HSC Chemistry 2024

[2 marks]. A metal plus an acid gives a salt plus hydrogen gas. The two products are magnesium acetate (magnesium ethanoate, $\text{Mg(CH}_3\text{COO)}_2$) and hydrogen gas ($\text{H}_2$).

Marker's note. Name both products; the salt is magnesium acetate, not "magnesium acid" or "magnesium carbonate". The acetate anion comes from the acid losing its acidic hydrogen.

[2 marks]. Addition polymerisation opens the C=C double bond and joins monomers into a saturated chain. A six-carbon (three-monomer) section, with both ends shown continuing:

That is three repeat units of $\text{-CHF-CH}_2\text{-}$ (six carbons), with bonds shown continuing at both ends.

Marker's note. Show every bond and make sure each backbone carbon has four bonds. Use the continuation lines at both ends to signal a section of polymer, and place the fluorine on alternate carbons (the double bond is what opens during addition).

[3 marks]. Becoming more blue means the concentration of the blue $[\text{CoCl}_4]^{2-}$ ion (a product) has increased, so heating has shifted the equilibrium to the right, favouring the forward reaction.

The equilibrium expression is

Because the product concentration in the numerator has risen and the reactant concentrations in the denominator have fallen, $K_{eq}$ increases. (This tells us the forward reaction is endothermic.)

Marker's note. Link the blue colour to a rise in $[\text{CoCl}_4^{2-}]$, then to the forward reaction being favoured, then to a larger numerator and so a larger $K_{eq}$. If "left" and "right" are confusing, write "reactant" and "product".

(a) [3 marks]. Plot carbon number (1 to 4) on the x-axis and boiling point on the y-axis. Use a scale that fills the grid, for example $-10$ to 120 degrees C in 10-degree increments, plot all eight points, and add a labelled key with a different symbol for each series.

(b) [4 marks]. Within each series, boiling point rises with carbon number. A longer carbon chain has more electrons, so the temporary dipoles are larger and the dispersion forces between molecules are stronger, needing more energy to separate them.

Between the series, the alcohol always boils higher than the amine with the same carbon number. Both can hydrogen bond (both have an electronegative atom, O or N, bonded to H), but oxygen is more electronegative than nitrogen, so the O-H...O hydrogen bonds in alcohols are stronger than the N-H...N hydrogen bonds in amines. The stronger hydrogen bonding raises the alcohols' boiling points above the amines'. The gap narrows along the series because, as the chain lengthens, dispersion forces increasingly dominate over hydrogen bonding.

Marker's note. Hydrogen bonding requires N, O or F bonded to H; use electronegativity (O greater than N) to explain why alcohols sit higher. Explain the rise within a series by increasing dispersion forces with more electrons. On the graph, use an even scale that fills the page and clearly distinct symbols for the two series.

[4 marks]. Read the concentration of the diluted solution from the calibration graph at absorbance 0.64. This gives about 0.78 mg L$^{-1}$ of phosphate.

Reverse the dilution. The sample was diluted 1.00 mL to 1.000 L, a factor of 1000:

Convert to mol L$^{-1}$ using the molar mass of $\text{PO}_4^{3-}$:

So the waste water contains about $8.2 \times 10^{-3}$ mol L$^{-1}$ phosphate.

Marker's note. Use a ruler to interpolate the graph and check the scale. Convert mg to g and apply the 1000-fold dilution factor the right way (the original is more concentrated). Use the correct molar mass for the polyatomic phosphate ion, and label each step.

(a) [2 marks]. Start at $2.0\times10^{-5}$ mol L$^{-1}$ at $t=0$. The curve falls, steeply at first then more gently, reaching $1.0\times10^{-5}$ mol L$^{-1}$ (half) at $t=3$ min, then stays flat (horizontal) at $1.0\times10^{-5}$ mol L$^{-1}$ from $t=3$ to $t=6$ min.

(b) [3 marks]. At $t=0$ the $[\text{I}_2]$ is at its highest, so $\text{I}_2$ particles collide most frequently with HCN; with the most frequent successful collisions per second, the forward reaction rate is at its maximum. As $\text{I}_2$ is consumed, $[\text{I}_2]$ falls, collisions become less frequent and the rate slows. By $t=3$ min the system is at equilibrium: the forward and reverse rates are equal, $\text{I}_2$ is being made as fast as it is consumed, so $[\text{I}_2]$ stays constant at $1.0\times10^{-5}$ mol L$^{-1}$ to $t=6$ min.

Marker's note. Link a higher $[\text{I}_2]$ to more frequent successful collisions and a faster rate, and a falling $[\text{I}_2]$ to a slowing rate. Make clear that beyond $t=3$ min the rate has not stopped: forward and reverse rates are equal. This is collision theory and rate, not a Le Chatelier shift.

[4 marks]. When only barium ions are present, step 1 precipitates barium sulfate:

so a precipitate correctly identifies barium, and step 3 gives no precipitate, correctly showing no lead. The result is right.

When both ions are present, the problem is that lead(II) sulfate is also insoluble, so step 1 precipitates both barium sulfate and lead(II) sulfate ($\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s)$). Step 2 then filters out both solids, removing the lead from the solution. By step 3 there are essentially no lead(II) ions left in the filtrate to react with bromide, so no precipitate forms and the test wrongly concludes that lead is absent. This is a false negative for lead.

Marker's note. State that lead(II) sulfate is also insoluble and so precipitates in step 1 alongside barium sulfate, then is filtered off in step 2, leaving no lead for step 3. Include a balanced equation with states. The two white precipitates in step 1 also cannot be told apart by eye.

[3 marks]. Both solutions have pH 1.151, so for each:

For iodic acid the actual $[\text{H}^+]$ (0.0706) is less than the acid concentration (0.100 mol L$^{-1}$), and for sulfamic acid 0.0706 is less than 0.120 mol L$^{-1}$. If either acid had dissociated completely, a monoprotic acid would give $[\text{H}^+]$ equal to its concentration. Since both produce less $\text{H}^+$ than that, neither dissociates completely, so both are weak acids.

Both give the same $[\text{H}^+]$, but iodic acid does so from a lower starting concentration (0.100 versus 0.120 mol L$^{-1}$). To reach the same $[\text{H}^+]$ from less acid, iodic acid must ionise to a greater extent, so iodic acid is the stronger acid.

Marker's note. Use $[\text{H}^+] = 10^{-\text{pH}}$ and compare it with the stated acid concentration: a smaller $[\text{H}^+]$ proves incomplete dissociation. You are told both acids are monoprotic, so do not waste time deducing formulae. The stronger acid is the one giving the same $[\text{H}^+]$ from the lower concentration.

[4 marks]. Find the moles, then the excess.

The reaction is

Each mole of acid needs 2 moles of NaOH, so 0.010 mol H2SO4 consumes $2\times0.010 = 0.020$ mol NaOH. NaOH is in excess:

In 250 mL:

Marker's note. Use the 1:2 mole ratio from a balanced equation, identify NaOH as the excess reagent, and divide its excess moles by the total 250 mL volume. Go via pOH then pH. Show every step.

[4 marks]. Adding CO (a product) pushes the equilibrium left, so $[\text{H}_2\text{O}]$ falls. The change in $[\text{H}_2\text{O}]$ is $0.400 - 0.200 = 0.200$ mol L$^{-1}$ (it is consumed). With 1:1:1:1 ratios, set up an ICE table, letting $x$ be the moles of CO added (per litre):

$K_{eq}$ is unchanged at constant temperature:

So about 4.92 mol of CO was added (the container is 1 L).

Marker's note. Recognise that adding product shifts the system left and raises reactant concentrations. Add the unknown $x$ to the initial CO. Keq does not change at constant temperature, so substitute the equilibrium row and solve for $x$.

[3 marks]. The DMC process is preferable, on three grounds from the data:

• Atom economy: the DMC route has AE 48.4% versus 35.9% for phosgene, so a larger fraction of the starting material ends up as urea and less is wasted.
• Toxicity: DMC has low toxicity, whereas phosgene has high toxicity, making the DMC route safer for workers and the environment.
• Reactant quantity: the DMC route uses only 2 moles of ammonia per urea, against 4 moles for the phosgene route, so it consumes less reactant.

On all three measures, higher atom economy, lower toxicity and less ammonia, the DMC process is the better choice for urea production.

Marker's note. Use the figures given (48.4% versus 35.9%, low versus high toxicity, 2 versus 4 mol ammonia) rather than general green-chemistry statements, and make a clear comparative judgement that DMC is preferred.

[4 marks]. Write the dissolution equation and let $s$ mol of $\text{Cd}_3(\text{PO}_4)_2$ dissolve per litre:

So $[\text{Cd}^{2+}] = 3s$ and $[\text{PO}_4^{3-}] = 2s$. The solubility product is

Solve for $s$:

Therefore

Marker's note. Write the balanced equation and the matching $K_{sp}$ expression, expand $(3s)^3(2s)^2$ to $108s^5$ (watch the order of operations), solve for $s$, then multiply by 3 for the cadmium ion. Quote the final answer to 3 significant figures.

(a) [2 marks]. In acetone the central carbon is double bonded to oxygen and single bonded to two methyl groups: three regions of electron density, so the shape is trigonal planar. In propan-2-ol the central carbon has four single bonds (to OH, H and two CH3): four regions, so the shape is tetrahedral.

(b) [3 marks]. At the start, acetone has two carbon environments, the carbonyl carbon (C=O) and the two equivalent methyl carbons, giving two signals in the $^{13}$C NMR spectrum. As the reduction proceeds, the C=O carbon becomes an H-C-OH carbon at a different chemical shift, so a new signal appears for that product carbon while the original carbonyl signal shrinks and disappears. Following the loss of the carbonyl signal and the growth of the new alcohol-carbon signal tracks the reaction from start to finish.

Marker's note. Name both shapes in full (trigonal planar and tetrahedral, not "planar" and "bent"). For (b), explain the specific change in the carbon environment (C=O to C-OH) and how a signal appearing while another disappears lets you monitor progress.

[4 marks]. Before the equivalence point (up to about 4.5 mL), the added ammonia neutralises the hydrochloric acid:

This replaces highly conducting H$^+$ ions (relative conductivity 4.76) with poorly conducting NH4$^+$ ions (1.00), while the Cl$^-$ count is unchanged, so the conductivity falls. The minimum is at the equivalence point, where almost all the H$^+$ has been converted to NH4$^+$.

After the equivalence point, the excess ammonia is a weak base that ionises only slightly:

This adds a few extra NH4$^+$ and OH$^-$ ions, so the conductivity rises slightly. The rise is small because the NH4$^+$ already in solution suppresses this ionisation (Le Chatelier's principle), keeping $[\text{OH}^-]$ low.

Marker's note. Both balanced equations are needed. Explain the fall as H$^+$ (high conductivity) being replaced by NH4$^+$ (low conductivity), and the small rise as the weak ionisation of excess ammonia adding a little NH4$^+$ and OH$^-$, suppressed by common-ion NH4$^+$, not simply "OH$^-$ has low conductivity".

[7 marks]. Work through each clue.

X has no C=C. X does not react with bromine in the dark (no addition) and does not hydrate, so it has no carbon-carbon double bond. It must be the saturated acid. Structure 2 ($M=74.078$, propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$) is the only saturated carboxylic acid, so X is Structure 2.

Y and Z both have C=C (they add bromine and hydrate). Comparing them:

• 0.100 g of Y reacts with the same bromine as 0.200 g of Z, so for equal moles of C=C, Y has half the mass of Z. This means Z has double the molar mass of Y (consistent with $144.124 \approx 2 \times 72.062$).
• Hydration of Z gives only one product, so Z is symmetrical about its double bond; Y gives two products, so Y is not symmetrical.

Titration confirms the protons. For 0.100 g, X (monoprotic, $M=74.078$) gives $0.100/74.078 = 1.35\times10^{-3}$ mol, and Y (if $M=72.062$, monoprotic) gives $0.100/72.062 = 1.39\times10^{-3}$ mol, matching their near-equal titres. Z has double Y's molar mass but the same titre as Y for 0.100 g:

Z neutralises the same amount of NaOH as Y only because Z is diprotic (two COOH groups), which fits Structure 3 ($M=144.124$, a symmetrical diacid).

Conclusion: X is Structure 2 (saturated, propanoic acid), Y is Structure 1 ($M=72.062$, unsymmetrical unsaturated monoprotic acid), Z is Structure 3 ($M=144.124$, symmetrical diprotic unsaturated acid).

Marker's note. Use every piece of data: the bromine test for C=C, the mass ratio for the molar-mass relationship, the "one versus two hydration products" for symmetry, and the equal titres with double molar mass to show Z is diprotic. Sequence the reasoning and identify the exact double bond you mean.

[5 marks]. The reaction is

Find moles of each reactant:

The ratio is 1:1, so HCl (0.180 mol) is in excess and NaHCO3 (0.175 mol) is the limiting reagent.

Mass of solution for $q = mc\Delta T$: add the solid and the acid, then subtract the CO2 that escapes.

The solution cooled, so the reaction is endothermic. Per mole of NaHCO3:

Marker's note. Use density to convert the acid volume to mass, add both reactant masses, then subtract the escaping CO2 before computing $q$. Identify NaHCO3 as the limiting reagent and divide by its moles. The temperature fell, so $\Delta H$ is positive.

[3 marks]. The graph shows that as $\Delta G^\circ$ becomes large and negative, $K_{eq}$ becomes extremely large. A very large $K_{eq}$ means the products vastly outnumber the reactants at equilibrium, so the reactants are negligible and the reaction effectively goes to completion.

Calculate $\Delta G^\circ$ for this reaction:

This is large and negative (dominated by the very exothermic $\Delta H^\circ$; the small $T\Delta S^\circ$ term barely matters). A $\Delta G^\circ$ this negative corresponds, from the graph, to an enormous $K_{eq}$, so the position of equilibrium lies so far towards products that the reaction effectively proceeds to completion.

Marker's note. Combine $\Delta H^\circ$ and $T\Delta S^\circ$ with the correct signs to get $\Delta G^\circ = -214$ kJ mol$^{-1}$, then read the graph: a large negative $\Delta G^\circ$ gives a very large $K_{eq}$, hence effectively complete reaction. The graph is $K_{eq}$ against $\Delta G^\circ$, not a reaction-versus-time plot.

[7 marks].

Identify the halogen and A and B. The mass spectrum of A has peaks at m/z 122 and 124 of roughly equal height: a pair separated by 2 with similar abundance is the signature of bromine (isotopes $^{79}$Br and $^{81}$Br). So X is Br and the compounds are $\text{C}_3\text{H}_7\text{Br}$. The peak at m/z 43 is $\text{C}_3\text{H}_7^+$ after loss of Br.

There are two isomers of $\text{C}_3\text{H}_7\text{Br}$: 1-bromopropane and 2-bromopropane. Substitution with hydroxide gives propan-1-ol (a primary alcohol) and propan-2-ol (a secondary alcohol). Only the secondary alcohol, propan-2-ol, oxidises to a ketone (propanone). Since D oxidises to a ketone, D is propan-2-ol, so B is 2-bromopropane; C oxidises but not to a ketone, so C is propan-1-ol and A is 1-bromopropane.

• Compound A: $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$ (1-bromopropane).
• Compound B: $\text{CH}_3\text{CHBrCH}_3$ (2-bromopropane).

Identify E from the $^1$H NMR. E is the ester of 3-methylbutanoic acid with C or D.

• The peak at 4.0 ppm (2H, triplet) is a $\text{CH}_2$ bonded to the ester oxygen ($\text{-O-CH}_2$): a triplet integrating for 2 means E is a propyl ester (-OCH2CH2CH3), so the alcohol was C (propan-1-ol), not the isopropyl that propan-2-ol would give.
• The rest of the propyl group gives the 1.7 ppm $\text{CH}_2$ multiplet (2H) and the 0.95 ppm $\text{CH}_3$ triplet (3H).
• The 2.2 ppm doublet (2H) is the $\text{CH}_2$ next to the carbonyl, the 2.1 ppm multiplet (1H) is the branch CH, and the 0.96 ppm doublet (6H) is the two equivalent CH3 groups of the isopropyl end, confirming the 3-methylbutanoyl group.

So E is propyl 3-methylbutanoate, $\text{(CH}_3\text{)}_2\text{CHCH}_2\text{COOCH}_2\text{CH}_2\text{CH}_3$.

Marker's note. Read the m/z 122/124 doublet as bromine and m/z 43 as loss of Br. Use the "ketone on oxidation" clue to place the bromine (2-bromopropane gives the secondary alcohol). For E, the 4.0 ppm triplet (OCH2) fixes a propyl, not isopropyl, ester, and the 6H doublet plus 1H multiplet identify the isopropyl end of the 3-methylbutanoyl group. Draw structures with the correct number of carbons and hydrogens.

[4 marks]. In the water layer the acid dissociates: $\text{BrCH}_2\text{COOH} \rightleftharpoons \text{H}^+ + \text{BrCH}_2\text{COO}^-$, with $[\text{BrCH}_2\text{COO}^-] = [\text{H}^+] = 9.18\times10^{-3}$ mol L$^{-1}$. Rearrange the $K_a$ expression for the undissociated acid:

Now use a mass balance. The initial 0.1000 mol L$^{-1}$ (equal volumes, so concentrations are directly comparable) is split between the undissociated acid in water, the bromoacetate ion in water, and the acid that has moved into the octan-1-ol:

Therefore

Marker's note. Use $K_a$ with $[\text{H}^+]=[\text{BrCH}_2\text{COO}^-]$ to get the undissociated aqueous acid (0.0654 mol L$^{-1}$). The amount in octan-1-ol is the initial concentration minus the undissociated aqueous acid minus the bromoacetate ion, then substitute into the partition $K_{eq}$ expression.