# 12.1: The Standard Gibbs Free Energy Change and Equilibrium in Ideal Gas Reactions

Figure 5. Cycle demonstrating the relationship between $$\Delta_r G$$ and $$\Delta_r G^o$$.

The relationship between $${\Delta }_rG$$ and $${\Delta }_rG^o$$ is evident from the cycle in Figure 5. Since we have shown that $${\Delta }_{sep}G\left(P_A,P_B,P_C,P_D\right)={\Delta }_rG\left(P_A,P_B,P_C,P_D\right)$$, we can consider the bottom equation in this cycle to represent the reaction occurring in a mixture while calculating its free energy change as the free energy difference between pure products and pure reactants. Since $${\Delta }_{cycle}G=0$$,

${\Delta }_{cycle}G={\Delta }_rG^o-{\Delta }_rG+RT\left({ \ln p^c_C\ }+{ \ln p^d_D\ }-{ \ln p^a_A\ }-{ \ln p^B_B\ }\right)=0$

which can be rearranged to the result obtained in §2:

${\Delta }_rG={\Delta }_rG^o+RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ }$

$${\Delta }_rG$$ is the Gibbs free energy change for one unit of the reaction occurring in a system whose composition is specified by $$P_A$$, $$P_B$$, $$P_C$$, and $$P_D$$. In this spontaneously reacting system, the molar Gibbs free energy of ideal gas $$A$$ is

${\overline{G}}_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln p_A\ }$

If the system is at equilibrium, $$P_A$$, $$P_B$$, $$P_C$$, and $$P_D$$ are equilibrium pressures; these values characterize an equilibrium state. Then $${\Delta }_rG$$ is the free energy change for a reaction occurring at equilibrium at constant pressure and temperature, and $${\Delta }_rG$$ is zero. The equation

$0={\Delta }_rG^o+RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ }$

is exact. We have, when the partial pressures are those for a system at equilibrium,

${\Delta }_rG^o=-RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ }$

Since $${\Delta }_rG^o$$ is a constant, it follows that

$\frac{p^c_Cp^d_D}{p^a_Ap^b_B}$

is a constant. It is, of course, the equilibrium constant. We have

$K_P=\frac{p^c_Cp^d_D}{p^a_Ap^b_B}$

and ${\Delta }_rG^o=-RT{ \ln K_P\ }$

or, solving for $$K_P$$

$K_P = \mathrm{exp}\left( + \frac{{\mathrm{\Delta }}_rG^o}{RT}\right)$

Note that the value of the equilibrium constant is calculated from the Gibbs free energy change at standard conditions, not the Gibbs free energy change at equilibrium, which is zero.