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28 May, 16:18

The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this reaction increase as the temperature rises from 231 to 293 oC?

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Answers (2)
  1. 28 May, 16:40
    0
    The rate of the reaction increased by a factor of 1012.32

    Explanation:

    Applying Arrhenius equation

    ln (k₂/k₁) = Ea/R (1/T₁ - 1/T₂)

    where;

    k₂/k₁ is the ratio of the rates which is the factor

    Ea is the activation energy = 274 kJ/mol.

    T₁ is the initial temperature = 231⁰C = 504 k

    T₂ is the final temperature = 293⁰C = 566 k

    R is gas constant = 8.314 J/Kmol

    Substituting this values into the equation above;

    ln (k₂/k₁) = 274000/8.314 (1/504 - 1/566)

    ln (k₂/k₁) = 32956.4589 (0.00198-0.00177)

    ln (k₂/k₁) = 6.92

    k₂/k₁ = exp (6.92)

    k₂/k₁ = 1012.32

    The rate of the reaction increased by 1012.32
  2. 28 May, 16:48
    0
    By a factor of 2.25.

    Explanation:

    Using the Arrhenius equations for the given conditions:

    k1 = A * (exp^ (-Ea / (RT1))

    k2 = A * (exp^ (-Ea / (RT2))

    T1 = 231°C

    = 231 + 273.15 K

    = 574.15 K

    T2 = 293°C

    = 293 + 273.15 K

    = 566.15 K

    Ea = 274 kJ mol^-1

    R = 0.008314 kJ/mol. K

    Now divide the second by the first:

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

    = 0.444

    2.25k2 = k1
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