2024 HSC Chemistry Exam Paper Exemplar Answers & Solutions

Can't wait for the solutions to the 2024 HSC Chemistry exam paper solutions to be released? Are you finding the publicly available solutions and sample answers inadequate?

Not to worry! The state-ranking tutors at Cognito have your back! 💪

Section I: Multiple Choice

MCQ Answer Key

1. A
2. B
3. D
4. A
5. D
6. B
7. C
8. C
9. B
10. B
11. D
12. C
13. C
14. B
15. A
16. C
17. C
18. B
19. A
20. D

Question 1

A homologous series refers to a set of compounds with the same set of functional groups. Option A shows two alcohols, so the answer is A.

Question 2

The process of leaching refers to the dissolution of toxins in the water to wash them away. This is not a chemical reaction, so the answer cannot be A.

The decomposition of food has little to do with the removal of toxins from plant items, so the answer cannot be C.

Digestion and toxicity are two separate properties of food items, so the answer cannot be D.

Although certain toxins in foods detoxified by Aboriginal and Torres Strait Islander Peoples are highly soluble (such as cycasin in cycad fruits), leaching over multiple days was performed for toxins of lower solubility. Hence, the answer is B.

Question 3

An Arrhenius base refers to a substance which dissociates to produce hydroxide ions in water. Only sodium hydroxide contains hydroxide ions, so is the only valid option for an Arrhenius base.

Question 4

There is a strong, broad trough at the wavenumber, 3250 - 3550 cm^-1, so the organic compound must be an alcohol. Although the argument can be made that the trough is very broad, the lack of a visible trough at the wavenumber, 1680 - 1750 cm^-1, indicates a lack of C=O bond, which immediately rules options B and D out.

Question 5

2-methylpropan-1-ol is a primary alcohol, whereas 2-methylpropan-2-ol is a tertiary alcohol.

As neither contain a carbon-carbon double bond, neither compound would produce a positive result in the bromine water test.

Potassium and nitrate ions are soluble ions with no notable reactions with either of these alcohols.

Sodium carbonate can react with acids to produce effervescence. The acidity of alcohols is negligible, and neither alcohol would produce effervescence in the presence of sodium carbonate.

Using acidified potassium permanganate, the primary alcohol would oxidise (and hence turn the pink permanganate ions into colourless manganese ions), whereas the tertiary alcohol would not oxidise (resulting in the pink permanganate ions remaining pink).

Question 6

Since pH + pOH = 14 (assuming the temperature is 25ºC), pOH = 14 - 11 = 3.

Hence, [OH^-] = 10^-pOH = 10^-3 molL^-1

Question 7

Increasing the volume would result in a decrease in pressure, by PV = nRT. To minimise the disturbance by Le Chatelier’s Principle, the equilibrium would shift left, which increases the total number of gas particles present (as per the stoichiometric gas ratio of 3:2). This would decrease the yield of SO₃(g).

Increasing the temperature would result in the system favouring the reverse, endothermic reaction by Le Chatelier’s Principle as well. Similarly, this decreases the yield of SO₃(g).

Removing the product as it is formed reduces the [SO₃] in the reaction chamber. To minimise the disturbance, the equilibrium would shift right to minimise the disturbance, by Chatelier’s Principle. This results in a net greater amount of SO₃(g) being produced, and thus a higher yield of SO₃(g). Hence, the answer is C.

Keeping the temperature and the volume constant would not result in an equilibrium shift, as no disturbance was made. This would result in the yield of SO₃(g) remaining unchanged.

Question 8

Anions and certain cations such as Ag⁺ and Mg²⁺ release UV light during electron relaxation in a flame. As such, the flame colour remains unchanged for options A, B, and D.

In the case of option C, Cu²⁺ ions produce a blue-green flame, whereas Ca²⁺ ions produce a brick-red flame, rendering the answer as C.

Question 9

Ethanamine has a molar mass of approximately 45 gmol^-1, so the parent peak is expected at m/z = 45. The only mass spectrum showing the parent peak at this value is the one in option B.

Question 10

When a catalyst is added, the activation energy for both the forward and the reverse reaction decreases by the same extent, resulting in their rates increasing by the same extent as well. As a result, the position of equilibrium doesn’t shift, and the yield of propan-2-one remains the same.

Question 11

To calculate solubility of lead(II) iodide in a 0.1 molL^-1 solution of NaI, we can let the molar solubility of lead(II) iodide be x.

PbI₂(s) ⇌ Pb²⁺(aq) + 2I^-(aq)

‍

Then, K_sp = [Pb²⁺][I^-]² = (x)(0.1 + 2x)² = 9.8 x 10^-9

Since the K_sp of PbI₂ is small, x must be small. Thus, applying the small change assumption, 0.1 + 2x ≈ 0.1. 

Hence, (x)(0.1)² = 9.8 x 10^-9 ⇒ x = 9.8 x 10^-9 / (0.1)²

Question 12

The carbon chain on which the amide is present is four carbons long, so this is a form of butanamide.

Adding on the ethyl chain on the N, we have N-ethylbutanamide.

Adding on the Cl atom attached to locant 3 (since locant 1 is occupied by the amide group), we have 3-chloro-N-ethylbutanamide.

Question 13

To find the molar mass of the fuel, we can consider the ratio of the enthalpies of combustion in the differing units, which provides the ratio between the mass and molar quantity of the fuel.

MM = (-2057.8 kJmol^-1) / (-48.9 kJg^-1) ≈ 42.0817996 = 42.1 gmol^-1

This corresponds to propene.

Question 14

The silk polymer contains a methyl group periodically, yet glycine does not have a methyl side group. As such, the methyl side group must be originating from the other monomer.

The only polymer which is an amino acid (as the amine and carboxyl group is required for the condensation polymer to propagate) and contains a methyl side group is the monomer provided in option B.

This is demonstrated by the following diagram, where the portion of the polymer originating from the glycine is illustrated in red, and the monomer shown in option B is illustrated in blue:

Question 15

The K_eq for this equilibrium system is [O₂(g)], as the remaining reactants / products are solid, which are excluded from equilibrium expressions.

As both containers, P and Q are at the same temperature, the K_eq = [O₂(g)] is equal in each container. Hence, the ratio of [O₂(g)] across the two containers is 1:1.

Question 16

When a strong acid (represented as H₃O⁺(aq)) is added to a buffer solution, it would react with the basic portion of the buffer mixture. In this case, acetate is the base in the buffer conjugate pair. Hence, the correct reaction is the reaction between acetate and hydronium ions. This is best demonstrated by option C.

Question 17

The acid being neutralised must be diprotic, where each of the two steep changes in pH corresponds to the complete neutralisation of each proton from the acid. As such, let the acid be H₂X, where X is a placeholder anion.

As such:

H₂X(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HX^-(aq), K_a1 = [H₃O⁺][HX^-] / [H₂X]

HX^-(aq) + H₂O(l) ⇌ H₃O⁺(aq) + X^2-(aq), K_a2 = [H₃O⁺][X^2-] / [HX^-]

When an acid is neutralised halfway to an equivalence point, it is called the half-equivalence point.

At the first half-equivalence point at the 10 mL mark, half of H₂X has been neutralised to HX^-. That is, [H₂X] = [HX^-] at 10 mL. Thus, at 10 mL, K_a1 = [H₃O⁺]. Taking the negative logarithm, pK_a1 = pH = 1.91.

At the first equivalence point, the remainder of the H₂X has been neutralised, leaving only HX^- in the solution.

At the second half-equivalence point at the 30 mL mark, half of HX^- has been neutralised to X^2-. That is, [HX^-] = [X^2-] at 30 mL. Thus, at 30 mL, K_a2 = [H₃O⁺]. Taking the negative logarithm, pK_a2 = pH = 6.30.

Question 18

Based on the provided expression for the reaction quotient (Q), the equilibrium must be:

N₂O₄(g) ⇌ 2NO₂(g)

If the rate of the forward reaction is initially greater than the rate of the reverse reaction Q would be less than K_eq. As a result [NO₂] will increase and [N₂O₄] will decrease until equilibrium is reached, at which point Q becomes a constant and is equal to K_eq.

Hence, it follows that Q will gradually increase until it reaches a constant.

However, as the question specifies that both N₂O₄(g) and NO₂(g) is present initially, Q must be a non-zero value initially. Hence, the answer is B.

Question 19

To produce two doublets, there must be two distinctly different hydrogen environments with one neighbouring hydrogen each. This is only exhibited by the compound shown in option A, and the two environments in question are highlighted in red and blue respectively.

Question 20

n(ascorbic acid)_added = 0.05 g / 176.125 gmol^-1 = 0.000283889283 mol

Since V(KOH)_{required for solution A} / V(KOH)_{required for solution A + 50.00 mg of ascorbic acid} = n(ascorbic acid)_{solution A} / n(ascorbic acid)_{solution A + 50.00 mg of ascorbic acid}, it follows that:

17.50 / 33.10 = n(ascorbic acid)_{solution A} / (n(ascorbic acid)_{solution A} + 0.000283889283 mol)

⇒ 0.528700906(n(ascorbic acid)_{solution A} + 0.000283889283) = n(ascorbic acid)_{solution A}
⇒ n(ascorbic acid) = (0.000283889283 x 0.528700906) / (1 - 0.528700906) = 0.000318465541 mol

Hence, [ascorbic acid] = 0.000318465541 mol / 0.025 L = 0.0127386216 = 1.274 x 10^-2 molL^-1

Section II: Written Responses

Question 21

Exemplar Answer

• Magnesium ethanoate or magnesium acetate
• Hydrogen gas

Marker’s Notes

“Magnesium acetate” is not the preferred IUPAC name of Mg(CH₃COO)₂.

Question 22

Exemplar Answer

Question 23

Exemplar Answer

When the solution is heated, the temperature of the solution increases. By Le Chatelier’s Principle, the system will shift to decrease the temperature by favouring the reaction which absorbs heat, i.e. the endothermic reaction.

Since the mixture becomes more blue, this indicates that the concentration of [CoCl₄^2-](aq) increases when temperature increases. As a result, the system must have shifted to the right.

Temperature is the only factor which can change the equilibrium constant (K_eq) for a reaction, and since the position of equilibrium shifts to the right, the concentration of products would increase whilst that of reactants will decrease. Hence, K_eq = [[CoCl₄]^2-] / [[Co(H₂O)₆]²⁺][Cl^-]⁴ would increase overall.

Question 24(a)

Exemplar Answer

Marker’s Notes

Responses could have also included a curve or line of best fit.

Question 24(b)

Exemplar Answer

Both primary alcohols and primary amines are polar with an O–H and N–H bond respectively. As such, they can form dispersion forces and dipole-dipole forces including hydrogen bonding.

As the strength of intermolecular forces increases, more energy is required to overcome these forces and separate the molecules, resulting in a higher boiling point.

(1) It is firstly observed that as the number of carbon atoms per molecule increases, the boiling point of both primary alcohols and primary amines increase. This is due to a larger number of electrons per molecule resulting in stronger momentary dipoles and hence stronger dispersion forces.

(2) The boiling point of the primary alcohol is consistently higher than the boiling point of the primary amines of the same number of carbon atoms per molecule. This is because although the equal number of carbon atoms per molecule ensures that the dispersion forces are similar in strength, the O-H bond in alcohols is more polar than the N-H bond in amines due to oxygen having a larger electronegativity. This results in stronger hydrogen bonding in alcohols. Additionally, the trigonal pyramidal geometry around the carbon atoms in primary amines also results in poor packing.

(3) As the number of carbons increases, the boiling points of primary alcohols and primary amines converge. This is because as the number of carbon atoms per molecule increases, the stronger dispersion forces now account for a greater proportion of the intermolecular forces. As a result, the differing hydrogen bonding strength between primary alcohols and primary amines has a lesser effect on the boiling points as the number of carbon atoms per molecule increases.

Marker’s Notes

The third trend is likely unnecessary for a question worth 4 marks, but is included for the sake of completeness.

Question 25

Exemplar Answer

By interpolating the graph provided at the absorbance of 0.64, [PO₄^3-] = 0.78 mgL^-1 = 0.00078 gL^-1

[PO₄^3-]_{in 1 L} = 0.00078 gL^-1 / (30.97 + 4 x 16) gmol^-1 = 8.21311 x 10^-6 molL^-1

Then, to find the [PO₄^3-] in the original 1.00 mL sample, n(PO₄^3-)_{1.00 mL} = n(PO₄^3-)_{1 L} 

Hence, by C₁V₁ = C₂V₂:
[PO₄^3-]_{in 1 L} x 1 L = [PO₄^3-]_{in 1 mL} x 0.001 L

∴ [PO₄^3-]_{in 1 mL} = 8.21311 x 10^-6 x (1 / 0.001) = 0.0082 molL^-1 (2 s.f.)

Marker’s Notes

Students should have shown their interpolation on the graph provided in the question.

Question 26(a)

Exemplar Answer

Question 26(b)

Exemplar Answer

Collision theory states that for a successful chemical reaction to occur, particles must collide with sufficient energy greater than the activation energy and the correct molecular orientation. The rate of the forward and reverse reactions thus depends on the number of successful collisions per unit time and will increase with a higher concentration of reactants and products respectively.

t = 0 min:
When 2.0 x 10^-5 molL^-1 of I₂(aq) is added to the system there will be a non-zero forward rate of reaction. 

t = 0 min - t = 3 min:
As the forward reaction occurs the concentration of I₂(aq) will decrease, this results in a decrease in the frequency of collisions between reactant particles and as a result the forward rate of reaction decreases overtime. However, as the concentration of products increases the reverse rate of reaction would also increase.

t = 3 min - t = 6 min:
When the forward and reverse rate of reactions become equal the concentration of I₂(aq) remains constant as it is produced and consumed at the same rate and thus dynamic equilibrium is established.

Question 27

Exemplar Answer

When only Ba²⁺(aq) ions are present:
The addition of Na₂SO₄(aq) results in a precipitate with Ba²⁺(aq) ions:

Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)

which is a white precipitate. Hence, the formation of a white precipitate after step 1 confirms the presence of Ba²⁺(aq) ions. Subsequently, adding NaBr(aq) in step 3 would not result in a precipitate forming due to the absence of Pb²⁺(aq) ions indicating that only Ba²⁺(aq) ions are present.

When both Ba²⁺(aq) and Pb²⁺(aq) ions are present:
The addition of of Na₂SO₄(aq) would precipitate out both Ba²⁺(aq) and Pb²⁺(aq) ions as white precipitates:

Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)
Pb²⁺(aq) + SO₄^2-(aq) → PbSO₄(s)

Although the experimenter would correctly identify that Ba²⁺(aq) ions are present, all Pb²⁺(aq) ions will have already precipitated out as well in this process. As a result, adding NaBr(aq) in step 3 would not result in a precipitate forming, and the experimenter will come to the inaccurate conclusion that Pb²⁺(aq) ions are not present.

Question 28

Exemplar Answer

If the acids were both strong monoprotic acids, they would both ionise to completion. Letting the acid be HX(aq), where X^-(aq) is the conjugate base:

HX(aq) + H₂O(l) → H₃O⁺(aq) + X^-(aq)

In this case, the initial concentration of HX(aq) would equal the measured concentration of H₃O⁺(aq).

For both acids, [H₃O⁺] is 10^-1.151 = 0.0706 molL^-1, since pH = -log₁₀[H₃O⁺]. As this concentration is lower than the initial concentration of either acid, the acids must be weak and not dissociate completely in water.

Since iodic acid has a lower concentration of 0.100 molL^-1, compared to the concentration of 0.120 molL^-1 of the sulfamic acid, yet it reaches the same [H₃O⁺], iodic acid must have a higher degree of ionisation, and thus be the stronger acid.

Question 29

Exemplar Answer

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

n(NaOH) = 0.2 x 0.15 = 0.03 mol
n(H₂SO₄) = 0.1 x 0.1 = 0.01 mol

Assume all NaOH reacts. Then, n(H₂SO₄)_required = 0.03 / 2 = 0.015 mol (< n(H₂SO₄)_present)

∴ H₂SO₄ is the limiting reagent

The excess NaOH(aq) will dissociate completely:

NaOH(aq) → Na⁺(aq) + OH^-(aq)

n(OH^-) = n(NaOH)_{in excess} = 0.03 - 0.01 x 2 = 0.01 mol
[OH^-] = 0.01 / 0.25 = 0.04 molL^-1

pH = 14 - pOH = 14 - (-log₁₀[OH^-]) = 14 + log₁₀(0.04) = 12.60

Question 30

Exemplar Answer

Let x be the initial increase in [CO] due to the addition of CO(g):

K_eq = [CO][H₂O] / [CO₂][H₂] 

⇒ 1.6 = (0.2)(1.8 + x) / (1.2)(0.7)
⇒ 1.8 + x = 6.72
⇒ x = 4.92 molL^-1

Hence, n(CO)_added = 4.92 x 1 = 4.92 = 5 mol (1 s.f.)

Question 31

Exemplar Answer

Atom economy:
The second process of producing urea has a greater atom economy than the first process (48.4% > 35.9%). As a result, a lower amount of waste products would be produced, which is favourable in the industry as less reagents are required to produce the same amount of urea. Furthermore, there would be a lesser cost expenditure from disposing the waste products responsibly. Hence, the profit margin for the second process is likely higher.

Toxicity:
The first process uses phosgene, which is a highly toxic substance, whereas the second process uses DMC, which is less toxic. Hence, the second process is more favourable as prolonged exposure to toxic substances may lead to negative health effects on both the workers at the manufacturing plant and local residents in the event of a gas leakage.

Therefore, the second process is preferable for urea production.

Question 32

Exemplar Answer

Cd₃(PO₄)₂(s) ⇌ 3Cd²⁺(aq) + 2PO₄^3-(aq) 

Let x be the molar solubility of the salt. Then:

K_sp = [Cd²⁺]³[PO₄^3-]² = 2.53 x 10^-33

⇒ (3x)³(2x)² = 2.53 x 10^-33
⇒ 108x⁵ = 2.53 x 10^-33
⇒ x = √(2.53 x 10^-33) / 108
⇒ x = 1.1856… x 10^-7 molL^-1

Hence, [Cd²⁺]_eq = 3x = 3 x (1.1856… x 10^-7) = 3.56 x 10^-7 molL^-1 (3 s.f.)

Question 33(a)

Exemplar Answer

• Acetone: trigonal planar
• Propan-2-ol: tetrahedral

Question 33(b)

Exemplar Answer

Acetone would produce 2 signals on a ¹³C NMR spectrum, as it has 2 unique carbon environments. The carbon in the carbonyl group would produce a signal between 190 - 220 ppm, whereas the 2 carbon atoms in the methyl groups would produce a signal between 20 - 50 ppm, as they are adjacent to a carbonyl group.

Propan-2-ol would also produce 2 signals on a ¹³C NMR spectrum, as it has 2 unique carbon environments. However, the carbon bearing the hydroxyl group would produce a signal between 50 - 90 ppm, whereas the 2 carbon atoms in the methyl groups would also produce a signal between 5 - 40 ppm.

Hence, the progress of the reaction can be monitored by observing the signal integral of the signal at 190 - 220 ppm and 50 - 90 ppm on the ¹³C NMR spectrum. As the concentration of acetone decreases, the signal integral for the signal at 190 - 220 ppm would decrease proportionally, whereas the signal at 50 - 90 ppm would increase proportional to the increasing concentration of propan-2-ol. A similar analysis can be performed with the 20 - 50 ppm signal and 5 - 40 ppm signal respectively.

When the signal at 190 - 220 ppm reaches a signal integral of 0, this indicates that the reaction has gone to completion and all of the acetone has been reduced.

Question 34

Exemplar Answer

Initially, only HCl(aq) is present in the conical flask. HCl(aq) is a strong acid, so ionises completely as per:

HCl(aq) → H⁺(aq) + Cl^-(aq)

As H⁺(aq) and Cl^-(aq) have a high relative conductivity of 4.76 and 1.04 respectively, the overall conductivity of the solution is initially very high.

As NH₃(aq) is added to the conical flask, it reacts with the H⁺(aq) ions, as per:

H⁺(aq) + NH₃(aq) → NH₄⁺(aq)

Hence, H⁺(aq) ions are replaced with less conductive NH₄⁺(aq) ions, with a relative conductivity of 1.00. Hence, the overall conductivity of the solution decreases, until all of the H⁺(aq) ions from the hydrochloric acid is neutralised. This is the equivalence point.

At the equivalence point, the conductivity remains non-zero, as NH₄⁺(aq) and Cl^-(aq) ions are still present in solution.

As more NH₃(aq) is added after the equivalence point, it ionises partially as it is a weak acid:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)

This results in the conductivity increasing, as the production of NH₄⁺(aq) and OH^-(aq) ions with a relative conductivity of 1.00 and 2.70 respectively, contributes to the conductivity of the solution. The increase in conductivity is not as steep, as the ionisation of the weak acid is minimal and OH^-(aq) ions have a lower relative conductivity than the H⁺(aq) ions. 

Question 35

• Structure 1: Acid Y
• Structure 2: Acid X
• Structure 3: Acid Z

‍Bromine Water Test & Hydration:
Since both acids Y and Z provide a positive result with the bromine water test and can undergo a hydration reaction, they must be unsaturated organic compounds capable of undergoing addition reactions.

In contrast, the negative results from acid X indicates that it is saturated. Structure 2 is the only unsaturated acid of the three, so structure 2 must be acid X.

Both acids Y and Z have a C=C bond, matching the unsaturated nature of structures 1 and 3. However, since twice the mass of acid Z compared to acid Y is required to react with the same amount of Br₂ it must have double the molar mass of acid Y. Since the molar mass of structure 3 is double that of structure 1, structure 3 must be acid Z and structure 1 must be acid Y.

Titration with NaOH:
n(X) = 0.1 / 74.078 = 0.0013 mol
n(NaOH)_{required to react with acid X} = 0.0617 x 0.02188 = 0.0013 mol

Since n(NaOH) required to react with acid X is equal to n(X), this indicates that they react in 1:1 ratio, and that the acid is thus monoprotic. This is consistent with the structure of acid X, as structure 2 has 1 carboxyl functional group.

n(Y) = 0.1 / 72.062 = 0.0014 mol
n(NaOH)_{required to react with acid Y} = 0.0617 x 0.02249 = 0.0014 mol

Similarly as with acid X, acid Y must be monoprotic based on n(NaOH) matching n(Y). This is also consistent with the structure of acid Y, as structure 1 has 1 carboxyl functional group.

n(Z) = 0.1 / 144.124 = 0.0007 mol
n(NaOH)_{required to react with acid Z} = 0.0617 x 0.02249 = 0.0014 mol

Since n(NaOH):n(Z) = 2:1, this indicates that acid Z is diprotic, which is consistent with the structure of acid Z, structure 3 has 2 carboxyl functional groups.

Possible Products of Hydration:
Since 2 possible products are possible when acid Y undergoes hydration, this further confirms that acid Y must match structure 1:

Additionally, since only 1 possible product can be produced when acid Z undergoes hydration, this confirms structure 3, as the molecule is symmetrical around the double bond.

Question 36

Exemplar Answer

NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g) 

n(NaHCO₃) = 14.7 / 84.008 = 0.1749833… mol
n(HCl) = 1.5 x 0.12 = 0.18 mol 

Since NaHCO₃(aq) and HCl(aq) react in 1:1 ratio, NaHCO₃ is the limiting reagent.

Hence, n(H₂O)_produced = 0.1749833… mol (since n(NaHCO₃):n(H₂O) = 1:1) 

m(HCl) = ⍴V = 1.02 x 120 = 122.4 g

n(CO₂)_lost = 0.1749833… mol (since n(NaHCO₃):n(CO₂) = 1:1)
m(CO₂)_lost = 0.1749833 x 44.01 = 7.7010… g

Hence, m(reaction solution) = m(HCl) + m(NaHCO₃) - m(CO₂)_lost = 122.4 + 14.7 - 7.7010 = 129.39… g

Thus, q = mcΔT = (129.39 g)(3.80 Jg^-1K^-1)(11.5 - 21.5 K) = -4917.16… J = -4.91716… kJ

Hence, ΔH = - q / n(H₂O)_produced = 4.91716 kJ / 0.1749833 mol = 28.1 kJmol^-1

Question 37

Exemplar Answer

From the graph, the more negative the ∆Gº is, the larger the equilibrium constant (K_eq) for the reaction. The larger the K_eq, the more products that will be formed, since K_eq = [SO₂] / [O₂]. Hence, reactions which go to completion have a very large K_eq.

For the above reaction: ∆G° = ∆H° - T∆S° = (-217) - (-3) = -214 kJmol^-1

Since this reaction has a very negative ∆G°, the reaction will have a very large K_eq. Hence, it can be concluded that the reaction only reaches equilibrium once it goes effectively to completion.

Question 38

Exemplar Answer

Compound A

Compound B

Compound E

‍

Mass Spectrum (Presence of Br atom):
The mass spectrum shows two parent peaks at m/z = 122 and 124 in a 1:1 ratio. There is also another set of peaks in a 1:1 ratio separated by 2 m/z at m/z = 107 and 109. This suggests the presence of bromine in the molecule, as stable isotopes of bromine occur with a difference in atomic mass by approximately 2 amu (79 and 81 amu), with a relative abundance ratio of roughly 1:1.

The molar mass of C₃H₇Br is 122.986 gmol^-1, which is consistent with the m/z values of the two parent peaks, as the molar mass of bromine is a weighted average of the molar mass of its two stable isotopes. Thus, the two isotopes must be 1-bromopropane and 2-bromopropane.

The mass spectrum of compound A has a base peak of m/z = 43. This corresponds to the molecule with the Br atom fragmented out (i.e. C₃H₆⁺), as 122 - 79 = 124 - 81 = 43.

Flowchart (Determination of A & B):
As compound D oxidises into a ketone, it must be a secondary alcohol, namely propan-2-ol. As the bromine atom is substituted with a hydroxyl group to form compound D, compound B must be 2-bromopropane.

This suggests that compound A is 1-bromopropane and compound C is propan-1-ol via the same substitution reaction. Propan-1-ol is a primary alcohol, and does not oxidise into a ketone. Instead, it would oxidise to form propanal, then propanoic acid.

¹H NMR Spectrum (Determination of E):
The ester produced by refluxing 3-methylbutanoic acid with alcohol C (propan-1-ol) is shown below. Let this be compound X:

The ester produced by refluxing the same acid with alcohol D (propan-2-ol) is shown below. Let this be compound Y:

Compound X has 6 unique hydrogen environments, whereas compound Y has 5 unique hydrogen environments. Since the ¹H NMR spectrum data shows 6 signals, compound E must be compound X.

To confirm the identity of compound E, the splitting pattern based on the n +1 rule and ¹H chemical shift data can be used:

• H₁ has one neighbouring hydrogen, so it produces a doublet. As this environment contains 6 hydrogens, this signal has an integration of 6, matching the characteristics of the signals at 0.96 ppm. Additionally the chemical shift value is consistent with that expected of the hydrogens in a –CH₃ group of an alkyl chain which should be between 0.7 - 2.1 ppm as seen on the ¹H chemical shift data.
• H₂ has one hydrogen in its environment with eight neighbouring hydrogens and thus produces the multiplet of 9 with an integration of 1 at 2.1 ppm. This signal is also within the expected 0.7 - 2.1 ppm range of a hydrogen in a –CH group.
• H₃ has two hydrogens in its environment with one neighbouring hydrogen, so it produces the doublet at 2.2 ppm with an integration of 2. This aligns with the ¹H chemical shift data provided, as hydrogens in H₂C–CO groups produce a signal from 2.2 - 2.6 ppm.
• The 2 hydrogens in H₄ have two neighbouring hydrogens and thus produce the triplet at 4.0 ppm with an integration of 2. This aligns with the ¹H chemical shift data provided, as the H₂–CO groups produce a signal from 3.2 - 5.0 ppm.
• H₅ has 2 hydrogens in its environment with five neighbouring hydrogens so it produces the multiplet of 6 with an integration of 2 at 1.7 ppm. This signal is also within the expected 0.7 - 2.1 ppm range of a H in a –CH₂– group of an alkyl chain.
• H₆ has 3 hydrogens in its environment with two neighbouring hydrogens, producing the triplet at 0.95 ppm with an integration of 3. The chemical shift value is consistent with that expected of the hydrogens in a –CH₃ group of an alkyl chain which should be between 0.7 - 2.1 ppm.

Question 39

Exemplar Answer

BrCH₂COOH(aq) + H₂O(l) ⇌ BrCH₂COO^-(aq) + H₃O⁺(aq)

[BrCH₂COO^-]_eq = [H₃O⁺]_eq = 9.18 x 10^-3molL^-1, since they form in a 1:1 ratio.

K_a = [BrCH₂COO^-][H₃O⁺] / [BrCH₂COOH(aq)]

∴ [BrCH₂COOH(aq)]_eq = [BrCH₂COO^-][H₃O⁺] / K_a = (9.18 x 10^-3)² / 1.29 x 10^-3 = 0.06532744186 molL^-1

Since [BrCH₂COOH(aq)]_initial = [BrCH₂COOH(octan-1-ol)]_eq + [BrCH₂COOH(aq)]_eq + [BrCH₂COO^-]_eq = 0.1 molL^-1, it follows that:

[BrCH₂COOH(octan-1-ol)]_eq = 0.1 - (0.06532744186 + 9.18 x 10^-3) = 0.02549255814 molL^-1

K_eq = [BrCH₂COOH(octan-1-ol)]_eq / [BrCH₂COOH(aq)]_eq = 0.02549255814 / 0.06532744186 = 0.390 (3 s.f.)

---

The small classes and exam-based approach at Cognito allows our talented, state-ranking tutors to personally help you get from where you're at to producing top-quality responses like these. Try out our 2-week obligation-free trial today and supercharge your HSC marks!