chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution

    4. Solve

    viii. The rate constant for the first order reaction is given by log10 k = 14.34 – 1.25 × 104 T. Calculate activation energy of the reaction. (239.3 kJ/mol)

    Answer:- 

    The given rate constant equation is

    log10k = 14.34 – 1.25 × 104 T ...(1)

    Arrhenius equation is

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

    Therefore, In k = In A - (Ea)/(2.303 RT) ...(2)

    Comparing (1) and (2),

    (Ea)/(2.303 RT) = 1.25 × 104 T

    Therefore, (Ea)/(2.303 R) = 1.25 × 104

    Therefore, (Ea)/(2.303 × 8.314) = 1.25 × 104

    Therefore, Ea = 1.25 × 104 × 2.303 × 8.314 = 239339 K mol-1 = 239.3 kJ mol-1

    The energy of activation of the reaction is 239.3 kJ mol-1

    Question 4.7
    Question 4.8
    Question 4.9

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