Chemistry - Thermodynamics - II Question with Solution | TestHub

(a) $2.303\log K=-\frac{\Delta H°}{RT}+\frac{\Delta S°}{R}$

This equation is obtained by combining

$\Delta G°=\Delta H°-T\Delta S°$

with

$\Delta G°=-RT\ln K$

(when $\ln K$ is converted to $\log K$, the factor 2.303 appears).

(b) $\Delta G°=-2.303RT\log K$

This is the standard relation between Gibbs free energy change and the equilibrium constant.