Q24 / 75 · JEE Main 2026 (02 April Shift 1) · Chemistry
For reaction $A \rightarrow P$, rate constant $k = 1.5 \times 10^3$ s$^{-1}$ at $27°$C. If activation energy for the above reaction is $60$ kJ mol$^{-1}$, then the temperature (in $°$C) at which rate constant, $k = 4.5 \times 10^3$ s$^{-1}$ is _______. (Nearest integer)
Given : $\log 2 = 0.30$, $\log 3 = 0.48$, $R = 8.3$ J K$^{-1}$ mol$^{-1}$, $\ln 10 = 2.3$
Given : $\log 2 = 0.30$, $\log 3 = 0.48$, $R = 8.3$ J K$^{-1}$ mol$^{-1}$, $\ln 10 = 2.3$
Correct answer: 41
Solution
Using the Arrhenius equation:
$\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.3 R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)$
Given:
$k_1 = 1.5 \times 10^3 \text{ s}^{-1}$
$k_2 = 4.5 \times 10^3 \text{ s}^{-1}$
$T_1 = 27^\circ\text{C} = 300 \text{ K}$
$E_a = 60 \text{ kJ mol}^{-1} = 60000 \text{ J mol}^{-1}$
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
Substituting the values into the equation:
$\log \left( \dfrac{4.5 \times 10^3}{1.5 \times 10^3} \right) = \dfrac{60000}{2.3 \times 8.3} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$
$\log 3 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$
$0.48 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$
$\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{0.48 \times 19.09}{60000}$
$\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{9.1632}{60000} = 1.5272 \times 10^{-4}$
$\dfrac{1}{T_2} = \dfrac{1}{300} - 1.5272 \times 10^{-4}$
$\dfrac{1}{T_2} = 3.3333 \times 10^{-3} - 0.1527 \times 10^{-3} = 3.1806 \times 10^{-3} \text{ K}^{-1}$
$T_2 = \dfrac{1}{3.1806 \times 10^{-3}} \approx 314.4 \text{ K}$
Converting to Celsius:
$T_2 = 314.4 - 273 = 41.4^\circ\text{C}$
The nearest integer is $41$.
Answer: $41$
$\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.3 R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)$
Given:
$k_1 = 1.5 \times 10^3 \text{ s}^{-1}$
$k_2 = 4.5 \times 10^3 \text{ s}^{-1}$
$T_1 = 27^\circ\text{C} = 300 \text{ K}$
$E_a = 60 \text{ kJ mol}^{-1} = 60000 \text{ J mol}^{-1}$
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
Substituting the values into the equation:
$\log \left( \dfrac{4.5 \times 10^3}{1.5 \times 10^3} \right) = \dfrac{60000}{2.3 \times 8.3} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$
$\log 3 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$
$0.48 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$
$\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{0.48 \times 19.09}{60000}$
$\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{9.1632}{60000} = 1.5272 \times 10^{-4}$
$\dfrac{1}{T_2} = \dfrac{1}{300} - 1.5272 \times 10^{-4}$
$\dfrac{1}{T_2} = 3.3333 \times 10^{-3} - 0.1527 \times 10^{-3} = 3.1806 \times 10^{-3} \text{ K}^{-1}$
$T_2 = \dfrac{1}{3.1806 \times 10^{-3}} \approx 314.4 \text{ K}$
Converting to Celsius:
$T_2 = 314.4 - 273 = 41.4^\circ\text{C}$
The nearest integer is $41$.
Answer: $41$
Hence, the correct answer is 41.