According to the Arrhenius equation, $k = A e^{-E_a/RT}$. Taking the natural logarithm on both sides, we get $\ln k = \ln A - \frac{E_a}{RT}$. This equation is of the form $y = mx + c$, representing a straight line where $y = \ln k$, $x = \frac{1}{T}$, slope $m = -\frac{E_a}{R}$, and intercept $c = \ln A$. Therefore, a plot of $\ln k$ versus $\frac{1}{T}$ gives a straight line. By determining its slope, the activation energy ($E_a$) can be calculated.