Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter Chemical Kinetics. This page offers a step-by-step solution to the specific question from Exercise 2, Question 8: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
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.
(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+ k3 / 3
= 1.911 × 10 - 2 + 1.957 × 10 - 2 + 2.075 × 10 - 2 / 3
= 1.981 x 10-2 s - 1
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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