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# The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s– 1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

4.23   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 pre-exponential factor.

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Given that,
k = $2.418 \times 10 ^{-5} s ^{-1}$
$E_{a}$= 179.9 KJ/mol
T(temp) = 546K

According to Arrhenius equation,

$k=Ae^{-E_{a}/RT}$
taking log on both sides,
$\log k = \log A - \frac{E_{a}}{2.303 RT}$

$\log A =\log k + \frac{E_{a}}{2.303 RT}$

$=\log (2.418\times 10^{-5}) + \frac{179.9\times 10^{3}}{2.303 \times 8.314 \times 546}$

= (0.3835 - 5) + 17.2082
= 12.5917
Thus A = antilog (12.5917)
A = 3.9 $\times 10^{12}$ per sec (approx)

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