Edexcel A-Level Chemistry Paper 1, June 2017: Question 7

9 marks · Medium difficulty · Calculations

Calculate the temperature change when hydrochloric acid reacts with aqueous ammonia, and comment on the relative enthalpy changes of neutralisation for various acids.

Practise this question

Question

Part (a) gives an enthalpy change equation and asks to calculate the expected temperature change when mixing hydrochloric acid and aqueous ammonia, given specific volumes, concentrations, density, and specific heat capacity values. Part (b) presents a table of enthalpy changes of reaction for 1 mol of different acids (hydrochloric, nitric, sulfuric, and ethanoic) neutralised by sodium hydroxide solution, and asks to comment on the relative values using data and relevant equations.
Question text

7 In acid-base neutralisation reactions, there is a temperature change.

(a) The enthalpy change when hydrochloric acid reacts with aqueous ammonia is –53.4 kJ mol–1.

HCl(aq) + NH3(aq) r NH4Cl(aq)

Calculate the temperature change you would expect when

25.0 cm3 of 1.00 mol dm–3 hydrochloric acid is mixed with

25.0 cm3 of 1.00 mol dm–3 aqueous ammonia.

Give your answer to an appropriate number of significant figures.

Assume: the density of the solution is 1.00 g cm–3

[ the specific heat capacity of the solution is 4.18 J g–1 qC–1]

(3)

*(b) The table shows the enthalpy changes of reaction when 1 mol of different acids

are neutralised by sodium hydroxide solution, at 298 K.

Enthalpy change of reaction

Acid –1

for 1 mol of acid / kJ mol

hydrochloric acid, HCl –58

nitric acid, HNO3 –58

sulfuric acid, H2SO4 –115

ethanoic acid, CH3COOH –56

Comment on the relative enthalpy changes of reaction, using the data from the

table and including any relevant equations.

(6)

… 16

… *P48058A01628*

(Total for Question 7 = 9 marks)

Mark scheme

Show the mark scheme The mark scheme provides step-by-step guidance for calculating moles, energy, and temperature change for part (a), including significant figures. For part (b), it outlines indicative marking points relating to strong/weak acids, diprotic nature, ionisation, and neutralisation equations, along with criteria for structure and lines of reasoning.

Question

Answer Additional Guidance Mark

Number

7(a) Example of calculation (3)

calculation of moles used (1) moles used = 25.0 x 1.00/1000 = 0.0250

calculation of energy for that number of moles (1) energy released = 0.025 x 53.4 = 1.335 (kJ)

/ 1335 (J)

TE on moles used

Ignore sign

calculation of temperature change temperature change = 1335/(50.0 x 4.18)

and = 6.3876

gives answer to 2 SF (because a school = 6.4 (oC / K)

thermometer cannot measure to 3 SF) (1) TE on moles and energy

Allow final answer to 3 SF 6.39 (oC / K)

Ignore units

Correct answer with no working scores (3)

Question

Answer Additional Guidance Mark

Number

7(b)* This question assesses a student’s ability to show a Guidance on how the mark scheme should

coherent and logically structured answer with linkages and be applied:

fully-sustained reasoning. The mark for indicative content should be

added to the mark for lines of reasoning.

Marks are awarded for indicative content and for how the For example, an answer with five indicative

answer is structured and shows lines of reasoning. marking points that is partially structured

with some linkages and lines of reasoning

The following table shows how the marks should be scores 4 marks (3 marks for indicative

awarded for indicative content. content and 1 mark for partial structure

Number of Number of and some linkages and lines of reasoning).

indicative marks If there are no linkages between points, the

marking awarded for same five indicative marking points would

points seen indicative yield an overall score of 3 marks (3 marks

in answer marking for indicative content and no marks for

points linkages).

5–4 3

3–2 2

(6)

The following table shows how the marks should be

awarded for structure and lines of reasoning.

Number of marks In general it would be expected that 5 or 6

awarded for indicative points would get 2 reasoning

structure of marks. 3 and 4 indicative points would get

answer and 1 mark for reasoning and 0, 1 or 2

sustained line of indicative points would score zero marks for

reasoning reasoning.

Answer shows a coherent and 2

logical structure with linkages

and fully sustained lines of

reasoning demonstrated

throughout.

Answer is partially structured 1

with some linkages and lines of

reasoning.

Answer has no linkages between 0

points and is unstructured.

Comment: Look for the indicative marking points first,

then consider the mark for the structure of answer and

sustained line of reasoning.

Indicative content Allow correct formulae for names

throughout the answer

Hydrochloric acid and nitric acid

(same value for) hydrochloric acid and nitric acid as Ignore sulfuric acid as strong(est) acid

they are strong / completely dissociated into ions (in

solution)

reaction taking place is H+ + OH― → H O / Allow HCl + NaOH → NaCl + H O and

H O+ + OH― → 2H O HNO + NaOH → NaNO + H O

32 3 3 2

Allow hydrochloric acid and nitric acid are

both monoprotic / monobasic / provide 1

mol H+ / produce 1 mol H O

Sulfuric acid Allow

sulfuric acid is diprotic / dibasic H2SO4 + 2NaOH → Na2SO4 + 2H2O

or

(1 mol of) sulfuric acid provides 2 mol H+ /

produces 2 mol H2O

so value is (almost) twice that of hydrochloric acid /

nitric acid or reverse argument

Ethanoic acid

ethanoic acid is weak /partially dissociated into ions (in Allow ethanoic acid is the weakest acid

solution) / CH COOH ⇌ CH COO― + H+ /

CH COOH + H O ⇌ CH COO― + H O+

32 3 3

some energy is needed to break (O-H) bond(s) to Allow some energy is needed to ionise

release H+ ions (so enthalpy change of neutralisation is

ethanoic acid

less than for a strong acid)

or

enthalpy change of neutralisation includes the enthalpy

of dissociation of ethanoic acid so it is less exothermic

(Total for Question 7 = 9 marks)

How to answer it

Enthalpy Changes of Neutralisation

📋 What this question tests

This question assesses your understanding of thermochemistry and calorimetry calculations (q = mcΔT) combined with a deep conceptual analysis of acid-base neutralisation. You will be tested on handling moles in solution, applying specific heat capacities, explaining the enthalpy of neutralisation for strong vs. weak acids, and accounting for monoprotic versus diprotic behaviour.

Part (a): Neutralisation Calorimetry Calculation

Calculate the expected temperature change (3 marks)

📐 Step-by-Step Calculation

  • Step 1: Calculate moles of acid used
    Moles = (25.0 / 1000) × 1.00 = 0.0250 mol
  • Step 2: Calculate energy released (q)
    Energy = Moles × Enthalpy change = 0.0250 × 53.4 = 1.335 kJ = 1335 J
  • Step 3: Calculate mass of solution (m)
    Total volume = 25.0 cm³ + 25.0 cm³ = 50.0 cm³. Given density = 1.00 g cm⁻³, mass = 50.0 g .
  • Step 4: Rearrange q = mcΔT to find temperature change
    ΔT = q / (m × c) = 1335 / (50.0 × 4.18) = 6.3876 °C

✅ Final Answer & Significant Figures

6.4 °C (or K)

Mark Breakdown:
• Mark 1: Calculation of moles.
• Mark 2: Calculation of energy released.
• Mark 3: Correct temperature change given to 2 significant figures with unit.

🧠 Exam Technique

Always pay close attention to significant figure instructions. Examiners award the final mark for using 2 SF because standard school laboratory thermometers can only reliably measure to the nearest 0.5 or 0.1 °C, making 3 SF unjustified.

❌ Common Errors

  • Using the wrong total mass (forgetting to add the acid and ammonia volumes together to get 50 g).
  • Forgetting to convert kJ into J when equating q to mcΔT .
  • Rounding to 3 SF and losing the final accuracy/formatting mark.

Part (b): Explaining Relative Enthalpy Changes

Comment on relative enthalpy changes using data and equations (6 marks)

💡 Key Knowledge Required

  • Strong acids (HCl, HNO₃): Fully dissociate into ions in aqueous solution.
  • The ionic equation for neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l)
  • Diprotic acids (H₂SO₄): Release 2 moles of H⁺ ions per mole of acid, producing double the energy.
  • Weak acids (CH₃COOH): Partially dissociate; energy is absorbed to break covalent bonds (or dissociate) prior to neutralisation.

✅ Indicative Content & Marking Points (6 Points Total)

  • Point 1: HCl and HNO₃ have the same value (-58 kJ mol⁻¹) because they are strong acids and fully dissociate.
  • Point 2: The reaction taking place is the same: H⁺ + OH⁻ → H₂O .
  • Point 3: H₂SO₄ is diprotic (provides 2 moles of H⁺ per mole of acid).
  • Point 4: Consequently, sulfuric acid's enthalpy change is approximately twice that of HCl/HNO₃ (-115 kJ mol⁻¹).
  • Point 5: Ethanoic acid is a weak acid and only partially dissociates ( CH₃COOH ⇌ CH₃COO⁻ + H⁺ ).
  • Point 6: Energy is required to ionise the weak acid, meaning the overall enthalpy of neutralisation is less exothermic.
Marking Structure: Marks are awarded based on how many indicative points you make (up to 4 marks), combined with a separate 2-mark assessment on the clarity and logical structure of your reasoning.

🧠 Examiner Commentary & Top-Level Responses

To secure top marks, students must explicitly link data values from the table to chemical properties. For instance, stating that sulfuric acid is dibasic/diprotic and thus liberates double the moles of hydrogen ions directly justifies the -115 value compared to -58.

❌ Common Errors

Students often lose marks by writing generic descriptions without including explicit equations or failing to mention that ethanoic acid requires an endothermic bond-breaking/dissociation step which reduces the overall exothermic output.

Topics

Physical Chemistry · Topic 8: Energetics I · Topic 12: Acid-base Equilibria

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.