To solve this problem, we need to understand how activation energy is determined using graphs.$\newline$The Arrhenius equation is given by:$\newline k = A e^{-\frac{E_a}{RT}} \newline$where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy,$\newline R$ is the gas constant, and $T$ is the temperature in Kelvin.$\newline$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:$\newline y = \ln k, \ m = -\frac{E_a}{R}, \ x = \frac{1}{T}, \ c = \ln A \newline$Thus, a plot of $\ln k$ vs. $\frac{1}{T}$ will yield a straight line with a slope of $-\frac{E_a}{R}. \newline$The activation energy $E_a$ can be determined from the slope as follows:$\newline E_a = -$slope$\times R \newline$Analyzing the options:$\newline$• Option 1: $\frac{\ln k}{T}$ vs. $T$ is incorrect as it doesn't relate to the Arrhenius equation.$\newline$• Option 2: $\ln k$ vs. $\frac{1}{T}$ is correct as it directly relates to the Arrhenius equation.$\newline$• Option 3: $\frac{T}{\ln k}$ vs. $\frac{1}{T}$ is incorrect as it doesn't relate to the Arrhenius equation.$\newline$• Option 4: $\ln k$ vs. $T$ is incorrect as it doesn't relate to the Arrhenius equation.$\newline$Therefore, the correct option is Option 2.