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 1: . 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.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N2O(g) Rate = k[NO]2
(ii) H2O2 (aq) + 3 I - (aq) + 2 H+→ 2 H2O (l) + I3- Rate = k[H2O2][I - ]
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl]
(i) Given rate = k [NO]2
Therefore, order of the reaction = 2
Dimension of k = Rate / [NO]2
= mol L-1 s-1 / (mol L-1)2
= mol L-1 s-1 / mol2 L-2
= L mol-1 s-1
(ii) Given rate = k [H2O2][I - ]
Therefore, order of the reaction = 2
Dimension of k = Rate / [H2O2][I - ]
= mol L-1 s-1 / (mol L-1) (mol L-1)
= L mol-1 s-1
(iii) Given rate =k [CH3CHO]3/2
Therefore, order of reaction = 3/2
Dimension of k = Rate / [CH3CHO]3/2
= mol L-1 s-1 / (mol L-1)3/2
= mol L-1 s-1 / mol3/2 L-3/2
= L½ mol-½ s-1
(iv) Given rate = k [C2H5Cl]
Therefore,order of the reaction = 1
Dimension of k = Rate / [C2H5Cl]
= mol L-1 s-1 / mol L-1
= 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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