chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution

    4. Solve

    iii. The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 0C is 3.7 × 10-5 s-1. What is the rate constant at 300C? (R = 8.314 J/K mol)
    (7.4 × 10-5)

    Answer:- 

    Given:

    Activation energy (Ea) = 104 kJ mol-1 = 104 × 103 J mol-1

    Rate constant (k1) = 3.7 × 10-5 s-1

    Temperatures; T1 = 25 + 273 = 298 K, T2 = 30 + 273 = 303 K

    R = 8.314 J K-1 mol-1

    To find:

    Rate constant (k2) at 30°C

    Formula:

    log10 (k2/k1) = Ea / (2.303R) × ((T2 - T1) / (T2T1))

    Calculation:

    log10 (k2 / (3.7 × 10-5 s-1)) = (104 × 103 J mol-1) / (2.303 × 8.314 J K-1 mol-1) × ((303 K - 298 K) / (303 K × 298 K))

    ⇒ log10 (k2 / (3.7 × 10-5 s-1)) = 0.301

    ⇒ k2 / (3.7 × 10-5 s-1) = antilog(0.301) = 2.00

    ⇒ k2 = 2.00 × 3.7 × 10-5 s-1 = 7.4 × 10-5 s-1

    The rate constant of the reaction is 7.4 × 10-5 s-1.

    Q 4.2
    Q 4.3
    Q 4.4

    chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution page 137