The decomposition of \(H_2O_2\) in alkaline medium, in presence of \(I^-\) ion occurs as follows:
\(2H_2O_2 \overset{I^-, \text{alk. medium}}{\longrightarrow}2H_2O + O_2\)
If the reaction occurs in the following two steps:
\(\text{Step 1: }H_2O_2 + I^- \underset{slow}{\longrightarrow} H_2O + IO^-\)
\(\text{Step 2: }H_2O_2 + IO^- \underset{fast}{\longrightarrow} H_2O + I^- + O_2\)
The correct rate law expression for the reaction is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(r = k[H_2O_2][I^-]\)
When a reaction takes place in steps, the slowest step decides the rate. Each step is an elementary reaction, so its rate law follows from the species that collide in it. The slow step here is bimolecular, with one \( \ce{H2O2} \) and one \( \ce{I-} \):
\( \text{Rate} = k[\ce{H2O2}][\ce{I-}] \)
Why the other options are incorrect
Option (1) \(r = k[H_2O_2]^2[I^-]\) — Incorrect. Takes the exponent 2 from the balanced equation; only one \( \ce{H2O2} \) takes part in the slow step.
Option (3) \(r = k[H_2O_2]^2[IO^-]\) — Incorrect. \( \ce{IO^{-}} \) is an intermediate formed in the slow step; it cannot appear in the rate law.
Option (4) \(r = k[H_2O_2]^2[I^-][OH^-]\) — Incorrect. \( \ce{OH^{-}} \) only provides the alkaline medium and is not in the slow step; the \( \ce{H2O2} \) exponent is also wrong.
NCERT (2026-27 edition):
Molecularity of a Reaction – “… Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium.
\( \ce{2H2O2} \xrightarrow{\ce{I-},\ \text{alkaline medium}} \ce{2H2O + O2} \)
The rate equation for this reaction is found to be
\( \displaystyle \text{Rate} = \frac{-d[\ce{H2O2}]}{dt} = k[\ce{H2O2}][\ce{I-}] \)
This reaction is first order with respect to both H2O2 and I–. Evidences suggest that this reaction takes place in two steps
(1) \( \ce{H2O2 + I- -> H2O + IO-} \)
(2) \( \ce{H2O2 + IO- -> H2O + I- + O2} \)
Both the steps are bimolecular elementary reactions. … The first step, being slow, is the rate determining step. Thus, the rate of formation of intermediate will determine the rate of this reaction.”