Edexcel A-Level Chemistry AS Paper 2, June 2016: Question 4
10 marks · Medium difficulty · Calculations
Calculate the number of molecules, write the standard enthalpy of formation equation, calculate the enthalpy of combustion using mean bond enthalpies, complete a reaction profile, and explain the difference between calculated and data book values for ethanol.
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Question text
4 Ethanol, C2H5OH, is a member of the homologous series of alcohols.
(a) Calculate the number of molecules in 55.2kg of ethanol.
[Avogadro Constant = 6.02 × 1023 mol–1]
(2)
(b) Write the equation to represent the standard enthalpy change of formation of
ethanol. Include state symbols.
(2)
(c) Ethanol burns completely in excess oxygen.
C2H5OH(l) + 3O2(g) o 2CO2(g) + 3H2O(l)
(i) The table shows some mean bond enthalpy data.
Bond C C C H C O O H O O C O
Mean bond enthalpy
−1 347 413 358 464 498 805
/ kJ mol
Calculate the enthalpy change, in kJ mol−1, for the complete combustion of
1 mol of ethanol.
(3)
(ii) Complete the reaction profile diagram for the combustion of ethanol and fully
label the diagram.
(2)
enthalpy
C2H5OH(l) + 3O2(g)
*P49838A01128*
reaction pathway
(iii) A data book value for the standard enthalpy change of combustion of ethanol
is −1367.3 kJ mol−1.
Give the main reason why the value you calculated in (b)(i) is different from
this data book value.
(1)
(Total for Question 4 = 10 marks)
Mark scheme
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How to answer it
Study Guide: Ethanol Thermochemistry & Calculations
What this question tests
This question assesses core physical chemistry concepts required at AS Level: mole calculations involving mass conversions and the Avogadro constant, writing standard enthalpy changes of formation equations with correct state symbols, calculating enthalpy changes using mean bond enthalpies, interpreting reaction profile diagrams for exothermic reactions, and explaining limitations of mean bond enthalpy values compared to data book values.
Calculating Number of Molecules from Mass
📐 Step-by-Step Calculation
- Convert mass to grams: 55.2 kg = 55,200 g
- Calculate moles of ethanol (C₂H₅OH):
Molar mass = (2 × 12.0) + (6 × 1.0) + 16.0 = 46.0 g mol⁻¹
Moles = 55,200 / 46.0 = 1200 mol - Calculate number of molecules:
Molecules = 1200 × 6.02 × 10²³ = 7.224 × 10²⁶
❌ Common Errors
- Forgetting to convert kilograms to grams (using 55.2 g directly), resulting in a power of 10 error ( 7.224 × 10²⁰ ).
- Using an incorrect molar mass for ethanol.
Standard Enthalpy Change of Formation
✅ Correct Answer
2C(s, graphite) + 3H₂(g) + 0.5O₂(g) → C₂H₅OH(l)
💡 Key Knowledge
The standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
❌ Common Errors & Examiner Pitfalls
- Failing to include correct state symbols (must include (s) or (s, graphite) , (g) , and (l) ).
- Multiplying the equation to give 2 moles of ethanol (formation equations must produce exactly 1 mole of the product). Fractions like 0.5O₂ or ½O₂ are required.
Enthalpy Change of Combustion from Mean Bond Enthalpies
📐 Step-by-Step Calculation
Equation: C₂H₅OH(l) + 3O₂() → 2CO₂(g) + 3H₂O(l)
- Bonds broken in reactants:
1 × (C—C) = 347
5 × (C—H) = 5 × 413 = 2065
1 × (C—O) = 358
1 × (O—H) = 464
3 × (O=O) = 3 × 498 = 1494
Total energy in = 4728 kJ - Bonds made in products:
2 moles of CO₂ contain 4 × (C=O): 4 × 805 = 3220
3 moles of H₂O contain 6 × (O—H): 6 × 464 = 2784
Total energy out = 6004 kJ - Calculate ΔH:
ΔH = Energy in − Energy out = 4728 − 6004 = -1276 kJ mol⁻¹
🧠 Exam Technique
Always structure your bond enthalpy calculations by clearly separating energy required to break bonds (positive) and energy released when making bonds (negative). Always include the negative sign in your final enthalpy change answer!
Reaction Profile Diagram
💡 What to Draw
- A curved reaction pathway starting at C₂H₅OH(l) + 3O₂is and rising to a peak representing activation energy ( Eₐ ).
- A product energy level line drawn lower than the reactants to reflect an exothermic reaction ( ΔH is negative).
- An arrow labelled ΔH pointing downwards from the reactant level to the product level.
- An arrow or curve labelled Eₐ from the reactants to the top of the curve.
❌ Common Errors
- Drawing products higher than reactants (treating combustion as endothermic).
- Drawing straight lines instead of a smooth curve for the activation energy peak.
- Omitting or mislabelling the ΔH and Eₐ arrows.
Comparing Calculated vs. Data Book Values
✅ Acceptable Answers
Mean bond enthalpies are average values derived from a range of different compounds, rather than exact values specific to ethanol and water in their exact physical states (liquid states vs. gaseous state assumption).
🧠 Examiner Tip
Keep your answer concise. The examiners are strictly looking for the key limitation of mean bond enthalpies: they are averages across different molecules, and bond enthalpies strictly apply to substances in the gaseous state (whereas ethanol and water are liquids in standard combustion equations).
Topics
Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics I
Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2016. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.