VCE Chemistry 2024

Every question from the 2024 VCE Chemistry examination, the first under the current study design. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle.

How to use this page

• Questions are from the 2024 VCE Chemistry examination, copyright Victorian Curriculum and Assessment Authority (VCAA). Each is summarised briefly here; open the official examination PDF for the full wording, diagrams, spectra and the Data Book.
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2024 Chemistry external assessment report (Word document), and every Section B calculation was redone and compared with the report. Both files are listed on the VCAA Chemistry examinations page.
• Percentages after multiple-choice answers are the share of students who chose the correct option. In Section B, "From the report" notes summarise what the examiners said students got wrong.

Structure and timing

The paper is 120 marks in 150 minutes (plus 15 minutes reading time), with a scientific calculator and the Data Book. That is 1.25 minutes per mark.

• Section A (30 marks): 30 multiple-choice questions. Aim for about 35 minutes.
• Section B (90 marks): 9 questions (8, 8, 9, 8, 11, 13, 12, 12 and 9 marks). Show working, give states in equations and use sensible significant figures.

Section A: Multiple choice

Q1

Photosynthesis is what kind of reaction? Answer: B - an endothermic redox reaction: it is the reverse of combustion, and carbon is reduced while oxygen is oxidised. It is not a condensation reaction. (54%)

Q2

Which statement is correct? Answer: B - glucose releases about 16 kJ g⁻¹ ($\tfrac{2840}{180}$) on combustion, far less than hydrogen's $\tfrac{286}{2} = 143$ kJ g⁻¹. (40%)

Q3

0.50 mol of ethane and 100.0 g of air (22.0% m/m O₂) are burned. Which reactant is in excess, and by how much? Answer: B - $n(\text{O}_2) = \tfrac{22.0}{32.0} = 0.6875$ mol, but 0.50 mol of ethane needs $3.5 \times 0.50 = 1.75$ mol. So O₂ is limiting; it reacts with $\tfrac{0.6875}{3.5} = 0.196$ mol of ethane, leaving 0.30 mol of ethane in excess. (45%)

Q4

Oxidising agent and a conjugate redox pair for the dichromate/iodide reaction. Answer: A - Cr₂O₇²⁻ is reduced, so it is the oxidising agent, and the pair written oxidant/reductant is I₂/I⁻. (49%)

Q5

A Pt | Sn⁴⁺, Sn²⁺ half-cell (A) is connected to Pb | Pb²⁺ (B). When operating? Answer: C - a galvanic cell converts stored chemical energy into electrical energy, so chemical energy decreases. Sn⁴⁺ (the strongest oxidant) is reduced in half-cell A and Pb is oxidised in B, so electrons flow to half-cell A. (54%)

Q6

An acidified iodate/iodine half-cell ($E^0 = +1.20$ V) with an O₂/alkaline half-cell. Which statement is correct? Answer: C - IO₃⁻ is reduced at the cathode, so the O₂ half-cell is the negative electrode (anode), where 4OH⁻ → O₂ + 2H₂O + 4e⁻, and [OH⁻] decreases there. (43%)

Q7

Mass of Cu deposited per 0.50 mol of electrons. Answer: A - Cu²⁺ + 2e⁻ → Cu, so 0.25 mol of Cu: $0.25 \times 63.5 \approx 15.9$ g $= 1.6 \times 10$ g. (63%)

Q8

Porous electrodes reduce efficiency (I); a higher ratio of hydrogen supplied to electricity produced means higher efficiency (II). Answer: D - neither: porous electrodes improve efficiency, and needing more hydrogen for the same electricity means lower efficiency. (51%)

Q9

Hydroxide ions reacting with 1 mol of ethanol at the anode of an alkaline ethanol fuel cell. Answer: D - balancing in alkaline conditions: C₂H₅OH + 12OH⁻ → 2CO₂ + 9H₂O + 12e⁻, so 12 mol. (29%)

Q10

Statements about bioethanol and hydrogen peroxide in fuel cells. All four options were accepted after VCAA's review of the question.

Q11

Making iron with hydrogen produced from fossil fuels fits which idea? Answer: A - a linear economy: finite raw materials in, product out, with no reuse loop. (73%)

Q12

Why is the combustion of butane considered irreversible? Answer: C - the reverse reaction is so improbable (and slow) that it can be ignored. Being exothermic does not make a reaction irreversible, and combustion is not an equilibrium, so option D does not apply. (32%; another 32% chose D.)

Q13

Energy profile of an uncatalysed reaction with a catalyst available. Which statement is correct? Answer: C - removing a catalyst increases the activation energy. Catalysts do not change the energies of reactants or products, and removing one lowers the rate. (82%)

Q14

Conditions for the greatest yield of NO₂ from N₂O₄ ⇌ 2NO₂ ($\Delta H > 0$). Answer: C - high temperature favours the endothermic forward reaction, and low pressure favours the side with more gas particles: 100 kPa and 100 °C. (58%)

Q15

SO₃ decomposes in a sealed container: 2SO₃ ⇌ 2SO₂ + O₂. What is true at equilibrium at all temperatures? Answer: A (D was also accepted) - SO₂ and O₂ form in a 2 : 1 mole ratio, and SO₂ is twice as heavy per mole, so its mass is 4.0 times that of O₂. VCAA also accepted D because of how its wording could be read. (43% chose D; 17% chose A.)

Q16

NaSCN is added to an Fe³⁺/SCN⁻/FeSCN²⁺ equilibrium at constant temperature. Which is true? Answer: C - just after the addition $Q < K$, so the forward rate exceeds the reverse rate at $t_3$ until equilibrium returns. $K$ is unchanged (same temperature) and [Fe³⁺] ends up lower than before. (49%)

Q17

Electrolytic refining of blister copper containing Ag, Fe and Ni. Answer: D - Fe and Ni are oxidised at the blister-copper anode instead of some Cu, while Cu²⁺ is still reduced at the cathode, so [Cu²⁺] in the electrolyte falls. The blister copper is the anode, not the cathode. (38%)

Q18

Which compound with $M = 88$ g mol⁻¹ has the highest boiling point? Answer: B - butanoic acid: carboxylic acids form strong hydrogen bonds and dimers. (48%)

Q19

Cyclohexane and benzene. Answer: A - both have non-cyclic structural isomers (for example hexenes for C₆H₁₂ and hexadiynes for C₆H₆). Their average C-C bond strengths differ, they are not in the same homologous series, and each carbon has four valence electrons in both. (34%)

Q20

Aldehyde isomers of C₅H₁₀O. Answer: C - there are four (pentanal, 2-methylbutanal, 3-methylbutanal and 2,2-dimethylpropanal), and two have a single methyl branch. (51%)

Q21

Skeletal structure of 3-chlorobutyl butanoate. Answer: D - a four-carbon acid part (including the C=O) joined through O to a four-carbon chain with Cl on its third carbon. (42%)

Q22

A triglyceride and methanol give glycerol and Compound J, C₁₉H₃₀O₂. Formula of the triglyceride? Answer: D - J is a methyl ester, so the fatty acid is C₁₈H₂₈O₂. Then $3(\text{C}_{18}\text{H}_{28}\text{O}_2) + \text{C}_3\text{H}_8\text{O}_3 - 3\text{H}_2\text{O} = \text{C}_{57}\text{H}_{86}\text{O}_6$. (49%)

Q23

Should water be added when making biodiesel from a triglyceride and methanol? Answer: B - no: water would hydrolyse the esters back towards fatty acids, interfering with the desired reaction. (59%)

Q24

Volume of 0.100 M oxalic acid to react with 10.0 mL of 0.100 M KMnO₄. Answer: D - MnO₄⁻ takes 5e⁻ and oxalic acid gives 2e⁻, so the ratio is 2 : 5. $n(\text{MnO}_4^-) = 0.00100$ mol, $n(\text{C}_2\text{H}_2\text{O}_4) = 0.00250$ mol, $V = 25.0$ mL. (26%)

Q25

Retention order (lowest first) for hexane, hexan-1-ol and hexan-2-one on a non-polar stationary phase with a polar mobile phase. Answer: D - the most polar compound elutes first: hexan-1-ol, then hexan-2-one, then hexane. (50%)

Q26

How to find the hexane concentration accurately. Answer: A - run standards of known concentration under the same conditions to make a calibration curve. (63%)

Q27

Which statement about analytical techniques is correct? All four options were accepted after VCAA's review of the question.

Q28

Functional groups in the capsicum compound. Answer: A - alkenyl (C=C), amide (-NH-CO-) and hydroxyl (-OH). There is no carboxyl, carbonyl (ketone or aldehyde) or amine group. (63%)

Q29

A reaction between sulfur atoms in proteins. Answer: C - it forms covalent (disulfide) bonds that help hold the tertiary structure. (57%)

Q30

Poster sections for "a 50 mL solution of 0.1 M HCl was placed in a flask" (I), "the average is within the claimed range" (II) and "the table shows the titres" (III). Answer: C - method, discussion, results. (81%)

Section B

Question 1 (8 marks)

a

How can melting point determination check the identity and purity of solid consumer products? (2 marks)

b

Linalool, C₁₀H₁₈O ($M = 154$ g mol⁻¹), has two C=C double bonds. i. How many chiral centres does it have? (1 mark)

ii

A 15.0 g mixture of linalool and hexane reacts with excess iodine; 0.042 mol of I₂ reacts. Find the percentage by mass of linalool. (3 marks)

iii

Identify two processes used to obtain purified linalool from crushed lavender. (2 marks)

Show worked solution

a. [2 marks]

A pure compound melts over a narrow range; impurities make the range broader and lower. The identity is checked by comparing the melting point with the literature value for the pure compound (or by mixing with a pure sample: if the melting point is unchanged, it is the same compound).

b. i. [1 mark]

One (the carbon bearing the OH group, which has four different groups).

ii. [3 marks]

Each C=C reacts with one I₂, and linalool has two C=C (hexane has none):

$$n(\text{linalool}) = \frac{0.042}{2} = 0.021 \text{ mol}, \qquad m = 0.021 \times 154 = 3.23 \text{ g}$$

$$\%(\text{linalool}) = \frac{3.23}{15.0} \times 100\% \approx 22\%.$$

iii. [2 marks]. An extraction step, such as steam distillation or solvent extraction, then a purification step, such as fractional distillation (or HPLC).

From the report. Part a (11% full marks) was weak: the range of the melting point shows purity, and closeness to the literature value suggests identity; a difference is not always just due to impurity. In b.iii, recrystallisation was not accepted because linalool is an oil.

Question 2 (8 marks)

A pathway: hydrocarbon E reacts with Cl₂ in UV light to give 1-chloropentane; with NaOH(aq) this gives L; L is oxidised by acidified dichromate to pentanoic acid. Propene, C₃H₆, reacts with steam (H₃PO₄ catalyst) to give M, a primary alcohol. Pentanoic acid and M react with concentrated H₂SO₄ to give N.

a. i

Molecular formula of E. (1 mark)

ii

Type of reaction forming 1-chloropentane. (1 mark)

b

Semi-structural formula of L. (1 mark)

c

IUPAC name of M. (1 mark)

d. i

Give a qualitative test, with the observation, for the functional group in pentanoic acid. (2 marks)

ii

Draw the skeletal structure of N. (2 marks)

Show worked solution

a. i. [1 mark]

C₅H₁₂ (pentane).

ii. [1 mark]

Substitution.

b. [1 mark]

CH₃CH₂CH₂CH₂CH₂OH (or CH₃(CH₂)₄OH), pentan-1-ol.

c. [1 mark]

Propan-1-ol.

d. i. [2 marks]

For example: add sodium carbonate (or NaHCO₃) and observe bubbles of CO₂. Alternatives: a pH meter reading below 7, or universal indicator turning the acid colour.

ii. [2 marks]

N is the ester propyl pentanoate, CH₃CH₂CH₂CH₂COOCH₂CH₂CH₃:

From the report. In b, "OHCH₂CH₂..." is an incorrect way to write a terminal hydroxy group. In d.i, a limewater test alone was not accepted; the acid must first react with a carbonate and the gas then be tested. Group 1 metals were not accepted, since they react with any labile hydrogen.

Question 3 (9 marks)

100 g of bread contains 9.5 g protein, 3.4 g fat and 47.3 g carbohydrate.

a

Using Data Book item 12, calculate the energy available in 100 g of bread. (1 mark)

b

Gluten exorphin C is a short peptide. i. Name the bond joining its monomers. (1 mark)

ii

Name the type of reaction that forms this bond. (1 mark)

iii

Name a monomer that formed the circled section (with an isobutyl side chain and a CH₂OH side chain). (1 mark)

c

Oleic and stearic acid (Data Book item 21). i. Identify a structural difference. (1 mark)

ii

Write the equation for the complete combustion of oleic acid (no states needed). (2 marks)

iii

Predict which acid has the higher molar heat of combustion, using bond enthalpies (no calculation required). (2 marks)

Show worked solution

a. [1 mark].

$$9.5 \times 17 + 3.4 \times 37 + 47.3 \times 16 = 1044 \text{ kJ} \approx 1.0 \times 10^3 \text{ kJ}.$$

b. i. [1 mark]

Peptide (amide) bond.

ii. [1 mark]

Condensation.

iii. [1 mark]

Leucine (isobutyl side chain) or serine (CH₂OH side chain).

c. i. [1 mark]

Oleic acid (C₁₈H₃₄O₂) has one C=C double bond, whereas stearic acid (C₁₈H₃₆O₂) has no C=C (it is saturated), so oleic acid has two fewer hydrogens.

ii. [2 marks]

$$2\text{C}_{18}\text{H}_{34}\text{O}_2 + 51\text{O}_2 \rightarrow 36\text{CO}_2 + 34\text{H}_2\text{O}$$

iii. [2 marks]. Stearic acid. Where oleic acid has one C=C (614 kJ mol⁻¹), stearic acid has a C-C (346) and two extra C-H bonds (414 each), and burning those extra hydrogens uses another half O₂ (498) but forms an extra water molecule with two O-H bonds (463 each). Roughly, stearic acid needs $346 + 2(414) + \tfrac12(498) - 614 \approx 809$ kJ more to break bonds but releases $2(463) = 926$ kJ more on forming bonds, so its combustion releases more energy per mole.

From the report. In b.iii, many could not identify either amino acid. In c.i the answer needed a direct comparison (one C=C against none). In c.iii (9% full marks) most compared C=C with C-C or C-H, but few included the extra O=O broken or O-H formed.

Question 4 (8 marks)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), $\Delta H = -92.3$ kJ, iron catalyst. At 400 °C, $K_p = 1.60 \times 10^{-8}$ kPa⁻². At equilibrium, $p(\text{H}_2) = 88.7$ kPa and $p(\text{N}_2) = 43.4$ kPa.

a

Write the expression for $K_p$. (1 mark)

b

Find the equilibrium partial pressure of NH₃. (1 mark)

c

At constant temperature, a change makes $Q = 1.20 \times 10^{-9}$ kPa⁻². Identify a possible change and explain how the system responds. (3 marks)

d

Explain how the iron catalyst resolves the conflict between rate and temperature, referring to one green chemistry principle (Data Book item 26.ii). (3 marks)

Show worked solution

a. [1 mark].

$$K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2) \times p(\text{H}_2)^3}$$

b. [1 mark].

$$p(\text{NH}_3)^2 = 1.60 \times 10^{-8} \times 43.4 \times 88.7^3 = 0.4846, \qquad p(\text{NH}_3) = 0.696 \text{ kPa}.$$

c. [3 marks]. $Q$ ($1.20 \times 10^{-9}$) is less than $K_p$ ($1.60 \times 10^{-8}$). Possible changes: removing NH₃, adding N₂ or H₂, or decreasing the volume (which raises all partial pressures). For example, if NH₃ is removed, the system shifts forward (net production of NH₃), partially replacing the lost NH₃ and increasing $Q$ until $Q = K_p$ again.

d. [3 marks]. The forward reaction is exothermic, so a low temperature gives a high equilibrium yield but a very slow rate, while a high temperature gives a fast rate but a low yield. The iron catalyst lowers the activation energy, so an acceptable rate is reached at a lower temperature. This fits the green chemistry principle of catalysis (catalysts increase rates while using less energy): less energy is needed to heat the reactor and the yield is better.

From the report. Part b (34%) was not well answered. In c, answers had to link a specific stress to the forward shift and its effect; temperature changes were wrong because the question said constant temperature. In d, most linked a principle but did not explain the rate-yield conflict.

Question 5 (11 marks)

a

Burning 0.580 g of butane heats 100.0 g of water by 35.6 °C. i. Using Data Book item 13, calculate the energy released. (2 marks)

ii

Calculate the energy absorbed by the water. (1 mark)

iii

Calculate the percentage energy efficiency. (2 marks)

b

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq), $\Delta H = -49.0$ kJ, in a solution calorimeter with calibration factor 470.0 J °C⁻¹, starting at 25.0 °C. i. Why must a calorimeter be calibrated before measuring enthalpy changes? (2 marks)

ii

Find the maximum temperature reached if 0.048 mol of KI reacts. (3 marks)

iii

Repeated trials were very consistent but the average maximum temperature was higher than expected. Give one reason. (1 mark)

Show worked solution

a. i. [2 marks]

$n(\text{C}_4\text{H}_{10}) = \dfrac{0.580}{58.0} = 0.0100$ mol, so $E = 0.0100 \times 2880 = 28.8$ kJ.

ii. [1 mark]

$q = 100.0 \times 4.18 \times 35.6 = 14\,881$ J $\approx 14.9$ kJ.

iii. [2 marks]

$\dfrac{14.9}{28.8} \times 100\% = 51.7\%$ (three significant figures).

b. i. [2 marks]

The energy released heats not only the solution but also the calorimeter itself (its walls, stirrer and thermometer), so $q = mc\Delta T$ for the water alone would underestimate it. The calibration factor measures the energy needed to raise the temperature of the whole calorimeter and its contents by 1 °C, so a temperature change can be converted into an accurate energy.

ii. [3 marks]

2 mol of KI release 49.0 kJ, so 0.048 mol release

$$\frac{0.048}{2} \times 49.0 = 1.176 \text{ kJ} = 1176 \text{ J}.$$

$$\Delta T = \frac{1176}{470.0} = 2.50 \text{ °C}, \qquad T_{\max} = 25.0 + 2.5 = 27.5 \text{ °C}.$$

iii. [1 mark]. Any one: the KI solution was more concentrated than 0.960 M; more KI was added than intended; the volume in the calorimeter was smaller than during calibration; or the calibration factor used was smaller than the true value.

From the report. Energy released should be positive: $-28.8$ kJ and a negative temperature change were common errors. Using the density of water from the old Data Book was inappropriate. In b.ii, errors included ignoring the 2 : 1 stoichiometry and mixing J with kJ. Part b.iii was answered correctly by only 13%.

Question 6 (13 marks)

Compound X, C₅H₁₁ClO, has an IR spectrum with a broad band centred near 3350 cm⁻¹, a mass spectrum with a base peak and a peak at m/z = 73, a ¹³C NMR spectrum with four peaks, and a ¹H NMR spectrum with four singlets at 4.16, 3.44, 3.39 and 0.96 ppm with relative integrations 1 : 2 : 2 : 6.

a

Using Data Book item 22, identify the bond with the highest wave number. (1 mark)

b. i

m/z of the base peak (56)? (1 mark)

ii

Formula of the species at m/z = 73. (1 mark)

c

What do four ¹³C peaks say about the arrangement of the five carbons? Justify. (3 marks)

d. i

What feature of a ¹H NMR peak does its integration curve represent? (1 mark)

ii

In general, what do the relative sizes of the integration curves show about structure? (1 mark)

iii

What do the relative sizes for X indicate? (1 mark)

iv

What do the shift and size of the 0.96 ppm peak indicate (Data Book item 24)? (2 marks)

e

Draw the structure of X. (2 marks)

Show worked solution

a. [1 mark]

O-H (alcohol), the broad absorption around 3200 to 3550 cm⁻¹.

b. i. [1 mark]

m/z = 56.

ii. [1 mark]

[C₄H₉O]⁺, that is [(CH₃)₂CCH₂OH]⁺: the molecular ion (122 with ³⁵Cl) minus a CH₂Cl fragment (49).

c. [3 marks]

Four peaks means four carbon environments for five carbons, so two carbons are in identical environments. The molecule must therefore have a branch (it is not a straight chain) and some symmetry, such as two identical CH₃ groups on the same carbon.

d. i. [1 mark]

The area of the peak (signal) for that hydrogen environment.

ii. [1 mark]

The relative number of hydrogen atoms in each environment.

iii. [1 mark]

The four hydrogen environments contain hydrogens in the ratio 1 : 2 : 2 : 6 (11 hydrogens in total).

iv. [2 marks]

A shift near 0.9 to 1.0 ppm indicates R-CH₃ hydrogens, and an integration of 6 means two equivalent CH₃ groups. Being a singlet, they sit on a carbon with no hydrogens.

e. [2 marks]

3-chloro-2,2-dimethylpropan-1-ol, HOCH₂C(CH₃)₂CH₂Cl: a central carbon carrying two CH₃ groups, a CH₂OH and a CH₂Cl. Every hydrogen environment is a singlet (no neighbouring H), with OH (1H), CH₂OH (2H), CH₂Cl (2H) and 2 × CH₃ (6H). Shown here in skeletal form (a full structural diagram shows every C and H):

From the report. In a, the answer had to name O-H (alcohol) as in the Data Book. Write m/z values without "m/z" as a unit. Parts d.i and d.iii were answered correctly by only 20%; many repeated the same answer for d.i and d.ii, or did not relate the ratio back to X's structure.

Question 7 (12 marks)

a

1 M NaNO₃(aq) is electrolysed at 3 V with an iron electrode on the positive terminal and a platinum electrode on the negative terminal. i. Which electrode is the anode? (1 mark)

ii

Write the half-equation at the Pt electrode. (1 mark)

iii

Why does no reaction occur when the power is off? (1 mark)

b

On recharging a lithium-ion cell: electrode X, LiCoO₂ → CoO₂ + Li⁺ + e⁻; electrode Y, 6C + Li⁺ + e⁻ → LiC₆. i. Polarity of X during recharging. (1 mark)

ii

Mass of LiC₆ produced by recharging at 1.25 A for 1.00 hour. (4 marks)

c

Artificial photosynthesis in acidic conditions makes H₂ from water. i. Overall energy transformation. (1 mark)

ii

Direction of electron flow (water side on the left, H₂ side on the right). (1 mark)

iii

How is charge balanced if electrons flow only one way? (2 marks)

Show worked solution

a. i. [1 mark]. The iron electrode (connected to the positive terminal, where oxidation occurs).

ii. [1 mark]. Water is the strongest oxidant present at the cathode (Na⁺ is far weaker):

$$2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$$

iii. [1 mark]

Any one: the H₂ produced bubbles away and is not in contact with the electrode to react; or no oxidant present is strong enough to react spontaneously with iron (water, at $-0.83$ V, lies below Fe²⁺/Fe at $-0.44$ V).

b. i. [1 mark]

Positive. Oxidation occurs at X, making it the anode, and in an electrolytic (recharging) cell the anode is positive.

ii. [4 marks]

$$Q = It = 1.25 \times 3600 = 4500 \text{ C}, \qquad n(e^-) = \frac{4500}{96\,500} = 0.04663 \text{ mol}$$

$n(\text{LiC}_6) = n(e^-) = 0.04663$ mol, and $M(\text{LiC}_6) = 6.9 + 6(12.0) = 78.9$ g mol⁻¹, so

$$m = 0.04663 \times 78.9 \approx 3.68 \text{ g}.$$

c. i. [1 mark]

Light (solar) energy to chemical energy.

ii. [1 mark]

From left to right: from the water-oxidation side (anode) to the H₂-producing side (cathode).

iii. [2 marks]

Water is oxidised at the anode, producing H⁺ ions and electrons. The H⁺ ions migrate through the electrolyte to the cathode, where they are reduced to H₂ using the electrons. The flow of protons to the cathode balances the flow of electrons.

From the report. Part a.ii (25%) was marked for states and a single-headed arrow; many wrote equations for iron or platinum reacting. Part a.iii (6%) needed a specific reason, not just "non-spontaneous". Part b.ii was answered very well, but units were required. In c.iii, answers using hydroxide ions were not accepted because the conditions were acidic.

Question 8 (12 marks)

Lucia ferments 5.0 g of glucose in 100 mL of water at 35 °C with 2.0 g of yeast. Method A: the flask's mass on a balance (0.1 g display) falls from 282.5 g to 282.4 g in 30 minutes. Method B: gas is collected over water in an inverted burette, from 0.5 mL to 32.5 mL in 30 minutes. A later graph compares 10.0 g and 20.0 g of glucose per 100 mL: both produce gas at the same rate for the first 15 minutes, after which the 10.0 g curve levels off near 240 mL while the 20.0 g curve continues to about 500 mL.

a. i

Write a balanced equation for producing bioethanol from glucose. (1 mark)

ii

Why is boiling water not used (not for safety reasons)? (1 mark)

b. i

Amount of gas in Method A ($M = 44.0$ g mol⁻¹). (1 mark)

ii

Amount of gas in Method B at SLC. (1 mark)

iii

State an assumption behind b.ii. (1 mark)

iv

Compare the resolution of the data in Methods A and B. (2 marks)

c. i

Identify the independent variable. (1 mark)

ii

Does the graph support the hypothesis that a higher glucose concentration increases the rate? (2 marks)

d. i

Identify one limitation of Method B (different from b.iii). (1 mark)

ii

Which poster section should include it? (1 mark)

Show worked solution

a. i. [1 mark].

$$\text{C}_6\text{H}_{12}\text{O}_6(aq) \rightarrow 2\text{C}_2\text{H}_5\text{OH}(aq) + 2\text{CO}_2(g)$$

ii. [1 mark]

High temperatures would denature the enzymes in the yeast that catalyse fermentation.

b. i. [1 mark]

Mass lost $= 0.1$ g of CO₂: $n = \dfrac{0.1}{44.0} \approx 2.3 \times 10^{-3}$ mol.

ii. [1 mark]

$V = 32.5 - 0.5 = 32.0$ mL $= 0.0320$ L, so $n = \dfrac{0.0320}{24.8} = 1.29 \times 10^{-3}$ mol.

iii. [1 mark]

Any one: all the CO₂ is collected (none dissolves in the water); the only gas in the burette is CO₂; the collected gas is at 25 °C and 100 kPa; the gas behaves ideally.

iv. [2 marks]

Method A's balance has a resolution of 0.1 g (its last displayed digit). Method B's volumes are recorded to the nearest 0.5 mL, which means the burette was read from 1 mL graduations, so its resolution is 1 mL.

c. i. [1 mark]

The concentration of the glucose solution.

ii. [2 marks]

No. For the first 15 minutes the two curves overlap, with the same gradient (the same volume of gas at each time), so the rate did not increase with concentration. The higher concentration only produced more gas in total, over a longer time.

d. i. [1 mark]

Any one, for example: some CO₂ dissolves in the water as it bubbles through; the gas pressure in the burette may not equal atmospheric pressure; gas may escape before the stopper is fitted; the gas is not at 25 °C (it starts at 35 °C); a 50 mL burette cannot hold the 500 mL of gas shown on the graph.

ii. [1 mark]

The discussion.

From the report. In a.ii, "the yeast would be killed" was not accepted: the enzyme is the key component. In b.ii many forgot to subtract the initial reading or convert mL to L. In b.iii, "at SLC" alone was insufficient because the reaction runs at 35 °C. Part b.iv was answered fully by only 1%: resolution needs units, and a burette read to 0.5 mL has 1 mL graduations. In c.ii, answers had to refer directly to the graph.

Question 9 (9 marks)

Paracetamol (C₈H₉NO₂, 151 g mol⁻¹) is made from 4-aminophenol (109 g mol⁻¹). Reaction P1 uses ethanoic anhydride (102 g mol⁻¹), giving paracetamol and ethanoic acid (60 g mol⁻¹): 87% yield, 72% atom economy. Reaction P2 uses ethanoic acid, giving paracetamol and water (18 g mol⁻¹); 1.4 g of 4-aminophenol gives 0.80 g of paracetamol.

a. i

Calculate the percentage yield and atom economy of P2 and compare them with P1. (4 marks)

ii

Which reaction better aligns with green chemistry? Refer to the relevant principle (Data Book item 26.ii). (2 marks)

b

Discuss P1 and P2 in terms of the principle of prevention of waste. (3 marks)

Show worked solution

a. i. [4 marks].

$$n(\text{4-aminophenol}) = \frac{1.4}{109} = 0.01284 \text{ mol}, \qquad m_{\text{theoretical}} = 0.01284 \times 151 = 1.94 \text{ g}$$

$$\%\text{ yield} = \frac{0.80}{1.94} \times 100\% \approx 41\%, \qquad \%\text{ atom economy} = \frac{151}{109 + 60} \times 100\% \approx 89\%.$$

P1 has the higher percentage yield (87% against 41%), but P2 has the higher atom economy (89% against 72%).

ii. [2 marks]. The relevant principle is atom economy (percentage yield is not itself a green chemistry principle). P2 is better aligned, because more of the reactant atoms end up in the desired product.

b. [3 marks]. P1 produces ethanoic acid as a by-product, which becomes waste unless it is recovered (it could be recycled, for example as the reactant for P2). P2 produces only water as a by-product, which is harmless. However, P2's low yield (41%) means much of the reactant is left unconverted, potentially creating more waste than P1 with its 87% yield, unless the unreacted materials are recovered and reused.

From the report. In a.i some calculated only one of yield or atom economy. In a.ii the mark depended on naming atom economy as the principle. In b, the first two marks were for identifying each reaction's waste (ethanoic acid for P1, water for P2).

General advice from the 2024 report

• Energy released is positive. Writing $-28.8$ kJ, or using $-49$ kJ to get a negative temperature change, loses marks.
• Resolution needs a unit and depends on the graduations of the equipment, not just on how a reading is written.
• Balance half-equations in the stated conditions, including alkaline media (Section A Q9).
• Put answers where the question asks (the boxes for Questions 2d.ii and 7c.ii), and give units when the stem does not.
• Use a sketch of the cell to work out anodes, cathodes and the species that react (Question 7a).