Understanding the Gibbs free energy calculator — concept and principle
The Gibbs free energy change ΔG tells you whether a reaction at constant temperature and pressure can proceed on its own, and how far it goes before reaching equilibrium. Enter the enthalpy change ΔH, the entropy change ΔS and the temperature, and the calculator returns ΔG, a yes/no spontaneity verdict, the equilibrium constant K and the crossover temperature where the verdict flips. The defaults are the ammonia synthesis (Haber) reaction, N₂ + 3H₂ → 2NH₃, at 25 °C.
How it is calculated
- Gibbs energy: ΔG = ΔH − TΔS (ΔS is divided by 1000 to convert J to kJ)
- Equilibrium constant: K = exp(−ΔG / RT), with R = 8.314462618 J/(mol·K) and ΔG in J/mol
- Crossover temperature: T = ΔH / ΔS, where ΔG = 0
ΔH in kJ/mol, ΔS in J/(mol·K), T in kelvin. When standard values (ΔH°, ΔS°) are entered, the result is the standard ΔG° and the standard equilibrium constant.
The principle
The second law says the total entropy of system plus surroundings must increase for a spontaneous change. At constant temperature and pressure, the heat −ΔH released into the surroundings raises their entropy by −ΔH/T. Adding the system's own ΔS and multiplying by −T gives ΔG = ΔH − TΔS, so a negative ΔG means total entropy increases. The link to equilibrium, ΔG° = −RT ln K, follows from how Gibbs energy depends on concentrations.
Worked example
- Inputs: ΔH = −92.2 kJ/mol, ΔS = −198.7 J/(mol·K), T = 298.15 K.
- TΔS = 298.15 × (−0.1987) = −59.2424 kJ/mol.
- ΔG = −92.2 − (−59.2424) = −32.957595 kJ/mol, so the answer is yes (ΔG < 0).
- −ΔG / RT = 32957.6 / (8.314462618 × 298.15) = 13.2949; K = e^13.2949 = 594,183.8403.
- Crossover: −92.2 × 1000 / −198.7 = 464.0161047 K (about 191 °C).
| ΔH (kJ/mol) | ΔG (kJ/mol) |
|---|---|
| -23.1 | 36.192405 |
| -46.1 | 13.142405 |
| -69.2 | -9.907595 |
| -92.2 | -32.957595 |
| -115 | -56.007595 |
| -138 | -79.057595 |
| -184 | -125.157595 |
How to read the result
ΔG below zero means products are favoured at equilibrium; above zero, reactants are favoured. K above 1 corresponds to negative ΔG°. The signs of ΔH and ΔS decide how temperature acts. With both negative, as here, the reaction is favoured at low temperature and stops being favoured above the crossover temperature. Industrial ammonia plants nevertheless run at roughly 400 to 500 °C with a catalyst, because at room temperature the reaction is far too slow, and high pressure is used to push the equilibrium back towards ammonia.
Common mistakes
- Mixing kJ and J: ΔH is usually in kJ/mol, ΔS in J/(mol·K).
- Reading "spontaneous" as "fast"; ΔG says nothing about the rate.
- Applying the crossover formula when ΔH and ΔS have opposite signs; then there is no crossover.
Frequently asked questions
Does a negative ΔG mean the reaction goes to completion?
No. It means products are favoured; the exact composition at equilibrium is given by K.
Where do I find ΔH° and ΔS° for a reaction?
Combine tabulated standard enthalpies of formation and standard molar entropies of products minus reactants, for example from the NIST Chemistry WebBook.
Sources: P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press · NIST Chemistry WebBook, NIST Standard Reference Database 69