June 2026
The 2025 VCE Chemistry exam showed that Section B is where chemical understanding becomes most visible.
Multiple-choice questions can test precision, but Section B requires students to build the answer themselves. They must write equations, show working, use correct states, justify conclusions, explain mechanisms and connect data to chemical principles.
That is why Section B often separates students who recognise Chemistry from students who can use Chemistry.
In 2025, students had to explain combustion efficiency, limiting reagents, electroplating, equilibrium yield, organic pathways, spectroscopic evidence, titration data and biochemical fuel comparisons. These were familiar topics, but the questions were demanding because they required sequence and control.
What equation applies?
Which species is limiting?
Which reagent is in excess after reaction?
What does the half-equation show?
What does the equilibrium quotient mean?
Which spectrum feature supports the conclusion?
How does the calculated result compare with the hypothesis?
The strongest responses answered the chemical task directly.
Thermochemical equations needed to be complete
Question 1a asked students to write a balanced thermochemical equation for the complete combustion of propane.
A strong response needed:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) ΔH = −2220 kJ
The report noted several common errors: students omitted the enthalpy value, used incorrect states or failed to balance the equation correctly.
This is a useful Section B lesson.
A thermochemical equation is not only a balanced equation. It is a complete chemical statement. The states matter because the enthalpy value corresponds to those states. The coefficient ratio matters because the enthalpy value corresponds to the balanced amount of reaction shown.
Writing H₂O(g) instead of H₂O(l) is not a minor formatting issue if the enthalpy value is for liquid water at SLC.
In Section B, every part of the chemical equation matters.
Calculation answers needed working, units and interpretation
Question 1b asked students to calculate the energy released from the combustion of 9.0 g of propane.
The working was:
n(C₃H₈) = 9.0 ÷ 44.0 = 0.2045 mol
Energy released = 0.2045 × 2220 = 454 kJ
The report noted that some students omitted units or included the negative sign while saying the energy released was −454 kJ.
This distinction matters.
The enthalpy change is negative because the reaction is exothermic. But the amount of energy released is expressed as a positive quantity.
A high-scoring answer communicates the chemistry clearly:
454 kJ of energy is released
or
ΔH = −454 kJ for the amount combusted
Those two statements are not identical in wording, and the sign must match the meaning.
Energy efficiency required two different energy values
Question 1c asked students to calculate the energy absorbed by water and then determine the efficiency of the heating process.
The water absorbed:
q = mcΔT
q = 166.0 × 4.18 × 66.0 = 45.8 kJ
The propane released:
454 kJ
So the efficiency was:
45.8 ÷ 454 × 100 = 10%
This question required students to keep two energy values separate.
The fuel released energy through combustion. The water absorbed only part of that energy. The efficiency compared useful energy absorbed by the water with total energy released by the fuel.
Students who confuse these values cannot explain why efficiency is low.
In the practical set-up, much of the energy is lost to the surroundings, saucepan and air rather than being transferred to the water. Section B rewarded students who understood the experiment behind the numbers.
Limiting reagent questions required the amount remaining
Question 1d was a strong discriminator.
Students had to determine the mass of the reactant in excess after combusting 9.0 g of propane in a sealed container containing air.
The oxygen amount required careful conversion:
0.125 m³ = 125 L
The air contained 21.0% v/v O₂, so:
V(O₂) = 0.210 × 125 = 26.25 L
At SLC:
n(O₂) = 26.25 ÷ 24.8 = 1.058 mol
From the propane mass:
n(C₃H₈) = 0.2045 mol
The combustion equation requires five moles of oxygen per mole of propane:
n(O₂ needed) = 5 × 0.2045 = 1.0225 mol
Oxygen was therefore in excess:
n(O₂ excess) = 1.058 − 1.0225 = 0.0355 mol
m(O₂ excess) = 0.0355 × 32.0 = 1.1 g
The report noted two major errors: students struggled with converting cubic metres to litres, and some did not calculate the amount of excess reagent left after reaction.
That second point is crucial.
The question did not ask which reactant was in excess. It asked for the mass of the reactant in excess. That means students had to determine what remained after the limiting reagent was consumed.
In Section B, wording like in excess, remaining, consumed and initial must be read carefully.
Fuel comparison needed direct chemical reasoning
Question 1e asked why different masses of ethanol and propane were required to produce the same temperature change.
The report showed that this question was poorly answered.
The expected reasoning was that ethanol has a lower energy content than propane:
ethanol: 29.7 kJ g⁻¹
propane: 40.5 kJ g⁻¹
Because ethanol releases less energy per gram, a larger mass of ethanol is needed to produce the same temperature increase in the water.
A stronger response could also explain that ethanol is already partially oxidised due to the oxygen atom in its structure, so complete combustion releases less energy per gram than propane.
The key was comparison.
Students needed to explain why the two fuels behaved differently, not simply state that they are different fuels.
Electroplating required oxidation number language
Question 2a asked students to explain, using oxidation numbers, why chromium electroplating involves reduction.
The half-equation was:
Cr³⁺(aq) + 3e⁻ → Cr(s)
Chromium changes from an oxidation number of +3 to 0.
A decrease in oxidation number is reduction.
This question is a reminder that Section B often specifies the required type of explanation. Because the question said using oxidation numbers, a response needed to include the oxidation number change. Simply writing “chromium gains electrons” was relevant, but not sufficient if the oxidation number reasoning was missing.
The answer must match the instruction.
Electroplating explanations needed practical cell logic
Question 2b asked why effective electroplating requires the same ion to be involved in both the anode and cathode reactions.
At the cathode, metal ions are reduced and deposited onto the object. At the anode, metal atoms are oxidised to replenish the metal ions in the electrolyte.
This maintains the concentration of the metal ion in solution and allows continuous plating.
For chromium electroplating, chromium ions are reduced to chromium metal at the cathode, while chromium metal at the anode can be oxidised to replace chromium ions in the electrolyte.
The response needed to connect the half-reactions to the purpose of electroplating.
It was not enough to say that “the same metal is needed”. Students had to explain why the process depends on replenishing the ions being deposited.
Data Book references had to be used chemically
Question 2c asked students to use the Data Book to justify why the amount of sulfuric acid in the chromium electroplating cell must be carefully managed.
This required students to consider competing reduction reactions. Hydrogen ions can be reduced to hydrogen gas under some conditions. If too much acid is present, hydrogen gas production may compete with chromium deposition at the cathode.
That would reduce plating efficiency and affect the quality of the chromium coating.
The question specifically told students to use the Data Book. A generic answer about acid being dangerous or corrosive would not meet the chemical demand. The relevant issue was electrochemical competition between reducible species.
When Section B points students to a data source, the response should use that data to support a chemical conclusion.
Faraday calculations had to follow the half-equation
Question 2d asked students to calculate the current required to electroplate 7.80 g of chromium over 2.50 hours.
The method was:
n(Cr) = 7.80 ÷ 52.0 = 0.150 mol
From:
Cr³⁺ + 3e⁻ → Cr
n(e⁻) = 0.150 × 3 = 0.450 mol
Q = n(e⁻)F = 0.450 × 96500 = 43 425 C
t = 2.50 × 60 × 60 = 9000 s
I = Q ÷ t = 43 425 ÷ 9000 = 4.83 A
This calculation shows the standard Section B expectation: every step has a chemical purpose.
Mass gives moles of metal. The half-equation gives moles of electrons. Faraday’s constant gives charge. Time gives current.
Students who jump straight to a formula without the half-equation risk using the wrong electron ratio.
Same mass did not mean same current
Question 2e asked why less current would be required to plate the same mass of rhodium compared with chromium.
The answer was not primarily about electrode potential.
Both metal ions involved a 3+ charge, so three moles of electrons were needed per mole of metal atoms plated. However, rhodium has a much higher molar mass than chromium. The same mass of rhodium contains fewer moles of atoms than the same mass of chromium. Therefore, fewer moles of electrons are needed, less charge must pass, and less current is required for the same time period.
This was a reasoning question.
The student needed to recognise which quantity mattered. The relevant comparison was moles of metal and moles of electrons, not voltage.
Section B often tests whether students choose the right chemical data, not just whether they can use data.
Equilibrium yield required rate-yield separation
Question 3b asked about methanol synthesis:
CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH < 0
Students were told that increasing temperature can increase the reaction rate, then asked to explain the impact on yield.
A strong answer needed both sides.
Increasing temperature increases rate because particles have greater kinetic energy and more collisions exceed activation energy. However, the forward reaction is exothermic, so increasing temperature favours the reverse endothermic reaction. The equilibrium shifts left and the yield of methanol decreases.
This question rewarded students who did not confuse rate with yield.
In industrial Chemistry, conditions are compromises. A change that improves rate may reduce equilibrium yield.
Kc and Q required a conclusion
Question 3c asked students to determine what needed to happen for a methanol system to return to equilibrium.
The expression was:
Kc = [CH₃OH] ÷ ([CO][H₂]²) = 1.21
At the given concentrations:
Q = 5.60 ÷ (2.00 × 1.25²) = 1.79
Since Q > Kc, the system had too much product relative to equilibrium. The reverse reaction needed to be favoured. Methanol concentration would decrease, while carbon monoxide and hydrogen concentrations would increase.
This question shows that Section B calculations must end with interpretation.
Calculating Q was not enough. Students needed to state what the comparison meant chemically.
Equilibrium tables required stoichiometric control
Question 3d asked students to calculate the initial amount of hydrogen added to a sealed container.
The reaction was:
CO(g) + 2H₂(g) ⇌ CH₃OH(g)
If equilibrium methanol concentration was 1.30 M in a 1.00 L container, then 1.30 mol of methanol was formed. That means 1.30 mol of CO and 2.60 mol of H₂ were consumed.
Students then needed to use the Kc expression to find the equilibrium hydrogen concentration and add back the amount consumed.
The most important feature was the 2:1 coefficient for hydrogen.
In equilibrium calculations, the equation controls the changes. Students who ignore coefficients lose the chemical logic of the table.
Organic pathways needed conditions and reactants
Question 4a asked students to write a balanced equation for producing C₃H₇Cl from C₃H₈, including reaction conditions.
This required substitution of propane with chlorine under ultraviolet light:
C₃H₈ + Cl₂ → C₃H₇Cl + HCl
with UV light.
The same question asked for two reactants required to form propan-1-amine in a single substitution reaction. Students needed a suitable haloalkane, such as 1-chloropropane, and ammonia.
This question rewarded students who knew organic pathways as reactions, not just product names.
Conditions matter.
Reactants matter.
A named product is not enough.
Spectroscopy explanations needed evidence, not guesses
Question 4 included IR, mass spectrometry and ¹³C NMR analysis.
Students needed to identify functional groups, base peaks, molecular ion peaks, isotope patterns and possible structures. These questions required evidence-based reasoning.
For example, a sample could be classified as an amine or carboxylic acid based on characteristic IR absorptions. A chlorine-containing compound could be recognised through isotope patterns. Isomers could share the same molecular ion peak because they have the same molecular formula.
A strong Section B response does not say “it is this compound” without evidence.
It points to the spectral feature that supports the conclusion.
Titration calculations required the whole analytical pathway
Question 6 involved oxalate ions in spinach and titration with permanganate.
The report noted that common errors involved failing to account for dilution or reaction stoichiometry.
A full analytical calculation required students to:
- use the average titre
- calculate moles of permanganate
- apply the redox mole ratio to determine moles of oxalate
- scale from aliquot to total solution
- calculate the mass of oxalate
- compare the result with the hypothesis
This is exactly what Section B demands.
The student must follow the sample from preparation to aliquot to titre to final conclusion.
A titre is not the endpoint of the reasoning. It is the measured evidence that begins the calculation.
Hypothesis evaluation needed comparison with prediction
The spinach question also asked whether the result supported the hypothesis that boiling would remove 75% of oxalate compounds.
The report indicated that the calculated reduction was 60%, which did not support the hypothesis.
This is a scientific reasoning point.
A hypothesis is not supported merely because there was some reduction. It is supported only if the result matches the predicted reduction closely enough. Since 60% is below the predicted 75%, the hypothesis was not supported.
In Section B, students need to make that evaluative step.
Calculate first.
Then interpret.
Why Section B causes avoidable mark loss
Section B marks are often lost because students stop one step too early.
They calculate energy but omit units.
They identify excess oxygen but do not calculate the amount remaining.
They calculate Q but do not state the direction of shift.
They name a catalyst but do not explain activation energy.
They identify a spectrum but do not cite the evidence.
They calculate a titre result but do not compare it to the hypothesis.
The 2025 report repeatedly showed that high-scoring responses completed the chain.
Equation → calculation → chemical meaning.
Observation → evidence → conclusion.
Condition → principle → effect.
Data → comparison → judgement.
That is Section B Chemistry.
What future Chemistry students should learn from 2025
The 2025 VCE Chemistry exam shows that Section B preparation needs to focus on complete responses.
Students should practise:
- writing thermochemical equations with states and enthalpy values
- showing calculation working clearly
- using units at every stage
- distinguishing ΔH from energy released
- calculating limiting reagents and excess remaining
- explaining fuel comparisons using energy content
- using oxidation numbers when asked
- applying half-equations in electroplating calculations
- interpreting Kc and Q chemically
- setting up equilibrium changes with correct stoichiometry
- writing organic reactions with conditions
- using spectroscopy evidence to justify structures
- completing titration calculations with dilution and mole ratios
- evaluating hypotheses against calculated results
These are the skills that turn topic knowledge into marks.
How ATAR STAR approaches Section B in Chemistry
At ATAR STAR, Section B is taught as structured chemical reasoning.
Students learn how to build answers that move from chemical principle to working to conclusion. They practise writing equations, interpreting data, explaining mechanisms and using evidence precisely, so that each response matches the command and mark allocation.
The 2025 Examination Report confirms why this matters. High-scoring students did not simply recognise the topic.
They completed the chemistry.
That is what Section B rewards.