chem yr 12

Heinemann

Heinemann Chemistry 2 Student Workbook Heinemann Chemistry 2 Fourth Edition Teacher

Resource and Assessment Book Heinemann Chemistry 2 Fourth Edition Teacher Lounge Heinemann Chemistry 2 Fourth Edition Enhanced Pearson Reader

Student book — key features:

978 1 7408 1721 9

978 1 4425 3755 2

978 1 7408 1726 4

Exam Café Chemistry VCE Units 3&4 Exam Café is an interactive package that assists students with learning, studying, revising and preparing for tests and examinations. www.pearsonplaces.com.au Key features:

Pearson Places is the online destination that allows you to access current educational content, download lesson material, use rich media and connect with students, educators and professionals around Australia. With over 30 000 resources online and constantly evolving, Pearson Places is the only place you’ll need for your digital solutions.

Bob Hogendoorn • Chris Commons • Susan Jarrett • Carol McKenzie • Warren Moseley • Maria Porter • Mark Williamson

Answers to all text questions A complete electronic copy of the student book A comprehensive glossary ‘How to’ sections on study and revision, and on planning and writing exam answers Practice exams and answers Interactive quizzes Exam-style questions with samples of higher-order and lower-order answers Access to interactive learning and revision modules with instantaneous feedback at www.ChemCAL.com.au Weblinks

Enhanced

978 1 4425 3752 1

4th edition

Stimulating content that provides an exact match to the structure and sequence of the Study Design A focus on new and emerging technologies Clearly marked extension and enrichment material Explanations of concepts in easy-to-understand language Extensive use of illustrative material, visual communication and graphic design Relevant, high-interest material that draws on student interest and experience, and features current applications of chemistry and the issues surrounding these Plenty of review questions, further questions and examtype questions

Heinemann

chemistry 2

The fourth editions of Heinemann Chemistry 1 and Heinemann Chemistry 2 have been updated to support the current accredited Chemistry Study Design, which has been extended to 2014. The new and enhanced Heinemann Chemistry 2 is presented as a student pack consisting of a student book and an Exam Café CD.

Also available:

chemistry 4th edition

chemistry 2

Heinemann

Enhanced

VCE Units 3 & 4

www.pearsonplaces.com.au

Bob Hogendoorn Chris Commons Susan Jarrett Carol McKenzie Warren Moseley Maria Porter Mark Williamson

PEARSON

Pearson Australia (a division of Pearson Australia Group Pty Ltd) 20 Thackray Road, Port Melbourne, Victoria 3207 PO Box 460, Port Melbourne, Victoria 3207 www.pearson.com.au Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) First published 2010 by Pearson Australia 2013 2012 2011 10 9 8 7 6 5 4 3 2 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/ or communicated by any educational institution for its educational purposes provided that that educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact Copyright Agency Limited (www.copyright.com.au). Reproduction and communication for other purposes Except as permitted under the Act (for example any fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All enquiries should be made to the publisher at the address above. This book is not to be treated as a blackline master; that is, any photocopying beyond fair dealing requires prior written permission. Publisher: Malcolm Parsons and Misal Belvedere Editor: Elaine Cochrane and Catherine Greenwood Text Designer: Ruth Comey, Sarah Hazell and Ben Galpin Typesetters: Palmer Higgs and Sunset Digital Cover designer: Ruth Comey and Ben Galpin Copyright & Pictures Editor: Julia Weaver Project Editor: Sarah Flockhart Production Controller: Claire Henry Cover art: Photolibrary Illustrator/s: Guy Holt and Wendy Gorton Printed in China National Library of Australia Cataloguing-in-Publication entry Heinemann Chemistry. 2 enhanced : VCE units 1 & 2 /authored by Bob Hogendoorn...[et al.] ; editor, Catherine Greenwood. Edition: Enhanced 4th ed. ISBN: 9781442537552 (pbk.) Notes: Includes index. Target Audience: For secondary school age. Subjects: Chemistry--Textbooks. Other Authors/Contributors: Hogendoorn, Bob; Greenwood, Catherine. Dewey Number: 540 Pearson Australia Group Pty Ltd

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Acknowledgements We would like to thank the following for permission to reproduce copyright material. The following abbreviations are used in this list: t = top, b = bottom, l = left, r = right, c = centre. AAP: p.313, Australian Associated Press Pty Ltd/Andrew Brownbill, p.187; Australian Associated Press Pty Ltd/Paul Crock, p.366; Australian Associated Press Pty Ltd/Gavin Blue, p.371; Australian Associated Press Pty Ltd/Bill Bachman, p.373. The Age: ‘Council Acts on Toxic Waste’ Martin Boulton/Courtesy of The Age; p.311. Alamy Ltd: Alamy Limited/John Glover, p.5; Alamy Limited/Jim West , p.172; Alamy Limited/blickwinkel/Dolder, p.191; Alamy Limited/Mira, p.203; Alamy Limited/ INTERFOTO, p.223; Alamy Limited/TRG, p.230; Alamy Limited/Vanessa Miles, p.252bl; Alamy Limited/David R. Frazier Photolibrary, Inc, p.434t; Alamy Limited/ Greenshoots Communications, p.459. Apple Computer, Inc: Courtesy of Apple, p.456. Archiv der Max-Planck-Gesellschaft, Berlin-Dahlem: p.322t. Australian Associated Press Pty Ltd: ‘US backs Australia’s uranium plan’ from AAP, published in The Age on 11 January 2006. Australian Broadcasting Corporation: ‘Sydney Harbour commercial fishing halted’, first published on ABC Online on 24 January, 2006, is reproduced by permission of the Australian Broadcasting Corporation and ABC Online. (c) 2006 ABC. All rights reserved, p.309.

The Australian Synchrotron Project: p.117. Corbis Australia Pty Ltd: ©Arthur Rackham Estate / Bridgeman Art Library /CORBIS, p.82bl; © Bettmann/CORBIS, p.192; © David Woods/CORBIS, p.247; © Patrik Giardino/CORBIS, p.249; © Charles Rotkin/CORBIS, p.322b; © Bettmann/CORBIS, p.329; © James L. Amos/CORBIS, p.331b; © Jonathan Blair/CORBIS, p.377t; © Igor Kostin/Sygma/CORBIS, p.386; © Brooks Kraft/CORBIS, p.389; © Bettmann/CORBIS, p.401b. CSIRO Publishing: Archives, p.81. Dreamstime: pp.1, 3, 50, 79, 144, 152br, 155r, 178, 221, 252l, 253, 257, 264, 295, 301b, 317, 328, 330, 349r, 382, 392t, 445 Fairfax Photo Sales: Glenn Campbell, p.7; Craig Abraham, p.4; Glen McCurtayne, p.278; Wayne Taylor, p.312; James Davies, p.351; Craig Abraham, p.377b; Joe Armao, p.456. Fergus Photography: p.146. Getty Images: Getty Images Australia Pty Ltd/Megan Fizell, Tres Jolie Studios, p.319; Getty Images Australia Pty Ltd/Marcel Mochet, p.392b. iStockphoto: © iStockphoto.com/elusivemuse, p.163; © iStockphoto.com/borsheim, p.222. John Innes Centre and Institute of Food Research: ‘Nitrogen-fixing crops might be produced more easily’ p.325. Lonely Planet Images: Diana Mayfield, p.285l. Los Angeles Times: Los Angeles Times: ‘European Parliament OKs Rules on Chemical Safety’ by Marla Cone, 18 November 2005. p.312. Malcolm Cross: pp.62, 339, 339bl MFB (Metropolitan Fire and Emergency Services Board): C.Maxwell/MFB Photographic Management Unit, p.356t NASA: pp.400, 408, 427, 437; NASA Photo/Tom Tschida, p.390. National Archives of Australia: National Archives of Australia: R32, SUNDRY 2/54, p.296. Newspix: © Newspix/News Ltd/3rd Party Managed Reproduction & Supply Rights/ Glenn Barnes, p.11; © Newspix/News Ltd/3rd Party Managed Reproduction & Supply Rights/Jane Ollerenshaw, p.48; © Newspix/News Ltd/3rd Party Managed Reproduction & Supply Rights/Amos Aikman, p.303. New Scientist Magazine: Forensic expert slates Maguire analysis, New Scientist Magazine, Issue 1722, 1990, p.73; Engineered enzyme washes whiter than white, New Scientist Magazine, Issue 1671, 1989, p.201. Orica Limited: p.401t Pearson Australia: pp.170, 213tl; Alice McBroom, pp.20, 28, 40, 41, 37, 133, 135, 150bl, 167, 252tl, 275, 285r. Photolibrary Pty Ltd: Photolibrary Pty Ltd/Gusto Gusto p.39; Photolibrary Pty Ltd/ ALM p.53; Photolibrary Pty Ltd/MEHAU KULYK/SPL, p.61; Photolibrary Pty Ltd/ Scott Bauer, p82tl; Photolibrary Pty Ltd/Dr Jurgen Scriba, p.115; Photolibrary Pty Ltd/Robert Clark, p.116; Photolibrary Pty Ltd/Science Photo Library, pp.130tl, 173, 181l, 228, 321, 331t, 385l, 413, 414; Photolibrary Pty Ltd/Yoav Levy, p.150tl; Photolibrary Pty Ltd/AJ Photo/Hop Americain, p.154; Photolibrary Pty Ltd/ Dr P Marazzi, pp.162, 181r; Photolibrary Pty Ltd/Biophoto Associates, p.186l; Photolibrary Pty Ltd/Vincent Serventy, p.188; Photolibrary Pty Ltd/Pasieka Pasieka, p.193; Photolibrary Pty Ltd/CNRI CNRI, p.204; Photolibrary Pty Ltd/A Barrington Brown, p.205l; Photolibrary Pty Ltd/Inc-Photo, p.205r; Photolibrary Pty Ltd/Corbin O’Grady Studio, p.210; Photolibrary Pty Ltd/Manfred Kage, p.211tr; Photolibrary Pty Ltd/Walter Reinhart, p.211br; Photolibrary Pty Ltd/Simon Fraser/SPL, p.213br; Photolibrary Pty Ltd/Pascal Goetgheluck, p.213tr, 217b; Photolibrary Pty Ltd/David Parker/SPL, p.217tl; Photolibrary Pty Ltd/Andy Crump/TDR/WHO/SPL, p.227br; Photolibrary Pty Ltd/Robert Lambiase, p.227bl; Photolibrary Pty Ltd/Pasquale Sorrentino/SPL, p.255; Photolibrary Pty Ltd/Susumu Nishinaga/SPL, p.276b; Photolibrary Pty Ltd/Dirk Wiersma, p.276t; Photolibrary Pty Ltd/Paul Rapson, p.301; Photolibrary Pty Ltd/Jeremy Burgess/SPL, p.318; Photolibrary Pty Ltd/Dr Keith Wheeler, p.323; Photolibrary Pty Ltd/David Taylor, p.339br; Photolibrary Pty Ltd/ Bowater Peter, p.344; Photolibrary Pty Ltd/Warwick Kent; Photolibrary Pty Ltd/ Patrick Landmann, p.385r; Photolibrary Pty Ltd/Sam Ogden/SPL, pp.444, 448; Photolibrary Pty Ltd/Maximilian Stock Ltd/SPL, p.447. Qenos Pty Ltd: pp.348, 352, 356b. Science Photo Library: Erich Schrempp, cover. Shutterstock: pp.2, 10, 14, 27, 36, 47, 66, 76, 100, 110, 129, 130tc,136, 151,152tl, 155l, 166, 177,183, 186r, 236, 245, 246, 262, 284, 291, 302t, 337, 349l, 350, 404, 428t, 428b, 432, 434b, 455, 457. Every effort has been made to trace and acknowledge copyright. However, should any infringement have occurred, the publishers tender their apologies and invite copyright owners to contact them. Disclaimer/s The selection of internet addresses (URLs) provided for this book was valid at the time of publication and was chosen as being appropriate for use as a secondary education research tool. However, due to the dynamic nature of the internet, some addresses may have changed, may have ceased to exist since publication, or may inadvertently link to sites with content that could be considered offensive or inappropriate. While the authors and publisher regret any inconvenience this may cause readers, no responsibility for any such changes or unforeseeable errors can be accepted by either the authors or the publisher.

Contents

unit

area of study 01

3

area of study 02

Chemical analysis Chapter 1.1 1.2 1.3

1

Organic chemical pathways

What is chemical analysis?

Chapter

Chemical analysis Quality control Analytical techniques Chapter review

Chapter

2

3 4 6 9

Analysis by mass

2.1 Measuring water content 2.2 Finding the composition of a compound 2.3 Calculating masses of reactants and products 2.4 Finding the composition of a mixture Chapter review

Chapter

3

11 16 19 22 25

Volumetric analysis

3.1 Concentration 3.2 Volumetric analysis Chapter review

Chapter 4.1 4.2 4.3 4.4

Analysing acids and bases

5

37 39 39 43 45

Analysing oxidants and reductants

What is a redox reaction? Oxidation numbers: classifying redox reactions Writing half equations Volumetric analysis Chapter review

Chapter 6.1 6.2 6.3

28 30 34

Acid–base chemistry revisited pH Indicators Back titration Chapter review

Chapter 5.1 5.2 5.3 5.4

4

6

49 51 54 56 58

Chromatography

Chromatography Thin-layer chromatography Column chromatography Chapter review

Chapter

7

62 62 65 72

Spectroscopy

7.1 The electromagnetic spectrum 7.2 Analysis of atoms 7.3 Atomic absorption spectroscopy 7.4 UV–visible spectroscopy 7.5 Infrared spectroscopy 7.6 Nuclear magnetic resonance spectroscopy The Cutting Edge NMR and the structure of membrane peptides Chapter review

Chemical pathways

77 78 81 85 89 97

9

Compounds of carbon

9.1 Bonding in carbon compounds 9.2 Homologous series 9.3 Functional groups Chapter review

Chapter

10

Organic reactions: pathways

to new products

10.1 Reactions of alkanes 10.2 Reactions of alkenes 10.3 Reactions of functional groups 10.4 Esters: sweet flavours and strong aromas 10.5 Reaction pathways 10.6 Principles of fractional distillation The Cutting Edge Free radicals: a highly reactive species! Chapter review

Chapter

11

12

13

14

167 167 174 176

178 182 183 188 194 197 200

DNA

13.1 The biological importance of DNA 13.2 The structure of DNA 13.3 How does DNA control biochemical processes? 13.4 Forensic applications: DNA profiling Chapter review

Chapter

162 164

Pathways to biomolecules

12.1 Fats 12.2 Condensation polymerisation 12.3 Carbohydrates 12.4 Proteins 12.5 Enzymes: biological catalysts 12.6 Proteins as markers for disease Chapter review

Chapter

145 146 149 152 156 160

Biochemical fuels

11.1 Fossil fuels 11.2 Biochemical fuels The Cutting Edge Two approaches to using biofuels for public transport Chapter review

Chapter

130 131 137 143

204 204 210 212 220

From organic molecules to medicines

8.1 Mass spectrometry 8.2 Combined techniques 8.3 Assessing analytical techniques Chapter review

111 115 119 122

14.1 Aspirin: a modern painkiller from a herbal remedy 222 The Cutting Edge Computer-aided drug design 226 14.2 Penicillin: a chance discovery 228 14.3 Proteins as medicines 230 14.4 A cleaner, greener future? 231 14.5 Developing a new drug 233 The Cutting Edge Designer medicines: molecules of the future 236 Chapter review 238

124

Chapter

8

102 106

Combining and choosing analytical

techniques

Area of Study review


239

iii v

Contents

unit

4

area of study 02

area of study 01

Supplying and using energy

Industrial chemistry Chapter

15

Chapter

16

17

18

Chemical industry: risks and rewards

18.1 Australia’s chemical industry 18.2 A global enterprise 18.3 Inside a chemical industry 18.4 Working safely 18.5 Waste management and pollution 18.6 Community safety 18.7 Location of a plant Chapter review

Chapter

19

20

21

iv

22

329 331 333 335 336

Production of sulfuric acid

21.1 Uses of sulfuric acid 21.2 The contact process 21.3 Waste management 21.4 Health and safety Chapter review

Chapter

318 320 324 326 327

Production of nitric acid

20.1 Uses of nitric acid 20.2 The Ostwald process 20.3 Waste management 20.4 Health and safety Chapter review

Chapter

296 297 298 301 306 311 312 315

Production of ammonia

19.1 Uses of ammonia 19.2 Production of ammonia 19.3 Waste management 19.4 Health and safety Chapter review

Chapter

286 289 293

338 340 343 345 346

Production of ethene

22.1 Uses of ethene 22.2 Production of ethene 22.3 Waste management 22.4 Health and safety Chapter review

349 351 354 355 358

360


23

Fossil fuels

23.1 Energy sources today 23.2 Energy converters 23.3 Fossil fuels Chapter review

Chapter 263 266 271 276 280

Equilibria involving acids and bases

17.1 Acidity of solutions 17.2 Acidity constants Chapter review

Chapter

247 248 251 259

Controlling the yield of reactions

16.1 A problem for industry: incomplete reactions 16.2 The equilibrium law 16.3 Changing the equilibrium position of a reaction 16.4 Do all reactions reach equilibrium? Chapter review

Chapter

Chapter

Fast and slow chemistry

15.1 Fast and slow chemistry 15.2 Chemical energy 15.3 Making reactions go faster Chapter review

Chemistry at work

24

Alternative energy sources

24.1 Nuclear energy 24.2 Renewable energy sources The Cutting Edge Polymer solar cells 24.3 A time for decisions Chapter review

Chapter

25

26

27

28

405 406 410

414 417 420 422 423 426

Cells and batteries

27.1 Portable power 27.2 Primary cells 27.3 Rechargeable cells and batteries 27.4 Fuel cells The Cutting Edge Solid oxide fuel cells Chapter review

Chapter

402

Electricity from chemical reactions

26.1 Galvanic cells 26.2 The electrochemical series The Cutting Edge Making unnatural products by natural means 26.3 Potential difference 26.4 Predicting direct redox reactions Chapter review

Chapter

383 388 394 396 398

Energy from chemical reactions

25.1 Thermochemical equations 25.2 The connection between energy and temperature change 25.3 Measuring the heat released during a reaction Chapter review

Chapter

367 369 370 380

428 429 432 436 440 442

E lectrolysis: driving chemical reactions by electricity

28.1 Chemical energy from electrical energy 28.2 Electroplating 28.3 Faraday’s laws 28.4 Competition at electrodes 28.5 Useful chemicals from electrolysis Chapter review

445 445 448 451 453 460


463

Appendices

470

Answers

475

Glossary

481

Index

491

Periodic table

496

How to use this book Heinemann Chemistry 2 4th Edition Enhanced Student Book The enhanced fourth edition of Heinemann Chemistry 2 has been developed to support the 2008–2014 Chemistry Study Design. The content provides an exact match to the study design and the layout and design incorporates many features that assist student learning. The authors have updated the student book with the very latest developments and applications of chemistry, presented in an Australian context. The book is divided into four Areas of Study corresponding to the Study Design and these are further divided into chapters. The following features will assist students to grasp the key concepts. · Each chapter opening includes key knowledge statements and outcomes. These assist students to unpack the Study Design and expand on what they are expected to know and be able to do. · Each chapter is further divided into clear-cut sections, which finish with a set of summary points and key questions that will assist students to Analysis by mass consolidate the key points and concepts of that section.

·

· ·

·

chemfact If the uncertainty of the balance used for weighing is ±0.02 g, ‘constant mass’ would include a mass increase or decrease by up to 0.02 g.

02

chapter 02

key knowledge

• gravimetric analysis • calculations including amount of solids, liquids and gases, and volume and pressure of gases.

chapter outcomes

·

After completing this chapter, you should be able to:

·

·

·

• calculate the water content of a material • describe the technique of gravimetric analysis • identify the characteristics of a suitable precipitate for gravimetric analysis • write balanced equations for precipitation reactions • use the relationship between amount, in mole, and mass, in gram • use the general gas equation to determine the amount in moles of a gas, or its volume, temperature or pressure • determine the empirical and molecular formula of a compound • use stoichiometry to calculate quantities of a reactant or product, including calculations involving gases and excess reactants.

Chapter review questions are found at the end of each chapter. These test students’ ability to apply the knowledge gained from the chapter. Chemistry in Action boxes contain chemistry in an applied situation or relevant context. These include the nature and practice of chemistry, applications of chemistry and the associated issues, and historical development of concepts and ideas. ChemCAL boxes indicate when to go to the ChemCAL website which is accessed through Exam Café Online. ChemCAL is a collaboration between the Chemistry Education Association, the School of Chemistry at the University of Melbourne and Pearson. It allows students to revise key ideas and skills and extend their knowledge of applications and examples. Students can monitor their progress through interactive quizzes throughout, which provide instant feedback on progress.

Extension boxes contain material that goes beyond the core content of the Study Design. These are intended for students who wish to expand their depth of understanding in a particular area. The material may be conceptual or contextual. The Area of Study Review includes a large range of exam-style questions both multiple choice and extended response. The Cutting Edge spreads are written by practising Australian scientists. New and emerging technologies are explained and discussed and help bring modern chemistry to life, while addressing this vital area of the Study Design. Chemfacts are snippets of information that add interest and relevance to the text.

·

The glossary at the end of the book can be used to check the meaning of important words. A comprehensive index is included and appendices include important support material.

Exam Café CD Features of the Heineman Chemistry 2 Exam Café CD include: · student book worked solutions · electronic student book · glossary · practice exams and answers · course content checklist · general advice • a link to Exam Café online.

The Heinemann Chemistry 4th Edition package Don’t forget the other components that will assist teachers and help engage and excite students in Chemistry: Heinemann Chemistry 2 Student Workbook Heinemann Chemistry 2 Teacher Resource and Assessment Book Heinemann Chemistry 2 Teacher Lounge Heinemann Chemistry 2 Enhanced Pearson Reader Chemistry VCE Units 3 & 4 Exam Café Online

Stoichiometry • The ideal gas equation

v How to use this book

Series features The enhanced fourth edition of Heinemann Chemistry 2 has been developed to support the 2008–2014 Chemistry Study Design. The content provides an exact match to the study design and the layout and design incorporates many features that assist student learning.

Heinemann Chemistry 2 4th Edition Enhanced Student Book The book is divided into four Areas of Study corresponding to the Study Design and these are further divided into chapters. • Each chapter opening includes key knowledge statements and outcomes. These assist students to unpack the Study Design and expand on what they are expected to know and be able to do. • Each chapter is further divided into clear-cut sections, which finish with a set of summary points and key questions that will assist students to consolidate the key points and concepts of that section. • Chapter review questions are found at the end of each chapter. These test students’ ability to apply the knowledge gained from the chapter. • Chemistry in Action boxes contain chemistry in an applied situation or relevant context. These include the nature and practice of chemistry, applications of chemistry and the associated issues, and historical development of concepts and ideas. • ChemCAL boxes indicate when to go to the ChemCAL website which is accessed through Exam Café Online. ChemCAL is a collaboration between the Chemistry Education Association, the School of Chemistry at the University of Melbourne and Pearson. It allows students to revise key ideas and skills and extend their knowledge of applications and examples. Students can monitor their progress through interactive quizzes throughout, which provide instant feedback on progress.

vi vi

•

• •

• • •

Extension boxes contain material that goes beyond the core content of the Study Design. These are intended for students who wish to expand their depth of understanding in a particular area. The material may be conceptual or contextual. The Area of Study Review includes a large range of exam-style questions, both multiple choice and extended response. The Cutting Edge spreads are written by practising Australian scientists. New and emerging technologies are explained and discussed and help bring modern chemistry to life while addressing this vital area of the study design. Chemfacts are snippets of information that add interest and relevance to the text. The glossary at the end of the book can be used to check the meaning of important words. A comprehensive index is included and appendices include important support material.

Exam Café CD The Heinemann Chemistry 2 Exam Café CD comes with the student book and includes: • worked solutions to all student book questions • a complete electronic copy of the student book • a comprehensive glossary • practice exams and answers * course content checklist • ‘How to’ sections on study and revision, and planning and writing exam answers • a link to Exam Café Online.

Heinemann Chemistry 2 Student Workbook Heinemann Chemistry 2 Student Workbook provides outstanding support and guidance for students studying VCE Chemistry. Designed to be used in conjunction with the student book, it includes: • highly illustrated study notes covering the main points of each Area of Study • multiple intelligence worksheets that cater for a range of learning styles • sample assessment tasks • practical activities and investigations.

Heinemann Chemistry 2 Teacher’s Resource and Assessment Book (includes CD) Heinemann Chemistry 2 Teacher’s Resource and Assessment Book consists of the teacher’s resource book and CD. It includes: • a vast range of experiments, demonstrations and exercises all with comprehensive teacher notes • course advice • risk management advice and templates • answers and worked solutions to all textbook questions • PowerPoint slides • a sample work program.

Heinemann Chemistry 2 Teacher Lounge

Exam Café Online Chemistry VCE Units 3 & 4 Exam Café Online is an interactive package that assists students with learning, studying, revising and preparing for tests and examinations. Features include: • ChemCAL: www.ChemCAL.com.au includes interactive learning and revision modules with instantaneous feedback • interactive quizzes • exam-style questions with samples of higher order and lower order answers • practice exams and answers • useful weblinks • general advice • scaffolded questions with answers.

Heinemann Chemistry 2 Teacher Lounge is the one location for free teacher support material—course advice, practical notes and advice, ICT support for activities and more! The Teacher Lounge can be accessed through the Pearson Places website.

Pearson Places Pearson Places is your online destination that is constantly evolving to give you current educational content on the web. Visit Pearson Places to access educational content, download lesson material, use rich media and connect with students, educators and professionals around Australia. www.pearsonplaces.com.au

Heinemann Chemistry 2 Pearson Reader More than an eBook, Pearson Reader provides unique online student books that allow teachers and students to harness the collective intelligence of all who participate. Search for a unit of work and contribute by adding links and sharing resources. The Heinemann Chemistry 2 Pearson Reader contains media-rich resources, including interactive lessons, animations, videos and teacher support material in one location.

vii Series features

About the authors The Chemistry Education Association was formed in 1977 by a group of chemistry teachers from both secondary and tertiary institutions. It aims to promote the teaching of chemistry, particularly in secondary schools. The CEA has established a tradition of providing up-to-date text and electronic material and support resources for both students and teachers and professional development opportunities for teachers. CEA offers scholarships and bursaries to students and teachers to further their interest in chemistry. CEA supports STAV with sponsorship for the Chemistry Conference, Science Drama Awards and The Science Talent search. The first editions of Heinemann Chemistry 1 and Heinemann Chemistry 2 were published in 1990 and 1991. The fourth editions along with the Teacher Support material continue this commitment to chemistry education. Bob Hogendoorn has taught chemistry in Victoria for over 35 years. He has been actively involved in the development of senior chemistry courses including membership of the exam panels and chemistry subject committees. He is the manager of the VCE ChemCAL project, a collaborative project between the University of Melbourne and Chemistry Education Association, which has developed interactive web-based learning modules for VCE Chemistry. He has also written and edited publications in support of chemistry and science education in this state and has presented at Chemistry Teacher workshops and in-services. Bob was principal and assistant principal at a number of secondary colleges. Bob is a member of the executive of the Chemistry Education Association. Chris Commons was the editor of the first three editions of Heinemann Chemistry 2. He is a senior member of staff at Scotch College, Hawthorn. Chris was awarded a PhD for his research in transition metal chemistry at the University of Melbourne. He has been actively involved in teaching, curriculum development and assessment and was Chief Examiner for VCE Chemistry for several years. Chris has co-authored several chemistry text books and has had extensive

viii

experience in government and independent schools and on secondment to the Australian Council for Educational Research and the University Melbourne as Lecturer in Chemistry Method. In 2004, he received the Chemistry Educator of the Year award from the Chemistry Education Association. Susan Jarrett has had extensive experience teaching chemistry and science in Britain, Hong Kong, USA and Australia. She has also had much involvement with curriculum development at the national and international level. Her coordination and teaching of the International Baccalaureate has given additional insight into her teaching of the VCE. Carol McKenzie has worked as a research chemist in areas of environmental and drug analysis. She has taught chemistry and general science and has an ongoing interest in curriculum development and effective teaching and learning. Presently, she is the Academic Development Director for Study Group Higher Education Division in Australasia. Warren Moseley has been teaching for over 25 years, specialising in chemistry, science and IT. He has been a VCE Coordinator and Curriculum Coordinator, and has served on several committees for the Victorian Board of Studies and the Victorian Curriculum and Assessment Authority. Warren is currently Director of Information, Communication and Learning Technologies at Whitefriars College in outer Melbourne. Maria Porter has taught mathematics and science in Victorian schools since 1972. She is currently head of Senior Science at Wesley College, Prahran. As well as chemistry, Maria teaches biology and general science. Mark Williamson is senior chemistry teacher with extensive experience in teaching and natural resource management. He has completed research in science education with a focus on concept development and mapping. Mark coordinated the Dandenong Ranges Innovation and Excellence project and continues to work with others to improve our natural and social environment. Mark is currently Performance and Development Manager at Monbulk College.

unit

3

areas of study 01 chemical analysis On completion of this unit the student should be able to evaluate the suitability of techniques and instruments used in chemical analyses.

02 organic chemical pathways On completion of this unit the student should be able to identify and explain the role of functional groups in organic reactions and construct reaction pathways using organic molecules.

Chemical pathways

0

chapter 01

What is chemical analysis?

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • explain what is meant by the term ‘chemical analysis’ • explain the term ‘quality control’ • distinguish between qualitative and quantitative analysis • give examples of chemical analyses • list a range of techniques used in chemical analysis • recall that analytical techniques utilise particular physical or chemical properties of the substance under investigation.

1.1

Chemical analysis Think about what you had for breakfast today. The packaging on any processed food you ate, including cereal, milk, bread and yogurt, has a label that provides you with nutrition information so you can make healthy diet choices. This information comes from chemical analysis. Suppose you ate a boiled egg and a banana instead of the cereal and milk. There is no label with nutrition information there, but analytical chemists have tested a sample of eggs and bananas for residues of harmful chemicals so you can be confident that your breakfast is safe to eat. Your toothpaste, too, will have a label that says something like ‘active ingredient 0.7% w/w sodium monofluorophosphate’. An analytical chemist would have checked that this amount of active ingredient was correct. Analytical chemists are employed in a range of industries, from food manufacturing to environmental monitoring and forensic sciences and many more. They use a range of techniques and technologies to determine the identity, quality and quantity of the chemicals that make up our world. The composition of consumer goods is of great importance. For example, residues of pesticides such as dieldrin and DDT have been detected in meat. Nutrionists urge us to eat fish on a regular basis to take advantage of the beneficial fatty acids that many fish contain. But fish can also contain heavy metals, such as mercury. Governments have placed restrictions on the amount of pesticides and heavy metals allowed in foods. For example, the acceptable daily intake of DDT is 0.002 milligram per kilogram of body weight per day (0.002 mg/kg/day) while that of dieldrin is 0.0001 mg/kg/day. Chemical analysis is used to answer many other important industrial and environmental questions, for example: Is the amount of mineral in this ore body high enough to make the mine profitable? What mineral deficiencies does this soil have? What type of fertiliser is needed and how much should be added to address this problem? Is the sugar content of my grapes high enough to make a quality wine? What additives have been added to this food? Does the amount comply with those specified in the food standards? Chemists can answer these questions by analysing foods and other substances to determine their composition.

Figure 1.1 Chemical analysis plays a central role in determining the suitability and quality of consumer goods.

chemfact Acceptable daily intake, ADI, is the amount of pesticide or other contaminant that can be taken into the body on a daily basis over a lifetime and not cause any injury of any kind. The highest amount that can be taken daily and cause no effect is referred to as the no observable adverse effect limit, NOAEL. To provide a margin of safety, a pesticide’s ADI is set at significantly less than its NOAEL. In practice, zero levels are often not feasible.

chemistryy in action What is in the water? Arsenic leaked into river page one headline, The Age, 22 August 2005

According to the Age report, people have been fishing for years in a stretch of the Maribyrnong River that is heavily contaminated with a range of toxic chemicals. Groundwater entering the river contains: Arsenic—up to 20 000 times safe environmental limit Copper—up to 154 000 times safe environmental limit Zinc—up to 5000 times safe environmental limit Lead—up to 250 times safe environmental limit

Ammonia—up to 33 500 times safe level for human contact. ‘Most of the fishermen here eat everything they catch,’ says Colin Wilkinson, who has fished in the area for 50 years. ‘No one has ever warned us about eating the fish.’

Contaminated groundwater is leaching from a site on which a chemical, fertiliser and acid plant formerly stood. The final cost of decontamination of the land is likely to exceed $10 million.

3 What is chemical analysis?

chemistryy in action What is in the water? (continued) b

c

How the contamination occurs

RAY RD

Contaminated fill

YA R

WH

ITE

HA LL S T R AV ILL E Maribyrnong River

FOOTSC

Coode Island

Maribyrnong River

a

Polluted ground water enters here Volcanic layer Sand/silt layers

Former Pivot site

ver a Ri Yarr N

0

Water table line 500 m

Silt

Figure 1.2 a Fishing on the Maribyrnong River. b Location of the contaminated region. c How the contamination occurs.

1.2

Quality control In most processes that convert raw materials into a manufactured product, information is needed about the composition of the raw materials and also about the final product. A biscuit manufacturer, for example, would purchase batches of raw materials such as flour, sugar, fats and dried fruits. The quality of ingredients such as flour and fruits might vary with each batch. These variations must be monitored and the manufacturing process adjusted to ensure consistent quality in the final product. Chemists have developed methods to determine the composition of the raw materials and the quality of the final product. These are methods of quality control. Chemical analysis of the chemical composition of a substance seeks to answer two questions: What chemicals are present in the substance? The process of finding this information is known as qualitative analysis. How much of each chemical is present? The process of finding this information is called quantitative analysis. In the chapters that follow, you will learn about the different techniques that chemists have developed for qualitative and quantitative analysis. But let us look first at an example of the use of each type of analysis.

Qualitative analysis: identification of chemicals in plant extracts The fragrant oils of many eucalypt species were first extracted by Australian settlers in the 1800s. These pioneers found that the oils could be used as perfumes and disinfectants, and they are still widely used for those purposes

4 Chemical analysis C

today. The identification of substances present in plant oils is one example of qualitative analysis. Chemists are now investigating some of the plants used in the bush medicines of Aboriginal people and early settlers (Figure 1.3). The drug curare was once extracted from plants by South American native peoples and used as poison on the tips of arrows. It is now used widely as a muscle relaxant in open-heart surgery. Another substance, vincristine, obtained from the Madagascar periwinkle, can be used to cure about 60% of childhood leukaemia cases. Qualitative analysis can identify the types of chemicals present in a substance. It can also give information about the structure of newly discovered compounds such as potential new drugs. Analytical techniques are used to separate, identify and determine the composition and structure of biologically active compounds that may be used as drugs or medicines. Once the active agent has been identified chemists can devise procedures to make the drug synthetically. It is estimated that only 1% of plants have been tested to see if they have medicinal properties. Not surprisingly then, the rapid destruction of the world’s forests to provide timber and farming land is being viewed with alarm.

Quantitative analysis: aflatoxins in peanuts A common mould that grows on nut and cereal crops produces substances known as aflatoxins. Aflatoxins are among the most powerful cancer-causing substances known. It is believed that liver cancer, in particular, can be caused by high levels of these substances. Peanuts are particularly prone to the development of the mould, depending upon the conditions under which the nuts are grown and stored. In Australia, the National Health and Medical Research Council has set a maximum permitted level of aflatoxins of 15 μg/kg in peanuts and products such as peanut butter. Since the mould that produces aflatoxins cannot usually be seen, rigorous testing is required to ensure that contaminated products do not reach the consumer. Analytical techniques such as gas chromatography, liquid chromatography and mass spectrometry are used to measure the level of aflotoxins in peanuts. Single or muliple techniques may be used.

summary

Figure 1.3 Crinum lilies (Crinum pedunculatum). Rubbing the lily on the sting of a box jellyfish eases the pain. The active component in the lily has been determined by chemical analysis.

chemfact Your body has the ability to handle small amounts of many poisonous compounds. Trace amounts of chemicals believed capable of causing cancer and mutations are found in many foods, including honey, celery, mushrooms, chocolate, eggs, red wine and cooked meat, to name just a few.

1.1–1.2

Ch Chemical i l analysis l i is i usedd to t determine d t i the th chemical h i l composition and chemical structure of substances. Qualitative analysis is used to identify the chemicals present in a substance.

Quantitative Q tit ti analysis l i provides id iinformation f ti about b t the th amountt off chemicals present in a substance.

key question 1 Classify the following analyses as qualitative or quantitative. a an analysis that showed that Vegemite contains 0.55 mg of vitamin B1 and 0.80 mg of riboflavin per 5 g b detection of toxic materials in the gaseous discharge from a factory c confirmation of the presence of lead in a sample of blood d comparison of the concentrations of trace elements in glass fragments removed from the skin of a hit-and-run accident victim with those in the glass fragments from the broken headlights of a suspect’s car.


1.3

Analytical techniques

! The analytical techiques described here are not the only ones available to the analytical chemist. Just a few important techniques not dealt with here are electrophoresis, used in many biological applications, for example for the analysis of DNA collected from crime scenes (Chapter 13); ion-specific electrodes for analysing such chemicals as fluorine in drinking water; and immunoassay techniques that can identify specific proteins in venoms.

Chemists have developed a number of techniques for identifying chemicals. Apart from their use in analysing the substances present in consumer goods, these techniques are used in many other areas, including the diagnosis of disease, to monitor environmental pollution, and in police forensic work. Some of these analytical techniques will be explained in Chapters 2–8. The most appropriate chemical technique to use depends on factors including the nature of the sample, the nature of the chemical under analysis, the concentration of that chemical, the accuracy required, the time available, and, not least, the costs and benefits in carrying out the analysis. Many different chemical and physical properties can be used as a basis for qualitative analysis (Table 1.1). These properties are then compared with the properties of known substances. In some cases, the basis for identification might be physical properties such as colour, mass, solubility, melting point or response to electromagnetic radiation. In other instances, we might consider chemical properties, such as reactions with acids or oxidants. You will meet a number of analytical methods in the following chapters; these and the properties on which they are based are listed in Table 1.1. Chemists prefer to use techniques that are convenient and give unambiguous results. As you might imagine, the expertise of the chemist, availability of equipment and chemicals, and expense can also affect the choice of analysis method.

TABLE 1.1 Examples of some analytical techniques. Each technique utilises a physical or chemical property of the substance under investigation. Technique

Physical or chemical property

Examples of analysis

Mass of chemical, isolated from a mixture or precipitated in a reaction

Percentage of water in wheat stored in a silo Concentration of salt in a jar of baby food

Acid–base titrations (Chapter 4)

Neutralisation reactions by acid or base

Concentration of ammonia in window cleaner

Redox titrations (Chapter 5)

Reactant as oxidant or reductant

Chromatographic analysis (Chapter 6) Paper Thin-layer Gas High performance liquid Spectroscopic analysis (Chapter 7) Flame tests Atomic emission (AES) Atomic absorption (AAS) Ultraviolet–visible (UV) Infrared (IR) Nuclear magnetic resonance (NMR) Mass spectrometry (Chapter 8)

Adsorption onto a stationary phase and solubility in a mobile phase

Concentration of hydrogen peroxide in hair bleach Analysis of codeine in a tablet Concentration of alcohol in the blood Presence of DDT in minced meat

Analysis by mass Gravimetric analysis (Chapter 2) Volumetric analysis


Absorption or emission of part of the electromagnetic spectrum

Monitoring levels of calcium in the blood (AES) Concentration of copper in a ‘gold’ bracelet (AAS) Amount of aspirin in a cold remedy (UV) Determination of structure of a new drug (IR, NMR)

Mass/charge ratio of ionised compound

Identification of lead isotope ratios in meteors Identification of structure of potential new drug

It may be possible to analyse the same chemical by a range of chemical techniques, each adding a different piece of information to complete the final ‘jigsaw’. The analysis of ethanol by a number of techniques is described in Chapter 8. A number of the techniques listed in Table 1.1 are often combined in a single composite instrument to give a more sensitive or faster analysis. Some examples of combined techniques such as gas chromatography with mass spectrometry (GC-MS) are described in Chapter 8. The techniques that you meet in Chapters 3–5 are described as ‘wet chemical techniques’. These can be carried out by you using simple equipment in a chemistry laboratory. Chapters 6–8 go on to look at instrumental methods of chemical analysis.

extension Chemical analysis prevents a catastrophe

Figure 1.4 The likelihood of catastrophic engine failure can be reduced by analysing ‘wear metals’ in engine oil.

H often How ft ddo you hhear off a plane l crashh bbeing i bl blamedd on engine i failure? It may surprise you to learn that chemical analysis of engine oil can be used to predict possible engine problems. Lubricating oil carries away the tiny particles that flake way from metal surfaces as they rub against each other. The type of ‘wear metal’ present (qualitative analysis) and the amount (quantitative analysis) can give information about the health of various parts of the engine, since different engine parts are composed of different metal alloys. In most cases atomic emission spectrometry (AES, p. 80) is used to analyse for the metals present. If the sample volume is extremely small or if very accurate results are required, atomic absorption spectroscopy (AAS) or inductively coupled plasma (ICP) spectroscopy are used (p. 81).

The Th concentrations t ti off more th than 20 elements l t att a level l l off parts t per million (ppm) are measured: wear metals, e.g. Al, Cr, Cu, Sn,Ti, Ni additives to the lubricant, e.g. Ca, Ba, Zn, P, Mg, B, Mo contaminants, e.g. Si, Na, K. Other analytical techniques are used to check the oil for chemical breakdown and contamination. For example, initial analysis of the lubricant oil from an engine of an F15 aircraft showed no change in the levels of Fe, Ag, Al, Cr, Ni and Ti in the oil. However, after 430 hours of operation the iron content jumped from 4 to 11 ppm. A ‘teardown’ of the aircraft was ordered so the dismantled engine could be examined for problems. The ground crew found a bearing case broken in pieces. Oil analysis had prevented an imminent malfunction and possible catastrophic failure.


extension Chemical analysis prevents a catastrophe (continued)

Monitor, manage and control

Diagnosis: Silicon high, Check air filt er Fe 102 ppm Al 3 ppm Cr 12 ppm Cu 19 ppm Si 37 ppm Pb 55 ppm TAN 6 Visc. 12 ...........

Maintenance recommendations

Oil samples

Laboratory Analysis, diagnosis and recommendations

Management reports Monthly Report Monthly Equipm ent rtMonthly Repo t enpo Re Equipm rt Equipment

Machine condition

Figure 1.5 Analysis of wear metals can be a useful part of predictive maintenance programs.

questions

Lubricant condition

?

E1 Explain Expllaiin hhow ow quali qualitative litative and quantitative analysis of metals in oil are used to identify wear of engine components. components E2 What analytical techniques are used to determine the level of metals in oils?

summary

1.3

A variety i t off ttecniques i are usedd tto id identify tif andd quantify tif chemicals.

The Th physical h i l andd chemical h i l properties ti off the th substance bt under d investigation determine the analytical technique to be used.

key question 2 Look at Table 1.1. List the techniques from the table that are based on the: a physical properties of the substance being tested b chemical properties of the substance being tested.


01 atomic absorption spectroscopy (AAS) atomic emission spectrometry (AES) chemical analysis chromatography

Quantitative and qualitative analysis 3 Think back to when you first got up this morning. List every product you were in contact with during the first 30 minutes you were awake. a How many of these products are made from materials that have been processed? b How many are unprocessed? c Is there any evidence that any products you have listed were subject to some process of quality control? 4 The ingredients of a bottle of jam are listed on the label as ‘sugar, strawberries, pectin and food acid (citric acid)’. a Is this an example of qualitative or quantitative analysis? b Rewrite this label so that it is an example of analysis of the other type. 5 Which of the following represents example(s) of quantitative analysis? I A sample of ‘fries seasoning’ contains salt, chicken extract, rice flour, paprika and chilli. II Dry air contains 78% nitrogen vol/vol. III A sample of iron ore weighs 23.5 kg. IV A chocolate biscuit contains 2.3 g protein, 6.6 g fat and 24.4 g carbohydrate per 100 g serve. A IV only B I and IV C II and IV D II, III and IV 6 Analytical chemists use a variety of techniques. Give an example of the information provided by each of the following techniques. a gravimetric analysis b volumetric analysis c chromatography d spectroscopy

key terms concentration qualitative analysis quantitative analysis

HIGH LEVELS OF MERCURY FOUND IN FISH Canada restricts phosphate levels in detergent

Apples a forbidden fruit for school children

Nitrates contamin ate drinking water in western France

Pesticide spray link to cot death

in ox in s pr es en t St ud y sh ow s di ies pp na le di sp os ab

Insecticide found in organic food

D M IU M LE AD AN D CA IN CARCINOGEN IN BOT TLED WATER: D RE VE D IS CO RECALL ORDERED BA BI ES ’ FO O D Figure 1.6 These headlines reflect community concerns about chemical contamination of consumer goods.

8 Many of the foodstuffs we purchase are subject to quality control and a summary of their contents is often printed on the label. a Give three examples of substances that may be found in food and which are analysed by chemical means. b How can the information provided in a list of ingredients be used by the consumer for their benefit? 9 Obtain several brands of spring water and compare the information provided on the labels. a What evidence is there that the water has been subject to quality control? b How might this control have been carried out? c Compare the ions that that each contains. Are all samples of spring water the same? How do they differ?

Connecting the main ideas 7 Research a consumer good whose purity is currently making news. Find out how this product is analysed.


0

chapter 02

Analysis by mass

key knowledge • gravimetric analysis • calculations including amount of solids,

liquids and gases, and volume and pressure of gases.

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • calculate the water content of a material • describe the technique of gravimetric analysis • identify the characteristics of a suitable precipitate for gravimetric analysis • write balanced equations for precipitation reactions • use the relationship between amount, in mole, and mass, in gram • use the general gas equation to determine the amount in moles of a gas, or its volume, temperature or pressure • determine the empirical and molecular formula of a compound • use stoichiometry to calculate quantities of a reactant or product, including calculations involving gases and excess reactants.

The quantitative analysis of substances based on the measurement of mass is called gravimetric analysis. The use of sensitive electronic balances has enabled a high degree of accuracy to be achieved in the determination of the composition of compounds and mixtures. Gravimetric analysis is also used to determine the chemical formula of a compound.

2.1

Measuring water content Water is a component in many consumer products, including foodstuffs. In some cases water occurs naturally in the food, and in others it is added as part of the manufacturing process. When buying cordial in your local supermarket, you may be confronted with a bewildering array of brands. Which bottle of cordial represents the best value? One factor that might affect your choice is the number of glasses of diluted cordial you could make from each bottle. Some brands are less concentrated than others and contain a higher water content. Cheapest or biggest doesn’t always give best value for money because the mass of water in the sample may disguise the actual amount of product being purchased. Farmers need to check the water content of the grain they harvest. Grain buyers specify a maximum acceptable water content because grain that is too moist can sprout. If the weather is damp at harvest time, the farmer may use special driers to reduce the water content to an acceptable level. The water contents of various foods are shown in Table 2.1. The percentage of water in a sample can be determined by heating the sample at 110°C until there is no change in mass. This is referred to as heating to constant mass. Suppose you wish to find the percentage of water in a brand of canned soup. You would need to perform the steps shown in Figure 2.2.

TABLE 2.1 Water content of various foods Food

Water content (% by mass) 96

Lettuce Lemonade

90

Egg

75

Apple

65

Orange

65

Grilled steak

59

Sardines

49

Bread

40

Honey

23

Plain cake

20

Breakfast cereal (plain)

3

Weigh a sample.

Heat the sample in an oven at 110°C.

Allow the sample to cool in a desiccator.

The cycle must be repeated until the mass is constant.

Reweigh.

Figure 2.1 Which of these foods contains the most water? Table 2.1 may help you decide.

Figure 2.2 Procedure for determining the amount of water in a sample.

11 Analysis by mass

Worked example 2.1 An analytical chemist wanting to find the percentage of water in a can of soup obtains the following results. What is the percentage of water in the soup? Initial mass of soup Second mass, after heating at 110°C

33.6 g

Third mass, after more heating

24.3 g

Fourth mass, after more heating

24.3 g

Solution

chemfact

The third and fourth masses are the same, since all the water has been driven off. Mass of water in the soup = 223.1 g − 24.3 g = 198.8 g 198.8 g × 100 % water in the soup by mass = 223.1 g = 89.1%

If the uncertainty of the balance used for weighing is ±0.02 g, ‘constant mass’ would include a mass increase or decrease by up to 0.02 g.

summary

223.1 g

2.1

AAnalysis l i based b d on the th determination d t i ti off mass is i called ll d gravimetric analysis. Substances or mixtures containing a volatile component, such as water, are heated to drive off the volatile material and then the mass of the solid product left behind is measured.

The Th substance bt mustt be b heated h t d until til a constant t t mass is i obtained to ensure that all the volatile material has been driven off.

key questions 1 a Why was the soup sample in Worked Example 2.1 heated to 110°C? b Why was it necessary to weigh the sample four times? 2 Some laboratories use microwave ovens in place of conventional ovens to dry samples. What advantage could this have? 3 Soy sauce weighing 74.6 g was heated in an oven to constant mass. The final mass was 14.2 g. What percentage of water did the sauce contain? 4 Three brands of dog food were heated and dried to constant mass. The data recorded are shown in Table 2.2.

TABLE 2.2 Determination of water content of dog food Dog food tested Phydeaux Deluxe K9 Budget Fresh meat—buffalo mince


Mass of dog food sample (g) 19.8

Final mass (g) 3.9

7.4

1.9

15.0

3.8

a Which brand contained the highest percentage of water? b Do you consider that water content is a good guide to the relative value of different dog foods? What other factors might be important? 5 A student determined the water content of a sample of jam. The following measurements were obtained. What was the percentage, by mass, of water in the jam? Mass of evaporating dish

20.22 g

Mass of jam and evaporating dish Before heating

30.95 g

After heating

27.22 g

After more heating

26.50 g

After more heating

26.49 g

rreview

The mole Th Before looking at other ways that chemists analyse substances by weighing, we will recall some of the concepts that were covered in Units 1 and 2 of this chemistry course. One of the most useful units that chemists use is the mole. The mole is a unit used for counting particles. Just as a farmer might find it more convenient to count large numbers of eggs by the dozen, chemists count the enormously large numbers of particles in a substance in mole (which can be abbreviated to mol). You can count any type of particle you wish in mole, including atoms and molecules; even people can be counted in mole. But it is important to state clearly what you are counting so that there can be no confusion. For instance, 3 mol of H atoms is quite different from 3 mol of H2 molecules. One mole is a standard amount of substance, which contains the same number of specified particles as there are atoms in exactly 12 g of carbon-12. The symbol for amount of substance is n. Because atoms are so small, chemists have defined the mole so that it represents a very large number. From experiment, scientists have found that there are about 6.02 × 1023 particles in one mole. This number is called Avogadro’s number and has the symbol NA.

STOICHIOMETRY • Using Avogadro’s number • Number of particles and the mole • The mole and molar mass

Counting by weighing At first glance, it might seem strange that one mole contains 6.02 × 1023 particles. Why not another number? Why not just 1023? In fact, our definition of the mole has been selected with some care. The relative isotopic mass of carbon-12 is set at 12 exactly, and we have defined one mole of carbon-12 as weighing 12 g. Consider the mass of one mole of chlorine atoms. It will be greater than the mass of one mole of carbon atoms, since chlorine atoms are heavier than carbon atoms. Chlorine has a relative atomic mass of 35.5, so one mole must therefore weigh 35.5 g. In general, the mass of one mole (molar mass) of an element or compound is the relative mass of the element or compound expressed in grams. The molar mass of a substance is represented by the symbol M and has the unit g mol–1. To count one mole of particles, all you need to do is weigh out the relative mass of those particles in grams. As long as you know the mass of a pure substance and its molar mass, you can work out the amount of the substance. The amount of substance (n), measured in mole, can be calculated from the mass (m) and the molar mass (M ) using the formula: m n= M

Worked example a Calculate the amount, in mol, of H2O molecules in 9.0 g of water.

Solution The relative molecular mass of H2O = 18. So, molar mass, M(H2O) = 18 g mol−1 m 9.0 g Since n = , n(H2O) = = 0.50 mol M 18 g mol–1 So, 9.0 g of water contains 0.50 mol of H2O.

13 Analysis by mass

rreview (continued)

Amounts of gases Gases: spread to fill the volume available have low densities are easily compressed mix together rapidly. These properties are explained by regarding the particles of a gas as being widely separated and moving rapidly in random, straight-line motion. This model for explaining the behaviour of gases is known as the kinetic molecular theory of gases. It is not convenient to measure the amounts of gases by weighing. For example, when a cylinder for a gas barbecue is filled or air tanks for a scuba diving trip are prepared, the gas is forced into the tank until the contents are at a particular pressure. When the internal volume of a gas container is known, it is possible to calculate the amount of substance (n), measured in mole, by using the general gas equation: pV = nRT

Figure 2.3 The amount of air in a diver’s tank is indicated by measuring the pressure.

where p is the gas pressure, V is the volume, and T is the temperature. R is called the general gas constant and has the value 8.31 J K−1 mol−1 when pressure is measured in kilopascals (kPa), volume is measured in litres (L), and the temperature is measured in kelvin (K). Once the amount of gas, n, has been found, its mass can be calculated using the m formula n = . M

TABLE 2.3 Useful conversion factors for units Quantity

Units

Pressure

1 atmosphere = 760 mmHg = 101 325 Pa = 101.3 kPa 1 hectopascal (hPa) = 100 Pa = 0.1 kPa 1 bar = 100 kPa 1000 mL = 1 L 1 mL = 1 cm3 1 L = 1 dm3 1000 L = 1 m3 temperature in K = temperature in °C + 273

Volume

Temperature

Molar volume of a gas STOICHIOMETRY • The ideal gas equation


The volume of one mole of a gas depends on the gas temperature and pressure. At standard temperature and pressure (STP, which is 0°C and 1 atmosphere) the molar volume of a gas is 22.4 L mol−1. At standard laboratory conditions (SLC, which is 25°C and 1 atmosphere) the molar volume is 24.5 L mol−1. The formula: V n= Vm may be used to calculate the amount of a gas, measured in mole, from the volume of the gas and the molar volume (Vm). V and Vm are measured at the same temperature and pressure.

rreview (continued) Worked example b A steel cylinder with a volume of 30.0 L is filled with nitrogen gas to a pressure of 2.00 atm at 25.0°C. What mass of nitrogen does the cylinder contain?

Solution Since pV = nRT pV n(N2) = RT (2.00 × 101.3) kPa × 30.0 L = 2.45 mol 8.31 × (25.0 + 273) K m(N2) = n(N2) × M(N2) = 2.45 × 28.0 = 68.7 g So, the cylinder contains 68.7 g of nitrogen gas. =

Worked example c Calculate the amount, in mol, of sulfur dioxide in 10.0 L of the gas, measured at STP.

Solution Since n =

V Vm

10.0 L = 0.446 mol 22.4 L mol–1 So, 10.0 L of sulfur dioxide, measured at STP, contains 0.446 mol of SO2 molecules. n(SO2) =

summary The Th amountt off substance bt is i measuredd in i mole l (n). ( ) The number of particles in one mole is called Avogadro’s number (NA ), where NA = 6.02 × 1023 particles. The relationship between amount of substance (n), mass (m) m and molar mass (M ) is given by n = . M.

The Th relationship l ti hi between b t volume l (V )), pressure ((p )), temperature (T ) and amount of a gas in mol (n) is given by the general gas equation pV = nRT. At STP one mole of a gas occupies 22.4 L and at SLC one mole of gas occupies 24.5 L. The number of mole of a gas can be determined from the V volume (V ) and the molar volume Vm : n = . Vm

key questions 6 Calculate the amount (in mole) of: a NaCl in 5.85 g of salt b Fe atoms in 112 g of iron c CO2 molecules in 2.2 g of carbon dioxide d Cl– ions in 13.4 g of nickel chloride (NiCl2) e O2– ions in 159.7 g of iron(III ) oxide (Fe2O3) 7 Calculate the mass of: a 3.0 mol of oxygen molecules (O2)

b 1.2 mol of aluminium chloride (AlCl3) c 2.0 mol of nitrogen atoms 8 A small oxygen cylinder carried by an ambulance has an internal volume of 1.42 L. What mass of oxygen is present at a pressure of 15 000 kPa and temperature of 15.0°C? 9 Calculate the mass of the following gases: a 3.5 L of argon at SLC b 250 mL of ammonia (NH3) at STP

15 Analysis by mass

2.2

Finding the composition of a compound Empirical formulas Chemical analysis enables us to determine the empirical formula of a compound. This shows the simplest whole-number ratio of the atoms present in the compound. For example, ethene, which is made up of molecules of C2H4, has an empirical formula CH2 because the simplest ratio of the number of carbon atoms to number of hydrogen atoms is 1 : 2. The empirical formula of glucose, which is composed of C6H12O6 molecules, is CH2O. A crystal of table salt, containing billions of sodium ions and an equal number of chloride ions, has the formula NaCl. Empirical formulas are determined experimentally, usually by finding the mass of each element in a given mass of compound.

Worked example 2.2a Chemical analysis of an oxide of sulfur present in the gaseous emissions from a factory shows that it contains 40.0% sulfur. Find its empirical formula.

Solution A 100 g sample of this compound will contain 40.0 g of sulfur and 60.0 g of oxygen. The ratio of the amounts of sulfur and oxygen is: m(S) m(O) n(S) : n(O) = : M(S) M(O) 40.0 g 60.0 g : = 32.0 g mol–1 16.0 g mol–1 = 1.25 : 3.75 Simplify by dividing both values by the smaller number (i.e. 1.25): =1:3 So, the empirical formula of this oxide of sulfur is SO3. The empirical formula of a compound may also be determined from the quantities of the products formed by the compound in a chemical reaction.

Worked example 2.2b Find the empirical formula of an alkane extracted from natural gas if, upon complete combustion, a sample of the alkane produces 7.75 L of carbon dioxide, measured at SLC, and 7.59 g of water.

Solution This information allows us to calculate the amounts of carbon and hydrogen that were present in the alkane: V 7.75 L = = 0.316 mol n(CO2) = Vm 24.5 L mol–1 There is only one carbon atom in each CO2 molecule, so the amount of carbon in the sample is also 0.316 mol.


The amount of water, formed from hydrogen atoms in the alkane, is: m(H2O) 7.59 g = 0.422 mol n(H2O) = = M(H2O) 18.0 g mol–1 Since each water molecule has two hydrogen atoms, there was 0.422 × 2 = 0.844 mol of hydrogen atoms in the sample. The ratio n(C) : n(H) = 0.316 : 0.844 Divide both values by the smaller number (i.e. 0.316): = 1.00 : 2.67 This is not a whole number ratio, so the decimals must be converted to fractions: 2 = 1 : 23 8 =1:3 Multiply both numbers by the lowest common denominator (i.e. 3): =3:8 The alkane has the empirical formula C3H8 (propane).

Worked example 2.2c After heating of 2.95 g of crystals of hydrated magnesium sulfate (MgSO4.xH2O), the mass of the residue is found to be 1.44 g. Find the value of x in the formula of the compound.

Solution The mass of water lost from the sample is 2.95 g – 1.44 g = 1.51 g. The residue is anhydrous magnesium sulfate (MgSO4). m(MgSO4) m(H2O) : So, the ratio n(MgSO4) : n(H2O) = M(MgSO4) M(H2O) 1.44 g 1.51 g : = 120.4 g mol–1 18.0 g mol–1 = 0.0120 : 0.0839 Divide each value by the smaller number (i.e. 0.0120): =1:7 Since the ratio n(MgSO4) : n(H2O) is equal to 1 : x, the formula of the hydrated salt is MgSO4.7H2O.

review Anhydrous means ‘without water’; hydrated means ‘with water’. The water molecules in a hydrated salt are incorporated into the crystal structure and are referred to as water of crystallisation.

! Many materials are sensitive to atmospheric moisture. Sachets of chemicals such as silica gel that absorb water are often placed inside the packaging of moisture-sensitive goods. Substances that absorb atmospheric moisture are described as hygroscopic. Can you suggest why some people put rice in salt shakers?

Worked example 2.2d An organic compound formed by fermentation of sugar is known to contain only the elements carbon, hydrogen and oxygen. It is analysed by burning a 1.00 g sample in air. If 1.91 g of carbon dioxide and 1.17 g of water are produced, determine the empirical formula of the compound.

Solution m(CO2) 1.91 g = 0.0434 mol = M(CO2) 44.0 g mol–1 Since each carbon dioxide molecule has one carbon atom, there is 0.0434 mol of carbon atoms in the sample. m(H2O) 1.17 g n(H2O) = = 0.0650 mol = M(H2O) 18.0 g mol–1 Since each water molecule has two hydrogen atoms, there is 0.0650 × 2 = 0.130 mol of hydrogen atoms in the sample. To find the amount in mol of oxygen in the sample, the mass of the oxygen must first be deduced. This is found by subtracting the mass of the other two elements, hydrogen and carbon, from the total mass of the sample. n(CO2) =

17 Analysis by mass

m(O) = m(sample) – (m(C) + m(H)) = 1.00 – ((n(C) × M(C)) + (n(H) × M(H)) = 1.00 – ((0.0434 × 12.0) + (0.130 × 1.0)) = 1.00 – (0.521 + 0.130) = 0.349 g m(O) 0.349 g = = 0.0218 mol n(O) = M(O) 16.0 g mol–1 The ratio n(C) : n(H) : n(O) = 0.0434 : 0.130 : 0.0218 Divide each value by the smallest number (i.e. 0.0218): =2:6:1 The compound has the empirical formula C2H6O.

Molecular formulas STOICHIOMETRY • Percentage composition • Empirical and molecular formulas

A molecular formula gives the actual number of atoms of each element present in a molecule of a compound. The molecular formula of a compound is not necessarily the same as the empirical formula. The molecular formula is always a whole-number multiple of the empirical formula. For example, the empirical formula of glucose is CH2O and its molecular formula is C6H12O6, six times larger. A molecular formula can be deduced from an empirical formula if the molar mass is known. In modern laboratories, mass spectrometers are used to determine the molar mass (see Chapter 8).

Worked example 2.2e A compound has the empirical formula CH and a molar mass of 78 g mol–1. What is its molecular formula?

Solution The molecule must contain a whole number of CH units. M(CH) = 13 g mol–1 molar mass of compound 78 g mol–1 number of CH units in a molecule = =6 = molar mass of one unit 13 g mol–1 The molecular formula of the compound is therefore C6H6.

summary

2.2

Th The empirical i i l formula f l indicates i di t the th simplest i l t whole-number h l b ratio of atoms present in a compound. Empirical formulas can be determined by gravimetric analysis. The analysis of combustion products of an organic compound can be used to determine its empirical formula.

A molecular l l formula f l gives i the th actual t l number b off atoms t off eachh element present in a molecule of the compound. The molecular formula can be determined from an empirical formula if the molar mass is known.

key questions 10 Determine the percentage composition of the following compounds: a lead(IV ) oxide (PbO2) b sodium carbonate (Na2CO3) 11 A gaseous hydrocarbon that is used as a fuel for hightemperature welding of metals contains 92.3% carbon. a Determine its empirical formula. b If the molar mass of the hydrocarbon is 26 g mol–1, find its molecular formula.


12 When 1.66 g of tungsten (W) is heated in excess chlorine gas, 3.58 g of tungsten chloride is produced. Find the empirical formula of tungsten chloride. 13 A sample of blue copper(II ) sulfate crystals weighing 2.55 g is heated and decomposes to produce 1.63 g of anhydrous copper(II) sulfate. Show that the formula of the blue crystals is CuSO4.5H2O.

2.3

Calculating masses of reactants and products The mole is particularly useful for calculating the quantities of substances consumed or produced in chemical reactions. Consider the reaction between solutions of lead nitrate and potassium iodide. When the solutions are mixed, a bright yellow precipitate of lead iodide is formed. The equation for this reaction is:

review A precipitate is a solid formed when two solutions are mixed.

Pb(NO3)2(aq) + 2KI(aq) –—h PbI2(s) + 2KNO3(aq)

The coefficients indicate the relative numbers of particles or moles. We can write mole ratios as: n(Pb(NO3)2) 1 n(Pb(NO3)2) 1 n(KI) 2 = and = and = 2 1 n(KI) n(PbI2) n(PbI2) 1 Mole ratios can be used to calculate masses of reactants and products.

Worked example 2.3a Calculate the mass of lead iodide that can be made from 30.0 g of potassium iodide.

Solution Step 1: Write a balanced equation. Pb(NO3)2(aq) + 2KI(aq)) → PbI2(s) + 2KNO3(aq) Step 2: Calculate the amount, in mol, of potassium iodide consumed. m(KI) 30.0 g n(KI) = = 0.181 mol = M(KI) 166 g mol−1 Step 3: Use the ratio of amounts of substances to calculate the number of mole of lead iodide formed. From the equation, 1 mol of PbI2 is produced from 2 mol of KI. n(PbI2) 1 = So, the ratio n(KI) 2 1 n(PbI2) = × n(KI) 2 1 = × 0.181 mol = 0.0905 mol 2 Step 4: Calculate the mass of lead iodide. m(PbI2) = n(PbI2) × M(PbI2) = 0.0905 mol × 461 g mol–1 = 41.7 g So, the reaction can produce 41.7 g of lead iodide from 30.0 g of potassium iodide.

Worked example 2.3b The active ingredient of an antacid powder is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g sample of the antacid produced 714 mL of carbon dioxide, measured at 22.0°C and 101.3 kPa pressure. Calculate the percentage of magnesium carbonate in the powder.

19 Analysis by mass

!

Solution

Remember that pressure must be in kPa, volume in L and temperature in K.

Step 1: The reaction to produce the CO2 is: MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l) Step 2: Since pV = nRT pV 101.3 × 0.714 n(CO2) = = = 0.0295 mol RT 8.31 × (273 + 22) Step 3: Hence n(MgCO3) = n(CO2) = 0.0295 mol Step 4: m(MgCO3) = n(MgCO3) × M(MgCO3) = 0.0295 mol × 84.3 g mol–1 = 2.49 g Therefore, percentage MgCO3 in the 3.50 g sample 2.49 g × 100 = 71.1% = 3.50 g So, the antacid contains 71.1% magnesium carbonate.

STOICHIOMETRY • Stoichiometry and chemical formula

Calculations involving excess reactants Calculations become more complex if the reactants are not present in their stoichiometric ratio. In these cases you must determine which reactant is completely consumed in the reaction (called the limiting reactant) and which one is present in excess. The amount of limiting reactant determines how much product is formed. Figure 2.5 show a series of test-tubes in which 3.0 mL of lead nitrate solution has different amounts of potassium iodide added to it. The height of the lead iodide precipitate gives a good indication of the amount of PbI2 formed. In each test-tube the same amount of lead nitrate is present (0.0010 mol), and progressively increasing amounts of potassium iodide solution are added to it (0.0005 mol, 0.0010 mol, 0.0015 mol, 0.0020 mol, 0.0025 mol, 0.0030 mol). In the first three test-tubes, a–c, the limiting reactant is KI and so the amount of precipitate formed increases steadily as the amount of KI increases. In the last two test-tubes, e and f, Pb(NO3)2 is the limiting reactant and so the same amount of precipitate forms in both cases. In test-tube d, the two reactants are present in their stoichiometric ratio. The height of PbI2 precipitate is graphed against amount of KI added (Figure 2.4).

Height of PbI2 precipitate (mm)

25

a

b

c

d

e

f

20 15 10 5

0.001 0.002 Amount of KI added (mol)

0.003

Figure 2.4 Graph of the height of PbI2 precipitate formed versus the amount of KI added to Pb(NO3)2.


Figure 2.5 The height of lead iodide formed when various amounts of potassium iodide are added to lead nitrate solution.

Worked example 2.3c Calculate the mass of silver bromide that can be formed if a solution containing 15.0 g of silver nitrate is allowed to react with a solution containing 10.0 g of calcium bromide.

Solution Step 1: 2AgNO3(aq) + CaBr2(aq) → 2AgBr(s) + Ca(NO3)2(aq) Step 2: We can calculate the amounts of both reactants. m(AgNO3) 15.0 n(AgNO3) = = = 0.0883 mol M(AgNO3) 169.9 m(CaBr2) 10.0 = = 0.0500 mol n(CaBr2) = M(CaBr2) 199.9 Since the equation shows that 2 mole of AgNO3 reacts with 1 mole of CaBr2, all the AgNO3 will be consumed. AgNO3 is the limiting reactant, CaBr2 is present in excess. Step 3: Use a mole ratio involving silver nitrate (not calcium bromide) to calculate the amount of silver bromide formed. n(AgBr) 1 = n(AgNO3) 1 n(AgBr) = n(AgNO3) = 0.0883 mol Step 4: m(AgBr) = n(AgBr) × M(AgBr) = 0.0883 × 187.8 = 16.6 g So, the reaction can produce 16.6 g of silver bromide. The steps in this calculation can be summarised as shown below. 2AgNO3(aq) 15.0 g ↓ 0.0833 mol

Initially Reacting (same ratio as equation) Finally

summary

0.0833 mol –

+

CaBr2(aq) 10.0 g ↓ 0.0500 mol

→

2AgBr(s)

0.0417 mol ↓ 0.0083 mol excess

→

0.0833 mol ↓ 0.0833 × 187.8 = 16.6 g

+

Ca(NO3)2(aq)

0.0417 mol 0.0417 mol

2.3

The stoichiometry off a chemical reaction can be used to determine the amount of product formed f rmed or the amount of reactant consumed. consumed

key questions 14 Magnesium reacts with hydrochloric acid according to the equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) If 10.0 g of magnesium reacts completely, calculate: a the mass of magnesium chloride that forms b the mass of hydrogen that forms. 15 Iron metal is extracted in a blast furnace by a reaction between iron(III ) oxide and carbon monoxide: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

To produce 1000 kg of iron, calculate: a the mass of iron(III ) oxide required b the volume of carbon dioxide produced at SLC (25°C and 101.3 kPa). 16 A solution containing 10.0 g of silver nitrate is mixed with a solution containing 10.0 g of barium chloride. What mass of silver chloride precipitate is likely to be produced? 2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)

21 Analysis by mass

2.4

Finding the composition of a mixture Most commercial products are mixtures. An analytical chemist may seek to find the percentage of one component in such a mixture. Many ions can be analysed by a technique called gravimetric analysis. This involves forming a suitable precipitate with the ion and using the amount of precipitate to calculate the amount of the ion in the sample. The precipitate should: have a known formula have low solubility be stable when heated (so it can be dried easily) not form precipitates with other ions that are likely to be present. A number of elements routinely analysed by precipitation and gravimetric analysis are listed in Table 2.4.

TABLE 2.4 Precipitates formed for gravimetric analysis Element to be analysed

Precipitate

Compound name

Chlorine

AgCl

Silver chloride

Bromine

AgBr

Silver bromide

Iodine

AgI

Silver iodide

Iron

Fe2O3

Iron(III ) oxide

Phosphorus

Mg2P2O7

Magnesium pyrophosphate

Magnesium

Mg2P2O7

Magnesium pyrophosphate

Sulfur

BaSO4

Barium sulfate

Barium

BaSO4

Barium sulfate

For example, to find the proportion of chloride in a sample of peanut butter, a known mass of the peanut butter would be diluted with water, blended in a food processor and filtered. The amount of chloride in the resulting solution may then be determined by gravimetric analysis (Figures 2.6 and 2.7). The steps in the analysis are summarised in Figure 2.7. In practice, no solid is absolutely insoluble, or infinitely soluble. All ionic solids are soluble to some extent. However, if excess Ag+ ions are added to a solution containing sodium chloride, virtually all the Cl− ions will be precipitated (Table 2.5).


TABLE 2.5 Solubilities of common ionic compounds

Weigh the peanut butter sample, blend and filter.

High solubility

Low solubility

Compounds containing the following ions are generally soluble in water: Na+, K+, NH4+, NO3−, CH3COO− Cl−, Br−, I− (unless combined with Ag+ or Pb2+) SO42− (except PbSO4 and BaSO4; Ag2SO4 and CaSO4 are slightly soluble)

Compounds containing the following ions are generally insoluble, unless combined with Na+, K+ or NH4+: CO32−, PO43−, S2−, OH− (unless combined with Ba2+ or Sr2+; Ca(OH)2 is slightly soluble)

Add an excess of silver nitrate solution to precipitate Cl– ions as silver chloride.

Filter the white precipitate of silver chloride.

Wash the precipitate with a small volume of water.

Dry the precipitate in an oven at 100°C.

Weigh the AgCl precipitate.

Vacuum pump

1 Weighing the sample

2 Forming a precipitate

Repeat these steps until the mass is constant.

Figure 2.7 Steps in the analysis of the chloride content in peanut butter.

3 Filtering the solution 4 Weighing the dry precipitate

Figure 2.6 Key steps in gravimetric analysis.

Worked example 2.4a Infant cereals should not contain more than 0.3% sodium chloride (common salt). A 7.802 g sample of baby cereal was blended with water and filtered. The solution obtained was mixed with excess silver nitrate solution, causing silver chloride to precipitate. The precipitate was collected by filtration, dried and weighed. A mass of 0.112 g was obtained. What is the percentage of sodium chloride in the baby food, assuming that all the chloride is present as sodium chloride?

Solution The full equation for the reaction is: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) Alternatively, this may be written as an ionic equation: Ag+(aq) + Cl−(aq) → AgCl(s) Calculating the amount in mol of AgCl present in the precipitate: m(AgCl) 0.112 g = n(AgCl) = = 0.000 781 mol M(AgCl) 143.4 g mol−1 From the equation, 1 mole of NaCl yields 1 mole of AgCl. n(NaCl) 1 = n(AgCl) 1 n(NaCl) = n(AgCl) = 0.000 781 mol m(NaCl) = n(NaCl) × M(NaCl) = 0.000 781 mol × 58.5 g mol−1 = 0.0457 g

! Note that this analysis calculates only the amount of chloride present. Other tests would be required to demonstrate that it was all sodium chloride and not potassium chloride or some other salt.

chemfact When a reaction occurs in solution, only ions that are involved in the reaction appear in an ionic equation. Other ions present are called spectator ions and are omitted. To convert a full equation to an ionic equation, first rewrite it with the soluble ionic compounds dissociated into ions. Ions that remain in the solution and are unchanged at the end of the reaction are then deleted from both sides of the equation. For the reaction in Worked Example 2.4a: Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq) simplifies to Ag+(aq) + Cl−(aq) → AgCl(s)

23 Analysis by mass

STOICHIOMETRY • Relative atomic and molecular mass • Relative molecular mass • The mole • Balancing chemical equations • Interpreting stoichiometric calculations • Reaction between barium and water • Reduction of iron(III) oxide • Oxidation of hydrogen sulfide • Burning hexane

So the 7.802 g sample contains 0.0457 g of NaCl. 0.0457 g × 100 Therefore, % NaCl = = 0.5857% 7.802 g So, the percentage of sodium chloride by mass in the baby cereal is 0.586%, which is well above the regulatory limit of 0.3%. Sometimes the substance being measured needs to be converted to a more convenient form for analysis.

Worked example 2.4b The content of saccharine (C7H7NO3S) in diet sweetener tablets can be determined by oxidising the sulfur to sulfate and precipitating it as barium sulfate (BaSO4). A 0.607 g sample yields 0.3196 g barium sulfate. What was the percentage of saccharine in the sample?

review

Solution

The accuracy of an answer is limited to the same number of significant figures as there are in the least accurate piece of data used in the calculation. In this question, the least accurate data is the mass of the precipitate, which has three significant figures. The answer is therefore accurate only to three significant figures.

! Notice that no equation has been written for this reaction. Apart from the fact that it would not be easy to write an equation using the information supplied, all we really need to know is the mole ratio between C7H7NO3S and BaSO4.

summary

m(BaSO4) 0.3196 g = = 0.001 369 mol M(BaSO4) 233.4 g mol−1 1 mole of C7H7NO3S yields 1 mole of BaSO4 (as the number of sulfur atoms in each compound must be the same). n(C7H7NO3S) 1 So, the ratio = n(BaSO4) 1 n(C7H7NO3S) = n(BaSO4) = 0.001 369 mol m(C7H7NO3S) = n(C7H7NO3S) × M(C7H7NO3S) = 0.001 369 mol × 185.2 g mol−1 = 0.2535 g So, the 0.607g sample contains 0.2535 g of C7H7NO3S. Therefore, 0.2535 g × 100 % C7H7NO3S = = 41.76% 0.607 g The percentage of saccharine in the tablets is 41.8% (to three significant figures). n(BaSO4) =

2.4

Th The mass off precipitate i it t fformedd in i a precipitation i it ti reaction ti can be used to determine the purity or composition of some substances. The solid precipitate is heated to constant mass to ensure that it is dry.

The Th composition iti off a mixture, it th the purity it off a compoundd andd empirical formula can be determined by gravimetric analysis. Gravimetric analysis makes use of chemical ideas such as the mole concept, general gas equation and the degree of solubility of precipitates.

key questions 17 A chemist determined the salt content of a sausage roll by precipitating chloride ions as silver chloride. If an 8.45 g sample of sausage roll yielded 0.636 g of precipitate, calculate the percentage of salt in the food. Assume that all the chloride is present as sodium chloride.


18 An impure sample of iron(II ) sulfate, weighing 1.545 g, was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate was 0.315 g, calculate the percentage of iron in the sample.

02 amount of substance anhydrous Avogadro’s number empirical formula general gas equation

key terms gravimetric analysis heating to constant mass ionic equation limiting reactant molar mass

molar volume of a gas mole molecular formula precipitate spectator ions

The mole 19 Find the amount in mol of: a Ca atoms in 60.0 g of calcium b NH3 molecules in 22 g of ammonia c H2O molecules in 20.0 g of CuSO4.5H2O d Cl− ions in 34 g of FeCl3 20 Find the mass of: a 0.30 mol of zinc atoms b 0.16 mol of iron(III ) oxide (Fe2O3) c 1.5 mol of ammonium phosphate ((NH4)3PO4) 21 A brand of toothpaste contains 0.22% by mass sodium fluoride (NaF). Calculate the mass of fluoride ions in a tube containing 120 g of the paste.

27

28

General gas equation 22 6.00 g of helium gas was blown into a fairground balloon. On the day, the temperature was 28.0°C and the pressure inside the balloon was 103.4 kPa. Assuming it is infinitely elastic, to what volume did the balloon inflate? 23 Calculate the volume of the following gases: a 1.50 mol of oxygen at STP b 28.0 g nitrogen at STP c 17 g of sulfur dioxide at SLC d 1.23 × 1022 atoms of helium at SLC 24 A 1.22 g sample of pure gas extracted from the gases from a car exhaust occupied 991 mL at 24.0°C and 1.00 atmosphere pressure. a Calculate the amount of gas, in mol, present in the sample. b What is the molar mass of the gas? c Suggest the identity of the gas.

29

30

Reacting quantities 25 Solutions of silver nitrate and potassium chromate react to produce a red precipitate of silver chromate: 2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq) If 0.778 g of precipitate is formed in a reaction, find: a the mass of potassium chromate that reacted b the mass of silver nitrate that reacted. 26 Magnesium in distress flares burns in air according to the equation: 2Mg(s)+ O2(g) → 2MgO(s)

31

standard temperature and pressure (STP) standard laboratory conditions (SLC) water content

If 10.0 g of magnesium bums in air, calculate: a the mass of magnesium oxide produced b the mass of oxygen that reacts. Lithium peroxide may be used as a portable oxygen source for astronauts. Calculate the volume of oxygen gas, measured at 25°C and pressure of 101.3 kPa, that is available from the reaction of 0.500 kg of lithium peroxide with carbon dioxide according to the equation: 2Li2O2(s) + 2CO2(g) → 2Li2CO3(s) + O2(g) A power station burns coal at 45.5 tonnes per hour (1 tonne = 106 g). Assuming the coal is pure carbon and that all the coal is oxidised completely to carbon dioxide gas on combustion, what volume of carbon dioxide is released to the atmosphere per hour when the atmospheric pressure is 758 mmHg and the temperature is 19.0°C? Silicon steel is an alloy of the elements iron, carbon and silicon. An alloy sample was reacted with excess hydrochloric acid and the following reaction occurred: Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) The carbon and silicon in the alloy did not react with the acid. If an alloy sample with a mass of 0.160 g produced 62.0 mL of hydrogen gas, measured at SLC, calculate the: a amount of hydrogen evolved in the reaction b mass of iron that reacted to produce this amount of hydrogen c percentage of iron in the alloy. Explain the meanings of the following terms: a relative atomic mass b relative molecular mass c mole d Avogadro’s number e molar mass f precipitate g gravimetric analysis h ionic equation 16.0 g of hydrogen sulfide is mixed with 20.0 g of sulfur dioxide and they react according to the equation: 2H2S(g) + SO2(g) → 2H2O(l) + 3S(s) a What mass of sulfur is produced? b What mass of reactant is left after the reaction?

25 Analysis by mass

32 Calculate the volume of carbon dioxide gas produced, at SLC, when 5.00 g of calcium carbonate is added to a solution containing 5.00 g of nitric acid.

Empirical and molecular formulas 33 The following compounds are used in fertilisers as a source of nitrogen. Calculate the percentage of nitrogen, by mass, in: a ammonia (NH3) b ammonium nitrate (NH4NO3) c urea (CO(NH2)2) 34 Find the empirical formula of: a a compound that contains 65.2% scandium and 34.8% oxygen by mass b an oxide of copper that contains 89% copper by mass c a polymer used to make plastic drain pipes, which contains 38.4% carbon, 4.84% hydrogen and 56.7% chlorine by mass. 35 Gypsum is hydrated calcium sulfate (CaSO4.xH2O). A residue of 5.65 g of anhydrous calcium sulfate is obtained by heating 7.15 g of gypsum. Determine the empirical formula of gypsum. 36 A 2.203 g sample of an organic compound that contains oxygen, Cx Hy Oz , was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidised to 3.23 g of carbon dioxide. a Find the empirical formula of the compound. b Another sample was analysed in a mass spectrometer. The mass spectrum produced showed that the molar mass of the compound was 60.0 g mol−1. What is its molecular formula?

Gravimetric analysis 37 What mass of barium chloride (BaCl2) will remain after a 15.0 g sample of the hydrated salt BaCl2.2H2O is heated to drive off all of the water? 38 A student is given solutions of lead(II ) nitrate, copper(II ) chloride and barium hydroxide. a Using Table 2.5, name the precipitates that could be made by mixing together pairs of solutions. b Write full and ionic equations for each of the reactions. 39 Design a flowchart to show how the salt content of a savoury spread could be determined by gravimetric analysis. 40 The iodide ions in a solution containing 0.300 g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed? 41 A precipitate of Fe2O3, of mass 1.43 g, was obtained by treating a 1.5 L sample of bore water. What was the concentration of iron, in mol L−1, in the water? 42 The chlorine in a 0.63 g sample of a chlorinated pesticide, DDT (C14H9Cl5), is precipitated as silver chloride. What mass of silver chloride is formed? 43 A 0.693 g sample of a silver alloy used to make cutlery is dissolved completely in nitric acid. Excess sodium chloride

26 62 Chemical analysis C

solution is added to produce a precipitate of silver chloride. The precipitate is filtered, dried and found to weigh 0.169 g. a Find the percentage of silver in the alloy. b If the precipitate was not completely dry when weighed, what effect would this have on the answer for part a? 44 A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium sulfate (Al2(SO4)3) were dissolved in a volume of water and excess barium chloride was added to precipitate barium sulfate. What was the total mass of barium sulfate produced? 45 Water pollution can result from the phosphates added to washing powders to improve the stability of their suds. The phosphorus in a 2.0 g sample of washing powder is precipitated as Mg2P2O7. The precipitate weighs 0.085 g. a What is the percentage, by mass, of phosphorus in the washing powder? b Suppose you were in charge of an advertising campaign to promote the washing powder. Would you advertise the percentage of phosphorus or phosphate in the product? Explain.

Connecting the main ideas 46 A 2.10 g sample of a commercial antacid powder is treated with excess hydrochloric acid. The volume of carbon dioxide evolved is 430 mL, measured at 21.0°C and 109.6 kPa pressure. If magnesium carbonate is the active ingredient in the antacid, calculate the percentage of magnesium carbonate in the sample. 47 When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of phosphorus is produced. The molar mass of the oxide is 284 g mol−1. a Determine the empirical formula of the phosphorus oxide. b Determine the molecular formula of the phosphorus oxide. 48 Excessive salt intake in the diet can contribute to high blood pressure and heart disease. The salt content of a 14.96 g sample of powdered chicken soup was determined by dissolving it in water to make a volume of 250.0 mL. A 20.00 mL volume of this stock solution was pipetted into a conical flask and excess silver nitrate was added. The silver chloride precipitate that formed was then filtered, washed and dried. Its mass was 0.246 g. a Write an ionic equation for the formation of the silver chloride precipitate. b Calculate the amount, in mol, of silver chloride that was produced. Assume all the chloride in the powdered soup came from sodium chloride (common salt). c Determine the amount of sodium chloride in the 20.00 mL volume of stock solution in the conical flask. d Calculate the amount of sodium chloride in 250.0 mL of the stock solution. e What mass of sodium chloride was in the sample?

0

chapter 03

Volumetric analysis

key knowledge • volumetric analysis: simple titrations • calculations including concentration

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • discuss the principle of volumetric analysis • calculate the concentration of a solute dissolved in solution, in mol L−1 or g L−1 • list the characteristics of a primary standard and give examples of primary standards • describe how a standard solution is prepared from a primary standard • describe the steps in volumetric analysis • distinguish between end point and equivalence point • identify the equipment used in volumetric analysis.

STOICHIOMETRY • Molar concentrations

This chapter describes a method of analysis that involves chemical reactions between substances in aqueous solution. In volumetric analysis the amount, in mol, of the substance being analysed is determined by measuring the volume of the test solution required to react completely with a volume of solution of known concentration. This method is usually faster and more convenient than gravimetric analysis.

3.1

Concentration Figure 3.1 A memory aid showing the relationship between concentration c in mol L−1, the amount in moles (n ) and the volume (V ) in litres. n n n=c×V V= c= V c

Concentration is a measure of the amount of solute in a specified volume of solvent. Chemists often express concentration as molar concentration, referred to as molarity and defined as the amount of solute, in mole, dissolved in one litre of solution.

or

amount of solute, in mol concentration = volume of solution, in L n c = V

The unit of molarity is mol L−1, which is given the symbol M. For example, 1.00 L of a 2.00 M solution of NaCl contains 2.00 mol of dissolved NaCl. Such a solution is commonly referred to as 2.00 molar. Another concentration measure is mass per unit volume, e.g. grams of solute dissolved in each litre of solution (g L−1), easily calculated by dividing the mass of solute, in grams, by the solution volume, in litres. To convert a concentration measured in mol L−1 to g L−1, multiply the molarity by the molar mass of the solute. A 2.00 M NaCl solution has concentration of 2.00 mol L−1 × (23.0 + 35.5) g mol−1 = 117 g L−1.

chemistryy in action Units of concentration Other ways of expressing concentration include: parts per million (ppm), i.e. gram per million gram parts per billion (ppb), i.e. gram per billion gram percentage mass/volume (%w/v), i.e. gram per 100 mL percentage by mass (%w/w), i.e. gram per 100 g percentage by volume (%v/v), i.e. mL per 100 mL. Parts per million (ppm) and parts per billion (ppb) are very low concentrations. They are used for measures of air pollution or effluent contamination. For example, the amount of nitrogen dioxide in the air at a particular site might be given as 0.034 ppm or 34 ppb. Percentage mass/volume (%m/v or %w/v), percentage mass (%m/m or %w/w) and percentage volume (%v/v) are common units for commercial products. Figure 3.2 The label on White King Premium Bleach says ‘Active ingredient: Sodium hypochlorite 42 g/L (Available chlorine 4.0% m/v) At use by date: Available chlorine 2.0% m/v. Sodium hydroxide 9 g/L.’ The label on PineOCleen Disinfectant says ‘Active ingredient: Benzalkonium chloride 1.5% w/w.’


!

Standard solutions A solution with an accurately known concentration is called a standard solution. Preparing a standard solution is not a simple matter of dissolving a measured mass of a substance in water to give a known volume of solution. In practice, this is possible for only a few of the chemicals encountered in the laboratory. Many chemicals are impure because they decompose or react with chemicals in the atmosphere. Substances that are so pure that the amount of substance, in mole, can be calculated accurately from their mass are called primary standards. A primary standard should: be readily obtainable in a pure form have a known formula be easy to store without deteriorating or reacting with the atmosphere have a high molar mass to minimise the effect of errors in weighing be inexpensive. Examples of primary standards are: bases: anhydrous sodium carbonate (Na2CO3), sodium borate (Na2B4O7.10H2O) acids: hydrated oxalic acid (H2C2O4.2H2O) and potassium hydrogen phthalate (KH(C8H4O4)). A standard solution can be prepared from a primary standard by dissolving an accurately known mass to make up an accurately known volume of solution. Precise volumes of solutions are obtained using volumetric flasks (Figure 3.4). The common strong acids and bases cannot be made directly into standard solutions. The concentrations of commercial supplies of hydrochloric, nitric and sulfuric acids are not accurately specified, and strong bases such as sodium hydroxide and potassium hydroxide are usually impure as they absorb water and react with carbon dioxide in the air.

Suppose you wanted to make one litre of an aqueous solution of sodium chloride. You would weigh out the required amount of sodium chloride and place it in a volumetric flask, then add distilled (de-ionised) water to make the volume up to one litre. You make the volume up to one litre rather than add one litre of water to allow for volume changes caused by the attractions between solute and solvent particles.

50 mL distilled water

50 mL ethanol

98 mL

100 mL measuring cylinder

Figure 3.3 Ethanol and water molecules are attracted together by hydrogen bonding, dipole–dipole and dispersion forces. Look at the result!

STOICHIOMETRY • Volumetric analysis: equipment • Volumetric analysis: calculations • Flow chart

1 Place weighed sample in volumetric flask.

2 Half fill with water. Shake to dissolve the sample.

3 Add water to the calibration line. Shake again.

Figure 3.4 Steps in preparing a standard solution. A known mass of solute is used to make a known volume of solution.

29 Volumetric analysis

summary

3.1

C Concentration t ti is i a measure off the th amountt off solute l t dissolved di l d in i a specified volume of solution. Molarity (M) is one measure of concentration. It is the amount, in mol, of solute dissolved in one litre of solution. The n concentration, c mol L−1, is given by c = , where n = number V of mole and V = volume in litres.

A primary i standard t d d has h a high hi h level l l off purity, it enabling bli the number of moles to be determined from its mass. A standard solution has a known concentration and is prepared by dissolving a known mass of a primary standard in an accurately known volume of solution.

key questions 1 Calculate the molarity of these solutions: a 1.5 mol of HCl dissolved in 3.0 L of solution b 0.64 g of H2SO4 dissolved in 500 mL of solution c 2.1 g of NaHCO3 dissolved in 1.00 L of solution 2 Calculate the amount, in mol, of solute present in: a 20.00 mL of 0.255 M KOH solution b 2.50 L of 1.05 M sucrose (C12H22O11) solution c 25.00 mL of 0.0200 M AgNO3 solution 3 Calculate the mass of solute present in these solutions: a 100.0 mL of 0.50 M NaOH b 20.00 mL of 1.50 M CuSO4 c 10.00 mL of 0.10 M KSCN 4 Calculate the concentration, in g L−1, of the following solutions: a sports drink containing 6.0 g of sucrose in each 100 mL (i.e. 6.0% w/v sucrose) b mouthwash containing 1.50 mg of benzoic acid per mL c vinegar that is a 0.95 M solution of ethanoic acid (CH3COOH)

5 Calculate the molarity of solutions with the following concentrations: a 5.884 g L−1 K2Cr2O7 b 11.6 g L−1 KCl c 1.50 mg of benzoic acid (C6H5COOH) per mL of solution 6 For 250 mL of a 0.20 M solution of potassium sulfate, K2SO4, calculate the amount, in mol, of: a potassium ions, K+ b sulfate ions, SO42− c oxygen atoms 7 The seal enclosure at the Melbourne Zoo contains 455 000 L of salt water. The water in the pool is maintained at a sodium chloride concentration of 0.048 M. What total mass of sodium chloride is present in the water when the pool is completely full? 8 During the preparation of the standard solution shown in Figure 3.4, why is water added to the level of the calibration mark on the flask after the solid has dissolved, rather than before?

3.2

Volumetric analysis There are many situations where we want to find the amount of a substance in solution, for example the chemicals present in industrial effluent, the composition of antacid tablets or the acidity of polluted rainwater. Volumetric analysis involves reacting a measured volume of a standard solution with a measured volume of the solution of unknown concentration. Ideally, the solutions are mixed until they have just reacted completely, in the mole ratio indicated by the stoichiometric equation. This process is known as performing a titration. The steps involved in an acid–base titration are shown in Figure 3.5. Volumetric analysis (titrations) involve the following steps: 1 A known volume of one of the solutions is measured using a pipette and transferred into a conical flask. The volume measured out by the pipette is called an aliquot. You should always draw the liquid into a pipette using a safety filler. Do not use your mouth.


Acid

Conical flask Indicator

1 Burette is filled with one reactant.

2 Pipette delivers a known volume of the other reactant into flask.

3 Indicator has been added.

4 The end point has been reached.

Figure 3.5 Steps in an acid–base titration.

2 Th The other h solution l is d dispensed d slowly l l into the h titration flask k from f a burette. 3 The equivalence point is the point during the titration when the solutions have been mixed in the mole ratio shown by the reaction equation. The volume of solution delivered by the burette is known as the titre. 4 To minimise errors, the titration is repeated several times and the average titre found. For example, in the reaction: HCl(aq) + NaOH(aq) –—h NaCl(aq) + H2O(l)

the equivalence point is reached when the reactants have been mixed in the mole ratio given by the equation (the stoichiometic ratio). At this point n(HCl) = n(NaOH) = 1. Similarly, in the reaction:

Graduation line

2HCl(aq) + Ca(OH)2(aq) –—h CaCl2(aq) + 2H2O(l)

the equivalence point is reached when the mole ratio n(HCl(aq) ) : n(Ca(OH)2(aq) ) = 2 : 1. In this case, the equivalence point is reached when n(HCl(aq) ) = 2n(Ca(OH)2(aq) ).

Identifying the equivalence point

Figure 3.6 A pipette must always be filled using a safety pipette filler. When the pipette is filled correctly, the meniscus sits on the graduation line.

If the solutions being titrated are colourless, as is often the case in acid–base titrations, an indicator must be added to the reaction mixture to detect the equivalence point (Chapter 4). The indicator must be chosen carefully to ensure that the point during the titration at which the indicator changes colour, called the end point, closely matches the equivalence point of the reaction. With redox reactions (Chapter 5), the end point can often be identified by a colour change in one of the reactants. To minimise the effect of errors, titrations are repeated several times and the results averaged. Once the unknown concentration has been determined by volumetric analysis, the solution can be used as a standard solution for other titrations.


How accurate are our measurements? 19

20

21

Figure 3.7 The volume of solution in a burette is read going down from the top of the burette.

!

The accuracy with which the volumes of the aliquot and titre are measured depends on the calibration of the pieces of equipment used. Analytical pipettes deliver fixed, accurate volumes of liquid. There are always errors associated with measurements made during experimental work. Typical uncertainties associated with volumetric analysis are: 20 mL pipette ±0.05 mL burette ±0.02 mL for each reading 250.0 mL volumetric flask ±0.3 mL. Burettes are calibrated in intervals of 0.1 mL. The meniscus is usually read to the bottom curve and the volume can be estimated to the nearest 0.02 mL. If the meniscus lies exactly on a line, it should be recorded to the second decimal place (e.g. 18.40 mL) to indicate this. It is usual to keep repeating titrations until three concordant titres are obtained, that is, three titres that differ by a maximum of 0.10 mL from highest to lowest. This is because the volume of one single drop from a burette is about 0.05 mL, so taking the average of three results assumes that they will be no more than one drop over and one drop under the accurate titre. Look at the set of sample results in Table 3.1.

TABLE 3.1 Sample titration results

The last figure recorded as your burette reading should be the one where you are uncertain— you have to estimate where the meniscus falls between calibration lines.

Titration number

1

2

3

4

5

Final burette reading (mL)

20.20

40.82

20.64

41.78

21.86

Initial burette reading (mL)

0.00

21.00

1.00

22.00

2.00

20.20

19.82

19.64

19.78

19.86

Titre (mL)

The first reading was obviously out of range and was just a rough reading to give the person doing the titration an idea of approximately where the end point was going to fall. The three titres shown in bold are the concordant titres. Difference between the highest and lowest readings is 19.86 − 19.78 = 0.08 mL, i.e. within the acceptable range for concordant results. The mean titre to use in calculations is: 19.82 + 19.78 + 19.86 = 19.82 mL 3

extension Measurement errors Wh conducting When d ti analysis l i chemists h i t often ft measure suchh quantities titi as temperature, mass and volume. There are errors associated with nearly all measurements carried out in the laboratory, and these need to be considered in any calculations. Experimental error can be either systematic errors or random errors. Systematic errors result from an equipment fault, for example an incorrect calibration on a measuring cylinder.


Wh you repeatt a measurementt severall titimes, you may When notice that not all the measurements are the same and that they vary slightly. These variations are called random errors. They may arise because of difficulty in judging the fraction between two scale markings on a burette, for example. By making repeated measures the effects of random errors can be reduced.

question

?

E1 A student sttuddentt nnoted ottedd that her 50 mL burette was dirty and that droplets of liquid stuck to the inside surface of the tube tube. She decided to investigate if this would affect the burette’s accuracy. She filled the burette to the zero mark and drained a 50.00 mL titre into a flask that had previously been weighed. The flask was weighed again and the mass of water calculated. The student then calculated the volume of water by assuming that the density of water is 1.000 00 g mL−1. This procedure was repeated another four times. The masses of water for the five trials were 49.95 g, 49.93 g, 49.90 g, 49.93 g and 49.92 g. a What is the average mass of the five water samples? b Calculate the average volume of these samples. c Is the use of a dirty burette a source of random error or systematic error? d How can the source of this error be removed? The burette was cleaned and the experiment repeated. This time the following masses of water were obtained: 50.02 g, 50.00, 49.98 g, 49.99 g, 50.01 g. e Calculate the average volume of these samples. f Does the variation in volumes indicate a random error or systematic error? Another burette was selected. Before testing it was cleaned thoroughly. The same procedure was used and the average titre was found to be 49.50 g rather than the expected 50.00 g. The error was thought to be due to a poor standard of calibration. g Is this a systematic error or random error? h How should the measurements taken from this burette be adjusted?

summary

3.2

IIn a titration, tit ti a measuredd volume l off a standard t d d solution l ti is reacted with a measured volume of the solution whose concentration is to be determined. The equivalence point of a reaction occurs when the reagents have been mixed in the mole ratio shown by the reaction equation. An indicator may be used to identify the equivalence point of a reaction.

The Th stage t att which hi h an indicator i di t changes h colour l is i called ll d the th end point. For accurate analysis the indicator end point must closely match the reaction equivalence point. Concordant titres vary within narrowly specified limits. Three concordant titres are usually obtained during a titration.

key questions 9 Explain the difference between the following: a standard solution and primary standard b equivalence point and end point c burette and pipette d aliquot and titre. 10 The following paragraph describes an acid–base titration. Some of the key words are missing. Use the list of words in the box to fill in the gaps. pipette measuring cylinder beaker volumetric flask primary standard indicator base standard solution burette indicator titre desiccator aliquot

A sample of anhydrous sodium carbonate of approximately 2 g is weighed accurately. (The solid must be dry if it is to be .) The solid is tipped into used as a a and shaken with about 50 mL of distilled water until the solid dissolves. More water is added to make the solution to a volume of exactly 100.0 mL. A 20.00 mL of the solution is taken by using a and placed in a conical flask. A few drops of methyl orange are added and the mixture is titrated against dilute hydrochloric acid.


03

key terms aliquot average titre burette concentration concordant titres

conical flask end point equivalence point indicator molarity

pipette primary standard standard solution titration titre

uncertainties volumetric analysis volumetric flask

STOICHIOMETRY • Molar concentrations • Barium hydroxide solution • Sodium carbonate solution

Co oncentration Concentration 11 Calculate the molarity of: a 250.0 mL solution containing 1.42 g of potassium chloride (KCl) b 100.0 mL solution containing 0.63 g of anhydrous sodium carbonate (Na2CO3 ) c 250.0 mL solution containing 0.400 g of sodium oxalate (Na2C2O4 ) 12 What mass of solute is required to prepare the following standard solutions? a 250 mL of 0.500 M sodium oxalate (Na2C2O4 ) b 100 mL of 0.250 M potassium hydrogen phthalate (KH(C8H4O4 )) 13 40.4 g of Fe(NO3)3.9H2O is dissolved in sufficient water to make up one litre of solution. Find the concentration of each of the following ions in the solution: a iron(III ) ions b nitrate ions 14 a Convert the following molar concentrations to units of g L−1: i 0.0024 M NaCl solution ii 6.3 × 10−5 M Pb(NO3)2 solution b Convert the following concentrations in g L−1 to units of mol L−1: i 0.068 g L−1 dissolved O2 in tap water ii 0.32 g L−1 cadmium ions in seawater c Convert the following molar concentrations to units of ppm: i 0.0036 M Ca(OH)2 solution ii 2.9 × 10−6 M Cd(NO3)2 solution 15 A carton of orange juice lists among its contents: Total carbohydrates: 8.5 g/100 mL Vitamin C: 40 mg/100 mL


16

17 18

19

20

21

a Express the concentration of each solute in g L−1. b Why is it also possible to express the concentration of vitamin C as a molarity, but not possible to do so for the carbohydrates? Each 200 mL of an electrolyte solution designed for treating dehydration contains 0.47 g of sodium chloride (NaCl), 0.30 g of potassium chloride (KCl) and 3.56 g of glucose (C6H12O6). a Calculate the concentration of each compound, in g L−1. b Calculate the molarity of each compound. c Calculate the concentration of potassium ions in the solution, in mol L−1. d Calculate the concentration of chloride ions in the solution, in mol L−1. To what volume of water must 10 mL of 8.0 M HCl be added in order to prepare a 0.50 M HCl solution? The label on a laundry stain remover indicates that it contains 50 g/L hydrogen peroxide (H2O2). What is the molarity of the hydrogen peroxide? Schools normally purchase concentrated (14 M) nitric acid and then dilute it for use. What volume is required to prepare 2.0 L of 0.15 M acid? A student wishes to prepare 500 mL of a standard solution of any base of concentration 0.2500 M. a Would it be better to prepare the solution using solid sodium hydroxide or anhydrous sodium carbonate? Explain. b Write instructions to the student for making up the solution. Fluorine compounds are added to Melbourne’s supplies of drinking water to give a concentration of fluoride ions of about 0.90 ppm (1 ppm = 1 g in 106 g). a What amount, in mol, of fluoride is present in 1.0 g of Melbourne water? b How many fluoride ions would you swallow if you drank a 200 mL glass of Melbourne water? (1 mL of water has a mass of 1 g.)

Volumetric analysis 22 Sketch a pipette, showing the calibration line. Clearly indicate the positions of the curved surface (meniscus) of a liquid in the pipette before and after the liquid is delivered. 23 A student is to perform an analysis of sodium hydroxide solution by titrating it with standard hydrochloric acid, as shown in Figure 3.5. Before beginning, the student rinses the glassware that is to be used in the analysis. However, the student does not wish to wait until the glassware has dried before using it. For each of the following apparatus, a, b and c, state if it should be rinsed with: i de-ionised water only ii sodium hydroxide solution only iii hydrochloric acid only. a pipette b burette c conical flask

Connecting the main ideas 24 The concentration of a solution of sodium hydroxide was found by titration with a standard solution of hydrochloric acid. The following steps (not in the correct order) were used when carrying out the titration: I The burette was filled with hydrochloric acid. II The conical flask was rinsed with water. III The burette was rinsed with hydrochloric acid. IV An aliquot of sodium hydroxide was placed in the conical flask. V The level of acid in the burette was read. VI The level of acid in the burette was read. VII A pipette was rinsed with sodium hydroxide solution.

VIII Hydrochloric acid was discharged from the burette until the end point was reached. IX An indicator was added to the sodium hydroxide solution. The titration was repeated several times and the following values were obtained: 25.32, 25.30, 25.34, 25.90, 25.30 mL. a What is meant by the term ‘standard’ in ‘a standard solution of hydrochloric acid’? b Explain why sodium hydroxide solution should not be left standing in a glass burette. c From the list above, list the steps required to complete a titration in a correct order. d A ‘rough’ titration is usually carried out before the first accurate titration. Explain what is meant by a rough titration and why it is used. e Titrations are repeated until concordant results are obtained. Explain what is meant by the term ‘concordant’ in this context. f What is the best value to use for the titre of acid, considering the values given above? g For the titration values 25.32, 25.30, 25.34 mL, what sort of error explains the differences—random or systematic? Explain your answer. 25 Anhydrous sodium carbonate is used as a primary standard in determining the concentration of hydrochloric acid by volumetric analysis. a What criteria are used to determine whether or not a substance is suitable for use as a primary standard? b What mass of anhydrous sodium carbonate is required to prepare 200 mL of a standard solution of sodium carbonate that has a concentration of 0.20 M? c How would you prepare this standard solution?


0

chapter 04

Analysing acids and bases

key knowledge • volumetric analysis, simple and back

titrations, acid–base titrations

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • write balanced chemical equations for acid–base reactions • identify a substance acting as an acid or a base in a given reaction • recall that pH is a measure of the strength of an acid or a base • explain the reasons for the selection of a particular acid–base indicator • perform stoichiometric calculations for volumetric analyses involving acid–base reactions • perform stoichiometric calculations for volumetric analyses that involve back titrations.

chemistryy in action Antacids: curing indigestion Over-eating or eating too quickly may cause the pain in the stomach or chest that signals you are suffering from indigestion. In response, glands in the walls of the stomach produce excessive quantities of acidic gastric juices. Taking antacids, such as those shown in Figure 4.1, can relieve the discomfort caused by these excessive quantities of acid. The stomach has thick, muscular walls that enable food to be churned into a mushy fluid before it passes into the small intestine for further digestion. Gastric juice contains pepsin, a chemical that speeds the breakdown of proteins in food. Pepsin works best in acidic conditions and so gastric juice also contains large amounts of hydrochloric acid. To counter the potential damage that protein-digesting pepsin (an enzyme) and H3O+ ions could inflict on the stomach wall, secretions from other glands produce a thick mucus layer containing basic HCO3– ions to line the stomach. Conditions that produce an imbalance in this process may cause the lining to be eaten away and a crater-like sore, called an ulcer, can develop. Commercial antacid tablets and powders commonly contain bases such as magnesium hydroxide, sodium hydrogen carbonate and aluminium hydroxide. These bases react with (or neutralise) the acid in the stomach. The tablets usually also contain fillers, such as starch or sugar, to bind the powders together and mask the bitter taste of the bases. How can you decide which of the many brands of antacid to buy? In order to compare the effectiveness of the antacids in neutralising acids, you could perform a test to find out how much hydrochloric acid reacts with each brand. In this chapter we will look at methods for performing analyses that involve acid–base reactions.

Figure 4.1 The effectiveness of antacids can be assessed using acid–base reactions.

! Antacids should be used strictly as indicated on the label. A doctor must be consulted if the symptoms persist.

4.1

Acid–base chemistry revisited You will recall from your studies of Unit 2 that acids and bases can be defined in the following ways: Acids are proton donors. Bases are proton acceptors. Acid–base reactions involve the transfer of a proton from an acid to a base. This definition was proposed by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry in 1923.

37 Analysing acids and bases

Hydrochloric acid is a strong acid that ionises almost completely in water, forming H3O+ and Cl– ions: HCl(aq) + H2O(l) –—h H3O+(aq) + Cl–(aq) acid base

This is an example of an acid–base reaction. Notice that during the reaction, a proton (H+) is transferred from the HCl to the H2O. Proton transfer occurs in all acid–base reactions. Substances such as HCl and Cl– that differ by only one proton are called a conjugate acid–base pair. In the above reaction, H2O and H3O+ are also a conjugate acid–base pair. Some acids have more than one proton that can be donated. Sulfuric acid, H2SO4, can donate one or two protons, and is called a diprotic acid. Phosphoric acid, H3PO4, a triprotic acid, can donate up to three protons. Strong acids, such as hydrochloric acid, are acids that readily donate protons. They react almost completely with water to form ions (‘ionise’). Weak acids, such as ethanoic (acetic) acid, ionise only slightly: CH3COOH(l) + H2O(l)

chemfact The double arrow ( ) indicates that a reaction can go backwards as well as forwards. When the forwards and backwards reactions proceed at the same rate, the reaction is said to be in equilibrium (Chapter 16).

summary

CH3COO–(aq) + H3O+(aq)

At any instant, most ethanoic acid molecules are not ionised. Similarly, strong bases are bases that readily accept protons and weak bases accept protons only to a limited extent. Strength of acids and bases is discussed in more detail in Chapter 17. The concentration of acids and bases can be determined experimentally by volumetric analysis. Acid–base titrations utilise the neutralisation reaction between acids and bases to form a salt plus water. The reaction between acids and carbonates to form a salt, water and carbon dioxide is also used in some acid–base titrations.

4.1

AAcids id are proton t donors; d bases b are proton t acceptors. t Acid–base reactions involve a transfer of protons from the acid to the base. The conjugate acid of a base has one more proton than the base. The conjugate base of an acid has one less proton than the acid.

An A acid id will ill reactt with ith a base b to t form f a saltlt plus l water. t When the base reacting with an acid is a carbonate or hydrogen carbonate, the products are carbon dioxide as well as a salt and water.

key questions 1 Antacid tablets should normally be chewed before they are swallowed. Why? 2 A laboratory test to determine how much hydrochloric acid is neutralised by a brand of antacid does not give a complete picture of its effectiveness in the stomach. What other factors might be important? 3 Examine the range of antacids shown in Figure 4.1. If you were to choose one of these from all the others, what features, apart from its ability to neutralise acid, might influence your choice? 4 Many antacids fizz when dissolved in a glass of water. One such brand lists among its ingredients sodium hydrogen


carbonate and citric acid. Write an ionic equation for the reaction between HCO3–(aq) and H3O+(aq) responsible for the ‘fizz’. 5 In each of the following equations: i identify the acids and bases ii name the conjugate acid–base pairs. a NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq) b HSO4–(aq) + H2O(l) → H3O+(aq) + SO42–(aq) c NH4+(aq) + S2–(aq) → NH3(aq) + HS–(aq) d CH3COO–(aq) + H3O+(aq) → H2O(l) + CH3COOH(aq)

4.2 SO3

_

_ SO3

pH The concentration of H3O+ ions in a solution is referred to as the solution’s acidity. For convenience, acidity is measured using a logarithmic scale, called the pH scale. The definition of pH is:

N

N

N

N

N+

N

pH = – log [H3O+] +

+

–1

where [H3O ] is the concentration of H3O ions measured in mol L . Neutral solutions have a pH = 7 at 25°C. Acidic solutions have a pH < 7. Basic solutions have a pH > 7. pH is covered in more detail in Chapter 17.

CH3

CH3

H Acid form

Indicators An indicator is used during acid–base titration to identify the equivalence point of the reaction. An acid–base indicator is a substance whose colour depends on the concentration of H3O+ ions in solution. Indicators are weak acids with their acid form being one colour and their conjugate base being another (Figure 4.2). The colours of common acid–base indicators and the pH range over which they change colour are given Table 4.1.

TABLE 4.1 Common indicators Indicator

CH3

CH3

4.3

Colour of base form Pink

pH range

Phenolphthalein

Colour of acid form Colourless

Methyl orange

Pink

Yellow

3.2–4.4

Bromothymol blue

Yellow

Blue

6.0–7.6

Base form

Figure 4.2 The acid–base forms of methyl orange indicator.

chemfact Universal indicator is mixture of indicators that changes colour as the concentration of H3O+ ions changes (Figure 4.3). It is often used for measuring the pH of solutions.

8.2–10.0

The indicator must be chosen carefully to ensure that the point during the titration where the indicator changes colour, the end point, closely matches the equivalence point of the reaction. Most of the common acid– base indicators change colour within the range of pH values 3–11. If we follow the pH changes that take place during the titration of a solution of a strong base, such as a 0.1 M solution of sodium hydroxide, with a solution of a strong acid, such as 0.1 M hydrochloric acid, we would see a plot similar to that shown in Figure 4.4a. At the end point, addition of a very small volume of strong acid produces a large change in pH. This is referred to as a sharp end point. In the case of the reaction between a strong acid and a strong base the equivalence point lies close to pH 7. When a weak base is titrated with a strong acid, or a strong base is titrated with a weak acid, there is a much more gradual change of pH around the end point (Figure 4.4b). In this situation, only indicators whose colour changes within the end point range can be used. When a strong acid reacts with a weak base, the equivalence point occurs between pH 3 and pH 7. Methyl orange changes colour over this pH range and would be a suitable indicator.

Figure 4.3 Universal indicator colours over the pH range 1 to 12.


a

b

pH p 14

pH p 14

Pink

Pink

Phenolphthalein changes colour

Phenolphthalein changes colour

Colourless

7

Equivalence point

} Sharp end point

Volume of acid

Colourless

7

Equivalence point

Sharp end point with methyl orange

Broad end point with phenolphthalein

Volume of acid

Figure 4.4 pH curves showing change of pH during a titration of a a strong base with a strong acid, and b a weak base with a strong acid. Phenolphthalein, which changes colour in the pH range 8.2–10, gives a sharp end point in a but a broad end point in b. Methyl orange, which changes colour between pH 3.1 and 4.5, would be a more suitable indicator for the second titration.

The equivalence point for a reaction between a weak acid and a strong base occurs between pH 7 to 11. A suitable indicator for these reactions would be phenolphthalein. In the case of reactions between weak acids with weak bases the pH change is so gradual that this combination cannot be analysed by a simple, direct titration. Instead we use instruments such as a pH meter to find the equivalence point.

Worked example 4.3a The concentration of ethanoic acid (CH3COOH) in a brand of white vinegar was determined by titration with standard sodium hydroxide solution. A 25.00 mL aliquot of vinegar was pipetted into a flask and several drops of phenolphthalein indicator added. Using a burette, 0.995 M sodium hydroxide solution was slowly added until the indicator turned permanently pink. The volume of sodium hydroxide solution required to reach this point was 21.56 mL. Calculate the concentration of acid in the vinegar. The equation for the reaction is: Figure 4.5 The sour-tasting ingredient in vinegar is ethanoic acid. Various types of vinegar are sold, including rice wine, malt and balsamic. They contain ethanoic acid in approximately 1 M concentration.

review The final result is rounded to three significant figures, corresponding to the smallest number of significant figures in the original data. Four significant figures are used throughout the calculations to avoid introducing errors by rounding off too soon.


CH3COOH(aq) + NaOH(aq) –—h CH3COONa(aq) + H2O(l)

Solution Calculate the amounts, in moles, of the substances involved in the reaction. In 21.56 mL of NaOH solution: n(NaOH) = c(NaOH) × V (NaOH) = 0.995 mol L–1 × 0.021 56 L = 0.021 45 mol Since 1 mol of CH3COOH reacts with 1 mol of NaOH, we can write the ratio: n(CH3COOH) 1 = n(NaOH) 1 n(CH3COOH) = n(NaOH) = 0.021 45 mol n(CH3COOH) 0.021 45 mol c(CH3COOH) = = = 0.8580 mol L–1 V (CH3COOH) 0.025 00 L The concentration of ethanoic acid in the vinegar is 0.858 M (to three significant figures).

It is often necessary to dilute a solution before carrying out a titration in order to obtain more convenient concentrations. This is seen in the following example.

Worked example 4.3b A commercial concrete cleaner (Figure 4.6) contains concentrated hydrochloric acid. A 25.00 mL volume of cleaner was diluted to 250.0 mL in a volumetric flask. A 20.00 mL aliquot of 0.4480 M sodium carbonate solution was placed in a conical flask. Methyl orange indicator was added and the solution was titrated with the diluted cleaner. The methyl orange indicator changed permanently from yellow to pink when 19.84 mL of the cleaner was added. Calculate the concentration of hydrochloric acid in the concrete cleaner.

Solution 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) In 20.00 mL of Na2CO3 solution: n(Na2CO3) = c (Na2CO3) × V (Na2CO3) = 0.4480 mol L–1 × 0.020 00 L = 0.008 960 mol From the equation, 2 mol of HCl reacts with 1 mol of Na2CO3. n(HCl) 2 So, the ratio = n(Na2CO3) 1 n(HCl) = 2 × n(Na2CO3) = 2 × 0.008 960 = 0.017 92 mol n(HCl) c(HCl) = V(HCl) 0.017 92 mol = 0.019 84 L = 0.9032 mol L–1 The concentration of HCl in the diluted cleaner is 0.9032 M. Since the original concrete 250.0O cleaner had been diluted by a factor of (i.e. 10.00 times), the concentration of acid 25.00 in the cleaner is given by: 250.0O c(HCl) = 0.9032 × = 9.032 mol L–1 25.00 The concrete cleaner has an HCl concentration of 9.032 M.

Figure 4.6 Concrete cleaner, used to remove excess concrete from walls and paths, works through an acid–base reaction.

review The ionic equation for the reaction in Worked Example 4.3b is: 2H+(aq) + CO32–(aq) → H2O(l) + CO2(g) The H+(aq) ion actually exists as H3O+(aq), so this equation could also be written as: 2H3O+(aq) + CO32–(aq) → 3H2O(l) + CO2(g).

During your chemistry course you will probably perform other volumetric analyses using acid–base titrations. This analytical method is a convenient way to determine the amount of active ingredient in products such as cleaning agents containing ammonia, antacid tablets or aspirin.

summary

4.2–4.3

pH H is i a measure off the th concentration t ti off H3O+ ions i in i an aqueous solution. The concentration of acidic or basic solutions can be determined by volumetric analysis.

The Th indicator i di t usedd in i an acid–base id b titration tit ti mustt provide id a sharp end point. The end point in an acid–base titration must be at or close to the equivalence point.


key questions 6 The graphs in Figure 4.7 show the pH curves for titrations involving combinations of acids and bases of various strengths. You have a choice of phenolphthalein and methyl orange indicator. Phenolphthalein changes colour over a pH range 8.2

a

pH

14 1

Equivalence point

7

to 10.0. Methyl orange changes colour between pH 3.2 and 4.4. Decide which indicator(s) would be suitable to identify the equivalence point for each reaction. Provide reasons for your selections.

b

14 1

pH

7 Equivalence point

Volume of acid

c

Volume of acid

14 1

d

14 1

Equivalence point pH pH

7

Equivalence point

7

Volume of acid

Volume of acid

Figure 4.7 Change in pH during a titration of: a a strong acid with a strong base; b a strong acid with a weak base; c a weak acid with a strong base; d a weak acid with a weak base.

7 The ethanoic acid content of white vinegar was determined by titrating a 20.00 mL aliquot of the vinegar with 0.9952 M sodium hydroxide solution. The phenolphthalein indicator changed permanently from colourless to pink when 20.34 mL of sodium hydroxide solution was added from the burette. a Write an equation for the reaction. b Calculate the amount of sodium hydroxide, in mol, used in the titration. c Calculate the amount of ethanoic acid, in mol, used in the titration. d Calculate the concentration of ethanoic acid in the vinegar. 8 The ammonia content of cloudy ammonia was determined by first diluting a 25.0 mL sample to 250 mL in a volumetric flask.


A 20.0 mL aliquot of this solution was titrated with 0.0987 M hydrochloric acid. The volume of the acid used was 22.18 mL. The equation for the reaction is: NH3(aq) + HCl(aq) → NH4Cl(aq) a Calculate the amount of hydrochloric acid, in mol, used in the titration. b Calculate the amount of ammonia, in mol, used in the titration. c Calculate the concentration of ammonia in the diluted solution. d Calculate the concentration of ammonia in the original sample.

4.4

Back titration Some acids and bases are so weak that they do not produce a sharp colour change at the end point of a titration. A technique known as back titration is used to overcome this problem. The procedure for a back titration has two parts: 1 If the substance to be analysed acts as a weak acid, it is mixed with an excess of a strong base. The original amount of strong base is known. All of the weak acid reacts in stoichiometric proportions, leaving some of the strong base unused. 2 The unused strong base is titrated as normal with a standard solution of a strong acid. Knowing the original amount of strong base and the amount of strong base left unused, it is possible to work back to find the amount of weak acid present. The same procedure can be used to determine the concentration of a weak base by adding an excess of a strong acid.

Determining the concentration of ammonium ions in lawn fertiliser

1 Fertiliser added to standard NaOH solution.

2 Mixture boiled

Nitrogen-containing ions, such as ammonium ions, are required by plants to synthesise materials needed for healthy growth. For this reason, lawn fertilisers often contain a high proportion of ammonium sulfate or ammonium nitrate. The ammonium ion (NH4+) acts as a very weak acid, and a direct titration with a base does not give a sharp end point. Instead back titration is used. 1 A measured mass of fertiliser is added to a solution containing an excess amount of sodium hydroxide. The ammonium ions react with hydroxide ions: NH4+(aq) + OH−(aq) –—h NH3(aq) + H2O(l)

The mixture is boiled to ensure complete reaction of the ammonium ions. Some of the hydroxide ions are consumed by this reaction but others remain unreacted. 2 The quantity of excess hydroxide ions is found by titrating the solution with a standard solution of hydrochloric acid: HCl(aq) + NaOH(aq) –—h NaCl(aq) + H2O(l)

Indicator added 3 Excess NaOH titrated with HCI solution.

Figure 4.8 Steps in a back titration used to analyse the concentration of NH4+ ions in fertiliser.

The procedure is shown in Figure 4.8. Knowing the original quantity of sodium hydroxide and the amount in excess after reaction with the fertiliser, we can calculate, by subtraction, the quantity of hydroxide ions that reacted with the fertiliser. The concentration of ammonium ions in the fertiliser can then be calculated by working backwards.

Worked example 4.4 A 1.50 g sample of lawn fertiliser was boiled with 25.00 mL of 0.9987 M sodium hydroxide solution. When no further ammonia gas was evolved from the mixture, it was cooled and titrated with 0.2132 M hydrochloric acid, using phenolphthalein as an indicator. A titre of 19.78 mL was required. Calculate the percentage of ammonium ions in the fertiliser.


Solution n (NaOH) supplied

–

n (NaOH) remaining

n (NaOH) reacting with NH4+

n (NH4+)

m (NH4+) in sample

% (NH4+) in sample

Figure 4.9 Steps in the calculation of ammonium content in a fertiliser.

Step 1: Find the original amount of NaOH used: n(NaOH) = c(NaOH) × V (NaOH) = 0.9987 mol L−1 × 0.025 00 L = 0.024 97 mol Step 2: Find the amount of NaOH that did not react with NH4+ ions: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) n(HCl) = c(HCl) × V (HCl) = 0.2132 mol L−1 × 0.019 78 L = 0.004 217 mol From the equation, 1 mol of HCl reacts with 1 mol of NaOH. n(NaOH) 1 = So, the ratio n(HCl) 1 n(NaOH) = 0.004 217 mol Step 3: The amount of NaOH that reacted with NH4+ ions is given by: n(NaOH) reacting = n(NaOH) used originally − n(NaOH) not reacting with NH4+ ions = 0.024 97 − 0.004 217 = 0.020 75 mol Step 4: Find the amount of NH4+ in the fertiliser sample: NH4+(aq) + OH−(aq) → NH3(aq) + H2O(l) From the equation, 1 mol of NH4+ reacts with 1 mol of NaOH. n(NH4+) 1 = So, the ratio n(NaOH) 1 n(NH4+) = 0.020 75 mol Step 5: The percentage of NH4+ in the fertiliser sample can now be found: m(NH4+) = n(NH4+) × M(NH4+) = 0.020 75 mol × 18.01 g mol−1 = 0.3737 g + Percentage NH4 in the fertiliser mass of NH4+ in the sample × 100 = mass of sample 0.3737 g × 100 = 24.91% = 1.50 g The fertiliser contains 24.9% ammonium ions (three significant figures). A flow chart summarising the steps in this calculation is shown in Figure 4.9.

summary

4.4

Back titrations are used sed to analyse anal se solutions sol tions of weak eak acids or bases where indicators do not produce a sharp end point.

In a back titration, titration an excess of a strong acid or base is added to the chemical that is to be analysed and the amount of excess is then determined by titration.

key question 9 A 0.4376 g aspirin tablet was heated with 50.00 mL of 0.196 M sodium hydroxide solution. The aspirin reacts according to the equation: C6H4(OCOCH3)COOH(aq) + 2NaOH(aq) → C6H4(OH)COONa(aq) + CH3COONa(aq) + H2O(l) aspirin After cooling, the resulting solution was titrated against 0.298 M hydrochloric acid to determine the amount of unreacted sodium hydroxide. A titre of 18.64 mL was obtained. Calculate: a the amount, in mol, of sodium hydroxide initially added to d the amount of sodium hydroxide that reacted with the the aspirin aspirin b the amount of hydrochloric acid used e the amount of aspirin in the tablet c the amount of sodium hydroxide in excess after reaction f the percentage, by mass, of aspirin in the tablet. with the aspirin


04 acid acid–base indicator acid–base reaction acidity

key terms back titration base conjugate acid–base pair ionise

neutralisation pH pH curve strong acid

strong base weak acid weak base

STOICHIOMETRY • Volumetric analysis: calculations • More stoichiometric calculations

Acid–base reactions 10 Write full equations for these acid–base reactions: a nitric acid is added to sodium hydroxide solution b sulfuric acid is added to potassium hydroxide solution c hydrochloric acid is added to ammonia solution d ethanoic acid solution is added to potassium hydroxide solution. 11 Write ionic equations for the reactions in Question 10.

Volumetric analysis of acids and bases 12 What mass of sodium sulfate is produced when 25.0 mL of 0.100 M sulfuric acid is added to 20.0 mL of 0.15 M sodium hydroxide solution? 13 What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide? 14 The cleaning agent in a household window cleaner is an aqueous solution of ammonia. Draw a flow chart to show how you would perform a titration to find the concentration of ammonia in the window cleaner. 15 Read the description of the analysis of concrete cleaner on page 41 again. What safety instructions would you give to a student intending to perform an analysis such as this? 16 The concentration of sodium hydroxide in waste water from an alumina refinery was found by titrating 20.00 mL aliquots of waste water against 0.150 M hydrochloric acid, using phenolphthalein as indicator. The average titre of several titrations was 11.40 mL. a Why is an indicator used? b Write an equation for the reaction that occurred. c What was the molarity of NaOH in the waste water? d What mass of NaOH would be present in 100 L of the waste water?

17 A 42.7 mL volume of a hydrochloric acid solution is required to react completely with 20.0 mL of 0.612 M sodium carbonate solution. a Write an equation for the reaction. b Calculate the concentration of the HC l, in mol L−1. 18 A 1.20 g antacid tablet contains 80.0% by mass of Mg(OH)2 as the active ingredient. What volume of 0.1500 M HC l could the antacid tablet neutralise? 19 Washing soda is added to hard water to allow soap to lather. A certain brand of washing soda contains partially hydrated sodium carbonate solid. A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloric acid. a What mass of sodium carbonate was present? b Calculate the percentage by mass of sodium carbonate in the washing soda. 20 A 50.0 mL sample of vinegar was diluted to 250.0 mL in a volumetric flask. A 20.00 mL aliquot of this solution required the addition of 27.98 mL of 0.134 M sodium hydroxide solution in order to be neutralised. a Write an equation for the neutralisation reaction. b What is the molarity of ethanoic acid in the original vinegar? c Express your answer to part b in g L−1. 21 A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher than, lower than or unchanged from the actual value if the student had previously washed with water, but not dried, the following apparatus: a the pipette used to deliver the aliquot of sodium carbonate solution b the flask containing the aliquot c the burette.


22 In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00 mL aliquots of a standard solution of sodium carbonate. Methyl orange indicator was used to identify the end point of the reaction: 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) The sodium carbonate solution had been prepared by dissolving 1.236 g of anhydrous Na2CO3 in water and making the solution up to 250.0 mL in a volumetric flask. The titres recorded were 21.56 mL, 20.98 mL, 20.96 mL and 21.03 mL. a What value for the titre of hydrochloric acid solution should the student use in the calculation of the acid concentration? Explain your answer. b What is the molarity of the sodium carbonate solution? c Calculate the concentration of the HC l, in mol L−1. 23 A solution of a metal carbonate with the formula M2CO3 was prepared by dissolving 2.80 g of the solid in 250 mL of water. 20.00 mL aliquots of this solution were then titrated with 0.150 M H2SO4, using methyl orange indicator. The average titre was 10.8 mL. a Write an equation for the reaction between the metal carbonate and sulfuric acid. b What amount of sulfuric acid, in mol, was needed to reach the end point of the titration? c What amount of the metal carbonate was present in each 20.00 mL aliquot? d What amount of metal carbonate was present in the original sample? e Determine the molar mass of the metal carbonate. f What is the identity of the metal M? 24 Sketch the pH curve that would be obtained when a strong base is titrated with a weak acid.

Back titration 25 A 50 mL volume of 0.10 M nitric acid is mixed with 60 mL of 0.10 M calcium hydroxide solution. What volume of 0.050 M sulfuric acid is required to neutralise the mixture? 26 Lawn fertiliser contains ammonium ions (NH4+). A 1.234 g sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide solution. The flask was heated until the reaction: NH4+(aq) + OH−(aq) → NH3(aq) + H2O(l) was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added. Calculate:


a the amount, in mol, of HC l used in the titration b the amount of NaOH in excess after reaction with the fertiliser c the amount of NaOH that reacted with the NH4+ ions d the amount of NH4+ ions in the 1.234 g fertiliser sample e the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

Connecting the main ideas 27 The drain cleaner Drainol contains 13% m/v NaOH. The original solution is diluted, by taking 10 mL and making it up to 500 mL with distilled water. a What is the molarity of the original solution? b Concentrated solutions of NaOH are not used in a titration. Give two reasons why. c The concentration of the diluted NaOH is determined by titration. Give a suitable reactant and indicator to use in the titration with NaOH. d Standard NaOH solutions cannot be prepared directly from the solid. Explain why. 28 An impure sample of limestone, mainly calcium carbonate, was analysed by back titration. Approximately 1 g of the finely powdered limestone was weighed accurately and transferred to a conical flask. An excess of HC l, exactly 50.00 mL, was added to the limestone. The mixture was stirred for 15 minutes with a magnetic stirrer to allow the reaction to be completed. The mixture was then titrated with a standard solution of NaOH and the following results were obtained: Mass of watchglass

8.7954 g

Mass of limestone and watchglass

9.8460 g

Concentration of standard NaOH solution

0.0489 M

Titration value of NaOH obtained

22.32 mL

Concentration of HC l

0.395 M

a Write balanced equations for the two reactions that occur. b Determine the moles of HC l in excess after the reaction with the limestone. c Calculate the total moles of HC l added to the limestone. d How many moles of HC l reacted with the limestone? e Calculate the number of moles of calcium carbonate in the limestone. f What is the percentage of calcium carbonate in the limestone? g In this experiment the whole sample of limestone was used in one titration. How could the precision of the titration have been improved?

0

chapter 05

Analysing oxidants and reductants

key knowledge • volumetric analysis: redox titrations

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • recognise a redox reaction • identify chemicals undergoing oxidation and reduction in a redox reaction • write half equations for redox reactions • combine half equations to produce overall equations • assign oxidation numbers to elements in redox reactions and hence identify oxidants and reductants • describe appropriate procedures for volumetric analyses that involve redox titrations of some common consumer products • perform stoichiometric calculations for volumetric analysis involving redox reactions • perform calculations for volumetric analysis that involve back titrations.

chemistryy in action Alcohol and the road toll

Figure 5.1 The blood alcohol content of motorists can be analysed using a breathalyser.

TABLE 5.1 Typical ethanol contents of selected alcoholic beverages Drink ‘Light’ beer Beer Alcoholic cider Wines (including sparkling wines) Port, sherry Spirits (brandy, bourbon, gin, rum, vodka, whisky etc.) Pre-mixed vodka-based drinks

Ethanol content (%v/v) 2.5 5.0 10 12 18 37 9

Drinks such as wine, beer, vodka and bourbon contain ethanol (CH3CH2OH). Ethanol acts as a depressant, slowing the functioning of the brain. When consumed in excess, the intoxicating qualities of alcoholic drinks can lead to antisocial behaviour and can damage a person’s health. The ethanol content of alcoholic drinks varies, as shown in Table 5.1. Regulations require that the ethanol content of alcoholic drinks be specified on their labels, since the content determines the quantity of drink that can be consumed without adverse effects. Because ethanol slows down a person’s reaction time, it seriously affects driving skills. It is estimated that alcohol has contributed to nearly 40% of all road accidents. Governments have responded by introducing penalties, such as fines and licence disqualification, for drivers who register a concentration of alcohol in the blood exceed-ing a certain level—typically 0.05% (w/v). Probationary or P-plate drivers are required to have a zero blood alcohol level. The introduction of penalties, a policy of randomly testing motorists for blood alcohol levels, and various advertising campaigns have all helped to increase public awareness of the link between alcohol consumption and road accidents. Figure 5.2 shows how the road toll in Victoria changed during the period when these various measures designed to discourage drink-driving were introduced. Chemists were involved in the invention of the ‘breathalyser’, an instrument designed for use by police to estimate blood alcohol content. Rather than analysing samples of blood, this instrument measures the concentration of alcohol in a person’s breath, which is closely related to the concentration of alcohol in their blood. If this screening test indicates a driver is over 0.05, more accurate measurements are taken either in a ‘booze bus’ or at a police station.The blood alcohol content may then be confirmed using instrumental techniques such as infrared spectroscopy (Chapter 7) or gas–liquid chromatography (Chapter 6). An alcohol fuel cell sensor (Chapter 27) is also used by some police departments. The first breathalysers operated by detecting the colour change that occurs when ethanol reacts with an acidified solution of potassium dichromate (K2Cr2O7), forming Cr3+ ions and ethanoic acid. The reaction can be represented by the equation:

2Cr2O72−(aq) + 3CH3CH2OH(g) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l) orange green This reaction can also be used to determine the ethanol content of drinks by volumetric analysis and is an example of a particular class of reactions known as oxidation–reduction (or redox) reactions. We will now look at redox reactions in more detail and examine how they are used in chemical analysis.


chemistryy in action Alcohol and the road toll (continued) 10 9

0.05% law introduced

Fatality rate per 10 000 vehicles

8 50 km/h limit in residential areas introduced

7 6 5

Zero blood alcohol for probationary drivers

Increased blood alcohol penalties

4 3

40 km/h around schools

Random breath-testing introduced

2

3 km/h tolerance on speeding replaces 10% tolerance

Increased penalties

1 0

62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 00 02 04 06 08 10 Year

Figure 5.2 Deaths on Victorian roads, 1960–2009

5.1

_

Mg

2e

What is a redox reaction? You may remember from your study of Unit 2 (Heinemann Chemistry 1, Chapter 16) that redox reactions involve the transfer of electrons from one chemical to another. They can often be considered as occurring in two parts. In these reactions: One of the reactants loses electrons—this process is called oxidation. One of the reactants gains these electrons—this is called reduction. Oxidation and reduction occur simultaneously. Redox reactions occur when, for example, fireworks are triggered. These substances usually contain magnesium metal, which burns with an intense light to produce a white, ionic solid—magnesium oxide. During the reaction, magnesium atoms lose two electrons to become Mg2+ ions (Figure 5.3):

Mg

O

Mg2+

O

O2

_

_

O2

Mg2+

_

2e

Electron transfer 2Mg(s) O2(g) + Reductant Oxidant (is reduced) (is oxidised)

2MgO(s)

Figure 5.3 A representation of the electron transfer reaction between magnesium and oxygen.

Mg –—h Mg2+ + 2e−

This part of the overall reaction is oxidation. At the same time, oxygen molecules in the air (O2) each gain four electrons that were lost by the magnesium atoms and form O2− ions: O2 + 4e− –—h 2O2−

This part of the overall reaction involves reduction.

review The words ‘OIL RIG’ can be used to remember that ‘Oxidation Is Loss of electrons’ and ‘Reduction Is Gain of electrons’. Redox reactions are described in further detail in Heinemann Chemistry 1, Chapter 16.

49 Analysing oxidants and reductants

The number of electrons produced during the oxidation is the same as the number of electrons consumed in the reduction. To write an overall equation for the reaction we add the half equations, making sure that the number of electrons lost is equal to the number of electrons gained during the reaction. Overall equation: 2(Mg –—➤ Mg2+ + 2e−) O2 + 4e− –—➤ 2O2− 2Mg(s) + O2(g) –—➤ 2MgO(s)

Substances that cause oxidation to occur are called oxidising agents, or oxidants. In the reaction between magnesium and oxygen above, oxygen acts as the oxidant. Note that the oxidant causes oxidation but itself will always be reduced. Substances that cause reduction to occur are called reducing agents, or reductants. Magnesium is the reductant in our example. Reductants cause reduction but are always themselves oxidised. Redox reactions occur as part of many everyday processes, such as the bleaching of hair, the corrosion of metals, extraction of metals from their ores, respiration, photosynthesis and film-based photography.

chemistryy in action Photochromic sunglasses

Figure 5.4 The lenses of these photochromic glasses darken in sunlight as a consequence of redox reactions involving silver chloride.


Some people wear glasses fitted with photochromic lenses, which darken in bright sunlight and become more transparent when the light intensity drops (Figure 5.4). Use of these glasses means that a pair of sunglasses is not required, as the photochromic lenses can decrease the amount of transmitted light by up to 80%. Many silver compounds, including silver chloride, are sensitive to light. Tiny crystals of silver chloride are incorporated into the complex silicate-based structure of the glass used to make the photochromic lenses. On exposure to ultraviolet light, which is present in sunlight, the chloride ions are oxidised to chlorine atoms: Cl− → Cl + e− Electrons produced from this reaction are transferred to silver ions, causing the silver ions to be reduced to metallic silver: Ag+ + e− → Ag The silver metal formed causes light to be reflected, the lenses to darken, and the intensity of light reaching the eyes of a person wearing the glasses to be reduced. To prevent the silver metal and chlorine atoms forming silver chloride again immediately, copper(I) chloride is also added to the glass in tiny amounts. It reacts with the chlorine atoms, reducing the rate at which silver chloride can be re-formed: Cl + Cu+ → Cl− + Cu2+ The darkening process must be reversible for the glasses to be effective. In the absence of strong sunlight, silver ions are re-formed by a redox reaction involving the silver metal and Cu2+ ions: Cu2+ + Ag → Cu+ + Ag+ As a consequence, the lenses of the glasses recover their transparency.

summary

5.1

O Oxidation id ti involves i l a loss l off electrons. l t Reduction involves a gain of electrons. Oxidation and reduction reactions occur simultaneously.

An A oxidant id t causes oxidation id ti andd is i itself it lf reduced. d d A reductant causes reduction and is itself oxidised.

key questions 1 Name the chemicals that undergo oxidation in the following reactions. a 2Zn(s) + O2(g) → 2ZnO(s) b Ca(s) + Cl2(g) → CaCl2(s)

c 2AgBr(s) → 2Ag(s) + Br2(g) 2 Identify the oxidants and reductants in each of the reactions in Question 1.

5.2

Oxidation numbers: classifying redox reactions We have seen that the reaction of magnesium with oxygen is a redox reaction. Similar reactions occur when other substances burn in oxygen. During a traditional barbecue, carbon in charcoal burns to form carbon dioxide: C(s) + O2(g) –—➤ CO2(g)

If the air supply is limited, such as where charcoal is burning strongly in the centre of the fire, carbon monoxide may be formed: 2C(s) + O2(g) –—➤ 2CO(g)

Carbon monoxide is flammable and burns with a clean, blue flame in the outer regions of the fire: 2CO(g) + O2(g) –—➤ 2CO2(g)

These reactions are so similar to the combustion reaction of magnesium that chemists have agreed they should also be classified as redox reactions. However, since the products in the reactions are molecular rather than ionic, they cannot be classified as redox reactions according to our earlier definitions of oxidation and reduction as involving the transfer of electrons. This difficulty is overcome in a somewhat artificial manner. To determine whether a reaction is a redox reaction, oxidation numbers are assigned to elements involved in the reaction. These numbers are determined by applying a set of rules that treat molecular substances as if they were composed of ions. Oxidation numbers have no physical meaning. Oxidation numbers are also known as oxidation states.

! The oxidation number of an element is written above its symbol. The plus or minus sign precedes the number and so distinguishes the oxidation number from the charge on an −2

ion. The O 2− ion, for example, has a charge of 2− and an oxidation number of −2.


review Electronegativity is a measure of the electronattracting ability of an element. The order of electron-attracting power is: F > O > Cl > N > other elements Metals generally have low electronegativities. Electronegativities of the elements are given in the periodic table at the end of this book.

! Remember that oxidation numbers have no physical meaning. In the case of water, the hydrogen is assigned a value of +1 and the oxygen a value of −2, as if the compound were made of H+ and O2− ions. (Of course, water really consists of H2O molecules.) This is done so that reactions involving covalent substances, as well as those involving ionic substances, can be classified as redox reactions.

Oxidation number rules Oxidation numbers are determined using the following rules: Free elements have an oxidation number of zero. 0

0

0

e.g. Na(s), C(s), Cl2(g) In ionic compounds composed of simple ions, the oxidation number is equal to the charge on the ion. +1 −1

+2

−1

+3

−2

e.g. NaCl, CaCl2, Al2O3 Some elements have oxidation numbers that are regarded as fixed, except in a few exceptional circumstances. − Oxygen usually takes −2 in compounds. In peroxides such as H2O2 and BaO2 it has −1. − Hydrogen takes +1 in compounds, except in metal hydrides such as NaH and CaH2 where it has −1. − Electronegative elements such as F, Cl and O take numbers equal to the charges on their simple ions (−1, −1 and −2, respectively) when part of a compound, provided that they are the most electronegative element present in the compound. The sum of the oxidation numbers in a neutral compound is zero. In a polyatomic ion the sum of the oxidation numbers is equal to the charge on the ion. The most electronegative element in a compound has the negative oxidation number.

Worked example 5.2a The sum of the oxidation numbers in the neutral compound HNO3 equals zero. H is fixed as +1 and O is fixed as −2. Find the oxidation number of nitrogen.

Solution (oxidation number of H) + (oxidation number of N) + 3 × (oxidation number of O) = 0 So, (+1) + x + 3(−2) = 0 x = +5 +1 +5 −2

The oxidation numbers are H N O3.

Worked example 5.2b For CO32−, the sum of the oxidation numbers equals the ionic charge of 2−. The oxidation number of O is fixed as −2. Find the oxidation number of carbon.

Solution (oxidation number of C) + 3 × (oxidation number of O) = −2 So, x + 3(−2) = −2 x = +4 +4 −2 The oxidation numbers are C O32−.

Worked example 5.2c The sum of the oxidation numbers of F2O is zero. Find the oxidation number for each atom.

Solution F is always −1 and O is normally −2 in compounds. In this case F is more electronegative, so O becomes +2. As a result, the sum of the oxidation numbers is zero.


Variable oxidation number In all their compounds the group 1 metals only have an oxidation number of +1 and the group 2 metals only have an oxidation number of +2. The oxidation numbers of the transition metals vary depending upon the compound. The different oxidation states of the transition metals often have characteristic colours (Figure 5.5). The oxidation numbers in non-metallic elements may also vary from compound to compound. The oxidation number of nitrogen in NH3 is −3, in NO it is +2, in NO2 and N2O4 it is +4, while in HNO3 and N2O5 it is +5.

chemfact Oxygen is transported through the bodies of crustaceans and molluscs by haemocyanin instead of the haemoglobin used in mammals. Haemocyanin is a blue compound containing copper ions. The transfer of oxygen from a crustacean’s blood to its cells employs an equilibrium reaction involving copper(I ) and copper(II ) ions: Cu2+ + e− Cu+

Figure 5.5 The test tubes contain solutions of vanadium compounds that have oxidation numbers from left to right of +5, +4, +3 and +2.

Using oxidation numbers to name chemicals There are two compounds that can be called iron chloride: FeCl2 and FeCl3. To distinguish between the two iron chlorides, roman numerals representing the appropriate oxidation number are inserted in the name: FeCl2 is named iron(II) chloride. FeCl3 is named iron(III) chloride.

Changing oxidation numbers Once oxidation numbers are assigned to the elements involved in a chemical reaction, we can look for a change in these numbers during a reaction. An increased oxidation number means the element has been oxidised. A decreased oxidation number means the element has been reduced. Both processes always occur in a redox reaction.

Worked example 5.2d Let us apply the rules for assigning oxidation numbers to the reactions that occur in a barbecue. 0

0

+4 −2

i C(s) + O2(g) → C O2(g) Since the oxidation number of carbon increases from 0 in elemental carbon to +4 in carbon dioxide, the carbon is oxidised. The oxidation number of oxygen decreases from 0 to −2, so oxygen is reduced. Carbon is the reductant and oxygen is the oxidant in this reaction.


0

0

+2 −2

+2 −2

0

+4 −2

ii 2C(s) + O2(g) → 2C O(g) Carbon is oxidised; oxygen is reduced. Carbon is the reductant and oxygen is the oxidant. iii 2 C O + O2(g) → 2C O2(g) Carbon in carbon monoxide is oxidised. (We often say the carbon monoxide is oxidised.) Molecular oxygen is reduced. Carbon is the reductant and oxygen is the oxidant.

summary

5.2

O Oxidation idation numbers n mbers can be used sed to identify identif species that ha havee been oxidised or reduced. Oxidation numbers have no physical meaning.

An element has been oxidised o idised if its oxidation o idation number n mber has increased. An element has been reduced if its oxidation number has decreased.

key questions 3 State the oxidation number of carbon in: a CO b CO2 c CH4 d C (graphite) e HCO3− 4 Which one or more of the following substances contain manganese in the +6 oxidation state: MnCl2, MnCl3, MnO2, K2MnO4, KMnO4? 5 Find the oxidation numbers of each element in the following compounds or ions. Hint: For ionic compounds, use the charge on each ion to help you.

a CaO b CaCl2 c HSO4− − d MnO4 e F2 f SO32− g NaNO3 h K2Cr2O7 6 Assign oxidation numbers to each element in these equations, and hence identify the oxidant and reductant: a Mg(s) + Cl2(g) → MgCl2(s) b 2SO2(g) + O2(g) → 2SO3(g) c Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) d 2Fe2+(aq) + H2O2(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)

5.3

Writing half equations Half equations that involve atoms or simple ions can be written with little difficulty. For instance, knowing that magnesium metal is oxidised to form Mg2+ ions, you can quickly write the half equation as: Reactants

Products

Mg –—➤ Mg2+ + 2e−

Half equations involving polyatomic ions may be less obvious. The anaesthetic nitrogen(I) oxide or laughing gas (N2O), can be prepared by the reduction of nitric acid in an acid solution: 2NO3−(aq) + 10H+(aq) + 8e− –—➤ N2O(g) + 5H2O(l)

Figure 5.6 Remember, in balanced half and overall equations: the number of atoms of each element is equal on both sides the total charge on each side is equal.


Such equations can be deduced using the following procedure. Let us take the reduction of the nitrate ion as an example. 1 Balance all elements except hydrogen and oxygen in the half equation. 2NO3− –—➤ N2O

2 Balance the oxygen atoms by adding water. (Oxygen atoms react to form water in acidic solution.) 2NO3− –—➤ N2O + 5H2O

3 Balance the hydrogen atoms by adding H+ ions (which are present in acidic solution). 2NO3− + 10H+ –—➤ N2O + 5H2O

4 Note that the total charge on the left-hand side of this incomplete equation is (2−) + (10+), which equals 8+. The total charge on the righthand side is 0. Balanced equations should have the same total charge on each side. Balance the charge on both sides of the equation by adding electrons. Add states. 2NO3−(aq) + 10H+(aq) + 8e− –—➤ N2O(g) + 5H2O(l)

Worked example 5.3 When a pale green solution containing Fe2+ ions is mixed with a purple-coloured solution of MnO4− ions, the purple colour disappears. Fe3+ and Mn2+ ions are formed. Write a balanced equation for this reaction.

Solution The half equation for the reaction involving Fe2+ ions is written easily: Fe2+(aq) → Fe3+(aq) + e− We can work out the equation involving MnO4− ions by following the steps described above. 1 MnO4− → Mn2+ 2 MnO4− → Mn2+ + 4H2O 3 MnO4− + 8H+ → Mn2+ + 4H2O 4 MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) To write an overall equation, the half equations are multiplied so that the number of electrons in each is the same. They are then added together and simplified if required: 5 × (Fe2+(aq) → Fe3+(aq) + e−) MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

summary

5.3

AAn oxidation id ti or a reduction d ti reaction ti can be b represented t d by b a half equation. An oxidation half equation and a reduction half equation can be combined by multiplying the half equations so that there is

an equall number b off electrons l t in i eachh half h lf equation. ti Th The hhalflf equations can then be added.

key questions 7 Write ionic half equations for the: a reduction of MnO2 to Mn2+ b reduction of MnO4− to MnO2 c reduction of SO42− to H2S d oxidation of SO2 to SO42− e oxidation of H2S to S f oxidation of SO32− to SO42− 8 When copper(II ) sulfate solution is stored in a steel (iron) container, the container gradually corrodes. Write ionic half equations and a balanced ionic equation to represent the reaction if the products of the reaction are copper and iron(II ) sulfate solution.

9 When zinc powder is sprinkled into an acidified solution of potassium dichromate, a reaction occurs that leaves zinc ions and chromium(III ) ions in solution. a Write the oxidation half equation for the reaction. b Write the reduction half equation for the reaction. c Use your answers to parts a and b to write a balanced ionic equation for the overall reaction. 10 Write the ionic half equations and the balanced overall ionic equation for the reaction in which: a a solution containing iron(II ) ions is oxidised by an acidified solution containing dichromate ions (Cr2O72−). The products include iron(III ) and chromium(III ) ions.

55 5 Analysing oxidants and reductants

key questions (cont.) b a solution containing sulfite ions (SO32−) reacts with an acidified solution of permanganate ions (MnO4−) to produce a colourless solution containing sulfate ions and manganese(II ) ions. c manganese dioxide (MnO2) reacts with concentrated hydrochloric acid to form chlorine gas and a solution containing manganese(II ) ions. 11 The following equations are not balanced. i Identify the species that has been reduced and the species that has been oxidised.

chemfact Selecting a suitable indicator can be more difficult for redox titrations than for acid–base titrations. Redox indicators must behave as oxidants or reductants after the equivalence point has been reached and a small excess of the solution from the burette is present. They must also be highly coloured in either oxidised or reduced form. For example, the permanganate ion often used in redox titrations has a purple colour while the manganese(II ) ion is colourless: MnO4−(aq) + 8H+(aq) + 5e− purple → Mn2+(aq)+ 4H2O(l) colourless

a b c d

ii Write balanced half equations for the oxidation and reduction reactions. iii Combine the half equations to write a balanced equation. Ce4+(aq) + H2S(g) → Ce3+(aq) + S(s) + H+(aq) NO3−(aq) + H+(aq) + Cu(s) → NO(g) + H2O(l) + Cu2+(aq) H2O2(aq) + Br−(aq) + H+(aq) → Br2(l) + H2O(l) MnO2(s) + H+(aq) + S(s) → Mn2+(aq) + H2O(l) + SO2(g)

5.4

Volumetric analysis The concentration of solutions of substances that can undergo redox reactions can be found by volumetric analysis, the technique described in Chapters 3 and 4. Instead of reacting an acid with a base, a redox titration involves reaction of an oxidant with a reductant. As for acid–base titrations, one solution is usually pipetted into a conical flask and the other is dispensed into the flask from a burette. For some reactions, for example those involving the permanganate ion (MnO4−), the equivalence point will be indicated by a colour change in the reacting solutions, whereas for others an indicator must be added in order to detect the equivalence point. Volumetric analysis involving redox reactions can be used to determine the composition of many substances, including fruit juice, wine and bleach. Table 5.2 gives details of how the oxidants and reductants in some everyday substances are analysed. In some cases back titrations are used (p. 43).

TABLE 5.2 Analysis of oxidants and reductants in common products Item Wine

Ingredient for analysis Ethanol

Wine Fruit juice Household bleach

Sulfur dioxide Vitamin C (ascorbic acid) Hypochlorite ion

Hair bleach

Hydrogen peroxide

Titrate with Iron(II) solution, after reaction with an excess of potassium dichromate solution Iodine solution Iodine solution Sodium thiosulfate solution, after reaction with an excess of acidified potassium iodide solution Potassium permanganate solution

Worked example 5.4 A 10.0 mL sample of white wine was placed in a volumetric flask and water was added to make 100.0 mL of solution. Then 20.0 mL aliquots of the diluted wine were titrated against 0.100 M acidified potassium dichromate (K2Cr2O7) solution. The mean titre was 24.61 mL. Calculate the concentration of ethanol in the sample of white wine. 2Cr2O72−(aq) + 3CH3CH2OH(aq) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)


Solution n(Cr2O72−) = c(Cr2O72−) × V(Cr2O72−) = 0.100 mol L−1 × 0.024 61 L = 0.002 461 mol 2− From the equation, 2 mol of Cr2O7 reacts with 3 mol of CH3CH2OH. n(CH3CH2OH) 3 So, the ratio = n(Cr2O72−) 2 3 n(CH3CH2OH) = × 0.002 461 = 0.003 692 mol 2 The amount of ethanol in the 20.0 mL volume of diluted wine is 0.003 692 mol. Since 100.0 this volume of wine was taken from a total volume of 100.0 mL, there would be 20.0 (or 5.000 times) this amount in the original 10.0 mL sample. 100 mol n(CH3CH2OH) in original 10.0 mL sample = 0.003 692 × 20 = 0.018 46 mol n(CH3CH2OH) c(CH3CH2OH) = V(CH3CH2OH) 0.018 46 mol = 0.100 L = 1.846 mol L−1 The concentration of alcohol in the wine is 1.85 M.

summary

chemfact Ethanol reacts only slowly with dichromate ions at room temperature. To overcome this problem, the ethanol content of wine is usually found by performing a back titration, rather than by using the method described here. Excess dichromate ions are first added to the wine and the mixture is heated so that reaction occurs. The amount of dichromate ions in excess is then found by titration with a solution of Fe2+ ions.

5.4

VVolumetric l t i analysis l i can be b usedd to t determine d t i the th concentration t ti or mass off substances bt that th t undergo d redox d reactions. ti

key questions 12 Potassium permanganate reacts with hydrogen peroxide: 2MnO4−(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) 25.0 mL of 0.02 M KMnO4 solution is reduced by 20.0 mL of H2O2 solution. What is the concentration of the hydrogen peroxide solution? 13 An artist uses 10.0 mL of 15.0 M HNO3 to etch a design into a sheet of copper: Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g) What mass of copper will have reacted with the acid?


05

key terms half equation oxidant oxidation

oxidation number oxidation state oxidising agent

redox reaction redox titration reducing agent

reductant reduction

STOICHIOMETRY • Volumetric analysis: calculations • More stoichiometric calculations

Oxidation Oxiidation numbers 14 What is the oxidation number of sulfur in each of the following compounds? a SO2 b H2S c H2SO4 d SO3 d Na2SO3 f Na2S2O3 15 Which of the following may be regarded as redox reactions? Give reasons for each of your answers. a BaCl2 + H2SO4 → BaSO4 + 2HCl b 2Ag + Cl2 → 2AgCl c 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4 d ZnCO3 → ZnO + CO2 d HPO32– + I2 + OH– → H2PO4– + 2I– f 2Cu+ → Cu2+ + Cu g CaF2 → Ca2+ + 2F– h P4 + 6H2 → 4PH3 16 Copper bowls and trays can be decorated by etching patterns on them using concentrated nitric acid. The overall reaction is: Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) a What is the oxidation number of copper: i before reaction? ii after reaction? b What is the oxidation number of nitrogen: i before reaction? ii after reaction? c Name the oxidant and the reductant in this process.

Writing and combining half equations 17 Aluminium metal can reduce the hydrogen ions in a solution of hydrochloric acid to hydrogen gas, according to the equation: 2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g) a Write half equations for the oxidation and reduction processes involved in this reaction. b Use these half equations to explain why six hydrogen ions are reduced for every two aluminium atoms oxidised. 18 As a result of a traffic accident, residents in a Melbourne suburb had to be evacuated when toxic fumes leaked from


a container of sodium dithionite (Na2S2O4). The dithionite ion reacts with water according to the equation: 2S2O42–(aq) + H2O(l) → S2O32–(aq) + 2HSO3–(aq) a State the oxidation number of sulfur in the following ions: i S2O42– ii S2O32– iii HSO3– b Write the ionic half equations for the oxidation and reduction reactions that occur when sodium dithionite is mixed with water. 19 During each of the following analyses of a substance, redox reactions occurred. Write half equations for the oxidation and reduction reactions. Use these half equations to write an overall equation for each reaction. a Zinc was analysed by reacting it with a solution of Pb2+ ions. Lead metal was precipitated and Zn2+ ions were formed. b Fe2+ ions in iron tablets were determined by oxidising them to Fe3+ ions, using an acid solution of MnO4– ions, which were themselves reduced to Mn2+ ions during the reaction. c Sulfur dioxide (SO2), a preservative in dried fruit, was determined by oxidation to SO42– using a solution of I2. Iodide (I–) ions were produced. d An acidic solution of bleach, which contains OCl– ions, was titrated against a solution of I– ions. The reaction products included Cl– and I2.

Calculations involving redox reactions 20 In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed. a Write half equations, and hence an overall equation, for the reaction. b Calculate the mass of zinc that would be needed to react completely with 8.0 g of MnO2 in a dry cell. 21 Find the mass of silver metal that will react with 2.000 L of 10.00 M nitric acid, according to the equation: Ag(s) + 2HNO3(aq) → AgNO3(aq) + H2O(l) + NO2(g)

22 The thermite process can be used to weld lengths of railway track together. A mould placed over the ends of the two rails to be joined is filled with a charge of aluminium powder and iron(III) oxide. When the mixture is ignited, a redox reaction occurs to form molten iron, which joins the rails together. a Write a half equation for the conversion of iron(III) oxide to metallic iron. b Is the half equation you wrote for part a an oxidation or a reduction process? c Write the overall equation for the thermite process. d What mass of iron(III) oxide must be present in the charge if each joint requires 3.70 g of iron to weld it together?

Volumetric analysis 23 A vitamin C tablet with a mass of 1.306 g was crushed and dissolved in de-ionised water. The solution was titrated against 0.0500 M iodine solution, using starch solution as indicator, to determine the vitamin C (C6H8O6) content of the tablet. The reaction can be represented by the equation: C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I–(aq) The end point occurred when 28.40 mL of iodine solution had been added. a Find the mass of vitamin C in the tablet. b Calculate the percentage of vitamin C in the tablet by mass. c Suggest the function of the other substances that make up the remainder of the mass of the tablet. 24 The iron content in a 0.200 g sample of fencing wire was determined by dissolving the wire in dilute sulfuric acid and making up the resulting pale green solution of Fe2+ ions to 25 mL. The solution was titrated with 0.0300 M potassium permanganate (KMnO4) solution, which is purple in colour. A titre of 20.22 mL was obtained. The solution of Mn2+ and Fe3+ ions produced by the reaction was almost colourless. a Write an overall equation for the titration reaction. b Calculate the amount, in mol, of Fe2+ ions in the 25 mL volume of solution. c Calculate the percentage, by mass, of iron in the wire. d An indicator was not required for this titration. Why not? e Briefly describe two safety precautions that should be observed when carrying out this titration. 25 A food and drugs authority analysed a sample of light beer to see if it conformed with the regulation of a maximum of 2% by mass of alcohol (ethanol). The alcohol content was determined by volumetric analysis according to the reaction:

dichromate (K2Cr2O7). Three separate titrations gave titres of 9.20 mL, 9.16 mL and 9.22 mL. Calculate: a the amount, in mol, of Cr2O72– present in the average of the titres b the amount, in mol, of ethanol present in each 20.00 mL aliquot c the amount, in mol, of ethanol in the original 10.00 mL sample of beer d the mass of ethanol in the original sample e the percentage by mass of alcohol in the beer, if the density of light beer is 1.10 g mL–1. Would this product conform with the regulations for low-alcohol beer? 26 Megavit multivitamin tablets contain iron(II) sulfate. The iron content of a sample of Megavit tablets was analysed. Ten tablets were crushed and ground into a paste with a little dilute sulfuric acid. All of the paste was carefully transferred into a 250.0 mL volumetric flask and the solution was made up to the mark with more dilute sulfuric acid. 20.00 mL aliquots of the solution were titrated with 0.0200 M potassium permanganate solution and a mean titre of 12.95 mL was obtained. a Write the half equation for the oxidation of Fe2+ to Fe3+. b Write the half equation for the reduction of purple MnO4– to colourless Mn2+ in acidic solution. c Write a balanced ionic equation to represent the overall reaction. d Calculate the amount, in mol, of Fe2+ present in each aliquot. e Calculate the amount of Fe2+ in the 250.0 mL volumetric flask. f Calculate the mass of iron(II) sulfate present in each tablet. g The label says that each tablet contains 250 mg FeSO4. How do these results compare with the manufacturer’s specification? List two possible sources of error that could account for any discrepancy. 27 The amount of vitamin C in fruit juice can be determined by titration with a standard 0.0100 M iodine solution: C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I–(aq)

2Cr2O72–(aq) + 3C2H5OH(aq) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)

If the maximum concentration of vitamin C is likely to be 0.000 50 g mL–1, describe how you would perform the analysis. You should mention: i the volume of fruit juice used ii the maximum titre of iodine you would expect to obtain. 28 The alcohol content of an imported brandy was found by taking 10.0 mL and diluting it to 500 mL. 20.00 mL aliquots of this solution were then titrated against 0.100 M acidified potassium dichromate (K2Cr2O7) solution. The mean titre was 17.98 mL.

The beer was tested by taking a 10.00 mL sample and making it up to 250 mL in a standard flask. 20.00 mL aliquots were titrated against a 0.0500 M solution of potassium

Calculate the concentration of ethanol in the sample of brandy.

2Cr2O72–(aq) + 3CH3CH2OH(aq) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)


29 The active ingredient in bathroom mould-killers is the bleach sodium hypochlorite (NaOCl). The concentration of this chemical in a 20.00 mL sample was determined by adding an acidified solution containing an excess of I– ions to the sample. This reacted according to the equation: OCl–(aq) + 2I–(aq) + 2H+(aq) → I2(aq) + Cl–(aq) + H2O(l) The iodine formed by this reaction was titrated using 0.750 M sodium thiosulfate (Na2S2O3) solution. I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O62–(aq) 25.10 mL of the thiosulfate solution was required to reach the end point. Calculate: a the amount, in mol, of I2 reacting with the S2O32– ions b the amount of NaOCl in the sample c the percentage of NaOCl, by mass, in the mould-killer. (Assume the density of the solution is 1.0 g mL–1.)

Connecting the main ideas 30 What is the oxidation number of the underlined element in each of the following ions? a Fe2+ b Cr3+ c CrO42– d Cr2O72– 2– 3– – e SO3 f PO4 g NO3 h MnO4– 2+ – i VO j OCl 31 Use oxidation numbers to determine whether each of the following equations represents a redox reaction and, if it is a redox reaction, identify the oxidant and reductant. a 3Mg(s) + Fe2O3(s) → 3MgO(s) + 2Fe(s) b AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) c Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) d Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(aq) 32 For each of the following reactions write: i the oxidation half equation ii the reduction half equation iii the balanced ionic equation. a Chlorine gas is bubbled into sodium iodide solution to form a solution of sodium chloride and iodine. b A piece of copper wire is dipped into silver nitrate solution, producing silver crystals and copper(II) nitrate solution. 33 Potassium permanganate is used in many redox titrations. No indicator is required as the permanganate ion is purple while the manganese(II) ion is colourless. MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) Potassium permanganate cannot be used as a primary standard because it is slightly unstable. Prior to use for analytical purposes a potassium permanganate solution must be standardised. Sodium oxalate, Na2(COO)2, can be used as a


primary standard for this purpose. During the analysis oxalate ions are oxidised to CO2: (COO)22–(aq) → 2CO2(g) + 2e–

! Oxalic acid, (COOH)2, is a toxic diprotic acid found in rhubarb leaves.

a Write a balanced equation eq for the reaction between the permanganate and oxalate ions. b A solution containing 0.161 g of sodium oxalate, Na2(COO)2, reacted with 26.7 mL of a potassium permanganate solution. Determine the concentration of this solution. 34 Liquid household bleach contains sodium hypochlorite (NaOCl) as its active ingredient. The manufacturer maintains quality control by performing volumetric analysis on random samples taken from the end of the production line. In the analytical laboratory a 25.00 mL sample was diluted to 250.0 mL using distilled water. 20.00 mL aliquots were then pipetted into clean conical flasks and excess acidified potassium iodide solution was added to each flask to cause the reaction: ClO–(aq) + 2I–(aq) + 2H+(aq) → Cl–(aq) + I2(aq) + H2O(l) The free iodine was then titrated with 0.1000 M sodium thiosulfate solution (Na2S2O3), using starch indicator solution. This indicator gives a blue-black colour in the presence of iodine. The titration reaction is represented by: 2S2O32–(aq) + I2(aq) → S4O62–(aq) + 2I–(aq) The average titre was 20.42 mL. a What is the oxidation state of chlorine in: i ClO–? ii Cl–? b Explain whether sodium hypochlorite acts as an oxidant or reductant when it is mixed with potassium iodide solution. c What colour change is observed at the end point of the titration? d What amount of sodium thiosulfate, in mol, was used in the titration with the free iodine? e What amount of iodine (I2), in mol, was released by the reaction of the sodium hypochlorite? f Deduce the amount, in mol, of sodium hypochlorite present in each 20.00 mL aliquot of diluted bleach solution. g What amount of sodium hypochlorite, in mol, was present in the original 25.00 mL sample of bleach? h What mass of sodium hypochlorite was present in the original 25.00 mL sample of the household bleach? i What is the concentration of sodium hypochlorite in g L–1?

0

chapter 06

Chromatography key knowledge • principles and applications of chromatographic techniques and interpretation of qualitative and quantitative data from thin-layer chromatography (TLC), high performance liquid chromatography (HPLC) and gas chromatography (GC)

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the equipment and procedures necessary for thin-layer chromatography • identify an appropriate analytical procedure for a particular sample • calculate Rf values from a chromatogram and use them to identify the components of a sample • describe the equipment and procedures necessary for high performance liquid chromatography and gas chromatography • explain the chemical or physical basis of each analytical technique and relate this to the nature of samples that are appropriate for analysis by the technique • determine the concentration of samples from data obtained by instrumental analyses • use retention times from HPLC and GC analyses to determine the components present in a sample • explain how chromatograms from HPLC and GC analyses are used to determine the relative amounts of components present in a sample.

6.1

Chromatography

Figure 6.1 A simple chromatogram. One end of this chalk was dipped in black ink before being placed in the beaker of water. The black ink separates into its different coloured components as it rises up the chalk.

! The term ‘chromatography’ comes from the Greek khromatos, meaning ‘of colour’, and graphos, meaning ‘written’. The technique was developed by the Russian botanist Mikhail Tsvet to separate plant pigments.

! Adsorption occurs when a substance forms a bond with a surface. The breaking of this bond is called desorption.

Chromatography is a technique that is used to separate the substances present in a mixture. It is also widely used to determine the identity of a substance. Chromatography is used mainly for mixtures of organic compounds. Its applications include the identification of: drugs present in blood sugars in fruit juice hydrocarbons in oil pollutant gases in exhaust fumes pesticides in water and soil. You can perform a simple chromatography experiment by dipping the end of a stick of chalk into water-soluble black ink and then standing the chalk in a beaker containing a small amount of water (Figure 6.1). As the water carries the ink up the chalk, you will see that it separates into bands of different colours. Each band contains one of the substances that are present in the ink.

How does chromatography work? All methods of chromatography have: a stationary phase a moving or mobile phase. In our simple exercise using chalk and ink, the stationary phase is the chalk and the mobile phase is the water. As the components in the ink are swept forward over the stationary phase by the solvent, they undergo a continual process of adsorption onto the solid stationary phase, followed by desorption back into the liquid mobile phase. The rate of movement of each component depends mainly upon: how strongly it adsorbs onto the stationary phase how readily it dissolves in the mobile phase. Since each component in the ink undergoes these two processes to a different degree, the components separate. The blue dye in the ink is more soluble in the mobile phase than the red dye, and bonds less strongly with the stationary phase than the red dye does. The blue dye in the ink has moved faster up the piece of chalk than the red dye, resulting in their separation.

6.2

Thin-layer chromatography You will probably have conducted paper chromatography experiments in previous years. In paper chromatography a piece of high-quality absorbent paper, similar to filter paper, is used as the stationary phase. Thin-layer chromatography (TLC) is very similar to paper chromatography. In this case the stationary phase is a thin layer of a fine powder such as alumina (aluminium oxide) spread on a glass or plastic plate. Paper and thin-layer chromatography are useful for qualitative analysis. Similar procedures are used in both techniques. In thin-layer chromatography a solution of the sample to be analysed is made up and as small a spot as possible of this solution is placed onto one end of the


chromatography plate. The position of this spot is called the origin. The plate is then placed in a container with the edge of the plate below the spot submerged in a solvent. It is important to have the sample spot, the origin, above the level of the solvent so that it can be transported up the plate and not dissolve into the liquid in the beaker. A similar technique is used in paper chromatography. Figure 6.2 shows thin-layer chromatography of three different samples in progress. As the solvent rises up the plate the components of each sample separate.

Watchglass used as a cover

Solvent front

Thin-layer plate coated with alumina (stationary phase) x A

Interpreting chromatograms of thin-layer and paper chromatography A chromatogram is the pattern of bands or spots formed on the plate in thin-layer chromatography or on the paper in paper chromatography. The identity of the chemicals in the mixture can be identified in two ways: 1 running standards of known chemicals on the same chromatogram as the unknown sample 2 calculating the Rf value of the sample. In Method 1 it is necessary to have some idea of the chemical that you are looking for in the sample. For example, if you wish to find out whether a vitamin tablet contains vitamins A and D, a sample of the vitamin tablet can be run on the same chromatogram as a pure sample of each of vitamins A and D. If spots from the sample of tablet move the same distance from the origin on the chromatogram as the spots from the pure samples, then the tablet is likely to contain vitamins A and D. In Figure 6.2 it is likely that the green dye in sample C is the same green dye as in sample A because the green spots occur at the same distance from the origin on the chromatogram. The sample and standard must be run on the same chromatogram because the distances moved from the origin will depend on the distance moved by the solvent front. The further the solvent front is allowed to travel, the further the spots travel and the greater the separation between the spots. In Method 2, the components of a mixture can be identified by the distance they travel along the stationary phase compared to the distance travelled by the solvent front. This is expressed as an Rf value for a component: Rf =

distance moved from origin by component distance moved from origin by solvent

x B

x C

Origin Solvent (mobile phase)

Figure 6.2 Thin-layer chromatography of three different food colours (A, B and C). A number of chemicals once used as food colours are now regarded as hazardous and can be detected by this method.

Solvent front

10 cm 7 cm

3 cm X Origin 3 Rf (blue component) = 10 = 0.3 Rf (red component) = 7 = 0.7 10

Figure 6.3 Calculation of the Rf values of components on a chromatogram.

You can see from Figure 6.3 that: Rf values will always be less than one the component most strongly adsorbed onto the stationary phase moves the shortest distance and has the lowest Rf value. Each component has a characteristic Rf value for the conditions under which the chromatogram was obtained. By comparing the Rf values of components of a particular mixture with the Rf values of known substances under identical conditions, the compounds present in a mixture can be identified. Although in chromatograms of plant pigments and food dyes the components can be seen easily, most compounds are colourless and must be made visible. Many organic compounds fluoresce and appear blue when viewed under ultraviolet light. With other compounds, the chromatogram can be sprayed with a chemical that reacts to form coloured or fluorescent compounds. In Method 2 the distance moved by the solvent front is no longer critical as the proportion of the distance moved from the origin (the Rf value) stays the same provided the conditions under which the chromatogram is run are the same. This means the Rf values of unknown spots can be compared

63 Chromatography

chemfact For a particular combination of stationary phase and mobile phase, many different chemicals may move similar distances from the origin, relative to the stationary phase, and have similar Rf values. Paper and thin-layer chromatography are only a guide to the identity of a chemical. They may be followed up by gas chromatography or high performance liquid chromatography if a more certain identification is needed.

summary

against a data table of Rf values of common materials. However, changes in the temperature, the type of stationary phase, the amount of water vapour around the plate or paper and the type of solvent will all change the Rf value for a particular chemical. The choice between paper and thin-layer chromatography depends upon the sample being analysed. Advantages of each method are given in Table 6.1.

TABLE 6.1 A comparison of paper and thin-layer chromatography Paper chromatography Cheap Little preparation More efficient for polar and water-soluble compounds Easy to handle and store A wide range of stationary phases is available.

Thin-layer chromatography Faster Detects smaller amounts Better separation of less polar compounds Corrosive materials can be used

6.1–6.2

Ch Chromatography t h is i a technique t h i commonly l usedd to t separate t andd identify the components in a mixture. All chromatographic techniques involve a mobile phase and a stationary phase.

Components C t iin a mixture i t have h diff differing i affi ffinities iti ffor the th mobile bil and stationary phases. Comparison with known standards and determination of Rf values are used in paper and thin-layer chromatography to identify components in a mixture.

key question 1 An extract from a plant was analysed using thin-layer chromatography with a non-polar solvent. The chromatogram obtained is shown in Figure 6.4.


TABLE 6.2 R f values of some plant materials Chemical Xanthophyll β-carotene Chlorophyll a Chlorophyll b Leutin Neoxanthin

Solvent front

Origin

Table 6.2 gives the Rf values of some chemicals commonly found in plants.

Figure 6.4 Chromatogram for Question 1.

Rf 0.67 0.82 0.48 0.35 0.39 0.27

a Measure and record the distance from the origin to the centre of each band, and the distance of the solvent front from the origin. b Calculate the Rf value of each band. c Compare Rf values for the bands to the Rf values in Table 6.2 and name the chemicals present in the extract. d If water had been used as the solvent, would the chromatogram be likely to have a similar appearance? Explain.

6.3

Column chromatography Another chromatographic technique, column chromatography, can be used to separate the components in a mixture (Figure 6.5). The stationary phase is a solid, or a solid that has been thinly coated in a viscous liquid, and packed into a glass column. The sample is applied carefully to the top of the packing and a solvent, which acts as the mobile phase, is dripped slowly onto the column from a reservoir above. A tap at the bottom of the column allows the solvent, which is called the eluent, to leave the column at the same rate as it enters it at the other end. Two instrumental chromatographic techniques based on column chromatography, high performance liquid chromatography (HPLC) and gas chromatography (GC), are commonly used for the separation and identification of very complex mixtures of similar compounds, such as drugs in blood and hydrocarbons in oil samples.

Sample originally placed here

Solvent Component 1: most strongly adsorbed

Component 2 Solid stationary phase

Component 3: least strongly adsorbed

Porous barrier

High performance liquid chromatography High performance, or high pressure, liquid chromatography (HPLC) makes possible extremely sensitive analysis of a wide range of compounds. It is now used routinely for pharmaceutical and industrial analyses. For example, it can be used to detect barbiturates (sleeping tablets) in the blood, a procedure that is useful in hospitals in cases where it is suspected that a patient’s medical condition is caused by overdose. Figure 6.6 shows the apparatus used for HPLC. There are many ways in which this technique differs from traditional chromatography. For instance, in HPLC: the particles in the solid used in the column are often 10–20 times smaller than in a column chromatography the very small size of the solid particles allows for more frequent adsorption and desorption of the components, giving much better separation of similar compounds the small particle size creates a considerable resistance to the flow of the mobile phase and so the solvent is pumped through under high pressure—up to about 14 000 kPa a range of solids is available for use in HPLC columns, some with chemicals specially bonded to their surfaces to improve the separation of particular classes of compounds. In HPLC, the components are usually detected by passing the eluent stream through a beam of UV light. Many organic compounds absorb UV light, so when an organic compound passes in front of the beam of light, a reduced signal is picked up by a detector. The amount of light received by the detector is recorded on a chart that moves slowly at a constant speed. The resulting trace is called a chromatogram (Figure 6.8). The time taken for a component to pass through the column is called the retention time, Rt, and is characteristic of the compound for the conditions of the experiment. It is analogous to the Rf value in paper and thin-layer chromatography. The retention times are used to identify the components associated with the peaks on a chromatogram. The relative amounts of each component in a mixture may be determined by comparing the areas under each peak with areas under peaks for standard samples.

Eluent

Figure 6.5 Using column chromatography to separate the components in a mixture.

Syringe for injecting sample

Liquid mobile phase

Pump Column Solid stationary phase

Detector Recorder Waste

Figure 6.6 The construction of a high performance liquid chromatograph.

65 Chromatography

chemistryy in action Making the grade by Robin Meadows

Gourmet cooks value olive oil for its distinctive taste and aroma. Nutritionists favour it because it is rich in mono-unsaturated triglycerides, which are believed to lower blood cholesterol levels and reduce the risk of heart disease. But some of the companies that process and distribute olive oil succumb to the temptation to mix their olive oil with less expensive oils like corn, peanut and soybean. A European company sent a shipment of olive oil to the United States. A routine sample was taken and forwarded to chemist Richard Flor at the US Customs Service laboratory in Washington, DC. The sample looked and poured like olive oil, but it didn’t taste quite right. Because taste is a subjective test and doesn’t hold up in court, Flor and his colleague Le Tiet Hecking developed an analytical test based on olive oil’s unique composition. Flor used an instrument called the ‘high pressure liquid chromatograph’ (HPLC) to separate the oils into their component triglycerides. Flor and Hecking found that the oil in the suspect sample had too much of the polyunsaturated triglycerides, confirming their suspicion that the imported sample was diluted with corn oil.

Good olive oil

Suspect shipment

Absorbance (concentration)

Figure 6.7 Pure olive oil is highly regarded due to its aroma and taste, and because it is thought to reduce blood cholesterol levels.

Figure 6.8 Peaks present in the chromatogram of the suspect oil indicate the presence of compounds found in corn oil, rather than olive oil.

10

20

30

Time (minutes)

10

20

30

Time (minutes)

Gas chromatography The most sensitive of the chromatographic techniques is gas chromatography (GC). It is capable of detecting as little as 10-12 g of a compound. However, it is limited to compounds that can be readily vaporised without decomposing. Such compounds usually have relative molecular masses less than 300. High performance liquid chromatography, on the other hand, can separate compounds with relative molecular masses of 1000 or more. The extreme sensitivity of gas chromatography makes it ideal for the analysis of trace contaminants in samples or for the detection of tiny amounts


of very potent compounds. For example, urine samples are routinely taken from athletes competing in major events to ensure that the athletes are not benefiting from the use of illegal, performance-enhancing drugs. These samples are analysed by GC. There are two types of gas chromatography, gas–liquid chromatography (GLC) and gas–solid chromatography (GSC). Both gas chromatographic techniques operate in a similar way as outlined below. A generalised diagram of the equipment used in GC is shown in Figure 6.9. Sample injection port

Detector

Chart recorder Oven Column

Carrier gas (usually nitrogen)

Figure 6.9 A gas chromatograph.

Gas chromatography has the following features: The mobile phase is a gas, generally nitrogen, called the carrier gas. A small amount of sample is injected into the top of the column through an injection port. The injection port is heated to a temperature sufficient to instantly vaporise the sample, which is then swept into the column by the carrier gas. The column is a loop, or series of loops, of glass that has an internal diameter of about 4 mm and is 2–3 m long in total. In gas–liquid chromatography, the column is packed with a porous solid that has been coated with a liquid hydrocarbon or ester with a high boiling point. This liquid acts as a liquid stationary phase. In gas–solid chromatography, the column is packed with an adsorbant solid such as silica gel or alumina. The solid acts as a solid stationary phase. The column is mounted in an oven and heated. The components of the sample repeatedly pass into and out of solution with the stationary phase. The least soluble are swept out first by the gas into the detector. One of the most useful detectors is the flame ionisation detector, invented by an Australian, Ian McWilliam. In this detector, organic compounds leaving the column are burnt in a hydrogen–oxygen flame. Ions produced in the flame are attracted to electrodes and cause a current to flow. This current is used to indicate the presence of the compound.

67 Chromatography

Porous solid, e.g. silica

B

B

B B

B B

A A

A B

Stationary phase, e.g. high boiling point liquid hydrocarbon

B B

B

A

B

A

A A

A

B A

A A

A A

B

A

B

A

B

B

Component A is more soluble in the stationary phase. Component B is swept away faster

A

A

A

A A

A Mobile phase, e.g. nitrogen

Figure 6.10 Section through a GLC column.

Interpreting the chromatograms The chromatogram shows the response of the detector against time. The response of the solvent is constant and gives a flat baseline. As components pass through the detector, the response changes and a peak is formed. The sample is injected into the column at time zero. Different chemicals show differing strengths of attraction for the stationary phase and the mobile phase, and so pass through the column at different rates. The time a component takes to pass through the column is called the retention time, Rt. The same compound will give the same retention time if the conditions (temperature, mobile phase, stationary phase, flow rate, pressure etc.) remain the same. Each component forms one peak in the chromatogram. However, it is possible for a number of peaks to coincide and so be indistinguishable. Figure 6.11 shows the gas chromatogram of a petrol sample. A number of components of the petrol, peaks marked 1–5, have been identified. The lighter components are more soluble in the solvent used and so are eluted more quickly and have a lower retention time. The peaks can be identified (qualitative analysis) by running a sample containing known pure compounds under exactly the same conditions as the sample. A compound can also be identified by ‘spiking’ the sample with a known compound. Figure 6.13 shows the petrol sample ‘spiked’ with benzene. Note that peak 4 is much bigger in the spiked sample. The concentration of a component in the sample (quantitative analysis) can be determined by running a series of standards of known concentration under the same conditions as the sample. The area under the peak is proportional to the amount of the substance present.


2

Recorder response

1

3

4 5

Figure 6.11 Gas chromatogram of a petrol sample. 0

5

10

15

20

Time (min) Number of carbon atoms 4

5

2

7

3

Recorder response

1

6

4 5

1 butane 2 2-methylbutane 3 hexane 4 benzene 5 2-methylhexane

Figure 6.12 Chromatogram of a reference sample containing a mixture of butane, 2-methylbutane, hexane, benzene and 2-methylhexane. 0

5

10

15

20


5

7 4

2

Recorder response

1

6

3 5

Figure 6.13 Petrol sample spiked with benzene. 0

5

10

15

20


5

6

7

Example 6.3 The concentration of benzene in a sample of petrol was determined by gas chromatography. A series of standards with an accurately known concentration of benzene were run under the same conditions as the sample. The chromatogram of the sample and the standards is shown in Figure 6.14.

Figure 6.14 Chromatogram of standard samples of benzene and an unknown sample. 0.2%

0.4%

0.8%

1.2%

Sample

69 Chromatography

The areas of the peaks were measured and a calibration curve was plotted from the data (Figure 6.15). A line can be drawn on the calibration graph corresponding to the peak area of the sample and the concentration read off from the graph. The concentration of benzene in the petrol sample is 0.6%.

Peak area (cm2)

8.0

6.0

4.0

Sample 2.0

0.2

0.4

0.6

0.8

1.0

1.2

Concentration of benzene (%)

Both gas chromatography and high performance liquid chromatography can be combined with another analytical technique, mass spectrometry (MS). Together the techniques allow chemists to determine smaller and smaller quantities and identify a wider range of materials, as well as gain valuable information about the structures of the compounds. Mass spectrometry and the ‘hyphenated techniques’, GC-MS and HPLC-MS, are described in Chapter 8.

Figure 6.15 Calibration graph for benzene in petrol.

extension The case of the tainted potato chip a

b

p-Cresol

Skatole Indole

Retention time

Figure 6.16 Gas chromatograms of extracts of potato chips with a an acceptable flavour, and b an offflavour. The amount of each component can be found by comparing the area under the peaks with those of standards for each component.


The flavour of a food is determined by its chemical composition and the response these chemicals evoke in the taste-buds of the tongue and the odour receptors of the nose. The chemicals that affect the sense of taste are usually non-volatile compounds such as inorganic salts, organic acids and sugars; those affecting the sense of smell are generally volatile organic compounds of many different structures. It is these volatile compounds that are essentially responsible for the characteristic flavours associated with different types of food and that provide the consumer with the means of assessing the quality of fresh and processed foods. There are only five primary tastes: sweet, sour, salt, savoury and bitter. However, there are potentially many hundreds of distinctive aromas. Thus when consumers refer to a food as having a distinctive flavour, they are usually describing the aroma of its volatile organic compounds. Dr Frank Whitfield and Kevin Shaw of CSIRO Division of Food Research outline some of the work they have been doing in this area. ‘The potato (Solanum tuberosum), when served as a boiled food, is quite bland; however, when it is baked or fried, compounds are produced on the surface tissue which give it an appetising, roasted flavour. Furthermore, this flavour is sometimes enhanced by the absorption of compounds from the heated cooking oils. ‘When so much is known about the development of flavour in potato chips, could anything possibly go wrong? Consumers and processors discovered that the answer was “yes”. Packets of frozen chips were found to contain the occasional chip which possessed a disagreeable off-flavour, described as “cheese-like” in less severe cases, and “faecal” or “nauseous” in the worst examples. Examination of the volatile components isolated from normal and off-flavoured chips shows that the latter contained two compounds which were not detectable in the normal chips; a third compound, which was detectable in the normal chips, was present at a far higher concentration in the tainted product. Sniffing the column effluent during the gas chromatographic analysis demonstrated that these three compounds had odours similar to the off-flavour of the product. The two additional peaks were identified as p-cresol and skatole, and the third compound was found to be indole.

extension The case of the tainted potato chip (continued) ‘All of these compounds could be produced by bacteria from certain amino acids which occur naturally in potato tubers. To find their origin, the factory producing the potato chips was inspected; in one section of a storage shed, some potatoes were found that were infected with “soft rot”. The rotting tissue smelled very strongly of the three off-flavour compounds, whose presence was confirmed by gas chromatography. In addition, tubers not affected by the rot but stored near the rotting potatoes were also tainted to varying degrees.’

questions

?

E1 Solids Solid lids are nott iinjected njjected into the inlet port of a gas chromatograph. chromatograph How might the chromatograms of the chips have been obtained? E2 What method was used to confirm that new peaks in the chromatogram of the off-flavoured chips could be attributed to the substances that caused the disagreeable flavour? E3 What recommendations would you make to overcome the problem at the factory?

summary

6.3

Hi Highh performance f liliquid id chromatography h t h (HPLC) andd gas chromatography (GC) are very sensitive techniques and are used for qualitative and quantitative analysis. In HPLC the mobile phase is liquid under pressure. An inert gas is used as the mobile phase in GC.

In I high hi h performance f liliquid id chromatography h t h andd gas chromatography, retention time is used to identify substances. GC and HPLC can be linked to other instruments (e.g. mass spectrometers) to enhance analysis and identification of components.

key question 2 Australian wines are routinely tested for ethanol content. A quick and reliable method is by gas–liquid chromatography. The peak areas produced by a sample of wine and a number of standard solutions of ethanol are shown in Table 6.3.

TABLE 6.3 Peak areas from GLC analysis of a wine sample and standards Wine Standard 1 Standard 2 Standard 3 Standard 4

% ethanol

Relative ethanol peak area

? 4.00 8.00 12.0 16.0

82 400 31 200 62 900 94 200 125 700

a Plot a calibration curve of concentration of ethanol against peak area. b Determine the percentage of ethanol in the wine sample. c Why is it necessary to measure the peak areas produced by a number of standards?

71 Chromatography

06

key terms adsorption calibration graph carrier gas chromatogram chromatography column chromatography desorption eluent flame ionisation detector

gas chromatography (GC) gas–liquid chromatography (GLC) gas–solid chromatography (GSC) high performance (or pressure) liquid chromatography (HPLC) injection port

Principles of chromatography 3 Write a definition of each of the following terms: adsorption, desorption, mobile phase, stationary phase, eluent, retention time, carrier gas. 4 There are several types of chromatography, including thin-layer and paper, gas and high performance liquid chromatography. What features are common to all kinds of chromatography? 5 a Use the terms ‘adsorbed’ and ‘absorbed’ correctly in each of the sentences below. i Water was by the towel as the wet swimmer dried himself. ii A thin layer of grease onto the cup when it was washed in the dirty water. b Explain the difference between the two terms ‘adsorbed’ and ‘absorbed’.

Thin-layer chromatography 6 Phenacetin was once an ingredient in analgesic drugs, but it is not used now because it causes liver damage. It is soluble in chloroform. A chemist wishes to analyse a brand of analgesic using thin-layer chromatography to determine whether it contains phenacetin. Outline the steps in the analysis. (Assume that a sample of pure phenacetin is available to the chemist.) 7 A sample of brown dye from a lolly is placed at the origin on a chromatography plate. The solvent front moves 9.0 cm from the origin. A blue component of the dye moves 7.5 cm and a red component 5.2 cm in the same time. Calculate the Rf values of the two components. 8 Consider the diagram of thin-layer chromatography of three food colours in Figure 6.2. a Why must the level of the solvent be lower than the origin where spots of the mixture are originally placed? b Why are Rf values always less than one? c How many different components have been used to make colour A? d Which components present in colours B and C are also in colour A? Explain.


liquid stationary phase mobile phase origin retention time Rf value solid stationary phase solvent solvent front

standards stationary phase thin-layer chromatography (TLC)

e Which component of colour A is most strongly adsorbed on the stationary phase? f Calculate the Rf values of each component of colour C. 9 Thin-layer chromatography showed that the black dye used in a brand of writing ink contained blue, red, orange and yellow components. The Rf values of these substances using ethanol as solvent are 0.59, 0.32, 0.80 and 0.19, respectively. a How far apart would the blue and yellow components be after the solvent front had moved 8.0 cm from the origin? b When the red component had travelled 6.0 cm from the origin, how far would the orange component have travelled? c Sketch the chromatogram of the ink to scale after the solvent front had moved 15 cm from the origin. 10 The amino acids present in a sample of fruit juice can be detected by thin-layer chromatography. Rf values of some amino acids using two separate solvents are given in Table 6.4.

TABLE 6.4 Amino acid Lysine Leucine Proline Valine 2-Aminobutyric acid Threonine Hydroxyproline β-Phenylamine Isoleucine Alanine Serine Glutamic acid Glycine Arginine Taurine Tyrosine

Solvent 1 R f 0.12 0.58 0.39 0.40 0.28 0.21 0.21 0.50 0.57 0.24 0.19 0.25 0.20 0.13 0.12 0.38

Solvent 2 R f 0.55 0.82 0.88 0.74 0.58 0.49 0.67 0.86 0.81 0.55 0.34 0.33 0.40 0.60 0.33 0.62

Forensic expert slates Maguire analysis

To achieve better separation of the complex mixture of substances present in the juice, a ‘two-way’ chromatogram was prepared. The first step in this procedure was to run a chromatogram using Solvent 1. The results of this chromatogram are shown in Figure 6.17.

A leading forensic scientist . . . criticised the scientific evidence in the so-called Maguire case, in which seven people were convicted in 1976 for running an IRA bomb factory. The surviving defendants (one died in jail) have, since their release in 1985, sought to clear their names. The forensic tests were the main evidence against the Maguire Seven. Government scientists told the 1976 court hearing that the seven had been ‘kneading’ the nitroglycerine [used in the bombing]. An 18-year-old trainee at the Royal Armament Research and Development Establishment (RARDE) found sizeable amounts of the explosive nitroglycerine in a thin-layer chromatography analysis of hand swabs from the accused. All seven were strongly positive. No other traces of explosives were found in the house.

Front using Solvent 1

Distance (cm)

50 40 A

30

B 20

C

10

D E Original spot

0 X

Figure 6.17

a Calculate the Rf value of each spot on the chromatogram. b Try to identify the amino acids responsible for each spot. The paper was then turned around so that it lay at a right angle to the original and a second chromatogram was run using Solvent 2. Figure 6.18 shows the appearance of the paper after some time. Front using Solvent 2

40 30 20 10

A

B

C

Front using Solvent 1

Distance (cm)

50

E

! At the time the Maguire Seven were arrested, civil unrest in Northern Ireland had spread to England to include a program of bombing civilian targets such as pubs. By the time convictions for the Maguire Seven were overturned in 1991 a peace deal had been worked out. In February 2005 the British Prime Minister, Tony Blair, issued a public apology to the Maguire family for the miscarriage of justice.

12 Refer to Figure 6.19. Solvent front

D

A = leucine B = taurine C = valine D = glycine

Original placement of spot X

0 50

40

30

20

10

X = medicine

0

Figure 6.18

c Use the table of Rf values to identify each component in the mixture. d What is the advantage of a two-way chromatogram? 11 Read the following article from a 1990 British magazine and answer the questions. a Nitroglycerine forms colourless solutions. How might it be made visible in thin-layer chromatography? b In the mid 1970s, scientific policy was to use up all the sample when performing such tests. Suggest a reason for this. c On what grounds might the forensic scientist be criticising the scientific evidence?

A

B

X

C

D

Figure 6.19

73 Chromatography

a The following materials are commonly used in chromatography: water, ethanol, paraffin wax, glass sheets, paper strips, powdered alumina, nitrogen gas and hexane. Choose a suitable stationary phase and mobile phase from the list to use in the analysis of amino acids by thin-layer chromatography. b Calculate the distance moved by the sample of leucine if the solvent front moves 13 cm from the origin. c The separation of taurine and glycine on the chromatogram is not very great. State what would happen to the Rf values of the two amino acids if the separation was carried out for a longer period so the solvent front moved 10 cm instead of 5 cm from the origin. Explain your answer. d Write a short paragraph explaining how leucine is separated from the other amino acids.

High performance liquid chromatography 13 What are the advantages of HPLC for the analysis of drugs compared to an analysis technique based on column chromatography? 14 The organophosphorus insecticide parathion has been widely used in mosquito-prone areas. An empty drum of the insecticide was found close to a major reservoir. The EPA was asked to analyse the water to determine whether it was a threat to human health. Levels above 0.01 mg L–1 in water are a threat to human health. Parathion has an LD50 value (lethal dose to 50% of test animals exposed to this concentration) of 8 mg kg–1. Parathion standards in water of 0 ppm, 10 ppm, 20 ppm and 30 ppm parathion were prepared and analysed by HPLC (Figure 6.20). An undiluted sample of the reservoir water was run on the column under the same conditions.

a Complete the table below. Standard Standard 0 ppm parathion Standard 10.0 ppm parathion Standard 20.0 ppm parathion Standard 30.0 ppm parathion Reservoir water b Construct a calibration curve for the analysis of parathion. c Determine the concentration of parathion in the water sample. (Assume that peak height is a measure of the concentration of the insecticide.) d Is the reservoir water within the legal limits for safe drinking? Demonstrate your answer with a calculation. e What volume of water would a laboratory mouse, mass 150 g, need to drink to reach the LD50 dose? f Give possible reasons to explain the small peak observed with the 0 ppm standard of parathion.

Gas chromatography 15 In gas–liquid chromatography: a what is the usual mobile phase? b what is the stationary phase? c why is the column packed with very fine particles? d why is the injection port heated? 16 Explain how gas chromatography can be used for: a qualitative analysis b quantitative analysis 17 A mixture of four alkanes, decane, heptane, hexane and octane, were separated by gas–liquid chromatography (Figure 6.21). The solubility of the hydrocarbons in the liquid phase on the column was directly proportional to the relative molecular mass of the alkane.

A 0 ppm

10 ppm

20 ppm

30 ppm

Figure 6.20 Chromatograms produced for suspected contaminated water sample and parathion standards.

Chemical analysis C

B

C

D

Sample

Standards

74 47

Peak height (cm)

Figure 6.21 Chromatogram of four hydrocarbons.

a Identity the hydrocarbons A, B, C and D. b Why does decreased solubility of a chemical in the stationary phase decrease the retention time?

c Why do the peaks become broader with increased retention time? d List the factors that can increase the retention time of a particular chemical in GLC. 18 The organic compounds dimethyl ether (CH3OCH3) and ethanol (CH3CH2OH) have the same molecular formula, C2H6O, and the same molar mass; however they have different structural formulas. A sample containing both ethanol and dimethyl ether is analysed by GLC using a flame ionisation detector. a Will the two compounds produce peaks with the same retention time? Explain your answer. b If the sample has the same concentration of both chemicals, will the peaks produced have the same area? Explain your answer.

Connecting the main ideas

Absorbance

19 A herbal tea extract was analysed using HPLC. The chromatogram obtained is shown in Figure 6.22. a Explain what information chemists can obtain from this chromatogram. b How many components are evident? c Briefly explain how the components are separated by the HPLC technique.

0

1

2 3 Time (minutes)

4

5

Figure 6.22

20 Various forms of heart disease including angina are very successfully treated using nitroglycerine patches. The small colourless plastic patches are stuck to the skin and slowly release nitroglycerine into the blood at a rate of 0.4 mg/h. The amount of nitroglycerine in the patches must be controlled carefully to prevent an overdose.

A nitroglycerine patch was dissolved in approximately 20 mL ethanol and then diluted to 100 mL. 10.0 mL of this solution was diluted again to 100 mL with water. 20 μL volumes of the diluted sample and 20 μL of each of the prepared standards of nitroglycerine were injected onto a 1.5 m GLC column and analysed using a nitrogen-specific detector. Nitroglycerine standards Standard 5 μg/mL Standard 10 μg/mL Standard 15 μg/mL Diluted sample solution

Peak area (mm2) 7.2 14.6 22.0 10.8

a Construct a calibration curve and determine the concentration of nitroglycerine in the diluted sample. b What mass of nitroglycerine was in the patch? c The patch has a total mass of 0.50 g. What is the percentage concentration (w/w) of nitroglycerine in the patch? 21 Aspirin is a drug widely used in headache and cold preparations. Organic compounds such as aspirin in a formulation can be analysed by a variety of methods, for example paper chromatography, thin-layer chromatography, gas–liquid chromatography or high performance liquid chromatography. Final choice of a method of analysis involves consideration of the facilities available in the laboratory as well as problems inherent in each technique. A range of problems that can occur with some of these techniques is listed. Complete the table below using the letter codes to indicate the problem or problems associated with each technique. Problems may be used more than once. A Difficult or impossible to get quantitative data B Requires large amounts of solvent to operate C Samples must be stable to heat D Samples must be able to be dissolved in solvent E Expensive equipment needed F Samples cannot have a high molecular mass Technique Paper chromatography

Problem(s) (give letter)

Column chromatography Thin-layer chromatography Gas–liquid chromatography High performance liquid chromatography

75 Chromatography

0

chapter 07

Spectroscopy key knowledge • principles and applications of spectroscopic techniques and interpretation of qualitative and quantitative data from atomic absorption spectroscopy (AAS), infrared (IR) spectroscopy, nuclear magnetic resonance (NMR) spectroscopy and visible and ultraviolet (visible–UV) spectroscopy

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • identify the parts of the electromagnetic spectrum associated with the following techniques: flame tests, visible–ultraviolet spectroscopy, atomic emission spectroscopy, atomic absorption spectroscopy, infrared spectroscopy and nuclear magnetic resonance spectroscopy • suggest appropriate techniques for the analysis of elements (atoms) and molecular compounds • describe the equipment and procedures necessary for the techniques • identify an appropriate analytical procedure for a particular sample • explain the chemical or physical basis of each analytical technique and relate this to how the nature of samples makes them appropriate for analysis by the technique • make qualitative and quantitative interpretations of the data from these spectroscopic techniques • determine the concentration of samples from data obtained by instrumental analyses.

7.1

The electromagnetic spectrum Visible

Radiofrequency 106

Frequency (Hz) Wavelength L (m) Spectroscopic technique

107

Microwave

108

103

109 1

Nuclear magnetic resonance (NMR)

1010

1011

Infrared 1012

Ultraviolet

1013

1014

10-3

1015

1016

10-6 Infrared (IR)

X-rays 1017

1018

G-rays 1019

1020

10-9 Ultraviolet (UV)

X-ray

Atomic absorption (AAS)

Figure 7.1 The electromagnetic spectrum. Different types of spectroscopy use radiation from different parts of the electromagnetic spectrum.

All forms of spectroscopy use a part of the electromagnetic spectrum to give us information about the materials around us. Electromagnetic radiation interacts with atoms and molecules. The nature of this interaction depends upon the energy of the electromagnetic radiation. In this chapter you will learn about the effect of radiation from different parts of the electromagnetic spectrum on atoms or molecules, and how different spectroscopic techniques utilise these effects to provide us with information about the structure and composition of substances. The various spectroscopic techniques provide us with information about: the type of atom or molecule that is present (qualitative analysis) how much of a particular atom or molecule is present (quantitative analysis) the structure and bonding of the molecule. Spectroscopic techniques utilise the fact that: atoms or molecules absorb and emit electromagnetic radiation of specific energies atoms and molecules undergo a change when they absorb electromagnetic radiation different parts of the electromagnetic spectrum affect different parts of the atom or molecule. Radiation from each portion of the electromagnetic spectrum has a specific frequency, wavelength and energy associated with it. Ultraviolet light is very short wave radiation with high energy, while radio waves have long wavelengths and low energy. The energy of the radiation determines which part of an atom or molecule is affected. In each of the spectroscopic techniques we will look at in this chapter, the atom or molecule absorbs a specific quantum of energy which causes the atom or molecule to move to a higher energy level. With atoms, we look at the movement of electrons to higher energy levels. With molecules, as well as the movement of electrons to higher energy levels, we observe the

Wavelength, L

Wave A L

Wave B Distance travelled (m)

Figure 7.2 Wavelength measures the distance travelled by a wave during one cycle. Frequency refers to the number of waves that pass a particular point every second.

chemfact Some of the properties of light are explained by a wave model while other properties are explained by a particle model. The relationship between the frequency (ν) and wavelength (λ) of electromagnetic radiation is given by the equation c = λν where c is the speed of light. Different colours of light have different wavelengths. Light is emitted in discrete packets of energy called photons. The energy (E ) of a photon of light depends upon its frequency (ν): E = hν where h is called the Planck constant and has a value of 6.63 × 10–34 J s.

77 Spectroscopy

Increasing energy Electronic energy

H

Vibrational energy

H

O H

O H

O H

H Rotational energy

Translational energy

O H

H

O H

H

Figure 7.3 A water molecule has different types of energy.

movement of molecules to higher vibrational, rotational and nuclear spin energy levels. Like the electronic energy levels, all of these energy levels are quantised; that is, they have certain fixed values. In each case the energy difference between the quantum levels is given by the equation ΔE = hν.

TABLE 7.1 Spectroscopic techniques make use of the way electromagnetic radiation interacts with atoms and molecules Spectroscopic technique

Part of the electromagnetic spectrum Ultraviolet

Wavelength range (cm) (approx) 4 × 10–5 to 10–7

Visible

7 × 10–5 to 4 × 10–5

Atomic absorption (AAS) and atomic emission spectroscopy (AES); flame tests Infrared (IR) spectroscopy

Visible

7 × 10–5 to 4 × 10–5

Infrared

0.01 to 7 × 10–5

Nuclear magnetic resonance (NMR) spectroscopy

Radio

> 10

Ultraviolet (UV) spectroscopy Colorimetry

Part of atom or molecule affected Electrons in molecules Valence electrons in molecules Valence electrons in atoms

Bending and stretching of bonds in molecules Nuclear spin states

7.2 TABLE 7.2 Flame colours* Metal Sodium Strontium Copper Barium Lithium Calcium Potassium

Flame colour Yellow Scarlet Green Yellow-green Crimson Red Lilac

*Some common metal ions, including iron, silver, tin, aluminium, zinc and magnesium, do not produce flame colours. They emit radiation in the ultraviolet (UV) region of the electromagnetic spectrum.


Analysis of atoms We will start by looking at three techniques that use radiation from the visible region of the electromagnetic spectrum to give us information about the elements present in a sample: flame tests, atomic emission spectroscopy, and atomic absorption spectroscopy. While the details of the instruments are different, the chemistry behind the techniques is similar.

Flame tests We are all familiar with the colourful effects that are created in fireworks displays (Figure 7.4). Small quantities of different metal compounds are incorporated in fireworks to create different colours. For example, strontium compounds produce an eye-catching scarlet, and sodium compounds are added to produce yellow. If you have ever spilt salt into the flame of a gas stove, you might have noticed this same yellow colour. Chemists use the fact that some metals produce particular colours when they are heated as a convenient method of analysis. The metallic elements present in a compound can often be determined simply by inserting a sample of the compound into a non-luminous Bunsen burner flame (Figure 7.5). The metal is identified by comparing the flame colour with a list of the characteristic colours produced by metals (Table 7.2). To understand how the colour is produced we must look at the structure of the atom. A simple model of an atom is shown in Figure 7.6. In this model a central nucleus is surrounded by electrons moving in regions called shells.

Flame colour Sample adhering to wire

Burner

Figure 7.5 Performing a flame test. A moist wire has been dipped in the sample and then placed in the flame. A fine spray of the solution from a spray bottle could be used instead.

Figure 7.4 Metal ions incorporated in the fireworks are responsible for the colours in this display.

These shells, also called energy levels, and electrons in these shells, are located at different distances from the nucleus. The negatively charged electrons are attracted to the positively charged protons in the nucleus by electrostatic attraction. Electrons in energy levels close to the nucleus have the lowest energies and experience the strongest attraction to the nucleus. An electron can jump to a higher energy level if it absorbs energy that corresponds exactly to the difference in energy between the lower energy level and the higher energy level. When electrons jump to a higher energy level they move to a higher energy subshell in a different shell. For some metal atoms, the heat of a Bunsen flame provides sufficient energy for this to happen. Higher energy levels are unstable so the excited electrons quickly return to lower energy levels, emitting the energy they had absorbed. If the energy emitted falls within the band of energies visible to the eye, it can be seen as coloured light (Figure 7.7). Radiation emitted from other parts of the electromagnetic spectrum, such as the ultraviolet region, is detected using instruments. The energy of a photon of light emitted by an excited electron has a fixed value, equal to the difference in energy between the higher energy level and the lower energy level to which it returns. The light consequently has a specific wavelength and colour. Electrons in an atom can be excited to

Nucleus: contains protons and neutrons 1st shell 2nd shell 3rd shell

Figure 7.6 A simplified representation of the arrangement of electrons in a sodium atom. Higher energy shells that are not occupied by electrons are not shown. (In Unit 1 you learned that electrons are now thought to be distributed in a more complex fashion than shown here. Within each shell, electrons are arranged in energy levels called subshells.)

Energy emitted

Energy from the flame promotes an electron to a higher energy level.

The electron quickly returns to a lower energy level, emitting light of a particular colour.

Figure 7.7 During a flame test, electrons ‘jumping’ from a high to a low energy level lose energy by emitting coloured light.

79 Spectroscopy

review For more detail on electron shells, see Heinemann Chemistry 1, Chapter 2.

a number of higher energy levels, so light emitted from a sample placed in a flame is likely to be a mixture of several different colours (and energies). Violet light, the highest energy form of visible light, has a relatively small wavelength. Conversely, red light has low energy and a longer wavelength.

Atomic emission spectroscopy

ATOMIC STRUCTURE E • Visible spectrum of hydrogen • Emission spectra

Flame tests can give us only limited qualitative information about the likely elements present in a sample. Only a few elements give a coloured flame in a Bunsen burner flame, and the colours of some are alike (Table 7.2). In impure samples a fainter colour may be masked by a stronger one. Figure 7.5 shows the sample giving a reddish colour to the flame. Is this the scarlet of strontium, crimson from lithium, or red due to calcium? Two modifications greatly improve the usefulness of the technique: using a hotter flame, so that sufficient energy is available to excite electrons in a wider range of elements passing the light through a prism, as shown in Figure 7.8. The different energies in the light emitted by a heated sample are separated into a series of coloured lines, called an emission spectrum. Hot sample

Slit

Figure 7.8 Essential elements of an atomic emission spectrometer.

Prism

Emission spectrum

Because each element has a different number of protons in the nucleus, the attraction of the nucleus for electrons will vary from element to element. No two elements will therefore have energy levels of exactly the same energy, so a spectrum is characteristic of a particular element. It may be used as a ‘fingerprint’ to identify the elements present in a substance (Figure 7.9).

a

b

c

d

Figure 7.9 The emission spectra of a calcium, b sodium, c mercury and d cadmium.


Each line in the emission spectrum corresponds to radiation of a specific wavelength, frequency and energy exactly equal to the difference in energy of the electron energy levels. Because the energy levels have a specific value—are quantised, with fixed values—discrete lines appear rather than a continuous spectrum. Separating the light from the flame into its individual wavelengths—the atomic emission spectrum—allows us to use atomic emission spectroscopy (AES) to obtain a more certain qualitative analysis. Figure 7.9a shows the lines formed from our red flame in Figure 7.5—the characteristic fingerprint of calcium.

summary

7.1–7.2

Th The manner in i which hi h electromagnetic l t ti radiation di ti interacts i t t with ith matter depends upon the energy of the radiation. Electrons are promoted to higher energy levels when they absorb energy from radiation in the ultraviolet and visible regions of the electromagnetic spectrum.

Radiation R di ti off a particular ti l frequency f andd wavelength l th is i emitted itt d when electrons fall from a high energy level to a lower energy level. Use is made of this in atomic emission spectroscopy and in analysis based on flame colour.

key questions 1 Potassium chloride can be used instead of salt by people suffering from high blood pressure. Suppose, while cooking, someone spilt some potassium chloride into the flame of a gas stove. a What colour would the flame be? b Suggest why the presence of chloride ions has no effect on the flame colour.

2 Barium and calcium are both group 2 elements. Account for the observation that samples of barium and calcium compounds produce different colours when they are held in a flame. 3 Why does an emission spectrum contain a number of lines of different colours? 4 Explain why atomic emission spectroscopy is regarded as a superior method of analysis to flame tests.

7.3

Atomic absorption spectroscopy Very few elements are excited by even the hottest laboratory flame, and flame atomic emission spectroscopy is therefore useful for identifying only a limited number of metals, particularly the group 1 and 2 elements. In the early 1950s, the Australian scientist Alan Walsh was working on the measurement of small concentrations of metals at the CSIRO using atomic emission spectroscopy. One Sunday morning, while working in his garden, he had the idea of looking at the light absorbed by atoms, rather than at the light emitted by them. This flash of inspiration proved to be the key to a method of analysing metals that is much more sensitive and accurate than atomic emission spectroscopy, and that can be used to identify and measure a much wider range of metals. Next morning, Walsh was able to test and confirm his idea in his laboratory, using some makeshift pieces of equipment. His method is called atomic absorption spectroscopy (AAS).

Figure 7.10 Australian scientist Alan Walsh in 1952 with the atomic absorption spectrometer he developed.

81 Spectroscopy

chemistryy in action A life-saving technique It was not long before Alan Walsh’s exciting new technique was able to prove its worth. A young boy, who had suffered severe burns and was fighting for his life in a Sydney hospital, had shown no signs of improvement and was beginning to have violent convulsions.

Figure 7.11 A modern atomic absorption spectrometer. A computer allows rapid analysis of many samples for a wide range of applications.

Figure 7.12 ‘The Mad Hatter’s Tea Party’, one of Arthur Rackham’s illustrations for Alice in Wonderland. The expression ‘mad as a hatter’ has its origin in the days when mercury was used in the treatment of felt for hats. Workers who treated the felt suffered from mercury poisoning; their bodies would shake and they would experience muscular tremors and personality changes.


Analysis of his blood by atomic absorption spectroscopy revealed an acute magnesium deficiency, which was promptly treated and the boy’s life was saved.

Atomic absorption spectroscopy is: one of the most widely used of modern instrumental techniques an Australian invention that is now used all around the world and has earned millions of export dollars for Australia very versatile, being capable of detecting over 70 elements extremely sensitive, detecting concentrations of elements at part per million (ppm) levels or, in some cases, part per billion (ppb) levels. One of the many uses of atomic absorption spectroscopy (AAS) is to measure the concentration of mercury in fish. Perhaps you are hoping to have fish and chips for dinner tonight. Fish is an excellent source of protein, low in fat and cholesterol—nothing could be better … or could it? Fish, particularly large predatory fish such as tuna and shark, can sometimes contain high concentrations of mercury. Mercury is an extremely toxic metal that accumulates in the liver, kidneys and brain tissue of living creatures. It occurs naturally in the environment in small quantities, but increased levels can result from the discharge of industrial wastes. The mercury present in some Australian waters is thought to result from use of that element to extract gold during the gold rush era of the nineteenth century. Mercury builds up as it passes along the food chain. Organisms such as shellfish concentrate mercury 100 000 times above the concentration in surrounding water. Sharks are at the top of a long food chain; some have been analysed for mercury by AAS and found to have levels as high as 29 ppm. This is considerably higher than the limit of 0.5 ppm that is recommended as the maximum mercury level fit for human consumption. Analysis of magnesium in blood and mercury in fish are just two examples of the large range of applications of AAS. Others include: urine and blood analysis to detect an excess or deficiency of metals detection of toxic metals such as lead or copper in blood analysis of toxic metals in food and drink environmental sampling—testing for metal ion pollution of air, soil and water assays of mineral and soil samples to determine the type and amount of metals present analysis of metals present in engine oil to predict the possibility of engine failure (Chapter 1).

How atomic absorption spectroscopy works Alan Walsh’s breakthrough was to recognise that atoms will absorb light if the energy of the light is exactly that required to promote an electron from its normal energy level to a higher energy level. Because each element has a unique absorption spectrum, each element to be analysed requires its own light source that will emit light of the correct wavelength. The light is provided by a special lamp called a hollow source cathode lamp. The

cathode is composed of the element of interest, and when this is vaporised it produces light of the correct wavelength required for the analysis. A solution of the sample to be analysed is sprayed into a flame where it is converted into an atomic vapour (Figure 7.13). Light containing the chosen wavelength is passed through the flame. Atoms of the element being analysed that are present in the flame absorb some of the radiation. The light beam is then passed through a filter (monochromator) to select the light of the chosen wavelength, and its intensity is measured by an electronic detector. The amount of light absorbed indicates the quantity of the element present in the original sample.

Atomic vapour in flame

Lamp emits light of wavelength required to excite electrons

Light beam is chopped into pulses

Solution of sample is sprayed into flame

Detector measures pulsed light

Monochromator and slit select light of a particular wavelength

Figure 7.13 Essential elements of an atomic absorption spectrometer. Absorbance 10

Interpreting atomic absorption spectra Worked example 7.3 A batch of oysters was suspected of containing dangerous levels of lead. The following is an account of their analysis. One of the oysters was removed from its shell and chopped finely in a food processor. A 1.50 g portion was heated on a hotplate with 10 mL of nitric acid. This mixture was filtered and then sprayed into the flame of an atomic absorption spectrometer. A hollow cathode lamp that emitted light with a wavelength that is absorbed by lead atoms was in place. A significant amount of light was absorbed, indicating that there was lead in the oyster. To find the concentration of lead ions in the sample, the absorbance of different solutions containing known concentrations of lead ions was also measured. The results are graphed in Figure 7.14. The recommended limit of lead in oysters for human consumption is 2.5 ppm. Are these oysters fit to eat?

8 6 Sample 4 2

Absorbance of standard solutions 5

10

15

20

Concentration of lead (ppm)

Figure 7.14 The variation of absorbance against concentration, showing measurements for the standard solutions of lead ions and for the oyster sample.

Solution From the graph, the concentration of lead in the solution prepared from the oyster sample is 9.0 ppm, which is equivalent to 9.0 μg mL–1. So, in the 10 mL solution there is 9.0 × 101 = 90 μg of lead. Since 90 μg lead was present in the 1.50 g portion of oyster: concentration in the oyster = 90 μg/1.50 g = 60 μg g–1, or 60 ppm Since the recommended limit of lead in oysters for human consumption is 2.5 ppm, you would be wise not to eat the oysters that were the subject of this test!

! In a solid sample: 2.5 ppm = 2.5 μg per g of sample = 2.5 mg per kg sample In an aqueous sample: 2.5 ppm = 2.5 μg per mL of sample = 2.5 mg per L of sample

83 Spectroscopy

extension ICP-AES Even with atomic absorption and the use of very hot flames (a mixture of oxygen and ethyne gas gives the hottest flame, up to 3400 K), there are still many elements that cannot be analysed by AAS or that give only a poor response. The most modern instruments now can analyse most elements through the use of an inductively coupled plasma (ICP) instead of a flame to excite the atoms. The ICP generates very high temperatures of 7000–10 000 K to create the plasma. At these temperatures virtually all the atoms in the sample are excited and able to emit electromagnetic radiation as they relax (return to their unexcited ground state). The advantages of ICP-AES over AAS include:

questions

• ICP-AES can be used to identify most elements • it is suitable for all concentrations, down to very low part per billion (ppb) levels for some elements • it can identify many elements at once as the emission spectra are resolved by comparison to a computerised database of spectral lines, whereas AAS can determine only one element at a time • it is very fast: a multi-element analysis of 70 elements takes 2 minutes.

! The ICP can also be used to vaporise the sample that goes into a mass spectrometer (Chapter 8).

?

E1 What are the advantages of using inductively coupled plasma rather than a flame to excite atoms in an atomic absorption spectrometer?

summary

E2 Plasma is often referred to as the fourth state of matter. Conduct a web search to find out how the plasma state differs from the other states of matter.

7.3

IIn AAS, AAS a hollow h ll cathode th d lamp l is i usedd that th t emits it the th particular ti l wavelength required to excite the atoms of the element being analysed.

The Th concentration t ti off metal t l ions i is i determined d t i d by b measuring i the th amount of radiation of a particular wavelength that is absorbed.

key question 5 Iron is essential to our health. To determine the iron content in a flavoured milk drink, a 5.0 mL sample was diluted to 50.0 mL. The absorbance of the diluted solution and of several standard solutions was measured using AAS. The results are shown in Table 7.3.

TABLE 7.3 Atomic absorption spectroscopy measurements Solution concentration (ppm) 0.00 1.00 2.00 3.00 4.00 Sample


Absorbance 0.010 0.080 0.150 0.220 0.290 0.190

a Plot a graph of absorbance against concentration of iron. b What is the concentration of iron, in ppm, in the diluted drink? c Calculate the concentration of iron, in ppm, in the undiluted drink. d What mass of iron would you consume in a 250 mL glass of drink? e The recommended daily allowance (RDA) of iron for people over the age of 11 years is 18 mg. What percentage of your daily needs does a 250 mL glass of this drink provide? f The 0.00 ppm standard, which contained no added iron, gave a small absorbance reading. Suggest an explanation for this. g Suggest why the sample of the drink was diluted in order to measure its absorbance.

7.4

UV–visible spectroscopy UV–visible spectroscopy makes use of the fact that many substances absorb light of characteristic wavelengths. The wavelengths of the light absorbed by compounds can be used for their identification. UV–visible spectra arise because photons in this region of the electromagnetic spectrum have sufficient energy to promote electrons from low energy levels to higher energy levels. Such electron transitions may occur in atoms, ions or molecules. Because different substances have many different energy levels, the energy and wavelength of the light required to promote the electrons vary. The spectrum can therefore be used to assist in identifying a substance. When a substance absorbs visible light, it appears coloured. The colour observed is the complement of the absorbed colour because this is what remains to reach our eyes (Table 7.4). For example, chlorophyll appears green because its molecules absorb violet and red light. A solution of copper(II) sulfate is blue because of the absorption of energy from the orange region of the visible spectrum.

chemfact The relationship between absorbance and concentration is described by the Beer– Lambert Law: A = εcl where A is absorbance, c concentration in mol L–1, l is the length of the cell in cm and ε is constant. If the same cell size is used, the absorbance A can be used to measure the concentration of the analyte.

TABLE 7.4 Colours of visible light and complementary colours Wavelength (nm) 380–420 420–440 440–470 470–500 500–520 520–550 550–580 580–620 620–680 680–780

Colour absorbed Violet Violet–blue Blue Blue–green Green Yellow–green Yellow Orange Red Purple

Colour observed Green–yellow Yellow Orange Red Purple Violet Violet–blue Blue Blue–green Green

Although it can be used for qualitative analysis, UV–visible spectroscopy is mainly used for determining the concentration of a substance in a sample. This procedure involves recording the spectrum of the pure substance and selecting a wavelength at which the substance absorbs strongly but other components in the sample do not. The absorbance of the sample is then measured at this wavelength and compared to the absorbance of a series of standard solutions. UV–visible spectroscopy has a wide range of applications, including: clinical analysis, measuring the concentrations of specific substances in body fluids such as urine or blood. Hospital pathology laboratories use UV–visible spectroscopy to determine the haemoglobin content and sugar levels in blood determining the amount of coloured dye in plastics identifying the presence of metal ions; even if the metal ion itself is not coloured, it may be possible to analyse it by converting it into a coloured compound determining the levels of nutrients, additives and contaminants in water and foods in quantitative analysis of DNA and proteins in the field of molecular biology.

! The wavelength of visible light is measured in nanometres (nm). 1 nm = 10–9 m. The human eye detects electromagnetic radiation as visible light in the range from 400 nm (violet) to 700 nm (red). Radiation in the ultraviolet (< 400 nm) and infrared (> 700 nm) regions cannot be detected by the human eye.

85 Spectroscopy

The basic features of a simple spectrophotometer are shown in Figure 7.15. The radiation source provides ultraviolet and visible light of all wavelengths. The monochromator selects a particular wavelength from those emitted by the source, and the detector measures the intensity of the light that passes through the sample. Narrow beam

Light detector

Light source Slit

Monochromator (wavelength selector)

Recorder Sample solution

Absorbance

Figure 7.15 Essential elements of a simple UV–visible spectrophotometer.

300

400

500

600

700

Wavelength (nm)

Figure 7.16 Visible spectrum of chlorophyll. Chlorophyll absorbs strongly in the violet (420 nm) and red (660 nm) regions of the spectrum, so it appears green in colour.

Usually the substance under investigation is in solution. It is placed in the spectrophotometer in a special cell made of quartz or fused silica transparent to ultraviolet and visible light. With a simple spectrophotometer, a reference reading is first taken with a cell containing only pure solvent. This is used to compensate for any reflection, scattering or absorbance of the light by the cell and the solvent. The reference cell is then replaced with a cell containing a solution of the sample. The absorbance by the sample is found by comparing the two readings. By measuring the absorbance at various wavelengths a graph, or spectrum, for the sample can be obtained. The characteristic spectrum of chlorophyll is shown in Figure 7.16. Modern analytical laboratories now use the more sophisticated double beam scanning spectrophotometer (Figure 7.17). In this instrument the beam of light is rapidly ‘chopped’ and passed alternately through the reference and sample cells. This compensates for changes in the output of the radiation source and enables the absorbance to be measured accurately. Semitransparent mirror

M

ot

or

Rotating mirror (beam chopper)

Detector

Radiation source

Monochromator

Sample cell

Chart recorder

Slit

Mirror

Mirror Reference cell

Figure 7.17 Essential elements of a double beam scanning spectrophotometer. The light beam is passed alternately through the sample and the reference cells by the rotating mirror (‘beam chopper’).

Interpreting ultraviolet–visible spectra In ultraviolet–visible spectroscopy, the absorbance of visible or ultraviolet light by a series of standard solutions is measured and a calibration graph is drawn. By measuring the absorbance of light of an unknown solution, its concentration can be found directly from the graph.


Worked example 7.4 The concentration of iron in a sample of dam water was determined by UV spectroscopy. Iron present as Fe2+ ions was reacted to form an orange–yellow complex, iron(II)-1,10phenanthroline. The absorbance of a series of standards and a sample of dam water is shown in Table 7.5. Determine the concentration of iron in the dam water in μg mL–1 (ppm).

TABLE 7.5 Concentration Fe2+ (μg mL–1) Sample 4.0 8.0 12.0 16.0

Absorbance 0.38 0.16 0.31 0.47 0.63

Solution The first step is to construct a calibration graph with the data from Table 7.5. See Figure 7.18. The concentration of iron is plotted on the horizontal axis and the absorbance on the vertical axis. 0.7 0.6

Absorbance

0.5 0.4 0.3 0.2 0.1 0 0

2.0

4.0 6.0 8.0 10.0 12.0 14.0 16.0 Concentration Fe2+ μg/mL

Figure 7.18 Calibration graph for iron in a sample of dam water.

The observed absorbance of the sample is located on the vertical axis and a horizontal line drawn to intercept the calibration graph. A vertical line is dropped to the concentration axis and the concentration in the sample from the dam water is read from the scale. The dam water was not diluted before being treated and analysed so the concentration of iron in the dam water was 9.9 μg mL–1 or 9.9 ppm Fe2+.

summary

7.4

Th The unique i absorption b ti spectrum t off compounds d in i the th UV–visible region can be used to identify unknown substances.

UV–visible UV i ibl spectroscopy t can also l be b usedd to t determine d t i concentration by measuring absorbance of a sample at a selected wavelength and comparing this to the absorbance at the same wavelength of a series of standard solutions.

87 Spectroscopy

key questions

400

500 600 Wavelength (nm)

ultraviolet violet

blue

Figure 7.19 Absorption spectrum of a dye.

visible green

700

infrared yellow

red

Absorbance

11 The absorption spectrum of benzene is shown in Figure 7.20.

180

200

220

240

260

280

Wavelength (nm)

Figure 7.20 Absorption spectrum of benzene.

a In what part of the elctromagnetic spectrum does benzene absorb? b Explain why humans see benzene as colourless. 12 The phosphate content of a detergent may be analysed by UV–visible spectroscopy. In one analysis, a 0.250 g sample of detergent powder was dissolved in water and the solution made up to 250 mL. The solution was treated to convert any phosphate present to a blue-coloured molybdenum phosphorus compound. The absorbance of the solution at a wavelength of 600 nm was measured as 0.17. The absorbances of five standard phosphate solutions were measured in a similar fashion and the calibration graph in Figure 7.21 was obtained. Absorbances of standard solutions

Absorbance

6 Two samples of copper sulfate solution, with concentrations of 0.080 M and 0.30 M, were analysed using a spectrophotometer. a Which sample would allow most light to pass through to the detector? b Which sample would show the strongest absorbance of the light? 7 Why would red light, rather than blue light, be used in a colorimeter to measure the concentration of a blue copper sulfate solution? 8 Compare the analytical techniques of AAS and UV–visible spectroscopy. In what ways are they: a similar? b different? 9 The absorption spectrum of chlorophyll is shown in Figure 7.16. a At what wavelengths is there maximum absorbance of light? b What wavelength would you select if you were required to determine the concentration of chlorophyll in a leaf extract using UV–visible spectroscopy? Provide an explanation for your answer. 10 The absorption spectrum of a commercial dye is shown in Figure 7.19. What is the colour of the dye?

0.20

0.10

0.00 0

20 40 Concentration of P (mg L-1)

Figure 7.21

a What is the concentration of phosphorus in the 250 mL detergent solution? b Determine the percentage by mass of phosphorus in the detergent powder. c Why was a wavelength of 600 nm selected for this analysis?


7.5 chemfact

Infrared spectroscopy Infrared radiation is lower in energy and of a longer wavelength than visible and ultraviolet light. The energy from infrared radiation is not enough to promote electrons to a higher energy level, but it is enough to cause changes to the bonds in molecules. Covalent bonds can be compared to springs that can undergo specific amounts of bending or stretching. As the atoms in the bonds change position due to bending or stretching, the molecule vibrates. Just as electrons can only occupy discrete electronic energy levels, so molecules are only able to occupy discrete vibrational energy levels. A molecule will absorb a discrete quantum of energy to move from one vibrational energy level to the next. The amount of energy required is provided by radiation in the infrared region of the electromagnetic spectrum. Infrared spectroscopy therefore exploits the fact that molecules are flexible structures that can bend and stretch. Infrared spectroscopy is a powerful analytical technique because, apart from a few homonuclear molecules such as O2 and N2, all molecules absorb infrared radiation. For a molecule to absorb infrared radiation, the bending or stretching vibrations must change the overall dipole moment of the molecule.

Cancerous tumours can sometimes be detected using medical digital infrared thermal imaging. The growth of new blood vessels in cancerous tumours generates a ‘hot spot’ that can emit more infrared radiation than the surrounding tissue.

Energy levels

H O H Molecule in lowest vibrational energy level—the ground state

O

O

O

hν Energy levels

H H H

H

H

H

Figure 7.22 Bending (scissoring) in water molecules.

H O H

Diatomic molecule

Triatomic molecule

Nonlinear triatomic molecule

Stretc Stretch asymmetrical symmet ettrical Molecule gets more vibrational energy and moves to a higher energy level

Stretch St Stretch symme symmetrical al

Figure 7.24 As the molecule moves to a higher vibrational energy level the frequency of the stretching vibration increases. Bending, g, sciss scissoring Wagging, twisting and rocking

Figure 7.23 Stretching and bending motions in diatomic and triatomic molecules.

chemfact Some gases in the atmosphere, such as water, carbon dioxide and methane, will not absorb the higher energy ultraviolet and visible radiation reaching the Earth from the Sun, but they will absorb and ‘trap’ the lower energy infrared radiation emitted back into space by the Earth.

89 Spectroscopy

chemfact How does a microwave oven work? Microwave radiation is slightly lower in energy than infrared radiation. There is sufficient energy to cause changes to the rotational energy levels of the molecules. Molecules with a strong dipole such as water absorb microwave energy and move to higher rotational energy levels. As the molecules rotate faster and bump into each other, the temperature of the water and the food in which it is contained rises.

TABLE 7.7 The energy of absorption depends upon the mass of the atom bonded to carbon Bond C–H C–C C–O C–Cl C–Br

Typical absorption frequency (cm–1) 3000 1200 1100 750 600

chemfact The frequencies given in Tables 7.6 and 7.7 are typical values only. The absorption actually occurs over a band of frequencies because the vibrations of every bond in a molecule interact.

Generally, the range of energies absorbed depends on the strength of the bonds. A single C–C bond absorbs lower-energy infrared radiation than a triple C≡C bond (Table 7.6).

TABLE 7.6 The energy needed to excite the molecule to a higher vibrational energy level is dependent on the strength of the bond Bond

Bond energy (kJ mol–1)

C–C C=C C≡C

346 614 839

The dipole moment of a polar molecule is determined by the size of the charge separation and the distance between its atoms. As the molecule vibrates, the distance between the charges is changing and so the dipole moment is changing. Homonuclear diatomic molecules such as O2 and N2 do not absorb infrared radiation as they have no dipole moment to change as they vibrate.


Increasing energy

The mass of the atoms attached by the bond also affects the energy of the infrared radiation absorbed: the higher the mass, the lower the energy of the radiation absorbed. Table 7.7 compares atoms of different masses all bonded to carbon with a single bond. These characteristics of the infrared spectrum make it a powerful tool for identifying the bonds in molecules. It is particularly important in the pharmaceutical industry where knowledge of the molecular structure of existing drugs can help in the design of new medicines. You will recall that frequency and wavelength are related by the expression c = λν. Since c, the speed of light, is constant, frequency is inversely proportion to wavelength. In infrared spectroscopy frequency is measured in wavenumber, which is the reciprocal of wavelength (1/λ) and has units of cm–1. The relationship between frequency, wavelength and wavenumber is shown in Figure 7.25. 4000

2.5

1.0 x 1014

chemfact

Infrared absorption, stretch (cm–1) ~880 ~1720 ~2300

3000

2000

1000

Wavenumber (1/λ, cm–1)

10

Wavelength (λ, μm)

2.5 x 1013

Frequency (ν, Hz)

Figure 7.25 The relationship between wavenumber, wavelength and frequency.

Instrumentation Features of a dispersive infrared spectrophotometer are: a source of infrared radiation a sample and reference cell or disc made out of NaCl, KBr or similar; glass and plastic cannot be used as they absorb IR radiation and so are opaque to IR radiation a wavelength selector an infrared detector. The infrared radiation from the source is split into two beams and passed separately through the sample cell and a reference cell. The reference cell is used to discount the effect of the material of the sample cell, the solvent, and any interference from water or carbon dioxide in the atmosphere and from the background temperature of the room. The difference in transmittance, or transmitted radiation, between the sample and reference cell is due to the absorption of certain frequencies by the molecules of the sample. These absorptions result in changes in vibrational energy in the molecule under examination.

Sample cell for solution of sample

Monochromator Infrared source (electrically heated filament)

chopper

Computer Reference cell for solvent only

Figure 7.26 Essential features of a dispersive IR spectrophotometer.

Infrared detector

Interpreting the infrared spectrum chemfact

Qualitative analysis

Functional groups are atoms or groups of atoms that determine the properties and reactions of an organic molecule.

Infrared spectroscopy is most important for the information it can give us about the functional groups in a molecule. (You will learn more about functional groups in Chapters 9 and 10.) This can be used to identify a known substance or clarify the structure of an unknown material. It is a very powerful tool for biochemists investigating the structure of DNA and other complex biochemicals. The energy of radiation absorbed and released depends on: the kind of bonds in the molecule. If bonds are thought of as springs, then the strength of the spring will determine the energy of the wavelength absorbed. Each type of bond absorbs infrared radiation over a typical narrow range of wavelengths. For example, it takes a different amount of energy and so a different range of wavelengths of radiation to stretch a C–H bond compared to a C–C bond. The infrared spectrum of gaseous ethanol and some of the bond vibrations that cause the absorption bands are shown in Figure 7.27.

Transmittance (%)

100 C–H stretching 3010–2850 cm–1

80

60

H O-H stretch

H C

40

20

C–O stretching 1050 cm–1

O–H stretching 3670 cm–1

C O

H C-H stretch 3000

2000 Wavenumber (cm-1)

C-O stretch

H

1000

the environment or other bonds in the molecule. The wavelength absorbed depends upon the entire molecular structure. For example, propanoic acid and methyl ethanoate have the same molecular formula (C3H6O2), but their structures are different (Figure 7.28). Both contain a C=O group. The infrared spectra of these compounds have some features in common and some differences (Figure 7.29). Both spectra have an absorption band at about 1700 cm–1 due to the C=O bond. In methyl ethanoate the absorption just below 3000 cm–1 is due to the C–H bond. In propanoic

H H

Figure 7.27 Bond vibrations in gaseous ethanol and its infrared spectrum.

91 Spectroscopy

acid the broad absorption band from 2700 cm–1 to 3600 cm–1 is due to the O–H bond. This broad absorption band partly masks the absorption due to C–H bonds. The C=O stretch in propanoic acid produces a peak at a different wavenumber from the C=O stretch in methyl ethanoate.

a

a

O CH2

90

C

80 O

H

Transmittance (%)

CH3

Propanoic acid

b

O CH3

100

C O

70 60 50 O

40

20

CH3

10

Methyl ethanoate

C

0 4000

Figure 7.28 Structure of a propanoic acid and b methyl ethanoate.

H

30

H

C

3000

O

2000

1500

1000

500

1000

500

Wavenumber (cm–1)

b

100 90

Transmittance (%)

80 70 60 C

50

H

40 30 20 10

C

0 4000

H

H C

C

N

As the structure of a compound becomes more complex the number of vibrational modes increases. Acrylonitrile (propenenitrile, Figure 7.30) is used in the manufacture of synthetic rubber. Its infrared spectrum is shown

Figure 7.30 Structure of acrylonitrile

CH rock

H ~0.2% H2O impurity

C

Combination band

C

H C

H

C N stretch

C

C C C stretch

C H stretch

C

C

4000 3600 3200 2800 2400 2000

1800

1600 1400 Wavenumber (cm–1)

C

C

C CH wag

Figure 7.31 Infrared spectrum of acrylonitrile (propenenitrile).

Chemical analysis C

H

H

1200

H

C C stretch

CH2 rock H

H CH2 deformation

92

O


Figure 7.29 Infrared spectra of a propanoic acid and b methyl ethanoate.

C

H

3000

1000

C

C

H C

C

H C

H C twist

H CH2 wag 800

600

400

in Figure 7.31. The types of bond distortions (see Figures 7.22 and 7.23) that produce particular absorption bands are shown. No two molecular compounds are identical in bonds and bonding environment, and so infrared spectroscopy can give us a characteristic fingerprint for each compound. Some of the peaks in an IR spectrum are narrow. This usually means that the peak corresponds to one particular type of vibration of a bond. Other peaks are broad. These broad bands are combinations of peaks due to a number of vibrational changes that have overlapping energies. Peaks are described as strong, medium or weak. Strong peaks are observed when a high proportion of the molecules absorb at the given wavenumber. The infrared spectrum above 1000 cm–1 is used to identify functional groups. The spectrum at wavenumbers less than 1000 cm–1 is characteristic of a particular compound. This region is largely a result of bending and stretching of the entire molecule and so is unique to a particular compound. For this reason it is called the fingerprint region. So if a known substance and an unknown sample have the same absorption spectrum below 1000 cm–1 they are almost certainly the same compound. Chemists use tables showing the wavenumbers of the main functional groups in a variety of compounds. These can give broad clues to the functional groups present in the molecule, but for positive identification the entire spectrum will be compared to a computer library of spectra. Libraries containing tens of thousands of spectra are readily available. High energy

! An infrared spectrum is traditionally shown as a plot of transmittance, which is the amount of light coming through. The base line—100% transmittance of radiation—is at the top of the chart, and the absorption of radiation corresponds to a trough (inverted peak). An IR spectrum is upside down compared to a UV–visible spectrum. The peaks point down. This is because while the IR spectrum measures transmittance, UV spectroscopy measures absorbance, or the amount of light blocked.

Low energy Wavenumber (cm–1)

4000

3500

3000

2500

O H N H C H

C C C N

2000

1500 C C C O

1000

500

Figure 7.32 Typical regions of absorption in the infrared spectrum.

Fingerprint region C C C O

The absorbances can be summarised as shown in Figure 7.32 and can be used to identify the bonds that produce particular peaks in an infrared spectrum.

TABLE 7.8 Characteristic infrared absorbance bands Bond

Location

C–Cl C–C C–O C=O C=C O–H C–H C–H N–H O–H

Chloroalkanes Alkanes Alkanols, esters Carboxylic acids, esters Alkenes Carboxylic acids Alkanes Alkenes Amines (primary) Alkanols

Wavenumber (cm–1) 600–800 750–1100 1000–1300 1680–1750 1600–1700 2500–3300 2850–3100 3000–3100 3100–3500 3200–3600

Worked example 7.5 The structural formula and infrared spectrum of ethanoic acid are shown in Figure 7.33. The bond types present in ethanoic acid can be identified from the structural formula. Use Table 7.8 to identify the bond types corresponding to peaks A, B and C. It is useful to summarise this in a table.

chemfact Tables such as Table 7.8 indicate the wavenumber regions where a bond absorbs radiation. Note that the precise wavenumber of the absorption depends upon the molecular structure of the compound.

! Refer to the VCE Chemistry Data Book for a listing of infrared absorption data. The data book is available on the VCAA website.

93 Spectroscopy

100

For a wide range of IR, NMR and mass spectra available to download free of charge, visit the website via hi.com.au. (Be sure to acknowledge your source if you use one of these spectra in a report.)

90 80 70 Transmittance (%)

!

60 50 40 30 A

H H

10 C 0 4000

3000

O

H

B 20


C

C

O H Ethanoic acid

1000

500

Figure 7.33 Infrared spectrum of ethanoic acid.

Bond

Absorption range (cm–1) 2850–3100 2500–3300 1680–1750

C–H O–H C=O

Solution The bond type or functional group producing each peak in the spectrum can now be identified. Peak A is a broad peak with maximum absorption at 3100 cm–1. This falls within the characteristic absorption range for an O–H group in a carboxylic acid. The peak at 2990 cm–1 produced by the C–H bond is partly masked by the O–H peak and is of limited use. Peak C at 1750 cm–1 is indicative of a C=O bond. The set of peaks below 1000 cm–1 is more difficult to interpret. This set of peaks is unique to ethanoic acid and is known as the fingerprint region. IR spectrometry is a very powerful tool that can be applied to a huge range of samples. It is often used in conjunction with other techniques such as UV spectroscopy, chromatography, mass spectroscopy and nuclear magnetic resonance spectroscopy to determine the structure of a substance.

chemistryy in action Forensic use of spectroscopy

Nylon 6

Nylon 6,6

1400


Figure 7.34 IR spectrum of two fibres.


Infrared spectroscopy is an important tool in the armoury of the forensic chemist. Serious crimes such as rape and murder have been successfully prosecuted on the basis of IR spectroscopic analysis of paint, dye, glass and fibre samples. Trace fibre evidence from hair, clothing, carpet or upholstery can be exchanged between the perpetrator and the victim at a crime scene. An optical microscope often cannot differentiate between synthetic fibres left at the crime scene, but IR spectroscopy can distinguish between fibres from two very similar samples. Look at the spectra in Figure 7.34. While these are similar at first glance, there are subtle differences in the region from 1200 to 1000 cm–1. • The amide (–CONH–) band around 1270 is shifted by 10 cm–1. • Additional bands are present in the top spectrum.

chemistry in action Forensic use of spectroscopy (continued) a

H N

O (CH2)5

C

H N

(CH2)5

H

b

O N

C

C

N

O

O (CH2)4

C

H

O N

(CH2)6

H

N

C

H

Figure 7.35 Molecular structure of a nylon 6 and b nylon 6,6.

By comparison with a library of reference spectra, the two fibres were identified as nylon 6 and nylon 6,6 (Figure 7.35).

Quantitative analysis If one peak in the spectrum is selected, the absorbance increases as the concentration of the molecule in the sample increases. This means IR spectroscopy can be used for quantitative analysis. A strong sharp peak in the spectrum of the molecule is chosen. As with chromatography and the other spectroscopic techniques, a calibration curve is constructed by using standards of known concentration. The concentration of the sample is found by comparison with the calibration curve. Unfortunately calibrations are often significantly curved rather than linear. Quantitative analysis of toxic gases in the atmosphere such as sulfur dioxide, hydrogen cyanide and carbon disulfide is often carried out by IR spectroscopy.

chemistryy in action IR spectroscopy and breath tests If a roadside breath test (which usually uses a fuel cell) shows that a driver is over the legally allowed blood alcohol limit, the driver may be asked to undergo a second test at the police station or in a booze bus. In that second test, the driver breathes into the cell of an IR spectrophotometer that can give a printout that can be used

as evidence in court. The absorbance at 2950 cm–1 is measured to determine the ethanol concentration. If the reading is above the legal limit or the driver is unable to give a breath sample, a blood or urine sample may be taken for testing by gas chromatography.

95 Spectroscopy

7.5

summary

IInfrared f d spectroscopy t can be b usedd to t identify id tif molecular l l structure and verify the identity of unknown compounds. The covalent bonds between atoms vibrate due to stretching and bending motions; these cause a change in the dipole moment of the molecule. Molecules have discrete vibrational energy levels. The absorption of infrared radiation causes molecules to move to higher vibrational energy levels.

Each E h covalent l t bond b d absorbs b b infrared i f d radiation di ti over a characteristic range of frequencies (wavenumbers), enabling the covalent bond and the structure of the compound to be identified. Each compound has a unique absorbance pattern in the fingerprint region of the infrared spectrum and this can be used to identify the compound.

key questions 13 Consult Table 7.8. What bond is undergoing a vibrational change in ethanol at 2950 cm–1? 14 Look at the IR spectrum and structural formula of propanone (Figure 7.36). Use Table 7.8 to identify Peak A and Peak B.

a Explain how these IR spectra confirm that both samples abelong to the alkanol group b of compounds. b Are A and B the same compound? Explain. 100

Sample A (liquid film)

100

Transmittance (%)

80

Transmittance (%)

90 Peak A

70 60 50

H3C

40 30

C H3C

O

50

0 4000

20

2000 Wavelength (cm–1)

100

1000

Figure 7.36 Infrared spectrum of propanone.

15 Ethanol belongs to a group of organic compounds called alkanols (alcohols). Its IR spectrum is shown in Figure 7.27. The broad absorption between 3200 cm–1 and 3600 cm–1 is due to the O–H group of atoms common to all alcohols. The absorption just below 3000 cm−1 is caused by the C–H stretch while the C–O stretch is at 1050 cm–1. IR spectra of two liquids, A and B, are shown in Figure 7.37.

Transmittance (%)

3000

Chemical analysis C


1000

500

1000

500

Peak B

10

96

3000

Propanone

Sample B (liquid film)

50

0 4000

3000


Figure 7.37 IR spectra of sample A and sample B.

7.6

Nuclear magnetic resonance spectroscopy Nuclear magnetic resonance spectroscopy (NMR) is one of most powerful techniques for determining the structure of complex biochemical molecules. NMR uses energy in the radio frequency range of the electromagnetic spectrum. The energy of the radiation is too low to cause electronic, vibrational or rotational transitions. The radio waves used in NMR cause a change in the ‘spin’ of particles in the nucleus. Protons, neutrons and electrons can be thought of as spinning on their axes in either an ‘up’ (↑) or a ‘down’ (↓) direction. In many nuclei, the orientations of the spins of all the nucleons (protons and neutrons) are paired and so cancel out, but in atoms with an odd number of nucleons, such as 1H or 13C, the nucleus always has an overall spin. A spinning charged particle creates a tiny magnetic field along its axis. In the absence of an external magnetic field, the two states of the spinning nucleon have the same energy. When an external magnetic field is applied, the tiny magnetic field of a spinning nucleon aligns either with the field (lower energy) or against the field (higher energy) (Figure 7.38). Slightly more nucleons will be in the lower energy state, aligned with the external magnetic field. The difference in energy between the two states, ΔE = hν, is in the radio region of the electromagnetic spectrum. A nucleon can absorb a quantum of energy in the radio region and be promoted to the higher spin energy level. When it relaxes back to the lower energy level, the same radio frequency is emitted (Figure 7.39). The most common form of NMR is based on the hydrogen-1 (1H), nucleus, or proton, and so is called proton NMR. It can give information about the structure of any molecule containing hydrogen atoms. Because of the large number of carbon atoms in complex biochemical molecules, NMR using the 13C isotope of carbon is often also used. The difference in energy of the two spin states depends on: the identity of the nucleus, e.g. 1H is different from 13C the chemical environment surrounding the nucleus. The actual magnetic field experienced by the nucleus is not the same as the applied external magnetic field. Electrons around each nucleus are spinning and so have an associated magnetic field that shields the nucleus from the applied magnetic field. The amount of nuclear shielding depends on the other atoms surrounding the nucleus. This in turn affects the amount of energy needed for the nucleus to change its spin. The energy emitted when the nucleus relaxes to the lower spin state depends on the amount of nuclear shielding experienced. For example hydrogen in –CH2– will absorb at a slightly different frequency from hydrogen in –CH3 or hydrogen in –OH. For instrumental reasons it is difficult to compare frequencies from one NMR machine to another. To standardise measurements on different instruments and under different experimental conditions, a reference sample, usually tetramethylsilane (TMS), (CH3)4Si, is used (Figure 7.40). TMS has a symmetrical tetrahedral structure and therefore all the H atoms in the molecule are in an identical environment. A single peak is formed well away from the peaks that most chemists are interested in. TMS is also chemically inert so can be added to a sample without causing a chemical reaction.

! When two magnetic fields align, the north pole of one magnet points to the south pole of the other.

Aligned against field N

N

S N

S

S

N

E = hν S

No external field

S N

Aligned with field Energy

External magnetic field

Figure 7.38 The alignment of small magnets at two energy levels within a large magnetic field.

Absorbs energy Direction of magnetic field

Releases energy

+ Oν

– Oν

Figure 7.39 Absorption of energy and change in spin.

! Other nuclei used in NMR include 15N, 19F and 31P.

CH3 Si H3C

CH3 CH3

Figure 7.40 Structure of tetramethylsilane (TMS).

97 Spectroscopy

The difference in energy needed to change spin state in the sample is compared to TMS and is called the chemical shift, and is measured in ppm. The chemical shift of TMS is defined as zero. The symbol δ is often used to represent the chemical shift.

TABLE 7.9 Some characteristic chemical shifts in proton NMR relative to TMS (protons highlighted) (R is any alkyl group; see Chapter 9) Proton type R

CH3

R

CH2

R′ R″ R

CH

Chemical shift (ppm) 0.7–1.6 R

or

O C

R

1.2–1.4

R3CH

1.6–2.9

O

or

C

CH3

CH2

2.0–2.9

R

OH O

1.0–5.5 CH3 or

R

CH2

R

C

R

C

R

CH

O

CH2

Cl

R

3.3–4.3 3.0–4.0

O 8.0–10.0

H O

! A more detailed listing of chemical shifts is available in the VCE Chemistry Data Book, which can be downloaded from the VCAA website.

R

O

11.0–12.0

H CH

R

CH

CH

CH3

CH

CH

CH2

4.5–6.0 1.7 R

2.3

NH2

0.5–6.5

H H

H

H

H

6.0–9.0

H


Instrumentation

Sample tube

Main features of a basic NMR spectrometer include: a radio transmitter coil that produces a short powerful pulse of radio waves a powerful magnet that produces strong magnetic fields the sample is placed in a glass tube that spins so the test material is subject to a uniform magnetic field. Solid samples are dissolved in a solvent that will not give a signal, for example D2O, CD2Cl2 or CD3COCD3. Deuterium (D or 2H), 16O and 12C do not have a magnetic dipole moment as the magnetic moments of their protons and neutrons are paired and cancel out a radio receiver coil that detects the radio frequencies emitted as nuclei relax to a lower energy level a computer that analyses and records the data.

Radio frequency transmitter

Pole of magnet

Detector Recorder

Figure 7.41 Schematic diagram of a nuclear magnetic resonance (NMR) spectrometer.

Proton NMR Interpreting the NMR spectrum (low and high resolution spectra)

chemfact A hydrogen nucleus is a proton, and NMR spectroscopists tend to use the words ‘hydrogen’ and ‘proton’ interchangeably.

Proton NMR is used to identify the number of chemically distinct hydrogen ‘environments’ there are in a molecule. In low resolution proton NMR spectra, the number of peaks is equal to the number of different bonding environments experienced by the hydrogen nuclei in the molecule. In propanone, CH3COCH3, all six hydrogen atoms are in –CH3, and both CH3 are bonded to the C = O. They are all in the same environment so there is only one peak in the NMR spectrum (Figure 7.42a). In dimethoxymethane, CH3OCH2OCH3, a total of six hydrogen atoms is in the two –CH3 groups and both are attached to the O. Two hydrogen atoms are in a –CH2– group. There are two peaks in the NMR spectrum as there are two different environments for hydrogen atoms (Figure 7.42b). The area under the –CH3 peak is three times bigger because there are six hydrogen atoms in that environment compared with two hydrogens in the –CH2– environment.

a H H H H H

b

Absorption by CH3 protons

H H H

C C C

O

C

H

H

H C

C

H H


O O Dimethoxymethane


Propanone

H

TMS

TMS

8.0

7.0

6.0

5.0 4.0 3.0 Chemical shift (ppm)

2.0

1.0

0δ

8.0

7.0

6.0

5.0 4.0 3.0 Chemical shift (ppm)

2.0

1.0

0δ

Figure 7.42 a Low resolution NMR spectrum of propanone. b Low resolution NMR spectrum of dimethoxymethane.

In ethanol, CH3CH2OH, there are three hydrogen atoms in CH3, two hydrogen atoms in CH2 and one hydrogen in OH. The low resolution NMR spectrum thus has three peaks (Figure 7.43a). The area under each peak is proportional to the number of hydrogen atoms in each environment.

99 Spectroscopy

At high resolution the NMR spectrum of ethanol shows more detail. The peak due to –CH3 protons is split into three fine peaks, while the peak due to –CH2– protons is spit into four fine peaks (Figure 7.43b). Peak area

a

2

:

1

:

3

b 3H

CH3 H H H

—CH3

OH C

C

—CH2—

CH2

H H

H

Ethanol

2H

O

HO— TMS

1H

5

4

3 2 Chemical shift (ppm)

1

0

5

4

3 2 Chemical shift (ppm)

1

0

Figure 7.43 a Low resolution NMR spectrum of ethanol. b High resolution spectrum of ethanol showing peak splitting.

chemfact Peak splitting acts over small distances only. The interacting hydrogens must be bonded to adjacent atoms. For example, CH3OCH2OCH3 shows no spin–spin coupling and peak splitting because the hydrogens are not joined to adjacent atoms; they are separated by an oxygen atom.

! The H in OH groups doesn’t split the peaks of adjacent H atoms, nor is its own peak split, for reasons that are beyond this course.

The number of peaks caused by splitting equals n + 1, where n is the number of H atoms on the neighbouring atoms, i.e.: CH splits the signal from hydrogens attached to adjacent atoms into two peaks CH2 splits the signal from hydrogens attached to adjacent atoms into three peaks CH3 splits the signal from hydrogens attached to adjacent atoms into four peaks. In summary: chemical shift provides information about the functional group to which the hydrogen H is attached peak splitting gives information about the neighbouring protons peak areas give information about the number of equivalent hydrogen atoms.

chemistryy in action Magnetic resonance imaging NMR is used in medicine in the form of magnetic resonance imaging (MRI). The MRI scanner is effectively an NMR machine; the patient takes the place of the sample and is rolled into the opening of the huge magnet. The body is an abundant source of protons. MRI measures the amount of time it takes for excited protons to relax back to the lower energy state. This time is different for protons in fat, muscle, tissue and bone. Normal tissue gives a different response from a tumour. A powerful computer is used to analyse the data and present a coloured image of the patient’s body. The MRI scan shows repeated sections through the body as the patient is slowly moved deeper into the machine. MRI can be carried out on a patient who is conscious and can react to stimuli. This allows the technique to show a unique picture of the working brain as the patient is shown images, hears sounds, carries out mental arithmetic tasks, and the like. Figure 7.44 An MRI scan of a human head.


Interpretation of NMR spectra Worked example 7.6 The proton NMR spectrum of a compound, molecular formula C4H8, is shown in Figure 7.45. Peak set B (relative area = 3)

Peak set A (relative area = 1) TMS

7

6

5

4 3 2 Chemical shift (ppm)

1

0

Figure 7.45 NMR spectrum of C4H8.

From the spectrum, record the following data in a table for each peak: • the number of peak sets, and hence the number of different environments of 1H • the chemical shift of the major peak • the relative number of protons in each peak set (from the relative peak area) • the fine structure—the number of fine peaks each major peak is split into. From this information, deduce the type of proton that each peak represents and predict the structure of the compound.

Solution Peak set A Peak set B

Relative area 1 3

Peak split into 4 2

Chemical shift 5.6 1.7

There are two peak sets and so two different types of H environments in the compound. Peak set A has four fine peaks so the adjacent carbon, peak set B, must have three hydrogens in the group. Peak set B is therefore due to CH3. Peak set B is split into two, so the adjacent carbon, peak set A, must have one hydrogen in the group. Peak set A is therefore due to CH. For every one hydrogen in peak set A there are three hydrogens in peak set B, so for every CH group there is a CH3 group. We need to refer to Table 7.9 to use chemical shift to identify the group. Chemical shift for peak set B, 1.7, is consistent for CH3 attached to a =CH group. Chemical shift for peak set A, 5.6, is consistent with CH attached to a CH group by a double bond. The likely structure for the compound C4H8 is CH3CH=CHCH3, but-2-ene (Figure 7.46). (Note. There are two structures that correspond to this formula.)

H

CH3 C

C

H

CH3

H

H C

H3C

C CH3

Figure 7.46 Likely structures for the compound C4H8.

101 Spectroscopy

the cutting edge key terms

NMR structure The and pagethe A header ofgoes membrane here peptides by Frances Separovic ur laboratory is using powerful solidstate NMR methods to study the relationships between structure and function of biomolecules. We are employing these techniques to determine the structure and dynamics of polypeptides within model membranes. The structures of many biological systems cannot be obtained by traditional methods. Many systems are simply too large for solution-state NMR or have not been crystallised for X-ray diffraction.

O

Figure 7.47 The location of the amyloid Aβ (1–42) in a model cell membrane as deduced from our solid-state NMR work.

Solid-state NMR experiments have been designed for determining the structure of molecular systems that do not lend themselves to solution-state NMR and crystallography. For example, structures of powder samples can be determined and compared to single crystal X-ray structures of the same compound. Similarly, colloidal and membrane dispersions and protein complexes that precipitate out of solution can be studied using solid-state NMR.

Figure 7.48 Professor Separovic with an NMR machine at the Bio21 Molecular Science and Biotechnology Institute, University of Melbourne.

102 102 C Chemical Chemical analysis analysis

bio o We have used novel solid-state NMR methods to determine the molecular structure of the antibiotic gramicidin A, which is derived from soil bacteria, and the toxin in bee stings, melittin, within the membranes of cells. Both gramicidin A and melittin form membrane ion channels, which are small pores in the membrane through which ions move in and out of the cell. The techniques used to study these polypeptides are being extended to other membrane-active peptides and membrane proteins.

We are using solid-state NMR to study biological ion channels, amyloid peptides, toxins and antibiotics in membranes, as well as polymer coatings and resins, silk fibres, and protein–sugar complexes with pharmaceutical applications. In collaboration with other multidisciplinary research groups, we are providing insight into the biophysical chemistry and the relationship between structure and function of peptides and proteins active in membranes and relevant to disease states and treatments.

Professor Frances Separovic

Figure 7.49 Our NMR structure of a spider toxin in a membrane environment.

Frances Separovic is a biophysical chemist who specialises in NMR spectroscopy at the University of Melbourne. Frances has developed solid-state NMR techniques for determining the structure and dynamics of molecules within cell membranes, in particular peptide antibiotics and toxins. Membrane polypeptides are studied both as dry powders and in hydrated lipid membranes for biomolecular engineering applications. Frances is the Head of the School of Chemistry and Associate Dean of the Faculty of Science at the University of Melbourne.

103 Spectroscopy

TABLE 7.10 Carbon-13 chemical shifts Carbon type R–CH3 R–CH2–R R3–CH R–CH2–(F,Cl,Br,I) R3C–NH2 R3–CH2–OH RC≡CR C≡N R2C=CR2

Chemical shift (ppm) 8–27 27–50 37–60 12–78 35–70 50–88 77–98 115–133 108–150

RCOOH

13

C NMR spectroscopy

Carbon-13 is a naturally occurring isotope of carbon that has nuclear spin. It is used in NMR spectroscopy to identify different carbon atom environments within a molecule. The chemical shift from the TMS reference is dependent on a carbon atom’s environment within a molecule. Chemical shifts range from 0 ppm to 200 ppm (Table 7.10). The peaks in the spectrum are a single line produced for each different carbon atom environment. The 1H NMR and 13C NMR spectra for a particular compound are different. The 1H NMR spectrum for ethanol, CH3CH2OH, was given in Figure 7.43b; the 13C NMR spectrum is shown in Figure 7.50. The proton NMR shows three sets of peaks representing the three different hydrogen environments. The carbon-13 NMR shows two peaks respresenting the two carbon atom environments. —CH2—

CH3—

160–220

CH3

CH2

OH

TMS

! Refer to the VCE Chemistry Data Book for a full listing of 13C chemical shifts. 200

180

160

140


60

40

20

0

Figure 7.50 Carbon-13 NMR spectrum for ethanol.

summary

7.6

At Atomic i nuclei l i produce d a magnetic ti field ld as a resultlt off proton t andd neutron spin. The nucleus will always have an overall spin when the number of protons plus neutrons is an odd number. The interaction of an external magnetic field with the magnetic field of the nucleus of an atom is used in nuclear magnetic resonance spectroscopy to determine the identity and structure of a compound. The magnetic fields generated by nuclei align with or against an external field. Nuclei aligned against the external field are at a higher energy level than those aligned with the external field. Electromagnetic radiation in the radio frequency range causes nuclei to move to a higher energy level. The external magnetic field experienced by an atom within a molecule is modified by the magnetic field of neighbouring atoms. The energy required to excite a nucleus of an atom to a higher energy level depends on the arrangement of atoms within a molecule or the atom’s ‘environment’.


Each E h peakk in i an NMR spectrum t represents t a different diff t chemical h i l environment. The area under the peaks indicates the proportion of atoms in each environment. The frequency at which a nucleus absorbs radiation relative to the absorption of the TMS standard is called the chemical shift. The chemical shift is characteristic of an atom’s environment. The peaks in high resolution proton NMR spectra may be split into a cluster of peaks. The amount of splitting provides information about the number of hydrogens attached to adjacent carbon atoms. The interaction between protons on adjacent atoms that causes the peaks to split is called spin–spin coupling. Analysis of an NMR spectrum involves consideration of: − the number of peak sets − the chemical shift of each peak set − the relative area under each peak set − peak splitting.

key questions 16 Explain how radio waves can interact with the nucleus of an atom. How is this phenomenon applied to NMR? 17 a Samples analysed by nuclear magnetic resonance spectroscopy are not dissolved in ordinary water but instead dissolved in CD2Cl2 or D2O or similar solvents. Explain why ordinary water is not used. b Explain why tetramethylsilane is used as a standard to set the zero point in an NMR spectrum. c What is meant by the term ‘chemical shift’? How are chemical shifts used in NMR? d What does the area under each peak indicate?

18 Ethyl ethanoate is a solvent commonly used in nail varnish remover. The proton NMR spectrum and semi-structural formula of ethyl ethanoate are shown in Figure 7.51. To answer this question you may need to refer to the chemical shift data in Table 7.9. a How many hydrogen or proton environments are there in ethyl ethanoate? b Refer to Table 7.9 and identify the expected chemical shift for the groups present in the ethyl ethanoate molecule. c What is the relative number of protons in each peak set A, B and C? d Explain why: i peak set A is split into a quartet (four peaks) ii peak set B is a single peak iii peak set C is a triplet. e Identify the protons responsible for each peak set and the carbon atoms to which they are bonded. f How many peaks would you expect in the 13C NMR spectrum of ethyl ethanoate?

3 Ethyl ethanoate 1H NMR O H3C

C O

CH2

CH3 B

3

C

2 A

10

9

8

7


TMS

3

2

1

0

Figure 7.51 The proton NMR spectrum of ethyl ethanoate.

105 Spectroscopy

07

key terms dipole moment electromagnetic spectrum electronic energy levels emission spectrum excited electrons fingerprint frequency functional group infrared spectroscopy

Flame tests 19 Figure 7.5 shows a flame test being performed. a What colour would the flame be if copper were present in the sample? b Before performing the test it is necessary to heat the wire strongly for several minutes. Why? c Why would copper wire be unsuitable for use in flame tests? d Why are flame tests rarely used for qualitative analysis by modern chemists? 20 When electrons are excited they lose energy by jumping from the shell they are in to a lower energy shell. Examine the spectrum of calcium in Figure 7.9a. Which line represents the electron jump of the largest energy? Explain.

nuclear magnetic resonance spectroscopy (NMR) nuclear shielding photon proton NMR quantum of energy reference cell rotational energy levels sample cell

shells spectrum transmittance ultraviolet (UV) UV–visible spectroscopy vibrational energy levels wavenumber wavelength

d What is the percentage by mass of lead in the milk powder? e The zero standard, distilled water, gave an absorbance reading that was not zero. Suggest a reason for this. 0.35 0.30 0.25 Absorbance

absorbance atomic absorption spectroscopy (AAS) atomic emission spectroscopy (AES) calibration graph carbon-13 NMR chemical shift colorimeter colorimetry

0.20 0.15 0.10

Atomic absorption and atomic emission spectroscopy 21 Atoms will emit as well as absorb light of a characteristic wavelength when they are sprayed into a flame. Use Figure 7.13 to explain how the emitted light is prevented from interfering with the absorption measurement. 22 Explain the difference between inductively coupled plasma atomic emission spectroscopy and atomic absorption spectroscopy in terms of the origin of the light being measured and the changes in energy levels. 23 Lead exposure can cause permanent brain damage in infants and young children, even at very low levels. The amount of lead in infant milk formula can be measured by atomic absorption spectroscopy. A 2.5 g sample of milk powder was dissolved in 50 mL of distilled water. A very small volume of this solution was analysed and gave an absorbance of 0.130. Four standard solutions were analysed in the same way and the calibration graph in Figure 7.52 was obtained. a Determine the concentration of lead (ng mL–1) in the diluted milk powder solution. b Calculate the concentration of lead (ng g–1) in the dried milk powder. c Express your answer to part b in ppm.


0.05

0

1.0

2.0 3.0 4.0 5.0 Lead concentration (ng mL–1)

6.0

Figure 7.52

24 The mineral cobaltite is mined for the production of cobalt. Ore containing cobaltite may also contain trace quantities of nickel. A sample of ore was analysed by AAS to determine the concentration of nickel present. 5.0 g of the ore was dissolved in 25 mL concentrated nitric acid, then diluted to 100 mL. A concentrated stock solution containing 1000 ppm nickel was also prepared. 10 mL of the ore sample solution was pipetted separately into four 100 mL flasks and 1, 2, 4 and 6 mL of the concentrated stock nickel solution was added to the flasks. All five flasks were then made up to the mark. A ‘standard addition’ calibration curve was prepared using 10, 20, 40 and 60 ppm added nickel, giving the standards listed in the table.

Standard 1 Standard 2 Standard 3 Standard 4 Sample

Concentration of nickel (ppm) x + 10 x + 20 x + 30 x + 40 x

Absorbance 0.25 0.34 0.44 0.53 0.15

The absorbance readings for the standards and sample were determined using light of wavelength 325.4 nm.The value of x, the concentration of nickel in ppm, was found from the intercept of the calibration curve with the x-axis at absorbance 0. a Copy Figure 7.53 and plot the absorbance values for the standards and sample. Insert a line of best fit through the points and determine the concentration of nickel in the sample solution. b What mass of nickel was present in 5.0 g of ore? c What is the percentage by mass of nickel in the ore sample? d Why is it preferable to use a standard addition calibration technique instead of preparing the standards in distilled water? e Arsenic is present in equal concentrations to that of cobalt in cobaltite. Why does it not interfere in the calculation of the cobalt or nickel? Absorbance 0.6 0.5

26 a Name a substance, other than copper sulfate solution, whose concentration can be determined directly by UV– visible spectrometry. b Describe the purpose of the monochromator in the spectrometer (Figure 7.17). c Refer to Table 7.4. What wavelength of light would be best used to analyse a sample of copper sulfate solution? d Describe the changes on the atomic level that occur during UV spectroscopy.

Infrared spectroscopy 27 a Which of the following molecules will not absorb infrared radiation: water, methane, oxygen or carbon dioxide? b Explain why the molecule you have chosen doesn’t absorb infrared radiation. 28 Show, using labelled drawings of the water molecule: a symmetrical stretching b assymetrical stretching c scissoring. 29 Examine the structure and infrared spectrum of butan-2-ol (Figure 7.54). Which bonds correspond to to the absorption peaks labelled A and B? 100

80 Transmittance (%)

Standard

A 60 H H

40

H H

20 B

0.4

3000

0.3

H C

C

H C

H H

C

OH H butan-2-ol

2000 Wavenumber (cm–1)

1000

Figure 7.54 Structure and IR spectrum of butan-2-ol.

0.2 0.1

20

10

x

x+10

x+20

x+30

x+40

Nickel concentration (ppm)

Figure 7.53 Calibration curve for nickel.

UV–visible spectroscopy 25 The wavelength selector on a UV–visible spectrophotometer is adjusted to give light of wavelength 450 nm. a What is the colour of the light that is passed through the sample? b What colour would you expect the sample solution to be? Explain your answer.

30 The infrared spectrum of the nitrile (C≡N) group shows an absorbance at 2300 cm–1. Calculate: a the wavelength of this absorbance (in m) b the frequency of this absorbance (Hz). 31 Order the following molecules from highest to lowest energy required to cause stretching of the bond: HCl, HI, HBr.

Nuclear magnetic resonance spectroscopy 32 Figure 7.55 shows the proton NMR spectrum of ethanol (CH3CH2OH). a How many hydrogen environments are there in this compound? b What is the relative number of protons at each peak set? c Identify the groups that produce the peak sets labelled A, B and C.

107 Spectroscopy


d Explain why: i peak A is split into a quadruplet ii peak C is split into a triplet. 3H C

2H A 1H

TMS

B

3.7 2.6 1.2 0 Chemical shift (ppm)

Figure 7.55 High resolution NMR spectrum of ethanol.

33 The semi-structural formulas of propane, propan-1-ol and tartaric acid are shown in Figure 7.56. ∗ CH3CH2CH3

Propane

b

∗ CH3CH2CH2OH

Propan-1-ol

c

COOH

Tartaric acid

HO

C

H∗

HO

C

H

0.3

COOH

Figure 7.56 Semi-structural formulas of a propane, b propan-1-ol and c tartaric acid.

Each of the samples is analysed by 1H NMR. For each molecule state: a the number of different types or environments of protons in the molecule b the number of peaks into which the signal due to the proton marked with (*) would be split. Molecule Propane Propan-1-ol Tartaric acid


Number of types of 1H

Number of peaks H(*) spilt into

Absorbance

a

34 Place in order of the energy required to cause a change in quantum energy level (lowest to highest): vibrational change in a molecule; electronic change in an atom, change of spin orientation in 1H (proton). 35 Write a definition of each of the following terms that you have met in this chapter: emission spectrum, quantum of energy, complementary colour, homonuclear molecule, opaque, transmitted light, chemical shift, wavenumber. 36 Atomic absorption spectroscopy (AAS) and UV–visible spectroscopy both involve absorption of light. Both can be used to determine the amount of copper in a solution. a What species absorbs the light when copper nitrate is analysed by i UV–visible spectroscopy? ii AAS? b Which technique would be simplest for the analysis of an approximately 0.5 M copper nitrate solution? Explain your answer. c How is the light of the required wavelength selected in i UV–visible spectroscopy? ii AAS? 37 A copper mine can produce copper at a profit if the average concentration of ore is 0.4% w/w copper or greater. A 2.0 g sample of ore was ground, the copper dissolved in 10.0 mL aqua regia (a mixture of concentrated hydrochloric and nitric acids) and diluted to 200 mL with distilled water. Copper standards containing 2.0, 4.0. 6.0 and 12.0 μg/mL copper were also prepared and the UV-visible absorbance measured (Figure 7.57).

Sample solution 0.2

0.1

2.0

4.0

6.0

8.0

10.0

12.0

[Cu] (μg/mL)

Figure 7.57 Calibration curve for the analysis of copper.

a What was the concentration of copper in the sample solution? b What was the mass of copper in the 2.00 g sample of ore? c What mass of copper would be found in a tonne of ore based on the concentration in this sample? Is it commercially viable to extract the copper from this ore?

38 Carrots contain the pigment carotene. The absorption spectrum of carotene is shown in Figure 7.58.

A

1 H NMR Compound A

(1H) (1H)

Absorbance

X

visible

ultraviolet

Y

infrared TMS

260 300

340 380

420 460 500 540 580 620 660 700 Wavelength (nm)

ultraviolet

green

10

infrared

visible violet blue

yellow

red

9

8

7

4 3 5 6 Chemical shift (ppm)

2

1

0

(3H)

B 1

H NMR Compound B

Figure 7.58 Absorption spectrum of carotene.

Z

a Explain why carrots have an orange colour. b What wavelength should be used to determine the concentration of carotene in carrot juice using a UV–visible spectrometer? c Draw a flow chart summarising the steps involved in analysing the concentration of carotene in carrot juice using UV–visible spectroscopy. 39 Compare and contrast the techniques of infrared spectroscopy and ultraviolet spectroscopy under the following headings: Energy changes that occur in the sample Main features of the spectrometer Examples of samples analysed with the technique 40 Ethyl methanoate, propanoic acid and methyl ethanoate have the same molecular formula, C3H6O2. The structural formulas of these compounds are shown in Figure 7.59. The proton NMR spectra of these compounds, labelled A, B and C, are shown in Figure 7.60. The peak sets are labelled X, Y and Z.

(2H) (1H) X

10

C

9

Y

8

7

TMS


2

1

0

(3H)

1 H NMR Compound C

Z

(2H)

Y TMS

(1H) X

O

I

E Ethyl methanoate

H

12

C O

CH2

Propanoic acid P

CH3

CH2

CH3

8 7 6 5 4 Chemical shift (ppm)

3

2

1

0

Figure 7.60 Proton NMR spectra of compounds A, B and C. H

O M Methyl ethanoate

9

C O

III

10

CH3 O

II

11

C O

CH3

Figure 7.59 Semi-structural formulas of I ethyl methanoate, II propanoic acid and III methyl ethanoate.

a For each NMR spectrum, indicate: i the the number of proton environments in the molecule ii the relative number of hydrogens in each environment iii the approximate value of the chemical shifts for each peak set iv the protons responsible for each peak set and the carbon to which they are bonded. b Use the information from part a to match each compound with its NMR spectrum.

109 Spectroscopy

0

chapter 08

Combining and choosing analytical techniques

key knowledge • principles and applications of spectroscopic techniques and interpretation of qualitative and quantitative data from mass spectrometry • matching analytical techniques to a particular task

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the instrumentation and uses for mass spectrometry • explain the chemical and physical basis of mass spectrometry and relate this to the nature of samples analysed • make qualitative interpretations of mass spectra data • describe instrumental methods of analysis that combine a number of techniques • discuss the factors to be considered when choosing an analytical or instrumental technique • identify appropriate analytical techniques for a range of samples.

In the previous chapters you were introduced to a number of chemical and physical techniques that allow us to analyse a range of materials. A chemist presented with a sample will usually have more than one technique to call upon. For example, acetylsalicylic acid, the active ingredient in aspirin, can be analysed by volumetric analysis, UV–visible spectroscopy, high performance liquid chromatography, infrared spectroscopy and nuclear magnetic resonance spectroscopy. Each technique will give different information. One technique may be better suited to determine the concentration of the substance, while another gives information about the structure. A number of instruments have been developed that combine two techniques and so can give more information, more quickly, in a single analysis. In this chapter you will look at: mass spectrometry combined techniques, such as gas chromatography–mass spectrometry (GC-MS), which combine two analytical techniques in one instrument the range of data that can be determined about a sample from a number of techniques the synchrotron—the ultimate combined analytical technique the factors chemists consider when choosing an analytical technique.

8.1

ATOMIC STRUCTURE E • Mass of the atom

review In Unit 1 you learned about the use of mass spectrometry as a tool to measure the mass and percentage abundance of isotopes. See Heinemann Chemistry 1, Chapter 4. You used this data to calculate the relative atomic mass of an element. In this chapter we will look at the use of mass spectrometry to investigate molecular rather than atomic substances.

!

Mass spectrometry One of the most common and useful analytical tools used in combination with other techniques is mass spectrometry (MS). The mass spectrometer can be used for: quantitative analysis—as a sophisticated and very sensitive detector that can measure how much of a substance is present qualitative analysis—to provide a unique fingerprint of a substance; this can be used to identify the substance from an on-line database or to give information about the structure of a new or unknown compound. All of this information can be determined from the same analysis.

A mass spectrometer does not use electromagnetic radiation like the spectroscopic techniques outlined in Chapter 7. The mass spectrometer gives very accurate information about the mass of positive ions formed in the instrument.

Detector to measure ion current

Ion collector

Instrumentation The key principle of mass spectrometry is that a charged particle passing through a magnetic field is deflected along a circular path of radius proportional to the mass-to-charge ratio, m/e. A schematic diagram of a mass spectrometer is shown in Figure 8.1. Its operation is as follows. The sample, as a gas, enters the evacuated tube. Positive ions are formed in the ionisation chamber when an electron beam dislodges electrons from the sample atoms or molecules. The positive ions are accelerated by an electric field. The ions enter a magnetic field perpendicular to their path. This causes the ions to move in a curved path with a radius that depends upon the mass-to-charge ratio (m/e) of the ions. Only ions moving in a curved path of a particular radius, corresponding to a fixed m/e ratio, will reach the collector. Particles of different m/e ratio are able to reach the collector through adjustments to the accelerating voltage or the strength of the magnetic field. The collector measures the current due to the ions reaching the detector and the data is recorded as a mass spectrum.

Magnetic field at right angles to plane of diagram Charged plates to provide electric field Sample Ionisation chamber

Filament to provide bombarding electrons To vacuum pump

Figure 8.1 The construction of a mass spectrometer.

111 Combining and choosing analytical techniques

How the spectrum is formed A molecular substance can give a range of peaks in the spectrum. Two factors cause the many peaks in the spectrum: the fragmentation of the molecules into a large number of different positive ions the occurrence of different isotopes of the atoms that make up the molecules.

Fragmentation The high energy electron beam can knock just one electron from the molecule, M, to form a positive ion M+: M +

chemfact Radicals or free radicals are atoms or molecules with an unpaired electron. Radicals are highly reactive. The unpaired electron on a free radical is shown by a dot.

–—➤ M+ + 2e– e– bombarding electron

M+ is called the molecular ion or parent molecular ion. The parent molecular ion is a radical, with one unpaired electron. It is chemically unstable and so will often break into smaller fragments consisting of more positive ions and uncharged free radicals. For example, ethanoic acid, CH3COOH, will ionise as follows: CH3COOH + e– –—➤ CH3COOH+ + 2e– parent molecular ion; m/e = 60

CH3COOH+ will fragment: CH3COOH+ –—➤ CH3+ + •COOH fragment uncharged ion free radical m/e = 15

The CH3+ may fragment further: CH3+ –—➤ CH2+ + H• fragment ion m/e = 14

A mass spectrum of ethanoic acid showing the peaks caused by fragment ions is shown in Figure 8.3. When an ion fragments into two smaller parts, one ion will retain the electron to become the uncharged free radical and the other is positively charged.There are always two possibilities: either fragment could theoretically become the charged ion. M+ –—➤ A+ + B• or M+ –—➤ A• + B+

The molecular ion can break into almost every type of fragment, down to individual atoms. Only the positive ions reach the detector. The uncharged radicals are evacuated by the vacuum pump.

Isotope effects In the same spectrum additional peaks can be formed due to the occurrence of different isotopes of an element. Chlorine, for example, has two isotopes: 35 Cl (76% abundance) and 37Cl (24% abundance). As a result, the mass spectrum of chloromethane, CH3Cl, shows a peak at about m/e = 50 due to the molecular ion CH335Cl+ and another peak at m/e = 52 due to CH337Cl+. There are two molecular ions, one due to the molecule with 35Cl and the


other to the molecule with 37Cl. Dichloromethane, CH2Cl2, has three molecular ion peaks: one at approximately m/e = 84 due to the molecule of CH2Cl2 containing two 35Cl atoms, another at 88 corresponding to the molecule with two 37Cl atoms, and one at 86 corresponding to one 35Cl atom and one 37Cl atom (Figure 8.2).

! Dichloromethane is found in a wide range of household products, including paints, polishes, air fresheners, hairsprays, household insecticides, shoe polish, pet flea and tick products, and waterproofing compounds.

CH2Cl2 + e– –—➤ CH2Cl2+ + 2e– parent molecular ion 49

100

84

Relative abundance

80 CH2Cl+

CH2Cl2+

60

40

20

Figure 8.2 Mass spectrum of dichloromethane. 10

20

30

40

50 60 Mass/charge

70

80

90

100

110

Interpreting the mass spectrum 43 CH3CO+ 45 base peak COOH+

100 90

Relative intensity

80

O (m = 45)

60 CH3COOH+ M+, parent molecular ion

70 60

+C O

50

Loss of CH3

40 Molecular ion

30 20 10 10

H

15 CH3+

15

42 CH2CO+

29 COH+ 20

25

30

35

CH3

C O

45

40

+ O

50

55

60

65

70

H

m = 60

m/e Loss of COOH

Figure 8.3 Simplified mass spectrum of ethanoic acid.

Loss of OH CH3CO+ (m = 43)

CH3+ (m = 15)

Loss of H

The height of the peak in the mass spectrum gives the relative concentration of the ions present. In Figure 8.3, the highest peak, the base peak, has a fragment mass of 43 and is assigned an intensity of 100%. The intensities of all other peaks are measured relative to the base peak. In some spectra the peak with the highest intensity is also the parent molecular ion. In other spectra, such as that of dichloromethane, the peak corresponding to a smaller fragment may be the base peak. Smaller ions formed by fragmentation are usually present in smaller amounts. The relative intensities of the ions depend on: the energy of the bombarding electrons the stability of the ion fragments formed the ease with which ions can lose atoms.

CH2CO+ (m = 42)

Figure 8.4 Fragmentation of ethanoic acid. Only the fragments corresponding to the major peaks of the mass spectrum are shown.


! It is possible for doubly charged ions to form in the ionisation chamber. Consideration of mass spectra involving these ions is beyond the scope of this course.

summary

The mass spectrum is unique to the compound and so can be used to identify a substance. Computer data banks store thousands of mass spectra that chemists can compare with their spectra. The peak with the greatest mass will often give the relative molecular mass of the molecule. However, in many cases the peak from the parent molecular ion may be so weak as to be unnoticeable. The difference in mass between the fragment ions can give information about the structure of the molecule. The mass spectrum of ethanoic acid (Figure 8.3) indicates that the parent molecular ion has a mass of 60. The fragment corresponding to the peak at mass 45 is 15 mass units lighter than the parent molecular ion, indicating that CH3COOH+ has lost a CH3 group of atoms to form the COOH+ fragment, which has a mass of 45 (Figure 8.4). Mass spectrometry can be used to analyse many organic molecules. Applications include the analysis of pharmaceutical samples, environmental samples such as contaminants in water, and geological samples such as oil and gas, and medical screening.

8.1

A mass spectrometer t t measures the th mass-to-charge t h ratio ti ((m/e) /) of atomic and molecular ions. The peak with the highest m/e ratio is used to determine the molecular mass of a compound.

The Th fragmentation f t ti pattern tt off a molecule l l is i usedd to t determine d t i its structure. Each compound has a unique mass spectrum that can be used to identify the compound.

key questions 1 Use the data to calculate the relative atomic mass of lead. Relative isotopic mass 203.97 205.97 206.98 207.98

Abundance (%) 1.48 23.6 22.6 52.3

2 What is the relative molecular mass of the parent molecular ion of dichloromethane due to: a CH235Cl37Cl? b CH237Cl2? 3 What ions cause the peaks at m/e 49 and 35 in the mass spectrum of dichloromethane? 4 Complete the table below. Ion fragment CH3+ OH+ CO+ CH3CH2+ Cl+ C2H4+ COCH3+ COOH+ C6H5+


m/e 15

5 A compound of oxygen and hydrogen is analysed and the following fragments are measured.

m/e 1 16 17 18 19 20

Relative intensity (%) < 0.1 1.0 21 100 0.08 0.22

Possible fragment H+

a Complete the column ‘Possible fragment’ in the table above. b What is the m/e of the base peak? c What is the m/e of the parent molecular ion? d Draw the structure of the parent molecular ion. e Draw the structure of the ion at m/e 20. Note: There are two isotopes of hydrogen, 1H and 2H. There are also two isotopes of oxygen, 16O and 18O.

8.2

Combined techniques Figure 8.5 Mass spectrometers are now small enough to sit on a benchtop. This one is linked to a gas chromatograph that is used to separate the components of a mixture. The mass spectrum of each component is recorded.

!

M Many iinstruments combine bi two techniques h i to provide id more detailed d il d and d rapid information about a sample. The most commonly used combined techniques are gas chromatography–mass spectrometry (GC-MS) and high performance liquid chromatography–mass spectrometry (HPLC-MS).

Electrons

Source

DC+RF

Other combined techniques include the mass spectrometer linked to an inductively coupled plasma atomic emission spectrometer (ICP-MS), and nuclear magnetic resonance spectroscopy linked to liquid chromatography (high performance liquid chromatography–nuclear magnetic resonance spectroscopy, HPLC-NMR).

DC+RF

Detector

Figure 8.6 A quadropole mass spectrometer. This machine is much lighter and cheaper than the traditional mass spectrometer and is used in combined MS instruments. The beam of ions passes between two pairs of parallel rods. An electromagnetic field is generated to allow ions of only one M+ to reach the detector; the other ions collide with the rods and are evacuated. The frequency and voltage are varied to change the electromagnetic field to allow the whole spectrum to be measured.

Chromatography and mass spectrometry The advantage of the techniques of gas chromatography–mass spectrometry (GC-MS) and high performance (or high pressure) liquid chromatography– mass spectrometry (HPLC-MS) is that the chromatography can separate a complex sample into any number of components and each one can be positively identified through mass spectrometry. GC-MS and HPLC-MS have become the essential techniques for forensic analysis. A number of cases in which these techniques have played a central role are described in the following Chemistry in Action. As the individual components of the sample leave the chromatography column, they enter the ionisation chamber of the mass spectrometer where they are bombarded with a stream of electrons. The positive ions formed are accelerated into a quadropole mass spectrometer. The MS cycles through


the range of mass-to-charge ratios to form a scan for each component. The mass spectrometer can scan up to 50 times per second, allowing hundreds of components to be recorded from a single sample. Unknown mass spectra can be compared with an on-line library containing hundreds of thousands of different spectra.

chemistryy in action Some analytical case studies Sniffing for explosives The destruction of the Twin Towers in New York on 11 September 2001 led to a vastly increased awareness of the danger of terrorist action. Massively increased security surrounding airports and many public places has followed. Many of the devices used for the identification of explosive material are based on GC-MS. Hand-held devices can be used to ‘sniff’ air from around bodies or bags, and can detect amounts of less than a nanogram. This is much less than the amount that remains on the hands or suitcase of a person who has handled explosives even after the person has cleaned their hands.

Drug testing

Figure 8.7 Massively increased security followed the attacks of 9/11.

Since competitive sports began, people have sought to gain a competitive advantage by whatever means are available. The Romans and Greeks athletes used ‘drugs’ to improve their performance. In horse racing, the use of illegal ‘go-fast’ or ‘stopper’ drugs is now relatively rare because of their ease of detection by GC-MS. Today, drug problems in sport usually result from the inappropriate use of therapeutic substances and manipulation of substances such as hormones and electrolytes that occur naturally in the animal’s or athlete’s system.

Synchrotron—the ultimate combined technique A synchrotron is a giant collection of spectrometers linked via circular channels and covering an area equal to half a football field. Australia’s first synchrotron, costing $206 million, has been built adjacent to Monash University, Clayton, Victoria, and opened for business in 2007. A synchrotron accelerates and circulates electrons to close to the speed of light. The electrons are accelerated by magnets in a linear accelerator and in a booster ring and then transferred to an outer storage ring. As the electrons’ paths are bent through magnetic fields, the electrons generate electromagnetic radiation across the spectrum from infrared to X-ray. The radiation is channeled down ‘beam lines’ where it is used as the source radiation for a range of chemical instruments. Synchrotrons have a number of advantages over conventional spectrometers: the electromagnetic radiation is hundreds of thousands of times more intense a wide energy spectrum from infrared to X-rays is produced the radiation can be very finely tuned to select precise frequencies the radiation is highly polarised short intense pulses of radiation lasting less than a nanosecond are produced. Most of the thirteen beamlines and spectroscopic techniques used in the synchrotron use X-ray radiation to perform a range of analyses. However, two of the beamlines use radiation in the region which we have considered in this book, infrared and ultraviolet radiation.


Radio-frequency accelerator cavity

Beamline

Experimental stations

Booster ring Storage ring

Linear accelerator electron gun

Bending magnets

Figure 8.9 Australia’s first synchrotron, located in Melbourne.

Figure 8.8 Schematic diagram of the first Australian synchrotron.

A fantastic f i range off analyses l can be b carried i d out using i a synchrotron: h The development of the anti-influenza drug Relenza depended on the identification of the structure of a crucial influenza enzyme by Australian scientists working on synchrotrons overseas. Chocolate manufacturers used a synchrotron to investigate the optimum conditions of heating, cooling and stirring to form the best cocoa crystals for chocolate making. Metal and ceramic surfaces used in jet aircraft engines are under investigation using the synchrotron to devise components that can run hotter and for longer. Researchers in Chicago were able to take X-ray videos using a synchrotron and made new discoveries regarding the breathing patterns of beetles.

! For more information about the Australian synchrotron, visit their website.

Combining analyses to learn more In many cases chemists faced with an unknown sample will employ a number of analytical techniques to get more information about the substance. Some techniques will provide similar information, while others may help to confirm a guess or provide another piece of the jigsaw. As an example we will look at the various techniques that can be used to gain information about a familiar compound, ethanol. In practice chemists would employ multiple techniques to investigate much more complex organic and biochemical molecules.

Analysis of ethanol You have already looked at the analysis of ethanol by a number of techniques in this book. In Chapter 5 (p. 56), the redox reaction between ethanol and potassium dichromate is described. In this reaction the red-orange coloured


dichromate ion, Cr2O72–, is reduced by ethanol to the green chromium(III) ion, Cr3+. In the early breathalyser kits a tiny colorimeter was used to compare the difference in colour between the sample that the motorist breathed into and a tube of the unreacted mixture. The greater the concentration of ethanol, the greater the colour change. The same redox reaction is the basis of volumetric analysis of ethanol in wine. You may have performed an experiment in the laboratory to find the percentage of ethanol (v/v) in a sample of wine after first reacting the ethanol in the wine with potassium dichromate. This technique is relatively easy, cheap and accurate, when dealing with concentrations of ethanol in the range of 5–8%. However, volumetric analysis would not be suitable to determine the amount of ethanol in a blood sample or in the motorist’s breath. If a motorist is suspected of being above the legal limit, a second test may be performed. In some of these the suspect breathes into the cell of an infrared spectrophotometer. The intensity of radiation absorbed at 2950 cm–1 is measured in order to determine the concentration of ethanol in the breath. The IR spectrum also yields information about the functional groups in the molecule. The very strong broad band at about 3300–3400 cm–1 is characteristic of the O–H functional group, while the sharper band at 1050 cm–1 is characteristic of the C–O stretch. 100 90 80 70 Transmittance %

Acid electrolyte material

Breath

60 50 O-H stretch 40 CH2 bend

30

CH3 bend

20

+

Platinum electrodes

–

0.000

10 0 4000

Relative intensity

45

46 27 29 15 15

20

25

30 m/e

35

40

Figure 8.12 Major peaks in the mass spectrum of ethanol.


C-O stretch 2000 Wavenumber

1500

1000

Figure 8.10 Infrared spectrum of liquid ethanol.

31

10

3000

Readout

Figure 8.11 Fuel cell used to detect ethanol in breath.

100 90 80 70 60 50 40 30 20 10

C-H stretch

45

Another way to measure alcohol in the breath is to use a fuel cell (Chapter 27). The breathalyser fuel cell consists of two platinum electrodes with a porous acid electrolyte. The ethanol in the cell is oxidised to ethanoic acid at one electrode, and the oxygen in the air is reduced at the other. Transfer of electrons through the circuit creates an electric current proportional to the amount of ethanol in the breath. The fuel cell provides a more sensitive and specific test for concentration of ethanol than IR spectroscopy. For even more accuracy and specificity, chemists turn to gas chromatography–mass spectroscopy (GC-MS). Ethanol is easily volatilised and so the separation of ethanol from the sample is easily achieved with gas chromatography. The amount of ethanol present is determined by comparison of the sample with standards of known concentration. Most common GC detectors can be used for ethanol. A mass spectrometer gives additional definitive information about the structure and identity of the compound.

summary

Absorbance

One of the most useful pieces of data from this spectrum is that the m/e of the parent molecular ion is 46. This data, along with the knowledge from the IR spectrum that the substance is an alcohol, identifies the sample as ethanol. This is confirmed by the fragmentation pattern shown in the mass spectrograph. For large and complex organic and biochemical molecules, the mass spectrum and infrared spectrum may be too complicated to determine their structure fully. The nuclear magnetic resonance spectrum can also be used to give valuable information about the environment of the carbon and hydrogen atoms in the compound. The NMR spectrum in Figure 8.13 was generated using spin in the nucleus 13 C. You will recall that 13C is an isotope of carbon. The spectrum is very simple. Two peaks of equal intensity show an equal number of carbon atoms in two different environments. The number of hydrogen atoms present in each environment can be confirmed by proton (1H) NMR as described in Chapter 7, Figure 7.43 (p. 100).

TMS

160 140 120 100 80 60 40 20 ppm

0

Figure 8.13 C NMR spectrum of ethanol.

13

8.2

C Combined bi d tech techniques i ues su suchh as gas chromatography–mass hromatograph mass spectrometry (GC-MS) and high performance liquid chromatography–mass spectrometry (HPLC-MS) are used in analytical laboratories. A complex sample can be separated into a number of components and each one can be identified.

A variety va iety off qualitative and a d quantative quantati e analytical an lyti l techniques tech i ues are used to identify a particular substance. The synchrotron is a sophisticated spectrometer that produces a wide spectrum of electromagnetic radiation from infrared to X-rays.

key questions 6 Infrared spectrophotometers are sometimes used to determine blood alcohol content. What is the functional group that is used as the reference peak for quantitative analysis of ethanol? 7 Blood alcohol content can also be determined by using a fuel cell. Write half equations describing the oxidation of ethanol and reduction of oxygen in an acidic electrolyte.

8 The mass spectrum of ethanol is shown in Figure 8.12. Identify the fragment ions causing the peaks marked on the ethanol spectrum.

8.3

Assessing analytical techniques Many of the techniques you have studied in the preceding chapters could be used to analyse a chemical. In the study of ethanol above, we looked at six techniques that could all give quantitative information about ethanol. What factors do chemists consider when choosing the analytical technique or techniques to use? Factors to consider fall into two main types: chemical considerations and economic factors.


TABLE 8.1 A summary of analytical techniques Technique Gravimetric analysis

Typical analytes H2O; NaCl; Ag

Typical samples Foods; water

Advantages Very cheap—only basic laboratory equipment needed; easy to perform

Disadvantages Limited analytes and samples; suitable for high concentrations only

Mass spectrometry

Any elements and compounds that can be volatilised

Acids/bases; oxidants/ reductants, e.g. sodium hydroxide, vitamin C Dyes, amino acids

Very low detection limits; unique fingerprints formed; huge range of analytes; readily automated Very cheap—only basic laboratory equipment needed; easy to perform

Expensive instrument; trained technician needed to operate

Volumetric analysis

Huge range of elements and organic molecules; must be able to be volatilised Cleaning products, soft drinks

Drug preparations; dyes in foodstuffs

Very cheap–only basic laboratory equipment needed; easy to perform High sensitivity and precision; small sample size; readily automated High sensitivity and precision; small sample size; readily automated

TLC

GC

HPLC

Atomic emission spectroscopy

Low molecular mass organic compounds, e.g. acetone, aspirin Medium to high molecular mass organic compounds, e.g. pesticides, enzymes Many metals, e.g. Ca, Na, Mg

Water, foods, drugs, biological samples Foods, drugs, biological samples

Generally low viscosity solutions, e.g. waste water

Atomic absorption spectroscopy

Most metals, e.g. Cu, Fe, Zn

Generally lowviscosity solutions, e.g. waste water

UV–visible

Low molecular mass organic molecules, e.g. aspirin, quinine Organic molecules, e.g. ethanol Organic molecules

Liquid and gas samples, e.g. sunscreens, soft drinks Solids, liquids or gases

IR

NMR


Generally liquid or solid

Suitable for relatively high concentrations only; relatively large sample size needed

Comments Analyte must be easily separated from the other materials with which it is found (e.g. by heating or precipitation and separation) and the residue weighed. Used in quantitative analysis as a detector for AES, GC and LC; most useful for qualitative analysis to determine structure and identity of a compound Suitable for routine quality control of samples

Poor precision and accuracy

Samples need to be coloured or visible under UV light

Moderately expensive instrument; trained technician to operate

Samples must be volatilised without breaking down

Moderately expensive instrument; trained technician needed to operate

Samples must be soluble in a suitable solvent

High sensitivity and precision; rapid, multielement analysis; readily automated High sensitivity and precision; readily automated

Moderately expensive instrument; trained technician needed to operate Moderately expensive instrument; trained technician needed to operate

ICP-AES can determine most elements; simpler machines are more limited

Relatively simple to operate; readily automated

Not suitable for very low concentrations

Huge range of analytes and samples; small sample size High sensitivity and precision; small sample size

Moderately expensive; trained technician needed to operate Very expensive to buy and operate

Solid samples can be analysed by graphite furnace AAS in which the viscous liquid or solid is atomised in a tiny carbon cup As a detector for GC and HPLC can readily determine ppm levels of analytes Most useful for qualitative analysis—fingerprint or elucidation of structure Technique mainly used to determine structures rather than routine chemical analysis

!

Chemical considerations

The substance that is being analysed is sometimes called the analyte.

Analyte Consider whether the analyte is a metal or non-metal. Is it coloured or transparent? What functional groups are present? Is the analyte volatile? Is it soluble in water or other solvents?

Sample Consider the amount of sample available, the concentration of the analyte in the sample, the physical state of the sample, and possible interference from other compounds in the sample.

Economic considerations Time The time taken for an analysis depends on whether a single analysis is to be performed per day or if hundreds of samples are required to be analysed per day, and if continuous monitoring, multi-element determination, or complex sample preparation are needed.

Cost of equipment Many instruments cost tens or even hundreds of thousands of dollars. The cost of running analytical instruments and the cost of consumables also varies. Some NMR machines use expensive liquid helium as a coolant.

Expertise Are specialist staff required to operate the instruments and interpret the data?

Quality data Costs are generally greater for high quality data.

summary

8.3

VVarious i ttechniques h i are usedd iin chemical h i l analysis. l i The technique selected depends upon the chemical and physical properties of the analyte, the nature of the sample and

i considerations id ti suchh as time, ti t off purchasing h i economic costs and operating equipment, salary costs of staff, and the level of accuracy and precision required.

key questions 9 Name one or more analytical technique that could be used to: a determine the concentration of aspirin in a headache remedy b find the concentration of the large organic molecule vitamin A

c find the blood alcohol content of a motorist d determine the molecular structure of an organic compound. 10 Modern laboratories use a range of instrumental techniques to analyse substances. Briefly outline the advantages of using instrumental techniques.


08

key terms gas chromatography–mass spectrometry (GC-MS) high performance liquid chromatography–mass spectrometry (HPLC-MS)

Mass spectrometry 11 Compare and contrast the techniques of mass spectrometry and infrared spectroscopy under the following headings: a The energy source with which the sample interacts b The chemical or physical basis of the analysis c The type of sample that can be analysed 12 The mass spectrum and molecular structure of 1,2-ethanediol are shown in Figure 8.14 and Table 8.2. a Identify the molecular ion peak in the spectrum. b Identify the base peak in the spectrum. c What is the structure of the base peak ion? d Find out the common name for 1,2-ethanediol. e What is the best way to analyse quantitatively for this substance in a sample of waste water?

H H

80 Relative intensity

43

100

42

80

60 41 40 2729

20

0 10

15

20

25

30

57

35

40 45 T/L

50

55

72

60 65

70 75

31

100

Figure 8.15 Mass spectrum for a hydrocarbon.

H C

H

C

OH

OH

TABLE 8.3 Data for mass spectrum of unknown hydrocarbon

60

40

33

20

29

15 10

mass spectrometry (MS) molecular ion parent molecular ion synchrotron

13 The mass spectrum for a hydrocarbon is shown in Figure 8.15. Table 8.3 gives the relative intensities of the peaks.

Relative intensity

base peak beamlines combined techniques fragmentation free radicals

15

20

25

30

43 35 40 T/L

Figure 8.14 Mass spectrum for 1,2-ethanediol.

TABLE 8.2 Data for 1,2-ethanediol mass spectrum m/e

Relative intensity

15.0 29.0 31.0 33.0 48.0 62.0

9.6 13.2 100.0 34.9 8.3 3.5


45

62 50

55

60

m/e

Relative intensity

27.0 29.0 41.0 42.0 43.0 57.0 72.0

22.2 20.3 50.5 78.3 100.0 20.6 17.8

a What is the formula of the hydrocarbon? b Copy and complete the following table for the peaks shown.

m/e

Possible ion

c What fragment was lost to form m/e 57? d Write an equation illustrating this fragmentation. e Write an equation showing the formation of the base peak.

Analytical techniques 14 Name one or more analytical technique that could be used to: a analyse for low concentrations of lead in blood b determine the concentration of a silver nitrate solution c verify the identity of a sample of an organic compound d identify the components in a mixture of alkanes.

Connecting the main ideas 15 a Distinguish between thin-layer chromatography and atomic absorption spectroscopy in terms of the following features: i the type of sample analysed, for example metals, inorganic compounds, organic compounds, anions, cations etc. ii whether atoms or compounds are determined iii whether separation of the components of the sample occurs iv quantitative or qualitative analysis v cost of the test. b Thin-layer chromatography and atomic absorption spectroscopy involve very different chemical principles. Below are four key terms that can be used in an explanation of each of the techniques. For each of the four terms, identify whether it applies to thin-layer chromatography or atomic absorption spectroscopy and use the key term in a sentence to explain its meaning. i desorption ii energy levels iii excited electron iv solvent front 16 A sample of the energy drink High Caff was analysed for the concentration of caffeine it contained. Details on the pack indicated that it should contain 12 mg caffeine per 100 mL. A chemist decided to analyse a sample of the drink, without dilution, by HPLC. 100 mL of 10.0 mg mL–1 stock standard solution of caffeine was prepared from pure caffeine tablets. The chemist decided to prepare standards of the following concentration: 5, 10 and 20 mg/100 mL caffeine. Available were 10 mL and 20 mL pipettes and 100 mL and 200 mL volumetric flasks. a Describe the dilutions the chemist would need to carry out in order to prepare the standard solutions using the equipment available. b The chemist could have used a 1 mL pipette and 100 mL volumetric flask to obtain a 10 mg/100 mL standard from the stock solution with only one dilution. Why is this a less accurate method?

17 Condy’s crystals, a traditional remedy composed of potassium permanganate, can act as a fungicide. The amount of potassium permanganate in a sample of Condy’s crystals can be determined by a number of analytical techniques. In the table below, for each technique indicate: a the chemical species being analysed b any chemical reagents other than water and the sample needed to carry out the analysis c whether or not a calibration curve is needed d the sensitivity: low sensitivity is used when samples are concentrated; high sensitivity is used when samples are very dilute. Technique

a Species analysed

b Other reagents

c Calibration curve

d Sensitivity

UV–visible Volumetric analysis Atomic absorption spectroscopy 18 Low concentrations of gold can be extracted economically by dissolving the tiny gold particles in the rock with sodium cyanide. An equation for the reaction is: 4Au(s) + 8CN–(aq) + 2H2O(l) + O2(g) → 4Au(CN)2–(aq) + 4OH–(aq) a Is this an acid–base or redox reaction? Give an example of an acid/conjugate base or an oxidant/conjugate reductant redox pair in this equation to support your answer. b What mass of gold could be theoretically extracted using 1.00 L of 0.540 M sodium cyanide solution? c What technique can be used to analyse gold solutions in the concentration range mg Au per litre? 19 Nuclear magnetic resonance spectrometers and mass spectrometers both employ a magnet. Describe the function of the magnet in each instrument. 20 High performance liquid chromatography and mass spectrometry both involve the movement and separation of the sample components in the instrument. Describe and explain how, for each: a the components of the sample move through the instrument b the components of the sample are separated.


3

unit

area of study 1

review

Chemical analysis

multiple-choice questions 1 A sample of nitrogen (Mr = 28.0) has a mass of 10 g and a volume of 8 L. Which of the following values can be calculated for the nitrogen sample from these data? I moles II % w/v III density A I only B I and II only C I and III only D I, II and III 2 Which concentration of sodium chloride is much higher than the others? B 1% m/v NaCl A 1 g L–1 NaCl C 1000 ppm NaCl D 0.017 M NaCl 3 What volume of 2.0 M stock sodium iodide solution is needed to prepare 500 mL of 0.3 M sodium iodide solution? A 25 mL B 75 mL C 150 mL D 300 mL 4 A 20 cm3 sample of which solution contains the greatest number of chloride ions? A 0.2 M NaCl solution B 0.1 M KCl solution C 0.2 M CaCl2 solution D 0.1 M AlCl3 solution 5 A conjugate acid–base pair for this reaction is: HSO3–(aq) + OH–(aq) → SO32–(aq) + H2O(l) A HSO3– and SO32– B HSO3– and OH– C OH– and SO32– D SO32– and H2O 6 Potassium hydroxide reacts with carbon dioxide as follows: 2KOH(aq) + CO2(g) → K2CO3(aq) + H2O(l) Which expression correctly represents the volume of 0.1 M KOH solution that will react with 56 mL of CO2 at STP? 22.4 × 0.1 2 × 0.056 A L B L 2 × 0.056 22.4 × 0.1 0.1 × 0.056 C L D 2 × 0.056 × 22.4 × 0.1 L 2 × 22.4 7 Consider the following reaction. K2Cr2O7(aq) + 4H2SO4(aq) + 6HCl(aq) → Cr2(SO4)3(aq) + K2SO4(aq) + 3Cl2(g) + 7H2O(l) Which statement about this reaction is not true? A Dichromate ions have been reduced. B Chloride ions have been oxidised. C Hydrogen ions have been reduced. D Sulfate ions have not been oxidised or reduced.


8 In gas–liquid chromatography the liquid stationary phase is coated on very small particles of solid silica or alumina. Small particles are used in preference to large particles to: A make the flow of gas through the column easier B reduce the overall mass of the column C increase the available surface area of the liquid phase D make it easier to fill the column. 9 The retention time can be used in gas chromatography to determine: A the identity of a chemical B the concentration of a chemical C the amount of a chemical in the sample D all of the above. 10 The concentration of a substance is most accurately determined in HPLC by measuring: B peak height A peak area C retention time D Rf value. 11 Chromatographic analysis of an unknown sample of an angina preparation containing nitroglycerine resulted in a narrow peak of height 25.3 cm. A standard solution containing 9.5 μg/mL nitroglycerine gives a narrow peak of 10.5 cm. What is the concentration of nitroglycerine in the sample? A 23 μg/mL B 27 μg/mL C Cannot be determined as peak area is the only acceptable measure of concentration. D Cannot be determined as a range of standards must be tested to determine concentration. 12 Sodium salts give a distinctive yellow colour in a flame test. The colour is due to: A electrons near the nucleus of the sodium atom absorbing energy B excited electrons returning to lower energy levels and releasing energy C sodium atoms losing an electron to form sodium ions D sodium ions being converted to sodium atoms in the reducing flame. 13 UV and atomic absorption are forms of spectroscopy that involve: A a prism or grating to separate the light B a light that is passed through the sample C emission of light by the sample D a flame to excite the atoms.

14 The function of the flame in atomic absorption spectroscopy is: I to evaporate the solvent II to convert metal ions to atoms III to excite the metal atoms. A I B I and II C I, II and III D II and III 15 The concentration of low levels of the heavy metal cadmium in a sediment sample is best determined by: A gas chromatography B high pressure liquid chromatography C colorimetry D atomic absorption spectroscopy. 16 More energy is required to vibrate: A C–C compared to C=C B C–Cl compared to C–Br C C–O compared to C–H D C=C compared to C≡C. 17 The number of peaks expected in the 1H NMR and 13C NMR spectra of methyl ethanoate, CH3COOCH3, are: 1

A B C D

H NMR spectrum (low resolution) 2 3 6 1

13

C NMR spectrum

3 2 3 3

18 The following events occur in a mass spectrometer but not in the order indicated. I acceleration of particles in an electric field II vaporisation of the sample III deflection of particles in a magnetic field IV bombardment of the sample with high energy electrons V production of an electric current when charged particles enter the detector The correct sequence of events is: A IV, II, I, III, V B II, I, IV, III, V C IV, I, III, II, V D II, IV, I, III, V 19 Which of the following techniques does not use the mass spectrometer as a detector? A gas chromatography B high performance liquid chromatography C nuclear magnetic resonance spectroscopy D atomic absorption spectroscopy

short-answer questions 20 Methyl salicylate, a compound of hydrogen, carbon and oxygen, is commonly known as oil of wintergreen. It is often used in medical rubs. If a 20.00 g sample of methyl salicylate was burned in air and it produced 9.46 g of water and 46.30 g of carbon dioxide, show that the empirical formula of this compound must be C8H8O3. 21 A student wishes to prepare a standard solution of a base for an acid–base titration. She chooses to weigh the required amount of anhydrous sodium carbonate, dissolve the solid in water and make up the volume in a volumetric flask. The 0.0500 M standard sodium carbonate solution is then used to determine the amount of hydrochloric acid in a sample of concrete cleaner. 20.00 mL of the diluted concrete cleaner is pipetted into the conical flask. After several titrations a 23.45 mL titre is obtained and the concentration of the hydrochloric acid is calculated. a Distinguish between the terms ‘primary standard’ and ‘standard solution’. b List three features that make anhydrous sodium carbonate a suitable choice of solid to use to prepare the standard solution. c What is the concentration of hydrochloric acid in the diluted concrete cleaner? d Determine the effect of the following errors on the final calculated value of the hydrochloric acid concentration. i The burette was rinsed with water. ii The conical flask was rinsed hydrochloric acid. 22 The active ingredient in a certain brand of antacid is magnesium carbonate. A 1.30 g tablet was crushed and placed in a conical flask. A small volume of water was added, the mixture stirred, and methyl orange indicator added. The indicator changed colour permanently after 27.75 mL of 1.00 M hydrochloric acid had been added. a Describe the changes that would be observed in the colour of the indicator. b Write an equation for the reaction that occurs during the titration. c Calculate the amount, in mol, of hydrochloric acid used to reach the end point. d Calculate the percentage by mass of magnesium carbonate in the tablet. e What is the function of the substances that make up the remaining mass of the tablet? f Why do people often burp after taking an antacid tablet?

125 Area of study review

23 Health experts are encouraging us to consume unsaturated oils, rather than saturated oils (or fats) in our diet. Unsaturated oils contain one or more double carbon-to-carbon bonds. Iodine reacts readily with these double bonds. Chemists measure the degree of unsaturation by determining the ‘iodine number’ of the oil (see table). The iodine number is the number of grams of iodine that reacts with 100 g of oil. The oil is put in a stoppered flask, a known excess of iodine is added, and the flask is shaken to allow the iodine to react. The remaining iodine is determined by titration with sodium thiosulfate solution. The reaction of iodine with an unsaturated oil is illustrated. C

C

b Write half equations and an overall equation for the reaction between iodine and sodium thiosulfate, given thiosulfate ion (S2O32–) is oxidised to S4O62– and I2 is reduced to I–. c Calculate the number of moles of iodine in the 10.00 mL aliquot. d Calculate the number of grams of iodine that reacted with the 10 g of oil. e What is the identity of the oil in the sample? f Why is a back titration sometimes used instead of a direct titration? 24 HA is a monoprotic acid and MOH is a base. The titration curve shown is obtained when 20.0 mL of 0.200 M HA is titrated with MOH. 14

C

C

12

I2 H

10

H H

C

8 pH

C

6

C

H

C I

I

4 2 0

Reaction of iodine with an unsaturated oil.

Oil or fat Peanut oil Butter fat Sunflower oil Canola oil Olive oil Coconut oil

Iodine number 93 36 126 99 81 10

10 g of an unknown oil sample is shaken with 100 mL of solution containing 15 g of iodine. The mixture is shaken vigorously and 10.00 mL of the unreacted iodine solution is removed and titrated with 0.250 M sodium thiosulfate solution. A titre of 21.70 mL of the sodium thiosulfate is required. a Consider the table, which shows iodine numbers of a range of oils. Using the iodine number, list the oils in the table from most saturated to least saturated.


25.0 Volume MOH (mL)

Titration curve for HA and MOH.

a Is HA a strong or a weak acid? Explain your answer. b What is meant by the term ‘monoprotic acid’? c Methyl orange changes colour in the pH range 3.1 to 4.5 while phenolphthalein changes colour between pH 8.2 and 10. Which indicator would be the most suitable to use for this titration? Give reasons for your answer. d Determine the concentration of the base. 25 a The amount of alcohol, CH3CH2OH, in your breath can be determined by blowing into a tube containing acidified potassium dichromate, K2Cr2O7. Which substance has been oxidised and which has been reduced? b Explain how an IR spectrometer may be used to determine the amount of alcohol in a driver’s breath.

26 The structures of propan-2-ol and propanone are shown below.

100

H3C

C

CH3

Transmittance (%)

OH

propan-2-ol

H

50

H3C C

O

propanone

0 4000

* 3000*

2000 1500 Wavenumber

H3C

The infrared spectra of these compounds are shown below and are labelled IR Spectrum A and IR Spectrum B.

a Identify the bonds that produced the IR absorption marked with a (*) in each spectrum. b Determine which spectrum represents which molecule. c The 1H NMR spectrum of propanone is shown below.

*

3000

*

2000 1500 Wavenumber

1000

500

Absorbance

Transmittance (%)

0 4000

500

IR spectrum B.

100

50

1000

TMS

IR spectrum A. 10

9

8

7

6

5

4

3

2

1

0

1

H NMR spectrum of propanone.

Explain why there is only a single peak although there are six hydrogen atoms in the molecule.


d Explain why there are two peaks in the 13C NMR spectrum of propanone, shown below, when the molecule contains three carbon atoms.

i What is the mass of the molecular ion? ii What information does this provide about the molecule? iii Suggest a formula for the molecular fragment that produces the peak at m/e 43. 27 The diagram shows a thin-layer chromatogram of amino acids in a medicine.

Absorbance

Solvent front

A = leucine B = taurine C = valine D = glycine

TMS

X = medicine 200

180

160

140

120

100

80

60

40

20

0

C NMR spectrum of propanone.

13

e How many sets of peaks would you expect in the 1H NMR and 13 C NMR spectra of propan-2-ol? Provide an explanation for your answers. f The mass spectrum of propanone is shown below. 100

A

B

X

C

D

TLC of some amino acids.

90 Relative intensity

80 70 60 50 40 30 20 10 0

0

15

20

25

Mass spectrum of propanone.

128 Chemical analysis

30

35 40 m/e

45

50

55

60

a What amino acids are in the medicine? b Amino acids are colourless. What technique could be used to visualise the amino acid spots? c Calculate the Rf value of taurine. d Which amino acid is bonded least strongly to the stationary phase? 28 Indicate how each of the parts of the electromagnetic spectrum listed below interacts with matter. For each part of the electromagnetic spectrum listed, name a spectroscopic technique that utilises this interaction. a radio frequency b infrared c visible–UV

0

chapter 09

Compounds of carbon

key knowledge • structure and systematic nomenclature of alkanes, alkenes, amines, chloroalkanes, alkanols and carboxylic acids up to C10

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • recall that all living matter is composed of carbon compounds • explain how carbon compounds can be synthesised • provide reasons for the existence of so many carbon compounds • explain the nature of the chemical bonds formed between carbon atoms • explain the nature of chemical bonds formed between carbon atoms and the atoms of other elements • identify functional groups and explain their influence on the physical and chemical properties of carbon compounds • explain and provide examples of structural isomerism • name a range of carbon compounds using systematic nomenclature protocols.

Figure 9.2 Friedrich Wöhler.

chemfact It used to be thought that organic molecules could only be made by processes taking place in living organisms. This was disproved in 1828 by the German chemist Friedrich Wöhler, who produced urea, a substance naturally present in the urine of all living mammals, simply by heating a solution of the inorganic compound ammonium cyanate. NH4CNO(aq) → (NH2)2CO(s) urea

chemfact Carbon is a member of group 14 in the periodic table. See Heinemann Chemistry 1, Chapter 3. In older texts you might find that the group containing carbon was named group IV.

Figure 9.1 Carbon compounds are the basis for a huge range of materials from fuels, plastics, fibres and paints through to the chemicals that make up living organisms.

Over seven million compounds containing carbon are known, with more being discovered or created constantly. Carbon compounds exhibit an enormous range of properties, allowing us to develop materials for applications as diverse as fuels, fibres, plastics, detergents, dyes, paints, pharmaceuticals, perfumes and insecticides and a myriad of other substances that we have come to take for granted. Carbon compounds also form the basis of every cell in all living organisms. Because so many carbon compounds are found in plants and animals, the study of carbon compounds is sometimes called organic chemistry. Substances containing carbon, with the exception of carbon dioxide, carbon monoxide, cyanides and carbonates, are known as organic compounds. The element hydrogen is usually present in organic molecules; atoms of carbon and hydrogen can be assembled to form many thousands of different hydrocarbon molecules of various sizes and shapes. When atoms of other elements are also incorporated in these molecules, the number of possible arrangements becomes almost limitless.

9.1

Bonding in carbon compounds To explain why there are so many carbon compounds we need to understand the structure of the carbon atom and the way carbon forms bonds. You will recall from your studies in Unit 1 that carbon’s electronic structure is 1s22s22p2, and that it is found in period 2, group 14 of the periodic table. This means that it is a small atom with four outer-shell electrons. The carbon atom forms covalent bonds by sharing these electrons with other atoms. Carbon forms strong covalent bonds with hydrogen atoms, and nearly all organic molecules contain hydrogen bonded to carbon. Carbon also forms strong covalent bonds with other carbon atoms, forming long chains and rings. Single, double or triple bonds can form between carbon atoms.

130 Organic chemical pathways O

Covalent bonds can also be formed between carbon and atoms of other elements such as oxygen, nitrogen, sulfur, phosphorus and the halogens. When four single bonds are formed around a carbon atom, the pairs of electrons in each bond act as charge cloud. These charge clouds repel each other so that they spread out as far apart as possible. This arrangement is called a tetrahedral distribution because the four single bonds or charge clouds are pointing to the corners of a tetrahedron. Molecules consisting of long chains of carbon atoms joined by single bonds are often referred to as ‘straight chain’ molecules. In fact, the tetrahedral distribution of the bonding pairs of electrons means that the chain is bent.

C

C C

C C

C

C

Figure 9.3 A tetrahedron has four triangular faces. The electrons in four single bonds around a carbon atom repel so that the bonds point to the corners of a tetrahedron. In this arrangement the angle between the bonds is 109.5°.

CHEMICAL BONDING • Dispersion forces • Tetrahedral arrangement

Figure 9.4 Tetrahedral arrangement of bonds in a carbon chain.

9.2

Homologous series The study of organic chemistry can be simplified by grouping the molecules into families or homologous series where the compounds have similar structures and chemical properties and the same general formula. Each member of a homologous series differs by a –CH2– group from the previous member.

review You met the homologous series of the alkanes and alkenes in Unit 1 of this chemistry course. See Heinemann Chemistry 1, Chapter 8.

Alkanes The simplest series of hydrocarbons is composed of molecules with the general molecular formula CnH2n + 2 where n = 1, 2, 3, 4… The alkanes are called saturated hydrocarbons because they do not react further with hydrogen. Only single carbon–carbon bonds are present in these molecules. Table 9.1 gives the names and formulas of the alkanes up to 10 carbon atoms. The molecules can be represented by a structural formula. The structural formula of butane is shown in Figure 9.5. This can also be written as a semi-structural formula: CH3–CH2–CH2–CH3 or as CH3CH2CH2CH3 or as CH3(CH2)2CH3 H

H

H

H C

H

C

C H H

H

C

H

Figure 9.5 Structural formula of butane. H

TABLE 9.1 The alkane homologous series, general formula CnH2n + 2 n 1 2 3 4 5 6 7 8 9 10

Name

Formula

Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane

CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22

131 Compounds of carbon

Cyclic molecules A chain of carbon atoms can also be bent round to form a ring. The molecules formed in this way are called cyclic molecules.

a CHEMICAL BONDING • Forming a covalent bond • Covalent bonding within molecules • Describing molecular shape • Predicting molecular shape • Dispersion forces • Questions about dispersion forces

b CH2

H2C

CH2 H2C

CH2

H2C

CH2

CH2

CH2

Cyclopropane

Cyclohexane

Figure 9.6 a Cyclopropane, C3H6, and b cyclohexane, C6H12, are examples of cyclic molecules.

Sometimes, as in the special case of benzene, three of the four outershell electrons from each carbon atom form normal covalent bonds but the fourth electron is shared (delocalised) around the six-membered carbon ring. In all, there are six delocalised electrons in the benzene molecule that are effectively shared by the six carbon atoms. Compounds like this with rings of delocalised electrons are known as arenes. The ring of delocalised electrons in benzene, C6H6, is often represented in the shorthand form shown in Figure 9.7b.

a

C

H

c

H C

C

C H

b

H

C C

H

H The shaded ring represents the 6 delocalised electrons.

The delocalised electrons are represented by a circle inside the hexagon.

Figure 9.7 a Benzene has delocalised electrons. b A shorthand representation of the benzene molecule. c The delocalised electrons are sometimes represented by these structures.

Isomers Molecules of alkanes that contain less than four carbon atoms per molecule have only one possible structure. From butane, C4H10, onwards, there are progressively more molecules that have different arrangements of atoms but share the same molecular formula. Molecules having the same molecular formula but different structural formulas are known as structural isomers. Structural isomers have different physical and chemical properties. Structural isomers of C4H10 are shown in Figure 9.8. There are three structural isomers of pentane, C5H12. As molecules become larger, the degree of isomerism increases rapidly. There are 75 structural isomers for decane, whereas C20H42 has over 350 000 possible isomers! Each of these is a different alkane. You will develop a better idea of isomers if you make models of some of these molecules.


a

H

/

b

H

H

/

H C

H

H

C

H

H

*

C

C H

/

H

/

/

* *

/

* /

/ /

/

Figure 9.8 Structural isomers of butane: a butane (b.p. –0.5°C) b methylpropane (b.p. –12°C).

Naming alkanes Alkanes are named in a systematic way. The first part of the name indicates the number of carbon atoms and the last part ends in ane. The first ten members of the homologous series of alkanes are shown in Table 9.1. The number of possible isomers with side chains increases as the number of carbon atoms increases. The side chain can be considered as an alkane molecule that has lost a hydrogen atom. The fragment of the alkane that forms the side chain is called an alkyl group. The alkyl groups and the alkanes from which they are derived are shown in Table 9.2.

chemfact The different molecular shapes of structural isomers affect the intermolecular forces, resulting in variations in physical properties such as melting and boiling points.

chemistryy in action A systematic way of naming carbon compounds The over-the-counter antiseptic Dettol contains as an active ingredient the organic compound 4-chloro-3,5-dimethylphenol, C8H9ClO. It is one of the thousands of organic chemicals. Because of its complexity, the name of this compound has been written in a systematic way so that other chemists can quickly work out the compound’s structure. Systematic names are required to keep track of the many natural and synthetic organic compounds. Systematic nomenclature helps international communication between chemists, but common names are also used for well-known chemicals.

a

b

OH

CH3

H3C Cl

Figure 9.9 a The structure of 4-chloro-3,5-dimethylphenol, C8H9ClO, which is the antiseptic agent in Dettol b.


TABLE 9.2 The alkane homologous series and alkyl groups. Alkane Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane

CH3

H3C 5

CH2

CH2

CH

CH3

4

3

2

1

Figure 9.10

Formula CH4 CH3CH3 CH3CH2CH3 CH3(CH2)2CH3 CH3(CH2)3CH3 CH3(CH2)4CH3 CH3(CH2)5CH3 CH3(CH2)6 CH3 CH3(CH2)7CH3 CH3(CH2)8CH3

Alkyl Methyl Ethyl Propyl Butyl Pentyl Hexyl Heptyl Octyl Nonyl Decyl

There is a systematic way to name isomers in a manner that describes their structure. Systematic names for alkanes are made up of: the name of the longest alkane hydrocarbon chain the name and number of any alkyl side chains the position of the side chain on the main carbon chain. The longest unbranched carbon chain is chosen as the stem for naming the molecule. Where necessary, a number can be used to show where the side chain is attached to the longest chain. Carbon atoms are numbered from the start to the end of the longest continuous chain in the compound. The prefix of the name is derived from the alkyl group forming the side chain. The systematic name usually contains a prefix, a stem and a suffix. The name of the hydrocarbon shown in Figure 9.10 can be determined as follows: 1. Identify the longest chain of carbon atoms. The carbon atoms in this chain are numbered. In this case the longest chain contains five carbon atoms so the stem must be pent. 2. Because all the bonds are single, the compound must be an alkane and its name will end in ane. 3. Identify the side chain and the number of the carbon atom on the longest chain to which it is attached. As a methyl group is attached to the second carbon atom, the prefix will be 2-methyl. Numbering always starts from the end of the chain closest to a side chain. The name of this hydrocarbon is 2-methylpentane. Remember that the bonds between carbon atoms have a tetrahedral orientation, although for convenience they are often shown at right angles. The longest chain in the compound shown in Figure 9.11 has nine carbon atoms. So the stem of its name will be non. It is an alkane, so the suffix will be ane. Methyl groups are attached to the third and sixth carbons, and an ethyl group is also attached to the sixth. The prefix will be 3,6-dimethyl6-ethyl. The complete systematic name of this compound is 3,6-dimethyl6-ethylnonane.

H3C

CH2

CH2

9

8

7

CH3

1

CH3

CH2

2

CH2

C 6

CH3 Figure 9.11 Can you name this compound?


Formula CH3– CH3CH2– CH3(CH2)2– CH3(CH2)3– CH3(CH2)4– CH3(CH2)5– CH3(CH2)6– CH3(CH2)7– CH3(CH2)8– CH3(CH2)9–

CH2

CH2

CH

5

4

3

CH3

extension Octane ratings Petrol is a complex mixture of C6 to C12 hydrocarbons, and contains many isomers. Refineries carefully control the exact blend of hydrocarbons. A major factor determining the efficiency of a particular blend of petrol is the petrol’s undesirable tendency to ignite in an uncontrolled manner in the engine cylinder, causing ‘knocking’. Straight-chain hydrocarbons are more likely to cause knocking than branched hydrocarbons. The tendency of a hydrocarbon to cause knocking is measured in terms of the octane rating. Heptane has an octane rating of zero. The highly branched eight-carbon alkane 2,2,4-trimethylpentane has an octane rating of 100. Branched-chain alkanes are incorporated in petrol blends because they have better antiknock properties than straight-chain alkanes and so improve the fuel’s octane rating. Petrol that contains 95% 2,2,4-trimethylpentane and 5% heptane has an octane rating of 95. Refineries convert long hydrocarbon chains into smaller molecules by catalytic cracking. These smaller molecules are then reformed into hydrocarbons such as benzene that have a higher octane rating than alkanes.

questions

Figure 9.12 The octane rating of unleaded petrol.

?

E1 Write Writite the the structural sttructtural formula of the heptane and 2,2,4-trimethylpentane 2 2 4 trimethylpentane molecules. molecules E2 What is the name of the straight-chain hydrocarbon that has same molecular formula as 2,2,4-trimethylpentane?

Alkenes Because of their relatively small size, it is also possible for two adjacent carbon atoms to share two of their outer shell electrons to form double covalent bonds. Hydrocarbons with double covalent bonds between carbon atoms are called alkenes. Hydrocarbons that contain double or triple bonds between carbon atoms are described as unsaturated hydrocarbons. The additional density of electrons held in bonds between the carbon atoms makes these substances much more reactive than the alkanes. The shape of alkenes in the region around the double bond is planar, as there are in effect three charge clouds around each carbon atom. The angles between these bonds is 120°. Those carbon atoms not involved in the double bond still have tetrahedral orientations. The first ten members of the alkene homolgous series are shown in Table 9.3. The general formula of alkenes is CnH2n where n = 2, 3, 4, 5, … For butene and alkenes higher in the homologous series, the double bond can be located in various positions in the carbon chain. The names of alkenes end in ene. The stem of the systematic name of alkenes is based on the longest carbon chain that contains the double bond. The position of the double bond is indicated by the number of the first carbon atom involved in the double bond. By convention, numbering starts from the end nearest the double bond. The rules for naming any side chains are similar to those for the alkanes. For example, the compound CH2=CHCH2CH3 is called but-1-ene (showing that the double bond starts from the first carbon atom in the chain) to distinguish it from its isomer but-2-ene, CH3CH=CHCH3, where the double bond starts at the second

chemfact Carbon atoms can also share three electrons with a neighbouring carbon atom to form a triple bond. The homologous series containing triple covalent bonds between carbon atoms is known as the alkynes. The alkyne ethyne (also called acetylene) is mixed with oxygen in oxyacetylene torches used to weld or cut iron.

H 120˚

H

120˚ C 1

C 2

H

H

3C

H

H

Figure 9.13 Structural formula of propene. The bonds around carbon atoms 1 and 2 are planar while the bonds around carbon atom 3 are arranged tetrahedrally.


TABLE 9.3 The alkene homologous series, general formula CnH2n n 1 2 3 4 5 6 7 8 9 10

Name – Ethene Propene Butene Pentene Hexene Heptene Octene Nonene Decene

Formula – C2H4 C3H6 C4H8 C5H10 C6H12 C7H14 C8H16 C9H18 C10H20

carbon atom. A third isomer contains three carbons in a chain with a methyl side chain and is called 2-methylpropene, (CH3)2C=CH2. The structural formulas of these isomers of butene are shown in Figure 9.14.

a

b H

c

H

C C

H

H

H H

H

C C

C C

H H

H

But-1-ene CH2=CHCH2CH3

H C

H H

H

H

H

C C H

But-2-ene CH3CH=CHCH3

H

C H H

C

H

C

H H

H

2-Methylpropene (CH3)2C=CH2

Figure 9.14 Isomers of butene: a but-1-ene, b but-2-ene, c 2-methylpropene

! CHEMICAL BONDING • Diatomic Lewis structures • Multiple covalent bonds

Some books use the name 1-butene rather than but-1-ene. Both names are valid.

chemistryy in action Does a rotten apple spoil the whole barrel?

Figure 9.15 Bananas are harvested and transported while they are green, and ripened using ethene gas.


Ethene is a natural by-product of the chemical changes that take place during the last stages of ripening in fruits such as apples and bananas. Not all fruits produce ethene when they ripen, but many are sensitive to its presence. Citrus fruits, such as oranges and lemons, that have been picked while they are still green ripen quickly if they are exposed to the gas. The response of fruits to ethene is used to enable fruit growers to meet market demands at all times of the year. Ripe bananas are very easily damaged during picking and transportation so bananas are harvested while they are still green and hard, then kept cool to slow the ripening process. When they reach their destination the bananas can be exposed to ethene, causing them to ripen within a few days, in time for their sale. Tomatoes can be treated in a similar way. A reverse process is used to maintain the freshness of fruits such as apples. Under normal storage conditions, apples stay fresh for only a few weeks after picking. However, under controlled storage conditions, apples are cooled and the carbon dioxide levels in the surrounding atmosphere are increased slightly. This makes the apples less sensitive to their own naturally produced ethene and slows the ageing process so that they can be available to the consumer throughout the year. A side-effect of this storage process is that the apples lose some of the chemicals that give the fresh-picked flavour, even though the apples look good. When apples are bruised or damaged, they produce extra ethene, causing nearby fruit to spoil rapidly. So it is true; one rotten apple can spoil the whole barrel!

summary

9.1–9.2

Carbon C b atoms t form f stable t bl covalent l t bonds b d with ith eachh other th and with other non-metallic atoms such as hydrogen, oxygen, nitrogen, phosphorus and chlorine. Single, double or triple bonds can be formed between carbon atoms. Single bonds involving carbon atoms have a tetrahedral orientation; double bonds have planar shape. Double bonds give a planar shape. Organic compounds that are members of the same homologous series have similar structures and properties. Each member differs by –CH2– from the previous member. Structural isomers are molecules having the same molecular formula but a different structural formula.

Alkanes Alk have h the th generall formula f l CnH2n + 2 andd there th are only l single bonds between carbon atoms. An alkyl group is derived by removing a hydrogen from an alkane. The names of alkanes end in ane. The stem of the name is derived from the number of carbons in the longest chain. The prefix indicates the position and number of atoms in any side chains. Alkenes have the general formula CnH2n. There is a double bond between two carbon atoms. The names of alkenes end in ene. The stem is based on the longest chain that contains the double bond. The position of the double bond is indicated by the number of the first carbon involved in the double bond.

key questions

1 Give the meaning of the following terms: a homologous series b structural isomers c structural formula d semi-structural formula e saturated f unsaturated 2 Identify the homologous series to which each of the following belongs: a C3H8 b C2H4 c C5H10 d C8H18 e CH3(CH2)5CH3 f CH3CH=CHCH2CH3

3 Draw the structural formula of: a ethane b propene c butane d methylbutane e 3-ethyloctane f 2,3-dimethylhexane 4 Explain why there is only one compound corresponding to the formula C3H8 while there are over 70 compounds corresponding to the molecular formula C10H22. 5 Describe and explain the difference in bonding and structure between alkanes and alkenes. 6 Why does the alkene homologous series begin with ethene, C2H4, while the alkanes start with methane, CH4?

9.3

Functional groups So far we have looked at compounds containing only carbon and hydrogen. However, carbon can also form covalent bonds with other atoms or groups of atoms. These are called functional groups. A functional group is an atom or a group of atoms that gives a characteristic set of chemical properties to a molecule containing that group. Many compounds can be produced by substituting a hydrogen atom in an alkane molecule with different functional groups. You will recall that the remaining fragment of the alkane, which has one hydrogen atom removed, is called an alkyl group, represented by R (Table 9.2). The bonds formed between carbon atoms and between carbon and hydrogen atoms, C–C and C–H bonds, are strong and relatively unreactive. They do not react with common reagents, such as acids and bases. As a consequence, the saturated hydrocarbon skeleton of an organic molecule has less influence on its properties than the functional groups.

TABLE 9.4 Common functional groups in organic compounds Name Chloro Hydroxyl Carboxyl

Formula –Cl –OH O R

C O

Amino

H

–NH2

Ester

O C O

Amide

R O

R

C NH

Ether

R

O

R′

R′


! The double bond in alkenes can be considered as a functional group.

To a large extent, the chemistry of many organic compounds is dominated by reactions that involve these functional groups. Knowing the characteristic properties of particular functional groups allows us to classify the vast number of organic chemicals systematically and to understand and predict their properties. Some of the common functional groups are listed in Table 9.4. Functional groups located in different positions in a molecule result in structural isomers. Where necessary, a number can be used to show where the functional group is attached to the carbon chain. Thus, there are two isomers of C3H7OH: propan-1-ol and propan-2-ol.

Chloroalkanes CHEMICAL BONDING • Dipoles • Dipole–dipole attraction • Polar bonds

a

b

Cl

H C H

H H

H

H

d

H

Cl

C

Cl H

H 1-Chloropropane CH3CH2CH2Cl

C C

H

H H

H

C C

H

Cl H

Chloroethane CH3CH2Cl

H

H

C

H

Chloromethane CH3Cl

c

H C

C

H H

H 2-Chloropropane CH3CHClCH3

Figure 9.16 Four chloroalkanes. Cl H

H C

H C

C

C H

H H

H

H

C H

H

H

Figure 9.17 Can you name this chloroalkane?

chemfact The group 17 elements, F, Cl, Br, I and At, are also known as the halogens.


It is possible to replace a hydrogen atom in an alkane with a chlorine atom to form a chloroalkane. The systematic name for a chloroalkane starts with chloro followed by the name of the alkane from which it is derived. Chloroalkanes can be represented by the general formula RCl where R represents an alkyl group. The first two members of the chloroalkane homologous series are chloromethane, CH3Cl, and chloroethane, CH3CH2Cl. In a three-carbon chain it is possible to place a chlorine atom on the end carbon atom or on the middle carbon atom. Numbers are used to indicate the position of the chlorine atom. The two isomers of chloropropane are named 1-chloropropane and 2-chloropropane. The names of some chloroalkanes and their structural and semi-structural formulas are shown in Figure 9.16. The names and positions of any alkyl side chains are also included in the name. In Figure 9.17 the longest chain contains four carbon atoms and is derived from an alkane. The chlorine atom is on the second carbon atom and a methyl side chain is on the third carbon atom. The name of this compound is 2-chloro-3-methylbutane. More than one hydrogen atom in an alkane can be replaced with a chlorine atom as illustrated in Figure 9.18. The other halogens, fluorine, bromine and iodine, form similar homologous series to the chloroalkanes.

a

b

H

c H

C H

Cl Cl

Cl Dichloromethane CH2Cl2

H C

C

Cl

H H

H C

Cl H

H

1,1-Dichloroethane CHCl2CH3

C

Cl H

1,2-Dichloroethane CH2ClCH2Cl

! Figure 9.18 Some dichloroalkanes.

Trichloromethane, CHCl3 or chloroform, was once used as an anaesthetic.

Alkanols Ethanol, found in alcoholic drinks, is a member of the alkanol homologous series. Alkanols is the systematic name for the group of chemicals commonly called alcohols. Alkanols have the general formula ROH. The hydroxyl functional group, –OH, has been substituted for one hydrogen atom in an alkane molecule. Hydrogen bonds involving the polar bond of the –OH group influence the properties of alcohols. Compared to alkanes, alcohols are more soluble in water. They have higher boiling points than alkanes of similar molecular mass. Alkanols are named by dropping the ‘e’ at the end of the hydrocarbon name and replacing it with ‘ol’. The alkanol CH3CH2OH has two carbon atoms and is called ethanol. In higher members of this homologous series the –OH functional group may be placed at the end or in the middle of the carbon chain. Numbers are used to indicate the position of the functional group. For example, propan-1-ol tells us there is a hydroxyl group on the first carbon atom in the chain (Table 9.5). Similarly, propan-2-ol indicates a hydroxyl group on the second carbon atom.

TABLE 9.5 Representations of propan-1-ol, propan-2-ol and 2-methylpropan-1-ol Name

Semi-structural formula

P ropan-1-ol

CH3CH2CH2OH

Alcoholic beverages contain varying amounts of ethanol. Beer contains about 5% ethanol by volume, wine about 12% and spirits such as whisky about 50%.

chemfact The hydroxyl functional group in alcohols can form hydrogen bonds with water molecules. This allows methanol and ethanol to dissolve readily in water but larger alcohol molecules become progressively less soluble. Why?

! The isomers of propanol can also be named 1-propanol and 2-propanol.

Structural formula H

H H

C

C C

H H

CH3CHOHCH3

OH H

H

Propan-2-ol

chemfact

H

H OH

C

C C

H H

H H

H

2-Methylpropan-1-ol

(CH3)2CHCH2OH

H

H

H C

C

C

H H

OH

H C H

H

H


chemistryy in action Antifreeze H

H H

C

C H OH

HO

Figure 9.19 Structural formula of ethylene glycol.

! Propane-1,2,3-triol or glycerol, HOCH2CHOHCH2OH, is used as a sweetener in confectionery and as a moisturising agent in hand lotions.

Carboxylic acids

Oδ− δ+

C O

H

δ−

δ+

Figure 9.20 Structure of the carboxylic acid functional group.

O

H H H

C

C

C H

C C

H

OH H

H

If you inspect the liquid used as coolant in a car radiator, you will probably see that it is coloured and obviously not pure water. The coolant is a solution of different chemicals, including substances that prevent corrosion of metals in the engine’s cooling system. One substance also present is ‘antifreeze’. Antifreeze solutions are made with ethane-1,2-diol, HOCH2CH2OH, more commonly known as ethylene glycol. Pure ethylene glycol is a viscous, colourless liquid. It is soluble in water as a result of the two polar hydroxyl groups in the molecule (Figure 9.19), which form hydrogen bonds with water molecules. Aqueous solutions of ethylene glycol have excellent properties for use in car radiators, having lower freezing temperatures and higher boiling temperatures than pure water. Importantly, they do not promote corrosion of iron. Before World War II ethylene glycol was the major chemical made from ethene. With the subsequent growth of the plastics industry, it has been overtaken by polyethene, vinyl chloride and styrene. Ethylene gycol is converted to toxic oxalic acid in the body. Propylene glycol, HOCH2CH(OH)CH3, is also used as antifreeze and is metabolised to the non-toxic lactic acid. Propylene glycol is also used in foods and cosmetics. Isn’t it amazing how a small difference in structure can result in a big difference in properties!

H H

Figure 9.21 Structure of 3-methylbutanoic acid.

Carboxylic acids contain the carboxyl functional group (Figure 9.20). The general formula of carboxylic acids is RCOOH, where R is an alkyl group. Vinegar is a dilute solution of the carboxylic acid ethanoic acid (CH3COOH), which is commonly known as acetic acid. The characteristic odour of ants is due to methanoic acid. The acid properties of this group of organic compound is due to the ionisation of the COOH functional group. Bonds within the COOH functional group are polarised, resulting in the formation of hydrogen bonds between molecules or between carboxylic acid molecules and water. Low molecular mass carboxylic acids are soluble in water. You may remember from Unit 2 that carboxylic acids are weak acids (see Heinemann Chemistry 1, Chapter 14). The ionisation of ethanoic acid in aqueous solution can be shown as: CH3COOH(l) + H2O(l)

The stem of the systematic name of carboxylic acids is determined from the number of carbon atoms in the longest chain that contains the COOH functional group, and oic acid is added at the end of the name. The names and semi-structural formulas of some carboxylic acids are given in Table 9.6.

TABLE 9.6 Some carboxylic acids Number of carbon atoms 1 2 3 4 5



Semi-structural formula HCOOH CH3COOH CH3CH2COOH CH3(CH2)2COOH CH3(CH2)3COOH

Name Methanoic acid Ethanoic acid Propanoic acid Butanoic acid Pentanoic acid

Branched carboxylic acids are named in a similar way to the other branched organic compounds. Figure 9.21 shows the structure of 3-methylbutanoic acid, an isomer of pentanoic acid. Citric acid is found in lemons and other citrus fruits. A molecule of citric acid contains three carboxylic acid functional groups. The semi-structural formula is shown in Figure 9.22.

Amines N CH3

CH2

H

H

H H

Methanamine (aminomethane) CH3NH2

Ethanamine (aminoethane) CH3CH2NH2

COOH

C

COOH

H2C

COOH

Figure 9.22 Citric acid.

NH2

N

H3C

HO

H2C

CH2 H3C

CH3 CH

Butan-2-amine (2-aminobutane) CH3CH2CH(NH2)CH3

Figure 9.23 Some amines.

Organic compounds that contain the –NH2 functional group are called amines. The general formula for the amine homologous series is RNH2. The amines are named by replacing the ‘e’ at the end of the name of the alkane with amine as shown in Figure 9.23. Around the nitrogen atom, there are three pairs of electrons involved in bond formation and one lone pair of electrons. The bonds around the nitrogen atom are polar due to the high electronegativity of nitrogen. You will recall that this is similar to the bonding around the nitrogen atom in ammonia. Hence, the amino functional group is also able to function as a base, with the lone pair of electrons accepting a proton. (You may wish to revise your understanding of Brønsted–Lowry acids and bases from Heinemann Chemistry 1, Chapter 14.)

chemfact Like ammonia, the smaller amines have a strong odour. The characteristic odour of decaying flesh is caused by the diamines putrescine, H2N(CH2)4NH2, and cadaverine, H2N(CH2)5NH2. δ−

O δ+

δ− δ+

N

H δ−

Hδ+

O Hδ+

δ−

Hδ+

CH3–CH2–NH2(aq) + H2O(l) –—➤ CH3–CH2–NH3+(aq) + OH–(aq) ethanamine ethyl ammonium ion

The smaller amines are very soluble in water due to the formation of hydrogen bonds. This solubility decreases as the size of the alkyl group becomes larger. Amino acids are the basic building block for proteins. They contain the amine and carboxylic acid functional groups. The structural formula of alanine, an amino acid, is shown in Figure 9.25. The systematic name for alanine is 2-aminopropanoic acid.

Hδ+

H

CH3 Hδ+ O Hδ+

Figure 9.24 Hydrogen bonding between methylamine and water.

! The amines can also be named by adding -amine to the alkyl stem; giving methylamine, ethylamine.

CH3 O C C

H2N H

OH

Figure 9.25 The amine and carboxylic acid functional groups in alanine.


chemistryy in action Changing names The original names for some of the organic compounds that have been known for centuries tell you something about their origins. For example, the compound we now call ethanoic acid, CH3COOH, used to be known as acetic acid from the Latin word acetum—vinegar. Some of these old names are so well known that people still use them. As more and more substances were identified, remembering their names became rather complicated. An international committee called the International Union of Pure and Applied Chemistry (IUPAC) devised the system of nomenclature (naming) that is based on the number of carbon atoms.

summary

There is also a system for naming the homologous series. For example, the structure of the alcohols is based on an alkane chain with an alcohol functional group so they are known as the alkanols. Similarly the carboxylic acids should be called the alkanoic acids. However, the similarity of the names based on the hydrocarbon backbone becomes rather confusing and chemists often use the names such as alcohols and carboxylic acids that are familiar to them.

9.3

C Carbon b is i able bl to t form f stable t bl bonds b d with ith functional f ti l groups suchh as Cl, –OH, –COOH and –NH2. Functional groups largely determine the physical and chemical properties of carbon compounds. Chloroalkanes have a general formula RCl where R is an alkyl group. The general formula of alkanols is ROH. They contain the hydroxyl (–OH) functional group.

Carboxylic C b li acids id contain t i the th carboxyl b l (–COOH) ( COOH) functional f ti l group. Their general formula is RCOOH. Compounds that have the general formula RNH2 are called amines. They contain the amino (–NH2) functional group. Numbers are used to indicate the position of functional groups within a molecule of an organic compound.

key questions 7 Give the systematic names for: a CH3OH b HCOOH c CH3Cl d CH3NH2 8 Write the systematic names of: a CH3CH2CH2Cl b CH2ClCH2CH3 c CH3(CH2)3CH2OH d CH3(CH2)3CHOHCH2CH3 e CH3CH2CH2NH2 9 Write the semi-structural formulas of: a 2-hexanol b 1-chloropentane c butan-1-amine d 2-methylhexane e 4-nonanol 10 Why don’t we use the names 1-chloroethane or 3-propanol?


11 The semi-structural formulas of some organic compounds are given below. For each compound: i identify the homologous series to which it belongs ii give its systematic name. a CH3(CH2)5CH2Cl b CH3(CH2)2CHCl(CH2)2CH3 c CH3CHOH(CH2)4CH3 d CH3(CH2)4COOH e CH3(CH2)2CH(NH2)CH3 f (CH3)2CHCH2CH3 g (CH3)2C=CH2 12 For each of the substances whose molecular formulas are given below: i draw the structural formula ii name one structural isomer. a C5H11Cl b C4H9OH c C2H5COOH d C3H7COOH e C5H11NH2

09

key terms

alcohol alkanes alkanoic acid alkanol alkenes alkyl group alkyne

amine amino acid arenes carboxylic acid chloroalkane cyclic molecules functional group

Structural and semi-structural formulas and systematic names 13 Give the systematic names for: a CH3(CH2)7CH2Cl b CH3CH2COOH 14 Write the semi-structural formulas of: a 2-chloroheptane b butan-2-ol c decan-3-amine d butanoic acid 15 Draw the structural formulas of: a butan-2-ol b pentanoic acid c 2-chloropentane d propene e hexan-3-amine f octane 16 Write semi-structural formulas and systematic names for the following substances. a H H

C

OH C

C H

O

H

H

b

OH

H

H

C C

C H H

c

H

H O

C H

OH

H

homologous series hydrogen bonds polar bond saturated hydrocarbons semi-structural formula structural formula structural isomer

systematic name unsaturated hydrocarbons

Isomers 17 Draw and name all possible isomers of: a C3H7Cl b C5H10 (an alkene) c C5H12 18 Draw the structural formulas of five isomers of C5H11Cl.

Properties of chloroalkanes, alkanols, carboxylic acids and amines 19 Table 9.7 gives the molecular masses and boiling points of butane, propan-1-ol and chloroethane. Explain why the boiling points differ even though the compounds have similar molecular masses.

TABLE 9.7 Compound Butane Propan-1-ol Chloroethane

Molecular mass 58 60 65

Boiling point (°C) –0.5 97.0 12.5

20 Explain why alkanols such as methanol and ethanol are soluble in water in all proportions whereas the alkanols higher in the homologous series such as 1-octanol are insoluble. 21 Write an equation for the ionisation of methanoic acid in water. 22 With the aid of a diagram explain why methylamine is soluble in water. 23 Why would ethanamine be more soluble in water than butanamine?

Connecting the main ideas 24 Write equations for the reaction between ethanoic acid solution and: a sodium hydroxide solution b solid sodium carbonate c magnesium 25 Prepare a poster that summarises the rules for the systematic naming of carbon compounds. 26 Prepare notes for a PowerPoint presentation that outlines the reasons why carbon is able to form so many different compounds.


1

chapter 10

Organic reactions: pathways to new products

key knowledge • common reactions of organic compounds: addition reactions of alkenes; substitution reactions of alkanes and primary chloroalkanes; oxidation of primary alkanols; esterification • organic reaction pathways including the production of esters from alkenes • principles of fractional distillation

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe substitution reactions in alkanes and recall the names of products formed by these reactions • write equations illustrating substitution reactions • describe and write equations for addition reactions in alkenes and recall the products of these reactions • distinguish between substitution and addition • explain the importance of functional groups • describe the typical reactions of chloroalkanes, alkanols and carboxylic acids • write equations for these reactions and name the compounds formed • identify the names and structures of the alkanols and carboxylic acids used in the preparation of esters • identify organic reaction pathways for the synthesis of organic compounds from given starting compounds • recall the factors involved in the planning of an organic synthesis • calculate the yield of a reaction • describe the process of fractional distillation.

10.1

Reactions of alkanes An alkane molecule such as ethane has a ‘backbone’ consisting of a chain of single carbon-to-carbon bonds (Figure 10.1). This backbone is surrounded by carbon-to-hydrogen bonds. The alkane molecule is non-polar because all the carbon atoms have the same electronegativity (2.5), the electronegativity of hydrogen (2.1) is very similar to that of carbon and the hydrogen atoms are arranged symmetrically around the carbon chain. The alkanes are insoluble in water but are soluble in non-polar solvents. There are weak dispersion forces between molecules as evidenced by their low melting and boiling points. The stability of the carbon–carbon bonds and the non-polar nature of the molecules means that alkanes are very resistant to reaction. Most reactions involving alkanes are either combustion or substitution.

H

a

H C

H H

C

H H

b

Combustion Alkanes are used as fuels. Combustion reactions involving alkanes release large amounts of heat energy. Methane is the major component in natural gas, and octane is an important component of petrol. The combustion equations for these two alkanes are:

Figure 10.1 a Structural formula of ethane. b Space filling model of large alkane.

CH4(g) + 2O2(g) –—➤ CO2(g) + 2H2O(g) + energy 2C8H18(g) + 25O2(g) –—➤ 16CO2(g) + 18H2O(g) + energy

Substitution reactions In substitution reactions, one or more of the hydrogen atoms in an alkane is replaced by a different atom or functional group. This involves breaking the carbon-to-hydrogen bonds and making new bonds with the substituting atom or group. Chloroethane (CH3CH2Cl) is a gas at room temperature and is used as a local anaesthetic spray. It can be produced by heating mixtures of chlorine and ethane. Alternatively, the reaction can be initiated by exposing the gas mixture to ultraviolet light. In both circumstances, the reaction conditions cause the chlorine molecule, Cl2, to break into separate chlorine atoms. Because these are unstable with only seven outer-shell electrons, the chlorine free radicals attack the carbon-to-hydrogen bonds in the ethane. This results in a mixture of chlorinated alkanes, including CH3CH2Cl, CH3CHCl2, CH3CCl3 and CH2ClCH2Cl. The amounts of each product depend on the relative starting amounts of chlorine and ethane. Chloroethane can be separated from the other reaction products by fractional distillation (see section 10.6, page 160). We can write a simplified reaction for the manufacture of chloroethane: heat or light

CH3CH3(g) + Cl2(g) —–––—➤ CH3CH2Cl(g) + HCl(g)

The chlorination of alkanes can be represented by the general equation: RH + Cl2 –—➤ RCl + HCl

where R represents an alkyl group (p. 134). A reaction like this, in which one atom is replaced by another atom or group of atoms, is called a substitution reaction. In this instance, a chlorine atom has replaced a hydrogen atom in ethane.

145 Organic reactions: pathways to new products

10.1

summary

Alkanes undergo ndergo combustion comb stion and substitution s bstit tion reactions. reactions Carbon atoms are oxidised in combustion reactions and heat energy is produced.

In substitution s bstit tion reactions, reactions one or more hydrogen h drogen atoms is replaced by other atoms or groups of atoms.

key questions 1 Write balanced equations to represent the formation of all possible products of the reactions of chlorine with methane.

H

H

C

10.2

Reactions of alkenes

C

H

2 Draw the structural formulas and name all the possible products that can be formed by the reaction of chlorine with ethane. Which substances are isomers?

H

Figure 10.2 The structure of the ethene molecule.

Ethene, or ethylene as it is known in industry, is the first member of the homologous series of alkenes. Ethene (C2H4): is unsaturated is a non-polar molecule is insoluble in water is a flammable gas participates in addition reactions polymerises to produce polyethene. Since ethene is a small, non-polar molecule, the only attractive forces between its molecules are dispersion forces and ethene therefore has a very low boiling temperature: –104°C.

Addition reactions of alkenes Figure 10.3 Testing for unsaturation. Adding a few drops of red-coloured bromine to hexane (left) produces no reaction. The colour in bromine added to sunflower oil (right) disappears almost immediately because molecules in sunflower oil contain carbon–carbon double bonds that undergo addition reactions with bromine to produce colourless brominated compounds.

The double covalent bond in the ethene molecule has a significant effect on its chemical properties. Ethene reacts more readily, and with more chemicals, than ethane (CH3CH3), which contains only single bonds. The reactions of ethene usually involve addition of a small molecule to produce a single product. For example, it reacts with bromine solution (Br2 dissolved in an organic solvent) as shown in Figure 10.4. This reaction, in which the red-brown colour of bromine disappears as it reacts with the alkene, is used as a general test for unsaturation (Figure 10.3). This type of reaction is known as an addition reaction. During addition reactions the double carbon–carbon bond is converted to a single bond. H

H

C

Figure 10.4 Addition of bromine to ethene.


+ Br

C

Br

H

H Ethene

H

H H C

Br Bromine

C

H Br

1,2-Dibromoethane

An addition reaction between ethene and hydrogen chloride forms chloroethane: H2C=CH2(g) + HCl(g) –—➤ CH3CH2Cl(g)

Ethanol can be produced by an addition reaction of ethene and water using a catalyst to speed up the reaction. In industry, large amounts of ethanol are made by the addition reaction of steam and ethene using a phosphoric acid catalyst (Figure 10.5).

H

H

C

+ H2O

C

H

C

C H

300˚C H

H

Some equations in this chapter are not balanced. When writing equations for reactions involving organic chemistry, it is common for chemists to omit all but the organic reactants and products.

H

H

H3PO4 catalyst

chemfact

Figure 10.5 The formation of ethanol from ethene.

O–H

H

The reaction of ethene with steam is often described as a hydrolysis reaction. Hydrolysis reactions involve water as a reactant. The reaction occurs when steam and ethene are passed over a solid phosphoric acid catalyst. It is used for the commercial manufacture of ethanol because it is a one-step process that uses little energy, apart from initial heating. Ethene is an extremely useful substance because of its ability to undergo addition reactions. You may learn about its manufacture on a very large scale in Unit 4, Chapter 22. The other alkenes undergo similar addition reactions to produce alkanols and chloroalkanes.

Addition polymerisation H

H

! H

H

H

H

C

C

C

C

H

H

H

H

Ethene is also called ethylene. The polymer made from ethene, polyethene, is also known as polyethylene or Polythene, an ICI (UK) trademark.

Catalyst

n

C

C

H

H Ethene

(CH2

Figure 10.6 The formation of polyethene by addition polymerisation.

CH2)n

Polyethene

A type of addition reaction of ethene is involved in making polyethene (Figure 10.6). The number n in this reaction is very large—several thousand or more. A molecule made by linking a large number of small molecules, such as ethene, is called a polymer (meaning ‘many units’); the small molecule, ethene, is called a monomer (‘one unit’). This type of reaction is known as addition polymerisation. When the polymer is being formed, the ethene molecules add to the end of growing polymer chains. Imagine making a necklace of beads. If you think of each bead as representing an ethene molecule and imagine adding one bead at a time to the growing necklace, you have a picture of the way the polymer molecule forms. Of course, you would make an enormous necklace if you used as many beads as the number of ethene molecules used to make most polymer chains. Ethene is also used to synthesise other monomers which are used to manufacture addition polymers, for example PVC and polystyrene.

review See Heinemann Chemistry 1, Chapter 8, for a more detailed discussion of addition polymerisation.


review The empirical formula of an addition polymer is the same as that of the monomer. The monomers add to the ends of the growing polymer chain. Addition polymerisation was discussed in Heinemann Chemistry 1, Chapter 8.

PVC is the abbreviated name for the polymer polyvinyl chloride. PVC has many uses and often replaces materials such as glass, ceramics, rubber and steel. Additives that modify the properties of PVC have enabled it to become one of our most versatile polymeric materials, and it is used in an enormous range of articles. You will be familiar with a number of these, such as insulation around electrical cables and many kinds of packaging. Polyvinyl chloride is manufactured by an addition polymerisation reaction from the monomer vinyl chloride (chloroethene) (Figure 10.7). The vinyl chloride monomer is made from ethene. H

H

H

H

C

C

Catalyst

n

C

C

n

Figure 10.7 Production of polyvinyl chloride from vinyl chloride.

Cl

H

H

Vinyl chloride

Cl

Polyvinyl chloride

Polystyrene is another polymer for which there is a wide range of applications, including packaging. It is made from the monomer styrene (Figure 10.8), which in turn is made from ethene. Styrene is also mixed with acrylonitrile and butadiene to make ABS plastic, which is used to make hard hats and bicycle helmets. ABS is an example of a copolymer—a polymer made from more than one monomer. H

H

H

H

C

C

Catalyst

n

C

C

n H

Figure 10.8 Polymerisation of styrene to form polystyrene.

H Styrene

Polystyrene

chemistryy in action Customised polymers Other polymers that have particular properties and specific applications have been developed. Polyethenol dissolves in water and is used in hospital laundry bags. A bag containing soiled linen is placed in the washing machine, where the bag then dissolves, letting the linen out. Thus laundry workers do not have to touch contaminated linen.

summary

There is a triple bond between the carbon atoms in ethyne, HC≡CH. Ethyne can undergo addition polymerisation to form polyethyne. There are isomeric forms of polyethyne that glow in the dark—one isomer glows red while the other glows blue.

10.2

Alk Alkenes participate ti i t in i addition dditi reactions. ti The carbon–carbon double bond is converted to a single bond during addition reactions.


Alkenes Alk reactt with ith chlorine hl i to t form f chloroalkanes. hl lk Alkenes react with water to form alkanols. Alkenes undergo addition polymerisation reactions.

key questions 3 Name the products formed when ethene reacts with: a hydrogen chloride b chlorine c water d hydrogen 4 Use structural formulas to write equations and name the products for the reactions of: a but-1-ene with chlorine b but-2-ene with hydrogen bromide 5 Polyethenol is used to make soluble hospital laundry bags. The structure of the monomer ethenol is shown in Figure 10.9. Draw the structure of a section of the polymer.

H

H C

C

H

O–H

Figure 10.9 Structure of ethenol.

10.3

Reactions of functional groups The influence of functional groups on the chemistry of organic molecules may be seen by studying their reactions. When we consider how new substances can be made, we should think about how the structure of its functional group determines the way a particular molecule reacts and the conditions needed for the reaction to occur. H

Reactions of chloroalkanes Once a more electronegative atom such as chlorine has been substituted for a hydrogen atom in an alkane, the molecule becomes polar. Electrons in the carbon–chlorine bond are attracted towards the more electronegative chlorine atom. This makes the carbon atom at the other end of the bond susceptible to attack by negatively charged ions (anions). For example, chloromethane is converted to methanol when it is reacted with hydroxide ions, as shown Figure 10.11. The chlorine atom is substituted by an OH functional group to form methanol. H O– +

H

C

Cl

H

Chloromethane

H C

H H

Cl H

Chloroethane

+

H

Figure 10.10 Structural formula of chloroethane. Hydrogen has electronegativity of 2.1, carbon 2.5 and chlorine 3.2.

Figure 10.11 Reaction of chloromethane with hydroxide ions.

Methanol

H

H

C

CIδ−

Cl–

+

C

O

H

H

H

Cδ+

H

δ−

δ+

C

H

H

H

H

H

H

Catalyst O

C

H

+

C

H+(aq)

+

Cl–(aq)

O–H H

H

H Ethanol

Figure 10.12 Reaction of chloroethane with water.


chemfact Amines behave as weak bases: CH3CH2NH2(aq) + H2O(l) CH3CH2NH3+(aq) + OH–(aq)

The carbon–chlorine bond is also susceptible to attack by the negatively charged end of a polarised molecule. Ethanol is formed in this way when chloroethane reacts with water. Chloroalkanes will also react with ammonia to form amines: δ−

RCl + NH3 ––—➤ RNH2 + HCl

The chloroalkanes higher in the homologous series undergo similar substitution reactions.

Reactions of alkanols

Figure 10.13 Alkanols are a constituent of many consumer products.

As you discovered above, alkanols can be produced by addition reactions of alkenes or substitution reactions of chloroalkanes. Ethanol is the most widely used alkanol. Ethanol is added to some types of petrol to reduce our dependence on oil reserves. The use of ethanol as a biochemical fuel is discussed in Chapter 11. Ethanol has very different properties from ethane or chloroethane. It is a liquid at room temperature. It is widely used as a solvent in cosmetics and pharmaceuticals, as well as being the active ingredient in alcoholic drinks such as wine, beer and spirits. It acts as a depressant on the human body, slowing reactions and responses. Excess ethanol consumption also blocks the production of antidiuretic hormones, increasing urination and resulting in dehydration. Ethanol is soluble in water as a consequence of its highly polar OH group, which readily forms hydrogen bonds with water molecules. Alkanols can undergo substitution reactions. The amino functional group can be introduced into the alkane chain by a substitution reaction between ammonia and an alkanol. Ethylamine is formed by passing ammonia and ethanol vapour over alumina (aluminium oxide) heated to about 400°C: alumina

CH3CH2OH(g) + NH3(g) –––—➤ CH3CH2NH2(g) + H2O(g) 400°C

Alkanols can be oxidised to form carboxylic acids: O2(g)

CH3CH2OH(aq) ––—➤ CH3COOH(aq) Figure 10.14 Most wines contain added antioxidants. Antioxidants are chemicals that slow the oxidation of ethanol to ethanoic acid. This has to be shown on the label. Preservative 220 is sulfur dioxide.

chemfact This redox reaction involves marked colour changes. If a purple-coloured solution of MnO4– ions is used, the reaction mixture becomes almost colourless due to the formation of Mn2+ ions. If orange-coloured Cr2O72– solution is used, a green Cr3+ solution is formed.


In the laboratory, a carboxylic acid such as ethanoic acid is manufactured more rapidly and with more control, using an oxidant such as an aqueous solution of acidified potassium permanganate or acidified potassium dichromate. MnO4−(aq) or Cr2O72−(aq)

CH3CH2OH(aq) ––—–—————➤ CH3COOH(aq) H+(aq)

Not all alkanols will oxidise to form carboxylic acids. The position of the OH functional group in an alkanol determines the oxidation product. Carboxylic acids are synthesised from the oxidation of primary alkanols. In primary alkanols the OH functional group is attached at the end of a chain of carbon atoms or at the end of a side chain, i.e primary alkanols have a –CH2OH grouping. Methanol, CH3OH, is also classed as a primary alkanol.

a

H

b

H H

H H

H C C

O–H

H

H

c

H

O–H

H

H

C

C

H C

C

C

H

H

C

C

H

H

C C

H

H

H

H

H

H

C O–H

H

H

H H

H

Figure 10.15 Structures of the isomers of C4H9OH: a a primary alkanol, butan-1-ol b a secondary alkanol, butan-2-ol c a tertiary alkanol, 2-methylpropan-2-ol.

Alkanols react with carboxylic acids to form esters. The formation of esters is discussed in detail later in this chapter.

Reactions of carboxylic acids All carboxylic acids are weak acids, reacting with water to form a weakly acidic solution: CH3COOH(aq) + H2O(l)


Ethanoic (acetic) acid is classed as a weak acid, as in a 0.1 M solution only 1% of the ethanoic acid molecules ionise. Vinegar is an aqueous solution of ethanoic acid and other chemicals; the concentration of ethanoic acid is about 1 M. An uncorked bottle of wine will soon develop a sour taste as ethanol is oxidised to ethanoic acid. Some wineries take advantage of this process and produce wine vinegar in commercial quantities by allowing poor quality wine to react with oxygen in the air, with the aid of the microorganism Acetobacter. Carboxylic acids such as ethanoic acid react with bases, reactive metals and carbonates. Carboxylic acids will also react with alkanols to form esters as discussed in section 10.4.

summary

Figure 10.16 Wines are increasingly being sold with screw caps as the traditional cork is more likely to be porous to air, causing the ethanol in the wine to oxidise to ethanoic acid. This makes the wine taste ‘vinegary’. Poor quality cork can also introduce off-flavours to the wine.

10.3

Chl Chloroalkanes lk reactt with ith water t to t form f alkanols. lk l The hydroxyl group, OH, in alkanols is polar and forms hydrogen bonds with water.

The Th OH group in i alkanols lk l can be b substituted b tit t d by b other th functional f ti l groups such as NH2. Alkanols are oxidised to form carboxylic acids. Carboxylic acids are weak acids.

key questions 6 Draw molecular structures and refer to the electronegativities given in the periodic table at the end of this book to work out whether each of the following substances has a dipole or is non-polar. a CH3F b CH3CCl3 c CCl3CH2CCl3 7 Write unbalanced equations for the production of: a methanol from chloromethane b chloromethane from methane c propan-1-ol from 1-chloropropane.

8 Write a sequence of reactions that shows the production of: a ethanoic acid from ethane b ethanoic acid from ethene. 9 Label each step in Question 8 to show whether it is an addition, substitution or oxidation reaction. 10 Write balanced equations for the reaction between ethanoic acid and: a sodium hydroxide solution b magnesium metal c sodium carbonate solution.


10.4

Esters: sweet flavours and strong aromas Esters are a group of organic compounds responsible for some of the natural and synthetic flavours and smells in ice-creams, lollies, flowers and fruits (Table 10.1). Your favourite perfume or aftershave may well contain certain esters responsible for its distinctive and appealing odour.

TABLE 10.1 Ester smells and flavours.

Figure 10.17 Esters are responsible for many fruity smells and odours. Over a thousand synthetic and natural flavours are used as food additives.

Ester Pentyl propanoate Ethyl butanoate Octyl ethanoate 2-Methylpropyl methanoate Ethyl methanoate Pentyl ethanoate

Smell or flavour Apricot Pineapple Orange Raspberry Rum Banana

Esters composed of small molecules are volatile and have distinctive odours. They have low boiling points that allow them to evaporate easily and reach your nose. Esters of larger molecular size are oils and waxes (Figure 10.18). Esters are made by a condensation reaction between a carboxylic acid and an alkanol. Reactions that involve the combination of two reactants and the elimination of a small molecule, such as water, are called condensation reactions. For example, gently heating a mixture of ethanol and pure ethanoic acid with a trace amount of sulfuric acid added produces an ester (ethyl ethanoate) and water. The sulfuric acid acts as a catalyst (Figure 10.19).

Figure 10.18 Bees’ wax is largely composed of an ester, CH3(CH2)14COO(CH2)29CH3, called tracontyl palmitate.


Ester functional group H

H

O

H

+

C

C

H–O

C

Catalyst

C

O

+

C

H2O

H H

CH3

O–H

H H

H+

C

H H

O

H

CH3

Ethyl ethanoate is commonly known as ethyl acetate. As well as being used as an artificial flavouring, it is used as a solvent in some paints, adhesives and nail varnish. The general equation for the esterification reaction involving a carboxylic acid and an alkanol is shown in Figure 10.20. Esters have two-part names, with the first part derived from the name of the alkanol from which it is made, where ‘yl’ replaces ‘anol’. The second part comes from the carboxylic acid, where ‘ic acid’ is replaced by the suffix ‘ate’. Thus the ester formed from ethanol and ethanoic acid is ethyl ethanoate (Figure 10.21).

C

chemfact Although the first part of an ester’s name is derived from the alkanol from which it is formed, when we are writing the formula the alkanol section is normally placed second. For example: methyl ethanoate CH3COOCH3 propyl methanoate HCOOCH2CH2CH3

O

O

O R

Figure 10.19 Production of the ester ethyl ethanoate from ethanol and ethanoic acid.

+ H–O

R´

+ H2O

C

R

O–H

CH3C Ethanoate

R´

O

Ethyl OCH2CH3

Figure 10.21 Naming esters.

Figure 10.20 A general equation for ester formation. The OH group in the acid is replaced with the –OR′ of the alkanol.

extension Polyesters Polyesters are copolymers made from alkanol and carboxylic acid monomers bonded together by ester bonds formed between the hydroxyl and carboxyl functional groups of the monomers (Figure 10.22). O

O HO

C

C

OH + HO

Diacid

O

OH + HO

O

O

C

C

OH + HO

OH

Diol

O

O

C

C + H2O

O

O

O

O

C + H2O

C + H2O

O

O

Figure 10.22 Polyesters are copolymers. Monomers with two hydroxyl functional groups react in a condensation polymerisation reaction with molecules that have two carboxylic acid groups to form a polymer.

Polyester

Polyesters are an extremely versatile group of polymers. They are used in fabrics under trade names such as Dacron, Terylene and Crimplene; clothes made from these fabrics can be ‘permanently pressed’ by applying heat to soften the fibres. They are also used to make drink bottles, for coating surfboards, and constructing car bodies and swimming pools.


extension Polyesters (continued) Soluble polyesters are used to stitch wounds and in controlled release medications. A biodegradable polymer made from the monomers lactic acid and glycolic acid is used in absorbable stitches for wounds (Figures 10.23 and 10.24). While the wound is healing, the stitches are slowly hydrolysed to harmless products that can be processed by the body’s biochemical pathways. CH 3

HO

CH2

+

COOH

HO

CH

Glycolic acid Condensation

Hydrolysis

O O

COOH Lactic acid

CH2

C

CH3 O

CH

O C

O

+ H2O

Figure 10.23 Condensation polymerisation of lactic acid and glycolic acid to form a soluble polyester. The reaction is reversed when ester bonds are slowly hydrolysed by water.

Figure 10.24 Surgeons use sutures made from a polyester thread that is slowly hydrolysed in the body to harmless compounds after the wound has healed.


extension Polyesters (continued) A condensation reaction between propane-1,3-diol and terephthalic acid produces a dimer that can polymerise to form the polyester polymer polypropylene terephthalate (Sorona), which is used to make fibres for clothing (Figure 10.25). One of the monomers, propane-1,3-diol, is usually synthesised from petroleum. The DuPont Chemical Company have developed a new, greener, reaction pathway using genetically engineered Escherichia coli bacteria to convert glucose (obtained from corn starch) into propane-1,3-diol. This biological method of making propane-1,3-diol is cheaper and more environmentally friendly than the synthesis of propane-1,3-diol from petroleum. CO2 + H2O

+

sunlight (photosynthesis)

O

O

OCH2CH2CH2 O C

C

n

+

n H2 O

Glucose

E. coli fermentation

HO CH2 CH2 CH2OH

+

HOOC

COOH

propane-1,3-diol Terephthalic acid

Figure 10.25 Pathway to a greener polymer.

question

?

E1 A cond condensation densatition polymer used in medical sutures is made from glycolic acid monomer. The semi-structural formula of glycolic acid is HOCH2COOH. a What is meant by the term ‘condensation polymer’? b Name the functional groups present in glycolic acid.

summary

c d

Write an equation to represent the formation of a section of the polymer. Draw the structural formula for a section of the polymer and circle the ester bond.

10.4

C Carboxylic b li acids id reactt with ith alkanols lk l to t form f esters t andd water. t The reaction between carboxylic acids and alkanols is called a condensation reaction.

The Th name off an ester t indicates i di t the th alkanol lk l andd acid id from f which hi h it derived. The general formula of esters is RCOOR′.


key questions 12 Write equations that describe the formation of an ester that smells like: a pineapple (ethyl butanoate) b rum (ethyl methanoate).

11 Draw the structural formulas of the following esters and name them. a CH3COOCH3 b CH3COOCH2CH3 c CH3CH2COOCH2CH3

10.5

Reaction pathways Organic chemists are highly skilled at devising molecular structures that have exactly the right properties needed for a particular purpose. They then have to devise an efficient method for converting a readily available starting material—often an alkene or an alkane—into the product they want. The reaction pathway selected needs to take into account the yield and purity of the products and also minimise any unwanted side-products and waste materials. Time and cost factors also need to be considered. There is a lot of current interest in working out greener synthetic routes—ones that minimise waste, use more environmentally friendly solvents, require less energy and help to preserve the world’s resources. Figure 10.26 shows the main chemical reaction pathways that can be used to form compounds based on ethane and ethene. Pathways for other alkanes can be constructed by replacing a CH3 group in ethane with another alkyl group.

O–H

Cl

H Cl2(g)

CH3 H Ethane

H+(aq)

C

C

C H

O Cr2O72-(aq)

NaOH(aq)

CH3

H

CH3

H

H Chloroethane

C

H Ethanol

HCl(g)

Ethanoic acid ROH H2SO4

H2O(g) H3PO4(s)

H

O

H C

C

C R

H

Ethene

H

Figure 10.26 Reaction pathways based on ethane and ethene.


CH3

HO

O

CH3

Alkyl ethanoate (where R is an alkyl group, e.g.

CH3)

Example: synthesis of ethyl propanoate

O

Suppose you wanted to form ethyl propanoate using alkanes or alkenes as starting materials. A close look at the structure of the compound (Figure 10.27) indicates that it is an ester produced by the condensation reaction between propanoic acid and ethanol. Ethanol is a two-carbon compound that can be synthesised directly from ethene, or from ethene via the intermediate product chloroethane (Figure 10.28). In this case we will select the more direct route.

CH3

CH2

C CH2

O

CH3

Figure 10.27 Structure of ethyl propanoate.

Pathway 1 H

H

H

H

C

C

H

H

H

C

C

H

H

Ethene

H

OH−

HCl

H

H

Cl

C

C

H

H O–H

Chloroethane (an intermediate)

chemfact

Ethanol

Pathway 2 H

H C

H H2O

C

H

H

H

Adding a hydrogen halide such as HCl to an alkene larger than ethene results in the formation of the secondary haloalkane. When these compounds are treated with hydroxide solution they form the secondary alkanol. ‘Secondary’ means that the carbon atom bonded to the functional group is also bonded to 2 carbon atoms. Oxidation of secondary alkanols results in the formation of another type of compound, the alkanones (ketones). It is also possible to form tertiary alkanols in which the carbon bonded to the OH functional group is bonded to three other carbon atoms. The tertiary alkanol group is not oxidised. See Figure 10.15.

H C

C

H

H

O–H

Ethene

Ethanol

Figure 10.28 Synthetic pathways for ethanol.

Propanoic acid is a carboxylic acid containing three carbon atoms. It is prepared by the oxidation of the primary alkanol propan-1-ol. This in turn can be formed by the reaction of 1-chloropropane with sodium hydroxide. 1-chloropropane can be prepared by reacting propane with chlorine. A number of chlorine-substituted products will be formed. These are separated by fractional distillation. This reaction pathway for the synthesis of propanoic acid is summarised in Figure 10.29. H H

H C

C

H

H3C

H Cl2

H

UV light

H C

C

H3C

H Propane

H

OH− H

H

H

C

C

H3C

Cl

H H

Cr2O72− H+

O

H

Propan-1-ol

1-Chloropropane

H

O C

C

H3C

O

H

Propanoic acid

Figure 10.29 Synthesis of propanoic acid.

The substitution reaction of propane is chosen rather than an addition reaction of propene because the addition of HCl to propene will result in the formation of unwanted 2-chloropropane. Having synthesised ethanol and propanoic acid, we can now prepare the ester, ethyl propanoate, using a condensation reaction (Figure 10.30). O

O CH3

CH2

+

C O

H

CH2 H

O

CH3

H2SO4

CH3

CH2

+ H2O

C O

CH2

CH3

Figure 10.30 Synthesis of ethyl propanoate.


The reaction pathways for the preparation of ethanol, propanoic acid and ethyl propanoate can be summarised as shown in Figure 10.31. CH2

CH2

Ethene

H2O catalyst

CH3CH2OH Ethanol H2SO4

CH3CH2COOCH2CH3 Ethyl propanoate

CH3CH2CH3 Propane

Cl2

NaOH

UV light + fractional distillation

CH3CH2CH2Cl 1-Chloropropane

CH3CH2CH2OH

Cr2O72− H+

Propan-1-ol

CH3CH2COOH Propanoic acid

Figure 10.31 Reaction pathway for the preparation of ethyl propanoate.

The ester can be separated from the reaction mixture and purified by fractional distillation (p. 160). Its identity can then be verified using instrumental analysis such as IR, NMR or mass spectroscopy (Chapters 7 and 8).

Considerations in devising a synthesis

chemfact % yield = actual mass of product obtained × 100% theoretical mass of product

The above example illustrates some of the processes involved in planning and carrying out a synthesis. When planning a reaction pathway, the structure of the required compound or target molecule is studied and the functional group(s) are identified. A synthetic pathway is devised using knowledge of the reactions of functional groups. The synthesis may require the preparation of a number of intermediate compounds. More than one possible pathway may need to be considered as a desired product may be synthesised via number of pathways. For example, ethanol may be synthesised from ethene using several possible pathways (Figure 10.28). The formation of isomers and other by-products also needs to be considered; for example the substitution reaction of propane with chlorine produces a mix of the isomers 1-chloropropane and 2-chloropropane. Various compounds containing two or more chlorine atoms are also formed. In the above example only 1-chloropropane was required as an intermediate in the synthesis of the ester. The methods of separation of the desired intermediate and final product from isomers and other by-products must be determined. The final product then needs to be purified and the purity evaluated. The yield must also be taken into account, as not all of the reactants are necessarily converted to product. For example, in the reaction: CH3CH2Cl + OH– –—➤ CH3CH2OH + Cl–

you would expect 6.45 g of chloroethane to produce 4.60 g of ethanol if all the chloroethane was converted to ethanol. If only 2.30 g of ethanol was obtained then the yield would be 50%. Where there are a number of intermediate steps involved in a synthesis, the yield for each step must be taken into account. A low yield in one of the intermediate reactions can have a dramatic effect on the overall yield.


Worked example 10.5 In a particular synthesis, the yield of A → B is 80% and the yield of B → C is 70%. Calculate the overall percentage yield for the preparation of C from A.

Solution

80 of the theoretical mass of B is formed. 100 70 Then when B forms C, only 70% or of C is formed. 100 70 80 × × 100% Hence, the overall yield of C = 100 100 = 56%

When A reacts to form B, only 80% or

summary

10.5

O Organic i compounds d are synthesised th i d via i reaction ti pathways th th thatt take into account the reactions of functional groups. Organic reaction pathways can involve addition, substitution, oxidation and condensation reactions. Organic reaction pathways may involve a number of intermediate steps, and a number of separation and purification processes.

The Th yield i ld off a product d t is i given i bby: actual mass of product obtained % yield = × 100% theoretical mass of product Alkenes are the starting compounds for a number of organic syntheses. A number of factors, including yield, influence the choice of an appropriate reaction pathway.

key questions 13 Devise reaction pathways for the following reactions: a ethylamine from ethene b butan-1-ol from butane c pentanoic acid from 1-chloropentane 14 Name these esters and provide the systematic name of the alkanols and carboxylic acids used to form them. a CH3CH2CH2CH2COOCH3 b CH3CH2COOCH2CH2CH2CH2CH2CH3 15 Propose the shortest reaction pathway that will synthesise the following: a methanoic acid from methane b ethanol from ethene c propyl methanoate from propane and methane

16 Describe a reaction pathway to synthesise 1-butyl ethanoate using the appropriate alkanes or alkenes as starting materials. 17 Calculate the percentage yield if 5.0 g of ethanol is oxidised to produce 4.8 g of ethanoic acid. 18 Compound D can be synthesised using a reaction pathway that involves a number of intermediate steps. The yield for each step is shown. A ⎯→ B ⎯→ C ⎯→ D 70%

50%

90%

a Determine the overall yield for the preparation of compound D from compound A. b How would the overall yield be affected if the yield for B → C was only 10%?


10.6

Principles of fractional distillation review Simple distillation is outlined in Heinemann Chemistry 1, page 225.

Fractional distillation is a technique used to separate liquids that have different boiling points. It is commonly used in the laboratory to separate volatile liquids from a reaction mixture. Industrial applications of fractional distillation include: separation of the fractions from crude oil (p. 374) production of oxygen and nitrogen by the fractional distillation of liquid air (Heinemann Chemistry 1, p. 346) extraction of ethanol in the fermentation of sugar (p. 169). The apparatus used in fractional distillation in the laboratory is shown in Figure 10.32. The column is packed with glass beads or has glass shelves, providing a large surface area upon which the vapours condense. There is a temperature gradient up the fractionating column; the column is cooler at the top than at the bottom. Thermometer Water out Condenser

Glass beads Vent to fume cupboard Fractionating column

Water in Receiving flask

Distillate

Distillation flask Mixture of liquids Heating mantle

Figure 10.32 Fractional distillation apparatus used in the laboratory.

The principle on which fractional distillation is based is that if a mixture of volatile liquids is heated, the vapour contains a higher concentration of the lower boiling point components. In fractional distillation the components of a mixture of volatile liquids are separated by what can be considered to be a succession of simple distillations. When the mixture of liquids is heated in the distillation flask, the vapours rise up the fractionating column. These vapours contain a higher concentration of the more volatile component than the liquid in the distillation flask. Eventually the vapours reach a height in the fractionating column where the temperature is low enough for condensation to occur. As the condensed liquid trickles back down the column it is re-heated by vapours rising from


the distillation flask. As a result, some of the condensed liquid evaporates, and the resulting vapour has an even higher concentration of the low boiling point component. This process of evaporation and condensation is repeated many times throughout the length of the column, and the concentration of the more volatile substance in the vapour increases in each evaporation– condensation cycle. At the same time, the concentration of the less volatile (higher boiling point) substances in the distillation flask will increase. When the vapour reaches the top of the fractionating column it will ideally consist of only the more volatile component. When the relatively pure component reaches the top of the fractionating column the temperature remains relatively stable. The material that condenses over a small temperature range near the boiling point of the substance of interest is collected. It is not always possible to achieve a complete separation. Ethanol produced by the fermentation of sugar can be separated from water by fractional distillation but the distillate will always contain about 5% water.

Using fractional distillation Referring back to our example from section 10.5, the ester, ethyl ethanoate, is prepared by reacting ethanol with ethanoic acid in the presence of concentrated sulfuric acid, which acts as a catalyst. CH3CH2OH(l) + CH3COOH(l) –—➤ CH3COOCH2CH3(l) + H2O(l)

Pure ethyl ethanoate can be extracted from the reaction mixture by fractional distillation. The boiling points of the reactants and products are given in Table 10.2. The reaction mixture is heated in the distillation flask. The vapour rises up the fractionating column. The temperature at the top of the column slowly increases until it stabilises at about 57°C, which is the boiling point of ethyl ethanoate. The fraction condensing over a small range of temperatures, 55–59°C, near the boiling point of ethyl ethanoate is collected.

summary

TABLE 10.2 Boiling points of components in the reaction mixture Component CH3COOCH2CH3

Boiling point 57°C

CH3CH2OH

78°C

H2O

100°C

CH3COOH

118°C

10.6

FFractional ti l distillation di till ti is i usedd to t separate t the th components t in i a mixture containing liquids that have different boiling points. The vapours over a mixture of liquids will have a higher proportion of the more volatile (low boiling point) component than is in the liquid. Fractional distillation involves a series of evaporation– condensation cycles. The concentration of the more volatile component in the vapour increases after each cycle.

The Th fraction f ti condensing d i over a smallll temperature t t range near the boiling point of the compound of interest is collected. Industrial applications of fractional distillation include refining crude oil, extraction of oxygen and nitrogen from air, and the separation of ethanol produced by the fermentation of sugar.

key questions 19 Briefly describe the principles of fractional distillation. 20 Methane undergoes a series of substitution reactions with bromine to form a number of compounds with the following boiling points: bromomethane (3.6°C), dibromomethane

(97°C), tribromomethane (150°C) and tetrabromomethane (190°C). Describe how you would obtain pure samples of each compound from the mixture.


thecutting cutting edge the edge chapter review Free a highly10 Theradicals: page A header reactive species! key terms

goes here by Uta Wille

F

ree radicals are chemical species with one (or more) unpaired electrons. In 1929, Friedrich Adolf Paneth and Wilhelm Hofeditz prepared the free, highly reactive methyl radical, CH3• (the dot indicates the unpaired electron) for the first time, by heating tetramethyl lead [Pb(CH3)4]. During World War II (1939–1945), synthetic rubber was produced using chain polymerisation of alkenes induced by free radicals. Radicals can have the unpaired electron on atoms other than carbon, for example on oxygen, sulfur and nitrogen atoms. The unique properties of free radicals are causing great interest in discovering new types of reactions. One of my research interests is the unusual reactions of oxygen-centred radicals, for example thiyl–peroxyl radicals, RSOO•. These free radicals can be very easily produced in the laboratory, using photochemical (light) or electrochemical (potential) techniques,

Figure 10.33 Cumulative oxidative damage caused by free radicals may contribute to diseases associated with ageing, such as cataracts.

162 162 Organic Organic chemical chemical pathways pathways

where we first generate sulfur-centred thiyl radicals, RS•, which are then rapidly trapped by the oxygen in the air. We have discovered that RSOO• radicals are excellent reagents for the oxidation of alkynes to carbonyl compounds. Whereas ‘traditional’ procedures require the use of harsh conditions and often toxic heavy metal compounds, we can use thiyl radicals to ‘activate’ molecular oxygen, which is the cheapest oxidising agent we have. We are surrounded by free radicals; some occur naturally while others are atmospheric pollutants. All the chemistry that happens in the atmosphere—even where we are, in the lower troposphere— is caused by free radicals, and many of these are oxygen-centred. For example, the hydroxyl radical, HO•, controls the degradation during the day of natural and synthetic chemicals released into the atmosphere. At night the nitrate radical, NO3•, takes over this role from HO•.

bio o Free radicals are also responsible for the ageing of human cells. Our research is focused on understanding the impact of atmospheric free radicals on human health. For example, it is believed that ageing results from random oxidative damage to tissues by reactive oxygen and nitrogen species and other radicals. These radicals are formed by both metabolic processes and atmospheric pollutants, including radiation and ozone. The body is equipped with a very complex defence system to fight against free radicals, but when it is overrun, the consequence is extensive damage. This is called oxidative stress, and some diseases are believed to be a consequence of oxidative stress. Such diseases include Alzheimer’s and Parkinson’s disease, cancer, cataracts and other eye diseases, multiple sclerosis and others. When we are walking outside at night, NO3• may come into our airways as we breathe. Does NO3• lead to any damage to the molecules sitting on the outside of the cell membranes that are exposed to the

environment? Nobody knows this yet. Using simplified chemical models we are studying how NO3• reacts with biologically important molecules such as proteins, carbohydrates, lipids and DNA, and we have found that the reaction of NO3• with aromatic amino acids leads to products that are associated with respiratory tract diseases, such as asthma. We analyse the reaction products and investigate the rate of these reactions. This helps us gain an understanding of the processes that may lead to oxidation, cell damage and, ultimately, cell death. Recently, our work has expanded through the newly established ARC Centre of Excellence for Free Radical Chemistry and Biotechnology. The centre focuses on the important issues involved in defending Australia and its population against oxidative and other free-radical–induced damage. It also aims to broaden our existing theoretical understanding of radical chemistry and develop the free-radical technology of the future.

Dr Uta Wille Uta Wille received her PhD in Physical Chemistry at the University of Kiel, Germany. After spending some time at Basel University, Switzerland, she completed a ‘habilitation’ (which is an additional qualification required for an academic career in Germany) in Organic Chemistry at the University of Kiel. In 2003 she accepted an academic position at the University of Melbourne, where she is currently a Senior Lecturer in Chemistry and Molecular Science and Biotechnology.

Figure 10.34 The reaction of NO3• with aromatic amino acids leads to products that are associated with respiratory tract diseases, such as asthma.


10

key terms addition polymerisation addition reaction alkanol alkyl group biodegradable by-product carboxyl group

carboxylic acid chloroalkane combustion condensation reaction copolymer ester esterification reaction

Reactions of alkanes, alkenes and functional groups 21 Write unbalanced equations for the production of: a chloroethane from ethane b ethanol from ethene c ethanol from chloroethane d ethanoic acid from ethanol. 22 Classify the reactions in your answers to Question 21 as addition, substitution or oxidation. 23 Write the structural formulas of the products of the following reactions: a H H C

fractional distillation functional group hydrolysis hydroxyl group monomer polyester polymer

c butan-1-ol and acidified potassium dichromate solution d propan-1-ol and acidified potassium permanganate solution e ethane and oxygen. 25 Write chemical equations in the form of a flow diagram (similar to Figure 10.31) to describe the preparation of propanoic acid from an: a alkane b alkene. 26 Consider the following compounds: H I

H2O

C

C

H3PO4 CH3

b

reaction pathway substitution reaction synthesis yield

H

H

H

H

H

NaOH(aq)

H

H

II

Cl

C

C H

C H

H

H

Cl

H

H

H

c

III

H

H

H

H

C C

C H

H

H

IV

C

C

O

OH

C

+

H2SO4(l)

H

C C

H H

O

H

24 Write an unbalanced chemical equation for the reaction between: a butane and chlorine b 1-chlorobutane and sodium hydroxide solution


H H

H

d

CH3OH

C

H

O–H

H

H

C H

+ H (aq)

H

C

Cr2O72–(aq)

H

H

O–H

V

H C

H

Fractional distillation

H

31 In an experiment the reaction of chlorine with ethane resulted in a mixture containing a number of products with the following boiling points: chloroethane (12°C), 1,1-dichloroethane (57°C), 1,2-dichloroethane (84°C), and 1,1,2-trichlorethane (97°C). Describe how you could obtain a pure sample of 1,2-dichlorethane from this mixture.

C

H C H

H

a Write the names of each of these compounds. b Which of these compounds is a member of the alkane homologous series? c Give one important difference in chemical behaviour between compounds I and V. d What important chemical is manufactured from compound II? e Which of these compounds dissolves in water to form an acidic solution?

Connecting the main ideas 32 Complete the missing structural formulas and name the compounds A, B and C. H

H C

HCl A

C

H

H

H H

Esters 27 Complete the following reactions: O a

H

H Reactant B

C

C

O

H

H

Cr2O72– C H+(aq)

CH3OH

+

C OH

CH3CH2 O

b

33 Give the semi-structural formulas for substances P and Q and the systematic names for substances R and S in the following equation: H+(aq)

+

C CH3

c

CH3(CH2)3CH2OH

OH

R

O

+

C H

CH3CH2COOH + P

(CH3)2CHOH

OH

28 a Write a chemical equation to describe the preparation of propyl ethanoate from an alkanol and a carboxylic acid. b What name is given to this type of reaction? c Sulfuric acid would be present in the reaction mixture. Why? 29 Write equations to show the production of: a methyl ethanoate b ethyl propanoate. 30 Write the formula of the alkanol and carboxylic acid used to make the following esters: a CH3COOCH2CH3 b CH3CH2COOCH2CH3

warm

CH3CH2COOCH3 + Q S

34 Write a reaction pathway for the synthesis of: a ethyl propanoate from ethene and propene b ethyl ethanoate from an alkene. 35 Fractional distillation is an important laboratory technique. a What is fractional distillation used for? b Fractional distillation has been described as a series of simple distillations. Explain.


1

chapter 11

Biochemical fuels

key knowledge • biochemical fuels including fermentation of sugars to produce ethanol

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • discuss the environmental impact of burning hydrocarbon fuels • explain the advantages of using biochemical fuels • describe how ethanol is produced • write equations for the production of ethanol • describe, with the aid of appropriate equations, the production of biodiesel.

11.1

Fossil fuels Today most of our energy needs are met by burning fossil fuels, such as coal, petroleum and natural gas. Coal is used to generate electricity. Products derived from crude oil, such as petrol and diesel, are used as transport fuels. Other products derived from crude oil are used to manufacture a diverse range of products, from plastics to pharmaceuticals. While there are large reserves of coal, petroleum deposits are limited.

Figure 11.1 Our current way of life involves the burning of large amounts of fossil fuels.

Peak oil ‘Peak oil’ is the time when worldwide oil extraction can no longer keep up with increasing demand. The greatest amount of oil discovered in any one year was in 1964, and since then the new reserves found have gone down in size each year. Meanwhile demand for oil has increased as the world population has increased and lifestyle expectations have risen. The actual timing of ‘peak oil’ is debatable. We may be there now! The situation provides motivation to devise renewable and sustainable sources of the carbon compounds, both for fuels and to provide feedstock for the organic chemical industry.

11.2

Biochemical fuels We are making enormous demands on the planet in terms of providing sufficient fuel resources. We are increasingly aware of the environmental effects of pollution of the atmosphere and global warming. The latter is caused in part by increased amounts of carbon dioxide produced by the burning of carbon-based fossil fuels.

167 Biochemical fuels

Projection 6.0 5.5 5.0 4.5

Temperature change in ˚C

4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 13.7

Figure 11.2 Average surface temperature of the Earth from the year 1000 projected to 2100.

–0.5 –1.0 1000

! Cellulose is the structural material for the cell walls of plants, while starch is found in large amounts in grain where it acts as a form of energy storage for the germinating seed. It is also found in some roots, such as potato.

! Polysaccharides are large molecules consisting of many simple sugar units such as glucose joined together. You will learn more about polysaccharides in Chapter 12.

chemfact Grains as a raw material for fermentation are particularly important in the US, which is a major producer of maize (corn). The grain used to make biochemical fuel can be of second-class quality and therefore unsuitable for food products.

1100

1200

1300

1400

1500

1600 Year

1700

1800

1900

2000

Governments around the world are now acknowledging that the effect of global warming, with its associated rise in sea levels and changing weather patterns is becoming noticeable. The limited nature of crude oil reserves is also a concern.Consequently there is increased motivation to ensure that the new sources of energy are clean (produce much less pollution) and green (come from sustainable and renewable sources). Biochemical fuels (biofuels) are fuels derived from plant materials such as grains (maize, wheat, barley or sorghum), sugarcane or vegetable waste, and vegetable oils. The two main biofuels are ethanol and biodiesel. They are used either alone or blended with fossil fuels such as petrol and diesel. Biochemical fuels are not considered to contribute to an increase in atmospheric carbon dioxide. The plant materials used in the generation of biochemical fuels are produced by photosynthesis, which removes carbon dioxide from the atmosphere and produces glucose: photosynthesis

6CO2(g) + 6H2O(l) ––———➤ C6H12O6(aq) + 6O2(g) sunlight glucose

The plants convert the glucose into cellulose and starch (polysaccharides). The carbon dioxide is then released when the biochemical fuel is burned. Biochemical fuels are therefore considered to be carbon neutral.

Ethanol For millennia, humans have employed the enzymes in yeasts to convert starches and sugars to ethanol. Enzymes are used to catalyse the breakdown of the polysaccharide starch in grains to glucose, and then the fermentation process uses other enzymes from yeast organisms to convert small sugar molecules such as glucose to ethanol and carbon dioxide: enzymes

C6H12O6(aq) –––—➤ 2CH3CH2OH(aq) + 2CO2(g)

This is an anaerobic process—it takes place in the absence of oxygen.


2100

grain

2

feed hopper

4

1 3

ethanol for use as a fuel

7

5

6 add 5% petrol CO2(g)

carbonated drinks solid waste processed for use as stock feed

1 Grain feedstock is crushed and milled 2 Water is added and mixed to form a slurry 3 Heat to between 85°C and 105°C with added enzymes to hydrolyse the starch content — converts starch molecules to glucose in 2 to 4 hours 4 Cool with heat exchangers to around 32°C 5 Fermentation — enzymes complete the breakdown of starch and glucose to ethanol and carbon dioxide 6 Steam distillation of product from the fermentation process increases the ethanol content to 95% 7 Dehydration (removal of water content) produces 99.7% pure ethanol

Figure 11.3 The production of ethanol from grain.

The fermentation stops when the ethanol content is between 10 and 20%, at which point the yeasts and their enzymes can no longer function. The fermented mixture is pumped to an evaporation plant where steam is added to cause the (lower boiling point) ethanol to evaporate off. When cooled, the resulting liquid contains about 95% ethanol and 5% water. This is dehydrated—the water is removed—leaving ethanol that is 99.7% pure. In Australia, the final ethanol is then ‘denatured’ (poisoned) by adding up to 5% petrol to make it unsuitable for consumption as drinking alcohol. The process falls under the control of Australian excise laws. Care is taken to avoid waste and pollution. The carbon dioxide produced in the fermentation step can be sold to manufacturers of carbonated drinks. The waste water and cooling water can be used for irrigating crops, and the protein-rich remains from the fermentation can be sold as animal feed. When molasses, a residue from sugar refining, is used as the feedstock, as it is in Queensland, some of the by-products can be recycled as a fertiliser for sugarcane and banana plantations. The yeasts that are used to ferment material such as molasses and grains contain enzymes that act on glucose, C6H12O6 (Figure 11.4). However, the fibrous waste material from sugar production, called bagasse, is made of a different type of sugar molecule, called xylose. This is a pentose, a sugar that contains five carbon atoms per molecule, and is not recognised by the usual yeasts. An Australian development has been to selectively breed yeasts whose enzymes are able to digest the bagasse by-product. The development of these yeasts will greatly increase the amount of ethanol produced from a given quantity of sugarcane.

a

CH2OH O H

HO

b

H

H

OH

H

H

OH

OH

CH2OH O H H

HO

H OH

OH

H

Figure 11.4 Structure of a glucose, and b xylose.

! Enzymes are special proteins that control the rate of reactions occurring in living systems.


Figure 11.5 E10 petrol is sold at many service stations.

Biodiesel

chemfact Modern car engines can function normally on E10 petrol/ethanol mixtures. A higher proportion of ethanol can cause corrosion problems in the fuel tank and fuel lines due to ethanol’s ability to absorb water from the atmosphere.

O CH2

O

C

(CH2)14CH3

O CH

O

C

(CH2)14CH3

O CH2

O

The cellulose found in biomass such as woody plants can also serve as a raw material for ethanol production. Genetically engineered E. coli bacteria have been developed that convert the polysaccharides in materials like forestry waste, corn stalks and sugarcane residue into ethanol. Aside from producing a valuable chemical, the process reduces the need to dispose of these waste materials in landfill or by burning. In Canada, a different process has been developed using a special patented enzyme to convert the plant cellulose into glucose. This is then fermented, in the same sort of way as shown in Figure 11.3, to make ethanol. It can make more than 300 L of ethanol from every tonne of plant fibre. The lignin (the non-cellulose parts of the plant fibre) is burned to make electricity to drive the process, and the emissions are recycled. The roll-out of E10 petrol (containing 10% ethanol) has begun across Australia. The speed of this is partly limited by the availability of supplies of ethanol. This mix can be used by most modern car engines and will extend the availability of petrol while helping to reduce production of additional carbon dioxide. Research also suggests that the use of E10 petrol reduces the quantity of particles such as unburnt carbon and gases such as oxides of nitrogen that are emitted by exhausts and that contribute to air pollution.

C

Biodiesel is a mix of esters produced by a chemical reaction between vegetable oil and an alcohol such as methanol. The chemical and physical properties of the esters in biodiesel are very similar to those of diesel (petrodiesel). In Europe, biodiesel containing 5% esters has been available since 1995. It is possible to run diesel vehicles on 100% ester fuel. The usual raw material is fresh vegetable oil from sources such as soybeans, canola or palm oil, but recycled vegetable oil or animal fats can also be used. Fats and oils are triglycerides with a molecular structure consisting of three hydrocarbon chains attached by ester functional groups to a backbone of three carbon atoms, as shown in Figure 11.6. The triglyceride is hydrolysed by warming it with methanol and potassium hydroxide solution. The potassium hydroxide solution acts as a catalyst in this hydrolysis reaction so that overall the triglyceride breaks down to three molecules of fatty acid esters, plus glycerol (Figure 11.7).

(CH2)14CH3 H

Figure 11.6 A triglyceride found in vegetable oil. Can you identify the ester functional groups?

H

C

O O

C

H (CH2)14CH3

O H

C

O

C

+ 3CH3OH

! The esters in biodiesel are known as FAMEs— fatty acid methyl esters.

C

O

C

Organic chemical pathways O

OH

H

C

OH

H

C

OH

+ 3CH3OOC(CH2)14CH3

H

(CH2)14CH3

H Triglyceride (palm oil)

Methanol

Figure 11.7 The hydrolysis of triglycerides to fatty acids and glycerol.

170

C

(CH2)14CH3

C H

H

Glycerol

Fatty acid methyl esters (biodiesel)

Ethanol and sodium hydroxide can be used as alternative reagents but there are some practical issues involved that make them less useful for large-scale production. The esters produced are less viscous (less thick) than vegetable oil. When burnt they release a similar amount of energy to petrodiesel. Biodiesel can also be used in a mixture with petrodiesel; 5% and 10% mixtures of biodiesel in petrodiesel are known as B5 and B10, respectively. These mixtures can be used in normal diesel engines although a slow transition to the mixture is recommended for older engines. This is because the biodiesel acts as a better solvent for any contamination that has built up in the engine and hence can cause blockages in filters. 100% biodiesel, known as B100, can also be used. Biodiesel is biodegradable, non-toxic and produces fewer pollutants in the vehicle emissions. Like ethanol as a biofuel, it is not adding to the overall amount of carbon dioxide in the atmosphere as it is simply recycling the carbon dioxide already present rather than releasing carbon stored in fossil fuels. Another form of biodiesel, called SunDiesel, is being trialled in Germany. It uses a patented process called Carbo-V that starts with biomass (waste plant material such as straw left after crop production) to make ‘syngas’, a mixture of carbon monoxide and hydrogen. Syngas can be used as a combustible gas to generate heat, steam and electricity, or it can be used as feedstock to make SunDiesel. This involves passing the syngas mixture over an inorganic catalyst to produce large hydrocarbon molecules of the same general size as the heavier fractions from normal oil fractionation. These large molecules are then cracked to give hydrocarbons of the required size. This process gives rise to the same sort of molecules as found in petrodiesel, but the additional processing required means that SunDiesel is significantly more costly than conventional diesel. There have been some feasibility studies on the production of oil for biodiesel from algae. It appears that this would give a higher yield than plant sources and could be developed on land where it is not possible to grow crops, or in coastal seawater.

chemfact The ethanol molecule has a non-polar hydrocarbon tail but the hydroxyl functional group is polar. This allows it to mix with non-polar substances such as petrol and polar solvents such as water.

chemfact Backyard biodiesel! Some motoring enthusiasts convert the used oil from fish and chip shops into biodiesel for use in their cars. Would the exhaust fumes smell of fish and chips?

chemfact In the drive to produce increasing amounts of sustainable fuel supplies, the use of animal fat (tallow) from abattoirs is being investigated. This fat can be converted to biodiesel. It is estimated that the fat from one sheep will provide enough biodiesel to drive the family car a distance of 105 km. The fat from a pig will provide enough fuel for a 140 km trip, a cow 450 km, while a chicken will only drive your car for 600 m. Weekend Australian, 12 August 2006

Biogas Biogas consists mainly of carbon dioxide and methane in roughly equal quantities and is generated when organic material decays in the absence of oxygen. It can be used for heating and to power homes and farms. Biogas generators are particularly suited to farms—there are over seven million generators in China. The waste from a biogas generator can be used as a rich fertiliser. Manure and farm waste

Valve

Valve

Valve Gas outlet

Rotting manure and waste vegetable matter

Methane

Methane storage tank

Valve Solid waste

Figure 11.8 A biogas digestor.


Figure 11.9 Pipes buried in this rubbish tip collect biogas which is burnt to supply thermal energy for the generation of electricity.

In the future, it is likely we will obtain more energy from biogas generated at sewerage works, chicken farms, piggeries and food-processing plants. Your local rubbish tip also has the potential to supply biogas. The gas can be used directly for small-scale heating or used to generate electricity.

Why develop biochemical fuels? chemfact Classing biofuels as carbon-neutral ignores the non-renewable resources consumed in growing the raw materials and their subsequent processing. The environmental impact can vary enormously with the nature and origin of the raw material and the method used to convert it to fuel.


Our current use of fossil fuels is releasing carbon into the environment, mainly in the form of atmospheric carbon dioxide. This carbon has been stored in the Earth for millions of years in the form of coal, oil and natural gas. Gases such as water vapour and carbon dioxide absorb heat energy from the Sun and act as a sort of ‘blanket’ to trap this heat in the atmosphere. Life on Earth could not survive without this greenhouse effect, but the enormous additional amount of carbon dioxide produced by burning fossil fuels is causing further heating effects, which is believed to be triggering consequential shifts in weather patterns and climate, causing droughts in some parts of the world and floods in other regions. There is growing evidence that the effects are extending throughout nature, with some plant and animal species already being driven into extinction. Biochemical fuels are seen as a useful alternative to the continued exclusive use of fossil fuels. Although their combustion does produce carbon dioxide, they are produced from plant materials made by photosynthesis, which consumes an equal amount of carbon dioxide. Some energy is used in the processing needed to ferment and distill the industrial ethanol, but the process is much closer to being ‘carbon neutral’ than the use of conventional petroleum products. While the use of land to produce sufficient food for the world is another problem to be faced, the production of bio ethanol at present generally uses waste material left over after food production. Furthermore, the production of ethanol uses resources that are available in Australia. This means that production in Australia is not likely to be influenced by problems in importing raw materials whose cost can vary dramatically due to political circumstances. It also provides employment at many different levels.

The future plans for our fuel and energy supplies in Australia and the world require a complex balancing of the extent of current resources, their environmental advantages and disadvantages, whether sources are sustainable and renewable, and cost.

chemistryy in action Rudolf Diesel: the earliest greenie? Rudolf Diesel (1858–1913) was a German engineer who invented the ‘oil engine’ that was named after him. His prototype engine first ran in 1893 and was powered by peanut oil. Diesel was well aware of the potential value of an engine that could run on renewable fuel. He demonstrated his engine at the World Exhibition in Paris in 1900 and was awarded the Grand Prix—the highest prize. In the years just before his death in 1913 he stated: ‘The diesel engine can be fed with vegetable oils and would help considerably in the development of agriculture of the countries which use it.’ He also predicted that: ‘The use of vegetable oils for engine fuels may seem insignificant today. But such oils may become, in course of time, as important as petroleum and the coal tar products of the present time.’

Figure 11.10 Rudolf Diesel.

summary

11.1–11.2

Li Limited it d crude d oilil resources andd environmental i t l concerns have h promoted research into biochemical fuels. Various plant materials are fermented to produce ethanol.

Biodiesel Bi di l is i produced d d from f vegetable t bl oils. il The use of biochemical fuels can be considered as carbon dioxide neutral.

key questions 1 Write a balanced equation for the combustion of hydrogen in air. 2 a Why can ethanol for use as a biofuel be produced from a variety of different plant sources? b Write a balanced equation, including subscripts, for the combustion of ethanol in a good air supply. c Explain why the use of ethanol as fuel can be considered to be carbon neutral.

3 It is possible to make biodiesel from animal fat as well as from vegetable oils. a Assuming that the animal fat consists only of tristearin, which can be hydrolysed to make stearic acid, C17H35COOH, write a balanced equation to represent how this substance can react with methanol to form an ester. b Write a balanced equation to represent the combustion of this biodiesel in a plentiful air supply.


the cutting edge Two approaches to using biofuels for public transport eoff Peel, Fleet Manager of InvictaBus, and Andrew Cornwall, Managing Director of Ventura Bus Lines, are on a mission to make a positive contribution to improving the environment. Both are well aware that transport cannot continue to rely on diesel fuel from petrochemical sources, and both want their companies to be leaders in environmental change. It is interesting that they have tried different biochemical fuels. Here are their stories.

G

by Geoff Peel

Geoff Peel

The company has always been environmentally concerned. As a member of ‘Greenhouse Challenge’, a governmentsponsored organisation, I was thinking about what we could do for the environment as part of the Greenhouse Challenge when I was contacted by a fuel company representative to see if I was interested in trialling biodiesel. The fuel we tried was 100% biodiesel, made from canola oil. We ran the fuel through the older buses in our fleet of vehicles so we that we could gauge the effects on different types of bus engines in a variety of situations.

Figure Fig Figure 11.11 In using I the th trial, t i l buses b i biodiesel bi di l ran cleaner than those using petrodiesel.

174 Organic chemical pathways

Generally the buses performed well. There was a slight drop in performance but it was only really noticeable in the vehicles with smaller engines, and there was about a 5% rise in fuel consumption, so it did cost a bit more. However, the engines ran ‘cleaner’, with no black smoke coming out of the exhaust as the vehicles pulled away from bus stops, and the buses had a different smell. Research also showed a drop in emissions of carbon monoxide, carbon dioxide and oxides of nitrogen. The drivers explained to our passengers what was going on and the complimentary phone calls soon started. In turn this led to contacts from the media, radio interviews across Australia and even contacts from a US Senator. I am in touch with environmental scientists at CSIRO and RMIT and have spoken at overseas conferences about our work. We had to stop this first trial because of fuel supply problems caused by the drought. However, we are not giving up. Once we have some logistics sorted out we will look at using B20 (a 20% biodiesel, 80% petrodiesel mix). This will reduce problems with vehicle warranties, and our newest buses will be far in advance of Australian emission standards. In the immediate future, I believe that biofuels will be used to extend the availability of carbon-based fuels while helping to protect the environment. In the longer term, we will probably be using u hydrogen.

bio o by Andrew Cornwall think it is very important for transport operators to demonstrate to the public that we are not major polluters. If we can entice people out of their cars, it will be a lot better for the environment. The three Ventura buses that run on 100% ethanol have purpose-built engines that run on hydrated ethanol. We use them on routes that cover university areas to help capture an aware market. When you are evaluating an alternative fuel, it is important to look at the emissions produced by manufacturing the fuel as well as measuring any vehicle emissions from the exhaust pipe. Currently our ethanol is from CSR Distilleries in Queensland, using molasses, a waste product left after processing sugarcane. This means that the life cycle of the carbon dioxide emissions is very low. I made a point of finding a source of hydrated ethanol as this means that energy is not used by the manufacturer to dehydrate the fuel. You couldn’t use it in E10 mixes for conventional cars because of corrosion problems, but in purpose-built engines it is fine.

I

I am in touch with research done by CSIRO and we are helped by the Australian Greenhouse Office and the Department of Sustainability, so everyone is working together to try to produce the best outcomes. The ethanol-powered buses use more fuel, but there is a government energy credit that currently makes them overall cheaper to run than our diesel-powered vehicles. By the time the energy credit is phased out, diesel may be a lot more expensive anyway. We must always make sure that we comply with Australian emissions standards. These follow the European standards so we know that in the future we may have to have at least 20% of vehicles operating on a renewable source of fuel. We have recently purchased an electric drive bus so that we can gather first-hand data on its performance and emissions, including those produced at the fuel source, and costs.

Andrew Cornwall Andrew is the managing director of the Ventura and National Bus Company, Victoria’s largest bus operator. He first saw ethanol-powered buses when in Stockholm, Sweden, and decided to adapt similar ethanol buses for the Australian climate. Andrew’s ongoing commitment to environmental sustainability has led to the purchase of three successful ethanol buses, with Ventura Bus lines winning the ‘Greenhouse Challenge Aware 2001’ Award.

Figure 11.12 The ethanol buses use a purpose-built engine that runs on hydrated ethanol, which is not suitable for blending with petrol.


11

key terms B100 biochemical fuel biodiesel biofuel

cellulose diesel E10 ethanol

fatty acid feedstock fermentation hydrolysis

polysaccharides renewable sustainable syngas


Fossil fuels 4 The world has become very dependent on the products of the petrochemical industry but the raw materials of crude oil and natural gas are likely to be virtually exhausted by 2100. Assuming that current production remains unchanged and no alternative sources are available, suggest the impact of the lack of raw materials on our lifestyle. 5 The fact that global warming is taking place is now generally accepted. Use the Internet to research some of the consequences. Give one example each of the effect on: a the polar ice caps b changing weather patterns c crop production d extinction of plant or animal species.

Biochemical fuels 6 a Explain what is meant by E10 petrol. b How will the introduction of E10 help with the potential shortage of crude oil? 7 What advantages does biodiesel have over petrodiesel? Are there any disadvantages to its use? 8 Syngas contains a mixture of carbon monoxide and hydrogen. Write a balanced equation for the combustion in a good air supply of: a carbon monoxide b hydrogen 9 Assuming that SunDiesel consists of molecules such as C18H38, write a balanced equation for the combustion of this substance in a good air supply. 10 What might be the advantages and disadvantages of using SunDiesel in place of petrodiesel? 11 The reactions that produce ethanol can be summarised as: A

B

CO2 + H2O → C6H12O6 → CH3CH2OH What are the names of the reactions represented by the letters A and B?


12 Currently hydrogen is produced by steam reforming of alkanes such as methane: CH4(g) + H2O(g) → CO(g) + 3H2(g) a Why is hydrogen produced by this method not considered a renewable source of energy? b What more sustainable methods could be used to produce hydrogen? c List some of the advantages and disadvantages of using hydrogen as a fuel now. 13 Ethanol is produced industrially by reacting ethene with water using a phosporic acid catalyst at 300°C: C2H4(g) + H2O(g) → CH3CH2OH(g) a Is ethanol produced by this method a biochemical fuel? Explain. b Describe how ethanol classified as a biochemical fuel can be produced. 14 Draw a concept map that includes the following terms: biochemical fuel, biodiesel, ethanol, fermentation, sustainable 15 a Write suitable balanced equations to represent the: i use of atmospheric carbon dioxide in photosynthesis ii fermentation of glucose to produce ethanol iii combustion of ethanol in a good air supply. b Explain how these reactions support the statement that the use of ethanol as a fuel is carbon neutral.

1

chapter 12

Pathways to biomolecules

key knowledge • organic reaction pathways, condensation and polymerisation reactions that produce large biomolecules • primary, secondary and tertiary structure of proteins and the function of protein catalysts (enzymes) • use of proteins as markers for disease

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the molecular structure of common carbohydrates, lipids and proteins • use functional groups to describe the condensation reactions of monosaccharides to form disaccharides and polysaccharides, amino acids to form proteins, and glycerol and fatty acids to form saturated and unsaturated triglycerides • describe the primary, secondary and tertiary structures of a protein and their impact on a protein’s molecular shape and characteristics • explain the action of enzymes as catalysts in biochemical reactions • describe the denaturation of proteins by heat and changing acidity • describe the enzyme-catalysed hydrolysis of proteins, fats and carbohydrates • describe how protein markers are used in the diagnosis of disease.

Biomolecules are molecules such as fats and oils, carbohydrates, proteins and nucleic acids that are found in all living things. They have an essential role in the supply of energy to the body, the growth and repair of organs and tissue, the movement of muscles, the activity of nervous and hormonal systems, and the elimination of waste. Many biomolecules are polymers. Cotton, wool and silk are naturally occurring polymers that have been in use for thousands of years. They are characterised by the size of the molecules, each one often containing tens of thousands of atoms, that make up the material. In this chapter you will study some of the chemistry of biomolecules that constitute lipids, carbohydrates and proteins. The chemistry of nucleic acids, another class of biomolecules, is covered in Chapter 13.

12.1

Fats Sources of fat Humans can make most of their own fat internally. However, this is usually not necessary in our society as more than enough fat is available in most diets. Typically, fat makes up about a fifth of the solid matter in the food we eat. It is present in meat, fish, dairy products, vegetable oils and all fried foods.

What are fats? Figure 12.1 We tend to enjoy eating fatty foods. They remain in the stomach longer than other foods, giving us the sensation of being satisfied.

‘Fat’ is a name used to describe a large number of organic compounds belonging to an even larger class of biological molecules called lipids. Fats and oils are the best known types of lipids. The oils found in foods have quite different structures and properties from the hydrocarbon oils produced by petroleum refining. Compounds such as waxes and steroids (which include cholesterol) are also members of the lipid family. Lipids are based mainly on carbon and hydrogen. They also contain small amounts of oxygen and, occasionally, other elements. Most lipids are essentially non-polar and so they are insoluble in water. However, since nonpolar substances dissolve in non-polar liquids, fats and oils are one of our sources of the non-polar vitamins A, D, E and K. Fats and oils have very similar chemical structures. They are simply distinguished on the basis of their physical states at normal temperatures: Fats are solids. Oils are liquids. In Chapter 10 you read about condensation reactions used to make esters. As Figure 12.2 shows, producing an ester requires a molecule containing Ester functional group H

H

O

Figure 12.2 Ester links form when the carboxyl group reacts with a hydroxyl group. A water molecule is also produced.


C

+

C

HO

H

C

OH

CH3

H H

+

Catalyst

H H

O

H H

C

C

O

C

H H

+ H2O

CH3

a carboxyl group, COOH, to be reacted with one with a hydroxyl group, OH. The ester group, –COO–, provides a covalent link between the two molecules. A water molecule, H2O, is also produced. Most fats and oils are formed by a condensation reaction between a single molecule of glycerol and three molecules of fatty acids, as shown in Figure 12.3. Fatty acids contain a carboxyl functional group which reacts with the hydroxyl groups in the glycerol. The –COO– groups in fat are ester functional groups or ester linkages. Three molecules of water are also produced in this reaction. Fats and oils formed from this process are called triglycerides. Most fats have two or three different fatty acids attached to the glycerol molecule. Triglycerides are large, non-polar molecules and therefore insoluble in aqueous solutions. O H

H HO

H

C

OH

H

C

OH

C

+

HO

C

O

C17H35

O C17H35

H

C

OH HO

H 1 Glycerol molecule

C

Three ester linkages

H

C

O

C O

C17H35

H

C

O

C O

C17H35

H

C

O

C

C17H35

O C17H35

3 Stearic acid molecules

! Glycerol is an example of an alkanol; fatty acids are carboxylic acids. Fatty acids are so-called because they consist of a long hydrocarbon chain connected to a carboxyl group, which has acid properties.

+

3H2O

H Tristearin (a fat)

3 Water molecules

Figure 12.3 Glycerol reacts with three molecules of fatty acids to form a fat.

Types of fats Fats are distinguished on the basis of the fatty acids from which they are made (Table 12.1). Saturated fats are made from fatty acids, such as palmitic acid, that contain only single carbon–carbon bonds. Saturated fats are generally unreactive and occur as waxy solids at room temperature. Mono-unsaturated fats are made from fatty acids, such as oleic acid, that contain one carbon–carbon double bond. Polyunsaturated fats are made from fatty acids, such as linoleic acid, that contain more than one carbon–carbon double bond. Polyunsaturated fats have lower melting points than saturated fats, and often occur as liquids (oils) at room temperature. They are more reactive than saturated fats.

Table 12.1 Examples of different types of fatty acids Type

Semi-structural formula

Molecular formula

Name

Saturated

CH3(CH2)14COOH

C16H32O2

Palmitic acid

Mono-unsaturated

CH3(CH2)7CH=CH(CH2)7COOH

C18H34O2

Oleic acid

Polyunsaturated

CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH

C18H32O2

Linoleic acid

chemfact Antioxidants and stabilisers are added to margarine because the carbon–carbon double bonds in polyunsaturated fats are reactive. The antioxidants are oxidised in preference to the double bond in the unsaturated fats. These additives prevent the polyunsaturated fats from reacting with oxygen in the air and producing substances with unpleasant odours whose presence leads to a condition known as rancidity.

! Notice that, on the basis of molecular formulas, in saturated fats the H : C ratio is 2 : 1 whereas for unsaturated fats it is less than 2 : 1.

The different physical states of saturated and polyunsaturated fats are thought to arise because molecules of saturated fats can pack more closely together, resulting in stronger dispersion forces between the molecules. The arrangement of the carbon chain sections at the double bonds sites of polyunsaturated fats do not permit such close packing. Consequently saturated fats have higher melting points than unsaturated fats.

179 Pathways to biomolecules

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chemfact Human body fat typically contains about 32% saturated fats, 48% mono-unsaturated fats and 10% polyunsaturated fats.

! An emulsion is a suspension of tiny droplets of one liquid suspended in another liquid.

! In a hydrolysis reaction, water reacts with another substance.

Both saturated and polyunsaturated fats are present in foods that contain fat. In general, animal fats contain higher proportions of saturated fats, while vegetable oils are richer in polyunsaturated fats (Table 12.2).

TABLE 12.2 Composition of fats and oils Fat or oil

Saturated fat (%) 52 66 92 51 18 15 14 13 11 9

Beef Butter Coconut Palm Peanut Soybean Olive Corn Sunflower Safflower

Polyunsaturated fat (%) 4 4 2 10 34 61 9 62 69 78

Mono-unsaturated fat (%) 44 30 6 39 48 24 77 25 20 13

You will have seen the word ‘polyunsaturated’ used to describe the fats in margarine (Figure 12.4). Unlike butter, which is made from saturated animal fats, margarine is usually made from vegetable oils that contain more polyunsaturated fats. Vegetable oils from plants such as sunflower, palm, soybean and coconut are blended together to make margarine. In order to convert the unsaturated oils to solid form, they are partially saturated by reacting them with hydrogen, in the presence of a nickel catalyst.

What happens to fat in the body? Chemical digestion of fat does not start until it reaches the small intestine. There, it is mixed with bile from the gall bladder. Bile contains surfactants that convert fats into an emulsion of smaller fat globules. This increases the surface area of the fat, allowing it to react more rapidly. Secretions from the pancreas and the walls of the small intestine contain enzymes that catalyse the hydrolysis of fat into fatty acids and glycerol, reversing the condensation reaction from which the fat was made. Unlike the enzymes in the stomach, which require acidic conditions to be effective, enzymes present in the small intestine operate best under alkaline conditions. As food from the stomach enters the intestine, it is neutralised by alkaline secretions. Once the fatty acids and glycerol have been absorbed into the body, they are reassembled into triglycerides via condensation reactions (Figure 12.5).

Digestion (using enzymes and water)

Fats

Glycerol and fatty acids

After absorption into body (using enzymes)

Fats

Figure 12.5 When fats are digested they are first hydrolysed and then re-formed by condensation reactions for transport to the adipose tissue.


extension Fats and good health Coronary heart disease is one of the major causes of death in Australia. It is a disease of the arteries that supply blood to the heart that begins with the build-up of fatty deposits on the inner walls of the arteries. This leads to a condition called atherosclerosis, or ‘hardening of the arteries’. As the deposits increase, arteries become narrowed and the flow of blood is restricted (Figure 12.6). Eventually, if hardening of the arteries continues, an artery supplying blood to some part of the heart may become completely blocked. Without oxygen, that part of the heart muscle cannot function and a heart attack results. One indication of possible danger is the amount of cholesterol in the blood. For all Australians, the National Heart Foundation has the following advice. To avoid coronary heart disease: • exercise regularly • maintain a balanced diet • do not smoke.

a

b

Figure 12.6 a A healthy artery. b An artery showing the effect of atherosclerosis.

TABLE 12.3 Maximum recommended levels of cholesterol and triglycerides in the blood Age (years) Above 18 Below 18

Blood cholesterol (mol L–1) 0.0050 0.0045

Blood triglycerides (mol L–1) 0.0020 0.0020

HO

HO

Cholesterol

Vitamin D

OH

Cholesterol is a lipid that occurs naturally in all animals. It is a white, fatty substance that is used by the body in the manufacture of hormones, bile and vitamin D (Figure 12.7). Its role in the body is therefore important. As it is made only by animals, cholesterol is not found in fruit, vegetables or other plant products. The small amount of cholesterol available from the animal products in a balanced diet is not enough to meet our daily requirements. The body synthesises most of its own cholesterol (3 to 5 grams per day) in the liver. High blood cholesterol is caused by the body making more than it needs. Why the body does this is as yet unknown and is the subject of much current medical and scientific research.

question

HO

Oestrogen (female sex hormone)

OH

O

Testosterone (male sex hormone)

Figure 12.7 Cholesterol and three substances made from it. The substances have very similar molecular structures. (Carbon and hydrogen atoms in the molecules have been omitted for clarity. The lines represent carbon–carbon bonds; there are carbon atoms at each line junction.)

?

E1 Since the amount of polyunsaturated fats (PUFA) in your diet may affect your health, it is worthwhile to look closely at the contents of several different foods that you might use to prepare sandwiches. Use the Internet and food labels to investigate the PUFA content of butter, several different brands of margarine, and avocados. a Which have the highest total fat content? b What proportion of the fat content is unsaturated? c Does the labelling distinguish between the amount of saturated, mono-unsaturated and polyunsaturated fats? d Is information about the source of these fats provided on the label?


12.1

summary

• The functioning f nctioning of large biomolecules biomolec les is essential to life processes. • Fats and oils belong to the class of biomolecules called lipids. • Fats and oils are formed in a condensation reaction between glycerol and fatty acids. • Three ester linkages are formed in the condensation reaction between glycerol and fatty acids to form triglycerides.

• Fatty Fatt acids can be sat saturated, rated mono-unsaturated mono nsat rated and polyunsaturated. • Fats and oils are broken down into glycerol and fatty acid molecules via a hydrolysis reaction in the digestive system.

key questions 1 Which two functional groups are directly involved in the formation of a lipid? 2 a Write an equation showing the formation of the triglyceride glyceryl trioleate from the reaction of glycerol and oleic acid, CH3(CH2)7CH=CH(CH2)7COOH. b Draw the structure of glyceryl trioleate, showing the ester bonds. 3 A fat present in a vegetable oil has the structure shown in Figure 12.8. O CH2

O

C

(CH2)14CH3

O CH

O

C

(CH2)14CH3

O CH2

O

C

(CH2)14CH3

Figure 12.8 A fat found in a vegetable oil.

a Circle an ester functional group. b Is this fat saturated or unsaturated? Explain. c The fat is broken down into two different substances in the stomach and small intestine. i Write the semi-structural formulas of the reaction products. ii What chemical, apart from the fat, is a reactant in this process? iii What name is given to this type of reaction? 4 Describe the types of fatty acids that combine with glycerol to form: a saturated fats b mono-unsaturated fats c polyunsaturated fats. 5 Sunflower oil is a liquid and can be used to make margarine. How are polyunsaturated oils, such as sunflower oil, converted to a solid form?

12.2

Condensation polymerisation review You will recall from Unit 1 that polymers are molecular substances containing many recurring smaller units covalently bonded together.

The monomers used to make a polymer by a condensation polymerisation reaction have a pair of functional groups that are able to react together, producing a new functional group that links the monomers and also producing a small molecule such as water. Synthetic polymers such as nylon and polyester are formed in this way, as are many biologically important molecules, including proteins, DNA, cellulose and starch. Table 12.4 lists the functional groups in monomers that react to form some condensation polymers.

TABLE 12.4 Features of monomers used to produce different types of polymers Type of polymer Polyester Polyamide (polypeptide) Polysaccharide


Functional group in monomer(s) Carboxyl, –COOH, and hydroxyl, –OH Carboxyl, –COOH, and amine, –NH2 Hydroxyl, –OH, and hydroxyl, –OH

12.3

Carbohydrates Have you ever wandered through a forest? As you gaze round you are looking at the most abundant organic compound on Earth. The main structural component of all plants is a polymer called cellulose. Over half the world’s organic carbon is bound up in trees and other plants in the form of cellulose. Cellulose belongs to a class of compounds called carbohydrates. As much as 75% of a plant’s dry mass is cellulose or other carbohydrates. Green plants, in the presence of sunlight, convert carbon dioxide and water into a number of substances including the carbohydrate glucose. Carbohydrates are a source of energy in our diets. Carbohydrates are made from the elements carbon, hydrogen and oxygen, and usually have the formula Cx(H2O)y where x and y are whole numbers. Carbohydrates range in size from small molecules, with relative molecular masses between 100 and 200, to very large polymers, with molecular masses greater than one million.

Monosaccharides: the simple sugars The smallest carbohydrates are the monosaccharides. They are white crystalline solids that are highly soluble in water. Most monosaccharides have a sweet taste. Monosaccharides, together with another group of carbohydrates called disaccharides, are often called sugars (Table 12.5).

Figure 12.9 Most visible plant material in this forest is made from carbohydrates.

! Although the formula of a carbohydrate can be written as Cx(H2O)y there are no water molecules actually present in carbohydrates. The name carbohydrate was chosen for these substances because it was once thought that they were made of carbon (‘carbo-’) and water (‘-hydrate’), since the ratio of hydrogen to oxygen is the same as that in water molecules.

Glucose CH2OH

TABLE 12.5 Important carbohydrates Carbohydrate Monosaccharides

Composition

Glucose Fructose Galactose Disaccharides

C6H12O6 C6H12O6 C6H12O6 Formed from:

Fructose

Where found

H HO

Fruit juices Fruit juices, honey Not found naturally in its free form

HOCH2 O

O H

H HO

OH

H

H

OH

OH

H

HO CH OH 2 H

HO

Galactose CH2OH HO

Maltose Sucrose Lactose Polysaccharides

Glucose + glucose Glucose + fructose Glucose + galactose Polymer of:

Germinating grain Sugarcane, sugar beet Milk

Glycogen

Glucose

Starch

Glucose

Cellulose

Glucose

Energy store in animals—in the liver and muscles Energy store in plants—abundant in wheat, potatoes, corn etc. Plant fibre—in plant cell walls

The most abundant monosaccharide is glucose. It has the molecular formula C6H12O6 and the atoms in the molecule are arranged in a ring, as shown in Figure 12.10. The structures of two other important monosaccharides, fructose and galactose, are also shown. All three molecules contain a number of polar OH groups, enabling them to form hydrogen bonds with water. As a result, monosaccharides are highly soluble in water.

H

H

O H

H

OH

H

H

OH

OH

Figure 12.10 Structural formulas of glucose, fructose and galactose. The lines represent carbon–carbon bonds; carbon atoms in the rings have been omitted for clarity.

chemfact Glucose, fructose and galactose are structural isomers. They have the same molecular formula but different molecular structures. The differences in their structures appear slight but they are of great importance to living organsims.


chemfact Some sugars such as ribose (C5H10O5) and deoxyribose (C5H10O4) contain five carbon atoms. These monosaccharides are found in nucleic acids such as ribonucleic acid (RNA) and deoxyribonucleic acid (DNA). The structure of DNA is discussed in Chapter 13.

Glucose is found in all living things, especially in the juice of fruits, the sap of plants and in the blood and tissue of animals. Glucose functions as the key energy source in most forms of life and it is also a major component of most of the larger carbohydrates. Both glucose and its polymer, starch, are more rapidly digested than other forms of food. They are the main sources of energy in most diets and our bodies use them for energy in preference to fats and proteins. Fructose and galactose are not as abundant as glucose. In fact, galactose is not found in nature as a free monosaccharide. However, it occurs frequently as a component of larger carbohydrates. Fructose is found in many fruit juices and honey. It is the sweetest sugar known, being 1.7 times sweeter than table sugar! Although fructose is a component of other large biological molecules, its main role in the body is as an energy source, where it is used in much the same way as glucose.

chemistryy in action Why are sugars sweet? An O–H or N–H group

350 pm

300 pm

An oxygen or nitrogen atom

1 pm = 10–12 m

550 pm

A non-polar group e.g. –CH3, –C6H5

Figure 12.11 It is thought that this ‘triangle of sweetness’ must be present in a molecule for it to taste sweet.

Figure 12.12 Aspartame—food additive 951—is used as an artificial sweetener in diet soft drinks and sugar-substitute tablets.

The fact that some substances taste sweet is believed to be due to the shape of their molecules. To taste sweet, molecules seem to need three sites that can form bonds with taste buds on the surface of the tongue. Provided the bonds are of the correct type and the sites are certain distances apart, our brain recognises a substance as tasting sweet. Two of these bonds to the taste-buds are hydrogen bonds, and the third is formed with a non-polar site. The so-called ‘triangle of sweetness’ that must be present in the molecule is shown in Figure 12.11. Some molecules have many such triangles that can ‘lock onto’ the taste-buds for long periods. Substances that are several thousand times sweeter than glucose exist. Table sugar (sucrose) is twice as sweet as glucose. Use of artificial sweeteners has proved a popular way of reducing the amount of sugar added to foods. One of the most common sweeteners is aspartame, which is sold under the trade names NutraSweet and Equal. Although aspartame contains about the same amount of energy as an equal mass of sucrose, it can be used in much smaller quantities than sugar because it is nearly 200 times sweeter (Figure 12.12). Neotame, another artificial sweetener, is 50 times sweeter than aspartame!

O

H2N CH

HO C

CH2

C

CH2 CH

NH

O

O C

CH3

O

Disaccharides A disaccharide is the carbohydrate formed when two monosaccharides undergo a condensation reaction. Like monosaccharides, disaccharides also dissolve in water, taste sweet and are called sugars. A disaccharide called maltose is formed when two glucose molecules react, with the elimination of a water molecule (Figure 12.13a). Note how the hydroxyl functional groups react to form the disaccharide and water. The two rings are joined via an oxygen atom. This linkage is called a glycosidic (or ether) linkage.


Two other important disaccharides are lactose and sucrose. Lactose is a disaccharide made by the condensation of galactose and glucose. It is not as sweet as glucose. Lactose is synthesised in the mammary glands of mammals and is the main carbohydrate present in milk. In contrast, sucrose is widely used as a sweetener because of its intense taste. It is formed from the condensation of fructose and glucose (Figure 12.13b). Sucrose is found in the sap of some trees and the juices of many fruits. Table sugar is produced commercially by extracting sucrose from sugarcane or sugar beet.

a HOCH2 H H HO

HOCH2 O

H OH H

H

H

+

OH

H

H OH

HO

OH

H

Glucose

HOCH2

HOCH2 O H

H OH

H HO

OH

O

O

H H H

H OH

H

+ H 2O

OH

H

H

OH

O H

OH

Glucose

Maltose

H OH

Glycosidic linkage

Water

b HOCH2 H H HO

H OH H

Glycosidic linkage

HOCH2 O

H

H OH

Glucose

OH

O

CH2

H

+ HO

H

HO

HO H

CH2OH

H HO

O

O

HOCH2

H

H

H OH

+ H2O H

H

HO

HO

H

O H

OH

Fructose

CH2OH

Sucrose

Water

Figure 12.13 The condensation of a glucose molecules to form maltose, and b glucose and fructose molecules to form sucrose. Glycosidic linkages are highlighted.

Polysaccharides: the complex carbohydrates As the name implies, polysaccharides are polymer carbohydrates made by linking monosaccharides into a chain. Polysaccharides are polymers of glucose molecules linked together in different ways by condensation reactions. Polysaccharides are generally insoluble in water and have no taste. The three most important polysaccharides biologically are: glycogen starch cellulose. Glycogen is found in animals, where it acts as a glucose storage molecule. As glucose can be oxidised to produce energy more rapidly than fat, all animals store some glucose for use when energy is required quickly. Excess glucose is polymerised to form glycogen, which is stored in the liver and in muscle tissue. When no more glycogen can be stored in these places, additional glucose is converted into fat. As energy is required, the glycogen is hydrolysed to yield glucose for respiration. Starch is the glucose storage molecule in plants; the plant equivalent of glycogen. The starch is stored and used at night to meet the plant’s ongoing energy requirements when glucose production from photosynthesis has ceased. Figure 12.15 represents the digestion of starch and glycogen in the body. During digestion, the polymers must be broken down to release the glucose monomers. The polymers are first hydrolysed. This hydrolysis is catalysed

*HYIVO`KYH[LZ

Monosaccharide

:[HYJO

Disaccharide

.S`JVNLU

Polysaccharides

*LSS\SVZL

Figure 12.14 Starch and glycogen are polymers of glucose. The glucose molecules are joined together the same way, but glycogen is more highly branched than starch. Hydrogen bonds between parallel chains are formed in cellulose, producing strong, water-insoluble fibres.


chemfact Iodine can be used to test for the presence of starch. If starch is present the brown colour of iodine changes to dark blue. Starch or glycogen

Digestion

Further digestion

(using enzymes and water)

(using enzymes and water) Maltose

Glucose

! Chew on a piece of bread for a few minutes. It begins to taste sweet. Enzymes in your saliva catalyse the conversion of some of the starch to the sugars maltose and glucose.

a

CH2OH O OH

O

b O

CH2OH O O

OH

O

OH

OH

OH

CH2OH O

OH O CH2OH

O

CH2OH O

OH

O OH

OH O

OH

OH

OH O CH2OH

Figure 12.16 Sections of a starch and b cellulose polymers. Note the different orientations of the –CH2OH groups on adjacent glucose monomers.

O

Figure 12.15 During digestion, enzymes catalyse the breakdown of starch into maltose and then into glucose.

by enzymes; every second glucose is hydrolysed to produce maltose. Another enzyme catalyses the hydrolysis of maltose into glucose, which is absorbed into the body. These reactions can be regarded as the reverse of the condensation reactions from which maltose and starch or glycogen are formed from glucose. Cellulose is the main structural material in plants. It is a large linear polymer of glucose, with a molecular mass varying from 50 thousand to 2.5 million. Its chemical structure differs from that of starch and glycogen because of slight differences in the way the glucose monomers are joined. As you can see from Figure 12.16, the CH2OH groups on adjacent glucose monomers in starch are on the same side of the polymer chain, while in cellulose they are on alternating sides. In plants, individual cellulose molecules are organised into bundles of parallel chains. The bundles are held together by hydrogen bonding between adjacent chains. These bundles are used to form the cell walls which surround and support all plant cells (Figure 12.17).

Figure 12.17 Strands of cellulose in the wall of a plant cell. This photograph was taken using an electron microscope.

! Bread and potatoes contain starch and plenty of fibre, so they make us feel full, without supplying much energy. These foods are often wrongly considered to be particularly fattening.


Figure 12.18 Sheep obtain much of their energy from cellulose. Bacteria living in the animal’s gut are responsible for hydrolysing the cellulose to glucose.

chemistryy in action Lactic acid: causing the pain an athlete experiences The athletes turn the final bend in a 400-metre race. Gasping for breath, hearts racing, they strive to maintain or increase their speed. Their legs feel like lead and, when they finish, they experience severe cramping pain in their muscles. The experience of these athletes is a consequence of their muscles running out of oxygen to oxidise glucose. Despite the accelerated beating of their hearts and their rapid breathing, the athletes reach a point during the race when they experience ‘oxygen debt’: their blood can no longer provide oxygen at a sufficiently fast rate. From this stage onwards their bodies obtain energy anaerobically (‘without oxygen’). Unlike aerobic (‘with oxygen’) energy production, in which the products of glucose oxidation are water and carbon dioxide, anaerobic energy production involves the breakdown of glucose molecules into lactic acid (Figure 12.20), yielding much less energy. The accumulation of lactic acid in the muscles causes much of the pain and fatigue that the athletes experience. Once the race is over the discomfort eventually disappears as the lactic acid is oxidised to carbon dioxide and water. The training of top-class athletes involves conditioning to use energy supplies as efficiently as possible. CH2OH H HO

O H

COOH H 2

OH

H

H

OH

C

OH CH3

HO Glucose

summary

H Lactic acid

Figure 12.20 The anaerobic production of energy involves formation of two lactic acid molecules from each glucose molecule.

Figure 12.19 The fatigue experienced by a runner is partly due to the formation of lactic acid produced by the anaerobic break-down of glucose molecules in muscle cells.

12.2–12.3

C Carbohydrates b h d t are a source off energy in i our diets. di t Starch, glycogen and cellulose are condensation polymers of glucose.

The Th condensation d ti reactions ti that th t produce d polysaccharides l h id involve the formation of glycosidic linkages as hydroxyl functional groups on adjacent monosaccharide molecules interact. During digestion, some complex carbohydrates are hydrolysed into monosaccharides.

key questions 6 Glucose is the most abundant monosaccharide. Draw a structural formula for glucose and use it to explain why it is highly soluble in water. 7 Glucose is described as a monosaccharide, maltose as a disaccharide and glycogen as a polysaccharide. a What is the difference between these three types of carbohydrate? b What type of reaction is involved in converting glucose to maltose and then to glycogen? c Which functional group in the molecules is involved in these reactions?

8 The structure of lactose is shown in Figure 12.21. Lactose undergoes hydrolysis in the digestion process. a What is meant by ‘hydrolysis’? b Circle a hydroxyl and a glycosidic functional group in the molecule.

CH2OH H

O

H

H CH2OH O

HO H H

OH

H

H

OH

OH

H

H

OH

O

OH

H

Figure 12.21 The structure of lactose.


key questions (cont.) c Draw the structures of the hydrolysis products. d Lactose is a carbohydrate. Why does 10 g of lactose provide more energy for our bodies than 10 g of the main carbohydrate found in celery? 9 a i Write a chemical equation for the condensation reactions between two glucose molecules to form a maltose molecule. ii Draw structural formulas for all molecules involved in this reaction and circle the glycosidic linkage in maltose. b i Write a chemical equation for the condensation reactions between a glucose molecule and a fructose molecule to form a sucrose molecule.

ii Draw structural formulas for all molecules involved in this reaction and circle the glycosidic linkage in sucrose. 10 Starch, glycogen and cellulose are all polymers that contain glucose monomers. Explain why the human body can completely digest the starch and glycogen but cannot digest cellulose to any great extent. 11 Write a chemical equation for: a the hydrolysis of sucrose b the hydrolysis of tristearin (see Figure 12.3)

12.4

Proteins The biological importance of proteins When you look at yourself in a mirror, you are looking only at proteins. Proteins are found in every cell and are fundamental to cell structure and operation. There are thousands of different types of proteins, each with its own specific purpose. Some of these are listed in Table 12.6.

TABLE 12.6 Types of protein

Figure 12.22 Milking venom from a snake. Snake venom is made of protein.

H C

O Glycine

H

Function Protection, support, movement

Enzymes Hormones Transport

Biological catalysts Regulation of body functions Movement of compounds between and within cells Defence Attack Energy source (only in extreme circumstances)

Protective Toxins Most proteins

C

H2N

Type Structural

Examples Skin, bone, cartilage, ligaments, tendons, muscle, hair, teeth, feathers, beaks, cocoons, insect exoskeletons Digestive enzymes Insulin Haemoglobin Antibodies Snake and spider venoms

OH

Amino acids CH3 Alanine

O C C

H2N H

OH

Figure 12.23 Two of the simpler amino acids.


Proteins are polymers built up from small monomer molecules called amino acids. Two of the simpler amino acids are glycine and alanine. The structures of these two substances are shown in Figure 12.23. Every amino acid has an amino group (NH2) and a carboxyl group (COOH). The twenty amino acids commonly found in proteins in the human body are listed in Table 12.7. They have the general formula H2N–CHZ–COOH.

TABLE 12.7 The 20 amino acids that make up proteins in the body. Some amino acids are essential to our diet, but others can be made by our body. Essential amino acids are shown by *. Amino acid Alanine

Abbreviation ala

Structure

arg

HN

CH2

C

O C C

H2N H

CH3

CH

CH2

OH

COOH

NH2

Arginine

Z

CH2

CH

NH

COOH

Figure 12.24 General structural formula of an α-amino acid. Different amino acids have different groups of atoms as the Z group.

NH2

NH2

Asparagine

asn

H2N

C

CH2

CH

O

Aspartic acid

NH2

HOOC

asp

COOH

CH2

CH

COOH

NH2

Cysteine

cys

HS

CH2

CH

COOH

NH2

Glutamic acid

glu

HOOC

CH2

CH2

CH

COOH

NH2

Glutamine

gln

H2N

C

CH2

CH2

CH

O

Glycine

COOH

NH2

gly

H

CH

COOH

NH2

Histidine*

Isoleucine*

his

ile

CH2 HN

N:

H3C

CH2

CH NH2

CH

CH

H3C

Leucine*

leu

H3C

COOH

CH

COOH

NH2 CH2

CH

COOH

H3C NH2

Lysine*

lys

H2N

(CH2)4

CH

COOH

NH2

Methionine*

met

H3C

S

(CH2)2

CH

COOH

NH2

Phenylalanine*

phe CH2

CH

COOH

NH2

Proline

pro + N H

COOH H


TABLE 12.7 (Continued) Amino acid Serine

Abbreviation ser

Structure HO

CH2

CH

COOH

NH2

Threonine*

thr

H3C

CH

HO

Tryptophan*

CH

COOH

NH2

trp

CH2

CH

COOH

NH2

N H

Tyrosine

tyr HO

CH2

CH

COOH

NH2

Valine*

val

H3C

CH

CH

COOH

H3C NH2

Z H H H

+

N

O

C C H

–

O

Figure 12.25 General structure of a zwitterion.

The major difference between one amino acid and another is the collection of atoms that make up the side chain known as the Z group. These amino acids are known as 2-amino acids or α-amino acids because the amino, carboxyl and Z groups are all attached to the second carbon atom (Figure 12.24). As a consequence of the polar amino and carboxyl functional groups, amino acids are soluble in water. In solution, the amino group can act as a base and the carboxyl group can act as an acid. As a result, an amino acid molecule in a solution at a particular pH will usually be in the form + H3N–CHZ–COO–. Such a molecule is called a zwitterion or dipolar ion (Figure 12.25). A proton has been lost from the acidic carboxyl group and the basic amino group has gained a proton. The pH at which an amino acid exists as a zwitterion depends on the structure of the Z group. The dual acidic and basic nature of amino acids means that several different forms of an amino acid can be in equilibrium in a solution. The predominant form depends upon the acidity of the solution: +

H3N–CHZ–COO–(aq) + H3O+(aq)

+

+

H2N–CHZ–COO–(aq) + H2O(l)

–

–

H3N–CHZ–COO (aq) + OH (aq)

H3N–CHZ–COOH(aq) + H2O(l)

In acidic solutions the +H3N–CHZ–COOH form is most abundant, whereas the H2N–CHZ–COO– form is most abundant in alkaline solutions. The ability of amino acids to react with both acids and bases means that they can act as buffers (p. 290), minimising the effect that the addition of H+ or OH– ions to a solution would have on acidity. Their buffering action is of great importance in cells as biochemical processes can only operate correctly if the pH is maintained within a narrow range.


extension Essential amino acids Plants can manufacture all the amino acids they require from simple inorganic ingredients, such as water, carbon dioxide and nitrates, obtained from the atmosphere and soil. However, animals cannot do this. Humans are able to assemble 11 of the 20 amino acids they require, using the organic compounds available in their diets. However, the other nine must be provided directly in the food we eat. These are known as essential amino acids (Table 12.7). A healthy diet should therefore include proteins that contain these essential amino acids. The human body does not store amino acids so a balanced intake of protein is required each day. Proteins from animal sources contain all the essential amino acids. Vegetable proteins, however, are often deficient in one or more of the essential amino acids (Table 12.8). Maize, for example, is too low in lysine and tryptophan to support normal growth in young children. An unsupplemented diet based on maize will result in diseases such as kwashiorkor that are caused by protein deficiency. Vegetarians must ensure that their diets contain complementary proteins. Instead of eating plant protein from a single source, a variety of plant foods must be eaten so that deficiencies of amino acids in one plant are supplemented by the amino acids in another. Rice and lentils, for example, contain complementary proteins. In Asia, they have been part of the staple diet for thousands of years. The traditional Mexican meal of beans and maize also provides complementary proteins. In developed countries, the need to eat complementary proteins creates few problems as only in extreme circumstances do people eat food from one source.

TABLE 12.8 Amino acid deficiencies in various foods Food Legumes Soybeans Wheat Maize Nuts

Figure 12.26 This child is suffering from kwashiorkor or protein–energy malnutrition.

Amino acid deficiency Methionine Methionine Lysine Lysine, tryptophan Methionine

questions

! Vegetarians also need to ensure they get adequate amounts of vitamin B12 because this vitamin is found mainly in animal-derived foods.

?

E2 Why must vegetarians ensure that their diet contains complementary proteins? E3 A healthy diet includes proteins that include essential amino acids. Why should essential amino acids be included in a healthy diet? E4 Kwashiorkor or protein–energy malnutrition is a disease found in some traditional African societies. Find out more about this disease and its causes.

Protein structure When a molecule that contains a carboxyl group, COOH, combines with a molecule containing an amine group, NH2, a condensation reaction occurs to form an amide functional group, –CONH–, that links the two molecules (Figure 12.27). A water molecule, H2O, is also formed. O O

H

C

N O

H

C

N

+ H2O

H

Figure 12.27 An amide link forms when the carboxyl group reacts with an amine group; a water molecule is also produced.

H Carboxylic acid

Amine

Amide group

Water


chemfact During the synthesis of a protein in the body, the sequence of amino acids in the protein is controlled by a complex polymer called deoxyribonucleic acid (DNA) that is found in the nuclei of cells. Sections called genes along the DNA polymer chain contain the instructions for building all the thousands of different proteins found in any organism. The structure and function of DNA are discussed in Chapter 13.

H2N

H

O

C

C

OH

+

H

H

H

O

N

C

C

H

CH3

OH

Alanine

Glycine

Amide or peptide linkage

H2N

H

O

C

C

H

H

O

N

C

C

H

CH3

OH

+

H 2O

A dipeptide

Water

Figure 12.28 The condensation reaction between two amino acids. Note how the carboxyl and amine groups react in forming the dipeptide and water.

chemfact A shorthand notation is often used represent the sequence of amino acids in a polypeptide chain. The sequence of amino acids (valine, serine, cysteine, alanine) in the section of the polypeptide chain shown in Figure 12.28 is represented as: val ser cys ala

H

O

C

C

(CH3)2CH

H

O

N

C

C

H

CH2 OH

Valine

Serine

H

O

N

C

C

H

CH2

H

O

N

C

C

H

CH3

N H

SH Cysteine

Alanine

Figure 12.29 A section of a polypeptide chain, showing peptide linkages.

Proteins are polymers formed by condensation reactions between amino acids; during these reactions the amino acids join and form long unbranched chains. The amine group of one amino acid reacts with the carboxyl group of an adjacent amino acid. A covalent bond is formed and a molecule of water is eliminated, as shown in Figure 12.28. Molecules made from amino acids are often called peptides. When two amino acid molecules react together a dipeptide is formed, and when three molecules react together a tripeptide is formed. A polymer made from amino acids is known as a polypeptide, and polypeptides built up from more than 50 amino acids are usually called proteins (Figure 12.29). The group of atoms –CONH– that links the constituent amino acids is an amide group, referred to as a peptide linkage in proteins. Proteins differ from one another in the number, type and sequence of their constituent amino acids. Each protein has a precise chemical composition, amino acid sequence, and three-dimensional shape. There can be more than 500 amino acid units in a large protein. Determining the structure of these complex materials has provided challenging problems for chemists. Figure 12.30 Professor Dorothy Hodgkin and her research group discovered the three-dimensional structure of one of the smallest proteins, insulin, in 1969. It contains 51 amino acid units and acts as a hormone, controlling the use of glucose in the body. In 1964, Hodgkin was awarded the Nobel Prize in Chemistry for using X-ray crystallography to determine the structure of vitamin B12.


The role that any protein fulfils in an organism depends on its shape. This is regarded as being made up of a primary, secondary and tertiary structure, as described in Table 12.9.

TABLE 12.9 Three levels of structure of protein molecules Primary structure Cys

Val

Leu

His

His

Glu

A sequence of amino acids Secondary structure H

C

N

C

O

H

H

O

C

O

N

C

O

N H

N

Hydrogen bonds between sections of the keratin chains in hair and wool give these proteins a helical structure. Tertiary structure

A ribbon diagram representing the tertiary structure of a protein, Factor VIII, that helps blood clot after an injury.

summary

The order of the amino acids is known as the protein’s primary structure. The entire shape of a protein is a consequence of the precise order in which its amino acids are joined together.

Parts of the chain can attract each other, creating particular threedimensional shapes. Such coiling, folding or pleating of sections of a protein is described as the protein’s secondary structure. For example, the helical structure of keratins in fibres of hair and wool result from hydrogen bonding between neighbouring peptide links in the polypeptide chain, as shown on the left. The Z groups in the amino acid units of a protein molecule influence the overall three-dimensional shape of the molecule. As Table 12.7 shows, the Z groups in some amino acids are relatively large and their size can cause distortions in the protein structure. Furthermore, some of these side groups are capable of forming bonds with side groups elsewhere on the protein chain. As a result, sections of the chains may be folded back on each other in intricate and unique shapes. Four types of attractions are important in chain folding: • hydrogen bonds between polar Z groups such as those containing –O–H or –N–H bonds • ionic interactions between a Z group that contains –NH3+ and another group that contains –COO– • covalent cross-links between chains, particularly where the amino acid cysteine (containing an –SH group) is present. Two neighbouring chains with cysteine units can react, forming a disulfide link (–S–S–). Disulfide bridges are found in many proteins, including insulin. • dispersion forces between non-polar Z groups, such as valine and phenylalanine. As a consequence of such bonds, a variety of protein shapes exists. Some proteins resemble flat sheets, others are long and helical, and others still are compact and globular. The overall three-dimensional shape adopted by a protein is called its tertiary structure.

12.4

AAmino i acids id contain t i the th amino i andd carboxyl b l functional f ti l groups. Proteins are condensation polymers of 2-amino acids. The primary structure of proteins is determined by the sequence of amino acids. Secondary structure of proteins involves hydrogen bonding.

Various V i ttypes off attractions tt ti between b t NH andd C — O groups influence the tertiary structure of proteins. Attractions such as ionic interactions, covalent cross-links between chains, hydrogen bonding and dispersion forces determine the tertiary structure of proteins.


key questions 12 What two functional groups are present in all amino acids? 13 Draw the structural formula of glycine as it is likely to exist in: a an acidic solution b an alkaline solution c neutral solution. 14 With the aid of Table 12.7: a draw structural formulas of serine and cysteine b write an equation to show the formation of a dipeptide from these amino acids c name the type of reaction in part b. 15 Why is protein more important in your diet as a source of raw materials for building other molecules, rather than as a source of energy? 16 Explain what is meant by the primary, secondary and tertiary structures of a protein. Clearly outline the bonding type involved at each level of protein structure. (Include labelled diagrams.)

review A catalyst increases the rate of a reaction without being consumed by it. It allows the reaction to proceed via a new pathway with a lower activation energy. Catalysts are discussed in more detail in Chapter 15.

chemfact Defects in the glucokinase gene, located on chromosome 7, have long been suspected of contributing to genetic susceptibility to non insulin-dependent (type II) diabetes.


17 Draw the structure of the two amino acids produced when the dipeptide in Figure 12.31 undergoes hydrolysis.

H

H

O

N

C

C

H

H

H

O

N

C

C

H

CH2

O

H

OH

Figure 12.31

18 Use the information in Table 12.7 to sketch the structural formula of the tripeptide that would be represented as ala-ser-val.

12.5

Enzymes: biological catalysts Many chemical reactions occur more rapidly when a catalyst is present. The human body can be likened to a complex chemical factory. Thousands of chemical reactions are involved in sustaining life. These reactions occur in a highly organised, sequential fashion. Catalysts control almost all the chemical reactions happening in our bodies. The biological catalysts that accelerate the rate of chemical reactions in living things are a type of protein called enzymes. Enzymes control the manufacture of complex substances, such as skin and blood, as well as the breaking down of chemicals to provide energy. Enzymes make life possible. Enzymes are generally more efficient than inorganic catalysts, because they allow reactions to occur rapidly within the narrow band of temperatures in which life can survive. A biological reaction often has a large number of stages, each one controlled by its own enzyme. Compared to inorganic catalysts: Enzymes produce much faster reaction rates. They can increase the rate of a chemical reaction by as much as 1010 times. This is like taking 1 second to do a task that normally takes 300 years. Enzymes operate under much milder conditions. Even with a catalyst, the industrial production of ammonia requires temperatures around 500°C and a pressure of 250 atmospheres. Enzymes found in bacteria in the roots of leguminous plants, such as peas and clover, perform a similar reaction at normal temperatures and pressures. Enzymes are more sensitive. The catalytic activity of many enzymes is destroyed when they are heated strongly because their delicate structure breaks down. The enzyme is then described as being denatured.

Enzymes are very selective. For example, platinum metal catalyses many reactions. On the other hand, the enzyme glucokinase catalyses the first step in the oxidation of sugar in human liver cells (glycolysis). This is the only function that glucokinase has, and no other enzyme can perform this function. A typical living cell contains 200 different enzymes, each one responsible for controlling a particular reaction. Hundreds of enzymes have been isolated in pure form, ranging in size from molecules with relative molecular masses of 10 000 to molecules with relative molecular masses of several million. The catalytic activity of an enzyme depends on its tertiary structure. A slight change in its three-dimensional shape can render an enzyme inoperative. The active site of an enzyme is usually a flexible hollow or cavity within the molecule. Some enzymes have small, non-protein parts called cofactors, such as vitamins or metal atoms, associated with the active site. These cofactors are necessary for the catalytic effect. A reactant molecule, known as the substrate, is manoeuvred into this site and it is there at the surface of the enzyme that reaction takes place. The steps in the action of an enzyme are shown in Figure 12.32. They are as follows: 1 The reactant (substrate) enters the active site. 2 Bonds formed between the enzyme and substrate weaken bonds within the substrate, lowering the reaction’s activation energy. 3 The substrate breaks or rearranges into new products and these products are released. Typically, reactions occur at the rate of one thousand per second on an active site. During respiration the enzyme carbonic anhydrase catalyses the decomposition of 600 000 carbonic acid molecules (H2CO3) into carbon dioxide and water each second! In many cases, the enzyme and substrate bind together because part of the substrate and the active site are non-polar and so dispersion forces are significant. In other cases, the substrate is held in place by attraction between positive and negative charges, attraction of a metal ion in the enzyme to a negative dipole on the substrate, or hydrogen bonding between enzyme and substrate. Figure 12.33 shows the interactions at the active site of the enzyme (ACE) responsible for catalysing a biochemical reaction in which a dipeptide consisting of leucine and histidine is chopped off the end of a short polypeptide chain (angiotensin I). ACE active site

NH

CH2

Angiotensin I

N H

CH

H

CH2 C

N

δ− O

H

Zn2+

CH

C

CH3

N

δ− O

H

CH

Substrate

The substrate enters the active site in the enzyme

b

Bonds are formed with the enzyme

Bonds break in the substrate The enzyme is regenerated

c

New products are released

!

CH2 C

Enzyme E

Figure 12.32 a-c Steps in the action of an enzyme.

CH3

N

a

C

O–

+

NH3

The polypeptide product of the reaction catalysed by ACE is an important factor in controlling blood pressure.

O

H

chemfact N

ACE catalyses hydrolysis at this point

It takes energy to break bonds so a chemical reaction can take place. The amount of energy required is called the activation energy. Activation energy will be discussed in more detail in Chapter 15.

Figure 12.33 A model of the active site of angiotensin-converting enzyme (ACE).

195 19 Pathways to biomolecules

The selectivity of enzymes is one of their most important features. Although thousands of different reactions are possible in a cell, the presence of enzymes ensures that particular reactions occur rapidly and that others proceed at insignificant rates. In this way, order is maintained in living cells. This selectivity arises because the shape and functional groups in the active site of the enzyme allow it to bind only with certain substrates. The enzyme and substrate are often described as fitting together ‘like a lock and key’. Figure 12.33 shows some of the types of bond that can form between enzyme and substrate: δ– 1 ion–dipole interactions (between Zn2+ and O=C) δ+

δ–

2 hydrogen bonds (between N–H and O=C) 3 ionic interactions (O– and +NH3). Dispersion forces and dipole–dipole interactions may also be involved as an enzyme–substrate complex forms.

Denaturation of enzymes

Denaturation

Coagulation

Rate

Figure 12.34 When a protein is denatured the chains unravel and then often clump together to form a solid mass.

20˚

40˚ Temperature

Figure 12.35 Effect of temperature on enzyme activity.


60˚

We have seen that the structure of a protein is important to its correct functioning. Since the bonds responsible for holding the protein chains in particular shapes are often relatively weak, slight changes in conditions can disrupt the attractions between parts of the chain, causing the chain to unfold and rendering the protein ineffective (Figure 12.34). A change that destroys the biological activity of a protein is called denaturation. Denaturation may result because of increased temperature, a change in pH, or the addition of various chemicals. Enzyme activity is dependent on temperature. As temperature increases, the rate of the reaction catalysed by the enzyme increases because of increased collisions between enzyme and substrate, until an optimum temperature, about 40°C, is reached. As the temperature increases further the enzyme is denatured and the reaction rate decreases rapidly (Figure 12.35). Enzymes operate effectively within a small pH range (Figure 12.36). Not all enzymes have the same optimum pH. Pepsin, an enzyme that breaks down protein in the stomach, is most efficient at a pH of 1.5. The optimum pH of salivary amylase, an enzyme that catalyses the hydrolysis of starch in the mouth, is 7.4. Lipase, which breaks down fats in the small intestine, has an optimum pH of 8–9. You will recall that the ionisation of amino acids is dependent on pH. The bonds that determine the tertiary structure of the enzyme are altered as changes in pH alter the ionisation of the amino acid residues in the protein.

summary

Salivary amylase

Pepsin

Lipase

Rate

Once a protein has been denatured, the unfolded chains tend to form randomly looped structures which come into close contact. The chains become entangled and bond with each other, often by disulfide bridges, so that large clumps of protein molecules are formed. This process is called coagulation (Figure 12.34). You will be familiar with a number of instances of denaturation: When an egg is cooked, the ovalbumin protein in the egg white changes from a clear liquid to a white solid. Heating above 60°C causes the protein to denature and coagulate. Milk curdles when vinegar is added to it. Denaturation and coagulation of the casein protein in milk is caused by the change in acidity. A similar effect is observed when milk sours due to the production of lactic acid by bacteria called Lactobacillus that are present in the milk.

2

4

6 pH

8

10

Figure 12.36 Effect of pH on enzyme activity.

12.5

EEnzymes are special i l proteins t i that th t actt as catalysts t l t ffor reactions ti in living systems.

Enzymes E are specifi ific ffor a particular ti l reaction. ti Enzymes are denatured by heat or by a change in pH.

key questions 19 a Find the names and describe the functions of two enzymes in the body. b Explain why the action of enzymes justifies the statement ‘Enzymes make life possible’. c Why is the action of an enzyme often described as operating like a ‘lock and key’? 20 What are the main differences between enzymes and inorganic catalysts? 21 The enzyme carbonic anhydrase catalyses the decomposition of carbonic acid molecules to carbon dioxide and water in the

lungs. When heated to more than 60°C, the enzyme becomes denatured. a What is meant by the term ‘denatured’? b Describe the events that usually occur to the structure of an enzyme when it is denatured. c Does the primary structure of the carbonic anhydrase enzyme change during this process? d Why is the functioning of the enzyme closely related to its tertiary structure?

12.6

Proteins as markers for disease New and improved techniques for the separation, identification and measurement of small amounts of substances have enabled researchers to identify proteins that indicate the presence of disease. Analytical techniques such as mass spectrometry (Chapter 8), infrared and NMR spectroscopy (Chapter 7) and advanced chromatographic techniques (Chapter 6) as well as two-dimensional electrophoresis are used to identify these protein markers. The body’s natural defences produce proteins called antibodies in the fight against infection caused by bacteria or viruses. These antibodies are specific to a particular disease.

! Electrophoresis is the movement of electrically charged particles under the influence of an electric field. Different particles move at different speeds, so electrophoresis can be used for analysis.


The cells in diseased or damaged body organs may release specific proteins that are unique to that organ. A raised level of these marker proteins in a patient’s blood or tissue can be used to: identify a disease at early or advanced stages of development monitor the progress of the disease measure the effectiveness of treatment test for the recurrence of the disease. Protein markers are now used in the diagnosis of a number of diseases including heart disease and several different types of cancer. In many cases a raised level of a particular protein marker may indicate the presence of disease before physical symptoms are evident. However, there are often other causes for raised levels of protein markers. Doctors consider the evidence from a variety of tests and observations when making a diagnosis. The continuing search for new, specific, protein markers for a range of diseases is an exciting field of research. The study of protein structure and function is called proteomics and it has an increasingly important role in the identification of disease.

Heart attack Lactate dehydrogenase Troponin Creatine phosphokinase

Enzyme level increase above normal

30×

20×

10× 6× 2× 1 2

Onset of chest pain

4

6

8

10

12

Time (days)

Figure 12.37 Changes in the levels of cardiac enzymes following a heart attack.


14

Heart attack is the most common cause of death in Australia. Each year about 50 000 people suffer from a heart attack. Half of these are fatal. A heart attack is caused when the blood supply to a part of the heart muscle is blocked by a clot in one of the cardiac arteries. The cardiac muscle cells are starved of oxygen and nutrients and are consequently injured or die. The cells in the damaged part of the heart muscle release various cardiac enzymes into the blood stream: troponin (TnI, TnT) creatine phosphokinase (CPK) lactate dehydrogenase (LDH). Blood levels of these enzymes increase within a few hours after a heart attack and drop back to normal after several days. The changes in the levels of specific cardiac enzymes are indicated in Figure 12.37. While troponin is found only in heart muscle, the other enzymes are also elsewhere in the body. Forms of the CPK enzyme are found in heart muscle, skeletal muscles and in the brain. LDH is also found in cells in the liver, kidneys, brain and lungs. An injury to these organs also causes a rise in the blood level of the enzymes. A rise in the blood level of one cardiac enzyme does not necessarily indicate that a heart attack has occurred. For example vigorous exercise, muscle injury or muscle inflammation can raise the total CPK level in the blood. An increase in LDH can be caused by diseases such as anaemia, leukaemia, pneumonia or a blood clot in the blood vessels in the lung as well as by a heart attack. LDH is a marker for injury to several organs and is not used as a specific marker for a heart attack. Troponin I and a specific form of creatine phosphokinase, CPK-MB, are only released by damaged heart muscle, so these enzymes can be used as specific markers for heart attack and to determine the extent of damage to heart muscle. Doctors diagnose a heart attack by interpreting data from a variety of sources, including physical examination, medical history and the electrical activity of the heart muscle measured by an electrocardiogram (ECG), as well as the level of cardiac enzymes in the blood.

Prostate cancer The prostate is a small organ located at the base of the bladder in males. It surrounds the urethra and produces a milky fluid that is added to sperm from the testes to produce semen. Cancer is an uncontrolled growth of cells. Cancer of the prostate is a common form of cancer in men. About 3500 men in Victoria are diagnosed with prostate cancer each year. Prostate cancer can interfere with the ability to urinate and may spread to other organs. It is initially diagnosed by digital examination of the rectum and by measuring the level of a protein, prostate specific antigen (PSA), in the blood. The cells in the prostate gland produce PSA, which prevents semen from coagulating. Normally a small amount is present in the bloodstream. Abnormal cancerous or non-cancerous cell growths block the ducts in the prostate, forcing more PSA into the bloodstream. The concentration of PSA in the blood is used as a marker for possible prostate cancer. There is a greater risk of cancer with high levels of PSA, but there are other possible reasons for an elevated PSA level. If the PSA concentration in the blood is a cause of concern to a doctor, follow-up tests are conducted to determine if this is due to a cancerous or non-cancerous growth. These tests may include an ultrasound and a biopsy (removal of a small amount of prostate tissue for examination). Treatment for prostate cancers may include surgery, chemotherapy and radiation therapy. The PSA test can be used as screening test to determine if cancer might be present before symptoms are noticed. Treatment is generally more successful if an early diagnosis is made.

summary

:LTPUHS ]LZPJSL

7YVZ[H[L NSHUK

*V^WLY»Z NSHUK

7LUPZ

,WPKPK`TPZ

;LZ[PZ

Figure 12.38 The male urinogenital system.

12.6

Di Diseasedd or injured i j d tissue ti produces d specifi ific proteins t i th thatt can be used in diagnosis.

Modern M d analytical l ti l techniques t h i are usedd to t identify id tif protein t i markers.

key questions 22 Explain why LDH is not used as a specific marker for heart attack. 23 A competitor collapses during a race. A blood test showed a raised level of total CPK enzyme. a Does the test indicate that the competitor has had a heart attack?

b What other enzyme tests would confirm a heart attack? 24 Explain why doctors sometimes rely on information from a number of tests and observations to make a diagnosis rather than relying solely on the presence of a protein marker.


12

key terms α-amino acid 2-amino acid active site carbohydrate cellulose cholesterol coagulation cofactor complementary proteins condensation reaction denaturation dipolar ion disaccharide disulfide link

enzyme essential amino acids ether link fat fatty acid glycerol glycogen glycosidic linkage hydrolysis lipid monosaccharide mono-unsaturated fat oil peptide linkage

Lipids 25 Explain the difference between: a a fatty acid and a fat b a polyunsaturated fat and a saturated fat. 26 By referring to Figure 12.3, write reactions that show: a the hydrolysis of the fat tristearin b the complete oxidation of the products from part a. 27 The structure of a fatty acid, linoleic acid, is shown in Figure 12.39. CH3(CH2)4CH

CHCH2CH

CH(CH2)7COOH

Figure 12.39 The structure of linoleic acid.

a Name the chemical needed to convert linoleic acid into a lipid. b Draw the structure of the lipid made from linoleic acid. c Name and circle the functional group in the lipid molecule. d Classify the lipid as saturated, mono-unsaturated or polyunsaturated. e Is this fat likely to be a solid or liquid at room temperature? Explain.

polypeptide polysaccharide polyunsaturated fat primary structure protein saturated fat secondary structure steroid substrate sugar tertiary structure triglyceride Z group zwitterion

b Apart from water and maltose, what other substance is required for the hydrolysis of maltose? 29 Why is glycogen sometimes called ‘animal starch’? 30 The human brain relies almost exclusively on glucose as an energy source. Normal functioning of the brain depends on a constant supply of glucose from the blood. Explain how your body maintains a fairly constant concentration of glucose in the blood even though you don’t eat foods that supply glucose constantly throughout the day.

Proteins 31 Explain why body builders and weightlifters eat a large amount of protein. 32 When alanine and glycine react, two different dipeptides can be formed. a Write the formula for each dipeptide. b How many tripeptides can be formed from three different amino acids? c Proteins are formed using 20 different amino acids and can be hundreds of amino acid units in length. On the basis of your answers to parts a and b, what can you say about the number of different proteins possible? 33 Aspartic acid is one of the non-essential amino acids (Figure 12.40). It exists as a zwitterion at pH 2.80.

Carbohydrates 28 a By referring to Figure 12.13, draw structural formulas to show: i the formation of a disaccharide from two fructose molecules ii the hydrolysis of maltose


CH2COOH H2N

C

COOH

H

Figure 12.40 The structure of aspartic acid.

a Name: i the acidic functional group ii the basic functional group. b Draw the structure of the molecule as it is most likely to exist in a solution at: i pH much more than 2.8 ii pH much less than 2.8 iii pH 2.8. c Aspartic acid is needed to make important proteins in the body, yet it is described as a non-essential amino acid. Explain. 34 a Circle the peptide linkages in the part of the protein chain shown in Figure 12.41. /

6

*

*

*/

/

6

5

*

*

/

*/

/

6

5

*

*

/

/

:/

/

6

5

*

*

/

*/

5 /

6/

Figure 12.41 A section of a protein molecule.

35 36 37

38

39

b Write the formulas of the amino acids that made up this part of the chain. c Where in the body does hydrolysis of proteins to form amino acids take place? Children need proportionately more protein in their diets than adults. Suggest why. Draw a structure for an amino acid that could not be classified as a 2-amino acid (α-amino acid). Scientists have spent many years trying to develop a form of the protein insulin that can be given to diabetics orally rather than by injection. Suggest an aspect of the chemistry of proteins that has so far prevented insulin from being taken orally. Proteins are often described in terms of their primary, secondary and tertiary structures. What is the difference between these levels of structure and how does each contribute to the role of the protein? Mutations can cause the wrong amino acid to be inserted into a protein chain when it is being assembled. In some instances such a mistake has no effect on the function of the protein, but in others it disables the protein. Explain how a single change in a protein’s primary structure could cause such dramatically different results.

Enzymes 40 Enzymes allow some reactions to proceed far more rapidly than they would if inorganic catalysts were used. Find out why enzymes are not used widely in industrial processes. 41 Jellied pineapple dessert cannot be made using gelatine and fresh pineapple because an enzyme in the pineapple causes molecules in the gelatine to break down instead of setting. Suggest how jellied pineapple might be prepared. 42 Digestion of proteins occurs in both the stomach and the small intestine. Trypsin is an enzyme that catalyses the hydrolysis of proteins. Trypsin is secreted into the small intestine, where the pH is approximately 8. Would you expect this same enzyme to be secreted into the stomach, where the pH is approximately 2? Explain your reasoning. 43 a The steps in the action of an enzyme involve, in particular, an active site and a substrate. With the use of a diagram describe in detail the action of an enzyme. b The forces of attraction that enable a substrate to bind to an active site can vary. Describe four such forces. 44 Read the magazine article and answer the following questions.

Engineered enzyme washes whiter than white Until Danish company Novo-Nordisk came along no one had developed a commercially available enzyme that would break down fats. In its hunt for a suitable enzyme, NovoNordisk investigated a range of enzymes called lipases. Molecular biologists at the company identified a fungus which will produce a specific type of lipase that is good at breaking down the fats found in human foods. Although the fungi produce these useful lipases naturally, they do not produce enough to be financially viable. So the geneticists identified the sequence of nitrogen bases which make up the DNA of the fungus and cloned that sequence. They then spliced the section of the DNA which holds the gene that codes for the lipase into a new host organism, a fungus. [They] found that it will produce the lipase in sufficient quantities. Novo complains bitterly at the length of time it has had to wait for a licence to produce Lipolase in Denmark … The same process took only four weeks in Japan.

a This article reflects research in the development of enzymes for commercial use. What are the advantages of enzymes compared to inorganic catalysts?


b What tests would you require the Novo-Nordisk company to perform on the enzyme Lipolase before it was released commercially? c Why might governments be cautious about licensing the use of genetically engineered organisms to produce enzymes? Discuss the issues involved in this question with other class members.

Connecting the main ideas 45 What functional groups are required to react to form each of the following, and with which biological polymers are each of these linkages associated? a an ester linkage b a glycosidic linkage c a peptide linkage 46 a Explain the difference between condensation reactions and hydrolysis reactions. b What is the importance of these reactions in the human body? c Write equations for condensation reactions between: i two glucose molecules ii a glycerol molecule and three palmitic acid molecules (see Table 12.1) iii two alanine molecules 47 a By referring to Figure 12.33 on page 195, draw the structure of the dipeptide produced when the ACE enzyme catalyses the conversion of angiotensin I to its active form. b Identify the amino acids formed when the amide functional group is hydrolysed. c List the types of bonds involved at the active site of ACE when it is catalysing this reaction. d The other molecule produced is known to stimulate the constriction of blood vessels, which can lead to an increase in blood pressure. It is a small polypeptide called

202 202 Organic chemical pathways O

angiotensin II that contains the amino acid sequence asp-arg-val-tyr-ile-his-pro-phe. What is the number of each of the following types of functional groups in angiotensin II? i amide (peptide) ii carboxyl. 48 A fatty acid that was extracted from sunflower oil is analysed in a laboratory in a number of ways. a Using IR spectroscopy, it was deduced that the fatty acid is unsaturated. Explain what feature of the IR spectrum would lead to such a conclusion. b The unsaturated fatty acid was reacted quantitatively with hydrogen gas to determine the number of double bonds per molecule. A 0.0102 mol sample of the fatty acid reacted with exactly 500 mL of hydrogen gas, under SLC conditions, for the complete conversion of the fatty acid sample to a saturated fatty acid. i What type of reaction occurred between the unsaturated fatty acid and hydrogen gas? ii Calculate the amount, in moles, of hydrogen gas that reacted with the fatty acid. c Determine the number of double bonds per molecule. d Use the data in Table 12.1 to identify the unsaturated fatty acid. 49 From many thousand proteins in a human tissue sample, a research scientist needed to separate and identify individual proteins that might be used as markers for particular diseases. Briefly describe how each of these analytical techniques could be used in such a process. a HPLC b proton NMR spectroscopy c IR spectroscopy d mass spectroscopy

1

chapter 13

DNA key knowledge • the structure and bonding of DNA and its application in forensic analysis • organic reaction pathways, condensation and polymerisation reactions that produce large biomolecules

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • identify the phosphate, deoxyribose and base groups that combine to form DNA • describe the nature of the bonds in DNA • explain the pairing of the bases • describe the primary, secondary and tertiary structure of DNA • explain the basic principles of DNA profiling and its application in forensic analysis • recall the principles of electrophoresis.

13.1

The biological importance of DNA ! Your red blood cells do not have a nucleus. The chromosomes in your sex cells (sperm or ova) are unpaired and therefore these cells contain only half the number of chromosomes found in normal body cells.

Within the nucleus of almost every one of your cells, 46 DNA (deoxyribonucleic acid) molecules or chromosomes contain approximately 20 000 to 25 000 genes that act as a blueprint directing which proteins are to be synthesised by the cell (Figure 13.1). By controlling cell functioning, DNA is able to determine the characteristics of the entire person—your eye colour, build, skin type and susceptibility to some diseases. DNA is unique among molecules because its structure allows it to manufacture another molecule, identical to itself, and in doing so transmit its genetic code from parents to their children and from one cell to another as new cells are formed.

Figure 13.1 The set of 46 DNA molecules of a human male. Referred to as chromosomes, they have been arranged in pairs, numbered 1 to 22, and the sex chromosomes, X and Y. This man inherited one chromosome of each pair from his mother and one of each pair from his father.

a N

C C

H

N

C C N

H

The structure of DNA

H

b

H

H

C C

N

C H

13.2

C N

C N

H

Figure 13.2 a Purine. b Pyrimidine.


The quest to identify the structure of DNA was one of the most exciting scientific sagas of the twentieth century and, without doubt, its discovery has been among the most important and far-reaching ever made in science. DNA is a condensation polymer constructed from four monomers known as nucleotides. Each nucleotide consists of a phosphate group, a deoxyribose sugar group, and one of the four bases adenine (A), cytosine (C), guanine (G) and thymine (T). Adenine and guanine are derivatives of the organic compound purine (Figure 13.2a), and as a result are referred to as purine bases. Cytosine and thymine are derivatives of another organic compound, pyrimidine (Figure 13.2b), and are referred to as pyrimidine bases.

a

!

b

James Watson, Francis Crick and Maurice Wilkins shared the Nobel Prize in Chemistry in 1962 for the discovery of the structure of DNA. Rosalind Franklin’s X-ray analysis of DNA crystals and Erwin Chargaff’s stoichiometric investigations of DNA had also provided crucial information. Since 1962, it has often been argued that Rosalind Franklin’s evidence was worthy of her inclusion in that Nobel Prize. However, Franklin died in 1958, and the Nobel Prize cannot be given posthumously. In an interview for the magazine Scientific American, Watson himself suggested that it might have been better to to have given Wilkins and Franklin the Nobel Prize in Chemistry, and he and Crick the Nobel Prize in Physiology or Medicine.

Figure 13.3 a James Watson and Francis Crick with their model of the DNA double helix. b Rosalind Franklin, who obtained the X-ray diffraction pattern.

The structure of each of these components of the nucleotide and the way in which the three groups are bonded together are shown in Figures 13.4 and 13.6. The base forms a covalent bond to the carbon atom labelled C1 in the deoxyribose mulecule. The phosphate group bonds to the carbon atom labelled C5. A water molecule is produced when each bond is formed.

a

b OH

–O

5′ HOCH2 4′ C

O

P

OH

O

C 1′ H

H

C

C

OH

H

H

H 3′

2′

OR Phosphate

Deoxyribose

c

d

NH2 N

H

O N

C C

N

C

C

H

C N

N

*

C C

C C N

H

*

f

O C

H

C

*

C C

C N

NH2

H

N

C H

NH2

Guanine

Adenine

H3C

C N

H

H

e

H N

N

C O

H

C N

*

H

H

Thymine

Cytosine

O

Figure 13.4 Structure of a phosphate group, b deoxyribose sugar group, and bases c adenine, d guanine, e thymine, and f cytosine. * indicates the site where the base bonds to the C1 deoxyribose carbon atom.

205 DNA

OH

5′ end O

O

P

O–

OH

H

H3C

N

O

O

T

O

P

O– H

H3C CH2

O

N

N

O

O

CH2

O

O

N

+ H2O

O H

O

OH

O O

O

P

O–

O

N

P

O–

N

H

H

O

N

O

N

G

CH2

O

N

N

CH2

NH2

O

N

N

NH2

+ H2O O O

NH2

O

O–

P

N

H OH

N

O NH2

O

P

O–

CH2

N

O

N

N

N

O

A

CH2

O

+ H2O

N

N

O NH2

O O

P

O– N

O

H

CH2

OH

O

N

O

NH2

O

P

O– N

O CH2

O

N

C

OH

O

3′ end

OH

Figure 13.5 A simplified presentation of the condensation reaction between nucleotides to form a section of DNA.


O

OH

O

N

O–

P

OH O

H

O

N

O–

P

N

H N

O

H

Phosphate

H

O

O N

N CH2 5′

O

4′

OH

H

Guanine

+ 2H2O

NH2 CH2 5′

1′ 3′

4′

2′

N

N

NH2

1′ 3′

OH

Deoxyribose

O

2′

OH

Guanosine monophosphate nucleotide

Figure 13.6 Formation of a nucleotide. Skeletal formulas are used for simplicity.

Primary structure of the polymer Covalent bonds are responsible for producing the primary structure of DNA. A covalent bond is formed between the 5′-phosphate group on one nucleotide and the 3′-hydroxy group on the deoxyribose of another nucleotide. One end of the DNA strand has a hydroxyl group and is called the 3′ end. The end of the polymer with the phosphate group is called the 5′ end. In this way, nucleotides undergo condensation polymerisation to form a long chain of nucleotides (Figure 13.5). A base is also covalently bonded to the sugar group at the C1 site via a condensation reaction (Figure 13.6). It is the sequence of bases along the sugar–phosphate backbone that forms the primary structure and is the basis for the genetic code.

Secondary structure of the polymer As is the case in proteins, hydrogen bonding is responsible for maintaining the secondary structure of DNA. Erwin Chargaff’s stoichiometric studies of the mole amounts of the four bases in a wide variety of living things had led to the discovery that: 1 for a given species, the percentage of each of the four bases is the same in all cells and is characteristic of the organism 2 the mole amounts of thymine (a purine) and adenine (a pyrimidine) are equal 3 the mole amounts of cytosine (a pyrimidine) and guanine (a purine) are equal. By considering the structures of the base pairs adenine and thymine, it can be seen that they are able to fit together, rather like the pieces of a jigsaw. It is possible for two hydrogen bonds to form between the δ– on nitrogen and oxygen atoms and the δ+ on hydrogen atoms on the adjacent base (Figure 13.7). Similarly with the base pairs cytosine and guanine, three hydrogen bonds can form between δ– on nitrogen and oxygen atoms of one base and the δ+ on hydrogen atoms on the adjacent base (Figure 13.7). As shown in Figure 13.8, the secondary structure of DNA is a pair of DNA polynucleotide strands held together by hydrogen bonding between the bases adenine–thymine and cytosine–guanine. The structure twists around to form a right-handed double helix—the shape of a spiral staircase—where the sugar–phosphate backbone forms the handrails and the pairs of bases are the steps. The pairing of bases, A–T and C–G, is referred to as complementary base pairing and results in ‘steps’ of a constant width.

chemfact The bases in the hen are 29% each adenine and thymine and 21% each cytosine and guanine. In contrast, the marine crab has approximately 47% each adenine and thymine and only 3% each cytosine and guanine.

207 DNA

a

CH3 Oδ– Hδ+ N

H N

N

Bond to DNA chain

Hδ+

or T

N

Nδ–

O

N

N

A

Bond to DNA chain

H

b

N Hδ+ Oδ–

Nδ–

N Bond to DNA chain

Hδ+

N

N

Oδ– Hδ+ or C

N

N

N

G

Bond to DNA chain

H

Figure 13.7 Hydrogen bonding between a thymine and adenine, and b guanine and cytosine.

a

b

Phosphate group

A

Deoxyribose sugar group 5′

P

3′ Nucleotide

Nitrogen base

S

S

S P

G

G G

S

S

A

C P

T

C C

T

P

Sugar-phosphate Sugar-ph backbone

P S

S P

G

C

P

T

A C

A

Hydrogen bonds between bases

A T

S

S 3′

G T

P

G A

5′

Sugar-phosphate backbone

Figure 13.8 a Hydrogen bonding between complementary bases produces a ladder-like structure. Note that the strands run in opposite directions. b The polynucleotide ‘ladder’ twists to form a right-handed double helix.


The DNA strands align in opposite directions. The orientation of one strand is 5′→3′ while the orientation of the other is 3′→5′.

Tertiary structure of the polymer The phosphate groups in the backbone of the double helix give DNA molecules a negative charge, and this enables the molecules to interact with a group of proteins called histones. The DNA molecules wrap around histones and become super-coiled. This allows the DNA to be very tightly and efficiently packaged, each molecule forming a structure known as a chromosome.

chemfact Histones contain a high proportion of the amino acids lysine and arginine, whose Z-groups are positively charged at the pH of body cells. As a result, ionic interactions occur between DNA and histones.

DNA (2 nm in diameter)

Initial DNA NA double helix elix

Histones Active site e of DNA polymerase merase

AT GC GC T A A T G C C G

G

C G S P C T A A A T C G S P T C G A C G S P S P T A T GC G C S P C T A T A P A T A T S T A T A Free C G C G nucleotides des G C G C G C G C Free A T A T nucleotides nuc n T A T GC Template e G C C G C G strand A T A T T New T G

S P

Replication The division of plant and animal cells involves a mechanism that generates an exact copy of their DNA. The DNA double helix partially unwinds, as hydrogen bonds between the two DNA strands are broken. Enzymes catalyse this process. The bases exposed on the separated strands then act as a template to which new nucleotides attach by hydrogen bonds between complementary base pairs, C and G, A and T. These bases then undergo a condensation polymerisation reaction catalysed by the enzyme DNA polymerase to form two exact copies of the original DNA double helix.

T A G C C G AT

strand

C

Figure 13.9 Coiling and super-coiling of DNA.

TA GC CG AT

TA C G TA

Pair of identical DNA double helices

Figure 13.10 Replication of DNA is possible because of the hydrogen bonds between complementary bases (A–T, C–G) that break and then re-form.

209 DNA

chemistryy in action Fred Sanger—two Nobel Prizes in Chemistry

Figure 13.11 Fred Sanger.

Fred Sanger (Figure 13.11) has been awarded two Nobel Prizes in Chemistry: for work on protein sequencing, and for work on DNA sequencing. Born in England in 1918, the son of a doctor, Sanger became very interested in nature and science. He decided not to study medicine because he felt that a career in science would give him a better chance to become a problem solver. Sanger was a conscientious objector during World War II (his family were Quakers, who opposed conflict on religious grounds). After completing his degree at Cambridge in 1939, he studied amino acid metabolism for his PhD and then started working on identifying the primary structure of insulin. He was the first person to obtain a protein sequence and won his first Nobel Prize in Chemistry in 1958 for his work on protein structure. By this stage he had moved on to study DNA-related problems, in collaboration with other eminent scientists including Francis Crick and John Kendrew. Solving the problem of DNA sequencing became a natural extension of his work in protein sequencing. Sanger won a second Nobel Prize in Chemistry in 1980, for research concerned with determining the base sequences in nucleic acids.

13.3 chemfact In the body, the amino acid phenylalanine is converted to another amino acid, tyrosine. This reaction is catalysed by an enzyme, phenylalanine hydroxylase, coded for on chromosome 12. In people with phenylketonuria (PKU), a fault in both copies of the gene means the enzyme does not function. High levels of phenylalanine can build up, damaging the brain and nerve cells. A blood test soon after birth will detect the problem and damage can be avoided by excluding phenylalanine from the diet.

How does DNA control biochemical processes? Within a population, alternative sequences of nucleotides can occur at specific positions on a given chromosome. These alternative sequences are referred to as alleles. Alleles give rise to most of the variation we see between people, but even the smallest change in the nucleotides in a gene can have dramatic consequences. You learned in Chapter 12 that the function of a protein is determined by the sequence of amino acids or its primary structure. The mechanism of protein synthesis is complex and a detailed study is beyond the scope of this course. Briefly, however, a sequence of three DNA bases codes for a particular amino acid. This is known as the triplet code. For example, the base sequence CTT codes for glutamic acid while the base sequence AAA codes for phenylalanine. Another nucleic acid, ribonucleic acid (RNA), decodes the information on DNA in the nucleus of the cell and transfers this information to the part of the cell where protein synthesis takes place. Another type of RNA controls the selection and sequencing of the amino acids that form proteins. transcription

translation

DNA —–––—➤ RNA —–––—➤ proteins

A tiny change in DNA can lead to a variation in the sequence of amino acids in a protein, resulting in major health problems.


Case study: sickle-cell anaemia

a

Haemoglobin is a protein assembled from four polypeptide chains, each of which is able to combine reversibly with one oxygen molecule. It is the pigment in red blood cells, and its function is to transport oxygen to active tissues. Sickle-cell anaemia is a heritable disease in which the red blood cells deform into a sickle shape when oxygen concentration is low (Figure 13.12). These conditions are encountered in blood vessels near active tissues. The sickle shape can impede the free movement of red blood cells in the bloodstream and dramatically reduces the red blood cells’ life span. The most common form of the disease occurs because the gene that codes for one of the haemoglobin polypeptide chains has just one nucleotide in the gene incorrect! At one point in the DNA sequence on chromosome 11 that codes for the polypeptide chain β-globin, the base thymine is replaced by adenine. A person who has sickle-cell anaemia will have the same defective gene on both chromosome 11s. As a result, the sixth amino acid in the β-globin polypeptide chains in haemoglobin is always valine instead of glutamic acid and the protein does not function correctly.

Case study: cystic fibrosis Cystic fibrosis is a heritable disease caused by a tiny error in the DNA sequence of a coding allele on chromosome 7. The gene codes for a protein needed to transport chloride ions across cell membranes. A person with cystic fibrosis is unable to regulate chloride ion concentrations in cells lining the lungs, intestine, pancreas and sweat glands. The sufferer is prone to chronic lung infections and produces excessive amounts of thick mucus in the lungs, and has very salty sweat and impaired digestion. Of the 6100 nucleotides in the gene that codes for this protein, a person with the defective allele has a sequence of just three nucleotides missing. This results in one amino acid, phenylalanine, being omitted from the chloride transport polypeptide chain that is manufactured in their cells. The protein folds incorrectly and cannot perform its normal function.

summary

b

Figure 13.12 a Normal red blood cells. b Red blood cells from a person with sickle-cell anaemia.

13.1–13.3

DNA directs di t the th synthesis th i off proteins t i in i cells. ll DNA transmits the genetic code from each generation to the next. DNA is able to replicate itself. DNA is a condensation polymer consisting of four different monomers called nucleotides. A nucleotide consists of a phosphate group, a deoxyribose sugar group and a base group. These groups are covalently bonded together to form a nucleotide via condensation reactions that also produce water molecules. There are four base groups: adenine (A), cytosine (C), guanine (G) and thymine (T). The nucleotides are bonded together by a condensation polymerisation reaction that involves the formation of covalent

bbonds d between b t the th phosphate h h t group in i one nucleotide l tid andd the th deoxyribose sugar group in another. This polynucleotide strand is the primary structure of DNA. The secondary structure of DNA is a pair of polynucleotide strands bonded together by hydrogen bonds formed between the bases. The structure twists to form a double helix. Two hydrogen bonds can form between the base pairs adenine and thymine. Three hydrogen bonds form between cytosine and guanine. This is called complementary base pairing. The tertiary structure involves coiling and twisting of the double helix around special proteins called histones to form the chromosomes. The sequence of nucleotides in DNA directs the sequence of amino acids in proteins.

211 DNA

key questions 1 Copy the structural formula of deoxyribose sugar. Now indicate where covalent bonds form to link this group to: a a phosphate group b a base group. 2 Give the sequence of bases that will pair with the sequence 5′-AAACCTGAACGA-3′. 3 a What structural similarities do the purine bases adenine and guanine share? b What structural similarities do the pyrimidine bases cytosine and thymine share? c In terms of the structure of these bases, explain how the two polymer chains of phosphate and deoxyribose sugar in

the double helix bring about a constant separation along the length of the DNA molecule. 4 a i What structural change occurs in the DNA of a person who suffers from sickle-cell anaemia? ii How does this change affect the polypeptide chains of their haemoglobin molecules? b i What structural change occurs in the DNA of a person who suffers from cystic fibrosis? ii How does this change affect the polypeptide chains of their transmembrane chloride-transport molecules?

13.4

Forensic applications Electrophoresis Electrophoresis is an analytical technique used in creating a DNA fingerprint. It is a little like chromatography and is the best method for the separation and qualitative analysis of very large molecules that have an overall charge, such as DNA and proteins. A solution of the sample to be analysed is usually placed in a well cut into a block of a special type of gel, as shown in Figure 13.13. The gel contains an electrolyte so it can conduct an electric current. Positive and negative electrodes are connected to opposite ends of the gel, causing the ions in the sample to migrate towards the oppositely charged electrode. Smaller ions move faster than larger ones. All fragments of DNA polymer chains have a negative charge due to the phosphate groups. They must be introduced into wells at the end of the gel near to the negative electrode because they will migrate towards the positive electrode. The rate at which they move depends on their molecular mass, which is proportional to the number of base pairs in the fragment. The distance the DNA copies have travelled through the gel is compared with the distance travelled by known standards. The separated components are made visible by using various methods, including: adding a dye that binds to them and fluoresces (glows) in UV light adding radioactive probes that bind to them; the radiation is then used to expose a photographic plate shining lasers onto fragments that have a fluorescent dye incorporated into their structure.


a

b

c

d DNA fragments from test samples

Molecular weight standards Size (bp)

1

2

3

1500 1000 900 800 700

4 Wells in agarose gel

600 500 400 300 200

100

Figure 13.13 Gel electrophoresis of DNA. a The gel is made in a mould. The two combs shown leave two rows of wells for samples and standards. b The gel is placed in an electrophoresis bath where it is covered with buffer solution (an electrolyte). A power source is attached to the electrophoresis bath and switched on. c DNA fragments in the gel can be observed when the gel is placed in an ultraviolet light box. d The diagram shows how DNA fragments from four different samples and a set of standards have migrated. The molecular mass of the test fragment can be determined by comparison with the standards.

DNA profiling Scientists have identified the sequence of bases in human DNA (the genome) and are now decoding the genes. You will recall from Chapter 12 that proteins catalyse the many biochemical reactions occurring in our cells and serve as hormones, oxygen carriers and structural components. Because they contain the blueprint for the proteins we all need, the sections of DNA that code for proteins tend to be relatively similar in all humans. Only about 5% of the nuclear DNA carries the code for proteins that are so important for biochemical processes. Those regions of DNA that are not used for the coding of proteins, the ‘non-coding’ regions, are most useful to the forensic scientist. It is these sections of the polymer that can vary dramatically from person to person, enabling people to be identified on the basis of differences in their DNA.

chemfact The human genome project was completed in 2003. It determined the sequence of the 6 billion base pairs found in human DNA and determined that there were an estimated 20 000 to 25 000 protein-coding genes.

213 DNA

In many non-coding regions a sequence of bases may be repeated a number of times. For example, on human chromosome 5 the sequence AGAT may be repeated between seven and 15 times. The number of times this sequence is repeated varies between individuals. In DNA profiling, the number of repeats at ten locations across various chromosome pairs is considered to provide sufficient information to identify an individual accurately. A DNA profile or fingerprint can be used to identify suspects if they leave behind a biological sample, perhaps skin or blood under the nails of a victim. Just as importantly, DNA fingerprinting can be used to clear an innocent suspect. Forensic scientists also use DNA profiling data to: provide experimental data in court cases involving disputed paternity and identity help in identifying bodies recovered after natural disasters, such as the tsunami triggered by a massive earthquake under the Indian Ocean on 26 December 2004, and terrorist attacks, such as the 2002 Bali bombing identify the species of a plant or animal, e.g. birds being smuggled out of the country or cheap fish being substituted for more expensive species. DNA fingerprinting is a technology that provides a unique identification of individuals with the exception of identical twins, based on individual differences in base sequences present in the non-coding parts of DNA. It uses electrophoresis to produce a unique series of bands on a gel, each band being a fragment of DNA of a different length. The series of bands produced from DNA collected from a crime scene, for example, is compared with the bands produced from the DNA obtained from suspects. The degree of matching of the bands is used as part of the evidence to prove that a suspect was or was not at the scene of the crime.

Forensic techniques DNA profiling involves the isolation of a sample of DNA and treating this with restriction enzymes. Restriction enzymes act like molecular scissors, cutting DNA molecules everywhere a particular base sequence is encountered (Figure 13.14). Each restriction enzyme recognises a specific sequence of bases. Restriction enzyme Recognition site DNA sequence

5′

~ C–G–G A–A–T–T–C–T –C ~ G–C–C–T–T–A–A G–A

Two DNA fragments

3′

– G 3′

~ ~

Figure 13.14 The restriction enzyme EcoR1 cuts DNA polymer chains when it recognises the base 5′-CTTAAG-3′ pair sequence . 3′-GAATTC-5′

5′

Restriction enzyme cuts here

5′ A – A – T–T A–A– T–T –C– T 3′– G – – C A – ~ 3′ G~ 5′ 5′ ~ C – G – G 3′ 3′ ~ G – C – C – T – T – A – A 5′

Thousands of DNA fragments with varying sizes are produced. The fragments are then: 1 duplicated using a technique known as polymerase chain reaction 2 separated using gel electrophoresis (Figure 13.13) 3 blotted onto a suitable medium, such as a nitrocellulose or nylon filter 4 labelled with a radioactive piece of DNA or fluorescent dye that will bond to its complementary base sequence in a fragment 5 exposed to an X-ray film or UV light.


The pattern of bands on the DNA fingerprint can then be compared with the pattern of DNA profiles from known sources. For example, the determination of a paternity case relies on the fact that a child will inherit DNA from both parents as intact chromosomes. The child’s DNA fingerprint would therefore be a composite of those of each parent. Bands present in the child’s DNA fingerprint must be present in either the mother’s or the father’s profile.

chemistryy in action Immigration status challenged 1985 saw the first case to be decided on the basis of evidence provided from analysis of DNA. It involved a dispute between the immigration authorities in the United Kingdom and a teenage boy. The boy, Andrew, claimed that he had returned to his mother Christiana, in Britain, after a long stay in Ghana with his father. The immigration department suspected that the boy was Christiana’s nephew, rather than her son, and were disputing his right to remain in the United Kingdom. The legal aid centre that was providing assistance to the family asked Sir Alec Jeffreys, the scientist who pioneered the technique of using DNA to prove paternity, to conduct DNA tests on Andrew, Christiana and Christiana’s three acknowledged children. The results are shown in Figure 13.15, as well as the DNA fingerprint of an unrelated individual. Sister 1

Sister 2

Mother

Andrew

Brother

! Professor Alec Jeffreys discovered that noncoding DNA base pairs repeated a number of times while he was conducting research aimed at finding the DNA code of myoglobin, a protein found in muscle tissue.

Unrelated

Figure 13.15 DNA fingerprints of Andrew and his family.

It was not possible to obtain a DNA fingerprint from Andrew’s father or from Christiana’s sister, whom the immigration department suspected of being Andrew’s mother. A comparison of bands in the DNA fingerprints from Andrew and Christiana revealed that 25 were the same, providing strong evidence that they were mother and son. This match could only occur in 1 of 600 000 cases were Christiana Andrew’s aunt rather than his mother. Further comparisons with the other children’s DNA fingerprints only strengthened the family’s argument, and Andrew was allowed to stay in Britain with his mother, brother and sisters.

215 DNA

child

suspect father 1

suspect father 2

Electrophoresis

mother

Figure 13.16 Who is the child’s father? Bands present in the child’s DNA fingerprint must be present either in the mother’s or the father’s DNA profile.

Polymerase chain reaction (PCR)

! When amplified fragments are separated, each band consists of DNA fragments that have the same length and base sequence and hence molecular mass.

chemfact Analysis of repeating base sequences on the Y chromosome can be used to determine biological relationships between males.


The polymerase chain reaction (PCR) is a technique that takes advantage of DNA’s ability to produce an identical replica molecule (p. 209). By alternately heating and cooling a sample containing a DNA molecule, millions of copies of a segment of the molecule can be made in a matter of hours. Forensic chemists use this property to generate thousands of new copies of a DNA sample recovered from a crime scene, victim or suspect. PCR has enabled forensic scientists to work with extremely small samples of DNA, such as might be extracted from a single hair or a spot of dried blood left at a crime scene, or when they are analysing human remains that have been buried for many years. PCR is used to replicate DNA fragments. The PCR cycle involves three stages: 1 Denaturation. The sample is heated to 95°C for 1–5 minutes. Hydrogen bonding between the DNA strands is broken. 2 Annealing. The mixture is cooled to 55°C for about 1 minute. Primers bond to the start and end of the sequence to be copied. 3 Elongation. The mixture is heated to 72°C for about 1 minute. Complementary base pairs are added to the single DNA strands to form a double strand of DNA. This reaction is catalysed by the enzyme polymerase and involves the formation of hydrogen bonds. Each heating and cooling cycle doubles the number of copies of each DNA fragment, so that after 35 cycles there will be 236 or approximately 68 billion copies of each DNA fragment. A fluorescent dye is added to the DNA fragment during this process. The fragments are separated into bands according to their length by using gel electrophoresis. The position of each band is identified using computer-controlled lasers. Today DNA profiling is automated and it is possible to analyse base sequences at a number of chromosome locations simultaneously. A computer printout of a PCR analysis is shown in Figure 13.18. The technique is so sensitive that samples of DNA in the order of 1 nanogram may be used. The technique also enables damaged DNA, for example from decomposed bones, to be used.

!

1 DNA is extracted from blood, semen, hair roots and other body tissues.

Visit a site with a simulation of PCR via hi.com.au

2 Segments of DNA are synthesised using a process called the polymerase chain reaction (PCR). Each segment is bonded to groups of atoms which fluoresce.

Annealing

Denaturation

PCR Elongation

3 The DNA segments are separated into bands during electrophoresis.

Gel DNA sample

Segments move according to size to form bands

Electric current is passed through gel

4 In an automated process, the positions of the DNA segments are detected by their fluorescence when exposed to light of various energies.

Light source

Figure 13.17 Making a DNA profile using PCR.

217 DNA

Plots - ST1 16.11.00 GS2 FDI GFa 120 120

A base sequence is repeated 17 times at the same location on both chromosome 3s.

140 140

s

160 160

VWA

260 260

280 280

300 300

1500 1000 500

FGA

18

21

D21S11

1500 1000 500

D18S51

D8S1179 30

12 14

13

17

Chromosome and site identifier.

11 Yellow

D5S818

Figure 13.18 A computer-generated DNA profile showing base sequence repeats at 10 locations.

240 240

11 Green

Amelogenin

s

220 220

16 18

17

X Y

200 200

11 Blue

D3S1358

s

A base sequence is repeated 11 times on one chromosome 5 and 12 times at the same site on the other chromosome 5.

180 180

11 12

D13S317

11

D7S820

1500 1000 500

8 10

extension Mitochondrial DNA

chemfact Whereas the average number of base pairs in human nuclear DNA is approximately 6 billion, mitochondrial DNA (mtDNA) contains fewer than 20 000 base pairs.

chemfact Tsar Nicholas II of Russia and his family were executed in 1918 during the Russian Revolution. The analysis of mitochondrial DNA was used to confirm that the identity of skeletons found in a mass grave 75 years later were those of the tsar, his wife and three of their daughters. The same technology was used to prove that a woman who, throughout her life, claimed to be the tsar’s youngest daughter, Anastasia, was an impostor.

questions

As well as the 46 DNA molecules that are the chromosomes in the nucleus of your cells, there are smaller DNA molecules in cell organelles called mitochondria. The mitochondria are found outside the nucleus, and control energy-releasing reactions within the cell. Although there are only a few locations in mitochondrial DNA (mtDNA) that can be used for profiling, this additional DNA can be a valuable source of material for the forensic scientist for a number of reasons: 1 Some cells that are commonly the only available biological material, such as red blood cells and the cells in shafts of hairs, do not have a nucleus. 2 Each cell may contain many thousands of copies of the mtDNA molecule whereas it contains only one copy of each of the 46 DNA molecules. This means that if the sample has deteriorated, as may occur if a body part is recovered after many years or an archaeological specimen is examined, there is a better chance of extracting enough mtDNA, rather than nuclear DNA, to use. Great care must be taken when assessing data collected using this technique. It is highly prone to being contaminated by mtDNA from an external source. Mitochondrial DNA was first used in the United States to convict a man of the brutal rape and murder of a four-year-old child in 1986. A hair found on the body of the child provided the source of his mitochondrial DNA. All people who share the same mother, maternal grandmother, etc. will have the same mtDNA profile because mtDNA is inherited only via the egg cell at conception.

?

E1 Why would the hair sample found on the murdered child not be suitable for producing a DNA profile similar to the one shown in Figure 13.18? E2 Suggest the ways in which mitochondrial DNA might become contaminated from an external source before analysis. E3 Discuss the similarities and differences in mtDNA of Tsar Nicholas, his wife, their daughters and the imposter.


chemistryy in action DNA databases In the New South Wales town of Wee Waa in 2000, the search for the criminal responsible for the sexual assault of an elderly lady led to a call for all males between the ages of 18 and 45 in the town to be DNA fingerprinted. After police had taken samples from 500 men, the culprit confessed his crime, even before his sample was analysed! The first national DNA database was established in the United Kingdom in 1995, and by 2000 they had a register that included over one million people, mostly convicted criminals or people suspected of committing a crime. In Victoria a similar database was commenced by 1997, and a common Australian database was set up in 1998.

summary

13.4

RRegions i off DNA that th t do d nott code d for f proteins t i are usedd in i forensic analysis. Sections of non-coding DNA consist of particular base sequences that are repeated a number of times. The number of repeats varies between individuals. The base sequence at a number of sites (loci) is determined to increase accuracy.

The Th polymerase l chain h i reaction ti (PCR) technique t h i produces d millions of duplicate copies of DNA fragments. Individuals can be identified by the separation pattern of DNA fragments produced by electrophoresis and comparison with fragments from a known source. Electrophoresis is the separation of charged particles under the influence of an applied electric field.

key question 5 Make a flow chart of the steps involved in producing DNA profiles such as the one shown in Figure 13.15 or 13.16.

219 DNA

13

key terms adenine annealing base pairs chromosome complementary base pairing condensation polymerisation

cytosine denaturation deoxyribose deoxyribonucleic acid DNA database DNA profile double helix electrophoresis

DNA structure and function 6 a Draw structural formulas for the bases adenine and thymine, showing clearly how hydrogen bonds form during complementary base pairing. b Draw structural formulas for the bases cytosine and guanine, showing clearly how hydrogen bonds form during complementary base pairing. 7 Sketch a single nucleotide that includes the base thymine. 8 A segment of DNA includes the base sequence GATTATCAA. List the base sequence on the complementary strand of the molecule. 9 a List three diseases that result from changes in the base sequence on a gene. b By referring to the heritable disease sickle-cell anaemia, explain how a small change in the nucleotide sequence in a protein-coding region of a DNA molecule can have serious biochemical consequences.

DNA profiling 10 a What is a repeating base sequence? b Give an example of a repeating base sequence, including the number of the chromosome on which it is found. c Explain the significance of repeating base sequences to forensic analysis. 11 Describe the ways in which those sections of DNA used for forensic analysis can differ from individual to individual. 12 Explain why a DNA molecule is able to produce a replica of itself. 13 How are DNA databases useful for forensic analysis? 14 Why might the reliability of DNA fingerprinting be questioned and a profile be withdrawn as evidence in a court case? 15 ‘A database of DNA fingerprints of everyone in the country should be established.’ Discuss the merits and disadvantages of this proposal with your class members. You might consider the following questions in your discussion. a Should such a database contain both coding and non-coding DNA sequences?


elongation gene genetic code genome guanine histone mitochondrial DNA nucleotides

phosphate group polymerase chain reaction (PCR) polynucleotide purine base pyrimidine base restriction enzyme thymine

b Who should have access to a DNA database? police and security services? private organisations such as insurance companies? 16 a How does mitochondrial DNA (mtDNA) differ from nuclear DNA? b Give an example to illustrate the usefulness of mtDNA in forensic analysis.

Connecting the main ideas 17 Distinguish between the primary, secondary and tertiary structures of DNA. 18 a Outline the main steps in the forensic use of DNA. b Why is PCR such an important tool in forensic analysis? 19 Figure 13.19 shows a small segment of a strand of DNA. a Identify that part of the strand that represents a nucleotide. b Name the three species that have combined to form the nucleotide. c What type of reactions are involved in the formation of nucleotides? O O

P

NH2

O–

N N

O CH2

O

N

N

O O

P

NH2

O–

N

O CH2

O

OH

Figure 13.19 Small section of DNA.

N

O

1

chapter 14

From organic molecules to medicines

key knowledge • function of organic molecules in the design and synthesis of medicines including the production of aspirin from salicylic acid

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe how drugs are synthesised using organic reaction pathways • name the reagents used in the synthesis of aspirin and write an equation for this reaction • describe how aspirin works • list the therapeutic uses of aspirin and the side effects associated with its use • understand how changes in the molecular structure of a drug affect its effectiveness • describe the factors that influence the choice of reaction pathway used to synthesise drugs • recall that some proteins are used for therapeutic purposes and that these proteins are manufactured using genetic engineering technologies • describe the processes that lead to the discovery of drugs • describe how modern analytical techniques are used to determine the structure of drugs.

One very useful application of organic reaction pathways is in the design and synthesis of medicines. The main ingredients of medicines are drugs— substances that affect how the body functions. A drug is any substance that alters a chemical process in the body. We take some drugs for medicinal purposes, e.g. painkillers, antibiotics, sedatives. Some drugs such as caffeine are found in the foods and drinks we consume. It is legal for people over 18 to use drugs such as nicotine or alcohol even though they have harmful effects. Other drugs such as the narcotic heroin or the hallucinogen LSD are illegal to use.

14.1

! Many Aboriginal bush medicines contain biologically active compounds. For example, there is a type of convolvulus (botanical name Ipomoea pes-caprae) growing on the edge of the sandy coastline of the Northern Territory that has been found to contain a substance antagonistic to (acts to block the action of) box jellyfish venom. The locals also use the heat-softened leaves to treat insect bites, to drain pus from infected sores, and to stop bleeding. A number of plants used as traditional Aboriginal medicines are being investigated to find sources of new medicines.

Aspirin: a modern painkiller from a herbal remedy

Figure 14.1 Ipomoea pes-caprae is a traditional bush medicine used in coastal regions of the Northern Territory for its soothing antihistamine and anti-inflammatory properties.

chemfact The name ‘salicylic acid’ comes from salix, the Latin for willow. The systematic chemical name for salicylic acid is 2-hydroxybenzoic acid.

! An infusion is made by soaking or boiling plant material in water and then straining off the resulting solution, just like making tea. You drink the solution.


Pharmaceutical products are often developed from substances found in a plant that has been used as a traditional medicine. Aspirin is one such substance. Its origins are from a naturally occurring substance called salicin found in the leaves and bark of willow trees and in the herb meadowsweet. As long ago as 400 BCE, the Greek physician Hippocrates recommended ‘an infusion of willow leaves and bark to relieve aches, pains, inflammation and fever’. It was even thought to help with the pain of childbirth. The active ingredient in willow bark, salicin, was isolated in 1829. It is now known that the body converts salicin into salicylic acid and this is the active substance that helps to reduce fever and acts as a painkiller. Salicylic acid is a more effective painkiller (analgesic) than salicin, and was prepared from salicin a few years later. By 1870, doctors were prescribing salicylic acid directly, but the acidic nature of the substance gave it a very unpleasant taste and irritated the lining of the mouth and stomach. In 1897, Felix Hoffmann, a chemist with the Bayer Company in Germany, synthesised an improved modification of salicylic acid. He was motivated to do this to help his father, who suffered severe rheumatoid arthritis but could not tolerate taking salicylic acid to help his pain. Hoffmann’s starting material

was an extract from coal tar—the black, sticky substance that is produced when coal is heated in the absence of air. Once he had made salicylic acid he replaced the hydroxyl functional group with an ester functional group to form acetylsalicylic acid (Figure 14.4). This is the compound known commercially as aspirin. The production of a substance to be used as a medicine usually requires a number of chemical steps, known as a synthetic pathway. You should recall that the general chemical reaction to form an ester is: carboxylic acid + alcohol –—➤ ester + water

This reaction could be used to form acetylsalicylic acid. Interestingly, it is the hydroxyl group in the salicylic acid molecule that functions as the alcohol in a reaction with ethanoic acid (acetic acid) to form acetylsalicylic acid and water. This is a slow reaction. Also the yield is low, as the water formed tends to drive the reaction backwards. In an alternative reaction pathway, which is faster and produces higher yields, the ethanoic acid is replaced with ethanoic anhydride (acetic anhydride). This is the preferred pathway for the synthesis of aspirin (Figures 14.3, 14.4). O–H CH3

C

H–O

O

+

C

O

CH3

CH3

C

C

O

O

CH3 + H2O

O

Ethanoic acid

Ethanoic acid

Water

Ethanoic anhydride

OH

O

O–H

O

C O

C

+

O

C

CH3

O

O–H

+

CH3

C O

C O Salicylic acid

Figure 14.3 Formation of ethanoic anhydride.

OH

O CH3

C

Figure 14.2 Felix Hoffmann, who first synthesised aspirin in a chemically pure form in 1897.

CH3

Ethanoic anhydride

Acetylsalicylic acid (aspirin)

Ethanoic acid (acetic acid)

Figure 14.4 Structural equation for the preparation of acetylsalicylic acid (aspirin) from salicylic acid and ethanoic (acetic) anhydride. Unreacted ethanoic anhydride is converted to ethanoic acid by the addition of water.

The products, acetylsalicylic acid and acetic (ethanoic) acid, have to be separated and the product purified before it can be put into tablet form and packaged for sale. Acetylsalicylic acid is virtually tasteless and is much less irritating to the stomach than salicylic acid itself. It is not until after it has passed through the stomach that it reacts with water (hydrolyses) in the alkaline conditions in the small intestine and returns to the more effective salicylic acid.

! Ethanoic anhydride is prepared by combining two molecules of ethanoic acid (Figure 14.3). A water molecule is also produced.

chemfact During World War I, Britain was unable to obtain aspirin from its German manufacturer. In 1915, the British government offered a reward of £20 000 to anyone who could develop a workable manufacturing process for aspirin so that the country could make its own supply. The reward was won by a Melbourne pharmacist, George Nicholas, who went on to market aspirin under the name ‘Aspro’. The name ‘Aspro’ comes from Nicholas Products.

COOH

COOH O O

C

+ H2O CH3

Acetylsalicylic acid (aspirin)

OH– OH + CH3COOH in small intestine

Salicylic acid

Ethanoic acid

Figure 14.5 Acetylsalicylic acid is hydrolysed in the small intestine to salicylic acid.

223 From organic molecules to medicines

Soluble aspirin

COO–Na+ O

Although the molecule has a –COOH functional group, pure acetylsalicylic acid (aspirin) is not very soluble in water. However, converting the carboxylic acid functional group into the sodium salt changes the molecule into an ion and makes it much more soluble. It is used in many headache and cold remedies in this form. Buffered aspirin provides further protection against stomach irritation.

C

O

CH3

Figure 14.6 Sodium salt of acetylsalicylic acid.

PolyAspirin: a polymer to prevent pain

!

A recent development is to make a polymer structure using a condensation reaction between salicylic acid and 1,8-octanedioic acid (suberic acid).

PolyAspirin was accidently discovered by Dr Kathryn Urich, who was trying to develop new drugs for the treatment of brain cancers.

O

O H

O

C

O O

O

H

C O

n

H

H +

H

O

C

CH2

CH 2

CH 2

CH 2

CH 2

CH 2

C

O

H

O

+

Suberic acid (a diacid)

Salicylic acid

O

O

O O

C

O

O C

O

C

CH2

CH2

CH2

CH2

CH2

CH2

C

O

+

2n H2O

PolyAspirin n

Figure 14.7 Formation of PolyAspirin.

chemfact In salicylic acid and acetyl salicylic acid, some of the hydrogen atoms on the benzene ring have been replaced by other functional groups. Benzene undergoes substitution reactions where hydrogen atoms are replaced rather than addition reactions where the double bonds are broken.


This larger molecule passes through the stomach to the small intestine, where it enters the bloodstream before it is hydrolysed to salicylic acid. This means that it doesn’t irritate the stomach. PolyAspirin will need to undergo clinical trials before it is released for general use. PolyAspirin has a number of potential advantages and applications: It can be used as a controlled-release painkiller because the polymer breaks down slowly. Because it is a polymer with a similar molecular structure to polyesters, it can be made into thread and used to stitch cuts or wounds together. As it breaks down slowly, it can fight pain and reduce inflammation exactly where it is needed. It has the potential to be used as a plastic coating for an injured bone or joint. Polymers of this type have been shown to promote bone growth so it could help the healing process while easing pain.

extension How does aspirin work? Aspirin is an example of a mild analgesic. It acts to relieve pain and fever by blocking the enzyme-controlled production of prostaglandins. These substances are a type of hormone that controls body temperature by causing blood vessels to constrict. This results in less heat escaping from the body tissues into the blood. Prostaglandins also increase the permeability of capillaries, allowing water to pass from blood capillaries into tissues where it causes swelling and pain. By lowering the concentration of prostaglandins, aspirin acts to reduce inflammation, pain and fever. Aspirin has been found to have a much wider range of applications than simply reducing pain and fever. Aspirin can prevent the formation of blood clots and has been used in the treatment of patients who have suffered certain types of strokes and after a heart attack. It may also be useful in preventing further heart attacks and strokes and in protecting against certain types of colon cancer and the onset of Alzheimer’s disease. Aspirin has some negative side effects as it can irritate the stomach. Some people who are allergic to aspirin can suffer a life-threatening reaction called anaphylactic shock. Children or adolescents infected with the influenza or chickenpox virus should not be given aspirin because they can develop a reaction called

questions

Reye’s syndrome, which can cause fatal brain and liver damage. Its effect on blood clotting is sometimes desirable and sometimes dangerous. Aspirin belongs to a class of pharmaceuticals called nonsteroidal anti-inflammatory drugs (NSAIDs). These substances are thought to work by inhibiting the cyclo-oxygenase (COX) enzymes that are needed to make prostaglandins. However, there are two different forms of this enzyme. Cyclo-oxygenase 1 (COX-1) is present in the stomach and produces prostaglandins that protect the stomach lining. Cyclo-oxygenase 2 (COX-2) is present only at the sites of inflammation. As aspirin inhibits the production of both COX-1 and COX-2 it can cause stomach irritation as well as treat pain, fever and inflammation. You may recognise the names of painkillers such as Ibuprofen and Naprosyn. Like aspirin, these drugs inhibit COX-1 and COX-2 and so can cause stomach problems, and also affect blood clotting, although less strongly. An ideal painkiller would inhibit only COX-2, but in 2004 Vioxx, one of the first COX-2 inhibitors, was withdrawn from use as there was concern that long-term use significantly increased the risk of heart attack or stroke.

?

E1 List the ailments that can be treated by using acetylsalicylic acid (aspirin). E2 Why should someone with stomach ulcers not take aspirin?

summary

14.1

D Drugs, including i l di medicinal di i l drugs, d are chemical h i l substances bt that alter the way the body functions. Drugs are synthesised using reaction pathways involving organic molecules. Aspirin is synthesised from salicylic acid and ethanoic anhydride.

Aspirin A i i has h a number b off therapeutic th ti uses as wellll as some undesirable side effects. The effectiveness of a drug can be altered by changes to its molecular structure.

key question 1 a Draw a flow chart showing the reaction pathways for the preparation of aspirin from salicylic acid using: i ethanoic acid ii ethanoic anhydride.

b Explain why the pathway using ethanoic anhydride is preferred over the one using ethanoic acid. c Which pathway could be considered greener because it produces less waste?



Computer-aided The page A header drug design goes here by Jonathan Baell am a practising medicinal chemist. A lot of people of ask me what medicinal chemistry actually is; my reply is that it is the science of organic chemistry applied to the discovery of new pharmaceuticals. It was only towards the end of second year in university that I began to realise that organic chemistry was ‘my thing’ and so I pursued it regardless of whether there was a career afterwards or not. Fortunately, of course, there is. I think it’s important to follow what you enjoy and are good at, and the rest will generally follow. My Institute’s drug discovery facility has only been going since around 2002. It is involved in the very first stages of the process of discovering new drugs. We run high-throughput screens on our in-house library of 100 000 different compounds to find those that bind most tightly to our target. We can visualise this interaction

I

Figure 14.8 The crystal structure of the enzyme trypanothione reductase (TR) from the parasite Trypanosoma cruzi. Only backbone atoms are shown and coloured according to secondary structure.

226 226 O Organic Organic chemical chemical pathways pathways

using protein X-ray crystallography. Computer-aided drug design can help us develop improved compounds from our screening hits. Our aim is to develop compounds that have possible use as drugs, then to license this intellectual property to a pharmaceutical company that can fund clinical trials to test whether our compound is therapeutically useful. Our current main focus is on cancer, multiple sclerosis, and (sub)tropical diseases such as malaria and leishmaniasis, all terrible diseases in dire need of better treatment. It’s early days as yet, but we’re optimistic we will ultimately play an important role in better treatment of these diseases. The importance of this is perhaps the reason why I’ve yet to catch that common Western disease—Monday-morning-itis. The primary knowledge required in our research is organic chemistry, but we also

bio o need the skill to be able to turn synthetic compounds into good pharmaceuticals. These are ones that you can take as a pill, that reach the target tissue where they need to act, and that don’t have unacceptable side-effects. It sounds simple, but the body has evolved numerous ways, such as metabolism in the liver, to remove substances that aren’t recognised as being useful. So I try to optimise synthetically how a molecule ‘fits’ or binds to the target receptor that may be responsible for a disease state. The compound needs to have high polarity so it is sufficiently soluble in water, but if it is too polar it will not be able to penetrate the cellular membrane. The compound also needs to be metabolically stable and non-toxic. This research is part of the reason why pharmaceuticals cost so much. Not only does it cost a large amount of money to test compounds in clinical trials, but before this can take place, a huge amount of chemistry is required to find the best compound that is safe enough for people to ingest in clinical trials. It has been estimated that a new drug appearing on the market today represents in

Figure 14.9 Malaria parasite entering a red blood cell. Malaria image by Drew Berry, the Walter and Eliza Hall Institute of Medical Research.

excess of $US1 billion worth of investment, and drug companies need to recoup that before the life of their patent runs out. I and others are interested in using smart medicinal chemistry to reduce the attrition rate that leads to such costs. Figure 14.8 shows the enzyme trypanothione reductase (TR), from the parasite Trypanozoma cruzi that causes Chagas’ disease, with an estimated 100 million people at risk. TR inhibitors potentially could treat this and any other disease caused by trypanosome parasites. These diseases include African sleeping sickness, which is transmitted by the tsetse fly and can also cause a disease in cattle, called ‘nagana’. We found compounds that inhibit this enzyme by high-throughput screening of 100 000 diverse compounds, and knowing the crystal structure could help to optimise our hits. Fortunately, there are now charities that could help make any resulting TR inhibitor drug affordable to those people in poorer countries who need it.

Dr Jonathan Baell Jonathan Baell was born in Tanzania in 1965 but grew up mainly in England and then Tasmania. He obtained his BSc(Hons) in organic chemistry from the University of Tasmania and his PhD in 1992 from the Victorian College of Pharmacy in Melbourne. He then joined the CSIRO (1991–95, Sydney, then Melbourne), following this by a position as Senior Research Scientist in the Biomolecular Research Institute (Clayton, Victoria) until 2000, at which time he took up a position as a Special Fellow at the Walter and Eliza Hall Institute of Medical Research (WEHI), where he is presently based at the WEHI Biotechnology Centre in the La Trobe R&D Park at Bundoora. Here he is Co-Head of Laboratory and has a role in directing medicinal chemistry research with a focus on cancer, autoimmune diseases, and subtropical and tropical diseases. He is an Honorary Senior Research Fellow at the University of Melbourne and Group Leader, CRC for Cancer Therapeutics.

Figure 14.10 A person afflicted with the parasitic disease leishmaniasis.


14.2

! Fleming, Florey and Chain were awarded the Nobel Prize in Medicine for their ‘discovery of penicillin and its curative effect in various infectious diseases’.

Penicillin: a chance discovery The discovery and development of penicillin is a very good example of drugs resulting from chance observations in the laboratory, with chemists having sufficient insight to follow up unexpected events. Alexander Fleming was a British doctor who was conducting research on substances capable of overcoming bacterial infections. In August 1928, he returned from a holiday to find an unusual mould growing in a dish in which he had a culture of harmful staphylococcus bacteria. In the area around the mould, the bacteria appeared to be dying off. The mould turned out to be an unusual strain of penicillium, one that was different from the usual blue ‘fur’ that you see on mouldy bread or jam. However, when an extract was made from the mould, the first trials did not give very promising results so Fleming put aside this line of research. In 1940, Howard Florey, an Australian, and Ernst Chain, both working at Oxford University, decided to investigate the mould further. By this time improved processes has been developed that allowed them to extract larger samples of penicillin. Initially the substance was tested on mice for its antibiotic effects, and then on humans. In the developmental work, it was found that a number of different chemical substances were present in the extract obtained from the culture. Some were more stable and clinically effective than others. The actual chemical structure was not finally resolved until 1945, by Dorothy Hodgkin using X-ray crystallography.

chemistryy in action Early use of penicillin

Figure 14.11 Fleming’s culture plate. The mould on the bottom (Penicillium notatum) is overcoming the growth of the smaller staphylococci.


The first patient to receive a purified sample of penicillin was an Oxford policeman, Albert Alexander. He had developed severe streptococcal and staphylococcal infections from a sore on his mouth. He was desperately ill and all other possible treatments had been exhausted. On 12 February 1941, Alexander was given an intravenous injection of 200 mg of penicillin, followed by further 100 mg injections every three hours. His condition improved dramatically within 24 hours, and after five days he appeared to be on the way to recovery. However, the supply of penicillin was very limited. The research team even tried to recycle unused penicillin from the patient’s urine but there was not enough to maintain the treatment. Alexander suffered a relapse and died on 15 March of septicaemia (the effects of bacterial toxins and large numbers of bacteria in his bloodstream). You can see why the instructions on prescriptions for antibiotics always say that you must take the entire course of medication. You should not stop taking the medication as soon as the symptoms have disappeared because there might be a few remaining bacteria that will start multiplying again. The unfortunate first trial was soon followed by successes. By June 1941, five more seriously ill patients had been treated and four of them recovered completely. The only failure was a four-year-old boy who had severely infected spots from measles. The penicillin cured the bacterial infection in the spots but the boy died of neurological (brain and nerve) damage caused by the measles virus.

H

H

H

!

S CH2

C

N

C

CH3

C

C CH3

O

C

C

N

O

C OH

O

H

Pharmacological and non-systematic (generic) names, rather than the systematic chemical names, are used for many compounds throughout this chapter. The systematic chemical names for these complex substances are themselves very large and complex and are not widely used.

Penicillin G H

H

H

N

C

C

C

N

C

S CH3

CH2

CH

CH

CH2

C

CH3 CH3

O

C

O

C OH

O

H

Penicillin F H

H

H

N

C

C

S O

CH2

C

CH3 C CH3

O

C

N

C

O

C OH

O

H

Penicillin V

Figure 14.12 The molecular structures of penicillins F, G and V. The structures were worked out by Dorothy Hodgkin (Figure 12.29), who also worked out the structures of insulin and vitamin B12.

H

S C

The first extracts of penicillin isolated by Florey and his co-workers contained mainly penicillin F with small amounts of penicillin G. Changing the culture medium allowed greater amounts of the more stable (and hence more effective) penicillin G to be formed. Adding ‘side-chain precursors’ to the culture medium resulted in the addition of new side chains to the basic penicillin molecule. A range of related molecules could now be produced and tested for suitability as an antibiotic. Out of these forms, penicillin V (phenoxymethylpenicillin) was found to be more stable in the presence of stomach acid. Although it was not as potent as penicillin G, it became the preferred antibiotic for general use as doctors could prescribe tablets for patients to take orally, instead of requiring a series of injections. In 1958, another penicillin molecule, 6-aminopenicillanic acid (Figure 14.13), was extracted from the culture medium. You can see that this substance lacks the side chains of the earlier forms of penicillin. It was an important breakthrough because chemists could use this molecule as the basis for preparing any other penicillin structure that they wanted to test. Even in 1940, while the various forms of penicillin were being introduced to fight bacterial infections, bacteria evolved to develop resistance. (Bacteria can reproduce every half an hour or so and hence evolve much faster than animals such as humans.) Knowledge of the chemical structure has allowed hundreds of related structures to be prepared and tested. Some of these substances, such as oxacillin (Figure 14.14) have structures with fairly bulky side chains. These are not recognised by the bacteria and so help prevent the development of bacterial resistance.

H

H2N

CH3

C

C CH3

C

C

N

O

C OH

O

H

Figure 14.13 The structure of 6-aminopenicillanic acid.

H

H

C

C

S

C N

C

NH

CH3 C CH3

O

C CH3

C

N

C

O

C OH

O

H

Figure 14.14 Structure of oxacillin.


chemistryy in action How do medicines work? There are two main ways in which a pharmacologically (biologically) active substance might work in the body: by attaching to a receptor site such as a nerve ending and blocking or changing the way in which it functions by interfering with a biochemical mechanism, often by inhibiting (stopping) the functioning of the enzymes that are needed to maintain some process. An example of the second mechanism is the way in which penicillin works. It stops the growth of new bacteria by inhibiting an enzyme needed to construct new cell walls. The way in which the substance acts in the body is often related to the overall shape of the molecule and to the presence of particular functional groups. The molecular structure is capable of attaching to a receptor site using some form of intermolecular bonding such as hydrogen bonding or dipole–dipole forces. Hence, once the structure of a molecule that acts as a useful medicine has been determined, other similar molecules can be made (synthesised) and their effectiveness tested. If you look at the structures of various forms of penicillin (Figures 14.12, 14.13, 14.14), you can see that one part of the molecular structure is always the same.

! There are two forms of diabetes. Insulindependent, or type I, diabetes often first appears in childhood and is caused by the body’s inability to manufacture insulin. Insulinresistant, or type II, diabetes usually appears in older people; their bodies produce insulin but do not respond to it properly. People who are inactive or obese are at higher risk of developing type II diabetes than active people of normal weight.

Figure 14.15 People with insulin-dependent diabetes selfinject insulin to manage the disease.


14.3

Proteins as medicines The treatment for a number of diseases involves the use of proteins as medicines. As you have discovered in Chapter 12, proteins are large molecules that have a complex structure. We do not yet have the technology to use chemical reaction pathways to manufacture therapeutic proteins. Pharmaceutical and biotechnology companies use recombinant DNA technology to manufacture a required medicinal protein using bacteria, yeast, fungi or animal cells that have been genetically modified. Insulin is a protein that controls the storage and use of glucose in the body. Insulin-dependent diabetes is a disease caused by the body’s inability to produce insulin, resulting in a build up of high levels of glucose in the blood (hyperglycaemia). If untreated this can lead to diabetic coma and to the sufferer developing complications such as heart, circulatory system and kidney disease, as well as blindness and gangrene. People with insulin-dependent diabetes require regular injections of insulin. In the past insulin was obtained from the pancreas of pigs and cattle. Today cloning or genetic engineering techniques are used to manufacture human insulin. The insulin gene is inserted into the DNA of a bacteria cell. The genetically modified bacteria are grown in vats containing a nutrient mixture. The cells multiply and the insulin is then extracted from the bacteria and purified.

Some bacterial cells, unlike human cells, contain plasmids—tiny circular pieces of DNA which are able to pass between cells

Plasmid

The human gene responsible for insulin production can be built up from insulin mRNA using viruses

Other bacterial DNA (in reality this is much bigger than the plasmid)

Plasmids can be extracted and cut with restriction enzymes which break a sugar—phosphate link in the DNA ‘backbone’

Human gene Other enzymes re-form sugar—phosphate links and ‘splice’ the human gene into the plasmid The modified plasmid is put back into bacterial cells

! A new technique for treating insulin-dependent diabetes is being trialled by Australian doctors. Insulin-producing cells coated with seaweed extract are injected into the body. Insulin seeps into the bloodstream while the seaweed coating prevents the body’s immune system from destroying the foreign insulin-producing cells. It is hoped that this technique will reduce the number of insulin injections diabetics need to have.

The cells multiply in the fermenter

The modified bacteria produce human insulin ( ) from the insulin gene

The protein is extracted and modified if necessary to give the final product Waste bacterial cells are destroyed

Figure 14.16 Steps in cloning for the production of insulin.

Other biosynthetic proteins that are used as drugs include: Etanercept, used in the treatment of rheumatoid arthritis, a crippling and painful disease of the joints. Etanercept blocks the enzyme that causes inflammation of the joints. Clotting factor VIII: People with haemophilia can suffer from uncontrolled bleeding because they do not produce sufficient quantities of the protein that causes blood to clot. Injections of clotting factor VIII are used to treat this heritable disease. Erythropoietin (EPO), a protein that controls the production of red blood cells. Oxygen and carbon dioxide are transported throughout the body by haemoglobin found in red blood cells. People who have a low level of red blood cells in their body suffer from anaemia. Synthetic erythropoietin is used to treat this disease by stimulating the production of red blood cells. It is also used to treat some kidney diseases.

! EPO has been used illegally by some athletes to increase the concentration of red blood cells in their blood and thus its oxygen-carrying capacity in order to enhance their performance in endurance events. High pressure gas liquid chromatography (HPLC) and mass spectrometry (MS) are just two examples of techniques used to detect the use of drugs in sport.

! Biotechnology companies are investigating the use of genetically modified plants such as tobacco, rice and corn to produce proteins for medical use. This is sometimes referred to as biopharming.

14.4

A cleaner, greener future? When chemists first developed synthetic pathways for making useful drugs in large amounts, their initial concern was just to get a good yield of pure product. Some of the methods used had many steps in the process and generated large amounts of by-products that resulted in a disposal problem. The processes were very wasteful in terms of many of the atoms of the reactants not being incorporated into the product and ending up in the wasted by-products. This is referred to as poor atom economy or poor atom utilisation.

review The idea of atom economy was explained in Chapter 17 of Heinemann Chemistry 1. It is a means of tracking all the atoms used in a reaction. Percentage atom economy is given by: molar mass of atoms used in the products × 100 molar mass of all reactants


H

H3C H3C

O

CH3

O

H3C 2-methylpropyl benzene

O Step 1

AICI3

O Step 1

O

H3C

H3C

O

HF

O CH3

H3C

O H CH3

CH3

COOC2H5

Cl

CH3

H

Step 2 NaOC2H5 O

H3C CH3

H3C

CO2C2H5 H3C Step 2

H2

Green pathway

Brown pathway H3O+

Step 3

CH3 O

OH CH3 CH3

CH3

H3C

H3C

NH2OH

Step 4 CH3 N

CH3

CO

OH

Pd H3C

Step 3 CH3 CH3 H3C

H COOH

Ibuprofen

Step 5

H O

H

O

H

Step 6

CH3

CH3 C

N

H3C

Figure 14.17 Comparison of two reaction pathways for the synthesis of ibuprofen. Atoms not utilised in the final product are shown in red.

Typical figures in the old (brown) processes would be 40% of reactant atoms ending up in the product and 60% being wasted. When this is translated into masses of substance, for example 70 million kilograms of the common analgesic ibuprofen (enough for 25 billion tablets) produced in the US alone in one year, the concern about over 100 million kilograms of useless, wasted by-products becomes obvious. However, as we have become aware of the need to make good use of all of our resources, chemists are beginning to work out greener reaction pathways that are more efficient in terms of using fewer steps in the synthesis, greater yields and better atom utilisation. Applying these principles to the commercial production of ibuprofen has improved the atom economy from 40% in the old, six-step brown method to 77% in the three-step green method. This figure can be considered to rise to 99% for the green method if the recycling of ethanoic acid by-product is


taken into account. This has made an enormous savings both in dollar terms and in environmental impact. A comparison of the two synthesis pathways is shown in Figure 14.17. The ‘green’ pathway is shown on the left.

! The principles of green chemistry are detailed in Chapter 18 p. 310.

14.5

Developing a new drug Substances that can potentially be developed into drugs are called ‘lead’ compounds. Some drugs have been developed from herbal or traditional remedies. The development of aspirin is an example. Other drugs such as penicillin were discovered by accident. Today pharmaceutical companies screen thousands of chemicals for biochemical activity. Some of these chemicals may be derived from plants and animals, or they may be known drugs that are tested for new applications. A study of the venom of Brazilian Arrowhead viper led to the development of captopril, a drug used to treat high blood pressure. The anti-AIDS drug, AZT, was developed from a chemical extracted from sponges found in the Caribbean. Taxol, a cancer medication, is derived from Californian yew trees. Other chemicals for screening for potential use as a drug are developed using an automated chemical process called combinatorial chemistry that makes thousands of molecules that might show pharmacological activity. Computer modelling has led to the development of specifically designed drugs that match the size and shape of active sites in specific target proteins. The anti-inflammatory drug Celebrex and the Australian-designed antiinfluenza drug Relenza (Figure 14.18) are two examples of drugs purposely designed to treat specific diseases. The development of these designer drugs depends on an understanding of the biology of the specific diseases and the identification of the three-dimensional structure of the target protein. Converting the starting material into the final drug requires one or more steps in a synthetic pathway. For each step the chemist has to deduce what reagents and reaction conditions to use. It is vital to pay attention to the principles of equilibrium to choose the best reaction conditions to maximise the yield. You will learn more about this in Unit 4. A suitable formulation has to be devised for administering the medicine. This involves practical issues such as whether the required amount of active ingredient should be delivered in a tablet or as a liquid; if it is to be swallowed, inhaled or injected, and, if it is to be injected, if it is to be injected into the muscles or intravenously. Solubility and taste play a part in making these decisions. The potential medicine is tested on cell cultures and on animals to establish effectiveness, suitable doses and potential side effects. If the results look promising, the medicine then goes through a number of clinical trials using volunteers and patients. Care has to be taken to allow for variability in the responses from different individuals, as well as investigator bias. A great deal of use is made of statistics to help overcome these problems. Finally the results are submitted to the government’s Therapeutic Goods Administration for approval before the new drug is made available to the public. The development of a drug is expensive. About one in ten thousand compounds screened for pharmacological activity is actually developed into a drug for use. The identification, research, development and approval of a new drug can take 10–20 years and costs in excess of $A450 million.

OH C O

CH2 HC

CH

C

OH

CH

CH

HN

OH

CH2

C O

C

O

HO

NH CH3

NH C NH2

Figure 14.18 The structure of the anti-influenza drug Relenza. This drug blocks the function of a protein in the virus neuramidase that controls the spread of the virus in the body.


Analysis in action When a potential new drug molecule has been synthesised in the laboratory, tests have to be performed on a pure sample of the compound to determine whether the structure is exactly as intended. Instrumental methods such as mass spectroscopy, infrared spectroscopy, nuclear magnetic resonance spectroscopy and X-ray spectroscopy all contribute information that together confirms the exact structure of the substance. HO

O

Infrared spectroscopy

C O

Look carefully at the molecular structure of acetylsalicylic acid shown in Figure 14.19. There are two carbonyl, C=O, functional groups present in the molecule. You can see from the structural formula in Figure 14.19 that one is part of the ester functional group, the other is found in the carboxyl group. These correspond to the two absorptions at 1760 and 1770 cm–1 in the IR spectrum.

C

O

CH3

Transmittance (%)

Figure 14.19 Molecular structure of acetylsalicylic acid showing the carbonyl functional groups.

1770

4000

3000

2000

1760

1600

1200

800

Wavenumber (cm–1)

Figure 14.20 Infrared spectrum of acetylsalicylic acid.

O COOH H

O—CC

H

Nuclear magnetic resonance (NMR) spectroscopy

H H H

H H TMS

14

12


2

Figure 14.21 Proton NMR spectrum of acetylsalicylic acid.


0

You may remember from Chapter 7 that proton NMR shows the proportions of hydrogen atoms (protons) in different environments in the molecule. Look again at Figure 14.19 and count the number of hydrogen atoms in each environment. You should be able to identify four Hs around the benzene ring, three Hs in the CH3 group and one H on the OH group. The four hydrogen atoms attached to the benzene ring (shown in red) give the four separate peaks lying between 7.0 and 8.5 ppm. The three hydrogen atoms in the CH3 group (blue) all experience the same environment so they appear as the large peak at 2.3 ppm. The single hydrogen atom in the COOH group (green) would give a small peak; this is not shown in the spectrum.

Mass spectroscopy The peak at 180 is the molecular ion peak, Mr(C9H8O4) = 180.0. The masses of the various fragments are used to confirm the structural formula. 120

10 000

Abundance

8000 6000

138

92

4000

42 64

2000

93

180

Figure 14.22 Mass spectrum of acetylsalicylic acid.

0 0

20

summary

40

60

80 100 Mass/charge

120

140

160

180

14.2–14.5

Choice of reaction pathway path a to synthesise s nthesise a drug dr g is determined by considerations of rate, yield and waste minimisation. Drugs are discovered via a variety of means including traditional remedies, accident, screening for biological activity, and by specific design. The molecular structure of a drug is altered when groups of atoms are added or removed. The effectiveness of a drug can be modified by changing the molecular structure.

Penicillin is an antibiotic originally originall extracted e tracted from mould. mo ld Various types of penicillin have differing molecular structures. Some proteins such as insulin are used as drugs to treat disease. Genetic engineering technologies are used to manufacture medicinal proteins. Modern analytical techniques such as IR, NMR, mass spectroscopy and X-ray spectroscopy are used to determine the molecular structure of drugs.

key questions 3 Why do the instructions on prescription antibiotics tell you to complete the whole course of tablets? 4 Why were antibiotics like penicillin V preferred to penicillin G? 5 What information about the molecular structure of a drug does each of these analytical techniques provide? a infrared spectroscopy b nuclear magnetic resonance spectroscopy c mass spectroscopy

6 Medicinal proteins such as insulin are administered by injection. Suggest why they cannot be given orally as tablets or capsules.



Designer medicines: The page A header molecules goes hereof the future by Jenny Martin Recent developments in drug biotechnology mean it is now possible to design molecules that have a very specific biological activity. Molecules designed in this way are aimed at producing drugs with fewer side effects and less chance of toxicity. My research at the Institute for Molecular Bioscience at the University of Queensland focuses on protein structure and its application to drug design. When we know more about biological molecules and how they interact, function and malfunction, we can begin to develop new medicines, including for diseases that are currently untreatable. The process of developing a drug from a chemical on the laboratory bench to a product on the pharmacy shelves has many stages. Estimates are that the chances of a drug making it through the process are 1 in 10 000, costs are on the order of

Figure 14.23 Foxglove has been used as a medicine since ancient times. The active ingredient, digitalis, is still used today as a heart stimulant.

236 236 O Organic Organic chemical chemical pathways pathways

$US300–500 million, and the time required is 10–20 years. However this may change in the future. The later stages of drug development are regulated by the Therapeutic Goods Administration, to ensure that drugs making it to market are safe and effective. It is not possible to reduce the costs or time at these stages, but there are ways to reduce them at the early stages, specifically at the drug discovery step such as the identification of ‘lead compounds’. Lead compounds is the name researchers give the small molecules that serve as the starting point for the development of drugs. (The term should not be confused with compounds of the metal lead.) Historically, we have identified lead compounds in a variety of ways: • traditional use of natural products

bio o • a chance observation (for example penicillin, p. 228) • random screening and testing of tens of thousands of chemicals until one is found with the desired biological activity. We can be smarter about the way in which we discover candidates for new drugs, using a technique called structurebased design. This reduces the odds of finding a drug from 1 in 10 000 to 1 in 300. There are certain prerequisites for structure-based design to work: the disease biology must be understood, a target protein must be identified, and its threedimensional structure must be known. There are currently only a small proportion of diseases that meet these criteria. Protein structures are most often determined using X-ray crystallography. When the structure is known, it allows us to visualise hot spots in the protein, active sites where the enzyme does its work. Once we know what the hot spot looks like, we

can design drugs that match the size, shape and physical and chemical properties of the hot spot. We are only now starting to see the first structure-based designed drugs making their way into the market. Structure-based drug design will provide us with the tools for designing better drugs, and to combat diseases that are currently untreatable. These methods herald a new era in drug discovery. A recent example of structure-based drug design is Relenza, an Australiandesigned drug to treat influenza, in a collaboration between research groups of Peter Colman, Mark von Itzstein and Graeme Laver. My work has focused on the adrenaline-synthesising enzyme recently. We solved the structure of the enzyme and now use structure-based methods to develop lead compounds. Access to synchrotron radiation is critical to the success of this research.

Professor Jenny Martin Dr Jenny Martin is a pharmacist who followed a research career path. She has a Master of Pharmacy from the Victorian College of Pharmacy and a PhD from the University of Oxford. After postdoctoral research at Rockefeller University in New York, she returned to Australia to establish the first protein crystallography laboratory in Queensland. Her research at the Institute for Molecular Bioscience in the University of Queensland focuses on protein structure and drug design. In 2009 she was awarded a prestigious Australian Research Council Australian Laureate Fellowship.

Figure 14.24 Structure of the adrenaline-synthesising enzyme (green) showing how drug-like molecules (yellow, cyan) interact at hot spots.


14

key terms acetylsalicylic acid analgesic antibiotic aspirin culture

drug formulation genetic engineering insulin

non-steroidal anti-inflammatory drugs (NSAIDs) penicillin prostaglandin

13 Codeine, morphine and heroin are drugs derived from the opium poppy. The molecular structure of these drugs is shown in Figure 14.26. Identify the part of each molecular structure that is common to all three drugs. How does the structure of these molecules differ?

Aspirin 7 Why is the sodium salt of acetylsalicylic acid more soluble than normal aspirin? 8 Briefly explain how aspirin is able to reduce pain and fever. 9 Acetaminophen (paracetamol) is the active agent in some over-the-counter painkillers. It can be synthesised from p-hydroxyaniline (4-hydroxyaminobenzene) and acetic anhydride. An incomplete equation for this reaction is shown in Figure 14.25. a What by-product is formed in this reaction? b Determine the atom economy of this reaction. OH

receptor site salicylic acid synthetic pathway therapeutic protein

O CH2

CH2 OH

CH3

CH2

N

CH3

N

+

?

Acetic anhydride

p-Hydroxyaniline

N H

CH3

O

N

CH2

O

OCH3

CH3COCCH3

NH2

CH2

O O

Codeine +

OCCH3

OH

O

OH

O O

CH2

OH Morphine

OCCH3 Heroin

Figure 14.26 Molecular structures of codeine, morphine and heroin. Caution—it is not possible to convert codeine into morphine or heroin by any simple laboratory process.

C

! Acetaminophen

CH3

Figure 14.25 Synthesis of acetaminophen.

10 a The drugs celecoxib and rofecoxib were specifically designed to be COX-2 inhibitors. They do not affect the production of COX-1. Aspirin inhibits both COX-1 and COX-2. What are the benefits of these COX-2 inhibiting drugs over aspirin? b Find out why rofecoxib, marketed as Vioxx, was withdrawn from use by the pharmaceutical company Merck. 11 From the description of the use of the Australian native plant with the Aboriginal name of bunaangu, what active ingredient might be present to give it these medicinal properties? ‘You only eat the sap of the bunaangu. It has a bitter taste. You boil it up to drink. It helps toothache and a bad chest.’

12 As well as aspirin, analgesics include acetaminophen, ibuprofen, naproxen sodium, ketoprofen. Use the web to find out: a the names under which they are sold b their side effects c their chemical structure d how they are synthesised.


All three drugs, codeine, morphine and heroin, relieve pain and are sedatives. These molecules bind to the chemical receptors for pain located in the brain. By attaching to these sites, these strong analgesics stop the transmission of pain. Morphine and heroin are both addictive. Morphine is used under strict medical supervision. The medical use of heroin is banned in many countries due to its addictive nature. Medications containing codeine are available over the counter.

Proteins as medicines 14 Suggest some advantages and disadvantages of using genetically modified plants to produce therapeutic proteins. 15 Use the web to find out about the diseases that are treated by the following medicinal proteins: a growth hormone b interferon c monoclonal antibodies

Connecting the main ideas 16 Illustrate, using aspirin as an example, how biochemists modify the structure of drugs to eliminate undesirable side effects. 17 Explain why pharmaceutical companies manufacture drugs such as aspirin using chemical reactions, while biotechnological processes are used to produce drugs such as insulin.

3

unit

area of study 2

review

Organic chemical pathways

multiple-choice questions 1 The smallest molecule in which homologous series must contain two carbon atoms is: A an alkane B an alkanol C an alkene D an alkanoic acid. 2 The systematic name for CH3CH2CH2CH(CH3)2 is: A 1,1-dimethylbutane B 2-methylpentane C 2-methylpentene D propyldimethylmethane 3 Which compound would be expected to have the highest boiling point? A CH3CH2CH2OH B CH3CH2CH2CH3 C CH3CH2CH2Cl D CH3CH2CH3 4 Which of the following statements are true of the homologous series of primary alkanols? I The members differ by one CH2 unit. II They are all strong bases. III They can be oxidised to form carboxylic acids. A I and II B II and III C I and III D I, II and III 5 What is the product formed by the reaction of CH2CH2 with Br2? A CH2BrCH2Br B CH2CHBr C CH3CH2Br D CHBrCHBr 6 When ethanol is heated under reflux with an acidified solution of potassium dichromate, the final product has the formula: A CH3CH2CH2OH B CH3COOCH3 C CH3COOH D CH3CH2CH3 7 Consider the following reaction pathway: I II III ethene –—➤ chloroethane –—➤ ethanol –—➤ ethanoic acid The reactions that occur in steps I, II and III of the pathway are: A cracking, addition, hydrolysis B chlorination, substitution, addition C addition, hydrolysis, oxidation D addition, reduction, hydrolysis

8 Acetylsalicylic acid could be formed from salicylic acid by warming it with: A ethanol B methanol C ethanoic acid D methanoic acid COOH

COOH

OH

O

C

CH3

O Salicylic acid

Acetylsalicylic acid

(Chemically inactive hydrogen atoms have been omitted for clarity.)

9 Consider the following molecular structures of substances that all function as drugs. (Chemically inactive hydrogen atoms have been omitted for clarity.) CH3

O

O

CH3

N H3C Cl

C N

C

C

C

N C

N O

N

H

N

CH3 I

II H

O

N –O

N+

N

O

III

The substances that are expected to have similar pharmacological effects are: A I and II B II and III C I and III D I, II and III

239

239 Organic chemical pathways

Area of study review

10 The ester methyl ethanoate could be made by reacting together: A CH3CH2OH and CH3COOH B CH3CH2OH and HCOOH C CH3OH and CH3CH2COOH D CH3OH and CH3COOH 11 The correct systematic name for this substance is: CH

CH3

CH2

CH3

CH

CH3

CH2

CH3

A 2,4-dimethylhexane B 2-methyl-4-ethylpentane C 2-ethyl-4-methylpentane D 1,1,3-trimethylpentane 12 A carboxylic acid with five carbon atoms per molecule could have the formula: A C5H11COOH B CH3(CH2)4COOH C CH3CH2CH2CH2COOH D C3H7COOCH3 13 Which statement about compounds I and II is not true? H

H

H

H

H

H

C H

C C

H

C O

H

H

H

C C

H

H H

H

O

Compound I

Compound II

H

A The 13C NMR spectrum of compound I has three peaks, and that of compound II has two peaks. B The low resolution 1H NMR spectrum of I has four peaks, and that of II has three peaks. C The mass spectrum of both will show a peak at a mass-tocharge ratio of 60. D The fingerprint region of the IR spectra will be identical for both compounds.


Questions 14 and 15 refer to the following information. The following list gives the formulas of a number of important biochemical molecules. I C6H12O6 II CO(NH2)2 III C2H5NO2 IV C3H7NO2 V C17H33COOH VI C3H8O3 VII CO2 VIII C17H35COOH 14 From the list, identify two molecules that would react together to form a dipeptide. A II and III B III and IV C IV and VI D IV and II 15 When a saturated fat undergoes hydrolysis, two products that might be formed are: A VI and VIII B I and III C V and VI D II and VIII 16 When starch is formed from glucose, the condensation reaction that occurs involves: A a carboxyl functional group and an amino functional group B two hydroxyl functional groups C a hydroxyl functional group and a carboxyl functional group D a hydroxyl functional group and an amino functional group. 17 Olive oil contains a high proportion of oils made from a fatty acid called oleic acid. The formula of oleic acid is C18H34O2. The number of carbon-to-carbon double covalent bonds in oleic acid is: A 0 B 1 C 2 D 3 18 When a protein such as insulin is formed, the condensation reaction that occurs involves: A a carboxyl functional group and an amino functional group B two hydroxyl functional groups C a hydroxyl functional group and a carboxyl functional group D a hydroxyl functional group and an amino functional group. 19 Protein molecules are formed by condensation reactions between amino acids. The secondary structure of many proteins is a helical shape. The type of chemical bond responsible for maintaining this shape is: A hydrogen bonds B ion–dipole bonds C ionic bonds D covalent bonds.

short-answer questions 20 a What feature of the bonding of alkenes makes them so useful? b The presence of an alkene can be detected by shaking a sample with a dilute aqueous solution of bromine. i Describe what changes you would observe if an alkene was present. ii Write a balanced equation for the reaction of aqueous bromine solution with ethene. c i Give the name of an instrumental method that could be used to confirm that the compound was an alkene. ii Briefly outline what observation would be expected in the output of the method if the substance was an alkene. d Under suitable conditions chloroethene can be converted into a polymer. i What type of polymerisation process would you expect chloroethene to undergo? ii Write an unbalanced equation to represent the polymerisation reaction. iii Give the systematic name of the product. 21 a What is meant by the terms: i renewable? ii sustainable? b Assume that petrol is made entirely of octane. i Write a balanced equation to represent the combustion of octane in an abundant supply of air. ii Explain why the use of petrol as a fuel for vehicles is of environmental concern. c The use of ethanol as an alternative biochemical fuel to petrol is increasing. i Write a balanced equation for the combustion of ethanol in a good air supply. ii Why is the use of biofuel ethanol seen as being beneficial for the environment? 22 a Crude oil consists mainly of alkanes of varying molecular sizes. What feature of the intermolecular forces of alkanes allows the alkanes to be separated into fractions at an oil refinery?

b Ethanol is an alternative biochemical fuel to petrol. i Draw the structural formula of ethanol. ii Why does ethanol have an affinity for water whereas the alkanes in petrol do not? c E10 petrol contains 10% ethanol. Outline the environmental advantages of using E10 petrol. 23 A student attempted to prepare a sample of acetylsalicylic acid (aspirin) starting from salicylic acid. COOH

COOH

OH

O

C

CH3

O Salicylic acid

Acetylsalicylic acid

Structural formulas of salicylic acid and acetylsalicylic acid.

a Apart from salicylic acid (2-hydroxybenzoic acid), what other reagent is needed? b Write a balanced equation for the reaction. You may use either structural or molecular formulas. c Salicylic acid and acetylsalicylic acid are both white crystalline solids. The student collected and purified the product of the reaction before sending it off for analysis by proton NMR spectroscopy. How would the spectrum produced by this technique allow the student to identify whether or not any acetylsalicylic acid had been formed in the reaction? d A sample of the product was also sent off for analysis by mass spectroscopy. In the spectrum returned there was a peak at m/e = 138 and a peak at m/e = 180. What does this tell you about the sample? 24 Traditional medicine used infusions of willow bark to help reduce pain and fever. Analysis has shown that the infusion contains the substance salicin. It is now known that digestive enzymes are able to convert salicin into salicylic acid. A student proposes to synthesise a soluble form of aspirin from salicin via the reaction pathway shown on the next page.


25 Aspirin, acetylsalicylic acid, C9H8O4, is made by reacting salicylic acid, C7H6O3 (Mr = 138.0), with ethanoic anhydride, C4H6O3 (Mr = 102.0), according to the equation:

CH2OH CH2OH O O

hydrolysis

OH

HO Salicin

Salicyl alcohol

COOH

COO–Na+

COOH

A

2C7H6O3 + C4H6O3 –—➤ 2C9H8O4 + H2O

OH

OH

C

B OH

OCCH3 O Acetylsalicylic acid (aspirin)

Salicylic acid

CH2OH

OCCH3 O Sodium acetylsalicylate (soluble aspirin)

a Name the type of reactions represented by the letters: i A ii B iii C b Copy the two structures (salicin and salicylic acid) shown below and circle their common features. CH2OH

CH2OH

COOH

O O

OH

HO

OH

OH

Salicin

Salicylic acid

c Acetylsalicylic acid (aspirin) has the structure shown below: COOH

O

CCH3

a Determine which is the limiting reagent if 10.0 g salicylic acid is reacted with 10.0 g ethanoic anhydride. b Find the maximum mass of acetylsalicylic acid that could be produced by this reaction. c Explain why sufferers of pain and fever would prefer to take acetylsalicylic acid (aspirin) rather than salicylic acid. Are there any situations where aspirin should not be given? 26 a Draw the structural formula for: i hexanoic acid ii 2-methyl-butan-2-ol iii pent-2-ene b Draw the structural formula and provide the systematic name of four isomers of C4H9Cl. c Give systematic names for: i CH3CH2CH2OH ii CH3CH2CH2COOH iii CH3CH2CH2COOCH2CH3 27 Ethyl hexanoate is an ester commonly used as artificial pineapple essence. a What starting materials are required to prepare a sample of this ester by one reaction? b Describe the reaction conditions that you would use in this preparation. Explain the purpose of any additional reagents. c Write a balanced equation to represent the reaction. 28 The boxes below represent structures that make up a polymer of biological importance. Deoxyribose (sugar)

O Acetylsalicylic acid Phosphate

i Circle the part of this molecular structure that would not contribute to its pharmacological activity. ii Acetylsalicylic acid is hydrolysed to salicylic acid in the body. Explain why it is preferable to take acetylsalicylic acid rather than salicylic acid.


Adenine A

Guanine G

Cytosine C

Amine bases

Thymine T

a What is the name of this biologically important polymer? b Draw a simple diagram using boxes to show the arrangement of the components of a nucleotide containing cytosine. c Nucleotides can bond together to form a long chain. What name is given to this type of reaction? What molecule is also produced in this reaction? d Draw a diagram to show the arrangement of nucleotide monomers in a single strand of DNA. e Draw a simple diagram to show how two single polynucleotide strands are aligned in DNA. Show any bonds that are formed between two single strands of DNA. f The structure of DNA is often described as a double helix. Explain what is meant by this term. 29 a Explain how a DNA profile can be obtained. b How is a DNA profile used in forensic analysis? 30 Cysteine is an amino acid that forms part of the polypeptide chain in albumin, a protein that occurs in eggs. The Z-group of cysteine has the structure –CH2SH. a Draw the structure of the amino acid cysteine. b On your diagram, label and name the two functional groups that are common to all amino acids. c Draw the structure of cysteine at low pH. d Draw the structure of the two possible dipeptides that can form when a molecule of cysteine reacts with a molecule of alanine. (Alanine has a Z-group of –CH3.) 31 Condensation reactions are responsible for the formation of many important biomolecules, e.g. cellulose, proteins, starch, lipids. a Draw the structure of the products of the condensation reactions between the following pairs of compounds. i CH2OH H C HO

C

O

H OH

H

C

C

H

OH Glucose

C OH

O

HOH2C

H +

C HO

H

H

OH

C

C

OH

H

C CH2OH

ii H

O

H

C

OH

H

C

OH

H

C

OH

3 CH3(CH2)14C

+

O

H

Palmitic acid

H Glycerol

iii OH H H

O N

C

H

+

C OH

CH2

H N

C

H

O C OH

H

H

Glycine

Serine

b Name the functional group that has been produced as a result of: i the reaction between glucose and fructose ii the reaction between glycerol and palmitic acid iii the reaction between glycine and serine 32 Compound X contains hydrogen, carbon and oxygen. 4.50 g of X was burnt in excess air. 2.70 g of water was produced. 3.67 L of carbon dioxide measured at 25°C and 1.00 atm pressure was also produced. a Determine the percentage composition of the compound. b What is the empirical formula of the compound? c The mass spectrum of compound X is shown below. Deduce the molecular formula of X. d The molecular formula of compound X suggests that it is either a carboxylic acid or an ester. Write the structural formula and give the name of the two isomers that correspond to the molecular formula of X. e Use the IR spectrum of X below to explain why the compound is not an acid.

Fructose


100 90

Relative intensity

80 70 60 50 40 30 TMS

20 10 0

0

15

20

25

30

35 40 m/e

45

50

55

60

9

8

7


3

2

1

0

1

Mass spectrum of X.

H NMR spectrum of compound X.

33 The DNA profiles below were used as evidence in two separate court cases in which women (M) claimed particular men (F) were the fathers of their children (C).

90 80 Transmittance (%)

10

70

Case 1

60

M

C

Case 2 F

M

C

F

50 40 30 20 10 0 4000

3000


1000

500

IR spectrum of X.

f Explain why there are only two peaks in the 1H NMR spectrum of compound X shown below. g Identify compound X. a What do you think should have been the verdict in each case? b Each band in these profiles represents a repeating base sequence region. Why are multiple repeating base sequence regions compared in DNA profiling?


unit

4

areas of study 01 industrial chemistry On completion of this unit the student should be able to analyse the factors that determine the optimum conditions used in the industrial production of the selected chemical.

02 supplying and using energy On completion of this unit the student should be able to analyse chemical and energy transformations occurring in chemical reactions.

Chemistry at work

1

chapter 15

Fast and slow chemistry

key knowledge • collision theory and factors that affect the rate of a reaction including activation energy • energy profile diagrams and the use of ΔH notation

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the energy changes occurring during a chemical reaction in terms of breaking and forming of chemical bonds • compare the relative amounts of energy involved in bond breaking and bond forming in endothermic reactions and exothermic reactions • use ΔH values to identify reactions as endothermic or exothermic • write thermochemical equations for a reaction, including the appropriate values for ΔH • sketch and interpret energy profiles for reactions • describe chemical reactions in terms of the collision theory of particles • provide explanations for the increase in reaction rate that occurs when solute concentration is increased, particle size of solid reactants is decreased, temperature is increased and a catalyst is added.

15.1

Fast and slow chemistry This chapter examines the factors that influence the rate of reactions and the changes in energy that occur during chemical reactions.

chemistryy in action Saved by a very fast chemical reaction Imagine the scene. An eighteen-year-old borrows his parent’s car to take his girlfriend for a drive to celebrate gaining his driver’s license. Roof down, enjoying the beautiful afternoon and the countryside, the driver rounds a corner to find the road wet. The car begins to slide on the wet surface. In his inexperience the driver brakes; the car starts to spin. Suddenly, the car is leaving the road and heading straight for a large tree. Then, bang! Later, the car was estimated to have been travelling at 60 km/h when it hit the tree. The collision was a ‘head-on’, with the front and passenger side taking most of the impact. Yet the girl in the passenger seat suffered just a chipped tooth, and her boyfriend sustained only minor bruising. This is the true story of a lucky escape, thanks to a very rapid chemical reaction. As the collision took place, airbags were inflating and then deflating as the travellers were slammed forward towards the windscreen. The driver described it as being ‘all over in a flash’ and had no clear recollection of the airbags going off. Hidden in the car’s steering wheel, dashboard and windscreen pillars, special nylon bags fill with gas within 30 milliseconds of impact. As a consequence, the car occupants are prevented from smashing their heads against the steering wheel, dash, windscreen or front pillars, all within the blink of an eye. As the head and body strike the airbags, the cushion of gas is forced out of the bag through tiny vents, and within 100 milliseconds the bag has completely deflated. Air bags contain a mixture of crystalline solids—sodium azide (NaN3), potassium nitrate (KNO3) and silica (SiO2)—stored in a canister. Sensors in the front of the car are able to detect the difference between a bump and life-threatening impact. When a response is required, an electronic impulse ‘ignites’ the sodium azide. Sodium metal and hot nitrogen gas are the products of this energy-releasing redox reaction: 2NaN3(s) –—➤ 2Na(l) + 3N2(g)

The pulse of hot nitrogen gas released from this reaction commences inflating the nylon bag. The molten sodium metal immediately reduces the potassium nitrate, generating more nitrogen gas, as well as sodium oxide and potassium oxide, which are white powdery solids. The equation for this reaction is: 10Na(l) + 2KNO3(s) –—➤ K2O(s) + 5Na2O(s) + N2(g)

A filtration system prevents any of the oxides from entering the nylon bag, while a third reaction ‘captures’ them to produce a harmless glassy solid. In this reaction they combine with silica: K2O(s) + 5Na2O(s) + SiO2(s) –—➤ alkaline silicate (‘glass’)

Figure 15.1 Testing air bags. Equipped with 30 internal sensors, the dummy provides over 4000 readings during the course of a halfsecond crash.

! Although widely used in the United States since the 1970s, airbags were initially not seen as necessary in Australia because of our compulsory use of seatbelts. However, ongoing concerns about road safety led to their adoption in most Australian cars during the late 1990s. Dual front and side airbags are now available as standard equipment in most new cars.

chemfact The oxidation number of the nitrogen atoms in the azide ion (N3–) is rather interesting: –1/3. You can regard the ion as being made up of two nitrogen atoms with an oxidation number of –1, and one with an oxidation number of +1. The structure of the azide ion can be represented as: –1

+1

–1

N

N

N

Chemical reactions do save lives!

247 Fast and slow chemistry

chemfact Remember, chemical reactions do not produce or consume matter. Every atom present before the reaction will still be there after it is over. Only the atoms are rearranged as bonds are broken and new bonds are formed.

During chemical reactions, particles (atoms, molecules or ions) collide and undergo change during which atoms are rearranged to produce new particles. The transfer of electrons from one particle to another, as in redox reactions, or the transfer of protons, as in acid–base reactions, are just two examples of many different types of reactions. It is important to appreciate that collisions between reactants do not always result in chemical change. For example, while a car’s fuel tank is being filled with petrol the hydrocarbon gas molecules in the fuel are colliding with oxygen molecules in the air without a reaction taking place. An ignition source, such as a spark from a spark plug, is needed to initiate a combustion reaction involving the two substances.

15.2

Chemical energy What is chemical energy?

a

Petrol is useful to us because it contains large amounts of chemical energy. All substances have chemical energy. The chemical energy of a substance is the sum of its potential energy (stored energy) and kinetic energy (energy of movement). These energies result from such things as: attractions between electrons and protons in atoms repulsions between nuclei repulsions between electrons movement of electrons vibrations of and rotations around bonds. The chemical energy of a substance is sometimes called its heat content or enthalpy, and is given the symbol H.

Exothermic reactions Ex

Energy

Reactants Energy released (ΔH negative) Products

b

Endothermic reactions E

Energy

Products

Reactants

Energy absorbed (ΔH positive)

Figure 15.2 Possible energy changes for a chemical reaction.

Energy changes during chemical reactions During a chemical reaction, the atoms in the reactants are rearranged into products with different chemical energies. As Figure 15.2 shows, two situations arise, depending upon the relative energies of the reactants and products: The total chemical energy of the products might be less than the energy of the reactants (Figure 15.2a). Since energy is never lost, the difference in energy between the reactants and products is released into the environment—often as heat energy. A reaction that releases energy, such as the combustion of petrol, is known as an exothermic reaction. The chemical energy of the products might be greater than the energy of the reactants (Figure 15.2b). Energy must be absorbed from the environment around the reactants in order for the reaction to occur. A reaction that absorbs energy is known as an endothermic reaction (Figure 15.3). The energy released or absorbed during a chemical reaction is called the heat of reaction. Since the heat of reaction is equal to the difference in enthalpy between the products and the reactants, it is given the symbol ΔH (‘delta H’), where: ΔH = H(products) – H(reactants)

248 Industrial In n chemistry

You can see that: for exothermic reactions, H(products) is less than H(reactants), so values of ΔH will be negative; i.e. ΔH < 0 for endothermic reactions, H(products) is greater than H(reactants), so values of ΔH will be positive; i.e. ΔH > 0. Fuels such as wood, coal, oil and natural gas provide us with energy by undergoing exothermic combustion reactions. Food can also be considered to be a type of fuel; it undergoes exothermic reactions inside our bodies. For example, the exothermic oxidation of glucose in muscles is used to power movement. Indeed, most of the reactions we commonly encounter are exothermic.

Thermochemical equations One of the most important endothermic reactions is photosynthesis. Plants absorb energy from sunlight and use it to change water and carbon dioxide into glucose and oxygen. Plants then use glucose for a number of purposes such as making structural material for growth and as an energy source. The thermochemical equation for this reaction is: 6CO2(g) + 6H2O(l) –—➤ C6H12O6(aq) + 6O2(g); ΔH = +2803 kJ mol–1

In this reaction, 2803 kilojoules (kJ) of energy is absorbed when 6 moles of carbon dioxide reacts completely with 6 moles of water, producing one mole of glucose (C6H12O6) and 6 moles of oxygen. Thermochemical equations show the energy released or absorbed during a chemical reaction. Energy is measured in joules (J) or kilojoules (kJ). The heat of reaction, ΔH, has the units J mol–1 or kJ mol–1, to indicate that the amount of energy corresponds to the mole amounts specified by the equation. Oxidation of glucose involves the opposite chemical change to photosynthesis:

Figure 15.3 An endothermic chemical reaction provides the lowered temperature used by some types of cold pack to reduce inflammation in sporting injuries.

chemfact About 40 000 J of energy is needed to boil the water for a cup of tea or coffee. 1 kJ = 1000 J.

C6H12O6(aq) + 6O2(g) –—➤ 6CO2(g) + 6H2O(l); ΔH = –2803 kJ mol–1

As you might expect, this process results in the release of 2803 kJ of energy for each mole of glucose oxidised, and so the sign of ΔH is negative. Another example of a thermochemical equation is the combustion of methanol (CH3OH): 3 CH3OH(l) + O2(g) –—➤ CO2(g) + 2H2O(g); ΔH = –726 kJ mol–1 2

This reaction is used as the source of power for some racing cars. For each mole of methanol fuel that reacts with oxygen to produce carbon dioxide and steam, 726 kJ of energy is released. Note that if the equation is written for the reaction of 2 moles of methanol, then twice as much energy would be released and the value of the ΔH is doubled. In this case the thermochemical equation is: 2CH3OH(l) + 3O2(g) –—➤ 2CO2(g) + 4H2O(g); ΔH = –1452 kJ mol–1

Activation energies: getting a reaction started Consider the combustion of natural gas. Since natural gas is mainly methane, we can represent the reaction as: CH4(g) + 2O2(g) –—➤ CO2(g) + 2H2O(g); ΔH = –890 kJ mol–1

It is an exothermic reaction, so the energy of the reactants is higher than the energy of the products. Why then doesn’t natural gas burst immediately into flame and release energy when it contacts air? To start a gas oven or heater, why must you use a match or electric spark?

chemfact Light energy can provide the activation energy for some reactions. For instance, light from the Sun allows plants to convert carbon dioxide and water into oxygen and glucose during photosynthesis.


Energy

Energy absorbed as bonds in reactants break Activation energy Reactants Energy released as bonds form in products

ΔH

Products

Figure 15.4 Energy profile of an exothermic reaction.

Energy

Activation Products energy ΔH

Reactants

Figure 15.5 Energy profile of an endothermic reaction.

EQUILIBRIUM • Energy changes

summary

To answer these questions, we need to recall what happens to chemical bonds during the course of a reaction. The bonds between atoms in the reactants must first be broken. For this to occur, energy must be absorbed. Then new bonds form as the products are created. Energy is released as this happens. In the combustion of methane, energy is used to break covalent bonds within the methane and oxygen molecules. Energy is then released when covalent bonds form as carbon dioxide and water are produced. Since the combustion of methane results in an overall release of thermal energy, the energy absorbed when the bonds break must be less than the energy released when new bonds form. The heat of reaction, ΔH, is the net result of the energy absorbed in breaking bonds and the energy released by making them. Energy changes during the course of the reaction are shown in Figure 15.4. A diagram like this is referred to as an energy profile. The top of the curve represents an intermediate step in the reaction, referred to as an activation or transition complex. At this stage, the bonds in the reactants are partially broken and the bonds in the products are partially formed. The energy required to break the bonds of reactants so that a reaction can proceed is called the activation energy of the reaction. In the combustion of natural gas there is insufficient energy at room temperature to break the bonds in oxygen or methane. You can see why natural gas must be lit. The match or spark supplies the energy required to overcome the activation energy ‘barrier’. Like exothermic reactions, endothermic reactions also involve the breaking and forming of bonds. An activation energy barrier is present in both exothermic and endothermic reactions (Figure 15.5).

15.1–15.2

Ch Chemical i l reactions ti involve i l the th collision lli i off reactant t t particles. ti l During a chemical reaction the breaking of bonds in reactants absorbs energy, while the formation of bonds in products releases energy. The energy required to break bonds is called the activation energy. All matter contains chemical energy. This comprises potential energy from the electrostatic attractions and repulsions of subatomic particles and kinetic energy arising from translational, vibrational and rotational motion. Chemical energy is called heat content or enthalpy and is represented by the symbol H.

The Th enthalpy th l change h (ΔH) ((ΔH H) off a chemical h i l reaction ti iis given i bby: ΔH = enthalpy of products – enthalpy of reactants In an exothermic reaction the enthalpy of the reactants is greater than the enthalpy of the products. ΔH is negative as energy is lost to the surroundings from the reaction system. In an endothermic reaction the enthalpy of the reactants is less than the enthalpy of the products. ΔH is positive as the reaction system gains energy from the surroundings. Energy profile diagrams can be used to describe the energy changes occuring during a chemical reaction.

key questions 1 Natural gas begins to burn when lit with a match. Why does it continue to burn when the match is taken away? 2 From the data given, identify each of the following reactions as either exothermic or endothermic. a CO2(g) + C(s) –—➤ 2CO(g); ΔH = +161 kJ mol–1


b N2(g) + 3H2(g) –—➤ 2NH3(g); ΔH = –91 kJ mol–1 c 6CO2(g) + 6H2O(l) –—➤ C6H12O6(aq) + 6O2(g); ΔH = +2803 kJ mol–1 + – –— d H (aq) + OH (aq) ➤ H2O(l); ΔH = –57 kJ mol–1

15.3

Making reactions go faster The rate at which chemical reactions occur is an important consideration for industrial chemists and chemical engineers. Some reactions take less than 10–11 seconds, while others take many years. For the manufacture of most products to be profitable, the reactions involved need to occur rapidly. Considerable effort is directed towards maximising reaction rates in industry. In your experience, too, there will be situations where the rate of a chemical reaction is of interest or concern. At times you might have considered the rate at which: wood on a camp fire was burning bathroom tiles were being cleaned a cake was baking a tomato plant was growing. To understand how to speed up or slow down a chemical process, we must visualise what happens to particles during a reaction and understand the energy changes that take place.

Collision theory For a chemical reaction to occur, the particles involved must collide with each other with sufficient energy to overcome the activation energy ‘barrier’ to the reaction. This way of visualising reactions is known as collision theory. It is an extension of the kinetic theory of matter and covers the detail of what happens at the particle level during a chemical reaction. It makes sense that the greater the number of collisions between reactant particles, the greater the rate of reaction. But this doesn’t completely explain what we observe. Most chemical reactions involve enormous numbers of reactant particles. The frequency of collision of these reactant particles is also enormous, so why isn’t every reaction an explosion? The answer is that in most chemical reactions the majority of collisions do not result in reaction. In these collisions the energy involved is less than that required to break bonds in the reactants, i.e. the energy is less than the activation energy. Only ‘successful’ collisions, where the energy of collision is greater than the activation energy, allow a chemical reaction to progress. So, the rate of a chemical reaction is also dependent on the proportion of collisions that are successful.

review The kinetic molecular theory of particle behaviour is described in Heinemann Chemistry 1, Chapter 21.

chemfact When talking about successful collisions, chemists use a variety of terms. Other terms you might encounter that mean the same thing are fruitful, effective and productive.

Factors that affect rates We can understand the factors that affect the rate of a reaction better if we understand how these factors influence the frequency of collision and the proportion of successful collisions. There are four main ways in which reaction rates can be increased: increasing the surface area of solids increasing the concentration of reactants in solution (or pressure of gaseous reactants) increasing the temperature adding a catalyst.

Increasing the surface area of solids In a solid, only those particles that are at the surface can be involved in reaction. Crushing a solid into smaller parts means that more particles are


present at the surface. As a consequence of the greater number of exposed particles, the frequency of collisions between these particles and reactant particles increases, and so reaction occurs more rapidly. Sherbet (Figure 15.6) provides an example of how a reaction can be speeded up by using fine-grained powders. It is a mixture of a very fine powder of a weak organic acid, malic acid (C4H6O5), and sodium hydrogen carbonate. On contact with the saliva in the mouth, these chemicals quickly dissolve and react, releasing carbon dioxide gas and providing the fizzy sensation. In simplified form, the equation can be written: C4H6O5(aq) + 2NaHCO3(aq) –—➤ Na2C4H4O5(aq) + 2H2O(l) + 2CO2(g)

Fireworks manufacturers modify the surface area of solid reactants to control the rate at which the fuels in fireworks are oxidised, using this to create different effects. For example, finely divided aluminium confined in a shell explodes violently. If larger pieces of aluminium are used, the reaction is slower and sparks are seen as pieces of burning metal are ejected.

Figure 15.6 Sherbet ‘fizzes’ on the tongue as the finely powdered mixture of a weak acid and sodium hydrogen carbonate reacts rapidly to form carbon dioxide gas.

! Malic acid is also known as ‘apple acid’. Can you suggest why?

CHEMICAL EQUILIBRIUM RIUM • Factors affecting rates of reactions— concentrations • Factors affecting rates of reactions— temperature • Factors affecting rates of reactions— particle size

Figure 15.7 Particle size can be used in fireworks to control the rate of reaction and create different effects.

Increasing the concentration of reactants

Figure 15.8 Air pollution, which can cause relatively high concentrations of acids in rainwater, has caused the rapid recent decay of this marble statue.


Reactions involving molecules or ions dissolved in solution occur faster if the concentration of the dissolved particles is increased. With more particles moving randomly in a given volume of solution, the frequency of collisions is increased and so more successful collisions occur. The relatively rapid deterioration of marble statues in locations prone to acid rain provides an example of the effect of concentration on rate (Figure 15.8). Most rainwater is slightly acidic as a result of the presence of carbonic acid formed by carbon dioxide gas dissolved in the water. However, oxides of nitrogen and sulfur released by cars and industry also dissolve, raising the concentration of acids in rainwater. As a result, the rate of the reactions that disfigure marble statues are also increased. Acid rain also has some serious environmental effects. Increasing the pressure of gases raises the concentration of gas molecules, causing more frequent collisions. High gas pressures are therefore often

employed by engineers in their design of chemical plants that employ gasphase reactions. As you will learn later in this Unit, there are occasions when this approach needs to be carefully balanced against other constraints.

Increasing the temperature We use refrigerators to slow down the rate of chemical reactions that lead to food spoilage (Figure 15.9). On the other hand, a gardener might use a hot house to accelerate the rate of biochemical reactions in tomato plants to obtain a crop of tomatoes in a shorter time. As temperature increases, the average speed and average kinetic energy of the particles increases as well. Hence, as temperature increases so does the rate of reaction. There are numerous everyday examples of the effect of temperature on reaction rate, including: rapid cooking of food in a pressure cooker—at the high pressures involved, the boiling temperature of water rises above 100°C and rates of reactions associated with cooking are increased the use of water to extinguish fires—soaking burning trees or buildings in water lowers the temperature as energy is absorbed by the water. The lower temperature reduces the rate of the combustion reaction.

Figure 15.9 We keep perishable foods in a refrigerator because the rates of reactions that lead to food spoilage are much lower at low temperatures.

Extending collision theory

Number of particles

Number of particles

The effect of temperature on reaction rate cannot be simply explained by the increased frequency of collisions. A temperature increase of just 10°C causes the rate of many reactions to double, but collisions have increased by only about one-fiftieth of this amount! Clearly, we need to look further than the simple version of collision theory we have used so far to understand such a dramatic increase in the reaction rate. Figure 15.10 shows the distribution or spread of kinetic energies of particles at a particular temperature. At any given temperature particles have a wide spread of kinetic energies and velocities. Consequently, temperature is a measure of the average kinetic energy. If we increase the temperature of the particles, more particles will move at higher speeds and have higher kinetic energies. Figure 15.11 shows how the distribution of kinetic energies changes as you increase the temperature. There is still a wide spread of energies, but at a higher temperature more particles have higher kinetic energies.

Kinetic energy

Figure 15.10 The distribution of energies of particles at a particular temperature.

CHEMICAL EQUILIBRIUM RIUM • Collision and chemical reactions in a gas • Collision and chemical reactions • Rates of reactions—an explanation

chemfact The spread of energies of particles at a particular temperature has a characteristic shape called a Maxwell–Boltzmann distribution.

chemfact

20°C 30°C

Kinetic energy

Many animals maintain a constant body temperature to ensure the rates of biological reactions remain constant. Cold-blooded animals, such as lizards and snakes, are more responsive to temperature variations and become more active when the weather is warm.

Figure 15.11 The distribution of energies of particles at two temperatures. Raising the temperature increases the proportion of particles with higher energies.


Number of particles

20°C 30°C

,A

Kinetic energy

Figure 15.12 The energies of particles at two temperatures. EA represents the activation energy for the reaction. The shaded area under the graph represents the proportion of particles with sufficient energy to react.

a

b

A reaction can take place only if the particles colliding have more energy than the activation energy of the reaction. At any instant only a fraction of the particles present have sufficient energy to participate in successful collisions and react. In Figure 15.12, EA represents the activation energy of a reaction. The proportion of particles with kinetic energies in excess of the activation energy is represented by the shaded area under the graph. As the temperature increases, the shaded area under the graph increases. This means the proportion of particles with more energy than EA increases. A temperature increase of 10°C almost doubles the number. A small increase in temperature can lead to a big increase in the proportion of particles with enough energy to react. This factor has a greater impact on rate than the increase in the number of collisions at higher temperatures. In many reactions, the position and angle of the reactant molecules at the point of collision will determine if a reaction occurs. Figure 15.13 represents the reaction: H2(g) + I2(g) –—➤ 2HI(g)

Unfavourable collision orientation

Favourable collision orientation

Less likelihood of successful reaction

The orientation of the hydrogen and iodine molecules as shown in Figure 15.13a is unlikely to lead to the formation of hydrogen iodide. The orientation of H2 and I2 as shown in Figure 15.13b will result in a reaction.

Figure 15.13 Effect of orientation on collision.

chemistryy in action The catalysts in cars: towards cleaner air

chemfact Modern cars burn ‘unleaded’ petrol and use a catalytic converter in the exhaust system to help minimise pollution. Before exhaust pollution was recognised as a problem, lead was added to petrol to help it burn efficiently. However, lead in exhaust fumes is not only toxic to humans, it renders the catalyst in car exhaust systems ineffective (‘poisoned’).

All new cars sold in Australia since 1986 have a catalyst fitted between the engine and the exhaust pipe. The purpose of the catalyst is to clean up the exhaust gases from the engine and reduce the air pollution that could be caused if these gases entered the atmosphere (see Heinemann Chemistry 1, Chapter 19). Catalyst coated on a metal framework

Catalyst Carbon dioxide, water and nitrogen

Figure 15.14 A catalyst in the exhaust system of a car reduces the emission of gases that form smog.

Exhaust gases: unburnt fuel, nitrogen oxides, carbon monoxide and air

Catalysts in cars are used to convert carbon monoxide and nitrogen oxide, formed in the engine, to the non-toxic gases carbon dioxide and nitrogen: 2CO(g) + 2NO(g) –—➤ 2CO2(g) + N2(g) The catalyst is a mixture of platinum and rhodium metals and aluminium oxide, and it is mounted on a honeycomb-shaped support. There are millions of tiny pores in the metals, providing a large surface area. Exhaust gases from the engine enter the catalyst chamber, pass quickly over the metals, and leave the exhaust pipe in a purified condition. The catalyst is unchanged by the reaction and can be used without replacement for many years.


Catalysts The catalyst in car exhaust systems is remarkable. How does it stimulate the gases to react without being used itself? Why does the reaction between carbon monoxide and nitrogen oxide not happen anyway, without the catalyst? Many reactions occur more rapidly in the presence of particular elements or compounds. These substances, known as catalysts, are not consumed during the reactions and therefore do not appear as either reactants or products in reaction equations. The chemical industry uses catalysts extensively. Without catalysts, many reactions would be too slow for products to be obtained at an economical rate. Table 15.1 lists some important industrial processes and the catalysts used in them. Every year, millions of tonnes of chemicals are produced using these processes.

chemfact A catalyst will become ‘poisoned’ and lose its functionality if it is exposed to chemicals that bind to and block the active site. In most cases this process is irreversible and the poisoned catalyst has to be replaced.

TABLE 15.1 Some industrial processes that involve the use of catalysts Product Polyethene Ammonia Sulfuric acid Gasoline Nitric acid Margarine Wine

Name of process Low-pressure polymerisation Haber process Contact process Catalytic cracking Ostwald process Hydrogenation Fermentation

Catalyst Titanium/aluminium compound Iron Vanadium(V) oxide Zeolite Platinum/rhodium Nickel Enzyme in yeast

Many catalysts have been discovered by simple trial and error. For example, over 2500 catalysts have been tried for the Haber process, an industrial process that converts nitrogen and hydrogen into ammonia (Chapter 19). Once chemists discover how a particular catalyst speeds up a reaction, they start to look for other similar substances that might be even more effective. Sometimes small modifications to a catalyst’s composition produce outstandingly successful results. There are two types of catalysts: Homogeneous catalysts are those in the same state as the reactants and products. For example, in the outer atmosphere, chlorine atoms in the gaseous state act as homogeneous catalysts in the decomposition of ozone gas into oxygen gas. The chlorine atoms may have come from chlorofluorocarbons (CFCs) released into the atmosphere from refrigerators or air-conditioners. Heterogeneous catalysts are in different states from the reactants. The solid catalyst in car exhaust systems is of this type. Chemists prefer to use heterogeneous catalysts for industrial processes as they are more easily separated from the products of a reaction.

Figure 15.15 Some industrial catalysts that are used to speed up reactions.

chemfact The iron catalyst used in the Haber process is made by melting magnetite (Fe3O4) with small amounts of potassium and aluminium oxides. The mixture is ground and heated with hydrogen. One gram of the catalyst has the remarkable surface area of 50 m2.

review Enzymes are special proteins that control the rate of reactions in biological systems (p. 194).

How do catalysts work? Details of how many catalytic effects occur are not well understood. In general, particles at the surface of a solid of high surface energy tend to adsorb (form a bond with) gas molecules that strike the surface, lowering the surface energy of the solid. Adsorption distorts bonds in the gas molecules or may even break them completely, allowing a reaction to proceed more easily than it would if the solid were absent.

CHEMICAL EQUILIBRIUM RIUM • Factors affecting rates of reactions— catalysts


Nitrogen molecule

Hydrogen molecules

Iron catalyst surface

A powdered or sponge-like form of the solid catalyst is used to provide the greatest possible surface area. The larger the surface area, the more reactant molecules that can be adsorbed, and the faster the reaction. One catalysed reaction that has been studied a great deal and is now quite well understood is the Haber process. This industrial process is of great commercial importance as it produces ammonia gas, from which fertilisers, nylon, explosives and some pharmaceuticals can be manufactured. In the process, hydrogen and nitrogen gas are converted to ammonia using a catalyst of powdered iron. The reaction is represented by the equation: N2(g) + 3H2(g)

New bonds form to make ammonia molecules

Figure 15.16 The catalyst in the Haber process allows the bonds in the nitrogen and hydrogen molecules to be broken more easily.

The reaction that takes place at the iron surface is shown in Figure 15.16. Hydrogen and nitrogen molecules both adsorb onto the iron surface. As they attach themselves to the surface, the bonds within their molecules break. The hydrogen and nitrogen atoms now readily combine to form ammonia molecules and move away from the iron surface. The catalyst remains unaltered by the reaction. Without a catalyst, temperatures greater than 3000°C are needed for both the nitrogen and hydrogen molecules to break apart sufficiently and significant reaction to occur. The catalyst allows the manufacture of ammonia to proceed at an economical rate using a temperature of about 500°C and a pressure of 250 atmospheres. The energy needed to break bonds (activation energy) is less if a catalyst is used

Without catalyst

With catalyst Energy

Molecules are adsorbed on the catalyst and bonds break

2NH3(g); ΔH = –91 kJ mol–1

Nitrogen and hydrogen Ammonia

a Number of particless

Figure 15.17 Energy changes in the uncatalysed and the catalysed reaction of nitrogen and hydrogen to form ammonia. No catalyst

,A Kinetic energy Number of particles es

b

Catalyst present

,A Kinetic energy

Figure 15.18 A higher proportion of particles have sufficient energy to react when a catalyst is added b, compared with no catalyst a.


The catalyst Th l provides id an alternative l i reaction i pathway h that h d dramatically i ll reduces the activation energy ‘barrier’—the energy needed to break the covalent bonds in the nitrogen and hydrogen molecules (Figure 15.16). Consequently, a much higher proportion of reactant particles collide with sufficient energy to overcome the activation energy barrier (Figure 15.18). So, providing such a pathway increases the proportion of successful collisions and therefore the rate of reaction increases. You could liken this situation to that of two people travelling from Melbourne to Canberra. One person climbs over the top of Mount Kosciuszko (the uncatalysed route) in order to reach their destination. The other person gets to Canberra much faster by travelling only over the foothills of the mountains (the catalysed route) (Figure 15.19). Notice that the relative energies of the reactants and products shown in Figure 15.17 are unaffected by the presence of the catalyst (in the same way as both our travellers arrive at the same destination). This means the heat of reaction, ΔH, is not changed. Adding a catalyst only changes the activation energy of the reaction.

Mt KO

SCIUSZKO

THE FOOT HILLS

Melbourne

Canberra

Figure 15.19 Travelling from Melbourne to Canberra by two routes. The slower route can be likened to the progress of an uncatalysed reaction, the faster route to the progress of a catalysed reaction.

chemistryy in action Zeolites: tailor-made catalysts A group of catalysts called zeolites is becoming ever more important. With zeolites, the catalytic effect takes place inside the catalyst. Zeolites are a naturally occurring group of crystalline substances composed of aluminium, silicon and oxygen. They have a three-dimensional honeycomb structure made up of channels of a specific size (Figure 15.20). The honeycomb skeleton is negatively charged, while the channels contain mobile cations that maintain the electrical neutrality of the zeolite. Chemists have been able to make a variety of synthetic zeolites with channels of different sizes. Appropriately sized molecules can fit into these channels. Highly charged sites inside the channels then catalyse the breakup of the molecules and new products are formed.

Figure 15.20 This diagram shows the position of atoms and the channels in the skeleton of a zeolite.

extension Solving Australia’s looming oil crisis? M Many scientists i ti t believe b li that th t world ld oilil production d ti is i peaking ki andd that th t production d ti will ill soon begin to decline unless there are significant new discoveries. Worldwide demand for oil has been growing steadily for decades. Recently, growth in consumption has begun to accelerate, mainly because of increasing demand from the rapidly expanding economies of China and India. What does this mean? We know that oil reserves are being depleted; the question is how fast. There is much debate, but a number of experts predict that reserves will dwindle quickly and run out in as little as twenty years. The main concern is for transport fuels, especially petrol. In contrast, supplies of natural gas are abundant. If only we could convert natural gas into petrol. Well, with the use of a catalyst, we can. In the 1960s, Mobil Oil in the United States developed a synthetic zeolite known as ZMS5, which allows methanol to be converted to petrol (Figure 15.21). The process starts with the conversion of natural gas to methanol. Methane first reacts with steam in the presence of a nickel catalyst to form ‘synthesis gas’, a mixture of carbon monoxide, carbon dioxide and hydrogen. The synthesis gas is cooled, compressed, reheated and passed over a catalyst to produce methanol. The methanol is passed through

Figure 15.21 A zeolite catalyst used for converting methanol to petrol. Each bead of catalyst is about half the size of a pea.

257 2 5 Fast and slow chemistry

extension Solving Australia’s looming oil crisis? (continued)

! Issues related to fuel resources and their use are discussed in Chapters 11, 23 and 24.

Figure 15.22 Steps involved in the manufacture of petrol from natural gas.

CH4 + H2O

Natural gas

questions

the h microscopic i i channels h l iin the h ZMS5 catalyst l at 350 350–450°C 450°C to yield i ld high-octane hi h petroll and water (Figure 15.22). In the 1980s Mobil teamed with the New Zealand government to prove the viability of the process by constructing a plant at Waitara on New Zealand’s North Island, using methane piped from nearby gas fields. The ‘Synfuel’ plant was built to reduce the impact that soaring oil prices in the early 1980s had on the New Zealand economy. However, the price of oil declined steadily through the late 1980s and 1990s and the lower world oil prices ruined the economic viability of Synfuel. The conversion of methanol to petrol at Waitara ceased in 1997, although methanol production from natural gas continues. Now the worldwide price of oil is more than double what it was in the early 1980s. Perhaps it is time for Australia, with its huge reserves of natural gas, to consider Synfuel once more?

Nickel mesh catalyst

Synthesis gas

C8H18 + H2O

CH3OH

CO + H2

Pellets of copper and zinc oxide catalyst

Methanol Zeolite catalyst

Octane (a component of petrol)

?

E1 What advantage does this source of petrol offer? E2 Certain zeolites containing sodium cations are used to soften water. Suggest why this is possible.

summary questions ques qu ess

15.3 ?

E1 What thist source At anyadvantage particular ti l does temperature, t particles tiof lpetrol have h offer? a spreadd off ti l collide llid with ith enoughh energy to t overcome the th − particles E2 Certain zeolites containing sodium cations are used to soften water. Suggestactivation why this energy is possible. velocities and kinetic energies. barrier The proportion of particles having a high kinetic energy − particle collide with a favourable orientation. increases with increasing temperature. The rate of reaction can also be increased by: − increasing the surface area At higher temperatures more particles have energy greater than − increasing the concentration or pressure, and the activation energy, leading to an increase in the rate of reaction. − by using a catalyst. A reaction will occur if: − particles collide A catalyst lowers the activation energy of a reaction.

key questions 3 Consider the examples of reactions mentioned on page 251— wood burning on a camp fire, bathroom tiles being cleaned, a cake baking, and a tomato plant growing. a How would you speed up the rates of these reactions? b Explain why the methods you suggested would produce an increase in the reaction rate. 4 Explain the following observations in terms of the behaviour of particles. a There have been many explosions in coal mines.


b Refrigeration slows down the browning of sliced apples. c Bushfires often start during lightning storms. d Iron anchors from shipwrecks can show little corrosion after years in the sea. e A burning match is used to light a candle, but the candle continues to burn when the match is extinguished. 5 Some reactions are called explosions. What are the features of these reactions?

15

key terms

activation energy adsorption alternative reaction pathway catalyst chemical energy collision theory

endothermic energy profile enthalpy exothermic heat content heat of reaction (ΔH )

heterogeneous catalysts homogeneous catalysts reaction rate thermochemical equation zeolite

CHEMICAL EQUILIBRIUM • Factors affecting rates of reactions—problems

Ch ical energy Chemical

2C4H10(g) + 13O2(g) –—➤ 8CO2(g) + 10H2O(l); ΔH = –5772 kJ mol–1

a How does the overall energy of the bonds in the reactants compare with the energy of the bonds in the products? b Draw an energy profile diagram for the reaction, labelling ΔH and the activation energy. c Describe how your diagram would change for the equation: C4H10(g) +

13 O (g) –—➤ 4CO2(g) + 5H2O(l) 2 2

10 Many major car makers have unveiled hydrogen-powered cars. In the engines of these cars hydrogen reacts with oxygen from the air to produce water. 2H2(g) + O2(g) –—➤ 2H2O(g)

Energy changes for the reaction are shown in Figure 15.23. a What is the magnitude of the activation energy of this reaction? b What is ΔH for this reaction?

1370 Energy (kJ mol-1)

6 Decide, giving reasons for your answers, whether the following processes are endothermic or exothermic: a burning of wood b melting of ice c recharging of a car battery d decomposition of plants in a compost heap 7 The activation energy for the reaction A + B –—➤ C is greater than the activation energy for the reverse (opposite) reaction C –—➤ A + B. Is the reaction A + B –—➤ C exothermic or endothermic? Explain. 8 Hydrogen reacts explosively with oxygen to form water. a What chemical bonds are broken in the reaction? b What chemical bonds are formed? c Explain how the energy changes during bond-breaking and bond-forming affect the energy change for the reaction. d Why is there no reaction until the reaction mixture is ignited? 9 The combustion of butane gas in portable stoves can be represented by the equation:

0 2H2(g) +O2(g)

_

572

2H2O(g)

Figure 15.23 Energy changes for the reaction of hydrogen and oxygen.

c Several groups of scientists have claimed to have split water into hydrogen and oxygen using a molybdenum catalyst: Mo

2H2O(g) ––—➤ 2H2(g) + O2(g)

Sketch energy change graphs for this reaction with and without the presence of a catalyst. d What is the value of ΔH for this water-splitting equation? 11 When one mole of methane gas burns completely in oxygen, the process of bond breaking uses 3380 kJ of energy and 4270 kJ of energy is released as new bonds form. a Write a balanced chemical equation for the reaction. b Calculate the value of the heat of reaction, ΔH, for the reaction. c Draw a diagram to show the changes in energy during the course of the reaction. 12 The formation of hydrogen iodide from its elements is represented by the equation: H2(g) + I2(g) –—➤ 2HI(g)

This endothermic reaction has an activation energy of 167 kJ mol–1 and the heat of reaction, ΔH, is +28 kJ mol–1. What is the activation energy for the reverse reaction, the decomposition of two mole of hydrogen iodide?


13 Account for the following observations with reference to the collision model of particle behaviour. a Surfboard manufacturers find that fibreglass plastics set within hours in summer but may remain tacky for days in winter. b A bottle of fine aluminium powder has a caution sticker warning that it is ‘highly flammable, dust explosion possible’. c A potato cooks much more slowly in a billy of boiling water on a trekking holiday in Nepal than a potato boiled in a similar way in the Australian bush. Hint: At high altitudes pressure is lower so water boils at a lower temperature than at sea level. 14 a Explain why surface properties are important to the operation of catalysts. b Many industrial catalysts are made into porous pellets. What is the reason for this? 15 Explain the meaning of the terms: a catalyst b activation energy 16 If a sugar cube is held in the flame of a candle, the sugar melts and browns but does not burn. However, the cube will burn if salt is first rubbed into it, even though the salt does not react. Explain the effect of the salt on the activation energy of this combustion reaction. 17 Perform a literature search using an Internet search engine to find details about three new catalysts that are being developed.

b Draw a graph of number of particles versus kinetic energy that shows the effect of a catalyst on a reaction. c Use the diagram you have drawn in part b to explain in terms of collision theory how a catalyst increases the rate of a reaction. 19 Lumps of limestone, calcium carbonate, react readily with dilute hydrochloric acid. Four large lumps of limestone, mass 10.0 g, were reacted with 100 mL 0.100 M acid. a Write a balanced equation to describe the reaction. b Which reactant is in excess? Use a calculation to support your answer. c Describe a technique that you could use in a school laboratory to measure the rate of the reaction. d 10.0 g of small lumps of limestone will react at a different rate from four large lumps. Will the rate of reaction with the smaller lumps be faster or slower? Explain your answer in terms of collision theory. e List two other ways in which the rate of this reaction can be altered. Explain your answer in terms of collision theory. 20 The reaction of hydrogen and iodine to form hydrogen iodide: H2(g) + I2(s)

2HI(g)

is shown on the energy level diagram Figure 15.25.

,ULYN`

Rates of reaction

18

5\TILYVMWHY[PJSLZ

Connecting the main ideas (

;PTL

)

Figure 15.25 Energy level diagram for the production of hydrogen iodide. 2PUL[PJLULYN`

Figure 15.24 Energy profiles at 40°C and 60°C.

a Figure 15.24 shows the kinetic energy profile of particles at two different temperatures, 40°C and 60°C. Indicate the temperatures represented by graph A and graph B.

260 062 IIndustrial In n chemistry

a Copy Figure 15.25 and label the following: H2(g) and I2(s); HI(g); ΔH; activation energy. b Is the reaction endothermic or exothermic? c Draw on the diagram the energy profile that would result if a catalyst was used in the reaction.

21 Read the following article and answer the questions below it.

Exploding iron In 1996, while the Turkish ship MV B. Onal was riding at anchor in Delaware Bay, near Philadelphia in the USA, a 2 tonne hatch cover suddenly blew off. As the ship was carrying a cargo of iron, the surprised crew asked themselves, ‘Can iron explode?’ As you may be aware, traditionally iron oxide (Fe2O3) is reduced to molten iron in a blast furnace [...] A new process which uses less energy has been developed. Iron oxide is converted directly to solid iron without having to heat the reactants to the melting point of iron. Iron oxide is heated to 550°C in the presence of carbon monoxide and hydrogen gas. The iron oxide is reduced to iron by both gases with the formation of carbon dioxide or water. Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O (g)

(1) (2)

The pellets of pure iron that are formed are extremely porous and full of many tiny holes, in contrast to the solid formed when the molten iron from a blast furnace cools. Under the right conditions the iron pellets can be oxidised back to iron oxide. In most cases, iron is oxidised slowly by oxygen back to iron oxide and the resulting heat can readily escape. If the pellets are more than one metre deep, as in the hold of a ship, the heat cannot escape quickly enough and the temperature rises. This speeds up the reaction rate. If the temperature increases sufficiently and water is present, another reaction occurs and the oxidation rate is speeded up 100-fold, with the release of more heat: Fe(s) + H2O(g) → FeO(s) + H2(g)

a What is the main reason the new reduction process uses less energy than the old process? b Write equations showing the oxidation of iron by oxygen to form iron(II) oxide and iron(III) oxide. c If water is present the oxidation reaction is speeded up 100-fold. Is water acting as a catalyst? Explain your answer. d Is the reaction shown in equation 3 endothermic or exothermic? e List the factors that increased the rate of reaction in equation 3. f Firefighters were not able to use water to put out the fire in the cargo hold. Why not? Suggest how they could put out the fire. 22 Chemical reactions in the body normally take place at 37°C. Explain how the rate of chemical reactions in the body can account for the following facts. a The body often responds to illness by an increase in temperature, accompanied by a higher pulse rate and faster breathing. b People rescued from drowning after 20 or 30 minutes in freezing water can sometimes survive and recover with no brain damage. 23 The first step in most toffee recipes is to dissolve about three cups of sugar in one cup of water. Although sugar is quite soluble in water, this step could be time-consuming. Use your knowledge of reaction rates to suggest at least three things you could do to increase the rate of dissolution without ruining the toffee.

(3)

Any spark or fire will set off an explosion of hydrogen gas, and that is what happened on the MV B. Onal.


1

chapter 16

Controlling the yield of reactions

key knowledge • reversible reactions: homogeneous equilibria and the equilibrium law, Le Chatelier’s principle and factors that affect the position of equilibrium

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the features of dynamic equilibrium • describe how rates of forward and back reactions change as a reaction approaches equilibrium • compare the rates of forward and back reactions at equilibrium • write an expression for the equilibrium law for a homogeneous reaction • explain the significance of the magnitude of the equilibrium constant for a reaction • use the equilibrium law to decide whether a particular reaction mixture is at equilibrium • perform calculations involving an equilibrium constant and concentration data • predict the effect on an equilibrium constant if temperature is increased or decreased • define and use Le Chatelier’s principle to predict the direction in which the equilibrium position shifts in response to change • recall that when a catalyst is added to a reaction, the value of the equilibrium constant and the equilibrium position are unchanged.

16.1

Volume = 1 L Temperature = 400˚C

A problem for industry: incomplete reactions 1.0 mol N2

Ammonia ranks as one of the top six chemicals produced in the world. As with many other products of the chemical industry, only a small proportion of the ammonia manufactured is sold directly to the public. Most of the ammonia is used in the production of fertilisers, plastics, explosives, pharmaceuticals, and other important chemicals. Ammonia is prepared from nitrogen and hydrogen gas. The reaction can be written as:

3.0 mol H2

Initially

N2(g) + 3H2(g) –—➤ 2NH3(g)

Suppose you mixed 1 mol of nitrogen and 3 mol of hydrogen gas in a sealed container. You might expect from the equation that 2 mol of ammonia would eventually be formed. In fact no matter how long you wait, the reaction seems to ‘stop’ when much less than 2 mol of ammonia is present (Figure 16.1). The stage when the quantities of reactants and products in the reaction remain unchanged is called chemical equilibrium. The fact that this reaction does not proceed to completion has serious consequences for companies manufacturing ammonia. From their point of view, the presence of large amounts of unreacted starting materials in reaction mixtures is wasteful and costly. The profitability of the industry is dependent upon the reaction yield—the extent of conversion of reactants into products (that is, how ‘far’ the reaction will go). As you might anticipate, this is not the only reaction that behaves this way. There are many reactions of this type. Some are employed in industrial processes; others are important in biological systems. To maximise the yield from such reactions, the following questions are vital: Why do some reactions reach equilibrium? How can the amount of product from a reaction that reaches equilibrium be increased?

Why are some reactions incomplete? Reversible reactions

0.7 mol N2 2.1 mol H2 0.6 mol NH3

After a long time (equilibrium)

Figure 16.1 When 1 mol of nitrogen and 3 mol of hydrogen are mixed, the reaction to form ammonia seems to stop before the reactants are consumed.

chemfact The yield of a chemical reaction is equal to the fraction of reactants converted into a product. It can be represented as a percentage. Percentage yield = actual amount of product × 100% theoretical amount of product

Some physical and chemical changes can be reversed. For example: An ice block in a drink melts, forming water. If the drink is placed in a freezer the water freezes. We can write these changes as: H2O(s)

H2O(l)

Water melting is represented by the forward equation: H2O(s) –—➤ H2O(l)

CHEMICAL EQUILIBRIUM RIUM • Incomplete chemical reactions • Incomplete chemical reactions and reaction rates

Water freezing is represented by the reverse reaction: H2O(l) –—➤ H2O(s)

A double arrow ( ) is used to indicate that this is a reversible reaction. Nickel–metal hydride (NiMH) batteries are used as portable energy sources in portable DVD players, digital cameras and calculators (Figure 16.2). A chemical reaction provides electrical energy when the

chemfact Note that equations for reversible processes may also be written in reverse. In this case the equation could have been written as: H2O(s) H2O(l)

263 Controlling the yield of reactions

batteries are in use. They are conveniently recharged, using an adaptor plugged into the mains electricity supply, and can be reused hundreds of times. During recharging the reverse reaction takes place, regenerating chemicals that had been consumed. The reactions involved are: discharging NiOOH(s) + MH(s) recharging

Figure 16.2 Ni–metal hydride batteries are used to power this digital camera. The chemical reaction in the batteries that provides electricity is reversed in order to recharge them.

Rate N 2

+

2

2N H

+3

CuSO4(s) + 5H2O(l)

The reaction between nitrogen and hydrogen to form ammonia, described above, can also be regarded as a reversible reaction, since ammonia decomposes into nitrogen and hydrogen when it is heated. The equation for this reaction is written as: 2NH3(g)

Equilibrium explained

H2

3

Equilibrium first established

Time

Figure 16.3 The variation of the rates of the forward and back reactions with time when nitrogen and hydrogen are mixed.

Concentration (M)

H2

2

1

CuSO4.5H2O(s)

2NH 3 N2

3

where M represents metals such as V, Ti, Cr and Co. You may have heated blue-coloured hydrated copper(II) sulfate (CuSO4.5H2O) in a test-tube. A white solid (CuSO4) forms and droplets of water appear on the walls of the test-tube. When water is mixed with the white powder, the blue compound is re-formed. This reaction is represented by the equilibrium equation:

N2(g) + 3H2(g)

3H

Ni(OH)2(s) + M(s)

N2 NH3 Time N2 and H2 added to container

Equilibrium established

Figure 16.4 Changes in the concentrations of N2, H2 and NH3 as a mixture of nitrogen and hydrogen gas reacts. As indicated by the coefficients of the equation for the reaction, for every amount of N2 that reacts, three times as much H2 reacts and twice as much NH3 is produced.


We can use the idea that processes can be reversed to understand why some reactions reach an equilibrium. Chemists have shown that in these reactions the forward and reverse reactions occur simultaneously. Suppose nitrogen gas and hydrogen gas were added to a sealed container at a constant temperature, as in Figure 16.1. The nitrogen and hydrogen start to react immediately, forming ammonia. The following sequence of events then occurs: As the forward reaction proceeds, the concentrations of nitrogen and hydrogen decrease, so the rate of the ammonia production, that is, N2(g) + 3H2(g) –—➤ 2NH3(g), decreases. At the same time as ammonia is being formed, some ammonia molecules react to re-form nitrogen and hydrogen. The rate of this reverse (or ‘back’) reaction, 2NH3(g) –—➤ N2(g) + 3H2(g), increases as the concentration of ammonia increases. Eventually the forward and back reactions proceed at the same rate (Figure 16.3). When this situation is reached, ammonia is formed at exactly the same rate as it is breaking down. The concentrations of ammonia, nitrogen and hydrogen will then remain constant. At the equilibrium position no further change will take place in the rate of either the forward reaction or the back reaction. The reaction has reached a point of balance—an equilibrium, i.e. N2(g) + 3H2(g) 2NH3(g). Figure 16.4 shows the changes in the concentrations of the chemicals. Equilibrium is a dynamic state, since the forward and back reactions have not ceased. Instead they occur simultaneously at the same rate. During dynamic equilibrium: the amounts and concentrations of chemical substances remain constant the total gas pressure is constant (if gases are involved) the temperature is constant the reaction is ‘incomplete’ (all of the substances are present in the equilibrium mixture).

extension Investigating dynamic equilibrium Chemists have used radioactive isotopes to monitor the behaviour of processes in dynamic equilibrium. Radioactive isotopes behave chemically in the same way as non-radioactive atoms of the same element, but have the advantage that their presence and location can be easily determined using a radiation detector. When solid sodium iodide (NaI) is added to water, it dissolves readily at first. As more and more of the salt is added the concentration of dissolved sodium iodide increases until a saturated solution forms and no further solid dissolves. When some solid sodium iodide, containing radioactive iodide ions, is added to a saturated solution, the subsequent movement of these ‘labelled’ ions can be traced. To the naked eye there is no sign that any of this added solid dissolves, and analysis shows that the concentration of sodium iodide in the solution does not increase. However, the solution becomes radioactive (Figure 16.6). The radioactivity of the solution shows that some of the radioactive sodium iodide has dissolved. Since the concentration remains constant, particles that were not radioactive must crystallise from the solution at the same rate as the radioactive solid dissolves (Figure 16.5). Even though we see nothing happening, there must be continual activity at the surface of the solid. The process is a dynamic equilibrium.

Ions leave the surface of the solid as they dissolve

Dissolved ions return to the surface of the solid at the same rate

Figure 16.5 The behaviour of ions in a saturated solution.

Add radioactive sodium iodide

Saturated sodium iodide solution

Radiation detector

Solid sodium iodide

No radiation

Solution becomes radioactive

Figure 16.6 This experiment shows a dynamic equilibrium is present in a saturated solution.

questions

?

E1 Write an equation to show the equilibrium that exists between NaI(s) and Na+(aq) and I–(aq). E2 a Sketch a graph of the change in the radioactivity of the solution over time. b On the same axes, sketch a graph of the change in the radioactivity of the solid over time.


How far do equilibrium reactions go? Do all reactions proceed to the same extent? We can answer this question with a simple experiment. Both hydrogen chloride (HCl) and ethanoic acid (CH3COOH) react with water in a hydrolysis reaction to form ions according to the equations: HCl(aq) + H2O(l) H3O+(aq) + Cl–(aq) CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq)

Solutions of hydrogen chloride and ethanoic acid will conduct electricity because they contain mobile ions. By measuring the electrical conductivity of solutions of the same concentration, we can compare the extent to which each compound ionises in water. Figure 16.7 shows the results obtained. You can see that the solution formed when hydrogen chloride dissolves in water (called hydrochloric acid) is a much better conductor than the ethanoic acid solution. The difference in conductivity arises because these reactions occur to remarkably different extents. At equilibrium in a 0.10 M solution, at 25°C, almost all the hydrogen chloride molecules are ionised, whereas only approximately 1% of the ethanoic acid molecules have reacted. Different reactions proceed to different extents. Battery

A

High conductivity reading

A

Low conductivity reading

+ H3O (aq) CH3COO-(aq) CH3COOH(aq)

+ H3O (aq) Cl (aq)

Figure 16.7 This experiment compares the electrical conductivity of 1 M solutions of hydrogen chloride and ethanoic acid. 1 M HCl(aq)

1 M CH3COOH(aq)

16.2

The equilibrium law So many reactions of importance involve equilibria that chemists have tried to find mathematical relationships they can apply to them. Consider the concentration of substances in the equilibrium: N2(g) + 3H2(g)

2NH3(g)

An unlimited number of different equilibrium mixtures of nitrogen, hydrogen and ammonia can be prepared. Table 16.1 shows data for four different equilibrium mixtures at 400°C. You can see that there is no obvious relationship between individual concentrations of N2, H2 and NH3.


TABLE 16.1 Concentrations of different equilibrium mixtures for 2NH3(g) N2(g) + 3H2(g) Equilibrium mixture

[N2] (M)

[H2] (M)

[NH3] (M)

A B C D

0.25 0.55 0.0025 0.0011

0.75 0.65 0.0055 0.0011

0.074 0.089 4.6 × 10–6 2.7 × 10–7

[NH3] [N2][H2]

[NH3]2 [N2][H2]3

0.39 0.25 0.33 0.23

0.052 0.052 0.051 0.051

[NH3] for each equilibrium mixture also [N2][H2] gives values for which no relationship can be seen. Many other fractions could be evaluated with a similar result. However, Table 16.1 shows that [NH3]2 the fraction gives an almost constant value for each mixture. This [N2][H2]3 concentration fraction is known as the reaction quotient for the reaction. While the concentration fraction can be calculated for any mixture of reactants and products, it is only at equilibrium that it has a constant value. We write: Evaluation of the fraction

K=

chemfact Units for equilibrium constants can be determined by substituting the unit M, or mol L–1, for each concentration into the concentration fraction. In this case the unit is given by: M2 = M–2 (or mol–2 L2) M × M3

[NH3]2 [N2][H2]3

where K is known as the equilibrium constant. The equilibrium constant for the reaction at 400°C is 0.052 M–2. Now consider another reaction. Table 16.2 gives concentrations in different equilibrium mixtures at 80°C for the reaction described by the equation: N2O4(g)

2NO2(g)

chemfact A concentration fraction can be calculated for any time during a chemical reaction and has the same expression as the equilibrium constant. When the reaction reaches equilibrium, the value of the concentration fraction becomes equal to K.

TABLE 16.2 Concentrations of substances in different equilibrium 2NO2(g) mixtures for N2O4(g) Equilibrium mixture

[N2O4] (M)

[NO2] (M)

[NO2]2 [N2O4]

A B C D

1.1 × 10–4 3.4 × 10–4 8.6 × 10–4 1.3 × 10–3

2.2 × 10–2 3.9 × 10–2 6.2 × 10–2 7.6 × 10–2

4.5 4.5 4.5 4.4

In this case the concentration fraction or reaction quotient that gives a [NO2]2 . The equilibrium constant, K, for constant value at equilibrium is [N2O4] this reaction has a value of 4.5 M at 80°C. From similar studies of equilibria, chemists have found that: different chemical reactions have different values of K for a particular reaction, K is constant for all equilibrium mixtures at a fixed temperature. By comparing the expressions for the reaction quotients with the corresponding equations, you will see that the index of each concentration in the expression is the same as the coefficient for the substances in the equation for the reaction.


For the equation aW + bX then: K=

EQUILIBRIUM • Principles of equilibrium • Concentration fraction and K • Concentration fraction • Equilibrium law (extension) • Determining concentration fractions

! K1, K2 and K3 are related mathematically: K1 =

1 = K32 K2

cY + dZ at a particular temperature,

[Y]c[Z]d [W]a[X]b

where K is a constant. This is known as the equilibrium law. Note that the equilibrium law depends upon the equation used for the reaction. For example, the equilibrium between the gases N2O4 and NO2 can be represented by several equations, including: N2O4(g)

2NO2(g), for which K1 =

[NO2]2 [N2O4]

2NO2(g)

N2O4(g), for which K2 =

[N2O4] [NO2]2

1 N O (g) 2 2 4

NO2(g), for which K3 =

[NO2] [N2O4]1/2

It is therefore important that the equation be specified when an equilibrium constant is quoted. You can see that if one equation is the reverse of another the equilbrium constants will be the inverse or reciprocal of each other. Doubling the coefficients of an equation causes the value of K to be squared.

What does an equilibrium constant tell us? Since the value of an equilibrium constant is based on the equilibrium concentrations of products divided by the equilibrium concentrations of reactants, it gives an indication of the extent of reaction, i.e. how far the forward reaction proceeds before equilibrium is established. For values of K that are between about 10–4 and 104, there will be significant amounts of reactants and products present at equilibrium. The reactions in Tables 16.1 and 16.2 are examples of this. For values of K that are very large (> 104), the equilibrium mixture consists mostly of products, with relatively small amounts of reactants. For example: HCl(aq) + H2O(l)

H3O+(aq) + Cl–(aq); K = 107 at 25°C

For values of K that are very small (< 10–4), the equilibrium mixture consists mostly of reactants, with relatively small amounts of products. For example: CH3COOH(aq) + H2O(l) Δ/

;

H3O+(aq) + CH3COO–(aq); K = 1.8 × 10–5 at 25°C

2

Effect of temperature on equilibria Exothermic (–)

Endothermic (+)

Figure 16.8 Effect of temperature on K for exothermic and endothermic reactions.


By experiment, it has been shown that the value of the equilibrium constant, K, for a particular reaction depends only upon temperature. It is not affected by actions such as addition of reactants or products, changes in pressure, or the use of catalysts. The effect of a change in temperature on an equilibrium depends on whether the reaction is exothermic or endothermic. As temperature increases: for exothermic reactions, the amount of products decreases and so the value of K decreases for endothermic reactions, the amount of products increases and so the value of K increases (Figure 16.8).

Since the value of K depends on temperature, it is important to specify the temperature at which an equilibrium constant has been measured.

Calculations using equilibrium constants The following examples deal with equilibrium reactions in which reactants are not completely consumed.

Worked example 16.2a Calculate the value of the equilibrium constant for the reaction represented by the equation: H2(g) + I2(g)

2HI(g)

at 460°C, if a 2.00 L vessel contains an equilibrium mixture of 0.0860 mol of H2, 0.124 mol of I2 and 0.716 mol of HI.

Solution [H2] =

n(H2) 0.0860 = = 0.0430 M V(H2) 2.00

[I2] =

0.124 = 0.0620 M 2.00

[HI] =

0.716 = 0.358 M 2.00

K=

[HI]2 0.3582 = = 48.1 [H2][I2] 0.0430 × 0.0620

The equilibrium constant has a value of 48.1. (Note that in this example the constant has no units as the number of particles in the numerator and denominator of the concentration fraction are equal.)

Worked example 16.2b The equilibrium constant for the reaction described by the equation: N2O4(g)

2NO2(g)

is 4.5 M at 80°C. In an equilibrium mixture at this temperature, what is the concentration of NO2 if the concentration of N2O4 is 0.0012 M?

Solution K=

[NO2]2 = 4.5 M [N2O4]

[NO2]2 = 4.5 0.0012 [NO2]2 = 4.5 × 0.0012 = 0.0054 Taking the square root of both sides, [NO2] = 0.0054 = 0.073 M The concentration of NO2 in the equilibrium mixture is 0.073 M. Therefore

Worked example 16.2c At a particular temperature 0.0500 mol of SO2, 0.0100 mol of O2 and 0.1500 mol of SO3 were mixed in a 2.00 L vessel and allowed to reach equilibrium according to the equation: 2SO2(g) + O2(g)

2SO3(g)

Analysis showed that 0.1400 mol of SO3 was present in the gas mixture at equilibrium. Calculate the value of the equilibrium constant at this temperature.


Solution Because the amount of SO3 has decreased, a net back reaction must have occurred. The amount of SO3 that has reacted is 0.1500 – 0.1400 = 0.0100 mol. From the equation, 2 mol of SO3 decomposes to 2 mol of SO2 and 1 mol of O2. Therefore, 0.0100 mol of SO3 will form 0.0100 mol of SO2 and 0.005 00 mol of O2. So at equilibrium: n(SO2) = 0.0100 + 0.0500 = 0.0600 mol n(O2) = 0.005 00 + 0.0100 = 0.0150 mol n(SO3) = 0.1400 mol n(SO2) 0.600 [SO2] = = = 0.0300 M V(SO2) 2.00 0.0150 [O2] = = 0.007 50 M 2.00 0.1400 [SO3] = = 0.0700 M 2.00 [SO3]2 K= [SO2]2[O2] 0.07002 = 726 M–1 0.03002 × 0.007 50 The equilibrium constant has a value of 726 M–1. =

Worked example 16.2d The equilibrium constant for the reaction N2O4(g) 2NO2(g) is 4.5 M at 80°C. A gas mixture in a 2.0 L vessel at 80°C contained 0.20 mol of N2O4 and 0.30 mol of NO2. Decide if the reaction is at equilibrium and, if it is not, predict the direction the reaction will shift to reach equilibrium.

chemfact When writing an equilibrium law, the concentration of solids is assigned as 1, since the concentration of a solid does not depend on how much of it is present. For example, K = [CO2] for the equation: CaCO3(s) CaO(s) + CO2(g) This is a heterogeneous reaction: a reaction involving different phases. Most reactions discussed in this chapter involve just one phase, either the gaseous state or aqueous solution. Such reactions are described as homogeneous.

summary

Solution n(N2O4) 0.20 = = 0.10 mol V(N2O4) 2.0 n(NO2) 0.30 [NO2] = = = 0.15 mol V(NO2) 2.0 [NO2]2 0.0225 The reaction quotient = = 0.23 M [N2O4] 0.10 [N2O4] =

Since the reaction quotient is not equal to K, the reaction is not in equilibrium. For an equilibrium to be established and the reaction quotient to equal K, the concentration of NO2 must increase and the concentration of N2O4 must decrease, i.e. a net forward reaction will occur.

16.1–16.2

EEquilibria ilib i arise i because b chemical h i l reactions ti can be b reversed. d At equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. Equilibria obey a mathematical relationship involving the concentrations of the reactants, known as the equilibrium law. The equilibrium constant, K, is a constant at a particular


ttemperature t andd its it value l provides id a measure off the th extent t t off reaction. The concentrations of reactants and products can be calculated using the equilibrium law. As temperature increases, the value of K increases for endothermic reactions and decreases for exothermic reactions.

key questions 1 a By referring to the equilibrium: H2O(l)

H2O(g)

explain what is meant by the ‘dynamic nature’ of chemical equilibrium and why wet clothes in a closed laundry bag do not dry. b When the bag in part a is opened the clothes begin to dry. Is this due to an equilibrium process? Explain your answer. 2 Write an expression for the equilibrium constant for the following equations. a 2NH3(g) N2(g) + 3H2(g) b 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) c Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq) 2+ d Ni (aq) + 6NH3(aq) Ni(NH3)62+(aq) 3 At 255°C the equilibrium constant is 100 M2 for the reaction: 2NH3(g)

N2(g) + 3H2(g)

a Write an expression for the equilibrium constant for the equation: N2(g) + 3H2(g)

2NH3(g)

b Calculate the equilibrium constant for the equation in part a. c Write an expression for the equilibrium constant for the equation: NH3(g)

1 3 N2(g) + H2(g) 2 2

d Calculate the equilibrium constant for the equation in part c. e Use your answers to parts a to d to state the effect on the value of an equilbrium constant when: i the equation is reversed ii the coefficients of the equation are halved. 4 A chemist investigated three different reactions and determined the value of the equilibrium constant for each. In which of the reactions are the products favoured over the reactants? Explain.

a Reaction 1: K = 0.0057 b Reaction 2: K = 2.5 × 109 c Reaction 3: K = 3.1 × 10–3 5 State whether the equilibrium constants for each of the following would be increased, decreased or unchanged by a rise in temperature: a 2NH3(g) N2(g) + 3H2(g); ΔH = +91 kJ mol–1 b 4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g); ΔH = –113 kJ mol–1 c H2(g) + CO2(g) H2O(g) + CO(g); ΔH = +42 kJ mol–1 d 2CO(g) + O2(g) 2CO2(g); ΔH = –564 kJ mol–1 6 For each of the equilibria in Question 5, would the amounts of products increase, decrease or remain unchanged as temperature increases? 7 Calculate the equilibrium constant for the reaction represented by the equation: N2O4(g)

2NO2(g)

if an equilibrium mixture in a 3.0 L container was found to consist of 0.120 mol of N2O4 and 0.500 mol of NO2. 8 The equilibrium constant for the reaction given by the equation: 2HI(g)

H2(g) + I2(g)

is 48.8 at 455°C. An equilibrium mixture in a 2.0 L vessel at this temperature contains 0.220 mol of H2 and 0.110 mol of I2. a Calculate the concentration of HI in this mixture. b Another mixture was prepared by placing 4.0 mol of HI in a 2.0 L vessel at 330°C. At equilibrium 0.44 mol of H2 and 0.44 mol of I2 were present. Calculate the value of the equilibrium constant at this temperature. c A third mixture consisted of 1.0 mol of HI, 0.24 mol of H2 and 0.32 mol of I2 in a 2.0 L container at 330°C. Decide if the mixture is at equilbrium and, if not, predict the direction the reaction will shift to reach equilibrium.

16.3

Changing the equilibrium position of a reaction Because economics is one of the driving forces of the chemical industry, the dependence of the composition of equilibrium mixtures on reaction conditions is of great importance to industrial chemists. By careful control of conditions, chemists can maximise the yield of a desired product. The equilibrium position (the relative amounts of reactants and products) of a reaction may be changed by: adding or removing a reactant or product changing the pressure by changing the volume (equilibria involving gases)


dilution (for equilibria in solution) changing the temperature. Let us consider the effect of each of these changes in turn.

Adding extra reactant or product

Concentration (not to scale)

Suppose a vessel contains an equilibrium mixture represented by the equation: N2

N2(g) + 3H2(g)

H2

NH3

Initial equilibrium Nitrogen added

Time New equilibrium established

Figure 16.9 A representation of changes in concentrations that occur when additional nitrogen gas is added to the equilibrium 2NH3(g). N2(g) + 3H2(g)

chemfact The effect of adding more nitrogen can also be predicted using the equilibrium law: K=

[NH3]2 [N2][H2]3

When extra nitrogen is added, the concentration fraction will be momentarily less than K. The mixture is then no longer in equilibrium. A net forward reaction will occur, increasing the amount of products, until the concentration fraction again becomes equal to K.

! Henri Louis Le Chatelier (1850–1936) published ‘A General Statement of the Laws of Chemical Equilibrium’ in the Journal of the French Academy of Sciences in 1884. Le Chatelier also developed the oxyacetylene welding torch, and the thermocouple for accurate temperature measurement.

CHEMICAL EQUILIBRIUM RIUM • Le Chatelier’s principle—concentration


2NH3(g)

If extra nitrogen gas were added to the vessel without changing the volume or temperature, the mixture would momentarily not be in equilibrium. The following events occur as the composition of the mixture adjusts to return to equilibrium: the increased concentration of nitrogen gas causes the rate of the forward reaction to increase and more ammonia is formed as the concentration of ammonia increases, the rate of the back reaction to re-form N2 and H2 increases ultimately the rates of the forward and back reaction become equal again and a new equilibrium is established (Figure 16.9). Note that when equilibrium is re-established the concentrations of all substances have changed. The overall effect of adding nitrogen is to increase the concentration of ammonia at equilibrium—a net forward reaction. For similar reasons, adding extra amounts of the other reactant, H2, also increases the concentration of ammonia. It is important to note, however, that even though the concentration of the added reactant decreases as the system moves to establish the new equilibrium, its final concentration is higher than in the original equilibrium. Addition of more product, NH3, increases the rate of the back reaction. When equilibrium is re-established the concentrations of the reactants, N2 and H2, are increased. The concentration of NH3, while decreasing as the system moves to the new equilibrium, is higher than at the original equilibrium. It is sometimes possible to arrange to remove a reactant or product from an equilibrium mixture. As you would anticipate, this has the opposite effect on an equilibrium from addition of the substance. To summarise: addition of a reactant leads to the formation of more products (a net forward reaction) addition of a product leads to the formation of more reactants (a net back reaction). Notice that, in all cases, a net reaction occurs that acts to partially counteract the effect of the change. A useful generalisation called Le Chatelier’s principle will help you to remember this: If an equilibrium system is subjected to a change, the system will adjust itself to partially oppose the effect of the change. This principle can be used to predict the consequences of adding substances to or removing substances from an equilibrium mixture. It can also be employed to predict the effect of changes of pressure and temperature on equilibria.

chemistryy in action Why do chickens lay eggs with thinner shells in the summer? Chickens, like dogs, do not perspire. Therefore in hot weather they must resort to panting to try to maintain a healthy temperature. This means that they exhale more carbon dioxide gas than when they are breathing normally. The following series of equilibria are affected: CO2(g) CO2(aq) CO2(aq) + H2O(l) H2CO3(aq) H2CO3(aq) H+(aq) + HCO3–(aq) – HCO3 (aq) H+(aq) + CO32–(aq) 2+ 2– CO3 (aq) + Ca (aq) CaCO3(s) eggshell Removing CO2(g) shifts each equilibrium, in turn, to the left. So, in summer, chickens lay eggs with thinner shells! This means eggs are more easily broken, at great economic cost to farmers and supermarkets. How to solve the problem? Ted Odorn, of the University of Illinois, found a bubbly solution: give chickens carbonated water to drink. Apparently the chickens like it, and they will produce stronger eggs. Another victory for chemistry and Le Chatelier.

! Acid rain is thought to be making the shells of birds’ eggs thinner in Britain. Acid rain reduces the calcium content of leaf litter consumed by the worms that are the staple diet of some birds.

Changing the pressure By changing volume The pressure of gases in an equilibrium mixture can be changed by increasing or decreasing the volume of the container while keeping the temperature constant. Consider the effect of increasing the pressure on different equilibrium systems.

Example 16.3a During sulfuric acid manufacture, one step involves sulfur dioxide reacting with oxygen to form sulfur trioxide gas: 2SO2(g) + O2(g) 3 gas particles

2SO3(g) 2 gas particles

In this equilibrium the forward reaction involves a reduction in the number of particles of gas, causing a reduction in pressure. The back reaction involves an increase in the number of gas particles, causing an increase in pressure. Applying Le Chatelier’s principle, an equilibrium will respond to an increase in pressure by adjusting to reduce the pressure. A net forward reaction will occur, increasing the amount of sulfur trioxide present at equilibrium (Figure 16.10). The effect on the pressure change on concentrations is shown in Figure 16.11.

Example 16.3b Colourless dinitrogen tetroxide gas and brown nitrogen dioxide gas exist in equilibrium: N2O4 1 gas particle

2NO2(g) 2 gas particles

When an equilibrium mixture of the gases is compressed it is observed that, after an initial darkening because of the higher concentration of the brown gas, the colour of the gas mixture fades. Some nitrogen dioxide has converted to dinitrogen tetroxide.


Concentration

Increased pressure

SO2 molecules:

5

SO2 molecules:

1

O2 molecules:

3

O2 molecules:

1

SO3 molecules:

1

SO3 molecules:

5

Total

9

Total

7

Figure 16.10 A representation of the effect of increased pressure on the equilibrium 2SO2(g) + O2(g)

2SO3(g).

O2 SO2 SO3 Time Initial equilibrium Pressure increase

New equilibrium established

Figure 16.11 The effect of increased pressure on the equilibrium 2SO2(g) + O2(g) 2SO3(g).

Concentration

The system adjusts to the increased pressure by undergoing a net back reaction: the equilibrium moves in the direction that produces fewer particles and reduces the pressure (Figure 16.12). Note that the concentration of NO2 in the new equilbrium will be higher than in the initial equilibrium, but not as high as it would be if there had not been a net back reaction (Figure 16.13).

N2O4 NO2

Time Initial equilibrium Pressure increase


Figure 16.12 The effect of increased pressure on the equilibrium N2O4(g) 2NO2(g).

Before compression (initial equilibrium)

Immediately after compression

At new equilibrium

Figure 16.13 When the pressure on an equilibrium mixture of N2O4 and NO2 is increased, the colour of the gas initially darkens but then fades as the reaction 2NO2(g) N2O4(g) occurs.

In general the effect of a change of pressure, by changing the container volume, depends on the relative number of gas particles on both sides of the equation.

By adding an inert gas CHEMICAL EQUILIBRIUM RIUM • Le Chatelier’s principle—pressure • Le Chatelier’s principle—inert gas • Le Chatelier’s principle—dilution

The total pressure of an equilibrium mixture of gases may also be changed, without changing the volume of the container, by adding a non-reacting gas such as helium, neon or argon. The impact of this change is very different to the previous case where the pressure is altered by changing the container’s volume. Despite the increase in total pressure that occurs when a non-reacting gas is added, the concentrations of the individual chemicals involved in the equilibrium are not affected by the presence of the extra gas. The system therefore stays in equilibrium and there is no net forward or back reaction.

Dilution For equilibria occurring in solution, the effect of diluting the solution by adding water is similar to changing the volume in gaseous equilibria. Where possible, a net reaction occurs in the direction that produces the greater number of particles.


For example, dilution of a solution containing the equilibrium system: –

Fe (aq) + SCN (aq) 2 particles in solution

2+

Fe(SCN) (aq) 1 particle in solution

results in an increase in the amount of Fe3+ and SCN– ions. In terms of Le Chatelier’s principle, a net back reaction increases the total concentration of particles in solution, offsetting the effect of dilution. Despite the net back reaction, the concentration of Fe3+ and SCN– at the new equilibrium will be lower than their concentration prior to dilution (Figure 16.14) as the equilibrium shift only partially offsets the change.

Changing the temperature Earlier we saw that the effect of a temperature change on an equilibrium reaction depends upon whether the reaction is exothermic or endothermic. Le Chatelier’s principle can also be used to determine the effect of heating on equilibrium mixtures. The conversion of brown nitrogen dioxide gas to colourless dinitrogen tetroxide gas is exothermic, and therefore could (but wouldn’t usually) be written as: 2NO2(g)

N2O4(g) + energy

Heating increases the energy of the substances in the mixture. Applying Le Chatelier’s principle, the reaction ‘opposes’ an increase in energy by removing energy—that is, a net backward reaction occurs (Figure 16.16). When a new equilibrium is attained, there is less dinitrogen tetroxide and more nitrogen dioxide present (Figure 16.15).

SCN– Concentration

3+

Fe3+ FeSCN2+

Time Initial equilibrium Dilution


Figure 16.14 Graph showing the effect of dilution on the equilibrium Fe3+(aq) + SCN–(aq) Fe(SCN)2+(aq).

chemfact FeSCN2+ has a dark red colour. It has been used in theatrical productions to make fake blood, formed by breaking separate plastic bags containing pale-coloured Fe3+ ions and colourless SCN– ions together.

Concentration

Figure 16.15 Heating an equilibrium mixture of NO2 and N2O4 favours the formation of brown NO2 gas.

N2O4

NO2

Time

For endothermic reactions, net forward reactions occur as the temperature rises. We can summarise this as follows. An increase in temperature in an equilibrium mixture results in: a net backward reaction (less products) for exothermic reactions a net forward reaction (more products) for endothermic reactions.

The influence of catalysts Catalysts increase the rate of reactions (Chapter 15). It has been shown experimentally that they increase the rate of forward and back reactions equally. As a consequence, the presence of a catalyst does not change the position of an equilibrium. A catalyst therefore has no effect on the equilibrium yield of a reaction. A catalyst may, however, greatly increase the rate at which an equilibrium is attained and it is for this reason that catalysts are employed in many industrial and biological systems.

Initial equilibrium Temperature increase


Figure 16.16 The effect of heating on the equilibrium 2NO2(g) N2O4(g).

CHEMICAL EQUILIBRIUM RIUM • Le Chatelier’s principle—temperature


16.4

Do all reactions reach equilibrium? Lime (CaO) is used to make mortar and plaster and to improve soils. It is made by heating limestone (calcium carbonate) strongly (Figure 16.17). Carbon dioxide gas is a by-product of the process. CaCO3(s) –—➤ CaO(s) + CO2(g)

Figure 16.17 Limestone is heated to 1000°C in this kiln to convert it to lime. If the process were performed in a sealed container an equilibrium mixture would form. Instead, the carbon dioxide produced is allowed to escape so that conversion into lime is complete. The limestone takes approximately 30 hours to travel the vertical length of the kiln, before emerging as lime.

extension Equilibria in the bloodstream Figure 16.18 Red blood cells photographed using an electron microscope.

chemfact A haemoglobin molecule contains four heme groups. An oxygen molecule binds to each heme group. Thus up to four oxygen molecules can be bonded to a haemoglobin molecule. (In the equation Hb is used to represent a heme group.) Hb4(O2)4 Hb4 + 4O2


After limestone is heated for some time, no trace of calcium carbonate remains. This reaction has not reached an equilibrium, since no reactants are present. The carbon dioxide gas produced during this process is allowed to pass into the atmosphere. From Le Chatelier’s principle, loss of product causes a net forward reaction to occur. Since carbon dioxide escapes continually, the reaction is driven forward until it is complete (no reactant remains). Many reactions can be regarded as continuing until they are complete. These include: reactions that produce products such as gases that escape from the reaction mixture as they are formed. Continual loss of products drives these reactions forward. reactions that form equilibria in which only minute quantities of reactants are present. The reaction of hydrogen chloride gas with water, which was described earlier, is such a case.

extension Equilibria in the bloodstream (continued)

Oxygen transport A large protein molecule called haemoglobin is responsible for the ability of blood to carry oxygen. Haemoglobin is the pigment in red blood cells. It combines with oxygen to form an equilibrium with oxyhaemoglobin: haemoglobin + oxygen

Atmospheric pressure of oxygen at 25°C

0.1

0.2

0.3

Partial pressure of oxygen (atm)

oxyhaemoglobin

When we inhale, oxygen from the air combines with haemoglobin present in the small blood vessels in the lining of the lungs. Using Le Chatelier’s principle, we can see that relatively high concentrations of oxygen in this environment will cause a net forward reaction. Most haemoglobin is converted to oxyhaemoglobin in the lungs (Figure 16.19). Oxyhaemoglobin is transported to the tissues through the blood circulatory system (Figure 16.20). In the tissues the concentration of oxygen is low. Using Le Chatelier’s principle again, we can see that the back reaction is favoured, releasing oxygen for use by cells. Thus, this important life-enabling process is achieved through the forward and backward movement of a chemical equilibrium as it responds to changes in the concentration of oxygen.

Carbon dioxide transport Carbon dioxide is produced by cells during respiration. An equilibrium between carbon dioxide gas and dissolved carbon dioxide is established. CO2(g)

100

Conversion to oxyhaemoglobin (%)

O bbodies Our di obtain bt i th the energy needed d d tto sustain t i lif life bby respiration, i ti a process in i which hi h glucose obtained from food reacts with oxygen. We obtain the oxygen from the air we breathe. The reactions that are involved in the process of respiration occur in cells throughout the body. One of the products is carbon dioxide; it is eliminated when we breathe out. The bloodstream transports oxygen around the body and carries away the carbon dioxide produced. The transport of both oxygen and carbon dioxide in the body involve chemical equilibria.

Figure 16.19 The fraction of oxyhaemoglobin in the blood depends on the pressure of oxygen.

Head and arms Carotid artery Pulmonary artery

Lung

Lung Pulmonary vein Right atrium Vena cava Right ventricle

Left atrium Liver Intestine

CO2(aq)

Aorta

Kidney

The relatively high concentration of carbon dioxide gas drives this reaction forward. The forward reaction is assisted by reaction of dissolved carbon dioxide with water: CO2(aq) + H2O(l) H2CO3(aq) H2CO3(aq) H+(aq) + HCO3–(aq)

In this sequence of reactions, carbon dioxide forms carbonic acid, a weak acid, which in turn is in equilibrium with the hydrogen carbonate ion. Almost 95% of carbon dioxide is transported back to the lungs as HCO3–. In the lungs, the low concentration of carbon dioxide gas in air causes these reactions to occur in reverse, releasing the carbon dioxide gas, which is exhaled.

Carbon monoxide poisoning Carbon monoxide is a colourless, odourless and tasteless gas present in cigarette smoke and in the exhaust gases from car engines. It forms when carbon or carbon compounds burn in a limited supply of air. It is toxic because, like oxygen, it reacts with haemoglobin: haemoglobin + carbon monoxide

Left ventricle

Legs

Figure 16.20 The circulatory system for blood.

chemfact The decomposition of carbonic acid is normally slow, but a protein catalyst or enzyme called carbonic anhydrase present in the body ensures that decomposition is sufficiently rapid for carbon dioxide to be expelled by the lungs.

carboxyhaemoglobin


extension Equilibria in the bloodstream (continued)

Figure 16.21 Police officers on traffic duty must be careful to limit their exposure to carbon monoxide gas from vehicle exhausts.

Carbon monoxide bonds much more strongly to the haemoglobin molecule than does oxygen. The equilibrium constant for this reaction is nearly 20 000 times greater than for the reaction between oxygen and haemoglobin. Even small concentrations of carbon monoxide cause the position of equilibrium to go well to the right. This has a critical effect on the equilibrium between haemoglobin and oxygen: haemoglobin + oxygen

oxyhaemoglobin

When carbon monoxide is available, the formation of carboxyhaemoglobin reduces the concentration of haemoglobin. This in turn causes the back reaction of oxyhaemoglobin. In extreme cases almost no oxyhaemoglobin is left in the blood and ‘carbon monoxide poisoning’ occurs. Symptoms of carbon monoxide poisoning include drowsiness, dizziness, headaches, shortness of breath and loss of intellectual skills. Loss of consciousness and death can result from carbon monoxide concentrations as low as 200 ppm. Large cities have average carbon monoxide concentrations of 7 ppm and this can rise to 120 ppm at busy intersections (Figure 16.21). Faulty car exhaust systems that allow carbon monoxide gas to enter the vehicle have caused numerous accidents and deaths. Similarly, using faulty gas and kerosene heaters in poorly ventilated rooms can be fatal. The reactions of oxygen and carbon monoxide with haemoglobin are described as competing equilibria, since both oxygen and carbon monoxide ‘compete’ for the same substance, haemoglobin. The equilibrium with the larger equilibrium constant has a significant effect on the position of equilibrium of the other reaction.

questions

?

E3 Death D ath De th by by carbon carbbon monoxide poisoning has been likened to suffocation. suffocation Explain why why. E4 Suggest how a person suffering from carbon monoxide poisoning should be treated.


summary

16.3–16.4

Th The arrows show h how h the th position iti off an equilibrium ilib i will ill shift hift in i response to t changes. h indicates a shift ‘to the right’ (a net forward reaction) indicates a shift ‘to the left’ (a net backward reaction) reactants

products

Add reactants: Add products: Increase pressure:

e.g. 2 reactant particles

1 product particle

e.g. 1 reactant particle

2 product particles

Increase temperature: exothermic reactions: endothermic reactions: Add a catalyst:

does not alter the extent of a reaction

The position of an equilibrium may be changed by adding or removing reactant or product, changing the pressure by changing volume (for equilibria involving gases), dilution (for equilibria in solution) and changing the temperature. The position of equilibrium is not affected by catalysts. Catalysts affect the forward and reverse reactions equally. If an equilibrium system is subject to change, the system will adjust itself to partly oppose the change. This is known as Le Chatelier’s principle and may be used to predict the effect of a change on an equilibrium.

key questions 9 Use Le Chatelier’s principle to predict the effect of adding more hydrogen gas to the following equilibria: a H2(g) + I2(g) 2HI(g) b 2NH3(g) N2(g) + 3H2(g) c H2(g) + CO2(g) H2O(g) + CO(g) 10 Predict the effect of the following changes on the position of each equilibrium. a addition of SO3 to the equilibrium: 2SO2(g) + O2(g)

2SO3(g)

–

b removal of CH3COO from the equilibrium: CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq)

c halving the volume (doubling the pressure) of the equilibrium: N2(g) + 3H2(g)

2NH3(g)

d increasing the pressure of the equilibrium: H2(g) + I2(g)

11 Consider the following equilibria. i H2(g) + CO2(g) H2O(g) + CO(g);

ΔH = +42 kJ mol–1 ii N2O4(g) 2NO2(g); ΔH = +58 kJ mol–1 iii H2(g) + F2(g) 2HF(g); ΔH = –536 kJ mol–1 How would you alter: a the temperature of each equilibrium mixture in order to produce a net forward reaction? b the volume of each mixture in order to produce a net forward reaction? 12 An equilibrium mixture consists of the gases N2O4 and NO2: N2O4(g)

2NO2(g)

The volume of the container was increased at constant temperature and a new equilbrium was established. Predict how each of the following quantities would change at the new equilibrium compared with the initial equilibrium: a concentration of NO2 b mass of NO2

2HI(g)

e increasing the temperature of the endothermic equilibrium: N2(g) + O2(g)

2NO(g)


16

key terms chemical equilibrium competing equilibria concentration fraction dynamic equilibrium

equilibrium constant equilibrium law equilibrium position extent of reaction

Le Chatelier’s principle reaction quotient reaction yield reversible reaction

EQUILIBRIUM • Le Chatelier’s principle—quantitative problems

Thh equilibrium ilib i llaw The 13 The magnitude of an equilibrium constant changes with temperature. Explain why it is referred to as a constant. 14 The equilibrium constant is 10–10 at 25°C for the reaction Sn4+(aq) + 2Fe2+(aq)

Sn2+(aq) + 2Fe3+(aq)

a Write an expression for the equilibrium constant for this reaction. b Would significant reaction occur when solutions of tin(IV) chloride and iron(II) chloride are mixed? Explain. c Determine the value of the equilibrium constant for the reaction: 2+

3+

Sn (aq) + 2Fe (aq)

4+

2+

Sn (aq) + 2Fe (aq)

d Would significant reaction occur when solutions of tin(II) chloride and iron(III) chloride are mixed? Explain. 15 Write balanced equations for the reactions with the following equilibrium laws: [H ]2[CO] a K= 2 [CH3OH] [H S]2 b K= 2 2 [S2][H2] [N O ]1/2 c K= 2 4 [NO2] 16 Acetone (C3H6O) is used to remove nail polish. It can be prepared from propan-2-ol using a copper–zinc catalyst, according to the equation: C3H8O(g)

PCl3(g) + Cl2(g)

A 3.00 L vessel contained 6.00 mol of PCl3, 4.50 mol of PCl5 and 0.900 mol of Cl2 at equilibrium at 250°C. a Write an expression for the equilibrium constant for this reaction.


PCl3(g) + Cl2(g)

PCl5(g)

18 The reaction used to manufacture ammonia is represented by: N2(g) + 3H2(g)

2NH3(g)

The equilibrium constant for the reaction is 0.052 M–2 at 400°C. Each of the following gas mixtures is contained in a 1.0 L vessel at 400°C. Decide if each mixture is in equilibrium. If not, predict the direction the reaction will shift to reach equilibrium. a 0.20 mol of N2, 0.20 mol of H2, 0.20 mol of NH3 b 1.00 mol of N2, 1.00 mol of H2, 0.25 mol of NH3 c 0.050 mol of N2 and 0.050 mol of H2 only. 19 The equilibrium constant for the reaction CO(g) + 3H2(g)

CH4(g) + H2O(g); ΔH = +208 kJ mol–1

is 0.667 M–2 at a particular temperature. A mixture of 0.100 M CO, 0.200 M H2, 0.300 M CH4 and 0.400 M H2O is heated to this temperature. Copy the table below and mark with a tick what would happen to the concentration of each chemical as the reaction approaches equilibrium.

C3H6O(g) + H2(g)

If an equilibrium mixture of these gases consists of 0.018 mol of propan-2-ol, 0.082 mol of acetone and 0.082 mol of hydrogen in a 20 L vessel, calculate the value of the equilibrium constant. 17 Consider the equilibrium: PCl5(g)

b Calculate the equilibrium constant for the reaction at 250°C. c Another equilibrium mixture contains 0.0020 M PCl5 and 0.0010 M PCl3 at 250°C. What is the concentration of chlorine in this mixture? d Determine the equilibrium constant at 250°C for the reaction:

Concentration increases

Concentration does not change

Concentration decreases

CO H2 CH4 H2O 20 At one step during the synthesis of nitric acid, dinitrogen tetroxide is in equilibrium with nitrogen dioxide: N2O4(g)

2NO2(g)

0.540 mol of N2O4 was placed in a 2.00 L vessel. When equilibrium was achieved, 0.280 mol of NO2 was present. Calculate the value of the equilibrium constant at this temperature. 21 5.89 mol of N2 and 8.23 mol of H2 were mixed in a 5.0 L vessel and reacted according to the equation: N2(g) + 3H2(g)

2NH3(g)

When equilibrium was reached, it was found that 0.48 mol of NH3 had been formed. Calculate the value of the equilibrium constant. 22 Consider the reaction: A + 3B

2C + D

Analysis of an equilibrium mixture in a 2.0 L container shows that 2.0 mol of A, 0.50 mol of B and 3.0 mol of D are present. If the equilibrium constant of the reaction is 0.024 M–1, calculate: a the concentration of A, B and D at equilibrium b the concentration of C in the equilibrium mixture c the amount of C, in mol, in the equilibrium mixture 23 2.0 mol of sulfur dioxide gas and 2.0 mol of nitrogen dioxide gas are placed in a 1.0 L vessel, which is then sealed and heated. After equilibrium has been reached, analysis of the mixture shows that 1.3 mol of nitrogen dioxide is present. The reaction that occurred is described by the equation: SO2(g) + NO2(g)

SO3(g) + NO(g)

a What amount of sulfur trioxide is present in the equilibrium mixture? b What amount of sulfur dioxide is present in the equilibrium mixture? c Calculate the value of the equilibrium constant for the reaction under these conditions.

Changing the equilibrium position 24 How will the concentration of hydrogen gas in each of the following equilibrium mixtures change when the mixtures are heated and kept at constant volume? a N2(g) + 3H2(g) 2NH3(g); ΔH = –91 kJ mol–1 b CH4(g) + H2O(g) CO(g) + 3H2(g); ΔH = +208 kJ mol–1 25 The following equations represent reactions that are important in industrial processes. Predict the effect on the equilibrium position if each reaction mixture was compressed at constant temperature. a C3H8O(g) C3H6O(g) + H2(g) b CO(g) + 2H2(g) CH3OH(g) c N2(g) + O2(g) 2NO(g) 26 Some elderly people, especially women, become very susceptible to bone breakages. It is thought that as people

age they absorb Ca2+ ions from food inefficiently, reducing the concentration of these ions in body fluids. An equilibrium exists between calcium phosphate in bone and calcium ions in body fluids: 3Ca2+(aq) + 2PO43–(aq)

Ca3(PO4)2(s)

Explain why inefficient absorption of Ca2+ ions could cause weakness in bones. 27 Carbon monoxide is used as a fuel in many industries. It reacts according to the equation: 2CO(g) + O2(g)

2CO2(g)

In a study of this exothermic reaction, an equilibrium system is established in a closed vessel of constant volume at 1000°C. a Predict what will happen to the equilibrium position as a result of the following changes: i decrease in temperature ii addition of a catalyst iii addition of more oxygen b What will happen to the equilibrium constant as a result of each of the changes above? c If carbon monoxide can be used as a fuel, comment on the magnitude of the equilibrium constant for the reaction. 28 Sealed bottles of fizzy drinks, such as lemonade and sparkling wines, contain carbon dioxide gas in equilibrium with dissolved carbon dioxide: CO2(g)

CO2(aq)

The forward reaction is exothermic. a Use Le Chatelier’s principle to explain why bubbles appear when the bottles are opened. b Why are the drinks usually cooled before they are carbonated? 29 A step during nitric acid production is the oxidation of nitrogen oxide to nitrogen dioxide. 2NO(g) + O2(g)

2NO2(g); ΔH = –114 kJ mol–1

Nitrogen dioxide is a brown coloured gas and nitrogen oxide and oxygen are colourless. An equilibrium mixture was prepared in a 1 L container at 350°C. Copy the table below, and for each of the following changes indicate if the reaction mixture would become darker or lighter, giving a reason for your choice. Colour change lighter / darker

Explanation

a The temperature is increased to 450°C at constant volume. b The volume of the container is increased at constant temperature. c A catalyst is added at constant volume and temperature.


C2H6(g)

C2H4(g) + H2(g); ΔH = +138 kJ mol–1

a Copy the table below and place a tick in the appropriate box to indicate what will happen to the equilibrium percentage yield of ethene when each of the following changes is made. Increased ethene yield

No change in ethene yield

Decreased ethene yield

i The volume is reduced at constant temperature. ii More hydrogen gas is added at constant temperature and volume. iii The temperature is increased at constant volume. iv A catalyst is added. v Argon gas is added at constant temperature and volume. b How will each of the changes in part a affect the rate at which the reaction achieves equilibrium? 31 Methanol is manufactured for use as a fuel for racing cars. It can be made by reaction between carbon monoxide and hydrogen: CO(g) + 2H2(g)

CH3OH(g); ΔH = –103 kJ mol–1

a What conditions of temperature and pressure would be required for: i a fast reaction rate? ii a high equilibrium yield of methanol? b Will a compromise be needed in the choice of temperature or pressure? c Suggest another method that could be employed at a manufacturing plant to increase the rate of methanol production. d As part of an investigation of this process, the concentrations of a mixture of CO, H2 and CH3OH were monitored continuously. The mixture was initially at equilibrium at 400°C and constant volume. After 10 minutes additional CO was added to the mixture, as shown in Figure 16.22.

Concentration (not to scale)

d More oxygen is added at constant volume and temperature. 30 Ethene gas is produced from ethane gas in an endothermic reaction represented by the equation:

H2

CO CH3OH

10

Time (min)

Figure 16.22 Graph of concentration versus time.

i Sketch a graph to show how concentrations would change as a consequence of the addition of CO. ii Following the addition of the CO the mixture again reaches equilibrium. Sketch a second graph to show the effect on the concentrations if the temperature were then increased to 450°C. 32 During the contact process for sulfuric acid manufacture, sulfur dioxide is converted to sulfur trioxide at temperatures of 400–500°C: 2SO2(g) + O2(g)

2SO3(g); ΔH = –197 kJ mol–1

a What would be the effect of increasing the pressure on: i reaction rate? ii equilibrium yield? b In practice, this step is usually performed at atmospheric pressure. Suggest why. c During the process, sulfur trioxide is removed from the reaction mixture by converting it to sulfuric acid. The remaining gases are then recycled to the reaction vessel. Explain the reason for recycling the gases. d What factors would have influenced the choice of the reaction temperature?

Connecting the main ideas 33 Write a paragraph about chemical equilibrium that uses the terms ‘forward reaction’, ‘reverse reaction’, ‘equilibrium law’, ‘equilibrium constant’, and ‘Le Chatelier’s principle’. 34 Explain the difference between the terms ‘concentration fraction’ and ‘equilibrium constant’. 35 a The equilibrium constant is 0.67 M–2 at a particular temperature for the reaction: CO(g) + 3H2(g)

CH4(g) + H2O(g)

A mixture of 0.100 M CO, 0.200 M H2, 0.300 M CH4 and 0.400 M H2O is heated to this temperature. Copy the table below and indicate with a tick what would happen to the concentration of each chemical as the reaction approaches equilibrium.


Concentration does not change

Concentration decreases

CO H2 CH4 H2O b When the temperature of the reaction mixture is increased by 10°C, the equilibrium constant for the reaction becomes 0.71 M–2. What conclusion can you make about the enthalpy change of this reaction? 36 Carbon disulfide gas (CS2), which is used in the manufacture of rayon, can be made using an endothermic gas-phase reaction between sulfur trioxide gas (SO3) and carbon dioxide. Oxygen gas is also produced in the reaction. a Write a balanced chemical equation for the reaction. b Write an expression for the equilibrium constant of the reaction. c An equilibrium mixture of these gases was made by mixing sulfur trioxide and carbon dioxide. It consisted of 0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2 in a 20 L vessel. Calculate the value of the equilibrium constant at that temperature. d Predict how each of the following changes to an equilibrium mixture would affect the yield of CS2. i removal of O2 (at constant total volume) ii increasing the temperature iii adding a catalyst iv increasing the pressure by decreasing the volume of the reaction vessel (at constant temperature) v increasing the pressure by introducing argon gas into the reaction vessel (at constant volume)

37 Sulfur dioxide gas and oxygen gas were mixed at 600°C to produce a gaseous equilibrium mixture: 2SO2(g) + O2(g)

2SO3(g)

A number of changes were then made, including the addition of a catalyst, resulting in the formation of new equilibrium mixtures. Figure 16.23 shows how the concentrations of the three gases changed. SO2 0.100

Concentration (M)

Concentration increases

O2 0.050 SO3

0.000

10

20

30

Time (minutes)

Figure 16.23 Graph of concentration versus time.

a Write an expression for the equilibrium constant, K, of the reaction. b During which time intervals was the reaction at equilibrium? c Calculate the value of K at 18 minutes. d At what time was the catalyst added? Explain your reasoning. e What change was made to the system at 20 minutes?


1

chapter 17

Equilibria involving acids and bases

key knowledge • pH as a measure of strength of acids and bases: Kw, Ka for weak acids

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • write an expression for the ionisation constant of water • explain the difference between acidic, basic and neutral solutions in terms of the relative concentrations of hydronium and hydroxide ions • define pH as a measure of acidity • perform calculations involving pH and concentrations of hydronium and hydroxide ions • describe the effect of temperature on the pH of water • write an expression for the acidity constant of an acid • perform calculations involving acidity constants and concentration data • describe the function of a buffer.

chemistryy in action The equilibria in a swimming pool

Figure 17.1 Chemical equilibria involving acids and bases keep the water in a backyard swimming pool hygienic and safe for use.

! HOCl molecules diffuse through the cell walls of micro-organisms and react with the enzyme required for digestion of glucose, thus cutting off the micro-organism’s energy supply.

O a hhot summer’s On ’ dday, there’s h ’ probably b bl nothing hi more refreshing f hi than h a long l dip di in i the h cool, clear water of a swimming pool. It turns out that there is some interesting chemistry at the heart of keeping swimming pools clean, clear and safe to swim in. The water in swimming pools is used again and again. Even though it is filtered, the water can quickly become contaminated with microscopic algae and bacteria. Swimming pools are ‘chlorinated’ to prevent the growth of harmful micro-organisms. Chlorination produces hypochlorous acid (HOCl), which is a very efficient antibacterial agent and algicide. Commercially available ‘pool chlorine’ powder (Figure 17.2) is composed of calcium hypochlorite (Ca(OCl)2), which dissolves in water to release hypochlorite ions. The hypochlorite ions then react with hydronium ions in the water to form hypochlorous acid:

Figure 17.2 Commercial ‘pool chlorine’ contains the hypochlorite ion (OCl–).

Ca(OCl)2(s) Ca2+(aq) + 2OCl–(aq) – + OCl (aq) + H3O (aq) HOCl(aq) + H2O(l)

The H3O+ is available from the self-ionisation of water, a reaction that occurs between water molecules, forming H3O+ and OH– ions:

10

80

20

70

30

60

40

50

50

40

60

H3O+(aq) + OH–(l) –

+

The relative amounts of HOCl, OCl ions and H3O ions in a swimming pool need to be controlled carefully. This is done by monitoring the pH of the swimming pool and adding either more pool chlorine or acid as needed to maintain a pH in the range 7.2 to 7.8. Figure 17.3 shows the relationship between the three substances. Remember, pH measures the concentration of H3O+ ions. As pH increases, the concentration of H3O+(aq) decreases. Le Chatelier’s principle tells us the position of equilibrium will move to the left, consuming some of the HOCl. If the pH rises above about 7.8, the concentration of HOCl will be insufficient to control the growth of bacteria and algae. On the other hand, as pH falls the concentration of H3O+(aq) increases. Le Chatelier’s principle tells us the position of equilibrium will move to the right and more HOCl will be formed. Although pH values below around 7.2 result in greater amounts of HOCl in the pool, swimmers experience eye and skin irritation if the pool is too acidic. Thus, maintaining a pool so that it is hygienic and comfortable for swimmers involves carefully using pH to maintain an optimum equilibrium position in the reaction mixture.

30

70

20

80

10

90 4

5 6 7 8 9 10 pH Desirable pH range for swimming pools

%OCI-

0

90

%HOCI

H2O(l) + H2O(l)

100

100

Figure 17.3 The effect of pH on the proportion of HOCl and OCl– in water at 20°C.

285 Equilibria involving acids and bases

17.1

Acidity of solutions We have studied acid–base reactions previously in some detail in Units 2 and 3. We can now re-examine some aspects of acids and bases in light of our new understanding of equilibrium. The Brønsted–Lowry theory defines an acid as a proton donor and a base as a proton acceptor. An acid–base reaction involves the transfer of one or more protons from an acid to a base. Substances that donate protons are referred to as acids, and substances that accept protons are referred to as bases. Some substances have the ability to donate or accept protons depending on the circumstances. These substances are described as amphiprotic. Water is a good example.

The ionisation constant of water In Unit 2 we saw that water will react with itself in a process called selfionisation: H2O(l) + H2O(l)

H3O+(aq) + OH–(aq)

[H3O+][OH–] . [H2O]2 In aqueous solutions, water molecules are usually far more abundant than any other substance present. The concentration of water in aqueous solutions is virtually constant at about 56 M. We therefore write the equilibrium law for the ionisation of water as:

At equilibrium, K =

[H3O+][OH–] = K × [H2O]2 = a constant

This expression can also be written as: Kw = [H3O+][OH–]

where Kw is called the ionisation constant of water. This equilibrium law applies to both pure water and all aqueous solutions. In pure water at 25°C, chemists have found that the concentration of both H3O+ and OH– ions is 10–7 M. The value of Kw at 25°C can be calculated: Kw = [H3O+][OH–] = 10–7 × 10–7 = 1.0 × 10–14 M2

Notice that the concentrations of ions in pure water are extremely small. For each H3O+ ion present in a glass of water, there are 560 million H2O molecules! There are ions there, though, so, although pure water is regarded as a poor conductor of electricity, it does conduct slightly.

Acidic and basic solutions In solutions of acidic substances, H3O+ ions are formed by reaction of the acid with water, as well as from self-ionisation of water. So the concentration of H3O+ ions will be greater than 10–7 M at 25°C. Since the product [H3O+][OH–] remains constant, the concentration of OH– ions in an acidic solution at 25°C must be less than 10–7 M. The opposite is true for basic solutions. The concentration of OH– ions in a basic solution is greater than 10–7 M and that of H3O+ ions is less than 10–7 M. In summary, at 25°C: in pure water and neutral solutions: [H3O+] = [OH–] = 10–7 M in acidic solutions: [H3O+] > 10–7 M and [OH–] < 10–7 M


in basic solutions: [H3O+] < 10–7 M and [OH–] > 10–7 M. The higher the concentration of H3O+ ions in a solution, the more acidic the solution is. Concentrations of H3O+ range from about 10–14 M in strongly basic solutions, such as 1 M sodium hydroxide solution, to 10 M in concentrated strong acids, such as 10 M hydrochloric acid. Indicators that change colour at specific concentrations of H3O+ ions can be used to measure the acidity of solutions.

pH: a convenient way to measure acidity The range of H3O+ concentrations in solutions is so great that a convenient scale, called the pH scale, has been developed to measure acidity. pH is defined as:

14 1 M NaOH 13 Drain cleaner, oven cleaner

pH = –log10 [H3O+]

Detergents for 12 dishwashing machines

Alternatively, this expression can be rearranged to give: Base: [H3O+] < [OH-]

[H3O+] = 10–pH

For pure water at 25°C, [H3O+] = 10–7 M, so pH = –log10 10–7 = 7. For acidic solutions, [H3O+] > 10–7 M, so pH < 7. For example, a solution of 0.020 M HCl at 25°C has [H3O+] = 0.020 M so pH = –log10 0.020 = 1.7. For basic solutions, [H3O+] < 10–7 M; so pH > 7. For example, a solution of 0.300 M NaOH at 25°C has [OH–] = 0.300 M. Since [H3O+][OH–] = 1.0 × 10–14 M2, [H3O+] × 0.30 M = 1.0 × 10-14 M2 1.0 × 10-14 M2 [H3O+] = = 3.33 × 10-14 M 0.30 M So pH = –log10 (3.33 × 10–14) = 13.5 Using the pH scale, the most acidic solutions have pH values slightly less than 0 and the most basic solutions have values of about 14 (Figure 17.4).

11 Laundry detergents 10 Sea water 9 Pancreatic juice 8

Neutral (at 25˚C): [H3O+] = [OH-]

Eggs, toothpaste Blood

7 Pure water Milk 6 Rain water 5 Black coffee Tomatoes 4 Oranges

Acid: [H3O+] > [OH-]

3 Soft drinks 2

Vinegar Lemons

1 Gastric juice 0 1 M HCl

Worked example 17.1a

–1 10 M HCl

Calculate the concentration of hydrogen ions in blood if the pH of blood is 7.4. Figure 17.4 The pH values of some common solutions.

Solution [H3O+] = 10–pH = 10–7.4 = 4.0 × 10–8 M (using a calculator) The concentration of H3O+ ions in blood is 4.0 × 10–8 M.

Worked example 17.1b 40.0 mL of 0.200 M hydrochloric acid reacts with 20.0 mL of 0.100 M sodium hydroxide solution. Calculate the pH of the resulting solution.

Solution The reaction can be represented by the equation: HCl(aq) + NaOH(aq)

NaCl(aq) + H2O(l)

Reactions between strong acids and strong bases can be regarded as being virtually complete, so: n(HCl) = c × V = 0.200 × 0.0400 = 0.008 00 mol n(NaOH) = 0.100 × 0.0200 = 0.002 00 mol From the equation, n(HCl) reacted with NaOH = n(NaOH) = 0.002 00 mol. So, n(HCl) unreacted = 0.008 00 – 0.002 00 = 0.006 00 mol.


Since 1 mol of HCl gives 1 mol of H3O+, n(H3O+) = 0.006 00 mol. The total volume of the mixture is 40.0 + 20.0 mL = 60.0 mL. n(H3O+) 0.006 mol = 0.100 M c(H3O+) = = V(H3O+) 0.0600 L So, pH = –log10 0.100 = 1.00

How is pH affected by temperature? Earlier, we defined the ionisation constant for water as: Kw = [H3O+][OH–] = 1.0 × 10–14 M2 at 25°C

We can use this relationship to calculate either the [H3O+] or [OH–] at 25°C in different solutions. But what happens if the temperature is not 25°C? From experimental data we know that the self-ionisation reaction for water is endothermic:

TABLE 17.1 Variation of Kw and pH with temperature. Temperature (°C)

Kw (M2)

0 5 15 25 35 45 55

1.14 × 10–15 1.85 × 10–15 4.51 × 10–15 1.00 × 10–14 2.09 × 10–14 4.01 × 10–14 7.29 × 10–14

summary

pH of pure water 7.47 7.37 7.17 7.00 6.83 6.70 6.57

H2O(l) + H2O(l)

H3O+(aq) + OH–(aq); ΔH = +57 kJ mol–1

In Chapter 16, we saw that equilibrium constants are temperature dependent. If a reaction is endothermic, the equilibrium constant increases as the temperature increases and decreases as temperature decreases. As the temperature of a solution rises above 25°C, the position of equilibrium will favour the endothermic forward reaction; [H3O+] and [OH–] increase, causing Kw to rise. Consequently, pH will decrease (Table 17.1). The opposite is true when a solution cools below 25°C. The position of equilibrium favours the exothermic back reaction; [H3O+], [OH–] and Kw all decrease and the pH increases. The pH of pure water only equals 7.00 at 25°C. At other temperatures, even though the pH of pure water is not equal to 7.00, it can still be described as neutral because the concentrations of H3O+ and OH– ions are equal.

17.1

W Water t self-ionises lf i i according di tto H2O(l) + H2O(l) H3O+(aq) + OH–(aq). The ionisation constant of water is: Kw = [H3O+][OH–] = 10–14 M2 at 25°C. In an acidic solution the concentration of H3O+(aq) is greater than the concentration of OH–(aq). In a basic solution the concentration of OH–(aq) is greater than the concentration of H3O+(aq).

In I neutral t l solution l ti the th concentrations t ti off H3O+((aq)) andd OH–((aq)) are equal. Acidity is measured using the pH scale: pH = –log10 [H3O+]. At 25°C the pH of a neutral solution is 7, the pH of an acidic solution is less than 7 and the pH of a basic solution is greater than 7. The pH of pure water decreases as temperature increases because the self ionisation of water is endothermic.

key questions 1 Calculate the pH of the following solutions: a 0.000 10 M HNO3(aq) b 0.050 M HCl(aq) c 0.000 10 M KOH(aq) d 0.020 M NaOH(aq) 2 Calculate the concentration of H3O+ and OH– ions at 25°C in: a pancreatic juice of pH 8.0 b Coca-Cola of pH 3.0 c urine of pH 6.0


3 Copy and complete the following table. Solution

[Solute] (M)

HCl(aq)

0.010

HNO3(aq)

[OH–] (M)

0.030

NaOH(aq) Ca(OH)2(aq)

[H3O+] (M)

0.020 0.0010

Is the solution acidic, basic or neutral?

17.2

Acidity constants Most acid–base reactions in water can be considered as equilibrium reactions. Hydrochloric acid, for example, ionises in water according to the equation: HCl(aq) + H2O(l)

H3O+(aq) + Cl–(aq)

The expression for the equilibrium constant can be written as: K=

[H3O+][Cl–] [HCl][H2O]

In this reaction, the Cl– ion can be described as the conjugate base of HCl. In any acid–base reaction, the conjugate base is the substance formed from the acid when it donates a proton (H+). Similarly, HCl is referred to as the conjugate acid of Cl–. Since water is the solvent in aqueous solutions and its concentration is virtually constant, we can write the equilibrium expression as: K[H2O] =

[H3O+][Cl–] = Ka [HCl]

The quantity Ka is known as the acidity constant. The value of Ka for hydrochloric acid is 107 M at 25°C. This means that in hydrochloric acid solutions, most of the hydrochloric acid has been converted to H3O+ and Cl–. This is why hydrochloric acid is classified as a strong acid. In contrast, ethanoic acid has an acidity constant of 1.75 × 10–5 M at 25°C. In a solution of ethanoic acid, the position of equilibrium favours the reactants and there is a relatively small amount of products. CH3COOH(aq) + H2O(l)


[H3O+][CH3COO–] K= [CH3COOH][H2O] [H3O+][CH3COO–] Ka = [CH3COOH] Ethanoic acid is classified as a weak acid because at equilibrium only a small proportion of the acid has been ionised. The acidity constant can be used as a measure of an acid’s strength. The acidity constants of several common acids are given in Table 17.2. From the table you can see, for example, that while phosphoric acid is stronger than citric acid, it is weaker than nitric acid.

Calculations involving acidity constants The following example shows how an acidity constant can be used to calculate the pH of a solution of a weak acid.

TABLE 17.2 K a values for some common acids at 25°C Acid

Ka (M)

Ethanoic acid Citric acid* Phosphoric acid* Nitric acid Hydrochloric acid Sulfuric acid*

1.75 × 10–5 7.2 × 10–4 7 × 10–3 2 × 101 107 109

*In the case of the polyprotic acids citric acid, phosphoric acid and sulfuric acid, the Ka value quoted involves the loss of the first proton.

! Refer to the Acidity Constant table in the VCE Chemistry Data Book. This is available on the VCAA website.

Worked example 17.2 Calculate the pH and percentage hydrolysis of a 0.50 M ethanoic acid solution, given that the Ka for ethanoic acid is 1.75 × 10–5 M.

Solution The equation for the ionisation of ethanoic acid is: CH3COOH(aq) + H2O(l) +

Ka =


–

[H3O ][CH3COO ] = 1.75 × 10–5 M [CH3COOH]


From the equation for the reaction we can see that for every mole of ethanoic acid that ionises, one mole of H3O+ and one mole of CH3COO– will be formed. So we can say:

review Since an HCl molecule can donate only one proton in its reaction with water, it is described as monoprotic. Polyprotic acids can donate more than one proton per molecule. H2SO4, with two protons per molecule to donate, is a diprotic acid, and H3PO4, with three protons it can donate, is a triprotic acid.

[H3O+] = [CH3COO–]

If we substitute this into the expression for Ka, we get: [H3O+]2 = 1.75 × 10–5 M [CH3COOH] We also know from the equation for the reaction that in a 0.50 M solution of ethanoic acid: [CH3COOH] + [CH3COO–] = 0.50 M

However, ethanoic acid is a weak acid with a very small Ka, so we know it will only ionise to a very small extent. As the small value of Ka indicates that little CH3COO– is formed by ionisation, we can approximate the above expression to: [CH3COOH] ≈ 0.50 M

chemfact Chemists define a corresponding quantity, the basicity constant, Kb, to measure the strength of bases. Taking ammonia as an example: NH4+(aq) + OH–(aq) NH3(aq) + H2O(l) [NH4+][OH–] Kb = [NH3] For a weak acid and its conjugate base, Ka × Kb = Kw.

If we substitute this into the revised expression for Ka, we get: [H3O+]2 = 1.75 × 10–5 M 0.50 M [H3O+]2 = 8.75 × 10–6 [H3O+] = 2.96 × 10–3 pH = –log [2.96 × 10–3] = 2.5 We can measure the extent of the reaction by calculating the percentage hydrolysis (or percentage ionisation), the fraction of the acid that is ionised. Percentage hydrolysis is given by the expression: [CH3COO–] × 100 2.96 × 10–3 = × 100 [CH3COOH] 0.50 = 0.59% This means that in a 0.50 M solution just six in every thousand CH3COOH molecules are hydrolysed at any instant—no wonder it’s classified as a weak acid! These ideas can be generalised to solutions of any weak acid represented by HA: HA(aq) + H2O(l)

A–(aq) + H3O+(aq)

[H3O+] = [A–] and [HA] does not change during the ionisation (hydrolysis).

Buffers: using equilibrium to resist change /(

/HX (¶ Z[H`ZYV\NOS` JVUZ[HU[ILJH\ZL

7SLU[`VM/([V THRLTVYL/HX PMZVTL\ZLK\WI` HSRHSP[OH[NL[ZHKKLK

7SLU[`VM(¶[V JVTIPUL^P[OHU` /HX[OH[NL[Z HKKLK

Buffers are solutions that can absorb the addition of acids or bases with little change of pH. They can be made most easily by mixing a weak acid and a salt of its conjugate base. To understand how buffers work, let us consider one made from a mixture of ethanoic acid and sodium ethanoate. The resulting solution will contain an equilibrium mixture of CH3COOH, H3O+ and CH3COO–. CH3COOH(aq) + H2O(l)

Figure 17.5 How the equilibrium between a weak acid, HA, and its conjugate base, A–, keeps the pH constant.



The important feature of this solution is that it contains significant amounts of both the weak acid and its conjugate base. If a solution of a strong acid such as HCl is added to the equilibrium mixture of ethanoic acid/ethanoate ion, the pH will decrease, but much less than one would expect. The addition of the HCl solution disturbs the equilibrium. Le Chatelier’s principle tells us that the system will respond to oppose the change and restore equilibrium. The addition of H3O+ causes a net back reaction, consuming H3O+ and creating more CH3COOH.

The addition of a strong base consumes H3O+, causing a net forward reaction to produce more H3O+. In both cases, the overall effect on [H3O+] is small and so the change in pH is minimal. Buffers are especially important in environmental and living systems where they help maintain delicate chemical balances essential to life.

extension Buffering action of the oceans Th pH The H off the h oceans iis about b 8. 8 ItI is i maintained i i d at this hi level l l by b a series of reactions as summarised in Figure 17.6. A change in the equilibrium position of one reaction will cause a shift in the equilibrium position of the others. The interaction between these reactions means that the oceans act as a giant buffer. Gradual changes in the concentration of atmospheric carbon

dioxide di id over a period i d off thousands h d off years hhave not resulted l d in i any great change to the pH of the oceans. This happy situation may not last. Climate scientists are concerned that the increase in atmospheric carbon dioxide in the last two hundred years has happened too quickly for the buffer system in the oceans to deal with, and that the ocean pH will decrease in consequence.

CO2(g) CO2(aq)

If the HCO3–(aq) concentration falls, this equilibrium will shift to the right

CO2(aq) + H2O(l)

If oceans start becoming too acidic, HCO3–(aq) ions remove H+(aq) ions as this equilibrium shifts to the left H+(aq) + HCO3–(aq)

CO32–(aq) + H+(aq)

HCO3–(aq)

CaCO3(s) Ca2+(aq) + CO32–(aq) If carbonate ions react with H+(aq) ions, then calcium carbonate will dissolve to restore this equilibrium Shells and limestone in the sea are a reservoir for the anions which can remove H+(aq) ions from the oceans

Figure 17.6 Sea water contains an equilibrium mixture of CO2, H2CO3, HCO3– and CO32– ions. The mixture acts as a buffer, preventing changes of pH detrimental to marine life.

Figure 17.7 Coral reefs and shellfish will be particularly vulnerable to changes in the equilibrium position in the buffering system of the oceans.

questions

?

E1 UUse se tthe he iinformation nformati f tion provided in Figure 17.6 17 6 to explain why gradual changes in the concentration of atmospheric carbon dioxide have not resulted in a change in the pH of the oceans. E2 The shells of marine animals consist of calcium carbonate, which is in equilibrium with dissolved carbonate ion in the oceans. Why would some scientists have suggested that the increased concentration of dissolved carbon dioxide could have an adverse effect on marine organisms before there is a measurable change in the pH of the oceans?


pH in the body

Blood pH Death

7.8

A number of reactions that occur in the human body involve acid–base reactions. Without a means of controlling acidity, the pH of body fluids could fluctuate from extremely basic to extremely acidic. Biological systems cannot function in such extreme environments. Blood, for example, is maintained within a narrow pH range of 7.35–7.45. Severe respiratory illnesses such as pneumonia and emphysema can cause this value to drop to potentially lethal levels, a condition called acidosis. Untreated diabetes can have a similar effect. On the other hand, hyperventilation, caused by rapid breathing, increases the pH value and causes alkalosis (Figure 17.8). The presence of buffers maintains pH values within narrow limits in the body. Control of pH in the blood is achieved using different buffers. One of the important buffers is made up of carbonic acid (H2CO3) and the hydrogen carbonate ion (HCO3–):

Alkalosis 7.45 7.35 Acidosis

7.0

Death

Figure 17.8 Effects of changes in blood pH.

H2CO3(aq) + H2O(l)

chemfact Another buffer system in blood involves haemoglobin. Haemoglobin is a weak acid that loses a proton to form a weak conjugate base.

summary

HCO3–(aq) + H3O+(aq)

The action of this buffer system becomes clear if we consider what happens when an acid or base is added to blood: If H3O+ ions are added, a net back reaction occurs, removing most of these ions. If OH– ions are added, they react with H3O+ ions. A net forward reaction results, producing more H3O+ ions. The OH– ions have effectively reacted with the carbonic acid.

17.2

Th The equilibrium ilib i expression i for f the th ionisation i i ti reaction ti off a monoprotic acid in water: HA(aq) + H2O(l)

can be written as Ka =

H3O+(aq) + A–(aq)

[H3O+][A–] [HA]

Ka is i known k as the th acidity idit constant t t andd is i a measure off the th strength of an acid. Buffers absorb the effects of the addition of acids or bases with little change in pH. • A buffer solution contains a weak acid and its conjugate base. Buffers have an important role in maintaining pH in many chemical and biochemical processes.

key questions 4 Write Ka expressions for each of the following acids: a NH4+ b HCOOH c HCN 5 Chloroacetic acid is a weak monoprotic acid with a Ka of 1.3 × 10–3 M. For a 1.0 M solution of chloroacetic acid, calculate: a the pH b the percentage hydrolysis 6 The following equilibria are involved in the transport of carbon dioxide throughout the body: In the lungs: CO2(g)

CO2(aq)

In body tissues: CO2(aq) + H2O(l)

H2CO3(aq) H+(aq) + HCO3–(aq)

Buffers in the blood prevent these reactions from causing large changes in pH.


a Use Le Chatelier’s principle to explain what would otherwise happen to the pH of blood: i in the lungs, where CO2(g) is breathed out ii in the tissues, where CO2(aq) is produced by reactions occurring in the cells. b Hyperventilation results from rapid breathing. i When might a person experience hyperventilation? ii Use the above equations to explain what effect hyperventilation has on blood pH. iii Find out how to treat a person suffering from this condition. c During cardiac arrest the heart stops beating, but other cellular functions continue. Doctors sometimes quickly inject a sodium hydrogen carbonate solution directly into the heart muscle before restarting the heart. Why?

17

key terms

acid acidity acidity constant base buffer

conjugate acid conjugate base ionisation constant of water neutral percentage hydrolysis

Acids, bases and pH 7 Identify the conjugate acid–base pairs in the following equations. H3O+(aq) + NO3–(aq) a HNO3(aq) + H2O(l) b HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) – c HCO3 (aq) + H2O(l) H2CO3(aq) + OH–(aq) 3– d PO4 (aq) + H2O(l) HPO42–(aq) + OH–(aq) 8 Perchloric acid (HClO4) is a strong monoprotic acid. For a 0.0100 M perchloric acid solution, calculate: a [OH–] b pH 9 100 mL of hydrochloric acid of pH 4.0 is mixed with 100 mL of hydrochloric acid of pH 5.0. What is the pH of the solution formed? 10 Calculate the pH of the solution formed when 7.30 g of hydrogen chloride is dissolved in water to make up 2.00 L of solution. 11 10.0 mL of lemon juice of pH 2.00 was diluted so that the final volume was 1000 mL. Calculate the pH of the resulting solution. 12 What mass of hydrogen chloride gas must be dissolved in 1.00 L of hydrochloric acid to change the pH from 2.0 to 1.0, if it is assumed that no change in volume occurs? 13 Find the pH of the solution formed when: a 100 mL of 0.0100 M hydrochloric acid is added to 20 mL of 0.0100 M potassium hydroxide solution b 50 mL of 0.0100 M hydrochloric acid is added to 50 mL of 0.0200 M nitric acid c 50 mL of 0.0100 M hydrochloric acid is added to 50 mL of 0.0100 M sodium hydroxide solution 14 In pure water at 25°C, the concentration of both H3O+ and OH– ions is 10–7 M due to the reaction described by the equation: 2H2O(l)

H3O+(aq) + OH–(aq)

The forward reaction is endothermic. Suppose some pure water is heated. a What happens to the value of Kw? b What happens to the concentration of H3O+ and OH– ions? c What happens to the pH? d Does the water become acidic, neutral or basic? Explain.

percentage ionisation pH self-ionisation of water strong acid weak acid

15 Solutions of the acid–base indicator methyl red contain an equilibrium mixture of the red-coloured acid form of the indicator and a yellow-coloured conjugate base form. The equilibrium can be represented by: HMr(aq) red

H+(aq) + Mr–(aq) yellow

The equilibrium constant for the reaction is 10–5 M at 25°C. a Write an expression for the equilibrium constant for the reaction. [Mr–] b Calculate the value of the ratio when the indicator is [HMr] dissolved in a solution of pH 2. c Predict the colour of the indicator in the solution in part b. d What colour would the indicator be in solutions of: i pH 5? ii pH 10?

Acidity constants 16 A 0.100 M solution of hypoiodous acid (HOI) has a pH of 5.8. a Write an equation for the ionisation of hypoiodous acid. b Write an expression for Ka for hypoiodous acid. c Calculate [H3O+] in the solution. d Calculate Ka for hypoiodous acid. 17 A student investigating an equilibrium mixture produced by reacting methanoic acid (HCOOH) and water finds that the pH of the solution is 2.4 and the concentration of HCOOH is 0.10 M. Calculate the Ka of methanoic acid. 18 Ethanoic acid, with a Ka of 1.75 × 10–5 M, is the principal acidic constituent in vinegar. A particular brand of vinegar is found to have a pH of 3.0. Calculate the concentration of ethanoic acid in the vinegar. 19 Benzoic acid is added in small amounts to some foods to act as a preservative. A 1.0 M solution of benzoic acid is 1.4% hydrolysed. a Calculate Ka for benzoic acid. b What percentage of benzoic acid molecules will be hydrolysed in a 0.10 M solution? 20 A buffer made from sodium hydrogen carbonate (NaHCO3) and carbonic acid (H2CO3) is sometimes used in swimming pools. a What is the function of a buffer?


b Using this buffer as an example, explain how buffers operate. c Use the information on page 285 to explain the advantages of a buffer in pool water.

Connecting the main ideas 21 a Write ionic equations to explain why: i ammonium chloride has a pH less than 7 ii sodium sulfide has a pH greater than 7 b Sodium carbonate is readily soluble in water but calcium carbonate is insoluble. A solution of sodium carbonate was found to have a pH of 8. A quantity of calcium nitrate was added to the solution. Using ionic equations, explain the effect the addition of calcium nitrate would have on the pH of the solution. 22 Propanoic acid (abbreviated to HPr) is a weak monobasic acid used to retard the growth of mould in foods. A 0.100 M solution was prepared. The acidity constant for the acid is Ka = 1.30 × 10–5 M. a Assuming x moles of HPr dissociate, complete the following table. Initial concentration (M) Change in concentration (M) Equilibrium concentration (M)

[HPr]

[H3O+]

[Pr–]

0.100

0

0

–x

0.100 – x ≈ 0.100

b Determine [H3O+]. c Determine the pH of the solution. d Determine the percentage dissociation of HPr.


23 Butanoic acid (C4H8O2), a weak monoprotic acid, is responsible for the unpleasant smell of curdled milk and parmesan cheese. Butanoic acid is known to have a Ka of 1.5 × 10–5 M. 0.881 g of butanoic acid is mixed with water to make up 100 mL of solution. a Write an equation for the ionisation of butanoic acid with water. b Write an expression for Ka for butanoic acid. c Is butanoic acid a stronger or weaker acid than ethanoic acid? d Calculate the pH of the butanoic acid solution. e Would you expect the pH of a similar solution of ethanoic acid to be higher or lower?

1

chapter 18

The chemical industry: risks and rewards

key knowledge • principles of waste management and occupational health and safety used in the chemical industry

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the relevance of the chemical industry in our society • discuss issues raised by the activities of the chemical industry • describe ways in which the yields of chemical products can be optimised • explain the significance of energy requirements, safety considerations, economics and pollution for the chemical industry • understand the risk assessment process and the role of labels and Material Safety Data Sheets for safe working with chemicals • identify sources and impacts of chemical waste • describe, compare and apply waste management principles, strategies and options • describe factors considered when a new chemical industry site is established • describe interrelationships between different chemical industries.

18.1

Australia’s chemical industry chemistryy in action The beginnings of an industry Phosphorus is an essential plant nutrient. It is often lacking in Australian soils, so farmers supply it in the form of superphosphate. The active ingredient in superphosphate is calcium dihydrogen phosphate. Pouring sulfuric acid onto sheep bones was the way to make it in 19th century Footscray and Yarraville, near Melbourne. In earlier times, bones were simply ground up and spread over farms to release their phosphorus content. Sheep may have been in short supply in Europe, where the bones of men and horses were collected from the batttlefields and shipped to England for use as fertiliser! Sulfuric acid is now used in Australia to manufacture superphosphate fertiliser from imported phosphate rock: Figure 18.1 A worker discharging phosphate to Cuming Smith & Co., Yarraville, 1911.

2Ca5(PO4)3F(s) + 7H2SO4(l) + 3H2O(l) –—➤ 7CaSO4(s) + 3Ca(H2PO4)2.H2O(s) + 2HF(g)

Despite some reduced demand for locally manufactured superphosphate since the 1980s, it is still added in huge amounts to fertilise Victoria’s nutrient-poor soils.

Paints and inks Explosives 10% 3% Soaps, detergents and cosmetics 12%

Industrial gases and inorganics 9% Resin and industrial organics 15%

Fertilisers and pesticides 8%

Rubber, plastics products and packaging 43%

Figure 18.2 Australian chemical industry products: turnover by sector.


In addition to its use in making superphosphate, sulfuric acid is increasingly used to produce ammonium phosphate fertiliser and for extracting nickel from ores. Demand for rubber, plastics and packaging drives Melbourne’s chemical industry today. Other local products are shown in Figure 18.2. Australia is about twentieth on the list of world exporters of chemicals, with annual trade worth around US$3 billion. Over 40 000 Australians are employed directly in the industry, with many thousands more employed indirectly supplying goods and services. Together, this represents about one person in every ten employed in manufacturing industries. If we were to include the large petroleum and metal extraction industries as well, the chemical industry would be the largest manufacturing group in the country. A range of different skills and qualifications are required by industrial chemists. Chemical engineers constantly supervise the operation of the plant (factory), solve production problems, develop modifications to the existing plant and assist in the design of new projects. Analytical chemists are concerned with quality control, providing analyses of raw materials, products and emissions and problems, develop modifications to the existing plant, and assist in the design of new projects. Research chemists undertake research on new processes and new materials and develop new analytical techniques.

Raw materials (minerals, plants, atmospheric gases)

Chemical industry

Other manufacturing industry

Consumers

Figure 18.3 The chemical industry plays a key, but often hidden, role in the production of consumer goods.

summary

18.1

Th The chemical h i l industry i d t produces d chemicals h i l that th t are usedd in i the th manufacture of materials that contribute to our lifestyle.

Chemists Ch i t perform f a variety i t off roles l in i the th chemical h i l industry. i d t

key questions 1 Compare the tasks of chemical engineers, analytical chemists and research chemists by filling in the Venn diagram (Figure 18.4) with the following characteristics: scientist, problem-solver, detective, discoverer, communicator, mathematician, production supervisor, quality controller, accountable, specialist. The example shows ‘chemist’ applying to all three roles. 2 Would you expect to find qualified chemists working in the sales or management sections of a chemical company? Explain how chemical skills could be useful in such commercial roles.

Research chemist

Chemical engineer Chemist

Analytical chemist

Figure 18.4 The role of the industrial chemist.

18.2

A global enterprise A large, healthy chemical industry is vital to the well-being of our society, and demand for chemicals is increasing steadily. Consumers purchase very few chemicals as raw materials. Perhaps this is one of the reasons why the public knows so little about the chemical industry and often fails to recognise its importance. Most of its products are used by other manufacturing industries, such as the pharmaceutical, agriculture and textile industries, to make more sophisticated consumer products.

297 The chemical industry: risks and rewards

25 20 15 10 5 0

Su lfu ric

ac id Eth en e Ph os ph Lime ori ca cid Am mo nia Pr op e ne So Ch di u l or m i n hy So dro e di u xid m e ca r Ch bon a l or oe te the ne

Annual production (Gt)

45 40 35 30

Substance

Figure 18.5 The top ten chemicals produced in the US.

! Chemistry is big business! Multinational chemical companies earn billions of dollars every year in their biggest market, the US. For example, Dow Chemical has annual sales of about US$40 billion and around 43 000 employees. Dow supplies more than 3300 products through 165 manufacturing sites in 37 countries, including an Australian plant at Altona near Melbourne. The company has customers in 175 countries and supplies many markets including food, transport, health and medicine, personal and home care, and building and construction. About half of Dow’s income is generated by plastic products.

Research

Laboratory < 1 kg Review + modifications

Scaling up

Pilot plant ~ 1000 kg Review + adjustments

Scaling up

Production plant (tonnes)

Figure 18.6 The search for optimum conditions requires several stages before the chemical plant is built.


Chemical industries convert raw materials, such as plant material, minerals or atmospheric gases, into useful and profitable products. Figure 18.5 shows the top ten chemicals produced in the United States, which dominates the global chemical industry. The chemicals shown in Figure 18.5 are used in many different industries. For example, uses of sodium hydroxide include: manufacturing soap and detergents, cosmetics, pharmaceuticals, household bleaches, paper, rayon and cellophane, dyes, and cleaning compounds refining vegetable oil and petroleum engraving and etching peeling fruit and vegetables electroplating. Companies producing chemicals vary from small businesses with only a handful of employees producing specialised materials, to large multinational operations. Since the products of one company are often the raw materials for others, chemical industries tend to be concentrated in large complexes, such as those at Footscray and Altona in Victoria and Botany in New South Wales. Multinational chemical companies such as Dow Chemical, H.B. Fuller, Lubrizol, Orica, PPG Industries, Rohm & Haas and Sigma Aldrich all have offices or plants in Australia, many in Victoria. Europe and Asia are also major forces in the global chemical industry, again with many links to the Australian economy.

18.3

Inside a chemical industry A chemical industry must operate efficiently if it is to make and market a product that will provide company shareholders with an adequate return on their investment. At the same time, companies must operate in a responsible manner that minimises hazards to employees and damage to the environment. The needs to be efficient and to be environmentally responsible are often conflicting. As a result, industries do a great deal of research and development and frequently modify processes and work practices. The development of a new industrial plant for the manufacture of a chemical involves a number of stages. It is important to build a plant that produces the maximum amount of product at the lowest possible cost. Initially small-scale experiments are carried out in the laboratory. Reaction conditions are then tested on a larger scale in a pilot plant, which is a smaller version of the the final industrial plant. It is only after these tests have been completed satisfactorily that a full scale industrial plant is built. The production of most industrial chemicals occurs in a number of stages. These are summarised in Figure 18.7.

Energy in or out

Excess energy recycle loop

Products Feedstock (reactants)

REACTION Temperature, pressure, catalyst

Feedstock preparation

Recycle loop

Separation

Unused feedstock

Co-products (and by-products)

Input or removal of energy may be required at any stage

Figure 18.7 Sequence of operations in a chemical plant.

Yields As we saw in Chapter 16, the extent and rate of conversion of raw materials to product is influenced by reaction conditions such as temperature, pressure and concentration. To avoid waste and reduce costs, conditions are carefully selected to ensure that acceptable yields of product are obtained. Chemists therefore try to ensure that: the reaction rate is fast a high proportion of reactants is converted to products at equilibrium. As you can see from Table 18.1, selection of an appropriate temperature and pressure will be difficult for some processes, since conditions that favour rapid reaction will sometimes give low yields of product. There is often a need to compromise between between rate and equilibrium considerations. In Chapters 19 to 22, we examine some important industrial processes, using our knowledge of how reaction rate and the position of equilibrium can be controlled by modifying reaction conditions.

Batch and continuous processing There are two main ways that reactions are controlled in industry.

Batch processing

TABLE 18.1 Conditions that favour fast reaction rates and high equilibrium yields For fast rates High concentrations/pressures High temperatures High surface area of solids Use of a catalyst For high equilibrium yields Pressures depend on the relative numbers of reactant and product particles Low temperatures for exothermic reactions; high temperatures for endothermic reactions Addition of excess reactant Removal of product as it forms

Batch processing is the method used when fixed amounts of reactants are mixed, giving fixed amounts of products. It is used when there are low costs or negligible delays in starting or stopping production, for one-off production and small quantities of product.

Continuous flow processing In continuous flow processing, reactants are mixed continuously, providing a constant supply of products. This technique uses smaller, less expensive equipment than batch processing and is the preferred method when there are high costs or long delays in starting or stopping production. However, it is only cost effective if there is sufficient demand for the large quantities of product obtained. Continuous flow processing allows greater control over reaction conditions, especially temperature, than batch processing. Reactants may be added or products removed to increase yield.


Costs Other factors also influence the final choice of reaction conditions. Expense, for example, may influence the selection of: catalyst—a less expensive catalyst may be preferable to a more effective one pressure—high-pressure reaction vessels are costly to build and operate temperature—fuel costs rise rapidly as operating temperatures increase choice of reactant to use in excess safety and waste minimisation and disposal—the safest option is not always the cheapest! For an industry to be economic, expenses must be carefully estimated and controlled. Aside from the capital required to build a new plant, raw materials usually represent the largest single cost, with other expenses for labour, energy, marketing, transport, waste disposal and depreciation. It may take a company many years to recover its expenses for the research and development of new products. Income

+ Revenue

Research and development

Production

Construction

Time (years)

–

Expenditure

Figure 18.8 Establishment and operation costs for a chemical plant during the development, construction and operational phases.

Energy In the laboratory, we tend to use Bunsen burners without thinking about the cost of the gas. Industries cannot afford to do this—even simple drying processes can become expensive when performed on a large scale. Costs associated with energy use are closely monitored. For industries such as aluminium extraction that use electrical energy to bring about reactions, energy costs represent a major part of the total expenditure. Energy use can sometimes be reduced by careful choice of processes. For example, copper and zinc minerals in the ore mined at Broken Hill are separated from other worthless minerals by floating the ore in large vats of special solutions. This process very effectively concentrates the minerals and employs little energy. With efficient management, heat produced in one part of a plant may be utilised elsewhere. If sufficient thermal energy is generated, it may be possible to convert it to electrical energy for use in the plant and, perhaps, sale to the electricity grid supply grid. A good example of this is sulfuric acid manufacture, where the energy produced by the process can be more valuable to the operating company than the sulfuric acid that it makes!


summary

18.2–18.3

Ch Chemical i l industries i d t i use either ith a batch b t h process that th t produces d a fixed amount of product at a time or a continuous process that produces a continous supply of product. Ideally the production of an industrial chemical should occur at a fast rate with a high equilibrium yield. In reality there is often a compromise between rate and yield as industry selects the conditions that produce a product at the most economical rate.

Reaction R ti conditions diti suchh as temperature, t t pressure andd use off a catalyst also need to be taken into account. Other factors that industry must consider are availability and cost of feedstock, and energy costs, waste management, and health and safety issues.

key questions 3 Is a typical school chemistry experiment batch or continuous processing? Use an example to explain your answer. 4 Platinum is a useful catalyst. Explain how its rising cost (platinum has outstripped gold over the last few years largely in response to demands from industry) is related to the size of the global chemical industry. 5 How might new legislation to enforce stricter safety and pollution control measures affect the price of the next product

you buy? Who should pay this cost—the manufacturer or the consumer? 6 Which chemist would you call on to improve yields in a chemical manufacturing process—a chemical engineer, a research chemist or an analytical chemist? Explain. 7 How would the price of a chemical product affect what yields are acceptable to the manufacturer?

18.4

Working safely Occupational health and safety The area of employee safety is usually called occupational health and safety, or OHS. Australia has laws that enforce a duty of care by organisations that supply or use chemicals. For instance, Dangerous Goods (Storage and Handling) Regulations 2000 sets out severe penalties on people who fail to discharge their safety duties. Penalties for negligence that results in someone’s death or serious injury can include hefty fines or imprisonment. Chemical industries also self-regulate their activities. Members of the Australian Plastics and Chemical Industries Association must sign and follow a set of ‘Responsible Care’ policies that includes six codes covering all aspects of the life cycle of the chemical: community right to know manufacturing process safety environment protection storage and transport safety employee health and safety product stewardship.

Figure 18.9 OHS at a chemical plant—a safety notice at Qenos Olefins, Altona.

PETROLEUM GASES LIQUEFIED UN No.

1075 HAZCHEM

2WE

FLAMMABLE GAS 2

Dangerous goods and hazardous substances Dangerous goods have the potential to pose immediate danger to people, property or the environment. The risk typically arises during transport, storage or handling. Examples of dangerous goods include petrol, pool

Figure 18.10 A HAZCHEM placard for a tank containing liquefied petroleum gas (LPG).


Figure 18.11 HAZCHEM signs.

EXPLOSIVE

1.1

EXPLOSIVE

1.3

1.6 EXPLOSIVE

Figure 18.12 Explosive hazard label diamonds.


chlorine and ammonium nitrate fertiliser. Dangerous goods are assigned a specific United Nations (UN) identity number and classified as: Class 1: Explosive Class 2: Gases (flammable, non-flammable, toxic) Class 3: Flammable liquids Class 4: Flammable solids (spontaneously combustible or emit flammable gases when wet) Class 5: Oxidising Class 6: Toxic or infectious Class 7: Radioactive Class 8: Corrosive Class 9: Miscellaneous dangerous substances. Hazardous substances are classified on the basis of their short- and longterm effects on the health or safety of people, typically in the workplace. Many are also classed as dangerous goods, but others such as asbestos may cause illness or disease some time after exposure. Health effects caused by hazardous substances may range from nausea and headaches to birth defects and major organ damage. These effects depend upon the substance involved and the degree of exposure. Over 2000 Australians die each year as a result of exposure to hazardous substances. Disinfectant, glue, paint and pesticides can be hazardous. Other hazardous substances found in the workplace range from acids and caustic substances (bases) to heavy metals, solvents and petroleum products. Warning signs about dangerous goods are required at any site where they are used or stored, for example factories, hospitals and storage depots. Vehicles transporting dangerous goods must also carry appropriate signs. Some farms also require signage for agricultural chemicals. HAZCHEM signs inform emergency services about firefighting requirements, violent reactions, protective equipment and the need for evacuation. Classes and subclasses of dangerous goods are identified by safety ‘diamonds’ (Figure 18.12).

Material Safety Data Sheets and risk assessment Everything we do poses some risk. Doing nothing at all is also a risk— never getting out of bed is likely to give you pressure sores and lead to an early death! A risk assessment is a formal way of identifying risks and assessing potential harm from a hazardous chemical in a particular situation. Effective risk management reduces exposure to hazardous chemicals to at least acceptable levels. Risk assessments can be performed for lifestyles as well as industrial production and school chemical experiments. Physical, social and legal factors are usually considered. For example, benzene and nicotine are carcinogens found in cigarettes. The risk of cancer increases with exposure. Ways of managing this hazard include banning sales to people under 18 and persuading smokers to quit. Banning smoking in confined spaces reduces the risk of exposure for non-smokers. Victoria’s Occupational Health and Safety (Hazardous Substances) Regulations 1999 require all work places, including schools, to perform a risk assessment for each activity using hazardous chemicals. A risk assessment should include the following steps: 1 identification of hazardous chemicals, including products as well as reactants (refer to a published or web-based list; not all chemicals are hazardous) 2 assessment of the risk (refer to container labels and Material Safety Data Sheets (MSDS), identify routes of entry into the body and other risks from how the chemical will be used) 3 control of the risk (procedures and equipment used to reduce the risk, safe disposal) 4 documentation (record the risk assessment and its author and date—a risk assessment lasts for five years). Legislation requires manufacturers to supply Material Safety Data Sheets (MSDS) for hazardous substances. An MSDS provides detailed information on a chemical or product and its hazards, such as: company name (manufacturer) and contact details scientific and trade names, including products made of several substances, and ingredients; for instance ‘Araldite’ is a glue made of bisphenol A epoxy resin and biethyl phthalate major uses and applications physical data such as appearance, colour, boiling point, melting point, solubility, vapour pressure (the higher the vapour pressure the more likely inhalation will occur) and vapour density (very dense vapour sinks in air), specific gravity, reactivity with water, flammability, flashpoint (lowest temperature at which a spark will ignite the vapour), autoignition temperature (lowest temperature for spontaneous combustion) health effects including the effects of swallowing and inhalation or exposure to skin and eyes treatment including first aid and medical advice precautions for use including short- and long-term exposure limits, ventilation, flammability and personal protection safe handling including storage, transport, spills, disposal, fire and explosion hazards date of publication: an MSDS must be less than five years old. Figure 18.14 shows a condensed sample MSDS for acetone. (More detail is found in a complete MSDS, typically several pages long.) Codes, acronyms and abbreviations are common. Some manufacturers use plainer English than others.

Figure 18.13 Warning label on a packet of cigarettes.


MATERIAL SAFETY DATA SHEET: ACETONE 1. Manufacturer: Kool Kemikals Pty Ltd, 1 Polymer Place, Factorytown, Australia 2. Date of publication: May 2007 3. Chemical name(s) and formula: acetone, beta-ketopropane, dimethyl ketone, dimethylformaldehyde, dimethylketal, methyl ketone, propanone, 2-propanone, pyroacetic acid, (CH3)2CO. 4. Physical properties: Colourless liquid, sweet fragrant odour, melting point –95°C, boiling point 56°C, vapour density 2.0, vapour pressure 181 mm Hg at 20°C, specific gravity/density 0.79 g/cm3, flash point: –18°C, explosion limits 2.6% to 13.0%, autoignition temperature 538°C. 5. Stability: Stable. Incompatible with halogen compounds, strong bases, strong oxidising agents, caustics, amines and ammonia, chlorine and chlorine compounds, strong acids, nitrosyl compounds. Highly flammable. Readily explosive in air. 6. Toxicology: Harmful by ingestion, inhalation or skin absorption. Irritant to skin, lungs and eyes. Liquid may cause permanent eye damage. Vapours may cause drowsiness and dizziness. Long-term exposure may cause liver damage. Typical TLV (threshold limit value) 750 ppm, UNR-MAN LDLO (unreported lowest published lethal dose for man) 1159 mg kg–1, ORL-RAT LD50 (oral lethal dose for 50% rats) 5800 mg kg–1, IHL-RAT LC50 (inhaled lethal concentration for 50% rats) 50 100 mg m–3 / 8h, ORL-MUS LD50 (oral lethal dose for 50% mice) 3000 mg kg–1, IHL-MUS LC50 (inhaled lethal concentration for 50% mice) 44 g m–3 / 4h, EYE-HMN (irritant to human eye) 500 ppm, SKN-RBT (mild irritant to rabbit skin) 395 mg open mld, EYE-RBT (severe irritant to rabbit eye) 20 mg sev, R11 R20 R21 R22 R36 R37 R38 R66 R67 7. Ecotoxicology: LC50 Fish (lethal concentration for 50% fish) 96 h 5540 mg/L. 8. Transport and storage: UN No 1090. Hazard class 3.0. Transport category 2. EMS No 2.0-06. Packing group III. Keep lid tightly closed. Keep away from heat, sparks and open flame. 9. Personal protection: Safety glasses, Nitrile/chemical resistant gloves, respirator or effective ventilation. Remove sources of ignition. S9 S16 S23 S26 S33.

Figure 18.14 A condensed sample MSDS for acetone, a common solvent.


International codes and common acronyms exist for toxicology, risks and safety. For instance an LD50 is the dose that kills 50% of a group (often based on rodent studies). The risk code R11 means ‘highly flammable’ and R36 is ‘irritating to eyes’. Risk can also be posed to the environment; although not specified for acetone, R57 means ‘toxic to bees’. The safety code S9 means ‘keep container in a well-ventilated place’ and S23 is ‘do not breathe vapour’.

! Risk and safety codes are found on the student CD.

Academia Heights College Risk Assessment Form—to be completed for all chemistry experiments Activity

Solubility testing

Date of activity

28 October 2011

Teacher

A Teacher

Class/level

Year 12

Hazardous substances

MSDS date

Risk phrases

Safety phrases

acetone

August 2005

R11 Highly flammable R20 Harmful by inhalation R21 Harmful in contact with skin R22 Harmful if swallowed R36 Irritating to eyes R37 Irritating to respiratory system R38 Irritating to skin R66 Repeated exposure may cause skin dryness or cracking R67 Vapours may cause drowsiness and dizziness

S9 Keep container in a well ventilated place S16 Keep away from sources of ignition—No Smoking! S23 Do not breathe gas/fumes/ vapour/spray S26 In case of contact with eyes, rinse immediately with plenty of water and seek medical advice S33 Take precautionary measures against static discharges

General risk control (all chemical experiments)

All students must wear safety goggles, gloves, laboratory coats and closed shoes for all chemistry experiments. Chemical labels, instructions for mixing and disposing chemicals and Science Safety Rules must be followed at all times.

Disposal

Collect for disposal. Do not pour down sink.

Other comments (optional)

Use nitrile gloves. Keep away from flames. Use in a fume cupboard or well-ventilated space. Keep lid on container.

Author (signed)

Andy Teacher

Date

28 October 2011

Figure 18.15 A sample risk assessment form for a classroom activity using acetone.

summary

18.4

Chemicals are classifi classified ed according to their hazardous properties.

Appropriate signage and Material Safety S fety Data Sheets provide information about the hazardous nature of dangerous substances and assist in risk assessment.


key questions 8 We communicate in different ways. Some of us are better readers and listeners, and others may have excellent visual communication skills. This means a multi-media approach to hazard warnings may be appropriate. The age or training of a message’s target audience is also relevant to effective communication. a Draw a cartoon or warning poster (e.g. ‘Read the label, Mabel!’) to communicate a threat from a dangerous good or hazardous substance to schoolchildren. b Write a jingle or rap to remind people to read warning signs and labels (e.g. write new lyrics for an existing tune). c Who is the target audience for a chemical label? a HAZCHEM sign? 9 a The ‘2WE’ HAZCHEM code for liquefied petroleum gas means that water fog should used for fighting any fire (‘2’), full protective clothing should be worn (‘W’) and evacuation may be required (‘E’). The code for acetone, another flammable substance, is ‘2YE’. Explain the difference. Hint: refer to Figure 18.11. b Translate the hazard label diamonds in Figure 18.12 into plain English. Are the meanings clear from the graphics alone? c Design a hazard label diamond for a new class of dangerous goods that freeze living tissue on contact. Test your design on a partner—is it effective?

10 a Complete a risk assessment for your next chemistry experiment. b How might the age of the students, the concentration of a chemical or its amount affect a risk assessment for a school experiment? 11 Despite obvious hazards, tobacco and its products are generally exempt from the requirement for an MSDS. However it is possible to find MSDS for some components of tobacco products. Some MSDS data for nicotine includes: Poison: May be fatal if inhaled/swallowed/absorbed through skin. May cause liver/kidney/respiratory/reproductive/ cardiovascular/nervous damage/disease/injury. Adverse delayed/cumulative effects. Toxicity: Oral-rat LD50 50 mg kg–1, Skin-rat LD50 140 mg kg–1, Intravenous-mouse LD50 0.8 mg kg–1 Risk codes: R23 R24 R25. Safety codes: S23 S24 S25 S37 S38 S39 a What do these risk and safety codes mean? (Refer to the student CD.) b Use the above information and the label from a pack of cigarettes (Figure 18.13) to complete a risk assessment for smoking in a public place. c What legal controls have been imposed in response to the health hazards associated with harmful chemicals in tobacco and tobacco smoke?

18.5

Waste management and pollution What is a waste? What wastes are hazardous? Photosynthesis (and therefore life on Earth) requires carbon dioxide, but too much CO2 contributes to excessive global warming. Sulfuric acid is a valued raw material for industrial production but it can kill forests as acid rain. Life-saving DDT, an insecticide used to combat malaria, is bioaccumulative in food chains with destructive consequences for bird populations (Figure 18.16). A waste can be considered as an unwanted or unvalued substance that is discarded or discharged. Wastes occur as contaminated soil and solids, organic liquids of low volatility, wastes containing highly volatile liquids, solid and semi-solid process materials, aqueous wastes, sludges, spillages, and products past their use-by date. Hazardous waste may be toxic, radioactive or contaminated by disease-causing organisms. Chemical waste is generated from many sources, as shown in Figure 18.17. It includes corrosives, acids and bases, solvents and flammable materials, heavy metals, poisons, pesticides, fertilisers, explosives, and products of combustion. The waste can be in solid, liquid or gaseous forms.


pre-1947 thickness mean

DDT enters agriculture

chemfact DDT was introduced in Australia as an agricultural insecticide in 1947. A thinning of the egg shells of a number of species of predatory birds at the top of the food chain has been linked to DDT.

Thickness index

2.0

1.8

20% thinning

1.6

1.4

Figure 18.16 The thickness of egg shells of the Australian Peregrine Falcon decreased by between 11 and 38% after the introduction of DDT as an agricultural insecticide.

1.2

1890

1910

1930

1950

1970

Some chemical wastes may not pose any hazard, but others may cause subtle damage that appears only after long-term exposure. Different chemicals can affect animals and humans at different stages of development. Negative effects from the past or current use of some widely occurring chemicals are well documented (Table 18.2).

2

3

1

Residential

cturing Manufa treated t effluen

are eathc

H

4

Mu

Septic

nic

Le

Le

ge

ha

l se

1

Treated effluent, untreated storm overflow, system failure

wa

ge

te Aquifer

3

Reuse

ipa

aka

ac

Leach ponds

Sludge

Treated effluent ‘Straightpiping’

ring

ufactu

Man

6

Agriculture runoff

G recround har wa ge ter

5

Household chemicals, medicines, domestic cleaners, etc. to sewer

2

Releases of treated/untreated hospital wastes to domestic sewage systems

3

Discharge of regulated/controlled industrial manufacturing waste streams

4

Sewage treatment plants discharged to surface waters or re-injected into aquifers (recharge)

5

Transfer of sewage solids (’biosolids’) to land (e.g. soil amendment/fertilisation). Release from agriculture: spray drift from tree crops (e.g. antibiotics)

6

Disposal to landfills via domestic refuse, medical wastes, and other hazardous wastes Leaching from defective (poorly engineered) landfills and cemeteries

7

Release to open waters from aquaculture (medicated feed and resulting excreta)

Leaching Orchard

7 Aquaculture

Farmland

Figure 18.17 Origin and fate of chemicals in the environment.


TABLE 18.2 Some negative effects of exposure to some common chemicals Chemical Asbestos

Negative impact Fatal lung disease

Benzene

Leukaemia

Polychlorinated biphenyls (PCBs) and dioxins

Cancer and toxic effects to immune, endocrine, nervous and reproductive systems Damaged reproductive ability

Dichlorodiphenyltrichloroethane (DDT)

Beijing apologizes for chemical spill China today offered Russia apologies for the chemical spill that will soon flow into Russian rivers in eastern Siberia.

Radio Free Europe Radio Library 26 November 2005

Murky dealing surrounds slick The Age, 28 November 2005 Figure 18.18 Pollution does not recognise national boundaries.

Use (past or current) Brake linings, building material, noncombustible thermal and electrical insulation Solvent, raw material for polymers, dyes, detergents, pesticides and explosives, component/product of petrol Insulating fluids and coolants, electrical components (PCBs); dioxin is a waste product Insecticide

Some benzene is emitted from cars running on unleaded petrol. PCBs, dioxins and DDT are persistent, i.e. they have long half lives and take a long time to break down. They are also bioaccumulative, being retained in body tissues and resulting in increased concentrations as they move through the food chain. This is a scary situation if these complex issues are not well managed, particularly since many wastes readily cross local and international boundaries. Governments, industry and the general community all have roles and responsibilities in waste management. There are international agreements to tackle waste. The 2002 World Summit on Sustainable Development set a target date of 2020 for ensuring that chemicals are used and produced in ways that minimise significant adverse effects on human health and the environment. The international Globally Harmonised System (GHS) is a common system for describing chemical hazards and information on labels and safety data sheets, and has been in use in Australia from 2006. Ideally, chemical waste management should be part of a sound risk management process for the global community, and follow the principles of green chemistry (Table 18.3, p. 310) to minimise or eliminate hazardous substances. In Victoria and locally, the Environment Protection Authority (EPA) and local governments regulate waste disposal.

Waste treatment Industrial waste treatment technologies now go beyond landfill and dumping at sea, and include high-temperature incineration and vitrification (sealing in molten slag). Other options, some still under development, include: dispersion in controlled amounts in water or air removal of pollutants from waste gases and liquids storage in sealed drums in secure locations catalysed dechlorination ionic replacement of halogens in halogenated organic compounds high-temperature steam and water treatments. The physical form of the waste, rather than the chemical constituents, often determines the choice of treatment technology to be used. At the industrial


level, chemical waste management needs to consider the particular hazardous chemical. Considerations include quantity and physical form, threats to people and the environment, and economically, socially and environmentally acceptable options for disposal. High temperature incineration (HTI) takes place at temperatures over 1100°C in a series of combustion chambers, followed by filtering and wet scrubbing to reduce the threat of gaseous emissions. The mostly inert ash residues are dumped in landfills. Under optimal conditions, high temperature incineration results in the almost complete combustion of organic compounds to carbon dioxide, water, and oxides of sulfur and phosphorus. Other products include hydrochloric, hydrobromic and hydrofluoric acids, iodine, and metal oxides. Responsible corporations practise sound waste management. Chemical industries aim to deal with wastes in the following strategic order of priority: 1 Prevention 4 Recycling 2 Elimination 5 Treatment 3 Reduction 6 Disposal Waste management also applies in the school laboratory, preferably as part of a risk assessment for activities with hazardous chemicals. General good practice includes collecting chemical wastes for safe disposal in accordance with the relevant MSDS. Laboratories should have purpose-built storage that separates incompatible substances. Containers should be labelled, closed, and compatible with their contents.

chemistryy in action Nasty dioxins Over 200 polychlorinated organic compounds are classified as dioxins. The colourless and odourless 2,3,7,8-tetrachlorodibenzop-dioxin (TCDD), C12H4Cl4O2,, is extremely toxic. Dioxins are formed as by-products of combustion, spread through airborne emissions, and contaminate soil where they land. Dioxins are persistent and accumulate in fatty tissues. Exposure to 2,3,7,8-TCDD can cause chloracne (skin lesions), cancer and birth defects. It can interfere with reproduction, development, and immune system function in humans and animals. High levels of dioxin have been found in bream and prawns in Sydney Harbour; a three-month ban on commercial fishing has been put in place and could be extended. The Minister for Primary Industries, Ian Macdonald, says while there are no national levels of risk for dioxins in Australia, the levels found in some harbour fish were above World Health Organization standards.

‘The levels have not changed, the levels of dioxin in the fish from the tests that I’ve seen in 1996 and now have not significantly changed,’ he said. ‘What has changed is the international assessment of the levels of risk of dioxin so it’s got a higher bar in recent years than it did in previous years.’

from ABC NewsOnline, 24 January 2006 High temperature incineration greatly reduces dioxin emissions compared to low temperature burns. Despite this advantage, HTI may not completely remove dioxin emissions. Greenpeace and other opponents of HTI have pointed out concerns with the ability of operators to maintain suitable operating temperatures, burn time, reactant mixing and excess oxygen. Dioxins can be formed at temperatures under 900°C. Despite government attempts to set up HTI plants and hazardous waste storage faclities in Australia, local opposition has been strident and politically effective.

Green chemistry Green chemistry aims to promote chemical technologies to minimise or eliminate hazardous substances in the design, manufacture or use of chemical products. The principles of green chemistry are summarised in Table 18.3.

review The principles of green chemistry were introduced in Chapter 17 of Heinemann Chemistry 1.

309 30 The chemical industry: risks and rewards

TABLE 18.3 The 12 principles of green chemistry 1 2 3 4

5

6

7

8 9 10 11 12

Prevent waste It is better to design chemical processes to prevent waste than to treat waste or clean it up after it is formed. Design safer chemicals and products Design chemical products to be fully effective, yet have little or no toxicity. Design less hazardous chemical syntheses Methods should be designed that use and generate substances with little or no toxicity to humans and the environment. Use renewable raw materials Use starting materials that are derived from renewable resources such as plant material rather than those such as from fossil fuels that will eventually run out. Use catalysts, not stoichiometric reagents Minimise waste by using catalysts in small amounts that can carry out a single reaction many times. They are preferable to stoichiometric reagents, which are used in excess and work only once. Avoid chemical derivatives Avoid using blocking or protecting groups or any temporary modifications if possible. Derivatives use additional reagents and generate waste. Maximise atom economy Design syntheses so that the final product contains the maximum proportion of the starting materials. There should be few, if any, wasted atoms. Use safer solvents and reaction conditions Avoid using toxic solvents to dissolve reactants or extract products. Increase energy efficiency Energy requirements should be minimised. Run chemical reactions at room temperature and pressure whenever possible. Design for degradation Chemical products should be designed to break down to harmless substances after use so that they do not accumulate in the environment. Analyse in real time to prevent pollution Include continuous monitoring and control during process to minimise or eliminate the formation of by-products. Minimise the potential for accidents Design chemicals and their forms (solid, liquid or gas) to minimise the potential for chemical accidents including explosions, fires and releases to the environment.

Source: Paul Anastas and John Warner, Green Chemistry: Theory and Practice (Oxford University Press: New York 1998)

summary

18.5

Ch Chemical i l industries i d t i are required i d to t manage th theiri wastes t in i a way that does not have a detrimental effect on health, safety or the environment. A variety of waste management options exist, and the choice of which to use depends on the nature of the waste product.

Some S chemical h i l industries i d t i are adopting d ti practices ti consistent it t with the principles of green chemistry in an effort to reduce waste and the impact of waste on the environment.

key questions 12 a What properties of some organic compounds make them long-term threats to the environment? b Would DDT or dioxins present the same hazards if these substances had short half-lives? (The half-life is the time taken for half the mass of a pesticide to break down.) Explain.


13 a List at least three options for disposing of excess acetone left over from a science experiment in class. What is the best disposal method? Explain. b Are leftover chemicals or by-products always a waste? Explain.

key questions 14 a With reference to Figure 18.17, suggest at least three ways to prevent marine pollution. b Heavy metals and some persistent organic toxins can contaminate land, making it unfit for habitation or food production. Suggest what safeguards could be made to reduce the chances of building a new home on land affected by chemical contamination from past uses. c Why might prevention be technically and economically more desirable than clean up and rehabilitation in the above cases?

15 Match each of the twelve principles of green chemistry listed above with none, one or more of the three Rs of ‘reduce, re-use and recycle’. 16 Apply the relevant principles of green chemistry to your next school chemistry experiment. Why would application of the principles usually be more relevant to large-scale commercial chemical manufacture than a school science activity? 17 Why might waste disposal be considered as a last resort in dealing with hazardous substances?

18.6

Community safety chemistryy in action Safety news Nobody wants a dump in their backyard We all want to dispose of wastes, but who wants to live near a toxic waste site?

Council acts on toxic waste by Martin Boulton Pressure is mounting on the State Government to pick a site for a new toxic waste dump after the Greater Dandenong Council said that it wants to stop hazardous waste being dumped at the Lyndhurst landfill in Melbourne’s south-east. Legal opinion seen by The Age shows that the tip owner, SITA Environmental Solutions, could be breaching its permit by accepting thousands of tonnes of proscribed industrial waste each year.

The 1990 planning permit states that rubbish dumped at the 54-hectare site must be ‘domestic garbage, solid commercial refuse and solid industrial wastes’ and not include soluble chemical, hazardous or liquid waste. However, legal advice from a Queen’s counsel shows that industrial waste ‘in the more hazardous categories’, including asbestos, hydrochloric acid, lead and mercury, has been dumped at Lyndhurst for several years. ‘Many types of waste . . . at the point of receipt are risky or perilous,’ the advice said …

The Age, 24 November 2005


chemistryy in action Safety news (continued) pesticides—would have to undergo basic toxicity testing . . . the law could cost US industries that export products to Europe billions of dollars …

More expensive than poison Are strict safety benefits worth the cost?

What’s your poison? Europe makes it clear by Marla Cone, London The European Parliament has approved legislation requiring safety testing of thousands of compounds widely used in everyday products. The legislation aims to improve the ways people are protected from toxic chemicals … The standards would be the world’s strictest, eclipsing US laws, and could lead to global bans on some compounds. Chemicals found in many products—such as computers, cosmetics, cars, furniture, detergent and

The new European law was prompted by discoveries that chemicals are amassing in human bodies, particularly breast milk, as well as in wildlife. In most cases, the potential dangers are unknown. Roughly 70 000 to 100 000 chemicals are in commerce today, and experts say that more than 90 per cent have not been subjected to basic toxicity testing for health and ecological effects. ‘These new rules will make a huge difference in protecting people’s health … Companies will have to show that the chemicals they produce or import are safe.’

Los Angeles Times, quoted in The Age, 19 November 2005

As the newspaper articles above show, the public is concerned about potential dangers presented by chemical industries to local communities. Such dangers were highlighted by the accident in 1984 in Bhopal, India, in which 2500 people died when the malfunction of a pesticide manufacturing plant released the chemical methyl isocyanate. In 1976 a dioxin spill from a chemical plant in the Italian city of Seveso resulted in 37 000 people being exposed to the chemical. The faces of some children were disfigured from contact with the chemical. Eighty thousand contaminated farm animals had to be destroyed and the produce from large areas of polluted agricultural land was declared unfit for human use. Occasional fires and accidental releases of toxic materials by local industries also remind us of hazards. Responsible companies recognise and accept their obligation to provide a working environment for their employees and nearby residents that is as free of hazard as possible. No human activity is free of risk and, as you would expect, no chemical industry is accident-free. Of the accidents that do occur, only a few are directly related to chemical processing. The majority are trips, falls or accidents with tools—the kinds of accidents that also happen in other industries.

18.7 Figure 18.19 The Coode Island storage depot near the Port of Melbourne. On 21 August 1991 a series of explosions occurred at Coode Island. The resulting fire sent a spectacular plume of smoke over the city of Melbourne, forcing thousands indoors to avoid the toxic fumes, and leaving a damage bill of $30 million. The advantages and disadvantages of relocating the storage depot further from the city have been vigorously debated since the fire.


Location of a plant Once it has been decided to construct a plant to manufacture a particular chemical, the company must decide on the location. For the plant to be profitable, it must have ready access to raw materials and markets. Chemical companies interact as both customers and suppliers. This is demonstrated by the links that Qenos has for the production of olefins at Altona in Victoria (Figure 18.20).

ESSO/BHPP ESSO/BHPP MOBIL/SHELL

Ethane

P Propylene

Natural gas

E Ethylene

LPG

QENOS OLEFINS

Phenol, Acetone HUNTSMAN

Fu Fuel Gas Ga

QENOS PLASTICS Polyethylene

Gasoil (HVGO) Polypropylene ESSO/BHPP GIPPSLAND CRUDE IMPORT REFINERY FEEDSTOCKS

QENOS RESINS

MOGAS MOBILE ALTONA REFINERY

QENOS ELASTOMERS

Butadiene

Quench oil

Polyethylene

BR

TABLE 18.4 Factors considered in locating a plant Availability of raw materials Proximity to markets Proximity to other, interacting, chemical manufacturers Transport costs Availability and cost of energy Labour supply Environmental concerns Waste disposal facilities Water supply Legal, social and political constraints Cost of land

Latex DOW/ BASF CABOT AUSTRALASIA CBFS

Carbon black

Figure 18.20 Qenos feedstocks, products and customers at the Altona petrochemical complex in Victoria. Chemical companies interact as both customers and suppliers.

extension Siting an integrated steelworks

Figure 18.21 The Newcastle steelworks.

The sites selected for Australia’s iron and steelworks demonstrate the influence of some of the factors discussed above. One of Australia’s earliest surveys to identify a site for a chemical industry was commissioned by the BHP company in 1912. BHP had decided to extend its existing interests from mining iron ore to the production of iron and steel. At that time, over two tonnes of coal were needed for each tonne of iron ore smelted, and the cost of transporting these raw materials was the major constraint.


extension Siting an integrated steelworks (continued)

Port Hedland

Whyalla

Newcastle Port Kembla

Major iron ore deposits Coal mines for making iron Steelworks Iron making

Newcastle, in New SSouth Wales, was chosen because of its good port facilities, existing labour force, access to quality coal (for making coke), and competitive coal prices. The Newcastle iron and steelworks was officially opened in 1915. The Newcastle site was closed in 1999 following changing worldwide patterns in demand for steel and globalisation of the iron and steel industry. The efficiency with which today’s large tankers can transport raw materials such as coal and iron ore leads to new integrated steelworks being sited close to deep-water harbours and in reasonable proximity to large markets, thereby reducing the cost of distributing the finished product. By choosing a coastal site, the potential for importing raw materials and exporting products is favoured. In general, however, the enormous cost involved in establishing a new steelworks means that much effort is directed towards expanding or improving the efficiency of existing facilities.

Figure 18.22 The location of Newcastle steel works and sources of coal and iron ore.

question

?

E1 Is it llikely ikely l that thatt a steelworks will be built near Melbourne, Melbourne Victoria, Victoria in the next decade? Explain the reasons for your answer. answer

summary

18.6–18.7

The location of chemical industries near residential or environmentally sensitive areas poses some risk and is at times a cause of community concern.

key questions 18 a Why are many of Melbourne’s chemical industries sited in Footscray, 5 km from central Melbourne? b Do you think this is a good site? Why, or why not? What would be the features of a better site? Explain your choice. 19 Residential areas have grown to encircle the industries at Footscray. Some residents near the industries are concerned about their safety and want the factories relocated.


a What would the residents consider to be the main risks? b Why do you think that residential areas have been allowed to develop so close to these industries? c What do you think should be done?

18 batch processing bioaccumulative continuous flow processing dangerous goods green chemistry

key terms hazardous chemical HAZCHEM high temperature incineration Material Safety Data Sheet (MSDS) occupational health and safety

persistent risk assessment waste wet scrubbing yield

Chemical industry 20 In what ways is the industrial production of a chemical such as sulfuric acid likely to differ from the way it is made in the laboratory? 21 What methods might be used by a chemical industry to save on the cost of energy used for their processes? 22 Suppose a major chemical industry, employing about 200 people, wishes to build a large plant in a country town. List: a five potential benefits to residents b five potential disadvantages to residents. 23 The Australian chemical industry operates at a disadvantage compared with some overseas chemical industries. What do you think these disadvantages could be? 24 Only a few companies in the world are prepared to develop a new substance, such as an insecticide, and produce it commercially. Suggest reasons for this.

Location of chemical industries 25 Give an example from an industrial process to show the significance of the following factors in the chemical industry: a concern for the environment b energy costs c availability of raw materials d transportation costs 26 a Name a process in which the choice of raw materials varies, depending upon where the plant is located. b List three factors, apart from location, that influence the choice of raw materials for an industrial process.

Safety 27 a Match the caution symbols shown in Figure 18.23 with the following hazard descriptions: corrosive, oxidising, flammable, radioactive, explosive. b State which of these hazard descriptions you would expect to find on: i a bottle of concentrated sulfuric acid ii a cylinder of ethene gas

Figure 18.23 Hazard warning symbols – can you crack the code?

28 a Obtain Material Safety Data Sheets for HCl, NaOH, CH3COOH and Na2CO3. (You might be able to download these from the Internet.) b Compare the R and S codes and determine the two most and two least hazardous chemicals to use in a junior science experiment that involves mixing an acid with a base. c Do a risk assessment for a titration of 0.1 M HCl with 0.1 M NaOH. Would you take the same precautions when titrating 0.01 M ethanoic acid with 0.01 M Na2CO3? Explain. d Suggest basic safety precautions you might take when handling any chemical in a laboratory.


29 Why are you unlikely to find ORL-HUM LD50, the lethal dose for 50% of a human population, on the MSDS for most substances? 30 a Why is it poor practice to store HCl in a metal container? b How should the products of a reaction between 50 mL 0.1 M HCl and 50.0 mL 0.1 M AgNO3 be disposed of? (Hint: refer to MSDS on the Internet.) 31 a Carbon dioxide is produced in huge quantities by coalburning in the power industry. Is the carbon dioxide a waste? Explain. b Once a waste is released into the environment, the industry it came from has little, if any, control of it. Debate the proposition: That the cost of managing wastes should be borne by chemical manufacturers. Give at least three arguments for and against this proposition.

(Electrical energy from imported coal) Port

Mountains Lake Hydroelectric power station Capital city (Population: 10 000)

B D Port

Connecting the main ideas 32 a A company intends to locate an aluminium smelter at one of the four sites, A, B, C or D, on the island shown in Figure 18.24. Large amounts of electricity are used to extract aluminium from its ore, bauxite. Which site would you choose? Discuss the reasons for your choice with a partner. b A company intends to locate a high temperature incineration plant at one of the four sites, A, B, C or D, on the island shown in Figure 18.24. Large amounts of organic compounds and medical wastes will be treated. The predominant winds are westerly. Which site would you choose? Discuss the reasons for your choice with a partner. c Investigate answers to parts a or b by performing a role play with members of the class representing four competing companies, each owning one of the sites A, B, C or D. Other class members can represent the the local communities, industrial producers or conservation groups. A panel of judges, representing the government, will hear each case and reach a decision based on the arguments presented. 33 Acetone in used as a solvent in the production of fibreglass, a material widely used for boats, canoes and surfboards. a Why is it a poor idea for workers to use acetone in an enclosed space? b What special precautions might be wise for smokers in a speedboat factory? c What are the risks of autoignition in the boat factory? d Why might polyester overalls, plastic shoes and wool carpet be a particularly bad idea in a speedboat factory? (Exclude fashion statements from your answer!)


Bauxite mine and treatment plant

C

A Road Railway line

Port

National park

High-current power line Deep-water port (Population: 10 000)

50 km

Figure 18.24 Map of the island.

34 Although the following scenarios appear far-fetched in developed countries, use an example to predict what might happen to workers in a chemical plant (or others in society) in each of the following scenarios: a There are no Occupational Health and Safety laws requiring employers to apply a duty of care. b Chemicals are stored in unlabelled containers. c Material Safety Data Sheets are unavailable, ignored or out of date. d Wastes are disposed by flushing down the sink or into waterways. e The national government chooses to ignore international agreements to minimise waste production. 35 What would be the best method of disposing of the following wastes from a chemistry laboratory class? The laboratory has a sink connected to a neutralising tank, a rubbish bin and aqueous waste container and an organic waste container. a 150 mL of ethyl ethanoate b 100 mL of 0.05 M AgNO3 solution c 5 g of CuSO4 precipitate d 100 mL of 0.1 M HCl

1

chapter 19

Production of ammonia

key knowledge • the industrial production of ammonia − factors affecting the production including rate and equilibrium position, catalysts, temperature, pressure − waste management including generation, treatment and reduction − health and safety − uses of ammonia

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to:

VCE Chemistry students only need to study the industrial production of one of the following chemicals: ammonia, nitric acid, sulfuric acid or ethene (Chapters 19–22).

• name the raw materials used in the production of ammonia • describe the reaction steps by which ammonia is manufactured • explain how the principles of equilibrium and reaction rates play a significant role in determining reaction conditions • explain the reasons for choice of reaction conditions such as pressure, temperature and catalyst • describe waste management procedures • describe health and safety issues involved in the production of ammonia • describe how the production of ammonia relates to the principles of green chemistry • recall the major uses of ammonia.

N

N

Figure 19.1 The nitrogen molecule contains a strong triple bond. The strength of this bond causes nitrogen gas to be relatively unreactive.

19.1

Uses of ammonia Properties of ammonia Some important properties of ammonia are listed in Table 19.1.

chemfact In liquid ammonia, hydrogen bonds exist between ammonia molecules, causing ammonia to have a higher boiling temperature (–33°C) than most substances with molecules of similar size. Since it can be liquefied relatively easily, ammonia is transported in liquid form in refrigerated vehicles.

TABLE 19.1 Properties of ammonia Properties Polar molecule

Gas at room temperature and pressure Acts as a weak base

Explanation Due to the large difference in electronegativity (electron attracting ability) between nitrogen and hydrogen, the hydrogen atoms in the molecule are slightly positive and the nitrogen atom is slightly negative. The intermolecular forces between NH3 molecules are not strong enough for it to exist as a liquid or solid at these conditions. It melts at –78°C and boils at –33°C. NH3 reacts slightly with water: NH3(aq) + H2O(l)

Extremely soluble in water

NH4+(aq) + OH–(aq)

In addition to reacting with water, the polar NH3 molecules form hydrogen bonds with H2O molecules.

Fertilisers

Figure 19.2 Rhizobium nodules on the roots of legume plants fix atmospheric nitrogen.


One of the substances continually needed by plants is the element nitrogen, which is used to make proteins and other organic compounds essential for growth. Despite an abundance of nitrogen gas (N2) in the atmosphere, few plants are able to use it directly. Most absorb soluble nitrogen compounds from the soil. Life on Earth depends upon the activities of a few species of microorganism that are able to convert nitrogen gas into nitrogen compounds. The conversion of atmospheric nitrogen into soluble nitrogen compounds that can be used by plants is called nitrogen fixation. These micro-organisms use the enzyme nitrogenase to catalyse the reduction of nitrogen gas to ammonia (NH3) or ammonium ions (NH4+). These organisms include Rhizobium bacteria which invade the roots of certain plants such as legumes (peas, beans, clover, lucerne and wattle), forming nodules such as those shown in Figure 19.2. Rhizobium lives in close association with its host plant, helping the plant to survive by providing it with nitrogen in a form the plant can use. Since plant growth is dependent on the supply of fixed nitrogen, farmers can enhance crop yields by planting legumes such as clover. The legumes can then be dug into the soil before the next crop is sown or, alternatively, a crop can be sown at the same time (Figure 19.3). Adding nitrogen-containing manure to the soil has similar benefits. Conventional agricultural methods rely upon the use of synthetic fertilisers, such as urea and ammonium nitrate. By applying these fertilisers to soil, farmers obtain higher yields than those that are otherwise possible. Almost all of the major fertilisers are manufactured from ammonia.

Figure 19.3 Tick beans, a legume, are planted between rows of dormant grape vines in order to fix nitrogen in the soil. At the end of winter, just before the vine leaves burst from their buds, the legume plants are rotaryhoed into the ground. N

Farmers may fertilise crops with ‘anhydrous ammonia’. They use cylinders containing pure ammonia, stored as a liquid under pressure. Specially designed machinery injects ammonia directly into the soil, about 15 cm below the surface. Although ammonia is a gas at atmospheric pressure, very little is lost from moist soil because ammonia is highly soluble in water. Its solubility results from the ability of polar ammonia molecules to form hydrogen bonds with polar water molecules (Figure 19.5). Ammonia also reacts with water, acting as a base by accepting a proton: NH3(aq) + H2O(l)

H

Figure 19.4 Structural formula of the ammonia molecule.

Oδ– H

NH4+(aq) + OH–(aq)

Because only a small proportion of ammonia molecules react, ammonia is a weak base. The ammonium ion (NH4+) that is formed can be absorbed by the root systems of plants. The advantage of this method of fertilising is that ammonia has a very high nitrogen content (85% by mass) and anhydrous ammonia is conveniently transported and distributed. However, many farmers find it unsuitable because the equipment has a high initial cost and a large amount of ammonia is lost from dry, sandy soils. World production of ammonia has grown strongly from 62 million tonnes in 1974 to more than 156 million tonnes in 2008. More ammonia is produced industrially per mole than any other chemical. Over one third of the production is in the high growth centres of China and India. Over 80% of the ammonia produced is consumed for the manufacture of ammonium salts used as fertilisers.

H

H

Hδ+

Nδ– Hδ+ H

O

Hδ+

Hδ+

δ–

Hδ+

Figure 19.5 The ability of ammonia molecules to form hydrogen bonds with water molecules causes ammonia to be soluble in water. It is the most soluble of all gases—740 mL dissolves in 1 mL of water at 25°C and one atmosphere pressure.

319 Production of ammonia

The reactions involved can be summarised: –—h ammonium nitrate, NH4NO3 + nitric acid –—h ammonium sulfate, (NH4)2SO4 ammonia + sulfuric acid + phosphoric acid –—h ammonium phosphate, (NH4)3PO4 Zinc and nickel extraction (7%)

Explosives (3%)

Ammonia acts as a base in all these reactions. It is also converted into the important fertiliser urea by reaction with carbon dioxide:

Fibres, plastics, refrigeration (10%)

2NH3(aq) + CO2(g) –—h (NH2)2CO(aq) + H2O(l)

Additional uses

Fertilisers (83%)

Figure 19.6 Uses of ammonia.

The behaviour of ammonia as a weak base is employed in household cleaning solutions such as ‘cloudy ammonia’. Ammonia is also used for commercial refrigeration because a considerable amount of energy is absorbed when liquid ammonia evaporates, as hydrogen bonds between molecules are disrupted. Ammonia is used to make a number of economically important chemicals, including: nitric acid (see Chapter 20) fibres, such as nylon explosives, such as ammonium nitrate pharmaceuticals, such as sulfonamides.

chemistryy in action Don’t mix ammonia cleaners with bleach! Don’t be tempted to ‘improve’ on household chemicals by mixing them. For example, mixing bleach with ammonia produces irritating chloroamine vapour, not a better cleaner: NH3(aq) + OCl–(aq) –—h OH–(aq) + NH2Cl(aq) NH3(aq) + 2OCl–(aq) –—h 2OH–(aq) + NHCl2(aq) NH3(aq) + 3OCl–(aq) –—h 3OH–(aq) + NCl3(aq)

These compounds are potentially toxic and burn the membranes in the eyes and throat. They are the chemicals we normally associate with the smell of ‘chlorine’ in swimming pools. In this case they are formed when the chlorine or hypochlorite compound added to the pool to sterilise the water reacts with urea in the water. (Just why there is urea in swimming pool water is best not discussed!)

19.2

Production of ammonia Raw materials ! Kwinana is a large integrated industrial complex adjacent to a deep-water port south of Perth. The complex processes oil, gas, minerals and a variety of chemicals, including fertilisers such as ammonia.

Ammonia is formed from the reaction of nitrogen with hydrogen. Air is always the source of nitrogen used in making ammonia. The choice of raw materials used to obtain hydrogen largely depends on their availability and cost to a particular plant. Fossil fuels are usually used, with natural gas being the simplest, least expensive and most efficient source. For example Wesfarmers CSBP uses the reaction between natural gas and steam known as steam reforming to produce hydrogen at their plant in Kwinana, Western Australia. The initial reaction produces a mixture of carbon monoxide and hydrogen, called synthesis gas: CH4(g) + H2O(g)


CO(g) + 3H2(g)

Nickel is usually used as the catalyst for reforming reactions. Since carbon dioxide and carbon monoxide act as poisons to the catalyst used for ammonia synthesis, preventing it from acting efficiently, the carbon monoxide is converted into carbon dioxide: CO(g) + H2O(g)

CO2(g) + H2(g)

The mixture of gases is then treated to remove carbon dioxide. Various treatment methods to remove carbon dioxide are available, including physical methods and chemical treatment using bases such as amines or hot potassium carbonate solutions. Coal might be expected to become the dominant source of hydrogen in the future, based on consumption rates and reserves of fossil fuels. On the other hand, at the zinc refinery in Risdon, Tasmania, hydrogen is obtained by electrolysis of water: 2H2O(l) –—h 2H2(g) + O2(g)

An important factor in the latter process is the availability of relatively cheap hydroelectricity in Tasmania.

chemfact In 1901, several years before Haber developed his process for ammonia production, Le Chatelier attempted to make ammonia from nitrogen and hydrogen at 60°C at a pressure of 200 atm using an iron catalyst. A violent explosion resulted from the accidental inclusion of air in the mixture. In his old age Le Chatelier wrote, ‘I let the discovery of the ammonia synthesis slip through my hands. It was the greatest blunder of my scientific career.’

! Electrolysis involves the passage of an electric current through a conducting liquid; see Chapter 27.

The Haber process Ammonia is manufactured from nitrogen and hydrogen gas using the Haber process, which employs an iron/iron oxide catalyst. The reaction is exothermic: 2NH3(g); ΔH = –91 kJ mol–1

By using Le Chatelier’s principle you can predict the way in which temperature and pressure will alter the equilibrium yield of ammonia: If the pressure exerted by the equilibrium mixture is increased, the system moves to decrease the pressure by moving the position of equilibrium to the side with fewer particles. Since the number of gas particles on the right side of the equation (two NH3 molecules) is less than on the left (one N2 and three H2 molecules), increased pressures will lead to an increase in the amount of ammonia in equilibrium. In practice, the pressure used is limited by the cost of compressing the gas mixture. Since the forward reaction is exothermic, the position of equilibrium will lie further to the right and the equilibrium yield of ammonia will increase at low temperatures. That is, the lower the reaction temperature, the greater will be the equilibrium constant for the reaction. The rate of the reaction will be faster: as temperature increases as pressure increases if a catalyst is employed. Experimental data shown in Figure 19.8 supports these predictions. The reaction will occur most rapidly at high temperatures. There is a conflict between the low temperature needed for a high equilibrium yield and the high temperature that gives economically viable reaction rates. By using a catalyst, it is possible to achieve a compromise and obtain acceptable reaction rates at temperatures that give acceptable equilibrium yields. The operating conditions in the converters of modern ammonia plants are: the nitrogen and hydrogen gas enter in a stoichiometric 1 : 3 ratio temperatures ranging from 350–550°C and pressures of 100–250 atm are required a porous iron catalyst made from iron oxide (Fe3O4) is usually employed.

Figure 19.7 The German chemist Fritz Haber (1868–1934) invented the process for producing ammonia. He was awarded the 1918 Nobel Prize in Chemistry.

100

Fraction of ammonia in the equilibrium mixture (%)

N2(g) + 3H2(g)

80

200°C 300°C 400°C

60

500°C

40

600°C

20

0 200

400

600

800

1000

Pressure (atm)

Figure 19.8 Fraction of ammonia present in a mixture of nitrogen and hydrogen (mole ratio 1 : 3) at equilibrium.


Unreacted nitrogen and hydrogen recycled to reactor

1 : 3 nitrogen and hydrogen

Nitrogen Compressor

Converter

Ammonia, nitrogen and hydrogen

Condenser

Hydrogen Liquid ammonia

Figure 19.9 A flow diagram of the Haber process.

Figure 19.10 Haber’s experimental apparatus for the synthesis of ammonia.

At each pass of the gases through the converter, the yield of ammonia is about 20%. During commercial ammonia production, sufficient time is not allowed for the gas mixture to reach equilibrium. The gas mixture leaving the converter is cooled to liquefy the ammonia so it can be removed. The remaining gases, containing unreacted nitrogen, hydrogen and some unrecovered ammonia gas, are then continuously recycled to improve the ultimate yield to about 98% overall. Heat released by the exothermic reaction between nitrogen and hydrogen must be removed or the temperature would rise above the optimum value. This waste heat is used to heat up the incoming gas mixture, providing an important cost saving. After production, ammonia may be piped to a plant nearby for use in the manufacture of other chemicals, or liquefied for storage or transport. The site used for ammonia production is often connected directly to the sites of plants synthesising other chemicals, such as urea and nitric acid, thus minimising the hazards and costs associated with transport of the chemical.

Figure 19.11 Ammonia is transported as the anhydrous liquid. At this plant in the US a 34 kL tanker (left) is contrasted with a monster 190 kL tanker (right).


chemistryy in action Improving the catalyst for ammonia synthesis The effectiveness of the catalyst in the converter in the Haber process is critical to its efficiency. Fritz Haber and his colleague, Carl Bosch, tested nearly 2000 different materials as catalysts in developing their process for ammonia synthesis. By good fortune, the iron ore that they tested was from Sweden and contained traces of group 1 metal compounds which act as ‘promoters’, increasing the efficiency of the catalyst. A promoter creates many small pores in the catalyst, exposing iron crystals and providing a greater surface area and more sites for reaction. The catalyst widely used today is not very different—iron oxide with a potassium hydroxide promoter. An alternative catalyst is ruthenium metal supported on high-surface-area carbon. This material began to attract considerable interest from industry in the 1990s. While this catalyst is much more expensive, it is 45 times more active per square metre of surface, allowing the use of a lower operating temperature of about 400°C and 40 atm pressure. It has been installed in a number of ammonia plants. Even more recently, a Danish company has announced the discovery of a new catalyst based on metal nitrides such as Fe3Mo3N3 and Co3Mo3N. This catalyst is two to three times as effective as commercial iron oxide catalysts but less expensive than the rutheniumbased ones. A dream for many chemists has been to devise a catalyst with the efficiency of nitrogenase, the enzyme found in nitrogen-fixing bacteria in legumes, which converts atmospheric nitrogen to ammonia without the need for the high pressures and temperatures of the industrial process. Nitrogenase contains clusters of iron, molybdenum and sulfur and research is continuing to develop similar artificial clusters which could allow low-cost ammonia synthesis at normal temperatures and pressures.

summary

Figure 19.12 Sections through root nodules from broad beans, stained to show the nitrogen-fixing bacteria.

19.1–19.2

AAmmonia i is i a major j industrial i d t i l chemical h i l that th t iis usedd in i the th manufacture of fertilisers and nitric acid. Nitrogen, from air, and hydrogen, from fossil fuels or the electrolysis of water, are the raw materials used in the production of ammonia. The reaction is exothermic and higher equilibrium yields are favoured by the use of low temperatures and high

ti rates t are favoured f d by b th hi h pressures. FFastt reaction the use off high temperatures and pressures. The conditions actually chosen for ammonia manufacture are a compromise between achieving rapid rates, high equilibrium yields, and cost. In practice, ammonia is manufactured at moderate temperatures (350–550°C) and pressures (100–250 atm) in a converter using an iron/iron oxide catalyst.

key questions 1 a Write balanced equations for each of the reactions of ammonia with acids to make fertilisers described on page 320. b Suppose each ammonium salt was sold at the same price per kilogram. Which fertiliser would provide nitrogen most cheaply? c Apart from the result of your calculations in part b, what else might you consider when choosing the best fertiliser for a particular application?

d Why do farmers need to know the exact composition of the fertilisers they use? 2 Construct a flow diagram to show each of the major steps in the production of ammonia from natural gas. Write important equations that occur beside each step in the process.


19.3

Waste management

140

Mass of ammonia (106 t)

120

100

80

60

40

20

0 1920

1940

1960 Year

1980

Figure 19.13 Annual world production of ammonia.


2000

2008

There are sulfur compounds present in natural gas. These sulfur-containing compounds need to be removed (desulfurisation) to avoid environmentally damaging sulfur dioxide emissions and to prevent the poisoning of catalysts. One of the main sources of potential gas emissions is flue gases from the steam reforming process of industrial plants that use hydrogen derived from fossil fuels. These emissions are mainly carbon dioxide, nitrogen oxides NOx, and small amounts of sulfur dioxide and carbon monoxide. The levels of these emissions are carefully monitored and controlled. Carbon dioxide is the main by-product of such plants. It can be liquefied and sold in limited quantities to the food and beverage markets, as at Wesfarmers CSBP’s Kwinana plant. At Incitec’s Brisbane plant it is used for manufacturing the fertiliser urea. However not all the carbon dioxide produced worldwide can be used in these ways, and excess gas is released into the atmosphere. Chemical removal of carbon dioxide from hydrogen by treatment with amines results in liquid waste. Whereas this waste was once processed and dumped in landfill sites, in modern plants it is usually disposed of by burning, which results in the production of CO2, H2O and NOx. The heat energy is used elsewhere in the plant. The Haber process maximises the conversion of nitrogen and hydrogen into ammonia by recycling unreacted gases back into the converter for further passes over the catalyst. In this way, almost complete conversion is achieved. Because the hydrogen is recycled the amount of hydrogen feedstock required from the hydrogen generation process is reduced. Consequently the amount of raw materials, energy and waste materials involved in the production of the hydrogen used in the Haber process is reduced. In some sections of the production process, aqueous solutions of ammonia are produced when gases being released to the atmosphere are purified by passing them through water. These solutions can also be used in the manufacture of urea. Careful attention is given to energy efficiency, resulting in lower operating costs and higher yields. Examples include: using heat released from exothermic reactions to generate steam to drive turbines, compressors and other machines using waste heat to preheat reaction gas mixtures using cold gases from the refrigerator section of the plant to cool gas mixtures before compression computer control of all plant operations. While most ammonia production plants produce relatively large quantities of the chemical, recent improvements in technology such as those described above have meant that small-scale profitable plants can be constructed to target local markets. The Haber process has been used for the synthesis of ammonia for over 80 years and may be regarded as technologically mature, with perhaps only limited scope for future improvements. Nevertheless, energy costs and emissions continue to be reduced and operating efficiencies increased. In part this arises from an increased awareness of environmental issues and responsibility, coupled with other legislative, technological and economic

influences. The emerging acceptance of green chemistry philosophies could lead to: reduced use of ammonia for the industrial synthesis of other chemicals reduced use of ammonia as optimum application rates of fertilisers to soils are better understood design of more efficient catalysts for the Haber process modifications to the genetic structure of agricultural crops to make them less dependent on fertilisers.

extension An end to nitrogenous fertilisers?

Ni Nitrogen-fi fixing i crops might i h b be produced more easily Scientists have reported the discovery of a plant gene that is essential in controlling the interactions between plants and micro-organisms that enable them to establish intimate associations of benefit to both partners. Published in Nature, the report’s findings suggest that it may be easier than previously imagined to design green plants that are able to make their own nitrogenous fertiliser. The discovery was made by scientists at the Sainsbury Laboratory in Norwich, UK. The roots of many plants are able to form intimate relationships with particular fungi living in the soil. These so-called arbuscular mycorrhizal associations are a symbiosis—a partnership of direct benefit to both partners . . . The fungi are very efficient at absorbing nutrients, especially phosphate, from the soil. This is

questions

exchanged in sugars h d with i h the h plant l i return for f plant l that are absorbed and used by the fungus . . . ‘Scientists had always imagined that the nitrogenfixing symbiosis between legumes and rhizobia bacteria was a unique relationship, so the discovery that it actually uses some of the same genes that control the very common mycorrhizal association of plant roots with fungi, is really exciting’ said Dr Martin Parniske, the Project Leader. ‘This suggests that evolution of the nitrogen-fixing symbiosis used some of the genes that were controlling the plant– fungal partnerships that are widespread in the plant kingdom. ‘So we know that part of the genetic blueprint needed to establish a symbiotic relationship with nitrogen-fixing bacteria is present in all major plant types, including crop species such as wheat and rice. Consequently, relatively few genetic changes might enable breeders to produce a wide range of plants that can establish symbiotic relationships with nitrogen fixing bacteria, and perhaps manufacture their own nitrogen fertiliser’, he added.

?

E1 Explain Expllaiin the the meaning of the terms ‘nitrogen fifixing’ xing’, ‘arbuscular mycorrhiza’ and ‘symbiosis’. ‘symbiosis’ E2 Legume plants such as peas form a symbiosis with nitrogen-fixing bacteria. What are the benefits of this relationship to: a the plant? b the bacteria? E3 What would be the environmental benefits of using genetically engineered plants to add nitrogen to the soil compared with the use of nitrogenous fertilisers such as ammonia? E4 Is the addition of nitrogen to the soil using plants a green chemistry process? Explain your answer.


19.4

AMMONIA ANHYDROUS

Health and safety

TOXIC GAS

UN No.

1005

2

HAZCHEM

CORROSIVE

2RE Figure 19.14 Hazard sign for storing bulk ammonia. Ammonia is a hazardous chemical, and health and safety are paramount.

summary

Ammonia is a toxic gas that can be harmful by all means of exposure. In gaseous form, it is an extreme irritant to the eyes, respiratory system and other parts of the body. Appreciable exposure can be fatal. Since ammonia boils at –33°C, liquid ammonia can cause frost bite and severe burning. Companies therefore ensure areas are well ventilated, and that workers involved with liquid ammonia storage and transport wear impervious gloves, face shields, and rubber boots and aprons. Breathing apparatus must be available, and employees are drilled in emergency procedures to protect themselves and the local community in the event of a major accident. In the event of an ammonia spill, because ammonia is a gas at room temperature and lighter than air, it can usually be allowed to disperse after ensuring the safety of people who are downwind. Other major hazards associated with the manufacture of ammonia involve fires and explosions in the sections of the plants where hydrogen is produced from fossil fuels. Historically, these explosions have been small and are minimised by careful design and operation of the plant. The carbon monoxide gas produced during reforming is toxic and exposure to this gas must also be carefully monitored. As a consequence of these potential hazards, modern ammonia plants pay careful attention to operating, safety and maintenance routines.

19.3–19.4

Unreacted hydrogen and nitrogen in the gases leaving the converter are recycled and the ammonia is isolated by liquefaction. Various methods are employed to remove unwanted chemicals and isolate useful compounds.

Ammonia is a harmful gas and care must be taken to minimise exposure. Fire and explosions also constitute a risk to workers. Companies follow strict governmental and voluntary guidelines designed to protect the environment and the health and safety of employees and the community.

key questions 3 Using the information about waste management in the text, construct a table that lists the main by-products of ammonia production from natural gas and describes what becomes of each one. 4 The principles of green chemistry can be used to evaluate the environmental impact of a chemical process. Construct a table with two columns headed ‘Principles’ and ‘Practice’, as shown below. In the first column, list the twelve principles of green chemistry (Table 18.3 p. 310). In the second column, indicate the ways a modern plant using the Haber process could be regarded as complying with these principles.


Principles 1. Prevent waste . . . . . . . 12. Minimise the potential for accidents

Practice

19 converter desulfurisation Haber process

key terms hydrogen bonds nitrogenase nitrogen fixation

The Haber process 5 Fritz Haber has been credited with developing a process that has helped to feed millions of people. He has also been regarded as responsible for killing 5000 other people. Haber’s work before the start of World War I allowed Germany to produce sufficient ammonia for its agricultural needs and for production of nitric acid, which was needed to make explosives. Find out more about his eventful life. 6 During ammonia production the nitrogen and hydrogen gas are thoroughly purified before they are admitted to the reactor chamber. Explain the reason for this. 7 At 200°C, nearly twice as much ammonia forms when a 3 : 1 hydrogen/nitrogen mixture under 250 atm pressure reaches equilibrium as is formed at 400°C. Why are temperatures above 400°C used to manufacture ammonia? 8 In the commercial production of ammonia, nitrogen is reduced by hydrogen gas. a Write a balanced chemical equation for the reaction. b Write an expression for the equilibrium constant for the reaction. c Predict the reaction conditions that would favour the maximum equilibrium yield of ammonia. Explain your reasoning. d How would these conditions affect the rate of reaction? e Discuss the conflict that arises in the choice of reaction conditions for the Haber process and the strategies used to deal with this conflict. 9 During the Haber process, as gas passes through the catalyst in the converter its temperature increases. a Why does the temperature of the gas rise? b Why is the gas leaving the converter cooled? c What side benefit is obtained from the need to cool gases? 10 Ammonia is normally liquefied to isolate it from the reaction mixture obtained from the Haber process. a Suggest another way that ammonia could be separated from nitrogen and hydrogen gas.

steam reforming synthesis gas waste heat

b What are the advantages and disadvantages of the method you have suggested? 11 Suppose that Fritz Haber did not invent a method to synthesise ammonia and that no economical method had ever been discovered. What do you think the world would be like today? 12 Describe two ways that expenses are minimised in the Haber process.

Connecting the main ideas 13 Under the title ‘Ammonia: providing fertilisers to feed the world’, write a paragraph which uses correctly the words: fertilisers, urea, Haber process, steam reforming, synthesis gas, converter, equilibrium, reaction rate and condenser. 14 A citizens’ action group calls a meeting of residents living near a major ammonia manufacturing plant with the intention of lobbying the government to have the plant moved because of health, safety and environmental concerns. a Design a flyer advertising the first public meeting of the group, highlighting the main issues of concern. b The public relations manager of the ammonia company has been invited to speak at the meeting. List the points that the manager might wish to make to the concerned residents. 15 Construct a concept map that includes the following terms: ammonia, hydrogen bonds, fertilisers, Haber process, converter, condenser, waste heat, desulfurisation, steam reforming and synthesis gas. 16 Design a safety notice that summarises: a the effects to a person’s health that may result from exposure to ammonia fumes b the first aid treatment that should be used to treat exposure to ammonia c the personal protective equipment that someone working with ammonia should use or have access to.


2

chapter 20

Production of nitric acid

key knowledge • the industrial production of nitric acid − factors affecting the production including rate and equilibrium position, catalysts, temperature, pressure − waste management including generation, treatment and reduction − health and safety − uses of nitric acid



• name the raw materials used in the production of nitric acid • describe the reaction steps by which nitric acid is manufactured • explain how the principles of equilibrium and reaction rates play a significant role in determining reaction conditions • explain the reasons for choice of reaction conditions, such as pressure, temperature and catalyst • describe waste management procedures • describe how the production of nitric acid relates to the principles of green chemistry • describe health and safety issues involved in the production of nitric acid • recall the major uses of nitric acid.

Nitric acid ranks about thirteenth in the order of world chemical production. Approximately 50–60 million tonnes of the acid are produced worldwide each year. Incitec Pivot has nitric acid plants at Kooragang Island near Newcastle, New South Wales. Even though nitric acid is made in great quantities, the average person rarely sees it. It is used mainly for making other materials.

Miscellaneous (10%)

Explosives (15%)

20.1

Uses of nitric acid Fertilisers Plants obtain nitrogen in the form of soluble nitrogen compounds present in soil. Since plants are able to absorb nitrate ions through their roots, nitrates can be used as fertilisers. The main use of nitric acid is, in fact, the production of the important fertiliser ammonium nitrate. Consequently, nitric acid plants are often adjacent to ammonia production facilities to minimise transport costs. The growth in production of nitric acid has slowed in recent years because urea, CO(NH2)2, has increasingly been used as a nitrogen fertiliser in preference to ammonium nitrate. Reaction between nitric acid and ammonia produces a solution of ammonium nitrate: HNO3(aq) + NH3(g) –—h NH4NO3(aq)

Heat released during this acid–base reaction causes the solution to boil so that a very concentrated solution forms. By spraying this solution from a tall tower, solid pellets are formed as the droplets fall through the air and lose water by evaporation. Since ammonium nitrate decomposes explosively at high temperatures, the process requires elaborate safety measures.

Fertilisers (75%)

Figure 20.1 Uses of nitric acid.

chemfact Ammonium nitrate has been used by terrorists to make bombs. As a consequence, access to fertilisers containing large amounts of this chemical has been restricted by law in Australia. Urea, ammonium sulfate and ammonium nitrate/ammonium sulfate blends are now used as substitutes.

Figure 20.2 In 1921 this ammonium nitrate factory in Germany exploded, killing 600 people and leaving 7000 homeless.

329 Production of nitric acid

Other common fertilisers made from nitric acid are potassium nitrate, sodium nitrate, calcium nitrate and cobalt nitrate. The fertiliser chosen by a farmer depends upon the nature of the soil and type of crop or pasture.

Explosives Most chemical explosives, such as ammonium nitrate, trinitrotoluene (TNT) and nitroglycerine, are nitrogen compounds made from nitric acid (Figure 20.3). When they decompose they release large amounts of energy and gaseous products, including nitrogen gas.

CH3

a O2N

b

CH2ONO2

NO2 CHNO2 CH2ONO2 NO2

Figure 20.3 Structural formulas of a TNT and b nitroglycerine.

Figure 20.4 Nitric acid is used in the manufacture of many explosives.

Ammonium nitrate, for example, decomposes above 300°C, forming nitrogen, oxygen and water vapour: 2NH4NO3(s) –—h 2N2(g) + O2(g) + 4H2O(g)

In a confined space enormous force is generated as large quantities of hot gas are formed.

Nitrate salts

chemfact Recent research studies have questioned the use of dinitrogen monoxide as a base component in anaesthetics. Patients treated with this gas were slower to recover and likely to stay in hospital longer than those given other anaesthetic mixtures. Dinitrogen oxide has been used as an anaesthetic since the 1840s.


Nitric acid is a strong oxidant. It reacts with most metals to form solutions of the corresponding metal nitrate salt. Metal nitrates have a variety of applications, including: silver nitrate—in silver plating and medicine potassium nitrate—as a fertiliser lead nitrate—to make special explosives, such as lead azide strontium nitrate—red flares and fireworks (pp. 78–9) sodium nitrate—as a food additive.

Laughing gas At temperatures between 200°C and 260°C, ammonium nitrate decomposes to produce dinitrogen monoxide gas: NH4NO3(s) –—h N2O(g) + 2H2O(g)

Nitrogen(I) oxide is widely used in anaesthetics used in dentistry and surgery. Since small quantities of the gas cause involuntary laughter, it has been called ‘laughing gas’.

20.2

The Ostwald process Until early last century, nitric acid was manufactured by the addition of concentrated sulfuric acid to sodium nitrate. The sodium nitrate was obtained from deposits occurring naturally in Chile. Nitric acid is now made from ammonia in a three-step process involving the successive oxidation of compounds of nitrogen: NH3 –—h NO –—h NO2 –—h HNO3

This method of synthesis of nitric acid from ammonia is known as the Ostwald process, after the German chemist Wilhelm Ostwald (1853–1932) who discovered how to convert ammonia to nitrogen(II) oxide. Although the Ostwald technique may seem indirect, it avoids direct oxidation of nitrogen to nitrogen(II) oxide. Even at very high temperatures such as 3000°C, direct conversion of nitrogen into nitrogen(II) oxide gives possible yields of only a few per cent. Water

Figure 20.6 The German chemist Wilhelm Ostwald (1853–1932) was awarded the Nobel Prize in Chemistry in 1909 for his work on catalysis, chemical equilibria and rates of reaction.

Air

Hot air

Air

Ammonia Nitrogen(II) oxide

chemfact

Nitrogen(lV) oxide

Gauze catalyst

Nitric acid Converter

Cooling chamber

Absorption tower

The oxidation states of nitrogen in these compounds are, in order, –3, +2, +4 and +5. Nitrogen atoms undergo oxidation in each step of this process.

Figure 20.5 A flow diagram of the Ostwald process.

Step 1: Catalytic oxidation of ammonia Air is preheated and mixed with ammonia, and the gas mixture is then passed through a converter. Oxygen, from air, oxidises the ammonia to nitrogen(II) oxide: 4NH3(g) + 5O2(g) –—h 4NO(g) + 6H2O(g); ΔH = –907 kJ mol–1

The converter in which this reaction takes place contains a catalyst made of many layers of a thin gauze woven from an alloy of 90% platinum and 10% rhodium (Figure 20.7). The incoming gas mixture must be filtered to remove solid particles that could poison the catalyst. The composition of the incoming gas mixture, the time the gas is in contact with the catalyst and the temperature are all controlled to ensure that the production of nitrogen(II) oxide occurs more rapidly than two alternative reactions—the conversion of ammonia to nitrogen or to nitrogen(I) oxide. 4NH3(g) + 3O2(g) –—h 2N2(g) + 6H2O(g) 2NH3(g) + 2O2(g) –—h N2O(g) + 3H2O(g)

The production of nitrogen(II) oxide is maximised and alternatives minimised by the use of high temperatures (820–930°C) and high pressures (about

Figure 20.7 A gauze made from platinum and rhodium acts as the catalyst for nitric acid manufacture. The catalyst is eroded by the flow of hot gases and needs to be replaced every 6 to 10 weeks.

chemfact Nitrogen(II) oxide, NO, is also known as nitric oxide or nitrogen monoxide.


chemfact Cells on the inside of blood vessels release nitrogen(II) oxide (NO), activating an enzyme to relax muscle cells and lower blood pressure. Drugs such as nitroglycerine and amyl nitrite produce nitrogen(II) oxide that relaxes the blood vessels, allowing an improved flow of blood to the heart. NO plays a similar role in the act of sex. Sexual arousal in a man can send a stimulus to the muscle in his penis, which releases NO, relaxing the muscle so that blood enters the tissues and creates an erection. Drugs such as Viagra enhance this effect and provide treatment for men suffering from erectile dysfunction.

1100 kPa or 11 atm). The flow rate of incoming gas is also adjusted so that it is in contact with the catalyst for an extraordinarily short period of about 0.003 seconds. The catalyst itself is selected so that it promotes the desired reaction in preference to the competing ones. The yield of nitrogen(II) oxide from this step is about 95%.

Step 2: Oxidation of nitrogen monoxide The nitrogen monoxide produced in the converter is cooled and mixed with air, oxidising the nitrogen monoxide to nitrogen(IV) oxide: 2NO(g) + O2(g)

2NO2(g); ΔH = –114 kJ mol–1

This reaction is the slow step in the Ostwald process and is highly dependent on temperature and pressure. Applying Le Chatelier’s principle, the equilibrium yield of nitrogen(IV) oxide is increased by: using low temperatures: since the reaction is exothermic the system will compensate for the low temperature by favouring the direction that releases energy, i.e. the forward reaction using high pressures: the system reacts to high pressure by favouring the direction that will result in a decrease in pressure. Since there are more gas particles on the left-hand side of the reaction (one O2 and two NO molecules) than on the right (two NO2 molecules), the forward reaction will tend to decrease pressure and is thus favoured. Unlike many reactions, this reaction is unusual in that its rate increases with decreasing temperature. In this instance no conflict exists between the conditions needed for high equilibrium yield and economic reaction rate. In practice, the incoming gas mixture is cooled to the lowest temperature conveniently achieved with the available cooling water, about 30°C, and it is sometimes compressed. Rapid formation of nitrogen dioxide occurs under these conditions. The brown-coloured nitrogen(IV) oxide gas is in equilibrium with colourless dinitrogen tetroxide gas: 2NO2(g)

N2O4(g)

Step 3: Absorption of nitrogen dioxide Water is mixed with the nitrogen dioxide gas in absorption towers. The following reaction occurs: 3NO2(g) + H2O(l) –—h 2HNO3(aq) + NO(g)

The absorption towers are packed with inert granular material to increase the contact between the gases and water. The nitrogen(II) oxide formed in this reaction reacts with a stream of air which is also introduced into the tower. It is oxidised to nitrogen(IV) oxide, which then reacts with water as before. Solutions of 45–50% nitric acid are obtained. Further distillation yields nitric acid solutions of 68% concentration. Industries and schools usually purchase nitric acid of this concentration.


summary

20.1–20.2

Th The main i use for f nitric it i acid id is i in i the th production d ti off fertilisers, f tili including ammonium nitrate. It is also used in the manufacture of explosives. The raw materials used to make nitric acid are ammonia, water, and oxygen from the air. The Ostwald process used to manufacture nitric acid involves three main steps: ammonia is oxidised to nitrogen(II) oxide, which is further oxidised to nitrogen(IV) oxide, followed by reaction of the nitrogen(IV) oxide with water.

The Th conditions diti actually t ll chosen h for f the th conversion i off ammonia i (NH3) to nitrogen(II) oxide (NO) involve low pressures and the reactant gases are sometimes compressed. A platinum–rhodium catalyst is employed. The oxidation of nitrogen(II) oxide (NO) to nitrogen(IV) oxide (NO2) in the converter is exothermic and higher equilibrium yields are also favoured by the use of low temperatures and high pressures. Fast reaction rates are also favoured by the use of low temperatures and high pressures.

key questions 1 TNT is one of the products made from nitric acid. Combustion of TNT (C7H5N3O6) is rapid and releases large amounts of energy. a Write a balanced equation for this reaction if the products are carbon dioxide, water and nitrogen. b By referring to the equation from part a, explain why TNT is used as an explosive.

2 Prepare a flow chart to show the sequence of processes used to convert ammonia to concentrated nitric acid. Write equations for significant chemical reactions beside the appropriate section of the flow chart.

20.3

Waste management The main emissions from nitric acid manufacture come from the absorption tower and are mainly NO and NO2 (NOx) and trace amounts of nitric acid. Since these gases contribute to photochemical smog there is further treatment to reduce the levels that are emitted into the atmosphere. One approach involves heating these gases with a fuel such as natural gas, naphtha or hydrogen, over a catalyst, so that the NOx is reduced to N2. Alternatively, the absorption tower may be modified by increasing its size or operating pressure, or an additional absorption tower may be incorporated, with the advantage that more NOx is converted into nitric acid. Other means by which waste is minimised in accordance with the principles of green chemistry include: optimising reaction conditions in the converter, cooling chamber and absorption tower to give high yields with efficient energy use and to minimise waste emissions careful filtration of the gas mixture entering the converter because contaminated catalyst not only requires more frequent replacement but increases the pressure needed to force the gas through the catalyst bed, increasing energy expenditure using heat exchangers to remove heat released by the reaction in the converter and then using it to heat incoming gases or generate electricity. Modern plants have been described as ‘power plants that also make acid’—in some cases the surplus energy can be more valuable to a company than the acid itself

review Photochemical smog is discussed in Heinemann Chemistry 1, p. 332.

! Heat exchangers are devices used to transfer heat from one material to another. An example is a car radiator, in which the hot coolant is cooled by the air flow over the radiator surface.


recovering the costly platinum and rhodium metals from the catalyst. At the high temperatures and pressures used in the converter, the catalyst slowly vaporises and is lost. Gases leaving the converter are passed through a filter to recover the metals trialling methods such as the use of an additional catalyst in the converter to decompose the gas nitrogen(I) oxide, N2O, which is a significant greenhouse gas. Without this additional catalyst, about 1.5% of the ammonia in the converter is converted into nitrogen(I) oxide as an unwanted by-product.

extension Green chemistry solutions Th iinflfluence off green chemistry The h i may reduce d the h ddemandd ffor nitric acid in the future. For example, almost two billion kilograms of the chemical adipic acid are used to make nylon, polyurethane, plasticisers and lubricants. Food quality adipic acid is used as a setting agent in gelatine. It is also used in beverages, and in processed foods such as relishes, frozen dairy desserts and snack foods. Adipic acid is usually made by a process involving oxidation of benzene, a known carcinogen, using nitric acid. The greenhouse gas nitrogen(I) oxide is a significant and unwanted by-product O H2

O2/150°C

Ni catalyst

10 atm

(Figure 20.8). Due to the (Fi 20 8) D h massive i scale l off adipic di i acid id production, d i 10% of the annual increase in nitrogen(I) oxide levels in the atmosphere has been attributed to this process. A process described in 1994 by Dr Karen Draths and Professor John Frost of Michigan State University, the Draths–Frost synthesis, uses genetically altered bacteria, called biocatalysts, to convert glucose into adipic acid at relatively low temperatures in aqueous solution without the use of nitric acid (Figure 20.9). Because glucose is derived from plants, it is the ultimate renewable raw material— you can simply grow as much plant material as needed.

OH

+

HNO3

HO2C(CH2)4CO2H + N2O Adipic acid

Benzene

Figure 20.8 The conventional process for adipic acid production.

C6H12O6 Glucose

Genetically modified bacteria

HO2C

CH

CH

CH

CH

CO2H

H2 Pt catalyst

HO2C(CH2)4CO2H Adipic acid

Figure 20.9 The Draths–Frost synthesis for adipic acid.

questions

?

E1 Adipic Adipic i acid id (hexanedioic (hexanedioic acid) acid), HOOC(CH2)4COOH COOH, and 1,6-diaminohexane, H2N(CH2)6NH2, are monomers used to manufacture nylon 6,6. a Draw a section of the repeating units of this polymer. b What other substance is formed in this polymerisation reaction?


c What name is given to the bond formed between the monomer units? d Which polymer of biological importance has the same linkage between its monomers? E2 Why is the synthesis pathway developed by Draths and Frost considered to be a green chemistry alternative for the production of adipic acid?

20.4

NITRIC ACID

Health and safety

UN No.

Concentrated nitric acid is corrosive and causes severe burns to the skin and eyes. The fumes are harmful if inhaled. As a result, there is careful monitoring in nitric acid plants for leaks and spills, and employees must be familiar with appropriate procedures to handle them if they do occur. Full protective equipment and breathing apparatus must be available. Stainless steel pipes and reaction vessels are used in sections of the plant because of the corrosive nature of the acid. Spills of the acid can be treated by isolating the acid with sand or earth and carefully neutralising it with calcium hydroxide (‘slaked lime’) or sodium carbonate. Levels of nitrogen(IV) oxide gas must also be carefully monitored since inhalation at relatively low concentrations may cause lung oedema (fluid in the lungs) and excessive exposure may be fatal. The ratio of ammonia to air in the gas entering the converter is continuously measured and controlled to ensure it does not reach explosive conditions.

summary

2031

CORROSIVE

HAZCHEM

2PE Figure 20.10 Signage for storing bulk nitric acid.

20.3–20.4

UUnreacted t d ammonia i andd nitrogen it in i the th gases leaving l i the reactor are recycled and the ammonia is isolated by liquefaction. Various methods are employed to limit NOx emissions, maximise conversion efficiency, and minimise loss of energy.

Nitric Nit i acid id is i a corrosive i liquid li id andd nitrogen( it (IVV) oxide id is i toxic. t i As a result, manufacturers follow strict guidelines designed to protect the environment and the health and safety of employees and the community.

key questions 3 Using the information in the text, construct a table that lists the main wastes from nitric acid production, how they are treated, and how the quantities produced are reduced. 4 The principles of green chemistry can be used to evaluate the environmental impact of a chemical process. Construct a table with two columns headed ‘Principles’ and ‘Practice’, as shown below. In the first column, list the twelve principles of green chemistry (Table 18.3, p. 310). In the second column, indicate the ways a modern plant using the Ostwald process could be regarded as complying with these principles.

Principles 1. Prevent waste . . . . . . 12. Minimise the potential for accidents

Practice


20

key terms converter heat exchangers

Ostwald process

The Ostwald process 5 The Arab alchemist Jabir ibin Hayyan is credited with the discovery of nitric acid in about 800 AD. He made it by heating a mixture of sodium nitrate powder with sulfuric acid and condensing the gas produced. The process can be written as: 2NaNO3(s) + H2SO4(aq) –—h Na2SO4(aq) + 2HNO3(aq)

a Use this equation to calculate the percentage atom economy for nitric acid production by this process. b In the Ostwald process the overall reaction can be represented by: NH3(g) + 2O2(g) –—h HNO3(aq) + H2O(l)

6

7

8

9

Use this equation to calculate the percentage atom economy for the Ostwald process. c Comment on the significance of the difference between the percentage atom economies of each method for producing nitric acid. During the Ostwald process, nitrogen monoxide is made from ammonia at about 900°C and then cooled to 30°C before being reacted with air to make nitrogen(IV) oxide. Both these reactions are exothermic. Why are the temperatures used for these reactions so different? As the gas passes through the catalyst bed in the converter during nitric acid manufacture, its temperature increases. The gas must be cooled before it is mixed with air. a Why does the temperature of the gas rise? b Why is it necessary to cool the gas? c What side benefit is obtained from the need to cool gases? In the commercial production of nitric acid, nitrogen(II) oxide is oxidised to nitrogen(IV) oxide by oxygen gas. a Write a balanced chemical equation for the reaction. b Write an expression for the equilibrium constant for the reaction. c Predict the reaction conditions that would favour maximum equilibrium yield of nitrogen dioxide. Explain your reasoning. d How do these conditions affect the rate of reaction? A number of different oxidation states of nitrogen are involved in the industrial production of nitric acid from ammonia. a Give the equation for the reaction in which nitrogen in a compound is oxidised from the +2 to the +4 oxidation state.


b Give the equation for the reaction in which nitrogen in a compound is oxidised from the –3 to the +2 oxidation state. c Give the equation for the reaction in which nitrogen in a compound is oxidised from –3 to the +5 oxidation state. 10 Nitrogen(II) oxide is present in the gaseous emissions from nitric acid plants. a Explain why the levels of these emissions need to be restricted. b What other human activities cause nitrogen(II) oxide to be produced? 11 If you were designing a new nitric acid plant, briefly list important factors that you would consider when deciding: a whether to provide further treatment for waste gases containing NOx from the absorption tower b which catalyst to use, platinum–rhodium or powdered iron c whether to use air or oxygen for the reaction in the converter 12 Describe two ways that expenses are minimised in the Ostwald process.

Connecting the main ideas 13 Under the title ‘Nitric acid: providing fertilisers to feed the world’, write a paragraph that uses the words: fertilisers, Ostwald process, converter, cooling chamber, absorption tower, equilibrium, reaction rate, and catalytic oxidation. 14 Imagine a serious acid spill occurred at a nitric acid plant. As a consequence, the local council invited the plant manager to speak to the local citizens about their health and safety concerns. a Design a flyer advertising the meeting and highlighting the main issues of concern. b List the points that the manager might wish to make to the concerned residents. 15 Construct a concept map that includes the following terms: nitric acid, explosives, fertilisers, Ostwald process, converter, absorption tower, cooling chamber, oxidation, reaction rate, equilibrium yield, and catalyst. 16 Design a safety notice that summarises: a the effects to a person’s health that may result from contact with nitric acid b the first aid treatment that should be used to treat contact with nitric acid c the personal protective equipment that someone working with nitric acid should use or have access to.

2

chapter 21

Production of sulfuric acid

key knowledge • the industrial production of sulfuric acid − factors affecting the production including rate and equilibrium position, catalysts, temperature, pressure − waste management including generation, treatment and waste reduction − health and safety − uses of sulfuric acid



• name the raw materials used in the production of sulfuric acid • describe the reaction steps by which sulfuric acid is manufactured • explain how the principles of equilibrium and reaction rates play a significant role in determining reaction conditions • explain the reasons for choice of reaction conditions, such as pressure, temperature and catalyst • describe waste management procedures • describe health and safety issues involved in the production of sulfuric acid • describe how the production of sulfuric acid through the contact process relates to the principles of green chemistry • recall the major uses of sulfuric acid.

! One of the first chemical industries established in Melbourne was a plant to manufacture sulfuric acid (p. 296).

Sulfuric acid is used so extensively by industry that the quantity of sulfuric acid manufactured by a country provides an indicator of the country’s industrial activity. It is produced in greater quantities than any other chemical in both Australia and the world. Annual worldwide production is estimated at about 200 million tonnes. In Australia our production meets almost all of our requirements. In future years it is anticipated that Australia will become a major exporter of the chemical. As transport and storage are both hazardous, a high proportion of the acid is used close to the site of manufacture. Major Australian producers of sulfuric acid are at Kalgoorlie in Western Australia, Roxby Downs in South Australia, Risdon in Tasmania and Mt Isa in Queensland. These sulfuric acid plants are located near smelting and refining industries that produce waste sulfur dioxide, a raw material for the production of sulfuric acid.

21.1

Uses of sulfuric acid

Fertilisers (75%)

Other chemicals manufacture (15%)

Metal production (5%)

Why is sulfuric acid so useful to us? Figure 21.1 shows the main uses of sulfuric acid. Most sulfuric acid is used to make fertilisers such as superphosphate, ammonium sulfate and ammonium phosphate. The amount of acid manufactured therefore mainly depends upon the demand for fertilisers and the size of the market held by local fertiliser producers. Sulfuric acid is also used in the manufacture of paper, household detergents, pigments, dyes and drugs. It is the electrolyte in car batteries (Chapter 27) and is used in petroleum refining and metallurgical processes.

Petroleum refining (5%)

Figure 21.1 The uses of sulfuric acid.

! The South Pacific island of Nauru was once a major source of rock phosphate for the Australian superphosphate industry. The phosphate was formed over thousands of years from the excretions and bodies of sea birds. Over 80% of the island has been mined, resulting in significant environmental damage and changes to the islanders’ lifestyle and culture. In 1997 the Australian Government paid Nauru $75 million as compensation for the environmental damage.


Superphosphate An adequate supply of phosphorus is vital for plant growth. Many Australian soils are seriously deficient in this element and it must be added if the land is to be farmed. During superphosphate manufacture, insoluble calcium phosphate Ca3(PO4)2 contained in rock phosphate is converted to a soluble form that plants can absorb. Superphosphate is manufactured by adding sulfuric acid to finely powdered rock phosphate. Over a period of several weeks the following reaction occurs: Ca3(PO4)2(s) + 2H2SO4(l) + 4H2O(l) –—➤ Ca(H2PO4)2(s) + 2CaSO4.2H2O(s) superphosphate

The final mixture is superphosphate. It is crushed into a powder and bagged for easy distribution. Although Queensland has deposits of phosphate rock, these have not yet been exploited to any major extent. The company Incitec Pivot manufactures superphosphate at Portland in Victoria, using rock phosphate that comes from North Africa, where it is cheap and readily available.

Other uses As a strong acid Pure sulfuric acid is a viscous liquid that reacts with water in two steps: H2SO4(l) + H2O(l) –—➤ HSO4–(aq) + H3O+(aq) HSO4–(aq) + H2O(l) SO42–(aq) + H3O+(aq)

It acts as a diprotic acid, since it donates two protons to water. Although the first step is virtually complete, the Ka for the second step is much smaller. A large amount of heat is evolved during this process. For this reason, if you are preparing a sulfuric acid solution you must always take care to add acid to water slowly with continuous stirring (Figure 21.2). If water is added to pure acid, sufficient heat can be generated to cause the water to boil. The steam that is formed can cause the acid to splatter violently. Sulfuric acid is used extensively as a general-purpose strong acid. One example is the ‘pickling’ of iron and steel. Before sheet iron is galvanised or electroplated a surface layer of iron(III) oxide must be removed. Treatment with sulfuric acid converts the oxide into water and soluble iron(III) sulfate.

As a dehydrating agent

Figure 21.2 Considerable heat is evolved when sulfuric acid is diluted. Note the safety precautions being used. In the event of an acid spill on the skin, flood the area with running water. (For a large volume of acid, the bulk of the acid should first be quickly wiped away.)

Concentrated sulfuric acid is a powerful dehydrating agent. For example, sugar is dehydrated (Figure 21.3) according to the equation: H2SO4(l)

C12H22O11(s) ––––—➤ 12C(s) + 11H2O(l)

The action of concentrated sulfuric acid on hydrated copper sulfate (Figure 21.4) provides a good visual demonstration of dehydration as the blue hydrated solid changes to the white anhydrous salt, according to the equation: H2SO4(l)

CuSO4.5H2O(s) ––––—➤ CuSO4(s) + 5H2O(l) blue white

The dehydrating ability of sulfuric acid is often utilised in laboratories to dry gas mixtures that are being prepared or analysed. The method of bubbling the gas mixture through concentrated sulfuric acid is suitable for drying gases such as air, carbon dioxide and nitrogen. It is not suitable for bases, such as ammonia, or reductants, such as hydrogen sulfide, because these will react with the acid. Figure 21.3 Concentrated sulfuric acid dehydrates sugar. The reaction is exothermic, producing steam and a porous pillar of carbon.

chemfact

Figure 21.4 Sulfuric acid behaving as a dehydrating agent. Blue crystals of CuSO4.5H2O are converted to white CuSO4 when treated with concentrated sulfuric acid.

Always be extremely cautious when you use concentrated sulfuric acid. It can cause serious burns if spilled or splashed on skin or other body tissues. The main reason for this is its ability to dehydrate and, in the process, decompose the organic molecules you are made of.

339 Production of sulfuric acid

As an oxidant Concentrated sulfuric acid is a strong oxidant, especially when hot. Sulfuric acid can be reduced to sulfur dioxide (SO2), sulfur (S), or hydrogen sulfide (H2S), depending on the temperature, the strength of the reductant involved, and the mole ratio of the reactants. For example, the following reactions can occur when zinc is added to sulfuric acid: Zn(s) + 2H2SO4(aq) –—➤ ZnSO4(aq) + 2H2O(l) + SO2(g) 3Zn(s) + 4H2SO4(aq) –—➤ 3ZnSO4(aq) + 4H2O(l) + S(s) 4Zn(s) + 5H2SO4(aq) –—➤ 4ZnSO4(aq) + 4H2O(l) + H2S(g)

Like other strong acids, dilute sulfuric acid reacts with zinc to produce hydrogen gas: Zn(s) + H2SO4(aq) –—➤ ZnSO4(aq) + H2(g)

In this last reaction it is the H+(aq) ion that acts as the oxidant.

21.2

The contact process Raw materials for making sulfuric acid Sulfuric acid is manufactured in stages from sulfur dioxide. These involve oxidation of sulfur dioxide to sulfur trioxide; followed by conversion to the acid. The process can be summarised:

! The contact process was developed in 1831 by Peregrine Phillips.

SO2 (from various sources) –—➤ SO3 –—➤ H2SO4

Figure 21.5 shows the main stages involved in the contact process for producing sulfuric acid from sulfur.

Concentrated sulfuric acid

Molten sulfur Sulfur dioxide Air

Air

Unreacted sulfur dioxide

Burner

Absorption tower

Catalyst bed

Oleum diluted to form sulfuric acid

Hot gas cooled and returned

Figure 21.5 The contact process for manufacturing sulfuric acid.


Converter

Sulfur trioxide

The sulfur dioxide used to produce sulfuric acid is obtained from two principal sources: combustion of sulfur recovered from natural gas and crude oil sulfur dioxide formed during the smelting of sulfide ores of copper, zinc or lead. A third source is from the mining of underground deposits of elemental sulfur by a method known as the Frasch process, but this is little used in Australia due to the ready availability of sulfur from the other two processes. The sources of sulfur dioxide used in Australia are attractive from an environmental viewpoint as they use a by-product of other industries and limit the amount of sulfur dioxide emitted into the atmosphere. Increasingly, less and less imported sulfur recovered from fossil fuels is being used to make Australia’s sulfuric acid, with most sulfur dioxide derived from smelters. Several smelters and refineries in Australia act as sources of sulfur dioxide or sulfur for conversion to sulfuric acid.

chemfact ‘Smelting’ is a technique used to extract metals from their ores. It involves heating the ore in a furnace with a reducing agent. The smelting of copper, lead and zinc ores, in particular, is an important source of sulfur dioxide.

Step 1: Burning sulfur If sulfur is used as a raw material for making sulfuric acid, the first stage of the manufacturing process involves spraying molten sulfur under pressure into a furnace, where it burns in air to produce sulfur dioxide gas. The high surface area of the sulfur spray allows combustion to be rapid. The reaction is represented by the equation: S(l) + O2(g) –—➤ SO2(g); ΔH = –297 kJ mol–1

Temperatures up to 1000°C may be reached. The sulfur dioxide gas is then cooled for the next step in the process.

Step 2: Catalytic oxidation of sulfur dioxide Sulfur dioxide gas is oxidised to sulfur trioxide gas by oxygen, using vanadium(V) oxide as a catalyst: 100

2SO3(g); ΔH = –197 kJ mol–1

This step is performed in a reaction vessel called a converter. Sulfur dioxide is mixed with air and passed through trays containing loosely packed porous pellets of catalyst (catalyst beds). The converter contains several catalyst beds and the gas mixture passes over each in succession. Because the reaction is exothermic, it is necessary to cool the gas mixture as it passes from one tray of catalyst to another to maintain the desired reaction temperature. The temperature in the converter is maintained between 400°C and 500°C and the gas pressure is close to 1 atmosphere. Nearly complete conversion of sulfur dioxide to sulfur trioxide is achieved. Knowing the effect of reaction conditions on equilibrium yield and reaction rate, you can appreciate the reasons behind this arrangement. Using Le Chatelier’s principle, the equilibrium yield of sulfur trioxide will increase: as temperatures decrease. Since the reaction is exothermic, the equilibrium system will compensate for the loss of energy associated with the temperature decrease by favouring the direction that releases energy, i.e. the forward reaction (Figure 21.6) as pressure increases. The equilibrium system reacts to high pressure by favouring the direction that will result in a decrease in pressure. Since there are more gas particles on the left-hand side of the equation

Conversion to sulfur trioxide at equilibrium (%)

2SO2(g) + O2(g)

80

60

40

20

400

600

800

1000

Temperature (˚C)

Figure 21.6 The effect of temperature on the equilibrium yield of sulfur trioxide for a typical mixture of sulfur dioxide and oxygen.


(one O2 and two SO2 molecules) than on the right-hand side (two SO3 molecules), the forward reaction will result in a decreased pressure. if excess reactants (oxygen or sulfur dioxide) are added. The rate of the reaction will be faster: as temperature increases as pressure increases if a catalyst is employed. A conflict exists because a high equilibrium yield is favoured by low temperatures, whereas a fast rate of production of sulfur trioxide is favoured by high temperatures. By using a catalyst it is possible to use lower temperatures in the converter and still achieve an acceptable reaction rate. The effectiveness of the catalyst is improved by the large surface area obtained by spreading it over trays. The most effective catalyst is not necessarily used, since cost and the chance of the catalyst being ‘poisoned’ must also be considered. Vanadium(V) oxide (V2O5) is chosen, and dust particles that could poison the catalyst are removed by passing the air and sulfur dioxide through electrostatic precipitators before they enter the converter. The desired reaction temperature is maintained by cooling the gas mixture as it passes from one catalyst tray to another. The equilibrium yield of sulfur trioxide is also improved by using an excess of the cheaper reactant, oxygen, in the form of air. Since high yields can be achieved at atmospheric pressure, expensive high pressure equipment is not usually employed.

Step 3: Absorption of sulfur trioxide Sulfur trioxide reacts with water to form sulfuric acid: SO3(g) + H2O(l) –—➤ H2SO4(aq); ΔH = –130 kJ mol–1

chemfact The Persian alchemists Jabir ibn Hayyan and Ibn Zakariya al-Razi in the eighth and ninth centuries heated hydrated copper(II) sulfate or iron(II) sulfate to produce the metal oxide and SO3 and H2O, which then combined to form a dilute solution of sulfuric acid H2SO4. In the seventeenth century Johan Glauber prepared sulfuric acid by burning sulfur with potassium nitrate. The sulfur is oxidised stepwise to sulfur trioxide by the nitrate ion and then by oxygen and various oxides of nitrogen.


However, direct reaction with water is not used, because so much heat evolves when sulfur trioxide is added to water that a fine mist of acid is produced, which is difficult to collect. Instead, sulfur trioxide gas is passed into concentrated sulfuric acid in an absorption tower. The reaction can be regarded as occurring in two steps: 1 The sulfur trioxide gas dissolves almost totally in the acid to form a liquid known as oleum (H2S2O7): SO3(g) + H2SO4(l) –—➤ H2S2O7(l)

2 Oleum obtained from the absorption tower is then carefully mixed with water to produce sulfuric acid: H2S2O7(l) + H2O(l) –—➤ 2H2SO4(l)

In practice, many plants trickle in the additional water, together with the concentrated sulfuric acid, to meet the incoming sulfur trioxide. This results in two moles of sulfuric acid being produced for every one introduced into the absorption tower.

summary

21.1–21.2

SSulfuric lf i acid id is i produced d d in i greater t quantities titi than th any other th chemical. The main use for sulfuric acid is to make fertilisers, including superphosphate. Sulfur or sulfur dioxide, water, and oxygen from air are the main raw materials for sulfuric acid production. Sulfur may be obtained from the processing of fossil fuels, and sulfur dioxide from metal smelting operations. The contact process is the main method employed for manufacturing the acid. It involves converting sulfur dioxide into sulfur trioxide, followed by absorption into oleum and dilution with water.

The Th reaction ti in i the th converter t is i exothermic th i andd higher hi h equilibrium yields are favoured by the use of low temperatures and high pressures. Fast reaction rates are favoured by the use of high temperatures and pressures. The conditions actually chosen for sulfuric acid manufacture are a compromise between achieving rapid rates, high equilibrium yields, and cost. In practice, it is manufactured at moderate temperatures (400–500°C) and atmospheric pressure using a vanadium( V ) oxide catalyst.

key questions 1 Examine Figure 21.2, which shows concentrated sulfuric acid being diluted. a List the safety precautions in use. b Explain why each of these precautions is used. c For large spills of sulfuric acid on the skin, why is it often recommended that the bulk of the acid should be quickly wiped away before washing thoroughly with water? (Note: Small-scale laboratory spills should be immediately washed with water.)

2 Prepare a flow chart to show the sequence of processes used to convert sulfur into concentrated sulfuric acid. Write equations for significant chemical reactions beside the appropriate section of the flow chart. Indicate how the conflict between rate and equilibrium considerations is dealt with.

21.3

Waste management From an environmental perspective, an attractive feature of sulfuric acid plants is that they use sulfur or sulfur dioxide that is a by-product from other industries such as metal smelters and petroleum refining. Manufacturers of sulfuric acid must maximise the conversion of sulfur dioxide to sulfur trioxide because strict limits are set for sulfur dioxide emissions into the atmosphere to minimise the formation of acid rain (p. 252). Most plants built after 1970 now use a double absorption process. The gas mixture, after several passes over the catalyst beds in the converter, is passed through the absorption tower to remove the sulfur trioxide that has formed. Unreacted gases are then recycled to the converter for one or two more passes over the catalyst before being returned for absorption. In this way the percentage of sulfur dioxide converted can be increased from 98% to better than 99.6%. Emissions are continuously monitored for sulfur dioxide, and manual sampling and chemical analysis measure levels of sulfur trioxide and sulfuric acid. The amount of sulfuric acid mist emitted from the process is minimised by controlling the operating temperature of the absorber, gas flow rates, and concentrations. Improvements in conversion have also been made possible by adding small amounts of caesium to the vanadium(V) oxide catalyst to increase its


efficiency and allow it to operate at lower temperatures. Caesium-doped catalysts are about three times more expensive than the usual vanadium(V) oxide catalyst. Because sulfuric acid has a relatively high boiling temperature of 290°C, it has a low vapour pressure and there is no appreciable air pollution problem with its storage, handling and shipping. There is relatively little solid waste produced from the sulfuric acid process. After recovery of the mildly toxic vanadium from spent catalyst, the catalyst is disposed of in landfill sites. Cooling water is usually recycled. All three major steps: burning sulfur, catalytic conversion to sulfur trioxide and absorption, are exothermic. As a consequence, manufacturers have a ready supply of energy for other operations in the plant, and modern plants use this energy to operate efficiently and minimise costs. Since more energy is produced than is used in the process, sulfuric acid plants are often used as the energy source for the production of other chemicals. Some sulfuric acid plants use the energy to generate electricity. Since no carbon dioxide is produced, the energy can be described as ‘green’ compared with other forms of energy production (Figure 21.7).

Green chemistry and sulfuric acid production Figure 21.7 A sulfuric acid manufacturing plant. Heavy insulation is used to reduce heat loss. The small proportion of sulfur dioxide that is not converted into sulfuric acid is released from the chimney.

The contact process for the manufacture of sulfuric acid existed well before the concept of green chemistry emerged, and the scope for further improvements in the process seems limited. None the less, developments in efficiency and reduction in emissions as a result of environmental, legislative, technological and economic pressures may be viewed as conforming to some of the principles of green chemistry. The contact process is being modified to enable it to make use of the very low concentrations of sulfur dioxide emissions that are emitted from some industries, and already some plants use dilute waste sulfuric acid and hydrogen sulfide gas as their feedstock.

extension Reducing the need for sulfuric acid L Large excess quantities i i off sulfuric lf i acid id are usedd in i the h iindustrial d i l nitration of organic compounds. These nitration reactions involve the addition of a NO2 group to an organic compound, using concentrated sulfuric acid as a catalyst. The reactions are employed in the manufacture of synthetic rubber, pharmaceuticals, pesticides, and dyes. Large quantities of sulfuric acid catalyst are required because the acid is diluted by the water that is formed as a by-product during the nitration reactions. The separation of the acid from the reaction mixture is difficult and energy consuming.

questions

A green chemistry h i t alternative lt ti has h recently tl been b designed. d i d This Thi uses a solid catalyst called silica-sulfuric acid, and eliminates the drawbacks of the conventional process. In this catalyst the sulfuric acid molecule is covalently bonded to the surface of silica gel. This may represented as SiO2–O–SO3H. The nitration of organic compounds using silica-sulfuric acid requires milder reaction conditions and simpler procedures, and produces less waste than the traditional method while still producing good yields.

?

E1 The The w water ater t by-product by product d formed during the nitration process slows the reaction down by diluting the sulfuric acid acid. Why does dilution of the sulfuric acid slow down the reaction? E2 What are the advantages of reducing the amounts of sulfuric acid needed during the nitration procedures? E3 The researchers investigating the use of silica-sulfuric acid state that ‘a new feature of the reaction is the fact that it is heterogenous. This could be worthwhile in an industrial setting.’ Name one advantage of using a heterogeneous reaction system.


21.4

Health and safety One of the most significant risks associated with the industry is in the transportation of sulfuric acid. Procedures are highly regulated and companies have strict protocols for the loading and unloading of the acid because there is a serious risk to personnel if the liquid is mishandled. Sulfuric acid is highly corrosive and can burn skin and eyes severely. It can cause blindness and third degree burns on contact. Exposure to sulfuric acid mist may also cause other health problems, including a build-up of fluid in the lungs (pulmonary oedema). Workers in sulfuric acid plants can also be exposed to the acid through breathing air contaminated with emissions containing oxides of sulfur. Both sulfur dioxide and sulfur trioxide are respiratory irritants, damage plants, and contribute to a major extent to acid rain (p. 252). Oleum is a highly corrosive oily liquid that produces sulfur trioxide fumes. Strict safety procedures, including adequate methods to trap the fumes, are required to minimise risks to workers and the environment in the case of accidental release. Work areas must be well ventilated and employees wear protective clothing. Acid spills are contained using materials such as earth, clay or sand, and then slowly diluted with water before being neutralised with a base such as limestone (CaCO3) or sodium carbonate.

summary

SULFURIC ACID UN No.

1830

CORROSIVE

HAZCHEM

2P Figure 21.8 Signage for storing and transporting sulfuric acid.

21.3–21.4

A process off double absorption bsorption is used us d to increase i crease conversion conversi n of sulfur dioxide to sulfur trioxide and minimise gas emissions. Sulfuric acid and the sulfur oxides are harmful and care must be taken to minimise exposure.

Companies C mpani s follow f llow strict stri t governmental government l and a d voluntary v luntary protocols protocol designed to protect the environment and the health and safety of employees and the community.

key questions 3 Use the information about waste management in the text to construct a table that lists the main by-products of sulfuric acid production from sulfur dioxide and describes how these problems are treated or reduced. 4 The principles of green chemistry can be used to evaluate the environmental impact of a chemical process. Construct a table with two columns headed ‘Principles’ and ‘Practice’, as shown. In the first column, list the twelve principles of green chemistry (Table 18.3 p. 310). In the second column, indicate the ways a modern plant using the contact process could be regarded as complying with these principles.

Principles 1. Prevent waste

Practice

. . . . . . . . 12. Minimise the potential for accidents


21

key terms absorption tower contact process converter dehydrating agent

diprotic acid double absorption oleum

The contact process 5 In the mid-eighteenth century, sulfuric acid was made from sulfur dioxide by heating it with potassium nitrate and then mixing the gases that were formed with water. The process can be written as:

8

4KNO3(s) + 4SO2(g) + 4H2O(l) –—➤ 2K2O(s) + O2(g) + 4NO(g) + 4H2SO4(aq)

a Use this equation to calculate the percentage atom economy for sulfuric acid production by this process. b In the contact process there are three main reactions involved: 2SO2(g) + O2(g) SO3(g) + H2SO4(l) H2S2O7(l) + H2O(l)

2SO3(g) H2S2O7(l) 2H2SO4(l)

i By adding these equations, write one overall equation that shows the production of H2SO4 from SO2. (Hint: Before adding, you will need to double the coefficients of some equations and cancel identical reactants and products from both sides of the overall equation.) ii Use this overall equation to calculate the percentage atom economy for the contact process. iii Comment on the significance of the difference between the percentage atom economies of the two methods for producing sulfuric acid. 6 The sulfur dioxide gas used as the raw material for manufacturing sulfuric acid can come from either sulfur extracted from processing fossil fuels or from smelting metal ores. Write balanced equations for the production of sulfur dioxide from: a elemental sulfur b smelting iron(II) sulfide. 7 In the commercial production of sulfuric acid, sulfur dioxide is oxidised to sulfur trioxide by oxygen gas. a Write a balanced chemical equation for the reaction. b Write an expression for the equilibrium constant for the reaction. c Predict the reaction conditions that would favour maximum equilibrium yield of sulfur trioxide. Explain your reasoning. d How would these conditions affect the rate of reaction?


9

10

11

12 13

e Discuss the conflict that arises in the choice of reaction conditions for the contact process and the strategies used to deal with this conflict. During the contact process, the temperature of the gas increases as it passes through each catalyst bed in the converter. The gas must be diverted for cooling before returning for another pass. a Why does the temperature of the gas rise? b Why is it necessary to cool the gas? c What side benefit is obtained from the need to cool the gas? Although water reacts directly with sulfur trioxide to form sulfuric acid, this reaction is not employed when sulfuric acid is made. a Explain why not. b Describe the process that is used. c Write equations for the reactions that occur in the process described in part b above. A number of different oxidation states of sulfur are involved in the industrial production of sulfuric acid from elemental sulfur. a Give the equation for the reaction in which sulfur is oxidised from the +4 to the +6 oxidation state. b Give the equation for the reaction in which sulfur is oxidised from the 0 to the +4 oxidation state. c Give the equation for the reaction in which a gaseous compound containing sulfur in the +6 oxidation state reacts with a liquid. If you were designing a new sulfuric acid plant, briefly list important factors that you would consider when deciding: a whether to use powdered sulfur from a local supplier or molten sulfur transported a greater distance b if sulfur were to be burnt in air or oxygen c whether to obtain sulfur dioxide by burning sulfur or from the waste gases of a copper smelter d whether to use air or oxygen for the reaction in the converter. Air is thoroughly purified before it is used in the contact process. Explain the reason for this. Sulfur dioxide is present in emissions from sulfuric acid plants. a Why is the release of large amounts of sulfur dioxide in the atmosphere undesirable?

b What is done to limit the extent of these emissions in sulfuric acid plants? c What other industries emit sulfur dioxide?

Connecting the main ideas 14 Under the title ‘Sulfuric acid: the world’s most wanted chemical’, write a paragraph that uses the words: fertilisers, contact process, smelting, converter, absorption tower, equilibrium, reaction rate and catalytic oxidation. 15 Describe two ways that expenses are minimised in the contact process. 16 Imagine a major accidental emission of sulfur dioxide gas occurred at a sulfuric acid plant. As a consequence, the local council invited the plant manager to speak to the local citizens about their health, safety and environmental concerns.

a Design a flyer advertising the meeting and highlighting the main issues of concern. b List the points that the manager might wish to make to the concerned residents. 17 Construct a concept map that includes the following terms: sulfuric acid, fertilisers, oxidation, contact process, converter, absorption tower, oxidation, reaction rate, equilibrium yield and catalyst. 18 Design a safety notice that summarises: a the effects to a person’s health that may result from contact with sulfuric acid b the first aid treatment that should be used to treat contact with sulfuric acid c the personal protective equipment that someone working with sulfuric should use or have access to.


2

chapter 22

Production of ethene

key knowledge • the industrial production of ethene − factors affecting the production including rate and equilibrium position, catalysts, temperature, pressure − waste management including generation, treatment and reduction − health and safety − uses of ethene



• recall that petroleum deposits consist of a mixture of hydrocarbon compounds • describe the main steps in the production of ethene from crude oil and natural gas, including fractional distillation and steam cracking • identify the steps in the production of ethene where equilibrium plays a role in determining reaction conditions • describe issues associated with management of wastes during ethene production, including their generation and treatment and reducing the quantity of waste • describe health and safety issues associated with the production of ethene • describe how the process complies with the principles of green chemistry • recall and explain the physical and chemical properties of ethene • describe the main uses of ethene.

22.1

Uses of ethene

Polyethene (50%)

Ethene is the raw material used for the production of an enormous range of useful materials, including plastics, dyes and pharmaceuticals. The presence of a double covalent bond in the ethene molecule has a significant effect on its chemical properties. Ethene reacts more readily, and with more chemicals, than ethane (CH3CH3), which contains only single bonds. The reactions of ethene usually involve an addition reaction with a small molecule to produce a single product (Chapter 10). Figure 22.1 shows the main synthetic chemicals made from ethene.

Polystyrene and copolymers Antifreeze (10%) (10%) Fibres (5%)

Polymers The products shown in Figure 22.2 were introduced to Australia in the 1980s. You will recognise them as familiar household items.

Miscellaneous (includes ethanol) (10%)

PVC (15%)

Figure 22.1 Uses of ethene.

review The chemistry of ethene is discussed in Chapter 10, pp. 146–8.

140

90

Figure 22.2 Products made from ethene introduced in the 1980s.

Microwave polyethene film has flexibility, stretch and very low permeability to water vapour, making it a versatile cover for foods during microwave cooking. Plastic milk bottles are easily manufactured and stacked for transport, and are light and unbreakable. Large wheeled rubbish bins provide an efficient and aesthetic means of collecting household refuse. These products are all made from polyethene, an addition polymer produced from ethene gas (p. 147). Figure 22.3 shows the dramatic increase in the amount of ethene synthesised between 1950 and 2005. Over 400 000 tonne of ethene is produced in Australia each year. That represents 20 kg per head of population. Approximately 50% of the ethene produced is used to make various forms of polyethene. Polyethene was first made in 1933 at the ICI laboratories in Cheshire, UK. As happened with many other processes and materials, World War II provided a stimulus for the research and development of new applications for polyethene—it was used as insulation for radar cables. Since the war the market for polyethene has seen spectacular growth. Other important addition polymers made from ethene include PVC and polystyrene (p. 148).

World ethene production (millions of tonnes)

57

33 21

0.7

3

1950 1960 1970 1980 1990 2000 2010 (est.)

Figure 22.3 World production of ethene is currently growing at about 5% per year.

349 Production of ethene

Ethanol For industrial purposes ethanol is manufactured from ethene for use as a reagent and as a solvent in cosmetics, pharmaceuticals and inks (p. 147).

extension Synthetic oil: improving on nature

chemfact The name ‘polyalphaolefin’ derives from ‘poly’ (many identical units joined together), ‘alpha’ (carbon atoms joined without sidebranches in the monomer chain) and ‘olefin’ (the industrial name for an alkene).

Figure 22.4 Synthetic lubricants such as Mobil 1 are ideal for cars that operate under challenging conditions.

CH2

CH2

Ethene

T di i l lubricants Traditional l bi for f cars, trucks k andd other h types off machinery hi are obtained b i d from f fractions derived from crude oil. However, the properties of these lubricants do not always match up to the demands of equipment operating under extreme conditions, such as at very high temperatures or during continuous operation. Modern synthetic lubricating oils provide better performance in these circumstances, making an important contribution to energy conservation (Figure 22.4). Lubricants derived directly from oil fractions contain traces of organic compounds of oxygen, sulfur and nitrogen, as well as small amounts of inorganic substances. Under extreme operating conditions these substances can catalyse unwanted reactions in equipment, leading to corrosion and wear. Synthetic lubricants are derived from ethene and propene, which currently are obtained from oil fractions. The synthetic oil that we are familiar with for use in car engines is actually a polyalphaolefin. It is manufactured by polymerising ethene into decene, an alkene with 10 carbon atoms. This acts as a basic building block to make a range of polyalphaolefins (Figure 22.5). The decene in turn is polymerised in the presence of a catalyst to form dimers (two decene molecules joined together), trimers (three molecules), tetramers (four molecules) and some heavier fractions including pentamers (five molecules). Most synthetic oils are based on the trimer and tetramer molecules, although dimers and pentamers are used to obtain the very low or very high viscosities required for some specialised applications. The oil is then reacted with hydrogen to remove the remaining carbon–carbon double bonds. This improves its stability to heat and oxidation. Of course, all of this processing contributes to the cost of the product. Synthetic oils are more expensive than traditional mineral oils, but the improvements they bring to the automobile or machinery by way of performance and protection are very worthwhile. As petroleum reserves become depleted, scientists believe that it will be possible and economically viable to make synthetic lubricants starting from coal, oil-shale or even from oils derived from plants. Polymerisation catalyst

_ C8H17 CH

CH2

1-Decene

Polymerisation, H2 catalysts

C8H17 _ CH

CH2 _ CH

H X

CH3

C8H17

‘Polyalphaolefin’

Figure 22.5 Production of synthetic oil from ethene.

questions

?

E1 Use Use your kknowledge nowlleddge of chemical bonding to explain why synthetic oils made from trimer molecules are less viscous than oils made from pentamer molecules. E2 Suggest a suitable catalyst for use in the reaction of the unsaturated oil with hydrogen.


22.2

Production of ethene Petroleum deposits are found in the form of natural gas and crude oil. Natural gas and crude oil are mixtures of alkanes. Physical separation and chemical processing are used to produce ethene from crude oil. This involves two main steps: fractional distillation, and steam cracking. Figure 22.7 shows the main stages involved in the production of ethene from petroleum. Fractional distillation Quench tower Distillation and waste treatment

Furnace

Figure 22.6 Oil production in Bass Strait.

Crude oil or natural gas Ethene Steam Cold water

Hot water

Figure 22.7 Steps in the production of ethene from petroleum. Heat energy recovered from the cracked gases in the quench tower can be used in distillation and to produce high pressure steam.

review The distillation of crude oil into petroleum products is discussed in Chapter 23.

Stage 1: Fractional distillation A process called fractional distillation, discussed in Chapters 10 and 23, is used to separate the components of crude oil. The process is performed in a fractionating tower (Figure 23.13 p. 375) where the oil is separated into several fractions, each containing a number of hydrocarbons with similar boiling points.

Stage 2: Steam cracking When alkanes are subjected to sufficiently high temperatures, carbon–carbon bonds can be broken, resulting in the production of smaller molecules. The molecules formed may be small alkanes as well as unsaturated molecules such as ethene, propene and butadiene. Hydrogen gas may also be produced. This decomposition by heating is called cracking. The process can be brought about in two main ways: 1 catalytic cracking, to increase the amount of lighter fractions recovered from crude oil (pp. 375–6) 2 thermal cracking, to produce unsaturated hydrocarbons, such as ethene and propene for use in the petrochemical industry. Unsaturated hydrocarbons such as ethene are primarily produced by a type of thermal cracking called steam cracking. The feedstocks for steam cracking are either ethane or propane from natural gas or naphtha or gas-oil fractions obtained from fractional distillation of crude oil.


Figure 22.8 A view of the cracking furnace (left) and a distillation tower (centre) of the Qenos Company at Altona, Victoria. Emissions of carbon dioxide, hydrocarbons and nitrogen dioxide are carefully monitored and minimised.

Steam and the feedstock are passed through coiled metal tubes in a furnace at temperatures between 750°C and 900°C, as shown in Figure 22.7. Conditions are adjusted to produce the desired product from the feedstock being used. Typical reactions occurring during the cracking process are: C2H6(g) C3H8(g)

chemfact The process that occurs in the furnace is complex, involving many different reactions, and equilibrium is not achieved. Computer simulations designed to represent this process have included hundreds of individual reactions in their models. Most of the reactions involve free radicals (atoms or groups of atoms with an unpaired electron). For example, when ethane is heated it can split into two methyl radicals: CH3CH3(g) –—➤ 2CH3•

chemfact In 2008, US Department of Energy scientists announced the development of a membrane that continuously removes hydrogen from a stream of ethene gas at high temperatures. In the future, ethene might be manufactured using this membrane instead of from steam cracking, avoiding the production of substantial amounts of unwanted products.

C2H4(g) + H2(g); ΔH = +138 kJ mol–1 C2H4(g) + CH4(g); ΔH = +81 kJ mol–1

Since these reactions are endothermic, a high equilibrium yield of ethene is favoured by the use of high temperatures, and hydrocarbon gas pressures of less than one atmosphere are used to drive the reaction forward. (Le Chatelier’s principle predicts that low pressure favours the reaction that produces the greatest number of molecules.) The use of steam allows a high overall gas pressure to be employed while keeping the concentrations of the hydrocarbons low. The complex series of reactions that take place means that applying the principles of equilibrium to this reaction system has its limitations, as the reactions do not end with the production of ethene. In practice the gas is in the furnace for less than one second to prevent continued cracking of the ethene, which would form unwanted products such as carbon (coke). Some modern cracking furnaces employ such high gas velocities that the hydrocarbons are in the furnace for only a few milliseconds, in order to improve the yield of the desired products. The gas leaving a furnace using ethane feedstock typically contains about 50% ethene, 35% uncracked ethane, and a mixture of other gases. Ethane is a particularly good feedstock for steam cracking if ethene is the main product required. As the size of the hydrocarbons in the feedstock increases, the products from the cracking process become increasingly diverse and the yield of ethene decreases, with corresponding increases in the yields of larger alkene molecules such as propene, butene and butadiene. Apart from allowing the concentrations of hydrocarbons to be relatively low, the presence of the steam in the furnace helps to prevent the ethene produced from undergoing further cracking to produce coke, which would prevent the hydrocarbon gases from reaching the correct reaction temperature and lower the yield. Instead, any coke produced is converted into carbon monoxide and hydrogen gas: C(s) + H2O(g) –—➤ CO(g) + H2(g)


Stage 3: Quenching The gas leaving the furnace is quickly cooled to about 30°C using water in a quench tower. This is to prevent the cracking reaction from continuing to produce unwanted coke.

Stage 4: Distillation and waste treatment

chemfact

The gas is then cooled to below –100°C by progressively compressing and expanding it, and ethene is separated from the other components in the gas by distillation. The compounds other than ethene that are generated from the steam cracking process are extracted or treated at this stage. Unreacted ethane is recycled back into the feedstock.

When a gas is compressed and then allowed to expand rapidly, the temperature of the gas drops. This is known as the Joule–Thompson effect and is the basis for the operation of a refrigerator.

extension Crack a candle T cracking Try ki hhydrocarbons d b at home. h You Y can crackk the h hhydrocarbon d b molecules in paraffin wax, one of the materials obtained from the residue from fractional distillation of crude oil. First light a candle. Allow the flame to burn for a few moments until there is a pool of molten paraffin wax under the flame. Now light another match and blow out the candle. Quickly bring the match to a point about three centimetres above the wick.

questions

?

E3 What Whatt is is the the purpose of a candle wick?

summary

Did you see the h flame jump j back b k to the h wick, i k relighting li h i the h candle? It was not just paraffin vapour that ignited—small gaseous hydrocarbons formed by cracking of the paraffin were also present in the vapour.

E4 When a candle burns, burns what is reacting with oxygen?

22.1–22.2

Th The main i chemical h i l made d from f ethene th is i polyethene. l th It iis made d from the monomer by an addition polymerisation reaction. Other important chemicals derived from ethene are PVC, polystyrene and ethanol. Ethene is produced from either natural gas or the gas-oil fraction obtained from the fractional distillation of crude oil.

Ethene Eth production d ti involves i l a process called ll d steam t cracking, ki in i which lighter unsaturated hydrocarbons are made from heavier hydrocarbons. The cracking reactions are endothermic and higher yields are favoured by the use of high temperatures and low pressures. The gas from the cracking furnace is cooled quickly to prevent undesirable secondary reactions and reduced yield.

key questions 1 The fractions from a petroleum fractionating tower may be subjected to either steam or catalytic cracking. a Explain the different purposes of the two cracking processes. b What are the differences in the two cracking methods? 2 Prepare a simple flow chart showing the main steps involved in producing polyethene from crude oil.

3 Write a balanced equation to represent the steam cracking of C14H30, from the gas-oil fraction of a fractionating tower, to form ethene and hydrogen as the only products. 4 List the conditions used by industry for steam cracking: C2H6(g)

C2H4(g) + H2(g); ΔH = +138 kJ mol–1

Explain how each condition helps to maximise the yield.


22.3 chemfact A substance that restricts a catalyst’s normal action is described as a catalytic poison.

Waste management The sulfur compounds present in crude oil and natural gas are removed in a process known as desulfurisation. This is performed during the initial processing of the petroleum, before the cracking process commences. It is important for several reasons: to minimise emissions of the pollutant sulfur dioxide (SO2) when fuels are burned (p. 252) to prevent poisoning of the catalysts used in refinery processes as feedstock for the manufacture of sulfuric acid and other useful chemicals. Australian natural gas contains only small amounts of sulfur, in the form of hydrogen sulfide, but its removal is important for environmental reasons. Natural gas from some overseas sources contains significant amounts of hydrogen sulfide. Recovery of sulfur in these cases is economically worthwhile. Apart from ethene, chemicals in the gas leaving the furnace in the steam cracking process include unreacted hydrocarbon feedstock, ethyne, methane, propene, butadiene, hydrogen, carbon dioxide and hydrogen sulfide. These chemicals may be treated in the following ways. Unreacted feedstock in the product mixture is recycled back to the cracking furnace. Ethyne (C2H2) is converted to additional ethene by reacting it with hydrogen using a transition metal catalyst, such as Pd or an Fe/Ni mixture. metal catalyst

C2H2(g) + H2(g) ––––––––—➤ C2H4(g)

Carbon dioxide and hydrogen sulfide are both weakly acidic gases and are removed by treatment with dilute sodium hydroxide solution: 2NaOH(aq) + H2S(g) –—➤ Na2S(aq) + 2H2O(l) 2NaOH(aq) + CO2(g) –—➤ Na2CO3(aq) + H2O(l)

review Photochemical smog is discussed in Heinemann Chemistry 1, Chapter 13.


Unreacted sodium hydroxide solution is then neutralised by treatment with acid. One Australian company uses the carbon dioxide gas present in flue gases from its furnaces for this purpose. Apart from ethene, several other useful compounds are produced as by-products of this process. The most important is propene, which is extracted and used for making polypropene. Butadiene is also obtained from the process, and is used to make synthetic rubber. Hydrogen and methane are extracted and used as a fuel gas for furnaces. Coke that builds up on the walls of the cracking furnace must be removed periodically because it degrades performance. The coke can be used as a fuel. Modern cracking furnaces can run for several months between de-cokings. Heat released during the quenching process is used to generate more high-pressure steam for the steam cracking process. The water used in the quench tower is recycled. The volatile organic compounds that are involved in ethene production can react with nitrogen oxides in the atmosphere in the presence of sunlight to form photochemical smog. As a consequence, the level of these emissions is controlled by government legislation and, in most cases, even stricter voluntary limits set by the companies involved. Australian manufacturers have achieved substantial reductions in emissions in recent years.

Government legislation and voluntary limits also apply to solid and liquid wastes. As a result, the wastes are carefully analysed and monitored. When they are not recycled, they may be taken off site for incineration or chemical treatment, often ultimately ending up in landfill sites or the sewer system.

chemistryy in action Greening the production of ethene If ethanol were able to be produced in much greater quantities by fermentation, commercial synthesis of ethene by the dehydration of ethanol may become feasible:

The scope for dramatic improvements in the production of ethene by steam cracking is limited. Over the past 40 years there have been substantial improvements in the efficiency of the process and in the reduction and management of emissions. Aside from further developments of this type, the demand for ethene produced from fossil fuels may decrease in the future as a result of a number of other developments.

In recent studies, nearly 100% yields of ethene have been achieved in pilot studies using a continuous reactor.

Ethene from biological sources

Ethene from recycled materials

A large proportion of the ethanol manufactured for industrial use is produced by reacting ethene with steam. This reaction has a theoretical atom economy of 100%, but some toxic by-products are produced and the alcohol is not used for human consumption. Ethanol can also be produced by the fermentation of sugars and chemists are investigating the potential of this reaction as an alternative to the petrochemical-based process. While crops such as maize and sugarcane can provide significant quantities of sugars, the amount from these sources may not be sufficient to meet the demand for ethanol. An alternative raw material for ethanol production is cellulose, which is found widely in biomass such as woody plants. Recent advances in biotechnology have resulted in the development of genetically engineered E. coli bacteria that convert materials such as forestry waste, maize stalks and sugarcane residue into ethanol. Aside from producing a valuable chemical, the process reduces the need to dispose of these chemicals in landfill or by burning.

The demand for ethene may be reduced in the future as the extent of recycling of polymers increases. Recycling saves raw materials and reduces the cost and energy needed to make articles from polymers, as well as minimising the pollution that can occur if a plastic article is discarded or incinerated. On the other hand, every time a polymer, such as polyethene, is recycled, the molecules are broken as it is melted and remoulded, weakening it and restricting the market for this material. Some polyethene is recycled into cracking furnaces to make new chemicals.

C2H5OH(g) –—➤ C2H4(g) + H2O(g)

Replacement materials Chemists are developing a range of polymers that can replace the more traditional polymers derived from ethene in many applications. For example, the polymer Biopol, produced from genetically modified plants, can be used for moulded goods, as a film and as a packaging agent. Importantly, Biopol is completely biodegradable. A number of bioplastics are now available commercially and new polymers of this type are still under development.

22.4

Health and safety Short-term exposure to ethene is generally considered not to be harmful. It can act as an asphyxiant and, at moderate to high concentrations, can cause dizziness, headaches and nausea. Butadiene gas emissions are carefully monitored because high exposures may cause damage to the nervous system, and other illnesses. The biggest risk to workers in the petrochemical industry is usually considered to be explosions and fire. Like other hydrocarbons, ethene can readily form explosive mixtures with air. As a consequence, extensive


Figure 22.9 Hazchem symbol for ethene.

Figure 22.10 Fire-fighting drills are performed regularly in petrochemical plants to ensure employees are well prepared in case of a major accident.

fire prevention and fire fighting strategies are employed by the industry (Figure 22.10). Since both high temperature and low temperature stages are involved in ethene production, special attention is also given to the prevention of burning and freezing injuries. To ensure the well-being of employees and the community, the companies producing ethene follow strict procedures for the storage, handling and transport of chemicals. These procedures include appropriate training, signage, protective clothing and first-aid services.

extension Case study: the Altona petrochemical complex

Figure 22.11 The Altona chemical complex is Australia’s largest centre for production of plastics, petrochemicals and synthetic rubbers.


extension Case study: the Altona petrochemical complex (continued) A Australia’s li ’ major j ethene h producers d are O Orica i at BBotany iin N New SSouthh Wales, and Qenos at Altona in Victoria (Figure 22.11). At Altona, ethene is derived from two sources.

1 Natural gas Bass Strait petroleum is transported to a processing plant at Longford near Sale, 217 km east of Melbourne, using undersea pipelines more than 600 km long. It is then transferred via a 285 km pipeline to the Esso–BHP plant at Long Island Point on Western Port Bay for fractional distillation. Ethane from this plant is delivered to Qenos using a pipeline that runs underneath Port Phillip Bay. At Qenos the ethane undergoes steam cracking to produce ethene.

2 Crude oil Crude oil, also from Bass Strait, is separated by fractional distillation at the Mobil Altona Refinery. The gas-oil fraction is then piped to the nearby Qenos Olefins plant. Steam cracking yields propene, butadiene, and small amounts of alkanes.

questions E5 a b

The off products Th gaseous mixture i d iis separatedd bby fractional f i l distillation. Sulfur compounds are removed and the unsaturated products are piped to nearby companies that manufacture polyethene, polypropene, polystyrene and polybutadiene rubber (Figure 18.20, p. 313). The alkanes are recycled for further processing. The hazards and costs involved in transporting ethene, propene and butadiene are major factors in the establishment of complexes such as Altona. Although the companies on the site are independent, they benefit from exchanging materials and services. Since Altona is close to Melbourne, there is a ready supply of labour and markets are close by. The site allows access to a nearby port and other transport facilities, and operates 24 hours a day throughout year. One potential disadvantage of a site close to populated areas is the risk of pollution from the plants. Industries must be highly conscious of the need to protect the health and environment of local residents. The industries have pollution monitoring equipment and operate to standards set by the Environment Protection Authority of Victoria.

?

Summarise Summarise i tthe he main reasons why the clustering of companies in the Altona complex is commercially advantageous advantageous. What compromises would be involved in a decision to organise a group of industries in such a manner?

summary

22.3–22.4

VVarious i ttreatments t t off the th furnace f gas are employed l d tto remove unwanted chemicals and isolate other useful compounds. Unreacted hydrocarbon feedstock is recycled back into the furnace.

Fire Fi andd explosions l i constitute tit t the th greatest t t risk i k to t workers k in i the th petrochemical industry. Companies follow strict governmental and voluntary guidelines designed to protect the environment and the health and safety of employees and the community.

key questions 5 The principles of green chemistry can be used to evaluate the environmental impact of a chemical process. Construct a table with two columns headed ‘Principles’ and ‘Practice’, as shown below. Principles 1. Prevent waste

In the first column, list the twelve principles of green chemistry (Table 18.3 p. 310). In the second column, indicate the ways a modern plant using the steam cracking process could be regarded as complying with these principles.

Practice

. . . . . 12. Minimise the potential for accidents


22

key terms addition polymerisation addition reaction alkane alkene

catalytic cracking cracking desulfurisation fractional distillation

Industrial production of ethene 6 Ethene can also be produced by the catalytic dehydration of ethanol. This process was done on an industrial scale to produce ethene in the nineteenth century. There is renewed interest in this reaction as an alternative green source of ethene. The process can be written as: C2H5OH(g) –—➤ C2H4(g) + H2O(g)

a Use this equation to calculate the percentage atom economy for ethene production by this process. b Suppose we represent the cracking reaction for the production of ethene from propane by: C3H8(g) –—➤ C2H4(g) + CH4(g)

Use this equation to calculate the percentage atom economy for this process. c Comment on the significance of the percentage atom economies of each method for producing ethene. d What are some of the issues that would be associated with the widespread adoption of the production of ethene from ethanol? 7 In order to obtain ethene from crude oil, the oil is treated by fractional distillation. a Why is the fractional distillation of oil necessary? b Name three fractions obtained from fractional distillation of crude oil and describe one use for each fraction. 8 Unsaturated hydrocarbons, such as ethene and propene, are produced by steam cracking of natural gas or petroleum fractions. a Explain what is meant by the term ‘cracking’. b Why is steam used in this process? c Briefly describe the conditions employed for steam cracking and explain why these are chosen. d Write equations for two different reactions that could occur during steam cracking of heptane (C7H16). 9 In 2005 a Dutch chemist reported he had developed a solid containing oxygen that reacts with hydrogen gas to produce water. The solid is referred to as an ‘oxygen sponge’. By referring to reactions that occur during the cracking of ethane gas, explain why it is thought that the oxygen sponge may allow higher yields to be obtained in ethene production.


fractionating tower free radicals monomer petroleum

quench tower steam cracking thermal cracking

10 Consider the flow chart for ethene production below. Ethane C2H6

Steam Cracker

Mixture

Fractional distillation

Cool

Ethene Other gases

a Explain why ethane, the feedstock used in the steam cracking process, is present among the products, and suggest what would be done with this substance. b Name one of the ‘other gases’. c What physical property of the gases does fractional distillation depend on? 11 Explain, using a diagram of the energy changes that occur during the course of a reaction (energy profile diagram), the difference between steam and catalytic cracking. 12 Under the title ‘Crude oil: An important natural resource’, write a paragraph that uses correctly the words: fractional distillation, fraction, cracking, saturated, unsaturated, alkane, ethene, polymer.

Connecting the main ideas 13 Use the information in the text about waste management to construct a table that lists the main by-products of steam cracking of ethane, and describe what becomes of each one. 14 Consider the reaction scheme shown below. Substance A

Br2 Thermal Substance B Cracking

Substances C,D C2H4

Ethanol

H2 and Ni

Substance E

a Write the chemical formulas and names of substances A, B and E.

b Name substances C and D needed to produce ethanol from ethene. c Describe the conditions used when converting substance B to ethene. 15 When the present supplies of crude oil and natural gas are depleted, an alternative source of ethene may be the production of ethanol by fermentation of plant material, followed by the dehydration of ethanol: C2H5OH(g)

C2H4(g) + H2O(g); ΔH = +45 kJ mol–1

You have been asked to design a pilot plant to test the feasibility of this process. Describe how you might maximise the yield of ethene. 16 A simplified flow chart involving ethene is shown below. Liquid A

, eam

Crude oil

Process X

Mixture Z

St

Process Y Ethene

st

aly

cat

High temperature, high pressure, catalyst

Solid B

Oxy

gen

Gas C and water

a b c d

Describe the composition of crude oil. Name processes X and Y. Describe the composition of mixture Z. Name another raw material from which ethene can be manufactured. e Identify substances A, B and C and write equations for the reactions that produce them. f What type of reaction is involved in the production of substances A and B? 17 The world has become very dependent on the products of the petrochemical industry, but the raw materials of crude oil and natural gas are likely to be virtually exhausted by 2100. Assuming the current production process of ethene stays unchanged, suggest the impact of the lack of raw materials on our lifestyle. 18 A group calling itself the ‘Concerned Residents Group’ has been formed in the suburb next to a major petrochemical plant with the intention of lobbying the state government to have the plant moved. a Design a flyer advertising the first public meeting of the group to highlight the main issues of concern. b The public relations manager of the petrochemical company has been invited to speak at the meeting. List the points that the manager might wish to make to the concerned residents. 19 Construct a concept map that includes the following terms: addition polymerisation, alkanes, alkenes, cracking, fractional distillation, fractionating tower, monomers, petroleum and polymers.


4

unit

area of study 1

review

Industrial chemistry

multiple-choice questions 1 Nickel metal reacts with carbon monoxide according to the equation: Ni(s) + 4CO(g) –—➤ Ni(CO)4(s)

Which one of the following would decrease the activation energy of the reaction? A addition of a catalyst B increasing the temperature C increasing the pressure of CO D crushing the nickel into a fine powder Questions 2 and 3 refer to the following information. An energy profile diagram shown below is for the reaction: CO2(g) + NO(g) –—➤ CO(g) + NO2(g)

390 kJ mol–1 260 kJ mol–1

From a comparison of the enthalpy change, ΔH, it can be deduced that: A the activation energy of equation I > activation energy equation II B the activation energy of equation I < activation energy equation II C the activation energy of equation I = activation energy equation III D no information about activation energy can be deduced from ΔH. 5 A reaction is found to have an activation energy of 960 kJ mol–1 and an increase in enthalpy of 120 kJ mol–1. The reaction: A requires a source of energy to start B will occur spontaneously at room temperature C requires a catalyst to occur D can be used as a source of energy. 6 Which of the following statements about a catalyst is true? A Only a small amount of a catalyst is consumed in a reaction. B A catalyst can occur as a reactant or product in the overall equation. C A catalyst decreases the ΔH of the reaction. D The proportion of molecules with sufficient energy to react is increased by a catalyst. 7 Water reacts with chlorine according to the equation: 2H2O(g) + 2Cl2(g)

2 The ΔH of the forward reaction, in kJ mol–1, is: A –170 B +130 C +230 D +260 3 The activation energy of the reverse reaction, in kJ mol–1, is: A 40 B 130 C 230 D 360 4 Consider the following reactions: I A(g) + B(g) –—➤ 2C(g); ΔH = +180 kJ mol–1 II D(g) + 3E2(g) –—➤ 2F(g); ΔH = –90 kJ mol–1 III 2G(g) –—➤ H(g) + I(g); ΔH = –180 kJ mol–1

4HCl(g) + O2(g)

At a particular temperature the value of the equilibrium constant for this reaction is 4.0 × 10–4. At the same temperature, the value of the equilibrium constant for the reaction: 1 2HCl(g) + O2(g) H2O(g) + Cl2(g) 2 is equal to: A 2 × 10–4 B 2 × 10–2 C 2.5 × 103 D 50 8 Consider the following equation: A + 3B

2C + 4D

The correct expression for the equilibrium constant is: [A][3B]3 A [2C]2[4D]4

360 Industrial chemistry In

A decreased in amount and decreased in concentration B increased in amount and decreased in concentration C decreased in amount and increased in concentration D increased in amount and increased in concentration. 12 The reaction between nitrogen monoxide and ozone is:

[C]2[D]4 B [A][B]3 C

[A][B]3 [C]2[D]4

D

[2C][4D] [A][3B]

NO(g) + O3(g)

9 The concentration–time diagram shown could represent the concentrations of A and B in the equation:

Concentration (mol L–1)

1.0 0.8 0.6 B

0.4

A

0.2 0.0

Time

A A –—➤ B B B –—➤ A C 2A –—➤ B D A –—➤ 2B 10 In a sealed vessel, nitrogen oxide, oxygen and nitrogen dioxide form the equilibrium: 2NO(g) + O2(g)

2NO2(g); ΔH = –114 kJ mol–1

Which one of the following sets of conditions is likely to lead to the highest yield of nitrogen dioxide gas? A 200°C and 1 atm pressure B 200°C and 2 atm pressure C 300°C and 1 atm pressure D 300°C and 2 atm pressure 11 A sample of NOCl is allowed to come to equilibrium according to the following equation: 2NOCl(g)

2NO(g) + Cl2(g)

The volume of the mixture is halved and allowed to come to a new equilibrium, the temperature remaining constant. At the new equilibrium the chlorine has:

NO2(g) + O2(g); K = 6 × 1034 at 25°C

Equal amounts of NO and O3 are used. Which of the following statements cannot be inferred from the magnitude of the equilibrium constant? A [NO2][O2] >> [NO][O3] B The equilibrium of the reaction lies well to the right. C The reaction has a low activation energy. D Very little ozone will remain at equilibrium. 13 Carbon dioxide gas dissolves to a small extent in water, forming carbonic acid in an exothermic reaction: CO2(g) + H2O(l) H2CO3(aq); ΔH negative H2CO3(aq) + H2O(l) HCO3–(aq) + H3O+(aq)

This is the reaction involved in forming fizzy drinks. Which of the following strategies would not be effective at increasing the amount of dissolved carbon dioxide? A Decreasing the pH of the solution. B Decreasing the temperature of the solution. C Increasing the concentration of carbon dioxide in the gas. D Increasing the pressure of the carbon dioxide gas. 14 The pH of a tomato juice drink is 5.3 at 25°C. The hydroxide ion concentration in the drink, in mol L–1, is: A 105.3 B 10–5.3 C 108.7 D 10–8.7 15 The self-ionisation of water is affected by temperature: H2O(l) + H2O(l)

Temperature (°C) 5 15 25 35

H3O+(aq) + OH–(aq); Kw = [H3O+][OH–]

Ionisation constant for water, Kw 1.85 × 10–15 4.51 × 10–15 1.00 × 10–14 2.09 × 10–14


From the data it can be inferred that: A the pH of pure water at 35°C is less than 7 B self-ionisation of water is an exothermic reaction C equilibrium for the reaction lies well to the left with mainly reactants present D [OH–] of pure water at 15°C is less than [H3O+]. 16 The dihydrogen phosphate ion, H2PO4–(aq), is an amphiprotic ion formed after the first dissociation step of phosphoric acid in water. The equilibrium constants of the dihydrogen phosphate ion acting as an acid and as a base at 25°C are:

Hydrochloric acid

short-answer questions 17 A student investigated the factors affecting the rate of reaction between a solution of sodium thiosulfate and hydrochloric acid: Na2S2O3(aq) + 2HCl(aq) –—➤ 2NaCl(aq) + SO2(g) + S(s) + H2O(l)

The reaction was carried out in a conical flask placed on top of a piece of white paper with a dark cross marked on it. The rate of reaction was determined by measuring the time taken for the cross to be masked by the suspension of sulfur formed during the reaction, as shown.

Cross marked on sheet of paper

Sodium thiosulfate solution (colourless)

H2PO4–(aq) + H2O(l) –—➤ HPO42–(aq) + H3O+(aq); K = 6.3 × 10–8 M – H2PO4 (aq) + H2O(l) –—➤ H3PO4(aq) + OH–(aq); K = 1.6 × 10–7 M

From these data, dihydrogen phosphate will: A form an acidic solution in water B accept a proton and donate a proton to an equal extent in water C more readily accept a proton than donate a proton in water D readily lose two protons in water.

Cross not visible

Cross visible

At the start

Cloudy

After t seconds

The rate was determined for different concentrations of sodium thiosulfate and for different temperatures. The volume of each solution and the concentration of hydrochloric acid was kept constant. The results are summarised in the table below. Experiment no.

[Na2S2O3(aq)] (M)

Temperature (°C)

1 2 3

0.1 0.2 0.1

20 20 25

Time taken for the cross to be masked (s) 36 20 28

a Explain in terms of collision theory why the rate in Experiment 2 is faster than the rate in Experiment 1. b Explain in terms of collision theory why the rate in Experiment 3 is faster than the rate in Experiment 1. c What other factors, other than the two investigated in this experiment, may affect the rate of a reaction? 18 The activation energy for the decomposition of hydrogen peroxide to oxygen was measured under two different conditions. 2H2O2(aq) –—➤ 2H2O(l) + O2(g)

I The activation energy for the decomposition reaction when an enzyme was added was found to be 36.4 kJ mol–1. The temperature of the reaction mixture increased. II When platinum was added to another sample of the hydrogen peroxide solution the activation energy was 49.0 kJ mol–1. The temperature of the reaction mixture increased.


a What is the function of the enzyme and platinum in each these reactions? b Sketch, on the same energy profile diagram, the energy profiles for the decomposition of hydrogen peroxide with the enzyme and the decomposition using the platinum. c Which reaction system, I or II, would be faster? Explain your answer. 19 Dinitrogen tetroxide, N2O4, and nitrogen(IV) oxide, NO2, coexist according to the equilibrium reaction: N2O4 2NO2. A concentration–time graph for the system coming to equilibrium is shown.

Change Increase T Decrease T Add I2(g) Add CO2(g) Double volume Remove CO Add catalyst Add inert gas Ar

Quantity K amount I2O5(s) K amount I2(g) concentration CO2 amount CO2 I2(g) K

Effect

21 The reaction between hydrogen and iodine in the gaseous phase to produce hydrogen iodide is described by the equation: A

B

0.05

0.0

Time

a Identify A and B. b Write an expression for the equilibrium constant of the decomposition of N2O4. c On the graph, sketch the line showing the effect on A if a catalyst were added to the mixture. d Calculate K for the reaction at equilibrium according to this concentration graph. 20 Carbon monoxide and iodine pentoxide react to form iodine and carbon dioxide in the equilibrium reaction: 5CO(g) + I2O5(s)

I2(g) + 5CO2(g); ΔH = –1175 kJ

Use your knowledge of Le Chatelier’s principle to predict the effect of the change (column 1) on the designated quantity (column 2). Complete the effect on the equilibrium (column 3) by using the terms decrease, increase, no change. Assume that the change listed is the only one taking place, e.g. if I2 is added, the volume and the temperature are kept constant.

2HI(g); ΔH negative

H2(g) + I2(g)

The diagram shows the change in concentration of gaseous hydrogen, iodine and hydrogen iodide as the equilibrium is reached and disturbed.

Concentration

Concentration (mol L–1)

0.10

Hl I2 H2

A

Time

B

C

a At point A, a sudden change occurs to the system. What has occurred? b At point B, another change has occurred. What has happened? c Mark with an X on the time axis, a point when the system as at equilibrium. d Extend the graph past point C, until equilibrium is reestablished, to show what would happen to the concentration of the gases if the volume of the reaction vessel was doubled. Assume the temperature remains constant.


22 Hydrocyanic acid, HCN, is a weak acid. It ionises in water as follows: HCN(aq) + H2O(l)

CN–(aq) + H3O+(aq)

A 0.1 M HCN solution has a pH of 5.15 at 25°C. a Write an expression showing how pH is related to the concentration of H3O+(aq). b Write an expression for the acidity constant, Ka, of hydrocyanic acid. c What is the concentration of H3O+(aq) and CN–(aq) ions in a 0.1 M HCN solution? d What assumption can you make about the equilibrium concentration of HCN(aq)? e Determine the value of the acidity constant, Ka, of hydrocyanic acid. f Explain why a 0.1 M solution of HCl has a pH of 1 while a 0.1 M solution of HCN has a pH of 5.15. 23 A mixture of 0.040 M Na2HPO4 and 0.040 M KH2PO4 acts as a buffer, maintaining a pH of approximately 7. Use equations to show how the mixture can resist change of pH when a small amount of acid or base is added. 24 Both thymol blue and methyl orange are acid–base indicators that are red in the acidic form and yellow in the basic form. The acidity constants, Ka, of the two indicators are: thymol blue Ka = 0.020 M, methyl orange Ka = 0.000 40 M. Both indicators are weak monobasic acids. The protonated form of the indicator can be represented as HIn and the unprotonated form as In–: HIn + H2O(l) –—➤ H3O+(aq) + In–(aq)

a Determine the pH at which [In–] = [HIn] for each indicator. b A weak acid and strong base are titrated and found to have an equivalence point of pH ≈ 4.3. Which of the indicators would be more suitable to use in this titration? Justify your answer. c Calculate the value of the ratio [In–]/[HIn] when thymol blue is in a solution of pH = 5. What is the colour of the indicator in this solution? 25 A solution of barium hydroxide, Ba(OH)2, is found to have a pH of 12.60 at 25°C. Assuming that it fully ionises in water, what was the concentration of the barium hydroxide solution?


26 Methanol is used as a fuel for some racing cars. The reaction conditions used to manufacture methanol must be carefully adjusted in order to ensure efficient production. The synthesis of methanol from methane involves three major steps: I reaction of methane with steam to yield carbon monoxide and hydrogen II an exothermic reaction between carbon monoxide and hydrogen to produce methanol III separation of the methanol from the reaction mixture. a Draw a flowchart of this process, showing the chemicals involved in each stage. b Write equations for the two reactions that are described above. c Suggest how methanol might be separated from other gases in the reaction mixture. In practice the reaction of carbon monoxide with hydrogen is performed at about 250°C and 100 atm pressure. Copper, zinc oxide and alumina are also present and about 10% of the reactants are converted to methanol as the reactant gases pass through the reactor. d i What is the likely function of the copper, zinc oxide and alumina in the reactor? ii How do the copper, zinc oxide and alumina allow the process to operate more efficiently? e A much higher conversion of carbon monoxide to methanol than 10% is obtained in industry. Suggest how this is achieved. f For the reaction between carbon monoxide and hydrogen, what would be the effect on the yield of methanol if you were to: i increase the pressure? ii increase the temperature? g With reference to reaction rates and equilibrium, discuss why the particular temperatures and pressures described above would have been chosen. h Suggest two waste management strategies that might be involved in this method of methanol production. i Suggest two health and safety issues that might be involved in this method of methanol production.

27 Select one of the following chemicals of industrial importance: ammonia, ethene, sulfuric acid or nitric acid. a Name and write the formulas of the raw materials from which the chemical is made. b Write chemical equations for the main reactions that occur during its production. c Sketch and label a flow diagram of the process, indicating the reactants and products at each stage. d The synthesis of each chemical involves a step in which conditions must be carefully controlled in order for the production process to be efficient. i What pressures and temperatures would be used in this step if a high equilibrium yield were desired? ii What pressures and temperatures would be used if a high reaction rate were desired? iii State the pressures and temperatures actually used for this step and, where appropriate, explain why they vary from those given for parts d i and d ii. e Apart from careful selection of pressure and temperature, describe one other way that the efficiency of the production of the chemical is improved. f Name three wastes generated in the production process and where appropriate describe how these wastes are treated and minimised. g Name three health and safety issues associated with the process and describe the precautions taken. h List three commercial uses for the chemical. 28 Ethanol is a major product of the chemical industry, with many applications as a solvent and as a raw material for other products. With reference to a MSDS for ethanol (this can be downloaded from the Internet): a Explain why large storage tanks should not be located next to residential areas. b What is the LD50 for rats, and what does this term mean? c What steps should you take in making a risk assessment for using ethanol in a school laboratory experiment? d What personal protection should be used when working with ethanol? e Should waste ethanol be flushed down the sink? Explain.

29 ‘Money makes the world go round’, or so the saying goes. Explain the impact of finances on the production and use of chemicals in the following situations. a The effect of high retail prices on the acceptable yield of a product. b The effect of a carbon tax on the cost of producing electricity by burning coal. c The effect of toxicology testing for all chemical products. d The effect of rehabilitating polluted land charged against the emitter of the pollutant. e The effect of waste disposal charged against the manufacturer of a product. f The effect of costly start-up procedures on a decision to produce small quantities of a chemical by batch or continuous processing. 30 Victoria has struggled to find a suitable site for a proposed high temperature incinerator (HTI) to treat chemical wastes, with the state government and locals repeatedly failing to reach agreement on where it should be located. a List two considerations for each of the following factors involved in locating a HTI (or a chemical production plant such as an oil refinery): i technical factors ii social factors iii environmental factors iv economic factors b Explain how the following decision-making tools could assist in finding a suitable location for a high temperature incinerator (or a chemical production plant such as an oil refinery): i cost–benefit analysis ii environmental effects statement c Which of the above decision-making tools is most likely to give the fairest answer? Explain.


2

chapter 23

Fossil fuels key knowledge • comparison of energy sources: types, uses and sustainability, including brown coal, natural gas, nuclear fission and biochemical fuels

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe types of fossil fuels, their uses and sustainability • distinguish between renewable and non-renewable energy sources • use social, economic and environmental factors to compare common energy sources • assess the suitability of various energy sources for particular situations • explain how energy conservation and consumption affect the sustainability of energy sources.

23.1

Energy sources today Human history becomes more and more a race between education and catastrophe. H.G. Wells, The Outline of History (1920) vol. 2, ch. 41, part 4

How much energy do we use? Energy is measured in units called the joule, J. Larger amounts of energy are commonly expressed in kilojoules, kJ (1 kJ = 1000 J), or megajoules, MJ (1 MJ = 1 000 000 J). A petajoule, PJ, is a useful unit for discussing national energy needs: 1 PJ = 1 × 1015 J. A person in a hunter–gatherer society uses approximately 10 MJ of energy per day. Despite being less active than hunter–gatherers, each person in our modern industrial society uses, on average, nearly 1000 MJ per day. This figure is one hundred times greater than the body’s basic requirement. The bulk of our energy use is for transport, heating, and domestic purposes (Figure 23.1). The national energy budget is somewhat higher than your personal tally. Australia uses about 2000 PJ of energy per year for mining, manufacturing, construction, commercial services, farming, fishing, forestry, and residential use. World energy consumption is around 4 × 1020 joules per year. Some nations could be considered gluttons in terms of their energy consumption. The United States consumes a quarter of the world energy pie. Australia is just under the US on consumption per head of population (Figure 23.2). The world’s poorest nations are starved of food and energy.

chemfact In our society, about 10 J of energy is expended on production, processing and distribution to provide 1 J of food.

Transportation (47%)

Heating (40%) Cooking (2%) Lighting and motors (7%)

Other – including refrigeration, electrolysis (4%)

Figure 23.1 Use of energy in Australia.

Consumption per capita Tonnes oil equivalent (toe)

toe per capita 0–1.5 1.5–3.0 3.0–4.5 4.5–5.0 > 6.0

Figure 23.2 Energy consumption per capita (2004). Energy use in the world is far from equal. Source: BP Statistical Review of World Energy, June 2005.

367 Fossil fuels

New technologies Nuclear energy Natural gas

Energy

Oil

1860

Coal Hydraulic power, wood, wind

1880

1900

1920

1940

1960

1980

2000

2020

Figure 23.3 World energy consumption, 1860 to 2005, with projections to 2020.

Meeting our energy needs Burning wood (and in some countries, dung) was the dominant method of obtaining energy up to the middle of the nineteenth century. Wood supplies once seemed inexhaustible and, like fossil fuels today, satisfied most of the demands of the time. The clearing of forests and an increasing demand for fuel to operate machines led to the transition from wood as a source of energy to coal and oil. The fossil fuels coal, oil and natural gas provide nearly 90% of the world’s energy needs. As members of a society that is heavily dependent on fossil fuels as a source of energy, it is sometimes hard for us to imagine obtaining energy from elsewhere. Feasible alternatives to fossil fuels are not always easy to find. In Chapter 24 we briefly consider some existing alternatives and look at what the future may hold in terms of our energy supplies. The dominance of fossil fuels as an energy source occupies only a brief part of human history. There is growing pressure to reduce our future dependence on these fuels for several reasons, of which the main ones are: increasing concern over the effects of gases released into the atmosphere as a by-product of the combustion of fossil fuels. In particular, carbon dioxide contributes to atmospheric warming (the greenhouse effect) and sulfur dioxide is known to cause acid rain (p. 252) appreciation that fossil fuels are non-renewable: world reserves are limited and will eventually run out demand for petroleum as a raw material for polymers, paints and other petrochemical products. The pressure for change was first felt strongly during the ‘oil crisis’ of the early 1970s, when several Middle Eastern oil exporters restricted production for political reasons. The effect of their actions on the price of oil is shown in Figure 23.4. The rising cost of oil during the 1970s prompted greater interest in other energy sources. Nuclear energy became more attractive economically. Research into alternative sources of energy such as solar, wind, tidal power and biochemical energy was encouraged, and more efficient ways of using oil were developed. Much of the research and development resulting from the oil crisis is ongoing. The 1991 and 2003 Gulf Wars in the Middle East and

368 Supplying Su u and using energy

Russian oil exports begin

140 130 120

Discovery of Spindletop, Texas

Sumatra production begins

Pennsylvanian oil boom

Loss of Iranian supplies

East Texas field discovered

Fears of shortage in USA

Global Netback pricing economic introduced Asian financial crisis crisis Iranian revolution Iraq Invasion invaded of Iraq Yom Kuwait Kippur war

Post-war reconstruction

Growth of Venezuelan production

Suez crisis

110 100 90 80

?

$ 2006

70

?

60 ?

50

?

40 30 20

Figure 23.4 Changes in the price of oil.

10 -0 9

-9 9

00 20

-8 9

90

80

19

-7 9 19

-6 9

70 19

-5 9

60

50

19

-4 9 19

-3 9

40 19

-2 9

30

20

19

-1 9 19

10

-0 9 19

-9 9

00

90

19

-8 9 80

18

-7 9 18

70 18

18

51

-6 9

0

the subsequent political instablity in this region, where most of our planet’s oil reserves are found, has continued to force up the price of crude oil. Economic forces pushed crude oil prices above US$140 a barrel before the economic global recession in late 2008. The market price dropped below US$50 a barrel in 2009.

23.2

chemfact One litre of petrol provides about 34 200 kJ of energy when it is burnt in the engine of a car.

Energy converters Conversions of energy from one form to another involves ‘losses’, usually in the form of thermal energy.

33 J lost as heat through exhaust gases

Device

Energy transformation

Steam turbine Gas burner Galvanic cell (battery) Fuel cell Generator Solar cell Hot-water heater (electric) Electric motor Petrol engine Hydrogen engine Fuel cell powered vehicle Photosynthesis in leaf

Thermal to mechanical Chemical to thermal Chemical to electrical Chemical to electrical Mechanical to electrical Light to electrical Electrical to thermal Electrical to mechanical Chemical to mechanical Chemical to mechanical Chemical to electrical to mechanical Light to chemical

Typical efficiency (%) 45 85 60–90 50 90 10–14 Nearly 100 60–90 20–25 25–50 20–30 1

The engine in a car is an energy converter. It converts the chemical energy in petrol to thermal energy, by combustion. Petrol is mostly octane (C8H18). The combustion of octane is a redox reaction: 2C8H18(l) + 25O2(g) –—➤ 16CO2(g) + 18H2O(g)

100 J of chemical energy in petrol

TABLE 23.1 Efficiency of energy converters

30 J lost as heat through engine cooling water

5 J lost by pumping combustion air 7 J lost to engine friction 5 J accessories 20 J to propel the car

Figure 23.5 Energy transformations in a car. Only a small proportion of the chemical energy in petrol is converted into motion; most is converted into heat.

369 Fossil fuels

Only about 20–25% of the energy released by the combustion of petrol is transformed into mechanical energy and used to move the car. In other words, more than 4 J of chemical energy is needed to obtain 1 J of mechanical energy. Most of the waste is transferred through the exhaust and the cooling system to the outside air as waste heat (Figure 23.5).

23.3

Fossil fuels

100

Percentage

80

60

40

20

0 S. Korea Japan Canada 2001 TWh: 281 1033 588

Nuclear Oil

USA 3864

OECD Europe 3258

Gas Coal

UK 393

Aust 217

Hydro and other

Non-renewable resources are those that are used at a rate faster than they can be replaced. Fossil fuels and uranium are fuels of this type. There are finite reserves of these which can eventually be exhausted. There is obvious conflict between using non-renewable resources and sustainability. Sustainability is all about meeting the long-term ongoing needs of society, as well as immediate and short-term demands. This concept includes care of the environment as well as the economic and social wellbeing of future generations. In a United Nations report, sustainable development was defined as ‘meeting the needs of the present without compromising the ability of future generations to meet their own needs’ (Our Common Future (Brundtland Report), Oxford University Press, 1987). About 80% of Australia’s electricity is generated from burning coal. Gas-fired and hydroelectric plants generate about 10% each. The sources of electricity generated in Victoria are shown in Table 23.2. Despite enormous uranium reserves, Australia does not have nuclear power plants. The Victorian government rejected a proposal in the 1960s to build a large nuclear plant on French Island in Western Port Bay. Increasing world oil prices and pressure to reduce greenhouse gas emissions may lead to a rethink on nuclear options for Australia. Relative electricity generation by fuel source in several developed countries is shown in Figure 23.6.

TABLE 23.2 Victoria’s electricity sources (2006) Figure 23.6 Electricity production by fuel source. The size of the block represents a region’s energy use. Energy use in the world is far from equal. Source: ‘Nuclear Energy Prospects in Australia’, Nuclear Issues Briefing Paper 44, April 2005, Uranium Information Centre Ltd.

Source Brown coal Gas Hydro Wind

Megawatt 6395 1505 2325 238

Percentage 62 14 22 2

Cost per megawatt hour ($) 31–40 42–47 45–55 70–80

The Age, 5 August 2006

Coal, oil and natural gas formed from primitive plants and animals. When we burn these fuels we are burning the remains of plants and animals that were alive millions of years ago. The plant and animal material has undergone complex changes while buried under tonnes of mud and sand, but it still retains the chemical energy the plants originally accumulated by carrying out photosynthesis. Chemical energy in fossil fuels can be considered as trapped solar energy.


Coal

Decaying vegetation

Increasing quality as fuel

Peat 60% carbon Heat released 25 kJ g–1

Brown coal 70% carbon Heat released 30 kJ g–1

Black coal 90% carbon Heat released 35 kJ g–1

Increasing time, pressure, temperature

As wood and other plant material turn into coal, some gradual chemical changes occur. Wood is about 50% carbon. As it is converted into coal, the carbon content increases and the proportion of hydrogen and oxygen decreases. The wood progressively becomes peat, brown coal (also called lignite) and then black coal or anthracite (Figure 23.7). Coal is a mixture of large molecules made from carbon, hydrogen, nitrogen, sulfur and other elements. The relative molecular masses of these molecules can be as large as 3000. The amount of water in coal decreases as these changes occur. When coal is burned, energy is used to vaporise this water, reducing the amount of heat released. Black coal, which contains least water, is therefore a better fuel than brown coal or peat. Although black coal is usually buried further underground, its higher heat value often makes mining worthwhile. Large brown coal deposits are located in the Latrobe Valley in Victoria. The power stations located next to the open-cut mines burn brown coal to generate electricity.

Uses of coal Over three-quarters of Australia’s electricity is produced in coal-fired power stations. You may indirectly be burning coal when you switch on an electric heater, use a hair dryer or watch television. Rather than cart coal to every factory, office and household, the chemical energy in coal is converted to electrical energy at a power station. Electricity is easily transmitted and is a convenient and versatile form of energy, but it is no cleaner than the energy source used to generate it. The reaction that occurs when coal burns can be written as: C(s) + O2(g) –—➤ CO2(g)

This equation is a simplification of the actual reactions that occur, since coal is not pure carbon. Significant quantities of ash are formed, as well as water vapour and sulfur dioxide. Oil and natural gas, by comparison, are cleaner burning fuels, producing less smoke and fewer pollutants.

Figure 23.7 Steps in the formation of coal. Values of the carbon content and heat released upon combustion are for dried coal.

chemfact Peat is not an important fuel in Australia. It is still dug by hand from the surface of boggy ground in parts of Ireland and Scotland where the traditional right to dig peat is retained within families. A smoky peat fire was an essential energy source for some of our ancestors, and is still required for the manufacture of whisky.

chemfact The generation of electricity by power stations using brown coal generates 1.7 tonnes of CO2 per megawatt hour of electricity produced. Power stations fired by black coal produce 0.9 tonne of CO2 per MWh, while those that use natural gas produce 0.6 tonnes of CO2 per MWh.

Figure 23.8 Mining brown coal in the Latrobe Valley.

371 Fossil fuels

Research is being conducted into ways of reducing carbon dioxide emissions from coal-fired power stations. These include: • geosequestration or storage of carbon dioxide deep underground • densification processes where water is removed from brown coal by heating. The volatile hydrocarbons that are also driven off are burnt to produce electricity, producing about the same level of carbon dioxide emissions as natural gas. Coke that is also produced can be used for steel-making. • chemical capture. Carbon dioxide is absorbed into a solvent, e.g. amino alkanols.

H O H H H

NH2

C C

C H

H H

1-Aminoethan-2-ol (an aminoalkanol)

Figure 23.9 Aminoalkanols absorb carbon dioxide.

Coal-fired power stations A number of energy transformations are involved in a coal-fired power station (Figure 23.9). • Coal is burnt—chemical energy in coal is converted to thermal energy. • Heat from the burning coal is used to boil water—thermal energy from burning coal becomes thermal energy in steam. • Steam is passed through a turbine—thermal energy in the steam becomes mechanical energy as the turbine spins. • Electricity is produced from a generator that is driven by the turbine— mechanical energy is converted to electrical energy.

! A pilot plant has been constructed at a Latrobe Valley power station where carbon dioxide is dissolved in a solvent. The captured carbon dioxide reacts with calcium salts in the ash slurry from the boilers to produce calcium carbonate and reduces the pH of the slurry.

Flue gas (a) a

35% of the coal's chemical energy is converted to electricity

10% of the coal's chemical energy is lost as heat in the chimney gases

50% of the coal's chemical energy is lost as heat in steam

Steam

Chimney

Electricity Generator Turbine 5% of the coal's chemical energy is lost in miscellaneous ways in the power plant

Boiler

Condenser Pump Cooling tower Pump

Coal (b)

b

Chemical energy in coal

Thermal energy of burning coal

Figure 23.10 a How a coal-fired power station works and where energy is lost. b Energy conversions in the power station.


Thermal energy of steam

Mechanical energy of turbine

Electrical energy from generator

Figure 23.11 Water vapour rising from steam-cooling towers at Victoria’s Loy Yang power station shows that energy conversions are less than 100% efficient.

The most inefficient transfer of energy occurs when thermal energy in the steam is transformed into mechanical energy by the turbine. Overall efficiencies of about 30–40% are achieved. While this figure does not sound impressive, it is greater than the 25% efficiency of the car engine.

Coal reserves As you can see from Table 23.3, there are large reserves of coal remaining in the world. Estimates of reserves are continually changing as new deposits are located and mining of less accessible deposits becomes economically viable. Australia is fortunate to have large deposits of coal, and exports more than any other nation.

TABLE 23.3 Estimated global fossil fuel reserves and their lifetime at current production and use rates (2004) Fossil fuel Coal Oil Natural gas

Size of reserves (tonnes) 1 × 1012 1 × 1011 1 × 1011

Life of reserves (years) 250 40–100 45–75

Coal represents about one quarter of the fossil fuels consumed around the world and it is anticipated that this fraction will increase greatly as reserves of oil and natural gas diminish. Coal is likely to become more widely used than oil in the future. A number of methods for converting coal to liquid and gaseous fuels have been demonstrated. In fact, coal-to-oil conversion was employed in Germany during World War II when the Germans were unable to import oil. A process known as flash pyrolysis has been developed by CSIRO for converting coal to oil. Pulverised coal is heated to 600°C. A tar forms that is reacted with hydrogen to give a type of crude oil. These processes may well offer economic alternatives in the future. With growing global awareness of global warming, Victoria’s brown coal industry has faced the challenge of reducing emissions of the greenhouse gas CO2. Victoria has set up a research centre to develop cleaner burning technologies.

Figure 23.12 Australian coal-producing basins.

chemfact There have been proposals to store carbon dioxide produced by power stations and other industries deep underground in depleted oil and gas basins. This process of isolating the CO2 is often described as sequestering the CO2 or carbon sequestration. A trial sequestration was conducted in south-west Victoria in 2007.

373 Fossil fuels

Oil and gas basins Resources are shown as a percentage of total resources. estimated Australian resources as at 1 January 2003 Gas = 167,285 PJ Liquids = 32,601 PJ (Geoscience Australia 2004)

Bonaparte Basin 18.6%

DARWIN

16.9% Browse Basin

Coal basins

16.0%

Uranium mineral deposit

18.3%

Laura Basin

Operating uranium mine Oil basin: producing Oil basin: not producing

Galilee Basin

Carnarvon Basin 49.7%

Bowen Basin Amadeus Basin

55.7%

0.1%

Gas basin: producing

0.1%

Gas basin: not producing

Cooper/Eromanga Basin

Arekaringa Basin

Maryborough Basin Tarong Basin

Bowen/Surat Basin

Coal basin: producing Perth Basin Perth Basin

Callide Basin

Surat Basin Adavalle Basin

Leigh Creek

Moreton Basin BRISBANE

Gunnedah Basin

0.8% PERTH

Gloucester Basin

0.6% Collie

Sydney Basin ADELAIDE Oaklands Basin

SYDNEY WOLLONGONG PORT KEMBLA Bacchus Latrobe Marsh Valley Anglesea Gippsland Basin MELBOURNE 10.9% Bass Otway Basin Basin 5.1% 0.2% Fingal 1.89% 1.0% HOBART 0.4%

Figure 23.13 Australian oil, gas, coal and uranium fields.

Oil Crude oil (petroleum) is a mixture of hydrocarbon molecules that are members of the homologous series of alkanes. Crude oil itself is of no use as a fuel, but many of the compounds in it are. The relative amounts of alkanes in crude oil from different sources vary. Oil from Bass Strait, for instance, contains relatively few of the larger molecules needed to form lubricants and bitumen. As a consequence of this, Australia imports oil to blend with the oil it extracts locally.

Fractional distillation of crude oil Separation of the components of crude oil is accomplished by a process called fractional distillation, which is performed in a fractionating tower (Figure 23.14). The oil is separated into several fractions, each containing a number of hydrocarbons that have similar boiling temperatures. Oil entering the fractionating tower is first heated to about 400°C so that it becomes a hot mixture of vapour and liquid. On entering the tower the liquid and vapour immediately separate and the vapour rises upwards through the tower. The temperature in the tower decreases gradually with increasing height. Within the tower there are horizontal trays, each containing hundreds of bubble caps. These impede the upward movement of gases (Figure 23.14). As the vapour rises it forces the caps up and it bubbles through condensed liquid in the trays. Those substances in the vapour that have boiling temperatures almost equal to the temperature of the liquid in the trays condense and are collected. Consequently, fractions collected from trays higher in the tower will be those with lower boiling temperatures.


Refinery gas (C1–C4) 110°C

Bubble cap

Gasoline (C5–C6) 150°C 180°C

Crude oil

260°C

Naphtha (C6–C10) Kerosene (C10–C14) Gas oil (C14–C20)

340°C Heater

Fractionating column

Bitumen (larger than C20)

Figure 23.14 Fractional distillation of oil.

The boiling temperature of molecular compounds depends on the strength of intermolecular forces. Attractions between non-polar alkane molecules arise from weak dispersion forces, the strengths of which increase with molecular mass. As a result, each fraction consists of alkanes within a specific mass range. Lighter alkanes condense near the top of the tower, whereas heavier ones condense near the bottom. The composition, boiling range and use of each fraction is summarised in Figures 23.14 and 23.15. The naphtha fraction that boils (and condenses) between 110°C and 180°C, for example, consists of alkanes containing between 6 and 10 carbon atoms, that is, C6H14 to C10H22. The liquid remaining in the bottom of the tower contains hydrocarbons of high molecular mass. This residue is normally distilled again at reduced pressure to separate its components. The exact operating conditions and fractions separated in the fractional distillation process vary from one refinery to another, depending on both the source of the crude oil used and market demands. Fractionating towers operate continuously and may be designed to accommodate a number of different feedstocks, such as Bass Strait crude oil or crude oil from the Middle East, which is richer in the heavy fractions.

Catalytic cracking: producing lighter hydrocarbon fractions The process that gives rise to more small alkane molecules is catalytic cracking, using a zeolite catalyst made of aluminium, oxygen and silicon. A catalyst is a substance that changes the rate of a reaction. The presence of the catalyst allows cracking to be performed at relatively low temperatures (450–500°C) and provides good control of the range of products formed. The catalyst is a fine powder which is in a fluid-like suspension in the flow of the hydrocarbon gases. Higher quality fuels can be produced by this process. A typical reaction occurring during catalytic cracking is:

chemfact Zeolites are substances that occur naturally in volcanic rocks. They have a sponge-like structure, with minute channels running through them. By modifying the size of these channels in synthetic zeolites, chemists have developed useful catalysts (Chapter 15).

C29H60(g) –—➤ C8H18(g) + C8H16(g) + C13H26(g)

The products from fractional distillation of crude oil do not necessarily match the fuel needs of society. They are often deficient in the proportion of lighter fractions needed for transport fuels, and contain much greater proportions of the residue fractions required for production of paraffin wax, lubricating oils and bitumen.

375 Fossil fuels

Size of molecules increases

Supply in oil

Demand for product

Refinery gas: LPG, feedstock for chemical industry Gasoline: Petrol

Naphtha: Cracking to petrol; chemical industry Kerosene: Jet fuel

Crude oil

Gas oil: Diesel; cracking to petrol

Bitumen: Lubricatiing oil; bitumen; fuel oil

Figure 23.15 Supply and demand for different fractions of crude oil.

Cracking breaks up (cracks) larger alkane molecules into smaller alkanes as well as producing small alkene molecules, such as ethene, together with hydrogen. The alkenes are such useful substances that they are deliberately produced in large amounts. Chapter 22 discusses ethene production in detail. Molecules such as octane (C8H18) are a major component of petrol. The process also yields larger alkanes and alkenes that can be used as a feedstock for thermal cracking furnaces to produce more ethene. As well as being used for fuels and lubrication, some of the hydrocarbons are used as solvents, and significant proportions are converted into alkenes to be used as the starting point for the enormous range of polymers, plastics, dyes and pharmaceuticals that we rely on. When investigating the sources of fuels for the future, we must also think about how to obtain molecules that can be converted into all of these other useful substances.

Uses of crude oil The various products obtained from the fractional distillation of oil include fuels such as petrol, kerosene, diesel and liquefied petroleum gas (LPG). Other products obtained from crude oil are used as raw materials in the manufacture of a range of products from plastics to pharmaceuticals. Liquefied petroleum gas (LPG) has replaced petrol as the fuel in over 500 000 vehicles in Australia. Most of these vehicles are taxis or trucks, or are fleet-operated. LPG is predominantly a mixture of propane and butane. It is expected that it will be available for at least the first half of this century, as it is obtained in significant quantities from local oil fields. The major impediments to greater use of LPG are: the need for large gas tanks, which reduce space in the boot of a car the cost of engine conversion to allow its use actual and perceived risks of explosions from leaking gas limited availability in remote areas. Conversion to LPG is likely to remain attractive as long as the price is significantly less than that of petrol. In 2006 the federal government provided a financial incentive for owners to convert their cars to LPG in response to community concern about rising petrol prices.


Oil reserves Estimates vary, but world oil reserves are much smaller than those of coal and may last only for several decades (Table 23.3). More than half of these reserves are situated in the politically sensitive Middle East. Australia’s relatively small oil reserves are likely to be exhausted early this century (Figure 23.13). Importation of large amounts of oil would have a significant impact on Australia’s economy.

Figure 23.16 A geologist examines a sample of oil shale from a drill core. Vast amounts of this rock, which contains an oil-like material, are found in Australia and overseas. In future it may become economic to mine the rock and extract this oil.

Natural gas Natural gas is mainly methane (CH4) together with small amounts of other hydrocarbons such as ethane (C2H6) and propane (C3H8). Natural gas is a popular fuel for home heating and cooking. As world oil prices increase, natural gas may increasingly replace petrol in cars and, in future, diesel fuel in many buses and trucks. The 510 MW Newport power station in Melbourne uses natural gas to generate electricity. Its city location near the mouth of the Yarra River provides water for cooling and minimises transmission lines to distribute the electricity. Unlike coal-fired power stations, the energy from a gas or oil-fired power station can be increased or decreased quickly to meet changing demands for electricity over a daily period. A gas-fired power station operates in a different way from a coal or nuclear powered plant. In coal and nuclear power stations, heat is used to produce steam that drives turbines that turn generators. However, steam is not used for spinning turbines in a gas-fired power station. Instead, hot gases from combustion expand air in a combustion turbine, spinning propellorlike blades and an attached generator. Other differences are the source of heat energy, how the energy is released, and how wastes are managed. About one third of the energy released from combustion drives the electric generator. Gas is simply burnt in air to to release mainly CO2 and H2O and heat energy. The amount of CO2 released per MW of electricity by gas-fired plants is half that of coal-fired plants. Unlike coal in Victoria, the combustion of gas produces fewer pollutants such as smoke particles and sulfur dioxide, a precursor of acid rain. Smoke particles can be a significant pollution problem because particles smaller than 10 µm are small enough to enter our lungs, increasing the risk of heart and lung disease including cancer.

Figure 23.17 Newport Power Station, Melbourne.

377 Fossil fuels

Air

Exhaust

Common shaft for turbine/generator

Generator

Pressurises air Fuel

Spins turbine and generator

Figure 23.18 Expanding gases drive the blades of a gas-fired turbine.

chemistryy in action Environmental impact of Newport power station Local fishermen know that the warm water in Hobson’s Bay next to the 100–500 megawatt Newport power station increases their chances of coming home with a catch. The gas-fired power station draws water used for cooling and pumps it back into the shipping channel near the mouth of the Yarra River, causing an increased temperature of several degrees. The fish are usually black bream, mullet and tailor. Mulloway and ‘pinky’ snapper are occasionally found in ‘the hotties’, as some locals call the spot.

Victoria’s Environment Protection Authority (EPA) specifies limits for temperature, flow, acidity, iron, ammonia, chlorine, carbon, nitrate, turbidity and dissolved oxygen in the cooling water released from the Newport power plant. The plant operators expressed concern with dredging proposed in 2005 to deepen the channel to allow the passage of deep-water ships. Sediment containing high levels of sulfide could corrode tubes in the power plant, forcing the station off-line with consequent power blackouts in Melbourne.

Reserves of natural gas About 20% of the world’s energy is obtained from natural gas. It supplies a similar proportion of Australia’s energy needs. As with oil, known reserves of natural gas are severely limited (Table 23.3).

Fossil fuels may cost the Earth Apart from the limited reserves, there are other concerns about the use of fossil fuels. Chemicals extracted from fossil fuels are used as the raw materials for many of the products of the chemical industry. Many of the materials that contribute to our current lifestyle, such as plastics, fibres and pharmaceuticals, have originated from fossil fuels. Can we afford to use fossil fuels in transport? Can we afford to burn such valuable raw materials? Both wood and fossil fuels release carbon dioxide and water when they burn. Increasing atmospheric levels of carbon dioxide and some other gases that trap infrared radiation contribute to the phenomenon known as the greenhouse effect that is increasing world temperatures. The warming of our planet, associated rises in ocean levels and disrupted climates may be the long-term consequences of using fossil fuels.


summary

2.5 2.0 Tonnes

Governments and industry are exploring a number of possible energy sources for the future: production of biochemical fuels such as ethanol and biodiesel production of hydrogen from natural gas solar-generated electrolysis of water to produce hydrogen and oxygen solar cells as a direct source of electrical energy wind power geothermal power nuclear energy. Future scenarios predict that there will be no one single replacement fuel. Instead there will be a range of sources of energy, and the use of each will shift as appropriate technology is developed. Different countries have greater or less access to the resources needed, and being able to ensure secure access to the resources is of great economic and political importance. In Chapter 24 we examine some of the possibilities.

1.5 1.0 0.5 0.0

Mar Jun Sep Dec Mar 06 06 06 06 07 Your greenhouse gas emissions from this account were 1.19966 tonnes

Figure 23.19 Household electricity bills may shock the consumer with feedback on the amount greenhouse gas produced as a result of their electricity consumption.

23.1–23.3

FFossil ssil fuels fuel are derived deri d from the remains remai s off plants pl nts and animals ani ls that died millions of years ago. There are large reserves of coal but the reserves of oil and natural gas are limited. Coal is mainly used in the generation of electricity. Crude oil is a mixture of alkanes. The components in crude oil are separated by fractional distillation into fractions with similar boiling points.

Heavier H avi r fractions f actions can be b cracked cra k d to produce pr duce more of the lighter li hter fractions that are used as petroleum. Products derived from crude oil include petroleum, LPG, ethene, and other chemicals used in the manufacture of consumer products such as plastics and pharmaceuticals. Natural gas is used for heating and in electricity generation.

key questions 1 a The water content of freshly mined brown coal is about 60–70%. What implications does this high water content have for the energy released from burning brown coal? b What pre-treatment could raise the energy content per gram of brown coal consumed in a power station? c If some electricity produced from burning brown coal was used to reduce its water content, then how could this energy cost be minimised? d What impact do impurities such as sulfur have on generating electricity from brown coal? 2 In Australia, which resource is likely to last longer before it is depleted, coal, oil or natural gas? Explain your choice. 3 Wood from forests is a renewable resource that supplied global energy needs for thousands of years. a Why is wood no longer sustainable as the major energy source for today’s society? b Is it possible to have a non-renewable and sustainable energy source? Explain.

c What can be done to increase the sustainability of our planet’s energy resources? In your opinion, who should make decisions regarding practices that affect sustainability? Discuss your answer with your classmates. 4 Describe two major problems associated with the use of fossil fuels. 5 a Why is it necessary to treat crude oil by fractional distillation? b What is the function of the bubble caps in the fractionating tower? c Name three fractions obtained from fractional distillation of crude oil and describe one use for each fraction. 6 a Why is catalytic cracking performed on some of the fractions obtained from crude oil? b Give an equation that represents the sort of reaction that might take place in a catalytic cracker. c Name any homologous series that can be produced by this process.

379 Fossil fuels

23

key terms black coal brown coal bubble cap catalytic cracking energy converter

feedstocks fossil fuels fractional distillation fractionating tower fractions

greenhouse effect joule natural gas non-renewable oil

peat sustainability waste heat

Oil

Coal 7 a A simplified chemical equation for the combustion of coal can be written as: C(s) + O2(g) –—➤ CO2(g)

Sketch an energy profile for the combustion of coal. Label on your sketch: i the activation energy ii the energy required to break the bonds in the reactants iii the energy released in forming bonds in the products iv the enthalpy change for the reaction b Suggest two advantages and two disadvantages of using brown coal to generate electricity. c Write the equation for the combustion of methane, the main component of natural gas. d Write the equation for the combustion of hydrogen gas. e Explain which of natural gas, hydrogen or coal will produce the most (and least) carbon dioxide. 8 The brown coal power stations in Victoria generate a total of over 30 000 GWh of electrical energy each year. (1 GWh = 3.6 × 1012 J.) a Calculate how many joules of energy are produced by these power stations each day. b One tonne of brown coal yields about 8.7 × 109 J of energy. On the basis of the current rate of use, calculate how long you expect Victoria’s estimated 40 000 Mt of brown coal to last. (1 Mt = 106 t.) c In view of predicted patterns of energy use, would you expect the figure calculated in part b to be higher or lower than the actual length of time before brown coal supplies are exhausted? Give reasons for your answer. d A household uses about 1.7 × 1010 J in a year. Calculate the mass of coal that is burnt to provide this energy. 9 The Hazelwood power station in the Latrobe Valley consumes about 13 million tonnes of coal in one year. The coal used in the power station is composed of approximately 25% carbon. Calculate the volume of the greenhouse gas carbon dioxide released each year by the power station. (1 mol of carbon dioxide occupies 24.5 L at 25°C and 1 atm; 1 t = 106 g.)


10 Explain why: a energy from oil is very suitable for transport b electrical energy is very suitable for domestic lighting and heating 11 Crude oil consists mainly of saturated hydrocarbons. a What is meant by the term ‘saturated hydrocarbons’? b One of the first steps in processing crude oil is fractional distillation. What is the purpose of this step? c Some of the fractions obtained from fractional distillation are subjected to catalytic cracking. What is the purpose of catalytic cracking? 12 Crude oil is a mixture of hydrocarbons including C15H32, C3H8, C9H20, C30H62 and C5H12. a In what order would these hydrocarbons liquefy as a gaseous mixture moves up the fractionating tower? Explain. b Copy and complete the table below to show what fraction each of these compounds belongs to and its main uses. Hydrocarbon C15H32

Fraction

Use

C3H8 C9H20 C30H62 C5H12 13 The US has 5% of the world’s population, but consumes about 25% of the total oil produced in the world. The nation consumes nearly twice as much oil as it produces. a Account for this high rate of oil usage in the US. b What political and economic problems could result from this rate of oil consumption? c Do you think that nations such as the US that consume large amounts of energy have special responsibilities to the rest of the world? d Suppose that the US government desired to reduce the nation’s dependence on oil. What steps might be most effective in the i short term? ii long term?

14 Dmitri Mendeleev, the Russian chemist who developed the periodic table in 1869, said that burning petroleum ‘would be akin to firing up a kitchen stove with banknotes’. Why do you think he believed petroleum was too valuable to burn?

Connecting the main ideas 15 Use the Internet to find out which nations are the top ten consumers of energy. 16 When natural gas burns, carbon dioxide and steam are formed. An advertisement for the South Australian Gas Company Limited claimed that, ‘If you use natural gas in your home for cooking, water heating and space heating, you’ll cut carbon dioxide emissions by up to 60% compared to other forms of energy.’ Explain why the company could make this claim. 17 Rather than spending money to reduce gas emissions from their power stations further, the Dutch government decided to spend $35 million on reduction of sulfur emissions in nearby Poland and $300 million in tropical countries to plant new trees in areas where trees have been felled. Why might the government have made this decision?

18 The 1997 Kyoto Protocol was an international response to global warming in which the developed nations were asked to restrict their carbon dioxide emissions to set amounts. Australia was asked to restrict its emission of greenhouse gases to 108% of the level in 1990. a Australia has huge reserves of coal. What impact would adoption of the Kyoto Protocol have on this country? b Suggest why an increase of 108% of emissions, rather than a decrease, was set as a target for Australia. 19 There is an emerging concern that CO2 emissions could lead to increasing acidification of our oceans, with consequent impacts upon marine life and, in turn, ours. a What chemical reaction is likely to be the basis of this concern? b With reference to equilibrium, is such a change and its effects likely to be permanent? c What is the role of the scientific community in checking such a claim? d If true, how could marine acidification influence a choice between coal, petrol and other forms of energy? e How would the successful development of geosequestration (capture and ‘burial’ of emitted CO2) change your answer to part c?

381 Fossil fuels

2

chapter 24

Alternative energy sources

key knowledge • comparison of energy sources: types, uses and sustainability, including brown coal, natural gas, nuclear fission and biochemical fuels

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the range of alternative energy sources available to society, including nuclear fission and fusion and biochemical fuels • identify some advantages and disadvantages in the use of nuclear energy and biochemical fuels • distinguish between renewable and non-renewable energy sources • use social, economic and environmental factors to compare common energy sources • assess the suitability of various energy sources for particular situations • explain how energy conservation and consumption affect the sustainability of energy sources.

20 000

Million tonnes oil equivalent

As we saw in Chapter 23, the world’s dependence on fossil fuels poses two main problems: resources are finite, so use of fossil fuels is not sustainable burning fossil fuels releases carbon dioxide, leading to global warming and climate change. Consumption of energy is predicted to increase, both in developed countries and with increasing industrialisation of developing countries such as India and China (Figure 24.1). This means that problems of supply and environmental impact are also predicted to increase unless alternative sources of energy are found. This chapter explores some of the possible alternatives.

15 000

10 000

5000

2000 Coal Oil

2010

Gas Nuclear

2020

2030

Hydro Other renewable

Figure 24.1 Projected changes in the world’s consumption of energy.

24.1

Nuclear energy !

Fission reactors Large quantities of energy are released when the nucleus of the uranium isotope 235 92 U splits in response to being bombarded by neutrons. The nuclei split in a process called nuclear fission. New elements are formed in this process, so it is described as a nuclear reaction. The neutrons produced in this process then split more uranium atoms, so producing more energy and more neutrons. This process continues, causing a chain reaction that yields considerable energy (Figure 24.2).

93 36

1 0

n

Kr

1 0

n 1 0

n

Neutron = 10n

93 36

140 56

Kr 1 0

n

1 0

n

235 92

93 36

U

Ba 1 0

n

Kr

1 0

1 0

1 0

1 0

n

U 140 56

chemfact Uranium deposits contain two isotopes, 99.3% uranium-238 and 0.7% uranium-235. The uranium-235 isotope undergoes fission readily. Uranium-238 decays progressively to produce stable lead-206. Fuel for nuclear power stations is enriched so that it contains 2–3% uranium-235. Weapons-grade uranium contains about 90% uranium-235.

n

n

235 92

Chemical reactions involve rearrangements of atoms. Elements present in the reactants are always present in the products. Nuclear reactions involve rearrangements of protons and neutrons in atomic nuclei to form different elements.

235 92

140 56

U

n

Ba

93 36

Ba

chemfact

Kr 1 0

n

235 92

U

1 0

n

140 56

Ba

1 0

n

Uranium-235 is naturally radioactive, with a half-life of 7 million years. This spontaneous decay is a different process from the fission caused by bombardment with neutrons in a nuclear reactor.

Figure 24.2 A nuclear fission chain reaction.

383 Alternative energy sources

Steam Electricity

Turbine Generator

Control rods Boiler

Reactor vessel

To cooling towers

Condenser

Pump

Uranium fuel

Figure 24.3 Elements of a nuclear power station.

Pump

Circulating reactor coolant

Concrete containment building

TABLE 24.1 Half-lives of some radioactive isotopes produced by nuclear power stations Radioactive isotope Radon-222 Iodine-131 Krypton-85 Hydrogen-3 (tritium) Strontium-90 Caesium-137 Americium-241 Radium-226 Plutonium-239 Thorium-230

Half-life 3.8 days 8 days 10 years 12 years 29 years 30 years 443 years 1622 years 24 000 years 80 000 years


The energy released from the fission of uranium-235 atoms can be used as the heat source for generating electricity in a power station. One kilogram of uranium produces about the same electrical energy as 2500 tonnes of coal. Thermal energy is used to produce steam. The steam drives a turbine connected to an electricity generator, as shown in Figure 24.3. Over 30 countries, including France, the US, countries from the former USSR, Japan and the UK, are operating a total of more than 440 nuclear power stations. Several other countries are planning and building reactors. Some nations, particularly in Asia, are expanding their nuclear power. In contrast, nuclear plants are closing in parts of western Europe. Some countries are completely dependent on nuclear power for electricity generation. Others, such as Australia, do not use nuclear power stations, although a small research reactor operates at Lucas Heights, near Sydney, producing medically and industrially important isotopes. Reactor designs vary. In some nuclear reactors, control rods containing neutron absorbers such as boron are inserted between fuel rods containing uranium to control the reaction rate. The reactor vessel is in water, which cools it and also absorbs neutrons to moderate the chain reaction, below. 1 0n

–—➤ 140 + 235 92U 56Ba +

93 36Kr

+ 3 10n + energy

Other fission reactions for uranium-235 include: 1 0n 1 0n 1 0n

–—➤ 144 + 235 92U 56Ba + 235 + 92U –—➤ 141 56Ba + –—➤ 130 + 235 92U 52Te +

90 36Kr 92 36Kr 94 40Zr

+ 2 10n + energy + 3 10n + energy + 3 10n + energy

In these fission reactions a small amount of mass is converted to energy. The amount of energy (E) obained when mass (m) is converted to energy is given by E = mc2, where c is the speed of light. Most isotopes directly formed from the fission of uranium have atomic masses spread around 95 and 135. Most are radioactive and undergo further decay until stable isotopes are formed. Table 24.1 gives the half-lives of some radioactive products from uranium-fuelled nuclear power stations.

Nuclear waste The presence of radioactive isotopes in spent fuel obviously demands longterm waste disposal solutions. A typical nuclear power station might produce 30 tonnes of spent fuel each year. Some of this spent fuel is uranium and plutonium that can be recycled, but there is also about 1 tonne of unwanted, highly radioactive material produced each year. This material can be solid, liquid or gaseous, depending upon the design of the particular reactor from which it comes. Steel and concrete containers are used to transport nuclear waste by road and rail. As a consequence of its radioactivity, nuclear waste gives off a considerable amount of heat and must be stored for about five years to allow short-lived radioactive isotopes to decay and the material to cool down (Figure 24.4). The waste containing the longer-lived isotopes must then be stored safely for several thousand future generations. (The Great Pyramid of Egypt was built only around 200 generations ago!) Methods used to store waste include: burial in sealed tanks or stockpiling in deep mines in geologically stable regions stockpiling in air-conditioned warehouses dumping sealed containers of waste in deep ocean trenches sealing it in a special type of glass. Fission is not the only source of troublesome radioactive by-products. Instead of undergoing fission, uranium-238 in the fuel rods can sometimes capture a neutron generated by the fission of uranium-235 to form plutonium. This reaction is exploited in breeder reactors. Plutonium is used in some power stations and in nuclear weapons, but as a component of nuclear waste it is particularly troubling as it has a half-life of 24 000 years.

Figure 24.4 Where should we put all the nuclear waste? This spent fuel is stored in a ventilated vault at the Wylfa nuclear power station in Wales.

238 1 –— ➤ 239 92U + 0 n 92U 239 239 –— U ➤ Np + –10e 92 93 239 0 –—➤ 239 93Np 94 Pu + –1e

chemfact A ceramic called Synroc was used to trap plutonium waste stockpiled in the United States during the Cold War. Synroc is made from several natural minerals and is able to incorporate into its crystal structure nearly all of the elements found in highlevel radioactive waste. It is claimed to be the most effective and durable material for immobilising various high-level radioactive wastes for disposal, and was developed in 1978 by Professor Ted Ringwood at the Australian National University.

Figure 24.5 These storage tanks in Washington State in the US store high-level liquid waste. The tanks were buried in earth after construction.


Advantages and disadvantages of nuclear power

Figure 24.6 The nuclear power station at Chernobyl. Some scientists predicted that thousands of people would die of cancer from radiation released during the accident, an estimate confirmed by Greenpeace International in 2006.

A hazard associated with nuclear power is the release of radioactive substances by accident, hijacking or sabotage. Major accidents at Three Mile Island in the US in 1979 and Chernobyl in the former USSR in 1986 both released radioactive particles in dangerous amounts (Figure 24.6). On the positive side, accident rates are very low in nuclear power plants. There is also concern that radioactive material designed for peaceful uses could be diverted and used in the manufacture of weapons. In contrast to fossil-fuel power stations, nuclear power stations are not usually sources of air pollution. Nuclear power stations offer us the chance to conserve fossil fuels, and consequently reduce acid rain and the production of greenhouse gases. They also have some drawbacks, including a long construction time, often about a decade or more from planning to commission, and a limited working life before they need to be decommissioned. Greenhouse gases are produced in the mining, transport and refining of uranium and in the construction of nuclear power stations. There is also low-level nuclear waste produced in mining uranium ore, and in by-products from refining the ore. In summary, the greatest problems associated with nuclear energy are: the disposal of highly radioactive waste materials, many with long halflives and the potential to enter the food chain the risk and consequences of serious accidents or terrorist attacks involving nuclear reactors.

extension Nuclear trade

US backs b k A Australia’s l ’ uranium plan l The United States has backed a plan by Australia to sell uranium to China for nuclear power as long as there are strict safeguards to stop it falling into the hands of terrorists. US Energy Secretary Sam Bodman . . . said countries around the globe were facing an increase of about 50 per cent in demand for electricity in the next 20 years and would have to look at developing nuclear power in response. While Australia remains cagey about whether it will adopt nuclear power, Mr Bodman said the US had no problem with the federal government selling uranium to China. ‘We don’t object to that,’ Mr Bodman told reporters.

questions

‘However we would encourage both the Australians as well as the Chinese to make sure there are adequate safeguards in place … there is concern over the potential access of terrorists to nuclear material.’ Australia’s Industry Minister Ian Macfarlane said Australia was yet to have a proper debate based on facts rather than emotion about adopting nuclear power in Australia. ‘Whether or not Australia goes down that [nuclear] path or looks at a different mix of renewable energy along with zero emission coal it really needs to be debated.’ Source: AAP, in The Age, 11 January 2006

?

E1 AAustralia usttralilia aand ndd C China hina are both signatories to the Treaty for the Non Proliferation of Nuclear Weapons hi Weapons. Use the Internet to find find out about the purposes of this treaty. E2 Identify some of the issues involved in the mining of uranium and its sale to other countries.


Advocates for nuclear power argue that: environmental damage associated with nuclear reactors is less than that caused by the burning of fossil fuels costs and supply of imported oil have been unreliable while nuclear energy is readily available another energy form is required to replace diminishing fossil-fuel reserves switching to nuclear power stations will reduce greenhouse gas emissions and global warming. The world’s reserves of uranium are estimated at 1 × 1015 MJ, sufficient for more than 100 years at current consumption rates. Australia has between one-quarter to one-third of the world’s known reserves of uranium ore, U3O8. Some of Australia’s uranium is exported, although there is vigorous debate about whether it should be mined and sold.

Nuclear fusion At extremely high temperatures, hydrogen nuclei can combine to form helium nuclei, with no radioactive by-products. Such a process, in which a larger nucleus is formed from smaller ones, is called nuclear fusion. Nuclear fusion occurs in the Sun and is the source of solar energy. The step-wise nuclear reaction can be summarised as:

! 0 1

e represents a positron—a small, positively charged particle the size of an electron.

4 11H –—➤ 42He + 2 01e + energy

Nuclear fusion also occurs when a hydrogen bomb is detonated, but such a rapid release of energy is unsuitable for a power station. Controlled nuclear fusion, if it were possible, would offer clean and safe power. One kilogram of fusionable material may yield the energy available from 10 000 000 kg of fossil fuel. Two main problems have been encountered in the development of this energy source: sustaining the temperatures of 100 000 000°C required for the reaction containing material at such high temperatures. Recent experiments involve the use of ‘containers’ made from magnetic fields. There is a prospect of nuclear fusion becoming a useful energy source at some future date. France, in partnership with the European Union, Japan, Russia, India, South Korea, China and the United States, has recently announced the proposed construction of a nuclear fusion demonstration power plant at Cadarache, near Marseille. This quest for controllable fusion has immeasurable rewards for us all if it can be achieved.

summary

24.1

D During i nuclear l reactions, ti protons t andd neutrons t in i the th nuclei l i off atoms are rearranged to form new elements. In a fission reaction the nucleus of a heavy element is split, forming the nuclei of lighter elements. In a fusion reaction the nuclei of lighter elements combine to form the nucleus of a heavier element. 235 92 U is used as the fuel in a nuclear reactor. The generation of electricity using nuclear power does not produce CO2, a greenhouse gas.

There Th are large l reserves off uranium i readily dil available il bl in i Australia. The wastes from nuclear power stations remain radioactive for a long time, posing a problem for their long-term storage and disposal. There is community concern about the risk of accidents involving nuclear power stations. Radioactive waste contains materials that can be used to manufacture nuclear weapons.


key question 1 Australia exports most of the uranium it mines rather than using it for domestic nuclear power stations. Respond to each of the following questions, giving the reasons for your response in point form.

a Should Australia take back the waste products from nuclear power stations fuelled by Australian uranium? b What possible changes might lead to a greater use of nuclear power in Australia?

24.2

! You can find out more about Australia’s nonrenewable and renewable energy resources by following the link from hi.com.au.

Renewable energy sources Renewable energy sources are those that are continually being replaced by natural processes. These sources include rivers, tides, waves, rubbish tips, plant crops, hot underground water, and the Sun. Some of the most exciting recent developments in science are in the area of renewable energy sources. Rapid technological advances mean that many alternative energy sources viewed as exotic or fanciful a decade or two ago are now economically viable.

Biochemical fuels H

O

H

C

O

C O

C17H29

H

C

O

C O

C15H31

H

C

O

C

C17H31

+

3CH3OH

H

enzymes

C6H12O6(aq) ––––—➤ 2CH3CH2OH(aq) + 2CO2(g)

NaOH H CH3OOC17H29 + CH3OOC15H31 + CH3OOC17H31 Mixture of methyl esters of fatty acids (biodiesel)

The use of crops such as sugar to produce ethanol for use as a petrol additive and canola to produce biodiesel was discussed in Chapter 11. These are considered to be a renewable energy source as natural processes, the growth of plants, can be used to produce them. They are also considered by some to be carbon dioxide neutral in that carbon dioxide is consumed in the photosynthesis reaction that occurs in plants. You will recall that ethanol for use as a fuel can be produced by the fermentation of glucose:

H

C

OH

H

C

OH

H

C

OH

+

H Glycerol

Figure 24.7 A triglyceride containing different fatty acid residues undergoes a transesterification reaction with methanol. Once the glycerol is separated, the mixture of the methyl esters of the fatty acids is sold as biodiesel.


Research into the conversion of cellulose and other fibrous plant wastes into ethanol may result in an increase in the amount of ethanol produced from plant material. Blends of 10% ethanol with petrol, known as E10, can be used in petrol engines. Biodiesel is derived from the triglycerides in vegetable oils and animal fats that undergo a transesterification reaction with methanol to produce a mixture of methyl esters of the fatty acids (see Figure 24.7). The glycerol and esters are separated. The mixture of esters of the fatty acids is sold as biodiesel. Waste plant material can be converted into the mixture of carbon monoxide and hydrogen known as syngas, which can be used to generate heat and electricity. It can also be converted into diesel. Animal waste and plant material can also be used as a source of biogas, by using a biogas digestor, as described in Chapter 11. Biogas is a popular form of energy in many developing countries where people rely on burning biomass, wood, stubble or animal dung for their cooking and heating needs.

chemistryy in action Hydrogen: a fuel of the future? Hydrogen gas can be produced simply from the electrolysis of water, using a range of renewable or non-renewable energy sources to drive the process. A future means of production that is still in the laboratory developmental stage is to obtain hydrogen from the green chloroplasts in spinach using an enzyme from the bacterium Escherichia coli. Other lines of research involve the use of different bacteria and also using light to reduce carbon dioxide to the methanoate ion, HCOO–, followed by the use of enzymes to convert this to hydrogen and carbon dioxide. Research is also under way into electrolysis by photobiological methods whereby microbes exposed to sunlight split water to produce hydrogen and oxygen, and photoelectrolysis by photoelectric semiconductors placed in water. The overall reaction is:

! The E. coli bacterium is not usually such a helpful organism. It is commonly found in raw sewage and can cause food poisoning. However, it has been well studied for many years and is often used as a basis for genetic engineering.

2H2O –—➤ 2H2(g) + O2(g); ΔH = 571 kJ mol–1

This reaction is reversed when hydrogen is burnt, say in a motor vehicle, releasing 142 kJ per gram of hydrogen consumed.

Photovoltaic power

d.c. electricity

Water

Figure 24.8 Hydrogen production in the future.

Rectifier

Oxygen gas Electrolyser

Wind power Hydrogen gas a.c. electricity

Hydroelectricity

Compressor Compressed hydrogen gas

Solar thermal electricity: the collectors concentrate the sunlight to heat oil, which is used to generate steam to drive turbines

Pipeline to user

About 95% of the hydrogen produced today is extracted from methane using hightemperature steam (steam methane reforming, see Chapter 19) in a two-step sequence: CH4(g) + H2O(g) –—➤ CO(g) + 3H2(g) CO(g) + H2O(g) –—➤ CO2(g) + H2(g)

Hydrogen is a contender for the top motor fuel in the next few decades (Figure 24.9). All major car manufacturers are developing engines that are able to use hydrogen as a fuel. The big advantage of hydrogen is its non-polluting combustion, although greenhouse gases are released if coal or other fossil fuels are used to produce the hydrogen in the first place. It is an extremely light and highly flammable gas, and problems remain with its safe use and setting up a distribution infrastructure. Hydrogen’s low energy density at standard temperature and pressure make it difficult to store in a small fuel tank. One of the most promising developments is in improved storage technology in hydrogen-powered motor vehicles, giving them a range comparable to that of petrol-powered cars.

Figure 24.9 A hydrogen fuel-cell vehicle fills up at the first retail hydrogen dispenser in the US.


chemistryy in action Hydrogen: a fuel of the future? (continued) Fuel cells, which are described in Chapter 27, convert chemical energy into electrical energy. They offer the promise of clean and safe energy. Heat and water are the only waste products from hydrogen–oxygen fuel cells. Fuel cells can be up to 80% efficient. At this rate, they can produce electricity much more efficiently than it can be produced by fuel combustion in a conventional power station. Fuel cells in hydrogen-powered vehicles are currently more expensive and short-lived than other fossil-fuelled engines, but fuel cells have been used successfully in the United States’ Space Shuttle. Ongoing technological improvements are paving the way for the large-scale commercial use of fuel cells.

Wide eaves to protect windows from summer sun

Summer sun Solar water heater

Winter sun

Insulation to prevent loss of heat in winter

Large windows to admit winter sun Solid building materials to store heat

Solar energy The energy in fossil fuels, and most of the other forms of energy that are described in this chapter, originate from energy supplied by the Sun. Solar energy can also be used directly. The Sun provides the Earth with about 1020 J of energy per hour. Solar-efficient design of buildings dramatically reduces heating, cooling and lighting costs. Solar water-heating systems can provide hot water for domestic or industrial purposes such as heating homes, water for domestic use and for use in swimming pools.

Figure 24.10 Features of a solar-efficient house.

Solar collector Hot water Cold water Pump Filter

Figure 24.11 A solar pool heater. Solar water-heating can provide more than 60% of the energy required to heat a swimming pool.


Figure 24.12 This solar-powered aircraft has a 63 m wingspan. It is designed to carry scientific instruments to heights of up to 30 000 m.

Figure 24.13 Rows of dishes form part of a solar power station serving the remote township of White Cliffs in New South Wales.

! It has been estimated that there is sufficient roof space on Australian homes to meet Australia’s current electrical energy needs using photovoltaic cells. Less than 1% of Australia’s energy needs are currently met by solar energy.

TABLE 24.2 Average daily solar energy received for some Australian cities Electricity can also be generated by using parabolic collectors or dishes to focus the Sun’s energy (Figure 24.13). A liquid is used to transfer the heat from this point to a turbine and generator, as in a coal-fired power station. The disadvantages of generating electricity using solar cells are that: they are expensive they are less effective in cloudy weather and do not operate at night large areas of land are needed for collection because of their relatively low efficiency as energy converters.

Hydroelectricity: a solar-powered water wheel Hydroelectricity is obtained by using the energy of falling water to drive turbines connected to electricity generators. In a sense, hydroelectric power stations are solar powered. Solar energy evaporates water, transferring it from low to higher altitudes in the form of rain. Gravitational potential energy is converted to kinetic energy as rainwater in rivers flows downhill (Figure 24.14). Hydroelectric power stations supply about 25% of the world’s and 10% of Australia’s electricity. Further development of this energy resource is possible in Australia, but will be restricted because of: a limited number of suitable sites concerns about the environmental impact of building large dams. To dam or not to dam? In order to provide a reliable flow of water in all seasons of the year it is often necessary to build a storage dam in a river. Land must be flooded in order to build a dam and environmental issues associated with the flooding of land are frequently difficult to resolve.

Wind power

City

Average solar energy levels (MJ m–2) 18.0 17.9 17.5 21.3 13.8 14.7 18.6 16.4 18.9

Adelaide Brisbane Canberra Darwin Hobart Melbourne Mildura Sydney Perth

Dam wall Dam

Pipeline

Turbine and generator

Figure 24.14 The operation of a hydroelectric power station.

Windmills—wind operated water pumps—are iconic features of the Australian outback. Wind-powered electricity generators are a more recent development. They are usually sited near the coast to take advantage of higher wind speeds. Average wind speeds exceeding 5 m s–1 are regarded as suitable for electricity generation. Wind-powered generators have been used successfully to supply electricity to towns in Europe and the United States (Figure 24.15), and a 14-turbine wind farm is planned for the south-western coast of Victoria. Other wind turbines are appearing elsewhere in Victoria.


Figure 24.15 Wind turbines dot the landscape.

The disadvantages of wind-powered electricity generation are that: the turbines are highly visible and their effect on the landscape must be considered the moving blades of the turbines produce some noise so they must be sited away from populous areas electricity production is dependent on wind speeds and is no more reliable than the wind very large numbers of turbines are needed to equal the output of a coalfired power station. The siting of wind farms is often a contentious issue. The construction of a wind farm was blocked by a federal minister in 2006 on the basis of threats to the endangered orange-bellied parrot, a bird with indeterminate flightpaths that had been reported in the area of the proposed wind farm. The decision was reversed later that year when revised plans were prepared locating the wind farm a little further from the coast. Despite these problems, wind power is a promising and non-polluting way of harnessing a free form of energy.

Tidal power: power from the Moon

Figure 24.16 A power station at La Rance, France, produces electricity from tides.


A Melbourne company is installing wave-power generation units in the ocean near Portland. The power generated will be sold to electricity companies. Each generator produces about 20 kilowatts of electricity, which is sufficient to power 10–20 houses. Average wave heights of 2.5 metres off the Australian coast are ideal for wave-energy production. A tidal power station at La Rance in France generates up to 240 megawatts of electricity by capturing the incoming tide in sluice gates and then releasing the water via turbines (Figure 24.16). Tidal power stations are best located where there is a large drop between high and low tides. There are several potential sites in Australia, particularly in north-west Western Australia, where the difference between high and low tides can be as much as 11 metres.

Geothermal power In volcanic regions of the world, heat flowing from hot underground rocks to underground water may cause superheated steam to reach the surface. The energy from the steam can be harnessed to produce electricity. Geothermal power stations have been built in New Zealand and at other locations. This is a proven, cheap and relatively clean form of power. Like tidal power, geothermal power generation is restricted to suitable sites. Away from the more active volcanic areas, underground water at temperatures of 30–80°C can be used to heat buildings, domestic water and for industrial processes. The city of Portland in Victoria is located near a volcanic hotspot. There, hot water from a well is used to heat a swimming pool and six major buildings, reducing the city’s annual fuel bill by several hundred thousand dollars.

Conservation: the hidden energy source You will appreciate by now that choosing energy sources for the future is a complex process that requires us to weigh up the advantages and disadvantages of each source. There are two main ways in which we can extend the lives of our nonrenewable energy reserves: develop alternative energy sources to produce more energy conserve our existing energy reserves by using less energy. The government has encouraged and demanded energy conservation through rebates for householders who install solar heating and solar panels, and legislation to enforce new building codes for energy-efficient housing. You can practise some energy conservation measures as an individual; others require people to cooperate and work as a group. Some measures, such as turning off the television instead using the standby feature, will barely affect your lifestyle; others are more drastic. The selection and use of the best energy sources requires us to make decisions. The decisions that we make as individuals and those made by our governments will determine the quality of our lives in the future.



The page A header Polymer solar cells goes here by Gerry Wilson

Figure 24.17 Construction of a typical organic photovoltaic device. The active layer (red) can be a bilayer of vapor-deposited molecules, or an interpenetrating network of polymers with C60 or an inorganic quantum structure.

394 394 Su Supplying Supplying andand using using energy energy

One of the reasons solar energy has attracted so much interest is that the Sun provides the Earth with more energy per hour (4.3 × 1020 J) than all the energy consumed by the planet in a year (4.1 × 1020 J). However, converting even a small fraction of this into useable electricity remains challenging, both scientifically and technically. Current solar panels are made of crystalline silicon, which have practical efficiencies of about 12%. They are very expensive to produce as their manufacture often requires high-temperature and highvacuum techniques. By making sandwichlike devices from silicon, cadmium, telluride, indium, gallium and arsenic, people have been able to generate higher efficiencies (30%); however, the costs of such devices are prohibitive ($US50 000 per square metre) and many of the components are toxic. We are developing a new generation of electroactive polymers that will allow us to make large-area, low-cost solar panels that have efficiencies greater than 10% (Figure 24.17). Converting solar energy into electricity involves the following steps: 1 absorbing as much light as possible across the solar spectrum, from the ultraviolet through to the infrared 2 transferring that energy to a ‘reactive site’ in the device 3 using the energy to generate charges (i.e. move an electron from one molecule to another, thereby creating positive and negative molecules—a so-called exciton) 4 separating these charges in space before they can recombine

5 moving these charges to separate electrodes and finally 6 moving the charge through an external circuit. The overall solar cell efficiency depends on the efficiencies of each of these steps. Polymers are best known as insulators; however in the late 1980s Heeger, McDiarmid and Shirakawa discovered that some polymers could become conducting— and for this work they received the 2000 Nobel Prize in Chemistry. So in principle we now have molecular wires along which we can move electrical charge. Over the past decade we have developed techniques that give precise control over both the polymer chain length and the polymer ‘architecture’—the sequence of monomers in a copolymer. We have made controlled polymer structures composed of various dye molecules that: • absorb in different parts of the solar spectrum • transfer this energy with high efficiency (> 95%) to a reaction site • produce an exciton, again with high efficiency (Figure 24.19). We are now integrating these polymers with conducting polymers of the type shown in Figure 24.18 and with other additives such as C60 (a hollow ball of carbon atoms) to make prototype solar cells. The aim is to develop a low-cost, roll-to-roll printing technique to produce large-area solar cells that have high efficiencies.

bio o O

)n

OMe O

( O

Figure 24.18 A typical conjugated polymer (MDMO-PPV) and a soluble derivative of C60 (PCBM) used in an all-polymer solar cell.

Dr Gerry Wilson

Figure 24.19 Construction and performance of a star shaped, light-harvesting polymer having 24 coumarin monomers (in blue) and 36 acenaphthalene monomers (in green) per arm. The reaction site is the core of the molecule, which is based on a ruthenium bipyridine complex (in yellow). The solid black arrow indicates the direction of energy transfer and the red arrow indicates electron transfer from the ruthenium core to a methyl viologen molecule (in red).

Dr Gerry Wilson received his PhD from the Australian National University and an undergraduate degree in Chemistry from University College, Dublin. Following research fellowships at the universities of New South Wales and Melbourne, he joined CSIRO to work on security features for high-security documents. His inventions appear on several currencies around the world.

Figure 24.20 Gerry Wilson at work in his laboratory.


24.3

A time for decisions TABLE 24.3 Some advantages and disadvantages of energy sources available in Australia Source Coal

Advantages Large Australian reserves Easily mined

Disadvantages Non-renewable Causes greater pollution than oil and natural gas

Low cost Provides export income Oil

Natural gas

Limited Australian reserves Easily transported

Non-renewable Limited world supplies

Many uses

Causes pollution

Limited Australian reserves Efficient source of domestic heat

Non-renewable Causes pollution, although less than coal and oil

Moderate cost Nuclear fission

Large Australian reserves High energy output per unit mass of fuel

Non-renewable Radioactive waste disposal, long half-life of nuclear wastes

Low accident rate

Accidents have great impact

Provides export income

Security risks

Nuclear fusion

Large global fuel reserves High energy output per unit mass of fuel

Unproven high-cost technology Radioactive waste disposal

Solar

Renewable Low running costs Efficient heating Causes little pollution Renewable Causes little pollution Low running costs Renewable Causes little pollution Low running costs Renewable Causes little pollution Low running costs Renewable Biogas uses wastes Low CO2 emission per unit mass of fuel Production via photosynthesis absorbs CO2

Dependent on weather Large collectors needed

Hydroelectricity

Wind

Tidal/wave Biomass (wood, biogas, ethanol, biodiesel)


Restricted sites Dam destroys local habitat Dependent on weather Restricted sites Dependent on tides/waves Restricted sites Limited rate of supply Low energy output per unit mass of fuel

How do we choose what energy sources to use? How much, how quickly, for which purpose and in what ways should we use energy? What are the risks and how can we manage them? What will future demands be? These are not easy questions to answer. To reach a decision we must carefully balance the benefits and costs (and risks) of each option. The advantages and disadvantages. of various energy options are summarised in Table 24.3.

summary

24.2–24.3

Th The lives li off non-renewable bl energy resources can be b extended t d d by conserving existing reserves and by developing alternative energy sources. Renewable energy resources are continuously being replaced by natural processes.

Biochemical Bi h i l fuels f l suchh as ethanol, th l bi biodiesel di l andd biogas bi are produced from plant materials. Other renewable sources of energy include solar, wind, hydro, geothermal and tidal power.

key questions 2 Bagasse, the fibrous waste from sugarcane mills, is burnt to satisfy about 2% of Australia’s energy needs. Firewood satisfies a similar amount. Should we use more or less of these renewable resources? Justify your answer. 3 a Select a form of biochemical energy and one other form of renewable energy. For your chosen energy sources:

i draw up a table that summarises their advantages and disadvantages ii comment on their usefulness in supplying the future energy needs of your local community. b Repeat part a for two non-renewable forms of energy.


24

key terms biogas chain reaction geothermal power

hydroelectricity nuclear fission nuclear fusion

Nuclear energy 4 a Write an overall equation for the formation of plutonium239 from a neutron hitting uranium-238 in a nuclear reactor. Hint: use electrons (beta particles, –10 β) in your equation. b Plutonium can be used to make atomic weapons. What implications does this have for global security when a nation decides to introduce nuclear power? Suggest some safeguards to minimise nuclear proliferation. 5 Research the characteristics of some radioactive isotopes in spent nuclear fuel that make them particularly hazardous to human health. 6 Australia has large reserves of uranium. At present, it has no nuclear power stations and exports only a limited amount of this source of energy. Discuss with other members of your class the benefits and costs associated with the following proposals: a that Australia should build nuclear power stations b that mining of uranium should be banned 7 The coal versus nuclear power debate has undergone a recent revival in Australia. a Given the choice, would you prefer to live near a coal-fired power station or a nuclear power station? Why? b Is there just a choice between coal and nuclear power as they now operate? What other options exist?

Renewable energy sources 8 Oil, coal and natural gas are described as non-renewable sources of energy. List three examples of renewable sources of energy. 9 The use of solar energy in Australia is increasing. About 6% of dwellings have solar water-heaters, and around 30 000 new solar water-heaters are being produced each year. a What situations are most suited to the use of solar waterheaters? b Why do you think that the use of solar water-heaters is not more widespread? c Solar cars can be constructed and they are cheap to run. Why then do most people prefer petrol-driven cars?


renewable solar energy tidal power

wind power

Energy conservation 10 a Summarise the energy that you are supplied with from outside sources in a single day. Identify the energy source (e.g. coal, gas, etc.) and its application (e.g. transport, hot water, etc.) in your answer. b How could you change your day’s activities so that your energy use is minimised? Can you make a difference? 11 The ever-increasing use of computer-controlled systems offers opportunities for increasing our society’s efficiency in using energy. Describe how energy can be saved by the computer control of: a engines in cars b traffic systems c household lights and heating d industrial production systems

Connecting the main ideas 12 Changes in energy consumption over time are shown in Figure 23.3 (p. 368). a In which decade was nuclear energy introduced? What events led to the introduction of this form of energy? b Why has wood not shown the same increase in consumption as other fuels? c Suggest why the rate of use of coal was outstripped by the rate of use of oil. d Describe four ways in which the use of energy in the daily life of a person living in an Australian city in 1860 would have differed from that of a person living now. 13 Tasmania obtains nearly all of its power supplies from hydroelectricity, whereas Victoria obtains over three-quarters of its electricity from brown coal. Account for the difference between the two states in methods of energy production. 14 ‘Most forms of energy available to us originated from energy supplied by the Sun.’ a Justify this statement by explaining the origins of wind energy, oil and hydroelectricity. b List the energy transformations involved in generating electricity from biogas.

15 The Yallourn power station in Victoria has a maximum capacity of 1450 MW. The average daily rate at which Melbourne receives solar energy is 1.0 kW m–2. a What area of solar cells, operating at 20% efficiency, would produce the same amount of power as the Yallourn station? b If a typical residential block of land in the city has an area of 750 m2, how many of these blocks does your answer in part a represent? 16 Is there an energy source that may be used without any environmental impact? Explain your answer. 17 Describe how you expect energy to be provided in the year 2050 for each of the following purposes. Give reasons for your answers. a domestic hot water b domestic lighting c private transport d industrial heating

18 Compare the diagrams showing the operation of a coal-fired power station (Figure 23.9), nuclear power station (Figure 24.3) and hydroelectric power station (Figure 24.13). a In what ways are the power stations similar? b How do they differ? c Name an advantage and a disadvantage of each form of energy production. 19 Coal, oil, nuclear energy and solar energy are all important sources of energy today. Give your opinion of the likely importance of each source of energy in: a 10 years time b 100 years time c 1000 years time


2

chapter 25

Energy from chemical reactions

key knowledge • application of calorimetry to measure energy changes in chemical reactions in solution calorimetry and bomb calorimetry

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • write and interpret thermochemical equations • perform calculations involving thermochemical equations • compare the specific heat capacity of common materials • perform calculations involving specific heat capacities • describe the procedure for the electrical calibration of a calorimeter and calculate a calibration factor from experimental data • explain how solution and bomb calorimetry can be used to determine energy changes of reactions • use calorimetry data to calculate ΔH and heat of combustion for reactions • discuss factors that may affect the selection of a particular fuel for an application.

chemistryy in action Explosives: a blast of chemical energy

Figure 25.1 An explosives manufacturing vehicle. Most explosives contain the element nitrogen. Nitrogen compounds tend to react to form the more stable nitrogen molecule N2. The explosive mixture is prepared on site and pumped directly into the borehole.

Th Chi The Chinese were th the firstt tto mix i saltpetre, lt t sulfur lf andd charcoal h l ttogether th iin 919 919. Th They quickly realised that there were many uses for this mixture, which later became known as gunpowder. It was put to military use and eventually led to the development of the bombs, cannons and guns which humanity uses so effectively against itself. Today, explosives are an essential tool for mining and other engineering works, such as road construction, tunnelling and building (Figure 25.1). Whereas early mining operations involved thousands of labourers, modern explosives allow millions of tonnes of ore to be set free with a single blast. All along our highways we can see where cuts have been made in the landscape to make way for the road; these cuts have also been created using small quantities of explosives. Explosives are compounds that transform chemical energy into large quantities of thermal energy. Thermal energy is also released when fuels such as petrol and natural gas burn, but the rate of combustion in such reactions is limited by the availability of oxygen gas to the fuel. The compounds making up an explosive, on the other hand, contain sufficient oxygen for complete, or almost complete, reaction to occur. An explosion involves a very rapid reaction. In the case of the explosive nitroglycerine (Figure 25.2), the reaction can be written as:

CH2ONO2

CHONO2

CH2ONO2

Figure 25.2 Nitroglycerine.

4C3H5N3O9(l) –—➤ 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g); ΔH = –1456 kJ mol–1

Notice that large amounts of gas are produced in this reaction—29 mol of gas are produced from 4 mol of nitroglycerine. At atmospheric pressure, the reaction products would expand to fill a volume more than 10 000 times larger than the volume of the nitroglycerine. During a blast, this gas is usually produced within a small hole where the explosive has been placed, creating huge pressures that shatter the surrounding rock. Nitroglycerine can detonate unexpectedly and is too unstable for practical use. During the nineteenth century, a Swedish chemist, Alfred Nobel, recognised its potential and worked to produce a safer material. In the course of this work, Nobel’s brother was killed in an explosion. Fearful local residents forced him to continue his experiments on a barge in the middle of a lake. By absorbing nitroglycerine in a type of earth, Nobel produced a material called dynamite that was much safer to handle but which exploded with almost the same power as nitroglycerine when detonated. Despite Nobel’s concern for human welfare, his invention made him unpopular, linking him to death and destruction. In his will he left funds for the establishment of annual prizes. These prizes are awarded for the most significant research in the fields of chemistry, physics, medicine or physiology, literature, and peace. The Nobel Prizes, which have been awarded each year since 1901, have become the most highly esteemed awards in these fields. A sixth prize, for economics, was added after Nobel’s death.

Figure 25.3 Alfred Nobel (1833–96) invented dynamite. Money earned from this discovery was used to fund the prizes named after him.

401 Energy from chemical reactions

25.1

Thermochemical equations chemfact The term enthalpy change is sometimes used to describe the energy change.

An explosion is an exothermic reaction in which the chemical energy in the explosive is transformed into thermal energy as chemical bonds are broken and new ones are formed. In Chapter 15 we saw that the heat of reaction, ΔH, is equal to the difference in chemical energy, or enthalpy, between reactants and products: ΔH = Hproducts – Hreactants

review Energy profile diagrams and activation energy were discussed on p. 250.

chemfact In warfare, the size of a bomb is described by the mass of explosive it carries.

review

The heat of reaction, ΔH, is negative when there is an overall release of energy (exothermic reaction) and positive when heat is absorbed (endothermic reaction). As you would expect, the size of an explosion depends upon the amount of explosive. The gunpowder in a Christmas cracker releases a puny bang compared to the massive push given to a missile by the explosive packed inside it. Similarly, you know that 10 litres of petrol in a car allows you to travel twice as far as 5 litres. The amount of petrol in the tank is a direct measure of the energy available. We can put this in another way—the enthalpy change obtained by burning petrol molecules is directly proportional to the number of mole of those molecules. In general, for any reaction, ΔH is directly proportional to the amount of substance. Octane is a major component of petrol. Complete combustion of one mole of octane molecules to form carbon dioxide and steam releases 5054 kJ. This information can be written as a thermochemical equation: 1

C8H18(g) + 12 2 O2(g) –—➤ 8CO2(g) + 9H2O(g); ΔH = –5054 kJ mol–1

This reaction is exothermic because energy is released (ΔH negative); reactions that involve the absorption of energy are described as endothermic (ΔH positive).

C 3 CH H2C C 2 CH 1

Energy changes for C8H18 + 122– O2

H2C C 2 CH H2C

Energy

CH2 H3C

Transition complex O O H C OH O C C H H Energy absorbed O O C H to break bonds O

8CO2 + 9H2O

Energy released as new bonds made

O O

H

–5054 kJ mol–1 H O

C

H O

∇

Products Time

Figure 25.4 In an exothermic reaction, less energy is used to break bonds than is released when new bonds form.


If you were to burn twice as much octane, twice the energy (10 108 kJ) would be released. This may be written as the thermochemical equation: 2C8H18(g) + 25O2(g) –—➤ 16CO2(g) + 18H2O(g); ΔH = –10 108 kJ mol–1

When working with thermochemical equations you should be aware that: The coefficients of the reactants indicate the amounts, in mole, of each substance that react to give the specified heat energy change. For example, the last thermochemical equation indicates that 10 108 kJ of energy is released by the reaction of two moles of octane with 25 moles of oxygen. States of reactants and products must be specified, since energy changes occur when solids are converted to liquids or liquids to gases, and vice versa. The evaporation of water, for example, absorbs energy. It can be written as a thermochemical equation: H2O(l) –—➤ H2O(g); ΔH = +44 kJ mol–1

For this reason, the thermochemical equation for the combustion of octane to form water has a different ΔH value from the earlier equation for the combustion of octane to form steam: 1

C8H18(g) + 12 2 O2(g) –—➤ 8CO2(g) + 9H2O(l); ΔH = –5450 kJ mol–1

If a reaction occurs in reverse, it has the same magnitude of ΔH but the opposite sign. The condensation of one mole of steam, for instance, releases the same amount energy as the energy absorbed when one mole of water evaporates. H2O(l) –—➤ H2O(g); ΔH = +44 kJ mol–1 H2O(g) –—➤ H2O(l); ΔH = –44 kJ mol–1

chemfact The difference between the ΔH value for the reaction when steam is formed (–5054 kJ mol–1) and this reaction when water is formed is –5054 – (–5450) = +396 kJ mol–1. This difference in the values of ΔH arises from the change in state of 9 mol of water from liquid to gas.

chemfact Different structural isomers of octane have slightly different heats of combustion.

Calculations using thermochemical equations Worked example 25.1a Calculate the heat energy released when 50.00 mL of 0.200 M sodium hydroxide reacts with excess dilute hydrochloric acid. H+(aq) + OH–(aq) –—➤ H2O(l); ΔH = 57.2 kJ mol–1

chemfact The reaction between a dilute acid and a dilute base is essentially a reaction between H+(aq) and OH–(aq) ions.

Solution From the equation: 1 mol of NaOH releases 57.2 kJ of energy. n(NaOH) = 0.200 × 0.0500 = 0.0100 mol Let 0.0100 mol of NaOH release x kJ. x 0.0100 = By proportion: 57.2 1 x = 0.0100 × 57.2 = 0.572 kJ The energy released is 0.572 kJ.

Worked example 25.1b Calculate the energy released when 250.0 g of petrol burns completely in a car engine. Assume petrol is mainly octane (C8H18) and burns according to the equation: 2C8H18(g) + 25O2(g) –—➤16CO2(g) + 18H2O(g); ΔH = –10 108 kJ mol–1

Solution From the equation: 2 mol of C8H18 release 10 108 kJ of energy. 250.0 g = 2.193 mol 250.0 g = 114 g mol–1


Let 2.193 mol of C8H18 release x kJ. x 2.193 By proportion: = 10 108 2 10 108 × 2.193 x= = 11 083 kJ 2 The energy released when 250.0 g of petrol burns is 1.108 × 104 kJ.

Worked example 25.1c What volume of methane, measured at standard laboratory conditions, is burnt to form carbon dioxide and water in order to provide 4.00 × 104 kJ? CH4(g) + 2O2(g) –—➤ CO2(g) + 2H2O(l); ΔH = –890 kJ mol–1

Solution From the equation, 1 mol of CH4 releases 890 kJ of energy. Let x mol of CH4 release 4.00 × 104 kJ of energy. x 4.00 × 104 = = 44.9 mol By proportion: 890 1 Since 1 mol of gas occupies a volume of 24.5 L at SLC, the volume occupied by 44.94 mol = 44.94 × 24.5 = 1.10 × 103 L.

chemistryy in action Energy in a stick can light up your life!

Figure 25.5 A scuba diver uses a chemical light stick for a night dive. The light is released as a consequence of an exothermic chemical reaction.

A cold plunge into wintry ocean waters may not appeal to everyone. But some people consider night-time scuba dives with a group of friends to check out a sunken ship a lot of fun! Waterproof torches are part of the gear, but a hands-free light that can be seen from all directions is needed for spotting your buddies in the inky blackness. The answer is a chemical light stick: a handy plastic tube that glows for hours when the chemicals inside it react (Figure 25.5). The chemistry of the light stick is fairly straightforward. The aqueous reactants mix when the translucent plastic light stick is bent and a glass tube inside it is broken. Energy is released from the exothermic redox reaction between hydrogen peroxide and the ester diphenyl oxalate (Figure 25.6). Instead of the energy from this reaction being released solely as heat, a carrier molecule transfers the energy to a chemoluminescent dye which is also in the light stick. The electrons in the dye are excited to higher energy levels. Light is emitted as these electrons return to their original, lower, energy levels. The glow from the light stick is simply the emission spectrum of the dye molecule.

O

Figure 25.6 The reaction that occurs in a light stick.

O

O

C

C

O

Diphenyl oxalate

+

H2O2

2

OH + 2CO2 + Energy

Hydroxybenzene (phenol)

Light sticks are useful additions to emergency kits for bushwalkers and rescue teams. The amount of light released depends on the size of the light stick. You can also buy ornamental chemoluminescent hoops, earrings and necklaces, just for fun or to create a romantic glow in the evening light. Glow-worms and fireflies apply similar chemical principles for their glow-in-the-dark bioluminescence.


summary

25.1

Th The energy change h produced d d in i a chemical h i l reaction ti is i the th difference between the energy required to break bonds in the reactants and the energy released when new bonds are formed in the products. Energy required to break the bonds in reactants is called the activation energy.

Energy E change h is i represented t d by b the th symbol b l ΔH. ΔH Energy is released in an exothermic reaction, so ΔH is negative. Energy is absorbed in an endothermic reaction, so ΔH is positive. The amount of energy released or absorbed during a reaction depends on the number of mole of reactants.

key questions 1 Sodium nitrate dissolves readily in water according to the equation: NaNO3(s) –—➤ NaNO3(aq); ΔH = +21.0 kJ mol–1

a Determine the energy change when 1.0 g of solid NaNO3 is dissolved in water. b Would you expect the temperature to rise or fall when sodium nitrate dissolves in water? Explain your answer.

2 Calculate the energy released when the following quantities of ethane gas burn according to the equation: 2C2H6(g) + 7O2(g) –—➤ 4CO2(g) + 6H2O(l); ΔH = –3120 kJ mol–1

a 3.00 mol b 100 g c 10.0 L at SLC 3 What volume of ethane, measured at STP, must be burnt according to the equation in Question 2 in order to yield 100 kJ of heat energy?

25.2

The connection between energy and temperature change Suppose you heat a saucepan of oil and a saucepan containing an equal volume of water on a stove, using the same amount of heat energy. As the liquids absorb energy, their temperatures will increase. However, their temperatures rise at different rates—oil heats up much more rapidly than water. The increase in temperature when a given amount of energy is absorbed depends upon the material’s ability to store thermal energy in its bonds. The amount of energy required to raise the temperature of one gram of a substance by 1°C is called the specific heat capacity of the substance. Specific heat capacities for different materials are listed in Table 25.1. The higher the specific heat capacity, the more effectively a material will store heat energy. The following example demonstrates the use of specific heat capacities in a calculation.

TABLE 25.1 Heat capacities of common materials Substance Water Oil, cooking Ethanol Iron Copper Aluminium Sand

Specific heat capacity (J g–1 °C–1) 4.184 ≈ 2.2 2.4 0.473 0.389 1.97 0.48

review Water is distinguished by its very high heat capacity, which is a consequence of the hydrogen bonding between its molecules. The special properties of water are described in Chapter 10 of Heinemann Chemistry 1.

Worked example 25.2 Calculate the energy required to heat 120 mL of water for a cup of coffee to boiling point if the initial water temperature is 20.0°C.


!

Solution

A useful formula to use in calculations such as this is: Energy (J) = 4.184 × mass of water (g) × temperature rise (°C) or Energy (J) = 4.184 × volume of water (mL) × density of water × temperature rise (°C). The density of water varies with temperature. The density of water at 5°C is 1.0000 g mL–1.

Since the density of water is 1.00 g mL–1, the mass of 120 mL of water is 120 g. The energy required to raise the temperature of 120 g of water by 1°C = 4.184 × 120 = 502.0 J. Since the temperature rises by (100.0 – 20.0) = 80.0°C, the total energy required = 502.0 × 80 = 40 160 J. The energy required to raise the temperature of the water to 100°C is 40.2 kJ.

25.3

Measuring the heat released during a reaction Bomb and solution calorimetry Enthalpy changes are measured directly using an instrument called a calorimeter. Figure 25.7a shows the components of a ‘bomb’ calorimeter used for reactions that involve gaseous reactants or products. The reaction vessel in a bomb calorimeter is designed to withstand the high pressures that may be created during reactions. A calorimeter employed for reactions in aqueous solutions is shown in Figure 25.7b. Both calorimeters are insulated to reduce loss or gain of energy to or from the outside environment. When a reaction takes place in a calorimeter, the heat change causes a rise or fall in the temperature of the contents of the calorimeter. Before the calorimeter can be of use, we must first determine how much energy is required to change the temperature within a calorimeter by 1°C. This is known as the calibration factor of the calorimeter. Electric heater for calibrating the calorimeter

a

Electric heater to ignite sample

b Stirrer

Thermometer Stirrer

Electric heater for calibrating the calorimeter

Thermometer

Pressurised vessel Oxygen under pressure

Water

Sample in crucible

Insulated container Insulated container

Glass bulb containing second reactant

Figure 25.7 Calorimeters used for measuring energy changes in reactions. a A bomb calorimeter used for reactions that involve gases. b A solution calorimeter. Breaking the glass bulb starts the reaction.


Solution of one reactant

The calorimeter is calibrated by using an electric heater to release a known quantity of thermal energy and measuring the resultant rise in temperature. The thermal energy released when an electric current passes through the heater can be calculated from the formula: Energy = voltage (volts) × current (amps) × time (seconds), or E = VIt

Once the calorimeter has been calibrated, we can measure the temperature change caused by a reaction occurring in the calorimeter. The calibration factor is then used to determine what energy change is responsible for this temperature change.

Worked example 25.3 A bomb calorimeter was calibrated by passing 1.50 A through the electric heater for 50.1 s at a potential difference of 6.05 V. The temperature of the water in the calorimeter rose by 0.387°C. ΔH for the equation: CH4(g) + 2O2(g) –—➤ CO2(g) + 2H2O(l)

was determined by burning 8.58 × 10–4 mol of methane gas in the calorimeter. The temperature rose from 20.241°C to 20.891°C.

Solution Step 1: Determine the calibration factor of the calorimeter. The thermal energy released by the heater in the calorimeter is found from the expression: E = VIt = 6.05 V × 1.50 A × 50.1 s = 454.7 J Since this energy made the temperature rise by 0.387°C, the energy required to raise the 454.7 = 1175 J °C–1 temperature by 1°C = 0.387 Step 2: Calculate the energy change during the reaction. During the reaction the temperature rose by 20.891 – 20.241 = 0.650°C. The energy change in the calorimeter = calibration factor × temperature change = 1175 J °C–1 × 0.650°C = 763.8 J Step 3: Calculate ΔH for the equation. From the equation, we require the heat change for reaction of 1 mol of CH4. Combustion of 8.58 × 10–4 mol of CH4 caused an energy change of 763.8 J. By proportion, 1 mol of methane would cause an energy change of

! You may prefer to calculate the calibration factor of the calorimeter directly from the formula: calibration factor (J °C–1) energy released in the calorimeter (J) = temperature rise (°C) VIt = ΔT

763.8 = 890 210 J 8.58 × 10– 4

So, the ΔH for the equation is –890 kJ mol–1. (Note that the negative sign indicates the reaction released energy, causing the temperature to rise.)

Heat of combustion The heat of combustion of a substance is defined as the energy released when a specified amount (e.g. 1 g, 1 L, 1 mol) of the substance burns completely in oxygen. Many substances, including wood, coal and kerosene, are mixtures of chemicals and do not have a specific chemical formula or molar mass. The energy released when they burn is therefore measured per gram or per litre, rather than per mole. Heats of combustion are measured using a calorimeter (Table 25.2).


TABLE 25.2 Heats of combustion of various fuels and pure substances. Volumes of gases are measured at SLC and combustion products are CO2(g) and H2O(l). Substance

Wood, dried Peat, dried Brown coal, dried Black coal, dried Carbon (graphite) Hydrogen, liquid (–253°C)

Heat of combustion (kJ g–1) ≈ 18 ≈ 20 ≈ 30 ≈ 33 33 143

Substance

Petrol, standard Petrol, super Kerosene Diesel fuel Heating oil Ethanol LPG

extension Which fuel is best? A wide range of fuels are used. Factors that need to be considered when selecting a fuel for a particular purpose include: • energy released per unit mass or unit volume • availability and cost of fuel • technology to convert chemical energy to useful work • ease of transport • hazards to people associated with its use or waste products • hazards to the environment associated with its use or waste products.

Figure 25.8 Liquid hydrogen was used to propel the Saturn V, used in the Apollo program and the most powerful rocket ever built.


Heat of combustion (kJ L–1) ≈ 33 800 ≈ 34 200 ≈ 36 500 ≈ 38 000 ≈ 38 500 ≈ 23 418 ≈ 93

Substance

Natural gas (methane) Ethane Propane Butane Octane Methanol Ethanol Hydrogen

Heat of combustion (kJ mol–1) 890 1560 2220 2886 5450 727 1367 286

extension Which fuel is best? (continued) Many forms of transport, such as cars, ships and aeroplanes, carry their fuel with them. The mass of the fuel as well as its energy content needs to be considered when selecting the most appropriate energy source. The energy of a fuel per unit mass is called energy density. The energy density of some common fuels is shown in Table 25.3.

questions

TABLE 25.3 Energy density of some common fuels Fuel

Energy density (MJ kg–1) 45.8 45.3 41.6 29.7 32.8 143.0

Petrol Petroleum diesel Biodiesel Ethanol Anthracite (black coal) Hydrogen

?

E1 E2 E3 E4

Suggest Suggestt why whhy liquid liquid i hydrogen was chosen to fuel the Saturn rockets used in the American space program program. What are the limitations of using hydrogen as fuel for cars? Why is petrol preferred to coal as a fuel in cars? Biodiesel and ethanol have been promoted as being renewable energy resources and as being carbon dioxide neutral. Which would drive your car further, one kilogram of ethanol or one kilogram of biodiesel? E5 Calculate the energy density of octane, C8H18, in MJ kg–1 if the heat of combustion of octane is –5450 kJ mol–1.

summary

25.2–25.3

SSpecifi ific hheatt capacity it is i the th amountt off energy required i d to t raise i the temperature of 1 gram of a substance by 1°C. Energy change is measured in a calorimeter.

A calorimeter l i t mustt be b calibrated lib t d bbefore f use. Th The amountt off energy required to raise the temperature of a calorimeter by 1°C is called the calibration factor. The heat of combustion is the energy released when 1 mole of a substance is burnt.

key questions 4 Calculate the energy needed to heat: a 100 mL of water from 20.0°C to 80.0°C b 250 mL of water from 25.0°C to 100.0°C c 1.5 kg of water from 20.0°C to 30.0°C d 300 g of water from 18.0°C to 100.0°C e 300 g of cooking oil from 18.0°C to 100.0°C 5 A calorimeter containing 100 mL of water is calibrated by passing a 3.00 A current through the instrument for 36.0 s at a potential difference of 3.50 V. The temperature rises by 0.82°C. Potassium hydroxide weighing 0.654 g is added to the calorimeter and dissolved rapidly by stirring. The temperature rises from 20.82°C to 22.23°C. a Calculate the calibration factor for the calorimeter and water.

b Determine ΔH for the equation: H2O(l)

KOH(s) –––—➤ KOH(aq)

6 A temperature rise of 1.78°C was observed when 1.00 × 10–3 mol of propane gas was burnt in a calorimeter. The calibration factor of the calorimeter was previously determined to be 1250 J °C–1. a Write an equation for the combustion of propane. b Calculate the energy change during the reaction. c Calculate the heat of combustion for propane in kJ mol–1. d Write a thermochemical equation for the combustion of propane.


25

key terms calibration factor calorimeter energy density

heat of combustion specific heat capacity thermochemical equation

Thermochemical equations 7 In a steelworks, carbon monoxide present in the exhaust gases of the blast furnace can be used as a fuel elsewhere in the plant. It reacts according to the equation: 1

CO(g) + 2 O2(g) –—➤ CO2(g); ΔH = –283 kJ mol–1

a Which has the greater heat content: 1 mol of CO(g) and 0.5 mol of O2(g), or 1 mol of CO2(g)? b Write the value of ΔH for the reactions: i 2CO(g) + O2(g) –—➤ 2CO2(g) ii 2CO2(g) –—➤ 2CO(g) + O2(g) 8 Calculate the energy released when the following quantities of hydrogen gas burn according to the equation: 2H2(g) + O2(g) –—➤ 2H2O(g); ΔH = –572 kJ mol–1

a 1.00 mol b 1.00 kg c 100 L at SLC 9 What volume of hydrogen, measured at STP, must be burnt according to the equation in Question 8 in order to yield 100 kJ of heat energy? 10 Butane is used as the fuel in pocket cigarette lighters. It is a liquid when stored under pressure in the lighter, but vaporises when the valve is opened. Combustion of butane is represented by the equation: 2C4H10(g) + 13O2(g) –—➤ 8CO2(g) + 10H2O(l); ΔH = –5772 kJ mol–1

a How much energy is evolved when 10.0 g of butane burns completely? b How much energy is evolved when 0.100 L of butane, measured at SLC, burns completely? c Calculate the volume of butane, measured at 15°C and 108 kPa, that must be burnt to yield 1.00 kJ of energy. 11 A hand-warmer purchased by campers and skiers consists of a sealed outer plastic bag and an inner perforated plastic bag containing powdered iron, water, salt and sawdust. When the inner bag is taken out and shaken it becomes hot because the following reaction occurs: 4Fe(s) + 3O2(g) –—➤ 2Fe2O3(s); ΔH = –823 kJ mol–1


The bag contains 15.0 g of iron that reacts completely. a Draw and label a diagram to show the energy changes that occur during the course of the reaction (an energy profile diagram). b Calculate the energy released by the hand-warmer. c Suggest why salt is included in the contents of the bag. 12 An oxy-acetylene welding torch uses ethyne (acetylene) at the rate of 500 mL per minute, measured at STP. The reaction is described by: 2C2H2(g) + 5O2(g) –—➤ 4CO2(g) + 2H2O(l); ΔH = –2599 kJ mol–1

a What is the heat of combustion of ethyne in kJ mol–1 and kJ g–1? b Calculate the rate of energy production by the torch in kJ per minute. 13 Calculate the heat evolved when 25.00 mL of 0.100 M sulfuric acid reacts with excess sodium hydroxide solution. H+(aq) + OH–(aq) –—➤ H2O(l); ΔH = –57.2 kJ mol–1

14 Glucose is produced by photosynthesis according to the following equation: 6CO2(g) + 6H2O(l) –—➤ C6H12O6(aq) + 6O2(g); ΔH = +2803 kJ mol–1

a Write a thermochemical equation for the reaction of glucose with oxygen. b How much energy is released when 5.00 g of glucose is oxidised? 15 Recycled ammonia has been proposed as a new source of hydrogen for electricity generation. In a closed system, solar energy is used to dissociate ammonia into hydrogen and nitrogen gas. The reverse reaction releases heat for power generation. Under standard conditions, the reaction is: 2NH3(l)

N2(g) + 3H2(g); ΔH = 92.4 kJ mol–1

a Is ammonia produced or consumed when power is being generated? b What mass of hydrogen gas is required to power a 100 W light bulb for an hour? (Hint: 1 W = 1 J s–1) c How would the temperature and pressure affect the equilibrium position of the reaction?

Energy and temperature change 16 Water evaporating from the skin after swimming causes us to feel cool. Account for the fact that during this process the heat content of the water is increasing. 17 On a hot day a swimmer leaves the water, walks across the sand and, when entering a car, sits on the metal buckle of the seat belt. Explain how the data in Table 25.1 are related to the swimmer’s experiences. 18 The moist brown coal used in coal-fired power stations in Victoria is not as good a fuel as the black coal used in power stations in New South Wales. Only about one-third of the substances present in brown coal are combustible. What other factor reduces the heat energy available from this fuel? 19 Sand and water are both used to smother fires. Which material would absorb more heat from a fire? Why? 20 By referring to Table 25.1, calculate the energy required to raise the temperature of: a 1.00 kg of iron from 25.0°C to 100.0°C b 500 g of water from 25.0°C to 100.0°C c 0.500 kg of ethanol from 50.0°C to 100.0°C d 500 g of aluminium from 50.0°C to 100.0°C 21 A 500 mL volume of water in an electric kettle was heated using 136 kJ of electrical energy. If the water was initially at 15°C, calculate its final temperature. 22 An ‘instant’ cold pack sold for the treatment of sporting injuries contains solid ammonium nitrate and a weak inner plastic bag full of water. Punching the pack breaks the inner bag and causes the ammonium nitrate to dissolve. NH4NO3(s) –—➤ NH4+(aq) + NO3–(aq); ΔH = +25 kJ mol–1

A bag containing 25.0 g of ammonium nitrate was punched and dropped into a 2.00 L container of water at a temperature of 19.00°C. a Calculate the lowest temperature that the water could reach. b Why would this temperature not be achieved in practice? 23 Surf on, play on! Super-cool ‘hotsuits’ are a new Australian invention that use heat-releasing liquid gel packs in a wet suit. The packs are placed in pockets near parts of the body where there is substantial flow of blood under the skin. Surfers can get over an hour more playtime in cold conditions. The packs contain supercooled sodium ethanoate in water. Sodium ethanoate solidifies at 54°C but it can be supercooled and stay as a stable liquid gel at much lower temperatures until disturbed by vigorous motion. The gel packs can be re-used by boiling them and allowing them to cool. a Would you expect the reaction to be exothermic or endothermic? Explain. b The gel packs crystallise as they release heat. Write a feasible reaction for this process.

c Footballers sometimes wear ice vests to cool them down in warm weather. Suggest a thermochemical alternative to ice. 24 The equation for dissolving crystals of ammonium chloride in water at SLC is: NH4Cl(s) + H2O –—➤ NH4+(aq) + Cl–(aq); ΔH = 15.15 kJ mol–1

a Is this reaction endothermic or exothermic? b Will the temperature of the solution increase or decrease as the reaction proceeds? Explain. c Which is greater in this reaction, the energy required to break bonds in the reactants or the energy released when new bonds are formed? Explain. d Calculate the energy absorbed when 2.00 g of NH4Cl is dissolved in: i 100 mL water ii 1000 mL water e Predict the expected temperaure change for each case in part d.

Calorimetry 25 What is the calibration factor of a calorimeter if a current of 1.20 A at a potential difference of 5.67 V raises its temperature by 1.25°C in 60.0 seconds? 26 When 1.01 g of solid potassium nitrate was dissolved in a calorimeter, the temperature dropped by 0.672°C. When the calorimeter was calibrated, a current of 1.50 A applied at a potential difference of 5.90 V for 60.0 seconds caused a temperature rise of 0.456°C. a Write an equation for the dissolution of potassium nitrate. b Determine the calibration factor of the calorimeter. c Calculate the energy change during the reaction. d Write the thermochemical equation for the reaction. 27 When a current of 1.50 A passed through the electric heater of a calorimeter for a time of 25.1 s at a potential difference of 8.05 V, the temperature of the calorimeter rose by 0.399°C. 7.62 × 10–4 mol of ethene gas was then burnt in the calorimeter. The temperature rose from 20.12°C to 21.54°C. a Write an equation for the combustion of ethene. b Calculate ΔH for the equation in part a. 28 A calorimeter is used to measure energy changes that occur during chemical reactions. a Why should a calorimeter be well insulated? b Why is it necessary to calibrate a calorimeter? c Try to devise a method for calibrating a calorimeter that does not involve the use of an electric heater. d The combustion of a 3.00 g piece of wood in a bomb calorimeter results in an observed temperature rise of 1.6°C. The calibration factor of the calorimeter was 30.0 kJ °C–1. Calculate the heat of combustion of the wood.


29 Like a calorimeter, your classroom requires a certain amount of energy to raise its temperature by one degree. a Explain how you could determine the ‘calibration factor’ of your classroom. b Could a classroom be used as a calorimeter to measure the energy released by students during a lesson? Why, or why not? 30 A camper uses a butane stove to heat water in a metal cup for a drink of coffee. The camper boils 150 mL of water that was initially at 20.0°C. 2C4H10(g) + 13O2(g) –—➤ 8CO2(g) + 10H2O(l); ΔH = –5718 kJ mol–1

a Calculate the minimum volume of butane, measured at SLC, required to boil the water. b Why would more butane be required in practice than your answer to part a? 31 A mixture of 0.0100 mol hydrogen gas and an unknown amount of methane was burnt completely in a calorimeter to produce a temperature rise of 5.75°C. The calibration factor of the calorimeter was 1500 J °C–1. If the heats of combustion of hydrogen gas and methane are 286 kJ mol–1 and 890 kJ mol–1 respectively: a write the thermochemical equations for the two combustion reactions b calculate the amount of methane in the mixture.

Heat of combustion 32 Methane and propane are used as alternative transport fuels. Given the heats of combustion of methane and propane are 890 kJ mol–1 and 2220 kJ mol–1, respectively, calculate the energy released per gram by each fuel. 33 Ethanol produced from sugar is used as a substitute for petrol in Brazil. Heats of combustion of ethanol (CH3CH2OH) and octane (C8H18), a major component of petrol, are listed in Table 25.2. a Write equations to represent the combustion reactions of ethanol and octane. b Calculate the heats of combustion of each fuel in units of kJ g–1. c If the fuels were the same price per gram, which fuel would represent the better value? d Ten per cent blends of ethanol in petrol are available from some outlets in Australia. Would cars using a 10% blend of ethanol in petrol travel a greater or smaller distance per tank than those using unblended petrol?

412 214 SSupplying u and using energy

e Comment on the prospects of replacing petrol with ethanol produced from sugar or cellulose and using it as fuel for cars. 34 300 mL of LPG is burnt to heat 400 mL of water from 18.0°C to 34.7°C. What is the heat of combustion of the LPG in kJ L–1?

Connecting the main ideas 35 A 0.254 g sample of black coal was burnt in a bomb calorimeter. The calorimeter contained 300 mL of water which rose from 18.25°C to 24.92°C. a Using the specific heat capacity of water, calculate the heat of combustion of the coal. b Is your answer likely to be higher or lower than the actual value? Explain. 36 A particular car travels 20 000 km in a year and operates at a 20% efficiency in transforming the chemical energy in fuel into mechanical energy (motion). The car has a fuel consumption of 10 L per 100 km and its petrol tank holds 50 L. Using data from Table 25.2, calculate: a the amount of energy supplied by a full tank of petrol b the amount of energy released as waste heat during the year c the cost of the fuel that is converted to waste heat, if petrol costs $1.40 cents per litre 37 a Use data from Table 25.2 to compare the energy supplied by one gram of black coal with that supplied by one gram of octane (petrol). Which is the better fuel on this basis? b Describe one application in which coal is preferred to petrol and another in which petrol is preferred. Explain the reasons behind the choice of fuel in each case. 38 A foam cup containing 200 mL of black coffee was heated from 20°C to 100°C in a microwave oven that operates at 1200 W. (1 W (watt) is equivalent to an energy consumption for 1 J s–1.) a Suppose that there was no heat loss. Calculate the time required to heat the coffee. b Calculate the calibration factor of the foam cup when it contained 200 mL of solution.

2

chapter 26

Electricity from chemical reactions

key knowledge • the use of the electrochemical series in predicting the products of redox reactions and writing half equations • limitations of predictions made using the electrochemical series

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • use the electrochemical series to predict redox reactions in aqueous solutions • write half and full equations for redox reactions • describe the limitations of using the electrochemical series.

26.1

Galvanic cells chemistryy in action A technological leap beginning with a frog’s leg

Figure 26.1 Luigi Galvani.

! In the early days of electrochemistry, intensity of pain was used to measure a cell’s power.

The history of electrochemistry began in 1791 when an Italian biologist, Luigi Galvani, and his assistant were working in their laboratory. While experimenting with dissected frogs, they were startled to see a frog’s leg hanging on a copper hook twitch when it touched an iron rail. The muscles of the frog were stimulated by an electric shock. Galvani had just discovered how to generate such a current. His ‘error’ was in thinking that the current was some sort of life force, a perfectly reasonable idea for the time. Other scientists set out to investigate the possibility that the metals were somehow involved in this phenomenon. After several years’ work, in 1800 Alessandro Volta developed a device that used a chemical reaction to produce an electric current. The device, now called a ‘voltaic pile’, consisted of a stack of alternating silver and zinc disks separated by cardboard soaked in salt water. In 1836, an English chemist, John Daniell, invented a device that could supply a useful electric current. The device, which became known as the Daniell cell, was used almost exclusively to power the early English and American telegraph systems because of its reliable output.

We can produce a model of the Daniell cell using simple laboratory apparatus (Figure 26.2). An electric current flows through the wire and light globe. This part of the cell is called the external circuit. The globe converts the electrical energy of the current into light and heat. The current flows because a chemical reaction is taking place in the cell. Initially, there is little indication that a reaction is taking place in the beakers.

Low-power light globe

Salt bridge: a strip of filter paper soaked in potassium nitrate solution Zinc

Copper

Zinc nitrate solution

Copper(II) sulfate solution

Figure 26.2 A model of the Daniell cell constructed using simple laboratory equipment.


However, if we leave the cell with a light globe connected for several hours a number of changes can be observed: the zinc metal corrodes the copper metal becomes covered with a furry brown-black deposit the blue copper(II) sulfate solution loses some of its colour. These changes provide evidence of a chemical reaction. If the light globe is replaced with a galvanometer (an instrument for detecting electric current), the galvanometer indicates that electrons flow from the zinc electrode, through the wire, and to the copper electrode. Current flows only if the salt bridge is present. These observations allow us to decide what is happening in the cell. The reaction in the cell is a redox reaction, as electrons are being produced and consumed. The zinc electrode is eaten away, forming zinc ions in solution: Zn(s) –—➤ Zn2+(aq) + 2e–

The oxidation of the zinc metal releases electrons, which flow through the wire to the copper electrode. Electrons are accepted by copper ions in the solution when the ions collide with the copper electrode: Cu2+(aq) + 2e– –—➤ Cu(s)

Copper atoms are insoluble and deposit on the electrode producing a brown-black coating. The salt bridge forms an essential part of the cell. Figure 26.3 summarises the processes occurring during the operation of the cell. The equation for the overall reaction in the cell, found by adding the two half equations, is: Cu2+(aq) + Zn(s)

chemfact The same changes occur when a strip of Zn is immersed in an aqueous solution of Cu2+ ions, say CuSO4(aq). However, when the Cu2+(aq) and Zn are in contact, the energy transformation is chemical to thermal.

Cu(s) + Zn2+(aq)

Copper ions act as the oxidant and zinc metal acts as the reductant.

review

Electron flow

A chemical that oxidises another substance is called an oxidant and is itself reduced in the process. Similarly, a substance that reduces another substance is called a reductant. Reductants are oxidised in redox reactions.

Salt bridge Zinc

Copper e–

e–

e–

e–

Zn

K+

NO3– Cu2+

Cu Zn

Zn2+ Cu2+

Zn Cu2+

Zn(s)

Zn2+(aq) + 2e–

Cu2+(aq) + 2e–

Cu(s)

Figure 26.3 A representation of the electrode reactions in a Daniell cell.

415 Electricity from chemical reactions

Why is electrical energy released?

chemfact An oxidant and its corresponding reduced form are known as a conjugate redox pair.

Half cells

Silver electrode

Ag+(aq)

Ag+(aq)/Ag(s) half cell

Platinum electrode

Fe3+(aq) and Fe2+(aq)

Fe3+(aq)/Fe2+(aq) half cell

Platinum electrode Glass tube

H2(g)

A galvanic cell is designed so that half reactions occur in two separate compartments of the cell. Because the oxidant and reductant do not come into direct contact with each other, electrons can only be transferred by travelling through an external circuit connecting the negative and positive electrodes. This flow of electrons creates an electric current. Chemical energy of the reactants is transformed into electrical energy.

H+(aq)

H+(aq)/H2(g) half cell

Figure 26.4 Some typical half cell configurations.

review In balanced half and overall equations: • The number of atoms of each element is equal on both sides. • The total charge on each side is equal.

The Daniell cell in Figure 26.3 can be regarded as consisting of two half cells. Each half cell contains an electrode in contact with a solution. In the Daniell cell, one half cell contains Cu and Cu2+, the other contains Zn and Zn2+. The species present in each half cell form a conjugate redox pair. In general, if one member of the conjugate pair in a half cell is a metal, it is usually used as the electrode. However, if no metal is present an inert electrode such as platinum or graphite is used (Figure 26.4). In some half cells, one of the conjugate pairs may be a gas; in such cases, a special ‘gas electrode’ is used. Half cells usually contain other species not involved in the reaction, such as spectator ions and the particles that make up a solvent. The electrode at which oxidation occurs is called the anode. In a galvanic cell the anode, where electrons are produced, is described as the negative terminal. The electrode at which reduction occurs is called the cathode. The cathode, where electrons are consumed, is the positive terminal in a galvanic cell. In the Daniell cell the zinc electrode is the anode and the copper electrode is the cathode.

The purpose of the salt bridge If no salt bridge were present, the solution in one half cell would accumulate negative charge and the solution in other half cell would accumulate positive charge as the reaction proceeded. Such accumulations of charge would very quickly prevent further reaction, and hence prevent production of electricity. The salt bridge contains ions that are free to move so that they can balance charges formed in the half cells. Cations (positive ions) migrate to the cathode and anions (negative ions) migrate to the anode as shown in Figure 26.3. This part of the cell is called the internal circuit.

Writing half cell equations The half equation that involves a conjugate redox pair containing an atom and a simple ion can be written with little difficulty. For instance, knowing that the conjugate redox pair is Zn2+ ions and Zn, you can quickly write the half cell equation as: Zn2+(aq) + 2e– –—➤ Zn(s)

Half cell equations involving polyatomic ions may, however, be more complex to write. The equation for a half cell containing the dichromate ion (Cr2O72–) and Cr3+ ion redox pair is: Cr2O72–(aq) + 14H+(aq) + 6e– –—➤ 2Cr3+(aq) + 7H2O(l)

Such half equations can be deduced using the procedure outlined in Chapter 5 (p. 54).


26.2

The electrochemical series Why does the zinc metal in the Daniell cell react with Cu2+ ions, rather than the copper metal react with Zn2+ ions? Although both zinc and copper metals have a tendency to lose electrons and form ions, they differ in their tendencies to do so. Zinc loses its electrons much more readily than copper— we often simply say that zinc is more reactive than copper. When Zn2+/Zn and Cu2+/Cu half cells are connected, electrons flow from the zinc metal in one half cell to the copper ions in the other. Zinc is described as a stronger reductant than copper. Since zinc loses electrons more readily, it follows that the oxidised form of zinc, Zn2+, will not accept electrons as readily as Cu2+ ions. Cu2+ is described as a stronger oxidant than Zn2+. Figure 26.5 shows a cell constructed from Cl2/Cl– and Cu2+/Cu half cells. Notice that in this cell the copper electrode is negative—Cu is oxidised to Cu2+—whereas in the Daniell cell Cu2+ is reduced to Cu and the copper electrode is positive. Since electrons flow from the Cu2+/Cu half cell to the Cl2/Cl– half cell: copper is a stronger reductant than Cl– ions Cl2 is a stronger oxidant than Cu2+ ions. Galvanometer Electron flow

Platinum cathode

+

Copper anode

_

Chlorine gas

Cl2(g) + 2e–

Cl– solution

2Cl–(aq)

Cu2+ solution

Cu(s)

Cu2+(aq) + 2e–

Figure 26.5 A cell constructed from Cl2/Cl– and Cu2+/Cu half cells.

This information can be combined with our knowledge of the relative redox strength of Zn and Zn2+ ions in the Daniell cell to produce Table 26.1, which lists the half cell equations in order of their tendency to occur as reduction reactions.

TABLE 26.1 Relative oxidising and reducing strengths of Cl2/Cl–, Zn2+/Zn and Cu2+/Cu Cl2(g) + 2e– Cu2+(aq) + 2e– 2+

Zn (aq) + 2e

–

2Cl–(aq) Cu(s) Zn(s)

increasing reducing strength

Reductants g–—–––––

––––––––—h

increasing oxidising strength

Oxidants

review The reactivity of zinc allows it to be used to protect iron from rusting. Galvanised iron has a surface coating of zinc that corrodes in preference to the iron. Copper is much less reactive and is useless for this purpose (Heinemann Chemistry 1, Chapter 16).

By constructing cells from various combinations of half cells, chemists have developed the larger table shown in Table 26.2. Such a list is called an electrochemical series.


Notice the figure of 0.00 given for hydrogen. All other E ° values are given relative to this arbitrary standard.

! In a galvanic cell, the stronger reductant is in the half cell with the negative electrode (anode). The stronger oxidant is in the half cell with the positive electrode (cathode).

! Another way to predict the electrode reactions is to remember that the higher of the two half reactions in the electrochemical series goes forward and the lower one is reversed. The higher half reaction occurs at the positive electrode; the lower reaction occurs at the negative one.


E ° (V)

Reductants F2(g) + 2e–

2F–(aq)

H2O2(aq) + 2H+(aq) + 2e–

2H2O(l)

MnO–4(aq) + 8H+(aq) + 5e–

Mn2+(aq) + 4H2O(l)

Cr2O72–(aq) + 14H+(aq) + 6e–

2Cr3+(aq) + 7H2O(l)

Cl2(g) + 2e– O2(g) + 4H+(aq) + 4e–

2Cl–(aq) 2H2O(l)

Br2(aq) + 2e–

2Br–(aq)

Ag+(aq) + e–

Ag(s)

Fe3+(aq) + e–

Fe2+(aq)

O2(g) + 2H+(aq) + 2e–

H2O2(aq)

I2(s) + 2e– O2(g) + 2H2O(l) + 4e–

2I–(aq) 4OH–(aq)

Cu2+(aq) + 2e–

Cu(s)

Sn4+(aq) + 2e–

Sn2+(aq)

S(s) + 2H+(aq) + 2e–

H2S(g)

2H+(aq) + 2e–

H2(g)

Pb2+(aq) + 2e–

Pb(s)

Sn2+(aq) + 2e–

Sn(s)

Ni2+(aq) + 2e–

Ni(s)

Fe2+(aq) + 2e–

Fe(s)

Zn2+(aq) + 2e–

Zn(s)

2H2O(l) + 2e–

H2(g) + 2OH–(aq)

Al3+(aq) + 3e–

Al(s)

Mg2+(aq) + 2e–

Mg(s)

Na+(aq) + e–

Na(s)

Ca2+(aq) + 2e–

Ca(s)

K+(aq) + e–

K(s)

Li+(aq) + e–

Li(s)

increasing reducing strength

chemfact

Oxidants

g–—–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

• Strong reductants donate electrons more readily than weak ones. • Strong oxidants accept electrons more readily than weak ones. • Strong reductants have weak conjugate oxidants, and vice versa.

TABLE 26.2 The electrochemical series. Strongest oxidants are at top left of the table and the strongest reductants are at bottom right of the table.


chemfact

The electrochemical series is strictly valid only for the conditions under which it was determined. The series in Table 26.2 applies at: a temperature of 25°C pressures of 1 atm 1 M concentration of solutions. These are known as standard laboratory conditions or standard conditions.

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––—h

! Under different conditions, the order of the half cells may change.

2.87 1.78 1.49 1.36 1.36 1.23 1.09 0.80 0.77 0.68 0.54 0.40 0.34 0.15 0.14 0.00 –0.13 –0.14 –0.23 –0.41 –0.76 –0.83 –1.71 –2.38 –2.71 –2.87 –2.92 –3.05

The electrochemical series can be used to predict what will happen when two specific half cells are combined to form a cell.

Worked example 26.2 Use the electrochemical series to predict the behaviour of a cell composed of Ag+/Ag and Fe2+/Fe half cells. Ag+(aq) + e– –—➤ Ag(s); 0.80 V Fe2+(aq) + 2e– –—➤ Fe(s); –0.41 V

–—–––➤


––––—➤

Solution increasing reducing strength

Since Ag+ is higher on the left side of the table than Fe2+, it is the stronger oxidant. Iron, being lower on the right side of the table than silver, is a stronger reductant. The strongest oxidant will react with the strongest reductant, so we predict that the two half equations will be:

Voltmeter Electron flow –

+ Silver

Iron

Ag+(aq) + e– Ag(s) Fe(s) Fe2+(aq) + 2e–

To write the overall cell equation, multiply the Ag+/Ag half-cell equation by two so that the number of electrons in each half cell is the same, and then add the two equations together. The overall equation for the reaction will be: 2Ag+(aq) + Fe(s)

Fe2+(aq) 1 M

2Ag(s) + Fe2+(aq)

The silver electrode will be the cathode and the iron electrode will be the anode. Figure 26.6 indicates our predictions. It is important to appreciate that this is only a prediction. Since the order of half cells in Table 26.2 applies for standard conditions, different conditions might affect the order of half cells, altering the cell reaction. Predictions should be checked by experiment.

summary

Fe(s)

+ Fe2+(aq) + 2e–

Ag+(aq) 1 M Ag+(aq) + e–

Ag(s)

Figure 26.6 A cell constructed from Ag+/Ag and Fe2+/Fe half cells.

26.1–26.2

O Oxidation id ti involves i l the th loss l off electrons, l t andd reduction d ti involves i l the gain of electrons. In galvanic cells the oxidation and reduction reactions are separated and the transfer of electrons occurs through an external circuit. The movement of ions from the salt bridge into the half cell maintains electrical neutrality. Oxidation occurs at the anode, which is the negative terminal of a galvanic cell.

Reduction R d ti occurs att the th cathode, th d which hi h is i the th positive iti terminal t i l of a galvanic cell. In the electrochemical series, chemicals are arranged in order of their strengths as oxidants. The order within the electrochemical series is based on the assumption that the standard conditions of 25°C, pressure of 1 atmosphere and a concentration of 1 M apply. In a galvanic cell the stronger oxidant will be reduced and the stronger reductant will be oxidised.

key questions 1 Draw a labelled diagram of a cell formed from Cl2/Cl– and Sn2+/Sn half cells. Use the electrochemical series to indicate: a half cell reactions b the anode and the cathode c direction of electron flow d electrode polarities (which electrode is positive and which is negative) e the overall reaction.

2 Repeat Question 1 for the cells formed from the following half cells: a Fe3+/Fe2+ and H+/H2 b Cl2/Cl– and Pb2+/Pb


the cutting edge Making The unnatural page A header products by natural means goes here

key terms

by Steven J. Langford y group’s interests are in utilising nature’s concepts in the laboratory— making unnatural products by natural means! Most of your chemistry studies so far have involved looking at interactions between atoms to form discrete molecules. We work in the area of supramolecular chemistry, dealing with the interactions between molecules in large arrangements on a nanometre scale. In our work we think of molecules as being like Lego blocks. This situation is not too different from the way nature approaches life. On the cellular or sub-cellular level, ‘simple’ molecular building blocks are usually assembled into larger, more complicated systems with a high degree of specificity. For example, in order to translate the genetic code so protein synthesis can occur, messenger RNA, ribosomes and transfer RNA must all combine into a molecular machine that synthesises proteins specific to the genetic code. The systems we design are thermodynamically stable, like nature’s, meaning they are typically the lowest energy form possible. The simple molecular building blocks we use are constructed using covalent bonds or ionic bonds. The larger assemblies, consisting of these molecular building blocks, are held together using a mixture of non-covalent interactions such as hydrogen bonding, dispersion forces and other dipole–dipole interactions. The number of interactions as well as their different arrangements lead to the level of stability and selectivity observed. This is rather like assembling a structure from Lego. Our research focuses mainly on two processes—photosynthesis and translation. Photosynthesis is the process of converting solar energy into chemical energy, leading to the synthesis of glucose and oxygen from carbon dioxide and water.

M

CH

CH2 CH3

H3C

CH2CH3 N

N Mg

N O C

H H3C CH2CH2 H

C20H39O

N CH3 H O

C

O CH3O

Figure 26.7 Structure of chlorophyll.


6CO2(g) + 6H2O(l) –—➤ C6H12O6(aq) + 6O2(g)

Why the interest? Well, our society is becoming ever more dependent on nonrenewable energy sources. The recent global increase in fuel costs and its implications to our cost of living are encouraging consumers and business groups to look for alternative methods of producing energy that are cheaper and sustainable. Effective conversion of electromagnetic radiation (as solar energy) into electrical energy may be important as a source of renewable energy. The conversion of solar energy into electrical energy also has application to a number of molecular and nano-based technologies such as molecular computing, acting as molecularscale batteries, and ‘powering’ light-driven mechanical processes on a molecular level. This rationale is bio-inspired as nature adopts these very same design principles to power photosynthesis in plants and bacteria, and has done so from very early on in the evolution process. Photosynthetic cells function through large arrays of highly colourful molecules such as chlorophylls. These natural pigments are called porphyrins. The structure of chlorophyll consists of a magnesium atom in the centre of large open ring structure (Figure 26.7). In other porphyrins the magnesium atom is replaced by iron, as in heme (part of the haemoglobin molecule), and by cobalt, as in vitamin B12. Porphyrin molecules collect absorbed visible light energy and deliver it to the primary reaction centre in the cell via energy transfer (EnT) mechanisms. Once there, the energy is converted to chemical potential via a relay of electron transfer (ET) mechanisms within the reaction centre with an efficiency greater than 99%.

bio o Our laboratories are dedicated to finding out more about the processes of ET and EnT, using simple model systems that employ building blocks similar to those used by nature. Translation is a genetic process used to convert the information stored within the genetic code in DNA into proteins. Not all

proteins are good, and those that are not may lead to disease states and in some cases the death of a cell. My group is very interested in motor neuron disease and the implications of death-inducing proteins in triggering the degeneration of nerve cells. This, however, is a story we will have to leave for another time.

Professor Steven J. Langford

Figure 26.8 We are studying the effects of alkali metal ions on the ability to harvest light and transfer energy of molecules like this, where a porphyrin is linked to C60. This may lead to better solar energy devices and a more sustainable future.

Figure 26.9 My research aims to use the same approach as plants to focus the Sun’s energy through natural pigments (porphyrins). Here we show four different views of the molecule formed when porphyrins are linked to form a cyclic macromolecule with dimensions on the nanometre scale.

Steven J. Langford received his BSc (Hons I) and PhD from the University of Sydney following secondary schooling at Liverpool Boys High School, NSW. He was awarded a Ramsay Memorial Fellowship (one is awarded annually world-wide) in 1994 to work in the area of supramolecular chemistry, an area he still researches today. In 1996 he returned to Australia as an Australia Research Council Postdoctoral Fellow and was appointed as a Lecturer in Organic Chemistry in 1998 at Monash University, Melbourne. In 2006 he was appointed as Professor of Organic Chemistry. His research and teaching contributions have been recognised through many awards. In 2006 he was awarded a prestigious Young Investigator Award by the Society of Porphyrins and Phthalocyanines (an international award for scientists under 40), a Carrick Citation for Outstanding Contributions to Student Learning, and the Centenary of Federation Tertiary Teaching Award of the Royal Australian Chemical Institute. He is currently Head, School of Chemistry and Deputy Dean for the Faculty of Science.


26.3

Potential difference ! The volt is named after Italian physicist Alessandro Volta in recognition of his contribution to our understanding of electricity and electrochemical cells.

Voltmeter Electron flow

0.41 V + H2(g)

Platinum electrode

H+(aq)

Iron electrode

Fe2+(aq)

Both cells at standard conditions

Figure 26.10 Measuring the standard half-cell potential of a Fe2+(aq)/Fe(s) half cell. From the voltmeter reading and negative polarity of the iron electrode, the E° of the half cell is 0.41 V.

_

A current flows in a galvanic cell because one half cell has a greater tendency to push electrons into the external circuit than the other half cell. Chemists say that a potential difference exists between the two half cells. The potential difference of a cell is sometimes called the emf. The potential difference of a cell is measured in volts (symbol V) and is determined using a voltmeter. The potential difference measures the amount of energy supplied by a fixed amount of charge flowing from a galvanic cell. A Daniell cell has a potential difference of about one volt. A rough indication of cell voltage can be obtained from the electrochemical series. Large potential differences are obtained when a half cell containing a strong oxidant is connected to a half cell with a strong reductant. A potential difference of 3.7 V is obtained by connecting a Mg2+/Mg half cell to a Cl2/Cl– half cell at standard conditions. The reactions in these two half cells are spaced far apart in the electrochemical series. Potential differences of cells at standard conditions can be calculated using the E° values listed in Table 26.2. These standard half-cell potentials give a numerical measure of the tendency of a half cell reaction to occur as a reduction reaction. Reduction reactions are most likely to occur in half cells with high E° values, such as F2(g)/F–(aq) (E° = 2.87 V), whereas oxidation reactions are most likely to occur in half cells with low E° values, such as Li+(aq)/Li(s) (E° = –3.05 V). It is impossible to measure the potential difference of an isolated half cell since both oxidation and reduction must take place for a potential difference to exist. E° values for half cells are measured by connecting the cells to a reference half cell. A hydrogen half cell, H+(aq)/H2(g), at standard conditions, known as the standard hydrogen electrode, is used for this purpose and its E° value is arbitrarily assigned as zero. The standard potentials of other cells are then measured as shown in Figure 26.10. The potential difference of a cell at standard conditions is the difference between the E° values of its two half cells, i.e. cell potential difference = higher half-cell E° – lower half-cell E°

For the cell composed of Ag+/Ag and Fe2+/Fe half cells as shown in Figure 26.6, the cell voltage at standard conditions can be calculated as follows: cell potential difference = higher half-cell E° – lower half-cell E° = E° (Ag+(aq)/Ag(s)) – E° (Fe2+(aq)/Fe(s)) = 0.80 – (–0.41) = 1.21 V

Different values for the cell voltage are obtained under non-standard conditions.


summary

26.3

Th The standard t d d half h lf cellll potential t ti l is i the th voltage lt a half h lf cellll generates when connected to the standard hydrogen electrode.

The Th electrochemical l t h i l series i can predict di t the th potential t ti l difference diff e generated by a galvanic cell. This is equal to the difference between the higher half cell E° and the lower half cell E°.

key question 3 Show that under standard conditions, according the electrochemical series, the Daniell cell (Figure 26.2) should have a cell potential difference of 1.10 V.

26.4

!

Predicting direct redox reactions In a galvanic cell, the higher half reaction in the electrochemical series occurs in the forward direction (as reduction) and the lower reaction occurs in reverse (as oxidation). This principle applies equally to redox reactions that occur when reactants are mixed directly. Thus, for appreciable reaction to occur, a chemical on the left of the electrochemical series must react with a chemical on the right that is lower in the series (Figure 26.11). Since an electrochemical series applies to standard conditions, predictions should be checked by experiment.

Worked example 26.4 Consider the following equations that appear in the order shown in the electrochemical series: Br2(aq) + 2e– Ni2+(aq) + 2e– Mg2+(aq) + 2e–

2Br–(aq) Ni(s) Mg(s)

If the contents of the half cells of a galvanic cell were mixed, the reactants would react directly. Energy would be released as heat rather than as electrical energy. Reactions that occur in galvanic cells or when chemicals are directly mixed are described as naturally occurring reactions, or spontaneous reactions.

Oxidants A2+

Reductants A2+ is reduced to A

reac t

A

s wit

h

B is oxidised to B2+ B2+

B

A2+ reacts with B. No reaction occurs for the other combinations: A and B, A2+ and B2+, A2+ and A, A and B2+, B2+ and B.

Figure 26.11 Oxidants only react significantly with those reductants that are lower in the electrochemical series.

Predict the effect of mixing: a Br2(aq) and Mg2+(aq) b Mg2+(aq) and Ni(s) c Ni2+(aq) and Mg(s)

Solution a Br2(aq) + 2e– 2Br–(aq) Mg2+(aq) + 2e– Mg(s) No reaction occurs because both Br2 and Mg2+ are oxidants. b Ni2+(aq) + 2e– Ni(s) Mg2+(aq) + 2e– Mg(s) No reaction because the oxidant Mg2+ is below the reductant Ni in the electrochemical series. c Ni2+(aq) + 2e– Ni(s) Mg2+(aq) + 2e– Mg(s)


A reaction occurs because the oxidant, Ni2+, is higher in the electrochemical series than the reductant, Mg. The higher equation occurs in the forward direction: Ni2+(aq) + 2e–

Ni(s)

The lower equation occurs in reverse: Mg(s)

Mg2+(aq) + 2e–

The overall reaction equation is: Ni2+(aq) + Mg(s)

Ni(s) + Mg2+(aq)

Limitations of predictions The standard half cell potentials are measured under standard conditions. These standard conditions stipulate a temperature of 25°C, one molar concentration of the solutions, and one atmosphere pressure of gases. As you would expect, the half cell potentials are likely to vary if you move away from the standard conditions. If reaction conditions remain close to the standard conditions, the standard half cell potentials will provide a guide to reactivity. However, when conditions change significantly, the order of half reactions in the electrochemical series will also change and the standard half cell potentials will not be relevant. It is also important to bear in mind that the electrochemical series gives no information about the rate at which reactions occur. The significance of the rate of reaction is seen in the reaction of hydrogen peroxide molecules (H2O2) with each other. Since hydrogen peroxide is found in the electrochemical series as an oxidant and also as a reductant at a lower position in the series, its molecules would be predicted to react with each other (to form water and oxygen gas). H2O2(aq) + 2H+(aq) + 2e– –—➤ 2H2O(l); E° = 1.78 V O2(g) + 2H+(aq) + 2e– –—➤ H2O2(aq); E° = 0.68 V

Overall reaction: 2H2O2(aq) –—➤ 2H2O(l) + O2(g)

In fact the rate of reaction is so slow that no reaction is apparent. As you may be aware, in practice solutions of hydrogen peroxide can be stored for years at room temperature, unless a catalyst such as manganese(IV) oxide is added. Hydrogen peroxide is stored in darkened bottles to further slow the rate of self-decomposition.

extension Cell potentials and non standard conditions

! Walther Nernst (1864–1941) was awarded the 1920 Nobel Prize in Chemistry.


Th lilimitations The i i off the h use off standard d d hhalflf cellll potentials i l were outlined li d above. b BBasically i ll their h i use only applies to standard conditions; concentrations of 1 M, 1 atmosphere pressure and 25°C. The German chemist Walther Nernst carried out experimental investigations to determine how variations in concentration, temperature and pressure affected cell potential. He developed a mathematical model, now known as the Nernst equation, that provides an accurate prediction of cell potential under non standard conditions. Consider the cell shown in Figure 26.12. (Notice that the ion concentrations are less than 1 M.)

extension Cell potentials and non standard conditions (continued) The Nernst Th N t equation ti ffor this thi reaction ti att 25°C 25 C can be b represented t d as: Ecell = E°Ag+ / Ag – E°Fe3+ / Fe2+ – 0.059 log

–

[Ag(s)][Fe3+] [Ag+(aq)][Fe2+(aq)]

The concentration of a solid is assigned a value of 1. The equation simplifies to: Ecell = E°Ag+ / Ag – E°Fe3+ / Fe2+ – 0.059 log = 0.80 – 0.77 – 0.059 log

1 × [Fe3+] [Ag (aq)][Fe2+(aq)] +

+

Salt bridge

–

+ Ag

Pt

0.001 0.1 × 0.1

= 0.80 – 0.77 – 0.059 log 0.1 = 0.80 – 0.77 – 0.059 × (–1) = 0.090 volt

[Fe3+] = 0.001 M [Fe2+] = 0.10 M

[Ag+] = 0.10 M

Figure 26.12 Cell for the reaction Ag+(aq) + Fe2+(aq) –—➤ Ag(s) + Fe3+(aq).

question

?

E1 The The equ equation atition for for the the Daniell cell shown in Figure 26.2 26 2 is: Zn(s) + Cu (aq) –—➤ Zn (aq) + Cu(s) 2+

2+

The Nernst equation for this reaction is: Ecell = E°Cu2+ / Cu – E°Zn2+ / Zn –

summary

Calculate the cell potential if: a [Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 1.0 M b [Zn2+(aq)] = 10–3 M and [Cu2+(aq)] = 1.0 M c [Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 10–3 M

0.059 [Zn2+(aq)][Cu(s)] log 2 [Zn(s)][Cu2+(aq)]

26.4

A spontaneous reaction will occur if the th oxidant o id nt of the conjugate redox pair higher in the electrochemical series is added to the reductant of the conjugate redox pair lower in the electrochemical series.

Predictions P di tions based b sed on the th electrochemical el ctro hemi l series seri s will ill only o ly apply under standard conditions, i.e. 1 M concentration, 1 atmosphere pressure and a temperature of 25°C. The electrochemical series provides no information about the rate at which a reaction might occur.

key question 4 The following equations appear in the order shown in the electrochemical series: Cl2(g) + 2e– Ag+(aq) + e– Zn2+(aq) + 2e–

2Cl–(aq) Ag(s) Zn(s)

i Predict whether a redox reaction would occur in the following mixtures. Assume 1 M and 1 atm.

For mixtures where you predict a reaction will occur: ii write separate half equations for the oxidation and reduction reactions iii write a balanced equation. a Cl2(g) and Zn(s) b Ag+(aq) and Ag(s) c Ag+(aq) and Zn(s) d Zn2+(aq) and Cl–(aq)


26

key terms anode cathode conjugate redox pair electrochemical series emf external circuit

galvanic cell half cell internal circuit oxidant potential difference reductant

You will need to refer to the electrochemical series on page 418 to answer some of the following questions.

Electrochemical series 5 Iron nails are placed into 1 M solutions of CuSO4, MgCl2, Pb(NO3)2 and ZnCl2. In which solutions would you expect a coating of a metal other than iron to appear on the nail? Explain your answer. 6 Use the electrochemical series to predict whether a reaction will occur in each of the following situations. Write an equation for each reaction that you predict will occur. a Copper(II) sulfate solution is stored in an aluminium container. b Sodium chloride solution is stored in a copper container. c Silver nitrate solution is stored in a zinc container. d An iron nail is placed in 1 M hydrochloric acid solution. e A plumber uses hydrochloric acid to clean copper pipes. 7 Explain the difference between the terms: a oxidant and reductant b anode and cathode c conjugate redox pair and conjugate acid–base pair d external circuit and internal circuit 8 The following equations form part of the electrochemical series. They are ranked in the order shown. Ag+(aq) + e– Pb2+(aq) + 2e– Fe2+(aq) + 2e– Zn2+(aq) + 2e– Mg2+(aq) + 2e–

Ag(s) Pb(s) Fe(s) Zn(s) Mg(s)

a Which species is the strongest oxidant and which is the weakest oxidant? b Which species is the strongest reductant and which is the weakest reductant? c Lead rods are placed in solutions of silver nitrate, iron(II) sulfate and magnesium chloride. In which solutions would you expect to see a coating of another metal form on the lead rod? Explain.


salt bridge spontaneous reaction standard laboratory conditions standard half-cell potential

d Which of the metals silver, zinc or magnesium might be coated with lead when immersed in a solution of lead(II) nitrate? 9 Use the electrochemical series to determine whether: a elemental iodine is an oxidant or reductant b calcium metal is a strong or weak reductant c the permanganate ion is a strong or weak oxidant d nickel is a better reductant than silver e Cu2+(aq) is a better oxidant than Ag+(aq) f Fe2+(aq) can act as an oxidant and a reductant 10 a Use the electrochemical series to predict what might be expected to occur if hydrogen gas were bubbled through a solution containing Fe3+ ions. b Write an equation for the predicted reaction. c When the reactants were mixed in an experiment, no reaction was observed. Suggest possible reasons for this.

Connecting the main ideas 11 A student was told that the redox pair Mn2+(aq)/Mn(s) is lower in the electrochemical series than the Fe2+(aq)/Fe(s) pair and higher than Al3+(aq)/Al(s). Explain how the student could experimentally determine the exact location of the manganese pair in the series on page 418. 12 Account for the following observations. a Bromine reacts with iodide ions in solution but does not react with chloride ions. b Hydrogen peroxide (H2O2) solution can spontaneously decompose to form water and oxygen. c Tin metal is added to solutions of tin(II) chloride to prevent oxidation of the tin(II) ions by oxygen in air. d Blocks of zinc are attached to the iron hulls of ships to reduce corrosion. 13 A student working in the laboratory spilled an iodine solution over the bench, causing a dark brown stain to form. Suggest how the student could remove the iodine stain. 14 What are the limitations that need to be considered when using the electrochemical series to predict whether or not a certain reaction will occur?

2

chapter 27

Cells and batteries

key knowledge • the construction and operation of simple galvanic primary and secondary cells • the construction and operation of fuel cells and the advantages and disadvantages of fuel cells compared to conventional energy sources

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the main features of a typical galvanic cell and explain how it produces electrical energy • describe the construction and functioning of common primary, secondary and fuel cells • explain the recharging process in common secondary cells • compare the energy efficiency of a fuel cell to the corresponding combustion process • explain how fuel cells operate, describe their advantages and limitations and give examples of their use.

27.1

Portable power A mobile phone, a notebook computer, an MP3 player and a hearing aid all depend on small portable sources of electricity: cells and batteries. In this chapter we will examine a range of common commercial galvanic cells to see how they use chemical reactions to produce electricity.

Figure 27.1 A laptop computer is a popular use of electrical energy.

chemistryy in action A case study: lithium ion rechargeable cells

Figure 27.2 Lithium ion power supply in a modern camera.

Electron flow

CoO2(s) + Li+ + e– LiCoO2(l) +

Favourites Music Photos Games

> > > >

Lithium ion cell

– Li(s)

Lithium ion cells and batteries have become the power source of choice for many portable electronic products. They are light and powerful, and unlike ordinary alkaline cells, lithium ion cells and batteries are not discarded when flat. Instead, they can be recharged and reused. Lithium-based cells and batteries are common—they power most of the latest mobile phones, notebook computers, personal organisers and digital cameras. Sony released the first commercial lithium ion cells in 1991. Lithium’s light weight and high reactivity had long been recognised as potentially useful in cells and batteries—the chemist Gilbert Lewis pioneered the concept of a lithium cell in 1912. However, while lithium’s reactivity might be ideal, it also makes practical applications difficult. Lithium reacts with many everyday materials and can react explosively with water. As a consequence, the development of a lithium-based cell using a non-aqueous electrolyte has been a significant scientific as well as technological achievement. Lithium ion cells are designed principally to overcome the safety concerns surrounding the use of a strong reactant such as metallic lithium. Lithium ion cells do not contain lithium in the metallic state; rather, the lithium is adsorbed into the lattice structure of a graphite electrode. To find out how a lithium ion cell works, let us consider what happens when it is connected into an electronic appliance such as a mobile phone. A redox reaction occurs within the cell. Electrons flow from the negative terminal of the cell, through the appliance, to the cell’s positive terminal (Figure 27.3). At the negative electrode: The oxidation of lithium generates electrons: Li(s) –—➤ Li+ + e–

Li+ + e–

Electronic appliance

Figure 27.3 Electrons flow from the negative to the positive terminal of a galvanic cell.


Once formed, Li+ ions migrate into the electrolyte. At the positive electrode: The electrode material, which is usually a metal oxide such as cobalt oxide, accepts electrons. The product combines with Li+ ions from the electrolyte and precipitates. For example: CoO2(s) + Li+ + e– –—➤ LiCoO2(s)

The equation for the overall cell reaction is formed by adding the two half equations: Li(s) + CoO2(s) –—➤ LiCoO2(s)

chemistryy in action A case study: lithium ion rechargeable cells (continued) A lithium ion cell delivers a steady 3.6 volts as it discharges. If you have used lithium ion cells, you will have noticed that they become flat suddenly. When a lithium ion cell is discharging, the products of the reaction remain available within the cell. By supplying the discharged cell with electrical energy, the electrode reactions can be reversed and the reactants regenerated.

Cathode Anode

Separator

LiCoO2(s) –—➤ Li(s) + CoO2(s)

As a cell is discharged and recharged, lithium ions migrate back and forth through the electrolyte, giving rise to the name ‘lithium ion’ cell. Although their complex internal structure makes them more expensive to manufacture, lithium ion cells have increased in popularity because of their high energy density and light weight. They also have a low rate of self-discharge and do not suffer from ‘memory’ effects like some other types of rechargeable cell. The only notable disadvantage of the lithium ion cell is its comparatively short three-year working life. Still, in many applications a lithium ion cell will outlast the appliance that it is used to operate.

! A prototype electric car, the Eliica or electric lithium ion car, reached a speed of 370 km/h. The downside is that it takes 10 hours to recharge the banks of lithium ion batteries.

Casing

Header

! Lithium ion cells are very different from lithium metal cells. Lithium metal cells have only recently become widely available. They are also high energy and light, but they are not rechargeable.

27.2

Primary cells Batteries such as dry cells, alkaline cells and button cells have one important feature in common: they are non-rechargeable. They go flat when the cell reaction reaches equilibrium, and you have to buy a replacement. Cells that cannot be recharged are called primary cells. In primary cells, the products slowly migrate away from the electrodes and are consumed by side reactions occurring in the cell.

The zinc–carbon dry cell The first mass-produced and widely used small-scale source of electrical energy was the zinc–carbon dry cell. It was an updated commercial version of a cell first developed by the French scientist Georges Leclanché in 1866. Figure 27.5 shows what you would see if you cut a dry cell open. An electrolyte composed of a moist paste of zinc chloride and ammonium chloride plays the same role as the salt bridge in the cells we discussed in Chapter 26.

Figure 27.4 When a lithium ion cell or battery is recharged, the electrical energy from the recharger forces the reverse reaction to regenerate the reactants.

Metal cap (+)

Mixture of carbon and manganese dioxide

Ammonium chloride and zinc chloride electrolyte

Cathode: carbon rod

Anode: zinc case (–)

Figure 27.5 The construction of a zinc–carbon dry cell is simple and has changed little in the last 100 years.

429 Cells and batteries

! A high performance version of the zinc–carbon dry cell has been produced using a thicker zinc case, more manganese dioxide and a zinc chloride electrolyte. Consequently the cell would last longer, produce higher currents and be less susceptible to polarisation. These ‘heavy duty’ dry cells have largely been replaced by alkaline cells.

! Removal of the products of a reaction favours formation of a greater amount of products—an application of Le Chatelier’s principle.

Metal cap (+) Cathode: outer steel case Powdered zinc Mixture of manganese dioxide and carbon Potassium hydroxide electrolyte Anode: steel or brass

Metal base (–)

Figure 27.6 Construction of an alkaline cell.

! An alkaline cell needs less electrolyte than a dry cell, allowing more reactant to be included. A typical D size alkaline cell contains about 40 g of manganese dioxide, compared with 25 g in a dry cell of equivalent size.

At the anode (–): Oxidation of the zinc case produces electrons: Zn(s) –—➤ Zn2+(aq) + 2e–

At the cathode (+): Manganese dioxide is reduced in a complicated reaction that is thought to be: 2MnO2(s) + 2NH4+(aq) + 2e– –—➤ Mn2O3(s) + 2NH3(aq) + H2O(l)

A new cell produces about 1.5 volts, but this diminishes significantly during use. To maintain a net forward reaction, the soluble reaction products must migrate away from the electrodes. During use, the build-up of products around the electrodes slows and can even stop the forward reaction. This effect, known as polarisation, is more marked during prolonged operation. If a cell is allowed to rest, the products migrate away from the electrodes and the cell ‘recovers’. Warming the cell with the heat of your hand can often increase the rate of recovery. However, once the reaction in the cell finally reaches equilibrium, the cell will be ‘flat’, and cannot be used again. You can still buy zinc–carbon cells and batteries. While some minor improvements have been made to the cell, the technology is basically the same as it was in the first dry cells over a century ago. Zinc–carbon dry cells now tend to be restricted to applications that require a cheap portable source of energy for infrequent low-drain use. Their inability to perform in high-drain, high-use applications has seen zinc–carbon cells largely replaced by alkaline cells.

Alkaline cells The alkaline cell was invented in the late 1800s, but it was not until after World War II (1939–45) that the basic idea was developed seriously. Since that time, an expanding range of appliances has appeared that require highcapacity, small power sources. With earlier types of cell unable to meet these demands, the alkaline cell appeared on the market in the late 1960s. Although the chemical reaction that occurs within an alkaline cell has much in common with the zinc–carbon dry cell, the construction is completely different (Figure 27.6). The alkaline cell is optimised for performance and longevity. At the anode (–): Zinc powder around the central metal rod is oxidised: Zn(s) –—➤ Zn2+(aq) + 2e–

Once formed, Zn2+ reacts immediately with OH– ions in the electrolyte to form zinc hydroxide. The overall reaction at the anode is therefore written as: Zn(s) + 2OH–(aq) –—➤ Zn(OH)2(s) + 2e–

At the cathode (+): Manganese dioxide is reduced: 2MnO2(s) + H2O(l) + 2e– –—➤ Mn2O3(s) + 2OH–(aq)

The technological improvements incorporated in the alkaline cell’s construction give it about five times the life of a zinc–carbon dry cell. Although alkaline cells are slightly more expensive than dry cells, they offer better value for money and are especially cost-efficient in torches, flashguns and motorised toys, where high currents are needed intermittently.


Button cells Button cells are used in very small devices such as watches, some calculators and PDAs, and remote car locks. There are two main types: silver–zinc cells and lithium cells. Both are relatively expensive because of their small size and the materials used to make them. Lithium button cells produce about 3 volts during discharge, and silver– zinc cells give an almost constant 1.6 volts. A variant of the lithium cell employing iodine as the oxidant is used to power heart pacemaker implants and is designed to last for fifteen years.

chemfact The first button cells used the reaction between zinc and mercury(II) oxide: Zn(s) + 2OH– (aq) –—➤ Zn(OH)2(s) + 2e– HgO(s) + 2H2O(l) + 2e– –—➤ Hg(l) + 2OH–(aq) Mercury and its compounds are toxic, so the disposal of these mercury cells created health and environmental safety concerns. Cells such as the silver–zinc cell have replaced mercury cells.

Worked example 27.2a The overall reaction that occurs in the silver–zinc button cell is: Zn(s) + Ag2O(s) + H2O(l) –—➤ Zn(OH)2(s) + 2Ag(s)

Write equations for the anode and cathode reactions.

Solution Oxidation occurs at the anode. Zinc is oxidised to Zn(OH)2: Zn(s) + 2OH–(aq) –—➤ Zn(OH)2(s) + 2e–

Silver oxide is reduced to silver at the cathode: Ag2O(s) + H2O(l) + 2e– –—➤ 2Ag(s) + 2OH–(aq)

The construction of a lithium button cell is shown in Figure 27.7. The most common type of lithium button cell uses manganese dioxide as the oxidant.

Metal cap (–) Lithium metal

Worked example 27.2b

Electrolyte

The following reactions occur in the lithium button cell:

Cathode: outer container of nickel or steel (+)

Li –—➤ Li+ + e– MnO2 + Li+ + e– –—➤ LiMnO2

Manganese(IV) oxide

a Which reaction occurs at the negative electrode? b Write an overall equation for the reaction occuring in the lithium cell.

Figure 27.7 Construction of a lithium button cell.

Solution a Electrons are produced at the negative electrode. The oxidation of lithium metal to lithium ions produces electrons. The reaction is Li –—➤ Li+ + e–. b The equations must be combined in such a way that there are equal numbers of electrons on both sides. In this case the equations can simply be added: Li + MnO2 + Li+ + e– –—➤ Li+ + e– + LiMnO2

This simplifies to: Li + MnO2 –—➤ LiMnO2

summary

27.1–27.2

G Galvanic l i cells ll that th t cannott be b recharged h d are known k as primary i cells. The products produced by the redox reactions in primary cells migrate away from the electrodes and are consumed by side reactions.

The Th dry d cell, ll alkaline lk li cellll andd the th silver–zinc il i andd lithium lithi button b tt cells are examples of primary cells.


key questions 1 One major disadvantage of the dry cell was that the zinc casing would begin to disintegrate if it was left operating in an appliance for several weeks, potentially causing damage to the appliance. Explain why this occurs. 2 Although very small alkaline cells can be made, they are not used in watches. Why are button cells preferred for this application? 3 Write overall equations for the cell reactions in the dry cell, alkaline cell, and lithium and silver–zinc button cells using the anode and cathode reactions given in the text. 4 Describe the energy transformations that occur when a galvanic cell discharges.

5 A galvanic cell is constructed by connecting a Cl2/Cl– half cell to a Zn2+/Zn half cell using a salt bridge at standard conditions. You will need to refer to Table 26.2 (p. 418). a Write an overall equation for the cell. b What is the polarity (positive or negative) of the zinc electrode? c Calculate the potential difference of the cell. d What happens to the concentrations of Cl– and Zn2+ ions when electric current is drawn from the cell? e Explain the purpose of the salt bridge in the cell.

27.3

Rechargeable cells and batteries ! A battery is a combination of cells connected in series. The potential difference of a battery is equal to the sum of the potential differences of the individual cells. A battery made from two 1.5 V dry cells, for example, has a potential difference of 3 V.

Rechargeable cells, such as lithium ion cells, are known as secondary cells or accumulators. To recharge a cell, the products of the reaction must be converted back into the original reactants: the cell reaction must occur in reverse. This is done by connecting the cell to a ‘charger’, a source of electrical energy, which has a potential difference greater than the potential difference of the cell. Electrical energy supplied by the charger is converted into chemical energy in the cell. In order for it to be possible to regenerate the reactants, the products formed in the cell during discharge must remain in contact with the electrodes in a convertible form. Rechargeable cells and batteries have become very popular with costconscious consumers. Although initially more expensive, rechargeable cells and batteries become cheaper each time they are recharged and re-used. Most types of rechargeable cells are able to undergo at least 500 recharges. Lithium ion and nickel–metal hydride cells and batteries are relatively new, but many of the early developments in rechargeable cells date back over a hundred years.

Car batteries

Figure 27.8 Lead–acid batteries are used to start cars, trucks and motorcycles.


Lead–acid batteries are the most widely used type of secondary cell. They are relatively cheap and reliable, provide high currents, and have a long lifetime. Each year over 300 million are manufactured around the world and there are well over ten million currently in use across Australia. Most people know the lead–acid battery simply as a car battery (Figure 27.8). It is used to start the car’s engine and to operate the car’s electrical accessories when the engine is not running. Once the engine starts, an alternator, which is run by the engine, provides electrical energy to operate the car’s electrical system. The alternator also provides electrical energy to recharge the battery (Figure 27.9).

Ignition electronics (controls the engine and switches the current to each spark plug in turn) Ignition coil (produces high voltage needed to operate the spark plugs)

Figure 27.9 Part of the electrical system of a car. The car battery supplies over 100 amps to the starter motor for several seconds. It is recharged by the alternator when the engine is running.

12 V battery Earth

Spark plug (ignites petrol in engine)

Starter motor (operated by battery to start engine)

Alternator (recharges battery; provides electricity when engine is running)

Lead–acid batteries are also used for emergency light and power systems, for small-scale energy storage and to power some electric vehicles like golf buggies and small fork-lifts. Although a modern lead–acid battery looks like a single unit, it is actually six separate cells connected together in series. In a typical car battery each cell contains three positive electrodes sandwiched between four negative electrodes. Contact between the electrodes is prevented by the presence of a porous separator. The positive electrodes consist of a lead grid packed with lead(IV) oxide (PbO2), while the negative electrodes consist of a lead grid packed with powdered lead. A solution of sulfuric acid, of about 4 M concentration, acts as the electrolyte (Figure 27.10). Each cell has a potential difference of just over 2 volts. A car battery has six of these cells connected in series, giving a total potential difference of about 12 volts. At the anodes (–): The reaction when the battery is supplying electricity is described by the equation: Pb(s) + SO42–(aq) –—➤ PbSO4(s) + 2e–

In this reaction, lead is oxidised to Pb2+ ions. These ions combine with sulfate ions from the electrolyte to form a coating of lead(II) sulfate on the electrodes (the white chemical that you sometimes see on battery terminals). At the cathodes (+): The reaction is: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e– –—➤ PbSO4(s) + 2H2O(l)

In this reaction, lead(IV) oxide is reduced to Pb2+ ions. As at the anodes, these ions combine with sulfate ions from the electrolyte to form a lead(II) sulfate coating on the electrodes. The two half equations can be combined as an overall equation: Pb(s) + PbO2(s) + 2SO42–(aq) + 4H+(aq) –—➤ 2PbSO4(s) + 2H2O(l)

The product of both electrode reactions, lead(II) sulfate, forms as a solid on the surface of the electrodes. This enables the battery to be recharged.

Porous separator Lead plates impregnated with lead(IV) oxide (+)

Sulfuric acid Lead plates (–)

Figure 27.10 The construction of a car battery.

! The first working lead–acid cell was invented by a French scientist, Gaston Planté, in 1859. Full-scale production of the cell had begun by the late 1880s. Like the dry cell, the basic technology of the lead–acid cell is over 100 years old.

chemfact During the recharging of cheaper models of lead–acid battery, some of the water in the electrolyte is electrolysed to form hydrogen gas and oxygen gas. While this is normal, it means the electrolyte must be periodically topped up with distilled water. Explosions have occurred when gases produced from charging batteries have been ignited by a spark. In some cases serious injuries have resulted.

433 43 Cells and batteries

To recharge the battery, the electrode reactions are reversed.The alternator, with a potential difference of about 14 V, is used to force electrons into the battery’s negative terminal and draw them out at the positive terminal, in effect driving the spontaneous reaction backwards. The recharging process converts electrical energy into chemical energy. The overall equation for the reaction as the battery recharges is: 2PbSO4(s) + 2H2O(l) –—➤ Pb(s) + PbO2(s) + 2SO42–(aq) + 4H+(aq)

Nickel-based cells and batteries

Figure 27.11 A hydrometer is used to measure the density of a battery’s electrolyte, and hence the charge on the battery. As a battery discharges, sulfuric acid is consumed and water is formed, so the density of the electrolyte decreases.

! Nickel–cadmium rechargeable batteries are still preferred in some applications because of their superior recharge life—over 1000 recharge cycles.

The use of nickel compounds in rechargeable cells dates back to at least 1899, when a Swedish engineer, Waldemar Jungner, invented the nickel–cadmium cell. The nickel–cadmium cell first went into mass production in the early 1950s. Although still widely used today, it is steadily being overtaken by its better performing and more environmentally friendly relative the nickel– metal hydride cell. The two types of cell are similar in many respects. Both consist of a coiled sandwich of anode, porous separator and cathode immersed in a concentrated potassium hydroxide electrolyte. When these cells are connected to an electronic appliance, the following half reactions are observed. At the anode (–): In a nickel–cadmium cell, oxidation of cadmium generates electrons and the Cd2+ ions react immediately with OH– ions in the electrolyte and precipitate as solid cadmium hydroxide. Cd(s) + 2OH–(aq) –—➤ Cd(OH)2(s) + 2e–

In a nickel–metal hydride cell, the negative electrode is made of a special hydrogen-absorbing metal alloy (MH). At the surface of the electrode, absorbed hydrogen reacts with OH– ions from the electrolyte, forming water and releasing electrons. MH(s) + OH–(aq) –—➤ M(s) + H2O(l) + e–

At the cathode (+): Electrons are accepted by nickel ions in the nickel hydroxide and they are reduced from an oxidation state of +3 to +2. The half equation is: NiOOH(s) + H2O(l) + e– –—➤ Ni(OH)2(s) + OH–(aq)

Figure 27.12 A variety of sizes of nickel–cadmium and nickel–metal hydride cells and batteries are sold. When flat, they are recharged using a battery charger.

chemfact Electrodes made of nanocrystals of nickel and metal hydride are being investigated as a way of improving battery performance. The large surface area obtained using nanoparticles will mean that the battery will need to be recharged less often and will last longer.


The electrode reactions are fully reversible and this enables the reactants to be regenerated when the cell is recharged. Most brands of nickel–cadmium cells are capable of 1000 discharge–recharge cycles. Because cadmium is toxic, old nickel–cadmium cells have to be recycled properly rather than just thrown away. The newer nickel–metal hydride cells will do around 500 cycles. The development of nickel–metal hydride batteries reduced the weight and extended the operating times of mobile phones, lap-top computers and many other portable electronic devices.

extension A clever combination U il recently, Until l llarge amounts off electricity l i i could ld not bbe storedd efficiently. Most of the electricity we use to power our modern society is created as we use it in big coal-fired power stations. This is referred to as base load power because it is required continually. The intermittent nature of solar and wind power limits their use in providing base load power. However, a new type of a room-sized battery is poised to transform the way baseload power is supplied. A 2-volt cell the size of a soft-drink can could allow intermittent power sources to be used for baseload power. The NaS cell, which uses sodium and sulfur electrodes, is efficient, has a high energy density and a very long life. When assembled into large batteries it is ideal for utility-grade power storage. The cell consists of a sodium anode, sulfur cathode and solid alumina electrolyte. During discharge, sodium ions are created and pass through the electrolyte to form sodium sulfide at the cathode. Recharging the cell regenerates the reactants. Power companies in both the USA and Japan have successfully coupled NaS batteries to wind farms to create baseload power sources.

questions

Figure 27.13 An artist’s impression of a battery that stores excess electrical energy generated by wind turbines or solar panels.

?

E1 Write Writite equ equations atitions for for the reactions that occur at the anode and cathode of a NaS cell as it discharges. discharges E2 Write an overall equation for the cell reaction during: a discharge b recharge. E3 To operate efficiently, the cells require a temperature above 270°C. Why do you think this is necessary? E4 Provided the cells are discharged and recharged every day, they remain at working temperatures. Suggest the reason for this. E5 Suggest a meaning for the term ‘energy density’. Why is energy density an important factor in the construction of most types of cells and batteries? E6 Explain how the use of NaS batteries enables an intermittent source such as wind power to be used for baseload power.

summary

27.3

G Galvanic l ani cells c ll that th t can be b recharged rech rg d are known as secondary seco dary cells. Secondary cells are recharged by reversing the reaction that occurs when the cell discharges. This can occur because the products produced when the cell is discharged remain in contact with the electrodes.

An A external source of electricity lectri ity is i used to force electrons el ctrons into the cell’s negative terminal and draw out electrons at the cell’s positive terminal, reversing the half reactions that occur when the cell is discharged. The lead–acid accumulator, lithium ion cell, nickel–cadmium and nickel–metal hydride cells are examples of secondary cells.


key questions 6 The pH of the electrolyte in a lead–acid battery varies according to the state of charge. Describe how the pH would change as a battery discharged and was recharged. 7 A lead–acid battery uses 50 g of lead to provide the electrical energy needed to start a typical car. Calculate the mass of lead oxide consumed at the positive electrode and the total mass of lead sulfate produced in the battery while starting the car.

8 A mechanic wishes to recharge a car battery using a battery charger. a Which battery terminals should the positive and negative terminals of the battery charger be connected to? b What would be the effect on the battery if the charger were connected incorrectly?

27.4

Fuel cells

chemfact Some types of breathalyser use fuel cell technology to measure the blood alcohol content. In this fuel cell, ethanol is oxidised and the electricity generated provides an indication of blood alcohol level.

chemfact Fuel cells are found in nature. The electric eel, which grows to more than one metre in length, has a battery made of specialised fuel cells within its body. The fuel for each cell is the food consumed by the eel and oxygen is the oxidant. Each cell generates about 1.5 V, yielding a total voltage from the head of the eel to its tail of more than 300 V!

The major limitation of the cells that we have examined so far is that they contain relatively small amounts of reactants. When the reaction reaches equilibrium, the cell must be discarded or recharged. Cells can be constructed in which the reactants are supplied continuously, allowing constant production of electrical energy. Such devices are called fuel cells. Fuel cells transform chemical energy directly into electrical energy, enabling efficient use to be made of the energy released by spontaneous redox reactions. Energy losses such as those that occur in a coal-fired power station are avoided. They are up to 80% efficient, compared with 30–40% for thermal power stations. In addition, modern designs for fuel cells employ the waste heat that they produce to make steam. This steam can be used for heating or to operate a turbine, thus raising the efficiency of the cells even more. The basic principles behind the operation of a fuel cell were discovered in 1839 and the first working example was created by Sir William Grove in the 1840s. However, it was not until the 1950s that fuel cells were used for small-scale power production. Fuel cells were the main on-board power supply units during the Apollo program that put humans on the Moon (Figure 27.15). In fact, an explosion in a fuel cell was responsible for the failure of the Apollo 13 mission. Fuel cells are also used as the main on-board power systems in the Space Shuttle. The fuel cell used in the Apollo program used pure oxygen and hydrogen gas as reactants (Figure 27.14). Potassium hydroxide solution was used as the electrolyte and the cell operated at about 250°C. Because of the use of the potassium hydroxide electrolyte, it is commonly referred to as an alkaline fuel cell. Alkaline fuel cells are still used in the space program today. At the anode (–): Hydrogen gas reacts with hydroxide ions from the electrolyte. H2(g) + 2OH–(aq) –—➤ 2H2O(l) + 2e–

At the cathode (+): O2(g) + 2H2O(l) + 4e– –—➤ 4OH–(aq)

The overall equation for the reaction is: 2H2(g) + O2(g) –—➤ 2H2O(l)


–

+

Electrolyte

Oxygen gas inlet

Hydrogen gas inlet

Porous cathode

Porous anode

Water outlet

Figure 27.14 An alkaline fuel cell.

Each cell produces about one volt. Higher voltages are obtained by connecting a number of fuel cells in series to form a battery or fuel cell stack. The only by-products are water and heat. The nature of the electrodes is crucial to successful operation of the cell as they function as catalysts for the reaction, and the size of the current depends on their surface area. Advances in fuel-cell technology have led to the development of the acid fuel cell. This cell uses concentrated phosphoric acid as the electrolyte and air as the source of oxygen; it uses hydrogen but does not require the hydrogen to be pure. The overall reaction is the same as in the alkaline fuel cell, but the specific electrode reactions are different. Since the first 11 MW experimental unit was installed at Goi, Japan, in 1991, over 400 phosphoric acid fuel cells have been put into use around the world, including one in Sydney. Scientists are striving to reduce the overall costs of cells and to improve the current that can be drawn by increasing the rate of reaction at the electrodes. Variations on the basic concept have been developed and tested, with several types reaching limited trials. Some of these are summarised in Table 27.1. Fuel cells are named according to the electrolyte and catalysts used in them. The various electrolytes operate most efficiently over a set temperature range, and the nature of the electrolyte influences the use of the fuel cell.

Figure 27.15 This fuel cell was used to supply electricity for the Apollo space program some 40 years ago. Fuel cells are still used in the space program.

chemfact The electrode reactions in an acid fuel cell are: anode: H2(g) –—➤ 2H+(aq) + 2e– cathode: O2(g) + 4H+(aq) + 4e– –—➤ 2H2O(l)

Transport Most major vehicle manufacturers are investigating the use of fuel cells as an alternative to the internal combustion engine. In Perth, buses using hydrogen-powered fuel cells have been in operation for a number of years. This use of fuel cells improves fuel efficiency and reduces greenhouse gases and other emissions as well as our reliance on oil.


TABLE 27.1 Fuel cell comparisons Fuel cell

Electrolyte

Efficiency (%)

Alkaline fuel cell

Potassium hydroxide

60

Operating temperature (°C) 90–100

Application

Equations

Space vehicles Military

Anode H2 +2OH– –—➤ 2H2O + 2e– Cathode O2 + 2H2O + 4e– –—➤ 4OH–

Methanol fuel cell

Molten carbonate fuel cell

Phosphoric acid fuel cell

Polymer membrane

40

Molten lithium, sodium or potassium carbonates

50–60

Liquid phosphoric acid

40–45

50–100

600–1000

Mobile phones Lap-top computers

Anode CH3OH + H2O –—➤ CO2 + 6H+ + 6e–

Electricity utility

Anode H2 + CO32– –—➤ H2O + CO2 + 2e–

Cathode 4H+ + O2 + 4e– –—➤ 2H2O

Cathode O2 + 2CO2 + 4e– –—➤ 2CO32– 175–200

Electricity utility Transportation

Anode H2 –—➤ 2H+ + 2e– Cathode O2 + 4H+ + 4e– –—➤ 2H2O

Proton exchange membrane fuel cell

Solid polyperfluoro sulfonic acid

60

60–100

Portable power Transportation

Anode H2 –—➤ 2H+ + 2e– Cathode O2 + 4H+ + 4e– –—➤ 2H2O

Solid oxide fuel cell

Solid zirconium oxide

50–60

600–1000

Electricity utility

Anode H2 + O2– –—➤ H2O + 2e– or 4O2– + CH4 –—➤ CO2 + 2H2O + 8e– or 3O2– + CH3OH –—➤ CO2 + 2H2O + 6e– Cathode O2 + 4e– –—➤ 2O2–

Source: Adapted from Fuel Cells: Green Power. Los Alamos National Laboratory

Electricity generation Stationary fuel cell systems are used to generate electricity for domestic, commercial and industrial purposes. A 100 kW fuel cell will provide electricity for domestic purposes, while units having a capacity of several megawatt provide enough electricity for some industries or a large commercial building.

Small appliances A methanol-powered fuel cell has been developed for use with portable electronic equipment such as mobile telephones and lap-top computers. To ‘recharge’ the appliance, you just swap a small replaceable fuel pack.


Advantages and disadvantages of fuel cells Fuel cells have a number of advantages over other energy sources. Fuel cells convert chemical energy directly to electrical energy. This is more efficient than the series of energy conversions: chemical energy –—➤ heat energy –—➤ mechanical energy –—➤ electrical energy

that takes place in power stations that burn fossil fuels. Hydrogen fuel cells produce water as a by-product. No greenhouse gases such as carbon dioxide are released into the atmosphere by the reactions taking place in these types of fuel cell. Fuel cells will generate electricity as long as the fuel is supplied. In conventional batteries electricity production ceases once the reaction reaches equilibrium. The batteries then need to be recharged or replaced. Fuel cells can use a variety of fuels. Electricity can be generated on-site and users are not reliant on connection to an electricity grid. There are currently some disadvantages associated with the use of fuel cells. The fuel cells require a constant fuel supply. They are expensive. Fuel cell technology is still developing. Fuel cells are only manufactured in limited numbers and as yet there are no economies of scale associated with high volume production. Some types of fuel cell use expensive electrolytes and catalysts, leading to high purchase price and maintenance costs. Fuel cells generate a direct current (DC), electrical appliances used in the home and in industry rely on an alternating current (AC). An inverter is required to change DC to AC at the appropriate voltage. The effectiveness of some fuel cells, particularly those operating at lower temperatures, is affected by impurities in the hydrogen fuel. The use of fuel cells in transport is limited by the lack of facilities for hydrogen storage and distribution. The use of hydrogen fuel cells to power transport will require an extensive network of hydrogen filling stations before it can become widespread.

summary

! Hydrogen fuel cells produce water as a waste product. Regenerative fuel cell systems are under development that use electricity from solar or wind power to electrolyse this water waste into hydrogen and oxygen, which are then used as fuels in the cell to generate more electricity.

27.4

PPrimary i andd secondary d cells ll contain t i a limited li it d amountt off reactants. These cells stops producing electricity when equilibrium is reached. Fuel cells have a continuous supply of reactants and will produce electricity as long as the reactants are supplied.

Fuel F l cells ll produce d electricity l t i it more effi fficiently i tl th than coal-fi l firedd power stations. Hydrogen fuel cells produce only water as a by-product. Current fuel cell technology is expensive.

key questions 9 Why is large-scale production of electricity using fuel cells suggested as part of a solution to the greenhouse problem?

10 The use of hydrogen-powered fuel cells in public transport is being considered in many cities around the world. What are the advantages and disadvantages of using fuel cells for this purpose?


the cutting edge The page A header Solid goesoxide here fuel cells

key terms

by Karl Föger here is growing demand for energy across the globe. Yet the existing supplies may not cope with this demand, and new generation systems are needed that also cut carbon emissions. Ceramic Fuel Cells Limited (CFCL) is developing products based on proprietary solid oxide fuel cell technology, to provide low-emission electricity. CFCL was formed in 1992 and is based in Melbourne, with offices in Germany and the United Kingdom, and all its technology has been developed in house and is wholly owned. CFCL’s first products will be installed in homes and other buildings to generate electricity and heat from widely available natural gas. Later products can use LPG and renewable fuels such as biogas and ethanol. Fuel cells generate electricity and heat through an electrochemical reaction.

T

Figure 27.16 Ceramic Fuel Cell Limited’s fuel cell.

Solid oxide fuel cells operate at high temperatures ( >700°C) and very high efficiencies. Figure 27.16 shows a diagram of CFCL’s fuel cell. A single fuel cell is a low-voltage, high-current device (0.8 V at >20 A) and cells need to be linked to form the fuel cell stack. Figure 27.17 shows the stack and its building blocks. CFCL has achieved an electrical efficiency of 60% from its 2 kW power generator—far higher than any other microgeneration technology. The product recovers the heat to provide hot water or space heating, for total system efficiency of 85%. There are negligible emissions of oxides of nitrogen and sulfur (NOx, SOx ) and no noise or vibrations. This makes them ideal to be installed as ‘mini power stations’ in homes and other buildings.

Steam

Pre-reforming– removal of ethane, propane and butane

Sulfur removal

Natural gas

Recoverable high-temperature exhaust heat

Methane and steam at approx. 800°C Internal reforming of methane into hydrogen and carbon monoxide

Methane fuel input

Reaction with oxygen ions generates electricity and forms water vapour and carbon dioxide exhaust.

Anode

–

O2–

O2–

O2– Electrolyte

2–

O

O2–

O2–

2–

O2–

O

O2–

Electrical load

O2– +

Cathode

Air input


Oxygen

Oxygen ions

Air exhaust

bio o

Cell

Stack

Sealing frame

Interconnector

Figure 27.17 Ceramic Fuel Cell Limited’s stack and its building blocks.

With very high electrical efficiency (see Figure 27.18), CFCL’s products can reduce carbon emissions by up to two-thirds—and cut water use by up to 95%—compared to existing brown coal power stations. In 2009, CFCL developed a modular generator called Blue GenTM, which can produce up to 2 kW of electricity and up to 200 litres of hot water a day. A Blue Gen unit could be installed at your house, probably next to where the gas hot water unit now sits. BlueGen plugs into the existing natural gas, water and power grid connections. The unit is about the size of a dishwasher. It converts the gas into power which supplies your house and heat for hot

water. BlueGen runs constantly, generating more than twice the power needed by the average home during the year. The excess can be sold back to the power grid. During peak times of the day, the home would import power from the grid; the rest of the time the home exports power (similar to a solar photovoltaic system, except that BlueGen uses gas, makes a lot more power and its output can be controlled and turned up and down remotely over the Internet). Ceramic Fuel Cells has sold early BlueGen units in Australia, Germany, Japan and The Netherlands. As the BlueGen product matures, the company will expand into other large global markets.

70

1200

60

1000

50

800

40

600

30

400

20

200

10

0 0

500 1000 1500 AC export power (W)

Figure 27.18 Energy efficiency of Blue Gen.

0 2000

Net electrical efficiency (%)

Thermal power (W)

Thermal and electrical performance 1400

Karl Föger Karl Föger is Chief Technology Officer and one of the initiators of solid oxide fuel cell technology in Australia. He has an international reputation in research and development in the energy and environmental fields. He has a PhD in Physical Chemistry and is a Fellow of the Royal Australia Chemical Institute and an Adjunct Professor at RMIT University in Melbourne. Dr Föger held various research and management roles at CSIRO, culminating in his appointment as a Chief Research Scientist, He has been involved with the company since its inception in 1992 and has published a number of papers on solid oxide fuel cell technology.

Figure 27.19 Ceramic Fuel Cell Limited’s modular generator Blue Gen.


27

key terms accumulator alkaline cell battery button cell dry cell fuel cell

galvanic cell lead–acid battery lithium ion cell nickel–cadmium cell nickel–metal hydride cell non-rechargeable

You will need to refer to the electrochemical series on page 418 to answer some of the following questions.

Principles of galvanic cells 11 When constructing a galvanic cell in the laboratory, two separate containers are generally used for the half cells. However, in commercial cells such as the dry cell, all the reactants are placed in the one container. a How are commercial cells constructed so that the use of two containers is avoided? b Design a cell that could be constructed in one container in the laboratory and which uses the same cell reaction as in the Daniell cell (p. 415). 12 Explain the difference between the terms: a positive electrode and negative electrode b cell and battery c primary cell and secondary cell d discharge and recharge 13 What feature enables secondary cells such as lithium ion cells to be recharged? 14 Explain what happens to the oxidant, reductant, anode and cathode when a secondary cell is switched from discharge to recharge.

Commercial cells and batteries 15 Both dry and alkaline cells use powdered graphite mixed with a manganese dioxide oxidant. From your knowledge of the properties of graphite, suggest what its role might be in these cells. 16 Describe an application where each of the following would be used: a a nickel–metal hydride cell b a lead–acid accumulator c a lithium button cell d a dry cell e an alkaline cell 17 A manual car with a flat battery can often be started by pushing, but this is not done with an automatic car. Find out why a manual car can be started in this way but an automatic cannot. 18 Mains electricity costs about 5 cents per MJ of energy, or less, depending on the tariff. The cost of the same amount


primary cell rechargeable cell secondary cell silver–zinc cell zinc–carbon dry cell

of electrical energy produced by a cell is far more—about $1300 for a dry cell and even more for a button cell. Why are people prepared to use cells and pay such relatively high prices for electricity?

Fuel cells 19 A car has been powered using an experimental ten-cell aluminium–air battery. Each of the ten cells contains aluminium plates and sodium hydroxide solution, and air is fed into the battery. New aluminium plates and water must be added regularly to the battery. a Write half equations for the anode and cathode reactions in the battery. b Draw a diagram to show how each cell might be constructed. c Give two reasons why this battery weighs less than a similar lead–acid battery. d Why can this battery be described as a fuel cell? 20 Preliminary roadside breath tests are often performed by measuring the voltage produced when a motorist breathes into a small fuel cell. The cell operates like a hydrogen–oxygen fuel cell, except that ethanol replaces hydrogen. Carbon dioxide is one of the products formed when the ethanol is oxidised at the anode. a Predict the anode, cathode and overall reactions in the cell. (Assume the cell electrolyte is acidic.) b Would you expect the fuel cell to generate a voltage if natural gas were blown into the breath inlet? Explain your answer. 21 An experimental fuel cell that uses methanol as the fuel has the half equations: CH3OH(g) + H2O(l) + 6OH–(aq) –—➤ CO2(g) + 6H2O(l) + 6e– O2(g) + 2H2O(1) + 4e– –—➤ 4OH–(aq)

a b c d

Write the equation for the overall cell reaction. Which reaction occurs at the positive electrode of the cell? Suggest a suitable electrolyte for the cell. When the cell begins to produce electricity, the pH of the electrolyte near the cathode increases and eventually reaches a constant value. Explain why this occurs.

e Electricity could be obtained from thermal energy produced by combustion of methanol. What is the main advantage of using a fuel cell to produce electricity?

Connecting the main ideas 22 Many of the ‘alkaline cells’ on the market contain zinc electrodes in contact with an electrolyte containing hydroxide ions. The half-cell reaction might be represented as: Zn(s) + 4OH–(aq) –—➤ Zn(OH)42–(aq) + 2e–

of electron drift through the external circuit is shown in Figure 27.21. The electrode reactions are: O2 + 4e– –—➤ 2O2– CH4 + 4O2– –—➤ CO2 + 2H2O + 8e–

Waste gases (CO2, H2O) Electrons

To investigate if the standard electrode potential (E °) of these half cells is the same as that of a zinc electrode in contact with a zinc nitrate electrolyte, for which the electrode reaction would be Zn(s) –—➤ Zn2+(aq) + 2e–

Electrolyte O2–

a student is provided with the two half cells shown in Figure 27.20, as well as a Cu2+/Cu half cell. Zn

Zn

Depleted air

Cu Anode Fuel, CH4

Cathode Air (oxidant)

Figure 27.21 Methane fuel cell.

Zn(OH)42–(aq)

Zn2+(aq)

Cu2+(aq)

Figure 27.20 The three half cells

a Write a net equation for the reaction in a galvanic cell in which the ‘alkaline zinc half cell’ is connected to the Cu2+/Cu half cell. b Carefully explain how the student could use the half cells that were provided to determine whether the two different half cells containing zinc had the same E° value. Include fully labelled diagrams with your answer and explain how the results could be interpreted. 23 In the quest for power plants with lower emissions of carbon dioxide, a Victorian company is developing solid oxide fuel cells (SOFC). Also known as ceramic fuel cells, the units have a ceramic electrolyte through which oxide ions, O2–, can move. On the top and bottom of the electrolyte is a thin layer of electrode material. Fuel cells are designed so that the fuel and air are directed separately onto the appropriate electrode. The units are arranged in stacks so that high voltages can be generated. The diagram represents a fuel cell that is using methane gas, CH4, as the energy source. The direction

a Write an overall equation for the reaction that occurs in this SOFC. b On the diagram, clearly label the positive and negative electrodes. c Explain the claim that this method of generating electricity will reduce carbon dioxide emissions, by comparison with a power plant that burns natural gas (containing methane) and uses the heat released to boil water and then generate electricity. d Why is it necessary to aim for reduced emissions of carbon dioxide from power-generating facilities? 24 The current generation of hybrid cars such as the Honda Insight and the Toyota Prius use nickel–metal hydride batteries. Nickel–metal hydride technology has replaced lead–acid batteries previously used in many developmental hybrid and electric cars. Yet the ‘holy grail’ for both hybrid and electric vehicles remains lithium ion technology. a What aspects of lead–acid battery technology have resulted in their replacement in hybrid cars with nickel–metal hydride batteries? b What features of lithium ion batteries make them an attractive option for future hybrid and electric vehicles? c What is currently the main impediment to the use of lithium ion batteries in hybrid and electric vehicles?


2

chapter 28

Electrolysis: driving chemical reactions by electricity

key knowledge • the construction and operation of simple electrolytic cells: comparison of electrolytic cells • application of Faraday’s laws in electrochemistry

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • describe the main features and chemical reactions of a typical electroplating cell • use the electrochemical series to predict reactions in electrolytic cells, and understand the limitations of these predictions • apply Faraday’s laws to calculate electric current, electric charge, amounts of substance or time involved in the production of chemicals in typical electrolytic cells • describe the use of electrolysis in the industrial production of chemicals such as aluminium, chlorine, copper, sodium and sodium hydroxide • list and explain the factors that influence the construction of the commercial electrolytic cells used to produce these substances.

28.1 Electrolytic cell

Chemical energy from electrical energy

Electrical energy

Chemical energy Galvanic cell

In Chapters 26 and 27 we saw that spontaneous redox reactions can be used as a source of electrical energy. In this chapter we will see that the passage of electrical energy from a power supply, such as a battery, through a conducting liquid causes redox reactions to occur. This process is called electrolysis and has a variety of applications. During electrolysis, electrical energy is converted into chemical energy. The reactions that occur in electrolytic cells are essentially the opposite to those occurring in galvanic cells (Figure 28.1). Reactions in electrolytic cells would not normally happen without the application of electrical energy, and so they are called non-spontaneous reactions. Chemicals formed by electrolysis are often difficult to obtain by other means. Many useful materials are manufactured using this process. Important applications of electrolysis include: plating of a thin film of metal on surfaces of other metals to improve appearance or prevent corrosion (Figure 28.2) extraction of reactive metals such as sodium and aluminium from their ores industrial production of sodium hydroxide, copper, chlorine and hydrogen recharging of car batteries and other rechargeable cells, such as lithium batteries increasing the thickness of the surface oxide layer on aluminium metal to improve its resistance to corrosion.

Figure 28.1 Energy conversions in electrolytic and galvanic cells.

Figure 28.2 Tin cans are used as packaging for a wide range of products. The tin coating on the can prevents it from corroding.

28.2

Electroplating

+

We tend to take the existence of tin cans for granted. They are used extensively for packaging food items as varied as soups, baked beans, peaches and fish. Although we call them tin cans, they are mainly composed of steel, with a thin surface coating of tin only a fraction of a millimetre thick that has been applied by electrolysis. Tin corrodes very slowly and prevents contact between the iron, moisture and air; it therefore prevents the iron rusting. The deposition of a layer of metal on the surface of another metal by electrolysis is called electroplating. Electroplating is performed in an electrolytic cell such as the simplified one used for tin plating shown in Figure 28.3. The object to be plated is connected by a wire to the negative terminal of a power supply. This object is regarded as the negative electrode in the cell. It is immersed in a solution, such as tin nitrate solution, which contains ions of the metal that is to form the plating. This solution is referred to as the electrolyte. In tin plating, an electrode of tin metal is connected to the positive terminal of the power supply. When the cell is in operation the power supply acts as an ‘electron pump’, pushing electrons onto the negative electrode and removing electrons from the positive electrode.

Power supply

–

Electron flow +

–

Tin rod

Sn2+

NO3–

Object being plated

Tin nitrate solution (electrolyte)

Figure 28.3 A cell used for electroplating tin.

445 Electrolysis: driving chemical reactions by electricity

! Romans used tin plating to decorate objects made from bronze and copper. The tin was applied by rubbing the object with a piece of the metal. Techniques for plating steel cans by electrolysis were developed commercially in the 1930s.

e–

Tin rod

Tin ions accept electrons and become tin metal

e–

e–

chemfact A cathode is defined as an electrode at which reduction occurs. Note that in electrolytic cells the cathode is the negative electrode, whereas we saw in Chapter 26 that the cathode is positive in galvanic cells.

Sn Sn2+ e–

Object being plated

Sn2+ Sn

Sn Sn2+

Sn2+(aq) + 2e–

Sn

Sn(s)

Sn(s)

Sn2+(aq) + 2e–

chemfact An anode is defined as an electrode at which oxidation occurs. In electrolytic cells the anode is positive, whereas in galvanic cells the anode is the negative electrode.

Figure 28.4 A representation of the reaction at the cathode in the process of tin plating.

Figure 28.5 A representation of the reaction at the anode in the process of tin plating.

At the negative electrode: The negative terminal of the power supply pushes electrons along the wire to the electrode. Tin ions are attracted to the electrode, accept electrons, and are converted to tin metal: Sn2+(aq) + 2e– –—➤ Sn(s)

A coating of tin forms on the object. Since reduction of the tin ions, Sn2+, has occurred, the negative electrode is acting as a cathode. The process is shown in Figure 28.4. At the positive electrode: The positive terminal of the power supply withdraws electrons from the tin electrode. A reaction that releases electrons must occur. Tin metal slowly dissolves as Sn2+ ions are formed: Sn(s) –—➤ Sn2+(aq) + 2e–

This reaction replaces the Sn2+ ions in solution that were consumed by the reaction at the negative electrode. The overall concentration of tin ions therefore remains constant. Since an oxidation reaction is involved, the positive electrode is acting as an anode. The process at this electrode is shown in Figure 28.5. The excess of positive charge caused by the production of Sn2+ ions at the anode will be balanced by movement of NO3– ions in the electrolyte to that region and by movement of Sn2+ ions away from it. At the cathode, loss of positive charge causes cations to migrate towards it and anions away from it. You can see that there is an internal electric current through the cell electrolyte involving movement of cations towards the cathode and anions towards the anode (Figure 28.3).


extension Tin plating in practice

Figure 28.6 An industrial tin plating cell. Before entering the cell, the steel sheet is cleaned to remove grease and surface deposits of iron oxide. The tin plate produced from the cell is coiled into large rolls before being transported to the can manufacturers.

The cell that we have described for tin plating is simpler than that used in practice. Our simple cell would produce a tin coating that adhered poorly and was unattractive because of irregularities in its thickness. In all likelihood it would look like a black sludge. The precise voltage, current, temperature, electrode positions, and the concentration and identity of the electrolyte, can be critical in achieving a smooth, tightly bonded coating that prevents oxygen and water from reaching the steel. The conditions that are most suitable for industrial electroplating are largely determined by trial and error. In practice the tin coating is applied to the steel sheet before the can is made, using a series of large cells similar to the one shown in Figure 28.6. The iron sheet is continuously plated at speeds of up to 600 m min–1 using more than 50 000 litres of electrolyte. Until recently, the most popular electrolyte used by the plating industry incorporated a strong organic acid, phenol sulfonic acid, to allow the use of high currents and therefore provide rapid rates of plating. This acid is made from phenol, and both chemicals are hazardous. Following a major leak of phenol sulfonic acid from an American electroplating plant in the 1980s, causing substantial environmental damage to a local river, chemical companies have been investigating alternative, more environmentally friendly electrolytes. To be accepted by the industry, a new electrolyte needs to work in existing plating cells. Inorganic acids, such as H2SO4, HNO3 and HCl, are too corrosive to equipment and suffer from other drawbacks. After extensive research, companies have developed and patented several electrolytes, including one based on methane sulfonic acid (Figure 28.7). While its highly acidic nature means it is still hazardous, it is much less likely to cause environmental harm than phenol sulfonic acid. It is increasingly being used in industrial tin-plating cells.

questions

O O

S

OH

CH3

Figure 28.7 Methane sulfonic acid is one of the electrolytes developed for tin plating to address the requirement for a less environmentally hazardous process.

?

E1 Use Use the the IInternet nternet t t or another reference to find find out why phenol (and (and, if possible possible, phenol sulfonic acid) is regarded as a hazardous material. E2 Before the steel plate enters the tin plating cell it is cleaned thoroughly. Why is this cleaning process essential? E3 Why does the presence of a strong acid in the electrolyte allow higher rates of tin plating than a weak acid?


Other examples of electroplating

Figure 28.8 Chromium plating prevents rusting and gives objects a shiny appearance.

Iron is widely used because it is cheap and strong. Objects made from iron are often electroplated to improve their resistance to corrosion (Figure 28.8). Electroplating can also be used to make an object more attractive. If you browse through a jeweller’s shop you will see many shiny ‘silver’ articles such as bracelets, necklaces, cutlery, trays and jugs. The main metals present in many of these articles will be less expensive metals such as copper, zinc and nickel; the silver is a surface coating that has been applied by electrolysis. The cell used for silver plating is similar to the one used for tin plating. Silver is used instead of tin as the anode and the electrolyte contains silver ions. Chromium is decorative, non-tarnishing and hard-wearing. Chromium plating is used extensively on items such as bicycle handlebars, wheel rims and taps. Chromium plating is performed using an acidic electrolyte containing chromium ions. Unlike the previous metal-plating processes, chromium metal is not used for the anode in a chromium-plating cell because it dissolves in the acid. An unreactive anode made of a lead–tin alloy is used instead and chromium(VI) oxide, CrO3, is continually added to the electrolyte to replace the chromium ions that have been consumed.

28.1–28.2

summary

El Electrolysis t l i involves i l the th conversion i off electrical l t i l energy into i t chemical energy. Reactions in electrolytic cells are nonspontaneous.

In I an electrolytic l t l ti cellll the th anode d is i positive iti andd the th cathode th d is i negative. Oxidation occurs at the anode and reduction occurs at the cathode.

key questions 1 A student wishes to copper-plate her iron locker key. a Design a suitable electroplating cell. Identify the anode and cathode, electrode polarities and the electrolyte to be used.

b If the student connects the cell to the wrong terminals of the power supply, what happens to the locker key? 2 Draw a diagram to show the migration of ions in a nickelplating cell containing nickel sulfate solution.

Power supply +

Ammeter (for current measurement)

–

28.3 Variable resistor

A

Faraday’s laws The first law of electrolysis

Electroplating cell

Figure 28.9 Experimental apparatus used to investigate electroplating.

448 8 Supplying Su u and using energy

If you wished to electroplate an object, you might ask yourself the following questions: How can I determine how much metal is being plated? How long should I leave the object being plated in the electroplating cell? What size electric current should be used? Suppose we constructed the electroplating cell shown in Figure 28.9. A variable resistor is included in the circuit to allow the current entering the cell to be kept constant. We can use this cell to study the relationship between the mass of metal formed at the cathode and the quantity of charge that passes through the cell. While the mass of metal formed is easily measured by weighing the cathode, electric charge is determined indirectly.

charge (coulombs) = current (amps) × time (seconds) Q=I×t

Consider the experimental results obtained from a silver-plating cell given in Table 28.1. The results are hardly surprising: the more charge that passes through the cell, the more metal is formed at the cathode. This relationship is shown graphically in Figure 28.10. Even if we perform this experiment in different silver-plating cells and vary conditions such as temperature, voltage, electrolyte concentration and current, the same graph is obtained as long as silver is the only product at the cathode.

Mass of silver deposited at cathode (g)

Electric charge, symbol Q, is measured using the unit coulomb. The electric charge passing through a cell may be calculated from measurements of the current, I, through the cell and the time, t, for which the current flows. The relationship is:

2.5 2.0

1.5 1.0 0.5

500

Time (s) 200 400 600 800 1000

Charge (C) Q = I × t 400 800 1200 1600 2000

! Michael Faraday (1791–1867) was an English chemist and physicist who, apart from working out the laws of electrochemistry, discovered the connection between magnetism and electricity and devised the first electric motor.

We can repeat the previous experiment using electroplating cells for metals other than silver. Results from experiments with cells using electrolytes containing Cu2+, Sn2+, Pb2+, Cr3+ and Ag+ ions are shown graphically in Figure 28.11. While the data for each metal obey Faraday’s first law, at first inspection there seems little relationship between the graphs for each metal. However, if we graph the amount of each metal, measured in moles, against charge (Figure 28.12), the results are more interesting. All data lie along just three lines. We see that one mole of silver is produced by about 96 500 coulombs of charge, whereas one mole of copper, tin or lead is produced by 193 000 (which is 2 × 96 500) coulombs. An underlying pattern is now evident. In order to deposit one mole of silver from a solution of Ag+ ions, just one mole of electrons is required:

Mass of metal deposited at cathode (g)

These experimental results may be summarised by saying that the mass of metal produced at the cathode is directly proportional to the quantity of electricity passed through the cell. This relationship is known as Faraday’s first law of electrolysis, after Michael Faraday who discovered it. It may be written symbolically as:

The second law of electrolysis

2000

Figure 28.10 The variation with charge of mass of silver formed at the cathode passing through a silverplating cell.

Mass of silver formed (g) 0.45 0.90 1.34 1.79 2.24

m∝Q

1500

Charge (C)

TABLE 28.1 Experimental data from a silver-plating cell Current (A) 2.0 2.0 2.0 2.0 2.0

1000

2.5 Silver Lead

2.0 1.5

Tin 1.0 Copper 0.5

Chromium

500

1000 1500 Charge (C)

2000

Figure 28.11 Variation with charge of the mass of metal formed at the cathode for different metalplating cells.

Ag+(aq) + e– –—➤ Ag(s)

The charge on one mole of electrons must be 96 500 C. The charge on one mole of electrons is known as a faraday, and given the symbol F. We can therefore write: 1 F = 96 500 C mol–1

chemfact A more accurate value for one faraday, obtained by experiment, is 96 487 C mol–1. The value of 96 500 C mol–1 will be sufficiently accurate for our purposes.


The charge on a given number of moles of electrons, n(e–), may be calculated readily. Two moles of electrons, for example, would have a charge of 193 000 C. We can express this mathematically by: Q = n(e–) × F or

Q = n(e–) × 96 500

The electrode reactions when copper, tin or lead are electroplated are:

Amount of metal deposited at cathode (mol)

Cu2+(aq) + 2e– –—➤ Cu(s) Sn2+(aq) + 2e– –—➤ Sn(s) Pb2+(aq) + 2e– –—➤ Pb(s)

For one mole of each of these metals to be deposited, two moles of electrons are required. Two faradays of charge (193 000 C) must therefore be passed through the cell. To deposit one mole of chromium, three faradays of charge (289 500 C) are required, since the equation for the electrode reaction is:

4 Silver 3

2

Copper, tin, lead

1 Chromium

96 500

193 000 =2× 96 500

289 500 =3× 96 500

Charge (C)

Figure 28.12 Variation with charge of amount of metal (mol) formed at the cathode for different metalplating cells.

Cr3+(aq) + 3e– –—➤ Cr(s)

This is just what we observe in our graph (Figure 28.12). The three different lines on the graph arise because of the different charges on the metal ions. These observations may be summarised by saying that in order to produce one mole of a metal, one, two, three, or another whole number of moles of electrons must be consumed. This is a modernised statement of Faraday’s second law of electrolysis. Although we have developed Faraday’s laws by studying the deposition of metal at the cathode in electroplating cells, the laws apply just as well to the production or consumption of any substance at the electrodes of either electrolytic cells or galvanic cells. Calculations are based on the relationships: Q = I × t or Q = n(e–) × F

The following examples show how such calculations are performed.

Worked example 28.3a chemfact Faraday’s laws may be restated more generally as follows: The mass of any substance deposited, evolved or dissolved at an electrode in an electrochemical process is directly proportional to the amount of electricity passed through the cell. For one mole of a substance to be deposited, evolved, or dissolved at an electrode, the passage of one, two, three, or another whole number of mole of electrons is required.

A silver-plating cell operates with a steady current of 30.0 A for 20.0 minutes. What mass of silver is plated on the object at the cathode?

Solution Quantity of charge passed through the cell: Q =I×t = 30.0 A × (20.00 × 60) s = 36 000 C (Note that time must be expressed in seconds.) Amount, in mol, of electrons passed through the cell: Q n(e–) = F 36 000 C = 0.3731 mol = 96 500 C mol–1 Since the cathode reaction is: Ag+(aq) + e– –—➤ Ag(s), 1 mol of electrons will deposit 1 mol of silver metal, so n(Ag) = n(e–) = 0.3731 mol Mass of silver plated at the cathode: m(Ag) = n × M = 0.3731 mol × 107.9 g mol–1 = 40.25 g The mass of silver plated on the cathode is 40.3 g.

Worked example 28.3b How long would it take to deposit 50.0 g of copper at the cathode of a copper-plating cell operating at a current of 8.00 A?


Solution Amount of copper to be deposited: m n(Cu) = M 50.0 g = 0.787 mol = 63.5 g mol–1 Since the cathode reaction is Cu2+(aq) + 2e– –—➤ Cu(s), 2 mol of electrons will deposit 1 mol of copper metal, so: n(e–) = 2 × n(Cu) = 2 × 0.787 = 1.57 mol Quantity of charge passed through the cell: Q = n(e–) × F = 1.57 mol × 96 500 C mol–1 = 152 000 C Time required: Q t= I 152 000 C = 19 000 s = 8.00 A = 5.27 hours

summary

28.3

Th The mass off a chemical h i l produced d d or consumedd att th the electrode l t d of an electrolytic cell is directly proportional to the quantity of electricity passed through the cell.

For F one mole l off a substance bt to t be b produced d d or consumedd att an electrode, the passage of a whole number of moles of electrons is required.

key questions 3 Calculate the quantity of electric charge, in coulomb, represented by: a a current of 6.00 A flowing for 25.0 minutes b a current of 25 mA flowing for 2.0 days 4 The electrolyte in an electrolytic cell used to chromium-plate bicycle frames contains Cr3+ ions. If the cell operates at 30.0 A for 25.0 minutes, calculate: a the quantity of charge that passes through the cell

b the amount, in mol, of electrons that pass through the cell c the mass of metal deposited on the frames 5 A 0.300 g mass of silver is to be used to plate a sporting trophy. How long should the trophy be left in a silver-plating cell using a current of 6.00 A? 6 A solution used to gold-plate bathroom fittings contains Au3+ ions. What current would be required to plate 5.00 g of gold onto a tap in 30.0 minutes?

28.4

Competition at electrodes Knowing that an oxidation reaction occurs at the anode and a reduction reaction occurs at the cathode, it is often possible to predict the electrode reactions in a particular electrolytic cell. Usually there are several chemicals present at each electrode and you must decide which of the possible reactions have the greatest tendency to occur. Even the material used for the electrodes may react. The electrochemical series (p. 418) is a useful guide to likely reactions.

chemfact During electrolysis: • The highest reaction in the electrochemical series that can occur in the forward direction is likely to occur at the cathode (that is, the strongest oxidant will usually react at the cathode). • The lowest reaction in the series that can occur in reverse is likely to occur at the anode (that is, the strongest reductant will usually react at the anode).


Worked example 28.4 +

Power supply

–

Predict the products of electrolysis of 1 M nickel sulfate solution using copper electrodes.

Solution Copper anode

Copper cathode

Copper corrodes

Nickel deposit

Blue Cu2+ ions form

Nickel sulfate solution

Figure 28.13 Electrolysis of nickel sulfate solution using copper electrodes.

The chemicals present at the electrodes are Ni2+(aq), SO42–(aq), and H2O(l). The electrodes are made of copper. By referring to the electrochemical series we can see that a number of reactions need to be considered as the cell contains two oxidants, Ni2+ and H2O, and two reductants, Cu and H2O. O2 + 4H+ + 4e– –—➤ 2H2O 1.23 V Cu2+ + 2e– –—➤ Cu 0.34 V Ni2+ + 2e– –—➤ Ni –0.23 V 2H2O + 2e– –—➤ H2 + 2OH– –0.83 V At the cathode (–) we see that the reduction reactions that could occur are: Ni2+(aq) + 2e– –—➤ Ni(s) 2H2O(l) + 2e– –—➤ H2(g) + 2OH–(aq) We predict that nickel metal will plate onto the electrode, since the highest reduction reaction in the electrochemical series is favoured; that is, the strongest oxidant, Ni2+, will be reduced. At the anode (+), oxidation reactions that could occur are: 2H2O(l) –—➤ O2(g) + 4H+(aq) + 4e– Cu(s) –—➤ Cu2+(aq) + 2e– The copper anode is therefore likely to dissolve as the oxidation reaction lowest in the series is favoured; that is, the strongest reductant, Cu, will be oxidised. Our predictions are confirmed by the results of the experiment shown in Figure 28.13. The overall equation is: Ni2+(aq) + 2e– –—➤ Ni(s) Cu(s) –—➤ Cu2+(aq) + 2e– Ni2+(aq) + Cu(s) –—➤ Ni(s) + Cu2+(aq) Note that experiments are the only sure way to find out what products form in a particular instance. In practice, reactions are affected by factors such as electrolyte concentration, gas pressures, current, voltage, and even the nature of the electrodes. Some electrodes permit particular reactions to occur more readily than others.

summary

28.4

Th The electrochemical l t h i l series i can be b usedd to t predict di t electrode l t d reactions in electrolytic cells where more than one reaction is possible.

The Th strongest t t oxidant id t will ill usually ll reactt att the th cathode th d andd the th strongest reductant will usually react at the anode.

key question 7 Predict the products at each electrode during electrolysis (using unreactive electrodes) of the following 1 M solutions. The nitrate ion is not involved in any of the reactions. a copper(II) bromide


b c d e

sodium iodide lead(II) nitrate zinc chloride aluminium nitrate

28.5

Useful chemicals from electrolysis Chemical industries tend to avoid using electrolysis for the manufacture of chemicals because of the relatively high cost of electrical energy. However, electrolysis brings about some chemical reactions that would not ordinarily occur. By converting electrical energy into chemical energy, we can perform non-spontaneous reactions in which reactive chemicals are products, rather than reactants. As examples of this, we will look at how sodium, aluminium and chlorine and are produced and how copper is purified. All four elements are in great demand by manufacturing industry. To be commercially viable, these electrolysis processes must: be located in places with access to cheap, abundant and reliable electrical power use electrolytes that are cheap and readily available where possible, use electrodes made from inexpensive materials, such as carbon or iron be sited in locations that allow relatively inexpensive transport of materials and products operate continuously—it should not be necessary to stop the cells in order to add electrolyte or remove products involve low electrolyte temperatures to reduce heating costs and minimise wear on equipment consider the environmental impact of the materials used in the construction and operation of the electrolytic cell.

Commercial cells using molten electrolytes Electrolysis is a useful method for the production of highly reactive elements such as the metals sodium, potassium, calcium and aluminium. The electrolytes used in the production of these metals are molten salts, rather than solutions of the salt in water. The electrochemical series shows that water is a stronger oxidant than the metal ions Na+, K+, Ca2+ and Al3+. If water were present in the cells, it would be reduced preferentially at the cathode and no metal would form. In comparison to electrolytic processes that use aqueous electrolytes, the use of a molten electrolyte usually involves greater energy expenditure and greater wear on the components of the cell. In many cells, once the cell becomes operational it is not necessary to heat the electrolyte because the passage of electric current through the electrolyte keeps it molten.

Sodium

VCE Chemistry students should understand the principles of operation of electrolytic cells. You will not be expected to know details of the construction of individual industrial cells.

chemfact

Sodium metal is a starting material for the manufacture of the bleach sodium peroxide and for the production of a range of organic chemicals, including dyes. The metal is also used in liquid form in some nuclear reactors to transfer heat from the reactor core. The heat energy carried by the sodium is then used to boil water for subsequent production of electrical energy in steam turbines. Sodium is extracted from molten sodium chloride by electrolysis. Before examining the details of the cell used for the industrial production of sodium,

Sodium is one of the most reactive metals. Few substances are capable of reducing Na+ ions to the metal. Because of this, sodium was not isolated until 1807 when Humphry Davy electrolysed molten sodium hydroxide. Davy was also the first to isolate potassium, calcium, barium, magnesium and strontium, using electrolysis.


+

Power supply

–

Electron flow

Cathode

Anode

consider the simple cell shown in Figure 28.14. The electrodes are made of an unreactive conducting material such as graphite or platinum. At the cathode (–): The power supply pushes electrons towards this electrode. Positive sodium ions accept electrons and become sodium atoms: Na+(l) + e– –—➤ Na(l)

Na+

Sodium is solid at normal temperatures but is liquid at the temperatures required to melt sodium chloride. It is less dense than molten sodium chloride and floats to the top of the cell. At the anode (+): Electrons are withdrawn from this electrode by the power supply. Since the electrode is made of unreactive material, the electrode itself will not participate in an oxidation reaction to supply electrons. Instead, Cl– ions in the electrolyte give up electrons and form chlorine atoms. These atoms quickly form molecules of Cl2 and bubbles of chlorine gas form at the electrode. The equation for the process is:

Sodium metal forms

Cl–

Chlorine gas forms

Molten sodium chloride (electrolyte)

Figure 28.14 Electrolysis of molten salt.

2Cl–(l) –—➤ Cl2(g) + 2e–

An overall equation may be written for the cell reaction by adding the half equations: 2(Na+(l) + e– –—➤ Na(l)) 2Cl–(l) –—➤ Cl2(g) + 2e– 2Na+(l) + 2Cl–(l) –—➤ 2Na(l) + Cl2(g)

! Heat energy is produced when an electric current passes through a resistance. Electric radiators and the heating elements in electric stoves are applications of this principle. In the Downs cell, the electrolyte acts as the resistance.

This reaction is the reverse of the reaction that occurs spontaneously between sodium and chlorine to form sodium chloride. Electrical energy from the power supply has caused a non-spontaneous reaction to occur. Since Cl2 is a strong oxidant and Na is a strong reductant, there must be no contact between them; otherwise they will re-form sodium chloride and the products from the electrolysis will be lost. This will certainly occur in the simple cell shown in Figure 28.14. In order to produce sodium and chlorine commercially, a modified cell, known as the Downs cell, has been developed to minimise contact between the two products (Figure 28.15). A screen is used to prevent contact between the chlorine gas formed at the anode and the sodium formed at the cathode.

Chlorine gas

!

Sodium chloride added

Up to 40 cells may be connected in series, each operating at about 7 V at currents of 30 000 A.

Molten sodium

– Cylindrical iron cathode

Figure 28.15 The Downs cell for the production of sodium.

Molten sodium chloride/ calcium chloride mixture

Carbon anode

Iron mesh screen +


Aluminium

Figure 28.16 Boats made from aluminium are light, strong and resistant to corrosion.

The low density and high strength of aluminium makes it a particularly useful metal for many applications, including cooking foil, drink cans, car engines, gutters, caravans, aeroplanes, boats and window frames (Figure 28.16). The properties of pure aluminium metal are often modified to suit these applications by alloying the metal with other elements such as copper, magnesium and silicon. Aluminium is much more reactive than you might imagine considering how widely it is used. By referring to the electrochemical series on p. 418, you can see that it is only slightly less reactive than magnesium. However, aluminium reacts rapidly with oxygen in the air to form a thin, unbroken layer of aluminium oxide that prevents further reaction between the metal and oxygen. As a result of this oxide layer, aluminium resists further corrosion and other chemical attack. The first samples of aluminium were prepared by direct reaction of aluminium chloride with metals that are stronger reductants than aluminium: potassium or sodium. Today, aluminium metal is obtained from aluminium oxide (alumina, Al2O3) by electrolysis. Pure alumina is extracted from the mineral bauxite, which is found in large deposits in Weipa in northern Queensland, Gove in the Northern Territory and the Darling Ranges in Western Australia. Alumina melts at 2050°C, which inhibits its use as the electrolyte. However, it is soluble in molten cryolite, a chemical with the formula Na3AlF6. By dissolving alumina in cryolite, electrolysis can be performed at 950–1000°C. Both the anodes and cathodes of the electrolytic cell used for the extraction of aluminium metal are made from carbon (Figure 28.17). Many cells are connected in series, each operating at about 5 V using currents of up to 350 000 A (Figure 28.18). The heat released by the passage of the current through the electrolyte provides energy to keep the electrolyte molten. At the cathode (–): Electrons from the power supply convert Al3+ ions from the alumina into aluminium metal:

chemfact Such was the difficulty in extracting aluminium metal that Napoleon III proudly displayed a small bar of the metal with his crown jewels in 1855.

! Charles Hall, an American, developed the electrolytic process in 1886 at the age of 23. The same process was discovered independently by Paul Héroult, a Frenchman, just a few weeks later.

Al3+(in cryolite) + 3e– –—➤ Al(l)

The aluminium, which is molten at the temperatures used, sinks to the bottom of the cell and is periodically removed by siphoning. At the anode (+): Electrons are removed from the oxide ions of the alumina: 2O2–(in cryolite) –—➤ O2(g) + 4e–


! Australia produces about 60 million tonnes of bauxite a year and has one of the largest identified reserves of the ore. At the current rate of extraction these reserves will last 90 years. Only a small fraction is converted to aluminium in Australia. Most of the ore is converted to alumina and shipped overseas.

Reaction of oxygen with the carbon in the electrodes yields carbon dioxide gas. The electrode reaction may therefore be written as: C(s) + 2O2–(in cryolite) –—➤ CO2(g) + 4e–

The carbon anodes must therefore be replaced at regular intervals. The overall equation for the extraction of aluminium is: 2Al2O3(in cryolite) + 3C(s) –—➤ 4Al(l) + 3CO2(g) Alumina in

Carbon dioxide

Alumina dissolved in molten cryolite

Figure 28.17 The Hall–Héroult cell used to extract aluminium from alumina.

+

Molten aluminium –

+

Molten aluminium is collected by vacuum siphon

Carbon anode

Carbon cathode

Figure 28.18 A row of electrolytic cells for aluminium extraction.

Anodising A di i

Figure 28.19 The case of this popular portable multimedia player is made from anodised aluminium.


We have seen that the oxide layer on the surface of aluminium metal restricts further reaction of the metal with other chemicals. The corrosion resistance and hardness of this layer can be further increased by a process known as anodising. In this process, the oxide layer is made about 1000 times thicker by connecting the metal to the anode in an electrolytic cell with sulfuric acid as the electrolyte. The thick oxide film that forms is easily dyed to give an attractive appearance. Coloured aluminium utensils such as saucepans and cake tins, window frames and the cases of electronic goods are manufactured this way (Figure 28.19).

Commercial cells using aqueous electrolytes Compared with cells using molten electrolytes such as used to produce sodium and aluminium, electrolytic cells using aqueous electrolytes tend to use less energy and have lower operating costs. We will look at two examples of these cells: the membrane cell used for the production of chlorine and sodium hydroxide, and the electrorefining cell used for purifying copper metal.

Chlorine and sodium hydroxide

chemfact

Since chlorine is a strong oxidant, relatively few compounds are capable of oxidising Cl– ion to Cl2. Electrolysis therefore provides a convenient method for the industrial manufacture of chlorine. Chlorine ranks about tenth in order of world chemical production. Most of the chlorine manufactured around the world is used in the production of other chemicals, such as polyvinyl chloride (PVC), pesticides, and chemicals for paper bleaching and water treatment (Figure 28.20). Although chlorine is produced in the Downs cell during the extraction of sodium, most of the world’s supply of chlorine is produced by electrolysis of concentrated salt solution (brine). Sodium hydroxide, which ranks about eighth in the order of chemical production, is also obtained using this process. You might find sodium hydroxide (caustic soda) in your home: it is used in drain cleaners, paint strippers and oven cleaners. Since it is highly corrosive, it must be treated with caution; gloves must be worn when handling it, and it must be kept out of the reach of children. It is used in a diverse range of industries, including the manufacture of soap, paper and aluminium. Before looking at the electrolytic cell used by industry, first consider the simple cell shown in Figure 28.21. At the cathode (–): The power supply pushes electrons towards this electrode, resulting in a reduction reaction. The reaction that will occur will be the one that has the greatest natural tendency to do so. The electrode itself will not participate in the reaction as it is made of unreactive material. The chemicals present at the cathode are H2O(l), Na+(aq) and Cl–(aq). Using the electrochemical series (p. 418), you can see that there are two reduction reactions that might occur:

Chlorine is a green, poisonous gas. Chemicals used in domestic swimming pools are often labelled as chlorine, but such labels are misleading. These chemicals are actually compounds of chlorine such as calcium hypochlorite or sodium hypochlorite (Ca(OCl)2 or NaOCl).

Figure 28.20 A water treatment plant. Chlorine can be used to reduce the number of bacteria in drinking water.

+

2H2O(l) + 2e– –—➤ H2(g) + 2OH–(aq) Na+(aq) + e– –—➤ Na(s)

Reactions higher in the series involve stronger oxidants and have the greater tendency to occur as reduction reactions. We would therefore predict that the first reaction will be observed. As predicted, the products at the cathode of this cell are found by experiment to be hydrogen gas and hydroxide ions. At the anode (+): The power supply causes electrons to be withdrawn from this electrode, so an oxidation reaction occurs. The chemicals present at the cathode are H2O(l), Na+(aq) and Cl–(aq). Using the electrochemical series, you can see that two oxidation reactions are possible: 2Cl–(aq) –—➤ Cl2(g) + 2e– 2H2O(l) –—➤ O2(g) + 4H+(aq) + 4e–

Reactions lower in the series involve stronger reductants and have the greater tendency to occur as oxidation reactions. Although the above

Anode

Power supply

–

Cathode

Hydrogen gas forms Chlorine gas forms

Sodium chloride solution

Figure 28.21 Electrolysis of concentrated salt solution using unreactive electrodes.


! Salt-water chlorinators used in some domestic swimming pools sterilise water using this process. Salt is added to the pool and a small electric current through the water generates chlorine. Periodic additions of acid may be required to neutralise the hydroxide ions that are also formed.

chemfact Other industrial cells used to manufacture chlorine and sodium hydroxide include the diaphragm cell. Instead of a membrane, this cell uses a porous asbestos barrier and the solution obtained from the cathode compartment is contaminated by unreacted sodium chloride. Membrane cells do not use hazardous asbestos, are more efficient and use less electricity, and are therefore cheaper to operate.

! Cells are connected in series and operate at voltages of 3.5 to 5 V using currents from 1000 to 80 000 A. The manufacture of 1 tonne of sodium hydroxide uses 2200 kWh of electrical energy.

Figure 28.22 A membrane cell. The membrane is selective in the ions that can pass through it. Chloride and hydroxide ions are unable to move through the membrane.

reactions are located very close to each other in the series, we might expect the second reaction to be observed. In fact, it is possible for either reaction to occur when sodium chloride solution is electrolysed. Recall that the electrochemical series was determined for reactions at standard conditions (concentrations of 1 M, gas pressures of 1 atm and at 25°C). Variations in conditions and even the nature of the electrodes can cause one reaction to be favoured over another. In practice, at very low concentrations of sodium chloride the products at the anode of this cell are oxygen and H+ ions, whereas at high concentrations chlorine is formed. At intermediate concentrations both reactions can occur simultaneously. So, for the electrolysis of concentrated sodium chloride solution the electrode reactions are: anode (+): 2Cl–(aq) –—➤ Cl2(g) + 2e– cathode (–): 2H2O(l) + 2e– –—➤ H2(g) + 2OH–(aq) The overall reaction is: 2H2O(l) + 2Cl–(aq) –—➤ H2(g) + 2OH–(aq) + Cl2(g)

or 2NaCl(aq) + 2H2O(l) –—➤ Cl2(g) + H2(g) + 2NaOH(aq)

Like all electrolysis reactions, the reaction when brine is electrolysed is nonspontaneous and the products formed can react if they come in contact with each other. Industrial cells are carefully designed to avoid this. The membrane cell (Figure 28.22) is a modern industrial cell designed to prevent contact between chlorine and sodium hydroxide. A specially developed polymer sheet, or cation exchange membrane, prevents the mixing of the chlorine and hydrogen gas formed and allows only Na+ ions to move from one electrode chamber to the other.

Chlorine

+

–

Hydrogen

Brine in Na+

+

Power supply

–

Cl–

Impure copper anode

Pure copper cathode

OH– Pure sodium hydroxide solution

Spent brine out Semipermeable plastic membrane

Cu2+

Pure copper forms

Sulfuric acid ‘Mud’: contains silver and gold

Figure 28.23 Purifying copper by electrorefining.


Electrorefining of copper Copper metal is extracted from its ores by a process called smelting. Australia’s largest mining and copper smelters are at Olympic Dam in South Australia and Mount Isa in Queensland. The process involves several stages in which the copper ore is heated strongly in air to produce impure copper metal. The metal that is produced in this process is cast into blocks and left to cool. As hot gases escape, its surface becomes blistered and it is commonly known as ‘blister copper’. It contains about 2% of impurities such as sulfur, iron, antimony, silver and gold. Blister copper is purified by electrolysis. Sheets of blister copper are placed in a large tank of sulfuric acid and thin sheets of pure copper are positioned between them. An external power source is connected so that

the blister copper acts as a positive electrode (anode) and the pure copper acts as a negative electrode (cathode), as in Figure 28.23. At the anode (+): Electrons are drawn away from the blister copper anode to the positive terminal of the power source. Copper and impurities such as nickel and zinc that are more reactive (stronger reductants) than copper are oxidised and enter the solution as ions: Cu(s) –—➤ Cu2+(aq) + 2e– Ni(s) –—➤ Ni2+(aq) + 2e– Zn(s) –—➤ Zn2+(aq) + 2e–

Impurities less reactive than copper, such as silver, gold and platinum, are not oxidised and simply fall from the anode, collecting at the bottom of the tank. This anode residue is sometimes called ‘anode mud’. The precious metals are later recovered from the valuable mud. At the cathode (–): At the pure copper cathode, metal is deposited as electrons from the power source are accepted by metal ions from solution. Since copper ions are the best electron acceptors (strongest oxidants) present in solution, copper is the only metal formed. The Ni2+ and Zn2+ ions stay in solution.

Figure 28.24 Plates of freshly-deposited copper are removed from an electrolysis cell.

Cu2+(s) + 2e– –—➤ Cu(s)

Copper of 99.99% purity is obtained this way. Electrorefining uses a lot of electrical energy, so a refinery must be located close to a source of abundant, cheap power. Because of this, copper produced in the smelter at Mount Isa is shipped to Townsville for refining. Australia has the third largest reserves of copper in the world and each year extracts about 900 000 tonnes of metal (10% of world production).

summary

28.5

El Electrolysis t l i iis usedd commercially i ll ffor electroplating l t l ti to t preventt corrosion and enhance the appearance of metals. Electrolysis is also employed in the production of a range of useful chemicals, such as chlorine, copper, sodium, sodium hydroxide and aluminium.

The Th electrolytic l t l ti cells ll usedd in i industry i d t are designed d i d to t preventt mixing of the reaction products. They use electrodes and electrolytes that are as inexpensive as possible and minimise energy use.

key question 8 From the industrial electrolytic cells described in the text, select one that uses a molten electrolyte and another that uses an aqueous electrolyte. a Write the half reactions that occur at each electrode. b What materials are used for the anode and cathode in each cell? c Explain the reasons for the choice of the electrode materials.

d Why is an aqueous electrolyte more energy efficient than a molten electrolyte? e Explain why aqueous solutions are not employed in the cell using a molten electrolyte. f What factors may be important in deciding where to site an industrial cell of each type?


28

key terms anodising coulomb Downs cell electrolysis

electrolyte electrolytic cell electroplating faraday

Faraday’s laws Hall–Héroult cell membrane cell

Faraday’s laws 9 a Explain what is meant by a faraday of charge. b State the number of faradays of charge needed to produce 1 mole of: i silver atoms from silver nitrate solution ii zinc atoms from zinc nitrate solution iii chlorine molecules from molten potassium chloride iv hydrogen molecules from water 10 A car battery is charged at a current of 2.0 A for 9.00 hours. Calculate the mass of lead deposited at the cathode if the electrode reaction is represented by the equation:

17

18

PbSO4(s) + 2e– –—➤ Pb(s) + SO42–(aq)

11 Sodium chloride solution is electrolysed in a diaphragm cell to manufacture chlorine. The cell operates at 100 000 A. For one day’s operation, calculate: a the volume of chlorine produced, measured at SLC b the mass of sodium chloride consumed 12 Iron is plated with tin in an electrolytic cell containing K2Sn(OH)6 as the electrolyte. A cell operates for 5.00 hours at a current of 25.0 A. a What is the charge on the tin ions in the electrolyte? b How many faradays of charge are required to produce 1.00 mole of tin? c Calculate the mass of tin deposited during this period. 13 A charge of 0.400 faradays was passed through 1.00 L of 1.00 M copper(II) sulfate solution using carbon electrodes. a Write equations for the reactions at each electrode. b Calculate the concentration of the copper ions in solution after electrolysis. 14 Three electrolytic cells containing silver nitrate solution, copper(II) sulfate solution and chromium(III) sulfate solution, respectively, were connected in series so that the same amount of electric charge passed through each cell. Metal was deposited at the cathode of each cell. If 10.0 g of silver was obtained from one cell, what mass of metal would be obtained from each of the other cells? 15 Electrolysis of concentrated sodium chloride solution produced 15.0 L of chlorine gas, measured at SLC. What mass of silver would be produced if the same amount of electricity were passed through a silver-plating cell? 16 A nickel teapot, with a surface area of 900 cm2, is to be silver plated.


19

20

non-spontaneous reaction

a Which electrode should the teapot be connected to? b What is the polarity of this electrode? c If the plating is to be 0.005 00 cm thick and the density of silver is 10.5 g cm–3, how long should the pot be put in a silver-plating cell with a current of 0.500 A? Electrolysis of a molten ionic compound with a current of 0.50 A for 30.0 minutes yielded 0.700 g of a metallic element at the cathode. If the element has a relative atomic mass of 150, calculate the charge on the metal ions. A student wishes to copper-plate a nickel medallion and sets up an electrolytic cell using 250 mL of 1.00 M copper(II) sulfate solution and a large copper anode. The student runs a current of 10.0 A through the cell for 20 minutes. a What mass of copper will be plated onto the medallion? b What will be the concentration of copper(II) sulfate remaining in solution? The student replaces the copper anode with an inert graphite electrode and runs the same current through the cell for a further 20 minutes. c What gas will be produced at the anode? d What will be the concentration of copper(II) sulfate remaining in solution? A number of important metals are produced by electrolysis of molten salts. In one such process a current of 25 000 A produces 272 kg of metal in 24 hours. If the cation in the salt has a 2+ charge, what is the metal? Two electrolytic cells are connected in series as shown in Figure 28.25. A current of 5.00 A flows for 15.0 minutes. I = 5.00 A +

A C

–

A

B

C Fe

1.0 M Al2(SO4)3 solution

Figure 28.25 Electrolytic cells in series.

Cu

D Fe

1.0 M H2SO4 solution

a Calculate the charge flowing through each cell. b Write half equations for the reactions occurring at each electrode (A–D) when the current commences to flow. c Calculate the change in mass of electrode C after 15 minutes. d Calculate the volume of gas, measured at SLC, that is formed at electrode B after 15.0 minutes. e As the current flows through the cells, how does the reaction at electrode D change?

Electrolysis 21 For electrolytic and galvanic cells, compare: a the polarity of the anode and cathode b the direction of electron flow c the energy transformation occurring in the cells d the tendency of the cell reaction to occur spontaneously 22 Sodium and chlorine are manufactured by passing direct current through molten sodium chloride. a Why would alternating current be unsuitable? b Why is it necessary for the electrolyte to be molten? 23 Using the electrochemical series, complete the table by predicting the initial products of each of the following electrolysis experiments. Experiment a

b

c

d

e

Electrolyte

Electrodes

Molten potassium iodide Copper(II) sulfate solution Potassium bromide solution A mixture of copper(II) nitrate and nickel(II) nitrate solutions Lithium fluoride

Platinum

Cathode Anode reaction reaction

Platinum

Copper

Carbon

Platinum

24 Early attempts to produce aluminium by electrolysis of aqueous solutions of aluminium compounds were unsuccessful. Use the electrochemical series to explain why. 25 A student attempts to electroplate magnesium using a cell made of a magnesium anode, iron cathode and magnesium nitrate solution as electrolyte. The student is dismayed to discover that no magnesium has plated onto the cathode,

despite careful checks of all the electrical connections in the electrolytic cell. a Why didn’t magnesium plate onto the cathode? b What reaction occurred at the iron cathode?

Industrial applications 26 Sodium and chlorine are produced by electrolysis of molten sodium chloride in the Downs cell. A typical Downs cell runs at 600°C, 7 V and with a current of 30 kA. In a 24-hour period: a what mass of sodium metal will be produced? b what volume of chlorine gas (at SLC) will be produced? 27 Account for the fact that: a most chlorine is produced by electrolysis of sodium chloride solution, rather than by electrolysis of the molten salt b fluorine was first isolated from fluorine compounds by electrolysis c the concentration of Sn2+ ions in a tin-plating cell (Figure 28.3) remains constant as the cell operates d although calcium chloride is present in the electrolyte in the Downs cell, calcium metal is not formed at the cathode 28 Magnesium and zinc are both extracted from their compounds by electrolysis. Sketch and label cells that you consider would be suitable for commercial extraction of these metals. 29 An electrolytic cell for the extraction of aluminium operates at a current of 150 000 A. In order to produce 1.00 tonne (106 g), calculate: a how long the cell must operate b the mass of carbon consumed at the anodes c the volume of carbon dioxide gas produced, measured at STP 30 Commercial operation of an aluminium smelter depends upon the availability of cheap electric power. a Calculate the electric charge required to produce 1.00 tonne of aluminium from alumina (Al2O3) by electrolysis. b If the cell operates at 5.00 V, calculate the energy needed to make 1.00 tonne of aluminium. (1 volt = 1 joule per coulomb.) c Calculate the cost of the electricity required to make 1.00 tonne of aluminium if the smelter purchased electricity at the domestic rate of 12 cents per kilowatt hour. (1 kilowatt hour = 3 600 000 J.) d Apart from the cost of electricity, name three other important costs involved in the production of aluminium. 31 Chemists are investigating the possibility of replacing the carbon anodes in the electrolytic cell for aluminium extraction with anodes made of an unreactive material. a Write an equation for the anode reaction using unreactive electrodes.


b What would be the advantages of using unreactive anodes? 32 The Portland aluminium smelter in Victoria produces about 1000 tonnes of aluminium a day. What volume of carbon dioxide gas (measured at SLC) is produced by the smelter each day? 33 Impure copper, called ‘blister copper’ is produced by the reduction of copper ore in a blast furnace. Blister copper can be purified by the electrolytic process called electrorefining. Use the description of this process in the text to help you answer the following questions. a To which electrode is the blister copper attached? b What is the nature of the other electrode? c What is the composition of the electrolyte? d Write the half equations for the reactions involving copper that occur at the two electrodes. e What happens to impurities in the blister copper that are: i more reactive than copper, such as zinc? ii less reactive than copper, such as gold?

462 264 SSupplying u and using energy

Connecting the main ideas 34 Construct a concept map that includes the terms: electrolysis, electrolytic cell, chemical energy, electrical energy, anode, cathode, reductant, oxidant, reduction, oxidation, electrolyte and non-spontaneous reactions. 35 Under the title ‘Using electrical energy to produce a useful chemical’, write one or two paragraphs that explain to a Year 11 Chemistry student how a chemical of your choice can be produced industrially by electrolysis. 36 When constructing a galvanic cell in the laboratory, why are two half cells usually used whereas the reactants of an electrolysis cell are often placed in a single container?

4

unit

area of study 2

review

Supplying and using energy

multiple-choice questions 1 Which one of the following lists includes only renewable energy sources? A ethanol, coal, natural gas B animal waste, the Sun, crude oil C uranium, ‘hot rocks’, hydrogen D hot underground springs, biochemical fuels, wind 2 Two gaseous fuels that have been investigated as possible alternative energy sources are hydrogen gas, H2, and biogas, which consists mostly of methane, CH4. The table shows the energy released by the combustion of 1 mol of hydrogen and 1 mol of methane. Fuel

Hydrogen Methane

Equation for combustion reaction 1 H2(g) + O2(g) –—➤ H2O(l) 2 CH4(g) + 2O2(g) –—➤ CO (g) + 2H O(l) 2 2

Energy released per mole of fuel (kJ mol–1)

The following information relates to Questions 4 and 5. As a response to rising petrol prices the Australian government in 2006 offered car owners a $2000 incentive to convert their cars from petrol to LPG. Propane is one of the gases in the mixture sold as LPG. The thermochemical equation for the complete combustion of propane is: C3H8(g) + 5O2(g) –—➤ 3CO2(g) + 4H2O(l); ΔH = –2220 kJ mol–1

4 Which energy profile diagram best represents the energy changes that take place during this reaction? A B 2220 kJ mol–1 –1

2220 kJ mol

286 890

Consider the following situations that may apply to the use of the fuels. I Limiting greenhouse gas emissions is very important. II Equal masses of the two gases are to be stored. III Equal volumes of the gases are to be stored (at the same temperature and pressure). Hydrogen gas would be the preferred fuel in situations: A I, II and III B I and II C II and III D I only 3 Some eminent politicians and scientists have recently proposed that we should consider utilising our reserves of uranium to generate electricity for our cities, instead of the current situation where most electricity is supplied by coal-fired power stations. An argument in favour of greater reliance on nuclear power, rather than coal, could be based on: A whether the energy source is renewable or non-renewable B the relative amounts of greenhouse gases produced by the power stations C the safe disposal of waste materials from the power plant D the number of energy conversions involved between the energy source and the electrical energy that is generated

C

2220 kJ mol–1

D

2220 kJ mol–1

5 When 100 g of propane undergoes complete combustion: A 5.04 × 103 kJ of energy is absorbed B 5.04 × 103 kJ of energy is released C 3.92 × 101 kJ of energy is absorbed D 3.92 × 101 kJ of energy is released. 6 Which one of the following sources of power provides the highest quantity of energy per gram of fuel? A coal B natural gas C nuclear fission D biochemical fuels

463

463 Supplying and using energy

Area of study review

7 A temperature rise of 2.0°C occurred when 1.50 × 10–3 mol of ethane gas burnt in a bomb calorimeter.

piece of paper towel that had been soaked in a potassium nitrate solution. The cell is shown in the diagram.

2C2H6(g) + 7O2(g) –—➤ 4CO2(g) + 6H2O(l); ΔH = –3120 kJ mol–1

V Salt bridge (paper towel)

The calibration constant of the calorimeter, in J °C–1, is: A 1170 B 3120 C 2340 D 4680 8 The following equations involve ions of the transition metal vanadium. The half cell potential E° (V) is shown for each equation. VO2+(aq) + 2H+(aq) + e– –—➤VO2+(aq) + H2O(l); E ° = +1.00 V 2+ + – –— 3+ VO (aq) + 2H (aq) + e ➤ V (aq) + H2O(l); E ° = +0.36 V V3+(aq) + e– –—➤ V2+(aq); E ° = –0.25 V

Of the ions listed, the strongest oxidant and the strongest reductant are: Strongest oxidant Strongest reductant A VO2+ V2+ + B VO2 VO2+ 3+ C V VO2+ 2+ D V VO2+ 9 Use the electrochemical series to determine which one of the following would not be expected to occur to an appreciable extent: A 2H+(aq) + Fe(s) –—➤ Fe2+(aq) + H2(g) B 2Ag+(aq) + Ni(s) –—➤ 2Ag(s) + Ni2+(aq) C Br2(aq) + 2Fe2+(aq) –—➤ 2Br–(aq) + 2Fe3+(aq) D 2I–(aq) + Pb2+(aq) –—➤ I2(s) + Pb(s) 10 Which of the following best describes the features of an anode in a galvanic cell? Polarity Electrode reaction A positive oxidation B positive reduction C negative oxidation D negative reduction 11 An electrochemical cell was made by dipping a copper rod into a solution of 1 M CuSO4 in one beaker and dipping a nickel rod into a solution of 1 M NiSO4 in another beaker. The metals were connected with wire and the two solutions were connected by a

464 Supplying and using energy SSu

+

–

Cu

Ni

NiSO4

CuSO4

Beaker 1

Beaker 2

The solution in beaker 1 was initially coloured blue, owing to the presence of Cu2+ ions. The solution in beaker 2 was initially coloured green because of the presence of Ni2+ ions. After the galvanic cell has been discharging for a period of time it might be possible to detect the following changes. A The green colour in beaker 2 has faded and the mass of the copper electrode has increased. B The blue colour in beaker 1 has faded and the mass of the nickel electrode has increased. C The green colour in beaker 2 has faded and the mass of the copper electrode has decreased. D The blue colour in beaker 1 has faded and the mass of the nickel electrode has decreased. The following information relates to Questions 12 and 13. A car manufacturer has developed a prototype of an electric car that uses hydrogen fuel cells to generate electricity. 12 A reaction that might occur at the cathode of a fuel cell using hydrogen and oxygen as reactants is: A 2H2(g) + O2(g) –—➤ 2H2O(g) B O2(g) + 2H2O(g) + 4e– –—➤ 4OH–(aq) C H2(g) –—➤ 2H+(aq) + 2e– D H2(g) + 2OH–(aq) –—➤ 2H2O(g) + 2e– 13 Consider the following features of the fuel-cell–powered car. I Two half-cell reactions occur spontaneously at two separated locations (the electrodes). II Oxygen behaves as an oxidant in the reaction that supplies energy. III Hydrogen gas stores 143 kJ g–1 of fuel.

The fuel cell powered car differs from a conventional petrol-driven car with respect to: A I, II and III B I and II C I and III D II and III The following information relates to Questions 14 and 15. The diagram shows an electroplating cell that was devised by a student to place a silver coating onto a key. An electric current is passed through a solution of 1 M silver nitrate, AgNO3, using an inert carbon electrode as the anode. Power source

e–

e–

Ag+

short-answer questions

– Ag+

Key (cathode)

A Cr > Cu > Ag B Ag > Cu > Cr C Cr > Ag > Cu D the same in all three beakers 17 A series of experiments is described below. I Zinc granules are added to a solution of tin(II) chloride. II Copper turnings are added to tin(II) chloride solution. III Electrolysis of molten tin(II) chloride. IV Electrolysis of an aqueous solution of tin(II) chloride. The experiments that would result in the production of metallic tin, Sn, are: A all of them B I, III and IV C I and II D only III

Carbon block (anode) Silver nitrate solution

14 The reaction that occurs at the electrode connected to the negative terminal of the external power supply is: A Ag+(aq) + e– –—➤ Ag(s) B 2H2O(l) + 2e– –—➤ H2(g) + 2OH–(aq) C 2Cl–(aq) –—➤ Cl2(g) + 2e– D 2H2O(l) –—➤ O2(g) + 4H+(aq) + 4e– 15 If 0.055 g silver is deposited on the surface of the key when a current is passed through the cell for 7.0 minutes, what current is required? A 0.12 A B 0.75 A C 7.0 A D 49 A 16 Three beakers contain solutions of 1.0 M chromium(III) nitrate, 1.0 M copper(II) nitrate and 1.0 M silver nitrate. Each solution has 0.60 F of electric charge passed through it, using carbon electrodes and a power pack. In each beaker a different metal is deposited onto the cathode. In decreasing order, the amount of metal (in moles) deposited onto the cathode in each beaker will be:

18 a During the 20th century and continuing to the present day, atmospheric carbon dioxide concentrations have increased significantly, due mainly to the burning of fossil fuels. i Describe two undesirable atmospheric pollutants, other than carbon dioxide emissions, produced by the burning of fossil fuels. ii What is thought to be the environmental impact of increased levels of atmospheric carbon dioxide? iii Identify one fossil fuel and briefly discuss the sustainability of its continued use into the future. Your response should include a statement about the extent of the fossil fuel’s reserves and outline one advantage and one disadvantage of its continued use. b The main use of coal is in the generation of electricity. Electricity can also be generated using non-fossil fuel alternatives. Name one of these alternatives and identify one advantage (other than less CO2 emission) and one disadvantage compared to burning fossil fuels for electricity generation. c Some countries use nuclear fission instead of fossil fuels for electricity production. i Explain what the term ‘nuclear fission’ means. ii What energy transformation occurs when 235U undergoes fission? d There has been an increased awareness of the use of biochemical fuels as an alternative to fossil fuels.


i How does a biochemical fuel differ from a fossil fuel? ii Give two examples of biochemical fuels and describe briefly how they are produced. iii Why are biochemical fuels considered by some to be carbon dioxide neutral? 19 The heat of combustion of brown coal was measured in a bomb calorimeter that had been calibrated by passing a current of 3.00 amperes at 5.00 volts through the electrical heater for 2.00 minutes. The temperature rose by 1.1°C. When a 0.50 g sample of freshly crushed brown coal was completely burned in the calorimeter, the temperature rose by 3.0°C. a Calculate the value of the calorimeter constant. b Determine the heat of combustion of brown coal in kJ g–1. c Another sample of brown coal was heated at 100°C for 2 hours. A sample weighing 0.5 g was then burned completely in the calorimeter. Would this sample release more or less energy per gram than the freshly crushed sample? Explain your answer. d A brown-coal fired power station produces 1.7 tonnes of carbon dioxide gas for every megawatt of electricity generated. A black-coal fired power station releases 0.9 tonne of carbon dioxide into the atmosphere per megawatt of electricity. Explain the difference in the amount of carbon dioxide released to produce the same amount of electricity. Assume that both types of power stations operate at the same level of efficiency. 20 One of the boilers at a power station burns 75.6 kg of brown coal every second. Each kilogram of brown coal produces 9.82 MJ of energy. The electrical output of the generator connected to this boiler is 300 MJ s–1. a Calculate the amount of energy produced by burning 75.6 kg of coal each second. b Give two reasons why the answer to part a is much greater than the output of the generator. c Over three-quarters of the electricity generated in Victoria comes from brown coal, a non-renewable energy source. Explain what is meant by the term non-renewable energy source. d Describe briefly how one renewable energy source can be used to generate electricity.


21 When a Ni–Cad cell is converting chemical energy to electrical energy (discharge), the electrode reactions are best described as follows: positive electrode: NiOOH(s) + H2O(l) + e– –—➤ Ni(OH)2(s) + OH–(aq) negative electrode: Cd(s) + 2OH–(aq) –—➤ Cd(OH)2(s) + 2e–

a Which metal, nickel or cadmium, is at the anode of this cell as it is discharging? b Give the formula of a salt that might be expected to be found in the electrolyte paste of a Ni–Cad cell. Ni–Cad cells are superior to alkaline cells as they can be recharged. c Explain why cells such as the Ni–Cad can be recharged. d Write equations for the half reactions and the overall cell reaction occurring when a Ni–Cad cell is recharged. e Describe the energy transformation that occurs when a Ni-Cad cell is recharged. 22 Zinc–air cells have been developed for a range of different uses, including electric vehicles, heart pacemakers and lap-top computers. They offer a favourable alternative to other batteries because of their non-toxic contents and high energy density. In the cells, zinc metal is oxidised at the negative electrode and oxygen gas is reduced to water at the positive electrode. A zinc–air cell used for a pacemaker operates with a steady current of 3.5 × 10–5 A. a Write a half equation for the half reaction at the negative electrode. b If the cell will continue operating until 1.5 g of zinc in the negative electrode has been used up, determine how long the pacemaker will last before it needs replacing. 23 An ‘alcometer’ is a small, hand-held device that can be used to give reliable readings of blood alcohol levels when a person blows air from their lungs into the instrument. It relies on the equilibrium established between ethanol dissolved in the blood and gaseous ethanol in the lungs: C2H5OH(aq) (in blood)

C2H5OH(g) (in lungs)

which has equilibrium constant K = 4.35 ×10–4 at 37°C.

The alcometer is a fuel cell and air from the lungs is directed onto the anode of the cell where ethanol, C2H5OH, is oxidised to carbon dioxide, CO2. It consists of a porous glass plate containing a phosphoric acid electrolyte and coated with layers of platinum metal on the top and bottom. The fuel cell’s voltage depends on the concentration of ethanol in the air blown onto the anode, so the instrument can be calibrated to switch on a red light if blood alcohol concentration is greater than 0.05%, or a green light for zero blood alcohol concentration. Meter +

– O2

Air from lungs

Phosphoric acid electrolyte

d Give two different functions of the layers of platinum metal in the fuel cell. 24 An experiment was carried out as follows to compare the effectiveness of two different methods of generating energy from liquid methanol (CH3OH) as a source of fuel. a The heat of combustion of methanol was determined in a bomb calorimeter which had a calibration factor of 8.26 kJ °C–1. When 2.0 g of methanol was completely burned in excess oxygen the temperature rose from 20.0°C to 25.5°C. i How much energy is released when 2.0 g of methanol is burned in excess oxygen? ii How much energy is released when 1.0 mol of methanol is burned in excess oxygen? iii What is the value of ΔH for the reaction: 2CH3OH(l) + 3O2(g) –—➤ 2CO2(g) + 4H2O(l)

O2

Layers of platinum metal

The net reaction that occurs in this fuel cell is given by the equation: C2H5OH(g) + 3O2(g) –—➤ 2CO2(g) + 3H2O(l)

a By assigning oxidation numbers to carbon in both ethanol and carbon dioxide, verify that ethanol is oxidised in the fuel cell reaction. b The half reaction that occurs at the cathode is: O2(g) + 4H+(aq) + 4e– –—➤ 2H2O(l)

What is the equation for the half reaction that occurs at the anode in the alcometer? c On the diagram, place an arrow to show the direction of electron movement through the external circuit.

b The methanol fuel cell generates electricity by consuming methanol and oxygen. The equations for the reactions occurring at the electrodes are: CH3OH(l) + H2O(l) –—➤ 2CO2(g) + 6H+(aq) + 6e Equation 1 O2(g) + 4H+(aq) + 4e– –—➤ 2H2O(l) Equation 2

i Which equation represents the reaction occurring at the anode? ii Write an equation that represents the overall cell reaction. iii Calculate the electric current generated by the methanol fuel cell when 0.64 g of methanol is consumed in 10 minutes. iv How much energy in kJ is delivered per mole of methanol from a fuel cell that generates 1.26 V? c Methanol can also be used as a fuel for an internal combustion engine driving an electrical generator. i Give one reason why the fuel cell would produce electricity more efficiently than a generator driven by an internal combustion engine. ii Give one reason why fuel cells are not used to generate electricity on a large scale.


25 a Complete the half equations for the reactions predicted when each of the following aqueous solutions undergoes electrolysis, using the electrodes described in the table. The process is carried out in a U-tube, as shown in the diagram. e– +

+

–

+ Pt electrode

– Pt electrode

e–

–

MgCl2 and SnCl2

Solution

Electrode materials

KI(aq)

Positive electrode carbon

Negative electrode carbon

PbCl2(aq)

carbon

carbon

AlCl3(aq)

copper

copper

Half equation for the reaction at the Positive Negative electrode electrode

b For one of the solutions described in the table, explain how you might verify that the predicted products were being formed at each electrode. 26 A 100 mL solution containing a mixture of magnesium chloride, MgCl2, and tin(II) chloride, SnCl2, is prepared by dissolving 0.025 mol of each salt in water. Platinum electrodes are placed in the solution and a small current is passed through it, as shown in the diagram.


a Write half equations for the electrode reactions occurring just after the electrolysis is started: i at the anode ii at the cathode. b After electrolysis has been occurring for a considerable period of time, all of the metal that was first plated on the cathode will have been used up and a new electrode reaction will occur at the cathode. Write a half equation for the next electrode reaction that occurs at the cathode. c What major difference would occur in the cell if the solution that was used had been a saturated solution of tin(II) chloride? 27 In an electrolysis experiment, a student is provided with a solution of nickel(II) nitrate, Ni(NO3)2. The electrodes to be used were a carbon rod, as the positive electrode, and a metal spatula, as the negative electrode. a Will the nickel coating appear on the carbon rod or the metal spatula during the experiment described? b Write equations for the half reactions that occur at each electrode. c A current of 2.5 A was passed through this nickel-plating cell for 15 minutes. Calculate the mass of nickel that could be plated on the cathode. 28 Lithium metal is prepared by electrolysis of a molten mixture of lithium chloride and potassium chloride. a Write a half equation for the reaction occurring at the cathode in this cell. b Write a half equation for the reaction occurring at the anode in the cell.

c Why is it not possible to produce lithium metal by the electrolysis of an aqueous solution of lithium chloride? d Suggest why a mixture of lithium chloride and potassium chloride is used, rather than pure lithium chloride, in this electrolytic process. e Assuming that lithium reacts in a similar way to sodium, describe two precautions that would need to be taken in this preparation. 29 This question involves both electrochemical cells and electrolytic cells. a Compare the energy transformation occurring in an electrochemical cell with that occurring in an electrolytic cell. b Explain why the negative electrode of an electrochemical cell is the site of oxidation, whereas the negative electrode of an electrolytic cell is the site of reduction. 30 Some industrially important chemicals are produced by the electrolysis of a molten electrolyte while others are produced by electrolysis of an aqueous solution.

a Name one chemical produced by the electrolysis of: i a molten electrolyte ii an aqueous solution. b For each chemical that you have selected: i write an equation for the reaction occurring at the anode ii write an equation for the cathode reaction iii describe one feature of the electrolytic cell used to manufacture of each selected chemical. c Explain why some chemicals can be produced by the electrolysis of a molten electrolyte but not from the electrolysis of an aqueous solution. 31 Write equations to represent the anode and cathode reactions that occur when an electric current is passed through: a a dilute solution of sodium chloride b a concentrated solution of sodium chloride c molten sodium chloride.


Appendices A Ap pp pe end Appendix 1 Significant figures and standard form Significant figures The number of significant figures a piece of data has indicates the precision of a measurement. For example, compare the following data: A jogger takes 20 minutes to cover 4 kilometres. A sprinter takes 10.21 seconds to cover 100.0 metres. The sprinter’s data has been measured more precisely than that of the jogger. This is indicated by the greater number of significant figures in the second set of data.

Which figures are significant? A significant figure is an integer or a zero that follows an integer. In the data above: the distance ‘4 kilometres’ has one significant figure the time ‘20 minutes’ has two significant figures (the zero follows the integer 2) the 10.21 seconds and 100.0 metres each have four significant figures. A zero that comes before any integers, however, is not significant. For example: the value 0.0004 has only one significant figure, whereas the value 0.0400 has three significant figures. The zeros that come before the integer 4 are not significant, whereas those that follow the integer are significant.

Using significant figures In chemistry you will often need to calculate a value from a set of data. It is important to remember that the final value you calculate is only as precise as your least precise piece of data.

470 Appendices

Example An athlete takes 3.5 minutes to complete four laps of an oval. What is the average time taken for one lap? 3.5 minutes = 0.875 minutes Average time = 4 Because the data (3.5 minutes) has only two significant figures, the answer will need to be adjusted to two significant figures so that it has the same degree of precision as the data. (Note: The ‘four’ is taken to indicate a precise number of laps and so is considered to have as many significant figures as the calculation requires. This applies to values that describe quantities rather than measurements.) When adjusting the number of significant figures, if the integer after the last significant figure is equal to or greater than 5, then the last significant integer is rounded up. Otherwise, it is rounded down. Therefore, in this example, 0.875 minutes is adjusted to 0.88 minutes to two significant figures.

Standard form A value written in standard form is expressed as a number equal to or greater than 1 and less than 10 multiplied by 10x, where x is an integer. For example, when written in standard form: 360 becomes 3.6 × 102 0.360 becomes 3.60 × 10−1 0.000 045 6 becomes 4.56 × 10−5. Sometimes you need to use standard form to indicate the precision of a value.

Appendix 2 Symbols, units and fundamental constants TABLE 1 Units and symbols based on the SI system Quantity

Symbol for physical quantity

Mechanics Length Area Volume Mass Density Time Force Pressure Energy Electricity Electric current Electric charge Electric potential difference Nuclear and chemical quantities Atomic number Neutron number Mass number Amount of substance Relative atomic mass Relative molecular mass Molar mass Molar volume Concentration Thermal quantities Temperature Molar heat capacity

Corresponding SI unit

Synbol for SI unit

Definition of SI unit

/ A V m ρ t F p E

metre square metre cubic metre kilogram

fundamental unit

second newton pascal joule

m m3 kg kg m−3 s N Pa J

I Q V

ampere coulomb volt

A C V

fundamental unit As J A–1 s–1

Z N A n Ar Mr M Vm c

– – – mole – – – – –

– – – mol – – – – –

– – Z+N fundamental unit – – kg mol–1 m3 mol–1 mol m−3

T C

kelvin –

K –

fundamental unit J K−1 mol−1

2

fundamental unit fundamental unit kg m s–2 N m–2 Nm

Units listed in red are the arbitrarily defined fundamental units of the SI system.

TABLE 2 Some fundamental constants Description Avogadro’s constant Faraday constant Gas constant Ionic product of water Molar volume of an ideal gas at 273 K, 101.3 kPa at 298 K, 101.3 kPa Planck constant Speed of light in a vacuum Specific heat capacity of water

Symbol NA F R Kw Vm

h c

Value 6.02 × 1023 mol–1 96 500 C mol–1 8.31 J K–1 mol–1 1.00 × 10–14 mol2 L–2 at 298 K 22.4 L mol–1 24.5 L mol–1 6.63 × 10–34 J s–1 3 × 108 m s–1 4.18 J g–1 K–1

TABLE 3 Summary of the main decimal fractions and multiples Fraction –1

10 10–2 10–3 10–6 10–9 10–12 10–15

Prefix Symbol deci centi milli micro nano pico femto

d c m μ n p f

Multiple 1

10 102 103 106 109 1012

Prefix Symbol deca hecto kilo mega giga tera

da h k M G T

471 Appendices

Appendix 3 Table of relative atomic masses* Name

Actinium Aluminium Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium Caesium Calcium Californium Carbon Cerium Chlorine Chromium Cobalt Copper Curium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold

Symbol

Atomic number

Ac Al Am Sb Ar As At Ba Bk Be Bi B Br Cd Cs Ca Cf C Ce Cl Cr Co Cu Cm Dy Es Er Eu Fm F Fr Gd Ga Ge Au

89 13 95 51 18 33 85 56 97 4 83 5 35 48 55 20 98 6 58 17 24 27 29 96 66 99 68 63 100 9 87 64 31 32 79

Relative atomic mass – 26.9815 – 121.75 39.948 74.9216 – 137.34 – 9.01218 208.9806 10.81 79.904 112.40 132.9055 40.08 – 12.011 140.12 35.453 51.996 58.9332 63.54 – 162.50 – 167.26 151.96 – 18.9984 – 157.25 69.72 72.59 196.9665

Name

Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium

Symbol

Atomic number

Hf He Ho H In I Ir Fe Kr La Lr Pb Li Lu Mg Mn Md Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr

72 2 67 1 49 53 77 26 36 57 103 82 3 71 12 25 101 80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59

Relative atomic mass 178.49 4.00260 164.9303 1.0080 114.82 126.9045 192.22 55.847 83.80 138.9055 – 207.2 6.94 174.97 24.305 54.9380 – 200.59 95.94 144.24 20.179 – 58.71 92.9064 14.0067 – 190.2 15.9994 106.4 30.9738 195.90 – – 39.102 140.9077

* Based on the atomic mass of 12C = 12. The values for relative atomic masses given in the table apply to elements as they exist in nature, without artificial alteration of their isotopic composition, and, further, to natural mixtures that do not include isotopes of radiogenic origin.

472 Appendices

Name

Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium

Symbol

Atomic number

Pm Pa Ra Rn Re Rh Rb Ru Sm Sc Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W U V Xe Yb Y Zn Zr

61 91 88 86 75 45 37 44 62 21 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40

Relative atomic mass – – – – 186.2 102.9055 85.467 101.07 150.4 44.9559 78.96 28.086 107.868 22.9898 87.62 32.06 180.947 – 127.60 158.9254 204.37 232.0381 168.9342 118.69 47.90 183.85 238.029 50.941 131.30 173.04 88.9059 65.37 91.22

Appendix 4 Chemical bonding overview All bonding results from electrostatic attraction between positive and negative particles. Atoms form bonds by losing, gaining or sharing electrons. In the process they often achieve the more stable noble gas electron configurations in their outer shells. The strengths of bonds vary enormously. Bonds that require more than 100 kJ mol–1 of energy to be broken are classified as strong bonds; bonds that can be broken by energy of less than 100 kJ mol–1 are called weak bonds.

these covalent network substances are very hard and have high melting temperatures. Rather than forming covalent network lattices, non-metallic atoms usually form smaller clusters of atoms held together by covalent bonds. The clusters are known as molecules. Whilst the bonds within molecules are strong, bonds between molecules are weak. Molecular substances are therefore often soft and easy to melt.

Weak bonds Strong bonds There are three types of strong bonds: metallic, ionic and covalent.

Weak bonds are classified as dispersion forces, dipole–dipole attractions, hydrogen bonds or ion– dipole bonds.

Metallic bonds

Dispersion forces

Metallic elements and alloys consist of a regular, three-dimensional lattice of positively charged metal ions, called cations, surrounded by delocalised electrons from the outer shells of the metal atoms. The strong attraction between the cations and the electrons holds the lattice together and is called metallic bonding.

These weak forces are present between all particles and are often only evident when no significantly stronger forces exist. Dispersion forces are responsible for holding the molecules together in non-polar molecular substances such as polyethene, wax, oil and grease. Because of the weakness of dispersion forces, such substances tend to be soft and melt at low temperatures. Dispersion forces increase in strength as the number of electrons in molecules increases.

Ionic bonds Ionic bonds are present in compounds formed from a metal and one or more non-metal elements. Electrons are transferred from the metal atoms to the non-metal atoms. The metal atoms become positively charged cations and the non-metallic elements become negatively charged anions. A regular, three-dimensional lattice is formed, in which the cations are surrounded by the oppositely charged anions and the anions are surrounded by cations. Strong electrostatic attraction between the cations and anions holds the lattice together and is called ionic bonding.

Covalent bonds Atoms of non-metallic elements share electrons and form covalent bonds in order to obtain outer shells of higher stability. In some substances, including diamond, silica (SiO2) and graphite, atoms are linked together by a continuous three-dimensional network of covalent bonds. As a consequence

Dipole–dipole attractions When different non-metallic atoms form a molecule, one or more of the atoms may draw the electrons in the covalent bonds more strongly towards them than the other atoms. Partial negative or positive charges are created on the atom. In the case of a molecule of HCl, for example, the chlorine atom has a partial negative charge and the hydrogen atom has a partial positive charge. HCl molecules are said to be polar. The partial charges in the HCl molecule can be represented as: δ+ δ– H–Cl The molecules tend to arrange themselves so that the oppositely charged ends of the molecules are next to each other. This type of bonding between molecules is called dipole–dipole attraction.

473 Appendices

Hydrogen bonds Dipole–dipole attractions are observed to be much stronger than usual in compounds containing molecules in which hydrogen is bonded to one of the three most electronegative elements: fluorine, nitrogen or oxygen. These highly electronegative elements have lone pairs of electrons that are attracted to the highly polarised hydrogen. In these compounds the dipole–dipole attractions between the molecules are usually referred to as hydrogen bonds. Water is an important example of a compound containing hydrogen bonding. The partial positive

charge on the hydrogen atoms on one side of a water molecule tends to attract the unpaired electrons from the oxygen atoms (that are carrying a partial negative charge) on neighbouring molecules. Due to these hydrogen bonds between the molecules, water has a higher melting and boiling temperature than substances composed of molecules of similar size.

Ion–dipole bonds Attraction between an ion and a polar molecule is referred to as ion–dipole attraction.

An overview of chemical bonding is also provided in the ChemCAL module ‘Bonding’.

474 Appendices

Answers Detailed answers and fully worked solutions are contained in the Heinemann Teacher’s Resource and Assessment Kit.

Chapter 1 1 a quantitative b qualitative c qualitative d quantitative E2 atomic emission spectroscopy, atomic absorption spectroscopy 5 C

Chapter 2 2 Microwave ovens dry samples more rapidly than conventional ovens. 3 81.0% 4 a Phydeaux Deluxe 80%, K9 Budget 74%, Fresh meat—buffalo mince 75% 5 % (H2O) = 41.6% 6 a 0.100 mol b 2.01 mol c 0.050 mol d 2.07 × 10–1 mol e 3.000 mol 7 a 96 g b 1.6 × 102 g c 28 g 8 285 g 9 a 5.7 g b 0.190 g 10 a %Pb = 86.6%, %O = 13.4% b %Na = 43.4%, %C = 11.3%, %O = 45.3% 11 a CH b C2H2 12 WCl6 14 a 39.2 g b 0.823 g 15 a 1430 kg b 659 000 L 16 8.44 g 17 3.07% 18 14.3% 19 a 1.50 mol b 1.3 mol c 0.401 mol d 0.63 mol 20 a 20 g b 26 g c 2.2 × 102 g 21 0.12 g 22 36.3 L 23 a 33.6 L b 22.4 L c 6.5 L d 0.49 L 24 a 0.0407 mol b 30.0 g mol–1 c NO 25 a 0.455 g b 0.797 g 26 a 16.6 g b 6.58 g 27 133 L 28 9.10 × 107 L 29 a 2.53 × 10–3 mol b 0.141 g c 88.1% 31 a 22.6 g b 4.94 g 32 0.972 L 33 a 82.4% b 35.0% c 46.7% 34 a Sc2O3 b Cu2O c C2H3Cl 35 CaSO4.2H2O 36 a CH2O b C2H4O2 37 12.8 g 38 a lead(II) chloride, lead(II) hydroxide, copper(II) hydroxide b Pb(NO3)2(aq) + CuCl2(aq) — PbCl2(s) + Cu(NO3)2(aq); Pb2+(aq) + 2Cl–(aq) — PbCl2(s); Pb(NO3)2(aq) + Ba(OH)2(aq) — Pb(OH)2(s) + Ba(NO3)2(aq); Pb2+(aq) + 2OH–(aq) — Pb(OH)2(s); CuCl2(aq) + Ba(OH)2(aq) — Cu(OH)2(s) + BaCl2(aq); Cu2+(aq) + 2OH–(aq) — Cu(OH)2(s) 40 0.470 g 41 0.012 mol L–1 42 1.3 g 43 a 18.4% b The result would be too high. 44 1.85 g 45 a 1.2% 46 77.4% 47 a P2O5 b P4O10 48 a Ag+(aq) + Cl–(aq) — AgCl(s) b 0.001 72 mol c 0.001 72 mol d 0.0215 mol e 1.25 g

Chapter 3 1 a 0.50 M b 0.013 M c 0.025 M 2 a 5.10 × 10–3 mol b 2.63 mol c 5.00 × 10–4 mol 3 a 2.0 g b 4.79 g c 0.097 g 4 a 60 g L–1 b 1.5 g L–1 c 57 g L–1 5 a 0.020 00 mol L–1 b 0.155 M c 0.0123 M 6 a 0.10 mol b 0.050 mol c 0.20 mol 7 1.3 × 103 kg E1 a 49.93 g b 49.93 mL c systematic error d Cleaning the burette will remove any error caused by dirty glass. e 50.00 mL f random error g systematic error h add 0.50 mL for every 50 mL or 0.01 mL for every 1 mL delivered by the burette 11 a 0.0762 M b 0.059 M c 0.0119 M 12 a 16.7 g b 5.10 g 13 a 0.100 mol b 0.300 M 14 a i 0.14 g L–1 ii 0.021 g L–1 b i 0.0021 M ii 0.0029 M c i 270 ppm ii 0.69 ppm 15 a 0.40 g L–1 16 a 2.4 g L–1, 1.5 g L–1, 17.8 g L–1 b 0.041 M, 0.020 M, 0.0989 M c 0.020 M d 0.061 M 17 0.15 L 18 1.5 M 19 21 mL 21 a 4.7 × 10–8 mol g–1 b 5.7 × 1018 23 a ii b iii c i 25 b 4.24 g

Chapter 4 4 HCO3–(aq) + H3O+(aq) — CO2(g) + 2H2O(l) 5 a i H2O; NH3 ii NH4+, NH3 and H2O, OH– b i HSO4–; H2O ii HSO4–, SO42– and H3O+, H2O c i NH4+; S2– ii NH4+, NH3 and HS–, S2– d i H3O+; CH3COO– ii CH3COOH, CH3COO– and H3O+, H2O 7 a CH3COOH(aq) + NaOH(aq) — CH3COONa(aq) + H2O(l) b 0.020 24 mol c 0.020 24 mol d 1.012 M 8 a 2.19 × 10–3 mol b 2.19 × 10–3 mol c 0.109 M d 1.09 M 9 a 0.009 80 mol b 0.005 55 mol c 0.005 55 mol d 0.004 25 mol e 0.002 13 mol f 87.6% 10 a HNO3(aq) + NaOH(aq) — NaNO3(aq) + H2O(l) b H2SO4(aq) + 2KOH(aq) — K2SO4(aq) + 2H2O(l) c HCl(aq) + NH3(aq) — NH4Cl(aq) d CH3COOH(aq) + KOH(aq) — CH3COOK(aq) + H2O(l) 11 a H+(aq) + OH–(aq) — H2O(l)

b H+(aq) + OH–(aq) — H2O(l) c H+(aq) + NH3(aq) — NH4+(aq) d CH3COOH(aq) + OH–(aq) — CH3COO–(aq) + H2O(l) 12 0.21 g 13 0.134 L 16 a An indicator is needed to detect the equivalence point as both the reactants and products are colourless. b NaOH(aq) + HCl(aq) — NaCl(aq) + H2O(l) c 0.0855 M d 342 g 17 a 2HCl(aq) + Na2CO3(aq) — 2NaCl(aq) + H2O(l) + CO2(g) b 0.573 M 18 220 mL 19 a 0.265 g b 88.3% 20 a CH3COOH(aq) + NaOH(aq — CH3COONa(aq) + H2O(l) b 0.937 M c 56.3 g L–1 21 a higher b unchanged c lower 22 a 20.99 mL b 0.046 66 M c 0.088 92 M 23 a H2SO4(aq) + M2CO3(aq) — M2SO4(aq) + H2O(l) + CO2(g) b 0.001 62 mol c 0.001 62 mol d 0.0203 mol e 138 g mol–1 f potassium 25 70 mL 26 a 0.001 13 mol b 0.001 13 mol c 0.000 915 mol d 0.0114 mol e 13.0% 27 a 3.25 M 28 a 2HCl(aq) + CaCO3(s) — CaCl2(aq) + H2O(l) + CO2(g), HCl(aq) + NaOH(aq) — NaCl(aq) + H2O(l) b 0.001 09 mol c 0.0198 mol d 0.0187 mol e 0.009330 mol f 88.9%

Chapter 5 1 a Zn b Ca c Br– ions in AgBr 2 (oxidant is listed first) a O2; Zn b Cl2; Ca c Ag+ in AgBr; Br – in AgBr 3 a +2 b +4 c –4 d 0 e +4 4 K2MnO4 5 a Ca: +2; O: –2 b Ca: +2; Cl: –1 c H: +1; S: +6; O: –2 d Mn: +7; O: –2 e F: 0 f S: +4; O: –2 g Na: +1; N: +5; O: –2 h K: +1; Cr: +6; O: –2 6 (oxidant is listed first) a Cl2; Mg b O2; SO2 c Fe2O3; CO d H2O2; Fe2+ 7 a MnO2(s) + 4H+(aq) + 2e– — Mn2+(aq) + 2H2O(l) b MnO4–(aq) + 4H+(aq) + 3e– — MnO2(s) + 2H2O(l) c SO42–(aq) + 10H+(aq) + 8e– — H2S(g) + 4H2O(l) d SO2(g) + 2H2O(l) — SO42–(aq) + 4H+(aq) + 2e– e H2S(g) — S(s) + 2H+(aq) + 2e– f SO32–(aq) + H2O(l) — SO42–(aq) + 2H+(aq) + 2e– 8 Cu2+(aq) + 2e– — Cu(s), Fe(s) — Fe2+(aq) + 2e–, Cu2+(aq) + Fe(s) — Cu(s) + Fe2+(aq) 9 a Zn(s) — Zn2+(aq) + 2e– b Cr2O72–(aq) + 14H+(aq) + 6e– — 2Cr3+(aq) + 7H2O(l) c 3Zn(s) + Cr2O72–(aq) + 14H+(aq) — 3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l) 10 a Fe2+(aq) — Fe3+(aq) + e–, Cr2O72–(aq) + 14H+(aq) + 6e– — 2Cr3+(aq) + 7H2O(l), Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) — 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq) b SO32–(aq) + H2O(l) — SO42–(aq) + 2H+(aq) + 2e–, MnO4–(aq) + 8H+(aq) + 5e– — Mn2+(aq) + 4H2O(l), 5SO32–(aq) + 2MnO4–(aq) + 6H+(aq) — 5SO42–(aq) + 2Mn2+(aq) + 3H2O(l) c MnO2(s) + 4H+(aq) + 2e– — Mn2+(aq) + 2H2O(l), 2Cl–(aq) — Cl2(g) + 2e–, MnO2(s) + 4H+(aq) +2Cl–(aq) — Mn2+(aq) + 2H2O(l) + Cl2(g) 11 a i Ce4+ is reduced to Ce3+ and H2S is oxidised to S ii Ce4+(aq) + e– — Ce3+(aq), H2S(aq) — S(s) + 2H+(aq) + 2e– iii 2Ce4+(aq) + H2S(aq) — 2Ce3+(aq) + S(s) + 2H+(aq) b i NO3– is reduced to NO and Cu is oxidised to Cu2+ ii NO3–(aq) + 4H+(aq) + 3e– — NO(g) + 2H2O(l), Cu(s) — Cu2+(aq) + 2e– iii 2NO3–(aq) + 8H+(aq) + 3Cu — 2NO(g) + 4H2O(l) + 3Cu2+(aq) c i H2O2 is reduced to H2O and Br – is oxidised to Br2 ii H2O2(aq) + 2H+(aq) + 2e– — 2H2O(l), 2Br –(aq) — Br2(l) + 2e– iii H2O2(aq) + 2H+(aq) + 2Br –(aq) — 2H2O(l) + Br2(l) d i MnO2 is reduced to Mn2+ and S is oxidised to SO2 ii 2MnO2(s) + 4H+(aq) + 2e– — Mn2+ + 2H2O(l), S(s) + 2H2O(l) — SO2(g) + 4H+(aq) + 4e– iii MnO2(s) + 4H+(aq) + S(s) — 2Mn2+ + 2H2O(l) + SO2(g) 12 0.06 M 13 2.38 g 14 a +4 b –2 c +6 d +6 e +4 f +2 15 b, c, e, f and h are redox reactions as the elements in the reactions undergo changes in oxidation number b Ag from 0 to +1; Cl from 0 to –1 c Fe from +3 to +2; Sn from +2 to +4 e P from +3 to +5; I from 0 to –1 f Cu from +1 to +2; Cu from +1 to 0 h P from 0 to –3; H from 0 to +1 16 a i 0 ii +2 b i +5 ii +4 c oxidant HNO3; reductant Cu 17 a Al(s) — Al3+(aq) + 3e–, 2H+(aq) + 2e– — H2(g) 18 a i +3 ii +2 iii +4 b S2O42–(aq) + 2H2O(l) — 2HSO3–(aq) + 2H+(aq) + 2e–, S2O42–(aq) + 2H+(aq) + 2e– — S2O32–(aq) + H2O(l) 19 a Zn(s) — Zn2+(aq) + 2e–, Pb2+(aq) + 2e– — Pb(s), Zn(s) + Pb2+(aq) — Zn2+(aq) + Pb(s) b Fe2+(aq) — Fe3+(aq) + e–, MnO4–(aq) + 8H+(aq) + 5e– — Mn2+(aq) + 4H2O(l), 5Fe2+(aq) + MnO4–(aq) + 8H+(aq) — 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

475 Answers

c SO2(aq) + 2H2O(l) — SO42–(aq) + 4H+(aq) + 2e–, I2(aq) + 2e– — 2I–(aq), SO2(aq) + 2H2O(l) + I2(aq) — SO42–(aq) + 4H+(aq) + 2I–(aq) d OCl–(aq) + 2H+(aq) + 2e– — Cl–(aq) + H2O(l), 2I–(aq) — I2(aq) + 2e–, OCl–(aq) + 2H+(aq) + 2I–(aq) — Cl–(aq) + H2O(l) + I2(aq) 20 a Zn(s) — Zn2+(aq) + 2e–, 2MnO2(s) + 2H+(aq) + 2e– — Mn2O3(s) + H2O(l); Zn(s) + 2MnO2(s) + 2H+(aq) — Zn2+(aq) + Mn2O3(s) + H2O(l) b 3.0 g 21 1079 g 22 a Fe2O3(s) + 6e– — 2Fe(l) + 3O2–(s) b reduction c Fe2O3(s) + 2Al(s) — 2Fe(l) + Al2O3(s) d 5.29 g 23 a 0.250 g b 19.1% c fillers, binders, sweeteners, flavours and colouring 24 a 5Fe2+(aq) + MnO4–(aq) + 8H+(aq) — 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) b 0.003 03 mol c 84.7% d The end point is recognised by the first permanent tinge of pink caused by excess permanganate ions. e wear safety glasses and a laboratory coat; hold pipette close to the top when fitting pipette filler 25 a 0.000 460 mol b 0.000 690 mol c 0.008 62 mol d 0.396 g e 3.60% 26 a Fe2+(aq) — Fe3+(aq) + e– b MnO4–(aq) + 8H+(aq) + 5e– — Mn2+(aq) + 4H2O(l) c 5Fe2+(aq) + MnO4–(aq) + 8H+(aq) — 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) d 0.001 30 mol e 0.0162 mol f 0.246 g g Low results, due to loss of vitamin paste during transfer, or aliquots being less than the correct volume. 28 6.74 M 29 a 0.009 41 mol b 0.009 41 mol c 3.5% 30 a +2 b +3 c +6 d +6 e +4 f +5 g +5 h +7 i +4 j +1 31 a redox reaction; oxidant Fe2O3; reductant Mg b not a redox reaction c redox reaction; oxidant H+; reductant Zn d redox reaction; oxidant Cl2; reductant Br– 32 a i 2I–(aq) — I2(aq) + 2e– ii Cl2(g) + 2e– — 2Cl–(aq) iii 2I–(aq) + Cl2(g) — I2(aq) + 2Cl–(aq) b i Cu(s) — Cu2+(aq) + 2e– ii Ag+(aq) + e– — Ag(s) iii Cu(s) + 2Ag+(aq) — Cu2+(aq) + 2Ag(s) 33 a 2MnO4–(aq) + 16H+(aq) + 5(COO)22–(aq) — 2Mn2+(aq) + 8H2O(l) + 10CO2(g) b 0.0180 M 34 a i +1 ii –1 b oxidant c blue-black to colourless d 0.002 042 mol e 0.001 021 mol f 0.001 021 mol g 0.012 76 mol h 0.9506 g i 38.02 g L–1

b Rf

Propane

2

7

Propan-1-ol

4

3

Tartaric acid

3

1

Protons that are chemically equivalent do not split their own signal. For example ClCH2CH2Cl does not show peak splitting. 34 spin change of proton < vibrational change of molecule < electronic change in an atom 36 a i Cu(H2O)62+ ii Cu atoms 37 a 7.2 µg mL–1 Cu b 1.4 × 10–3 g Cu c 0.07% Cu; yes, commercially viable 38 a Carotene absorbs light in the violet–blue region of the visible spectrum and transmits light in the yellow–red region resulting in an orange colour. b Carotene absorbs strongly at about 425 nm. This wavelength should be selected. 40 Compound A: a i 2 ii X : Y = 1 : 1 iii Y = 2.0, X = 3.7 b III; Compound B: a i 3 ii X : Y : Z = 1 : 2 : 3 iii X = 8.0, Y = 4.2, Z = 1.3 b I; Compound C: a i 3 ii X : Y : Z = 1 : 2 : 3 iii X = 11.8, Y = 2.4, Z = 1.1 b II

Chapter 8 1 207.18 2 a 86 b 88 3 Peak at 49 = CH235Cl+; Peak at 35 = 35Cl+ 4 Ion fragment

m/e 17

CO+

28

CH3CH2+

29

Cl+

35 or 37

C2H4+

28

c Compound

COCH3+

43

COOH+

45

C6H5+

77

20

0.33

Chlorophyll b

Dark green

27

0.45

Chlorophyll a

Orange

40

0.67

Xanthophyll

Yellow

50

0.83

β-carotene

Solvent front

60

–

–

2 b 10.4% 7 blue: 0.83; red: 0.58 8 c 3 e blue f 0.63, 0.13 9 a 3.2 cm b 15 cm 10 a A: 0.59; B: 0.49; C: 0.38; D: 0.20; E: 0.13 12 b 7.5 cm 14 c 13 ppm d no e 92 mL 15 a nitrogen 17 a A hexane; B heptane; C octane; D decane 18 a no b no 19 b 4 20 a 8.0 µg/mL b 8.0 mg c 1.6% w/w

Chapter 7 1 a lilac 5 b 2.57 ppm c 25.7 ppm d 6.4 × 10–3 g e 36% 6 a 0.080 M solution b 0.30 M solution 9 a There are two strong absorption peaks, at 420 nm and 660 nm. b Either 660 nm or 420 nm could be used as chlorophyll absorbs strongly at both wavelengths. 10 violet–blue 11 a Benzene absorbs in the ultraviolet region of the electromagnetic spectrum. b The human eye only detects electromagnetic radiation in the visible region of the electromagnetic spectrum. 12 a 32 mg L–1 b 3.2% c orange light of wavelength 600 nm 13 C–H stretching 14 Peak A C–H stretching; Peak B C=O stretching 18 a 3 c A : B : C = 2 : 3 : 3 f 4 19 a green 23 a 2.45 ng mL–1 b 49 ng g–1 c 0.049 ppm d 4.9 × 10–6 % 24 a 16 ppm b 16 mg c 0.32% w/w 25 a blue b orange; blue light of wavelength 440–470 nm is most strongly absorbed by samples of its complementary colour, orange 26 a Any coloured solution can be chosen, for example chlorophyll solution. b The monochromator is used to select the optimum wavelength for absorption by the sample. c Electrons in the atom, ion or molecule are excited by the absorption of UV light, 580–620 nm. d Electrons move from a lower energy level to a higher energy level and subsequently return to the ground state. 27 a oxygen

Answers

Number of peaks H(*) spilt into

15

Light green

476

Number of types of 1H

CH3

1 a b c a Distance from origin (mm)

Molecule

OH+

Chapter 6 Band

29 a C–H peak = 2970 cm–1 b O–H peak = 3600 cm–1 30 a 4.4 × 10–6 m b 6.9 × 1014 Hz 31 HCl > HBr > HI 32 a 3 b A : B : C = 2 : 1 : 3 33 a and b

+

5 a

m/e

Relative abundance

1

<0.1

16

1.0

17

21

18

100

Possible fragment H

1 +

O

16 +

H O

1 16 +

H216O+

1

19

0.08

18

20

0.22

1

O 1H+ or 1H2H16O+ H2 18O+ or 2H216O+

b 18 c 18 d H2O e  H2 O or H2 O 7 CH3CH2OH + H2O — CH3CHOOH + 4H+ + 4e–; O2 + 4H+ + 4e– — 2H2O 8 1

m/e

18

2

16

Possible fragment

15

CH3+

27

C2H3+

29

C2H5+

31

CH3O+

45

C2H5O+

46

C2H6O+

9 a HPLC, UV–visible b HPLC c IR spectroscopy, fuel cell d mass, IR and NMR spectroscopy

11 Mass spectrometry

Infrared spectroscopy

‘Source’ with which sample interacts

The sample is bombarded with a beam of electrons

The sample is irradiated with infrared light

Chemical or physical basis for analysis

Different molecules yield ions with a different mass/charge ratio

Different molecules have bonds of differing strengths requiring differing amounts of energy to cause vibration.

Samples analysed

A huge range of elements and organic molecules. Samples must be able to be vaporised e.g. identification of an explosive in a bomb

A very wide range of organic and other molecules including solids, liquids and gases e.g. determination of the composition of soils or food materials

Chapter 9 E2 C8H18 octane 2 a alkane b alkene c alkene d alkane e alkane f alkene 7 a methanol b methanoic acid c chloromethane d methanamine 8 a 1‑chloropropane b 1‑chloropropane c pentan‑1‑ol d heptan‑3‑ol e propan-1-amine or 1-propylamine 9 a CH3CHOHCH2CH2CH2CH3 b CH3CH2CH2CH2CH2Cl c CH3CH2CH2CH2NH2 d CH3CH(CH3)CH2CH2CH2CH3 e CH3CH2CH2CHOHCH2CH2CH2CH2CH3 13 a 1‑chlorononane b propanoic acid 14 a CH3CHCl(CH2)4CH3 b CH3CHOHCH2CH3 c CH3CH2CHNH2CH2CH2CH2CH2CH2CH2CH3 or CH3CH2CHNH2(CH2)6CH3 d CH3(CH2)2COOH 16 a CH3CH2COOH; propanoic acid b CH3CHOHCH3; propan‑2‑ol c HCOOH; methanoic acid 21 HCOOH(l) + H2O(l) HCOO–(aq) + H3O+(aq) 23 The highly polar amino group can form hydrogen bonds with water molecules but as the non-polar hydrocarbon chain grows longer, as in butanamine, the solubility is reduced. 24 a CH3COOH(aq) + NaOH(aq) — CH3COONa(aq) + H2O(l) b 2CH3COOH(aq) + Na2CO3(s) — 2CH3COONa(aq) + CO2(g) + H2O(l) c 2CH3COOH(aq) + Mg(s) — (CH3COO)2Mg + H2(g)

Chapter 10

m/e

Possible ion

27

C2H3+

29

C2H5+

41

C3H5+

1 CH4(g) + Cl2(g) — CH3Cl(g) + HCl(g), CH3Cl(g) + Cl2(g) — CH2Cl2(g) + HCl(g), CH2Cl2(g) + Cl2(g) — CHCl3(g) + HCl(g), CHCl3(g) + Cl2(g) — CCl4(g) + HCl(g) 3 a chloroethane b 1,2‑dichloroethane c ethanol d ethane 6 a polar b polar c non-polar OH– 7 a CH3Cl(g) —— CH3OH(aq) Cl2 b CH4(g) —— CH3Cl(g) light OH– c CH3CH2CH2Cl(l) —— CH3CH2CH2OH(l)

42

C3H6+

8 a CH3CH3(g)

43

C3H7+

57

C4H9+

72

C5H12+

12 a m/e = 62 b m/e = 31 c CH2OH+ d anti-freeze e gas chromatography or HPLC 13 a C5H12 b

Thin-layer chromatography

Atomic absorption spectroscopy

i

organic compounds especially coloured components

atoms, principally metals

ii

compounds are analysed

atoms of particular elements in compounds are analysed

Cr2O72– Cl2 OH– —— CH3CH2Cl(g) —— CH3CH2OH(aq) —–—   H+ heat or UV light CH3COOH(aq) H2O Cr2O72–  b CH2CH2(g) —–— CH3CH2OH(g) —–— CH COOH(aq) 3 H+ H3PO4 catalyst 9 a substitution, substitution, oxidation b addition, oxidation 10 a CH3COOH(aq) + NaOH(aq) — CH3COONa(aq) + H2O(l) b 2CH3COOH(aq) + Mg(s) — (CH3COO)2Mg(aq) + H2O(l) c 2CH3COOH(aq) + Na2CO3(aq) — 2CH3COONa(aq) + CO2(g) + H2O(l) H2SO4 12 a CH3CH2CH2COOH(l) + CH3CH2OH(l) —–— CH3CH2CH2COOCH2CH3(l) + H2O(l) H2SO4 b HCOOH(l) + CH3CH2OH(l) —–— HCOOCH2CH3(l) + H2O(l) 14 a methyl pentanoate; from methanol and pentanoic acid b hexyl propanoate; from hexanol and propanoic acid 17 74% 18 a 32% b 6.3% Cl2 21 a CH3CH3(g) —— CH3CH2Cl(g) UV light

iii

components are separated

components not separated

iv

primarily qualitative analysis

qualitative and quantitative analysis

b CH2CH2(g)

v

cheap

more expensive

c CH3, methyl group, was lost to form ion m/e = 57 d CH3CH2CH2CH2CH3 — CH3CH2CH2CH2+ + CH3 e base peak = m/e 43 CH3CH2CH2+, CH3CH2CH2CH2CH3 — CH3CH2CH2+ + CH2CH3 14 a AAS b gravimetric analysis, volumetric analysis, AAS c IR, mass or NMR spectroscopy d gas chromatography 15 a

18 a Redox reaction Au/Au+ or O2/OH– b 53 g c AAS Unit 3 Area of Study 1 Review 1 B 2 B 3 B 4 C 5 A 6 B 7 C 8 C 9 A 10 A 11 D 12 B 13 A and B 14 B 15 D 16 C 17 A 18 D 19 D 20 C8H8O3 21 c 0.117 M 22 a yellow to pink b 2HCl(aq) + MgCO3(aq) — MgCl2(aq) + H2O(l) + CO2(g) c 0.0278 mol d 90.0% e binder and flavouring 23 a coconut oil, butterfat, olive oil, peanut oil, canola oil, sunflower oil b I2(aq) + 2e– — 2I–(aq); 2S2O32–(aq) — S4O62–(aq) + 2e–; I2(aq) + 2S2O32–(aq) — 2I–(aq) + S4O62–(aq) c 0.002 71 mol d 8.12 g e olive oil 24 d 0.160 M 26 a spectrum B propan‑2‑ol: peak at ~3350 cm–1 due to OH, peak at ~3000 cm–1 due to CH; spectrum A propanone: peak at ~3000 cm–1 due to CH, peak at ~1700 cm–1 due to C=O b Spectrum B is propan‑2‑ol; Spectrum A is propanone f i 58 ii Molar mass of propanone is 58. iii The peak at m/e 43 indicates that a fragment of molar mass 15, CH3, has been removed leaving a CH3CO+ ion having mass of 58 – 15 = 43. 27 c 0.12 d leucine

H2O —–—  CH3CH2OH(g) H3PO4 catalyst

OH– c CH3CH2Cl(g) —— CH3CH2OH(aq) Cr2O72–  CH3COOH(aq) d CH3CH2OH(aq) —–— H+ 22 a substitution b addition c substitution d oxidation Cl2 24 a CH3CH2CH2CH3(g) —— CH3CH2CH2CH2Cl(l) UV light OH– b CH3CH2CH2CH2Cl(l) —— CH3CH2CH2CH2OH(aq) Cr2O72–  CH3CH2CH2COOH(aq) c CH3CH2CH2CH2OH(aq) —–— H+ – MnO4  CH3CH2COOH(aq) d CH3CH2CH2OH(aq) —–— H+ e 2C2H6(g) + 7O2(g) — 4CO2(g) + 6H2O(g) 26 a I methane, II chloroethane, III propane, IV ethanoic acid, V propene b I, III c Compound V will decolourise bromine almost immediately, whereas compound I will not. In general, compound I participates in substitution reactions (e.g. chlorination by chlorine gas to form chloromethane) whereas compound V participates in addition reactions (e.g. addition of HCl across the double bond to form chloropropane). d ethanol e IV 27 a CH3CH2COOCH3(l) + H2O(l) b CH3COO(CH2)4CH3(l) + H2O(l) c HCOOCH(CH3)2(l) + H2O(l) Answers

477

H2SO4 28 a CH3COOH(l) + CH3(CH2)2OH(l) — —h CH3COO(CH2)2CH3(l) + H2O(l) b condensation and esterification c catalyst H2SO4 29 a CH3COOH(l) + CH3OH(l) — —h CH3COOCH3(l) + H2O(l) H2SO4 b CH3CH2COOH(l) + CH3CH2OH(l) — —h CH3CH2COOCH2CH3(l) + H2O(l) 30 a CH3CH2OH, CH3COOH b CH3CH2OH, CH3CH2COOH 33 P = CH3OH; Q = H2O; R = propanoic acid; S = methyl propanoate

Chapter 11 1 2H2(g) + O2(g) —h 2H2O(g) 2 b C2H5OH(l) + 3O2(g) —h 2CO2(g) + 3H2O(g) 3 a C17H35COOH(l) + CH3OH(l) —h C17H35COOCH3(l) + H2O(l) b 2C17H35COOCH3(l) + 55O2(g) —h 38CO2(g) + 38H2O(g) 6 a a mixture of 10% ethanol and 90% conventional petrol 8 a 2CO(g) + O2(g) —h 2CO2(g) b 2H2(g) + O2(g) → 2H2O(g) 9 2C18H38(l) + 55O2(g) —h 36CO2(g) + 38H2O(g) 11 A: photosynthesis. B: fermentation. Carbon dioxide and water are converted to glucose and oxygen by photosynthesis. During fermentation glucose is converted to ethanol and carbon dioxide.

Chapter 12 1 hydroxyl, carboxyl 3 c i HOCH2CHOHCH2OH and CH3(CH2)14COOH ii water iii hydrolysis 7 b condensation c hydroxyl functional groups 10 Humans lack the enzyme (cellulase) that catalyses the hydrolysis of cellulose to form glucose. 12 amino, carboxy 14 b H2NCH(CH2OH)COOH + H2NCH(CH2SH)COOH —h H2NCH(CH2OH)CONHCH(CH2SH)COOH + H2O c condensation 27 a glycerol d polyunsaturated 28 b enzyme 32 a H2NCHCH3CONHCH2COOH; H2NCH2CONHCHCH3COOH b 6 (using each amino acid once in each peptide) c very large numbers 33 a i carboxy ii amino 34 b H2NCHCH3COOH; H2NCH(CH2SH)COOH; H2NCH2COOH; H2NCH(CH2OH)COOH c stomach and small intestine 45 a carboxyl and hydroxyl groups; lipids b two hydroxyl groups; carbohydrates c carboxyl and amino groups; proteins 47 b histidine, leucine c ion–dipole interaction, hydrogen bonds, ionic interaction d 7 amide, 1 terminal carboxyl group, 1 carboxyl group on the Z group of asp. 48 a Absorbance of 1600–1700 cm–1 indicative of C=C bond b i addition ii n = 0.0204 mol c 2 double bonds d linoleic acid

Chapter 13 2 TTTGGACTTGCT 3 a They are derivatives of purine. b They are derivatives of pyrimidine. c The width of the double helix is constant because in each ‘rung’ one purine base is hydrogen-bonded to one pyrimidine base. 4 a i Thymine is replaced by adenine at one point in the DNA sequence on chromosome number 11. ii The sixth amino acid in the β-globin polypeptide chains is valine, instead of glutamic acid. b i A sequence of just three nucleotides is missing on chromosome 7. ii One phenylalanine molecule is omitted from the sequence of amino acids used to make the chloride transport polypeptide chain. 8 CTAATAGTTA 9 a cystic fibrosis, PKU, sickle-cell anaemia

Chapter 14 9 a ethanoic acid, CH3COOH b 58% Unit 3 Area of Study 2 Review 1 C 2 B 3 A 4 C 5 A 6 C 7 C 8 C 9 C 10 D 11 A 12 C 13 D 14 B 15 A 16 B 17 B 18 A 19 A 20 b i Brown bromine solution becomes colourless. ii C2H4(g) + Br2(aq) —h CH2BrCH2Br(l) c i IR spectroscopy ii characteristic absorption of C=C bond at 2070 to 2250 cm–1 d i addition polymerisation ii nCH2CH2 —h ~(CH2CH2)n~ iii polyethene 21 b i 2C8H18(l) + 25O2(g) —h 16CO2(g) + 18H2O(g) c i C2H5OH(l) + 3O2(g) —h 2CO2(g) + 3H2O(g) 23 a ethanoic anhydride b 2C7H6O3 + C4H6O3 —h 2C9H8O4 + H2O 25 a salicylic acid b 13.0 g 26 c i propan-1-ol ii butanoic acid iii ethylbutanoate 27 a ethanol, hexanoic acid b Concentrated sulfuric acid is added as a catalyst. c CH3CH2OH(l) + CH3(CH2)4COOH(l) —h CH3(CH2)4COOCH2CH3(l) + H2O(l)

478 Answers

28 a deoxyribonucleic acid, DNA c condensation polymerisation, water 30 c +NH3CH(CH2SH)COOH 31 b i glycoside link ii ester link iii peptide bond 32 b CH2O c C2H4O2 g methyl methanoate

Chapter 15 2 a endothermic b exothermic c endothermic d exothermic 6 a exothermic b endothermic c endothermic d exothermic 7 endothermic 10 a 1370 kJ mol–1 b –572 kJ mol–1 d +572 kJ mol–1 11 a CH4(g) + 2O2(g) —h CO2(g) + 2H2O(g) b –890 kJ mol–1 12 139 kJ mol–1 19 a CaCO3(s) + 2HCl(aq) —h CaCl2(aq) + CO2(g) + H2O(l) b CaCO3 is in excess 20 b endothermic 21 a The temperature of the new process (550°C) is much lower than the temperature of 1800°C used in the blast furnace. b 2Fe(s) + O2(g) —h 2FeO(s), 4Fe(s) + 3O2(g) —h 2Fe2O3(s) c No. The reaction of iron with water is different from the reaction of iron with dry oxygen and proceeds at a different rate. d exothermic

Chapter 16

[N2][H2]3 [N2]2[H2O]6 bK= [NH3]2 [NH3]4[O2]3 [Cu(NH3)42+] [Ni(NH3)62+] [NH3]2 d K = 2+ 3aK= b 0.01 M–2 cK= [N2][H2]3 [Cu2+][NH3]4 [Ni ][NH3]6 [N2]1/2[H2]3/2 cK= d 10 M 5 a increase b decrease c increase d decrease [NH3] 6 a increase b decrease c increase d decrease 7 0.69 M 8 a 0.011 M b 0.020 9 a net forward reaction b net back reaction c net forward reaction 10 a net back reaction b net forward reaction c net forward reaction d no effect e net forward reaction 11 a i increase ii increase iii decrease b i cannot cause forward reaction [Sn2+][Fe3+]2 ii increase iii cannot cause forward reaction 14 a K = 4+ 2+ 2 b No; K is very [Sn ][Fe ] small. c 1010 d Yes; K is very large (provided the rate is sufficiently fast) 15 a CH3OH(g) 2H2(g) + CO(g) b S2(g) + 2H2(g) 2H2S(g) c NO2(g) N2O4(g) [PCl3][Cl2] 16 0.019 M 17 a K = b 0.400 M c 0.80 M d 2.5 M–1 20 0.0980 M [PCl5] –2 21 0.0024 M 22 a 1.5 M b 0.016 M c 0.032 mol 23 a 0.7 mol b 1.3 mol c 0.29 24 a increase b increase 25 a net back reaction b net forward reaction c no effect 27 a i net forward reaction ii no change iii net forward reaction b i increase ii no change iii no change c likely to be large 30 b i increase ii increase iii increase iv increase v no change 31 a i high temperature; high pressure ii low temperature; high pressure b temperature c use a catalyst 32 a i increase ii increase 36 a 2SO3(g) + CO2(g) CS2(g) + 4O2(g) [CS2][O2]4 bK= c 1.6 × 10–3 M2 d i increase ii increase iii no effect [SO3]2[CO2] [SO3]2 iv decrease v no effect 37 a K = b 15–20 minutes; 25–30 minutes; [SO2]2 [O2] –1 35–40 minutes c 1.56 M at 20 minutes 3.53 M–1 at 30 minutes and 0.32 M–1 at 40 minutes d 10 minutes; before 10 minutes the concentrations were changing slowly but the catalyst caused equilibrium to be reached more rapidly e Increased pressure by reducing volume. Also, because the value of K at 25–30 minutes is larger than at 15–20 minutes and this is an exothermic reaction, a decrease in temperature must have occurred. E1 NaI(s)

Na+(aq) + I–(aq) 2 a K =

Chapter 17 1 a 4 b 1.3 c 10 d 12.3 2 a 1.0 × 10–6 M b 1.0 × 10–11 M c 1.0 × 10–8 M [NH3][H3O+] [HCOO–][H3O+] [CN–][H3O+] 4 a Ka = b Ka = c Ka = 5 a 1.4 b 3.6% + [NH4 ] [HCOOH] [HCN] 7 a HNO3 and NO3–, H3O+ and H2O b HCN and CN–, H3O+ and H2O c H2CO3 and HCO3–, H2O and OH– d HPO42– and PO43–, H2O and OH– 8 a 1 × 10–12 M b 2 9 4.3 10 1.00 11 4.00 12 3.3 g 13 a 2.2 b 1.8 c 7 14 a As the temperature increases the forward endothermic reaction is favoured and the [H3O+] and [OH–] will both increase resulting in an increase in Kw. b both increase c decreases [H+][Mr –] d neutral; since [H3O+] = [OH–] 15 a K = b 10–3 c red d i orange [HMr] ii yellow

16 a HOI(aq) + H2O(l) H3O+(aq) + OI–(aq) [H3O+][OI–] b Ka = c 1.6 × 10–6 M d 2.6 × 10–11 M 17 1.6 × 10–4 M [HOI] 18 0.057 M 19 a 2.0 × 10–4 M b 4.4% 22 b 1.14 × 10–3 M c 2.94 d 1.14% 23 a C4H8O2(l) + H2O(l) —h C4H7O2–(aq) + H3O+(aq) [C4H7O2–][H3O+] b Ka = d 2.9 [C4H8O2]

Chapter 18 8 c All users (label); emergency services personnel (HAZCHEM sign) 9 a the ‘Y’ means ‘use breathing apparatus’. b Hazard class 1.1 is ‘most explosive’, 1.6 is ‘least explosive’ and 1.3 is in between. 11 a R23 Toxic by inhalation, R24 Toxic in contact with skin, R25 Toxic if swallowed, S23 Do not breathe gas/fumes/vapour/ spray, S24 Avoid contact with the skin, S25 Avoid contact with eyes, S37 Wear suitable gloves, S38 In case of insufficient ventilation wear suitable respiratory equipment, S39 Wear eye/face protection b Responses should cover identification of hazardous chemicals (e.g. nicotine); assessment of the risk; control of the risk and documentation. c Compulsory health warning labels, restrictions on age of purchaser, restrictions on venues where smoking is permitted. 27 a from top to bottom: explosive, flammable, oxidising (left), radioactive (right), corrosive b i corrosive ii flammable 28 b CH3COOH and Na2CO3 are less hazardous than HCl and NaOH 30 a HCl reacts with many metals. b Collect for disposal (silver can be recycled). 33 a Acetone is a respiratory irritant that can cause drowsiness, dizziness or death. b Acetone is highly flammable. c the risk of autoignition at 538°C is low. d Sparks from static electricity may ignite acetone vapours. 35 a organic waste b aqueous waste c aqueous waste d sink or aqueous waste

Chapter 19 1 a NH3(aq) + HNO3(aq) NH4NO3(aq); 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq); 3NH3(aq) + H3PO4(aq) (NH4)3PO4(aq) b ammonium nitrate 8 a N2(g) + 3H2(g) 2NH3(g) [NH3]2 bK= 12 Recycling of nitrogen and hydrogen gases; use of heat [N2][H2]3 exchangers to recover energy released by the reaction between nitrogen and hydrogen.

Chapter 20 1 a 4C7H5N3O6(l) + 21O2(g) —h 28CO2(g) + 10H2O(g) + 6N2(g) E1 a [~NH(CH2)6NHCO(CH2)4CO~]n b water molecule c amide bonds NHCO [NO2]2 5 a 42% b 77.7% 8 a 2NO(g) + O2(g) 2NO2(g) b K = [NO]2[O2] 2NO2(g) 9 a 2NO(g) + O2(g) b 4NH3(g) + 5O2(g) —h 4NO(g) + 6H2O(g) c NH3(g) + 2O2(g) —h HNO3(aq) + H2O(l)

Chapter 21 5 a 68% b i 2SO2(g) + O2(g) + 2H2O(l) 2H2SO4(l) ii 100% 6 a S(l) + O2(g) —h SO2(g) b Cu2S(s) + O2(g) —h 2Cu(s) + SO2(g) [SO3]2 7 a 2SO2(g) + O2(g) —h 2SO3(g) b K = [SO2]2[O2] 10 a 2SO2(g) + O2(g) —h 2SO3(g) b S(s) + O2(g) —h SO2(g)

8 d for example: C7H16(g) C2H4(g) + C5H12(g), C7H16(g) C3H8(g) + C4H8(g) 10 a Ethane is present in the products because the cracking process involves equilibrium reactions. The ethane could be recycled. b hydrogen c differences in boiling points of the gases 14 a A: C2H4Br2, 1,2-dibromoethane; B: for example C2H6, ethane; E: C2H6, ethane b water and a catalyst (phosphoric acid) c Steam and the hydrocarbon fraction are passed through coiled metal tubes in a furnace at 750°C to 900°C. 15 Use low pressure, high temperature and a suitable catalyst, and employ a continuous flow process so products are continually removed. 16 a Crude oil is a mixture of hydrocarbons. Most are alkanes, with formulas ranging from CH4 to C70H142. b X: fractional distillation; Y: thermal cracking c Mixture Z contains saturated hydrocarbons of similar masses and similar boiling temperatures. d natural gas H3PO4 e A: ethanol; C2H4(g) + H2O(g) —–—h CH3CH2OH(g); catalyst B: polyethene; nC2H4(g) —––—h (CH2CH2)n(s); C: carbon dioxide; C2H4(g) + 3O2(g) —h 2CO2(g) + 2H2O(g) f addition reactions Unit 4 Area of Study 1 Review 1 A 2 D 3 B 4 D 5 A 6 D 7 D 8 B 9 C 10 B 11 C 12 C 13 A 14 D 15 A 16 C 19 a A is NO2 and B, N2O4 21 a Iodine vapour has been added to the mixture. b the temperature of the mixture has been increased. 22 a pH = –log10 [H3O+(aq)] [CN–(aq)][H3O+(aq)] b Ka = c 7 × 10–6 M d 0.1 M e 4.9 × 10–10 M f 1, 5.15 [HCN(aq)] 24 a pH = 1.7, pH = 3.4 c yellow 25 2.0 × 10–2 M 27 for ammonia: a nitrogen (N2) from air and hydrogen (H2) b N2(g) + 3H2(g) 2NH3(g); for nitric acid: a ammonia (NH3), oxygen (O2) from air and water (H2O) b 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g); 2NO(g) + O2(g) 2NO2(g); 3NO2(g) + H2O(l) ) 2HNO3(aq) + NO(g); for sulfuric acid: a sulfur (S) or sulfur dioxide (SO2), oxygen (O2) from air and water (H2O) b S(l) + O2(g) SO2(g); 2SO2(g) + O2(g) 2SO3(g); SO3(g) + H2SO4(l) H2S2O7(l); H2S2O7(l) + H2O(l) 2H2SO4(l); for ethene: a ethane (C2H6) or propane (C3H8) from natural gas or naphtha (e.g. C8H18) or gas-oil (e.g. C16H34) fractions from crude oil b for example C3H8(g) C2H4(g) + CH4(g) 28 a Ethanol is flammable and toxic. b LD50 = 5860 mg kg–1, the lethal dose for 50% rats (based on laboratory tests) c identify hazardous chemicals, assess risk, control risk, document risk d safety glasses, gloves, respirator or ventilation, remove sources of ignition 29 a low yields become more profitable —h increased production b increased cost of electricity —h reduced production and use c increased cost of products —h reduced production and use d increased cost use —h reduced production and use e increased cost of disposal —h reduced production and use f Costly start-up favours continuous production but small quantities favour batch processing.

Chapter 23 6 b C29H60(g) —h C8H18(g) + C8H16(g) + C13H26(g) c alkanes and alkenes 7 c CH4(g) + 2O2(g) —h CO2(g) + 2H2O(g) d 2H2(g) + O2(g) —h 2H2O(g) e coal > natural gas > hydrogen 8 a 3.0 × 1014 J b 3.2 × 104 years d 2.0 tonnes 9 6.6 × 1012 L 12 b Hydrocarbon

Fraction

Use

c SO3(g) + H2SO4(l) —h H2S2O7(l)

C15H32

gas oil

diesel, heating oil, cracked to produce petrol

12 to prevent poisoning of the catalyst

C3H8

refinery gas

LPG, feedstock for the production of organic chemicals

Chapter 22

C9H20

naphtha

E2 nickel or platinum E4 paraffin wax vapour and smaller hydrocarbons 3 C14H30(g) —h 7C2H4(g) + H2(g) 4 high temperatures increase the value of K, low pressures cause a net forward reaction, short reaction time restricts the extent of secondary reactions 6 a 61% b 64%

feedstock for the production of organic chemicals, cracked to produce petrol

C30H62

residue, bitumen

fuel oil, lubricating oil, asphalt

C5H12

gasoline

petrol

19 a CO2(g) + H2O(l) —h H2CO3(aq) → H+(aq) + HCO3–(aq)

479 Answers

Chapter 24 239 0 4 a 10n + 238 92U —h 94Pu + 2–1β 11 a Improved engine tune wastes less petrol. b Reduced travel times waste less petrol. c Energy for heat and light is required for shorter periods. d Continuous monitoring of raw materials and products can provide feedback to maintain optimum energy consumption for industrial processes. 15 a 1.5 × 107 m2 b 1.9 × 104 blocks

Chapter 25 1 a 0.25 kJ 2 a 4.68 × 103 kJ b 5.19 × 103 kJ c 637 kJ 3 1.44 L E5 47.72 MJ kg–1 4 a 25.1 kJ b 78.5 kJ c 62.8 kJ d 103 kJ e 54.1 kJ 5 a 461 J °C–1 b –55.8 MJ mol–1 6 a C3H8(g) + 5O2(g) —h 3CO2(g) + 4H2O(l) b 2.23 kJ c –2.23 MJ mol–1 d C3H8(g) + 5O2(g) —h 3CO2(g) + 4H2O(g); ΔH = –2.23 MJ mol–1 7 a 1 mol of CO(g), 0.5 mol of O2(g) b i –566 kJ mol–1 ii +566 kJ mol–1 8 a 286 kJ b 143 MJ c 1.17 MJ 9 7.83 L 10 a 497 kJ b 11.8 kJ c 7.68 mL 11 b 55.3 kJ 12 a –49.97 kJ g–1 b 29.0 kJ min–1 13 286 J 14 a C6H12O6(aq) + 6O2(g) —h 6CO2(g) + 6H2O(l); ΔH = –2803 kJ mol–1 b 77.9 kJ 15 a Ammonia is produced. b 23.4 g 19 Water, because it has a higher specific heat (4.184 J °C–1 g–1) than sand (0.48 J °C–1 g–1). 20 a 35.5 kJ b 157 kJ c 60.0 kJ d 49.3 kJ 21 80°C 22 a 18.07°C 23 a exothermic b Na+(aq) + CH3COO–(aq) CH3COONa(s); ΔH negative c an endothermic reaction, similar to that in a cold pack (e.g. NH4NO3(s) NH4+(aq) + NO3–(aq); ΔH = 25 kJ mol–1) 24 a endothermic b decrease d i 0.566 kJ ii 0.566 kJ e i 1.35°C ii 0.135°C H2O(l) 25 327 J °C–1 26 a KNO3(s) —–—h K+(aq) + NO3–(aq) b 1.16 kJ °C–1 c 783 J d KNO3(s) —h K+(aq) + NO3–(aq); ΔH = +78.3 kJ mol–1 27 a C2H4(g) + 3O2(g) —h 2CO2(g) + 2H2O(l) b ΔH = –1416 kJ mol–1 28 16 kJ g–1 30 a 430 mL 31 a 2H2(g) + O2(g) —h 2H2O(g); ΔH = –572 kJ mol–1, CH4(g) + 2O2(g) —h CO2(g) + 2H2O(g); ΔH = –890 kJ mol–1 b 6.48 × 10–3 mol 32 CH4 –55.6 kJ g–1,C3H8 –50.5 kJ g–1 33 a CH3CH2OH(g) + 3O2(g) —h 2CO2(g) + 3H2O(l); ΔH = –1367 kJ mol–1, 25 —h 8CO2(g) + 9H2O(l); ΔH = –5450 kJ mol–1 C8H18(g) + — 2 O2(g) b –47.7 kJ g–1 c octane d shorter 34 93.2 kJ L–1 35 a 33.0 kJ g–1 36 a 1.7 × 106 kJ b 5.5 × 107 kJ c $1120 37 a octane 47.7 kJ g–1, black coal 33 kJ g–1; octane is the better fuel. 38 a 56 s b 837 J °C–1

Chapter 26 3 1.10 V E1 a 1.10 V b 1.2 V c 1.01 V 4 a i Reaction will occur. ii Zn(s) —h Zn2+(aq) + 2e– (oxidation); Cl2(g) + 2e– —h 2Cl–(aq) (reduction) iii Zn(s) + Cl2(g) —h Zn2+(aq) + 2Cl–(aq) b i no reaction c i Reaction will occur. ii Zn(s) —h Zn2+(aq) + 2e– (oxidation); 2Ag+(aq) + e– —h 2Ag(s) (reduction) iii 2Ag+(aq) + Zn(s) —h 2Ag(s) + Zn2+(aq) d i no reaction 6 a 3Cu2+(aq) + 2Al(s) —h 3Cu(s) + 2Al3+(aq) b no reaction c 2Ag+(aq) + Zn(s) —h 2Ag(s) + Zn2+(aq) d 2H+(aq) + Fe(s) —h H2(g) + Fe2+(aq) e no reaction 8 a Ag+(aq); Mg2+(aq) b Mg(s); Ag(s) c A coating of silver will form on the lead when it is placed in silver nitrate solution because Ag+ ions are stronger oxidants than Pb2+ ions. d zinc and magnesium 9 a oxidant b strong reductant c strong oxidant d true e false f true 10 a predict Fe2+(aq) and H+(aq) are formed b 2Fe3+(aq) + H2(g) —h 2Fe2+(aq) + 2H+(aq)

Chapter 27 3 dry cell: Zn(s) + 2MnO2(s) + 2NH4+(aq) —h Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l); alkaline cell: Zn(s) + 2MnO2(s) + H2O(l) —h Zn(OH)2(s) + Mn2O3(s); lithium cell: Li + MnO2 —h LiMnO2; silver–zinc cell: Zn(s) + Ag2O(s) + H2O(l) —h Zn(OH)2(s) + 2Ag(s) 5 a Zn(s) + Cl2(g) —h Zn2+(aq) + 2Cl–(aq) b negative c 2.12 V E1 anode reaction Na(l) —h Na+(l) + e–; cathode reaction S(l) + 2e– —h S2–(l) E2 2Na(l) + S(l) —h S2–(l) + 2Na+(l); S2–(l) + 2Na+(l) —h 2Na(l) + S(l) 7 58 g, 1.5 × 102 g

480 Answers

19 a anode: Al(s) —h Al3+(aq) + 3e–; cathode: O2(g) + 2H2O(l) + 4e– —h 4OH–(aq) 20 a anode: CH3CH2OH(g) + 3H2O(l) —h 2CO2(g) + 12H+(aq) + 12e–; cathode: O2(g) + 4H+(aq) + 4e– —h 2H2O(l); overall: CH3CH2OH(g) + 3O2(g) —h 2CO2(g) + 3H2O(l) 21 a 2CH3OH(g) + 3O2(g) —h 2CO2(g) + 4H2O(l) b reduction of O2(g) c NaOH(aq) or KOH(aq) 22 a Zn(s) + 4OH–(aq) + Cu2+(aq) —h Zn(OH)42–(aq) + 2Cu(s) 23 a CH4(g) + 2O2(g) —h CO2(g) + 2H2O(l)

Chapter 28 3 a 9.00 × 103 C b 4.3 × 103 C 4 a 4.50 × 104 C b 0.466 mol c 8.08 g 5 44.7 s 6 4.08 A 7 a Br2(aq); Cu(s) b I2(aq); H2(g) and OH–(aq) c O2(g) and H+(aq); Pb(s) d O2(g) and H+(aq) (Cl2(g) is formed in practice); Zn(s) e O2(g) and H+(aq); H2(g) and OH–(aq) 9 a A faraday is the charge on 1 mole of electrons, 96 500 C. b i 1 F ii 2 F iii 2 F iv 2 F 10 70 g 11 a 1.10 × 106 L b 5.24 tonnes 12 a 4+ b 4.00 F c 139 g 13 a 2H2O(l) —h 4H+(aq) + O2(g) + 4e–; Cu2+(aq) + 2e– —h Cu(s) b 0.800 M 14 Cu 2.95 g; Cr 1.61 g 15 132 g 16 a cathode b negative c 23.5 h 17 2+ 18 a 3.95 g b 1.00 M c O2(g) d 0.75 M 19 magnesium 20 a 4500 C b A: 2H2O(l) —h O2(g) + 4H+(aq) + 4e–; B: 2H2O(l) + 2e– —h H2(g) + 2OH–(aq); C: Cu(s) —h Cu2+(aq) + 2e–; D: 2H+(aq) + 2e– —h H2(g) c 1.48 g d 0.571 L e Cu2+(aq) + 2e– —h Cu(s) 23 (cathode reaction; then anode reaction) a K+(l) + e– —h K(l); 2I–(l) —h I2(g) + 2e– b Cu2+(aq) + 2e– —h Cu(s); 2H2O(l) —h O2(g) + 4H+(aq) + 4e– c 2H2O(l) + 2e– —h H2(g) + 2OH–(aq); Cu(s) —h Cu2+(aq) + 2e– d Cu2+(aq) + 2e– —h Cu(s); 2H2O(l) —h O2(g) + 4H+(aq) + 4e– e 2H2O(l) + 2e– —h H2(g) + 2OH–(aq); 2H2O(l) —h O2(g) + 4H+(aq) + 4e– 25 b 2H2O(l) + 2e– —h H2(g) + 2OH–(aq) 26 a 6.2 × 102 g b 3.3 × 105 L 29 a 19 h 51 min b 333 kg c 6.23 × 105 L 30 a 1.07 × 1010 C b 5.36 × 1010 J c $1780 31 a 2O2– (in cryolite) —h O2(g) + 4e– 32 6.811 × 105 L 33 a anode b a thin sheet of pure copper c sulfuric acid d anode Cu(s) —h Cu2+(aq) + 2e–; cathode Cu2+(aq) + 2e– —h Cu(s) Unit 4 Area of Study 2 Review 1 D 2 B 3 B 4 C 5 B 6 C 7 A 8 A 9 D 10 C 11 D 12 B 13 C 14 A 15 A 16 B 17 B 19 a 1636 J °C–1 b 9.82 kJ g–1 20 a 742 MJ s–1 21 a cadmium b KOH (or NaOH) d positive electrode: Ni(OH)2(s) + OH–(aq) —h NiOOH(s) + H2O(l) + e–; negative electrode: Cd(OH)2(s) + 2e– —h Cd(s) + 2OH–(aq); overall: 2Ni(OH)2(s) + Cd(OH)2(s) —h 2NiOOH(s) + 2H2O(l) + Cd(s) e electrical energy —h chemical energy 22 a Zn(s) —h Zn2+(aq) + 2e– b 4.0 years 23 a Ethanol: the CH3 carbon is in the –3 oxidation state and the CH2OH carbon is in –1 oxidation state (attributing zero effect to the C–C bond); carbon dioxide: carbon is in the +4 oxidation state. Carbon is oxidised. b C2H5OH(g) + 3H2O(l) —h 2CO2(g) + 12H+(aq) + 12e– 24 a i 45.43 kJ ii 727 kJ iii 1454 kJ mol–1 b i Oxidation occurs at the anode. The anode reaction is: CH3OH(l) + H2O(l) —h CO2(g) + 6H+(aq) + 6e– ii 2CH3OH(l) + 3O2(g) —h CO2(g) + 4H2O(l) iii 19.3 A iv 730 kJ mol–1 26 a i 2H2O(l) —h O2(g) + 4H+(aq) + 4e– ii Sn2+(aq) + 2e– —h Sn(s) b 2H2O(l) + 2e– —h H2(g) + 2OH–(aq) c 2Cl–(aq) —h Cl2(g) + 2e– 27 b spatula (negative): Ni2+(aq) + 2e– —h Ni(s); carbon rod (positive): 2H2O(l) —h O2(g) + 4H+(aq) + 4e– c 0.68 g 28 a Li+(l) + e– —h Li(l) b 2Cl–(l) —h Cl2(g) + 2e– 30 a i sodium, aluminium ii chlorine, sodium hydroxide, copper 31 a anode: 2H2O(l) —h O2(g) + 4H+(aq) + 4e–; cathode: 2H2O(l) + 2e– —h H2(g) + 2OH–(aq) b anode: 2Cl–(aq) —h Cl2(g) + 2e–; cathode: 2H2O(l) + 2e– —h H2(g) + 2OH–(aq) c anode: 2Cl–(aq) —h Cl2(g) + 2e–; cathode: Na+(l) + e– —h Na(l)

Glossary absorbance The proportion of incident electromagnetic

alkanoic acid Systematic name for a carboxylic acid. An

radiation absorbed by the sample. absorption tower A reaction vessel in the contact process in which sulfur trioxide is converted into oleum. absorption spectrum A plot of the proportion of electromagnetic radiation absorbed by a substance over a range of wavelengths (or frequencies). accumulator A secondary or rechargeable electrochemical cell. acetylsalicylic acid An analgesic commonly known as aspirin. acid A substance that donates hydrogen ions (protons). acid rain Rainwater that has reacted with acidic emissions from industry and has a pH less than 5.5. acid–base indicator A substance whose colour depends on the concentration of H3O+ ions in solution. Indicators are weak acids and bases; their acid form is one colour and their conjugate base is another. acid–base reaction A reaction involving the transfer of a hydrogen ion (proton) from an acid to a base. acidic oxide An oxide that either reacts with water to form an acid or reacts with hydroxide ions. acidic solution Aqueous solution in which [H3O+] > [OH–]. At 25°C, pH < 7. acidity The concentration of H3O+ ions in an aqueous solution. Acidity is measured using the pH scale. acidity constant An equilibrium constant, Ka, which measures the degree of ionisation, or strength, of an acid. activation energy The energy required by reactants for reaction to occur. This energy is needed to break the bonds between atoms in the reactants to allow products to form. active site The location on an enzyme’s surface at which reaction occurs. It is usually a flexible hollow or cavity in the molecule. addition polymerisation Reaction of a large number of unsaturated monomers to form a polymer that has the same empirical formula as the monomers. addition reaction A reaction in which a molecule binds to an unsaturated hydrocarbon, forming a single carbon-tocarbon bond. In this process two reactant molecules become one. adenine A base found in nucleic acids such as DNA. It is the complementary base of thymine. adsorption The attraction of one substance to the surface of another. alcohol An organic compound containing the hydroxyl functional group, e.g. ethanol (CH3CH2OH). aliquot A volume of liquid measured by a pipette. alkaline cell A commercial electrochemical cell with an alkaline electrolyte that is a moist paste rather than a solution. An organic compound that has the carboxyl functional group attached to an alkane chain, e.g. CH3COOH0. alkanes Hydrocarbons with only single bonds between carbon atoms, which have the general formula CnH2n + 2.

organic compound that has the carboxyl functional group attached to an alkane chain, e.g. CH3COOH. alkanol Systematic name for an alcohol. An organic compound that has the hydroxyl functional group attached to an alkane chain, e.g. ethanol, CH3CH2OH. alkenes Hydrocarbons with one double bond between carbon atoms, which have the general formula CnH2n. alkyl group A group obtained by removing a hydrogen atom from an alkane, with general formula CnH2n + 1, e.g. methyl (CH3). alkyne Hydrocarbon with one triple bond between carbon atoms, which have the general formula CnH2n – 2. alternative reaction pathway A product may be formed by more than one reaction pathway. Each alternative pathway involves a different sequence of intermediate reactions, each having a different activation energy. amide A compound containing the CONH functional group. This group forms the link between amino acids in proteins. See peptide linkage. amine An organic molecule containing the NH2 functional group. amino acid A molecule that contains both an amino group and a carboxy group. α-amino acid An amino acid that has its amino, carboxy and Z groups attached to one central carbon atom. Also known as a 2-amino acid. 2-amino acid See α-amino acid. amount of substance A measure used by chemists for counting particles; the unit is the mole. amphoteric oxide An oxide that can act as an acid or a base. analgesic A drug that relieves pain. anhydrous An anhydrous solid does not contain water molecules in its lattice structure. annealing The stage in the polymerase chain reaction (PCR) where primers form hydrogen bonds to their complementary base sequences at the end of a DNA strand. anode An electrode at which an oxidation reaction occurs. anodising An electrolytic process that increases the thickness of the oxide layer on aluminium, making it harder and more corrosion resistant. antibiotic A chemical used to treat infectious disease in humans and animals. arenes Cyclic (ring-shaped) molecules in which electrons are delocalised around the ring, e.g. benzene. aspirin A common drug used to treat pain, fever and inflammation, and to reduce blood clotting. Also known a acetylsalicylic acid. atomic absorption spectroscopy (AAS) An analytical technique that uses light absorption to measure the concentration of a metal in a sample. atomic emission spectrometry (AES) An analytical technique that uses light emission to measure the concentration and identity of a metal in a sample.

481 Glossary

atomic number Number of protons in the nucleus of an atom.

average (mean) titre The value to use in calculating the result of a volumetric analysis. Found by taking the average of three concordant titres. Avogadro’s number The number of particles in a mole; symbol NA. NA = 6.022 × 1023.

buffer A solution composed of a weak acid and its conjugate base that resists changes of pH.

burette An item of volumetric equipment that delivers variable volumes of solution accurately.

button cell A small galvanic cell used in watches, cameras, calculators etc.

by-product A substance that is produced at the same time as an intended product in the course of a chemical reaction.

B100 100% biodiesel fuel. back titration A technique used when an acid or base is too weak to be analysed by direct titration. The process involves adding an excess of a strong base or acid to the original substance and then determining, by titration, how much of the added chemical remains unreacted. base A substance that accepts hydrogen ions (protons). base pairs Pairs of complementary bases found in DNA, i.e. (i) adenine (A) and thymine (T), or (ii) cytosine (C) and guanine (G). base peak The highest intensity peak in a mass spectrum; it is assigned an intensity of 100%. basic solution A solution where [H3O+] < [OH–]. At 25°C, pH > 7. basic oxide An oxide of a metal which either reacts with water to form hydroxide ions or reacts with H3O+ ions. batch processing A method of performing reactions in which fixed amounts of reactants are mixed, giving fixed amounts of products. battery A combination of cells connected in series. beamlines Instrumentation in the synchrotron that transports the photon beam from the source and selects the wavelength of radiation emitted. benzene ring A six-carbon ring, C6H6, containing delocalised electrons. Forms compounds by replacing hydrogens with functional groups. bioaccumulative Describes the tendency of a substance to increase in concentration as it moves through a food chain. biochemical fuel (biofuel) A fuel produced from biomass, i.e. from a plant or animal source or from micro-organisms. biodegradable Able to be broken down by biological processes. biodiesel A biochemical fuel made from vegetable oils or animal fats that can be used in place of diesel from petroleum. Biodiesel molecules contain the ester functional group so are also known as bioesters. biofuel A fuel derived from plant, animal or microbial material, e.g. bioethanol, biodiesel. Biofuels are considered to be renewable energy resources. biogas Combustible gas that is produced from the decay of organic matter in the absence of oxygen. biomass The mass of living matter in a population of organisms. black coal A fuel made from fossilised primitive plant material, with a higher carbon content than brown coal. brown coal A fuel made from fossilised primitive plant material with a carbon content between that of peat and black coal. bubble caps Caps fitted in the horizontal trays of a crude oil fractionating tower to impede the flow of gases through the tower, giving better separation of hydrocarbons.

482 Glossary

calibration factor A measure of the amount of energy required to raise the temperature of a calorimeter and its contents by 1°C. calibration graph A plot of data involving two variables that is used to determine values for one of the variables. calorimeter An instrument designed to measure energy changes in a reaction. carbohydrate A naturally occurring condensation polymer containing carbon, hydrogen and oxygen that has the general formula Cx(H2O)y. carbon-13 NMR Nuclear magnetic resonance utilising the interaction of the 13C nucleus with an external magnetic field and electromagnetic radiation. Used to determine the chemical environment of carbon atoms in compounds. carboxyl (or carboxy) group The functional group COOH. carboxylic acid An organic molecule containing the carboxyl functional group, e.g. ethanoic acid (CH3COOH). carrier gas The gas used as the mobile phase in gas chromatography. catalyst A substance that increases the rate of a reaction but is not consumed in the reaction. catalytic cracking The production of small hydrocarbon molecules by heating and breaking down larger hydrocarbons in the presence of a catalyst. cathode An electrode at which a reduction reaction occurs. cellulose A carbohydrate found in the cell walls of plants. It is a polysaccharide formed by a condensation reaction of glucose. chain reaction A self-sustaining reaction involving a rapid series of steps initiated by a single reaction step. chemical analysis Identification of the chemicals present in a substance or the measurement of their concentration. chemical energy The sum of the chemical potential energy and molecular kinetic energy in a substance. chemical equilibrium The point in a chemical reaction when the rate of the forward reaction equals the rate of the back reaction. chemical shift Frequency of the signal in the NMR spectrum, relative to the TMS standard. chloro group The Cl functional group. chloroalkane A molecule derived from an alkane, containing a Cl functional group. cholesterol A lipid found in animals that is involved in the synthesis of a number of hormones. High levels of cholesterol are linked to cardiovascular disease. chromatogram The trace or plot that is obtained when the components of a mixture are separated by chromatography. chromatography A technique for separating the components of a mixture. The components are carried by a mobile phase (gas or liquid) over the adsorbent surface of a stationary phase, which can be solid or liquid.

chromosome Threadlike strand of DNA, found in the

conjugate redox pair An oxidant and its corresponding

nucleus of most cells. coagulation Clumping of denatured protein molecules to form a suspension. cofactor A small non-protein part of an enzyme which is essential for its ability to act as a catalyst. collision theory A theoretical model that accounts for the rate of chemical reactions in terms of collisions between particles during a chemical reaction. colorimeter An instrument that enables the determination of the concentration of a coloured solution by measuring the amount of light passing through a sample. colorimetry An analytical technique that uses the amount of light of a particular wavelength absorbed by a sample to determine concentration. column chromatography A chromatographic technique in which the stationary phase is contained in a column, as for example in gas chromatography. combined techniques A combination of two or more analytical techniques in the same instrument, e.g. gas chromatography–mass spectrometry. combustion Rapid reaction with oxygen accompanied by the release of large amounts of heat; burning. competing equilibria Equilibria that have a common reactant; each equilibrium can be regarded as competing for this reactant. The equilibrium with the larger equilibrium constant has a significant effect on the position of equilibrium of the other reaction. complementary base pairing The formation of hydrogen bonds between the base pairs adenine (A) and thymine (T), or cytosine (C) and guanine (G), in the two strands of the DNA double helix. complementary colour When a substance absorbs a particular energy from white light, it appears coloured. The colour that is observed is not the same as the colour of the light absorbed. The two colours are said to be complementary. complementary proteins Proteins that in combination provide the essential amino acids that are required for a healthy diet. concentration A measure of how much solute is dissolved in a unit quantity of solution. concentration fraction The ratio of concentrations of reactants and products, as expressed in the equilibrium law for a reaction. The concentration fraction is also known as the reaction quotient. concordant titres Titres that vary within narrowly specified limits. condensation polymerisation Formation of a polymer by condensation reactions. condensation reaction A reaction in which two molecules link together, eliminating a small molecule such as water. conical flask A flask with a broad base and narrow neck, used in titrations. conjugate acid The conjugate acid of a base contains one more hydrogen ion (proton) than the base. conjugate acid–base pair An acid and its conjugate base. conjugate base The conjugate base of an acid contains one less hydrogen ion (proton) than the acid.

reduced form, e.g. Cu2+/Cu. contact process An industrial process used to manufacture sulfuric acid. continuous flow processing A method of performing reactions in which reactants are mixed continuously to provide a constant supply of products. converter A vessel in which reaction occurs in an industrial process. copolymer A polymer formed from more than one monomer. coulomb The unit of charge; symbol C. cracking The breaking of carbon-to-carbon bonds in hydrocarbons to form smaller molecules, some of which are unsaturated. culture Medium for growing bacteria. current The flow of charged particles through a conductor; symbol I. The unit of current is the ampere, A. cyclic molecule A ring-shaped molecule. cytosine A base found in nucleic acids such as DNA. It is the complementary base of guanine.

dangerous goods Materials that are potentially harmful to people, property or the environment.

dehydrating agent A substance that removes water from other substances. denaturation A change to the structure and shape of a protein that prevents it from functioning. deoxyribose A five-carbon sugar that is a constituent of DNA. deoxyribonucleic acid (DNA) A nucleic acid that carries genetic information in the cells of plants and animals. Each strand is formed by a condensation polymerisation reaction between nucleotide monomers. Hydrogen bonds between complementary bases shape pairs of DNA molecules into a double helix structure. desorption The breaking of the bonds between a substance and the surface to which the substance is adsorbed. desulfurisation The removal of sulfur compounds from petroleum during refining. diesel A fuel consisting of hydrocarbon molecules of chain length C13 to C25. dilution The addition of solvent to make a solution less concentrated. dipolar ion A particle carrying both a positive and negative charge, e.g. +H3NCH2COO–. dipole moment A measure of the strength and direction of charge separation in a molecule. diprotic acid Substance capable of donating two hydrogen ions. disaccharide A carbohydrate made from a condensation reaction between two monosaccharide molecules. discharge reaction The spontaneous reaction that occurs in an electrochemical cell, during which chemical energy is converted into electrical energy. disulfide link A bond between two sulfur atoms that forms a link between different parts of a protein chain, influencing the tertiary structure of the protein.

483 Glossary

DNA database

A collection of DNA fingerprints or profiles from a large group of individuals in a particular population DNA profile A series of bands formed by DNA fragments of different length, uniquely identifying an individual. Also referred to as DNA fingerprint. double absorption A step in the contact process for the manufacture of sulfuric acid in which unreacted gases are recycled into the converter to prevent the release of harmful emissions. double helix The secondary structure of DNA, in which the two strands wind around each other. Downs cell A commercial electrolytic cell that is used to produce sodium and chlorine. drug A substance that affects how the body functions. dry cell A commercial electrochemical cell with an electrolyte that is a moist paste rather than a solution. dynamic equilibrium A point in a chemical reaction when the rate of the forward reaction is equal to the rate of the back reaction.

E10 A fuel consisting of 10% ethanol and 90% petrol. electrochemical cell A device that converts chemical energy into electrical energy.

electrochemical series A list of chemical species arranged in order of their strength as oxidants.

electrode A solid conductor in a half cell at which oxidation or reduction reactions occur. electrolysis The production of a non-spontaneous redox reaction by the passage of electrical energy from a power supply through a conducting liquid. electrolyte A solution that contains ions that are free to move and so conduct electricity. electrolytic cell A cell in which electrolysis can occur. electromagnetic spectrum All possible frequencies of electromagnetic radiation shown in wavelength order. electronegativity A measure of the electron-attracting power of an element. electronic energy levels Definite fixed energies that electrons in an atom or molecule can take. electrophoresis The process using an electric field to separate DNA fragments or proteins on the basis of their size and charge. electroplating The use of electrolysis to deposit a layer of metal on the surface of another material. element Matter containing atoms of the same atomic number. elongation Addition of complementary base pairs to a single strand of DNA. eluent A liquid used as the mobile phase in chromatography. emf Electromotive force; a measure of the energy given to electrons in a circuit. emission spectrum A display of the wavelengths of light emitted by a substance. For many metals this appears as coloured lines on a black background. empirical formula The formula of a compound that indicates the simplest ratio of atoms of each element in the compound. end point The point during a titration when the indicator changes colour.

484 Glossary

endothermic An endothermic reaction is one in which the chemical energy of the reactants is less than that of the products (ΔH > 0). energy conservation The reduced or more efficient use of energy resources. energy converter A device that changes energy from one form to another. energy density Amount of energy per unit mass or unit volume. energy profile A graphical representation of the energy changes occurring during a chemical reaction. energy transfer efficiency The percentage of energy transformed to its required form. enthalpy Heat content, symbol H; the sum of the chemical potential and molecular kinetic energies in the substance. environmental effects statement A decision-making tool that documents the environmental impact of a proposed development project and typically involves consultation between community, government and corporate stakeholders. enzyme A protein molecule that acts as a catalyst for reactions in living things. equilibrium constant The value of K in the equilibrium law. equilibrium law The equilibrium law for the chemical equation aW + bX ↔ cY + dZ is given by the expression K = [Y]c[Z]d/[W]a[X]b, where K is a constant. equilibrium position The relative amounts of reactants and products in a reaction at equilibrium. equivalence point A point during a titration when the solutions have been mixed in the mole ratio shown by the reaction equation. essential amino acids Nine amino acids that cannot be manufactured by the human body and are required for protein formation. They must be provided directly from food. ester An organic compound containing the ester functional group –COO–. esterification reaction The chemical reaction between an alcohol (alkanol) and a carboxylic (alkanoic) acid to form an ester as the main product. ethanol A two-carbon member of the alkanol homologous series; found in alcoholic drinks. ether linkage A functional group that consists of an oxygen linking two hydrocarbons or sugars. excess reactant A reagent that is not completely consumed in a reaction. excited electrons Electrons that have absorbed particular quantities of energy and moved from their ground states to higher energy levels. excited state The state of an atom, ion, etc. in which one or more electrons, nuclei or bonds have absorbed energy and are not in their lowest energy levels. exothermic An exothermic reaction is one in which the chemical energy of the reactants is greater than that of the products (ΔH < 0). extent of reaction The relative amounts of products compared with reactants. The extent of a reaction is indicated by the value of the equilibrium constant. external circuit The section of an electrochemical cell circuit in which electrons move, e.g. wires.

faraday The charge on one mole of electrons, symbol F. Faraday’s first law of electrolysis The mass of any substance deposited, evolved or dissolved at an electrode in an electrochemical process is directly proportional to the quantity of electricity passed through the cell. Faraday’s second law of electrolysis For one mole of a substance to be deposited, evolved or dissolved at an electrode, the passage of one, two, three or another whole number of moles of electrons is required. fats A nutrient group composed of triglyceride molecules formed from the reaction of glycerol and fatty acids. The term ‘fats’ may be used to describe triglycerides that are solids, rather than liquids, at room temperature. fatty acid A carboxylic acid composed of molecules with long hydrocarbon chains. feedstock The raw materials that are introduced (fed) into an industrial chemical process. fermentation The breakdown of sugar solutions, by the action of enzymes in yeasts, into ethanol and carbon dioxide. fingerprint In chemical analysis, a spectrum or analysis that is unique to a particular chemical. flame ionisation detector A device used to identify organic compounds leaving a GC column. The compounds are burnt in a hydrogen–oxygen flame and are attracted to electrodes, creating an electric current. formulation The particular form in which a drug is administered, e.g. as tablets, as an injection, etc. fossil fuels Coal, oil and natural gas; fuels derived from prehistoric plant and animal matter. fraction A group of molecules with similar boiling temperatures, separated from a mixture by fractional distillation. fractional distillation The separation of the components in a mixture on the basis of their boiling temperatures. fractionating tower A tower used in oil refining to separate oil into its fractions. fragmentation The break-up of a molecule into a number of different smaller fragments in a mass spectrometer, the mass of which can be used to identify the molecule. free radicals An atom or group of atoms with at least one unpaired electron; this tends to make them highly reactive. frequency (ν) The number of complete waves passing a particular point per second. fuel cell An electrochemical cell in which the reactants are supplied continuously, allowing continuous production of electrical energy. functional group Atoms or groups of atoms in an organic molecule that largely determine the molecule’s properties and reactions.

galvanic cell An electrochemical cell; a device that converts chemical energy into electrical energy.

gas chromatography (GC) A form of chromatography in which the mobile phase is a gas. It includes the techniques of gas–liquid chromatography and gas–solid chromatography. gas chromatography–mass spectrometry (GC-MS) A combined analytical technique in which the mass spectrometer replaces the conventional detector in the gas chromatograph.

gas–liquid chromatography (GLC) A very sensitive form of chromatography that is limited to separating compounds that can be vaporised readily without decomposing. It uses an inert gas as the mobile phase and a viscous liquid as the stationary phase. gas–solid chromatography (GSC) A chromatographic technique in which the mobile phase is a gas and the stationary phase an adsorbent solid such as alumina or activated charcoal. gene A segment of DNA that provides the code for a polypeptide. general gas constant The constant, R, in the general gas equation (R = 8.31 J mol–1 K–1). general gas equation The equation that describes the behaviour of a gas: pV = nRT. genetic code The way in which the nucleotide sequence of DNA provides information for the production of proteins. genetic engineering The deliberate alteration of part of the DNA of an organism so as to introduce desired characteristics. genome All of the genetic information carried on the DNA of an individual a particular species. geothermal power A form of power generation using steam or hot water coming from within the Earth’s interior. glycerol An alcohol containing three OH groups. A glycerol molecule can react with three molecules of fatty acids to form a fat molecule. glycogen A carbohydrate that is a means of storing glucose in animals. glycosidic (ether) linkage A C–O–C link formed when two monosaccharides combine to form a disaccharide. gravimetric analysis A type of analysis that involves precipitation and weighing. green chemistry Chemical technology to minimise or eliminate hazardous substances in the design, manufacture or use of chemical products. greenhouse effect An effect involving absorption of infrared radiation by gases such as carbon dioxide and water vapour in the atmosphere, which results in the trapping of heat by the Earth’s atmosphere. ground state A term used to describe atoms in which the electrons or nuclei are in their lowest possible energy levels. guanine A base found in nucleic acids such as DNA. It is the complementary base of cytosine.

Haber process The industrial process used to manufacture ammonia from nitrogen and hydrogen.

half cell Half an electrochemical cell containing an oxidant and the conjugate reductant. When two half cells are combined a galvanic cell is formed. half equation An equation involving either an oxidation or a reduction process. The electrons lost or gained in this process are shown in the equation. Hall–Héroult cell An electrolytic cell used for the industrial production of aluminium. hazardous substance A substance with the potential to harm the health or safety of people. HAZCHEM A code for identifying hazardous chemicals. Used in storage and transport of hazardous materials.

485 Glossary

heat content Chemical energy, symbol H; the sum of the chemical potential energy and molecular kinetic energy in a substance. heat exchangers Devices used to transfer heat from a substance to either its surroundings or another substance. heat of combustion The energy released when a specified amount (1 g, 1 L, 1 mol) of a substance burns completely in oxygen. heat of reaction (ΔH) The energy released or absorbed during a chemical reaction. heating to constant mass The technique of repeatedly heating a sample to 100°C until the same mass is recorded on successive weighings, to ensure that all water has been removed. heterogeneous catalysts Catalysts that are in a different state from the reactants.

high performance (or pressure) liquid chromatography (HPLC) A very sensitive form of chromatography in which the mobile solvent phase is pumped through the stationary phase under high pressure.

high performance liquid chromatography–mass spectrometry (HPLC-MS) A combined analytical technique in which the mass spectrometer replaces the conventional detector in HPLC. high temperature incineration The incineration of materials, typically above temperatures of 1100°C. histone A type of protein. DNA wraps around histones to form chromosomes. homogeneous catalysts Catalysts that are in the same state as the reactants. homologous series Organic compounds with the same functional groups in which each member of the groups differs from the previous member by CH2. hydroelectricity Electricity generated by the flow of water through turbines. hydrogen bonds A strong dipole–dipole attraction that forms between molecules when hydrogen is bonded to one of the electronegative elements oxygen, nitrogen or fluorine. hydrolysis A reaction in which water is a reactant. hydroxyl group The OH functional group.

indicator A substance that has different colours in acidic and basic solutions.

infrared spectroscopy An analytical technique that uses the infrared part of the electromagnetic spectrum.

injection port The part of an analytical instrument into which a sample is injected. insulin A hormone that regulates the level of glucose in the body. This small protein is composed of 51 amino acids. internal circuit The part of an electrochemical cell in which ions move, e.g. solutions and salt bridge. ion–dipole bonding Attraction between an ion and a dipolar molecule. ionic equation An equation that contains only the ions that are involved in the reaction. ionisation constant of water The equilibrium constant Kw, where Kw = [H3O+][OH–]. At 25°C, Kw = 1.0 × 10–14 M2. ionise Form a ion. isomers See structural isomers.

486 Glossary

joule The unit of energy, symbol J. law of conservation of energy When one form of energy is transformed, the total amount of energy remains unchanged. law of conservation of mass During a chemical reaction the total mass of the reactants is equal to the total mass of the products. Le Chatelier’s principle If an equilibrium system is subjected to a change, the system will adjust itself to partially oppose the effect of the change. lead–acid battery A secondary cell or accumulator used to start most cars. limiting reactant The reagent that is completely consumed in a reaction and which determines the amounts of products formed. lipids A class of nutrients made up of naturally occurring fats and oils that are esters of fatty acids. liquid stationary phase A viscous liquid used as the stationary phase in chromatography. The components of a mixture undergo adsorption to this liquid as they are carried along by the mobile phase. lithium ion cell A galvanic cell that generates electricity from the oxidation of lithium embedded in the lattice structure of a graphite rod and the reduction of a metal oxide, the reduction product combining with the lithium ions.

mass spectrometer An instrument designed to measure the mass-to-charge ratio of particles.

mass spectrometry (MS) An analytical technique that uses the mass-to-charge ratio of atoms, molecules and fragments of molecules to identify substances. mass spectrum A graph of data produced from a mass spectrometer indicating the abundance of each particle and the mass-to-charge ratio. Material Safety Data Sheet (MSDS) Documented description of a substance that includes its identity, chemical and physical properties, health hazard information, uses, precautions for use, disposal and safe handling information. membrane cell An electrolytic cell, used to produce sodium hydroxide, hydrogen and chlorine, that has a special plastic membrane to prevent the mixing of products. mitochondrial DNA Non-nuclear DNA found in mitochondria, the energy-producing part of a cell. mobile phase The phase that moves over the stationary phase in chromatography. molar mass The mass of one mole of a substance, symbol M. It is equal to the relative mass of the substance expressed in grams. molar volume of a gas The volume occupied by one mole of gas at a particular pressure and temperature; symbol Vm. At standard laboratory conditions, the molar volume of a gas is approximately 24.5 L mol–1; at standard temperature and pressure it is 22.4 L mol–1. molarity The amount of solute (mole) dissolved in one litre of solution. mole The amount of substance that contains the same number of fundamental particles as there are atoms in 12 g of 126C; symbol n.

molecular formula A formula that indicates the actual numbers and kinds of atoms in a molecule. molecular ions A molecule with either more or fewer electrons than it has protons so it carries an overall charge; in mass spectrometry, the ions formed when electrons are knocked out of a molecule or molecule fragment. monochromator A device that selects a narrow range of wavelengths of light from a light source. monomers Small molecules that are able to react to form long chains of repeating units, called polymers. monoprotic An acid which is able to donate only one hydrogen ion. monosaccharide The simplest type of carbohydrate; formula C6H12O6. mono-unsaturated fat A fat made from fatty acids that contains only one carbon-to-carbon double bond.

natural gas Hydrocarbons that are gases under the conditions they are extracted from the Earth.

neutral In a neutral solution, [H3O+] = [OH–]. At 25°C, pH = 7.

neutralisation The reaction of an acid with a base to form a salt and water.

neutron A neutral particle found in the nucleus of an atom. nickel–cadmium cell A rechargeable cell that generates an electric current by the oxidation of cadmium and the reduction of NiOOH. nickel–metal hydride cell A rechargeable cell that generates an electric current by the oxidation of nickel metal hydride and the reduction of NiOOH. nitrogen fixation The conversion of atmospheric nitrogen into soluble nitrogen compounds that can be used by plants. nitrogenase An enzyme that catalyses the fixation of nitrogen. non-rechargeable cell A primary electrochemical cell. non-renewable Non-renewable energy sources are those that are being used at a rate faster than they can be replaced. non-spontaneous reaction Reactions that would not normally occur without the application of electrical energy. non-steroidal anti-inflammatory drugs (NSAIDs) A type of pharmaceutical that acts as a mild, non-addictive analgesic (pain killer). nuclear fission A nuclear reaction involving the splitting of the nucleus of an atom and the formation of smaller nuclei. nuclear fusion A nuclear reaction where two small nuclei join together to form a larger nucleus. nuclear magnetic resonance (NMR) spectroscopy A technique used to analyse materials using the interaction of the nucleus of particular atoms, usually 1H or 13C, with an external magnetic field and electromagnetic radiation. nuclear shielding Modification of the effect on an atom of an external magnetic field by the magnetic field of surrounding electrons. nucleons The protons and neutrons in the nucleus of an atom. nucleotides The monomers of DNA, consisting of a phosphate group, deoxyribose sugar and a base (adenine, cytosine, guanine or thymine) bonded together.

nucleus The centre of an atom, which contains most of the mass of the atom and around which electrons move.

occupational health and safety Workplace practices to ensure employee safety.

oil A mixture of hydrocarbons that is used as a source of chemicals and fuel. The term ‘oil’ is also used for a triglyceride that is a liquid at room temperature. oilfield A site where crude oil is extracted from the Earth. oleum The compound H2S2O7 that forms when sulfur trioxide reacts with sulfuric acid. origin The point at which a small spot of a mixture is placed so that it can be separated by thin-layer chromatography. Ostwald process The industrial process used to manufacture nitric acid. oxidant A substance that causes another substance to be oxidised; in the process the oxidant is reduced. oxidation The loss of electrons by a chemical in a reaction. oxidation number Numbers that are assigned to elements to determine if a reaction can be classified as a redox reaction. Although they have no physical meaning, the numbers are assigned on the basis of a set of rules assuming that all molecular compounds are composed of ions. oxidation state See oxidation number. oxidising agent An oxidant. oxidising strength The ability of a substance to accept electrons and so oxidise another.

parent molecular ion The ion formed when one electron is knocked out of a molecule in a mass spectrometer.

peat A fuel made from prehistoric plant material; it has a lower carbon content than brown coal.

penicillin A group of anti-bacterial drugs developed from moulds and which all contain a particular structural unit.

peptide linkage The CONH functional group between amino acid units in polypeptides and proteins. Also called an amide group. percentage hydrolysis A measure of how readily a substance reacts with water. percentage ionisation A measure of how readily a substance undergoes an ionisation reaction. persistent pesticide A chemical that is used to kill insects and that does not break down rapidly. petrochemical A chemical derived from natural gas and crude oil. petrochemical industry Industries that use crude oil and natural gas to produce other chemicals. petroleum A mixture of hydrocarbons found underground. pH A measure of acidity; pH = –log10[H3O+]. pH curve A plot showing change in pH during a volumetric titration. phosphate group A phosphorus atom bonded to four oxygen atoms that is also bonded to deoxyribose in DNA. photon A ‘packet’ of energy that has a particular wavelength. A light beam is made up of photons. photosynthesis The conversion by green plants of carbon dioxide and water, in the presence of sunlight and chlorophyll, to glucose and oxygen.

487 Glossary

pipette An item of volumetric equipment used to deliver fixed volumes of liquid accurately. plasma A state of matter composed of electrons and bare nuclei that occurs at temperatures above 10 000°C. polar bond A covalent bond in which the electrons are not shared equally between the two atoms. polyester A condensation polymer formed from the reaction of certain carboxylic acids and alcohols. polymer A long-chain molecule that is formed by the reaction of large numbers of repeating units (monomers). polymerase chain reaction (PCR) A technique for making millions of copies of a segment of a DNA molecule. polynucleotide A polymer made from nucleotides, i.e. DNA or RNA. polypeptide A molecule made from amino acids. polyprotic acid An acid that can donate more than one hydrogen ion. polysaccharides Carbohydrates formed by a condensation reaction involving monosaccharide molecules. polyunsaturated fat A fat made from fatty acids that contain more than one carbon-to-carbon double bond. potential difference The emf, or ‘electrical pressure’, between two points in a circuit, such as the electrodes of an electrochemical cell. precipitate A solid formed during a reaction in which two or more solutions are mixed. primary cell A galvanic cell that is non-rechargeable because the products of the reaction migrate away from the electrodes. primary standard A substance that is so pure that its amount (in mole) can be determined accurately from its mass. primary structure The sequence of amino acids in a protein. prostaglandin A hormone that controls body temperature by causing blood vessels to constrict. It also increases the permeability of capillaries, allowing fluid build-up in tissues. protein Naturally occurring condensation polymer formed from amino acids. proton A positively charged particle found in the nucleus of an atom. proton NMR Nuclear magnetic resonance using the differences in the magnetic properties in 1H atoms in the compounds. Often written as 1H NMR. purine base A derivative of purine, such as adenine or guanine. pyrimidine base A derivative of pyrimidine, such as cytosine or thymine.

qualitative analysis An analysis to determine the identity of the chemicals present in a substance.

quality control Analysis by the manufacturer of the composition of the raw materials or the final product of an industrial process to ensure they meet a particular standard. quantum of light A photon. quantitative analysis A determination of how much of each chemical is present in a substance. quantum of energy An indivisible ‘packet’ of energy. quench tower A tower used to cool hot gases rapidly.

488 Glossary

reaction pathway A series of chemical reactions that convert a starting material into a product in a number of steps.

reaction quotient The ratio of concentrations of reactants and products, as expressed in the equilibrium law for a reaction. The reaction quotient is also known as the concentration fraction. reaction rate The speed at which a reaction occurs. reaction yield The proportion of product recovered from a reaction mixture, usually expressed as a percentage of the maximum amount possible if the reaction were to proceed to completion. receptor site A molecule, usually a protein, that accepts and responds to a chemical such as a drug. rechargeable cell A secondary electrochemical cell. Recharging can occur because the products formed in the cell during discharge remain in contact with the electrodes in a convertible form. redox reaction A reaction that involves the transfer of electrons from one chemical to another. redox titration A titration involving a redox reaction. reducing agent A reductant. reducing strength The ability of a substance to donate electrons and so reduce another. reductant A substance that causes another substance to be reduced; in the process the reductant is oxidised. reduction The acceptance of electrons by a chemical in a reaction. reference cell A cell that contains a reference material, such as pure solvent, and is used for the purpose of comparison in an instrumental analysis. relative atomic mass The weighted mean of the relative isotopic masses of an element on the scale where carbon-12 is 12 exactly; symbol Ar. relative isotopic mass The mass of an atom of an isotope on the scale where carbon-12 is 12 exactly. relative molecular mass The relative mass of a molecule on the scale where carbon-12 is 12 exactly; symbol Mr. It is equal to the sum of the relative atomic masses of the atoms in the molecule. renewable Resources are renewable if they are not finite, e.g. wind power, or can be replenished, e.g. biochemical fuels. restriction enzyme An enzyme that cuts a DNA molecule at each occurrence of a particular sequence of bases. retention time The time taken for a component to pass through a chromatography column. reversible reaction A chemical reaction in which the products can be converted back to the reactants. Rf value The ratio of the distance a component in a chromatogram has moved from the origin to the distance the solvent has moved from the origin. risk assessment A formal way of identifying risks and assessing potential harm from a hazardous chemical in a particular situation. rotational energy levels Different fixed energies that a molecule with a dipole moment can have as a result of rotation around a bond.

salicylic acid A chemical used to make acetylsalicylic acid (aspirin).

salt bridge An electrical connection between the two half

stationary phase A solid, or a solid that is coated in a

cells in an electrochemical cell; it is usually made from a material saturated with electrolyte solution. sample cell A cell in an analytical instrument that contains the sample to be analysed. saturated hydrocarbon Hydrocarbons composed of molecules with only single carbon-to-carbon covalent bonds. saturated fat A fat derived from fatty acids that contain only single carbon-to-carbon bonds. secondary cell An accumulator or rechargeable cell. Recharging can occur because the products formed in the cell during discharge remain in contact with the electrodes in a convertible form. secondary structure The coiling, folding or pleating of proteins due to the formation of weak bonds, e.g. hydrogen bonds, between parts of the protein chain. self-ionisation of water The reaction of water with itself to form H3O+ and OH– ions. semi-structural formula A formula that summarises the structural formula without giving the arrangement of atoms in space. shells Major energy levels around the nucleus of an atom in which electrons move; shells are numbered 1, 2, 3 etc. silver–zinc cell A galvanic cell generating a current when zinc is oxidised and silver oxide is reduced. solar cell A cell that converts solar energy into an electric current. solar energy Electromagnetic energy radiated from the Sun. solid stationary phase An adsorbent solid such as alumina or activated charcoal which is used as the stationary phase in chromatography. The components of a mixture undergo adsorption to this solid as they are carried along by the mobile phase. solvent A liquid in which other substances dissolve. solvent front The point reached by the mobile phase as it moves along the stationary phase in thin-layer chromatography. specific heat capacity The energy required to raise the temperature of one gram of a substance by 1°C. spectator ions Ions that remain in solution and are unchanged in the course of a reaction. Spectator ions are not included in ionic equations. spectrum A display of the wavelengths of light that are either emitted or absorbed by a substance. spontaneous reactions Reactions that occur naturally, either in galvanic cells or when chemicals are mixed directly. standard half-cell potential The emf that is measured when a half cell, at standard laboratory conditions, is connected to a standard hydrogen half cell. The standard half cell potential (E°) gives a numerical measure of the tendency of a half cell reaction to occur as a reduction reaction. standard laboratory conditions (SLC) Conditions at which gas pressure is 1.0 atm (101.3 kPa), the concentrations of dissolved species are 1.0 M and the temperature is 25°C. standard solution A solution of known concentration. standard temperature and pressure (STP) 0°C and 1 atm pressure. standards Chemicals of known composition used for comparison.

viscous liquid, used in chromatography. The components of a mixture undergo adsorption to this phase as they are carried along by the mobile phase. steam cracking An industrial process in which hydrocarbons are diluted with steam and heated to produce smaller hydrocarbon molecules. steam reforming A process for the production of hydrogen from natural gas and steam. steroids Molecules derived from cholesterol that perform a particular function, such as the sex hormones. strong acid An acid that readily donates hydrogen ions (protons). Strong acids have a large acidity constant, Ka. strong base A base that readily accepts hydrogen ions (protons). structural formula A formula that represents the threedimensional arrangement of atoms in a molecule. structural isomers Molecules that have the same molecular formula but different structures. substitution reaction A reaction that involves the replacement of an atom or group of atoms by another atom or group of atoms. substrate The substance that is acted upon by an enzyme. sugars Carbohydrates that have a sweet taste. sustainable Able to support energy and resources into the future without depletion. synchrotron A particle accelerator in which the magnetic field and electric field are carefully synchronised with the accelerated electrons to produce intense electromagnetic radiation. syngas A gaseous mixture of carbon monoxide and hydrogen produced from the processing of biomass. synthesis The process of forming a new chemical substance from simpler ones. synthesis gas A gaseous mixture of carbon monoxide and hydrogen produced from the reaction of natural gas and steam. synthetic pathway The series of chemical reactions required to produce a wanted material, often starting from simpler substances. systematic name The name of a molecule as determined by a set of rules involving the number of carbon atoms and the location and type of functional groups present.

tertiary structure The three dimensional shape of a protein.

therapeutic protein A protein used to treat a medical condition, e.g. insulin given to people with insulin-dependent diabetes because their bodies are unable to make their own insulin. thermal cracking Production of small hydrocarbon molecules by heating larger hydrocarbons in the absence of a catalyst. thermal pollution The adverse effect on plants and animals of heat energy released into the environment. thermochemical equation A chemical equation that includes the enthalpy change of the reaction, ΔH. thin-layer chromatography (TLC) Chromatography using as the stationary phase a thin layer of fine powder spread on a glass or plastic plate.

489 Glossary

thymine A base found in nucleic acids such as DNA. It is the complementary base of adenine. tidal power Electricity generated by the tidal movement of sea water. titration A process used to determine the concentration of a reagent in which one solution is added from a burette to a known volume or mass of another substance. titre The volume of liquid, measured by a burette, used in a titration. total solids The total mass of solid material in a substance. transmittance The amount of light that passes through the sample. triglyceride A fat or oil molecule formed when three fatty acid molecules react with a glycerol molecule.

ultraviolet (UV) A region of the electromagnetic spectrum with wavelengths smaller than that of violet light.

uncertainties Errors associated with measurements made during experimental work.

waste An unwanted or unvalued substance that is discarded or discharged.

waste heat Heat energy that cannot be converted to useful forms of energy.

water content The amount of water in a substance, usually expressed as a percentage by mass.

wavelength The distance between successive peaks in a wave.

wavenumber Inverse of the wavelength of the radiation, measured in cm–1, used as a measurement of frequency in infrared spectroscopy. weak acid An acid that donates protons to a limited extent. Weak acids have a small value of the acidity constant, Ka. weak base A base that accepts hydrogen ions to a limited extent. wet scrubbing A pollution-control technique used by industry to remove dust and pollutants from gases prior to their release into the atmosphere. wind power Electricity generated by the movement of air.

unsaturated hydrocarbons Hydrocarbons composed of molecules with at least one carbon-to-carbon double or triple covalent bond. UV–visible spectroscopy An analytical technique involving measurement of the UV–visible light absorbed by a substance.

yield The proportion of product in a reaction mixture,

vibrational energy levels Different fixed energies that

aluminium, silicon and oxygen that have a three-dimensional honeycomb structure of channels in which reactants lodge and readily undergo reaction. zinc–carbon dry cell A commonly used galvanic cell that generates an electric current by the oxidation of zinc and the reduction of manganese dioxide at a carbon electrode. zwitterion A dipolar ion formed when the proton on the carboxy group of an amino acid molecule is transferred to the amino functional group, as in +H3NCH2COO–.

molecules can have as a result of bending and stretching of bonds. volumetric analysis Analysis using measurement of volumes, e.g. titration. volumetric flask A flask used to measure precise volumes of solutions.

490 Glossary

usually expressed as a percentage of the maximum amount possible.

Z group The side chain of an amino acid. zeolite A class of crystalline substances composed of

Index A absorbance 83 and concentration 85 absorption tower 342 accumulators 432 acetone, MSDS 304 acetylsalicylic acid 223, 234–5 acid–base indicators 39–40 acid–base reactions 37–41 acid–base titrations 6, 31, 39–41 acid rain 252 acidic solutions 39, 286–7 acidity constants (Ka) 289–90 acidity of solutions 39, 286–8 acids 37, 286 activation energies 249–50 active site (enzymes) 195 addition polymerisation, alkenes 147–8, 349 addition reactions, alkenes 146–7, 349 adenine 204 adsorption 62, 255 aerobic energy production 187 aflatoxins in peanuts 5 air bags 247 alcohol and the road toll 48 alcohols 139 aliquot 30 alkaline cells 430 alkanes 131–5, 351 combustion 145, 249 cyclic molecules 132 homologous series 131, 134 reactions 145–6 semi-structural formula 131 structural formula 131 structural isomers 132–3 substitution reactions 145 systematic names 133, 134 alkanoic acids 142 alkanols 139, 147 reactions 150–1 alkenes 135–6, 352 addition polymerisation 147–8 addition reactions 146–7 homologous series 135, 136 hydrolysis reactions 147 reactions 146–8 structural isomers 135–6 systematic names 135–6 see also ethene alkyl groups 133, 134, 137, 138, 145 alkynes 135 alleles 210 -amino acids 190 alternative energy sources 379, 383–97 nuclear energy 383–7

renewable energy sources 388–97 Altona petrochemical complex 356–7 aluminium metal, production 455–6 amide functional group 137, 191 amines 141, 150 amino acids 141, 188–90 condensation reactions 191–2 dual acidic and basic nature of 190 essential 189–90, 191 amino functional group 137, 141, 188, 199 ammonia catalytic oxidation 331 fertiliser use 318–20 in chemical manufacture 320 properties 318 ammonia production 263, 319, 320–3 equilibrium 264, 266–7 Haber process 256, 321–3 health and safety 326 raw materials 320–1 waste management 324–5 ammonium ions 319 in lawn fertiliser 43–4 amount of substance 13 anaerobic energy production 187 analgesics 222, 225 analysis by mass 6, 11–24 analytical techniques 6–7 assessing 119–21 chemical considerations 121 combined 115–18 economic considerations 121 see also specific types anhydrous solids 17 annealing 216 anode 416, 446 anodising 456 antacids 37 antibiotics 228 antibodies 198 antifreeze 140 aqueous electrolytes, uses 457–9 arenes 132 aspirin 222–5 atherosclerosis 181 atom economy 231 atomic absorption spectra 83 atomic absorption spectroscopy (AAS) 7, 78, 81–4, 120 atomic analysis 78–81 atomic emission spectrometry (AES) 7, 78, 80–1, 120

average titre 31 Avogadro’s number 13

B

B100 (100% biodiesel) 171 back titrations 43–4 bagasse 169 base pairs (DNA) 207 base peak 113 baseload power 435 bases (alkalis) 37, 286 bases (DNA) 204 basic solutions 39, 286–7 batch processing 299 batteries lead–acid 432–4 nickel-based 434 beamlines 116 Beer–Lambert law 85 bioaccumulation 306, 308 biochemical fuels (biofuels) 167–72, 388–90 for public transport 174–5 reasons for developing 172 biodegradable polymers 154 biodiesel 170–1, 388 biogas 171–2, 388 biosynthetic proteins 230 black coal 371 bloodstream, equilibrium in 276–7 Blue Gen 441 bomb calorimetry 406–7 bond strength 90 bonding in carbon compounds 130–1 breathalysers 48, 117, 118 Brønsted–Lowry theory 37, 286 brown coal 371 bubble caps 374 buffers 190, 290–1, 292 burettes 31, 32 button cells 431

C

calibration factor (calorimeters) 406 calibration graphs 86 calorimeters 406–7 car batteries 432–4 carbohydrates 183–7 carbon compounds bonding 130–1 changing names 142 functional groups 137–41, 149–55 homologous series 131–7 reaction pathways 155, 156–9 reactions 144–55 carbon dioxide transport 277 carbon monoxide poisoning 277–8 carbon-13 NMR 104, 119

carboxylic acid functional group 137, 140, 141, 188 carboxylic acids 140–1, 150 reactions 151 cardiac enzymes 198 carrier gas 67 catalysts 255–7 effect on equilibria 275 effect on rate of reaction 255–6 in ammonia synthesis 323 in cars 254 in oxidation of sulfur dioxide 341–2 catalytic cracking 351, 375–6 cathode 416, 446 cell potentials see standard half-cell potentials cells fuel 118, 436–41 primary 429–31 rechargeable 428–9, 432–4 see also batteries cellulose 168, 170, 183, 185, 186 Ceramic Fuel Cells 440–1 chain reaction 383 chemical analysis 3 chemical energy 248–50 chemical equilibrium see equilibrium chemical industry aluminium metal production 455–6 ammonia production 318–26 as a global enterprise 297–8 chlorine production 457–8 community safety 311–12 copper electrorefining 458–9 ethene production 349–57 in Australia 296–7 nitric acid production 329–35 production issues 298–300 sodium hydroxide production 455, 456 sodium metal production 453–4 sulfuric acid production 338–45 waste management and pollution 306–10 working safely 301–5 chemical plant, location 312–14 chemical reactions activation energies 249–50 energy changes 248–9 rate of 251–4 thermochemical equations 249 chemical shift 98 chemical wastes 306–7 see also waste management

491 Index

chlorine manufacture 457–8 chloro functional group 137 chloroalkanes 138, 147 reactions 149–50 chloroethane 145, 149 chlorophyll 420 cholesterol 178 chromatograms gas 68–70 TLC 63–4 chromatographic analysis 6, 62–71 chromatography 62 column 65–70 gas 5, 65, 66–8 high performance liquid 65–6 phases 62 thin-layer 62–4 chromium plating 448 chromosomes 204 clotting factor VIII 231 coagulation 197 coal 371 uses of 371–2 coal-fired power stations 372–3 coal reserves 373–4 cofactors 195 collision theory 251, 253–4 colorimetry 78 column chromatography 65–70 combined analytical techniques 115–19 combustion alkanes 145, 249 sulfur 341 community safety 311–12 competing equilibria 278 complementary base pairing 207 composition of a compound 16–18 of a mixture 22–4 compounds empirical formula 16–18 molecular formulas 18 computer-aided drug design 226–7 concentration 6, 28–30 concentration fraction 267 concentration of reactants, and reaction rates 252–3 concordant titres 32 condensation polymerisation 154, 182, 207 condensation reactions 152–3, 155, 178–9, 192, 206 conical flasks 30 conjugate acid 289 conjugate acid–base pair 38 conjugate base 289 conjugate redox pair 416 contact process absorption of sulfur trioxide 342

492 Index

burning sulfur 341 catalytic oxidation of sulfur dioxide 341–2 raw materials 340–1 continuous flow processing 299 converters 321, 331 copolymers 148, 153 copper electrorefining 458–9 coronary heart disease 181 coulomb 449 covalent bonds 130–1, 135, 207, 473 cracking 351 crude oil 374 catalytic cracking 375–6 fractional distillation 374–5 reserves 377 uses of 376 cyclic molecules 132 cystic fibrosis 211 cytosine 204

D

dangerous goods 301–2 Daniell cell 414–15, 416 dehydrating agents 339 delocalised electrons 132 denaturation 196–7, 216 deoxyribonucleic acid see DNA deoxyribose sugar group 204 desorption 62 desulfurisation 324, 354 diatomic molecules 89 diesel 168 Diesel, Rudolf 173 digestion of fats 180 of starch or glycogen 186 dilution, effect on equilibrium 274–5 dioxins 309 dipolar ions 190 dipole–dipole attractions 473 dipole moment 89, 90 diprotic acids 38, 339 disaccharides 183, 184–5 dispersion forces 473 distillation 353 see also fractional distillation disulfide links 193 DNA biological importance 204 control of biochemical processes 210–12 formation 204–7 replication 209 DNA databases 219 DNA profiling/fingerprinting 212, 213–15 forensic techniques 214 polymerase chain reaction 216–18 DNA structure 204–7 primary structure 207

secondary structure 207–8 tertiary structure 208–9 double absorption process 343 double helix 207 Downs cell 456 drug design computer-aided 226–7 structure-based 236–7 drug development 233–5 analytical procedures 234–5 drug synthesis, greener reaction pathways 231–3 drug testing 116 drugs 222 aspirin 222–5 function 230 penicillin 228–9 proteins as 230–1 dry cells 429–30 dynamic equilibrium 264–5

E E10 (petrol containing 10% ethanol) 170 electricity generation 436 electricity sources 370 electrochemical series 417–19 predicting direct redox reactions 423–4 electrodes, competition at 451–2 electrolysis 445 aqueous electrolyte use 457–9 chemicals from 453–9 first law of 448–9 molten electrolyte use 453–6 second law of 449–51 electrolytes 445 aqueous 457–9 molten 453–6 electrolytic cells 445 competition at electrodes 451–2 electromagnetic spectrum 77–8 electron shells 79 electronic energy levels 78, 79–80 electrophoresis 212–13 electroplating 445–8 electrorefining of copper 458–9 elongation 216 eluent 65 emf 422–3 emission spectra 80 empirical formulas 16–18 end point 31, 39 endothermic reactions 248, 249, 268, 288, 402 energy profile 250 energy, and temperature change 405–6

energy changes during chemical reactions 248–50, 402 energy converters 369–70 energy density 409 energy levels (electrons) 79 energy needs, meeting 368–9 energy profiles 250 energy sources advantages/disadvantages 396 alternative sources 379, 383–97 choosing 397 fossil fuels 167, 368, 370–8 nuclear energy 383–7 renewable energy sources 388–97 energy usage 367–8 engine oil, analysis 7 enthalpy (H) 248, 402, 406 enzyme–substrate bonding 196 enzymes 194–6 action of 195 denaturation 196–7 selectivity 196 temperature and pH effects 196–7 equilibrium 263, 264–5 in bloodstream 276–8 in swimming pools 285 involving acids and bases 285–92 reaching equilibrium 276 temperature effects 268–9 equilibrium constant 267, 268 calculations 269–70 equilibrium law 266–70 equilibrium position of a reaction 271–5 adding extra reactant or product 272 catalyst effects 275 dilution effects 274–5 pressure change by adding inert gas 274 pressure change by changing the volume 273–4 summary 279 temperature effects 275 equilibrium reactions 265–6 equivalence point 31, 39 erythropoietin 231 essential amino acids 189–90, 191 esterification reactions 153 esters 137, 151, 152–3 condensation reactions 152–3, 178–9 in biodiesel 170–1 naming 153 polyesters 153–5 smells and flavours 152 etanercept 231 ethane 145, 352

ethanoic acid 38, 140, 289 ethanol 139 analysis 56–7, 117–18 as biofuel 168–70 as fermentation product 168–9 synthesis from ethene 350 ethene 136, 146 addition polymerisation 147–8 addition reactions 146–7, 349 hydrolysis reactions 147 in ethanol synthesis 147, 350 in polymer synthesis 148, 349 in synthetic oil production 350 uses 349–50 ethene production 351–3 distillation and waste treatment 353 fractional distillation 351 green chemistry 355 health and safety 355–6 quenching 353 steam cracking 351–2 waste management 354–5 ethers 137 ethyl ethanoate 153, 161 ethyl propanoate, synthesis 157–8 ethylene glycol 140 ethyne 354 excess reactants, calculations involving 20–1 excited electrons 79 exothermic reactions 248, 249, 268, 288, 402 energy profile 250 explosives 116, 329, 401 extent of reaction 268 external circuit 414

F

faraday 449 Faraday, Michael first law of electrolysis 448–9 second law of electrolysis 449–51 fast chemical reactions 247 fats 178–82 and good health 181 composition 180 digestion of 180 sources of 178 types of 179–80 what are they? 178–9 fatty acids 170, 179, 180 feedstocks 375 fermentation 168–9, 350 fertilisers 318–20, 329 fission reactors 383–4 flame ionisation detector 67 flame tests 78–80

forensic use of DNA 212–15 of spectroscopy 94 formulation 233 fossil fuels 167, 368, 370–6 concerns about 378–9 see also coal; crude oil; natural gas fractional distillation 160–1, 351, 374–5 fractionating tower 374 fragmentation of molecules 112 free radicals 112, 162–3 fructose 183, 184 fruit ripening 136 fuel cells 118, 436–41 advantages/disadvantages 438–9 comparison 438 uses 437–8 fuels factors in choosing 408–9 heat of combustion 408 functional groups 91, 133, 134, 137–41, 145 in monomers 182 reactions of 149–51 fundamental constants 471 fusion reactors 387

G

galactose 183, 184 galvanic cells 414–16, 423, 428 primary cells 429–31 gas chromatography 5, 65, 66–70, 120 gas chromatography–mass spectrometry (GC-MS) 115, 118 gas-fired power stations 377–8 gas–liquid chromatography (GLC) 67 gas pressure change by adding inert gas to equilibrium mixture 274 volume change effects on equilibrium 273–4 gas–solid chromatography (GSC) 67 gases amounts of 14 molar volume 14–15 general gas equation 14 genes 204 genetic code 204 genetic engineering 230 geothermal power 393 global warming 167–8 glucose 169, 183, 184 glycerol 179, 180 glycogen 185 glycosidic linkages 184 gravimetric analysis 6, 11–24, 120

key steps 23 precipitates formed 22 green chemistry 155, 308, 309–10, 334, 344, 355 greener reaction pathways, drug synthesis 231–3 greenhouse effect 368 guanine 204

hydrogen gas, as fuel 389–90 hydrogen-powered fuel cells 437 hydrolysis reactions 147, 170, 180 hydroxyl functional group 137

H

ibuprofen synthesis 232–3 ICP-AES (inductively coupled plasma–AES) 84 incomplete reactions 263–6 indicators 31, 39–41 industrial chemical production 298 batch and continuous processing 299 costs 300 energy use 300 yields 299 see also chemical industry industrial chemists 296–7 inert gas, addition to equilibrium mixture of gases 274 infrared absorbance bands 93 infrared spectroscopy (IR) 78, 89–96, 118, 120 drug development use 234 fingerprint 92 forensic use 94 instrumentation 90–1 infrared spectrum qualitative analysis 91–4 quantitative analysis 95 injection port 67 insulin 230–1 internal circuit 416 ion–dipole attraction 474 ion–dipole bonds 474 ionic bonding 473 ionic bonds 473 ionic compounds, solubilities 23 ionic equations 23 ionic half equations 50, 54–5 ionisation constant of water (Kw) 286, 288 ionise 38 isomers see structural isomers

Haber process 256, 321–2 catalyst effects 323 waste management 324–5 haemoglobin 211, 276–8 half cell equations 416 half cells 416 half equations 50, 54–5 half-lives, radioactive isotopes 384 Hall–Héroult cell 456 hazardous chemicals 303 adverse effects 308 hazardous substances 302 hazardous wastes 306 HAZCHEM signs 302 health and safety 301–5 ammonia production 326 ethene production 355–6 nitric acid production 335 sulfuric acid production 345 heart attack, diagnosis 198 heat content 248 heat exchangers 333 heat of combustion 407–8 heat of reaction (ΔH) 248–51, 402 heat released during a reaction 406–9 heating to constant mass 11 heterogeneous catalysts 255 high performance liquid chromatography (HPLC) 65–6, 120 high performance liquid chromatography–mass spectrometry (HPLC-MS) 115 high temperature incineration 309 histones 209 homogeneous catalysts 255 homologous series 131 alkanes 131–5 alkanols 139 alkenes 135–6 amines 141 carboxylic acids 140 chloroalkanes 138 human genome project 213 hydrocarbon fractions 374, 375 hydrochloric acid 37, 289 hydroelectricity 391 hydrogen bonds 140, 141, 183, 193, 196, 207, 208, 319, 474

I

J joule 369

L lactic acid 187, 197 lactose 185 laughing gas 330 Le Chatelier’s principle 272, 321, 332, 341–2 lead–acid batteries 432–4 light sticks 404 limiting reactants 20 lipids 178 liquefied petroleum gas (LPG) 376

493 Index

liquid stationary phase (GLC) 67 lithium button cells 431 lithium ion rechargeable cells 428–9, 432

M

magnetic resonance imaging 100 maltose 184 mass spectrometry 5, 111–14, 120 drug development use 235 instrumentation 111 mass spectrum formation 112–13 fragmentation 112 interpreting 113–14 isotope effects 112–13 Material Safety Data Sheets (MSDS) 303–4 measurement errors, titrations 32 membrane cells 458 metal ions, flame colours 78 metallic bonds 473 methane 377 combustion 249 methanol combustion 249 conversion to petrol 257–8 methanol-powered fuel cells 438 mitochondrial DNA 218 mixture, composition 22–4 mobile phase 62 molar mass 13 molar volume of a gas 14–15 molarity 28 mole 13 molecular formulas 18 molecular ions 112 molten electrolytes, uses 453–6 monomers 147, 182 functional groups 182 monoprotic acids 292 monosaccharides 183–4 mono-unsaturated fats 179

N

NaS cell 435 natural gas 377 combustion 249 conversion to petrol 257–8 reserves 368 neutral solutions 39 neutralisation reactions 38 Newport power station, environmental impact 378 nickel–cadmium cells 434 nickel–metal hydride cells 432, 434 nitrate salts 330 nitric acid explosives use 329 fertiliser use 329

494 Index

green chemistry solutions 334 uses 329–30 nitric acid production health and safety 335 Ostwald process 331–2 waste management 333–4 nitrogen(II) oxide, oxidation 332 nitrogen(IV) oxide, absorption 332 nitrogen fixation 318, 325 nitrogenase 318 non-rechargeable cells 429–35 non-renewable energy resources 368–9, 370–9 conservation strategies 393 non-spontaneous reactions 445 non-steroidal antiinflammatory drugs 225 nuclear energy 383–7 nuclear fission 383–4 nuclear fusion 387 nuclear magnetic resonance (NMR) spectroscopy 78, 97–105, 120 13 C NMR 104, 119 and peptide structure 102–3 drug development use 234 instrumentation 99 proton NMR 97, 99–101, 234 nuclear magnetic resonance spectrum, interpretation 99–101 nuclear power, advantages/ disadvantages 386–7 nuclear power stations 384 nuclear shielding 97 nuclear waste 385 nucleotides 204–7

O occupational health and safety 301 see also health and safety oceans, buffering action in 291 octane ratings 135 oil reserves 377 oils 178, 179, 180 see also crude oil oleum 342 olive oil, HPLC analysis 66 origin (chromatography plate) 63 Ostwald process 331–2 absorption of nitrogen(IV) oxide 332 catalytic oxidation of ammonia 331–2 oxidation of nitrogen(II) oxide 332 oxidants (oxidising agents) 50, 340, 415, 417, 418

oxidation numbers 51 changing 53–4 rules 52 to name chemicals 53 variable 53 oxidation–reduction reactions 48, 49–54 oxidation states 51 oxygen transport 277

P

paper chromatography 62, 63–4 parent molecular ions 112 peat 371 penicillin 228–9 peptide linkages 192 peptide structure 102–3 peptides 192 percentage hydrolysis 290 percentage ionisation 290 petrochemical industry 349 pH 39, 287–8 effect on enzymes 196–7 in the body 292 temperature effects 288 pH curves 40 phosphate group (DNA) 204 phosphoric acid 38 photochromic sunglasses 50 photons 79 photosynthesis 249, 420 pipettes 30, 32 plant extracts, chemicals in 4–5, 7 polar bonds 140, 141 polar groups 140, 183, 193 PolyAspirin 224 polyesters 153–5 polyethene 349 polymer solar cells 394–5 polymerase chain reaction (PCR) 216–18 polymers 147, 154, 349 greener synthesis 155 polynucleotides 207 polypeptides 192 polysaccharides 168, 183, 185–6 polystyrene 147, 148, 349 polyunsaturated fats 179, 180 porphyrins 420, 421 potato chips, tainted 70–1 potential difference 422–3 precipitates 19, 22 predicting direct redox reactions 423–4 limitations of predictions 424 primary cells 429–35 primary standards 29 primary structure (proteins) 193 propan-1,3-diol, synthesis 155 propanol isomers 138, 139 prostaglandins 225 prostate cancer 199

prostate serum antigen 199 proteins 188–93 amino acids in 188–91 as disease markers 197–9 as medicines 230–1 biological importance 188 enzymes 194–7 structure 191–3 synthesis 192 types of 188 proton NMR 97, 99–101, 234 chemical shifts 98 public transport, alternative fuels 174–5, 437 purine bases 204 PVC (polyvinyl chloride) 147, 148, 349 pyrimidine bases 204

Q

qualitative analysis 4–5 quality control 4–5 quantitative analysis 4, 5, 7 quantum of energy 77 quench towers 353

R

radioactive isotopes, half-lives 384 rates of reactions see reaction rates reactants and products adding extra in equilibrium mixtures 272–3 calculating masses of 19–20 calculations involving excess reactants 20–1 reaction pathways 155, 156–9 reaction quotient 267 reaction rates catalyst effects 255–7 collision theory 251, 253–4 concentration of reactant effects 252–3 increasing surface area of solids effects 251–2 speeding up 251–4 temperature effects 253 reaction yield 263 receptor site 230 rechargeable cells 428–9, 432–4 redox reactions 48, 49–51 classifying 51–4 half equations 50, 54–5 predicting direct 423–4 redox titrations 6, 56–7 reductants (reducing agents) 50, 415, 417, 418 reference cell 86 relative atomic masses 472 renewable energy sources 167, 388–97 biochemical fuels 167–72, 174, 388–9 geothermal power 393 hydroelectricity 391

solar energy 390–1 tidal power 392–3 wind power 391–2 respiration 277 restriction enzymes 214 retention time (Rt) 65 reversible reactions 263–4 Rf value 63 risk assessment 303 risk assessment form 305 RNA (ribonucleic acid) 210 rotational energy levels 90

S

salicylic acid 222 salt bridge 415, 416 sample cell 86 saturated fats 179, 180 saturated hydrocarbons 131 secondary cells 432 secondary structure (proteins) 193 self-ionisation of water 285 semi-structural formulas 131, 138, 139, 140, 141 shells 79 sickle-cell anaemia 211 significant figures 470 silver plating 448 silver–zinc cells 431 sodium hydroxide production 455, 456 sodium metal, production 453–4 solar energy 390–1, 394–5, 420, 421 solid oxide fuel cells 440–1 solid stationary phase (GSC) 67 solubilities of ionic compounds 23 soluble aspirin 224 solution calorimetry 406–7 solvent 63 solvent front 63 specific heat capacity 405 spectator ions 23 spectroscopic analysis 6, 7, 78, 80–105 see also specific types, e.g. atomic absorption spectroscopy spontaneous reactions 423 standard form 470 standard half-cell potentials 422, 424 and non-standard conditions 424–5 standard laboratory conditions (SLC) 14, 418 standard solutions 29 standard temperature and pressure (STP) 14 standards (chromatograms) 63 starch 185, 186 stationary phase 62, 67 steam cracking 351–2

steam reforming 320 steel works, location 313–14 steroids 178 strong acids 38, 39–40, 289, 339 strong bases 38, 39–40 strong bonds 475 structural formulas 131, 138, 139, 140, 141 structural isomers alkanes 132–3 alkenes 135–6 chloroalkanes 138 monosaccharides 183 structure-based drug design 236–7 styrene 148 substitution reactions, alkanes 145 substrate 195 sucrose 184, 185 sugars 183–4 sweetness of 185 sulfur combustion 341 sulfur dioxide 345 catalytic oxidation 341–2 sulfur trioxide 345 absorption 342 sulfuric acid 38 as dehydrating agent 339 as oxidant 340 as strong acid 339 health and safety 345 superphosphate use 338 sulfuric acid production contact process 340–2 green chemistry 344 health and safety 345 raw materials 340–1 waste management 343–4 SunDiesel 171 superphosphate 338 surface area of solids, and reaction rates 251–2 sustainable energy resources 167, 372 swimming pools, equilibrium in 285 symbols and units (SI) 471 synchrotron 116–17 synthesis propan-1,3-diol 155 considerations 158–9 ethyl propanoate 157–8 synthesis gas (syngas) 171, 320 synthetic oils 350 synthetic pathways 223 greener reaction pathways 231–3 systematic names 133, 142 alkanes 133, 134 alkanols 139 alkenes 135–6 amines 141 carboxylic acids 140 chloroalkanes 138

T

temperature and reaction rates 253 effect on enzymes 196 effect on equilibrium 268–9, 275 effect on pH 288 temperature change, and energy 405–6 tertiary structure (proteins) 193 tetramethylsilane (TMS) 97–8 therapeutic proteins 230 thermal cracking 351–2 thermochemical equations 249, 402–3 calculations 403–4 thin-layer chromatography (TLC) 62–4, 120 thymine 204 tidal power 392–3 tin plating 447 titrations 30–2 acid–base 6, 31, 39–41 back 43–4 redox 6, 56–7 titres 31 transmittance 90 triatomic molecules 89 triglycerides 170, 179 triprotic acids 38

weak acids 38, 39, 289 weak bases 38, 39–40 weak bonds 473 wet scrubbing 309 wind power 391–2 working safely 301–5, 326, 335, 345, 355–6

X

xylose 169

Z

Z groups 190, 193 zeolites 257 zinc–carbon dry cell 429–30 zwitterions 190

U

uncertainties (volumetric analysis) 32 unsaturated hydrocarbons 135 uranium 386 UV–visible spectra 86–7 UV–visible spectroscopy 78, 85–6

V

vibrational energy levels 89, 90 vinyl chloride 148 volumetric analysis 6, 28, 30–3, 56–7, 120 accuracy of measurements 32 volumetric flasks 32

W

waste heat 322 waste management ammonia production 324–5 ethene production 354–5 nitric acid production 333–4 sulfuric acid production 343–4 technologies 308–9 wastes 306–7 water content, measuring 11–12 wavenumber 90

495 Index

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

17

18

p BLOCK 1.0

1

4.0

2

–252.7 –259.2 2.1

H

–268.9 –269.7 –

He

Hydrogen

Helium

1

1s

6.9

3

9.0

4

1330 180.5 1.0

Li

1s2

10.8

5

2770 1277 1.5

Be

12.0

6

3900 (2030) 2.0

B

14.0

7

4830 3727g 2.5

C

16.0

8

–195.8 –210 3.0

N

19.0

9

–183 –218.8 3.5

O

–188.2 –219.6 4.0

F

20.2

10

–246 –248.6 –

Ne

Lithium

Berylliu m

Boron

Carbon

Nitrogen

Oxygen

Fluorin e

Neon

1s22s1

1s22s2

1s22s22p1

1s22s22p2

1s22s22p3

1s22s22p4

1s22s22p5

1s22s22p6

23.0

11

Na

892 97.8 0.9

24.3

12

Mg

Sodium

Magnesium

[Ne]3s1

[Ne]3s2

39.1

19

K

40.1

20

760 63.7 0.8

d BLOCK 45.0

21

Sc

47.9

22

2730 1539 1.3

Ti

50.9

23

3260 1668 1.5

V

52.0

24

3450 1900 1.6

2665 1875 1.6

Cr

54.9

25

Mn

55.8

26

2150 1245 1.5

3000 1536 1.8

Fe

58.9

27

Co

58.7

28

2900 1495 1.8

2730 1453 1.8

Ni

63.5

29

Cu

65.4

30

2595 1083 1.9

Zn

28.0

14

2450 660 1.5

Al

1440 838 1.0

Ca

27.0

13

1107 650 1.2

Si

31.0

15

2680 1410 1.8

P

32.0

16

280w 44.2w 2.1

S

35.5

17

444.6 119.0 2.5

–34.7 –101.0 3.0

Cl

39.9

18

–185.8 –189.4 –

Ar

Aluminium

Silicon

Phosphorus

Sulfur

Chlorine

Argon

[Ne]3s23p1

[Ne]3s23p2

[Ne]3s23p3

[Ne]3s23p4

[Ne]3s23p5

[Ne]3s23p6

69.7

31

906 419.5 1.6

Ga

72.6

32

2237 29.8 1.6

Ge

74.9

33

2830 937.4 1.8

As

79.0

34

613 817 2.0

Se

79.9

35

685 217 2.4

Br

83.8

36

58 –7.2 2.8

–152 –157.3 –

Kr

Potassium

Calcium

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Gallium

Germanium

Arsenic

Selenium

Bromine

Krypton

[Ar]4s1

[Ar]4s2

[Ar]3d14s2

[Ar]3d24s2

[Ar]3d34s2

[Ar]3d54s1

[Ar]3d54s2

[Ar]3d64s2

[Ar]3d74s2

[Ar]3d84s2

[Ar]3d104s1

[Ar]3d104s2

[Ar]3d104s24p1

[Ar]3d104s24p2

[Ar]3d104s24p3

[Ar]3d104s24p4

[Ar]3d104s24p5

[Ar]3d104s24p6

85.5

37

Rb

87.6

38

688 38.9 0.8

Sr

88.9

39

1380 768 1.0

2927 1509 1.3

Y

91.2

40

Zr

3580 1852 1.4

92.9

41

Nb

3300 2468 1.6

95.9

42

Mo

(98)

43

5560 2610 1.8

Tc

101.1

44

– 2140 1.9

Ru

102.9

45

4900 2500 2.2

Rh

106.4

46

4500 1966 2.2

Pd

107.9

47

3980 1552 2.2

Ag

2210 960.8 1.9

112.4

48

Cd

114.8

49

765 320.9 1.7

In

118.7

50

2000 156.2 1.7

Sn

121.8

51

2270 231.9 1.8

Sb

127.6

52

1380 630.5 1.9

Te

126.9

53

989.8 449.5 2.1

I

131.3

54

183 113.7 2.5

Xe

–108.0 –111.9 –

Rubidium

Strontium

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Indium

Tin

Antimony

Tellurium

Iodine

Xenon

[Kr]5s1

[Kr]5s2

[Kr]4d15s2

[Kr]4d25s2

[Kr]4d4s1

[Kr]4d55s1

[Kr]4d55s2

[Kr]4d75s1

[Kr]4d85s1

[Kr]4d105s0

[Kr]4d105s1

[Kr]4d105s2

[Kr]4d105s25p1

[Kr]4d105s25p2

[Kr]4d105s25p3

[Kr]4d105s25p4

[Kr]4d105s25p5

[Kr]4d105s25p6

132.9

55

Cs

137.3

56

690 28.7 0.7

57

1640 714 0.9

Ba

138.9 3470 920 1.1

La

72

178.5

Hf

180.9

73

5400 2222 1.3

Ta

183.9

74

5425 2996 1.5

W

186.2

75

5930 3410 1.7

Re

190.2

76

5900 3180 1.9

Os

192.2

77

5500 3000 2.2

Ir

195.0

78

5300 2454 2.2

Pt

197.0

79

4530 1769 2.2

Au

2970 1063 2.4

200.6

80

Hg

204.4

81

357 –38.4 1.9

Tl

207.2

82

1457 303 1.8

Pb

208.0

83

1725 327.4 1.8

Bi

(210)

84

1560 271.3 1.9

Po

(210)

85

– 254 2.0

At

(222)

86

– (302) 2.2

(–61.8) (–7.1) –

Rn

Caesium

Barium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Thallium

Lead

Bismuth

Polonium

Astatine

Radon

[Xe]6s 1

[Xe]6s 2

[Xe]5d 16s2

[Xe]4f145d26s2

[Xe]4f145d46s2

[Xe]4f145d46s2

[Xe]4f145d56s2

[Xe]4f145d66s2

[Xe]4f145d76s2

[Xe]4f145d96s1

[Xe]4f145d106s1

[Xe]4f145d106s2

[Xe]4f145d106s26p1

[Xe]4f145d106s26p2

[Xe]4f145d106s26p3

[Xe]4f145d106s26p4

[Xe]4f145d106s26p5

[Xe]4f145d106s26p6

(223)

87

Fr

– (27) 0.7

(226)

88

– 700 0.9

Ra

89

(227) – 1050 1.1

104 (261) 105 – – –

(262)

Db

(263)

106

– – –

107

– – –

Sg

(262)

Bh

(265)

108

– – –

Hs

(266)

109

– – –

110

– – –

Mt

(271)

111

– – –

Ds

(272) – – –

Rg

(285)

112

– – –

Ac

Rf

Francium

Radium

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium *

[Rn]7s 1

[Rn]7s 2

[Rn]6d 17s2

[Rn]5f146d27s2

[Rn]5f146d37s2

[Rn]5f146d47s2

[Rn]5f146d57s2

[Rn]5f146d67s2

[Rn]5f146d77s2

[Rn]5f146d87s2

[Rn]5f146d97s2

[Rn]5f146d107s2

*temporary names RELATIVE ATOMIC MASS ( ) indicates most stable isotope

ATOMIC NUMBER

12

24.3 1107 650 1.2

SYMBOL

Mg

ELECTRON STRUCTURE

Magnesium [Ne]3s2

BOILING POINT °C MELTING POINT °C ELECTRONEGATIVITY

{

58

140.1

Ce

59

3468 795 1.1

140.9 3127 935 1.1

Pr

144.2

60

Nd

3027 1024 1.2

61

(147) – (1027) –

Pm

62

150.4

Sm

63

1900 1072 1.2

152.0

Eu

1439 826 –

64

157.3

Gd

Cn

(284)

113

– – –

Uut Ununtrium*

114

(289) – – –

Uuq

(288)

115

– – –

Uup

Ununquadium* Ununpentium*

[Rn]5f146d107s27p1 [Rn]5f146d107s27p2 [Rn]5f146d107s27p3

116

(292)

(293)

118

– – – Element 117 has – – been reported but – was not confirmed Ununhexium* Ununoctium* at time of [Rn]5f146d107s27p4 [Rn]5f146d107s27p6 publication.

Uuh

Uno

f BLOCK 65

3000 1356 1.1

158.9

Tb

66

2800 1356 1.2

162.5

Dy

2600 1407 –

67

164.9

Ho

2600 1461 1.2

167.3

68

2900 1497 1.2

Er

69

168.9

Tm

1727 1545 1.2

173.0

70

Yb

175.0

71

1427 824 1.1

Lu

3327 1652 1.2

Cerium

Praseodymium

Neodymium

Promethium

Samarium

Europium

Gadolinium

Terbium

Dysprosium

Holmium

Erbium

Thulium

Ytterbium

Lutetium

[Xe]4f25d06s2

[Xe]4f35d06s2

[Xe]4f45d06s2

[Xe]4f55d06s2

[Xe]4f65d06s2

[Xe]4f75d06s2

[Xe]4f75d16s2

[Xe]4f95d06s2

[Xe]4f105d06s2

[Xe]4f115d06s2

[Xe]4f145d06s2

[Xe]4f125d06s2

[Xe]4f145d06s2

[Xe]4f145d16s2

90

232.0

Th

3850 1750 1.3

91

(231)

Pa

– (1230) 1.5

238.0

92

U

3818 1132 1.7

93

(237)

Np

– 637 1.3

94

(242)

Pu

3235 640 1.3

95

(243)

Am

– – 1.3

96

(247)

Cm

– – –

97

(247)

Bk

– – –

(249)

98

Cf

– – –

99

(254)

Es

– – –

100

(253)

Fm

– – –

101

(256)

Md

– – –

102

(254)

No

– – –

103

Lr

(257) – – –

Thorium

Protactinium

Uranium

Neptunium

Plutonium

Americium

Curium

Berkelium

Californium

Einsteinium

Fermium

Mendelevium

Nobelium

Lawrencium

[Rn]5f06d27s2

[Rn]5f26d17s2

[Rn]5f36d17s2

[Rn]5f46d17s2

[Rn]5f66d07s2

[Rn]5f76d07s2

[Rn]5f76d17s2

[Rn]5f96d07s2

[Rn]5f106d07s2

[Rn]5f116d07s2

[Rn]5f126d07s2

[Rn]5f136d07s2

[Rn]5f146d07s2

[Rn]5f146d17s2

The relative atomic masses in the table are given to 1 decimal place. More accurate values are given in Appendix 3, page 472.

A periodic table of the elements

s BLOCK

16

chem yr 12

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