The activation energy of one of the reactions | vedclass

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(A) Using the Arrhenius equation: $\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{E_a}{R} \left(\frac{1}{T_{300}} - \frac{1}{T_{310}}\right)$ $\ln \

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(A) Using the Arrhenius equation: $\ln \left(\frac{k_{310}}{k_{300}}ight) = \frac{E_a}{R} \left(\frac{1}{T_{300}} - \frac{1}{T_{310}}ight)$ $\ln \left(\frac{k_{310}}{k_{300}}ight) = \frac{532611}{8.3} 	imes \left(\frac{310 - 300}{310 	imes 300}ight)$ $\ln \left(\frac{k_{310}}{k_{300}}ight) = \frac{532611}{8.3} 	imes \frac{10}{93000} = 64170 	imes \frac{10}{93000} \approx 6.9$ Since $\ln 10 = 2.3$,then $6.9 = 3 	imes 2.3 = 3 	imes \ln 10 = \ln 10^3$. Therefore,$\frac{k_{310}}{k_{300}} = 10^3$,which means $k_{300} = 10^{-3} \, k_{310}$. Comparing this with $k_{300} = x 	imes 10^{-3} \, k_{310}$,we get $x = 1$.

Similar Questions

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