Unit 1–9 Quick ReviewUnit 1–9 速览复习

~45 min

Learning goals

  • Check the core of all nine units with active recall and find the units that need another look
  • Plan review time by unit weight, putting Units 3 and 8 first
  • Practice using two or three units together in one question

Before you start

  • AP Units 1–9
  • Exam format and scoring (4.1.1)

How to review effectively

Before the exam, active recall (saying the answer yourself before checking it) works far better than rereading: wherever you get stuck while recalling is where you’d lose points on the exam.

  1. Test first, then read: go through the cards below unit by unit and mark anything you can’t answer, or get wrong, as “missed.”
  2. Split time by weight: Unit 3 (18–22%) and Unit 8 (11–15%) carry the most points; the other units are 7–9% each.
  3. Go back to the lessons: open each unit with misses, reread that lesson’s examples, and redo its practice.
  4. Test again a few days later: repeat the same cards after two or three days until you can answer them all.

The core of the nine units

Active recall: think of the answer first, then reveal it and rate yourself honestly

U1 Atomic Structure and Properties 7–9%

  1. In a mass spectrum, what do the horizontal position and relative height of each peak show?

  2. Across a period from left to right, how do atomic radius and first ionization energy change, and why?

  3. In a photoelectron spectrum (PES), what do a peak’s position and height show?

  4. What are the steps to find an empirical formula from the mass of each element?

Exam questions often span units

AP questions rarely stay inside one unit. Long FRQs often combine stoichiometry (Unit 4), gases (Unit 3), thermochemistry (Unit 6) and equilibrium (Unit 7) in a single scenario. Each example below links three units.

Example 1: stoichiometry + gases (Units 1, 3, 4)

0.243 g of magnesium reacts with excess hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). What volume of H₂ forms at 25 °C and 0.980 atm? (Mg 24.31 g/mol)

Step-by-step solution
  1. Moles of magnesium (Unit 1).

    n(Mg)=0.24324.31=1.00×10−2 moln(\ce{Mg}) = \frac{0.243}{24.31} = 1.00 \times 10^{-2}\ \text{mol}
  2. Convert with the equation (Unit 4). n(H₂) : n(Mg) = 1 : 1, so n(H₂) = 1.00 × 10⁻² mol.

  3. Ideal gas law (Unit 3). With T = 298.15 K:

    V=nRTP=(0.0100)(0.08206)(298.15)0.980=0.250 L\begin{aligned} V &= \frac{nRT}{P} = \frac{(0.0100)(0.08206)(298.15)}{0.980} \\ &= \boxed{0.250\ \text{L}} \end{aligned}

Example 2: equilibrium + acids + thermodynamics (Units 7, 8, 9)

At 25 °C, HF has Ka = 6.4 × 10⁻⁴. Find ΔG° for the ionization of HF, and the pH of 0.10 M HF.

Step-by-step solution
  1. ΔG° (Unit 9). ΔG° = −RT ln K:

    ΔG∘=−(8.314)(298.15)×ln⁡(6.4×10−4)=1.82×104 J/mol=+18.2 kJ/mol\begin{aligned} \Delta G^\circ &= -(8.314)(298.15) \\ &\quad \times \ln(6.4 \times 10^{-4}) \\ &= 1.82 \times 10^{4}\ \text{J/mol} \\ &= \boxed{+18.2\ \text{kJ/mol}} \end{aligned}

    ΔG° > 0 matches K < 1: only a small fraction of HF ionizes in water.

  2. Try the approximation first (Unit 8).

    x≈Kac=6.4×10−4×0.10=8.0×10−3x \approx \sqrt{K_a c} = \sqrt{6.4 \times 10^{-4} \times 0.10} = 8.0 \times 10^{-3}
  3. Check it. 8.0 × 10⁻³ ÷ 0.10 = 8%, more than 5%, so the approximation isn’t good enough; solve the quadratic from the ICE table (Unit 7):

    x20.10−x=6.4×10−4x=7.7×10−3 M\begin{gathered} \frac{x^2}{0.10 - x} = 6.4 \times 10^{-4} \\ x = 7.7 \times 10^{-3}\ \text{M} \end{gathered}
  4. pH. pH = −log(7.7 × 10⁻³) = 2.11\boxed{2.11}.

Practice: true or false?
  1. A catalyst increases a reaction’s equilibrium constant K
  2. Across a period from left to right, atomic radius decreases
  3. When a weak acid is titrated with a strong base, the equivalence pH is 7
  4. An endothermic reaction can never be favorable at any temperature
  5. The higher the bond order, the shorter and stronger the bond
  6. A first-order half-life doesn’t depend on the starting concentration
  7. Lowering the temperature always shifts an equilibrium forward
  8. Real gases deviate clearly from ideal behavior at high pressure and low temperature

Common mistakes

Mistake 1: “Rereading the book cover to cover counts as review.” Rereading makes everything feel familiar, but the exam asks you to retrieve. Test yourself first, then read what you couldn’t answer.

Mistake 2: “Spend the same time on every unit.” Unit 3 is worth more than twice as much as most units, and Unit 8 clearly more too. Split your time by weight and by your own weak spots.

Mistake 3: “Mastering each unit separately is enough.” Exam questions often span units. In a combined question, you need to recognize which unit’s method each step uses (for example, stoichiometry first, then the gas law).

Quick check

Which unit carries the most weight on the AP Chemistry exam?

Which review method works best?

For a reaction with ΔG° = +18.2 kJ/mol, how does K compare with 1?

After approximating x ≈ √(Ka·c), when must you switch to the quadratic?

Which pair of topics is most likely to appear in the same question?

Of CH₄, NH₃ and H₂O, which has the highest boiling point?

AP-style practice

AP-style free response

Ammonia synthesis: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Data at 25 °C (OpenStax Appendix G), with ΔfH° in kJ/mol and S° in J/(mol·K):

SubstanceΔfH°S°
N₂(g)0191.6
H₂(g)0130.7
NH₃(g)−45.9192.8

(a) Draw the Lewis structure of NH₃, give its shape, and say whether it is polar.

(b) Calculate ΔH° and ΔS° for the reaction.

(c) Calculate ΔG° at 25 °C. Under standard conditions at this temperature, is the forward reaction thermodynamically favored?

(d) Industry runs this reaction at about 700 K. How does raising the temperature affect K and the reaction rate?

Model answer and scoring

(a) N forms a single bond to each of three H atoms and has one lone pair; with four electron domains, the molecule is trigonal pyramidal. The N–H bonds are polar and the pyramidal shape keeps the dipoles from cancelling, so NH₃ is polar.

(b) ΔH° = 2(−45.9) − 0 = −91.8 kJ/mol; ΔS° = 2(192.8) − [191.6 + 3(130.7)] = 385.6 − 583.7 = −198.1 J/(mol·K).

(c) ΔG° = −91.8 − 298.15 × (−0.1981) = −32.7 kJ/mol < 0, so it is thermodynamically favorable under standard conditions at 25 °C.

(d) The forward reaction is exothermic, so raising the temperature makes K smaller (the equilibrium shifts left); but more molecules then have energy above the activation energy, so the rate increases. Industry uses a higher temperature (and a catalyst) so the reaction is fast enough.

Scoring: (a) Lewis structure, trigonal pyramidal, polar; (b) values and signs of ΔH° and ΔS°; (c) ΔG° and the judgment; (d) K decreases and the rate increases, each with a reason.

Summary

  • Review with active recall: test yourself, go back to the units you missed, and test again a few days later.
  • Split time by weight, with Units 3 and 8 first.
  • In combined questions, identify which unit’s method each step uses, and check approximations, units and signs.

AP tip: In the last two weeks, a daily set of questions that span units (such as one long FRQ) is closer to the real exam than reviewing a single unit at a time.

Sources: unit weights from the College Board AP Chemistry Course and Exam Description; Ka of HF from OpenStax Chemistry 2e Appendix H; ΔfH° and S° from Appendix G of the same book.

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