For the reaction C_2H_5I + OH^- rightarrow C_ | vedclass

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(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$Given: $K_1 = 0.325$,$K_2 =

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$20260$

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(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} ight]$Given: $K_1 = 0.325$,$K_2 = 6.735$,$T_1 = 30 + 273 = 303 \ K$,$T_2 = 60 + 273 = 333 \ K$,$R = 1.987 \ cal \ K^{-1} \ mol^{-1}$.Substituting the values:$\log \left( \frac{6.735}{0.325} ight) = \frac{E_a}{2.303 	imes 1.987} \left[ \frac{1}{303} - \frac{1}{333} ight]$$\log(20.723) = \frac{E_a}{4.575} \left[ \frac{333 - 303}{303 	imes 333} ight]$$1.3164 = \frac{E_a}{4.575} 	imes \frac{30}{100899}$$E_a = \frac{1.3164 	imes 4.575 	imes 100899}{30} \approx 20260 \ cal \ mol^{-1}$.

For the reaction $C_2H_5I + OH^- \rightarrow C_2H_5OH + I^-$,the rate constants at $30^\circ C$ and $60^\circ C$ are $0.325$ and $6.735 \ L \ mol^{-1} \ s^{-1}$ respectively. The value of activation energy $(E_a)$ is .......... calories.

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