The big idea: Whether a reaction happens on its own — without being driven — depends on two things: the energy change (ΔH) and the disorder change (ΔS). The quantity that combines them is the Gibbs energy change, ΔG.
The rule is simple: a reaction is spontaneous (it is thermodynamically feasible — it can go forward on its own) when
ΔG < 0.
If ΔG > 0 the reaction is non-spontaneous in the forward direction, and if ΔG = 0 the system is at equilibrium.
What 'spontaneous' really means: Spontaneous does not mean fast. It only tells you the reaction is energetically allowed — whether it actually goes, and how quickly, is a question of kinetics (activation energy), not Gibbs energy.
The rusting of iron and the conversion of diamond to graphite are both spontaneous (ΔG < 0), yet one is slow and the other almost immeasurably slow.
Why two factors, not one: Many spontaneous reactions are exothermic (ΔH < 0) — releasing energy favours a reaction. But some endothermic reactions still happen, because an increase in disorder (ΔS > 0) can drive them.
Dissolving ammonium nitrate in water is endothermic (the beaker gets cold), yet it happens spontaneously because the ions spread out and disorder rises. ΔG captures both drivers at once.
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The Gibbs equation is given in the data booklet, so you never have to recall it — but you do have to use it carefully.
- Gibbs energy change (kJ mol⁻¹)
- enthalpy change (kJ mol⁻¹)
- absolute temperature (K)
- entropy change — convert to kJ K⁻¹ mol⁻¹
The single biggest mistake: ΔH is given in kJ mol⁻¹, but ΔS is given in J K⁻¹ mol⁻¹. The units do not match.
Before you put them into ΔG = ΔH − TΔS you must convert ΔS to kJ by dividing by 1000:
ΔS (kJ K⁻¹ mol⁻¹) = ΔS (J K⁻¹ mol⁻¹) ÷ 1000.
Forgetting this makes the TΔS term 1000× too big, which usually flips the sign of ΔG — and the wrong conclusion.
Also: T must be in kelvin (K = °C + 273).
Worked example — is the reaction spontaneous at 298 K?
For the reaction N2(g) + 3H2(g) → 2NH3(g), ΔH = −92.0 kJ mol⁻¹ and ΔS = −199 J K⁻¹ mol⁻¹. Determine whether the reaction is spontaneous at 298 K.
Solution
- Formula first — the given Gibbs equation:
- Fix the units — convert ΔS from J to kJ by dividing by 1000:
- Substitute (T = 298 K):
- Work out the TΔS term, then combine:
- Apply the rule: ΔG < 0, so the reaction is spontaneous at 298 K.
Final answer
ΔG = −32.7 kJ mol⁻¹. Because ΔG < 0, the reaction is spontaneous at 298 K.
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You can often predict spontaneity without calculating by looking at the signs of ΔH and ΔS. There are exactly four cases.
Read it straight off the equation: In ΔG = ΔH − TΔS, the entropy term carries a minus sign and is multiplied by T:
- A negative ΔH pushes ΔG down (favours spontaneity). - A positive ΔS pushes ΔG down too (because −TΔS becomes negative).
So when ΔH and ΔS pull in the same direction the answer is fixed; when they disagree, temperature decides, because it weights the ΔS term.
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneous? |
|---|---|---|---|
| negative (exothermic) | positive (more disorder) | always negative at all T | Yes — always spontaneous |
| positive (endothermic) | negative (less disorder) | always positive at all T | No — never spontaneous |
| negative (exothermic) | negative (less disorder) | negative only at low T | spontaneous at low temperature |
| positive (endothermic) | positive (more disorder) | negative only at high T | spontaneous at high temperature |
How to read the two temperature-dependent cases: ΔH < 0, ΔS < 0 (exothermic but more ordered): the −TΔS term is positive and grows with T. At low T it is small, so ΔG stays negative (spontaneous); at high T it overwhelms ΔH and ΔG turns positive.
ΔH > 0, ΔS > 0 (endothermic but more disorder): the −TΔS term is negative and grows with T. At low T the positive ΔH wins (non-spontaneous); at high T the −TΔS term wins and ΔG becomes negative (spontaneous).
Memory hook: Same signs on ΔH and ΔS ⇒ temperature decides. Opposite signs ⇒ the answer is the same at every temperature (ΔH<0, ΔS>0 always go; ΔH>0, ΔS<0 never go).
How this is tested: For the two temperature-dependent cases, IB Paper 2 asks: at what temperature does the reaction become spontaneous (or stop being spontaneous)?
At that switch-over point the reaction is exactly on the boundary — neither favoured nor disfavoured — so:
ΔG = 0.
Set ΔG = 0 in the given equation and rearrange for the crossover temperature T.
- crossover temperature (K)
- enthalpy change (use kJ mol⁻¹)
- entropy change (use kJ K⁻¹ mol⁻¹ for matching units)
Same units, again: ΔH and ΔS must be in matching units before you divide. The cleanest way is to put both in kJ (convert ΔS by ÷1000), so T comes out directly in kelvin. If you leave ΔH in kJ and ΔS in J the answer is 1000× wrong.
Worked example — find the crossover temperature
For the decomposition CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Calculate the temperature above which this reaction becomes spontaneous.
Solution
- At the crossover the reaction is on the boundary, so set ΔG = 0 — formula first:
- Fix the units — convert ΔS to kJ so it matches ΔH:
- Substitute and divide:
- Work it out — the unit is kelvin:
- Both ΔH and ΔS are positive, so the reaction is spontaneous above this temperature (the −TΔS term wins once T is large enough).
Final answer
T ≈ 1.11 × 10³ K (about 1106 K). The decomposition becomes spontaneous above this temperature.