The time required for 10% completion of | Class 12 Chemistry Chapter Chemical Kinetics, Chemical Kinetics NCERT Solutions

Question:

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 1010 s-1. Calculate k at 318 K and Ea.

Answer:

For a first order reaction,

t = 2.303 / k log/ a - x

At 298 K,

t = 2.303 / k log 100  / 90

= 0.1054 / k

At 308 K,

t' = 2.303 / k' log 100  / 75

= 2.2877 / k'

According to the question,

t = t'

0.1054 / k  =  2.2877 / k'

⇒ k' k  = 2.7296

From Arrhenius equation,we obtain

 To calculate k at 318 K,

It is given that, A = 4 x 1010 s-1, T = 318K

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10-2 s -1

    


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Comments

  • Dhruv
  • 2019-11-15 11:17:09

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  • Ashu
  • 2019-10-16 14:07:25

Helps me a lot


  • Not important
  • 2019-01-13 00:12:51

Thanks for clearing my doubts


  • Aynal Hoque
  • 2018-10-23 12:02:27

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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 29: The time required for 10% completion of a first order reaction at 298 K is equal to that required fo....