VCE Chemistry 2021

Walkthrough of the 2021 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

The VCAA Chemistry Units 3 & 4 examination (120 marks in 150 minutes) covers:

Unit 3 - How can chemical processes be designed to optimise efficiency: rates of reaction and chemical equilibrium (Le Chatelier's principle, the equilibrium constant $K_c$ ), galvanic cells, fuel cells and electrolysis, and the production of chemicals.
Unit 4 - How are organic compounds categorised and synthesised: organic structure, nomenclature, functional groups and reaction pathways; analytical techniques including mass spectrometry, infrared, and NMR spectroscopy; volumetric analysis (titration); and energy content of food/fuels including calorimetry.

The exam combines multiple choice with calculation-heavy short and extended response. Markers expected balanced equations with state symbols, the chosen relationship written before the arithmetic, significant figures consistent with the data, and clear reasoning in qualitative items.

Structure and timing

120 marks in 150 minutes is a rate of 1.25 minutes per mark.

Section A - Multiple choice (~30 marks). Target about 30-35 minutes; do scratch working for calculation MCQs.
Section B - Short and extended response (~90 marks). Target about 105-110 minutes at ~1.25 min/mark.

A workable plan: ~35 minutes Section A, ~105 minutes Section B, with a final 10-minute buffer to check balancing, state symbols, units and significant figures.

Worked practice questions (exam-style)

Worked example

Question 1 (5 marks): A $20.00\ \text{mL}$ sample of vinegar (ethanoic acid solution) is titrated against $0.250\ \text{mol L}^{-1}$ NaOH, requiring an average titre of $16.8\ \text{mL}$ . Calculate the concentration of ethanoic acid in the vinegar.

Balanced equation (1:1 mole ratio):

\text{CH}_3\text{COOH}(aq) + \text{NaOH}(aq) \rightarrow \text{CH}_3\text{COONa}(aq) + \text{H}_2\text{O}(l)

Moles of NaOH:

n(\text{NaOH}) = c \times V = 0.250 \times 0.01680 = 4.20 \times 10^{-3}\ \text{mol}

From the 1:1 ratio, $n(\text{CH}_3\text{COOH}) = 4.20 \times 10^{-3}\ \text{mol}$ .

Concentration:

c = \frac{n}{V} = \frac{4.20 \times 10^{-3}}{0.02000} = 0.210\ \text{mol L}^{-1}\ (3\ \text{sig figs})

Worked example

Question 2 (4 marks): A propane fuel cell operates by the overall reaction $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$ . Write the half-equation occurring at the anode (use $\text{H}^+$ ) and state whether it is oxidation or reduction.

At the anode the fuel (propane) is oxidised. Carbon in propane (oxidation state $-\tfrac{8}{3}$ average) is oxidised to carbon in $\text{CO}_2$ (+4), releasing electrons. Balancing the half-equation in acidic conditions:

\text{C}_3\text{H}_8(g) + 6\text{H}_2\text{O}(l) \rightarrow 3\text{CO}_2(g) + 20\text{H}^+(aq) + 20e^-

Check: 3 C and 8+12 = 20 H on the left; 3 C and 20 H (as $\text{H}^+$ ) on the right; oxygen 6 left, 6 right; charge $+20 - 20 = 0$ left, $0$ right. This is the anode (oxidation) half-equation because electrons are released.

Worked example

Question 3 (6 marks): For the equilibrium $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ , $\Delta H = +57\ \text{kJ mol}^{-1}$ , a sealed flask at equilibrium is (a) compressed to half its volume and (b) then heated. Predict and justify the effect on the amount of $\text{NO}_2$ in each case.

(a) Compression raises the total pressure. By Le Chatelier's principle the system shifts toward the side with fewer gas moles to oppose the increase: the left has 1 mol, the right has 2 mol, so it shifts left, decreasing the amount of $\text{NO}_2$ (the driver is the increase in pressure/concentration). $K_c$ is unchanged because temperature is constant.

(b) The forward reaction is endothermic ( $\Delta H > 0$ ). Heating adds energy, which the system opposes by favouring the endothermic (forward) direction, shifting right and increasing the amount of $\text{NO}_2$ (the driver is the added heat). Here $K_c$ increases because temperature has changed.

Common errors students made

The 2021 VCAA examiners' report noted that students omitted state symbols, mixed up oxidising and reducing agents, quoted Le Chatelier shifts without naming the driver, and drew energy diagrams without an activation-energy marker. Adding to those:

Confusing $K_c$ changes with position shifts - claiming concentration changes alter $K_c$ , when only temperature does.
Half-equation balancing errors - failing to balance charge with electrons, or O and H with water and $\text{H}^+$ .
Significant figures quoted beyond the precision of the data.
Spectroscopy misreads - assigning an IR or NMR signal to the wrong functional group or environment without justification.

How to use this paper

Sit Section A in a strict ~35-minute block, then Section B in a single ~105-minute run under exam conditions, keeping a 10-minute buffer to check balancing, state symbols, units and significant figures. Mark every calculation against the official VCAA assessment report and answers at the authority page linked in the frontmatter, noting where method marks sit - they are recoverable. Re-attempt any calculation that lost a method mark on a blank page, writing the balanced equation and the chosen relationship before any arithmetic. Keep a one-page error log of recurring slips and re-test those item types after a week.

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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