Explain why the reaction of nitrogen gas with oxygen gas to form nitrogen monoxide, N₂(g) + O₂(g) → 2NO(g), is non-spontaneous at standard conditions, yet becomes spontaneous at very high temperatures.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The standard enthalpy change for the reaction N₂(g) + O₂(g) → 2NO(g) is +180 kJ mol⁻¹. The standard entropy change for the same reaction is approximately +25 J K⁻¹ mol⁻¹.
Model answer (4 marks)
1. ΔG° = ΔH° – TΔS°.
2. ΔH° = +180 kJ mol⁻¹ is large and positive, so at 298 K ΔG° ≈ +180 kJ – 298 K×0.025 kJ K⁻¹ = +174 kJ, positive → non‑spontaneous.
3. ΔS° is +25 J K⁻¹ mol⁻¹ (≈+0.025 kJ K⁻¹ mol⁻¹) because two moles of gaseous reactants give two moles of gaseous products – a small increase in disorder.
4. As T rises, the TΔS term grows; when T > ΔH/ΔS (≈180 kJ ÷ 0.025 kJ K⁻¹ ≈ 7200 K) ΔG becomes negative, so the reaction becomes spontaneous at very high temperatures.
2. ΔH° = +180 kJ mol⁻¹ is large and positive, so at 298 K ΔG° ≈ +180 kJ – 298 K×0.025 kJ K⁻¹ = +174 kJ, positive → non‑spontaneous.
3. ΔS° is +25 J K⁻¹ mol⁻¹ (≈+0.025 kJ K⁻¹ mol⁻¹) because two moles of gaseous reactants give two moles of gaseous products – a small increase in disorder.
4. As T rises, the TΔS term grows; when T > ΔH/ΔS (≈180 kJ ÷ 0.025 kJ K⁻¹ ≈ 7200 K) ΔG becomes negative, so the reaction becomes spontaneous at very high temperatures.
Examiner tips
- Use the ΔG = ΔH – TΔS formula explicitly; show the sign of each term.
- State the numerical values and the sign of ΔG at 298 K.
- Explain how a positive ΔS makes the TΔS term favourable as T increases.
- Mention the crossover temperature where ΔG changes sign.
Common mistakes
- Saying the reaction is spontaneous at all temperatures; ignoring the large positive ΔH.
- Confusing ΔS sign or giving the wrong magnitude; forgetting to convert J to kJ.
- Not showing the calculation of ΔG at 298 K or the crossover temperature.
Mark scheme (4 marks)
- The reaction is non-spontaneous at standard conditions because ΔG is positive, arising from a large positive ΔH that outweighs the TΔS term at low temperatures.
- The entropy change is positive (small but positive) because two moles of reactant gases produce two moles of product gases — there is a slight increase in disorder/the number of accessible microstates.
- As temperature increases, the TΔS term increases in magnitude, because ΔS is positive; therefore ΔG = ΔH − TΔS decreases.
- Above a critical/crossover temperature (where T > ΔH/ΔS), ΔG becomes negative and the reaction is spontaneous.
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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