OCR A-Level Chemistry AS Breadth in chemistry (01), June 2018: Question 13

1 mark · Medium difficulty · Multiple Choice

Deduce the changes in pressure and temperature required to shift the equilibrium position towards the products for the given reversible reaction.

Practise this question

Question

Multiple-choice question 13 showing a reversible reaction: 2SO2(g) + O2(g) = 2SO3(g) with delta H = -197 kJ mol^-1. A table lists four options (A, B, C, D) with various combinations of pressure and temperature changes (Decrease/Increase), and an answer box at the bottom right worth 1 mark.
Question text

13 The reversible reaction below is at equilibrium.

2SO (g) + O (g) 2SO (g) ∆H = −197 kJ mol−1

22 3

Which changes in pressure and temperature would shift the equilibrium position towards the

products?

Pressure Temperature

A Decrease Decrease

B Decrease Increase

C Increase Decrease

D Increase Increase

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 13 indicating the correct answer is C, worth 1 mark.

13 C 1

How to answer it

Shifting Equilibrium Position (Pressure & Temperature)

OCR AS Level Chemistry

What this question tests

This question assesses your understanding of Le Chatelier's Principle. Specifically, you need to predict how changes in external conditions (pressure and temperature) affect the position of a dynamic equilibrium in a gas-phase reaction, taking into account stoichiometric gas moles and the enthalpy change (ΔH).

Question 13 Analysis

Evaluating Pressure and Temperature Effects on 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

✅ Correct Answer: C

Pressure: Increase

Temperature: Decrease

Awarded 1 mark for selecting option C.

💡 Key Knowledge

  • Le Chatelier's Principle: If a factor is changed, the position of equilibrium shifts to counteract that change.
  • Effect of Pressure: Increasing pressure shifts equilibrium towards the side with fewer moles of gas.
  • Effect of Temperature: A negative ΔH means the forward reaction is exothermic. Decreasing temperature shifts equilibrium in the exothermic direction to release heat.

🧠 Exam Technique

  • Step 1 (Pressure): Count the balancing numbers for gas molecules on both sides. Left = 2 + 1 = 3 moles. Right = 2 moles. To favour products (fewer moles), you must increase the pressure.
  • Step 2 (Temperature): Look at the sign of ΔH (−197 kJ mol⁻¹). This tells you the forward reaction is exothermic. To promote the exothermic reaction, you must decrease the temperature.
  • Step 3: Match your two conclusions (Increase pressure, Decrease temperature) to the multiple-choice options to find option C.

❌ Common Errors

  • Confusing endothermic and exothermic directions based on the sign of ΔH.
  • Miscounting total moles of gaseous reactants (forgetting to sum coefficients, e.g., missing the unwritten '1' in front of O₂).
  • Assuming a change in pressure alters the value of the equilibrium constant (Kc) rather than just shifting the position of equilibrium.

Topics

Module 3: Periodic table and energy · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.