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 .

Question

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.

in progress 0
Eirian 4 years 2021-08-02T05:19:38+00:00 1 Answers 10 views 0

Answers ( )

  1. Calantha
    0
    2021-08-02T05:20:54+00:00
    Reply

    Answer:

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )