VCE Chemistry Unit 3+4 practice questions: The 2026 guide

Study guide16 min readUpdated 7 Oct 2026

Quick answer

VCE Chemistry Unit 3-4 exam practice requires mastery of multiple calculation types (calorimetry, Kc and ICE tables, the reaction quotient, Faraday's laws, redox titration, percentage yield, IUPAC nomenclature) with three significant figures and units throughout; the 120-mark single paper (30 multiple choice + 90 short and extended response) is built on the VCAA Study Design (Units 3 and 4 from 2024) across all five areas of study, and a six-week practice routine moving from key-knowledge review through targeted drills to full timed papers secures top scores.

Jump to a section

What this guide is for
Exam structure
Question types
Marking criteria
Sample questions by Area of Study
Practice routine
Check your knowledge

What this guide is for

VCE Chemistry Unit 3-4 is examined through SACs and one end-of-year exam. The exam is worth 50 percent of the study score and SACs the other 50 percent (Unit 3 is 20 percent and Unit 4 is 30 percent). This guide covers exam structure, question types, marking criteria, and a six-week preparation routine.

Exam structure

Duration: 2 hours 30 minutes plus 15 minutes reading time.
Total marks: 120.
Section A: 30 multiple-choice questions, 1 mark each. 30 marks.
Section B: Short and extended response. 90 marks.

The exam draws on all the Unit 3 and Unit 4 key knowledge and the key science skills. The student-designed investigation poster and logbook are school-assessed, but the exam can still ask about investigation design, evidence and communication.

Question types

Multiple choice (Section A): Knowledge recall, simple calculations, structure identification.
Short answer (Section B): 2-5 mark items. Specific calculations or short explanations.
Extended response (Section B): 5-10 mark items. Multi-step calculations, longer explanations, sometimes incorporating data analysis.

Marking criteria

VCAA rewards:

Correct calculations with appropriate working.
Significant figures consistent with the data (typically 3 sig fig).
Units (always include).
Chemical correctness (equations balanced, formulas correct).
Communication (clear scientific writing in explanations).

Top band requires excellence in all five.

Sample questions by Area of Study

Unit 3 AoS 1 (Energy)

Sample. Calculate the energy released when 5.0 g of methane is combusted completely. $\Delta H_c$ of methane = -890 kJ/mol.

Solution. Moles methane = 5.0 g / 16.04 g/mol = 0.312 mol. Energy = 0.312 × 890 = 278 kJ.

Reference reaction profile for the exothermic combustion of methane:

\Delta H = -890

kJ mol

^{-1}

sits between reactant and product plateaus, with

E_a

marking the activation energy.

Unit 3 AoS 2 (Rates, equilibrium and electrolysis)

Sample. A reaction A + B ⇌ C + D has $K_c = 4.0$ at a given temperature. At equilibrium, [A] = 0.20, [B] = 0.10, [C] = ?. [D] = 0.40.

Solution. $K_c = [C][D] / [A][B]$ . $4.0 = [C](0.40) / (0.20)(0.10) = 20[C]$ . $[C] = 0.20$ M.

Sample. For $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at a temperature where $K_c = 50$ , a 1.0 L vessel holds 0.10 mol $H_2$ , 0.10 mol $I_2$ and 0.50 mol HI. Predict the direction of net change.

Solution. $Q_c = [HI]^2 / ([H_2][I_2]) = (0.50)^2 / (0.10 \times 0.10) = 25$ . Because $Q_c < K_c$ , the reaction proceeds in the forward direction (to the right) until $Q_c = K_c$ .

Unit 4 AoS 1 (Organic)

Sample. Identify the IUPAC name of (CH3)2CHCH2CH(CH3)CH2OH.

Solution. Longest chain with -OH: 5 carbons starting from OH end. OH on C1 (after numbering to give OH lowest locant). Branches: methyl on C2 (or higher; check both directions). Final: 2,4-dimethylpentan-1-ol.

Unit 4 AoS 1 and 2 (Organic synthesis, analysis and medicines)

Sample. A 0.500 g sample of an iron tablet is dissolved in acid and titrated with $0.0200$ M $MnO_4^-$ . The endpoint is reached at 18.40 mL. Calculate the percentage by mass of iron(II). The reaction is $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$ .

Solution. $n(MnO_4^-) = 0.0200 \times 0.01840 = 3.68 \times 10^{-4}$ mol. $n(Fe^{2+}) = 5 \times 3.68 \times 10^{-4} = 1.84 \times 10^{-3}$ mol. Mass of iron $= 1.84 \times 10^{-3} \times 55.85 = 0.103$ g. Percentage by mass $= 0.103 / 0.500 \times 100 = 20.6$ percent. (Redox titrations, mass spectrometry, IR and NMR data analysis, and the distinction between saturated and unsaturated fats, measured with iodine, are all Unit 4 AoS 2 key knowledge.)

Practice routine

Weeks 1-2 of revision: Key knowledge review. Use VCAA Study Design as checklist.
Weeks 3-4: Calculation drills. Calorimetry, Kc and Qc, Faraday's laws, redox titrations, percentage yield.
Week 5: Past papers under timed conditions. Mark against assessor's reports.
Week 6: Weak-area focus. Identify specific knowledge gaps; drill those.

Common errors

Significant figures: 3 sig figs is the convention. Reporting 5 sig figs from 3-sig-fig data signals carelessness.
Units missing: Every numerical answer needs units.
Equations not balanced: Check before writing.
Direction confusion: Markovnikov for HX addition to alkene; oxidation of primary vs secondary alcohols; equilibrium shifts under Le Chatelier.
Forgetting the data book: Look up the electrochemical series, constants and heats of combustion rather than guessing.

Check your knowledge

A broad VCAA-style set spanning Units 3 and 4 (equilibrium, electrochemistry, calorimetry, organic, analytical, medicinal chemistry). Attempt under exam conditions, then check against the solutions block.

State Le Chatelier's principle and use it to predict the effect of increasing pressure on the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ . (3 marks)
A galvanic cell is built from a $Zn/Zn^{2+}$ half-cell and a $Ag/Ag^+$ half-cell at standard concentrations and 25 degrees C. ( $E^\circ(Zn^{2+}/Zn) = -0.76 \ \text{V}$ , $E^\circ(Ag^+/Ag) = +0.80 \ \text{V}$ .) (a) Identify the anode, the cathode, and write the overall cell equation. (b) Calculate $E^\circ_{cell}$ . (c) Calculate the change in mass of the silver electrode after a current of 0.50 A flows for 30.0 minutes ( $A_r(Ag) = 107.9$ ). (6 marks)
For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , the initial concentrations in a 2.0 L sealed vessel are $[SO_2] = 0.40 \ \text{M}$ , $[O_2] = 0.20 \ \text{M}$ , $[SO_3] = 0 \ \text{M}$ . At equilibrium, $[SO_3] = 0.30 \ \text{M}$ . (a) Construct an ICE table. (b) Calculate $K_c$ at this temperature. (5 marks)
A 5.000 g sample of an iron supplement is dissolved in dilute sulfuric acid and made up to 250.0 mL. A 25.00 mL aliquot is titrated with $0.0200 \ \text{M}$ $KMnO_4$ , and the endpoint is reached at 18.40 mL. (a) Write the balanced equation for the reaction ( $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$ ). (b) Calculate the amount of $Fe^{2+}$ in the aliquot and in the whole sample. (c) Calculate the percentage by mass of iron in the supplement ( $A_r(Fe) = 55.85$ ). (5 marks)
The combustion of methane is exothermic, $\Delta H = -890 \ \text{kJ mol}^{-1}$ . (a) Calculate the volume of methane (at SLC, $V_m = 24.79 \ \text{L mol}^{-1}$ ) needed to heat 2.50 L of water from 18.0 degrees C to 95.0 degrees C, assuming 65 percent heat-transfer efficiency and $c_{water} = 4.18 \ \text{J g}^{-1} \text{K}^{-1}$ . (b) Comment on whether the assumption of complete combustion is realistic in a gas hot-water system. (5 marks)
An unknown organic compound $X$ has the empirical formula $C_3H_6O$ and $M = 58 \ \text{g mol}^{-1}$ . (a, 2) Determine the molecular formula. (b, 4) The $^1H$ NMR of $X$ shows a single peak. Its IR shows a strong absorption at $1715 \ \text{cm}^{-1}$ and no broad O-H. Identify $X$ and justify your reasoning. (c, 2) Predict the mass spectrum base peak and account for it. (8 marks)
For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ , $K_c = 50$ at a given temperature. A 1.0 L vessel holds 0.10 mol $H_2$ , 0.10 mol $I_2$ and 0.50 mol $HI$ . (a) Calculate $Q_c$ and predict the direction of net change. (b) State the effect of adding a catalyst on the value of $K_c$ and on the rate at which equilibrium is reached. (c) The forward reaction is exothermic. Explain the conflict between rate and yield when choosing the operating temperature. (6 marks)
(a, 2) Define saturated and unsaturated fatty acids at the molecular level. (b, 3) Olive oil is roughly 75 percent oleic acid ( $C_{18}H_{34}O_2$ , one C=C). Calculate the mass of bromine ( $Br_2$ , $M = 159.8$ ) that would react completely with 5.00 g of oleic acid via addition across the double bond. (c, 2) Comment on why olive oil's iodine value is a useful measure for assessing oil oxidative stability. (7 marks)

Solutions

Q1: Le Chatelier's principle: if a system at dynamic equilibrium is disturbed, the system shifts in the direction that minimises the disturbance. For the Haber equilibrium, increasing pressure favours the side with fewer moles of gas. There are $1 + 3 = 4$ moles of gas on the reactant side and $2$ moles on the product side, so the equilibrium shifts to the right (toward $NH_3$ ), increasing the yield of ammonia.
Q2: (a) Zn is more readily oxidised (more negative $E^\circ$ ), so Zn is the anode: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$ . Ag is the cathode: $Ag^+(aq) + e^- \rightarrow Ag(s)$ . Overall: $Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$ . (b) $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.76) = +1.56 \ \text{V}$ . (c) Charge $Q = It = 0.50 \times 30.0 \times 60 = 900 \ \text{C}$ . Moles of $e^- = 900 / 96\,500 = 9.33 \times 10^{-3}$ . Moles of Ag deposited $= 9.33 \times 10^{-3}$ (1:1 ratio). Mass gained $= 9.33 \times 10^{-3} \times 107.9 = 1.01 \ \text{g}$ . The silver electrode gains 1.01 g.
Q3: (a) ICE table (concentrations in mol L $^{-1}$ ):

Species	$SO_2$	$O_2$	$SO_3$
Initial	0.40	0.20	0
Change	$-0.30$	$-0.15$	$+0.30$
Equilibrium	0.10	0.05	0.30

(b) $K_c = \dfrac{[SO_3]^2}{[SO_2]^2[O_2]} = \dfrac{(0.30)^2}{(0.10)^2 (0.05)} = \dfrac{0.090}{5.0 \times 10^{-4}} = 1.8 \times 10^{2} \ \text{M}^{-1}$ . The equilibrium constant at this temperature is $1.8 \times 10^{2} \ \text{M}^{-1}$ , indicating products are strongly favoured.

Q4: (a) $5Fe^{2+}(aq) + MnO_4^-(aq) + 8H^+(aq) \rightarrow 5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_2O(l)$ . (b) $n(MnO_4^-) = 0.0200 \times 0.01840 = 3.68 \times 10^{-4} \ \text{mol}$ . $n(Fe^{2+})$ in the aliquot $= 5 \times 3.68 \times 10^{-4} = 1.84 \times 10^{-3} \ \text{mol}$ . The aliquot is one tenth of the 250.0 mL solution, so the whole sample contains $1.84 \times 10^{-2} \ \text{mol}$ . (c) Mass of iron $= 1.84 \times 10^{-2} \times 55.85 = 1.03 \ \text{g}$ . Percentage by mass $= 1.03 / 5.000 \times 100 = 20.6$ percent.
Q5: (a) Energy needed to heat water: $Q = mc\Delta T = 2500 \times 4.18 \times (95.0 - 18.0) = 8.047 \times 10^{5} \ \text{J} = 804.7 \ \text{kJ}$ . At 65 percent efficiency, fuel energy required $= 804.7 / 0.65 = 1238 \ \text{kJ}$ . Moles methane $= 1238 / 890 = 1.391 \ \text{mol}$ . Volume at SLC $= 1.391 \times 24.79 = 34.5 \ \text{L}$ . (b) In a well-tuned domestic gas heater combustion is close to complete, but the air-to-gas ratio fluctuates and some CO is produced. The complete-combustion assumption typically overstates the heat output by a few percent.
Q6: (a) Empirical mass $= 3(12.01) + 6(1.01) + 16.00 = 58.08 \ \text{g mol}^{-1}$ , matching $M$ . The molecular formula equals the empirical formula: $C_3H_6O$ . (b) A single $^1H$ NMR peak means all the H environments are equivalent. The IR carbonyl absorption at $1715 \ \text{cm}^{-1}$ is in the ketone/aldehyde region; absence of broad O-H rules out a carboxylic acid. Six equivalent H and a C=O fits propan-2-one (acetone, $CH_3COCH_3$ ): two equivalent methyl groups on a central carbonyl. (c) Cleavage of one methyl-acyl bond gives the acylium cation $[CH_3CO]^+$ at $m/z = 43$ , the well-known base peak of acetone (with the parent ion at $m/z = 58$ ).
Q7: (a) $Q_c = [HI]^2 / ([H_2][I_2]) = (0.50)^2 / (0.10 \times 0.10) = 25$ . Since $Q_c < K_c$ , the reaction proceeds in the forward direction (to the right) until $Q_c = 50$ . (b) A catalyst provides an alternative pathway with lower activation energy. It does not change $K_c$ or the position of equilibrium; it increases the rate of both the forward and reverse reactions, so equilibrium is reached sooner. (c) Because the forward reaction is exothermic, raising the temperature shifts the equilibrium to the left (and lowers $K_c$ ), reducing the yield; but a lower temperature makes the reaction too slow. The operating temperature is a compromise between acceptable rate and acceptable yield, which is why catalysts are used to allow a moderate temperature.
Q8: (a) Saturated fatty acids contain only single C-C bonds in the hydrocarbon tail; unsaturated fatty acids contain one or more C=C double bonds (mono- or poly-unsaturated). (b) Moles of oleic acid $= 5.00 / 282.5 = 1.770 \times 10^{-2} \ \text{mol}$ (using $M(C_{18}H_{34}O_2) = 282.5$ ). One $Br_2$ adds per C=C, so moles $Br_2 = 1.770 \times 10^{-2}$ . Mass $Br_2 = 1.770 \times 10^{-2} \times 159.8 = 2.83 \ \text{g}$ . (c) The iodine (or bromine) value measures the degree of unsaturation in a fat or oil. Polyunsaturated oils with high iodine values are more prone to free-radical oxidation at the C=C sites (rancidity), so iodine value is a direct proxy for oxidative stability and shelf life.