Calculate ΔG, explain the role of temperature and distinguish a favourable energy change from a rapid reaction.
Combine enthalpy and entropy in compatible units
At constant temperature and pressure, ΔG = ΔH − TΔS. A negative ΔG indicates a thermodynamically favourable forward change under the conditions to which the value applies. A positive value favours the reverse direction; ΔG = 0 describes the boundary for that process.
T is absolute temperature in kelvin. If ΔH is in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹, divide ΔS by 1000 before multiplying by T. TΔS then has units kJ mol⁻¹, allowing subtraction from ΔH. Converting after an inconsistent subtraction cannot rescue the calculation.
Standard ΔG° uses standard-state quantities. Its sign is not a promise of complete conversion at every composition; real reaction mixtures have composition-dependent driving forces. Within OCR calculations, state the conditions and use the supplied standard-data model rather than making an absolute claim that products must form instantly.
Worked example: explain the temperature threshold
For an original illustrative reaction, ΔH = +48.0 kJ mol⁻¹ and ΔS = +120 J K⁻¹ mol⁻¹ = +0.120 kJ K⁻¹ mol⁻¹. At 298 K, ΔG = 48.0 −298(0.120) = +12.24 kJ mol⁻¹, so it is not favourable in the forward direction under this data model.
Set ΔG = 0: T = ΔH/ΔS = 48.0/0.120 = 400 K. Above 400 K the favourable entropy contribution outweighs the endothermic cost; at 450 K, ΔG = −6.0 kJ mol⁻¹. The threshold itself is the boundary, not a strictly negative-ΔG condition.
This estimate assumes ΔH and ΔS remain approximately constant across the range and that no phase change changes the process. State this limitation when interpreting a long extrapolation. On a plot of ΔG against T, the intercept is ΔH and the gradient is −ΔS.
Derive each sign combination
A negative ΔH favours the forward process. A positive ΔS makes −TΔS negative and increasingly favourable as T rises. Use those two contributions to derive the four cases instead of memorising disconnected labels.
If both ΔH and ΔS are negative, heating makes the positive −TΔS term larger, eventually outweighing the favourable enthalpy. For example ΔH = −60.0 kJ mol⁻¹ and ΔS = −0.150 kJ K⁻¹ mol⁻¹ give ΔG < 0 below 400 K, not above it. The same positive ratio ΔH/ΔS can therefore describe an upper or lower temperature boundary; inspect the signs.
| ΔH | ΔS | When ΔG is negative |
|---|---|---|
| Negative | Positive | At all positive temperatures in this model |
| Positive | Negative | At no positive temperature in this model |
| Positive | Positive | Above the threshold |
| Negative | Negative | Below the threshold |
Feasible does not mean fast
Thermodynamics compares the initial and final states. Rate depends on the pathway and its activation barrier. A process with negative ΔG can be extremely slow if few particles cross a large barrier; a catalyst provides a faster pathway without changing ΔH, ΔS or the equilibrium position for the same reaction.
A strong explanation separates the two questions: is there a favourable thermodynamic change under these conditions, and is there an accessible kinetic pathway on the timescale observed? Heating can improve the rate yet make the equilibrium yield worse for an exothermic process. Industrial decisions often require that distinction.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.
Q1. Calculate ΔG at 300 K if ΔH = −35.0 kJ mol⁻¹ and ΔS = −80.0 J K⁻¹ mol⁻¹.Show answer
Convert ΔS to −0.0800 kJ K⁻¹ mol⁻¹. ΔG = −35.0 −300(−0.0800) = −11.0 kJ mol⁻¹. The forward change is favourable within the stated model.
Q2. Find the threshold for those data and decide which side is favourable.Show answer
T = (−35.0)/(−0.0800) = 437.5 K. Because ΔS is negative, ΔG increases as T rises, so the forward direction is favourable below this threshold.
Q3. A plot of ΔG in kJ mol⁻¹ against T in K has gradient −0.0850. Find ΔS.Show answer
Gradient = −ΔS, so ΔS = +0.0850 kJ K⁻¹ mol⁻¹ = +85.0 J K⁻¹ mol⁻¹. The conversion factor applies to the entropy unit.
Q4. Why can a reaction with ΔG < 0 remain unobservably slow?Show answer
A high activation energy can make the fraction of effective collisions very small. Negative ΔG concerns the energy balance, not the speed of the pathway.
Q5. Does a catalyst lower the Gibbs energy of the products?Show answer
No. It changes the pathway and activation barrier. The initial and final state energies, the reaction ΔG under the same conditions and the equilibrium position remain unchanged.
Sources
Sources and examiner guidance (reviewed 6 October 2026)
- OCR A H432 specification — version 3.1 — 5.2.2, printed pp. 48; outcomes and additional guidance, with relevant Module 1 practical skills.
- Chemrevise — OCR A 5.2.2 — Pages 3–6; secondary coverage cross-check. Lesson explanations, data and questions are original Finesse material.
- OCR H432/01 mark scheme — June 2025 — Q17(a–b); printed pp. 16–17. Question-specific evidence, not universal marking rules.
- OCR H432/01 examiner report — June 2025 — Q17(a–b); printed pp. 26–29. Read with the corresponding question context.
- OCR H432/01 question paper — June 2025 — Q17(a–b); context for the assessment references, not reproduced questions.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.