AQA A-Level Chemistry Paper 3, 2019: Question 9

1 mark · Easy difficulty · Multiple Choice

Identify which change to the reaction mixture shifts the equilibrium to produce a higher concentration of sulfur trioxide.

Practise this question

Question

Question 09 shows the reversible reaction: 2 SO2(g) + O2(g) ⇌ 2 SO3(g) with ΔH = -188 kJ mol⁻¹. The question asks: 'Which change leads to a higher concentration of SO3 in this equilibrium mixture?' The multiple choice options are: A higher concentration of O2, B higher temperature, C lower pressure, D use of a catalyst.
Question text

09 Which change leads to a higher concentration of SO3 in this equilibrium mixture?

2 SO (g) + O (g) ⇌ 2 SO (g) ∆H = −188 kJ mol−1

22 3

[1 mark]

A higher concentration of O2

B higher temperature

C lower pressure

D use of a catalyst

Mark scheme

Show the mark scheme Mark scheme table showing question number 9 with the correct answer as 'A' and the mark value as 1.

9 A 1

How to answer it

Equilibrium Shifts & Le Chatelier's Principle

📌 What this question tests

This question assesses your ability to apply Le Chatelier’s Principle to predict how changes in conditions affect the position of dynamic equilibrium and the concentrations of products in a gaseous system.

  • Concentration: Adding a reactant shifts equilibrium towards the products.
  • Temperature: Effect of enthalpy change (ΔH) on equilibrium position.
  • Pressure: Relationship between gas volumes (moles of gas) and pressure changes.
  • Catalysts: Understanding their effect on rate vs position of equilibrium.
Question 09

Shifting the Equilibrium to Maximise [SO₃]

Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)  |  ΔH = -188 kJ mol⁻¹

✅ Correct Answer

A: higher concentration of O₂

Award 1 mark for selecting option A

🧠 Exam Technique: Systematic Elimination

For multiple choice equilibrium questions, test each variable against Le Chatelier's Principle (the system opposes any change imposed):

  • A: Add reactant → opposes change by consuming reactant → shifts Right.
  • B: Increase temp → opposes change by absorbing heat (endothermic / reverse) → shifts Left.
  • C: Lower pressure → opposes change by shifting towards more gas moles (3 vs 2) → shifts Left.
  • D: Catalyst → affects rate only → No shift.

📐 Option-by-Option Breakdown

  1. Option A (higher concentration of O₂):
    O₂ is a reactant. Increasing [O₂] causes the equilibrium to shift to the right to oppose the change by consuming the added O₂. This produces more SO₃, resulting in a higher equilibrium concentration of SO₃.
  2. Option B (higher temperature):
    The forward reaction is exothermic ( ΔH = -188 kJ mol⁻¹ ), meaning heat is released. Increasing the temperature shifts equilibrium in the endothermic direction (to the left) to absorb added heat. This decreases [SO₃].
  3. Option C (lower pressure):
    Count the moles of gas on each side:
    • Left-hand side: 2 + 1 = 3 mol of gas
    • Right-hand side: 2 mol of gas
    Decreasing pressure causes the equilibrium to shift to the side with more gas moles (the left) to increase the pressure. This decreases [SO₃].
  4. Option D (use of a catalyst):
    A catalyst increases the rate of both forward and reverse reactions equally by providing an alternative pathway with a lower activation energy. It allows equilibrium to be reached faster, but does not alter the position of equilibrium or concentrations.

💡 Key Knowledge: Le Chatelier's Rules

  • Increase [Reactant]: Shifts Right (makes more products).
  • Increase Temperature: Favours the endothermic reaction (shifts towards positive ΔH direction).
  • Decrease Temperature: Favours the exothermic reaction (shifts towards negative ΔH direction).
  • Increase Pressure: Favours side with fewer moles of gas.
  • Decrease Pressure: Favours side with more moles of gas.
  • Add Catalyst: No change in yield or equilibrium position.

❌ Common Student Pitfalls

  • Catalyst confusion: Believing a catalyst increases the yield of products because it speeds up the reaction. It only speeds up rate, not position of equilibrium!
  • Sign of ΔH: Forgetting that a negative ΔH ( -188 kJ mol⁻¹ ) denotes an exothermic forward reaction, so heat acts like a product.
  • Counting moles incorrectly: Overlooking that O₂(g) has a coefficient of 1, leading to an incorrect total of 3 moles of gas on the reactant side.

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.