Lesson 20: Full AP Exam ReviewAP

This final lesson is a map of the whole exam and a rehearsal of the skills that convert knowledge into marks. The AP Chemistry exam has two sections, each worth 50%. Section I: 60 multiple-choice questions in 90 minutes — about 90 seconds each, no penalty for wrong answers, so never leave a blank. Section II: 7 free-response questions in 105 minutes — three long questions (10 points each) and four short ones (4 points each). A calculator is permitted throughout the exam (a change from older practice), and you are given a periodic table, a sheet of equations and constants, and reference tables including standard reduction potentials and Ka/Kb values. Everything in this course maps onto the College Board's nine CED units; the twenty lessons you have worked through cover each unit at least once, and this lesson's job is to make the connections between them automatic.

The habits that earn marks

Show the setup, not just the answer. On free-response questions, the points live in the substitution line: an equation with numbers in it earns the point even if the arithmetic slips. A bare wrong answer earns nothing. Carry units through calculations and report answers with units and a sensible number of significant figures — usually three — and never round intermediate values. Obey the verb. "Calculate" wants a number with work; "explain" and "justify" want a claim tied to evidence (a value, a trend, a principle), typically two or three sentences that name the particles involved; "compare" wants both quantities addressed; "identify" wants a name, not a paragraph. Draw particle diagrams carefully — they are scored on counts, charges, relative sizes and which species touch: if a question says a solution contains Na⁺ and Cl⁻ ions and water, your drawing must show separated, hydrated ions, not NaCl pairs. Answer every part — later parts of a long question often reuse results from earlier ones, and the scoring guidelines usually award follow-on credit if your earlier value is used consistently.

Worked example 1: a complete titration walk-through (Units 4 + 8)

A 50.0 mL sample of 0.100 M HF (Ka = 6.8 × 10⁻⁴) is titrated with 0.100 M NaOH. Key points of the curve, calculated the way the exam expects: Before any base: weak-acid ICE: [H₃O⁺] = √(6.8 × 10⁻⁴ × 0.100) = √(6.8 × 10⁻⁵) = 8.2 × 10⁻³ M; pH = 2.08. At half-equivalence (25.0 mL of base): half the HF has become F⁻, so [HF] = [F⁻] and pH = pKa = −log(6.8 × 10⁻⁴) = 3.17. At equivalence (50.0 mL of base): all the acid is now 5.00 × 10⁻³ mol of F⁻ in 100.0 mL, so [F⁻] = 0.0500 M. F⁻ hydrolyses: Kb = Kw/Ka = 1.0 × 10⁻¹⁴ ÷ 6.8 × 10⁻⁴ = 1.5 × 10⁻¹¹; [OH⁻] = √(1.5 × 10⁻¹¹ × 0.0500) = √(7.4 × 10⁻¹³) = 8.6 × 10⁻⁷ M; pOH = 6.07; pH = 7.93. Three different calculations — ICE, Henderson–Hasselbalch, hydrolysis — for three different regions of one curve. Examiners design titration FRQs so that each region is worth its own points, and students lose most of them by using the wrong region's method, not by bad arithmetic.

Worked example 2: one spectrum, three units (Units 1 + 2 + 3)

An unknown element has a PES spectrum with peaks (binding energy MJ/mol, relative height): 117 (1), 9.2 (1), 6.8 (3), 1.0 (1), 0.7 (3). Convert the heights to electron counts using the subshell capacity pattern (an s subshell holds 2 electrons, a p subshell holds 6): the heights 1, 1, 3, 1, 3 scale by 2 to give 2, 2, 6, 2, 6 electrons, and the peaks assign as 1s² (117), 2s² (9.2), 2p⁶ (6.8), 3s² (1.0), 3p⁶ (0.7). Total = 18 electrons, so a neutral atom has Z = 18: argon. The habit to build: always add the electrons and check the subshell pattern before naming the element — a peak list ending in 3p⁴ with 16 electrons in total would be sulfur, not argon. Now the cross-unit payoffs: its configuration is [Ne] 3s² 3p⁶, so it has 8 valence electrons and a complete octet, explaining its inertness (Unit 1/2); its chloride-free chemistry means questions about its compounds do not exist, but its neighbour chlorine's PES first peak would sit at *higher* binding energy because of the extra proton (Unit 1); and liquid argon is used as an inert atmosphere for welding reactive metals precisely because there are no low-energy reactions available to it. A single PES figure routinely anchors questions across three units — read every figure for all it contains.

Traps that are set on purpose

Ksp is not on this exam. Solubility equilibria and solubility-product calculations were removed from the AP curriculum; precipitation questions are handled with net ionic equations and, occasionally, Q versus a *given* threshold. If you find yourself wanting Ksp, reread the question. Significant figures in pH are inverted: the digits *after the decimal point* of a pH are the significant figures of the concentration. [H₃O⁺] = 1.6 × 10⁻³ M (2 sf) has pH = 2.80 (2 decimal places) — not 2.8. Entropy and enthalpy tables mix units by design (kJ for ΔH, J for ΔS) — convert before combining, every time. "Spontaneous" is not "fast", a catalyst never moves an equilibrium, and an intermediate never appears in a rate law: these three confusions are tested so reliably that they should be read as free marks. Oxidation numbers of the peroxide/hydride exceptions and removing 4s electrons before 3d complete the standard trap list.

In the lab and in the real world

The exam's lab questions draw on a short list of investigations: gravimetric analysis, titration (including back-titration), calorimetry, kinetics by spectrophotometry, chromatography, Beer–Lambert concentration finding, equilibrium by colour, electroplating by Faraday's law, and the identification of unknowns by precipitation and flame tests. For each one, be able to state four things cold: the measured quantity, the conversion to the target quantity, one controlled variable, and the direction of one plausible error. That four-part fluency — practised across Lessons 8, 9, 15, 18 and 19 of this course — is what the free-response lab questions reward, and it transfers wholesale to any first-year university laboratory.

Common mistakes (the final checklist)

• Blank answers on multiple choice. With no guessing penalty, an unanswered question is a free mark thrown away.
• Answers without units or with absurd magnitudes left unchecked. A molar mass of 0.018 g/mol or a pH of 21 should trigger an immediate recheck, not submission.
• Using the Henderson–Hasselbalch equation at the equivalence point, where one member of the pair is absent — the most common titration error of all.
• Assuming equilibrium means equal concentrations. Equilibrium means equal *rates* and Q = K; the concentrations can be wildly unequal, and usually are.
• Stopping a mechanism test after checking the sum. A mechanism must also reproduce the experimental rate law; both checks, every time.
• Defining "strong" as "concentrated". Strength is degree of ionisation; concentration is mol/L. A 6 M solution of acetic acid is concentrated and weak; 10⁻⁵ M HCl is dilute and strong.

AP Exam Experience

Question 1. A reaction at 298 K has ΔG° = −30. kJ/mol. The equilibrium constant K is closest to

• (A) 2 × 10⁻³
• (B) 1
• (C) 2 × 10⁵
• (D) 2 × 10¹⁰

Question 2. A first-order reaction has a half-life of 20 minutes. Starting with 0.80 mol of reactant, how much remains after 60 minutes?

• (A) 0.40 mol
• (B) 0.20 mol
• (C) 0.10 mol
• (D) 0.05 mol

Question 3. Which of the following changes will increase the rate of a reaction but leave the equilibrium constant unchanged?

• (A) Increasing the temperature for an endothermic reaction
• (B) Adding a catalyst
• (C) Increasing the concentration of a reactant
• (D) Decreasing the volume of the container

Answer key. 1. (C) — log K = −ΔG° ÷ (2.303 RT) = 30,000 ÷ (2.303 × 8.314 × 298) = 30,000 ÷ 5706 = 5.26, so K ≈ 10^5.26 ≈ 2 × 10⁵. 2. (C) — 60 minutes is three half-lives: 0.80 → 0.40 → 0.20 → 0.10 mol. 3. (B) — a catalyst changes only kinetics (both rates rise together), so K is untouched. Raising the temperature changes K for any reaction with ΔH ≠ 0; changing concentration or volume changes positions and rates transiently but the question's clean, universal answer is the catalyst — in particular, for an endothermic reaction choice A increases K, eliminating it.

Free-response question (long format). Calcium carbonate decomposes on heating: CaCO₃(s) ⇌ CaO(s) + CO₂(g). In a sealed vessel at 1000 K, the equilibrium pressure of CO₂ is 0.40 atm. (a) Write the Kp expression for the reaction and state its value at 1000 K. (b) Calculate ΔG° for the reaction at 1000 K. (R = 8.314 J/mol·K.) (c) Additional CO₂ is injected into the vessel at 1000 K, raising P_CO₂ momentarily to 1.5 atm. State, with reference to Q, what happens as the system returns to equilibrium. (d) A student adds more solid CaCO₃ to the equilibrium vessel at constant temperature and volume. State the effect, if any, on the equilibrium pressure of CO₂, and justify your answer. (e) At 1200 K, Kp for the reaction is 2.5. Determine whether the decomposition is endothermic or exothermic, justifying your answer with the two Kp values, and calculate ΔG° at 1200 K. Scoring guideline.

• (a) 1 point for Kp = P_CO₂ (solids omitted) = 0.40.
• (b) 1 point for ΔG° = −RT ln Kp = −8.314 × 1000 × ln(0.40) = −8314 × (−0.916) = +7.6 kJ/mol.
• (c) 1 point for recognising Q (= 1.5) now exceeds Kp (0.40), so the reaction runs in reverse — CO₂ is consumed and CaCO₃ forms — until P_CO₂ returns to 0.40 atm.
• (d) 1 point for "no effect": the solid does not appear in the Kp expression, so Q and the position of equilibrium in terms of P_CO₂ are unchanged (provided some of each solid is present).
• (e) 1 point for endothermic: Kp increases as T rises (0.40 → 2.5), which for a rise in temperature means the equilibrium shifts toward products, the endothermic direction; 1 point for ΔG° = −8.314 × 1200 × ln(2.5) = −9977 × 0.916 = −9.1 kJ/mol — its sign has flipped from part (b), as it must once K passes 1.