AP Chemistry Unit 5 Kinetics: Learning Reaction Mechanisms and Deriving Rate Laws

Elementary Reactions

What an elementary reaction is

An elementary reaction (also called an elementary step) is a single molecular event that occurs exactly as written. That means the reactant particles shown in that step collide (or rearrange) in one “move” to form the products of that step.

This matters because kinetics is about how reactions happen, not just what the overall balanced equation looks like. For many real reactions, the overall equation hides the actual sequence of events. Elementary steps are the building blocks of that sequence.

Molecularity: counting who participates in a single step

For an elementary step, the number of reacting particles in that step is called the molecularity:

Unimolecular: 1 reactant particle breaks apart or rearranges.
Bimolecular: 2 reactant particles collide.
Termolecular: 3 reactant particles collide simultaneously (rare in real systems).

A key idea: molecularity is defined only for an elementary step, not for an overall reaction.

Why elementary steps are special: their rate laws come “straight from” their equations

For an elementary step, the rate law exponents match the stoichiometric coefficients of the reactants in that step.

Examples:

1) Unimolecular elementary step

A \rightarrow products

Rate law:

rate = k[A]

2) Bimolecular elementary step

A + B \rightarrow products

Rate law:

rate = k[A][B]

3) Bimolecular with two of the same reactant

2A \rightarrow products

Rate law:

rate = k[A]^2

This is a huge contrast with overall reactions, where you generally cannot write the rate law from the balanced equation.

Connecting to collision model (conceptually, not mathematically)

Elementary steps fit naturally with collision ideas: the rate depends on how often the required particles meet. A unimolecular step depends on how many A molecules exist (so [A]). A bimolecular step depends on how often A meets B, which scales with [A][B].

Energy perspective: one step, one barrier

Each elementary step has:

a transition state (a high-energy arrangement along the path), and
an activation energy for that step.

A multi-step mechanism therefore has multiple “humps” on a potential energy diagram—one per elementary step.

Worked example: identifying the rate law for an elementary step

Suppose a proposed elementary step is:

2NO_2 \rightarrow 2NO + O_2

Because it is stated to be elementary, the rate law would be:

rate = k[NO_2]^2

If an experiment instead finds rate = k[NO_2], that would be strong evidence the reaction is not elementary as written.

Exam Focus

Typical question patterns:
- You’re told a step is “elementary” and asked to write the rate law directly from its reactant coefficients.
- You’re shown a proposed step (uni- vs bi- vs termolecular) and asked to classify its molecularity.
- You’re asked why you cannot generally use the overall balanced equation to write the rate law.
Common mistakes:
- Treating the overall reaction as elementary and copying coefficients into a rate law.
- Calling an overall reaction “bimolecular” or “termolecular” (molecularity applies to steps).
- Thinking “termolecular” is common; simultaneous three-particle collisions are unlikely.

Introduction to Reaction Mechanism

What a reaction mechanism is

A reaction mechanism is a proposed sequence of elementary steps that adds up to the overall chemical reaction. It’s your best molecular-level story for how reactants become products.

A mechanism is not just a random set of steps—it must satisfy two major constraints:

1) It must reproduce the overall balanced equation when you add the steps.
2) It must be consistent with the experimentally determined rate law.

If either fails, the mechanism is not acceptable (or at least incomplete).

Why mechanisms matter in kinetics

Kinetics experiments often give you a rate law like:

rate = k[A]^m[B]^n

The exponents m and n tell you something about the underlying molecular events. Mechanisms are how chemists interpret those exponents in terms of collisions, intermediates, and slow steps.

Mechanisms also explain:

why catalysts change rates,
why intermediates might build up or stay low,
why temperature changes can affect some steps more than others.

Elementary steps, intermediates, and catalysts

When you add elementary steps to get the overall reaction, some species may cancel. These are the key players to recognize:

Intermediate: produced in one step and consumed in a later step. It does not appear in the overall reaction.
Catalyst: consumed in an early step and regenerated in a later step. It appears in the mechanism but not in the net reaction.

A practical way to spot them:

Write all steps.
Add reactants and products across steps.
Cancel species that appear on both sides.

If a species cancels and was formed then used, it’s an intermediate. If it was used then re-formed, it’s acting like a catalyst.

Rate-determining step: the “bottleneck” idea

Many mechanisms contain a rate-determining step (often abbreviated RDS): the slowest elementary step. If one step is much slower than the others, it controls how fast product can form—like a bottleneck in a factory line.

When an RDS exists and is clearly the slow step, the overall rate law often matches the rate law of that slow elementary step (with an important warning: sometimes that slow step contains an intermediate, and then you must eliminate it using earlier steps).

Energy diagram intuition for multi-step mechanisms

On a potential energy diagram:

each elementary step corresponds to a peak (a transition state),
the tallest peak (largest activation energy) often corresponds to the slow step.

You don’t usually need detailed diagrams to do AP Chemistry mechanism problems, but the “tallest hump is slowest step” intuition helps you make sense of why a mechanism could have a bottleneck.

Worked example: identifying intermediates in a mechanism

Consider the mechanism:

Step 1:

NO_2 + NO_2 \rightarrow NO_3 + NO

Step 2:

NO_3 + CO \rightarrow NO_2 + CO_2

Add the steps and cancel species:

Left side total: 2NO_2 + CO

Right side total: NO_3 + NO + NO_2 + CO_2

Cancel NO_3 (produced then consumed) and cancel one NO_2 (consumed then regenerated):

Net reaction:

NO_2 + CO \rightarrow NO + CO_2

Here NO_3 is an intermediate because it is formed in Step 1 and used up in Step 2.

Exam Focus

Typical question patterns:
- You’re given a multi-step mechanism and asked to identify intermediates and/or catalysts.
- You’re asked to add steps to show the mechanism sums to the overall reaction.
- You’re given an energy diagram with multiple peaks and asked which step is slow.
Common mistakes:
- Calling any species that cancels a “catalyst” (you must check whether it is consumed first or produced first).
- Forgetting that intermediates do not appear in the overall equation.
- Assuming the step with the largest stoichiometric coefficients is slow (rate depends on activation energy, not coefficients).

Reaction Mechanism and Rate Law

The core goal: connect experimentally measured rate laws to molecular steps

In kinetics, the rate law is determined by experiment. A mechanism must explain that law.

A general experimental rate law might look like:

rate = k[A]^m[B]^n

where:

rate is the reaction rate (often in M s^{-1}),
k is the rate constant (units depend on overall order),
[A] and [B] are molar concentrations,
m and n are reaction orders (not necessarily integers).

A mechanism is consistent with the rate law if, after you derive the rate expression from its steps (and eliminate intermediates), you obtain the same dependence on reactant concentrations.

Case 1: slow step is first and contains only reactants

If the slow step is the first step and involves only species that appear in the overall reactants, the derivation is straightforward: the overall rate law is the rate law for that slow elementary step.

Example mechanism:

Step 1 (slow):

A + B \rightarrow I

Step 2 (fast):

I \rightarrow products

Because Step 1 is elementary and slow:

rate = k[A][B]

This is the simplest scenario.

Case 2: slow step contains an intermediate (you must substitute)

More interesting mechanisms have a slow step that depends on an intermediate. Since intermediates are not measurable in a typical rate law, you must rewrite the rate law in terms of reactants.

A common AP Chemistry approach is fast pre-equilibrium when the first step is reversible and fast.

Fast pre-equilibrium method (common in AP problems)

If an early step is fast and establishes equilibrium:

A + B \rightleftharpoons I

then at equilibrium:

K = \frac{[I]}{[A][B]}

So:

[I] = K[A][B]

If a later step is slow and uses I:

I + C \rightarrow products

then the rate is controlled by the slow step:

rate = k[I][C]

Substitute [I] = K[A][B]:

rate = kK[A][B][C]

Often you combine constants into a new observed rate constant:

rate = k_{obs}[A][B][C]

where k_{obs} = kK.

Worked example (classic pattern): mechanism that produces a squared reactant order

Overall reaction:

2NO + Br_2 \rightarrow 2NOBr

Proposed mechanism:

Step 1 (fast equilibrium):

NO + Br_2 \rightleftharpoons NOBr_2

Step 2 (slow):

NOBr_2 + NO \rightarrow 2NOBr

Step 1 (equilibrium relationship):

K = \frac{[NOBr_2]}{[NO][Br_2]}

So:

[NOBr_2] = K[NO][Br_2]

Step 2 (rate-determining step):

rate = k[NOBr_2][NO]

Substitute the expression for [NOBr_2]:

rate = kK[NO]^2[Br_2]

So the predicted rate law is:

rate = k_{obs}[NO]^2[Br_2]

This kind of derivation explains why experimental orders can be 2, 1, or even fractional—because they come from how intermediates depend on reactant concentrations.

Checking a mechanism: two necessary tests

When AP problems ask whether a mechanism is “valid,” they typically mean:

1) Does it add to the overall reaction?

When you add steps, all intermediates should cancel.

2) Does it match the experimental rate law?

Derive the rate law from the mechanism (using RDS and substitutions).
Compare exponents and reactants appearing.

Failing either test means the mechanism is inconsistent.

What can and cannot be concluded from a rate law

A common misconception is that reaction order “tells you the balanced equation.” It doesn’t. The rate law can suggest features of a mechanism, but multiple mechanisms can sometimes produce the same overall rate law. In AP Chemistry, you usually treat “matches the rate law and sums correctly” as strong evidence the mechanism is acceptable.

Exam Focus

Typical question patterns:
- Given a mechanism, derive the rate law (identify the slow step; eliminate intermediates).
- Given an experimental rate law and a proposed mechanism, decide if they are consistent.
- Given a mechanism with a fast equilibrium followed by a slow step, use the equilibrium relationship to substitute for an intermediate.
Common mistakes:
- Writing the rate law from the overall balanced equation instead of from the mechanism.
- Forgetting to eliminate intermediates (leaving [I] in the final rate law).
- Treating a “fast equilibrium” step like it has no mathematical consequence (it provides the crucial substitution relationship).

Steady-State Approximation

What the steady-state approximation is

The steady-state approximation is a method for handling intermediates in a mechanism when you cannot (or are not told to) assume a simple fast equilibrium. It assumes that the concentration of an intermediate stays approximately constant during most of the reaction.

In plain language: the intermediate is formed and consumed quickly, so it never builds up much.

Mathematically, for an intermediate I, steady-state means:

\frac{d[I]}{dt} \approx 0

That does not mean [I] = 0. It means the _rate of formation_ of I approximately equals the _rate of consumption_ of I.

Why it matters

Many real mechanisms involve reactive intermediates (radicals, activated complexes, short-lived species). You often can’t measure them directly, but they influence the observed rate law. The steady-state approximation is a bridge between the mechanism steps and an observable rate law written only in terms of stable reactants.

It’s also a flexible tool: it can handle mechanisms where intermediates are consumed in multiple ways, or where the first step is not a clean equilibrium.

How to apply it (the workflow)

When you see a mechanism with an intermediate and you’re asked to derive a rate law, you can follow this logic:

1) Write the rate expression from the rate-determining step (or from the step that forms product at the relevant rate).
2) If that rate expression contains an intermediate, write a steady-state equation for that intermediate:

\frac{d[I]}{dt} = \text{(formation terms)} - \text{(consumption terms)} \approx 0

3) Solve for [I] in terms of reactant concentrations and rate constants.
4) Substitute back into the overall rate expression.

A subtle but important point: formation/consumption terms must come from the elementary-step rate laws. Each step has its own rate constant (often written k_1, k_{-1}, k_2, etc.).

Worked example: steady-state gives a non-simple rate law

Consider this mechanism:

Step 1 (fast):

A + B \rightarrow I

Step 2 (fast):

I \rightarrow A + B

Step 3 (slow):

I + B \rightarrow products

Here I is an intermediate: it is formed in Step 1 and consumed in Steps 2 and 3.

Assign rate constants:

Step 1 forward: k_1
Step 2: k_{-1} (a common notation because it “undoes” Step 1)
Step 3: k_2

Step A: write the rate law from the slow (rate-determining) step.

Step 3 is elementary, so:

rate = k_2[I][B]

But [I] is an intermediate, so we use steady-state.

Step B: write the steady-state equation for I.

Formation of I comes from Step 1:

\text{formation rate of } I = k_1[A][B]

Consumption of I comes from Step 2 and Step 3:

\text{consumption rate of } I = k_{-1}[I] + k_2[I][B]

Steady-state condition:

\frac{d[I]}{dt} = k_1[A][B] - k_{-1}[I] - k_2[I][B] \approx 0

Step C: solve for [I].

Factor out [I] from the consumption terms:

k_1[A][B] = [I](k_{-1} + k_2[B])

So:

[I] = \frac{k_1[A][B]}{k_{-1} + k_2[B]}

Step D: substitute into the rate expression.

rate = k_2[B] \left(\frac{k_1[A][B]}{k_{-1} + k_2[B]}\right)

So:

rate = \frac{k_1k_2[A][B]^2}{k_{-1} + k_2[B]}

This result is valuable because it shows why some reactions do not follow a simple power-law order across all conditions. For example:

If k_{-1} dominates (or [B] is small so k_2[B] is small), the denominator is approximately constant and the rate behaves roughly like [A][B]^2.
If k_2[B] dominates (large [B]), the denominator is approximately proportional to [B] and the rate behaves roughly like [A][B].

You are not always asked to do those limiting-case interpretations in AP Chemistry, but they help you see the physical meaning: at high [B], the pathway consuming I via Step 3 becomes very significant compared to the “back reaction” Step 2.

Where students often go wrong with steady-state

Two common pitfalls are worth calling out explicitly:

1) Setting [I] = 0 instead of setting its rate of change to zero. The intermediate exists; it just stays low and roughly constant.
2) Forgetting one of the consumption pathways. If I is consumed in two different steps, both must appear in the steady-state balance.

Steady-state vs fast pre-equilibrium

Both methods eliminate intermediates, but they rely on different assumptions:

Fast pre-equilibrium assumes a reversible step reaches equilibrium quickly, giving an equilibrium constant relationship.
Steady-state assumes the intermediate concentration stays roughly constant, giving a formation-equals-consumption relationship.

In AP Chemistry, problems often cue “fast equilibrium” explicitly. If they do not, or if the intermediate has multiple consumption routes, steady-state is often the appropriate tool.

Exam Focus

Typical question patterns:
- You’re given a mechanism with an intermediate and asked to derive a rate law using \frac{d[I]}{dt} \approx 0.
- You’re asked to write the steady-state expression by summing formation and consumption terms for an intermediate.
- You’re asked to eliminate an intermediate from a rate law so the final expression contains only reactants.
Common mistakes:
- Writing the steady-state equation but forgetting to include every step that creates or destroys the intermediate.
- Mixing up which step’s rate expression corresponds to product formation (using the wrong step as “the rate”).
- Algebra errors when solving for [I], especially factoring [I] out of multiple consumption terms.