Derive the relation between the equilibrium constant $K$,reaction quotient $Q_{c}$,and Gibbs energy change $\Delta G$.

(N/A) The thermodynamic relationship between Gibbs energy change and the reaction quotient is given by the equation:
$\Delta G = \Delta G^{\circ} + RT \ln Q \quad (Eq. I)$
Where:
$\Delta G^{\circ} = \text{Standard Gibbs energy change}$
$\Delta G = \text{Gibbs energy change at any state}$
$R = \text{Gas constant} = 8.314 \ J \ mol^{-1} \ K^{-1}$
$T = \text{Temperature in Kelvin}$
$Q = \text{Reaction quotient}$
At equilibrium,the system reaches a state where $\Delta G = 0$ and $Q = K_{c}$. Substituting these values into $(Eq. I)$:
$0 = \Delta G^{\circ} + RT \ln K$
$\Delta G^{\circ} = -RT \ln K$
$\ln K = -\frac{\Delta G^{\circ}}{RT} \quad (Eq. II)$
Taking the antilog of both sides,we get:
$K = e^{-\Delta G^{\circ} / RT} \quad (Eq. III)$
Significance of $\Delta G^{\circ}$:
If $\Delta G^{\circ} < 0$,then $-\Delta G^{\circ} / RT$ is positive,so $K > 1$,indicating a spontaneous reaction where products predominate.
If $\Delta G^{\circ} > 0$,then $-\Delta G^{\circ} / RT$ is negative,so $K < 1$,indicating a non-spontaneous reaction where reactants predominate.