4.32 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?

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    Payal Gupta | Contributor-Level 10

    4 months ago

    4.32 Given,

    k = 2.418 * 10-5 s-1

    T = 546 K

    Ea = 179.9 kJ mol-1 = 179.9 * 103J mol-1

    The Arrhenius equation is given by k = Ae-Ea/RT Taking natural log on both sides,

    Ln k = ln A- (Ea/RT) Substituting the values,

    ln (2.418 * 10-5 ) = ln A-179.9/ (8.314 * 546)

    ln A = 12.5917

    A = 3.9 * 1012 s-1 (approximately)

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