Ocr a2 Chemistry Teacher Guide

Teacher Guide

Teacher Guide OCR

A2 Chemistry Philip Allan Updates, an imprint of Hodder Education, an Hachette UK company, Market Place, Deddington, Oxfordshire OX15 0SE

Orders Bookpoint Ltd, 130 Milton Park, Abingdon, Oxfordshire OX14 4SB tel: 01235 827720 fax: 01235 400454 e-mail: [email protected] Lines are open 9.00 a.m.–5.00 p.m., Monday to Saturday, with a 24-hour message answering service. You can also order through the Philip Allan Updates website: www.philipallan.co.uk First published 2009 ISBN 978-1-84489-336-2 © Philip Allan Updates 2009 Printed by Marston Book Services Ltd, Didcot

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© Philip Allan Updates

OCR A2 Chemistry

P01405

The paper on which this title is printed is sourced from managed, sustainable forests.

Contents Introduction .........................................................................................................................iv

Unit 4 Rings, polymers and analysis Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8

Arenes and phenols .............................................................................................2 Carbonyl compounds...........................................................................................5 Carboxylic acids and esters..................................................................................9 Amines ..............................................................................................................14 Amino acids and chirality ..................................................................................17 Polyesters and amides .......................................................................................20 Synthesis...........................................................................................................23 Analysis.............................................................................................................28

Unit 5 Equilibria, energetics and elements Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16

How fast? ..........................................................................................................33 How far? ...........................................................................................................41 Acids, bases and buffers ....................................................................................44 Lattice enthalpy.................................................................................................50 Enthalpy and entropy ........................................................................................54 Electrode potentials and fuel cells .....................................................................57 Transition elements ...........................................................................................66 Practical skills ...................................................................................................70

Answers to chapter summary worksheets Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15


Arenes and phenols ...........................................................................................74 Carbonyl compounds.........................................................................................78 Carboxylic acids and esters ................................................................................81 Amines ..............................................................................................................86 Amino acids and chirality ..................................................................................89 Polyesters and amides .......................................................................................93 Synthesis...........................................................................................................96 Analysis.............................................................................................................98 How fast? ........................................................................................................101 How far? .........................................................................................................104 Acids, bases and buffers...................................................................................106 Lattice enthalpy...............................................................................................108 Enthalpy and entropy ......................................................................................109 Electrode potentials and fuel cells....................................................................110 Transition elements .........................................................................................113

OCR A2 Chemistry

iii

Introduction This Teacher Guide accompanies the OCR A2 Chemistry textbook by Mike Smith and John Older, published by Philip Allan Updates. It provides answers to questions that appear throughout OCR A2 Chemistry, as well as in the Chapter Summary Worksheets on the CD-ROM provided with the book. The answers to questions are not model answers. They are designed to help students understand what is required. Examiner notes are included after some answers, indicated by the icon e, and are additional points to help students: G understand what is required in examinations G avoid some common errors. A-level examiners are instructed not to be too literal in following mark schemes. The important point is that students’ responses are correct and answer the question. Although it is important to encourage students to write answers that are logical and in good English, too much emphasis should not be placed on providing responses that follow the exact wording suggested here. Nonetheless, students who contradict themselves in an answer will inevitably lose marks — for example, writing that ionic compounds have a giant molecular structure. Numerical questions can often be tackled in various ways and teachers are free to choose their own approach. However, the answers presented in this guide are usually consistent with the approach taken in the textbook. Students often ask whether it is necessary to include all the steps in a calculation, especially if they believe them to be obvious. It is strongly recommended that, even if an examiner may take a more lenient view, all steps should be given. Examiners are often surprised that candidates offer numerical answers that simply cannot be true. Although it may not always be possible to check an answer carefully in the heat of an examination, this is not the case with class or homework exercises. Students should be encouraged to assess how likely a numerical answer is to be correct. If students cannot detect where an error has been made, they should be instructed to indicate that their answer is unlikely. This will give them a better general understanding of physical reality, and perhaps stop them from suggesting answers such as an element with an atomic mass of 900 g or a solution with a concentration of 2000 mol dm−3. Relative atomic mass values, characteristic absorptions and chemical shift values are not given in the questions because students are expected to consult the data booklet supplied with each examination. The textbook contains a periodic table and spectral data to enable students to attempt the questions within the book. The introduction of the A* grade means that examination papers now contain questions that are designed to ‘stretch and challenge’ the most able students. It is with this in mind that at the end of each chapter there is an additional reading section to challenge such students and to fire their imaginations. The additional reading sections are accompanied by more demanding questions, some of which go beyond the specification. These questions will not suit everyone but they are the sort of questions that potential A* grade candidates should be able to do. This guide includes a CD-ROM, designed so that answers to individual questions and the examiner comments can be cut and pasted and issued to students.


OCR A2 Chemistry

iv

Introduction Specific terms used in examination questions State… This means that no explanation is required, so no further marks are obtained by giving extra information.

Name… If the name of a substance is asked for, examiners may accept either the name or the formula, but if both are given, both must be correct. In general it is better to answer such questions directly — if asked to name a compound then simply give the name and not the formula.

Identify… The answer here can also be a name or a formula. If the formula is required, only the formula and not the name will score the mark.

Molecular formula Only the correct numbers of each element in the compound are required. For example, C2H4O2 is the correct molecular formula for ethanoic acid. CH3COOH would not be accepted.

Structural formula The formula provided must be unambiguous. For example, the simplest acceptable structural formula for ethanoic acid is CH3COOH and similarly for ethanal is CH3CHO (not CH3 COH). However, it is better to show the bonds clearly. For example, ethene should be written as H2C = CH2.

Displayed formula A displayed formula gives more information than molecular or structural formulae and should show both the relative position of atoms and the number of bonds between them. For example, phenylethene can be shown as: H

H

C

C

or

H

H

C

C H

H

Skeletal formula A skeletal formula is used to show a simplified organic formula by removing hydrogen atoms from alkyl chains, leaving a carbon skeleton and associated functional groups, as illustrated below for butanal and ethylbutanoate. O O O Butanal

Ethylbutanoate

Observations Questions sometimes ask students to state what they observe when a reaction takes place. This is not meant to test their interpretation of the reaction, simply what they would see on performing the reaction. They should remember to state the colour before as well as after the reaction. For example, bromine water changes from brown to colourless (not ‘clear’) and acidified potassium dichromate(VI) changes from orange to green, not just ‘goes green’. Stating that a gas is evolved is not an observation (it is a deduction). Bubbles of gas or effervescence are the correct observations.


OCR A2 Chemistry

v

Introduction Signs It is usually unnecessary to indicate with a plus sign (+) that a number is positive, as this can be taken for granted. However, a plus (+) or minus sign (−) should be included in answers involving ΔH or oxidation numbers.

Significant figures It is usual for a least one numerical question on an examination paper to ask for the answer to be quoted to a specified number of significant figures. It is good practice to ensure that answers do not exceed the accuracy of the information given. For example, a mass of 3.40 g is accurate to three significant figures, whereas 3.4 g has two significant figures. Examination questions are always constructed so that the expected number of significant figures is clear. If data are given with a variable number of significant figures, it is the figure with the lowest number that determines the precision of the answer. Be careful to check that significant figures are not muddled with number of decimal places — a common mistake. A further concern arises with questions that are broken into several steps (e.g. titration calculations). Although it is correct to give the answer to each step to the appropriate number of significant figures, students should be advised that the full number obtained should be used for subsequent calculations, otherwise accuracy may be lost. For example, in a two-step calculation, if step 1 involves 6.8/5.0, the answer is 1.36 — which to two significant figures is 1.4. If step 2 involves multiplying the answer from step 1 by 40, the answer using 1.36 is 54 (to two significant figures). However, using 1.4 gives the answer 56, which is clearly inaccurate. In practice, examiners usually require only the final answer to be reported to a specific number of significant figures and intermediate answers are ignored.


OCR A2 Chemistry

vi

Unit

4

Rings, polymers and analysis

Unit 4 Rings, polymers and analysis

Chapter 1 Arenes and phenols 1 a

A

B

C

CH3

CH3

CH3

CH3

CH3 1,2-dimethylbenzene

1,3-dimethylbenzene

H3C

CH3 1,3,5-trimethylbenzene

D

E

F

CH3

OH

OH

NO2

Cl

Cl

Cl 2-nitromethylbenzene

CH3

2,4,6-trichlorophenol

4-methylphenol

b C7H8O c C4H5 d 1,4-dimethylbenzene and ethylbenzene CH3

C2 H 5

CH3 1,4-dimethylbenzene

Ethylbenzene

e C9H12 + 12O2 → 9CO2 + 6H2O f Two reactions are required. Credit should be given regardless of the order. H2SO4 (i) C6H6 + HNO3 ⎯⎯→ C6H5NO2 + H2O 60°C AlCl3 (ii) C6H5NO2 + CH3Cl ⎯⎯⎯⎯⎯⎯→ CH3C6H4NO2 + HCl Halogen carrier


OCR A2 Chemistry

2

Chapter 1 Arenes and phenols Reaction (i) is straight out of the specification. Reaction (ii) requires adaptation of the chlorination reaction from the specification to alkylation. 2 NaOH can behave as either a base or a nucleophile. Phenol is acidic; benzene and ethanol are not acidic. Hence, phenol will undergo an acid–base reaction with NaOH to form a salt (sodium phenoxide) and water. 3 a

Benzene

Phenol

Cyclohexene

Type of reaction

Electrophilic substitution

Electrophilic substitution

Electrophilic addition

Reagents and

Bromine plus a halogen

Bromine; no special conditions

Bromine; no special

conditions (if any)

carrier, such as AlBr3 or Fe

Organic product and

Bromobenzene, C6H5Br;

2,4,6-tribromophenol; white

1,2-dibromocyclohexane;

any observations

bromine is decolourised

precipitate; bromine is decolourised

bromine is decolourised

conditions

b (i) Bromine reacts faster with phenol because one of the lone pairs of electrons on the oxygen in the OH group becomes delocalised into the ring. This increases the electron density, which polarises the Br–Br bond generating an electrophile. This attacks the ring. e Accept e An

the reverse argument explaining why the reaction with benzene is slower.

alternative approach would be to use a diagram showing the sideways overlap of one of the p-orbitals on the

oxygen with the six p-orbitals of the ring, so that the delocalisation incorporates the oxygen.

(ii) Bromine reacts faster with cyclohexene because the C=C double bond has a high electron density. This polarises the Br–Br bond generating an electrophile. This attacks the C=C double bond. e Accept

4

the reverse argument explaining why the reaction with benzene is slower. CH 3

O2 N

NO 2

+ 5 –14 O2 → 7CO2(g) + 2 –12 H2O(l) + 1 –12 N2(g)

NO 2

C7H5N3O6 From the equation, 1 mol TNT produces 7 + 1–12 = 8.5 mols of gas (this assumes that H2O(l) was produced) molar mass of TNT = 227.0 g mol−1 moles of TNT used = 9.08/227.0 = 0.0400 mol moles of gas = 8.5 × 0.0400 = 0.340 mol volume of gas = 0.340 × 22.4 = 7.62 dm3 = 76 200 cm3


OCR A2 Chemistry

3

Chapter 1 Arenes and phenols 5 a

O − Na +

OH Cl

Cl

Cl

Cl

+ NaOH →

+ H2O

Cl

Cl

moles of NaOH = 0.400 × 25.0/1000 = 0.0100 = moles of TCP in 100 cm3 solution molar mass of TCP = 197.5 g mol−1 mass of TCP in 100 cm3 solution = 0.0100 × 197.5 = 1.975 = 1.98 g (3 s.f.) b concentration of TCP = 0.0100 × 10 = 0.100 mol dm−3

Stretch and challenge a

H3 C 109° 28’

CH 3 CH 120°

All the bond angles in the ring are 120°. b H2 C H3 C CH3 CH3 CH

H3 C

CH +

CH3 CH

H

H3 C

CH3 CH

H+ +

+

H+

c Carbonium ions are unstable and often do not exist long enough to collide and react. Anything that stabilises the carbonium ion increases the chance of it existing long enough to collide and react. Carbonium ions are stabilised by the inductive effect of adjacent alkyl groups that release electrons along the σ-bond. The primary carbonium ion is stabilised by the adjacent ethyl (C2H5) group. The secondary carbonium ion is stabilised by both adjacent methyl (CH3) groups. This makes the secondary carbonium ion more stable, and therefore more likely to exist long enough to collide and react. Inductive effect H3C

CH2

CH2

H H3 C

C

CH3

Inductive effects Primary carbonium ion


Secondary carbonium ion

OCR A2 Chemistry

4


Chapter 2 Carbonyl compounds 1 a

A

CH3

H3C

CH

H C

B

CH3

H3C

CH

O

C

C

CH2

O

H

O

2-methylpropanal

3-methylbutanone or 3-methylbutan-2-one

O

D

C

CH3

CH3

C

E

CH3

H3C

C

F

H

O

C

O

O

CH3 Phenylethanone

Phenylethanal

2,2-dimethylpropanal

Cyclohexane-1,4-dione

b A: aldehyde B: ketone C: aldehyde D: ketone E: aldehyde F: ketone c C6H8O2 d C from 2-phenylethanol, C6H5CH2CH2OH D from 1-phenylethanol, C6H5CH(OH)CH3 e Any three of the following: O O O

O

O Pentanal

Pentan-2-one

2-methylbutanal

Pentan-3-one

3-methylbutanal

2 Carbonyl compound: butanone, CH3COCH2CH3 Reagents and conditions: NaBH4, water as solvent Balanced equation: CH3COCH2CH3 + 2[H] → CH3CH(OH)CH2CH3 δ

H

δ−

δ+

−

H

C CH 3

Butanone


δ− O

H

O−

O C 2H 5

+

O

H −

C 2H 5

C CH 3

H

C 2H 5

C

H

+

O H

CH 3 Butan-2-ol

OCR A2 Chemistry

5

Chapter 2 Carbonyl compounds 3 a O CHBr

CHBr

C

CH

CH

CH2 OH

H

Reagent — NaBH4

Br2

O CH

CH

C OH

Tollens’ reagent

O CH

CH

C OH

b

O

O CH

CH

Tollens' reagent

C H

CH

CH

C OH

+ [O]

In the reaction with Tollens' reagent the Ag+ ion will be reduced to metallic Ag(s)

Ag+ + e− → Ag c

H

H C

H

CHO

C

C

C

and CHO

Z (or cis) isomer

H

E (or trans) isomer

Stretch and challenge 1 a

H3 C

H3 C C

O

+

H2 N

H Ethanal


C

N

+

H2O

H Loss of H2O

OCR A2 Chemistry

6

Chapter 2 Carbonyl compounds b

H3 C

H3 C C

O

+

H2 N-OH

C

H

+

H2O

Loss of H2O

H

H3 C C

O

+

H2 N-NH2

+

O

H

H

H3 C

C

C CH3

H

N

N

+ 2H2O

C CH3

Loss of H2O

Ethanal

2

OH

H

Ethanal

c

N

Oxidation number =?

Oxidation number = +1

Oxidation number = −2

CH3CHO — molecular formula is C2H4O

There are four hydrogens, giving a total of +4 and one oxygen (−2). Hence, the contribution from the carbon atoms must add up to −2, which means that the theoretical oxidation number of each carbon is −1. Oxidation number =?

Oxidation number = +1

Oxidation number = −2

CH3COOH — molecular formula is C2H4O2

There are four hydrogens, giving a total of +4 and two oxygens (−4). Hence, the contribution from the carbon atoms must add up to 0, which means that the theoretical oxidation number of each carbon is 0. The oxidation number of each carbon changes from −1 to 0. Hence two electrons are lost: CH3CHO → CH3COOH + 2e− This can be balanced by including H2O. The half-equation for the oxidation of ethanal to ethanoic acid is: H2O + CH3CHO → CH3COOH + 2e− + 2H+ The half-equation for the reduction of the dichromate is: 14H+ + Cr2O72− + 6e– → 2Cr3+ + 7H2O This means that the half-equation for the oxidation of ethanal has to be multiplied by 3, thus: 3H2O + 3CH3CHO → 3CH3COOH + 6e− + 6H+ Adding together the two half-equations: 3H2O + 3CH3CHO → 3CH3COOH + 6e− + 6H+ 14H+ + Cr2O72− + 6e− → 2Cr3+ + 7H2O gives: 3H2O + 3CH3CHO + 14H+ + Cr2O72– + 6e− → 3CH3COOH + 6e− + 6H+ + 2Cr3+ + 7H2O This simplifies to: 3CH3CHO + 8H+ + Cr2O72− → 3CH3COOH + 2Cr3+ + 4H2O Therefore, the fully balanced equation is: 3CH3CHO + 4H2SO4 + K2Cr2O7 → 3CH3COOH + Cr2(SO4)3 + 4H2O + K2SO4


OCR A2 Chemistry

7

Chapter 2 Carbonyl compounds 3

δ−

Step 1

O

O−

δ+

H

C

H

Step 2

O

O−

δ− O

H

O

C

H

O

C

H

−

H

O−

O

δ+

H

C

H

O

C

H

+

Step 3

O O

H

+


H

+

O C

C

H

C

−

O H

C

OH O−

H

C

H

+

OCR A2 Chemistry

8


Chapter 3 Carboxylic acids and esters 1 a

A

B

O C

H

H3C

C

H

O

CH3

O

Methanoic acid

O

C

CH2

OH

Methyl propanoate

O

D

C

O

C

F

CH2

CH3

OH

CH3

CH3

O

C O

CH2

Propyl methanoate

O

E

CH2

C O

Ethyl benzoate

Benzoic acid

Phenyl ethanoate

b C7H6O2 c C4H4O d Molecular formula = C9H10O2 Hence molar mass = (108.0 + 10.0 + 32.0) = 150.0 % carbon = (108.0/150.0) × 100 = 72% e O Methanoic acid is soluble in δ+ H C H water because it can form δ+ δ− δ− H-bonds with water molecules O H O H

f

δ+

Acid catalyst

O H

+

C

HO

CH 2

CH 2

CH3

O H

OH

O

Methanoic acid e Esterification

+ H2 O

C

Propan-1-ol

CH 2

CH 2

CH 3

Propyl methanoate

is a reversible reaction and it is better to use reversible arrows,

, rather than a single forward

arrow, →.

g (i)

O H

O

C

+ Na O

(ii)

H

C

+

e Students

O

C

+ NaHCO3 O

H2

O− Na+

H

O H

1 2

H

H

C

+ H2O

+

CO2

O− Na+

often try to write equations using molecular formulae and the above equation is often written as:

CH2O2 + NaHCO3 → CHO2−Na+ + H2O + CO2 This would probably be awarded the marks in an exam. However, by doing this, students make it more difficult for themselves and often make mistakes. They should be encouraged to use displayed or structural formulae, not molecular formulae. © Philip Allan Updates

OCR A2 Chemistry

9

Chapter 3 Carboxylic acids and esters 2 a Refluxing is continuous evaporation and condensation or boiling in such a way that volatile components cannot escape. e Students

find this difficult to put into words. A simple sketch would suffice and would score the marks.

b

O H3C

O

+ NaOH

C O

H3 C

+

C

Sodium ethanoate

3

H H

O C

Propan-1-ol

O

C C

O H

O

C

O

C

H C H

O

1 COOH

9

18

CH2 CH2 CH3

O − Na+

CH2 CH2 CH3

4

HO

12

Both C=C double bonds are in the cis-form, so the molecule takes the shape: 9

COOH

12

5 Z (cis) fatty acids have a distinctive ‘kinked’ shape. E (trans) fatty acids, the ‘kink’ is straightened out.

Oleic acid E-form (trans) Z-form (cis)

Z-form has the distinctive ‘kink’

E-form has lost the ‘kink’ and forms a straight chain

Straight-chain fatty acids such as E (trans) fatty acids can stack together. This can lead to a build up of plaque on the inside of the artery walls. 6 a CH3COOH + CH3OH CH3COOCH3 + H2O The H2SO4 is a catalyst. b molar mass CH3COOH = 60.0 g mol−1 molar mass CH3COOCH3 = 74.0 g mol−1 moles of CH3COOH used = 12.0/60.0 = 0.200 moles of CH3COOCH3 produced = 3.70/74.0= 0.0500


OCR A2 Chemistry

10

Chapter 3 Carboxylic acids and esters % yield = (0.0500/0.200) × 100 = 25.0% c atom economy = (mass of desired product/mass of all products) × 100 = (74.0/92.0) × 100 = 80.4% d The reaction is reversible/equilibrium may have not been reached. e The

yield will depend on the value of the equilibrium constant, K. It is probable that students will not be aware of

this until later in the course, but potential A* candidates might be expected to read around the topic of equilibria.

7 a

OH O

CH 2 CH

OH

+− Na O

C

CH 2 OH O − + O Na

C

+− Na O

C O

b

OH CH 2 CH

O OH

C

− + O Na

CH 2 OH

Propane-1,2,3-triol is saturated because it contains only single carbon-to-carbon bonds. The fatty acid is referred to as a saturated fatty acid because the hydrocarbon chain contains only single bonds. O +− Na O

C

This molecule is referred to as an monounsaturated fatty acid because the hydrocarbon chain contains one C=C double bond. +− Na O

C O

This molecule is referred to as a polyunsaturated fatty acid because the hydrocarbon chain contains more than one C=C double bond.


OCR A2 Chemistry

11

Chapter 3 Carboxylic acids and esters Stretch and challenge 1 a C6H5COOH > (C6H5)2CHCOOH > C6H5CH2COOH most acidic least acidic The aryl group, C6H5, is electron deficient and pulls electrons into the ring. This withdraws electrons in the O–H bond in the COOH group and weakens the O–H bond. This makes it easier to donate a proton and therefore increases the acidity. Intervening alky groups (CH2 and CH) reduce this effect. e Candidates

tend to write about the ‘negative inductive effect’ of the benzene ring. This is not entirely correct. They

should be advised to relate electron movement to the electron deficiency of the ring.

b CCl3COOH > CH2ClCOOH > CH3COOH most acidic least acidic Cl is electronegative and pulls electrons towards it. This has the ‘knock-on’ effect of reducing the electron density in the O–H bond and hence makes it easier to donate a proton and behave as an acid. 2 a

O

−

O

O

O

O

−

−

−

−

The negative charge (lone pair) on the oxygen is dispersed throughout the ring. In the benzoate ion, C6H5COO− the negative charge is dispersed over the COO group. The p-orbitals on the O–C–O− in the carboxylic acid can overlap creating a delocalised electron cloud above and below the group: O C

O –

O

C

– O

Sideways overlap of adjacent p-orbitals

b Both phenol and benzoic acid are weak acids and therefore do not ionise completely: OH

O

O

−

OH C

+ H+

O−

O C

+ H+

Benzoic acid is the stronger acid, which suggests that the benzoate ion, C6H5COO−, is more stable than the phenoxide ion, C6H5O−, because the reverse reaction in the phenol equilibrium is more favoured than the reverse reaction in the benzoic acid equilibrium.


OCR A2 Chemistry

12

Chapter 3 Carboxylic acids and esters 3 2,4-dinitrophenylhydrazine reacts with C=O, but the double bond does not exist in a carboxylic acid because the negative charge in the carboxylate ion, RCOO− is spread over the COO group. The p-orbitals on the O–C–O− in the carboxylate ion can overlap creating a delocalised electron cloud above and below the group. The carboxylate ion COO−, is probably best represented as: C

O – O

O

rather than

C O–

Since there is no genuine C=O group, 2, 4-dinitrophenylhydrazine cannot react with a carboxylic acid.


OCR A2 Chemistry

13


Chapter 4 Amines 1 a

A

B

CH3

H3C

CH

CH2

NH2

C

CH3

H3C

CH

CH

CH3 H3C

NH2

C

D

2-amino-3-methylbutane

H3C

C

2-amino-3-methylbut-2-ene

E

CH3 CH2

F H3C

NH2

NH2

CH3

CH3 2-methylpropylamine or 1-amino-2-methylpropane

C

CH2

NH2

CH2

NH2

CH3 2,2-dimethylpropylamine or 1-amino-2,2-dimethylpropane

b

A

4-aminomethylbenzene or 4-methylphenylamine B

NH 2

2-phenylethylamine or 1-amino-2-phenylethane C

NH 2

NH 2

c C7H9N d CH3CH(CH3)CH2Cl + NH3 → CH3CH(CH3)CH2NH2 + HCl NH3 is dissolved in ethanol. e First, benzene has to be nitrated and alkylated to produce CH3C6H4NO2. Two reactions are required and credit should be given whichever order is used. H2SO4 (i) C6H6 + HNO3 ⎯⎯⎯→ C6H5NO2 + H2O 60 °C AlCl3 (ii) C6H5NO2 + CH3Cl ⎯⎯⎯⎯⎯→ CH3C6H4NO2 + HCl Halogen carrier Reaction (i) is straight out of the specification. Reaction (ii) requires adaptation of the chlorination reaction from the specification to alkylation. Second, the NO2 group is reduced to NH2 by heating with tin and concentrated HCl as the reducing mixture: Sn/conc. HCl CH3C6H4NO2 + 6[H] ⎯⎯⎯⎯⎯→ CH3C6H4NH2 + 2H2O Heat


OCR A2 Chemistry

14

Chapter 4 Amines 2

Amine

Diazonium ion

CH3

CH3 NH2

+ N

HNO2 + HCl

Phenol

Azo dye

− NCl

CH3

OH

N

+

<10 °C

N OH

OH

NH2 O HNO2 + HCl

HO

<10 °C

Cl − NCl +

+ N

S

Cl O

Cl HO

S

N

N

OH

O

O

Cl

S O

e In

O OH

an exam, substitution of the diazonium compound in any position on the phenol would be allowed.

Stretch and challenge 1 a The reason is uncertain and either of the following arguments is valid: G When ammonia reacts with chloroalkanes, it behaves as a nucleophile. It is the lone pair of electrons on the nitrogen that enable ammonia to behave as a nucleophile. Ammonia reacts with water to produce NH4+ and OH− ions. When NH3 forms NH4+ the nitrogen loses its lone pair of electrons, so it can no longer behave as a nucleophile. G NH3 reacts with water to produce NH4+ and OH− ions and the following equilibrium is set up: NH3 + H2O NH4+ + OH− The hydroxide ion is also a nucleophile and reacts more readily than NH3. This forces the equilibrium to the right and generates more OH− ions, so NH3 is unable to behave as a nucleophile. b NH4OH is a weak base and exists in an equilibrium: NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) If NH4Cl(aq) is added to the aqueous ammonia there will be a large excess of NH4+(aq) ions which will force the equilibrium to the left and enable ammonia to exist as :NH3, i.e. it can behave as a nucleophile. c Step 1: H H

δ+ C

Cl δ

H

−

H

H

N H

H

H

C

N

H

H

+

H

+

Cl

−

H

Step 2:

H

H

H

C

+ N

H

H

H H

H

NH 2

+

NH4+

H N H


C

H H

OCR A2 Chemistry

15

Chapter 4 Amines d (i) Use a large excess of NH3 (ii) Use a large excess of chloroethane 2 (CH3CH2)3N > CH3CH2NH2 > C6H5CH2NH2 > NH3 > C6H5NH2 > (C6H5)3N Most basic Least basic Basicity depends on the ability to accept a proton. The lone pair of electrons on the nitrogen forms a dative covalent bond with H+. The greater the electron density on the nitrogen, the easier it is to accept protons. Alkyl groups (such as CH3CH2–) are electron releasing; aryl groups (such as C6H5–) are electron withdrawing. The more alkyl groups attached to the nitrogen the more basic the compound and vice versa. e Candidates

tend to write about the ‘negative inductive effect’ of the benzene ring. This is not entirely correct. They

should be advised to relate electron movement to the electron deficiency of the ring or to talk about the possibility of the lone pair of electrons on the nitrogen being delocalised into the ring.


OCR A2 Chemistry

16


Chapter 5 Amino acids and chirality 1 a

A

H

H2N

C

COOH

B

H

H2N

C

CH3

2-aminopropanoic acid

C

H

H2N

C

COOH

D COOH

H

H2N

C

CH2

CH2

(CH2)4

COOH

OH

NH2

2-aminobutane-1,4-dioic acid

2-amino-3-hydroxypropanoic acid

COOH

2,6-diaminohexanoic acid

b H

H

C

C

HOOC

NH 2

H2 N

COOH

CH 3

c

CH 3

NH2

A

OH

B O

O O

d

A

H

H3+N

C

NH2

OH

OH

COO–

C

H

H3+N

C

COO–

CH2

CH3

OH

e

pH = 2

B

H

H3+N

C

e Students

COOH

pH = 12

D

H

H3+N

C

COOH

B

H

H2N

C

COO–

D

H

H2N

C

COO–

CH2

(CH2)4

CH2

(CH2)4

COOH

+NH

COO–

NH2

3

usually deduce the correct charges on the amine and carboxylic acid groups. However, they often forget

that acidic or basic R-groups are also affected. Common responses are: pH = 2

B

H

H3+N

C

COOH

CH2 COOH ✔


pH = 12

D

H

H3+N

C

COOH

(CH2)4 NH2 ✗

B

H

H2N

C

COO–

CH2 COOH ✗

D

H

H2N

C

COO–

(CH2)4 NH2 ✔

OCR A2 Chemistry

17

Chapter 5 Amino acids and chirality f (i) 4 (ii) ala–ser; ser–ala; ala–ala; ser–ser (iii) Any two from the following:

H2N

H

O

C

C

CH3

H COOH

H2N

H

O

C

C

N

C

H

CH2

CH2

OH

OH

Ala–Ser

H2N

H

O

C

C

CH3

H N

C

H

CH3

Ser–Ala

H N

C

H

CH3

COOH

H2N

H

O

C

C

CH2

H N

C

H

CH2

OH Ala–Ala e Students

COOH

COOH

OH Ser–Ser

often assume that only two different dipeptides are possible and that they are formed only between the

two different amino acids.

g

H

H

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

O

H

H

O

N

C

C

N

C

C

N

C

C

N

C

C

N

C

C

N

C

C

N

C

C

CH3

A

CH2

CH2

COOH

COOH

B

B

CH3

CH2

CH3

OH A

C

(CH2)4 NH2

A

D

2 Proteins can be hydrolysed to amino acids by refluxing the protein (or peptide) with 6.0 mol dm−3 hydrochloric acid. Hydrolysis can also be achieved by refluxing with a base. Hydrolysis results in the formation of α-amino acids. Since hydrolysis is either acid or base catalysed, the amino acids will form the corresponding ions.

Stretch and challenge 1 a (i) H

H

C* H2 N


COOH CH(CH3)2

*C HOOC

NH 2

(CH3)2HC

OCR A2 Chemistry

18

Chapter 5 Amino acids and chirality (ii) Threonine has two chiral centres, which makes drawing the optical isomers more difficult. COOH H

C*

NH2

H3C

C*

H

OH H H2N

H

H COOH

C*

HOOC

H2N

OH

H

H3C

HOOC

OH

Non-superimposable mirror images = one pair of enantiomers

CH3

HO

H

CH3

NH2

*C *C

C*

HO

H3C

H COOH

C*

*C

C* H

*C

NH2

H


b 2,3-dimethyl-2,3-diphenylbutanedoic acid also has two chiral centres. There is internal symmetry and therefore only three stereoisomers. COOH C6H5

C*

CH3

H3C

C*

C6H5

COOH C6H5 H3C

C*

C6H5 COOH

HOOC

*C

H3C

*C

C* C6H5

C6H5

CH3

COOH

COOH Internal mirror image

C* H3C

C6H5

HOOC CH3

H3C

C*

COOH C6H5


2 Polymer A: the CH3 groups are all on the same side of the chain, so this is an isotactic polymer Polymer B: the CH3 groups occur alternately on either side of the chain, so this is a syndiotactic polymer Rotation is not possible because each of the asterisked carbons is asymmetric. H3C

H

H

C* CH2


CH3 H3C C*

CH2

H

H

C* CH2

CH3 H3C C*

CH2

H C*

CH2

CH2

OCR A2 Chemistry

19


Chapter 6 Polyesters and amides 1 a

Simplest repeat unit H (CH2)6

N

H

O

N

C

(CH2)8

O

H

C

N

(CH2)6

H

O

N

C

O (CH2)8

C

b 0

0

C

C N

N

H

H

H

H

N

N

C

C

0

0

c

0

0

C

C

CH2

0

0

0

0 C

CH2

0 C CH2

0 CH2 e Students

usually draw polymers as linear structures. This is accepted by examiners and would score full marks.

2 a

O CH2

b

CH2

CH2

O C

CH2

CH CH3


CH2

O

H

O

N

C

C

O (CH2)3

C

O O

CH2

CH2

CH2

CH2

O

C

O (CH2)3

C

O

H CH2

CH

N

CH3

OCR A2 Chemistry

20

Chapter 6 Polyesters and amides 3

Acid hydrolysis will give: O HO

O

C

C

CH

CH

CH3

CH3

0H

and

(CH2)6

HO

0H

Base hydrolysis (with sodium hydroxide) will give: O

O +Na–O

C

CH

CH

CH3

CH3

0–Na+

C

and

(CH2)6

HO

0H

Stretch and challenge 1

H

H

H

H

C δ+ O

H

H

C δ+

δ−

O

H

H

δ−

H

C

H

H

C δ+ δ−

O

H

H

C

H

O

H

H

C

O

H C δ+

δ−

O

H

C

O

H

C δ+

δ−

H C

O

O

O

2 a Nylon-4,6: 1,4-diaminobutane and hexane-1,6-dioic acid (or hexane-1,6-dioyl chloride) b Nylon-6,4: 1,6-diaminohexane and butane-1,4-dioic acid (or butane-1,4-dioyl chloride) c Nylon-6,10: 1,6-diaminohexane and decane-1,10-dioic acid (or decane-1,10-dioyl chloride) 3 Nylon-6, made from caprolactam, has the structure: O CH2

C

CH2

CH2

CH2

CH2

H

O

N

C

H CH2

CH2

CH2

CH2

CH2

N

The structure of nylon-6,6 is: H (CH2)6

N

H

O

N

C

(CH2)4

O

H

C

N

(CH2)6

H

O

N

C

O (CH2)4

C

Nylon-6 has five CH2 groups between each consecutive pair of amide links; in nylon-6,6 there are alternately six CH2 groups and four CH2 groups. In nylon-6, the amide link is always orientated the same way; in nylon-6,6 the amide link alternates –(CO)NH– and –NH(CO)–. 4 a

CH3 O n H

O

C H

C

CH3 O O

H

C H

C

+ H2O

O n

The polymer contains ester links, which would react with the water, so it would be broken down immediately to the monomer.


OCR A2 Chemistry

21

Chapter 6 Polyesters and amides b

CH3

H

O

C

O

C

H + H

H

O

C

O

CH

CH3 O

CH3 O

C

O

O

C

C

O

+ 2H2O

H

H

CH

O

CH3

c

CH3 O

CH O

C

C

O

n

C H

CH

O

CH3 O

CH3 O C

O

C H

C

O n

CH3

When the dimer polymerises, water is not formed. Hence the possibility of hydrolysis of the ester linkages is reduced. d

CH3 O C H


C

CH3 O O

C

C

O

H

OCR A2 Chemistry

22


Chapter 7 Synthesis 1 For each of these multi-stage syntheses there are several possible routes. a For the conversion of propene to propanone, the simplest way is: OH

Step 1: H3C

CH

CH2 + H2O

≈300°C

H3C

CH3

C

6 MPa H

Step 2:

OH

H3C

O

Oxidising agent = H+/Cr2O72– CH3

C

H3C

+ [O]

CH3

C

+ H2O

Heat H

b For the conversion of 3-chloropropan-1-ol to 3-hydroxypropene, the simplest way is: Loss of water H

Step 1:

H

OH

H

H

Acid catalyst Cl

Step 2: Cl

e It

C

C

C

H

H

H

H

H

C

H

H

H

C

C

H

H

HO

+ H2O

C H

H

NaOH as catalyst + H2O

C

Cl

H

H

C

Heat

C

C

H

H

+ HCl

C H

is also possible to react 3-chloropropan-1-ol first with NaOH(aq) to produce propan-1,3-diol and then

to dehydrate this to form 3-hydroxypropene. Full credit would be given for this, despite the fact that 3-hydroxypropene would only be formed if partial dehydration occurred.


OCR A2 Chemistry

23

Chapter 7 Synthesis c For the conversion of phenylethanone to poly(phenylethene), the simplest way is: Step 1:

CH3

0

CH3

H0

C

CH

+ 2[H]

Step 2:

Reducing agent = NaBH4(aq)

CH3

H0

CH2

CH

CH Dehydration by heating with an acid catalyst + H2O

CH2

Step 3:

Two repeat units H

CH2

CH

CH

C

H

H

C

C

C

H

Polymerisation

n

H

H

n

2 a (i) In stereoisomerism, the isomers have the same molecular formula, the same structural formula but a different spatial arrangement (different 3-D arrangement) (ii) Compound X exhibits both E/Z isomerism and optical isomerism: G E/Z isomerism — each carbon in the C=C double bond is bonded to two different atoms or groups CH3 H

H

H

CH 0

C

C

C

CH3

C

C H

CH

H

0 HO

C

C (cis)

HO

E (trans)

H G

Optical isomerism — there is an asymmetric carbon atom atom/chiral centre CH3

CH3

*C

C* H

CH C

CH H

HO

O

H O

CH C

CH H

OH

Each optical isomer can exist as either an E- or Z-isomer, giving a total of four isomers.


OCR A2 Chemistry

24

Chapter 7 Synthesis b (i) When compound X reacts with hot NaBH4(aq), it is reduced to a primary alcohol: CH3 HO

CH

CH

CH OH

H2C

(ii) When compound X reacts with hot Na2Cr2O7 in the presence of H2SO4, it is oxidised to a carboxylic acid: CH3 HO

CH

CH

CH O

C HO

(iii) When compound X reacts with Tollens’ reagent, it is oxidised to a carboxylic acid: CH3 HO

CH

CH

CH O

C HO

(iv) Both the phenol and alkene groups react with bromine. The phenol groups reacts by electrophilic substitution; the alkene group reacts by electrophilic addition: Br Br

Br CH3

HO

CH

CH

CH O

C H

Br

3 a C13H10N2O4 e Variation

in the sequence of the elements in the molecular formula is allowed; C13H10N2O4, C13H10O4N2 and

C13O4N2H10 would all be credited.

b (i) When thalidomide reacts with an excess of hot NaBH4(aq), all the carbonyl groups are reduced to alcohol groups: OH

CH

H2C N

CH

CH OH e There

CH2 CH

CH

OH

NH

HO

are four ketone groups, each of which should be reduced — partial marks would be awarded for partial

reduction. Asking for a balanced equation would be interesting and challenging.


OCR A2 Chemistry

25

Chapter 7 Synthesis (ii) When thalidomide reacts with HCl(aq), both amine groups are protonated: O C

CH2

H2C +

NH

CH

C

+

C

O

NH2

O

O

c

C

O C

H2C N

*CH

C O

CH2 C

C

O

NH

O

d Only one of the optical isomers has the correct shape to be pharmaceutically active and it is essential to separate this from the other isomer. It is possible that the other isomer could have adverse side effects or make the active isomer less effective. It is, therefore, important that a pharmaceutical manufacturer isolates the effective isomer. e Techniques include: G chiral chromatography with a solid stationary phase that has either an enzyme or a chiral stationary phase as the active site. One optical isomer absorbs more strongly to the stationary phase than the other isomer does, so separation can be achieved. G synthesis using naturally occurring enzymes or bacteria, which results in the formation of a single optical isomer. Enzymes can achieve this because their active sites have specific shapes that promote specific reactions. G chiral catalysis.


OCR A2 Chemistry

26

Chapter 7 Synthesis Stretch and challenge 1

CH3

CH3

CH3Cl with a halogen carrier AlCl3

HNO3/H2SO4 Sn/ conc. HCl, heat

CH3

NO2

CH3

CH3 HNO2/ NH2 HCl Temp. < 10°C

H2O/ warm N2 + HCl +

+

OH

N

Cl–

N

This reaction involves five separate steps. Each step will introduce impurities — for example, in the second step (the nitration of methylbenzene), it is likely that the 2, 4 and 6 positions will be nitrated and that a mixture of mono-, di- and tri-nitro products will be produced. 2

O

O

OH H

O

C

C

NaOH H3C

C

H + H3C

C

H

H3C

C

H

H H 3-hydroxybutanal Dehydrated by heating with a concentrated acid O H3C

C

C

H

H

C

H

But-2-enal


OCR A2 Chemistry

27


Chapter 8 Analysis 1 a Mobile phase: the phase that moves over the stationary phase. It is usually a liquid or a gas that carries the components over the stationary phase. b Stationary phase: the phase that does not move. It is usually a solid or a liquid on the surface of an inert solid. c Adsorption: the interaction between the components in the mobile phase and the stationary phase. d Partition: the ratio of the solubilities of a component in two immiscible liquids or the relative distribution of the component between the mobile and the stationary phases: concentration of solute in mobile phase = constant concentration of solvent in stationary phase e Rf value =

distance moved by spot/solute distance moved by solvent

f Retention time: the time from the injection of the sample to a component leaving the column. 2

After

Before

Solvent front

Baseline

Mixture of amino acids

Known samples of pure separate amino acids

Mixture of amino acids

Known samples of pure separate amino acids

The mixture of amino acids is spotted on the base line on the plate and known samples of pure amino acids are spotted alongside. The plate is then put into a suitable solvent and left. When the solvent has moved almost to the end of the plate, the plate is removed and the position of the solvent front is recorded. The plate is dried and developed using ninhydrin. The positions of the spots from the mixture are compared with the positions of the spots from the reference samples. e It

is also possible to identify the amino acids by cross referencing the Rf values against those in data books. If

students use this argument, it is important to recognise that Rf values are solvent dependent.

3 Any four from the following: G Environmental analysis: GC–MS can be used to identify most organic compounds. The detection of pesticides and herbicides is both sensitive and effective. G Forensic science: GC–MS can be used to analyse particles from a human body to help link a criminal to a crime and to detect and identify small amounts of narcotics. G Airport security: GC–MS is used in airports to detect explosives.


OCR A2 Chemistry

28

Chapter 8 Analysis Food and drink analysis: GC–MS is used extensively in the analysis of compounds that occur in food and drink. These include esters, fatty acids and alcohols. It is also used to measure contamination by pesticides and herbicides. Medicine: GC–MS is used widely in medicine for the analysis of pharmaceuticals and of metabolic compounds labelled with 13C. Astrochemistry: GC–MS has provided much information from planets, including Mars and Saturn.

G

G

G

4 a

1H NMR detects the two proton

13C NMR detects the three carbon environments labelled C1, C2 and C3

environments labelled Ha and Hb O Ha

O Ha

C C

Ha Hb

C C Hb

C

C1 C2

Ha

C3

Hb

C2 C3

Hb

b C1 in the range 190– 220 ppm C2 in the range 20– 30 ppm (allow 5–55 ppm) C3 in the range 5–55 ppm c Ha in the range 2.0– 2.9 ppm Hb in the range 0.7–1.6 ppm 5 a molar mass of CH3CH(OH)COOH (molecular formula = C3H6O3) = 90.0 % carbon = (36.0/90.0) × 100 = 40.0% % hydrogen = (6.0/90.0) × 100 = 6.7% % oxygen = (48.0/90.0) × 100 = 53.3% b (i) The infrared spectrum would contain peaks for both carboxylic acid and alcohol groups: G COOH group would give rise to a peak in the range 1680–1750 cm−1 due to C=O and a broad peak in range 2500–3300 cm−1 due to O–H. G OH would give rise to a peak in the range 3230–3550 cm−1 due to O–H. The two O–H peaks would overlap and not be easy to distinguish. (ii) The mass spectrum would have M peak at 90 due to the molecular ion CH3CH(OH)COOH+(g) and also fragment peaks, including CH3CH+COOH(g) at 73, CH3CH(OH)+(g) at 45 and CH3+(g) at 15.

Transmittance/%

6 Infrared spectrum: 100 80 60 40 C=O 20 0 4000

3000

2000

1500

1000 500 Wavenumber/cm–1

The infrared spectrum confirms the presence of a carbonyl group.


OCR A2 Chemistry

29

Chapter 8 Analysis e Infrared

is used to identify the C=O group. However, the region 800–1300 cm−1 can be used as a fingerprint region.

Each peak should match with the value for a corresponding sample in a databank. It is difficult to assign the C–O

Transmittance/%

type because there are so many peaks in this region. 100 80 60 40 C=O 20 C–O region

0 4000

3000

2000

1500

1000 500 Wavenumber/cm–1

Relative intensity

Mass spectrum: 100

Peak at m/z = 43 likely to be due to CH3CO+(g)

80 60

Molecular ion at m/z = 72, hence molar mass = 72

Peak at m/z = 29 likely to be due to CH3CH2+(g)

40 20

CH3+(g)

0 20

10

30

40

50

60

70

80

90 m/z

The mass spectrum indicates the presence of two fragments CH3CO+(g) and CH3CH2+(g) that add up to the molecular ion. e It

is not necessary to identify all the peaks in the mass spectrum, but the peak at 57 would be consistent with the

formation of the fragment ion CH3CH2CO+(g). The peak at 27 is more difficult to assign. Absolute confirmation can only be attributed by comparison with a data bank. Each peak should match like a fingerprint. 13C

NMR spectrum: All in the ranges 5–55 ppm indicating that these three carbons are in the alkyl groups and not attached to either a halogen, oxygen or nitrogen

C=O in range 190–220 ppm

200 13C

180

160

140

120

100

80

60

40

20

0 ppm

NMR confirms the presence of four different carbon environments, one of which is C=O.


OCR A2 Chemistry

30

Chapter 8 Analysis 1H

NMR spectrum: B

C A 11

10

9

8

7

6

5

4

3

2

1

0 ppm

1H

NMR confirms the presence of three different hydrogen environments. Peaks A and C show classic quartet and triplet splitting, which confirms the presence of a CH3 group next to a CH2 group. It also confirms that the carbon on the other side of the CH2 has no hydrogens attached. Peak B is a singlet, which confirms that it is bonded to a carbon that has no hydrogens attached. Conclusion: compound X is butanone, CH3CH2COCH

Stretch and challenge 1 CH4 is most shielded; H2O is the least de-shielded. The oxygen is atom is highly electronegative and draws the electron cloud towards itself, so that the electron density around the attached protons is decreased. This is known as de-shielding. Alkyl groups, such as CH3, are electron releasing (by the inductive effect) and increase the electron density around the protons. This is known as shielding. 2 Spin–spin splitting of the CH3 group gives rise to a triplet with intensity 1:2:1. Spin–spin splitting for the CH2 group gives rise to a septet with intensity 1:6:15:20:15:6:1. 3 The number of peaks in the 13C NMR spectrum of each of compounds A–F is: 1

A

CH3 1

2 2

3

1 CH3

B

2

CH3

5

2 4

4

4 peaks D

3 1

1

3 peaks


E

1 3

2

3

2

8

2

9

7

3 2

3 peaks

1 CH3 3

OH

5

10 peaks

2

5

HO

OH 1

3

2 4

4

F

1

6

10

3

3 1 CH3

5 peaks

1 2

2 3

4 3

4

1 CH3

C

5

1 2

4

5 peaks

OCR A2 Chemistry

31

Unit

5

Equilibria, energetics and elements

Unit 5 Equilibria, energetics and elements

Chapter 9 How fast? 1 Rate has units of mol dm−3 s−1 and concentrations are measured in mol dm−3. a Adding units to both sides of the equation, rate = k[A][B][C] gives: mol dm−3 s−1 = k × (mol dm−3) (mol dm−3)( mol dm−3) units of k = =

mol dm−3 s−1 (mol dm−3)( mol dm−3)

(mol dm−3)

s−1 (mol dm−3)

(mol dm−3)

= mol−2 dm6 s−1 = dm6 mol−2 s−1 The units should be written as dm6 mol−2 s−1. b If the rate equation, rate = k[A]2[B] is re-written as: rate = k[A][A][B] then it can be seen that this problem is similar to part a. Therefore, the units of k are the same: dm6 mol−2 s−1. e Units

for the rate constant cause students a lot of trouble — perhaps because they are not usually immediately

obvious. In addition, they can look rather strange! For many students, this means that they need practice. Although it is better to deduce units using the method shown above, the answer can be obtained using the formula: k has units of (concentration)(1−n)(time)−1 where n is the total order for the reaction In both examples above the total order is 3. Therefore, the units are mol−2 dm6 s−1 or dm6 mol−2 s−1.

2 a When the concentration of NOCl is doubled from 0.1 mol dm−3 to 0.2 mol dm−3, the initial rate of the reaction increases from 4 × 10−10 mol dm−3 s−1 to 1.6 × 10−9 mol dm−3 s−1, which is a four-fold increase (remember that 1.6 × 10−9is the same as 16 × 10−10). This means that the order with respect to NOCl is 2. e The

same conclusion could be reached by noticing that doubling concentration of NOCl from 0.2 mol dm−3 to 0.4 mol

dm−3, the

rate also increases four-fold.

If you have the opportunity it is always worth checking that what you have done is consistent with any other information you have been given.

b The rate equation is: rate = k[NOCl]2 Using the first set of results gives 4 × 10−10 = k(0.1)2 Therefore, k =

4 × 10−10 (0.1)2

=

4 × 10−10 0.01

= 4 × 10−8 mol−1 dm3 s−1


OCR A2 Chemistry

33

Chapter 9 How fast? e The

rate equation gives rate = k[NOCl]2

Therefore, k =

units of k =

rate [NOCl]2

mol−1 dm3 s−1 (mol dm−3)2

=

s−1 mol dm−3

= mol−1 dm3 s−1

c The initial rate of reaction is obtained by substituting the concentration 0.15 mol dm−3 into the rate equation. rate = 4 × 10−8 × (0.15)2 = 9 × 10−10 mol dm−3 s−1 3 a (i) When the concentration of (COO)22− is doubled from 0.020 mol dm−3 to 0.040 mol dm−3 the initial rate of the reaction increases from 0.48 × 10−4 mol dm−3 min−1 to 1.92 × 10−4 mol dm−3 min−1, which is a four-fold increase. This means the order with respect to (COO)22− is 2. (ii) When the concentration of HgCl2 is halved from 0.080 mol dm−3 to 0.040 mol dm−3 the initial rate of reaction halves from 1.92 × 10−4 mol dm−3 min−1 to 0.96 × 10−4 mol dm−3 min−1. This means the order with respect to HgCl2 is 1. b The rate equation is: rate = k[(COO)22−]2[HgCl2] Using the first set of results gives 0.48 × 10−4 = k(0.020)2 × 0.080 Therefore, k =

= e Check

0.48 × 10−4 (0.020)2 × 0.080 0.48 × 10−4 = 1.5 mol−2 dm6 s−1 3.2 × 10−5

that you understand how to obtain the units.

4 a The order with respect to Br− can be obtained from the second and third sets of results. When the Br−concentration is doubled from 0.1 mol dm−3 to 0.2 mol dm−3 the rate doubles from 1.64 × 10−3 mol dm−3 s−1 to 3.28 × 10−3 mol dm−3 s−1. This means that the order with respect to Br− is 1. From the first and third sets of results, it can be seen that when the BrO3− concentration is doubled, the rate doubles. Therefore, the order with respect to BrO3− is 1. From the second and fourth sets of results, it can be seen that when the concentration of H+ is doubled, the rate increases four-fold. Therefore, the order with respect to H+ is 2. The rate equation for the reaction is: rate = k[Br−][BrO3−][H+]2 e There

are other ways of processing these results, which are equally valid, providing that they lead to the correct

orders of reaction.

b Using the rate equation and substituting the values from the first set of results: 1.64 × 10−3 = k × 0.1 × 0.2 × (0.1)2 Therefore, k =

=

1.64 × 10−3 0.1 × 0.2 × (0.1)2 1.64 × 10−3 = 8.2 mol−3 dm9 s−1 2 × 10−4

c rate = 8.2 × 0. 25 × 0.25 × (0.25)2 = 0.032 mol dm−3 s−1


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34

Chapter 9 How fast?

Concentration/mol dm–3

5 There are two ways of showing that this reaction is first order with respect to B: G For a first-order reaction, over fixed time intervals the concentration of B will always fall by the same proportion. In the first 20 s, the concentration of B falls from 0.800 mol dm−3 to 0.640 mol dm−3. This is a drop of (0.640/0.800) × 100 = 80%. In the second 20 s, the drop is from 0.640 mol dm−3 to 0.512 mol dm−3. This is a drop of (0.512/0.640) × 100, which is also 80%. This is repeated throughout the set of results proving that the reaction is first order. It has, in effect, a constant ‘0.8’ life. G The second method perhaps fits the specification better but takes more time. First, a graph of concentration against time must be drawn. The graph is then used to establish that the reaction has a constant half-life. 1.0

0.8

0.6

0.4

0.2

0

0

20

40

60

80

100 Time/s

For example, it takes 60 s for the concentration of B to be reduced from 0.800 mol dm−3 to 0.400 mol dm−3 and 60 s for the concentration of B to fall from 0.600 mol dm−3 to 0.300 mol dm−3. This also shows that the reaction must be first order. e Either

method is equally valid if all that is required is to show that the reaction is first order. The second method has

the advantage when, as is often the case, the question asks for the half-life to be quoted.


6 a This can be solved by plotting a graph of concentration against time. 0.6 0.5 0.4 0.3

0.2 0.1 0

0

30

60

90

120

150 Time/s

The graph is a straight line. This indicates that the reaction is zero order with respect to X. In practice, the regular drop in concentration of X every 30 s can be seen directly in the table of data. This tells you that the graph will be a straight line and the reaction is zero order with respect to X. © Philip Allan Updates

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35

Chapter 9 How fast?


b The concentration of Y does not change linearly with time and therefore the reaction is not zero order with respect to Y. To find out if the reaction is first order with respect to Y, you could use either of the two methods given in question 5. The first two results show that in the first 30 s the concentration of Y drops form 1.00 mol dm−3 to 0.70 mol dm−3, i.e. to 70% of the original value. This is repeated in the next 30 s, when the concentration falls from 0.70 mol dm−3 to 70% of this value, i.e. 0.49 mol dm−3. So the reaction is first order with respect to Y. However, a glance at part c should suggest that it would be better to plot the concentration–time graph because both the initial rate and the half-life are required to answer this part-question. 1.2 1.0 0.8 0.6

0.4 0.2 0

0

30

60

90

120

150 Time/s

It takes 58 s (strictly 58.3 s) for the concentration of Y to fall from 1.00 mol dm−3 to 0.50 mol dm−3, and 58 s for the concentration of Y to fall from 0.70 mol dm−3 to 0.35 mol dm−3. This also shows that the reaction must be first order with respect to Y. e Whenever

a value is deduced from a graph, examiners allow some leeway in the value quoted because it is difficult

to read a graph precisely.


c (i) rate = k[X]0[Y]1 = k[Y] (ii) The initial rate of the reaction is obtained by drawing a tangent to the graph at time = 0 s and measuring its gradient. 1.2 1.0 0.8 0.6

0.4 0.2 0 –0.2 0

30

60

90

120

150 Time/s

The initial rate is 0.012 mol dm−3 s−1.


OCR A2 Chemistry

36

Chapter 9 How fast? (iii) The rate constant is obtained by using the initial rate calculated in part (ii). 0.012 = k × 1.00 Therefore, k = 0.012 s−1 e If

you have looked at the ‘Additional reading’ section of Chapter 9 you will have learned that the half-life, t1/2, is

related to k by the equation kt1/2 = ln 2. This gives a means of checking the value of k: k × 58.3 = 0.693 so, k =

0.693 58.3

= 0.012 s−1

7 The results show clearly that the reaction is first order since it is possible to see that the half-life is constant. Concentration of hydrogen peroxide/mol dm−3

Half-life/s

0.1

5210

0.8

6889 − 1678 = 5211

0.6

9052 − 3841 = 5211


8 a This is very similar to question 6 above and can also be solved by either of two methods. It is clear that the reaction is not zero order as the fall in the concentration of SO2Cl2 is not linear with time. To check if the reaction is first order, the first method is to see whether, over a fixed time interval, the concentration of SO2Cl2 drops by the same proportion. In the first 30 min, the concentration drops to 75% of its initial value. This is repeated in the next 30 min when the concentration falls from 0.750 mol dm−3 to 75% of this value, i.e. 0.750 × 75/100 = 0.563 mol dm−3. This is the value given in the results table, thus proving that the reaction is first order. (A further check shows that 75% of 0.563 is 0.422 (the 90 min value), which acts as confirmation). The second method involves plotting a concentration–time graph. 1.2 1.0 0.8 0.6

0.4 0.2 0

0

30

60

90

120

150 Time/s

Measuring half-lives from the graph it can be seen that it takes approximately 72 min for the concentration of SO2Cl2 to fall from 1.00 mol dm−3 to 0.500 mol dm−3 and 72 min for the concentration to fall from 0.75 mol dm−3 to 0.375 mol dm−3. Therefore, the reaction is first order.


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Chapter 9 How fast?


b The rate constant requires a rate of reaction to be established for a known concentration. Taking the initial concentration of 1.00 mol dm−3, a tangent is drawn to the graph. Its gradient gives the initial rate of reaction. 1.2 1.0 0.8 0.6 0.4 0.2 0 –0.2 –0.2 0

30

60

90

120

150 Time/s

The initial rate of reaction is 0.0096 mol dm−3 min−1. So, 0.0096 = k[1] Therefore, k = 0.0096 min−1 e As

with question 6, the ‘Additional reading’ section of Chapter 9 shows that the half-life, t1/2, is related to k by the

equation kt1/2 = ln 2. This gives a means of checking the value of k: k × 72 = 0.693 so, k =

0.693 72

= 0.0096 min−1

9 a The rate equations are obtained from the rate-determining steps. For mechanism X: rate = k[O2][NO]2 For mechanism Y: rate = k[NO][O2] For mechanism Z: rate = k[O2] b (i) From the first two results, it can be seen that reducing the pressure from 50 kPa to 20 kPa (a factor of 2/5) while keeping the pressure of NO constant reduces the rate of reaction from 0.100 kPa h−1 to 0.040 kPa h−1. Since this is also a factor of 2/5, it means that the order of reaction with respect to O2 is 1. Using the second and third results, the initial pressure of O2 is increased three-fold from 20 kPa to 60 kPa while the initial pressure of NO is doubled from 50 kPa to 100 kPa. Since the order of reaction with respect to O2 is 1, the threefold increase in pressure will triple the rate, i.e. it will increase from 0.040 to 3 × 0.040 = 0.120 kPa h−1. However, the rate has actually increased to 0.48 kPa h−1, which is a four-fold increase. This must be because of doubling the concentration of the NO. This indicates that the order of reaction with respect to NO is 2. (ii) Therefore, the overall rate equation is: rate = k[O2][NO]2 This indicates that mechanism X is correct. © Philip Allan Updates

OCR A2 Chemistry

38

Chapter 9 How fast? Stretch and challenge a volume of alcohol in the beer = 568 × (4.1/100) = 23.29 cm3 mass of alcohol = volume × density = 23.29 × 0.789 = 18.37 g molar mass of ethanol = 46.0 g mol−1 amount (in moles) of ethanol = 18.37/46.0 = 0.40 mol b (i) Solving this graphically, by plotting a graph of moles against time, gives a constant half-life. Therefore the reaction is first order with respect to ethanol. e Given

that the half-life can be obtained directly from the results once it has been established that the reaction is

first order, it is easy to prove that it is first order by noting that in any fixed time interval the percentage drop in moles is the same — for example, the percentage drop from 0.40 mol to 0.34 mol is to (0.34/0.40) × 100 = 85% of the original value. The same percentage drop occurs from 0.34 mol to 0.29 mol, i.e. (0.29/0.34) × 100 = 85% of 0.34 mol.

b (ii) Note from the table of results that the time taken to reduce the amount in moles from 0.34 mol to 0.17 mol is 4 min. This is easier than using the graph. Either way gives the value for the half-life as 4 minutes. b (iii) The gradient at t = 0 gives the initial rate. In this case, it is 0.067 mol min−1 Therefore, using the initial amount in moles, the rate equation gives 0.067 = k × 0.4 Therefore, k = 0.17 min−1 e If

you understand the mathematical derivation, you could establish the value of k by noting that kt1/2 = ln 2 and,

therefore, k = 0.693/4 = 0.17 min−1.

c The alcohol transferred to the body fluids can be calculated from the amount remaining in the stomach. The concentration is (0.40 − x)/40, where x is the amount in moles left in the stomach — for example, for the measurement after 1 min the concentration in the body fluid including the blood is (0.40 − 0.34)/40 = 0.0015 mol dm−3. The table is shown below. Concentration of ethanol/mol dm−3

0.00

0.0015

0.0028

0.0040

0.0058

0.0081

0.0097

0.0098

Time/min

0

1

2

3

5

10

20

30

d (i) rate constant = 0.2 g dm−3 h−1 = 0.2/60 g dm−3 min−1 molar mass of ethanol = 46.0 g mol−1 rate constant = (0.2/60)/46.0 = 7.25 × 10−5 mol dm−3 min−1 e Strictly,

the data given in the questions only allow the answer to be given to 1 s.f.

(ii) Every minute the concentration of ethanol in the stomach is reduced by 7.25 × 10−5 mol dm−3. As the reaction is zero order, this is a constant rate of removal. Therefore, the ethanol remaining in the body fluids is the concentration listed in the table above minus (7.25 × 10−5)t where t is the time in minutes — for example, after 3 min the concentration is 0.0040 − (3 × 7.25 × 10−5) = 0.00378 = 0.0038 mol dm−3. Concentration of ethanol/mol dm−3

0

0.00143 0.00261

0.00378

0.00538 0.00738 0.00820 0.00767

Time/min

0

1

3

5


2

10

20

30

OCR A2 Chemistry

39

Chapter 9 How fast?

Ethanol concentration/mol dm–3

e Plotting the graph of concentration of ethanol against time gives something similar to the curve below, depending on the accuracy with which the calculations are carried out. 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

1

2

3

5

10

20 30 Time/min

The answer is therefore in the region of 20 min. f From the graph, the maximum concentration is 0.0082 mol dm–3. This corresponds to 0.00082 mol per 100 cm3 of blood, so the mass of ethanol in 100 cm3 of blood is 0.00082 × 46.0 = 0.038 g. This is 38 mg per 100 cm3 of blood. If all the assumptions in this calculation are accepted, it would be legal to drive. However, this does not mean that it would be wise to do so.


OCR A2 Chemistry

40


Chapter 10 How far? 1 a Kc =

[NO2(g)]2 [N2O4(g)]

b Kc =

[N2(g)]2[O2(g)] [N2O(g)]2

c Kc =

[CO(g)]2 [O2(g)]

e Remember,

carbon is a solid, so it has no concentration and therefore does not appear in the expression for the

equilibrium constant.

d Kc =

[CO2(g)]2 [CO(g)]2[O2(g)]

2 a Kc =

[PCl3(g)][Cl2(g)] [PCl5(g)]

b The units are mol dm−3 e In

this expression, the units of [PCl3] cancel with the units of [PCl5]. So the units of Kc are those of [Cl2].

c molar mass of PCl5(g) is 208.5 g mol−1 so, 2.085 g is 0.0100 mol PCl5(g) PCl3(g) + Cl2(g) so, [PCl5(g)] = 0.01 mol dm–3 At equilibrium, if the amount (in moles) of Cl2 is x mol: x×x Kc = 1.9 = 0.01 x2 = 0.019 x = 0.138 mol concentration of Cl2 = 0.138 × 71.0 = 9.8 g dm−3 3 There is 80 mg (0.08 g) of ethanol in 100 cm3 of blood. So, concentration of ethanol = 0.8 g dm−3 In mol dm−3, this is 0.8/46.0 = 0.017 mol dm−3 [C2H5OH(g)] Kc = [C2H5OH(blood)] 4.5 × 10−4 =

[C2H5OH(g)] 0.017

Concentration of ethanol in the vapour in equilibrium with the blood = 0.017 × 4.5 × 10−4 = 7.8 × 10−6 mol dm−3 or 3.6 × 10−4 g dm−3


OCR A2 Chemistry

41

Chapter 10 How far? 4 a Before equilibrium is established the amounts (in moles) are: 2N2(g) + O2(g) 2N2O(g) 1.00 mol 0 mol 0 mol Once equilibrium is established, there is 0.100 mol of N2O present and therefore 0.900 mol of the N2O has been converted to N2 and O2. The equation indicates that 1.00 mol of N2O reacts to produce 1.00 mol of N2 and 0.500 mol of O2. So, 0.900 mol of N2O will produce 0.900 mol of N2 and 0.450 mol of O2. Since the volume of the container is 1 dm3, the concentration of N2 is 0.900 mol dm−3and the concentration of O2 is 0.450 mol dm−3. b To summarise, at equilibrium: 2N2O(g) 2N2(g) + O2(g) 0.100 mol 0.900 mol 0.450 mol Kc =

=

[N2(g)]2[O2(g)] [N2O(g)]2 (0.900)2 × 0.450 (0.100)2

= 36.5 mol dm−3 c The value of Kc will not change. e It

is easy to be trapped into redoing the calculation and changing the concentrations to half their previous values to

obtain a new value for Kc. However, changing the volume of the container changes the amount of moles of each component in such a way that the value of Kc stays the same. Remember, only a change in temperature causes a change in the value of Kc.

5 a Kc =

[Cl]2 [Cl2]

If the value of Kc increases when the temperature is raised, this must be because more Cl atoms are produced. Using Le Chatelier’s principle, this must mean that the forward reaction of the equilibrium Cl2(g) 2Cl(g) is endothermic. e This

is perhaps predictable because heat is necessary to break the Cl–Cl bond in Cl2.

b (i) The reason is that the pressure in the stratosphere is very much lower and, by Le Chatelier’s principle, a lower pressure will cause the equilibrium to move to the side of higher volume (greater number of moles). e The

tempting answer is to suggest that it is because of an increased temperature. However, the temperature in the

stratosphere is considerably lower than that at the Earth’s surface.

(ii) Chlorine radicals attack the ozone layer by the mechanism covered in Unit 322 of the AS course. Cl + O3 → ClO + O2 ClO + O → Cl + O2


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42

Chapter 10 How far? Stretch and challenge a mass (in g) = volume (in cm3) × density (in g cm−3) amount (in moles) = mass/molar mass amounts (in moles) are: C2H5OH = (20.0 × 0.79)/46.0 = 0.343 = 0.34 mol CH3COOH = (20.0 × 1.05)/60.0 = 0.350 = 0.35 mol CH3COOC2H5 = (30.0 × 0.92)/88.0 = 0.314 = 0.31 mol H2O = (10.0 × 1.0)/18.0 = 0.556 = 0.56 mol b amount (in moles) of sodium hydroxide in 14.50 cm3 = (14.50 × 0.250)/1000 = 0.003625 mol This will have neutralised both the sulphuric acid catalyst and the ethanoic acid. The equation for the neutralisation of sulphuric acid is: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) 10.0 cm3 of 0.500 mol dm−3 of sulphuric acid contains 0.005 mol, which will be contained in the 80.0 cm3 of equilibrium mixture. However, only 1.00 cm3 was titrated. The amount (in moles) contained in 1.00 cm3 is 0.005/80 = 6.25 × 10−5 mol. Hence, this will react with 6.25 × 10−5 × 2 = 1.25 × 10−4 mol NaOH. The equation for the neutralisation of ethanoic acid is: CH3COOH(aq) + NaOH(aq) → CH3COO−Na+(aq) + H2O(l) 1 mol of NaOH neutralises 1 mol of CH3COOH amount (in moles) of ethanoic acid in 1.00 cm3 = 0.003625 − 1.25 × 10−4 = 0.00350 mol c amount (in moles) in 80.0 cm3 = 80.0 × 0.00350 = 0.280 mol d Using the equation: C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) amount of ethanoic used up in creating the equilibrium = 0.350 − 0.280 = 0.070 mol moles of ethanol present in the equilibrium mixture = 0.343 − 0.070 = 0.273 mol moles of ethyl ethanoate present in the equilibrium mixture = 0.314 + 0.070 = 0.384 mol moles of water present in the equilibrium mixture = 0.556 + 0.070 = 0.626 mol e Kc =

[CH3COOC2H5(l)][H2O(l)] [CH3COOH(l)][C2H5OH(l)]

so, Kc =

0.384 × 0.626 = 3.14 0.280 × 0.273

f The total number of molecules on either side of the equation is the same and therefore the volume would cancel out in the equilibrium expression. g ΔH = (−485.8 − 285.9) − (−277.7 − 484.5) = −9.5 kJ mol−1 h The difference in enthalpy between the reactants and products is quite small. This means that neither reactants nor products are likely to predominate within the mixture. e This

does not mean the that the equilibrium will necessarily be established quickly, as this depends on the

activation energy.

i The overall reaction (to form ethyl ethanoate and water) is exothermic, so a decrease in temperature would favour their formation. Therefore, at 20°C the value of the equilibrium constant would be higher. So the difference between the value obtained and the value quoted in data books is not explained by the experiment being carried out at 20°C. The difference is more likely to result from either a failure of the mixture to reach true equilibrium or to experimental error.


OCR A2 Chemistry

43


Chapter 11 Acids, bases and buffers 1 In parts a–c, the conjugate acid–base pairs are identified beneath the equations — for example, in part a one conjugate acid–base pair is H2O and OH–. a HCO3− + H2O H2CO3 + OH− Base 2 Acid 1 Acid 2 Base 1 b HCO3− + OH− Acid 1 Base 2

H2O + CO32− Acid 2 Base 1

c HCO3− + HCOOH Base 2 Acid 1

HCOO− + H2O + CO2 Base 1 The mixture of CO2 and H2O can be regarded as Acid 2

2 a pH = −log 0.15 = 0.82 [H+(aq)][CN−(aq)] b Ka = [HCN(aq)] [H+(aq)] = [CN−(aq)] so, Ka =

[H+(aq)]2 [HCN(aq)]

4.8 × 10−10 =

[H+(aq)]2 0.15

[H+(aq)] = 0.72 × 10–10 = 8.5 × 10−6 mol dm−3 pH = –log (8.5 × 10–6) = 5.1 c [H+(aq)][OH−(aq)] = 1 × 10−14 [H+(aq)] × 0.15 = 1 × 10−14 [H+(aq)] = 6.67 × 10−14 mol dm−3 pH = −log (6.67 × 10−14) = 13.2 This calculation can also be carried out using pH + pOH = 14 So pH −log (0.15) = 14 pH – 0.82 =14 pH = 13.2 d Na2SO4(aq) is a salt of a strong acid and a strong base and is neutral. Therefore the pH is 7. 3 When 20 cm3 of 1.00 mol dm−3 HCl is added to 10 cm3 of 1.00 mol dm−3 NaOH a reaction occurs and the NaOH is neutralised to produce neutral NaCl. 10 cm3 of the 1.00 mol dm−3 HCl remain un-neutralised. However, this has been diluted to the total volume of the mixture (30 cm3). This means that the concentration of the HCl is now (10/30) × 1.00 = 0.333 mol dm−3. Therefore, pH = −log 0.333 = 0.48 4 a [H+(aq)][OH−(aq)] = 1 × 10−14 [H+(aq)] × 0.5 = 1 × 10−14 [H+(aq)] = 2 × 10−14 mol dm−3 b pH = 4.0, so [H+(aq)] = 1 × 10−4 mol dm−3 (0.0001 mol dm−3)


OCR A2 Chemistry

44

Chapter 11 Acids, bases and buffers c pH = 2.7, so [H+(aq)] = 1 × 10−2.7 = 2.0 × 10−3 mol dm−3 d pH = 11.2 so [H+(aq)] = 1 × 10−11.2 = 6.3 × 10−12 mol dm−3 5 pH of the milk is 6.3, so [H+(aq)] = 1 × 10− 6.3 = 5.0 × 10−7 mol dm−3 6 pH of the magnesium hydroxide solution is 10.5 so, [H+(aq)] = 1 ×10−10.5 = 3.16 × 10−11 mol dm−3 so, [OH−(aq)] × ( 3.16 × 10−11) = 1 × 10−14 mol dm−3 [OH−(aq)] = 3.16 × 10−4 mol dm−3 When magnesium hydroxide dissolves it breaks into its ions: Mg(OH)2 → Mg2+(aq) + 2OH−(aq) There is one Mg2+(aq) ion produced for every two OH−(aq) ions. So [Mg2+] = 1.58 × 10−4 mol dm−3 7 a [H+(aq)] = 1 × 10−2.45 = 3.55 × 10−3 mol dm−3 b Assuming the equilibrium: H3X → 3H+(aq) + X3−(aq) the concentration of citrate ions will be (3.55 × 10–3)/3 = 1.18 × 10−3 mol dm−3 8 a Since the acid in apple juice is monobasic, 1 mole of it will be neutralised by 1 mole of sodium hydroxide. 22.90 cm3 of 0.120 mol dm–3 sodium hydroxide contains (22.90/1000) × 0.120 = 0.002748 mol. This neutralises 25.0 cm3 of the apple juice, so the concentration of the acid in the apple juice is (1000/25.0) × 0.002748 = 0.110 mol dm−3 b In the apple juice, [H+(aq)] = 1 × 10−3.5 = 3.2 × 10−4 mol dm−3 c If the anion of the acid in the apple juice is A−, then: [H+(aq)][A−(aq)] Ka = [HA(aq)] and since [H+(aq)] = [A−(aq)] [H+(aq)]2 Ka = [HA(aq)] =

(3.16 × 10−4)2 = 9.1 × 10−7 mol dm−3 0.11

9 a When potassium ethanoate is dissolved in water it ionises fully: CH3COO−K+ → CH3COO− + K+ The water in which it is dissolved will be slightly ionised: H2O H+ + OH− The presence of the hydrogen ions from the water is sufficient to create an equilibrium involving a weak acid: CH3COO− + H+ CH3COOH The effect is small, but the removal of hydrogen ions to form molecular CH3COOH is sufficient to mean that a solution of potassium ethanoate is very slightly alkaline with a pH above 7. e The

interaction of equilibria in aqueous solution often causes small effects such as that described above.

An alternative way of considering this is to say that the equilibrium: H2O + CH3COO−

CH3COOH + OH−

is set up and lies slightly to the right-hand side. The two arguments are essentially the same.


OCR A2 Chemistry

45

Chapter 11 Acids, bases and buffers b The extent of the removal of hydrogen ions discussed in part a depends on the strength of the acid. The weaker the acid, the greater is the formation of the molecular form. For HCN, the HCN lies further to the right-hand side than does the equilibrium equilibrium CN− + H+ − + CH3COO + H CH3COOH, so a greater number of hydrogen ions are removed from the water equilibrium. K+CN− produces a solution with a higher pH than CH3COO−K+ indicating that HCN is a weaker acid. c When ammonium chloride is dissolved in water it ionises fully: NH4+Cl− → NH4+ + Cl− The water in which it is dissolved will be slightly ionised: H2O H+ + OH− The presence of the hydroxide ions from the water is sufficient to create the equilibrium: NH4+ + OH− NH3 + H2O This results in the removal of some OH− ions from the water, leaving an excess of hydrogen ions. This makes the solution slightly acidic with a pH between 5 and 6.5. e An

alternative way of considering this is to say that the NH4+ ion is a stronger acid than H2O and therefore the

equilibrium: NH4+ + H2O

NH3 + H+

is set up and lies slightly to the right-hand side.

10 a Ka =

[H+(aq)][CH3CH2COO−(aq)] [CH3CH2COOH(aq)]

In the buffer solution, the concentrations of CH3CH2COO–(aq) and of CH3CH2COOH(aq) are approximated to their initial concentrations. When the two solutions are mixed, they dilute each other. So the concentrations are: [CH3CH2COO−(aq)] = 0.05 mol dm−3 [CH3CH2COOH(aq)] = 0.05 mol dm−3 Using the equilibrium constant above gives: [H+(aq)] × 0.05 1.3 × 10–5 = 0.05 Therefore [H+(aq)] = 1.3 × 10–5 mol dm−3 pH = 4.9 b The addition of potassium propanoate would cause some of the CH3CH2COO−(aq) ions to react with H+(aq) ions in the buffer solution to form more CH3CH2COOH(aq). As a result the pH will rise. 11 a pH of the acid is 2.70, so [H+] = 10–2.70 = 2.0 × 10−3 mol dm−3 In the solution of ethanoic acid, [H+(aq)] = [CH3COO−(aq)] [H+(aq)]2 so, Ka = [CH3COOH(aq)] Therefore 1.7 × 10−5 =

(2.0 × 10−3)2 [CH3COOH(aq)]

so, [CH3COOH(aq)] = (4.0 × 10−6)/1.7 × 10−5 = 0.235 = 0.24 mol dm−3 b For a buffer of pH 4.0, [H+(aq)] = 1 × 10−4 mol dm−3 The concentration of the ethanoic acid stays the same, so: 1 × 10−4 × [CH3COO−(aq)] 1.7 × 10−5 = 0.235


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Chapter 11 Acids, bases and buffers [CH3COO−(aq)] = (0.235 × 1.7 × 10−5)/10−4 = 0.04 mol dm−3 molar mass of sodium ethanoate = 82.0 g mol−1 mass required is 82.0 × 0.04 = 3.3 g (3.28 g) 12 a The ‘neutral’ colour of the indicator occurs when [HIn] = [In−]. This is at pH 4.1. b Since bromophenol blue changes colour at a relatively low pH value, it would be suitable for the titration of a strong acid with either a strong base or a weak base. 13 a At the mid-point of the reaction, pH = pKIn so, pH = −log (6.31 × 10−7) = 6.20 b Writing the indicator as HIn, the indicator equilibrium is: HIn H+ + In− yellow red In acid solution, the added H+ ions push this equilibrium to the left and, therefore, the indicator appears yellow. In alkaline solution, the added OH− ions react with the H+ ions from the indicator to produce water. This causes the indicator equilibrium to adjust by creating more In− ions and, therefore, it appears red. c (i) 0.0001 mol dm−3 has a pH of 4, so the indicator appears yellow. (ii) Pure water is pH 7 so the indicator colour will be changing to its red colour. To the eye, the solution will appear orange because the conversion to the HIn form will be incomplete. 14 a The equation for the neutralisation reaction is: CH3CH2COOH(aq) + NaOH(aq) → CH3CH2COO−(aq) + Na+(aq) + H2O(l) amount of propanoic acid used in the titration is (25.0/1000) × 0.020 = 5.0 × 10−4 mol concentration of sodium hydroxide = 0.025 mol dm−3 volume required = 5.0 × 10–4/0.025 dm3 = 0.020 dm3 = 20 cm3 eA

little thought might in fact allow you to see that the volume required must be 20 cm3, without the need for a

formal calculation.

b Ka =

[H+(aq)][CH3CH2COO−(aq)] [CH3CH2COOH(aq)]

since [H+(aq)] = [CH3CH2COO−(aq)] [H+(aq)]2 Ka = [CH3CH2COOH(aq)] so, 1.3 × 10−5 = [H+(aq)]2/0.020 [H+(aq)] = (2.6 × 10–7 = 5.1 × 10−4 mol dm−3 pH = −log (5.1 × 10−4) = 3.3 c [H+(aq)][OH−(aq)] = 1 × 10−14 [H+(aq)] × 0.025 = 1 × 10−14 [H+(aq)] = 4.0 × 10−13 mol dm−3 pH = 12.4


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Chapter 11 Acids, bases and buffers d

pH

14 12 10 8 6 4 2 0

eA

0

5

10

15 20 25 30 Volume of 0.025 mol dm–3 NaOH/cm3

rough sketch is fine in an answer to a question of this sort. However, the marking points would require a starting

pH of 2.8, the end point close to pH 7 when 20.0 cm3 of sodium hydroxide has been added and a final pH close to 12.4. (Since the solution is diluted at this stage, the final pH will be slightly less than this).

e There are a number of suitable indicators for a weak acid and strong base titration. Phenolphthalein is a suitable example. The requirement is that the indicator change is in the pH range 8–10. 15 a The equation for the reaction is: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) amount (in moles) of sulphuric acid used is (50.0/1000) × 1.00 = 0.0500 mol amount (in moles) of sodium hydroxide used = 0.0500 mol The equation indicates that twice the amount in moles of sodium hydroxide is required, so the sulphuric acid must be in excess. b The heat, q, produced by the reaction = total mass of the solution × specific heat capacity × temperature rise. Therefore, q = 100.0 × 4.18 × (23.6 − 19.0) = 1922.8 J 1922.8 J is produced by the reaction of 0.0500 mol of sodium hydroxide. Therefore the enthalpy change of neutralisation is: ΔH = 1922.8/0.0500 = 38.5 kJ mol−1

Stretch and challenge 1 a At pH 12, [OH−(aq)] = 0.01 mol dm−3 At pH 11.5, [H+(aq)] = 10−11.5 = 3.16 × 10−12 mol dm−3 This means that at pH 11.5, [OH−(aq)] = 10−14/(3.16 × 10−12) = 0.00316 mol dm−3 Therefore, the change in concentration is 0.00684 mol dm−3. b The neutral point is at pH 11.5. If the indicator equilibrium is written as NPBH KIn = [H+] Therefore KIn = 10−11.5 = 3.16 × 10−12 mol dm−3 c [H+][NPB−] = 3.16 × 10−12 If pH = 10.8, it follows that [H+] = 10−10.8 10−10.8 × [NPB−] Hence, = 3.16 × 10−12 [NPBH] © Philip Allan Updates

H+ + NPB−, this is when [NPBH] = [NPB]− and

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Chapter 11 Acids, bases and buffers [NPB−] 3.16 × 10−12 = [NPBH] 10−10.8 = 3.16 × 10−1.2 = 0.20 2 a If the acid equilibrium for aspirin is written as: HA(aq) H+(aq) + A−(aq) [H+(aq)][A−(aq)] = 3.0 × 10−4 [HA(aq)] If pH = 1, H+ = 10−1 = 0.1 mol dm−3 0.1 × [A−(aq)] = 3 × 10−4 [HA(aq)]

Therefore,

[A−(aq)] = 3 × 10−3 [HA(aq)]

This means that the aspirin is largely in its molecular form, HA. This would be likely to dissolve in the lipids in the stomach lining. Hence, bleeding might be a problem. b At pH 7.4, [H+(aq)] = 4.0 × 10−8 mol dm−3 [A−(aq)] 3.0 × 10−3 = = 7.5 × 104 [HA(aq)] 4.0 × 10−8 The aspirin is now largely ionised and is present mostly as the anion. c relative formula mass of Ca(OH)2 is 74.1 concentration of Ca(OH)2 = 0.0100 mol dm−3 Calcium hydroxide is a strong base, so it ionises fully: [OH−(aq)] = 0.0200 mol dm–3, so [H+(aq)] = 10–14/0.0200 mol dm–3 pH = 12.30 At pH 12.30 there would be substantial hydrolysis of the aspirin. d (i) C6H4(OCOCH3)CO2H + H2O C6H4(OH)CO2H + CH3COOH Since the equilibrium constant has no units the calculation can be carried out in moles rather than mol dm−3. molar masses of the components are: G C6H4(OH)CO2H — 138.0 g mol−1 G CH3COOH — 60.0 g mol−1 G C6H4(OCOCH3)CO2H — 180.0 g mol−1 G H2O —18.0 g mol−1 initial amount (in moles) of C6H4(OCOCH3)CO2H = 0.900/180.0 = 5.00 × 10−3 mol initial amount (in moles) of H2O = 0.100/18.0 = 5.55 × 10−3 mol equilibrium mass of CH3COOH = 0.117 g equilibrium moles of CH3COOH = 0.117/60.0 = 1.95 × 10−3 mol amounts (in moles) of the other components at equilibrium are: G C6H4(OH)COOH — 1.95 × 10−3 mol G C6H4(OCOCH3)CO2H — (5.00 – 1.95) × 10−3 = 3.05 × 10−3 mol G H2O — (5.55 − 1.95) × 10−3 = 3.60 × 10−3 mol [C H (OH)COOH][CH3COOH] (1.95 × 10−3)(1.95 × 10−3) = = 0.346 Ka = 6 4 [C6H4(OCOCH3)COOH][H2O] (3.05 ×10−3)(3.60 × 10−3) (ii) % of aspirin hydrolysed = © Philip Allan Updates

1.95 × 10−3 × 100 = 39.0% 5.00 × 10−3

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Chapter 12 Lattice enthalpy 1 a

Na+(g) + Cl(g) 1st ionisation enthalpy of Na

1st electron affinity of Cl

Na(g) + Cl(g) Na+(g) + Cl–(g) Enthalpy of atomisation of Cl 1 Na(g) + –Cl 2 2(g)

ΔH

Enthalpy of atomisation of Na

(Lattice enthalpy of Na+Cl–)

1 Na(s) + –Cl 2 2(g)

Enthalpy of formation of Na+Cl–(s)

b

Na+Cl–(s)

Na+(g) + Cl(g) 500kJ

–364 kJ

Na(g) + Cl(g) Na+(g) + Cl–(g) 122 kJ Na(g) +

1 –Cl 2 2(g)

108 kJ

ΔH

1 Na(s) + –Cl 2 2(g)

–411 kJ Na+Cl–(s)

Therefore 108 + 122 + 500 + (− 364) + ΔH = −411 ΔH = −411 − 108 − 122 − 500 + 364 = −777 kJ mol−1 e There

are several acceptable ways of drawing Born–Haber cycles and you should not worry if your diagram is not

exactly the same as the one above. The important thing is that you provide a complete enthalpy cycle that relates all the terms together. However, you should be aware that when a Born–Haber cycle is drawn on an exam paper it is usual for it to look similar to the diagram above.


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Chapter 12 Lattice enthalpy 2 a

2K+(g) + O2–(g)

2K+(g) + O(g) 2 × 420 kJ

+790.8 kJ –141.4 kJ 2K+(g) + O–(g)

2K(g) + O(g) 249.4 kJ 2K(g) +

ΔH

1 –O 2 2(g)

2 × 89.5 kJ 1 2K(s) + –O 2 2(g)

–361.5 kJ K2+O2–(s)

b Therefore (2 × 89.5) + 249.4 + (2 × 420) − 141.4 + 790.8 + ΔH = −361.5 ΔH = −179 − 249.4 − 840 + 141.4 − 790.8 − 361.5 = –2279.3 kJ mol−1 e This

is quite a complicated diagram. It is worth emphasising again that your version does not have to look the same

as the diagram above.

3 a

Ag+(g) + F(g) –332.6 kJ 730 kJ Ag+(g) + F–(g) Ag(g) + F(g)

79.1 kJ 1 Ag(g) + –F 2 2(g)

–943.0 kJ 286.2 kJ 1 Ag(s) + –F 2 2(g)

ΔH Ag+F–(s)

Therefore, the enthalpy of formation of silver fluoride is: ΔH = 286.2 + 79.1 + 730.0 − 332.6 − 943.0 = −180.3 kJ mol−1 The Born–Haber cycle for AuF is similar, so the enthalpy of formation of AuF is: ΔH = 369.6 + 79.1 + 890.1 − 332.6 − 772.0 = +234.2 kJ mol−1 b AgF has an exothermic enthalpy of formation and is, therefore, thermodynamically stable with respect to the elements silver and fluorine. The enthalpy of formation of AuF is endothermic and, therefore, it is thermodynamically unstable. e It

is possible that if AuF could be created then it might remain kinetically stable. It should be emphasised that

enthalpy differences only give one view of stability. In fact, gold forms a fluoride of formula AuF3.


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Chapter 12 Lattice enthalpy You may wonder how the lattice enthalpy of AuF can be given if the lattice does not exist. It is possible to predict a lattice enthalpy based on theoretical calculations related to the lattice structure of another compound. In this case, the numerical value is estimated based on the lattice structure of sodium chloride. The value has a degree of uncertainty but estimates of this kind have turned out to be very useful.

4 a The enthalpy cycle is: ΔH1

AgCl(s) ΔH2

Ag+(aq) + Cl–(aq) ΔH3

Ag+(g) + Cl–(g)

where ΔH1 is the enthalpy of solution, ΔH2 is the lattice enthalpy of silver chloride and ΔH3 is the sum of the enthalpies of hydration of Ag+(g) and Cl−(g) ΔH2 + ΔH1 = ΔH3 enthalpy of solution of silver chloride, ΔH1 = ΔH3 − ΔH2 = −464.4 − 384.1 − (− 890) = +41.5 kJ mol−1 b For silver iodide, the enthalpy of solution ΔH1 = ΔH3 − ΔH2 = −464.4 − 306.7 − (− 867) = +95.9 kJ mol−1 e In

both cases, the enthalpies of solution are endothermic and, of course, neither silver chloride nor silver iodide

dissolves in water to any extent. The figures also predict correctly the relative solubilities with silver iodide being considerably less soluble than silver chloride.

Stretch and challenge a (i) The fluoride ion is a very small anion and, therefore, has a high charge density. This means that it is likely to form stable ionic compounds. (ii) One factor that might encourage an ionic compound to form is the ease with which an element is able to form a cation. Xenon atoms, as the largest stable noble gas atoms, would lose an electron most readily and xenon is therefore the best possibility. b The enthalpy change is 1170 + 79.1 − 332.6 = +916.5 kJ mol−1 c The lattice enthalpies of fluorides, M+F−, do not have particularly high exothermic values. The values for alkali metals, are shown in the table. Fluoride

Lattice enthalpy/kJ mol−1

LiF

–1030

NaF

–915

KF

–813

RbF

–780

CsF

–735

As expected, the strongest lattice occurs with the smallest cation, Li+. However, Xe+ will be quite a large cation. Xe is in the same period as Rb (period 5) but, as atomic radii decrease across a period, Xe+ would probably be smaller than Rb+. However, it would not be as small as K+ in period 4, so it is unlikely that the formation of a XeF lattice could offset the 916.5 kJ mol−1 required to create the gaseous ions. d The use of UV to initiate the reaction between xenon and fluorine suggests that the bond in the fluorine molecule has been broken homolytically (rather than heterolytically to form ions). The low boiling point of XeF2 indicates that the bonding is simple molecular.


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Chapter 12 Lattice enthalpy e XeF4 has four covalent bonds, one to each of the four fluorine atoms; xenon has two further nonbonding pairs of electrons. Therefore its shape is square planar with bond angles of 90°: •• • •

•×

•• • •

F

F

•× ×

••

Xe ×

• •

• • ×

×

••

•×

F

• •

••

×

•

Square planar with lone pairs of electrons above and below the plane F F •• Xe •• F F

• • There are four bonded pairs and two lone pairs of electrons •• around the central Xe atom F

f 6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2 (i) The oxidation number of xenon in XeF4 is +4; in XeO3, it is +6. The oxidation number of the element xenon is 0. This is, therefore, a redox process with XeF4 being both oxidised and reduced. The oxidation number of oxygen in H2O is −2; in O2, it is 0. So oxygen has also been oxidised in the reaction. A fairly lively movement of electrons has occurred! (ii) It could be argued that this is an acid–base reaction because the H2O has donated protons to the XeF4 to create HF. However, the transfer mechanism is complicated. g It is likely that each of the bonds from the Xe to the three oxygen atoms is dative covalent. This leaves one non-bonding pair of electrons on the xenon. × ×

•• • •

O

× ×

••

Xe ××

• •

Pyramidal •• Xe

•• × ×

O ••

• •

O

O O

• • There are three bonded pairs and one lone pair of electrons •• around the central Xe atom O

The shape of XeO3 would be pyramidal. h Ignoring the small peaks at 162 and 164, the molecular mass is: (1.9 × 166) + (26.4 × 167) + (4.1 × 168) + (21.2 × 169) + (26.9 × 170) + (10.4 × 172) + (8.9 × 174) 99.8 This gives a relative molecular mass of 169.4. (Since 0.2% has been ignored, you might have chosen to divide by 100 instead of 99.8. In this case, the answer is 169.1) i The two fluorine atoms contribute 38.0 to the mass of the XeF2, so the relative atomic mass of xenon is 169.4 − 38.0 = 131.4 (or 131.1, if dividing by 100)


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Chapter 13 Enthalpy and entropy 1 e For

these questions it is a matter of deciding whether the reactant particles or the product particles have the greater

freedom to move. If the product particles are more restricted then ΔS will be negative; if the product particles have greater freedom then ΔS will be positive.

a b c d

ΔS will be negative (particles in ice are more restricted) ΔS will be positive (the particles in an aqueous solution have more freedom than in the solid) ΔS will be negative (the oxygen molecules lose their freedom to move) ΔS will be negative (the reduction in overall volume as the reaction takes place reduces freedom)

2 Substances with particles with the least freedom of movement tend to have the lowest entropy values. Of the three substances here, one is a solid, one is a liquid and the other is a gas. Solid iodine has particles with the least freedom to move. This identifies it as substance B, with an entropy value of 58.4 J mol−1 K–1. Methanol is a liquid, so it will have the next highest entropy value. Therefore, methanol is substance C with an entropy value of 127.2 J mol−1 K–1. Ammonia is a gas and has particles with the greatest freedom to move. Therefore, ammonia is substance A with an entropy value of 192.5 J mol−1 K–1. e Although

the argument above is basically sound, it should be mentioned that it is not always possible to judge from

the size of an entropy value whether a substance is a solid, liquid or gas.

3 The equation for the reaction is: 2Na(s) + –12 O2(g) → Na2O(s) The entropy change, ΔS, is: 72.8 − (2 × 51.0 + –12 × 204.9) = −131.7 J mol−1 K−1 e When

doing entropy calculations, it is worth checking that the sign of the answer is what you might expect because

this could prevent you from making careless mistakes. In this case, there is restricted movement of the particles in the product and therefore you would expect ΔS to be negative.

4 The equation for the reaction is: C2H6(g) + 3–12 O2(g) → 2CO2(g) + 3H2O(g) The total entropy of the products is (2 × 213.8) + (3 × 70.0) = 637.6 J mol−1 K−1 The total entropy of the reactants is 229.5 + (3–12 × 204.9) = 946.65 J mol−1 K−1 Therefore ΔS = 637.6 − 946.65 = −309.1 J mol−1 K−1 5 a The reaction is: C2H4(g) + H2(g) → C2H6(g) free energy change, ΔG = −32.8 − (68.1 + 0) = −100.9 kJ mol−1 b The reaction is: C2H4(g) + H2O(g) → C2H5OH(l) free energy change, ΔG = −174.9 − (68.1 − 228.6) = −14.4 kJ mol−1 6 a C6H6(l) + 3H2(g) → C6H12(l) (i) ΔH = −156.2 − 49.0 = −205.2 kJ mol−1 (ii) ΔS = 204.4 − (172.8 + (3 ×130.6)) = −360.2 J mol−1 K−1


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Chapter 13 Enthalpy and entropy b ΔG = ΔH −TΔS so, ΔG = −205.2 − 298 × (−0.3602) = −312.5 kJ mol–1 e Remember

that the enthalpies and entropies given are standard and are therefore measured at 298 K. Don’t forget

that entropy is measured in joules whereas enthalpy is measured in kilojoules.

7 C(s) + H2O(g) → CO(g) + H2(g) ΔH = −110.5 − (−241.8) = +131.3 kJ mol−1 ΔS = (197.9 + 130.6) − (5.7 + 188.7) = +134.1 J mol−1 K−1 Equilibrium will be achieved when ΔG = 0 Since ΔG = ΔH − TΔS, this will be when ΔH = TΔS The temperature at which equilibrium is reached is 131.3/0.1341 = 979 K or 706°C.

Stretch and challenge a (i) The equation for the formation of carbon monoxide is: C(s) + –12 O2(g) → CO(g) There is an increase in volume as –12 mol O2(g) produces 1 mol CO(g). This means that there is an increase in entropy as the reaction takes place and ΔS is therefore positive. ΔG = ΔH − TΔS and, although ΔH does not alter significantly as the temperature rises, TΔS will increase steadily and this makes ΔGf become more negative. (ii) The equation for the formation of carbon dioxide is: C(s) + O2(g) → CO2(g) There is no change in the volume as the reaction takes place. Therefore, there will be very little change in the value of the entropy of the reaction. This means that ΔGf is unlikely to change much as the temperature rises. e The

table shows that ΔGf becomes slightly more negative, which suggests that there is a small positive change in the

entropy as the reaction occurs. Data books give the values of the standard entropies for C as 5.7 J mol−1 K−1, O2 as 204.9 J mol−1 K−1 and CO2 as 213.6 J mol−1 K−1. This confirms the entropy change as 213.7 − (5.7 + 204.9) = 3.1 J mol−1 K−1.

(iii) As the oxides form, the oxygen gas is converted into the solid oxide. Therefore ΔS will be negative. Hence as the temperature rises, TΔS will become increasingly negative and ΔH − TΔS will become steadily more positive. b ΔG = ΔH − TΔS At 298 K, ΔGf = −137.2 kJ mol−1 and, assuming that ΔH does not change as the temperature rises, the change in the value of ΔG will be: −(change in temperature) × ΔS for the reaction −(2000 − 298) × (89.8) = −152.8 kJ mol−1 ΔGf at 2000 K = −137.2 − 152.8 = 290.0 kJ mol−1 c At 500 K, ΔGf(CO2) = −395.1 kJ mol−1, ΔGf(CO) = −155.4 kJ mol−1 Free energy change for the reaction: CO + –12 O2 → CO2 is ΔGf(CO2) − ΔGf(CO) = −395.1 − (−155.4) = −239.7 kJ mol−1 1 CO(g) + –O 2 2(g)

ΔG = ?

ΔGf (CO) = –155.4

CO2(g) ΔGf (CO2) = –395.1

C(s) + O2(g)


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Chapter 13 Enthalpy and entropy d Since ΔG = ΔH − TΔS At 500 K: − 239.7 = ΔH − 500ΔS At 750 K: − 218.1 = ΔH − 750ΔS subtracting gives: −21.6 = 250ΔS ΔS = −0.0864 kJ mol−1 K−1 or −86.4 J mol−1 K−1 Substituting into the first equation gives: −239.7 = ΔH − 500 × (−0.0864) so ΔH = −239.7 + (−43.2) = −282.9 kJ mol−1 e Equilibrium will be achieved when ΔH = TΔS, i.e. when −282.9 = T(−0.0864), which gives T = 3274 K f The formation of iron from its ore requires very high temperatures for reduction of the oxide. The table indicates that the free energies of formation at various temperatures of copper(II) oxide have much smaller negative values than those of iron(III) oxide. This means that copper(II) oxide is less stable with respect to reduction than is iron(III) oxide. The reduction is therefore possible at lower temperatures and this means it can be achieved without the use of modern furnaces. This is presumably what made the production of copper possible for ancient civilizations. g At elevated temperatures carbon is oxidised to carbon monoxide or carbon dioxide. The reduction of iron(III) oxide to iron therefore requires the free energy of the reaction: 3CO(g) + Fe2O3(s) → 2Fe(s) + 3CO2(g) to be negative for the reaction to take place. As the temperature rises, the reduction of Fe2O3 to Fe becomes easier (as shown in the table of values, which gives figures for the reverse reaction to form Fe2O3). However, the calculations in parts c and e and the value given in part d show that the oxidation of CO to CO2 is less favoured as the temperature is increased. (At 500 K, ΔG = −239.7 kJ mol−1; at 750 K, it is −218.1 kJ mol−1; at 3274 K, it is 0 kJ mol−1). The best temperature to operate the furnace depends on the balance between these two factors; a temperature of around 1850 K has been found to be optimum. e You

will probably not have considered all these aspects (and this is not a question that you would expect to find in

any exam without further guidance) but it does illustrate the complexity of what appear to be simple chemical reactions.

h Within a furnace, the presence of air means that carbon is oxidised to carbon monoxide or carbon dioxide. The only possible way, therefore, that aluminium metal could be formed from aluminium oxide would be if carbon monoxide was the reducing agent. Even at 3500 K, the table of ΔGf values indicates that Al2O3 is very stable with respect to its elements. (ΔGf = −580 kJ mol− 1). By 3274 K, the free energy change for the oxidation of CO to CO is 0 kJ mol−1. Therefore, it is 2 not possible to reduce aluminium oxide to aluminium using carbon monoxide. The only possible way of achieving the production of aluminium is by electrolysing molten aluminium compounds. Electricity was not discovered until the end of the eighteenth century. Following its discovery, many of the most reactive metals were discovered in quick succession, including aluminium in 1808. e Davy

also discovered sodium and potassium using electrolysis.


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Chapter 14 Electrode potentials and fuel cells 1 a CO32−(aq) + 2H+(aq) → H2O(l) + CO2(g) b CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g) e In

this reaction, the calcium carbonate should be included as shown because both the ions change — the Ca2+ from a

solid to an aqueous ion.

Ca2+(aq) + CO32−(aq) → CaCO3(s) OH−(aq) + H+(aq) → H2O(l) Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s) ZnO(s) + 2H+(aq) → Zn2+(aq) + H2O(l)

c d e f

+0 +5 +5 +6 +1 +6 +3 +5

2 a b c d e f g h

i j k l m n o p

+4 +7 +3 +5 +4 +3 +3 +2

3 e For

each of the part-questions a–g, the half-equations for the two reactions must be completed by adding the

appropriate number of electrons. The two half-equations should then be combined in such a way that the number of electrons produced in one half-reaction is the same as the number required by the other half-reaction.

a Fe2+(aq) → Fe3+(aq) + e− 2I−(aq) → I2(s) + 2e− Multiplying the first half-equation by 2 gives: 2Fe2+(aq) → 2Fe3+(aq) + 2e− The number of electrons generated (2) is the number required by the second half-equation. The overall equation is: 2Fe3+(aq) + 2I−(aq) → I2(s) + 2Fe2+(aq) b

MnO4−(aq) + H+(aq) → Mn2+(aq) + H2O(l) V2+(aq) → V3+(aq) (V is vanadium) In this case, the first half-equation must be completed taking into account that the oxidation number of Mn in MnO4– is +7 and in Mn2+ it is +2. The same result can be achieved by balancing the equation with H+ and e− to make sure that the total charge of the species on the left-hand side of the equation is the same as that on the right-hand side. 5e− + MnO4−(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l) V2+(aq) → V3+(aq) + e− Multiplying the second half-equation by 5 and adding the two half-equations together gives: MnO4−(aq) + 8H+(aq) + 5V2+(aq) → 5V3+(aq) + Mn2+(aq) + 4H2O(l)

e Always

check that the total electrical charge of the species on the left-hand side of the equation is the same as that

on the right-hand side. In this case, both are +17.


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Chapter 14 Electrode potentials and fuel cells c The balanced half-equations are: 5e− + MnO4−(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l) Sn2+(aq) → Sn4+(aq) + 2e− Multiplying the first half-equation by 2 and the second by 5 gives 10 e− in each half-equation: 10e− + 2MnO4−(aq) + 16H+(aq) → 2Mn2+(aq) + 8H2O(l) 5Sn2+(aq) → 5Sn4+(aq) + 10e− The balanced overall equation is: 2MnO4−(aq) + 16H+(aq) + 5Sn2+(aq) → 5Sn4+(aq) + 2Mn2+(aq) + 8H2O(l) e The

total electrical charge of each side of the equation is +24.

d The balanced half-equations are: 5e− + MnO4−(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l) V2+(aq) + 3H2O(l) → VO3−(aq) + 6H+(aq) + 3e− Multiplying the first half-equation by 3 and the second by 5 gives 15 e− in each half- equations: 15e− + 3MnO4−(aq) + 24H+(aq) → 3 Mn2+(aq) + 12H2O(l) 5V2+(aq) + 15H2O(l) → 5VO3−(aq) + 30H+(aq) + 15e− The balanced overall equation is: 3MnO4−(aq) + 24H+(aq) + 5V2+(aq) + 15H2O(l) → 5VO3−(aq) + 30H+(aq) + 3Mn2+(aq) + 12H2O(l) This can be simplified by removing the H+ and H2O that appear on both sides of the equation to give: 3MnO4−(aq) + 5V2+(aq) + 3H2O(l) → 5VO3−(aq) + 6H+(aq) + 3Mn2+(aq) e The

total electrical charge on each side of the equation is +7.

e The balanced half-equations are: 6e− + Cr2O72−(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l) 2Br−(aq) → Br2(aq) + 2e− Multiplying the second half-equation by 3 and combining the two half-equations gives: Cr2O72−(aq) + 14H+(aq) + 6Br−(aq) → 3Br2(aq) + 2Cr3+(aq) + 7H2O(l) e The


f The balanced half-equations are: 6e− + Cr2O72−(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l) SO2(aq) + 2H2O(l) → SO42−(aq) + 4H+(aq) + 2e− Multiplying the second half-equation by 3 and combining the two half-equations gives: Cr2O72−(aq) + 14H+(aq) + 3SO2(aq) + 6H2O(l) → 3SO42−(aq) + 12H+(aq) + 2Cr3+(aq) + 7H2O(l) This can be simplified to: Cr2O72−(aq) + 2H+(aq) + 3SO2(aq) → 3SO42−(aq) + 2Cr3+(aq) + H2O(l) e The

total electrical charge of each side of the equation is 0.

g The balanced half-equations are: 3e− + 4H+(aq) + NO3−(aq) → NO(g) + 2H2O(l) Cu(s) → Cu2+(aq) + 2e− Multiplying the first half-equation by 2 and the second by 3 gives: 6e− + 8H+(aq) + 2NO3−(aq) → 2NO(g) + 4H2O(l) 3Cu(s) → 3Cu2+(aq) + 6e− So the balanced overall equation is: 8H+(aq) + 2NO3−(aq) + 3Cu(s) → 3Cu2+(aq) + 2NO(g) + 4H2O(l) e The



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Chapter 14 Electrode potentials and fuel cells 4 e For

each of part-questions a–e, the cell potentials must be combined to make sure that:

G

one half-cell gives electrons and the other receives them

G

the maximum positive potential is achieved E °/V

Reaction Mg2+(aq) + 2e−

Mg(s)

−2.37

Zn2+(aq) + 2e−

Zn(s)

−0.76

Sn4+(aq) + 2e−

Sn2+(aq)

+0.15

I2(aq) + 2e−

2I−(aq)

+0.54

Fe3+(aq) + e−

Fe2+(aq)

+0.77

Br2(aq) + 2e−

2Br−(aq)

+1.09

a Mg(s) → Mg2+(aq) + 2e− E° = +2.37 V Zn2+(aq) + 2e− → Zn(s) E° = −0.76 V Therefore, the overall cell potential is +1.61 V. b Fe3+(aq) + e− → Fe2+(aq) E° = +0.77 V Sn2+(aq) → Sn4+(aq) + 2e− E° = −0.15 V Therefore, the overall cell potential is +0.62 V. e Remember,

you do not need to worry about balancing the number of electrons when combining electrode potentials.

So in this example, don’t be tempted to multiply the cell potential of the first equation by 2 because only one electron is released and the second reaction requires two electrons. Simply combine the two potentials as shown.

c Br2(aq) + 2e− → 2Br−(aq) E° = +1.09 V 2I−(aq) → I2(aq) + 2e− E° = – 0.54 V Therefore, the overall cell potential is +0.55 V. d Zn(s) → Zn2+(aq) + 2e− E° = +0.76 V I2(aq) + 2e− → 2I−(aq) E° = +0.54 V Therefore, the overall cell potential is +1.30 V. e Br2(aq) + 2e− → 2Br−(aq) E° = +1.09 V Sn2+(aq) → Sn4+(aq) + 2e− E° = −0.15 V Therefore, the overall cell potential is +0.94 V. 5 The aluminium foil sets up an electrical cell with the amalgam: Al(s) → Al3+(aq) + 3e− E° = +1.66 V Either of the two components of the amalgam is able to receive the electrons using saliva as the electrolyte. The cell potentials under standard conditions are as follows: Al(s) →Al3+(aq) + 3e– E° = +1.66 V + − Ag + e → Ag (amalgam) E° = +0.85 V This has an overall potential of +2.51 V. Al(s) → Al3+(aq) + 3e− E° = +1.66 V Sn2+ + 2e− → Sn(amalgam) E° = −0.13 V This has an overall potential of +1.53 V. Although the conditions in the mouth are, of course, not standard conditions these electric cells are enough to cause a current to flow and produce a nasty shock.


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Chapter 14 Electrode potentials and fuel cells 6 e For

each of part-questions a–g the electrode potentials must be combined to see if the net potential is positive, in

which case a reaction will occur, or negative, in which case a reaction under standard conditions is not possible. To answer these part-questions, there is no need to calculate the overall potential if you can see that it is positive or negative. Here, it has been done to confirm the prediction. E °/V

Reaction Co2+(aq) + 2e−

Co(s)

−0.28

Ni2+(aq) + 2e−

Ni(s)

−0.25

Pb2+(aq) + 2e−

Pb(s)

−0.13

2I−(aq)

+0.54

I2(aq) + 2e− Fe3+(aq) + e−

Cr2O72−(aq) + 14H+(aq) + 6e− Cl2(aq) + 2e−

+ 0.77

Fe2+(aq) 2Cr3+(aq) + 7H2O(l)

+1.36

2Cl−(aq)

MnO4−(aq) + 8H+(aq) + 5e−

+1.33

Mn2+(aq) + 4H2O(l)

+1.51

a The reaction to be considered is: Co(s) + Pb2+(aq) → Co2+(aq) + Pb(s) The relevant potentials are: Co(s) → Co2+(aq) + 2e− E° = +0.28 V Pb2+(aq) + 2e− → Pb(s) E° = −0.13 V net potential = 0.28 − 0.13 = +0.15 V The net potential is positive, so the reaction is energetically possible. b The reaction to be considered is: Pb(s) + Ni2+(aq) → Pb2+(aq) + Ni(s) The relevant potentials are: Pb(s) → Pb2+(aq) + 2e− E° = +0.13 V Ni2+(aq) + 2e− → Ni(s) E° = −0.25 V net potential = 0.13 − 0.25 = −0.12 V The net potential is negative, so the reaction is not possible energetically. c The reaction to be considered is: I2(aq) + Ni2+(aq) → 2I−(aq) + Ni(s) The relevant potentials are: I2(aq) + 2e− → 2I−(aq) E° = +0.54 V Ni2+(aq) + 2e− → Ni(s) E° = −0.25 V This reaction cannot occur since both I2 and Ni2+ require electrons. e Don’t

forget that for a reaction to occur one reagent must supply electrons and the other must receive them.


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Chapter 14 Electrode potentials and fuel cells e In

questions similar to parts d, e and f, it is unnecessary to provide a balanced equation unless one has been

requested specifically. Only the direction in which the reaction might occur and the size of the electrode potentials need to be considered.

d The reaction to be considered is: Cl−(aq) + MnO4−(aq) + H+(aq) → Cl2(aq) + Mn2+(aq) + H2O The relevant potentials are: MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) E° = +1.51 V 2Cl−(aq) → Cl2(aq) + 2e− E° = −1.36 V net potential =1.51 − 1.36 = +0.15 V The net potential is positive, so the reaction is energetically feasible. e The reaction to be considered is: I−(aq) + Cr2O72−(aq) + H+(aq) → I2(aq) + Cr3+(aq) + H2O(l) The relevant potentials are: Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) E° = +1.33 V 2I−(aq) → I2(aq) + 2e− E° = −0.54 V net potential = 1.33 – 0.54 = +0.79 V The net potential is positive, so the reaction is energetically feasible. f The reaction to be considered is: Cl−(aq) + Cr2O72−(aq) + H+(aq) → Cl2(aq) + Cr3+(aq) + H2O(l) The relevant potentials are: Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) E° = +1.33 V 2Cl−(aq) → Cl2(aq) + 2e− E° = −1.36 V net potential = 1.33 − 1.36 = −0.03 V The net potential is negative, so the reaction is not energetically feasible. This negative potential is small, so it might be possible to induce the reaction by using nonstandard conditions. g The reaction to be considered is: Cr3+(aq) + H2O(l) + MnO4−(aq) + H+(aq) → Cr2O72−(aq) + H+(aq) + Mn2+(aq) + H2O The relevant potentials are: 2Cr3+(aq) + 7H2O(l) → Cr2O72–(aq) + 14H+(aq) + 6e– E° = −1.33 V − + − 2+ MnO4 (aq) + 8H (aq) + 5e → Mn (aq) + 4H2O(l) E° = +1.51 V net potential = −1.33 + 1.51 = +0.18 V The net potential is positive, so the reaction is energetically feasible. e It

is worth reminding students that if the net potential is positive it indicates that the products have a lower energy

than the reactants but gives no indication that the reaction will definitely occur. If the activation energy for the reaction is high, the reaction may not take place.

7 The relevant electrode potentials are: Fe3+(aq) + e− → Fe2+(aq) E° = +0.77 V − − 2I (aq) → I2(aq) + 2e E° = −0.54 V Therefore, a reaction will occur between Fe3+(aq) and I−(aq). Hence, any attempt to make iron(III) iodide would result in iodide ions being oxidised to iodine and iron(III) ions being reduced to iron(II). Iron(III) chloride will, however, be stable because the relevant potential is: 2Cl−(aq) → Cl2(aq) + 2e− E° = −1.36 V Therefore, iron(III) ions cannot oxidise chloride ions to chlorine.


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Chapter 14 Electrode potentials and fuel cells 8 a Zn(s) → Zn2+(aq) + 2e− O2(g) + 2H2O(l) + 4e− → 4OH−(aq) b The first half-equation must be multiplied by 2 so that the correct number of electrons are supplied to the second half-equation: 2Zn(s) + O2(g) + 2H2O(l) → 4OH−(aq) + 2Zn2+(aq) c Zn2+(aq) + 2e− → Zn(s) E° = −0.76 V − − O2(g) + 2H2O(l) + 4e → 4OH (aq) E° = +0.40 V so for the reaction: 2Zn(s) + O2(g) + 2H2O(l) → 4OH−(aq) + 2Zn2+(aq) net potential = +0.76 + 0.40 = +1.16 V d The cell is not operating under standard conditions. The oxygen concentration and the zinc ion concentration will be substantially less than 1 mol dm−3 and the temperature of operation will be normal body temperature of around 37°C. 9 a Pb(s) + HSO4−(aq) → PbSO4(s) + H+ (aq) + 2e− b To obtain the overall equation: Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4−(aq) → 2PbSO4(s) + 2H2O(l) the reaction at the cathode must involve the reduction of PbO2 with the formation of 2H2O. This suggests Pb is also formed and gives the half-equation as: PbO2(s) + 4H+ (aq) + 4e− → Pb(s) + 2H2O(l) c At the anode: Zn(s) + 2OH−(aq) → Zn(OH)2(s) + 2e− At the cathode: Ag2O(s) + H2O(l) + 2e− → 2Ag(s) + 2OH−(aq) 10 a From the table, since: Fe3+(aq) + e− → Fe2+(aq) E° = +0.77 V 2+ 3+ − Cr (aq) → Cr (aq) + e E° = +0.41 V the reaction that occurs is: Fe3+(aq) + Cr2+(aq) → Fe2+(aq) + Cr3+(aq) giving an overall voltage of +1.18 V The other electrode potentials quoted do not give a combined positive potential. b As the reaction takes place, in the compartment containing the Fe3+(aq) ions, the ions are reduced to Fe2+(aq) and as a result the net positive charge decreases. In the compartment containing the Cr2+(aq) ions, oxidation to Cr3+(aq) means that the net positive charge is increasing. The passage of chloride ions from the compartment containing the iron ions to the compartment containing the chromium ions resolves the imbalance of charge and allows the cell to keep operating.

Stretch and challenge 1 a The standard electrode potential requires the concentration of oxygen to be 1 mol dm−3, the concentration of hydrogen ions to be 1 mol dm−3, the temperature to be 25°C and the pressure to be 101 kPa. The first two of these conditions do not occur in soil; the second two may not because of the atmospheric conditions. Typically, soils have a pH in the range 6–8 and the oxygen supply from the air is of limited concentration.


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Chapter 14 Electrode potentials and fuel cells b The electrode potential of the half-cell, O2(g) + 4H+(aq) + 4e− 2H2O(l) is reduced because low concentrations of hydrogen ions and oxygen encourage a movement of the equilibrium from right to left. c Iron is kept oxidised as Fe3+(aq) by the combination of the half-cells: O2(g) + 4H+(aq) + 4e− → 2H2O(l) E° = +0.80 V 2+ 3+ − Fe (aq) → Fe (aq) + e E° = −0.77 V The typical E° value of +0.80 V is only just sufficient to maintain iron as Fe3+. If, as a result of the soil becoming compacted or waterlogged, the concentration of oxygen falls further (and its pressure is perhaps reduced as well) the value may well fall below the 0.77 V required. In this case, the Fe2+ will not be oxidised. This is indicated by the soil having the greenish colour indicative of iron(II) compounds. d 10H+ + SO42− + 8e− → H2S + 4H2O 4H+ + CH3COOH + 4e− → CH4 + 2H2O 2 a The two redox half-equations combine and the hydrogen peroxide decomposes to oxygen and water: H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l) E° = +1.77 V + − H2O2(aq) → O2(g) + 2H (aq) + 2e E° = −0.69 V Since the net potential is 1.77 − 0.69 = +1.08 V, the overall reaction can take place as follows: 2H2O2(aq) → O2(g) + 2H2O(l) b (i) The reaction, Fe3+(aq) + e− → Fe2+(aq) (E° = +0.77 V) would allow the reaction H2O2(aq) → O2(g) + 2H+(aq) + 2e− to take place. Fe3+(aq) + e− → Fe2+(aq) E° = +0.77 V H2O2(aq) → O2(g) + 2H+(aq) + 2e− E° = −0.69 V The net potential for the reaction: H2O2(aq) + 2Fe3+(aq) → O2(g) + 2H+(aq) + 2Fe2+(aq) is positive (0.77 − 0.69 = +0.08 V). (ii) Fe3+(aq) might be a catalyst for the process because the Fe2+(aq) produced can be oxidised back to Fe3+(aq): H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l) E° = +1.77 V 2+ 3+ − Fe (aq) → Fe (aq) + e E° = −0.77 V The net potential for the reaction: H2O2(aq) + 2H+(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l) is positive (1.77 − 0.77 = +1.00 V). c (i) molar mass of hydroquinone = 110.0 g mol−1 amount (in moles) of 10 g hydroquinone = 10/110.0 = 0.091 mol total volume of solution = 100 cm3 concentration of 10% solution of hydroquinone = 0.91 mol dm−3 (ii) molar mass of hydrogen peroxide is 34.0 g mol−1 amount (in moles) of 25 g hydrogen peroxide = 25/34.0 = 0.735 mol total volume of solution = 100 cm3 concentration of 25% solution of hydrogen peroxide = 7.4 mol dm−3 d (i) amount (in moles) of hydrogen peroxide contained in 0.1 cm3 = (0.1 /1000) × 7.4 = 0.00074 mol The equation for the redox reaction is: H2O2(aq) + 2Fe3+(aq) → O2(g) + 2H +(aq) + 2Fe2+(aq) so, amount (in moles) of oxygen produced = 0.00074 mol


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Chapter 14 Electrode potentials and fuel cells so, volume of oxygen at 100°C = 0.00074 × 30 100 = 22 cm3 This volume of oxygen produced suddenly is sufficient to propel the beetle’s boiling spray some distance. (ii) Heating hydrogen peroxide causes it to decompose, thus: 2H2O2(aq) → O2(g) + 2H2O(l) so, 0.5 mol of oxygen is formed from 1 mol of hydrogen peroxide so, the volume of oxygen produced is 0.5 × 22 = 11 cm3 e Quinhydrone can be produced by the following redox reaction: HO

OH

→ quinhydrone + 2H+(aq) + 2e−

H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l) Quinone is then produced by the redox reaction:

E° = −0.63 V E° = +1.77 V

O

+ 2H+(aq) + 2e−

Quinhydrone →

E° = −0.75 V

O

H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l)

E° = +1.77 V

3 e At

first glance, this question seems as if it should be easy. However, it must be remembered that 1 volt is unit of 1

joule per coulomb: it is not a straightforward energy term in joules. Therefore, the values given cannot be simply added or subtracted.

Suppose 1 coulomb of electricity is supplied to make the conversion Fe3+(aq) + e− → Fe2+(aq). From the electrode potential, it is clear that 0.77 joules would be released. A further 2 coulombs are required to convert the Fe2+ formed to Fe, since the process is Fe2+ (aq) + 2e− → Fe(s). These 2 coulombs would require 0.88 joules because the electrode potential indicates that the energy is −0.44 V (joules per coulomb). Therefore, to convert Fe3+ to Fe requires 3 coulombs. The energy for this process is 0.77 − 0.88 = −0.11 J. Electrode potentials are expressed in volts i.e. joules per coulomb. Therefore, Fe3+(aq) + e− → Fe2+(aq) has an electrode potential of −0.11/3 = −0.037 V 4 a The reaction: 2Cu2+(aq) + 2I− (aq) → 2Cu+(aq) + I2(s) does not appear to be feasible because the net potential is +0.15 − 0.54 = −0.39 V b It is necessary to appreciate that the net potential of a reaction gives only the position of the equilibrium between the reactants and products. A high positive potential for a reaction indicates that the conversion to the products is substantial; a negative potential indicates that the reactants are favoured. In this case, the potential of −0.39 V shows that the equilibrium mixture contains very little Cu+ and I2. However, some Cu+ ions are present and, in addition, copper(I) iodide is extremely insoluble — sufficiently insoluble to cause a precipitate of copper(I) iodide to form as it reacts with the I− present. This disrupts the equilibrium and more Cu+ is generated to replace that which


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Chapter 14 Electrode potentials and fuel cells has been precipitated. This continues until all the Cu2+ has been used up. The result is a precipitate of copper(I) iodide and iodine. Another way of looking at this is to appreciate that Cu+ is never present in solution, let alone a solution of concentration 1 mol dm−3. The use of the standard electrode potential is therefore misleading in this case.


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Chapter 15 Transition elements 1 a b c d e

[Zn(NH3)4(H2O)2]2+ [Fe(CN)6]4− [Co(NH3)5Cl]2+ [Co(C2O4)3]4− [Cr(CH3COO)2(H2O)2]+

2 a The change in colour suggests that a ligand exchange has taken place. The aqueous copper(II) sulphate solution is light blue because of the presence of the [Cu(H2O)6]2+ ion. 1,2-diaminoethane is a stronger ligand than water and the change in colour to a darker blue is due to the formation of a copper(II) 1,2-diaminoethane complex ion. b When hydrochloric acid is added, the hydrogen ions form a stronger bond with the 1,2-diaminoethane than does the Cu2+ ion. H2C H2N

CH2 NH2

H2C

+ 2H+ H3N +

CH2 NH3 +

The copper(II) 1,2-diaminoethane complex ion therefore breaks down and the copper(II) recomplexes with water molecules so the original light blue colour is restored. 3 a The aqueous cobalt(II) ion, [Co(H2O)6]2+, is pink. Chloride ions in high concentration replace the water molecules, forming the blue cobalt chloride complex ion, [CoCl4]2−. b In the presence of larger amounts of water, the process in part a is reversed and pink [Co(H2O)6]2+ ions are re-formed. The equilibrium: [Co(H2O)6]2+ + 4Cl− [CoCl4]2− + 6H2O moves readily to one side or the other depending on the concentration of chloride ions and water molecules present. c When aqueous silver nitrate is added, the silver ions react with chloride ions to produce a precipitate of silver chloride, AgCl. The blue cobalt chloride complex ion, [CoCl 4]2–,then decomposes and pink [Co(H2O)6]2+ ions are formed once again, i.e. the Ag+ ion removes the Cl− ion from the equilibrium: [Co(H2O)6]2+ + 4Cl− [CoCl4]2− + 6H2O pink blue so that the equilibrium moves to the left and the solution turns pink. d [CoCl4]2− is soluble in propanone. It is extracted into that layer and, therefore, a blue colour is obtained. 4 a In complex B, chloride forms a complex cation that does not release the chloride ion when aqueous silver nitrate is added. Only the chloride present as an anion forms a precipitate when aqueous silver nitrate is added. Compound A has three Cl− ions; compound B has only two Cl− ions. Therefore, compound A produces 3/2 times as much precipitate of silver chloride.


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Chapter 15 Transition elements b Compounds C and D must only have one Cl− ion available as an anion. The other two Cl− ions must be present in a complex cation with the four ammonia ligands. This suggests that C and D are cis–trans isomers. +

NH3 H3N

Co

Cl

NH3 Cl

NH3 cis

+

NH3 Cl Co

NH3 Cl

H3N NH3 trans

5 The half-equation for the reduction of MnO4−(aq) is: MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) The half-equation for the oxidation of VO2+(aq) is: VO2+(aq) + 2H2O(l) → VO3– (aq) + 4H+(aq) + e− Multiplying the second half-equation by 5, the overall equation for the reaction is: MnO4−(aq) + 8H+(aq) + 5VO2+(aq) + 10H2O(l) → 5VO3− (aq) + 20H+(aq) + Mn2+(aq) + 4H2O(l) or MnO4−(aq) + 5VO2+(aq) + 6H2O(l) → 5VO3− (aq) + 12H+(aq) + Mn2+(aq) amount (in moles) of MnO4–(aq) in 23.30 cm3 of 0.0150 mol dm−3 solution = (23.30/1000) × 0.0150 = 3.495 × 10−4 mol From the equation : amount (in moles) of VO2+(aq) used in the titration = 5 × 3.495 × 10−4 = 1.7475 × 10−3 mol so, concentration of VO2+(aq) = (1000/25.0) × 1.7475 × 10−3 = 0.0699 mol dm−3 6 a Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) Sn2+(aq) → Sn4+(aq) + 2e− Multiplying the second half-equation by 3 and adding the two equations together gives the overall equation: Cr2O72−(aq) + 14H+(aq) + 3Sn2+(aq) → 3Sn4+(aq) + 2Cr3+(aq) + 7H2O(l) b amount (in moles) of Cr2O72–(aq) in 20.00 cm3 of 0.0175 dichromate solution is (20.00/1000) × 0.0175 = 3.50 × 10−4 mol From the equation: amount (in moles) of Sn2+(aq) used in the titration = 3 × 3.500 × 10–4 = 1.05 × 10−3 mol so, concentration of Sn2+(aq) = (1000/25.0) × 1.05 × 10−3 = 0.0420 mol dm−3 c Ar of tin = 118.7 mass of tin in 0.0420 mol = 0.0420 × 118.7 = 4.985 g percentage by mass of tin in the solder = (4.985/10.00) × 100 = 49.9% 7 a MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (COO)22−(aq) → 2CO2(g) + 2e− Multiplying the first half-equation by 2 and the second half-equation by 5 and adding the two together gives the overall equation: 2MnO4−(aq) + 16H+(aq) + 5(COO)22−(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l) b amount (in moles) of MnO4−(aq) in 23.90 cm3 of 0.0200 mol dm−3 KMnO4 solution = (23.90/1000) × 0.0200 = 4.78 × 10−4 mol From the equation: amount (in moles) of (COO)22−(aq) used in the titration = 5/2 × 4.78 × 10−4 = 1.195 × 10−3 mol so, amount (in moles) of (COO)22−(aq) in 250 cm3 = (250/25.0) × 1.195 × 10−3 = 0.0120 mol


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Chapter 15 Transition elements c Mr of ethanedioic acid is 90.0 mass of ethanedioic acid in four rhubarb leaves = 90.0 × 0.0120 = 1.08 g d 24 g ethandioic acid is a fatal dose, suggesting that the number of rhubarb leaves required is (24/1.08) × 4 = 89 leaves. e This

calculation assumes the rhubarb leaves are of uniform size. Long before someone had munched their way

through 89 leaves they would no doubt be very ill!

8 The relevant equations are: Br2(aq) + 2I−(aq) → I2(aq) + 2Br−(aq) I2(aq) + 2e− → 2I−(aq) 2S2O32−(aq) → S4O62−(aq) + 2e− The overall equation for the titration is: 2S2O32−(aq) + I2(aq) → S4O62−(aq) + 2I−(aq) amount (in moles) of S2O32−(aq) in 24.75 cm3 of 0.800 mol dm−3 sodium thiosulphate solution = (24.75/1000) × 0.800 = 0.0198 mol From the equation: amount (in moles) of I2 produced by the addition of KI(aq) to 25.0 cm3 of the bromine solution = 1/2 × 0.0198 = 0.00990 mol From the first equation, 1 mol of Br2 produces 1 mol of I2 so, amount (in moles) of bromine in the 25.0 cm3 = 0.00990 mol so, concentration of the aqueous bromine = (1000/25.0) × 0.00990 = 0.396 mol dm−3 9 The overall equation for the titration is: 2S2O32−(aq) + I2(aq) → S4O62−(aq) + 2I−(aq) amount (in moles) of S2O32−(aq) in 10.25 cm3 of 1.500 × 10−3 mol dm−3 sodium thiosulphate solution = (10.25/1000) × 1.500 × 10−3 = 1.5375 × 10−5 mol From the equation: amount (in moles) of I2 produced by the addition of KI(aq) to 500.0 cm3 of the swimming pool water = 0.5 × 1.5375 × 10−5 = 7.688 × 10−6 mol Cl2(aq) + 2I−(aq) → I2(aq) + 2Cl–(aq) so, amount (in moles) of chlorine in 500.0 cm3 water = 7.688 × 10−6 mol Mr of Cl2 = 71.0 so, mass of chlorine in 500.0 cm3 water = 7.688 × 10−6 × 71.0 = 5.458 × 10−4 g mass of chlorine in 1000 dm3 = (1000/0.5) × 5.458 × 10−4 = 1.09 g The water should be safe to swim in. eA

titration in which 500.0 cm3 of one solution is balanced by 10.25 cm3 of another (very dilute) solution is subject to

significant error, so there must be doubt about the reliability of the result.

10 25.0 cm3 of NaXO3 contains (25.0/1000) × 0.0500 = 1.25 × 10−3 mol The overall equation for the titration is: 2S2O32−(aq) + I2(aq) → S4O62−(aq) + 2I−(aq) Mr of sodium thiosulphate, Na2S2O3, = 158.2 so, concentration of Na2S2O3 = 13.63/158.2 = 0.08616 mol dm−3 amount (in moles) of S2O32−(aq) in 29.00 cm3 Na2S2O3 solution = (29.00/1000) × 0.08616 = 2.50 × 10–3 mol so, amount (in moles) of iodine produced = 1.25 × 10–3 mol Therefore, in the reaction, 1 mol of XO3− produces 1 mol of I2. Since 2I–(aq) → I2(aq) + 2e–, the oxidation state of XO3− must be reduced by 2, from +5 to +3.


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Chapter 15 Transition elements Stretch and challenge e The

answers to the following questions are reasonably straightforward once the method has been understood. EGTA

forms complexes with the calcium and magnesium ions; some EGTA is left uncomplexed. The thymolphthalexone then forms a complex with magnesium ions. When the lead ions are added, they first form a complex with the excess EGTA and then displace magnesium ions from the complex with thymolphthalexone. The end point occurs when the magnesium has just been displaced completely. Throughout the titration, the calcium–EGTA complex remains unchanged. e In

answers a (i) to a (v) ‘>’ means ‘has a greater a stability than’

a (i) Ca–EGTA > Ca–H2O (ii) Pb–EGTA > Pb–H2O As the lead ions are added in the titration, they react with the free EGTA. (iii) Pb–thymolphthalexone > Mg–thymolphthalexone During the titration, the magnesium ions are displaced from the thymolphthalexone complex and are replaced by lead ions. (iv) Mg–thymolphthalexone > Mg–EGTA > Mg–H2O This follows the sequence, as the ligands are added to the magnesium prior to the titration. (v) Ca–EGTA > Pb–EGTA > Mg–EGTA The Ca–EGTA remains at the end of the titration but during the course of the titration the lead ions form a complex with EGTA displacing the magnesium ions. b The end point of the titration is colourless. Therefore, thymolphthalexone must be colourless. c At the end of the titration, magnesium must be present as the Mg–H2O complex because the lead has formed complexes with the thymolphthalexone and any EGTA in the solution. The calcium ions have remained bound to the EGTA throughout the titration and are present as Ca–EGTA. Lead ions are present as Pb–EGTA together with a small amount of Pb–thymolphthalexone, which has been formed during the titration. d At the end of the titration, the EGTA added is present either as Ca–EGTA or as Pb–EGTA. Therefore (in moles): amount of calcium present = (total amount of EGTA added) – (amount present in the lead–EGTA complex) Since all the lead ions added are complexed with EGTA at the end point, this is the same as: (total amount of EGTA added) – (amount of lead ions added) There are two assumptions made: G The complexes formed between calcium and EGTA, and between lead and EGTA, are in a 1:1 ratio. G The small amount of lead that is complexed with thymolphthalexone at the end point can be ignored.


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Chapter 16 Practical skills 1 Error 1: the water supply to the apparatus is connected incorrectly. As illustrated, the water would not surround the inner tube of the reflux condenser adequately and provide suitable cooling. Error 2: there should not be a stopper in the top of the reflux condenser. It would be blown off when the apparatus was heated and either the air inside expanded or the liquid boiled and vaporised into the condenser. 2 a Aqueous sodium chloride is colourless but aqueous copper sulphate is a blue solution. e Confusing

the words ‘clear’ and ‘colourless’ is a common error.

b When sodium chloride is dissolved in water the white solid forms a colourless solution. When aqueous silver nitrate solution is added, a white precipitate is formed. e When

3 a

a white solid dissolves to form a colourless solution, it is easy to forget to include this as an observation.

Titration Final burette

reading/cm3

Initial burette

reading/cm3

Volume used/cm3

e Care

Rough

1

2

3

28.00

27.80

27.35

27.25

0.00

0.15

0.20

0.00

28.00

27.65

27.15

27.25

is needed to make sure that two decimal places are used consistently for burette readings: either ‘ .00’ or ‘ .05’.

b The mean value for the titration is 27.20 cm3 The mean value should be obtained from the consistent titres 27.15 cm3 and 27.25 cm3. 0.06 4 a % error for the pipette = × 100 = 0.24% 25.0 b In this case, two readings are taken, each with a maximum error of 0.05 cm3. 2 × 0.05 % error for the burette = × 100 = 0.4% 25.0 5 a Two temperature measurements are taken, each with a maximum error of 0.5°C. 2 × 0.5 % error for the thermometer = × 100 = 29% 3.5 b Two temperature measurements are taken, each with a maximum error of 0.1°C. 2 × 0.1 % error for the thermometer = × 100 = 5.7% 3.5 6 a (i) 34.76 (ii) 34.8 (iii) 35 b (i) 0.00438 (ii) 0.0044 c (i) 0.50 (ii) 0.5 d 1570


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Chapter 16 Practical skills 7 If the balance were completely accurate, the mass of solid would be 68.67 − 66.78 = 1.89 g. However, each reading has a maximum error of 0.005 g. 2 × 0.005 Therefore, the maximum % error in the mass of the solid = × 100 = 0.53% 1.89 8 a heat produced = mass of solution × specific heat capacity × temperature rise = 50 × 4.18 × (23.5 − 18.0) = 1.1495 kJ (ignoring significant figures at this stage) b Using the extremes of the errors and ignoring significant figures at this stage: (i) heat produced = 49 × 4.18 × (23.0 − 18.5) = 0.92169 kJ (ii) heat produced = 51 × 4.18 × (24.0 − 17.5) = 1.38567 kJ c The calculations indicate the wide range of uncertainty that might exist in the answer. It is acceptable to quote the answer to part a to two significant figures, i.e. 1.1 kJ. e There G

are two points to note:

In an assessed task, you should note the quantity that has been measured to the least number of significant figures and use that to supply the correct number of significant figures to use in the final answer.

G

The maximum error in the mass of the calcium carbonate would become important if the enthalpy change per mole of the reaction was calculated.

9 a (i) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) or Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (ii) ZnCl2(aq) + Na2CO3(aq) → ZnCO3(s) + 2NaCl(aq) or Zn2+(aq) + CO32−(aq) → ZnCO3(s) Either type of equation is acceptable, but ionic equations are preferable. b amount (in moles), of hydrochloric acid used = (25/1000) × 2.00 = 0.050 mol c Mr of zinc carbonate = 65.4 + 12.0 + 48.0 = 125.4 amount (in moles) of zinc carbonate precipitated = 0.32/125.4 = 0.00255 mol The same amount (in moles) of zinc must have been formed. so, mass of zinc in the brass = 0.00255 × 65.4 = 0.167 = 0.17 g d % of zinc in the brass = (0.167/0.50) × 100 = 33% e amount (in moles) of zinc present = 0.00255 mol amount (in moles) of hydrochloric acid used = 0.050 mol The equation in part a (i) indicates that 1 mol of zinc requires 2 mol of hydrochloric acid for reaction. Therefore, 0.00255 mol of zinc requires 0.0051 mol of hydrochloric acid. So the hydrochloric acid is substantially in excess. The instructions also state that the aqueous sodium carbonate is in excess. Therefore the precise volumes of the solutions used are unimportant. The accuracy of the balance is the important factor in obtaining a more accurate result. 10 The important precautions that should be taken are: G heat the crucible to constant mass G cool the crucible in a dry atmosphere (e.g. use a desiccator)

Stretch and challenge The equation for the reaction is: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Reason 1: 2.00 g of marble chips contains 2.00/100.1 = 0.02 mol This requires a minimum of 0.04 mol of hydrochloric acid for complete reaction.


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Chapter 16 Practical skills The amounts (in moles) of hydrochloric acid used in the experiments are: G 50.0 cm3 of 4.00 mol dm−3 = 0.200 mol G 50.0 cm3 of 2.00 mol dm−3 = 0.100 mol G 50.0 cm3 of 1.00 mol dm−3 = 0.050 mol G 50.0 cm3 of 0.500 mol dm−3 = 0.025 mol G 50.0 cm3 of 0.250 mol dm−3 = 0.0125 mol G 50.0 cm3 of 0.125 mol dm−3 = 0.00625 mol Therefore, only the first three experiments have sufficient hydrochloric acid to allow the reaction to go to completion Reason 2: The use of marble chips is unsatisfactory because each sample will have a different surface area. Reason 3: A fundamental difficulty stems from the use of 1/t to measure the rate of the reaction. 1/t is an approximation to the rate, which makes the assumption that over the time, t, the reaction proceeds at a constant rate. This is acceptable if the time measurements are short enough for the reaction to have occurred to only a small extent. As time passes, the reaction slows down as the acid is being used up, so the approximation becomes progressively less valid. The results obtained by allowing the reaction to run its full course will almost certainly be meaningless. This applies particularly to the third reaction in which there is only just sufficient acid and the reaction will proceed very slowly as it nears completion. However, it also applies to the first two reactions.


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OCR

A2 Chemistry Answers to chapter summary worksheets

Answers to chapter summary worksheets Chapter 1 Arenes and phenols Structure of benzene 1 a C = 92.3/12 = 7.69; H = 7.7/1.0 = 7.7 C:H ratio = 7.69 : 7.7 = 1:1 empirical formula is CH mass of empirical unit = 13.0 Therefore, there are six empirical units in the molecular unit (78.0/13.0 = 6) so, molecular formula = C6H6 b (i) Benzene does not decolourise bromine readily. When hydrogenated, the ΔH value is not three times the value of that for the hydrogenation of cyclohexene. The carbon-to-carbon bond lengths are the same and are intermediate in length between a C–C single bond and a C=C double bond. (ii) The six hexagonally arranged carbon atoms and six hydrogen atoms are in a single plane. Each carbon atom has an electron in a p-orbital, which is at right angles to the plane. The electrons in the p-orbitals overlap with adjacent p-orbitals, so that the electrons are delocalised across all six carbons. This generates a p-electron cloud above and below the hexagonal plane. The bond angles in benzene are all 120°.

Each C atom has a p-orbital at right angles to the plane of atoms

c (i)

π-delocalised ring above and below the plane

Skeletal formula of benzene

H 109° 28’ H

C

H 120°

(ii)

H O

Approx. 104°

Cl

Cl 120°

Cl

(iii)

H H

120° C


C H

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Answers to chapter summary worksheets Electrophilic substitution of arenes 2 a (i) Electrophile: electron pair acceptor (ii) Substitution: replacement of an atom or group by another atom or group b (i) Reagents: benzene and nitric acid; sulphuric acid as a catalyst (ii) Conditions: both acids are concentrated; temperature must be about 60°C (iii) Balanced equation: C6H6 + HNO3 → C6H5NO2 + H2O c Step 1: formation of the electrophile H2SO4 + HNO3 → HSO4− + H2NO3+ → H2O + (NO2)+ or 2H2SO4 + HNO3 → 2HSO4− + H3O+ + (NO2)+ Step 2: electrophilic attack at the ring +

H

NO2

NO2

NO2

+

+ H+

Step 3: regeneration of the H2SO4 catalyst H+ + HSO4− → H2SO4 d Step 1: formation of the electrophile Cl

δ+ Cl

Cl + AlCl3

δ– Cl AlCl3

_

AlCl4 + Cl+

The type of bond fission is heterolytic. Step 2: electrophilic attack at the ring + Cl

H

Cl

Cl

+

+ H+

Step 3: regeneration of the catalyst H+ +AlCl4− → AlCl3 + HCl e (i) Balanced equation benzene: Br

+ Br2

+ HBr

cyclohexene: Br + Br2 Br

(ii) Conditions benzene: halogen carrier, AlBr3; anhydrous cyclohexene: no specific conditions


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Answers to chapter summary worksheets (iii) Type of reaction benzene: electrophilic substitution cyclohexene: electrophilic addition (iv) Type of bond fission in the Br2 molecule benzene: heterolytic cyclohexene: heterolytic (v) Electron densities benzene: low electron density and therefore cannot polarise the Br–Br bond cyclohexene: high electron density in the C=C double bond and therefore polarises the Br–Br bond (vi) The C=C double bond in an alkene has high electron density and can polarise the Br–Br bond. Polarised Br2 is attracted to the double bond, leading to electrophilic addition. The electron density in benzene is much lower and is not sufficient to polarise the Br–Br bond, so Br2 is not attracted to benzene. A halogen carrier catalyst is needed to polarise Br2. In addition, benzene is stable and the overall reaction is electrophilic substitution.

Phenols 3 a (i) OH + Na+OH–

O–Na+ + H2O

OH + Na+OH–

O2N

O–Na+ + H2O

O2N

Cl

Cl

OH + 2Na+OH–

HO

Na+O–

O–Na+ + 2H2O

Cl

Cl

(ii) Phenol is an acid. b 1 O–Na+ + –H 2 2

OH + Na

O–Na+

OH

1 + –H 2 2

+ Na

O–Na+

OH

HO

CH3 + 3Na

OH


Na+O–

1 CH3 + 1–H 2 2

O–Na+

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Answers to chapter summary worksheets c (i)

Br

OH + 3Br2

Br

0H + 3HBr

Br

(ii) Unlike benzene, phenol does not require a halogen carrier and reacts instantly with bromine. This is explained by the activation of the ring by one of the lone pairs of electrons on the oxygen in the OH group, which is delocalised into the ring. This increases the electron density in the ring so that a dipole is induced in the Br–Br bond thereby generating an electrophile, Brδ+. The increased electron density of the ring then attracts the electrophile more readily leading to a rapid reaction at room temperature. eA

common mistake is for candidates to refer to the movement of the lone pair of electrons into the ring as an

‘inductive effect’. Inductive effect refers to the movement of electrons along a σ-bond and is usually associated with alkyl groups.

d Aqueous solutions of phenol are used as antiseptics because they are effective antibacterial agents; they are also effective against fungi and many viruses. They are also used as disinfectants. Examples of phenol-containing products are lysol, pine-sol, cresi-400 and environ. Chlorophenols, for example 2,4,6-trichlorophenol (TCP), are also used widely as antiseptics and disinfectants.


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Answers to chapter summary worksheets Chapter 2 Carbonyl compounds 1 C

δ+

O

C

δ–

O

The p-orbitals overlap to form a π-bond

The bond angles are approximately 120°. The carbonyl group is C=O. Like the C=C double bond, the bond is formed by the sideways overlap of adjacent p-orbitals. Like the alkenes, molecules of carbonyl compounds are unsaturated. However, unlike the alkenes, carbonyl compounds are polar due to the difference in the electronegativity of the carbon and the oxygen. 2

Carbonyl

Alcohol used

Propanal

Alcohol used to make propanal

H

H

H

C

C

H

H

O C

H H

H

H

H

C

C

C

H

H

H

OH

Structural formula = CH3CH2CHO

Structural formula = CH3CH2CH2OH

Phenylethanone

Alcohol used to make phenylethanone O

H

C

C

H

H Structural formula = C6H5COCH3

OH

H

C

C

H

H

H

Structural formula = C6H5CH(OH)CH3

3 a Oxidising agent: (potassium or sodium) dichromate(VI), Cr2O72− e When

asked to ‘state’ the oxidising agent it is safer to give the answer in words, rather than using a formula. Sodium

dicromate would be awarded the mark even though dichromate is spelt incorrectly; Cr2O7− would always be marked wrong because the formula is incorrect.

Conditions: acidified/H+ ; heated/refluxed or distilled Observations: colour change from orange to green b Propanal: CH3CH2CH2OH + [O] → CH3CH2CHO + H2O Phenylethanone: C6H5CH(OH)CH3 + [O] → C6H5COCH3 + H2O


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Answers to chapter summary worksheets 4 Water out

Water in

Propan-2-ol and acidified potassium dichromate

Heat

Refluxing involves continuous evaporation and condensation, so that the volatile components cannot escape. 5 When refluxing, the volatile components cannot escape, so the aldehyde formed would be oxidised to a carboxylic acid. Therefore, distillation is the method used. 6 a Oxidising agent: (potassium or sodium) dichromate(VI), Cr2O72− Conditions: acidified/H+ ; heated/refluxed Observations: colour change from orange to green b CH3CH2CHO + [O] → CH3CH2COOH 7 a Reducing agent: NaBH4, sodium tetrahydridoborate(III) Conditions: aqueous/warm b (i) Propanal: CH3CH2CHO + 2[H] → CH3CH2CH2OH (ii) Phenylethanone: C6H5COCH3 + 2[H] → C6H5CH(OH)CH3 8 a Nucleophile: an electron pair donor b An addition reaction is a reaction between two molecules/particles that results in the formation of a single compound. c δ– O

H3C

δ+ C

CH2

H

d δ+

O C2H5

H

δ–

δ+

H Propanal


–

H +

H

O– – + H2O H

C

δ–

O

C2H5

C H

O H

O OH

C2H5

C

H

H Propan-2-ol

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Answers to chapter summary worksheets e On

the exam paper, there is likely to be at least one question requiring a mechanism, and the use of ‘curly arrows’ is

essential. Examiners are looking for accuracy: O –

C2H5

H

C H

would be awarded 2 marks for the two curly arrows, but the following would fail to score, for the reasons shown on the diagram. This curly arrow should start at the double bond and move to the O O –

C2H5

H

C H

This curly arrow should go to the C and not to the bond

9 a Reagent: add excess 2,4 dinitrophenylhydrazine solution Observations: orange/yellow precipitate b Purify the derivative by re-crystallisation. Dry the pure crystals and determine the melting point. Check the melting point against values in a data book. 10 a Reagent: AgNO3 dissolved in ammonia/Ag(NH3)2+ Conditions: warm water bath Observations: silver mirror is produced with aldehydes but not with ketones Method: Tollens’ reagent is prepared by adding a few drops of NaOH solution to a solution of silver nitrate. This immediately produces a brown precipitate of silver oxide, Ag2O(s). Dilute ammonia is then added until the Ag2O(s) just re-dissolves. This creates an ammoniacal solution of silver nitrate, Ag(NH3)2+, which acts as the oxidising agent. In this reaction, Ag(I) is reduced to silver metal and the aldehyde is oxidised to a carboxylic acid. b (i) [Ag (NH3)2]+(aq) + e− → Ag(s) + 2NH3(aq) (ii) Propanal: CH3CH2CHO + [O] → CH3CH2COOH Phenylethanal: C6H5CH2CHO + [O] → C6H5CH2COOH


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Answers to chapter summary worksheets Chapter 3 Carboxylic acids and esters Properties of carboxylic acids 1 a Ethanoic acid b Propanoic acid c 2,2-dimethylpropanoic acid 2 a Reagents: ethanol: potassium dichromate(VI), Cr2O72− Conditions: acidified/H+; refluxed Apparatus: heat under reflux Observations: colour change from orange to green Equation: CH3CH2OH + 2[O] → CH3COOH + H2O e Many

candidates lose marks when attempting to write a balanced equation for the formation of a carboxylic acid by

the oxidation of a primary alcohol. The most common error is to write, for example: CH3CH2OH + [O] → CH3COOH + H2 This appears to be balanced and seems to work. However, hydrogen is not a product of this reaction — water is always formed.

b Reagents: ethanal; potassium dichromate(VI), Cr2O72− Conditions: acidified/H+; refluxed Apparatus: heat under reflux Observations: colour change from orange to green Equation: CH3CHO + [O] → CH3COOH c Reagents: phenylmethanol (benzyl alcohol); potassium dichromate(VI), Cr2O72− Conditions: acidified/H+; refluxed Apparatus: heat under reflux Observations: colour change from orange to green Equation: C6H5CH2OH + 2[O] → C6H5COOH + H2O d Reagents: benzaldehyde; potassium dichromate(VI), Cr2O72− Conditions: acidified/H+; refluxed Apparatus: heat under reflux Observations: colour change from orange to green Equation: C6H5CHO + [O] → C6H5COOH e O H3C

C δ+ δ– O

δ–

Hδ

+

δ+

H

O δ– Hδ

+

Hydrogen bond

The ability to form hydrogen bonds and to undergo dipole–dipole interactions explain why methanoic acid and ethanoic acid are soluble in water. Both form hydrogen bonds with water as shown in the diagram. 3 a An acid is a proton (H+) donor. b A weak acid dissociates partially into its ions. c A salt is the species formed when the H+ in an acid is replaced by either a metal ion or an ammonium (NH4+) ion.


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Answers to chapter summary worksheets d (i) CH3COOH + NaOH → CH3COO−Na+ + H2O (ii) HCOOH + NaHCO3 → HCOO−Na+ + CO2 + H2O (iii) 2CH3CH2CH2COOH + Mg → (CH3CH2CH2COO−)2Mg2+ + H2 (iv) 2C6H5COOH + CaCO3 → (C6H5COO−)2Ca2+ + H2O + CO2 e (i) HCOO−Na+ (ii) CH3COO−K+ (iii) (CH3COO−)2Ca2+ (iv) (CH3CH2CH2COO−)2Mg2+ e Candidates

struggle over the correct formulae for the salts of magnesium and calcium. It is essential to go back to

basics and deduce each formula from the charge on the ions. Calcium forms 2+ ions and ethanoate ions have a charge of 1−. Hence, we need two ethanoate ions for each calcium ion, giving the formula (CH3COO−)2Ca2+. A similar argument holds true for magnesium.

4 a Reagents: ethanol and ethanoic acid Conditions: heat in the presence of a concentrated acid/conc. H2SO4 e It

is essential to use concentrated acid as the catalyst. Mention of dilute or (aq) will lose the mark because this

would result in the equilibrium moving to the left-hand side, reducing esterification.

Equation: CH3COOH + C2H5OH b (i)

CH3COOC2H5 + H2O

Displayed formula

Skeletal formula O

O H3C

C CH3

O

(ii)

O

Displayed formula

Skeletal formula O

CH2 H3C

O

C

CH2

O

CH2

O

CH3

(iii)

Displayed formula

Skeletal formula O C

O CH3

O

CH

O

CH3

c

Compound

Carboxylic acid

Alcohol

A

Propanoic acid

Propan-2-ol

B

Phenylethanoic acid

1-phenylethanol

C

4-methylpentanoic acid

Butan-2-ol

D

6-phenylhexanoic acid

Methanol


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Answers to chapter summary worksheets d Using a symmetrical acid anhydride, such as those in reactions E and F, means that it is only possible to form a single ester and a single carboxylic acid. E

O

H3C

C

O O

O

C

CH3 + C2H5OH

H3C

C

O O

C2H5 + HO

Ethyl ethanoate O

F H3C

CH2

O

CH3

Ethanoic acid O

O

C

C

CH2

C

CH3 + C2H5OH

H3C

CH2

O O

C

C2H5 + HO

Ethyl propanoate

CH2

C

CH3

Propanoic acid

Using an unsymmetrical acid anhydride, such as that in reaction G, means that two different esters and two different carboxylic acids can be produced, depending on which C–O bond is broken in the acid anhydride functional group. O

G H3C

CH2

O

C

O

O

C

CH3 + C2H5OH

H3C

C–O bond breaking O H3C

CH2

CH2

O

C

CH3 + C2H5OH

H3C

C

O

C2H5 + HO

C

CH3

Ethyl propanoate

Ethanoic acid

O

O

O

C

O

CH2

C

OH + C2H5

Propanoic acid

O

C

CH3

Ethyl ethanoate

Properties of esters e

H

O H3C

O Hot H2SO4(aq)

CH2

C O

CH

H3C

CH2

CH3 + HO

C OH

CH3

CH CH3

CH3 HO

I O

O Hot NaOH(aq)

CH2

C O

J

C

CH2 CH

+ O–Na+

CH3

O

O Hot KOH(aq)

O

CH3 CH

O–K+

+ HO

f In perfumes; in artificial flavourings


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Answers to chapter summary worksheets 5 a A triglyceride is a triester of propane-1,2,3-triol. It has the general structure: O H2C

O R

C

O

O

C

R’

O

C

R’’

CH H2C

O R is a long-chain hydrocarbon

b Fatty acids are naturally occurring carboxylic acids. The smallest is butanoic acid. Many fatty acids have a long hydrocarbon chain.

Butanoic acid CH3(CH2)2COOH e The

Octadecanoic acid (stearic acid)

CH3(CH2)16COOH

diagrams are to illustrate the answer. They would not be required in an exam.

c (i) A saturated fatty acid: COOH

(ii) A monounsaturated fatty acid: COOH

(iii) A polyunsaturated fatty acid: COOH

d A molecule of a monounsaturated fatty acid, for example oleic acid, CH3(CH2)7CH=CH(CH2)7COOH, has one C=C double bond. In the Z (cis) isomer, the two hydrogen atoms attached to the two carbon atoms in the double bond are on the same side of the double bond.

Z-oleic acid (cis)

In the E (trans) isomer the two hydrogen atoms attached to the two carbon atoms in the double bond are on opposite sides.

E-oleic acid (trans) — this has lost the ‘kink’ and forms a straight chain e The

diagrams are to illustrate the answer. They would not be required in an exam.


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Answers to chapter summary worksheets e E (trans) fatty acids can stack together. This can lead to a build up of plaque that thickens the walls of arteries. Saturated fats, and possibly E (trans) fatty acids, can also lead to increased levels of cholesterol and low-density lipoproteins, LDL, in the blood. There are two categories of lipoprotein: G High-density lipoprotein, HDL, which carries about one third of the cholesterol and is known as ‘good cholesterol’ as it carries cholesterol away from the arteries and back to the liver G Low-density lipoprotein, LDL, which is known as ‘bad cholesterol’ because it can lead to a build up of plaque and to atherosclerosis. Recent studies show a positive correlation between the amount of E (trans) fatty acids and LDL.


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Answers to chapter summary worksheets Chapter 4 Amines Nomenclature 1 a b c d e

Ethylamine; aminoethane Propylamine; 1-aminopropane 2-methyl-2-propylamine; 2-amino-2-methylpropane Phenylamine; aminobenzene Phenyl-1,4-diamine; 1,4-diaminobenzene

2 a An acid is a proton (H+) donor. b A base is a proton (H+) acceptor. c A nucleophile is an electron pair donor. d H + Dative bond H

H3C

N

H

H3C

H

+

N

H

H

3 a CH3CH2NH2 + HCl → CH3CH2NH3+Cl− b CH3CH2CH2NH2 + HCl → CH3CH2CH2NH3+Cl− c (CH3)3CNH2 + HCl → (CH3)3CNH3+Cl− H

d

N

+

+ HCl

NH3Cl–

H H

H

e

N

N + 2HCl

H

Cl–H3N

+

+

NH3Cl–

H

4 a

Bond angle = 109°28’ because the C atom has four bonded pairs around it

H H

C

H

H

H

b H H

Bond angle = 107° because the N atom has three bonded pairs and one lone pair around it

N

Bond angle = 109°28’ because the C atom has four bonded pairs around it H Cl– +

C

N

H

H

H Bond angle = 109°28’ because the N atom has four bonded pairs around it

Preparation of amines 5 a Reagents: CH3CH2Cl and NH3 Conditions: ethanol as the solvent; heated under pressure (often described as heated in a sealed tube) Type of reaction: nucleophilic substitution Equation: CH3CH2Cl + NH3 → CH3CH2NH2 + HCl © Philip Allan Updates

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Answers to chapter summary worksheets b Stage 1: Reagents: HNO3, H2SO4 and C6H5CH3 Conditions: concentrated acids. Benzene requires a temperature of around 60°C. Compounds like methylbenzene are more reactive than benzene and require a lower temperature. Type of reaction: electrophilic substitution (nitration) Equation: C6H5CH3 + HNO3 → (CH3)C6H4NO2 + H2O Mechanism: 2H2SO4 + HNO3 → 2HSO4− + H3O+ + NO2+ CH3

CH3

CH3

+ H+

+ +

O2N

NO2

H

NO2

H+ + HSO4− → H2SO4 e Candidates

are good at showing the intermediate when substitution occurs in position 1 (at the top of the ring) but

often make mistakes when substitution occurs at any other position. In the formation of 4-nitromethylbenzene from methylbenzene it is common to see: CH3

+

O2N

H

The correct formula is: CH3

+ O2N

H

Stage 2: Reagents: Sn, HCl Conditions: reflux; concentrated acid Type of reaction: reduction Equation: (CH3)C6H4NO2 + 6[H] → (CH3)C6H4NH2 + 2H2O

Synthesis of azo dyes 6 a Reagents Phenylamine; nitrous acid, HNO2, (which is made from NaNO2 and HCl); HCl Conditions: temperature below 10°C; excess HCl Equation: Cl– +N

NH2

+ HNO2 + HCl

Temp. <10°C

N

+ 2H2O

Benzenediazonium chloride


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Answers to chapter summary worksheets e It

is essential to show that the ‘+’ charge is on the nitrogen that is attached directly to the ring. Candidates often

float the ‘+’ charge between the nitrogen atoms in the hope that they will be awarded the mark. This is not the case. The structure shown below would not be given credit. +

N

N

b Reagent: a phenol Conditions: alkaline Equation: Cl– +N

N

OH

N +

OH + HCl

Alkaline/OH–

N

Benzenediazonium chloride

Observations: formation of a coloured dye c Diazonium ion

Phenol

Azo dye

O

O C

O

CH3

C

O

CH3

+

H3C

N

H3C OH

N

Cl

N OH

N

Cl

H3C

H3C +

N

N

OH N

Cl

OH

N Cl

H3C

H3C

+

OH

N N


N N

OCR A2 Chemistry

OH

88

Answers to chapter summary worksheets Chapter 5 Amino acids and chirality 1 NH2CHRCOOH 2 a Alanine: H H2N

C

COOH

CH3

b Valine: H H2N

C

COOH

CH H3C

CH3

c Aspartic acid: H H2N

C

COOH

CH2 COOH

d Asparagine: H H2N

C

COOH

CH2 C H2N

O

3 In an α-amino acid, the NH2 group and the COOH group are attached to the same carbon atom. 4 a Stereoisomers are compounds that have the same molecular formula and the same structure, but a different 3-dimensional/spatial arrangement. b A chiral centre is a centre that has no symmetry. It is usually a carbon atom bonded to four different atoms or groups. c

CH2

H2C

C

C

H2N

COOH H

e Drawing

NH2

HOOC H

3-dimensional structures can cause problems. A solid line (—) is used to represent a bond in the plane of

the paper, a dotted line (——-) to represent a bond behind the plane of the paper and a wedge shape

to

represent a bond in front of the plane of the paper. It is essential that any two (—) bonds are adjacent to each other and are not separated by a dotted or a wedge-shaped bond. For example, in the formula below the two (—) bonds are adjacent and not separated:


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Answers to chapter summary worksheets R C H2N

COOH H

This formula would score the mark. However, in the formula below the two(—) bonds are not adjacent and are separated: R C NH2

HOOC H

This forumla would not score the mark.

d They can be distinguished by using plane-polarised light. One of the optical isomers will rotate the plane of plane-polarised light to the left (L-isomer) and the other will rotate it to the right (Disomer). 5 a +NH3CHRCOO− b Transfer of a H+ from COOH to NH2 H

H +

H2N

C

COOH

R

H3N

C

COO–

R

c The isoelectric point is the pH at which the zwitterion exists. d (i) H +

H3N

C

COOH

H pH = 2.0

(ii)

H +

H3N

COO–

C H pH = 5.8

(iii)

H H2N

C

COO–

R pH = 10.0

e (i) Glutamic acid at pH 12: H H2N

C

COO–

CH2 COO–


OCR A2 Chemistry

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Answers to chapter summary worksheets (ii) Lysine at pH 2.0: H +

H3N

C

COOH

(CH2)4 NH3 +

6 a

H H2N

H

C

b H2N

COOH + NaOH

H

H

H COOH + 2NaOH

H2N

CH2

CH2

COOH

COO–Na+ H

C

+

Cl– H3N

COOH + HCl

C

CH3

d

COO–Na+ + 2H2O

C

H H2N

COO–Na+ + H2O

C

H

C

c

H2N

CH3

H H2N

H

C

+

Cl– H3N

COOH + 2HCl

CH2

C

COOH

CH2

C

C NH2

O

7 a

COOH

+

NH3 Cl–

O

H C

N

O

b When two different amino acids react it is possible to form four different dipeptides. It is the sequence in which they link that gives rise to the different dipeptides. If glycine (gly) reacts with alanine (ala) it is possible to form a dipeptide designated as gly–ala:

H

H

H

O

N

C

C

O

H

H

O

N

C

C

Loss of water

H Gly

H

H

H

O

H

CH3 Ala

H

H

O

H

H

O

N

C

C

N

C

C

O

H

+ H2O

CH3

H Gly-Ala


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Answers to chapter summary worksheets It is also possible to form ala–gly:

H

H

H

O

N

C

C

O

H

H

O

N

C

C

Loss of water

CH3 Ala

H

H

H

O

H

H Gly

H

H

O

H

H

O

N

C

C

N

C

C

O

H

+ H2O

H

CH3 Ala-Gly

Using glycine and alanine the four possible dipeptides are: gly–gly, gly–ala, ala–gly and ala–ala e The

diagrams are to illustrate the answer, rather than being a requirement.

c (i) H

H

H

O

H

H

O

N

C

C

N

C

C

CH2

O

H

O

H

CH CH3

H3C Phe-Val

(ii) H

H

H

O

H

H

O

N

C

C

N

C

C

CH2

CH H3C

CH3 Val-Phe

d gly–ala–leu–gly–phe–lys e Reflux with 6.0 mol dm−3 HCl for approximately 8 hours.


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Answers to chapter summary worksheets Chapter 6 Polyesters and amides Addition polymers 1 a (i)

W n

X C

C

Y

Z

W

X

C

C

Y

Z n

(ii) CH3 H

H

C

C

C

C

H

H

H

H

Propene

(iii)

Phenylethene

CH3 H

CH3 H

C

C

C

C

H

H

H

H

(iv) H

b

H

C

C

C

C

H

H

H

H

Isomer of C4H8

Name

C2H5 H

But-1-ene

Two repeat units of the polymer C2H5 H

C2H5 H

C

C

C

C

C

C

H

H

H

H

H

H

CH3 CH3

But-2-ene

CH3 CH3 CH3 CH3

C

C

C

C

C

C

H

H

H

H

H

H

CH3 H C

C

CH3 H

Methylpropene

CH3 H

CH3 H

C

C

C

CH3 H

C

CH3 H

But-2-ene exists in both the E (trans) and the Z (cis) forms. This influences the polymer formed. c (i) Addition polymers are unreactive because they have strong, non-polar, single covalent bonds throughout. (ii) It is difficult to dispose of addition polymers because they are unreactive and non-biodegradable; they produce toxic fumes when burnt. d They are poor conductors since there are no mobile charge carriers. e Any three from: HCl, dioxins, acrolein, aldehydes, ketones.


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Answers to chapter summary worksheets Condensation polymers Polyesters 2 a (i)

HO

O C

C

CH2

HO

O

OH Benzene-1,4-dioic acid

(ii)

OH

CH2

Ethane-1,2-diol

O

O C

C

O

O

CH2

CH2

Polyamides b (i)

H H

O

H

N

(CH2)6

N

HO

H

O

C

C

(CH2)4

OH

Hexanedioic acid

1,6-diaminohexane

(ii)

Simplest repeat unit H (CH2)6

N

H

O

N

C

(CH2)4

O

H

C

N

H

O

N

C

(CH2)6

O (CH2)4

C

n

c (i) H

H

H

N

N

H

HO

1,4-diaminobenzene

O

O

C

C

OH

Benzene-1,4-dioic acid

(ii)

Simplest repeat unit H

H

O

O

H

H

O

O

N

N

C

C

N

N

C

C

n

d The ester link and the peptide (amide) link are polar and can, therefore, be hydrolysed. e


H

O

N

C

C

H

H

N

H Amide

H

O

C

C

CH3

Ester

O

H

H

C

C

H

H

O

O

CH2

C

C

N

H

Ester

O

H

C

C

N

H Amide

H

H

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94

Answers to chapter summary worksheets Monomers:

H

N H

H

O

C

C

OH

N

H

H

H

H

O

C

C

OH

CH3

Glycine

Alanine

HO

H

H

C

C

H

H

Ethane-1,2-diol

OH

HO

O

CH2

C

C

N

H

H

H

Phenylalanine

Poly(lactic acid), PLA, and poly(glycolic acid), PGA 3

a PLA: O

H

O

C

C

O

CH3

PGA: O

O

C

C

CH3

H

O

C

C

H

H

O

H

O

C

C

H

b (i) The monomers can be obtained from plants, which can be replaced by growing more plants. e Some

candidates find it difficult to explain what is meant by ‘renewable’, often confusing it with ‘reusable’. A

common question is to ask candidates to explain why ethanol could be thought of as a renewable fuel. Candidates often state that this is because ‘when it has been used, it can be used again’. Obviously, once it has been burnt, it can’t be burnt again! The correct response is that it can be replaced easily. The feedstock for ethanol is plant-based and plants can be grown relatively easily.

(ii) They can be broken down by bacteria.


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Answers to chapter summary worksheets Chapter 7 Synthesis 1 a (i) A : arene , alkene , aldehyde (carbonyl) B: alkene, ketone (carbonyl) C: primary alcohol, alkene b (i) Bromine (or other reagents capable of addition to the double bond e.g. H2) (ii) Br Br H2C

CH

H2C

Br CH

CH

CH2

CH CH2

CH3

C Br O

(or equivalent addition product) c Acidified potassium or sodium dichromate(VI) d Tollens’ reagent (Ag+ dissolved in NH3) e Na, PCl5, RCOOH 2 a (i) 1-phenylethanol (ii) Reagents: NaBH4 Conditions: warm, aqueous solution Equation: O

OH

H C

CH3

C

CH3

+ 2[H]

(iii) Reagent: H2SO4 Conditions: concentrated acid; temperature about 170°C Equation: OH

H C

H

CH3

C

CH2

Conc. H2SO4

+

H2O

b (i) Nitrobenzene (ii) Reagents: HNO3 and H2SO4 Conditions: both acids are concentrated; temperature about 60°C Equation: NO2

+ HNO3


+

H2O

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Answers to chapter summary worksheets (iii) Reagents: Sn; HCl Conditions: concentrated acid; reflux Equation: NO2

NH2

+ 6[H]

+

2H2O

c (i) Propan-1-ol (ii) Reagents: H2O Conditions: heat with either dilute HCl or dilute NaOH Equation: CH3COOCH2CH2CH3 + H2O → CH3COOH + CH3CH2CH2OH (iii) Reagents: acidified potassium dichromate(VI) Conditions: distillation to prevent oxidation to propanoic acid Equation: CH3CH2CH2OH + [O] → CH3CH2CHO + H2O

Chirality in pharmaceutical synthesis 3 a Chiral centre — a centre that has no symmetry, usually a carbon atom bonded to four different atoms or groups b Racemic mixture — a 50 : 50 mixture of two optical isomers c Non-superimposable — two optical isomers do not ‘fit’ on top of each other d Stereoselectivity — produces only one optical isomer; shapes have to match e Chiral catalyst — a catalyst that has the correct shape to promote the formation of one optical isomer over another 4 The preparation of a single chiral compound in the laboratory is difficult. The synthetic mixture would probably contain 50% of the isomer that rotates the plane of plane-polarised light to the right, and 50% of the isomer that rotates the plane of plane-polarised light to the left. By contrast, synthesis using naturally occurring enzymes or bacteria results in the formation of a single optical isomer. 5 Laboratory synthesis produces a mixture of isomers, only one of which has the correct shape to be pharmacologically active. The other may have adverse side-effects. The two isomers are difficult and expensive to separate. 6 Enzymes and bacteria produce only the pharmacologically active isomer. Nowadays, single optical isomers can also be produced by using starting materials such as L-amino acids or L-sugars or by using a chiral catalyst.


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Answers to chapter summary worksheets Chapter 8 Analysis Chromatography 1 a (i) Mobile phase: the solvent that moves over the plate Stationary phase: silica gel (SiO2) or alumina (Al2O3 (ii) Mobile phase: carrier gas that is unreactive or inert, such as nitrogen or one of the noble gases Stationary phase: either a liquid or a solid adsorbed onto the surface of an inert solid in the column distance moved by spot/solute b (i) Rf = distance moved by solvent (ii) Retention time is the time taken from injection of the sample for a component to leave the column. c G Similar compounds have similar Rf values and retention times. G Retention times vary with temperature, flow rate and from one GC machine to another. d (i) The limitations of GC are overcome; GC and MS together provide a powerful analytical tool. GC separates the components and MS allows identification of each component by enabling checking of the fragmentation peaks against a database. Different compounds have different fragmentation patterns and provide a ‘fingerprint’ of the molecule. This fingerprint can then be cross-matched against a large computer database to identify a specific compound. (ii) Uses: environmental, medical, forensic, astrochemistry, airport security, pharmaceutical and others

Nuclear magnetic resonance spectroscopy, NMR 2 a Radio wave b Tetramethylsilane, TMS, is used a reference; all other peaks are measured relative to TMS. The chemical shift of TMS is assigned δ = 0 ppm. TMS is used because: G it is chemically inert and does not react with the sample G it is volatile and easy to remove afterwards G it absorbs at a higher frequency than other organic compounds, so its peak does not overlap with those from the sample c Both 13C and 1H have an odd number of nucleons (number of nucleons = number of protons + number of neutrons) d Deuterium is an isotope of hydrogen (D = 2H). It is not detected in the same region as 1H so absorptions from the solvent do not overlap with absorptions from the sample. e (i) 2 peaks (ii) 4 peaks (iii) 2 peaks (iv) 4 peaks (v) 6 peaks (vi) 3 peaks e 13C-NMR

spectroscopy is new to the specification. The key to answering this question is the identication of the

number of different carbon environments. Similarly, with 1H-NMR spectroscopy the identification of different hydrogen environments is essential.


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Answers to chapter summary worksheets f C1: 35–60 ppm C2: 160–185 ppm C3: 5–55 ppm C4: 50–70 ppm g (i) Chemical shift is the frequency at which 1H or 13C absorb energy in the radio wave part of the spectrum. However, the frequency of the absorption varies depending on the surrounding atoms. Therefore, the exact absorption frequency depends on the chemical environment. It is always measured relative to a standard, typically TMS. (ii) The hydrogen atoms attached to one carbon atom influence the hydrogens on adjacent carbon atoms. This is called spin–spin coupling. The easiest way to predict the splitting pattern is to count the number of hydrogens on the adjacent carbon atoms and then use the ‘n + 1’ rule, where n = the number of hydrogens on the adjacent carbon atoms. If the adjacent carbon is a CH2, the split will be (2 + 1) = triplet. (iii) The relative peak area is proportional to the number of hydrogens in that environment. (iv) When some polar organic compounds are dissolved in water there is a rapid exchange between the protons in the functional groups (known as labile protons) and the protons in the water. h Alcohol; carboxylic acid; amine i Chemical environment CH3 in CH3CH2CH3 CH3 in C6H5CH3 CH3 in CH3CHO CH3 in CH3COOH δ/ppm (range)

0.7–1.6

2.3–2.7

2.0–2.9

2.0–2.9

Chemical environment

H in CH3CH2OH

H in CH3COCH2CH3

H in CH3COOH

H in C6H5OH

δ/ppm (range)

1.0–5.5 (but can be

2.0–2.9

11.0–12.0 (but can be

4.5–10.0 (but can be

outside this range)

outside this range)

outside this range) Chemical environment

CH2 in CH3CH2CH3

CH2 in CH3CH2OH

CH2 in CH3COOCH2CH3

CH2 in CH3CH2CHO

δ/ppm (range)

1.2–1.4

3.3–4.3

3.3–4.3

2.0–2.9

Chemical environment

H in CH3CHO

CH3 in CH3OH

H in C6H6

H in CH3CONH2

δ/ppm (range)

9.0–10.0

3.3–4.3

6.5–8.0

5.0–12.0 (but can be outside this range)

e Identifying

the correct or most appropriate chemical-shift range is the most difficult part of 1H-NMR spectroscopy.

You need lots of practice prior to the exam. There will almost certainly be a question on 1H-NMR spectroscopy on the paper.

j (i) Ha

Ha

Hb

OHc

He

C

C

C

C

Ha

Hb

Hd

He

Relative peak area

e *The

He

Splitting

Chemical shift, δ/ppm (range)

Ha

3

3 peaks

0.7–1.6

Hb

2

5 peaks*

1.2–1.5

Hc

1

1 peak

1.0–5.3

Hd

1

6 peaks*

1.2–1.5

He

3

2 peaks

0.7–1.6

splitting is not as predicted by the simple (n + 1) rule and it is unlikely that you will be asked this. Several

peaks appear in the same region and can be difficult to assign.


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Answers to chapter summary worksheets (ii)

O Ha

C

O

Hb

Hc

Hd

C

C

C

Hb

Hc

Hd

Relative peak area

e *The


Splitting

Ha

1

1 peak

2.0–2.9

Hb

2

3 peaks

3.3–4.3

Hc

2

6 peaks*

1.2–1.4

Hd

3

3 peaks

0.7–1.6

splitting is not as predicted by the simple (n + 1) rule and it is unlikely that you will be asked this.

(iii)

Ha

Ha

Ha

C

Ha

Hb

C

Hb

Ha

Hb

C

C

Ha

Hb

Hb

Ha

C

C

C

Hc

Hb

Ha

Relative peak area

e *The

Hd

Ha

Splitting


Ha

9

3 peaks

0.7–1.6

Hb

6

5 peaks*

1.2–1.4

Hc

1

7 peaks*

1.2–1.4

splitting is not as predicted by the simple (n + 1) rule and it is unlikely that you will be asked this.

(iv) Ha

Ha

Hb

Hc

Hb

Ha

C

C

C

C

C

Ha

Hc Relative peak area

e *The

Splitting


Ha

2 (4)

2 peaks

4.5–6.0

Hb

1 (4)

5 peaks*

1.2–1.4

Hc

1 (2)

2 peaks

4.5–6.0

splitting is more complex and it is unlikely that you will be asked this. Note that the ratio should always be

written in the simplest form, in this case, 2:1:1.


OCR A2 Chemistry 100

Answers to chapter summary worksheets Chapter 9 How fast? 1 a A rate equation is an equation that shows how the overall rate of a reaction depends on the concentrations of the reactants, for example: rate = k[A]a[B]b e It

is hard to give a clear explanation of rate equation using words alone. In an exam it is best to give an example of

the equation.

b The order of a reaction is the power to which an individual reactant concentration is raised in the rate equation — for example, in the rate equation: rate = k[A]a[B]b the orders are the superscripted a and b. e In

an exam, the best way to answer this is to give an example, as shown above.

c The rate constant is the constant of proportionality in the rate equation and is represented by k. The rate constant, k, is temperature dependent. 2 a s−1 b mol−1 dm3 s−1 c mol−2 dm6 s−1 e The

order in which you write these units is unimportant. For example, it is perfectly acceptable to write

dm3 mol−1 s−1,

rather than mol−1 dm3 s−1.

4 a

Rate of reaction

3 The rate-determining step is the slowest step in a reaction mechanism.

b

Rate of reaction

Concentration of reagent

Concentration of reagent



Answers to chapter summary worksheets Concentration

5 a

Time Concentration

b

1.0 0.8 0.6 0.4 0.2 0

e If

0

30

60

90

120

150 Time/s

you are asked to draw this curve it is important to give a reasonable indication that it possesses a constant half-

life. Including a few numbers on the axes would help to indicate this.

6 a When the initial concentration of ICl is held constant at 0.2 mol dm−3 and the concentration of H2 is increased 1.5-fold from 0.2 mol dm−3 to 0.3 mol dm−3, the rate increases 1.5-fold from 6.4 × 10−3 mol dm−3 s−1 to 9.6 × 10−3 mol dm−3 s−1. Therefore, the order with respect to H2 is 1. When the initial concentration of H2 is held constant at 0.3 mol dm−3 and the concentration of ICl is increased 2.5-fold from 0.2 mol dm−3 to 0.5 mol dm−3, the rate increases 2.5-fold from 9.6 × 10−3 mol dm−3 s−1 to 2.4 × 10−2 mol dm−3 s−1. (Remember, 2.4 × 10−2 = 24 × 10−3). Therefore, the order with respect to ICl is 1. b rate = k[H2][ICl] Substituting the values from the first set of results gives: 6.4 × 10−3 = k[0.2][0.2] k = (6.4 × 10−3)/0.04 = 0.16 mol−1 dm3 s −1 7 a Either of two methods can be used. Method 1: In the first 20 min, the concentration of sucrose falls from 2.00 mol dm−3 to 1.70 mol dm−3. This is a percentage drop in concentration to (1.70/2.00) × 100 = 85% of the original value. In the next 20 min, the concentration falls from 1.70 mol dm–3 to 1.45 mol dm−3. The percentage drop is to (1.45/1.70) × 100 = 85.3% of 1.70 mol dm–3 (the value at 20 min). Since the percentage drop over the same time period is the same the reaction must be first order. (This could be checked using other results).



Answers to chapter summary worksheets

Sucrose concentration/mol dm–3

Method 2: A graph is drawn of concentration against time and checked to see that the half-life is constant. 2.5

2.0

1.5

1.0

0.5

0

0

20

40

60

80

100 120 Time/min

The half-life determined from 2.00 mol dm−3 to 1.00 mol dm−3 is 85 min. The half-life determined from 1.50 mol dm−3 to 0.75 mol dm−3 is also 85 min. Therefore the reaction is first order. e You

can measure the half-life from any concentration to half of that value. In this case, the half-life should always

be 85 min.

Sucrose concentration/mol dm–3

b The second method provides a means of calculating the rate constant. The initial rate can be obtained by drawing a tangent to the curve at t = 0. 2.5

2.0

1.5

1.0

0.5

0

0

20

40

60

80

100 120 Time/min

From the gradient of the tangent, the initial rate = 0.0167 mol dm−3 min−1 This is the rate for a concentration of sucrose of 2.00 mol dm−3. Therefore, 0.0163 = k × 2.00 k = 0.00815 = 8.2 × 10−3 min−1 e Tangents

are difficult to draw. In an exam, the tolerance allowed would be substantial; the accepted value of k could

be from 8.0 × 10−3 to 9.0 × 10−3.



Answers to chapter summary worksheets Chapter 10 How far? 1 A chemical reaction is in dynamic equilibrium when, in a closed system, the rate of the forward reaction is equal to the rate of the reverse reaction. e There

are a number of acceptable ways of wording this — the key elements are that it is a closed system and that the

rates of forward and reverse reaction have become equal. The amount of each component of the reaction remains constant but reaction occurs continuously in both directions.

2 a Kc =

[HCl(g)]2[Br2(g)] [HBr(g)]2[Cl2(g)]

Kc does not have units. [CO(g)]2 b Kc = [CO2(g)] The units of Kc are mol dm−3. e Carbon

is excluded from the equilibrium expression because it is a solid.

3 a (i) Increasing the pressure favours a shift in the equilibrium to the side of lower volume, i.e. towards the carbon dioxide. Therefore, less carbon monoxide is produced. (ii) Increasing the temperature favours the endothermic reaction. The forward reaction is endothermic and so more carbon monoxide is produced. b (i) Increasing the pressure has no effect on the equilibrium constant. (ii) Increasing the temperature means that more carbon monoxide is formed. Therefore, the equilibrium constant is increased. 4 If the equilibrium constant decreases with a rise in temperature, this means that the equilibrium position shifts to produce more of the reactants. Applying Le Chatelier’s principle, the forward reaction must be exothermic. 5 a The number of moles is the same on both sides of the equation. Therefore, Kc has no units. [ICl(g)]2 b Kc = = 81 [I2(g)][Cl2(g)] At equilibrium, [I2(g)] = [Cl2(g)] = 0.10 mol e The

volume is irrelevant as the number of moles is the same.

[ICl(g)]2 = 81 0.10 × 0.10

Kc =

[ICl(g)] =

81 × 0.10 × 0.10 = 0.90 mol

6 a CO(g) + 3H2

CH4(g) + H2O(g) CO(g)

3H2(g)

CH4(g)

H2O(g)

Initial amounts (in moles)

1.0

1.0

0

0

Equilibrium amounts (in moles)

1.0 − x

1.0 − 3x

x

x

From the data in the question, x = 0.24 mol G CH4 = 0.24 mol e Every

time 1 molecule of steam is formed, 1 molecule of methane is also formed. So, the amount (in moles) of

methane is the same as the amount in moles of steam. G


CO = 1.0 − 0.24 = 0.76 mol


Answers to chapter summary worksheets e Every

time 1 molecule of steam is formed, 1 molecule of carbon monoxide is used up. So the amount (in moles) of

carbon monoxide in the equilibrium mixture is 1.0 − 0.24 mol.

H2 = 1.0 − (3 × 0.24) = 0.28 mol

G e Every

time 1 molecule of steam is formed, 3 molecules of hydrogen are used up. So the amount (in moles) of

hydrogen in the equilibrium mixture is 1 − (3 × 0.24) mol

Kc =

(b)

=

[CH4(g)][H2O(g)] [CO(g)][H2(g)]3 0.24 × 0.24 0.76 × 0.283

= 3.45 = 3.5 mol−2 dm6 e Care

must be taken to ensure that the units are included in the answer.



Answers to chapter summary worksheets Chapter 11 Acids, bases and buffers 1 The Brønsted–Lowry theory defines an acid as a species that donates a proton to another species in a reaction; a base is a species that accepts a proton from an acid in a reaction. The acids and bases are indicated below the equation: CH3NH2 + H2O CH3NH3+ + OH− Base 2 Acid 1 Acid 2 Base 1 2 HNO3 is a strong acid. In solution, it ionises completely: HNO3(aq) → H+(aq) + NO3−(aq) HCOOH is a weak acid. In solution, it ionises partially and exists in equilibrium with its ions: HCOOH(aq) H+(aq) + HCOO−(aq) 3 a Ka =

[HCOO−(aq)][H+(aq)] [HCOOH(aq)]

b pKa = −log Ka c pH = −log [H+(aq)] e Remember

that it is not necessary to try to put these definitions into words — just state the expressions.

4 a Since HCl is a strong acid: pH = −log [HCl(aq)] = −log 0.035 = 1.46 [CH3COO−(aq)][H+(aq)] b Ka = 1.7 × 10−5 = [CH3COOH(aq)] [H+(aq)] = [CH3COO−(aq)] [H+(aq)]2 [H+(aq)]2 = so, 1.7 × 10−5 = [CH3COOH(aq)] 0.035 Therefore, [H+(aq)]2 = 1.7 × 10−5 × 0.035 [H+(aq)] = 7.7 × 10–4 mol dm−3 pH = 3.11 5 [H+] × 0.08 = 10−14 Therefore, [H+] = 10−14/0.08 = 1.25 × 10−13 mol dm−3 pH = −log(1.25 × 10−13) = 12.9 6 a A buffer solution is a solution that resists a change in pH when small quantities of either acid or alkali are added. b If hydrogen ions are added, the reserve of butanoate ions from the sodium butanoate in the mixture combine with the hydrogen ions to make more undissociated butanoic acid, so there is only a small change in the overall pH. If hydroxide ions are added, they react with the existing hydrogen ions present to form water. The reserve of butanoic acid then ionises to replace the lost hydrogen ions, so the pH changes very little. 7 Substituting into the expression for Ka 0.01 × [H+] 6.4 × 10−5 = 0.02 Therefore, [H+] = 2 × 6.4 × 10−5 = 1.28 × 10−4 mol dm−3 pH = −log (1.28 × 10−4) = 3.89



Answers to chapter summary worksheets 8 In water, HCO3−(aq), forms the following equilibrium system: H+(aq) + HCO3–(aq) H2CO3(aq) CO2(aq) + H2O(l) + Addition of H (aq) pushes the first equilibrium position to the right and the weak acid H2CO3(aq) is formed. This then decomposes into CO2(aq), which causes the removal of H+(aq). Addition of OH−(aq) ions causes the removal of H+(aq) and the reverse procedure takes place, forming HCO3− (aq) and replacing the H+(aq) ions. 9

pH

14 12 10 8 6 Equivalence point 4 2 0

e The

0

5

10

15

20

25 30 35 40 Volume of alkali added/cm3

end point of the reaction depends on the concentrations of the acid and the alkali. If asked to sketch this curve,

you should give some indication of the likely pH values.

10 a HIn(aq) H+(aq) + In−(aq) Yellow Blue In acid solution, the equilibrium position moves to the left and HIn dominates. Therefore the indicator appears yellow. In alkaline solution, H+ ions are removed. The equilibrium position moves to the right and In− dominates. Therefore, the indicator appears blue. At the midpoint, the indicator is likely to appear green because there is a mixture of HIn and In−. b The indicator could be used for the titration of a strong acid with either a strong base or a weak base. 11 Enthalpy of neutralisation is defined as the enthalpy change that occurs when 1 mole of water is produced in the reaction between an acid and an alkali. The enthalpy is ‘standard’ if the enthalpy change is measured at 1 atm pressure and 298 K. 12 Neutralisation occurs when the reaction: H+(aq) + OH−(aq) → H2O(l) takes place. A strong acid is ionised fully, but a weak acid ionises during the course of the neutralisation to replace the H+(aq). This process requires some energy, and therefore the enthalpy of neutralisation of a weak acid is less than that for a strong acid.



Answers to chapter summary worksheets Chapter 12 Lattice enthalpy 1 a 2K+(g) + O2−(g) → K2O(s) b –12 Br2(l) → Br(g) c O−(g) + e– → O2−(g) excess H O

2 d Mg2+(g) — → Mg2+(aq) 2+ e MgCl2(s) → Mg (aq) + 2Cl−(aq)

2 a

Ca2+(g) + 2Cl(g) 2nd ionisation enthalpy of Ca Ca+(g) + 2Cl(g)

2 × 1st electron affinity of Cl

1st ionisation enthalpy of Ca Ca(g) + 2Cl(g) Ca2+(g) + 2Cl–(g) 2 × enthalpy of atomisation of Cl Ca(g) + Cl2(g) ΔH

Enthalpy of atomisation of Ca

(Lattice enthalpy of CaCl2)

Ca(s) + Cl2(g) Enthalpy of formation of CaCl2(s)

CaCl2(s)

b Using the figures supplied and the diagram above, the lattice enthalpy of calcium chloride, ΔH, is obtained from: 108 + (2 × 122) + 590 + 1100 + (2 × – 364) + ΔH = −795 ΔH = −795 − 108 − 244 − 590 − 1100 + 728 = −2109 kJ mol−1 3 a The magnesium ion is smaller than the sodium ion and therefore has a greater charge density. This means that the magnesium chloride lattice will be stronger than the sodium chloride lattice. Therefore, magnesium chloride has the larger numerical value for lattice enthalpy. b The calcium ion is smaller than the barium ion and therefore has a greater charge density. This means that the calcium chloride lattice will be stronger than the barium chloride lattice. Therefore, calcium chloride has the larger numerical value for lattice enthalpy. c The chloride ion is smaller than the bromide ion and therefore has a greater charge density. This means that calcium chloride lattice will be stronger than the calcium bromide lattice. Therefore, calcium chloride has the larger numerical value for lattice enthalpy. 4 The enthalpy cycle is: ΔH1

BaCl2(s) ΔH2

Ba2+(aq) + 2Cl–(aq) ΔH3

Ba2+(g) + 2Cl–(g)

The enthalpy of solution, ΔH1 = ΔH3 − ΔH2 = [−1272.8 + (−384.1 × 2)]− (−1958) = −83.0 kJ mol−1



Answers to chapter summary worksheets Chapter 13 Enthalpy and entropy 1 a When a solid is melted, its entropy increases. b The precipitation of a solid results in a decrease in entropy. c The entropy decreases as ammonia is formed in the Haber process. e There

is a decrease in the total number of moles as the reaction takes place.

d As a gas condenses its entropy decreases. 2 The entropy change is {2 × 70.0} − {(2 × 130.6) + 204.9} = −326.1 J mol−1 K−1 3 ΔGf° = ΔHf° − TS° = −769.9 − (298 × 0.1134) = 803.7 kJ mol−1 4 total free energy of products = {−27.8 + (−237.2)} = −265 kJ mol−1 total free energy of reactants = {(−174.9 + (−53.2)} = −228.1 kJ mol−1 ΔG° = −265 − (−228.1) = −36.9 kJ mol−1 5 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) total free energy of products = (3 × −394.6) + (4 × −237.2) = −2132.6 kJ mol−1 total free energy of reactants = −23.5 + (5 × −61.1) = −329.0 kJ mol−1 ΔG° = −2132.6 − (−329.0) = −1803.6 kJ mol−1 6 A reaction is spontaneous if ΔG is negative. For an endothermic reaction, ΔH is positive. However, since ΔG = ΔH − TΔS, ΔG can achieve a negative value if TΔS is sufficiently positive for it to become numerically greater than ΔH. This may occur if ΔS is very large or if the temperature is high. 7 ΔH = TΔS 8 Equilibrium can only be achieved if the relationship ΔH = TΔS applies. This is not possible if ΔH and ΔS have different signs. 9 For the reaction: 2NO2(g) → N2O4(g) ΔH = 9.2 − (2 × 33.2) = −57.2 kJ mol−1 ΔS = 304.2 − (2 × 240) = −175.8 J mol−1 K−1 At equilibrium, ΔH = TΔS so, T = 57.2/0.1758 = 325 K (52°C)



Answers to chapter summary worksheets Chapter 14 Electrode potentials and fuel cells 1 a b c d e

Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s) OH−(aq) + H+(aq) → H2O(l) Ca2+(aq) + CO32–(aq) → CaCO3(s) Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l)

e Remember,

symbols and charges have to balance.

2 a b c d e f g h i j

O2: 0 Al(OH)3: +3 BaCrO4: +6 Na2SO4: +6 Ca(NO3)2: +5 Na2O2: −1 Na2Zn(OH)4: +2 Na2Cr2O7: +6 MgH2: −1 KClO4: +7

3 a b c d

Oxidation is the loss of electrons by a species in a reaction. Reduction is the gain of electrons by a species in a reaction. An oxidising agent promotes oxidation by readily gaining electrons from another species. A reducing agent promotes reduction by readily losing electrons to another species.

4 a b c d e f

Mg(s) → Mg2+(aq) + 2e− 2Br−(aq) → Br2(aq) + 2e− Cr2+(aq) → Cr3+(aq) + e− MnO4–(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) ClO3−(aq) + 6H+(aq) + 6e− → Cl−(aq) + 3H2O(l) SO2(g) + 2H2O(l) → SO42−(aq) + 4H+(aq) + 2e−

e Remember,

symbols and charges have to balance.

5 a Mg(s) + Br2(l) → Mg2+(aq) + 2Br−(aq) b 5Cr2+(aq) + MnO4−(aq) + 8H+(aq) → 5Cr3+(aq) + Mn2+(aq) + 4H2O(l) e Cr2+(aq)

→ Cr3+(aq) + e− must be multiplied by 5 to provide the correct number of electrons for the reduction of

−(aq).

MnO4

Always remember to check the total charge on each side of the final equation to help avoid mistakes. In this case, it is 17+.

c 5SO2(g) + 10H2O(l) + 2MnO4−(aq) + 16H+(aq) → 5SO42−(aq) + 20H+(aq) + 2Mn2+(aq) + 8H2O(l) This simplifies to: 5SO2(g) + 2H2O(l) + 2MnO4−(aq) → 5SO42−(aq) + 4H+(aq) + 2Mn2+(aq) e SO2(g)

+ 2H2O(l) → SO42−(aq) + 4H+(aq) + 2e− must be multiplied by 5 and MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq)

+ 4H2O(l) by 2 to balance the electrons. The total charge on each side of the final equation is 2–.



Answers to chapter summary worksheets 6 Salt bridge H2(g)

Solution containing equal concentrations in mol dm–3 of Fe2+(aq) and Fe3+(aq)

1 mol dm–3 H+(aq)

Platinum electrodes

7 a The cell reactions are: Zn(s) → Zn2+(aq) + 2e− Co2+(aq) + 2e− → Co(s) So, the cell voltage is +0.48 V. b The cell reactions are: Zn(s) → Zn2+(aq) + 2e− I2(aq) + 2e− → 2I−(aq) So, the cell voltage is +1.30 V. c The cell reactions are: Fe3+(aq) + e− → Fe2+(aq) Co(s) → Co2+(aq) + 2e− So, the cell voltage is +1.05 V.

E° = +0.76 V E° = −0.28 V

E° = +0.76 V E° = +0.54 V

E° = +0.77 V E° = +0.28 V

Sn2+(aq) → Sn4+(aq) + 2e− E° = −0.15 V Cl2(aq) + 2e− → 2Cl−(aq) E° = +1.36 V net potential = −0.15 + 1.36 = +1.21 V The net potential is positive, so the reaction is feasible. b Br2(aq) + 2e− → 2Br−(aq) E° = +1.09 V 2Cl−(aq) → Cl2(aq) + 2e− E° = −1.36 V net potential = 1.09 − 1.36 = −0.27 V The net potential is negative, so the reaction is not feasible. c Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) E° = +1.33 V 2Br−(aq) → Br2(aq) + 2e− E° = −1.09 V net potential = 1.33 − 1.09 = +0.24 V The net potential is positive, so the reaction is feasible.

8 a

9 The cell reactions are: Li(s) → Li+(aq) + e− E° = +3.03 V − − F2(g) + 2e → 2F (aq) E° = +2.87 V voltage produced under standard conditions = 3.03 + 2.87 = +5.90 V 10 Advantages of the use of hydrogen fuel cells are: G They are alternative to fossil fuels (petrol or diesel), which will ultimately be depleted. G They avoid the production of polluting products such as carbon monoxide, carbon dioxide and oxides of nitrogen. G They are relatively lightweight and are more efficient than engines based on fossil fuels.



Answers to chapter summary worksheets Disadvantages of the use of hydrogen fuel cells are: G If hydrogen is to be used directly it has to be carried on board the vehicle, which is potentially dangerous as hydrogen is explosive. G The transportation of hydrogen under pressure is potentially hazardous. G If the hydrogen is absorbed into a metal, then the metal adds considerably to the weight. G If hydrides are used, they have to be very hot before the hydrogen is released.



Answers to chapter summary worksheets Chapter 15 Transition elements 1 a b c d e f

V: 1s2 2s2 2p6 3s2 3p6 4s2 3d11 3d21 3d31 or 1s2 2s2 2p6 3s2 3p6 3d3 4s2 Cu: 1s2 2s2 2p6 3s2 3p6 4s1 3d12 3d22 3d32 3d42 3d52 or 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Mn2+: 1s2 2s2 2p6 3s2 3p6 3d11 3d21 3d31 3d41 3d51 or 1s2 2s2 2p6 3s2 3p6 3d5 Co2+: 1s2 2s2 2p6 3s2 3p6 3d12 3d22 3d31 3d41 3d51 or 1s2 2s2 2p6 3s2 3p6 3d7 Cr3+: 1s2 2s2 2p6 3s2 3p6 3d11 3d21 3d31 or 1s2 2s2 2p6 3s2 3p6 3d3 Ni2+: 1s2 2s2 2p6 3s2 3p6 3d12 3d22 3d32 3d41 3d51 or 1s2 2s2 2p6 3s2 3p6 3d8

2 A transition element is a d-block element that can form an ion with at least one incompletely filled d-orbital. 3 There are several possible answers to this question. Any three from: G they conduct electricity well G they have high density G they have high melting and boiling points G they have variable oxidation states G they form coloured aqueous ions G they can often be used as catalysts G they form complex ions. 4 Aqueous sodium hydroxide reacts with the pale green aqueous iron(II) sulphate to produce a dark green precipitate of iron(II) hydroxide: Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s) The green precipitate turns brown slowly as it is oxidised to iron (III) hydroxide: 5 A bidentate ligand uses two different atoms with lone pairs of electrons bound to a cation in two positions. 6 The coordination number of a complex ion is the number of coordinate bonds attaching the ligands to the central metal ion. 7

2+

M

e In

2+

M

the diagram, the loops represent a bidentate ligand attaching to a central metal ion (M). The complex ion would

have a charge that depends on the charge on M and the charge on the ligand.

8 There are several acceptable answers — for example: [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ The hydrated copper ion is blue. On adding concentrated ammonia solution, ligand substitution occurs creating an intensely dark blue solution of the copper(II) tetraammine ion. Before the copper(II) tetraammine ion forms, a light blue precipitate of copper(II) hydroxide may be observed. [CoCl42−] 9 Kstab = [Co(H2O)62+][Cl−]4



Answers to chapter summary worksheets 10 a VO2+(aq) + 2H+(aq) + e− → VO2+(aq) + H2O(l) E° = +1.00 V Zn(s) → Zn2+(aq) + 2e− E° = +0.76 V + net potential of the reaction of Zn(s) and VO2 (aq) = 1.00 + 0.76 = +1.76 V The net potential is positive, so the reaction is energetically feasible. b The equation for the reaction that occurs is: 2VO2+(aq) + 4H+(aq) + Zn(s) → 2VO2+(aq) + 2H2O(l) + Zn2+(aq) 11 a The equation for the reaction is: 5H2O2(aq) + 2MnO4−(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) amount (in moles) of MnO4−(aq) in 26.35 cm3 is (26.35/1000) × 0.0200 = 5.27 × 10−4 mol From the equation: amount (in moles) of H2O2(aq) used in the titration = 5/2 × 5.27 × 10−4 = 1.318 × 10−3 mol concentration of H2O2(aq) = (1000/25.0) × 1.318 × 10−3 = 0.0527 mol dm−3 b In the reaction, 5H2O2(aq) produces 5O2(g) amount (in moles) of oxygen produced during the titration = 1.318 × 10−3 mol volume of oxygen = 1.318 × 10−3 × 24 000 = 31.6 = 32 cm3



Ocr a2 Chemistry Teacher Guide

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