Question:
The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from $27^{\circ} \mathrm{C}$ to $42^{\circ} \mathrm{C}$. Its energy of activation in $\mathrm{J} / \mathrm{mol}$ is______________. (Take $\ln 5=1.6094$; $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
Solution:
$(84297.48)$
$\because k=A \mathrm{e}^{-E_{a} / R T}$
$\ln \frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \Rightarrow \ln 5=\frac{E_{a}}{R}\left(\frac{1}{300}-\frac{1}{315}\right)$
$\Rightarrow E_{a}=\frac{1.6094 \times 8.314 \times 300 \times 315}{15}=84297.48 \mathrm{~J} / \mathrm{mol}$