Question

The rate of a particular reaction doubles when temperature changes from ${27}^{o}C$ to ${37}^{o}C$. Calculate the energy of activation for such reaction. $[R=8.314J{K}^{-1}mole]$

Answer 1

$log\frac{K_2}{K_1}=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$log\frac{2K}{K}=\frac{E_a}{2.303\times 8.314}\times [\frac{1}{300}-\frac{1}{310}]$

$\frac{0.3010\times 2.303\times 8.314\times 300\times 310}{10}=E_a$

$E_a=53.603KJ/mol$