If the rate constants of a reaction at 500 K | vedclass

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Question

(A) The Arrhenius equation is given by $\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 \ R} \left(\frac{T_2 - T_1}{T_1 \ T_2}\right)$.Given: $k_

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$49.49 \ kJ \ mol^{-1}$

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(A) The Arrhenius equation is given by $\log \left(\frac{k_2}{k_1}ight) = \frac{E_a}{2.303 \ R} \left(\frac{T_2 - T_1}{T_1 \ T_2}ight)$.Given: $k_1 = 0.002 \ s^{-1}$,$T_1 = 500 \ K$,$k_2 = 0.06 \ s^{-1}$,$T_2 = 700 \ K$,$R = 8.314 \ J \ mol^{-1} \ K^{-1}$.Substituting the values:$\log \left(\frac{0.06}{0.002}ight) = \frac{E_a}{2.303 	imes 8.314} \left(\frac{700 - 500}{700 	imes 500}ight)$$\log(30) = \frac{E_a}{19.147} \left(\frac{200}{350000}ight)$$1.477 = \frac{E_a}{19.147} 	imes \frac{2}{3500}$$E_a = \frac{1.477 	imes 19.147 	imes 3500}{200} \approx 49.49 \ kJ \ mol^{-1}$.

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

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