2025 HSC Chemistry Exam Paper Exemplar Answers & Solutions

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Section I: Multiple Choice

MCQ Answer Key

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

Question 1

A strong acid is one that completely ionises in solutions. Since the solution contains only H⁺ and A^- ions, this suggests that the acid is fully ionised (HA(aq) → H⁺(aq) + A^-(aq)) and therefore must be a strong acid.

Since there are very few acid molecules in the solution, this suggests the solution is dilute and not concentrated.

Therefore, option B is the correct answer.

Question 2

The reaction illustrates the conversion of butan-2-ol (a secondary alcohol) into butan-2-one (a ketone). This is performed via an oxidation reaction.

Therefore, option B is the correct answer.

Question 3

1,2-dibromopentane is the name of an alkane with a 5-carbon chain that has a bromine substituent on carbon 1 and 2. 

This corresponds to option A.

Question 4

The addition of I^– to the solution containing Pb²⁺ ions will result in a yellow precipitate forming: Pb²⁺(aq) + 2I^–(aq) → PbI₂(s). Adding I^– to the solution containing Na⁺ will not produce a precipitate. This allows the two solutions to be identified. Therefore, option A is the correct answer.

The addition of NH₄⁺, NO₃^–, and CH₃COO^– to both solutions will not result in any visible reaction, as no precipitate forms with any of the cations. Therefore, the addition of NH₄⁺, NO₃^- and CH₃COO^- cannot be used to distinguish between the two solutions.

Question 5

Since there is initially some Cl_2, the concentration of Cl₂ is initially non-zero. This eliminates options B and D. 

The concentration of Cl₂ will decrease until it reaches a constant non-zero value when dynamic equilibrium is established.

Therefore, option C is the correct graph.

Question 6

HCl is a strong acid: HCl(aq) + H₂O(l)  → H₃O⁺(aq) + Cl^-(aq)

∴ [H₃O⁺] = 0.25 molL^-1 (1:1 stoichiometric ratio)

pH = -log₁₀0.25 = 0.60

Question 7

Since the two complexes are different in colour, UV-visible spectrophotometry is most suitable for distinguishing between the two complexes as they will absorb a different peak wavelengths in the visible region of the electromagnetic spectrum.

Therefore, option C is the correct answer.

Question 8

The molecular ion peaks of X and Y correspond to the molar mass of X and Y, respectively.

By the Law of Conservation of Mass: MM(X) + MM(unknown substance) = MM(Y)

∴ MM(unknown substance) = MM(Y) - MM(X) = 92 - 72 = 20 gmol^-1

20gmol^-1 ≈ MM(HF)

Therefore, option D is the correct answer.

Question 9

As saponin has both water-soluble and fat-soluble components, it will be able to act as a cleaning agent, similarly to soaps. This is because the fat-soluble component will interact with any grease, while the water-soluble component will interact with water molecules. Upon agitation, micelles will form around the grease, producing an emulsion which can be washed away, leaving the surface clean.

Therefore, the answer is A.

Question 10

Butanoic acid is a covalent molecule, so it can exhibit dispersion forces. It is also polar due to the presence of a carboxyl group, allowing it to form dipole-dipole interactions. Since the acidic proton is a hydrogen bond donor, while the lone pairs of electrons on the oxygen atoms are hydrogen bond acceptors, butanoic acid can also form hydrogen bonds.

Therefore, the correct answer is D.

Covalent bonding is an intramolecular force. It is not present between molecules of butanoic acid, but instead it is present within butanoic acid, holding the atoms together.

Question 11

A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. X is propanoate, the weak conjugate base of propanoic acid and therefore can accept a proton, acting as a Brønsted-Lowry base. Y is ethanamine, a weak base and therefore is also capable of accepting a proton and acts as a Brønsted-Lowry base.

Therefore, A is the correct answer.

Question 12

K_eq = [products] / [reactants] = [Fe(NO₃)₃] / ([AgNO₃]³[FeCl₃])

Chemicals in a liquid or solid state are omitted from the equilibrium expression, so AgCl(s) is not included.

Therefore, option B is the correct answer.

Question 13

Combustion of octane above 100°C: 2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

Note that water is gaseous as the reaction is occurring above 100°C.

Combustion is an exothermic reaction. Hence, the reaction has a negative enthalpy change,i.e. ∆H < 0. Since there are significantly more gaseous molecules on the products side than the reactants side, entropy increases in this process, i.e. ∆S > 0.

Therefore, option C is the correct answer.

Question 14

Experiment 1: K_eq = [H₂][I₂] / [HI]² = 0.01² / 0.04²

Experiment 2: K_eq =  [H₂][I₂] / [HI]² = 0.02² / 0.04²

By comparing the expressions for K_eq, K_eq(experiment 2) > K_eq(experiment 1). This eliminates options C and D.

Since K_eq(experiment 2) is larger, this implies that the system shifts to the right when the temperature is changed from experiment 1. Since the forward reaction is endothermic, the temperature at which experiment 2 is performed must be higher than experiment 1 so that the equilibrium will shift to the right by Le Chatelier’s Principle to decrease temperature.

Therefore, option B is the correct answer.

Question 15

When prop-2-en-1-ol reacts with hydrogen gas, it undergoes hydrogenation, an addition reaction, to produce propan-1-ol.

Upon oxidation, since propan-1-ol is a primary alcohol, it will oxidise into propanoic acid. The identity of propanoic acid is further confirmed, as it effervesced upon the addition of sodium carbonate due to the production of carbon dioxide.

When propanoic acid is reacted under reflux with metanol and concentrated sulfuric acid, it undergoes an esterification reaction to form an ester, i.e. methyl propanoate.

Hence, X is propan-1-ol, Y is propanoic acid, and Z is methyl propanoate, and their structures are correctly illustrated by option D.

Question 16

These are two ways of approaching this question.

Method 1

MM(3-hydroxypropanoic acid) = 3 x 16.00 + 3 x 12.01 + 6 x 1.008 = 90.078 gmol^-1

3-hydroxypropanoic acid undergoes homocondensation polymerisation. Therefore, when n monomer units react, (n-1) water molecules are produced.

∴ MM(strand) = MM(monomers)_used - MM(H₂O)_condensed

= 1000 x  90.078 - 999 x (2 x 1.008 + 16.00) = 72080 gmol^-1

Method 2

Repeating unit: –OCH₂CH₂CO– 

MM(repeating unit) = 2 x 16 + 3 x 12.01 + 4 x 1.008 = 72.062 gmol^-1

MM(strand) = 1000 x MM(repeating unit) + MM(H₂O) = 72.062 x 1000 + (2 x 1.008 + 16) = 72080 gmol¹

Here, the molar mass of water was added once in the expression to account for the H and OH on either end of the chain, which did not condense. 

Question 17

The distinct chemical environments for each compound are provided below:

As can be seen, the only option where the number of carbon chemical environments equals the number of proton chemical environments is option A.

Question 18

This question is effectively describing a Mohr’s method titration.

Cl^–(aq) + Ag⁺(aq) → AgCl(s)

CrO₄^2-(aq) + 2Ag⁺(aq) → Ag₂CrO₄(s)

For potassium chromate to act as a suitable indicator, it must precipitate out after the analyte (Cl^– ions) is fully precipitated out. Hence, the more soluble salt (the one that precipitates out later) is Ag₂CrO₄.

In addition, the indicator must undergo a colour change to indicate the endpoint of the titration. Since it is given that chromate ions are yellow in the question, the only valid colour change is yellow to red.

Thus, option B is the correct answer.

Question 19

These are two ways of approaching this question.

Method 1 (More methodological)

The following reaction takes place and occurs to completion, as HCl(aq) is a strong acid:

CH₃COO^–(aq) + HCl(aq) → CH₃COOH(aq) + Cl^–(aq)

n(CH₃COO^–) = 0.1 mol
n(HCl) = 0.500 L x 0.1 molL^-1 = 0.05 mol

Since the two species react in a 1:1 ratio, HCl(aq) is the limiting reagent.

After the reaction,

n(CH₃COO^–)_{after neutralisation} = n(CH₃COO^–)_initial - n(CH₃COO^–)_reacted
= n(CH₃COO^–)_initial - n(HCl)_added
= 0.1 mol - 0.05 mol
= 0.05 mol

n(CH₃COOH)_{after neutralisation} = n(HCl)_added
= 0.05 mol

The following equilibrium is then established:

CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO^–(aq) + H₃O⁺(aq)

Hence, constructing an ICE table:

Hence, K_a = [CH₃COO^–][OH^–] / [CH₃COOH]
= (0.1 + 10^–4.8)(10^–4.8) / (0.1 – 10^–4.8)
= 1.585395 x 10^-5

Then, considering the dilution:

Substituting and solving, we have:

K_eq = x(0.05 + x) / (0.05 – x) = 1.585395 x 10^–5

As K_eq is small, x is small and hence 0.5 + x ≈ 0.5 – x ≈ 0.5 (small change assumption)
⇒ x = 1.585395 x 10^–5 

Thus, pH = -log₁₀[H₃O⁺] = -log₁₀(1.585395 x 10^-5) = 4.79986252 = 4.8 (1 d.p.)

Method 2 (Faster)

Since exactly half of the CH₃COO^–(aq) are neutralised by the HCl(aq) to CH₃COOH(aq), this represents a buffer solution / half-equivalence point, where:

[CH₃COO^–]_initial = [CH₃COOH]_initial

Hence, making the small change assumption that [CH₃COO^–]_initial ≈ [CH₃COO^–]_eq and [CH₃COOH]_initial ≈ [CH₃COOH]_eq:

K_a = [CH₃COO^–][H₃O⁺] / [CH₃COOH]  = [H₃O⁺] = 10^-4.8

In other words, [H₃O⁺] is equal to the K_a and independent of the actual concentration of the acid or the volume of water present.

Thus, the pH does not change and remains at 4.8, which corresponds to option C.

Question 20

[Ag⁺]_diluted = 0.11 x 10^-6 molL^-1 (by interpolation)

Since it is given that [Ag⁺]_saturated / [Ag⁺]_diluted = 2000, it follows that:

[Ag⁺]_saturated = 2000 x [Ag⁺]_diluted
= 2000 x (0.11 x 10^-6 molL^-1)
= 2.2 x 10^-4 molL^-1

Consider the actual dissolution of silver oxalate:

Ag₂C₂O₄(s) ⇌ 2Ag⁺(aq) + C₂O₄^2-(aq)

Since Ag⁺(aq) and C₂O₄^2-(aq) are produced in 2:1 ratio,

[C₂O₄^2-]_saturated = ½ x [Ag⁺]_saturated
= ½ x (2.2 x 10^-4 molL^-1)
= 1.1 x 10^-4 molL^-1

Hence, K_sp = [Ag⁺]²[C₂O₄^2-]
= (2.2 x 10^-4)²(1.1 x 10^-4)
= 5.3 x 10^-12

Section II: Written Responses

Question 21

Exemplar Answer

Reaction condition X: UV light

IUPAC name of organic product: 2-bromobutane

Question 22

Exemplar Answer

Qualitative analysis

Qualitative analysis is used to identify chemical species in the water, which may be environmental contaminants such as NO₃^-(aq), which can cause eutrophication or Pb²⁺(aq) ions, which are toxic and can bioaccumulate and biomagnify.

Quantitative analysis

Quantitative analysis allows the concentration of each chemical species to be calculated. This allows chemists to monitor the concentration of species to ensure they are within safe limits and implement management strategies when they trend above the safe limits.

Question 23

Exemplar Answer

1. The mass of the watch glass and filter paper have not been measured separately from the BaSO₄ precipitate. This results in a systematic error, where the mass of the washglass and filter paper inflates the mass of the precipitate and reduces the accuracy of the % w/w calculated. These pieces of equipment should be weighed separately and subtracted from the mass measured in step 7.
2. The precipitate should be weighed in multiple 20-minute intervals throughout the drying process until a constant mass is obtained. This would ensure that all of the moisture is removed, which could otherwise inflate the mass of the precipitate and act as a systematic error that reduces accuracy.

Answers Could Include

• A higher concentration of BaCl₂(aq) could be added to ensure that the sulfate ions are precipitated out completely.
• The drying of the precipitate could be performed in a more controlled environment in the lab, such as a dessicator or a drying oven. This would minimise the likelihood of impurities such as pollens or dust settling on the watch glass, or some of the precipitate being lifted away by the wind. These random errors would otherwise reduce the reliability of the % w/w measured.

Question 24(a)

Exemplar Answer

Structural formula

Shape of molecule: Linear

Question 24(b)

Exemplar Answer

n(C₂H₂) = m(C₂H₂) / MM(C₂H₂)
= 65 g / 26.04 gmol^-1
= 2.49654325 mol

Since n(C₂H₂)_used = n(C₂H₃Cl)_produced = 1:1

n(C₂H₃Cl)_produced = n(C₂H₂)_used = 2.49654325 mol

Hence, m(C₂H₃Cl)_produced = n(C₂H₃Cl)_produced x MM(C₂H₃Cl)
= 2.49654325 mol x 62.50 gmol^-1
= 156.033953 g
= 156 g (3 s.f.)

Question 25

Exemplar Answer

Propan-1-ol and butanoic acid react in 1:1 ratio.

Since n(propan-1-ol) = 0.267 mol < n(butanoic acid) = 0.298 mol, propan-1-ol is the limiting reagent.

Since n(propan-1-ol)_used:n(propyl butanoate)_{theoretically produced} = 1:1,

n(propyl butanoate)_{theoretically produced} = n(propan-1-ol)_used
= 0.267 mol

m(propyl butanoate)_{theoretically produced} = 0.267 mol x 130.2 gmol^-1 = 34.7634 g

Now, calculating the mass actually produced:

m(propyl butanoate)_{actually produced} = 12.2 mL x 0.873 gmL^-1 = 10.6506 g

Hence, % yield = (m(propyl butanoate)_{actually produced} / m(propyl butanoate)_{theoretical produced}) x 100%
= (10.6506 g / 34.7634 g) x 100%
= 30.6373945%
= 30.6% (3 s.f.)

Question 26(a)

Exemplar Answer

Marker’s Notes

• A better way to represent unrelated data points like this is using a column graph, but we have decided to use a line graph above, as a column graph is unprecedented for the HSC.

Question 26(b)

Exemplar Answer

As halogens are taken from further down in Group 17, the pK_a of the haloethanoic acid increases. Since K_a = 10^-pKa, the increase in pK_a corresponds to a decrease in K_a. As K_a = [X–CH₂COO^–][H₃O⁺] / [X–CH₂COOH] decreases, the concentration of the ionised acid at equilibrium decreases.

Since the greater the degree of ionisation of an acid, the stronger the acid is, the strength of the acid decreases as the halogen in haloethanoic acid becomes larger.

Question 27(a)

Exemplar Answer

The combustion of petrol in engines releases greenhouse gases such as CO₂(g) and / or CO(g): C₇H₁₆(l) + 11O₂(g) → 7CO₂(g) + 8H₂O(l).

These greenhouse gases contribute to the greenhouse effect, which prevents sunlight from escaping into space after being reflected by the Earth. This results in a rise in the climate temperature (global warming).

This results in increasing frequency and severity of bushfires, as well as the melting of ice caps and a consequent rise in sea level. In either case, natural habitat is lost, which is a detrimental implication for the environment.

Question 27(b)

Exemplar Answer

Question 27(c)

Exemplar Answer

Question 28(a)

Exemplar Answer

Question 28(b)

Exemplar Answer

The stronger the intermolecular forces, the greater the amount of energy required to separate the polymer chains, resulting in a greater difficulty in pulling them apart, as well as an increasing melting point. 

Kevlar is a condensation polymer featuring amide linkages. Thus, it possesses a polar amide functional group throughout the polymer, resulting in strong polar hydrogen bonding between Kevlar molecules. 

Meanwhile, Polystyrene is an addition polymer without polar functional groups. Thus, it is non-polar and can only form weak dispersion forces between each polymer molecule. In addition, its large benzene ring side chain prevents efficient stacking and creates chain chain-stiffening effect, which increases the brittleness of the polymer. 

Hence, Kevlar has stronger intermolecular forces than Polystyrene, as well as being less brittle. Thus, Kevlar is harder to pull apart than polystyrene, and it possesses a higher melting point.

Marker's Notes

Since the question specifies physical properties in plural, more than one physical property should be discussed.

Question 29

Exemplar Answer

Since the system is initially at equilibrium, Q = [N₂O₄] / [NO₂]²= K_eq. When argon gas is added to the system, as the volume is fixed, the concentrations of NO₂ and N₂O₄ do not change. Therefore, Q  = K_eq after the addition of argon gas, and the system does not shift in either direction, remaining at equilibrium.

As there are no macroscopic changes to a system at dynamic equilibrium, the temperature will remain constant.

Question 30(a)

Exemplar Answer

As phosgene is a toxic gas, it should be handled in a fume cupboard to prevent its diffusion and inhalation. Gas masks may be worn to further prevent inhalation.

Question 30(b)

Exemplar Answer

Excess carbon monoxide

When an excess of carbon monoxide is injected into the system. By Le Chatelier's Principle, the position of equilibrium will shift right to decrease the carbon monoxide concentration. Consequently, the yield of phosgene will increase, which is desirable in industry.

Use of a catalyst

The use of a catalyst provides an alternate reaction pathway with a lower activation energy. This results in a greater success rate of collisions, as a greater proportion of particles can collide with sufficient energy above the activation energy barrier. Therefore, the rate at which phosgene is synthesised will increase, increasing the yield of phosgene within a set timeframe, which is also desirable in industry.

Question 31(a)

Exemplar Answer

By Gay-Lussac’s Law of Combining Volumes, the volume ratio of gases in a reaction matches the stoichiometric ratio.

N₂H₄(g) + 3O₂(g) → 2NO₂(g) + 2H₂O(g)

Using N₂H₄ as the molecular formula for hydrazine gives a stoichiometric ratio of n(hydrazine):n(O₂):n(NO₂):n(H₂O) = 1:3:2:2, which matches the volume ratio provided of V(hydrazine):V(O₂):V(NO₂):V(H₂O) = 1 L : 3 L : 2 L : 2 L.

Question 31(b)

Exemplar Answer

K_a(N₂H₅⁺) = K_w / K_b(N₂H₄) = 10^-14 / (1.7 x 10^-6) = 5.88235 x 10^-9

N₂H₅⁺(aq) + H₂O(l) ⇌ N₂H₄(aq) + H₃O⁺(aq) 

Let x be [H₃O⁺]_eq:

K_a = [H₃O⁺] [N₂H₄] / [N₂H₅⁺] 

⇒ x² / (0.2 - x) = 5.88235 x 10^-9 

As K_a is small, x is small and hence 0.20 - x ≈ 0.20 (small change assumption)

⇒ x² / 0.2 = 5.88235 x 10^-9
⇒ x = 3.429971 x 10^-5 = [H₃O⁺]

pH = –log₁₀[H₃O⁺]
= –log₁₀(3.429971 x 10^-5)
= 4.46 (2 d.p.)

Question 32

Exemplar Answer

As all sodium and nitrate salts are soluble, the only precipitates that can form are Mg(OH)₂(s) and MgCO₃(s).

As the K_sp of Mg(OH)₂ = 5.61 x 10^-12 is significantly smaller than the K_sp of MgCO₃ = 6.82 x 10^-6, Mg(OH)₂ will precipitate out first. 

[Mg²⁺] = 0.006 mol / 1 L = 0.006 molL^-1
[OH^–] = 0.010 mol / 1 L = 0.01 molL^-1

Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH^–(aq)

Q_sp = [Mg²⁺][OH^–]² = 0.006 x 0.01² = 6 x 10^-7
K_sp(Mg(OH)₂) = 5.61 x 10^-12

As Q_sp > K_sp, Mg(OH)₂ will form a precipitate. 

Assuming the precipitation reaction goes to completion (as K_sp is very small):

Mg²⁺(aq) + 2OH^–(aq) → Mg(OH)₂(s)

n(Mg²⁺) = 0.006 mol
n(OH^–) = 0.01 mol

Since ½ x n(OH^–) = 0.005 mol < n(Mg²⁺), OH^– is the limiting reagent (1:2 stoichiometric ratio) 

n(Mg²⁺)_excess = 0.006 mol - (0.01 mol / 2) = 0.001 mol
[Mg²⁺]_excess = 0.001 mol / 1 L = 0.001 molL^-1 

[CO₃^2-] = 0.002 mol / 1 L = 0.002 molL^-1 

MgCO₃(s) ⇌ Mg²⁺(aq) + CO₃^2-(aq) 

Q_sp = [Mg²⁺][CO₃^2-] = 0.001 x 0.002 = 2 x 10^-6
K_sp(MgCO₃) = 6.82 x 10^-6

As Q_sp < K_sp, MgCO₃ will not precipitate. 

Hence, the only precipitate that forms is Mg(OH)₂.

Question 33

Exemplar Answer

V(KOH) = (7.20 + 7.10 + 7.15) / 3 = 7.15 mL = 0.00715 L (ignoring 7.80 mL as an outlier) 

n(KOH) = 0.1 molL^-1 x 0.00715 L
= 0.000715 mol

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l) 

n(HCl)_titrated = 0.000715 mol (based on 1:1 stoichiometric ratio) 

n(HCl)_excess = 0.000715 mol x (100.0 mL / 20 mL)
= 0.003575 mol 

n(HCl)_initial = 0.1 L x 0.550 molL^-1
= 0.0550 mol

n(HCl)_{reacted with CaCO3} = n(HCl)_initial - n(HCl)_excess
= 0.0550 mol - 0.003575 mol
= 0.051425 mol

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) 

n(CaCO₃) = 0.051425 / 2 (based on 1:2 stoichiometric ratio)
= 0.0257125 mol

m(CaCO₃) = 0.0257125 mol x (40.08 + 12.01 + 3 x 16.00) gmol^-1
= 2.573564125 g 

%w/w(CaCO₃) = (2.573564125 / 3) x 100%
= 85.78547%
≈ 85.5%

Therefore, Brand X was used.

Question 34(a)

Exemplar Answer

Method 1 (Half-equivalence point)

The equivalence point occurs at the point where 0.024 L of NaOH has been added.

The half-equivalence point must thus occur at the point where exactly half of that amount of NaOH has been added, at V(NaOH) = 0.024 / 2 = 0.012 L, which corresponds to a pH of 4.4.

At the half-equivalence point, [HX] = [X^–], so K_a = [X^–][H⁺] / [HX] = [H⁺], and pK_a = pH = 4.4.

Hence, K_a = 10^-pKa = 10^–4.4 ≈ 4.0 x 10^-5 (2 s.f.)

Method 2 (Initial pH)

V(NaOH)_required = 0.024 L
n(NaOH)_required = 0.024 L x 0.10 molL^-1 = 0.0024 mol

HX(aq)+ NaOH(aq) → NaX(aq) + H₂O(l)

Since HX(aq) and NaOH(aq) react in 1:1 ratio,

n(HX) = n(NaOH) = 0.0024 mol

Hence, [HX]_initial = 0.0024 mol / 0.010 L = 0.24 molL^-1

Based on the initial pH on the titration curve, [H⁺]_eq = 10^-2.5 molL^-1

HX(aq) ⇌ H⁺(aq) + X^–(aq)

Thus, K_a = [X^–][H⁺] / [HX]
= (10^-2.5)² / (0.24 - 10^-2.5)
= 0.000042223
≈ 4.2 x 10^-5 (2 s.f.)

Question 34(b)

Exemplar Answer

At the equivalence point, when 0.024 L of NaOH is added, all of the acid (HX) is neutralised:

HX(aq) + NaOH(aq) → NaX(aq) + H₂O(l)

As NaOH is further added after the equivalence point, it is diluted by the 0.010 L (acid) + 0.024 L (base) = 0.034 L of salt solution present. This results in the overall concentration of NaOH always being less than the 0.10 molL^-1 NaOH solution added. Since pOH = -log₁₀[OH^-], the pOH of the solution will always be > 1. Further, since pH = 14 - pOH (at 25°C), the pH of the solution will always be less than 13 and never reach 13.

Question 35

Exemplar Answer

The reaction is exothermic in the forward direction, as the products have a lower enthalpy than the reactants.

Le Chatelier's Principle

Le Chatelier’s Principle states that when a system at dynamic equilibrium is disturbed the equilibrium will shift to minimise the disturbance until a new equilibrium is established. 

When the solution is cooled from 80ºC to 0ºC, by Le Chatelier's Principle, the system will shift to increase the temperature by favouring the forward exothermic reaction, which releases heat. Therefore, the equilibrium shifts to the right and increases the concentration of [Co(H₂O)₆]²⁺, which is pink. Hence, the solution turns pink.

Collision theory

Collision theory states that for a chemical reaction to occur, chemicals must collide with sufficient energy (above the activation energy) with the correct molecular orientation.

When the solution is cooled to 0ºC, the collisional frequency of all particles decreases, and the proportion of particles that can overcome the activation energy barriers decreases, which results in the success rate of collisions also decreasing. Thus, both reaction rates decrease. However, as the reverse, endothermic reaction has a greater activation energy (E_a2) than that of the forward, exothermic reaction (E_a1), the reverse reaction experiences a greater decrease in the proportion of particles that can now overcome the activation energy barrier. Thus, the reverse reaction rate decreases to a greater extent, and the forward reaction is favoured, shifting the equilibrium to the right and increasing the concentration of [Co(H₂O)₆]^2+, which turns the solution pink.

Question 36

Exemplar Answer

IR Spectrum

A very broad, strong trough will be observed in the wavenumber range of 2500 - 3000 cm^-1 due to the presence of an acidic O–H bond in the carboxyl group. A sharp, strong trough will be observed in the wavenumber range of 1680 - 1750 cm^-1 due to the presence of the C=O bond in the carboxyl group.

¹³C NMR

3 signals would be present on the ¹³C NMR spectrum, as the compound has 3 distinct carbon environments, as labelled below:

• C₁ produces a signal between 160 – 185 ppm as it is the carbon in the carbonyl group of a carboxylic acid. This peak is significantly downfield due to the electronegative oxygen atoms deshielding C₁ as they draw electron density away from it.
• C₂ produces a signal from 20 – 50 ppm as it is the carbon atom adjacent to the carbonyl group.
• C₃ produces a signal from 5 – 40 ppm, as it is not in proximity to any electronegative elements and produces a peak corresponding to C–C carbons. This signal will be more upfield than that produced by C₂, as it is further away from the electronegative oxygen atom and hence is more shielded.

Proton NMR 

3 signals would be present on the proton NMR spectrum as the compound has 3 distinct hydrogen environments, as labelled below:

• H₁ would produce the most downfield signal as it is directly bonded to an electronegative oxygen atom in the carboxyl group, resulting in the hydrogen atom being highly deshielded. As this hydrogen is directly bonded to oxygen, it produces a singlet in the spectrum.
• The two hydrogens in H₂ would produce a signal from 2.1 – 4.5 ppm, as it corresponds to a CH₂–CO– group on the carboxylic acid. The two hydrogens in H₂ are neighboured by the 3 hydrogens in H₃ and would hence produce a quartet by the n+1 rule of spin-spin coupling.
• The three hydrogens in H₃ would produce a signal from 0.7 – 2.1 ppm as they correspond to –CH₃ protons as seen on the data sheet. These three hydrogens are neighboured by the 2 hydrogens in H₂ and would hence produce a triplet by the n+1 rule of spin-spin coupling.

The signal integral ratio of the three signals produced by H₁, H₂, and H₃ environments would be 1:2:3, corresponding to the number of hydrogen atoms present in each environment.

Mass Spectroscopy

The molecular ion peak would be observed at 74 m/z, as propanoic acid has a molar mass of approximately 74 gmol^-1. A smaller peak would likely appear at 75 m/z due to the presence of isotopes such as ¹³C.

Potential fragments that would produce peaks on the mass spectrum are summarised in the table below:

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Marker’s Notes

The HSC has mistakenly termed mass spectrometry as mass spectroscopy in this question.

Question 37

Exemplar Answer

C₅H₁₀ matches the chemical formula of a 5-carbon alkene. It undergoes a hydration reaction (H⁺ / H₂O) to produce compounds B and C. Thus, B and C must be alcohols.

As compound B does not react with acidified dichromate (H⁺ / Cr₂O₇^2-), this means it cannot oxidise and thus must be a tertiary alcohol. 

As compound C does react with acidified dichromate, this means compound C can be oxidised, and as a result, must be a primary or secondary alcohol.

Compound D, the product of the oxidation of compound C, does not react with Na₂CO₃. This means compound D cannot be a carboxylic acid. As primary alcohols oxidise into carboxylic acids, compound C must therefore be a secondary alcohol, as compound D would be a ketone, which does not react with Na₂CO₃.

Thus, compound A is a 5-carbon alkene that hydrates to produce both secondary and tertiary alcohols. This means one of the carbon atoms in the C=C bond must be bonded to three other carbon atoms, while the other must be bonded to two.

Therefore, compound A is 2-methylbut-2-ene. 

Marker’s Notes

• It should be noted that compound B is 2-methylbutan-2-ol, compound C is 3-methylbutan-2-ol, and compound D is 3-methylbutan-2-one.
• The name of compound A does not need to be provided to score full marks.

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