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

Answer:$ 84\;kJmol^{-1}$

Taking natural logarithms of the Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$ we get:

$ln\; k=ln A-\large\frac{E_a}{RT}$ which we can compare to $y=mx+c$

$\Rightarrow m = -\large\frac{E_a}{R}$

$\Rightarrow E_a = -mR = - (-1.01 \times 10^{4}\;K) \times (8.314\; J/K·mol) = 83971\;Jmol^{-1} = 84\;kJmol^{-1}$

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