2024 AP Chemistry FRQ Answers Explained

Parts (a)–(c)3 points

(a) Circle the H on the carboxylic acid group (the O–H of COOH); the C–H hydrogens are not acidic.

(b) 10.22 g ÷ 40.00 g/mol = 0.2555 mol NaOH in 0.500 L, so 0.511 M. The molarity guide covers this.

(c) pKa is the pH at the half-equivalence point: about 3.9 (3.7 to 4.0 accepted).

Part (d)4 points

(i)–(ii) The particle diagram shows more lactic acid molecules than lactate ions, so the mixture is before the half-equivalence point: an X anywhere from 3 mL to just under 8 mL of NaOH. Before half-equivalence, [HA] > [A⁻].

(iii) With NaOH twice as concentrated, half the volume reaches equivalence: about 8 mL instead of 16 mL. The new curve starts at the same pH, rises sharply at 8 mL, and ends near the same final pH. One point is for the equivalence volume, one for the shape. See AP Chemistry titration.

Part (e)3 points

100.0 mL of 0.500 M lactic acid and 100.0 mL of 0.500 M NaOH, both at 20.0 °C, reach 23.2 °C (density 1.00 g/mL, c = 4.2 J/g·°C).

(i) q = (200.0 g)(4.2 J/g·°C)(3.2 °C) = 2700 J (2.7 × 10³ J, two significant figures). (ii) Moles reacting = (0.100 L)(0.500 M) = 0.0500 mol, so ΔH = −2.7 kJ ÷ 0.0500 mol = −54 kJ/mol_rxn. (iii) Agree. Heat lost to the air gives a smaller temperature rise, so q and the magnitude of ΔH come out smaller than the true value.

Use the combined 200.0 g, not 100.0 g. Calorimetry sign rules are in the calorimetry guide.

Part (a)2 points

(i) 0.0114 mol × 44.01 g/mol = 0.502 g CO₂. (ii) V = nRT/P = (0.0114)(0.08206)(293 K) ÷ 1.25 atm = 0.219 L. Try it in the gas law calculator.

Part (b)3 points

(i) Grinding into powder increases the surface area. (ii) Shorter. More surface means more frequent collisions between water and the solid particles, so the solids dissolve faster. (iii) Equal. The same amounts of reactants give the same moles of CO₂, and at the same temperature and pressure the same volume. Surface area changes the rate, not the yield.

Parts (c)–(e)3 points

(c) NaHCO₃ is limiting in trial 3: 1.251 g ÷ 84.01 g/mol = 0.01489 mol NaHCO₃, which gives 0.01489 mol CO₂ (matching the data), while 1.543 g of maleic acid could make 0.02659 mol CO₂. Trial 4 confirms it: more NaHCO₃ gave more CO₂. See the limiting reactant calculator.

(d) The reaction makes 2 moles of gas from no gaseous reactants, and gases are far more dispersed than solutions and liquids, so entropy increases.

(e) Disagree. With ΔH > 0 and ΔS > 0, ΔG = ΔH − TΔS is negative only at high enough temperatures, not at all temperatures. See Gibbs free energy.

Parts (f)–(g)2 points

(f) pK_a2 = −log(8.5 × 10⁻⁷) = 6.07. (g) 7.00 = 6.07 + log([C₄H₂O₄²⁻]/[HC₄H₂O₄⁻]), so the ratio = 10^0.93 = 8.5.

The buffer uses the second dissociation, so use K_a2, not K_a1. See the Henderson-Hasselbalch equation.

Parts (a)–(c)5 points

(a) Ag(s) is 0; in Ag₂S, silver is +1 (sulfide is −2). See oxidation numbers.

(b)(i) Ag (165 pm) and Cu (145 pm) have similar radii, so copper atoms replace silver atoms in the lattice: a substitutional alloy. Interstitial alloys need a much smaller atom. (ii) Silver’s valence electrons are in the fifth shell, copper’s in the fourth, so silver’s are farther from the nucleus and less strongly attracted.

(c) Mass of Ag₂S removed = 409.21 − 398.94 = 10.27 g. Then 10.27 g ÷ 247.80 g/mol × 2 mol Ag/mol Ag₂S = 0.08289 mol Ag. One point is for the mass difference, one for the moles.

Part (d)3 points

(i) Rhodium is reduced and water is oxidized to O₂; balancing 12 electrons: 4 Rh³⁺(aq) + 6 H₂O(l) → 4 Rh(s) + 3 O₂(g) + 12 H⁺(aq). Check it with the balancing equations calculator.

(ii) E°_cell = 0.80 V − 1.23 V = −0.43 V. (iii) A negative E°_cell means ΔG° is positive, so the reaction is not thermodynamically favorable and needs an external power source.

Part (e)2 points

2.8 g ÷ 102.9 g/mol = 0.0272 mol Rh; × 3 = 0.082 mol e⁻; × 96,485 C/mol = 7.9 × 10³ C; ÷ 2.0 C/s = 3.9 × 10³ s.

Rh³⁺ needs three electrons per atom. The full method is in electrolysis and Faraday’s law.

Parts (a)–(d)4 points, 1 each

(a) Read the thermometer to one estimated digit beyond the smallest marking: 38.5 °C.

(b) The water warms, so its molecules move faster on average: draw arrows that are longer, on average, than in the Before diagram.

(c) c = q ÷ (mΔT) = 2940 J ÷ [(98.1 g)(61.5 °C)] = 0.487 J/g·°C, using the metal’s drop from 100.0 °C to 38.5 °C.

(d) Aluminum’s specific heat (0.897 J/g·°C) is larger, so the same 2940 J changes the temperature of the same mass less: about 33.4 °C instead of 61.5 °C. The calorimetry guide has more practice.

Parts (a)–(b)4 points, 1 each

(a) K_c = [HI]² ÷ ([H₂][I₂]).

(b)(i) The diagram shows 3 H₂ and 2 I₂. For Q = 0.67: [HI]² = 0.67 × 3 × 2 = 4, so draw 2 HI molecules.

(ii) The moles of HI rise after time t without adding HI, so the change could be lowering the temperature (favoring the exothermic forward reaction) or adding H₂ or I₂.

(iii) Remains the same. There are 2 moles of gas on each side, so changing the volume changes every concentration by the same factor and leaves Q equal to K. See Le Chatelier’s principle and the reaction quotient.

Parts (a)–(c)4 points, 1 each

(a) The plot of 1/[NO₂] against time is linear, which is the signature of a second-order reaction. See integrated rate laws.

(b) O₂ forms at half the rate NO₂ disappears (1 : 2 coefficients): 6.52 × 10⁻⁷ ÷ 2 = 3.26 × 10⁻⁷ M/s.

(c)(i) NO₂⁺ has 16 valence electrons: O=N=O with two lone pairs on each O and none on N. (ii) Agree. NO₂ has an unpaired electron on N, giving a bent shape near 120°, while NO₂⁺ has only two electron domains on N and is linear at 180°. See Lewis structures and the VSEPR chart.

Parts (a)–(c)4 points

(a) (0.340 mol/L)(0.1000 L)(58.44 g/mol) = 1.99 g NaCl (1 point).

(b) Step 2: transfer the NaCl into the 100.0 mL volumetric flask with some distilled water. Step 4: add distilled water up to the 100.0 mL calibration mark (1 point each). A graduated cylinder or beaker is not precise enough for a 0.340 M solution.

(c) The separation would be smaller. In 15 minutes the solvent front travels a shorter distance, so the spots have less distance to separate (1 point). See AP Chemistry chromatography.