Question

The rate constant for a reaction at ${500}^{o}C$ is ${1.6\times 10}^3{M}^{-1}{s}^{-1}$. What is the frequency factor of the rection if its energy of activation is 56 kJ ${mol}^{-1}$?

( ${9.72\times 10}^6{M}^{-1}{s}^{-1}$)

Answer 1

We know that,

$k=A\times e^{\frac{{-E}_a}{RT}}$

Given $E_a=56\text{kJ}{mol}^{-1}$

$k=1.6\times {10}^3{M}^{-1}{s}^{-1}$

$T=500+273=773K$

Substituting the values,

$1.6\times {10}^3=A\times e^{\frac{56\times 1000}{8.314\times 773}}$

$A=0.26s^{-1}$