The rate constant for the decomposition of hydrocarbons is 2.418 × 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.
k = 2.418 × 10−5 s−1
T = 546 K
Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1
According to the Arrhenius equation,
$k=\mathrm{Ae}^{-E_{a} / \mathrm{R} T}$
$\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$
$\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{R} T}$
$\Rightarrow \log \mathrm{A}=\log k+\frac{E_{a}}{2.303 \mathrm{RT}}$
$=\log \left(2.418 \times 10^{-5} \mathrm{~s}^{-1}\right)+\frac{179.9 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1} \times 546 \mathrm{~K}}$
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 1012 s−1 (approximately)