Chemistry · Year 12 · Module 6 · Lesson 3
HSC Exam Practice
Enthalpy of Neutralisation: Practical & Theory
Short answer
1.Short answer
Define enthalpy of neutralisation (ΔHn) and state its sign convention for an exothermic reaction.
Write the net ionic equation for the reaction between hydrochloric acid and sodium hydroxide in aqueous solution, and state the approximate ΔHn value.
Distinguish between the molar enthalpy of neutralisation for a strong acid + strong base combination and for a weak acid + strong base combination. Refer to the concept of ionisation in your answer.
Identify two specific sources of systematic error in a foam-cup calorimetry experiment. For each, describe the effect on the recorded temperature change (ΔT) and the resulting effect on the calculated ΔHn.
Explain why the total mass of the combined acid and base solution, rather than the mass of only one reactant, must be used in the formula q = mcΔT.
Outline how a cooling-curve extrapolation technique can be used to improve the accuracy of the measured Tmax in a calorimetry experiment.
Data response
2.Data response — multi-step calculation
A student at a school in Wollongong mixes 30.0 mL of 2.00 mol/L nitric acid (HNO3) with 50.0 mL of 1.00 mol/L sodium hydroxide (NaOH) in a foam-cup calorimeter. The initial temperature of both solutions is 19.5°C. The maximum temperature recorded is 25.8°C. Assume density = 1.00 g/mL and c = 4.18 J g−1°C−1.
(a) Identify the limiting reagent and calculate n(H2O) formed.
(b) Calculate q (in J) using the total mass of solution.
(c) Calculate ΔHn in kJ/mol.
(d) The theoretical ΔHn for HNO3 + NaOH is −57 kJ/mol. Identify one specific reason why the experimental value is less negative, and suggest one improvement to the procedure.
3.Data response — graph interpretation
The graph below shows temperature vs time data for two neutralisation experiments performed under identical conditions (50.0 mL of 1.00 mol/L acid + 50.0 mL of 1.00 mol/L base; Tinitial = 20.0°C; m = 100.0 g; mixing at t = 0). Curve A uses HCl + NaOH; Curve B uses acetic acid (CH3COOH) + NaOH.
(a) State the measured Tmax for each curve and calculate ΔT for each experiment.
(b) Calculate ΔHn (kJ/mol) for each experiment. Show full working for Curve A.
(c) Account for the difference in ΔT between Curve A and Curve B, using the theory of ionisation.
Extended response
4.Extended response
Assess the reliability of a foam-cup calorimetry experiment as a method for determining the molar enthalpy of neutralisation (ΔHn) of a strong acid with a strong base. In your response, refer to the sources and direction of systematic error, the role of cooling-curve extrapolation, and the difference between the concepts of accuracy and reliability.
Chemistry · Year 12 · Module 6 · Lesson 3
Answer Key & Marking Guidelines
Section 1 · Short answer · 2 marks · Band 3
Sample response. The enthalpy of neutralisation (ΔHn) is the heat energy released per mole of water formed when an acid reacts with a base. Because neutralisation is exothermic (heat is released to the surroundings), ΔHn is always negative by sign convention.
Marking notes. 1 mark for “heat energy released per mole of water formed in a neutralisation reaction”; 1 mark for “ΔHn is negative because the reaction is exothermic.”
Section 1 · Short answer · 2 marks · Band 3
Sample response. Net ionic equation: H+(aq) + OH−(aq) → H2O(l). The approximate ΔHn value is −57 kJ/mol (of H2O formed).
Marking notes. 1 mark for correct net ionic equation including states; 1 mark for −57 kJ/mol (accept −56 to −58 kJ/mol).
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. For a strong acid + strong base (e.g. HCl + NaOH), both species are fully ionised before the reaction. The only reaction is H+(aq) + OH−(aq) → H2O(l), so ΔHn ≈ −57 kJ/mol. For a weak acid + strong base (e.g. CH3COOH + NaOH), the weak acid is only partially ionised in solution. During neutralisation, additional energy must be absorbed to drive further ionisation of the weak acid; this energy reduces the net heat released, so |ΔHn| < 57 kJ/mol.
Marking notes. 1 mark for strong+strong gives −57 kJ/mol with reference to full ionisation; 1 mark for weak acid only partially ionised before reaction; 1 mark for ionisation during reaction is endothermic, reducing net heat → |ΔHn| < 57 kJ/mol.
Section 1 · Short answer · 4 marks · Band 4
Sample response. Source 1: Heat loss through the walls and open top of the foam cup to the surrounding atmosphere. Effect: Tmax recorded is lower than the true Tmax → ΔT is smaller → q is smaller → |ΔHn| is underestimated (less negative). Source 2: Heat absorbed by the thermometer or measuring probe. Effect: the thermometer absorbs heat from the solution, so the solution loses some heat that is not measured → smaller ΔT → smaller q → |ΔHn| underestimated.
Marking notes. 2 marks per source: 1 for naming the source specifically (not “human error”); 1 for correctly tracing effect on ΔT → q → ΔHn. Accept any two valid sources from: heat loss through cup/lid opening, heat absorbed by thermometer, density/SHC approximation, incomplete mixing, thermometer precision.
Section 1 · Short answer · 2 marks · Band 3
Sample response. All heat released by the reaction is absorbed by the combined acid + base solution, not just by one component. Using only the mass of the acid would underestimate the total heat capacity of the solution, giving a q value that is too small and therefore a ΔHn value that is incorrectly calculated.
Marking notes. 1 mark for “heat is absorbed by the entire combined solution”; 1 mark for correct consequence of using wrong mass on the calculated q or ΔHn.
Section 1 · Short answer · 3 marks · Band 4
Sample response. A data logger records temperature continuously during and after the reaction. After the reaction peak, the temperature decreases as heat is lost to the surroundings (cooling phase). The linear cooling trend from the post-peak region is extrapolated back to the moment of mixing (t = 0 for the reaction). The temperature at the intersection of the extrapolated line and the mixing time gives the true Tmax — the temperature that would have been reached if no heat had been lost. This corrected Tmax gives a more accurate ΔT and therefore a more accurate q and ΔHn.
Marking notes. 1 mark for data logger records temperature continuously; 1 mark for extrapolating the cooling portion of the graph back to the moment of mixing; 1 mark for the extrapolated Tmax gives a more accurate ΔT because it accounts for heat loss during the rise phase.
Section 2 · Data response · 7 marks · Band 4–5
(a) Limiting reagent (2 marks): n(HNO3) = 2.00 mol/L × 0.0300 L = 0.0600 mol H+. n(NaOH) = 1.00 mol/L × 0.0500 L = 0.0500 mol OH−. OH− is the limiting reagent (0.0500 mol < 0.0600 mol). n(H2O) = 0.0500 mol.
(b) q (2 marks): Total volume = 30.0 + 50.0 = 80.0 mL → m = 80.0 g. ΔT = 25.8 − 19.5 = 6.3°C. q = 80.0 × 4.18 × 6.3 = 2106.7 J.
(c) ΔHn (2 marks): ΔHn = −2106.7 J / 0.0500 mol = −42 134 J/mol ÷ 1000 = −42.1 kJ/mol.
(d) Source + improvement (1 mark): Source: heat loss through the walls and open top of the foam-cup calorimeter to the surrounding atmosphere reduces the recorded Tmax, giving a smaller ΔT and therefore a less negative ΔHn than the theoretical −57 kJ/mol. Improvement: add a tight-fitting lid to the foam cup, or use a data logger with cooling-curve extrapolation to recover the true Tmax.
Section 2 · Data response · 8 marks · Band 4–5
(a) Tmax and ΔT (2 marks): Curve A (HCl + NaOH): Tmax = 27.4°C; ΔT = 27.4 − 20.0 = 7.4°C. Curve B (CH3COOH + NaOH): Tmax = 25.8°C; ΔT = 25.8 − 20.0 = 5.8°C.
(b) ΔHn (3 marks — full working for A required): Curve A: q = 100.0 × 4.18 × 7.4 = 3093.2 J; n(H2O) = 0.0500 mol; ΔHn = −3093.2/0.0500/1000 = −61.9 kJ/mol (slightly above −57 due to rounding/graph reading; accept −58 to −62). Curve B: q = 100.0 × 4.18 × 5.8 = 2424.4 J; ΔHn = −2424.4/0.0500/1000 = −48.5 kJ/mol.
(c) Account for difference (3 marks): HCl (strong acid) is fully ionised in solution before the reaction, so all H+ ions are immediately available. The only reaction is H+(aq) + OH−(aq) → H2O(l), releasing the full ≈ −57 kJ/mol as heat, producing a larger ΔT. CH3COOH (acetic acid) is a weak acid: only approximately 0.4–1% is ionised before the reaction. When NaOH is added, OH− reacts with the available H+, pulling the equilibrium CH3COOH ⇌ H+ + CH3COO− further to the right. This additional ionisation step is endothermic — it absorbs energy that would otherwise be released as heat — reducing the net heat available to warm the solution. The result is a smaller ΔT and a less negative ΔHn than for the strong acid combination.
Section 3 · Extended response · 7 marks · Band 5–6
Sample response. The foam-cup calorimeter is a simple and relatively reliable method for determining ΔHn, but it is not highly accurate due to a consistent pattern of systematic error. Reliability refers to the reproducibility of results when the experiment is repeated under the same conditions; accuracy refers to how close those results are to the true (theoretical) value. The foam-cup method is moderately reliable — repeated trials with the same solutions and equipment tend to give similar results — but it is systematically inaccurate because all sources of error act in the same direction, producing experimental |ΔHn| values that are consistently less negative than the true value of −57 kJ/mol. The dominant sources of systematic error are: (1) heat loss through the walls and open top of the foam cup to the surrounding atmosphere, which means Tmax recorded is lower than the true Tmax, reducing ΔT, q, and |ΔHn|; (2) heat absorbed by the thermometer or probe, which further reduces the apparent temperature rise of the solution; (3) assumptions that solution density = 1.00 g/mL and specific heat capacity = 4.18 J g−1°C−1 introduce small but systematic errors in the calculated mass and q. Because all errors act in the same direction (all reduce |ΔHn|), a student can never obtain a result more negative than the true value — systematic error is unidirectional in this experiment. Cooling-curve extrapolation substantially improves accuracy. By recording temperature continuously with a data logger, the linear cooling trend after the peak can be extrapolated back to the moment of mixing, recovering the Tmax that would have been reached if no heat had been lost. This corrected Tmax gives a more accurate ΔT and therefore a less biased ΔHn. Even with extrapolation, the remaining errors (thermometer heat capacity, density/SHC assumptions) mean the method is still not as accurate as a bomb calorimeter or a calibrated adiabatic calorimeter. In summary: the foam-cup method is reliable (reproducible) but systematically inaccurate; the degree of inaccuracy can be reduced but not eliminated by cooling-curve extrapolation and careful experimental technique.
Marking notes. 1 mark — defines reliability vs accuracy correctly. 1 mark — states that all systematic errors are unidirectional (all reduce |ΔHn|). 1 mark — names at least two specific sources of systematic error with correct effect on ΔT and ΔHn. 1 mark — explains the cooling-curve extrapolation procedure (data logger, post-peak cooling region, back-extrapolation to mixing time). 1 mark — explains why extrapolation improves accuracy (“recovers Tmax that would have been reached without heat loss”). 1 mark — acknowledges that remaining errors still prevent full accuracy even with extrapolation. 1 mark — reaches an explicit assessment: foam-cup method is reliably reproducible but systematically inaccurate; adequate for ranking experiments but not for precise absolute ΔHn determination.