Lesson 5 of 5

Gibbs energy, temperature and limits

Decide whether a process is favourable under the stated conditions.

A-Level 9476 (2026-2027)

I can use the standard Gibbs-energy equation.Syllabus
Syllabus 9476, 7(j). Use the standard Gibbs-energy equation.
I can infer spontaneity from the sign of Gibbs energy.Syllabus
Syllabus 9476, 7(k). Infer spontaneity from the sign of Gibbs energy.
I can explain limits of standard Gibbs-energy predictions.Syllabus
Syllabus 9476, 7(l). Explain limits of standard Gibbs-energy predictions.
I can predict how temperature changes spontaneity.Syllabus
Syllabus 9476, 7(m). Predict how temperature changes spontaneity.

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 exampleAn 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.

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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.

Answer

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.

Try it step by step

Work it out in steps: delta S in kJ mol^-1 K^-1, then T delta S, then delta G = delta H - T delta S.

Problem 1. For N₂(g) + 3H₂(g) -> 2NH₃(g), delta H = -92.0 kJ mol^-1 and delta S = -199 J mol^-1 K^-1. Find delta G at 298 K in kJ mol^-1 to 3 significant figures.
Problem 2. For CaCO₃(s) -> CaO(s) + CO₂(g), delta H = +178 kJ mol^-1 and delta S = +161 J mol^-1 K^-1. Find delta G at 298 K in kJ mol^-1 to 3 significant figures.
Problem 3. For H₂O(l) -> H₂O(g) at its boiling point, delta H = +40.7 kJ mol^-1 and delta S = +109 J mol^-1 K^-1. Find delta G at 373 K. Give delta S to 3 significant figures, and T delta S and delta G to 1 decimal place.

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Problem 1.

delta S in kJ mol^-1 K^-1: delta S = -199 / 1000 = -0.199 kJ mol^-1 K^-1. Convert J to kJ before mixing it with delta H.

T delta S: T delta S = 298 x (-0.199) = -59.3 kJ mol^-1.

delta G = delta H - T delta S: delta G = -92.0 - (-59.3) = -32.7 kJ mol^-1. Negative, so the reaction is feasible at 298 K.

Problem 2.

delta S in kJ mol^-1 K^-1: delta S = +0.161 kJ mol^-1 K^-1.

T delta S: T delta S = 298 x 0.161 = +48.0 kJ mol^-1.

delta G = delta H - T delta S: delta G = 178 - 48.0 = +130 kJ mol^-1. Positive, so the decomposition is not feasible at 298 K.

Problem 3.

delta S in kJ mol^-1 K^-1: delta S = +0.109 kJ mol^-1 K^-1.

T delta S: T delta S = 373 x 0.109 = +40.7 kJ mol^-1.

delta G = delta H - T delta S: delta G = 40.7 - 40.7 = 0.0 kJ mol^-1. At the boiling point, liquid and vapour are in equilibrium, so delta G is zero.

Spontaneous does not mean rapid, complete or condition-independent

Use a sign table for temperature trends, then state what the calculation cannot decide.

Make a guess. You are not marked.

A reaction has a negative delta H and a negative delta S. What does raising the temperature do to how feasible it is?

No change, because delta H is negative
It becomes less feasible, as -T delta S grows
It becomes more feasible, as particles have more energy

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It becomes less feasible, as -T delta S grows

With delta S negative, the term -T delta S is positive and grows with T. At a high enough temperature, delta G turns positive.

Predictions when enthalpy and entropy are approximately constant
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.

Limits of a standard Gibbs-energy prediction
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.

Check your understanding

A conversion has a negative delta G, yet no change is seen over a whole lesson. Is the free-energy calculation wrong?

Yes. A negative delta G means the reaction is spontaneous, so it should be seen happening at once.
Not necessarily. delta G depends on the amount used, so a small sample shows no change.
Not necessarily. A high activation energy can make a feasible reaction far too slow to notice.

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Not necessarily. A high activation energy can make a feasible reaction far too slow to notice.

A negative delta G says the reaction is feasible, not how fast it goes. A large kinetic barrier can make the rate tiny, as with diamond turning to graphite. A catalyst or other conditions can change the rate.