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SACE Chemistry Stage 2 — Booklet 2

SACE Chemistry Stage 2 — Booklet 2 — Free Online Pack 0

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SACE Stage 2 Examination - Booklets 1 and 2 2026 Edition - Pack 0 v2.0
Updated 30 Sep 2026

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Subject Outline alignment: Stage 2 Chemistry Subject Outline — reissued 2025; accredited from 2018.

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2026 exam timetable reference View SACE 2026 timetable
  • Written examination Wed 4 Nov 2026 9:00 am Upcoming9 days left

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SACE Chemistry Stage 2 — Booklet 2

4 questions

61 marks

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Covered in this pack

  • Current SACE Stage 2 Chemistry subject outline - Topic 1 - Analytical chemistry and monitoring 20 marks · 1 question
  • Current SACE Stage 2 Chemistry subject outline - Topic 1 - Environmental monitoring and atmospheric chemistry 18 marks · 1 question
  • Current SACE Stage 2 Chemistry subject outline - Topic 2 - Optimising chemical processes 21 marks · 1 question
  • Current SACE Stage 2 Chemistry subject outline - Topic 3 - Biological chemistry and biomolecules 15 marks · 1 question
  • Current SACE Stage 2 Chemistry subject outline - Topic 3 - Organic structure and reactions 14 marks · 1 question
  • Current SACE Stage 2 Chemistry subject outline - Topic 4 - Electrochemistry and redox applications 16 marks · 1 question
  • Current SACE Stage 2 Chemistry subject outline - Topic 4 - Materials, resources and sustainability 16 marks · 1 question

Read SACE Chemistry Stage 2 — Booklet 2 online

Skill Align

Skill Align SACE Stage 2 Chemistry - Free Online Pack 0 Booklet 2

An independently authored SACE Chemistry examination booklet with worked answers, atomic marking guidance and diagnostic review prompts.

Paper
Booklet 2
Reading
No separate reading time
Writing
Approximately 65 minutes
Assessment
61 marks

Approved calculators may be used. A Skill Align-formatted data sheet and periodic table based on the current SACE Chemistry reference material are supplied in this booklet.

Booklet 2 - Questions 4 to 7

Answer every question and every part. Show working for calculations, include units and use the supplied chemical evidence where required.

Question 4

14 marks
Stimulus

An algal oil is converted to a diol, HO-R-OH, where R is the algal-derived carbon skeleton between the two hydroxyl groups. The diol reacts with butanedioic acid, HOOC-CH₂-CH₂-COOH, to form a polyester. The repeat unit contains ester links. A 12.0 g batch yields 8.64 g purified polymer; the theoretical mass is 10.8 g.

Use the supplied algal polyester route evidence to answer all parts of this question.
(a) 4 marks
Represent a repeat unit from the algal-derived diol and butanedioic acid. Identify the ester links, the origin of each ester oxygen and the water removed during chain growth.
(b)(i) 2 marks
Identify the functional groups on the algal-derived diol and butanedioic acid that react to form the polyester.
(b)(ii) 2 marks
Explain how condensation between those functional groups creates ester links in the polymer backbone.
(c) 2 marks
Calculate the percentage yield of purified polyester from the actual and theoretical masses.
(d)(i) 2 marks
State what your calculated isolated yield establishes, then explain why that result alone cannot establish environmental superiority.
(d)(ii) 2 marks
Propose a controlled performance or end-of-life comparison between the algal polyester and a conventional polyester. Name the measurement and the variable that must be controlled.

Question 5

15 marks
Stimulus

A grape juice contains 180 g L⁻¹ glucose. Yeast converts glucose according to C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g). A 0.500 L batch consumes 75.0% of its glucose. M(glucose) = 180.16 g mol⁻¹.

Use the supplied grape-sugar fermentation evidence to answer all parts of this question.
(a)(i) 2 marks
State one structural feature of glucose and the mole ratio between glucose and ethanol in the supplied fermentation equation.
(a)(ii) 2 marks
Explain how the enzyme-catalysed anaerobic conditions and balanced equation describe glucose fermentation.
(b) 5 marks
Calculate the amount, in moles, and mass, in grams, of ethanol formed when 75.0% of the glucose in the batch is consumed.
(c) 3 marks
Use the balanced fermentation equation to trace glucose carbon into ethanol and carbon dioxide, and explain how yeast enzymes affect rate without changing the overall mole ratio.
(d)(i) 2 marks
A batch recovers less ethanol than the glucose stoichiometry predicts. Give one result that would support incomplete glucose conversion and one different route by which ethanol recovery could fall after glucose was consumed.
(d)(ii) 1 mark
Set up matched replicate fermentation batches. Specify how ethanol, residual glucose and carbon-dioxide measurements would test whether the predicted ethanol was formed and recovered.

Question 6

16 marks
Stimulus

A flow cell uses V²⁺/V³⁺ at the negative electrode and VO₂⁺/VO²⁺ in acidic solution at the positive electrode during discharge. The reduction half-equation is VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O. A current of 2.50 A flows for 38.6 min; use F = 96485 C mol⁻¹.

Use the supplied vanadium flow-cell evidence and question-specific visual to answer all parts of this question.
Diagram Preview
Use the labelled half-cell species to determine oxidation, reduction, and electron flow. external wire ion-selective membrane V²⁺(aq) | V³⁺(aq) VO₂⁺(aq), H⁺(aq) | VO²⁺(aq)
(a)(i) 1 mark
Write the vanadium oxidation half-equation at the negative electrode during discharge.
(a)(ii) 3 marks
Use the supplied positive-electrode half-equation and your oxidation half-equation to account for acid consumption, charge balance and the direction of electron flow during discharge.
(b) 4 marks
Trace one discharge through the two vanadium tanks: identify the species oxidised and reduced, match electron release to consumption, and give the external electron-flow direction. Leave current-to-amount arithmetic for Part (c).
(c) 4 marks
Calculate the charge transferred and the amount, in moles, of VO₂⁺ reduced during the 38.6-minute discharge.
(d) 4 marks
The reported discharge charge is smaller than the theoretical inventory charge. Propose a measurement sequence that separates active-species crossover from incomplete utilisation and explain how it would change an operating decision.

Question 7

16 marks
Stimulus

A wastewater plant precipitates struvite, MgNH₄PO₄·6H₂O, from a stream containing 0.0200 mol phosphate. The trial recovers 85.0% of the phosphate. One mole of struvite contains one mole of phosphorus; M(struvite) = 245.4 g mol⁻¹. The recovered struvite is intended for use as a land-applied fertiliser if it meets the required purity, contaminant and agronomic-quality specifications.

Use the supplied phosphorus-recovery trial evidence to answer all parts of this question.
(a) 3 marks
Explain how phosphate is transferred into struvite, identify the one-to-one phosphate-to-product relationship, and state the chemical or operational factors that affect recovery.
(b) 5 marks
Explain how magnesium availability, phosphate speciation and pH control struvite precipitation, then outline how the solid is separated and checked for purity. No unsupported net-ionic equation is required.
(c) 4 marks
Calculate the mass of struvite recovered and the amount, in moles, of phosphate left unrecovered.
(d) 4 marks
Make a justified resource or treatment choice for the phosphorus-recovery trial using the stated functional outcome, then evaluate one omitted lifecycle, quality, or field-performance factor.

SACE examinations are administered by the SACE Board of South Australia. Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by or endorsed by the SACE Board or the Government of South Australia.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank. Every question, dataset, visual, answer and marking description in this pack is original Skill Align content.

Worked Solutions And Marking Guide

General marking principles

  • Award each listed raw mark independently when the required chemical evidence is demonstrated.
  • Accept chemically correct equivalent equations, representations, working and wording.
  • A conclusion or Science as a Human Endeavour judgement must be supported by relevant evidence and acknowledge material limitations or trade-offs.

Question 4

(a) - From the supplied HO-R-OH, a suitable repeat unit is [-O-R-O-C(=O)-CH₂-CH₂-C(=O)-]ₙ, where R is the defined algal-derived carbon skeleton and ester links occur in the backbone. - The single-bond ester oxygen comes from a hydroxyl group of the algal-derived diol. - The carbonyl oxygen remains with the carboxylic-acid-derived acyl group. - For each condensation linkage, the acid's hydroxyl oxygen and an alcohol hydroxyl hydrogen leave together as H₂O.

(b)(i) - A diol has two hydroxyl functional groups. - Butanedioic acid has two carboxyl functional groups.

(b)(ii) - Condensation between -OH and -COOH forms an ester link. - Water is eliminated for each ester linkage formed.

(c) - Yield ratio = 8.64/10.8 = 0.800. - Percentage yield = 0.800 × 100 = 80.0%.

(d)(i) - The 80.0% result establishes the purified polymer mass relative to the stated theoretical mass. - The isolated yield alone does not establish environmental superiority or complete reaction.

(d)(ii) - Compare equal-thickness algal and conventional polyester samples under the same temperature, light and exposure time. - Measure strength retention, mass change or degradation products after the controlled exposure.

Detailed marking criteria

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: From the supplied HO-R-OH, a suitable repeat unit is [-O-R-O-C(=O)-CH₂-CH₂-C(=O)-]ₙ, where R is the defined algal-derived carbon skeleton and ester links occur in the backbone.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The single-bond ester oxygen comes from a hydroxyl group of the algal-derived diol.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The carbonyl oxygen remains with the carboxylic-acid-derived acyl group.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: For each condensation linkage, the acid's hydroxyl oxygen and an alcohol hydroxyl hydrogen leave together as H₂O.

Part (b)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: A diol has two hydroxyl functional groups.

Part (b)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Butanedioic acid has two carboxyl functional groups.

Part (b)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Condensation between -OH and -COOH forms an ester link.

Part (b)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Water is eliminated for each ester linkage formed.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Yield ratio = 8.64/10.8 = 0.800.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Percentage yield = 0.800 × 100 = 80.0%.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The 80.0% result establishes the purified polymer mass relative to the stated theoretical mass.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The isolated yield alone does not establish environmental superiority or complete reaction.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Compare equal-thickness algal and conventional polyester samples under the same temperature, light and exposure time.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Measure strength retention, mass change or degradation products after the controlled exposure.

Question 5

(a)(i) - Glucose is a monosaccharide containing multiple hydroxyl groups. - The supplied equation gives a one-to-two glucose-to-ethanol mole ratio.

(a)(ii) - Fermentation is enzyme-catalysed under anaerobic conditions. - C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g) is balanced for carbon, hydrogen and oxygen atoms.

(b) - Initial glucose mass = 180 × 0.500 = 90.0 g. - Consumed glucose mass = 0.750 × 90.0 = 67.5 g. - n(glucose) = 67.5/180.16 = 0.3747 mol. - n(ethanol) = 2 × 0.3747 = 0.749 mol. - Using M(ethanol) = 46.07 gives 34.5 g ethanol.

(c) - The required reaction representation is C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g). - The carbohydrate glucose is converted into the alcohol ethanol and carbon dioxide while conserving six carbon atoms. - Fermentation includes coupled oxidation and reduction steps mediated by yeast enzymes.

(d)(i) - Residual glucose would support incomplete conversion of the supplied glucose rather than complete fermentation. - Ethanol lost during collection or carbon diverted to other products could lower recovered ethanol even if glucose was consumed, so recovery alone cannot select a cause.

(d)(ii) - Run replicate batches at fixed temperature, pH and yeast loading, measure ethanol concentration by gas chromatography or a validated density method, and compare measured yield with residual glucose and carbon-dioxide data.

Detailed marking criteria

Part (a)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Glucose is a monosaccharide containing multiple hydroxyl groups.

Part (a)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The supplied equation gives a one-to-two glucose-to-ethanol mole ratio.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Fermentation is enzyme-catalysed under anaerobic conditions.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g) is balanced for carbon, hydrogen and oxygen atoms.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Initial glucose mass = 180 × 0.500 = 90.0 g.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Consumed glucose mass = 0.750 × 90.0 = 67.5 g.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: n(glucose) = 67.5/180.16 = 0.3747 mol.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: n(ethanol) = 2 × 0.3747 = 0.749 mol.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Using M(ethanol) = 46.07 gives 34.5 g ethanol.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The required reaction representation is C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g).

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The carbohydrate glucose is converted into the alcohol ethanol and carbon dioxide while conserving six carbon atoms.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Fermentation includes coupled oxidation and reduction steps mediated by yeast enzymes.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Residual glucose would support incomplete conversion of the supplied glucose rather than complete fermentation.

Part (d)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Ethanol lost during collection or carbon diverted to other products could lower recovered ethanol even if glucose was consumed, so recovery alone cannot select a cause.

Part (d)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Run replicate batches at fixed temperature, pH and yeast loading, measure ethanol concentration by gas chromatography or a validated density method, and compare measured yield with residual glucose and carbon-dioxide data.

Question 6

(a)(i) - The oxidation half-equation is V²⁺ → V³⁺ + e⁻.

(a)(ii) - The reduction half-equation consumes two H⁺ per electron. - Adding the half-equations conserves charge and electrons. - Electron flow is from the V²⁺ half-cell to the VO₂⁺ half-cell.

(b) - V²⁺ is oxidised to V³⁺ at the negative electrode during discharge. - VO₂⁺ is reduced to VO²⁺ at the positive electrode during discharge. - One mole of electrons is released per mole of V²⁺ oxidised and one mole of electrons is consumed per mole of VO₂⁺ reduced. - Electrons flow from the negative electrode through the external circuit to the positive electrode.

(c) - Charge Q = It = 2.50 × (38.6 × 60) = 5790 C. - n(e⁻) = 5790/96485 = 0.0600 mol. - The one-electron half-reaction gives 0.0600 mol VO₂⁺ reduced. - Time must be converted from minutes to seconds.

(d) - Before and after the same discharge, measure each tank's volume, total vanadium concentration and oxidation-state distribution separately to establish compartment-specific species inventories. - A loss of active vanadium from one compartment accompanied by a gain or misplaced vanadium species in the other supports membrane crossover. - Usable V²⁺ and VO₂⁺ remaining in their correct compartments at the discharge cutoff, without a corresponding transfer between tanks, support incomplete utilisation. - Membrane repair or replacement follows a crossover diagnosis; residual usable reactants instead support testing flow, mixing or discharge-cutoff settings.

Detailed marking criteria

Part (a)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The oxidation half-equation is V²⁺ → V³⁺ + e⁻.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: The reduction half-equation consumes two H⁺ per electron.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Adding the half-equations conserves charge and electrons.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Electron flow is from the V²⁺ half-cell to the VO₂⁺ half-cell.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: V²⁺ is oxidised to V³⁺ at the negative electrode during discharge.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: VO₂⁺ is reduced to VO²⁺ at the positive electrode during discharge.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: One mole of electrons is released per mole of V²⁺ oxidised and one mole of electrons is consumed per mole of VO₂⁺ reduced.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Electrons flow from the negative electrode through the external circuit to the positive electrode.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Charge Q = It = 2.50 × (38.6 × 60) = 5790 C.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: n(e⁻) = 5790/96485 = 0.0600 mol.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The one-electron half-reaction gives 0.0600 mol VO₂⁺ reduced.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Time must be converted from minutes to seconds.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Before and after the same discharge, measure each tank's volume, total vanadium concentration and oxidation-state distribution separately to establish compartment-specific species inventories.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: A loss of active vanadium from one compartment accompanied by a gain or misplaced vanadium species in the other supports membrane crossover.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Usable V²⁺ and VO₂⁺ remaining in their correct compartments at the discharge cutoff, without a corresponding transfer between tanks, support incomplete utilisation.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Membrane repair or replacement follows a crossover diagnosis; residual usable reactants instead support testing flow, mixing or discharge-cutoff settings.

Question 7

(a) - Struvite precipitation transfers dissolved phosphate into a solid product. - The one-to-one formula links phosphate moles to struvite moles. - Magnesium availability, pH, phosphate speciation, mixing and competing ions affect struvite recovery.

(b) - Precipitation occurs when the ionic product exceeds the solubility threshold. - Increasing suitable magnesium ion concentration can promote nucleation. - Controlled pH changes phosphate speciation and struvite formation. - Filtration separates the solid from treated water. - Wash the recovered solid and assay its phosphorus, magnesium and contaminant content before claiming struvite purity.

(c) - Recovered phosphate = 0.850 × 0.0200 = 0.0170 mol. - The stoichiometric struvite amount is also 0.0170 mol. - m(struvite) = 0.0170 × 245.4 = 4.17 g. - Unrecovered phosphate = 0.0200-0.0170 = 0.00300 mol.

(d) - Adopt struvite recovery only when the separated solid meets phosphorus-availability, contaminant and purity requirements for land application. - Use plant-available phosphorus delivered to soil as the functional unit. - Measure heavy metals, pathogens, crop uptake and runoff in a field trial. - Include magnesium reagent, aeration, pumping and dewatering in the lifecycle comparison.

Detailed marking criteria

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Struvite precipitation transfers dissolved phosphate into a solid product.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The one-to-one formula links phosphate moles to struvite moles.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Magnesium availability, pH, phosphate speciation, mixing and competing ions affect struvite recovery.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Precipitation occurs when the ionic product exceeds the solubility threshold.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Increasing suitable magnesium ion concentration can promote nucleation.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Controlled pH changes phosphate speciation and struvite formation.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Filtration separates the solid from treated water.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Wash the recovered solid and assay its phosphorus, magnesium and contaminant content before claiming struvite purity.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Recovered phosphate = 0.850 × 0.0200 = 0.0170 mol.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The stoichiometric struvite amount is also 0.0170 mol.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: m(struvite) = 0.0170 × 245.4 = 4.17 g.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Unrecovered phosphate = 0.0200-0.0170 = 0.00300 mol.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Adopt struvite recovery only when the separated solid meets phosphorus-availability, contaminant and purity requirements for land application.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Use plant-available phosphorus delivered to soil as the functional unit.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Measure heavy metals, pathogens, crop uptake and runoff in a field trial.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Include magnesium reagent, aeration, pumping and dewatering in the lifecycle comparison.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Topic 3 - Biological chemistry and biomolecules Q5 15 ___ Review carbohydrates, triglycerides, amino acids, proteins, hydrolysis, condensation and biological structure-function links.
Topic 3 - Organic structure and reactions Q4 14 ___ Review functional groups, naming, structure-property relationships, reactions, synthesis and polymer chemistry.
Topic 4 - Electrochemistry and redox applications Q6 16 ___ Review oxidation states, half-equations, cell conventions, electron flow, electrolysis and redox applications.
Topic 4 - Materials, resources and sustainability Q7 16 ___ Review energy, water, soil, materials, resource recovery, functional units and evidence-based sustainability judgements.

What is included

Booklet 1 questions (59 marks)

Booklet 2 questions (61 marks)

Worked solutions and marking guidance shown online

Diagnostic checklist shown online

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2026 exam timetable reference View SACE 2026 timetable
  • Written examination Wed 4 Nov 2026 9:00 am Upcoming9 days left

Times are South Australian local time. Check your school's day-by-day instructions and any special arrangements.