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27 January, 09:48

If a reaction has a rate constant of 3.7 x 10-3 s-1 at 25oC, and an activation energy of 43.6 kJ/mol, what will be the rate constant for this reaction at 75oC?

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  1. 27 January, 10:05
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    The rate constant for the reaction at 75 °C = 4.6 * 10^ (-2) s^ (-1)

    Explanation:

    By Arrhenius equitation

    k = Ae^ (-Ea / (RT))

    Therefore

    3.7 x 10-3 s-1 = A*e^ (43600J/mol / (8.3145*298))

    Therefore A = 160900.320

    Therefore at 75 °C T = 348.15 K and k = 160900.320*e^ (-43600 / (8.314*348.15)

    k = 4.6 * 10^ (-2) s^ (-1)

    the rate constant for this reaction at 75 °C is k = 4.6 * 10^ (-2) s^ (-1)

    Which shows that the rate of reaction increases as temperature increases that is the proportions of effective collisions with energy equal to or greater than the activation energy increases.
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