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13 July, 10:06

The rate constant of a chemical reaction increased from 0.100 s-1 to 3.10 s-1 upon raising the temperature from 25.0 ∘C to 51.0 ∘C.

Part A

Calculate the value of (1/T2-1/T1) where T1 is the initial temperature and T2 is the final temperature.

Express your answer numerically.

Part B

Calculate the value of ln (k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Express your answer numerically.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

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Answers (1)
  1. 13 July, 10:30
    0
    A. (1/T₂ - 1/T₁) = - 2.69 x 10⁻⁴ K⁻¹

    B. ln (k₁/k₂) = - 3.434

    C. E = 106.13 KJ/mol

    Explanation:

    Part A:

    we have:

    T₁ = 25°C = 298 K

    T₂ = 51°C = 324 K

    (1/T₂ - 1/T₁) = (1/324 k - 1/298 k)

    (1/T₂ - 1/T₁) = - 2.69 x 10⁻⁴ K⁻¹

    Part B:

    ln (k₁/k₂) = ln (0.1/3.1)

    ln (k₁/k₂) = - 3.434

    Part C:

    The activation energy can be found out by using Arrhenius Equation:

    ln (k₁/k₂) = E/R (1/T₂ - 1/T₁)

    where,

    E = Activation Energy

    R = General Gas Constant = 8.314 J/mol. k

    Therefore,

    -3.434 = (E/8.314 J/mol. k) (-2.69 x 10⁻⁴ K⁻¹)

    E = 106134.8 J/mol = 106.13 KJ/mol
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