Year 12 ChemistryModule 6, Acid/Base Reactions⏱ ~45 minLesson 3 of 19IQ1
Enthalpy of Neutralisation, Practical & Theory
In 1883, Danish chemist Julius Thomsen published the first systematic calorimetric measurements of neutralisation, reporting that strong acid + strong base reactions consistently released 57.3 kJ per mole of water formed, regardless of which strong acid or base was used. It took another decade before Wilhelm Ostwald explained why: when both reactants are fully ionised, the only reaction is H⁺ + OH⁻ → H₂O, and the heat is the bond energy of that single reaction.
🌡️
Today's hook, In 1883, Julius Thomsen measured 57.3 kJ released when strong acids neutralise strong bases, the same number every time, regardless of which acid or base he used. Why is the heat the same? And why is it smaller when the acid or base is weak?
0/5QUESTS
1
You’re here
Get oriented
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.
A pharmacist is comparing three antacid products for a patient with chronic acid reflux. Product A contains sodium hydroxide solution. Product B contains sodium hydrogen carbonate (baking soda). Product C contains magnesium hydroxide suspension. All three neutralise excess hydrochloric acid in the stomach. The pharmacist measures the temperature change in a simulated stomach acid solution after adding each antacid. Product A produces the largest temperature rise. Products B and C produce noticeably smaller temperature rises, and Product B also produces vigorous bubbling.
Question 1: Why do you think Product A releases the most heat? What is special about NaOH compared to NaHCO₃ and Mg(OH)₂?
Question 2: What is the bubbling from Product B, and why does a neutralisation reaction produce a gas at all?
📐 Formulas & Patterns
q = mcΔT
q = heat released by reaction (J)m = total mass of solution (g), use acid + base combined; assume 1.00 g/mL for dilute solutionsc = specific heat capacity = 4.18 J g⁻¹ °C⁻¹ (assume dilute aqueous)ΔT = Tfinal − Tinitial (°C), positive when temperature rises
n = c × V
n = moles of limiting reagent (mol)c = concentration (mol/L)V = volume in litres, convert mL ÷ 1000n(H₂O) formed = n(limiting reagent H⁺ or OH⁻)
ΔHn = −q / n(H₂O) [units: kJ/mol]
ΔHn = molar enthalpy of neutralisation (kJ/mol)Negative sign: exothermic, system releases heat, so ΔH is negativeConvert J → kJ: divide q by 1000 before dividing by n
Valid only when BOTH acid and base are fully ionised (strong)Strong acid + strong base: ΔHn ≈ −57 kJ/mol | Weak acid or base involved: |ΔHn| < 57 kJ/mol
Know
Key facts
The formulas q = mcΔT, n = cV, and ΔHn = −q/n
The net ionic equation for strong + strong neutralisation: H⁺ + OH⁻ → H₂O
ΔHn ≈ −57 kJ/mol for strong acid + strong base
Four main sources of error in neutralisation calorimetry
Understand
Concepts
Why ΔHn is constant for all strong + strong combinations (same net ionic equation)
Why weak acid or base gives |ΔHn| < 57 kJ/mol (energy consumed in ionisation)
Why experimental |ΔHn| is always less than theoretical (heat loss)
Why total mass (not just acid mass) is used in q = mcΔT
Can do
Skills
Calculate q, n(H₂O), and ΔHn from experimental data
Identify the limiting reagent when volumes and concentrations differ
Explain discrepancies between experimental and theoretical ΔHn values
Suggest specific, named improvements to the calorimetry procedure
Enthalpy of neutralisation (ΔHn)
The heat energy released when one mole of water is formed in a neutralisation reaction.
Standard ΔHn for strong acids/bases
Approximately −57 kJ mol⁻¹ for any strong acid + strong base combination.
Calorimetry
A technique to measure heat released or absorbed in a reaction using q = mcΔT.
Specific heat capacity (c)
The energy required to raise 1 g of a substance by 1°C; for dilute aqueous solutions ≈ 4.18 J g⁻¹ °C⁻¹.
Heat lost to surroundings
A source of systematic error in calorimetry; addressed by insulating the calorimeter.
Weak acid ΔHn
Less exothermic than strong acid ΔHn because energy is also consumed to fully ionise the weak acid.
2
Why neutralisation releases heat
Core Content
Cross-lesson links: The −57 kJ mol⁻¹ benchmark from Thomsen's 1883 measurements is revisited in L10 where you compare calorimetry data for strong vs weak acid neutralisations. The q = mcΔT method practised here reappears in every calorimetry question. The reason weak acids give less heat connects directly to Ka (L05) and ionisation energy.
1
Why Neutralisation Releases Heat, The Thermochemical Basis
Strong + strong always gives −57 kJ/mol · Same net ionic equation regardless of which acid/base
Before any formula is applied, the key question is physical: why does mixing an acid and a base always release heat, and why does the amount of heat depend on whether the acid and base are strong or weak?
Neutralisation is exothermic because forming a covalent O–H bond in liquid water releases more energy than is required to break the bonds that precede it. At the most fundamental level, every neutralisation between a strong acid and a strong base in aqueous solution is the same reaction at the ionic level, regardless of which specific acid or base is used.
HCl, HNO₃, and H₂SO₄ are all fully ionised in solution before the reaction begins, contributing H⁺(aq). NaOH, KOH, and Ca(OH)₂ are fully dissociated, contributing OH⁻(aq). The only reaction that occurs is:
H⁺(aq) + OH⁻(aq) → H₂O(l) ΔHn ≈ −57 kJ/mol
The spectator ions (Na⁺, K⁺, Cl⁻, NO₃⁻) do nothing, they don't react, don't change, and don't contribute to or absorb any energy. Because the reaction is always identical at the ionic level, the enthalpy released is always the same: approximately −57 kJ per mole of water formed, regardless of which strong acid and strong base are combined.
Net ionic equation
H⁺ + OH⁻ → H₂O
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
H⁺ + NH₃ → NH₄⁺
CH₃COOH + NH₃ → CH₃COO⁻ + NH₄⁺
ΔHn (kJ/mol H₂O)
≈ −57
More positive than −57 (less heat)
More positive than −57 (less heat)
Most positive (least exothermic)
Critical Rule: The net ionic equation H⁺ + OH⁻ → H₂O is valid ONLY when both acid and base are strong, fully ionised before the reaction begins. For weak acids or bases, the net ionic equation must include the ionisation step, which consumes energy and reduces the net heat released.
Common Error: Students write ΔHn ≈ −57 kJ/mol for ALL neutralisation reactions. This value applies ONLY to strong acid + strong base combinations. If either the acid or the base is weak, the measured |ΔHn| will be less than 57 kJ/mol. Using −57 kJ/mol for a weak acid + strong base calculation produces the wrong answer and demonstrates a fundamental misconception.
Strong acid + strong base neutralisation has a constant net ionic equation H⁺(aq) + OH⁻(aq) → H₂O(l), giving ΔHn ≈ −57 kJ/mol regardless of which specific strong acid or base is used, because spectator ions do not react; if either species is weak, |ΔHn| < 57 kJ/mol.
Pause, copy the highlighted definition into your book before moving on.
+5 XP, Which combination gives the LEAST negative ΔHn?
3
Why weak acids give less heat
2
Why Weak Acids and Bases Give Lower |ΔHn|, The Ionisation Energy Explanation
Partial ionisation before reaction → energy consumed → net heat reduced
A weak acid or base is not fully ionised in solution before the reaction begins, and the energy needed to complete that ionisation during the neutralisation reaction comes directly from the heat that would otherwise be released, reducing the net enthalpy measured.
When a weak acid such as acetic acid (CH₃COOH) is mixed with a strong base such as NaOH, the neutralisation cannot simply be described as H⁺ + OH⁻ → H₂O. In a 0.1 M CH₃COOH solution, only about 1.3% of the acetic acid molecules have ionised, the other 98.7% remain as intact CH₃COOH molecules.
When NaOH is added, the OH⁻ ions rapidly react with the small available supply of H⁺ ions, driving the acetic acid equilibrium further to the right: CH₃COOH ⇌ H⁺ + CH₃COO⁻. This additional ionisation is endothermic, energy must be absorbed to break the O–H bond in CH₃COOH. This energy comes directly from the heat that the H⁺ + OH⁻ → H₂O reaction releases. The net heat measured is therefore:
(heat from H⁺ + OH⁻ → H₂O) − (energy consumed to ionise CH₃COOH) = |ΔHn| < 57 kJ/mol
The same logic applies to weak bases: NH₃ must accept a proton from the reaction rather than already having it dissociated, which requires energy input. The weaker the acid or base, the less ionised it is before the reaction, and the more energy the ionisation step consumes, and therefore the less heat is released overall.
Ionisation before reaction
Strong acid: 100%, fully ionised
Weak acid: ~1–5% at typical concentrations
Strong base: 100%, fully dissociated
Weak base: ~1% at typical concentrations
Energy consequence for ΔHn
Strong acid: No ionisation energy needed, full −57 kJ/mol released
Weak acid: Energy absorbed to complete ionisation, |ΔHn| < 57 kJ/mol
Strong base: No dissociation energy needed, full −57 kJ/mol released
Weak base: Energy absorbed for protonation, |ΔHn| < 57 kJ/mol
Energy budget: weak acid ionisation consumes energy that would otherwise be released as heat
high-mark Response: In any extended response asking you to explain why weak acid + strong base gives lower |ΔHn|, your answer must explicitly state: (1) weak acid is only partially ionised before the reaction; (2) energy must be absorbed to complete the ionisation during the reaction; (3) this energy reduces the net heat released. All three points are required, stating only "weak acid releases less energy" earns minimal marks.
Common Error: "Weak acids are less acidic, so they release less energy." This is circular and chemically vague. The correct explanation is mechanistic, it links the degree of ionisation before the reaction (partial vs complete) to the energy budget of the reaction. "Less acidic" is not an explanation; "requires additional energy input to fully ionise during the reaction" is.
Real-World Link: Antacids containing weak bases (Mg(OH)₂, Al(OH)₃, CaCO₃) are preferred over strong bases like NaOH for treating stomach acid. Mg(OH)₂ is a weak base, it neutralises HCl with a lower |ΔHn|, producing less heat in the stomach. NaOH is a strong base, it would release the full −57 kJ/mol and could cause thermal burns to stomach lining. The lower heat release of weak-base antacids is a safety feature, not a limitation.
When a weak acid reacts with a strong base, |ΔHn| < 57 kJ/mol because energy is consumed to complete the partial ionisation of the weak acid during the reaction, the three-step explanation is: (1) weak acid only ~1–5% ionised beforehand; (2) endothermic ionisation step during reaction; (3) net heat released is reduced; weak + weak gives the least exothermic neutralisation.
Add the highlighted point to your notes before the check below.
+5 XP, Why is |ΔHn| less than 57 kJ/mol when a weak acid reacts with NaOH?
4
The practical, measuring ΔHn
3
The Practical, Measuring ΔHn Using a Calorimeter
q = mcΔT → n = cV → ΔHn = −q/n · Use total mass, not just acid or base
We just saw that weak acid neutralisations release less heat due to the ionisation energy cost. That raises a question: How do we actually measure ΔHn in the lab? This card answers it → the calorimetry method uses q = mcΔT to find heat from the temperature rise, then converts to kJ/mol using n(H₂O) formed.
The experimental measurement of enthalpy of neutralisation relies on a simple energy transfer principle, all heat released by the reaction is assumed to be absorbed by the surrounding solution, and that absorbed heat is calculated from the temperature change.
In a school calorimetry experiment, a polystyrene (foam) cup is used as an insulated vessel. The procedure is:
Safety
Run this only under teacher supervision with a risk assessment. Wear safety glasses and gloves: dilute HCl and NaOH used here are corrosive and the mixture warms as it reacts. Use dilute solutions (about 1.0 mol/L or less), add base to acid slowly while swirling, and keep the working volume small. Never seal the cup. Rinse any splashes on skin with plenty of cool water. If your class is not equipped for the practical, use the supplied secondary temperature data instead, the calculation skill (q = mcΔT → ΔHn) is identical.
1
Measure and record the volume of each solution (acid and base) in mL.
2
Record the initial temperature of both solutions (Tinitial), ideally they should be at the same temperature, or average the two if slightly different.
3
Add the base to the acid (or vice versa), swirl gently, and record the maximum temperature reached (Tfinal) with a thermometer or temperature probe.
4
Calculate: total mass m = (Vacid + Vbase) × 1.00 g/mL; ΔT = Tfinal − Tinitial; q = mcΔT (in joules).
5
Calculate moles of water formed: n(H₂O) = n(limiting reagent). Use n = c × V for each reactant; the one with fewer moles is the limiting reagent.
6
Calculate: ΔHn = −q/n (convert J → kJ by dividing by 1000). The negative sign is applied because the reaction is exothermic.
Heat released
What you calculate: q (in joules)
Formula: q = mcΔT
Key assumption: All heat absorbed by combined solution; density = 1.00 g/mL
Moles of H₂O
What you calculate: n(H₂O)
Formula: n = c × V
Key assumption: Use limiting reagent; 1:1 ratio H⁺ : H₂O at equivalence
Molar enthalpy
What you calculate: ΔHn (kJ/mol)
Formula: ΔHn = −q/n × (1/1000)
Key assumption: Negative = exothermic
A foam cup calorimeter reduces heat loss, but the cup, lid opening, and thermometer still create systematic error.
Critical: The mass used in q = mcΔT is the TOTAL mass of solution, acid plus base combined. A very common error is to use only the mass of the acid or only the mass of the base. All heat produced is absorbed by the entire combined solution, so the entire combined mass must be used.
Common Error: Students calculate ΔT as Tinitial − Tfinal (subtracting in the wrong order), getting a negative ΔT, then become confused about signs. Always: ΔT = Tfinal − Tinitial. For an exothermic neutralisation, Tfinal > Tinitial, so ΔT is positive. The negative sign in ΔHn = −q/n is then applied deliberately to show the reaction is exothermic.
To measure ΔHn: m = (V_acid + V_base) × 1.00 g/mL (total mass); ΔT = T_final − T_initial; q = mcΔT (c = 4.18 J g⁻¹ °C⁻¹); n(H₂O) = n(limiting reagent) = c × V; ΔHn = −q/n ÷ 1000 kJ mol⁻¹, always use combined volume for mass, and apply the negative sign to show exothermic.
Pause, write the highlighted equation sequence into your book.
+5 XP, 30 mL of acid and 50 mL of base are mixed. What mass should be used in q = mcΔT?
5
Sources of error and improvements
4
Sources of Error and Improvements in Neutralisation Calorimetry
Every error makes ΔHn less negative, always in the same systematic direction
We just saw the five-step calculation method for ΔHn from calorimetry data. That raises a question: Why do experimental results always come out less exothermic than the theoretical −57 kJ/mol? This card answers it → heat losses to the foam cup, lid, thermometer, and surroundings all reduce the measured ΔT in the same systematic direction.
Every school calorimetry experiment gives a ΔHn value that is slightly less exothermic than the theoretical value, understanding exactly why, and what can be done about it, is a standard HSC extended response target.
The fundamental assumption is that all heat released by the reaction is absorbed by the solution. In reality, heat is lost to several other pathways. All of these errors produce the same systematic effect: the experimental |ΔHn| is always less than the theoretical value never more.
Source of error
Effect on ΔT
Effect on |ΔHn|
Improvement
Heat loss through calorimeter walls and open top
Smaller ΔT (Tmax not fully reached)
Underestimated (less negative)
Add a lid; nest two foam cups; use Dewar flask
Heat absorbed by thermometer and cup
Slightly smaller ΔT
Slightly underestimated
Use a thin temperature probe with low heat capacity
Density/SHC assumption differs from pure water
Small error in m or c
Small systematic error
Measure actual density; use tabulated SHC values
Incomplete mixing
Smaller ΔT (not all reacted)
Underestimated
Swirl thoroughly; use magnetic stirrer
Thermometer precision (±0.1°C or ±0.5°C)
Random error in ΔT
Random error in ΔHn
Use a calibrated digital probe (±0.01°C); repeat and average
Advanced improvement, cooling curve extrapolation: Use a temperature probe and data logger to record temperature continuously before and after mixing. Plot temperature vs time; extrapolate the cooling portion of the curve back to the moment of mixing to find the true Tmax that would have been reached without heat loss. This is more accurate than simply reading the highest temperature from the thermometer.
Reading only the highest measured temperature underestimates the true temperature rise. Extrapolation improves the value of ΔT and therefore ΔH.
HSC Response Structure: In an extended response on sources of error, always link: specific source of error → effect on ΔT → effect on q → effect on ΔHn. For example: "Heat is lost through the walls of the foam cup to the surroundings → Tmax recorded is lower than true Tmax → ΔT is smaller → q is smaller → ΔHn is less negative than the true value." This three-step chain earns full marks; a vague statement earns partial marks.
Common Error: "Human error" is not an acceptable source of error in an HSC response, it is too vague and cannot be improved systematically. Name the specific source (e.g. "heat loss through the walls of the foam cup") and the specific improvement ("add a lid and/or use a better-insulated vessel such as a Dewar flask"). Generic answers earn zero marks.
All systematic errors in neutralisation calorimetry make |ΔHn| smaller (less negative), never larger, primarily because heat lost through the foam cup walls reduces ΔT; name the specific source and improvement (e.g. "heat loss through cup walls → add a lid and use cooling-curve extrapolation to find T_max"); "human error" is never acceptable in HSC responses.
Add the highlighted point to your notes before the check below.
+5 XP, A student measures ΔHn = −53.2 kJ/mol for HCl + NaOH (theoretical: −57 kJ/mol). What is the most likely explanation?
Deep dive, Extension. Identifying sources of error and proposing improvements is core. Cooling-curve extrapolation to correct for heat loss, and high-resolution temperature-probe treatment, go beyond the Module 6 core.
6
Worked examples and apply
🌍 Measuring Antacid Neutralising Capacity
Antacid Calorimetry: Why Gaviscon, Mylanta, and Rennie Feel Different
Gaviscon (sodium alginate + sodium bicarbonate), Mylanta (aluminium hydroxide + magnesium hydroxide), and Rennie (calcium carbonate + magnesium carbonate) all neutralise excess HCl in the stomach, but they feel different because they release heat at different rates and with different magnitudes. This is direct evidence of ΔHn differences between acid-base combinations.
Rennie contains CaCO₃, an acid + carbonate reaction (Pattern 2 from L02). This is not a simple H⁺ + OH⁻ → H₂O reaction; it involves CaCO₃ dissolving and CO₃²⁻ accepting protons from HCl, which produces CO₂ gas (the fizzing). The enthalpy is distributed across multiple bond-breaking and bond-forming steps, the net |ΔHn| is lower than for a strong + strong combination. The same experiment you perform in the lab, measuring ΔT with a foam cup, can be used to compare the neutralising capacity and enthalpy of each antacid product, giving quantitative data that a pharmacist or pharmaceutical chemist would use to compare products.
!
⚠️ Common Misconceptions, Module 6 Lesson 3
✗
"ΔHn ≈ −57 kJ/mol for all neutralisation reactions." This value applies only to strong acid + strong base. Any combination involving a weak species gives |ΔHn| < 57 kJ/mol because energy is consumed in the ionisation step during the reaction.
✗
"Use only the mass of the acid in q = mcΔT." The total mass of solution (acid + base combined) must be used. All heat produced is absorbed by the entire combined solution, not just the acid component.
✗
"ΔT = Tinitial − Tfinal." Always ΔT = Tfinal − Tinitial. For an exothermic reaction, Tfinal > Tinitial, so ΔT is positive. The negative sign in ΔHn = −q/n is applied separately and deliberately.
✗
"The experimental ΔHn can be more negative than the theoretical value." All systematic errors in this experiment (heat loss, thermometer heat capacity, etc.) make ΔHn less negative than theoretical, never more negative. If your experimental value is more negative than −57 kJ/mol, check your calculation for errors.
✗
"Human error caused the discrepancy." This is not an acceptable scientific answer. Name the specific source: "heat loss through the walls of the foam cup calorimeter to the surroundings" → specific improvement: "add a lid and use a better-insulated vessel."
✏️ Worked Examples
Worked Example 1, Calculating ΔHn from experimental data
50.0 mL of 1.00 mol/L HCl is mixed with 50.0 mL of 1.00 mol/L NaOH in a foam cup calorimeter. Initial temperature: 21.5°C. Maximum temperature: 28.0°C. Calculate (a) q, (b) n(H₂O), and (c) ΔHn in kJ/mol.
Comparison: theoretical = −57 kJ/mol. Experimental value is less negative, consistent with heat loss to surroundings from the foam cup calorimeter during the experiment.
Worked Example 2, Comparing ΔHn values and explaining the difference
The experiment from Example 1 is repeated but 50.0 mL of 1.00 mol/L acetic acid (CH₃COOH) replaces HCl. Tmax = 26.8°C; Ti = 21.5°C. (a) Calculate ΔHn. (b) Explain why this value differs from the HCl + NaOH result.
a
ΔT = 26.8 − 21.5 = 5.3°C. m = 100.0 g. q = 100.0 × 4.18 × 5.3 = 2215.4 J. n(H₂O) = 0.0500 mol.
(1) In 1.00 mol/L CH₃COOH solution, only approximately 0.4% of acetic acid molecules are ionised before the reaction (Ka = 1.8 × 10⁻⁵). The other ~99.6% exist as intact CH₃COOH molecules.
(2) When NaOH is added, OH⁻ reacts with the available H⁺, driving the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ further to the right. This further ionisation is endothermic, energy must be absorbed from the surroundings to break the O–H bond in CH₃COOH.
(3) This energy comes directly from the heat released by H⁺ + OH⁻ → H₂O, reducing the net heat available to warm the solution. The result: a smaller ΔT, smaller q, and a less negative ΔHn (−44.3 vs −54.3 kJ/mol).
Answers: (a) ΔHn = −44.3 kJ/mol (b) CH₃COOH is partially ionised before reaction; energy consumed to complete ionisation during the reaction reduces net heat released → less negative ΔHn than HCl + NaOH.
Worked Example 3, Unequal volumes: limiting reagent, error analysis (high-mark)
(7 marks) 30.0 mL of 2.00 mol/L HNO₃ is mixed with 50.0 mL of 1.00 mol/L NaOH. Ti = 20.0°C; Tmax = 26.4°C. (a) Identify the limiting reagent. (b) Calculate q. (c) Calculate ΔHn in kJ/mol. (d) Theoretical value is −57 kJ/mol. Identify one specific source of error and one improvement.
Source of error: Heat loss through the walls and open top of the foam cup calorimeter to the surrounding atmosphere. As the solution warms, heat transfers from the warm solution to the cooler surroundings before the thermometer reading reaches the true maximum. This means the recorded Tmax is lower than the true Tmax → smaller ΔT → smaller q → less negative ΔHn.
Improvement: Add a tight-fitting lid to the foam cup to reduce heat loss through the top surface. A more significant improvement: use a temperature probe with a data logger to record temperature every second and extrapolate the cooling curve back to the moment of mixing, recovering the true Tmax without depending on the observed peak temperature.
Answers: (a) NaOH is limiting (0.0500 mol OH⁻); n(H₂O) = 0.0500 mol (b) q = 2140 J (c) ΔHn = −42.8 kJ/mol (d) Heat loss through foam cup walls → lower Tmax → smaller ΔT → less negative ΔHn. Improvement: add lid + use data logger with cooling curve extrapolation.
Apply, Predict Then Reveal
🔬Predict, Then Reveal+8 XP
A student mixes 50 mL of 1.0 mol/L HCl with 50 mL of 1.0 mol/L NaOH at 20°C and measures a temperature rise of 6.3°C. Predict whether the experimental ΔH_neut will be higher or lower than the theoretical −57.3 kJ/mol, and explain why.
Your predictionExpert answerCompare
The experimental value will be lower in magnitude (less negative) than −57.3 kJ/mol. Heat losses to the surroundings, the calorimeter, and the thermometer mean not all the heat released by the reaction is captured in the temperature rise. Additionally, the specific heat capacity assumption (using 4.18 J/g·°C for the solution rather than measuring it) introduces systematic error. Both effects cause the calculated ΔH_neut to be less exothermic than the true thermodynamic value.
Good comparison. Keep the reasoning that matched the expert answer.
Re-read the expert answer and identify the missing reasoning step.
Complete the Learn phase to unlock Practice.
✓
Practice questions
Activities
ACTIVITY, CALCULATE + INTERPRET
Neutralisation Calorimetry Data Analysis
A student performs three neutralisation experiments using identical equipment. All solutions are at 20.0°C initially. Results are shown below. For each experiment: calculate q, n(H₂O), and ΔHn (kJ/mol). Then answer the interpretation questions.
Experiment
Acid (50.0 mL, 1.00 mol/L)
Base (50.0 mL, 1.00 mol/L)
Tmax (°C)
q (J)
n(H₂O) (mol)
ΔHn (kJ/mol)
1
HCl (strong)
NaOH (strong)
26.6
Calculate
Calculate
Calculate
2
CH₃COOH (weak)
NaOH (strong)
25.8
Calculate
Calculate
Calculate
3
HCl (strong)
NH₃ (weak)
25.1
Calculate
Calculate
Calculate
Rank the three experiments from most to least exothermic. Does this match your prediction from the theory?
All three experimental ΔHn values are less negative than −57 kJ/mol. For Experiment 1 (strong + strong), identify the most likely reason.
Explain in three steps why Experiment 2 gives a less negative ΔHn than Experiment 1.
01
Multiple Choice
+5 XP
A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct
✍️ Short Answer
SA
Short Answer
UnderstandBand 3
6. Explain why the molar enthalpy of neutralisation is approximately −57 kJ/mol for any strong acid + strong base combination, regardless of which specific acid and base are used. Write the net ionic equation to support your explanation. (3 marks)
SA
Short Answer
ApplyBand 4
7. 25.0 mL of 2.00 mol/L HCl is mixed with 25.0 mL of 2.00 mol/L NaOH. Ti = 19.5°C; Tmax = 32.7°C. Calculate: (a) q in joules, (b) n(H₂O) formed, (c) ΔHn in kJ/mol. Comment on how your value compares to the theoretical −57 kJ/mol and suggest one reason for the difference. (5 marks)
SA
Short Answer
EvaluateBand 5
8. Real-World Application, Antacid Comparison: A pharmacist tests three antacids against 50.0 mL of 1.00 mol/L HCl (Ti = 20.0°C): • Product A: NaOH (strong base), Tmax = 26.8°C • Product B: Mg(OH)₂ (weak base), Tmax = 24.1°C • Product C: NaHCO₃ (acid + carbonate), Tmax = 23.3°C with vigorous bubbling
(a) Explain why Product A releases the most heat. Write the net ionic equation for this reaction. (2 marks) (b) Explain why Product B releases less heat than Product A, despite both being bases that neutralise HCl. (2 marks) (c) What is the gas produced by Product C, and why does this reaction produce less heat? (2 marks)(6 marks)
Exp 2 less than Exp 1 (3 steps): (1) CH₃COOH is only ~0.4% ionised at 1.00 mol/L before the reaction, most exists as intact molecules. (2) When NaOH is added, the OH⁻ drives the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ further right; this endothermic ionisation step absorbs energy from the solution. (3) The energy consumed in ionisation reduces the net heat available to raise the temperature → smaller ΔT → smaller q → less negative ΔHn.
Short Answer Model Answers
Q6 (3 marks): Strong acids and strong bases are both fully ionised in solution before the reaction occurs, all H⁺ and OH⁻ ions are already present as free ions [1]. Regardless of which strong acid or base is used, the only species that react are H⁺ and OH⁻: H⁺(aq) + OH⁻(aq) → H₂O(l) [1]. The spectator ions (e.g. Na⁺, Cl⁻, K⁺, NO₃⁻) do not participate, the same bond is formed and the same energy is released each time, giving ΔHn ≈ −57 kJ/mol for any strong + strong combination [1].
Q7 (5 marks): (a) m = 50.0 g; ΔT = 32.7 − 19.5 = 13.2°C; q = 50.0 × 4.18 × 13.2 = 2758.8 J [1]. (b) n(HCl) = 2.00 × 0.0250 = 0.0500 mol; n(NaOH) = 0.0500 mol; equal, so n(H₂O) = 0.0500 mol [1]. (c) ΔHn = −2758.8/0.0500 = −55 176 J/mol ÷ 1000 = −55.2 kJ/mol [1]. The value (−55.2 kJ/mol) is less negative than the theoretical −57 kJ/mol [1]. This is consistent with heat loss from the foam cup calorimeter to the surroundings, the measured Tmax is lower than the true maximum, making q and therefore |ΔHn| smaller [1].
Q8 (6 marks): (a) NaOH is a strong base, fully dissociated in solution. All OH⁻ ions are immediately available to react with H⁺ from HCl [1]. Net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l), no ionisation energy is consumed; the full ΔHn ≈ −57 kJ/mol is released [1]. (b) Mg(OH)₂ is a sparingly soluble weak base, it does not fully dissociate in solution before reacting [1]. Energy must be consumed to dissolve and further ionise Mg(OH)₂ during the reaction; this energy comes from the heat released by H⁺ + OH⁻ → H₂O, reducing the net heat available and giving a less negative ΔHn [1]. (c) The gas is CO₂ [1]. NaHCO₃ + HCl → NaCl + H₂O + CO₂ (acid + hydrogen carbonate reaction). CO₂ is produced when the intermediate H₂CO₃ immediately decomposes. This reaction involves multiple bond-breaking and bond-forming steps beyond simple H⁺ + OH⁻ → H₂O, consuming additional energy, the net heat released per mole of acid neutralised is lower than for a simple strong base neutralisation [1].
C
Calculation practice
Written calculation questions in the style and mark range of the HSC paper. Set your working out in full, because method marks are awarded for the steps as well as the final value.
ApplyBand 4(4 marks) In a calorimetry experiment, 50.0 mL of 1.00 mol/L hydrochloric acid is mixed with 50.0 mL of 1.00 mol/L sodium hydroxide solution. The temperature of the mixture rises from 21.2 °C to 27.9 °C. Assuming the density of the mixture is 1.00 g/mL and its specific heat capacity is 4.18 J/g/°C, calculate the enthalpy of neutralisation in kJ/mol.
ApplyBand 4(3 marks) A student mixes 100.0 mL of 0.500 mol/L nitric acid with 100.0 mL of 0.500 mol/L potassium hydroxide and records a temperature rise of 3.4 °C. Calculate the heat released, then explain why the value obtained is likely to be smaller in magnitude than the accepted enthalpy of neutralisation.
✓
Retrieve and reflect
Check what actually stuck
Review, Flashcard Drill
Key Concept Flashcards
Flashcard 1: State the calculation sequence for finding ΔHn from experimental calorimetry data.
1) m = (Vacid + Vbase) × 1.00 g/mL (total mass). 2) ΔT = Tfinal − Tinitial. 3) q = mcΔT (joules); c = 4.18 J g⁻¹ °C⁻¹. 4) n(H₂O) = n(limiting reagent) = c × V. 5) ΔHn = −q/n × (1/1000) in kJ/mol.
Flashcard 2: Why is ΔHn ≈ −57 kJ/mol for all strong + strong combinations?
Both strong acid and strong base are fully ionised before the reaction, all H⁺ and OH⁻ are free ions. The net ionic equation is always identical: H⁺(aq) + OH⁻(aq) → H₂O(l). The same bond forms each time, releasing the same energy. Spectator ions (Na⁺, Cl⁻, etc.) don't participate → ΔHn is constant at ≈ −57 kJ/mol regardless of which specific acid/base is chosen.
Flashcard 3: Explain in three steps why weak acid + strong base gives |ΔHn| < 57 kJ/mol.
(1) Weak acid is only partially ionised (~1–5%) before the reaction, most molecules remain intact. (2) As the reaction proceeds, the OH⁻ drives the weak acid equilibrium further right, completing this ionisation is endothermic (absorbs energy). (3) This endothermic ionisation energy is taken from the heat of the H⁺ + OH⁻ → H₂O reaction, reducing the net heat released → |ΔHn| < 57 kJ/mol.
Flashcard 4: List two specific sources of error in foam cup calorimetry and their improvements.
1) Heat loss through foam cup walls and open top → lower Tmax → smaller ΔT → less negative ΔHn. Improvement: add a tight-fitting lid; nest two foam cups; use Dewar flask. 2) Heat absorbed by thermometer → slightly smaller ΔT. Improvement: use a thin digital probe with low heat capacity. Advanced: data logger + cooling curve extrapolation to recover true Tmax.
Flashcard 5: A student reports ΔHn = −58.3 kJ/mol for HCl + NaOH. Is this value possible? What does it suggest?
This value is more negative than the theoretical −57 kJ/mol. All systematic errors make |ΔHn| smaller (less negative), never larger. A value of −58.3 kJ/mol is suspicious, it suggests a calculation error. Most likely causes: ΔT was calculated as Ti − Tf with wrong sign, giving an artificially large magnitude; incorrect mass (using half the volume); or a unit conversion error (e.g. not dividing by 1000 to convert J → kJ consistently). The student should check every step of the calculation.
Look Back at Your Initial Answers
Go back to your Think First predictions at the top of this lesson. Recall Thomsen's 1883 finding: 57.3 kJ mol⁻¹ for strong acid + strong base, every time. Weak bases like Mg(OH)₂ release less because ionisation consumes some of the energy first.
Q1: You predicted why Product A (NaOH) releases the most heat. Was your answer in terms of full ionisation, the net ionic equation, or something else? Can you now write the three-step mechanism that explains why NaOH releases more heat than Mg(OH)₂?
Q2: You predicted what the bubbling from Product B (NaHCO₃) is. Can you now write the balanced equation and explain why CO₂ is produced?
