Introduction to reaction mechanisms — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Elementary reactions, molecularity, reaction intermediates, rate-determining step (RDS), and how to write and validate rate laws from proposed reaction mechanisms per AP Chemistry CED learning objectives.

You should already know: Definition of rate law and reaction order; how to determine rate laws from experimental data; net reaction balancing.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Introduction to reaction mechanisms?

A reaction mechanism is the step-by-step sequence of elementary bond-breaking and bond-forming events that describes how an overall net chemical reaction occurs at the molecular level. Per the AP Chemistry CED, this topic makes up approximately 20% of Unit 5 (Kinetics) exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, typically 1-2 MCQ questions and one multi-part FRQ subquestion. Standard notation conventions separate each elementary step as a distinct reaction, with single-headed arrows for irreversible steps and double equilibrium arrows for reversible steps. The overall net reaction is obtained by adding all elementary steps and canceling species that appear on both sides of the reaction arrow. Unlike the overall reaction, which only shows starting reactants and final products, mechanisms reveal intermediate species that are produced in one step and consumed in a later step, so they never appear in the net reaction. This topic is the core connection between microscopic molecular behavior and macroscopic measured reaction kinetics.

2. Elementary Reactions and Molecularity

An elementary reaction is a single step in a mechanism that describes one actual molecular collision or rearrangement; it cannot be broken down into smaller discrete steps. A key rule specific to elementary reactions: the rate law for an elementary reaction can be written directly from its stoichiometry, because it describes a single molecular event. This is not true for overall net reactions, which require experimental data to get a rate law. Molecularity is defined as the number of reactant particles that participate in an elementary step. There are three common classifications:

1. Unimolecular: One reactant particle reacts (e.g., $A\to \text{products}$), rate law: $\text{rate}=k[A]$, first order overall.
2. Bimolecular: Two reactant particles collide and react (either $A+B\to \text{products}$ or $2A\to \text{products}$), rate laws: $\text{rate}=k[A][B]$ (first order in A, first order in B, second order overall) or $\text{rate}=k[A{]}^2$ (second order overall).
3. Termolecular: Three reactant particles collide simultaneously (very rare, due to the low probability of three particles colliding with correct energy and orientation at the same time), rate law: $\text{rate}=k[A][B][C]$, third order overall.

Worked Example

Write the rate law for each elementary reaction below, and state the molecularity of each: (a) $C_4{H}_8(g)\to 2C_2{H}_4(g)$ (b) $NO(g)+O_3(g)\to N{O}_2(g)+O_2(g)$

1. For reaction (a): Count the number of reactant molecules: there is only 1 molecule of $C_4{H}_8$, so molecularity is unimolecular.
2. For elementary reactions, the order of each reactant equals its stoichiometric coefficient. The exponent for $[C_4{H}_8]$ is 1, so the rate law is $\text{rate}=k[C_4{H}_8]$.
3. For reaction (b): There are two different reactant molecules, so molecularity is bimolecular.
4. Each reactant has a stoichiometric coefficient of 1, so exponents are both 1, giving the rate law $\text{rate}=k[NO][O_3]$.

Exam tip: On the AP exam, you can never write a rate law for an overall reaction from stoichiometry—only for explicitly labeled elementary reactions. If the problem does not state the step is elementary, you must use experimental data or the mechanism's rate-determining step to get the rate law.

3. Reaction Intermediates and Catalysts

When adding elementary steps to get the overall net reaction, two types of species do not appear in the final net reaction: reaction intermediates and catalysts. It is critical to be able to distinguish between the two for AP exam questions. A reaction intermediate is produced in one elementary step and consumed in a subsequent later step. Intermediates are not present at the start of the reaction, and they never appear in the overall rate law, because their concentration is dependent on the concentration of the starting reactants. A catalyst is consumed in an early elementary step and regenerated in a later elementary step. Catalysts are present at the start of the reaction, speed up the reaction by changing the mechanism, and are not consumed overall. Like intermediates, they do not appear in the net overall reaction. To distinguish between the two: track the order of appearance: intermediate = product first, reactant second; catalyst = reactant first, product second.

Worked Example

The decomposition of hydrogen peroxide in the presence of bromide ion follows this mechanism: Step 1: $H_2{O}_2(aq)+H^{+}(aq)+B{r}^{-}(aq)\to HOBr(aq)+H_{2}O(l)$ Step 2: $HOBr(aq)+H_2{O}_2(aq)\to H^{+}(aq)+B{r}^{-}(aq)+O_2(g)+H_{2}O(l)$ Identify the intermediate and catalyst in this mechanism.

1. List all species by their position in each step: $B{r}^{-}$ is a reactant in step 1 (early step) and a product in step 2 (late step).
2. $HOBr$ is a product in step 1 (early step) and a reactant in step 2 (late step).
3. By definition: any species that is consumed first (reactant early) and produced later (product late) is a catalyst, so $B{r}^{-}$ is the catalyst.
4. Any species produced first (product early) and consumed later (reactant late) is an intermediate, so $HOBr$ is the reaction intermediate.

Exam tip: Always check for $H^{+}$ or other common ions that are regenerated—they are often catalysts that students misclassify as intermediates because they don't track their order of appearance.

4. Rate-Determining Step and Rate Law Derivation

The overall rate of a reaction mechanism is limited by the slowest step in the mechanism, called the rate-determining step (RDS). This is analogous to how the time to drive across a city is limited by the slowest stretch of highway: the RDS sets the maximum speed of the entire reaction, so the overall rate law is exactly the rate law of the RDS. If the RDS is the first step in the mechanism, and there are no intermediates in it, you can write the rate law directly from the RDS elementary step stoichiometry. If the RDS comes after one or more fast reversible steps, the RDS will usually contain an intermediate (from the fast step), so you need to substitute the intermediate concentration using the pre-equilibrium approximation: the fast reversible step reaches equilibrium quickly, so the rate of the forward step equals the rate of the reverse step, so you can solve for the intermediate concentration in terms of starting reactants. The formula for pre-equilibrium is: $$k_{forward}[reactants]=k_{reverse}[intermediate]⟹[intermediate]=\frac{k_{forward}}{k_{reverse}}[reactants]$$

Worked Example

A reaction has this proposed mechanism: Step 1 (fast, reversible): $B{r}_2(g)\rightleftharpoons 2Br(g)$ (intermediate $Br$) Step 2 (slow): $Br(g)+H_2(g)\to HBr(g)+H(g)$ (intermediate $H$) Step 3 (fast): $H(g)+B{r}_2(g)\to HBr(g)+Br(g)$ Derive the rate law for this overall reaction.

1. The RDS is the slow step (step 2), so write its elementary rate law: $\text{rate}=k_2[Br][H_2]$.
2. $Br$ is an intermediate from the fast pre-equilibrium step 1, so substitute it using the pre-equilibrium approximation: $k_1[B{r}_2]=k_{-1}[Br{]}^2⟹[Br]={\left(\frac{k_1}{k_{-1}}[B{r}_2]\right)}^{1/2}$.
3. Substitute $[Br]$ into the RDS rate law: $$\text{rate}=k_2{\left(\frac{k_1}{k_{-1}}[B{r}_2]\right)}^{1/2}[H_2]=k[B{r}_2{]}^{1/2}[H_2]$$ where $k=k_2{\left(\frac{k_1}{k_{-1}}\right)}^{1/2}$ is the overall rate constant.

Exam tip: When asked to confirm if a proposed mechanism matches an experimental rate law, always start with the RDS, substitute intermediates, then compare. AP exam graders require this order of working to award full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing a rate law for the overall reaction directly from its stoichiometry (e.g., for $2NO+O_2\to 2N{O}_2$, writing $\text{rate}=k[NO{]}^2[O_2]$ just from stoichiometry). Why: Students confuse the rule for elementary reactions with the rule for overall reactions, since the rate law often matches by coincidence in examples. Correct move: Only write rate law from stoichiometry if the step is explicitly stated to be elementary; for all overall reactions, rate law must come from experiment or the mechanism's RDS.
• Wrong move: Classifying a catalyst as an intermediate because neither appears in the overall reaction. Why: Students only remember that both are canceled from the net reaction, so they mix up the classification. Correct move: Always track order of appearance: produced first then consumed = intermediate; consumed first then produced = catalyst.
• Wrong move: Leaving an intermediate concentration term in the final overall rate law. Why: Students forget that intermediates are not starting reactants, so their concentration is dependent on reactant concentrations. Correct move: Always substitute any intermediate that appears in the RDS using the pre-equilibrium approximation to replace it with reactant concentration terms.
• Wrong move: Assuming the RDS must be the first step in the mechanism. Why: Most introductory examples have RDS as the first step, so students get used to this pattern. Correct move: Always use the labeled slow step as the RDS, regardless of its position in the mechanism.
• Wrong move: Eliminating a mechanism with a termolecular step on MCQ because termolecular steps are rare. Why: Students overgeneralize the "rare" label to "impossible". Correct move: Termolecular steps are rare but possible; only eliminate the mechanism if it conflicts with the experimental rate law.
• Wrong move: Adding the exponents from all elementary steps to get the overall reaction order. Why: Students assume all steps contribute equally to the overall rate. Correct move: Only the RDS exponents (after substituting intermediates) determine the overall rate order.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

The experimental rate law for the reaction $2A+B\to C$ is $\text{rate}=k[A{]}^2[B]$. Which of the following proposed mechanisms is consistent with this rate law? A) Step 1 (slow): $2A\to D$; Step 2 (fast): $D+B\to C$ B) Step 1 (fast): $A+B\rightleftharpoons D$; Step 2 (slow): $D+A\to C$ C) Step 1 (slow): $A+B\to D$; Step 2 (fast): $D+A\to C$ D) Step 1 (fast): $2A\rightleftharpoons D$; Step 2 (fast): $D+B\to C$

Worked Solution: We derive the rate law for each option and compare it to the given experimental rate law. For A, the RDS is step 1, giving rate = $k[A{]}^2$, which is second order overall, inconsistent with the third order experimental rate law. For B, the RDS is step 2, so rate = $k_2[D][A]$. From pre-equilibrium: $k_1[A][B]=k_{-1}[D]⟹[D]=\frac{k_1}{k_{-1}}[A][B]$. Substituting gives rate = $\frac{k_1{k}_2}{k_{-1}}[A{]}^2[B]=k[A{]}^2[B]$, which matches. For C, the RDS is step 1, giving rate = $k[A][B]$, which is second order, inconsistent. For D, there is no labeled slow step, so the mechanism has no defined RDS and is inconsistent. The correct answer is B.

Question 2 (Free Response)

The reaction of nitrogen monoxide and oxygen is $2NO(g)+O_2(g)\to 2N{O}_2(g)$. A proposed mechanism is: Step 1 (fast, reversible): $NO(g)+NO(g)\rightleftharpoons N_2{O}_2(g)$ (intermediate $N_2{O}_2$) Step 2 (slow): $N_2{O}_2(g)+O_2(g)\to 2N{O}_2(g)$

(a) Identify the intermediate in this mechanism. (b) Write the overall net reaction and confirm it matches the given overall equation. (c) Derive the rate law for this mechanism.

Worked Solution: (a) $N_2{O}_2$ is produced in step 1 and consumed in step 2, so it is the reaction intermediate. (b) Add the two steps: $NO+NO+N_2{O}_2+O_2\to N_2{O}_2+2N{O}_2$. Cancel the intermediate $N_2{O}_2$, giving $2NO(g)+O_2(g)\to 2N{O}_2(g)$, which matches the given overall equation. (c) The RDS is the slow step 2, so its rate law is $\text{rate}=k_2[N_2{O}_2][O_2]$. Substitute the intermediate $N_2{O}_2$ using pre-equilibrium: $k_1[NO{]}^2=k_{-1}[N_2{O}_2]⟹[N_2{O}_2]=\frac{k_1}{k_{-1}}[NO{]}^2$. Substitute into the RDS rate law: $\text{rate}=k_2\left(\frac{k_1}{k_{-1}}[NO{]}^2\right)[O_2]=k[NO{]}^2[O_2]$, where $k=\frac{k_1{k}_2}{k_{-1}}$.

Question 3 (Application / Real-World Style)

The acid-catalyzed iodination of acetone has an experimental rate law of $\text{rate}=k[acetone][H^{+}]$. A proposed mechanism for the reaction is: Step 1 (fast, reversible): $acetone+H^{+}\rightleftharpoons protonatedacetone(intermediate)$ Step 2 (slow): $protonatedacetone\to enol+H^{+}$ Step 3 (fast): $enol+I_2\to iodinatedacetone$ In an experiment, the initial rate is measured as $2.1\times 10^{-6}M/s$ when $[acetone]=0.25M$ and $[H^{+}]=0.10M$. Calculate the rate constant $k$, and confirm if the mechanism is consistent with the experimental rate law.

Worked Solution: First, derive the rate law from the proposed mechanism. The RDS is step 2, which is elementary, so rate = $k_2[protonatedacetone]$. Substitute the intermediate from pre-equilibrium: $[protonatedacetone]=\frac{k_1}{k_{-1}}[acetone][H^{+}]$, so overall rate = $k[acetone][H^{+}]$, which matches the experimental rate law. Solve for $k$: $k=\frac{\text{rate}}{[acetone][H^{+}]}=\frac{2.1\times 10^{-6}M/s}{(0.25M)(0.10M)}=8.4\times 10^{-5}{M}^{-1}{s}^{-1}$. This mechanism is consistent with the experimental rate law, so it is a plausible mechanism for the acid-catalyzed iodination of acetone.

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the foundation for all further work in kinetics and reaction dynamics. Immediately after this, you will study how catalysts modify reaction mechanisms to lower activation energy, a frequent topic on both AP MCQ and FRQ. Without mastering how to identify intermediates and extract rate laws from mechanisms, you will not be able to correctly explain catalytic behavior or validate proposed mechanisms, a common high-weight FRQ task on the AP exam. This topic also connects to equilibrium in Unit 6, as you use equilibrium principles to approximate the concentration of intermediates in fast pre-equilibrium steps, and to organic reaction mechanisms in later units of the course.

• Catalysis and reaction energy profiles
• Experimental rate law determination
• Arrhenius equation and activation energy