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27 April, 13:47

If a temperature increase from 12.0 ∘C to 21.0 ∘C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction

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  1. 27 April, 15:11
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    53.7kJ/mol

    Explanation:

    Using Arrhenius equation

    Given

    T1 = 12°C = 273 + 12 = 285K

    T2 = 21°C = 273 + 21 = 294K

    k = A exp (-Ea/RT)

    Where k = Rate constant

    A = the pre-exponential factor

    Ea = the activation energy

    R = the Universal Gas Constant = 8.314J/kmol

    T = the temperature

    Taking logarithms of both sides of the Arrhenius Equation.

    ln (k) = ln (A) - Ea/RT

    If there are the rates at two different temperatures, we can derive the expression to be;

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

    The reaction doubles the rate constant

    So, k2/k1 = 2 (Given)

    Then we have

    ln (2) = Ea/8.314 (1/285 - 1/294)

    ln (2) * 8.314 = Ea * (1/285 - 1/294)

    6.9314E-1 * 8.314 = Ea * (1/285 - 1/294)

    5.7628 = Ea * (1/285 - 1/294)

    5.7628 = Ea*1.0741E-4

    Ea = 5.7628 / 1.074E-4

    Ea = 53657.35567970204J

    Ea = 53.7kJ/mol
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