How is the equilibrium constant related to Gibbs free energy?

Question

Serenity Allen · Accepted Answer

The equation links the change in Gibbs Free Energy for any given reaction under standard conditions to the equilibrium constant for that reaction. #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_(eq)# at this temperature is a simple matter of calculating #DeltaG^0# for reaction as the difference in Gibbs Free Energies of the products and reactants, and then using the equation above with #T=298.15K#. Sometimes we need to calculate #K_(eq)# at a different temperature, and this involves a somewhat more complicated calculation: First, calculate #DeltaH^0# for the reaction, taking the difference in standard enthalpies of formation of the products and reactants.  Then calculate #DeltaS^0# by taking the difference in entropies of products and reactants.  The #DeltaG^0# for reaction can then be calculated approximately from the equation #DeltaG^0=DeltaH^0-TDeltaS^0# Here, we can use any value of #T# because #DeltaH^0# and #DeltaS^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_(eq)#. Note that we cannot simply change #T# in the first equation because #DeltaG^0# is strongly dependent on temperature, as shown in the second equation.

Jose Ayala · Answer

undefined The relationship is given by the equation ΔG = -RT ln(K), where ΔG is the Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant.