OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 7

1 mark · Medium difficulty · Multiple Choice

Calculate the enthalpy change of neutralisation when sulfuric acid reacts with potassium hydroxide, given the volume, concentration, and energy given out.

Practise this question

Question

Multiple choice question 7 presents the chemical equation for the reaction of sulfuric acid with potassium hydroxide: H2SO4(aq) + 2KOH(aq) -> K2SO4(aq) + 2H2O(l). It states that 25 cm3 of 1.00 mol dm-3 H2SO4 is reacted with excess KOH, and the energy given out is 2.8 kJ. Four options are provided: A (-56), B (-70), C (-112), and D (-224).
Question text

7 The equation for the reaction of sulfuric acid with potassium hydroxide is shown below.

H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)

25 cm3 of 1.00 mol dm–3 H SO is reacted with excess KOH.

The energy given out is 2.8 kJ.

What is the enthalpy change of neutralisation, in kJ mol–1?

A –56

B –70

C –112

D –224

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme for question 7 indicates the correct answer is option A, worth 1 mark, with an instruction to allow -56 as the correct numerical answer.

7 A 1 ALLOW -56 (correct numerical answer)

How to answer it

Calculating Enthalpy Change of Neutralisation

OCR A-Level Chemistry • Multiple Choice Question

What this question tests

This question assesses your understanding of enthalpy change of neutralisation (ΔH_neut), stoichiometry, and reacting mole ratios. Specifically, it tests whether you can scale enthalpy changes per mole of water formed versus moles of limiting reagent used, while correctly incorporating sign conventions for exothermic reactions.

Question 7 Breakdown

Enthalpy of Neutralisation Calculation

✅ Correct Answer

Option A: −56

Marks awarded: 1 / 1 for selecting A.

💡 Key Knowledge

  • Definition: Enthalpy change of neutralisation is the enthalpy change when an acid and alkali react to form one mole of water ( H₂O ) under standard conditions.
  • Sign Convention: Neutralisation is always exothermic, meaning the value must be negative ( − ).
  • Stoichiometry matters: 1 mole of H₂SO₄ produces 2 moles of H₂O .

🧠 Exam Technique

  • Do not blindly divide energy by moles of the acid without checking the balanced equation.
  • Always double-check if the question asks for enthalpy change per mole of reaction or per mole of water formed.
  • Watch out for distractor options caused by stoichiometry traps (e.g., Option C results from forgetting to account for the 2:1 water-to-acid ratio).

❌ Common Errors

  • The 1:1 Ratio Trap: Assuming 1 mole of H₂SO₄ yields 1 mole of water, leading to the incorrect calculation of 2.8 / 0.025 = −112 kJ mol⁻¹ (Option C).
  • Sign Omission: Forgetting to include the negative sign for an exothermic energy change.

📐 Step-by-Step Calculation

  1. Calculate the moles of sulfuric acid ( H₂SO₄ ) reacted:
    Moles = Concentration × Volume (in dm³) = 1.00 × (25 / 1000) = 0.0250 mol
  2. Determine the moles of water ( H₂O ) produced using the stoichiometric ratio:
    From the equation H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O , 1 mole of H₂SO₄ produces 2 moles of H₂O .
    Moles of H₂O = 0.0250 × 2 = 0.0500 mol
  3. Calculate the enthalpy change per mole of water ( ΔH ):
    ΔH = Energy released / Moles of water = 2.8 kJ / 0.0500 mol = 56 kJ mol⁻¹
  4. Apply the correct exothermic sign:
    Since energy is "given out", the enthalpy change is −56 kJ mol⁻¹ .

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 2.1 Atoms and reactions · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.