Gibbs Free Energy Calculator
This Gibbs free energy calculator works out ΔG = ΔH − TΔS from the enthalpy change, the entropy change and the temperature. For the vaporization of water (ΔH = 44.01 kJ/mol, ΔS = 118.8 J/(mol·K)) at 298 K it gives +8.61 kJ/mol – not spontaneous at room temperature.
Example: Enthalpy change ΔH 44.01 kJ/mol · Entropy change ΔS 118.8 J/(mol·K) · Temperature 298 · Temperature unit: Kelvin (K) → ΔG: ΔG = 8.61 kJ/mol – not spontaneous (ΔG > 0). Source: OpenStax Chemistry 2e, 16.4 Free Energy – ΔG = ΔH − TΔS, Table 16.3 (spontaneity), four sign cases, T = ΔH/ΔS at ΔG = 0, ΔG° = −RT ln K with R = 8.314 J/(K·mol), Table 16.4; Examples 16.7 (vaporization of water, 8.6 kJ at 298 K), 16.8 (HgO), 16.11 (boiling point estimate 370.5 K), 16.13 (Ksp of AgCl) (read 10.10.2026). Updated: .
How it is calculated
The change in Gibbs free energy is ΔG = ΔH − TΔS. If ΔG is negative the process is spontaneous, if it is positive it is not, and ΔG = 0 means equilibrium. This is how the textbook OpenStax Chemistry 2e introduces the free energy of J. Willard Gibbs: at constant temperature and pressure, ΔG combines the enthalpy change ΔH and the entropy change ΔS of the system. T is the absolute temperature in kelvin and can only be positive. ΔS is usually given in J/(mol·K), ΔH in kJ/mol – so TΔS must be divided by 1000 before subtracting.
Gibbs free energy examples
| ΔH (kJ/mol) | ΔS (J/(mol·K)) | T (K) | ΔG (kJ/mol) |
|---|---|---|---|
| 44.01 | 118.8 | 298 | 8.61 |
| 44.01 | 118.8 | 373 | −0.30 |
| 90.46 | 107.4 | 298.15 | 58.44 |
| −100 | 50 | 298 | −114.90 |
| 50 | −100 | 298 | 79.80 |
| −50 | −100 | 298 | −20.20 |
| −50 | −100 | 600 | 10.00 |
The first row is OpenStax Example 16.7, the vaporization of water: 44.01 kJ − 298 K × 118.8 J/K ÷ 1000 = 44.01 − 35.40 = 8.61 kJ (the book rounds to 8.6 kJ), so boiling is not spontaneous at 25 °C. The third row is the decomposition of yellow mercury(II) oxide (Example 16.8); the book prints 58.45 kJ/mol, exact arithmetic gives 58.44. The last four rows use round numbers to show the four sign cases (calculated).
Spontaneity and the crossover temperature
- ΔH negative, ΔS positive: spontaneous at all temperatures.
- ΔH positive, ΔS negative: not spontaneous at any temperature.
- Both positive: spontaneous only at high temperature; both negative: only at low temperature.
When ΔH and ΔS have the same sign, ΔG changes sign at T = ΔH/ΔS. For water: 44,010 J ÷ 118.8 J/K = 370.5 K = 97.3 °C. OpenStax uses this to estimate the boiling point and compares it with the accepted 373.2 K – the gap comes from using data measured at 298 K. With standard values, ΔG° = −RT ln K links ΔG to the equilibrium constant (R = 8.314 J/(K·mol)): ΔG° = 55.7 kJ/mol for dissolving AgCl at 298.15 K gives K = 1.74 × 10⁻¹⁰.
The temperature dependence of cell potentials is handled by the Nernst equation calculator; pressure, volume and temperature of gases are related by the ideal gas law calculator.
Frequently asked questions
What is the formula for Gibbs free energy?
ΔG = ΔH − TΔS, with ΔH the enthalpy change, T the absolute temperature in kelvin and ΔS the entropy change. Convert ΔS from J to kJ (÷ 1000) before subtracting.
When is a reaction spontaneous?
When ΔG is negative. ΔG > 0 means not spontaneous in the forward direction, ΔG = 0 means the system is at equilibrium.
At what temperature does a reaction become spontaneous?
If ΔH and ΔS have the same sign, at T = ΔH/ΔS (ΔH in J). Above that temperature when both are positive, below it when both are negative.
How do I get K from ΔG?
With standard values: K = e^(−ΔG°/RT), R = 8.314 J/(K·mol). A negative ΔG° gives K > 1 (products favoured), a positive ΔG° gives K < 1.
Why does the water example give 97.3 °C instead of 100 °C?
ΔH and ΔS were taken from tables at 298 K and treated as constant. OpenStax notes that values measured closer to the boiling point give a better result.
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