Section 1.1 Section 1.2 Section 1.3 - The Student Room

SECTION 1 

Section 1.1 

1 a 2.0 b 5.3 c 1.3 d 10.0 e 5.0 

f 50.2 

2 a 144 b Neodymium 

3 

Mass of Amount of Number of 

sample/g sample/mol atoms 

– – 6.02 ¥ 10 23 

– 2.00 – 

56.0 – 6.02 ¥ 10 23 

80.0 – 12.04 ¥ 10 23 

63.5 0.50 – 

4 a 1 b 0.5 c 0.25 d 0.1 e 0.25 

f 0.5 g 0.25 h 0.1 i 2 j 5 

Atoms of copper are approximately twice as heavy as 

atoms of sulphur. Thus the same mass contains only half 

as many moles of copper as it does of sulphur. 

5 Black copper(II) oxide (CuO) contains equal numbers of 

copper and oxygen particles (Cu 2+ and O 2– ions). Red 

copper(I) oxide (Cu 2 

O) contains twice as many copper 

particles as oxygen particles (Cu + and O 2– ions). 

6 a The mass of the sample is needed to be sure that 

iodine and oxygen are the only elements in the 

compound. 

b The relative number of moles of iodine and oxygen. 

c To change the relative number of moles into the ratio 

of moles of oxygen relative to 1 mole of iodine. 

d In order to produce a ratio involving whole numbers 

e I 2 

O 5 

, I 4 

O 10 

, I 6 

O 15 

, etc. 

f The molar mass is needed. 

7aH 2 

O b CO c CS 2 

d CH 4 

e Fe 2 

O 3 

f CuO g CaO h SO 2 

i MgH 2 

8a92.3 b 7.7 c CH 

9aSiH 4 

b CO c CO 2 

d MgO e C 2 

H 6 

O 

f CaCO 3 

g HClO 3 

h NaHCO 3 

10 a CH 2 

b P 2 

O 3 

c AlCl 3 

d BH 3 

e C 4 

H 5 

f C 3 

H 4 

g CH 2 

O h C 12 

H 22 

O 11 

11 a H 2 

O 2 

b CO c C 2 

H 2 

d C 6 

H 6 

e C 6 

H 12 

12 a 2 b 11 c 2 d 10 e 2 

13 a 30 b 78 c 130 d 100 e 158 

f 242 g 132 

14 a 2 b 4 c 10 d 0.02 e 5 

f 1 ¥ 10 6 


1 a 2Mg + O 2 

Æ 2MgO 

b 2H 2 

+ O 2 

Æ 2H 2 

O 

c 2Fe + 3Cl 2 

Æ 2FeCl 3 

d CaO + 2HNO 3 

Æ Ca(NO 3 

) 2 

+ H 2 

O 

e CaCO 3 

+ 2HCl Æ CaCl 2 

+ CO 2 

+ H 2 

O 

f H 2 

SO 4 

+ 2NaOH Æ Na 2 

SO 4 

+ 2H 2 

O 

g 2HCl + Ca(OH) 2 

Æ CaCl 2 

+ 2H 2 

O 

h 2Na + 2H 2 

O Æ 2NaOH + H 2 

i CH 4 

+ 2O 2 

Æ CO 2 

+ 2H 2 

O 

j 2CH 3 

OH + 3O 2 

Æ 2CO 2 

+ 4H 2 

O 

2 a 2Ca + O 2 

Æ 2CaO 

b Ca + 2H 2 

O Æ Ca(OH) 2 

+ H 2 

c C + CO 2 

Æ 2CO 

d N 2 

+ 3H 2 

Æ 2NH 3 

e C 3 

H 8 

+ 5O 2 

Æ 3CO 2 

+ 4H 2 

O 

3 a Zn(s) + H 2 

SO 4 

(aq) Æ ZnSO 4 

(aq) + H 2 

(g) 

b Mg(s) + 2HCl(aq) Æ MgCl 2 

(aq) + H 2 

(g) 

c MgCO 3 

(s) Æ MgO(s) + CO 2 

(g) 

d 2C 2 

H 6 

(g) + 7O 2 

(g) Æ 4CO 2 

(g) + 6H 2 

O(l) 

e BaO(s) + 2HCl(aq) Æ BaCl 2 

(aq) + H 2 

O(l) 


1 a All the magnesium reacts. 

b So that we know the number of moles of each 

substance involved in the reaction. 

c Mass of 1 mole of magnesium oxide. 

d Because 2 moles of magnesium oxide are produced. 

e To find the mass of magnesium oxide produced from 

1 g of magnesium. 

f 80/48 would be multiplied by 50 rather than by 6. 

2 20 g 

3 a 2.8 g b 3.1 g c 2.5 g 

4 3667 g (3.667 kg) 

5 a C 8 

H 18 

+ 12.5O 2 

Æ 8CO 2 

+ 9H 2 

O 

b 175 kg 

c 154 kg 

6 a 56 tonnes 

b S + O 2 

Æ SO 2 

c 64 g 

d 64 tonnes 

e 2 tonnes 

f 112 tonnes 

7 a 217 tonnes 

b 0.58 tonnes 

8 a Fe 2 

O 3 

+ 3CO Æ 2Fe + 3CO 2 

b 160 g 

c 1.43 g 

d 1.43 tonnes 

e 2.86 tonnes 

f 11.9 tonnes 

162

SECTION 2 

Section 1.4 

1 The particles in a gas are much further apart than in a 

liquid or solid. In a gas, therefore, the volume of the 

particles is a very small part of the total volume and does 

not significantly affect it. In a liquid or solid the particles 

are close together and their volumes must be taken into 

account when deciding on the total volume. 

2 a CH 4 

(g) + 2O 2 

(g) Æ CO 2 

(g) + 2H 2 

O(g) 

b Volume of oxygen is twice that of methane. 

c The volume of water vapour formed is twice the 

volume of methane burnt. 


1 a i 0.02 dm 3 ii 1.5 dm 3 

b i 220 000 cm 3 ii 1600 cm 3 

2 There is 0.4 mole of sodium hydroxide dissolved in every 

dm 3 of the solution. 

3 a 0.5 b 0.4 c 1 d 0.2 e 0.05 f 0.002 

4 a 2 b 2 c 5 d 0.2 e 4 f 0.2 

5 a 0.25 b 0.2 c 5 d 0.4 e 0.5 f 0.125 

6 a 40 g b 4g c 20 g d 0.4 g e 800 g f 1g 

7 a 117 g b 3.95 g c 1.4 g d 9930 g 

e 0.0024 g f 2.385 g g 0.0126 g h 0.1825 g 

i 25 .0 g j 13.9 g 

8 

Concentration/g dm –3 Concentration/mol dm –3 

– 5.15 

– 1.51 

31.5 – 

13.4 – 

– 0.174 

– 0.065 

0.6 – 

3 a H 2 

(g) + Cl 2 

(g) Æ 2HCl(g) 

b Volumes of hydrogen and chlorine are the same. 

Volume of hydrogen chloride is twice the volume of 

hydrogen or chlorine. 

4 a 2H 2 

(g) + O 2 

(g) Æ 2H 2 

O(l) b 5cm 3 

5 a C 3 

H 8 

(g) + 5O 2 

(g) Æ 3CO 2 

(g) + 4H 2 

O(l) 

b 500 cm 3 c 300 cm 3 

6 a 3 b 3 c 4 d 2 e C 3 

H 4 

7 0.414 dm 3 (414 cm 3 ) 

8 a 0.25 b 2 c 48 dm 3 d 240 dm 3 e 30 dm 3 

9 Na + Cl – 

Na + 2– 

CO 3 

Ag + – 

NO 3 

Mg 2+ Br – 

H + SO 2– 4 

(or HSO – 4 

) 

10 a 1moldm –3 b 0.02 mol dm –3 

c 0.3 mol dm –3 d 0.4 mol dm –3 

11 a 0.0019 mol b 0.0019 mol 

c 0.076 mol d 0.076 mol dm –3 or 2.77 g dm –3 

12 a 0.0022 mol b 0.0044 mol 

c 0.176 mol d 0.176 mol dm –3 

13 a 7.75 ¥ 10 –4 (0.000 775) mol dm –3 b 0.0574 g dm –3 

14 a 0.0025 b 0.0025 c 25 cm 3 

15 a 0.02 mol b 0.04 mol c 1dm 3 d 500 cm 3 

e 20 cm 3 


1 3 a A r 

(Br) = 80.0 

Isotope Symbol Atomic Mass Number of 

b A r 

(Ca) = 40.1 

number number neutrons 

2 

carbon-12 

12 

6 

C 6 12 6 

carbon-13 

13 

6 

C 6 13 7 

oxygen-16 

16 

8 

O 8 16 8 

strontium-90 

90 

38 

Sr 38 90 52 

131 

iodine-131 

53I 53 131 78 

123 

iodine-123 

53I 53 123 70 

a 

b 

c 

d 

Protons Neutrons Electrons 

35 44 35 

35 46 35 

17 18 17 

17 20 17 

4 a 100 – x 

b 193x 

c 191(100 – x) 

d 193x + 191(100 – x) 

e [193x + 191(100 – x)] ∏ 100 

f 60% iridium-193, 40% iridium-191 

5 37.5% antimony-123, 62.5% antimony-121 

6 23.5% rubidium-87 

163

SECTION 2 


1 a 238 234 

94Pu Æ 

92 U + 4 2 He 

b 221 217 

87Fr Æ 

85 At + 4 2 He 

c 230 226 

90Th Æ 

88 Ra + 4 2 He 

2 a 

38 90 90 

Sr Æ 

39 Y + –1 0e 

b 131 

53 I Æ 131 

54 Xe + –1 0e 

c 231 231 

90Th Æ 

91 Pa + –1 0e 

3 a 

3 7Li + 1 1 p Æ 24 2 He 

b 14 7 N + 0 1n Æ 14 6 C + 1 1 p 


1 No. Isotopes have the same number of protons and the 

same number of electrons. 

2 

core 

nucleus plus 

filled shell 1 

3 a 2.1 b 2.8.5 c 2.8.8.2 

4 Electronic shell configuration Group Period 

– 3 2 

– 6 3 

2.4 – – 

2.8.4 – – 

– 1 2 

– 1 3 

– 1 4 

5 Elements A, C and E are in the same group. 


outer shell 

with seven electrons 

1 s block: metals 

p block: mixture of metals and non-metals 

d block: metals 

f block: metals 

2 a The electron is in the first electron shell. 

b The electron is in an s type orbital. 

c There are two electrons in this orbital. 

3 a s block b p block 

c p block 

d f block 

e d block 

f p block 

g f block 

h s block 

c 14 7 N + 2 4He Æ 17 8 O + 1 1 p 

d 

13 27Al 

+ 2 4 30 

He Æ 

15 P + 0 1n 

4 226 

88 

222 218 

Ra Æ 

86Rn Æ 

a a 84 a 

Po Æ 

214 

82 Pb 

5 232 

90 Th – 64 2He – 4–1 0 e Æ 208 

82 Pb 

6 a 5g b 0.625 g c 4.5 ¥ 10 –4 s d 0.039 g 

7 b Approx. 4.3 g 

c Approx. 185 days 

6 a X is in Group 1. 

b Y is in Group 0. 

Group 1 elements have a single electron in their outer 

shell, which they lose readily. Noble gases in Group 0 

have 2 or 8 electrons in their outer shell and it is 

difficult to remove one of these electrons. 

7 a lst ionisation Ca(g) Æ Ca + (g) + e – 

2nd ionisation Ca + (g) Æ Ca 2+ (g) + e – 

3rd ionisation Ca 2+ (g) Æ Ca 3+ (g) + e – 

b Once an electron has been removed the remaining 

electrons are held more tightly. Hence it is more 

difficult to remove a second electron. 

c Second ionisation enthalpy involves removal of an 

electron from shell 4 but third involves removal of an 

electron from shell 3 which is closer to the nucleus. 

8 The second ionisation enthalpy for sodium is high 

because removing a second electron involves removing 

an electron from the full second shell. This requires 

much more energy than removing the second electron 

from the third shell of magnesium which is further from 

the nucleus. 

4 Z = 16. The element is sulphur. 

5 a Chlorine b Potassium 

c Titanium d Tin 

6 a 1s 2 2s 2 2p 1 

b 1s 2 2s 2 2p 6 3s 2 3p 3 

c 1s 2 2s 2 2p 6 3s 2 3p 5 

d 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 

e 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 


1 a, c, e 2.8.8 

3 a 

b, d 2.8 

f 2.8.14 

g 2.8.13 

b 

h 2.8.17 

2 K + , Ca 2+ , (Sc 3+ ), Cl – , S 2– , P 3– 

Li 

K 

+ – 

X 

H 

+ 

XX 

F 

X 

XX 

X 

X 

– 

164

SECTION 3 

c 

d 

e 

f 

Mg 

Ca 

Ca 

Na 

2+ 

2+ 

2+ 

+ 

X 

O 

X 

XX 

XX 

Cl 

X 

XX 

X 

S 

X 

XX 

Na 

X 

X 

X 

X 

X 

X 

+ 

2– 

– 

2– 

XX 

Cl 

X 

XX 

X 

S 

X 

XX 

X 

X 

X 

X 

– 

2– 

5 a In a normal covalent bond, each atom supplies a single 

electron to make up the pair of electrons involved in 

the bond. In a dative covalent bond one atom supplies 

both electrons. 

b 

+ 

H 

6 

H 

H 

X 

N 

X 

X 

H 

X 

O 

X 

H 

H 

H 

+ 

H 

X 

N 

X 

H 

X 

H 

g 

h 

Na 

Al 

+ 

3+ 

X 

X 

Na 

F 

X 

XX 

+ 

X 

X 

– 

Na 

X 

X 

F 

X 

XX 

+ 

X 

X 

– 

X 

X 

N 

X 

X 

X 

X 

X 

F 

X 

XX 

3– 

X 

X 

– 

7 

H 

H 

X 

X 

N 

X 

H 

X 

X 

XX 

X 

X F 

X 

X 

X XX 

B 

X 

F 

X 

X 

X XX 

X 

X F 

X 

X 

XX 

4 a 

b 

c 

d 

e 

f 

g 

X 

X 

H 

H 

H 

X 

X 

XX 

Cl X Cl 

XX 

XX 

Cl 

X 

XX 

X 

H 

X 

C 

X 

X 

X 

X 

H 

XX 

S 

X X 

XX 

XX 

X 

Br 

XX 

XX 

X 

X Br 

X 

Al 

XX X 

X X 

X Cl 

XX X 

H 

H 

X 

X 

Si 

H 

H 

XX 

X X 

X Cl X 

X X 

X Cl X 

XX 

C 

X 

X 

X 

X 

X 

X 

XX 

X 

Br 

XX 

X 

X 

XX 

X X 

Cl X 

XX 

C 

X 

X 

H 

H 

8 Chlorine has a stronger electron pulling power. Thus in a 

covalent bond between carbon and chlorine, the 

electron pair forming the bond is more strongly attracted 

by the chlorine atom than by the carbon atom. 

9 a H—F b N∫N 

d+ d– non-polar 

c H—Cl 

d Cl—F 

d+ d– d+ d– 

e H—I 

f S=C=S 

d+ d– non-polar 

10 a Three 

b The delocalised electrons form a pool of electrons 

which are free to move around the positive ions in the 

metal. When a potential difference is applied across 

the metal, electrons move towards the positive 

terminal, ie a current flows. 

11 a Metallic b Covalent c Covalent 

d Ionic e Ionic f Covalent 

Silicon tetrafluoride contains polar covalent bonds. 

12 a NaCl 

b MgCl 2 

c FeCl 3 

d Al 2 

O 3 

e NH 4 

Cl 

f NaOH 

g K 2 

CO 3 

h MgSO 4 

h 

i 

H 

H 

X 

X 

C 

H 

X 

X 

X 

C 

H 

X 

X 

X 

O 

X 

X C H 

X 

X 

X 

H 

13 a NH 3 

b H 2 

S 

c CO 2 

d HCl 

e CO 

f SO 2 

g N 2 

O 

h SO 3 

165

SECTION 3 


1 a Li(g) because it has an extra electron shell. 

b Li + (aq) because the ion is surrounded by water 

molecules. 

c Cl – (g) because it has an extra electron. 

d Cl – (aq) because the ion is surrounded by water 

molecules. 

2 a K + (g) because it has an extra electron shell. 

b Br – (g) because it has an extra electron shell. 

c Na + (g) because it has one less proton in the nucleus 

and therefore attracts the same number of electrons 

less strongly. 

d Fe 2+ (g) because it has one more electron. 

3 a K + (aq) 

b K + (aq) 

c K + (aq) 

4 a Na + 

b Na + Cl – 

c Na + Cl – 

5 a i 11, 12, 13, 15, 16, 17 

ii Na + , Mg 2+ , Al 3+ 1s 2 2s 2 2p 6 

P 3– , S 2– , Cl – 1s 2 2s 2 2p 6 3s 2 3p 6 

b i 

ii 

Same electronic configuration. 

Different number of protons in the nucleus 

leading to different ionic charges. 

c Contraction of ion size as the number of protons 

relative to the number of electrons increases. 

d P 3– is the biggest and Cl – is the smallest because of the 

increasing attraction on the same number of electrons 

of an increasing number of protons in the nucleus. 

e They each contain an extra shell of electrons. 


1 a 

b 

c 

d 

e 

f 

g 

2 a 

b 

H 

H 

H 

H 

H 

H 

X 

X 

Si 

X 

H 

XX 

S 

X 

X X 

XX 

XX 

X X 

X 

X 

P 

H 

H 

H 

H 

H 

X 

X 

109° 

O X C O 180° 

XX 

X 

S X 

XX 

Approx. 109° 

Approx. 109° 

F F Approx. 109° 

Cl X Cl 

X 

B 120° 

X 

Cl 

C 

X 

H 

C 

H 

X 

X 

X 

X 

X 

X 

C 

X 

H 

180° 

H 

C H 

H 109∞ 

3 

d 

e 

f 

g 

F 

H 

H 

H 

C 

109∞ 

H 

Cl 

Cl 

X 

C 

F 

N 

H 

X 

X 

N 

XX 

H 

C 

H 

C 

120∞ 

O 

F 

a Tetrahedral 

b Pyramidal 

4 a Tetrahedral 

b Planar 

c Linear 

5 a i 

H 

X 

O 

H 

H 

H 

180∞ 

C 

N 

H 

+ 

X 

C 

X 

H 

All bond angles 120° 

All bond angles approx. 109° 

H 

C 

O 

H 


ii Planar (trigonal) 

c 

H 

H 

H 

C 

H 

N 

H 

H 


b 

i 

H 

H 

– 

X 

X X 

C 

XX 

H 

ii Pyramidal (tetrahedral with respect to electron 

pairs) 

166

SECTION 3 

6 a 

H 

XX 

X X 

N 

XX 

H 

– 

About 109° 

b 

H 

X 

XX 

N 

X 

H 

+ 

About 120° 


1 a C 6 

H 14 

, C 6 

H 12 

, not isomers 

b C 4 

H 9 

Cl, C 4 

H 9 

Cl, isomers 

c C 3 

H 8 

O, C 3 

H 6 

O, not isomers 

d C 7 

H 8 

O, C 7 

H 8 

O, isomers 

e C 3 

H 9 

N, C 3 

H 9 

N, isomers 

2 Isomers of C 5 

H 12 

: 

C C C C C 

C 

pentane 

4 Isomers of C 8 

H 10 

(containing a benzene ring): 

CH 3 CH 3 

CH 3 

CH 3 

CH 3 

CH 2 CH 3 

CH 3 

C C C C 

2-methylbutane 

5 

OH 

C C C C OH 

C C C C 

C 

C C C 

2,2-dimethylpropane 

C 

3 There are 4 isomers of C 4 

H 9 

Br: 

C C C C Br C C C C 

Br 

C 

C C C OH 

C O C C C 

C 

C C OH 

C 

C 

C C C Br 

C 

C C C 

Br 

C 

C O C C 

C C O C C 

167

SECTION 3 


1 F F 

4 a F 

2 

H 

F 

H 

H 3 C 

H 

H 3 C 

H 

C 

C 

C 

C 

C 

C 

C 

C 

H 

H 

F 

H 

CH 2 CH 3 

CH 2 CH 3 

H 

cis-1,2-difluoroethene 

trans-1,2-difluoroethene 

trans-pent-2-ene 

cis-pent-2-ene 

F 

C 

C 

H 

H 

b No 

c There must be the same two atoms or groups of atoms 

on each C atom attached to the double bond. 

Geometric isomers are not possible if two groups on 

one side of the double bond are the same. 

5 a Cis–trans isomers 

b 2 moles of H 2 

c Citronellol is a partially hydrogenated form of nerol 

(or geraniol). 

d Structural isomers of the cis- and trans-isomers. 

6 

3 a 

Cl 

H 

C 

C 

Cl 

H 

cis-1,2-dichloroethene 

cis-poly(ethyne) 

n 

Cl 

H 

C 

C 

H 

Cl 

trans-1,2-dichloroethene 

trans-poly(ethyne) 

n 

b Cis-1,2-dichloroethene 

c The cis-isomer has a dipole moment whereas the 

trans-isomer does not. The permanent 

dipole–permanent dipole intermolecular forces in the 

cis-isomer make it more difficult to separate the 

molecules. 

7 a 

H 

H 

X 

X 

C 

X 

H 

X 

N 

X 

X 

b H 3 C CH 3 

N 

N 

XX 

N 

X 

H 

X 

C 

X 

H 

X 

H 

H 3 C 

N 

N 

cis form 

CH 3 

trans form 


1 a A chiral centre. 

b iii 

2 a Br 

b 

3 a b 

c 

CH 3 

H 3 C 

H 

CH 

CH 2 CH 3 

CH 3 

CH CH 2 C * 

3 CH 2 CH 2 CH 3 

H 

Br 

C 

H 3 C 

H 3 CH 2 CH 2 C 

C 

CH 2 CH 3 

H 

CH 2 CH 3 

CH 3 CH 2 

Br 

C 

H 

H 3 CH 2 C 

4 a b 

CH 2 OH 

NH 2 COOH 

C * 

CH 2 CH 2 CH 3 

H 

c d H 

H 

C 

C 

CH 3 

HOOC NH 2 

H 2 N COOH 

CH 2 OH 

HOH 2 C 

D-isomer 

L-isomer 

(Use the CORN rule to name the isomers.) 

5 L-Cysteine is readily available as a hydrolysis product of 

CH 3 

proteins. 

D-Cysteine has to be made synthetically. It is probably 

C 

made together with L-cysteine from which it must be 

H 

separated. 

168

SECTION 4 

6 a 

H 

H 

H 

H 

b 

H 3 C 

H 3 C 

C 

C C 

* H C C 

H H 

CH 3 

H 3 C 

C 

C C 

* H C C 

H 

CH 3 

H 3 C 

CH 

CH 3 

H 

C 

H 

C 

C 

H 

C 

H 

C 

C 

H 

H 

H 

H 

H 

O 

limonene 

carvone 

c 

No: the carbon which was chiral is 

now bonded in the same way in both 

directions around the ring. 


1 a Standard enthalpy change of combustion is the 

enthalpy change when 1 mole of the compound is 

burnt completely in oxygen, under standard 

conditions (ie the compound and the products in their 

most stable states at 1 atmosphere pressure and at a 

stated temperature, often 298 K). 

b Standard enthalpy change of formation is the 

enthalpy change when 1 mole of a compound is 

formed from its elements, with both the compound 

and its elements being in their standard states (ie their 

most stable state at 1 atmosphere pressure and at a 

stated temperature, often 298 K). 

2 The formation of a compound from its elements may be 

an exothermic reaction (DH f 

negative) or an 

endothermic reaction (DH f 

positive). However, energy is 

liberated whenever a substance burns, so combustion 

reactions are always exothermic (DH c 

negative). 

3 

Enthalpy 

1 

1 

2 H 2 (g) + 2 Cl 2 (g) 

DH = –92.3 kJ mol –1 

HCl(g) 

Progress of reaction 

e 6C(s) + 6H 2 

(g) + 3O 2 

(g) Æ C 6 

H 12 

O 6 

(s) 

f C 6 

H 12 

O 6 

(s) + 6O 2 

(g) Æ 6CO 2 

(g) + 6H 2 

O(l) 

5 DH c 

!(C) is enthalpy change when 1 mole of carbon 

burned completely under standard conditions, ie 

C(s) + O 2 

(g) Æ CO 2 

(g) 

DH f 

!(CO 2 

) is enthalpy change when 1 mole of carbon 

dioxide is formed from its elements with both the carbon 

dioxide and its constituent elements in their standard 

states, ie 

C(s) + O 2 

(g) Æ CO 2 

(g) 

6 a Thermometer, measuring cylinder, gas meter. 

b Volume of water used, temperature rise of water, 

volume of gas used. 

c Cooling losses, impurity of the gas, etc. 

7 a DH = –667 kJ mol –1 

b Much heat was lost to the surroundings in the 

experiment whereas the accurate DH value in the data 

book would have been determined using a calorimeter 

in which very little heat is lost. 

8 a M r 

(C 7 

H 16 

) = 100 

b i 481.7 kJ released ii 481 700 kJ released 

(Assumed combustion is complete and CO 2 

and H 2 

O 

are the only products. Also, that combustion is carried 

out under standard conditions.) 

c Density of heptane 

9 a H 2 

(g) + O 2 

(g) Æ H 2 

O(l) 

b DH f, 

! 298 

(H 2 

O) = –286 kJ mol –1 

c –143 kJ (assuming combustion takes place under 

standard conditions) 

d +286 kJ mol –1 

Enthalpy 

1 

1 

2 

2 H 2 (g) + I 2 (s) 

HI(g) 

DH = +26.5 kJ mol –1 


10 a Enthalpy change of formation of propane. 

b Enthalpy change of combustion of 3 moles of carbon 

and 4 moles of hydrogen. 

c Enthalpy change of combustion of propane. 

d DH 1 

+ DH 3 

= DH 2 

e DH 1 

= DH 2 

– DH 3 

= 3(–393) kJ mol –1 + 4(–286) kJ mol –1 

– (–2220) kJ mol –1 

= –103 kJ mol –1 

11 a 4C(s) + 5H 2 

(g) Æ C 4 

H 10 

(g) 

b 

DH 

4C(s) + 5H 2 

(g) 

1 

C 4 

H 10 

(g) 

4 a 2C(s) + 3H 2 

(g) + O 2 

(g) Æ C 2 

H 5 

OH(l) 

b C 2 

H 5 

OH(l) + 3O 2 

(g) Æ 2CO 2 

(g) + 3H 2 

O(l) 

c 4C(s) + 5H 2 

(g) Æ C 4 

H 10 

(g) 

d C 4 

H 10 

(g) + 6 O 2 

(g) Æ 4CO 2 

(g) + 5H 2 

O(l) 

+ 6 O 2 

(g) DH 2 

DH 3 

+ 6 O 2 

(g) 

4CO 2 

(g) + 5H 2 

O(l) 

169

SECTION 4 

CH 3 

CHO(l) +2 O 2 

(g) Æ 2CO 2 

(g) + 2H 2 

O(l) 

 

c DH 1 

+ DH 3 

= DH 2 

13 a 

DH 1 

= DH 2 

– DH 3 

b 

= –890 kJ mol –1 

= 4(–393) kJ mol –1 + 5(–286) kJ mol –1 

– (–2877) kJ mol –1 

CH 3 

CHO(l) + 2 O 2 

(g) 

DH 1 

2CO 2 

(g) + 2H 2 

O(l) 

= –125 kJ mol –1 

12 a Enthalpy change of combustion of methane. 

DH 2 

DH 3 

b Enthalpy change of formation of methane. 

c Enthalpy change of combustion of carbon or enthalpy 

2C(s) + 2H 2 

(g) + 30 2 

(g) 

change of formation of carbon dioxide; the enthalpy of 

combustion of 2 moles of hydrogen or enthalpy 

c Standard enthalpy change of combustion of ethanol 

change of formation of 2 moles of water. 

= DH 1 

= – DH 2 

+ DH 3 

d DH 1 

= – DH 2 

+ DH 

= – (–192) kJ mol –1 + 2(–393) kJ mol –1 + 2(–286) kJ mol –1 

3 

e DH 1 

= – (–75) kJ mol –1 + (–393) kJ mol –1 

= –1166 kJ mol –1 

+ 2(–286) kJ mol –1 


1 a CH 4 

(g) + 2O 2 

(g) Æ CO 2 

(g) + 2H 2 

O(g) 

b H 

H 

C 

H 

H + 2O O O C O + 2H O H 

c 4 (C–H) 

2 (O=O) 

d 2 (C=O) 

4 (O–H) 

e +2648 kJ mol –1 (or +2737 kJ mol –1 depending on the 

value used for E(C–H)) 

f –3466 kJ mol –1 

g –818 kJ mol –1 (or –729 kJ mol –1 ) 

(The value of the standard enthalpy change of 

combustion at 298 K is for H 2 

O(l) and hence that value 

will be more exothermic than the value obtained here.) 

2 a C 3 

H 8 

(g) + 5O 2 

(g) Æ 3CO 2 

(g) + 4H 2 

O(g) 

b H H H 

H 

C 

H 

C 

H 

C 

H 

c 2 (C–C) 

8 (C–H) 

5 (O=O) 

d 6 (C=O) 

8 (O–H) 

e +6488 kJ mol –1 

f –8542 kJ mol –1 

g –2054 kJ mol –1 

H + 5 O O 3 O C O + 4 H 

O 

H 

3 a CH 3 

OH(l) + 1 O 2 

(g) Æ CO 2 

(g) + 2H 2 

O(g) 

b H 

H 

C 

H 

c 3 (C–H) 

1 (C–O) 

1 (O–H) 

l.5 (O=O) 

d 2 (C=O) 

4 (O–H) 

e +2786 kJ mol –1 

f –3466 kJ mol –1 

g –680 kJ mol –1 

O H + 1 1 2O 

O O C O + 2H O H 

4 N 2 

(g) + 3H 2 

(g) Æ 2NH 3 

(g) 

Bond enthalpies/kJ mol –1 : 

(+945) + 3(+436) Æ 2(3 ¥ +391) 

DH ! = –93 kJ mol –1 

5 N 2 

H 4 

(g) + O 2 

(g) Æ N 2 

(g) + 2H 2 

O(g) 


(+158) + 4(+391) + (+498) Æ (+945) + 4(+464) 

DH ! = –581 kJ mol –1 

6 C 2 

H 4 

(g) + Br 2 

(g) Æ C 2 

H 4 

Br 2 

(g) 


(+612) + 4(+413) + (+193) Æ 4(+413) + 2(+290) 

+ (+347) DH ! = –122 kJ mol –1 


1 a Increase b Decrease c Increase 

d Increase e Decrease f Decrease 

2 a Molten wax (Liquids have higher entropies than solids.) 

b Br 2 

(g) (Gases have higher entropies than liquids.) 

c Brass (Mixtures have higher entropies than the 

pure substances.) 

d Octane (Complex molecules have higher 

entropies than simpler molecules.) 

3 1 in 256 

4 a Ar molecule larger than He. 

b Gases have higher entropies than liquids. 

c Cl 2 

molecule larger than F 2 

. 

5 a Greater. A gas is formed as a product. 

b Smaller. Number of molecules of gas in product 

smaller than the number of molecules of gas in 

reactants. 

c Smaller. One reactant is a gas; product only solid. 

170

SECTION 4 


1 The entropies increase for the first four alkanes as the 

molecules become heavier and composed of more atoms 

(the number of energy levels increases with the number 

of atoms). Pentane is a liquid and so has a lower entropy 

than butane. 

2 a Entropy decrease, because the number of moles of gas 

is reduced by half as reaction proceeds. 

b Entropy increase; the number of moles of gas doubles 

during the reaction and a solid has much lower 

entropy than a gas. 

c Entropy decrease; 2 moles of gaseous reactants are 

replaced by 1 mole of solid. 

d Entropy decrease; 5 moles of gaseous oxygen are 

removed, the only product is a solid. 

e Entropy increase; 5 moles of gaseous product are 

formed. 


1 a When 1 mole of sodium fluoride is formed from 1 mole 

of Na + (g) and 1 mole of F – (g), 915 kJ mol –1 of energy are 

released. This is the lattice enthalpy of sodium fluoride. 

b The lattice enthalpy becomes more negative as the 

ionic radii decrease. 

2 a LiF; Li + has a smaller radius than Na + and attracts F – 

ions more strongly. 

b Na 2 

O; Na + has a smaller radius than Rb + and attracts 

O 2– more strongly. 

c MgO; Mg 2+ is smaller and more highly charged than 

Na + , and attracts O 2– more strongly. 

d KF; F – has a smaller radius than Cl – and attracts K + 

more strongly. 

3 a SrF 2 

; Sr 2+ is smaller and more highly charged than Rb + 

and will attract F – more strongly. 

b By the same arguments as in a, BaO should have the 

more exothermic lattice enthalpy. 

c Cu 2+ is more highly charged than Cu + , so CuO should 

have the more exothermic lattice enthalpy. 

4 a Li + attracts water molecules more strongly than Na + 

because of its smaller size. 

b Mg 2+ attracts water molecules more strongly than 

Ca 2+ because of its smaller size. 

c Ca 2+ and Na + have similar sizes, but Ca 2+ is more highly 

charged and so attracts water molecules more strongly. 

5 a The ions in the lattice attract each other less strongly 

as the size of the anion increases from F – to Cl – . 

b DH hyd 

becomes less exothermic as the anion becomes 

bigger and attracts water molecules less strongly. 

c 

AgF(s) + aq 

–DH LE 

DH solution 

Ag + (g) + F – (g) + aq 

Ag + (aq) + F – (aq) 

DH hyd 

(Ag + ) 

+ 

DH hyd 

(F – ) 

3 Students’ answers should be based on the following 

deductions. 

a 

b 

c 

d 

e 

DS sys 

DS surr 

Explanation 

/J K –1 mol –1 /J K –1 mol –1 

+203 –44 Spontaneous: total 

entropy change positive 

+63 +329 Spontaneous: total 


+25 –604 Not spontaneous: total 

entropy change negative 

+209 +416 Spontaneous: total 


–4 –6.7 Not spontaneous: total 

entropy change negative 

4 Values for both DS sys 

and DS surr 

are negative. Therefore 

DS total 

must always be negative, whatever value of T is 

chosen, and the process can never be spontaneous. 

Enthalpy 

AgCl(s) + aq 

–DH LE 

DH solution 

Ag + (g) + Cl – (g) + aq 

Ag + (aq) + Cl – (aq) 

DH hyd 

(Ag + ) 

+ 

DH hyd 

(Cl – ) 

Both silver halides have endothermic enthalpy 

changes of solution and the enthalpy level diagrams 

will have the form shown below. 

–DH LE 

AgX(s) + aq 

Ag + (g) + X – (g) + aq 

Ag + (aq) + X – (aq) 

DH solution 

d DH solution 

= –DH LE 

+ DH hyd 

(Ag + ) + DH hyd 

(X – ) 

DH solution 

(AgF) = +6 kJ mol –1 ; 

DH solution 

(AgCl) = +95 kJ mol –1 

DH hyd (Ag + ) 

+ 

DH hyd (X – ) 

e AgF may be soluble in water. AgCl will be insoluble. 

6 a Mg(OH) 2 

+ 152 kJ mol –1 

Ca(OH) 2 

+ 7 kJ mol –1 

b Ca(OH) 2 

: enthalpy change of solution much less 

endothermic. 

c Entropy changes of the processes. 

171

SECTION 4 


1 a i Li(s) + Cl 2 

(g) Æ LiCl(s) 

ii Li(s) Æ Li(g) 

iii Cl 2 

(g) Æ Cl(g) 

iv Li(g) Æ Li + (g) + e – 

v Cl(g) + e – Æ Cl – (g) 

vi Li + (g) + Cl – (g) Æ LiCl(s) 

Enthalpy/kJ mol –1 

+800 

+700 

Li + (g) + Cl(g) 

b 

Li(s) + Cl 2 

(g) 

DH 1 

LiCl(s) 

DH EA (Cl) 

+600 

DH 2 

DH 4 

Li(g) +Cl(g) 

DH 3 

Li + (g) + Cl – (g) 

+500 

DH i (1)(Li) 

DH 1 

= DH 2 

+ DH 3 

+ DH 4 

DH! LE 

(LiCl) = DH 4 

= DH 1 

– DH 2 

– DH 3 

= (– 409 kJ mol –1 – (+ 159 kJ mol –1 

+ 121 kJ mol –1 ) – (+519 kJ mol –1 

– 355 kJ mol –1 ) 

DH! LE 

(LiCl) = – 853 kJ mol –1 

 

c See Graph in facing column 

2 a i Mg(s) + Cl 2 

(g) Æ MgCl 2 

(s) 

ii Mg(s) Æ Mg(g) 

iii Cl 2 

(g) Æ Cl(g) 

iv Mg(g) Æ Mg + (g) + e – 

v Mg + (g) Æ Mg 2+ (g) + e – 

vi Cl(g) + e – Æ Cl – (g) 

vii Mg 2+ (g) + 2Cl – (g) Æ MgCl 2 

(s) 

b 

Mg(s) + Cl 2 

(g) 

DH 1 

MgCl 2 

(s) 

+400 

+300 

+200 

+100 

0 

Li(g) + Cl(g) 

DH at (Cl) 

1 

Li(g) + 2 Cl 2 (g) 

DH at (Li) 

Li + (g) + Cl – (g) 

1 

Li(s) + 2 Cl 2 (g) 

DH 2 

DH 4 

Mg(g) + 2Cl(g) 

DH 3 

Mg 2+ (g) + 2Cl – (g) 

–100 

DH LE (LiCl) 

DH 1 

= DH 2 

+ DH 3 

+ DH 4 

DH! LE 

(MgCl 2 

) = DH 4 

= DH 1 

– DH 2 

– DH 3 

= (– 641 kJ mol –1 ) – (+ 147 kJ mol –1 

+ 2 (+ 121 kJ mol –1 )) 

– (+744 kJ mol –1 + 1457 kJ mol –1 

+2 (– 355 kJ mol –1 )) 

DH! LE 

(MgCl 2 

) = – 2521 kJ mol –1 

–200 DH f (LiCl) 

–300 

–400 

LiCl(s) 

–500 

As Group 1 is ascended, the enthalpy changes of formation of 

the chlorides become less negative and the first ionisation 

enthalpies of the elements become more positive. 

172

SECTION 4 

3 a 

DH 

Ca(s) + Cl 2 

(g) 

1 

DH 

CaCl(s) 

Ca(s) + Cl 2 

(g) 1 

CaCl 2 

(s) 

DH 2 

DH 4 

DH 2 

DH 4 

Ca(g) +Cl(g) 

DH 3 

Ca + (g) + Cl – (g) 

Ca(g) +2Cl(g) 

DH 3 

Ca 2+ (g) + 2Cl – (g) 

DH! f 

(CaCl) = DH 1 

= DH 2 

+ DH 3 

+ DH 4 

DH! f 

(CaCl 2 

) = DH 1 

= DH 2 

+ DH 3 

+ DH 4 

= (+178 kJ mol –1 + 121 kJ mol –1 ) 

+ (+596 kJ mol –1 –355 kJ mol –1 ) 

+ (–711 kJ mol –1 ) 

DH! f 

(CaCl) = – 171 kJ mol –1 

= (+178 kJ mol –1 + 2 (+121 kJ mol –1 )) 

+ (+596 kJ mol –1 + 1152 kJ mol –1 ) 

+ 2 (–355 kJ mol –1 ) +(–2237kJ mol –1 ) 

DH! f 

(CaCl 2 

) = – 779 kJ mol –1 

b 

2DH EA (Cl) 

DH i (2)(Ca) 

+800 

+600 

DH i (1)(Ca) 

DH EA (Cl) 

DH i (1)(Ca) 

DH LE (CaCl 2 ) 

+400 

+200 

Enthalpy/kJ mol –1 1 

+2200 

Ca 2+ (g) + 2Cl(g) 

+2000 

+1800 

+1600 

+1400 

Ca 2+ (g) + 2Cl – (g) 

+1200 

+1000 

Ca + (g) + Cl(g) 

Ca + (g) + Cl – (g) 

DH at (Cl) 

DH LE (CaCl) 

2DH at (Cl) 

0 

–200 

DH at (Ca) 

DH f (CaCl) 

Ca(s) + Cl 2 (g) 

2 

CaCl(s) 

DH at (Ca) 

Ca(s) + Cl 2 (g) 

–400 

DH f (CaCl 2 ) 

–600 

–800 

CaCl 2 (s) 

c The enthalpy change of formation of CaCl 2 

is much more negative than 

that of CaCl. CaCl 2 

has lower energy and is more stable relative to the 

elements Ca and Cl 2 

, and so it is more likely to be formed. 

173

SECTION 5 


1 a Ca 2+ (aq) + 2OH – (aq) 

b Mg 2+ (aq) + SO 2– 4 

(aq) 

c 2Na + (aq) + O 2– (aq) 

d K + (aq) + OH – (aq) 

e Ag + (aq) + NO – 3 

(aq) 

f 2Al 3+ (aq) + 3SO 2– 4 

(aq) 

2 a NaBr 

b Mg(OH) 2 

c Na 2 

S 

d BaO 

e CaCO 3 

f Ca(NO 3 

) 2 

g K 2 

CO 3 

3 a Ba 2+ (aq) + SO 2– 4 

(aq) Æ BaSO 4 

(s) 

b Mg 2+ (aq) + 2OH – (aq) Æ Mg(OH) 2 

(s) 

c Ca 2+ (aq) + CO 2– 3 

(aq) Æ CaCO 3 

(s) 

d Ba 2+ (aq) + CrO 2– 4 

(aq) Æ BaCrO 4 

(s) 


1 The small size of the carbon atom makes it possible for 

carbon to form double bonds with oxygen to produce 

discrete covalent molecules. Silicon, on the other hand, 

bonds to 4 oxygen atoms to form single bonds and 

hence a covalent network structure. The attractive forces 

between the molecules of carbon dioxide 

(intermolecular forces) are weak so that little energy is 

needed to separate the individual molecules. Strong 

covalent bonds exist throughout the SiO 2 

covalent 

network structure so a lot of energy is needed to melt it. 

2 a Any attractive forces between the solvent and the 

atoms in the covalent network structure are too weak 

to overcome the strong covalent bonds holding the 

network together. 

b At room temperature, the kinetic energy of atoms and 

molecules is small but it may be enough to overcome 

the weak attractive forces between simple molecules 

or isolated atoms. 

3 a Diamond structure with silicon and carbon atoms 

alternating. 

4 2Na + (aq) + SO 2– 4 

(aq) + 10H 2 

O(l) Æ Na 2 

SO 4 

.10H 2 

O(s) 

5 a H + (aq) + OH – (aq) Æ H 2 

O(l) 

b Zn(s) + 2H + (aq) Æ Zn 2+ (aq) + H 2 

(g) 

c CuO(s) + 2H + (aq) Æ Cu 2+ (aq) + H 2 

O(l) 

d CaCO 3 

(s) + 2H + (aq) Æ Ca 2+ (aq) + H 2 

O(l) + CO 2 

(g) 

6 a i Cubic 

ii Eight 

iii Eight 

b The attractions between the oppositely charged 

caesium and chloride ions greatly outweigh the 

repulsions between ions with the same charge. The 

net attractive force is very high. The solid is hard and 

has a high melting point as this strong attraction has to 

be overcome to separate the particles. 

b The silicon and carbon atoms are held together by 

strong covalent bonds in an extended network 

structure. This makes the substance very hard. 

4 a Graphite structure with alternating boron and nitrogen 

atoms. 

b In graphite, each carbon atom has 4 bonding electrons. 

Only 3 are needed to form the layer structure. The 

fourth electron is ‘free’ and causes graphite to be a good 

electrical conductor. Both boron and nitrogen have only 

3 bonding electrons. All 3 are needed to form the 

structure of the layer. There are no electrons available to 

become delocalised and conduct electricity. 

5 It was assumed that buckminsterfullerene was a covalent 

network structure and would be insoluble like graphite 

and diamond, but solution in benzene showed it, 

surprisingly, to be a molecular form of carbon. 

Attractive forces between the solvent molecules and 

the carbon atoms are sufficient to overcome the weak 

intermolecular forces between the buckminsterfullerene 

molecules. 


1 a Kr, Xe (higher b.p.) 

b C 6 

H 14 

, C 8 

H 18 

(higher b.p.) 

c CH 4 

, CCl 4 

(higher b.p.) 

d CH and 

CH 3 

CH 3 

CH 3 CH 3 CH 2 CH 2 

CH 3 

(higher b.p.) 

3 

pentane 

e 

CH 3 

methylbutane 

CH 3 CH CH CH 3 

CH 3 

and 

174 

CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 (higher b.p.) 

2 In the solid or liquid state, noble gas atoms are held 

together by weak instantaneous dipole–induced dipole 

forces. It takes very little energy to break these attractions 

and this results in very low melting and boiling points. 

dimethylpropane

SECTION 5 

a Pentane has the strongest intermolecular forces and 

hence the highest boiling point. Dimethylpropane has 

the weakest intermolecular forces and hence the 

lowest boiling point. 

b The molecules of pentane, because it is a straight 

chain alkane, can approach closely to each other 

which increases the opportunities for 

instantaneous–induced dipole interactions and hence 

stronger intermolecular attractions. Methylbutane has 

one methyl side chain and so it is more difficult for 

these molecules to approach each other and 

instantaneous–induced dipole interactions are weaker. 

Dimethylpropane has two methyl side-chains and so it 

is even more difficult for molecules of this compound 

to approach each other. 

4 The strength of instantaneous dipole–induced dipole 

forces between molecules increases as the relative 

molecular masses of the molecules increase. To be sure 

that the higher boiling point of the polar substance is 

due only to the increased strength of dipole–dipole 

attractions it will be necessary to ensure that the 

instantaneous dipole–induced dipole forces in both 

polar and non-polar substances are of similar strength. 

This can be done by comparing substances of similar 

molecular mass. 

5 a A and D; C and G. 

b A and G have the stronger intermolecular forces 

compared to D and C, respectively. 

6 Polarity will occur as follows: 

C-F; H-Cl; H-N; C-O with charges in the order d+ d– in 

each case. 

7 CHCl 3 

, CH 3 

OH, (CH 3 

) 2 

CO, cis-1,2-difluoroethene and 

1,2-dichlorobenzene possess dipoles. 

8 a i Eighteen 

ii The attractions will be similar. 

iii H 2 

S has a permanent dipole. It is a bent molecule 

with two lone pairs. SiH 4 

does not have an overall 

permanent dipole as it is a symmetrical molecule. 

b Both compounds have similar instantaneous 

dipole–induced dipole forces. However, H 2 

S also has 

permanent dipole–permanent dipole attractions so its 

boiling point is higher than that of SiH 4 

. 

9 a Instantaneous dipole–induced dipole. 

b Instantaneous dipole–induced dipole. 

c Instantaneous dipole–induced dipole, permanent 

dipole–permanent dipole. 

d Instantaneous dipole–induced dipole. 

e Instantaneous dipole–induced dipole, permanent 


f Instantaneous dipole–induced dipole. 

g Instantaneous dipole–induced dipole. 

h Instantaneous dipole–induced dipole, permanent 



1 a As the temperature rises, solids and liquids expand. 

The temperature increase raises the kinetic energy of 

the particles present. In solids, the rotational and 

vibrational energy increases. In liquids, rotational, 

vibrational and translational energy increases. The 

increases in vibrational and translational energy 

increase the volume occupied by the particles. As they 

occupy an increasing volume, the density of the solid 

or liquid decreases. 

b i When ice melts, much of the open, hydrogenbonded 

structure collapses. This enables the 

molecules to occupy less space so the density 

increases on melting. 

ii The boiling point of water is higher than 

expected, as more energy is needed to break the 

hydrogen bonding. 

iii The specific heating capacity of water is higher 

than expected, as more energy is absorbed by the 

water to break hydrogen bonds in the liquid. 

2 a i H 2 

O instantaneous dipole–induced dipole 

permanent dipole–permanent dipole 

hydrogen bonding. 

H 2 

S instantaneous dipole–induced dipole 

permanent dipole–permanent dipole. 

H 2 

Se instantaneous dipole–induced dipole 


H 2 

Te instantaneous dipole–induced dipole 


ii Intermolecular forces must be overcome when a 

liquid boils. 

Hydrogen bonding present between molecules of 

H 2 

O – but not between those of H 2 

S, H 2 

Se or H 2 

Te. 

Hydrogen bonding forces are much stronger than 

other intermolecular forces and so the boiling point of 

water is higher than that of the other hydrides. 

b The strength of instantaneous dipole–induced dipole 

and permanent dipole–permanent dipole attractions 

in a substance gets weaker as its relative molecular 

mass gets smaller. This produces a lower boiling point. 

The boiling point of H 2 

O should be lower than that of 

H 2 

S but it is in fact much higher. This suggests that, 

compared to H 2 

S, a different and much stronger type 

of intermolecular bonding exists in H 2 

O. 

c ii All have instantaneous dipole–induced dipole 

forces. 

iii The shape of the graph is nearly a straight line 

with positive slope. There is no hydrogen 

bonding between the hydride molecules of 

Group 4. The increase in boiling points down the 

group is a result of their regularly increasing 

molecular masses. 

3 A, D, E and F 

4 a Hydrogen bonding will be present in NH 3 

, CH 3 

OH, 

and HF. 

b 

N H CH 3 

F 

H H O 

H 

H 

H 

H 

H CH 3 F 

F 

N 

O 

H H 

H 

175

SECTION 5 

5 a 

H 

H 

H 

c 

H 

b 

H 

O 

H 

C 

H 

H 

C 

N 

H 

O 

H 

H 

O 

H H 

O C C H 

H 

O 

H 

O 

O 

H 

O 

H 

O 

H H H 

H 

H 

H 

C 

H 

H 

H 

C 

H 

O 

H 

H 

H 

O 

H 

H 

H 

6 a Instantaneous dipole–induced dipole, permanent 

dipole–permanent dipole, hydrogen. 

b Instantaneous dipole–induced dipole. 

c Instantaneous dipole–induced dipole, permanent 

dipole–permanent dipole, hydrogen. 

d Instantaneous dipole–induced dipole. 

e Instantaneous dipole–induced dipole, permanent 


Section 5.5, Part 1 (Addition polymers) 

1 a c 

CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 

H 3 C 

H 

b 

CH 2 CH CH 2 CH CH 2 CH 

C 

C 


H 

CO OCH 3 

2 CH 2 CH CH 2 CH CH 2 CH 

3 a 1 

b 4 

4 a 

b 

c 

5 a 

b 

H 

H 

H 

H 3 C 

C 

C 

C 

Cl Cl Cl 

C 

C 

C 

H 

CN 

H 

H 

H 

H CO OCH 3 

H 

C 

H 

H 

C 

H 

C 

CN 

H 

C 

H 

6 But-1-ene and propene 

7 H H H H 

C C C C 

H 

Cl 

H 

O C CH 3 

O 

8 a Poly(ethene), poly(propene), poly(chloroethene) 

b Poly(ethene) instantaneous dipole–induced 

dipole 

Poly(propene) instantaneous dipole–induced 

dipole 

Poly(chloroethene) instantaneous dipole–induced 

dipole 

permanent dipole–permanent 

dipole 

c Poly(chloroethene) 

9 a Isotactic polymer is a stereoregular polymer, that is, 

the side-chains all have the same orientation. In an 

atactic polymer the side-chains are randomly 

orientated. 

b The regular arrangement of the side-chains in an 

isotactic polymer allows the polymer chains to pack 

together closely, which means intermolecular forces 

are stronger. The polymer chains do not slide past 

each other easily, making the polymer stronger and 

less flexible than the atactic form of the same polymer. 

176

SECTION 5 

Section 5.5, Part 2 (Addition and condensation polymers) 

1 a 

O O 

O O 

O 

CH 2 

CH 2 

O 

C 

CH 2 

C 

O 

CH 2 

CH 2 

O 

C 

CH 2 

C 

b 

O 

O 

O 

O 

N 

CH 2 

CH 2 

N 

C 

CH 2 

C 

N 

CH 2 

H H 

CH 2 

CH 2 

N 

C 

CH 2 

CH 2 

C 

H 

H 

2 a 

O 

H 2 N 

(CH 2 ) 5 

C 

OH 

b 

O 

O 

HO CH 2 CH 2 CH 2 OH and HO C 

C OH 

c 

CH 3 

O 

HO 

CH 

C 

OH 

3 Polarity increases from –CH 3 

to –Cl to –CN so the 

intermolecular attractions become stronger which leads 

to an increase in T g 

. 

4 a i Instantaneous dipole–induced dipole. 

ii Instantaneous dipole–induced dipole. 

Permanent dipole–permanent dipole. 

b (CH 2 

O) n 

would have a higher T g 

because of the 

permanent dipole–permanent dipole interactions 

between the chains. 


1 a Ionic lattice 

b i Ne monatomic 

ii 

S 

H H 

simple molecular (covalent) 

iii 

H 

H 

C 

H 

H 

simple molecular (covalent) 

iv O C O simple molecular (covalent) 

c i See Figure 6, page 92, Chemical Ideas. 

ii Covalent network or giant covalent. 

d The diagram should be similar to Figure 15, page 39, 

Chemical Ideas. There will be 2+ charges on the 

positive ions as the metal is magnesium. The outer 

electrons, two from each magnesium atom, contribute 

to a ‘pool’ of electrons which move randomly through 

the lattice of positive ions. Each positive ion is 

attracted to the negatively charged delocalised 

electrons and vice versa. 

5 There is hydrogen bonding in poly(caprolactam). 

In poly(caprolactone) there are only permanent 

dipole–permanent dipole and instantaneous 

dipole–induced dipole attractions. 

6 a Model building should show that the T m 

values reflect 

the alignment of hydrogen bonding between the 

polymer chains for nylon-6. There are fewer 

opportunities for hydrogen bonding in nylon-11. 

2 

b Intermolecular hydrogen bonding is more extensive in 

nylon-6,6. The chains in nylon-6,10 can slide past one 

another more easily. 

a 

b 

c 

d 

e 

f 

g 

h 

Name State at Solubility Electrical 

room in water conductivity 

temperature 

sodium solid soluble conducts 

iodide 

when molten 

or in aqueous 

solution 

carbon gas insoluble does not conduct 

monoxide 

diamond solid insoluble does not conduct 

tetrachloro- liquid insoluble does not conduct 

methane 

ethanol liquid soluble does not conduct 

copper(II) solid soluble conducts 

chloride 

when molten 

or in aqueous 

solution 

vanadium solid insoluble conducts when 

solid or liquid 

poly(propene) solid insoluble does not conduct 

3 a Isolated atoms 

b Metallic; giant lattice 

c Covalent network (or giant covalent); giant lattice 

d Macromolecular; covalent molecular 

e Ionic; giant lattice 

f Simple molecular; covalent molecular 

4 A Macromolecular; covalent molecular 

B Metallic; giant lattice 

C Ionic; giant lattice 

D Metallic; giant lattice 

E Simple molecular; covalent molecular 

177

SECTION 6 

5 a In ionic substances, the charge-carriers (ions) are held 

in the solid lattice and are not free to move. In metals, 

the charge-carriers (electrons) are delocalised and are 

free to move throughout the lattice. 

b When an ionic melt conducts electricity, the ions move 

to opposite electrodes where they are discharged, 

leading to decomposition. When a metal conducts, 

electrons move towards the more positive terminal 

and are replaced at an equal rate at the more negative 

terminal. 

c Any attractive forces between the solvent and the 

atoms in the giant covalent lattice are too weak to 

overcome the strong covalent bonds holding the 

lattice together. 

d At room temperature, the kinetic energy of atoms and 

molecules is small. It is not enough to overcome the 

energy of attraction between particles in giant or 

macromolecular structures, but it may be enough to 

overcome the weak intermolecular forces between 

simple molecules. 

178 


1 a 7.22 ¥ 10 –28 J 

b 6.6 ¥ 10 –17 J 

c 9.2 ¥ 10 10 

2 5.5 ¥ 10 13 Hz 

3 a The four lowest energy lines in the Balmer Series arise 

from transitions to level 2 from levels 3, 4, 5 and 6. 

b 


1 

a 

b 

c 

d 

3 2 

4 2 5 2 6 2 

frequency 

The lines converge towards high frequency. The lines 

in the Balmer Series are at lower frequency than the 

lines in the Lyman Series. 

Energy of Frequency/Hz Type of Type of 

photon radiation energy 

emitted or 

change 

absorbed/J 

in molecule 

4.6 ¥ 10 –17 6.9 ¥ 10 16 u.v. electronic 

2.3 ¥ 10 –20 3.5 ¥ 10 13 i.r. vibrational 

2.1 ¥ 10 –22 3.2 ¥ 10 11 microwave rotational 

5.5 ¥ 10 –19 8.3 ¥ 10 14 visible electronic 

2 a 5.43 ¥ 10 –20 J 

b 8.19 ¥ 10 13 Hz; infrared 

c 3.66 ¥ 10 –6 m 

3 

+214 kJ mol 

a E = ¥ 1000 

6.02 ¥ 10 23 mol –1 = + 3.55 ¥ 10 –19 J 

b E = hν 

3.55 ¥ 10 –19 J = 6.63 ¥ 10 –34 J Hz –1 ¥ ν Hz 

ν = 5.36 ¥ 10 14 Hz 

c This is in the visible region of the electromagnetic 

spectrum. 


1 F, OH, NO 2 

and CH 3 

are radicals. 

2 a i Photodissociation 

ii Homolytic 

b Reaction A initiation 

Reaction B ⎞ 

⎬ 

Reaction C ⎠ 

propagation 

c H. and HO2 . 

d i 2O 3 

Æ 3O 2 

ii Catalyst 

e The rates of radical reactions depend on the 

concentrations of the radicals involved. 

f Termination 

c The line must go from a lower energy level to a 

higher level. 

4 a Ground state 

b i DE = hν 

= 6.63 ¥ 10 –34 J Hz –1 ¥ 3.27 ¥ 10 15 Hz 

= 2.17 ¥ 10 –18 J 

ii One photon of frequency 

3.27 ¥ 10 15 Hz provides this energy. 

For 1 mole of hydrogen atoms 

DE = 6.02 ¥ 10 23 mol –1 ¥ 2.17 ¥ 10 –18 J 

= 1310 kJ mol –1 . 

Data book value for the ionisation enthalpy of 

hydrogen is 1318 kJ mol –1 . 

4 a 1.89 ¥ 10 4 J 

b 1.62 ¥ 10 –24 J 

c 0.978 J 

d 19 330 moles of photons 

5 a CO 2 

absorbs infrared radiation of specific frequencies 

corresponding to transitions between vibrational 

energy levels. (Vibrational energy is quantised.) The 

specific frequencies absorbed make the molecules 

vibrate in particular ways. The vibrational energy 

increases. 

b The molecules which have absorbed radiation have 

more kinetic energy. (A more complete answer could 

include energy being subsequently transferred to 

other molecules in the air by collision.) 

3 a Oxidation of N 2 

in internal combustion engines. 

b Exothermic 

c i O 3 

+OÆ O 2 

+ O 2 

ii Catalyst 

iii DH = –292 kJ mol –1 

d It catalyses breakdown of ozone, thus removing it 

from the stratosphere. 

4 a Initiation: reaction 1 

Propagation: reactions 2, 3 and 4 

Termination: reactions 5 and 6 

b i Endothermic: reaction 1 (C–C bond broken) 

ii Exothermic: reaction 6 (C–C bond formed)

c CH 3 

. methyl radical 

C 2 

H 5 

. ethyl radical 

H. hydrogen radical 

5 a Cl 2 

+ hν Æ Cl. + Cl . initiation 

CH 4 

+ Cl. Æ CH 3 

. + HCl ⎞ 

⎬ propagation 

CH 3 

. + Cl2 Æ CH 3 

Cl + Cl. ⎠ 

Cl. + Cl . Æ Cl ⎞ 

2 termination 

CH 3 

. + CH3 . Æ C2 H 6 

⎬ 

⎠ 

SECTION 6 

b The chloromethane formed can also react with Cl. 

radicals. 

CH 3 

Cl + Cl. Æ . CH2 Cl + HCl 

. CH2 Cl + Cl 2 

Æ CH 2 

Cl 2 

+ Cl. 

Similarly, CH 2 

Cl 2 

can react with Cl. , and so on, until 

CCl 4 

is produced. The chlorinated products react with 

Cl. radicals more quickly than CH4 does, so a mixture 

of products is always obtained. 


1 a 4.24 mm 

b 7.08 ¥ 10 13 Hz 

2 

Absorption/cm –1 Bond 

3660 O–H 

3060 C–H 

(arene) 

3 a 

OH 

CH 3 CH CH 3 

butan-2-ol 

cA butan-2-ol 

B butan-2-one 

4 a 

Compound Absorption/cm –1 Bond 

C 3580 O-H 

2990 C-H (alkane) 

1775 C=O 

D 3670 O-H 

2950 C-H (alkane) 

E 2990 C-H (alkane) 

1770 C=O 

CH 2 

CH 3 CH 3 

b 

CH 3 CH 2 

O 

C CH 3 

butan-2-one 

Compound Absorption/cm –1 Bond 

A 3660 O–H 

2970 C–H (alkane) 

B 2990 C–H (alkane) 

1730 C=O 

5 

bC A carboxylic acid 

D An alcohol 

E An ester 

Bond Absorption/cm –1 

O-H (phenol) 3600–3640 

C-H (arene) 3000–3100 

C=O (ester) 1735–1750 


b 

[CH 3 

] + 

1 a 78, 72, 106 

15 

b Two isotopes of chlorine leading to C 3 

H 35 7 

Cl (78) and 

C 3 

H 37 7 

Cl (80) 

2 43, [C 3 

H 7 

] + 

43, [CH 3 

CO] + 

77, [C 6 

H 5 

] + 

3 a, b A Ethanoic acid 

B Ethanol 

A 43, [CH 3 

CO] + 

45, [COOH] + 

60, [CH 3 

COOH] + 

B 31, [CH 3 

O] + or [CH 2 

OH] + 

46, [C 2 

H 5 

OH] + 

6 a 

4 a 88, molecular ion 

43, [CH 3 

CO[ + 

b [CH 3 

COOCH 2 

CH 3 

] + Æ [CH 3 

CO] + + OCH 2 

CH 3 

5 a Mass of peak 

Possible fragment 

58 [C 4 

H 10 

] + 

43 [C 3 

H 7 

] + 

29 [C 2 

H 5 

] + 

CH 3 

CH 2 

CH 2 CH 3 CH CH 3 

c Adopting a ‘Lego’ approach to these fragment ions, 

the full structural formula of the hydrocarbon must be: 

CH 3 CH 2 

CH 2 CH 3 

Fragmentation of this by breaking C–C bonds leads to 

the four ions in the table. 

The branched isomer would not produce the C 2 

H 5 

+ 

ion by breaking C–C bonds. However, the other three 

ions will also appear in the spectrum of this isomer. 

CH 3 

CH 2 

H 

C 

b 58 – 43 = 15; CH 3 

c i 58 – 57 = 1; H 

ii 57 – 29 = 28; CO or C 2 

H 4 

d i [CH 3 

CO + ] 

ii 28 [CO] + or [C 2 

H 4 

] + 

29 [CH 3 

CH 2 

] + 

57 [CH 3 

CH 2 

CO] + 

eCis CH 3 

COCH 3 

O 

D is CH 3 

CH 2 

CHO 

CH 3 

O 

C CH 3 

179

SECTION 6 

7 Accurate atomic masses give C 4 

H 8 

O as the formula. 

Peaks to be identified: 

Mass of peak 

Possible fragment 

15 [CH 3 

] + 

29 [CH 3 

CH 2 

] + 

43 [CH 3 

CO] + 

57 [CH 3 

CH 2 

CO] + 

72 [CH 3 

CH 2 

COCH 3 

] + 

8 The sketch should show two molecular ion peaks of 

equal height at 108 [C 2 

H 79 5 

Br] + and 110 [C 2 

H 81 5 

Br] + . 

Also peaks at 93 [CH 79 2 

Br] + and 95 [CH 81 2 

Br] + . 

(Actual data have 29 as the base peak (100%) and peaks 

at 108 and 110 (94%). The next largest peak is 27 (70%). 

Peaks at 79 and 81 are very small (3%).) 

The compound has the structure 

O 

CH 3 CH 2 C CH 3 


1 a 3 

b 

Chemical shift Relative number of protons 

1.0 3 

2.1 3 

2.4 2 

6 


1.3 3 

2.1 3 

4.1 2 

O 

CH 3 

cd CH 3 

– CH 2 

– CO – CH 3 

CH 3 

(1.3), CH 3 CO (2.1), C O CH 2 R (4.1) 

1.0 2.4 2.1 

O 

2 a b Tartaric acid 3 signals 1 : 1 : 1 

CH 2 

Succinic acid 2 signals 2 : 1 

CH 3 C O CH 2 CH 3 

Citric acid 3 signals 1 : 4 : 3 

H is ethyl ethanoate. 

3 a 


7 



4.0 3 

2.6 1 

6.4 1 

3.7 2 

7.0 1 

b CH 3 

(1.2), OH (2.6), CH 2 

(3.7) 

7.4 2 

c CH 3 

–CH 2 

–OH 

9.8 1 

E is ethanol. 

4 a 

CH 3 O (4.0), 



OH (6.4), 

2.0 1 

H 

b CH 3 

(1.3), OH (2.0) 

c CH 3 

(7.0–7.4), 

CH 3 C OH 

H H 

CH 3 

CHO (9.8) 

F is 2-methylpropan-2-ol. 

8 a 

Protons in Expected Relative numbers 

5 


the molecule chemical of each type 

shift 

of proton 


2.7 2 

CH 3 

O ca 3.7 3 

7.0–7.4 5 

C 6 

H 4 

6.0–9.0 2 

H H 

It is good enough to identify two types of proton (CH 3 

and C 6 

H 4 

), in the ratio of 3 : 2. However, there are 

CH 

actually two different environments for the protons on 

3 (1.2), 

CH 2 R (2.7), H (7.0–7.4) 

the aryl group, one being adjacent to the ester group, 

and the better answer would be 3 : 1 : 1. 

H H 

180 

G is ethylbenzene.

SECTION 7 


1 a Green 520 nm–580 nm approx. 

b Red 620 nm–700 nm approx. 

2 Blue and blue-green; approx. 440 nm – 520 nm 

3 It will appear green. The sketch should show two peaks 

with λ max 

at about 640 nm and 410 nm. 


1 a Spectrum (b) b Violet 

2 a i 

ii 

Intensity of 

absorption 

300 500 700 

l / nm 

4 a 6.97 ¥ 10 14 Hz 

b 8.57 ¥ 10 14 Hz 

c 4.29 ¥ 10 14 Hz 

b A reflectance spectrum of a black pigment would show 

a low percentage of reflected light for all wavelengths 

in the visible region. 

3 a Spectrum (a) λ max 

= 440 nm (approx.) 

Spectrum (b) λ max 

= 420 nm (approx.) 

b Spectrum (b) corresponds to haemoglobin. 

4 a Red ochre. 

b It absorbs violet, blue and green light 

(λ = 400 nm–530 nm) but reflects yellow and red light 

(λ = 530 nm–700 nm). 

Intensity of 

absorption 

300 500 700 

l / nm 


1 a A chromophore is the part of a dye molecule 

responsible for its colour. It contains unsaturated 

groups such as C=O and –N=N– which are often part 

of an extended delocalised electron system involving 

arene rings. 

b 

N N 

2 a Cyanidin contains an extended delocalised 

(conjugated) electron system. The electrons in such 

systems require less energy to excite them than those 

in single bonds or in isolated double bonds. The 

excitation energy corresponds to the visible region. 

b Molecule absorbs orange light; the diagram should 

show a smaller excitation energy. 

c An extended delocalised system of electrons. 


1 a Rate of evaporation = rate of condensation. 

b It is a closed system, in which there is a dynamic 

equilibrium between water vapour and liquid. 

c No longer a closed system, H 2 

O(g) escapes. 

d i Towards H 2 

O(l) 

ii Towards H 2 

O(g). 

2 a Equilibrium lies to the reactants’ side because at 

equilibrium there is a greater concentration of 

reactants than products. 

b Equilibrium is reached at the point where the graphs 

become horizontal. 

3 B 

4 a left Æ right b right Æ left c no change 

d left Æ right e left Æ right 

5 a The concentrated hydrochloric acid moves the 

position for the equilibrium to the left, the bismuth 

trichloride is predominantly present as BiCl 3 

(aq). 

b The equilibrium would move to the right due to the 

large amount of water, hence a white precipitate of 

BiOCl(s) would be seen. 

6 a There would be no change to fizziness as the 

concentration of CO 2 

(g) is not influenced by the 

amount of air present since the system had already 

come to equilibrium. 

b The increase in concentration of CO 2 

(g) would make 

the first equilibrium move to the right and form more 

CO 2 

(aq) which in turn would make the second 

equilibrium move to the right and increase the 

concentration of H + (aq). 

c Dilute alkali would react with H + (aq) and the 

reduction in concentration of H + (aq) would cause the 

second equilibrium to move to the right. The resulting 

reduction in the concentration of CO 2 

(aq) would 

cause the first equilibrium to move to the right and 

more CO 2 

(g) would dissolve and so the equilibrium 

pressure of carbon dioxide would decrease. 

181

SECTION 7 


1 a 0.09 

5 a 2H 2 

(g) + O 2 

(g) D 2H 2 

O(g) 

0.08 

[H 

b K c 

= 2 

O(g)] 2 

[H 2 

(g)] 2 [O 2 

(g)] 

0.07 

c i Equilibrium moves towards reactants (reaction is 

0.06 

exothermic) 

ii Equilibrium moves towards products (fewer 

0.05 

gaseous molecules). 

0.04 

d i Decreases 

ii No effect 

0.03 

0 1 2 3 

[PCl 3 

(g)][Cl 2 

(g)] 

Pressure CO 6 a K c 

= 

2 (g)/atm 

[PCl 5 

(g)] 

b [CO b 0.196 mol dm –3 

Gradient = 2 

(aq)] (0.09–0.03) mol dm –3 

= 

p CO2(g) 

(2.6–0.8) atm 

[H 

7 a K c 

= 2 

S(g)] 2 

= 0.033 mol dm –3 atm –1 

[H 2 

(g)] 2 [S 2 

(g)] 

= equilibrium constant at 292 K 

Concentration CO 2 (aq)/mol dm –3 

c Drawing a ‘best fit’ line eliminates errors more 

effectively than taking an average. 

[NO 

2 a K c 

= 2 

(g)] 2 

mol –1 dm 

[NO(g)] 2 [O 3 

2 

(g)] 

[C 2 

H 4 

(g)][H 2 

(g)] 

b K c 

= mol dm –3 

[C 2 

H 6 

(g)] 

c K c 

= 

d K c 

= 

[H 2 

(g)][I 2 

(g)] 

[HI(g)] 2 

(no units) 

[HCO 3 – (aq)][H + (aq)] 

[CO 2 

(aq)][H 2 

O(l)] 

[In 

e K c 

= 2 

(aq)] 3 

mol 2 dm – 6 

[In 6 

(aq)] 

(no units) 

0.442 2 ⎞ 

b At equilibrium, [S 2 

(g)] = mol dm –3 

9.4 ¥ 10 5 ¥ 0.234 2 ⎠ 

= 3.80 ¥ 10 –6 mol dm –3 

8 a Products 

[NO(g)] 2 [O 

b K c 

= 

2 

(g)] 

[NO 2 

(g)] 2 

c 0.083 mol dm –3 

d Yes. The equilibrium concentration of NO 2 

(g) is much 

lower than the concentrations of NO(g) and O 2 

(g). 

9 a K c 

= 

[CH 3 

CH(OC 2 

H 5 

) 2 

(l)][H 2 

O(l)] 

[C 2 

H 5 

OH(l)] 2 [CH 3 

CHO(l)] 

⎞ ⎠ 

f K c 

= 

[CH 3 

COOC 3 

H 7 

(l)][H 2 

O(l)] 

[CH 3 

COOH(l)][C 3 

H 7 

OH(l)] 

3 2SO 2 

(g) + O 2 

(g) D 2SO 3 

(g) 

[NH 3 

(g)] 2 

4 a K c 

= 

[N 2 

(g)] [H 2 

(g)] 3 

b K c 

= 2.09 mol –2 dm 6 

(no units) 

b Reactants 

c Low 

d 0.074 mol –1 dm 3 

10 Likely to be low. 


p H2 O 2 

1 a K p 

= atm 

p –1 

f K 2 p 

= (no units) 

H2 p O2 p N2 p O2 

p NO 

2 

p CH3 OH 

b K p 

= 2 atm 

p –2 

CO p H2 

p H2 p CO2 

c K p 

= (no units) 

p H2 O p CO 

p C2 H 5 OH 

d K p 

= atm 

p –1 

C2 H 4 

p H2 O 

2 

p SO3 

e K p 

= 2 atm 

p –1 

SO2 p O2 

3 

p CO p H2 

2 a K p 

= 

p CH4 p H2 O 

b i K p 

increases 

ii K p 

unchanged 

iii K p 

unchanged 

c i A larger proportion of reactants (fewer gaseous 

molecules) 

ii A larger proportion of products (reaction is 

endothermic) 

iii No effect 

182

SECTION 7 

3 a K p 

= 

p O 

2 

p O2 

b 3 ¥ 10 –8 atm 

The position of the equilibrium is much further to the 

right at this altitude compared to lower regions of the 

atmosphere because of the higher intensity of the 

ultraviolet radiation with the right frequency to cause 

this dissociation. 

p NH3 

2 

4 a K p 

= 

3 

p N2 p H2 

b 1.7 ¥ 10 –4 atm –2 (The calculation follows the steps in 

the worked example on page 179.) 

c 23.7 atm 

d 100 atm, 24% 


1 Experiment number K 

1 87.1 

2 87.6 

3 87.3 

An average value of 87.3 

2 a Partition coefficient = [butanedioic acid] ethoxyethane 

[butanedioic acid] water 

= 0.148 

b [butanedioic acid] ethoxyethane 

= 0.148 ¥ 0.036 mol dm –3 

= 0.0053 mol dm –3 

[DDT(octan-1-ol)] 

3 a K ow 

= 

[DDT(aq)] 

b K ow 

is greater than 1 

c Lower 


1 The position of equilibrium would move toward the 

products, ie Ca 2+ ions will displace Mg 2+ ions from the 

clay surface. 

2 The equilibrium involved is 

R–H + (s) + Na + (aq) D R–Na + (s) + H + (aq) 

Washing with acid moves the equilibrium back to the left 

(the regenerated form). A high volume of concentrated 

acid is used to ensure that the equilibrium moves as far 

as possible to the left so that regeneration is as near 

complete as possible. 

3 a resin –OH – (s) + Cl – (aq) Æ resin –Cl – (s) + OH – (aq) 

b One might expect I – (aq) to be held more strongly by 

the resin than Cl – (aq), because I – has fewer 

surrounding water molecules. 

c H + (aq) reacts with OH – (aq) to form H 2 

O(l) 

4 The caesium, as Cs + ions, was deposited by rain. In the 

soil, the caesium ions exchanged with other cations such 

as hydrogen ions on the surface of the clays. 

Soil–H + (s) + Cs + (aq) Æ Soil–Cs + (s) + H + (aq) 

The caesium is thus held in the soil rather than being 

rapidly leached away. Like any nutrient cation, it is slowly 

released from the soil by exchange with hydrogen ions in 

subsequent rainfall. The initial ion exchange reaction is 

reversed 

Soil–Cs + (s) + H + (aq) Æ Soil–H + (s) + Cs + (aq) 

The rate of release will be greater in areas of higher 

rainfall. This will lead to a greater concentration in grass 

and a consequently higher level in livestock. This higher 

level is likely to exceed that acceptable in food. 


1 a X = 0.75; Y = 0.25 

b Compared to compound X, compound Y has a greater 

affinity for the stationary phase and/or a lower affinity 

for the mobile phase. 

c To maintain constant conditions in which the space 

around the thin layer is saturated with solvent vapour. 

2 a Retention time depends on the affinity of the 

compound for the stationary phase compared to its 

affinity for the carrier gas. More volatile compounds 

usually have shorter retention times. Other factors 

include the length and packing of the column, the 

flow rate of the gas and the temperature. 

b The extent to which a compound distributes itself 

between the mobile phase and the stationary phase, 

and hence the retention time for that compound, 

depends on the temperature, so this must be constant 

throughout the column. Most columns are kept above 

room temperature to give reasonably short retention 

times. The temperature (and other conditions) must 

be recorded and kept constant in order to obtain 

reproducible results. If the temperature varies, the 

rates of elution of the substances will vary and results 

will be inconsistent between experiments. 

c Compounds with very high boiling points would have 

very long retention times. Compounds which 

decompose on heating may break down into smaller 

compounds in the column. 

3 a Ratio of cis : trans isomers was 1 : 1.4. 

b They are likely to be similar because they are isomers. 

4 a Methylpropane 

b The more carbon atoms in the molecule, the longer 

the compound takes to travel through the column. 

Larger molecules are less volatile, spend more time 

dissolved in the stationary liquid phase and less time 

in the gas phase. 

c All the times would be longer and the peaks would be 

further apart. 

183

SECTION 8 


1 a i AgI(s) D Ag + (aq) + I – (aq) 

ii BaSO 4 

(s) D Ba 2+ (aq) + SO 4 2– (aq) 

iii PbI 2 

(s) D Pb 2+ (aq) + 2I – (aq) 

iv Fe(OH) 3 

(s) D Fe 3+ (aq) + 3OH – (aq) 

b i K sp 

= [Ag + (aq)] [I – (aq)] mol 2 dm –6 

ii K sp 

= [Ba 2+ (aq)] [SO 4 2– (aq)] mol 2 dm –6 

iii K sp 

= [Pb 2+ (aq)] [I – (aq)] 2 mol 3 dm –9 

iv K sp 

= [Fe 3+ (aq)] [OH – (aq)] 3 mol 4 dm –12 

c The reactant is a solid and so its concentration can be 

regarded as constant. 

2 a 

Product of ion 

K sp 

concentrations 

/ mol 2 dm –6 

i 

ii 

iii 

iv 


4.0 ¥ 10 –12 5.0 ¥ 10 –13 mol 2 dm –6 

9.0 ¥ 10 –12 1.0 ¥ 10 –10 mol 2 dm –6 

2.0 ¥ 10 –9 1.6 ¥ 10 –8 mol 2 dm –6 

2.0 ¥ 10 –6 5.0 ¥ 10 –9 mol 2 dm –6 

b A precipitate will form in i and iv as in these cases the 

product of the ion concentrations is greater than the 

solubility product. 

3 a [Ag + (aq)] = 0.005 mol dm –3 ; 

[Cl – (aq)] = 0.005 mol dm –3 

b [Ca 2+ (aq)] = 0.001 mol dm –3 

4 a [Tl + (aq)] = 0.5 ¥ 7.0 ¥ 10 –3 mol dm –3 . 

(Remember the concentration will be halved on 

mixing with an equal volume of sodium chloride.) 

1 Acid donates hydrogen ions/protons and the base 

accepts hydrogen ions/protons. 

2 a HNO 3 

+ H 2 

O Æ H 3 

O + – 

+ NO 3 

acid base 

b NH 3 

+ H 2 

O Æ NH + 4 

+ OH – 

base acid 

c NH + 4 

+ OH – Æ NH 3 

+ H 2 

O 

acid base 

d SO 2– 4 

+ H 3 

O + Æ HSO – 4 

+ H 2 

O 

base acid 

e H 2 

O + H – Æ H 2 

+ OH – 

acid base 

f H 3 

O + + OH – Æ 2H 2 

O 

acid base 

g NH 3 

+ HBr Æ NH + 4 

+ Br – 

base acid 

h H 2 

SO 4 

+ HNO 3 

Æ HSO – + 

4 

+ H 2 

NO 3 

acid base 

i CH 3 

COOH + H 2 

O Æ CH 3 

COO – + H 3 

O + 

acid base 

3 a Acid–base 

b Acid–base 

c Redox 

d Redox 

K sp 

= (0.5 ¥ 7.0 ¥ 10 – 3 

mol dm –3 ) ¥ [Cl – (aq)] 

1.75 ¥ 10 –4 mol 2 dm –6 

[Cl – (aq)] = 

0.5 ¥ 7.0 ¥ 10 –3 mol dm –3 

= 0.05 mol dm –3 

As with the thallium ion, the concentration of chloride 

ion will be halved on mixing. Hence the concentration 

of sodium chloride above which a precipitate will just 

form is 0.10 mol dm –3 . 

b The volume of the mixture becomes 200 cm 3 , therefore 

both ionic concentrations become 5 ¥ 10 –3 mol dm –3 . 

Hence [Ag + (aq)] ¥ [BrO – 3 

(aq)] = 2.5 ¥ 10 –5 mol 2 dm –6 

at 298 K, which is less than K sp 

so no precipitate would 

be observed. 

5 a [Ag + (aq)] = 1.41 ¥ 10 –5 mol dm –3 , 

[Cl – (aq)] = 1.41 ¥ 10 –5 mol dm –3 

b 2.02 ¥ 10 –3 gdm –3 

c The new [Cl – (aq)] = 5.0 ¥ 10 –1 mol dm –3 . 

(The Cl – ions from AgCl will be negligible in 

comparison.) The product of the concentrations of 

silver and chloride ions 

= (5.0 ¥ 10 –1 mol dm –3 ) ¥ (0.5 ¥ 1.41 ¥ 10 –5 mol dm –3 ) 

= 3.5 ¥ 10 –6 mol 2 dm –6 

which exceeds the K sp 

at this temperature and so a 

white precipitate of silver chloride would be observed. 

4 Conjugate pairs: acid base 

a 

+ 

NH 4 

NH 3 

H 3 

O + 

H 2 

O 

b H 2 

SO 4 

– 

HSO 4 

+ 

H 2 

NO 3 

HNO 3 

c HClO 4 

– 

ClO 4 

+ 

CH 3 

COOH 2 

CH 3 

COOH 

5 a CH 3 

COOH + OH – Æ CH 3 

COO – + H 2 

O 

Conjugate pairs: acid 

base 

CH 3 

COOH CH 3 

COO – 

H 2 

O OH – 

b HCO – 3 

+ HCl Æ H 2 

CO 3 

+ Cl – 


base 

HCl Cl – 

H 2 

CO 3 

– 

HCO 3 

c H 2 

O + HSO – 4 

Æ SO 2– 4 

+ H 3 

O + 


base 

– 

HSO 4 

2– 

SO 4 

H 3 

O + 

H 2 

O 

184

SECTION 8 


1 a [H + (aq)] = 1 ¥ 10 –2 mol dm –3 ; therefore pH = 2 

b [H + (aq)] = 2 ¥ 10 –1 mol dm –3 ; therefore pH = 0.7 

c [H + (aq)] = 4 ¥ 10 –1 mol dm –3 ; therefore pH = 0.4 

d [H + (aq)] = 4 ¥ 10 –1 mol dm –3 ; therefore pH = 0.4 

[H + (aq)] 2 

2 a 1.7 ¥ 10 –5 = 

1 ¥ 10 –1 

[H + (aq)] 2 = 1.7 ¥ 10 –6 mol 2 dm –6 

[H + (aq)] = 1.3 ¥ 10 –3 mol dm –3 

pH = 2.9 

[H + (aq)] 2 

b 1.7 ¥ 10 –5 = 

5 ¥ 10 –2 

[H + (aq)] 2 = 8.5 ¥ 10 –7 mol 2 dm –6 

[H + (aq)] = 9.2 ¥ 10 –4 mol dm –3 

pH = 3.0 

[H + (aq)] 2 

c 6.3 ¥ 10 –5 = 

1 ¥ 10 –3 

[H + (aq)] 2 = 6.3 ¥ 10 –8 mol 2 dm –6 

[H + (aq)] = 2.5 ¥ 10 –4 mol dm –3 

pH = 3.6 

[H + (aq)] 2 

d 1.6 ¥ 10 –4 = 

2.5 

[H + (aq)] 2 = 4.0 ¥ 10 –4 mol 2 dm –6 

[H + (aq)] = 2.0 ¥ 10 –2 mol dm –3 

pH = 1.7 

3 a The reaction of the acid with water goes to 

completion. 

b [H + (aq)] = [A – (aq)] 

[HA(aq)] at equilibrium = original [HA(aq)]. 

4 a Strong acid – hydrochloric acid 

Weak acid – nitrous acid (nitric(III) acid) 

5 

b The position of the equilibrium for the reaction of the 

strong acid with water is completely to the right: 

HCl(aq) + H 2 

O(l) Æ Cl – (aq ) + H 3 

O + (aq) 

The amount in moles of H + (aq) ions is equal to the 

amount in moles of HCl put into solution. Thus, 

[H + (aq)] = 0.01 mol dm –3 and pH = 2. 

The position of the equilibrium for the reaction of the 

weak acid with water is more to the left: 

HNO 2 

(aq) + H 2 

O(l) D NO 2 – (aq) + H 3 

O + (aq) 

The amount in moles of H + (aq) ions is very much less 

than the amount in moles of HNO 2 

put into solution. 

To obtain a solution with pH = 2, the nitrous acid 

solution must be more concentrated than the 

hydrochloric acid solution. 

a 

b 

c 

[OH – (aq)]/mol dm –3 [H + (aq)]/mol dm –3 pH 

1 1 ¥ 10 –14 14 

0.01 1 ¥ 10 –12 12 

0.2 5 ¥ 10 –14 13.3 

6 a In alkaline solution, the equilibrium shifts to the right 

as H + (aq) is removed by reaction with OH – (aq), so the 

indicator will be present as the pink In – form. 

b K a 

= [H+ (aq)][In – (aq)] 

[HIn(aq)] 

c K a 

= 5.01 ¥ 10 –10 mol dm –3 

= [H+ (aq)][In – (aq)] 

[HIn(aq)] 

At the end point [HIn(aq)] = [In – (aq)] 

Hence 5.01 ¥ 10 –10 mol dm –3 = [H + (aq)] 

pH at end point = 9.3 

7 a i K a 

= 5.0 ¥ 10 –10 mol dm –3 , pK a 

= 9.3 

ii K a 

= 1.3 ¥ 10 –10 mol dm –3 , pK a 

= 9.9 

iii K a 

= 4.8 ¥ 10 –4 mol dm –3 , pK a 

= 3.3 

b HF, HCN, phenol 


1 a 1.6 ¥ 10 –4 mol dm –3 = [H + 0.1 mol dm–3 

f In 750 cm 

(aq)] ¥ 3 of solution the concentrations of acid and 

0.1 mol dm –3 

salt are 

Therefore [H + (aq)] = 1.6 ¥ 10 –4 mol dm –3 

0.1 ¥ 250 

pH = 3.8 

[acid] = mol dm 

750 

–3 

b 6.3 ¥ 10 –5 mol dm –3 = [H + 0.03 mol dm–3 

(aq)] ¥ 

0.01 mol dm –3 

0.1 ¥ 500 


[salt] = mol dm –3 

750 

pH = 4.7 

0.1 ¥ 500 mol dm –3 

c The concentrations of acid and salt in the mixture will 

1.7 ¥ 10 –5 mol dm 3 = [H + (aq)] ¥ 

0.1 ¥ 250 moldm –3 

both be 0.05 mol dm –3 . This is equivalent to diluting 

the buffer in 1 a with an equal volume of water. The 

[H + (aq)] = 8.5 ¥ 10 –6 mol dm –3 

pH remains unchanged at 3.8. 

pH = 5.1 

d 1.3 ¥ 10 –5 mol dm –3 = [H + (aq)] ¥ 5.0 ¥ 10–3 mol dm –3 

1 ¥ 10 –1 mol dm –3 

g 1.7 ¥ 10 –5 mol dm –3 = [H + 0.1 mol dm –3 

(aq)] ¥ 


0.2 mol dm –3 

pH = 3.6 

[H + (aq)] = 3.4 ¥ 10 –5 mol dm –3 

2 ¥ 1.5 ¥ 10 –2 mol dm –3 

e 1.6 ¥ 10 –4 mol dm –3 = [H + pH = 4.5 

(aq)] ¥ 

2 ¥ 5 ¥ 10 –3 mol dm –3 The pH of the buffer solution, on adding a higher 

[H + (aq)] = 5.3 ¥ 10 –5 mol dm –3 

proportion of acid, has decreased. Changing the ratio 

pH = 4.3 

of [salt] : [acid] provides a way of ‘fine tuning’ the pH 

of a buffer solution. 

185

SECTION 9 

2 a Buffer solutions are made of either a weak acid and 

one of its salts or a weak base and one of its salts. 

b i Ethanoate ions from the salt react with the extra 

H + (aq) ions to form ethanoic acid and water and 

so prevent a fall in the pH. 

ii The addition of OH – (aq) ions removes H + (aq) 

but these are replaced by further dissociation of 

the ethanoic acid so the pH will remain constant. 

iii Addition of a small amount of water will change 

the concentration of the acid and the salt by the 

same factor which means the ratio of [salt] to 

[acid] will remain constant and the pH will 

remain constant. 

3 Ethanoic acid, because its K a 

is close to the required 

[H + (aq)] (pK a 

close to the required pH). 


1 a b i K Æ K + + e – oxidation 

ii H 2 

Æ 2H + + 2e – oxidation 

iii O+2e – Æ O 2– reduction 

iv Cu + Æ Cu 2+ + e – oxidation 

v Cr 3+ + e – Æ Cr 2+ reduction 

2 a Ag(+1) f N(–3) k S(+6), F(–1) 

b Br(0) g Mg(+2), Cl(–1) l S(+6), O(–2) 

c P(0) h C(+4), O(–2) m N(+5), O(–2) 

d H(+1) i P(+5), Cl(–1) n P(+5), O(–2) 

e H(–1) j Al(+3), O(–2) 

3 a Cl(0) Æ Cl(–1) reduced 

Fe(0) Æ Fe(+3) oxidised 

b Cl(0) Æ Cl(–1) 

reduced 

H(0) Æ H(+1) 

oxidised 

c Cl(0) Æ Cl(–1) 

reduced 

Fe(+2) Æ Fe(+3) oxidised 

d F(0) Æ F(–1) 

reduced 

O(–2) Æ O(0) 

oxidised 

4 a i Cl 2 

ii Fe 

b i Cl 2 

ii H 2 

c i Cl 2 

ii FeCl 2 

d i F 2 

ii H 2 

O 

5 a Cl(+5) Æ Cl(–1) reduced 

O(–2) Æ O(0) 

oxidised 

b S(+6) Æ S(+4) reduced 

Br(–1) Æ Br(0) 

oxidised 

c S(+6) Æ S(–2) 

reduced 

I(–1) Æ I(0) 

oxidised 

d I(0) Æ I(–1) 

reduced 

S(+4) Æ S(+6) oxidised 

6 a Cu 2 

O + 2H + Æ Cu 2+ + Cu + H 2 

O 

Cu: 2(+1) +2 0 oxidised and 

reduced 

O: –2 –2 no change 

H: 2(+1) 2(+1) no change 

b 3Br 2 

+ 6OH – Æ BrO – 3 

+ 5Br – + 3H 2 

O 

Br: 3(0) +5 5(–1) oxidised and 

reduced 

O: 6(–2) 3(–2) 3(–2) no change 

H: 6(+1) 6(+1) no change 

c 4IO – 3 

Æ 3IO – 4 

+ I – 

I: 4(+5) 3(+7) –1 oxidised and reduced 

O: 12(–2) 12(–2) no change 

7 a tin(II) oxide h nitrate(III) 

b tin(IV) oxide i nitrate(V) 

c iron(II) chloride j sulphate(IV) 

d iron(III) chloride k sulphate(VI) 

e lead(IV) chloride l manganate(VII) 

f copper(I) oxide m chromate(VI) 

g manganese(II) hydroxide n vanadate(V) 

8 a KClO 2 

b NaClO 3 

c Fe(OH) 3 

d Cu(NO 3 

) 2 


1 a Pb 2+ (aq) + 2e – Æ Pb(s) Zn(s) Æ Zn 2+ (aq) + 2e – 

b 6H + (aq) + 6e – Æ 3H 2 

(g) 2Al(s) Æ 2Al 3+ (aq) + 6e – 

c 2Ag + (aq) + 2e – Æ 2Ag(s) Cu(s) Æ Cu 2+ (aq) + 2e – 

d Cl 2 

(g) + 2e – Æ 2Cl – (aq) 2I – (aq) Æ I 2 

(aq) + 2e – 

e S(s) + 2e – Æ S 2– (s) Zn(s) Æ Zn 2+ (s) + 2e – 

2 a Mg(s) + Cu 2+ (aq) Æ Mg 2+ (aq) + Cu(s) 

b Zn(s) + 2Ag + (aq) Æ Zn 2+ (aq) + 2Ag(s) 

c 3Mg(s) + 2Au 3+ (aq) Æ 3Mg 2+ (aq) + 2Au(s) 

d 2Fe 2+ (aq) + Cl 2 

(g) Æ 2Fe 3+ (aq) + 2Cl – (aq) 

3 a i 3.16 V 

ii 1.10 V 

iii 0.18 V 

iv 0.32 V 

v 1.36 V 

b i Ag + (aq)/Ag(s) 

ii Cu 2+ (aq)/Cu(s) 

iii Ni 2+ (aq)/Ni(s) 

iv Fe 2+ (aq)/Fe(s) 

v MnO – 4 

(aq)/Mn 2+ (aq) 

186

SECTION 9 

c i 2Ag + (aq) + Mg(s) Æ 2Ag(s) + Mg 2+ (aq) 

ii Cu 2+ (aq) + Zn(s) Æ Cu(s) + Zn 2+ (aq) 

iii Ni 2+ (aq) + Fe(s) Æ Ni(s) + Fe 2+ (aq) 

iv Fe 2+ (aq) + Zn(s) Æ Fe(s) + Zn 2+ (aq) 

v 2MnO – 4 

(aq) + 5Sn 4+ (aq) + 16H + (aq) 

Æ 2Mn 2+ (aq) + 5Sn 2+ (aq) + 8H 2 

O(l) 

4 Cd 2+ (aq)/Cd(s) E! = –0.40 V 

5 Co 2+ (aq)/Co(s) E! = –0.28 V 

6 Pb 2+ (aq)/Pb(s) E! = –0.13 V 

7 Fe 3+ (aq)/Fe 2+ (aq) E! = + 0.77 V 

8 F 2 

(g)/2F – (aq) E! = +2.85 V 

9 K, Ce, Cd, Ni, Sn, Ag 

10 2H + (aq)/H 2 

(g) E! = 0 V by definition 

Pb 2+ (aq)/Pb(s) E! = –0.13 V 

Cd 2+ (aq)/Cd(s) E! = –0.40 V 

Ag + (aq)/Ag (s) E! = +0.80 V 

Cr 3+ (aq)/Cr(s) E! = –0.74 V 


1 i 2I – (aq) Æ I 2 

(aq) + 2e – ; Cl 2 

(g) + 2e – Æ 2Cl – (aq) 

overall: Cl 2 

(g) + 2I – (aq) Æ 2Cl – (aq) + I 2 

(aq) 

ii 2Br – (aq) Æ Br 2 

(aq) + 2e – ; 

MnO – 4 

(aq) + 8H + (aq) + 5e – 

Æ Mn 2+ (aq) + 4H 2 

O(l) 

overall: 2MnO – 4 

(aq) + 16H + (aq) + 10Br – (aq) 

Æ 2Mn 2+ (aq) + 8H 2 

O(l) + 5Br 2 

(aq) 

iii 2I – (aq) Æ 2I 2 

(aq) + 2e – ; Br 2 

(aq)+ 2e – Æ 2Br – (aq) 

overall: 2I – (aq) + Br 2 

(aq) Æ I 2 

(aq) + 2Br – (aq) 

2 a Yes c No 

b Yes 

d Yes 

3 I 2 

(aq) + 2e – Æ 2I – (aq) 

Cr 2 

O 2– 7 

(aq) + 14H + (aq)+6e – Æ 2Cr 3+ (aq) + 7H 2 

O(l) 

overall: Cr 2 

O 2– 7 

(aq) + 14H + (aq) + 6I – (aq) 

Æ2Cr 3+ (aq)+ 7H 2 

O(l) + 3I 2 

(aq) 


1 a i Cobalt(III) fluoride dissolves in water to form 

cobalt(III) ions, [Co(H 2 

O) 6 

] 3+ . E! for the 

[Co(H 2 

O) 6 

] 3+ /[Co(H 2 

O) 6 

] 2+ half-cell is more 

positive than E! for the O 2 

(g), H + (aq)/H 2 

O(l) 

half-cell. Electrons are supplied to the more 

positive half-cell, so cobalt(III) ions are reduced 

to cobalt(II) ions and water is oxidised to release 

oxygen. 

ii The overall equation is 

4[Co(H 2 

O) 6 

] 3+ (aq) + 2H 2 

O(l) 

Æ 4[Co(H 2 

O) 6 

] 2+ (aq) + O 2 

(g) + 4H + (aq) 

iii The hydrated Co 2+ ion is more stable than the 

hydrated Co 3+ ion. 

b Oxygen in the air is unable to oxidise the pink cobalt(II) 

ion as the E! for the oxygen half equation is less positive 

than that required for oxidising [Co(H 2 

O) 6 

] 2+ (aq). 

c The E! for the oxygen half equation is more positive 

than that needed to oxidise the yellow-brown 

[Co(NH 3 

) 6 

] 2+ (aq) to the dark brown [Co(NH 3 

) 6 

] 3+ (aq). 

The overall equation is 

O 2 

(g) + 4H + (aq) + 4[Co(NH 3 

) 6 

] 2+ (aq) 

Æ 2H 2 

O(l) + 4[Co(NH 3 

) 6 

] 3+ (aq) 

d [Co(NH 3 

) 6 

] 3+ is more stable than [Co(NH 3 

) 6 

] 2+ . 

4 b i No 

ii No 

iii Yes 

5 a Yes b No 

c Yes 

d Yes 

e Yes 

6 a,b i CH 4 

(g) + 2O 2 

(g) Æ CO 2 

(g) + 2H 2 

O(l) 

ii 2CH 3 

OH(aq) + O 2 

(g) Æ 2HCHO(aq) + 2H 2 

O(l) 

iii 2HCHO(aq) + O 2 

(g) Æ 2HCOOH(aq) 

2 b i As the E! for aqueous iron(III), [Fe(H 2 

O) 6 

] 3+ (aq), 

is more positive than that of the iodine half-cell, 

the iron(III) will oxidise the aqueous iodide ion 

to iodine. 


2[Fe(H 2 

O) 6 

] 3+ (aq) + 2I – (aq) 

Æ 2[Fe(H 2 

O) 6 

] 2+ (aq) + I 2 

(aq) 

E cell = +0.77 V – (+0.54 V) = +0.23 V 

ii As the E! for the iodine half-cell is more positive 

than that of hexacyanoferrate half-cell, iodine will 

oxidise the hexacyanoferrate(II) to 

hexacyanoferrate(III). 


I 2 

(aq) + 2[Fe(CN) 6 

] 4– (aq) 

Æ 2I – (aq) + [Fe(CN) 6 

] 3– (aq) 

E cell = +0.54 V – (+0.36 V) = +0.18 V 

c A major reason why many predicted reactions do not 

occur is because of high activation enthalpies, ie there 

is a kinetic barrier to the reaction despite it appearing 

thermodynamically feasible. Also, these calculations 

assume standard conditions: if conditions are not 

standard, the results may be different. 

187

SECTION 10 


1 

a b c 

A rate will increase with rate will increase with rate will increase with 

temperature temperature temperature temperature 

B rate of forward reaction not rate increases rate of forward reaction not 

total pressure of gas affected affected 

C increasing the concentration solutions not involved increasing the concentration of 

concentration of solution of acid will increase the rate peroxide will increase the rate 

D the more finely divided the the more finely divided the solids not involved 

surface area of solid magnesium, the faster the rate catalyst, the faster the rate 

2 Both the acid and the enzyme can act as catalysts for the 

hydrolysis of a protein. 

3 a The greater the concentration of reactants, the greater 

the rate of collisions and hence the faster the reaction 

proceeds. 

b A change of temperature has little effect. Most 

collisions result in a reaction. 

4 a B and C 

bAand D 

cD 

dB 

eB 

f D 


1 a A b A c B d Mainly B, with A to a minor extent. 

2 a This reaction has a high activation enthalpy that 

prevents it occurring at a significant rate at room 

temperature, but the reaction is exothermic, and once 

the spark has provided the energy needed to get it 

started, the reaction produces enough energy to 

sustain itself regardless of how much is present. 

b The platinum catalyst lowers the activation enthalpy to 

such an extent that it is close to the thermal energy of 

molecules at room temperature. 

3 Above a certain temperature, enzymes are denatured and 

become inactive. 

4 a The surface area of the coal is much greater in the 

powder than in the lump. Many more collisions with 

oxygen molecules are possible and the speed of 

reaction will be much greater. 

b Although the gas molecules are moving freely, the 

molecules have insufficient kinetic energy to 

overcome the activation enthalpy for reaction. 

c The particles in the solids are in fixed positions in their 

respective lattices. The only movement will be due to 

low energy vibrations or rotations about these fixed 

positions. The number of collisions is very low indeed. 

There is also unlikely to be sufficient energy available 

to overcome the activation enthalpy for reaction. 

d The fine flour dust allows maximum chances of 

collisions with oxygen molecules. A spark will cause 

instant ignition followed by a very rapid reaction 

amounting to an explosion. 

5 Catalytic converters catalyse redox reactions involving 

CO, NO x 

and oxygen from the air (see Developing 

Fuels for details). The catalyst lowers the activation 

enthalpies of these reactions, but the activation 

enthalpies are still high, and the reactions do not occur 

at a significant rate until the catalyst is hot. 

6 The added curve is above the original, with a greater 

slope at the start of the reaction but plateauing at the 

same final volume of hydrogen given off. 

7 a The area shaded is underneath the T 1 

curve and to the 

right of E a 

. 

b The area shaded a different colour is underneath the 

T 2 

curve and to the right of E a 

, encompassing the firstcoloured 

area. The T 2 

curve has a lower and broader 

maximum than the T 1 

curve and the maximum value is 

shifted to the right. It tails off above the T 1 

curve. 

188 


1 a The reaction is first order with respect to 

bromoethane and zero order with respect to 

hydroxide ion. 

b The reaction is first order with respect to methyl 

methanoate, zero order with respect to water and first 

order with respect to H + . 

c The reaction is first order with respect to urea, zero 

order with respect to water and first order with 

respect to urease. 

d The reaction is a single step in the mechanism. It is 

first order with respect to the methyl radical and first 

order with respect to the chlorine molecule. 

e The reaction is order with respect to carbon 

monoxide and first order with respect to chlorine. 

f The reaction is second order with respect to nitrogen 

dioxide. 

2 a Rate = k[CH 3 

CH 2 

CH 2 

CH 2 

Cl] [OH – ] 

b Rate = k[C 12 

H 22 

O 11 

] [H + ]

SECTION 11 

3 a i First order 

ii Second order 

b Rate = k[H 2 

][NO] 2 

c k = 0.384 mol –2 dm 6 s –1 

4 b All are ca 1150 s (Allow within 1100–1200 s.) 

First order. 

e Rate = k[N 2 

O 5 

] or –d[N 2 

O 5 

]/dt = k[N 2 

O 5 

] 

f 6.2 ¥ 10 –4 s –1 

(Method is difficult to use. Allow for an answer 

between 5 and 7 ¥ 10 –4 s –1 .) 

5 a Structural isomerism 

b 0.5 atm 

d t 

is constant and about 55 ¥ 10 3 s. First order. 


1 a Platinum on aluminium oxide (heterogeneous 

catalyst). 

b Zeolite (heterogeneous catalyst). 

c Platinum on aluminium oxide (heterogeneous 

catalyst). 

2 a 2CO(g) + 2NO(g) Æ 2CO 2 

(g) + N 2 

(g) 

b These gases are hazardous to health, produce acid rain 

and can take part in producing photochemical smog. 

c i A substance is adsorbed when it is bound to the 

surface of another substance. 

ii The mechanism should show CO and NO being 

adsorbed to the catalyst surface. 

O N N O C O 

The bonds in the molecules are weakened and 

new bonds form between the atoms. 

O N N 

catalyst surface 

Products are released from the surface. 

d The reaction between nitrogen and carbon monoxide 

on the surface of the catalyst is faster at high 

temperature. 

e i Lead poisons the catalyst in the converter. It is 

adsorbed strongly to the surface of the catalyst 

and prevents CO and NO being adsorbed. 

ii Use unleaded fuel. 

O 

C 

O 

catalyst surface 


1 a 

Enthalpy 

Platinum 

catalyst 

Reactants 

Section 11.1 

Enzyme 

catalyst 


+ 49 kJ mol –1 

+ 36 kJ mol –1 

DH 

Products 

1 a d block b p block c s block d f block 

2 a Peaks: carbon, silicon 

Troughs: helium, neon, argon 

b Elements in the same group occur at similar positions 

on graph eg Group 4 at peaks, Group 0 in troughs. 

3 a They are all in Group 1. 

b The molar atomic volumes for each period of elements 

vary in similar regular patterns. 

4 a LiCl, BeCl 2 

, BCl 3 

, CCl 4 

, NCl 3 

, OCl 2 

, FCl 

NaCl, MgCl 2 

, AlCl 3 

, SiCl 4 

, PCl 5 

, SCl 4 

, Cl 2 

c Second period: number of chlorine atoms same as 

group number for Groups 1–4. After Group 4, falls by 

b The rate will be faster for the enzyme catalysed 

reaction, as the activation enthalpy is significantly 

lower. This means that more pairs of colliding 

molecules have sufficient energy to react at room 

temperature. 

2 a The catalyst is in the same phase as the reactants. 

b i An intermediate is formed. In this example, the 

intermediate is the radical, ClO. 

ii The first peak represents the energy that must be 

supplied to enable bonds in the reactants (Cl and 

O 3 

) to stretch and break as new bonds form in 

the products (ClO and O 2 

). The trough 

represents the energy released when ClO and O 2 

are formed from Cl and O 3 

. The second peak 

represents the energy required for ClO and O to 

come together and allow the formation of Cl and 

O 2 

which are the products. 

one chlorine atom for each subsequent group. 

Similar pattern for third period, rise continues to 

Group 5 before falling. 

d Same formulae. 

5 a Li 2 

O, BeO, B 2 

O 3 

, CO 2 

, N 2 

O 5 

, F 2 

O 

Na 2 

O, MgO, Al 2 

O 3 

, SiO 2 

, P 2 

O 5 

, SO 3 

, Cl 2 

O 7 

c Second period: number of oxygen atoms per atom of 

element increases by a half up to Group 5 before falling. 

Third period, increases from Groups 1 through to 7. 

d Same formulae except for halogens where fluorine is 

unable to form the higher oxides. 

189

SECTION 11 


1 

Element Trend in reactivity Trend in thermal Trend in pH of Trend in solubility Trend in solubility 

with water stability of carbonate hydroxide with water of hydroxide of carbonate 

Mg 

Ca 

Sr 

Ba 

increases 

decompose 

at increasingly 

higher temperature 

increases 

increases 

increases 

2 Sulphate solubility decreases down group. 

3 a magnesium + steam 

Æ magnesium hydroxide + hydrogen 

Mg(s) + 2H 2 

O(g) Æ Mg(OH) 2 

(s) + H 2 

(g) 

b calcium oxide + hydrochloric acid 

Æ calcium chloride + water 

CaO(s) + 2HCl(aq) Æ CaCl 2 

(aq) + H 2 

O(l) 

c beryllium carbonate 

Æ beryllium oxide + carbon dioxide 

BeCO 3 

(s) Æ BeO(s) + CO 2 

(g) 

d barium hydroxide + sulphuric acid 

Æ barium sulphate + water 

Ba(OH) 2 

(aq or s) + H 2 

SO 4 

(aq) Æ BaSO 4 

(s) + 2H 2 

O(l) 

4 a Cs (or Fr) 

b Cs (or Fr) 

c +1 

di M 2 

O ii MOH iii M 2 

CO 3 

5 a lithium + water Æ lithium hydroxide + hydrogen 

2Li(s) + 2H 2 

O(l) Æ 2LiOH(aq) + H 2 

(g) 

b hydrochloric acid + sodium hydroxide 

Æ sodium chloride + water 

HCl(aq) + NaOH(aq) Æ NaCl(aq) + H 2 

O(l) 

c Little reaction even at high temperatures 

d sodium oxide + sulphuric acid 

Æ sodium sulphate + water 

Na 2 

O(s) + H 2 

SO 4 

(aq) Æ Na 2 

SO 4 

(aq) + H 2 

O(l) 


1 The triple bond in N 2 

is very strong (bond enthalpy = 

945 kJ mol –1 ), so a great deal of activation enthalpy must 

be supplied before N 2 

will react. P 4 

, on the other hand, 

needs only enough energy to break one of the P–P bonds 

(bond enthalpy = 198 kJ mol –1 ) to start it reacting. The 

reason that P does not form triple bonds like those in N 2 

is related to its larger size. 

2 a i +3 Æ +5 

ii +5 Æ 0 

iii –3 Æ –3 

iv 0 Æ –3 

v +2 Æ +4 

b i A 

ii B 

iii C 

iv B 

v A 

3 a +2 to +1 


1 a i Deep brown colour 

2I – (aq) + Cl 2 

(aq) Æ I 2 

(aq/s) + 2Cl – (aq) 

ii No change 

iii Light brown/red colour deepens in tone, 

becoming dark brown 

2I – (aq) + Br 2 

(aq) Æ I 2 

(aq/s) + 2Br – (aq) 

iv Light brown/red colour 

2Br – (aq) + Cl 2 

(aq) Æ Br 2 

(aq) + 2Cl – (aq) 

b i Ag + (aq) + Cl – (aq) Æ AgCl(s) 

ii Ag + (aq) + Br – (aq) Æ AgBr(s) 

iii Ag + (aq) + I – (aq) Æ AgI(s) 

b 2NO(g) + 2H + (aq) + 2e – Æ N 2 

O(g) + H 2 

O(l) 

4 i NO – 2 

(aq) + H 2 

O(l) Æ NO – 3 

(aq) + 2H + (aq) + 2e – 

ii NO – 3 

(aq) + 6H + (aq) + 5e – Æ N 2 

(g) + 3H 2 

O(l) 

iii NH + 4 

(aq) + OH – (aq) Æ NH 3 

(g) + H 2 

O(l) 

iv N 2 

(g) + 6H + (aq) + 6e – Æ 2NH 3 

(g) 

v NO(g) + H 2 

O(l) Æ NO 2 

(g) + 2H + (aq) + 2e – 

5 a The NO is oxidised by air as in reaction (2), to produce 

further NO 2 

which takes part in reaction (3). 

b Ammonium nitrate fertiliser. 

c i 1 mole of NH 3 

Æ mole of HNO 3 

1000 kg ¥ (63/17) ¥ = 2471 kg 

ii Incomplete reaction at each of the three stages. 

Loss of intermediates. 

d The reactants and intermediates include serious 

environmental pollutants, particularly NO x 

. Escape of 

these, even in small quantities, would lead to acid rain 

and direct effects on living things. The NO x 

could be 

absorbed in aqueous sodium hydroxide. 

c i 2Na(s) + Br 2 

(l/g) Æ 2NaBr(s) 

ii Mg(s) + Cl 2 

(g) Æ MgCl 2 

(s) 

iii 2K(s) + I 2 

(s/g) Æ 2KI(s) 

iv Ca(s) + Cl 2 

(g) Æ CaCl 2 

(s) 

2 a I +1, Cl –1 

b Br +3, F –1 

c Br +1 

d I +5 

e I +7 

190

SECTION 11 

3 a Cl 2 

(g) + 2NaOH(aq) 

Æ NaCl(aq) + NaClO(aq) + H 2 

O(l) 

b i Chlorine 

ii Cl from 0 in Cl 2 

to –1 in NaCl, ie reduction 

Cl from 0 in Cl 2 

to +1 in NaClO, ie oxidation 

c i 2NaClO(aq) Æ 2NaCl(aq) + O 2 

(g) 

ii Oxygen and chlorine. 

iii Oxygen from –2 in NaClO to 0 in O 2 

, ie oxidation 

Chlorine from +1 in NaClO to –1 in NaCl, ie 

reduction. 

4 a isotopes and 

210 At 

abundances 

melting point/K 575 

(allow 

480–600) 

boiling point/K 610 

(allow 

550–700) 

b 

i The orange aqueous bromine water turns a dark 

brown/black and a black precipitate may form. 

Br 2 

(aq) + 2At – (aq) Æ 2Br – (aq) + At 2 

(s) 

ii The orange-brown colour of iodine is replaced by 

a much darker colour and/or a black precipitate. 

I 2 

(aq) + 2At – (aq) Æ 2I – (aq) + At 2 

(s) 

iii The two colourless solutions produce a yellow 

precipitate. 

Ag + (aq) + At – (aq) Æ AgAt(s) 

iv The sodium burns with an orange flame and a 

white solid is produced. 

2Na(s) + At 2 

(g) Æ 2NaAt(s) 

solubility 

at 293 K/g per 

100 g water 

allow any 

low value 


1 

Element Electronic Element Electronic 

configuration 

configuration 

Sc [Ar] 3d 1 4s 2 Fe [Ar] 3d 6 4s 2 

Ti [Ar] 3d 2 4s 2 Co [Ar] 3d 7 4s 2 

V [Ar] 3d 3 4s 2 Ni [Ar] 3d 8 4s 2 

Cr [Ar] 3d 5 4s 1 Cu [Ar] 3d 10 4s 1 

Mn [Ar] 3d 5 4s 2 Zn [Ar] 3d 10 4s 2 

2 a i Cu 2+ [Ar] 3d 9 4s 0 

ii Cu + [Ar] 3d 10 4s 0 

iii Fe 3+ [Ar] 3d 5 4s 0 

iv V 3+ [Ar] 3d 2 4s 0 

v Cr 3+ [Ar] 3d 3 4s 0 

vi Ni 2+ [Ar] 3d 8 4s 0 

b Cu 2+ has a partially filled 3d sub-shell and behaves as a 

typical d block transition metal ion. 

Cu + has a filled 3d sub-shell and cannot show such 

properties. 

3 a The three transition metals atoms (Cr, Fe, Co) are the 

same size as each other, but are smaller than the 

atoms of sodium (Na) and magnesium (Mg). 

b The melting and boiling points of the transition metals 

are much higher than those of sodium and magnesium. 

c The three transition metals are much denser than 

sodium and magnesium. 

d Cr 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 

Fe 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 

Co 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7 4s 2 

Na 1s 2 2s 2 2p 6 3s 1 

Mg 1s 2 2s 2 2p 6 3s 2 

e The properties of elements are governed by the 

arrangement of electrons in the outermost incomplete 

shells. In the first row of the d block, the inner 3d 

orbital is being filled and most elements have two 

electrons in the outer 4s shell. (Chromium is an 

exception, having only one electron in the 4s shell.) 

Thus, transition elements have similar properties 

because they have essentially the same outer 

electronic arrangement as each other, in the same way 

as the elements in a vertical group. They differ only by 

the number of electrons in the inner incomplete 3d 

sub-shell. 

Metals in different groups in the periodic table have 

different numbers of outer electrons and hence 

different properties. 

4 a The E! for the O 2 

(g), 4H + (aq)/2H 2 

O(l) half-cell is more 

positive and receives electrons from the 

Fe 3+ (aq)/Fe 2+ (aq) half-cell. So overall O 2 

is reduced to 

water and the Fe 2+ (aq) is oxidised to Fe 3+ (aq). 

O 2 

(g) + 4H + (aq) + 4Fe 2+ (aq) 

Æ 2H 2 

O(l) + 4Fe 3+ (aq) 

E cell = +1.23 V – (+0.77 V) = +0.46 V 

The E! for the O 2 

(g), 4H + (aq)/2H 2 

O(l) half-cell is less 

positive than that for the Mn 3+ (aq)/Mn 2+ (aq) half-cell 

and so it is not oxidising enough to form Mn(III) from 

Mn(II). The Mn 3+ (aq)/Mn 2+ (aq) half-cell receives 

electrons from the O 2 

(g), 4H + (aq)/2H 2 

O(l) half-cell. In 

the presence of water, Mn 3+ (aq) is reduced to 

Mn 2+ (aq) and the water oxidised to oxygen. 

4Mn 3+ (aq) + 2H 2 

O(l) 

Æ 4Mn 2+ (aq) + O 2 

(g) + 4H + (aq) 

E cell = +1.56 V – (+1.23 V) = +0.33 V 

b i An acidified solution of iron(II) will be oxidised 

by air. 

ii An acidified solution of marganese(II) will not be 

oxidised by air. 

c The flow of electrons is to the half-cell with the most 

positive E! value. Cu + (aq) will be expected to be 

oxidised to Cu(s) and reduced to Cu 2+ (aq) 

(disproportionation). 

2Cu + (aq) Æ Cu 2+ (aq) + Cu(s) 

E cell = +0.52 V – (+0.16 V) = +0.36 V 

191

SECTION 12 


1 a 2 

b 4 

c 6 

d 6 

2 a [Mn(H 2 

O) 6 

] 2+ 

b [Zn(NH 3 

) 4 

] 2+ 

c [FeF 6 

] 3– 

d [Cr(H 2 

O) 5 

OH] 2+ 

3 a +1 

b +2 

c +3 

d +3 

4 a Hexaaquavanadium(III) ion 

b Hexacyanoferrate(II) ion 

c Tetrachlorocobaltate(II) ion 

d Diamminesilver(I) ion 

e Tetraaquadichlorochromium(III) ion 

5 a TiO 2 

(s) contains titanium with a 3d 0 electron 

configuration. It is white as no 3d electron transitions 

are possible. 

b Sc 3+ [Ar] 

Zn 2+ [Ar] 3d 10 no 3d electron transitions 

Cu + [Ar] 3d 10 possible, hence colourless 

6 a 6 

b +4 

c Hexachlorotitanate(IV) ion 

d 

Cl 

Cl 

Cl 

Ti 

Cl 

7 A five-membered chelate ring seems to lead to a more 

stable complex than a six-membered one. 

8 a Monodentate 

b Bidentate 

c Polydentate 

d Bidentate 

9 A complex of a metal ion with edta contains six 

5-membered chelate ring systems (see Figure 32 on page 

270 of Chemical Ideas). The complex is much more 

stable than the corresponding complexes with NH 3 

or 

H 2 

O which contain no chelate rings. The extra stability is 

due to the large increase in entropy when edta 4– 

displaces six ligands. 

Cl 

Cl 

2– 


1 

Empirical formula Molecular formula M r 

7 a C 7 

H 16 

b C 8 

H 18 

c C 10 

H 22 

C 3 

H 8 

C 3 

H 8 

44 8 a H H 

CH 2 

C 12 

H 24 

168 

X X 

CH C 6 

H 6 

78 

H X C 

X 

X C X H 

H 

C H C H 282 

X X 

H H 

10 21 

CH 2 

20 42 

C 5 

H 10 

70 

CH C 2 

H 2 

26 

C 5 

H 4 

C 10 

H 8 

128 

b H H H 

2 a CH 2 

b C 2 

H 4 

3 a CH b C 6 

H 6 

4 a, b C : H 

Ratio by mass 82.8 : 17.2 

Ratio by moles 6.9 : 17.2 

Simplest ratio 1 : 2.5 

Whole number ratio 2 : 5 

\ Empirical formula is C 2 

H 5 

But M r 

(C 2 

H 5 

) = 29 

So, the molecular formula is C 4 

H 10 

. 

5 a 0.085 g C 0.014 g H 

b 0.0071 mol C 0.014 mol H 

c CH 2 

d C 6 

H 12 

e 

(or a hexene such as ) 

c 

H 

H 

H 

X 

X 

X 

H 

C 

X 

C 

X 

H 

X 

X 

X 

C 

X 

C 

X 

H 

X 

X X 

X X 

X 

X 

X 

X 

H 

C 

X 

X 

X 

C 

X 

X 

H 

H 

H 

H 

9 a Cyclooctane b Methylcyclopentane 

c Cycloheptane 

d Butylcyclohexane 

10 a b c 

d 

H 

H 

H 

H 

C 

H 

C 

H 

H 

C 

H 

C 

H 

C 

H 

H 

C 

H 

H 

C 

H 

C 

H 

H 

H 

H 

H 

6 a B D E 

bA C F 

e 

f 

192

SECTION 12 

11 butane 

methylpropane 

12 a C 5 

H 12 

+ 8O 2 

Æ 5CO 2 

+ 6H 2 

O b C 5 

H 12 

+ 5 O 2 

Æ 5CO + 6H 2 

O 

13 a For ‘cat’ cracking, high temperature (500 °C) and zeolite catalyst. 

For steam cracking, high temperature (900 °C) with short 

residence time and steam as dilutent. 

b True: (ii), (iv), (v) 

c Three from: 

+ 

+ 

+ 

+ 

14 a A B C 

D 

E 

bD c A d B e Cand E 

f eg C or E could have been produced by cracking D 

+ 

gA 


1 a 

b 

c 

3 a 

b 

c 

d 

e 

d 

2 a propene 

b hept-2-ene 

c 2-methylbut-2-ene 

d hept-1,4-diene 

e 2,4-dimethylpent-2-ene 

f cyclopentene 

4 

e 

H 

H CH 3 H H 

C C C C C 

H CH 3 H H 

H 

193

SECTION 12 

5 a 

H 

H 

H 

Br 

Br 

+ 2Br 2 Br 

Br 

H C C C + Br 2 H C C C H 

H H 

H 

H H H 

+ H 2 One of the carbon atoms has become positively 

b H 

H 

H H H 

Br 

H C C C + H 

8 a Bromine; room temperature; non polar solvent 

2 H C C C H 

b Steam; catalyst, phosphoric acid on silica; 300 °C; high 

H H 

H 

H H H 

pressure (60 atm) 

c Hydrogen; catalyst, nickel; 150 °C; 5 atm pressure; 

6 H Br H 

H H Br 

(or platinum at lower temperatures and atmospheric 

pressure) 

H C C C H H C C C H 

d Hydrogen; catalyst, nickel; 150 °C. 

H H H 

H H H 

9 a H 

H 

H 

H 

7 a 

C C H C 

+ 

C 

+ Br 2 

H 

H d+ H 

H 

H 

Br 

Br d– 

Br 

b 

Br 

H 

H 

H H 

+ Br 2 (aq) 

H C 

+ 

C 

H C C H 

c 

OH 

H 

H 

H Br 

+ HBr 

– 

Br 

Br 

b The hydrogen bromide is polarised with the hydrogen 

d + 2H atom being slightly positively charged. The hydrogen 

2 

atom behaves as an electrophile and reacts with the 

e 

double bond. 

charged and so it reacts with the bromide ion to form 

a C–Br covalent bond. 

10 Shake the compound with about 1 cm depth of bromine 

water in a stoppered test-tube. If the compound is 

unsaturated, the bromine will be decolorised. The 

reaction occurs quickly at room temperature. 

f 

Br 

Br 

– 


1 a i 1,4-dimethylbenzene 

ii 1-ethy1-3-methylbenzene 

iii 1,2,4-trimethylbenzene 

d 

Cl 

e 

b i ii iii 

Cl 

2 a 1,3,5-trimethylbenzene 

b 1-methyl-3-propylbenzene 

c bromobenzene 

d 1,3-dinitrobenzene 

e 4-methylphenylamine 

f 2,6-dimethylphenol 

3 a b c OH 

4 a 0.154 nm The ring is not a regular 

hexagon, but has alternating 

0.134 nm long and short bonds. 

b 

c All the bonds in the ring are the same and so only one 

isomer is possible. 

CH 3 

CH 3 

194

SECTION 12 


1 a Product: Cl 

2 a CH 3 

b 

Cl 

CHClCH 2 Cl 

b Reagents and conditions: Br 2 

; AlBr 3 

or Fe; reflux 

c Product: NO 2 

CH 3 

d Reactant: 

3 a Difference in electronegativity between I and Cl 

b 

I 

+ I Cl 

+ HCl 

e Product: 

H 3 C 

CH 3 

CH 

CH 3 

c As the molecule is permanently polarised, a catalyst is 

not needed. 

d– d+ 

Cl—I 

d Formation of chlorobenzene requires a chlorine with a 

d+ charge. 

4 a NO 2 

f Product: 

O 

C CH 2 CH 3 

g Reagents and conditions: CH 3 

COCl; AlCl 3 

; reflux 

h Product: SO 2 OH 

b 

+ HNO 3 

c. H 2 SO 4 

< 55 ∞C 

+ H 2 O 

SO 2 OH 

+ H 2 SO 4 + H 2 O 

5 a The benzene ring is resistant to hydrogenation 

because this destroys the stable delocalised electron 

system. The reaction has a high activation enthalpy. 

6 a Electrophilic substitution 

b Electrophilic addition 

c Radical substitution 

d Radical addition 


1 a trichloromethane 

b 2-chloropropane 

c 1,1,1,–trichloro–2,2,2–trifluoroethane 

d 2–chloro–1,1,1–trifluoropropane 

e 2,2–dibromo–3–chlorobutane 

2 d 

Cl 

F 

4 a 

H 

b 

H Cl 

C C Cl 

H Cl 

F F H 

H C C C H 

e 

F 

Cl 

F 

F F H 

c Cl H H Br H 

H C C C C C H 

Br 

Br 

3 a CH 3 

CH 2 

CH 2 

Cl(l) + NaOH(aq) 

Æ CH 3 

CH 2 

CH 2 

OH(aq) + NaCl(aq) 

b The chlorine atom in 1-chloropropane has been 

replaced by a hydroxyl group, –OH. 

c XX 

– 

H 

X O X 

XX 

5 a D 

bA 

cA 

dE 

e 

Cl 

H 

OH 

H 

H 

H 

cyclohexanol 

195

SECTION 13 

6 a 

H 

H 

H 

H 

H 

C 

C 

I 

+ OH – H C C 

OH 

+ I – 

H 

H 

H 

H 

b 

H 

H 

H 

H 

H 

C 

C 

Br 

+ CN – H C C 

CN 

+ Br – 

H 

H 

H 

H 

c 

Cl + OH – OH + Cl – 

d 

H 

H 

H C H H C 

H H 

H 

H 

H 

H 

C 

C 

C 

H 

+ H 2 O H C C C H + HCl 

H 

Cl 

H 

H 

OH 

H 

e 

H 

H 

H 

H 

Br 

C 

C 

Br 

+ 2OH – HO C C 

OH 

+ 2Br – 

H 

H 

H 

H 

f 

H 

H 

H 

H 

H 

H 

H 

C 

Br 

+ H C C O – 

H C O C C H + Br – 

g 

H 

H 

H 

H 

H 

H 

CH 3 

H 

C 

O 

H 

Cl 

H 

O 

H 

O 

H 

H 

C 

C 

C 

H 

+ H C O – 

H C C C H + Cl – 

3 C 

H 

H 

H 

H 

H 

H 

7 a CH 3 

CH 2 

Br + NH 3 

Æ CH 3 

CH 2 

NH 2 

+ H + +Br – 

b H H 

H 

H 

H 

H C C Br 

H C C 

N + 

H + 

Br 

– 

H 

H 

H 

H 

H 

H N H 

H 

H 

H 

H 

H 

H 

H 

H C C 

N + 

H H C C 

N 

H + H + 

196 

H 

H 

H 

c d Explanation along similar lines to that on pages 302–303 in Chemical Ideas. 

Ammonia is the nucleophile because of its lone pair. 

8 a 1-iodopentane (or bromo or chloro); aqueous NaOH; reflux. 

b 2-iodopropane (or bromo or chloro): concentrated aqueous solution of NH 3 

(heat under pressure in a sealed tube). 

c 1-iodopropane (or bromo or chloro); alcoholic solution of KCN (or NaCN); reflux. 

hv 

9 a CH 3 

I æÆ CH 3 + I . 

hv 

CH 3 

Cl æÆ CH 3 

. + Cl . 

b Bond enthalpy of C–I bond is less than that of C–Cl bond. C–I bond can be broken 

by ultraviolet radiation in the troposphere although this has relatively less energy 

than the ultraviolet radiation in the stratosphere. 

H 

H

SECTION 13 


1 a pentan-1-ol e 2-methylbutan-2-ol 

b heptan-3-ol 

f ethoxypropane 

c butane-2,3-diol g decanol 

d cyclohexanol 

2 O 

methoxybutane 

O 

ethoxypropane 

Some students may give this isomer: 

O 

ethoxy-2-propane 

3 a Hydrogen bonding between ethanol and water 

molecules (see Chemical Ideas, page 306). As the 

hydrocarbon chain becomes longer, the importance of 

the –OH group relative to that of the alkyl group 

becomes less and hexanol is unable to mix with water. 

b No hydrogen bonding occurs. 

4 a B D F 

bA 

cC 

dE 

eAand B; C and F 

f A 

gE 

5 a Ethanol has hydrogen bonds between molecules, ethane 

does not. Hydrogen bonds require more energy to be 

broken than the weak attractive forces between ethane 

molecules, so boiling point of ethanol is higher. 

b Water forms more hydrogen bonds than ethanol so its 

boiling point is higher. 

c Both have –OH group and form hydrogen bonds 

between molecules. Boiling point increases down a 

homologous series as M r 

increases. Hence butan-1-ol 

has a higher boiling point than ethanol. 

d Butan-1-ol forms hydrogen bonds, ethoxyethane does 

not. Hence boiling point of butan-1-ol is higher. 


O 

CH 3 

CH 2 

1 a methanoic acid 

d 

b pentanoic acid 

CH 

c 2-methylbutanoic acid 

3 CH 2 C 

CH 3 CH 2 C 

CH 3 

2 a butanoic acid 

O 

b octanoic acid 

CH 3 CH 2 C 

c pentanedioic acid 

O 

d benzene-1,2-dicarboxylic acid 

e O 

O 

3 a H H H H H 

O 

C CH 2 CH 2 CH 2 CH 2 C 

H C C C C C C 

Cl 

Cl 

H H H H H O H 

b O H H H H 

O 

5 

O 

C C C C C C 

C 

propyl ethanoate 

H O 

O 

H H H H O H 


c O O H 

O 

C 

H C 

butyl methanoate 

O CH 2 CH 2 CH 2 CH 3 

O 

C 

O 

O 

H 

CH 3 CH 2 C 

ethyl propanoate 

O CH 2 CH 3 

O CH 3 

CH 3 CH 2 CH 2 C 

methyl butanoate 

4 a 

O 

CH 3 CH 2 C 

O 

O CH 3 

b Cl O 

C 

Student may also draw isomers with branched chains, 

for example 

O 

CH 3 

C 

2-propyl ethanoate 

c 

O 

O CH CH 3 

NH 2 

197

SECTION 13 


1 a A butan-1-ol 

B butan-2-ol 

C 2,4-dichlorophenol 

b D 

E 

F 

CH 3 

CH 3 

CH 2 CH 2 

CH 2 CH 2 

O 

O 

C 

H 

O 

C 

O 

H 

b 

5 a 

CH 3 

b 

O 

C 

O – Na + 

O 

CH 2 CH 2 C 

O – Na + 

CH 3 

2-methylpropanal 

CH CHO 

O 

cyclohexanone 

3 


G 

CH 

CH 3 CH 2 C CH 2 CH 3 

c 

OH 

COOH 

2-hydroxybenzoic acid 

(salicylic acid) 

OH 

2 a B b D 

cC 

dF 

eG 

f C 

gE 

hA and D 

3 a OH O – Na + 

6 a A F 

bC 

cD E 

dB D E 

eA 

7 a H 

O 

H 

+ NaOH + H 2 O 

H C O C C H 

b CH 3 

CH 2 

COOH + KOH Æ CH 3 

CH 2 

COO – K + + H 2 

O 

c 2CH 3 

CH 2 

CH 2 

COOH + Na 2 

CO 3 

Æ 2CH 3 

CH 2 

CH 2 

COO – Na + + CO 2 

+ H 2 

O 

H 

b methyl ethanoate 

c H H 

H 

O 

d CH 3 

COOH + NaHCO 3Æ 

CH3 COO – Na + + CO 2 

+ H 2 

O 

e 2CH 3 

CH 2 

COOH + Mg 

Æ (CH 3 

CH 2 

COO – ) 2 

Mg 2+ + H 2 

4 a O – K + O – 

O – K + 

O – 

H C C O C H 

H 

H 

d ethyl methanoate 

e ethanol; methanoic acid 

or 

198

SECTION 13 


CH 3 C OH 

CH 3 CH 2 OH 

1 a methyl propanoate 

b propyl ethanoate 

c ethyl propanoate 

d methyl methanoate 

e methyl butanoate 

2 a and b 

Ester Alcohol Acid 

ethyl methanoate ethanol methanoic acid 

H C O CH 2 CH 3 

H C OH 

CH 3 CH 2 OH 

O 

O 

3–methylbutyl ethanoate 3–methylbutanol ethanoic acid 

CH 3 C O CH 2 CH 2 CHCH 3 

CH 3 CHCH 2 CH 2 OH 

O 

CH 3 

CH 3 

O 

ethyl 2–methylbutanoate ethanol 2–methylbutanoic acid 

CH 3 CH 2 CH C O CH 2 CH 3 

CH 3 CH 2 CH C OH 

CH 3 O 

CH 3 O 

phenylmethyl ethanoate phenylmethanol ethanoic acid 

O 

CH 2 OH 

CH 2 O C CH 3 

CH 3 C OH 

O 

c They are structural isomers. 

3 a 

CH 3 

CH 

CH 3 

CH 3 CH 2 CH 3 CH 

O 

O 

OH 

+ 

C 

OH 

CH 3 

C CH 2 

CH 3 H 2 O 

O 

+ 

b 

HO 

O 

O 

O 

CH 2 CH 2 OH + 2CH 3 C OH CH 3 C O CH 2 CH 2 O C CH 3 

+ 2H 2 O 

4 

O 

O 

O 

O 

O CH 2 CH 2 O C 

C O 

CH 2 

CH 2 

O 

C 

C 

5 a The C–O bond next to the C=O in the ester is broken. c A hydrogen ion attaches itself to the oxygen of the 

O 

carbonyl group, and thus makes the carbon atom in 

the group more susceptible to attack by a nucleophile 

C CH 3 

(eg water). 

CH 3 CH 2 O 

O 

C CH 3 + H + 

b The C=O bond is very polar with a d+ charge on the 

carbon atom. Hydrolysis occurs by nucleophilic attack 

CH 3 CH 2 O 

on this carbon atom by a lone pair on the oxygen-18 of 

a water molecule and the C–O bond break. 

HO + 

CH 3 

CH 2 

O 

C 

CH 3 

A water molecule attacks the positive carbon atom, the 

CH 3 

–CH 2 

–O – group is displaced and combines with H + to 

form ethanol. 

199

SECTION 13 

6 a 

O 

O 

CH 3 

CH 2 OH + CH 3 C Cl CH 3 CH 2 O C CH 3 

+ 

HCl 

b 

O 

CH 3 

O 

CH 3 

CH 3 

C OH + CH 3 C Cl CH 3 

C CH 3 

CH 3 

C O 

CH 3 

+ 

HCl 

c 

O 

O O O 

CH 3 

OH + CH 3 C O C CH 3 CH 3 O C CH 3 

+ CH 3 C OH 

d 

O 

CH 3 O C CH 3 

+ H 2 O CH 3 OH + CH 3 C 

H + 

O 

OH 

e 

O 

O 

O 

C 

CH 3 

+ 2OH – + CH 3 C + H 2 O 

O – 

O – 

(or NaOH as reactant; sodium salts as products) 

H 

7 

OH 

8 a Phenylmethanol 

C O H 

CH 3 

4-methylphenol 

CH 3 

COCl (ethanoyl chloride) or CH 3 

CO–O–COCH 3 

(ethanoic anhydride) 

Mix the reagents carefully. If the anhydride is used, warm 

the mixture under reflux. 

b ethanoic acid, CH 3 

COOH 

ethanoyl chloride, CH 3 

COCl 

ethanoic anhydride (CH 3 

CO) 2 

O 

c H 2 

O, water 

HCl, hydrogen chloride 

CH 3 

COOH, ethanoic acid 

d 

O 

CH 2 O C 

CH 3 

H 

CH 2 OH 

+ H 2 O + CH 3 COOH 

e Becomes increasingly vinegary. 


1 a O 

b H O 

H C O 

R O C R¢ 

(R' can be H) 

` 

C 

O 

H C O C 

O 

H C O C 

CH 3 

CH 3 

CH 3 

H 

c 

H 

O 

H 

H C O 

C 

O 

H C O 

H 

O 

H C O C 

O 

CH 3 

+ 3NaOH H C O H 

CH 3 

+ 3CH 3 C 

O – + Na 

H C O C 

H 

CH 3 

H C O H 

H 

200

SECTION 13 

2 a, b 1 mole glycerol (propane-1,2,3-triol) 

2 moles oleic acid 

1 mole linoleic acid 

3 a Saturated fats are esters of fatty acids with no (or few) carbon double bonds. 

b Monounsaturated fats contain fatty acids with one carbon double bond (such as oleic acid). 

c Polyunsaturated fats contain a high proportion of fatty acid groups with two 

or more carbon double bonds (such as linoleic acid). 

4 a O 

O 

O 

O 

O 

O 

b 

O 

Na + 

– O 

c The hydrocarbon chain 

Instantaneous dipole–induced dipole forces 

d Soaps are sodium and potassium salts of long chain carboxylic acids. The ions 

in solution are readily hydrated. 

5 a 150 °C; pressure (5 atmospheres); nickel as catalyst 

b i M r 

= 882 

ii 1.13 ¥ 10 3 moles of oil react with 2.45 ¥ 10 3 moles of hydrogen 

iii 4 

iv 4 moles 

v To fully saturate 1 tonne of oil would require 

1.13 ¥ 10 3 ¥ 4 moles hydrogen = 4.52 ¥ 10 3 moles 

2.45 ¥ 10 3 

Percentage of double bonds hydrogenated = ¥ 100 = 54.2% 

vi Easier to spread (less hard); healthier 

primary 

CH 3 


1 a methanal 

c 

CH 3 O 

b propanal 

C 

c 4-methylpentanal 


CH 3 C C H aldehyde 

H 

2 a butan-2-one 

H 

3 

b pentan-3-one 

5 a H H H H 

a H H H H O 

O 

H C C C C OH 

H C C C C C H 

4.52 ¥ 10 3 

H 

H H H H 

H H H H 

b H OH H H 

b H O H H H 

O 

H C C C C H 

H C C C C C H 

H H H H 

H H H H 

4 a 

O CH 3 

c 

H 

CH 3 CH 2 C C CH 3 ketone 

H H C H OH H 

H 

H C C C C H 

b CH 3 O 

H H C H H H 

ketone 

secondary 

secondary 

201

SECTION 13 

6 a H H OH 

b 

H 

H 

OH 

H 

H 

C 

H 

C 

H 

C 

C 

H 

N 

H 

C 

H 

C 

H 

C 

C 

C 

H 

N 

H 

c Similar argument to Chemical Ideas pages 328–329. 


NH 2 NH 2 

1 a ethylamine d ethyldimethylamine 

b dimethylamine e cyclohexylamine 

c 2-aminopropane 

2 a CH 3 CH 2 CH 2 NH 2 

b 

NH 2 

c CH 3 CH 2 N CH 3 

CH 2 CH 3 

d CH 3 e f NH 2 

CH 3 CH 2 CH 2 CH 2 NCH 2 CH 3 

CH 3 CH 2 CH CH 2 CH 3 CH 2 CH CH 2 CH CH 3 

+ 

3 a NH 3 

b NH 

c 

CH 3 CH CH 3 Cl – CH 3 CH 

CH 3 

CO CH 3 

CH 

NH 

CH 2 CH 3 

+ HCl 

CH 3 

CH 3 

+ HCl 

4 a CH CH 2 CH 2 NH 2 + HCl CH 3 CH 2 CH 2 NH 3 + Cl – 

3 CH 2 

CH 2 

+ 

b 

c 

N 

+ CH 3 COCl + HCl 

NH 2 NHCOCH 3 

H + C 2 H 5 COCl N C C 2 H 5 

+ HCl 

O 

O 

d 

+ 

– 

NH 2 + H 2 SO 4 NH 3 HSO 4 

or 

NH 2 + H 2 SO 4 

2 

+ 

NH 3 SO 4 

2– 

e 

f 

CH 3 CH 2 NH 2 + CH 3 Br CH 3 CH 2 NH CH 3 + HBr 

CH 3 CH 2 CO NH CH 3 + NaOH CH 3 CH 2 COO – Na + + CH 3 NH 2 

or 

CH 3 CH 2 CO NH CH 3 + OH – CH 3 CH 2 COO – + CH 3 NH 2 

5 a Tertiary amines cannot react with acid chlorides 

because there is no H atom attached to the nitrogen. 

b H 

H 

C 4 H 9 

N 

H 

H 

O 

H 

6 a Reagents and conditions: HCl(aq) (moderately 

concentrated); reflux 

b Products: 

CH 3 

+ 

CH 3 CH 2 COOH + N H Cl – 

c Reactant: 

O 

CH 3 

O 

H 

Because of strong attraction (hydrogen bonding) 

between amine and solvent molecules. 

c [Cu(H 2 

O) 6 

] 2+ + 4C 4 

H 9 

NH 2 

blue 

Æ [Cu(C 4 

H 9 

NH 2 

) 4 

(H 2 

O) 2 

] 2+ + 4H 2 

O 

deep blue 

(cf [Cu(NH 3 

) 4 

(H 2 

O) 2 

] 2+ ) 

d Product: 

H 2 C 

H 2 C 

NH 

C 

O 

C 

O – Na + 

CH 2 

NH 2 

CH 3 

202

SECTION 13 


1 a H R 

O 

H C C O – 

N + 

H 

H 

b 

i 

CH 3 CH 3 

+ HCl Cl – H 3 N + CH COOH 

H 2 N 

CH 

COOH 

ii 

CH 2 OH 

CH 2 OH 

H 2 N 

CH 

COOH 

+ NaOH H 2 N CH COO – Na + 

+ H 2 O 

iii 

+ 

(CH 2 ) 4 NH 3 Cl – 

(CH 2 ) 4 NH 2 

+ 2HCl Cl – H 3 N + CH COOH 

H 2 N 

CH 

COOH 

iv 

CH 2 COOH 

H 2 N 

CH 

COOH 

CH 2 COO – Na + 

+ 2NaOH H 2 N CH COO – Na + + 2H 2 O 

2 

O 

H 2 N CH C OH 

CH 

CH 3 

O 

CH 3 

H 2 N CH C OH 

3 a 

O 

O 

O 

O 

CH 3 

H 2 N CH C N CH 2 C OH 

CH 3 

H 2 N CH C OH 

+ H 2 O 

C 

OH 

CH 3 

CH 2 

+ H 2 N 

H 

The salts will be formed in the presence of HCl(aq), for example 

O 

O 

+ 

Cl – H 3 N CH 2 C OH 

b 

OH 

CH 2 

O 

O 

H2N CH C N 

C OH 

H 

CH 2 

H 2 N CH C 

+ 2OH – 

OH 

CH 2 

CH 2 

O 

O 

C O – + H 2 N O – + H 2 O 

c Key points which should be included are: correct choice of solvent; the container used for 

the separation should have a lid; authentic samples of the amino acids used alongside the 

reaction mixture; R f 

values carefully measured. 

203

SECTION 14 


1 a 

+ 

– 

NH 3 HSO 4 

b 

+ 

– 

N N HSO 4 

c 

N N OH 

(strictly the sodium salt) 

2 a 

HO 

+ 

N N Cl – 

b 

i 

+ 

HO N N Cl – + OH HO N N OH + HCl 

ii 

HO 

+ 

HO N N Cl – + HO N N 

+ HCl 

HO 

iii 

+ 

HO N N Cl – + NH 2 HO N N NH 2 + HCl 

3 

HOO 2 S 

+ 

N N X – 

and 

N 

CH 3 

CH 3 

O 2 N 

+ 

N N X – 

and 

OH 

NH 2 

and 

X – N 

+ N 

N + N X – 

4 The diazonium salt is a relatively weak electrophile and will only 

react with particularly electron-rich activated benzene rings. 

SO 2 OH 

204 


1 a 73.9% 

b 73.1% 

2 Route I (Route I 40%; Route II 30%) 

3 16.2 g 

4 a (Selection of starting materials); reaction; extraction of 

product from reaction mixture; purification of product. 


1 a A solution of hydrogen bromide (HBr) in a polar 

solvent at room temperature. 

b Reflux with dilute aqueous sodium hydroxide solution 

b A small amount of product is never recovered from the 

reaction mixture. This is due to factors such as wetting 

of the walls of the reaction vessel, incomplete 

crystallisation of the product, loss on filter papers, loss 

by evaporation, etc. 

c Reflux with an acidified solution of potassium 

dichromate(VI) 

d Treat with hydrogen cyanide solution with a small 

amount of alkali.

SECTION 14 

2 a Addition; electrophilic 

b Substitution; nucleophilic 

c Oxidation 

d Addition; nucleophilic 

3 a Radical substitution 

b The mechanism is: 

Cl 

Cl 

hν 

2 Cl 

initiation 

H 

CH 3 

+ Cl 

C 

+ HCl 

H 

propagation 

H 

H 

C 

+ Cl 2 

C 

Cl 

+ Cl 

H 

H 

Cl Cl 2 

C C 

+ Cl 

H 

H 

H 

H 

C 

+ 

C 

H 

H 

H 

H 

termination 

H 

C 

H 

+ Cl C Cl 

H 

H 

c 

CH 2 Cl 

CH 2 NH 2 

c. NH 3 (aq) 

heat in sealed tube 

4 a 

CH 3 

CH 3 

Cl 

and 

Cl 

b Electrophilic substitution 

c AlCl 3 

helps to polarise the chlorine molecule, to produce the electrophile, Cl + 

d+ d– 

+ Cl Cl 

Cl 

Cl + + AlCl 4 

– 

+ AlCl 3 

+ + AlCl 3 

H 

Cl 

205

SECTION 14 

5 a i Reagents and condition: treat with a nitrating 

mixture (conc. HNO 3 

and conc. H 2 

SO 4 

); keep 

temperature below 55 °C. 

ii Product: Br 

iv 

H 

C 

O 

OH 

+ CH 3 

OH 

Br 

iii Reactant: CH 3 

Cl (or CH 4 

) 

iv Reactant: CH 3 

–CH 2 

–CH 2 

–X 

(X = Cl, Br, I) 

v Products: 

O 

O 

HO C 

C OH 

vi Products: 

O 

CH 2 CH 2 

+ HCl 

CH 3 

+ 2HCl 

v 

H 

H 

H 

O 

C 

+ H 2 O 

O CH 3 

H 

C C + H 2 CH 3 CH 3 

H 

b 

6 a i 

b 

C 

CH 

i Electrophilic substitution 

ii Electrophilic addition 

iii Radical substitution 

iv Nucleophilic substitution 

v Nucleophilic substitution 

vi Electrophilic substitution 

ii 

iii 

iv 

v 

COO – Na + 

CH 2 CH 3 

+ H 2 O 

CH 3 COO – Na + + NH 2 

+ H 2 O 

CH 3 CH 2 OH 

i Acid–base 

ii Hydrolysis 

iii Dehydration 

iv Esterification 

v Reduction 

+ 3H 2 

Pt 

c Examples may be: 

i CH 3 COOH + NaOH 

ii 

iii 

CH 3 

C NH CH 3 

O 

Al 2 O 3 

CH 3 CH 2 OH CH 2 CH 2 + H 2 O 

+ 

CH 3 COO – Na + + CH 3 NH 2 

CH 3 COO – Na + + H 2 O 

NaOH 

7 a React with hydrogen in presence of finely divided 

nickel at 150 °C and 5 atm. 

CH 3 CH 2 CH CH CH 2 COOH + H 2 

CH 3 CH 2 CH 2 CH 2 CH 2 COOH 

b Treat with a solution of HBr in a polar solvent at 

room temperature. 

CH 3 CH 2 CH CH CH 2 COOH HBr 

Br 

CH 3 CH 2 CH CH 2 

(+ isomer) 

Br 

CH 2 

+ 

CH 3 CH 2 CH CH 2 CH 2 COOH 

(prepared as in b) 

OH 

CH 3 CH 2 CH 

Then acidify with dilute acid. 

OH 

OH 

COOH 

c Hydrolyse by refluxing with a dilute aqueous solution 

of NaOH. 

2NaOH 

CH 2 CH COO – Na + 2 

+ NaBr + H 2 O 

CH 3 CH 2 CH CH 2 CH 2 COO – Na + + HCl 

CH3 CH 2 CH CH 2 CH 2 COOH + NaCl 

+ 

206

SECTION 14 

d Dehydrate by heating with Al 2 

O 3 

at 300 °C or by heating with concentrated sulphuric acid. 

OH 

CH 3 CH 2 

CH CH 2 CH 2 COOH CH 3 CH CH CH 2 CH 2 COOH 

(prepared as in c) 

(+ isomer) + H 2 O 

8 OH 

OH 

O 

H 2 (g) / Ni 

300∞C; 30 atm 

Cr 2 O 2– 7 / H + (aq) 

reflux 

cyclohexanol 

9 

COOH 

NO 2 

COOC 2 H 5 

NO 2 

COOC 2 H 5 

NH 2 

C 2 H 5 OH 

Sn + c. HCl 

c. H 2 SO 4 as catalyst 

reflux 

reflux 


a Continuous 

b Batch 

c Continuous 

d Continuous 


a 

Conditions: lower pressure, or less separation required. 

Feedstock: methanol can be obtained from a variety of 

feedstocks. 

Product: higher yield 

Co-products: far fewer. 

b 

Disadvantages: very expensive catalyst, and more than 

one stage to process. 


2 a i As 3 moles of gas form 1 mole of gas, increasing 

the pressure will cause more methanol to form. 

ii As the conversion of synthesis gas to methanol is 

exothermic, an increase in temperature will 

reduce the equilibrium yield of methanol. 

b i The pressure of 100 atm will increase the yield. 

The temperature is not too high, otherwise the 

yield is decreased. It is sufficient to provide a high 

rate of reaction with the catalyst used. It is 

carefully controlled to prevent the exothermic 

reaction raising the temperature and reducing the 

yield. 

ii The catalyst lowers the temperature required to 

give a high rate of reaction. However, some 

sacrifice in yield is made to maintain this rate. The 

high pressure also increases the reaction rate. 


a 

Thermal energy released from the reaction can be used 

to pre-heat, via a heat exchanger, the water which will be 

made into steam as one of the reactants. 

b 

The unreacted feedstock can be recycled again over the 

catalyst, saving costs of energy needed in the high 

temperature and pressure conditions. 

207

SECTION 15 


1 a Many of the fixed costs for a 200 tonne day –1 plant are 

less than double those of a 100 tonne day –1 plant: for 

example, the larger plant will require less than twice 

the area of land for building, and the cost of building 

access roads will be less than twice as great. 

b You have to pay the same fixed costs such as wages 

and land rental regardless of how much you are 

producing. But variable costs such as costs of raw 

materials and energy will double if you double output. 

2 a Germany 2.2 ¥ 10 5 US$ 

France 

3.3 ¥ 10 5 US$ 

UK 

2.1 ¥ 10 5 US$ 

Italy 

2.6 ¥ 10 5 US$ 

Belgium 

3.8 ¥ 10 5 US$ 

Spain 

2.5 ¥ 10 5 US$ 

The Netherlands 3.7 ¥ 10 5 US$ 

Switzerland 

3.7 ¥ 10 5 US$ 

Ireland 

7.9 ¥ 10 5 US$ 

Sweden 

2.7 ¥ 10 5 US$ 

b Ireland 

c The costs of operating in each country (general level 

of wages; workforce productivity; taxation policies); 

the age and inherent efficiency of the plant in each of 

the countries if similar products are being made; the 

type of product each country produces; some of these 

factors are controlled by political forces, others are the 

results of decisions and developments in the past. 

d It may be that a country’s chemical industry is biased 

towards products that need a larger number of 

employees. Thus, one should also look at other 

statistics such as profits per employee. 

208

Section 1.1 Section 1.2 Section 1.3 - The Student Room

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