Derive the relation between G^0 and equilibrium constant ( K ) for the reactionaA + bB cC + dD.

The free energy (G) of any substance at a temperature T is represented as A+B C+D G_A, G_B, G_C and G_D are standard free energies G_A=G_A^0+R T [A] G_B=G_B^0+R T [B] G_C=G_c^0+R T [C] G_D=G_D^0+R T [D] G= G_ product - G_ reactant = [G_C+G_D ]- [G_A+G_B ] = G_C^0+R T [C]+G_D^0+R T Ln[D] - G_A^0+R T [A]+G_B^0+R T [B] (G_C^0+G_D^0 )- (G_A^0+G_B^0 ) + R T [C]+R T [D]-R T [A]+R T [B] if G^0= (G_C^0+G_D^0 )- (G_A^0+G_B^0 ) G= G^0+(R T [C] [D]-R T [A] [B]) G= G^0+R T or G= G^0+R T Q Q= [ product ] [ reactant ] Q=K=[C] [D] [A] [B] Hence from above equation G= G^0+R T k since at equillibrium G=0 O= G^0+R T K G^0=-R T K or G^0=-2.303 R T _ 10 K

Derive the relation between G^0 and equilibrium constant ( K ) fo…

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Question

Derive the relation between $\Delta G^0$ and equilibrium constant $( K )$ for the reaction$aA + bB \rightleftharpoons cC + dD$.

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Answer

The free energy $(G)$ of any substance at a temperature $T$ is represented as
$A+B \rightleftharpoons C+D$
$G_A, G_B, G_C$ and $G_D$ are standard free energies
$G_A=G_A^0+R T \ln [A]$
$G_B=G_B^0+R T \ln [B]$
$G_C=G_c^0+R T \ln [C]$
$G_D=G_D^0+R T \ln [D]$
$\therefore \Delta G=\sum G_{\text {product }}-\sum G_{\text {reactant }}$
$=\left[G_C+G_D\right]-\left[G_A+G_B\right]$
$=\left\{G_C^0+R T \ln [C]+G_D^0+R T \operatorname{Ln}[D]\right\}-\left\{G_A^0+R T \ln [A]+G_B^0+R T \ln [B]\right\}$
$\left\{\left(G_C^0+G_D^0\right)-\left(G_A^0+G_B^0\right)\right\}+\{R T \ln [C]+R T \ln [D]-R T \ln [A]+R T \ln [B]\}$
if $\Delta G^0=\left\{\left(G_C^0+G_D^0\right)-\left(G_A^0+G_B^0\right)\right\}$
$\therefore \Delta G=\Delta G^0+(R T \ln [C] \times[D]-R T \ln [A] \times[B])$
$\Delta G=\Delta G^0+R T \ln$
or $\Delta G=\Delta G^0+R T \ln Q$
$Q=\frac{[\text { product }]}{[\text { reactant }]}$
$Q=K=\frac{[C] \times[D]}{[A] \times[B]}$
Hence from above equation
$\Delta G=\Delta G^0+R T \ln k$
since at equillibrium $\triangle G=0$
$\therefore O=\Delta G^0+R T \ln K$
$\therefore \Delta G^0=-R T \ln K$
or
$\therefore \Delta G^0=-2.303 R T \log _{10} K$