Rate constant of a reaction is given as log_{ | vedclass

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Question

(A) The Arrhenius equation in logarithmic form is $\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$.Comparing this with the given equation $\log_{10}

Vedclass · Accepted Answer

$9.21$

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(A) The Arrhenius equation in logarithmic form is $\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$.Comparing this with the given equation $\log_{10} K = -\frac{2000}{T} + 6.0$,we get $\frac{E_a}{2.303R} = 2000$.Given $R = 1.987 	imes 10^{-3} \ kcal \ K^{-1} \ mol^{-1} \approx 2 	imes 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$.$E_a = 2000 	imes 2.303 	imes 2 	imes 10^{-3} \ kcal \ mol^{-1}$.$E_a = 2000 	imes 2.303 	imes 0.002 = 9.212 \ kcal \ mol^{-1}$.Thus,the activation energy is $9.21 \ kcal \ mol^{-1}$.

Rate constant of a reaction is given as $\log_{10} K = -\frac{2000}{T} + 6.0$. Activation energy is ....... $kcal$.

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