OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the bond enthalpy for the O-H bond given the reaction enthalpy and other bond enthalpies.

Practise this question

Question

Multiple choice question 11. The equation given is 2H2(g) + O2(g) -> 2H2O(g) with enthalpy change delta r H = -486 kJ mol-1. A table lists bond enthalpies for H-H (+436 kJ mol-1) and O=O (+498 kJ mol-1). Four options A (+221), B (+355), C (+464), and D (+928) are provided for the bond enthalpy of the O-H bond.
Question text

11 Hydrogen and oxygen react as shown below.

2H (g) + O (g) → 2H O(g) ∆ H = −486 kJ mol−1

22 2 r

Bond enthalpies are shown in the table.

Bond H−H O=O

Bond enthalpy

−1 +436 +498

/ kJ mol

What is the bond enthalpy, in kJ mol−1, for the O−H bond?

A +221

B +355

C +464

D +928

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is C.

11 C 1 2.6

How to answer it

Calculating Bond Enthalpies from Enthalpy Change of Reaction

📌 What this question tests

This question assesses your understanding of enthalpy changes, specifically applying the relationship between enthalpy of reaction ( ΔrH ), bonds broken, and bonds formed. You are required to rearrange a standard thermochemical formula to calculate an unknown individual bond enthalpy.

Question 11 (Multiple Choice)

Determining the O-H Bond Enthalpy

✅ Correct Answer

C (+464)

The correct bond enthalpy for the O–H bond is +464 kJ mol⁻¹ .

💡 Key Knowledge

  • The Master Equation: ΔrH = Σ(Bonds broken) − Σ(Bonds formed)
  • Bonds broken (reactants) are always endothermic (positive values).
  • Bonds formed (products) are always exothermic (negative values in reality, but represented as positive bond enthalpy terms subtracted in the formula).

🧠 Exam Technique

Always write out the balanced equation with explicit structural awareness before substituting values:

2(H–H) + (O=O) → 4(O–H)

Carefully count every single bond according to the stoichiometric coefficients in the balanced equation.

❌ Common Errors

  • Stoichiometry Traps: Forgetting that a water molecule ( H₂O ) contains two O–H bonds, meaning 4 moles of O–H bonds are formed in total for 2 moles of H₂O . Option D ( +928 ) is a common distractor representing the sum for 2 bonds instead of dividing by 4.
  • Sign inversion errors when rearranging the equation for ΔrH .

📐 Step-by-Step Calculation

  1. Identify bonds broken (Reactants): 2 moles of H–H and 1 mole of O=O.
    Energy in = (2 × 436) + (1 × 498) = 872 + 498 = +1370 kJ mol⁻¹
  2. Set up the formula: ΔrH = Σ(Bonds broken) − Σ(Bonds formed)
    -486 = 1370 − Σ(Bonds formed)
  3. Rearrange for bonds formed: Σ(Bonds formed) = 1370 − (-486) = 1370 + 486 = 1856 kJ mol⁻¹
  4. Calculate single O–H bond enthalpy: Since the equation produces 2 moles of H₂O , and each water molecule has 2 O–H bonds, there are 4 moles of O–H bonds formed in total.
    O–H bond enthalpy = 1856 / 4 = +464 kJ mol⁻¹ (matches option C).
Mark Scheme Allocation: 1 mark for selecting C. (Assessment Objective: AO1 / AO2 - Applying chemical knowledge and quantitative skills).

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), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.