WBJEE · Chemistry · Chemical Equilibrium
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
- A \(4 \times 10^{-4}\)
- B \(4 \times 10^{-3}\)
- C \(4 \times 10^{-5}\)
- D \(2.5 \times 10^{-4}\)
Answer & Solution
Correct Answer
(A) \(4 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
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…
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