5.8.1 Rate-Determining Step and the Rate Law

AP Syllabus focus: ‘If the first step is rate limiting (or all steps are irreversible), the overall rate law is determined by the slowest elementary step.’

Reaction mechanisms connect microscopic steps to a macroscopic rate law. In many AP Chemistry situations, one slow step controls how fast products form, allowing the overall rate law to be inferred from that step.

Core idea: the slow step “sets” the pace

In a multistep mechanism, not every elementary step contributes equally to the observed reaction rate. If one step is much slower than the others, it throttles the overall process.

Rate-determining step (RDS): the slowest elementary step in a reaction mechanism; it limits the overall reaction rate because faster steps cannot proceed faster than this bottleneck allows.

A helpful analogy is an assembly line: the slowest station determines the throughput, even if all other stations are faster.

When the RDS determines the overall rate law (AP focus)

The AP-specified condition is:

If the first step is rate limiting, the overall rate law comes from that slow first elementary step.
Alternatively, if all steps are irreversible, the mechanism behaves like a one-way sequence where the slowest step controls the rate; under this condition, the rate law is likewise determined by the slowest elementary step.

This is a powerful shortcut because it avoids having to derive a complex law from multiple steps.

What you use from the slow elementary step

For an elementary step, the rate law is written directly from the reactant side of that step.

Particulate collision-orientation diagram for a bimolecular elementary step, contrasting an ineffective collision with a properly oriented, reactive collision. The figure emphasizes that the rate of an elementary step depends on the concentrations of the colliding reactants (e.g., a bimolecular step implies a form like $\text{Rate} = k[A][B]$ ). Source

Therefore, under the AP condition above, you can write the overall rate law directly from the reactants in the RDS.

Identify the slow step (often labelled “slow” in problems).
Use only the reactants that collide/react in that slow step.
Use their stoichiometric coefficients in that step as exponents in the rate law (because it is elementary).

$\text{Rate} = k\prod_i [\text{reactant}_i]^{m_i}$

$k$ = rate constant (units depend on overall order)

$[\text{reactant}_i]$ = molar concentration (M)

$m_i$ = reaction order with respect to reactant $i$ (unitless)

$\text{If the RDS is an elementary step: Rate} = k[\text{A}]^{a}[\text{B}]^{b}$

$[\text{A}],[\text{B}]$ = concentrations of reactants in the slow step (M)

$a,b$ = stoichiometric coefficients of A and B in the slow elementary step (unitless)

This approach is explicitly limited to the AP condition stated: first step rate limiting, or all steps irreversible, so the overall rate is controlled by the slowest elementary step.

Recognising what can (and cannot) appear in the rate law

Under the RDS shortcut, the species in the overall rate law must be consistent with the slow elementary step you selected.

Species that are not reactants in the RDS should not appear in the rate expression you write from that step.
The rate law is about the mechanism, not the overall balanced equation; overall stoichiometric coefficients do not automatically become rate-law exponents.

Common pitfalls to avoid

Do not “read” the rate law from the overall balanced equation unless you are explicitly told the reaction is elementary (rare for overall reactions).
Do not assume the last step controls the rate because it makes products; the RDS is the slowest step, regardless of where it occurs in the sequence.
Do not mix steps: the RDS-based rate law comes from one elementary step (the slow one), not from adding reactants across multiple steps.

How this shows up on AP-style prompts

You are typically asked to do one or more of the following:

Given a proposed mechanism with a labelled slow step, write the overall rate law.
Decide whether a proposed mechanism is consistent with an experimentally determined rate law, by comparing it to the rate law predicted by the slow step (under the AP condition).

FAQ

Because intermediates formed by the slow step cannot be produced faster than that step proceeds.

Fast subsequent steps can only process what is supplied, so they do not increase the overall throughput.

Not reliably.

When multiple steps have comparable rates, the observed rate may depend on more than one step, and the simple RDS shortcut may fail.

Yes, in principle.

If a catalyst participates in the slow step, it can appear in the rate law even though it cancels from the overall equation.

You can state that negligible reverse rates mean no significant back-reaction to establish equilibrium between steps.

Thus, the net forward progress is limited by the slowest forward elementary step.

A common clue is that the experimentally determined rate law matches the reactant concentrations from the first elementary step.

Additionally, changing concentrations of species only used in later steps has little to no effect on the measured rate.

Practice Questions

(2 marks) A proposed mechanism has a slow first step: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$ (slow). Write the predicted rate law.

Correct form: $\text{Rate} = k[\text{NO}_2][\text{F}_2]$ (1)
Correct first-order dependence on each reactant in the slow step (1)

(5 marks) A mechanism is proposed: Step 1: $\text{A} + \text{B} \rightarrow \text{C}$ (slow)
Step 2: $\text{C} + \text{D} \rightarrow \text{E}$ (fast)
(a) State which step is rate-determining. (1)
(b) Write the overall rate law predicted by this mechanism. (2)
(c) An experiment finds $\text{Rate} = k[\text{A}][\text{B}]$ and shows no dependence on $[\text{D}]$ . Explain whether the data support the mechanism. (2)

(a) Step 1 identified as rate-determining because it is slow (1)
(b) Correct rate law from slow elementary step: $\text{Rate} = k[\text{A}][\text{B}]$ (2: one mark for correct species, one for correct orders)
(c) States the mechanism is supported because predicted law matches experiment and $[\text{D}]$ does not appear (2)

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