2026 AP Chemistry FRQ Answers Explained

Part A(i)1 point

Write the electron configuration of K⁺.

1s² 2s² 2p⁶ 3s² 3p⁶, or [Ne] 3s² 3p⁶: potassium loses its single 4s electron. Check ions with the electron configuration calculator.

Part A(ii)1 point

Which is larger, K or K⁺?

The K atom. Its outermost electron is in the n = 4 shell, farther from the nucleus than the n = 3 electrons that are outermost in K⁺. Explaining that K has the same nuclear charge but one more electron also earns the point.

Part B1 point

How do the temperature readings show that dissolving is complete?

The temperature stops falling: it reaches a minimum and then begins to drift back up as the room warms the calorimeter.

Part C(i)1 point

Calculate the magnitude of q in joules (c = 3.95 J/(g·°C)).

Use the mass of the whole solution, water plus KCl: q = mcΔT = (97.5 g + 6.80 g)(3.95 J/(g·°C))(21.1 − 24.5 °C), so |q| = 1.4 × 10³ J. The point requires two significant figures, set by the temperature change of 3.4 °C. See calorimetry.

Part C(ii)1 point

Calculate the molar enthalpy of solution, with its sign, for 0.0912 mol KCl.

The solution lost 1,400 J, so the dissolving system gained it: ΔH_soln = +1,400 J ÷ 0.0912 mol = +15 kJ/mol. The positive sign is required; the temperature drop means dissolving is endothermic.

Part D1 point

With imperfect insulation, is the calculated magnitude greater than, less than, or equal to the true value?

Less than. Heat leaks in from the warmer room, so the temperature drops less than it should. A smaller ΔT gives a smaller q and a smaller ΔH_soln.

Part E1 point

With 6.80 g of RbCl instead, which has nearly the same molar ΔH_soln, is ΔT greater, less, or equal?

Less than. RbCl has a larger molar mass (120.92 vs 74.55 g/mol), so the same mass contains fewer moles. Fewer moles at the same molar enthalpy absorb less heat.

The salts have similar molar ΔH values, so the deciding factor is moles per gram: the justification has to compare molar masses.

Part F3 points, 1 each

(i) RbCl(s) ⇌ Rb⁺(aq) + Cl⁻(aq).

(ii) K_sp = [Rb⁺][Cl⁻] = s² = 57, so s = 7.5 M. RbCl is very soluble, which is why K_sp is so large.

(iii) Less than. The Cl⁻ already present from 1.0 M KCl is a common ion that shifts the equilibrium toward solid RbCl. Solving s(s + 1.0) = 57 gives s = 7.1 M, lower than 7.5 M. See Ksp and molar solubility.

Part A2 points, 1 each

(i) Chromium has four bonding domains and no lone pairs, so CrO₄²⁻ is tetrahedral. See the VSEPR chart.

(ii) Draw two Cr=O double bonds and two Cr–O single bonds. Each double-bonded O has two lone pairs (formal charge 0) and each single-bonded O has three (formal charge −1), keeping the same electron count. See formal charge.

Part B2 points, 1 each

(i) 2 CrO₄²⁻(aq) + 2 H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l). Writing H₃O⁺ with 3 H₂O is also accepted. The nitrate from HNO₃ is a spectator ion.

(ii) No. Chromium is +6 in both ions (x + 4(−2) = −2 and 2x + 7(−2) = −2), so no element changes oxidation number. See oxidation numbers.

Part C1 point

Is plating chromium from CrO₃ (E° = −1.32 V, 6 electrons) favorable? Justify with ΔG°.

ΔG° = −nFE° = −(6)(96,485 C/mol)(−1.32 V) = +764 kJ/mol, so the reaction is thermodynamically unfavorable, which is why it must be driven by an outside current. See Gibbs free energy.

Part D2 points

What mass of Cr plates out with 15.0 A for 3,250 s?

Moles of electrons (1 point): (15.0 A)(3,250 s) ÷ 96,485 C/mol = 0.505 mol e⁻. Mass (1 point): 0.505 mol e⁻ × (1 mol Cr / 6 mol e⁻) × 52.00 g/mol = 4.38 g Cr. See electrolysis and Faraday’s law.

Chromium here gains 6 electrons per atom (from +6 to 0), not 3; using 3 doubles the answer.

Part E1 point

How does the concentration-time graph show the reaction is first order in Cr₂O₇²⁻?

The concentration halves every 20 minutes whatever the starting point. A constant half-life is the signature of a first-order reaction. See integrated rate laws.

Part F1 point

Calculate k from the ln[Cr₂O₇²⁻] versus time plot.

For a first-order reaction, k is the negative of the slope. Reading two points from the line, such as ln[ ] = −0.70 at 0 min and −1.20 at 15 min, gives k = 0.50 ÷ 15 min = 0.033 min⁻¹. Using the half-life instead, k = 0.693 ÷ 20 min = 0.035 min⁻¹, which is also accepted. The half-life calculator converts between the two.

Part G1 point

Draw the ln plot for a second trial starting at 0.25 M.

Start at ln(0.25) ≈ −1.40 and draw a straight line parallel to the original. The rate constant, and so the slope, does not depend on concentration.

Part A1 point

Either pair, labeled: HNO₂ (acid) and NO₂⁻ (base), or H₃O⁺ (acid) and H₂O (base). See Brønsted-Lowry acids and bases.

Part B2 points, 1 each

(i) [H₃O⁺] = 10^−2.09 = 0.0081 M for the 0.125 M solution.

(ii) [HNO₂] = 0.125 − 0.0081 = 0.117 M, because each ionized molecule produces one H₃O⁺. See ICE tables.

Part C2 points, 1 each

(i) At 333 K, K_a = (0.0109)² ÷ 0.114 = 1.04 × 10⁻³.

(ii) Endothermic. K_a rose from 5.6 × 10⁻⁴ to 1.04 × 10⁻³ as the temperature increased, and raising the temperature favors the endothermic direction. See Le Châtelier’s principle.

Part D1 point

From the curve, the equivalence point is at about pH 8 (any value above 7 and up to 10 is accepted): titrating a weak acid gives a basic equivalence point, because NO₂⁻ is a weak base.

Part E1 point

Equivalence takes 50.0 mL of NaOH: (0.0500 L)(0.16 M) = 0.0080 mol NaOH = 0.0080 mol HNO₂, and 0.0080 mol ÷ 0.0350 L = 0.23 M. See titration.

Part F1 point

Place the X anywhere before 25 mL, the half-equivalence point. Until then, less than half the acid has been converted, so [HNO₂] > [NO₂⁻]. See titration curve calculations.

Part G1 point

The neutralization equals the K_a reaction plus the reverse of water’s autoionization, so K₂ = K_a × (1/K_w) = 5.6 × 10⁻⁴ ÷ 1.0 × 10⁻¹⁴ = 5.6 × 10¹⁰. The huge value shows the reaction goes essentially to completion.

Part H1 point

Disagree. Methyl orange changes color between pH 3.1 and 4.4, well before the equivalence point near pH 8–9. Thymol blue (8.0–9.6) would change color at the equivalence point.

Parts A–C4 points

A (1 point). P₄ has single bonds and P₂ a triple bond. Sharing more electron pairs pulls the nuclei closer, so the single bonds are longer. See bond order.

B (2 points). P₄ ⇌ 2 P₂: forming 0.630 atm of P₂ uses half as much, 0.315 atm, of P₄, leaving 0.470 − 0.315 = 0.155 atm (1 point). K_p = (0.630)² ÷ 0.155 = 2.56 (1 point). See Kc and Kp.

C (1 point). Agree. One mole of gas becomes two, so ΔS° > 0. Since ΔG° = ΔH° − TΔS° is negative only above 1,500 K, ΔH° must also be positive: the reaction is endothermic.

In Part B, subtracting the full 0.630 atm from 0.470 atm ignores the 1 : 2 ratio and gives a negative pressure.

Parts A–C4 points, 1 each

A. C–F, because the electronegativity difference between C and F is the largest of the three bonds. See electronegativity.

B. Br and Cl have larger electron clouds than F, so their bonding electrons repel each other more and spread the Br–C–Cl angle wider than the F–C–F angle.

C(i). CBr₄: London dispersion forces only (it is nonpolar). CBrClF₂: London dispersion and dipole-dipole forces.

C(ii). CBr₄ has a much larger, more polarizable electron cloud, so its dispersion forces are stronger than the dispersion and dipole-dipole forces of CBrClF₂ combined. See London dispersion forces and intermolecular forces.

Base the explanation on the size and polarizability of the electron cloud, not on mass, and compare all the forces present in each liquid.

Parts A–C4 points, 1 each

A. An absorbance of 0.08 is one-fourth of 0.32, and absorbance is proportional to concentration, so draw one-fourth as many ions as the 0.32 diagram: 2 ions.

B(i). c = A ÷ 4.0 M⁻¹ = 0.22 ÷ 4.0 = 0.055 M (0.053–0.057 M read from the graph is accepted). See the Beer-Lambert law.

B(ii). M₁V₁ = M₂V₂: M₁ = (0.055 M)(25.0 mL) ÷ 3.00 mL = 0.46 M. See dilution calculations.

C. Agree. Filling past the mark adds extra water, so the diluted solution is more dilute than assumed. Its absorbance and calculated concentration are too low, and so is the back-calculated original concentration.

Parts A–C4 points

A (1 point). Na and O₂ are elements in their standard states (ΔH_f° = 0), so −828 = 2x and ΔH_f°(Na₂O) = −414 kJ/mol. See Hess’s law and enthalpies of formation.

B (2 points). 18.4 g Na ÷ 22.99 g/mol = 0.800 mol Na, which needs only 0.200 mol O₂; 12.8 g O₂ is 0.400 mol, so Na is limiting (1 point). 0.800 mol Na × (828 kJ / 4 mol Na) = 166 kJ released (1 point). See limiting reactant.

C (1 point). Rb⁺ has a larger radius than Na⁺, so the ion centers in Rb₂O are farther apart. Coulombic attraction falls with distance, so less energy is needed to separate the ions. See lattice energy.

Using 12.8 g O₂ as the limiting amount gives 331 kJ, twice the correct heat.