CHEM 150 Midterm I

Engineering Chemistry · Camosun College · Units 17 Reference

Unit 1 Matter & Measurement

Classification, atomic structure, stoichiometry, nomenclature

Classification of Matter
  • Element pure substance; cannot be separated by chemical means
  • Compound two+ elements, fixed ratio; separated only by chemical means
  • Homogeneous mixture uniform (e.g., salt water, glass, air)
  • Heterogeneous mixture non-uniform (e.g., drywall, sandy water)
Tip: Physical changes (melting, dissolving) preserve chemical identity. Chemical changes (burning, rusting) produce new substances.
Dalton's Atomic Theory (1808)
  1. Matter composed of indivisible atoms
  2. All atoms of an element are identical
  3. Compounds = atoms in fixed whole-number ratios
  4. Chemical reactions = rearrangement of atoms; atoms not created/destroyed
Note: Isotopes violate postulate 2; nuclear reactions violate postulate 4.
Atomic Structure
ParticleChargeMass (u)Location
Proton+11.0073Nucleus
Neutron01.0087Nucleus
Electron10.000549Orbitals

Atomic number Z = # protons  |  Mass number A = protons + neutrons  |  Isotopes = same Z, different A

Average Atomic Mass
avg mass = Σ (fractional abundance × isotope mass) e.g., Cl: 0.7577×34.97 + 0.2423×36.97 = 35.45 u
Stoichiometry Essentials
n = m / M     (mol = g ÷ g/mol) N = n × N   (N = 6.022×10²³) M = mol × M   (g = mol × g/mol)

Limiting reagent: convert all reactants to same product; smallest amount is the limit.

% yield = (actual / theoretical) × 100% M = n / V   (mol/L)    MᵢVᵢ = MᶠVᶠ
Nomenclature Quick Reference
  • Ionic: cation name + anion name; variable-charge metals use Roman numerals (Fe² = iron(II))
  • Oxyanions: "Nick the Camel" NO nitrate, NO nitrite, SO² sulfate, SO² sulfite, ClO perchlorate, ClO chlorate, ClO chlorite, ClO hypochlorite
  • Molecular: Greek prefixes (mono, di, tri, tetra, penta )
  • Acids: H + anion; -ide hydro___ic acid; -ate ___ic acid; -ite ___ous acid
Metal Types & Periodic Table Groups

Diatomic elements: H, N, O, F, Cl, Br, I (mnemonic: HOFBrINCl)

Metals (left/centre): ductile, malleable, lustrous, conduct heat & electricity.  Non-metals (right): brittle, poor conductors.  Metalloids: along staircase (B, Si, Ge, As, Sb, Te).

Variable-charge metals: Cu (1+, 2+) Fe (2+, 3+) Co (2+, 3+) Mn (2+, 7+) Sn (2+, 4+)  | Fixed: Na, Mg², Al³, Ca², Ba², K

Unit 2 Chemical Bonding

Bond types, periodic trends, ionization energy, lattice energy, Born-Haber

Types of Chemical Bonds
  • Ionic complete electron transfer; metal + nonmetal; e.g., NaCl. Electrostatic attraction of oppositely charged ions.
  • Covalent electron sharing; nonmetal + nonmetal; e.g., H, CO. Can be polar or nonpolar.
  • Metallic metal cations in a delocalized "sea" of electrons; explains conductivity, malleability.
ΔEN rule: 0 = nonpolar covalent | 01.5 = polar covalent | >1.5 ionic character
Periodic Trends Summary
Ionization Energy (I)
Increases leftright across period
Decreases topbottom in group

Exceptions: B < Be (2p vs 2s); O < N (paired e repulsion in O)

Atomic Radius
Increases topbottom (more shells)
Decreases leftright (Zeff increases)
Electron Affinity
More negative going right and up

More negative = more exothermic = "wants" electrons more. Exceptions: N (half-filled 2p stable)

Effective Nuclear Charge & Ionic Radii
Slater's Rule (simplified)
Z_eff = Z S    (S = # core electrons shielding valence shell) Examples: C (Z=6, 2 core e) Z_eff = 4 N (Z=7, 2 core e) Z_eff = 5 B (Z=5, 2 core e) Z_eff = 3

Ionic radius: Cations are smaller than parent atom (lost electrons, more Zeff per electron). Anions are larger (gained electrons, less Zeff per electron).

Isoelectronic series (same # electrons): more protons smaller. E.g., Al³ < Na < F < O² (all have 10 electrons).

Lattice Energy

Energy released when gaseous ions form 1 mol of ionic crystal. Always negative (exothermic).

Factors: Lattice energy with higher ionic charges and smaller ions (shorter distance).

CompoundLattice E (kJ/mol)mp (°C)
LiF1049848
NaCl787801
KBr691734
MgO37952825
AlO159162054
Born-Haber Cycle (NaCl example)
Na(s) Na(g) +107 kJ (sublimation) Na(g) Na(g) + e +496 kJ (IE) ½Cl(g) Cl(g) +121 kJ (½ bond energy) Cl(g) + e Cl(g) 349 kJ (EA) Na(g) + Cl(g) NaCl(s) 787 kJ (lattice E) Na(s) + ½Cl(g) NaCl(s) 412 kJ (ΔH°f)

Use Hess's Law: ΔH°f = sublimation + IE + ½ bond + EA + lattice E

IE Trend Practice (memorize order)

Increasing IE:  Sr < Ca < Se < Br  (same group ordering + period position)

Increasing ionic size:  Ti < K < S² < Se²  (isoelectronic Ti, K, S² all have 18 e; Se² has 36 e = bigger shell)

Increasing atomic radius:  Br < Se < Ca < Sr

Unit 3 Lewis Structures

Drawing rules, formal charges, resonance, exceptions to octet

Lewis Structure Drawing Steps
  1. Count total valence electrons (VE). For anions add n e; for cations subtract n e.
  2. Choose central atom: usually least electronegative (but not H or F). In oxoanions, the non-O atom is central.
  3. Place one bond between central atom and each terminal atom.
  4. Complete octets on terminal atoms first using lone pairs.
  5. Place remaining electrons on central atom.
  6. If central atom is short of octet, convert lone pairs on terminals to double/triple bonds.
  7. Verify: total electrons used = VE counted in step 1.
Quick VE check: VE = (sum of group numbers) ± charge. E.g., NO = 5+6+6+1 = 18 VE
Formal Charge Formula
FC = (valence e) ½(bonding e) (lone pair e) e.g., in NNO (best NO structure): Left N: 5 ½(6) 2 = 0 Right N: 5 ½(6) 0 = +2? check each structure

Best Lewis structure:

  1. Lowest formal charges (ideally zero)
  2. Negative FC on most electronegative atom
  3. Same-sign FCs on adjacent atoms = very bad
Resonance

When two or more valid Lewis structures can be drawn, resonance exists. The true structure is the resonance hybrid intermediate between all structures.

Key rule: Only electrons move in resonance. Atoms never move. Use double-headed arrow () between structures.

Equivalent resonance: NO, NO, CO² (all structures identical by symmetry contribute equally)

Non-equivalent resonance: NO structures NOT identical; best (lowest FC) contributes most to hybrid.

Exceptions to the Octet Rule

1. Odd-electron species (radical)

  • NO, NO odd total VE, always one atom with 7e
  • Very reactive; paramagnetic

2. Incomplete octets

  • BF (6e on B), BeCl (4e on Be)
  • Lewis acids can accept a lone pair

3. Expanded valence shell

  • Only elements in period 3+ (Si, P, S, Cl, As, Se, Br, I, Xe)
  • PCl (10e), SF (12e), XeF (12e)
  • SO², ClO expanded shells minimize formal charges
Never give C, N, O, F more than 8 valence electrons.
Bond Dissociation Energies (D°, kJ/mol)
BondD° (kJ/mol)BondD° (kJ/mol)BondD° (kJ/mol)
HH436CC347NN946
CH414C=C611O=O498
CO360CC837HO464
C=O799CN305HN391
ΔH reaction using bond energies
ΔH = Σ D°(bonds broken) Σ D°(bonds formed) (bonds broken = reactants; bonds formed = products)

Unit 4 Polarity & VSEPR

Electronegativity, dipole moments, molecular geometry

Electronegativity & Bond Polarity

Electronegativity (EN) = ability of atom in molecule to attract shared electrons (Pauling scale).

Increases up and to the right on periodic table. F is most electronegative (4.0).

μ = Q × r    (dipole moment, debyes D) 1 D = 3.34×10³ C·m % ionic character = (measured μ / theoretical μ) × 100%
Polarity: Bond dipoles must add as vectors. CO has polar bonds but is nonpolar (linear, dipoles cancel). HO has polar bonds AND is polar (bent, net dipole).
VSEPR Algorithm
  1. Draw Lewis structure
  2. Count Electron Groups (EG) around central atom: each single bond, double bond, triple bond, and lone pair = 1 EG
  3. Determine Electron Group Geometry (EGG) from # EG
  4. Determine Molecular Shape from bonding pairs only
  5. Determine if molecule is polar (net dipole 0)
Lone pairs repel more than bonding pairs bond angles are compressed. LPLP > LPBP > BPBP repulsion.
VSEPR Shape Reference Table
EGEGGLPMolecular ShapeAnglesExamplePolar?
2Linear0Linear180°CO, HCNNo / Yes
3Trig. planar0Trigonal planar120°BF, SONo
3Trig. planar1Bent (120°)~117°SO, NOYes
4Tetrahedral0Tetrahedral109.5°CH, CClNo
4Tetrahedral1Trig. pyramidal107°NH, PHYes
4Tetrahedral2Bent (109°)104.5°HO, HSYes
5Trig. bipyramidal0Trig. bipyramidal90°/120°PCl, PFNo
5Trig. bipyramidal1See-saw~90°/120°SFYes
5Trig. bipyramidal2T-shaped~90°BrF, IFYes
5Trig. bipyramidal3Linear180°XeF, INo
6Octahedral0Octahedral90°SFNo
6Octahedral1Square pyramidal~90°BrFYes
6Octahedral2Square planar90°XeFNo
TBP lone pairs: Always go equatorial (120° from each other) because equatorial sites have fewer 90° neighbors than axial sites.

Unit 5 Intermolecular Forces

London dispersion, dipole-dipole, hydrogen bonding, effects on properties

Types of IMFs (weakest strongest)
ForceBetweenStrengthExample
London DispersionALL molecules (induced dipoles)Weakest; increases with molar mass & surface areaHe, Ar, CH, I
DipoleDipolePolar moleculesModerate; depends on polarityHClHCl, acetone
Hydrogen BondingH bonded to F, O, or N near another F/O/NStrong (1540 kJ/mol)HO, NH, HF, alcohols
IonDipoleIons + polar solventsStrongest of Van der WaalsNa in water
Remember: IMFs are 10100× weaker than covalent bonds. Boiling water breaks H-bonds between HO molecules NOT the OH bonds within them.
Effects on Physical Properties
  • Boiling point: with stronger/more IMFs. Same mass polar > nonpolar. Same polarity heavier (more surface area) = higher bp.
  • Vapour pressure: with stronger IMFs (harder to escape)
  • Viscosity: with stronger IMFs; with higher temperature
  • Surface tension: net inward force on surface molecules; with stronger IMFs
BP order: He(269°C) < O(183°C) < Cl(34°C) < acetone(56°C) (London disp. only) (London) (London) (London + dipole-dipole)
H-Bonding Special Cases
  • Ice floats: H-bonding creates rigid open lattice ice less dense than liquid water (4°C = maximum density)
  • Anomalous bp: HO (100°C), HF (19°C), NH (33°C) are all higher than expected from their group trend
  • DNA: H-bonds hold the double helix together
  • Kevlar: strength comes from H-bonding between amide groups
H-bonding requires: H directly bonded to F, O, or N. The H must be attracted to a nearby F, O, or N on another molecule.
"Like Dissolves Like" Solubility

Polar solvents dissolve polar/ionic solutes. Nonpolar solvents dissolve nonpolar solutes.

Soluble in water (polar):

  • NH H-bonds with water
  • HF, HCl polar, H-bond/dipole
  • SO polar molecule
  • CHOH OH group H-bonds
  • NaCl, ionic compounds (if lattice E < solvation E)

Insoluble in water:

  • CH (acetylene) nonpolar
  • CCl nonpolar (symmetric)
  • Octane, hexane nonpolar
  • S nonpolar
  • MgO lattice E too large

Unit 6 Solutions & Colligative Properties

Raoult's Law, Henry's Law, vapour pressure, osmosis

Raoult's Law Vapour Pressure Lowering
P_A = x_A × P°_A ΔP = x_solute × P°_A x_A = mole fraction of A = n_A / n_total

Example: 1 mol sucrose + 15 mol water. x_water = 15/16 = 0.9375. P = 0.9375 × 31.26 mbar = 29.3 mbar (pure water = 31.26 mbar).

Benzene/toluene mixture (0.5 mol each): Total P = x_benz×P°_benz + x_tol×P°_tol. Vapour is enriched in the more volatile (higher vapour pressure) component.

Henry's Law Gas Solubility
S_g = k × P_g S_g = solubility of gas (mol/L) k = Henry's constant (mol/(L·kPa) or mol/(L·atm)) P_g = partial pressure of gas above liquid

Key facts:

  • Gas solubility increases with pressure
  • Gas solubility decreases with temperature (carbonated drinks more bubbly when warm)
  • Larger molecules (stronger London forces) higher solubility
Lake Nyos (1986): Deep lake CO suddenly outgassed under pressure release 1700 deaths.
Colligative Properties (depend on # particles only)
PropertyFormulaNotes
Vapour pressure loweringΔP = x_solute × P°Raoult's Law
Boiling point elevationΔT_b = K_b × m × im = molality (mol/kg)
Freezing point depressionΔT_f = K_f × m × iAntifreeze, salting roads
Osmotic pressureπ = MRTM = molarity, R = 8.314, T in K

van 't Hoff factor i: # particles per formula unit. NaCl i=2; MgCl i=3; glucose i=1.

Osmotic pressure example
NaCl at 0.15 M, 25°C: π = 2 × 0.15 × 8.314×10³ × 298 = 0.744 kPa×10³... Actually: π = MiRT = (0.15)(2)(0.08206)(298) = 7.33 atm enormous!
Osmosis: Solvent moves from low solute concentration to high solute concentration across semipermeable membrane. Isotonic = no net flow; hypertonic = cells shrink (crenation); hypotonic = cells burst (haemolysis).

Unit 7 Gases

Gas laws, ideal gas law, Dalton's Law, KMT, real gases

Pressure Units & Conversions
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 14.7 psi 1 Pa = 1 N/m²     STP: 0°C (273.15 K), 1 atm, molar volume = 22.4 L/mol
Simple Gas Laws
Boyle's: PV = PV    (const T, n) Charles': V/T = V/T (const P, n; T in Kelvin!) Gay-Lussac: P/T = P/T (const V, n) Avogadro: V n (const T, P) Combined: PV/T = PV/T
Always convert T to Kelvin! T(K) = T(°C) + 273.15
Ideal Gas Law & Dalton's Law
Ideal Gas Law: PV = nRT R = 8.314 L·kPa/(mol·K) [or] R = 0.08206 L·atm/(mol·K) Dalton's Law: P_total = P + P + P + ... P_i = x_i × P_total (x_i = mole fraction) Molar mass from ideal gas: M = mRT / PV

Gas over water: P_gas = P_atm P_HO(vapour)

Kinetic-Molecular Theory (5 Postulates)
  1. Gases consist of a large number of molecules in continuous random motion
  2. Volume of individual molecules is negligible compared to total volume
  3. No attractive or repulsive forces between molecules
  4. Average kinetic energy is proportional to absolute temperature: KE_avg = ½mv² = (3/2)k_B T
  5. Collisions between molecules are perfectly elastic (energy is transferred but not lost)

At same T: all gases have the same average KE. Lighter molecules move faster (effuse/diffuse faster).

Effusion: escape through a tiny hole. Diffusion: spread through space. Both faster for lighter gases.

Real Gases van der Waals
[P + n²a/V²][V nb] = nRT a = correction for intermolecular attractions (reduces observed pressure) b = correction for molecular volume (reduces available volume)

Real gases deviate most from ideal at high pressure and low temperature (conditions that bring molecules closer together).

CO at 273K: ideal 756.6 kPa; van der Waals 727.1 kPa (a=364, b=0.0427)

Common Gas Calculation Tips
Units: Use R = 8.314 when P is in kPa; use R = 0.08206 when P is in atm. Never mix.

Limiting reagent + gas: Find limiting reagent first, then use moles of product to calculate volume with ideal gas law.

Mixture pressures: Calculate moles of each gas, find total moles, then P_i = n_i × RT/V for each component.

Molar mass of gas from density
M (g/mol) = ρ × R × T / P (ρ in g/L)

Past Exam to Textbook Crosswalk

Build exam pattern recognition from past tests + equivalent Petrucci questions

Strategy Loop (Use Before Every Term Test)
  1. Collect the latest past tests (target 3 to 5 papers).
  2. Label each question by topic (e.g., ionization energy, Lewis resonance, VSEPR, gases).
  3. Find 1 to 3 equivalent Petrucci exercise questions for each exam question.
  4. Mark textbook pages with test year + question number so repeats are easy to spot.
  5. Practice in cycles: past exam question -> textbook equivalent -> unseen variant.
Execution rule: prioritize repeated patterns first, not random chapter order.
Current Crosswalk Examples (From Your Chem Test Files)
Past Test QuestionClosest Petrucci MatchTopic
Chem TT 1B Q1 (increasing ionization energy)p.406 Q21; p.410 Q76Periodic trends
Chem TT 1B Q2 (increasing ionic size)p.406 Q15, Q18; p.410 Q78Ionic radius / isoelectronic
Chem TT 1B Q3 (partial charges, bond polarity)p.458 Q41, Q42Electronegativity / polarity
Chem TT 1B Q5 (N2O Lewis + resonance)p.459 Q49, Q51Lewis / resonance
Chem TT 1B Q7 (lattice energy estimate)p.572 Q86Ionic crystal energetics
Midterm Practice A Q14 (first ionization energy comparison)p.406 Q21; p.410 Q76Ionization energy
Midterm Practice A Q16 (electron affinity)p.410 Q77Electron affinity
Midterm Practice A Q17 (isoelectronic pair)p.406 Q18; p.410 Q78Isoelectronic trend
Midterm Practice A Q26/Q27/Q28 (VSEPR shape/polarity)p.460 Q59, Q60, Q67, Q68Molecular geometry
TT 1A Q4 (balanced equations)p.141 Q5-Q8; p.142 Q9-Q12Balancing reactions
Page note: these are printed textbook page numbers. If using the local PDF viewer, page indexes may be offset.
Term Test II Seed Bank (New Week 9 + Week 10 Files)
Source PatternClosest Petrucci MatchPriority
Thermochemistry: calorimetry (coffee-cup / bomb)Ch. 7 p.292-293: Q37, Q38, Q40High
Thermochemistry: Hess's Law and combustion enthalpyCh. 7 p.293-294: Q68, Q69, Q75, Q76, Q78High
Strong acid/base pH and pOH conversionsCh. 16 p.782-783: Q10, Q11, Q12, Q14, Q20, Q21, Q22, Q29, Q30High
Weak acid/base + percent ionizationCh. 16 p.784-785: Q39, Q40, Q44, Q59, Q60High
Polyprotic and salt-solution pHCh. 16 p.786-788: Q82, Q90, Q95, Q98, Q117, Q119Medium
Buffer prep + Henderson-Hasselbalch + titration curvesCh. 17 p.822-824: Q9, Q13, Q14, Q15, Q16, Q17, Q19, Q20, Q39-Q42, Q49, Q50High
Solution check layer: use the Solution Manual PDF after attempting textbook questions (Ch. 7 around PDF p.263-304; Ch. 16 around PDF p.721-785).
Term Test 2A Direct Mapping (Old Paper You Added)
TT2 SourceClosest Petrucci MatchTopic
Q2 (HNO3 ionization in water)p.782 Q9; p.782 Q10Strong acid ionization / pH setup
Q3 (NaCl dissolves spontaneously, +DeltaH)p.631 Q21; p.631 Q22; p.631 Q23DeltaG, DeltaH, DeltaS sign logic
Q4 (pH of 0.00112 M Ca(OH)2)p.782 Q10; p.782 Q12Strong base pH / hydroxide stoichiometry
Q5 (define closed system)p.247 Concept Assessment 7-1Thermodynamic system definitions
Q6 (combustion enthalpy + isomer stability)p.293 Q68, Q69; p.294 Q76Hess's Law / combustion and stability
Q7 (hydrogen electrode with metal-ion half-cell)p.912 Q20(c); p.917 Q92Voltaic cell setup, anode/cathode, half-reactions
Q8 (pH of 0.111 M NaNO2 with Ka(HNO2))p.785 Q60Weak base from conjugate-acid Ka
TT2 Exact Questions + Solved Textbook Parallels
TT2 exact wordingSimilar Petrucci questionSolution manual answer (textbook question)
Q1: Write a dissociation equation for K2SO4 in water. Ch. 16 p.782 Q10(c): pH of 0.00683 M NaOH (strong electrolyte dissociation first). Manual p.736 Q10(c): pOH = 2.166, pH = 11.83.
Q2: Write an ionization equation for HNO3 in water. Ch. 16 p.782 Q10(b): pH of 6.14 x 10^-4 M HNO3. Manual p.736 Q10(b): strong-acid ionization, pH = 3.21.
Q3: Sodium chloride dissolves spontaneously in water at room temperature. The standard enthalpy for NaCl(s) -> NaCl(aq) is +5 kJ/mol. What can you confidently say about the standard entropy? Ch. 13 p.660 Q22: external influence/spontaneity sign logic. Manual p.958-959 (Q16/Q22): nonspontaneous needs DeltaG > 0; if DeltaH > 0 and DeltaS > 0, process is favored only at high T.
Q4: Calculate the pH of 0.00112 M Ca(OH)2(aq). Ch. 16 p.782 Q12: saturated Ca(OH)2 has pH 12.35; find solubility. Manual p.736 Q12: complete dissociation of Ca(OH)2 gives 81 mg per 100 mL solubility.
Q5: As far as thermodynamics is concerned define a closed system. Write a sentence, then give one example. Ch. 7 p.293 Q57: fixed-quantity ideal gas, isothermal expansion (closed-system behavior). Manual p.280 Q57: gas does work; exchanges heat with surroundings; temperature constant; DeltaU = 0.
Q6: Calculate the enthalpy of combustion of propene (A) and cyclopropane (B) from provided thermodynamic data (kJ/mol per mole combusted). The two enthalpies differ; which compound is more stable and why? Ch. 7 p.293 Q69 and p.294 Q76 (Hess-law combustion/hydrogenation set). Manual p.282 Q69: DeltaH = -290 kJ. Manual p.283 Q76: hydrogenation DeltaH = -234.2 kJ.
Q7: Draw a hydrogen electrode connected to a gold half-cell; [H+] = 0.010 M, H2 = 0.1 atm, [Au3+] = 0.0111 M. Identify anode/cathode, write half-reactions, and compute Ecell. Ch. 19 p.912 Q20(c): Cr(s)|Cr2+(aq)||Au3+(aq)|Au(s) half-cell writing and Ecell. Manual p.1014 Q19(c): Ox Cr → Cr2+ + 2e-; Red Au3+ + 3e- → Au; net 3Cr + 2Au3+ → 3Cr2+ + 2Au; Ecell° = 2.42 V.
Q8: Calculate the pH of 0.111 M NaNO2(aq), Ka(HNO2) = 7.2 x 10^-4. Ch. 16 p.785 Q59: pH of 0.089 M NaOCl (weak-base anion hydrolysis). Manual p.757 Q59: [OH-] = 1.7 x 10^-4 M, pOH = 3.77, pH = 10.23.
Reference note: TT2 wording is taken from Answer_Key_Term_Test_2A.pdf (1).pdf (pages 1-4). Some solution-manual chapter numbering is offset vs the textbook print chapter labels.
14-Day Term Test Plan
  • Days 14-10: map 15 to 20 past questions into textbook equivalents.
  • Days 9-7: redo only weak patterns (timed, no notes).
  • Days 6-4: mixed set: 50% past exam style, 50% textbook equivalents.
  • Days 3-2: full mock (exam conditions), then error log only.
  • Day 1: one-page formula sheet + top 10 repeated traps.
When You Send New Potential Questions

I will cross-check each one into this format:

Past question: [source test + Q#] Core topic: [e.g., Lewis resonance] Petrucci equivalents: [page + question #] Difficulty tag: Core / Medium / Stretch

This keeps your study focused on repeatable test patterns and makes future course expansion easy (physics, math, etc.).

📝

Term Test 2A — 28 March 2024

8 questions · 36 marks · Thermodynamics, pH, Electrochemistry · Full worked solutions + Petrucci parallels

Source: Chem 150 Term Test 2A_2.pdf. Questions are identical to the earlier TT2A answer-key paper; this version includes full step-by-step solutions. Click Show Solution on each question to reveal the worked answer.
Provided Data Sheet (as given on the test)

Selected standard reduction potentials:

Half-ReactionE° (V)
Au3+(aq) + 3e− → Au(s)+1.52
O2(g) + 4H+(aq) + 4e− → 2H2O(l)+1.229
Fe3+(aq) + e− → Fe2+(aq)+0.771
Cu2+(aq) + 2e− → Cu(s)+0.340
2H+(aq) + 2e− → H2(g)0.000 (defined)
Fe2+(aq) + 2e− → Fe(s)−0.440
Mg2+(aq) + 2e− → Mg(s)−2.356

Selected standard enthalpies of formation:

SubstanceΔH°f (kJ/mol)
Propene C3H6(g)+20.42
Cyclopropane C3H6(g)+53.3
CO2(g)−393.5
H2O(l)−285.8
Key equations on the test
ΔG° = ΔH° − TΔS° E_cell = E°_cell − (0.0592 V / n) log Q Ka × Kb = Kw = 1.0 × 10⁻¹⁴ pH + pOH = 14
Q1 · 2 marks · Ionic dissociation
Write a dissociation equation for K2SO4 in water.
Show Solution

K2SO4 is a soluble ionic compound — it dissociates completely:

K₂SO₄(s) → 2 K⁺(aq) + SO₄²⁻(aq)

Both ions are spectators (K⁺ does not hydrolyse; SO₄²⁻ is the conjugate base of a strong acid H2SO4 and does not hydrolyse significantly). The solution is neutral.

Petrucci parallel: Ch. 16 p.782 Q10(c) — pH of 0.00683 M NaOH requires writing the dissociation of NaOH first.
Q2 · 2 marks · Strong acid ionization
Write an ionization equation for HNO3 in water.
Show Solution

HNO3 is a strong acid — it ionizes completely (one-way arrow):

HNO₃(aq) + H₂O(l) → H₃O⁺(aq) + NO₃⁻(aq)

Or equivalently: HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

Key point: strong acid = complete ionization, single arrow, no equilibrium.

Petrucci parallel: Ch. 16 p.782 Q10(b) — pH of 6.14 × 10⁻⁴ M HNO₃. pH = −log(6.14×10⁻⁴) = 3.21.
Q3 · 2 marks · Gibbs energy / entropy sign logic
Sodium chloride dissolves spontaneously in water at room temperature. The standard enthalpy for this process NaCl(s) → NaCl(aq) is +5 kJ/mol. What can you confidently say about the standard entropy for this process?
Show Solution

Given: ΔG < 0 (spontaneous), ΔH° = +5 kJ/mol (endothermic)

From ΔG° = ΔH° − TΔS°:

For ΔG° < 0 with ΔH° > 0: TΔS° > ΔH° → ΔS° > ΔH°/T At T = 298 K: ΔS° > 5000 J·mol⁻¹ / 298 K ≈ +16.8 J·mol⁻¹·K⁻¹

Answer: ΔS° must be positive (the entropy of the system increases) and must be large enough that TΔS° > 5 kJ/mol at 25°C. Physically: dissolving creates more disorder (ions in solution vs. ordered crystal lattice).

Petrucci parallel: Ch. 13 p.660 Q22 — spontaneity/sign logic for endothermic processes. Solution manual p.958–959: for ΔH > 0, spontaneous only when TΔS > ΔH (i.e., high T or large ΔS).
Q4 · 3 marks · Strong base pH
Calculate the pH of 0.00112 M Ca(OH)2(aq).
Show Solution

Ca(OH)2 is a strong base — dissociates completely:

Ca(OH)₂(aq) → Ca²⁺(aq) + 2 OH⁻(aq) [OH⁻] = 2 × 0.00112 M = 0.00224 M pOH = −log(0.00224) = −log(2.24 × 10⁻³) = 2.650 pH = 14.00 − pOH = 14.00 − 2.650 = 11.35

Key step: remember Ca(OH)2 gives two OH− per formula unit. Do not forget to multiply the concentration by 2 before taking the log.

Petrucci parallel: Ch. 16 p.782 Q12 — saturated Ca(OH)₂ with pH 12.35; same complete-dissociation logic.
Q5 · 3 marks · Thermodynamic system definition
As far as thermodynamics is concerned define a closed system. Write a sentence! Next give an example of a closed system.
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Definition: A closed system can exchange energy (as heat or work) with its surroundings, but cannot exchange matter.

Example: A sealed gas cylinder, a bomb calorimeter, a sealed water bottle — matter stays inside but heat can flow through the walls.

Contrast: an open system exchanges both energy and matter (e.g., a beaker). An isolated system exchanges neither (e.g., an ideal thermos).

Petrucci parallel: Ch. 7 p.293 Q57 — fixed-quantity ideal gas in isothermal expansion; DeltaU = 0, heat and work exchanged with surroundings but no matter transfer.
Q6 · 6 marks · Hess's Law / combustion enthalpy / isomer stability
Calculate the enthalpy of combustion of propene (A) and cyclopropane (B) using the ΔH°f data on the formula sheet. The final result should be in kJ/mol per mole combusted. The two enthalpies are different — which compound is more stable and how can you tell?
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Balanced combustion equation (both are C3H6):

C₃H₆(g) + 9/2 O₂(g) → 3 CO₂(g) + 3 H₂O(l)

Using ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants) and ΔH°f(O₂) = 0:

Compound A — Propene [ΔH°f = +20.42 kJ/mol]:

ΔH°_comb = [3(−393.5) + 3(−285.8)] − [20.42] = [−1180.5 + (−857.4)] − 20.42 = −2037.9 − 20.42 = −2058.3 kJ/mol

Compound B — Cyclopropane [ΔH°f = +53.3 kJ/mol]:

ΔH°_comb = [3(−393.5) + 3(−285.8)] − [53.3] = −2037.9 − 53.3 = −2091.2 kJ/mol

Stability conclusion: Propene (A) releases less energy on combustion (|−2058| < |−2091|), meaning propene has a lower chemical potential energy → propene is more stable. Cyclopropane has ~33 kJ/mol extra energy from ring strain in its triangular C–C–C skeleton, making it less stable.

Common error: using an unbalanced equation or multiplying through by an integer and forgetting to divide back to per-mole. Always check: 3C, 6H, sufficient O₂ → 3CO₂ + 3H₂O.
Petrucci parallels: Ch. 7 p.293 Q69 (combustion Hess's Law); p.294 Q76 (hydrogenation enthalpy and isomer stability). Solution manual p.282–283.
Q7 · 10 marks · Voltaic cell / Nernst equation
Draw a hydrogen electrode properly connected to a gold half-cell. Name all vital parts. The hydrogen ion concentration is 0.010 M, H2(g) pressure is 0.1 atm, and the gold(III) ion concentration is 0.0111 M. Identify anode/cathode, write half-reactions and overall cell reaction, and calculate Ecell.
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Half-reactions (from E° table):

Cathode (reduction — higher E°): Au³⁺(aq) + 3e⁻ → Au(s) E° = +1.52 V Anode (oxidation — lower E°, SHE reversed): H₂(g) → 2H⁺(aq) + 2e⁻ E° = 0.00 V

Balance electrons (LCM of 3 and 2 = 6):

Cathode ×2: 2Au³⁺ + 6e⁻ → 2Au Anode ×3: 3H₂ → 6H⁺ + 6e⁻ ──────────────────────────────────────── Overall: 2Au³⁺(aq) + 3H₂(g) → 2Au(s) + 6H⁺(aq) n = 6

Standard cell potential:

E°_cell = E°_cathode − E°_anode = 1.52 − 0.00 = +1.52 V

Reaction quotient Q:

Q = [H⁺]⁶ / ([Au³⁺]² × p(H₂)³) = (0.010)⁶ / ((0.0111)² × (0.1)³) = 1.000×10⁻¹² / (1.232×10⁻⁴ × 1.0×10⁻³) = 1.000×10⁻¹² / 1.232×10⁻⁷ = 8.12×10⁻⁶

Nernst equation (E_cell):

E_cell = E°_cell − (0.0592/n) log Q = 1.52 − (0.0592/6) × log(8.12×10⁻⁶) = 1.52 − (0.009867) × (−5.090) = 1.52 + 0.0502 = 1.57 V

Cell diagram: Pt(s) | H2(g, 0.1 atm) | H+(0.010 M) ‖ Au3+(0.0111 M) | Au(s)

Electrons flow: anode (Pt/H₂, left) → cathode (Au, right) through the external wire.

Vital parts to name in the drawing: anode, cathode, salt bridge, external wire with electron-flow arrow, half-cell solutions with concentrations, Pt electrode (inert), Au electrode.
Petrucci parallel: Ch. 19 p.912 Q20(c) — Cr|Cr²⁺ ‖ Au³⁺|Au cell. Solution manual p.1014: E°_cell = 2.42 V for that pair. Same Nernst procedure applies.
Q8 · 8 marks · Weak-base hydrolysis / salt pH
Calculate the pH of 0.111 M aqueous sodium nitrite NaNO2(aq). The acid ionization constant for nitrous acid (HNO2) is Ka = 7.2 × 10−4.
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Step 1 — Identify the active species:

NaNO2 → Na+ + NO2−. Na+ is a spectator. NO2− is the conjugate base of the weak acid HNO2, so it hydrolyses to give a basic solution.

Step 2 — Find Kb for NO2−:

Kb = Kw / Ka = (1.00×10⁻¹⁴) / (7.2×10⁻⁴) = 1.389×10⁻¹¹

Step 3 — Hydrolysis equilibrium:

NO₂⁻(aq) + H₂O(l) ⇌ HNO₂(aq) + OH⁻(aq) ICE: [NO₂⁻]: 0.111 − x ≈ 0.111 (x small) [OH⁻]: x [HNO₂]: x Kb = x² / 0.111 = 1.389×10⁻¹¹ x² = 1.542×10⁻¹² x = 1.242×10⁻⁶ M = [OH⁻] Check: x / 0.111 = 1.12×10⁻⁵ ≪ 5% ✓ approximation valid

Step 4 — Calculate pH:

pOH = −log(1.242×10⁻⁶) = 5.906 pH = 14.00 − 5.906 = 8.09

The solution is basic (pH > 7), as expected for the salt of a weak acid and strong base.

Petrucci parallel: Ch. 16 p.785 Q59 — pH of 0.089 M NaOCl (same weak-base anion hydrolysis pattern). Solution manual p.757: [OH⁻] = 1.7×10⁻⁴ M, pOH = 3.77, pH = 10.23.
Petrucci Textbook Parallels — TT2A (28 Mar 2024)
TT2A QuestionTopicPetrucci MatchPriority
Q1 — K₂SO₄ dissociationStrong electrolyte dissociationCh. 16 p.782 Q10(c)Core
Q2 — HNO₃ ionizationStrong acid ionizationCh. 16 p.782 Q9, Q10(b)Core
Q3 — NaCl dissolves, ΔH > 0 → ΔS?ΔG = ΔH − TΔS sign logicCh. 13 p.660 Q21, Q22, Q23Medium
Q4 — pH of 0.00112 M Ca(OH)₂Strong base with 2 OH⁻ per unitCh. 16 p.782 Q10, Q12Core
Q5 — Define closed systemThermodynamic system typesCh. 7 p.247 Concept Assessment 7-1; p.293 Q57Core
Q6 — Combustion enthalpy + isomer stabilityHess's Law / ΔH°_fCh. 7 p.293 Q68, Q69; p.294 Q76Stretch
Q7 — H₂/Au voltaic cell + NernstElectrochemistry + Nernst equationCh. 19 p.912 Q20(c); p.917 Q92Stretch
Q8 — pH of 0.111 M NaNO₂Weak-base hydrolysis (Kb = Kw/Ka)Ch. 16 p.785 Q59, Q60Core
Study order: Core questions (Q1, Q2, Q4, Q5, Q8) are highest frequency on CHEM 150 tests. Stretch questions (Q6, Q7) require multi-step setups — practice the sub-skills (Hess's Law, Nernst) separately first.

Key Formulas & Constants

Everything you need on exam day

Constants
N = 6.022 × 10²³ mol¹ 1 u = 1.660 × 10² kg R = 8.314 L·kPa/(mol·K) R = 0.08206 L·atm/(mol·K) STP molar volume = 22.4 L/mol 1 atm = 101.325 kPa = 760 torr
All Key Formulas
n = m/M MᵢVᵢ = MᶠVᶠ % yield = actual/theoretical × 100% FC = VE ½(BE) LP Z_eff = Z S (core electrons) μ = Qr (dipoles) ΔH = Σ D°(broken) Σ D°(formed) P_A = x_A × P°_A (Raoult) ΔP = x_solute × P°_A S_g = k × P_g (Henry) π = MRT (osmotic pressure) PV = nRT PV/T = PV/T P_total = Σ P_i P_i = x_i × P_total [P + n²a/V²][V nb] = nRT (van der Waals)
Common Ion Charges
IonChargeIonCharge
NH+1OH1
NO1NO1
SO²2SO²2
CO²2PO³3
ClO1ClO1
ClO1ClO1
MnO1CrO²2
Naming Acids Quick Chart
AnionAcid nameAnionAcid name
Cl (chloride)HCl hydrochloric acidNO (nitrate)HNO nitric acid
SO² (sulfate)HSO sulfuric acidNO (nitrite)HNO nitrous acid
PO³ (phosphate)HPO phosphoric acidClO (chlorate)HClO chloric acid
CO² (carbonate)HCO carbonic acidClO (hypochlorite)HClO hypochlorous acid
Stoichiometry Worked Example Limiting Reagent
Reaction: 2Nb + 5Cl 2NbCl Given: 22.2 g Nb (M=92.91) and excess Cl mol Nb = 22.2 / 92.91 = 0.2390 mol Nb mol NbCl = 0.2390 mol Nb × (2 mol NbCl / 2 mol Nb) = 0.2390 mol mass NbCl = 0.2390 × 270.17 g/mol = 64.6 g (theoretical yield) If actual yield = 55.4 g % yield = 55.4/64.6 × 100 = 85.7%