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AP Chemistry

AP Chemistry Practice Test: Practice Test 11

Practice Test 11 for AP Chemistry: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

In oxygen gas, O2\text{O}_2O2​, two identical oxygen atoms share electrons equally. What type of bond is present in O2\text{O}_2O2​?

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Question 1

In oxygen gas, O2\text{O}_2O2​, two identical oxygen atoms share electrons equally. What type of bond is present in O2\text{O}_2O2​?

  1. Nonpolar covalent (correct answer)
  2. Polar covalent
  3. Metallic
  4. Dipole–dipole
  5. Ionic

Explanation: This question tests the ability to identify the type of chemical bond in diatomic oxygen. In O₂, identical oxygen atoms share electrons equally with ΔEN = 0, forming a nonpolar covalent double bond. No charge separation occurs due to symmetry. Nonpolar covalent is standard for O₂. A tempting distractor is polar covalent, but it is incorrect for zero ΔEN, misconceptions stem from oxygen's high electronegativity in other contexts. Confirm nonpolar covalent in homonuclear diatomics by equal electronegativities.

Question 2

When heated, solid potassium chlorate decomposes to form solid potassium chloride and oxygen gas.

2KClO3(s)→2KCl(s)+3O2(g)2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g)2KClO3​(s)→2KCl(s)+3O2​(g)

Which statement best classifies this reaction?

  1. It is an acid-base reaction because potassium chlorate is a salt of a strong base and a strong acid.
  2. It is an oxidation-reduction reaction because the oxidation states of both chlorine and oxygen change. (correct answer)
  3. It is a precipitation reaction because one of the reactants is a solid.
  4. It is not an oxidation-reduction reaction because no aqueous ions are involved in the process.

Explanation: This decomposition is an oxidation-reduction reaction. In KClO3KClO_3KClO3​, the oxidation state of Cl is +5 and O is -2. In the products, the oxidation state of Cl in KClKClKCl is -1 (reduction), and the oxidation state of O in O2O_2O2​ is 0 (oxidation). Since the oxidation states of multiple elements change, it is a redox reaction.

Question 3

A student has 0.500 mol0.500\ \mathrm{mol}0.500 mol of pure ammonia, NH3(s)\mathrm{NH_3(s)}NH3​(s). The molar mass of NH3\mathrm{NH_3}NH3​ is 17.0 g mol−117.0\ \mathrm{g\ mol^{-1}}17.0 g mol−1. What is the mass of the ammonia sample?

  1. 34.0 g34.0\ \mathrm{g}34.0 g
  2. 8.50 g8.50\ \mathrm{g}8.50 g (correct answer)
  3. 17.0 g17.0\ \mathrm{g}17.0 g
  4. 0.0294 g0.0294\ \mathrm{g}0.0294 g
  5. 0.500 g0.500\ \mathrm{g}0.500 g

Explanation: This problem requires using moles and molar mass to convert from moles to mass. The molar mass of NH₃ is 17.0 g/mol, which tells us the mass of one mole. To find the mass of 0.500 mol, we multiply: 0.500 mol × 17.0 g/mol = 8.50 g. A student might mistakenly divide the molar mass by moles (17.0 ÷ 0.500 = 34.0), which gives twice the molar mass rather than half. The reliable method is dimensional analysis: start with moles, multiply by molar mass (g/mol), and verify that mol cancels to leave grams.

Question 4

Two buffers are prepared using equal volumes at the same temperature.

Buffer A: H3PO4/H2PO4−\mathrm{H_3PO_4/H_2PO_4^-}H3​PO4​/H2​PO4−​ with relatively large amounts of both components Buffer B: H3PO4/H2PO4−\mathrm{H_3PO_4/H_2PO_4^-}H3​PO4​/H2​PO4−​ with relatively small amounts of both components

The same small amount of strong base is added to each buffer.

Which statement best explains which buffer has greater capacity?

  1. Buffer B has greater capacity because a smaller amount of solution changes pH less when base is added.
  2. Buffer A has greater capacity because it contains more total moles of the weak acid and its conjugate base to neutralize added base. (correct answer)
  3. Buffer B has greater capacity because the weak acid is present, and only the acid matters when base is added.
  4. Both have equal capacity because the H3PO4/H2PO4−\mathrm{H_3PO_4/H_2PO_4^-}H3​PO4​/H2​PO4−​ pair fixes the pH regardless of amount.
  5. Buffer A has lower capacity because larger amounts of buffer components shift equilibrium more, causing larger pH changes.

Explanation: This question evaluates buffer capacity principles. Buffer capacity is determined by the total moles of weak acid and conjugate base available to neutralize added H⁺ or OH⁻, not by solution volume or other factors. Buffer A contains relatively large amounts of both H₃PO₄ and H₂PO₄⁻, while Buffer B has relatively small amounts of both. When strong base is added, the H₃PO₄ (weak acid) will react to neutralize the OH⁻ ions, and Buffer A has more moles of H₃PO₄ available for this reaction. The misconception in choice A is that smaller amounts somehow change pH less, but this confuses dilution effects with buffer capacity. The key principle is that higher total concentration of buffer components provides greater capacity to resist pH changes from added acid or base.

Question 5

A student mixes 10 mL10\,\text{mL}10mL of hexane (C6H14\text{C}_6\text{H}_{14}C6​H14​, nonpolar) with 10 mL10\,\text{mL}10mL of water (polar) in a test tube and shakes the mixture. After standing, two distinct layers are observed. Which statement is correct?

  1. No homogeneous solution forms because a polar liquid and a nonpolar liquid are immiscible. (correct answer)
  2. A homogeneous solution forms; both liquids act as solvents since both are liquids.
  3. A homogeneous solution forms; water is the solvent and hexane is the solute.
  4. No homogeneous solution forms because equal volumes cannot make a solution.
  5. A homogeneous solution forms; hexane is the solvent and water is the solute.

Explanation: This question tests the concept of miscibility between liquids of different polarities and the formation of homogeneous solutions. Mixing 10 mL of nonpolar hexane with 10 mL of polar water results in two distinct layers because nonpolar and polar substances are immiscible, following the 'like dissolves like' principle where similar intermolecular forces are required for mixing. The shaking temporarily disperses the liquids, but upon standing, they separate due to the inability of water's hydrogen bonding to interact effectively with hexane's London dispersion forces. No homogeneous solution forms as the substances do not mix at a molecular level. A tempting distractor is choice A, which suggests a homogeneous solution with water as solvent, but this is incorrect due to the misconception that equal volumes guarantee miscibility regardless of polarity differences. A transferable strategy is to assess the polarity of components to predict whether they will form a single phase or separate layers.

Question 6

Excess solid zinc sulfide, ZnS(s)\text{ZnS}(s)ZnS(s), is added to pure water at 25∘C25^\circ\text{C}25∘C and allowed to reach equilibrium. The KspK_{sp}Ksp​ of ZnS\text{ZnS}ZnS is 1.0×10−241.0\times10^{-24}1.0×10−24.

Dissolution: ZnS(s)⇌Zn2+(aq)+S2−(aq)\text{ZnS}(s)\rightleftharpoons \text{Zn}^{2+}(aq)+\text{S}^{2-}(aq)ZnS(s)⇌Zn2+(aq)+S2−(aq)

Assuming initial ion concentrations are zero, what is the molar solubility of ZnS\text{ZnS}ZnS?

ICE table (in M):

Zn2+\text{Zn}^{2+}Zn2+S2−\text{S}^{2-}S2−
I00
C+sss+sss
Essssss
  1. 1.0×10−12 M1.0\times10^{-12}\,\text{M}1.0×10−12M (correct answer)
  2. 1.0×10−24 M1.0\times10^{-24}\,\text{M}1.0×10−24M
  3. 5.0×10−13 M5.0\times10^{-13}\,\text{M}5.0×10−13M
  4. 1.0×10−8 M1.0\times10^{-8}\,\text{M}1.0×10−8M
  5. 2.0×10−12 M2.0\times10^{-12}\,\text{M}2.0×10−12M

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Zn²⁺ and S²⁻. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻²⁴, so s = √(1.0×10⁻²⁴) = 1.0×10⁻¹² M. Choice B (1.0×10⁻²⁴) incorrectly uses the Ksp value as the molar solubility without taking the square root. For extremely insoluble salts like ZnS, the molar solubility is still found by taking the square root of Ksp.

Question 7

Solid calcium carbonate decomposes according to the balanced equation:   CaCO3(s)→CaO(s)+CO2(g)\;\text{CaCO}_3(s)\rightarrow \text{CaO}(s)+\text{CO}_2(g)CaCO3​(s)→CaO(s)+CO2​(g). When 10.0 g10.0\ \text{g}10.0 g of CaCO3\text{CaCO}_3CaCO3​ decomposes completely, what mass of CO2\text{CO}_2CO2​ is produced? (Molar masses: CaCO3=100.0 g mol−1\text{CaCO}_3=100.0\ \text{g mol}^{-1}CaCO3​=100.0 g mol−1, CO2=44.0 g mol−1\text{CO}_2=44.0\ \text{g mol}^{-1}CO2​=44.0 g mol−1.)

  1. 44 g44\ \text{g}44 g
  2. 2.2 g2.2\ \text{g}2.2 g
  3. 4.4 g4.4\ \text{g}4.4 g (correct answer)
  4. 22 g22\ \text{g}22 g
  5. 8.8 g8.8\ \text{g}8.8 g

Explanation: This question assesses the skill of stoichiometry. The balanced equation CaCO3 → CaO + CO2 provides mole ratios where 1 mole of CaCO3 produces 1 mole of CO2. These ratios allow conversion from grams of CaCO3 to moles using its molar mass, then to moles of CO2, and finally to grams of CO2 using its molar mass. For 10.0 g of CaCO3 (0.100 mol), the 1:1 ratio yields 0.100 mol of CO2, or 4.4 g. A tempting distractor is 44 g, which could result from forgetting to convert grams to moles and directly using the molar mass without the ratio. Always start from the balanced equation and convert to moles before applying ratios.

Question 8

In a closed vessel at constant temperature, the reversible reaction occurs:

2SO2(g)+O2(g)⇌2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}2SO2​(g)+O2​(g)⇌2SO3​(g)

At a moment before equilibrium is reached, the following concentrations are measured:

SpeciesCurrent concentration (M)
SO2\mathrm{SO_2}SO2​0.05
O2\mathrm{O_2}O2​0.10
SO3\mathrm{SO_3}SO3​0.90

In which direction will the net reaction proceed to reach equilibrium?

  1. No net change
  2. Toward products
  3. Toward reactants (correct answer)
  4. Toward products because SO3\mathrm{SO_3}SO3​ has the largest concentration
  5. No net change because the reaction is reversible

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows a high concentration of SO₃ (0.90 M) compared to low concentrations of SO₂ (0.05 M) and O₂ (0.10 M), indicating that products are currently favored relative to equilibrium. With more product present, the reverse reaction rate is higher than the forward, driving the system to produce more SO₂ and O₂. Therefore, the net reaction proceeds toward reactants to restore balance by decreasing the product and increasing the reactants. A tempting distractor is D, which is incorrect due to the misconception that 'highest concentration determines direction,' but direction is based on overall imbalance relative to equilibrium, not just the largest individual concentration. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 9

A rigid 2.00 L flask contains 0.100 mol of an ideal gas at 27C. What is the pressure of the gas? (Use R=0.0821 L0˘0b7atm0˘0b7mol−10˘0b7K−1R = 0.0821\ \text{L\u00b7atm\u00b7mol}^{-1}\text{\u00b7K}^{-1}R=0.0821 L0˘0b7atm0˘0b7mol−10˘0b7K−1.)

  1. 1.23 atm (correct answer)
  2. 0.616 atm
  3. 2.46 atm
  4. 0.0554 atm
  5. 12.3 atm

Explanation: This question tests the application of the ideal gas law, PV = nRT, to calculate the pressure of a gas sample. To find the pressure, rearrange the ideal gas law to P = nRT/V, using the given values of n = 0.100 mol, V = 2.00 L, T = 27°C (which must be converted to 300 K), and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Substituting these values gives P = (0.100 × 0.0821 × 300) / 2.00 = 1.23 atm, matching choice A. The calculation relies on the ideal gas law assuming the gas behaves ideally under these conditions, with all units consistent with R. A tempting distractor is choice C, 2.46 atm, which results from mistakenly using V = 1.00 L instead of 2.00 L, reflecting a misconception of misreading the given volume. Always double-check unit conversions and given values before plugging into the ideal gas law equation.

Question 10

A student has a mixture of two volatile liquids, acetone (bp 56 ∘C56\,^{\circ}\text{C}56∘C) and toluene (bp 111 ∘C111\,^{\circ}\text{C}111∘C), that are miscible. The student needs to separate them into two fractions. Which method is best?

  1. Use simple distillation because the boiling points are far apart (correct answer)
  2. Use filtration because acetone molecules are smaller than toluene molecules
  3. Use a separatory funnel because acetone and toluene form two layers
  4. Use paper chromatography because volatile liquids separate best on paper
  5. Use evaporation because both liquids will remain after heating gently

Explanation: This question tests the skill of separating miscible volatile liquids with significantly different boiling points. Simple distillation is appropriate here because acetone (bp 56°C) and toluene (bp 111°C) have a large boiling point difference of 55°C, which is sufficient for effective separation using simple distillation. When the mixture is heated to around 56°C, acetone vaporizes while toluene remains largely in the liquid phase, allowing acetone to be collected as the first fraction. After most acetone is removed, the temperature can be raised to collect toluene. Choice C incorrectly suggests using a separatory funnel, but acetone and toluene are both organic solvents that are miscible and form only one layer. The transferable strategy is to use simple distillation when boiling points differ by more than 25°C, and fractional distillation only when boiling points are closer together.

Question 11

A sample of CO2_22​(g) in a piston is compressed isothermally at constant temperature from 2.0 L to 1.0 L. No phase change occurs. For the gas (the system), does entropy increase, decrease, or remain approximately constant?

  1. Entropy remains approximately constant because the number of moles is unchanged.
  2. Entropy decreases because the gas occupies a smaller volume and is less dispersed. (correct answer)
  3. Entropy increases because compression requires work to be done on the gas.
  4. Entropy increases because the pressure increases during compression.
  5. Entropy remains approximately constant because temperature is constant.

Explanation: This question tests the concept of entropy changes due to volume alterations in gases under isothermal conditions. During isothermal compression of CO₂ gas from 2.0 L to 1.0 L, the gas molecules are confined to a smaller volume, reducing their dispersal and the number of accessible microstates. This confinement leads to a decrease in entropy, as the system becomes more ordered with less space for random motion. No phase change occurs, so the entropy change is primarily driven by the volume reduction. A tempting distractor is choice C, which suggests entropy remains constant because temperature is unchanged, but this misconceives that temperature constancy implies entropy constancy, ignoring volume's role in gas entropy. When assessing entropy for gases, consider how changes in volume affect molecular dispersal, especially in isothermal processes.

Question 12

A reaction is monitored by measuring [E][\text{E}][E] as a function of time. Based on the slope of the curve, during which interval is the rate of disappearance of E\text{E}E the smallest (slowest)?

  1. 0–10 s0\text{–}10\ \text{s}0–10 s
  2. 10–20 s10\text{–}20\ \text{s}10–20 s
  3. 20–30 s20\text{–}30\ \text{s}20–30 s
  4. 30–40 s30\text{–}40\ \text{s}30–40 s
  5. 40–50 s40\text{–}50\ \text{s}40–50 s (correct answer)

Explanation: This question tests understanding of reaction rates by asking for the interval with the smallest (slowest) rate. The rate of disappearance corresponds to the slope of the concentration versus time curve - a gentler slope indicates a slower rate. In typical reactions, the rate decreases over time as reactant concentration decreases, so the slowest rate usually occurs in the latest time interval. The 40-50s interval would have the smallest rate because [E] is lowest at this point, resulting in fewer collisions and the gentlest slope. Students often confuse low concentration with low rate, but it's actually the low concentration that causes the low rate, not the concentration itself that is the rate. To find the slowest rate, look for the interval where the concentration changes least steeply, typically occurring late in the reaction.

Question 13

Two bonds are compared: the O−H\mathrm{O-H}O−H bond within a water molecule and the hydrogen bond between two water molecules. Which interaction requires more energy to break?

  1. The hydrogen bond between two water molecules
  2. The O−H\mathrm{O-H}O−H covalent bond within a water molecule (correct answer)
  3. They require the same energy because both involve hydrogen and oxygen
  4. The hydrogen bond, because intermolecular forces are stronger than covalent bonds
  5. The O−H\mathrm{O-H}O−H bond only if water is in the gas phase

Explanation: This question tests the distinction between intramolecular covalent bonds and intermolecular hydrogen bonds in water, within intramolecular forces and potential energy. The O-H covalent bond within a water molecule requires more energy to break because it is a strong intramolecular bond, while the hydrogen bond between molecules is a weaker intermolecular interaction. Covalent bonds involve electron sharing and have much higher dissociation energies than hydrogen bonds. The stimulus compares these interactions, emphasizing energy differences. Choice A is a tempting distractor, claiming the hydrogen bond requires more energy, which stems from the misconception that intermolecular forces are stronger than covalent bonds. To differentiate bond types, remember covalent bonds are intramolecular and stronger than intermolecular forces, regardless of phase.

Question 14

The following reaction energy profile diagram represents a three-step mechanism with intermediates I1 and I2 and transition states TS1, TS2, and TS3. The rate-determining step corresponds to the step with the greatest activation energy, measured from the energy of the species at the start of that step to the top of the corresponding transition state. Based on the diagram, which step is rate-determining?

Step 1: Reactants→I1\text{Reactants} \to \text{I1}Reactants→I1 Step 2: I1→I2\text{I1} \to \text{I2}I1→I2 Step 3: I2→Products\text{I2} \to \text{Products}I2→Products

  1. Step 3
  2. Step 1
  3. Step 2 (correct answer)
  4. Step 1
  5. Step 3

Explanation: This problem involves analyzing a multistep reaction energy profile. The rate-determining step corresponds to the step with the greatest activation energy, which is explicitly defined as the energy difference from the starting species of that step to the top of its transition state. For Step 1, this is reactants to TS1; for Step 2, it's I1 to TS2; and for Step 3, it's I2 to TS3. The correct answer being Step 2 means the vertical distance from I1 to TS2 is the largest among the three barriers. Students sometimes incorrectly identify the highest absolute peak as determining the rate, but it's the relative barrier height that matters. To solve these problems, measure each activation barrier separately and identify the largest one.

Question 15

A student measures the absorbance of the same solution in two different cuvettes at the same wavelength. In a 1.00 cm1.00\,\text{cm}1.00cm cuvette, the absorbance is A=0.30A = 0.30A=0.30. What absorbance should be measured in a 3.00 cm3.00\,\text{cm}3.00cm cuvette, assuming the solution and wavelength are unchanged?

  1. 0.100.100.10
  2. 0.600.600.60
  3. 0.900.900.90 (correct answer)
  4. 0.300.300.30
  5. 3.03.03.0

Explanation: This question tests understanding of the Beer-Lambert Law's proportionality to path length when concentration and molar absorptivity are constant. The Beer-Lambert Law shows A is directly proportional to ℓ, so changing path length scales absorbance linearly. For the original A = 0.30 at ℓ = 1.00 cm, the new absorbance at ℓ = 3.00 cm is 0.30 × (3.00 / 1.00) = 0.90. This corresponds to choice C. A tempting distractor is 0.30, which might come from assuming absorbance is independent of path length, a misconception that ignores the role of light path in absorption. A transferable strategy is to identify proportional relationships in the Beer-Lambert equation and use ratios to predict changes without recalculating fully.

Question 16

A student dissolves a sample of table salt that contains a small amount of insoluble anti-caking agent in water. The student wants a clear sodium chloride solution with the insoluble solid removed. Which technique is most appropriate?​​

  1. Filtration, because insoluble particles can be trapped while dissolved ions pass through (correct answer)
  2. Fractional distillation, because dissolved ions have different boiling points than water
  3. Paper chromatography, because ions separate into colored bands on the paper
  4. Evaporation to dryness, because the insoluble solid will evaporate before the salt
  5. Magnetic separation, because the anti-caking agent is attracted to magnets

Explanation: This question tests the skill of removing insoluble impurities from a solution using filtration. When table salt containing an insoluble anti-caking agent is dissolved in water, the sodium chloride dissolves completely while the anti-caking agent remains as suspended solid particles. Filtration effectively separates this heterogeneous mixture because the filter paper's pores trap the insoluble solid particles while allowing the clear sodium chloride solution to pass through as the filtrate. This produces a clear solution free from suspended solids, which was the student's goal. Choice D is incorrect because evaporation to dryness would leave both the salt and the anti-caking agent together as a solid residue, failing to separate them, and reflects the misconception that solids evaporate at different rates. To obtain clear solutions from mixtures containing insoluble impurities, use filtration to remove suspended particles while keeping dissolved substances in solution.

Question 17

Two hydrocarbons are compared: pentane, C5H12\mathrm{C_5H_{12}}C5​H12​, and neopentane, C5H12\mathrm{C_5H_{12}}C5​H12​ (same formula but more compact/branched shape). Both are nonpolar and have no H-bond donors/acceptors. Which statement best explains why pentane has a higher boiling point than neopentane?

  1. Pentane has stronger dipoledipole forces because it is less symmetric
  2. Pentane has a larger surface area, leading to stronger dispersion forces (correct answer)
  3. Neopentane can hydrogen bond more effectively due to branching
  4. Neopentane has fewer covalent bonds, so it vaporizes more easily
  5. Pentane is more polar because it has more C ⁣−H\mathrm{C\!- H}C−H bonds

Explanation: This question tests understanding of how molecular shape affects London dispersion forces in nonpolar molecules. Both pentane and neopentane are nonpolar hydrocarbons with the same molecular formula (C₅H₁₂) and thus the same number of electrons, but pentane has a linear, extended shape while neopentane is compact and spherical. The extended shape of pentane provides a larger surface area for contact between molecules, allowing for stronger London dispersion forces compared to the compact neopentane. Stronger intermolecular forces require more energy to overcome, resulting in pentane's higher boiling point. Neither molecule is polar or capable of hydrogen bonding since they contain only C-H and C-C bonds. A common misconception is that pentane is more polar (choice E), but all hydrocarbons containing only carbon and hydrogen are nonpolar. When comparing dispersion forces in molecules with the same formula, consider molecular shape and surface area—more extended shapes generally have stronger dispersion forces.

Question 18

Use Hess’s law to find ΔH\Delta HΔH for: CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}CO(g)+21​O2​(g)→CO2​(g). The following reactions are given:

  1. C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}C(s)+O2​(g)→CO2​(g) ΔH=−394 kJ\Delta H = -394\ \mathrm{kJ}ΔH=−394 kJ

  2. C(s)+12O2(g)→CO(g)\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)}C(s)+21​O2​(g)→CO(g) ΔH=−111 kJ\Delta H = -111\ \mathrm{kJ}ΔH=−111 kJ

  1. ΔH=+283 kJ\Delta H = +283\ \mathrm{kJ}ΔH=+283 kJ
  2. ΔH=−505 kJ\Delta H = -505\ \mathrm{kJ}ΔH=−505 kJ
  3. ΔH=−283 kJ\Delta H = -283\ \mathrm{kJ}ΔH=−283 kJ (correct answer)
  4. ΔH=+505 kJ\Delta H = +505\ \mathrm{kJ}ΔH=+505 kJ
  5. ΔH=−111 kJ\Delta H = -111\ \mathrm{kJ}ΔH=−111 kJ

Explanation: This question tests the application of Hess’s law to calculate the enthalpy change for the oxidation of CO to CO_2. To obtain the target CO(g) + ½O_2(g) → CO_2(g), reverse the second reaction to get CO(g) → C(s) + ½O_2(g) with ΔH = +111 kJ. Add the first reaction C(s) + O_2(g) → CO_2(g) with ΔH = -394 kJ. The overall enthalpy change is +111 kJ - 394 kJ = -283 kJ, with C and ½O_2 canceling. A tempting distractor is -111 kJ, from using the second reaction without reversing, misconceiving the reaction direction. A transferable strategy is to reverse reactions as necessary to align reactants and products with the target.

Question 19

A student compares the reaction of NO(g) with O2(g) to form NO2(g) under two conditions. The student uses the same temperature and explicitly assumes the reaction mechanism is unchanged. In Condition 1, the mixture is prepared with a higher initial concentration of NO(g) than in Condition 2, while the O2(g) concentration is the same in both. The formation of NO2(g) is faster in Condition 1. Which statement best explains the faster rate in Condition 1 using collision frequency or collision effectiveness?

  1. Condition 1 is faster because higher NO concentration increases the frequency of collisions between NO and O2 molecules, increasing the number of effective collisions per unit time. (correct answer)
  2. Condition 1 is faster because increasing NO concentration increases the temperature of the gas mixture, which increases collision effectiveness.
  3. Condition 1 is faster because the products are more stable when NO is more concentrated, so the reaction becomes faster.
  4. Condition 1 is faster because higher NO concentration causes the mechanism to change so that O2 is no longer required for the reaction to proceed.
  5. Condition 1 is faster because at lower NO concentration the molecules collide more gently and therefore stop colliding after a few attempts.

Explanation: This question assesses the collision model, which explains reaction rates based on the frequency and effectiveness of molecular collisions. In Condition 1, the higher concentration of NO increases the number of NO molecules per unit volume, raising the frequency of collisions with O2 molecules. This leads to more effective collisions that produce NO2 at a faster rate. Consequently, the reaction proceeds quicker due to these increased interactions. A tempting distractor is choice C, which wrongly implies product stability affects rate, confusing thermodynamics with kinetic collision factors. Faster reactions result from more frequent or more energetic effective collisions.

Question 20

Element J is in Group 16 and has atomic number Z=34Z = 34Z=34 (selenium). What is the most likely ionic charge for J in a simple ionic compound?

  1. 1−1-1−
  2. 2−2-2− (correct answer)
  3. 2+2+2+
  4. 4−4-4−
  5. 1+1+1+

Explanation: This question assesses the skill of valence electrons and ionic compounds. Selenium, in Group 16, has 6 valence electrons and gains 2 to form 2- ion with krypton's configuration. The charge is negative, reflecting electron gain in nonmetals. Valence electrons guide the number needed for octet. A tempting distractor is 1-, but that's for Group 17, not 16 needing 2. Selenium's configuration is [Ar] 4s2 3d10 4p4. Main-group ions form to achieve noble-gas configurations.

Question 21

A student uses a rechargeable lithium-ion battery to power a phone (discharging). During use, the battery provides electrical energy to the circuit without being connected to a charger. Which statement correctly classifies the process occurring during discharge and its defining feature?

  1. Electrolytic cell; an external power source is required to force a nonspontaneous redox reaction
  2. Galvanic (voltaic) cell; a spontaneous redox reaction produces electrical energy in the external circuit (correct answer)
  3. Galvanic (voltaic) cell; an external power source is required to force a nonspontaneous redox reaction
  4. Electrolytic cell; a spontaneous redox reaction produces electrical energy in the external circuit
  5. Galvanic (voltaic) cell; the separator membrane supplies electrons to the circuit

Explanation: This question tests your understanding of galvanic (voltaic) and electrolytic cells. In galvanic cells, spontaneous processes generate electrical energy, with natural electron flow. Electrolytic cells require external energy to drive nonspontaneous processes, forcing electron direction. This spontaneity-energy relationship distinguishes cells and determines electron movement. Choice C is a tempting choice, misclassifying it as galvanic needing power, based on the misconception that batteries always require external input even during discharge. To distinguish between cell types, remember: if electrical energy is produced, the cell is galvanic; if it must be supplied, the cell is electrolytic.

Question 22

A reaction-energy diagram is shown for a single-step reaction. Which statement correctly describes the effect of increasing temperature on the diagram quantities?

(Interpretation question only: the diagram shape is unchanged.)

  1. The peak shifts right, so EaE_aEa​ becomes larger.
  2. EaE_aEa​ and ΔH\Delta HΔH read from the diagram do not change when temperature increases. (correct answer)
  3. EaE_aEa​ increases because molecules move faster at higher temperature.
  4. ΔH\Delta HΔH changes sign because higher temperature favors products.
  5. EaE_aEa​ decreases and ΔH\Delta HΔH becomes more negative when temperature increases.

Explanation: This question tests the skill of understanding temperature effects on quantities in reaction energy diagrams. Increasing temperature does not alter the diagram's shape or values of Ea and ΔH, as these are intrinsic properties of the potential energy surface, independent of temperature. However, higher temperature increases the fraction of molecules overcoming Ea, thus accelerating the rate, but the diagram itself remains unchanged. Thus, option A correctly states that Ea and ΔH do not change with increasing temperature. A tempting distractor is option B, which claims Ea decreases and ΔH becomes more negative, but this is incorrect due to the misconception that temperature modifies the energy landscape. Remember that energy diagrams represent zero-temperature potential energies; temperature affects rates via Boltzmann distribution, not the diagram quantities themselves.

Question 23

A reaction is reported to have ΔG=0\Delta G = 0ΔG=0 at a particular temperature and pressure. Which statement best describes the thermodynamic favorability of the reaction as written under those conditions?

  1. Spontaneous under these conditions because ΔG\Delta GΔG equals zero.
  2. Nonspontaneous under these conditions because ΔG\Delta GΔG equals zero.
  3. At equilibrium under these conditions (neither thermodynamically favorable nor unfavorable). (correct answer)
  4. Spontaneous only at high temperature because ΔG\Delta GΔG equals zero.
  5. Spontaneous only if a catalyst is added because ΔG\Delta GΔG equals zero.

Explanation: This question assesses the skill of Gibbs free energy and thermodynamic favorability. The Gibbs free energy change, ΔG, determines spontaneity, with ΔG < 0 spontaneous, ΔG > 0 nonspontaneous, and ΔG = 0 at equilibrium. Here, ΔG = 0 means the reaction is at equilibrium, neither forward nor reverse is favored. No net change occurs under these conditions. A tempting distractor is choice A, which mistakenly equates ΔG = 0 with spontaneity, confusing equilibrium with favorability. Judge spontaneity by ΔG's sign, not catalysts which affect rate, not thermodynamics.

Question 24

A student heats solid potassium chloride until it melts. During melting, the temperature remains constant even though the hot plate continues to supply energy. Which statement correctly classifies the melting process?

  1. The process is exothermic; heat is released as bonds form and ΔH<0\Delta H<0ΔH<0.
  2. The process is endothermic; heat is absorbed to overcome attractions and ΔH>0\Delta H>0ΔH>0. (correct answer)
  3. The process is exothermic; constant temperature means no heat transfer so ΔH=0\Delta H=0ΔH=0.
  4. The process is endothermic; constant temperature means ΔH<0\Delta H<0ΔH<0.
  5. The process is endothermic; because the temperature is constant, ΔG>0\Delta G>0ΔG>0.

Explanation: This question tests understanding of phase transitions and heat flow during melting. When KCl melts at constant temperature despite continuous heating, the supplied energy is being absorbed to break the ionic bonds in the solid lattice structure. Melting is endothermic because energy must be absorbed to overcome the attractive forces holding the solid together, making ΔH > 0. The constant temperature during melting occurs because all the absorbed energy goes into increasing potential energy (breaking bonds) rather than kinetic energy (temperature). Choice A incorrectly identifies melting as exothermic, confusing it with freezing which releases heat as bonds form. During any melting process, heat is absorbed (endothermic) even when temperature remains constant.

Question 25

A student compares two buffers and then adds the same small amount of strong base to each.

Buffer A: HNO2/NO2−\text{HNO}_2/\text{NO}_2^-HNO2​/NO2−​ with 0.40 M0.40\,\text{M}0.40M HNO2\text{HNO}_2HNO2​ and 0.40 M0.40\,\text{M}0.40M NaNO2\text{NaNO}_2NaNO2​. Buffer B: HNO2/NO2−\text{HNO}_2/\text{NO}_2^-HNO2​/NO2−​ with 0.10 M0.10\,\text{M}0.10M HNO2\text{HNO}_2HNO2​ and 0.70 M0.70\,\text{M}0.70M NaNO2\text{NaNO}_2NaNO2​.

Which buffer will show the smaller pH change?

  1. Buffer B, because it contains more conjugate base
  2. Buffer A, because it contains more weak acid to neutralize added base (correct answer)
  3. They will change equally because the total concentration of buffer components is the same
  4. Buffer B, because buffers are most effective when one component is in large excess
  5. They will change equally because both use the same conjugate pair

Explanation: This question tests your understanding of buffer capacity. Buffer capacity indicates resistance to added base, relying on the total concentrations of weak acid and conjugate base to neutralize OH⁻. Higher weak acid concentrations better handle base additions. Capacity depends on these levels and the ratio's proximity to 1:1. A tempting distractor is that they will change equally because the total concentration of buffer components is the same, but Buffer A's balanced ratio and higher acid give better capacity. A transferable strategy is, for base additions, to seek higher weak acid concentration and balanced ratio; this means greater capacity to absorb added OH⁻.