Find the value of the equilibrium constant $(K)$ of a reaction at $300 \mathrm{~K}$, when standard Gibbs…

Find the value of the equilibrium constant $(K)$ of a reaction at $300 \mathrm{~K}$, when standard Gibbs free energy change is $-25 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ? (Consider $R=8.33 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ )
  1. $e^8$
  2. $e^9$
  3. $e^{10}$
  4. $e^{11}$

Solution

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

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more THERMODYNAMICS questions on Aicharya