Free practice final
General Chemistry 2 practice final (free, with answers)
A General Chemistry 2 final usually covers colligative properties, kinetics, chemical equilibrium, acids, bases and buffers, solubility, thermodynamics and electrochemistry, with most of the points in multi-step calculations. Set up each problem on paper with units before you open the answer, then use the explanation to find the exact step where your work went off course.
Topics
- Solutions and Colligative Properties
- Chemical Kinetics
- Chemical Equilibrium
- Acids and Bases
- Buffers and Titrations
- Solubility Equilibria
- Thermodynamics
- Electrochemistry
15 questions
Try each one before you open its answer.
1. 0.100 mol of CaCl₂ is dissolved in 1.00 kg of water. Assuming complete dissociation, what is the freezing point of the solution? (Kf for water = 1.86 °C·kg/mol; pure water freezes at 0.00 °C.)
- A −0.186 °C
- B −0.372 °C
- C +0.558 °C
- D −0.558 °C
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Answer: D. −0.558 °C
CaCl₂ dissociates into Ca²⁺ and 2 Cl⁻, so i = 3. ΔTf = i·Kf·m = 3 × 1.86 × 0.100 = 0.558 °C, and a solute always lowers the freezing point, so the solution freezes at −0.558 °C.
2. A first-order reaction has a rate constant of 0.0462 min⁻¹. How long does it take for 87.5% of the reactant to be consumed? Give three significant figures.
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Answer: 45.0 min
The half-life is ln 2 / k = 0.693/0.0462 min⁻¹ = 15.0 min. Leaving 12.5% of the reactant means halving it three times (100 → 50 → 25 → 12.5%), so t = 3 × 15.0 = 45.0 min.
3. For 2NO(g) + O₂(g) → 2NO₂(g), initial rates were measured. Trial 1: [NO] = 0.010 M, [O₂] = 0.010 M, rate = 2.5 × 10⁻⁵ M/s. Trial 2: [NO] = 0.020 M, [O₂] = 0.010 M, rate = 1.0 × 10⁻⁴ M/s. Trial 3: [NO] = 0.010 M, [O₂] = 0.020 M, rate = 5.0 × 10⁻⁵ M/s. What is the rate law?
- A Rate = k[NO][O₂]
- B Rate = k[NO]²[O₂]
- C Rate = k[NO][O₂]²
- D Rate = k[NO]²[O₂]²
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Answer: B. Rate = k[NO]²[O₂]
From trial 1 to 2, doubling [NO] with [O₂] fixed multiplies the rate by 4 = 2², so the reaction is second order in NO. From trial 1 to 3, doubling [O₂] doubles the rate, so it is first order in O₂.
4. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH° = −92 kJ. Which change increases the amount of NH₃ present at equilibrium?
- A Raising the temperature
- B Adding an iron catalyst
- C Decreasing the container volume
- D Adding argon at constant volume
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Answer: C. Decreasing the container volume
Shrinking the volume raises the pressure, and the system shifts toward the side with fewer moles of gas (2 mol versus 4 mol), making more NH₃. Heating shifts an exothermic reaction toward reactants, a catalyst only speeds up the approach to equilibrium, and argon at constant volume leaves every partial pressure unchanged.
5. For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 49.0 at a certain temperature. If 0.200 mol H₂ and 0.200 mol I₂ are placed in a 1.00 L flask at that temperature, what is the equilibrium concentration of HI? Give three significant figures.
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Answer: 0.311 M
With x as the amount of H₂ that reacts, Kc = (2x)²/(0.200 − x)² = 49.0, so taking the square root gives 2x/(0.200 − x) = 7.00. Solving, 9.00x = 1.40 and x = 0.1556 M, so [HI] = 2x = 0.311 M.
6. What is the pH of 0.200 M acetic acid? (Ka = 1.8 × 10⁻⁵.)
- A 2.72
- B 0.70
- C 4.74
- D 5.44
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Answer: A. 2.72
For a weak acid, x²/(0.200 − x) = 1.8 × 10⁻⁵, and since x is small, x ≈ √(1.8 × 10⁻⁵ × 0.200) = 1.9 × 10⁻³ M = [H₃O⁺]. pH = −log(1.9 × 10⁻³) ≈ 2.72; 0.70 would be the pH if acetic acid were a strong acid, and 4.74 is its pKa.
7. Which 0.10 M aqueous solution is basic?
- A NH₄Cl
- B KNO₃
- C NaHSO₄
- D NaF
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Answer: D. NaF
F⁻ is the conjugate base of the weak acid HF, so it takes protons from water and produces OH⁻. NH₄⁺ and HSO₄⁻ are acidic, while K⁺ and NO₃⁻ come from a strong base and a strong acid and leave the solution neutral.
8. What is the pH of a buffer that is 0.250 M NH₃ and 0.400 M NH₄Cl? (Kb of NH₃ = 1.8 × 10⁻⁵; Kw = 1.0 × 10⁻¹⁴ at 25 °C.) Give two decimal places.
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Answer: pH ≈ 9.05
The Ka of NH₄⁺ is Kw/Kb = 5.6 × 10⁻¹⁰, so pKa = 9.255. By the Henderson-Hasselbalch equation, pH = pKa + log([NH₃]/[NH₄⁺]) = 9.255 + log(0.250/0.400) = 9.255 − 0.204 ≈ 9.05.
9. 25.0 mL of 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵) is titrated with 0.100 M NaOH. What is the pH at the equivalence point? (Kw = 1.0 × 10⁻¹⁴.)
- A 7.00
- B 8.72
- C 4.74
- D 5.28
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Answer: B. 8.72
At equivalence, 2.50 × 10⁻³ mol of acetate sits in 50.0 mL, so [CH₃COO⁻] = 0.0500 M, and Kb = Kw/Ka = 5.6 × 10⁻¹⁰. [OH⁻] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ M, so pOH = 5.28 and pH = 8.72; the solution is basic because acetate is a weak base.
10. The Ksp of PbI₂ is 9.8 × 10⁻⁹. What is its molar solubility in pure water?
- A 1.35 × 10⁻³ M
- B 9.90 × 10⁻⁵ M
- C 7.00 × 10⁻⁵ M
- D 2.14 × 10⁻³ M
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Answer: A. 1.35 × 10⁻³ M
If s mol/L dissolves, [Pb²⁺] = s and [I⁻] = 2s, so Ksp = s(2s)² = 4s³. Then s = (9.8 × 10⁻⁹ / 4)^(1/3) = 1.35 × 10⁻³ M; forgetting to square the 2s gives 7.00 × 10⁻⁵ M.
11. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ/mol and ΔS° = +161 J/(mol·K). Assuming both stay constant, above what temperature is the reaction spontaneous under standard conditions?
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Answer: ≈ 1.11 × 10³ K (1106 K, about 832 °C)
The reaction becomes spontaneous when ΔG° = ΔH° − TΔS° drops below zero, which happens above T = ΔH°/ΔS°. Converting kJ to J, T = 178,000 J/mol ÷ 161 J/(mol·K) ≈ 1106 K.
12. A reaction has ΔG° = −10.0 kJ/mol at 298 K. What is its equilibrium constant, K? (R = 8.314 J/(mol·K).)
- A 1.77 × 10⁻²
- B 1.00
- C 56.6
- D 4.04
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Answer: C. 56.6
ΔG° = −RT ln K, so ln K = 10,000 J/mol ÷ (8.314 × 298) = 4.036 and K = e^4.036 ≈ 56.6. A negative ΔG° always means K > 1; 4.04 is ln K, not K, and 1.00 comes from leaving ΔG° in kJ.
13. Standard reduction potentials: Ag⁺ + e⁻ → Ag, E° = +0.80 V; Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V. What is E°cell for 2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq)?
- A +1.26 V
- B +0.46 V
- C +1.14 V
- D −0.46 V
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Answer: B. +0.46 V
E°cell = E°(cathode) − E°(anode) = 0.80 V − 0.34 V = +0.46 V, with silver reduced at the cathode and copper oxidized at the anode. Potentials are intensive, so the coefficient of 2 on Ag⁺ does not double the 0.80 V.
14. A current of 2.00 A passes through a Cu²⁺ solution for 30.0 minutes. What mass of copper metal is deposited? (F = 96,485 C/mol; Cu = 63.55 g/mol.) Give three significant figures.
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Answer: 1.19 g
Charge = 2.00 A × 1800 s = 3600 C, which is 3600/96,485 = 0.0373 mol of electrons. Each Cu²⁺ needs 2 electrons, so 0.0187 mol Cu forms, or 0.0187 × 63.55 ≈ 1.19 g.
15. The Ksp of AgCl is 1.8 × 10⁻¹⁰. What is the molar solubility of AgCl in 0.10 M NaCl?
- A 1.8 × 10⁻⁹ M
- B 1.3 × 10⁻⁵ M
- C 1.8 × 10⁻¹¹ M
- D 4.2 × 10⁻⁵ M
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Answer: A. 1.8 × 10⁻⁹ M
The chloride already in solution suppresses dissolving (the common-ion effect), so [Cl⁻] ≈ 0.10 M and Ksp = s × 0.10. That gives s = 1.8 × 10⁻¹⁰ / 0.10 = 1.8 × 10⁻⁹ M, far below the 1.3 × 10⁻⁵ M solubility in pure water.