Q21 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry
$\mathrm{A} \longrightarrow \mathrm{B}$ (first reaction)
$\mathrm{C} \longrightarrow \mathrm{D}$ (second reaction)
Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At $500 \mathrm{~K}, 50 \%$ of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is $\_\_\_\_$ $\times 10^{-1}$ hour $^{-1}$ (nearest integer).
$\mathrm{C} \longrightarrow \mathrm{D}$ (second reaction)
Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At $500 \mathrm{~K}, 50 \%$ of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is $\_\_\_\_$ $\times 10^{-1}$ hour $^{-1}$ (nearest integer).
Correct answer: 5
Solution
For $A \xrightarrow{K_1} B$
$\ln(2) = \dfrac{E_{a_1}}{R} \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$
$E_{a_1} = \dfrac{\ln 2 \times R \times 1500}{2}$
$E_{a_2} = \dfrac{E_{a_1}}{2} = \dfrac{\ln 2 \times R \times 1500}{4}$
$(K_1)_{\text{at 500 K}} = \dfrac{\ln 2}{2}$
$(K_2)_{\text{at 500 K}} = \ln 2$
Now for $C \xrightarrow{K_2} D$
$\ln \left[ \dfrac{(K_2)_{\text{at 500K}}}{(K_2)_{\text{at 300K}}} \right] = \left( \dfrac{\ln 2 \times R \times 1500}{4} \right) \times \dfrac{1}{R} \times \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$
$(K_2)_{\text{at 300 K}} = \dfrac{\ln 2}{\sqrt{2}} = 0.49$
$(K_2)_{\text{at 300 K}} = 4.9 \times 10^{-1}$.
So, answer is 5.
$\ln(2) = \dfrac{E_{a_1}}{R} \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$
$E_{a_1} = \dfrac{\ln 2 \times R \times 1500}{2}$
$E_{a_2} = \dfrac{E_{a_1}}{2} = \dfrac{\ln 2 \times R \times 1500}{4}$
$(K_1)_{\text{at 500 K}} = \dfrac{\ln 2}{2}$
$(K_2)_{\text{at 500 K}} = \ln 2$
Now for $C \xrightarrow{K_2} D$
$\ln \left[ \dfrac{(K_2)_{\text{at 500K}}}{(K_2)_{\text{at 300K}}} \right] = \left( \dfrac{\ln 2 \times R \times 1500}{4} \right) \times \dfrac{1}{R} \times \left[ \dfrac{1}{300} - \dfrac{1}{500} \right]$
$(K_2)_{\text{at 300 K}} = \dfrac{\ln 2}{\sqrt{2}} = 0.49$
$(K_2)_{\text{at 300 K}} = 4.9 \times 10^{-1}$.
So, answer is 5.
Hence, the correct answer is 5.