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9 August, 11:22

A reaction has a rate constant of 0.0117 s-1 at 400.0 k and 0.689 s-1 at 450.0 k. calculate the activation energy in kilojoules for this reaction.

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  1. 9 August, 12:58
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    By definition we have to:

    LOG (k2 / k1) = (-Ea/R) * (1/T1-1/T2)

    Where,

    k1 = 0.0117 s-1

    K2 = 0.689 s-1

    T1 = 400.0 k

    T2 = 450.0 k

    R is the ideal gas constant

    R = 8.314 KJ / (Kmol * K)

    Substituting

    ln (0.0117/0.689) = - Ea / (8.314) * ((1/400) - (1/450))

    Clearing Ea:

    Ea = 122 kJ

    answer

    the activation energy in kilojoules for this reaction is

    Ea = 122 kJ
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