In a first order reaction at 27,^oC and 47,^o | vedclass

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(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.At $T_1 = 300 \ K$ $(27\,^oC)$,$t_{1/2} = 30 \ \min$,so $

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$43.85 \ kJ \ mol^{-1}$

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(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.At $T_1 = 300 \ K$ $(27\,^oC)$,$t_{1/2} = 30 \ \min$,so $k_1 = \frac{0.693}{30} \ \min^{-1}$.At $T_2 = 320 \ K$ $(47\,^oC)$,$t_{1/2} = 10 \ \min$,so $k_2 = \frac{0.693}{10} \ \min^{-1}$.Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.$\ln(\frac{0.693/10}{0.693/30}) = \ln(3) = \frac{E_a}{8.314} (\frac{1}{300} - \frac{1}{320})$.$1.0986 = \frac{E_a}{8.314} (\frac{20}{96000})$.$E_a = \frac{1.0986 	imes 8.314 	imes 96000}{20} \approx 43857 \ J \ mol^{-1} = 43.85 \ kJ \ mol^{-1}$.

In a first order reaction at $27\,^oC$ and $47\,^oC$,$50\%$ of the reaction is complete in $30\,\min$ and $10\,\min$ respectively. Calculate the energy of activation $(E_a)$.

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