Define the complex reactions.

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Q. Define the complex reactions.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2025Subjective· 1mImportance★★★★★

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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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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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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.Define the complex reactions.

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