Open with the Haber–Bosch reaction — the industrial synthesis of ammonia that feeds roughly half the planet. Tell students: at 500 K a reactor is sampled and found to contain [N₂] = 0.20 M, [H₂] = 0.30 M, and [NH₃] = 0.15 M once the mixture is no longer changing. Ask them to work with a neighbor for 90 seconds and write down what Kc equals for N₂ + 3H₂ ⇌ 2NH₃. Do not solve it — collect two or three answers on the board and note the disagreements. Tell students the goal of today is that every one of them can walk out of the room and reliably produce that number, plus handle the harder case where only ONE equilibrium value is measured. Targets SP 5 (Mathematical Routines) and SP 6 (Argumentation) as students commit to a numerical claim.

1. 6m

   The K expression uses EQUILIBRIUM values, not initial ones

   Content

   The equilibrium constant expression is products over reactants, each raised to its stoichiometric coefficient, and every value plugged in must be the equilibrium concentration (for Kc) or equilibrium partial pressure (for Kp). For N₂ + 3H₂ ⇌ 2NH₃, Kc = [NH₃]² / ([N₂][H₂]³). Using the hook numbers: Kc = (0.15)² / ((0.20)(0.30)³) = 0.0225 / (0.20 · 0.027) = 0.0225 / 0.0054 ≈ 4.17. Notice that K is reported as a pure number — no M, no atm. AP treats every concentration as an activity, meaning [X] divided by a 1 M reference state, and every partial pressure as P divided by 1 atm, so the units cancel. This is why K is unitless even though the raw numbers you plug in carry units.

   Delivery

   Anchor the two non-negotiables from the very first slide: (1) only equilibrium values go in, (2) every exponent is a stoichiometric coefficient. Walk the Haber calculation on the board digit by digit so pace-setters see the arithmetic. Pre-empt the misconception that K has units — tell them explicitly that AP graders will not deduct for omitting units on K and will deduct if a student writes 'Kc = 4.17 M⁻²'. Ask a cold-call check: 'If I had given you the concentrations 30 seconds after mixing, could I have used those numbers?' Correct answer: no, they aren't equilibrium values. Targets SP 1 and SP 5.

2. 8m

   Building an ICE table when only ONE equilibrium value is measured

   Content

   In most AP problems you are given initial concentrations and only one measured equilibrium value; you must solve for everything else with an ICE table. Consider H₂(g) + I₂(g) ⇌ 2HI(g) with [H₂]₀ = 1.00 M, [I₂]₀ = 1.00 M, [HI]₀ = 0, and a measurement that [HI]eq = 1.56 M. The Change row uses the extent x scaled by coefficients: −x for H₂, −x for I₂, +2x for HI. Because HI started at 0 and ended at 1.56 M, 2x = 1.56, so x = 0.78. That gives [H₂]eq = 1.00 − 0.78 = 0.22 M and [I₂]eq = 0.22 M. Then Kc = (1.56)² / ((0.22)(0.22)) = 2.4336 / 0.0484 ≈ 50.3, which matches the textbook value for this reaction near 700 K. The critical move is that the coefficient of HI is 2, so its change row entry is 2x, not x — this is where most students drop points.

   Delivery

   Fill the ICE table live, one row at a time, thinking out loud. When you write the Change row, say aloud 'stoichiometric coefficient times x' for every species. Stop after the Change row and ask students to predict which value the measurement pins down; the +2x = 1.56 step is where you want them to see that x = 0.78, not 1.56. Explicitly flag the misconception: 'HI's change is +1.56 total, but x itself is only 0.78 because the coefficient is 2.' Then ask 'why is the sign on H₂ negative and the sign on HI positive?' — because reactants decrease and products increase when the reaction proceeds forward, not because 'change is always negative.' Targets SP 5 and SP 3.

3. 6m

   Kp works the same way — with partial pressures

   Content

   When gaseous reactions are described in pressure units, use Kp with the same product-over-reactant, coefficient-as-exponent structure — just substitute partial pressures for concentrations. Example: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Start with P(PCl₅)₀ = 1.00 atm and no products. At equilibrium, P(PCl₅) is measured to be 0.60 atm. Change row: −x for PCl₅, +x for PCl₃, +x for Cl₂. Since 1.00 − x = 0.60, x = 0.40 atm. So P(PCl₃)eq = P(Cl₂)eq = 0.40 atm. Then Kp = (0.40)(0.40)/(0.60) = 0.267. Kp and Kc are related by Kp = Kc(RT)^Δn where Δn is moles of gaseous product minus moles of gaseous reactant, but on this standard you're mostly just choosing the right form for the data given.

   Delivery

   Emphasize that the procedure is identical — same ICE structure, same coefficient rules, just pressures instead of concentrations. Draw an analogy: 'The pressure-vs-time trace on the slide flattens out just like a concentration-vs-time trace would; the equilibrium partial pressures are the y-values at the plateau.' Watch for students who try to convert pressures to concentrations before starting — that's unnecessary work here. Quick check question: 'For 2A ⇌ B, if you measure P_B at equilibrium, what is x?' Answer: x = P_B (because the coefficient on B is 1), which means the change for A is −2·P_B. Targets SP 1 and SP 4.

1. 29m

   Three-Station K Gallery: Kc from Concentrations, Kc from ICE, Kp/Absorbance from DataLab

   Set up three stations around the room. Groups of 3 rotate every ~9 minutes. Every group must produce a completed ICE table (where relevant) and a final K value with reasoning. Walk around and check that the Change row uses coefficient·x, not just ±x, at every station. Targets SP 3 (Representing Data), SP 5 (Mathematical Routines), and SP 6 (Argumentation). Station A — Kc directly from equilibrium concentrations (no ICE needed) Student handout: At 1000 K, a sealed vessel containing the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is analyzed and the following equilibrium concentrations are measured: - [SO₂] = 0.10 M - [O₂] = 0.050 M - [SO₃] = 0.30 M 1. Write the Kc expression for this reaction. 2. Calculate Kc. Show the substitution. 3. Claim–Evidence–Reasoning: A classmate writes Kc = 0.30 / (0.10 · 0.050) = 60. Identify their error and state the correct Kc. Rule you must follow: every exponent equals the stoichiometric coefficient. Station B — ICE table with one measured equilibrium value Student handout: The water–gas shift reaction is run in a 1.00 L reactor: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). The reactor is charged with 1.00 mol CO and 1.00 mol H₂O and no products. At equilibrium the [CO₂] is measured to be 0.55 M. Part 1 — Complete this ICE table: - I: [CO] = ______, [H₂O] = ______, [CO₂] = ______, [H₂] = ______ - C: [CO] = ______, [H₂O] = ______, [CO₂] = ______, [H₂] = ______ - E: [CO] = ______, [H₂O] = ______, [CO₂] = ______, [H₂] = ______ Part 2 — Calculate Kc. Show the expression, the substitution, and the numeric answer. Part 3 — Reasoning: In one sentence, explain why you could NOT have plugged the INITIAL concentrations into the Kc expression. Station C — Absorbance data → equilibrium concentration → Kc Student handout: Reaction: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). A Beer's law calibration for FeSCN²⁺ at 447 nm has already been run for you and gives A = 4400 · [FeSCN²⁺] (path length 1.00 cm, slope in units of M⁻¹). Procedure: 1. Using a graduated cylinder, add 5.00 mL of 0.00200 M Fe(NO₃)₃ (in 0.5 M HNO₃) to a clean 50 mL beaker. 2. Add 5.00 mL of 0.00200 M KSCN. Swirl gently. Total volume = 10.00 mL. 3. Transfer to a cuvette and measure absorbance at 447 nm. Record A = ______. Part 1 — Find [FeSCN²⁺]eq using the Beer's law equation above. Show the arithmetic. Part 2 — Initial concentrations after mixing (dilution to 10.00 mL total): - [Fe³⁺]₀ = ______ M - [SCN⁻]₀ = ______ M - [FeSCN²⁺]₀ = ______ M Part 3 — Complete the ICE table, using x = [FeSCN²⁺]eq from Part 1. - I: ______ ______ ______ - C: ______ ______ ______ - E: ______ ______ ______ Part 4 — Calculate Kc. Part 5 — Argumentation (SP 6): Suppose your absorbance reading were 10% too high because the cuvette had a fingerprint. Would your calculated Kc be too high, too low, or unchanged? Justify in one sentence. Safety: Wear goggles. Fe(NO₃)₃ is in dilute HNO₃ — rinse skin immediately if contacted. Do not pipette by mouth. Dispose of solutions in the labeled waste beaker.

   Materials

   • Colorimeter or Vernier SpectroVis (one per group) with cuvettes
   • 0.00200 M Fe(NO₃)₃ in 0.5 M HNO₃ (50 mL per group)
   • 0.00200 M KSCN (50 mL per group)
   • 10 mL graduated cylinders (2 per group)
   • Small labeled beakers (50 mL, 2 per group)
   • Distilled water squirt bottle
   • Calculators
   • Station handouts (content below)

   Example outputs

   • Station A: Kc = (0.30)² / ((0.10)²(0.050)) = 0.09 / 0.00050 = 180. The classmate forgot to square [SO₂] and [SO₃]; the exponents come from the coefficients, not the concentrations.
   • Station B: I row 1.00, 1.00, 0, 0. C row −x, −x, +x, +x with x = 0.55. E row 0.45, 0.45, 0.55, 0.55. Kc = (0.55)(0.55)/((0.45)(0.45)) = 0.3025/0.2025 ≈ 1.49. Initial values can't be used because the system had not yet reached equilibrium — rates weren't equal.
   • Station C with A = 0.44: [FeSCN²⁺]eq = 0.44/4400 = 1.00×10⁻⁴ M. Initial [Fe³⁺]₀ = [SCN⁻]₀ = 1.00×10⁻³ M (5.00 mL of 0.00200 M diluted into 10.00 mL). Change: −x, −x, +x with x = 1.00×10⁻⁴. E: 9.00×10⁻⁴, 9.00×10⁻⁴, 1.00×10⁻⁴. Kc = (1.00×10⁻⁴) / ((9.00×10⁻⁴)²) ≈ 123. A high absorbance would inflate [FeSCN²⁺]eq (numerator up) and shrink [Fe³⁺]eq and [SCN⁻]eq (denominator down), so Kc would be too high.

   No-equipment fallback

   Replace Station C's live measurement with a provided data table: 'Student groups measured absorbances of 0.44, 0.42, and 0.46 for three trials with the same initial concentrations. Use the average and proceed with Parts 1–5.' All other calculation steps remain identical.

1. For the reaction 2NO₂(g) ⇌ N₂O₄(g) at some temperature, a sealed flask at equilibrium contains [NO₂] = 0.10 M and [N₂O₄] = 0.40 M. What is Kc? A) 4.0 B) 40 C) 0.025 D) 4.0 M⁻¹

   multiple choiceB) 40. Kc = [N₂O₄]/[NO₂]² = 0.40/(0.10)² = 0.40/0.010 = 40. Answer A misses the square. Answer C inverts numerator and denominator. Answer D attaches units — Kc is reported unitless. Targets SP 5.

2. N₂O₄(g) ⇌ 2NO₂(g) is set up with an initial [N₂O₄] of 0.100 M and no NO₂. At equilibrium, [NO₂] is measured to be 0.040 M. Construct an ICE table and calculate Kc.

   calculationChange in NO₂ = +0.040 M, and because the coefficient on NO₂ is 2, 2x = 0.040 so x = 0.020. ICE table: - I: [N₂O₄] = 0.100, [NO₂] = 0 - C: [N₂O₄] = −x = −0.020, [NO₂] = +2x = +0.040 - E: [N₂O₄] = 0.080, [NO₂] = 0.040 Kc = [NO₂]²/[N₂O₄] = (0.040)²/(0.080) = 0.0016/0.080 = 0.020. The key move is recognizing that x = 0.020 (not 0.040) because the coefficient on NO₂ is 2. Targets SP 5 and SP 3.

3. A student solves an equilibrium problem and reports 'Kc = 1.5 × 10² M⁻¹.' Their arithmetic is correct. State whether the reported answer is fully correct as written, and justify your answer in one or two sentences using AP conventions.

   short answerNot fully correct: K should be reported unitless. AP treats each equilibrium concentration as an activity (concentration divided by the 1 M reference state), so units cancel out of the expression regardless of the exponents. The student should write 'Kc = 1.5 × 10².' Targets SP 6 (Argumentation).

• ☐Colorimeter or Vernier SpectroVis + cuvettes (1 per group)
• ☐0.00200 M Fe(NO₃)₃ in 0.5 M HNO₃ (~50 mL per group)
• ☐0.00200 M KSCN (~50 mL per group)
• ☐10 mL graduated cylinders (2 per group)
• ☐50 mL beakers (2 per group), labeled
• ☐Distilled water squirt bottle
• ☐Waste beaker labeled 'Fe/SCN waste'
• ☐Station handouts (A, B, C) printed for each group
• ☐Goggles for every student
• ☐Calculators
• ☐Whiteboard/markers for board work during hook and direct instruction