Chemistry • Year 11 • Module 4 • Lesson 4
Calorimetry — Dissolution of Ionic Substances
Apply dissolution calorimetry calculations, read temperature–time graphs, interpret experimental data, and reason about real-world Australian contexts.
1. Temperature–time graph: exothermic vs endothermic dissolution
The graph below shows temperature vs time for two dissolution experiments performed in identical polystyrene cup calorimeters. Experiment A uses NaOH; Experiment B uses NH4NO3. Answer the sub-questions that follow. 8 marks total
(a) Describe the trend shown by Experiment A after the solid is added. Identify whether this dissolution is exothermic or endothermic, and justify your answer using the graph. 2 marks
(b) Read from the graph: for Experiment A, ΔT = _______ K. For Experiment B, ΔT = _______ K. 2 marks
(c) Using m = 100.0 g (water only, for simplicity), calculate q in kJ for Experiment A. Show all working. 2 marks
(d) After t = 3 min, both curves show a slight drift back towards room temperature. Explain why this occurs and identify the systematic error it introduces. 2 marks
2. Data table — comparing dissolution experiments
A student dissolved four different ionic substances in 150.0 g of water and recorded the initial and final temperatures. The table below summarises the data. 7 marks total
| Substance | Molar mass (g mol−1) | Mass dissolved (g) | Tinitial (°C) | Tfinal (°C) |
|---|---|---|---|---|
| CaCl2 | 111.10 | 5.55 | 20.0 | 23.6 |
| NH4Cl | 53.49 | 5.35 | 20.0 | 16.2 |
| NaOH | 40.00 | 2.00 | 20.0 | 22.9 |
| KNO3 | 101.10 | 5.05 | 20.0 | 17.7 |
(a) Identify which substances produced an endothermic dissolution. Justify your identification using the data. 2 marks
(b) For CaCl2, calculate ΔHsoln in kJ mol−1. Use m = mass of water + mass of solute. Show all working. 3 marks
(c) The accepted value for ΔHsoln(CaCl2) is −81.9 kJ mol−1. Compare your answer to this value and identify one likely source of error that could account for the discrepancy. 2 marks
3. Cause-and-effect chain — cold pack in AFL first aid
Complete the cause-and-effect chain below by filling in the empty boxes. The chain traces what happens at the molecular level when an AFL trainer cracks a cold pack containing NH4NO3 and water. 5 marks (1 per effect)
| Cause | → | Effect (fill in) |
|---|---|---|
| The inner bag of NH4NO3 is broken and mixes with water | → | |
| Lattice energy must be absorbed to separate NH4+ and NO3− ions | → | |
| Hydration enthalpy is released as water surrounds the separated ions | → | |
| Net energy is absorbed from the water (lattice energy > hydration enthalpy) | → | |
| The pack surface temperature drops significantly | → |
4. Predict and justify
A physiotherapy clinic in regional Victoria uses magnesium sulfate heptahydrate (MgSO4·7H2O, Epsom salts) for warm-soak treatments. A new batch has been accidentally ordered as the anhydrous form (MgSO4, M = 120.37 g mol−1). A lab assistant claims: “It doesn’t matter — the same mass of either form gives the same molar enthalpy of dissolution.” 4 marks
Predict whether the lab assistant’s claim is correct. Justify your prediction with reference to moles and ΔHsoln calculations, and explain how substituting anhydrous MgSO4 for the heptahydrate could affect the temperature of the soak.
Q1(a) — Graph description and classification
After the solid is added (at ~0.5 min) the temperature of Experiment A rises sharply from 20.0°C to approximately 26.0°C, then reaches a plateau. The temperature rise indicates energy was released by the dissolution process to the solution, confirming exothermic dissolution. The solution gained heat, so ΔHsoln is negative.
Marking notes: 1 mark for describing the temperature increase and plateau; 1 mark for identifying exothermic and linking to temperature rise.
Q1(b) — ΔT values
Experiment A: ΔT = 26.0 − 20.0 = +6.0 K
Experiment B: ΔT = 16.5 − 20.0 = −3.5 K
Q1(c) — q for Experiment A
q = mcΔT = 100.0 × 4.18 × 6.0 = +2508 J = +2.508 kJ
Marking notes: 1 mark for correct substitution; 1 mark for correct answer with units.
Q1(d) — Drift after plateau
After the maximum/minimum temperature is reached, the solution and the surroundings (air, bench) exchange heat because the polystyrene cup is not a perfect insulator. For Experiment A (exothermic) the warm solution slowly loses heat to the cooler surroundings, so the temperature drifts downward. For Experiment B (endothermic) the cold solution gains heat from the warmer surroundings. This introduces a systematic error: the plateau temperature read from the graph underestimates the true peak/trough (for A) or overestimates it (for B), leading to an underestimate of |ΔT| and therefore an underestimate of |ΔHsoln|.
Marking notes: 1 mark for identifying heat exchange with surroundings after dissolution is complete; 1 mark for correctly describing the systematic error on |ΔHsoln|.
Q2(a) — Endothermic substances
NH4Cl (T fell from 20.0 to 16.2°C) and KNO3 (T fell from 20.0 to 17.7°C) produced endothermic dissolution. The temperature fall indicates the dissolving process absorbed energy from the water solution, consistent with positive ΔHsoln and lattice energy > hydration enthalpy.
Marking notes: 1 mark for identifying both substances; 1 mark for justifying using temperature fall data.
Q2(b) — ΔHsoln(CaCl2) calculation
ΔT = 23.6 − 20.0 = +3.6 K
m = 150.0 + 5.55 = 155.55 g
q = 155.55 × 4.18 × 3.6 = 2341.4 J = 2.341 kJ
n = 5.55 ÷ 111.10 = 0.04995 mol
ΔHsoln = −q ÷ n = −2.341 ÷ 0.04995 = −46.9 kJ mol−1
Marking notes: 1 mark for correct m and ΔT; 1 mark for correct q; 1 mark for correct n and ΔHsoln with sign and units.
Q2(c) — Comparison to accepted value
The experimental value (−46.9 kJ mol−1) is substantially less negative than the accepted value (−81.9 kJ mol−1). This is a significant underestimate of the magnitude of ΔHsoln. A likely source of error is heat loss to the surroundings: the polystyrene cup is not perfectly insulating, so some of the energy released by the exothermic dissolution escaped to the bench and air rather than warming the solution. This makes the measured ΔT (and thus q and ΔHsoln) less than the true value. Other valid answers: incomplete dissolution reducing effective n; solute not fully added before recording Tinitial.
Q3 — Cause-and-effect chain
1. The ionic lattice of NH4NO3 begins to break apart as water contacts the solid crystals (dissolution commences).
2. Energy is absorbed from the water molecules; the kinetic energy of water decreases, so the solution temperature begins to fall.
3. Some energy is released back as water molecules form ion–dipole bonds with the separated NH4+ and NO3− ions (partial compensation).
4. The solution temperature falls because energy absorbed by the lattice-breaking step exceeds the energy returned by ion hydration; ΔHsoln > 0.
5. The pack feels cold against skin; heat flows from the injured tissue into the pack, providing a cooling/analgesic effect on the ankle.
Marking notes: 1 mark per logically correct effect (award if the causal reasoning is sound; exact wording not required).
Q4 — Predict and justify (Epsom salts)
The lab assistant’s claim is incorrect. The molar enthalpy of dissolution is an intensive property per mole of solute, so ΔHsoln per mole is a fixed thermodynamic value — it does not depend on the mass used. However, the same mass of the two forms (anhydrous vs heptahydrate) represents very different numbers of moles: M(MgSO4) = 120.37 g mol−1; M(MgSO4·7H2O) = 246.47 g mol−1. For a given mass, the anhydrous form provides more moles (~2×), releasing or absorbing approximately twice as much total heat. This means the soak temperature would change significantly more with the anhydrous form — for exothermic dissolution this would make the water hotter, which could potentially cause discomfort. Additionally, when anhydrous MgSO4 dissolves, it undergoes a hydration reaction as it forms the heptahydrate in solution, releasing extra energy (enthalpy of hydration of the anhydrous salt), further complicating the temperature change.
Marking notes: 1 mark for stating claim is incorrect; 1 mark for correct comparison of moles from same mass; 1 mark for predicting the larger temperature change; 1 mark for referencing either the hydration energy term or correctly using n in the calculation context.