Gibbs Free Energy and the Equilibrium Constant

ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K

Worked example: K = 1.0e5 at 25 C → dG0 = -28.540 kJ/mol — press Try an example to run it live, then adjust anything.

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Grade 12Grade 12 Chemistry

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Gibbs Free Energy and the Equilibrium Constant explained

ΔG°KT

Thermodynamic tables and equilibrium tables describe the same chemistry in different currencies, and this equation is the exchange rate. A negative ΔG° means K > 1 and products dominate; a positive ΔG° means K < 1 and reactants win; ΔG° = 0 sits exactly at K = 1. At 25 °C the conversion factor RT is 2.479 kJ/mol, so RT ln 10 = 5.708 kJ/mol — every factor of ten in K is worth 5.7 kJ/mol of free energy. A reaction with K = 1.0 × 10⁵ therefore has ΔG° = −5.708 × 5 = −28.5 kJ/mol.

Because the relationship is exponential, small energy differences produce enormous equilibrium swings: a change of just 11.4 kJ/mol multiplies K by a hundred. That steepness is why enzyme designers and medicinal chemists chase a few kilojoules of binding energy so hard, and why ΔG° values quoted to the nearest kilojoule are already good enough for most predictions. Two cautions: the standard state matters (K must be written with the same reference concentrations and pressures the ΔG° was tabulated for), and this ΔG° is not the ΔG of an actual running mixture — the two differ by RT ln Q, and it is ΔG, not ΔG°, that must reach zero at equilibrium.

Gibbs Free Energy and the Equilibrium Constant formula

ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K
Where
  • ΔG∘\Delta G^{\circ}= Standard free energy change (kJ/mol)
  • KK= Equilibrium constant
  • TT= Absolute temperature (°C)