Question

The energy of activation of a reaction is $140$ $kJ$ ${mol}^{-1}$. If its rate constant at $400$ $K$ is $2.0\times {10}^{-6}{s}^{-1}$, what is the value at $500$ $K$?

Answer 1

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

$log\frac{K_2}{2\times {10}^{-6}}=\frac{140\times 1000}{2.303\times 8.314}\times [\frac{1}{400}-\frac{1}{500}]$

$K_2=9.05\times {10}^{-3}$