OCR A-Level Chemistry Unified chemistry (03), June 2024: Question 2

14 marks · Hard difficulty · Structured Questions

Calculate bond enthalpy, energy released in gas reactions, determine the formula of a chloride from titration data, and deduce molecular formula and structural features of an organic antiviral molecule.

Practise this question

Question

Exam question in three main parts. Part (a) provides a reversible reaction for the manufacture of hydrogen gas from methane and steam with enthalpy change, plus a table of average bond enthalpies (H-H +436, O-H +464, C triple bond O +1077 kJ mol^-1), followed by three sub-questions on bond enthalpies and energy calculations. Part (b) describes compound A as a chloride of a Period 3 element, giving titration details with silver nitrate solution, followed by writing an ionic equation and determining its formula. Part (c) displays a skeletal structure of an antiviral medicine 'compound B', followed by finding its molecular formula, counting chiral carbon centers, and calculating the difference in relative molecular masses when replacing nitrogen with phosphorus and oxygen with sulfur.
Question text

2 This question is about different areas of chemistry.

(a) Hydrogen gas is manufactured by the chemical industry from the reversible reaction of methane

and steam, shown below.

CH (g) + H O(g) 3H (g) + CO(g) ΔH = +195 kJ mol–1

42 2

Average bond enthalpies are shown in the table.

Bond H–H O–H C≡O

Average bond enthalpy

–1 +436 +464 +1077

/ kJ mol

(i) Why do all average bond enthalpies have a positive value?

… [1]

(ii) Determine the C–H bond enthalpy, in kJ mol–1, using the information above.

C–H bond enthalpy = … kJ mol–1 [3]

(iii) Hydrogen gas is being considered as a household fuel to replace methane.

The enthalpy change of formation, Δ H, for H O(l) is –285.8 kJ mol–1.

f 2

Determine the energy released when 60.0 m3 of hydrogen is used as a household fuel at RTP.

Give your answer to 3 significant figures and in standard form.

energy released = … kJ [2]

(b) Compound A is a chloride of a Period 3 element.

A student carries out the 2 steps below to find the formula of compound A.

Step 1 The student adds 5.00 × 10–4 mol of compound A to water.

A colourless solution is formed.

Step 2 The colourless solution reacts with exactly 60.0 cm3 of 2.50 × 10–2 mol dm–3 AgNO (aq)

to form a white precipitate.

(i) Write an ionic equation, with state symbols, for the reaction in Step 2.

… [1]

(ii) Determine the formula of compound A.

formula of A = … [3]

(c) Compound B, shown below, is an antiviral medicine.

HO OH

O

O

N O

HO

N N O

H

compound B

(i) What is the molecular formula of compound B?

… [1]

(ii) How many chiral carbon atoms are there in one molecule of compound B?

… [1]

(iii) A research chemist synthesises two related compounds, compound C and compound D, from

compound B.

• In compound C, the N atoms in compound B had been replaced by P atoms.

• In compound D, the O atoms in compound B had been replaced by S atoms.

What is the difference between the relative molecular masses of compound C and compound D?

difference = … [2]

Mark scheme

Show the mark scheme Detailed mark scheme showing acceptable answers, marking points, and common errors with specific deductions or ECF guidelines for each sub-question from 2(a)(i) to 2(c)(iii).

Question Answer Marks Guidance

2 (a) (i) Bonds are breaking 1 IGNORE ‘more energy needed to break bonds than

AND released in making bonds’

endothermic OR energy is required/needed ✓

Unclear whether response refers to bond breaking or

IGNORE ‘overcome’ for ‘break’ overall enthalpy change

(a) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE 3 COMMON ERRORS ECF for other numbers

IF bond enthalpy = (+)413 (kJ mol–1) award 3 marks

315.5 OR 316 → 2 marks Wrong sign for 195

---------------------------------------------------------------------- Bonds made = 2385 ✓

Energy for bonds made ( 3 H–H + 1 C O ) 1 mark –195 + 2385 – 928 = 1262

3 436 + 1 1077 = 315.5 OR 316 ✓ ECF

OR 1308 + 1077 877 → 2 marks Wrong sign for 928

OR 2385 (kJ) ✓ Bonds made = 2385 ✓

IGNORE sign 195 + 2385 + 928 = 3508

= 877 ✓ ECF

4C–H bond enthalpy correctly calculated 1 mark 364 → 1 mark Missing ∆H, (195)

4 C–H bond enthalpy = 195 + 2385 – (2 O–H) Potentially 2 errors: missing 195 and sign for 195

= 195 + 2385 – 2 464 Bonds made = 2385 ✓

= 195 + 2385 – 928 2385 – 928 = 1457

= 1652 (kJ mol–1) ✓

= 364/ 364.3 /364.25 NO ECF

IGNORE sign

779.5 OR 780 → 1 mark Wrong sign for 928 AND 195

C–H bond enthalpy correctly calculated 1 mark

Bonds made = 2385 ✓

**This mark is NOT available from TWO previous errors

–195 + 2385 + 928 = 3118

OR from ∆H = 195 not being used **

= 779.5 NO ECF

1652

C–H bond enthalpy = 4 529 → 2 marks 1 O–H instead of 2 O–H:

= (+)413 kJ mol–1 ✓ Bonds made = 2385 ✓

For the final answer, 195 + 2385 – 464 = 2116

DO NOT ALLOW value with a negative sign = 529 ✓ ECF

--------------------------------------------------- 181 → 2 marks 4 O–H instead of 2 O–H:

COMMON ERRORS Bonds made = 2385 ✓

–413 → 2 marks Wrong sign for answer 195 + 2385 – 1856 = 724

724/4 = 181 ✓ ECF

304 → 2 marks 2 mol of H2 instead of 3 mol:

2 436 + 1 1077 = 872 + 1077 = 1949

195 + 1949 – 928 = 1216 ECF ✓

1216/4 = 304 ECF ✓

(a) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE 2

IF energy released = 7.15 105 kJ, award 2 marks

---------------------------------------------------------------------- ALLOW ECF ONLY from incorrect n(H2) based on with

incorrect unit conversion from m3

60.0 103 e.g.

n(H2) = 24.0 = 2500 (mol) ✓ 60.0 102

n(H2) = 24.0 = 250 (mol)

Energy released = 2500 285.8 = 7.15 105 kJ ✓ 250 285.8 = 7.15 104 kJ ECF ✓

3SF AND standard form required So ALLOW 1 mark for:

±7.15 10x (unit conversion)

IGNORE sign i.e. ALLOW + OR – OR no sign 7.145 105 (not 3SF)

715000 (not standard form)

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

ALLOW use of ideal gas equation with a sensible

temperature (290–298K) and pressure (100/101/101325 kPa)

e.g.

e.g. At 293K and 100 kPa,

100 103 60.0

n(H2) = = 2463… (mol)

8.314 293

→ 2463 285.8 = 7.04 105 kJ

e.g. At 298K and 100 kPa,

100 103 60.0

n(H2) = = 2421.7… (mol)

8.314 298

→ 2421.7 285.8 = 6.92 105 kJ

ALLOW use of 8.31 for R (same answers)

293K → 2464.24 285.8 = 7.04 105 kJ

298K → 2422.89 285.8 = 6.92 105 kJ

(b) (i) Ag+(aq) + Cl–(aq) → AgCl(s) ✓ 1 ALL 3 state symbols required

(b) (ii) n(AgNO3) 1 mark 3

= 2.50 10–2 60.0/1000 = 1.5(0) 10–3 (mol) ✓

Essential mark

Formula 2 marks Check equation from 2b(i) at top of response

Ratio --------------------------------------------------------------------------

5.00 10–4 mol A contains 1.5(0) 10–3 mol Cl ALLOW 1:3 or 3:1 ratio seen anywhere, e.g. XCl3

OR

ratio A : Cl = 1.5(0) 10–3 ÷ 5.00 10–4 = 1 : 3 ✓ ALLOW ECF from formula of silver chloride in 2b(i)

Formula e.g. From AgCl2

= AlCl ✓ n(Cl) = 2 1.5(0) 10–3 = 3.(00) 10–3 (mol)

Automatically subsumes 1:3 ratio mark ratio = 1 : 6

ALLOW Al2Cl6 ALLOW PCl3 Formula = SCl6

(c) (i) C13H19N3O7 ✓ 1 ALLOW elements in formula in any order

e.g. C13H19O7N3

(c) (ii) 4 ✓ 1

(c) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE 2 ALLOW other approaches based on different atoms in

IF difference = 61.7, award 2 marks C and D,

---------------------------------------------------------------------- e.g. Difference = 7 (32.1 – 16) – 3 (31 – 14)

= 112.7 – 51 = 61.7 ✓

Mr of C = 380 OR Mr of D = 441.7 ✓

Check answer from 2c(i) at top of response for ECF

Correct difference = 441.7 – 380 = 61.7 ✓ ALLOW ECF from incorrect formula from 2c(i)

AWARD mark for correct answer of 61.7 only e.g. From C12H16N3O6

Mr of C = 349 OR Mr of D = 394.6 ✓ ECF

difference = 394.6 – 349 = 45.6 ✓ ECF

How to answer it

Enthalpy Changes, Amount of Substance & Organic Structures

What this question tests

This synoptic question tests your mastery across physical, inorganic, and organic chemistry. Key topics include: applying average bond enthalpies and enthalpy changes of formation, performing molar calculations with gas volumes and stoichiometry, writing ionic equations with correct state symbols, deducing chemical formulas using titration mole ratios, determining molecular formulas from complex organic structures, and identifying chiral centres.

Question 2(a) — Enthalpy and Bond Calculations

Part (i): Bond Enthalpy Definitions

✅ Correct Answer

Bonds are breaking AND endothermic (or energy is required/needed).

❌ Common Errors

Writing vague statements like "more energy needed to break bonds than released in making bonds". Examiners penalise this because it refers to overall reaction enthalpy rather than explaining why bond enthalpy values are strictly positive.

Part (ii): Determining C–H Bond Enthalpy

📐 Step-by-Step Calculation

  1. Recall the relationship: ΔH = Σ(Bonds broken) − Σ(Bonds made)
  2. Identify bonds broken (reactants): 4 × (C–H) + 2 × (O–H) [Note: 2 moles of H₂O contains 4 O–H bonds!]
  3. Identify bonds made (products): 3 × (H–H) + 1 × (C≡O)
  4. Substitute values: +195 = [4(C–H) + 2(464)] − [3(436) + 1(1077)]
  5. Simplify numerical terms: +195 = [4(C–H) + 928] − [1308 + 1077]
  6. +195 = 4(C–H) + 928 − 2385
  7. +195 = 4(C–H) − 1457
  8. Rearrange for C–H: 4(C–H) = 195 + 1457 = 1652
  9. C–H = 1652 / 4 = +413 kJ mol⁻¹

❌ Common Traps

  • Using 1 × O–H instead of 2 × O–H for H₂O.
  • Incorrect sign handling for ΔH (+195 vs -195).
  • Forgetting to multiply the entire water molecule bonds correctly.

🧠 Exam Technique

Always write out the full algebraic expression before plugging in numbers. If you make an arithmetic error later, clear working allows examiners to award ECF (Error Carried Forward) marks.

Part (iii): Energy Released from Hydrogen Fuel

📐 Step-by-Step Calculation

  1. Find moles of H₂: n(H₂) = Volume / Molar volume at RTP = 60.0 m³ × 10³ / 24.0 = 2500 mol. (Watch out for the volume unit conversion from m³ to dm³!).
  2. Calculate energy released: ΔH = 2500 mol × 285.8 kJ mol⁻¹ = 714,500 kJ = 7.145 × 10⁵ kJ.
  3. Apply formatting rules: Round to 3 significant figures and standard form: 7.15 × 10⁵ kJ .
Marks: [2] — Awarded for correct calculation and correct 3SF standard form.

Question 2(b) — Inorganic Analysis & Stoichiometry

Part (i): Ionic Equation

✅ Correct Answer

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

💡 Key Knowledge

Precipitation reactions involving silver nitrate require correct ionic species and complete state symbols (aq) and (s) to secure the mark.

Part (ii): Determining Formula of Compound A

📐 Step-by-Step Calculation

  1. Moles of AgNO₃ added: n(AgNO₃) = (60.0 / 1000) × (2.50 × 10⁻²) = 1.50 × 10⁻³ mol.
  2. Relate to chloride ions: Since 1 mol of Ag⁺ reacts with 1 mol of Cl⁻, moles of Cl⁻ in compound A = 1.50 × 10⁻³ mol.
  3. Find stoichiometric ratio: Ratio of compound A to Cl⁻ = (5.00 × 10⁻⁴ mol) : (1.50 × 10⁻³ mol) = 1 : 3.
  4. Deduce formula: Since A is a chloride of a Period 3 element with 3 chlorine atoms, the formula is AlCl₃ (or dimeric Al₂Cl₆ ).
Marks: [3] — Moles of reagent calculated, ratio determined, final correct formula stated.

Question 2(c) — Organic Structure Analysis

✅ Part (i): Molecular Formula

Counting carbons, hydrogens, nitrogens, and oxygens from the skeletal structure yields: C₁₃H₁₉N₃O₇ (Order of elements does not matter).

✅ Part (ii): Chiral Centres

Number of chiral carbon atoms (carbons bonded to 4 different groups) = 4 .

Part (iii): Difference in Relative Molecular Masses

📐 Step-by-Step Calculation

  1. Analyze structural substitutions:
    • Compound C replaces N atoms with P atoms.
    • Compound D replaces O atoms with S atoms.
  2. Check atom counts in Compound B: Contains 3 Nitrogen atoms and 7 Oxygen atoms.
  3. Calculate atomic mass differences:
    • Difference per N → P swap = Ar(P) - Ar(N) = 30.974 - 14.007 = 16.967 (or use integer values: 31 - 14 = 17 per atom; 3 atoms × 17 = 51).
    • Difference per O → S swap = Ar(S) - Ar(O) = 32.06 - 16.00 = 16.06 (or integer values: 32 - 16 = 16 per atom; 7 atoms × 16 = 112).
  4. Calculate overall Mᵣ difference: Mᵣ(C) - Mᵣ(D) or direct difference calculation: (7 × (32 - 16)) - (3 × (31 - 14)) = 112 - 51 = 61.7 (or using precise periodic table values leading to 61.7).

🧠 Examiner Insight

Top-level responses avoided recalculating the entire Mᵣ of complex molecules C and D from scratch, saving valuable exam time by directly calculating the net mass change resulting from atom exchanges.

Marks: [2] — Correct difference value calculated successfully.

Topics

Module 3: Periodic table and energy · Module 4: Core organic chemistry · Module 2: Foundations in chemistry · 3.1 The periodic table · 3.2 Physical chemistry · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.