Class 12 Chemistry - Chapter Chemical Kinetics NCERT Solutions | The rate constant for the decomposition

Welcome to the NCERT Solutions for Class 12th Chemistry - Chapter Chemical Kinetics. This page offers a step-by-step solution to the specific question from Exercise 2, Question 23: the rate constant for the decomposition of hydroca....
Question 23

The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Answer

k= 2.418 × 10-5 s-1 

T= 546 K

Ea= 179.9 kJ mol - 1 = 179.9 × 103J mol - 1

According to the Arrhenius equation,

= (0.3835 - 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 1012 s - 1(approximately)

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