§-Past paper

VICChemistry2024

VCE Chemistry 2024

Walkthrough of the 2024 VCE chemistry exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks: 120
Time: 150 min
Authority: VCAA
Updated: 2026-05-29

Open the official PDF ->Official marking notes

What this paper assessed

This VCAA Chemistry Units 3 & 4 examination is a 120-mark paper sat over 150 minutes (plus 15 minutes reading time). It assesses:

Unit 3 - How can design and innovation help to optimise chemical processes? Chemical energy (fuels, calorimetry, $\Delta H$ , energy profile diagrams), redox chemistry (galvanic cells, fuel cells, electrolysis, the electrochemical series), and rate and extent of reaction (collision theory, equilibrium and $K$ , Le Chatelier's principle).
Unit 4 - How are organic compounds categorised, analysed and used? The structure, nomenclature and reactions of organic families (functional groups, isomerism, reaction pathways such as substitution, addition, esterification, oxidation), analysis using instrumental techniques (mass spectrometry, IR, ¹H and ¹³C NMR, chromatography), and the chemistry of food (carbohydrates, proteins, fats, and titration/volumetric analysis).

The paper has Section A (multiple choice) and Section B (short and extended response) and is calculation-heavy. The 2024 paper leaned harder on application where the context was unfamiliar but the skill was core.

Structure and timing

120 marks in 150 minutes is 1.25 minutes per mark:

Section A (~30 MC marks): ~35-40 minutes.
Section B (~90 marks): ~110 minutes at ~1.25 min/mark. A 6-mark spectroscopy or multi-step stoichiometry item deserves ~7-8 minutes.
Reserve the final ~10-15 minutes for unit and significant-figure checks.

Use the 15-minute reading time to locate any structure-determination and multi-step calculation items.

Worked practice questions (exam-style)

Worked example

Question 1 (5 marks): 25.0 mL of a sodium hydroxide solution is exactly neutralised by 21.4 mL of 0.105 mol L⁻¹ sulfuric acid. Calculate the concentration of the NaOH.

This is a titration calculation. The key is using the balanced equation to relate the moles of acid to the moles of base via the stoichiometric ratio, then dividing by the volume of base to find the concentration.

Step 1: Write the balanced equation

H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l).

The ratio of acid to base is 1:2, which we will use in Step 3.

Step 2: Calculate moles of sulfuric acid

Use $n = cV$ , with volume in litres:

n(H_2SO_4) = 0.105 \times 0.0214 = 2.247\times10^{-3}\ \text{mol}.

Step 3: Find moles of NaOH using the mole ratio

The 1:2 stoichiometry means every mole of $H_2SO_4$ reacts with 2 moles of NaOH:

n(NaOH) = 2 \times 2.247\times10^{-3} = 4.494\times10^{-3}\ \text{mol}.

Step 4: Calculate the concentration of NaOH

Divide moles by the volume of NaOH solution (in litres):

c(NaOH) = \frac{n}{V} = \frac{4.494\times10^{-3}}{0.0250} = 0.180\ \text{mol L}^{-1}.

Final answer: the concentration of NaOH is $0.180\ \text{mol L}^{-1}$ (3 significant figures).

Worked example

Question 2 (4 marks): The complete combustion of 1.50 g of ethanol ( $C_2H_5OH$ , $M = 46.0\ \text{g mol}^{-1}$ ) in a calorimeter raises the temperature of 250 g of water by 30.5 °C. Calculate the molar heat of combustion. (Take the specific heat capacity of water $= 4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}$ .)

Heat absorbed by water: $q = mc\Delta T = 250 \times 4.18 \times 30.5 = 31\,872.5\ \text{J} \approx 31.9\ \text{kJ}$ .

Moles of ethanol: $n = \dfrac{m}{M} = \dfrac{1.50}{46.0} = 0.03261\ \text{mol}$ .

Molar heat of combustion:

\Delta H_c = \frac{q}{n} = \frac{31.87\ \text{kJ}}{0.03261\ \text{mol}} \approx 977\ \text{kJ mol}^{-1}.

Since combustion is exothermic, $\Delta H_c \approx \mathbf{-977\ kJ\ mol^{-1}}$ . (This is below the data-book value because calorimeters lose heat to the surroundings.)

Worked example

Question 3 (6 marks): A compound with molecular formula C₃H₆O₂ shows a strong IR absorption near 1740 cm⁻¹ and no broad O-H band. Its ¹H NMR shows three signals; its ¹³C NMR shows three carbon environments. Deduce a likely structure and justify each piece of evidence.

Degrees of unsaturation: $\dfrac{(2\times3+2)-6}{2} = 1$ - one C=O or ring.

IR: ~1740 cm⁻¹ is an ester C=O stretch (higher than a ketone, ~1715); the absence of a broad 2500-3300 cm⁻¹ band rules out a carboxylic acid. This points to an ester.

¹³C NMR (3 environments) and ¹H NMR (3 signals) are consistent with methyl ethanoate $CH_3{-}COO{-}CH_3$ : the carbonyl C, and two distinct CH₃ groups (one attached to C=O, one attached to O), giving three carbon environments and (allowing for the two CH₃ singlets plus the carbonyl carbon having no H) the observed proton signals.

Putting it together: a likely structure is methyl ethanoate (methyl acetate), $CH_3COOCH_3$ . Check formula C₃H₆O₂ ✓; ester carbonyl ✓; three carbon environments ✓.

Common errors students made

The 2024 VCAA examination report flagged that students lost marks by:

omitting state symbols and not balancing charge in ionic/redox equations;
mixing up oxidising and reducing agents (and the direction of electron flow in cells);
quoting Le Chatelier shifts without naming the driver (moles of gas, exo/endothermic direction);
drawing energy profile diagrams without labels or an activation-energy marker.

Further traps to avoid:

Confusing strength and concentration of acids.
Forgetting that calorimetry answers should be negative for combustion, and not accounting for heat loss.
Quoting the wrong significant figures - match the least precise data in the stem.
In spectroscopy, jumping to a structure without first calculating degrees of unsaturation and assigning each key peak (note ester C=O ~1740 vs acid/ketone ~1715).

How to use this paper

Sit Section A in ~35-40 minutes, then Section B in ~110 minutes, with calculator and data book ready.
Self-mark against the official VCAA assessment report and answers at the links in the frontmatter, awarding method marks line by line.
Re-attempt any calculation you lost marks on, writing the relationship first, then substituting, then evaluating with units and sig figs.
Build a one-page summary of characteristic IR/NMR/mass-spec values and the electrochemical series, and review it before each re-sit.

Use this paper well

Sit the paper under exam conditions (150 minutes, 120 marks).
Mark yourself against the official VCAA marking notes.
Compare against the Chemistry hub to find the syllabus dot points this paper tested.

What this paper assessed

Structure and timing

Worked practice questions (exam-style)

Common errors students made

How to use this paper

Use this paper well

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