ANSWERS TO EXERCISES AND REVISION QUESTIONS14.10Unit 15
Exercises
15A
Statement (b) is correct. Statement (a) is incorrect
because the double arrows show that the reaction is an
equilibrium reaction and complete conversion of
reactants to products does not occur under the
conditions of the experiment.
15B
(i)
Kc[Fe3 +(aq)] (^2) [Cl–(aq)] 2
[Fe^2 +(aq)]^2 [Cl the units are mol dm–^3
2 (g)]
(ii)
K [ClONO^2 (g)]
c[NO 2 (g)][ClO(g)] the units are mol–^1 dm^3
(iii)
K [NO^2 (g)]^2
c[N 2 O 4 (g)] the units are mol dm–^3
15C
(i)Reactions in Table 15.1 where Kc 103 are 2, 7, 8 (at
298 K) and 10.
(ii)The reaction between sodium ions and EDTA
Na+(aq)EDTA^4 – (aq) NaEDTA^3 – (aq)
does not go to completion. This means that the
EDTA^4 – will not complex all the sodium ions in a
mixture and the Na+(aq)–EDTA^4 – reaction is therefore
worthless in quantitative analysis. (See also
Exercise 12R)
15D
Yes. The equilibrium constant for the reverse reaction at
1100 K is 1/(7 10 –^9 )1.4 108. We conclude that
decomposition goes to completion.
15E
(i)Kc(average)45 at 490 °C.
K [HI(g)]^2
c(T)[H 2 (g)][I 2 (g)]
or,
[HI(g)] (Kc(T)[H 2 (g)][I 2 (g)])
(453.003.00)20 mol dm–^3
In the reaction between hydrogen and iodine,
1H 2 (g)1I 2 (g)2HI(g). Therefore, if 20 mol dm–^3 of HI
is produced, (20/2) 10 mol dm–^3 of each reactant is
used up. This means that initially there must have been
10 3.0013 mol dm–^3 of H 2 (g) and I 2 (g).
(ii)
K [HI(g)]^2 [19.1]^2
c(800 K) [H 2 (g)][I 2 (g)][3.50][2.50]41.7
This value is lower than the value at 490°C (45),
suggesting that the equilibrium constant decreases with
increasing temperature.
15F
[N 2 (g)][H 2 (g)]^3 0.04(0.03)^3
3 10 – (^3) mol (^2) dm 6
[NH 3 ]^2 (0.02)^2
This is not equal to the equilibrium constant at this
temperature. Therefore, the reaction is not at
equilibrium.
15G
One of the products, water vapour, is driven away (by
heating and by the flow of unused hydrogen gas). This
prevents the reaction reaching equilibrium. In an attempt
to restore the concentration of water, more iron(III) oxide
reacts with hydrogen. Eventually, all the Fe 2 O 3 is
consumed.
15H
15I
Raising the temperature increases the equilibrium
concentration of NO 2. This suggests that the formation
of NO 2 is endothermic since, for endothermic reactions,
Kcincreases with increasing temperature. (It is useful to
note that chemical decompositions are generally
endothermic.)
15J
(i)Equilibrium concentration of ethane (C 2 H 6 (g)) rises;
Kcis unaffected.
(ii)This increases the concentration of hydrogen
reactant. Therefore the equilibrium concentration of
ethane increases; Kcis unaffected.
(iii)No effect on composition or upon Kc.
(iv)Exothermic reaction, so Kcincreases. Thus the
concentration of ethane increases.
15K
Applying the equilibrium law,
Kc(T)[H[H^2 O(g)]
2 O(l)]
However, the concentration of a pure liquid is fixed at
that temperature. We can now write
Kc(T)[H 2 O(l)][H 2 O(g)]
By the ideal gas equation,
PcRT
wherecis the concentration of gas (n/V) and Pis its
partial pressure. Therefore,
p(H 2 O)Kc(T)[H 2 O(l)]RT
whereRis the universal gas constant. All the quantities
on the right-hand side of this equation are fixed at a
particular temperature. We conclude that the partial
pressure of water vapour in equilibrium with its liquid is
also fixed at a particular temperature.
Revision questions
15.1The chemical equation is missing.
15.2 (i) The completed table is
(Concentration/mol dm–3)# 10 –3)
Time(s) [H 2 (g)] [I 2 (g)] [HI(g)] Q
000 1.000 1.000 0.000 00.000
050 0.897 0.897 0.206 00.053
100 0.813 0.813 0.374 00.211
200 0.685 0.685 0.630 00.846
300 0.592 0.592 0.817 01.905
400 0.521 0.521 1.040 03.985
500 0.465 0.465 1.069 05.285
600 0.420 0.420 1.160 07.628
700 0.383 0.383 1.234 10.381
800 0.352 0.352 1.296 13.556
850 0.333 0.333 1.334 16.048
900 0.332 0.332 1.336 16.193
950 0.331 0.331 1.338 16.340
(ii)A plot of [H 2 (g)] and [HI(g)] against time:
The initial rate of consumption of hydrogen is (1.000
0.897) 10 –^3 /502.06 10 –^6 mol dm–^3 s–^1.
The initial rate of HI production is 0.206 10 –^3 /50
4.12 10 –^6 mol dm–^3 s–^1. This is double the rate at
which H 2 (g) is consumed, as expected by the reaction
equation.
(iii)A plot of Qagainst time reveals that equilibrium has
been reached at about 900 s. The equilibrium constant
is very roughly 16.
(iv)Figure 15.6 suggests a crude estimate of the
temperature as being about 1000–1100 K.
15.3 (i)The presence of water in the initial mixture
produces a lower yield of ester in the equilibrium mixture
in order to maintainKcat that temperature. (Le
Chatelier’s principle allows us to come to the same
conclusion.)
(ii) (a)
K [CH^3 COOC^2 H^5 (l)][H^2 O(l)] (4.48)^2
c(T)[CH 3 COOH(l)][C 2 H 5 OH(l)](2.67)(1.93)3.89
(b)According to the chemical equation for this reaction,
1 mol of ester (and 1 mol of water) are produced
using 1 mol of acid and 1 mol of ethanol. Therefore,
the initial concentration of ethanol was
2.674.487.15 mol dm–^3. Similarly, the initial
concentration of ethanoic acid was
4.481.936.41 mol dm–^3.
Time/s
0
0
10
20
Q
200 400 600 800 1000
Time / s
0
0
0.4
0.8
1.2
1.6
[HI(g)]
[H 2 (g)]
200 400 600 800 1000
46
61 667 K
764 K
667K
Kc
Time / s
Platinum
O CO OC O
(a) Starting materials
O CO OC O
(b) Adsorption
O COO C O
(c) Rearrangement
O COO C O
(d) Desorption
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‘Kc’
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