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Gibbs Free Energy Calculator

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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.

Gibbs free energy for the Haber processΔH = −92.2 kJ/mol plus −TΔS = +59.24 kJ/mol gives ΔG = −32.96 kJ/mol at 298.15 K; above 464 K the reaction is no longer spontaneous.Making ammonia: heat out vs. order lost0−92.2kJ/mol+59.24−32.96ΔG < 0spontaneousΔHheat released−TΔS298.15 × 0.1987ΔG < 0ΔG > 0298 K464 K crossoverAt 298 K: K = e^(32,958 / RT) ≈ 5.9 × 10⁵
The released heat beats the entropy penalty at room temperature, so ΔG = −32.96 kJ/mol.

How it is calculated

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

  1. Inputs: ΔH = −92.2 kJ/mol, ΔS = −198.7 J/(mol·K), T = 298.15 K.
  2. TΔS = 298.15 × (−0.1987) = −59.2424 kJ/mol.
  3. ΔG = −92.2 − (−59.2424) = −32.957595 kJ/mol, so the answer is yes (ΔG < 0).
  4. −ΔG / RT = 32957.6 / (8.314462618 × 298.15) = 13.2949; K = e^13.2949 = 594,183.8403.
  5. Crossover: −92.2 × 1000 / −198.7 = 464.0161047 K (about 191 °C).
How ΔG (kJ/mol) changes with ΔH (kJ/mol)How ΔG changes with ΔH-150-100-50050-23.1-46.1-69.2-92.2-115-138-184Example: -32.957595ΔH (kJ/mol)
Each point is the calculator re-run with only “ΔH (kJ/mol)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
ΔH (kJ/mol)ΔG (kJ/mol)
-23.136.192405
-46.113.142405
-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

Keep in mind: The calculation treats ΔH and ΔS as constant with temperature, which is only an approximation far from 298 K, and K applies to standard states. Real equilibria also depend on pressure, concentration and non-ideal behaviour.

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

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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