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.