# Reaction mechanism basics

> IB Chemistry HL · R3: Mechanisms of chemical change
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u6-reaction-mechanism-basics/

This module introduces core concepts of reaction mechanisms, the step-by-step sequence of elementary reactions that describes how overall chemical reactions proceed. You will learn to connect mechanism structure to experimental rate law data.

**Prerequisites:** [Rate expressions and reaction order](https://www.owlsprep.com/study/ib-chemistry-hl-u6-rate-laws-order-of-reaction/)

## Learning objectives

- Define core terms related to reaction mechanisms
- Distinguish between reaction intermediates and catalysts
- Identify the rate-determining step from a multi-step mechanism
- Propose mechanisms consistent with experimental rate law data

## Core Terminology of Reaction Mechanisms

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.

**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.

**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. $I_2 \rightarrow 2I^\bullet$
- 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. $O_3 \rightarrow O_2 + O$, 2. $NO + O_3 \rightarrow NO_2 + O_2$

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:

*Conclusion:* Step 1 is unimolecular, Step 2 is bimolecular.

> **Exam tip:** Molecularity only applies to elementary steps, never to overall balanced reactions.

## Intermediates and Catalysts in Mechanisms

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.

**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.

**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: $NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$, Step 2: $NO_{2(g)} + O_{(g)} \rightarrow NO_{(g)} + O_{2(g)}$. 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: $NO_2$ is produced in Step 1 and consumed in Step 2.
4. Catalysts are consumed first, then regenerated: $NO$ is consumed in Step 1 and regenerated in Step 2.

*Conclusion:* Intermediate = $NO_2$, Catalyst = $NO$

> **Exam tip:** Remember the order rule: Intermediate = Made then Used, Catalyst = Used then Made.

## The Rate-Determining Step (RDS)

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.

**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.

> **info**
>
> The rate of the overall reaction can never be faster than its slowest step, just like a bottleneck in a factory limits total output.

**Worked example:** Overall reaction: $2NO_2 + F_2 \rightarrow 2NO_2F$, experimental rate law: $rate = k[NO_2][F_2]$. Proposed mechanism: Step 1 (slow): $NO_2 + F_2 \rightarrow NO_2F + F$, Step 2 (fast): $NO_2 + F \rightarrow NO_2F$. 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 $NO_2$ 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.

## Proposing Mechanisms From Rate Data

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.

**Check your understanding**

Test your basic understanding

1. A reaction has rate law $rate = k[A]^2[B]$. 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

   *Why:* 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: $A + 2B \rightarrow C$, rate law: $rate = k[A][B]$. 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.

*Conclusion:* The proposed two-step mechanism is consistent with all given data.

## Common pitfalls

- **Wrong:** Assigning molecularity to the overall reaction instead of just elementary steps
  - Why it fails: Molecularity describes the number of particles in a single molecular event, which only applies to individual steps, not the overall balanced equation
  - Correct: Only use molecularity when referring to elementary steps, never the overall reaction
- **Wrong:** Confusing reaction intermediates with catalysts
  - Why it fails: Both do not appear in the overall reaction, but they are formed and consumed in opposite orders
  - Correct: Intermediate = produced then consumed; Catalyst = consumed then regenerated
- **Wrong:** Leaving a reaction intermediate in the final rate law
  - Why it fails: Intermediate concentrations are very low and cannot be measured experimentally, so they cannot appear in the final rate law
  - Correct: Substitute intermediate concentration using the equilibrium expression for the preceding fast step
- **Wrong:** Assuming the rate-determining step must always be the first step
  - Why it fails: The RDS can be any step in the mechanism, depending on the activation energy of each step
  - Correct: Always match RDS stoichiometry to the experimental rate law regardless of step order
- **Wrong:** Claiming a matching mechanism proves the reaction follows that pathway
  - Why it fails: Multiple different mechanisms can match the same rate law, so we can only confirm consistency not proof
  - Correct: You only need to propose or confirm a consistent mechanism, never prove it is the only possible pathway

## 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 |

## 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.

- [AHL: Stereoisomerism](https://www.owlsprep.com/study/ib-chemistry-hl-u6-ahl-stereoisomerism/)
- [AHL: Advanced organic reaction mechanisms](https://www.owlsprep.com/study/ib-chemistry-hl-u6-ahl-advanced-organic-reaction-mechanisms/)
- [AHL: Electrolytic cells and standard electrode potentials](https://www.owlsprep.com/study/ib-chemistry-hl-u6-ahl-electrolytic-cells-and-standard/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-chemistry-hl-u6-reaction-mechanism-basics/