Thermodynamics (enthalpy, entropy, Gibbs free energy)
The idea
Three state functions run chemical thermodynamics: enthalpy H tracks heat at constant pressure (ΔH < 0 exothermic), entropy S measures how widely energy and matter are dispersed (gases and more particles mean more entropy), and Gibbs free energy ties them together through ΔG = ΔH − TΔS. At constant temperature and pressure, ΔG < 0 marks a spontaneous process — one that can proceed without continuous outside work. You already know reaction heats from thermochemistry; the new power is predicting direction and the temperature at which direction flips.
Read the equation as a competition: enthalpy and entropy each vote, and temperature weights the entropy vote. When ΔH and ΔS share a sign, the reaction has a crossover temperature T = ΔH/ΔS where ΔG passes through zero — endothermic, entropy-driven reactions switch on above it; exothermic, order-creating reactions switch off. Unit discipline matters: ΔH usually arrives in kJ and ΔS in J/K, so convert before subtracting.
The essential correction: spontaneous does not mean fast. Thermodynamics ranks where a system wants to end up, never how quickly it gets there — diamond converting to graphite is spontaneous and takes geological eons. Rate questions belong to kinetics.
Worked example
For the decomposition CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178.3 kJ/mol and ΔS° = +160.5 J/(mol·K). Is the reaction spontaneous at 298 K? Assuming ΔH° and ΔS° stay constant, above what temperature does it become spontaneous?
- Convert entropy to matching units first: ΔS° = 160.5 J/(mol·K) = 0.1605 kJ/(mol·K). Skipping this step is the single most common source of nonsense answers.
- Evaluate ΔG° at room temperature: ΔG° = 178.3 − 298 × 0.1605 = 178.3 − 47.8 = +130.5 kJ/mol. Positive, so limestone does not decompose spontaneously at 298 K — which matches everyday experience.
- Both ΔH° and ΔS° are positive (heat must flow in, but a gas is created), so raising T strengthens the entropy term until it overcomes the enthalpy cost.
- Set ΔG° = 0 to find the crossover: T = ΔH°/ΔS° = 178.3/0.1605 = 1111 K, about 838 °C.
- Interpret: above roughly 1111 K the −TΔS° term exceeds +178.3 kJ/mol and decomposition becomes spontaneous — consistent with industrial lime kilns running near 900 °C.
Answer. At 298 K, ΔG° = +130.5 kJ/mol, so the decomposition is nonspontaneous; it becomes spontaneous above about 1111 K (roughly 838 °C).
Check your understanding
- Why does creating a gas product give a reaction such a large positive ΔS, and how could you estimate the sign of ΔS by inspection of any equation?
- What does a positive ΔG actually forbid — and why can coupled reactions in living cells still drive such steps forward?
- How would the crossover-temperature argument change for a reaction with ΔH < 0 and ΔS < 0, and what does that predict about cooling?
- How would you explain to a friend the difference between a spontaneous reaction and a fast one, with a concrete example of each combination?
Build the foundations first
Thermodynamics (enthalpy, entropy, Gibbs free energy) builds on these concepts. If any feel shaky, start there.