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Chemistry LibreTexts

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

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    204505
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    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.