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Module 4 · L3 of 13 ~35 min ⚡ +50 XP in Learn · +25 to complete

Calorimetry, Neutralisation

In 1840, Germain Hess published measurements showing that mixing aqueous HCl with aqueous NaOH released approximately 57 kJ mol⁻¹ of heat, close to the result for other dilute strong acid/strong base pairs such as HNO₃ + KOH. Hess's careful neutralisation calorimetry, conducted at the St Petersburg Academy of Sciences, was one of the key observations that led him to formulate the Law of Constant Heat Summation: for dilute strong acids and bases, the molar enthalpy of neutralisation per mole of water came out approximately the same.

Today's hook, In 1840, Germain Hess at the St Petersburg Academy of Sciences measured neutralisation heat for three different strong acid–base pairs and got the same answer every time: ~57 kJ mol⁻¹. That observation, the acid and base don't matter, only the moles of water formed, is the key to this lesson.
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Warm up, context and priorities

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Where this lesson fits

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Where this lesson fits
Syllabus requirement

Lesson question: when an acid and a base are mixed in a cup and it warms up, how do you turn that temperature rise into a molar enthalpy of neutralisation, and why does every dilute strong acid + strong base give about the same value?

  1. 1Explain the constant value. The net ionic reaction H⁺(aq) + OH⁻(aq) → H₂O(l) is the same for every strong acid + strong base, so ΔHn ≈ −57 kJ mol⁻¹.
  2. 2Set up the calorimetry. Use q = mcΔT with m = the total mass of both solutions, measured in an insulating polystyrene cup.
  3. 3Convert to molar enthalpy. ΔHn = −q / n, where n is the moles of water formed by the limiting reagent.

Quick prerequisite: neutralisation is exothermic (ΔHn < 0), and the heat released is found with q = mcΔT (from L02). If either idea is shaky, start with the Supported route in the pathways card below. Syllabus reference: investigate enthalpy changes using calorimetry and q = mcΔT, comparing experimental results with reliable secondary data and explaining differences (ACSCH037, ACSCH073).

Know what matters most

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Know what matters most

Must know

  • q = mcΔT with m = total mass of both solutions combined
  • ΔHn = −q / n, where n = moles of H₂O formed (the limiting reagent)
  • Strong acid + strong base: net ionic H⁺(aq) + OH⁻(aq) → H₂O(l), ΔHn ≈ −57 kJ mol⁻¹

Should know

  • Diprotic acids: n(H⁺) = 2 × n(H₂SO₄); always check the limiting reagent
  • Weak acids give a less negative ΔHn because ionising them consumes energy
  • Compare experimental ΔHn to the accepted value and explain the difference

Going deeper

  • Why spectator ions leave ΔH unchanged in the ideal dilute-solution model
  • H⁺(aq) is a hydrated proton; solvent reorganisation adds to the heat released
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Recall, mastery and key terms

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Recall, your gut answer first
+5 XP warm-up

A nurse mixes two clear solutions, hydrochloric acid and sodium hydroxide, in a polystyrene cup. Within seconds, the cup feels warm in her hand. No flame, no fuel, just ions rearranging.

What's releasing the energy? And here's the challenge: would it matter if she used sulfuric acid instead of hydrochloric acid, would you get a different amount of heat per mole of water formed?

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What you'll master, and the words for it

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What you'll master
Know

Key Facts

  • The solution (coffee cup) calorimeter setup for neutralisation
  • Why strong acid + strong base gives ΔHn ≈ −57 kJ mol⁻¹
  • The net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)
Understand

Concepts

  • Why m = total mass of both solutions combined
  • Why weak acid + strong base gives a less negative ΔHn
  • How neutralisation differs from combustion calorimetry in setup and formula
Can Do

Skills

  • Calculate q and ΔHn from neutralisation data
  • Calculate n using n = cV, including for diprotic acids
  • Compare experimental ΔHn to the accepted −57 kJ mol⁻¹
Beyond the syllabus. Module 4 names two calorimetry contexts for practical investigation, the combustion of a fuel and the dissolution of a solid, and those are the ones you will be assessed on. Neutralisation calorimetry is not a third one: it is the same q = mcΔT skill applied to a different reaction, a supporting application rather than a separate Core endpoint. Everything transferable here is still examinable, using the mass of the whole solution rather than the mass of the acid alone, converting q into a molar ΔH, and explaining heat lost to the surroundings. Practise those, and do not treat the neutralisation context itself as a mastery gate.
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Key terms
Enthalpy of neutralisation (ΔHn)
Heat change when one mole of water is produced in a neutralisation reaction.
Standard ΔHn for strong acid + strong base
Approximately −57 kJ mol⁻¹ for dilute monoprotic strong acid/strong base pairs, because the net reaction is H⁺ + OH⁻ → H₂O.
Calculating ΔHn
Measure ΔT of the mixed solution; q = mcΔT; then ΔHn = −q / n(H₂O formed) in kJ mol⁻¹.
Weak acid neutralisation
Less exothermic than strong acid neutralisation; additional energy is consumed to fully ionise the weak acid before neutralisation.
Polystyrene cup calorimeter
A simple calorimeter using an insulating cup; assumes all heat is absorbed by the solution; neglects heat capacity of cup.
Moles of water formed
n(H₂O) = n(limiting reactant); used to convert q to molar enthalpy; must use limiting amount.
Cross-lesson links: Neutralisation calorimetry uses the same q = mcΔT method as combustion calorimetry (L02) and dissolution calorimetry (L04), the only difference is what reaction is being measured. The net ionic equation concept introduced here (H⁺ + OH⁻ → H₂O) is a miniature Hess's Law argument: spectator ions cancel. You'll use that cancellation logic formally in L08 when you manipulate multi-step thermochemical equations.
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Strong acid + strong base enthalpy

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Why Strong Acid + Strong Base Always Gives the Same ΔHn
Must know core concept

The reason a dilute monoprotic strong acid neutralised by a dilute strong base gives approximately the same ΔHn ≈ −57 kJ mol⁻¹ under comparable conditions is that, to a good approximation, the net reaction is the same one in each case.

When a strong acid and strong base are dissolved in water, they fully dissociate into ions. For example:

HCl(aq) → H⁺(aq) + Cl⁻(aq)
NaOH(aq) → Na⁺(aq) + OH⁻(aq)

The net ionic equation for any strong acid + strong base neutralisation is always:

H⁺(aq) + OH⁻(aq) → H₂O(l)    ΔH ≈ −57 kJ mol⁻¹

The spectator ions (Na⁺, Cl⁻, K⁺, NO₃⁻, etc.) do not appear in the net ionic equation, so in the ideal dilute-solution model they do not affect ΔH. This is why dilute HCl + NaOH and HNO₃ + KOH give approximately the same molar enthalpy of neutralisation per mole of water.

Going deeper: real solutions and the diprotic exception

The "no effect" rule is the ideal dilute-solution model. In real solutions, ion–ion and solvent effects can shift the measured value by a small amount. Diprotic H₂SO₄ is also a special case: its second ionisation is not complete in all conditions, so its value per mole of water can deviate slightly from the −57 kJ mol⁻¹ standard.

Acid + Base combination Spectator ions Net ionic equation Expected ΔHn
HCl(aq) + NaOH(aq) Na⁺, Cl⁻ H⁺ + OH⁻ → H₂O(l) ≈ −57 kJ mol⁻¹
HNO₃(aq) + KOH(aq) K⁺, NO₃⁻ H⁺ + OH⁻ → H₂O(l) ≈ −57 kJ mol⁻¹
H₂SO₄(aq) + 2NaOH(aq) Na⁺, SO₄²⁻ H⁺ + OH⁻ → H₂O(l) ≈ −57 kJ mol⁻¹
CH₃COOH(aq) + NaOH(aq) Na⁺, CH₃COO⁻ Partial ionisation of weak acid < −57 kJ mol⁻¹
Real-World Anchor, Why the Nurse's Cup Gets Warm: When HCl and NaOH mix, the net ionic reaction H⁺(aq) + OH⁻(aq) → H₂O(l) occurs, a highly exothermic process (ΔH ≈ −57 kJ mol⁻¹ per mole of H₂O formed); here H⁺(aq) stands for the hydrated proton, and the energy comes from forming water plus the reorganisation of the surrounding solvent. The spectator ions Na⁺ and Cl⁻ remain in solution and do not appear in the net ionic equation. If the nurse used dilute H₂SO₄ instead, the same net reaction would occur and the cup would warm to approximately the same temperature per mole of water formed. You'll explore this in Short Answer Q3.
Weak acid + strong base → less negative ΔHn: A weak acid (e.g. CH₃COOH) only partially ionises in water. During neutralisation, energy is consumed to force its complete ionisation. This reduces the net heat released, giving a ΔHn less negative than −57 kJ mol⁻¹. The stronger the acid (the more fully it ionises), the closer ΔHn is to the strong acid standard.
Calculating n for diprotic acids: In the dilute strong-acid approximation, H₂SO₄ is treated as releasing 2 mol H⁺ per mole of acid (this assumes its second ionisation is complete, which holds well for dilute solutions). So n(H₂O formed) = 2 × n(H₂SO₄), provided enough base is present. Always check the stoichiometry and which reagent is limiting, how many H⁺ does each mole of acid release?

For strong acid + strong base, the net ionic equation is always H⁺(aq) + OH⁻(aq) → H₂O(l), ΔH ≈ −57 kJ mol⁻¹; in the ideal dilute-solution model spectator ions do not affect ΔH. Weak acids give a less negative ΔHn because energy is consumed to fully ionise them during neutralisation.

Pause, copy the highlighted definition into your book before moving on.

Mini-task: H₂SO₄ and HCl are both mixed separately with excess NaOH. State whether the ΔHn per mole of H₂O formed will be the same or different for each experiment, and explain why using the net ionic equation concept.

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Calorimetry comparison and support

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Combustion vs Neutralisation Calorimetry
Must know core concept

We just saw that strong acid + strong base always gives ΔHn ≈ −57 kJ mol⁻¹ via the same net ionic equation. That raises a question: how does measuring ΔHn in a polystyrene cup differ from the combustion calorimeter in L02? This card answers it → by comparing what ‘m’ and ‘n’ represent in each experimental context.

The same formula q = mcΔT applies to both, but what 'm' and 'n' represent, and the apparatus used, are fundamentally different.

Feature Combustion (L02) Neutralisation (L03)
Apparatus Copper calorimeter + spirit burner Polystyrene cup + thermometer
'm' in q = mcΔT Mass of water in calorimeter Total mass of combined solution (acid + base)
'n' in ΔH = −q/n Moles of fuel burned Moles of H₂O formed
Typical ΔT 5–15°C 5–10°C
Sign of ΔH Always negative (exothermic) Negative for strong acid + strong base
Main source of error Heat loss to atmosphere; incomplete combustion Heat loss through cup walls; incomplete mixing
Why apparatus chosen Copper: high conductivity → fast heat transfer to water Polystyrene: low conductivity → insulates reaction mixture
In any calorimetry question, identify the type first. Then ask: What is 'm'? What is 'n'? These two decisions determine everything else in the calculation.
Common error, wrong 'n' for neutralisation: Using n = moles of acid solution added (e.g. n = 0.050 mol L⁻¹ × 0.050 L) without checking stoichiometry. For diprotic acids, each mole of acid produces 2 mol H⁺, so n(H₂O) = 2 × n(acid). Always identify H⁺ moles, not just acid moles.
Comparison table of combustion vs neutralisation calorimetry apparatus and variables
CALORIMETRY CALCULATOR, INTERACTIVE Interactive
Switch between Combustion, Neutralisation, and Dissolution tabs. Adjust parameters to see how ΔH changes.

In neutralisation calorimetry: m = total mass of combined solution; n = moles of H₂O formed; polystyrene cup insulates the mixture. These calculations assume the dilute solution has the density and specific heat capacity of water (1.0 g mL⁻¹, 4.18 J g⁻¹ K⁻¹) and neglect the heat capacity of the cup, so they are approximations. In combustion: m = mass of water; n = moles of fuel; copper cup rapidly conducts heat from flame to water.

Add the highlighted point to your notes before the check below.

Explain it: A student is setting up a neutralisation calorimetry experiment using 50 mL HCl and 50 mL NaOH. They use a copper beaker instead of a polystyrene cup, and they measure m as 50 g (the mass of just the acid). Identify both errors and explain the correct approach.

Same skill, choose your support

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Same skill, choose your support
pick one

All three routes practise the same Must-Know skill: turn a temperature rise into a molar enthalpy of neutralisation. Do one; try another if you want more practice.

Supported word bank + worked frame start here if unsure

Word bank: total mass · 4.18 · H₂O formed · negative

50.0 mL of 1.00 mol L⁻¹ HCl is mixed with 50.0 mL of 1.00 mol L⁻¹ NaOH; the temperature rises 6.6 °C. Complete the frame:

"m = the ______ of both solutions = 100 g. q = m × ______ × 6.6 = 2.76 kJ. n = moles of ______ = 0.0500 mol. ΔHn = −q ÷ n, so the sign is ______."

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Show a hint
A temperature rise means the reaction is exothermic, so once you divide −q by n the answer must come out negative, close to −55 kJ mol⁻¹.
Core independent HSC-standard the exam standard

25.0 mL of 1.00 mol L⁻¹ H₂SO₄ is mixed with 50.0 mL of 1.00 mol L⁻¹ NaOH in a polystyrene cup; the temperature rises 6.8 °C. Calculate ΔHn per mole of water, showing m, q, n and the correct sign.

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Stretch reasoning in a new context push yourself

A student neutralises a weak acid with NaOH and calculates ΔHn = −49 kJ mol⁻¹, then a strong acid and gets −55 kJ mol⁻¹. Explain, using ionisation and the net ionic equation, why the weak-acid value is less negative, and state one experimental factor that would make both values less negative than the true ΔHn.

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Worked examples, key idea, formula reference and worksheet

Worked examples · reveal as you go

Worked example 1 +5 XP on full reveal

HCl + NaOH, equimolar. 50.0 mL of 1.00 mol L⁻¹ HCl is mixed with 50.0 mL of 1.00 mol L⁻¹ NaOH in a polystyrene cup. The temperature rises from 21.4°C to 28.0°C. Calculate the molar enthalpy of neutralisation.

1
Total mass of solution
m = 50.0 + 50.0 = 100.0 mL ≈ 100.0 g
Combine both volumes. Assume density ≈ 1 g mL⁻¹ for dilute aqueous solutions.
2
Temperature change
ΔT = 28.0 − 21.4 = 6.6°C = 6.6 K
Temperature rose, reaction is exothermic. ΔHn will be negative.
3
Heat absorbed by solution
$q = mc\Delta T = 100.0 \times 4.18 \times 6.6 = 2758.8 \text{ J} = \mathbf{2.759 \text{ kJ}}$
Use total solution mass (100.0 g). Convert J → kJ.
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Moles of H₂O formed
$n = cV = 1.00 \times 0.0500 = \mathbf{0.0500 \text{ mol}}$
HCl and NaOH are equimolar and equal volume, limiting reagent gives 0.0500 mol H₂O. (Check: NaOH also gives 1.00 × 0.0500 = 0.0500 mol, neither is in excess.)
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Molar enthalpy of neutralisation
$\Delta H_n = \dfrac{-q}{n} = \dfrac{-2.759}{0.0500} = \mathbf{-55.2 \text{ kJ mol}^{-1}}$
Answer: ΔHn = −55.2 kJ mol⁻¹ (experimental). Accepted: ≈ −57 kJ mol⁻¹.
Negative, confirms exothermic. Compare to accepted ≈ −57 kJ mol⁻¹. The small discrepancy (3.5%) reflects heat loss through the polystyrene cup walls.
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Worked example 2 +5 XP on full reveal

H₂SO₄ + NaOH, diprotic acid. 25.0 mL of 2.00 mol L⁻¹ H₂SO₄ is neutralised with 50.0 mL of 2.00 mol L⁻¹ NaOH. The temperature rises by 13.2°C. Calculate ΔHn. (Assume density of solution = 1.00 g mL⁻¹)

1
Moles of H⁺ from H₂SO₄ (diprotic)
n(H₂SO₄) = 2.00 × 0.0250 = 0.0500 mol
n(H⁺) = 2 × 0.0500 = 0.100 mol H⁺
H₂SO₄ is diprotic, each mole releases 2 mol H⁺. This is the step students most often miss.
2
Moles of OH⁻ from NaOH
n(OH⁻) = 2.00 × 0.0500 = 0.100 mol OH⁻
H⁺ and OH⁻ are equimolar, neither is in excess. n(H₂O formed) = 0.100 mol.
3
Total mass of solution
m = 25.0 + 50.0 = 75.0 g
Combined volume of acid + base solutions.
4
Heat absorbed by solution
$q = 75.0 \times 4.18 \times 13.2 = 4138 \text{ J} = \mathbf{4.138 \text{ kJ}}$
ΔT = 13.2°C given directly. Temperature rose → q positive → ΔHn negative.
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Molar enthalpy of neutralisation
$\Delta H_n = \dfrac{-4.138}{0.100} = \mathbf{-41.4 \text{ kJ mol}^{-1}}$
Answer: ΔHn = −41.4 kJ mol⁻¹. Note: diprotic acid → n(H₂O) = 2 × n(H₂SO₄).
This experimental value is about 27% below the accepted −57 kJ mol⁻¹, a gap too large to attribute to heat loss alone. Treat it as experimental data to evaluate: likely contributors include heat loss to the surroundings, the temperature not being recorded at its true maximum, the heat capacity of the cup being ignored, and the assumption that the second ionisation of H₂SO₄ is complete under these conditions. Report it as a measured estimate requiring evaluation, not a reliable ΔHn.
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Key idea

Fill the blanks +4 XP

Complete this neutralisation calorimetry calculation. Type each missing value then click Check answers.

50 mL of 1.0 mol L⁻¹ HCl is mixed with 50 mL of 1.0 mol L⁻¹ NaOH. Temperature rises by 6.8 °C. Assume density = 1.0 g mL⁻¹ and c = 4.18 J g⁻¹ K⁻¹.

Mass of solution: m = g
Heat released: q = 100 × 4.18 × 6.8 = J
Moles of water formed: n = mol
ΔH = −q ÷ n = −2842.4 ÷ 0.05 = J mol⁻¹

What you'll master, and the words for it

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Formula reference · this lesson
core formula
📐

Formula Reference, This Lesson

$q = mc\Delta T$
q = heat energy absorbed by solution (J) m = total mass of combined solution, acid + base (g) c = 4.18 J g⁻¹ K⁻¹ (dilute aqueous solutions ≈ water) ΔT = Tfinal − Tinitial (°C or K)
$\Delta H_n = \dfrac{-q}{n}$
ΔHn = molar enthalpy of neutralisation (kJ mol⁻¹) q = heat energy in kJ (÷ 1000 from J) n = moles of H₂O formed = moles of limiting reagent
$n = cV$
n = moles (mol) c = concentration (mol L⁻¹) V = volume in litres convert mL ÷ 1000
Key difference from L02:   m = mass of all solution (acid + base combined)  |  n = moles of H₂O formed (not moles of acid or base added separately)

Warm up and recall

WS
Printable worksheet

Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.

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Mass in calorimetry calculations

Common errors · the 3 traps that cost marks

1

"m is just the mass of the acid (or the base)"

Students use only one of the solution volumes as the mass in q = mcΔT, ignoring the other solution.

Fix: m = total mass of both solutions combined. Both the acid and the base are present in the cup and both are heated by the exothermic reaction. Using only one volume will underestimate q and give a ΔHn that is less negative than the true value.

2

"n = moles of acid added" (missing diprotic factor)

Students calculate n(acid) directly from cV without considering how many H⁺ each mole of acid releases. For diprotic H₂SO₄, this halves the moles used and doubles the calculated ΔHn.

Fix: n = moles of H₂O formed = moles of H⁺ available from the limiting reagent. For H₂SO₄: n(H⁺) = 2 × n(H₂SO₄). Always identify how many protons per mole of acid.

3

"Using H₂SO₄ gives a different ΔHn per mole than HCl"

Students expect that swapping one strong acid for another changes the molar enthalpy of neutralisation.

Fix: The net ionic equation is always H⁺ + OH⁻ → H₂O regardless of which strong acid or base is used. Spectator ions have no effect on ΔH. ΔHn per mole of H₂O is ≈ −57 kJ mol⁻¹ for all strong acid + strong base combinations.

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Drill, then revisit

Quick-fire practice · 5 reps +2 XP per reveal

1

50.0 mL of 0.500 mol L⁻¹ HNO₃ is mixed with 50.0 mL of 0.500 mol L⁻¹ KOH in a polystyrene cup. The temperature rises from 20.5°C to 23.8°C. Density = 1.00 g mL⁻¹.
(a) Calculate the total mass of solution and ΔT.
(b) Calculate q in kJ.
(c) Calculate n(H₂O formed) and ΔHn. Compare to the expected −57 kJ mol⁻¹.

2

40.0 mL of 1.50 mol L⁻¹ HCl is mixed with 60.0 mL of 1.00 mol L⁻¹ NaOH. (a) Identify the limiting reagent. (b) How many moles of H₂O will form?

3

A student mixes 25.0 mL of 1.00 mol L⁻¹ H₂SO₄ with 50.0 mL of 1.00 mol L⁻¹ NaOH. The temperature rises by 6.8°C. Calculate ΔHn. (Density = 1.00 g mL⁻¹)

4

An experiment shows that neutralising CH₃COOH (ethanoic acid, a weak acid) with NaOH gives ΔHn = −51.6 kJ mol⁻¹. The expected value for a strong acid is −57 kJ mol⁻¹. (a) Is this result consistent with theory? (b) Explain why the value is less negative.

5

A nurse neutralises 50.0 mL of 1.00 mol L⁻¹ HCl with 50.0 mL of 1.00 mol L⁻¹ NaOH (ΔHn ≈ −57 kJ mol⁻¹). She then repeats with 50.0 mL of 1.00 mol L⁻¹ H₂SO₄ + 50.0 mL of 1.00 mol L⁻¹ NaOH. Predict whether the temperature rise will be greater, less, or the same in the H₂SO₄ experiment. Justify using the net ionic equation concept and mole calculations.

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Revisit your thinking

Go back to your Think First response. Now that you've studied neutralisation calorimetry, revisit Germain Hess's 1840 observation at the St Petersburg Academy of Sciences:

  • Hess measured ~57 kJ mol⁻¹ for three different strong acid–base pairs. Using your understanding of net ionic equations, explain in one sentence why the identity of the acid and base doesn't matter. Which ions are the actual "reactants"?
  • What is releasing the energy? (Answer: the net reaction H⁺(aq) + OH⁻(aq) → H₂O(l), the formation of water together with the reorganisation of the surrounding solvent, which is highly exothermic.) Connect this to the bond energy principle you'll study in L06.
  • Were you right about H₂SO₄ vs HCl? Using H₂SO₄ with the same volume gives twice the moles of H⁺, so twice the moles of H₂O (if enough base), but ΔHn per mole of H₂O is the same ≈ −57 kJ mol⁻¹. Hess observed exactly this, the molar value is unchanged.
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Interactive Tool, Enthalpy & Calorimetry Open fullscreen ↗
Use the Calorimetry Calculator. Heating 200 g of water by 5.0°C (c = 4.18 J/g·°C) requires how much heat?
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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next.

01b
Misconceptions to fix before short answer

Wrong: In calorimetry, m in q = mcΔT is the mass of the fuel burned.

Right: m in q = mcΔT is always the mass of the water (or solution) being heated, not the fuel. The fuel mass is used later to calculate ΔH per mole: ΔH = −q/n where n is moles of fuel. Using the wrong mass is the most common calorimetry error.

Wrong: For neutralisation, n = moles of acid solution added (ignoring stoichiometry).

Right: n = moles of H₂O formed = moles of H⁺ that react with OH⁻. For diprotic acids (H₂SO₄), each mole provides 2 mol H⁺, so n(H₂O) = 2 × n(H₂SO₄). Always trace H⁺ moles, not just acid moles.

Wrong: Using H₂SO₄ instead of HCl will change ΔHn per mole of H₂O formed.

Right: ΔHn per mole of H₂O is ≈ −57 kJ mol⁻¹ for all strong acid + strong base combinations. The net ionic equation H⁺ + OH⁻ → H₂O is always the same; spectator ions don't contribute to ΔH.

02
Short answer
UnderstandBand 3

Q6. In a neutralisation calorimetry experiment, explain why: (a) the total volume of both solutions is used as the mass in q = mcΔT, and (b) polystyrene is used instead of copper for the calorimeter. 3 MARKS

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ApplyBand 4

Q7. 40.0 mL of 1.50 mol L⁻¹ HCl is mixed with 60.0 mL of 1.00 mol L⁻¹ NaOH in a polystyrene cup. The temperature rises from 19.0°C to 24.5°C. (a) Identify the limiting reagent. (2 marks) (b) Calculate ΔHn. (3 marks) 5 MARKS

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EvaluateBand 5

Q8. Real-World Application: A nurse measures the heat produced when 50.0 mL of 1.00 mol L⁻¹ HCl is neutralised with 50.0 mL of 1.00 mol L⁻¹ NaOH (ΔHn ≈ −57 kJ mol⁻¹). She then repeats the experiment using 50.0 mL of 1.00 mol L⁻¹ H₂SO₄ with 50.0 mL of 1.00 mol L⁻¹ NaOH.

(a) Predict whether the temperature rise will be greater, less, or the same for the H₂SO₄ experiment compared to the HCl experiment. Justify using the net ionic equation concept. (2 marks)
(b) Calculate the expected temperature rise for the H₂SO₄ experiment given ΔHn = −57 kJ mol⁻¹. Note: H₂SO₄ is diprotic; check whether NaOH is sufficient to neutralise all H⁺. (3 marks) 5 MARKS

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Comprehensive Answers
Show comprehensive answers ▼

Short Answer Model Answers

Q6 (3 marks): (a) Both solutions are mixed in the cup and both are heated by the exothermic neutralisation reaction, so both contribute thermal mass that absorbs heat. Using only one volume would underestimate m, making q appear smaller and ΔHn less negative than the true value [2]. (b) Polystyrene is used instead of copper because polystyrene is a poor thermal conductor (low thermal conductivity), it insulates the reaction mixture, reducing heat loss to the surrounding air. Copper's high conductivity would allow heat to escape rapidly to the bench and atmosphere, making this setup impractical for neutralisation [1].

Q7 (5 marks): (a) n(HCl) = 1.50 × 0.0400 = 0.0600 mol; n(NaOH) = 1.00 × 0.0600 = 0.0600 mol. Both are equal, neither is in excess. Limiting reagent: both equally limiting [1 for each calculation, 1 for correct conclusion = 2 marks]. (b) m = 40.0 + 60.0 = 100.0 g; ΔT = 24.5 − 19.0 = 5.5°C; q = 100.0 × 4.18 × 5.5 = 2299 J = 2.299 kJ [1]; n(H₂O) = 0.0600 mol [1]; ΔHn = −2.299 ÷ 0.0600 = −38.3 kJ mol⁻¹ [1]. (Note: this is much lower than −57 kJ mol⁻¹, suggesting significant heat loss or an error in the experimental setup.)

Q8 (5 marks): (a) The temperature rise will be approximately the same [1]. Although H₂SO₄ is diprotic and 50 mL of 1.00 mol L⁻¹ H₂SO₄ supplies 2 × 0.0500 = 0.100 mol H⁺ (versus 0.0500 mol H⁺ from HCl), only 0.0500 mol NaOH is available, so NaOH is the limiting reagent in both experiments. The same 0.0500 mol of H₂O therefore forms and the same total heat is released; the extra H⁺ from H₂SO₄ is left unreacted. ΔHn per mole of H₂O is unchanged at ≈ −57 kJ mol⁻¹ [1]. (b) n(H⁺) = 2 × 1.00 × 0.0500 = 0.100 mol; n(OH⁻) = 1.00 × 0.0500 = 0.0500 mol, NaOH is limiting; n(H₂O) = 0.0500 mol [1]. q = 0.0500 × 57 = 2.85 kJ [1]. m = 100.0 g; ΔT = q ÷ (mc) = 2850 ÷ (100.0 × 4.18) = 6.8°C [1]. (Note: with NaOH limiting, only half the H₂SO₄ is neutralised, same temperature rise as the HCl experiment, not double.)