Mark it yourself: HSC Chemistry calculations and extended responses

How to use this pack

HSC Chemistry Section II rewards precise, complete working more than any other science. This pack covers four question types that appear every year. For each you get the official NESA criteria pattern (quoted from the 2025 HSC Chemistry marking guidelines), a mark-by-mark breakdown, an original practice question with our own top-mark sample answer (written by ExamExplained, not an official NESA exemplar), the errors that cost marks, and a self-marking prompt.

How the Chemistry marking guidelines work

• Calculations are marked by steps. Criteria read "Provides the main steps of the calculation" (partial marks) and "Correctly calculates" (full marks). A wrong final answer with clear, correct steps still earns most of the marks. A right answer with no working is risky.
• The equation is often a separate requirement. In 2025, Question 31(b) (4 marks) required "Correctly calculates the pH" together with "Provides a correct chemical equation".
• Significant figures can be the last mark. Question 25 (3 marks) was "Provides the correct calculations, with correct significant figures"; correct calculations with incorrect significant figures earned 2.
• "Explain" means linked cause and effect. Question 30(b) gave 3 marks for explaining both features, and only 2 for an answer that "Describes how BOTH features affect the reaction but does NOT link cause and effect".
• Big questions use a depth ladder. Question 36 (7 marks) moved from "extensive understanding" (7) to "thorough" (6), "sound" (4 to 5) and "basic" (2 to 3).

Type 1: the multi-step calculation (3 to 4 marks)

Official pattern (2025, Question 31(b), 4 marks): "Correctly calculates the pH" and "Provides a correct chemical equation" (4); "Provides the main steps of the calculation" (3); "Provides some steps of the calculation" OR "Provides a correct chemical equation" (2); "Provides some relevant information" (1).

Practice question (ours). Calculate the pH of a 0.10 mol/L solution of propanoic acid at 25 °C. $K_a$ = 1.3 x 10⁻⁵. Include a relevant chemical equation. (4 marks)

Our sample answer (full marks).

CH₃CH₂COOH(aq) + H₂O(l) ⇌ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

$K_a = \dfrac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}$

Let [H₃O⁺] = [CH₃CH₂COO⁻] = x. Because $K_a$ is small, assume [CH₃CH₂COOH] ≈ 0.10 mol/L.

x² = 1.3 x 10⁻⁵ x 0.10 = 1.3 x 10⁻⁶, so x = 1.14 x 10⁻³ mol/L

pH = -log(1.14 x 10⁻³) = 2.94

Check: 1.14 x 10⁻³ / 0.10 = 1.1% ionised, so the assumption is valid.

Other common mark-losing errors: no equation; an equation with → instead of ⇌ for a weak acid; forgetting the square root; quoting pH to five decimal places; mixing up $K_a$ and p$K_a$.

Type 2: the two-part "explain" question (3 marks)

Official pattern (2025, Question 30(b), 3 marks): "Explains the use of a catalyst" AND "Explains the use of a large excess of CO" (3); "Describes how BOTH features affect the reaction but does NOT link cause and effect" OR "Explains how one feature affects the reaction" (2); "Provides any relevant information" (1).

The lesson: a list of effects is a description. For 3 marks each feature needs "this, BECAUSE that, SO this".

Practice question (ours). When a student prepares an ester in the laboratory, she adds a few drops of concentrated sulfuric acid and heats the mixture under reflux. Explain the purpose of these two features. (3 marks)

Our sample answer (full marks).

Concentrated sulfuric acid is a catalyst: it provides an alternative reaction pathway with a lower activation energy, so a greater proportion of collisions are successful and the slow esterification reaches equilibrium faster. It also absorbs the water produced, and removing a product shifts the equilibrium to the right (Le Chatelier's principle), increasing the yield of ester.

Heating increases the rate because more particles collide with energy greater than the activation energy. Heating under reflux means a vertical condenser cools the vapours of the volatile alcohol, acid and ester and returns them to the flask, so the mixture can be heated for a long time without losing reactants or products, and flammable vapours are not released into the room.

Other common mark-losing errors: saying the catalyst "increases the yield" (a catalyst alone does not change the equilibrium position); explaining only one feature; describing the equipment without its purpose.

Type 3: determining a structure from spectra (4 marks)

Official pattern (2025, Question 37, 4 marks): "Correctly determines the structure of compound A" and "Correctly justifies the structure using all the data provided" (4); correct structure justified using "some of the data provided" OR "Correctly interprets all of the data provided" (3); "Correctly interprets some of the data provided" (2); relevant information (1). The phrase to notice is all the data: every spectrum given must be used.

Practice question (ours). Compound X has the molecular formula C₃H₆O₂. Use the data to identify X, justifying your answer. (4 marks)

Our sample answer (full marks).

X is propanoic acid, CH₃CH₂COOH.

• Mass spectrum: M⁺ at 74 matches C₃H₆O₂ (3 x 12 + 6 x 1 + 2 x 16 = 74).
• IR: the very broad 2500 to 3000 cm⁻¹ band is O-H of a carboxylic acid, and 1710 cm⁻¹ is C=O. Together they indicate a COOH group, which rules out the isomeric esters methyl ethanoate and ethyl methanoate (no O-H).
• ¹³C NMR: three carbon environments, matching three different carbons. 181 ppm is the carbonyl carbon of an acid; 28 and 9 ppm are alkyl carbons.
• ¹H NMR: three hydrogen environments in a 1 : 2 : 3 ratio. 11.7 ppm (1H, singlet) is the COOH proton. The quartet (2H) is a CH₂ next to a CH₃ (3 neighbours, n + 1 = 4), and the triplet (3H) is a CH₃ next to a CH₂ (2 neighbours). This is an ethyl group attached to the COOH.

Other common mark-losing errors: reading the n + 1 rule backwards (a quartet means 3 neighbouring H, not 4); ignoring the integration ratio; naming the compound without drawing or writing its structure.

Type 4: the extended response linking two theories (5 to 7 marks)

Official pattern (2025, Question 35, 5 marks): "Shows how both Le Chatelier's principle AND collision theory account for the colour change, with reference to the energy profile" (5); "a sound understanding of how both collision theory AND Le Chatelier's principle account for the colour change" (4); sound understanding of collision theory OR Le Chatelier's principle (3); "some understanding" of either (2); relevant information (1). The 7-mark Question 36 used the ladder "extensive", "thorough", "sound", "basic".

So the top band needs both theories, applied to the specific system, using any stimulus given. One theory done well caps you at about 60%.

Practice question (ours). Ammonia is made industrially by the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ/mol. Plants typically operate at about 400 to 450 °C and around 200 atm with an iron catalyst. Using collision theory and Le Chatelier's principle, explain why these conditions are chosen. (6 marks)

Our sample answer (top band).

The conditions are a compromise between a high equilibrium yield and a fast rate.

Temperature: the forward reaction is exothermic, so by Le Chatelier's principle a lower temperature shifts the equilibrium to the right and increases the yield of NH₃. However, at low temperatures few N₂ and H₂ molecules collide with energy greater than the high activation energy of this reaction (the N≡N triple bond is very strong), so the rate is too slow. About 400 to 450 °C gives a reasonable rate while keeping an acceptable yield.

Pressure: there are 4 mol of gas on the left and 2 mol on the right. Increasing the pressure causes the equilibrium to shift towards fewer gas molecules, to the right, increasing the yield. Higher pressure also increases the concentration of the gases, so collisions are more frequent and the rate increases. Pressures much above about 200 atm are not used because the equipment and energy costs rise steeply and the extra yield does not justify them.

Catalyst: iron provides an alternative pathway with lower activation energy, so a greater fraction of collisions are successful at 400 to 450 °C and equilibrium is reached faster. It does not change the position of equilibrium or the yield. Ammonia is also removed by cooling and liquefying it, which lowers [NH₃] and keeps shifting the equilibrium to the right, while unreacted N₂ and H₂ are recycled.

Other common mark-losing errors: using only Le Chatelier's principle; saying the catalyst "shifts the equilibrium"; not using the ΔH or mole numbers given in the question; writing "pressure increases collisions" without "because concentration increases".

Your self-marking prompt

Paste this into NSWEduChat, ChatGPT or another AI tool, with the criteria for the matching type above (or the official criteria for a past-paper question).

You are an experienced HSC Chemistry marker. Mark my answer strictly
against the marking criteria below. Do not award a mark the criteria do
not allow, and do not reward length.

Question ([number] marks): [paste the question, data and any equation]

Marking criteria:
[paste the criteria, one line per mark level]

My answer:
[paste your answer, including all working]

1. Redo the calculation or reasoning yourself first and state the correct
   answer, with the balanced equation if one is needed.
2. For each criterion, quote the part of my answer that meets it, or say
   that it is missing (for example: no equation, sig figs wrong, cause and
   effect not linked, one spectrum not used).
3. Give my mark and name the criteria level it matches.
4. Tell me the one change that would gain the next mark. Do not rewrite
   my whole answer.

Next steps

• Practise Section I with the harder exam-style HSC Chemistry quiz.
• Revise equilibrium and acid-base theory in the equilibrium and acid-base guide, and spectroscopy and synthesis in the organic chemistry guide.
• Then write a full NESA past paper under time and mark it with that year's official guideline.