Electrochemistry - Result Question 55
####57. The correct relationship between free energy and equilibrium constant $K$ of a reaction is [1996]
(a) $\Delta G^{\circ}=-R T \ln K$
(b) $\Delta G=R T \ln K$
(c) $\Delta G^{\circ}=R T \ln K$
(d) $\Delta G=-R T \ln K$
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Solution:
- (a) The relation between free energy change and equilibrium constant is given by Nernst equation
$E _{\text{cell }}=E^{o}-\frac{R T}{n F} \ln Q$
At equilibrium, $E _{\text{cell }}=0$ and $Q=K_C$
$\therefore E^{o}=\frac{R T}{n F} \ln K$
Again $\Delta G^{\circ}=-n F E^{\circ}$
put in (i)
$\frac{-\Delta G^{o}}{n F}=\frac{R T}{n F} \ln K \quad ; \quad \Delta G^{o}=-R T \ln K$
