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For an elementary chemical reaction, the Arrhenius plot is given below.
If the energy of activation is 6.64 kJ mol$^{-1}$ and R = 8.3 J K$^{-1}$ mol$^{-1}$, the temperature at which the rate constant becomes $e^2$ min$^{-1}$, is
If the energy of activation is 6.64 kJ mol$^{-1}$ and R = 8.3 J K$^{-1}$ mol$^{-1}$, the temperature at which the rate constant becomes $e^2$ min$^{-1}$, isDetailed Solution
$\ln k=\ln A-\frac{E_a}{RT}$. From the graph, $\ln A=6$ (intercept). For $k=e^2$: $2=6-\frac{6.64\times1000}{8.3\times T} \Rightarrow \frac{6640}{8.3\times T}=4 \Rightarrow T=200$ K.