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JEE MAIN 2021
25-02-2021 S2
Question
The rate constant of a reaction increases by five times on increase in temperature from $27^{\circ} \mathrm{C}$ to $52^{\circ} \mathrm{C}$. The value of activation energy in $\mathrm{kJ} \mathrm{mol}^{-1}$ is $\_\_\_\_$ (Rounded-off to the nearest integer)
$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
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Solution
$$ \begin{aligned} \mathrm{T}_1=300 \mathrm{~K}, \mathrm{~T}_2 & =325 \mathrm{~K}, \mathrm{~K}_2=5 \mathrm{~K}, \\ \ln \frac{\mathrm{~K}_2}{\mathrm{~K}_1} & =\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right] \\ \text { or, } \quad \ln 5 & =\frac{\mathrm{Ea}}{8.314}\left[\frac{1}{300}-\frac{1}{325}\right] \\ \text { or, } \mathrm{Ea} & =0.7 \times 2.303 \times 8.314 \times 12 \times 325 \\ & =52271 \mathrm{~J}=52.271 \mathrm{~kJ} \end{aligned} $$ Nearest integer answer will be 52 kJ
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