SCH4U Chemistry — Final Exam

Cumulative Summative Assessment — Strands A through F
30% of Final Mark
Duration: 3 hours  |  Total: /100 marks  |  Show all work. Calculator and periodic table permitted. F = 96485 C/mol; R = 8.314 J/mol·K; Kw = 1.0×10⁻¹⁴.
Knowledge & Understanding
/25
Thinking
/25
Communication
/25
Application
/25
Part A: Knowledge & Understanding [25]
Q1 Strand B [3]
Provide IUPAC names: (a) CH₃CH₂CH(CH₃)CH₂CH₃; (b) CH₂=C(CH₃)CH₂CH₃; (c) CH₃CH₂COOCH₂CH₃.
Q2 Strand C [3]
For SO₃ give: (a) Lewis structure; (b) molecular shape; (c) hybridization at S; (d) is it polar?
Q3 Strand D [3]
Define ΔHf°. Why is ΔHf°(O₂(g)) = 0?
Q4 Strand E [4]
Write Kc, conjugate acid–base pair, and predict pH (acidic / basic / neutral) for: NH₃(aq) + H₂O ⇌ NH₄⁺ + OH⁻.
Q5 Strand F [4]
Determine oxidation numbers: (a) Cl in HClO₃; (b) Mn in MnO₂; (c) C in CO₃²⁻; (d) S in SO₃²⁻.
Q6 Strand C [4]
Write electron configurations of: (a) Cl⁻; (b) Fe³⁺; (c) Cu; (d) Mn.
Q7 Strand B [4]
Identify the functional group: (a) CH₃CONH₂; (b) CH₃CH₂CHO; (c) CH₃COOH; (d) CH₃OCH₃.
Part B: Thinking & Investigation [25]
Q8 Strand B [5]
Organic synthesis pathway. Show the sequence of reactions to convert ethene → ethanol → ethanal → ethanoic acid → ethyl ethanoate. State reagents/conditions.
Q9 Strand E [6]
Equilibrium calculation (ICE). 2.00 mol of HI is placed in a 1.00 L flask at 700 K and decomposes via 2 HI(g) ⇌ H₂(g) + I₂(g) with Kc = 0.0185. Find equilibrium concentrations of all species.
Q10 Strand F [6]
Redox titration. 25.00 mL of acidified Fe²⁺ solution is titrated with 0.0250 M KMnO₄, requiring 18.40 mL to reach the endpoint. Determine [Fe²⁺] and the mass of Fe²⁺ in the original 25.00 mL sample (M = 55.85).
Q11 Strand D [4]
Use Hess: (1) C(s)+½O₂→CO ΔH=−110.5; (2) CO+½O₂→CO₂ ΔH=−283.0. Find ΔH for C(s)+O₂→CO₂.
Q12 Strand E [4]
Find pH of 0.20 M HCOOH (Ka = 1.8×10⁻⁴).
Part C: Communication [25]
Q13 Strand C [6]
Structure–property explanation. Compare and explain the relative b.p. of CH₄ (−161 °C), NH₃ (−33 °C), and H₂O (100 °C) using molecular structure, polarity, and IMFs.
Q14 Strand D [5]
Sketch and explain a Maxwell-Boltzmann curve. Show how a catalyst and a temperature increase each affect the proportion of molecules with KE > Ea.
Q15 Strand E [5]
Sketch the titration curve for 25.00 mL of 0.10 M acetic acid titrated with 0.10 M NaOH. Label initial pH, half-equivalence, equivalence (pH and volume), and indicator choice.
Q16 Strand F [5]
Draw a labelled diagram of a galvanic cell using Zn|Zn²⁺ and Cu|Cu²⁺ half-cells. Show e⁻ flow, ion flow, salt bridge, anode/cathode, and net reaction.
Q17 Strand B [4]
Compare addition vs condensation polymerization with one polymer example each, and state environmental implications of each.
Part D: Application [25]
Q18 Strand D / STSE [5]
A bomb calorimeter (C = 8.50 kJ/°C) burns 0.500 g of glucose; ΔT = 4.62 °C. Find ΔHcombustion per mole (M(C₆H₁₂O₆)=180.16). Compare with calorie content (1 cal = 4.184 J).
Q19 Strand E / STSE [5]
Haber process: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ. Argue why industrial conditions of ~450 °C and 200 atm represent a thermodynamic-kinetic compromise. Address Le Chatelier and rate.
Q20 Strand F / STSE [5]
An Al smelter passes 1.50×10⁵ A through molten Al₂O₃ for 24.0 h. Calculate the mass of Al produced and energy consumed (assume 4.5 V cell). Discuss two environmental impacts.
Q21 Strand B [5]
Aspirin (acetylsalicylic acid) is synthesized from salicylic acid + acetic anhydride. Write the equation and identify the functional group(s) introduced. Suggest why aspirin is unstable in moist air.
Q22 Strand C/E [5]
Hard water contains Ca²⁺, Mg²⁺. Explain (i) why these ions decrease soap effectiveness, (ii) how Ca(OH)₂ and Na₂CO₃ "soften" water using Ksp/precipitation, and (iii) the structure–property reason that Mg²⁺ is more strongly hydrated than Ca²⁺.
Complete Worked Answer Key

Q1. (a) 3-methylpentane; (b) 2-methylbut-1-ene; (c) ethyl propanoate.

Q2. SO₃: trigonal planar; sp²; nonpolar (symmetric, dipoles cancel); resonance among three S=O / S–O structures.

Q3. ΔHf° = enthalpy change to form 1 mol of compound from its elements in standard states. Elements in standard state have ΔHf° = 0.

Q4. Kc = [NH₄⁺][OH⁻]/[NH₃]. Pairs: NH₃/NH₄⁺ and H₂O/OH⁻. Solution is basic (OH⁻ produced).

Q5. (a) +5; (b) +4; (c) +4; (d) +4.

Q6. Cl⁻: [Ar]; Fe³⁺: [Ar]3d⁵; Cu: [Ar]4s¹3d¹⁰; Mn: [Ar]4s²3d⁵.

Q7. (a) Amide; (b) Aldehyde; (c) Carboxylic acid; (d) Ether.

Q8. CH₂=CH₂ +H₂O/H⁺ → CH₃CH₂OH. CH₃CH₂OH +oxidation (PCC or limited K₂Cr₂O₇/H⁺) → CH₃CHO. CH₃CHO +KMnO₄ → CH₃COOH. CH₃COOH + CH₃CH₂OH /H₂SO₄ → CH₃COOCH₂CH₃ + H₂O (esterification).

Q9. [HI]₀=2.00. Let x = [H₂]=[I₂] formed; [HI]=2.00−2x. Kc = x²/(2.00−2x)² = 0.0185 → x/(2.00−2x) = 0.136 → x = 0.272 − 0.272x ⇒ x(1.272)=0.272 ⇒ x=0.214. So [H₂]=[I₂]=0.214 M; [HI]=2.00−0.428=1.57 M.

Q10. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. n(MnO₄⁻)=0.0250×0.01840=4.60×10⁻⁴; n(Fe²⁺)=5×4.60×10⁻⁴=2.30×10⁻³ mol. [Fe²⁺]=2.30×10⁻³/0.02500=0.0920 M; mass = 2.30×10⁻³ × 55.85 = 0.128 g.

Q11. ΔH = (−110.5) + (−283.0) = −393.5 kJ/mol.

Q12. x² ≈ Ka·C = 1.8×10⁻⁴ × 0.20 = 3.6×10⁻⁵ → x=6.00×10⁻³ → pH = 2.22.

Q13. CH₄: nonpolar, only London dispersion. NH₃: polar with H-bonding (one H-bond donor effective). H₂O: polar with two H-bond donors and two acceptors → strongest H-bonding network. B.p. order matches.

Q14. Curve: KE on x-axis, # molecules on y-axis; threshold Ea shown. ↑T: curve broadens / shifts right, more area beyond Ea. Catalyst: lowers Ea position to the left → more molecules exceed it. Both increase rate but via different mechanisms.

Q15. Initial pH ≈ 2.87. Half-eq (12.5 mL): pH = pKa = 4.74. Equivalence (25.0 mL): pH ≈ 8.7 (basic — acetate hydrolyzes). Indicator: phenolphthalein.

Q16. Anode (Zn|Zn²⁺) on left: Zn → Zn²⁺ + 2e⁻. Cathode (Cu²⁺|Cu) on right: Cu²⁺ + 2e⁻ → Cu. e⁻ flow Zn→Cu via wire; salt-bridge cations move to Cu side, anions to Zn side. E°cell = +1.10 V; net: Zn + Cu²⁺ → Zn²⁺ + Cu.

Q17. Addition: alkene → polyethylene; nonbiodegradable, persistent waste. Condensation: diacid+diol → polyester (or diamine+diacid → nylon); releases water; many polyesters are recyclable.

Q18. q = CΔT = 8.50 × 4.62 = 39.27 kJ. n(glucose)=0.500/180.16=2.775×10⁻³ mol → ΔHcomb = −39.27/2.775×10⁻³ = −14150 kJ/mol ≈ −14.2 MJ/mol. Per gram: 39.27/0.500 = 78.5 kJ/g (≈ 18.8 kcal/g) — close to the 4 kcal/g often cited (note: bomb calorimeter measures heat of complete oxidation; food values use Atwater factors).

Q19. ΔH<0 → ↓T favours forward (more NH₃). High kinetic barrier → ↑T needed for adequate rate. ↑P shifts forward (4 mol → 2 mol gas) and improves yield. Compromise: ~450 °C (rate good, eq still favourable enough), ~200 atm (yield + manageable equipment), Fe catalyst speeds attainment without shifting K.

Q20. Q = It = 1.50×10⁵ × 86 400 = 1.296×10¹⁰ C. n(e⁻) = 1.343×10⁵ mol → n(Al)=4.477×10⁴ mol → m = 4.477×10⁴ × 26.98 = 1.21×10⁶ g ≈ 1.21 t. Energy = QV = 1.296×10¹⁰ × 4.5 = 5.83×10¹⁰ J = 1.62×10⁴ kWh. Impacts: huge electricity demand (often coal/gas → CO₂); fluoride/perfluorocarbon emissions from cryolite bath.

Q21. C₆H₄(OH)COOH + (CH₃CO)₂O → C₆H₄(OCOCH₃)COOH + CH₃COOH. New ester (–OCOCH₃) replaces phenolic OH. Aspirin's ester hydrolyzes in moist air, releasing salicylic acid + acetic acid (vinegar smell).

Q22. (i) Ca²⁺/Mg²⁺ react with stearate to form insoluble scum, reducing surfactant action. (ii) Ca(OH)₂ + Mg²⁺ + HCO₃⁻ → CaCO₃(s) + Mg(OH)₂(s) (Ksp very low → ions removed); Na₂CO₃ + Ca²⁺ → CaCO₃(s). (iii) Mg²⁺ has smaller ionic radius → higher charge density → polarizes water more strongly → larger hydration enthalpy.