JEE Main 2026 — 28 January, Evening Shift — Question 66

For $\mathrm{A} \xrightarrow{\mathrm{K}_{1}} \mathrm{~B}$ $\ln (2)=\frac{E_{a_{1}}}{R}\left[\frac{1}{300}-\frac{1}{500}\right]$ $\mathrm{E}_{\mathrm{a}_{1}}=\frac{\ln 2 \times \mathrm{R} \times 1500}{2}$ $\mathrm{E}_{\mathrm{a}_{2}}=\frac{\mathrm{E}_{\mathrm{a}_{1}}}{2}=\frac{\ln 2 \times \mathrm{R} \times 1500}{4}$ $\left(\mathrm{K}_{1}\right)_{\text {at } 500 \mathrm{~K}}=\frac{\ln 2}{2}$ $\left(\mathrm{K}_{2}\right)_{\mathrm{at} 500 \mathrm{~K}}=\ln 2$ Now for $\mathrm{C} \xrightarrow{\mathrm{K}_{2}} \mathrm{D}$ $\ln \left[\frac{\left(\mathrm{K}_{2}\right)_{\text {at } 500 \mathrm{~K}}}{\left(\mathrm{~K}_{2}\right)_{\text {at } 300 \mathrm{~K}}}\right]=\left(\frac{\ln 2 \times \mathrm{R} \times 1500}{4}\right) \times \frac{1}{\mathrm{R}} \times\left[\frac{1}{300}-\frac{1}{500}\right]$ $\left(\mathrm{K}_{2}\right)_{\mathrm{at} 300 \mathrm{~K}}=\frac{\ln 2}{\sqrt{2}}=0.49$ $\left(\mathrm{K}_{2}\right)_{\mathrm{at} 300 \mathrm{~K}}=4.9 \times 10^{-1}$ Ans is 5.