Chemistry for the IB Diploma
Marking scheme for AHL Worksheet – Chapter 5
1 a 6C(s) + 3H
2
(g)

C
6
H
6
(l) b 2C(s) + 2H
2
(g) + 1 /
2
O
2
(g)

CH
3
CHO(l) c 3Li(s) + 1 /
2
N
2
(g)

Li
3
N(s)
2 a 3C(s) + 4H
2
(g)

CH
3
CH
2
CH
3
(g)

H f
= 3
 –394 + 4  –286 + 2220

H f
= –106 kJ mol
–1 b C(s) + 2H
2
(g) + 1 /
2
O
2
(g)

CH
3
OH(l)

H f
= –394 + 2

–286 + 715

H f
= –251 kJ mol
–1 c C(s) + H
2
(g) + 1 /
2
O
2
(g)

HCHO(g)

H f
= –394 – 286 + 561

H f
= –119 kJ mol
–1 d 2C(s) + 2H
2
(g) + O
2
(g)

CH
3
COOH(l)

H f
= 2

–394 + 2

–286 + 876

H f
= –484 kJ mol
–1
3 a C
6
H
12
(l) + 9O
2
(g)

6CO
2
(g) + 6H
2
O(l)

H = (6

–394 + 6

–286) – (–156)

H f
= –3924 kJ mol
–1 b C
6
H
5
OH(s) + 7O
2
(g)

6CO
2
(g) + 3H
2
O(l)

H = (6

–394 + 3

–286) – (–163)

H f
= –3059 kJ mol
–1 c CH
2
CHCHCH
2
(g) + 11 /
2
O
2
(g)

4CO
2
(g) + 3H
2
O(l)

H = (4

–394 + 3

–286) – (112)

H f
= –2546 kJ mol
–1 d (CH
3
)
2
O(g) + 3O
2
(g)

2CO
2
(g) + 3H
2
O(l)

H = (2

–394 + 3

–286) – (–185)

H f
= –1461 kJ mol
–1
4 a (–246 – 597) – (–297 – 444)
–102 kJ mol –1 b (6

–314 + 3

–242) – (3

80 + 10

–46)
–2390 kJ mol –1
Copyright Cambridge University Press 2011. All rights reserved.
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Page 1 of 3

Chemistry for the IB Diploma
5 –276 = ( 
H f
+82) – (230+2

90)

H f
(ClNO) = +52 kJ mol
–1
6 a Mg + (g)

Mg 2+ (g) + e
– b O(g) + e
– 
O
–
(g) c MgF
2
(s)

Mg 2+ (g) + 2F
–
(g)
7
[1]
[1]
[1]
[1]
[1]
4 marks for all correct, lose 1 mark for each mistake

H latt
= +1214 + 193 + 590 + 1150 + 2

79 + 2

–348

H latt
= 2609 kJ mol
–1
[4]
[1]
[1]
8 Lattice enthalpies for the group 1 fluorides are lower than for the group 2 oxides. [1]
Group 2 oxides contain 2+ and 2– ions as opposed to 1+ and 1– ions in the group 1 fluorides [1] and there are stronger electrostatic attractions between 2+ and 2– ions, so the lattice enthalpies are higher.
Lattice enthalpy decreases from LiF to NaF to KF and from MgO to CaO to SrO the size of the positive ions increases so there are weaker electrostatic attractions between the positive and negative ions.
[1]
[1]
[1]
[1]
9 a decrease decrease in number of moles of gas b increase increase in number of moles of gas
[1]
[1]
[1]
[1]
Copyright Cambridge University Press 2011. All rights reserved. Page 2 of 3

Chemistry for the IB Diploma
10 a

S = (308 + 223) – (248 + 167)

S = 116 J K
–1
mol
–1 b

S = (192 + 2

188) – (121 + 205)

S = 242 J K
–1
mol
–1
11

G =

H – T

S

G = –102 – 298

116
1000

G = –137 kJ mol
–1 reaction is spontaneous as

G is negative
12 a

G o = (2

66) – (2

87) b

G o = –42 kJ mol
–1

S is negative for this reaction (decrease in number of moles of gas) as T increases

G becomes more positive as – T

S is positive reaction is less spontaneous at 500 K than at 300 K
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Copyright Cambridge University Press 2011. All rights reserved. Page 3 of 3