# Reaction Mechanism and Rate Law

> AP Chemistry · AP Chemistry 2024-2029
> Source: https://www.owlsprep.com/study/ap-chemistry-u5-reaction-mechanism-and-rate-law/

This module connects multi-step reaction mechanisms to experimentally measured rate laws, covering elementary steps, rate-determining steps, pre-equilibrium approximations, and validation rules for proposed mechanisms.

**Prerequisites:** [Experimental rate law determination from initial rate data](https://www.owlsprep.com/study/ap-chemistry-u5-initial-rate-methods/); [Basic collision theory for reaction kinetics](https://www.owlsprep.com/study/ap-chemistry-u5-collision-theory/)

## Learning objectives

- Deduce the rate law from a proposed reaction mechanism using the slow, rate-determining step
- Identify valid reaction intermediates and catalysts across elementary steps
- Verify if a proposed mechanism matches an experimentally derived overall rate law
- Justify why elementary reaction molecularity directly maps to its rate law order

## Elementary Reactions and Molecularity

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.

**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 | $rate = k[A]$ |
| Bimolecular | A + B → Products | $rate = k[A][B]$ |
| Bimolecular | A + A → Products | $rate = k[A]^2$ |
| Termolecular | A + B + C → Products | $rate = k[A][B][C]$ |

**Worked example:** Write the rate law for the elementary step: $NO_2 + CO → NO + CO_2$

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. $$rate = k[NO_2][CO]$$

**Check your understanding**

Test your understanding of elementary step rate laws:

1. 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

   *Why:* The stoichiometric coefficient of X is 2, so its order is 2, and Y has order 1.

## Rate-Determining Step Basics

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.

> **Simplest Case Rule**
>
> If the RDS is the first step in the mechanism, no pre-equilibrium substitution is required, and the rate law can be written directly from the RDS stoichiometry.

**Worked example:** Given the two-step mechanism below, write the overall rate law: Step 1 (slow): $NO_2 + NO_2 → NO_3 + NO$, Step 2 (fast): $NO_3 + CO → NO_2 + CO_2$

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.

**Exam command terms**

AP exam questions use specific phrasing to signal what justification you need to provide:

- **Show the mechanism is consistent with the rate law** — You must explicitly derive the rate law from the RDS, sum the steps to get the overall reaction, and confirm no intermediates appear in the final rate law *(Award 2 points for full derivation, 1 point for only identifying the RDS)*

## Pre-Equilibrium Approximation

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:** 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: $rate = k_2[I]$
4. Substitute the intermediate expression into the RDS rate law to eliminate [I]

*Conclusion:* Final rate law: $rate = k_{overall}[A][B]$, where $k_{overall} = k_2 K$

**Worked example:** Derive the rate law for this mechanism: Step 1 (fast, reversible): $NO + O_2 ⇌ NO_3$, Step 2 (slow): $NO_3 + NO → 2NO_2$

1. Write the equilibrium expression for Step 1: $K = \frac{[NO_3]}{[NO][O_2]}$
2. Rearrange to isolate the intermediate concentration: $[NO_3] = K[NO][O_2]$
3. Write the rate law from the slow Step 2: $rate = k_2[NO_3][NO]$
4. Substitute the intermediate expression to eliminate $NO_3$: $rate = k_2 K [NO]^2[O_2] = k_{overall}[NO]^2[O_2]$

## Identifying Intermediates and Catalysts

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.

**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: $Ce^{4+} + Mn^{2+} → Ce^{3+} + Mn^{3+}$, Step 2: $Ce^{4+} + Mn^{3+} → Ce^{3+} + Mn^{4+}$, Step 3: $Mn^{4+} + Tl^+ → Mn^{2+} + Tl^{3+}$

1. Trace species across steps: $Mn^{2+}$ is consumed in Step 1 and regenerated in Step 3, so it is a catalyst.
2. Trace remaining transient species: $Mn^{3+}$ and $Mn^{4+}$ 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: $2Ce^{4+} + Tl^+ → 2Ce^{3+} + Tl^{3+}$

## Common pitfalls

- **Wrong:** Using overall reaction stoichiometry to write the rate law directly
  - Why it fails: Only elementary steps have rate laws matching their stoichiometric coefficients; overall multi-step reactions never follow this rule
  - Correct: Derive rate law exclusively from the slow rate-determining elementary step
- **Wrong:** Including reaction intermediates in the final reported rate law
  - Why it fails: Intermediates are not present at the start of the reaction, so they cannot appear in the experimentally measurable rate law
  - Correct: Substitute intermediate concentration using the fast pre-equilibrium constant expression
- **Wrong:** Classifying a catalyst as a reaction intermediate
  - Why it fails: Catalysts are consumed early and regenerated later, while intermediates are formed early and consumed later
  - Correct: Trace all species across steps to confirm their appearance and disappearance order
- **Wrong:** Assigning a fractional molecularity to an elementary step
  - Why it fails: Molecularity counts discrete collision events, so it can only be 1, 2, or 3
  - Correct: Reject any proposed mechanism with an elementary step of non-integer molecularity
- **Wrong:** Forgetting to sum elementary steps to confirm they match the overall reaction
  - Why it fails: A mechanism is automatically invalid if it does not add up to the net balanced reaction
  - Correct: Cancel all intermediates and catalysts across steps before confirming mechanism validity

## 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 |

## 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.

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