Question

421000 Chemistry Integrated B
U3, L7: Energy In Chemical Reactions II
Apply
sample work

Answer 1

Below are five “Apply”–level sample problems with full worked answers covering calorimetry, Hess’s Law, bond enthalpies, energy‑profile interpretation, and experimental error analysis. These are the kind of sample work students would produce for U3 L7: Energy in Chemical Reactions II.

1) Calorimetry — combustion of ethanol (experimental)
- Data: 150.0 g of water in calorimeter, initial T = 21.3 °C, final T = 33.8 °C (ΔT = 12.5 °C). Mass of ethanol burned = 0.450 g. Specific heat of water = 4.18 J·g−1·°C−1. M(MC2H5OH) = 46.07 g·mol−1.
- Step 1: Heat absorbed by water (qwater) = m c ΔT
q = 150.0 g × 4.18 J·g−1·°C−1 × 12.5 °C = 7,837.5 J
- Step 2: Heat released by combustion (qreaction) = −qwater (assume all heat goes to water)
qreaction = −7,837.5 J
- Step 3: Moles ethanol burned = 0.450 g / 46.07 g·mol−1 = 0.00977 mol
- Step 4: Molar enthalpy of combustion ΔHcomb ≈ qreaction / n
ΔHcomb = −7,837.5 J / 0.00977 mol = −8.02 × 10^5 J·mol−1 = −802 kJ·mol−1 (to three sig figs)
- Comment: This value is lower (less exothermic) than literature (≈ −1367 kJ·mol−1 for ethanol) because of heat losses and incomplete combustion (see error analysis below).

2) Calorimetry — neutralization (check for accuracy)
- Data: 50.0 mL 1.00 M HCl mixed with 50.0 mL 1.00 M NaOH. Assume density ≈ 1.00 g·mL−1 and c = 4.18 J·g−1·°C−1. Initial T = 20.0 °C, final T = 26.6 °C (ΔT = 6.6 °C).
- Step 1: Total mass ≈ 100.0 g (100.0 mL × 1.00 g·mL−1)
- Step 2: q = m c ΔT = 100.0 g × 4.18 J·g−1·°C−1 × 6.6 °C = 2,758.8 J
- Step 3: Moles limiting reagent = 0.0500 mol (0.0500 L × 1.00 M)
- Step 4: Molar enthalpy ΔH = −q / n = −2,758.8 J / 0.0500 mol = −55,176 J·mol−1 = −55.2 kJ·mol−1
- Compare to theoretical ΔHneutralization ≈ −57.3 kJ·mol−1 (for strong acid + strong base). Percent error = |−55.2 − (−57.3)|/57.3 ×100% ≈ 3.6% (good agreement; smaller errors than the ethanol experiment because reaction was done in solution and heat losses are smaller).

3) Hess’s Law — combustion of methane
- Given standard enthalpies of formation:
ΔHf° (CO2(g)) = −393.5 kJ·mol−1
ΔHf° (H2O(l)) = −285.8 kJ·mol−1
ΔHf° (CH4(g)) = −74.8 kJ·mol−1
- Reaction: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
- Use ΔHrxn = Σ ΔHf°(products) − Σ ΔHf°(reactants)
ΔH = [−393.5 + 2(−285.8)] − [−74.8 + 0]
= [−393.5 − 571.6] − (−74.8)
= −965.1 + 74.8 = −890.3 kJ·mol−1
- Final: ΔHcombustion(CH4) = −890.3 kJ·mol−1 (exothermic)

4) Bond enthalpy estimate — H2 + Cl2 → 2 HCl
- Given average bond energies: H–H = 436 kJ·mol−1, Cl–Cl = 243 kJ·mol−1, H–Cl = 432 kJ·mol−1
- ΔH ≈ (bonds broken) − (bonds formed)
Bonds broken: one H–H + one Cl–Cl = 436 + 243 = 679 kJ·mol−1
Bonds formed: two H–Cl = 2 × 432 = 864 kJ·mol−1
ΔH ≈ 679 − 864 = −185 kJ·mol−1
- Final: Approximate ΔH = −185 kJ·mol−1 (exothermic). Note: this is an estimate using average bond enthalpies; values differ from standard enthalpies of formation.

5) Energy‑profile interpretation (conceptual)
- Example diagram data (relative to reactants energy = 0 kJ): top of activation peak = +250 kJ; product energy = −150 kJ.
- (a) Activation energy (forward) Ea,fwd = 250 kJ·mol−1 (energy difference reactants → peak).
- (b) Reaction enthalpy ΔH = Eproducts − Ereactants = −150 − 0 = −150 kJ·mol−1 (exothermic by 150 kJ).
- (c) Activation energy (reverse) Ea,rev = Ea,fwd − ΔH = 250 − (−150) = 400 kJ·mol−1 (or calculate as peak − product = 250 − (−150) = 400 kJ·mol−1).
- Interpretation: Reaction is exothermic but the reverse reaction has a larger activation barrier (kinetically hindered).

6) Short experimental‑error analysis (why the ethanol result was much less exothermic)
- Typical sources of error that make measured |ΔH| too small:
- Heat lost to calorimeter, air, flame, and container (not all heat went to the measured water).
- Incomplete combustion of ethanol (so less heat released per mass burned).
- Heat absorbed by the calorimeter vessel and thermometer (unless corrected using calorimeter constant).
- Evaporation of fuel or water during experiment.
- Improvements:
- Use a bomb calorimeter or apply a calorimeter constant (calibrate with a known reaction).
- Insulate the calorimeter better; reduce heat losses.
- Ensure complete combustion (or use combustion bomb to assure complete oxidation).

If you want, I can:
- Convert any of these samples into a student‑rubric with point‑breakdown,
- Generate additional practice problems (with varying difficulty), or
- Create a simple worksheet you can print for class. Which would you like?