AP Chemistry Mock ExamMock Exam 1
A complete AP Chemistry practice exam with the same question types, number of questions, timing, and scoring as the current AP Chemistry Exam. Every question is original to this site. Afterward you score your free responses against model answers to get a composite score and an estimated AP score; missed questions go into your mistake notebook automatically.
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Exam format
| Section | Question type | Questions | Time | Weight |
|---|---|---|---|---|
| Section I | Multiple choice (four options) | 60 | 90 min | 50% |
| Section II | Free response: 3 long (10 points each) and 4 short (4 points each) | 7 | 105 min | 50% |
Instructions
- You won’t see whether answers are right while you work; answers and explanations come after you submit.
- In Section I you can change answers at any time and flag questions to come back to. When time runs out, the section is submitted automatically.
- In Section II, write your answer in the box under each part; what you type is saved on this device. After submitting, you score yourself point by point against model answers and scoring points.
- You may use a calculator (College Board’s current policy allows one on both sections), plus the equations and constants and the periodic table below.
- If you close the page, you can pick up where you left off next time; the clock runs only while the page is open.
Equations and constants
Constants
- R = 8.314 J/(mol·K) = 0.08206 L·atm/(mol·K) = 62.36 L·mm Hg/(mol·K)
- F = 96,485 C/mol (charge of 1 mol of electrons)
- NA = 6.022 × 10²³ mol⁻¹
- h = 6.626 × 10⁻³⁴ J·s, c = 2.998 × 10⁸ m/s
- Kw = 1.0 × 10⁻¹⁴ at 25 °C
- 1 atm = 760 mm Hg = 760 torr = 101.3 kPa; 0 °C = 273.15 K
Atomic structure and light
- E = hν, c = λν
Gases, liquids, and solutions
- PV = nRT; PA = Ptotal × XA (XA = mole fraction); Ptotal = PA + PB + …
- n = m/M; M = dRT/P; KE per molecule = ½mv²
- M = moles of solute / liters of solution; A = εbc (Beer–Lambert law)
Kinetics
- Zero order: [A]t − [A]0 = −kt
- First order: ln[A]t − ln[A]0 = −kt; t½ = 0.693/k
- Second order: 1/[A]t − 1/[A]0 = kt
Equilibrium and acids and bases
- aA + bB ⇌ cC + dD: Kc = [C]c[D]d / ([A]a[B]b); Kp uses partial pressures
- pH = −log[H⁺], pOH = −log[OH⁻]; pH + pOH = 14 at 25 °C
- Ka × Kb = Kw; pK = −log K
- pH = pKa + log([A⁻]/[HA])
Thermodynamics and electrochemistry
- q = mcΔT; ΔH° = ΣnΔH°f(products) − ΣnΔH°f(reactants)
- ΔS° = ΣnS°(products) − ΣnS°(reactants); ΔG° = ΔH° − TΔS°
- ΔG° = −RT ln K = −nFE°
- I = q/t; E°cell = E°(cathode) − E°(anode)
Periodic table
The table is wide: swipe sideways to see all of it.
Click any element in the table.
Section I submitted
The real exam has a short break between sections. When you’re ready, start Section II: 7 free-response questions in 105 minutes.
Score your free responses
Compare with the model answers and scoring points, and check each point your answer earns (1 point each). When you’re done, use the button below to see your result.
- 1Multiple choice
Which sample contains the greatest total number of atoms? (H 1, He 4, N 14, O 16)
ExplanationConvert to moles, then multiply by atoms per particle: H₂O 1 mol × 3 = 3 mol of atoms; He 1 mol × 1 = 1 mol; O₂ 0.5 mol × 2 = 1 mol; N₂ 1 mol × 2 = 2 mol.
Review: U1.1 Moles and Molar Mass
- 2Multiple choice
CH₄, NH₃, and H₂O have similar molar masses (16, 17, and 18 g/mol). What is their order of increasing boiling point?
ExplanationWith similar molar masses, dispersion forces are comparable. Nonpolar CH₄ has only dispersion forces and boils lowest (−161 °C); NH₃ hydrogen-bonds (−33 °C); in H₂O, O is more electronegative and each molecule has 2 H atoms and 2 lone pairs, so it forms more and stronger hydrogen bonds and boils highest (100 °C).
Review: U3.1 Intermolecular Forces
- 3Multiple choice
A sealed container holds 4 H₂ molecules and 3 O₂ molecules, which react completely by 2H₂ + O₂ → 2H₂O. Which box shows the molecules after the reaction?
ExplanationFour H₂ need only 2 O₂, so H₂ runs out. That makes 4 H₂O and leaves 3 − 2 = 1 O₂. Atoms balance: 8 H and 6 O before, and 4 H₂O plus 1 O₂ also give 8 H and 6 O.
- 4Multiple choice
A compound is 40.0% C, 6.7% H, and 53.3% O by mass. What is its empirical formula?
ExplanationPer 100 g: C 3.33 mol, H 6.65 mol, O 3.33 mol. Dividing by the smallest, 3.33, gives 1 : 2 : 1, so CH₂O.
Review: U1.4 Composition of Mixtures
- 5Multiple choice
Which pair of atoms forms the bond with the greatest ionic character?
ExplanationThe bigger the electronegativity difference, the more ionic the bond. K sits low on the left (very low electronegativity) and F at the top right (the highest), so their difference is largest.
Review: U2.1 Types of Chemical Bonds
- 6Multiple choice
A rigid 2.0 L steel cylinder holds argon at 300 K and 3.0 atm. What is the pressure after it is heated to 400 K?
ExplanationWith volume and moles fixed, P is proportional to T in kelvins: P₂ = 3.0 × 400/300 = 4.0 atm. Temperatures must be in kelvins; 2.3 atm inverts the ratio.
Review: U3.4 Ideal Gas Law
- 7Multiple choice
The table gives initial-rate data for A + B → C. What is the rate law?
[A]₀ (M) [B]₀ (M) Initial rate (M/s) 0.10 0.10 2.0 × 10−3 0.20 0.10 8.0 × 10−3 0.10 0.20 2.0 × 10−3 ExplanationRuns 1 and 2: doubling [A] quadruples the rate, so second order in A. Runs 1 and 3: doubling [B] leaves the rate unchanged, so zero order in B. Rate = k[A]².
Review: U5.2 Introduction to Rate Law
- 8Multiple choice
Solutions of Pb(NO₃)₂ and KI are mixed, and a yellow precipitate forms. What is the net ionic equation?
ExplanationWrite only the particles that change: Pb²⁺ and I⁻ combine into insoluble PbI₂. K⁺ and NO₃⁻ are spectators, KNO₃ is soluble and doesn’t precipitate, and PbI doesn’t balance charge.
Review: U4.2 Net Ionic Equations
- 9Multiple choice
The mass spectrum of an element is shown (mass number on the horizontal axis, with the relative abundance of each isotope). The element is most likely which of the following?
ExplanationAverage atomic mass ≈ 69 × 0.601 + 71 × 0.399 = 69.8, between 69 and 71 and closer to the more abundant 69, which matches gallium’s 69.72.
- 10Multiple choice
A white solid melts at 801 °C, does not conduct as a solid but conducts when molten, and dissolves readily in water. It is most likely what kind of solid?
ExplanationA high melting point, no conduction as a solid but conduction when molten (ions free to move), and ready solubility in water all point to an ionic solid. Metals conduct as solids, molecular solids melt low, and covalent networks don’t conduct when molten and don’t dissolve in water.
Review: U3.2 Properties of Solids
- 11Multiple choice
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ/mol. How much heat is released when 0.25 mol of CH₄ burns completely?
ExplanationCH₄ has a coefficient of 1, so each mole releases 890 kJ: 0.25 × 890 = 223 kJ (about 220 kJ).
- 12Multiple choice
Of the possible Lewis structures for the cyanate ion, OCN⁻ (C in the center), which is best according to formal charges?
ExplanationThe best structure keeps formal charges near zero and puts negative charge on the most electronegative atom. In O–C≡N only O carries −1, and O is the most electronegative; O=C=N puts −1 on the less electronegative N, so it ranks second; the +1/−2 structure is worst; and the all-single-bond structure leaves C without an octet.
Review: U2.6 Resonance and Formal Charge
- 13Multiple choice
At 25 °C, what is the pH of a solution with [OH⁻] = 1.0 × 10⁻³ M?
ExplanationpOH = 3.00, so pH = 14.00 − 3.00 = 11.00, a basic solution.
- 14Multiple choice
A gas has a density of 1.25 g/L at 0 °C and 1.00 atm. Which gas is it most likely to be? (R = 0.0821 L·atm/(mol·K))
ExplanationRearranging PV = nRT gives M = dRT/P = 1.25 × 0.0821 × 273 ÷ 1.00 ≈ 28 g/mol, matching N₂ (28).
Review: U3.4 Ideal Gas Law
- 15Multiple choice
What is the Kp expression for C(s) + H₂O(g) ⇌ CO(g) + H₂(g)?
ExplanationC(s) is a pure solid and is left out; H₂O is a gas here and stays in. The partial pressures of the products CO and H₂ go on top and that of the reactant H₂O on the bottom.
- 16Multiple choice
This photoelectron spectrum shows every electron in one atom of an element (binding energy decreases from left to right; peak height is proportional to the number of electrons). Which element is it?
ExplanationThe five peaks are 1s, 2s, 2p, 3s, and 3p with 2, 2, 6, 2, and 3 electrons, 15 in all: 1s²2s²2p⁶3s²3p³, phosphorus. N has only two shells (three peaks), Al has 1 electron in 3p, and S has 4.
- 17Multiple choice
4.0 mol of Al reacts completely with 4.0 mol of Cl₂ by 2Al + 3Cl₂ → 2AlCl₃. What is the greatest amount of AlCl₃ that can form?
ExplanationAl: 4.0 ÷ 2 = 2.0; Cl₂: 4.0 ÷ 3 ≈ 1.33, so Cl₂ is limiting. n(AlCl₃) = 4.0 × 2/3 ≈ 2.7 mol. Basing it on Al gives 4.0 mol, but there isn’t enough Cl₂.
Review: U4.5 Stoichiometry
- 18Multiple choice
After NaCl dissolves in water, how are the water molecules around Na⁺ and Cl⁻ oriented?
ExplanationIn water, O carries a partial negative charge and H a partial positive charge. Opposite charges attract: the O end faces the cation Na⁺ and the H end faces the anion Cl⁻ (ion–dipole attractions).
- 19Multiple choice
A first-order reaction consumes 75% of its reactant in 60 min. What is its half-life?
Explanation25% remaining is 1/4 = (1/2)², so 2 half-lives have passed. A first-order half-life is constant, so t½ = 60 ÷ 2 = 30 min.
Review: U5.3 Concentration Changes Over Time (Integrated Rate Laws, Half-Life)
- 20Multiple choice
At 25 °C, a 0.10 M solution of a monoprotic weak acid HA has pH = 3.0. What is Ka for HA, approximately?
Explanation[H⁺] = [A⁻] = 10⁻³·⁰ = 1.0 × 10⁻³ M and [HA] ≈ 0.10. Ka = (1.0 × 10⁻³)² ÷ 0.10 = 1.0 × 10⁻⁵. 1.0 × 10⁻³ just takes [H⁺] as Ka; 1.0 × 10⁻² is the fraction ionized.
- 21Multiple choice
The potential energy curve for H₂ has its minimum at an internuclear distance of 74 pm, at −436 kJ/mol (zero being two far-apart H atoms). Which statement is correct?
ExplanationThe distance at the minimum is the bond length, and the depth from there to zero is the bond energy: breaking the bond absorbs that much, forming it releases the same. Closer than that, nuclear repulsion makes the energy rise steeply.
- 22Multiple choice
Blocks of copper and aluminum with equal masses and equal starting temperatures each absorb the same amount of heat (no phase change). Specific heats: Cu 0.385, Al 0.897 J/(g·°C). Which statement is correct?
ExplanationΔT = q ÷ (m·c); with equal q and m, the smaller specific heat gives the larger rise. ΔT(Cu) ÷ ΔT(Al) = 0.897 ÷ 0.385 ≈ 2.3. Conductivity affects how fast heat moves, not the temperature after a given amount is absorbed.
Review: U6.3 Heat Transfer and Thermal EquilibriumU6.4 Heat Capacity and Calorimetry
- 23Multiple choice
Samples of He and Ar are at the same temperature. Which statement about their atoms is correct?
ExplanationAverage kinetic energy depends only on temperature, so it is the same. Since KE = ½mv², the lighter He atoms must move faster on average.
Review: U3.5 Kinetic Molecular Theory
- 24Multiple choice
For which process is ΔS° positive?
ExplanationWhen hydrogen peroxide decomposes, 2 mol of liquid become 2 mol of liquid plus 1 mol of gas. Producing a gas raises the entropy: ΔS° = 2(70.0) + 205.2 − 2(109.6) ≈ +126 J/K. The others turn 3 mol of gas into 2, use up a gas to make a solid, or condense a gas, all lowering entropy.
Review: U9.1 Introduction to Entropy
- 25Multiple choice
At some temperature, K = 0.20 for N₂O₄(g) ⇌ 2NO₂(g). At some moment [N₂O₄] = 0.10 M and [NO₂] = 0.20 M. Which way does the reaction proceed?
ExplanationQ = [NO₂]² ÷ [N₂O₄] = 0.20² ÷ 0.10 = 0.40 > 0.20. There is too much product, so the reaction runs in reverse. Forgetting the square gives 2.0 (same conclusion, wrong Q).
- 26Multiple choice
The first ionization energy of oxygen is slightly lower than that of nitrogen. Which is the best explanation?
ExplanationN is 2p³ (one electron in each orbital); O is 2p⁴, with two electrons in one 2p orbital. Repulsion between the paired electrons makes one easier to remove, outweighing O’s larger nuclear charge. O actually has more protons and a smaller radius.
Review: U1.7 Periodic Trends
- 27Multiple choice
25.0 mL of H₂SO₄ solution is exactly neutralized by 40.0 mL of 0.100 M NaOH. What is the concentration of the H₂SO₄?
ExplanationH₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. n(NaOH) = 0.0400 × 0.100 = 4.00 × 10⁻³ mol, so n(H₂SO₄) = 2.00 × 10⁻³ mol and c = 2.00 × 10⁻³ ÷ 0.0250 = 0.0800 M. Ignoring the 1 : 2 ratio gives 0.160.
Review: U4.6 Introduction to Titration
- 28Multiple choice
Which mixture (each dissolved in 1.0 L of water) forms a buffer?
ExplanationReact first: 0.10 mol OH⁻ neutralizes half the acetic acid, leaving 0.10 mol each of CH₃COOH and CH₃COO⁻, a buffer. Equal amounts leave only CH₃COO⁻; HCl/NaCl has no weak conjugate pair; and NH₃ with excess HCl leaves strong acid.
- 29Multiple choice
Which of these liquids has the highest vapor pressure at 25 °C?
ExplanationThe weaker the intermolecular forces, the more easily molecules escape into the vapor and the higher the vapor pressure. Diethyl ether can’t hydrogen-bond with itself, so its forces are weakest; water, ethanol, and ethylene glycol all hydrogen-bond, and ethylene glycol, with two –OH groups, has the lowest vapor pressure.
Review: U3.3 Solids, Liquids, and Gases
- 30Multiple choice
The energy profile for a two-step reaction is shown (values are potential energies in kJ/mol). Which statement is correct?
ExplanationStep 1 has Ea = 50 kJ/mol and step 2 has 80 − 20 = 60 kJ/mol, so step 2 is higher, slower, and rate-determining. The products (−30) lie below the reactants (0), so the reaction is exothermic overall.
- 31Multiple choice
What is the molecular geometry of XeF₄?
ExplanationXe has 8 valence electrons: four bond to F, leaving 2 lone pairs, so there are 6 electron domains in an octahedral arrangement. The lone pairs sit opposite each other (180°), leaving the four F atoms in a plane: square planar.
- 32Multiple choice
Use average bond enthalpies to estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g). Bond enthalpies (kJ/mol): H–H 436, Cl–Cl 242, H–Cl 431.
ExplanationΔH ≈ Σbroken − Σformed = (436 + 242) − 2 × 431 = 678 − 862 = −184 kJ/mol. Reversing it (formed − broken) gives +184, and counting only one H–Cl gives +247.
Review: U6.7 Bond Enthalpies
- 33Multiple choice
At 300 K and 1 atm, which gas shows the largest deviation from ideal behavior?
ExplanationThe ideal-gas model assumes no attractions between molecules. NH₃ is polar and hydrogen-bonds, so its attractions are strongest and its real pressure falls furthest below the ideal value. He, H₂, and N₂ are nonpolar with very weak attractions.
- 34Multiple choice
2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0. Which conditions give the highest equilibrium yield of SO₃?
ExplanationThe forward reaction is exothermic, so cooling shifts it forward; 3 mol of gas on the left versus 2 on the right means higher pressure also shifts it forward. A catalyst doesn’t change the equilibrium yield. (Industry avoids very low temperatures to keep the rate up.)
- 35Multiple choice
10.0 mL of 0.10 M HCl is diluted with water to 100.0 mL. What is the pH after dilution?
ExplanationThe volume grows tenfold, so [H⁺] = 0.010 M and pH = 2.00. Diluting a strong acid tenfold raises the pH by 1.
- 36Multiple choice
What is the most likely formula of the ionic compound formed by calcium and nitrogen?
ExplanationCa has 2 valence electrons and forms Ca²⁺; N has 5 and gains 3 to form N³⁻. Balancing charge, 3 Ca²⁺ (+6) pair with 2 N³⁻ (−6): Ca₃N₂.
- 37Multiple choice
A galvanic cell is built from Zn/Zn²⁺ (E° = −0.76 V) and Cu/Cu²⁺ (E° = +0.34 V). Which statement is correct?
ExplanationZn, with the lower E°, is oxidized and is the anode. E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V. 0.42 just adds the two values.
Review: U9.7 Galvanic (Voltaic) and Electrolytic CellsU9.8 Cell Potential and Free Energy
- 38Multiple choice
What mass of NaCl (M = 58.44 g/mol) is needed to make 250.0 mL of 0.200 M solution?
Explanationn = 0.200 M × 0.2500 L = 0.0500 mol, m = 0.0500 × 58.44 = 2.92 g. 11.7 g would make 1 L, and 0.0500 is the number of moles, not grams.
Review: U3.7 Solutions and Mixtures
- 39Multiple choice
In 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O, what is oxidized?
ExplanationS goes from +4 in SO₃²⁻ to +6 in SO₄²⁻, losing electrons: oxidized. Mn goes from +7 to +2: reduced (MnO₄⁻ is the oxidizing agent). H stays +1 and O stays −2.
- 40Multiple choice
For A → products, a plot of ln[A] versus time is a straight line with a slope of −0.023 s⁻¹. Which statement is correct?
ExplanationFor first order, ln[A] = ln[A]₀ − kt, so ln[A] versus t is linear with slope −k, giving k = 0.023 s⁻¹ (k is always positive). Zero order makes [A] versus t linear; second order makes 1/[A] versus t linear.
Review: U5.3 Concentration Changes Over Time (Integrated Rate Laws, Half-Life)
- 41Multiple choice
Aqueous ammonia is titrated with HCl. What is the pH at the equivalence point, and why?
ExplanationAt equivalence all the NH₃ has become NH₄⁺ (Cl⁻ doesn’t affect pH), and NH₄⁺ donates H⁺, so the solution is acidic. pH = pKa(NH₄⁺) holds at the half-equivalence point.
Review: U8.5 Acid-Base Titrations
- 42Multiple choice
Steel, an interstitial alloy made by adding a little carbon to iron, is much harder than pure iron. Which is the best explanation?
ExplanationIn a pure metal the atoms line up neatly and layers slide easily, which makes it malleable. Carbon atoms are much smaller than iron atoms; sitting in the gaps, they disrupt the layers and resist sliding, so steel is harder and less malleable. Metallic bonding and mobile electrons remain.
- 43Multiple choice
A paper chromatography experiment separates the components of an ink. The paper (stationary phase) is polar and the solvent is nonpolar. Which component has the largest Rf?
ExplanationA large Rf means the component travels far with the solvent. The least polar component interacts more with the nonpolar solvent and less with the polar paper, so it moves farthest.
Review: U3.9 Separation of Solutions and Mixtures (Chromatography, Distillation)
- 44Multiple choice
How much heat is needed to melt 36 g of ice at 0 °C into water at 0 °C? (ΔHfus = 6.0 kJ/mol, M = 18 g/mol)
Explanationn = 36 ÷ 18 = 2.0 mol, so q = 2.0 × 6.0 = 12 kJ. The temperature stays the same during melting, but heat is still absorbed to overcome intermolecular attractions.
Review: U6.5 Energy of Phase Changes
- 45Multiple choice
At some temperature, K = 1.0 × 10⁻⁴ for A(g) ⇌ 2B(g). What is K for 4B(g) ⇌ 2A(g) at the same temperature?
ExplanationReversing takes the reciprocal: 1/(1.0 × 10⁻⁴) = 1.0 × 10⁴; doubling the coefficients squares it: (1.0 × 10⁴)² = 1.0 × 10⁸.
- 46Multiple choice
A reaction has ΔH° > 0 and ΔS° > 0. When is it thermodynamically favorable?
ExplanationΔG° = ΔH° − TΔS°: the positive ΔH° hurts and −TΔS° helps. Once T is large enough, TΔS° exceeds ΔH° and ΔG° < 0.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 47Multiple choice
Which of these oxyacids is the strongest?
ExplanationAll are H–O–X. The more electronegative X is, the more it pulls electron density from the O–H bond, the more easily H⁺ leaves, and the more stable XO⁻ is. Cl is the most electronegative of Cl, Br, and I.
- 48Multiple choice
Which substance is most soluble in hexane, C₆H₁₄?
Explanation“Like dissolves like”: hexane is nonpolar, and nonpolar I₂ interacts with it through dispersion forces, so it dissolves best. Ionic compounds and hydrogen-bonding methanol dissolve poorly in nonpolar solvents.
Review: U3.10 Solubility
- 49Multiple choice
What type of reaction is 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)?
ExplanationNa goes from 0 to +1 (oxidized) and H from +1 to 0 (reduced). Electrons are transferred, so it is a redox reaction (also a single replacement).
Review: U4.7 Types of Chemical Reactions
- 50Multiple choice
The mechanism for X₂ + 2Y₂ → 2XY₂ is: step 1 (fast, reversible) X₂ ⇌ 2X; step 2 (slow) X + Y₂ → XY₂ (occurs twice). Which rate law is consistent with the mechanism?
ExplanationThe slow step 2 gives rate = k₂[X][Y₂]. X is an intermediate and can’t appear in the rate law; the fast equilibrium K = [X]²/[X₂] gives [X] = (K[X₂])1/2. Substituting, rate = k[X₂]1/2[Y₂]. Writing k[X₂][Y₂]² from the overall coefficients is a common mistake.
- 51Multiple choice
Ag₂CrO₄ has a molar solubility of s mol/L in water. What is its Ksp in terms of s?
ExplanationAg₂CrO₄(s) ⇌ 2Ag⁺ + CrO₄²⁻, so [Ag⁺] = 2s and [CrO₄²⁻] = s. Ksp = (2s)²·s = 4s³.
- 52Multiple choice
A weak acid HA has pKa = 4.0. In a solution at pH 5.0, what is [A⁻] ÷ [HA]?
Explanationlog([A⁻]/[HA]) = pH − pKa = 1.0, so the ratio is 10. Above the pKa, the base form predominates.
Review: U8.7 pH and pKa
- 53Multiple choice
Absorbing which type of electromagnetic radiation mainly causes transitions between a molecule’s vibrational levels?
ExplanationInfrared excites molecular vibrations; lower-energy microwaves excite rotations; UV and visible light promote valence electrons; X-rays can eject core electrons.
- 54Multiple choice
What is ΔH for 2NO(g) + O₂(g) → 2NO₂(g)? ΔH°f (kJ/mol): NO(g) +90, NO₂(g) +33.
ExplanationΔH = 2(+33) − [2(+90) + 0] = 66 − 180 = −114 kJ/mol (O₂ has zero enthalpy of formation). +114 subtracts the wrong way, and −57 is the value per mole of NO.
Review: U6.8 Enthalpy of Formation
- 55Multiple choice
A redox reaction has E°cell = +0.30 V with n = 2. Which is correct about its ΔG° and K?
ExplanationΔG° = −nFE°, so a positive E° gives a negative ΔG°; ΔG° = −RT ln K, so a negative ΔG° gives K > 1.
Review: U9.8 Cell Potential and Free EnergyU9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)
- 56Multiple choice
A buffer contains a weak acid with pKa = 6.0, with [A⁻] ÷ [HA] = 0.10. What is the pH?
ExplanationpH = pKa + log([A⁻]/[HA]) = 6.0 + log 0.10 = 6.0 − 1.0 = 5.0.
- 57Multiple choice
In which solution is the molar solubility of AgCl lowest?
ExplanationCommon-ion effect: the more Cl⁻, the less AgCl dissolves. 0.10 M CaCl₂ gives [Cl⁻] = 0.20 M, more than 0.10 M NaCl. NaNO₃ has no common ion.
Review: U7.12 Common-Ion Effect
- 58Multiple choice
Using the same cuvette and wavelength, a 0.040 M solution of a colored compound has an absorbance of 0.60. What is the concentration of a solution of the same compound with an absorbance of 0.45?
ExplanationA = εbc with ε and b fixed, so A is proportional to c: c = 0.040 × 0.45/0.60 = 0.030 M.
Review: U3.13 Beer-Lambert Law
- 59Multiple choice
CuSO₄ solution is electrolyzed with a current of 1.93 A. About how long does it take to deposit 0.010 mol of Cu at the cathode? (F = 96,500 C/mol)
ExplanationCu²⁺ + 2e⁻ → Cu needs 0.020 mol of electrons: q = 0.020 × 96,500 = 1930 C, and t = q ÷ I = 1930 ÷ 1.93 ≈ 1000 s. Forgetting the 2 electrons gives 500 s.
- 60Multiple choice
Buffers X and Y have the same pH and are 100 mL each: X is 0.10 M in both HA and A⁻, and Y is 1.0 M in both. 0.010 mol of NaOH is added to each (ignore the volume change). Which statement is correct?
ExplanationX contains only 0.010 mol of HA, all neutralized by the 0.010 mol of OH⁻, leaving only A⁻: the buffer is exhausted and the pH jumps. Y has 0.10 mol of HA; afterward HA is 0.090 mol and A⁻ 0.11 mol, a small change in ratio, so the pH rises only about 0.09.
Review: U8.10 Buffer Capacity
Write your answer in the box under each part; key steps and equations are enough. Model answers and scoring points appear after you submit.
- 1Free response
Ozone (O₃) in the stratosphere can be destroyed in a reaction catalyzed by chlorine atoms. The overall reaction is O₃(g) + O(g) → 2O₂(g), and the proposed mechanism is: step 1 (slow) Cl + O₃ → ClO + O₂; step 2 (fast) ClO + O → Cl + O₂. Answer the following questions.
(a)Draw a Lewis diagram for O₃, including resonance structures, and compare the lengths of the two O–O bonds in O₃ with the O=O bond length in O₂.
(2 pts)
Model answerTwo resonance structures: O=O–O ↔ O–O=O, with one lone pair and a +1 formal charge on the central O and −1 on the singly bonded end O. Both O–O bonds are equal, each like “one and a half” bonds, and longer than the O=O double bond in O₂ (bond order 1.5 < 2).
Scoring points (1 point each)(b)Show that steps 1 and 2 add to give the overall reaction, and identify the catalyst and the intermediate in the mechanism.
(2 pts)
Model answerAdding the steps: Cl + O₃ + ClO + O → ClO + O₂ + Cl + O₂; Cl and ClO cancel, leaving O₃ + O → 2O₂. Cl is consumed in step 1 and regenerated in step 2, so it is the catalyst; ClO is made in step 1 and used in step 2, so it is the intermediate.
Scoring points (1 point each)(c)Write the rate law consistent with this mechanism.
(1 pt)
Model answerThe slow step 1 determines the rate, and its reactants are not intermediates: rate = k[Cl][O₃].
Scoring points (1 point each)(d)In an experiment with [Cl] held constant (in large excess), the concentration of O₃ was measured as shown. Explain how the data show first-order kinetics in O₃, and calculate the observed rate constant k′.
t (s) [O₃] (M) 0 4.0 × 10−6 40 2.0 × 10−6 80 1.0 × 10−6 120 5.0 × 10−7 (2 pts)
Model answer[O₃] halves every 40 s: a half-life that doesn’t depend on concentration is the mark of first-order kinetics. k′ = ln 2 ÷ t½ = 0.693 ÷ 40 s = 0.0173 s⁻¹.
Scoring points (1 point each)(e)On one energy diagram, how does the catalyzed pathway (through Cl) differ from the uncatalyzed one? Does ΔH for the overall reaction change?
(1 pt)
Model answerThe catalyzed path has two steps (two peaks with a valley for ClO), and its highest peak is lower than the uncatalyzed peak, so the activation energy is smaller. The reactant and product energies are unchanged, so ΔH is the same.
Scoring points (1 point each)(f)Calculate ΔH° for the overall reaction O₃(g) + O(g) → 2O₂(g) from standard enthalpies of formation: ΔH°f (kJ/mol) O₃(g) +142.7, O(g) +249.2. Is the reaction exothermic or endothermic?
(2 pts)
Model answerΔH° = 2 × 0 − (142.7 + 249.2) = −391.9 kJ/mol (O₂ is an element in its standard state). ΔH° < 0, so it is exothermic.
Scoring points (1 point each)ExplanationCommon losses: swapping the catalyst and the intermediate; writing the rate law from the overall coefficients; leaving off the units of k′; and dropping ΔH°f of O(g) (it isn’t zero, since O atoms are not oxygen’s standard state).
Review: U5.11 CatalysisU2.6 Resonance and Formal ChargeU5.8 Reaction Mechanism and Rate LawU5.3 Concentration Changes Over Time (Integrated Rate Laws, Half-Life)U6.8 Enthalpy of Formation
- 2Free response
Methanol can be made from CO and H₂: CO(g) + 2H₂(g) ⇌ CH₃OH(g). At some temperature, a rigid container is filled with CO and H₂ at initial partial pressures of 1.00 atm and 2.00 atm, with no CH₃OH. At equilibrium the total pressure is 2.20 atm (same temperature).
(a)Write the expression for Kp for this reaction.
(1 pt)
Model answerScoring points (1 point each)(b)Find the equilibrium partial pressure of each gas and calculate Kp.
(2 pts)
Model answerLet CO fall by x: CO 1.00 − x, H₂ 2.00 − 2x, CH₃OH x. Total = 3.00 − 2x = 2.20, so x = 0.40 atm. Equilibrium pressures: CO 0.60, H₂ 1.20, CH₃OH 0.40 atm. Kp = 0.40 ÷ (0.60 × 1.20²) = 0.463.
Scoring points (1 point each)(c)At constant temperature, the volume is suddenly halved. Use a comparison of Q and K to explain which way the equilibrium shifts.
(2 pts)
Model answerEvery partial pressure doubles. In Q = P(CH₃OH)/[P(CO)·P(H₂)²] the top is multiplied by 2 and the bottom by 2³ = 8, so Q becomes K/4 < K. There is too little product, so the reaction shifts forward, toward CH₃OH and fewer gas molecules.
Scoring points (1 point each)(d)ΔH°f (kJ/mol): CO(g) −110.5, CH₃OH(g) −201.0. Calculate ΔH°.
(1 pt)
Model answerΔH° = −201.0 − (−110.5) = −90.5 kJ/mol (H₂ is zero).
Scoring points (1 point each)(e)Without calculating, predict the sign of ΔS° and justify it. How does K change if the temperature is raised?
(2 pts)
Model answerΔS° < 0: 3 mol of gas become 1 mol, so there are fewer possible arrangements. The forward reaction is exothermic (ΔH° < 0), so raising the temperature shifts it in reverse and K decreases.
Scoring points (1 point each)(f)ΔS° = −219.2 J/K. Calculate ΔG° at 298 K and decide whether the reaction is thermodynamically favorable at 298 K under standard conditions.
(2 pts)
Model answerΔG° = ΔH° − TΔS° = −90.5 − 298 × (−0.2192) = −25.2 kJ/mol. ΔG° < 0, so it is thermodynamically favorable.
Scoring points (1 point each)ExplanationCommon losses: treating the total pressure as one gas’s pressure; forgetting H₂ changes by 2x; saying only “pressure increases” without comparing Q and K; and not converting ΔS° from J/K to kJ/K.
Review: U7.4 Calculating the Equilibrium ConstantU7.3 Reaction Quotient and Equilibrium ConstantU7.10 Reaction Quotient and Le Châtelier's PrincipleU6.8 Enthalpy of FormationU9.2 Absolute Entropy and Entropy ChangeU9.3 Gibbs Free Energy and Thermodynamic Favorability
- 3Free response
To identify a monoprotic weak acid HA, a student dissolves 0.500 g of the solid acid in 50.0 mL of water and titrates it with 0.100 M NaOH, recording the pH as shown (the equivalence point is at 41.00 mL).
Volume of NaOH added (mL) pH 0.00 2.64 10.00 3.71 20.50 4.19 30.00 4.63 41.00 8.42 45.00 11.62 (a)Calculate the moles of NaOH added at the equivalence point, and use it to find the molar mass of HA.
(2 pts)
Model answern(NaOH) = 0.100 M × 0.0410 L = 4.10 × 10−3 mol. For a monoprotic acid n(HA) = n(NaOH), so M = 0.500 g ÷ 4.10 × 10−3 mol = 122 g/mol.
Scoring points (1 point each)(b)Use the data to determine Ka for HA, and explain your method.
(2 pts)
Model answerAt 20.50 mL, half the equivalence volume, half the HA has become A⁻, so [HA] = [A⁻] and pH = pKa = 4.19. Ka = 10−4.19 = 6.5 × 10−5.
Scoring points (1 point each)(c)Calculate the pH at the equivalence point (25 °C).
(2 pts)
Model answerAt equivalence the solution holds 4.10 × 10−3 mol of A⁻ in 50.0 + 41.0 = 91.0 mL, so [A⁻] = 0.0451 M. Kb = Kw/Ka = 1.5 × 10−10, [OH⁻] = √(Kb·c) = 2.6 × 10−6 M, pOH = 5.58, pH = 8.42.
Scoring points (1 point each)(d)Choose an indicator for this titration, methyl red (pH 4.4–6.2) or phenolphthalein (8.3–10.0), and justify your choice.
(1 pt)
Model answerPhenolphthalein: the equivalence-point pH (about 8.4) falls within its range, while methyl red would change color well before equivalence.
Scoring points (1 point each)(e)After 30.00 mL of NaOH has been added, which is present at higher concentration, HA or A⁻? Explain.
(1 pt)
Model answerA⁻: 30.00 mL is past the half-equivalence point (20.50 mL), so more than half the HA has been converted; equivalently, the pH (4.63) is above the pKa.
Scoring points (1 point each)(f)If the solid acid the student weighed was not completely dry (it contained some water), would the calculated molar mass be too high, too low, or unaffected? Explain.
(2 pts)
Model answerToo high. Part of the 0.500 g is water, so there is less acid and less NaOH is needed (a smaller measured n(HA)), while the calculation still uses 0.500 g. With M = m ÷ n, the smaller denominator makes M too high.
Scoring points (1 point each)ExplanationCommon losses: giving pH 7 at equivalence; using 50.0 mL instead of the total volume for [A⁻]; and in the error analysis, saying only that there is error without giving the direction.
Review: U8.5 Acid-Base TitrationsU4.6 Introduction to TitrationU8.7 pH and pKaU8.3 Weak Acid and Base Equilibria (Ka, Kb)U8.4 Acid-Base Reactions and Buffers
- 4Free response
NH₃ (boiling point −33 °C), PH₃ (−88 °C), and NF₃ (−129 °C) all have trigonal pyramidal molecules.
(a)Draw a complete Lewis diagram for NF₃.
(1 pt)
Model answerN in the center, single-bonded to each of 3 F atoms, with one lone pair on N and three lone pairs on each F: 26 valence electrons, every atom with an octet.
Scoring points (1 point each)(b)Is the F–N–F bond angle in NF₃ larger or smaller than 109.5°? Explain.
(1 pt)
Model answerSmaller (about 102°). N has 4 electron domains (3 bonding, 1 lone pair) in a tetrahedral arrangement; the lone pair repels more strongly than the bonding pairs and squeezes the N–F bonds closer together.
Scoring points (1 point each)(c)PH₃ has a larger molar mass than NH₃ but a much lower boiling point. Explain.
(1 pt)
Model answerNH₃ molecules hydrogen-bond to each other (N is very electronegative, the N–H bond is very polar, and N has a lone pair); P is about as electronegative as H, so PH₃ can’t hydrogen-bond. Hydrogen bonding outweighs PH₃’s slightly larger dispersion forces, so NH₃ boils higher.
Scoring points (1 point each)(d)NF₃ (71 g/mol) is much heavier than NH₃, yet it boils at a lower temperature. Explain.
(1 pt)
Model answerNF₃ has no H bonded to N, O, or F, so its molecules can’t hydrogen-bond and have only weaker dipole–dipole and dispersion forces; NH₃ hydrogen-bonds, so it boils higher than the heavier NF₃.
Scoring points (1 point each)ExplanationCommon losses: thinking NF₃ can hydrogen-bond because it contains F (hydrogen bonding needs H bonded directly to N, O, or F); arguing only that heavier molecules boil higher without comparing types of forces; and giving 109.5° or 120° for the angle.
Review: U2.5 Lewis DiagramsU2.7 VSEPR and Bond HybridizationU3.1 Intermolecular Forces
- 5Free response
A nickel strip in 1.0 M Ni(NO₃)₂ and a copper strip in 1.0 M Cu(NO₃)₂ are connected by a KNO₃ salt bridge and a wire to form a galvanic cell. E°: Cu²⁺ + 2e⁻ → Cu, +0.34 V; Ni²⁺ + 2e⁻ → Ni, −0.26 V.
(a)Calculate E°cell.
(1 pt)
Model answerNi is oxidized (anode) and Cu²⁺ reduced (cathode): E°cell = 0.34 − (−0.26) = 0.60 V.
Scoring points (1 point each)(b)While the cell operates, which way do electrons flow in the wire, and which electrode gains mass?
(1 pt)
Model answerElectrons flow through the wire from the nickel electrode (anode) to the copper electrode (cathode); Cu deposits on the copper electrode, which gains mass.
Scoring points (1 point each)(c)Na₂S is added to the copper half-cell, precipitating CuS and greatly lowering [Cu²⁺]. How does the cell potential E change? Explain.
(1 pt)
Model answerE decreases. Cu²⁺ is a reactant in Ni + Cu²⁺ → Ni²⁺ + Cu, so lowering it raises Q = [Ni²⁺]/[Cu²⁺], and E = E° − (RT/nF) ln Q goes down.
Scoring points (1 point each)(d)The cell delivers an average current of 0.150 A for 2.00 h. What mass of copper deposits? (F = 96,485 C/mol, M(Cu) = 63.55 g/mol)
(1 pt)
Model answerq = 0.150 × 7200 = 1080 C; n(e⁻) = 1080 ÷ 96,485 = 0.01119 mol; n(Cu) = ÷ 2 = 0.00560 mol; m = × 63.55 = 0.356 g.
Scoring points (1 point each)ExplanationCommon losses: computing E°cell as anode − cathode (negative); not converting hours to seconds; and forgetting each Cu²⁺ needs 2 electrons.
Review: U9.7 Galvanic (Voltaic) and Electrolytic CellsU9.8 Cell Potential and Free EnergyU9.9 Cell Potential Under Nonstandard ConditionsU9.10 Electrolysis and Faraday's Law
- 6Free response
A student reacts a strip of magnesium with excess hydrochloric acid, Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g), and collects the hydrogen over water. At 25 °C, 38.2 mL of gas is collected, and the water level inside the tube matches the level outside. Atmospheric pressure is 757 mmHg, and the vapor pressure of water at 25 °C is 24 mmHg.
(a)Calculate the partial pressure of H₂ in the collected gas.
(1 pt)
Model answerWith the levels equal, the gas pressure equals atmospheric pressure. P(H₂) = 757 − 24 = 733 mmHg.
Scoring points (1 point each)(b)Calculate the moles of H₂ produced. (R = 0.08206 L·atm/(mol·K))
(1 pt)
Model answern = PV/RT = (733/760 atm)(0.0382 L) ÷ (0.08206 × 298 K) = 1.51 × 10−3 mol.
Scoring points (1 point each)(c)Calculate the mass of the magnesium strip. (M(Mg) = 24.31 g/mol)
(1 pt)
Model answern(Mg) = n(H₂) = 1.51 × 10−3 mol, so m = 1.51 × 10−3 × 24.31 = 0.0366 g.
Scoring points (1 point each)(d)If the student forgot to subtract the vapor pressure of water, would the calculated mass of magnesium be too high or too low? Explain.
(1 pt)
Model answerToo high. Taking all 757 mmHg as the H₂ pressure overstates n(H₂), and since n(Mg) = n(H₂), the mass of Mg comes out too high.
Scoring points (1 point each)ExplanationCommon losses: using atmospheric pressure as the H₂ pressure; not converting to liters and kelvins; and not converting mmHg to atm.
Review: U3.4 Ideal Gas LawU4.5 StoichiometryU3.5 Kinetic Molecular Theory
- 7Free response
Instant cold packs use ammonium nitrate dissolving in water: NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq), ΔH° = +25.7 kJ/mol, ΔS° = +108.7 J/(mol·K).
(a)Explain at the particle level why ΔS° > 0.
(1 pt)
Model answerThe ions go from fixed, ordered positions in the crystal to being dispersed and moving freely in solution, which greatly increases the number of possible arrangements, so entropy increases.
Scoring points (1 point each)(b)Calculate ΔG° for dissolving at 298 K.
(1 pt)
Model answerΔG° = 25.7 − 298 × 0.1087 = −6.7 kJ/mol.
Scoring points (1 point each)(c)Dissolving is endothermic, so why is it still thermodynamically favorable at 298 K?
(1 pt)
Model answerThe entropy gain makes TΔS° (32.4 kJ/mol) larger than ΔH° (25.7 kJ/mol), so ΔG° = ΔH° − TΔS° < 0: entropy outweighs the unfavorable enthalpy.
Scoring points (1 point each)(d)Assuming ΔH° and ΔS° don’t change with temperature, below what temperature would dissolving become unfavorable under standard conditions?
(1 pt)
Model answerSetting ΔG° = 0: T = ΔH° ÷ ΔS° = 25700 ÷ 108.7 ≈ 236 K. Below this, TΔS° < ΔH° and ΔG° > 0.
Scoring points (1 point each)ExplanationCommon losses: not converting ΔS° units; equating “endothermic” with “not spontaneous”; and mixing kJ and J when finding the temperature.
Review: U9.1 Introduction to EntropyU9.3 Gibbs Free Energy and Thermodynamic FavorabilityU7.14 Free Energy of DissolutionU6.2 Energy Diagrams
Result
| Unit | Correct |
|---|
Composite = Section I correct × 50/60 + Section II points × 50/46, so each section is worth 50%. The AP score is estimated from the conversion table published with the 2018 AP Chemistry practice exam (5: 75 and up, 4: 60–74, 3: 44–59, 2: 30–43, 1: below 30). Cutoffs change from year to year, so treat it as a rough guide.
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