6.7 Relationship between Equilibrium Constant K Reaction Quotient Q and Gibbs Energy G

The value of \(K_c\) for a reaction does not depend on the rate of the reaction. However, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, \(\Delta G\). If,

• \(\Delta G\) is negative, then the reaction is spontaneous and proceeds in the forward direction.
• \(\Delta G\) is positive, then reaction is considered non-spontaneous. Instead, as the reverse reaction would have a negative \(\Delta G\), the products of the forward reaction shall be converted to the reactants.
• \(\Delta G\) is 0, the reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:
\(
\Delta G=\Delta G^{\ominus}+\mathrm{RT} \ln Q \dots(7.21)
\)
where, \(G^{\ominus}\) is standard Gibbs energy.
At equilibrium, when \(\Delta G=0\) and \(Q=K_c\), the equation (7.21) becomes,
\(
\begin{aligned}
& \Delta G=\Delta G^{\ominus}+\mathrm{R} T \ln K=0 \\
& \Delta G^{\ominus}=-\mathrm{R} T \ln K \dots(7.22) \\
& \ln K=-\Delta G^{\ominus} / \mathrm{R} T
\end{aligned}
\)
Taking antilog of both sides, we get,
\(K=\mathrm{e}^{-\Delta G^{\ominus} / \mathrm{RT}} \dots(7.23)\)
Hence, using the equation (7.23), the reaction spontaneity can be interpreted in terms of the value of \(\Delta G^{\ominus}\).

• If \(\Delta G^{\ominus}<0\), then \(-\Delta G^{\ominus} / \mathrm{R} T\) is positive, and \(\mathrm{e}^{-\Delta \mathrm{G} / \mathrm{RT}}>1\), making \(K>1\), which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
• If \(\Delta G^{\ominus}>0\), then \(-\Delta G^{\ominus} / \mathrm{R} T\) is negative, and a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

Example 7.10: The value of \(\Delta G^{\ominus}\) for the phosphorylation of glucose in glycolysis is \(13.8 \mathrm{~kJ} / \mathrm{mol}\). Find the value of \(K_c\) at \(298 \mathrm{~K}\).

Answer:

\(
\begin{aligned}
& \Delta G^{\ominus}=13.8 \mathrm{~kJ} / \mathrm{mol}=13.8 \times 10^3 \mathrm{~J} / \mathrm{mol} \\
& \text { Also, } \Delta G^{\ominus}=-\mathrm{RT} \ln K_c \\
& \text { Hence, } \ln K_c=-13.8 \times 10^3 \mathrm{~J} / \mathrm{mol} \\
& \quad\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 298 \mathrm{~K}\right) \\
& \ln K_{\mathrm{c}}=-5.569 \\
& K_{\mathrm{c}}=\mathrm{e}^{-5.569} \\
& K_{\mathrm{c}}=3.81 \times 10^{-3} \\
\end{aligned} \\
\)

Example 7.11: Hydrolysis of sucrose gives,
Sucrose \(+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\) Glucose + Fructose
Equilibrium constant \(K_c\) for the reaction is \(2 \times 10^{13}\) at \(300 \mathrm{~K}\). Calculate \(\Delta G^{\ominus}\) at \(300 \mathrm{~K}\).

Answer: \(
\Delta G^{\ominus}=-\mathrm{R} T \ln K_c
\)
\(
\Delta G^{\ominus}=-8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K} \times \ln \left(2 \times 10^{13}\right)
\)
\(
\Delta G^{\ominus}=-7.64 \times 10^4 \mathrm{~J} \mathrm{~mol}^{-1}
\)