Reaction Mechanism and Rate Law

AP Chemistry· 12 min read

1. Elementary Reactions and Molecularity★★☆☆☆⏱ 10 min

Unlike overall balanced chemical reactions, which give no information about reaction pathway, elementary steps describe exactly what occurs at the molecular level during a single collision event. For elementary steps only, the rate law can be written directly from the stoichiometric coefficients of the step.

📘 Definition

Molecularity

The number of reactant particles participating in a single elementary collision event. Valid values are unimolecular (1 particle), bimolecular (2 particles), and termolecular (3 simultaneous, rare collisions).

Example:

A bimolecular step A + B → Products has rate = k[A][B]

Molecularity	Elementary Step	Rate Law
Unimolecular	A → Products	$r a t e = k [A]$
Bimolecular	A + B → Products	$r a t e = k [A] [B]$
Bimolecular	A + A → Products	$r a t e = k [A]^{2}$
Termolecular	A + B + C → Products	$r a t e = k [A] [B] [C]$

📐 Worked Example

Write the rate law for the elementary step: $N O_{2} + C O \to N O + C O_{2}$

1
Identify the molecularity of the step: two reactant molecules collide, so it is bimolecular.
2
Use the stoichiometric coefficients (both equal to 1) as the reaction orders for each reactant.
3
$rate = k[NO_2][CO]$

✓ Quick check

Test your understanding of elementary step rate laws:

Which of the following is a valid rate law for a termolecular elementary step 2X + Y → Products?
- rate = k[X][Y]
- rate = k[X]^2[Y]
- rate = k[X][Y]^2
- rate = k
Reveal answer
rate = k[X]^2[Y] —
The stoichiometric coefficient of X is 2, so its order is 2, and Y has order 1.

2. Rate-Determining Step Basics★★★☆☆⏱ 12 min

For any multi-step mechanism, the overall reaction rate can never be faster than the slowest elementary step, called the rate-determining step (RDS). All steps after the RDS are fast and do not impact the overall measured rate, as their reactants are supplied immediately from the preceding slow step.

📐 Worked Example

Given the two-step mechanism below, write the overall rate law: Step 1 (slow): $N O_{2} + N O_{2} \to N O_{3} + N O$ , Step 2 (fast): $N O_{3} + C O \to N O_{2} + C O_{2}$

1
Identify the slow RDS: Step 1, the first elementary step.
2
Write the rate law directly from the stoichiometry of Step 1, which is bimolecular for two NO2 molecules.
3
$rate = k[NO_2]^2$
4
Confirm no intermediates appear in the final rate law, as no intermediates are present in the RDS reactants.

3. Pre-Equilibrium Approximation★★★★☆⏱ 15 min

When the RDS is not the first step, a fast reversible pre-equilibrium step occurs before the slow RDS. The intermediate formed in the pre-equilibrium is consumed immediately in the slow step, so the pre-equilibrium maintains a constant ratio of reactant and intermediate concentrations.

🔬 Derivation

Goal:

Derive the rate law for a mechanism with a fast pre-equilibrium before the RDS

Starting from:

Fast pre-equilibrium: A + B ⇌ I, Slow RDS: I → Products

1
Write the equilibrium constant expression for the fast pre-equilibrium: $K = \frac{[ I ]}{[ A ] [ B ]}$
2
Rearrange to solve for the intermediate concentration: $[I] = K [A] [B]$
3
Write the rate law from the RDS: $r a t e = k_{2} [I]$
4
Substitute the intermediate expression into the RDS rate law to eliminate [I]

Result:

Final rate law: $r a t e = k_{o v er a l l} [A] [B]$ , where $k_{o v er a l l} = k_{2} K$

📐 Worked Example

Derive the rate law for this mechanism: Step 1 (fast, reversible): $N O + O_{2} ⇌ N O_{3}$ , Step 2 (slow): $N O_{3} + N O \to 2 N O_{2}$

1
Write the equilibrium expression for Step 1: $K = \frac{[ N O _{3} ]}{[ N O ] [ O _{2} ]}$
2
Rearrange to isolate the intermediate concentration: $[N O_{3}] = K [N O] [O_{2}]$
3
Write the rate law from the slow Step 2: $r a t e = k_{2} [N O_{3}] [N O]$
4
Substitute the intermediate expression to eliminate $N O_{3}$ : $r a t e = k_{2} K [N O]^{2} [O_{2}] = k_{o v er a l l} [N O]^{2} [O_{2}]$

4. Identifying Intermediates and Catalysts★★★☆☆⏱ 10 min

Intermediates and catalysts are both cancelled out when summing elementary steps to get the overall reaction, but they have distinct behavior that is frequently tested on the AP exam.

📘 Definition

Catalyst

A species consumed in an early elementary step and fully regenerated in a later step, which lowers the activation energy of the reaction without being used up

Example:

Cl atoms in the ozone depletion mechanism

📐 Worked Example

Identify all intermediates and catalysts in this 3-step mechanism: Step 1: $C e^{4 +} + M n^{2 +} \to C e^{3 +} + M n^{3 +}$ , Step 2: $C e^{4 +} + M n^{3 +} \to C e^{3 +} + M n^{4 +}$ , Step 3: $M n^{4 +} + T l^{+} \to M n^{2 +} + T l^{3 +}$

1
Trace species across steps: $M n^{2 +}$ is consumed in Step 1 and regenerated in Step 3, so it is a catalyst.
2
Trace remaining transient species: $M n^{3 +}$ and $M n^{4 +}$ are formed in early steps and consumed in later steps, so they are reaction intermediates.
3
Confirm no transient species appear in the overall balanced reaction: $2 C e^{4 +} + T l^{+} \to 2 C e^{3 +} + T l^{3 +}$

5. Common Pitfalls

Wrong move:

Using overall reaction stoichiometry to write the rate law directly

Why:

Only elementary steps have rate laws matching their stoichiometric coefficients; overall multi-step reactions never follow this rule

Correct move:

Derive rate law exclusively from the slow rate-determining elementary step

Wrong move:

Including reaction intermediates in the final reported rate law

Why:

Intermediates are not present at the start of the reaction, so they cannot appear in the experimentally measurable rate law

Correct move:

Substitute intermediate concentration using the fast pre-equilibrium constant expression

Wrong move:

Classifying a catalyst as a reaction intermediate

Why:

Catalysts are consumed early and regenerated later, while intermediates are formed early and consumed later

Correct move:

Trace all species across steps to confirm their appearance and disappearance order

Wrong move:

Assigning a fractional molecularity to an elementary step

Why:

Molecularity counts discrete collision events, so it can only be 1, 2, or 3

Correct move:

Reject any proposed mechanism with an elementary step of non-integer molecularity

Wrong move:

Forgetting to sum elementary steps to confirm they match the overall reaction

Why:

A mechanism is automatically invalid if it does not add up to the net balanced reaction

Correct move:

Cancel all intermediates and catalysts across steps before confirming mechanism validity

6. Quick Reference Cheatsheet

Species Type	First Appearance	Final Appearance	In Overall Reaction
Reactant	Consumed in Step 1	Never formed	Yes
Product	Never formed early	Formed in last step	Yes
Intermediate	Formed in early step	Consumed later	No
Catalyst	Consumed in Step 1	Regenerated later	No

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.

2025 · MCQ
Match mechanism to given rate law
2024 · FRQ Q3
Identify intermediate and RDS
2023 · FRQ Q2
Derive rate law with pre-equilibrium

What's Next

Mastering the link between reaction mechanisms and rate laws is a core requirement for 40-50% of AP Kinetics FRQ points, and it builds directly into your upcoming study of activation energy and Arrhenius equation relationships. You will use these skills to evaluate competing proposed mechanisms for complex atmospheric and industrial reactions, a common scenario in recent AP exam free-response questions. Before moving on, confirm you can quickly identify intermediates, derive a rate law for a mechanism with a fast pre-equilibrium, and reject invalid proposed mechanisms that do not match experimental rate data.