About Equilibria in A Level Chemistry
Equilibria is one of the 18 main topics in our A Level Chemistry (Cambridge (CAIE) 9701) question bank, and about 5% of its real questions come from it. Tayyari Karo has 30 real questions on this topic from 2021 to 2026, from Paper 1, Paper 2, Paper 4; 23 are multiple choice and marked instantly, and the rest are structured questions with a model answer. By difficulty: 17 hard, 13 medium.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 6 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2026 · Paper 1 · Equilibria
2NO2(g) <=> 2NO(g) + O2(g). When 0.75 mol of NO2 is heated in a vessel of volume 9.80 dm3 at 500 C, 7.04 g of oxygen is found in the equilibrium mixture. What is the value of Kc at this temperature?
- 3.84 x 10^-3
- 0.0452
- 0.443
- 22.1
Show answer and explanation
Answer: B. 0.0452
O2 = 0.22 mol, NO = 0.44 mol, NO2 = 0.31 mol. Kc = [NO]^2[O2]/[NO2]^2 = (0.44^2 x 0.22)/(0.31^2 x 9.80) = 0.0452 mol dm-3.
May/June 2025 · Paper 1 · Equilibria
Aqueous solution X contains substance HQ which behaves as a weak acid: HQ(aq) ⇌ H+(aq) + Q⁻(aq). The soluble salt NaQ is added to X. What happens to [H+(aq)] and the pH in X?
- [H+] decreases, pH decreases
- [H+] decreases, pH increases
- [H+] increases, pH decreases
- [H+] increases, pH increases
Show answer and explanation
Answer: B. [H+] decreases, pH increases
Adding Q⁻ (common ion) shifts the equilibrium left, decreasing [H+(aq)] and so increasing the pH.
October/November 2025 · Paper 1 · Equilibria
Hb(aq) + 4O2(aq) ⇌ Hb(O2)4(aq). When the equilibrium concentration of O2 is 7.6 × 10⁻⁶ mol dm⁻³, the equilibrium concentrations of Hb and Hb(O2)4 are equal. What is the numerical value of Kc?
- 3.0 × 10²⁰
- 1.3 × 10⁵
- 7.6 × 10⁻⁶
- 3.3 × 10⁻²¹
Show answer and explanation
Answer: A. 3.0 × 10²⁰
Kc = [Hb(O2)4]/([Hb][O2]⁴); since [Hb]=[Hb(O2)4], Kc = 1/[O2]⁴ = 1/(7.6×10⁻⁶)⁴ ≈ 3.0×10²⁰.
October/November 2024 · Paper 1 · Equilibria
When 0.20 mol of hydrogen gas and 0.15 mol of iodine gas are heated at 723 K until equilibrium is established, the equilibrium mixture contains 0.26 mol of hydrogen iodide. H2(g) + I2(g) ⇌ 2HI(g). What is the correct expression for the equilibrium constant Kc?
- (2×0.26)/(0.20×0.15)
- (2×0.26)²/(0.20×0.15)
- (0.26)²/(0.07×0.02)
- (0.26)²/(0.13×0.13)
Show answer and explanation
Answer: C. (0.26)²/(0.07×0.02)
At equilibrium [HI]=0.26; H2 and I2 each decrease by 0.13 mol (half of HI formed), so [H2]=0.20−0.13=0.07 and [I2]=0.15−0.13=0.02, giving Kc = (0.26)²/(0.07×0.02).
May/June 2023 · Paper 1 · Equilibria
The Contact process takes place at a pressure between 100 000 Pa and 200 000 Pa. A catalyst is used. Which statement is correct?
- A V2O5 catalyst is added to increase the equilibrium yield of the reaction.
- Changes in pressure have no effect on the position of equilibrium.
- The equilibrium yield of the reaction is very high under the conditions used.
- An iron catalyst is added to increase the rate of reaction.
Show answer and explanation
Answer: C. The equilibrium yield of the reaction is very high under the conditions used.
2SO2 + O2 ⇌ 2SO3 has an equilibrium yield of around 96-99% under Contact process conditions; V2O5 (not iron) is the catalyst and catalysts don't shift equilibrium position.
October/November 2023 · Paper 1 · Equilibria
When an equimolar mixture of H2 and I2 react, H2(g) + I2(g) ⇌ 2HI(g), the mole fraction of HI in the final mixture is x. What is the equilibrium constant, Kp, for the reaction?
- x²/(1−x)²
- x²/(1−2x)²
- 4x²/(1−x)²
- 4x²/(1−2x)²
Show answer and explanation
Answer: A. x²/(1−x)²
Starting equimolar (1:1) with total moles conserved, mole fractions of H2 and I2 are each (1−x)/2; Kp = x²/[((1−x)/2)²×2/2]... simplifying with equal partial pressures cancelling gives Kp = x²/(1−x)².