Question

Obtain the relationship between ∆G° of a reaction and the equilibrium constant.

Answer 1

1. Gibbs energy change for a chemical reaction is given by

ΔG = ΔG° + RT ln Q        …(1)

where, ΔG° is standard Gibbs energy change that is, the Gibbs energy change when the reactants and products in a reaction are in their standard states. Q is called reaction quotient. Q is analogous to that of the equilibrium constant and involves nonequilibrium concentrations or partial pressures in case of a gaseous reaction.

2. Consider the reaction, aA + bB → cC + dD

From equation (1),

ΔG = ΔG° + RT lnQ_C     or      ΔG = ΔG° + RT lnQ_P

= ΔG° + RT ln${\displaystyle \frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}}$   or   = ΔG° + RT ln ${\displaystyle \frac{{\text{P}}_{\text{C}}^{\text{c}}\times {\text{P}}_{\text{D}}^{\text{d}}}{{\text{P}}_{\text{A}}^{\text{a}}\times {\text{P}}_{\text{B}}^{\text{b}}}}$

3. When the reaction reaches equilibrium, ΔG = 0 and Q_C and Q_P become K_C and K_P, respectively.

Thus, 

∴ 0 = ΔG° + RT lnK_C    or    0 = ΔG° + RT lnK_P

∴ ΔG° = –RT ln K_C      or     ΔG° = –RT ln K_P

∴ ΔG° = –2.303 RT log₁₀K_C   or   ΔG° = –2.303 RT log₁₀K_P