AP Chemistry Mock ExamMock Exam 2
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
A 2.00 g sample contains only MgO and NaCl. Analysis shows that the sample contains 0.480 g of magnesium. What is the approximate mass percent of MgO in the sample? (Mg 24.31, O 16.00)
ExplanationMass of MgO = 0.480 g × 40.31 ÷ 24.31 = 0.796 g, so the mass percent is 0.796 ÷ 2.00 = 39.8%. 24.0% is the percent of Mg alone, 60.2% is the NaCl, and 15.8% counts only the oxygen in MgO.
Review: U1.4 Composition of Mixtures
- 2Multiple choice
The group 14 hydrides CH₄, SiH₄, and GeH₄ boil at −162 °C, −112 °C, and −88 °C. Which best explains the increase?
ExplanationAll three are tetrahedral with identical X–H bonds, so none of them has a net dipole: only dispersion forces act. From C to Ge the central atom has more shells, so the electron cloud is larger and more polarizable, the temporary dipoles are stronger, and the boiling point rises. H bonded to Si or Ge cannot hydrogen bond; the X–H bonds actually get weaker, and boiling does not break bonds within molecules anyway.
Review: U3.1 Intermolecular Forces
- 3Multiple choice
Solutions of NH₄Cl and NaOH are mixed and warmed, and a gas with a sharp odor is released. What is the net ionic equation?
ExplanationNH₄Cl, NaOH, and NaCl are all soluble strong electrolytes present as ions; Na⁺ and Cl⁻ are spectators. The actual change is NH₄⁺ transferring H⁺ to OH⁻, giving NH₃ gas and water. NaCl is soluble and does not precipitate.
Review: U4.2 Net Ionic Equations
- 4Multiple choice
Three equivalent resonance structures can be drawn for the carbonate ion, CO₃²⁻. Which statement about its C–O bonds is correct?
ExplanationThe real ion is an average of the three structures: one π bond is shared over three C–O bonds, so each has a bond order of 4/3. The bonds are equal, shorter than a single bond and longer than a double bond. Resonance structures are not different forms the ion switches between.
Review: U2.5 Lewis Diagrams
- 5Multiple choice
Initial-rate data for the reaction A + B → products are shown (concentrations in M, rates in M/s). What are the rate law and the rate constant, k?
Trial [A] [B] Initial rate 1 0.010 0.010 1.20 × 10−4 2 0.020 0.010 2.40 × 10−4 3 0.010 0.030 1.08 × 10−3 ExplanationTrials 1→2: doubling [A] at constant [B] doubles the rate, so first order in A. Trials 1→3: tripling [B] multiplies the rate by 9, so second order in B. From trial 1, k = 1.2 × 10⁻⁴ ÷ (0.010 × 0.010²) = 120 M⁻²·s⁻¹. The overall order is 3, so the units are M⁻²·s⁻¹.
Review: U5.2 Introduction to Rate Law
- 6Multiple choice
A sample of an ideal gas occupies 2.00 L at 300. K and 1.00 atm. It is compressed to 0.500 L and heated to 450. K. What is the new pressure?
ExplanationP₂ = P₁ × (V₁ ÷ V₂) × (T₂ ÷ T₁) = 1.00 × (2.00 ÷ 0.500) × (450 ÷ 300) = 6.00 atm. A quarter of the volume makes the pressure 4 times larger, and 1.5 times the temperature multiplies it by 1.5. 4.00 atm ignores the temperature; 2.67 atm inverts the temperature ratio.
Review: U3.4 Ideal Gas Law
- 7Multiple choice
What is the pH of 0.020 M Ba(OH)₂ at 25 °C?
ExplanationBa(OH)₂ is a strong base that releases two OH⁻ per formula unit: [OH⁻] = 0.040 M, pOH = −log 0.040 = 1.40, and pH = 14.00 − 1.40 = 12.60. 12.30 forgets the factor of 2; 1.40 is the pOH.
- 8Multiple choice
What is the ground-state electron configuration of Fe³⁺? (Fe has atomic number 26.)
ExplanationFe is [Ar]4s²3d⁶. Transition metals lose their outer 4s electrons before any 3d electrons, so Fe³⁺ loses both 4s electrons and one 3d electron: [Ar]3d⁵. [Ar]4s²3d³ removes the 3d electrons first, and [Ar]3d⁶ is Fe²⁺.
- 9Multiple choice
In an insulated container, 50.0 g of water at 80.0 °C is mixed with 100.0 g of water at 20.0 °C. What is the final temperature at thermal equilibrium?
ExplanationHeat lost by the hot water = heat gained by the cold water: 50.0 × c × (80.0 − T) = 100.0 × c × (T − 20.0), so T = 40.0 °C. There is twice as much cold water, so the final temperature is closer to 20 °C; 50.0 °C is a simple average that ignores the masses.
- 10Multiple choice
50.0 mL of 0.200 M CaCl₂ is mixed with 50.0 mL of 0.100 M NaCl, and the volumes are additive. Find [Cl⁻] in the mixture.
Explanationmol Cl⁻ = 0.0500 L × 0.200 × 2 + 0.0500 L × 0.100 = 0.0200 + 0.00500 = 0.0250 mol in 0.100 L, so [Cl⁻] = 0.250 M. 0.150 M forgets that each CaCl₂ gives two Cl⁻; 0.500 M ignores the larger volume after mixing.
Review: U3.7 Solutions and Mixtures
- 11Multiple choice
The equilibrium NH₄HS(s) ⇌ NH₃(g) + H₂S(g) is established in a closed container. More NH₄HS(s) is added at constant temperature. What happens to the partial pressure of NH₃?
ExplanationA pure solid has a constant “concentration” and is left out of the expression: Kp = P(NH₃)·P(H₂S). Adding solid doesn’t change Q, so Q still equals K and nothing shifts (the solid’s own volume is negligible). The equilibrium is dynamic; both reactions continue.
Review: U7.4 Calculating the Equilibrium ConstantU7.9 Introduction to Le Châtelier's Principle
- 12Multiple choice
What is the molecular geometry of ClF₃?
ExplanationCl has 7 valence electrons: 3 form bonds to F and 4 remain as 2 lone pairs, giving 5 electron domains in a trigonal bipyramidal arrangement. The lone pairs take equatorial positions (least repulsion), leaving the three F atoms in a T shape.
- 13Multiple choice
28.0 g of N₂ and 9.00 g of H₂ react by N₂ + 3H₂ → 2NH₃. What is the greatest mass of NH₃ that can form? (N₂ 28.02, H₂ 2.016, NH₃ 17.03 g/mol)
Explanationmol N₂ = 0.999 and mol H₂ = 4.46. The N₂ needs 3 × 0.999 ≈ 3.00 mol of H₂, so H₂ is in excess and N₂ limits: NH₃ = 2 × 0.999 mol × 17.03 = 34.0 g. 50.7 g is based on the H₂; 37.0 g adds the reactant masses, but the excess H₂ does not all react.
Review: U4.5 Stoichiometry
- 14Multiple choice
What is the approximate pH of 0.20 M formic acid (HCOOH, Ka = 1.8 × 10⁻⁴) at 25 °C?
ExplanationA weak acid ionizes only partly: [H⁺] ≈ √(Ka·c) = √(1.8 × 10⁻⁴ × 0.20) = 6.0 × 10⁻³ M (3% of 0.20, so the approximation holds), and pH = 2.22. 0.70 treats it as a strong acid, 3.74 is the pKa, and 4.44 forgets the square root.
- 15Multiple choice
Boiling points: HF 20 °C, HCl −85 °C, HBr −67 °C. Which statement explains both why HF has the highest boiling point and why HBr boils higher than HCl?
ExplanationF is small and very electronegative, so the H on one HF molecule hydrogen bonds to a lone pair on F in the next: HF is unusually high. HCl and HBr cannot hydrogen bond; Br has more electrons and a larger cloud, so dispersion forces win (HBr is actually less polar than HCl). Boiling does not break bonds within molecules.
Review: U3.1 Intermolecular Forces
- 16Multiple choice
For the reaction A → products, the concentration of A changes with time as shown. Which statement is correct?
Explanation[A] vs. time is a straight line: the same amount is used each second (slope −2.0 × 10⁻³ M/s) regardless of concentration, so the reaction is zero order. Falling from 0.50 to 0.10 M takes 0.40 ÷ 0.0020 = 200 s. A zero-order half-life, [A]₀/(2k), changes with the starting concentration; only first-order half-lives are constant.
Review: U5.3 Concentration Changes Over Time (Integrated Rate Laws, Half-Life)
- 17Multiple choice
Naturally occurring boron consists of ¹⁰B (10.01) and ¹¹B (11.01). The average atomic mass of boron is 10.81. What is the approximate abundance of ¹¹B?
ExplanationLet x be the fraction of ¹¹B: 10.01(1 − x) + 11.01x = 10.81, so x = 0.80, or 80%. The average is closer to 11.01, so ¹¹B is the more common isotope. 20% swaps the isotopes; 50% assumes equal amounts.
- 18Multiple choice
In a coffee-cup calorimeter, excess zinc powder is added to 50.0 mL of 0.200 M CuSO₄, and the temperature rises by 10.5 °C. Assume the solution has a mass of 50.0 g and a specific heat of 4.18 J/(g·°C), and that the calorimeter absorbs no heat. What is ΔH for Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), approximately?
ExplanationThe solution absorbs q = 50.0 × 4.18 × 10.5 = 2195 J. Cu²⁺ is 0.0500 L × 0.200 M = 0.0100 mol (zinc is in excess, so it all reacts). ΔH = −2.19 kJ ÷ 0.0100 mol ≈ −219 kJ/mol. The temperature rose, so the reaction is exothermic and ΔH is negative; −2.19 forgets to divide by the moles.
Review: U6.4 Heat Capacity and CalorimetryU6.6 Introduction to Enthalpy of Reaction
- 19Multiple choice
What is the energy of one photon of light with a wavelength of 500. nm? (h = 6.626 × 10⁻³⁴ J·s, c = 2.998 × 10⁸ m/s)
ExplanationE = hc ÷ λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (500. × 10⁻⁹ m) = 3.97 × 10⁻¹⁹ J. The wavelength must be in meters: 3.97 × 10⁻²⁸ J divides by 500 directly, and 1.99 × 10⁻²⁵ J is just hc.
Review: U3.12 Photoelectric Effect
- 20Multiple choice
At some temperature, PCl₅ is placed in a container at an initial concentration of 0.200 M and reacts by PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium [Cl₂] = 0.0500 M. What is K at this temperature?
ExplanationICE table: PCl₅ falls by 0.0500, so at equilibrium [PCl₅] = 0.150 M and [PCl₃] = [Cl₂] = 0.0500 M. K = (0.0500 × 0.0500) ÷ 0.150 = 0.0167. 0.0125 uses the initial 0.200; 0.333 uses only one product; 60.0 is the reciprocal.
Review: U7.7 Calculating Equilibrium Concentrations (ICE Tables)U7.4 Calculating the Equilibrium Constant
- 21Multiple choice
Propyne has the structure H₃C–C≡C–H. How many σ bonds and how many π bonds are in one molecule of propyne?
ExplanationEvery single bond is one σ bond, and a triple bond is one σ bond plus two π bonds. σ bonds: four C–H, one C–C, and one in the triple bond, six in all; the only π bonds are the two in the triple bond.
- 22Multiple choice
A 0.10 M solution of a weak acid is diluted with water to 0.010 M. Which statement is correct?
ExplanationDiluting lowers [H⁺], so the pH rises. For a weak acid [H⁺] ≈ √(Ka·c), so a tenfold dilution cuts [H⁺] only by about √10 and the pH rises by about 0.5. Percent ionization ≈ √(Ka/c) increases (adding water makes Q < Ka, shifting toward ionization). A rise of exactly 1 is what a strong acid does.
- 23Multiple choice
A NaOH solution is standardized with potassium hydrogen phthalate (KHP, a monoprotic acid, 204.22 g/mol): 0.5105 g of KHP requires exactly 25.00 mL of the NaOH solution. What is the concentration of the NaOH?
Explanationmol KHP = 0.5105 ÷ 204.22 = 2.500 × 10⁻³ mol. KHP is monoprotic and reacts 1 : 1 with NaOH, so [NaOH] = 2.500 × 10⁻³ mol ÷ 0.02500 L = 0.1000 M. 0.02042 divides the mass by the milliliters.
Review: U4.6 Introduction to Titration
- 24Multiple choice
A container holds 0.20 mol of N₂ and 0.30 mol of O₂ at a total pressure of 2.0 atm. What is the partial pressure of O₂?
ExplanationPartial pressure = mole fraction × total pressure. The mole fraction of O₂ is 0.30 ÷ 0.50 = 0.60, so P(O₂) = 0.60 × 2.0 = 1.2 atm. 0.80 atm is the N₂ pressure; 1.0 atm splits the total evenly.
Review: U3.4 Ideal Gas Law
- 25Multiple choice
For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ/mol and ΔS° = +161 J/(mol·K). Assuming these do not change with temperature, at what temperatures is the reaction thermodynamically favorable?
ExplanationΔG° = ΔH° − TΔS°. With ΔH° > 0 and ΔS° > 0, ΔG° is negative only when TΔS° exceeds ΔH°: T > 178,000 ÷ 161 ≈ 1106 K. At 298 K, ΔG° = 178 − 298 × 0.161 ≈ +130 kJ/mol, unfavorable. Convert ΔS° to kJ. Endothermic reactions can still be favorable at high temperature.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 26Multiple choice
The first four ionization energies (kJ/mol) of a period 4 element X are shown in the table. What is the most likely formula of the compound X forms with chlorine?
IE₁ IE₂ IE₃ IE₄ 590 1145 4912 6491 ExplanationIE₂ is less than twice IE₁, but IE₃ is more than four times IE₂, so X has two easily removed valence electrons; the third would come from the filled n = 3 shell. X forms X²⁺, which combines with two Cl⁻ as XCl₂.
Review: U1.7 Periodic TrendsU1.8 Valence Electrons and Ionic Compounds
- 27Multiple choice
For most reactions the rate increases noticeably with temperature. What is the main reason?
ExplanationAt a higher temperature the energy distribution spreads toward higher energies, so the fraction of molecules with energy ≥ Eₐ rises sharply and more collisions succeed. Collisions also happen a bit more often, but that is a minor effect. Eₐ and ΔH barely change with temperature; lowering Eₐ is what a catalyst does.
Review: U5.5 Collision Model
- 28Multiple choice
The graph shows the vapor pressure of two pure liquids, X and Y, as a function of temperature. Which liquid has the stronger intermolecular forces, and what is its normal boiling point (at 1 atm), approximately?
ExplanationAt every temperature Y has the lower vapor pressure, so its molecules escape into the gas phase less easily: Y has the stronger intermolecular forces. A liquid boils when its vapor pressure equals the outside pressure, and Y’s curve reaches 1 atm at about 90 °C. 40 °C is X’s normal boiling point.
Review: U3.3 Solids, Liquids, and GasesU3.1 Intermolecular Forces
- 29Multiple choice
(Same graph as the previous question.) On a high mountain the air pressure is only 0.50 atm. At about what temperature does liquid Y boil there?
ExplanationA liquid boils where its vapor pressure equals the outside pressure. On Y’s curve, 0.50 atm corresponds to about 71 °C, so a lower outside pressure lowers the boiling point. 22 °C reads X’s curve; 45 °C just halves the normal boiling point.
Review: U3.3 Solids, Liquids, and Gases
- 30Multiple choice
Given: S(s) + O₂(g) → SO₂(g), ΔH = −296.8 kJ/mol; 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197.8 kJ/mol. What is ΔH for S(s) + 3/2 O₂(g) → SO₃(g)?
ExplanationTarget = the first reaction + ½ × the second: ΔH = −296.8 + ½ × (−197.8) = −395.7 kJ/mol. −494.6 forgets to halve the second reaction.
Review: U6.9 Hess's Law
- 31Multiple choice
Which 0.10 M solution has the highest pH at 25 °C? (Ka: HNO₂ 4.6 × 10⁻⁴, CH₃COOH 1.8 × 10⁻⁵; Kb: NH₃ 1.8 × 10⁻⁵)
ExplanationThe anion of a weak acid makes a solution basic, and the weaker the acid, the larger the anion’s Kb = Kw/Ka. CH₃COOH is weaker than HNO₂, so CH₃COONa (pH ≈ 8.9) is more basic than NaNO₂ (pH ≈ 8.2). KNO₃ comes from a strong acid and a strong base, so it is neutral; NH₄⁺ makes NH₄NO₃ acidic (pH ≈ 5.1).
Review: U8.3 Weak Acid and Base Equilibria (Ka, Kb)U8.1 Introduction to Acids and Bases
- 32Multiple choice
Which of these molecules is polar (has a nonzero dipole moment)?
ExplanationNF₃ has a lone pair on N, so it is trigonal pyramidal and its three N–F bond dipoles do not cancel. BF₃ and SO₃ are trigonal planar and XeF₂ is linear; their bond dipoles are arranged symmetrically and cancel.
- 33Multiple choice
The equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) gives a blood-red solution. A small amount of solid Fe(NO₃)₃ is added (the volume barely changes). Which statement is correct?
ExplanationAdding Fe³⁺ makes the denominator of Q = [FeSCN²⁺] ÷ ([Fe³⁺][SCN⁻]) larger, so Q < K and the reaction runs forward, making more red FeSCN²⁺ and using up SCN⁻. K changes only with temperature. NO₃⁻ is a spectator, but the Fe³⁺ added with it takes part.
Review: U7.10 Reaction Quotient and Le Châtelier's Principle
- 34Multiple choice
In acidic solution: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. For each mole of Cr₂O₇²⁻ that reacts, how many moles of electrons does it gain, and which species is the oxidizing agent?
ExplanationCr goes from +6 to +3, gaining 3 electrons per Cr, and Cr₂O₇²⁻ has two Cr, so it gains 6 electrons; the six Fe²⁺ each lose one, which balances. Cr₂O₇²⁻ gains electrons (is reduced), so it is the oxidizing agent. 3 mol counts only one Cr; H⁺ does not change oxidation number.
- 35Multiple choice
A gas in a sealed, rigid container is heated. According to kinetic molecular theory, why does the pressure increase?
ExplanationTemperature measures the average kinetic energy. Hotter molecules move faster, so they hit the walls more often and transfer more momentum per hit, raising the pressure. In a sealed rigid container the number of molecules and the volume stay the same, and the molecules themselves do not get bigger.
Review: U3.5 Kinetic Molecular Theory
- 36Multiple choice
What are the molecularity and the rate law of the elementary step NO(g) + O₃(g) → NO₂(g) + O₂(g)?
ExplanationAn elementary step is a single collision, so its coefficients give the orders: one NO hits one O₃, a bimolecular step with rate = k[NO][O₃]. “Must be found by experiment” applies to overall reactions, not elementary steps, and rate laws contain reactants, not the products NO₂ and O₂.
Review: U5.4 Elementary Reactions
- 37Multiple choice
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is at equilibrium in a rigid container. Argon is added at constant temperature and volume, raising the total pressure. What happens to the equilibrium?
ExplanationWith the volume fixed, the amounts and concentrations of N₂, H₂, and NH₃ are unchanged, so their partial pressures and Q are too: Q still equals K. “Higher pressure favors fewer gas molecules” refers to compressing the container, which raises every partial pressure.
- 38Multiple choice
Which list puts these species in order of increasing radius?
ExplanationAll four species have 18 electrons in the same arrangement. The more protons, the more strongly the same electrons are pulled in, so the smaller the radius: Ca (20 protons) < K (19) < Cl (17) < S (16).
Review: U1.7 Periodic Trends
- 39Multiple choice
In cells, glucose + phosphate → glucose-6-phosphate + H₂O has ΔG° = +13.8 kJ/mol, and ATP + H₂O → ADP + phosphate has ΔG° = −30.5 kJ/mol. When the two are coupled, which is correct about the overall reaction glucose + ATP → glucose-6-phosphate + ADP?
ExplanationAdding the reactions cancels phosphate and H₂O and gives the overall reaction, so the ΔG° values add: +13.8 + (−30.5) = −16.7 kJ/mol < 0. A favorable reaction drives an unfavorable one this way. ATP is consumed, so it is not a catalyst.
Review: U9.6 Coupled Reactions
- 40Multiple choice
SiO₂ (quartz) melts at about 1700 °C, while CO₂ sublimes at −78 °C. Which is the best explanation?
ExplanationIn quartz every Si is bonded to four O and every O to two Si, so the crystal is one giant network with no separate SiO₂ molecules; melting it means breaking many covalent bonds. Solid CO₂ is made of separate molecules held by weak dispersion forces. Bond strength alone can’t explain it (C=O is actually stronger than Si–O).
Review: U3.2 Properties of Solids
- 41Multiple choice
A solid sample is heated at a constant rate. The graph shows its temperature against the heat added. Which statement is correct?
ExplanationThe melting plateau is at 10 °C and about 5 kJ long; the boiling plateau is at 80 °C and about 18 kJ long, so the same sample absorbs more heat to vaporize (the particles must be separated completely). On a plateau the temperature, and so the average kinetic energy, stays constant; the energy goes into overcoming intermolecular forces, raising potential energy. No covalent bonds break.
Review: U6.5 Energy of Phase ChangesU3.3 Solids, Liquids, and Gases
- 42Multiple choice
25.0 mL of 0.100 M weak acid HA (Ka = 1.4 × 10⁻⁴) is titrated with 0.100 M NaOH, giving the curve shown. At which point is the pH determined mainly by the reaction of A⁻ with water?
ExplanationR is the equivalence point (25.0 mL): all the HA has become A⁻, so the solution is mainly Na⁺ and A⁻, and A⁻ + H₂O ⇌ HA + OH⁻ gives pH ≈ 8.3 > 7. At P there is only HA; Q (12.5 mL) is a buffer with [HA] = [A⁻] and pH ≈ pKa; at S the excess OH⁻ from NaOH sets the pH.
Review: U8.5 Acid-Base Titrations
- 43Multiple choice
Of CaO, BaO, KF, and CsI, which has the greatest lattice energy (strongest attraction between its ions)?
ExplanationBy Coulomb’s law, higher charges and shorter distances give stronger attraction. CaO and BaO are +2/−2, far stronger than +1/−1 KF and CsI, and Ca²⁺ has two fewer shells than Ba²⁺, so its ions sit much closer together. CsI, with low charges and large ions, has the smallest lattice energy.
Review: U2.3 Structure of Ionic Solids
- 44Multiple choice
What is the total amount of ions in 250. mL of 0.200 M Al₂(SO₄)₃ solution?
Explanationmol Al₂(SO₄)₃ = 0.250 L × 0.200 M = 0.0500 mol. Each formula unit gives 2 Al³⁺ and 3 SO₄²⁻, five ions in all: 0.0500 × 5 = 0.250 mol. 0.150 mol counts only sulfate, and 0.100 mol only Al³⁺.
Review: U3.7 Solutions and MixturesU3.8 Representations of Solutions
- 45Multiple choice
0.100 mol of a weak acid HA (pKa = 4.20) and 0.050 mol of NaOH are dissolved in water to make 1.0 L of solution. What is the approximate pH?
ExplanationThe NaOH reacts completely: HA + OH⁻ → A⁻ + H₂O, leaving 0.050 mol HA and 0.050 mol A⁻, a buffer. pH = pKa + log(0.050/0.050) = 4.20. 3.90 treats all 0.100 mol as remaining HA; 2.60 is the pH of the acid before any NaOH is added.
Review: U8.9 Henderson-Hasselbalch EquationU8.4 Acid-Base Reactions and Buffers
- 46Multiple choice
For NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), which pair is a conjugate acid–base pair?
ExplanationA conjugate pair differs by exactly one H⁺. H₂O gives up H⁺ to become OH⁻, so H₂O/OH⁻ is a pair (the other is NH₄⁺/NH₃). NH₄⁺ and OH⁻ come from different reactants.
- 47Multiple choice
Early sulfuric acid plants used NO to speed up 2SO₂ + O₂ → 2SO₃. A proposed mechanism is: step 1 (slow) 2NO + O₂ → 2NO₂; step 2 (fast) NO₂ + SO₂ → NO + SO₃ (occurs twice). Which statement is correct?
ExplanationNO is used up in step 1 and regenerated in step 2, so it is the catalyst; NO₂ is made and then consumed, so it is an intermediate. The slow step sets the rate, and its reactants (NO and O₂) are not intermediates, so rate = k[NO]²[O₂]. Writing [SO₂]²[O₂] from the overall coefficients is a common mistake.
- 48Multiple choice
A galvanic cell is built from Mg/Mg²⁺ (E° = −2.37 V) and Pb/Pb²⁺ (E° = −0.13 V) half-cells, with KNO₃ in the salt bridge. Which is correct while the cell operates?
ExplanationMg, with the lower E°, is oxidized: it is the anode. Pb²⁺ is reduced at the Pb electrode, the cathode. E°cell = −0.13 − (−2.37) = 2.24 V. Electrons travel through the wire from Mg to Pb; Mg²⁺ builds up in the Mg half-cell, so NO₃⁻ anions move there to keep it neutral. Electrons never pass through the salt bridge; only ions move in it.
Review: U9.7 Galvanic (Voltaic) and Electrolytic CellsU9.8 Cell Potential and Free Energy
- 49Multiple choice
For the Mg–Pb cell in the previous question (Mg + Pb²⁺ → Mg²⁺ + Pb), if [Mg²⁺] is increased while [Pb²⁺] stays the same, how does the cell potential change?
ExplanationMg²⁺ is a product, so raising its concentration raises Q and lowers the driving force: from E = E° − (RT/nF) ln Q, a larger ln Q gives a smaller E. E° is only the standard-state value; the actual potential changes with concentration.
- 50Multiple choice
A colored solution has an absorbance of 0.80 in a cuvette with a 1.00 cm path length. It is diluted with water to twice its volume and measured at the same wavelength in a 2.00 cm cuvette. What is the absorbance, approximately?
ExplanationBy A = εbc, halving c and doubling b leave the product bc unchanged, so the absorbance stays about 0.80. 0.40 counts only the dilution, and 1.60 only the longer path.
Review: U3.13 Beer-Lambert Law
- 51Multiple choice
Compare the 1s peaks in the photoelectron spectra of Na and Mg atoms. Which statement is correct?
ExplanationMg has 12 protons and Na has 11, so the Mg nucleus holds its 1s electrons more tightly and more energy is needed to remove them: the peak sits at higher binding energy. Peak height reflects the number of electrons, and both atoms have two 1s electrons, so the peaks are equally tall.
- 52Multiple choice
Of HF, HCl, HBr, and HI, which has the greatest tendency to donate H⁺ (is the strongest acid), and why?
ExplanationDown the group the halogen gets larger, so the H–X bond gets longer and weaker and H⁺ leaves more easily: HF < HCl < HBr < HI. HF has the most polar bond, but it is also very strong, so HF only partly ionizes and is a weak acid. (In dilute water HCl, HBr, and HI all ionize almost completely; their differences show up in less basic solvents.)
- 53Multiple choice
At 25 °C, Ksp for BaSO₄ is 1.1 × 10⁻¹⁰ and Ksp for Ag₂CO₃ is 8.5 × 10⁻¹². Which statement about their molar solubilities in water is correct?
ExplanationBaSO₄: Ksp = s², so s = 1.05 × 10⁻⁵ M. Ag₂CO₃: Ksp = (2s)²·s = 4s³, so s = 1.29 × 10⁻⁴ M, more than ten times larger. When the formulas differ in type, compare solubilities, not Ksp values.
- 54Multiple choice
Pure silver is soft. Sterling silver is about 92.5% silver and 7.5% copper by mass; copper atoms (radius about 128 pm) occupy some of the lattice sites of silver atoms (about 144 pm). Which best explains why sterling silver is harder than pure silver?
ExplanationCu atoms replace some Ag atoms: a substitutional alloy (the radii are fairly close, and Cu is too big to fit in the gaps). In a pure metal all atoms are the same size, so whole layers slide easily under stress; the differently sized Cu atoms make the layers uneven and hinder sliding, so the alloy is harder. Both are metals held by metallic bonding, and the delocalized electrons are not reduced.
- 55Multiple choice
Use average bond enthalpies to estimate ΔH for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g). Bond enthalpies (kJ/mol): C–H 413, Cl–Cl 242, C–Cl 328, H–Cl 431.
ExplanationOnly the bonds that change matter: breaking one C–H and one Cl–Cl absorbs 413 + 242 = 655 kJ; forming one C–Cl and one H–Cl releases 328 + 431 = 759 kJ. ΔH = 655 − 759 = −104 kJ/mol. +104 subtracts the wrong way (formed − broken); the other three C–H bonds are unchanged.
Review: U6.7 Bond Enthalpies
- 56Multiple choice
Which process breaks covalent bonds within molecules?
ExplanationElectrolysis makes new substances, H₂ and O₂, so O–H covalent bonds must break: a chemical change. Boiling and melting only change how the molecules are arranged (overcoming hydrogen bonds and other intermolecular forces), and dissolved sucrose molecules stay intact, surrounded by water.
Review: U4.4 Physical and Chemical ChangesU4.1 Introduction for Reactions
- 57Multiple choice
A first-order reaction has k = 0.0231 min⁻¹. What percentage of the reactant remains after 90.0 min?
Explanationt½ = 0.693 ÷ k = 0.693 ÷ 0.0231 = 30.0 min, so 90.0 min is three half-lives: 100% → 50% → 25% → 12.5%. Equivalently, ln([A]/[A]₀) = −kt = −2.08, so [A]/[A]₀ = 0.125.
Review: U5.3 Concentration Changes Over Time (Integrated Rate Laws, Half-Life)
- 58Multiple choice
25.0 mL of 0.10 M CH₃COOH is titrated with 0.10 M NaOH. Which indicator is the best choice? (Its color-change pH range is in parentheses.)
ExplanationAt equivalence the solution is 0.050 M CH₃COONa, and CH₃COO⁻ makes it basic: pH ≈ 8.7. The indicator should change within the steep jump around that pH, so phenolphthalein is best. Methyl orange and methyl red change in the buffer region, and alizarin yellow only once NaOH is clearly in excess.
Review: U8.5 Acid-Base Titrations
- 59Multiple choice
Equal volumes of 0.010 M CaCl₂ and 0.010 M Na₂SO₄ are mixed. Ksp for CaSO₄ is 4.9 × 10⁻⁵. Will a precipitate form?
ExplanationMixing equal volumes halves each concentration: [Ca²⁺] = [SO₄²⁻] = 0.0050 M, so Q = 0.0050 × 0.0050 = 2.5 × 10−5, less than Ksp: no precipitate. 1.0 × 10⁻⁴ forgets the dilution on mixing. “Slightly soluble” is only qualitative; compare Q with Ksp.
- 60Multiple choice
Molten Al₂O₃ (dissolved in cryolite) is electrolyzed with a current of 10.0 A for 965 s. What mass of aluminum can form, in theory? (F = 96,485 C/mol, Al 26.98)
Explanationq = It = 10.0 × 965 = 9650 C, so mol e⁻ = 9650 ÷ 96,485 = 0.100 mol. Al³⁺ + 3e⁻ → Al gives 0.100 ÷ 3 = 0.0333 mol Al, or 0.0333 × 26.98 = 0.899 g. 2.70 g forgets that each Al needs 3 electrons; 1.35 g uses 2.
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
A car airbag is filled with nitrogen produced when sodium azide decomposes during a collision: 2NaN₃(s) → 2Na(s) + 3N₂(g). Answer the following questions.
(a)In the azide ion, N₃⁻, the three N atoms form a chain. Draw a valid Lewis diagram for it, showing all valence electrons and formal charges, and give the N–N–N bond angle.
(2 pts)
Model answer3 × 5 + 1 = 16 valence electrons. [N=N=N]⁻: two N=N double bonds, two lone pairs on each end N, none on the central N. Formal charges: −1 on each end and +1 in the center, totaling −1. The central N has two electron domains (two double bonds), so the ion is linear: 180°. (A resonance form N≡N–N is also acceptable.)
Scoring points (1 point each)(b)65.0 g of NaN₃ (molar mass 65.01 g/mol) decomposes completely. Calculate (i) the number of moles of N₂ produced and (ii) the volume of this N₂ at 25.0 °C and 1.05 atm.
(2 pts)
Model answer(i) mol NaN₃ = 65.0 ÷ 65.01 = 1.000 mol; mol N₂ = 1.000 × 3/2 = 1.500 mol. (ii) V = nRT ÷ P = 1.500 × 0.08206 × 298.15 ÷ 1.05 = 34.9 L.
Scoring points (1 point each)(c)The sodium metal produced is very reactive, so the airbag also contains KNO₃, which reacts by 10Na + 2KNO₃ → K₂O + 5Na₂O + N₂. What is the smallest mass of KNO₃ (101.1 g/mol) that would use up all the Na formed from the 65.0 g of NaN₃ in part (b)?
(1 pt)
Model answermol Na = mol NaN₃ = 1.000 mol; mol KNO₃ = 1.000 × 2/10 = 0.2000 mol; mass = 0.2000 × 101.1 = 20.2 g.
Scoring points (1 point each)(d)The standard enthalpy of formation of NaN₃(s) is +21.7 kJ/mol. Calculate ΔH° for 2NaN₃(s) → 2Na(s) + 3N₂(g), and state whether the reaction is endothermic or exothermic.
(2 pts)
Model answerNa(s) and N₂(g) are elements in their standard states, with ΔH°f = 0. ΔH° = 0 − 2 × (+21.7) = −43.4 kJ/mol of reaction. ΔH° < 0, so it is exothermic.
Scoring points (1 point each)(e)Right after inflation the N₂ in the bag is hot. It then cools to room temperature while the pressure inside stays about the same (the bag can shrink). How does the volume of the bag change? Explain at the particle level.
(2 pts)
Model answerThe volume decreases. As the gas cools, the N₂ molecules’ average kinetic energy falls, so they move more slowly and hit the walls less often and less hard. To keep the pressure the same, the molecules must hit the walls more often in a smaller space, so the volume shrinks (V is proportional to T).
Scoring points (1 point each)(f)At room temperature N₂ is a gas but Na is a solid. Explain in terms of the forces between particles.
(1 pt)
Model answerN₂ molecules are nonpolar and held together only by weak dispersion forces, so they separate easily. Na is a metal: Na⁺ cations are surrounded by a sea of delocalized valence electrons, and this metallic bonding holds all the atoms together, so much more energy is needed to melt it.
Scoring points (1 point each)ExplanationCommon losses: forgetting the extra electron from the azide’s charge; using °C in PV = nRT; inverting the Na : KNO₃ ratio; dropping the coefficient in ΔH°; and explaining the volume change as “things shrink when cold” without average kinetic energy and collisions.
Review: U4.5 StoichiometryU3.4 Ideal Gas LawU2.6 Resonance and Formal ChargeU2.7 VSEPR and Bond HybridizationU6.8 Enthalpy of FormationU3.5 Kinetic Molecular TheoryU3.2 Properties of Solids
- 2Free response
Crystal violet (CV⁺) is a purple dye that reacts in basic solution by CV⁺(aq) + OH⁻(aq) → CVOH(aq); the product is colorless. A student mixes crystal violet with NaOH so that [OH⁻] = 0.050 M (a large excess, so [OH⁻] stays nearly constant) and measures the absorbance every 60 s at 590 nm, where CV⁺ absorbs most strongly and nothing else absorbs, using a 1.00 cm cuvette. The data are shown.
Time (s) Absorbance, A ln A 0 0.870 −0.139 60 0.645 −0.439 120 0.477 −0.740 180 0.354 −1.038 240 0.262 −1.339 300 0.194 −1.640 (a)(i) Explain why the absorbance can be used to follow the concentration of CV⁺. (ii) At 590 nm, εb for CV⁺ is 8.7 × 10⁴ M⁻¹. Calculate [CV⁺] when the absorbance is 0.435.
(2 pts)
Model answer(i) By the Beer–Lambert law, A = εbc. At a fixed wavelength in the same cuvette, ε and b are constant, so the absorbance is proportional to [CV⁺]; only CV⁺ absorbs at 590 nm, so the absorbance tracks CV⁺ alone. (ii) c = A ÷ (εb) = 0.435 ÷ (8.7 × 10⁴) = 5.0 × 10−6 M.
Scoring points (1 point each)(b)From the data, find the reaction order in CV⁺, the observed rate constant k′ (with units), and the half-life of CV⁺ in this experiment.
(3 pts)
Model answerln A falls by the same amount each 60 s (about 0.30), so ln A vs. t is linear and the reaction is first order in CV⁺ (equivalently, A is multiplied by about 0.74 every 60 s). Slope = (−1.640 − (−0.139)) ÷ 300 s = −0.0050 s⁻¹, so k′ = −slope = 5.0 × 10−3 s⁻¹. For a first-order reaction t½ = 0.693 ÷ k′ ≈ 139 s.
Scoring points (1 point each)(c)In a second run with [OH⁻] = 0.100 M and everything else the same, k′ = 1.0 × 10⁻² s⁻¹. Write the rate law for the reaction, and calculate the rate constant k, with units.
(2 pts)
Model answerDoubling [OH⁻] doubles k′, so the reaction is first order in OH⁻: rate = k[CV⁺][OH⁻], with k′ = k[OH⁻]. k = 5.0 × 10−3 ÷ 0.050 = 0.10 M⁻¹·s⁻¹.
Scoring points (1 point each)(d)Someone proposes that the reaction occurs in a single step: one CV⁺ colliding directly with one OH⁻. Is this mechanism consistent with the rate law in part (c)? Explain.
(1 pt)
Model answerYes. That step is a bimolecular elementary reaction, so its rate law follows from its coefficients: rate = k[CV⁺][OH⁻], which matches the experimental rate law. (Agreement doesn’t prove the mechanism.)
Scoring points (1 point each)(e)If the first run were repeated at a higher temperature, would k′ be larger, smaller, or the same? Explain using the collision model.
(1 pt)
Model answerLarger. At a higher temperature the average kinetic energy is greater, so a larger fraction of collisions have energy at or above the activation energy (and collisions are a bit more frequent); more collisions succeed per second, so the rate constant increases.
Scoring points (1 point each)(f)The student later finds that the cuvette was scratched, so every reading was 0.020 higher than the true absorbance. Would the k′ calculated from these readings be too large, too small, or unaffected? Explain.
(1 pt)
Model answerToo small. Adding the same amount to every reading pulls the ratios between readings closer to 1: the true values fall from 0.870 to 0.194 by 300 s, but the readings fall only from 0.890 to 0.214. The fractional drop is smaller, the ln A slope is flatter (about −0.0047 s⁻¹), and the calculated k′ is too small.
Scoring points (1 point each)ExplanationCommon losses: calling it first order just because A decreases, without the linear ln A plot or a constant ratio; giving k′ units of M/s; treating k′ as k and not dividing by [OH⁻]; and in (f), assuming a constant offset can’t change the slope (it doesn’t for A vs. t, but it does for ln A vs. t).
Review: U5.3 Concentration Changes Over Time (Integrated Rate Laws, Half-Life)U3.13 Beer-Lambert LawU5.2 Introduction to Rate LawU5.4 Elementary ReactionsU5.5 Collision Model
- 3Free response
Milk of magnesia, an antacid, is a suspension of Mg(OH)₂(s) in water. At 25 °C the solubility equilibrium is Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). Answer the following questions.
(a)Write the expression for Ksp of Mg(OH)₂. A saturated solution of Mg(OH)₂ at 25 °C has pH = 10.42; calculate [OH⁻] in this solution and the value of Ksp for Mg(OH)₂.
(3 pts)
Model answerKsp = [Mg²⁺][OH⁻]². pOH = 14.00 − 10.42 = 3.58, so [OH⁻] = 10⁻³·⁵⁸ = 2.6 × 10−4 M. [Mg²⁺] = [OH⁻]/2 = 1.3 × 10−4 M, and Ksp = 1.3 × 10−4 × (2.6 × 10−4)² = 9.1 × 10−12.
Scoring points (1 point each)(b)Calculate the molar solubility of Mg(OH)₂ in a solution buffered at pH 12.00, and state whether it is greater or less than in pure water, and why.
(2 pts)
Model answerAt pH 12.00, [OH⁻] = 0.010 M, so s = [Mg²⁺] = Ksp ÷ [OH⁻]² = 9.1 × 10−12 ÷ (0.010)² = 9.1 × 10−8 M, far less than 1.3 × 10−4 M in pure water. The solution already contains a lot of OH⁻ (the common-ion effect), shifting the equilibrium toward the solid.
Scoring points (1 point each)(c)Give the net ionic equation for Mg(OH)₂ reacting with stomach acid (HCl), and use equilibrium reasoning to explain why Mg(OH)₂ dissolves in acid.
(2 pts)
Model answerMg(OH)₂(s) + 2H⁺(aq) → Mg²⁺(aq) + 2H₂O(l). H⁺ combines with OH⁻ to form water, lowering [OH⁻], so Q = [Mg²⁺][OH⁻]² < Ksp and the solubility equilibrium shifts right, dissolving more solid.
Scoring points (1 point each)(d)One dose of milk of magnesia contains 0.80 g of Mg(OH)₂ (58.32 g/mol). What is the greatest volume of 0.10 M HCl it can neutralize, in mL?
(1 pt)
Model answermol Mg(OH)₂ = 0.80 ÷ 58.32 = 0.0137 mol, which needs 0.0274 mol H⁺: V = 0.0274 ÷ 0.10 = 0.274 L ≈ 274 mL.
Scoring points (1 point each)(e)Some solid MgCl₂ is added to a saturated Mg(OH)₂ solution in contact with excess solid (constant temperature). After equilibrium is re-established, how have [OH⁻] and the mass of solid Mg(OH)₂ changed?
(1 pt)
Model answerThe added Mg²⁺ makes Q > Ksp, so the equilibrium shifts left: more Mg(OH)₂ precipitates, [OH⁻] decreases, and the mass of solid increases (Ksp is unchanged).
Scoring points (1 point each)(f)At 25 °C, is ΔG° for the dissolution of Mg(OH)₂(s) into Mg²⁺(aq) and OH⁻(aq) positive or negative? Justify your answer.
(1 pt)
Model answerPositive. ΔG° = −RT ln Ksp, and Ksp ≪ 1, so ln Ksp < 0 and ΔG° > 0: dissolving under standard conditions is not thermodynamically favorable, so only a little dissolves.
Scoring points (1 point each)ExplanationCommon losses: including the solid or dropping the square in Ksp; getting [OH⁻] from the pH without converting to pOH; setting [Mg²⁺] equal to [OH⁻]; splitting Mg(OH)₂ into ions in the net ionic equation; and missing the 1 : 2 ratio in the neutralization.
Review: U7.11 Introduction to Solubility Equilibria (Ksp)U7.12 Common-Ion EffectU7.13 pH and SolubilityU4.5 StoichiometryU7.14 Free Energy of Dissolution
- 4Free response
Acetic acid (CH₃COOH) and methyl formate (HCOOCH₃) both have the formula C₂H₄O₂. Acetic acid boils at 118 °C and methyl formate at 32 °C. Answer the following questions.
(a)Draw the Lewis diagram for acetic acid, showing all valence electrons.
(1 pt)
Model answer2 × 4 + 4 × 1 + 2 × 6 = 24 valence electrons. The CH₃ carbon has single bonds to three H atoms and to the other C. That carbon has a double bond to one O (two lone pairs) and a single bond to the other O, which is bonded to H (two lone pairs): CH₃–C(=O)–O–H.
Scoring points (1 point each)(b)Acetic acid has two carbon atoms. Compare the H–C–H angle at the methyl carbon with the O–C–O angle at the carbonyl carbon (the one bonded to both O atoms), using VSEPR theory.
(1 pt)
Model answerThe methyl carbon has four electron domains (four single bonds), tetrahedral, so H–C–H is about 109.5°. The carbonyl carbon has three domains (two single bonds and one double bond), trigonal planar, so O–C–O is about 120°. The angle at the carbonyl carbon is larger.
Scoring points (1 point each)(c)State how each of the two carbon atoms in acetic acid is hybridized, and count the σ and π bonds in one acetic acid molecule.
(1 pt)
Model answerThe methyl carbon is sp³ and the carbonyl carbon is sp². σ bonds: three C–H, one C–C, one in C=O, one C–O, and one O–H, seven in all; there is one π bond (in C=O).
Scoring points (1 point each)(d)The two compounds have the same molar mass. Explain, in terms of intermolecular forces, why acetic acid boils at a much higher temperature than methyl formate.
(1 pt)
Model answerAcetic acid has an O–H bond, with H attached to highly electronegative O, so its molecules hydrogen bond to each other. In methyl formate every H is bonded to C, so there are no hydrogen bonds between its molecules, only dipole–dipole and dispersion forces. With the same number of electrons, the dispersion forces are similar, and hydrogen bonding makes the attractions in acetic acid much stronger.
Scoring points (1 point each)ExplanationCommon losses: miscounting electrons or leaving lone pairs off O; giving both carbons the same hybridization; missing the O–H or C–O σ bond; claiming methyl formate hydrogen bonds to itself (its O can accept a hydrogen bond, but it has no O–H to donate); and describing boiling as breaking O–H covalent bonds.
Review: U2.5 Lewis DiagramsU2.7 VSEPR and Bond HybridizationU3.1 Intermolecular Forces
- 5Free response
A galvanic cell consists of an iron electrode in 1.0 M FeSO₄ and a nickel electrode in 1.0 M NiSO₄, connected by a salt bridge. Standard reduction potentials at 25 °C: Ni²⁺ + 2e⁻ → Ni, E° = −0.26 V; Fe²⁺ + 2e⁻ → Fe, E° = −0.45 V.
(a)Write the balanced equation for the overall cell reaction, and calculate E°cell.
(1 pt)
Model answerFe, with the lower E°, is oxidized: Fe(s) + Ni²⁺(aq) → Fe²⁺(aq) + Ni(s). E°cell = −0.26 − (−0.45) = +0.19 V.
Scoring points (1 point each)(b)Calculate ΔG° for the reaction, in kJ/mol (F = 96,485 C/mol).
(1 pt)
Model answerΔG° = −nFE° = −2 × 96,485 × 0.19 = −36.7 kJ/mol (n = 2).
Scoring points (1 point each)(c)Calculate the equilibrium constant, K, for this reaction at 25 °C.
(1 pt)
Model answerΔG° = −RT ln K, so ln K = 36664 ÷ (8.314 × 298.15) = 14.8, and K ≈ 2.7 × 106.
Scoring points (1 point each)(d)After the cell has run for a while, [Fe²⁺] has increased and [Ni²⁺] has decreased, and the voltmeter now reads less than E°cell. Explain why, and state what the cell potential eventually becomes.
(1 pt)
Model answerQ = [Fe²⁺]/[Ni²⁺] starts at 1 and keeps rising. By E = E° − (RT/nF) ln Q, E < E° once ln Q > 0. As the reaction continues, Q approaches K (2.7 × 106) and E keeps falling; when Q = K the system is at equilibrium, E = 0, and the cell can do no more work.
Scoring points (1 point each)ExplanationCommon losses: oxidizing the electrode with the higher potential; a positive ΔG° or n = 1; mixing kJ with R in J when finding K; and thinking a “dead” cell has run out of reactants rather than reached equilibrium (Q = K).
Review: U9.8 Cell Potential and Free EnergyU9.7 Galvanic (Voltaic) and Electrolytic CellsU9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)U9.9 Cell Potential Under Nonstandard Conditions
- 6Free response
A student determines the specific heat of an unknown metal. A 25.0 g block of the metal is heated to 99.5 °C in boiling water and quickly placed in 50.0 g of water at 21.0 °C in a coffee-cup calorimeter. After stirring, the water and metal reach 25.0 °C. The specific heat of water is 4.18 J/(g·°C); assume the calorimeter absorbs no heat. Specific heats of some metals are shown.
Metal Specific heat (J/(g·°C)) Aluminum 0.897 Iron 0.449 Copper 0.385 Lead 0.129 (a)Calculate the heat absorbed by the water.
(1 pt)
Model answerq = mcΔT = 50.0 × 4.18 × (25.0 − 21.0) = 836 J.
Scoring points (1 point each)(b)Calculate the specific heat of the metal, and use the table to identify the metal.
(1 pt)
Model answerThe heat lost by the metal equals the heat gained by the water: c = 836 ÷ [25.0 × (99.5 − 25.0)] = 0.449 J/(g·°C), so the metal is iron.
Scoring points (1 point each)(c)If some heat escaped from the calorimeter to the air during the experiment, would the calculated specific heat of the metal be too large, too small, or unaffected? Explain.
(1 pt)
Model answerToo small. Heat loss makes the measured final temperature too low, so the calculated heat absorbed by the water (the numerator) is too small and the metal’s temperature change, 99.5 − T (the denominator), is too large; both make c too small.
Scoring points (1 point each)(d)Describe at the particle level how heat is transferred after the metal is placed in the water, and why both end at the same temperature.
(1 pt)
Model answerThe metal atoms vibrate with a higher average kinetic energy than the water molecules have. Where they touch, collisions transfer energy from the more energetic metal atoms to the water molecules until their average kinetic energies, and so their temperatures, are equal; then there is no net transfer (thermal equilibrium).
Scoring points (1 point each)ExplanationCommon losses: using the water’s 4.0 °C as the metal’s temperature change; writing the units as J/°C; and in the error analysis, saying “heat was lost” without stating which way it moves the result.
Review: U6.4 Heat Capacity and CalorimetryU6.3 Heat Transfer and Thermal Equilibrium
- 7Free response
A student finds the molar mass of a volatile liquid this way: a few milliliters of the liquid are placed in a 0.250 L flask, the mouth is covered with foil pierced by a pinhole, and the flask is held in boiling water at 99.0 °C until all the liquid has vaporized and the excess vapor has escaped, leaving the flask full of vapor at 99.0 °C and 0.987 atm. The flask is then cooled so all the vapor condenses; the condensed liquid has a mass of 0.712 g.
(a)Calculate the number of moles of vapor in the flask.
(1 pt)
Model answern = PV ÷ RT = (0.987 × 0.250) ÷ (0.08206 × 372.15) = 8.08 × 10−3 mol.
Scoring points (1 point each)(b)Calculate the molar mass of the liquid.
(1 pt)
Model answerM = m ÷ n = 0.712 ÷ 8.08 × 10−3 = 88.1 g/mol.
Scoring points (1 point each)(c)Elemental analysis shows that the empirical formula of the liquid is C₂H₄O. What is its molecular formula?
(1 pt)
Model answerC₂H₄O has a formula mass of 44.05, and 88.1 ÷ 44.05 ≈ 2, so the molecular formula is C₄H₈O₂.
Scoring points (1 point each)(d)After taking the flask out of the water bath, the student did not dry the outside, so water droplets were still on it when it was weighed. Would the calculated molar mass be too large, too small, or unaffected? Explain.
(1 pt)
Model answerToo large. The droplets make the measured mass of condensed liquid too large, while the moles of vapor, calculated from P, V, and T, are unaffected, so M = m ÷ n is too large.
Scoring points (1 point each)ExplanationCommon losses: using 99.0 instead of 372.15 K; plugging in milliliters; giving the empirical formula as the molecular formula; and not saying which quantity (m or n) the error affects.
Result
| Unit | Correct |
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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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