Which equation correctly relates standard Gibbs free energy change to the equilibrium constant?

• A. ΔG° = −RT ln K
• B. ΔG° = RT ln K
• C. ΔG° = −RT
• D. ΔG° = −K

Answer: (A)

The main relationship being tested is how standard Gibbs free energy change connects to the equilibrium constant. At equilibrium, the driving force for the reaction vanishes, so the actual Gibbs energy change ΔG is zero. The general relation is ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. When the system sits at equilibrium, Q equals the equilibrium constant K, giving 0 = ΔG° + RT ln K. Rearranging yields ΔG° = -RT ln K.

This expression also explains the sign and magnitude. If K > 1, ln K is positive and ΔG° is negative, meaning the reaction favors products under standard conditions. If K < 1, ln K is negative and ΔG° is positive, favoring reactants. R is the gas constant (8.314 J/mol·K) and T is the absolute temperature in kelvin, so ΔG° is in joules per mole and K is dimensionless.

The other forms don’t fit this equilibrium relationship: ΔG° = RT ln K would give the opposite sign for cases where K > 1, and ΔG° = -K or ΔG° = -RT are not dimensionally or conceptually correct for relating ΔG° to K.