2024 HSC Chemistry Exam Paper Solutions

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2024 HSC Chemistry exam paper solutions

These are the responses to the 2024 Chemistry paper, which can be found here on the NESA website.

Section 1: Multiple choice questions

Question	Answer	Explanation
1	A	A homologous series is a family of compounds that can be represented by one general molecular formula, with each member in the series differing from the next member by a -CH₂– group.
2	B	Low solubility toxins exist in dynamic equilibrium inside the food: toxins(s) ⇌ toxins(aq) By leaching in flowing water, the concentration of aqueous toxins decreases and equilibrium shifts forward (Le Chatelier’s principle). Hence more of the solid low solubility toxins will dissolve into the water and be removed.
3	D	An Arrhenius base dissociates in water to produce hydroxide ions. Only sodium hydroxide in the options will dissociate to produce hydroxide ions.  NaOH(s)  ⟶  Na⁺(aq)  +  OH⁻(aq)
4	A	Broad signal between 3230–3550 cm-1 is characteristic of O–H bonds in alcohols.
5	D	2-methylpropan-1-ol is a primary alcohol and 2-methylpropan-2-ol is a tertiary alcohol. Primary alcohols can be oxidised by acidified potassium permanganate, resulting reduction of permanganate ions to manganese(II) ions, which is observed as a colour change from purple to colourless. Tertiary alcohols cannot be oxidised any further by acidified potassium permanganate.
6	B	pOH = 14 – pH = 14 − 11 = 3 [OH–] = 10-pOH = 10-3 mol L−1
7	C	Removal of SO3(g) as it forms decreases the concentration of SO3(g) in the system, hence will shift the equilibrium forward (Le Chatelier’s principle). This results in more SO3(g) production, hence yield increases. Increasing the volume shifts the equilibrium towards the reverse direction with more gas moles and increasing temperature shifts the equilibrium in the reverse endothermic direction.
8	C	Only metal cations will produce characteristic colours. Cu2+ will produce a green flame and Ca2+ will produce an orange-red flame.
9	B	The molecular ion peak (usually the peak with the highest mass-to-charge ratio) corresponds to the molar mass of the intact molecule. Ethanamine has the formula C2H7N which has a molar mass of 45.086 g mol−1. Only spectrum in option B has a signal at m/z 45.
10	B	Addition of a catalyst increases both the forward and reverse rates equally since the activation energy of both the forward and reverse reactions by the same amount. Since the rates are equal, there is no shift in the equilibrium. Hence yield remains the same.
11	D	[I−]initial from NaI = [NaI] = 0.1 mol L−1 Let solubility of lead(II) iodide be 𝑥 mol L−1 Concentration (mol L−1) PbI2(s) ⇌ Pb2+(aq) + 2I−(aq) Initial – 0 0.1 Change – +𝑥 +2𝑥 Equilibrium – 𝑥 0.1 + 2𝑥 Ksp = [Pb2+][I−]2= 9.8 × 10−9  Assuming [I−] does not change very much since Ksp is small.  Ksp = (𝑥)(0.1)2 = 9.8 × 10−9  𝑥 = 9.8 × 10−9 / 0.12
12	C
13	C
14	B	The polymer shown is a polyamide, a type of condensation polymer that is formed from difunctional monomers that possess a carboxylic acid and amine functional group.  During polymerisation, the carboxylic acid and amine combine to form the amide link with the simultaneous elimination of water. Thus the repeating unit(s) from the monomers occur between the amide links. The polymer shown has two different repeating units, the second repeating unit has a CH-CH3 in between the amide groups, hence the monomer would also have a CH-CH3  in between the carboxylic acid and amine groups.
15	A	Since the two mixtures were allowed to reach equilibrium at the same temperature, the equilibrium constant (Keq) must be the same for both containers.  For this heterogeneous system, Keq = [O2]. Thus [O2] in both containers will be the same, and hence ratio is 1:1.
16	C	A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers work by maintaining a relatively stable concentration of H3O+ or OH− ions.  When an acid is added, H3O+ ions will be added to the buffer. The weak base in the buffer system (which is the acetate ions CH3COO−), will react with the added H3O+, thereby preventing a large increase in the concentration of H3O+.  CH3COO−(aq) + H3O+(aq) ⟶ CH3COOH(aq) + H2O(l)
17	C	The titration curve depicts two equivalence points, hence the acid is diprotic and will have two pKa values. The pH at the half-equivalence points gives the pKa values for each dissociation. The first equivalence point occurs at 20 mL, hence the first half-equivalence point occurs at 10 mL. The pH here is 1.91, so pKa for the first dissociation is 1.91. The second equivalence point occurs after another 20 mL (total 40 mL) of NaOH is added, so the second half-equivalence point occurs after another 10 mL has been added (total 30 mL). The pH here is 6.30, so pKa for the second dissociation is 6.30.
18	B	Since Q = [products]/[reactants], the reversible reaction is  N2O4(g) ⇌ 2NO2(g) If the rate of the forward reaction is initially greater than the rate of the reverse reaction, the N2O4 is consumed faster than it can be produced. Hence concentration of N2O4 decreases over time. Conversely, NO2 will be produced faster than it can be consumed, hence NO2 concentration increases over time.  Hence Q increases over time, eliminating options C and D. Since the mixture initially contains both N2O4 and NO2, Q will be non-zero at time 0.
19	A	Two doublets suggest two proton environments that have 1 hydrogen in a different chemical environment on the immediately adjacent carbons. “a” and “d” hydrogens will produce doublets as the adjacent carbons both have 1 hydrogen (hydrogens b and c).
20	D	n(ascorbic acid) in 50.00 mg = 0.05000 / 176.124 = 2.838908951 × 10−4 mol Ascorbic and KOH react in a 1:1 ratio.  n(KOH) reacting with 50.00 mg of ascorbic acid = 2.838908951 × 10−4 mol V(KOH) required to react with 50.00 mg of ascorbic acid = 33.10 − 17.50 = 15.60 mL = 0.01560 L c(KOH) = 2.838908951 × 10−4/0.01560 = 0.0181981343 mol L−1 n(KOH) reacting with ascorbic acid in solution A = 0.0181981343 × 0.01750 = 3.184673502 × 10−4 mol n(ascorbic) in solution A = 3.184673502 × 10−4 mol c(ascorbic acid) = = 3.184673502 × 10−4/0.025 = 0.01274 mol L−1

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Section 2: Long response questions

Question 21

Magnesium ethanoate/magnesium acetate + hydrogen gas.

The products of an acid + metal reaction is salt + H_2(g)

Question 22

Question 23

When the system is heated, the equilibrium is disturbed. According to Le Chatelier’s Principle the system will favour the endothermic reaction to absorb some of the added heat to minimise the disturbance.

Since the mixture appears more blue when heated, it indicates the system is favouring the production of blue [CoCl₄]²⁻ and favouring the forward reaction.

As such the concentration of [CoCl₄]²⁻ will increase and the concentration of [Co(H₂O)₆]²⁺ and Cl⁻ will decrease when heated. Since \(K= \frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}][Cl^-]^4}\), this means K will increase when temperature increases.

Question 24 (a)

Question 24 (b)

For the same carbon chain length, amines have lower boiling points than alcohols. Both amines and alcohols can form hydrogen bonds. However, nitrogen is less electronegative than oxygen, so amines form weaker hydrogen bonds than alcohols. Thus, less thermal energy is required to break the intermolecular forces in amines compared to alcohols of the same carbon chain length.

As carbon chain length increases, the difference in boiling point between amines and alcohols of similar size becomes smaller. As chain length increases, dispersion forces become stronger, but hydrogen bonding remains the same.

Thus, the hydrogen bonding will have a smaller and smaller contribution to the total intermolecular force strength. This means the difference in strength of intermolecular forces between amines and alcohols of similar size becomes smaller as chain length increases.

Question 25

From the graph, a reading of 0.78 mg L⁻¹ of phosphate ions can be found in the diluted wastewater. 

Since the sample was diluted 1000× the original concentration must be 0.78 g L⁻¹.

\(c_1= \frac{c_2V_2}{V_1}= \frac{0.78 \times 1000}{1} = 780 mg L^{-1}= 0.78 g L^{-1}\)

To convert to mol L⁻¹:

0.78 g L⁻¹ ÷ MM(PO₄³⁻) = 0.78 ÷ (30.97 + 16 × 4) = 8.2 × 10⁻³ mol L⁻¹ (2 s.f.)

Question 26 (a)

Question 26(b)

Between 0 to 3 minutes, the concentration of I₂ decreases. As a result, the frequency of collisions between I₂ and HCN decreases, leading to a decrease in the rate of the forward reaction between 0 to 3 minutes.

Since the concentration of reactants decrease between 0 to 3 minutes, the concentration of products increases between 0 to 3 minutes. As a result, the frequency of collisions between product particles increases, leading to an increase in the rate of the reverse reaction between 0 to 3 minutes.

From 3 to 6 minutes, the concentration of I₂ is constant, indicating the frequency of collisions between the I₂ and HCN is equal to the frequency of collisions between ICN, I⁻ and H⁺, hence rate of forward reaction is equal to rate of reverse reaction between 3 to 6 minutes.

Question 27

The procedure is not correct when both barium ions (Ba²⁺) and lead(II) ions (Pb²⁺) are present because both ions precipitate with sulfate ions (SO₄²⁻). Although the K_sp value for BaSO₄ is smaller than PbSO₄, indicating BaSO₄ is less soluble and will preferentially be precipitated out, Pb²⁺ ions could still precipitate in step 1. This means when excess Na₂SO₄ solution is added, all the Pb²⁺ ions can precipitate out with the Ba²⁺ ions in step 1, following the equation: 

Pb²⁺_(aq) + SO₄²⁻_(aq) ⟶ PbSO_4(s)

This will give a false negative test for step 3 and an invalid procedure.

However, if only Ba²⁺ ions were present they would precipitate out in the presence of SO₄²⁻ ions in step 1 but will not precipitate out with bromide ions in step 3 resulting in a valid test.

Question 28

[H⁺] = 10^−pH = 10^−1.151 = 0.07063175543 mol L⁻¹

Both solutions have the same pH, hence the same H⁺ concentration of 0.0706 mol L⁻¹ (3 s.f.). Since the concentration of H⁺ is less than the concentration of both iodic acid and sulfamic acid, the acids must not have completely dissociated. 

Since a smaller concentration of iodic acid (0.100 mol L⁻¹) is required to produce the same concentration of hydrogen ions in larger concentration of sulfamic acid (0.120 mol L⁻¹), this means iodic acid is more ionised than sulfamic acid. A stronger acid is more ionised, so iodic acid is stronger. 

Alternatively:

Let iodic acid = HX

HX_(aq) ⇌ H^+_(aq)  + X⁻_(aq)

\(K_a(HX) = \frac{[H^+][X^-]}{[HX]}= \frac{(0.07063175543)^2}{0.100 – 0.07063175543} = 1.70 × 10^{−1} (3 s.f.)\)

Let sulfamic acid = HY

HY_(aq) ⇌ H+(aq) + Y−(aq)

\(K_a(HX) = \frac{[H^+][Y^-]}{[HY]}= \frac{(0.07063175543)^2}{0.120 – 0.07063175543} = 1.01 × 10^{−1} (3 s.f.)\)

Since K_a of iodic acid is larger, it is a stronger acid. 

Question 29

The neutralisation reaction between a strong acid and strong base can be simplified as a net ionic equation: H⁺_(aq) + OH⁻_(aq) ⟶ H₂O_(l)

n(OH⁻) = n(NaOH) = c × V = 0.20 mol L⁻¹ × 0.15 L = 0.03 mol

n(H⁺) = 2 × n(H₂SO₄) = 2 × c × V = 2 × 0.10 mol L⁻¹ × 0.100 L = 0.02 mol

Therefore n(H⁺) is the limiting reagent and n(OH⁻) is in excess.

n(OH⁻)excess = 0.03 − 0.02 = 0.01 mol 

V(total) = 150 mL + 100 mL = 250 mL = 0.25 L

c(OH⁻) = 0.01 ÷ 0.25  \( \frac{0.01}{0.25}\) = 0.04 mol L⁻¹

pH = 14 − pOH = 14 − (−log₁₀[OH⁻]) = 14 − (−log₁₀0.04) = 12.60 (2 sf) 

Question 30

\(K_{eq} = \frac{[H_2O][CO]}{[H_2][CO_2]}= \frac{(x – 0.200)(0.200)}{(1.200)(0.700)} = 1.600\)

\(x = \frac{1.600 × 1.200 × 0.700 – 0.04}{0.200}= 6.92\)

n(CO) initial = c × V = 2.000 mol L⁻¹ × 1 L = 2.000 mol

n(CO) after more CO added = c × V = 6.92 mol L⁻¹ × 1 L = 6.92 mol

n(CO) added = n(CO) after more CO added − n(CO) initial 

= 6.92 − 2.000 = 4.92 mol = 5 mol (1 sf)

Question 31

The second process reacting ammonia with DMC is preferred as the following reaction involves a less toxic reactant (DMC), whilst the first reaction involves phosgene a reactant with high toxicity.

Furthermore, since the second reaction has a higher AE of 48.4% in comparison to the reaction involving phosgene AE of 35.9%, it means there is a lower mass of waste product produced, and less toxic byproducts being released into the environment.

Although both byproducts ammonium chloride and methanol can be sold and used for other industrial processes, the primary product of interest is urea. Overall, the second reaction is safer for the environment and more efficient since less waste products are produced. 

Question 32

Cd₃(PO₄)_2(s) ⇌ 3Cd²⁺_(aq) + 2PO₄³⁻_(aq)

K_sp = [Cd²⁺]³[PO₄³⁻]² = 2.53 × 10⁻³³

Let solubility of Cd₃(PO₄)₂ = 𝑥 mol L⁻¹

[Cd²⁺] = 3𝑥 mol L⁻¹

[PO₄³⁻] = 2𝑥 mol L⁻¹

K_sp = (3𝑥)³(2𝑥)² = 2.53 × 10⁻³³

108𝑥⁵ = 2.53 × 10⁻³³

\(x = \sqrt[5]{\frac{2.53 \times 10^{-33}}{108}}= 1.185603 \times^{-7}\)

[Cd²⁺] = 3𝑥 mol L⁻¹ = 3 × 1.185603 × 10⁻⁷ mol L⁻¹= 3.56 × 10⁻⁷  mol L⁻¹ (3 s.f.)

Question 33(a)

Acetone has a trigonal planar shape around the central carbon atom (3 VSEPR groups, no lone pairs).

Propan-2-ol is tetrahedral around the central carbon (4 VSEPR groups, no lone pairs).

Question 33(b)

¹³C NMR spectroscopy can be used as both molecules have different functional groups that resonate at different frequencies in a magnetic field.

Acetone would produce 2 signals in the ¹³C NMR spectrum as it has 2 carbon environments. The signals would be between 190-220 ppm for the carbonyl carbon whilst the adjacent methyl carbons would produce a signal between 20-50 ppm.

Propan-2-ol would also produce 2 signals, however, the carbon bearing the hydroxyl group would produce a signal between 50-90 ppm whilst the adjacent methyl carbons would produce a signal between 5-40 ppm.

The progress of the reaction can be monitored by the transition of the carbonyl signal produced by acetone (190-220 ppm), eventually reaching a signal integral of 0, whilst a new signal between 50-90 ppm is formed to indicate the formation of the hydroxyl group indicating successful reduction of acetone.

Question 34

Conductivity is initially high as the HCl is present in the flask. HCl is a strong acid that is completely dissociated into H⁺ and Cl⁻, hence flask initially contains a large concentration of ions.

The conductivity initially decreases as ammonia is added because conductive H⁺ ions are replaced by less conductive NH₄⁺ ions, as shown in the neutralisation reaction between hydrochloric acid and ammonia:

H⁺_(aq) + Cl⁻_(aq) + NH_3(aq) ⟶ NH₄⁺_(aq) + Cl⁻_(aq) + H₂O_(l)

Additionally, the moles of ions in the reaction mixture stay the same, but the volume of the solution increases. Hence the concentration of the ions decreases resulting in a decrease in conductivity.

The conductivity reaches a minimum at the equivalence point when only NH₄⁺ and Cl⁻ are present. 

After the equivalence point, the conductivity is relatively stable as excess ammonia is added. Ammonia ionises in water as follows:

NH_3(aq) + H₂O_(l) ⇌ NH₄⁺_(aq) + OH⁻_(aq)

However, due to the presence of NH₄⁺ ions in the solution already, ammonia will not significantly ionise. As a result, the concentration of ions will remain relatively stable, resulting in a relatively stable level of conductivity.

Question 35

Both Y and Z can react with bromine, indicating the presence of a C=C bond in the structures, which is consistent with structures 1 and 3. 

Sample X cannot react with bromine, indicating the absence of a C=C bond in the structure. This means structure 2 must be X.

Both structure 1 and structure 3 have one C=C double bond. This means both react in a 1:1 mole ratio with Br₂, so for the same moles of Br₂, the moles X and Y would be the same. 

However, the mass of Z reacting with Br₂ is double that of Y, indicating molar mass of Z must be double that of Y (as mass = moles × molar mass). Since the molar mass of structure 3 is double the molar mass of structure 1, structure 3 must be Z and structure 1 must be Y.

Furthermore, Y and Z can both undergo hydration, supporting the presence of a C=C bond in the structures. Y hydrates to give two products, this is consistent for the hydration of structure 1, which is asymmetric, and hydrates as follows:

Z hydrates to only produce one product, this is consistent for the hydration of structure 3 which is symmetrical. 

Lastly, the volume of NaOH used to titrate the three acids is consistent with the proposed structures.

Structure 1 and structure 2 only have one carboxyl group, hence would be monoprotic and would react in a 1:1 mole ratio with NaOH. Structure 3 has two carboxyl groups, hence would be diprotic and would react in a 1:2 ratio with NaOH.

Question 36

NaHCO_3(s) + HCl_(aq) ⟶ NaCl_(aq)  + H₂O_(l) + CO_2(g)

n(NaHCO₃) = m / MM = 14.7 / 84.008 = 0.1749833349 mol

n(HCl) = c × V = 1.50 × 0.120 = 0.18 mol

Therefore NaHCO₃ is the limiting reactant

m(HCl) = 1.02 g mL⁻¹ × 120 mL = 122.4 g

m(CO₂) = n × MM = 0.1749833349 × (12.01 + 16 × 2) = 7.70101657 g

m(solution) = 122.4 + 14.7 − 7.70101657 = 129.3989834 g

q(solution) = mc∆T = 129.3989834 × 3.80 × (11.5 − 21.5) = −4917.16137 J = −4.91716137 kJ

∆H = −q / n = −(−4.91716137) / 0.1749833349 = 28.1 kJ mol⁻¹ (3 s.f.)

Question 37

Calculating ∆G for the following reaction using ∆G = ∆H-T∆S

∆G = −217 kJ mol⁻¹ − (−3 kJ mol⁻¹) = −214 kJ mol⁻¹

According to the graph shown, a more negative ∆G value corresponds to a larger equilibrium constant (K_eq). A larger K_eq indicates the system has a larger concentration of products in comparison to reactants, such that the reaction’s equilibrium lies further towards the right (towards the product side).

The equilibrium expression for this reaction is K_eq  = [SO₂] ÷ [O₂]. For this reaction, ∆G is very negative, hence K_eq is very large, such that [SO₂] is significantly larger than [O₂]. This means the equilibrium lies very far to the right and likely virtually proceeds to completion.

Question 38

Compounds A and B must have the formula C₃H₇Br.

The mass spectrum has two molecular ion peaks, at m/z 122 and 124, with similar relative intensities. This corresponds to the relative abundance of the two bromine isotopes, ⁷⁹Br and ⁸¹Br, which have approximately the same abundance. Additionally, the two molecular ion peaks have m/z values corresponding to the molecular mass of the molecular ion bearing the lighter and heavier isotope, respectively.

Molecular mass (C₃H₇⁷⁹Br]^•+) = (12 × 3) + (1 × 7) + 79 = 122

Molecular mass (C₃H₇⁸¹Br]^•+) = (12 × 3) + (1 × 7) + 81 = 124

Since compound D is oxidised to a ketone, D is a secondary alcohol. The only secondary alcohol for 3 carbon chain molecule is propan-2-ol. Propan-2-ol is produced from the substitution of compound B with OH⁻. Since this reaction substitutes the bromo group with a hydroxyl group, the position of the bromo group must be at carbon 2 in compound B. Hence compound B is 2-bromopropane.

Compound A must be 1-bromopropane, as it is the only isomer possible for 2-bromopropane. Substitution of 1-bromopropane forms propan-1-ol (compound C), which can also be oxidised to form a propanoic acid, which is not a ketone. This is consistent for information provided for compound C.  

Compound E is an ester, as it is produced from the condensation of a carboxylic acid and alcohol under reflux. 

When 3-methylbutanoic acid is condensed with alcohol C (propan-1-ol) or alcohol D (propan-1-ol), the following esters are produced.

The ¹H NMR spectrum for compound E has 6 signals. This indicates there are 6 unique proton environments in compound E. 

Propyl 3-methylbutanoate has 6 unique proton environments (labelled a-f) whereas propan-2-yl 3-methylbutanoate only has 5 unique proton environments (labelled A-E). Hence compound E must be Propyl 3-methylbutanoate. 

Additionally, the integration (which corresponds to the ratio of hydrogen nuclei in each environment) and the splitting (which indicates the number of hydrogens in a different chemical environment on immediately adjacent carbons) are consistent for each of the labelled hydrogen environments. 

Question 39

For the following ionisation reaction: BrCH₂COOH_(aq) ⇌ BrCH₂COO⁻_(aq) + H⁺_(aq)

\(K_a= \frac{[H^+][BrCH_2COO^-]}{[BrCH_2OOH]}\)

[BrCH₂COO⁻_(aq)] at equilibrium = [H⁺_(aq)] at equilibrium = 9.18 × 10⁻³ mol L⁻¹

\(K_a= 1.29 \times 10^{-3} = \frac{(9.18 \times 10^{-3})(9.18 \times 10^{-3})}{[BrCH_2OOH]}\)

Therefore [BrCH₂COOH_(aq)] at equilibrium 

\( = \frac{(9.18 \times 10^{-3})(9.18 \times 10^{-3})}{1.29 \times 10^{-3}}\)

= 0.06532744186 mol L⁻¹

Let [BrCH₂COOH_(aq)] initially before ionisation = 𝑥 mol L⁻¹

x = 0.06532744186 + 9.18 × 10⁻³ = 0.07450744186 mol L⁻¹

Suppose we begin with 1 L of 0.1000 mol L⁻¹ BrCH₂COOH:

n(BrCH₂COOH)_total = 0.1000 mol

n(BrCH₂COOH_(aq)) = 0.07450744186 mol 

n(BrCH₂COOH_(octan-1-ol))= 0.1000 − 0.07450744186 = 0.02549255814 mol

Since aqueous solution and octan-1-ol are of equal volume:

[BrCH₂COOH_(octan-1-ol)] = 0.02549255814 mol L⁻¹

K_eq = 0.0254925581 ÷ 0.06532744186 = 0.390 (3sf)