Chemistry • Year 11 • Module 4 • Lesson 3
Calorimetry — Neutralisation
Lock in the key vocabulary, the two calorimetry formulas, and the conceptual distinction between strong and weak acid neutralisation before attempting calculations.
1. Term–definition match
Match each term on the left to its correct definition. Write the letter of the definition in the right-hand column. Each definition is used once. 10 marks
| # | Term | Your answer (A–J) |
|---|---|---|
| 1.1 | Enthalpy of neutralisation (ΔHn) | |
| 1.2 | Net ionic equation for strong acid + strong base | |
| 1.3 | Spectator ions | |
| 1.4 | Polystyrene cup calorimeter | |
| 1.5 | m in q = mcΔT (neutralisation) | |
| 1.6 | n in ΔHn = −q/n | |
| 1.7 | Why ΔHn is negative for strong acid + strong base | |
| 1.8 | Why weak acid + strong base gives a less negative ΔHn | |
| 1.9 | Diprotic acid (e.g. H2SO4) | |
| 1.10 | Accepted ΔHn for any strong acid + strong base |
A — Ions that are present in solution but do not participate in the net reaction; their presence does not affect ΔH.
B — A simple insulating apparatus used for solution-phase reactions; assumes all heat is absorbed by the liquid mixture.
C — The heat change per mole of water produced when an acid and base neutralise.
D — H+(aq) + OH−(aq) → H2O(l)
E — Approximately −57 kJ mol−1, regardless of the specific strong acid or base used.
F — The reaction is exothermic — energy is released as H+ and OH− bond to form water.
G — An acid that releases two moles of H+ per mole; n(H2O) = 2 × n(acid).
H — The total combined mass of both the acid solution and the base solution.
I — Moles of H2O formed in the reaction (= moles of limiting reagent).
J — Energy is consumed to fully ionise the weak acid before neutralisation, reducing net heat released.
2. True or false — with correction
Circle T or F. If false, write the corrected statement on the line below. 10 marks (1 T/F + 1 correction each)
2.1 In a neutralisation calorimetry experiment, the value of ‘m’ in q = mcΔT is the mass of the acid solution only. T / F
2.2 The net ionic equation H+(aq) + OH−(aq) → H2O(l) applies to all strong acid + strong base reactions. T / F
2.3 HCl + NaOH and HNO3 + KOH give different ΔHn values because they have different spectator ions. T / F
2.4 Weak acid neutralisation releases more heat per mole of H2O than strong acid neutralisation. T / F
2.5 For H2SO4 + 2NaOH, one mole of acid produces two moles of H2O. T / F
3. Complete the cloze paragraph
Fill in each blank using one word or term from the word bank below. Each term is used once. 8 marks
When a strong acid and a strong base react in a solution calorimeter, the reaction is __________________ because heat is released as H+ and OH− bond together. The equation that describes this reaction is the __________________ equation: H+(aq) + OH−(aq) → H2O(l). The ions that are present but do not react are called __________________ ions. Because the fundamental reaction is always identical, ΔHn for all strong acid + strong base combinations is approximately __________________ kJ mol−1. In the laboratory, a __________________ cup is used as the calorimeter because it insulates the reaction mixture. The value of ‘m’ in q = mcΔT is the __________________ mass of both solutions. The value of ‘n’ in ΔHn = −q/n is the __________________. When a weak acid reacts, extra energy is consumed to fully __________________ the acid, making ΔHn less negative than for a strong acid.
4. Function recall
Answer each question in 1–2 sentences using precise lesson terms. 8 marks (2 each)
4.1 Why is polystyrene used instead of copper for a neutralisation calorimeter?
4.2 What is the physical significance of the negative sign in ΔHn = −q/n?
4.3 Why does a diprotic acid such as H2SO4 require special care when calculating n(H2O)?
4.4 A technician measures the temperature of an antacid reaction (Mylanta tablet dissolved in HCl) and records a ΔT of only 2.1 °C. A colleague neutralises pure NaOH with the same amount of HCl and records ΔT = 6.8 °C. State one reason for the difference, using lesson content.
5. Connect the concepts
Draw labelled arrows between the six terms in the box to show how they relate. Each arrow must carry a linking phrase (e.g. “determines”, “is used to calculate”, “depends on”). Aim for at least 6 labelled arrows. 6 marks
Terms: q = mcΔT • n(H2O formed) • ΔHn • limiting reagent • ΔT (temperature rise) • combined solution mass
6. Sequence the calculation steps
The steps below for calculating ΔHn from a neutralisation experiment are in the wrong order. Write the correct order (1–6) in the “Order” column. 6 marks
| Order | Step |
|---|---|
| Calculate ΔHn = −q(kJ) ÷ n(H2O) | |
| Identify the limiting reagent and calculate n(H2O formed) | |
| Record Tinitial of both solutions and Tfinal after mixing; calculate ΔT | |
| Calculate q = mcΔT and convert J to kJ (divide by 1000) | |
| Determine the combined mass of both solutions (assume density ≈ 1 g mL−1) | |
| Write the neutralisation equation and identify how many moles of H2O each mole of acid produces |
Q1 — Term–definition match
1.1 → C • 1.2 → D • 1.3 → A • 1.4 → B • 1.5 → H • 1.6 → I • 1.7 → F • 1.8 → J • 1.9 → G • 1.10 → E
Q2 — True / false
2.1 False. Correction: ‘m’ is the total combined mass of both the acid solution and the base solution. Using only the acid mass underestimates m and therefore underestimates q and ΔHn.
2.2 True.
2.3 False. Correction: spectator ions do not participate in the net ionic reaction and therefore do not affect ΔHn. HCl + NaOH and HNO3 + KOH give the same ΔHn ≈ −57 kJ mol−1.
2.4 False. Correction: weak acid neutralisation releases less heat per mole of H2O because energy is consumed to fully ionise the weak acid, reducing the net heat released.
2.5 True. H2SO4 + 2NaOH → Na2SO4 + 2H2O; one mole of diprotic acid produces two moles of water.
Q3 — Cloze paragraph (in order of blanks)
exothermic • net ionic • spectator • −57 • polystyrene • combined • moles of H2O • ionise
Q4 — Function recall
4.1 Polystyrene has low thermal conductivity, so it insulates the reaction mixture and reduces heat loss to the surroundings. Copper’s high conductivity would allow heat to escape rapidly, making temperature measurement unreliable.
4.2 The negative sign indicates that the reaction is exothermic — heat is released to the solution. Because q (heat absorbed by solution) is positive when temperature rises, −q assigns the correct negative sign to ΔHn, showing the system loses energy.
4.3 H2SO4 releases 2 mol H+ per mole, so n(H2O) = 2 × n(H2SO4). Using only n(acid) without multiplying by 2 would halve the calculated n and double the magnitude of ΔHn — a major error.
4.4 Antacid tablets contain weak bases (e.g. magnesium hydroxide, calcium carbonate) or partially ionised compounds that require energy to fully react with HCl. This reduces the net heat released compared to a fully dissociated strong base such as NaOH, giving a smaller ΔT.
Q5 — Concept map (sample arrows)
Accept any chemically correct linking phrases. Sample valid arrows:
- ΔT —is an input to→ q = mcΔT
- combined solution mass —is an input to→ q = mcΔT
- q = mcΔT —gives→ ΔHn (via divide by n)
- limiting reagent —determines→ n(H2O formed)
- n(H2O formed) —used to calculate→ ΔHn
- ΔT —depends on→ combined solution mass
Q6 — Correct order of calculation steps
1: Write the neutralisation equation and identify moles of H2O per mole of acid.
2: Record Tinitial and Tfinal; calculate ΔT.
3: Determine combined mass of both solutions.
4: Identify limiting reagent; calculate n(H2O formed).
5: Calculate q = mcΔT and convert to kJ.
6: Calculate ΔHn = −q(kJ) ÷ n(H2O).