Q9 / 155 · WBJEE 2021 · Chemistry
The equilibrium constant for the reaction $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$ is $4 \times 10^{-4}$ at $2000 \mathrm{~K}$. In presence of a catalyst the equilibrium is attained 10 times faster. Therefore, the equilibrium constant, in presence of the catalyst at $2000 \mathrm{~K}$ is
- Correct answerA$4 \times 10^{-4}$
- B$4 \times 10^{-3}$
- C$4 \times 10^{-5}$
- D$2.5 \times 10^{-4}$
Solution
As per Van't Hoff Equation
$\frac{\mathrm{d}\left(\ln \mathrm{K}_{\mathrm{eq}}\right)}{\mathrm{dT}}=\frac{\Delta \mathrm{H}_{\mathrm{T}}}{\mathrm{RT}^{2}}$
As catalyst don't affect enthalpy of reaction $\left(\Delta \mathrm{H}_{\mathrm{r}}\right)$ and Temperature is constant. Therefore, $\mathrm{K}_{\mathrm{eq}}$ (equilibrium constant) will remain constant i.e. $4 \times 10^{-4}$.
Hence, the correct answer is option: 'A'.