For reaction A rightarrow P , rate constant k =1.5 × 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 × 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

Chemistry · Chemical Kinetics

JEE Main 2026 — 2 April, Morning Shift — Question 59

For reaction $\mathrm{A} \rightarrow \mathrm{P}$ , rate constant $\mathrm{k}=1.5 \times 10^{3} \mathrm{~s}^{-1}$ at $27^{\circ} \mathrm{C}$ . If activation energy for the above reaction is $60 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , then the temperature (in ${ }^{\circ} \mathrm{C}$ ) at which rate constant, $\mathrm{k}=4.5 \times 10^{3} \mathrm{~s}^{-1}$ is $\_\_\_\_$ . (Nearest integer) Given : $\log 2=0.30, \log 3=0.48, \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ , $\ln 10=2.3$

Answer: 41

Numerical answer — enter this value.

Step-by-step solution

$\quad \ell \mathrm{n}\left(\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right]$

\begin{aligned} & \ln \left(\frac{4.5 \times 10^{3}}{1.5 \times 10^{3}}\right)=\frac{60 \times 10^{3}}{8.3}\left[\frac{\mathrm{~T}_{2}-300}{300 . \mathrm{T}_{2}}\right] & \ln 3=\frac{60 \times 10^{3}}{8.3}\left[\frac{\mathrm{~T}_{2}-300}{300 . \mathrm{T}_{2}}\right] & \Rightarrow \mathrm{T}_{2}=314.4 \mathrm{~K} & \Rightarrow \mathrm{t}=41.4^{\circ} \mathrm{C} \end{aligned}

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation