# What is the Gibbs free energy change?

Solves ΔG = ΔH − TΔS for the Gibbs free energy change, the enthalpy change, the entropy change or the temperature, says whether the process is spontaneous, and gives the temperature where ΔG = 0 and the equilibrium constant.

- Page: https://www.acalculator.org/chemistry/gibbs-free-energy-calculator
- JSON spec: https://www.acalculator.org/chemistry/gibbs-free-energy-calculator.json
- Version: 5d67ecc3d51e

## Default answer

Example with the default inputs (Find ΔG, ΔH, kJ/mol 44.01, ΔS, J/(mol·K) 118.8, Temperature 298 K): ΔG = 8.6076 kJ/mol: not spontaneous (ΔG > 0).

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | Which of ΔG, ΔH, ΔS and T to work out from the other three. |
| dh | ΔH, kJ/mol | The enthalpy change in kJ/mol; negative for an exothermic process. |
| ds | ΔS, J/(mol·K) | The entropy change in J/(mol·K); note joules, not kilojoules. |
| t | Temperature | The absolute temperature; above absolute zero. |
| dg | ΔG, kJ/mol | The Gibbs free energy change in kJ/mol; 0 at equilibrium. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The value worked out, to 6 significant figures. |
| spontaneity | Spontaneous? | Spontaneous when ΔG is below 0, not spontaneous above 0, at equilibrium at 0. |
| dG | ΔG (kJ/mol) | The Gibbs free energy change. |
| dH | ΔH (kJ/mol) | The enthalpy change. |
| dS | ΔS (J/(mol·K)) | The entropy change. |
| tK | T (K) | The temperature in kelvins. |
| tds | TΔS (kJ/mol) | The temperature times the entropy change, in kJ/mol. |
| trend | Spontaneous when | From the signs of ΔH and ΔS: at all temperatures, none, or above or below the temperature where ΔG = 0. |
| crossover | T where ΔG = 0 (K) | ΔH ÷ ΔS, when ΔH and ΔS have the same sign. |
| lnK | ln K | −ΔG ÷ RT, if ΔG is the standard free energy change ΔG°. |
| k | K | The equilibrium constant e^(−ΔG° ÷ RT), if ΔG is the standard free energy change. |

## Method

ΔG = ΔH − T × ΔS ÷ 1000 (ΔH and ΔG in kJ/mol, ΔS in J/(mol·K), T in K); the temperature where ΔG = 0 is T = 1000 × ΔH ÷ ΔS; K = e^(−1000 × ΔG ÷ (R × T)) with R = 8.31446261815324 J/(mol·K).

## Assumptions

- ΔH and ΔS are taken as constant over the temperature range.
- Spontaneous means ΔG < 0 at constant temperature and pressure; it says nothing about how fast the process is.
- K is shown only when ΔG is the standard change ΔG°; the page cannot tell which you typed.

## Worked examples

1. find = g, dh = 44.01, ds = 118.8, t = 298 gives dG = 8.6076, tds = 35.4024, spontaneity = not spontaneous (ΔG > 0). Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.7 Evaluation of ΔG° Change from ΔH° and ΔS°: TΔS° = 35.4 kJ, ΔG° = 8.6 kJ.
2. find = g, dh = 90.46, ds = 107.4, t = 298 gives dG = 58.4548. Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.8 (b): ΔG° = 90.46 − 298 × 0.1074 = 58.45 kJ/mol.
3. find = t, dh = 44.01, ds = 118.8, dg = 0 gives tK = 370.454545, result = T = 370.455 K (97.3045 °C). Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.11 Equilibrium Temperature for a Phase Change: 370.5 K (97.3 °C).
4. find = g, dh = 0, ds = 0, t = 298.15 gives dG = 0, k = 1, spontaneity = at equilibrium (ΔG = 0). Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Table 16.4: K = 1 when ΔG° = 0.
5. find = h, dg = 55.7, ds = 0, t = 298.15 gives dH = 55.7, k = 0. Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.13: Ksp = e^(−55,700 ÷ (8.314 × 298.15)) = 1.74 × 10⁻¹⁰ (the page uses R = 8.31446261815324).

## FAQ

### How do I calculate ΔG?

Use ΔG = ΔH − TΔS with T in kelvins. ΔS is usually given in J/(mol·K) and ΔH in kJ/mol, so divide TΔS by 1000 first. For water boiling at 298 K: 44.01 − 298 × 118.8 ÷ 1000 = 44.01 − 35.40 = 8.61 kJ/mol.

### How do I know if a reaction is spontaneous?

At constant temperature and pressure a process is spontaneous when ΔG is negative, not spontaneous when it is positive, and at equilibrium when it is 0. Water boiling at 25 °C has ΔG = +8.6 kJ/mol, so it does not happen on its own at that temperature.

### At what temperature does a reaction become spontaneous?

When ΔH and ΔS have the same sign, ΔG changes sign at T = ΔH ÷ ΔS (with ΔS in kJ/(mol·K)). If both are positive the process is spontaneous above that temperature; if both are negative, below it. If ΔH is negative and ΔS positive it is spontaneous at every temperature, and with the opposite signs at none.

### How is ΔG related to the equilibrium constant?

For the standard change ΔG°, ΔG° = −RT ln K, so K = e^(−ΔG° ÷ RT) with ΔG° in J/mol and R = 8.314 J/(mol·K). ΔG° below 0 gives K above 1 (products favored); ΔG° = 0 gives K = 1.

### Does spontaneous mean fast?

No. ΔG says whether a process can happen on its own, not how quickly. Diamond turning into graphite is spontaneous at room conditions, but its rate is extremely slow.

### Why must the temperature be in kelvins?

ΔG = ΔH − TΔS needs the absolute temperature: at 0 °C the TΔS term is not zero. The page reads °C and °F too and converts them (K = °C + 273.15).

## Sources

- OpenStax, Chemistry 2e, §16.4 Free Energy: ΔG = ΔH − TΔS, ΔG° = −RT ln K, the four sign cases of ΔH and ΔS, T = ΔH ÷ ΔS where ΔG = 0, Example 16.7 (water boiling at 298 K: TΔS° = 35.4 kJ, ΔG° = 8.6 kJ), Example 16.8 (HgO: 58.45 kJ/mol), Example 16.11 (boiling point 370.5 K) and Example 16.13 (AgCl, Ksp = 1.74 × 10⁻¹⁰), Table 16.4 (K = 1 when ΔG° = 0). https://openstax.org/books/chemistry-2e/pages/16-4-free-energy (retrieved 2026-10-19)
- OpenStax, Chemistry 2e, §16.1 Spontaneity: spontaneity is not correlated to the speed of a process (diamond to graphite). https://openstax.org/books/chemistry-2e/pages/16-1-spontaneity (retrieved 2026-10-19)
- NIST, CODATA 2022 value: molar gas constant R = 8.314 462 618… J/(mol·K), exact (N_A × k). https://physics.nist.gov/cgi-bin/cuu/Value?r (retrieved 2026-10-19)