Pre-Equilibrium Approximation
Connect reversible rate balance to the pre-equilibrium approximation used inside a mechanism.
AP Chemistry Topic 5.9 · CED printed p. 103 / PDF p. 110. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Balanced equations and stoichiometric ratios
- Molarity and concentration versus amount
- Bond rearrangement and energy conservation
- Slopes, logarithms and reciprocal graphs
A later slow step often consumes an intermediate. Its elementary law contains that intermediate, but the measured overall law should be written in terms of experimentally controlled reactants. Suppose 2NO ⇌ N2O2 is rapid and reversible, followed by N2O2 + O2 → 2NO2 as a slow step. In the pre-equilibrium approximation, the first step stays close to equilibrium because interconversion is much faster than removal by the slow step. Set its forward and reverse rates approximately equal: k1[NO]2 =k−1[N2O2]. Solve for the intermediate, [N2O2] ≈(k1/k−1)[NO]2. Substitute into the slow law r =k2[N2O2][O2] to get r ≈keff[NO]2[O2], where keff =k2k1/k−1. Now no unmeasured intermediate remains. This is not “just take the slow step’s coefficients”: you must justify the approximation and substitute. If the slow step removes intermediate too rapidly for the reversible step to re-equilibrate, this approximation may fail; a different kinetic treatment would be needed.
Let's work through it
For rapid A + B ⇌ I followed by slow I + C → P, derive the observed law under pre-equilibrium conditions.
- First-step forward and reverse rates are approximately equal: k1[A][B] ≈k−1[I].
- Solve [I] ≈(k1/k−1)[A][B]. The slow elementary step has r =k2[I][C].
- Substitute: r ≈(k2k1/k−1)[A][B][C]. This predicts first order in each reactant and third order overall in the modeled range. The shorthand labels describe a balanced route for the chemical units represented.
Try it from memory
Pause & retrieve
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Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Why not leave r =k2[I][C] as the final observed reactant law?
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Retrieval 1
Why not leave r =k2[I][C] as the final observed reactant law?
Worked answer and reason
I is an intermediate, not an independently controlled initial reactant. Use the fast-step relation to express it through A and B.
Eliminating the intermediate connects the mechanism to measurable concentration dependence.
Retrieval 2
What assumption permits equality of the fast forward and reverse rates?
Worked answer and reason
Rapid reversible interconversion maintains approximate equilibrium while the slow step removes relatively little intermediate.
The approximation is a statement about relative timescales.
Catch the tempting mistake
- If the second step is slow, copy the overall coefficients.
- Write the slow elementary law, then eliminate intermediates using a justified approximation.
- Pre-equilibrium means the whole reaction is at equilibrium.
- Only the fast reversible substep is near equilibrium; the net reaction can still form product.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 5 problem 16 · ap synthesis · symbolic
- Unit 5 problem 17 · ap synthesis · data
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For the teacher
- Require all three algebra steps: fast-rate equality, intermediate expression, substitution.
- If students equate the whole reaction’s rates, return to the distinction between a substep and net reaction.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 5.9.A
- Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting.
- 5.9.A.1
- If the first elementary reaction is not rate limiting, approximations (such as pre-equilibrium) must be made to determine a rate law expression.
Science references for this teaching
- College Board AP Chemistry Course and Exam Description, Fall 2024 — Exact unit learning requirements and exclusions; supplied PDF is pinned by the integration catalog.
- OpenStax Chemistry 2e: Reaction Mechanisms — Scientific check: intermediates, elementary molecularity, slow steps and rapid-equilibrium substitution.
Glossary and deeper reading · 4 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Stoichiometry
Stoichiometry uses a balanced chemical equation to relate amounts of reactants and products. Equation coefficients provide mole ratios rather than direct mass ratios. For example, forming water from hydrogen and oxygen consumes two moles of hydrogen molecules per mole of oxygen molecules.
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Molarity
Molarity is the amount of a solute in moles divided by the final solution volume in liters. It describes concentration rather than the total amount of solute. For example, 0.10 mole dissolved to make 0.50 liter of solution gives a molarity of 0.20 mole per liter.
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Reaction intermediate
A reaction intermediate is produced in one mechanism step and consumed in another. It cancels from the overall equation and differs from a catalyst, which is present before it is consumed and then regenerated. For example, a short-lived species formed in step one and used in step two can be an intermediate.
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Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.