Question

Find the value of the equilibrium constant $(K)$ of a reaction at $300 \,K$ when standard
Gibbs free energy change is $-25 \,kH\,mol ^{-1} ?$
(Consider $R =8.33\, J\, mol ^{-1}\, K ^{-1}$ )

• A. $e^{8}$
• B. $e^{9}$
• C. $e^{10}$
• D. $e^{11}$

Answer 1

Answer: (C)

Given,
Gibbs free energy $=-25 \,kJ \,mol ^{-1}$
Temperature $=300 \,K$
Equilibrium constant $(K)=?$
$\Delta G^{\circ}=-R T \ln \,K$
$-25 \,kJ \,mol ^{-1}=\left(-8.314\, J\,mol ^{-1} \,K ^{-1} \times 300 \,K \right) \ln k$
$\ln K=\left(-\frac{25\, kJ\,mol ^{-1}}{8.314 \,J\,mol ^{-1} \,K ^{-1} \times 300\, K }\right)$
$=\frac{25 \times 10^{3}\, J\,mol ^{-1}}{8.314\, J\,mol ^{-1} K ^{-1} \times 300\, K }$
$\ln\, K=-10.02, K=e^{10.02}$