About this tool
The Gibbs Free Energy Calculator is a free, in-browser thermodynamics tool built on ΔG = ΔH − T·ΔS. Choose which quantity to solve for — ΔG, ΔH, ΔS, temperature T, or the equilibrium constant K — and it returns that value together with a spontaneity verdict based on the sign of ΔG.
Watch the units, which are the most common source of error: enthalpy ΔH and free energy ΔG are entered in kJ/mol, but entropy ΔS is entered in J/mol·K. The tool converts ΔS by dividing by 1000, so it evaluates ΔG = ΔH − T·(ΔS ⁄ 1000) and the answer stays in kJ/mol. Solving for entropy rearranges to ΔS = (ΔH − ΔG)·1000 ⁄ T (back in J/mol·K), and solving for temperature gives T = (ΔH − ΔG)·1000 ⁄ ΔS.
Everything runs locally in your browser with no uploads. The equilibrium constant is linked through the standard free energy: ΔG° = −R·T·ln K, so K = exp(−ΔG°·1000 ⁄ (R·T)) with R = 8.314 J/mol·K and ΔG° in kJ/mol (multiplied by 1000 to match R's joules). A negative ΔG means a spontaneous (product-favoured) reaction, zero means equilibrium, and a positive ΔG means a non-spontaneous one.
Frequently asked questions
Why is ΔS in J but ΔG and ΔH in kJ?
By convention entropies are tabulated in J/mol·K while reaction enthalpies and free energies are in kJ/mol. The calculator handles the mismatch by dividing ΔS by 1000 inside ΔG = ΔH − T·(ΔS ⁄ 1000), so you can enter each in its usual unit.
How does ΔG determine spontaneity?
A negative ΔG marks a spontaneous, product-favoured process under the given conditions; ΔG = 0 means the system is at equilibrium; and a positive ΔG means the reaction is non-spontaneous forward and favoured in reverse. Spontaneity always refers to the sign of ΔG, not ΔH.
How is the equilibrium constant K calculated?
From the standard free energy: ΔG° = −R·T·ln K, rearranged to K = exp(−ΔG°·1000 ⁄ (R·T)). R = 8.314 J/mol·K, so ΔG° in kJ/mol is multiplied by 1000 to match, and T must be a positive temperature in kelvin.
Can a reaction switch between spontaneous and not?
Yes. Because ΔG = ΔH − TΔS, temperature can flip the sign when ΔH and ΔS have the same sign. Solving for T at ΔG = 0 gives the crossover temperature T = (ΔH ⁄ ΔS)·1000 where spontaneity changes.
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