Skill Align

Skill Align SACE Stage 2 Chemistry - Free Online Pack 0 Booklet 1

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

Paper: Booklet 1
Reading: No separate reading time
Writing: Approximately 65 minutes
Assessment: 59 marks

A calculator may be used. The SACE Chemistry data sheet and periodic table are reproduced as source-controlled reference pages in this booklet.

Booklet 1 - Questions 1 to 3

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

Question 1

18 marks

Stimulus

Three monitoring stations recorded afternoon ozone concentrations of 28, 46 and 71 ppb from the city centre to a downwind foothills site. NO₂ photolysis initiates ozone formation. The local advisory trigger is 65 ppb, and the final value has an uncertainty of ±4 ppb.

Use the supplied Adelaide ozone-transect evidence and question-specific visual to answer all parts of this question.

Graph Preview

(a)(i) 2 marks

State the principal measured pattern in the supplied Adelaide ozone-transect data and visual and identify the measurements used.

(a)(ii) 3 marks

Use the relevant values and units to interpret what the observations establish about the Adelaide ozone-transect, including any stated threshold, control or uncertainty.

(b) 2 marks

Calculate the percentage increase in ozone concentration from the city-centre station at 28 ppb to the downwind foothills station at 71 ppb.

(c) 5 marks

Write or use the environmental reaction supplied for the Adelaide ozone-transect, then connect the species-level process to the measured trend.

(d) 6 marks

Use the monitoring evidence to reach a justified conclusion about the Adelaide ozone-transect, propose an operational response, and state the uncertainty or sampling limitation that controls when that response is warranted.

Answer

(a)(i) - Ozone rises by 43 ppb across the transect, from 28 to 71 ppb. - The downwind result exceeds the 65 ppb trigger by 6 ppb.

(a)(ii) - The ±4 ppb interval is 67-75 ppb and remains above the trigger. - Transport time allows sunlight-driven reactions to continue downwind. - NO₂ absorbs light and dissociates to NO and an oxygen atom.

(b) - The absolute increase is 71-28 = 43 ppb. - The percentage increase is 43/28 × 100 = 154%, which is dimensionless.

(c) - The initiating step is NO₂ + light → NO + O. - The subsequent step is O + O₂ → O₃. - Volatile organic compounds convert NO back to NO₂ without consuming ozone. - This reaction cycle permits net ozone accumulation downwind. - Strong sunlight increases the photolysis rate and ozone-forming radical activity.

(d) - The evidence supports an afternoon ozone advisory at the foothills site. - The complete uncertainty interval remains above the operational trigger. - One afternoon cannot establish the frequency of high-ozone events. - Repeat measurements across seasons and comparable wind conditions. - Co-locate NOx and VOC measurements to test the proposed mechanism. - Public warnings reduce exposure while longer-term precursor controls address formation.

Question 2

20 marks

Stimulus

A 25.00 mL wine sample required 12.40 mL of 0.01000 mol L⁻¹ I₂. The reaction is I₂ + SO₃²⁻ + H₂O → 2I⁻ + SO₄²⁻ + 2H⁺. A second aliquot gave 12.46 mL. Assume the sample preparation converts all free sulfite to SO₃²⁻.

Use the supplied winery sulfite titration evidence and question-specific visual to answer all parts of this question.

Graph Preview

(a) 5 marks

Identify the analytical relationship, aliquot or calibration data, stoichiometric ratio, and stated assumption needed to interpret the supplied winery sulfite titration evidence.

(b) 5 marks

Calculate the mean of the two concordant iodine titres, then calculate the sulfite concentration in the 25.00 mL wine sample in mol L⁻¹.

(c)(i) 1 mark

Write or draw the chemical equation, structural formula, half-equation, equilibrium expression, or process representation relevant to the winery sulfite titration.

(c)(ii) 4 marks

Use that representation to explain the chemical behaviour and measured outcome in the winery sulfite titration.

(d) 5 marks

Assess whether the winery sulfite titration method supports the reported result. Identify a question-specific source of bias and propose a control, calibration, or repeat measurement that would test it.

Answer

(a) - The two titres differ by only 0.06 mL and are concordant. - The mean of both concordant titres is the appropriate analytical volume. - I₂ and SO₃²⁻ react in a one-to-one mole ratio. - Both titres refer to separate 25.00 mL aliquots under the same stated conditions. - The close agreement supports repeatability of the volume measurement.

(b) - Mean titre = (12.40 + 12.46)/2 = 12.43 mL. - n(I₂) = 0.01000 × 0.01243 = 1.243 × 10⁻⁴ mol. - The one-to-one equation gives 1.243 × 10⁻⁴ mol SO₃²⁻. - c(SO₃²⁻) = 1.243 × 10⁻⁴ / 0.02500 = 0.004972 mol L⁻¹. - The mean titre is converted from 12.43 mL to 0.01243 L before substitution.

(c)(i) - Sulfur is oxidised from +4 in sulfite to +6 in sulfate.

(c)(ii) - Iodine is reduced from 0 in I₂ to -1 in iodide. - The balanced equation conserves two iodine atoms and one sulfur atom. - Two electrons lost by sulfur match two electrons gained by iodine. - Acid is a product, consistent with the two H⁺ shown.

(d) - The method gives a precise estimate of free sulfite for this prepared sample. - Oxidation of sulfite by air before titration would bias the result low. - Analyse promptly in a closed vessel to reduce oxygen exposure. - A reagent blank can correct for iodine consumed by other reductants. - Certified sulfite reference material would test accuracy.

Question 3

21 marks

Stimulus

A pilot Haber loop uses N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ mol⁻¹ as written. It operates at high pressure and a moderate compromise temperature with an iron catalyst. A single pass converts 18.0% of 250 mol N₂. Ammonia is condensed before unreacted gases are recycled, and the hydrogen is supplied by electrolysis using contracted renewable electricity.

Use the supplied green-ammonia synthesis loop evidence to answer all parts of this question.

(a) 5 marks

Write the reaction or equilibrium representation relevant to the green-ammonia synthesis loop. Use collision, equilibrium, catalyst, or process chemistry to explain the effect of the stated conditions.

(b) 5 marks

Calculate the mass of ammonia, in kilograms, formed in one pass from the stated nitrogen feed and 18.0% conversion.

(c) 5 marks

Identify the reaction or process features established by the supplied green-ammonia synthesis loop evidence, including the relevant stoichiometry and effect of the stated operating conditions.

(d) 6 marks

Assess the stated operating choices for the green-ammonia synthesis loop. Explain one rate-yield or energy-separation trade-off and identify the additional plant evidence needed before changing conditions.

Answer

(a) - Kc = [NH₃]²/([N₂][H₂]³) for the stated gas reaction. - Removing ammonia shifts the equilibrium towards further ammonia formation by Le Chatelier's principle. - Compressing the mixture also favours the side with fewer gas moles. - Very low temperature would reduce production rate despite improving equilibrium yield. - The iron catalyst provides a lower-activation-energy pathway.

(b) - The theoretical ammonia amount is 2 × 45.0 = 90.0 mol. - M(NH₃) = 14.01 + 3(1.008) = 17.034 g mol⁻¹. - m(NH₃) = 90.0 × 17.034 = 1.53 × 10³ g. - The answer is 1.53 kg NH₃ to three significant figures. - Stoichiometry uses two moles NH₃ per mole N₂ consumed.

(c) - One pass reacts 45.0 mol N₂ because 0.180 × 250 = 45.0. - The equation predicts formation of 90.0 mol NH₃. - The reaction changes four moles of gas reactant to two moles of gas product. - High pressure favours ammonia because it favours fewer gas particles. - Lower temperature favours the exothermic forward reaction.

(d) - A moderate temperature balances acceptable rate against equilibrium yield. - High pressure improves yield but raises compression energy and equipment demands. - Condensing ammonia enables reactant recycle and shifts the equilibrium. - Renewable hydrogen lowers feed-related emissions only if electricity is genuinely low-carbon. - A functional comparison should use emissions per tonne of ammonia produced. - Plant decisions must consider safety, energy, capital cost and verified lifecycle emissions.

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 1

(a)(i) - Ozone rises by 43 ppb across the transect, from 28 to 71 ppb. - The downwind result exceeds the 65 ppb trigger by 6 ppb.

(b) - The absolute increase is 71-28 = 43 ppb. - The percentage increase is 43/28 × 100 = 154%, which is dimensionless.

Detailed marking criteria

Part (a)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Ozone rises by 43 ppb across the transect, from 28 to 71 ppb.

Part (a)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: The downwind result exceeds the 65 ppb trigger by 6 ppb.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: The ±4 ppb interval is 67-75 ppb and remains above the trigger.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Transport time allows sunlight-driven reactions to continue downwind.

Part (a)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: NO₂ absorbs light and dissociates to NO and an oxygen atom.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The absolute increase is 71-28 = 43 ppb.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The percentage increase is 43/28 × 100 = 154%, which is dimensionless.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The initiating step is NO₂ + light → NO + O.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The subsequent step is O + O₂ → O₃.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Volatile organic compounds convert NO back to NO₂ without consuming ozone.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: This reaction cycle permits net ozone accumulation downwind.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Strong sunlight increases the photolysis rate and ozone-forming radical activity.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: The evidence supports an afternoon ozone advisory at the foothills site.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: The complete uncertainty interval remains above the operational trigger.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: One afternoon cannot establish the frequency of high-ozone events.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Repeat measurements across seasons and comparable wind conditions.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Co-locate NOx and VOC measurements to test the proposed mechanism.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Public warnings reduce exposure while longer-term precursor controls address formation.

Question 2

(c)(i) - Sulfur is oxidised from +4 in sulfite to +6 in sulfate.

Detailed marking criteria

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The two titres differ by only 0.06 mL and are concordant.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The mean of both concordant titres is the appropriate analytical volume.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: I₂ and SO₃²⁻ react in a one-to-one mole ratio.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Both titres refer to separate 25.00 mL aliquots under the same stated conditions.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The close agreement supports repeatability of the volume measurement.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Mean titre = (12.40 + 12.46)/2 = 12.43 mL.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: n(I₂) = 0.01000 × 0.01243 = 1.243 × 10⁻⁴ mol.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The one-to-one equation gives 1.243 × 10⁻⁴ mol SO₃²⁻.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: c(SO₃²⁻) = 1.243 × 10⁻⁴ / 0.02500 = 0.004972 mol L⁻¹.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The mean titre is converted from 12.43 mL to 0.01243 L before substitution.

Part (c)(i) (1 mark)

Awards one mark for establishing this question-specific chemical point: Sulfur is oxidised from +4 in sulfite to +6 in sulfate.

Part (c)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Iodine is reduced from 0 in I₂ to -1 in iodide.

Part (c)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: The balanced equation conserves two iodine atoms and one sulfur atom.

Part (c)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Two electrons lost by sulfur match two electrons gained by iodine.

Part (c)(ii) (1 mark)

Awards one mark for establishing this question-specific chemical point: Acid is a product, consistent with the two H⁺ shown.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: The method gives a precise estimate of free sulfite for this prepared sample.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Oxidation of sulfite by air before titration would bias the result low.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Analyse promptly in a closed vessel to reduce oxygen exposure.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: A reagent blank can correct for iodine consumed by other reductants.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Certified sulfite reference material would test accuracy.

Question 3

Detailed marking criteria

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Kc = [NH₃]²/([N₂][H₂]³) for the stated gas reaction.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Removing ammonia shifts the equilibrium towards further ammonia formation by Le Chatelier's principle.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Compressing the mixture also favours the side with fewer gas moles.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: Very low temperature would reduce production rate despite improving equilibrium yield.

Part a (1 mark)

Awards one mark for establishing this question-specific chemical point: The iron catalyst provides a lower-activation-energy pathway.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The theoretical ammonia amount is 2 × 45.0 = 90.0 mol.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: M(NH₃) = 14.01 + 3(1.008) = 17.034 g mol⁻¹.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: m(NH₃) = 90.0 × 17.034 = 1.53 × 10³ g.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: The answer is 1.53 kg NH₃ to three significant figures.

Part b (1 mark)

Awards one mark for establishing this question-specific chemical point: Stoichiometry uses two moles NH₃ per mole N₂ consumed.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: One pass reacts 45.0 mol N₂ because 0.180 × 250 = 45.0.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The equation predicts formation of 90.0 mol NH₃.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: The reaction changes four moles of gas reactant to two moles of gas product.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: High pressure favours ammonia because it favours fewer gas particles.

Part c (1 mark)

Awards one mark for establishing this question-specific chemical point: Lower temperature favours the exothermic forward reaction.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: A moderate temperature balances acceptable rate against equilibrium yield.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: High pressure improves yield but raises compression energy and equipment demands.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Condensing ammonia enables reactant recycle and shifts the equilibrium.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Renewable hydrogen lowers feed-related emissions only if electricity is genuinely low-carbon.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: A functional comparison should use emissions per tonne of ammonia produced.

Part d (1 mark)

Awards one mark for establishing this question-specific chemical point: Plant decisions must consider safety, energy, capital cost and verified lifecycle emissions.

Diagnostic Checklist

Topic	Questions	Marks	Marks Lost	Action
Topic 1 - Analytical chemistry and monitoring	Q2	20	___	Review volumetric analysis, chromatography, atomic spectroscopy, calibration, uncertainty and fit-for-purpose monitoring.
Topic 1 - Environmental monitoring and atmospheric chemistry	Q1	18	___	Review atmospheric reactions, greenhouse and photochemical-smog chemistry, monitoring evidence and environmental interpretation.
Topic 2 - Optimising chemical processes	Q3	21	___	Review yield, rate, catalysts, energy, separation, recycling and process-condition trade-offs.

SACE Chemistry Stage 2 — Booklet 1 — Free Online Pack 0

Exam-pack paper structure

Read SACE Chemistry Stage 2 — Booklet 1 online

Skill Align SACE Stage 2 Chemistry - Free Online Pack 0 Booklet 1

Booklet 1 - Questions 1 to 3

Question 1

Question 2

Question 3

Worked Solutions And Marking Guide

General marking principles

Question 1

Detailed marking criteria

Question 2

Detailed marking criteria

Question 3

Detailed marking criteria

Diagnostic Checklist

What is included

Independent practice resource

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Questions about this exam pack