HSC Exam Practice

Sample response. The enthalpy of dissolution (ΔH_soln) is the heat energy absorbed or released when one mole of an ionic (or molecular) solid dissolves completely in a specified solvent under standard conditions. It can be positive (endothermic) or negative (exothermic).

Marking notes. 1 mark for ‘heat change per mole of substance dissolved’; 1 mark for noting it can be positive or negative (or equivalent acknowledgement that it is not always exothermic).

Sample response. In an exothermic dissolution, energy is released by the system (the dissolving process) to the surroundings (the solution and cup); the solution temperature rises; ΔH_soln is negative. In an endothermic dissolution, energy is absorbed from the surroundings by the system; the solution temperature falls; ΔH_soln is positive.

Marking notes. 1 mark for energy direction (exothermic = released to surroundings, temperature rises); 1 mark for energy direction (endothermic = absorbed from surroundings, temperature falls); 1 mark for correct sign for each (ΔH < 0 for exo, > 0 for endo).

Sample response. Step 1 (lattice dissociation): energy is absorbed (endothermic) to overcome the electrostatic forces of the ionic lattice and separate the ions into the gaseous state. This energy is the lattice energy and is always positive. Step 2 (hydration): energy is released (exothermic) as water dipoles form ion–dipole attractions around the separated gaseous ions. This is the hydration enthalpy and is always negative. If |hydration enthalpy| > |lattice energy|, the net ΔH_soln is negative (exothermic); if |lattice energy| > |hydration enthalpy|, ΔH_soln is positive (endothermic).

Marking notes. 1 mark for correctly identifying step 1 (lattice dissociation, endothermic); 1 mark for correctly identifying step 2 (hydration, exothermic); 1 mark for explaining how relative magnitudes determine the sign of ΔH_soln.

Sample response. In dissolution calorimetry: m = mass of water + mass of dissolved solid (the total solution mass, because the ionic solid becomes part of the solution); n = moles of ionic solid dissolved (n = m_solid / M_solid). In combustion calorimetry: m = mass of water in the calorimeter only (the fuel is outside, burning underneath); n = moles of fuel burned. The difference arises because in dissolution the solute becomes part of the solution and contributes to its heat capacity, whereas in combustion the fuel is a separate system that never becomes part of the water.

Marking notes. 1 mark for correct m in dissolution (water + solute); 1 mark for correct n in dissolution (moles of solid); 1 mark for a valid explanation of why combustion uses only the water mass.

Sample response. ΔT = 21.8 − 18.5 = +3.3 K m = 150.0 + 4.00 = 154.0 g q = mcΔT = 154.0 × 4.18 × 3.3 = 2122.6 J = 2.123 kJ n = 4.00 ÷ 111.10 = 0.03600 mol ΔH_soln = −q / n = −2.123 / 0.03600 = −58.9 kJ mol⁻¹

Marking notes. 1 mark for ΔT and correct total m; 1 mark for q in kJ with correct sign; 1 mark for n; 1 mark for correct ΔH_soln with sign and units (accept −58 to −60 kJ mol⁻¹).

Sample response. Heat loss from the solution to the surrounding air through the walls of the polystyrene cup (polystyrene is a good but imperfect insulator). For an exothermic dissolution, this makes the measured ΔT smaller than the true value, so the calculated q and hence |ΔH_soln| is an underestimate. For an endothermic dissolution, heat gained from the surroundings partially offsets the temperature drop, also making |ΔT| smaller and causing an underestimate of ΔH_soln.

Marking notes. 1 mark for identifying heat loss/gain through cup walls (or equivalent valid systematic error e.g. incomplete dissolution, heat absorbed by the thermometer); 1 mark for correct direction of effect on calculated |ΔH_soln| (underestimate).

Sample response. NaOH: T_i = 20.0°C, T_f = 25.5°C, ΔT = +5.5 K. NH₄NO₃: T_i = 20.0°C, T_f = 16.0°C, ΔT = −4.0 K. Accept readings within ±0.5°C of labelled values.

Marking notes. 1 mark for both NaOH values correct with positive ΔT; 1 mark for both NH₄NO₃ values correct with negative ΔT.

Sample response. After dissolution is complete and the temperature reaches its peak/trough, the polystyrene cup slowly exchanges heat with the surrounding air. The NaOH solution (warmer than air) gradually loses heat, so the temperature drifts down after t = 3 min. The NH₄NO₃ solution (colder than air) slowly gains heat, so the temperature drifts up after t = 3 min. Both drifts reflect the finite insulation of the polystyrene cup and the temperature difference between the solution and the surroundings.

Marking notes. 1 mark for identifying direction-specific heat exchange (NaOH loses heat to air; NH₄NO₃ gains heat from air); 1 mark for linking to the imperfect insulation of polystyrene or the temperature gradient between solution and surroundings.

Sample response (a) — q. ΔT = 25.5 − 20.0 = +5.5 K m = 100.0 + 4.00 = 104.0 g q = mcΔT = 104.0 × 4.18 × 5.5 = 2390.9 J = 2.391 kJ

Sample response (b) — ΔH_soln. n = 4.00 ÷ 40.00 = 0.1000 mol ΔH_soln = −q / n = −2.391 / 0.1000 = −23.9 kJ mol⁻¹

Sample response (c) — over/underestimate. The experimental value (−23.9 kJ mol⁻¹) is an underestimate of the magnitude of ΔH_soln compared with the accepted value (−44.5 kJ mol⁻¹). Heat lost from the warm solution to the surrounding air through the cup walls reduced the measured ΔT below the true value, giving a smaller q and thus a less negative ΔH_soln.

Marking notes. (a) 1 mark for total m; 1 mark for q in kJ. (b) 1 mark for n; 1 mark for ΔH_soln with sign and units. (c) 1 mark for underestimate + valid reason.

Sample response. The claim is incorrect on both counts: (1) the two compounds do not produce the same type of temperature change, and (2) they cannot be used interchangeably in cold packs. Being ionic is necessary but not sufficient to predict the sign of ΔH_soln — the outcome depends on the relative magnitudes of lattice energy and hydration enthalpy, both of which vary between compounds.

NH₄NO₃ has a large lattice energy relative to the combined hydration enthalpies of NH₄⁺ and NO₃⁻. More energy is absorbed breaking the lattice (step 1) than is released by hydration (step 2), so ΔH_soln is positive (+25.7 kJ mol⁻¹): an endothermic dissolution. The solution temperature falls, which is exactly what a cold pack requires. AFL trainers and sports physiotherapists across Australia use NH₄NO₃ cold packs precisely because squeezing the pack and mixing the contents produces a rapid, controllable temperature drop.

CaCl₂, by contrast, has a highly charged Ca²⁺ ion with a small ionic radius and therefore an exceptionally large hydration enthalpy. The energy released when water surrounds Ca²⁺ (step 2) far exceeds the lattice energy absorbed in step 1, giving ΔH_soln = −81.3 kJ mol⁻¹: a strongly exothermic dissolution. The solution temperature rises. A cold pack filled with CaCl₂ would get hot, providing the opposite effect — potentially burning rather than cooling an injured athlete’s ankle.

The claim also fails quantitatively: CaCl₂ releases ~81 kJ mol⁻¹ to the solution while NH₄NO₃ absorbs only ~26 kJ mol⁻¹. The thermal effects are opposite in sign and differ threefold in magnitude per mole. CaCl₂ is actually used in hot packs and for road de-icing in alpine NSW/VIC because its exothermic dissolution generates heat and lowers the freezing point of water simultaneously.

In conclusion, the claim is rejected: being ionic is not sufficient to predict the nature of dissolution. ΔH_soln is determined by the balance of lattice energy (always endothermic) and hydration enthalpy (always exothermic), which varies with ion charge and size. NH₄NO₃ and CaCl₂ produce opposite temperature effects and serve opposite purposes; they are not interchangeable in cold packs.

Marking notes.
1 mark — Correctly states the claim is false and links sign of ΔH_soln to relative magnitudes of lattice energy vs hydration enthalpy (not merely to being ionic).
2 marks — Correctly explains NH₄NO₃: lattice energy > hydration enthalpy → endothermic → temperature falls → suitable for cold packs (2 marks for full explanation; 1 mark for partial).
2 marks — Correctly explains CaCl₂: hydration enthalpy (especially Ca²⁺) > lattice energy → exothermic → temperature rises → unsuitable for cold packs, used in hot applications (2 marks for full; 1 mark for partial).
1 mark — Named Australian or real-world context for either compound (AFL cold packs / alpine road de-icing / physiotherapy hot packs).
1 mark — Explicit evaluative conclusion that rejects the claim with reference to evidence.