Which equation correctly relates standard Gibbs free energy change to the equilibrium constant at a given temperature?

Which equation correctly relates standard Gibbs free energy change to the equilibrium constant at a given temperature?

• A. ΔG° = -RT ln K
• B. ΔG° = RT ln K
• C. ΔG° = ΔH° - TΔS°
• D. ΔG° = -KRT

Answer: (A)

The relation between standard Gibbs free energy change and the equilibrium constant at a given temperature is ΔG° = -RT ln K. Here R is the gas constant (8.314 J/mol·K), T is the absolute temperature, and K is the equilibrium constant defined with standard states (activities). This means that when the reaction favors products (K > 1), ln K is positive and ΔG° is negative, signaling a spontaneous tendency toward products under standard conditions. If K < 1, ΔG° is positive and the reaction favors the reverse direction, while K = 1 gives ΔG° = 0 and no net tendency. This relationship comes from the broader expression ΔG = ΔG° + RT ln Q. At equilibrium, ΔG = 0 and Q = K, so 0 = ΔG° + RT ln K, which rearranges to ΔG° = -RT ln K. The other forms shown don’t correctly tie ΔG° to the equilibrium constant: one is a separate expression for ΔG° in terms of enthalpy and entropy, and the others use incorrect structure for the dependence on K (they lack the natural logarithm or have the wrong sign).

The relation between standard Gibbs free energy change and the equilibrium constant at a given temperature is ΔG° = -RT ln K. Here R is the gas constant (8.314 J/mol·K), T is the absolute temperature, and K is the equilibrium constant defined with standard states (activities). This means that when the reaction favors products (K > 1), ln K is positive and ΔG° is negative, signaling a spontaneous tendency toward products under standard conditions. If K < 1, ΔG° is positive and the reaction favors the reverse direction, while K = 1 gives ΔG° = 0 and no net tendency.

This relationship comes from the broader expression ΔG = ΔG° + RT ln Q. At equilibrium, ΔG = 0 and Q = K, so 0 = ΔG° + RT ln K, which rearranges to ΔG° = -RT ln K.

The other forms shown don’t correctly tie ΔG° to the equilibrium constant: one is a separate expression for ΔG° in terms of enthalpy and entropy, and the others use incorrect structure for the dependence on K (they lack the natural logarithm or have the wrong sign).