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Early and unofficial 2026 solutions

Early Solutions to the 2026 AP Chemistry FRQs | Step by Step

A complete walkthrough of all seven released FRQs, with balanced equations, particle-level reasoning, graph interpretation, calculations, units, signs, significant figures, and AP-ready final answers.

7 complete FRQs43 parts and subparts5 visual placeholdersAll calculations shown
Important: These are independent, early solutions prepared from the uploaded detailed-solution PDF and checked against the released 2026 question paper. They are not official College Board scoring guidelines. Equivalent chemically correct reasoning may also earn credit.

How to use these AP Chemistry FRQ solutions

For each part, start with the governing chemical principle, show the calculation or relationship, substitute the data with units, and then state the conclusion. For an “explain” or “justify” prompt, connect the observable result to particles, bonding, energetics, equilibrium, or experimental measurement.

Balance firstCheck every chemical equation for both atoms and total charge before using stoichiometric coefficients.
Track signs and unitsCooling means the solution loses heat while an endothermic dissolution absorbs it. “Heat released” corresponds to negative q for the reacting system.
Use the graph featureSupport a graphical claim with constant half-life, a linear slope, an inflection point, or a proportional relationship.
AP Chemistry past FRQs AP Chemistry score calculator AP study hub

Long free-response question

FRQ 1: Calorimetry, solution enthalpy, and solubility

A student dissolves KCl in a calorimeter, analyzes the temperature decrease, compares KCl with RbCl, and applies Ksp and the common-ion effect.

A(i)Ground-state electron configuration of K+

Task: Write the complete ground-state electron configuration for K+.

  1. Neutral potassium has atomic number 19 and the configuration 1s2 2s2 2p6 3s2 3p6 4s1.
  2. Forming K+ removes the outermost 4s electron, leaving 18 electrons.
AP-ready answerK+: 1s2 2s2 2p6 3s2 3p6

A(ii)Comparing the radii of K and K+

Task: Identify the larger species and explain using atomic structure.

The K atom has an occupied fourth energy level. K+ has lost that entire outer shell and has the Ar electron configuration. The same 19 protons also attract only 18 electrons in K+, increasing the attraction experienced by the remaining electrons.

AP-ready answerThe K atom has the larger radius. K+ loses its outermost shell, and its remaining electrons are pulled more strongly toward the nucleus.

Calorimetry data

KCl dissolution experiment
QuantityValue
Mass of water97.5 g
Mass of KCl(s)6.80 g
Initial temperature24.5°C
Final temperature21.1°C
Specific heat capacity of solution3.95 J g−1 °C−1

BDetermining when dissolution is complete

Task: Describe how the temperature readings show that dissolution is complete.

The endothermic dissolving process absorbs energy and lowers the solution temperature while KCl is still dissolving. When the reading reaches its minimum and stops changing because of dissolution, the process is complete.

AP-ready answerDissolution is complete when the temperature reaches its lowest value and then remains constant, aside from slow drift caused by heat exchange with the surroundings.

C(i)Magnitude of thermal energy transferred

Task: Calculate |q| in joules.

msolution = 97.5 g + 6.80 g = 104.3 gΔT = 21.1°C − 24.5°C = −3.4°Cqsolution = mcΔT = (104.3 g)(3.95 J g−1 °C−1)(−3.4°C) = −1.4007 × 103 J

The prompt requests the magnitude, so report the positive magnitude of the transfer.

AP-ready answer|q| = 1.40 × 103 J. The solution loses this energy, and the dissolution absorbs it.

C(ii)Molar enthalpy of solution

Task: Calculate ΔHsoln for 0.0912 mol KCl, including its sign.

qdissolution = −qsolution = +1.4007 kJΔHsoln = +1.4007 kJ0.0912 mol = +15.359 kJ mol−1
The positive sign is required: the falling temperature shows that the dissolution process absorbs heat and is endothermic.
AP-ready answerΔHsoln = +15.4 kJ mol−1

DEffect of imperfect insulation

Task: Compare the calculated magnitude of ΔHsoln with the accepted magnitude.

  1. The dissolving solution becomes colder than its surroundings.
  2. Heat leaks from the surroundings into the solution, partly offsetting the cooling.
  3. The measured |ΔT| and therefore mc|ΔT| are too small.
AP-ready answerThe calculated magnitude of ΔHsoln is less than the accepted magnitude.

EComparing |ΔT| for RbCl and KCl

Task: Equal masses of the salts have similar molar enthalpies. Compare their temperature changes.

nKCl = 6.80 g74.55 g mol−1 = 0.0912 molnRbCl = 6.80 g120.92 g mol−1 = 0.0562 mol

Because |q| = n|ΔHsoln|, fewer moles of RbCl absorb less energy. With equal total mass and specific heat capacity, |ΔT| = |q|/(mc) is smaller.

AP-ready answer|ΔT|RbCl < |ΔT|KCl. A numerical estimate gives about 2.1°C for RbCl versus 3.4°C for KCl.

F(i)Dissolution equation for RbCl

Task: Write the balanced net ionic dissolution equation with states and charges.

RbCl(s) ⇌ Rb+(aq) + Cl−(aq)
AP-ready answerRbCl(s) ⇌ Rb+(aq) + Cl−(aq)

F(ii)Molar solubility in pure water

Task: Calculate the molar solubility when Ksp = 57.

[Rb+] = s,   [Cl−] = sKsp = [Rb+][Cl−] = s2 = 57s = √57 = 7.5498 M
AP-ready answers = 7.55 M

F(iii)Solubility in 1.0 M KCl

Task: Compare the solubility with that in pure water and justify.

KCl supplies the common ion Cl−. Because Ksp = [Rb+][Cl−], beginning with a nonzero chloride concentration requires a smaller equilibrium concentration of dissolved Rb+ and shifts the dissolution equilibrium left.

s(1.0 + s) = 57  ⇒  s = −1 + √2292 = 7.07 M < 7.55 M
AP-ready answerThe molar solubility is less than the pure-water value because of the common-ion effect.

Long free-response question

FRQ 2: Chromate, electrochemistry, and kinetics

This question combines VSEPR and resonance, chromate-dichromate equilibrium, oxidation numbers, electrochemical thermodynamics, Faraday's law, and first-order kinetics.

A(i)Molecular geometry of CrO42−

Task: Predict the molecular geometry using VSEPR.

Chromium has four Cr-O bonding domains and no lone pairs in the resonance model. A single or double bond counts as one domain.

AP-ready answerCrO42− has tetrahedral molecular geometry, with ideal bond angles of approximately 109.5°.

A(ii)Resonance Lewis structure

Task: Draw a complete resonance structure with formal charge 0 on two O atoms and −1 on the other two.

Use two Cr=O double bonds and two Cr-O− single bonds. Each double-bonded O has two lone pairs; each single-bonded O− has three lone pairs. Enclose the structure in brackets with an overall 2− charge.

AP Chemistry 2026 FRQ 2 resonance Lewis structure of chromate with two chromium oxygen double bonds, two chromium oxygen single bonds, lone pairs, and an overall two minus charge
Replaceable placeholder for the completed CrO42− resonance Lewis structure.
FC(double-bonded O) = 6 − 4 − 4/2 = 0FC(single-bonded O) = 6 − 6 − 2/2 = −1
AP-ready answerTwo Cr=O bonds and two Cr-O− bonds give two O atoms with formal charge 0, two with formal charge −1, and total charge 2−.

B(i)Chromate-dichromate equilibrium in acid

Task: Write a balanced net ionic equation.

2 CrO42−(aq) + 2 H3O+(aq) ⇌ Cr2O72−(aq) + 3 H2O(l)
Equivalent H+ form: 2 CrO42−(aq) + 2 H+(aq) ⇌ Cr2O72−(aq) + H2O(l).
AP-ready answer2 CrO42−(aq) + 2 H3O+(aq) ⇌ Cr2O72−(aq) + 3 H2O(l)

B(ii)Is the reaction redox?

Task: Justify using the oxidation number of Cr.

Chromate: x + 4(−2) = −2  ⇒  x = +6Dichromate: 2x + 7(−2) = −2  ⇒  x = +6
AP-ready answerThe reaction is not redox. Chromium remains at oxidation number +6, so no oxidation number changes.

CThermodynamic favorability of chromium plating

Task: Use E° = −1.32 V for the six-electron half-reaction to calculate ΔG°.

CrO3(aq) + 6 H+(aq) + 6 e− → Cr(s) + 3 H2O(l)
ΔG° = −nFE°ΔG° = −(6 mol e−)(96,485 C mol−1)(−1.32 J C−1)ΔG° = +7.64 × 105 J mol−1 = +764 kJ mol−1
AP-ready answerΔG° = +764 kJ mol−1. Because ΔG° > 0, the half-reaction is thermodynamically unfavorable as written under standard conditions.

DMass of chromium plated

Task: Calculate the mass deposited by 15.0 A for 3250 s.

Q = It = (15.0 C s−1)(3250 s) = 4.875 × 104 Cne− = 4.875 × 104 C96,485 C mol−1 = 0.5053 mol e−nCr = 0.5053 mol e− × 1 mol Cr6 mol e− = 0.08421 mol CrmCr = (0.08421 mol)(52.00 g mol−1) = 4.38 g
AP-ready answer4.38 g Cr

EEvidence for first-order behavior

Task: Explain how the concentration-time data support first order in Cr2O72−.

A first-order reaction has a concentration-independent half-life. The concentration falls from 0.50 M to 0.25 M in about 20 min and from 0.25 M to 0.125 M in about another 20 min.

AP-ready answerThe approximately constant 20-minute half-life supports first-order behavior with respect to Cr2O72−.

FRate constant from the logarithmic plot

Task: Calculate k from the slope of ln[Cr2O72−] versus time.

ln[A]t = −kt + ln[A]0slope = −1.30 − (−0.70)18 min − 0 min = −0.0333 min−1k = −slope = 0.0333 min−1
AP-ready answerk = 3.33 × 10−2 min−1

GSecond logarithmic plot

Task: Draw the expected line when the initial concentration is 0.25 M instead of 0.50 M under identical conditions.

  1. The rate constant is unchanged, so the new line has the same slope, −k = −0.0333 min−1.
  2. The new intercept is ln(0.25) = −1.386.
  3. The line is parallel to the original and lower by ln 2 = 0.693.
ln[Cr2O72−] = −0.0333t − 1.386
AP Chemistry 2026 FRQ 2 graph showing parallel first-order plots of natural log dichromate concentration versus time for initial concentrations 0.50 M and 0.25 M
Replaceable placeholder for the completed second logarithmic kinetics plot.
AP-ready answerDraw a line through (0, −1.386) parallel to the original line, with slope −0.0333 min−1.

Long free-response question

FRQ 3: Weak-acid equilibrium and titration

Nitrous acid ionization, temperature dependence of Ka, a weak-acid/strong-base titration, neutralization equilibrium, and indicator choice.

Ionization reaction and given value

HNO2(aq) + H2O(l) ⇌ NO2−(aq) + H3O+(aq)

At 298 K, Ka = 5.6 × 10−4.

AConjugate acid-base pair

Task: Identify and label one conjugate pair.

HNO2 donates H+ and becomes NO2−; the two species differ by one proton.

AP-ready answerHNO2 is the acid and NO2− is its conjugate base. H2O/H3O+ is another valid pair.

B(i)Hydronium concentration from pH

Task: A 0.125 M HNO2 solution has pH 2.09. Calculate [H3O+].

[H3O+] = 10−pH = 10−2.09 = 8.13 × 10−3 M
AP-ready answer[H3O+] = 8.13 × 10−3 M

B(ii)Equilibrium concentration of HNO2

Task: Calculate [HNO2]eq.

Each mole of HNO2 that ionizes produces one mole of H3O+. Neglect the very small contribution from water.

[HNO2]eq = 0.125 M − 8.13 × 10−3 M = 0.11687 M
AP-ready answer[HNO2]eq = 0.117 M

C(i)Ka at 333 K

Task: Use [HNO2] = 0.114 M, [NO2−] = 0.0109 M, and [H3O+] = 0.0109 M.

Ka = [NO2−][H3O+][HNO2] = (0.0109)(0.0109)0.114 = 1.042 × 10−3
AP-ready answerKa(333 K) = 1.04 × 10−3

C(ii)Endothermic or exothermic ionization

Task: Compare Ka at 298 K and 333 K.

Ka increases from 5.6 × 10−4 to 1.04 × 10−3 when heated. Heating shifts equilibrium toward NO2− and H3O+, so heat acts as a reactant in the forward direction.

AP-ready answerThe ionization reaction is endothermic.

Titration of unknown HNO2

A student titrates 35.0 mL of HNO2 with 0.16 M NaOH. The equivalence point occurs at approximately 50.0 mL of NaOH added.

AP Chemistry 2026 FRQ 3 titration curve for nitrous acid with sodium hydroxide showing half-equivalence near 25 milliliters and equivalence near 50 milliliters at pH about 8.1
Replaceable placeholder for the annotated HNO2-NaOH titration curve.

DpH at the equivalence point

Task: Read the equivalence-point pH from the graph.

The equivalence point is the inflection point in the steep vertical region, at about 50.0 mL. The graph gives a pH near 8; conjugate-base hydrolysis gives approximately 8.1.

AP-ready answerpHeq ≈ 8.1

EMolarity of the original HNO2

Task: Use the 1:1 neutralization and 50.0 mL equivalence volume.

nOH− = (0.16 mol L−1)(0.0500 L) = 0.00800 mol[HNO2] = 0.00800 mol0.0350 L = 0.2286 M
AP-ready answer[HNO2] ≈ 0.23 M

FWhere [HNO2] > [NO2−]

Task: Place an X on a valid point on the titration curve.

At half-equivalence, [HNO2] = [NO2−]. Because equivalence is 50 mL, half-equivalence is 25 mL. Before 25 mL, more acid than conjugate base remains.

AP-ready answerPlace X anywhere on the curve where VNaOH < 25 mL.

GEquilibrium constant for neutralization

Task: Calculate K2 for HNO2 + OH− → NO2− + H2O.

Add acid ionization to the reverse of water autoionization. H3O+ and one H2O cancel, and equilibrium constants multiply.

K2 = Ka1Kw = 5.6 × 10−41.0 × 10−14 = 5.6 × 1010
AP-ready answerK2 = 5.6 × 1010, consistent with nearly complete neutralization by a strong base.

HChoosing an indicator

Task: Evaluate the claim that methyl orange is the best choice.

Available indicators
IndicatorpH rangeFit to equivalence region
Methyl orange3.1-4.4Changes before equivalence
Thymol blue8.0-9.6Overlaps steep region near pH 8.1
Clayton yellow12.2-13.2Changes after equivalence
AP-ready answerDisagree. Thymol blue is best. Methyl orange changes too early, making the measured endpoint volume and calculated acid concentration too low.

Short free-response question

FRQ 4: Phosphorus bonding and gas equilibrium

Compare P-P bond lengths, calculate Kp with an ICE table, and use ΔG° = ΔH° - TΔS° to explain high-temperature favorability.

AWhy P4 bonds are longer than P2

Task: Explain the 221 pm versus 189 pm bond lengths.

Each P-P connection in P4 is a single bond with bond order 1. P2 contains a P≡P triple bond with bond order 3. Higher bond order places more electron density between nuclei, producing stronger attraction and a shorter distance.

AP-ready answerP-P bonds in P4 are longer because they are single bonds, whereas the higher-bond-order P≡P bond in P2 is shorter.

BCalculating Kp

Task: For P4(g) ⇌ 2 P2(g), use PP4,initial = 0.470 atm and PP2,eq = 0.630 atm.

ICE table in atm
P4P2
Initial0.4700
Change−x+2x
Equilibrium0.470 − x2x
2x = 0.630 atm  ⇒  x = 0.315 atmPP4,eq = 0.470 − 0.315 = 0.155 atmKp = (PP2)2PP4 = (0.630)20.155 = 2.56
AP-ready answerKp = 2.56

CWhy high temperature favors decomposition

Task: Evaluate the claim that the reaction must be endothermic.

  1. The reaction changes one mole of gas into two, so ΔS°rxn > 0.
  2. In ΔG° = ΔH° - TΔS°, the entropy term becomes more negative as temperature rises.
  3. For ΔG° to be positive at low temperature and negative only above a threshold, ΔH° must be positive.
AP-ready answerAgree. ΔH°rxn > 0 and ΔS°rxn > 0. At sufficiently high temperature, the negative −TΔS° term outweighs the positive ΔH°, making ΔG° < 0.

Short free-response question

FRQ 5: Bond polarity, VSEPR, and intermolecular forces

Analyze CBrClF2 bond polarity and bond angles, then compare intermolecular forces and boiling points with CBr4.

AMost polar bond in CBrClF2

Task: Choose C-Br, C-Cl, or C-F and justify.

Bond polarity increases with electronegativity difference. Fluorine is more electronegative than chlorine or bromine and much more electronegative than carbon.

AP-ready answerThe C-F bond is most polar because it has the greatest electronegativity difference and largest bond dipole.

BExplaining unequal bond angles

Task: Explain F-C-F = 106.8° and Br-C-Cl = 112.3° using atomic structure and VSEPR.

Carbon has four bonding domains, so the basic geometry is tetrahedral. Fluorine pulls shared electrons away from central C, leaving less bonding electron density near C and weaker repulsion between the C-F domains. C-Br and C-Cl electron density remains closer to C, producing stronger repulsion around the central atom.

AP-ready answerWeaker repulsion between the C-F bonding domains compresses F-C-F to 106.8°, while stronger repulsion between the C-Br and C-Cl domains expands Br-C-Cl to 112.3°.

C(i)Intermolecular forces

Task: Identify all IMFs in pure liquid CBr4 and CBrClF2.

CBr4 is tetrahedral and symmetric, so bond dipoles cancel. CBrClF2 is asymmetric and polar. Neither contains H bonded to N, O, or F, so neither hydrogen bonds.

AP-ready answerCBr4(l): London dispersion forces only.
CBrClF2(l): London dispersion forces and dipole-dipole attractions.

C(ii)Explaining the boiling-point difference

Task: Explain why CBr4 boils at 463 K while CBrClF2 boils at 269 K.

CBr4 has four large bromine atoms and many more easily distorted electrons. Its highly polarizable electron cloud produces much stronger instantaneous dipoles and London dispersion forces.

AP-ready answerThe stronger London dispersion forces in CBr4 outweigh the additional dipole-dipole attractions in CBrClF2. More energy is needed to separate CBr4 molecules, so CBr4 has the higher boiling point.

Short free-response question

FRQ 6: Spectrophotometry and dilution

Use Beer-Lambert proportionality, read a calibration curve, apply M1V1 = M2V2, and trace the effect of overfilling a volumetric flask.

AParticle diagram at lower absorbance

Task: An equal-volume diagram has 8 V2+ ions at absorbance 0.32. Draw the diagram for absorbance 0.08.

c2c1 = A2A1 = 0.080.32 = 14;   8 × 14 = 2 ions
AP Chemistry 2026 FRQ 6 particle diagrams comparing eight vanadium two plus ions at absorbance 0.32 with two ions at absorbance 0.08 in equal volumes
Replaceable placeholder for the completed V2+ particle diagram.
AP-ready answerDraw 2 V2+ ions in the equal-volume circle.

Calibration curve

The line passes through the origin and approximately (0.090 M, 0.36), so absorbance is proportional to concentration with slope 4.00 M−1.

AP Chemistry 2026 FRQ 6 absorbance calibration curve for aqueous vanadium two plus with a linear fit and an absorbance of 0.22 corresponding to 0.055 M
Replaceable placeholder for the V2+ calibration curve.

B(i)Concentration of the diluted solution

Task: Determine concentration when absorbance is 0.22.

slope = 0.360.090 M = 4.00 M−1A = (4.00 M−1)cc = 0.224.00 M−1 = 0.0550 M
AP-ready answer[V2+]diluted = 0.055 M

B(ii)Concentration of the original solution

Task: A 3.00 mL sample is diluted to 25.0 mL. Calculate the original concentration.

M1V1 = M2V2Moriginal(3.00 mL) = (0.0550 M)(25.0 mL)Moriginal = 0.4583 M
AP-ready answer[V2+]original = 0.458 M ≈ 0.46 M

CEffect of overfilling the volumetric flask

Task: Evaluate the claim that overfilling makes the calculated original concentration too low.

  1. The actual final volume is greater than 25.0 mL, while transferred moles of V2+ are unchanged.
  2. The actual diluted concentration and measured absorbance are therefore lower.
  3. The student back-calculates with 25.0/3.00 even though the actual dilution factor is larger, so the calculated original concentration is below the true value.
Mtrue = MdilutedVactual3.00 > Mdiluted25.03.00 = Mcalculated
AP-ready answerAgree. Overfilling makes the back-calculated original concentration too low.

Short free-response question

FRQ 7: Enthalpy, limiting reactant, and lattice energy

Use a reaction enthalpy to find ΔH°f, identify the limiting reactant and heat released, and compare ionic lattice enthalpies with Coulomb's law.

Reaction and lattice data

4 Na(s) + O2(g) → 2 Na2O(s),   ΔH°rxn = −828 kJ molrxn−1
Lattice enthalpy data
CompoundLattice enthalpy
Na2O(s)2481 kJ mol−1
Rb2O(s)2163 kJ mol−1

AStandard enthalpy of formation of Na2O(s)

Task: Calculate ΔH°f using the balanced reaction.

ΔH°rxn = ΣnΔH°f,products − ΣnΔH°f,reactants−828 = 2ΔH°f[Na2O(s)] − [4(0) + 1(0)]ΔH°f[Na2O(s)] = −8282 = −414 kJ mol−1
AP-ready answerΔH°f[Na2O(s)] = −414 kJ mol−1

BHeat released and limiting reactant

Task: Calculate the heat released when 18.4 g Na reacts with 12.8 g O2.

nNa = 18.4 g22.99 g mol−1 = 0.800 molnO2 = 12.8 g32.00 g mol−1 = 0.400 mol

The stoichiometry requires 4 mol Na per 1 mol O2. The 0.800 mol Na requires only 0.200 mol O2, so O2 is excess and Na is limiting.

reaction extent = 0.800 mol Na × 1 molrxn4 mol Na = 0.200 molrxnq = (0.200 molrxn)(−828 kJ molrxn−1) = −166 kJ
AP-ready answer166 kJ of heat is released; equivalently, qsystem = −166 kJ.

CComparing lattice enthalpies

Task: Use Coulomb's law to explain why Rb2O has the smaller lattice enthalpy.

electrostatic attraction ∝ |q1q2|r

Ion charges are the same in both crystals: M+ and O2−. Rb+ has more occupied energy levels and a larger ionic radius than Na+, so the Rb+-O2− distance is larger and the attraction is weaker.

AP-ready answerThe larger ion-ion distance in Rb2O weakens Coulombic attraction, so less energy is required to separate its ions. Therefore Rb2O has the smaller lattice enthalpy, 2163 kJ mol−1 versus 2481 kJ mol−1.

Final AP Chemistry response checklist

  • Balance every requested equation for atoms and charge.
  • Include states of matter and ionic charges where required.
  • Match thermodynamic signs to the physical process.
  • Show formula, substitution, arithmetic, units, and final interpretation.
  • Use significant figures consistent with the provided measurements.
  • State an equilibrium expression before substituting.
  • Support qualitative conclusions with particle-level reasoning.
  • Use a visible graph feature such as slope, half-life, or inflection point.
  • Trace experimental error from procedure to measurement to calculated result.
  • Give only the number of examples or claims requested.

Continue your AP Chemistry review

Practice each released question again without the model response. Then compare your equations, units, signs, significant figures, and written justification with this walkthrough.

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Content basis: the user-supplied AP Chemistry 2026 FRQ Detailed Solutions PDF. Prompt wording, values, and figures were cross-checked against the released 2026 question paper. This is an independent educational guide, not an official scoring document.