HSC Chemistry 2023

[2 marks].

Butan-2-one (a ketone) and butanal (an aldehyde) share the formula C4H8O but have different functional groups, so they are functional group isomers. Butanal and 2-methylpropanal are both aldehydes with the same functional group but a different carbon skeleton (straight chain versus branched), so they are chain isomers.

Marker's note. Isomers share a molecular formula. Functional group isomers differ in the functional group present; chain isomers keep the same functional group but rearrange the carbon skeleton. Pair the compounds correctly and read all of the structures before choosing.

[4 marks]. Under the Arrhenius definition an acid ionises in water to produce H+ ions.

HCl produces H+ in water, so it is an Arrhenius acid. NH4Cl is a salt: in water it gives ammonium and chloride ions, not H+ directly, so Arrhenius does not classify it as an acid.

Under the Bronsted-Lowry definition an acid is a proton donor. HCl donates a proton to water and is a Bronsted-Lowry acid. The ammonium ion also donates a proton to water:

so NH4Cl behaves as a Bronsted-Lowry acid even though it is not an Arrhenius acid.

Marker's note. Classify both substances under both theories and include charges in every ionic equation. The key point is that NH4Cl is a salt: it is a Bronsted-Lowry acid (the ammonium ion donates a proton) but not an Arrhenius acid because it does not itself ionise to give H+.

[3 marks]. Both solutions start neutral with equal hydronium and hydroxide ions. In solution X there is nothing to resist the change, so the added OH- ions react with the small amount of hydronium present and then build up, sharply lowering [H3O+] and raising the pH from 7.00 to 12.00 (since $\text{pH}=-\log_{10}[\text{H}_3\text{O}^+]$).

Solution Y behaves as a buffer. It contains a weak acid and its conjugate base, so the added OH- ions are consumed by the weak acid, shifting that equilibrium and keeping [H3O+] almost constant. The pH therefore barely moves, from 7.00 to 7.02.

Marker's note. Identify that solution Y is a buffer that minimises pH change while solution X is not. Explain in terms of OH- being added: in X the hydronium ions fall and the pH rises, while in Y the buffer reacts with the OH- so [H3O+] and pH change little.

[2 marks]. An amphiprotic species can both accept and donate a proton. The hydrogen oxalate ion can accept a proton (acting as a base):

and donate a proton (acting as an acid):

Because it does both, HC2O4- is amphiprotic.

Marker's note. Show both behaviours with balanced equations and watch the charges. Accepting a proton gives neutral oxalic acid, while donating one gives the oxalate ion with a 2- charge.

(a) [3 marks]. Heat gained by the water:

This equals the heat released by the octan-1-ol. Find the moles burnt:

Then the mass:

(b) [2 marks]. Biofuels from plants are close to carbon neutral: the carbon dioxide released on combustion was recently absorbed from the atmosphere during photosynthesis as the plant grew, so burning the fuel adds little net carbon dioxide. Fossil fuels release carbon that was locked away for millions of years, increasing atmospheric carbon dioxide and the greenhouse effect.

Marker's note. In (a) use the mass of water (not octan-1-ol) in $q=mc\Delta T$, keep the units of q and delta H the same (both kJ), then divide to get moles before finding the mass. In (b) state a precise advantage and link it to its effect, rather than giving a generic statement.

(a) [2 marks]. In Reactor 2 there is the equilibrium $2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}$. In Reactor 3 the NO2 is consumed by the reaction with water, removing a reactant from this equilibrium. By Le Chatelier's principle the system shifts to oppose the loss of NO2, so the equilibrium moves to the left and N2O4 decomposes to replace the NO2. As NO2 keeps being removed, all of the N2O4 eventually decomposes.

(b) [3 marks].

1. The water produced at Separator 1 can be recycled into Reactor 3, where water is a reactant, instead of being sent to disposal. This conserves water and avoids the cost of supplying fresh water.
2. The heat released by the cooler/condenser after Reactor 1 can be captured and used in the emissions control (catalytic reduction) step, which needs energy input. This reduces overall energy consumption and running costs.

(Recycling the NO produced in Reactor 3 back into Reactor 2 is also acceptable.)

Marker's note. In (a) refer to the Reactor 2 equilibrium and state that removing NO2 shifts it left, decomposing the N2O4. In (b) name a specific improvement tied to the flow chart (for example, recycling Separator 1 water into Reactor 3) with a precise reason, not a generic "reduce waste".

[4 marks]. Mass of ethanol from the density:

Moles of ethanol:

From the equation the mole ratio of ethanol to carbon dioxide is 1 : 1, so $n(\text{CO}_2) = 3.17$ mol. Using the ideal gas law $PV=nRT$ with the data sheet value $R = 8.314$ J K-1 mol-1:

Marker's note. Use the density to get mass, then moles, then the 1 : 1 ratio to find moles of carbon dioxide. Take $PV=nRT$ and $R$ from the data sheet, keep pressure in kPa to give the volume in litres, and do not round until the end.

(a) [2 marks]. Add a few drops of orange-brown bromine water to a sample of Q in a test tube and shake. If Q is an alkene the bromine water is rapidly decolourised (turns colourless) as bromine adds across the carbon-carbon double bond. (Acidified potassium permanganate, decolourising from purple, is also acceptable.)

(b) [3 marks]. The molecular ion in the mass spectrum is at m/z = 114. Mass of carbon in R:

Non-carbon mass $= 114 - 36 = 78$. With three carbons and their hydrogens, and the addition of Cl2, this mass corresponds to two chlorine atoms, so R is C3H6Cl2. Since R forms by chlorine adding across the double bond of a propene, R is 1,2-dichloropropane:

Marker's note. A chemical test must involve a reaction with an observation (bromine water or permanganate decolourising), not a physical property or spectroscopy. In (b) use the molecular ion and the 32% carbon figure to get three carbons, deduce two chlorines, and place them on adjacent carbons consistent with addition of Cl2 across the double bond.

[3 marks]. As the molar mass increases the carbon chain gets longer, and the graph shows the solubility in water decreasing. Every alkan-1-ol has a hydroxyl (-OH) group that can hydrogen bond to water, which is what makes the short-chain alcohols soluble. As the chain lengthens, the non-polar alkyl part grows and its dispersion forces become more significant relative to the single hydrogen-bonding -OH group. The longer chain cannot hydrogen bond with water and instead disrupts the water structure, so the proportion of the molecule that dissolves well falls and solubility decreases.

Marker's note. Link the trend to named intermolecular forces. Strong answers state that the -OH allows hydrogen bonding with water (giving solubility) but that the growing alkyl chain increases dispersion forces and reduces solubility as chain length increases.

[4 marks]. Test in sequence so the carbonate is removed before testing for bromide:

1. Add dilute nitric acid to a sample. Effervescence (bubbles of a colourless gas that turns limewater milky) confirms carbonate, and the acid removes it ready for the next test:

2. To the remaining acidified solution add silver nitrate solution. A creamy (pale yellow) precipitate of silver bromide confirms bromide:

Acidifying first prevents silver carbonate forming and being mistaken for the bromide precipitate.

Marker's note. Give a logical sequence: test for and remove carbonate with acid first, then test for bromide with silver nitrate. Include states, colours of precipitates and gases, and two balanced equations.

[7 marks]. Plot absorbance (y-axis) against complex concentration (x-axis) from the calibration data and draw a straight line of best fit through the origin. The points lie close to a line of gradient about 14.4 (absorbance per mol L-1).

Reading the absorbance of 0.66 from the line of best fit gives an equilibrium complex concentration of about $0.046$ mol L-1.

Set up an ICE table (concentrations in mol L-1):

The lactic acid change is twice the complex change because of the 1 : 2 stoichiometry. The equilibrium constant:

Marker's note. Use a realistic axis scale, plot the calibration line and show on the graph how you read 0.66 off it. Then give the Keq expression and a clear ICE table, applying the 1 : 2 mole ratio to the lactic acid, before substituting to find Keq.

[5 marks]. Discard the first titre (22.65 mL) as an outlier and average the three concordant results:

Moles of HCl, equal to the excess NaOH in one 20.00 mL aliquot:

Scale up to the whole 250.0 mL flask:

Total NaOH added at the start:

NaOH that reacted with the ammonium ions:

The reaction is 1 : 1, so $n(\text{NH}_4^+) = 2.590\times10^{-2}$ mol. The mass:

Marker's note. Exclude the first titre as an outlier before averaging. Work from the excess HCl in the aliquot, scale by 250.0/20.00 to the whole flask, subtract from the total NaOH to get the NaOH that reacted with ammonium, then convert to a mass to four significant figures without rounding early.

(a) [2 marks]. Between 6 and 8 minutes the system is at equilibrium. The amounts of A2, B2 and AB2 stay constant because the forward and reverse reactions are occurring at equal rates, so there is no net change in the amount of any substance.

(b) [4 marks]. After 8 minutes AB2 decreases while A2 and B2 increase, so the equilibrium shifts towards the reactants. Two factors could cause this:

1. An increase in temperature. The forward reaction is exothermic, so adding heat shifts the equilibrium to the left to absorb the added heat. This also lowers the equilibrium constant K, so the reaction quotient Q now exceeds K and AB2 is consumed while A2 and B2 are produced until Q again equals K.
2. An increase in the container volume (decrease in pressure). The reactant side has three gas moles (1 + 2) and the product side has two, so the equilibrium shifts to the side with more gas molecules (the reactants) to oppose the pressure drop. Here K is unchanged but Q increases, so AB2 is consumed and A2 and B2 are produced until equilibrium is re-established.

Marker's note. In (a) explain (give a reason) rather than describe: equal forward and reverse rates, so amounts stay constant. In (b) the two valid factors are an increase in temperature (exothermic, so shift left to absorb heat) and an increase in volume (shift to the side with more gas moles). State the specific effect, not just the direction, and remember the y-axis shows amount in moles.

[5 marks]. The dissolution equilibrium is $\text{MgF}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{F}^-\text{(aq)}$.

Moles of MgF2 precipitated:

Initial moles of fluoride:

Each mole of MgF2 removes 2 mol of fluoride, so fluoride remaining:

Total volume is $0.125 + 0.175 = 0.300$ L, so:

Because Ksp is very small, almost all the magnesium has precipitated and the equilibrium [F-] is essentially this value. Rearranging $K_{sp}=[\text{Mg}^{2+}][\text{F}^-]^2$:

Marker's note. Provide the balanced equation and the correct Ksp expression. Find the leftover fluoride (subtract twice the moles of solid from the initial fluoride), convert to a concentration using the total combined volume, then solve for [Mg2+]. Work in mol L-1, not moles, inside the Ksp expression.

(a) [3 marks]. Find [H+] from the pH:

Set up an ICE table for $\text{CHCl}_2\text{COOH} \rightleftharpoons \text{H}^+ + \text{CHCl}_2\text{COO}^-$ (do not assume the acid is weak; here a significant fraction ionises):

(b) [3 marks]. A lower pKa means a stronger acid, so trichloroacetic acid (pKa 0.51) is much stronger than acetic acid (pKa 4.76). The strength is governed by delta G of ionisation, and the dominant difference between the two acids is the entropy term. Both delta S values are negative (an unfavourable, positive contribution to delta G), but acetic acid has a much more negative delta S (-91.6 J K-1 mol-1, giving -T delta S = +27.3 kJ mol-1) than trichloroacetic acid (-5.8 J K-1 mol-1, -T delta S = +1.7 kJ mol-1). This makes delta G for acetic acid (+27.2 kJ mol-1) far more positive (less favourable) than for trichloroacetic acid (+2.9 kJ mol-1). Ionisation is therefore much more favourable for trichloroacetic acid, which is why it is the stronger acid.

Marker's note. In (a) do not assume the acid is weak: use the measured [H+] and an ICE table so the acid concentration falls from 0.2000 to 0.1218 before substituting into the Ka expression. In (b) connect acid strength to pKa, then use the data coherently: the large difference in delta S (and so -T delta S and delta G) explains why trichloroacetic acid ionises more readily and is stronger.

[9 marks]

Compound A - alkene

A reacts with dilute H2SO4 to add water across a double bond, and it also adds H2 on heating, both of which are alkene reactions. Its 13C NMR has three peaks (three carbon environments), with the peak near 132 ppm characteristic of the alkene C=C carbons. A molar mass of 84.156 g mol-1 corresponds to C6H12 (a hexene), and having only three carbon environments for six carbons means the molecule is symmetrical: A is hex-3-ene, CH3CH2CH=CHCH2CH3.

Compound B - alcohol

B is the hydration product of the alkene, so it is an alcohol. Its IR spectrum shows a broad absorption near 3400 cm-1, the signature of the O-H stretch of a hydroxyl group. Adding water across the symmetrical hex-3-ene gives the secondary alcohol hexan-3-ol, CH3CH2CH(OH)CH2CH2CH3.

Compound C - ketone

Oxidising B with acidified KMnO4 gives C. A primary alcohol would oxidise to a carboxylic acid, but the 1H NMR of C has no peak in the 9.0 to 13.0 ppm region, so C is not a carboxylic acid (and B is therefore a secondary alcohol). C is a ketone. The five 1H NMR environments fit hexan-3-one, CH3CH2COCH2CH2CH3:

• 1.01 ppm triplet (area 3) and 1.05 ppm triplet (area 3): the two terminal CH3 groups, each next to a CH2;
• 1.65 ppm multiplet (area 2): a CH2 with several neighbouring hydrogens;
• 2.42 ppm triplet (area 2): the CH2 next to the C=O on the propyl side;
• 2.46 ppm quartet (area 2): the CH2 next to the C=O on the ethyl side, split by the adjacent CH3.

Marker's note. Identify the reactions (hydration then oxidation) and read each spectrum: the 132 ppm 13C peak and symmetry for the alkene A, the broad 3400 cm-1 IR band for the alcohol B, and the absence of a 9 to 13 ppm 1H peak (so a ketone, not an acid) for C. Draw structures consistent with the 84.156 g mol-1 molar mass and the splitting and integration data, with every carbon having four bonds.

(a) [2 marks]. Solid carbon is omitted from the expression. The reaction quotient:

Since $Q = K_{eq} = 10.00$, the system is at equilibrium.

(b) [3 marks]. At the new equilibrium [CO] = [CO2], so:

Build an ICE table from the original equilibrium (in a 1.00 L vessel, so amounts equal concentrations):

The CO concentration rises by $0.1000 - 0.0110 = 0.0890$ mol L-1. By the stoichiometry, making 0.0890 mol of CO requires consuming $0.0890/2 = 0.0445$ mol of CO2. The CO2 concentration also rises overall by $0.1000 - 0.00121 = 0.0988$ mol L-1.

The carbon dioxide added must supply both the net increase in CO2 and the CO2 consumed in making the extra CO:

Marker's note. In (a) leave the solid carbon out of the expression, calculate Q and compare with K. In (b) use [CO] = [CO2] in the Keq expression to find both equal 0.1000 mol L-1, then account for the CO2 added going both to raise [CO2] and to form the extra CO (via the 1 : 2 ratio) when adding up the total moles added.