Advanced Chemistry for You Answers

1 Atoms and electrons 1 a) Positive charge b) i) 2 electrons ii) 6 electrons iii) 10 electrons c) They spin in opposite d directions irections to minimise repulsion between the pair of electrons. electrons. 2 a) Alpha particles were fired at a very very thin sheet of gold foil and their their subsequent path detected detected by a radioactivity detector. b) If the Plum Pudding model model of the atom had been correct scientists would expect the positively charged alpha particles to pass straight through the gold foil, but some were deflected and a few rebounded from the foil. c) The Plum Pudding model model suggested that the the positive charge in an atom was spread through through the atom in a diffuse cloud of charge. To explain the results of this experiment, this was revised to propose a tiny positively charged nucleus at the centre of the atom into which all the positive charge was concentrated. d) Because it can be hammered into very thin gold foil (gold leaf) so the alpha particles could pass through and not be absorbed. electron transitions are specific fixed fixed ‘packets’ of energy, which 3 a) The energies absorbed or emitted in electron appear as lines on atomic spectra. This suggests that the electrons are moving between fixed energy levels. b) Continuous bands bands of light energy/electromagnetic energy/electromagnetic radiation radiation c) i) Principal quantum number of the electron shell or energy level ii) Electron falls from n = 3 into n = 1 iii) Electron falls falls from n = 4 into n = 1 iv) Electrons falling falling back into the n = 2 quantum shell mole of electrons from a mole of gaseous gaseous atoms 4 a) The energy needed to remove one mole b) log10I.E. 801 = 2.90; 2427 = 3.39; 3660 = 3.56; 25 026 = 4.40; 32 828 828 = 4.52 Graph – same axes as page 12 in Advanced Chemistry Chemistry for You You (2nd edition) c) The gradual rise in the first three three I.E.s occurs as the second second and third electrons are removed removed from the same quantum shell ( n = 2) but have to escape from increasingly positively charged ions. The large step up between the 3rd and 4th I.E.s on the graph arises as the fourth electron is removed from the n = 1 quantum shell, which is much nearer to the attractive force of the positively charged nucleus. d) i) The line gradually slopes upwards upwards until first large step up between the the 6th and 7th I.E.s, then slopes gradually upwards again until the second large step up between the 14th and 15th I.E.s., with 16th I.E. the highest point. ii) The first point is the lowest lowest with a large step up to the 2nd 2nd I.E. Then the line gradually gradually slopes upwards upwards until a second large step up between the 9th and 10th I.E.s, then slopes gradually upwards again until the third large step up between the 17th and 18th I.E.s, with 19th I.E. the highest point.

5 As shown at the bottom of page 11 in Advanced Chemistry Chemistry for You You (2nd edition) (without the top 4d sub-shell). 6 a) i) 1s22s22p3 ii) 1s22s22p63s23p5 iii) 1s22s22p63s23p64s2 iv) 1s22s22p63s23p63d24s2 v) 1s22s22p63s23p63d54s1 vi) 1s22s22p63s23p63d64s2 vii) 1s22s22p63s23p63d104s1 viii) 1s22s22p63s23p63d104s24p6 b) i) Ca – The The sub-shells fill up from from 1s to 3p then the the final two two electrons go into the 4s sub-shell sub-shell which is at a slightly lower energy level than the 3d sub-shell. ii) Cr – instead of the possible … 3d44s2 configuration Cr adopts the … 3d 54s1 alternative configuration which has the outer six electrons unpaired as this minimises the mutual repulsion between pairs of electrons in the same orbitals. iii) Cu – instead of the possible … 3d94s2 configuration Cu adopts the … 3d 104s1 alternative configuration which forms a complete 3d subshell. c) i) 1s22s22p6 ii) 1s22s22p6 Oxford University Press

Advanced Advan ced Chemis try for f or You, S econd Edition Editi on

© Ryan Books Limited 2015

1

2 Atoms and the nucleus 1 a) Atomic number = no. of protons in an atom Mass number = number of protons + neutrons Relative atomic mass = the weighted mean mass of atoms of an element, taking into account the relative proportions of its isotopes in a naturally occurring sample, on a scale which assigns carbon-12 a relative mass of exactly 12 b) Their numbers of neutrons differ. differ. c) Relative atomic masses are weighted means of the masses of naturally occurring isotopes so it is highly unlikely they will be whole numbers. d) Mass spectrometer 2

protons

neutrons

electrons

a)

27

32

27

b)

47

60

47

c)

92

143

92

d)

94

148

94

e) The proportion of neutrons neutrons compared with protons increases increases as the atoms get larger. larger. f) Their stability decreases and they become more susceptible to radioactive decay.

3 a) b)

(29.5

203)

×

(70.5

+

205)

×

100 (51.46

90)

×

(11.23

+

91)

×

204.4 (to 1 decimal place) or 204 (to 3 significant figures)

=

(17.11

+

92)

×

(17.40

+

94)

×

(2.80

+

96)

×

100

91.32 (to 2 decimal

=

places or 4 significant figures)

vaporised. 4 a) It is heated and vaporised. b) The atoms or molecules molecules are bombarded bombarded by electrons. electrons. The electrons knock electrons from the atoms or molecules, forming positively charged ions. c) The current flowing through through the coils is varied to vary the magnetic field – this ensures ensures that the range of ions with different m/z values enter the detector. d) ToF MS is quicker and less less expensive (also it is sometimes desirable desirable not to fragment the molecule molecule under test).

5 a)

(50.52

×

79)

(49.48

+

×

81)

79.99 (to 2 decimal places or 4 significant figures)

=

100 b) 158 peak is caused by two bromine-79 bromine-79 atoms in a Br2 ion. 160 peak is caused by a bromine-79 atom and and a bromine-81 atom in a Br2 ion. 162 peak is caused by two bromine-81 atoms in a Br2 ion. +

+

+

Oxford University Press



2

3 Calculations using the mole 1 a) b) c) d)

Molar mass Volume (in cm3) RT Concentration (in mol dm 3)

2 a) b) c) d)

6.02 × 1023 mol 1 The simplest ratio of the different elements elements in the compound compound The empirical formula multiplied multiplied by a whole whole number will give the actual formula. The relative relative molecular molecular (formula) (formula) mass For answers answers to questions 3, 4, 5, 6, 7, 8 and 9 see page page 451 in Advanced Chemistry Chemistry for You You (2nd edition).

10 a) b) c) d) e)

H2 + Br2 2HBr Fe2O3 + 2Al Al2O3 + 2Fe 2AgNO3 + Cu Cu(NO3)2 + 2Ag 2H2O2 2H2O + O2 CH4 + 2O2 CO2 + 2H2O For answers answers to questions 11, 12, 13, 14, 15 and 16 b)–e) see page page 451 in Advanced Chemistry Chemistry for You (2nd edition).

−

−

16 a) After all the iron(II) ions have been oxidised, the purple manganate( VII) ions become in excess and will start to colour the reaction mixture a permanent faint purple/pinkish colour at the end point. student data. 17 a)–d) Dependent on student e) No, as it the third most abundant abundant gas in the air, air, after nitrogen and oxygen. oxygen.




3

4 Ionic compounds 1 Ionic; non-metal; electrons; non-metal; positively; negatively; giant; lattice; ionic; electrostatic; oppositely; high; dissolve; molten; solid. 2 tin(II) bromide; sodium fluoride; iron( II) oxide; zinc chloride The substances with giant ionic structures are made up of metal ions and non-metal ions. electron transferred into outer outer shell of F atom (2,7); Li becomes becomes Li 3 a) Diagram of Li atom (2,1) showing electron (2) , and F becomes F (2,8) b) Diagram of two Na atoms (2,8,1) showing showing each one transferring transferring an electron into outer shell shell of O atom 2 2 (2,6); both Na’s become Na (2,8) , and O becomes O (2,8) +

+

−

+

4

+

−

−

I

NO3

SO42

PO43

LiI

LiNO3

Li2SO4

Li3PO4

NH4∙

NH4I

NH4NO3

(NH4)2SO4

(NH4)3PO4

Ba2∙

BaI2

Ba(NO3)2

BaSO4

Ba3(PO4)2

Cr3∙

CrI3

Cr(NO3)3

Cr2(SO4)3

CrPO4

∙

Li∙

∙

∙

∙

(short version of the Periodic Table Table in which Al is in Group Group 3) is equal to the positive 5 a) The group number (short charge on the metal ion. b) Short version of the Periodic Periodic Table: Table: 8 minus the group number number of the non-metal equals equals the charge on its negative ion. Long version version of the Periodic Table: 18 minus the group number number of the non-metal non-metal equals the charge on its negative ion. c) It requires too much energy for a carbon atom to either remove 4 electrons (to form form C 4 ions) or to gain 4 electrons (to form C4 ions). d) An H atom atom can gain gain 1 electron electron to form an H ion, gaining the electronic configuration of the noble gas helium; or it can lose its only electron to form an H ion with no electrons. +

−

−

+

6 a)

2+

−

F

Mg

−

F

or

Mg

b)

2+

+

Li

−

Br

or

Li

c)

+

2−

O

3+

Al

O

or

Al

2−

3+

Al

2−

O

or 3+

Al

3+

7 a) 8 b) 8 c) CsCl Oxford University Press



4

5 Covalent bonding 1 Sharing; covalent; pair; negatively; positively; four; melting; giant; simple; molecule; between; molecular; Waals’; London; temporary/instantaneous; polar; Hydrogen; hydrogen; electronegative; nitrogen/fluorine; nitrogen/fluorine; fluorine/nitrogen; fluorine/nitrogen; graphite 2 F

F

3 a) Allotropes b) Any fullerene/graphene fullerene/graphene c) Diamond is very hard and does does not conduct electricity electricity whereas graphite is soft and conducts conducts electricity. This is explained by diamond’s giant 3D network of carbon–carbon bonds, with each carbon atom forming 4 covalent bonds to their nearest neighbouring atoms. On the other hand, graphite’s giant structure is made up of sheets of carbon atoms arranged in hexagons, with each carbon atom forming 3 covalent bonds to its nearest neighbours. The forces between the layers are weaker than covalent bonds, as each carbon atom donates its fourth outer shell (2p) electron into bands of delocalised electrons located above and below the plane of the sheets. The weaker forces between sheets make it relatively easy for layers to slide over each other, making graphite soft (and slippery). The delocalised electrons can drift along the layers of carbon atoms and can flow as an electric current when a voltage is applied to graphite. On the other hand, diamond has no mobile electrons, with all its four outer shell electrons involved in its covalent bonding, so is an electrical insulator. 4 a) Single atoms b) van der Waals’ (London) (London) forces forces c) Instantaneous imbalances imbalances of electrons in atoms/molecules atoms/molecules result in temporary temporary dipoles that induce dipoles on neighbouring atoms/molecules, producing relatively weak forces of attraction between atoms/ molecules. d) As the atomic number increases increases (as the group is descended), descended), the boiling point increases. This This is because as the atoms get larger and have more electrons, the opportunities for the number and size of instantaneous dipole-induced dipole interactions increase so the forces between the atoms of the noble gases get stronger. 5 a) H

N

H

H

b) Dative or coordinate (covalent) bond c) H+ H

N

H

d) Tetrahedral




5

6 a) A small square b) Trigonal (planar) c)

2−

O

O

C

O

d) Trigonal (planar)

7 a) H

O

H

b) Its relatively low melting/boiling point c) Graph of melting points (vertical axis) v M r (horizontal axis, points join with line (or ‘stick bars’ for each hydride) d) Water e) 180–184 °C f) Although the permanent dipole–dipole attractions decrease from hydrogen sulfide to hydrogen telluride (as the dipole moment of the molecule decreases), the melting points rise as the molecules get larger and the greater number of electrons means that the opportunities for the number and size of instantaneous dipoleinduced dipole interactions increase. Therefore the van der Waals’ (London) forces between the molecules get stronger going from hydrogen sulfide to hydrogen telluride, outweighing the effect of the decreasing dipole moment. g) The presence of the relatively strong hydrogen bonds between the hydrogen and oxygen atoms in neighbouring water molecules. h) When the water molecules are fixed in position in ice crystals each water molecule forms two hydrogen bonds, in an open structure which occupies a greater volume than the same mass of liquid water.


Advanced Chemis try for You, Second Edition


6

6 Metals and structures Questions on metals 1 sea; giant; conduct; electrons; high; giant; density; close-/closely; sodium; body; alloys 2 a) Good conductors of electricity and heat, high melting and boiling points, malleable, ductile, high density b) E.g. mercury – low melting point – liquid at room temperature – used in thermometers c) They are good conductors of electricity (and heat). d) bulb

sample being tested

3 a) Fill the bottom of a tray with balls, then add layers to sit in the holes/indentations in successive layers of balls, leaving no gaps. b) Make a structure in which the third row of balls are directly above the first row balls or you could make a structure in which the first and fourth rows line up directly. c) An ‘aba’ or an ‘abc’-type structure respectively d) Body-centred cubic structure e) It is more open/has more space within the structure 4 a) The bonding in a giant metallic structure, in which each metal ion has donated electrons into a ‘sea’ of delocalised electrons. These mobile electrons bind the oppositely charge metal ions together. b) Because they have donated electrons into the ‘sea’ or conduction band of delocalised electrons, becoming positively charged ions (cations). c) They are made up of layers of ions of the same size which can slide over each other when subjected to forces. d) The differently sized atoms/ions disrupt the regular arrangement of the giant metallic structure, meaning that layers cannot slide past each other as easily – so the metal can withstand greater forces without changing shape. e) Melt the metals and stir them together; allow to cool and set to make the mixture. f) E.g. test which is harder – an alloy or pure metal – by placing a tube and ball-bearing in the surface of each; drop the same mass from the same height down the tube onto the ball-bearing; then measure diameter of indentation made in the surface of each metal. 5 a) Na – 1s22s22p63s1 Mg – 1s22s22p63s2 Al – 1s 22s22p63s23p1 b) Na 1 electron Mg 2 electrons Al 3 electrons c) Because aluminium has more electrons in the conduction band/more delocalised electrons per ion than Na or Mg in their giant structure. −




7

Questions on structure and bonding 1 covalent; ionic; giant; solid; dissolved; ions; non-metal; giant; high; molecules; low; graphite; inter; bonding; dipole; Waals’; induced; metallic 2 a) C because it conducts electricity when solid (as it contains mobile, delocalised electrons) b) B because it conducts no electricity when solid but does when molten (as the ions become mobile) c) A and D because they have low melting points (because of the weak forces between their individual molecules) d) D (as it melts at 114 °C , which is well above room temperature)) e) B as molten ionic compounds are decomposed by electrolysis as they conduct electricity f) Different intermolecular forces (factors affecting these could be the size of the molecules and any permanent dipole moments within molecules). 3 Zinc oxide (highest melting point), ammonia, methane Zinc oxide has ionic bonding so the forces between ions are strong as the electrostatic forces of attraction operate in all directions within their giant lattices. Then both ammonia and methane have simple molecular structures but ammonia has hydrogen bonding between its molecules – which is stronger than the van der Waals’ forces between the non-polar methane molecules.




8

7 The Periodic Table 1 number; groups; electrons; outer; periods; d-block; f-block; metalloids; heat; hammered; wires; high; cobalt/ nickel; nickel/cobalt; low; heat; electricity; repeating; periodicity; bonding; left; right 2 a) b) c) d) e) f) g) h) i) j)

hydrogen magnesium oxygen scandium bromine lithium phosphorus krypton strontium astatine

3 a) b) c) d) e) f) g) h) i) j)

1s22s1 1s22s22p5 1s22s22p63s23p4 1s22s22p63s23p64s2 1s22s22p63s23p63d14s2 1s22s22p63s23p63d54s1 1s22s22p63s23p63d74s2 1s22s22p63s23p63d104s1 1s22s22p63s23p63d104s2 1s22s22p63s23p63d104s24p6

4 a) b) c) d)

Silicon and germanium Carbon Tin and lead The 1st ionisation energy of lead is much smaller than that of carbon because lead’s outer electron is further from the attractive force of the nucleus and is shielded by more complete inner shells of electrons. These two factors outweigh the larger nuclear charge on the lead atom, making an electron in its outer shell easier to remove than one from carbon’s outer shell. e) Metals form positively charged ions in their reactions, so the lower 1st and 2nd ionisation energies of lead and tin compared to the rest of the Group 4(14) elements will make it easier for them to form positive ions in their reactions (and easier as elements to lose two outer electrons into the sea of delocalised electrons, forming cations in their giant metallic structures).

5 a) b) c) d) e)

Rises to a peak in Group 4(14), then falls away rapidly to the end of the period Increase steadily across a period to Group 7(17) Starts high on the left of the Periodic Table and drops abruptly on the right hand side Decreases steadily across a period General increase across a period (with small decreases at Group 3 and Group 6)

6 a) Giant metallic and giant covalent structures found in elements in Group 1 to 4(14), followed by the simple molecular structures of the non-metallic elements. b) The ability to attract the electrons in a covalent bond increases as the nuclear charge increases across a period and the size of the atoms decreases. c) The metals with their delocalised electrons are found on the left hand side of the Periodic Table whereas the non-metals, with their giant covalent or simple molecular structures, have no mobile electrons available for conduction. d) The nuclear charge increases across a period, so the electrons that enter the same shell successively are drawn slightly closer to the nucleus and the size of the atoms decreases.




1

e) As the nuclear charge increases across the period, the electrons become held more strongly to the nucleus, making it harder to remove them. (The slight fall at Group 3 is because the p sub-shell has an electron in it which is slightly further from the nucleus than the s sub-shell. The slight fall at Group 6 is the result of mutual repulsion between electrons in the spin pair in the outer p sub-shell.)

7 a) Rises to a peak in Group 4(14), then falls away rapidly to Group 5 and to the end of the period b) The periods start with giant metallic structures and the giant covalent structures of Group 4(14) elements carbon and silicon which need lots of energy to separate their particles to form a vapour. However, from Group 5(15) onwards the elements have simple molecular structures typical of many non-metallic elements in which relatively weak intermolecular forces can be easily overcome. 8 a) It is the 17th element in the Periodic Table – which is the number of protons (its atomic number) and also its number of electrons. In the short version of the Periodic Table, it is in Group 7 – which is the number of electrons in its outer quantum shell. It is in the 3rd period – which equals the number of quantum shells occupied or partially occupied by its electrons. b) They both have outer shell configurations of …s2 ..p5/same number of electrons in their outer quantum shell (highest energy level). 9 a) Alkali metals, as when the metals react with water they form alkaline solutions of their hydroxides. b) Lithium, as the smallest atom in the group has its outermost electron nearest to the attractive force of its nucleus, and this electron is also shielded less from the nuclear charge than the larger atoms, so it takes more energy to remove the outer electron to form a lithium ion. These two factors outweigh the fact that lithium, as the smallest atom in the group, has the smallest nuclear charge attracting its outer electron. c) rubidium + water rubidium hydroxide + hydrogen 2Rb(s) + 2H2O(l) 2RbOH + H2(g) d) Violent/vigorous/explosive reaction/flames (as the hydrogen released ignites) e) Low density, soft, low melting points for metals




2

8 Periodicity and chemical properties 1 Period; ionic; molecular; basic; amphoteric 2 a) sodium oxide – when added to water the oxide dissolves, reacting to form a solution of sodium hydroxide, NaOH(aq) – the alkaline solution has a pH of 14, so sodium oxide is basic. Sulfur trioxide gas dissolves when bubbled into water, reacting vigorously to form a solution of the acid, sulfuric acid (sulfuric( VI) acid), H2SO4(aq) – with a pH of 1, so it is an acidic oxide. b) E.g. aluminium oxide acting as a base: Al2O3(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2O(l) E.g. aluminium oxide as an acid: Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2NaAl(OH)4(aq) 3 a) E.g. 3rd period

Na

Mg

Al

Si

P

S

Cl

Na 2O

MgO

Al2O3

SiO2

P4O10

SO3/SO2

Cl2O7/ Cl2O

b) Argon does not form an oxide because it is so unreactive. c) Formula of oxide Na 2O MgO Al2O3 SiO2

P4O10

SO3/SO2

Cl2O7/ Cl2O

Formula of oxide

Moles of oxygen per mole of element

d) Student graph e) Formula of oxide Moles of oxygen per mole of element

0.5

1

1.5

2

2.5

3/2

3.5/0.5

Li 2O

BeO

B2O3

CO2

NO2

O2

OF2

0.5

1

1.5

2

2

1

0.5

f) The non-metal oxides of the third period can be oxidised to a greater extent than those in the second period because of the presence of the 3d sub-shell – which the elements can use to expand their octet, enabling them to bond to a larger number of oxygen atoms.

4 a) The initial rise peaks at Group 2 then decreases steadily to Group 6(16) and levels off to Group 7(17). b) The Group 1 and 2 chlorides are both high, followed by a steep fall and a levelling off at relatively low temperatures across to the rest of period 3. c) The ionic oxides and chlorides of Groups 1 and 2 have giant ionic structures with strong electrostatic forces of attraction between their oppositely charged ions, operating in all directions, explaining their high melting points. The large decrease takes place going from Group 2 to 3 in the chlorides, as aluminium chloride exists as simple molecules, whereas aluminium oxide has a giant ionic structure and silicon dioxide has a giant covalent structure, explaining why the steep fall occurs going from Group 4 to 5 in the oxides. The chlorides and oxides of Group 5(15) to 7(17) are all simple molecular structures so have relatively weak intermolecular forces, explain their low melting/boiling points. 5 a) Argon – because it has the highest nuclear charge in Period 3 whilst all Period 3 atoms are filling the same energy level/shell with electrons. b) The metal atoms lose electrons that can result in the loss of their outer shell so the positive ions are smaller than the original atoms. The non-metal atoms gain one or more electrons into their outer shell, increasing the mutual repulsion between electrons in that shell so increase in size. c) Silicon (because it takes too much energy to either gain 4 electrons or lose 4 electrons) and argon (because it has a complete outer octet of electrons which is a very stable electronic configuration).




3

6 a) b) c) d) e)

Sodium and fluorine Beryllium and aluminium Neon and argon Carbon and silicon Lithium and sodium

7 a) The first two chlorides dissolve in water whereas the remaining chlorides react, being hydrolysed by water, forming white fumes of hydrogen chloride gas and leaving acidic solutions. b) The chlorides of sodium and magnesium are purely ionic and dissolve in water, resulting in solutions of pH 7 and 6.5 respectively. Then from aluminium chloride, with some covalent character in its bonding and a simple molecular structure, onwards the chlorides are hydrolysed by water. c) The high charge density on the Al3 ion weakens the bonds in the water molecules surrounding the hydrated ion. When an O H bond breaks in a water molecule it releases an H (aq) ion into solution, making the solution acidic. d) SiCl4 has empty 3d orbital into which water molecules can donate pairs of electrons and initiate attack, resulting in the hydrolysis of silicon chloride. There is no equivalent 2d orbital in the carbon atom of CCl 4 so no attack occurs with water, and two immiscible layers form. +

+




4

9 Group 2 (s-block elements) 1 s; alkali; earth; more; more; flame; yellow; (brick) red; (crimson) red; barium; basic; alkaline; hydroxide/OH ; more; (thermally) stable −

2 a) Barium, Ba – because the electron removed from its outer shell is furthest away from the nucleus and is shielded from the attractive force of the nucleus by more inner shells full of electrons, despite it having the largest positive nuclear charge. b) Beryllium, Be c) Beryllium, Be d) Calcium, Ca e) Magnesium, Mg, and calcium, Ca f) Barium, Ba g) Barium, Ba h) Calcium, Ca i) Calcium, Ca j) Strontium, Sr k) Calcium, Ca l) Barium, Ba 3 a) Electrons gain energy from the flame and ‘jump’ into a higher energy level/shell (excited state); when they fall back to the lower energy level energy is released as light in the visible range of the electromagnetic spectrum. b) The energy released on the transition of an electron to a lower energy level does not correspond to any part of the visible range of the electromagnetic spectrum. c) CaO(s) + H2O(l) Ca(OH)2(aq) d) Calcium hydroxide/carbonate is used to raise the pH of acidic soil 4 a) b) c) d)

They lie diagonally next to each other in the Periodic Table. Na and Ca/K and Sr/Rb and Ba 2Mg(NO3)2(s) 2MgO(s) + 4NO2(g) + O2(g) 4LiNO3(s) 2Li2O(s) + 4NO2(g) + O2(g)

5 a) b) c) d) e) f) g) h)

Lower 7s 2+ i) RaO ii) Ra(OH)2 iii) RaCO3 iv) Ra(NO3)2 Violent/vigorous reaction with gas bubbles given off rapidly Ra(s) + 2H2O(l) Ra(OH)2(aq) + H2(g) i) Radium nitrate ii) radium carbonate i) Radium hydroxide will be more soluble than barium hydroxide/i.e. greater than 1.5 × 10 2 mol/100 cm3 of water) ii) Even less soluble than barium sulfate (effectively insoluble) −




5

10 Group 7(17) (p-block elements) 1 7/17; diatomic; Br2; lowest; most; strongest/most powerful; 1–; silver; white; ammonia; phosphoric 2 a) b) c) d) e) f)

Student table containing data as on page 126 Student graph of boiling points (y-axis) v atomic number (x-axis) 325–350 °C/598–623 K Black/dark grey At2 At

At

3 a) Sodium bromide b) Ag (aq) + Br (aq) AgBr(s) or Ag Br (s) c) Photographic film, as silver bromide decomposes in light to form grey deposits of silver metal. +

−

4 a) NaCl(s) b)

+

H2SO4(l)

NaHSO4(s)

+

−

HCl(g)

+

hydrogen chloride gas

sodium chloride + conc. H2SO4

c) Hydrogen chloride gas is too soluble in water. d) White, smoky fumes NH3(g) + HCl(g) NH4Cl(s) e) The hydrogen bromide or hydrogen iodide undergoes further reduction with concentrated sulfuric acid, forming bromine or iodine. f) Use concentrated phosphoric(V) acid instead of concentrated sulfuric acid.

5 a) A reaction in which the same species undergoes both reduction and oxidation. +1 b) 0 –1 Cl2(aq) + H2O(l) HCl(aq) + HClO(aq) or +1 0 –1 Cl2(aq) + 2NaOH(aq) 2NaCl(aq) + NaClO(aq) + H2O(l) or +5 0 –1 3Cl2(aq) + 6NaOH(aq) 5NaCl(aq) + NaClO3(aq) + 3H2O(l) c) i) 0 ii) –1 iii) +5 iv) +1 v) +7 6 a) A chemical species that accepts/gains electrons from another species in a reaction b) F2 + 2e 2F c) Because its outer shell, into which an incoming electron is transferred, is nearest to the attractive force of the nucleus and is shielded least by inner electrons, despite having the lowest positive nuclear charge. d) A chemical species that donates/gives/transfers electrons to another species in a reaction. e) Li Li + e f) A redox reaction g) F + 2e 2F −

−

+

−

−

2

2Li

2Li

+

+

2Li

−

F2

+


+

2e

−

2Li F (or 2LiF) +

−



6

11 d-block elements 1 Reactive; coloured; oxidation; Complex; ligands; catalysts 2 Any four transition metals and their uses, with relevant properties described. 3 a) i) 1s22s22p63s23p63d34s2 ii) 1s22s22p63s23p63d74s2 iii) 1s22s22p63s23p63d104s2 b) V and Co form ions with a partially filled d orbital in their electronic configuration whereas Zn does not (the Zn2 ion has a complete 3d sub-shell). c) A d-block element +

4 a) 1s22s22p63s23p63d54s1 b) 1s22s22p63s23p63d104s1 c) The chromium atom has a half-filled 3d sub-shell which minimises the mutual repulsion between electrons in orbitals compared with a configuration of [Ar]3d 44s2. Copper adopts the configuration that completes its 3d sub-shell. 5 a) Fe2 – 4 out of the 5 boxes for the 3d orbitals will each be occupied by 1 electron, with the other box contains 2 electrons. Fe3 – The 5 boxes for the 3d orbitals will be occupied by 1 electron. b) Cu – The 5 boxes for the 3d orbitals will each be occupied by 2 electrons. Cu2 – 4 out of the 5 boxes for the 3d orbitals will be occupied by 2 electrons, with 1 box containing a single electron. c) Mn2 – The 5 boxes for the 3d orbitals will each be occupied by 1 electron. Mn4 – 3 out of the 5 boxes for the 3d orbitals will be occupied by 1 electron. d) Fe3 is more stable than Fe 2 , as the 3 + ion has the half-filled 3d sub-shell, with less mutual repulsion between electrons. Cu is more stable than Cu2 , as the 1+ ion has the complete 3d sub-shell, with its symmetrical distribution of charge around the nucleus, giving it added stability. +

+

+

+

+ +

+

+

+

+

6 a) Because chromium has 6 electrons available to get involved in bonding – 1 4s electron and 5 3d electrons. b) The energy involved in removing 6 electrons would be too high for an energetically stable compound to be formed containing Cr 6 ions. c) Chromate(VI) ions, CrO42 , and dichromate(VI) ions, Cr2O72 +

−

7 a) b) c) d) e) f) g) h) i) j) k) l)

−

Manganese, +7 Zinc Iron Chromium and manganese Copper Copper Vanadium Chromium Cobalt Iron Copper Nickel

8 a) They donate one (or more) lone pair of electrons into empty orbitals on the central metal ion, forming dative (coordinate) covalent bonds. b) 4 c) Tetrahedral d) Because the complex carries a negative charge




7

e) E.g.

H

2+

H

H

H

O

O

H

O

H

O

H

Co2+ H

O O

H H

f) g) h) i)

H H

pink This complex is cationic. Linear It helps distinguish between the similarly coloured silver halide precipitates formed in positive tests for halide ions. The two ligands at the points of the octahedron are further from the central transition metal ion than the 4 ligands in the same plane as the metal ion. i) 3+ ii) 2+ iii) 3+

9 a) It forms only one coordinate (dative) covalent bond (or donates only one lone pair of electrons) to the central transition metal ion in a complex. b) E.g. H 3+ 2

C

H

H2C

H

N

H

H

H N

N CH2

H Cr

3+

H N

CH2

N

H H2C

c) i) 6

10 a) b) c) d) e) f)

C H2

H

H

H H

ii) hexadentate

iii) 2–

Fe Fe3 + 3e Cu2 + e Cu 2 3 V V + e 2H + VO2 + e V3 + H2O 4H + VO2 + 3e V2 + 2H2O 2H2O + Mn2 4H + MnO2 + 2e or 4OH + Mn2 2H2O + MnO2 + 2e 6H + CrO3 + 4e Cr2 + 3H2O H2O + MnO3 2H + MnO4 + e or 2OH + MnO3 H2O + MnO4 + e 14H + Mn2O7 + 10e 2Mn2 + 7H2O Reactions in b), d), e), g) and i) are reduction reactions because the reactant is gaining electrons, and there is a decrease in their oxidation state. Fe (in a)), V2 (in c)), Mn2 (in f)) and MnO3 (in h)) are acting as reducing agents as they all donate electrons (increase their oxidation state), and become oxidised themselves. V2 + VO2 + 2H 2V3 + H2O 2Cu2 + Mn2 + 2H2O 2Cu + MnO2 + 4H +

+

−

−

+

+

+

−

+

+

−

+

+

+

−

+

−

g) h)

N

+

+

−

+

+

−

−

+

+

−

−

i) j) k) l) m)

−

+

−

+

+

+

+

−


+

+

+

+

−

+

+

+



8

11 a) Homogeneous catalysis – the reactants and the catalyst are in the same phase. Heterogeneous catalysis – the reactants and the catalyst are in different phases. b) Homogeneous catalysis: 2 Fe (aq) e.g. 2I (aq) + S2O82 (aq) I2(aq) + 2SO42 (aq) Heterogeneous catalysis: +

−

−

−

MnO2(s)

e.g. 2H2O2(aq) 2H2O(l) + O2(g) c) It can help form transition intermediates by redox reactions with reactants (see enthalpy level diagram on page 155 in Advanced Chemistry for You (2nd edition)). d) E.g. Fe in Haber process in manufacture of ammonia, V2O5 in Contact process in manufacture of sulfuric acid, Ni in hydrogenation of oils in manufacture of margarine, Pt/Rh in manufacture of nitric acid.

12 a) b) c) d) e)

Fe2 Fe3 + e 8H + MnO4 + 5e Mn2 + 4H2O 5Fe2 + 8H + MnO4 5Fe3 + Mn2 + 4H2O MnO4 : Fe2 is 1 : 5 Adding potassium manganate(VII) from a burette to a conical flask containing a 25.0 cm3 portion of the original iron(II) sulfate solution until the solution just turned pale pink. f), g), h) and i) See answers on page 451 of Advanced Chemistry for You (2nd edition). +

+

+

−

+

−

−

+

−

+

−

+

+

+

13 a) [Cu(NH3)4(H2O)2]2 b) Tetraamminediaquacopper(II) ion c) NH +

3

H3N

2+

NH3 Cu2+

H3N

OH2 H2O

Z (or cis/ trans) isomerism d) E/ e) Square planar f) Z -dichlorodicyanonickelate(II) ion g) − − Cl

CN

2+

Ni2+

NC

−


Cl−



9

12 Introduction to organic chemistry 1 a) b) c) d)

Ketone Carboxylic acid Ester Alcohol

2 a) ethylamine, CH3CH2NH2, and H

H

H

C

C

H

H

H N H

b) ethanenitrile, CH3CN, or propanenitrile, CH3CH2CN, and

H H

C H

c) ethanal, CH3CHO, and

H H

C

N

or

H

H

H

C

C

H

H

C

N

O

C

C H

H

d) chloroethane, CH3CH2Cl, and H

e) ethanamide, CH3CONH2, and

H

H

C

C

H

H

H H

C

Cl

O C H N

H

H

f) ethanoic acid, CH3COOH, and

H H

C H

O C O

H

3 C3H6 4 a) CnH(2n b) C11H24

2)

+

5 a) H

b) H

H

H

H

C

C

C

H

H

H

H

H

H

C

C

C

H

H

H


H O

O C H



1

c) H

H

O

H

H

H

C

C

C

C

C

H

H

H

H

d) H

H

H

H

H

H

H

H

C

C

C

C

C

C

H

H

H

H

H

H

H

Br

Br

H

H

H

H

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

H

H

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

H

C

C

C

H

H

H

H N H

e) H

f) H

g) H

h) O

H

C

N

O C O

Cl

H

Cl

H

C

C

C

C

H

H

H

H

H

H

C

C

H

H

H

H

H

i)

H

H

j) F

6 a) b) c) d)

H

H

H

H

N

C

C

C

C

H

H

H

H

F

F

C

C

H

H

O C O

H

H

2-Chloropropane Propanal Methanoic acid 4-Chloro-2,2-difluoropentane




2

7 a) H

H

H

H

H

H

C

C

C

C

C

H

H

H

H

H H

H H

C

C H

H H H

H

C

C

C

H

H

H

H H

C

H H

C

C

C

C

H H

H H

H

H

H H

H

H

b) Cl

H

H

H

H

C

C

C

C

H

H

H

H

H

H

H

Cl

H

H

C

C

C

C

H

H

H

H

Cl H

H

c) Br

Br

Br

H

H

H

C

C

C

C

H

H

H

H

H

H

H

H

C

C

C

C

H

H

H

H

H

Br

Br

H

Br

H

H

C

C

C

C

H

H

H

H

Br

H

H

H

C

C

C

H H

H

H C

H

Br

H H

C

H H

C

C

C

H

H

H

Br

H

H

Br

H

C

C

C

C

H

H

H

H

H

Br

H

C

C

C

C

H H

H

C

C

C

H H

C

H H

H

H

H H

Br

H

Br

Cl

H

H

H H

H

H

H

H Br

C

H H

C

C

C

H

H

H

H

8 a), b) and d) See answers on page 451 of Advanced Chemistry for You (2nd edition) c) H H H H H

H

C

C

H

H

9 a) i)

O

and

C

O

C

H

H C

F

H

C Br

H

H

F

H

C

ii)

H

C

H

Br

Z -1-bromo-2-fluoroethene

E -1-bromo-2-fluoroethene

H3C

H

CH3 C

CH3

C

H

C H

H3C

Z -but-2-ene

C H

E -but-2-ene

b) i) CHBrClF because it contains a carbon atom bonded to four different groups/atoms (contains an asymmetric carbon atom or a chiral centre). ii) H H C

I

C Cl

Br


I

Cl Br



3

13 The alkanes 1 Saturated; C2H6; hexane; combustion; water; hydrogen; sunlight (U.V. light); incomplete; carbon; monoxide; sulfur; nitrogen; acid 2 a) CnH(2n b) C12H26 c) H

2)

+

H H

H

C

C

H

C

C

H

H

H

d) No, CnH2n e) The bond angles between carbon atoms in cyclobutane are about 90° (under strain), as they are much less than the 109.5° tetrahedral angles between the carbon atoms in butane (which minimise repulsion between bonding pairs of electrons).

3 a) Fractional distillation b) Petroleum gases; petrol/gasoline; kerosene; diesel; lubricating oil; fuel oil; bitumen c) The larger alkanes have stronger intermolecular forces/van de Waals’ forces between their molecules (as they contain more electrons within each molecule, therefore giving rise to more instantaneous dipoles and hence more induced dipoles on their neighbouring molecules, producing stronger forces of attraction between molecules). 4 a)

ice/water mixture (to condense water vapour) to water pump (to draw gases through the apparatus)

deposits of soot (carbon)

hexane

sand tray (for safety)

water droplets collect (tested by checking boiling point is 100ºC)

limewater turns milky (showing CO2 is produced)

b) Carbon dioxide – limewater turns cloudy/milky Water – check boiling point is 100 °C/white anhydrous copper sulfate turns blue/blue cobalt chloride turns pink c) CO bonds strongly to molecules of haemoglobin in the blood, taking up the sites that normally carry oxygen around the bloodstream, effectively starving cells of oxygen – this can result in losing consciousness and ultimately death. d) Carbon dioxide; carbon monoxide; carbon/soot particles; unburnt hydrocarbons, sulfur dioxide; nitrogen oxides e) They remove carbon monoxide (by oxidising it to carbon dioxide) and nitrogen oxides (by reducing them to nitrogen gas). f) No, as they can remove carbon monoxide and nitrogen oxides from exhaust emissions but the car still releases carbon dioxide gas. g) Premature ignition of fuel in an engine’s cylinders. h) They add more branched alkanes and arenes to the petrol mix.




4

5 a) b) c) d) e)

(free radical) substitution C2H6 + Br2 C2H5Br + HBr The Br–Br bond is broken (by energy from sunlight) Initiation/homolytic fission of the Br–Br bond) E.g. Br + C 2H6 C2H5 + HBr or C2H5 + Br2 C2H5Br + Br f) Two free radicals react to form a molecule. g) C2H5 + Br C2H5Br Br + Br Br2 C2H5 + C2H5 C4H10 h) C2H4Br2; C2H3Br3; C 2H2Br4; C 2HBr5; C 2Br6 ·

·

·

·

·

·

·

·

·

·

6 a) Any sensible ideas, e.g. landfill; advantages – cheap/fast process; disadvantages – unsightly/foul smelling/ uses up potentially useful land/destroys wildlife habitats (at least for the lifetime of the site until it is covered with soil) b) Any sensible ideas, e.g. incineration; advantages – does not use up valuable land as in landfill sites; energy released from burning can be put to good use, e.g. heating/generating electricity; disadvantages – can release toxic gases if temperature is not high enough/plant is unsightly/some sorting needed before incineration




5

14 The alkenes 1 double; reactive; high; double/pi; electrophiles; electrophilic; Bromine; colourless; addition 2 a)

H

H C

H

C

Cl

Cl C

Cl

C

Cl

H

Z (cis – trans/geometric) isomerism b) E/ c) There is no free rotation about the double bond in 1,2-dichloroethene. d) Cl Cl Cl H C

C

F

C H

F

C Cl

3 a) All three could produce carbon dioxide and water as combustion products but heptane would burn with the cleanest flame. b) Hept-1-ene and cycloheptene would decolourise the bromine water, but heptane would not react with it. 4 a) b) c) d)

C2H4 Decane, octane and ethene Cracking It converts larger, less useful alkanes into smaller, more useful alkanes.

5 a) b) c) d) e) f)

CH3CH2CH3 Propene, hydrogen, propane Nickel Finely divided to give a large surface area 140 °C It straightens the hydrocarbon chains in the oil molecules so they can pack together more effectively, increasing intermolecular (van der Waals’) forces between molecules which increase the melting point of the product compared with the oil, thus making the margarine a ‘spreadable’ solid.

6 a) b) c) d) e) f) g)

An electron-deficient species, capable of accepting a pair of electrons An organic ion containing a carbon atom carrying a positive charge A carbocation in which the carbon carrying the positive charge is bonded to one alkyl group A carbocation in which the carbon carrying the positive charge is bonded to three alkyl groups The ‘uneven’ breaking of a covalent bond, forming a positively charged ion and a negatively charged ion The reaction in which an electrophile bonds across the double bond in an alkene, forming a single product The electron-donating effect of alkyl groups which has the effect of reducing the charge density on the positively charged carbon atom in a carbocation

7 a) b) c) d)

C2H4Br2 C2H5Cl CH2FCHFCHFCH2F CH3CHBrCH3

8 a) (CH3)3C > CH 3C HCH3 > CH 3C H2 b) The tertiary carbocation has three alkyl groups donating electrons to its positively charged carbon atom, reducing the charge density and in effect spreading the positive charge around the whole ion, making it more stable. The secondary carbocation has one fewer alkyl groups and the primary has only one alkyl group exerting the inductive effect. +


+

+



6

15 The halogenoalkanes (haloalkanes) 1 nucleophiles; substitution; lone; electrons; carbon; alcohol, alkene; elimination; amine; nitrile; less; stronger 2 a)

H H

C

I

H

b) H

c) Cl

d) F

3 a) b) c) d) e) f)

Br

Br

H

H

H

C

C

C

C

C

H

H

H

H

H

Cl

H

H

C

C

C

Cl

H

H

H

H

F

Br

Br

H

C

C

C

C

H

H

H

H

H

CH3CH2CH2CH2CN + Cl ; 1-chlorobutane, pentanenitrile (CH3)2CHOH + Br ; 2-bromopropane, propan-2-ol C2H4 + H2O + Cl ; chloroethane, ethene CH3CN + I ; iodomethane, ethanenitrile CH3CH2NH2 + HBr; bromoethane, ethylamine (CH3CH2CH2)4N1Br ; 1-bromopropane, tetrapropylammonium bromide −

−

−

−

−

4 Spider diagram illustrating uses such as refrigerants, flame retardants, aerosol propellants, blowing agents for plastic manufacture; solvents, polymers, anaesthetics, paint strippers, dry cleaning 5 The larger relative formula mass, the stronger the intermolecular forces/van de Waals’ (London) forces between their molecules (as they contain more electrons within each molecule, therefore giving rise to more instantaneous dipoles and hence more induced dipoles on their neighbouring molecules, producing stronger forces of attraction between molecules) 6 a) b) c) d)

Heating under reflux To prevent the halogenoalkane reactant and the alcohol product evaporating off. A few anti-bumping granules CH3CH2CH2CH2Br + NaOH CH3CH2CH2CH2OH + NaBr or C4H9Br + NaOH C4H9OH + NaBr e) The OH ion acts as a nucleophile, attacking the partially positively charged carbon atom adjacent the bromine atom in 1-bromobutane. As the OH ion donates a pair of electrons to this carbon atom, forming a new covalent bond, the C Br bond weakens and breaks heterolytically, forming the alcohol as the Br ion breaks away from the halogenoalkane molecule. f) A tertiary halogenoalkane forms a carbocation first of all as its C Br bond breaks heterolytically, which is then attacked immediately by the OH ion to produce the tertiary alcohol. −

−

−

−




7

16 The alcohols 1 methanol; higher, hydrogen; hydrogen; sodium; ester; catalyse; aldehyde; carboxylic; ketones; Tertiary; nucleophilic; catalyst; dehydration/elimination; alkene 2 a) i) H

H

H

C

C

H

H

ii)

H

iii)

H

O H

H

O

H

C

C

C

H

H

H

H

H

H

H

O

O

H

C

C

C

H

H

H

H

b) Propane-1,2-diol > propan-2-ol > ethanol Because propane-1,2-diol can form hydrogen bonds from its two hydroxyl ( OH) groups as opposed to the one hydroxyl group on the other two alcohols so it has the strongest intermolecular bonds. Propan-2-ol has stronger intermolecular bonds than ethanol because it is a larger molecule than ethanol so in addition to its hydrogen bonding (present in both molecules), it has the stronger van der Waals’ (London) forces acting between its molecules. c) Yes, as they will form hydrogen bonds with water molecules, and the non-polar hydrocarbon chains are still small enough not to dominate the effects of this hydrogen bonding.

3 a) Oxidation b) First change – Add potassium dichromate solution, acidified with dilute sulfuric acid, dropwise to the ethanol and heat gently (distilling off the aldehyde as it forms). Second change – Add excess potassium dichromate solution, acidified with dilute sulfuric acid, and heat (under reflux). c) Ethanal, then ethanoic acid d) Propanone 4 a) b) c) d)

CH3CH2CH2O Na + ½ H2 CH3Cl + HCl CH3CHClCH3 + H2O 3CH3CH2I −

+

5 a) Propene and water (steam) b) Dehydration/elimination c) An ether 6 a) Fermentation b) C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) c) C2H4 + H2O C2H5OH Catalyst of conc. phosphoric(V) acid, 350 °C, 60 atm/6000 kPa d) Hydration of ethene by steam – ethene is made in the cracking of heavy fractions from crude oil, so the process uses a non-renewable resource as a raw material. Steam is a cheap raw material. The reaction is a reversible reaction but can be run continuously in industry. However, it needs a high temperature and pressure, again needing energy resources. Fermentation of glucose – fermentation occurs at relatively low temperatures and at atmospheric pressure, so is energy efficient but needs time to take place in a batch process. The glucose is extracted from plant material, e.g. sugar cane, sugar beet, which is a renewable resource.




8

17 Aromatic compounds 1 benzene; hexagonal; delocalised/six; plane; pi; electrophilic 2 a) Carbon dioxide and water b) C6H6 + 7.5O2 6CO2 + 3H2O or 2C6H6 + 15O2 12CO2 + 6H2O c) Smoky / dirty 3 a)

H

C

H C

C or

C

C C

H

H

H

b) Each of the six carbon atoms in the hexagonal ring forms 3 single/sigma bonds, to two other carbon atoms and a hydrogen atom. This leaves each carbon atom with 1 electron in a p atomic orbital. The lobes of these p orbitals overlap above and below the plane of the other atoms, and the 6 electrons become delocalised around the benzene ring. c) C6H6 + 3H2 y p l a h t n e

d) e)

–208 kJmol–1 C6H12

360 kJ mol

1

−

−

+ 3H2

y p l a h t n e

+ 3H2 –360 kJmol–1 –208 kJmol–1 C6H12

f) Higher g) 152 kJ mol

1

−

4 a) C6H5NO2 + H2O b) Nitration c) Nitronium ion/nitryl cation, NO2 ; reflux benzene and the nitrating mixture of concentrated nitric and concentrated sulfuric acids, not above 50 °C. d) HNO3 + 2H2SO4 NO2 + 2HSO4 + H3O +

+


−

+



1

5 a) C6H5Cl + HCl b) One of the Cl atoms in the Cl2 molecule donates a pair of electrons into the empty p orbital on the Al atom in AlCl 3. This draws electrons away from the other Cl atom, giving it a partial positive charge. (Alternatively, the bond between the 2 Cl atoms in Cl 2 which breaks heterolytically, forming a Cl ion.) This electron-deficient Cl acts as an electrophile attacking the benzene ring, forming a positively charged intermediate and leaving an [AlCl 4] ion. The AlCl3 is regenerated when the benzene ring’s delocalised pi-bonding system is re-established by releasing an H ion which accepts a pair of electrons from an Cl in the [AlCl 4] ion. +

−

+

−

6 a) C6H5CH2CH3 b) The HCl reacts with C2H4 in an addition reaction to form C 2H5Cl. Then: H –

+

CH3

C

Cl

AlCl3

CH3CH2

+

AlCl4

carbocation

H + AlCl4

+

CH2CH3 +

CH2CH3

–

CH2CH3

H


– +

H

CH2CH3

AlCl 4


+

HCl

+ AlCl3


2

18 Aldehydes and ketones 1 carbonyl; oxidised; Fehling’s; orange; oxidised; tetrahydridoborate(III); water/acid/alkali; primary; secondary; nucleophilic; dinitrophenylhydrazine; melting; condensation/addition-elimination 2 a)

O H

C H

b)

H H

c) H

H

H

H

H

C

C

C

C

C

C

H

O

H

H

H

H

H

H

H

H

C

C

C

C

H

H

H

H

H

O C H

3 In order of increasing melting points: Butane has the lowest melting point as it only has weak van der Waals’ (London) forces between its non-polar molecules. Propanone has stronger permanent dipole −dipole interactions between its molecules as well as temporary/ instantaneous dipole-induced dipoles (van der Waals’/ London forces) so has a higher melting point than butane. Then propan-1-ol comes next with its hydrogen bonding between molecules which can be thought of as stronger forms of dipole–dipole intermolecular forces. The highest melting point is ethanoic acid because its hydrogen bonding results in the formation of dimers (pairs of molecules) which effectively doubles its relative molecular mass resulting in much stronger van der Waals’/London forces. 4 a) Warm the ketone and the aldehyde mixed with Tollens’ reagent in separate boiling tubes in a water bath. The aldehyde will form a silver mirror on the inside of the boiling tube, the ketone will show no change. b) The aldehyde can be oxidised quite easily to a carboxylic acid by the silver ions present in Tollens’ reagent, reducing the silver ions to silver atoms – whereas the ketone cannot be readily oxidised. c) Add a few drops of the aldehyde or ketone to Brady’s reagent. Filter, wash and dry the yellow or orange precipitate formed and test its melting point. Check this against a table/database of known carbonyl derivatives of 2,4-dinitrophenylhydrazine to identify the specific aldehyde or ketone. d) NO CH3CH2

2

O + NH2NH

C

NO2

H

NO2

CH3CH2 C

N

NH

NO2 + H2O

H

5 a) b) c) d)

CH3CH2COOH CH3CH(OH)CH3 CH3CH2C(CN)(OH)CH2CH3 CH3CH2CH2HC NOH




3

6 a) Lithium tetrahydridoaluminate(III) in dry ether sodium tetrahydridoborate(III) in aqueous solution b) The secondary alcohol, propan-2-ol c) CH3COCH3 + 2[H] CH3CH(OH)CH3 d) H (from LiAlH4 or NaBH4) acts as a nucleophile, attacking the electron-deficient carbonyl carbon atom, forming a covalent bond by donating its pair of electrons to the carbon atom. Meanwhile, the oxygen atom in the carbonyl group draws a pair of electrons in the C O bond onto itself, gaining a negative charge. This negatively charged oxygen atom then forms an O H bond with one of the hydrogen atoms from a water molecule. −

CH3CH2C(CN)(OH)CH3 7 a) CH3CH2COCH3 + HCN b) Nucleophilic attack by the lone pair on the C atom in the CN ion on the carbonyl carbon attaches a CN group to it, while the carbonyl oxygen atom attracts a pair of electrons from the C O group onto itself, gaining a negative charge. Then this oxygen bonds to an H ion to form the hydroxyl group and 2-hydroxy-2-methyl-butanenitrile is formed. c) Butanal, because it only has one alkyl group donating electrons to the partially charged carbon atom of the carbonyl group, whereas butanone has the inductive effect of two alkyl groups. −

+




4

19 Carboxylic acids and related compounds 1 weak; large; molecules; hydrogen; dimer; higher; salt; carbonate; alcohol; strong; hydroxide; soap; anhydrides; water 2 a) Add lithium tetrahydridoaluminate(III)/LiAlH4 in dry ether, followed by water. b) Reduction c) Add methanol to propanoic acid in the presence of sulfuric acid and warm in a water bath/heat under reflux. d) Esterification e) Methanol CH3CH2CH2COCl + POCl3 + HCl 3 a) CH3CH2CH2COOH + PCl5 or CH3CH2CH2COOH + SOCl2 CH3CH2CH2COCl + SO2 + HCl b) i) CH3CH2COCl + H2O CH3CH2COOH + HCl ii) CH3COCl + CH3OH CH3CH2COOCH3 + HCl iii) CH3COCl + NH3 CH3CH2CONH2 + HCl or CH3COCl + 2NH3 CH3CH2CONH2 + NH4Cl iv) C6H5COCl + H3CH2NH2 C6H5CONHCH2CH3 + HCl

4 a) b) c) d) e)

CH3COCl + CH3COO Na (CH3CO)2O + NaCl (CH3CO)2O + H2O 2CH3COOH (CH3CO)2O + CH3OH CH3COOCH3 + CH3COOH (CH3CO)2O + NH3 CH3CONH2 + CH3COOH (CH3CO)2O + CH3CH2NH2 CH3CONHCH2CH3 + CH3COOH −

5 a) CH3COOCH2CH3 b)

+

+

NaOH

CH3COO Na −

+

+

CH3CH2OH

ethyl ethanoate + sodium hydroxide solution

heat (electric heating mantle could also be used as alternative heat source)

c)

RCOOCH2 RCOOCH

CH2OH +

3NaOH

RCOOCH2


3RCOO− Na+ +

CHOH CH2OH



5

6 a) CH3CH2COOH b) CH3CH2COOH c) O

+

H2O

H

CH3CH2COO

−

⇋

H3O+

+

O

CH3CH2C

CCH2CH3 O

H

O

d) Propanoic loses an H ion from its COOH group more easily than propan-1-ol loses an H ion from its OH group. This is because the presence of the extra O atom pulls electrons away from the O H bond in the COOH group, weakening it. Also, of the anions formed when an H ion is lost, the propanoate ion is less likely to attract H ions to act as a base and re-form the undissociated molecule, as the delocalisation of electrons around the COO group means that its negative charge is spread over all three atoms (and not concentrated on the end O atom as it is in the propoxide ion, CH 3CH2CO ). +

+

+

+

−

−




6

20 Organic nitrogen compounds 1 acids; ammonium; hydrogen; ammonia; reduction; reducing; tin; Diazonium; nitrous; 0; 5; amino; peptide/amide; hydrolysis; nucleotides; nitrogen 2 a) Ethylamine > ammonia > phenylamine b) The ability of the nitrogen atom to donate its lone pair of electrons and bond to an acidic H ion is enhanced by the inductive effect of the ethyl group in ethylamine, donating electrons to the N atom. On the other hand, the lone pair of the N atom in phenylamine is pulled towards the benzene ring as it becomes involved in extending the delocalised pi-bonding system; so the lone pair is less available to donate to an H ion than in ammonia or ethylamine. +

+

3 a) CH3CH2CH2NH2 + HBr b) A concentrated solution of ammonia in alcohol at 100 °C c) You do not just get the primary amine as the propylamine itself can react with bromopropane to give a secondary amine; then other substitution reactions can occur to form further substituted amines. d) Reduce nitropropane or propanenitrile with lithium hydridoaluminate(III)/LiAlH4 in dry ether, followed by water (or reduce with hydrogen gas over a nickel catalyst). e) Reduce nitrobenzene by refluxing with tin and concentrated hydrochloric acid in a boiling water bath. f) Step 1: C6H5NH2 + HNO2 + HCl C6H5N2 Cl + 2H2O Conditions: HNO2 made with chilled sodium nitrate( III) solution and concentrated hydrochloric acid, add slowly to chilled phenylamine − the temperature must not rise above 5 °C Step 2: C6H5N2+Cl + C 6H5O C6H5N2C6H4OH + Cl Conditions: Alkaline conditions/sodium hydroxide solution with phenol (to make phenoxide ion) added to the diazonium salt made in step 1 +

−

−

−

−

4 a) Add potassium cyanide/KCN and ethanol, then heat under reflux b) i) Propanoic acid ii) Propylamine c) i) Boil with dilute hydrochloric acid ii) Lithium hydridoaluminate(III)/LiAlH4 in dry ether, followed by water 5 a)

chiral

H

centre H2N

*C

COOH CH3

b) See label on 3D formula in part a) c) A racemic mixture/racemate




7

21 Polymers 1 polymers; monomers; poly(ethene); addition; condensation; any addition polymer, e.g. poly(ethene)/ poly(propene)/polystyrene/poly(phenylethene)/poly(chloroethene)/poly(tetrafluoroethene); any condensation polymer, e.g. nylon, polyester, polyamide, Kevlar, poly(lactic acid), DNA; Thermoplastics/Thermosoftening plastics; thermosets/thermosetting plastics; degradable; recycled 2 a) In a blocked cylinder of ethene b) Its polymer chains were highly branched and could not pack closely together. c) HDPE is made of straighter chain polymers than LDPE so its molecules can pack together more closely, resulting in stronger van der Waals’ forces. d) It is more rigid and has a higher softening temperature (so can be sterilised with boiling water). e) An impurity in a reaction vessel proved to be an essential component of the catalyst used. f) LDPE – e.g. plastic bags, toys HDPE – e.g. baby’s bottles, water pipes 3 a) temporary/instantaneous dipole-induced dipole interactions, resulting in van der Waals’/London forces b) hydrogen bonds (and van der Waals’/London forces) c) H H H O H H H C

N

C

C

N

H

O

H

H

H

H

H

O

N

C

C

C

C

H

O

H

C

C

N

C

C

H

O

H

H

H

H

N

C

N

C

C

H

H

O

H

d) Cross-links e) C f) C will be hard and rigid. B will have a greater tensile strength than A.

4 a) (C2H4)n b) Addition polymerisation c) H H H H H

H

C

C

C

C

C

C

H

H

H

H

H

H

d) Poly(propene) Isotactic – because all the methyl groups are on the same side of the chain e) H2C CHCH3 f) E.g. carpets, trawler nets, milk bottle crates

5 a) The reaction is started by a peroxide free radical initiator which reacts with ethene, breaking the double bond homolytically. A chain reaction follows as the alkyl free radicals bond with ethene molecules increasing the length of the free radical until a termination step occurs when two free radicals react together to complete a polymer chain. b) M CH CH M CH CH CH CH M CH CH CH CH CH CH 2

3

2

2

2

3

2

2

2

2

2

3

Ziegler catalyst

CH2


CH2

CH2

CH2

linear chain is produced with a Ziegler catalyst



8

6 a)

b) c) d e) f)

O

H

H

H

H

C

C

C

C

C

C

H

H

H

H

O

H

H

H

H

H

H

H

N

C

C

C

C

C

C

N

H

H

H

H

H

H

H

HOOCCH2CH2CH2CH2COOH or ClOCCH2CH2CH2CH2COCl and H 2NCH2CH2CH2CH2CH2CH2NH2 Water with the di-carboxylic acid or hydrogen chloride with the di-acyl chloride Condensation (or addition-elimination) Amide link Proteins

7 a) H2NCH2CH2CH2CH2CH2CH2NH2 and HOOCCH2CH2CH2CH2CH2CH2CH2CH2COOH or ClOCCH 2CH2CH2CH2CH2CH2CH2CH2COCl b) H H H H H H H H H H H H H H O H C

C

C

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

O

N

C

C

C

C

C

C

N

H

H

H

H

H

H

H

8 a) HOCH2CH2OH and HOOCC6H4COOH (or benzene ring drawn for C 6H4) b) Water c) O O

CH2

CH2

O

C

C n

O

d) Permanent dipole–dipole forces (and temporary/instantaneous dipole-induced dipole van der Waals’/ London forces) e) Drawing out the polymer from a narrow jet to strengthen the intermolecular forces as the chains align f) A thermosetting plastic (as a cross-link could form when two central OH groups from propane1,2,3-triol in repeating units of the adjacent polyester chains can react with a dicarboxylic acid monomer, joining the two chains together – this could happen at any points along the chains, forming a network of cross-links). g) It would be more rigid and heat resistant than Terylene.

9 a) H C

C

H

H

b) Poly(phenylethene) (or polystyrene) c) Addition d) If the bulky benzene rings are all aligned on the same side or arranged randomly, the polymer chains could be bent/curved compared with the straight chain with the benzene rings alternating regularly on one side then the other. The straighter chains could pack together more tightly, giving this plastic a higher density and a higher softening temperature.




9

10 a) E.g. any nylon b) Its linear polymer chains align next to each other with extensive hydrogen bonding holding them in position. c) E.g. bullet-proof vests, tennis rackets; fuel tanks; fire-proof protective clothing, racing ‘leathers’ d) The synthetic polymers have fewer types of monomer (usually two) giving very regular chains with a repeating unit compared with the different amino acids in a protein which can have very intricate shapes and join the numerous combinations (the primary structure of proteins show the sequence of amino acids so there is no repeating unit). e) They both polymerise via amide (peptide)/ CONH links and can from hydrogen bonds between amide links. 11 Discussion to include how chemists have made plastics that are biodegradable, photodegradable, soluble in water, recyclable/reusable, incineration using scrubbers to remove acidic gases and efficient, effective burners to ensure complete breakdown of plastics.




10

22 Organic synthesis 1 a) E.g. add concentrated hydrogen bromide to the ethene to form bromoethane. b) E.g. add concentrated hydrochloric acid to the ethene to form chloroethane. Then add excess ammonia solution to the chloroethane to form ethylamine. c) E.g. add steam in presence of conc. phosphoric acid catalyst to make ethanol. Then oxidise the ethanol with acidified (dil. sulfuric acid +) potassium dichromate solution dropwise, warm and distil off the ethanal formed. d) E.g. add steam in presence of conc. phosphoric acid catalyst to make ethanol. Then completely oxidise the ethanol by refluxing with excess acidified (dil. sulfuric acid +) potassium dichromate solution to make ethanoic acid. Add phosphorus( V) chloride to convert to ethanoyl chloride. e) E.g. add steam in presence of conc. phosphoric acid catalyst to make ethanol. Then completely oxidise the ethanol by refluxing with excess acidified (dil. sulfuric acid +) potassium dichromate solution to make ethanoic acid. Add phosphorus( V) chloride to convert to ethanoyl chloride, followed by addition of ammonia solution to make ethanamide. f) E.g. add concentrated hydrogen bromide to the ethene to form bromoethane. Then boil bromoethane under reflux with potassium cyanide dissolved in ethanol to form propanenitrile. Then hydrolyse the propanenitrile by boiling/heating under reflux with hydrochloric acid, to form ethanoic acid. This can then be reduced by lithium hydridoaluminate(III)/LiAlH4 in dry ether, followed by water to form ethanol in order to make propan-1-ol. g) E.g. add steam in presence of conc. phosphoric acid catalyst to make ethanol. Then completely oxidise the ethanol by refluxing with excess acidified (dil. sulfuric acid +) potassium dichromate solution to make ethanoic acid. Add phosphorus( V) chloride to convert to ethanoyl chloride, followed by addition of conc. ammonia solution to make ethanamide. Finally, add bromine and concentrated sodium hydroxide solution to remove a carbon atom and obtain methylamine. 2 a) A = sodium ethanoate, CH 3COO Na B = ethanoyl chloride, CH3COCl C = ethanamide, CH3CONH2 D = ethanoic acid, CH3COOH E = ethanol, CH3CH2OH F = ethyl ethanoate, CH 3COOCH2CH3 b) Ethyl ethanoate c) Hydrolysis d) i) Dehydration / elimination ii) Reduction / redox −

3 1 2 3

+

Reflux with red phosphorus and iodine/phosphorus( III) iodide = boil/reflux with hydrochloric acid = sodium hydroxide solution =

4 See answer on page 451 in Advanced Chemistry for You (2nd edition).




11

23 Organic analysis 1 empirical; mass; relative; molecular; molecular; mass; radiation/waves; visible; compared; database; finger-printing For answers to Q2 and 3 see page 451 in Advanced Chemistry for You (2nd edition). 4 a) b) c) d)

CH3COCH2CH3 [CH3COCH2CH3] It enables us to work out the molecular formula. The masses measured to a greater degree of accuracy for each atom mean that there are small, but detectable, differences in the relative molecular masses of different compounds that are unique to that compound (and any isomers of the compound). e) i) [CH3CO] ii) [CH3COCH2] iii) [CH3CH2] f) Peaks at m/z values of 72 and e.g. 14, 17, 27, 45, 55, 58 ∙

+

+

+

5 a) CnH(2n 2) b) 114 c) C8H18, with regularly-spaced peaks as the hydrocarbon chain fragments producing ions with different numbers of CH2 units +

6 a) Violet/indigo as the low end of the visible spectrum (red, orange, yellow and blue) is all absorbed by the compound b) The absorption bands are too broad for accurate ‘finger-printing’/not all organic compounds absorb energy in this range of the electromagnetic spectrum 7 a) Each organic compound has many characteristic peaks that can be used to finger-print compounds/each compound has a unique IR spectrum. b) They absorb energy/radiation that corresponds to the resonance frequency of their bonds − which causes the bonds to vibrate at their characteristic frequencies c) X = O H Y = C H d) IR spectra are most useful for determining the bonds/functional groups present so you need a combination of analytical techniques to be sure of the actual structure of a new compound. 8 a) A = CH3OCH3 B = CH3CH2OH b) In A we have peaks caused by [CH3] at 15, [CH3O] at 31, and [CH 3OCH2] at 45, as well as the molecular ion at 46. In B we have peaks caused by [CH3] at 15, [CH3CH2] at 29, and [CH 3CH2O] at 45, as well as the molecular ion at 46. c) In A all the H atoms/protons are equivalent so there is only one peak whereas in B there is a peak for the CH3, the CH2 and the OH H atoms/protons. d) The relative numbers of the H atoms/protons in different molecular/chemical environments. e) The first ( OH) peak at chemical shift 5p.p.m. would be a singlet, the next ( CH2) peak a quartet, and the largest peak ( CH3) would be a triplet. +

+


+

+

+


+


12

24 Enthalpy changes 1 exothermic; rises/increases; less; endothermic; fall/decrease; positive; 298; 100/101; combustion; change; oxygen; standard NaCl(aq) + H2O(l) 2 a) NaOH(aq) + HCl(aq) For answers to Q2 b), c) and d) see page 451 in Advanced Chemistry for You (2nd edition). e) data logger temperature probe

temp /˚C

thermometer

lid

lid time /s

polystyrene beaker

polystyrene beaker

50.0 cm3 of 1.0 mol dm–3 NaOH (sodium hydroxide) + 50.0 cm3 of 1.0 mol dm–3 HNO3 (nitric acid)

50.0 cm3 of 1.0 mol dm−3 NaOH (sodium hydroxide) + 50.0 cm3 of 1.0 mol dm−3 HNO3 (nitric acid)

f) Measurement uncertainty associated with volume of solutions (if volumes are measured out are actually lower than 100 cm3, the value calculated for enthalpy change will be too low and vice versa) and temperature change (if temperature change measurement is too high the value calculated for enthalpy change will be too high and vice versa). There is also the error in the temperature readings due to energy losses to the surroundings (this will produce a value that is too low; although the temperature change error can be adjusted using a graph as shown on page 317).

3 a)

thermometer

to water pump thermometer

stirrer water

copper calorimeter

water alcohol

block of wood

fuel

For answers to Q3 b), c) and d) see page 451 in Advanced Chemistry for You (2nd edition). e) Energy losses to the surroundings, as not all the energy from the burning fuel will be transferred to the water in the calorimeter (plus some incomplete combustion).

4 a) Enthalpy of atomisation of nitrogen b) Enthalpy of combustion of propane c) Enthalpy of formation of propane




1

5 a) See page 451 in Advanced Chemistry for You (2nd edition). b) 0

3C + 4H2 (+5O2)

C3H8 (+5O2) y p l a h t n e

3CO2 + 4H2O

reaction pathway

6 a) See page 451 in Advanced Chemistry for You (2nd edition). b) C(diamond) + O2

C(graphite) + O2

y p l a h t n e

–395.4 kJ mol–1 –393.5 kJ mol–1

CO2

reaction pathway

c) As graphite has the lower enthalpy value of the two allotropes, it is chosen as the standard form of the element carbon and given an enthalpy value of zero. For Q7 and Q8 see page 451 in Advanced Chemistry for You (2nd edition).

9 a) See page 451 in Advanced Chemistry for You (2nd edition). b) The bond enthalpy values are average values from a range of molecular environments so calculating an enthalpy change for a specific reaction will only provide an approximate answer (and calculation is only valid if all substances are in their gaseous state).




2

25 Lattice enthalpy 1 change; one; gaseous; ionic; negative; Haber; standard; Cl −(g); NaCl(s)/Na+Cl−(s); Na(s); ½ Cl2(g); ∆H 3; ∆H 2, formation; atomisation; ionisation; atomisation; affinity 2 a) Li+(g) + F−(g) b) Li + (g) + F −(g)

 H2

LiF(s)/Li+F−(s) 

 Hlattice

K + Br −(s)

 H3

Li(s) + ½F2(g)

c) Enthalpy of formation of lithium fluoride = −616 kJ mol−1 Enthalpy of atomisation of lithium = 159.4 kJ mol−1 Enthalpy of atomisation of fluorine = 79.0 kJ mol−1 1st I.E. of Li = 520 kJ mol−1 1st electron affinity of F(g) = −328 kJ mol−1 d) See page 452 in Advanced Chemistry for You (2nd edition).

3 a) See page 452 in Advanced Chemistry for You (2nd edition). b) i) The greater magnitude in the Born–Haber cycle values (more exothermic) than the theoretical values arise because of the polarisation of the iodide ion by the cations, giving rise to a degree of covalency in the bonding between the oppositely charged ions. ii) The Be2+ ion is much smaller than the Sr2+ ion so it has a greater charge density, which results in greater polarisation of the iodide ion in beryllium iodide compared with strontium iodide and so the greater the discrepancy between the actual and theoretical values for the lattice enthalpies. 4 a) See page 452 in Advanced Chemistry for You (2nd edition). b) There is some/a small degree of covalent character in the ionic bonding in sodium oxide.




3

26 Entropy and free energy 1 disorder; system/surroundings; surroundings/system; positive; increase; Gibbs; T; negative 2 The gases, with their random arrangement and motion, have the most entropy/disorder/ways in which their energy can be distributed. In liquids, there is less disorder, as shown by the values for steam and water, and the least disorder is found in the two solids, with their particles fixed in position, arranged in giant lattices with much more limited movement possible. 3 a) See page 452 in Advanced Chemistry for You (2nd edition). b) The entropy change of the system has a negative value because there are 3 moles of reactants in the gaseous state converted into only 2 moles of gaseous products so the disorder in the system decreases. c) and d) See page 452 in Advanced Chemistry for You (2nd edition). e) No, because the total entropy change for the reaction is negative. f) See page 452 in Advanced Chemistry for You (2nd edition).




4

27 Kinetics – rates of reaction 1 temperature; concentration; pressure; surface area; light; collision; concentration; volume; more; increases; large/great/double; collisions; reaction; effective/successful; Biological; protein; active; denatured; pH 2 a) Collect the gas in a gas syringe and measure the volume of carbon dioxide given off over time. b) Colorimeter, as the iodine formed causes the solution to become brown/coloured or measure change in conductivity over time as the number of ions in solution decreases as the reaction proceeds 3 a) CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) b) No c) Student graph, filling more than half the area of the graph paper, axes fully labelled, sensible scale chosen, points plotted accurately, smooth curves through points. d) Calcium carbonate was in excess, because the volume of CO2 given off was halved when the concentration of the hydrochloric acid was halved./If the calcium hydrochloric acid had been in excess, the volume of gas given off in each experiment would have been the same (assuming the mass of CaCO 3 was kept constant in both experiments). e) Steeper line, levelling off at 0.80 g of CO2 gas 4 a) number of particles

T ˚C

(T + 10) ˚C

the number of particles with energy greater than E a has roughly doubled for a 10 ˚C rise in temperature

activation energy

E a

energy

At the higher temperature the reacting particles have more energy and will move around faster, resulting in more frequent collisions but, more importantly, there will also be a much higher proportion of those collisions with energy greater than the activation energy, i.e. a much higher rate of successful/effective collisions (as shown by the area under the curve increasing at the higher temperature on the Maxwell−Boltzmann distribution). b) number of particles

all these extra particles in the area marked now have sufficient energy to react once the catalyst is present

new E a with catalyst

E a

energy

Using a catalyst lowers the activation energy, meaning that more of the collisions in any given time between reacting particles result in a reaction/are successful/effective (as shown by the area under the curve increasing when the value of Ea is decreased on the Maxwell −Boltzmann distribution). Hence the rate of reaction increases with a catalyst. c) Homogeneous catalysis occurs in a reaction in which the reactants and the catalyst are in the same state or phase. Fe2+(aq) e.g. 2I−(aq) + S2O82−(aq) I2(aq) + 2SO42−(aq) Heterogeneous catalysis occurs in a reaction in which the reactants and the catalyst are in different states or phases. MnO2(s) e.g. 2H2O2(aq) 2H2O(l) + O2(g) Oxford University Press



5

5 a) E.g. iron/iron(III) oxide to make ammonia; vanadium( V) oxide to make sulfuric acid/sulfur trioxide/sulfur( VI) oxide; Pt/Rh to make nitric acid b) The hydrogenation straightens the hydrocarbon chains in the oils, enabling them to pack together more closely with stronger van der Waals’ (London) forces, increasing the melting point of the oil and turning it into a ‘spreadable’ solid. c) They enable carbon monoxide and nitrogen oxides to react together on their honeycomb structure in the exhaust system of a vehicle, the carbon monoxide being oxidised to carbon dioxide and the nitrogen oxides being reduced to nitrogen. d) They convert carbon monoxide to the greenhouse gas carbon dioxide/do not remove carbon dioxide from exhaust gases. 6 a) b) c) d) e)

Proteins that act as biological catalysts. E.g. catalase breaking down toxic hydrogen peroxide in the liver The shape of their active site changes so substrate molecules can no longer bind to it and react. Fermentation C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)

7 a) Diagram of magnesium and sulfuric acid in a conical flask attached to a gas syringe/inverted measuring cylinder with gas collecting over water b) Hydrogen; a lighted splint will produce a squeaky ‘pop’ c) Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g) d) Student graph, filling more than half the area of the graph paper, axes fully labelled, sensible scale chosen, points plotted accurately, smooth curve through points. e) Steeper curve, levelling off at 80 cm3 f) The smaller pieces have the larger surface area, so more magnesium atoms are exposed to attack by the H+(aq) ions, resulting in more frequent collisions between them and a faster rate of reaction. g) Lower the temperature; use less concentrated acid; use a lump of magnesium with less surface area than the ribbon h) About 1.0 cm3 s−1 (+ or −0.1 cm3 s−1)




6

28 Rate equations 1 concentration; experiments; rate; k; varies; higher; increase; powers; 3; first/1st; second/2nd; zero; mechanism; slowest; determining; determining equation 2 a) The rate constant; mol−2 dm−6 s−1 b)−d) See page 452 in Advanced Chemistry for You (2nd edition). e) i) rate

[Y]

ii) rate

[Z]

3 a) Student graph, filling more than half the area of the graph paper, axes fully labelled (with ‘Initial rate/ mol dm−3 s−1 up the vertical (y) axis and concentration N 2O5/mol dm−3 along the horizontal ( x) axis), sensible scale chosen, points plotted accurately, straight line through points. b) As the concentration of N2O5 doubles, the reaction rate doubles. c), d) See page 452 in Advanced Chemistry for You (2nd edition). e) Substitute the values from one of the experiments to determine the initial rate into the rate equation. f) and g) See page 452 in Advanced Chemistry for You (2nd edition). h)

[N2O5] / 10–3 mol dm–3

time

i) The decay curve will have a constant half-life./The time for the [N2O5] to fall to half its value will be constant.




7

4 a) The brownish colour of the iodine will fade as the reaction proceeds. b) i) 1, 2 and 3 ii) 1, 6 and 7 iii) 1, 4 and 5 c), d) (Changing the concentration of iodine has no effect on the rate) and e) – See page 452 in Advanced Chemistry for You (2nd edition). f) No, because iodine must only be involved in fast steps in the mechanism of the reaction as it is zero order with respect to iodine g) See page 452 in Advanced Chemistry for You (2nd edition). 5 a) As the volume of the reacting mixture changes, the level of the liquid in the narrow-bore glass tube changes and can be measured on the scale next to the tube as time passes. b) See page 452 in Advanced Chemistry for You (2nd edition). 6 a) and b) See page 452 in Advanced Chemistry for You (2nd edition). c) Colorimeter, as the bromine formed causes the solution to become orange/yellow or measure change in conductivity over time as the number of ions in solution decreases as the reaction proceeds or titration, e.g. quenching reaction ,then finding concentration of acid at different times 7 a) The C Br bond in RBr breaks heterolytically, with Br taking both electrons to form a Br− ion, and leaving a positively charged tertiary carbocation. This carbocation is immediately attacked by an OH − ion to form the alcohol ROH. b) The C Br bond in RBr starts breaking as the OH− ion starts forming a bond by donating a pair of electrons to the partially positively charged C atom in RBr. This forms an intermediate in which the C Br bond breaks and the C O bond forms to make the alcohol ROH. c) See page 452 in Advanced Chemistry for You (2nd edition). d) The rate equation would be 1st order with respect to 2-methyl-2-bromopropane (a tertiary halogenoalkane), i.e. as we double its concentration the rate of reaction doubles. The formation of the carbocation from 2-methyl-2-bromopropane is the rate determining (slowest) step in the reaction mechanism. However, it would be zero order with respect to the hydroxide ions, as the hydroxide ions are involved in the fast step, attacking the carbocation rapidly as soon as it is formed.




8

29 Equilibrium mixtures 1 reversible; products; equilibrium; product; constant; forward/reverse; reverse/forward; dynamic; position; change; catalyst; established/achieved/reached; position 2 a) b) c) d)

Shift to the left/reactants, as this produces fewer moles/molecules of gas and will reduce the pressure Shift to the left/reactants, as this will reduced the concentration of hydrogen Shift to the right/products, as this will lower the temperature as the forward reaction is endothermic No effect on the position of equilibrium

3 a) Solution will turn orange as more dichromate( VI) ions are formed because the position of equilibrium shifts to the right to remove H (aq) ions. b) Solution will turn yellow as more chromate( VI) ions are formed because the position of equilibrium shifts to the left to replace H (aq) ions lost when they react with the OH (aq) ions added, forming water. +

+

−

4 a) The amount of solid increases as more PCl5 is formed in order to reduce the pressure of the system b) The enthalpy change of the forward reaction is endothermic as the position of equilibrium has shifted to the right when the temperature was increased. 5 a) b) c) d) e) f)

Position of equilibrium shifts to the left Position of equilibrium shifts to the left Position of equilibrium shifts to the right Position of equilibrium shifts to the right No effect on the position of equilibrium Position of equilibrium shifts to the left

6 a) nitrogen + hydrogen ⇋ ammonia N2(g) + 3H2(g) ⇋ 2NH3(g) b) Air for nitrogen and natural gas/methane/steam for hydrogen c) Iron (and iron(iii) oxide/other metal oxides) d) 1.5 × 104 − 3.0 × 104 kPa and 400−450 °C e) A high pressure will favour the production of more ammonia as there are fewer moles/molecules of gas on the right hand side of the equation. A low temperature will favour a higher yield of ammonia as the forward reaction exothermic, but quite a high temperature is used to speed up the rate of the reaction. 7

NH3 1 mole ?

+

HNO3

NH4NO3

1 mole ?

1 mole 2400 tonnes

Therefore mass of ammonia needed

=

2 400 000 000 (4 × 1.0) + (2 × 14.0) + (3 × 16.0) moles

=

2 400 000 000 moles 80

=

(

=

510 000 000 g =

=

000 000 (2 40080 )

=

1 890 000 000 g =

and mass of nitric acid needed


2 400 000 000 80

)

(14.0

×

+

3.0) g

510 tonnes

[(1.0

×

+

14.0

+

(3

×

16.0)] g

1890 tonnes



1

30 Equilibrium constants 1 constant; [G]c; [E]a; [F]b; gases; pressures; py(Z); ps(T); products; equilibrium; composition; K c; position; K c; K p; constant; decreased; catalyst; constant; unchanged/not affected

[Fe(SCN)]2 [Fe3 ] [SCN ] b) See page 452 in Advanced Chemistry for You (2nd edition). c) It lies well over to the right hand side/products. +

2 a) K c

3 a)

b) c) d)

=

+

−

p2(NO) K p = p(N2) × p(O2) p2(NO2) = K p p(N2O4) p2(HBr) K p = p(H2) × p(Br2) See page 452 in Advanced Chemistry for You (2nd edition).

4 [Ag(NH3)2] is more stable compared with [Ag(H 2O)2] . Because the stability constant is so large that in effect the reaction goes to completion (as the ammonia complex ion would be on the top line of the expression for the equilibrium constant given). +

5 a) K p

=

+

p2(NH3) p(N2) × p3(H2)

b) and c) See page 452 in Advanced Chemistry for You (2nd edition). d) It strongly favours the reactants/left hand side/nitrogen + hydrogen. e) The lower temperature will favour a higher yield of ammonia as the forward reaction is exothermic, but the rate at which ammonia would be formed at 298 K would be much too slow. [HI]2 [H2] × [I2] b) As there are the same number of moles of reactants and products so the units in the equilibrium expression cancel out completely. c) and d) See page 452 in Advanced Chemistry for You (2nd edition).

6 a) K c

=




2

31 Acid/base equilibria 1 hydrogen/H ; accept; conjugate; conjugate; Cl ; strong; dissociate/ionise; dissociation/equilibrium; weaker; concentration; −log[H (aq)]; indicator; titration; buffer; weak; salts; alkali +

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2 a)

Test

Result with aqueous ethanoic acid

Result with ethanoic acid in methylbenzene

Blue litmus paper

Turns red

Stays blue

Sodium carbonate

Gives off CO2 gas

No reaction

Magnesium ribbon

Gives off H2 gas

No reaction

Electrical conductivity Can conduct electricity

Does not conduct

b) Water enables ethanoic acid to display its acidic properties because it acts as a base, accepting an H ion from the carboxylic acid group/ COOH, so a small proportion of ethanoic acid molecules dissociate, forming ethanoate ions/CH 3COO (aq) and H3O (aq)/H (aq) ions. c) Any test, e.g. add sodium hydrogencarbonate/universal indicator/pH meter/pH sensor (probe) For questions 3 and 4 see answers on page 463 in Advanced Chemistry for You (2nd edition). +

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5 a) A small percentage of water molecules dissociate, according to the reaction: H2O(l) ⇋ H (aq) As K w of water is 1.0 × 10 14, = [H (aq)] [OH (aq)], the concentration of the H (aq) ions must be 1.0 × 10 7 mol dm 3. Therefore the pH of water at 25 °C is −log(1.0 × 10 7) = 7.0 b) In water the concentration of the OH (aq) is also 1.0 × 10 7 mol dm-3. The ionic product is the equilibrium constant for water given by the expression: K w = [H (aq)] [OH (aq)] 7 mol dm 3) × (1.0 × 10 7 mol dm -3) = (1.0 × 10 14 mol2 dm 3 = 1.0 × 10 c) See answers on page 452 in Advanced Chemistry for You (2nd edition). +

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OH (aq)

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6 a) The expression for the equilibrium constant when an acid is added to/dissolves in water. For example: HCN(aq) ⇋ H (aq) + CN (aq) Giving the acid dissociation constant, K a: [H (aq)] [CN aq)] [HCN] b) Hydrocyanic acid/HCN – because it has the smallest dissociation constant (so has the least proportion of its molecules dissociated in aqueous solution) c) See answers on page 452 in Advanced Chemistry for You (2nd edition). +

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7 a)

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strong acid / strong base

b)

strong acid / weak base

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pH

pH

7

7

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0

volume of base added / cm3





3

c)

d)

weak acid / strong base

weak acid / weak base

14

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pH

pH

7

7

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e) It is the pH of a solution where there is an equal balance of an indicator’s coloured species in a solution i.e. the point where it changes colour (although this appears to occur over a range of pH values). f) pK Ind gives us the pH at which an indicator has its equivalence point which should coincide a rapid change in the pH values in an acid −base pH curve if that indicator is suitable for determining the end point of the titration. For example, in a titration between a strong acid and a weak base, the rapid change in pH occurs over the region below pH 7 – so an indicator with a p K Ind value above 7, such as phenolphthalein, pK Ind = 9.4, will not be suitable for the titration, whereas methyl red, p K Ind = 5.1, will be. Alternatively, this can be shown diagrammatically on a pH curve. g) Strong acid with a weak base/b) (as well as a) where any acid–base indicator is suitable) 8 a) In the ammonia/ammonium chloride buffer we have the equilibrium: NH3(aq) + H (aq) ⇋ NH4 (aq) As ammonia is a weak base we have a high proportion of unprotonated ammonia molecules which will be available to accept H (aq) ions from the addition of acid to the buffer solution, minimising the lowering of the pH. The buffer also contains NH4 (aq), mainly from the ammonium chloride salt, which are ready to replace any H (aq) ions removed from the solution by the addition of a base (as the OH (aq) ions from the base react with H ions in the equilibrium mixture above to form water). The donation of H (aq) ions from the ammonium ions minimises the rise in pH on addition of the base. b) See answers on page 452 in Advanced Chemistry for You (2nd edition). +

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Advanced Chemistry for You Answers

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