New 21st Century Chemistry
Suggested answers to in-text activities and unit-end exercises
Topic 9 Unit 35
In-text activities
Checkpoint (page 71)
a)
b)
ΔHx = +90 kJ mol–1
ΔHy = –56 kJ mol–1
Checkpoint (page 79)
1 The following data is given in the question:
(1) N2O(g) à N2(g) + O2(g) ΔH1 = –82 kJ mol–1
(2) NO(g) + O2(g) à NO2(g) ΔH2 = –57 kJ mol–1
(3) N2(g) + O2(g) à 2NO(g) ΔH3 = +181 kJ mol–1
Looking at the target equation, we need 1 mole of N2O(g) as the reactant. So, keep equation (1) as it is.
We need 1 mole of NO2(g) as the reactant. Equation (2) has 1 mole of NO2(g), but it is on the product side. So, reverse the equation, giving equation (2)’.
By combining equations (1), (3) and (2)’, followed by collecting like items, we can obtain the target equation.
(1) N2O(g) à N2(g) + O2(g) ΔH1 = –82 kJ mol–1
(2)’ NO2(g) à NO(g) + O2(g) ΔH2 = –57 kJ mol–1
(3) N2(g) + O2(g) à 2NO(g) ΔH3 = +181 kJ mol–1
N2O(g) + NO2(g) à 3NO(g) ΔH4 = +156 kJ mol–1
∴ the enthalpy change for the reaction is +156 kJ mol–1.
2 a) Rhombic sulphur is more stable.
b) The following data is given in the question:
(1) S(rhombic) + O2(g) à SO2(g) ΔH = –296.0 kJ mol–1
(2) S(monoclinic) + O2(g) à SO2(g) ΔH = –296.4 kJ mol–1
By combining equations (1) and (2)’, followed by collecting like items, we can obtain the target equation.
(1) S(rhombic) + O2(g) à SO2(g) ΔH = –296.0 kJ mol–1
(2)’ SO2(g) à S(monoclinic) + O2(g) ΔH = +296.4 kJ mol–1
S(rhombic) à S(monoclinic) ΔH = +0.4 kJ mol–1
∴ the enthalpy change for the process is +0.4 kJ mol–1.
3 a)
b) By Hess’s Law
ΔHӨ f[ZnS(s)] = [(–348) + (–297) – (–441)] kJ mol–1
= –204 kJ mol–1
∴ the enthalpy change of formation of zinc sulphide is –204 kJ mol–1.
c)
Checkpoint (page 87)
a) i) Amount of heat released = 80 g x 4.18 J g–1 K–1 x 9.6 K
= 3 200 J
Number of moles of CuSO4(s) added =
= 0.050 mol
Enthalpy change of solution of CuSO4(s) =
= –64 000 J mol–1
= –64.0 kJ mol–1
ii) Amount of heat taken in = 80 g x 4.18 J g–1 K–1 x 2.0 K
= 670 J
Number of moles of CuSO4•5H2O(s) added =
= 0.0481 mol
Enthalpy change of solution of CuSO4•5H2O(s) =
= +14 000 J mol–1
= +14.0 kJ mol–1
b) i) ΔHr = ΔH1 – ΔH2
ii) Enthalpy change of the process ΔHr = ΔH1 – ΔH2
= [(–64.0) – (+14.0)] kJ mol–1
= –78.0 kJ mol–1
∴ the enthalpy change of the process is –78.0 kJ mol–1.
Checkpoint (page 94)
1 a) ΔH2 represents the enthalpy change of formation of C2H4(g).
ΔH3 represents the enthalpy change of combustion of C2H4(g).
b) ΔH1 represents the enthalpy change of combustion of C(graphite) and the enthalpy change of combustion of H2(g).
c) HӨ f[C2H4(g)] = 2 x ΔHӨ c[C(graphite)] + 2 x ΔHӨ c[H2(g)] – ΔHӨ c[C2H4(g)]
= [2(–394) + 2(–286) – (–1 411)] kJ mol–1
= +51 kJ mol–1
∴ the standard enthalpy change of formation of ethene is +51 kJ mol–1.
d)
e) Carbon and hydrogen do not react to produce ethene under normal conditions.
2 ΔHӨ f[C12H22O11(s)] refers to the standard enthalpy change of the following process:
12C(graphite) + 11H2(g) + O2(g) à C12H22O11(s)
ΔHӨ f[compound] = Σ ΔHӨ c[constituent elements] – ΔHӨ c[compound]
ΔHӨ f[C12H22O11(s)] = 12 x ΔHӨ c[C(graphite)] + 11 x ΔHӨ c[H2(g)] – ΔHӨ c[C12H22O11(s)]
= [12(–394) + 11(–286) – (–5 640)] kJ mol–1
= –2 230 kJ mol–1
∴ the standard enthalpy change of formation of sucrose is –2 230 kJ mol–1.
Checkpoint (page 100)
1 a)
b) ΔHӨ f[N2(g)] = 0
c) ΔHӨ r = Σ ΔHӨ f[products] – ΣΔHӨ f[reactant]
= 4 x ΔHӨ f[H2O(g)] + ΔHӨ f[Cr2O3(s)] – ΔHӨ f[(NH4)2Cr2O7(s)]
= [4(–242) + (–1 140) – (–1 810)] kJ mol–1
= –298 kJ mol–1
∴ the standard enthalpy change of the reaction is –298 kJ mol–1。
2 a) Amount of heat released when 1.00 g of butan-1-ol was burnt
= 200.0 g x 4.18 J g–1 K–1 x (44.0 – 22.0) K
= 18 400 J
= 18.4 kJ
Number of moles of butan-1-ol burnt =
= 0.0135 mol
Enthalpy change of combustion of butan-1-ol =
= –1 360 kJ mol–1
∴ the enthalpy change of combustion of butan-1-ol is –1 360 kJ mol–1.
b) ΔHӨ c[C4H9OH(l)] refers to the standard enthalpy change of the following process:
C4H9OH(l) + 6O2(g) à 4CO2(g) + 5H2O(l)
ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactant]
ΔHӨ c[C4H9OH(l)] = 4 x ΔHӨ f[CO2(g)] + 5 x ΔHӨ f[H2O(l)] – ΔHӨ f[C4H9OH(l)]
= [4(–394) + 5(–286) – (–327)] kJ mol–1
= –2 680 kJ mol–1
∴ the standard enthalpy change of combustion of butan-1-ol is –2 680 kJ mol–1.
c) Any two of the following:
• Heat loss to the surroundings.
• Incomplete combustion of butan-1-ol.
• Heat capacities of the container and the thermometer are ignored.
• Experiment not carried out under standard conditions.
• Loss of butan-1-ol / water via evaporation.
• The thermometer not precise enough.
STSE Connections (page 102)
1 ΔHӨ c[C8H18(l)] refers to the standard enthalpy change of the following process:
C8H18(l) + O2(g) à 8CO2(g) + 9H2O(l)
ΔHӨ c[C8H18(l)] = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= [8(–394) + 9(–286) – (–250)] kJ mol–1
= –5 480 kJ mol–1
ΔHӨ f[C2H5OH(l)] refers to the standard enthalpy change of the following process:
C2H5OH(l) + 3O2(g) à 2CO2(g) + 3H2O(l)
ΔHӨ c[C2H5OH(l)] = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= [2(–394) + 3(–286) – (–278)] kJ mol–1
= –1 370 kJ mol–1
2 Advantages and disadvantages of using ethanol as a fuel for cars
Advantages / Disadvantages· Ethanol can be produced from many different types of food crops. / · It requires a large area of land to produce enough food crops to meet the demand.
· Ethanol can burn directly as a fuel and does not require other additives to improve its efficiency. / · It may lead to diversion of investment from food production, resulting in increased food prices.
· Ethanol burns to form mainly carbon dioxide and water. It is a cleaner fuel than other fossil fuels. / · The toxic effluent from distilleries causes water pollution.
· Ethanol is a renewable resource. / · Ethanal () is also produced during combustion. Without adequate pollution control, it can harm vegetation, irritate the skin and eyes, and damage the lung at high concentrations.
· Ethanol can be easily transported to various places. / · Fuels are required to produce ethanol in distilleries.
· Distilleries are built for ethanol production. This creates jobs for people. / · Combustion of 1 kg of ethanol can give only 60% of the energy that can be released from 1 kg of petrol.
· Cars can be easily converted to run on pure ethanol. / · It may not be possible to obtain cheap ethanol in every part of the world.
Unit-end exercises (pages 105 – 115)
Answers for the HKCEE (Paper 1) and HKALE questions are not provided.
1 A possible concept map is shown below:
2
3 a) C8H18(l) + O2(g) à 8CO2(g) + 9H2O(l)
b)
c) ΔHӨ f[compound] = Σ ΔHӨ c[constituent elements] – ΔHӨ c[compound]
ΔHӨ f[C8H18(l)] = 8 x ΔHӨ c[C(graphite)] + 9 x ΔHӨ c[H2(g)] – ΔHӨ c[C8H18(l)]
= [8(–394) + 9(–286) – (–5 470)] kJ mol–1
= –256 kJ mol–1
4 a)
b) ΔHӨ f[H2(g)] = 0
c) ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
ΔHӨ f = ΔHӨ f[CO2(g)] – ΔHӨ f[CH4(g)] – 2 x ΔHӨ f[H2O(g)]
= [(–394) – (–74.8) – 2(–242)] kJ mol–1
= +165 kJ mol–1
5 a) None
ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= 3 x ΔHӨ f[CaO(s)] – ΔHӨ f[Fe2O3(s)]
Both ΔHӨ f[CaO(s)] and ΔHӨ f[Fe2O3(s)] are given in the question.
b) ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
=ΔHӨ f[CaO(s)] – ΔHӨ f[CuO(s)]
∴ the value of ΔHӨ f[CuO(s)] is required for the calculations.
c) ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= ΔHӨ f[Fe2O3(s)] – 3 x ΔHӨ f[CuO(s)]
∴ the value of ΔHӨ f[CuO(s)] is required for the calculations.
6 B ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= 3 x ΔHӨ f[CO(g)] – ΔHӨ f[Fe2O3(s)]
= [3(–110) – (–822)] kJ
= +492 kJ
7 C ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= 3 x ΔHӨ f[MgO(s)] + ΔHӨ f[KCl(s)] – ΔHӨ f[KClO3(s)]
= [3(–602) + (–437) – (–391)] kJ
= –1 852 kJ
8 B ΔHӨ r = Σ ΔHӨ f[products] – Σ ΔHӨ f[reactants]
= ΔHӨ f[Ca(OH)2(s)] – 2 x ΔHӨ f[H2O(l)]
∴ the extra information required is the enthalpy change of formation of H2O(l), i.e. the enthalpy change of combustion of H2(g).
9 A The following data is given in the question:
(1) H2(g) + O2(g) à H2O(l) ΔH = q
(2) S(s) + O2(g) à SO2(g) ΔH = r
(3) H2S(g) + O2(g) à H2O(l) + SO2(g) ΔH = s
Looking at the target equation, we need 1 mole of S(s) and 1 mole of H2(g) as the reactants. So, keep equations (1) and (2) as they are.
We need 1 mole of H2S(g) as the product. Equation (3) has 1 mole of H2S(g), but it is on the reactant side. So, we need to reverse the equation, giving equation (3)’.
By combining equations (1), (2) and (3)’, followed by collecting like items, we can obtain the target equation.
(1) H2(g) + O2(g) à H2O(l) ΔH = q
(2) S(s) + O2(g) à SO2(g) ΔH = r
(3)’ H2O(l) + SO2(g) à H2S(g) + O2(g) ΔH =– s
S(s) + H2(g) à H2S(g) ΔH = p = q + r – s
10 A (1) CuSO4(s) + 5H2O(l) à CuSO4 • 5H2O(s)
The above reaction cannot be properly carried out. Some CuSO4(s) may dissolve but may not be completely hydrated.
(2) C(s) + 2H2(g) à CH4(g)
Carbon and hydrogen do not react to produce methane under normal conditions.
(3) C6H12O6(s) + 6O2(g) à 6CO2(g) + 6H2O(l)
The enthalpy change of combustion of glucose can be measured directly: burn a fixed mass of glucose and measure the heat evolved.
11 —
12 a) 3C(graphite) + 2H2(g) à CH3C≡CH(g)
b)
ΔHӨ f[CH3C≡CH(g)] = 3 x ΔHӨ c[C(graphite)] + 2 x ΔHӨ c[H2(g)] – ΔHӨ c[CH3C≡CH(g)]
= [3(–394) + 2(–286) – (–1 938)] kJ mol–1
= +184 kJ mol–1
∴ the enthalpy change of formation of propyne is +184 kJ mol–1.
c) No
Carbon and hydrogen do not react to produce propyne under normal conditions.
13 a) N2(g) + O2(g) à N2O5(l)
b) Looking at equation (I), we need to eliminate the 2NO(g) on the product side. So, multiply equation (II) by 2, giving equation (II)’.
By combining equation (I), (II)’ and (III), followed by collecting like terms, we can obtain the equation for the formation of N2O5(g) from its constituent elements.
(I) N2(g) + O2(g) à 2NO(g) ΔH = +180 kJ
(II)’ 2NO(g) + O2(g) à 2NO2(g) ΔH = –114 kJ
(III) 2NO2(g) + O2(g) à N2O5(g) ΔH = –55 kJ
N2(g) + O2(g) à N2O5(g)
By Hess’s Law
ΔHӨ f[N2O5(g)] = [+180 + (–114) + (–55)] kJ mol–1
= +11 kJ mol–1
∴ the enthalpy change of formation of dinitrogen pentoxide is +11 kJ mol–1.
14 a) C6H12O6(s) + 6O2(g) à 6CO2(g) + 6H2O(l)
b)
ΔHӨ c[C6H12O6(s)] = 6 x ΔHӨ f[CO2(g)] + 6 x ΔHӨ f[H2O(l)] – ΔHӨ f[C6H12O6(s)]
= [6(–394) + 6(–286) – (–1 274)] kJ mol–1
= –2 806 kJ mol–1
c)
15 a)
ΔHӨ r = ΔHӨ f[Na2CO3(s)] + ΔHӨ f[CO2(g)] + ΔHӨ f[H2O(l)] – 2 x ΔHӨ f[NaHCO3(s)]
= [(–1 131) + (–394) + (–286) – 2(–951)] kJ
= +91 kJ
∴ the standard enthalpy change accompanying the thermal decomposition of 2 moles of NaHCO3 is +91 kJ。
b)
16 a) C3H8(g) + 5O2(g) à 3CO2(g) + 4H2O(l)
b) The standard enthalpy change of combustion of a substance is the enthalpy change when one mole of the substance
is completely burnt in oxygen under standard conditions.
c) i) Amount of heat produced = 200 g x 4.18 J g–1 K–1 x (68.3 – 18.0) K
= 42 100 J
= 42.1 kJ
ii) Number of moles of C3H8 burnt =
= 0.0227 mol
iii) Enthalpy change of combustion of C3H8 =
= –1 850 kJ mol–1
∴ the enthalpy change of combustion of C3H8 is – 1 850 kJ mol–1.
d) i) ΔHӨ f = [compound] = Σ ΔHӨ c[constituent] – ΔHӨ c[compound]
ΔHӨ f = [C3H8(g)] = 3 x ΔHӨ c[C(graphite)] + 4 x ΔHӨ c[H2(g)] – ΔHӨ c[C3H8(g)]
= [3(–394) + 4(–286) – (–2 219)] kJ mol–1
= –107 kJ mol–1
∴ the enthalpy change of formation of propane is –107 kJ mol–1.
ii) Carbon and hydrogen do not react to produce propane under normal conditions.
17 —
18 a) Amount of heat released = 600 J K–1 x 8.60 K
= 5 160 J
Number of moles of Mg reacted =
= 0.0100 mol
Enthalpy change of the reaction =
= –516 000 J mol–1
= –516 kJ mol–1
b) ΔHӨ f[Mg(OH)2(s)] refers to the enthalpy change of the following process: