# Pre-Equilibrium Approximation

> AP Chemistry · AP Chem 2024+
> Source: https://www.owlsprep.com/study/ap-chemistry-u5-pre-equilibrium-approximation/

This module explains how to derive valid rate laws for multi-step mechanisms where a fast reversible first step precedes a slow rate-determining step, eliminating unmeasurable intermediate concentrations to match experimental data.

**Prerequisites:** [Elementary reaction rate laws from stoichiometry](https://www.owlsprep.com/study/ap-chemistry-u5-elementary-reaction-rate-laws/); [Rate-determining step definition and properties](https://www.owlsprep.com/study/ap-chemistry-u5-rate-determining-step/)

## Learning objectives

- Distinguish the pre-equilibrium approximation from other kinetics simplifications for multi-step reactions
- Derive valid rate laws for mechanisms with a fast reversible first step followed by a slow rate-determining step
- Eliminate unmeasurable intermediate concentrations from final rate expressions
- Identify valid and invalid use cases for the approximation on AP exam questions

## Core Assumptions of the Pre-Equilibrium Approximation

For a mechanism where the first step is fast and reversible, and the second step is the slow RDS, the forward and reverse reactions of the first step will balance out to form a dynamic equilibrium before any significant amount of product is formed via the slow step.

**Pre-Equilibrium Approximation** — The ratio of the forward rate constant of the fast step to the reverse rate constant of the fast step equals the standard equilibrium constant for that pre-reaction.

*Notation:* K_{pre} = k_1 / k_{-1}

> **Validity Condition**
>
> This approximation only holds if the RDS is at least 100x slower than the reverse of the fast first step, so the intermediate has time to equilibrate before being consumed.

**Check your understanding**

Test your baseline understanding before proceeding:

1. Which species cannot appear in a final experimentally measured rate law?

   - Reactants
   - Products
   - Intermediates
   - Catalysts

   *Why:* Transient intermediates are present in too low concentration to measure reliably, so they must be eliminated from the final rate expression.

## Step-by-Step Derivation Workflow

1. Identify the slow RDS and write its raw rate law directly from elementary step stoichiometry
2. Write the equilibrium expression for the fast reversible step that occurs before the RDS
3. Rearrange the equilibrium expression to solve for the concentration of the intermediate species
4. Substitute the intermediate expression into the raw RDS rate law
5. Combine all rate constants into a single observed constant $k_{obs}$

**Worked example:** Derive the rate law for the mechanism: Step 1 (fast, rev): $2A \rightleftharpoons A_2$; Step 2 (slow, RDS): $A_2 + B \rightarrow C$

1. Step 1: Write raw rate law from the slow RDS
2. $$rate = k_2 [A_2][B]$$
3. Step 2: Write the pre-equilibrium expression for the fast first step
4. $$K_{pre} = \frac{k_1}{k_{-1}} = \frac{[A_2]}{[A]^2}$$
5. Step 3: Rearrange to solve for intermediate $[A_2]$
6. $$[A_2] = K_{pre} [A]^2$$
7. Step 4: Substitute back into the raw RDS rate law
8. $$rate = k_2 K_{pre} [A]^2 [B] = k_{obs} [A]^2 [B]$$

**Derivation:** Show that $k_{obs}$ aggregates all elementary constants

*Starting from:* rate = k_2 \times \frac{k_1}{k_{-1}} \times [A]^2 [B]

1. All three constants $k_1$, $k_{-1}$, and $k_2$ are measured at the same temperature
2. They can be combined into a single lumped constant with no loss of generality

*Conclusion:* Final observed rate law is second order in A, first order in B, and zero order in any products formed after the RDS

## Exam Phrasing and Scoring Expectations

**Exam command terms**

AP Chemistry uses specific command terms for mechanism questions that carry explicit scoring rules:

- **Show that the rate law is consistent with the mechanism** — You must explicitly show the intermediate substitution step, not just write the final rate law *(If you skip the substitution step, you will lose 50% of available points for the question)*

- **Justify the rate law** — You must state that the slow RDS determines overall rate, and the pre-equilibrium approximation eliminates the intermediate *(Stating "the slow step is rate determining" is required to earn full justification points)*

> **Exam Red Flag**
>
> Never write a rate law directly from the overall reaction stoichiometry on an AP exam. This automatic zero mistake is the single most common point loss on kinetics FRQs.

## Pre-Equilibrium vs Steady-State Comparison

**Comparing methods**

These two approximations are the only ones tested on the AP Chemistry exam:

- **Pre-Equilibrium Approximation** — Fast reversible step before slow RDS, intermediate concentration is defined by K
  - Pros: Simple 5-step workflow, no differential equations required
  - Cons: Only valid for mechanisms with a clear fast first step

- **Steady-State Approximation** — Rate of intermediate formation equals rate of intermediate consumption at all times
  - Pros: Works for any multi-step mechanism with no clear RDS
  - Cons: Requires more algebra, rarely tested on AP exams

## Common pitfalls

- **Wrong:** Writing the rate law directly from overall reaction stoichiometry
  - Why it fails: Only elementary steps have rate laws that match their stoichiometry; overall reactions never do
  - Correct: Always start your derivation from the slow rate-determining step first
- **Wrong:** Leaving an intermediate concentration in your final rate expression
  - Why it fails: Intermediates are unmeasurable, transient species that cannot appear in an experimentally observed rate law
  - Correct: Always substitute the intermediate away using the pre-equilibrium K expression
- **Wrong:** Including products of the slow RDS in the pre-equilibrium K expression
  - Why it fails: The fast equilibrium only applies to steps that occur before the RDS; products formed after the RDS cannot affect the pre-equilibrium
  - Correct: Only include species from the fast reversible first step in your K calculation
- **Wrong:** Applying pre-equilibrium to a mechanism where the slow step is first
  - Why it fails: If the first step is slow, no prior equilibrium can be established at all
  - Correct: If the first step is the RDS, its rate law is the full observed rate law with no substitutions needed
- **Wrong:** Setting the forward rate of the fast step equal to the rate of the slow RDS
  - Why it fails: The only two rates that balance at pre-equilibrium are the forward and reverse rates of the fast first step
  - Correct: Set $k_1 [reactants] = k_{-1} [intermediate]$ to derive your K expression

## Cheatsheet

| Step Number | Action | AP Exam Scoring Check |
| --- | --- | --- |
| 1 | Identify the slow RDS explicitly | Labeling RDS earns you 1 automatic point |
| 2 | Write raw rate law from RDS stoichiometry | Intermediate must appear at this stage |
| 3 | Write equilibrium expression for the fast pre-step | K = k_forward / k_reverse |
| 4 | Rearrange K to solve for intermediate | Eliminate all transient species |
| 5 | Combine constants into $k_{obs}$ | Final rate law only uses measurable reactants |

## What's next

Mastering the pre-equilibrium approximation is a critical milestone for AP Kinetics, as it accounts for nearly 10% of all Unit 5 free response points on recent exams. Once you can reliably derive rate laws for fast-first-step mechanisms, you will be ready to tackle more complex multi-step mechanisms that use the steady-state approximation, as well as reaction coordinate diagrams that map the relative energy of pre-equilibrium intermediates and the RDS transition state. You will also learn to test proposed reaction mechanisms against experimental rate data, a common extended response question that appears on almost every AP Chemistry exam administration. Practice applying this method to a full set of mechanism validation problems to solidify your understanding before moving on to advanced kinetics topics.

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