June 2026
The 2025 VCE Chemistry exam rewarded students who could combine chemical knowledge with disciplined execution.
That meant more than remembering formulas, reaction types or definitions. Students needed to read questions carefully, select the correct chemical principle, apply it to the scenario, carry units through calculations, and explain chemical reasoning clearly.
The paper covered familiar areas: fuels, calorimetry, electrochemistry, equilibrium, green chemistry, organic pathways, spectroscopy, experimental design and analytical techniques. But the strongest responses were not built on topic recognition alone. They showed precision.
Which species is oxidised?
Which electrode is the anode?
What moves through the electrolyte?
What is the limiting reagent?
What does the calibration factor measure?
What does the question mean by resolution?
Which functional group is actually present?
What does a broader and lower melting point range indicate?
These details shaped the exam.
In VCE Chemistry, marks are awarded for chemical reasoning in context.
Section A showed the importance of unit control
Question 2 asked students to calculate the energy content per 100 g of a snack bar. The snack bar contained carbohydrate, protein and fats, and students needed to use the energy values for each macronutrient.
The energy content of the 65 g bar was:
(16 × 45.0) + (17 × 10.0) + (37 × 10.0) = 1260 kJ
Converted to 100 g:
1260 × 100 ÷ 65 = 1938 kJ
That is 1.9 × 10⁶ J.
The report noted that the most common error was a unit error.
This is one of the clearest lessons from Chemistry. A student may do the chemical reasoning correctly but lose the answer by failing to convert kJ to J, mL to L, m³ to L, or by using the wrong unit in the final response.
In Chemistry, units are not decoration. They are part of the answer.
Energy questions required conceptual clarity
Question 3 asked which statement was true for all exothermic reactions.
The correct answer was that the energy required to break bonds in the reactants is less than the energy released when bonds form in the products.
This question tested a common misconception.
Breaking bonds requires energy. It is endothermic. Forming bonds releases energy. It is exothermic. A reaction is overall exothermic when more energy is released in bond formation than is required for bond breaking.
Students often remember that exothermic reactions “release energy”, but the question required the bond-level explanation.
This is the kind of precision that Chemistry rewards.
The label is not enough. The molecular reason matters.
Calorimetry rewarded formula discipline
Question 5 asked students to calculate the calibration factor of a solution calorimeter.
Oscar passed a current of 2.50 A through a heater at 12.0 V for 150 s. The water temperature increased from 22.3 °C to 28.4 °C, giving a temperature change of 6.1 °C.
The calibration factor was:
CF = VIt ÷ ΔT
CF = 12.0 × 2.50 × 150 ÷ 6.1 = 738 J °C⁻¹
This was answered well by many students, but it still shows an important Chemistry skill.
Students need to know not only the formula, but what each symbol represents and how the units behave. Voltage, current and time give energy in joules. Dividing by temperature change gives joules per degree Celsius.
A formula used without unit understanding is fragile.
Photosynthesis required reversing combustion logic
Question 6 asked what occurs to one mole of carbon dioxide during photosynthesis, given that the molar enthalpy of combustion of glucose is −2840 kJ mol⁻¹.
Combustion of glucose releases energy. Photosynthesis is the reverse process and absorbs energy.
For the formation of one mole of glucose, the enthalpy change is +2840 kJ. Since six moles of carbon dioxide are required to form one mole of glucose, one mole of carbon dioxide corresponds to:
2840 ÷ 6 = 473.3 kJ absorbed
This question rewarded students who could connect thermochemistry to stoichiometry.
It was not enough to remember that photosynthesis is endothermic. Students needed to scale the energy to one mole of carbon dioxide.
This is a recurring Chemistry demand: the concept and the mole ratio must both be right.
Electrochemistry was one of the clearest discriminators
The 2025 exam included several electrochemistry contexts: flexible metal–air batteries, fuel cells, nickel–cadmium recharge, electroplating and artificial photosynthesis.
Question 7 asked what migrates through the polymer or gel electrolyte when a flexible metal–air cell produces electrical energy.
The correct answer was metal ions flowing from the anode to the cathode.
This question was subtle because electrons also move from anode to cathode in a galvanic cell, but electrons move through the external circuit, not through the electrolyte. The question specifically asked about the polymer or gel electrolyte.
That wording mattered.
In electrochemistry, students must separate electron movement from ion movement. Electrons move through wires. Ions move through electrolytes or salt bridges to maintain charge balance.
A correct response depends on the path being described.
Metal–air cells required reduction potential reasoning
Question 8 asked which flexible metal–air cell had the highest potential difference. Students needed to use electrochemical data to compare the possible cells.
The report identified Circuit A as producing the highest potential difference, 3.60 V.
Question 9 then asked where unwanted side reactions with moisture were less likely. The correct answer was zinc–air cells, because zinc does not normally react with water, while sodium, magnesium and aluminium have the potential to spontaneously react with water.
This sequence shows how electrochemistry questions often require applied reasoning. Students are not only asked to identify an anode or cathode. They must use electrochemical data to compare feasibility, voltage, side reactions and chemical stability.
High-scoring students can interpret electrochemical information in context.
Fuel cell questions required charge-to-mole reasoning
Question 12 asked students to calculate the mass of oxygen consumed when a hydrogen fuel cell produces 36000 C of charge.
The calculation required Faraday’s constant:
n(e⁻) = Q ÷ F = 36000 ÷ 96500 = 0.373 mol
The oxygen half-equation involves four electrons per mole of oxygen:
O₂ + 4H⁺ + 4e⁻ → 2H₂O
So:
n(O₂) = 0.373 ÷ 4 = 0.09326 mol
m(O₂) = 0.09326 × 32.0 = 2.98 g
This is a classic electrochemistry calculation. The key step is not the arithmetic alone. It is connecting charge to moles of electrons, then using the balanced half-equation to find moles of oxygen.
Many Chemistry calculations follow this pattern.
Quantity → mole relationship → stoichiometric ratio → final value.
Skipping any step makes the answer vulnerable.
Equilibrium required mathematical and conceptual control
Question 15 asked students to determine the magnitude of Kc for a reaction derived from another equilibrium reaction.
For:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Kc was given as 2.86 × 10⁻¹ M⁻².
The second reaction was:
4NH₃(g) ⇌ 2N₂(g) + 6H₂(g)
This is the reverse reaction doubled. Therefore, the new Kc is:
(1 ÷ 0.286)² = 12.2
The correct answer was 1.22 × 10¹.
This question rewarded students who understood how equilibrium constants change when equations are reversed or multiplied.
Reversing a reaction takes the reciprocal of K. Multiplying coefficients raises K to that power.
This is not a memorised trick. It follows from the equilibrium expression.
Le Châtelier’s principle required graph interpretation
Question 16 asked students to interpret a temperature increase for the equilibrium:
2NO(g) + O₂(g) ⇌ 2NO₂(g), ΔH < 0
The forward reaction is exothermic. Increasing temperature favours the endothermic direction, which is the reverse reaction.
Therefore, the concentrations of NO and O₂ increase, while the concentration of NO₂ decreases. There is no instantaneous concentration change because temperature, not concentration, was changed.
This last point is important.
Students often expect a sudden jump in concentration whenever a disturbance occurs. But a temperature change does not instantly change the number of particles present. It changes the position of equilibrium, leading to gradual concentration changes as the system responds.
High-scoring equilibrium responses match the graph to the type of change.
Green chemistry questions required specific principles
The 2025 exam repeatedly assessed sustainability and green chemistry.
Question 4 asked how the circular economy of a bioethanol production process could be improved. The correct answer was developing a separate process that converts waste carbon dioxide into a useful product. This is because circular economy improves when waste materials are reused or recycled.
Question 21 asked for a key advantage of renewable feedstocks. The correct answer was that they reduce reliance on finite natural resources. Other effects, such as fewer by-products or lower energy requirements, may be possible, but they are not guaranteed in every scenario.
This distinction matters.
Green chemistry is not a set of generic “environmentally friendly” statements. Students need to connect the principle to the exact chemical process being described.
Renewable feedstock reduces reliance on finite resources. Circular economy reduces waste by keeping materials in use. Design for energy efficiency reduces energy loss or energy requirements.
The principle must fit.
Experimental design language mattered
Questions 19 and 20 focused on a vitamin C experiment involving a delayed colour change.
Question 19 asked students to order equipment by resolution. The report noted that students struggled with the term “resolution”. A higher-resolution instrument detects smaller changes and allows measurements in finer increments.
The 50 mL beaker had the lowest resolution. The 10 mL measuring cylinder had the highest resolution among the equipment shown.
Question 20 asked what caused variability in the graph. Because data points lay both above and below the line of best fit, the variation was best explained by random error. The most likely source was subjective judgement about when the colour change occurred.
This is important.
Chemistry practical questions do not only ask about apparatus. They assess how measurement quality affects data. Students need to understand resolution, random error, systematic error, variability and uncertainty.
Experimental language is part of Chemistry.
Organic chemistry required structural reasoning
The organic chemistry questions in Section A were highly instructive.
Question 22 asked for the major organic product when propan-1-ol reacts with butanoic acid in the presence of sulfuric acid. Students needed to recognise esterification and correctly assemble the ester structure.
Question 23 asked about protein hydrolysis. In hydrolysis, water is used to break amide links in proteins, producing smaller molecules such as amino acids.
Question 25 asked which technique would separate volatile polar compounds with similar boiling points from ethanol. The correct answer was fractional distillation, because it is suitable for separating volatile compounds with similar boiling points.
These questions rewarded students who could connect structure, reaction type and separation method.
Organic chemistry is not just naming functional groups. It is understanding how structure determines reactivity and physical properties.
Functional group tests needed specificity
Questions 26 and 27 used geranial and linalool, two natural compounds found in citrus essential oils.
Geranial has the molecular formula C₁₀H₁₆O and is an aldehyde. Linalool has the molecular formula C₁₀H₁₈O and is a tertiary alcohol.
The correct test to distinguish pure geranial from pure linalool was acidified dichromate. Geranial, as an aldehyde, is readily oxidised and causes the dichromate solution to change from orange to green. Linalool, as a tertiary alcohol, is not readily oxidised, so the solution remains orange.
This question was a strong reminder that tests must distinguish the compounds in the question.
Bromine solution would not distinguish them because both compounds contain carbon-carbon double bonds. Sodium hydrogen carbonate would not work because neither compound contains a carboxyl group.
A test is only useful if it produces different results for the two substances being compared.
Spectroscopy rewarded structural symmetry
Question 30 asked students to identify the ¹³C NMR spectrum of pentan-3-one.
Pentan-3-one is a symmetrical ketone. Because of its symmetry, it has only three distinct carbon environments and therefore three carbon signals in the ¹³C NMR spectrum. As a ketone, one signal must appear in the carbonyl region around 205–220 ppm.
This question rewarded students who understood symmetry and chemical environment.
Students often count carbons rather than unique carbon environments. In NMR, those are not the same thing. Equivalent carbons produce the same signal.
This is a major organic analysis skill.
A molecule’s structure determines its spectrum.
Section B calculations demanded full working
The Section B report comments reinforce the importance of clear working.
Question 1 involved propane combustion, energy absorbed by water, efficiency, limiting reagents in a sealed container and comparison with ethanol.
Students needed to write a balanced thermochemical equation with correct states and enthalpy value, calculate energy released from mass of propane, use q = mcΔT for the water, calculate efficiency, and then determine the excess reagent after combustion in a sealed container.
The report noted common errors such as failing to include an enthalpy value, using incorrect states, omitting units, including a negative sign when describing energy released, failing to convert m³ to L, and failing to determine the amount of excess reagent remaining.
This is Chemistry at its most procedural.
A student may understand combustion broadly, but the marks depend on balanced equations, correct states, stoichiometry, units and final interpretation.
Explanations needed chemical comparison
Question 1e asked why different masses of propane and ethanol were required to produce the same temperature change.
The key point was that ethanol has a lower energy content than propane, whether expressed in kJ g⁻¹ or kJ mol⁻¹. Therefore, a greater mass of ethanol was required to release enough energy to produce the same temperature change in the water.
The report noted that this question was poorly answered.
This is a valuable lesson.
Students often answer comparison questions too generally. A strong answer needed to compare the fuels directly and link the lower energy content of ethanol to the greater mass required.
The chemistry was not just “they are different fuels”. It was energy released per gram or per mole.
What the 2025 exam teaches future Chemistry students
The 2025 VCE Chemistry exam shows that high performance depends on controlled chemical reasoning.
Students need to:
- track units and conversions carefully
- distinguish energy released from enthalpy sign
- connect thermochemistry to stoichiometry
- separate electron flow from ion movement
- use half-equations in electrochemical calculations
- apply Le Châtelier’s principle to graphs correctly
- manipulate equilibrium constants mathematically
- connect green chemistry principles to specific processes
- understand resolution, random error and systematic error
- use functional group tests only when they distinguish substances
- interpret spectra using structure and symmetry
- show full working in multi-step calculations
The strongest Chemistry responses are not simply correct in principle.
They are chemically exact.
How ATAR STAR approaches VCE Chemistry
At ATAR STAR, Chemistry is taught as a subject of principles, calculations and evidence.
Students learn to move beyond formula recognition and into chemical reasoning. They practise tracking units, writing balanced equations, interpreting electrochemical systems, analysing organic structures, applying equilibrium principles and explaining experimental data in the language expected by VCAA.
The 2025 Examination Report confirms why this matters. High-scoring responses were precise, well-sequenced and grounded in chemical principles.
They did not just know Chemistry.
They used Chemistry accurately.