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

(A) Given:
$k = 2.418 \times 10^{-5} \, s^{-1}$
$T = 546 \, K$
$E_{a} = 179.9 \, kJ \, mol^{-1} = 179.9 \times 10^{3} \, J \, mol^{-1}$
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
Using the Arrhenius equation:
$k = A e^{-E_{a} / RT}$
Taking log on both sides:
$\log k = \log A - \frac{E_{a}}{2.303 RT}$
$\log A = \log k + \frac{E_{a}}{2.303 RT}$
Substituting the values:
$\log A = \log (2.418 \times 10^{-5}) + \frac{179.9 \times 10^{3}}{2.303 \times 8.314 \times 546}$
$\log A = (-4.6165) + 17.2082$
$\log A = 12.5917$
$A = \text{antilog}(12.5917) = 3.9 \times 10^{12} \, s^{-1}$