The rate of reaction decreased by 3.555 times when the temperature was changed from $40^{\circ} \mat

Question

The rate of 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; $R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ and $\ln 3.555=1.268$\

Professor PC Thomas Classes · Answer

(100): Given, $\frac{k_{T_{2}}}{k_{T_{1}}}=3.555$\ $T_{1}=30+273=303 \mathrm{~K}$,\ $T_{2}=40+273=313 \mathrm{~K}$\ $\ln \left(\frac{k_{T_{2}}}{k_{T_{1}}}ight)=\frac{E_{a}}{R}\left(\frac{1}{303}-\frac{1}{313}ight)$\ $\ln (3.555)=\frac{E_{a}}{8.314}\left(\frac{1}{303}-\frac{1}{313}ight)$\ $E_{a}=\frac{1.268 	imes 8.314 	imes 303 	imes 313}{10}$\ $=99980.7 \mathrm{~J} / \mathrm{mol}$\ $=99.98 \mathrm{~kJ} / \mathrm{mol} \approx 100 \mathrm{~kJ} / \mathrm{mol}$