Introduction to Reaction Mechanisms

AP Chemistry· AP Chemistry CED — Kinetics· 14 min read

1. Core Fundamentals of Reaction Mechanisms★☆☆☆☆⏱ 3 min

A reaction mechanism describes the step-by-step sequence of bond-breaking and bond-forming events that convert starting reactants to final products at the molecular level. Unlike the overall net reaction, which only shows starting and final species, mechanisms reveal the intermediate species that form and are consumed during the reaction.

2. Elementary Reactions and Molecularity★★☆☆☆⏱ 4 min

📘 Definition

Elementary Reaction

A single step in a reaction mechanism that describes one actual molecular collision or rearrangement, which cannot be broken down into smaller steps. For elementary reactions only, the rate law can be written directly from stoichiometry.

Molecularity is the number of reactant particles that participate in an elementary step. There are three common classifications:

Unimolecular: One reactant particle reacts, rate = $k [A]$ , first order overall
Bimolecular: Two reactant particles collide and react, rate = $k [A] [B]$ (or $k [A]^{2}$ for $2 A$ ), second order overall
Termolecular: Three reactant particles collide simultaneously (very rare), rate = $k [A] [B] [C]$ , third order overall

📐 Worked Example

Write the rate law for each elementary reaction below, and state the molecularity of each: (a) $C_{4} H_{8} (g) \to 2 C_{2} H_{4} (g)$ (b) $N O (g) + O_{3} (g) \to N O_{2} (g) + O_{2} (g)$

1
For reaction (a): Count the number of reactant molecules, there is only 1 molecule of $C_{4} H_{8}$ , so molecularity is unimolecular. For elementary reactions, the order of each reactant equals its stoichiometric coefficient, so the rate law is:
2
$text{rate} = k[C_4H_8]$
3
For reaction (b): There are two different reactant molecules, so molecularity is bimolecular. Each reactant has a stoichiometric coefficient of 1, so exponents are both 1, giving the rate law:
4
$text{rate} = k[NO][O_3]$

3. Reaction Intermediates and Catalysts★★☆☆☆⏱ 3 min

Two types of species do not appear in the final net reaction: reaction intermediates and catalysts. You must be able to distinguish between these two for AP exam questions.

📘 Definition

Reaction Intermediate

A species produced in one early elementary step and consumed in a subsequent later step. It is not present at the start of the reaction, and never appears in the final overall rate law.

📘 Definition

Catalyst

A species consumed in an early elementary step and regenerated in a later elementary step. It is present at the start of the reaction, speeds up the reaction by changing the mechanism, and is not consumed overall.

To distinguish between the two, track the order of appearance: intermediate = product first, reactant second; catalyst = reactant first, product second.

📐 Worked Example

The decomposition of hydrogen peroxide in the presence of bromide ion follows this mechanism: Step 1: $H_{2} O_{2} (a q) + H^{+} (a q) + B r^{-} (a q) \to H O B r (a q) + H_{2} O (l)$ Step 2: $H O B r (a q) + H_{2} O_{2} (a q) \to H^{+} (a q) + B r^{-} (a q) + O_{2} (g) + H_{2} O (l)$ Identify the intermediate and catalyst in this mechanism.

1
List all species by their position: $B r^{-}$ is a reactant in step 1 (early step) and a product in step 2 (late step). $H O B r$ is a product in step 1 and a reactant in step 2.
2
By definition, any species consumed first (reactant early) and produced later (product late) is a catalyst, so $B r^{-}$ is the catalyst.
3
Any species produced first (product early) and consumed later (reactant late) is an intermediate, so $H O B r$ is the reaction intermediate.

4. Rate-Determining Step and Rate Law Derivation★★★★☆⏱ 6 min

The overall rate of a reaction mechanism is limited by the slowest step in the mechanism, called the rate-determining step (RDS). The overall rate law is exactly equal to the rate law of the RDS.

If the RDS is the first step with no intermediates, you can write the rate law directly from the RDS stoichiometry. If the RDS comes after one or more fast reversible steps, the RDS will contain an intermediate from the fast step. You must substitute the intermediate concentration using the pre-equilibrium approximation: the fast step reaches equilibrium quickly, so the rate of the forward step equals the rate of the reverse step.

k_{forward}[reactants] = k_{reverse}[intermediate] \implies [intermediate] = \frac{k_{forward}}{k_{reverse}}[reactants]

📐 Worked Example

A reaction has this proposed mechanism: Step 1 (fast, reversible): $B r_{2} (g) ⇌ 2 B r (g)$ (intermediate $B r$ ) Step 2 (slow): $B r (g) + H_{2} (g) \to H B r (g) + H (g)$ (intermediate $H$ ) Step 3 (fast): $H (g) + B r_{2} (g) \to H B r (g) + B r (g)$ Derive the rate law for this overall reaction.

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The RDS is the slow step (step 2), so first write its elementary rate law:
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$text{rate} = k_2[Br][H_2]$
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$B r$ is an intermediate from the fast pre-equilibrium step 1, so substitute it using the pre-equilibrium approximation:
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$k_1[Br_2] = k_{-1}[Br]^2 \implies [Br] = \left(\frac{k_1}{k_{-1}}[Br_2]\right)^{1/2}$
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Substitute $[B r]$ into the RDS rate law to get the overall rate law:
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$text{rate} = k_2 \left(\frac{k_1}{k_{-1}}[Br_2]\right)^{1/2} [H_2] = k [Br_2]^{1/2}[H_2]$
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where $k = k_{2} (\frac{k _{1}}{k _{- 1}})^{1/2}$ is the overall rate constant.

📐 Worked Example

The reaction of nitrogen monoxide and oxygen is $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ . A proposed mechanism is: Step 1 (fast, reversible): $N O (g) + N O (g) ⇌ N_{2} O_{2} (g)$ (intermediate $N_{2} O_{2}$ ) Step 2 (slow): $N_{2} O_{2} (g) + O_{2} (g) \to 2 N O_{2} (g)$ (a) Identify the intermediate. (b) Confirm the net reaction matches. (c) Derive the rate law.

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(a) $N_{2} O_{2}$ is produced in step 1 and consumed in step 2, so it is the reaction intermediate.
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(b) Add the two steps: $N O + N O + N_{2} O_{2} + O_{2} \to N_{2} O_{2} + 2 N O_{2}$ . Cancel $N_{2} O_{2}$ to get $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ , which matches the given overall equation.
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(c) The RDS is step 2, so write its elementary rate law:
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$text{rate} = k_2[N_2O_2][O_2]$
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Substitute the intermediate using pre-equilibrium:
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$k_1[NO]^2 = k_{-1}[N_2O_2] \implies [N_2O_2] = \frac{k_1}{k_{-1}}[NO]^2$
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Substitute to get the final rate law:
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$text{rate} = k[NO]^2[O_2], \quad k = \frac{k_1k_2}{k_{-1}}$

✓ Quick check

Test your understanding: The experimental rate law for the reaction $2 A + B \to C$ is $rate = k [A]^{2} [B]$ . Which proposed mechanism is consistent with this rate law?

Which mechanism is consistent?
- A) Step 1 (slow): $2 A \to D$ ; Step 2 (fast): $D + B \to C$
- B) Step 1 (fast): $A + B ⇌ D$ ; Step 2 (slow): $D + A \to C$
- C) Step 1 (slow): $A + B \to D$ ; Step 2 (fast): $D + A \to C$
- D) Step 1 (fast): $2 A ⇌ D$ ; Step 2 (fast): $D + B \to C$
Reveal answer
1 —
Correct: Deriving the rate law for B gives $rate = k [A]^{2} [B]$ , which matches. Other options are inconsistent: A gives second order overall, C gives second order overall, D has no defined rate-determining step.

5. Common Pitfalls

Wrong move:

Writing a rate law for the overall reaction directly from its stoichiometry

Why:

Students confuse the rule for elementary reactions with overall reactions, and the rate law sometimes matches by coincidence, leading to incorrect assumptions

Correct move:

Only write a rate law from stoichiometry if the step is explicitly labeled elementary; for all overall reactions, the rate law must come from experiment or the mechanism's RDS

Wrong move:

Misclassifying catalyst as intermediate (or vice versa) because neither appears in the net reaction

Why:

Students only remember both are canceled from the net reaction, so they mix up the classification

Correct move:

Always track order of appearance: produced first then consumed = intermediate; consumed first then produced = catalyst

Wrong move:

Leaving an intermediate concentration term in the final overall rate law

Why:

Students forget intermediates are not starting reactants, so their concentration depends on starting reactant concentrations

Correct move:

Always substitute any intermediate that appears in the RDS using the pre-equilibrium approximation to replace it with reactant terms

Wrong move:

Assuming the RDS must be the first step in the mechanism

Why:

Most introductory examples have RDS as the first step, so students assume this pattern always holds

Correct move:

Always use the explicitly labeled slow step as the RDS, regardless of its position in the mechanism

Wrong move:

Adding exponents from all elementary steps to get overall reaction order

Why:

Students assume all steps contribute equally to the overall reaction rate

Correct move:

Only exponents from the RDS (after substituting all intermediates) determine the overall reaction order

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Elementary reaction rate law	Rate order = stoichiometric coefficient	Only applies to explicitly labeled elementary reactions
Unimolecular elementary reaction	$rate = k [A]$	1 reactant particle, first order overall
Bimolecular ( $A + B$ )	$rate = k [A] [B]$	2 particles, second order overall
Bimolecular ( $2 A$ )	$rate = k [A]^{2}$	2 particles, second order overall
Reaction intermediate	Produced early, consumed later	Never in net reaction or final rate law
Catalyst	Consumed early, regenerated later	Never in net reaction, not consumed overall
Rate-determining step rule	Overall rate = rate of slowest step	Always use this for rate law derivation
Pre-equilibrium approximation	$[in t er m e d ia t e] = \frac{k _{f or w a r d}}{k _{r e v er se}} [r e a c t an t s]$	Substitutes intermediates from fast pre-RDS steps

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.

2023 · MCQ
Identify intermediate in mechanism
2022 · FRQ
Derive rate law from mechanism

What's Next

This sub-topic is the foundation for all further work in kinetics and reaction dynamics. Immediately after mastering this material, you will study how catalysts modify reaction mechanisms to lower activation energy, a frequent topic on both AP MCQ and FRQ. Without understanding how to identify intermediates and extract rate laws from mechanisms, you will not be able to correctly explain catalytic behavior or validate proposed mechanisms, a common high-weight FRQ task on the AP exam. This topic also connects to equilibrium concepts in Unit 6 and organic reaction mechanisms in later units of the course.