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26 December, 21:22

If a temperature increase from 10.0 ∘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. 26 December, 21:41
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    According to this formula:

    ㏑ (K2/K1) = Ea/R (1/T1 - 1/T2)

    when K is the rate constant

    Ea is the activation energy

    R is the universal gas constant

    and T is the temperature K

    when K is doubled so K2: K1 = 2:1 & R = 8.314 J. K^-1. mol^-1

    and T1 = 10 + 273 = 283 k & T2 = 21 + 273 = 294 k

    So by substitution:

    ㏑2 = (Ea / 8.314) (1/283 - 1/294)

    ∴ Ea = 43588.9 J/mol = 43.6 KJ/mol
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