Chemistry • Year 11 • Module 4 • Lesson 3
Calorimetry — Neutralisation
Build HSC Band 5–6 extended-response technique on neutralisation calorimetry: evaluate experimental design, compare strong and weak acid systems, and synthesise quantitative and conceptual reasoning.
1. Stimulus-based extended response — soil neutralisation in NSW agriculture (Band 5–6)
8 marks
Background. Soil acidification is a significant problem in NSW cropping regions. Farmers spread agricultural lime (CaCO3) or quicklime (CaO) to neutralise acid soils. A soil scientist working near Dubbo compared two liming treatments over a 10-hectare paddock:
| Treatment | Reagent | Amount applied per ha | Initial soil pH | pH after 6 weeks | ΔT recorded (soil surface, °C) |
|---|---|---|---|---|---|
| A | Agricultural lime (CaCO3) | 2.0 t ha−1 | 4.8 | 6.0 | 0.4 |
| B | Quicklime (CaO) | 1.2 t ha−1 | 4.8 | 6.4 | 8.2 |
Hypothetical field data for illustrative purposes. CaO reacts with water to form Ca(OH)2 before neutralising soil acids; CaCO3 reacts with H+ to release CO2.
Q1. Analyse and evaluate the two treatments using neutralisation calorimetry principles. In your response you must:
- Explain why Treatment B (quicklime) produced a significantly larger ΔT than Treatment A, using the concept of enthalpy of neutralisation and the distinction between strong and weak base reactions.
- Identify one practical implication of the higher ΔT for farmers considering which treatment to apply near young seedlings.
- Compare the pH outcomes and assess which treatment provided more complete neutralisation of the acidic soil, using the data.
- Discuss one limitation of using ΔT as a measure of the neutralisation enthalpy in a field setting (as opposed to a controlled polystyrene cup experiment), and explain how this limitation affects the reliability of the field data.
- Reach a justified recommendation for which treatment a NSW farmer should choose when re-establishing canola after a fallow period.
2. Multi-step calculation + evaluation — experimental ΔHn vs accepted value (Band 5–6)
7 marks
Scenario. A Year 11 student at a school in the Hunter Valley is comparing ΔHn values for two reactions:
- Reaction P: 60.0 mL of 1.00 mol L−1 HCl + 60.0 mL of 1.00 mol L−1 NaOH in a polystyrene cup. Temperature rises from 20.2 °C to 26.4 °C.
- Reaction Q: 60.0 mL of 1.00 mol L−1 acetic acid (CH3COOH) + 60.0 mL of 1.00 mol L−1 NaOH in a polystyrene cup. Temperature rises from 20.2 °C to 25.1 °C.
Density of all solutions = 1.00 g mL−1; c = 4.18 J g−1 K−1.
(a) For Reaction P, calculate ΔHn in kJ mol−1. Show all steps clearly. 3 marks
(b) Calculate ΔHn for Reaction Q using the same method. 2 marks
(c) The accepted value for strong acid + strong base neutralisation is −57 kJ mol−1. Your calculated ΔHn for Reaction P will be less negative than this. Explain the most likely source of this systematic discrepancy (not random error) in a polystyrene cup experiment. 1 mark
(d) Using lesson content and your calculations, explain why Reaction Q gives a less negative ΔHn than Reaction P. 1 mark
Q1 — Sample Band 6 response (8 marks), annotated
Larger ΔT for Treatment B: CaO (quicklime) is a reactive oxide that first reacts exothermically with soil water to form Ca(OH)2: CaO + H2O → Ca(OH)2 ΔH < 0. The Ca(OH)2 formed is a strong base that fully ionises to release OH− directly, so the subsequent neutralisation follows the net ionic equation H+ + OH− → H2O with ΔHn ≈ −57 kJ mol−1. By contrast, CaCO3 reacts with soil H+ via the weaker reaction CaCO3 + 2H+ → Ca2+ + H2O + CO2, which is less exothermic per mole of H+ neutralised, similar in principle to a weak base reacting with an acid. The combined heat of CaO hydration plus strong-base neutralisation accounts for the much larger ΔT (8.2 vs 0.4 °C). [2]
Practical implication: The high ΔT for Treatment B (8.2 °C surface rise) could scorch or kill young canola seedlings whose roots are in the top few centimetres of soil. Agricultural lime (Treatment A) is safer to apply when seeds have just germinated or seedlings are emerging, because its lower heat release does not pose a thermal injury risk. [1]
pH comparison: Treatment B achieved a higher final pH (6.4 vs 6.0) using less material (1.2 vs 2.0 t ha−1), indicating more complete and rapid neutralisation. Treatment A raised pH by a smaller amount for more material applied. This is consistent with CaO being more reactive (higher enthalpy per mole) and therefore more effective at neutralising excess H+ in the same timeframe. [2]
Limitation of field ΔT: In a polystyrene cup, all heat is assumed to be absorbed by a known mass of solution, allowing accurate calculation of q via q = mcΔT. In a field setting, heat is conducted into the soil matrix, lost to the atmosphere, and absorbed by pore water — the thermal mass is indeterminate and varies with soil moisture. ΔT is therefore an unreliable proxy for ΔHn in the field; it can only provide a qualitative comparison, not an accurate molar enthalpy value. [2]
Recommendation: For re-establishing canola after fallow, I recommend Treatment A (agricultural lime). Although it is less reactive, the lower heat release reduces seedling scorching risk, it can be applied at sowing, and the 6-week pH improvement to 6.0 is within the acceptable range for canola (optimal pH 5.5–7.0). Treatment B would be appropriate for deep-ripping or inter-season application when no living crop is present. [1]
Marking criteria summary: 1 mark each for — CaO two-step mechanism (hydration + strong-base neutralisation vs CaCO3 weaker reaction) [1]; link to ΔHn ≈ −57 kJ mol−1 for strong base [1]; practical implication of high ΔT for seedlings [1]; comparative pH analysis using data (direction + at least one value) [1]; conclusion on more complete neutralisation [1]; limitation of field ΔT with mechanism (indeterminate thermal mass / heat loss to environment) [1]; how limitation affects reliability (cannot calculate q accurately) [1]; justified recommendation naming context (canola at sowing) [1].
Q2(a) — ΔHn for Reaction P (strong acid + strong base)
m = (60.0 + 60.0) × 1.00 = 120.0 g [1]
ΔT = 26.4 − 20.2 = 6.2 °C = 6.2 K
q = 120.0 × 4.18 × 6.2 = 3109.9 J = 3.110 kJ [1]
n(HCl) = 1.00 × 0.0600 = 0.0600 mol = n(H2O formed)
ΔHn = −3.110 ÷ 0.0600 = −51.8 kJ mol−1 [1]
Q2(b) — ΔHn for Reaction Q (weak acid + strong base)
m = 120.0 g (same combined mass)
ΔT = 25.1 − 20.2 = 4.9 °C
q = 120.0 × 4.18 × 4.9 = 2457.8 J = 2.458 kJ [1]
n(CH3COOH) = 1.00 × 0.0600 = 0.0600 mol = n(H2O)
ΔHn = −2.458 ÷ 0.0600 = −41.0 kJ mol−1 [1]
Q2(c) — Systematic discrepancy from accepted −57 kJ mol−1
The main systematic error is heat loss through the walls, base and open top of the polystyrene cup to the surrounding air during the experiment. Because the cup is not a perfect insulator, some of the heat released by the neutralisation escapes to the environment rather than being absorbed by the solution. This makes the recorded ΔT — and therefore q — smaller than the true value, giving a ΔHn that is less negative than −57 kJ mol−1. [1]
Q2(d) — Why Reaction Q gives a less negative ΔHn
Acetic acid (CH3COOH) is a weak acid and only partially ionises in water. During the neutralisation reaction, energy must be absorbed from the system to fully ionise the weak acid before the H+ can combine with OH− to form water. This energy cost reduces the net heat released, so ΔHn for Reaction Q is less negative than the −57 kJ mol−1 expected for a fully dissociated strong acid. [1]