Explain why the decomposition of calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), is non-spontaneous at 298 K but becomes spontaneous at temperatures above approximately 840 °C.

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 endothermic (ΔH>0), so the enthalpy change opposes spontaneity.
The entropy change is positive (ΔS>0) because a gas is produced from solid reactants, increasing disorder.
At 298 K the TΔS term is small, so ΔG=ΔH−TΔS remains positive and the reaction is non‑spontaneous.
At temperatures above ≈840 °C (≈1113 K) TΔS exceeds ΔH, giving ΔG<0 and the reaction becomes spontaneous; the crossover occurs when T=ΔH/ΔS.

Examiner tips

  • Use the ΔG=ΔH−TΔS equation explicitly. Mention the sign of ΔH and ΔS. Show the temperature at which ΔG changes sign. Keep the answer concise and to the point.

Common mistakes

  • Failing to state that ΔH>0 and ΔS>0. Confusing the sign of ΔG with spontaneity. Not mentioning the crossover temperature or the ΔH/ΔS ratio.

Mark scheme (4 marks)

  1. The reaction is endothermic (ΔH > 0), so the enthalpy change opposes spontaneity.
  2. Entropy increases (ΔS > 0) because a gas is produced from all-solid reactants, increasing the number/disorder of particles.
  3. At low temperatures, the TΔS term is small, so ΔG (= ΔH − TΔS) remains positive, meaning the reaction is non-spontaneous.
  4. At high temperatures, TΔS exceeds ΔH, making ΔG negative and the reaction spontaneous; the crossover occurs when T = ΔH/ΔS.

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