Topic 5 of 5
Gibbs energy, temperature and limits
Decide whether a process is favourable under the stated conditions.
A-Level 9476 (2026-2027)
Free energy combines the enthalpy and entropy contributions
Use Kelvin and match the energy units before subtracting.
ΔG° = ΔH° - TΔS°. T is absolute temperature in K. If ΔH is in kJ mol-1 and ΔS in J mol-1 K-1, divide the entropy by 1000 first. At constant temperature and pressure, a negative actual ΔG favours the forward process; zero corresponds to equilibrium; a positive value favours the reverse process.
The standard value ΔG° assesses the driving tendency from standard-state conditions. A negative standard value is often described as thermodynamically feasible or spontaneous under standard conditions. It does not promise that all reactant turns into product, that the change is fast, or that the same direction remains favoured at every composition.
Worked example
An endothermic process can be favourable
For a process, ΔH° = +36.0 kJ mol-1 and ΔS° = +120 J mol-1 K-1. Calculate ΔG° at 298 K and 350 K, treating these values as temperature-independent.
- Convert entropy to +0.120 kJ mol-1 K-1.
- At 298 K: delta G = 36.0 - 298(0.120) = +0.24 kJ mol-1.
- At 350 K: delta G = 36.0 - 350(0.120) = -6.00 kJ mol-1.
Slightly unfavourable at 298 K and favourable at 350 K under standard conditions. The larger entropy contribution at higher temperature outweighs the positive enthalpy change.
Spontaneous does not mean rapid, complete or condition-independent
Use a sign table for temperature trends, then state what the calculation cannot decide.
| ΔH | ΔS | When ΔG is negative |
|---|---|---|
| Negative | Positive | At all temperatures within the range where these signs and phases apply. |
| Positive | Negative | At no positive temperature under this approximation. |
| Negative | Negative | At sufficiently low temperature; the positive -T delta S term grows on heating. |
| Positive | Positive | At sufficiently high temperature; the negative -T delta S term becomes dominant. |
If ΔH and ΔS have the same sign, solve T = ΔH/ΔS for the crossover where ΔG = 0, using consistent units. For the +36.0 kJ mol-1, +120 J mol-1 K-1 example, T = 300 K. The direction of the inequality depends on the signs: positive/positive is favourable above the crossover; negative/negative below it.
| Limit | What to say |
|---|---|
| Kinetics | A large activation barrier can make a favourable reaction immeasurably slow; a catalyst changes the rate, not the endpoint free-energy difference. |
| Non-standard composition | Actual concentrations and partial pressures affect the driving tendency. Standard delta G is not the actual delta G at every composition. |
| Extent | A favourable forward tendency does not mean 100% conversion; equilibrium composition matters. |
| Temperature range and phase | Enthalpy and entropy vary with temperature, and a phase change can invalidate a simple extrapolation. |