Chemistry • Year 11 • Module 4 • Lesson 4

Calorimetry — Dissolution of Ionic Substances

Synthesise dissolution calorimetry data, evaluate claims with evidence, and construct extended responses that integrate quantitative and conceptual reasoning at Band 5–6.

Master · Band 5–6

Question 1 — Alpine road de-icing: choosing the right ionic compound

8 marks — ~15 lines of answer space

Background. The New South Wales and Victorian alpine regions (Snowy Mountains, Falls Creek, Mount Hotham) experience road icing in winter. Road authorities need a de-icing agent that can be spread from a truck and will melt ice rapidly. Two candidate ionic compounds are being compared: CaCl₂ and NH₄NO₃.

Table 1 summarises calorimetry data collected at 0°C (the approximate temperature when roads ice over).

Compound	M (g mol⁻¹)	ΔH_soln (kJ mol⁻¹)	Freezing-point depression (per mol solute in 1 kg H₂O)	Cost (relative units/kg)
CaCl₂	111.10	−81.3	−5.58°C (3 ions)	1.8
NH₄NO₃	80.05	+25.7	−3.72°C (2 ions)	0.9

Sources: ΔH_soln values from standard thermochemical tables; freezing-point depression constants: Atkins & de Paula (2014). Cost index is illustrative.

Extended response prompt. Using the data in Table 1 and your understanding of dissolution calorimetry and the Born–Haber cycle for dissolution, evaluate which compound is more suitable for alpine road de-icing. In your response you must:

Define ΔH_soln and explain how its sign relates to the temperature change observed when the compound dissolves on the road surface at 0°C;
Account for the difference in sign of ΔH_soln between CaCl₂ and NH₄NO₃ using the two-step (lattice energy + hydration enthalpy) model;
Use the data to compare the thermal and colligative (freezing-point depression) effects of each compound per kilogram of de-icing agent applied;
Reach an evidence-based judgement about which compound is more suitable, acknowledging at least one trade-off.

Structure: define → explain sign → two-step model → quantify per kg if possible → judge with trade-off.

Question 2 — Multi-step calculation and evaluation: three calorimetry experiments

7 marks across parts (a)–(d)

Scenario. A Year 11 student performs three calorimetry experiments on the same day using the same polystyrene cup calorimeter and the same balance (±0.01 g):

Experiment 1: Burns 0.46 g of ethanol (M = 46.07 g mol⁻¹) under a copper calorimeter containing 200.0 g of water; temperature rises 8.2°C.
Experiment 2: Mixes 50.0 mL of 1.00 mol L⁻¹ HCl with 50.0 mL of 1.00 mol L⁻¹ NaOH (assume density = 1.00 g mL⁻¹); temperature rises 6.5°C.
Experiment 3: Dissolves 4.01 g of NaOH (M = 40.00 g mol⁻¹) in 150.0 g of water; temperature rises 5.6°C.

(a) For each experiment, state clearly what ‘m’ and ‘n’ represent in the formulae q = mcΔT and ΔH = −q/n. Do not calculate — just identify. 3 marks (0.5 per identification × 6)

(b) Calculate ΔH_soln in kJ mol⁻¹ for Experiment 3 (NaOH dissolution). Show all working. 2 marks

(c) The accepted value for ΔH_soln(NaOH) is −44.5 kJ mol⁻¹. Determine the percentage error in the student’s result and explain one way the student could modify the procedure to reduce this error. 1 mark

(d) A classmate argues that Experiment 1 (combustion) will always give a higher percentage error than Experiment 3 (dissolution) when done in a school laboratory. Evaluate this claim. 1 mark

For (b), use the exact step sequence: find ΔT, find total m (water + solute), calculate q, find n from n = m/M, then ΔH = −q/n. For (c), percentage error = |experimental − accepted| ÷ |accepted| × 100.

Answers — Do not peek before attempting

Q1 — Alpine road de-icing evaluation (8 marks)

Criterion 1 — Define ΔH_soln and sign (1 mark): ΔH_soln is the enthalpy change when one mole of an ionic solid dissolves completely in water under standard conditions. A negative ΔH_soln (CaCl₂: −81.3 kJ mol⁻¹) means the dissolution releases heat to the surroundings — including the icy road surface — raising the local temperature and assisting melting. A positive ΔH_soln (NH₄NO₃: +25.7 kJ mol⁻¹) means dissolution absorbs heat from surroundings, which would cool the road surface further, actively opposing ice melting.

Criterion 2 — Born–Haber (two-step) model (2 marks): In both cases step 1 (lattice dissociation) is endothermic: energy is absorbed to overcome electrostatic forces and separate the ions. Step 2 (hydration) is exothermic: water dipoles stabilise the gaseous ions. For CaCl₂, the highly charged Ca²⁺ ion (charge density is high, radius small) generates an exceptionally large hydration enthalpy that exceeds the lattice energy, so the net ΔH_soln is negative. For NH₄NO₃, neither ion has particularly high charge density; the lattice energy exceeds the combined hydration enthalpies of NH₄⁺ and NO₃⁻, so the net ΔH_soln is positive.

Criterion 3 — Thermal + colligative comparison per kg (2 marks):

Moles per kg: CaCl₂: 1000/111.10 = 9.00 mol kg⁻¹; NH₄NO₃: 1000/80.05 = 12.49 mol kg⁻¹.
Heat released per kg: CaCl₂: 9.00 × 81.3 = 731.7 kJ (released to road); NH₄NO₃: absorbs 12.49 × 25.7 = 321 kJ (actively cools road — disadvantageous).
Freezing-point depression per kg applied: CaCl₂: produces 3 ions × 9.00 mol = 27.0 mol of particles → depression = 27.0 × 1.86 = 50.2°C; NH₄NO₃: 2 ions × 12.49 = 24.98 mol → depression = 24.98 × 1.86 = 46.5°C.
CaCl₂ provides both a greater thermal boost and a marginally larger freezing-point depression per kg applied.

Criterion 4 — Evidence-based judgement with trade-off (2 marks): CaCl₂ is clearly more suitable for alpine road de-icing: it releases substantial heat to warm the road surface (assisting immediate melting), produces 3 ions per formula unit (greater colligative effect than NH₄NO₃), and its exothermic dissolution works synergistically with the applied salt’s function. NH₄NO₃ is cheaper (0.9 vs 1.8 relative cost units per kg) and provides slightly more moles per kg, but its endothermic dissolution would cool the already-icy surface, counterproductively worsening the icing problem. The main trade-off for CaCl₂ is cost and potential environmental impact on alpine waterways. Overall, CaCl₂ is preferred unless cost is the overriding constraint.

Award: 1 mark each for criteria 1, 4; 2 marks each for criteria 2 and 3. Accept equivalent reasoning and calculation approaches.

Q2(a) — m and n identification

Experiment 1 (combustion): m = 200.0 g (mass of water in the copper calorimeter — the fuel is outside, heating the water); n = moles of ethanol burned = 0.46 ÷ 46.07 = 0.00998 mol.

Experiment 2 (neutralisation): m = (50.0 + 50.0) × 1.00 = 100.0 g (total mass of combined solution, using density = 1.00 g mL⁻¹); n = moles of H₂O formed = moles of limiting reagent = 1.00 × 0.0500 = 0.0500 mol.

Experiment 3 (dissolution): m = 150.0 + 4.01 = 154.01 g (total mass of solution = water + dissolved NaOH); n = moles of NaOH dissolved = 4.01 ÷ 40.00 = 0.1003 mol.

Q2(b) — ΔH_soln(NaOH) calculation

ΔT = +5.6 K
m = 150.0 + 4.01 = 154.01 g
q = 154.01 × 4.18 × 5.6 = 3601.2 J = 3.601 kJ
n = 4.01 ÷ 40.00 = 0.10025 mol
ΔH_soln = −3.601 ÷ 0.10025 = −35.9 kJ mol⁻¹

Marking notes: 1 mark for correct m, q calculation with units; 1 mark for correct n and final ΔH_soln with correct sign and units.

Q2(c) — Percentage error and improvement

Percentage error = |−35.9 − (−44.5)| ÷ 44.5 × 100 = 8.6 ÷ 44.5 × 100 = 19.3%.

One valid modification: place the polystyrene cup inside a second cup (or use a lid) to further reduce heat loss to the surrounding air, bringing the measured ΔT closer to the true value and reducing the percentage error.

Q2(d) — Percentage error comparison

The classmate’s claim is reasonable but not universally certain. Combustion experiments (Experiment 1) typically have larger percentage errors in a school lab because the copper calorimeter has significant heat capacity not accounted for in the basic q = mcΔT formula, and substantial heat is radiated to the air during combustion. Dissolution (Experiment 3) in a polystyrene cup is better insulated and has minimal radiation losses because the energy transfer is entirely within the solution. However, the extent of the difference depends on experimental technique, the specific calorimeter, and whether the student stirs the solution adequately. Accept any answer that identifies heat loss to the copper calorimeter and radiation as additional error sources in combustion that are absent in dissolution.