Download 8. Solution Guide to Supplementary Exercises

Topic
9
Chemical Reactions and Energy
Part A Unit-based exercise
Unit 34 Energy changes in chemical reactions
Fill in the blanks
1
specific heat capacity
2
heat capacity
3
kinetic energy; potential energy
4
enthalpy
5
a) exothermic
b) less
c) negative
6
a) endothermic
b) greater
c) positive
7
formation; elements; standard
8
combustion; completely; oxygen; standard
9
neutralization; water; standard
10 solution; solvent; standard
True or false
11 T
12 T
13 F
14 T
15 F
16 T
1
17 F
18 F
19 T
20 F
Multiple choice questions
21 C
22 D
23 B
24 D
25 A
26 B
27 D
28 A
29 D
30 B
31 A
32 A
33 A
34 C
35 A
36 A
37 C
38 B
39 B
2
40 D
41 C
42 C
43 D
44 A
45 C
46 B
47 D
48 D
49 B
50 B
51 C
52 B
53 D
54 A
55 D
56 C
57 C
58 B
59 A
60 C
61 A
62 A
63 D
3
64 C
65 C
66 D
67 A
Unit 35 Hess’s Law and its applications
Fill in the blanks
1
Hess’s; initial; final
2
a) hydrochloric acid
b) magnesium oxide; hydrochloric acid
c) water
3
constituent elements; compound
4
products; reactants
True or false
5
F
6
T
7
T
8
T
9
F
10 F
Multiple choice questions
11 C
12 A
13 B
14 C
4
15 A
16 A
17 D
18 B
19 C
20 A
21 D
22 C
23 C
24 B
25 B
26 B
27 D
28 C
29 A
30 A
31 C
32 B
33 A
34 C
35 B
36 C
37 A
38 B
5
39 C
40 B
41 C
42 B
43 A
44 A
45 D
46 C
47 D
48 D
49 D
50 A
51 D
52 D
53 B
54 C
55 B
6
Part B
Topic-based exercise
Multiple choice questions
1
D
2
B
3
B
4
B
5
A
6
B
7
D
8
B
9
B
10 C
11 D
12 C
13 C
14 D
15 D
16 D
17 B
18 A
19 B
20 C
21 C
7
22 D
23 B
24 B
25 A
26 C
27 A
28 B
29 A
30 B
31 D
32 A
33 B
34 C
35 B
36 A
37 D
38 D
39 C
40 D
41 A
42 C
43 C
44 D
8
Short questions
45 a) exothermic
(1)
b) endothermic
(1)
c) endothermic
(1)
d) exothermic
(1)
e) exothermic
(1)
46 Amount of heat required = 168 g x 0.400 J g–1 K–1 x (25.6 + 12.2) K
= 2 540 J
(1)
47 Let Tf °C be the final temperature.
Amount of heat released by copper = 44.9 g x 0.390 J g–1 K–1 x (99.8 – Tf) K
(1)
Amount of heat taken in by water = 152 g x 4.18 J g–1 K–1 x (Tf – 18.5) K
(1)
44.9 g x 0.390 J g–1 K–1 x (99.8 – Tf) K = 152 g x 4.18 J g–1 K–1 x (Tf – 18.5) K
Tf = 20.7 °C
(1)
∴ the final temperature is 20.7 °C.
2.51 g
–1
159.6 g mol
= 0.0157 mol
48 Number of moles of Fe2O3 reacted =
Amount of heat released = 24.8 kJ x 0.0157
= 0.389 kJ
(1)
(1)
∴ 0.389 kJ of heat is released.
49 a) standard enthalpy change of formation of H2O2(l)
(1)
b) standard enthalpy change of combustion of C2H5OH(l)
(1)
c) standard enthalpy change of neutralization of HCl(aq) and NaOH(aq)
(1)
d) standard enthalpy change of hydration of CuSO4(s)
(1)
Structured questions
50 a) • At a temperature of 25 °C (298 K).
• At a pressure of 1 atmosphere (1 atm).
(1)
(1)
• All substances involved are in their standard states, i.e. each in its most stable physical state at
25 °C and 1 atm.
(1)
b) C6H8(l) + 2H2(g)
C6H12(l)
(1)
9
c)
ΔH
C6H8(l) + 2H2(g) + 9O2(g)
O
O
C6H12(l) + 9O2(g)
O
O
ΔHc [C6H8(l)] + 2 x ΔHc [H2(g)]
ΔHc [C6H12(l)]
6CO2(s) + 6H2O(l)
(1)
ΔHO + ΔHOc [C6H12(l)] = ΔHOc [C6H8(l)] + 2 x ΔHOc [H2(g)]
ΔHO = ΔHOc [C6H8(l)] + 2 x ΔHOc [H2(g)] – ΔHOc [C6H12(l)]
= [(–3 584) + 2(–286) – (–3 924)] kJ mol–1
= –232 kJ mol–1
(1)
(1)
(1)
∴ the standard enthalpy change of hydrogenation of cyclohexa-1,3-diene is –232 kJ mol–1.
51 a) Amount of heat released during combustion = 200.0 g x 4.18 J g–1 K–1 x (64.1 – 28.0) K
= 30 200 J
= 30.2 kJ
(1)
(184.78 – 183.58) g
60.0 g mol–1
= 0.0200 mol
Number of moles of propan-1-ol burnt =
–30.2 kJ
0.0200 mol
–1
= –1 510 kJ mol
Enthalpy change of combustion of propan-1-ol =
(1)
∴ the enthalpy change of combustion of propan-1-ol is –1 510 kJ mol–1.
b) i) Evaporation of propan-1-ol
(1)
ii) Carbon
(1)
Less exothermic
c)
C3H7OH(l) +
O
(1)
O
c
ΔH [C3H7OH(l)]
9
O (g)
2 2
ΔHf [C3H7OH(l)]
3CO2(g) + 4H2O(l)
O
O
3 x ΔHf [CO2(g)] + 4 x ΔHf [H2O(l)]
3C(graphite) + 4H2(g) + 5O2(g)
(1)
ΔHOc [C3H7OH(l)] + ΔHOf [C3H7OH(l)] = 3 x ΔHOf [CO2(g)] + 4 x ΔHOf [H2O(l)]
ΔHOc [C3H7OH(l)] = 3 x ΔHOf [CO2(g)] + 4 x ΔHOf [H2O(l)] – ΔHOf [C3H7OH(l)]
= [3(–394) + 4(–286) – (–303)] kJ mol–1
= –2 020 kJ mol–1
10
∴ the standard enthalpy change of combustion of propan-1-ol is –2 020 kJ mol–1.
(1)
(1)
(1)
d) Heat is absorbed from the surroundings.
(1)
In the water / propan-1-ol system, the intermolecular attractions are mainly hydrogen bonds.
(1)
Water forms two hydrogen bonds per molecule on average while propan-1-ol forms only one.
(1)
The attraction between propan-1-ol and water is weaker than the average of water / water and
propan-1-ol / propan-1-ol attraction.
(1)
Thus when propan-1-ol and water are mixed, the enthalpy of the system will increase.
52 a) i) C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
(1)
ii) Respiration
(1)
3.60 g
180.0 g mol–1
= 0.0200 mol
b) Number of moles of glucose burnt =
(1)
–56.0 kJ
0.0200 mol
= –2 800 kJ mol–1
Standard enthalpy change of combustion of glucose =
(1)
∴ the standard enthalpy change of combustion of glucose is –2 800 kJ mol–1.
c) ΔHOf [C6H12O6(s)] refers to the standard enthalpy change of the following process:
6C(graphite) + 6H2(g) + 3O2(g)
C6H12O6(s)
O
6C(graphite) + 6H2(g) + 3O2(g) + 6O2(g)
O
ΔHf [C6H12O6(s)]
O
C6H12O6(s) + 6O2(g)
O
6 x ΔHf [CO2(g)] + 6 x ΔHf [H2O(l)]
ΔHc [C6H12O6(s)]
6CO2(g) + 6H2O(l)
(1)
ΔHOf [C6H12O6(s)] + ΔHOc [C6H12O6(s)] = 6 x ΔHOf [CO2(g)] + 6 x ΔHOf [H2O(l)]
ΔHOf [C6H12O6(s)] = 6 x ΔHOf [CO2(g)] + 6 x ΔHOf [H2O(l)] – ΔHOc [C6H12O6(s)]
(1)
ΔHOf [C6H12O6(s)] = [6(–394) + 6(–286) – (–2 800)] kJ mol–1
= –1 280 kJ mol–1
(1)
(1)
∴ the standard enthalpy change of formation of glucose is –1 280 kJ mol–1.
11
d)
Enthalpy (kJ)
6C(graphite) + 6H2(g) + 9O2(g)
–1 280 kJ
C6H12O6(s) + 6O2(g)
[6(–394) + 6(–286)] kJ
–2 800 kJ
6CO2(g) + 6H2O(l)
(1 mark for correct enthalpy levels; 1 mark for correct labels)
53 a) i)
(2)
ESBVHIUTDSFFO
UIFSNPNFUFS
TUJSSFS
MJE
DBMPSJNFUFSDPOUBJOJOH
XBUFS
DMBNQ
XBUFS
TQJSJUMBNQ
QFOUBOPM
(1 mark for basic set-up; 2 marks for correct labels; award 0 mark if the set-up is not workable) (3)
ii) Amount of heat released = 250.0 g x 4.18 J g–1 K–1 x 40.5 K
= 42 300 J
= 42.3 kJ
(1)
1.32 g
88.0 g mol–1
= 0.0150 mol
Number of moles of pentan-1-ol =
–42.3 kJ
0.0150 mol
= –2 820 kJ mol–1
Enthalpy change of combustion of pentan-1-ol =
(1)
∴ the enthalpy change of combustion of pentan-1-ol is –2 820 kJ mol–1.
b) i) The enthalpy change of a reaction depends on the initial and final states of the reaction
and is independent of the route by which the reaction may occur.
12
(1)
(1)
ii) ΔHOc [C5H11OH(l)] refers to the standard enthalpy change of the following process:
15
C5H11OH(l) +
O2(g)
5CO2(g) + 6H2O(l)
2
ΔHOc = ∑ ΔHOf [products] – ∑ ΔHOf [reactants]
ΔHOc = 5 x ΔHOf [CO2(g)] + 6 x ΔHOf [H2O(l)] – ΔHOf [C5H11OH(l)]
= [5(–394) + 6(–286) – (–354)] kJ mol–1
= –3 330 kJ mol–1
(1)
(1)
(1)
∴ the standard enthalpy change of combustion of pentan-1-ol is –3 330 kJ mol–1.
iii) Less negative
(1)
c) Any one of the following:
• Heat loss occurs in (a) / no heat loss in (b)
(1)
• Incomplete combustion of pentan-1-ol in (a)
(1)
54 a) The standard enthalpy change of combustion of decane refers to the enthalpy change of the following
process:
31
C10H22(l) +
O2(g)
10CO2(g) + 11H2O(l)
2
ΔHOc = 10 x ΔHOf [CO2(g)] + 11 x ΔHOf [H2O(l)] – ΔHOf [C10H22(l)]
(1)
–1
= [10(–394) + 11(–286) – (–301)] kJ mol
(1)
= –6 790 kJ mol–1
(1)
–6 790 x 1 000
kJ kg–1
142.0
= –47 800 kJ kg–1
b) Energy density of fuel =
108 000 kJ
–1
47 800 kJ kg
= 2.26 kg
(1)
c) Mass of decane =
d) To ensure complete combustion.
(1)
(1)
55 a) i) The standard enthalpy change of combustion of octane is the enthalpy change when one mole of
octane is completely burnt in oxygen under standard conditions.
(1)
25
C8H18(l) +
O2(g)
8CO2(g) + 9H2O(l)
(1)
2
ii) ΔHOc [C8H18(l)] = ∑ ΔHOf [products] – ∑ ΔHOf [reactants]
= 8 x ΔHOf [CO2(g)] + 9 x ΔHOf [H2O(l)] – ΔHOf [C8H18(l)]
(1)
–1
= [8(–394) + 9(–286) – (–250)] kJ mol
(1)
= –5 480 kJ mol–1
(1)
∴ the standard enthalpy change of combustion of octane is –5 480 kJ mol–1.
13
3.60 x 104 kJ
–1
5 480 kJ mol
= 6.57 mol
b) At 100.0% efficiency, number of moles of C8H18 used =
(1)
Since engine efficiency is 20.0%
∴ number of moles of C8H18 used = 6.57 mol x
100.0
20.0
= 32.9 mol
(1)
Mass of C8H18 used = 32.9 mol x 114.0 g mol–1
= 3 750 g
(1)
3 750 g
0.660 g cm–3
= 5 680 cm3
= 5.68 dm3
Volume of C8H18 used =
(1)
∴ 5.68 dm3 of petrol are used for the journey.
–5 480 kJ mol–1
114.0 g mol–1
= –48.1 kJ g–1
c) Enthalpy change of combustion of C8H18 =
(1)
–1
–1 368 kJ mol
46.0 g mol–1
= –29.7 kJ g–1
Enthalpy change of combustion of C2H5OH =
(1)
Enthalpy change of combustion of alternative fuel = [0.9(–48.1) + 0.1(–29.7)] kJ g–1
= –46.3 kJ g–1
(1)
The alternative fuel has a lower enthalpy change of combustion.
(1)
d) Any one of the following:
• Ethanol is an oxygen-containing compound. This makes it easier for the alternative fuel to undergo
complete combustion / less CO is produced / less particulates are formed.
(1)
• Ethanol is a renewable energy source. It can be obtained from crops.
(1)
• The cost for the production of ethanol is low in agricultural countries.
(1)
56 a) CH3(CH2)10COOH(s) + 17O2(g)
12CO2(g) + 12H2O(l)
(1)
1.00 g
200.0 g mol–1
= 5.00 x 10–3 mol
(1)
b) Number of moles of lauric acid burnt =
–36.9 kJ
5.00 x 10–3 mol
–1
= –7 380 kJ mol
Standard enthalpy change of combustion of lauric acid =
∴ the standard enthalpy change of combustion of lauric acid is –7 380 kJ mol–1.
14
(1)
c) The standard enthalpy change of formation of a substance is the enthalpy change when one mole of
the substance
(1)
is formed from its elements in their standard states.
(1)
d) ΔHfO [CH3(CH2)10COOH(s)] refers to the standard enthalpy change of the following process:
12C(graphite) + 12H2(g) + O2(g)
CH3(CH2)10COOH(s)
O
12C(graphite) + 12H2(g) + O2(g) + 17O2(g)
O
ΔHf [CH3(CH2)10COOH(s)]
O
CH3(CH2)10COOH(s) + 17O2(g)
O
12 x ΔHf [CO2(g)] + 12 x ΔHf [H2O(l)]
ΔHc [CH3(CH2)10COOH(s)]
12CO2(g) + 12H2O(l)
(1)
ΔHOf [CH3(CH2)10COOH(s)] + ΔHOc [CH3(CH2)10COOH(s)] = 12 x ΔHOf [CO2(g)] + 12 x ΔHOf [H2O(l)]
ΔHOf [CH3(CH2)10COOH(s)] = [12 x ΔHOf [CO2(g)] + 12 x ΔHOf [H2O(l)] – ΔHOc [CH3(CH2)10COOH(s)]
(1)
ΔHOf [CH3(CH2)10COOH(s)] = [12(–394) + 12(–286) – (–7 380)] kJ mol–1
= –780 kJ mol–1
(1)
(1)
∴ the standard enthalpy change of formation of lauric acid is –780 kJ mol–1.
57 a) Amount of heat required
= 750.0 g x 4.18 J g–1 K–1 x (100.0 – 28.5) K + 150.0 g x 0.900 J g–1 K–1 x (100.0 – 28.5) K
= 234 000 J
= 234 kJ
234 kJ
–1
1 368 kJ mol
= 0.171 mol
(1)
(1)
b) Number of moles of C2H5OH needed =
Mass of C2H5OH needed = 0.171 mol x 46.0 g mol–1
= 7.87 g
7.87 g
38.0%
= 20.7 g
(1)
(1)
c) Mass of C2H5OH needed in practice =
(1)
d) The greater the number of carbon atoms in a molecule of a straight-chain alcohol, the greater the
standard enthalpy change of combustion is.
(1)
The standard enthalpy change of combustion of each successive alcohol differs by the same amount. (1)
This is because the structure of each successive alcohol differs by a –CH2– unit.
(1)
As shown in the equations below, there is a constant difference in the number of bonds broken (2
3
C–H bonds, 1 C–C bond and
O=O bonds) and bonds formed (2 C=O bonds and 2 O–H bonds)
2
involved in the combustion of the alcohols.
(1)
15
CH3OH(l) +
3
O2(g)
2
CO2(g) + 2H2O(l)
CH3CH2OH(l) + 3O2(g)
CH3CH2CH2OH(l) +
2CO2(g) + 3H2O(l)
9
O2(g)
2
3CO2(g) + 4H2O(l)
58 The following data is given:
1
1
(1)
N2(g) +
O2(g)
NO(g)
2
2
(2) C(s) + O2(g)
CO2(g)
O
ΔHf [NO(g)] = +90.2 kJ mol–1
ΔHOf [CO2(g)] = –394 kJ mol–1
(3) 2NO(g) + O2(g)
2NO2(g)
ΔHO = –114 kJ
(4) 2CO(g) + O2(g)
2CO2(g)
ΔHO = –566 kJ
Looking at the target equation, we need 4 moles of CO(g) as the reactant. So, multiply equation (4) by 2,
giving equation (4)’.
(4)’ 4CO(g) + 2O2(g)
4CO2(g)
ΔHO = –1 132 kJ
We need 2 moles of NO2(g) as the reactant. Equation (3) has 2 moles of NO2(g), but it is on the product
side. So, reverse the equation, giving equation (3)’.
(3)’ 2NO2(g)
2NO(g) + O2(g)
ΔHO = +114 kJ
1
We need 1 mole of N2(g) as the product. Equation (1) has
mole of N2(g), but it is on the reactant side.
2
So, reverse the equation and multiply it by 2, giving equation (1)’.
(1)’ 2NO(g)
N2(g) + O2(g)
ΔHO = –180.4 kJ
By combining the three equations, followed by collecting like terms, we can obtain the target equation.
(1)’ 2NO(g)
(3)’ 2NO2(g)
N2(g) + O2(g)
2NO(g) + O2(g)
(4)’ 4CO(g) + 2O2(g)
4CO(g) + 2NO2(g)
4CO2(g)
4CO2(g) + N2(g)
ΔHO = –180.4 kJ
ΔHO = +114 kJ
ΔHO = –1 132 kJ
ΔHOr
ΔHOr = [(–180.4) + (+114) + (–1 132)] kJ
= –1 200 kJ
∴ the standard enthalpy change of the reaction is –1 200 kJ.
16
(2)
(1)
(1)
59 a) Place excess hydrochloric acid in a polystyrene cup. Record the initial temperature of the acid.
(1)
Add a known mass of solid magnesium carbonate to the acid.
(1)
Record the maximum temperature of the reaction mixture.
(1)
b) ΔHf[MgCO3(s)] refers to the enthalpy change of the following process:
3
Mg(s) + C(graphite) +
O2(g)
MgCO3(s)
2
Mg(s) + C(graphite) +
3
O (g) + 2HCl(aq)
2 2
ΔHf[MgCO3(s)]
MgCO3(s) + 2HCl(aq)
ΔH2 + ΔH3 + ΔH4
ΔH1
MgCl2(aq) + CO2(g) + H2O(l)
(2)
ΔHf[MgCO3(s)] + ΔH1 = ΔH2 + ΔH3 + ΔH4
ΔHf[MgCO3(s)] = ΔH2 + ΔH3 + ΔH4 – ΔH1
= [(–512) + (–394) + (–286) – (–89.9)] kJ mol–1
= –1 102 kJ mol–1
(1)
(1)
(1)
∴ the enthalpy change of formation of magnesium carbonate is –1 102 kJ mol–1.
c)
Enthalpy (kJ mol–1)
Mg(s) + C(graphite) +
3
O (g) + 2HCl(aq)
2 2
–1
[(–512) + (–394) + (–286)] kJ mol
–1 102 kJ mol–1
MgCO3(s) + 2HCl(aq)
MgCl2(aq) + CO2(g) + H2O(l)
–89.9 kJ mol–1
(1 mark for correct enthalpy levels; 1 mark for correct labels)
60 a) Si(s) + 2H2(g)
b) i) SiH4(g) + 2O2(g)
SiH4(g)
SiO2(s) + 2H2O(l)
ii) ΔHOc [SiH4(g)] = ΔHOf [SiO2(s)] + 2 x ΔHOf [H2O(l)] – ΔHOf [SiH4(g)]
= [(–910) + 2(–286) – (+34)] kJ mol–1
= –1 520 kJ mol–1
Assumption – Hess’s Law is followed.
(2)
(1)
(1)
(1)
(1)
(1)
(1)
17
61 a) (2) Initial temperature of KOH(aq)
b)
(1)
(3) Initial temperature of HCl(aq)
(1)
(4) Volume of HCl(aq)
(1)
(5) Final temperature of KCl(aq)
(1)
ΔH
KClO3(s) + 3Mg(s)
O
KCl(s) + 3MgO(s)
O
O
ΔHf [KClO3(s)]
K(s) +
O
ΔHf [KCl(s)] + 3 x ΔHf [MgO(s)]
1
3
Cl (g) +
O (g) + 3Mg(s)
2 2
2 2
(1)
ΔHOf [KClO3(s)] + ΔHO = ΔHOf [KCl(s)] + 3 x ΔHOf [MgO(s)]
ΔHO = ΔHOf [KCl(s)] + 3 x ΔHOf [MgO(s)] – ΔHOf [KClO3(s)]
= [(–437) + 3(–602) – (–391)] kJ
= –1 850 kJ
(1)
(1)
(1)
∴ the standard enthalpy change of the reaction is –1 850 kJ.
62 a) i) Heat released when BaCl2(s) dissolved = 50.0 g x 4.18 J g–1 K–1 x 1.20 K
= 251 J
–251 J
0.0300 mol
= –8 370 J mol–1
= –8.37 kJ mol–1
(1)
Enthalpy change of solution of BaCl2(s) =
ii) Heat taken in when BaCl2•2H2O(s) dissolved = 50.0 g x 4.18 J g–1 K–1 x 3.00 K
= 627 J
+627 J
Enthalpy change of solution of BaCl2•2H2O(s) =
0.0300 mol
–1
= +20 900 J mol
= +20.9 kJ mol–1
water
BaCl2(aq)
b) BaCl2(s)
ΔH1 = –8.37 kJ mol–1
water
BaCl2•2H2O(s)
BaCl2(aq)
ΔH2 = +20.9 kJ mol–1
(1)
(1)
(1)
The enthalpy change of the following process is required:
BaCl2(s) + 2H2O(l)
BaCl2•2H2O(s)
ΔH
Looking at the target equation, we need 1 mole of BaCl2(s) as the reactant. So, keep the first equation
as it is.
18
We need 1 mole of BaCl2•2H2O(s) as the product. The second equation has 1 mole of BaCl2•2H2O(s),
but it is on the reactant side. So, reverse the second equation.
ΔH = ΔH1 – ΔH2
= [(–8.37) – (+20.9)] kJ mol–1
= –29.3 kJ mol–1
(1)
(1)
∴ the enthalpy change of hydration of BaCl2(s) is –29.3 kJ mol–1.
c) It is difficult to measure the temperature of a solid.
(1)
63 a) Total volume of solution = (40.0 + 40.0) cm3
= 80.0 cm3
Amount of heat taken in = 80.0 cm3 x 4.18 J cm–3 K–1 x 1.0 K
= 334 J
Number of moles of NaHCO3 used = 2.00 mol dm–3 x
(1)
40.0
dm3
1 000
= 0.0800 mol
Number of moles of HCl used = 2.20 mol dm–3 x
40.0
dm3
1 000
= 0.0880 mol
According to the equation of Reaction 2, 1 mole of HCO3–(aq) would react with 1 mole of H+(aq).
i.e. 0.0800 mole of HCO3–(aq) would react with 0.0800 mole of H+(aq).
Hence H+(aq) was in excess. HCO3–(aq) was the limiting reactant.
+334 J
0.0800 mol
= +4 180 J mol–1
= +4.18 kJ mol–1
(1)
Standard enthalpy change of Reaction 2 =
b) i) ΔHOr =
=
=
=
∑ ΔHOf [products] – ∑ ΔHOf [reactants]
ΔHOf [H2O(l)] + ΔHOf [CO2(g)] – ΔHOf [HCO3–(aq)] – ΔHOf [H+(aq)]
[(–286) + (–394) – (–692)] kJ mol–1
+12 kJ mol–1
ii) Heat was taken in from the surroundings. Otherwise, the temperature drop could be more.
(1)
(1)
(1)
(1)
(1)
c) i) ΔHOr = ∑ ΔHOf [products] – ∑ ΔHOf [reactants]
ΔHOr = ΔHOf [H2O(l)] + ΔHOf [CO2(g)] – ΔHOf [CO32–(aq)] – 2 x ΔHOf [H+(aq)]
(1)
–2.3 kJ mol–1 = [(–286) + (–394)] kJ mol–1 – ΔHOf [CO32–(aq)]
ΔHOf [CO32–(aq)] = [(–286) + (–394) – (–2.3)] kJ mol–1
= –678 kJ mol–1
ii) ΔHOr =
=
=
=
∑ ΔHOf [products] – ∑ ΔHOf [reactants]
ΔHOf [CO32–(aq)] + ΔHOf [H2O(l)] + ΔHOf [CO2(g)] – 2 x ΔHOf [HCO3–(aq)]
[(–678) + (–286) + (–394) – 2(–692)] kJ
+26 kJ
(1)
(1)
(1)
(1)
(1)
19
64 a) i)
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(1 mark for correctly plotted points)
(1)
ii) Working on the graph to show temperatures at 3.5 minutes.
Calculated temperature change must be in the range 43.5 °C – 44.5 °C.
iii) To take into account the heat loss to the surroundings.
b) i) Low heat capacity / good insulator of heat / low mass / absorbs less heat
(1)
(1)
(1)
(1)
ii) To ensure uniform temperature.
(1)
iii) Burette / pipette / volumetric flask
(1)
3.17 g
65.4 g mol–1
= 0.0485 mol
c) Number of moles of Zn =
Number of moles of CuSO4 = 1.00 mol dm–3 x
(0.5)
50.0
dm3
1 000
= 0.0500 mol
(0.5)
According to the equation, 1 mole of Zn reacts with 1 mole of CuSO4. During the reaction, 0.0485
mole of Zn reacts with 0.0485 mol of CuSO4.
∴ CuSO4 is in excess.
(1)
d) Amount of heat released during the reaction = 50.0 g x 4.18 J g–1 K–1 x 43.8 K
= 9 150 J
= 9.15 kJ
–9.15 kJ
0.0485 mol–1
= –189 kJ mol–1
(1)
Enthalpy change of reaction =
∴ the enthalpy change of the reaction is –189 kJ mol–1.
20
(1)
e) Any two of the following:
• Put a lid on the polystyrene cup.
(1)
• Put the polystyrene cup in another polystyrene cup / in a beaker.
(1)
• Determine the heat capacities of the polystyrene cup and the thermometer. Take these values into
account in the calculations.
(1)
65 a) CaO(s) + H2O(l)
Ca(OH)2(aq)
(1)
b) Amount of heat released by 7.40 g of CaO(s) = 250.0 g x 4.18 J g–1 K–1 x 37.5 K
= 39 200 J
= 39.2 kJ
(1)
7.40 g
56.1 g mol–1
= 0.132 mol
Number of moles of CaO(s) used =
Amount of heat transferred to water by the reaction of 1.00 mole of CaO(s)
39.2 kJ
0.132 mol
= 297 kJ mol–1
=
(1)
∴ 297 kJ of heat are transferred to the water by the reaction of 1.00 mole of CaO(s).
c) i) It is hard to prevent the calcium hydroxide formed from dissolving.
(1)
ii) Allow a known mass of solid calcium oxide to react with excess hydrochloric acid. Record the
maximum temperature rise.
(1)
Calculate the enthalpy change for the reaction between solid calcium oxide and hydrochloric acid,
in kJ mol–1.
(1)
(1) CaO(s) + 2HCl(aq)
CaCl2(aq) + H2O(l)
ΔH1
Allow a known mass of solid calcium hydroxide to react with excess hydrochloric acid. Record the
maximum temperature rise.
(1)
Calculate the enthalpy change for the reaction between solid calcium hydroxide and hydrochloric
acid, in kJ mol–1.
(1)
(2) Ca(OH)2(s) + 2HCl(aq)
CaCl2(aq) + 2H2O(l)
ΔH2
Combine equation (1) and the reverse form of equation (2) (i.e. equation (2)’).
(1) CaO(s) + 2HCl(aq)
(2)’ CaCl2(aq) + 2H2O(l)
CaO(s) + H2O(l)
CaCl2(aq) + H2O(l)
Ca(OH)2(s) + 2HCl(aq)
Ca(OH)2(s)
ΔH1
–ΔH2
ΔH = ΔH1 – ΔH2
(1)
21
66 a) i) • At a temperature of 25 °C (298 K).
(1)
• At a pressure of 1 atmosphere (atm).
(1)
• All the substances involved are in their standard states, i.e. each in its most stable physical state
at 25 °C and 1 atm.
(1)
ii) C4H10(g) +
13
O2(g)
2
iii) ΔHOc [C4H10(g)] =
=
=
=
4CO2(g) + 5H2O(l)
∑ ΔHOf [products] – ∑ ΔHOf [reactants]
4 x ΔHOf [CO2(g)] + 5 x ΔHOf [H2O(l)] – ΔHOf [C4H10(g)]
[4(–394) + 5(–286) – (–125)] kJ mol–1
–2 880 kJ mol–1
iv)
Enthalpy (kJ mol–1)
C4H10(g) +
(1)
(1)
(1)
(1)
13
O (g)
2 2
–1
ΔHc [C4H10(g)] = –2 880 kJ mol
O
4CO2(g) + 5H2O(l)
(1)
–2 880 kJ mol–1
58.0 g mol–1
= –49.7 kJ g–1
b) i) Enthalpy change of combustion of butane =
ii) • A comparison of any two or three fuels by ΔHc in kJ g–1
(1)
(1)
e.g. butane gives out the greatest amount of heat per g.
• A comparison of any two or three fuels by ΔHc in kJ cm–3
(1)
e.g. C8H18 gives out the greatest amount of heat per cm3.
• A comparison of states, and consequence of state on use as fuel in motor vehicles
e.g. C4H10 is a gas; it needs a big fuel tank to be stored at high pressure.
or C2H5OH / C8H18 is a liquid; it needs a smaller fuel tank.
22
(1)
67 a) i) To obtain steady temperatures.
(1)
ii) Total volume of the solution mixture = 50.0 cm3 + 50.0 cm3
3
= 100.0 cm
Mass of the solution mixture = 100.0 g
Amount of heat released during neutralization = 100.0 g x 4.18 J g–1 K–1 x 13.5 K
= 5 640 J
= 5.64 kJ
(1)
Number of moles of HCl reacted = number of moles of KOH reacted
= 2.00 mol dm–3 x
50.0
dm3
1 000
= 0.100 mol
(1)
0.100 mole of HCl reacted with 0.100 mole of KOH to produce 0.100 mole of H2O.
–5.64 kJ
0.100 mol
= –56.4 kJ mol–1
Enthalpy change of neutralization =
(1)
∴ the enthalpy change of neutralization for the reaction is –56.4 kJ mol–1.
iii) Any one of the following:
• No heat was lost to the surroundings.
(1)
• The polystyrene cup and thermometer had negligible heat capacity.
(1)
• All the alkali was transferred from the beaker to the polystyrene cup.
(1)
b) Both Reactions 1 and 2 involve the neutralization between a strong monobasic acid and sodium
hydroxide solution.
The same chemical change occurs in both cases.
H+(aq) + OH–(aq)
(1)
H2O(l)
Hence Reactions 1 and 2 have the same enthalpy change of neutralization.
The enthalpy change for Reaction 3 is less because ethanoic acid is a weak acid.
(1)
Some energy is consumed when the acid dissociates to give hydrogen ions before neutralization.
(1)
23
68 a) Hydroxyl group
(1)
Carboxyl group
(1)
b)
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å5
5Jž$
5JNFNJO
i) Working on the graph to show temperatures at 3.0 minutes.
(1)
Maximum temperature rise of the mixture = 13.0 °C
(1)
ii) Enthalpy change = +25.0 g x 4.18 J g–1 K–1 x 13.0 K
= +1.36 kJ
(1)
+1.36 kJ
+45.2 kJ mol–1
= 0.0301 mol
iii) Number of moles of lactic acid in the cup =
0.0301 mol
25.0
dm3
1 000
= 1.20 mol dm–3
Concentration of the lactic acid solution =
c) Number of moles of NaOH = 1.60 mol dm–3 x
= 0.0320 mol
(
(1)
)
(1)
20.0
3
dm
1 000
(1)
According to the above equation, 1 mole of CH 3 CHOHCOOH requires 1 mole of NaOH for
neutralization.
24
i.e. number of moles of CH3CHOHCOOH = 0.0320 mol
Concentration of the lactic acid solution =
(
0.0320 mol
25.0
dm3
1 000
)
= 1.28 mol dm–3
(1)
d) Any two of the following:
• The specific heat capacity of the solution is greater than 4.18 J g–1 K–1.
(1)
• The density of the solution is greater than 1.00 g cm–3.
(1)
• The heat capacity of the apparatus is not taken into account.
(1)
69 a) The standard enthalpy change of solution of a substance is the enthalpy change when one mole of the
substance dissolves
(1)
in an infinite volume of solvent (or enough solvent so that further dilution has no additional effect)
under standard conditions.
(1)
b) Amount of heat taken in = 228 J K–1 x (28.6 – 25.3) K
= 752 J
(1)
2.55 g
80.0 g mol–1
= 0.0319 mol
Number of moles of NH4NO3 =
+752 J
0.0319 mol
–1
= +23 600 J mol
= +23.6 kJ mol–1
Enthalpy change of solution of NH4NO3 =
(1)
∴ the enthalpy change of solution of ammonium nitrate is +23.6 kJ mol–1.
c) Record the temperature of the water at intervals; then add the solid and stir; continue recording the
temperature.
(1)
Plot the temperature of the water against time.
(1)
Join the points before solid addition using a straight line and extrapolate to the time at which the solid
is added.
(1)
Join the points after solid addition using a straight line and extrapolate back to the time at which the
solid is added.
(1)
The separation of the lines at the time of solid addition corresponds to the maximum temperature
change for the process.
25
70 a) The heat capacity of a substance is the amount of heat required to raise the temperature of the
substance by 1 K (or 1 °C)
(1)
b) Number of moles of Pb(NO3)2 = 0.600 mol dm–3 x
50.0
3
dm
1 000
= 0.0300 mol
Number of moles of KI = 1.14 mol dm–3 x
(0.5)
50.0
dm3
1 000
= 0.0570 mol
(0.5)
According to the equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI. During the reaction, 0.0285
mole of Pb(NO3)2 reacted with 0.0570 mole of KI.
∴ Pb(NO3)2 was in excess.
(1)
Amount of heat released by the reaction = 49 000 J mol–1 x
0.0570
mol
2
= 1 400 J
(1)
1 400 J
3.10 K
= 452 J K–1
Heat capacity of calorimeter and its contents =
(1)
∴ the heat capacity of the calorimeter and its contents is 452 J K–1.
c) i)
Error
Calculated heat
capacity too low
(1) The concentration of the lead(II)
nitrate solution had been
incorrectly recorded and was
–3
actually 0.410 mol dm .
No effect on
calculated heat
capacity
✔
(2) The calorimeter was not placed
in its insulating jacket.
ii) Error (1)
Error (2)
26
Calculated heat
capacity too high
✔
Pb(NO3)2 was already in excess.
(1)
A lower temperature change leads to a higher heat capacity.
(1)
71 a)
7PMVNFPGOJDLFM**
DIMPSJEFTPMVUJPODN $IBOHFJOUFNQFSBUVSFž$
m
m
m
m
m
(1 mark for correctly plotted points)
(1)
b) Mark two straight lines on the graph, one through the first three points and (0,0) while the other
through the last three points and (100.0,0).
(1)
The two straight lines intersect at 67.0 cm3, i.e. 67.0 cm3 of nickel(II) chloride solution and 33.0 cm3 of
sodium carbonate solution would produce the maximum temperature fall when mixed.
(1)
c) i) Ni2+(aq) + CO32–(aq)
NiCO3(s)
(1)
ii) Number of moles of Na2CO3 = 1.00 mol dm–3 x
33.0
dm3
1 000
= 0.0330 mol
According to the equation, 1 mole of Na2CO3 reacts with 1 mole of NiCl2.
i.e. number of moles of NiCl2 = 0.0330 mol
0.0330 mol
Concentration of NiCl2(aq) =
67.0
dm3
1 000
(
)
= 0.493 mol dm–3
(1)
∴ the concentration of the nickel(II) chloride solution is 0.493 mol dm–3.
d) Amount of heat taken in = 100.0 g x 4.18 J g–1 K–1 x 4.50 K
= 1 880 J
= 1.88 kJ
+1.88 kJ
0.0330 mol
= +57.0 kJ mol–1
(1)
Enthalpy change of reaction =
(1)
∴ the enthalpy change of the reaction is +57.0 kJ mol–1.
27
–242 kJ mol–1
18.0 g mol–1
= –13.4 kJ g–1
72 a) Enthalpy change =
b) i) 4CH3NHNH2(l) + 5N2O4(l)
(1)
4CO2(g) + 12H2O(g) + 9N2(g)
(1)
ii) ΔH = 4 x ΔHOf [CO2(g)] + 12 x ΔHOf [H2O(g)] – 4 x ΔHOf [CH3NHNH2(l)] – 5 x ΔHOf [N2O4(l)]
= [4(–394) + 12(–242) – 4(+53) – 5(–20)] kJ
= –4 590 kJ
–4 590 kJ
[4(46.0) + 5(92.0)] g
= –7.13 kJ g–1
iii) Enthalpy change per gram of fuel mixture =
(1)
(1)
(1)
(1)
(1)
c) The propellant produces H2O(g) only.
(1)
d) CH3NHNH2(l) reacts with N2O4(l) on mixing. The propulsion can easily be started and restarted.
(1)
73 a) Amount of heat released = 300.0 g x 4.18 J g–1 K–1 x (76.0 – 4.0) K
= 90 300 J
= 90.3 kJ
(1)
b) ΔHOr = ΔHOf [Mg(OH)2(s)] – 2 x ΔHOf [H2O(l)]
= [(–925) – 2(–286)] kJ
= –353 kJ
(1)
(1)
(1)
c) Increase the rate of redox reaction. / Facilitate electron transfer.
(1)
d) i)
'PPEUFNQFSBUVSFž$
5JNFFMBQTFENJO
(1 mark for correctly plotted points; 1 mark for connecting the points smoothly)
28
(2)
ii) Within about 4.5 minutes after the release of water
(1)
iii) About 15 minutes
(1)
74 • Place excess dilute hydrochloric acid in a polystyrene cup with a lid.
• Measure the steady temperature of the acid.
(0.5)
• Add solid magnesium hydroxide of mass M1 to the polystyrene cup, with stirring.
(0.5)
• Measure the highest temperature reached.
(0.5)
• Repeat the above steps using magnesium of mass M2 instead of magnesium hydroxide.
(0.5)
Let
C1 be the heat capacity of the Mg(OH)2 / HCl system;
C2 be the heat capacity of the Mg / HCl system;
ΔT1 be the temperature rise of the Mg(OH)2 / HCl system; and
ΔT2 be the temperature rise of the Mg / HCl system.
Number of moles of Mg(OH)2(s) used =
M1
molar mass of Mg(OH)2
= n1
(0.5)
Enthalpy change of reaction between Mg(OH)2(s) and HCl(aq)
ΔH1 =
C1 x ΔT1
n1
(0.5)
Number of moles of Mg(s) used =
M2
molar mass of Mg
= n2
(0.5)
Enthalpy change of reaction between Mg(s) and HCl(aq)
ΔH2 =
C2 x ΔT2
n2
(0.5)
O
Mg(s) + 2HCl(aq) + H2(g) + O2(g)
ΔHf [Mg(OH)2(s)]
O
ΔH2 + 2 x ΔHf [H2O(l)]
Mg(OH)2(s) + 2HCl(aq)
ΔH1
MgCl2(aq) + 2H2O(l)
By Hess’s Law, ΔHOf [Mg(OH)2(s)] = ΔH2 + 2 x ΔHOf [H2O(l)] – ΔH1
(1)
(1)
(3 marks for organization and presentation)
29
75 • Place excess dilute hydrochloric acid in a polystyrene cup with a lid.
• Measure the steady temperature of the acid.
(0.5)
• Add solid sodium hydrogencarbonate of mass M1 to the polystyrene cup, with stirring.
(0.5)
• Measure the highest temperature reached.
(0.5)
• Repeat the above steps using solid sodium carbonate of mass M2 instead of sodium
hydrogencarbonate.
(0.5)
Let
C1 be the heat capacity of the NaHCO3 / HCl system;
C2 be the heat capacity of the Na2CO3 / HCl system;
ΔT1 be the temperature rise of the NaHCO3 / HCl system; and
ΔT2 be the temperature rise of the Na2CO3 / HCl system.
M1
molar mass of NaHCO3
Number of moles of NaHCO3(s) used =
= n1
(0.5)
Enthalpy change of reaction between NaHCO3(s) and HCl(aq)
C1 x ΔT1
n1
Number of moles of Na2CO3(s) used =
ΔH1 =
(0.5)
M2
molar mass of Na2CO3
= n2
(0.5)
Enthalpy change of reaction between Na2CO3(s) and HCl(aq)
ΔH2 =
C2 x ΔT2
n2
(0.5)
2NaHCO3(s) + 2HCl(aq)
2 x ΔH1
ΔH
Na2CO3(s) + H2O(l) + CO2(g) + 2HCl(aq)
ΔH2
2NaCl(aq) + 2H2O(l) + 2CO2(g)
By Hess’s Law, ΔH = 2 x ΔH1 – ΔH2
(1)
(1)
(3 marks for organization and presentation)
30