The rate constant (k) for a reaction was determined as a function of temperature. A plot of $ln\; k$ versus $\large\frac{1}{T}$ (in $K$) is linear and has a slope of $-1.01 \times 10^{4}\;K$. Calculate the activation energy for the reaction.

Question

$\begin{array}{1 1}84\;kJmol^{-1} \\ 48\;kJmol^{-1} \\ 24 \;kJmol^{-1} \\ 96\;kJmol^{-1} \end{array}$