AQA A-Level Chemistry Paper 2, 2023: Question 8
19 marks · Medium difficulty · State/Explain/Numerical
Biofuels and Energy Calculations This question involves completing a hydrolysis equation, calculating the enthalpy of combustion for palmitic acid, determining empirical formulas for a biofuel, identifying environmental problems of biofuel combustion, and calculating enthalpy changes using bond enthalpies and standard enthalpies of formation.
Practise this questionQuestion
Question text
08 This question is about biofuels.
Palmitic acid, CH3(CH2)14COOH, can be made by hydrolysis of the triester in palm oil
under acidic conditions.
Palmitic acid can be used as a biofuel.
08.1 Complete the equation for the hydrolysis of the triester in palm oil under
acidic conditions.
[2 marks]
08.2 Palmitic acid burns in air.
In a calorimetry experiment, combustion of 387 mg of palmitic acid increases the
temperature of 0.150 kg of water from 23.9 °C to 37.5 °C
Calculate a value, in kJ mol–1, for the enthalpy of combustion of
palmitic acid in this experiment.
Give your answer to the appropriate number of significant figures.
The specific heat capacity of water is 4.18 J K–1 g–1
[5 marks]
23 –1
Enthalpy of combustion kJ mol
08.3 State how the value calculated in Question 08.2 is likely to differ from
data book values.
Give one reason, other than heat loss, for this difference.
[2 marks]
Difference
Reason
08.4 A sample of a different biofuel, made from sewage sludge, is found to contain
37.08% carbon, 5.15% hydrogen and 24.72% oxygen by mass.
The rest of the sample is sulfur.
Calculate the empirical formula of this biofuel.
[3 marks]
Empirical formula
08.5 Complete combustion of the biofuel made from sewage sludge produces the
greenhouse gas carbon dioxide.
Suggest one other possible environmental problem with the complete combustion of
this biofuel.
State the formula of the pollutant responsible for this problem.
[2 marks]
Environmental problem
Formula
08.6 Ethanol is a biofuel that can be produced by the fermentation of glucose.
C6H12O6 → 2C2H5OH + 2CO2
Glucose has the structural formula shown.
Table 3 shows some mean bond enthalpy values.
Table 3
C–H C–C C–O C=O O–H
Mean bond enthalpy 412 348 360 805 463
/ kJ mol–1
Use the equation and the data in Table 3 to calculate an approximate value
of ΔH for the fermentation of glucose. For this calculation you should assume that all
the substances are in the gaseous state.
[3 marks]
25 –1
ΔH kJ mol
08.7 The carbon dioxide produced from fermentation can be reacted with steam to make
more ethanol.
*24* The equation for this reaction is
2CO2(g) + 3H2O(g) → C2H5OH(g) + 3O2(g)
Table 4 shows some standard enthalpies of formation.
Table 4
CO2(g) O2(g) C2H5OH(g) H2O(g)
∆ Hϴ / kJ mol–1 –394 0 –235 –242
f
Use the data in Table 4 to calculate a standard enthalpy change value for
this reaction.
[2 marks]
Standard enthalpy change kJ mol–1
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidelines Mark
M1 3 CH3(CH2)14COOH Penalise additional product(s) once 2
08.1 (2 x AO2)
M2 CH2(OH)CH(OH)CH2OH
M1 Mr = 256
0.387 -3
M2 n(CH3(CH2)14COOH) = = 1.51 × 10
M1
08.2 M3 Q = 150 × 4.18 × 13.6 = 8527.2 (J)
(5 x AO2)
M3
M4 ∆H = ÷ 1000 = (-)5641
M2
M5 ∆H = –5640 kJ mol–1 Must be negative and 3sf (allow ecf on M4)
M1 Less exothermic Allow Less negative (value) / Lower
08.3
(2 x AO2)
M2 Incomplete combustion Allow products of incomplete combustion
M1 % S = 33.05
C H O S
37.08 5.15 24.72 M1 = 33.05 M2 Calculation of moles 29
M2 ÷ Ar = 3.09 = 5.15 = 1.55 = 1.030 3
08.4 M3 Ratio of moles AND Empirical Formula (1 x AO1,
÷ = 3 = 5 = 1.50 = 1 2 x AO2)
If no Sulfur used ecf for M2 and M3
smallest
M2 3.09 : 5.15 : 1.55
M3 M3 C H O
6 10 3
Empirical formula = C6H10O3S2
M1 Acid rain Allow smog 2
08.5 (2 x AO3)
M2 SO2 Allow NOx
M1 Bonds broken = 9459 kJ mol–1 Allow if they cancel the common bonds
(5C-C + 7C-O + 7C-H + 5O-H) M1 4233
M2 Bonds formed = 9682 kJ mol–1 M2 4456 3
08.6
(3 x AO2)
(2C-C + 10C-H + 2C-O + 2O-H + 4C=O)
M3 ∆H = M1 – M2 = –223 kJ mol–1 M3 can be awarded as ecf from their M1 and M2
M1 ∆H = –235 – (2 × –394) – (3 × –242) 2
08.7 (2 x AO2)
M2 = +1279 kJ mol–1 If no sign assume positive
How to answer it
Step-by-Step Guide: Biofuels and Energetics
Question (a): Hydrolysis of Triester
Step 1: Write the Balanced Equation
The hydrolysis of the triester produces:
- 3 molecules of palmitic acid (CH3(CH2)14COOH)
- 1 molecule of glycerol (CH2(OH)CH(OH)CH2OH)
Equation:
Question (b): Enthalpy of Combustion
Step 1: Calculate Moles of Palmitic Acid
Use the formula: n = m / Mr .
Substitute: n = 0.387 / 256 = 1.51 × 10-3 mol
Step 2: Energy Released (Q)
Use the formula: Q = mcΔT .
Substitute:
Step 3: Enthalpy of Combustion
Convert to kJ mol-1:
Answer: -5640 kJ mol-1 (3 significant figures).
Question (c): Difference from Data Book Values
Step 1: State the Difference
The calculated value is less exothermic than the data book value.
Step 2: Reason
The difference is due to incomplete combustion , which produces products like CO or C instead of CO2.
Answer: Incomplete combustion.
Question (d): Empirical Formula
Step 1: Calculate Moles of Each Element
| Element | % by Mass | Divide by Ar | Ratio |
|---|---|---|---|
| C | 37.08 | 37.08 / 12 = 3.09 | 3 |
| H | 5.15 | 5.15 / 1 = 5.15 | 5 |
| O | 24.72 | 24.72 / 16 = 1.55 | 1.5 → 3 |
| S | Remaining | Divide to find 1 | 1 |
Empirical Formula: C6H10O3S2
Question (e): Environmental Problem
Problem: Acid rain due to sulfur dioxide emissions.
Formula: SO2
Question (f): Bond Enthalpy Calculation
Step 1: Bonds Broken
Total bonds broken:
Step 2: Bonds Formed
Total bonds formed:
Step 3: Calculate ΔH
ΔH = Bonds broken - Bonds formed:
Question (g): Standard Enthalpy Change
Step 1: Use Formation Values
Use the formula:
Substitute:
Evaluate:
Answer: +1279 kJ mol-1
Topics
Physical Chemistry · Organic Chemistry · 3.1.4 Energetics · 3.1.2 Amount of Substance · 3.3.9 Carboxylic Acids and Derivatives
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.