Unit 5 · Topic 5.9 · about 30 minutes · 7 parts · 11 scored checks, 2 of them multiple choice, plus 2 from earlier lessons

AP Chemistry 5.9: Pre-Equilibrium Approximation

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Unit 5: 7 to 9% of the multiple-choice section

What you'll be able to do

Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting, using the pre-equilibrium approximation.

Key vocabulary

Term	What it means	The exam trap
Pre-equilibrium approximation	Treating a fast, reversible step that comes before the slow step as if its forward and reverse rates were equal, so the intermediate's concentration can be written in terms of reactants.	The approximation is applied to the fast step before the slow one, not to the slow step itself.
Reversible step	An elementary step that can run in both directions, written with a double arrow, each direction having its own rate law and rate constant.	The two directions have different rate constants, so one k cannot stand for both.
Reaction intermediate	A species produced by some elementary steps and consumed by others, so it is present only while the reaction is occurring.	Check which species are intermediates before you finish, because none can stay in the final rate law.
Rate-limiting step	The slowest elementary step in a mechanism, which sets the rate of the overall reaction.	With a fast first step, the slow step's own rate law is only the starting point, not the answer.

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Part 1 Think first A holding lane at a car ferry

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

At a small car ferry dock, cars drive into a holding lane. Cars enter the lane quickly. Each car waiting in the lane has the same chance every minute of giving up and driving out, so the more cars are waiting, the more leave each minute, and the number in the lane soon settles at a steady level where as many leave each minute as arrive. The ferry loads cars from the lane only slowly. This is a teaching example. On a holiday weekend twice as many cars reach the dock each minute. About how does the steady number of cars in the holding lane change?

In short. When a fast, reversible step comes before the slow step, its forward and reverse rates are taken to be equal, and that balance gives the intermediate's concentration in terms of reactants.

The slow step's rate law contains an intermediate, which is not a reactant, so it has to be replaced before the rate law is finished.
Setting the fast step's forward rate equal to its reverse rate and solving for the intermediate lets you substitute it into the slow step's rate law.
A mechanism is consistent with the evidence only if the rate law this method gives matches the measured one.

Part 2 Learn it When the first step is fast

2a. The slow step needs an intermediate

In the last topic the slow step came first, so its rate law held only reactants. Now suppose a fast step comes first and the slow step uses what it makes.

Take 2 NO + O₂ → 2 NO₂, with this mechanism:

Step 1 (fast, reversible): NO + NO ⇌ N₂O₂
Step 2 (slow): N₂O₂ + O₂ → 2 NO₂

The double arrow means the step runs in both directions. Step 2 is the slow step, so it sets the rate: rate = k₂[N₂O₂][O₂]. But N₂O₂ is an intermediate. It is made and used up inside the reaction, so you cannot add a measured amount of it, and a rate law is written in terms of reactants. This rate law is not finished.

The first step cannot be ignored either. It is fast, so it makes and unmakes N₂O₂ many times before step 2 uses any up, and that back-and-forth decides how much N₂O₂ there is. Chemists handle this with an approximation, the pre-equilibrium approximation, which you will use in the next section.

Quick check · Multiple choice

A mechanism has two steps. Step 1 (fast, reversible): A + B ⇌ I. Step 2 (slow): I + C → P. I is an intermediate. This is a teaching example. A student proposes the rate law rate = k[I][C], the rate law of the slow step. Why is that rate law not finished?

2b. Forward rate equals reverse rate

A fast reversible step soon reaches a balance: the rate of its forward reaction equals the rate of its reverse reaction. Using that balance to find the intermediate is the pre-equilibrium approximation. It works because step 2 is slow. It takes N₂O₂ away so gradually that step 1 stays balanced.

Call the rate constants of the forward and reverse reactions of step 1 k₁ and k₋₁. Each direction is an elementary reaction with its own rate law:

forward rate = k₁[NO]²
reverse rate = k₋₁[N₂O₂]

Set them equal and solve for the intermediate:

k₁[NO]² = k₋₁[N₂O₂], so [N₂O₂] = (k₁/k₋₁)[NO]²

Put that into the slow step's rate law:

rate = k₂[N₂O₂][O₂] = k₂(k₁/k₋₁)[NO]²[O₂] = k[NO]²[O₂], where k = k₂k₁/k₋₁

The constants combine into one rate constant. The ratio k₁/k₋₁ is the equilibrium constant of the fast step, which a later unit treats in full. If the fast step makes two species, the reverse rate has both: for A ⇌ B + C it is k₋₁[B][C].

Quick check · Put it in order

Put the steps of a pre-equilibrium derivation in order.

2c. Choosing the method

Two kinds of mechanism now have rate laws you can find, and the first job is to tell which kind you have.

Which method a mechanism needs
The mechanism	What to do
The first step is the slow step	Write that step's rate law. It holds only reactants.
A fast, reversible step comes first, and the slow step uses its product	Write the slow step's rate law, then replace the intermediate using the fast step's balance.

Whichever method you use, the finished rate law is judged against the measured one, and it must hold no intermediate. A fast step that comes after the slow step does not change the rate law, because it only uses up what the slow step makes. And the approximation is only fair when the first step really is fast compared with the slow step after it.

Quick check · Sort it

Each card shows a two-step mechanism. Tap a card, then the bin for how its rate law is found.

Rate law straight from the slow step

Needs the pre-equilibrium approximation

Common mistakes

The mistake	Why it is wrong	Do this instead
Leaving the intermediate in the rate law.	An intermediate cannot be set or measured, so a finished rate law cannot contain one.	Replace it using the balance of the fast step before you stop.
Forgetting that the reverse rate has every product of the fast step in it.	A step that makes two species has a reverse rate that depends on both, so one of them can end up under the line.	Write the reverse rate law from the step's own products, then solve.
Using the coefficients of the overall equation as the exponents.	The exponents come from the slow step and the balance of the fast step, and they can differ from the coefficients.	Work the exponents out from the mechanism, or read them from the data.

Part 3 Watch it done Nitric oxide and bromine

Predict first, then watch each step, then finish the calculation yourself.

Nitric oxide reacts with bromine: 2 NO + Br₂ → 2 NOBr. Experiments find the rate is second order in NO and first order in Br₂. A proposed mechanism has two steps. Step 1 (fast, reversible): NO + Br₂ ⇌ NOBr₂. Step 2 (slow): NOBr₂ + NO → 2 NOBr. Does the mechanism predict the measured rate law?

Predict first

Before you open the first step, predict: will the finished rate law this mechanism gives contain NOBr₂, the intermediate?

Step 1 · Write the slow step's rate law
Step 2 is slow, so rate = k₂[NOBr₂][NO]. NOBr₂ is an intermediate, so it has to be replaced.
Step 2 · Use the balance of the fast step
Step 1 is fast and reversible, so its forward and reverse rates are equal: k₁[NO][Br₂] = k₋₁[NOBr₂]. Solving, [NOBr₂] = (k₁/k₋₁)[NO][Br₂].
Why this step?
Why can the forward and reverse rates of step 1 be set equal while the reaction is still going?
Step 1 is fast, so it keeps reaching balance, and step 2 is slow, so it removes NOBr₂ too gradually to upset that balance.
Step 3 · Substitute
rate = k₂(k₁/k₋₁)[NO][Br₂][NO] = (k₂k₁/k₋₁)[NO]²[Br₂]. This is second order in NO and first order in Br₂, as measured, so the mechanism is consistent with the experiment.
Your step · Calculate

Finish it. The rate law is rate = k[NO]²[Br₂] with k = k₂k₁/k₋₁. At one temperature k₁ = 4.0 M⁻¹·s⁻¹, k₋₁ = 2.0 s⁻¹ and k₂ = 0.50 M⁻¹·s⁻¹. These numbers are a teaching example. What is k, in M⁻²·s⁻¹? Give your answer to two significant figures.

Your answer M⁻²·s⁻¹

The rate constant of the reaction is not the constant of any one step. It combines the constants of three elementary reactions, the forward and reverse reactions of step 1 and step 2, and the first step's balance is why NO ends up squared even though only one NO is in the slow step.

Part 4 Try it Ozone and an oxygen atom

Ozone breaks down to oxygen: 2 O₃ → 3 O₂. A proposed mechanism has two steps:

Step 1 (fast, reversible): O₃ ⇌ O₂ + O
Step 2 (slow): O + O₃ → 2 O₂

Try it · Spot the error

A student derives the rate law for the ozone mechanism. Tap the line with the mistake.

Try it · Calculate

For the ozone mechanism above, by what factor does the rate change when [O₃] is doubled and [O₂] is doubled? Give a whole number.

Your answer

Part 5 Explain it Nitric oxide and hydrogen

On the exam this is handwritten, so answer on paper first. Nothing you type here leaves this page. Then check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like a short-answer question, which is worth 4 points.

Free-response practice · Short-answer style

The reaction 2 NO + 2 H₂ → N₂ + 2 H₂O is found by experiment to be second order in NO and first order in H₂. A proposed mechanism has three steps.

Step 1 (fast, reversible): NO + NO ⇌ N₂O₂

Step 2 (slow): N₂O₂ + H₂ → N₂O + H₂O

Step 3 (fast): N₂O + H₂ → N₂ + H₂O

(a) Identify the rate-limiting step, and write its rate law in terms of the species in that step.

(b) Determine the rate law of the reaction in terms of reactants only, using step 1.

(d) Explain why step 3 does not appear in the rate law.

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Model answer

(a) Step 2 is the rate-limiting step. Its rate law is rate = k₂[N₂O₂][H₂].

(b) N₂O₂ is an intermediate. Step 1 is fast and reversible, so its forward and reverse rates are equal: k₁[NO]² = k₋₁[N₂O₂]. Then [N₂O₂] = (k₁/k₋₁)[NO]², and rate = (k₂k₁/k₋₁)[NO]²[H₂].

(d) Step 3 comes after the slow step and is fast. It only uses up N₂O that step 2 makes, so it does not change how fast step 2 goes.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

(a) Identifies step 2 as rate limiting and gives rate = k₂[N₂O₂][H₂].
(b) Sets the forward and reverse rates of step 1 equal, k₁[NO]² = k₋₁[N₂O₂], and substitutes to get rate = (k₂k₁/k₋₁)[NO]²[H₂].
(c) Says it is consistent, because the derived rate law is second order in NO and first order in H₂, as measured.
(d) Says step 3 is fast and comes after the slow step, so it only uses up what step 2 makes and does not set the rate.

Part 6 Check yourself Nitric oxide and chlorine

Both case questions use the table.

The reaction 2 NO + Cl₂ → 2 NOCl is studied at one temperature, and the table gives the initial rate in three experiments. These numbers are a teaching example. Two mechanisms are proposed. Mechanism A: Step 1 (fast, reversible) NO + Cl₂ ⇌ NOCl₂, then Step 2 (slow) NOCl₂ + NO → 2 NOCl. Mechanism B: Step 1 (slow) NO + Cl₂ → NOCl + Cl, then Step 2 (fast) Cl + NO → NOCl.

Initial rates at one temperature
Experiment	[NO] (M)	[Cl₂] (M)	Initial rate (M/s)
1	0.100	0.200	6.00 × 10⁻³
2	0.200	0.200	2.40 × 10⁻²
3	0.100	0.400	1.20 × 10⁻²

Case question 1 · Multiple choice

Using the table: which mechanism is consistent with the rate data?

Case question 2 · Calculate

Using the table: what is the rate constant k for the rate law that mechanism A gives, in M⁻²·s⁻¹, from experiment 1? Give your answer to three significant figures.

Your answer M⁻²·s⁻¹

The rest stand on their own and do not use the case.

Check yourself · Match

Each pair is a fast, reversible first step and the concentration of its intermediate, I, when the step is in balance. Match each step to its expression.

Check yourself · Fill the statement

Complete the description of the pre-equilibrium approximation.

For a fast, reversible first step A + B ⇌ I, the forward rate k₁[A][B] and the reverse rate k₋₁[] are once the step is in balance. Solving for the gives [I] = (k₁/k₋₁)[A][B], which replaces it in the slow step's rate law.

Check yourself · Select all

Select the two true statements about the pre-equilibrium approximation.

Select 2 answers, then check.

Part 7 Keep it Bring back earlier ideas

This lesson leans on two earlier ideas: the slow step sets the rate of a reaction, and an intermediate is made by one step and used by a later one. Pulling them back up is what makes them last.

From 5.8 Reaction Mechanism and Rate Law · Multiple choice

A reaction has a two-step mechanism. Step 1 is slow and step 2 is fast. Which change would increase the overall rate the most?

A mechanism can have more than one intermediate. One proposed mechanism for 2 NO + 2 H₂ → N₂ + 2 H₂O has three steps:

Step 1: NO + NO → N₂O₂
Step 2: N₂O₂ + H₂ → N₂O + H₂O
Step 3: N₂O + H₂ → N₂ + H₂O

From 5.7 Introduction to Reaction Mechanisms · Select all

Select the two species that are intermediates in this mechanism.

Select 2 answers, then check.

In three lines

If a fast, reversible step comes before the slow step, the slow step's rate law holds an intermediate, which must be replaced before the rate law is finished.
Set the fast step's forward rate equal to its reverse rate, solve for the intermediate, and put the result into the slow step's rate law.
The finished rate law can differ from the coefficients of the overall equation, and it is consistent with a mechanism only if it matches the measured one.

Exam strategy for this topic

Section I, Multiple Choice: 60 questions in 90 minutes.
Section II, Free Response: 7 questions in 105 minutes.
A scientific or graphing calculator is recommended for use on both sections of the exam.

Write the slow step's rate law first. If it holds an intermediate, set the forward and reverse rates of the fast step equal, solve for the intermediate, and substitute. Check that the finished rate law has no intermediate in it, and compare it with the measured one.

Questions students ask

What is the pre-equilibrium approximation?

It treats a fast, reversible step that comes before the slow step as being in balance, so its forward and reverse rates are equal. That gives the concentration of the intermediate in terms of reactants.

Why can the rate law not contain an intermediate?

An intermediate is made and used up inside the reaction, so its concentration cannot be set or measured. The rate law has to be written in terms of reactants.

How do you find the rate law when the first step is fast and reversible?

Write the slow step's rate law, set the forward and reverse rates of the fast step equal, solve for the intermediate, and substitute it into the slow step's rate law.

Why can the rate law differ from the overall equation's coefficients?

The exponents come from the slow step and from the balance of the fast step, not from the coefficients of the overall equation, so they can be different.

What happens to a step that comes after the slow step?

It does not appear in the rate law. A fast step after the slow step only uses up what the slow step makes.

Course alignment, for teachers

AP Chemistry topic 5.9, Unit 5: Kinetics.

Skill 5.B: Identify an appropriate theory, definition, or mathematical relationship to solve a problem.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.