SCSA Chemistry Chemical synthesis

15 sample questions with marking guides and sample answers

Q3
2020
QCAA
Paper 2
9 marks
Q3

Ethanol can be produced by the fermentation of glucose or the hydration of ethene.

Q3a
3 marks

Describe the production of ethanol by fermentation of glucose by writing a balanced equation and indicating if a catalyst is required.

Reveal Answer

C6_6H12_{12}O6_6(aq) yeast\xrightarrow{\text{yeast}} 2CH3_3CH2_2OH(aq) + 2CO2_2(g)

Marking Criteria
DescriptorMarks

provides correct reactants and products

1

correctly balances the equation

1

indicates that yeast is required as a catalyst

1
Q3b
2 marks

Calculate the atom economy for the production of ethanol by fermentation of glucose.

Reveal Answer

Molar mass (ethanol) = 46.08 g
Molar mass (glucose) = 180.18 g
atom economy = 2×46.08180.18×100\frac{2 \times 46.08}{180.18} \times 100

atom economy = 2×46.08180.18×100=51.148%51%\frac{2 \times 46.08}{180.18} \times 100 = 51.148 \% \approx 51\%
Atom economy = 51%

Marking Criteria
DescriptorMarks

shows substitution correctly performed

1

determines atom economy

1
Q3c
2 marks

In terms of atom economy, determine which process for the production of ethanol (i.e. hydration of ethene or fermentation of glucose) is greener.

Reveal Answer

Hydration atom economy = 100%
Fermentation atom economy = 51%
Therefore, production of ethene by hydration is greener.

Marking Criteria
DescriptorMarks

determines atom economy for hydration reaction is 100%

1

identifies that hydration reaction is greener

1
Q3d
2 marks

Identify two principles of green chemistry, other than atom economy, that make the production of ethanol by fermentation greener than by hydration.

Reveal Answer

Use of renewable feedstocks
Design for energy efficiency

Marking Criteria
DescriptorMarks

provides use of renewable feedstocks

1

provides design for energy efficiency

1
Q38
2021
SCSA
12 marks
Q38
12 marks

Sulfuric acid is manufactured by the Contact process, the steps of which are outlined below.

Step One: Molten sulfur is burned in air at approximately 1000 °C:

S(l)+O2(g)SO2(g)+297kJS(l) + O_2(g) \rightarrow SO_2(g) + 297 kJ

Step Two: The resulting sulfur dioxide is converted to sulfur trioxide as shown in the following equilibrium reaction. It is conducted at a temperature of about 450 °C with a V2O5V_2O_5 catalyst at a pressure of between 100 and 200 kPa:

2SO2(g)+O2(g)2SO3(g)+198kJ2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g) + 198 kJ

Step Three: The resulting sulfur trioxide is absorbed into sulfuric acid, producing oleum (H2S2O7H_2S_2O_7). Water is added to the oleum, producing 18 mol L1^{-1} sulfuric acid:

SO3(g)+H2SO4(l)H2S2O7(l)SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l) H2S2O7(l)+H2O(l)2H2SO4(aq)H_2S_2O_7(l) + H_2O(l) \rightarrow 2 H_2SO_4(aq)

Use your understanding of collision theory and chemical equilibrium to discuss the reaction conditions for Steps 1 and 2 of the Contact process, given that the aim is to produce the greatest yield in the shortest time. In your discussion, also address economic concerns where appropriate.

Reveal Answer

High temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently. Also, more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases. The vanadium catalyst increases the rate of the forward reaction, and also the rate of the reverse reaction to an equal extent, as it provides an alternative pathway with a lower activation energy. Therefore, a greater proportion of the particles will have sufficient energy to react when they collide. High pressure or concentration has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases. As Step 1 is a combustion reaction, it essentially goes to completion at the high temperature and does not require a catalyst or high pressure. For Step 2, high temperature, high pressure and a catalyst would favour a high rate.

For equilibrium, which is only considered for Step 2, high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3_3(g) yield, which is not desired. A low temperature decreases the rate of reaction, which is also not desired. A high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3_3(g) yield which is desired.

Economically, high pressures are costly and dangerous.

Therefore, for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield. A compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.

Marking Criteria

Rates

DescriptorMarks

1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.

6

Equilibrium

DescriptorMarks

1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.

3

Economics

DescriptorMarks

States that high pressures are costly (and dangerous).

1

Compromise

DescriptorMarks

1 mark each for any of the following, up to a maximum of 2 marks: Identifies that for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield; Identifies that a compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.

2
Q2
2021
VCAA
1 mark
Q2
1 mark

Biodiesel and petrodiesel

A

have different viscosities.

B

have the same environmental impact.

C

contain molecules with no polar groups.

D

will flow easily through fuel lines in very cold climate conditions.

Reveal Answer
A

have different viscosities.

Correct Answer

This is correct because biodiesel contains polar ester groups that create stronger intermolecular forces, resulting in a higher viscosity compared to non-polar petrodiesel.

B

have the same environmental impact.

This is incorrect because biodiesel is derived from renewable biomass and generally produces fewer net carbon emissions and particulates compared to fossil-fuel-derived petrodiesel.

C

contain molecules with no polar groups.

This is incorrect because while petrodiesel consists of non-polar hydrocarbons, biodiesel consists of fatty acid methyl esters which contain polar ester groups.

D

will flow easily through fuel lines in very cold climate conditions.

This is incorrect because biodiesel has a higher cloud point than petrodiesel, meaning it is more prone to gelling and restricting flow in very cold climates.

Q1
2023
QCAA
Paper 2
2 marks
Q1
2 marks

Polylactic acid (PLA) and low-density polyethylene (LDPE) are both used to produce plastic wrapping film.

PlasticCompositionDensity (g/cm3^3)Tensile stress (MPa)Elongation (%)Degradation rate
PLAplant-based1.24606slow
LDPEpetrochemical-based0.9212148none

Analyse the data to discuss one advantage and one disadvantage of using PLA rather than LDPE to produce plastic wrapping film.

Advantage:

Disadvantage:

Reveal Answer

Advantage: An advantage is that PLA is plant based, therefore it uses (renewable) natural resources while LDPE is produced from non-renewable fossil fuels.
Disadvantage: PLA has less % elongation than LDPE, therefore PLA would stretch less.

Marking Criteria
DescriptorMarks

identifies an advantage of using PLA using data

1

identifies a disadvantage of using PLA using data

1
Q12
2021
QCAA
Paper 1
1 mark
Q12
1 mark

Green chemistry principles include the design of chemical synthesis processes that

A

use renewable raw materials and minimise unwanted products.

B

use renewable raw materials and minimise unwanted reactants.

C

use non-renewable raw materials and minimise unwanted products.

D

use non-renewable raw materials and minimise unwanted reactants.

Reveal Answer
A

use renewable raw materials and minimise unwanted products.

Correct Answer

Green chemistry principles explicitly advocate for the use of renewable feedstocks (Principle 7) and the prevention of waste by minimizing the formation of unwanted by-products (Principles 1 and 2).

B

use renewable raw materials and minimise unwanted reactants.

While using renewable materials is a core goal, the principles focus on minimizing waste (unwanted products) generated during the reaction, rather than minimizing reactants.

C

use non-renewable raw materials and minimise unwanted products.

Green chemistry promotes the transition to renewable raw materials (such as agricultural products) rather than relying on depleting non-renewable resources like petroleum.

D

use non-renewable raw materials and minimise unwanted reactants.

This option is incorrect because green chemistry aims to use renewable resources, and the primary objective regarding synthesis efficiency is the reduction of waste products.

Q4
2021
QCAA
Paper 1
1 mark
Q4
1 mark

The cleaning action of soap is impaired in hard water because the

A

hydrophilic end reacts with calcium ions to form insoluble salts.

B

hydrophobic end reacts with calcium ions to form insoluble salts.

C

hydrophilic end reacts with calcium ions to form insoluble fatty acids.

D

hydrophobic end reacts with calcium ions to form insoluble fatty acids.

Reveal Answer
A

hydrophilic end reacts with calcium ions to form insoluble salts.

Correct Answer

Hard water contains calcium ions (Ca2+Ca^{2+}) which react with the charged hydrophilic carboxylate group (COO-COO^-) of the soap to form insoluble calcium salts, commonly known as scum.

B

hydrophobic end reacts with calcium ions to form insoluble salts.

The hydrophobic end is the non-polar hydrocarbon chain that interacts with grease and oil; it does not react with the metal ions in the water.

C

hydrophilic end reacts with calcium ions to form insoluble fatty acids.

While the reaction occurs at the hydrophilic end, the product is an insoluble salt (calcium carboxylate), not a free fatty acid. Fatty acids are typically formed when soap reacts with an acid.

D

hydrophobic end reacts with calcium ions to form insoluble fatty acids.

The reaction occurs at the charged hydrophilic head, not the hydrophobic tail, and it results in the formation of insoluble salts rather than fatty acids.

Q23
2022
QCAA
Paper 1
4 marks
Q23
4 marks

Ibuprofen is manufactured using two different processes.

ProcessNumber of reagents usedReagentsReagentsIbuprofenIbuprofenWaste productsWaste products
  AtomsMr\text{M}_\text{r}AtomsMr\text{M}_\text{r}AtomsMr\text{M}_\text{r}
17C20H42NO10ClNa\text{C}_{20}\text{H}_{42}\text{NO}_{10}\text{ClNa}514.5C13H18O2\text{C}_{13}\text{H}_{18}\text{O}_2206.0C7H24NO8ClNa\text{C}_7\text{H}_{24}\text{NO}_8\text{ClNa}308.5
24C15H22O4\text{C}_{15}\text{H}_{22}\text{O}_4266.0C13H18O2\text{C}_{13}\text{H}_{18}\text{O}_2206.0C2H4O2\text{C}_2\text{H}_4\text{O}_260.0

Calculate the atom economy for each process and draw conclusions about the economic and environmental impact of each process.

Reveal Answer

Process 1: atom economy = 206.0/514.5×100=40.04%206.0/514.5 \times 100 = 40.04\%
Process 2: atom economy = 206.0/266.0×100=77.44%206.0/266.0 \times 100 = 77.44\%
Process 2 has 37.4% better atom economy than process 1

Economic impact: Process 2 has a better atom economy than process 1 (fewer reagents are required).
Environmental impact: Process 2 is greener than process 1 because fewer waste products (atoms) are produced.

Marking Criteria
DescriptorMarks

Calculates atom economy for Process 1 as 40%

1

Calculates atom economy for Process 2 as 77%

1

Concludes process 2 is cheaper as fewer reagent atoms are required

1

Concludes process 2 is greener as fewer waste atoms are produced

1
Q22
2024
VCAA
1 mark
Q22
1 mark

Use the following information to answer the question.

A triglyceride is reacted with methanol, CH3OH\text{CH}_3\text{OH}, in the presence of concentrated KOH(aq)\text{KOH}(\text{aq}). The products of this reaction are glycerol and Compound J.

The molecular formula of Compound J is C19H30O2\text{C}_{19}\text{H}_{30}\text{O}_2.

What is the molecular formula of the triglyceride?

A

C54H82O3\text{C}_{54}\text{H}_{82}\text{O}_3

B

C54H82O6\text{C}_{54}\text{H}_{82}\text{O}_6

C

C57H84O3\text{C}_{57}\text{H}_{84}\text{O}_3

D

C57H86O6\text{C}_{57}\text{H}_{86}\text{O}_6

Reveal Answer
A

C54H82O3\text{C}_{54}\text{H}_{82}\text{O}_3

This formula is incorrect because a triglyceride must contain 6 oxygen atoms, not 3, and the carbon and hydrogen counts do not balance the transesterification reaction.

B

C54H82O6\text{C}_{54}\text{H}_{82}\text{O}_6

This formula is incorrect because it fails to account for the 3 carbon atoms from the glycerol backbone, resulting in 54 carbons instead of 57.

C

C57H84O3\text{C}_{57}\text{H}_{84}\text{O}_3

This formula is incorrect because a triglyceride contains 3 ester groups, meaning it must have 6 oxygen atoms, not 3.

D

C57H86O6\text{C}_{57}\text{H}_{86}\text{O}_6

Correct Answer

This is correct. In the transesterification reaction, 1 Triglyceride + 3 Methanol (CH4O\text{CH}_4\text{O}) \rightarrow 1 Glycerol (C3H8O3\text{C}_3\text{H}_8\text{O}_3) + 3 Compound J (C19H30O2\text{C}_{19}\text{H}_{30}\text{O}_2). Balancing the atoms yields C57H86O6\text{C}_{57}\text{H}_{86}\text{O}_6.

Q11
2024
VCAA
1 mark
Q11
1 mark

Iron is produced from iron ore using heat and a reducing agent.

Using hydrogen, produced from fossil fuels, as the reducing agent to produce iron is consistent with

A

the concept of a linear economy.

B

the concept of a circular economy.

C

the United Nations Sustainable Development Goal 2: Zero hunger.

D

the United Nations Sustainable Development Goal 11: Sustainable cities and communities.

Reveal Answer
A

the concept of a linear economy.

Correct Answer

A linear economy follows a "take-make-dispose" model. Using finite fossil fuels to produce hydrogen relies on continuous resource extraction without renewing or recycling the resource, which perfectly aligns with this concept.

B

the concept of a circular economy.

A circular economy focuses on sustainability, recycling, and using renewable resources. Relying on non-renewable fossil fuels contradicts this closed-loop approach.

C

the United Nations Sustainable Development Goal 2: Zero hunger.

UN Sustainable Development Goal 2 focuses on ending hunger, achieving food security, and promoting sustainable agriculture, which is unrelated to industrial iron production.

D

the United Nations Sustainable Development Goal 11: Sustainable cities and communities.

UN Sustainable Development Goal 11 focuses on making cities inclusive, safe, resilient, and sustainable. Industrial processes reliant on fossil fuels do not support environmental sustainability.

Q39
2020
SCSA
12 marks
Q39

Fluorescent lights are glass tubes which are coated on the inside with rare earth metal phosphates (such as cerium, lanthanum and terbium phosphates) that provide light. Cerium, lanthanum and terbium are expensive, so are recovered once the fluorescent light is no longer functional.

The key steps in one method proposed for recovery of these rare earth metals are summarised below:

  • Step 1: Physical separation of the rare earth metal phosphates from the glass and any metallic components. This gives an impure powder consisting of cerium, lanthanum and terbium phosphates.
  • Step 2: Add excess solid sodium carbonate to the powder and heat, completely converting each rare earth metal phosphate to its corresponding oxide, as shown by the following balanced equations:
    2 LaPO4(s)+3 Na2CO3(s)La2O3(s)+2 Na3PO4(s)+3 CO2(g)2 \text{ LaPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{La}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}
    4 CePO4(s)+6 Na2CO3(s)+O2(g)4 CeO2(s)+4 Na3PO4(s)+6 CO2(g)4 \text{ CePO}_4\text{(s)} + 6 \text{ Na}_2\text{CO}_3\text{(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{ CeO}_2\text{(s)} + 4 \text{ Na}_3\text{PO}_4\text{(s)} + 6 \text{ CO}_2\text{(g)}
    2 TbPO4(s)+3 Na2CO3(s)Tb2O3(s)+2 Na3PO4(s)+3 CO2(g)2 \text{ TbPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{Tb}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}
  • Step 3: Wash the product from Step 2 with water.
  • Step 4: Add hydrochloric acid to the washed product from Step 3 to leach (dissolve) only the rare earth metal oxides.
  • Step 5: Use solvent extraction to separate the different rare earth metals from each other and create separate solutions of each of them.
  • Step 6: Add oxalic acid to the separated solutions to precipitate the rare earth metal ions as oxalate salts.
  • Step 7: Heat the oxalate salts to recover the rare earth metals as pure oxides, namely La2O3\text{La}_2\text{O}_3, Tb4O7\text{Tb}_4\text{O}_7 and CeO2\text{CeO}_2.

A chemist used the above procedure to determine the percentage by mass of lanthanum, terbium and cerium in some fluorescent lights and, after completing Step 1, had recovered 1.20 kg of the coating chemicals.

Q39b

The mass of the solid sent from Step 3 to Step 4 was 1.16 kg. This solid was leached with 6.00 mol L1^{-1} HCl at a solid to liquid ratio of 150 g per litre. Analysis of the solution at the end of leaching showed that it contained lanthanum, terbium and cerium, with its lanthanum concentration being 8.65×103 mol L18.65 \times 10^{-3} \text{ mol L}^{-1}.

Q39a
3 marks

At the completion of Step 2, the mass of the mixture had decreased by 11.3 g. Calculate the mass of sodium carbonate that reacted with the rare earth metal phosphates.

Reveal Answer

n(CO₂) = 11.3/44.01 = 0.257 mol

n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol

m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g

Marking Criteria
DescriptorMarks

n(CO₂) = 11.3/44.01 = 0.257 mol

1

n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol

1

m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g

1
Q39b
5 marks

Calculate the percentage, by mass, of lanthanum in the fluorescent light coating chemical, given that the leaching efficiency for lanthanum was 86%.

Note that the balanced equation for the leaching of lanthanum with hydrochloric acid is:
La2O3(s)+6 HCl(aq)2 LaCl3(aq)+3 H2O(l)\text{La}_2\text{O}_3\text{(s)} + 6 \text{ HCl(aq)} \rightarrow 2 \text{ LaCl}_3\text{(aq)} + 3 \text{ H}_2\text{O(l)}

Reveal Answer

Volume of HCl used: 1160/150 = 7.73 L

n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution

Taking into account the leaching efficiency,
n(La in the solid that was leached) = 0.0669/0.86 = 0.0778

m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g

% La in the coating chemical = (10.8/1200) × 100 = 0.900%

Marking Criteria
DescriptorMarks

Volume of HCl used: 1160/150 = 7.73 L

1

n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution

1

Taking into account the leaching efficiency, n(La in the solid that was leached) = 0.0669/0.86 = 0.0778

1

m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g

1

% La in the coating chemical = (10.8/1200) × 100 = 0.900%

1
Q39c
4 marks

Analysis of the cerium-containing solution produced in Step 5 showed that its cerium concentration was 0.146 mol L1^{-1}. This solution, which had a volume of 424 mL, was added to 110 mL of aqueous 1.15 mol L1^{-1} oxalic acid during Step 6, resulting in the precipitation of cerium oxalate, Ce(C2O4)2\text{Ce(C}_2\text{O}_4\text{)}_2. The balanced equation for this reaction is:

CeCl4(aq)+2 H2C2O4(aq)Ce(C2O4)2(s)+4 HCl(aq)\text{CeCl}_4\text{(aq)} + 2 \text{ H}_2\text{C}_2\text{O}_4\text{(aq)} \rightarrow \text{Ce(C}_2\text{O}_4\text{)}_2\text{(s)} + 4 \text{ HCl(aq)}

Did the chemist add enough oxalic acid solution to precipitate all of the cerium? Use calculations to support your answer.

Reveal Answer

n(cerium) = 0.146 × 0.424 = 0.0619 mol

n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol

n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol

comparison of the moles of oxalic acid shows that enough oxalic acid was added

Marking Criteria
DescriptorMarks

n(cerium) = 0.146 × 0.424 = 0.0619 mol

1

n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol

1

n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol

1

comparison of the moles of oxalic acid shows that enough oxalic acid was added

1
Q39
2023
SCSA
15 marks
Q39

Ethanol can be produced either from plant materials or from petrochemical sources.

Q39a

When ethanol is produced from plant sources, the material is ground up. The starches and cellulose in the material are then converted into sugars. Yeast or zymase is mixed with the sugars at 25 to 37 °C and a pH of between 3 and 5 at atmospheric pressure. The products of the fermentation process are ethanol and carbon dioxide.

Q39b

Ethanol can also be produced by the endothermic hydration of ethene. This is carried out at 250 to 300 °C and 6000 to 7000 kPa in the presence of an acid catalyst.

Q39a (i)
2 marks

Justify the conditions used for fermentation.

Reveal Answer

Zymase are enzymes, which are only effective in a narrow pH band and temperature band.

Marking Criteria
DescriptorMarks

Recognises that yeast or zymase are enzymes

1

Recognises that enzymes are only effective in a narrow pH band and temperature band, or recognises that without the enzymes, the reaction either does not proceed or is too slow to be viable

1
Q39a (ii)
2 marks

Write an equation for the fermentation process, using C6H12O6C_6H_{12}O_6 as the sugar. Use condensed structures in your equation.

Reveal Answer

C6H12O62 CH3CH2OH+2 CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2 \text{ CH}_3\text{CH}_2\text{OH} + 2 \text{ CO}_2

Marking Criteria
DescriptorMarks

Correct reactants and products

1

Correct balancing

1
Q39b (i)
3 marks

Write an equation for the hydration of ethene. Use condensed structures in your equation.

Reveal Answer

CH2CH2+H2OCH3CH2OH\text{CH}_2\text{CH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{OH}

Marking Criteria
DescriptorMarks

Correct reactants

1

Correct products

1

Uses condensed structures in equation

1
Q39b (ii)
5 marks

Justify the temperature and pressure used for the hydration of ethene.

Reveal Answer

high pressure increases rate of reaction as there are more particles per unit volume, therefore a greater frequency of collisions. It also increases yield due to it favouring the direction with the fewer number of gas particles which is the product side of the reaction. High temperature also means particles are moving more rapidly and collide more often, and more particles having sufficient energy for successful collisions, a greater proportion of collisions will be successful, increasing reaction rate.

Because hydration of ethene is endothermic, high temperature will favour the formation of the products.

High temperature favours both rate and yield, however, a moderate temperature will still produce an economical yield.

Marking Criteria
DescriptorMarks

Recognition that high pressure increases rate of reaction as there are more particles per unit volume, therefore a greater frequency of collisions

1

Recognition that high pressure also increases yield due to it favouring the direction with the fewer number of gas particles which is the product side of the reaction

1

Recognition that high temperature will increase rate as the particles are moving more rapidly and collide more often, as well as more particles having sufficient energy for successful collision, so greater proportion of collisions will be successful

1

Recognition that because hydration of ethene is endothermic, high temperature will favour the formation of the products

1

Recognition that, although high temperature favours both rate and yield, a moderate temperature will produce economical/safe yield

1
Q39c
3 marks

State three reasons why the fermentation process to produce ethanol is more common than the hydration of ethene.

Reveal Answer

Fermentation requires less energy input than hydration of ethene, fermentation costs less than hydration of ethene, and fermentation is a 'greener' process than hydration of ethene. Furthermore, fermentation uses a renewable feedstock while hydration of ethene does not.

Marking Criteria
DescriptorMarks

1 mark for each correct point (any 3 of):

  • Fermentation requires less energy input than hydration of ethene
  • Fermentation costs less than hydration of ethene
  • Fermentation is a 'greener' process than hydration of ethene
  • Fermentation uses a renewable feedstock while hydration of ethene does not
3
Q12
2022
VCAA
1 mark
Q12
1 mark

Enzymes are commonly not effective in acidic conditions because acids

A

change the charges on the enzymes.

B

react with the enzymes to form zwitterions.

C

esterify the enzymes into smaller molecules.

D

react with the carboxyl groups on the enzymes' amino acid residues.

Reveal Answer
A

change the charges on the enzymes.

Correct Answer

Acidic conditions increase the concentration of H+H^+ ions, which protonate amino acid side chains. This alters the enzyme's overall charge distribution, disrupting the ionic bonds that maintain its functional 3D structure.

B

react with the enzymes to form zwitterions.

Zwitterions are molecules with both positive and negative charges that net to zero, typically occurring at an amino acid's isoelectric point. In highly acidic conditions, amino acids become positively charged cations, not zwitterions.

C

esterify the enzymes into smaller molecules.

Acids do not esterify enzymes into smaller molecules. Breaking down an enzyme's protein chain into smaller molecules would involve the hydrolysis of peptide bonds, not esterification.

D

react with the carboxyl groups on the enzymes' amino acid residues.

While acidic conditions do protonate carboxylate groups into neutral carboxyl groups, this is only a partial explanation. The loss of enzyme effectiveness is due to the overall change in charges across all ionizable groups, which disrupts the active site.

Q4
2025
VCAA
1 mark
Q4
1 mark

The circular economy of a bioethanol production process could be improved by

A

increasing the fermentation temperature to accelerate the reaction rate.

B

using a new strain of yeast to produce a higher yield of ethanol from glucose.

C

developing a separate process that converts waste carbon dioxide into a useful product.

D

carrying out advanced distillation techniques to reduce energy consumption.

Reveal Answer
A

increasing the fermentation temperature to accelerate the reaction rate.

Increasing the fermentation temperature improves reaction kinetics but does not address waste reduction or resource recycling, which are the core principles of a circular economy.

B

using a new strain of yeast to produce a higher yield of ethanol from glucose.

While a higher yield improves process efficiency, it does not involve repurposing waste streams or closing material loops characteristic of a circular economy.

C

developing a separate process that converts waste carbon dioxide into a useful product.

Correct Answer

A circular economy focuses on eliminating waste by repurposing it; converting waste carbon dioxide into a useful product perfectly aligns with this principle by closing the material loop.

D

carrying out advanced distillation techniques to reduce energy consumption.

Reducing energy consumption improves the environmental footprint and sustainability of the process, but it does not represent the material recycling or waste valorization central to a circular economy.

Q12
2023
QCAA
Paper 1
1 mark
Q12
1 mark

Enzymes are classified as

A

carbohydrates.

B

proteins.

C

starches.

D

lipids.

Reveal Answer
A

carbohydrates.

Carbohydrates are organic compounds like sugars and fibers used primarily for energy and structure, not for catalyzing reactions.

B

proteins.

Correct Answer

Enzymes are biological catalysts composed of amino acid chains folded into specific shapes, classifying them as proteins.

C

starches.

Starches are complex carbohydrates (polysaccharides) used for energy storage in plants, whereas enzymes are the proteins that might break them down.

D

lipids.

Lipids include fats, oils, and waxes used for long-term energy storage and membrane structure, but they do not function as biological catalysts.

Q25
2022
VCAA
1 mark
Q25
1 mark

Consider the following statements regarding biodiesel and petrodiesel:

I Petrodiesel is more hygroscopic than biodiesel.
II Biodiesel forms crystals at a higher temperature than petrodiesel.
III Biodiesel and petrodiesel are derived from plants and animals.
IV The combustion of biodiesel releases less sulfur dioxide than petrodiesel.

Which of the statements above are correct?

A

I and IV only

B

III and IV only

C

I, II and III only

D

II, III and IV only

Reveal Answer
A

I and IV only

Statement I is incorrect because biodiesel contains polar ester groups, making it more hygroscopic (water-absorbing) than non-polar petrodiesel. Additionally, this option misses other correct statements.

B

III and IV only

While statements III and IV are correct, this option is incomplete. Statement II is also correct because biodiesel has a higher cloud point (forms crystals at a higher temperature) than petrodiesel.

C

I, II and III only

Statement I is incorrect due to biodiesel being more hygroscopic than petrodiesel. Furthermore, this option incorrectly excludes statement IV, which is a true statement.

D

II, III and IV only

Correct Answer

Statement II is correct because biodiesel has a higher cloud point. Statement III is correct as both fuels originate from organic matter (recent plants/animals for biodiesel, ancient marine life for petrodiesel). Statement IV is correct because biodiesel contains significantly less sulfur than petrodiesel.

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