AP Chemistry Practice Test 1
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 [Kr] 5s^2 4d^10 5p^2. What is the identity of the element?
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Correct: (B)
The configuration contains 36 + 2 + 10 + 2 = 50 electrons, so the neutral atom has atomic number 50, which is Sn.
Q2The photoelectron spectrum of an element shows four peaks, in order of decreasing binding energy, corresponding to 2, 2, 6, and 2 electrons. What is the element?
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Correct: (C)
The peaks correspond to 1s^2 2s^2 2p^6 3s^2, a total of 12 electrons, which is Mg.
Q3The ions Ca^2+, K+, Cl-, and S^2- are all isoelectronic with Ar. Which ion has the smallest radius?
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Correct: (A)
All have 18 electrons, but Ca^2+ has the greatest nuclear charge (20 protons), so it pulls the electron cloud in most strongly and is smallest.
Q4A sample of copper consists of two isotopes: Cu-63 with mass 62.93 amu and percent abundance 69.17%, and Cu-65 with mass 64.93 amu and percent abundance 30.83%. What is the average atomic mass of copper in the sample?
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Correct: (D)
(62.93 x 0.6917) + (64.93 x 0.3083) = 43.53 + 20.02 = 63.55 amu.
Q5The average atomic mass of neon is 20.18 amu. Neon is composed mainly of two isotopes with approximate masses of 20 amu and 22 amu. What is the percent abundance of the isotope with mass 20 amu?
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Correct: (B)
Let x be the fraction of the 20 amu isotope: 20x + 22(1 – x) = 20.18, so 22 – 2x = 20.18 and x = 0.91, or 91%.
Q6How many unpaired electrons are present in a ground-state Mn^2+ ion?
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Correct: (A)
Mn is [Ar] 4s^2 3d^5, and Mn^2+ loses the two 4s electrons first, leaving 3d^5 with one electron in each d orbital, for 5 unpaired electrons.
Q7Which element has the highest second ionization energy?
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Correct: (C)
Na+ already has the stable electron configuration of Ne, so removing a second electron from Na requires breaking into a filled core shell and takes far more energy.
🔗 Unit 2: Molecular & Ionic Compound Structure
Q8Which ionic compound has the greatest lattice energy?
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Correct: (D)
Lattice energy increases with ionic charge and decreases with ionic size. MgO combines 2+ and 2- charges with relatively small ions, giving the greatest lattice energy.
Q9In the Lewis structure of XeF4, how many lone pairs of electrons are on the central Xe atom?
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Correct: (B)
Xe contributes 8 valence electrons. Four are used in bonds to F atoms, leaving 4 electrons, which form 2 lone pairs on Xe.
Q10What is the formal charge on the nitrogen atom in NH4+?
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Correct: (A)
Formal charge = 5 valence electrons – (0 nonbonding electrons + 4 bonding electrons counted as 4) = 5 – 4 = +1.
Q11What is the molecular geometry of SF4?
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Correct: (C)
S in SF4 has 5 electron domains (4 bonds and 1 lone pair), giving a trigonal bipyramidal electron geometry and a seesaw molecular geometry.
Q12What is the hybridization of the sulfur atom in SF6?
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Correct: (D)
S in SF6 forms 6 bonds with no lone pairs, requiring 6 hybrid orbitals, which corresponds to sp^3d^2 hybridization.
Q13Which bond has the shortest bond length?
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Correct: (A)
Bond length decreases as bond order increases because higher bond order pulls the atoms closer together, so a triple bond is shorter than double and single bonds.
Q14What is the average N-O bond order in the nitrate ion, NO3-?
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Correct: (B)
Nitrate has 3 resonance structures with a total of 4 N-O bonds (one double and two single) shared over 3 positions, so the average bond order is 4/3 = 1.33.
💧 Unit 3: Intermolecular Forces & Properties
Q15Which substance can form hydrogen bonds with itself in the pure liquid?
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Correct: (A)
CH3NH2 has H atoms bonded directly to N, so it can both donate and accept hydrogen bonds; the others have no H bonded to N, O, or F.
Q16Which compound has the highest normal boiling point?
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Correct: (C)
CH3CH2OH forms hydrogen bonds, which are much stronger than the dipole-dipole and dispersion forces in the other compounds of similar size.
Q17Which solute is most soluble in hexane, C6H14?
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Correct: (D)
I2 is nonpolar, like hexane, so dispersion forces between them favor dissolving; the ionic and strongly polar solutes are favored by polar solvents instead.
Q18How much energy is required to vaporize 36.0 g of H2O(l) at 100°C? The molar enthalpy of vaporization of water is 40.7 kJ/mol.
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Correct: (B)
36.0 g / 18.0 g/mol = 2.00 mol, and 2.00 mol × 40.7 kJ/mol = 81.4 kJ.
Q19Which substance is a network covalent solid?
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Correct: (A)
In SiO2 (quartz) each Si is covalently bonded into a continuous three-dimensional network; ice is molecular, NaCl is ionic, and Cu is metallic.
Q20A 250.0 mL sample of 4.00 M KNO3 is diluted with water to a final volume of 1000. mL. What is the molarity of the diluted solution?
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Correct: (C)
Moles of solute are unchanged: 4.00 M × 0.2500 L = 1.00 mol, and 1.00 mol / 1.000 L = 1.00 M.
⚖ Unit 4: Chemical Reactions
Q21When the equation C3H8 + O2 → CO2 + H2O is balanced with the smallest whole-number coefficients, what is the coefficient of O2?
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Correct: (D)
The balanced equation is C3H8 + 5 O2 → 3 CO2 + 4 H2O, which balances C, H, and O.
Q22In the reaction Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s), which species is oxidized?
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Correct: (B)
Zn goes from oxidation number 0 to +2, losing electrons, so Zn is oxidized (and acts as the reducing agent).
Q23Solutions of BaCl2 and Na2SO4 are mixed and a white precipitate forms. What is the net ionic equation for the reaction?
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Correct: (A)
BaSO4 is the insoluble product; Na+ and Cl- are spectator ions and NaCl remains dissolved.
Q24Ammonia is made by N2(g) + 3 H2(g) → 2 NH3(g). If 6.0 mol of H2 reacts completely with excess N2, how many moles of NH3 are produced?
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Correct: (C)
6.0 mol H2 × (2 mol NH3 / 3 mol H2) = 4.0 mol NH3.
Q25Aluminum chloride is made by 2 Al + 3 Cl2 → 2 AlCl3. A mixture contains 4.0 mol of Al and 4.5 mol of Cl2. What is the maximum number of moles of AlCl3 that can form?
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Correct: (D)
4.0 mol Al would need 6.0 mol Cl2, so Cl2 is limiting; 4.5 mol Cl2 × (2 mol AlCl3 / 3 mol Cl2) = 3.0 mol AlCl3.
Q26A student carries out a preparation in which the theoretical yield of product is 25.0 g. She isolates 20.0 g of pure product. What is the percent yield?
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Correct: (B)
Percent yield = (20.0 g / 25.0 g) × 100% = 80.0%.
Q27A 25.00 mL sample of HCl(aq) of unknown concentration is titrated to the endpoint with 20.00 mL of 0.100 M NaOH(aq). The reaction is 1:1. What is the concentration of the HCl?
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Correct: (A)
Moles NaOH = 0.100 M × 0.02000 L = 0.00200 mol = moles HCl, so [HCl] = 0.00200 mol / 0.02500 L = 0.0800 M.
⏱ Unit 5: Kinetics
Q28For the reaction A + B → products, the following initial-rate data were collected. Experiment 1: [A] = 0.10 M, [B] = 0.10 M, initial rate = 0.020 M/s. Experiment 2: [A] = 0.20 M, [B] = 0.10 M, initial rate = 0.080 M/s. Experiment 3: [A] = 0.10 M, [B] = 0.20 M, initial rate = 0.040 M/s. What is the rate law for the reaction?
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Correct: (B)
Doubling [A] at constant [B] quadruples the rate (0.020 to 0.080), so the reaction is second order in A. Doubling [B] at constant [A] doubles the rate (0.020 to 0.040), so it is first order in B.
Q29A compound decomposes by a first-order process with a rate constant k = 4.62 × 10^-3 s^-1. What is the half-life of the decomposition?
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Correct: (C)
For a first-order reaction, t1/2 = 0.693 / k = 0.693 / (4.62 × 10^-3 s^-1) = 150 s.
Q30A proposed mechanism for the overall reaction A + 2 B → 2 D + E is: Step 1 (slow): A + B → C + D. Step 2 (fast): C + B → D + E. Which statement correctly identifies the intermediate and the rate law predicted by this mechanism?
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Correct: (A)
C is produced in Step 1 and consumed in Step 2, so it is an intermediate. The rate law is determined by the slow step, which involves A and B, giving Rate = k[A][B].
Q31Which of the following best explains, in terms of collision theory, why increasing the temperature increases the rate of a chemical reaction?
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Correct: (D)
A higher temperature raises the average kinetic energy, so a greater proportion of molecular collisions are energetic enough to exceed the activation energy and form the activated complex.
Q32For a particular reaction, the potential energy of the reactants is 40 kJ/mol, the potential energy of the activated complex is 110 kJ/mol, and the potential energy of the products is 15 kJ/mol. What is the activation energy for the forward reaction?
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Correct: (B)
The forward activation energy is the energy of the activated complex minus the energy of the reactants: 110 − 40 = 70 kJ/mol.
Q33Which statement about a catalyst is correct?
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Correct: (C)
A catalyst speeds up the reaction by providing a different mechanism with a lower activation energy. It is regenerated, does not change ΔH, and does not change the equilibrium position.
Q34A reaction is second order in reactant A, with k = 0.50 M^-1 s^-1. If the initial concentration of A is 0.40 M, what is the concentration of A after 10 s?
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Correct: (A)
Using the second-order integrated rate law, 1/[A] = 1/[A]0 + kt = 1/0.40 + (0.50)(10) = 2.5 + 5.0 = 7.5 M^-1, so [A] = 1/7.5 = 0.133 M.
🔥 Unit 6: Thermodynamics
Q35For the reaction 2 H2(g) + O2(g) → 2 H2O(l), ΔH = −572 kJ for the equation as written. How much heat is released when 0.50 mol of H2(g) reacts completely with excess O2(g)?
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Correct: (D)
The equation as written consumes 2 mol of H2 and releases 572 kJ. Therefore 0.50 mol of H2 releases 572 × (0.50 / 2) = 143 kJ.
Q36Given the following equations: (1) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g), ΔH = −460 kJ. (2) MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O(l), ΔH = −155 kJ. (3) H2(g) + 1/2 O2(g) → H2O(l), ΔH = −286 kJ. What is ΔH for the reaction Mg(s) + 1/2 O2(g) → MgO(s)?
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Correct: (B)
By Hess's law, combine equation (1), the reverse of equation (2), and equation (3): ΔH = −460 + 155 − 286 = −591 kJ.
Q37A student mixes 50.0 mL of 1.0 M HCl(aq) with 50.0 mL of 1.0 M NaOH(aq) in a coffee-cup calorimeter. The temperature of the resulting 100.0 g of solution rises by 6.8 °C. Assuming the specific heat capacity of the solution is 4.18 J/(g °C), what is the molar enthalpy of neutralization?
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Correct: (A)
Heat absorbed by the solution is q = mcΔT = (100.0)(4.18)(6.8) = 2842 J. The reaction forms 0.0500 mol of water, so ΔH = −2842 J / 0.0500 mol = −56,800 J/mol = −56.8 kJ/mol. The value is negative because neutralization is exothermic.
Q38Using the bond enthalpies H−H = 436 kJ/mol, Cl−Cl = 242 kJ/mol, and H−Cl = 431 kJ/mol, estimate ΔH for the reaction H2(g) + Cl2(g) → 2 HCl(g).
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Correct: (C)
Energy is required to break bonds and released when bonds form: ΔH = (436 + 242) − 2(431) = 678 − 862 = −184 kJ.
Q39Which of the following processes results in an increase in the entropy of the system?
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Correct: (D)
The decomposition produces a gas from a solid, greatly increasing the number of freely moving particles and the disorder of the system. Freezing, condensing, and compressing all decrease entropy.
Q40Given the standard enthalpies of formation: NH3(g) = −46 kJ/mol, NO(g) = +90 kJ/mol, H2O(g) = −242 kJ/mol, and O2(g) = 0 kJ/mol, calculate ΔH for the reaction 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g).
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Correct: (B)
Products: 4(90) + 6(−242) = 360 − 1452 = −1092 kJ. Reactants: 4(−46) + 5(0) = −184 kJ. ΔH = −1092 − (−184) = −908 kJ.
Q41Which of the following substances has a standard enthalpy of formation, ΔHf°, equal to zero?
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Correct: (A)
By definition, the standard enthalpy of formation of an element in its standard state, such as O2(g), is zero. Compounds and non-standard forms of elements have nonzero values.
⇆ Unit 7: Equilibrium
Q42For the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g), Kc = 85 at 1000 K. What is the value of Kp for this reaction at 1000 K? (R = 0.0821 L·atm/mol·K)
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Correct: (A)
Kp = Kc(RT)^Δn with Δn = 2 – 3 = -1, so Kp = 85 / (0.0821 × 1000) = 85 / 82.1 = 1.0.
Q43Hydrogen gas and iodine vapor react according to H2(g) + I2(g) ⇌ 2 HI(g), for which Kc = 54.3 at 430°C. A flask is filled with 0.50 M H2 and 0.50 M I2 and no HI. What is the equilibrium concentration of HI?
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Correct: (C)
If x mol/L of H2 reacts, [HI] = 2x and Kc = (2x)^2 / (0.50 – x)^2 = 54.3. Taking the square root gives 2x / (0.50 – x) = 7.37, so x = 0.39 and [HI] = 2x = 0.79 M.
Q44Ammonia is produced by N2(g) + 3 H2(g) ⇌ 2 NH3(g), for which ΔH = -92 kJ/mol. Which change will increase the amount of NH3 present at equilibrium?
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Correct: (D)
Increasing the pressure shifts the equilibrium toward the side with fewer moles of gas, which is the product side (2 mol versus 4 mol). A catalyst does not change the equilibrium position, and heating an exothermic reaction shifts it toward reactants.
Q45The Ksp of CaF2 is 3.9 × 10^-11 at 25°C. What is the molar solubility of CaF2 in pure water at this temperature?
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Correct: (B)
For CaF2(s) ⇌ Ca^2+(aq) + 2 F^-(aq), Ksp = (s)(2s)^2 = 4s^3 = 3.9 × 10^-11, so s^3 = 9.8 × 10^-12 and s = 2.1 × 10^-4 M. (Option D is [F^-] = 2s, not s.)
Q46For the reaction CO(g) + H2O(g) ⇌ CO2(g) + H2(g), Kc = 5.0 at a certain temperature. At one moment the concentrations are [CO] = 0.20 M, [H2O] = 0.20 M, [CO2] = 0.50 M, and [H2] = 0.50 M. Which statement correctly describes the system?
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Correct: (C)
Q = (0.50)(0.50) / (0.20)(0.20) = 6.25. Since Q = 6.25 is greater than Kc = 5.0, the reaction shifts toward reactants until Q decreases to Kc.
Q47The Ksp of PbCl2 is 1.6 × 10^-5 at 25°C. What is the molar solubility of PbCl2 in a 0.10 M NaCl solution?
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Correct: (A)
With the common ion Cl^- already at 0.10 M, Ksp = (s)(0.10 + 2s)^2 is approximated as s(0.10)^2 = 1.6 × 10^-5, giving s = 1.6 × 10^-3 M. (Option B is the solubility in pure water; the common ion lowers it.)
Q48Calcium carbonate decomposes according to CaCO3(s) ⇌ CaO(s) + CO2(g). Which expression gives Kc for this reaction?
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Correct: (D)
Pure solids do not appear in equilibrium constant expressions, so only the gas remains and Kc = [CO2].
🧫 Unit 8: Acids & Bases
Q49What is the pH of a 0.025 M solution of HCl?
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Correct: (B)
HCl is a strong acid, so [H^+] = 0.025 M and pH = -log(0.025) = 1.60.
Q50What is the pH of a 0.10 M solution of acetic acid, CH3COOH, for which Ka = 1.8 × 10^-5?
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Correct: (D)
[H^+] = (Ka × C)^1/2 = (1.8 × 10^-5 × 0.10)^1/2 = (1.8 × 10^-6)^1/2 = 1.3 × 10^-3 M, so pH = -log(1.3 × 10^-3) = 2.87.
Q51A buffer is prepared with 0.25 M CH3COOH and 0.15 M CH3COONa. The pKa of CH3COOH is 4.74. What is the pH of this buffer?
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Correct: (A)
pH = pKa + log([base] / [acid]) = 4.74 + log(0.15 / 0.25) = 4.74 + log(0.60) = 4.74 – 0.22 = 4.52.
Q52A 50.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HCl. The Kb of NH3 is 1.8 × 10^-5. What is the pH at the half-equivalence point of the titration?
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Correct: (C)
At the half-equivalence point, [NH3] = [NH4^+], so pOH = pKb = 4.75 and pH = 14.00 – 4.75 = 9.25.
Q53What is the conjugate base of H2PO4^-?
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Correct: (B)
A conjugate base is formed by removing exactly one proton: H2PO4^- – H^+ gives HPO4^2-. H3PO4 is its conjugate acid, and PO4^3- differs by two protons.
Q54Sulfuric acid is completely ionized in its first step, and Ka2 = 1.2 × 10^-2 for HSO4^-. What is the pH of a 0.10 M H2SO4 solution?
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Correct: (A)
The first ionization gives 0.10 M H^+ and 0.10 M HSO4^-. Solving x(0.10 + x) / (0.10 – x) = 0.012 for the second ionization gives x = 0.0099 M, so [H^+] = 0.110 M and pH = -log(0.110) = 0.96. (Option C wrongly assumes both protons ionize completely; option B ignores the second ionization.)
🔋 Unit 9: Applications of Thermodynamics
Q55A reaction has ΔH° = −120 kJ/mol and ΔS° = −250 J/(mol K). What is ΔG° for the reaction at 298 K, and is it spontaneous under standard conditions?
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Correct: (B)
ΔG° = ΔH° − TΔS° = −120 kJ/mol − (298 K)(−0.250 kJ/(mol K)) = −120 + 74.5 = −45.5 kJ/mol, and a negative ΔG° means spontaneous.
Q56Which of the following reactions has a positive ΔS° (an increase in entropy) at constant temperature?
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Correct: (C)
A solid produces a gas, so the products are much more disordered; the other choices form a more ordered solid or reduce the number of gas molecules.
Q57A galvanic cell is based on the reaction Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s). Given Cu^2+(aq) + 2 e− → Cu(s), E° = +0.34 V, and Zn^2+(aq) + 2 e− → Zn(s), E° = −0.76 V, what is E°cell?
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Correct: (A)
E°cell = E°(cathode) − E°(anode) = 0.34 V − (−0.76 V) = +1.10 V. Standard potentials are not multiplied by the number of electrons.
Q58A galvanic cell reaction transfers 3 mol of electrons per mole of reaction and has E°cell = +0.85 V. What is ΔG° for the reaction? Use F = 96,485 C/mol e−.
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Correct: (D)
ΔG° = −nFE° = −(3)(96,485 C/mol)(0.85 V) = −246,000 J/mol = −246 kJ/mol.
Q59A current of 5.00 A is passed through a Cu^2+(aq) solution for 965 s, depositing Cu(s) by Cu^2+(aq) + 2 e− → Cu(s). What mass of Cu is deposited? The molar mass of Cu is 63.55 g/mol and F = 96,485 C/mol e−.
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Correct: (B)
Charge = It = (5.00 A)(965 s) = 4,825 C, which is 4,825 / 96,485 = 0.0500 mol e−. This gives 0.0250 mol Cu, and (0.0250 mol)(63.55 g/mol) = 1.59 g Cu.
Q60A redox reaction at 298 K has E°cell = +0.56 V. Which statement about this reaction under standard conditions is correct?
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Correct: (A)
A positive E°cell gives a negative ΔG° through ΔG° = −nFE°, and a negative ΔG° corresponds to K greater than 1, so products are favored at equilibrium.
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