Explain why the reaction CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) is spontaneous at all temperatures, using both entropy and enthalpy considerations.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
1. The reaction is highly exothermic (ΔH°≈−802 kJ mol⁻¹), so the enthalpy term favours spontaneity.
2. The number of moles of gas is unchanged (3 → 3), so ΔS_sys≈0, but the large negative ΔH gives a large positive ΔS_surr (ΔS_surr=−ΔH/T).
3. ΔG=ΔH−TΔS_total; ΔH is strongly negative and ΔS_total (system + surroundings) is positive, giving ΔG<0.
4. Because ΔH is negative and ΔS_total is positive, the −TΔS term can never outweigh the negative ΔH, so ΔG remains negative at all temperatures, making the reaction spontaneous everywhere.
2. The number of moles of gas is unchanged (3 → 3), so ΔS_sys≈0, but the large negative ΔH gives a large positive ΔS_surr (ΔS_surr=−ΔH/T).
3. ΔG=ΔH−TΔS_total; ΔH is strongly negative and ΔS_total (system + surroundings) is positive, giving ΔG<0.
4. Because ΔH is negative and ΔS_total is positive, the −TΔS term can never outweigh the negative ΔH, so ΔG remains negative at all temperatures, making the reaction spontaneous everywhere.
Examiner tips
- Use the formula ΔG=ΔH−TΔS and explain the sign of each term; mention ΔS_surr explicitly.
- Show that ΔS_sys≈0 but ΔS_surr>0 due to exothermicity; this is key to the argument.
- Keep the answer concise – 4 marks, so one sentence per point is enough.
- Use correct units and symbols (ΔH, ΔS, ΔG).
Common mistakes
- Confusing ΔS_sys with ΔS_total; forgetting ΔS_surr.
- Claiming ΔS_sys>0 without justification.
- Using the wrong sign for ΔH or ΔS in ΔG expression.
Mark scheme (4 marks)
- The reaction is highly exothermic (ΔH° is large and negative / releases energy to surroundings), so the enthalpy term favours spontaneity.
- The entropy change of the system is positive (ΔS > 0) because the number of moles of gas increases from 3 to 3 — wait — correcting: moles of gas go from 3 mol gas reactants to 3 mol gas products, so ΔS_sys ≈ 0; however, the large negative ΔH means the entropy of the surroundings (ΔS_surr = −ΔH/T) is large and positive, favouring spontaneity.
- ΔG = ΔH − TΔS; because ΔH is strongly negative and ΔS_total (system + surroundings) is positive, ΔG is negative at all temperatures.
- Because ΔH is negative and ΔS_system is approximately zero (or slightly positive), the −TΔS term does not become large enough to make ΔG positive even at very high temperatures, so spontaneity is maintained across all temperatures.
Related
- All IB DP Chemistry Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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