OCR A-Level Chemistry AS Depth in chemistry (02), June 2025: Question 1
8 marks · Medium difficulty · Structured Questions
Define an alkali, write an ionic neutralisation equation, calculate the enthalpy change of neutralisation from temperature change data, and explain the effect of using a more concentrated solution on the temperature rise.
Practise this questionQuestion
Question text
1 A student is investigating the enthalpy change of neutralisation for a strong acid and a strong
alkali.
(a) Explain what is meant by an alkali.
… [1]
(b) Write the ionic equation for the neutralisation reaction between an acid and an alkali.
… [1]
(c) The student mixes 40.0 cm3 of 1.40 mol dm–3 HCl and 40.0 cm3 of 1.40 mol dm–3 NaOH.
The temperature rises by 9.40 °C
(i) Calculate the enthalpy change of neutralisation, in kJ mol–1.
Give your answer to 3 significant figures.
Assume that the density of all solutions and the specific heat capacity, c, of the reaction mixture
is the same as for water.
(ii) The student repeats the experiment using the same volume and concentration of NaOH but
using 20.0 cm3 of 2.80 mol dm–3 HCl instead of 40.0 cm3 of 1.40 mol dm–3 HCl .
The temperature rise is greater than 9.4 °C.
Explain why.
… [2]
Mark scheme
Show the mark scheme
Question Answer Mark Guidance
1 (a) releases OH–/hydroxide (ions in aqueous solution) 1 ALLOW containing/forms OH– (ions)
ALLOW ionises to give OH– (ions)
IGNORE mention of pH
1 (b) H+ + OH– → H O 1 IGNORE state symbols
1 (c) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 4 FULL ANNOTATIONS MUST BE USED
IF ΔH = –56.1 (kJ mol–1) award 4 marks -----------------------------------------------------
IF ΔH = (+)56.1 (kJ mol–1 ) award 3 marks ALLOW ECF throughout
Energy released in J OR kJ ALLOW 3140 (J) OR 3.14 (kJ)
Q = mc T = 80 × 4.18 × 9.4 DO NOT ALLOW < 3 SF
IGNORE any sign
= 3143.36 (J) OR 3.14336 (kJ)
IGNORE units i.e. ALLOW correctly calculated number in J OR
kJ OR no units
Calculates nH2O/nHCl/nNaOH
40 × 1.4 IGNORE trailing zeroes e.g ALLOW 0.056
= 0.056(0) (mol)
1000
H per mole H2O
3.14336 -1 Sign NOT needed
= ± 56.131428….(kJ mol )
0.056
OR
Common Errors
3143.36 -1
= ± 56,131.428…. (J mol ) -28.1 3 marks Use of 40g instead of 80g or 2 x 0.056 mol
0.056
-1690 3 marks +273 for temp change
ΔH in kJ mol–1 to 3 SF AND – sign
Final answer must have 3 SF AND – sign
H = –56.1 (kJ mol–1)
Question Answer 9 Mark Guidance
1 (c) (ii) 2 IGNORE responses related to changes in rate of reaction
Same number of moles (of HCl / H2O) ALLOW calculation which shows number of moles is
same i.e. 0.02 x 2.8 = 0.056 (ALLOW ECF from (i) for this
calculation)
Same q/ energy/heat (released)
AND
ALLOW H or Enthalpy change for ‘heat’
(spread over a) smaller (total) volume / mass
ALLOW calculation to give T = 12.5 C (using mass = 60
and same q value) (ALLOW ECF from (i) for this calculation)
How to answer it
Enthalpy of Neutralisation: Definitions, Calorimetry & Energy Conservation
This question assesses your foundational understanding of acid-base definitions, ionic equations, experimental thermochemistry calculations ( q = mcΔT and ΔH = -q / n ), correct sign conventions, significant figures, and the physical reasoning behind temperature changes when reaction volumes vary.
Part (a) Definition of an Alkali
Explain what is meant by an alkali. [1 Mark]
✅ Mark Scheme Answer
Releases OH⁻ / hydroxide ions (in aqueous solution / into water).
💡 Key Knowledge
- A base is any species that accepts protons (H⁺).
- An alkali is a specific type of base: a soluble base that dissolves in water to release OH⁻(aq) ions.
❌ Common Errors
- Stating "has a pH above 7" or "turns universal indicator purple" (describes properties, not what an alkali is).
- Omitting the word ions if naming hydroxide.
- Confusing an alkali with a general base (e.g. saying just "proton acceptor").
Part (b) Ionic Equation for Neutralisation
Write the ionic equation for the neutralisation reaction between an acid and an alkali. [1 Mark]
✅ Mark Scheme Answer
H⁺ + OH⁻ → H₂O
🧠 Exam Technique
- For any strong acid reacting with any strong alkali, spectator ions (e.g. Na⁺ and Cl⁻) cancel out completely.
- Always check atom balance (2 H and 1 O on each side) and charge balance (+1 - 1 = 0).
❌ Common Errors
- Writing the full molecular equation (e.g. HCl + NaOH → NaCl + H₂O ) instead of the ionic equation.
- Incorrect charges on ions, such as writing OH without a minus or H₂⁺ .
Part (c)(i) Calculating Enthalpy Change of Neutralisation
Calculate the enthalpy change of neutralisation, in kJ mol⁻¹. Give your answer to 3 significant figures. [4 Marks]
📐 Step-by-Step Calculation
- Total mass of reaction mixture ( m ):
Both solutions are mixed together:
m = 40.0 cm³ + 40.0 cm³ = 80.0 g (assuming density = 1.00 g cm⁻³) - Heat transferred ( q ):
q = m × c × ΔT
q = 80.0 × 4.18 × 9.40 = 3143.36 J ( 3.14336 kJ ) - Moles of water formed ( n ):
n(HCl) = (40.0 / 1000) × 1.40 = 0.0560 mol
n(NaOH) = (40.0 / 1000) × 1.40 = 0.0560 mol
Ratio is 1:1, so n(H₂O) = 0.0560 mol . - Enthalpy change per mole ( ΔH ):
ΔH = -q / n = -3.14336 kJ / 0.0560 mol = -56.131... kJ mol⁻¹ - Round to 3 SF with sign:
-56.1 kJ mol⁻¹
✅ Mark Breakdown (4 Marks Total)
- Mark 1: Correct energy released, q = 3143.36 J or 3.14 kJ .
- Mark 2: Correct moles of reactant/water, n = 0.0560 mol .
- Mark 3: Enthalpy value calculated: ±56.1... (evaluating q / n ).
- Mark 4: Correct final value to 3 SF AND negative sign: -56.1 kJ mol⁻¹ .
❌ Calculation Traps & Examiner Warnings
- Using m = 40.0 g instead of 80.0 g: Leads to -28.1 kJ mol⁻¹ (loses 1 mark). Always sum the volumes of all liquids in the cup.
- Adding 273 to the temperature change: A temperature rise of 9.40 °C is already a difference of 9.40 K! Do not convert ΔT.
- Forgetting the negative sign: Neutralisation is exothermic (temperature increased). A positive sign (+56.1) drops the 4th mark.
- Incorrect rounding: Writing -56 or -56.13 loses the final precision mark.
Part (c)(ii) Explaining the Higher Temperature Rise
The experiment is repeated using 20.0 cm³ of 2.80 mol dm⁻³ HCl and the same NaOH. Explain why the temperature rise is greater than 9.40 °C. [2 Marks]
✅ Mark Scheme Model Answer
- The same number of moles of HCl / H₂O react (therefore the same amount of heat energy is released). [1 mark]
- The heat energy is released into a smaller total volume / smaller mass of solution. [1 mark]
📐 Mathematical Comparison
| Feature | Experiment 1 | Experiment 2 |
|---|---|---|
| Moles of HCl | 0.040 × 1.40 = 0.056 mol | 0.020 × 2.80 = 0.056 mol |
| Total Volume ( m ) | 40 + 40 = 80.0 cm³ | 20 + 40 = 60.0 cm³ |
| Heat Released ( q ) | 3143 J | 3143 J (identical) |
| Predicted ΔT | q / (80 × 4.18) = 9.40 °C | q / (60 × 4.18) = 12.5 °C |
Since ΔT = q / (m × c) , reducing mass m with constant q directly increases ΔT .
❌ Major Misconception
Do NOT talk about rates of reaction!
Many students state: "The acid is more concentrated, so particles collide more frequently, making the reaction faster and hotter."
The mark scheme explicitly instructs: IGNORE responses related to changes in rate of reaction. Enthalpy and temperature changes are thermodynamic equilibrium quantities, completely independent of kinetics.
Topics
Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Practical Activity Groups · 1.1 Practical skills assessed in a written examination · 2.1 Atoms and reactions · 3.2 Physical chemistry · PAG 3: Enthalpy determination
Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.