Higher level
Equilibria mark scheme
Multiple choice
1 mark per question
1A
2C
3B
4A
5C
6D
7B
8A
9D
10 D
11 D
12 D
13 D
14 A
15 D
Structured answer questions
16 (a) A system that cannot exchange matter with its surroundings / OWTTE;
[1]
(b) Rate of forward reaction is equal to the rate of the reverse reaction; the concentration of reactants and
products remains steady
[2]
(c) If a change is imposed on a system in equilibrium it will respond in a way so as to oppose / minimize the
effect of the change / OWTTE;
[2]
17
(a) rate of the forward reaction is equal to the rate of the reverse reaction / forward and reverse reactions
occur and the concentrations of the reactants and products do not change / OWTTE;
[1]
(b)
[1]
(c) to move the position of equilibrium to the right/product side / increase the yield of biodiesel;
[1]
(d) no effect (on position of equilibrium);
increases the rate of the forward and the reverse reactions equally (so equilibrium reached
quicker) / it lowers Ea for both the forward and reverse reactions by the same amount / OWTTE;
No ECF for explanation.
[2]
18.
(a) (Kc = )[SO3]2 / [O2][SO2]2;
[1]
(b) yield (of ) SO3 increases / equilibrium moves to right / more SO3 formed;
3 gaseous molecules → 2 gaseous molecules / decrease in volume of gaseous molecules / fewer gaseous
molecules on right hand side; Do not allow ECF
[2]
(c) yield (of) SO3 decreases;
forward reaction is exothermic / reverse/backwards reaction is endothermic / equilibrium shifts to
absorb (some of) the heat; Do not accept “exothermic reaction” or “Le Châtelier’s Principle” without
further explanation. Do not allow ECF.
[2]
(d) rates of both forward and reverse reactions increase equally;
no effect on position of equilibrium;
no effect on value of Kc;
[3]
(e) kc decreases with increasing temperature ;
back reaction favoured and must be endothermic / forward reaction disfavoured and must be exothermic
OWTTE;
[2]
(f)
Initial
2SO2
1.5mol / 1.5dm3 = 1moldm-3
Change
-0.3moldm-3
Equilibrium 0.667moldm-3
O2
2mol / 1.5moldm-3 =
1.333moldm-3
-0.333 / 2 = -0.166moldm-3
1.17 moldm-3
2SO3
0moldm-3
0.5mol/1.5dm3 =
0.3moldm-3
0.3moldm-3
Kc = [SO3]2 / [SO2]2[O2]
= 0.3332 / 1.17 x 0.6672
= 0.213 dm3mol-1 /0.214 dm3mol-1
One mark for each of the equilibrium concentrations of SO2 and O2
One mark for correctly substituting values into Kc expression
One mark for answer
One mark for units
Allow ECF.
If 0.202 dm3 mol–1 is given award [4], this is obtained by premature rounding.
[5]
19
Kc = [N2][H2O]2/[NO]2[H2]2 = (0.019)(0.138)2 / (0.062)2(0.013)2 = 557;
One mark for the equilibrium concentrations of H2, N2 and H2O
One mark for correctly substituting values into the Kc expression and calculating a correct answer (no
units)
If units have been shown do not penalize
Award [4] for final correct answer .
Accept any value also in range 557–560.
Do not penalize significant figures.
[4]
20.
(a)
(i) shifts to the right/toward products / forward reaction favoured; to consume excess added; Do not
accept “due to Le Chatelier’s principle” without further explanation.
[2]
(ii) shifts to the left/toward reactants / reverse reaction favoured; NaOH reacts to consume / an increase in
the amount of H2O resulting from neutralization; Do not accept “due to Le Chatelier’s principle” without
further explanation.
[2]
(iii) no effect; catalyst increases the rate of the forward and backward reactions equally / lowers the
activation energy of both forward and backward reaction equally / lowers EA so rate of forward and
backward reactions increase equally;
[2]
(b) equilibrium constant decreases; forward reaction is exothermic/produces heat / reverse reaction is
endothermic/absorbs heat;
[2]
Total %
%
Level
12.0%
1
30.0%
2
36.0%
3
41.0%
4
58.0%
5
69.0%
6
79.0%
7