Question

Manufacture of $S{O}_3$ is given by the following reaction
$2S{O}_2\text{(g)}+O_2\text{(g)}\rightleftharpoons 2S{O}_3\text{(g)}$
If the equilibrium constant $K_c$ is $1.7\times {10}^2$ at $300\text{K}$, then the value of $\Delta r{G}^{\circ }$ at the same temperature will be

• A. $8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (3.4\times {10}^2)$
• B. $-8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (1.7\times {10}^2)$
• C. $-8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (2.7\times {10}^2)$
• D. $8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (2.7\times {10}^2)$

Answer 1

The correct option is B $-8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (1.7\times {10}^2)$

$\Delta rG=\Delta r{G}^{\circ }+RT\ln Q$
At equilibrium, $\Delta G=0,Q=K_{\text{eq}}$
So, $\Delta r{G}^{\circ }=-RT\ln K_{\text{eq}}$
$=-8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\times 300\text{K}\times \ln (1.7\times {10}^2)$