Question

The rates of most reactions double when their temperature is raised from $298K$ to $308K$. Calculate their activation energy.

Answer 1

$2.303\log \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$\frac{K_2}{K_1}=2;T_2=308K,T_1=298K$

$\therefore 2.303\log 2=\frac{E_a}{8.314}\times \frac{10}{308\times 298}$

$E_a=52.903\times {10}^{3}J$ or $E_a=52.903kJ$.