If the half lives of the first order reaction at $350 \mathrm{~K}$ and $300 \mathrm{~K}$ are 2 and 20 seconds respectively, the activation energy of the reaction in $\mathrm{kJ} \mathrm{mol}^{-1}$ is

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Aicharya · Accepted Answer

Given: $t_{1 / 2}(\mathrm{l})=20 \mathrm{~s}$ $$ \begin{aligned} & t_{1 / 2}(2)=2 \mathrm{~s} \quad\left[\because t_{1 / 2} \propto \frac{1}{k}ight] \ & T_1=300 \mathrm{~K}, T_2=350 \mathrm{~K} \ & \because \quad \log \frac{k_2}{k_1}=-\frac{E_a}{2.303 R}\left[\frac{1}{T_2}-\frac{1}{T_1}ight] \ & 	herefore \quad \log \frac{\frac{0.693}{2}}{\frac{0.693}{20}}=-\frac{E_a}{2.303 	imes 8.314}\left[\frac{1}{350}-\frac{1}{300}ight] \ & \Rightarrow \quad \log \frac{20}{2}=-\frac{E_a}{2.303 	imes 8.314}\left[\frac{-50}{350 	imes 300}ight] \ & 	herefore \quad E_a=\frac{2.303 	imes 8.314 	imes 350 	imes 300}{50} \ & E_a=40.209 \mathrm{~kJ} / \mathrm{mol} \ & \end{aligned} $$ ^

If the half lives of the first order reaction at $350 \mathrm{~K}$ and $300 \mathrm{~K}$ are 2 and 20…

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