The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The activation energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of the reaction is_____________ .Take; $\mathrm{R}=$ $8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \operatorname{In} 3.555=1.268$
(100) The Arrhenices equation is
$k=A e^{\frac{E_{\mathrm{a}}}{R T}}$
Assuming $A$ and $E_{a}$ to be independent of temperature
$\ln \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$
$\ln 3.555=\frac{E_{\mathrm{a}}}{8.314}\left(\frac{1}{303}-\frac{1}{313}\right)$
$\Rightarrow E_{\mathrm{a}}=\frac{1.268 \times 8.314 \times 303 \times 313}{10}$
$=99980.7=99.98 \mathrm{~kJ} / \mathrm{mol}$