HSC Exam Practice

Sample response. The enthalpy of neutralisation (ΔH_n) is the heat change per mole of water produced when an acid and a base react in aqueous solution under standard conditions.

Marking notes. 1 mark for identifying it as a heat/enthalpy change; 1 mark for specifying per mole of water produced in a neutralisation reaction.

Sample response. In combustion calorimetry, ‘m’ is the mass of the water in the calorimeter (the liquid being heated by the flame). In neutralisation calorimetry, ‘m’ is the total combined mass of both solutions — acid and base — because both liquids absorb the heat released.

Marking notes. 1 mark for combustion: mass of water in the calorimeter (accept ‘water being heated’). 1 mark for neutralisation: combined/total mass of both acid and base solutions.

Sample response. Both HCl and HNO₃ are strong acids, so they fully dissociate in water to give H⁺(aq) ions. Similarly, NaOH and KOH are strong bases that fully dissociate to give OH⁻(aq) ions. The only chemical reaction that actually occurs is the net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l). The other ions (Na⁺, K⁺, Cl⁻, NO₃⁻) are spectator ions — they do not participate in the reaction and therefore do not affect the enthalpy change. Because the net reaction is identical in both cases, ΔH_n is the same.

Marking notes. 1 mark for identifying that both acid + base combinations are strong electrolytes that fully dissociate. 1 mark for writing or identifying the net ionic equation H⁺ + OH⁻ → H₂O as common to both. 1 mark for stating that the remaining ions are spectators and do not affect ΔH.

Sample response. A weak acid only partially ionises in water. During the neutralisation, energy must first be absorbed to fully ionise the weak acid before the H⁺ can combine with OH⁻ to form water. This energy cost is subtracted from the heat released by the H⁺ + OH⁻ → H₂O step, so the net heat released — and therefore ΔH_n — is less negative than for a strong acid.

Marking notes. 1 mark for identifying that energy is consumed/needed to fully ionise the weak acid. 1 mark for concluding that this reduces the net heat released, making ΔH_n less negative.

Sample response. (1) Heat loss through the cup walls and open top to the surroundings: the solution loses heat to the environment before it can be measured, so ΔT is smaller than the true value, making q smaller and ΔH_n less negative than the true value. (2) Incomplete mixing / temperature gradient: if the acid and base are not fully mixed, the thermometer may record a temperature that does not represent the entire solution, leading to an inaccurate ΔT. A lower recorded ΔT makes ΔH_n less negative; a locally higher reading near the reaction site would overestimate it. Other valid sources: heat absorbed by the cup material itself (cup heat capacity ignored); solution density assumed to be 1.00 g mL⁻¹ when it may differ slightly.

Marking notes. 1 mark per source identified (max 2); 1 mark per correct effect on ΔH_n (max 2). Both a source and its effect needed for each pair of marks.

Sample response. Record temperature readings at regular intervals before and after mixing. After the maximum is reached, the temperature will fall as heat is lost to the surroundings — this cooling forms a linear trend. Extrapolate this cooling line back to the moment of mixing (time of acid addition); the y-intercept of the extrapolation is the corrected T_final. Calculate ΔT = T_corrected − T_initial. This gives a better result because the observed peak temperature is already reduced by heat loss that occurred during the mixing period; extrapolation corrects for that heat loss, recovering the temperature the mixture would have reached in a perfectly insulated system.

Marking notes. 1 mark for describing the procedure (regular temperature readings; identify cooling trend after peak). 1 mark for stating that the cooling trend is extrapolated back to the time of mixing. 1 mark for explaining why this is more accurate (heat is being lost even while the mixture is warming up, so the observed peak underestimates the true T_final).

Part (a) — 2 marks. m = 25.0 + 50.0 = 75.0 g [1]. ΔT = 31.7 − 18.5 = 13.2 °C [1].

Part (b) — 2 marks. q = mcΔT = 75.0 × 4.18 × 13.2 = 4138.2 J = 4.138 kJ [1 for formula with correct m; 1 for correct numerical answer with kJ conversion].

Part (c) — 3 marks. n(H⁺ from H₂SO₄) = 2 × 2.00 × 0.0250 = 0.100 mol [1]. n(OH⁻ from NaOH) = 2.00 × 0.0500 = 0.100 mol. Both equal → equimolar; n(H₂O) = 0.100 mol [1]. ΔH_n = −4.138 ÷ 0.100 = −41.4 kJ mol⁻¹ [1].

Part (d) — 1 mark. Percentage error = |(41.4 − 57)| ÷ 57 × 100 = 27.4%. The large discrepancy (much greater than typical 3–5%) most likely indicates significant heat loss through the polystyrene cup and possibly the thermometer and stirrer, or that the student did not use the extrapolation method to correct ΔT. [1 for calculation; credit also for correct explanation of source]

Part (a) — 2 marks. Mylanta (Mg(OH)₂) releases substantially more heat than Gaviscon (NaHCO₃): q = 1620 J vs 554 J [1], a difference of approximately 1066 J (about three times as much heat released) [1].

Part (b) — 3 marks. Mg(OH)₂ is a strong base that fully dissociates to release OH⁻ ions directly, so its neutralisation with HCl follows the net ionic equation H⁺ + OH⁻ → H₂O, releasing close to −57 kJ per mole of H₂O with no energy required for ionisation [1]. NaHCO₃ is a weak base (carbonate) that reacts with HCl via a different mechanism — NaHCO₃ + HCl → NaCl + H₂O + CO₂ — which involves partial ionisation of the carbonate and release of CO₂ gas [1]. Energy is consumed in these additional steps, reducing the net heat released per gram, so q (and therefore ΔT) is smaller for Gaviscon even when the same combined mass of solution is used [1].

Sample response.

The claim contains two statements, both of which are incorrect.

Error 1: “You can use either the acid mass or the base mass in q = mcΔT.” This is wrong. When acid and base are mixed in a polystyrene cup, the exothermic neutralisation reaction heats the entire combined solution — both liquids absorb the heat. Using only the acid mass or only the base mass underestimates m, which underestimates q and makes ΔH_n appear less exothermic (less negative) than it truly is. The correct value of m is the combined mass of both solutions.

Numerical example: If 50.0 mL of 1.00 mol L⁻¹ HCl is mixed with 50.0 mL of 1.00 mol L⁻¹ NaOH and ΔT = 6.6 °C, using only the acid mass (50.0 g) gives q = 50.0 × 4.18 × 6.6 = 1379 J = 1.379 kJ, and ΔH_n = −1.379 ÷ 0.0500 = −27.6 kJ mol⁻¹. Using the correct combined mass (100.0 g) gives q = 100.0 × 4.18 × 6.6 = 2759 J = 2.759 kJ, and ΔH_n = −2.759 ÷ 0.0500 = −55.2 kJ mol⁻¹, far closer to the accepted −57 kJ mol⁻¹.

Error 2: “HCl gives a bigger ΔH_n per mole of water than H₂SO₄, because sulfuric acid is stronger.” This is wrong. Both HCl and H₂SO₄ are strong acids that fully dissociate in water. The net ionic equation for both is H⁺(aq) + OH⁻(aq) → H₂O(l), with ΔH_n ≈ −57 kJ per mole of H₂O formed. The ‘extra strength’ of H₂SO₄ does not add extra energy; it simply means H₂SO₄ releases two moles of H⁺ per mole of acid, so the total heat per mole of acid is roughly double — but the molar enthalpy per mole of H₂O produced is still ≈ −57 kJ mol⁻¹ for both.

Marking criteria.

• 1 mark — Identifies Error 1: using only one solution’s mass is incorrect; the correct m is the combined mass of both solutions.
• 1 mark — Explains why: both solutions are heated by the reaction, so both must be included in the thermal mass for an accurate q.
• 1 mark — Provides a numerical example demonstrating the effect of using wrong vs correct m on the calculated ΔH_n.
• 1 mark — Identifies Error 2: ΔH_n per mole of H₂O is the same for HCl and H₂SO₄ (≈ −57 kJ mol⁻¹).
• 1 mark — Explains why: both are strong acids — both fully dissociate and produce the same net ionic equation; acid strength does not change ΔH per mole of water.
• 1 mark — Correctly distinguishes total heat per mole of acid (double for H₂SO₄ because diprotic) from ΔH_n per mole of H₂O (same for both).