AQA GCSE Chemistry Chemistry Paper 2 (Higher), November 2021: Question 10
17 marks · High Demand difficulty · Extended Answer
Questions on alkenes and alcohols including cracking, choosing compromise conditions for the industrial hydration of ethene, fermentation, mole calculations for fuel energy, and writing a balanced combustion equation.
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Question text
10 This question is about alkenes and alcohols.
Ethene is an alkene produced from large hydrocarbon molecules.
Large hydrocarbon molecules are obtained from crude oil by fractional distillation.
10.1 Name the process used to produce ethene from large hydrocarbon molecules.
[1 mark]
10.2 Describe the conditions used to produce ethene from large hydrocarbon molecules.
[2 marks]
10.3 Ethanol can be produced from ethene and steam.
The equation for the reaction is:
C2H4(g) + H2O(g) ⇌ C2H5OH(g)
The forward reaction is exothermic.
Explain how the conditions for this reaction should be chosen to produce ethanol as
economically as possible.
[6 marks]
10.4 Ethanol can also be produced from sugar solution by adding yeast.
Name this process.
[1 mark]
10.5 Butanol can be produced from sugar solution by adding bacteria.
Sugar solution is broken down in similar ways by bacteria and by yeast.
Suggest the reaction conditions needed to produce butanol from sugar solution
by adding bacteria.
[2 marks]
Ethanol and butanol can be used as fuels for cars.
10.6 A car needs an average of 1.95 kJ of energy to travel 1 m
Ethanol has an energy content of 1300 kilojoules per mole (kJ/mol).
Calculate the number of moles of ethanol needed by the car to travel 200 km
[3 marks]
Number of moles = mol
10.7 When butanol is burned in a car engine, complete combustion takes place.
Write a balanced equation for the complete combustion of butanol.
You do not need to include state symbols.
[2 marks]
Mark scheme
Show the mark scheme
Question 10
AO /
Question Answers Extra information Mark
Spec. Ref.
10.1 (steam / catalytic) cracking allow thermal decomposition 1 AO1
4.7.1.4
AO /
Spec. Ref.
10.2 high temperature allow a temperature in the range 1 AO1
300 – 900 °C 4.7.1.4
steam / catalyst 1
Question 10 continued
AO/
Question Answers Mark
Spec. Ref
10.3 Level 3: Relevant points (reasons/causes) are identified, given in 5–6 AO2
detail and logically linked to form a clear account. 4.6.1.3
4.6.2.4
Level 2: Relevant points (reasons/causes) are identified, and there 3–4 4.6.2.6
are attempts at logical linking. The resulting account is not fully 4.6.2.7
clear. 4.7.2.2
Level 1: Points are identified and stated simply, but their relevance 1–2
is not clear and there is no attempt at logical linking.
No relevant content 0
Indicative content
Rate
• higher temperature gives higher rate
• because more frequent collisions
• higher pressure gives higher rate
• because more frequent collisions
• a catalyst can be used to give a higher rate
• because the activation energy is reduced
Yield
• higher temperature gives lower yield
• because the reaction is exothermic
• higher pressure gives higher yield
• because there are more molecules on left hand side
Other factors
• higher temperatures use more energy so costs increase
• higher pressures use more energy so costs increase
• higher pressures require stronger reaction vessels so costs
24 increase
Compromise
• chosen temperature is a compromise between rate and yield
• chosen temperature is a compromise between rate and cost (of
energy used)
• chosen pressure is a compromise between rate and cost (of
energy used)
• chosen pressure is a compromise between yield and cost (of
energy used)
Question 10 continued
AO /
Spec. Ref.
10.4 fermentation allow ferment(ing) 1 AO1
4.7.2.3 25
AO /
Spec. Ref.
10.5 warm allow a value in the range 25 °C 1 AO2
to 45 °C 4.7.2.3
anaerobic (conditions) allow without oxygen / air 1
AO /
Spec. Ref.
10.6 (conversion) AO2
200 km = 200,000 m 1 4.7.2.3
200000 × 1.95 (mol) allow correct use of incorrect / 1
(moles = )
1300 no conversion for distance
= 300 (mol) 1
AO /
Spec. Ref.
10.7 C4H9OH + 6 O2 → 4 CO2 + 5 H2O allow C4H10O for C4H9OH 2 AO2
4.1.1.1
allow multiples 4.3.1.1
4.7.2.3
allow 1 mark for
C4H9OH + O2 → CO2 + H2O
with incorrect / no multipliers
ignore state symbols
Total 17
How to answer it
Alkenes, Alcohols, and Industrial Chemical Economics
This question assesses your understanding of organic chemistry and dynamic equilibria across AQA Topics 4.1, 4.6, and 4.7:
- Recall of hydrocarbon cracking methods and reaction conditions.
- Evaluating compromising industrial conditions (temperature, pressure, and catalyst) by balancing equilibrium position (yield), collision theory (rate), and monetary cost.
- Fermentation of sugars into alcohols and biological reaction conditions.
- Multi-step mole-energy calculations with metric unit conversion (km to m).
- Formulating and balancing complete combustion chemical equations for alcohols.
Process to Produce Ethene from Long Hydrocarbons
✅ Correct Answer
Cracking (or catalytic cracking / steam cracking / thermal decomposition)
❌ Common Errors
- Writing fractional distillation (this separates crude oil based on boiling points; it does not break down molecules).
- Spelling errors like "crackling".
Conditions Needed for Cracking
✅ Correct Answers (Any 2 points)
- High temperature (allow any value in the range 300 – 900 °C) [1 mark]
- Catalyst OR steam [1 mark]
💡 Key Knowledge
There are two types of cracking in GCSE Chemistry:
- Catalytic cracking: Vapour passed over a hot zeolitic/aluminium oxide catalyst (around 550 °C).
- Steam cracking: Vapour mixed with steam and heated to very high temperatures (over 800 °C).
Economic Choice of Conditions for Producing Ethanol
C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) (Forward reaction is exothermic)
🧠 How to Score Level 3 (5–6 Marks)
This question asks how conditions are chosen to produce ethanol as economically as possible. A top-band answer must clearly explain the conflict between Rate, Yield (Equilibrium), and Cost/Safety for both temperature and pressure, concluding with why a compromise condition is selected.
🌡️ 1. Temperature
- Yield: Forward reaction is exothermic, so a lower temperature shifts equilibrium right, giving a higher yield of ethanol.
- Rate: However, lower temperature means particles have less kinetic energy, causing fewer frequent successful collisions and a slow reaction rate.
- Cost: Very high temperatures require massive fuel/energy expenditure.
- Compromise: A moderately high temperature (approx. 300 °C) is chosen to give an acceptable yield at a reasonably fast rate.
💨 2. Pressure
- Yield: Left side has 2 moles of gas (1 C₂H₄ + 1 H₂O) and right side has 1 mole of gas (1 C₂H₅OH). Higher pressure shifts equilibrium towards fewer gas molecules (to the right), increasing yield.
- Rate: Higher pressure compresses gas molecules together, leading to more frequent collisions and a faster rate.
- Cost & Safety: Extremely high pressures require thick reinforced pipes/vessels and huge energy costs to compress gas, which is very expensive and risky.
- Compromise: A moderate/compromise pressure (approx. 60–70 atm) balances high yield/rate against plant construction and pumping costs.
⚡ 3. Catalyst
- A catalyst (phosphoric acid) provides an alternative pathway with a lower activation energy.
- Increases the rate without affecting the equilibrium position.
- Allows the reaction to run at a lower operating temperature, saving energy and production costs.
❌ Common Examiner Traps
- Forgetting to mention the word compromise.
- Saying "catalysts increase the yield" — False! Catalysts only increase the rate at which equilibrium is reached.
- Failing to state the number of gas molecules on both sides when discussing pressure.
Production of Ethanol from Sugar Solution
✅ Correct Answer
Fermentation (allow fermenting)
🧠 Quick Recall
Glucose → Ethanol + Carbon Dioxide
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
Reaction Conditions for Microbial Sugar Breakdown
✅ Correct Conditions
- Warm / suitable temperature (allow any value from 25 °C to 45 °C) [1 mark]
- Anaerobic conditions (without oxygen / without air) [1 mark]
❌ Why Conditions Matter
- Too cold (<20 °C): Enzymes work too slowly.
- Too hot (>45 °C): Bacterial/yeast enzymes become denatured.
- Presence of oxygen: Aerobic respiration would occur instead, or ethanol would oxidise into ethanoic acid (vinegar).
Mole-Energy Calculation for Car Journey
📐 Step-by-Step Calculation
- Step 1: Convert distance into metres
Car travels 200 km. There are 1000 m in 1 km.
Distance = 200 × 1000 = 200,000 m [1 mark] - Step 2: Calculate total energy required
Car uses 1.95 kJ per 1 m.
Total energy = 200,000 m × 1.95 kJ/m = 390,000 kJ - Step 3: Calculate moles of ethanol needed
Ethanol provides 1300 kJ per mole.
Moles = Total Energy ÷ Energy per mole
Moles = 390,000 kJ ÷ 1300 kJ/mol = 300 mol [2 marks for steps 2 & 3 combined]
❌ Common Trap
Forgetting to convert km to m! If you used 200 instead of 200,000, you would get (200 × 1.95) ÷ 1300 = 0.3 mol . The mark scheme allows an error-carried-forward (ECF) mark of 2/3 for 0.3 mol, but converting units correctly guarantees the full 3/3 marks.
Combustion Equation for Butanol
✅ Balanced Chemical Equation
C₄H₉OH + 6 O₂ → 4 CO₂ + 5 H₂O
(Also accepted: C₄H₁₀O + 6 O₂ → 4 CO₂ + 5 H₂O or correct multiples)
[1 mark] for correct balancing balancing coefficients. Total: [2 marks]
🧠 How to Balance Alcohols Systematically
- Carbons first: Butanol has 4 Carbons → 4 CO₂
- Hydrogens second: Butanol has 10 Hydrogens (9 + 1) → 5 H₂O
- Count Oxygens on right side: (4 × 2 in CO₂) + (5 × 1 in H₂O) = 8 + 5 = 13 Oxygens
- Balance Oxygens on left: Don't forget the 1 Oxygen already inside C₄H₉OH !
Need 13 - 1 = 12 Oxygens from O₂ → 12 ÷ 2 = 6 O₂ .
Topics
Chemistry · C3: Quantitative Chemistry · C6: The Rate and Extent of Chemical Change · C7: Organic Chemistry
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.