Chemistry • Year 11 • Module 4 • Lesson 3
Calorimetry — Neutralisation
Apply q = mcΔT and ΔHn = −q/n to real experimental data, interpret a temperature–time graph, and reason through real-world neutralisation scenarios.
1. Interpret a temperature–time graph of a neutralisation reaction
The graph below shows temperature measurements recorded during a calorimetry experiment in which 50.0 mL of 1.00 mol L−1 HCl was mixed with 50.0 mL of 1.00 mol L−1 NaOH in a polystyrene cup. Readings were taken every 30 seconds. The acid was added at t = 60 s. The dashed line is the extrapolated cooling trend used to correct for heat loss. 9 marks
Figure 1.1. Temperature–time profile for neutralisation of 50.0 mL of 1.00 mol L−1 HCl with 50.0 mL of 1.00 mol L−1 NaOH in a polystyrene cup. Hypothetical data consistent with Worked Example 1.
1.1 Read Tinitial and the corrected Tfinal from the graph. Calculate ΔT. 2 marks
1.2 Using your ΔT from 1.1, calculate q in kJ (m = 100.0 g, c = 4.18 J g−1 K−1). Show full working. 2 marks
1.3 Calculate n(H2O formed) and then ΔHn in kJ mol−1. Compare your answer to the accepted value of −57 kJ mol−1. 3 marks
1.4 Explain why the extrapolation method (dashed line) gives a more accurate result than simply using the highest recorded temperature. 2 marks
2. Compare antacid neutralisation data
A Year 11 student compared three antacid products by dissolving each in excess 1.00 mol L−1 HCl and measuring ΔT. She then calculated q and ΔHn per gram of antacid. The data are shown below. 8 marks
| Antacid | Mass used (g) | Combined solution mass (g) | ΔT (°C) | q (kJ) | ΔH per gram (kJ g−1) |
|---|---|---|---|---|---|
| Mylanta (Mg(OH)2) | 0.450 | 100.2 | 3.6 | 1.509 | −3.35 |
| Gaviscon (NaHCO3) | 0.420 | 100.2 | 1.2 | 0.503 | −1.20 |
| Quick-Eze (CaCO3) | 0.500 | 100.2 | 2.8 | 1.174 | −2.35 |
Hypothetical student data for illustrative purposes. q calculated using c = 4.18 J g−1 K−1.
2.1 Identify the antacid that releases the most heat energy per gram. State the value. 1 mark
2.2 Show how q was calculated for Gaviscon. Use q = mcΔT. 2 marks
2.3 Using the lesson’s explanation of weak vs strong neutralisation, suggest why NaHCO3 (Gaviscon) releases less heat per gram than Mg(OH)2 (Mylanta). 2 marks
2.4 The student’s teacher notes that using “mass of antacid only” instead of combined solution mass for m would overestimate |q|. Explain why using combined solution mass is correct. 3 marks
3. Case study — acid spill neutralisation at Port Kembla
In 2018, a minor sulfuric acid spill at a facility near Port Kembla (NSW) was treated by spreading powdered lime (Ca(OH)2) over the contaminated area. Safety engineers needed to estimate the heat generated during neutralisation to assess whether the ground surface would reach a temperature that could ignite organic material (>60 °C). They estimated 2.5 kg of 98% H2SO4 had spilled and reacted with excess Ca(OH)2. 7 marks
3.1 Write the balanced molecular equation for H2SO4 reacting with Ca(OH)2. Identify the spectator ion(s). 2 marks
3.2 Calculate n(H2SO4) in the spill. [M(H2SO4) = 98.09 g mol−1; assume 98% purity = 98% of mass is H2SO4.] 2 marks
3.3 Using ΔHn = −57 kJ mol−1 per mole of H2O and the fact that H2SO4 is diprotic, calculate the total heat (kJ) released in the spill neutralisation. 2 marks
3.4 Predict-and-justify: Would the engineers’ concern about surface temperature >60 °C be greater or less if the spill had involved acetic acid (weak acid) of the same mass and concentration? Justify using lesson content. 1 mark
Q1.1 — Read Ti, Tcorrected, ΔT
Tinitial = 21.4 °C (read from pre-mixing baseline). Tcorrected ≈ 28.4 °C (extrapolation line intersects t = 60 s). ΔT = 28.4 − 21.4 = 7.0 °C. Accept 6.9–7.1 °C based on graph reading. (1 for Ti and Tf; 1 for ΔT)
Q1.2 — Calculate q
q = mcΔT = 100.0 × 4.18 × 7.0 = 2926 J = 2.926 kJ. (1 for correct formula with combined mass; 1 for correct numerical answer with unit conversion)
Q1.3 — Calculate ΔHn
n(HCl) = 1.00 × 0.0500 = 0.0500 mol; n(NaOH) = 1.00 × 0.0500 = 0.0500 mol; equimolar → n(H2O) = 0.0500 mol [1]. ΔHn = −2.926 ÷ 0.0500 = −58.5 kJ mol−1 [1]. Comparison: −58.5 kJ mol−1 is approximately 2.6% more negative than the accepted −57 kJ mol−1 — within typical experimental error for a simple polystyrene calorimeter [1].
Q1.4 — Extrapolation method
After the acid is added, the solution begins losing heat to the surroundings even before the temperature has reached its true maximum [1]. The highest recorded temperature is therefore lower than the temperature the solution would have reached in a perfectly insulated system. Extrapolating the cooling trend back to the moment of mixing estimates the true maximum, giving a more accurate ΔT and therefore a more accurate ΔHn [1].
Q2.1 — Antacid with most heat per gram
Mylanta (Mg(OH)2) releases the most heat per gram: −3.35 kJ g−1.
Q2.2 — q for Gaviscon
q = mcΔT = 100.2 × 4.18 × 1.2 = 502.6 J = 0.503 kJ. (1 for correct m and formula; 1 for numerical answer)
Q2.3 — Why NaHCO3 releases less heat
NaHCO3 is a weak base (carbonate) that does not fully ionise in solution [1]. Energy is consumed to enable its complete reaction with HCl, reducing the net heat released per gram compared to Mg(OH)2, which reacts more directly and releases more of its neutralisation energy to the solution [1].
Q2.4 — Why combined mass is correct
When the acid and antacid react in solution, the exothermic neutralisation heats the entire mixed solution — both the acid solution and the dissolved antacid mixture [1]. Both volumes absorb heat, so both must be included in m for q = mcΔT [1]. Using only the mass of the antacid would underestimate the thermal mass, producing an artificially large q and an overestimated |ΔHn| [1].
Q3.1 — Balanced equation + spectator ions
H2SO4(aq) + Ca(OH)2(s) → CaSO4(s) + 2H2O(l) [1]. Note: CaSO4 is insoluble, so technically there are no spectator ions in this reaction — all ions participate. Award mark for balanced equation; accept also Ca2+ and SO42− as spectator if student assumes ionic reaction throughout [1].
Q3.2 — n(H2SO4)
Mass of pure H2SO4 = 2500 × 0.98 = 2450 g [1]. n(H2SO4) = 2450 ÷ 98.09 = 24.97 mol ≈ 25.0 mol [1].
Q3.3 — Total heat released
n(H2O) = 2 × 25.0 = 50.0 mol [1]. q = 50.0 × 57 = 2850 kJ released [1].
Q3.4 — Weak acid prediction
Less concern with acetic acid [1 for direction]. Weak acids require energy to fully ionise before neutralisation, so the net heat released per mole of H2O is less negative than −57 kJ mol−1. A same-mass acetic acid spill would release significantly less total heat, reducing the risk of the surface reaching 60 °C.