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AP Chemistry

AP Chemistry Practice Test: Practice Test 1

Practice Test 1 for AP Chemistry: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

For the reversible reaction A(g)+B(g)⇌C(g)\mathrm{A(g) + B(g) \rightleftharpoons C(g)}A(g)+B(g)⇌C(g) at a certain temperature, Kc=2.0K_c = 2.0Kc​=2.0. A reaction mixture is prepared with initial concentrations [A]=1.0 M[\mathrm{A}]=1.0\,\mathrm{M}[A]=1.0M, [B]=1.0 M[\mathrm{B}]=1.0\,\mathrm{M}[B]=1.0M, and [C]=0.50 M[\mathrm{C}]=0.50\,\mathrm{M}[C]=0.50M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ to KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

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Question 1

For the reversible reaction A(g)+B(g)⇌C(g)\mathrm{A(g) + B(g) \rightleftharpoons C(g)}A(g)+B(g)⇌C(g) at a certain temperature, Kc=2.0K_c = 2.0Kc​=2.0. A reaction mixture is prepared with initial concentrations [A]=1.0 M[\mathrm{A}]=1.0\,\mathrm{M}[A]=1.0M, [B]=1.0 M[\mathrm{B}]=1.0\,\mathrm{M}[B]=1.0M, and [C]=0.50 M[\mathrm{C}]=0.50\,\mathrm{M}[C]=0.50M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ to KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. the reaction will proceed toward reactants
  2. the reaction will proceed toward products (correct answer)
  3. the system is already at equilibrium
  4. the reaction will proceed toward reactants because the reactant concentrations are equal
  5. the reaction will proceed toward products until all reactants are consumed

Explanation: This question tests understanding of reaction quotient and equilibrium constant. To determine the direction of reaction, we calculate Q_c = [C]/([A][B]) = 0.50/((1.0)(1.0)) = 0.50/1.0 = 0.50. Since Q_c (0.50) < K_c (2.0), the system has too little product relative to equilibrium, so the reaction must shift toward products to increase Q_c until it equals K_c. Choice E incorrectly suggests the reaction will consume all reactants, but equilibrium reactions never go to completion - they reach a balance where both reactants and products are present. The key strategy is to calculate Q, compare it to K, and remember that when Q < K, the reaction shifts right (toward products) to reach equilibrium.

Question 2

For the weak acid HF in water, Ka=1.0×10−4K_a=1.0\times10^{-4}Ka​=1.0×10−4 at 25∘C25^\circ\text{C}25∘C. Which relationship between KaK_aKa​ and pKapK_apKa​ for HF is correct?

  1. pKa=log⁡(Ka)pK_a=\log(K_a)pKa​=log(Ka​)
  2. pKa=−log⁡(Ka)pK_a=-\log(K_a)pKa​=−log(Ka​) (correct answer)
  3. pKa=1KapK_a=\dfrac{1}{K_a}pKa​=Ka​1​
  4. pKa=log⁡(1Ka)+1pK_a=\log\left(\dfrac{1}{K_a}\right)+1pKa​=log(Ka​1​)+1
  5. pKa=−log⁡([H+])pK_a=-\log([\text{H}^+])pKa​=−log([H+])

Explanation: This question assesses the skill of pH and pK. The pKa is the negative logarithm of the acid dissociation constant Ka, reflecting the strength of the acid where lower pKa indicates stronger acids due to greater dissociation. Similarly, pH is -log[H+], so both are logarithmic measures that allow relative comparisons without full calculations. For HF with Ka=1.0×10^{-4}, pKa=4, illustrating how pKa inversely relates to Ka's magnitude. A tempting distractor is choice A, which incorrectly uses pKa=log(Ka) without the negative sign, leading to positive values that don't align with typical pKa ranges for weak acids. Remember as a transferable strategy that lower pKa means stronger acid because it corresponds to a larger Ka.

Question 3

A student is comparing electromagnetic radiation used in different spectroscopic methods. Which ordering lists the regions from lowest photon energy to highest photon energy?

  1. X-ray < ultraviolet < visible < infrared < microwave
  2. Microwave < infrared < visible < ultraviolet < X-ray (correct answer)
  3. Infrared < microwave < visible < ultraviolet < X-ray
  4. Microwave < visible < infrared < ultraviolet < X-ray
  5. Microwave < infrared < ultraviolet < visible < X-ray

Explanation: This question assesses understanding of spectroscopy and the electromagnetic spectrum. Photon energy in electromagnetic radiation increases with frequency across regions. The correct ordering from lowest to highest energy is microwave < infrared < visible < ultraviolet < X-ray, reflecting increasing frequency. Spectroscopic methods use these regions for different transitions, like IR for vibrations and UV for electronics. A tempting distractor is choice A, but it is incorrect because it reverses the order, starting with high-energy X-ray as lowest. To order spectrum regions by energy, remember that energy increases from radio to gamma rays.

Question 4

A liquid mixture contains 30.0 g30.0\ \text{g}30.0 g of acetone and 70.0 g70.0\ \text{g}70.0 g of water. What is the mass percent of acetone in the mixture?

  1. 30% (correct answer)
  2. 70%
  3. 40%
  4. 23%
  5. 3.0%

Explanation: This question tests the skill of calculating mass percent in a binary liquid mixture. Mass percent equals (mass of component / total mass) × 100%. The mixture contains 30.0 g acetone and 70.0 g water, giving a total mass of 100.0 g. The mass percent of acetone is (30.0 g / 100.0 g) × 100% = 30%. Students might incorrectly select 70% by confusing the mass of water with the mass percent of acetone, which represents a misconception about which value represents the component of interest versus the other component. When solving mass percent problems, always clearly identify which component's percentage is being requested.

Question 5

A student has a mixture of cooking oil and water in a beaker. The liquids form two layers after standing. The student wants to separate the two liquids with minimal cross-contamination. Which method is best?​​

  1. Use a separatory funnel to drain the bottom layer from the top layer (correct answer)
  2. Filter the mixture through filter paper to trap the oil while water passes through
  3. Use paper chromatography to separate the oil layer into fractions from the water layer
  4. Use fractional distillation to separate the layers based on density differences
  5. Evaporate the mixture to dryness to recover both liquids as separate solids

Explanation: This question tests the skill of separating immiscible liquids that form distinct layers. A separatory funnel is the best tool for this separation because it allows controlled drainage of the denser bottom layer (water) while retaining the less dense top layer (oil) in the funnel, minimizing cross-contamination between layers. The stopcock at the bottom provides precise control over the flow rate, allowing careful separation right at the interface between the two liquids. This method is specifically designed for liquid-liquid separations where density differences create distinct layers. Choice B is incorrect because filter paper cannot separate two liquids—both would pass through the filter since filtration only separates solids from liquids, demonstrating the misconception that immiscible liquids behave like solid-liquid mixtures. For separating immiscible liquids that form layers, use a separatory funnel for precise control, or careful decanting for less critical separations.

Question 6

A solution is measured in a 1.00 cm cuvette and has A=0.48A=0.48A=0.48. The molar absorptivity is ε=1.6×104 L mol−1cm−1\varepsilon=1.6\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}ε=1.6×104 L mol−1cm−1. What is the concentration?

  1. 3.0×10−4 M3.0\times10^{-4}\ \text{M}3.0×10−4 M
  2. 7.5×10−4 M7.5\times10^{-4}\ \text{M}7.5×10−4 M
  3. 7.5×10−5 M7.5\times10^{-5}\ \text{M}7.5×10−5 M
  4. 3.0×10−5 M3.0\times10^{-5}\ \text{M}3.0×10−5 M (correct answer)
  5. 4.8×10−5 M4.8\times10^{-5}\ \text{M}4.8×10−5 M

Explanation: This question tests the application of Beer-Lambert law to determine concentration from absorbance data. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.48/(1.6×10⁴ × 1.00) = 0.48/(1.6×10⁴) = 3.0×10⁻⁵ M. This matches the marked answer exactly. Students might choose option B (7.5×10⁻⁵ M) by making computational errors in the division. When working with Beer-Lambert calculations, double-check your arithmetic, especially when dividing by large numbers in scientific notation.

Question 7

A student compares two trials of the same reaction between a metal oxide solid and an acid solution. The mechanism is unchanged; only collision-related factors differ.

Condition 1: The solid metal oxide is present as large pellets in the acid. Condition 2: The same mass of the solid metal oxide is ground into a fine powder before being added to the acid.

Acid concentration, temperature, and stirring are the same. Which statement best explains why the reaction is faster in Condition 2 than in Condition 1, focusing on collision frequency at the solid surface?

  1. Grinding the solid increases the energy of the acid molecules, so collisions become effective more often even without changing surface area.
  2. Grinding the solid increases surface area, providing more sites where acid particles can collide with the solid per unit time, increasing the reaction rate. (correct answer)
  3. Grinding the solid shifts the reaction toward products, so the forward reaction rate increases to reestablish equilibrium.
  4. Grinding the solid changes the reaction mechanism to a faster one because freshly exposed atoms act as a catalyst.
  5. Grinding the solid decreases the number of reactant particles in solution, so fewer collisions occur but each collision forms more product.

Explanation: This question applies the collision model to surface area effects in heterogeneous reactions. Grinding the metal oxide into powder in Condition 2 creates vastly more surface area than the large pellets in Condition 1. Since the acid can only react with oxide atoms at the solid surface, this increased surface area provides many more sites where acid molecules can collide with the solid per unit time. The higher collision frequency at the expanded interface directly increases the reaction rate. Choice A incorrectly suggests grinding affects acid molecule energy—grinding only changes the solid's surface area, not the kinetic energy of solution particles. For reactions involving solids, remember that reaction rate is proportional to surface area because collisions can only occur at the interface.

Question 8

Two lasers emit photons of different frequencies. Laser X emits light at 5.0×1014 s−15.0\times10^{14}\ \text{s}^{-1}5.0×1014 s−1, and Laser Y emits light at 7.5×1014 s−17.5\times10^{14}\ \text{s}^{-1}7.5×1014 s−1. Which statement is correct?

  1. Laser X emits photons with greater energy because it has the lower frequency.
  2. Both lasers emit photons with the same energy because the speed of light is constant.
  3. Laser Y emits photons with greater energy because it has the higher frequency. (correct answer)
  4. Laser X emits photons with shorter wavelength because it has the lower frequency.
  5. Laser Y emits photons with lower energy because higher frequency means lower energy.

Explanation: This question assesses understanding of the properties of photons. The energy of a photon is directly proportional to its frequency, following E = hν, where h is Planck's constant. Frequency and wavelength are inversely related via ν = c/λ, so higher frequency means shorter wavelength and thus higher energy. Laser Y has a higher frequency (7.5×10^14 s⁻¹) than Laser X (5.0×10^14 s⁻¹), so its photons have greater energy. A tempting distractor is choice E, which wrongly claims higher frequency means lower energy, but the opposite is true since energy increases with frequency. Remember, higher frequency corresponds to higher energy.

Question 9

Solid silver acetate, AgC2H3O2(s)\text{AgC}_2\text{H}_3\text{O}_2(s)AgC2​H3​O2​(s), is added separately to two solutions at the same temperature: Solution 1 is pure water and Solution 2 is adjusted to pH 2 with strong acid. Compared with Solution 1, the solubility in Solution 2 will be

  1. smaller because added H+\text{H}^+H+ is a common ion that suppresses formation of Ag+\text{Ag}^+Ag+.
  2. greater because H+\text{H}^+H+ protonates acetate to form HC2H3O2\text{HC}_2\text{H}_3\text{O}_2HC2​H3​O2​, reducing [C2H3O2−][\text{C}_2\text{H}_3\text{O}_2^-][C2​H3​O2−​]. (correct answer)
  3. unchanged because acetate is not involved in any equilibrium other than dissolution.
  4. smaller because the acid buffers the acetate concentration at a higher value, shifting dissolution left.
  5. unchanged because KspK_{sp}Ksp​ does not change with pH and therefore solubility cannot change.

Explanation: This question tests understanding of how pH affects the solubility of salts containing weak acid conjugate bases. Silver acetate dissolves to produce Ag⁺ and C₂H₃O₂⁻ (acetate), where acetate is the conjugate base of the weak acid acetic acid (HC₂H₃O₂). In acidic solution (pH 2), H⁺ ions react with acetate to form acetic acid, reducing the concentration of free acetate ions and shifting the dissolution equilibrium to the right, increasing solubility. Choice E incorrectly states that solubility cannot change if Ksp doesn't change, but fails to recognize that the effective concentration of acetate is reduced by protonation. When dealing with salts of weak acids, acidic conditions will always increase solubility by converting the anion to its protonated form.

Question 10

The reaction of nitrogen and hydrogen to form ammonia is represented by the balanced equation N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}N2​(g)+3H2​(g)→2NH3​(g). If 0.60 mol0.60\ \text{mol}0.60 mol of H2\mathrm{H_2}H2​ reacts with excess N2\mathrm{N_2}N2​, how many moles of NH3\mathrm{NH_3}NH3​ are produced?

  1. 0.20 mol0.20\ \text{mol}0.20 mol
  2. 0.40 mol0.40\ \text{mol}0.40 mol (correct answer)
  3. 0.60 mol0.60\ \text{mol}0.60 mol
  4. 0.90 mol0.90\ \text{mol}0.90 mol
  5. 1.20 mol1.20\ \text{mol}1.20 mol

Explanation: This problem requires stoichiometry to determine the amount of ammonia produced from a given amount of hydrogen gas. The balanced equation N₂(g) + 3H₂(g) → 2NH₃(g) provides the mole ratio: 3 moles of H₂ produce 2 moles of NH₃. Starting with 0.60 mol H₂, we use the ratio (2 mol NH₃)/(3 mol H₂) to calculate: 0.60 mol H₂ × (2 mol NH₃)/(3 mol H₂) = 0.40 mol NH₃. A common mistake would be to use the coefficient 3 incorrectly, perhaps calculating 0.60 × 3 = 1.80 mol, which doesn't appear in the choices but shows confusion about which number goes in the numerator versus denominator. Always identify the mole ratio from the balanced equation, placing the desired substance in the numerator and the given substance in the denominator.

Question 11

A student has an aqueous solution containing two dissolved dyes, one blue and one yellow. The dyes have different polarities, and the student wants to determine whether the sample contains one dye or a mixture. Which technique best accomplishes this separation and identification task?

  1. Simple distillation, because dyes will vaporize at different temperatures
  2. Filtration, because dyes can be trapped by filter paper pores
  3. Paper chromatography, because components travel different distances based on interactions (correct answer)
  4. Decanting, because the denser dye settles out first in water
  5. Evaporation to dryness, because each dye crystallizes in separate layers

Explanation: This question tests the skill of identifying a chromatographic technique to separate and analyze dissolved components in a solution based on differences in polarity. Paper chromatography is best because the dyes have different polarities, causing them to interact differently with the stationary phase (paper) and mobile phase (solvent), resulting in separation into distinct bands for identification. If multiple spots appear, it indicates a mixture; a single spot would suggest one dye. This method allows visualization of whether the sample is pure or mixed without destroying the components. Simple distillation (choice A) is a tempting distractor but incorrect because dyes are nonvolatile solids with high boiling points, misconstruing that distillation separates based on volatility rather than polarity in solutions. For analyzing mixtures of solutes with similar physical properties but different intermolecular interactions, chromatography is a key strategy to achieve separation and identification.

Question 12

In a metabolic pathway, two reactions are coupled by sharing intermediate I\mathrm{I}I and occurring together. Reaction 1 is spontaneous: S→I\mathrm{S \rightarrow I}S→I with ΔG=−12 kJ mol−1\Delta G = -12\ \mathrm{kJ\,mol^{-1}}ΔG=−12 kJmol−1. Reaction 2 is nonspontaneous: I→T\mathrm{I \rightarrow T}I→T with ΔG=+12 kJ mol−1\Delta G = +12\ \mathrm{kJ\,mol^{-1}}ΔG=+12 kJmol−1. Based on thermodynamics, which statement best describes the overall coupled process?

  1. The coupled process is favorable because equal and opposite ΔG\Delta GΔG values make the pathway strongly spontaneous.
  2. The coupled process is favorable because the spontaneous step transfers spontaneity to the nonspontaneous step.
  3. The coupled process is not favorable because the net ΔG\Delta GΔG is zero, so there is no thermodynamic drive. (correct answer)
  4. The coupled process is favorable because coupling makes both steps have ΔG=0\Delta G = 0ΔG=0.
  5. The coupled process is not favorable because a catalyst is required to make ΔG\Delta GΔG negative.

Explanation: This question tests understanding of coupled reactions when ΔG values are equal and opposite. For reactions coupled through intermediate I, the overall Gibbs free energy change is: ΔG_total = ΔG₁ + ΔG₂ = -12 kJ/mol + 12 kJ/mol = 0 kJ/mol. When ΔG_total = 0, the system is at equilibrium with no net thermodynamic drive in either direction—the process is neither favorable nor unfavorable. This means there's no tendency for the overall reaction S → T to proceed spontaneously, though the system can exist with all species present at equilibrium. Choice B incorrectly suggests that spontaneity can be "transferred" between reactions, which misunderstands that only energy (not spontaneity itself) is coupled. For coupled reactions, always calculate the net ΔG; when it equals zero, the system is at equilibrium with no directional preference.

Question 13

A student adds a small amount of sodium chloride, NaCl(s), to distilled water and stirs until the solid is no longer visible. The resulting solution conducts electricity. Which statement best classifies the change that occurred to the NaCl?

  1. Chemical change, because the solution conducts electricity, indicating a new substance formed
  2. Physical change, because NaCl remains as neutral formula units dispersed in water
  3. Physical change, because Na+ and Cl− ions separate and become hydrated, but no new substances are formed (correct answer)
  4. Chemical change, because dissolving always produces new substances
  5. Chemical change, because the solid disappears, meaning it was consumed in a reaction

Explanation: This question tests understanding of physical and chemical changes. In a chemical change, particle identity changes as atoms rearrange to form new substances, while in a physical change, particles may separate or rearrange but maintain their fundamental identity. When NaCl dissolves, the ionic solid separates into Na+ and Cl- ions that become surrounded by water molecules (hydrated), but these are the same ions that existed in the solid—no new substances form. The tempting distractor A is incorrect because electrical conductivity results from mobile ions, not new substance formation. To determine if a change is physical or chemical, ask whether new substances with new compositions form—if the same particles exist before and after, just in different arrangements, it's physical.

Question 14

Consider the reaction A→ProductsA \rightarrow ProductsA→Products. The initial rate of reaction is measured at different initial concentrations of A. When the initial concentration of A is doubled from 0.10 M to 0.20 M, the initial rate of reaction also doubles.

Based on this information, what is the order of the reaction with respect to reactant A?

  1. Zero order
  2. First order (correct answer)
  3. Second order
  4. Third order

Explanation: The correct answer is B. The rate law is Rate = k[A]xk[A]^xk[A]x. When [A] is doubled, the new rate is k(2[A])x=2x(k[A]x)k(2[A])^x = 2^x(k[A]^x)k(2[A])x=2x(k[A]x). The problem states that the rate doubles, so 2x=22^x = 22x=2, which means x=1x = 1x=1. The reaction is first order with respect to A.

Question 15

Element N is a noble gas in Group 18. Based on valence electrons and stability, which statement best describes the tendency of N to form a monatomic ion under typical conditions?

  1. N most commonly forms N8+\text{N}^{8+}N8+ because it has 8 valence electrons.
  2. N does not commonly form a monatomic ion because it already has a stable valence shell. (correct answer)
  3. N most commonly forms N+\text{N}^+N+ because it is a nonmetal.
  4. N most commonly forms N2−\text{N}^{2-}N2− to complete an octet.
  5. N has 18 valence electrons and therefore forms N18−\text{N}^{18-}N18−.

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons determine an atom's stability and likelihood of forming ions. Noble gases in Group 18, like element N, have 8 valence electrons, completing their octet and making them stable without needing to gain or lose electrons. Thus, they do not commonly form monatomic ions under typical conditions. A tempting distractor is that N forms N^{2-} to complete an octet, but it is incorrect because noble gases already have a complete octet. Main-group ions form to achieve noble-gas configurations, but noble gases are already stable and rarely ionize.

Question 16

A student proposes the following mechanism for the net reaction 2SO2(g)+O2(g)→2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}2SO2​(g)+O2​(g)→2SO3​(g):

Step 1: SO2(g)+O2(g)→SO4(g)\mathrm{SO_2(g) + O_2(g) \rightarrow SO_4(g)}SO2​(g)+O2​(g)→SO4​(g) Step 2: SO4(g)+SO2(g)→2SO3(g)\mathrm{SO_4(g) + SO_2(g) \rightarrow 2SO_3(g)}SO4​(g)+SO2​(g)→2SO3​(g)

Which statement about the mechanism is correct?

  1. The mechanism is inconsistent because SO4(g)\mathrm{SO_4(g)}SO4​(g) is not included in the net reaction.
  2. The mechanism is consistent because SO4(g)\mathrm{SO_4(g)}SO4​(g) cancels when the steps are added to obtain the net reaction. (correct answer)
  3. The mechanism is inconsistent because O2(g)\mathrm{O_2(g)}O2​(g) appears only in the first step.
  4. The mechanism is consistent because each step produces exactly one mole of SO3(g)\mathrm{SO_3(g)}SO3​(g).
  5. The mechanism is inconsistent because SO2(g)\mathrm{SO_2(g)}SO2​(g) is used in two different steps.

Explanation: This question tests understanding of introduction to reaction mechanisms. To verify mechanism consistency, we add all steps and check if the sum equals the overall reaction after canceling species on both sides. Adding Step 1 and Step 2: SO₂(g) + O₂(g) + SO₄(g) + SO₂(g) → SO₄(g) + 2SO₃(g). The SO₄(g) appears on both sides and cancels, leaving 2SO₂(g) + O₂(g) → 2SO₃(g), which exactly matches the given net reaction. This confirms the mechanism is consistent. Choice A is incorrect because SO₄(g) is an intermediate that should cancel out and not appear in the net reaction—this is proper behavior, not an inconsistency. When evaluating mechanisms, always verify that intermediates cancel when steps are added, leaving only the overall reaction.

Question 17

A student adds 1.0 g of iodine (I2), a nonpolar molecular substance, to 50 mL of hexane (C6H14), a nonpolar solvent, and stirs. Which statement best describes the result?​

  1. A heterogeneous mixture forms because molecular solutes do not dissolve in molecular solvents.
  2. A homogeneous solution forms, with hexane as the solvent and iodine as the solute. (correct answer)
  3. A heterogeneous mixture forms because hexane is nonpolar and cannot dissolve any solute.
  4. A homogeneous solution forms, with iodine as the solvent and hexane as the solute.
  5. A homogeneous solution forms only if the solute is ionic, so no solution forms here.

Explanation: This question tests the application of "like dissolves like" for nonpolar substances. Iodine (I2) is a nonpolar molecular solid, and hexane (C6H14) is a nonpolar liquid solvent, so they are compatible and form a homogeneous solution. The nonpolar I2 molecules dissolve readily in hexane through London dispersion forces, creating a purple solution. Hexane is the solvent because it's the liquid medium present in larger amount (50 mL), while iodine is the solute being dissolved (1.0 g). Choice A incorrectly claims that molecular solutes don't dissolve in molecular solvents, when compatibility actually depends on matching polarities, not molecular versus ionic character. Remember that nonpolar solutes dissolve well in nonpolar solvents through dispersion forces.

Question 18

Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circΔHf∘​, given to calculate ΔHrxn∘\Delta H_{rxn}^\circΔHrxn∘​ for the reaction below.

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)CaCO3​(s)→CaO(s)+CO2​(g)

Given: ΔHf∘[CaCO3(s)]=−1207 kJ/mol\Delta H_f^\circ[\text{CaCO}_3(s)] = -1207\ \text{kJ/mol}ΔHf∘​[CaCO3​(s)]=−1207 kJ/mol, ΔHf∘[CaO(s)]=−635 kJ/mol\Delta H_f^\circ[\text{CaO}(s)] = -635\ \text{kJ/mol}ΔHf∘​[CaO(s)]=−635 kJ/mol, ΔHf∘[CO2(g)]=−394 kJ/mol\Delta H_f^\circ[\text{CO}_2(g)] = -394\ \text{kJ/mol}ΔHf∘​[CO2​(g)]=−394 kJ/mol.

  1. +178 kJ/mol (correct answer)
  2. -178 kJ/mol
  3. +1207 kJ/mol
  4. +241 kJ/mol
  5. -241 kJ/mol

Explanation: This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. Using ΔHrxn∘=Σ(ΔHf∘ products)−Σ(ΔHf∘ reactants)\Delta H^\circ_{\text{rxn}} = \Sigma(\Delta H^\circ_f \text{ products}) - \Sigma(\Delta H^\circ_f \text{ reactants})ΔHrxn∘​=Σ(ΔHf∘​ products)−Σ(ΔHf∘​ reactants), we calculate the enthalpy change for decomposition. For products: CaO(s) has ΔHf∘=−635\Delta H^\circ_f = -635ΔHf∘​=−635 kJ/mol and CO2_22​(g) has ΔHf∘=−394\Delta H^\circ_f = -394ΔHf∘​=−394 kJ/mol, giving a total of -1029 kJ/mol. For reactants: CaCO3_33​(s) has ΔHf∘=−1207\Delta H^\circ_f = -1207ΔHf∘​=−1207 kJ/mol. Therefore, ΔHrxn∘=−1029−(−1207)=+178\Delta H^\circ_{\text{rxn}} = -1029 - (-1207) = +178ΔHrxn∘​=−1029−(−1207)=+178 kJ/mol. A common mistake (choice B) is getting the sign wrong by subtracting in the wrong order, giving -178 kJ/mol. The positive value makes sense because decomposition reactions typically require energy input.

Question 19

In the carbonate ion, CO32−\mathrm{CO_3^{2-}}CO32−​, carbon is the central atom with three C–O bonding regions (each bond, whether single or double in any resonance form, counts as one electron domain) and no lone pairs on carbon. What is the molecular geometry around carbon?

  1. Linear
  2. Trigonal planar (correct answer)
  3. Tetrahedral
  4. Trigonal pyramidal
  5. Bent

Explanation: This question examines molecular geometry in polyatomic ions using VSEPR, considering resonance. In CO3^2-, carbon has three electron domains from bonds to oxygen (resonance averages to three equivalent domains) and no lone pairs, giving a trigonal planar electron-domain geometry. The molecular geometry is also trigonal planar, with bond angles of 120 degrees. This is consistent with AX3 notation in VSEPR. Choice B, tetrahedral, might attract those counting individual bonds in one resonance structure, a misconception ignoring resonance delocalization. For ions or resonant molecules, always consider the average structure or equivalent domains to predict geometry correctly.

Question 20

A sample of liquid water at 25∘C25^\circ\text{C}25∘C is converted to water vapor at the same temperature in an open container. Which statement correctly describes the enthalpy change for this phase change?

  1. The process is endothermic; heat is absorbed from the surroundings and ΔH>0\Delta H>0ΔH>0. (correct answer)
  2. The process is exothermic; heat is absorbed from the surroundings and ΔH<0\Delta H<0ΔH<0.
  3. The process is endothermic; heat is released to the surroundings and ΔH>0\Delta H>0ΔH>0.
  4. The process is exothermic; heat is released to the surroundings and ΔH<0\Delta H<0ΔH<0.
  5. The process is endothermic; because evaporation is spontaneous, ΔH<0\Delta H<0ΔH<0.

Explanation: This question tests understanding of phase changes and their enthalpy requirements. The conversion of liquid water to water vapor at constant temperature is vaporization, which requires breaking intermolecular hydrogen bonds between water molecules. This process is endothermic because energy must be absorbed from the surroundings to overcome these attractive forces, making ΔH > 0. Even though the temperature remains constant at 25°C, heat energy is still being absorbed and used to increase the potential energy of the molecules as they separate. Choice D incorrectly identifies vaporization as exothermic, confusing the energy requirement with condensation, which releases heat. Remember that all vaporization processes are endothermic (ΔH > 0) because energy is needed to overcome intermolecular forces.

Question 21

A proposed mechanism includes the following single step, which is explicitly identified as an elementary reaction:

Br(g)+H2(g)→HBr(g)+H(g)\mathrm{Br}(g)+\mathrm{H_2}(g)\rightarrow \mathrm{HBr}(g)+\mathrm{H}(g)Br(g)+H2​(g)→HBr(g)+H(g)

What is the rate law for this elementary step?

  1. Rate=k[Br][H2]2\text{Rate}=k[\mathrm{Br}][\mathrm{H_2}]^2Rate=k[Br][H2​]2
  2. Rate=k[Br][H2]\text{Rate}=k[\mathrm{Br}][\mathrm{H_2}]Rate=k[Br][H2​] (correct answer)
  3. Rate=k[HBr][H]\text{Rate}=k[\mathrm{HBr}][\mathrm{H}]Rate=k[HBr][H]
  4. Rate=k[H2]\text{Rate}=k[\mathrm{H_2}]Rate=k[H2​]
  5. Rate=k[Br]2[H2]\text{Rate}=k[\mathrm{Br}]^2[\mathrm{H_2}]Rate=k[Br]2[H2​]

Explanation: This question tests the skill of elementary reactions. The rate law for an elementary step mirrors the molecularity, with exponents matching the number of reactant molecules involved in the collision. Here, one Br and one H2 molecule react, indicating a bimolecular process, so the rate is first-order in each. This is because elementary reactions proceed via a single step, and the rate depends on the frequency of those specific collisions. A tempting distractor is choice E, Rate=k[Br]2[H2]\text{Rate} = k[\mathrm{Br}]^2[\mathrm{H_2}]Rate=k[Br]2[H2​], which is wrong because of the misconception of doubling the coefficient for Br without basis in the step. Always remember that only elementary steps allow coefficients to directly define rate laws; for overall reactions, experimental data is needed.

Question 22

The elements C, N, O, and F are consecutive elements across Period 2. Which element has the highest electronegativity?

  1. F (correct answer)
  2. N
  3. C
  4. O
  5. All have the same electronegativity because they are in the same period

Explanation: This question assesses understanding of periodic trends in electronegativity. Across a period from left to right, nuclear charge increases as protons are added to the nucleus. Shielding stays similar because electrons are filling the same energy level, resulting in a higher effective nuclear charge. This stronger pull makes atoms more effective at attracting electrons in bonds, increasing electronegativity, with F having the highest among C, N, O, and F. A tempting distractor is E, claiming all have the same electronegativity because they are in the same period, but this is wrong as the trend clearly rises across the period due to escalating effective nuclear charge. For electronegativity comparisons, always evaluate the effective nuclear charge, which intensifies from left to right in a period.

Question 23

A dissolution in water at 25∘C25^\circ\text{C}25∘C is observed to cool the solution (ΔHsoln>0\Delta H_{\text{soln}} > 0ΔHsoln​>0). The student also concludes that the overall entropy change is positive (ΔSsoln>0\Delta S_{\text{soln}} > 0ΔSsoln​>0). Which statement is correct about thermodynamic favorability at 25∘C25^\circ\text{C}25∘C?

  1. It is not favored because cooling means the process absorbs heat and therefore cannot be spontaneous.
  2. It is favored only if the dissolution happens rapidly, since rate determines spontaneity.
  3. It is favored only at low temperatures because endothermic processes prefer low TTT.
  4. It may be favored at 25∘C25^\circ\text{C}25∘C, because the positive ΔS\Delta SΔS can make ΔG\Delta GΔG negative even when ΔH\Delta HΔH is positive. (correct answer)
  5. It is not favored because dissolving always decreases entropy due to ordering of water.

Explanation: This question tests understanding of free energy of dissolution when ΔH > 0 and ΔS > 0. With the solution cooling (ΔH > 0, endothermic) and overall entropy increasing (ΔS > 0), the ΔG = ΔH - TΔS equation contains competing terms: positive ΔH opposes dissolution while positive ΔS favors it through the negative -TΔS term. At 25°C (298 K), if the magnitude of TΔS exceeds ΔH, then ΔG < 0 and dissolution is thermodynamically favored, which is often the case for salts like NH₄NO₃. Choice A incorrectly claims endothermic processes cannot be spontaneous, ignoring that entropy can drive spontaneity. The key principle is that endothermic processes can be thermodynamically favorable when the entropy increase is sufficient to overcome the unfavorable enthalpy change.

Question 24

Consider the bonds C−C\mathrm{C-C}C−C in ethane (H3C−CH3\mathrm{H_3C-CH_3}H3​C−CH3​) and C=C\mathrm{C=C}C=C in ethene (H2C=CH2\mathrm{H_2C=CH_2}H2​C=CH2​). Which bond is expected to be shorter?

  1. The C−C\mathrm{C-C}C−C bond in ethane
  2. The C=C\mathrm{C=C}C=C bond in ethene (correct answer)
  3. They are the same length because both are carbon-carbon bonds
  4. The C−C\mathrm{C-C}C−C bond in ethane because it has more hydrogens attached
  5. The C=C\mathrm{C=C}C=C bond in ethene only if ethene has stronger intermolecular forces

Explanation: This question assesses the relationship between bond length and bond order in the context of intramolecular forces and potential energy. The C=C bond in ethene is a double bond, which pulls the carbon atoms closer together due to the additional shared electrons, resulting in a shorter bond length compared to the single C-C bond in ethane. Bond length decreases as bond order increases because higher electron density between nuclei strengthens the attraction. The stimulus provides molecular formulas, highlighting the difference in bond multiplicity, which directly influences bond length. Choice A is a tempting distractor, claiming the C-C bond in ethane is shorter, based on the misconception that more attached hydrogens affect bond length rather than bond order. When evaluating bond lengths, focus on bond order rather than the number of surrounding atoms or intermolecular forces.

Question 25

At a given temperature, the equilibrium constant for CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq) is much less than 1 (K≪1K \ll 1K≪1). What does this imply about which side is favored at equilibrium?

  1. Products are favored; mostly ions are present at equilibrium.
  2. Reactants are favored; mostly molecular CH3COOH\mathrm{CH_3COOH}CH3​COOH is present at equilibrium. (correct answer)
  3. Neither side is favored; ions and molecules are present in comparable amounts at equilibrium.
  4. The reaction is fast, so products dominate at equilibrium.
  5. The reaction goes to completion, leaving no CH3COOH\mathrm{CH_3COOH}CH3​COOH at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant for acid dissociation. When K << 1 for the acid dissociation equilibrium, the equilibrium constant expression K=[H+][CH3COO−][CH3COOH]K = \frac{[\mathrm{H}^{+}][\mathrm{CH_3COO^{-}}]}{[\mathrm{CH_3COOH}]}K=[CH3​COOH][H+][CH3​COO−]​ has a very small value, which means the denominator (undissociated acid) is much larger than the numerator (ions). This indicates that at equilibrium, most of the acetic acid remains in its molecular form CH3COOH\mathrm{CH_3COOH}CH3​COOH, with very little dissociation into ions. The equilibrium position lies far to the left, favoring reactants. A common misconception (option D) is thinking that reaction rate determines equilibrium position - whether a reaction is fast or slow doesn't affect which side is favored at equilibrium. For weak acids with K << 1, remember that the molecular form predominates at equilibrium, not the ionized form.