Question

The rate constant for the decomposition of a hydrocarbon is $2.418\times {10}^{-5}se{c}^{-1}$ at $546$K. If the energy of activation is $179.9$ kJ/mol, what will be the value of pre-exponential factor?

Answer 1

The Arrhenius equation is
$k=A{e}^{-E_a/RT}$
It can also be written as
$\ln k=\ln A+\left(\frac{-E_a}{RT}\right)⟶\left(1\right)$
Given $k=2.418\times {10}^{-5}{sec}^{-1}{E}_a=179900J/molT=546KR=8.314J/mol.k$
$\therefore$ Equation (1) becomes$-$
$\ln (2.418\times 1^{-5})=\ln A+\left(\frac{-179900}{80314\times 546}\right)-10.63=\ln A+(-39.63)\ln A=39.63-63\ln A=29.63\log A=\frac{29.63}{2.303}=12.866A=antilog\left(12.866\right)$
$A=7.345\times {10}^{12}$ is the value of the pre-exponential factor.