For a reaction 2H₂O₂ xrightarrow[alkaline medium]I^- 2H₂O + O₂ the proposed mechanism is as given below: (1) H

Question

2H_2O_2 \xrightarrow[\text{alkaline medium}]{I^-} 2H_2O + O_2

(1)

H_2O_2 + I^- \rightarrow H_2O + IO^-

(slow)

(2)

H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2

(fast)

(i) Write rate law for the reaction.

(ii) Write the overall order of reaction.

(iii) Out of steps

(1) and (2), which one is rate determining step?

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Rate Determining Step

The Intuition: A Factory Assembly Line

Imagine a factory that makes bicycles. The process has three steps:

Frame is welded (takes 2 minutes)
Wheels are attached (takes 10 minutes)
Handlebars are fitted (takes 1 minute)

If you walk into the factory, how many complete bicycles come out per hour? The answer is not determined by the fast steps. Even if you weld frames instantly and fit handlebars in seconds, the bottleneck is the wheel-attaching step. That step takes 10 minutes, so at most 6 bicycles per hour leave the factory — no matter how fast the other steps are.

That bottleneck is the rate determining step.

The Precise Statement

In a chemical reaction that occurs through multiple elementary steps, the rate determining step (RDS) is the slowest elementary step. The overall rate of the reaction cannot exceed the rate of this step. All steps before it may build up intermediate, and all steps after it happen quickly once the RDS product is formed.

Important

The rate law for the overall reaction is determined only by the rate determining step and the steps that lead up to it. Steps that occur after the RDS do not appear in the rate law.

Why This Matters for Exam Problems

Consider a mechanism:

Step 1 (fast): $\ce{NO2 + NO2 -> NO3 + NO}$

Step 2 (slow): $\ce{NO3 + CO -> NO2 + CO2}$

The overall reaction is $\ce{NO2 + CO -> NO + CO2}$ . But the rate law is not simply $k[\ce{NO2}][\ce{CO}]$ . Because Step 2 is the RDS, the rate depends on $[\ce{NO3}]$ , which is an intermediate. You must use the fast equilibrium of Step 1 to express $[\ce{NO3}]$ in terms of $[\ce{NO2}]$ .

From Step 1 (fast equilibrium): $[\ce{NO3}] = K_{\text{eq}} [\ce{NO2}]^2 / [\ce{NO}]$

Then rate $= k_2 [\ce{NO3}][\ce{CO}] = k_2 K_{\text{eq}} \frac{[\ce{NO2}]^2 [\ce{CO}]}{[\ce{NO}]}$ …

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The rate law is determined by the slowest step in the mechanism. Since step (1) is slow and involves one molecule each of $H_2O_2$ and $I^-$ , the rate law is $r = k[H_2O_2][I^-]$ , making the overall order 2, and step (1) is the rate-determining step.

This problem is about connecting a proposed reaction mechanism to the observable rate law. In chemical kinetics, the rate law tells us how the reaction speed depends on reactant concentrations. When a mechanism has multiple steps, the slowest step — called the rate-determining step (RDS) — controls the overall rate. Any step faster than it doesn't affect the rate law directly.

Here, the decomposition of hydrogen peroxide is catalysed by iodide ions in alkaline medium. The mechanism has two steps: a slow one followed by a fast one. Let's work through it.

Identify the rate-determining step.

The problem states step (1) is slow and step (2) is fast. The slow step is always the RDS because the overall reaction cannot proceed faster than its slowest elementary step. So step (1) is the RDS.
Write the rate law from the RDS.

For an elementary step, the rate law is directly given by its molecularity — the coefficients in the step become the exponents in the rate law. Step (1) is:

H_2O_2 + I^- \rightarrow H_2O + IO^-

This is a bimolecular reaction involving one molecule of $H_2O_2$ and one of $I^-$ . Therefore, the rate law is:

r = k[H_2O_2][I^-]

where $k$ is the rate constant for step (1). No intermediates (like $IO^-$ ) appear because the RDS doesn't involve them as reactants.

Determine the overall order of the reaction.

The overall order is the sum of the exponents in the rate law. Here, exponent on $[H_2O_2]$ is 1, and on $[I^-]$ is 1. So overall order = $1 + 1 = 2$ . The reaction is second-order overall.
Confirm that step (1) is indeed the RDS. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 mark
$Q. Define the complex reactions.$
›Reveal solutionSolution

Complex reactions occur via a multi-step mechanism, in contrast to simple/elementary reactions which occur in exactly one step as written.

Most real chemical reactions with more than 2-3 reacting species do NOT occur exactly as their overall balanced equation suggests. Instead they proceed through a series of simpler elementary steps, each involving only 1, 2 (or rarely 3) molecules colliding - this sequence of steps is called the reaction mechanism, and such a multi-step reaction is called a complex reaction.

…

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CBSE 2022Set E1 markMCQ
$Q. The chemical reaction, 2O3 --> 3O2 proceeds as follows : O3 -> O2 + O ( fast ) ; O + O3 -> 2O2 (slow). Then the rate law expression of this reaction is$
(a) Rate = K[O3]2
(b) Rate = K[O3]2[O2]-1
(c) Rate = K[O3][O2]
(d) Rate = K[O3][O2]2
›Reveal solutionSolution
The slow (rate-determining) step gives Rate = k[O][O3]; substituting the O-atom concentration from the fast pre-equilibrium gives Rate = K[O3]^2[O2]^-1.

Mechanism:

Step 1 (fast, equilibrium): O3 ⇌ O2 + O

Step 2 (slow, rate-determining): O + O3 → 2O2

The overall rate is set by the slow step:

Rate = k2 [O][O3]

But [O] is a reactive intermediate; express it from the fast equilibrium. For step 1: …
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CBSE 2022Set ANNUAL1 mark
$Q. Rate of reaction controlled by slowest step called ............ .$
›Reveal solutionSolution

The slowest step of a multistep reaction mechanism controls the overall rate and is called the rate-determining (rate-limiting) step.

A reaction usually proceeds through a sequence of elementary steps. The overall rate cannot be faster than its slowest step.

…

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CBSE 2018Set ANNUAL1 mark
$Q. Fill in the blank: In mechanism of any reaction, the step going in slowest manner is called ___.$
›Reveal solutionSolution

In a multi-step reaction, the overall rate is governed by whichever elementary step is the slowest — this is the rate-determining step.

Most reactions occur through a sequence of elementary steps (the mechanism), not in a single step. Since the steps occur one after another, the overall reaction can proceed no faster than its slowest step. This slo …

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CBSE 2018Set ANNUAL1 mark
$Q. What is the rate determining step of a reaction?$
›Reveal solutionSolution

When a reaction occurs via several elementary steps, the overall rate is dictated entirely by whichever single step is slowest.

Most real chemical reactions do not occur in a single elementary step but proceed through a sequence (mechanism) of several elementary steps, each with its own rate.

Among these steps, one is invariably much slower than all the others — this step acts as a 'bottleneck': no matter how fast the other steps are, the overall reaction cannot proceed any faster than this slowest step allows.

…

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Q.For a reaction 2H2O2→alkaline mediumI−2H2O+O22H_2O_2 \xrightarrow[\text{alkaline medium}]{I^-} 2H_2O + O_22H2​O2​I−alkaline medium​2H2​O+O2​ the proposed mechanism is as given below:

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Q.For a reaction $2H_2O_2 \xrightarrow[\text{alkaline medium}]{I^-} 2H_2O + O_2$ the proposed mechanism is as given below:

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