Skill Align QCE Chemistry - Free Online Pack 0 Paper 1
An independently authored QCE Chemistry practice paper with worked answers and question-specific marking guidance.
- Paper
- Paper 1 Combined Independent Practice Book
- Reading
- 5 minutes perusal
- Writing
- 90 minutes
- Assessment
- 55 marks
The official QCAA Chemistry 2025 formula and data book is supplied separately as “snr_chemistry_25_formula_data_book.pdf”. A QCAA-approved scientific or graphics calculator may be used.
Section 1 - Multiple-choice questions
Questions 1-20 are multiple choice. Select the best answer for each question. Each question is worth 1 mark.
Question 1
1 mark- Kc = [N₂][H₂]³/[NH₃]²
- Kc = [NH₃]²/([N₂][H₂]³)
- Kc = 2[NH₃]/([N₂] + 3[H₂])
- Kc = [NH₃]/([N₂][H₂])
Question 2
1 mark- Net reverse reaction occurs until Qc equals zero.
- The equilibrium constant increases to 2.8.
- No change occurs because both values are positive.
- Net forward reaction occurs until Qc equals Kc.
Question 3
1 mark- towards NH₃ because that side has fewer gas particles.
- towards N₂ and H₂ because their concentrations decrease.
- nowhere because pressure never affects equilibrium.
- towards NH₃ because Kc increases.
Question 4
1 mark- HCO₃⁻ reacts with added H₃O⁺ to form H₂CO₃ and water.
- H₂CO₃ completely dissociates to make more H₃O⁺.
- Chloride precipitates as solid HCl.
- The value of Kw becomes zero.
Question 5
1 mark- 0.07813 mol L⁻¹
- 0.1250 mol L⁻¹
- 0.1000 mol L⁻¹
- 0.1563 mol L⁻¹
Question 6
1 mark- +4
- +7
- +2
- −1
Question 7
1 mark- +1.10 V
- −1.10 V
- +0.42 V
- +0.76 V
Question 8
1 mark- reduction occurs and electrons leave the circuit.
- anions are always produced.
- the more positive reduction half-equation occurs.
- oxidation occurs and electrons enter the external circuit.
Question 9
1 mark- 0.0100 mol
- 0.0400 mol
- 0.0200 mol
- 1.00 mol
Question 10
1 mark- water.
- hydrogen peroxide only.
- solid oxygen.
- chloride ions.
Question 11
1 mark- C₅H₁₂
- C₅H₈
- C₅H₁₂O
- C₅H₁₀
Question 12
1 mark- propanone.
- propanal.
- propanoic acid only, with no intermediate.
- propene.
Question 13
1 mark- an ester.
- an alcohol.
- an alkane.
- an amine.
Question 14
1 mark- 44
- 89
- 88
- 176
Question 15
1 mark- 1.67
- 2.4
- 0.40
- 0.60
Question 16
1 mark- the catalyst mass, even though the catalyst is not consumed.
- the reactant consumed first according to the balanced-equation mole ratio.
- the reactant with the largest starting mass in every case.
- the final product mass measured after purification.
Question 17
1 mark- 85.0%
- 17.6%
- 117.6%
- 6.30%
Question 18
1 mark- condense and eliminate water.
- oxidise to carboxylic acids.
- undergo addition polymerisation as C=C bonds open.
- form peptide bonds.
Question 19
1 mark- glycosidic bond.
- carbon–carbon double bond.
- metallic bond.
- peptide (amide) bond.
Question 20
1 mark- an addition reaction that consumes oxygen.
- a condensation reaction that forms water.
- electrolysis that releases hydrogen.
- a redox reaction that forms a metal.
Section 2 - Short-response questions
Questions 21-29 are short response. Answer all parts, show calculations and use the supplied chemical evidence where required.
Question 21
2 marksSulfur dioxide is oxidised according to 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Question 22
2 marksA reef-cleaning solution contains 3.20 × 10⁻³ mol L⁻¹ HCl at 25 °C.
Question 23
2 marksUnder standard conditions, a galvanic cell uses Mg/Mg²⁺ and Cu²⁺/Cu half-cells.
Question 24
4 marksEqual volumes of 4.0 × 10⁻⁴ mol L⁻¹ Ca(NO₃)₂ and 6.0 × 10⁻⁴ mol L⁻¹ NaF are mixed at 25 °C. For CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), Ksp = 3.9 × 10⁻¹¹.
Question 25
4 marksEthene is converted to ethanoic acid in two stages: CH₂=CH₂ → CH₃CH₂OH → CH₃COOH.
Question 26
3 marksAmino acids from a reef-protein hydrolysate are separated on chromatography paper using one mobile phase. The solvent front travels 8.0 cm; the glycine spot travels 2.4 cm and the leucine spot travels 6.0 cm.
Question 27
4 marksAt 20 MPa, the equilibrium percentage of NH₃ is 42% at 350 °C, 28% at 450 °C and 18% at 550 °C. The forward reaction N₂ + 3H₂ ⇌ 2NH₃ is exothermic.
Question 28
8 marksA 25.00 mL aquarium-water sample consumes H⁺ in a 1:1 amount relation to its acid-neutralising capacity. With 0.1000 mol L⁻¹ HCl, three titres are 18.62, 18.58 and 18.60 mL. The accepted mean volume has an uncertainty of ±0.04 mL; treat all other quantities as exact for this question. The equivalence point occurs at pH 5.0, while an indicator changes colour at pH 4.7. No individual reacting species is identified.
Question 29
6 marksA reef-monitoring connector is copper-plated from Cu²⁺ solution. A current of 2.50 A flows for 32.0 minutes at 92.0% current efficiency. Cu²⁺ + 2e⁻ → Cu(s). Use F = 96485 C mol⁻¹ and M(Cu) = 63.55 g mol⁻¹.
Worked Solutions And Marking Guide
General marking principles
- Award each listed mark independently when the required chemical evidence is demonstrated.
- Accept chemically accurate equivalent wording, balanced equations and logically equivalent calculations.
- Apply follow-through where a prior numerical error is used consistently and the subsequent chemical method is correct.
- Do not award conclusions that contradict the supplied data or rely on unstated evidence.
Section 1 Question 1
Answer: Kc = [NH₃]²/([N₂][H₂]³)
Concentration terms are raised to their stoichiometric coefficients.
Section 1 Question 2
Answer: Net forward reaction occurs until Qc equals Kc.
When Qc is less than Kc, products form by net forward reaction.
Section 1 Question 3
Answer: towards NH₃ because that side has fewer gas particles.
The system favours the side with fewer moles of gas; Kc is unchanged at constant temperature.
Section 1 Question 4
Answer: HCO₃⁻ reacts with added H₃O⁺ to form H₂CO₃ and water.
The conjugate base consumes much of the added hydronium ion.
Section 1 Question 5
Answer: 0.1000 mol L⁻¹
The 1:1 reaction uses 0.01600 × 0.1250 = 0.002000 mol acid; dividing by 0.02000 L gives 0.1000 mol L⁻¹.
Section 1 Question 6
Answer: +7
Four oxygen atoms contribute −8 overall, so manganese must be +7 to give a −1 ion.
Section 1 Question 7
Answer: +1.10 V
Subtracting the anode reduction potential from the cathode value gives E°cell = 0.34 − (−0.76) = +1.10 V.
Section 1 Question 8
Answer: oxidation occurs and electrons enter the external circuit.
Oxidation at the anode supplies electrons to the external circuit.
Section 1 Question 9
Answer: 0.0200 mol
Ag⁺ + e⁻ → Ag has a one-to-one electron-to-silver ratio.
Section 1 Question 10
Answer: water.
Combining hydrogen oxidation and oxygen reduction gives 2H₂ + O₂ → 2H₂O.
Section 1 Question 11
Answer: C₅H₁₀
Acyclic alkenes with one C=C bond have general formula CnH₂n.
Section 1 Question 12
Answer: propanal.
A primary alcohol is first oxidised to the corresponding aldehyde.
Section 1 Question 13
Answer: an ester.
An ester contains both C=O and C–O bonds without an O–H bond.
Section 1 Question 14
Answer: 88
For a singly charged molecular ion, m/z corresponds approximately to relative molecular mass.
Section 1 Question 15
Answer: 0.60
The retention factor is the alanine distance divided by the solvent-front distance: Rf = 3.6/6.0 = 0.60.
Section 1 Question 16
Answer: the reactant consumed first according to the balanced-equation mole ratio.
The limiting reactant is exhausted first under the equation stoichiometry and therefore fixes the theoretical product amount.
Section 1 Question 17
Answer: 85.0%
Percentage yield = 35.7/42.0 × 100 = 85.0%.
Section 1 Question 18
Answer: undergo addition polymerisation as C=C bonds open.
The alkene double bond opens and monomers link without a small-molecule by-product.
Section 1 Question 19
Answer: peptide (amide) bond.
The amino and carboxyl groups join to form an amide linkage and water.
Section 1 Question 20
Answer: a condensation reaction that forms water.
A glycosidic link forms with elimination of water.
Section 2 Question 21
(a) - Writes Kc = [SO₃]²/([SO₂]²[O₂]).
(b) - States that net forward reaction forms SO₃ until a new equilibrium is reached.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Writes Kc = [SO₃]²/([SO₂]²[O₂]).
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: States that net forward reaction forms SO₃ until a new equilibrium is reached.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 22
(a) - Uses pH = −log₁₀(3.20 × 10⁻³). - Obtains pH = 2.49 to appropriate precision.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Uses pH = −log₁₀(3.20 × 10⁻³).
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Obtains pH = 2.49 to appropriate precision.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 23
(a) - Identifies Mg as the anode where oxidation occurs. - States that electrons flow from Mg to Cu through the external circuit.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Identifies Mg as the anode where oxidation occurs.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: States that electrons flow from Mg to Cu through the external circuit.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 24
(a) - Accounts for dilution to obtain [Ca²⁺] = 2.0 × 10⁻⁴ mol L⁻¹ and [F⁻] = 3.0 × 10⁻⁴ mol L⁻¹. - Uses Qsp = [Ca²⁺][F⁻]². - Obtains Qsp = 1.8 × 10⁻¹¹.
(b) - Concludes no precipitate forms because Qsp < Ksp.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Accounts for dilution to obtain [Ca²⁺] = 2.0 × 10⁻⁴ mol L⁻¹ and [F⁻] = 3.0 × 10⁻⁴ mol L⁻¹.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Uses Qsp = [Ca²⁺][F⁻]².
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Obtains Qsp = 1.8 × 10⁻¹¹.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Concludes no precipitate forms because Qsp < Ksp.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 25
(a) - Identifies addition (hydration). - States steam/water with an acid catalyst.
(b) - Identifies oxidation of a primary alcohol. - States that an acidified oxidising agent and heating under reflux can produce ethanoic acid.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Identifies addition (hydration).
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: States steam/water with an acid catalyst.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Identifies oxidation of a primary alcohol.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: States that an acidified oxidising agent and heating under reflux can produce ethanoic acid.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 26
(a) - Calculates Rf(glycine) = 0.30. - Calculates Rf(leucine) = 0.75.
(b) - Infers that glycine is more strongly retained by the paper than leucine under these conditions.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Calculates Rf(glycine) = 0.30.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Calculates Rf(leucine) = 0.75.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Infers that glycine is more strongly retained by the paper than leucine under these conditions.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 27
(a) - States that equilibrium NH₃ percentage decreases as temperature increases.
(b) - Uses the exothermic forward reaction/endothermic reverse reaction. - Explains that higher temperature favours the endothermic reverse direction.
(c) - Explains that lower temperature gives a slower reaction rate and therefore lower throughput.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: States that equilibrium NH₃ percentage decreases as temperature increases.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Uses the exothermic forward reaction/endothermic reverse reaction.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Explains that higher temperature favours the endothermic reverse direction.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Explains that lower temperature gives a slower reaction rate and therefore lower throughput.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 28
(a) - Calculates mean titre = 18.60 mL. - Calculates n(H⁺) = 0.1000 × 0.01860 = 1.860 × 10⁻³ mol.
(b) - Divides the H⁺ amount by 0.02500 L. - Obtains 0.07440 mol L⁻¹ of H⁺ equivalents.
(c) - Uses 18.56 mL to obtain 0.07424 mol L⁻¹. - Uses 18.64 mL to obtain 0.07456 mol L⁻¹.
(d) - The equivalence point at pH 5.0 is where the H⁺ amount matches the neutralising capacity stoichiometrically. - The endpoint at pH 4.7 is the observed indicator colour change; it need not coincide exactly with equivalence.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Calculates mean titre = 18.60 mL.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Calculates n(H⁺) = 0.1000 × 0.01860 = 1.860 × 10⁻³ mol.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Divides the H⁺ amount by 0.02500 L.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Obtains 0.07440 mol L⁻¹ of H⁺ equivalents.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Uses 18.56 mL to obtain 0.07424 mol L⁻¹.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Uses 18.64 mL to obtain 0.07456 mol L⁻¹.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part d (1 mark)
Awards one mark for accurately establishing this question-specific point: The equivalence point at pH 5.0 is where the H⁺ amount matches the neutralising capacity stoichiometrically.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part d (1 mark)
Awards one mark for accurately establishing this question-specific point: The endpoint at pH 4.7 is the observed indicator colour change; it need not coincide exactly with equivalence.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Section 2 Question 29
(a) - Converts time to 1920 s. - Calculates total charge as 4800 C. - Applies 92.0% efficiency to obtain 4416 C.
(b) - Calculates n(e⁻) = 4416/96485 = 0.0458 mol. - Uses the 2:1 electron-to-copper ratio to obtain 0.0229 mol Cu. - Obtains approximately 1.45 g Cu.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Converts time to 1920 s.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Calculates total charge as 4800 C.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Applies 92.0% efficiency to obtain 4416 C.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Calculates n(e⁻) = 4416/96485 = 0.0458 mol.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Uses the 2:1 electron-to-copper ratio to obtain 0.0229 mol Cu.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Obtains approximately 1.45 g Cu.
Acceptable alternatives: Accept chemically equivalent wording or a logically equivalent calculation.
Do not credit by itself: Do not award an unsupported conclusion or a response that contradicts the supplied stimulus.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 3 Topic 1 - Chemical equilibrium systems | Q1, Q2, Q3, Q4, Q5, Q21, Q22, Q24, Q28 | 21 | ___ | Review equilibrium, Kc, Q, Ksp, acids, bases, buffers, indicators and volumetric analysis. |
| Unit 3 Topic 2 - Oxidation and reduction | Q6, Q7, Q8, Q9, Q10, Q23, Q29 | 13 | ___ | Review oxidation states, half-equations, galvanic and electrolytic cells, cell potentials and electrolysis calculations. |
| Unit 4 Topic 1 - Properties and structure of organic materials | Q11, Q12, Q13, Q14, Q15, Q25, Q26 | 12 | ___ | Review organic structure, nomenclature, isomerism, reactions, properties, spectra, chromatography and electrophoresis. |
| Unit 4 Topic 2 - Chemical synthesis and design | Q16, Q17, Q18, Q19, Q20, Q27 | 9 | ___ | Review Haber and contact processes, ethanol production, fuel cells, yields, polymers, peptides and carbohydrates. |