To solve this problem, we need to understand the relationship between the slope of an Arrhenius plot and the activation energy $(E_a). \newline$The Arrhenius equation is given by:$\newline k = A \cdot e^{-\frac{E_a}{RT}} \newline$Taking the natural logarithm of both sides, we get:$\newline \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \newline$This equation is in the form of a straight line $(y = mx + c)$ where:$\newline y = \ln k, \, m = -\frac{E_a}{R}, \, x = \frac{1}{T}, \, c = \ln A \newline$The slope of the plot of $\ln k$ vs. $\frac{1}{T}$ is $-\frac{E_a}{R}. \newline$Given that the slope is $-5 \times 10^3 \,$K$,$ we can write:$\newline -\frac{E_a}{R} = -5 \times 10^3 \newline$Solving for $E_a,$ we have:$\newline E_a = 5 \times 10^3 \times R \newline$Substitute the value of $R = 8.314 \,$J K$^{-1}$mol$^{-1}: \newline E_a = 5 \times 10^3 \times 8.314 \newline E_a = 41570 \,$J mol$^{-1} \newline$Convert $E_a$ to kJ mol$^{-1}: \newline E_a = \frac{41570}{1000} \,$kJ mol$^{-1} \newline E_a = 41.57 \,$kJ mol$^{-1} \newline$Therefore, the value of $E_a$ is approximately $41.5 \,$kJ mol$^{-1}. \newline$This corresponds to Option 1.