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Gibbs free energy and spontaneity
Spontaneous and non-spontaneous reactions, combining enthalpy and entropy in ΔG° = ΔH° − TΔS°, the four combinations of ΔH and ΔS, calculating ΔG° from standard data, the effect of temperature on spontaneity, and the crossover temperature.
Limestone sits in cliffs for millions of years without falling apart into lime and carbon dioxide. Put the same rock in a kiln at 900 °C and it decomposes steadily, as it has in lime kilns along the Australian coast since the first colonial settlements, where shells were burned to make mortar. Nothing about the reaction has changed except the temperature. This page explains exactly why heat flips the answer, and lets you predict the temperature at which it happens.
Spontaneous reactions
A spontaneous reaction is one that, once started, continues on its own without a continuous supply of energy from outside. Iron rusting, petrol burning, NH₄NO₃ dissolving and ice melting at room temperature are spontaneous. Photosynthesis is not: it stops the moment the light goes off. Electrolysing water is not: it stops when the power is switched off.
Two warnings about the word:
- Spontaneous does not mean fast. Diamond turning into graphite is spontaneous at room temperature, but its activation energy is so high that it will never happen in a human lifetime. Thermodynamics tells you whether a reaction can go; activation energy (4D) tells you how fast.
- Spontaneous does not mean it starts without help. Petrol needs a spark to overcome Eₐ, but once alight it keeps going.
Gibbs free energy
Whether a reaction is spontaneous depends on both drivers together: a fall in enthalpy (ΔH negative) and a rise in entropy (ΔS positive). Josiah Willard Gibbs combined them in one quantity, the Gibbs free energy change:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
- $\Delta G \lt 0$: the reaction is spontaneous in the forward direction.
- $\Delta G \gt 0$: the forward reaction is not spontaneous (the reverse reaction is).
- $\Delta G = 0$: the system is at equilibrium; neither direction is favoured.
$T$ is the temperature in kelvin. Because $\Delta S$ is in J K⁻¹ mol⁻¹ and $\Delta H$ is in kJ mol⁻¹, convert $\Delta S$ to kJ (divide by 1000) before substituting. The $T\Delta S$ term is the entropy driver expressed in energy units; multiplying by $T$ means entropy matters more and more as the temperature rises.
For NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq), ΔH° = +25.7 kJ mol⁻¹ and ΔS° = +108.7 J K⁻¹ mol⁻¹. Is dissolving spontaneous at 25 °C?
The four combinations
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneous? | Example |
|---|---|---|---|---|
| − | + | Always negative | At every temperature | Burning fuels, 2H₂O₂ → 2H₂O + O₂ |
| + | − | Always positive | Never | Photosynthesis (needs light) |
| − | − | Negative at low T | Below the crossover temperature | Freezing, Haber process, condensing |
| + | + | Negative at high T | Above the crossover temperature | Melting, boiling, CaCO₃ → CaO + CO₂ |
Temperature and the crossover point
When ΔH and ΔS have the same sign, the two drivers disagree, and temperature decides the winner. The switch happens when $\Delta G = 0$:
$$0 = \Delta H - T\Delta S \quad\Rightarrow\quad T = \frac{\Delta H}{\Delta S}$$
CaCO₃(s) → CaO(s) + CO₂(g). ΔH° = +178.3 kJ mol⁻¹, ΔS° = +160.6 J K⁻¹ mol⁻¹ (calculated in 4G and 4H). (a) Is the reaction spontaneous at 25 °C? (b) Above what temperature does it become spontaneous?
Use ΔfH° and S° data to decide whether 2SO₂(g) + O₂(g) → 2SO₃(g) is spontaneous at 25 °C and at 1200 K. Data: ΔfH° (kJ mol⁻¹) SO₂ −296.8, SO₃ −395.7; S° (J K⁻¹ mol⁻¹) SO₂ 248.2, O₂ 205.1, SO₃ 256.8.
- Choose Boiling water, work out ΔH°, ΔS° and the crossover yourself, then press Show all. Why is the crossover (370 K) so close to 373 K?
- Choose Smelting iron. At what temperature can carbon reduce iron(III) oxide? (Answer: 907 K.) Why do blast furnaces run much hotter than that?
- Choose N₂O₄ → 2NO₂ and move the temperature from 298 K to 400 K. When does ΔG change sign?
- Choose Methane burns. Its crossover is about 3660 K. Why do we call it spontaneous at all practical temperatures?
Driving non-spontaneous reactions
A reaction with a positive ΔG can still be made to happen, but only by coupling it to a continuous energy supply. Plants drive photosynthesis (ΔG about +2880 kJ per mole of glucose) with sunlight. Aluminium smelters drive the decomposition of Al₂O₃ with enormous electric currents, which is why smelters such as Tomago near Newcastle use about a tenth of NSW's electricity. Stop the energy supply and the reaction stops.
Test yourself
A reaction has ΔH° = +44 kJ mol⁻¹ and ΔS° = +119 J K⁻¹ mol⁻¹. It is spontaneous
T = 44 ÷ 0.119 = 370 K. With both positive, the TΔS term wins at high temperature.
For N₂O₄(g) → 2NO₂(g), ΔH° = +57.2 kJ mol⁻¹ and ΔS° = +175.9 J K⁻¹ mol⁻¹. ΔG° at 298 K is
57.2 − 298 × 0.1759 = 57.2 − 52.4 = +4.8 kJ mol⁻¹, slightly non-spontaneous; it becomes spontaneous above 325 K.
Diamond turning into graphite has ΔG° = −2.9 kJ mol⁻¹ at 298 K, yet diamonds last indefinitely. This is because
Spontaneous means favoured, not fast. The carbon–carbon bonds in diamond must break to rearrange, which needs a huge Eₐ.
Which pair of values gives a reaction that is spontaneous only at low temperatures?
Enthalpy favours it, entropy opposes it; the opposing −TΔS term grows with T, so it wins at high T.
Match each line to its spontaneity.
Classify each reaction from its ΔH and ΔS.
In ΔG = ΔH − TΔS, T must be in and ΔS must be converted to K⁻¹ mol⁻¹. A reaction is spontaneous when ΔG is . When ΔH and ΔS are both positive, the reaction is spontaneous at temperatures. At equilibrium ΔG equals .
Iron is extracted from haematite in a blast furnace. One possible reaction is Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g).
Data: ΔfH° (kJ mol⁻¹) Fe₂O₃ −824.2, CO −110.5. S° (J K⁻¹ mol⁻¹) Fe₂O₃ 87.4, C 5.7, Fe 27.3, CO 197.7.
(a) Calculate ΔH° and ΔS° for the reaction. (3 marks)
(b) Calculate the minimum temperature at which the reaction is spontaneous. (2 marks)
(c) Explain, in terms of enthalpy and entropy, why the reaction is not spontaneous at room temperature but is spontaneous in the furnace. (2 marks)
(a) ΔH° = [2(0) + 3(−110.5)] − [(−824.2) + 3(0)] = −331.5 + 824.2 = +492.7 kJ mol⁻¹.
ΔS° = [2(27.3) + 3(197.7)] − [87.4 + 3(5.7)] = 647.7 − 104.5 = +543.2 J K⁻¹ mol⁻¹.
(b) At the minimum temperature ΔG = 0, so T = ΔH/ΔS = 492.7 ÷ 0.5432 = 907 K (about 634 °C).
(c) The reaction is strongly endothermic, so the enthalpy driver opposes it. But three moles of gas are produced from solids, so the entropy increases greatly and the entropy driver favours it. At 298 K, TΔS = 298 × 0.5432 = 162 kJ, which is smaller than ΔH, so ΔG = +331 kJ mol⁻¹ and the reaction is not spontaneous. Above 907 K, TΔS exceeds 492.7 kJ, so ΔG becomes negative and the reaction is spontaneous, which is why it happens in the hot furnace.