A \to B The rate constants of the above reaction at 200 K and 300 K are 0.03 min ^-1 and 0.05 min ^-1 respectively. The activation energy for the reaction is \underline\hspace2cm J (Nearest integer) (Given : ln10=2.3R=8.3~J~K^-1~mol^-1\log5=0.70\log3=0.48\log2=0.30 )

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q23 with Solution

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Correct answer: 2520

For a reaction,

\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)

Given:

k_1=0.03\ \text{min}^{-1},\quad T_1=200\,\text{K}

k_2=0.05\ \text{min}^{-1},\quad T_2=300\,\text{K}

R=8.3\ \text{J K}^{-1}\text{mol}^{-1}

\ln\left(\frac{k_2}{k_1}\right)=\ln\left(\frac{0.05}{0.03}\right)=\ln\left(\frac{5}{3}\right)

Now,

\log\left(\frac{5}{3}\right)=\log 5-\log 3=0.70-0.48=0.22

Using $\ln x=2.3\log x$ ,

\ln\left(\frac{5}{3}\right)=2.3\times 0.22=0.506

\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{200}-\frac{1}{300}

=\frac{3-2}{600}=\frac{1}{600}

So,

0.506=\frac{E_a}{8.3}\times \frac{1}{600}

E_a=0.506\times 8.3\times 600

First,

8.3\times 600=4980

Then,

E_a=0.506\times 4980=2519.88\ \text{J mol}^{-1}

Nearest integer:

E_a\approx 2520\ \text{J mol}^{-1}

The activation energy is:

\boxed{2520}