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

8 PYQs, none HARD — a recall page. Which step decides the rate (the slowest), which species is the intermediate in a two-step mechanism (set four 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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Chemical KineticsMODERATE

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.

[Q72 · 26 April Shift II · 2025]

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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Chemical KineticsMODERATE

Identify the reaction intermediate of following reaction. (i)

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

; (ii)

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

; Overall:

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

[Q69 · 13th May Shift 1 · 2024]

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 4 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

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