HSC Chemistry 2025

[2 marks]. Condition X is UV (ultraviolet) light, which initiates the free-radical substitution of butane with bromine. The organic product (bromine on carbon 2) is 2-bromobutane.

Marker's note. State a reaction condition (UV light), not the reaction type ("substitution"), and use correct IUPAC nomenclature, naming the product from its structure rather than guessing.

[2 marks]. Qualitative analysis identifies which substances are present, for example detecting that a heavy-metal ion such as $\text{Pb}^{2+}$ or an excess of phosphate is in the water. Quantitative analysis then measures how much of that substance is present, for example the concentration in mg/L. Both matter because a contaminant must first be identified and then measured against the safe guideline level before the water can be judged safe.

Marker's note. Make the distinction explicit (qualitative identifies, quantitative measures amount) and tie it to a water-quality context, such as identifying a specific ion and then comparing its concentration to a safe limit.

[3 marks]. Two changes, each justified by its effect on accuracy:

1. Dry the precipitate to constant mass (reweigh after further drying until the mass stops changing) instead of drying for a fixed 20 minutes in the sun. Twenty minutes may leave trapped water, which adds to the measured mass of $\text{BaSO}_4$ and overstates the sulfate, so drying to constant mass removes that error.
2. Weigh the filter paper before use (or rinse the precipitate with distilled water during filtering). If the filter paper's mass is not subtracted, it is counted as precipitate; rinsing also washes off soluble salts that would otherwise crystallise on the solid and inflate the mass. Both give a truer mass of pure $\text{BaSO}_4$.

Marker's note. Each change must be paired with a reason tied to accuracy (not reliability). Read the given method carefully and target a genuine flaw, such as drying to constant mass, rather than suggesting unnecessary extra steps.

(a) [2 marks]. Ethyne is $\text{H-C}\equiv\text{C-H}$, with a triple bond between the two carbons and one hydrogen on each. The shape about each carbon is linear (180 degrees bond angle).

(b) [3 marks]. Ethyne adds one HCl across one of its bonds to give chloroethene:

Moles of ethyne: $n = \dfrac{65.0}{26.04} = 2.496$ mol.

The mole ratio $\text{C}_2\text{H}_2 : \text{C}_2\text{H}_3\text{Cl}$ is 1 : 1, and HCl is in excess, so $n(\text{C}_2\text{H}_3\text{Cl}) = 2.496$ mol.

Marker's note. In (a) draw the full structural formula (all bonds and atoms) and give the shape, which is distinct from the formula. In (b) write a balanced addition equation (not substitution), use the supplied molar masses, treat HCl as in excess so ethyne is limiting, and round to 3 significant figures.

[3 marks]. Mass of ester made: $m = V \times \rho = 12.2 \times 0.873 = 10.65$ g.

Moles of ester made: $n = \dfrac{m}{M} = \dfrac{10.65}{130.2} = 0.0818$ mol.

Esterification is 1 : 1, and propan-1-ol (0.267 mol) is the limiting reagent because butanoic acid (0.298 mol) is in excess, so the theoretical yield is 0.267 mol.

Marker's note. Use density and volume to get the actual mass, then moles; identify propan-1-ol as limiting so the theoretical yield is 0.267 mol; apply percentage yield (not percentage error or atom economy) and keep significant figures consistent.

(a) [3 marks]. Because the halogen X is categorical (F, Cl, Br, I), plot a column graph: the horizontal axis lists the halogens in order F, Cl, Br, I, and the vertical axis is pKa with a sensible scale (for example 0 to 3.5). Draw columns of height 2.6, 2.9, 2.9 and 3.2 with clear axis labels.

(b) [2 marks]. As the halogen X goes down Group 17 from F to I, the pKa rises (2.6 to 3.2), so the acid strength decreases (a higher pKa means a smaller Ka). Fluoroethanoic acid is the strongest and iodoethanoic acid is the weakest, because the more electronegative halogen withdraws more electron density and stabilises the conjugate base.

Marker's note. In (a) recognise the data are categorical, so a column graph (not a line graph) is correct, with accurate scale, labels and column heights. In (b) describe a trend rather than restating the numbers, and link the rising pKa to falling acid strength down the group.

(a) [2 marks]. Petrol is used as fuel in vehicle engines. Its combustion releases carbon dioxide, $\text{CO}_2$, into the atmosphere. The added $\text{CO}_2$ is a greenhouse gas that enhances the greenhouse effect and contributes to climate change.

(b) [3 marks]. Ethene undergoes acid-catalysed hydration: water adds across the double bond with dilute acid ($\text{H}^+$ / $\text{H}_2\text{O}$) as catalyst and reagent.

Drawn with full structural formulae, the C=C of ethene becomes a C-C single bond, with an H added to one carbon and an OH to the other to give ethanol.

(c) [3 marks]. At $t_1$ the concentration of ethanol jumps up vertically as the extra ethanol is added. The system is now disturbed, so the forward reaction is favoured and some of the added ethanol is consumed: the curve falls as a smooth concave decay, steepest just after $t_1$ and flattening as it approaches a new horizontal plateau. The new equilibrium concentration sits above the original equilibrium value (some, but not all, of the added ethanol reacts).

Marker's note. In (a) name a mixture (petrol), not a single compound, and link its use clearly to a specific impact, distinguishing the greenhouse effect from ozone depletion. In (b) use full structural formulae showing the C-O-H hydroxyl group and place the dilute acid catalyst on the arrow. In (c) show a vertical rise at $t_1$, a consistently concave fall and a straight horizontal new equilibrium line drawn higher than the original.

(a) [1 mark]. The missing monomer is the diamine 1,4-diaminobenzene (para-phenylenediamine), $\text{H}_2\text{N-}\text{C}_6\text{H}_4\text{-NH}_2$, whose two $-\text{NH}_2$ groups react with the two acid chloride groups to form the amide links of Kevlar.

(b) [3 marks]. Polystyrene chains have only dispersion forces between them, which are relatively weak, so the chains slide past one another easily. Kevlar chains carry N-H and C=O groups, so adjacent chains form many hydrogen bonds (as well as dispersion forces). Hydrogen bonds are much stronger than dispersion forces for similarly sized chains, so far more energy is needed to separate Kevlar chains, which is why Kevlar is strong and hard to pull apart while polystyrene is not.

Marker's note. In (b) identify the correct forces in both polymers and make clear it is the forces between whole chains (not between monomers, and not the covalent bonds within a chain) that are being overcome. Refer to the structural features given (the N-H and C=O of Kevlar).

[4 marks]. Adding argon at constant volume does not change the amounts of $\text{NO}_2$ or $\text{N}_2\text{O}_4$, and because the volume is fixed their concentrations (and partial pressures) are unchanged. Argon takes no part in the reaction, so the reaction quotient $Q$ still equals $K_{eq}$. The system is therefore not disturbed: the forward and reverse rates stay equal and the position of equilibrium does not shift. Because no net forward or reverse reaction occurs, no extra heat is released or absorbed, so the temperature of the system remains constant.

Marker's note. Recognise that an inert gas at fixed volume leaves the concentrations unchanged, so $Q = K_{eq}$ and the equilibrium is undisturbed. Avoid claiming a Le Chatelier shift; the key point is that nothing changes, so the temperature is unchanged.

(a) [2 marks]. Carry out the reaction inside a sealed system within a fume hood (or have workers wear suitable respiratory PPE). Because phosgene is a toxic gas, its main risk is being inhaled; a fume hood draws the gas away from workers and prevents it leaking into the workspace, so the precaution directly removes the inhalation hazard.

(b) [3 marks]. Excess carbon monoxide: by Le Chatelier's principle, adding more of a reactant shifts the equilibrium to the right, converting more chlorine into phosgene and so increasing the yield of the valuable product. Catalyst: a catalyst lowers the activation energy, increasing the rate of reaction so that equilibrium is reached faster. This does not change the yield, but it saves time and energy, making the industrial process more cost-effective. Together they give a high yield reached quickly.

Marker's note. In (a) name a specific precaution (fume hood or PPE) and link it to how it reduces the inhalation risk of a toxic gas. In (b) link each feature to its effect with cause and effect: excess CO shifts equilibrium right (more yield), the catalyst lowers activation energy and speeds the rate (saving time and cost) without changing yield.

(a) [2 marks]. By Avogadro's law, equal volumes of gas are equal moles, so the ratios are $\text{N}_x\text{H}_y : \text{O}_2 : \text{NO}_2 : \text{H}_2\text{O} = 1 : 3 : 2 : 2$:

Balancing nitrogen: the products hold $2$ N atoms, so $x = 2$. Balancing hydrogen: the products hold $2 \times 2 = 4$ H atoms, so $y = 4$. (Oxygen already balances: 6 on each side.) Therefore hydrazine is $\text{N}_2\text{H}_4$.

(b) [4 marks]. The hydrazinium ion is a weak acid that reacts with water:

Its acid constant comes from $K_a = \dfrac{K_w}{K_b} = \dfrac{1.0 \times 10^{-14}}{1.7 \times 10^{-6}} = 5.88 \times 10^{-9}$.

Let $x = [\text{H}_3\text{O}^+]$. With $[\text{N}_2\text{H}_5^+] \approx 0.20$ mol/L (x is small):

Marker's note. In (a) build the balanced equation from the volume ratios and use it to confirm both subscripts, not just assert the formula. In (b) write a correct acid equilibrium with a reversible arrow and $\text{N}_2\text{H}_5^+$ as the reactant, find $K_a$ from $K_w / K_b$, then solve the equilibrium expression with clear steps.

[5 marks]. Initial concentrations: $[\text{Mg}^{2+}] = 6 \times 10^{-3}$, $[\text{OH}^-] = 1 \times 10^{-2}$, $[\text{CO}_3^{2-}] = 2 \times 10^{-3}$ mol/L.

Test each possible magnesium salt against its $K_{sp}$ (data sheet: $K_{sp}(\text{Mg(OH)}_2) = 5.61 \times 10^{-12}$, $K_{sp}(\text{MgCO}_3) = 6.82 \times 10^{-6}$).

Magnesium hydroxide:

$Q \gg K_{sp}$, so $\text{Mg(OH)}_2$ precipitates. $\text{Mg(OH)}_2$ is far less soluble than $\text{MgCO}_3$, so it forms first.

Forming $\text{Mg(OH)}_2$ uses the hydroxide and removes some magnesium. The $\text{Mg}^{2+}$ is in excess of the hydroxide ($0.006$ vs the $0.005$ mol needed for $0.010$ mol $\text{OH}^-$), leaving about $[\text{Mg}^{2+}] \approx 6 \times 10^{-3} - 5 \times 10^{-3} = 1 \times 10^{-3}$ mol/L.

Magnesium carbonate with the leftover magnesium:

$Q < K_{sp}$ ($6.82 \times 10^{-6}$), so $\text{MgCO}_3$ does not precipitate.

So only $\text{Mg(OH)}_2$ forms.

Marker's note. Use solubility rules to spot the insoluble salts, compute the ionic product $Q$ for each and compare it to $K_{sp}$. The key insight is that $\text{Mg(OH)}_2$ forms first and consumes magnesium, so less is left for carbonate; with that lower $[\text{Mg}^{2+}]$, $\text{MgCO}_3$ stays in solution. Take care comparing magnitudes of powers of ten.

[7 marks]. This is a back titration: leftover HCl is titrated with KOH, so the HCl that reacted with the carbonate is found by difference.

Average titre (discard Trial 1 as an outlier):

Leftover HCl in one 20 mL aliquot ($\text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O}$, ratio 1 : 1):

The mixture made 100.0 mL, so there are five 20 mL aliquots:

HCl that reacted with the chalk:

Calcium carbonate ($2\text{HCl}(aq) + \text{CaCO}_3(s) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$, ratio 2 : 1):

This matches Brand X (85.5%).

Marker's note. Write the balanced neutralisation and the metal-carbonate reactions with correct states and irreversible arrows. Discard the first (outlier) titre, average the rest, and remember the 100 mL mixture gives five aliquots, so scale the moles by five. Show every step clearly, then compare the calculated percentage to the brands rather than just writing the nearest figure.

(a) [3 marks]. The gentle start and buffer region show the acid is weak and monoprotic, with equivalence at about 0.024 L NaOH.

At the half-equivalence point (0.012 L NaOH added), exactly half the acid is neutralised, so $[\text{HA}] = [\text{A}^-]$ and the equilibrium expression simplifies to $K_a = [\text{H}^+]$. Reading the curve at 0.012 L gives a pH of about 4.4, so:

A consistent alternative uses the starting pH (about 2.5, so $[\text{H}^+] = 3.16 \times 10^{-3}$) with the initial $[\text{HA}] = \dfrac{0.10 \times 0.024}{0.010} = 0.24$ mol/L, giving $K_a = \dfrac{(3.16 \times 10^{-3})^2}{0.237} = 4.2 \times 10^{-5}$.

(b) [2 marks]. After the equivalence point, the extra NaOH is the only source of hydroxide, but it is diluted by the acid solution and water already in the flask, so the hydroxide concentration of the mixture is well below 0.10 mol/L. Some hydroxide was also consumed neutralising the acid. A pH of 13 would need $[\text{OH}^-] = 0.10$ mol/L undiluted, which is never reached, so the pH stays below 13.

Marker's note. In (a) recognise the weak monoprotic acid, read the equivalence volume, then use that pH at half-equivalence equals $\text{p}K_a$ (or work from the start of the curve); annotate the graph and show all working. In (b) explain the two reasons (neutralisation and dilution) with correct cause-and-effect language, and note this is not a buffer beyond the equivalence point.

[5 marks]

The energy profile shows the forward activation energy $E_{a1}$ is smaller than the reverse $E_{a2}$, so the products lie lower in enthalpy and the forward reaction is exothermic (it releases heat).

Le Chatelier's principle

treat heat as a product of the forward reaction. Cooling removes heat, so the system shifts to oppose the change by favouring the forward (exothermic) direction, which replaces the lost heat. This produces more $\text{Co(H}_2\text{O)}_6^{2+}$, so the solution turns pink.

Collision theory (with the energy profile)

lowering the temperature lowers the average kinetic energy, so fewer particles exceed either activation energy and both forward and reverse rates slow. But because the reverse barrier $E_{a2}$ is higher, a greater fraction of reverse-reacting particles now fall below their activation energy, so the reverse rate falls more than the forward rate. The forward reaction is therefore temporarily faster, $[\text{CoCl}_4^{2-}]$ falls and $[\text{Co(H}_2\text{O)}_6^{2+}]$ rises, giving the pink colour.

Marker's note. Use the energy profile to establish the forward reaction is exothermic ($E_{a1} < E_{a2}$). Then use Le Chatelier (cooling favours the exothermic forward direction) and collision theory (cooling slows both rates, but the higher-barrier reverse reaction slows more) to explain the shift to pink. Do not claim activation energy changes with temperature.

[7 marks]. The molecule is propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$, which has a methyl group, a methylene group and a carboxylic acid group: three carbon and three proton environments.

Infrared (IR):

• A strong, broad O-H absorption of the carboxylic acid at roughly 2500 to 3300 cm$^{-1}$.
• A strong, sharp C=O (carbonyl) absorption at about 1700 to 1750 cm$^{-1}$.

Carbon-13 NMR: three signals (three carbon environments):

• $\text{CH}_3$ at about 5 to 40 ppm,
• $\text{CH}_2$ at about 20 to 50 ppm,
• the acid carbonyl carbon at about 160 to 185 ppm.

Proton NMR: three signals (three proton environments):

• $\text{CH}_3$ at 0.7 to 2.1 ppm, a triplet (split by the two adjacent $\text{CH}_2$ protons), integration 3,
• $\text{CH}_2$ at 2.1 to 4.5 ppm, a quartet (split by the three adjacent $\text{CH}_3$ protons), integration 2,
• $\text{COOH}$ at 9.0 to 13.0 ppm, a singlet, integration 1.

Mass spectroscopy: a molecular ion peak ($\text{M}^+$) at $m/z = 74$, the molar mass of $\text{C}_3\text{H}_6\text{O}_2$ (a fragment at 45 from $\text{COOH}^+$ or loss of OH at 57 is also typical).

Marker's note. Read the skeletal structure correctly as propanoic acid (not butanoic, ethanoic or a ketone/alcohol). For IR, distinguish the broad acid O-H plus C=O. For both NMRs, give three environments with the correct shift ranges, and for proton NMR include the triplet/quartet/singlet multiplicity and 3 : 2 : 1 integration. For MS, calculate the molar mass and give the molecular ion at 74.

[4 marks]. $\text{C}_5\text{H}_{10}$ has one degree of unsaturation; the reaction with $\text{H}^+$ / $\text{H}_2\text{O}$ (hydration) shows A is an alkene, adding water across the C=C to give two alcohols B and C (so the double bond is internal and not symmetric, giving two products).

• B does not oxidise with acidified dichromate, so B is a tertiary alcohol. A tertiary alcohol needs a carbon bonded to three other carbons, so A must have a branch.
• C oxidises with dichromate to D, so C is a primary or secondary alcohol. D does not react with $\text{Na}_2\text{CO}_3$, so D is not a carboxylic acid; it must be a ketone, which means C is a secondary alcohol.

For hydration of one alkene to give both a tertiary and a secondary alcohol, the double bond must lie between a branched (tertiary-forming) carbon and an internal (secondary-forming) carbon. This fits 2-methylbut-2-ene, $\text{(CH}_3)_2\text{C}{=}\text{CHCH}_3$:

• adding OH to the more substituted carbon gives 2-methylbutan-2-ol (tertiary, B),
• adding OH to the other carbon gives 3-methylbutan-2-ol (secondary, C), which oxidises to the ketone 3-methylbutan-2-one (D).

So compound A is 2-methylbut-2-ene.

Marker's note. Work through each clue: hydration shows an alkene; no oxidation of B means a tertiary alcohol (so a branch); oxidation of C to a non-acidic D means a secondary alcohol oxidising to a ketone. Combine these to place the double bond internally on a branched chain, giving 2-methylbut-2-ene, and draw the structure accurately.