AP Chemistry
College Board · 9 units · exam-weighted · high school
AP Chemistry: every unit with its share of the exam, as College Board publishes it, with practice questions per unit and a countdown to the exam.
Set up AP Chemistry → One set per unit, 9 of them pre-written and answer-checked. Free, no account needed.
Units and exam weighting
Unit 1: Atomic Structure and Properties 7–9% of the exam
Moles and molar mass, mass spectrometry, composition of mixtures, electron configuration, photoelectron spectroscopy, periodic trends, valence electrons and ionic compounds.
3 of 13 questions in this unit, with the worked answer
A sample of iron(III) oxide, Fe2O3, has a mass of 15.0 g. How many moles of Fe2O3 are present? (Molar mass Fe2O3 = 159.7 g/mol)
Answer. 0.0939 mol
Which 10.0 g sample contains the greatest number of atoms?
Answer. 10.0 g of Na
In a photoelectron spectroscopy (PES) spectrum, what do the relative heights of the peaks represent?
Answer. The relative number of electrons in that subshell
What people get wrong
- Forgetting to convert percent abundances to decimals before computing a weighted average atomic mass — always divide by 100 first, or the result will be off by roughly a factor of 100.
- Filling 3d before 4s when forming a transition metal cation — electrons are removed from the highest principal quantum number (4s) first, so Fe3+ is $[Ar]3d^5$, not $[Ar]4s^2 3d^3$.
- Confusing PES peak height with peak position — height tells you the number of electrons in a subshell, position tells you binding energy; mixing these up leads to misidentifying elements.
- Assuming ionization energy rises smoothly across every period — the Group 2→13 (e.g., Mg to Al) and Group 15→16 (e.g., N to O) dips are frequently tested and must be explained by subshell stability, not treated as errors.
Unit 2: Compound Structure and Properties 7–9% of the exam
Types of bonds, intramolecular force and potential energy, structure of ionic solids and metals, Lewis diagrams, resonance and formal charge, VSEPR and bond hybridisation.
3 of 14 questions in this unit, with the worked answer
Which type of bonding is characterized by valence electrons delocalized in an 'electron sea' surrounding a lattice of cations?
Answer. Metallic bonding
Rank NaCl, MgO, KBr, and CaS from largest to smallest lattice energy based on Coulomb's law.
Answer. MgO > CaS > NaCl > KBr
Rank the carbon-carbon bonds in ethane (C-C), ethene (C=C), and ethyne (C≡C) from shortest to longest.
Answer. C≡C < C=C < C-C
What people get wrong
- Forgetting to add or subtract electrons for the ion's overall charge when counting valence electrons for a polyatomic ion; always adjust the total electron count before drawing the skeleton.
- Treating one resonance structure as 'the' correct structure and assigning fixed bond lengths to it; instead recognize the resonance hybrid has bond lengths and orders intermediate between the individual structures.
- Confusing electron-domain geometry with molecular geometry when lone pairs are present on the central atom; state both separately, since lone pairs change the molecular shape (e.g., bent vs. linear) but not the electron-domain arrangement.
- Assuming formal charge of zero everywhere is always achievable; some structures require formal charges, and the goal is minimizing magnitude and placing negative charge appropriately, not forcing all charges to zero.
Unit 3: Properties of Substances and Mixtures 18–22% of the exam
Intermolecular forces, properties of solids, liquids and gases, ideal and non-ideal gases, kinetic molecular theory, solutions and mixtures, chromatography, solubility, spectroscopy and Beer-Lambert.
3 of 14 questions in this unit, with the worked answer
Which of the following intermolecular forces is present in all molecules and atoms, regardless of polarity?
Answer. London dispersion forces
NH3 has a significantly higher boiling point than PH3, even though PH3 has a larger molar mass. Explain why.
Answer. NH3 molecules form hydrogen bonds because H is bonded directly to N, while PH3 cannot hydrogen bond since P-H bonds aren't polar enough and P isn't N, O, or F.
Which of the following is NOT a correct assumption of the kinetic molecular theory of gases?
Answer. Average kinetic energy is directly proportional to the volume of the gas
What people get wrong
- Labeling any polar bond as 'hydrogen bonding' — hydrogen bonding only occurs when H is directly bonded to N, O, or F. Check for these specific bonds before invoking hydrogen bonding as the explanation.
- Assuming a heavier molecule always has stronger intermolecular forces — dispersion forces do scale with molar mass, but a lighter polar or hydrogen-bonding molecule can still have a higher boiling point than a heavier nonpolar one. Compare the strongest force present, not just mass.
- Using R with mismatched pressure units (plugging kPa into an equation set up for atm, or forgetting to convert Celsius to Kelvin) in PV=nRT. Always convert temperature to Kelvin and match R's units to the given pressure unit.
- Believing gas solubility increases with temperature the way solid solubility usually does. Gas solubility in liquids decreases as temperature rises because dissolved gas molecules gain enough kinetic energy to escape the solution.
Unit 4: Chemical Reactions 7–9% of the exam
Physical and chemical change, net ionic equations, representations of reactions, stoichiometry, titrations, types of reactions, acid-base and redox reactions.
3 of 14 questions in this unit, with the worked answer
A piece of dry ice (solid CO2) is left on a table and disappears, leaving behind CO2 gas. Is this a physical or chemical change, and why?
Answer. Physical change
Which observation is the strongest evidence that a chemical change has occurred?
Answer. Formation of a gas that was not present before, accompanied by a temperature change
Write the net ionic equation for the reaction between aqueous silver nitrate and aqueous sodium chloride.
Answer. Ag+(aq) + Cl-(aq) → AgCl(s)
What people get wrong
- Assuming every titration ratio is 1:1 — check the balanced equation first; polyprotic acids or bases with more than one reactive OH- change the ratio, as in H2SO4 + 2 NaOH.
- Forgetting to convert volumes from mL to L before using M = mol/L, which produces an answer off by a factor of 1000.
- Writing a net ionic equation without first checking solubility rules, so a spectator ion is left in (or a species that should dissociate is kept molecular) — always confirm which compounds are strong electrolytes before canceling ions.
- Confusing the oxidizing agent with the species that is oxidized — the oxidizing agent is the substance that gets reduced (gains electrons), even though it 'causes' oxidation in the other species.
Unit 5: Kinetics 7–9% of the exam
Reaction rates, rate laws, concentration changes over time, elementary reactions, collision model, reaction energy profiles, mechanisms, catalysis.
3 of 14 questions in this unit, with the worked answer
For the reaction 2A + B → 3C, the rate of disappearance of A is measured as 0.040 M/s. What is the rate of appearance of C?
Answer. 0.060 M/s
A reaction follows rate=k[A]^2[B]. If [A] is tripled and [B] is halved, by what factor does the rate change?
Answer. 4.5
Given: Exp1 [X]=0.050,[Y]=0.050,rate=2.5×10⁻4; Exp2 [X]=0.100,[Y]=0.050,rate=2.5×10⁻4; Exp3 [X]=0.100,[Y]=0.100,rate=1.0×10⁻3. What is the rate law?
Answer. rate = k[Y]^2 (zero order in X)
What people get wrong
- Assuming reaction orders match the coefficients of the overall balanced equation — this is only true for elementary steps. Always determine orders from experimental data for the overall reaction.
- Leaving a reaction intermediate in the final predicted rate law after using the rate-determining step. Instead, use the fast pre-equilibrium (or the step that produced the intermediate) to rewrite it in terms of original reactants.
- Confusing which linear plot corresponds to which order (mixing up ln[A] vs t with 1/[A] vs t). Instead, memorize: zero order is linear as [A] vs t, first order as ln[A] vs t, second order as 1/[A] vs t.
- Believing a catalyst changes ΔH or shifts the equilibrium position. Instead, remember a catalyst only lowers Ea for both directions and speeds up reaching the same equilibrium, without changing K or ΔH.
Unit 6: Thermochemistry 7–9% of the exam
Endothermic and exothermic processes, energy diagrams, heat transfer and thermal equilibrium, heat capacity and calorimetry, energy of phase changes, enthalpy of reaction, bond enthalpies, Hess's law.
3 of 13 questions in this unit, with the worked answer
In an energy diagram for an exothermic reaction, how does the energy of the products compare to the energy of the reactants?
Answer. The products are lower in energy than the reactants.
How much heat is released when 25.0 g of water cools from 80.0°C to 20.0°C? (c_water = 4.18 J/g·°C)
Answer. 6.27 kJ is released.
A 45.0 g piece of metal at 95.0°C is dropped into a coffee-cup calorimeter containing 100.0 g of water at 22.0°C. The mixture reaches a final temperature of 25.0°C. Calculate the specific heat capacity of the metal.
Answer. About 0.398 J/(g·°C).
What people get wrong
- Writing ΔH as positive for an exothermic process because energy is 'released.' Instead, remember the sign describes the system: exothermic reactions lose enthalpy, so ΔH is negative, even though the surroundings gain heat.
- In calorimetry, forgetting the sign flip between the two components (treating q_metal and q_water as equal instead of opposite). Instead, always start from $q_{lost} = -q_{gained}$ before substituting numbers.
- In bond-enthalpy calculations, computing (bonds formed) − (bonds broken) instead of the reverse. Instead, remember breaking bonds always costs energy (positive) and forming bonds always releases energy (subtracted), so it's always broken minus formed.
- When reversing or scaling an equation for Hess's law, changing the equation but forgetting to apply the identical operation to its ΔH. Instead, treat the equation and its ΔH as a single unit — whatever you do to one, do to the other.
Unit 7: Equilibrium 7–9% of the exam
Reversible reactions, the equilibrium constant, ICE tables, Le Chatelier's principle, reaction quotient, solubility equilibria and Ksp, free energy and equilibrium.
3 of 14 questions in this unit, with the worked answer
A reversible reaction has reached equilibrium in a sealed container. Which statement is correct?
Answer. The forward and reverse reaction rates are equal, so concentrations remain constant over time.
Write the Kc expression for 2NOCl(g) ⇌ 2NO(g) + Cl₂(g).
Answer. Kc = [NO]²[Cl₂]/[NOCl]²
Write the Kc expression for the heterogeneous equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g).
Answer. Kc = [CO₂]
What people get wrong
- Including pure solids or liquids in the K or Ksp expression. Instead, cross them out immediately after writing the balanced equation — only gases and aqueous species appear.
- Assuming the small-x approximation always works. Instead, check whether x is under about 5% of the initial concentration after solving; if not, redo the problem with the quadratic formula.
- Confusing Q with K by plugging in initial concentrations when the problem actually wants equilibrium concentrations, or vice versa. Instead, always identify whether the given concentrations are at equilibrium before deciding whether you're computing Q or K.
- Treating a catalyst or an inert gas added at constant volume as a stress that shifts equilibrium. Instead, remember a catalyst changes only the rate to reach equilibrium, and an inert gas at constant volume doesn't change any partial pressure of the reacting species, so K and the position of equilibrium are unaffected.
Unit 8: Acids and Bases 11–15% of the exam
pH and pOH, strong and weak acids and bases, acid-base reactions and titration curves, molecular structure and acid strength, buffers, Henderson-Hasselbalch.
3 of 14 questions in this unit, with the worked answer
Calculate the pH of a 0.025 M solution of HCl (a strong acid).
Answer. pH = 1.60
What is the pH of a 0.010 M NaOH solution?
Answer. pH = 12.00
A 0.20 M solution of weak acid HA has Ka = 1.0×10⁻⁵. Calculate its pH.
Answer. pH ≈ 2.85
What people get wrong
- Treating a weak acid like a strong acid by setting [H⁺] equal to the initial concentration — instead, build an ICE table and solve for x using Ka, since weak acids only partially dissociate.
- Assuming pH + pOH = 14 always holds — this depends on Kw = 1.0×10⁻¹⁴, which is only true at 25°C; at other temperatures Kw changes and so does the sum.
- Assuming the half-equivalence point always gives pH = pKa — this shortcut only applies to weak acid/strong base (or weak base/strong acid) titrations, not to strong-strong titrations which have no meaningful 'pKa'.
- Ranking acid strength purely by electronegativity — for binary acids down a group, decreasing bond strength (not increasing electronegativity) controls the trend, so HI is a stronger acid than HF.
Unit 9: Thermodynamics and Electrochemistry 7–9% of the exam
Entropy, Gibbs free energy and thermodynamic favourability, coupled reactions, galvanic and electrolytic cells, cell potential, Nernst equation, electrolysis and Faraday's laws.
3 of 14 questions in this unit, with the worked answer
Predict the sign of ΔS° for the reaction 2 NO2(g) → N2O4(g).
Answer. Negative
A reaction has ΔH° = −92.0 kJ/mol and ΔS° = −198 J/(mol·K). Is it thermodynamically favorable at 298 K?
Answer. Yes, ΔG° ≈ −33.0 kJ/mol
Which combination of ΔH° and ΔS° guarantees a reaction is thermodynamically favorable at every temperature?
Answer. ΔH° negative and ΔS° positive
What people get wrong
- Flipping a sign when a half-reaction is written as oxidation instead of reduction. Always look up the reduction potential as tabulated and use E°cell = E°cathode − E°anode without reversing any table values.
- Forgetting to convert ΔS° from J/(mol·K) to kJ/(mol·K) before combining it with ΔH° in kJ/mol in the Gibbs equation. Keep units consistent throughout, or convert ΔH° to joules instead.
- Using the 0.0592/n shortcut in the Nernst equation at a temperature other than 25°C. That constant already has T = 298 K built in; at any other temperature you must use the full RT/nF form.
- Multiplying E° values by the number of electrons when combining half-reactions. Cell potential is an intensive property and does not get multiplied even when you scale a half-reaction to balance electrons — only ΔG° scales with n.
How it works
Choosing this curriculum creates a course in Chalkset with one study set per unit and the exam date set to the start of the next testing window. Readiness is weighted the way the exam is, so a weak unit that carries 18% of the score pulls harder than one that carries 6%. Pre-written units come with worked explanations and an independently checked answer key; the rest are written by the agent when you open them.