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.

Given,

k = 2.418 × 10^{-5} s^{-1}

T = 546 K

E_{a} = 179.9 kJ mol^{-1} = 179.9 × 10^{3}J mol^{-1}

The Arrhenius equation is given by k = Ae^{-Ea/RT}

Taking natural log on both sides,

Ln k = ln A-(E_{a}/RT)

Substituting the values,

ln(2.418 × 10^{-5} ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10^{12} s^{-1}(approximately)

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