To solve this problem, we need to understand how activation energy can be calculated.$\newline$The activation energy of a chemical reaction can be determined using the Arrhenius equation:$\newline k = A \cdot e^{-\frac{E_a}{RT}} \newline$where:$\newline$• $k$ is the rate constant,$\newline$• $A$ is the pre-exponential factor,$\newline$• $E_a$ is the activation energy,$\newline$• $R$ is the universal gas constant, and$\newline$• $T$ is the temperature in Kelvin.$\newline$To find the activation energy $(E_a),$ we can use the rate constants at two different temperatures.$\newline$Taking the natural logarithm of the Arrhenius equation gives:$\newline \ln k = \ln A - \frac{E_a}{RT} \newline$For two different temperatures $T_1$ and $T_2$ with rate constants $k_1$ and $k_2,$ respectively, we have:$\newline \ln k_1 = \ln A - \frac{E_a}{RT_1} \newline \ln k_2 = \ln A - \frac{E_a}{RT_2} \newline$Subtracting these two equations gives:$\newline \ln k_2 - \ln k_1 = -\frac{E_a}{RT_2} + \frac{E_a}{RT_1} \newline \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \newline$Rearranging for $E_a$ gives:$\newline E_a = \frac{R \cdot \ln \left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \newline$Thus, knowing the rate constants at two different temperatures allows us to calculate the activation energy.$\newline$This corresponds to Option 2.