Q24 / 75 · JEE Main 2026 (22 January Shift 2) · Chemistry
Consider $\mathrm{A} \xrightarrow{\mathrm{k}_{1}} \mathrm{~B}$ and $\mathrm{C} \xrightarrow{\mathrm{k}_{2}} \mathrm{D}$ are two reactions. If the rate constant $\left(\mathrm{k}_{1}\right)$ of the $\mathrm{A} \longrightarrow \mathrm{B}$ reaction can be expressed by the following equation $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^{4}}{\mathrm{~T} / \mathrm{K}}$ and activation energy of $C \longrightarrow D$ reaction $\left(E a_{2}\right)$ is $\frac{1}{5}$ th of the $A \longrightarrow B$ reaction $\left(E a_{1}\right)$, then the value of $\left(E a_{2}\right)$ is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)
Correct answer: 57
Solution
From the Arrhenius equation:
$\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}$.
Comparing with given equation $\log_{10} k_1 = 14.34 - \frac{1.5 \times 10^4}{T}$, we get: $\frac{E_{a_1}}{2.303R} = 1.5 \times 10^4$ K
$E_{a_1} = 1.5 \times 10^4 \times 2.303 \times 8.314$
$= 1.5 \times 10^4 \times 0.01913$
$= 287$ kJ/mol
Since $E_{a_2} = \frac{1}{5} E_{a_1}$:
$E_{a_2} = \frac{287}{5} = 57.4 \approx 57$ kJ/mol
$\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}$.
Comparing with given equation $\log_{10} k_1 = 14.34 - \frac{1.5 \times 10^4}{T}$, we get: $\frac{E_{a_1}}{2.303R} = 1.5 \times 10^4$ K
$E_{a_1} = 1.5 \times 10^4 \times 2.303 \times 8.314$
$= 1.5 \times 10^4 \times 0.01913$
$= 287$ kJ/mol
Since $E_{a_2} = \frac{1}{5} E_{a_1}$:
$E_{a_2} = \frac{287}{5} = 57.4 \approx 57$ kJ/mol
Hence, the correct answer is 57.