# How is the equilibrium constant related to Gibbs free energy?

Sep 5, 2014

The equilibrium constant for any reaction is related to the change in Gibbs Free Energy for that reaction under standard conditions by the equation

K_(eq)=e^((-DeltaG^0)/(RT)

where R is the universal gas constant (8.314 J/mol-K) and $T$ is the absolute temperature in Kelvins.

Standard conditions means all reactants and products present in unit concentrations or pressures (e.g., 1 M, 1m or 1 bar) at the 'temperature of interest'. Most tables of thermodynamic values will give Gibbs Free Energy of formation for reactants and products at 298.15 K, so calculation of ${K}_{e q}$ at this temperature is a simple matter of calculating $\Delta {G}^{0}$ for reaction as the difference in Gibbs Free Energies of the products and reactants, and then using the equation above with $T = 298.15 K$.

Sometimes we need to calculate ${K}_{e q}$ at a different temperature, and this involves a somewhat more complicated calculation:

First, calculate $\Delta {H}^{0}$ for the reaction, taking the difference in standard enthalpies of formation of the products and reactants. Then calculate $\Delta {S}^{0}$ by taking the difference in entropies of products and reactants. The $\Delta {G}^{0}$ for reaction can then be calculated approximately from the equation

$\Delta {G}^{0} = \Delta {H}^{0} - T \Delta {S}^{0}$

Here, we can use any value of $T$ because $\Delta {H}^{0}$ and $\Delta {S}^{0}$ are not strongly dependent on temperature. Finally, use the first equation (with the same value of $T$ that you used in the second equation) to calculate ${K}_{e q}$.

Note that we cannot simply change $T$ in the first equation because $\Delta {G}^{0}$ is strongly dependent on temperature, as shown in the second equation.