AP Chemistry practice questions and exam guide

Which substance is most soluble in water?

1. Hexane (C₆H₁₄)
2. Carbon tetrachloride
3. Octane
4. Glucose (C₆H₁₂O₆)

Answer: D — Glucose (C₆H₁₂O₆)

'Like dissolves like.' Glucose has multiple -OH groups → strong H-bonding with water. A, C, B) All nonpolar hydrocarbons or nonpolar molecules — poorly soluble in polar water.

For the cell reaction Cu + 2 Ag⁺ → Cu²⁺ + 2 Ag with E°cell = +0.46 V, ΔG° is:

1. +89 kJ
2. −89 kJ (ΔG° = −nFE° = −2 × 96485 × 0.46 ≈ −88.8 kJ)
3. −44 kJ
4. +44 kJ

Answer: B — −89 kJ (ΔG° = −nFE° = −2 × 96485 × 0.46 ≈ −88.8 kJ)

ΔG° = −nFE° = −(2)(96485)(0.46) ≈ −88,800 J = −88.8 kJ. Negative confirms spontaneous. A) Sign error. C/D) Used n=1 incorrectly.

How many moles of CO₂ are in a 5.00 L flask at 2.00 atm and 300 K? (R = 0.0821 L·atm/mol·K)

1. 0.41 mol
2. 0.82 mol
3. 1.64 mol
4. 0.25 mol

Answer: A — 0.41 mol

n = PV/RT = (2.00)(5.00)/[(0.0821)(300)] = 10.0/24.63 ≈ 0.406 mol. B) Off by factor of 2. C) Off by factor of 4. D) Forgot a factor.

For 2 H₂O(l) → 2 H₂(g) + O₂(g) at 25°C, ΔS° is:

1. Negative — fewer moles
2. Approximately zero
3. Positive — gases produced from a liquid means greatly increased disorder
4. Cannot be determined without temperature

Answer: C — Positive — gases produced from a liquid means greatly increased disorder

Producing 3 moles of gas from 2 moles of liquid hugely increases disorder. Gases >> liquids in entropy. A) Wrong direction sign. B) Magnitudes are clearly nonzero. D) Sign of ΔS° doesn't depend on T.

Theoretical yield of NH₃ from 28.0 g N₂ and excess H₂ (N₂ + 3H₂ → 2NH₃)? (M(N₂)=28, M(NH₃)=17)

1. 17.0 g
2. 34.0 g
3. 28.0 g
4. 51.0 g

Answer: B — 34.0 g

28.0 g N₂ = 1.00 mol. Mole ratio: 1 N₂ → 2 NH₃, so 2.00 mol NH₃ = 34.0 g. A) Forgets the 2:1 ratio. C) Conserves mass naively (wrong — H₂ also reacts). D) Off-by-one mole multiplier.

Equivalence point of titrating 0.10 M NH₃ (Kb = 1.8 × 10⁻⁵) with HCl has pH:

1. > 7
2. = 14
3. = 7
4. < 7 (conjugate acid NH₄⁺ is weakly acidic)

Answer: D — < 7 (conjugate acid NH₄⁺ is weakly acidic)

At equivalence, only NH₄⁺ remains in solution; NH₄⁺ is a weak acid, making pH < 7. A) Reverse case (weak acid titrated with strong base). C) Strong-strong only. B) Pure NaOH.

For 2A + B → C, the rate law is rate = k[A]²[B]. Overall order?

1. 2
2. 3
3. 1
4. 4

Answer: B — 3

Sum of exponents: 2 + 1 = 3. Note: order is determined experimentally, not from stoichiometry. A) Order in A only. C) Order in B only. D) Confused with coefficient sum.

Which compound has the highest lattice energy?

1. NaCl
2. CsI
3. KBr
4. MgO (higher charges +2/-2 and small ionic radii give maximum Coulombic attraction)

Answer: D — MgO (higher charges +2/-2 and small ionic radii give maximum Coulombic attraction)

Lattice energy ∝ (q₁q₂)/r. MgO has the largest charges (2+ and 2-) and small ions. A) Singly charged. C/B) Singly charged with larger ions (weaker attraction).

A weak acid HA has Ka = 1.8 × 10⁻⁵ and initial concentration 0.10 M. Approximate pH?

1. 5.0
2. 2.9
3. 3.0
4. 1.0

Answer: C — 3.0

x² ≈ Ka × C = 1.8 × 10⁻⁶ → x = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³. pH = −log(1.34 × 10⁻³) ≈ 2.87 ≈ 2.9. C is closest at 3.0; B is also acceptable at 2.9. A) Forgets to take square root and divides. D) Treats as strong acid.

Which sample contains the most molecules? (Assume STP)

1. 22.4 L of CO₂
2. 22.4 L of H₂
3. 44.0 g of CO₂
4. All contain the same number

Answer: D — All contain the same number

At STP, 22.4 L of any ideal gas = 1 mole = 6.022 × 10²³ molecules. (A) 1 mol CO₂. (B) 1 mol H₂. (C) 44 g CO₂ / 44 g/mol = 1 mol. All three samples are 1 mole — same molecule count. Tests Avogadro's hypothesis.

Calculate ΔH using bond enthalpies: H₂ + Cl₂ → 2 HCl. (H-H = 436, Cl-Cl = 242, H-Cl = 431 kJ/mol)

1. +183 kJ
2. −247 kJ
3. +431 kJ
4. −184 kJ

Answer: D — −184 kJ

ΔH = Σ(bonds broken) − Σ(bonds formed) = (436 + 242) − 2(431) = 678 − 862 = −184 kJ. A) Sign error. C) Misses subtraction. B) Forgets the factor of 2.

For the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), at equilibrium [PCl₅] = 0.20 M, [PCl₃] = [Cl₂] = 0.10 M. Kc?

1. 0.050
2. 0.20
3. 0.10
4. 0.0050

Answer: A — 0.050

Kc = [PCl₃][Cl₂]/[PCl₅] = (0.10)(0.10)/0.20 = 0.0100/0.20 = 0.050. B) Just used [PCl₅]. C) Off by factor of 2. D) Off by factor of 10.

What does a strong-acid–strong-base titration curve look like?

Slow change, then sharp jump near equivalence (pH = 7), then slow change.

What's the difference between σ and π bonds?

σ: head-on overlap (single bonds, also one bond in multiples). π: side-by-side overlap of p orbitals (additional bonds in double/triple).

State Le Chatelier's Principle.

When a system at equilibrium is disturbed, it shifts to counteract the disturbance.

Why does a triple bond have higher energy than a single?

More shared electrons → stronger Coulombic attraction; multiple orbital overlaps.

What are the parts of an atom?

Protons (+, in nucleus), neutrons (neutral, in nucleus), electrons (−, in shells/orbitals).

Relation Ka × Kb?

For a conjugate acid-base pair: Ka × Kb = Kw = 1.0 × 10⁻¹⁴.