⚙️ Elementary Steps

Part 1 of 7 — Breaking Reactions into Steps

Most chemical reactions do not occur in a single step. Instead, they proceed through a series of simpler reactions called elementary steps. The collection of elementary steps that makes up an overall reaction is called the reaction mechanism.

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 Property

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.

Important

• 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

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

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

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

In reaction mechanisms, some species are produced in one step and consumed in another. These transient species are classified as either intermediates or catalysts, and understanding the difference is essential for AP Chemistry.

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

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

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

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

• 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

For an $n$-step mechanism:

• $n$ transition states (peaks)
• $n - 1$ intermediates (valleys)

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

In a multi-step mechanism, one step is usually much slower than the others. This rate-determining step (RDS) controls the overall rate of the reaction — just like the slowest person in a relay race determines the team's time.

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

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

On an Energy Diagram

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

Why It Matters

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.

Example

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

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)

Rate law from the slow step:

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

This is a bimolecular step, so exponents come from stoichiometry. The rate law is first order in NO₂ and first order in F₂ — overall second order.

Notice

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

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:

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

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

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[O₂]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

One of the most important skills on the AP Chemistry exam is deriving the predicted rate law from a proposed mechanism and comparing it with the experimentally observed rate law.

Strategy for Deriving Rate Laws

Step-by-Step Method

1. Identify the RDS (slow step)
2. Write the rate law for the RDS using its stoichiometry (it's an elementary step!)
3. Check for intermediates in the rate law
4. If intermediates present → eliminate them:
   • Use pre-equilibrium from a prior fast reversible step
   • Solve the equilibrium expression for [intermediate]
   • Substitute back into the rate law
5. Simplify — combine constants into $k_{\text{obs}}$

Result

The final rate law should contain only reactants (and possibly catalysts) — never intermediates.

Example 1: First Step Slow

Overall: $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$

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)

Derivation:

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!

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

Note: 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)

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

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)

Derivation:

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

NOBr₂ 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 = k₂[?][?]. 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

A proposed mechanism is a hypothesis — it must be tested against experimental evidence. In this part, you'll learn the two essential criteria for validating a mechanism and practice evaluating proposed mechanisms.

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.

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

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.

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

Important Caveat

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

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

Experimental rate law: Rate = $k[\text{NO}]^2[\text{O}_2]$

Proposed Mechanism A:

• Step 1: $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ (slow, one step)

Test Criterion 1: Sum = $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ ✓ Test Criterion 2: rate = $k[\text{NO}]^2[\text{O}_2]$ ✓ (termolecular step)

BUT: termolecular steps are extremely unlikely! Look for alternative.

Proposed Mechanism B:

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

Test Criterion 1: Sum = $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ ✓ Test Criterion 2: Rate = $k_2[\text{N}_2\text{O}_2][\text{O}_2]$ → $[\text{N}_2\text{O}_2] = K_{eq}[\text{NO}]^2$ → Rate = $k_{obs}[\text{NO}]^2[\text{O}_2]$ ✓

Mechanism B is preferred because it 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

Mistake 1: Confusing Order and Molecularity

• Order is experimental (can be 0, 1, 2, fractional)
• Molecularity is theoretical (must be 1, 2, or 3)
• They are equal ONLY for elementary steps

Mistake 2: Writing Rate Law from Overall Equation

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

Mistake 3: Leaving Intermediates in the Rate Law

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

Mistake 4: Forgetting to Check BOTH Criteria

A mechanism that gives the correct rate law but doesn't sum to the overall equation is INVALID (and vice versa).

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

This workshop focuses on the types of mechanism problems you'll see on the AP Chemistry exam: analyzing mechanisms, deriving rate laws, identifying species, and validating proposals.

Problem 1: Complete Mechanism Analysis

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

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

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 🧮

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

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

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

Note: I⁻ is not in the overall equation — it must be a catalyst.

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

This final part presents AP exam-level problems that integrate all mechanism concepts: elementary steps, intermediates, catalysts, rate-determining steps, rate law derivation, and mechanism validation.

Key Concepts Summary

Mechanism Fundamentals

• A mechanism is a series of elementary steps that sum to the overall reaction
• Molecularity (1, 2, or 3) = number of reactant particles in an elementary step
• For elementary steps only: rate law exponents = stoichiometric coefficients

Species Classification

• Intermediate: produced in one step, consumed in another (not in overall equation)
• Catalyst: consumed early, regenerated later (present at start and end)
• n steps → n transition states and n − 1 intermediates

Rate Law Derivation

• Rate law comes from the rate-determining step (slowest = highest $E_a$)
• If RDS rate law contains intermediates → use pre-equilibrium to eliminate
• Final rate law must contain only reactants (and catalysts)

Validation

• Steps must sum to the overall equation
• Derived rate law must match the experimental rate law
• A consistent mechanism is not proven — only not disproven

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[H₂]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 ✅