MHT-CET Chemistry · Chemical Kinetics

Reaction Mechanism, Intermediates and Rate-Determining Step

A complex reaction is a sequence of elementary steps; the slowest step sets the rate and its molecularity writes the rate law; a species made in one step and consumed in a later one is an intermediate and never appears in the overall equation.

All Chemical Kinetics notes MHT-CET Chemistry strategy

Why this matters

9 PYQs, none HARD — a recall page. Which step decides the rate (the slowest), which species is the intermediate in a two-step mechanism (set three times with the ClO⁻ and SO₂/NO pairs), which of four reactions is elementary, the rate law from a slow step, and one catalyst name (Fe–Cr for the water-gas shift). Ten minutes covers it.

Concept 1 of 2: The Rate-Determining Step Writes the Rate Law

A multistep reaction can go no faster than its slowest step, so that step's molecularity gives the rate law: for a slow step

\text{NO}_2\text{Cl} \to \text{NO}_2 + \text{Cl}

the rate is

k[\text{NO}_2\text{Cl}]

, whatever the fast step does afterwards.

Definition

The rate of a multistep reaction is the rate of the SLOWEST step — not the fastest, not an average.
Slow step $\text{NO}_2\text{Cl} \to \text{NO}_2 + \text{Cl}$ , fast step $\text{NO}_2\text{Cl} + \text{Cl} \to \text{NO}_2 + \text{Cl}_2$ : $r = k[\text{NO}_2\text{Cl}]$ , first order, though the overall equation is $2\text{NO}_2\text{Cl} \to 2\text{NO}_2 + \text{Cl}_2$ .
An elementary reaction happens in one step and its rate law follows its equation: $2\text{NO}_2 + \text{F}_2 \to 2\text{NO}_2\text{F}$ is the exam's example. Decompositions like $2\text{NO}_2\text{Cl} \to 2\text{NO}_2 + \text{Cl}_2$ and $\text{C}_2\text{H}_5\text{I} \to \text{C}_2\text{H}_4 + \text{HI}$ proceed through mechanisms.
Molecularity is defined for each elementary step, not for the overall reaction.

Rate from the slow step

\text{slow step } aA + bB \to \dots \ \Rightarrow\ r = k[A]^a[B]^b

Worked example

A mechanism is: (slow)

\text{A} + \text{B} \to \text{C}

; (fast)

\text{C} + \text{B} \to \text{D}

. Write the rate law and the overall reaction.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q72Moderate

Example 1 · Chemical Kinetics · Reaction Mechanism, Intermediates and Rate-Determining Step

A complex reaction takes place in following steps.

NO_{2}Cl_{(g)}\longrightarrow NO_{2(g)}+Cl_{(g)}

(slow)

NO_{2}Cl_{(g)}+Cl_{(g)}\longrightarrow NO_{2(g)}+Cl_{2(g)}

(fast) Identify rate law equation for this reaction.

Writing the rate law from the overall equation

2\text{NO}_2\text{Cl} \to 2\text{NO}_2 + \text{Cl}_2

suggests

k[\text{NO}_2\text{Cl}]^2

, option (D); the slow step is unimolecular and the law is first order.

Drill 2 more on the rate-determining step writes the rate law

Concept 2 of 2: Intermediates: Made in One Step, Used in the Next; Catalysts: Used, Then Regenerated

Add the steps of a mechanism and cancel what appears on both sides. A species that cancels because it is PRODUCED first and CONSUMED later is an intermediate; one that is consumed first and regenerated later is a catalyst. Neither appears in the overall equation.

Definition

$2\text{ClO}^- \to \text{ClO}_2^- + \text{Cl}^-$ , then $\text{ClO}_2^- + \text{ClO}^- \to \text{ClO}_3^- + \text{Cl}^-$ : $\text{ClO}_2^-$ is the intermediate; overall $3\text{ClO}^- \to \text{ClO}_3^- + 2\text{Cl}^-$ .
$2\text{SO}_2 + 2\text{NO}_2 \to 2\text{SO}_3 + 2\text{NO}$ , then $2\text{NO} + \text{O}_2 \to 2\text{NO}_2$ : NO is the intermediate (made in step i, used in step ii); NO $_2$ is consumed first and regenerated — it acts as a catalyst; overall $2\text{SO}_2 + \text{O}_2 \to 2\text{SO}_3$ .
Catalysts to know: Fe–Cr (iron–chromium oxide) for the water-gas shift $\text{CO} + \text{H}_2\text{O} \rightleftharpoons \text{CO}_2 + \text{H}_2$ ; platinised asbestos for SO $_2$ oxidation (contact process); MnO $_2$ for KClO $_3$ decomposition; Co–Th for Fischer–Tropsch.
A catalyst lowers the activation energy and raises $k$ ; it does not change the equilibrium position or the overall equation.

Cancelling species

\text{intermediate: produced then consumed;}\qquad \text{catalyst: consumed then regenerated}

Worked example

Mechanism: (i)

\text{H}_2\text{O}_2 + \text{I}^- \to \text{H}_2\text{O} + \text{IO}^-

; (ii)

\text{H}_2\text{O}_2 + \text{IO}^- \to \text{H}_2\text{O} + \text{O}_2 + \text{I}^-

. Identify the intermediate and the catalyst.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q69Moderate

Example 2 · Chemical Kinetics · Reaction Mechanism, Intermediates and Rate-Determining Step

Identify the reaction intermediate of following reaction: (i)

2SO_2(g) + 2NO_2(g) \rightarrow 2SO_3(g) + 2NO

; (ii)

2NO(g) + O_2(g) \rightarrow 2NO_2(g)

; overall:

2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)

Picking the product that appears in both steps

\text{Cl}^-

is formed in BOTH ClO

^-

steps and never consumed — it is a product, not an intermediate. The intermediate is the one that disappears again.

Drill 3 more on intermediates: made in one step, used in the next; catalysts: used, then regenerated

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The Rate-Determining Step Writes the Rate Law
Rate from the slow step
$\text{slow step } aA + bB \to \dots \ \Rightarrow\ r = k[A]^a[B]^b$
Intermediates: Made in One Step, Used in the Next; Catalysts: Used, Then Regenerated
Cancelling species
$\text{intermediate: produced then consumed;}\qquad \text{catalyst: consumed then regenerated}$

Watch out for (2)

Writing the rate law from the overall equation
→ The Rate-Determining Step Writes the Rate Law
Picking the product that appears in both steps
→ Intermediates: Made in One Step, Used in the Next; Catalysts: Used, Then Regenerated

Test yourself on Chemical Kinetics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Kinetics chapter test Drill all 9 questions

Related notes

Rate Law and Order — molecularity versus order →