Reaction Mechanisms and Intermediates

⚙️ Elementary Steps

Part 1 of 7 — Breaking Reactions into Steps

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

• Understanding the core concepts covered in Part 1
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📖 What Is an Elementary Step?

An elementary step (or elementary reaction) is a single molecular event — one collision or one molecular rearrangement. It describes exactly what happens at the molecular level.

🔑 Key Concept: For an elementary step, the rate law can be written directly from the stoichiometry of that step. This is NOT true for overall reactions.

Example

Overall: $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$

Proposed mechanism:

• Step 1: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$ (slow)
• Step 2: $\text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F}$ (fast)

Each step is an elementary reaction with its own rate law determined by its molecularity.

⚖️ Molecularity

Molecularity is the number of reactant particles (molecules, atoms, or ions) involved in an elementary step.

$\boxed{\text{Elementary step: Rate law exponents} = \text{stoichiometric coefficients}}$

⚠️ Warning: Molecularity applies only to elementary steps — never to overall reactions. Molecularity ≠ Order for overall reactions, but they ARE equal for elementary steps.

• Termolecular steps are extremely rare because three-body collisions are very unlikely
• Molecularity is always a positive integer (1, 2, or 3)
• Molecularity applies only to elementary steps, never to overall reactions
• Molecularity ≠ Order for overall reactions, but they ARE equal for elementary steps

Molecularity Quiz 🎯

📏 Rules for Valid Mechanisms

A proposed mechanism must satisfy two essential criteria:

Rule 1: Steps Must Sum to the Overall Reaction

When all elementary steps are added together, intermediates cancel, and the result must equal the overall balanced equation.

Rule 2: Rate Law Must Be Consistent

🔑 Key Concept: The rate law predicted by the mechanism must match the experimentally observed rate law.

Example Verification

Overall: $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$

Step 1: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$ Step 2: $\text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F}$

Sum: $2\text{NO}_2 + \text{F}_2 + \cancel{\text{F}} \rightarrow 2\text{NO}_2\text{F} + \cancel{\text{F}}$

$\boxed{2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}}$ ✓

Elementary Step Concepts 🔍

Practice: Analyzing Elementary Steps 🧮

Consider the mechanism:

• Step 1: $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{H}_2\text{O} + \text{IO}^-$ (slow)
• Step 2: $\text{H}_2\text{O}_2 + \text{IO}^- \rightarrow \text{H}_2\text{O} + \text{O}_2 + \text{I}^-$ (fast)

1) What is the molecularity of Step 1? (enter a number)

2) What is the molecularity of Step 2? (enter a number)

3) How many intermediates are there? (enter a number)

Exit Quiz — Elementary Steps ✅

🔎 Intermediates and Catalysts

Part 2 of 7 — Species That Appear and Disappear

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

• Understanding the core concepts covered in Part 2
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

⚗️ Reaction Intermediates

A reaction intermediate is a species that is:

• Produced in one elementary step
• Consumed in a subsequent step
• Not present in the overall balanced equation

How to Identify Intermediates

🔑 Key Concept: Any species that is produced in one step and consumed in a later step (cancelling out when steps are summed) is an intermediate.

1. Write out all elementary steps
2. Add them up to get the overall reaction
3. Any species that cancels out (appears on both sides) is an intermediate

Example

Step 1: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \textbf{F}$ (slow) Step 2: $\text{NO}_2 + \textbf{F} \rightarrow \text{NO}_2\text{F}$ (fast)

F (fluorine atom) is the intermediate — produced in Step 1, consumed in Step 2, not in the overall equation.

On an Energy Diagram

🔑 Key Concept: Intermediates sit in an energy valley (local minimum) between two transition state peaks.

⚙️ Catalysts in Mechanisms

A catalyst is a species that is:

• Consumed in an early step
• Regenerated in a later step
• Present at the beginning and end but not in the overall equation
• Not used up overall

How to Identify Catalysts

💡 Tip: A catalyst is consumed early and regenerated later — it’s present at both the start and end of the reaction.

1. Look for species present in the reactants of an early step that reappear in the products of a later step
2. The catalyst cancels when steps are added

Example: Ozone Decomposition (Cl-catalyzed)

Step 1: $\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2$ Step 2: $\text{ClO} + \text{O} \rightarrow \text{Cl} + \text{O}_2$

Overall: $\cancel{\text{Cl}} + \text{O}_3 + \cancel{\text{ClO}} + \text{O} \rightarrow \cancel{\text{ClO}} + \text{O}_2 + \cancel{\text{Cl}} + \text{O}_2$

$\boxed{\text{O}_3 + \text{O} \rightarrow 2\text{O}_2}$

• Cl = catalyst (consumed in Step 1, regenerated in Step 2)
• ClO = intermediate (produced in Step 1, consumed in Step 2)

Intermediate vs. Catalyst Summary

Identification Quiz 🎯

Consider this mechanism:

• Step 1: $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{H}_2\text{O} + \text{IO}^-$ (slow)
• Step 2: $\text{H}_2\text{O}_2 + \text{IO}^- \rightarrow \text{H}_2\text{O} + \text{O}_2 + \text{I}^-$ (fast)

Analyzing a Mechanism 🔍

Consider:

• Step 1: $2\text{NO} \rightarrow \text{N}_2\text{O}_2$ (fast, reversible)
• Step 2: $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$ (slow)
• Step 3: $\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}$ (fast)

📌 Intermediates on Energy Diagrams

For a multi-step mechanism, the energy diagram shows multiple peaks and valleys:

Two-Step Mechanism

$\text{Reactants} \xrightarrow{E_{a1}} \text{TS}_1 \rightarrow \text{Intermediate} \xrightarrow{E_{a2}} \text{TS}_2 \rightarrow \text{Products}$

• Peaks = transition states (one per elementary step)
• Valley = intermediate (local energy minimum)
• The tallest peak corresponds to the rate-determining step

Three-Step Mechanism

Has 3 peaks and 2 valleys (2 intermediates).

General Rule

🔑 Key Concept: For an $n$-step mechanism: $n$ transition states (peaks) and $n - 1$ intermediates (valleys).

$\boxed{n \text{ steps} \rightarrow n \text{ peaks (TS)}, \; (n-1) \text{ valleys (intermediates)}}$

Counting Species in Mechanisms 🧮

A mechanism has 4 elementary steps with the following species:

Step 1: A + B → C + D (slow) Step 2: C + B → E + F (fast) Step 3: E → G + H (fast) Step 4: H + D → P + A (fast)

1) How many intermediates? (count species that cancel)

2) Is A a catalyst, intermediate, or reactant? (enter: catalyst, intermediate, or reactant)

3) How many transition states on the energy diagram? (enter a number)

Exit Quiz — Intermediates & Catalysts ✅

🐢 Rate-Determining Step

Part 3 of 7 — The Bottleneck

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

• Understanding the core concepts covered in Part 3
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

⏱️ The Rate-Determining Step (RDS)

Definition

The rate-determining step is the slowest elementary step in a mechanism. It has the highest activation energy ($E_a$) of all the steps.

Key Principle

$\boxed{\text{Overall rate} \approx \text{Rate of the slowest step}}$

On an Energy Diagram

The RDS corresponds to the tallest peak (largest $E_a$ barrier) on the energy diagram.

🔑 Key Concept: The rate law for the overall reaction is determined by the rate-determining step. This is how we connect mechanisms to experimentally measured rate laws.

Rate-Determining Step Concepts 🎯

⏱️ Case 1: First Step Is Rate-Determining

This is the simplest case. When Step 1 is slow, the rate law comes directly from Step 1's elementary rate law.

Problem: Derive the rate law for $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$ given the mechanism below.

Mechanism:

• Step 1: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$ (slow — RDS)
• Step 2: $\text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F}$ (fast)

Solution: Write the rate law from the slow step (bimolecular elementary step):

$\boxed{\text{Rate} = k_1[\text{NO}_2][\text{F}_2]}$

This is a bimolecular step, so exponents come from stoichiometry. The rate law is first order in $NO_{2}$ and first order in $F_{2}$ — overall second order.

⚠️ Warning: The rate law only involves species from the slow step. The fast step has no effect on the rate law.

⚖️ Case 2: Second Step Is Rate-Determining (Pre-Equilibrium)

When a later step is rate-determining, the rate law from that step may contain an intermediate. Since intermediates cannot appear in the final rate law, we must eliminate them using the pre-equilibrium approximation.

The Pre-Equilibrium Method

💡 Tip: If a fast reversible step precedes the slow step, use the equilibrium expression to eliminate the intermediate from the rate law.

If Step 1 is fast and reversible, it establishes an equilibrium before the slow step:

$\text{Step 1 (fast, reversible):} \quad \text{A} + \text{B} \rightleftharpoons \text{C}$

$K_{eq} = \frac{[\text{C}]}{[\text{A}][\text{B}]}$

$\text{Step 2 (slow):} \quad \text{C} + \text{D} \rightarrow \text{products}$

Rate from slow step: $\text{Rate} = k_2[\text{C}][\text{D}]$

But C is an intermediate! Solve for [C] using equilibrium:

$[\text{C}] = K_{eq}[\text{A}][\text{B}]$

Substitute:

$\boxed{\text{Rate} = k_2 K_{eq}[\text{A}][\text{B}][\text{D}] = k_{\text{obs}}[\text{A}][\text{B}][\text{D}]}$

where $k_{\text{obs}} = k_2 K_{eq}$.

💡 Tip: An alternative to pre-equilibrium is the steady-state approximation, which assumes $d[\text{intermediate}]/dt \approx 0$. Both methods eliminate intermediates, but steady-state is more general and works even when the fast step is not fully reversible.

Pre-Equilibrium Practice 🧮

Mechanism:

• Step 1: $2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ (fast, reversible; $K_{eq}$)
• Step 2: $\text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\text{NO}_2$ (slow)

1) Write the rate law from the slow step. What is the intermediate? (enter the formula)

2) Express $[\text{N}_2\text{O}_2]$ in terms of $[\text{NO}]$ using $K_{eq}$. What power of $[\text{NO}]$ appears? (enter a number)

3) The final rate law is rate = k[NO]${}^{n}$[$O_{2}$]${}^{m}$. What are n and m? (enter as: n,m)

RDS Review 🔍

Exit Quiz — Rate-Determining Step ✅

🧮 Deriving Rate Laws from Mechanisms

Part 4 of 7 — From Steps to Predictions

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

• Understanding the core concepts covered in Part 4
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

🎯 Strategy for Deriving Rate Laws

📌 Step-by-Step Method

🔑 Key Rule: The final rate law must contain only reactants (and possibly catalysts) — never intermediates.

🧪 Example 1: First Step Slow

Problem: Derive the rate law for $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$ given the mechanism below.

Mechanism:

• Step 1: $\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}$ (slow)
• Step 2: $\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2$ (fast)

Solution:

Step 1 is the RDS (slow, bimolecular):

$\text{Rate} = k_1[\text{NO}_2][\text{NO}_2] = k_1[\text{NO}_2]^2$

No intermediates in this rate law → done!

$\boxed{\text{Rate} = k[\text{NO}_2]^2}$

💡 Tip: CO doesn't appear in the rate law even though it's a reactant in the overall equation! It reacts only in the fast step (after the RDS).

🧪 Example 2: Second Step Slow (Pre-Equilibrium Required)

Problem: Derive the rate law for $2\text{NO} + \text{Br}_2 \rightarrow 2\text{NOBr}$ using pre-equilibrium.

Mechanism:

• Step 1: $\text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2$ (fast, reversible)
• Step 2: $\text{NOBr}_2 + \text{NO} \rightarrow 2\text{NOBr}$ (slow)

Solution:

Step 2 is the RDS: $\text{Rate} = k_2[\text{NOBr}_2][\text{NO}]$

$NOBr_{2}$ is an intermediate! Eliminate it using Step 1 equilibrium:

$K_{eq} = \frac{[\text{NOBr}_2]}{[\text{NO}][\text{Br}_2]}$

$[\text{NOBr}_2] = K_{eq}[\text{NO}][\text{Br}_2]$

Substitute:

$\text{Rate} = k_2 \cdot K_{eq}[\text{NO}][\text{Br}_2] \cdot [\text{NO}]$

$\boxed{\text{Rate} = k_{\text{obs}}[\text{NO}]^2[\text{Br}_2]}$

where $k_{\text{obs}} = k_2 K_{eq}$.

Rate Law Derivation Quiz 🎯

Derivation Practice 🧮

Mechanism:

• Step 1: $\text{A} + \text{B} \rightleftharpoons \text{C}$ (fast, $K_{eq}$)
• Step 2: $\text{C} + \text{A} \rightarrow \text{D}$ (slow)

1) The rate law from the slow step is Rate = $k_{2}$[?][?]. Which species are in the rate law? (enter two formulas separated by a comma, alphabetically)

2) The intermediate is eliminated by writing [C] = Keq × [?] × [?]. Fill in the species. (enter two formulas separated by a comma, alphabetically)

3) The final rate law is Rate = k_obs[A]${}^{n}$[B]${}^{m}$. What are n and m? (enter as: n,m)

Mechanism → Rate Law Review 🔍

Exit Quiz — Deriving Rate Laws ✅

✅ Validating Mechanisms

Part 5 of 7 — Testing Proposed Mechanisms

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

• Understanding the core concepts covered in Part 5
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 The Two Essential Criteria

A valid mechanism must satisfy both of these conditions:

Criterion 1: Steps Sum to the Overall Reaction

When all elementary steps are added and intermediates/catalysts are cancelled, the result must equal the experimentally determined overall balanced equation.

$\boxed{\text{Step 1} + \text{Step 2} + \cdots = \text{Overall Reaction}}$

Criterion 2: Rate Law Matches Experiment

The rate law derived from the mechanism (using the RDS and pre-equilibrium as needed) must agree with the experimentally observed rate law.

$\boxed{\text{Rate law}_{\text{mechanism}} = \text{Rate law}_{\text{experimental}}}$

⚠️ Warning: Even if both criteria are met, the mechanism is not proven — it is only consistent with the data. Other mechanisms might also be consistent. We can disprove a mechanism but never definitively prove one.

🧪 Worked Example: Validating a Mechanism

Problem: Validate the proposed mechanisms for $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ given the experimental rate law Rate = $k[\text{NO}]^2[\text{O}_2]$.

Mechanism A: Single Termolecular Step

⚠️ Problem: Termolecular collisions (3 molecules simultaneously) are extremely unlikely. This mechanism is mathematically valid but physically implausible.

Mechanism B: Two-Step with Pre-Equilibrium

🔑 Conclusion: Mechanism B is preferred — it matches the experimental rate law AND avoids the improbable termolecular step.

Validation Quiz 🎯

Overall: $2\text{A} + \text{B} \rightarrow \text{C} + \text{D}$ Experimental: Rate = $k[\text{A}][\text{B}]$

Validating Mechanisms Practice 🔍

Overall: $\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}$ Experimental: Rate = $k[\text{H}_2][\text{I}_2]$

Mechanism X:

• Step 1: $\text{I}_2 \rightleftharpoons 2\text{I}$ (fast)
• Step 2: $2\text{I} + \text{H}_2 \rightarrow 2\text{HI}$ (slow)

Mechanism Y:

• Step 1: $\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}$ (one step, slow)

📌 Common AP Mistakes to Avoid

⚠️ Warning: Don't confuse order (experimental: can be 0, 1, 2, fractional) with molecularity (theoretical: must be 1, 2, or 3). They are equal ONLY for elementary steps.

⚠️ Warning: The rate law for an overall reaction must be determined experimentally. Only for elementary steps can you write the rate law from stoichiometry.

⚠️ Warning: The final rate law should contain only reactants (and catalysts). If your rate law has an intermediate, you need to eliminate it.

⚠️ Warning: A mechanism that gives the correct rate law but doesn't sum to the overall equation is INVALID (and vice versa). Always check both criteria.

Mechanism Validation Check 🧮

Overall: $\text{A} + 2\text{B} \rightarrow \text{C}$ Experimental: Rate = $k[\text{A}][\text{B}]$

Proposed mechanism:

• Step 1: $\text{A} + \text{B} \rightarrow \text{D}$ (slow)
• Step 2: $\text{D} + \text{B} \rightarrow \text{C}$ (fast)

1) Do the steps sum to A + 2B → C? (yes or no)

2) What is the predicted rate law from the RDS? (enter in form: k[X][Y] — use brackets)

3) Does the predicted rate law match the experimental rate law? (yes or no)

Exit Quiz — Validating Mechanisms ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Mechanism Analysis Practice

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

• Working through complete multi-step problems from start to finish
• Building problem-solving strategies you can apply on the AP exam
• Identifying which concepts to apply and in what order

⚙️ Problem 1: Complete Mechanism Analysis

Problem: Analyze the ozone decomposition mechanism and verify it matches the experimental rate law.

The reaction $2\text{O}_3 \rightarrow 3\text{O}_2$ has the experimental rate law:

$\boxed{\text{Rate} = k\frac{[\text{O}_3]^2}{[\text{O}_2]}}$

Proposed mechanism:

• Step 1: $\text{O}_3 \rightleftharpoons \text{O}_2 + \text{O}$ (fast, reversible)
• Step 2: $\text{O} + \text{O}_3 \rightarrow 2\text{O}_2$ (slow)

Problem 1 Analysis 🎯

Problem 2: Enzyme Kinetics Mechanism 🧮

Problem: Derive the rate law for the simplified enzyme-catalyzed reaction and identify all species.

An enzyme-catalyzed reaction has the mechanism:

• Step 1: $\text{E} + \text{S} \rightleftharpoons \text{ES}$ (fast, $K_{eq}$)
• Step 2: $\text{ES} \rightarrow \text{E} + \text{P}$ (slow)

where E = enzyme, S = substrate, ES = enzyme-substrate complex, P = product.

1) What is the intermediate? (enter formula)

2) What is the catalyst? (enter formula)

3) The derived rate law is Rate = $k_{obs}$[?][?]. Enter the two species. (separated by comma, alphabetically)

⚖️ Problem 3: Comparing Mechanisms

Problem: Compare mechanisms A and B for $H_{2}O_{2}$ decomposition and determine which matches the experimental rate law Rate = $k[\text{H}_2\text{O}_2][\text{I}^-]$.

Overall: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$

Experimental: Rate = $k[\text{H}_2\text{O}_2][\text{I}^-]$

💡 Tip: A negative-order dependence (rate ∝ 1/[product]) means adding that product shifts a pre-equilibrium backward, reducing the intermediate concentration.

Mechanism A:

• Step 1: $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{IO}^- + \text{H}_2\text{O}$ (slow)
• Step 2: $\text{IO}^- + \text{H}_2\text{O}_2 \rightarrow \text{I}^- + \text{H}_2\text{O} + \text{O}_2$ (fast)

Mechanism B:

• Step 1: $\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}$ (slow)
• Step 2: $\text{O} + \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2$ (fast)

Comparing Mechanisms A and B 🔍

Problem 4: Energy Diagram for a Mechanism 🧮

A two-step mechanism has:

• Step 1 (slow): $E_a = 80$ kJ/mol, $\Delta H_1 = -30$ kJ/mol
• Step 2 (fast): $E_a = 20$ kJ/mol, $\Delta H_2 = -10$ kJ/mol

If reactants start at energy = 0 kJ:

1) What is the energy of the first transition state? (in kJ)

2) What is the energy of the intermediate? (in kJ)

3) What is the energy of the products? (in kJ)

Exit Quiz — Mechanism Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Comprehensive Mechanism Problems

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

• Solving AP-style questions that integrate multiple concepts from this unit
• Writing clear, concise explanations using proper chemistry terminology
• Identifying and avoiding common AP exam traps and mistakes

📋 Key Concepts Summary

🧪 Mechanism Fundamentals

📌 Species Classification

💡 Counting rule: $n$ steps → $n$ transition states and $n - 1$ intermediates

🎯 Rate Law Derivation

$\text{Overall Rate} \approx \text{Rate of the slowest step (RDS)}$

✅ Validation Checklist

$\text{Valid mechanism} \iff \text{Steps sum correctly} \; \textbf{and} \; \text{Rate law matches experiment}$

AP Problem 1: Mechanism Analysis 🎯

The decomposition of hydrogen peroxide is catalyzed by iodide ion:

$2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$

Mechanism:

• Step 1: $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{IO}^- + \text{H}_2\text{O}$ (slow)
• Step 2: $\text{IO}^- + \text{H}_2\text{O}_2 \rightarrow \text{I}^- + \text{H}_2\text{O} + \text{O}_2$ (fast)

AP Problem 2: Full Mechanism Derivation 🧮

Reaction: $2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$

Mechanism:

• Step 1: $2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ (fast, $K_{eq}$)
• Step 2: $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$ (slow)
• Step 3: $\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}$ (fast)

1) How many intermediates? (number)

2) The rate law from the slow step contains an intermediate. After elimination, the rate law is Rate = k_obs[NO]${}^{n}$[$H_{2}$]${}^{m}$. What is n? (number)

3) What is m? (number)

AP Problem 3: Energy Diagram Interpretation 🎯

An energy diagram for a two-step mechanism shows:

• First peak is higher than the second peak
• A valley between the peaks (intermediate)
• Products are lower than reactants

Comprehensive Review 🔍

Challenge Problem 🧮

Reaction: $\text{A} + \text{B} + \text{C} \rightarrow \text{D} + \text{E}$

Experimental rate law: Rate = $k[\text{A}][\text{B}]^2$

A student proposes:

• Step 1: $\text{A} + \text{B} \rightarrow \text{F}$ (slow)
• Step 2: $\text{F} + \text{B} \rightarrow \text{G}$ (fast)
• Step 3: $\text{G} + \text{C} \rightarrow \text{D} + \text{E}$ (fast)

1) Does the mechanism sum to the overall reaction? (yes or no)

2) The rate law from Step 1 is rate = k[A][B]. Does this match the experimental rate law rate = k[A][B]${}^{2}$? (yes or no)

3) Is this proposed mechanism valid? (yes or no)

Final Exit Quiz — Reaction Mechanisms ✅