Question

At $407K$ the rate constant of a chemical reaction is $9.5\times 10^{-5}{s}^{-1}$ and at $420K$ , the rate constant is $1.9\times 10^{-4}{s}^{-1}$ . The frequency factor of the reaction is $x\times 10^5{s}^{-1}$ .The value of 'x' is.
Report your answer by rounding it up to nearest whole number.

Answer 1

Answer: 5

The Arrhenius equation is,
${\log }_{10}\frac{\frac{k_2}{k_1}}{k_1}=\frac{E_a}{2.303\times R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
Given $k_1=9.5\times 10^{-5}{s}^{-1};k_2=1.9\times 10^{-3}{s}^{-1}$
$R=8.314Jmo{l}^{-1}{K}^{-1}$
$T_1=407K$ and $T_2=420K$
Substituting the values in Arrhenius equation.
${\log }_{10}\frac{1.9\times 10^{-4}}{9.5\times 10^{-5}}=\frac{E_a}{2.303\times 8.314}\left[\frac{420-407}{420\times 407}\right]$
$E_a=75782.3Jmo{l}^{-1}$
Applying now $\log k_1=\log A-\frac{E_a}{2.303R{T}_1}$
$\log 9.5\times 10^{-5}=\log A-\frac{75782.3}{2.303\times 8.314\times 407}$
$\log \frac{\text{ A }}{9.5\times 10^{-5}}=\frac{75782.3}{2.303\times 8.314\times 407}=9.7246$
So, $A=5.0\times 10^5{s}^{-1}$