OCR A-Level Chemistry AS Depth in chemistry (02), June 2023: Question 4

16 marks · Hard difficulty · Structured Questions

Calculate the enthalpy of combustion of butan-1-ol, determine experimental uncertainties, evaluate combustion data, and calculate average bond enthalpy using bond data.

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Question

Question text

4 This question is about the enthalpy change of combustion of alcohols.

(a) Explain the term enthalpy change of combustion.

… [2]

(b) A student carries out an experiment to determine the enthalpy change of combustion, ΔcH,

of butan-1-ol, CH3CH2CH2CH2OH.

The student sets up the apparatus as shown below.

thermometer beaker

clamp

200g water

wick spirit burner

butan-1-ol

The student’s results are shown in the table below.

Initial temperature of water / °C 18.5

Final temperature of water / °C 49.5

Mass of burner before heating / g 212.38

Mass of burner after heating / g 211.07

(i) The thermometer had an uncertainty of ± 0.25 °C in each temperature reading.

Calculate the percentage uncertainty in the temperature change.

percentage uncertainty = … % [1]

(ii)* Use the student’s results to determine ∆ H of butan-1-ol in kJ mol–1.

Explain why this value of ΔcH is different from the data book value and suggest how

the experimental design could be modified to improve the accuracy of the ΔcH value

obtained. [6]

Additional answer space if required.

(iii) Another student carries out the experiment in 4(b) using 150 g of water in the beaker

instead of 200 g.

Calculate the mass of butan-1-ol that would produce the same temperature rise as in

the experiment in 4(b).

Assume the same heat losses.

mass of butan-1-ol = … g [1]

using average bond enthalpies.

H H

H C C O H(g) + 3 O=O(g) 2 O=C=O(g) + 3H O H(g) Δ H = –1276 kJ mol–1

H H

(i) Use this value of ΔcH and the average bond enthalpies below to calculate the average

bond enthalpy of C=O.

Average bond enthalpy /

Bond –1

kJ mol

C–H +413

C–C +347

C–O +358

O–H +464

O=O +498

(ii) Methoxymethane, CH3OCH3, is an isomer of ethanol.

On combustion, methoxymethane, in the gaseous state, produces carbon dioxide and

steam.

H H

H C O C H(g) + 3 O=O(g) 2 O=C=O(g) + 3H O H(g)

H H

ΔcH for methoxymethane is more negative than ΔcH for ethanol.

Explain why the ΔcH values are different, in terms of the bonds broken and the bonds–1

C=O bond enthalpy = … kJ mol [4]

formed.

… [2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

4 (a) (The enthalpy change) for complete combustion ✓ 2 AO1.1 DO NOT ALLOW energy required

× 2

IGNORE energy released

ALLOW combustion in excess oxygen/air

OR reacts in excess oxygen

OR reacts completely in oxygen

of 1 mol (of substance)✓ ALLOW element OR compound OR reactant

DO NOT ALLOW atoms

IGNORE standard states/conditions

(b) (i) 2 0.25 1 AO2.8 ALLOW 1 d.p. up to calculator value of 1.612903226

31 100 = 1.6 % ✓ correctly rounded

Question Answer Marks AO Guidance

element

4 (b)* (ii) Please refer to the marking instructions on page 4 of this mark 6 AO2.4 Indicative scientific points may include:

scheme for guidance on how to mark this question. × 2

Calculation of q and n

Level 3 (5–6 marks) AO3.1 • mass butan-1-ol burnt = 1.31 g

Calculates an acceptable value for ∆cH × 2 • T = 31 ºC

AND

• q = mc T = 200 × 4.18 × 31

Evaluates at least two differences from data book value AND AO3.3

suggests at least two suitable improvements. = 259(16) J OR 25.9(16) (kJ)

AO3.4 1.31

• n(butan-1-ol) = 74 = 0.0177(017…) (mol)

There is a well-developed line of reasoning which is clear and

logically structured. The information presented is relevant and

substantiated. Calculation of ∆cH

• H = 0.0177017..

Level 2 (3–4 marks)

Use of results to calculate moles of butan-1-ol AND an • H = (–)1460 to (–)1464 (kJ mol-1)

attempt to calculate q

AND 20 ALLOW answer in J mol-1 if units are given

Evaluate at least one difference from data book value AND

suggests at least one suitable improvement. ALLOW a single slip/rounding errors e.g. Mr = 73

OR IF no calculation seen check 4(b)(i) /page 10 for

Calculates an acceptable value for ∆cH any working.

There is a line of reasoning presented with some structure. Difference from data book value

The information presented is relevant and supported by some • Heat losses

evidence. • Incomplete combustion

• Data book uses standard values

Level 1 (1–2 marks) • Evaporation of alcohol from wick

Attempts to calculate moles of butan-1-ol OR attempts to • Evaporation of water from beaker

calculate q

AND Examples of improvements

Evaluate at least one difference from data book value OR • Burn in plentiful oxygen

suggests at least one suitable improvement. • Draft shield

21 • Copper can in place of beaker

OR • Use a bomb calorimeter

Correct use of results to calculate moles of butan-1-ol AND an

• Add lid to beaker

attempt to calculate q

• Place cap over wick when not burning

OR • Reduce gap between burner and beaker

Evaluate at least one difference from data book value AND • Use standard conditions

suggests at least one suitable improvement. • Use 3 DP balance OR digital thermometer

• Heat for longer to reduce % uncertainty in

mass/temperature measurements

The information is basic and communicated in an unstructured

way. The information is supported by limited evidence and the

relationship to the evidence may not be clear.

0 marks

No response or no response worthy of credit.

4 (b) (iii) Mass burnt = 0.9825 (g) ✓ 1 AO3.4 ALLOW any value from 0.98 to 0.99 g

ALLOW ECF from b(ii)

i.e. 0.75 × mass change calculated

q = 150 x 4.18 x 31 = 19437 J = 19.4(37) kJ

19.4(37)

Mass burnt = x 74

calculated H

element

4 (c) (i) FIRST, CHECK ANSWER ON ANSWER LINE 4 AO2.2 FULL ANNOTATIONS MUST BE USED

IF bond enthalpy = (+)805 (kJ mol–1) award 4 marks 4 -----------------------------------------------------------------

IF bond enthalpy = (+)3220 (kJ mol–1) award 3 marks ALLOW ECF throughout

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

Energy for bonds made: (6O–H) IGNORE signs for first 3 marking points

6 464

OR 2784 (kJ) ✓ 22 Alternative method

ALLOW (5O–H) from bonds made if O-H is omitted

Energy for bonds broken: from bonds broken.

(1 C–C + 5C–H + 1C–O + 1O–H + 3O=O) Made: 5 x 464 OR 2320 (kJ) ✓

347 + (5 413) + 358 + 464 + (3 498) Broken: 347 + (5 413) + 358 +(3 498)

OR 347 + 2065 + 358 + 464 + 1494 OR 4264 (kJ) ✓

OR 4728 (kJ) ✓ 4C=O bond enthalpy = 4264 – 2320 + 1276

= (+)3220 (kJ mol–1) ✓

4C=O bond enthalpy correctly calculated

–1276 = 4728 – (4C=O + 2784)

4C=O bond enthalpy = 4728 – 2784 + 1276

OR (+)3220 (kJ mol–1) ✓

C=O bond enthalpy correctly calculated

3220 –1

C=O bond enthalpy = 4 = (+)805 kJ mol ✓ DO NOT ALLOW – sign

COMMON ERRORS for 3 marks

-805 (Wrong Sign)

(+)718(.25) (Missing C-C)

(+)431.5 (Missing 3 x O=O)

(+)167 (Incorrect rearrangement)

methoxymethane ✓ × 2 ALLOW bonds in methoxymethane are weaker (than

in ethanol)

IGNORE calculations

Energy released in bond forming is same ✓ ALLOW Same bonds being formed

How to answer it

Enthalpy Changes of Combustion & Bond Enthalpies

What this question tests

This multi-step physical chemistry question assesses your understanding of enthalpy changes of combustion, calorimetric experimental design, percentage uncertainty calculations, multi-stage calculation of enthalpy changes from experimental data, and applying mean bond enthalpies to determine unknown bond energies. It also tests structural isomer comparisons.

Part (a)

Defining Enthalpy of Combustion

✅ Correct Answer

The enthalpy change when 1 mole of a substance
is completely burned in excess oxygen (or complete combustion).

❌ Common Errors

Using "energy required" instead of "enthalpy change".
Failing to state "1 mole" of the substance.
Mentioning "atoms" instead of "substance", "compound", or "element".

Marks: 2 marks (1 for complete combustion/excess oxygen, 1 for 1 mole of substance).

Part (b)(i)

Percentage Uncertainty in Temperature Change

📐 Calculation Steps

Find temperature change (delta T): 49.5 - 18.5 = 31.0 °C
Account for two readings: Initial and final readings both carry uncertainty, so total error = 2 × 0.25 = 0.5 °C
Calculate percentage uncertainty: (0.5 / 31.0) × 100 = 1.6129...%

Answer: 1.6% (or 1.61%)

🧠 Exam Technique

Always double the instrument uncertainty when a measurement involves a change (like initial and final burette or thermometer readings), because two separate readings were taken.

Marks: 1 mark.

Part (b)(ii)

Enthalpy of Combustion Calculation & Evaluation (Level of Response)

📐 Step-by-Step Calculation of Delta C H

Mass of alcohol burnt: 212.38 - 211.07 = 1.31 g
Moles of butan-1-ol (M_r = 74.0): 1.31 / 74 = 0.0177 mol
Energy transferred (q = mc delta T): 200 × 4.18 × 31 = 25916 J = 25.916 kJ
Enthalpy change (Delta H = -q / n): -25.916 / 0.0177 = -1464 kJ mol⁻¹ (Accept values between -1460 and -1464)

💡 Evaluation & Improvements

Why different from data book? Major heat losses to the surroundings, incomplete combustion of the alcohol, or evaporation of the alcohol wick/water.
How to improve: Add a draft shield, use a copper calorimeter instead of glass, add a lid, or use a digital balance reading to 3 decimal places.

Marks: 6 marks (Level of Response based on calculation accuracy, evaluation of differences, and suggested improvements).

Part (b)(iii)

Adjusting Mass for Changed Water Volume

📐 Calculation

New energy needed for 150 g water: q = 150 × 4.18 × 31 = 19437 J (19.437 kJ)
Moles required: 19.437 / 1464 = 0.01327 mol
Mass of butan-1-ol: 0.01327 × 74 = 0.9825 g (Allow 0.98 to 0.99 g )

🧠 Exam Technique

This question allows ECF (Error Carried Forward) from your calculated value in part (b)(ii). If your previous enthalpy value was wrong, scaling the energy proportionally will still gain full credit here.

Marks: 1 mark.

Part (c)(i)

Calculating C=O Bond Enthalpy

📐 Bond Enthalpy Cycle / Calculation

Energy for bonds made (O-H): 6 × 464 = 2784 kJ
Energy for bonds broken (C-C, C-H, C-O, O-H, O=O):
347 + (5 × 413) + 358 + 464 + (3 × 498) = 4728 kJ
Set up equation using Delta H_c (-1276 kJ mol⁻¹):
Delta H = (Bonds broken) - (Bonds made) - 4(C=O)
Rearrange for C=O:
4(C=O) = 4728 - 2784 - (-1276) = 3220 kJ
Divide by 4: 3220 / 4 = +805 kJ mol⁻¹

❌ Common Errors

Forgetting to include a positive sign (+805) or incorrectly inserting negative signs for bond enthalpies.
Miscounting the number of C-H bonds broken in ethanol (counting 6 instead of 5 due to the O-H bond).

Marks: 4 marks.

Part (c)(ii)

Comparing Isomer Combustion Enthalpies

✅ Correct Answer

Less energy is required to break bonds in methoxymethane (since bonds in methoxymethane are weaker than in ethanol).
The energy released in bond forming is the same (since the same products, CO₂ and H₂O, are formed).
Therefore, net enthalpy change is more exothermic (more negative).

🧠 Exam Technique

Always frame enthalpy comparisons around the energetic ledger:
Delta H = (Energy to break bonds) - (Energy released forming bonds). If fewer/weaker bonds require less energy input while output remains constant, the overall reaction becomes more exothermic.

Marks: 2 marks.

Topics

Module 3: Periodic table and energy · Module 4: Core organic chemistry · Practical Activity Groups · 3.2 Physical chemistry · 4.1 Basic concepts and hydrocarbons · PAG 3: Enthalpy determination

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.