Question

If the equilibrium constant $\left(K_C\right)$ for a reversible reaction is $5\times {10}^{13}$ at 300 K, then the value of $\Delta r{G}^o$ at the same temperature will be:

• A. $8.314J{\text{mol}}^{-1}{K}^{-1}\times 300K\times \text{ln}\left({10}^{13}\right)$
• B. $8.314J{\text{mol}}^{-1}{K}^{-1}\times 300K\times \text{ln}(3\times {10}^{13})$
• C. $-8.314J{\text{mol}}^{-1}{K}^{-1}\times 300K\times \text{log}(4\times {10}^{13})$
• D. $-8.314J{\text{mol}}^{-1}{K}^{-1}\times 300K\times \text{ln}(5\times {10}^{13})$

Answer 1

The correct option is D $-8.314J{\text{mol}}^{-1}{K}^{-1}\times 300K\times \text{ln}(5\times {10}^{13})$

${\Delta }_{r}G={\Delta }_r{G}^{\circ }+RT\text{ln}{Q}_c$
$\text{At equilibrium,}{Q}_C=K_{eq}$
$\text{and}{\Delta }_{r}G=0$
$\therefore 0=\Delta G^{\circ }+RT\text{ln}{K}_{eq}$
$\Delta G^{\circ }=-RT\text{ln}{K}_{eq}$
$=-8.314\times 300\text{ln}(5\times {10}^{13})$