The rate constant for the decomposition of hydrocarbons is
Question:

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.

Solution:

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)

 

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