Open with the ozone question. Say: 'The overall reaction destroying ozone in the stratosphere looks simple — 2 O₃ → 3 O₂ — but if you plug those coefficients into a rate law you get the WRONG answer. Meanwhile, the reaction NO₂ + CO → NO + CO₂, when written as an elementary step, gives a rate law of k[NO₂][CO] directly from the coefficients, and that matches experiment.' Ask students to turn-and-talk for 90 seconds: 'Why does the coefficient trick work for one and not the other?' Take 2-3 verbal responses. Do not resolve the puzzle — tell students that today they will learn the exact rule that separates these two situations. Targets SP 6 by asking for a claim before evidence.

1. 7m

   What makes a reaction 'elementary'

   Content

   An elementary reaction is a reaction that happens in a single microscopic event — one collision, one bond break, one rearrangement — exactly as the equation is written. There are no hidden steps and no reaction intermediates between reactants and products. Most equations students see in a textbook are NOT elementary; they are overall reactions, which are the net sum of two or more elementary steps. For example, 2 NO₂ + F₂ → 2 NO₂F looks like a single equation, but experiment shows it actually proceeds in two elementary steps: NO₂ + F₂ → NO₂F + F (slow), then F + NO₂ → NO₂F (fast). The fluorine atom is a reaction intermediate — it is produced in step 1 and consumed in step 2, and it never appears in the overall balanced equation.

   Delivery

   Emphasize the word 'microscopic' — elementary means 'this is literally what the particles do in one instant.' Ask students to point to the intermediate in the NO₂ + F₂ mechanism and explain how they know it's an intermediate (produced then consumed, not in overall equation). Pre-empt the misconception that every equation they see is elementary — most equations in their textbook are overall reactions in disguise. Check: 'If I write H₂ + Cl₂ → 2 HCl on the board, is that elementary? Answer: no, we cannot tell from the equation alone — we must be told or must have experimental evidence.'

2. 7m

   Molecularity: counting the colliders

   Content

   Molecularity is the number of particles that come together in a single elementary step. A unimolecular step has molecularity 1 — one particle spontaneously rearranges or decomposes, like N₂O₄ → 2 NO₂ or cyclopropane → propene. A bimolecular step has molecularity 2 — two particles collide, like NO + O₃ → NO₂ + O₂, or 2 NO₂ → NO₃ + NO. A termolecular step has molecularity 3 — three particles must collide simultaneously, like 2 NO + O₂ → 2 NO₂ if it occurred in one step. Termolecular steps are rare because the probability that three particles meet at the same point in space with the right orientation and energy at the same instant is very small. Almost all real mechanisms are built from unimolecular and bimolecular steps. Critically, molecularity is a property of a single elementary step, while reaction order is a property of an overall reaction determined by experiment — they are NOT the same thing, even though they can be numerically equal for a single-step reaction.

   Delivery

   Hammer the distinction between molecularity (theoretical, per-step, integer 1/2/3) and reaction order (experimental, overall, can be fractional or zero). This is the single biggest confusion in this topic. Ask: 'Why don't we see termolecular steps much?' — expect answers about collision probability; reinforce that a three-body simultaneous collision is geometrically rare. Give the analogy: two people bumping into each other in a hallway happens; three people all colliding at the same instant is much less likely.

3. 8m

   Writing the rate law — the rule and its limit

   Content

   For an elementary step, and only for an elementary step, the rate law can be written directly from the stoichiometric coefficients: the exponent on each reactant equals its coefficient in that step. Unimolecular A → products gives rate = k[A]. Bimolecular A + B → products gives rate = k[A][B]. Bimolecular 2 A → products gives rate = k[A]². Termolecular 2 A + B → products gives rate = k[A]²[B]. The physical reasoning: the rate of a single collision event is proportional to the frequency with which the required particles meet, and that frequency scales with the product of their concentrations raised to how many of each must show up. For an overall reaction, this rule collapses because the observed rate depends on the slow step of the mechanism, not on the net stoichiometry. Worked example: for the elementary step 2 NO + Br₂ → 2 NOBr, rate = k[NO]²[Br₂] — third order overall, and the molecularity is 3 (termolecular). But if 2 NO + Br₂ → 2 NOBr were an OVERALL reaction with an unknown mechanism, we could NOT write rate = k[NO]²[Br₂]; we would have to determine the rate law by an experiment such as the method of initial rates.

   Delivery

   Work the 2 NO + Br₂ example live on the board — say the coefficients, translate them to exponents, name the molecularity, state the overall order. Then contrast: 'If I hadn't told you this was elementary, could you write rate = k[NO]²[Br₂]?' — answer, no. This is the exact FRQ trap the AP exam sets. Pre-empt the misconception: students LOVE to write rate laws from overall equations because it feels systematic. Remind them: the coefficient rule requires the phrase 'elementary step' or explicit mechanism-step context.

1. 15m

   Classify-and-Write: Elementary Step Sort

   Hand each student the worksheet below. Give them 8 minutes to work individually, then 5 minutes in pairs to compare and revise, then 2 minutes to debrief items 6 and 8 as a class. Targets SP 1 (interpreting the step as a particle-level model), SP 4 (analyzing whether the coefficient rule applies), and SP 6 (justifying why it does or does not). Student handout: Classify-and-Write For each proposed elementary step below, (a) state the molecularity, (b) write the predicted rate law, and (c) note whether the step is plausible as a common mechanism step (unimolecular and bimolecular are common; termolecular is rare). You may ONLY use the coefficient rule because each item is labeled as an elementary step. Part 1 — Classify each elementary step 1. O₃ → O₂ + O (elementary) - Molecularity = ______ - Rate law: rate = ______ - Common or rare step? ______ 2. NO + O₃ → NO₂ + O₂ (elementary) - Molecularity = ______ - Rate law: rate = ______ - Common or rare step? ______ 3. 2 ClO → Cl₂ + O₂ (elementary) - Molecularity = ______ - Rate law: rate = ______ - Common or rare step? ______ 4. 2 NO + O₂ → 2 NO₂ (elementary) - Molecularity = ______ - Rate law: rate = ______ - Common or rare step? ______ 5. N₂O₅ → NO₂ + NO₃ (elementary) - Molecularity = ______ - Rate law: rate = ______ - Common or rare step? ______ Part 2 — Spot the trap 6. The OVERALL reaction 2 H₂ + 2 NO → N₂ + 2 H₂O has an experimentally determined rate law rate = k[H₂][NO]². A student writes rate = k[H₂]²[NO]² by applying the coefficient rule. - Is the student's rate law correct? ______ - In one sentence, explain what the student did wrong: ______________________________________ 7. The elementary step 2 NO₂ → NO₃ + NO is proposed as one step in a mechanism. - Rate law for THIS step: rate = ______ - Is this the rate law for the OVERALL reaction? ______ Why or why not? ______________ 8. A classmate says: 'Because item 4 is termolecular, this mechanism is probably wrong — three-body collisions are rare.' Do you agree? Give a one-sentence claim + reasoning response. Answer key (for teacher use): - 1: unimolecular; rate = k[O₃]; common. - 2: bimolecular; rate = k[NO][O₃]; common. - 3: bimolecular; rate = k[ClO]²; common. - 4: termolecular; rate = k[NO]²[O₂]; rare. - 5: unimolecular; rate = k[N₂O₅]; common. - 6: Incorrect — the coefficient rule only applies to elementary steps; this is an overall reaction whose rate law must be measured experimentally. - 7: rate = k[NO₂]²; no — this is only the rate law for that step, not the overall reaction, unless we know the mechanism and that this is the slow step. - 8: Reasonable claim — termolecular steps are geometrically improbable, so a proposed single termolecular step is a red flag; the mechanism likely proceeds through two bimolecular steps instead.

   Materials

   • Printed handout (one per student)
   • Pencil

   Example outputs

   • Item 2 correct response: bimolecular; rate = k[NO][O₃]; common — two particles colliding is the typical mechanism step.
   • Item 6 correct response: The student is wrong because the coefficient rule only works for elementary steps. 2 H₂ + 2 NO → N₂ + 2 H₂O is an overall reaction, so its rate law must come from experiment — which gives rate = k[H₂][NO]², not k[H₂]²[NO]².
2. 10m

   Bimolecular Collision Lab: N₂O₄ ⇌ 2 NO₂ demonstrationLab

   Use a sealed commercial NO₂/N₂O₄ demonstration tube. The tube stays SEALED at all times — NO₂ is highly toxic. Show students the tube at room temperature (medium brown), then submerge it in the ice bath for ~60 seconds (color fades toward pale/colorless as equilibrium shifts toward N₂O₄), then transfer to the hot bath for ~60 seconds (deep brown as it shifts toward NO₂). Targets SP 1 (particle-level model of a unimolecular decomposition and its reverse bimolecular recombination), SP 4 (interpret the color change as evidence of shifting particle populations), and SP 6 (argue from the observation which direction is unimolecular and which is bimolecular). After students observe the color changes, pose these questions and give 4 minutes for pair discussion + share-out: 1. Write the forward elementary step and its rate law. Answer: N₂O₄ → 2 NO₂, unimolecular, rate = k_f[N₂O₄]. 2. Write the reverse elementary step and its rate law. Answer: 2 NO₂ → N₂O₄, bimolecular, rate = k_r[NO₂]². 3. When we put the tube in ice, [NO₂] drops. Using rate laws, explain in terms of collision frequency why cooling shifts equilibrium toward N₂O₄. Expected answer: at lower temperature, k_f decreases more sharply than k_r for this system (or in Le Châtelier terms, the exothermic recombination is favored); either way, the reverse bimolecular collision rate exceeds the forward unimolecular decomposition, so N₂O₄ accumulates. 4. Could the reverse step be termolecular? Why is bimolecular more reasonable? Expected: two NO₂ molecules colliding is far more probable than three particles meeting at once; termolecular steps are rare. If the demo tube is unavailable, run the same discussion using the color-change photos in the auto-generated slide deck — the reasoning task is identical.

   Materials

   • Sealed glass tube or ampoule of NO₂/N₂O₄ equilibrium mixture (commercial demo tube)
   • Hot water bath (~60 °C) in a 400 mL beaker
   • Ice water bath in a 400 mL beaker
   • Tongs or gloves for handling the tube
   • Timer

   Example outputs

   • Forward step: N₂O₄ → 2 NO₂ is unimolecular; rate = k_f[N₂O₄]. One molecule breaks apart on its own — no collision required, so molecularity = 1.
   • Reverse step: 2 NO₂ → N₂O₄ is bimolecular; rate = k_r[NO₂]². Two NO₂ molecules must collide with the correct orientation and energy. A termolecular version would require a third body, which is much less probable — bimolecular is the reasonable choice.

   No-equipment fallback

   Use the deck's before/after color photos of the N₂O₄/NO₂ tube in cold and hot water. Have students do the same four-question analysis in pairs. The intellectual work (writing the two elementary rate laws and reasoning about collision frequency) does not require the physical demo.

1. The elementary step 2 NO(g) + Cl₂(g) → 2 NOCl(g) has what molecularity and rate law? A) Bimolecular; rate = k[NO][Cl₂] B) Termolecular; rate = k[NO]²[Cl₂] C) Termolecular; rate = k[NO][Cl₂]² D) Bimolecular; rate = k[NO]²[Cl₂]

   multiple choiceB. Termolecular; rate = k[NO]²[Cl₂]. Three reactant particles (2 NO + 1 Cl₂) collide in one step, so molecularity = 3. The coefficient rule applies to elementary steps, so exponents equal coefficients. Targets SP 5 (applying the correct routine) and SP 4 (model analysis).

2. The overall reaction 2 A + B → C has an experimentally measured rate law of rate = k[A][B]. A student claims that because the coefficients of A and B are 2 and 1, the rate law must be rate = k[A]²[B]. Explain in 2-3 sentences why the student's reasoning is wrong.

   short answerThe coefficient rule (exponents = coefficients) only applies to elementary steps, not overall reactions. Because 2 A + B → C is an overall reaction, its rate law can only be determined experimentally — which gave rate = k[A][B]. The overall reaction is almost certainly a multistep mechanism whose slow step involves one A and one B, not two A's. Targets SP 6 Argumentation.

3. Which of the following elementary steps would you expect to be LEAST common in real reaction mechanisms, and why? A) H₂O₂ → 2 OH B) NO + O₃ → NO₂ + O₂ C) 2 I + H₂ → 2 HI D) Cl₂ → 2 Cl

   multiple choiceC. 2 I + H₂ → 2 HI is termolecular (three particles must collide simultaneously), and termolecular steps are rare because the probability of a three-body collision with correct orientation and energy at the same instant is very low. Choices A and D are unimolecular, B is bimolecular — all common. Targets SP 4 Model Analysis.

4. Consider the two-step mechanism: Step 1 (slow): NO₂ + NO₂ → NO₃ + NO Step 2 (fast): NO₃ + CO → NO₂ + CO₂ Overall: NO₂ + CO → NO + CO₂ (a) State the molecularity of step 1 and write its rate law. (b) Identify any reaction intermediate. (c) A student writes the overall rate law as rate = k[NO₂][CO] by applying the coefficient rule to the overall equation. Is this correct? Justify using the mechanism.

   short answer(a) Step 1 is bimolecular; rate = k[NO₂]². (b) NO₃ is the intermediate — produced in step 1, consumed in step 2, not in overall equation. (c) No — the student's approach is invalid because the coefficient rule cannot be applied to overall reactions. However, by coincidence, the correct overall rate law is NOT k[NO₂][CO]. It is set by the slow step (step 1), which gives rate = k[NO₂]². The observed rate depends only on [NO₂], not [CO], because CO participates only in the fast step. Targets SP 4 and SP 6.

• ☐Classify-and-Write handout (one per student)
• ☐Sealed NO₂/N₂O₄ demonstration tube (commercial, sealed glass)
• ☐Hot water bath (~60 °C) in 400 mL beaker
• ☐Ice water bath in 400 mL beaker
• ☐Tongs or heat-resistant gloves
• ☐Timer or classroom clock
• ☐Pencils
• ☐Projector for auto-generated slide deck