OxfordAQA A Level Chemistry: Equilibria and Le Chatelier — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Equilibria and Le Chatelier revision notes

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Section A

1. Explain what is meant by a reversible reaction reaching a position of equilibrium. [2]

2. State the only factor that changes the value of an equilibrium constant. [1]

Section B

3. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ mol⁻¹:

(a) Write the expression for K_p. [2] (b) State and explain the effect on the equilibrium yield of increasing the temperature. [3] (c) State and explain the effect of increasing the total pressure. [3] (d) Explain why in the Contact process a pressure of only about 2 atm is used. [3]

4. At equilibrium a mixture contains 0.60 mol SO₂, 0.40 mol O₂ and 1.40 mol SO₃, at a total pressure of 200 kPa.

(a) Calculate the mole fraction and partial pressure of each gas. [4] (b) Calculate K_p, giving its units. [3]

5. Explain why adding an inert gas at constant volume does not shift the position of equilibrium. [3]

6. 0.60 mol of H₂ and 0.60 mol of I₂ are mixed in a 2 dm³ sealed container and allowed to reach equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium, 0.20 mol of I₂ remains.

(a) Use an ICE table to find the equilibrium amount, in moles, of each species. [3] (b) Calculate the equilibrium concentrations, and hence K_c, stating its units. [4]

7. Explain why increasing the pressure on the equilibrium in question 6 would have no effect on the position of equilibrium. [2]

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Answers

1. The forward and reverse reactions are proceeding at equal rates [1], so the concentrations of all species remain constant even though both reactions continue [1].

2. Temperature [1].

3. (a) K_p = p(SO₃)² ÷ (p(SO₂)² × p(O₂)) [1] [1]. (b) The yield decreases [1]. The forward reaction is exothermic [1], so the equilibrium shifts in the endothermic backward direction to oppose the temperature rise, and K_p falls [1]. (c) The yield increases [1]. There are 3 moles of gas on the left and 2 on the right [1], so the equilibrium shifts to the side with fewer gas molecules to reduce the pressure [1]. (d) The yield at 2 atm is already about 96%, so the extra yield from high pressure is very small [1]. High pressure requires thick-walled vessels and powerful compressors, which are expensive to build and run [1], so the small gain does not justify the cost [1].

4. (a) Total moles = 0.60 + 0.40 + 1.40 = 2.40 [1]. Mole fractions: SO₂ = 0.25, O₂ = 0.167, SO₃ = 0.583 [1]. Partial pressures: SO₂ = 50 kPa, O₂ = 33.3 kPa [1], SO₃ = 116.7 kPa [1]. (b) K_p = (116.7)² ÷ ((50)² × 33.3) [1] = 13 619 ÷ 83 250 = 0.164 [1]; units kPa⁻¹ [1].

5. The partial pressures of the reacting gases are unchanged, because the volume and the number of moles of each are unchanged [1]. Only the total pressure rises [1]. Since K_p depends on the partial pressures of the reacting species alone, the reaction quotient is still equal to K_p, so there is no shift [1].

6. (a) ICE table (moles): H₂ 0.60 → −x → 0.20 [1], so x = 0.40; I₂ 0.60 → −0.40 → 0.20 mol [1]; HI 0 → +2(0.40) → 0.80 mol [1]. (b) [H₂] = 0.20 ÷ 2 = 0.10 mol dm⁻³, [I₂] = 0.10 mol dm⁻³, [HI] = 0.80 ÷ 2 = 0.40 mol dm⁻³ [1]. K_c = [HI]² ÷ ([H₂][I₂]) = 0.40² ÷ (0.10 × 0.10) [1] = 0.16 ÷ 0.01 = 16 [1]; since there are 2 moles of gas on each side, the concentration units cancel and K_c has no units [1].

7. There are 2 moles of gas on the reactant side and 2 moles of gas on the product side [1], so increasing the pressure favours neither side, and the position of equilibrium is unchanged [1].

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Where marks are usually lost

• Saying K changes with pressure or with a catalyst.
• Forgetting to square the partial pressure of SO₃ and SO₂.
• Not deducing the units of K_p from the expression.
• Explaining the low Contact-process pressure by yield alone rather than by cost.
• Converting moles to concentration incorrectly in an ICE table calculation, or forgetting to divide by the volume before substituting into K_c.
• Assuming K_c always has units — when the total moles of gas are equal on both sides, as in question 6, the units cancel and K_c is dimensionless.

Using an ICE table — the method in one place

An ICE (Initial, Change, Equilibrium) table is the reliable way to turn a word problem into a K_c calculation: write the initial moles of every species, let the change in one species be x (using the reaction’s stoichiometric ratios to relate the changes in the others), then use whatever equilibrium information the question gives — as in question 6, where the equilibrium amount of I₂ is stated directly — to solve for x and complete the row. Only once every species’ equilibrium moles are known should you convert to concentration (dividing by the container’s volume) and substitute into the K_c expression; substituting moles directly, skipping the conversion, is one of the most common errors in this type of question. For the full treatment of K_p and K_c, together with the industrial compromise conditions for the Haber and Contact processes, see the Equilibria and Le Chatelier revision notes.