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

Question:

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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Comments

  • Aviral
  • 2019-12-28 09:40:03

What is the effect of temperature on the rate constant of reaction how can this temperature effect on the rate constant be represented


  • ansh d
  • 2019-08-12 21:12:15

yoyo


  • ansh d
  • 2019-08-12 21:11:53

yoyo


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 23: The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy ....