In general an equilibrium system is defined by writing an equilibrium equation for the reaction

\alpha A + \beta B \rightleftharpoons \sigma S + \tau T

In order to meet the thermodynamic condition for equilibrium, the Gibbs energy must be stationary, meaning that the derivative of G with respect to reaction coordinate (ΔG) must be zero. It can be shown that ΔG is, in fact, equal to the difference between the chemical potentials of the products and those of the reactants. Therefore, the sum of the Gibbs energies of the reactants must be the equal to the sum of the Gibbs energies of the products.

\alpha \mu_A + \beta \mu_B = \sigma \mu_S + \tau \mu_T \,

where μ is in this case a partial molar Gibbs energy, a chemical potential. The chemical potential of a reagent A is a function of the activity, {A} of that reagent.

\mu_A = \mu_{A}^{\ominus} + RT \ln\{A\} \,

Substituting expressions like this into the Gibbs energy equation:

\Delta G = Vdp-SdT+\sum_{i=1}^k \mu_i dN_i + \sum_{i=1}^n X_i da_i + \cdots \,

which at constant pressure and temperature becomes:

\Delta G =\sum_{i=1}^k \mu_i N_i

results in:

\Delta G = \sigma \mu_{S} + \tau \mu_{T} - \alpha \mu_{A} - \beta \mu_{B} \,

By substituting the chemical potentials:

\Delta G = ( \sigma \mu_{S}^{\ominus} + \tau \mu_{T}^{\ominus} ) - ( \alpha \mu_{A}^{\ominus} - \beta \mu_{B}^{\ominus} ) + ( \sigma RT \ln\{S\} + \tau RT \ln\{T\} ) - ( \alpha RT \ln\{A\} + \beta RT \ln \{B\} )

the relationship becomes:

\Delta G =\sum_{i=1}^k \mu_i^\ominus v_i + RT \ln \frac{\{S\}^\sigma \{T\}^\tau} {\{A\}^\alpha \{B\}^\beta}

At equilibrium \Delta G = 0 \, and therefore

\sum_{i=1}^k \mu_i^\ominus v_i + RT \ln \frac{\{S\}^\sigma \{T\}^\tau} {\{A\}^\alpha \{B\}^\beta} = 0

leading to:

\Delta G_m^{\ominus} = -RT \ln K

ΔGmO is the standard molar Gibbs energy change for the reaction and K is the equilibrium constant. Note that activities and equilibrium constants are dimensionless numbers.