ACS General Chemistry Practice Test 13
60 original exam-style multiple-choice questions with answers & explanations — covering every major topic tested on the ACS General Chemistry exam.
Note: These are original practice questions written in the ACS exam style — not reproduced exam questions. Use them to test your readiness for the real ACS General Chemistry exam.
📱 How to use: Try each question first, then click “Show Answer” to check. Covers: atomic structure, bonding, stoichiometry, gases, thermochemistry, kinetics, equilibrium, acids & bases, electrochemistry and more.
⚛ Atomic Structure & Periodic Trends
Q1A sample of copper contains two isotopes: Cu-63 (62.930 amu, 69.15% abundance) and Cu-65 (64.928 amu, 30.85% abundance). What is the atomic mass of copper?
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Correct: (B)
Weighted average: (0.6915 × 62.930) + (0.3085 × 64.928) = 43.52 + 20.03 = 63.55 amu.
Q2Which of the following orders correctly lists the elements from smallest to largest atomic radius?
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Correct: (A)
Atomic radius decreases left to right across a period due to increasing effective nuclear charge; the order across Period 2 is C > N > O > F, so smallest to largest is F < O < N < C.
Q3Which electron configuration corresponds to the element with the highest first ionization energy?
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Correct: (D)
The [Ne] 3s2 3p6 configuration is argon, a noble gas with a complete octet and the highest effective nuclear charge in its period, giving the greatest first ionization energy.
Q4Which species is expected to release the most energy when it gains one electron (highest electron affinity)?
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Correct: (B)
Although F is more electronegative, Cl has the highest electron affinity because F's small 2p orbitals cause greater electron-electron repulsion when the added electron enters.
Q5In which of the following pairs does the second species have a SMALLER radius than the first?
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Correct: (A)
A cation loses its outermost shell's electrons and experiences greater effective nuclear charge per remaining electron, so Na+ is smaller than Na. In all other pairs, the second species gains electrons or loses nuclear pull, making it larger.
Q6Which statement correctly explains why the atomic radius of Mg is smaller than that of Na?
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Correct: (B)
Na and Mg are in the same period with electrons in the same shell (n = 3), but Mg's extra protons increase the effective nuclear charge, pulling the electron cloud closer and shrinking the radius.
🔬 Chemical Bonding & Molecular Geometry
Q7What is the total number of valence electrons that must be accounted for in the Lewis structure of the phosphate ion, PO4 3-?
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Correct: (B)
P contributes 5, each O contributes 6 (24 total), and the 3- charge adds 3 electrons: 5 + 24 + 3 = 32 valence electrons.
Q8What is the molecular geometry of the XeF2 molecule?
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Correct: (B)
XeF2 has 5 electron domains around Xe (2 bonding pairs, 3 lone pairs); the 3 lone pairs occupy equatorial positions of a trigonal bipyramid, forcing the two F atoms into axial positions 180° apart, giving linear geometry.
Q9Which of the following bonds is the MOST polar?
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Correct: (C)
Bond polarity depends on electronegativity difference; oxygen (3.44) has the largest difference from H (2.20) among the choices (ΔEN = 1.24), making O-H the most polar.
Q10What is the hybridization of the nitrogen atom in the NO3- ion?
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Correct: (B)
N in NO3- is surrounded by 3 electron domains (3 sigma bonds to O, with the pi bond in resonance structures not counted), requiring 3 hybrid orbitals: sp2.
Q11Using molecular orbital theory, what is the bond order of the O2- ion?
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Correct: (B)
O2- has 13 valence electrons: (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)3, giving 8 bonding and 5 antibonding electrons; bond order = (8 – 5)/2 = 1.5.
Q12Which substance exhibits hydrogen bonding as its strongest intermolecular force?
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Correct: (C)
NH3 has H bonded directly to N, a highly electronegative atom with a lone pair, enabling hydrogen bonding. CH4 and CO2 are nonpolar (dispersion only), and CH3F has dipole-dipole forces but no H bonded to N, O, or F.
Q13Which statement correctly distinguishes intermolecular forces from intramolecular forces?
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Correct: (B)
Intermolecular forces (dispersion, dipole-dipole, hydrogen bonding) act between distinct molecules and are weaker than the intramolecular covalent or ionic bonds within a molecule; boiling point depends on intermolecular forces, and molecular shape depends on intramolecular bonding/VSEPR.
🧪 Stoichiometry & Aqueous Reactions
Q14How many moles of oxygen atoms are present in 2.50 mol of Al2(SO4)3?
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Correct: (C)
Each formula unit of Al2(SO4)3 contains 3 × 4 = 12 oxygen atoms, so 2.50 mol × 12 = 30.0 mol of oxygen atoms.
Q15Consider the reaction: 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). If 5.40 g of Al reacts with 14.2 g of Cl2, what is the limiting reactant?
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Correct: (B)
Moles Al = 5.40/26.98 = 0.200 mol; moles Cl2 = 14.2/70.90 = 0.200 mol. Required ratio Al:Cl2 = 2:3, so 0.200 mol Al needs 0.300 mol Cl2, but only 0.200 mol Cl2 is available; Cl2 is limiting.
Q16A student reacts 25.0 g of CaCO3 with excess HCl and isolates 9.50 g of CaCl2. What is the percent yield?
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Correct: (B)
Moles CaCO3 = 25.0/100.09 = 0.2498 mol; theoretical CaCl2 = 0.2498 mol × 110.98 g/mol = 27.7 g. Percent yield = (9.50/27.7) × 100% = 34.3%, approximately 34.4%.
Q17What volume of 0.150 M NaOH solution is required to neutralize 40.0 mL of 0.200 M H2SO4?
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Correct: (C)
Moles H2SO4 = 0.0400 L × 0.200 M = 0.00800 mol, giving 0.0160 mol H+ (diprotic). Moles NaOH needed = 0.0160 mol; V = 0.0160/0.150 = 0.107 L = 107 mL.
Q18A 25.0 mL sample of 0.500 M KNO3 is diluted to a final volume of 125 mL with water. What is the concentration of the diluted solution?
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Correct: (A)
Using M1V1 = M2V2: M2 = (0.500 M × 25.0 mL)/125 mL = 0.100 M.
Q19When aqueous solutions of Pb(NO3)2 and KI are mixed, a yellow precipitate forms. What is the net ionic equation for this reaction?
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Correct: (B)
The precipitate is PbI2; K+ and NO3- are spectator ions that appear unchanged on both sides, so the net ionic equation is Pb2+(aq) + 2 I-(aq) → PbI2(s).
Q20In the balanced redox reaction (acidic solution): MnO4- + C2O4 2- + H+ → Mn2+ + CO2 + H2O, what is the coefficient of H+?
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Correct: (B)
Balancing by half-reactions: MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O and C2O4 2- → 2 CO2 + 2 e-. Multiplying to equalize 10 e- gives 2 MnO4- + 5 C2O4 2- + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2O; the H+ coefficient is 16.
🌀 Gases & Intermolecular Forces
Q21A 2.00 L rigid container at 27°C holds a mixture of 0.40 mol N2 and 0.60 mol He. A small leak develops and 0.20 mol of gas escapes before it is sealed. Assuming the escaping gas has the same mole fraction as the original mixture, what is the final pressure in the container?
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Correct: (C)
0.80 mol of gas remains with unchanged composition. P = nRT/V = (0.80 mol)(0.0821 L·atm/mol·K)(300 K)/(2.00 L) = 9.9 atm.
Q22Which gas sample would show the greatest deviation from ideal behavior under identical conditions of high pressure and low temperature?
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Correct: (D)
NH3 has the strongest intermolecular attractions (hydrogen bonding) of the choices, so it deviates most from ideality at high pressure and low temperature.
Q23A balloon filled with an unknown gas effuses through a small pinhole 2.83 times faster than krypton gas under identical conditions. What is the molar mass of the unknown gas?
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Correct: (B)
Graham's law: 2.83 = √(83.8/M), so M = 83.8/(2.83)² = 83.8/8.01 = 10.5 g/mol.
Q24Rank the following substances in order of increasing boiling point: (I) CH3CH2CH2CH3, (II) CH3CH2CH2OH, (III) CH3OCH2CH3.
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Correct: (A)
Butane (I) has only London dispersion forces; dimethyl ether (III) adds dipole-dipole forces; 1-propanol (II) hydrogen bonds, giving the highest boiling point.
Q25Two identical flasks are connected by a stopcock. Flask A (1.0 L) contains O2 at 2.0 atm and flask B (3.0 L) contains N2 at 1.0 atm, both at the same temperature. The stopcock is opened and the gases mix. What is the partial pressure of O2 after mixing?
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Correct: (A)
Moles of O2 are unchanged, so P1V1 = P2V2: (2.0 atm)(1.0 L) = P(4.0 L), giving P = 0.50 atm.
🔥 Thermochemistry
Q26When 2.50 g of ethanol (C2H5OH, molar mass 46.1 g/mol) burns completely in a bomb calorimeter, the temperature of the calorimeter rises from 24.0°C to 31.5°C. The heat capacity of the calorimeter is 9.30 kJ/°C. What is the molar heat of combustion of ethanol?
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Correct: (B)
Heat released = CΔT = (9.30 kJ/°C)(7.5°C) = 69.75 kJ from 2.50/46.1 = 0.0542 mol, so ΔH_comb = −69.75/0.0542 = −1.29 × 10³ kJ/mol (negative because the reaction is exothermic).
Q27Given the following reactions:
(1) C(graphite) + O2(g) → CO2(g) ΔH = −394 kJ
(2) H2(g) + ½O2(g) → H2O(l) ΔH = −286 kJ
(3) C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(l) ΔH = −1562 kJ
What is ΔH for the reaction 2C(graphite) + 3H2(g) → C2H6(g)?
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Correct: (D)
By Hess's law, combine 2×(1) + 3×(2) − (3): ΔH = 2(−394) + 3(−286) − (−1562) = −788 − 858 + 1562 = −84 kJ.
Q28A 150.0 g sample of an unknown metal at 95.0°C is placed in 200.0 g of water at 22.0°C in an insulated container. The final equilibrium temperature is 27.4°C. What is the specific heat of the metal? (specific heat of water = 4.18 J/g·°C)
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Correct: (D)
Heat lost by metal = heat gained by water: (150.0)(c)(95.0 − 27.4) = (200.0)(4.18)(27.4 − 22.0), so c = 4514/10140 = 0.44 J/g·°C.
Q29Use bond enthalpies to estimate ΔH for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). Bond enthalpies (kJ/mol): C−H 414, O=O 498, C=O 799, O−H 463.
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Correct: (C)
Bonds broken: 4(414) + 2(498) = 2652 kJ. Bonds formed: 2(799) + 4(463) = 3450 kJ. ΔH = 2652 − 3450 = −798 kJ.
Q30Which process has a positive ΔH and a positive ΔS?
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Correct: (C)
Sublimation (solid → gas) requires heat input (ΔH > 0) and increases disorder (ΔS > 0); the other processes decrease entropy or release heat.
Q31Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), given ΔH_comb° = −728 kJ/mol for CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l); ΔHf°[CO2(g)] = −394 kJ/mol; ΔHf°[H2O(l)] = −286 kJ/mol.
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Correct: (A)
ΔH_comb = [ΔHf(CO2) + 2ΔHf(H2O)] − ΔHf(CH3OH), so −728 = [−394 + 2(−286)] − x, giving x = −966 + 728 = −238 kJ/mol.
⚡ Quantum Theory & Electronic Structure
Q32Light of wavelength 280 nm strikes a metal surface whose work function is 4.10 eV. What is the maximum kinetic energy of the ejected photoelectrons? (hc = 1240 eV·nm)
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Correct: (A)
Photon energy = hc/λ = 1240/280 = 4.43 eV, which exceeds the work function, so KE_max = 4.43 − 4.10 = 0.33 eV.
Q33Which set of quantum numbers (n, l, ml, ms) is NOT allowed for an electron in an atom?
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Correct: (B)
For l = 1, ml may only be −1, 0, or +1; ml = 2 is not permitted.
Q34The ground-state electron configuration of a neutral atom ends in …4s²3d⁶. Which element is it?
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Correct: (B)
[Ar]4s²3d⁶ corresponds to 26 total electrons, which is iron, Fe (Z = 26).
Q35Which statement about atomic orbitals is TRUE?
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Correct: (B)
A p orbital (l = 1) has exactly one angular (planar) node passing through the nucleus. A 3d orbital has 2 angular nodes and 0 radial nodes, and the n = 3 shell holds 9 orbitals.
⏱ Chemical Kinetics
Q36For the reaction 2NO(g) + O2(g) → 2NO2(g), the rate law is rate = k[NO]²[O2]. If the concentration of NO is tripled and the concentration of O2 is halved, by what factor does the reaction rate change?
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Correct: (C)
The rate changes by a factor of (3)² × (1/2) = 9 × 0.5 = 4.5.
Q37A first-order reaction has a half-life of 45 minutes. What fraction of the reactant remains after 3.0 hours?
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Correct: (C)
3.0 hours = 180 min, which is 180/45 = 4 half-lives, so the fraction remaining is (1/2)⁴ = 1/16.
Q38The rate constant of a reaction doubles when the temperature increases from 300 K to 310 K. What is the approximate activation energy?
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Correct: (D)
From the Arrhenius equation, ln 2 = (Ea/R)(1/300 − 1/310), so Ea = (0.693)(8.314 J/mol·K)/(1.075 × 10⁻⁴ K⁻¹) ≈ 5.4 × 10⁴ J/mol ≈ 53 kJ/mol.
Q39A proposed mechanism for a reaction is:
Step 1: A + B → C (slow)
Step 2: C + D → E + F (fast)
What is the rate law consistent with this mechanism?
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Correct: (D)
The slow step is rate-determining, so the rate law is determined by the reactants of Step 1: rate = k[A][B].
Q40For the second-order reaction 2HI(g) → H2(g) + I2(g), a plot of 1/[HI] versus time is linear with a slope of 0.0250 L/mol·s. If the initial [HI] is 0.400 M, what is [HI] after 180 s?
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Correct: (A)
Using the integrated second-order rate law, 1/[HI]t = kt + 1/[HI]0 = (0.0250)(180) + 1/0.400 = 4.50 + 2.50 = 7.00 M⁻¹, so [HI]t = 1/7.00 = 0.143 M.
Q41The first-order decomposition of a certain pesticide has a rate constant of 2.5 × 10^-3 s^-1 at 25°C and 1.1 × 10^-2 s^-1 at 50°C. What is the approximate activation energy for this reaction?
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Correct: (B)
Using ln(k2/k1) = (Ea/R)(1/T1 − 1/T2): ln(4.4) = 1.48, (1/298 − 1/323) = 2.60 × 10^-4 K^-1, so Ea = (8.314)(1.48)/(2.60 × 10^-4) ≈ 4.7 × 10^4 J/mol.
⇆ Chemical Equilibrium
Q42For the reaction A(g) ⇌ 2 B(g), Kc = 0.25 at a certain temperature. A 1.0 L vessel initially contains 0.50 mol of A and 0.20 mol of B. In which direction will the reaction proceed to reach equilibrium?
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Correct: (D)
Q = [B]^2/[A] = (0.20)^2/(0.50) = 0.080, which is less than Kc = 0.25, so the reaction proceeds forward to make more product.
Q43Consider the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), ΔH° = +178 kJ/mol. Which change will increase the equilibrium pressure of CO2?
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Correct: (A)
The reaction is endothermic, so raising the temperature shifts the equilibrium to the right, producing more CO2. Solids do not appear in the equilibrium expression, and at fixed T the equilibrium pressure of CO2 is independent of container volume.
Q44For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.045 at 250°C. A sealed flask initially contains 0.80 M PCl5 and no products. What is [PCl3] at equilibrium?
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Correct: (C)
With x = [PCl3] = [Cl2], x^2/(0.80 − x) = 0.045 gives x^2 + 0.045x − 0.036 = 0, so x = 0.17 M.
Q45At 700 K, a 2.0 L flask at equilibrium contains 0.30 mol SO2, 0.20 mol O2, and 0.90 mol SO3 for the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g). What is Kp at this temperature?
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Correct: (A)
Kc = (0.45)^2/[(0.15)^2(0.10)] = 90; Kp = Kc(RT)^Δn with Δn = −1 gives Kp = 90/[(0.0821)(700)] ≈ 1.6 atm^-1.
Q46For the reaction Cr2O7^2-(aq) + H2O(l) ⇌ 2 CrO4^2-(aq) + 2 H+(aq), the equilibrium-constant expression is
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Correct: (D)
Products over reactants, each raised to its stoichiometric coefficient; pure liquid water is omitted from the expression.
Q47A vessel containing 2 NO2(g) ⇌ N2O4(g) at equilibrium has its volume suddenly halved at constant temperature. After equilibrium is re-established, which statement is true?
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Correct: (B)
Halving the volume doubles the total pressure, shifting the equilibrium toward the side with fewer gas moles (N2O4); Kc depends only on temperature.
🧫 Acids, Bases & Buffers
Q48The Ka of propanoic acid (CH3CH2COOH) is 1.3 × 10^-5. What is the pH of a 0.10 M aqueous solution of propanoic acid?
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Correct: (B)
[H+] ≈ √(Ka · C) = √(1.3 × 10^-6) = 1.1 × 10^-3 M, so pH = 2.94.
Q4950.0 mL of 0.20 M HCN (Ka = 4.9 × 10^-10) is mixed with 50.0 mL of 0.30 M NaCN. What is the pH of the resulting solution?
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Correct: (A)
pH = pKa + log([CN-]/[HCN]) = 9.31 + log(0.30/0.20) = 9.31 + 0.18 = 9.49.
Q50At 25°C, Kb for pyridine (C5H5N) is 1.7 × 10^-9. What is Ka of the pyridinium ion (C5H5NH+)?
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Correct: (C)
For a conjugate pair, Ka · Kb = Kw, so Ka = 1.0 × 10^-14/1.7 × 10^-9 = 5.9 × 10^-6.
Q5125.0 mL of 0.20 M benzoic acid (Ka = 6.3 × 10^-5) is titrated with 0.20 M NaOH. What is the pH at the equivalence point?
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Correct: (B)
At equivalence, [C6H5COO-] = 0.10 M; Kb = Kw/Ka = 1.59 × 10^-10, [OH-] = √(1.59 × 10^-11) = 3.99 × 10^-6 M, giving pOH = 5.40 and pH = 8.60.
Q52A 0.50 L buffer contains 0.10 M CH3COOH and 0.10 M CH3COONa (Ka = 1.8 × 10^-5). After 0.010 mol of HCl is added to the buffer, what is the resulting pH? (Assume no volume change.)
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Correct: (D)
The added H+ converts 0.010 mol of acetate to acetic acid: [HA]/[A-] = 0.060/0.040, so pH = 4.74 + log(0.040/0.060) = 4.57.
Q53At 25°C, Kb for NH3 is 1.8 × 10^-5. What is the pH of a 0.10 M NH4Cl solution?
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Correct: (A)
Ka(NH4+) = Kw/Kb = 5.6 × 10^-10; [H+] ≈ √(5.6 × 10^-11) = 7.5 × 10^-6 M, so pH = 5.13.
🔋 Thermodynamics & Electrochemistry
Q54For which reaction is ΔS° expected to be negative?
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Correct: (A)
Only this reaction decreases the number of moles of gas (4 mol → 2 mol), reducing the disorder of the system.
Q55For the reaction 2 CO(g) + O2(g) → 2 CO2(g), ΔH° = −566 kJ/mol and ΔS° = −173 J/(mol·K). What is ΔG° at 298 K?
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Correct: (D)
ΔG° = ΔH° − TΔS° = −566 − (298)(−0.173) = −566 + 51.6 = −514 kJ/mol.
Q56What is E°cell for the galvanic cell Mg(s) | Mg2+(1 M) || Ag+(1 M) | Ag(s)? (E° Mg2+/Mg = −2.37 V; E° Ag+/Ag = +0.80 V)
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Correct: (B)
E°cell = E°cathode − E°anode = (+0.80) − (−2.37) = 3.17 V.
Q57For the cell Zn(s) | Zn2+(0.010 M) || Cu2+(1.0 M) | Cu(s) at 25°C, E°cell = 1.10 V. What is the measured cell potential?
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Correct: (C)
E = E° − (0.0592/2)log([Zn2+]/[Cu2+]) = 1.10 − (0.0296)(−2.00) = 1.16 V.
Q58How many grams of aluminum are deposited at the cathode when a current of 3.00 A flows for 2.00 h through molten Al2O3? (Molar mass of Al = 27.0 g/mol)
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Correct: (A)
Charge = (3.00)(7200) = 21,600 C = 0.224 mol e-; Al3+ requires 3 e- per atom, so (0.224/3)(27.0) = 2.01 g of Al.
☢ Nuclear & Descriptive Chemistry
Q59Phosphorus-32 has a half-life of 14.3 days. How long will it take for an 80.0 mg sample of phosphorus-32 to decay to 5.00 mg?
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Correct: (B)
80.0 → 5.00 mg is a factor of 16 = 2^4, so 4 half-lives elapse: 4 × 14.3 = 57.2 days.
Q60Which of the following oxides is the most acidic?
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Correct: (D)
Across Period 3, oxides become more acidic with increasing electronegativity of the element; SiO2 is an acidic oxide while Na2O and MgO are basic and Al2O3 is amphoteric.
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