Free this weekUnlimited AI marking for every account until Sunday 1 November. Type any practice answer, get a mark and the reason for every mark lost, in your subject's terms.Create a free account
…
Admin Owner (VCE) @calmfalcon-4b8c90248b4f1h ago

2026 HSC Chemistry exam: What to learn from last year's mistakes

Hi all, in preparation for the upcoming exam, here's what to learn from last year's mistakes. Exam: Friday 30 October 2026, 9:25 am to 12:30 pm. All times are NSW local time. The 2025 paper was 100 marks: 20 multiple choice and 80 in Section II. Questions worth redoing: Q23 (3 marks), Q26a–b (5), Q33 (7), Q35 (5) and Q36 (7). NESA does not publish question averages in this feedback. These are the specific calculation and explanation problems it identified. 1. Q33: Scale the aliquot before subtracting the leftover acid Where students went wrong Students included the initial outlier titre, missed that there were five aliquots, omitted the equations or reported a brand’s labelled percentage as their own calculation. What they should have done The accepted titres were 7.10, 7.20 and 7.15 mL: average 7.15 mL = 0.00715 L. HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l). n(KOH) = 0.10 × 0.00715 = 0.000715 mol. Leftover HCl in the full solution = 5 × 0.000715 = 0.003575 mol. Initial HCl = 0.550 × 0.1000 = 0.0550 mol. HCl consumed by the chalk = 0.0550 − 0.003575 = 0.051425 mol. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). n(CaCO₃) = 0.051425 ÷ 2 = 0.0257125 mol. Mass = 0.0257125 × 100.09 = 2.573564125 g. Percentage in the 3.00 g sample = 2.573564125 ÷ 3.00 × 100 ≈ 85.8%. That identifies Brand X. Keep intermediate precision and show the calculated percentage before matching the brand. 2. Q36: Identify propanoic acid before predicting spectra Where students went wrong Students misread the skeletal structure, missed carbon environments or gave proton NMR signals without integration and splitting. What they should have done For CH₃CH₂COOH: Mass spectrum: molecular ion at m/z 74. Carbon-13 NMR: three carbon environments, including the carboxyl carbon. Proton NMR: CH₃ triplet, integration 3; CH₂ quartet, integration 2; acid proton singlet, integration 1. IR: a strong carbonyl absorption near 1700 cm⁻¹ and the broad carboxylic-acid O–H absorption. Attach each signal to the part of the molecule producing it. For proton NMR, use the data sheet to give the relevant chemical-shift ranges too. Three signals alone does not address all four techniques. 3. Q35: Cooling does not change activation energy Where students went wrong Answers needed both collision theory and Le Chatelier’s principle. Some confused an energy profile with a concentration graph or claimed temperature changed activation energy. What they should have done The forward reaction was exothermic, so cooling favoured the forward direction and the pink complex. Immediately after cooling, both reaction rates decrease, but the reverse rate decreases more because its activation energy is higher. The forward rate is then greater until a new equilibrium is reached. Explain that temporary difference; at the new equilibrium the rates are equal again. 4. Q23: Improve accuracy, not just repeatability Where students went wrong Students proposed repeated trials without addressing the source of error in the gravimetric procedure, or suggested changes the procedure already included. What they should have done Identify two relevant procedural changes and justify each. Drying the precipitate to constant mass prevents retained water from being counted as sulfate-containing precipitate. Check the given procedure before choosing the other change, such as ensuring complete precipitation or avoiding loss during transfer. Explain the resulting measurement error, not merely that the result becomes more accurate. 5. Q26a–b: Use the graph type and logarithm correctly Where students went wrong Students used a line graph for categorical halogen data or listed pKa values without explaining the trend in acid strength. What they should have done Use a column graph with labelled axes and an appropriate scale. Since pKa = −log₁₀Ka, a higher pKa means a lower Ka and a weaker acid. Explain the trend across the supplied haloethanoic acids rather than treating a larger pKa as greater acid strength. Source: NESA 2025 Chemistry exam, marking guidelines and feedback. Use the Chemistry hub and past papers and marking guides alongside this.
0 repliesReply

Replies

Nothing yet
No replies yetBe the first to answer. Even a partial answer helps the next person.
Know the answer? Replies need a free account.Log in to reply →