AP® Chemistry Free-Response Questions Past Paper 2025 Solution

The total abundance must be 100%. Given Mg-24 is 79%, the remaining 21% is split equally between Mg-25 and Mg-26. Thus, each has an abundance of 10.5%. The completed spectrum shows lines at mass 25 and 26 with a relative abundance of 10.5% each.

Isotopes of an element have the same number of protons but a different number of neutrons. Both magnesium-25 and magnesium-26 have 12 protons. Magnesium-25 has 13 neutrons (25 - 12), while magnesium-26 has 14 neutrons (26 - 12). The additional neutron in magnesium-26 accounts for its greater mass.

i. The relative charge of the ions: According to Coulomb's law, the electrostatic force is directly proportional to the magnitude of the charges (F ∝ q₁q₂). The Mg²⁺ ion has a +2 charge, which is greater in magnitude than the +1 charge of the Na⁺ ion. This results in a stronger attraction between the Mg²⁺ ion and the partial negative charge on the oxygen atoms of water molecules.

ii. The relative radii of the ions: The Mg²⁺ ion is smaller than the Na⁺ ion because it has a greater nuclear charge (12 protons vs. 11) pulling on a similar number of electrons. Coulomb's law states that force is inversely proportional to the square of the distance (F ∝ 1/r²). The smaller radius of Mg²⁺ allows water molecules to approach more closely, decreasing the distance 'r' and resulting in a stronger force of attraction.

K_sp = [Mg²⁺][OH⁻]²

A precipitate will form.
Justification: The calculated reaction quotient, Q (2.07 × 10⁻¹¹), is greater than the given solubility product constant, K_sp (5.61 × 10⁻¹²). Since Q > K_sp, the solution is supersaturated with respect to Mg(OH)₂, and the equilibrium will shift to the left, causing solid Mg(OH)₂ to precipitate out of the solution until Q = K_sp.

The amount of undissolved Mg(OH)₂(s) will decrease.
Justification: The dissolution equilibrium is Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). Adding the strong acid HNO₃ introduces H⁺ ions, which neutralize the OH⁻ ions in the solution. This removal of a product causes the equilibrium to shift to the right, according to Le Châtelier's Principle, to produce more ions. This shift results in the dissolution of more solid Mg(OH)₂.

At the half-equivalence point (8.0 mL of NaOH added), pH = pKₐ. From the graph, at 8.0 mL, the pH is approximately 4.1.

Comparing Trial 1 and Trial 3, [I₃⁻] is held constant at 1.200 M while [HAsc] is doubled (0.450 M to 0.900 M). The initial rate also doubles (2.457×10⁻⁴ M/s to 4.914×10⁻⁴ M/s). Since doubling the concentration of HAsc doubles the reaction rate, the reaction is first order in [HAsc].

An ion-dipole force exists between the negatively charged I₃⁻ ion and the partial positive hydrogen atoms of the polar water molecules. This strong electrostatic attraction is not present between the nonpolar I₂ molecule and water, which explains the enhanced solubility of I₃⁻.

Each P atom has 5 valence electrons. In the tetrahedral P₄ molecule, each P atom forms 3 single bonds (using 3 electrons). To complete its octet, each P atom must have one lone pair of electrons (2 electrons). The completed diagram shows a lone pair drawn on each of the four phosphorus atoms.

(This 2D representation shows each P connected to the others, with one lone pair on each).

The reaction is P₄(s) + 5 O₂(g) → P₄O₁₀(s). The entropy of the system decreases (ΔS° is negative) because the reaction consumes 5 moles of gas (O₂), a state of high disorder, and combines it with a solid to produce only a solid product (P₄O₁₀), a state of much lower disorder. The reduction in the number of moles of gas leads to a decrease in the overall entropy of the system.

Yes, the student's claim is correct.
Justification: The thermodynamic favorability of a reaction is determined by the sign of the Gibbs free energy change, ΔG° = ΔH° - TΔS°. For the reaction to be favorable, ΔG° must be negative. We are given that the reaction is favorable, ΔH° is negative (exothermic), and ΔS° is negative. A negative ΔS° makes the -TΔS° term positive. Therefore, for ΔG° to be negative, the magnitude of the negative enthalpy term (|ΔH°|) must be greater than the magnitude of the positive entropy term (|-TΔS°|). This means the reaction is driven by the favorable enthalpy change, which overcomes the unfavorable entropy change.

ΔT for the second trial would be less than the value in the first trial.
Justification: The reaction is exothermic, and the amount of heat released (q) is directly proportional to the amount (moles) of the limiting reactant, P₄O₁₀, that reacts. In the second trial, less P₄O₁₀ was transferred to the calorimeter. Since less reactant is used, less heat will be released by the reaction. Because the mass of the water is the same, a smaller amount of heat released will result in a smaller temperature change (ΔT) for the solution.

The value of Kp will decrease.
Justification: The reaction (Equation 4) has a negative ΔH° (-88 kJ/mol), meaning it is an exothermic reaction. According to Le Châtelier's Principle, if the temperature of an exothermic reaction at equilibrium is increased, the equilibrium will shift to the left (toward the reactants) to absorb the added heat. A shift to the left decreases the concentration of products and increases the concentration of reactants, which results in a smaller value for the equilibrium constant, Kp.

The carbon atom in H₂CO is bonded to three other atoms (two H, one O) and has no lone pairs. It has three electron domains, which corresponds to sp² hybridization.

A dashed line should be drawn from the hydrogen atom of the -OH group on a CH₃OH molecule to the lone pair of electrons on the oxygen atom of an H₂CO molecule. This represents the hydrogen bond.

To be a liquid, the temperature must be below the substance's boiling point and above its melting point. For CH₃OH, the liquid range is 176 K to 338 K. For H₂CO, the liquid range is 181 K to 254 K. A temperature where both are liquid must be in the overlapping range. Therefore, any temperature between 181 K and 254 K is acceptable (e.g., 200 K).

The central Si atom in compound Y is bonded to four other atoms (one O, three C) and has no lone pairs. With four electron domains, the geometry around the Si atom is tetrahedral.

I agree with the student's claim.
Justification: London dispersion forces (LDFs) are stronger in molecules with more electrons and a larger, more polarizable electron cloud. Compound Y (90.2 g/mol) has a larger molar mass than compound X (74.1 g/mol), indicating it has more electrons. This leads to stronger LDFs in compound Y. Since both molecules are of similar polarity (both can hydrogen bond), the stronger LDFs in Y are the primary reason for its higher boiling point (98°C vs 82°C).

Compound X will have the higher vapor pressure.
Justification: Vapor pressure is the pressure of a gas in equilibrium with its liquid. Substances with weaker intermolecular forces (IMFs) evaporate more easily and thus have a higher vapor pressure at a given temperature. Compound X has weaker LDFs than compound Y (as established in part B), and therefore weaker overall IMFs. This allows its molecules to escape the liquid phase more readily, resulting in a higher vapor pressure.

The prompt states the Al(s) electrode decreases in mass, meaning it is being consumed. The anode is the site of oxidation. Therefore, the oxidation half-reaction is: Al(s) → Al³⁺(aq) + 3e⁻

Oxidation: Al(s) → Al³⁺(aq) + 3e⁻
Reduction (Zn²⁺ is reduced since Zn electrode increases in mass): Zn²⁺(aq) + 2e⁻ → Zn(s)
To balance electrons, multiply the oxidation reaction by 2 and the reduction reaction by 3.
2Al(s) → 2Al³⁺(aq) + 6e⁻
3Zn²⁺(aq) + 6e⁻ → 3Zn(s)
Overall: 2Al(s) + 3Zn²⁺(aq) → 2Al³⁺(aq) + 3Zn(s)

The Al electrode's mass changed the most.
Justification: The balanced equation shows that for every 6 moles of electrons transferred, 2 moles of Al (mass = 2 × 26.98 g = 53.96 g) are consumed, and 3 moles of Zn (mass = 3 × 65.38 g = 196.14 g) are produced. To compare the mass change per mole of electrons, we can calculate the mass change per 1 mole of electrons.
For Al: (53.96 g Al / 6 mol e⁻) = 8.99 g Al lost per mole of electrons.
For Zn: (196.14 g Zn / 6 mol e⁻) = 32.69 g Zn gained per mole of electrons.
However, the question asks which *electrode's* mass changed the most, which refers to the magnitude of the change. A better comparison is per mole of metal.
Let's assume 1 mole of Al reacts. Mass change = -26.98 g. This requires 3 moles of electrons. Those 3 moles of electrons will produce (3/2) = 1.5 moles of Zn. Mass change = 1.5 mol * 65.38 g/mol = +98.07 g.
Let's re-read the prompt. It says Zn electrode *increases* and Al *decreases*. This contradicts the standard reduction potentials (Al is more easily oxidized). Assuming the prompt's observation is correct for this specific cell:
Oxidation: Al(s) → Al³⁺(aq) + 3e⁻
Reduction: Zn²⁺(aq) + 2e⁻ → Zn(s)
For every 6 moles of e⁻ that flow, 2 moles of Al (53.96 g) are lost, and 3 moles of Zn (196.14 g) are gained. The change in mass of the zinc electrode (196.14 g) is greater than the change in mass of the aluminum electrode (53.96 g) for the same amount of charge passed. Therefore, the Zinc electrode's mass changed the most.

The atom that accepts the proton is the singly bonded oxygen atom that has a formal negative charge (the oxygen of the carboxylate group). This oxygen has available lone pairs and a negative charge, making it the most basic site on the ion. A circle should be drawn around this O atom in the Lewis diagram.

I disagree with the student's claim.
Justification: A catalyst is a substance that is consumed in an early step of a reaction mechanism and regenerated in a later step, so it does not appear in the overall balanced equation. In the proposed mechanism, H₃O⁺ is produced in Step 1 but is then consumed in Step 2. A substance that is produced and then consumed is a reaction intermediate, not a catalyst. Therefore, the student's claim is incorrect.