For the decomposition of azoisopropane t | Class 12 Chemistry Chapter Chemical Kinetics, Chemical Kinetics NCERT Solutions

Question:

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t (sec) P(mm of Hg)
0 35.0
360 54.0
720 63.0

Calculate the rate constant

Answer:

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, Pt = (Pº - p) + p + p

⇒ Pt = (Pº + p)

p = P - Pº

therefore, Pº - p = Pº  - P - Pº

= 2 Pº -  Pt

For a first order reaction,

k = 2.303/ Log  Pº / Pº  - p

=   2.303/ Log  Pº / 2 Pº  -  Pt

When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0

= 2.175 × 10 - 3 s - 1

When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0

= 2.235 × 10 - 3 s - 1

Hence, the average value of rate constant is

k = (2.175 × 10 - 3  + 2.235 × 10 - 3 ) / 2   s - 1

= 2.21 × 10 - 3 s - 1


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Comments

  • Gnkp
  • 2019-11-17 14:02:52

Approval hmmm!


  • Jishan
  • 2019-08-17 09:27:55

Thanx alot


  • Stuti
  • 2019-07-01 21:39:29

Thankyou !! Helped alot!!


  • Ayush
  • 2019-06-07 19:02:19

Thnx alot


  • Subhadeep Raj
  • 2019-03-06 07:59:09

Thanks for your help


  • Agrima
  • 2019-01-27 17:24:18

Thank u


  • Janhavi khalatkar
  • 2018-11-12 23:50:35

Ohhhh..thanks


  • Somya
  • 2018-09-25 17:47:50

Thanku


  • Arnav Shivam
  • 2018-07-26 14:55:59

ThnkZzzzzzzzx u....


  • Rajat garg
  • 2018-06-16 10:48:57

Thanks for your help


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 20: For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtai....