Question

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

• A. $2.9 \times 10^{-12} s^{-1} $
• B. $4.5 \times l0^{-5}s^{-1}m$
• C. $3.2 \times 10^{-4} s^{-1} $
• D. $3.9 \times 10^{12} s^{-1} $

Answer 1

Answer: (D)

According to Arrhenius equation
$ log A=log k +\frac{ E_{a}}{2.303 RT} $
log $ \left(2.418\times10^{-5}\right) +\frac{179.9\times10^{3}}{2.303\times8.314\times546} $
$\left(-5+0.3834\right)+ 17.2081=12.5924s^{-1} $
or $ A =Antilog\left(12.5924\right)s^{-1}=3.912 \times10^{12} s^{-1}$