To solve this problem, we need to understand the Arrhenius equation:$\newline k = A e^{-\frac{E_a}{RT}} \newline$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 $(m)$ is negative, indicating a downward slope.$\newline$Analyzing the options:$\newline$• Option 1: The plot shows an upward slope, which is incorrect.$\newline$• Option 2: The plot shows a downward slope, which is consistent with the Arrhenius equation.$\newline$• Option 3: The plot shows an upward slope, which is incorrect.$\newline$• Option 4: The plot shows an upward slope, which is incorrect.$\newline$Therefore, the correct option is Option 2.