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

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ 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.

12.1: The Standard Gibbs Free Energy Change and Equilibrium in Ideal Gas Reactions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.