According to the Arrhenius equation, $k = A e^{-E_a/RT}$. Taking the natural logarithm on both sides yields: $\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$ This equation is of the form $y = mx + c$, which represents a straight line. Here, $y = \ln k$, $x = 1/T$, slope $m = -\frac{E_a}{R}$, and intercept $c = \ln A$. Since the activation energy $E_a$ and the gas constant $R$ are positive quantities, the slope $-\frac{E_a}{R}$ is negative. Therefore, the plot of $\ln k$ versus $1/T$ is a straight line with a negative slope.