Question

The rate constant of a reaction increases by five times on increase in temperature from ${27}^{o}C$ to ${52}^{o}C$. The value of activation energy in ${\text{kJ mol}}^{-1}$ is:
(Rounded off to the nearest integer)
$[R=8.314J{K}^{-1}mo{l}^{-1}]$

Answer 1

Let, the rate constants be $k_{{52}^{0}C}$ and $k_{{27}^{0}C}$According to the question,
$\frac{k_{{52}^{0}C}}{k_{{27}^{0}C}}=5$
From Arrhenius equation,
ln$\left\{\frac{k_{T_2}}{k_{T_1}}\right\}=\frac{E_a}{R}\{\frac{1}{T_1}-\frac{1}{T_2}\}$
$\Rightarrow ln\left(5\right)=\frac{E_a}{R}\{\frac{1}{300}-\frac{1}{325}\}$
$\Rightarrow 2.303\log \left(5\right)=\frac{E_a}{R}\{\frac{1}{300}-\frac{1}{325}\}$
$\Rightarrow \frac{2.303\times 0.7\times 8.314\times 300\times 325}{25}=E_a$
$\Rightarrow E_a=52271.69\text{J /mol}$
$\Rightarrow E_a=52.272\text{kJ /mol}$