Which of the following reaction has fractional order with respect to one of the reactant?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl\)
The order of a reaction with respect to a reactant is the power of its concentration in the experimentally determined rate law. It need not equal the stoichiometric coefficient and can be 0, 1, 2, 3 or even a fraction.
For the chlorination of chloroform the experimental rate law is
Rate = k[CHCl3][Cl2]1/2
so the order with respect to \( \ce{Cl2} \) is 1/2 (3/2 in all).
Why the other options are incorrect
Option (1) \(2NO(g) + O_2(g) \rightarrow 2NO_2(g)\) — Incorrect. Rate = k[NO]2[O2]; orders 2 and 1, third order in all, no fraction.
Option (3) \(CH_3COOC_2H_5+H_2O \rightarrow CH_3COOH+C_2H_5OH\) — Incorrect. Acid hydrolysis of an ester with water in large excess; Rate = k[CH3COOC2H5], a pseudo first order reaction.
Option (4) \(CH_3COOC_2H_5+NaOH \rightarrow CH_3COONa+C_2H_5OH\) — Incorrect. Alkaline hydrolysis (saponification); Rate = k[ester][OH−], first order in each reactant, second order in all.
NCERT (2026-27 edition):
Rate Expression and Rate Constant – …
| Reaction | Experimental rate expression |
|---|---|
| 1. CHCl3 + Cl2 → CCl4 + HCl | Rate = k [CHCl3] [Cl2]1/2 |
| 2. CH3COOC2H5 + H2O → CH3COOH + C2H5OH | Rate = k [CH3COOC2H5]1 [H2O]0 |
“In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that: Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally.”