Question

The rate constant of a reaction is $1.5\times {10}^7{s}^{-1}$ at ${50}^{o}C$ and $4.5\times {10}^7{s}^{-1}$ at ${100}^{o}C$. Calculate the Arrhenius parameter $A$ and $E_a$.

Answer 1

${\log }_{10}\frac{k_2}{k_1}=\frac{E_a}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$
${\log }_{10}\frac{4.5\times {10}^7}{1.5\times {10}^7}=\frac{E_a}{2.303\times 8.314}(\frac{1}{323}-\frac{1}{373})$
$E_a=2.2\times {10}^{4}J/mol$
We know that,
${\log }_{10}k={\log }_{10}A-\frac{E_a}{2.303RT}$
${\log }_{10}(1.5\times {10}^7)={\log }_{10}A-\frac{2.2\times {10}^4}{2.303\times 8.314\times 323}$
$A=5.42\times {10}^{10}{s}^{-1}$