Explain why the reaction 2SO₂(g) + O₂(g) → 2SO₃(g) is spontaneous at low temperatures but becomes non-spontaneous at high temperatures.

IB DP Chemistry Higher Level (2023 syllabus) — R1.4 Entropy and spontaneity (HL only) · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

The reaction is exothermic (ΔH < 0), giving a negative contribution to ΔG and favouring spontaneity.
The entropy change ΔS is negative because 3 mol of gas are converted to 2 mol of gas, reducing disorder.
Using ΔG = ΔH − TΔS, at low temperatures the −TΔS term is small, so the negative ΔH dominates and ΔG < 0 – the reaction is spontaneous.
At high temperatures the −TΔS term (which is positive, since ΔS < 0) becomes large enough to make ΔG > 0, so the reaction is no longer spontaneous.

Examiner tips

  • Show the sign of ΔH and ΔS clearly; use ΔG = ΔH − TΔS to explain the temperature dependence.
  • Include the reasoning that a negative ΔS makes the TΔS term positive, which opposes the negative ΔH at high T.

Common mistakes

  • Confusing the sign of ΔS or forgetting that ΔS is negative for this reaction.
  • Failing to mention that the −TΔS term becomes positive and dominates at high temperatures.
  • Using the wrong equation (e.g., ΔG = ΔH + TΔS) or mis‑labeling the contribution of each term.

Mark scheme (4 marks)

  1. The reaction is exothermic (ΔH is negative), which gives a negative contribution to ΔG and favours spontaneity.
  2. The entropy change ΔS is negative because three moles of gas are converted to two moles of gas, reducing disorder.
  3. Using ΔG = ΔH − TΔS: at low temperatures the −TΔS term is small, so the negative ΔH dominates, giving ΔG < 0 (spontaneous).
  4. At high temperatures the −TΔS term (which is positive, since ΔS is negative) becomes large enough to make ΔG positive, so the reaction is no longer spontaneous.

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