AP Chemistry Practice Test 2
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
Q1An element has the ground-state electron configuration [Ne] 3s^2 3p^4. What is the identity of the element?
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Correct: (D)
The configuration contains 10 + 2 + 4 = 16 electrons, so the neutral atom has atomic number 16, which is S.
Q2The photoelectron spectrum of an element shows three peaks, in order of decreasing binding energy, with relative heights corresponding to 2, 2, and 4 electrons. What is the element?
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Correct: (A)
The peaks correspond to 1s^2 2s^2 2p^4, a total of 8 electrons, which is O.
Q3Which element has the largest atomic radius?
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Correct: (B)
Na, Mg, Si, and Cl are all in Period 3, and atomic radius decreases across a period as the nuclear charge increases. Na has the fewest protons of the four, so its electrons are held least tightly and it has the largest radius.
Q4A sample of bromine consists of two isotopes: Br-79 with mass 78.92 amu and percent abundance 50.69%, and Br-81 with mass 80.92 amu and percent abundance 49.31%. What is the average atomic mass of bromine in the sample?
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Correct: (C)
(78.92 x 0.5069) + (80.92 x 0.4931) = 40.00 + 39.90 = 79.91 amu.
Q5Silver has two naturally occurring isotopes with masses 106.905 amu and 108.905 amu, and its average atomic mass is 107.868 amu. What is the percent abundance of the isotope with mass 106.905 amu?
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Correct: (D)
Let x be the fraction of the 106.905 amu isotope: 106.905x + 108.905(1 – x) = 107.868, so 108.905 – 2x = 107.868 and x = 0.519, or 51.9%.
Q6How many unpaired electrons are present in a ground-state Zn^2+ ion?
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Correct: (B)
Zn is [Ar] 4s^2 3d^10, and Zn^2+ loses the two 4s electrons, leaving a completely filled 3d^10 subshell with 0 unpaired electrons.
Q7The successive ionization energies of an element, in kJ/mol, are 578, 1817, 2745, and 11577 for the first through fourth ionizations. How many valence electrons does the element have?
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Correct: (A)
The first three ionization energies increase gradually, but the fourth is much larger, showing the third electron was the last valence electron and the fourth comes from a core shell. The element has 3 valence electrons.
🔗 Unit 2: Molecular & Ionic Compound Structure
Q8Which ionic compound has the smallest lattice energy (least exothermic lattice formation)?
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Correct: (C)
CsI combines only 1+ and 1- charges with the largest cation and a large anion among the choices, so the electrostatic attraction is weakest and the lattice energy is smallest in magnitude.
Q9In the Lewis structure of BrF5, how many lone pairs of electrons are on the central Br atom?
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Correct: (D)
Br contributes 7 valence electrons. Five are used in bonds to F atoms, leaving 2 electrons, which form 1 lone pair on Br.
Q10In a Lewis structure of ozone, O3, with one single bond and one double bond, what is the formal charge on the central oxygen atom, which also has one lone pair?
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Correct: (B)
Formal charge = 6 valence electrons – (2 nonbonding electrons + 3 bonds) = 6 – 5 = +1.
Q11What is the molecular geometry of NH3?
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Correct: (A)
N in NH3 has 4 electron domains (3 bonds and 1 lone pair), giving a tetrahedral electron geometry, but the lone pair is not counted in the molecular geometry, which is trigonal pyramidal.
Q12What is the hybridization of the boron atom in BF3?
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Correct: (C)
B in BF3 forms 3 bonds with no lone pairs, requiring 3 hybrid orbitals, which corresponds to sp^2 hybridization.
Q13Which bond is the most polar?
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Correct: (D)
Polarity increases with the difference in electronegativity between the bonded atoms. F is the most electronegative element, so the H-F bond has the largest electronegativity difference and is the most polar.
Q14What is the C-C bond order in benzene, C6H6, which is represented by two resonance structures?
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Correct: (B)
In the resonance structures, each C-C position is single in one structure and double in the other, for a total bond order of 3 shared over 2 structures per position, giving an average bond order of 1.5.
💧 Unit 3: Intermolecular Forces & Properties
Q15Which substance experiences only London dispersion forces between its molecules?
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Correct: (B)
F2 is a nonpolar diatomic molecule, so it has no dipole-dipole forces and no hydrogen bonding; only dispersion forces remain.
Q16Which hydride has the highest normal boiling point?
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Correct: (A)
Although H2O is the lightest, hydrogen bonding between water molecules gives it a much higher boiling point than the heavier hydrides, which lack hydrogen bonding.
Q17A sealed bottle of carbonated water is opened and bubbles of gas rapidly form. This happens most directly because
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Correct: (C)
Gas solubility decreases as the pressure of the gas above the liquid decreases (Henry's law), so dissolved CO2 comes out of solution.
Q18How much heat is needed to raise the temperature of 50.0 g of H2O(l) from 25.0°C to 75.0°C? The specific heat capacity of water is 4.18 J/g°C.
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Correct: (D)
q = mcΔT = 50.0 g × 4.18 J/g°C × 50.0°C = 10,450 J, which is about 10.5 kJ.
Q19Which solid conducts electricity when molten but not when solid?
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Correct: (B)
MgCl2 is ionic; its ions are locked in place in the solid but are mobile in the melt, allowing conduction only when molten. Fe conducts in both states.
Q20A solution is prepared by dissolving 0.250 mol of glucose, C6H12O6, in enough water to make 500. mL of solution. What is the molarity of the solution?
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Correct: (A)
Molarity = 0.250 mol / 0.500 L = 0.500 M.
⚖ Unit 4: Chemical Reactions
Q21When the equation Fe2O3 + CO → Fe + CO2 is balanced with the smallest whole-number coefficients, what is the coefficient of CO?
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Correct: (C)
The balanced equation is Fe2O3 + 3 CO → 2 Fe + 3 CO2, which balances Fe, C, and O.
Q22What is the oxidation number of sulfur in H2SO4?
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Correct: (D)
With H at +1 and O at −2: 2(+1) + S + 4(−2) = 0, so S = +6.
Q23Hydrochloric acid is neutralized by potassium hydroxide: HCl(aq) + KOH(aq) → KCl(aq) + H2O(l). What is the net ionic equation?
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Correct: (B)
K+ and Cl- are spectator ions; the only chemical change is H+ combining with OH- to form water. Option (D) is the complete ionic equation, not the net ionic equation.
Q24Methane burns according to CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g). How many grams of CO2 are produced when 2.0 mol of CH4 burns completely? The molar mass of CO2 is 44.0 g/mol.
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Correct: (A)
2.0 mol CH4 produces 2.0 mol CO2 (1:1 ratio), and 2.0 mol × 44.0 g/mol = 88 g.
Q25Water is formed by 2 H2(g) + O2(g) → 2 H2O(g). A mixture contains 4.0 mol of H2 and 3.0 mol of O2. What is the maximum number of moles of H2O that can form?
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Correct: (C)
4.0 mol H2 needs only 2.0 mol O2, so H2 is limiting and O2 is in excess; 4.0 mol H2 forms 4.0 mol H2O (1:1 ratio).
Q26In a synthesis, the theoretical yield of a solid product is 18.0 g. A student recovers 13.5 g of the pure product. What is the percent yield?
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Correct: (D)
Percent yield = (13.5 g / 18.0 g) × 100% = 75.0%.
Q27What volume of 0.200 M KOH(aq) is required to neutralize 30.00 mL of 0.100 M HCl(aq)? The reaction is 1:1.
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Correct: (B)
Moles HCl = 0.100 M × 0.03000 L = 0.00300 mol = moles KOH needed, so volume = 0.00300 mol / 0.200 M = 0.0150 L = 15.0 mL.
⏱ Unit 5: Kinetics
Q28For the reaction D + E → products, the following initial-rate data were collected. Experiment 1: [D] = 0.30 M, [E] = 0.10 M, initial rate = 0.018 M/s. Experiment 2: [D] = 0.30 M, [E] = 0.30 M, initial rate = 0.162 M/s. Experiment 3: [D] = 0.60 M, [E] = 0.10 M, initial rate = 0.036 M/s. What is the rate law for the reaction?
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Correct: (C)
Tripling [E] at constant [D] increases the rate by a factor of 9 (0.018 to 0.162), so the reaction is second order in E. Doubling [D] at constant [E] doubles the rate, so it is first order in D.
Q29A first-order reaction has a half-life of 20 minutes. If the initial concentration of the reactant is 0.80 M, what concentration remains after 60 minutes?
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Correct: (D)
Sixty minutes is three half-lives. The concentration halves three times: 0.80 → 0.40 → 0.20 → 0.10 M.
Q30A proposed mechanism for the overall reaction 2 R + S → R2S is: Step 1 (slow): R + S → RS. Step 2 (fast): RS + R → R2S. Which statement correctly identifies the intermediate and the rate law predicted by this mechanism?
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Correct: (B)
RS is formed in Step 1 and consumed in Step 2, so it is an intermediate. The slow first step determines the rate law, giving Rate = k[R][S].
Q31Which of the following best explains why increasing the concentration of a reactant in solution usually increases the reaction rate?
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Correct: (A)
Collision theory states that reactions occur when particles collide. A higher concentration means particles are closer together and collide more often, increasing the rate. Concentration does not change the activation energy or the average kinetic energy.
Q32For a particular reaction, the potential energy of the reactants is 10 kJ/mol, the potential energy of the activated complex is 85 kJ/mol, and the potential energy of the products is 55 kJ/mol. What is the activation energy for the reverse reaction?
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Correct: (D)
The reverse activation energy is the energy of the activated complex minus the energy of the products: 85 − 55 = 30 kJ/mol. (The forward activation energy is 75 kJ/mol, and ΔH for the forward reaction is +45 kJ/mol.)
Q33A catalyst is added to a system that has reached chemical equilibrium. Which statement describes the effect of the catalyst?
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Correct: (C)
A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount, speeding up both equally. Equilibrium is reached faster, but the equilibrium constant and product yield are unchanged.
Q34The concentration of a reactant in a first-order reaction decreases from 0.80 M to 0.20 M in 40 s. What is the value of the rate constant, k?
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Correct: (B)
Using the first-order integrated rate law, k = ln([A]0 / [A]) / t = ln(0.80 / 0.20) / 40 = ln(4) / 40 = 1.386 / 40 = 0.035 s^-1.
🔥 Unit 6: Thermodynamics
Q35A solid salt is dissolved in 200. g of water in a calorimeter, and the temperature of the water decreases from 25.0 °C to 20.0 °C. Assuming the specific heat capacity of the water is 4.18 J/(g °C) and no heat is lost to the surroundings, what is the heat change for the dissolving process?
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Correct: (A)
The water loses q = mcΔT = (200.)(4.18)(5.0) = 4180 J = 4.18 kJ of heat. That heat is absorbed by the dissolving process, so the heat change for dissolving is +4.18 kJ, meaning the process is endothermic.
Q36Given the following equations: (1) C(s) + O2(g) → CO2(g), ΔH = −394 kJ. (2) 2 CO(g) + O2(g) → 2 CO2(g), ΔH = −566 kJ. What is ΔH for the reaction 2 C(s) + O2(g) → 2 CO(g)?
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Correct: (D)
By Hess's law, take twice equation (1) and subtract equation (2): ΔH = 2(−394) − (−566) = −788 + 566 = −222 kJ.
Q37Using the bond enthalpies for the nitrogen-nitrogen triple bond in N2 = 945 kJ/mol, H−H = 436 kJ/mol, and N−H = 391 kJ/mol, estimate ΔH for the reaction N2(g) + 3 H2(g) → 2 NH3(g).
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Correct: (C)
Bonds broken: 945 + 3(436) = 2253 kJ. Bonds formed: each NH3 has three N−H bonds, so 2 NH3 has six: 6(391) = 2346 kJ. ΔH = 2253 − 2346 = −93 kJ.
Q38A fuel sample is burned in a calorimeter containing 500. g of water. The temperature of the water increases by 8.0 °C. If 0.50 mol of the fuel was burned and the specific heat capacity of water is 4.18 J/(g °C), what is the molar enthalpy of combustion of the fuel, assuming all the heat released was absorbed by the water?
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Correct: (B)
Heat absorbed by the water is q = mcΔT = (500.)(4.18)(8.0) = 16,720 J = 16.7 kJ. Per mole of fuel: ΔH = −16.7 kJ / 0.50 mol = −33.4 kJ/mol. The value is negative because combustion is exothermic.
Q39Which of the following changes is accompanied by a decrease in entropy, ΔS < 0, for the system?
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Correct: (A)
Four moles of gas react to form two moles of gas, so the number of freely moving particles decreases and the entropy decreases. Melting, vaporizing, and producing a gas from a solid all increase entropy.
Q40Given the standard enthalpies of formation: CaCO3(s) = −1207 kJ/mol, CaO(s) = −635 kJ/mol, and CO2(g) = −394 kJ/mol, calculate ΔH for the reaction CaCO3(s) → CaO(s) + CO2(g).
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Correct: (C)
Products: −635 + (−394) = −1029 kJ. Reactants: −1207 kJ. ΔH = −1029 − (−1207) = +178 kJ, so the decomposition is endothermic.
Q41Which statement about an exothermic reaction is correct?
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Correct: (D)
In an exothermic reaction the products have lower enthalpy than the reactants, so ΔH is negative and the excess energy is released to the surroundings as heat.
⇆ Unit 7: Equilibrium
Q42For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.020 at 600 K. What is the value of Kp for this reaction at 600 K? (R = 0.0821 L·atm/mol·K)
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Correct: (B)
Kp = Kc(RT)^Δn with Δn = 2 – 1 = +1, so Kp = 0.020 × (0.0821 × 600) = 0.020 × 49.3 = 0.98.
Q43Iodine and chlorine react according to I2(g) + Cl2(g) ⇌ 2 ICl(g), for which Kc = 81 at a certain temperature. A mixture is prepared with 0.20 M I2 and 0.20 M Cl2 and no ICl. What is the equilibrium concentration of ICl?
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Correct: (D)
If x mol/L of I2 reacts, [ICl] = 2x and Kc = (2x)^2 / (0.20 – x)^2 = 81. Taking the square root gives 2x / (0.20 – x) = 9, so x = 0.164 and [ICl] = 2x = 0.33 M. (Option A is x, not 2x.)
Q44Carbon reacts with carbon dioxide according to C(s) + CO2(g) ⇌ 2 CO(g), for which ΔH = +172 kJ/mol. Which change will decrease the amount of CO present at equilibrium?
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Correct: (A)
Increasing the pressure shifts the equilibrium toward the side with fewer moles of gas, the reactant side (1 mol of gas versus 2 mol), so CO decreases. Heating this endothermic reaction, lowering the pressure, or adding CO2 would each increase CO.
Q45The Ksp of PbI2 is 7.1 × 10^-9 at 25°C. What is the molar solubility of PbI2 in pure water at this temperature?
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Correct: (C)
For PbI2(s) ⇌ Pb^2+(aq) + 2 I^-(aq), Ksp = (s)(2s)^2 = 4s^3 = 7.1 × 10^-9, so s^3 = 1.8 × 10^-9 and s = 1.2 × 10^-3 M. (Option B is [I^-] = 2s, not s.)
Q46For the reaction 2 NO2(g) ⇌ N2O4(g), Kc = 100 at a certain temperature. At one moment, [NO2] = 0.10 M and [N2O4] = 0.20 M. Which statement correctly describes the system?
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Correct: (B)
Q = [N2O4] / [NO2]^2 = 0.20 / (0.10)^2 = 20. Since Q = 20 is less than Kc = 100, the reaction shifts toward products until Q increases to Kc.
Q47The Ksp of AgCl is 1.8 × 10^-10 at 25°C. What is the molar solubility of AgCl in a 0.050 M AgNO3 solution?
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Correct: (D)
With the common ion Ag^+ already at 0.050 M, Ksp = (0.050)(s) = 1.8 × 10^-10, giving s = 3.6 × 10^-9 M. (Option A is the solubility in pure water; the common ion lowers it greatly.)
Q48Nitrosyl chloride decomposes according to 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g). At equilibrium in a 2.0 L flask there are 0.30 mol NOCl, 0.40 mol NO, and 0.20 mol Cl2. What is the value of Kc?
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Correct: (A)
The concentrations are [NOCl] = 0.15 M, [NO] = 0.20 M, and [Cl2] = 0.10 M, so Kc = (0.20)^2(0.10) / (0.15)^2 = 0.0040 / 0.0225 = 0.18. (Option B results from using moles instead of concentrations.)
🧫 Unit 8: Acids & Bases
Q49What is the pH of a 0.0040 M solution of NaOH?
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Correct: (C)
NaOH is a strong base, so [OH^-] = 0.0040 M and pOH = -log(0.0040) = 2.40, giving pH = 14.00 – 2.40 = 11.60. (Option A is the pOH, not the pH.)
Q50A 0.50 M solution of a weak acid, HA, has a pH of 2.50. What is the value of Ka for HA?
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Correct: (B)
[H^+] = 10^-2.50 = 3.2 × 10^-3 M, so Ka = (3.2 × 10^-3)^2 / (0.50 – 0.0032) = 1.0 × 10^-5 / 0.497 = 2.0 × 10^-5. (Option C is [H^+]^2 alone, before dividing by the acid concentration.)
Q51A buffer is prepared with 0.40 M NH3 and 0.20 M NH4Cl. The pKb of NH3 is 4.75. What is the pH of this buffer?
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Correct: (D)
pOH = pKb + log([acid] / [base]) = 4.75 + log(0.20 / 0.40) = 4.75 – 0.30 = 4.45, so pH = 14.00 – 4.45 = 9.55. (Option A is the pOH, not the pH.)
Q52A student mixes 50.0 mL of 0.10 M HCl with 25.0 mL of 0.10 M NaOH. What is the pH of the resulting solution?
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Correct: (A)
The mixture contains 0.0050 mol H^+ and 0.0025 mol OH^-, leaving 0.0025 mol H^+ in 75.0 mL, so [H^+] = 0.033 M and pH = -log(0.033) = 1.48.
Q53What is the conjugate acid of HPO4^2-?
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Correct: (C)
A conjugate acid is formed by adding exactly one proton: HPO4^2- + H^+ gives H2PO4^-. PO4^3- is its conjugate base, and H3PO4 differs by two protons.
Q54Hydrogen sulfide is a diprotic acid with Ka1 = 9.0 × 10^-8 and Ka2 = 1.0 × 10^-19. What is the pH of a 0.20 M H2S solution?
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Correct: (B)
Because Ka2 is far smaller than Ka1, only the first ionization matters: [H^+] = (Ka1 × C)^1/2 = (9.0 × 10^-8 × 0.20)^1/2 = (1.8 × 10^-8)^1/2 = 1.3 × 10^-4 M, so pH = -log(1.3 × 10^-4) = 3.87. (Option C is pKa1, not the pH.)
🔋 Unit 9: Applications of Thermodynamics
Q55A reaction has ΔH° = +48 kJ/mol and ΔS° = +160 J/(mol K). What is ΔG° for the reaction at 400 K, and is it spontaneous under standard conditions at that temperature?
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Correct: (D)
ΔG° = ΔH° − TΔS° = +48 kJ/mol − (400 K)(0.160 kJ/(mol K)) = 48 − 64 = −16 kJ/mol, and a negative ΔG° means spontaneous.
Q56Which of the following processes has a negative ΔS (a decrease in entropy)?
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Correct: (A)
Three moles of gas form two moles of gas, so disorder decreases; the other choices produce a gas or a liquid from a more ordered phase, increasing entropy.
Q57A galvanic cell is based on the reaction Mg(s) + Fe^2+(aq) → Mg^2+(aq) + Fe(s). Given Fe^2+(aq) + 2 e− → Fe(s), E° = −0.44 V, and Mg^2+(aq) + 2 e− → Mg(s), E° = −2.37 V, what is E°cell?
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Correct: (C)
E°cell = E°(cathode) − E°(anode) = −0.44 V − (−2.37 V) = +1.93 V.
Q58Silver is plated from a solution of Ag+(aq) by the half-reaction Ag+(aq) + e− → Ag(s). A current of 1.93 A is applied for 1,000 s. What mass of Ag is deposited? The molar mass of Ag is 107.87 g/mol and F = 96,485 C/mol e−.
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Correct: (B)
Charge = It = (1.93 A)(1,000 s) = 1,930 C, which is 1,930 / 96,485 = 0.0200 mol e−. Because Ag+ takes 1 e−, this gives 0.0200 mol Ag, and (0.0200 mol)(107.87 g/mol) = 2.16 g Ag.
Q59A galvanic cell reaction transfers 2 mol of electrons per mole of reaction and has E°cell = +1.56 V. What is ΔG° for the reaction? Use F = 96,485 C/mol e−.
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Correct: (D)
ΔG° = −nFE° = −(2)(96,485 C/mol)(1.56 V) = −301,000 J/mol = −301 kJ/mol.
Q60A reaction has ΔH° = +180 kJ/mol and ΔS° = +600 J/(mol K). Assuming ΔH° and ΔS° do not change with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous, and above or below that temperature is it spontaneous?
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Correct: (C)
At the crossover, ΔG° = 0, so T = ΔH° / ΔS° = 180,000 J/mol / 600 J/(mol K) = 300 K. Because ΔH° and ΔS° are both positive, the −TΔS° term dominates above 300 K, making ΔG° negative there.
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