If the half-lives of a first-order reaction a | vedclass

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(A) For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$,so $k \propto \frac{1}{t_{1/2}}$.Given: $T_1 = 300 \ K, t_{1/2}(1) = 20 \ s$ and $T_2 = 350

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$40.2$

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(A) For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$,so $k \propto \frac{1}{t_{1/2}}$.Given: $T_1 = 300 \ K, t_{1/2}(1) = 20 \ s$ and $T_2 = 350 \ K, t_{1/2}(2) = 2 \ s$.Therefore,$\frac{k_2}{k_1} = \frac{t_{1/2}(1)}{t_{1/2}(2)} = \frac{20}{2} = 10$.Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} [\frac{T_2 - T_1}{T_1 T_2}]$.Substituting the values: $\log(10) = \frac{E_a}{2.303 	imes 8.314 	imes 10^{-3}} [\frac{350 - 300}{350 	imes 300}]$.$1 = \frac{E_a}{2.303 	imes 8.314 	imes 10^{-3}} [\frac{50}{105000}]$.$E_a = \frac{2.303 	imes 8.314 	imes 10^{-3} 	imes 105000}{50} \approx 40.2 \ kJ \ mol^{-1}$.

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

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