Question

At $407$ $K$, the rate constant of a chemical reaction is $9.5\times {10}^{-5}{s}^{-1}$ and at $420$ $K$, the rate constant is $1.9\times {10}^{-4}{s}^{-1}$. Calculate the frequency factor of the reaction.

Answer 1

The Arrhenius equation is,
${\log }_{10}\frac{k_2}{k_1}=\frac{E_a}{2.303\times R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
Given: $k_1=9.5\times {10}^{-5}{s}^{-1};$ $k_2=1.9\times {10}^{-4}{s}^{-1};$
$R=8.314$ $J$ ${mol}^{-1}$ $K^{-1};$
$T_1=407$ $K$ and $T_2=420$ $K$
Substituting the values in Arrhenius equation
${\log }_{10}\frac{1.9\times {10}^{-4}}{9.5\times {10}^{-5}}=\frac{E_a}{2.303\times 8.314}\left[\frac{420-407}{420\times 407}\right]$
$E_a=75782.3$ $J$ ${mol}^{-1}$
Applying now $\log k_1=\log A-\frac{E_a}{2.303R{T}_1}$
$\log 9.5\times {10}^{-5}=\log A-\frac{75782.3}{2.303\times 8.314\times 407}$
or $\log \frac{A}{9.5\times {10}^{-5}}=\frac{75782.3}{2.303\times 8.314\times 407}=9.7246$
$A=5.04\times {10}^5{s}^{-1}$