Question

The rate constant of a reaction is $1.5\times {10}^7{S}^{-1}{at50}^o$ and $4.5\times {10}^7{S}^{-1}{at100}^o$, calculate value of activation energy (Ea) for the reaction. ${8.314JK}^{-1}{mol}^{-1}$

Answer 1

$log\frac{k_2}{k_1}$ = $\frac{E_a}{2.303\times R}(\frac{1}{T_1}-\frac{1}{T_2})$

Substituting the values, we get

$log3$ = $\frac{E_a}{2.303\times 8.314}(\frac{1}{323}-\frac{1}{373})$

$E_a=22.01kJ{mol}^{-1}$