Gibbs Free Energy
ΔG from ΔH, ΔS and temperature — with the J-to-kJ conversion handled and the crossover temperature.
at 298.15 K · K ≈ 5.941e+5
Spontaneous below this temperature, non-spontaneous above it (190.9 °C).
Gibbs free energy decides whether a reaction happens on its own: ΔG = ΔH − TΔS, and a negative ΔG means spontaneous. For ammonia synthesis with ΔH = −92.2 kJ/mol and ΔS = −198.7 J/(mol·K) at 298 K, ΔG is about −33 kJ/mol — spontaneous at room temperature.
Two competing tendencies
Reactions are pulled in two directions. Systems tend towards lower energy, which favours exothermic reactions with a negative ΔH. They also tend towards greater disorder, which favours a positive ΔS. Gibbs free energy combines the two into one number, weighting the entropy term by temperature, and the sign of that number settles the question.
Because temperature multiplies only the entropy term, it decides which tendency wins. At low temperature the TΔS term is small and enthalpy dominates; as temperature rises, entropy takes over. That is why ice melts above 0 °C and freezes below it — the same ΔH and ΔS, with the balance tipping at one temperature.
Spontaneous does not mean fast
This is the distinction the term hides. ΔG says whether a reaction is thermodynamically favourable, not whether it will happen in your lifetime. Diamond converting to graphite has a negative ΔG at room temperature and proceeds imperceptibly slowly, because the activation energy is enormous. Rate is kinetics; ΔG is thermodynamics, and neither predicts the other.
T in kelvin; ΔS is usually tabulated in J/(mol·K) while ΔH is in kJ/mol, so one of them must be converted
- 1 Convert ΔS into kilojoules. Entropy is tabulated in J/(mol·K), so −198.7 J/(mol·K) becomes −0.1987 kJ/(mol·K).
- 2 Put the temperature in kelvin. 25 °C is 298.15 K — the equation is meaningless with Celsius.
- 3 Work out the TΔS term. 298.15 × (−0.1987) = −59.24 kJ/mol.
- 4 Subtract it from ΔH. −92.2 − (−59.24) = −32.96 kJ/mol.
- 5 Read the sign. ΔG is negative, so the reaction is spontaneous in the forward direction at this temperature.
What the signs predict
Only the mixed cases depend on temperature; the other two are settled whatever it is.
| ΔH | ΔS | Spontaneous | Example |
|---|---|---|---|
| Negative | Positive | At every temperature | Combustion |
| Negative | Negative | At low temperature only | Freezing water |
| Positive | Positive | At high temperature only | Melting ice, boiling |
| Positive | Negative | Never | Nothing proceeds unaided |
The crossover temperature and the link to K
In the two mixed cases there is a temperature where ΔG passes through zero. Setting ΔG = 0 gives T = ΔH ÷ ΔS, and that is the point at which the reaction is at equilibrium — above or below it, one direction takes over. For the ammonia example the crossover is about 464 K, roughly 191 °C, which is why industrial synthesis has to trade yield against the higher temperature needed for a workable rate.
ΔG also connects to the equilibrium constant through ΔG° = −RT ln K. A negative ΔG° makes K greater than 1 and products dominate at equilibrium; a positive ΔG° makes K less than 1 and reactants dominate. The two are the same statement in different units, which is why the tool reports both. Note the standard-state symbol: ΔG° assumes 1 M concentrations and 1 bar pressure, while ΔG under real conditions shifts with them.