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Gibbs Free Energy Calculator

Find ΔG and reaction spontaneity.

Result

-32.70 kJ/mol
Spontaneous

How it is calculated

The Gibbs free energy change at a given temperature is ΔG = ΔH − T·ΔS. Enter enthalpy change ΔH in kJ/mol, temperature T in kelvin, and entropy change ΔS in J/(mol·K). Because ΔS is in joules and ΔH in kilojoules, the tool divides ΔS by 1000 first, so ΔG comes out in kJ/mol. Temperature must be zero or above. If ΔG is negative the reaction is spontaneous at that temperature, if positive it is non-spontaneous, and exactly zero means equilibrium. Spontaneous describes thermodynamic direction only; it says nothing about how fast the reaction is.

Worked example

ΔH = −92 kJ/mol, ΔS = −199 J/(mol·K), T = 298 K. Convert ΔS to kJ: −199 / 1000 = −0.199 kJ/(mol·K). T·ΔS = 298 × (−0.199) = −59.30. ΔG = −92 − (−59.30) = −32.70. The result is -32.70 kJ/mol, which is negative, so the reaction is Spontaneous at 298 K. Note that at a much higher temperature the −T·ΔS term would grow and ΔG would turn positive.

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Common mistakes

  • Entering ΔS in kJ instead of J/(mol·K). The tool divides it by 1000 itself, so the result would be wrong by a factor of 1000.
  • Entering the temperature in °C. T must be in kelvin; add 273.15.
  • Dropping the sign of ΔS. A negative entropy change makes the −T·ΔS term positive and pushes ΔG up.
  • Reading spontaneous as fast. A reaction with negative ΔG can still be extremely slow.
  • Using ΔH and ΔS at one temperature as constants over a wide range; they change somewhat with temperature.

Sources

About this calculator

This calculator finds Gibbs free energy change, ΔG = ΔH − T·ΔS, which combines a reaction's enthalpy change ΔH (heat absorbed or released), absolute temperature T, and entropy change ΔS into a single value that predicts spontaneity at constant pressure and temperature. A negative ΔG means the reaction is spontaneous (exergonic) and can proceed on its own; a positive ΔG means it's non-spontaneous (endergonic) and needs energy input; ΔG is also linked to the equilibrium constant K through ΔG = −RT·ln K.

Chemists use this to predict whether a reaction will run forward under given conditions, biochemists use it to explain energy-coupling in metabolism — such as ATP hydrolysis driving otherwise unfavorable reactions — and chemical engineers use it to assess the feasibility and yield of industrial processes before committing to them. Chemistry and biochemistry students, researchers, and process engineers all rely on this calculation to connect thermodynamics to real reaction behavior.

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Comments (1)

  • Editorial team

    When ΔH and ΔS have the same sign, the verdict flips at a crossover temperature where ΔG is zero: T = ΔH divided by ΔS, with matching units. For ΔH = -92 kJ/mol and ΔS = -199 J/(mol·K), that is about 462 K, spontaneous below it and not above it. Entering temperatures just under and just over that value should show ΔG changing sign, a handy test of your own numbers. Which reaction are you checking, and does the flip point look plausible to you?

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