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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

Use $n = \dfrac{m}{M}$: $n = \dfrac{15.0\text{ g}}{159.7\text{ g/mol}} = 0.0939\text{ mol}$. Always show the division with units to demonstrate dimensional analysis.

Which 10.0 g sample contains the greatest number of atoms?

Answer. 10.0 g of Na

Moles = mass/molar mass, so the element with the smallest molar mass produces the most moles from the same mass. Since each is monatomic, more moles means more atoms, and Na has the smallest molar mass of the four.

In a photoelectron spectroscopy (PES) spectrum, what do the relative heights of the peaks represent?

Answer. The relative number of electrons in that subshell

Peak position encodes binding energy, but peak height (area) is proportional to how many electrons occupy that particular subshell — this is how PES spectra reveal electron configuration.

All 13 questions for Unit 1, with the working →

What people get wrong

  1. 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.
  2. 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 $[Ar]4s^2 3d^3$.
  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.
  4. 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

In metals, valence electrons are not localized between specific atoms but move freely throughout the lattice, allowing metal atoms to slide past each other without breaking bonds. This explains conductivity and malleability.

Rank NaCl, MgO, KBr, and CaS from largest to smallest lattice energy based on Coulomb's law.

Answer. MgO > CaS > NaCl > KBr

Lattice energy scales with the product of ionic charges divided by the distance between ion centers. MgO and CaS both have 2+/2- ion pairs (larger charge product) versus 1+/1- for NaCl and KBr, so they dominate; MgO beats CaS because Mg2+ and O2- are smaller ions than Ca2+ and S2-, giving a shorter distance. Among the 1+/1- pairs, NaCl has smaller ions than KBr, so its lattice energy is larger.

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

Higher bond order pulls the bonded atoms closer together because more shared electron density increases attraction, so the triple bond is shortest and strongest, while the single bond is longest and weakest.

All 14 questions for Unit 2, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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.
  4. 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

Dispersion forces arise from temporary, instantaneous shifts in electron density that create momentary dipoles, so they exist in every atom or molecule. Dipole-dipole and hydrogen bonding require permanent polarity or specific N-H, O-H, or F-H bonds.

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.

Hydrogen bonding is a much stronger intermolecular force than the dispersion forces that dominate in PH3, so despite PH3's greater mass and stronger dispersion forces, NH3 requires more energy to separate its molecules and boils at a higher temperature.

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

KMT states that average kinetic energy is proportional to Kelvin temperature, not volume. The other three statements (negligible particle volume, elastic collisions, no intermolecular forces) are core KMT assumptions.

All 14 questions for Unit 3, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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.
  4. 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

Sublimation only changes the state of CO2 from solid to gas; the chemical identity of the substance (CO2) is unchanged, so no new substance is formed.

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

Gas evolution with an accompanying energy change indicates new bonds are forming and breaking, unlike a simple change of state or dissolving, which are physical.

Write the net ionic equation for the reaction between aqueous silver nitrate and aqueous sodium chloride.

Answer. Ag+(aq) + Cl-(aq) → AgCl(s)

The molecular equation is AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq). Dissociating all strong electrolytes and canceling the spectator ions Na+ and NO3- leaves only the ions that form the precipitate.

All 14 questions for Unit 4, with the working →

What people get wrong

  1. 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.
  2. Forgetting to convert volumes from mL to L before using M = mol/L, which produces an answer off by a factor of 1000.
  3. 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.
  4. 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

Using rate=-1/2·Δ[A]/Δt=1/3·Δ[C]/Δt, the reaction rate itself is 0.040/2=0.020 M/s, so Δ[C]/Δt=3×0.020=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

The factor is (3)^2×(0.5)^1=9×0.5=4.5, since rate depends on the square of [A] and the first power of [B].

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)

Doubling [X] (Exp1→2) with [Y] fixed causes no change in rate, so order in X is 0. Doubling [Y] (Exp2→3) with [X] fixed quadruples the rate, so order in Y is 2.

All 14 questions for Unit 5, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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.
  4. 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.

Exothermic reactions release energy from the system to the surroundings, so the system ends with less energy than it started with, giving a negative ΔH and a downward-sloping diagram.

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.

Use $q = mc\Delta T = 25.0(4.18)(20.0-80.0) = -6270\text{ J} = -6.27\text{ kJ}$. The negative sign shows the water loses heat, so 6.27 kJ is released to the surroundings.

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).

Heat gained by water: $q_{water}=100.0(4.18)(25.0-22.0)=1254\text{ J}$. By conservation, $q_{metal}=-1254\text{ J}=45.0\,c\,(25.0-95.0)$, so $c=1254/(45.0\times70.0)=0.398\text{ J/(g·°C)}$.

All 13 questions for Unit 6, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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.
  4. 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.

Equilibrium is dynamic: both reactions continue to occur, but at equal rates, so no net change in concentration is observed. Concentrations of reactants and products need not be equal to each other.

Write the Kc expression for 2NOCl(g) ⇌ 2NO(g) + Cl₂(g).

Answer. Kc = [NO]²[Cl₂]/[NOCl]²

Products go in the numerator and reactants in the denominator, each raised to its coefficient from the balanced equation. All species here are gases, so all appear in the expression.

Write the Kc expression for the heterogeneous equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g).

Answer. Kc = [CO₂]

Pure solids have an activity of 1 and are omitted from the equilibrium expression, so only the gaseous CO₂ concentration appears.

All 14 questions for Unit 7, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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.
  4. 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

HCl dissociates completely, so [H⁺] = 0.025 M. pH = −log(0.025) = 1.60.

What is the pH of a 0.010 M NaOH solution?

Answer. pH = 12.00

NaOH is a strong base and fully dissociates, giving [OH⁻] = 0.010 M, so pOH = −log(0.010) = 2.00. Then pH = 14.00 − 2.00 = 12.00.

A 0.20 M solution of weak acid HA has Ka = 1.0×10⁻⁵. Calculate its pH.

Answer. pH ≈ 2.85

Set up an ICE table: x²/(0.20 − x) ≈ x²/0.20 = 1.0×10⁻⁵ (approximation valid since Ka is small). Solving gives x = [H⁺] = 1.4×10⁻³ M, so pH = −log(1.4×10⁻³) = 2.85.

All 14 questions for Unit 8, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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'.
  4. 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

Two moles of gas combine to form one mole of gas, decreasing the number of independent gas particles and thus the dispersal of matter, so entropy decreases.

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

ΔG° = ΔH° − TΔS° = −92,000 J − (298 K)(−198 J/K) = −92,000 + 59,004 = −32,996 J/mol ≈ −33.0 kJ/mol, which is negative, so the reaction is favorable at this temperature despite the unfavorable entropy term.

Which combination of ΔH° and ΔS° guarantees a reaction is thermodynamically favorable at every temperature?

Answer. ΔH° negative and ΔS° positive

Since ΔG° = ΔH° − TΔS°, if ΔH° is negative and −TΔS° is also negative (ΔS° positive), every term is negative for any positive T, so ΔG° is always negative.

All 14 questions for Unit 9, with the working →

What people get wrong

  1. 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.
  2. 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.
  3. 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.
  4. 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.

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