Reaction mechanism basics

IB Chemistry HL· Topic 6: Chemical Kinetics· 15 min read

1. Core Terminology of Reaction Mechanisms★★☆☆☆⏱ 5 min

A reaction mechanism breaks an overall balanced chemical reaction into individual elementary steps, each describing a single molecular event like a collision between two particles. No reaction occurs in a single step unless it is an elementary reaction itself.

📘 Definition

Elementary reaction

A single step in a reaction mechanism that describes one individual molecular event, and cannot be broken down into simpler steps.

Example:

Collision of two NO₂ molecules to form NO and N₂O₅ is an elementary reaction.

📘 Definition

Molecularity

The number of reactant particles that participate in an elementary reaction, categorized by the count of reactant molecules.

Unimolecular: 1 particle decomposes or rearranges, e.g. $I_{2} \to 2 I^{∙}$
Bimolecular: 2 particles collide and react, the most common molecularity
Termolecular: 3 particles collide simultaneously, very rare due to low probability of collision

📐 Worked Example

Identify the molecularity of each elementary reaction: 1. $O_{3} \to O_{2} + O$ , 2. $N O + O_{3} \to N O_{2} + O_{2}$

1
Molecularity is determined by counting the number of reactant particles, not product particles, in the elementary step.
2
First elementary step has only 1 reactant molecule:
3
$O_3 \rightarrow O_2 + O$
4
Second elementary step has two separate reactant molecules:
5
$NO + O_3 \rightarrow NO_2 + O_2$
6
Final classification:

Exam tip:

Molecularity only applies to elementary steps, never to overall balanced reactions.

2. Intermediates and Catalysts in Mechanisms★★☆☆☆⏱ 5 min

In multi-step mechanisms, intermediate species and catalysts are both present in steps but do not appear in the overall reaction. However, they are formed and consumed in opposite orders.

📘 Definition

Reaction Intermediate

A species that is produced in one early elementary step, then consumed in a subsequent later step. It does not appear in the overall reaction or final rate law.

📘 Definition

Catalyst

A species that speeds up a reaction by providing an alternative mechanism with lower activation energy. It is consumed in an early step, then regenerated in a later step, so does not appear in the overall reaction.

📐 Worked Example

Given the two-step mechanism: Step 1: $N O_{(g)} + O_{3 (g)} \to N O_{2 (g)} + O_{2 (g)}$ , Step 2: $N O_{2 (g)} + O_{(g)} \to N O_{(g)} + O_{2 (g)}$ . Identify the intermediate and catalyst.

1
First add the two steps and cancel species that appear on both sides to get the overall reaction:
2
$O_{3(g)} + O_{(g)} \rightarrow 2O_{2(g)}$
3
Intermediates are produced first, then consumed: $N O_{2}$ is produced in Step 1 and consumed in Step 2.
4
Catalysts are consumed first, then regenerated: $N O$ is consumed in Step 1 and regenerated in Step 2.

Exam tip:

Remember the order rule: Intermediate = Made then Used, Catalyst = Used then Made.

3. The Rate-Determining Step (RDS)★★★☆☆⏱ 6 min

Not all steps in a multi-step mechanism proceed at the same rate. The slowest step has the highest activation energy, and limits the overall rate of the entire reaction. This step is called the rate-determining step.

📘 Definition

Rate-Determining Step (RDS)

The slowest elementary step in a reaction mechanism, which determines the overall rate of the reaction. The overall rate law matches the stoichiometry of the RDS.

📐 Worked Example

Overall reaction: $2 N O_{2} + F_{2} \to 2 N O_{2} F$ , experimental rate law: $r a t e = k [N O_{2}] [F_{2}]$ . Proposed mechanism: Step 1 (slow): $N O_{2} + F_{2} \to N O_{2} F + F$ , Step 2 (fast): $N O_{2} + F \to N O_{2} F$ . Confirm this mechanism matches the rate law.

1
The overall rate is determined by the slow step, so we write the rate law directly from the stoichiometry of the RDS.
2
Step 1 (the RDS) has 1 mole of $N O_{2}$ and 1 mole of $F_{2}$ as reactants, so the rate law becomes:
3
$rate = k[NO_2]^1[F_2]^1 = k[NO_2][F_2]$
4
This matches the experimentally observed rate law, so the mechanism is consistent.

Exam tip:

If an intermediate appears in the RDS, substitute its concentration using the equilibrium expression for the preceding fast step.

4. Proposing Mechanisms From Rate Data★★★★☆⏱ 7 min

Given an overall reaction and experimental rate law, we can propose a mechanism that is consistent with the observed data. The exponent of each reactant in the rate law equals its stoichiometric coefficient in the rate-determining step.

✓ Quick check

Test your basic understanding

A reaction has rate law $r a t e = k [A]^{2} [B]$ . What is the overall order, and how many moles of A are in the RDS?
- Overall order 2, 2 A particles
- Overall order 3, 2 A particles
- Overall order 3, 1 A particle
- Overall order 2, 1 A particle
Reveal answer
Overall order 3, 2 A particles —
Overall order is the sum of exponents: $2 + 1 = 3$ . The exponent of A equals the number of A particles in the RDS.

📐 Worked Example

Overall reaction: $A + 2 B \to C$ , rate law: $r a t e = k [A] [B]$ . Propose a consistent two-step mechanism.

1
The RDS has 1 A and 1 B, so we write the slow step first:
2
$A + B \rightarrow AB \quad \text{(slow, RDS)}$
3
The second step is fast, and uses the intermediate AB and the remaining 1 B to form product C:
4
$AB + B \rightarrow C \quad \text{(fast)}$
5
Check the overall reaction by adding the two steps and canceling the intermediate:
6
$A + 2B \rightarrow C$
7
Confirm the rate law matches the RDS: rate = k[A][B], which matches the given rate law.

5. Common Pitfalls

Wrong move:

Assigning molecularity to the overall reaction instead of just elementary steps

Why:

Molecularity describes the number of particles in a single molecular event, which only applies to individual steps, not the overall balanced equation

Correct move:

Only use molecularity when referring to elementary steps, never the overall reaction

Wrong move:

Confusing reaction intermediates with catalysts

Why:

Both do not appear in the overall reaction, but they are formed and consumed in opposite orders

Correct move:

Intermediate = produced then consumed; Catalyst = consumed then regenerated

Wrong move:

Leaving a reaction intermediate in the final rate law

Why:

Intermediate concentrations are very low and cannot be measured experimentally, so they cannot appear in the final rate law

Correct move:

Substitute intermediate concentration using the equilibrium expression for the preceding fast step

Wrong move:

Assuming the rate-determining step must always be the first step

Why:

The RDS can be any step in the mechanism, depending on the activation energy of each step

Correct move:

Always match RDS stoichiometry to the experimental rate law regardless of step order

Wrong move:

Claiming a matching mechanism proves the reaction follows that pathway

Why:

Multiple different mechanisms can match the same rate law, so we can only confirm consistency not proof

Correct move:

You only need to propose or confirm a consistent mechanism, never prove it is the only possible pathway

6. Quick Reference Cheatsheet

Term	Definition	Key Property
Elementary step	Single molecular reaction event	Molecularity defined here
Reaction intermediate	Made early, used later	Not in overall reaction
Catalyst	Used early, regenerated later	Lowers activation energy
Rate-determining step	Slowest step in mechanism	Determines overall rate law
Molecularity	Number of reactant particles	Uni=1, Bi=2, Termolecular=3

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Identify RDS from rate data
2021 · 1
Distinguish intermediate vs catalyst
2023 · 2
Propose consistent mechanism

Going deeper

syllabusIB Chemistry HL Official SyllabusTopic 6: Chemical Kinetics

What's Next

Mastering reaction mechanism basics is the foundation for all organic and inorganic mechanism topics in IB Chemistry HL. You will apply these core concepts to identify and distinguish between SN1 and SN2 nucleophilic substitution mechanisms, analyze electrophilic addition reactions of alkenes, and explain how different types of catalysts work in industrial and biological systems. The skill of connecting experimental rate data to proposed mechanism is also a frequent high-mark question in Paper 2, so practicing these concepts will prepare you for exam success.