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Equilibrium Constant Calculator

Find the equilibrium constant from ΔG.

Result

56.4965

How it is calculated

The engine applies K = exp(−ΔG / (R·T)) with R fixed at 8.314 J/(mol·K). Because R is in joules, ΔG must be entered in J/mol (not kJ/mol), and T in kelvin. The temperature has to be greater than zero, otherwise there is no result; any ΔG, positive, negative or zero, is accepted. K is shown to four decimals and has no unit; a K below 0.01 or at 1,000,000 and above is shown in scientific notation instead (ΔG = 30000 J/mol at 298.15 K gives 5.5454 × 10⁻⁶). ΔG = 0 gives K = 1, negative ΔG gives K above 1, positive ΔG gives K below 1. It treats ΔG as the standard value ΔG°; for a mixture away from standard state use the reaction quotient instead.

Worked example

Take ΔG = −10000 J/mol (that is −10 kJ/mol) at T = 298.15 K (25 °C). Step 1: R·T = 8.314 × 298.15 ≈ 2478.8 J/mol. Step 2: −ΔG / (R·T) = 10000 / 2478.8 ≈ 4.0342. Step 3: K = exp(4.0342). The result is 56.4965, so products are strongly favoured at equilibrium.

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Common mistakes

  • Entering ΔG in kJ/mol: typing −30 instead of −30000 gives K = 1.0122 instead of a value near 180,000.
  • Using Celsius for temperature; 25 must be entered as 298.15 K.
  • Flipping the sign of ΔG: a negative ΔG gives K above 1, so −10000 and +10000 give very different answers.
  • Using the value for a reaction not at standard conditions; the formula assumes the standard free energy change ΔG°.

About this calculator

This calculator finds the equilibrium constant K of a chemical reaction from its Gibbs free energy change using K = exp(−ΔG°/RT), the rearranged form of the thermodynamic relation ΔG° = −RT·ln(K), where R is the gas constant (8.314 J/mol·K) and T is the absolute temperature. This equation is the bridge between thermodynamics and chemical equilibrium: a large negative ΔG° (a thermodynamically favorable reaction) produces a K much greater than 1, meaning the equilibrium lies far toward products, while a positive ΔG° gives K less than 1, favoring reactants.

Chemistry students and researchers use this to connect free-energy data from tables to predicted reaction extent; chemical engineers use it when designing reactors to predict how far a reaction will proceed under given conditions; and biochemists use the same relationship to study equilibria in enzyme reactions, ligand binding, and protein folding.

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Comments (1)

  • Editorial team

    A handy sanity check: at 298 K, every 5.7 kJ/mol in the standard free energy shifts K by a factor of about ten. So entering minus 5708 J/mol should give a K near 10. A K below 0.01 prints in scientific notation, so a ΔG of 30000 J/mol at 298.15 K shows 5.5454 × 10⁻⁶ rather than a rounded zero. What reaction and temperature are you checking?

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