A Data Booklet may be used.
CommonConsider the reaction W + X ⇌ Y + Z. Which of the following statements is incorrect when a catalyst is added to the reaction mixture?
Source: AJC 2017 Promo, via Chemical Equilibria Topical Revision Booklet, Q1
A 0.0113 mol sample of ammonium nitrate is introduced into an evacuated vessel with a volume of 0.83 dm3. The vessel, of constant volume, is heated to a constant temperature such that the equilibrium below is established. NH4NO3(s) ⇌ NH3(g) + HNO3(g)
Given that the percentage dissociation of ammonium nitrate at equilibrium is 66.8%, what is the value of the equilibrium constant, Kc?
Source: MJC 2017 Promo, via Chemical Equilibria Topical Revision Booklet, Q7
The key stage in the manufacture of sulfuric acid is 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ mol−1. Which of the following statements are correct for an increase in temperature?
Source: MJC 2017 Promo, via Chemical Equilibria Topical Revision Booklet, Q8
In an experiment, 2 mol of compound P were placed in a vessel and heated. Compound P decomposed partially to give compounds Q and R. At equilibrium, the mixture was found to contain x mol of R and the total amount present was (2 + x). Which equilibrium represents this equilibrium? All compounds are gases.
Source: EJC 2022 JC1 Chemical Equilibria Tutorial, A2
Common0.200 mol of carbon monoxide and 0.200 mol of steam were placed in a reaction vessel and allowed to react at a temperature of 800 K. The percentage of carbon monoxide reacted was found to be 77.5%. CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Source: 2025 NJC J2 Prelim P2 Q4(c)
In the gaseous state, NO2 can form a dimer. 2NO2(g) ⇌ N2O4(g). NO2 is brown and N2O4 is colourless.
Source: EJC 2022 JC1 Chemical Equilibria Tutorial, A10(a) and (b)
Steam can react with carbon monoxide under appropriate conditions according to the following reversible reaction. H2O(g) + CO(g) ⇌ CO2(g) + H2(g) ΔH = −40 kJ mol−1
Source: RJC H2 Chemistry Tutorial 7 (Chemical Equilibria), Q4(a), (b) and (d), adapted
In the Haber process, a mixture of H2 and N2 reacts to form ammonia.
3H2(g) + N2(g) ⇌ 2NH3(g) ΔH = −92 kJ mol−1
Source: 2023 CJC J2 Prelim P3 Q2(b)(iii)
ChallengeIn the Contact process, SO2 is oxidised to SO3 using a vanadium(V) oxide catalyst. SO2(g) + ½O2(g) ⇌ SO3(g) ΔS = −93.5 J K−1 mol−1
Source: EJC 2022 JC1 Chemical Equilibria Tutorial, A11
BThe addition of a catalyst lowers the activation energy of the forward reaction but not that of the reverse reaction.[1]
A catalyst speeds up the rate of a reaction by providing an alternative pathway with a lower activation energy (A is correct). A catalysed and an uncatalysed reaction have the same enthalpy change (C is correct).
The activation energies for both the forward and backward reactions are lowered by the same amount (B is incorrect). Hence, a catalyst does not change the position of equilibrium (D is correct). Instead, it shortens the time to reach equilibrium.
B8.27 × 10−5[1]
At equilibrium, [NH3] = [HNO3] = 0.01130.83 × 66.8100 = 0.00909 mol dm−3
Kc = [NH3][HNO3] = (0.00909)(0.00909) = 8.27 × 10−5 mol2 dm−6
C2 and 3 only[1]
B2P ⇌ 2Q + R[1]
Let a and b be the stoichiometric coefficients of P and Q: aP ⇌ bQ + R. Equilibrium amounts: P (2 − ax), Q bx, R x.
At equilibrium, total amount = 2 − ax + bx + x = 2 + x, so 2 − ax + bx = 2. Therefore a = b, i.e. P and Q have the same stoichiometric coefficients.
Kc = [CO2][H2][CO][H2O][1]
Note[H2O] must be included, as this is a homogeneous system: all the species are gases.
| CO(g) | H2O(g) | CO2(g) | H2(g) | |
|---|---|---|---|---|
| initial / mol | 0.200 | 0.200 | 0 | 0 |
| change / mol | −0.2 × 0.775 | −0.2 × 0.775 | +0.2 × 0.775 | +0.2 × 0.775 |
| equilibrium / mol | 0.045 | 0.045 | 0.155 | 0.155 |
Kc = (0.155)2(0.045)2[1]
= 11.86 ≈ 11.9[1]
−18.4 × 103 = −(8.31)(600) ln Kc
ln Kc = 3.690, so Kc = 40.04 ≈ 40.0[1]
NoteConvert ΔG to J mol−1 first.
The dimerisation will be favoured by high pressure. At high pressure, the position of equilibrium will shift to decrease pressure by decreasing the total number of gaseous molecules present. Since there are two molecules of gas on the left-hand side of the equation and one on the right, the forward reaction is favoured, and the position of equilibrium shifts to the right to form more N2O4.[1]
The dimerisation will be favoured by low temperature. At low temperature, the position of equilibrium will shift to increase temperature. Since the forward reaction is exothermic, the forward reaction is favoured, and the position of equilibrium shifts to the right to form more N2O4.[1]
Tube in iced water: pale yellow. Tube in hot water: brown.[1]
In iced water, the low temperature favours the exothermic forward reaction, so the position of equilibrium shifts to the right to form more colourless N2O4. In hot water, the high temperature favours the endothermic backward reaction, so the position of equilibrium shifts to the left to form more brown NO2.[1]
Equilibrium amounts / mol: H2O 4 − x, CO 4 − x, CO2 x, H2 x. Let the volume of the vessel be V dm3.
Kc = (x/V)2((4 − x)/V)2 = 9[1]
Solving, x = 3 mol. Hence, the equilibrium amount of CO = 1 mol[1]
(i) At a lower temperature, the rate at which equilibrium is attained is slower. By Le Chatelier's Principle, since ΔH < 0, the position of equilibrium shifts to the right and produces heat so as to increase the temperature. Hence, the amount of CO2 is higher at equilibrium. Kc is also increased.[1]
(ii) A decrease in volume, which leads to an increase in concentration, favours neither the backward nor the forward reaction, since the number of molecules on both sides of the reaction is the same. The amount of CO2 produced is the same. The rate at which equilibrium is attained increases.[1]
(iii) In the presence of a suitable catalyst, the rate at which equilibrium is attained is faster. However, there is no effect on the amount of CO2 formed, and Kc remains the same.[1]
Advantage: by Le Chatelier's principle, a lower temperature favours the forward exothermic reaction, and hence the yield of ammonia increases.[1]
Disadvantage: if the temperature is too low, the reaction is too slow.[1]
ΔG = −2.303RT log Kc, so log Kc = (−71.14 × 1000) ÷ (−2.303 × 8.31 × 298) = 12.47[1]
Kc = 1012.47 = 2.95 × 1012 mol−½ dm3/2[1]
NoteR is in J K−1 mol−1, so ΔG must be in J mol−1, not kJ mol−1.