OCR A-Level Chemistry Unified chemistry (03), June 2022: Question 5

10 marks · Hard difficulty · Structured Questions

Calculate the enthalpy change of solution and hydration of calcium ions from experimental calorimetry data, and determine the energy released from the combustion of an octane-ethanol fuel blend.

Practise this question

Question

Question 5 consists of two parts. Part (a) is a 6-mark extended response question asking to determine the enthalpy of solution of CaCl2 and hydration of Ca2+ ions from experimental calorimetry data, requiring an energy cycle. It provides tables with literature values of lattice enthalpy of CaCl2 (-2223 kJ/mol) and hydration of Cl- (-378 kJ/mol), as well as mass readings of bottle, cup, and solution, and temperature changes from 21.0 to 39.5 °C. Part (b)(i) asks for the balanced equation for complete combustion of a 1:1 molar mixture of octane and ethanol (1 mark). Part (b)(ii) asks to calculate the energy released in kJ by complete combustion of 8.00 kg of this 1:1 fuel mixture given individual enthalpies of combustion (3 marks).
Question text

5 This question is about energy changes.

(a)* A student plans to determine the enthalpy change of hydration of calcium ions.

The student finds the information below from data tables.

Enthalpy change ∆H / kJ mol–1

Lattice enthalpy of calcium chloride –2223

Enthalpy change of hydration of chloride ions –378

The student carries out an experiment to find the enthalpy change of solution of calcium

chloride.

Student’s method:

• Weigh a bottle containing calcium chloride and weigh a polystyrene cup.

• Add water from a measuring cylinder to the polystyrene cup and measure its

temperature.

• Add the calcium chloride, stir the mixture, and measure the maximum temperature of

the final solution.

• Weigh the empty bottle and weigh the polystyrene cup with the final solution.

Mass readings

Mass of bottle + calcium chloride / g 27.45

Mass of empty bottle / g 18.17

Mass of polystyrene cup / g 21.24

Mass of polystyrene cup + final solution / g 127.84

Temperature readings

Initial temperature of water / °C 21.0

Maximum temperature of final solution / °C 39.5

Calculate the enthalpy change of solution of calcium chloride and determine the enthalpy

change of hydration of calcium ions.

Show your working, including an energy cycle linking the energy changes.

Assume that the density and specific heat capacity, c, of the solution are the same as for

water. [6]

Additional answer space if required.

(b) Internal combustion engines have historically used fuels obtained from crude oil as a source

of power.

The environmental effects of fossil fuel use can be reduced by blending petrol with biofuels

such as ethanol.

A fuel is being developed using a 1:1 molar ratio of octane and ethanol.

(i) Write the equation for the complete combustion of this fuel.

… [1]

(ii) Calculate the energy released, in kJ, by the complete combustion of 8.00 kg of this fuel.

∆ H(C H ) = –5470 kJ mol–1; ∆ H(C H OH) = –1367 kJ mol–1.

c 8 18 c 2 5

energy released = … kJ [3]

Mark scheme

Show the mark scheme Mark scheme for Question 5. Part (a) uses level of response criteria (6 marks) based on processing experimental data with q = mcΔT (q = 8.24 kJ), finding moles of CaCl2 (0.0835 mol), determining ΔsolH (-98.7 kJ/mol), and using an energy cycle to calculate ΔhydH(Ca2+) = -1566 kJ/mol. Part (b)(i) awards 1 mark for C8H18 + C2H5OH + 15.5 O2 -> 10 CO2 + 12 H2O (or integer multiple). Part (b)(ii) awards 3 marks for determining total molar mass of 1 mol octane + 1 mol ethanol = 160 g, calculating 50 mol of each in 8.00 kg, and computing total energy = 341850 kJ (or 342000 kJ).

Question Answer Marks AO Guidance

element

5 (a)* Please refer to the marking instructions on page 4 of this 6 AO3.1 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. ×4 1. Processing experimental data

Level 3 (5–6 marks) Energy change from mcΔT

Calculates correct enthalpy change AO3.2 • Energy in J OR kJ

with correct – sign for ∆ H (Ca2+),

hy ×2 = 106.6 × 4.18 × 18.5 = 8243.378 (J)

allowing for acceptable errors. OR 8.243378 (kJ)

3SF or more

There is a well-developed line of reasoning which is clear and Amount in mol of CaCl2

logically structured. 9.28

The information presented is relevant and substantiated. • n(CaCl2) = = 0.0835 … (mol)

111.1

Level 2 (3–4 marks) 0.08352835284 unrounded

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

Calculates a value of ∆solH (CaCl2(s)) from the:

2. ± value of ∆solH(CaCl2(s))

Energy change

AND 8.24….. –1

= ± 0.0835.… = ± 98.68957929 (kJ mol )

Amount in mol of CaCl2.

8.24

3 SF or more. From 3 SF: = 98.7

There is a line of reasoning presented with some structure. 0.0835

The information presented is relevant and supported by some ----------------------------------------------

3. CORRECT ∆ H(Ca2+) calculated with

evidence. hy

signs

Level 1 (1–2 marks) ∆ H(Ca2+) = L.E. + ∆ H(CaCl ) – 2 ∆ H(Cl–)

hy sol 2 hy

Processes experimental data to obtain the: = –2223 + (–98.7) – (2 × –378)

Energy change from mc∆T = –1566 (kJ mol–1)

OR 3SF or more with correct – sign

Amount in mol of CaCl2. From unrounded values, –1565.689579

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

There is an attempt at a logical structure with a line of See next page for examples of acceptable

reasoning. The information is in the most part relevant. errors

0 marks – No response or no response worthy of

credit.

24 element

Acceptable errors

ALLOW omission of trailing zeroes

ALLOW minor slips in rounding, transcription

errors, etc throughout

ALLOW one small error,

e.g. subtracting mass of CaCl2 for m

m = 106.60 – 9.28 = 97.32

q = 7.5257556 (kJ)

∆ H = 90.09821629 (kJ mol–1)

sol

∆ H(Ca2+) = –1557 (kJ mol–1)

hy

OR adding mass of CaCl2 for m

m = 106.60 + 9.28 = 115.88

q = 8.9610004 kJ

∆ H = 107.2809423 (kJ mol–1)

sol

∆ H(Ca2+) = –1574 (kJ mol–1)

hy

(b) (i) C8H18 + C2H5OH + 15½ O2 → 10 CO2 + 12 H2O 1 AO2.6 ALLOW multiples

e.g.

2 C8H18 + 2 C2H5OH + 31 O2

→ 20 CO2 + 24 H2O

ALLOW C10H24O for C8H18 + C2H5OH

Combining ethanol and octane!

Question Answer 25 Marks AO Guidance

element

(ii) FIRST CHECK ANSWER ON THE ANSWER LINE 3 AO2.2 IGNORE sign throughout

If answer = 341850 to 2 SF or more award 3 marks ×3

-------------------------------------------------------------------------- ALLOW approach based on mass for 2nd

M(C8H18) = 114 AND M(C2H5OH) = 46 mark

OR

1 mol C8H18 + 1 mol C2H5OH has mass of 160 g m(C8H18) = (114/160) × 8000 = 5700 g

AND

50 mol C8H18 OR 50 mol C2H5OH m(C2H5OH) = (46/160) × 8000 = 2300 g

OR

50 mol (C8H18 + C2H5OH) Energy = 5700/114 × 5470 + 2300/46 × 1367

OR = 341850 (kJ)

8.00 kg fuel contains 50 mol C8H18 + 50 mol C2H5OH ALLOW 2 SF or more correctly rounded

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

Energy = (50 × 5470) + (50 × 1367) Common errors

OR 50 × (5470 + 1367) OR 50 × 6837

OR 273500 + 68350 310800 → 2 marks

Use of equal masses (4 kg) of C8H18 & C2H5OH

= 341850 (kJ) (rather than equal moles)

Example

How to answer it

Enthalpy Changes: Hydration of CaCl₂ & Fuel Combustion

What this question tests

This question assesses mastery of quantitative energetics, experimental calorimetry, Born–Haber enthalpy cycles, and stoichiometry involving molar mixtures:

  • Calorimetry calculations: Processing experimental lab readings to determine mass of solution ( m ), temperature rise ( ΔT ), heat energy exchanged ( q = mcΔT ), and molar enthalpy of solution ( ΔsolH ).
  • Enthalpy cycle construction: Linking lattice enthalpy ( ΔLEH ), hydration enthalpy ( ΔhydH ), and enthalpy of solution using Hess's Law while respecting signs and stoichiometry (Ca²⁺ vs. 2Cl⁻).
  • Equation balancing: Writing balanced combustion equations for organic mixtures containing both hydrocarbons and alcohols.
  • Multi-step fuel stoichiometry: Calculating total combustion energy from a specified molar ratio (1:1 octane to ethanol) within a bulk mass (8.00 kg).
Level of Response • 6 Marks

Part (a)* Determining the Enthalpy Change of Hydration of Ca²⁺

Calorimetry calculation and energy cycle linking ΔsolH, ΔLEH, and ΔhydH

📐 Step-by-Step Calculation

  1. Mass of CaCl₂ added:
    27.45 g - 18.17 g = 9.28 g
  2. Moles of CaCl₂:
    M(CaCl₂) = 40.1 + (2 × 35.5) = 111.1 g mol⁻¹
    n(CaCl₂) = 9.28 / 111.1 = 0.083528... mol
  3. Mass of solution (m):
    127.84 g - 21.24 g = 106.6 g
  4. Temperature change (ΔT):
    39.5 °C - 21.0 °C = +18.5 °C
  5. Heat released (q):
    q = m × c × ΔT = 106.6 × 4.18 × 18.5 = 8243.38 J = 8.2434 kJ
  6. Enthalpy change of solution (ΔsolH):
    Reaction is exothermic (temperature increased):
    ΔsolH = -q / n = -8.2434 / 0.083528 = -98.7 kJ mol⁻¹
  7. Hydration enthalpy of Ca²⁺:
    Using the energy cycle relationship:
    ΔhydH(Ca²⁺) = ΔLEH + ΔsolH - 2ΔhydH(Cl⁻)
    ΔhydH(Ca²⁺) = -2223 + (-98.7) - 2(-378)
    ΔhydH(Ca²⁺) = -2223 - 98.7 + 756 = -1566 kJ mol⁻¹
    (Using unrounded values: -1565.7 kJ mol⁻¹)

💡 Hess's Law & Enthalpy Cycle

In OCR Chemistry, Lattice Enthalpy (ΔLEH) is defined as exothermic (formation of 1 mole of solid ionic lattice from gaseous ions):

Ca²⁺(g) + 2Cl⁻(g) → CaCl₂(s)   [ΔLEH = -2223 kJ mol⁻¹]

Ca²⁺(g) + 2Cl⁻(g) / \ ΔhydH(Ca²⁺) + \ ΔLEH = -2223 2ΔhydH(Cl⁻) \ / v v CaCl₂(s) Ca²⁺(aq) + 2Cl⁻(aq) / ^ / ΔsolH = -98.7 \_______________/ Hess's Law Route: Direct: Gaseous ions → Solution Indirect: Gaseous ions → Solid lattice → Solution ΔhydH(Ca²⁺) + 2ΔhydH(Cl⁻) = ΔLEH + ΔsolH

❌ Common Student Pitfalls

  • Wrong mass in q = mcΔT: Using 9.28 g (the solute mass) or 127.84 g instead of subtracting the cup weight ( 127.84 - 21.24 = 106.6 g ).
  • Missing the stoichiometric 2: Forgetting that CaCl₂ produces 2 moles of Cl⁻ ions, so hydration of chloride must be multiplied by 2 ( 2 × -378 ).
  • Lattice enthalpy sign confusion: OCR defines lattice enthalpy as formation (negative). If breaking the lattice, it is +2223 kJ mol⁻¹ .
  • Missing negative sign on ΔsolH: Since the temperature rose from 21.0 °C to 39.5 °C, the process is exothermic ( -98.7 kJ mol⁻¹ ).
Examiner Marking Criteria (Level 3, 5–6 marks): Full marks require processing data to get q , finding n(CaCl₂) , calculating ΔsolH , showing a clear energy cycle, and reaching ΔhydH(Ca²⁺) = -1566 kJ mol⁻¹ (or -1565 to -1574 depending on accepted experimental mass variances) with a correct negative sign and logical structure.
Short Answer • 1 Mark

Part (b)(i) Combustion Equation for 1:1 Molar Mixture

Balancing complete combustion for an equimolar fuel blend

✅ Correct Balanced Equation

C₈H₁₈ + C₂H₅OH + 15½ O₂ → 10 CO₂ + 12 H₂O

Alternative accepted balanced equations:

  • Doubled integers: 2 C₈H₁₈ + 2 C₂H₅OH + 31 O₂ → 20 CO₂ + 24 H₂O
  • Combined molecular formula (C₁₀H₂₄O): C₁₀H₂₄O + 15½ O₂ → 10 CO₂ + 12 H₂O

🧠 Exam Technique: Counting Atoms

  • Carbons: 8 (from octane) + 2 (from ethanol) = 10 C → gives 10 CO₂.
  • Hydrogens: 18 (from octane) + 6 (from ethanol) = 24 H → gives 12 H₂O.
  • Oxygens on RHS: (10 × 2) + 12 = 32 O atoms.
  • Oxygens from fuel: Ethanol already contains 1 O atom!
  • O₂ needed: (32 - 1) / 2 = 15.5 O₂ (or 31/2 O₂).
Structured Calculation • 3 Marks

Part (b)(ii) Energy Released by 8.00 kg of the Fuel Blend

Stoichiometry based on 1:1 molar ratio and combustion enthalpies

📐 Step-by-Step Calculation

  1. Calculate molar masses:
    M(C₈H₁₈) = (8 × 12.0) + (18 × 1.0) = 114 g mol⁻¹
    M(C₂H₅OH) = (2 × 12.0) + (6 × 1.0) + 16.0 = 46 g mol⁻¹
  2. Find mass of 1 mole of fuel blend (1 mol C₈H₁₈ + 1 mol C₂H₅OH):
    Mass per mole of blend unit = 114 + 46 = 160 g
  3. Calculate moles of blend in 8.00 kg (8000 g):
    n(blend) = 8000 g / 160 g mol⁻¹ = 50 mol
    This means the fuel contains exactly 50 mol C₈H₁₈ and 50 mol C₂H₅OH.
  4. Calculate total energy released:
    Energy from octane: 50 × 5470 = 273,500 kJ
    Energy from ethanol: 50 × 1367 = 68,350 kJ
    Total energy = 273,500 + 68,350 = 341,850 kJ

❌ The Major Trap: Equal Mass vs. Equal Moles

The question specifies a 1:1 molar ratio, NOT a 1:1 mass ratio!

Fatal error seen by examiners: Splitting 8.00 kg into 4.00 kg octane and 4.00 kg ethanol.

  • 4000 / 114 = 35.09 mol octane → 191,929 kJ
  • 4000 / 46 = 86.96 mol ethanol → 118,870 kJ
  • Incorrect total: 310,800 kJ (Capped at 2 marks maximum).

✅ Final Answer

341,850 kJ   (or 342,000 kJ to 3 SF)

Note: The question asks for "energy released", so a positive value is expected; negative sign is ignored by examiners.

Mark Breakdown:
• Mark 1: M(C₈H₁₈) = 114 and M(C₂H₅OH) = 46 OR mass of 1 mol blend = 160 g.
• Mark 2: Finding 50 mol of each component (or equivalent mass split: 5700 g octane + 2300 g ethanol).
• Mark 3: Correct total energy of 341850 kJ (allow standard 2 SF or more rounding, e.g. 342000 kJ).

Topics

Module 5: Physical chemistry and transition elements · Module 3: Periodic table and energy · Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Practical Activity Groups · 5.2 Energy · 3.2 Physical chemistry · 1.1 Practical skills assessed in a written examination · 2.1 Atoms and reactions · PAG 3: Enthalpy determination

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.