OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 23

8 marks · Medium difficulty · Structured Questions

State the features of dynamic equilibrium, calculate the S=O bond enthalpy in sulfur trioxide, and determine the equilibrium constant Kc.

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Question

Question 23 presents an equilibrium reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g) with ΔH = -196 kJ mol⁻¹. Part (a) asks to state two features of a dynamic equilibrium for 2 marks. Part (b) displays structural diagrams for SO2 (bent with two S=O bonds) and SO3 (trigonal planar with three S=O bonds), a table of bond enthalpies with O=O as +494 kJ mol⁻¹ and S=O in SO2 as +522 kJ mol⁻¹, and asks to calculate the S=O bond enthalpy in SO3 for 3 marks. Part (c) provides a table of equilibrium concentrations: [SO2] = 25.0 mol dm⁻³, [O2] = 108 mol dm⁻³, [SO3] = 42.0 mol dm⁻³, and asks for the Kc expression and calculated value to 3 significant figures in standard form for 3 marks.
Question text

23 This question is about equilibrium and bond enthalpies.

The chemical industry manufactures sulfur trioxide, SO3, by reacting sulfur dioxide, SO2, and

oxygen. This is a reversible reaction which can reach dynamic equilibrium.

2SO (g) + O (g) 2SO (g) ΔH = –196 kJ mol–1 Equilibrium 23.1

22 3

(a) State two features of a dynamic equilibrium.

1 …

2 …

[2]

(b) The displayed formulae of SO2 and SO3 are shown below.

O

S

O O S

O O

The bond enthalpies for the S=O bond in SO2 and SO3 are not the same.

The table shows bond enthalpies.

Bond Bond enthalpy / kJ mol–1

O=O +494

S=O in SO2 +522

Calculate the S=O bond enthalpy in SO3.

Use the information in the table and Equilibrium 23.1.

(c) The table shows equilibrium concentrations for Equilibrium 23.1.

[SO (g)] / mol dm–3 [O (g)] / mol dm–3 [SO (g)] / mol dm–3

22 3

25.0 108 42.0

For Equilibrium 23.1

• write the expression for the equilibrium constant, Kc

• calculate Kc to 3 significant figures and in standard form.

Bond enthalpy of S=O in SO = … kJ mol–1 [3]

K = … dm3 mol–1

c

[3]

Mark scheme

Show the mark scheme Mark scheme for question 23. (a) Awards 2 marks for any two of: rate of forward reaction equals rate of reverse reaction; concentrations of reactants and products remain constant; in a closed system. (b) Awards 3 marks: 1 mark for energy for bonds broken = (4 × 522) + 494 = 2582 kJ; 1 mark for total energy of 6 S=O bonds in SO3 = 2582 + 196 = 2778 kJ; 1 mark for dividing by 6 to give +463 kJ mol⁻¹. (c) Awards 3 marks: 1 mark for expression Kc = [SO3]² / ([SO2]² [O2]); 1 mark for substituting values (42.0)² / (25.0² × 108); 1 mark for evaluating to 2.61 × 10⁻² in standard form to 3 SF.

Question Answer Mark Guidance

23 (a) Two ( ) from: 2 IGNORE reactions take place together/reversible reaction

• rate of forward reaction = rate of reverse reaction ALLOW backward for reverse

• Concentrations (of reactants and products) do not DO NOT ALLOW concentration of reactants

change/are constant = concentration of products

• In a closed system/environment ALLOW ‘nothing can leave/enter’

Question Answer Answer Mark Guidance

23 (b) FIRST, CHECK THE ANSWER LINE, IF bond 3 FULL ANNOTATIONS MUST BE USED

enthalpy = (+)463 (kJ mol–1) award 3 marks -----------------------------------------------------------------

------------------------------------------------------------------------------

Energy for bonds broken (4 × S=O in SO2) + 1 ×

O=O)

(4 × 522) + (1 × 494) OR 2088 + 494 IGNORE sign

OR 2582 (kJ) (rearranging assessed below)

S=O bond enthalpy in 6 x SO3 correctly calculated

ECF for MP2: Ans to part 1 + 196

6 × S=O bond enthalpy in SO3 = 2582 + 196

= 2778 (kJ mol–1)

S=O bond enthalpy in SO3 correctly calculated

2778 ECF for MP3: Ans to part 2/6; IGNORE significant

S=O bond enthalpy in SO3 = 6

figures and rounding errors.

= (+)463 kJ

mol–1 COMMON ERRORS

Mark is for

answer Use of 2 S=O (reactants) and 3 S=O (product)

+578 → 1 mark

16 Wrong sign for 196

+398 → 2 marks (ans: 397.6...)

Use of 2 S=O (reactants) and dividing by 2

+867 → 1 mark

Addition of S=O (product) bonds, not subtraction

-463 → 2 marks

23 (c) Kc expression 3

[SO (g)]2

3 Square brackets required for K expression

(Kc =) 2 c

[SO2(g)] [O2(g)] IGNORE state symbols, even if wrong

DO NOT ALLOW

for first two marks

BUT ALLOW ECF to 3SF and standard form

i.e. 2.41 x 100

Calculation ALLOW 422 and 252 in substituted equation

42.02

(Kc =) ALLOW ECF for Calculation from inverted Kc

25.02 108 OR 0.02613 …

expression ONLY i.e. 3.83 × 101 for two marks (to 3

SF and standard form)

DO NOT ALLOW 0.0261 x 100

= 2.61 10–2 to 3 SF and standard form

ALLOW ECF from any incorrect Kc value for 3 SF and

standard form

IGNORE attempts at units

How to answer it

Equilibrium, Bond Enthalpy & Kc Calculations

Equilibrium 23.1: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)   ΔH = -196 kJ mol⁻¹

What this question tests

  • Dynamic Equilibrium: Defining the microscopic and macroscopic conditions required for a closed equilibrium system.
  • Mean Bond Enthalpy Calculations: Using stoichiometry, structural formulae, and Hess's law principles (ΔH = Σ(bonds broken) - Σ(bonds formed)) to solve for an unknown bond enthalpy.
  • Equilibrium Constant (Kc): Writing homogeneous equilibrium expressions correctly and formatting calculated values in standard index form to a stated degree of precision (3 significant figures).
Part (a) • 2 Marks

Features of a Dynamic Equilibrium

Core definitions and properties of reversible equilibrium systems

✅ Correct Answers (Any two)

  • Rate condition: The rate of the forward reaction equals the rate of the reverse (backward) reaction.
  • Macroscopic condition: The concentrations of reactants and products remain constant (do not change).
  • Environmental condition: It occurs in a closed system (nothing can enter or leave).

❌ Common Errors & Examiner Warnings

  • Fatal Error: Writing that the concentrations of reactants and products are equal. Concentrations are constant, not necessarily equal.
  • Vague assertions like "the reaction is reversible" or "reactants and products exist together" do not describe the dynamic balance.
  • Omitting the word "rate" when comparing forward and reverse processes (e.g. saying "forward reaction equals reverse reaction").
Mark Breakdown: 1 mark per correct feature stated up to a maximum of [2].
Part (b) • 3 Marks

Calculating Unknown Bond Enthalpy

Determining the S=O bond enthalpy in SO₃

📐 Step-by-Step Calculation

  1. Count and calculate bonds broken (reactants):
    • In 2 moles of SO₂: 2 × (2 × S=O) = 4 × (S=O) = 4 × 522 = 2088 kJ
    • In 1 mole of O₂: 1 × (O=O) = 494 kJ
    Total bonds broken = 2088 + 494 = +2582 kJ [1 mark]
  2. Set up the reaction enthalpy expression:
    ΔH = Σ(bonds broken) - Σ(bonds formed)
    Each SO₃ has 3 S=O bonds. For 2 moles of SO₃, there are 6 × (S=O) bonds.
    -196 = +2582 - 6(S=O in SO₃)
  3. Rearrange to find total energy of bonds formed:
    6(S=O in SO₃) = 2582 + 196 = 2778 kJ [1 mark]
  4. Solve for a single S=O bond in SO₃:
    Bond enthalpy = 2778 / 6 = +463 kJ mol⁻¹ [1 mark]

🧠 Exam Technique: Count the Bonds First

Always inspect the balanced equation together with displayed formulae:

  • 1 molecule of SO₂ has 2 S=O bonds → 2SO₂ has 4 bonds.
  • 1 molecule of SO₃ has 3 S=O bonds → 2SO₃ has 6 bonds.
  • Bond enthalpies are endothermic values, so quote your final answer as positive: +463 kJ mol⁻¹.

❌ Diagnostic Common Traps

  • Ignoring reaction balancing: Using only 2 S=O bonds broken and 3 S=O bonds formed yields an incorrect value of +578 kJ mol⁻¹ (scores only 1 mark).
  • Sign inversion on ΔH: Writing 2582 - 196 = 2386 instead of adding 196 leads to +398 kJ mol⁻¹ (scores 2 marks).
  • Negative final answer: Quoting -463 kJ mol⁻¹. Bond enthalpy definitions refer to breaking bonds and are strictly positive.
Mark Scheme Guidance: Full answer of +463 kJ mol⁻¹ on the answer line scores [3] marks immediately. Error-carried-forward (ECF) applies if arithmetic slips occur during rearrangement.
Part (c) • 3 Marks

Equilibrium Constant (Kc) Expression & Calculation

Constructing expressions and quoting standard form to 3 significant figures

💡 1. Writing the Kc Expression

For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g):

Kc = [SO₃(g)]² / ( [SO₂(g)]² [O₂(g)] )
  • Must use square brackets [ ] to represent equilibrium concentration. Round brackets ( ) are unacceptable.
  • State symbols are ignored even if omitted or incorrect, but reactants in denominator must be multiplied, never added.

📐 2. Substituting & Evaluating

Given equilibrium concentrations (in mol dm⁻³):
[SO₂] = 25.0   [O₂] = 108   [SO₃] = 42.0

  1. Substitute values into expression:
    Kc = (42.0)² / ( (25.0)² × 108 ) [1 mark]
  2. Calculate intermediate value:
    Kc = 1764 / (625 × 108) = 1764 / 67500 = 0.0261333...
  3. Convert to Standard Form & 3 Sig Figs:
    0.0261333... → 2.61 × 10⁻² [1 mark]

🧠 Exam Technique: Precision Checklist

  • Check your indices: Notice 0.0261 is 10⁻², not 10⁻³.
  • Significant figures: The question specifically asked for 3 significant figures (2.61). Writing 0.026 or 2.613 loses the final mark.
  • Standard form requirement: Leaving the answer as 0.0261 loses the format mark. Do not write 0.0261 × 10⁰.

❌ Common Examiner Traps

  • Adding denominators: Writing [SO₂]² + [O₂] in the formula is a severe algebra error and forfeits the first 2 marks completely.
  • Inverting expression: Putting reactants over products (yields 3.83 × 10¹). ECF can salvage marks for calculation and formatting, but loses the expression mark.
Units note: Units were already pre-printed on the paper ( dm³ mol⁻¹ ), so candidates did not lose marks for omitted unit derivations.

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 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.