QCAA Chemistry Chemical synthesis and design

15 sample questions with marking guides and sample answers

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
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.

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.

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
Q22
2023
QCAA
Paper 1
4 marks
Q22a
2 marks

Write a balanced chemical equation to describe how polytetrafluorethene (PTFE) is produced from its monomer.

Reveal Answer

nF2C=CF2[F2CCF2]nnF_2C=CF_2 \rightarrow \left[ F_2C-CF_2 \right]_n

Marking Criteria
DescriptorMarks

describes formulas for tetrafluorethene monomer and polytetrafluorethene polymer

1

provides a balanced equation

1
Q22b
2 marks

Determine whether polytetrafluorethene is an addition or condensation polymer. Explain your reasoning.

Reveal Answer

Addition polymer
The double bond in tetrafluorethene is broken to allow the monomers to join.

Marking Criteria
DescriptorMarks

identifies addition polymer

1

explains double bond in monomer is broken to allow the formation of polymer

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.

Q22
2025
SCSA
1 mark
Q22
1 mark

The main reason a vehicle powered by a hydrogen fuel cell has lower polluting emissions than a vehicle powered by an internal combustion engine is because

A

hydrogen fuel cells release carbon dioxide into the atmosphere; however, it is not produced by the burning of fossil fuels.

B

the only by-products of the hydrogen fuel cell are water and heat.

C

fuel cells convert chemical energy directly into electrical energy.

D

fuel cells will not work unless the reactants are constantly supplied.

Reveal Answer
A

hydrogen fuel cells release carbon dioxide into the atmosphere; however, it is not produced by the burning of fossil fuels.

Hydrogen fuel cells do not produce carbon dioxide (CO2CO_2) during operation; they generate electricity through an electrochemical reaction rather than combustion.

B

the only by-products of the hydrogen fuel cell are water and heat.

Correct Answer

The chemical reaction in a hydrogen fuel cell combines hydrogen (H2H_2) and oxygen (O2O_2) to produce only water (H2OH_2O) and heat, eliminating the harmful tailpipe emissions associated with burning fossil fuels.

C

fuel cells convert chemical energy directly into electrical energy.

While this describes the efficient energy conversion mechanism of a fuel cell, it does not explain the composition of the emissions or why they are less polluting.

D

fuel cells will not work unless the reactants are constantly supplied.

This statement describes the operational requirement of a fuel cell (continuous fuel supply) but is unrelated to the environmental impact or chemical composition of its emissions.

Q22
2024
QCAA
Paper 1
3 marks
Q22a
2 marks

Describe the type of reaction that occurs when amino acid monomers are joined to form polypeptides.

Reveal Answer

A condensation reaction occurs when amino acid monomers are joined to form a peptide bond and water is removed.

Marking Criteria
DescriptorMarks

Identifies condensation reaction

1

Describes that water is removed when the amino acids are joined to form polypeptides

1
Q22b
1 mark

Identify the bond formed when amino acid monomers join to form a dipeptide.

Reveal Answer

A peptide bond is formed.

Marking Criteria
DescriptorMarks

Identifies peptide bond

1
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.

Q14
2025
VCAA
1 mark
Q14
1 mark

The reaction to produce methanal, CH2OCH_2O, is shown below.

2CH3OH(g)+O2(g)catalyst2CH2O(g)+2H2O(g)2CH_3OH(g) + O_2(g) \xrightarrow{\text{catalyst}} 2CH_2O(g) + 2H_2O(g)

The primary role of the catalyst in the production of CH2OCH_2O is to increase the

A

speed of all particles.

B

number of collisions per unit time.

C

proportion of particles that react.

D

overall kinetic energy of the system.

Reveal Answer
A

speed of all particles.

A catalyst does not change the speed of the particles. Increasing the temperature of the system would increase particle speed.

B

number of collisions per unit time.

A catalyst does not increase the frequency of collisions. Factors like increased concentration, pressure, or temperature would increase the number of collisions per unit time.

C

proportion of particles that react.

Correct Answer

A catalyst provides an alternative reaction pathway with a lower activation energy, meaning a greater proportion of particles have sufficient energy to react upon collision.

D

overall kinetic energy of the system.

The overall kinetic energy of the system is determined by its temperature, not by the presence of a catalyst.

Q1
2020
VCAA
1 mark
Q1
1 mark

Glycogen breaks down into

A

glycerol.

B

amino acids.

C

triglycerides.

D

monosaccharides.

Reveal Answer
A

glycerol.

Incorrect. Glycerol is a structural backbone of lipids like triglycerides, not a breakdown product of the carbohydrate glycogen.

B

amino acids.

Incorrect. Amino acids are the fundamental building blocks of proteins, whereas glycogen is a complex carbohydrate.

C

triglycerides.

Incorrect. Triglycerides are a type of lipid (fat) composed of glycerol and fatty acids, unrelated to the breakdown of glycogen.

D

monosaccharides.

Correct Answer

Correct. Glycogen is a complex carbohydrate (polysaccharide) that breaks down into simple sugars called monosaccharides, specifically glucose.

Q23
2024
VCAA
1 mark
Q23
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.

When the triglyceride is reacted with methanol to produce glycerol and Compound J

A

H2O\text{H}_2\text{O} should be added to the KOH directly, because KOH is corrosive and causes skin burns.

B

H2O\text{H}_2\text{O} should not be added to the reaction mixture because it interferes with the desired reaction.

C

H2O\text{H}_2\text{O} should be added to separate the triglyceride from glycerol since the triglyceride is soluble in water.

D

H2O\text{H}_2\text{O} should not be added to the reaction mixture because it increases the frequency of successful collisions.

Reveal Answer
A

H2O\text{H}_2\text{O} should be added to the KOH directly, because KOH is corrosive and causes skin burns.

Adding water to a strong base like KOH\text{KOH} is highly exothermic and dangerous. Furthermore, water actually interferes with the transesterification reaction.

B

H2O\text{H}_2\text{O} should not be added to the reaction mixture because it interferes with the desired reaction.

Correct Answer

Water causes a competing saponification reaction (hydrolysis), which converts the triglyceride into soap instead of the desired methyl ester (Compound J).

C

H2O\text{H}_2\text{O} should be added to separate the triglyceride from glycerol since the triglyceride is soluble in water.

Triglycerides are nonpolar molecules and are insoluble in water, so adding water would not dissolve them.

D

H2O\text{H}_2\text{O} should not be added to the reaction mixture because it increases the frequency of successful collisions.

Adding water would dilute the reaction mixture, which decreases the concentration of reactants and thereby decreases, rather than increases, the frequency of successful collisions.

Q35
2023
SCSA
14 marks
Q35

The Ostwald process is used in the conversion of ammonia to nitric acid according to the equations below.

Equation 1:
4 NH3(g)+5 O2(g)4 NO(g)+6 H2O(g)ΔH=905.2 kJ mol14\text{ NH}_3(g) + 5\text{ O}_2(g) \rightarrow 4\text{ NO}(g) + 6\text{ H}_2O(g) \quad \Delta H = -905.2\text{ kJ mol}^{-1}

Equation 2:
2 NO(g)+O2(g)2 NO2(g)ΔH=114.0 kJ mol12\text{ NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{ NO}_2(g) \quad \Delta H = -114.0\text{ kJ mol}^{-1}

Equation 3:
3 NO2(g)+H2O()2 HNO3(aq)+NO(g)ΔH=117.0 kJ mol13\text{ NO}_2(g) + \text{H}_2O(\ell) \rightarrow 2\text{ HNO}_3(aq) + \text{NO}(g) \quad \Delta H = -117.0\text{ kJ mol}^{-1}

Q35a
8 marks

The reaction in Equation 1 is carried out with a platinum-rhodium catalyst at approximately 850.0 °C and 1500 kPa. Using collision theory, account for these conditions.

Reveal Answer

The platinum-rhodium catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. This means a greater proportion of collisions are successful.

The high temperature of about 850.0 °C increases the kinetic energy of the reacting particles. The particles move faster, so collisions occur more frequently, and a greater proportion of collisions have energy greater than the activation energy. Therefore, more collisions are successful.

The high pressure of 1500 kPa forces the gas particles closer together. This reduces the space between particles and increases the frequency of collisions. As a result, the number of successful collisions per second increases, so the reaction rate increases.

Marking Criteria

Catalyst

DescriptorMarks

Recognises that rate is increased as the platinum-rhodium catalyst provides an alternate reaction pathway with a lower activation energy

1

Recognises that there are an increased proportion of reacting particles colliding with energy greater than the activation energy (resulting in an increased frequency of successful collisions)

1

Temperature

DescriptorMarks

Recognises that increased temperature increases the average kinetic energy of particles and they collide more frequently

1

Recognises that the increased proportion of collisions will have energy higher than the activation energy (which has been lowered due to the catalyst)

1

Recognises that a greater proportion of collisions are therefore successful (leading to an increased rate of reaction)

1

Pressure

DescriptorMarks

Recognises that increased pressure reduces the space between reacting particles

1

Recognises that at high pressure particles collide more frequently

1

Recognises that this leads to a higher frequency of successful collisions

1
Q35b
6 marks

A nitric acid plant requires a production of 1095 tonnes of nitric acid by means of the Ostwald process each day. If the conversion of ammonia to nitric acid is 77.65% efficient, calculate the volume of ammonia at standard temperature and pressure (STP) that must be fed into the process each day. Give your answer to the appropriate number of significant figures.

Reveal Answer

m(HNO3)required=1.095×109 gm(\text{HNO}_3)\text{required} = 1.095 \times 10^9 \text{ g}

n(HNO3)required=1.095×10963.018=1.738×107 moln(\text{HNO}_3)\text{required} = \frac{1.095 \times 10^9}{63.018} = 1.738 \times 10^7 \text{ mol}

n(NH3)required 100%=128×1.738×107=2.606×107 moln(\text{NH}_3)\text{required } 100\% = \frac{12}{8} \times 1.738 \times 10^7 = 2.606 \times 10^7 \text{ mol}

n(NH3)required 77.65%=10077.65×2.606×107=3.356×107 moln(\text{NH}_3)\text{required } 77.65\% = \frac{100}{77.65} \times 2.606 \times 10^7 = 3.356 \times 10^7 \text{ mol}

v(NH3)STP=3.356×107×22.71=7.621×108 Lv(\text{NH}_3)\text{STP} = 3.356 \times 10^7 \times 22.71 = 7.621 \times 10^8 \text{ L}

Marking Criteria
DescriptorMarks

m(HNO3)required=1.095×109 gm(\text{HNO}_3)\text{required} = 1.095 \times 10^9 \text{ g}

1

n(HNO3)required=1.095×10963.018=1.738×107 moln(\text{HNO}_3)\text{required} = \frac{1.095 \times 10^9}{63.018} = 1.738 \times 10^7 \text{ mol}

1

n(NH3)required 100%=128×1.738×107=2.606×107 moln(\text{NH}_3)\text{required } 100\% = \frac{12}{8} \times 1.738 \times 10^7 = 2.606 \times 10^7 \text{ mol}

1

n(NH3)required 77.65%=10077.65×2.606×107=3.356×107 moln(\text{NH}_3)\text{required } 77.65\% = \frac{100}{77.65} \times 2.606 \times 10^7 = 3.356 \times 10^7 \text{ mol}

1

v(NH3)STP=3.356×107×22.71=7.621×108 Lv(\text{NH}_3)\text{STP} = 3.356 \times 10^7 \times 22.71 = 7.621 \times 10^8 \text{ L}

1

Answer to 4 significant figures

1
Q1
2022
QCAA
Paper 1
1 mark
Q1
1 mark

Identify the type of reaction that occurs when ethene undergoes polymerisation to form polyethene.

A

addition

B

elimination

C

substitution

D

condensation

Reveal Answer
A

addition

Correct Answer

This is an addition reaction because the carbon-carbon double bonds (C=CC=C) in the ethene monomers break to form single bonds, allowing the molecules to add together into a long chain without any by-products.

B

elimination

Elimination reactions involve removing atoms to form a double or triple bond, which is the opposite of this process where double bonds are consumed to form single bonds.

C

substitution

Substitution reactions involve replacing one atom or group with another, whereas this reaction involves the joining of monomers through the rearrangement of bonds.

D

condensation

Condensation polymerization involves the joining of monomers with the release of a small molecule (like water), but the polymerization of ethene produces no by-products.

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

Frequently Asked Questions

How many QCAA Chemistry questions cover Chemical synthesis and design?
AusGrader has 182 QCAA Chemistry questions on Chemical synthesis and design, all with instant AI grading and detailed marking feedback.

Ready to practise QCAA Chemistry?

Get instant AI feedback on past exam questions, aligned to the syllabus

Start Practising Free