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QLDChemistry2025

QCE Chemistry 2025

Walkthrough of the 2025 QCE chemistry exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks
110
Time
180 min
Authority
QCAA
Updated
Open the official PDF Official marking notes
Key fact

This paper was set on the 2019 QCAA syllabus. From the 2026 external assessments the 2025 syllabus applies, so some content and the exam format have changed. It is still useful practice, but check the current QCAA Chemistry syllabus and QCAA's 2026 key changes before relying on it. The mark and timing figures at the top of this page are a guide based on the 2024 papers (two papers, each with 90 minutes of working time), not QCAA's official structure for every year. Marks and question types vary, so use QCAA's own paper.

What this paper assessed

The QCAA Chemistry external assessment is the final 50% summative external, covering Units 3 and 4 of the senior syllabus. The 2025 paper drew on:

  • Unit 3 - Equilibrium, acids and redox reactions: chemical equilibrium and the equilibrium constant KeqK_{eq}, Le Chatelier's principle, acids and bases (pH, KaK_a, pKapK_a, buffers), and oxidation-reduction including electrochemical and galvanic cells.
  • Unit 4 - Structure, synthesis and design: the chemistry of carbon compounds (functional groups, isomers and nomenclature), reactions and reaction pathways of organic molecules, polymers, and macromolecules including proteins and carbohydrates.

You are provided with the QCAA data and formula sheet and periodic table. The paper rewards balanced equations with state symbols, calculations that state the relationship before the arithmetic, and answers reported to the correct number of significant figures. Several items blend theory with practical technique (titration, identification tests, spectroscopy).

Structure and timing

QCAA sets the external assessment as more than one paper, and the number of marks, the sections and the question types differ from year to year. Open the official paper for the exact structure of this year; the marks and time shown at the top of this page are a guide only.

  • Pace by marks. Share your working time in proportion to the marks on offer, so a 6-mark item gets about three times the time of a 2-mark item. Flag any multiple choice item that runs long and return to it.
  • Use perusal time. Scan for the data-heavy items and the longest calculations or extended responses so you can plan them early.
  • Subject habits. Use the QCAA data booklet rather than recalling constants. For any calculation chain, label the quantity at each step so method marks are visible even if the final number is wrong, and keep a few minutes at the end to check state symbols, balancing, units and significant figures.

Worked practice questions (exam-style)

Worked example

Question 1 (5 marks): pH of a 0.050 mol L−10.050\ \text{mol L}^{-1} methanoic acid solution with Ka=1.8×10−4K_a = 1.8\times10^{-4}

Methanoic acid (HCOOH) is a weak acid: it only partially ionises in water, so the hydrogen-ion concentration must be found from the equilibrium expression rather than assumed equal to the initial concentration.

Step 1: Write the ionisation equilibrium and set up the ICE table

Methanoic acid donates a proton to water. Let x=[H+]x = [\text{H}^+] at equilibrium.

HCOOH⇌HCOO−+H+\text{HCOOH} \rightleftharpoons \text{HCOO}^- + \text{H}^+

At equilibrium: [HCOOH]=0.050−x[\text{HCOOH}] = 0.050 - x, and [HCOO−]=[H+]=x[\text{HCOO}^-] = [\text{H}^+] = x.

Step 2: Write and simplify the KaK_a expression

Because ionisation is small compared with the initial concentration, approximate 0.050−x≈0.0500.050 - x \approx 0.050.

Ka=x20.050−x≈x20.050K_a = \frac{x^2}{0.050 - x} \approx \frac{x^2}{0.050}

This approximation is valid when xx is less than about 5%5\% of 0.050.

Step 3: Solve for x=[H+]x = [\text{H}^+]

x2=(1.8×10−4)(0.050)=9.0×10−6x^2 = (1.8\times10^{-4})(0.050) = 9.0\times10^{-6}

x=[H+]=3.0×10−3 mol L−1x = [\text{H}^+] = 3.0\times10^{-3}\ \text{mol L}^{-1}

Step 4: Convert to pH

pH=−log⁡10(3.0×10−3)=2.52\text{pH} = -\log_{10}(3.0\times10^{-3}) = 2.52

Treating it as a strong acid would wrongly give [H+]=0.050[\text{H}^+] = 0.050 and pH = 1.30. The equilibrium approach gives the correct, higher pH because the acid only partially dissociates.

Final answer: pH =2.52= \mathbf{2.52}.

Worked example

Question 2 (4 marks): For the equilibrium 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g), ΔH<0\Delta H < 0. Predict and justify the effect on the yield of SO₃ of (a) increasing pressure and (b) increasing temperature.

(a) Increasing pressure: the system shifts to partially oppose the change by moving to the side with fewer moles of gas. The left has 3 mol of gas (2 + 1); the right has 2 mol. The equilibrium shifts right, increasing the yield of SO₃.

(b) Increasing temperature: the forward reaction is exothermic (ΔH<0\Delta H<0), so heat behaves as a product. Adding heat shifts the equilibrium left (the endothermic direction) to absorb it, decreasing the yield of SO₃ and lowering KeqK_{eq}.

The mark for justification requires naming the driver in each case - moles of gas for pressure, the exothermic direction for temperature - not just the direction of the shift.

Worked example

Question 3 (6 marks): Describe how you could chemically distinguish samples of propan-1-ol, propanoic acid, and propene, including a balanced equation for one test.

1. Identify propene (the alkene)
add bromine water. Propene rapidly decolourises orange bromine water by addition across the C=C bond:
CH3CH=CH2+Br2→CH3CHBrCH2Br\text{CH}_3\text{CH}{=}\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3\text{CHBrCH}_2\text{Br}

The alcohol and acid do not decolourise bromine water quickly.
2. Identify propanoic acid
add sodium carbonate solution. The carboxylic acid effervesces, releasing CO₂, because it is acidic enough to react with carbonate; the neutral alcohol does not.
3. The remaining sample is propan-1-ol
Confirm by warming with acidified dichromate, Cr2O72−\text{Cr}_2\text{O}_7^{2-}: the orange solution turns green as the primary alcohol is oxidised. A complete answer names the observation and what it identifies for each test.

Common errors students made

The 2025 QCAA marker report flagged students omitting state symbols, mixing up oxidising and reducing agents, and quoting Le Chatelier shifts without naming the driver; energy diagrams were often unlabelled or drawn without an activation-energy marker. Further recurring traps:

  • Significant-figure errors - reporting more figures than the least-precise datum allows, or rounding mid-calculation.
  • Using the strong-acid approximation for a weak acid, badly overestimating [H+][\text{H}^+] and underestimating pH.
  • Confusing oxidation and reduction - remember OIL RIG (oxidation is loss, reduction is gain of electrons) and identify the agent by what it does to the other species.
  • Reaction pathways missing conditions - label reagents and conditions above each arrow, not just the products.

How to use this paper

Sit the paper under timed exam conditions, then mark strictly against the official marking guide and marker report using the QCAA links at the top of this page. Rework every calculation that lost more than half its marks, writing the governing relationship on its own line before substituting and re-checking sig figs at the end. Keep an error log of recurring slips (state symbols, sig figs, weak-vs-strong acid, oxidant/reductant) and re-test the exact question class in your next session.

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