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(C) From the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.Comparing this with the equation of a straight line $y = mx

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$10^{-4} \ s^{-1}$

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(C) From the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} ight)$.Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R}$.Given slope $= -4606 \ K$,so $\frac{E_a}{R} = 4606 \ K$.Using the integrated Arrhenius equation: $\ln \left( \frac{k_2}{k_1} ight) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$.Here,$T_1 = 400 \ K$,$k_1 = 10^{-5} \ s^{-1}$,$T_2 = 500 \ K$,and $k_2 = ?$.$\ln \left( \frac{k_2}{10^{-5}} ight) = 4606 	imes \left( \frac{1}{400} - \frac{1}{500} ight) = 4606 	imes \left( \frac{500 - 400}{200000} ight) = 4606 	imes \frac{100}{200000} = 4606 	imes \frac{1}{2000} = 2.303$.Since $\ln 10 \approx 2.303$,we have $\ln \left( \frac{k_2}{10^{-5}} ight) = \ln 10$.Therefore,$\frac{k_2}{10^{-5}} = 10$,which gives $k_2 = 10^{-4} \ s^{-1}$.

For a certain reaction,consider the plot of $\ln k$ versus $1/T$ given in the figure. If the rate constant of this reaction at $400 \ K$ is $10^{-5} \ s^{-1}$,then the rate constant at $500 \ K$ is:

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