At equilibrium, the reaction Gibbs free energy is zero:

$$\Delta G=0.$$

The thermodynamic reaction quotient has then reached its equilibrium value,

$$Q=K.$$

Substituting these conditions into

$$\Delta G=\Delta G^\circ+RT\ln Q$$

gives

$$0=\Delta G^\circ+RT\ln K,$$

so

$$\boxed{\Delta G^\circ=-RT\ln K}.$$

Equivalently,

$$\boxed{K=e^{-\Delta G^\circ/(RT)}}.$$

Here $K$ is the dimensionless thermodynamic equilibrium constant, built from equilibrium activities. Introductory concentration constants $K_c$ approximate this quantity for ideal dilute solutions when concentrations are normalized by the standard concentration; analogous pressure expressions use $p_i/p^\circ$ for ideal gases.

The equation links the thermodynamic preference encoded by a standard free-energy change to equilibrium composition:

• if $\Delta G^\circ<0$, then $K>1$ and products are favored relative to reactants;
• if $\Delta G^\circ>0$, then $K<1$ and reactants are favored;
• if $\Delta G^\circ=0$, then $K=1$.

Example

At $298,\mathrm K$, suppose

$$\Delta G^\circ=-10.0,\mathrm{kJ/mol}=-1.00\times10^4,\mathrm{J/mol}.$$

Then

$$\ln K=-\frac{\Delta G^\circ}{RT} =\frac{1.00\times10^4}{(8.314)(298)} \approx4.04.$$

Therefore

$$K=e^{4.04}\approx56.8.$$

The equilibrium is product-favored, but the finite value of $K$ means both reactants and products can still be present.

The equation also makes clear why an equilibrium constant belongs to a particular temperature. $\Delta G^\circ$ itself depends on temperature, so changing temperature generally changes $K$.