In ΔG = ΔH − TΔS, temperature multiplies only the entropy term. So the signs of $\Delta H$ and $\Delta S$ decide whether heating helps or hurts, and where the switch happens.
Key idea
When $\Delta H$ and $\Delta S$ have opposite effects, the process is favored on one side of a crossover temperature $T = \dfrac{\Delta H}{\Delta S}$ (with consistent units), where $\Delta G = 0$. Endothermic with $\Delta S > 0$: favored above it. Exothermic with $\Delta S < 0$: favored below it.
Why it works
At low temperature, $T\Delta S$ is small and $\Delta H$ decides. At high temperature, $T\Delta S$ dominates and $\Delta S$ decides. Setting $\Delta G = 0$ gives the balance point. This treats $\Delta H$ and $\Delta S$ as roughly constant over the range, which is a good first estimate.
| ΔH | ΔS | ΔG = ΔH − TΔS | Favored |
|---|---|---|---|
| − | + | always − | at every T |
| + | − | always + | never |
| − | − | − at low T | below $T = \Delta H/\Delta S$ |
| + | + | − at high T | above $T = \Delta H/\Delta S$ |
Worked examples
Worked example: Limestone to lime
$\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$ has $\Delta H^\circ = +179.2\ \text{kJ}$ and $\Delta S^\circ = +159.0\ \text{J/K}$. Above what temperature is it favored?
- Both positive: favored at high $T$.
- $T = \dfrac{179\,200\ \text{J}}{159.0\ \text{J/K}} = 1127\ \text{K}$, about $850\ ^\circ\text{C}$.
- Lime kilns run hotter than this, which is why heating limestone drives off $\mathrm{CO_2}$.
Answer: Above about 1130 K.
Worked example: A boiling point from tables
For $\mathrm{H_2O(l) \to H_2O(g)}$, $\Delta H^\circ = +44.0\ \text{kJ/mol}$ and $\Delta S^\circ = +118.8\ \text{J/(mol·K)}$. Estimate the boiling point.
- At the boiling point (1 bar), liquid and vapor are in equilibrium: $\Delta G = 0$.
- $T = \dfrac{44\,000}{118.8} = 370\ \text{K}$, close to the true $373\ \text{K}$. The small gap comes from using 298 K values of $\Delta H$ and $\Delta S$.
Answer: About 370 K.
Worked example: Favored only when cool
Ammonia synthesis has $\Delta H^\circ = -91.8\ \text{kJ}$ and $\Delta S^\circ = -198.1\ \text{J/K}$. Below what temperature is it favored?
- Both negative: favored at low $T$, where the $\Delta H$ term dominates.
- $T = \dfrac{-91\,800\ \text{J}}{-198.1\ \text{J/K}} = 463\ \text{K}$. Below about 463 K (190 °C) the reaction is favored under standard conditions.
Answer: Below about 463 K.
This is why the Haber process has to balance two needs: the reaction is favored only at lower temperatures, but it is fast enough only at higher ones (with a catalyst).
Common mistake
Thinking an endothermic reaction can never be favored. If $\Delta S$ is positive, heating makes $T\Delta S$ win: ice melts and limestone decomposes, both endothermic. Check the sign of $\Delta S$ before ruling a reaction out.