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10 June, 20:36

The activation energy of a certain reaction is 37.2 kJ/mol. At 20 ∘C, the rate constant is 0.0130 s-1. At what temperature would this reaction go twice as fast? Express your answer numerically in degrees Celsius

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  1. 10 June, 20:43
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    Answer: At 34°c

    Explanation:

    Using The Arrhenius Equation:

    k = Ae - Ea/RT

    k represents rate constant

    A represents frequency factor and is constant

    R represents gas constant which is = 8.31J/K/mol

    Ea represents the activation energy

    T represents the absolute temperature.

    By taking the natural log of both sides,

    ln k = ln A - Ea/RT

    Reactions at temperatures T1 and T2 can be written as;

    ln k1 = ln A - Ea/RT1

    ln k2 = ln A - Ea/RT2

    Therefore,

    ln (k1/k2) = - Ea/RT1 + Ea/RT2

    Since k2=2k1 this becomes:

    ln (1/2) = Ea/R*[1/T2 - 1/T1]

    Theefore,

    -0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]

    1/T2 - 1/293 = - 1.55 x 10^-4

    1/T2 = - 1.55 x 10^-4 + 34.13x 10^-4

    1/T2 = 32.58 x 10^-4

    Therefore T2 = 307K

    T2 = 307 - 273 = 34 °c
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