AQA A-Level Chemistry Paper 3, 2017: Question 32

1 mark · Medium difficulty · Multiple Choice

Determine the conditions under which an exothermic reaction with a negative entropy change is feasible.

Practise this question

Question

Question 32 asks: 'A reaction is exothermic and has a negative entropy change. Which statement is correct?' Four multiple-choice options are given: A: The reaction is always feasible; B: The reaction is feasible above a certain temperature; C: The reaction is feasible below a certain temperature; D: The reaction is never feasible.
Question text

32 A reaction is exothermic and has a negative entropy change.

Which statement is correct?

[1 mark]

A The reaction is always feasible

B The reaction is feasible above a certain temperature

C The reaction is feasible below a certain temperature

D The reaction is never feasible

Mark scheme

Show the mark scheme Mark scheme table row for question 32 showing the correct answer is option C.

32 C

How to answer it

AQA A-Level Chemistry • Thermodynamics

Temperature Dependence of Reaction Feasibility

What this question tests

This question assesses your ability to apply the Gibbs free energy relationship ( ΔG = ΔH - TΔS ) to predict the feasibility of a reaction given the signs of enthalpy change ( ΔH ) and entropy change ( ΔS ), and to determine how temperature alters feasibility.

Question 32 • Multiple Choice [1 Mark]

Analysis & Evaluation

A reaction is exothermic and has a negative entropy change. Which statement is correct?

✅ Correct Answer: C

The reaction is feasible below a certain temperature

Mark Scheme: Award 1 mark for option C.

💡 Key Knowledge

  • Feasibility condition: A reaction is feasible when ΔG ≤ 0 .
  • Gibbs equation: ΔG = ΔH - TΔS .
  • Exothermic: ΔH < 0 (negative).
  • Entropy decrease: ΔS < 0 (negative).
  • Since ΔS is negative, the term -TΔS becomes positive.

📐 Step-by-Step Derivation

  1. Substitute the signs into the Gibbs equation:
    ΔG = (-ΔH) - T(-ΔS) = (-|ΔH|) + T|ΔS|
  2. Analyze the competing terms:
    • The ΔH term is favourable (negative, drives ΔG < 0 ).
    • The -TΔS term is unfavourable (positive, drives ΔG > 0 ).
  3. Examine low vs. high temperatures:
    • At low T: The value of TΔS is small, so the negative ΔH dominates. Therefore, ΔG < 0 (feasible).
    • At high T: The positive T|ΔS| term becomes very large and outweighs ΔH . Thus, ΔG > 0 (not feasible).
  4. Conclusion: Feasibility is maintained only when T < ΔH / ΔS , meaning it is feasible below a certain temperature.

❌ Common Errors & Distractor Breakdown

  • Option A (Always feasible): Only true if ΔH < 0 and ΔS > 0 , where ΔG is always negative at all temperatures.
  • Option B (Feasible above a certain temperature): The opposite case! This applies to endothermic reactions with a positive entropy change ( ΔH > 0, ΔS > 0 ).
  • Option D (Never feasible): Only true if ΔH > 0 and ΔS < 0 , where ΔG is always positive regardless of temperature.

🧠 Exam Technique & Examiner Insight

  • Use Extreme Values: When solving these in multiple-choice, test extremes:
    • Let T → 0 K : ΔG ≈ ΔH . Since ΔH is negative, ΔG is negative (feasible at low T).
    • Let T → ∞ : -TΔS dominates. Since ΔS is negative, -TΔS is massively positive (not feasible at high T).
  • This immediate mental test rules out A, B, and D in under 15 seconds.

Topics

Physical Chemistry · 3.1.8 Thermodynamics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.