In a pseudo first order hydrolysis of es | Class 12 Chemistry Chapter Chemical Kinetics, Chemical Kinetics NCERT Solutions

Question:

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s 0 30 60 90

[Ester]mol L - 1

0.55 0.31 0.17 0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer:

(i) Average rate of reaction between the time interval, 30 to 60 seconds, = d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10 - 3mol L - 1s - 1

 

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]º / [R]

For t= 30 s, k1 = 2.303/ 30 log 0.55 / 0.31

= 1.911 × 10 - 2s - 1

For t = 60 s, k2 = 2.303/ 60 log 0.55 / 0.17

= 1.957 × 10 - 2s - 1

For t= 90 s, k3 = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 - 2s - 1

Then, average rate constant, k = k1 + k2+ k/ 3

= 1.911 × 10 - 2  + 1.957 × 10 - 2 + 2.075 × 10 - 2 / 3

= 1.981 x 10-2 s - 1


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Comments

  • Ankit kumar
  • 2018-07-22 11:05:53

Help me


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 8: In a pseudo first order hydrolysis of ester in water, the following results were obtained: ....