Chemistry • Year 12 • Module 6 • Lesson 3
Enthalpy of Neutralisation: Practical & Theory
Apply q = mcΔT and ΔHn = −q/n to real calorimetry data, interpret a temperature–time graph, and reason about Australian antacid products.
1. Interpret calorimetry data — antacid comparison
A Year 12 student in Sydney tests three antacid products against 50.0 mL of 1.00 mol/L HCl. Each antacid is adjusted so that it provides exactly 0.050 mol of base. All solutions start at 20.0°C. The combined solution mass in each experiment is 100.0 g. Results are shown below. 8 marks
| Experiment | Antacid product | Active base | Acid type | Tmax (°C) | ΔT (°C) | q (J) | ΔHn (kJ/mol) |
|---|---|---|---|---|---|---|---|
| A | Mylanta (simulated) | Mg(OH)2 — weak base | HCl — strong | 24.6 | Calculate | Calculate | Calculate |
| B | Gaviscon (simulated) | NaHCO3 — weak base | HCl — strong | 23.1 | Calculate | Calculate | Calculate |
| C | NaOH solution | NaOH — strong base | HCl — strong | 27.4 | Calculate | Calculate | Calculate |
1.1 Calculate ΔT, q, and ΔHn for each experiment. Show full working for Experiment A. 5 marks
1.2 Rank the three experiments from most to least exothermic. Explain why Experiments A and B give less negative ΔHn values than Experiment C, using lesson theory. 3 marks
2. Read and interpret the temperature–time graph
The graph below shows temperature vs time data recorded by a student using a data logger during a foam-cup calorimetry experiment (50.0 mL of 1.00 mol/L HCl + 50.0 mL of 1.00 mol/L NaOH). The dashed line is a cooling-curve extrapolation back to the moment of mixing. 9 marks
2.1 State the measured Tmax and the extrapolated Tmax from the graph. Calculate ΔT for each. 2 marks
2.2 Using the extrapolated Tmax, calculate q and ΔHn. Show all working. (n(H2O) = 0.0500 mol; m = 100.0 g; c = 4.18 J g−1°C−1; Ti = 20.5°C.) 3 marks
2.3 Explain why using the extrapolated Tmax instead of the measured Tmax gives a more accurate value of ΔHn. 2 marks
2.4 Even with extrapolation, the experimental |ΔHn| is still less than 57 kJ/mol. Identify one remaining source of systematic error. 2 marks
3. Cause-and-effect chain — tracing error through the calculation
Each cause box below describes a problem in a calorimetry experiment. In the effect box to its right, state the consequence on the final ΔHn value. The first row is done for you. 6 marks (1 per effect box + 1 for “overall direction”)
| Cause (problem in experiment) | Effect on ΔHn |
|---|---|
| Example: Foam cup has no lid — heat escapes from the top surface of the solution. | Given: Tmax recorded is lower than true Tmax → smaller ΔT → smaller q → |ΔHn| is underestimated (less negative). |
| The thermometer is thick glass and absorbs a significant amount of heat from the solution. | |
| The student uses only the volume of the acid (50.0 mL) instead of the total volume (100.0 mL) to calculate mass. | |
| The student does not swirl the mixture, so the acid and base do not completely mix before the temperature is recorded. | |
| The student reads Tmax as 0.5°C lower than its true value due to parallax error on the thermometer. | |
| The density of the solution is actually 1.03 g/mL but the student assumes 1.00 g/mL. | |
| Overall direction of ALL systematic errors in this experiment: They all make |ΔHn| (more negative / less negative) than the true value. Circle one and justify in one sentence. | |
4. Case study — Port Kembla acid spill neutralisation
In January 2024 a significant sulfuric acid spill occurred at the Port Kembla industrial complex near Wollongong, NSW. Emergency responders applied powdered calcium hydroxide (Ca(OH)2), a strong base, to neutralise the spill. The spill involved an estimated 2.0 kg of concentrated H2SO4 (molar mass = 98.1 g/mol). The net ionic equation for the neutralisation is: H+(aq) + OH−(aq) → H2O(l), ΔHn = −57 kJ/mol. 6 marks
4.1 Calculate the moles of H2SO4 in 2.0 kg of the acid. (M = 98.1 g/mol; H2SO4 produces 2 mol H+ per mol.) Hence calculate n(H2O) formed. 2 marks
4.2 Calculate the total heat released (in kJ) during the neutralisation. 1 mark
4.3 Responders wore full protective equipment despite Ca(OH)2 neutralising the acid. Using the calculated heat value, justify why heat release is a significant hazard at this scale. 2 marks
4.4 Predict and justify whether replacing Ca(OH)2 (strong base) with Mg(OH)2 (weak base) would change the heat released per mole of water formed. 1 mark
Q1 — Calorimetry data
Working for Exp A: ΔT = 24.6 − 20.0 = 4.6°C; q = 100.0 × 4.18 × 4.6 = 1922.8 J; n(H2O) = 0.050 mol; ΔHn = −1922.8/0.050/1000 = −38.5 kJ/mol
Exp B: ΔT = 3.1°C; q = 1295.8 J; ΔHn = −25.9 kJ/mol
Exp C: ΔT = 7.4°C; q = 3093.2 J; ΔHn = −61.9 kJ/mol (close to −57 kJ/mol; slight excess due to rounding).
1.2 Ranking: C (most exothermic) > A > B (least). Exp A uses Mg(OH)2, a weak base: energy must be absorbed to complete protonation of OH− from partially dissociated Mg(OH)2, reducing net heat. Exp B uses NaHCO3, which is an even weaker base and must first accept a proton before dissociating to water, further reducing the net heat released.
Q2 — Temperature–time graph
2.1: Measured Tmax = 26.8°C → ΔTmeas = 26.8 − 20.5 = 6.3°C. Extrapolated Tmax ≈ 27.9°C → ΔTextrap = 27.9 − 20.5 = 7.4°C.
2.2: q = 100.0 × 4.18 × 7.4 = 3093.2 J; ΔHn = −3093.2/0.0500/1000 = −61.9 kJ/mol (closer to theoretical −57 kJ/mol; remaining gap due to other systematic errors).
2.3: Heat is lost to the surroundings during the reaction, so the temperature continues to fall after the true peak. Extrapolating the cooling curve back to the mixing moment recovers the temperature that would have been reached if no heat loss had occurred, giving a more accurate ΔT and therefore a more accurate q and ΔHn.
2.4: Accept any one of: heat absorbed by the thermometer/probe (raises T less than expected); the assumption that density = 1.00 g/mL underestimates m; incomplete mixing means not all base has reacted at the moment of reading; heat loss through the cup walls still occurs even during the rapid rise phase.
Q3 — Cause-and-effect chain
Thick thermometer: absorbs heat → Tmax lower than true → smaller ΔT → smaller q → |ΔHn| underestimated.
Wrong mass (acid only): m is halved → q = mcΔT is halved → but n stays the same → ΔHn = −q/n is half the correct value → |ΔHn| is underestimated.
No swirling: not all acid+base reacts → fewer moles of H2O formed than expected → less heat produced → smaller ΔT → smaller q and n → |ΔHn| underestimated.
Parallax error (reads 0.5°C low): ΔT underestimated → smaller q → |ΔHn| underestimated.
Density 1.00 instead of 1.03 g/mL: m is underestimated → q = mcΔT is underestimated → |ΔHn| underestimated.
Overall direction: All systematic errors make |ΔHn| less negative than the true value, because they all reduce the calculated heat q below the actual heat released.
Q4 — Port Kembla case study
4.1: mass = 2000 g; n(H2SO4) = 2000/98.1 = 20.4 mol; n(H+) = 2 × 20.4 = 40.8 mol; n(H2O) formed = 40.8 mol.
4.2: q = 40.8 × 57 = 2326 kJ (approximately 2.3 MJ).
4.3: 2326 kJ is equivalent to over 2 MJ of heat — enough to boil approximately 9 L of water or cause severe thermal burns over a large area. At an industrial spill scale, even diluted heating of surrounding surfaces and water could ignite flammables or cause secondary steam explosions, justifying protective gear.
4.4: Using Mg(OH)2 (weak base) would produce less heat per mole of water formed (|ΔHn| < 57 kJ/mol), because energy must be absorbed to complete the ionisation/dissociation of the weak base during the reaction. This could actually be advantageous in a spill scenario, as less heat is released, reducing thermal hazard.