AP Chemistry Practice Test 10
60 original exam-style multiple-choice questions with answers & explanations — covering all nine units of the AP Chemistry course.
Note: These are original practice questions written in the AP Chemistry exam style — not reproduced College Board questions. Use them to test your readiness for the real AP Chemistry exam.
📱 How to use: Try each question first, then click “Show Answer” to check. Covers all 9 AP units: atomic structure, compound structure, intermolecular forces, reactions, kinetics, thermodynamics, equilibrium, acids & bases, and applications of thermodynamics.
⚛ Unit 1: Atomic Structure & Properties
Q1A ground-state atom of selenium, Se (Z = 34), has the electron configuration [Ar] 4s2 3d10 4p4. How many unpaired electrons are present in a ground-state Se atom?
Show Answer
Correct: (C)
The 4p4 subshell places one pair in the first 4p orbital and one unpaired electron in each of the other two, giving 2 unpaired electrons by Hund's rule.
Q2A bromide ion, Br−, is formed when a bromine atom (Z = 35) gains one electron. The ion is isoelectronic with which noble gas?
Show Answer
Correct: (B)
Br has 35 electrons, so Br− has 36 electrons, the same electron count as Kr.
Q3Which list orders the atoms from largest atomic radius to smallest atomic radius?
Show Answer
Correct: (A)
Rb is in period 5 and is largest; across period 4 the radius decreases from Ga to Se to Br as effective nuclear charge increases.
Q4A sample of copper consists of two isotopes: Cu-63 with mass 62.93 amu and 69.2% abundance, and Cu-65 with mass 64.93 amu and 30.8% abundance. What is the average atomic mass of copper in the sample?
Show Answer
Correct: (D)
(0.692 × 62.93) + (0.308 × 64.93) = 43.55 + 20.00 = 63.55 amu.
Q5What is the electron configuration of a Ga^3+ ion, given that a neutral Ga atom (Z = 31) is [Ar] 4s2 3d10 4p1?
Show Answer
Correct: (C)
Three electrons are removed first from 4p (1 electron) and then from 4s (2 electrons), leaving [Ar] 3d10.
Q6The first ionization energy of arsenic, As, is greater than that of selenium, Se, even though Se has a greater atomic number. Which statement best explains this observation?
Show Answer
Correct: (A)
As is 4p3 (half-filled and relatively stable), while Se is 4p4 with one paired orbital; repulsion makes that paired electron easier to remove.
Q7How many atoms are present in a 2.50 g sample of pure copper (molar mass 63.55 g/mol)?
Show Answer
Correct: (B)
2.50 g / 63.55 g/mol = 0.0393 mol, and 0.0393 mol × 6.022 × 10^23 atoms/mol = 2.37 × 10^22 atoms.
🔗 Unit 2: Molecular & Ionic Compound Structure
Q8In the best Lewis structure of carbon monoxide, CO, carbon and oxygen are joined by a triple bond and each atom carries one lone pair. What is the formal charge on the carbon atom?
Show Answer
Correct: (D)
Formal charge on C = 4 valence electrons − (2 nonbonding + 3 bonding electrons counted as half of 6) = 4 − 5 = −1.
Q9What is the molecular geometry of SO2, which has two bonding domains and one lone pair on the central sulfur atom?
Show Answer
Correct: (A)
Three electron domains give a trigonal planar electron geometry, but the lone pair leaves a bent molecular geometry.
Q10Nitrogen dioxide, NO2, is described by resonance structures in which one N-O bond is a double bond and the other is a single bond. What is the average N-O bond order in NO2?
Show Answer
Correct: (C)
Averaging one single and one double bond over the two resonance positions gives (1 + 2) / 2 = 1.5.
Q11What is the H-C-H bond angle in methane, CH4?
Show Answer
Correct: (B)
CH4 has four bonding domains and no lone pairs on carbon, giving a tetrahedral geometry with 109.5° bond angles.
Q12Which ionic compound has the lattice energy with the greatest magnitude, based on ionic charges and sizes?
Show Answer
Correct: (D)
ZnO contains Zn^2+ and O^2- ions; the product of the ionic charges (2 × 2) is larger than for the 1+/1− salts, giving the strongest lattice attractions.
Q13What is the hybridization of the nitrogen atom in ammonia, NH3, which has three N-H bonds and one lone pair?
Show Answer
Correct: (A)
Nitrogen has four electron domains (three bonds plus one lone pair), which corresponds to sp3 hybridization.
Q14Which compound contains both ionic bonds and covalent bonds within the same formula unit?
Show Answer
Correct: (C)
NH4Cl contains ionic attraction between NH4^+ and Cl− ions, and the N-H bonds inside NH4^+ are covalent.
💧 Unit 3: Intermolecular Forces & Properties
Q15Ammonia, NH3, has a much higher boiling point than methane, CH4, of similar molar mass, primarily because NH3 molecules experience which intermolecular force that CH4 molecules lack?
Show Answer
Correct: (B)
The N-H bonds and lone pair on nitrogen allow strong hydrogen bonding between NH3 molecules; CH4 has only dispersion forces.
Q16Which list orders the substances from lowest boiling point to highest boiling point?
Show Answer
Correct: (D)
CO boils near 82 K, CH4 near 112 K, NH3 near 240 K (hydrogen bonding), and SO2 near 263 K (strong dipole forces and larger electron cloud).
Q17Which gas is most soluble in water at room temperature, largely because it forms hydrogen bonds with water molecules?
Show Answer
Correct: (A)
NH3 hydrogen bonds extensively with water and also reacts slightly to form NH4^+ and OH−, making it highly soluble.
Q18How much heat is required to melt 18.0 g of ice at 0°C, given that the molar enthalpy of fusion of water is 6.01 kJ/mol and the molar mass of water is 18.0 g/mol?
Show Answer
Correct: (C)
18.0 g is 1.00 mol of water, so q = 1.00 mol × 6.01 kJ/mol = 6.01 kJ.
Q19A crystalline solid is a good conductor of electricity as a solid, is malleable, and has a high melting point. A sample of which substance fits this description?
Show Answer
Correct: (B)
Sn is a metal; delocalized electrons in a metallic solid account for conductivity and malleability.
Q20A nonvolatile solute is dissolved in a pure solvent. Compared with the pure solvent, the resulting solution will have
Show Answer
Correct: (A)
Nonvolatile solute particles reduce the tendency of solvent molecules to escape, lowering vapor pressure (it also lowers freezing point but raises, not lowers, boiling point).
⚖ Unit 4: Chemical Reactions
Q21When the equation Pb(NO3)2 + KI → PbI2 + KNO3 is balanced with the smallest whole-number coefficients, what is the sum of all the coefficients?
Show Answer
Correct: (D)
The balanced equation is Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3, and 1 + 2 + 1 + 2 = 6.
Q22What is the oxidation number of sulfur in SO2?
Show Answer
Correct: (C)
Each oxygen is −2, so S + 2(−2) = 0 gives S = +4.
Q23Solutions of AgNO3 and NaCl are mixed and a white precipitate of AgCl forms. Which equation is the correct net ionic equation for the reaction?
Show Answer
Correct: (B)
Only Ag^+ and Cl− combine to form the solid; Na^+ and NO3− are spectator ions and are omitted.
Q24A 2.50 g sample of zinc reacts completely with excess hydrochloric acid according to Zn + 2 HCl → ZnCl2 + H2. How many moles of H2 are produced (molar mass of Zn = 65.38 g/mol)?
Show Answer
Correct: (A)
2.50 g / 65.38 g/mol = 0.0382 mol Zn, and the 1:1 ratio gives 0.0382 mol H2.
Q25Copper metal and oxygen react according to 2 Cu + O2 → 2 CuO. A mixture contains 12.7 g of Cu (0.200 mol) and 1.60 g of O2 (0.0500 mol). Which reactant is the limiting reactant?
Show Answer
Correct: (D)
The 2:1 ratio means 0.200 mol Cu needs 0.100 mol O2; with only 0.0500 mol O2 present, O2 runs out first.
Q26A precipitation reaction that should theoretically produce 10.0 g of AgCl actually produces 8.50 g of AgCl. What is the percent yield?
Show Answer
Correct: (C)
Percent yield = (8.50 g / 10.0 g) × 100% = 85.0%.
Q27A 25.0 mL sample of 0.250 M acetic acid, CH3COOH, is titrated with 0.125 M NaOH. What volume of NaOH is required to reach the equivalence point?
Show Answer
Correct: (B)
Moles of acid = 0.0250 L × 0.250 M = 0.00625 mol; volume of base = 0.00625 mol / 0.125 M = 0.0500 L = 50.0 mL.
⏱ Unit 5: Kinetics
Q28For a reaction A + B → products, the following data were collected: with [A] = 0.10 M and [B] = 0.10 M the rate is 0.020 M/s; with [A] = 0.20 M and [B] = 0.10 M the rate is 0.080 M/s; with [A] = 0.10 M and [B] = 0.20 M the rate is 0.040 M/s. What is the rate law?
Show Answer
Correct: (A)
Doubling [A] quadruples the rate (second order in A), while doubling [B] doubles the rate (first order in B).
Q29A reaction has the rate law rate = k[NO2]^2. What are the units of the rate constant k, if rate is measured in M/s?
Show Answer
Correct: (D)
k = rate / [NO2]^2 = (M/s) / M^2 = M^-1 s^-1 for an overall second-order reaction.
Q30A first-order reaction has a rate constant k = 0.0693 min^-1. What is the half-life of the reaction?
Show Answer
Correct: (C)
For a first-order reaction, t1/2 = 0.693 / k = 0.693 / 0.0693 min^-1 = 10.0 min.
Q31A proposed mechanism for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) is: Step 1 (slow): NO2(g) + NO2(g) → NO3(g) + NO(g); Step 2 (fast): NO3(g) + CO(g) → NO2(g) + CO2(g). What is the rate law predicted by this mechanism?
Show Answer
Correct: (B)
The slow first step is rate-determining and involves two NO2 molecules, so the rate is second order in NO2 and independent of CO.
Q32A student increases the temperature of a gas-phase reaction from 300 K to 350 K and observes a large increase in the reaction rate. According to collision theory, the increase is best explained by which of the following?
Show Answer
Correct: (C)
Raising the temperature does not change the activation energy; it increases the fraction of molecular collisions energetic enough to overcome it.
Q33For a first-order analysis of a reaction, a plot of ln k versus 1/T gives a straight line with a slope of −7200 K. Using R = 8.314 J/(mol·K) and the Arrhenius relationship slope = −Ea/R, what is the activation energy of the reaction?
Show Answer
Correct: (A)
Ea = 7200 K × 8.314 J/(mol·K) = 59,900 J/mol, or about 59.9 kJ/mol.
Q34Chlorine atoms catalyze the decomposition of ozone in the stratosphere by the steps: Step 1: Cl(g) + O3(g) → ClO(g) + O2(g); Step 2: ClO(g) + O(g) → Cl(g) + O2(g). Which identification of the catalyst and the intermediate is correct?
Show Answer
Correct: (D)
Cl is consumed in Step 1 and regenerated in Step 2, making it a catalyst, while ClO is formed and then consumed, making it an intermediate.
🔥 Unit 6: Thermodynamics
Q35When a chemical process occurs in a coffee-cup calorimeter, 50.0 g of water warms from 25.0°C to 35.0°C. The specific heat capacity of water is 4.18 J/(g·°C). How much heat did the water absorb?
Show Answer
Correct: (B)
q = mcΔT = (50.0 g)(4.18 J/(g·°C))(10.0°C) = 2090 J = 2.09 kJ.
Q36Given: C(s) + O2(g) → CO2(g), ΔH = −393.5 kJ/mol, and CO(g) + 1/2 O2(g) → CO2(g), ΔH = −283.0 kJ/mol. What is ΔH for C(s) + 1/2 O2(g) → CO(g)?
Show Answer
Correct: (D)
By Hess's law, ΔH = −393.5 kJ/mol − (−283.0 kJ/mol) = −110.5 kJ/mol.
Q37Use the bond enthalpies H−H = 436 kJ/mol, Cl−Cl = 242 kJ/mol, and H−Cl = 431 kJ/mol to estimate ΔH for H2(g) + Cl2(g) → 2 HCl(g).
Show Answer
Correct: (A)
ΔH = bonds broken − bonds formed = (436 + 242) − 2(431) = 678 − 862 = −184 kJ/mol.
Q38An exothermic reaction is carried out in a calorimeter containing 200. g of solution with specific heat capacity 4.18 J/(g·°C). The temperature of the solution rises by 5.0°C, and 0.100 mol of the limiting reactant is consumed. What is ΔH for the reaction?
Show Answer
Correct: (C)
q(solution) = (200. g)(4.18)(5.0) = 4180 J = 4.18 kJ absorbed by the solution, so the reaction released 4.18 kJ for 0.100 mol, giving ΔH = −41.8 kJ/mol.
Q39Which of the following processes results in an increase in the entropy of the system?
Show Answer
Correct: (B)
Vaporization converts a liquid into a much more dispersed gas, greatly increasing the number of accessible microstates and the entropy.
Q40Given standard heats of formation: CH4(g) = −74.8 kJ/mol, CO2(g) = −393.5 kJ/mol, H2O(l) = −285.8 kJ/mol, and O2(g) = 0 kJ/mol, what is ΔH° for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)?
Show Answer
Correct: (A)
ΔH° = [−393.5 + 2(−285.8)] − [−74.8 + 0] = −965.1 + 74.8 = −890.3 kJ/mol.
Q41Given: N2(g) + O2(g) → 2 NO(g), ΔH = +180 kJ, and 2 NO(g) + O2(g) → 2 NO2(g), ΔH = −112 kJ. What is ΔH for N2(g) + 2 O2(g) → 2 NO2(g) as written?
Show Answer
Correct: (D)
Adding the two steps gives +180 kJ + (−112 kJ) = +68 kJ for the formation of 2 mol of NO2.
⇆ Unit 7: Equilibrium
Q42What is the equilibrium constant expression, Kc, for 2 SO2(g) + O2(g) ⇌ 2 SO3(g)?
Show Answer
Correct: (C)
Kc is the product concentrations raised to their coefficients divided by the reactant concentrations raised to their coefficients.
Q43For H2(g) + I2(g) ⇌ 2 HI(g), the initial concentrations of H2 and I2 are each 0.250 M with no HI present. At equilibrium, [HI] = 0.300 M. What is the value of Kc?
Show Answer
Correct: (B)
Forming 0.300 M HI consumes 0.150 M of each reactant, leaving [H2] = [I2] = 0.100 M, so Kc = (0.300)^2 / (0.100)(0.100) = 9.0.
Q44For N2(g) + 3 H2(g) ⇌ 2 NH3(g) at 350 K, what is the relationship between Kp and Kc?
Show Answer
Correct: (A)
The change in moles of gas is 2 − 4 = −2, so Kp = Kc(RT)^Δn = Kc(RT)^−2.
Q45The reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) is exothermic. Which change will increase the amount of SO3 present at equilibrium?
Show Answer
Correct: (D)
Removing a product shifts the equilibrium toward the products to replace it; heating an exothermic reaction and lowering the pressure both favor the reactants, and inert gas at constant volume changes nothing.
Q46The molar solubility of PbCl2 in pure water is 0.010 M. What is the Ksp of PbCl2?
Show Answer
Correct: (C)
Dissolution gives [Pb^2+] = 0.010 M and [Cl^-] = 0.020 M, so Ksp = (0.010)(0.020)^2 = 4.0 × 10^−6.
Q47For H2(g) + I2(g) ⇌ 2 HI(g), Kc = 50. at a certain temperature. At one moment, [HI] = 1.0 M and [H2] = [I2] = 0.10 M. In which direction will the reaction proceed to reach equilibrium?
Show Answer
Correct: (D)
Q = (1.0)^2 / (0.10)(0.10) = 100, which is greater than Kc = 50., so the reaction shifts toward the reactants.
Q48A saturated solution of PbCl2 is prepared. If solid Pb(NO3)2 is then added to the solution, what happens to the solubility of PbCl2?
Show Answer
Correct: (B)
The added Pb^2+ is a common ion, which shifts the dissolution equilibrium toward the solid and decreases the solubility of PbCl2; Ksp itself is unchanged.
🧫 Unit 8: Acids & Bases
Q49What is the pH of a 0.010 M solution of HCl(aq), a strong acid?
Show Answer
Correct: (A)
HCl ionizes completely, so [H^+] = 0.010 M and pH = −log(1.0 × 10^−2) = 2.00.
Q50Acetic acid has Ka = 1.8 × 10^−5. What is the pH of a 0.250 M solution of acetic acid?
Show Answer
Correct: (C)
[H^+] is approximately sqrt(Ka × C) = sqrt(1.8 × 10^−5 × 0.250) = 2.1 × 10^−3 M, so pH = 2.67.
Q51A buffer solution contains 0.100 M acetic acid and 0.250 M sodium acetate. The pKa of acetic acid is 4.74. What is the pH of the buffer?
Show Answer
Correct: (B)
pH = pKa + log([base]/[acid]) = 4.74 + log(0.250/0.100) = 4.74 + 0.40 = 5.14.
Q52A 25.0 mL sample of 0.100 M NH3(aq), a weak base, is titrated with 0.100 M HCl(aq). What is true of the pH at the equivalence point?
Show Answer
Correct: (D)
At the equivalence point the solution contains NH4^+, the conjugate acid of a weak base, which hydrolyzes to produce H^+ and makes the solution acidic.
Q53Which species is the conjugate acid of NH3?
Show Answer
Correct: (A)
A conjugate acid is formed by adding one H^+ to the base, so NH3 + H^+ gives NH4^+.
Q54A solution of the diprotic acid H2CO3 is titrated with NaOH(aq) to the first equivalence point. Which species is present in the highest concentration at that point, aside from water?
Show Answer
Correct: (B)
At the first equivalence point, essentially all of the H2CO3 has been converted to its singly deprotonated form, HCO3^−.
🔋 Unit 9: Applications of Thermodynamics
Q55A reaction has ΔH = −100 kJ/mol and ΔS = −200 J/(mol·K). What is ΔG for the reaction at 250 K?
Show Answer
Correct: (C)
ΔG = ΔH − TΔS = −100 kJ/mol − (250 K)(−0.200 kJ/(mol·K)) = −100 + 50 = −50 kJ/mol.
Q56A reaction has ΔH > 0 and ΔS > 0. Under which conditions will the reaction be spontaneous?
Show Answer
Correct: (D)
With both ΔH and ΔS positive, ΔG = ΔH − TΔS becomes negative only when T is large enough for the TΔS term to dominate.
Q57Using the standard reduction potentials Zn^2+ + 2 e^- → Zn, E° = −0.76 V, and Ag^+ + e^- → Ag, E° = +0.80 V, what is E°cell for the galvanic cell Zn(s) + 2 Ag^+(aq) → Zn^2+(aq) + 2 Ag(s)?
Show Answer
Correct: (A)
E°cell = E°(cathode) − E°(anode) = 0.80 V − (−0.76 V) = 1.56 V.
Q58A galvanic cell is built from a Pb electrode in Pb^2+(aq) and a Cu electrode in Cu^2+(aq), with E° = −0.13 V for Pb^2+/Pb and E° = +0.34 V for Cu^2+/Cu. Which half-reaction occurs at the anode?
Show Answer
Correct: (D)
Pb has the lower reduction potential, so Pb is oxidized at the anode while Cu^2+ is reduced at the cathode.
Q59Copper is deposited by electrolyzing a Cu^2+(aq) solution with a constant current of 2.00 A for 9650 s. The Faraday constant is about 96,500 C/mol e^-. What mass of Cu is deposited? (Molar mass of Cu = 63.55 g/mol)
Show Answer
Correct: (B)
Charge = (2.00 A)(9650 s) = 19,300 C = 0.200 mol e^-; since Cu^2+ requires 2 e^-, this deposits 0.100 mol Cu, or 0.100 × 63.55 = 6.35 g (using 63.5 g/mol gives 6.35 g).
Q60For the galvanic cell reaction Zn(s) + 2 Ag^+(aq) → Zn^2+(aq) + 2 Ag(s) with E°cell = 1.56 V, what is ΔG° for the reaction as written? (F = 96,500 C/mol e^-)
Show Answer
Correct: (C)
ΔG° = −nFE° = −(2)(96,500)(1.56) = −301,000 J/mol, or about −301 kJ/mol.
Want more practice?
All 20 AP Chemistry Practice Tests | AP Chemistry Practice Questions