OCR A-Level Chemistry AS Breadth in chemistry (01), June 2024: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the enthalpy change using the provided enthalpy cycle involving carbon, carbon monoxide, and carbon dioxide.

Practise this question

Question

An enthalpy cycle showing C(s) + 1/2 O2(g) reacting to form CO(g) with enthalpy change Delta_r H, which further reacts with 1/2 O2(g) to form CO2(g) with Delta_c H = -283 kJ mol^-1. Directly from C(s) + 1/2 O2(g) plus another 1/2 O2(g), CO2(g) is formed with Delta_c H = -393 kJ mol^-1. Four multiple-choice options for Delta_r H are given: A (+676), B (+110), C (-110), and D (-676).
Question text

11 An enthalpy cycle is shown below.

ΔrH

C(s) + ½O2(g) CO(g)

+ ½O2(g) + ½O2(g)

Δ H = –393 kJ mol–1 Δ H = –283 kJ mol–1

c c

CO2(g)

What is Δ H, in kJ mol–1, shown in the enthalpy cycle?

r

A +676

B +110

C –110

D –676

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 11 is C.

11 C 1

How to answer it

Calculating Enthalpy Change Using Hess's Law Cycles

What this question tests

This question assesses your understanding of Hess's Law and enthalpy cycles involving enthalpies of combustion ( ΔꭃH ). You must demonstrate the ability to apply conservation of energy principles to find an unknown reaction enthalpy ( ΔᵣH ) by routing correctly through an alternative pathway.

Question 11

Enthalpy Cycle Multiple Choice Question

✅ Correct Answer

C ( -110 kJ mol⁻¹)

💡 Key Knowledge

  • Hess's Law: The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
  • Combustion Arrows: In a standard combustion cycle, arrows point downwards from reactants/intermediates to combustion products (oxides).

🧠 Exam Technique

Trace routes from start to finish. Following the direction of a given arrow means you add that enthalpy change. Going against the direction of an arrow means you must reverse its sign (multiply by -1 ).

❌ Common Errors

Students frequently fall into the sign trap by forgetting to invert the value of ΔꭃH = -283 kJ mol⁻¹ when travelling backwards up the right-hand side arrow against its direction.

📐 Step-by-Step Calculation

  1. Identify the direct route: To get directly from C(s) + 1/2 O₂(g) to CO(g) , the enthalpy change is ΔᵣH .
  2. Identify the alternative indirect route: Go via the common combustion product ( CO₂(g) ).
    • First leg (with arrow): C(s) + 1/2 O₂(g) → CO₂(g) has ΔꭃH = -393 kJ mol⁻¹ .
    • Second leg (against arrow): CO(g) → CO₂(g) has ΔꭃH = -283 kJ mol⁻¹ , but since we are going backwards up this arrow, we change the sign to +283 kJ mol⁻¹ .
  3. Apply Hess's Law equation:

    ΔᵣH + (-283) = -393

    ΔᵣH = -393 - (-283) = -393 + 283 = -110 kJ mol⁻¹

Mark Allocation: [1 mark] awarded for selecting option C. This tests quick mental or rough-working execution under timed exam conditions.

Topics

Module 3: Periodic table and energy · 3.2 Physical chemistry

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