chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution

4. Solve

v. The rate constant of a reaction at 5000C is 1.6 × 103 M-1s-1. What is the frequency  factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M-1s-1)

Answer:- 

Given:

Rate constant (k) = 1.6 × 103 M-1 s-1,

Temperature (T) = 500 + 273 = 773 K,

Activation energy (Ea) = 56 kJ mol-1 = 56 × 103 J mol-1

To find:

Frequency factor (A)

Formula:

K = Ae(-Ea / (RT))

Calculation:

Substituting the given values

(1.6 × 103 M-1 s-1) / A = e((-56 × 103 J mol-1) / (8.314 J K-1 mol-1 × 773 K))

log((1.6 × 103 M-1 s-1) / A) = (-56,000) / (8.314 × 773 × 2.303)

((1.6 × 103 M-1 s-1) / A) = antilog(-3.7836)

((1.6 × 103 M-1 s-1) / A) = 1.646 × 10-4

A = ((1.6 × 103 M-1 s-1) / (1.646 × 10-4)) = 9.72 × 106 M-1 s-1

The frequency factor of the reaction is 9.72 × 106 M-1 s-1.

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