Chemical Thermodynamics & Equilibrium
CollegeSpontaneity, entropy and free energy
Thermodynamics asks not just how much heat a reaction releases, but whether it will happen at all without outside help — whether it is spontaneous. Enthalpy (ΔH) alone cannot answer this: some endothermic processes (ice melting above 0 °C) are spontaneous. The missing piece is entropy, S — a measure of the number of microscopic arrangements (microstates) consistent with a system's macroscopic state, popularly described as "disorder." The second law of thermodynamics states that the entropy of the universe (system + surroundings) increases for any spontaneous process.
Gibbs combined both quantities into one criterion evaluated using only the system:
A process is spontaneous (as written, at constant T and P) if ΔG < 0; non-spontaneous if ΔG > 0; at equilibrium if ΔG = 0. ΔG relates directly to the equilibrium constant:
Why cold packs work and why diamonds don't turn to graphite before your eyes
ΔG = ΔH − TΔS makes the interplay explicit. A cold pack's dissolving salt is endothermic (ΔH > 0, it should be unfavourable) yet spontaneous — because dissolving increases disorder sharply (ΔS > 0 is large), and at room temperature TΔS outweighs ΔH, making ΔG negative overall. Ice melting above 0 °C is the same story: ΔH > 0, but ΔS > 0 (liquid is more disordered than solid), and above 0 °C, TΔS wins.
Crucially, ΔG tells you whether a reaction is thermodynamically favourable, not how fast it happens — that is a separate question of kinetics (Chapter 19). Diamond converting to graphite has ΔG° < 0 at room temperature and pressure (graphite is the thermodynamically stable form of carbon), yet diamonds persist indefinitely because the activation energy for that conversion is enormous — the reaction is favourable but so slow it is, for all practical purposes, never observed. Thermodynamics decides whether; kinetics decides when.
Thermodynamics (ΔG) tells you if a reaction can happen and where equilibrium lies. Kinetics tells you how fast it gets there. A reaction can be thermodynamically favourable and kinetically frozen at the same time.
Worked example — predicting spontaneity across temperatures
A reaction has ΔH° = +58 kJ mol⁻¹ and ΔS° = +176 J mol⁻¹ K⁻¹. Is it spontaneous at 25 °C? At what temperature does it become spontaneous?
- Convert to consistent units: ΔS° = 0.176 kJ mol⁻¹ K⁻¹.
- At T = 298 K: ΔG° = ΔH° − TΔS° = 58 − (298 × 0.176) = 58 − 52.4 = +5.6 kJ mol⁻¹.
- Since ΔG° > 0, the reaction is non-spontaneous at 25 °C.
- Find the crossover temperature by setting ΔG° = 0: T = ΔH° / ΔS° = 58 / 0.176.
- Compute: T = 329.5 K ≈ 56.5 °C. Above this temperature, TΔS° exceeds ΔH° and the reaction becomes spontaneous — a case where heating drives a reaction forward purely through the entropy term.
Calculate ΔG° for a reaction at 500 K with K = 2.5 × 10³. (R = 8.314 J mol⁻¹ K⁻¹.)
ΔG° = −RT ln K = −8.314 × 500 × ln(2.5 × 10³) = −8.314 × 500 × 7.824 = −32 530 J mol⁻¹ ≈ −32.5 kJ mol⁻¹. Strongly negative, confirming K ≫ 1 (products favoured).
Why is ΔS positive when a solid dissolves into a solution?
In the solid lattice, particles are locked in fixed, highly ordered positions — few accessible microstates. Once dissolved, the same particles are free to move throughout the solvent in vastly more possible arrangements, so the number of accessible microstates — and hence entropy — increases sharply.
A reaction has ΔH° < 0 and ΔS° < 0. Under what temperature condition is it spontaneous?
ΔG° = ΔH° − TΔS°, with ΔH° negative (favourable) and −TΔS° positive (unfavourable, since ΔS° < 0). The reaction is spontaneous only when the favourable ΔH° term dominates, i.e. at low temperature — small T keeps the unfavourable TΔS° term small.
Why can a process with ΔG < 0 still appear not to happen at room temperature?
ΔG tells you the reaction is thermodynamically favourable and that equilibrium lies toward products — but it says nothing about the rate. If the activation energy is very high, the reaction can be immeasurably slow at room temperature even though it is thermodynamically "downhill," exactly as with diamond's slow conversion to graphite.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1A reaction has deltaH = +30 kJ and deltaS = +100 J/K. Is it spontaneous at 400 K?
- Convert deltaS to kJ: 100 J/K = 0.100 kJ/K.
- deltaG = deltaH - T deltaS = 30 - 400 x 0.100.
- deltaG = 30 - 40 = -10 kJ, which is negative.
Example 2Below what temperature does that same reaction become non-spontaneous?
- Set deltaG = 0: deltaH = T deltaS.
- T = deltaH / deltaS = 30 / 0.100.
- T = 300 K, so below 300 K deltaG is positive.
Now you try
Work each one out first, then tap to reveal the worked answer.