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Enthalpy, entropy and Gibbs free energy
The two drivers of chemical change: enthalpy and entropy. Predicting the sign of ΔS, calculating Gibbs free energy with ΔG = ΔH − TΔS, the temperature at which a reaction becomes spontaneous, and why reactions in closed systems stop at equilibrium.
Squeeze an instant cold pack and it gets cold enough to numb a sprained ankle. Inside, solid ammonium nitrate dissolves in water and absorbs heat. If reactions only happened when they released energy, this one shouldn't happen at all. It does, because energy isn't the only thing that matters. The other factor is entropy.
Enthalpy: the first driver
You met enthalpy, entropy and Gibbs free energy in Year 11 (Module 4, Drivers of reactions). This page revises them and uses them to explain equilibrium. $\Delta H$ is the heat absorbed or released at constant pressure.
- Exothermic ($\Delta H < 0$): products have lower enthalpy than reactants; heat is released to the surroundings.
- Endothermic ($\Delta H > 0$): products have higher enthalpy; heat is absorbed.
Systems tend towards lower enthalpy, so exothermic reactions are favoured. But plenty of endothermic changes happen on their own: ice melting on a warm day, salts like ammonium nitrate dissolving, water evaporating from a puddle. Something else must be pushing them.
Entropy: the second driver
Entropy ($S$) measures how dispersed, or spread out, the energy and particles of a system are. The more ways the particles and their energy can be arranged, the higher the entropy. Left alone, things tend to spread out: perfume fills a room, heat flows from hot to cold, a drop of ink colours a whole glass of water. Systems naturally move towards higher entropy.
The entropy change $\Delta S$ is measured in J K⁻¹ mol⁻¹. Note the joules, not kilojoules. You can usually predict its sign:
| Change | Sign of ΔS | Example |
|---|---|---|
| Solid → liquid → gas | Positive | H₂O(l) → H₂O(g) |
| More moles of gas formed | Positive | CaCO₃(s) → CaO(s) + CO₂(g) |
| Fewer moles of gas | Negative | N₂(g) + 3H₂(g) → 2NH₃(g) |
| Solid dissolving to form ions | Usually positive | NaCl(s) → Na⁺(aq) + Cl⁻(aq) |
| Precipitate forming from ions | Usually negative | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) |
| Raising temperature of a substance | Positive | Particles move faster, energy is spread over more states |
When gases are involved, counting moles of gas on each side is the quickest check. A change in the number of gas particles almost always dominates.
Gibbs free energy
A reaction is spontaneous if, once started, it continues without a continuous input of energy from outside. (Spontaneous says nothing about speed: diamond turning into graphite is spontaneous but immeasurably slow.) Whether a change is spontaneous depends on both drivers together, combined in the Gibbs free energy change:
$$\Delta G = \Delta H - T\Delta S$$
- $\Delta G < 0$: the forward reaction is spontaneous.
- $\Delta G > 0$: the forward reaction is not spontaneous (the reverse is).
- $\Delta G = 0$: the system is at equilibrium.
$T$ is the absolute temperature in kelvin (K = °C + 273). Because $\Delta H$ is usually in kJ mol⁻¹ and $\Delta S$ in J K⁻¹ mol⁻¹, divide ΔS by 1000 before using the equation. Forgetting this is the most common error in Gibbs questions.
For dissolving ammonium nitrate, $\Delta H = +25.7$ kJ mol⁻¹ and $\Delta S = +108.7$ J K⁻¹ mol⁻¹. Is dissolving spontaneous at 25 °C?
The temperature where a reaction "switches"
When $\Delta H$ and $\Delta S$ have the same sign, the two drivers pull in opposite directions and temperature decides the winner. Because $T$ multiplies $\Delta S$, entropy matters more at high temperature. Setting $\Delta G = 0$ gives the crossover temperature:
$$T = \frac{\Delta H}{\Delta S}$$
CaCO₃(s) → CaO(s) + CO₂(g); $\Delta H = +178.3$ kJ mol⁻¹, $\Delta S = +160.6$ J K⁻¹ mol⁻¹. Above what temperature is the decomposition spontaneous?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); $\Delta H = -92.2$ kJ mol⁻¹, $\Delta S = -198.7$ J K⁻¹ mol⁻¹. Find ΔG at 25 °C and at 450 °C, and comment.
- Choose Melting ice. Where does the line cross ΔG = 0? Why does that temperature make sense?
- Choose Haber process and move the temperature from 300 K to 700 K. What happens to ΔG, and why?
- Choose Burning methane and Photosynthesis. Explain why one is spontaneous at every temperature and the other at none.
- Make your own reaction with ΔH = +50 kJ mol⁻¹. What value of ΔS makes it spontaneous above 500 K?
Why reactions stop at equilibrium
If a reaction has a negative $\Delta G$, why doesn't it run to completion in a closed flask? Because the free energy of a mixture of reactants and products is lower than that of either pure reactants or pure products: mixing itself increases entropy. As the reaction proceeds, the total free energy falls until it reaches a minimum. At that point $\Delta G = 0$ for any further change in either direction, so there is no net driving force: the system is at equilibrium. A strongly negative $\Delta G$ puts the minimum close to the products (a large K); a strongly positive one puts it close to the reactants (a small K).
The open systems in 5A never reach this minimum. In a flame, hot gaseous products disperse into the room; in a leaf, a constant supply of light energy keeps pushing a reaction with $\Delta G > 0$ uphill.
Test yourself
Which change has the most positive ΔS?
Only this one produces gas from liquids.
A reaction has ΔH = −120 kJ mol⁻¹ and ΔS = −300 J K⁻¹ mol⁻¹. It is spontaneous
T = 120 ÷ 0.300 = 400 K. With ΔS negative, the −TΔS term becomes more positive as T rises.
At 298 K a reaction has ΔH = +40.0 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. ΔG is
40.0 − 298 × 0.100 = 40.0 − 29.8 = +10.2 kJ mol⁻¹.
Photosynthesis has ΔH > 0 and ΔS < 0. It occurs in plants because
ΔG is positive at every temperature, so it needs a continuous external energy input. A catalyst can't change ΔG.
Label each combination of ΔH and ΔS.
Predict the sign of the entropy change.
Entropy is a measure of the of energy and particles. In ΔG = ΔH − TΔS, T must be in . A reaction is spontaneous when ΔG is . At equilibrium, ΔG equals . A reaction with positive ΔH and positive ΔS becomes spontaneous at temperatures.
Water freezes: H₂O(l) → H₂O(s), ΔH = −6.01 kJ mol⁻¹, ΔS = −22.0 J K⁻¹ mol⁻¹. (a) Calculate ΔG at −10 °C and at +10 °C. (b) Use your answers and the concepts of enthalpy and entropy to explain why water freezes below 0 °C but not above it.
(a) ΔS = −0.0220 kJ K⁻¹ mol⁻¹. At −10 °C (263 K): ΔG = −6.01 − 263 × (−0.0220) = −6.01 + 5.79 = −0.22 kJ mol⁻¹. At +10 °C (283 K): ΔG = −6.01 + 6.23 = +0.22 kJ mol⁻¹.
(b) Freezing releases heat (ΔH negative), which favours it, but it turns a liquid into an ordered solid, so entropy decreases (ΔS negative), which opposes it. The entropy contribution, −TΔS, is positive and grows in proportion to temperature. Below the crossover temperature, T = ΔH/ΔS = 6.01 ÷ 0.0220 = 273 K, the enthalpy term wins and ΔG is negative, so water freezes spontaneously (as at −10 °C). Above 273 K the entropy term wins, ΔG is positive and ice melts instead. At exactly 0 °C, ΔG = 0 and ice and water are at equilibrium.