Chemistry - Chemical Kinetics Question with Solution | TestHub

We can calculate the Activation Energy by graphing ln k versus 1/T. When the $\ln k$ (rate constant) is plotted versus the inverse of the temperature (kelvin), the slope is a straight line. The value of the slope $(\mathrm{m})$ is equal to $-\mathrm{Ea}/\mathrm{R}$ where $\mathrm{R}$ is a constant equal to $8.314\mathrm{J}/\mathrm{mol}-\mathrm{K}$.