2024 VCE Chemistry Exam Solutions

2024 VCE Chemistry exam solutions

Want to see how you did in VCE Chemistry? Our Chemistry team has written the 2024 VCE Chemistry exam solutions, so read on to check your answers and see how you went! View the VCE Chemistry Study Design here.

2024 VCE Chemistry exam solutions: Section A

Question	Answer	Explanation
1	B	Photosynthesis is the process by which green plants and certain other organisms synthesise glucose and oxygen from carbon dioxide and water by absorbing energy from sunlight. 6CO2(g) + 6H2O(l) ⟶ C6H12O6(s) + 6O2(g) Since energy is absorbed, the process is endothermic. A redox reaction can be identified by determining the oxidation states of each element in the reactants and products. Element Oxidation state in reactants Oxidation state in products Carbon +4 0 Oxygen −2 (in both CO2 and H2O) −2 (in C6H12O6) 0 (in O2) Hydrogen +1 +1 Oxidation state of carbon decreases, so CO2 undergoes reduction. Oxidation state of oxygen increases, so H2O undergoes oxidation. This means photosynthesis is a redox reaction.
2	B	The oxidation of glucose in the body is the process of respiration. C6H12O6(s) + 6O2(g)  ⟶ 6CO2(g)  + 6H2O(l) This process goes to completion. Hence options A and C are incorrect. Combustion involves the reaction of a substance with oxygen. Hence oxygen does not undergo “combustion” and option D is incorrect. The molar enthalpy of combustion of glucose and hydrogen is provided in the data book. These values can be converted to energy per gram by dividing each by the molar mass of the substance.  ∆H (glucose) = −2840 kJ mol−1 = −2840 ÷ 180 = −15.8 kJ g−1 ∆H (hydrogen) = −286 kJ mol−1 = −286 ÷ 2 = −143 kJ g−1 Hence glucose provides less energy per gram than hydrogen during combustion.
3	B	The limiting reactant limits the amount of product formed. To determine the limiting reactant, determine the moles of product each reactant can produce.  n(C2H6) = 0.50 mol n(CO2) when all C2H6 is reacted = 2 × n(C2H6) = 2 × 0.50 = 1.0 mol m(O2) = 22% × 100 = 22 g n(O2) = m / M = 22 / 32 = 0.6875 mol n(CO2) when all O2 is reacted = (n(O2)/3.5) × 2 = (0.6875/3.5) × 2 = 0.392857 mol Since O2 produces less moles of product, it is the limiting reactant. This means C2H6 is in excess.  n(C2H6)used = n(O2) / 3.5 = 0.6875/3.5 = 0.1964285714 mol n(C2H6) remaining after the reaction = n(C2H6)total − n(C2H6)used = 0.50 − 0.1964285714 = 0.3035714 mol = 0.30 mol (2 s.f.)
4	A	The oxidising and reducing agents can be identified by determining the oxidation states of each element in the reactants and products. Element Oxidation state in reactants Oxidation state in products Chromium +6 +3 Oxygen −2 −2 Hydrogen +1 +1 Iodine −1 0 An oxidising agent causes another substance to be oxidised and itself is reduced (its oxidation state decreases). A reducing agent causes another substance to be reduced and itself is oxidised (its oxidation state increases). The oxidation state of chromium decreases so Cr2O72− undergoes reduction and is the oxidising agent. The oxidation state of iodine increases so I− undergoes oxidation and is the reducing agent. The conjugate oxidising agent of I− is I2.
5	C	A galvanic cell converts chemical energy to electrical energy, hence total chemical energy will decrease during operation.  In a galvanic cell, electrons flow from site of oxidation to site of reduction. The half-equation that is higher in the electrochemical series (from data book) is the reduction equation. Hence Sn4+ is reduced. Since reduction occurs in half-cell A, electrons flow to half-cell A.
6	C	The half equation involving O2 in an alkaline solution from electrochemical series is: O2(g) + 2H2O(l) + 4e− ⇌ 4OH−(aq) E° = +0.40 V A more positive electrode potential indicates a greater tendency to undergo reduction. Hence reduction occurs in half-cell 1 and oxidation occurs in half-cell 2. Half-cell 1 reduction half equation: IO3–(aq) + 6H+(aq) + 5e− ⟶ ½I2(aq) + 3H2O(l) Half-cell 2 oxidation half equation: 4OH–(aq) ⟶ O2(g) + 2H2O(l) + 4e− In a galvanic cell, the anode is the negative electrode and is the site of oxidation. As seen in the above oxidation half-equation, OH− ions are consumed, hence its concentration decreases at this electrode. The cathode is the site of reduction. As seen in above reduction equation, iodine and water are produced. Hence A is incorrect.  The oxidising agent is IO3–(aq) and reducing agent is OH–(aq), hence B is incorrect. The electrodes are inert, so their mass will be unchanged. Hence D is incorrect.
7	A	Half equation occurring in this galvanic cell: Cu2+(aq) + 2e− ⟶ Cu(s) n(Cu(s)) = ½ × n(e−) = ½ × 0.50 = 0.25 mol m(Cu(s)) = n × M = 0.25 × 63.5 = 15.875 g = 1.6 × 10 g
8	D	The use of porous electrodes increases the cell efficiency as it provides a high surface area and therefore more effective contact with reactants, allowing a larger current to be produced. The pores also allow the H2 fuel and O2 to easily diffuse through the electrode and come in contact with ions in the electrolyte, facilitating the redox reaction. As the ratio of hydrogen to amount of electricity produced increases, this efficiency is decreased as this indicates a larger quantity of hydrogen produces less electricity.
9	D	The negative electrode in a fuel cell is the cathode, which is the site of reduction.  The reduction half-equation for ethanol is: C2H5OH(l) + 12OH−(aq) ⟶ 2CO2(g) + 9H2O(l)+ 12e− So, 1 mole of ethanol will react with 12 moles of OH−.
10	B	Bioethanol is only theoretically carbon neutral. In reality, there is still a net release of CO2 because fossil fuels are used to provide energy for the other process, such as crop growing, harvesting, transport, milling, and distillation, involved in the production of bioethanol. H2O2 has a large electrode potential (from electrochemical series), hence has a high tendency to undergo reduction.  H2O2(aq) + 2H+(aq)+ 2e− ⇌ 2H2O(l)   E° = +1.77 V An oxidising agent causes another substance to be oxidised and itself is reduced. So H2O2 can act as an oxidising agent in fuel cells.
11	A	In a linear economy, natural resources are turned into products that will eventually become waste. A circular economy aims to create a ‘closed loop’ system that minimises the input of natural resources and the creation of waste. The iron is continually created using iron ore (natural resource) and not through recovering and recycling of iron products, hence would be consistent with the concept of linear economy.
12	C	When the position of equilibrium lies very close to the products, the reaction can be considered to be “irreversible” as the reverse reaction will not occur to any observable extent. A and B are incorrect because being exothermic in the forward direction, which results in less stable products, does not make the reaction irreversible. For example, the Haber process (the formation of ammonia from nitrogen and hydrogen) is exothermic in the forward direction but still reversible. D is incorrect as a stoichiometric amount of oxygen reacting with butane will still be irreversible.
13	C	A catalyst is a substance that increases the rate of a reaction by allowing the reaction to take an alternate reaction pathway with a lower activation energy. Hence removing it results in a reaction pathway with higher activation energy.
14	C	A lower pressure will shift the equilibrium towards the forward direction with more moles of gas and higher temperature shifts the equilibrium in the forward endothermic direction. This results in more NO2(g) production, hence yield increases.
15	A	The equilibrium is 2SO3(g) ⇌ 2SO2(g) + O2(g) Let initial moles of SO3 be W and the change in moles as reaction proceeds be signified by 𝑥: Moles  2SO3(g) ⇌ 2SO2(g) + O2(g) Initial W 0 0 Change −2𝑥 +2𝑥 +𝑥 Equilibrium W −2𝑥 2𝑥 𝑥 Since only SO3 is initially present in the container, the moles of SO2 and O2 present at equilibrium must be according to the stoichiometric ratio. As seen in the table, the moles of SO2 will be double that of O2.  Mass is determined using m = n × M: m(SO2) = 2𝑥 mol × 64.1 g mol−1 = (128.2 × 𝑥) g m(O2) = 𝑥 mol × 32.0 g mol−1 = (32.0 × 𝑥) g Hence m(SO2) is 4 times the m(O2).
16	C	As the concentration of SCN− increased at t2, more collisions would occur between Fe3+ and SCN−, resulting in a faster forward reaction. However, at t2 there is no change to the concentration of FeSCN2+, hence no initial change in the rate of the reverse reaction. This means the reverse reaction rate would be less than the forward reaction rate, until a new equilibrium is established at t4 where the rates become equal.
17	D	To purify blister copper using electrolysis, impure copper is connected as the anode and the cathode is made of pure copper. Oxidation occurs at the anode. According to the electrochemical series, iron and nickel are stronger reductants than copper, hence will oxidise with copper and form metal ions in solution. Silver metal will not oxidise.  Fe(s) ⟶ Fe2+(aq) + 2e− Ni(s) ⟶ Ni2+(aq) + 2e− Cu(s) ⟶ Cu2+(aq) + 2e− Hence Cu2+, Fe2+ and Ni2+ will be present in solution. Cu2+ is the strongest oxidant out of the three ions, so will preferentially reduce at the cathode, hence producing pure copper at the cathode.  Cu2+(aq) + 2e− ⟶ Cu(s) As copper ions are produced and consumed, its concentration in the electrolyte changes during electrolysis.
18	B	The structure for each compound is: For the same molar mass, carboxylic acid B has a higher boiling point than ester A, alcohol C and amine D.  The dominant intermolecular force exhibited between the ester molecules is dipole-dipole force which is weaker than the hydrogen bonding in carboxylic acid B, alcohol C and amine D. However, more hydrogen bonds can form between the carboxylic acid molecules than between the alcohol molecules and amine D would form weaker hydrogen bonds than carboxylic acid B as nitrogen is less electronegative than oxygen. Thus, more thermal energy is required to break the intermolecular forces in carboxylic acids to cause boiling.
19	A	Structural isomers are molecules that have the same molecular formula, but their atoms are arranged differently to give different structures.
20	C	Possible isomers of C5H10O that are aldehydes:
21	D
22	D
23	B	The reaction mixture needs to be as dry as possible to prevent the saponification side reaction (reaction of the free fatty acids to form soap).
24	D	Reduction half equation (from electrochemical series): MnO4−(aq) + 8H+(aq) + 5e− ⟶ Mn2+(aq) + 4H2O(l) Overall equation (combining oxidation and reduction half-equations):  5C2H2O4(aq) + 2MnO4−(aq) + 6H+(aq)⟶ 10CO2(g) + 2Mn2+(aq) + 8H2O(l) n(MnO4−) = c × V = 0.100 × 0.010 = 1 × 10−3 mol n(C2H2O4) = 5/2 × n(MnO4−) = 2.5 × 10−3 mol V(C2H2O4) = n / c = 2.5 × 10−3 / 0.100 = 0.025 L = 25 mL
25	D	The retention time (RT) is a measure of the time taken for a sample to pass through a chromatography column. It is calculated as the time from injection to detection. There will be a stronger attraction between polar molecules in the mixture with the polar solvent (mobile phase) than the non-polar stationary phase. The more polar the molecules in the mixture, the greater it’s attraction, and thus the more quickly the molecule can travel through the column with the solvent. This reduces the RT. Thus, the order of RT, from lowest to highest, corresponds to the order of decreasing polarity, which is hexan-1-ol, hexan-2-one, then hexane.
26	A	A calibration curve is required to determine the concentration of an analyte. To obtain a calibration curve, a series of standard solutions needs to be prepared and run through the HPLC column under the same conditions.
27	D	Acidified potassium permanganate oxidises both primary and secondary hydroxyl groups to give carbonyl compounds, hence 1H NMR would more definitively identify presence of a primary hydroxyl group.
28	A
29	C	When sulfur atoms (from the side groups in cysteine) chemically bond together, covalent bonding occurs. This forms disulfide linkages between sections of the polypeptide chain that contribute to the tertiary structure (three-dimensional shape) of the protein.
30	C	Statement I: describes a step taken in the analysis, which is reported in the method. Statement II: summarises the results, which is reported in the discussion.  Statement III: Presents the data measured, which is reported in the results.

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2024 VCE Chemistry exam solutions: Section B

Question 1 (a)

The measured melting point can be compared to the literature melting points of known compounds in the consumer product to help identify its components. If the measured melting point is similar, the components are likely to be present.

Pure substances have a sharp, well-defined melting point whereas impure substances usually have a melting point range that tends to be broad. Hence the purity can be checked by observing the measured melting point range. If the measured melting point range is broad, the product is likely impure.

Question 1 (b)(i)

There is one chiral centre. 

Question 1 (b)(ii)

Addition reaction will occur between I₂ and linalool. Since there is two carbon-carbon double bonds present in each linalool molecule, linalool will react in a 1:2 mole ratio with I₂.

C₁₀H₁₈O + 2I₂ d C₁₀H₁₈I₄O

n(C₁₀H₁₈O) = ½ × n(I₂) = ½ × 0.042 = 0.021 mol 

m(C₁₀H₁₈O) = n × M = 0.021 × (12.0 × 10 + 1.0 × 18 + 16) = 3.234 g

% m(C₁₀H₁₈O)/m(mixture) = (3.234/15.0) × 100 = 21.56% = 22% m/m (2 s.f.)

Question 1 (b)(iii)

Distillation

Solvent extraction using separating funnel

Other answers possible

Question 2 (a)(i)

Compound E – C₅H₁₂

Question 2(a)(ii)

Substitution reaction

Question 2(b)

Compound L – HOCH₂CH₂CH₂CH2₂CH₃

Question 2(c)

Compound M – propan-1-ol

Question 2(d)(i)

Carbonate test –Add a small amount of solid sodium carbonate to the organic liquid. Bubbles of CO₂ gas will be produced indicating presence of carboxyl functional group.

Question 2(d)(ii)

Question 3(a)

Energy = 9.5 g × 17 kJ g⁻¹ + 3.4 g × 37 kJ g⁻¹ + 47.3 g × 16 kJ g⁻¹ = 1044.1 kJ = 1.0 × 103 kJ (2 s.f.)

Question 3(b)(i)

Amide bond

Question 3(b)(ii)

Condensation reaction

Question 3(b)(iii)

Leucine or serine

Question 3(c)(i)
The hydrocarbon chain in stearic acid contains carbon-carbon single bonds, hence stearic acid is saturated. However, the hydrocarbon chain in oleic acid contains a carbon-carbon double bond so oleic acid is unsaturated.

Question 3(c)(ii)

C₁₈H₃₄O₂ + 51/2O₂ ⟶ 18CO₂ + 17H₂O

Question 3(c)(iii)

Stearic acid would have a higher molar heat of combustion. 

Less energy is required to break the bonds in stearic acid than oleic acid as the C−C bonds present in stearic acid have a lower bond energy of 346 kJ mol⁻¹, compared to that of the C=C bond present in oleic acid which has a bond energy of 614 kJ mol⁻¹. Additionally, one more mole of water is released per mole of stearic acid compared to oleic acid, hence more energy is produced in the formation of products.

Question 4(a)

\( K_p = \frac{(p(NH_3))^2}{(p(N_2))(p(H_2))^3}\)

Question 4(b)

\( 1.60 \times 10^{-8} =  \frac{(p(NH_3))^2}{(43.4)(88.7)^3}\)

\( p(NH_3) = \sqrt{1.60 \times 10^{-8} \times 43.4 \times 88.7^3} = 0.696 \) kPa (3 s.f.)

Questions 4(c)

Since Q < K, the change resulted in the system having more reactants than products. Hence the change could be the addition of H₂ or N₂ into the system (or removal of NH₃ from the system). 

The system would respond by shifting the equilibrium in the forward direction to consume some of the reactants and produce more products to counteract the change. This would increase the value of Q until Q = K and equilibrium is re-established.  

Question 4(d)

As the forward reaction is exothermic, the temperature used is a compromise between the yield of ammonia and the rate of reaction. A lower temperature would cause the equilibrium to shift in the forward exothermic direction to produce more heat and minimise the disturbance according to Le Chatelier’s principle. This increases the yield of ammonia. However, low temperatures would decrease the rate of reaction due to a lower proportion of collisions exceeding the activation energy. Therefore, a moderate temperature would be the most appropriate compromise to obtain a moderate yield at a moderate rate.

By using the iron catalyst, a faster rate could be achieved at a low temperature, so no compromise is required.  A catalyst provides an alternate reaction pathway of lower activation energy so a larger proportion of collisions will be successful. This follows the green chemistry principle “catalysis” as the use of the catalyst reduces energy consumption, making the process more economical and ultimately less polluting.

Question 5(a)(i)

n(C₄H₁₀) = m / M = 0.580 g / (12.0 × 4 + 1.0 × 10) g mol⁻¹ = 0.01 mol

q = −(∆H × n) = −(−2880 kJ mol⁻¹ × 0.01 mol) = 28.8 kJ (3 s.f.)

Question 5(a)(ii)

q(water) = mc∆T = 100.0 g × 4.18 J g⁻¹ °C⁻¹ × 35.6 °C = 14880.8 J = 14.8808 kJ = 14.9 kJ (3 s.f.)

Question 5(a)(iii)

% efficiency = (14.8808 kJ ÷ 28.8 kJ) × 100 = 51.7% (3 s.f.)

Question 5(b)(i)

Calibration allows the calibration factor for the calorimeter to be determined, which provides the joules of energy required to change the temperature of the contents in a specific calorimeter by 1 °C. This accounts for the heat that would be lost to the surrounding air and the other physical parts of the calorimeter, hence increasing the accuracy of the results.

Question 5(b)(ii)

From the thermochemical equation, 2 moles of KI release 49 kJ of energy. 

\( \frac{49 kJ}{2 mol} = \frac{x kJ}{0.048 mol} \)

\( x = \frac{49 kJ}{2 mol} \times 0.048 mol =\) 1.176 kJ = 1176 J (energy released from 0.048 mol of KI)

∆T = E / CF = 1176 J / 470 J °C⁻¹ = 2.50212766 °C

T_final = T_initial + ∆T = 25.0 + 2.50212766 = 27.5 °C (1 d.p.)

Question 5(b)(iii)

Incorrect calibration of the calorimeter, such that the calibration factor is higher than true value. This could result from a larger volume of solution used in the calibration than in the experiment. A larger volume would require more energy to cause 1 °C temperature change. 

Question 6(a)

O−H bond

Question 6(b)(i)
Base peak occurs at m/z = 56

Question 6(b)(ii)

[C₄H₉O]⁺

Question 6(c)

4 peaks in the ¹³C NMR indicates there are 4 unique carbon environments. Since there are 5 carbons in the molecule, the molecule must not be symmetrical in the middle and there must be two carbons in the molecule that share the same chemical environment. This means the 5 carbons cannot be arranged in a straight chain and must be a branched chain with either one or two methyl groups. 

Question 6(d)(i)

The area under the peak.

Question 6(d)(ii)

The relative size of the integration curves corresponds to the ratio of hydrogen nuclei in each hydrogen environment.

Question 6(d)(iii)

The ratio of hydrogen nuclei in each hydrogen environment:

Since there are only 11 hydrogens in the molecule, the ratio correlates to the number of hydrogens in each hydrogen environment. 

Question 6(d)(iv)

The peak at 0.96 ppm corresponds to hydrogens in a methyl group (R-CH₃). Since the integration of this peak indicates there are 6 hydrogens in this environment, compound X must have two CH₃ groups that share the same environment. 

Question 7(a)(i)

Iron (Fe) is the anode.

Question 7(a)(ii)

2H₂O_(l) + 2e⁻ ⟶ H_2(g) + 2OH⁻_(aq)

Question 7(a)(iii)

The redox reaction is non-spontaneous so needs to be driven by the input of external energy.

Question 7(b)(i)

Positive

Question 7(b)(ii)

Q = I × t = 1.25 × (60 × 60) = 4500 C

n(e⁻) = Q / F = 4500 C / 96500 C mol⁻¹ = 0.046632 mol

n(LiC₆) = n(e⁻) = 0.046632 mol

m(LiC₆) = n × M = 0.046632 mol × (6.9 + 12.0 × 6) g mol⁻¹ = 3.68 g (3 s.f.)

Question 7(c)(i)

Solar energy (or light energy from the sun) is transformed into chemical energy.

Question 7(c)(ii)

Question 7(c)(iii)

Oxidation: 2H₂O_(l)  ⟶  O_2(g) + 4H⁺_(aq) + 4e⁻

Reduction: 2H⁺_(aq) + 2e⁻  ⟶  H_2(g)

Net reaction: 2H₂O_(l) + 4H⁺_(aq) + 4e⁻  ⟶  O_2(g) + 4H⁺_(aq)+4e⁻ + 2H_2(g)

As seen in the above equations, the electrons produced from the oxidation of water will be consumed by the reduction of hydrogen ions. The hydrogen ions produced from the oxidation of water will flow from the anode to the cathode through the electrolyte and become reduced to hydrogen gas. Hence electrical charge is balanced. 

Question 8(a)(i)

Question 8(a)(ii)

Boiling water could kill the yeast and temperature of boiling water is too high for zymase activity so fermentation would not occur.

Question 8(b)(i) 

m(CO₂) = 282.5 g − 282.4 g = 0.1 g

n(CO₂) = m / M = 0.1 g / 44.0 g mol⁻¹ = 2.2727273 × 10⁻³ mol = 2 × 10⁻³ mol (1 s.f.)

Question 8(b)(ii)

V(CO₂) = 32.5 mL – 0.5 mL = 32.0 mL = 0.0320 L

n(CO₂) = V / V_m = 0.0320 L / 24.8 L mol⁻¹ = 1.29032258 × 10⁻³ mol = 1.29 × 10⁻³ mol (3 s.f.)

Question 8(b)(iii)

Assumption is the gas is at SLC.

Question 8(b)(iv)

Resolution of data in Method A is 0.1 g whereas the resolution of data in Method B is 0.1 mL or 0.0001 L. 

0.1 g of CO₂ would equate to 0.0564 L at SLC. Hence method B has higher resolution. This means method B has a smaller uncertainty and is more precise. 

Question 8(c)(i)

Independent variable is the concentration of glucose. 

Question 8(c)(ii)

No. Since the gradient of both graphs is the same, the rate of reaction is the same for 10 g of glucose/100 mL of water and 20 g of glucose/100 mL of water. 

Question 8(d)(i)

The method assumes all the CO_2(g) is transferred into the burette and is measured. However, some of the CO_2(g)  may be left in the conical flask or the delivery tube. 

Question 8(d)(ii)

Discussion section.

Question 9(a)(i)

% yield:

n(C₆H₇NO) = m / M = 1.4 g / 109 g mol⁻¹ = 0.0128440367 mol

n(C₈H₉NO₂)_theoretical = n(C₆H₇NO) = 0.0128440367 mol

m(C₈H₉NO₂)_theoretical = n × M = 0.0128440367 mol × 151 g mol⁻¹ = 1.9394495 g

% yield = (actual yield / theoretical yield) × 100

% yield = (0.80 g / 1.9394495 g) × 100 = 41 % (2 s.f.)

Atom economy:

% atom economy = (M(product) / M(reactants)) × 100 

% atom economy = (151 g mol⁻¹ / (109 + 60) g mol⁻¹)) × 100 = 89.3% (3 s.f.)

Comparison:

P2 has a lower yield compared to P1 (41% vs 87%)

P2 has a higher atom economy than P1 (89.3% vs 72%)

Question 9(a)(ii)

P2 is better aligned with green chemistry as has greater atom economy than P1 (89.3% vs 72%), hence better fulfils the green chemistry principle “atom economy”. =

Question 9(b)

Reaction P2 exhibits a higher atom economy than Reaction P1, meaning a smaller proportion of the reactants in Reaction P2 will be converted to waste. Moreover, the waste product of Reaction P2 is non-toxic water, which does not require expensive treatment or disposal. In contrast, Reaction P1 produces ethanoic acid as a waste product, which is hazardous due to its corrosive nature and necessitates neutralisation before disposal.

However, the yield of Reaction P2 is lower than that of Reaction P1. As a result, any unreacted reactants that are not recovered and reused will be considered waste. This would lead to a greater amount of waste generated by Reaction P2 in order to produce the same mass of desired product as Reaction P1.

Hence, Reaction P2 can be considered to be more aligned with green principle “prevention of waste” and hence is safer for the environment and more cost efficient, provided unused reactants are recycled.

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