New 21st Century Chemistry
Suggested answers to in-text activities and unit-end exercises
Topic 11 Unit 39
In-text activities
Checkpoint (page 4)
1 a) Reversible
b) Irreversible
c) Irreversible
2 a) A(g) and B(g)
b) C(g) and D(g)
c) A(g) + B(g) C(g) + D(g)
Checkpoint (page 9)
1 (a) and (c) are open systems. An equilibrium cannot be established.
b) A sealed bag containing a wet towel represents a system at equilibrium. This is a closed system in which no material can enter or escape.
2 a) The blue line indicates NH3(g).
The green line indicates N2(g).
The red line indicates H2(g).
b) During the first 10 minutes, the rate of the forward reaction is higher than the rate of the backward reaction.
c) At the 80th minute, the rate of the forward reaction is the same as the rate of the backward reaction.
3 a), b)
Checkpoint (page 15)
a) Kc =
b) Kc =
c) Kc =
d) Kc =
Checkpoint (page 21)
1 Kc =
=
= 957 dm3 mol–1
∴ the equilibrium constant, Kc, for the reaction is 957 dm3 mol–1.
2 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration mol dm–3 0 mol dm–3 0 mol dm–3
According to the equation, 2 moles of NOCl decompose to give 2 moles of NO and 1 mole of Cl2.
SpeciesNOCl(g) / NO(g) / Cl2(g)
Initial concentration
(mol dm–3) / / 0 / 0
Change in concentration
(mol dm–3) / – x 28% / + x 28% / + x x 28%
Equilibrium concentration
(mol dm–3) / 1.20 / 0.467 / 0.233
Kc =
=
= 0.0353 mol dm–3
∴ the equilibrium constant, Kc, for the reaction is 0.0353 mol dm–3.
3 a) The green line indicates A(g).
The red line indicates B(g).
The blue line indicates C(g).
b) y = 2
z = 3
The concentration of A(g) decreases by 0.50 mol dm–3 while that of B(g) decreases by 1.0 mol dm–3 and that of C(g) increases by 1.5 mol dm–3.
c) Kc =
=
= 6.9
∴ the equilibrium constant, Kc, for the reaction is 6.9.
4 a) 2A(g) + 3B(g) 4C(g)
Initial concentration mol dm–3 mol dm–3 0 mol dm–3
= 0.0480 mol dm–3 = 0.0720 mol dm–3
According to the equation, 2 moles of A(g) react with 3 moles of B(g) to give 4 moles of C(g).
SpeciesA(g) / B(g) / C(g)
Initial concentration
(mol dm–3) / 0.0480 / 0.0720 / 0
Change in concentration
(mol dm–3) / –0.0480 x
= –0.0320 / –0.0480 / –0.0320 x
= –0.0640
Equilibrium concentration
(mol dm–3) / 0.0160 / 0.0240 / 0.0640
b) Kc =
=
= 4 740 dm3 mol–1
c) Kc
=
=
Checkpoint (page 31)
1 Kc =
0.290 dm6 mol–2 =
[NH3(g)] = 5.94 x 10–3 mol dm–3
∴ the concentration of NH3(g) was 5.94 x 10–3 mol dm–3.
2 Let x mol dm–3 be the concentration of hydrogen ions in the wine.
RCOOH(aq) RCOO–(aq) + H+(aq)
Initial concentration mol dm–3 0 mol dm–3 0 mol dm–3
Equilibrium concentration mol dm–3 x mol dm–3 x mol dm–3
Kc =
1.00 x 10–3 mol dm–3 =
Rearranging the equation gives
(1.00 x 10–3)(0.0467 – x) = x2
x2 + 1.00 x 10–3x – 4.67 x 10–5 = 0
Solving the quadratic equation gives two solutions:
x = 6.35 x 10–3 or –7.35 x 10–3 (rejected)
pH of wine = –log10[H+(aq)] = –log10(6.35 x 10–3) = 2.20
∴ the pH of the wine is 2.20.
3 Let x% be the percentage decomposition of N2O4(g) at equilibrium.
N2O4(g) 2NO2(g)
Initial concentration mol dm–3 0 mol dm–3
= 0.125 mol dm–3
According to the equation, 1 mole of N2O4 decomposes to give 2 moles of NO2.
SpeciesN2O4(g) / NO2(g)
Initial concentration (mol dm–3) / 0.125 / 0
Change in concentration (mol dm–3) / –0.125x% / +2 x 0.125x%
Equilibrium concentration (mol dm–3) / 0.125(1 – x%) / 0.250x%
Kc =
1.51 mol dm–3 =
Rearranging the equation gives
x2 + 302x – 30 200 = 0
Solving the quadratic equation gives two solutions:
x = 79.0 or –381 (rejected)
∴ the percentage decomposition of N2O4(g) at equilibrium is 79.0%.
4 a) [N2(g)] = = 0.0500 mol dm–3
[H2(g)] = = 0.100 mol dm–3
[NH3(g)] = = 0.250 mol dm–3
Qc = =
= 1 250 dm6 mol–2
> 100 dm6 mol–2 (given Kc value)
Qc ≠ Kc, thus the system was not at equilibrium.
b) When Qc > Kc, the concentration of NH3(g) would decrease while those of N2(g) and H2(g) would increase until Qc = Kc.
Thus a net backward reaction would occur.
Checkpoint (page 37)
a) Kc =
b) Initial concentration of Fe3+(aq) ions
= mol dm–3
= 2.5 x 10–3 mol dm–3
c) 3.2 x 10–4 mol dm–3
d) Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq)
Initial concentration 2.5 x 10–3 mol dm–3 2.5 x 10–3 mol dm–3 0 mol dm–3
According to the equation, 1 mole of Fe3+ ions reacts with 1 mole of SCN– ions to give 1 mole of [Fe(SCN)]2+ ions.
As the concentration of [Fe(SCN)]2+ ions increased by 3.2 x 10–4 mol dm–3 when equilibrium was reached, those of both Fe3+ and SCN– ions would decrease by 3.2 x 10–4 mol dm–3.
SpeciesFe3+(aq) / SCN–(aq) / [Fe(SCN)]2+(aq)
Initial concentration
(mol dm–3) / 2.5 x 10–3 / 2.5 x 10–3 / 0
Change in concentration
(mol dm–3) / –3.2 x 10–4 / –3.2 x 10–4 / +3.2 x 10–4
Equilibrium concentration
(mol dm–3) / 2.2 x 10–3 / 2.2 x 10–3 / 3.2 x 10–4
Kc =
=
= 66 dm3 mol–1
∴ the equilibrium constant, Kc, for the reaction is 66 dm3 mol–1.
Unit-end exercises (pages 40 – 46)
Answers for the HKCEE (Paper 1) and HKALE questions are not provided.
1 A possible concept map:
2 a) A dynamic equilibrium is reached when the forward and backward reactions occur at the same rate.
b) Any two of the following:
• Equilibrium can only be established in a closed system with temperature and pressure held constant.
• Once equilibrium is reached, the composition of the system remains constant.
• The reaction has not stopped, rate of forward reaction = rate of backward reaction.
• Reactants and products will both be present.
• Equilibrium can be reached from either direction, starting with the substances on the left of the equation, or starting with the substances on the right.
• If the conditions (temperature, pressure and concentration) are changed, the equilibrium established may be affected.
3 a) The rate of the forward reaction is the same as that of the backward reaction.
b) Kc =
4 a) Kc =
b) Kc =
c) Kc =
d) Kc =
e) Kc =
5 a) Comparable concentrations of PCl3(g), Cl2(g) and PCl5(g) are present at equilibrium.
b) The reaction hardly proceeds at all.
6 C NO(g) + CO2(g) NO2(g) + CO(g)
Initial concentration mol dm–3 mol dm–3 0 mol dm–3 0 mol dm–3
According to the equation, 1 mole of NO reacts with 1 mole of CO2 to produce 1 mole of NO2 and 1 mole of CO.
i.e. as the amount of CO2 decreased by 0.800 mole at equilibrium, the amount of NO would decrease by the same amount, while the amounts of NO2 and CO would increase by the same amount.
SpeciesNO(g) / CO2(g) / NO2(g) / CO(g)
Initial concentration
(mol dm–3) / / / 0 / 0
Change in concentration
(mol dm–3) / – / – / + / +
Equilibrium concentration
(mol dm–3) / / / /
Kc =
=
= 2.00
7 B Kc =
13.0 dm3 mol–1 =
[SO2(g)]2 =
[SO2(g)]2 = 0.361 mol dm–3
8 B Option B — Kc =
The units of Kc are given by
i.e. dm3 mol–1
9 D K1 = 4.8 =
K2 = X =
X = =
10 A K1 = 5 x 10–3 =
K2 = = = = 4 x 104
11 K1 =
=
= 2.14 mol2 dm–6
12 a)
b) Ester
c) i)
Number of moles in the reaction mixtureHCOOH / C2H5OH / HCOOC2H5 / H2O
At start of experiment (mol) / 3.00 / 6.25 / 0.00 / 0.00
Change (mol) / –2.50 / –2.50 / +2.50 / +2.50
At equilibrium (mol) / 0.50 / 3.75 / 2.50 / 2.50
ii) Kc =
iii) Kc =
= 3.33
iv) Kc for this reaction has no units.
There are equal number of moles of substances on both sides of the equation.
13 a) X2(g) + Y2(g) 2XY(g)
According to the equation, 1 mole of X2 reacts with 1 mole of Y2 to give 2 moles of XY.
i.e. as the amounts of both X2 and Y2 decreased by 0.30 mole when equilibrium was reached, the amount of XY would increase by 0.60 mole.
∴ 0.60 mole of XY was present in the equilibrium mixture.
b) Kc =
=
= 9.0
c)
14 a) N2O4(g) 2NO2(g)
According to the equation, 1 mole of N2O4 dissociates to give 2 moles of NO2.
Number of moles of N2O4 dissociated = 1.00 x 57.0% mol
= 0.570 mol
Number of NO2 formed = 2 x 0.570 mol
= 1.14 mol
∴ 1.14 moles of NO2 were present in the equilibrium mixture.
b) Number of mole of N2O4 in the equilibrium mixture = (1.00 – 0.570) mol
= 0.430 mol
Kc =
=
= 1.51 mol dm–3
∴ the equilibrium constant, Kc, at the temperature of the experiment is 1.51 mol dm–3.
15 a) Number of moles of SO3 =
= 0.0100 mol
b) Number of moles of O2 formed =
= 1.75 x 10–3 mol
c) 2SO2(g) + O2(g) 2SO3(g)
According to the equation, 2 moles of SO3 decompose to give 2 moles of SO2 and 1 mole of O2.
i.e. as 1.75 x 10–3 mole of O2 was formed, 3.5 x 10–3 mole of SO2 would be present, and the amount of SO3 would decrease by 3.5 x 10–3 mole.
SpeciesSO3(g) / SO2(g) / O2(g)
Initial concentration
(mol dm–3) / / 0 / 0
Change in concentration
(mol dm–3) / – / + / +
Equilibrium concentration
(mol dm–3) / / /
Kc =
=
= 1.97 x 103 dm3 mol–1
∴ the equilibrium constant, Kc, for the reaction at the temperature of the experiment is 1.97 x 103 dm3 mol–1.
16 CH3OH(g) CO(g) + 2H2(g)
Kc =
0.44 mol2 dm–6 =
[CH3OH(g)] = 0.63 mol dm–3
∴ the equilibrium concentration of CH3OH(g) was 0.63 mol dm–3.
17 Let x mol dm–3 be the concentration of hydrogen ions in the acid.
HCOOH(aq) HCOO–(aq) + H+(aq)
Initial concentration 0.100 mol dm–3 0 mol dm–3 0 mol dm–3
Equilibrium concentration (0.100 – x) mol dm–3 x mol dm–3 x mol dm–3
Kc =
1.8 x 10–4 mol dm–3 =
Rearranging the equation gives
1.8 x 10–4(0.100 – x) = x2
x2 + 1.8 x 10–4x – 1.8 x 10–5 = 0
Solving the quadratic equation gives two solutions:
x = 4.2 x 10–3 or –4.3 x 10–3 (rejected)
pH of acid = –log10(4.2 x 10–3)
= 2.4
∴ the pH of the methanoic acid is 2.4.
18 —
19 2SO2(g) + O2(g) 2SO3(g)
Kc = 261 dm3 mol–1 =
a) 2SO3(g) 2SO2(g) + O2(g)
K1 = = = = 3.83 x 10–3 mol dm–3
b) SO2(g) + O2(g) SO3(g)
K2 = = =
c) SO3(g) SO2(g) + O2(g)
K3 = = =
20 2NO(g) N2(g) + O2(g) Kc1 = 2.40 x 10–18
NO(g) + Br2(g) NOBr(g) Kc2 = 1.40
N2(g) + O2(g) + Br2(g) 2NOBr(g) Kc3
Kc1 =
Kc2 =
Kc3 =
=
=
= 8.17 x 1017 dm3 mol–1
21 a) Qc =
b) Qc =
= 19.6 dm3 mol–1
Qc ≠ Kc, thus the system was not at equilibrium.
c) When Qc Kc, the concentration of SO3(g) would increase while those of SO2(g) and O2(g) would decrease until Qc = Kc.
Thus a net forward reaction would occur.
22 a) Qc =
=
= 7.20 x 10–3
Qc ≠ Kc, thus the system was not at equilibrium.
b) When Qc Kc, the concentration of NO(g) would decrease while those of N2(g) and O2(g) would increase until Qc = Kc.
Thus a net backward reaction would occur.
23 a) As a catalyst
b) To quench the reaction between ethanoic acid and ethanol.
c) NaOH(aq) reacts with CH3COOH(l) according to the following equation:
CH3COOH(l) + NaOH(aq) CH3COONa(aq) + H2O(l)
Number of moles of NaOH = number of moles of CH3COOH
0.250 mol dm–3 x dm3 = [CH3COOH(l)] x dm3
[CH3COOH(l)] = 8.50 mol dm–3
∴ concentration of CH3COOH(l) in the original mixture was 8.50 mol dm–3.
Volume of NaOH(aq) required to neutralize CH3COOH(l) in 1.00 cm3 of the mixture after reflux = (11.4 – 0.1) cm3 = 11.3 cm3
Number of moles of NaOH = number of moles of CH3COOH
0.250 mol dm–3 x dm3 = [CH3COOH(l)] x dm3
[CH3COOH(l)] = 2.83 mol dm–3
∴ concentration of CH3COOH(l) in the mixture after reflux was 2.83 mol dm–3.
Consider the equation for the esterification reaction:
CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
According to the equation, 1 mole of CH3COOH reacts with 1 mole of C2H5OH to give 1 mole of CH3COOC2H5 and 1 mole of H2O.