Pre-Equilibrium Approximation

AP Chemistry· 12 min read

1. Core Assumptions of the Pre-Equilibrium Approximation★★★☆☆⏱ 3 min

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.

📘 Definition

Pre-Equilibrium Approximation

K_{pre} = k_1 / k_{-1}

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.

✓ Quick check

Test your baseline understanding before proceeding:

Which species cannot appear in a final experimentally measured rate law?
- Reactants
- Products
- Intermediates
- Catalysts
Reveal answer
Intermediates —
Transient intermediates are present in too low concentration to measure reliably, so they must be eliminated from the final rate expression.

2. Step-by-Step Derivation Workflow★★★★☆⏱ 4 min

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

📐 Worked Example

Derive the rate law for the mechanism: Step 1 (fast, rev): $2 A ⇌ A_{2}$ ; Step 2 (slow, RDS): $A_{2} + B \to C$

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

Goal:

Show that $k_{o b s}$ 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

Result:

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

3. Exam Phrasing and Scoring Expectations★★★★☆⏱ 3 min

4. Pre-Equilibrium vs Steady-State Comparison★★★☆☆⏱ 2 min

Methods compared

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

5. Common Pitfalls

Wrong move:

Writing the rate law directly from overall reaction stoichiometry

Why:

Only elementary steps have rate laws that match their stoichiometry; overall reactions never do

Correct move:

Always start your derivation from the slow rate-determining step first

Wrong move:

Leaving an intermediate concentration in your final rate expression

Why:

Intermediates are unmeasurable, transient species that cannot appear in an experimentally observed rate law

Correct move:

Always substitute the intermediate away using the pre-equilibrium K expression

Wrong move:

Including products of the slow RDS in the pre-equilibrium K expression

Why:

The fast equilibrium only applies to steps that occur before the RDS; products formed after the RDS cannot affect the pre-equilibrium

Correct move:

Only include species from the fast reversible first step in your K calculation

Wrong move:

Applying pre-equilibrium to a mechanism where the slow step is first

Why:

If the first step is slow, no prior equilibrium can be established at all

Correct move:

If the first step is the RDS, its rate law is the full observed rate law with no substitutions needed

Wrong move:

Setting the forward rate of the fast step equal to the rate of the slow RDS

Why:

The only two rates that balance at pre-equilibrium are the forward and reverse rates of the fast first step

Correct move:

Set $k_{1} [r e a c t an t s] = k_{- 1} [in t er m e d ia t e]$ to derive your K expression

6. Quick Reference 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 $k_{o b s}$	Final rate law only uses measurable reactants

7. Frequently Asked

When do I use pre-equilibrium vs steady-state approximation?

Pre-equilibrium applies explicitly when the first step is fast and reversible, followed by a clearly defined slow RDS. Steady-state approximation is used when no single step is obviously rate-determining, or intermediates form in slow initial steps. AP exams almost exclusively test the pre-equilibrium framework for mechanisms with a fast first step.

Can pre-equilibrium produce fractional reaction orders?

No, unlike edge cases of the steady-state approximation, pre-equilibrium applied to elementary steps will always produce integer or zero reaction orders that match the stoichiometry of reactants in the fast equilibrium and slow RDS.

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
Rate law derivation for 2-step mechanism
2021 · FRQ
Intermediate elimination from rate expression
2019 · FRQ
Validate proposed mechanism against data

Going deeper

official_resourceAP Chemistry Kinetics CED ExcerptOfficial curriculum reference for reaction mechanism approximations

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.