Question

What is the value of rate constant k if the value of the activation energy Ea and the frequency factor A are 49 kJ / mol and 9 × 10^10 s^-1 respectively? (T = 313 K)

(a) 6 × 10^2  s^-1

(b) 9 × 10^2  s^-1

(c) 6 × 10^-2  s^-1

(d) 3 × 10^2  s^-1

This question was addressed to me during an internship interview.

I would like to ask this question from Chemical Kinetics topic in chapter Chemical Kinetics of Chemistry – Class 12

Answer 1

Right answer is (a) 6 × 10^2  s^-1

The best explanation: Given,

Ea = 49 kJ / mol^-1

T = 313 K

A = 9 × 10^2 s^-1

R = 8.314 J K^-1 mol^-1

log k = – Ea / (2.303 RT) + log A

log k = – 49000 / (2.303 × 8.314 × 313) + log 9 × 10^10

log k = 2.77843

k= antilog (2.77843)

k = 6 × 10^2 s^-1.