VCE Chemistry 2022

Every question from the 2022 VCE Chemistry examination. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle. This paper was set on the previous study design, so each question also says whether it still matches the course you are studying now.

How to use this page

• Questions are from the 2022 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 2022 Data Book (bound in at the back).
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2022 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.
• Data values are the ones in the 2022 Data Book (for example, octane 47.9 kJ g⁻¹ and propane 50.5 kJ g⁻¹). Some differ slightly in today's Data Book, so use the Data Book that matches the paper you are practising.
• 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 (7, 10, 10, 12, 10, 14, 8, 11 and 8 marks). Show working, give states in equations and use sensible significant figures.

Study design check for 2026

The 2022 exam was written for the 2017 to 2023 study design. The current Units 3 and 4 study design (from 2024) keeps fuels, calorimetry, galvanic and fuel cells, electrolysis (including green hydrogen), rates and equilibrium, organic reactions and spectroscopy, but it dropped most of the old "chemistry of food" area of study and a few other details. When you practise with this paper:

• Skip (no longer in the study design): Section A Q9 (omega fatty acids), Q21 (coenzymes) and Q25 (biodiesel against petrodiesel properties); Section B 4b (vitamin C as an essential vitamin and an antioxidant), 6d (glycaemic index) and 9a (flashpoint).
• Adapt: Section A Q5 (the current course says "repeatability", not "reliability") and Q17 (the sweetener options are old content, though glucose as a monomer is not); Section B 2a (gas calculations are now limited to SLC), 4a.ii (the idea of "essential" amino acids is gone, but hydrolysis and condensation are in), 6b (bomb calorimetry is no longer named; the calculation is the same) and 9b (Maxwell-Boltzmann curves are no longer named, though catalysts and activation energy are).
• Still fully examinable: everything else, including the green hydrogen electrolyser (Q2), equilibrium (Q3), spectroscopy (Q5), fuel cells (Q7) and the experimental design question (Q8).

Section A: Multiple choice

Q1

Which section of a scientific poster explains the reason for the investigation? Answer: C - the introduction. (93%)

Q2

A fuel is burned to heat water. Answer: A - the energy is absorbed by the water, and $\Delta H$ is negative because combustion is exothermic. (65%)

Q3

The equation for the incomplete combustion of ethanol. Answer: C - C₂H₅OH(l) + 2O₂(g) → 2CO(g) + 3H₂O(l): incomplete combustion gives CO (or C) and water, and this is the only balanced option that does. (78%)

Q4

The diagram showing the common design features of a fuel cell. Answer: A - fuel enters at the anode and air at the cathode, excess fuel leaves from the anode side, and the cell drives a load rather than drawing on a power source. D has the electrodes the wrong way round. (56%)

Q5

What does repeating trials with the same method and equipment improve? Answer: D - reliability (repeatability). Repeats do not remove bias or make an invalid method valid, and they do not fix a systematic error in accuracy. (67%) Study design: adapt. The current course uses "repeatability" and "reproducibility" rather than "reliability".

Q6

Galvanic cells and fuel cells have... Answer: B - the same energy transformations (chemical to electrical energy, with some heat) and both produce heat. Fuel cells are not reversible. (56%)

Q7

Hydrogen bonding in a protein is involved in forming... Answer: A - the secondary, tertiary and quaternary structures only. The primary structure is the sequence of amino acids held by covalent peptide bonds. (78%)

Q8

Unlike direct combustion, fuel cells... Answer: D - keep the oxidising and reducing agents apart, so electrons travel through an external circuit and chemical energy becomes electrical rather than thermal energy. (65%)

Q9

Arachidonic, linoleic, linolenic and stearic acids. Answer: B - linolenic acid, CH₃CH₂(CH=CHCH₂)₃(CH₂)₆COOH, is omega-3 (first C=C at the third carbon from the CH₃ end) with three C=C. Arachidonic (four C=C) and linoleic (two C=C) are omega-6; stearic acid is saturated. (81%) Study design: outside (omega fatty acids are no longer listed).

Q10

Producing 65.0 g of glycerol from tripalmitin. Answer: B - hydrolysis uses 3 H₂O per triglyceride: $n(\text{glycerol}) = \tfrac{65.0}{92.0} = 0.707$ mol, so $n(\text{H}_2\text{O}) = 2.12$ mol and $m = 2.12 \times 18.0 = 38.2$ g of water is required. (46%)

Q11

Which energy profile is the reverse of A + B ⇌ C ($\Delta H < 0$) without a catalyst? Answer: B - the reverse reaction is endothermic (products higher than reactants: Graphs 1 and 2), and the uncatalysed version has the higher peak: Graph 2. (63%)

Q12

Why are enzymes often ineffective in acidic conditions? Answer: A - acid protonates basic groups (such as -NH₂) on the side chains, changing the charges that hold the enzyme's shape and active site. Acids do not esterify enzymes or react with carboxyl groups, and a zwitterion needs both charges. (42%)

Q13

Which electrolyte gives a gas at the cathode with Pt electrodes? Answer: A - in KI(aq), water ($-0.83$ V) is a stronger oxidant than K⁺ ($-2.93$ V), so 2H₂O + 2e⁻ → H₂ + 2OH⁻. Molten NaCl and PbBr₂ give metals, and CuSO₄(aq) gives copper. (29%; 39% chose B.)

Q14

Vanadium redox battery during recharge. Answer: C - recharge reverses discharge, so VO²⁺ (V in +4) is oxidised to VO₂⁺ (V in +5): VO²⁺ is the reducing agent. H⁺ and H₂O do not change oxidation number, and VO₂⁺ is a product of recharge, not a reactant. (53%)

Q15

Cellular respiration with $\Delta H_c(\text{glucose}) = -2805$ kJ mol⁻¹. Answer: D - 6 mol of O₂ releases 2805 kJ, so 1 mol releases $\tfrac{2805}{6} = 467.5$ kJ. It is exothermic, and glucose loses electrons (is oxidised). (55%)

Q16

IUPAC name of CH₃CH₂CHClCHOHCH₃. Answer: B - 3-chloropentan-2-ol: number from the end nearer the -OH. (87%)

Q17

Which statement about sweeteners is correct? Answer: B - glucose acts as a monomer in the body, condensing to form glycogen. Glucose and fructose are structural isomers, not optical isomers; 2 mol glucose does not have the same energy as 1 mol sucrose; aspartame is about 200 times sweeter, not 200 times lower in energy. (56%) Study design: partly outside (sweeteners and aspartame are no longer listed).

Q18

Electrodes and solutions for a working Sn⁴⁺/Sn²⁺ and Ag⁺/Ag cell. Answer: C - the Sn⁴⁺/Sn²⁺ half-cell needs both ions and an inert (graphite) electrode, and the Ag⁺/Ag half-cell needs a silver electrode. A Sn electrode would itself be the reductant, and D has no Sn²⁺ to be oxidised. (44%)

Q19

Which compound has a chiral carbon? Answer: C - butan-2-ol: C2 carries H, OH, CH₃ and CH₂CH₃. (72%)

Q20

Zn metal is placed directly in a beaker of 1 M Co²⁺ and Mn²⁺ solution. Answer: B - Zn (the strongest reductant) reduces Co²⁺ (the strongest oxidant) by direct contact, so chemical energy becomes heat. There are no separate half-cells, so no voltage is produced; each Zn²⁺ formed replaces a Co²⁺, so the ion concentration does not rise. (34%; 34% chose A.)

Q21

Statements about enzymes and coenzymes. Answer: D - III (coenzymes can carry electrons) and V (they can change an enzyme's binding properties) are correct. Coenzymes can be changed during reaction, not all reactions need one, and they are organic molecules. (66%) Study design: outside (coenzymes are no longer listed).

Q22

Lithium ion battery: where does reduction occur during discharge and recharge? Answer: D - at the cathode in both, because the cathode is defined as the site of reduction. (65%)

Q23

Which statement about lithium ion batteries is correct? Answer: B - during discharge, cations (Li⁺) move towards the cathode, which is the positive electrode. LiCoO₂ is consumed during recharge, heat shortens battery life, and discharge is the galvanic process. (63%)

Q24

Retention times of Leu, Ala and Asn with a polar mobile phase and non-polar stationary phase. Answer: D - Asn (polar -CH₂CONH₂ side chain) stays with the polar mobile phase and elutes first; Leu (large non-polar side chain) is held longest by the stationary phase: Asn, Ala, Leu. (50%)

Q25

Statements about biodiesel and petrodiesel. Answer: D - II, III and IV: biodiesel's stronger intermolecular forces make it crystallise at a higher temperature; both come ultimately from plants and animals; petrodiesel contains more sulfur. Statement I is false: biodiesel, with polar ester groups, is the more hygroscopic. (33%; 31% chose B.) Study design: outside (this comparison is no longer listed).

Q26

Temperature-time graphs for the same reaction in two calorimeters calibrated with the same electrical energy. Answer: C - Calorimeter 1 rises 20 °C; Calorimeter 2's cooling line extrapolated back gives only about 12.5 °C. The same energy gives a smaller $\Delta T$, so $\text{CF} = \tfrac{E}{\Delta T}$ is higher for Calorimeter 2. (57%)

Q27

Highest atom economy for an organic product. Answer: C - addition of Br₂ to propene gives a single product, 100%. The iodine substitution gives about 57%, the dipeptide about 90%, and combustion gives no organic product. (46%)

Q28

A ¹³C spectrum with five peaks, one near 140 ppm and one near 115 ppm. Answer: D - pent-4-en-1-ol (CH₂=CHCH₂CH₂CH₂OH): five carbon environments, two of them C=C (110 to 150 ppm). A has no C=C, B has six environments, and C's alkyne carbons appear at 75 to 95 ppm. (42%; 32% chose B.)

Q29

Bond strengths in CH₃X for X = F, Cl, Br. Answer: A - C-Br is weakest and C-F strongest: bonds get longer, and weaker, as the halogen atom gets larger. (54%)

Q30

Four half-equations given out of order, with reaction data. Which cell has the highest potential difference? Answer: A - the data give the order (strongest oxidant first) HBrO/Br₂, BrO₃⁻/Br⁻, IO₃⁻/I₂, HCrO₄⁻/Cr³⁺, so pairing HBrO (strongest oxidant) with Cr³⁺ (strongest reductant) gives the largest voltage. (34%)

Section B

Question 1 (7 marks)

C₄H₈ reacts with HCl(g) to form two unbranched isomers, Compound A and Compound B; only Compound A can be converted into a primary alcohol.

a

Identify the type of reaction that converts C₄H₈ into Compound A. (1 mark)

b

Write the semi-structural formula of Compound B. (1 mark)

c

State the reagent(s) to convert Compound A into a primary alcohol. (1 mark)

d

Propan-1-ol reacts with methanoic acid. i. Identify the catalyst. (1 mark) ii. Write a balanced equation. (2 marks) iii. Name the organic product. (1 mark)

Study design: all parts are in the current course.

Show worked solution

a. [1 mark]

Addition. C₄H₈ here is but-1-ene; HCl adds across the C=C to give 1-chlorobutane (A) or 2-chlorobutane (B).

b. [1 mark]

CH₃CHClCH₂CH₃ (2-chlorobutane). Its -OH substitution product would be a secondary alcohol.

c. [1 mark]

NaOH(aq) or KOH(aq) (a hydroxide), which substitutes -OH for -Cl to give butan-1-ol.

d. i. [1 mark]

Concentrated sulfuric acid, H₂SO₄ (concentrated H₃PO₄ also accepted).

ii. [2 marks]

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{HCOOH} \rightleftharpoons \text{HCOOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}$$

iii. [1 mark]. Propyl methanoate.

From the report. In b (48% correct), a common error was giving 1-chlorobutane. In d.i (36% correct), the acid had to be concentrated, so "(aq)" was not accepted. In d.ii, some used the alcohol from the earlier parts instead of propan-1-ol. Names had to be spelt correctly.

Question 2 (10 marks)

A coal-fired power station produces CO₂.

a

5.17 × 10⁴ mol of captured CO₂ is stored at 28.0 °C in a sealed 20 000.0 L tank. Calculate the pressure in kPa. (2 marks)

b

An acidic electrolyser (cathode on the left giving H₂, anode on the right giving O₂, separated by a membrane) makes hydrogen. i. Show the direction of electron flow in the wire and justify it. (2 marks) ii. State two functions of the membrane. (2 marks)

c. i

Write the overall equation for the electrolyser at 80 °C. (1 mark) ii. How many moles of H₂ could 1625.0 A for 1.25 hours produce, at 100% efficiency? (3 marks)

Study design: b and c are in the current course (PEM electrolysis to make green hydrogen is named in it). Part a uses $pV = nRT$ away from SLC, which the current course no longer requires.

Show worked solution

a. [2 marks]. $T = 28.0 + 273 = 301$ K.

$$p = \frac{nRT}{V} = \frac{5.17 \times 10^4 \times 8.31 \times 301}{20\,000.0} = 6.47 \times 10^3 \text{ kPa}.$$

(With $V$ in L and $R = 8.31$ J mol⁻¹ K⁻¹, the pressure comes out in kPa.)

b. i. [2 marks]. Electrons flow from the anode to the cathode through the wire, so the arrow points along the wire towards the cathode (down into the H₂ electrode). Water is oxidised at the anode, releasing electrons, and H⁺ is reduced at the cathode, 2H⁺(aq) + 2e⁻ → H₂(g), which uses them. (In electrolysis the cathode is negative and the anode positive; saying otherwise contradicts the arrow.)

ii. [2 marks].

• It lets H⁺ ions (protons) pass from the anode side to the cathode side, completing the circuit.
• It keeps the products, H₂ and O₂, apart, so they cannot mix and react (explosively).

c. i. [1 mark].

$$2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)$$

ii. [3 marks].

$$Q = It = 1625.0 \times 1.25 \times 3600 = 7.31 \times 10^6 \text{ C}$$

$$n(e^-) = \frac{Q}{F} = \frac{7.3125 \times 10^6}{96\,500} = 75.8 \text{ mol}, \qquad n(\text{H}_2) = \frac{75.8}{2} = 37.9 \text{ mol}.$$

From the report. In a (47% full marks), some tried to find the mass of CO₂ instead. In b.i, answers that correctly said electrons flow from anode to cathode but then called the anode negative lost the second mark. Part b.ii was weak (6% full marks); the common error was saying the membrane separates the reactants.

Question 3 (10 marks)

4NH₃(g) + 3O₂(g) ⇌ 2N₂(g) + 6H₂O(g), $\Delta H < 0$, in a sealed container.

a. i

Write the equilibrium expression. (1 mark) ii. State its units. (1 mark)

b

The temperature is increased. i. How does this affect the equilibrium constant? Justify. (4 marks) ii. Compare the forward rate at the new and original equilibria, using collision theory. (3 marks)

c

For a different catalysed reaction reaching equilibrium at $t_1$, sketch the products curve without the catalyst. (1 mark)

Study design: all parts are in the current course.

Show worked solution

a. i. [1 mark].

$$K_c = \frac{[\text{N}_2]^2[\text{H}_2\text{O}]^6}{[\text{NH}_3]^4[\text{O}_2]^3}$$

ii. [1 mark]. M (mol L⁻¹): the units are $\text{M}^{8}/\text{M}^{7} = \text{M}$.

b. i. [4 marks].

• $K_c$ decreases.
• The forward reaction is exothermic, so the reverse reaction is endothermic.
• Raising the temperature favours the endothermic (reverse) reaction, so the equilibrium shifts left to partially oppose the change.
• At the new equilibrium, [NH₃] and [O₂] are higher and [N₂] and [H₂O] are lower, so the ratio that makes up $K_c$ is smaller.

ii. [3 marks].

• The forward rate is higher at the new equilibrium.
• At the higher temperature the particles move faster, so they collide more often (and the shift to the left has also increased the reactant concentrations).
• A greater proportion of collisions have energy at least equal to the activation energy, so there are more successful collisions per unit time.

c. [1 mark]. The products curve starts at zero, rises with a smaller gradient than the catalysed curve, and levels off at the same final concentration but later than $t_1$. (A catalyst changes the rate, not the position of equilibrium.)

From the report. In b.i (16% full marks), many did not say that both reactant and product concentrations change. In b.ii (26% full marks), collision theory needs "per unit time" (frequency) and the proportion of successful collisions; a direct comparison of the two rates was also needed.

Question 4 (12 marks)

Gelatin is made by boiling collagen (a protein) in water for many hours; both contain the same 18 amino acids, and gelatin is more easily metabolised.

a. i

Name the reaction that produces gelatin from collagen. (1 mark) ii. Explain how eating gelatin can help increase collagen in the body, naming the chemical processes. (3 marks)

b. i

What is the source of the vitamin C in the body? Justify. (2 marks) ii. How can vitamin C slow oxidative rancidity in foods? (1 mark)

c

The biofuel 2-methylpropan-1-ol has a heat of combustion of 36.1 kJ g⁻¹. i. Compare its energy content with octane's and explain the difference. (2 marks) ii. 2.36 g of it is burned under 500.0 g of water at SLC. Calculate the maximum temperature the water could reach. (3 marks)

Study design: a.i and c are in the current course. In a.ii, the hydrolysis and condensation reactions are in, but "essential" amino acids are not. Part b (vitamin C as an essential vitamin and antioxidant) is outside it.

Show worked solution

a. i. [1 mark]. Hydrolysis (of peptide bonds).

ii. [3 marks].

• Gelatin contains the amino acids, including essential ones, needed to build collagen; the essential ones cannot be made by the body and must come from the diet.
• In digestion, gelatin is hydrolysed (enzyme-catalysed) into its individual amino acids, which are absorbed.
• The body then joins these amino acids by condensation reactions, forming peptide bonds, to build new collagen.

b. i. [2 marks]

From the diet (food). Humans cannot synthesise vitamin C, so it is an essential nutrient that must be eaten.

ii. [1 mark]

Vitamin C is a reducing agent that is oxidised in preference to the fats and oils, so it reacts with the oxygen (or radicals) first and protects them.

c. i. [2 marks]

Octane releases more energy per gram (47.9 kJ g⁻¹ against 36.1 kJ g⁻¹). 2-methylpropan-1-ol contains oxygen, so it is already partially oxidised; less energy is released when it is fully oxidised to CO₂ and H₂O.

ii. [3 marks]

$$E = 2.36 \times 36.1 = 85.2 \text{ kJ}$$

$$\Delta T = \frac{E}{mc} = \frac{85.2 \times 10^3}{500.0 \times 4.18} = 40.8 \text{ °C}, \qquad T_{\max} = 25.0 + 40.8 = 65.8 \text{ °C}.$$

The water starts at 25.0 °C because it is at SLC.

From the report. In a.ii (14% full marks), a common error was saying gelatin simply condenses back into collagen; general words like "digestion" were not enough. In b.ii, "vitamin C is an antioxidant" did not explain how. In c.i (10% full marks), many wrongly claimed octane's longer chain gives more energy per gram; the Data Book shows heat of combustion per gram falls as chains get longer.

Question 5 (10 marks)

Molecule J contains chlorine. Its ¹³C NMR spectrum has three peaks (one at 168 ppm), its IR spectrum has a strong absorption near 1750 cm⁻¹, its mass spectrum shows peaks at m/z 108 and 110, and its ¹H NMR spectrum has two singlets (near 4.1 and 3.8 ppm). Its molecular mass is 108.5.

a

Identify one of the two possible carbon environments for the peak at 168 ppm. (1 mark)

b

Name the functional group, consistent with the IR spectrum, and justify. (2 marks)

c. i

Explain the peak at m/z 110. (1 mark) ii. Explain how the peaks relate to the molecular mass. (2 marks)

d

What do the two singlets tell you? (2 marks)

e

Draw a structural formula for J. (2 marks)

Study design: all parts are in the current course.

Show worked solution

a. [1 mark]

A carboxylic acid carbon, RCOOH (160 to 185 ppm), or an ester carbon, RCOOR' (165 to 175 ppm).

b. [2 marks]

Ester. The IR spectrum has a strong C=O absorption but no very broad acid O-H band between 2500 and 3500 cm⁻¹, so J cannot be a carboxylic acid.

c. i. [1 mark]

It is the molecular ion containing the heavier chlorine isotope, ³⁷Cl (the ³⁵Cl molecular ion is at 108).

ii. [2 marks]

The peaks at 108 and 110 are both molecular (parent) ions, one with ³⁵Cl and one with ³⁷Cl. Their heights are in about a 3 : 1 ratio, the natural abundance of the two isotopes, and the relative molecular mass is the weighted average: $0.75 \times 108 + 0.25 \times 110 = 108.5$.

d. [2 marks]

Two peaks mean two hydrogen environments. Both are singlets, so neither environment has hydrogens on the neighbouring carbon (no CH₃-CH₂ group, for example). The areas fit a 2 : 3 ratio.

e. [2 marks]

Methyl chloroethanoate, ClCH₂COOCH₃ (C₃H₅ClO₂, $M = 108.5$): the CH₂Cl (2H) and OCH₃ (3H) groups are separated by the C=O and the ester O, so both are singlets.

The isomer chloromethyl ethanoate, CH₃COOCH₂Cl, has the same formula and two singlets and was also accepted, but it fits the chemical shifts less well: a CH₃ next to an ester C=O is expected near 2.0 ppm (Data Book), not 3.8 ppm. Methyl chloroethanoate fits the data best.

From the report. Part c.ii was the hardest part of the paper (97% scored zero): most answers discussed the base peak or fragments instead of the two molecular ions and the weighted average. Only 15% explained the m/z 110 peak in c.i. In b, a formula instead of a name lost the mark; what a spectrum does not show (no broad acid O-H) is often the strongest evidence.

Question 6 (14 marks)

Air-popped popcorn contains, per 100 g, 10.7 g protein, 5.0 g fat and 78.7 g carbohydrate.

a

Calculate its energy content in kJ g⁻¹. (2 marks)

b

1.50 g burned in a bomb calorimeter (CF $= 6.54$ kJ °C⁻¹) raises the water from 22.2 °C to 25.7 °C. Calculate the energy released in kJ g⁻¹. (2 marks)

c

The calorimeter was accurately calibrated. State two factors that could explain the difference between a and b. (2 marks)

d

Rank amylose, amylopectin and cellulose by glycaemic index and explain each. (3 marks)

e

Picked corn loses sweetness as glucose forms starch. How does boiling and then freezing it keep it sweet? (2 marks)

f

Corn oil is 1% linolenic, 3% stearic, 13% palmitic, 31% oleic and 52% linoleic acid. Explain why it is liquid at SLC. (3 marks)

Study design: a, e and f are in the current course (food energy, enzyme denaturation and lowered activity, and melting point against unsaturation). Part b uses a bomb calorimeter, which is no longer named, although the calculation is the same, and c follows from it. Part d (glycaemic index) is outside it.

Show worked solution

a. [2 marks].

$$E = 10.7(17) + 5.0(37) + 78.7(16) = 181.9 + 185 + 1259.2 = 1.63 \times 10^3 \text{ kJ per } 100 \text{ g}$$

so the energy content is $\dfrac{1626}{100} = 16.3$ kJ g⁻¹.

b. [2 marks]. $\Delta T = 25.7 - 22.2 = 3.5$ °C.

$$E = \text{CF} \times \Delta T = 6.54 \times 3.5 = 22.9 \text{ kJ}, \qquad \frac{22.9}{1.50} = 15.3 \text{ kJ g}^{-1}.$$

c. [2 marks]. Any two that explain why b is lower than a (heat loss is ruled out by the accurate calibration), for example:

• the label gives average values, and popcorn is not uniform, so this 1.50 g sample may be less energy-dense;
• the popcorn may contain water, which does not burn;
• only one sample was tested, so the result may not be reliable;
• some mass may have been lost in transferring the sample from the balance;
• the calorimeter may have held more water than when it was calibrated.

d. [3 marks].

Rank	Carbohydrate	Explanation
Lowest GI	Cellulose	Humans lack the enzyme to hydrolyse it, so it releases essentially no glucose into the blood.
Middle	Amylose	A linear, tightly coiled chain: enzymes have few sites to attack, so it is hydrolysed to glucose slowly.
Highest GI	Amylopectin	Highly branched with many exposed ends, so enzymes hydrolyse it to glucose quickly.

e. [2 marks]. Boiling denatures the enzyme that catalyses the conversion of glucose to starch, so the reaction effectively stops. Freezing lowers the temperature, which slows the rate of any remaining reaction.

f. [3 marks].

• About 84% of the fatty acids (linolenic, oleic and linoleic) are unsaturated, with one or more C=C double bonds; only 16% are saturated.
• The cis double bonds put kinks in the chains, which prevent the triglyceride molecules packing closely together.
• The dispersion forces between the molecules are therefore weaker, so the melting point is below 25 °C and the oil is liquid at SLC.

From the report. Part c was very weak (1% full marks): "heat loss" and "poor insulation" were not accepted because the stem said the calorimeter was accurately calibrated. In e (16% full marks), both the denaturing of the enzyme and the slower rate at low temperature were needed.

Question 7 (8 marks)

LPG, mainly propane, is used in some cars.

a

Write a thermochemical equation for the complete combustion of propane gas. (2 marks)

b

A car uses 33.7 L of C₃H₈ (density 0.510 kg L⁻¹) to travel 270 km. Calculate the energy released. (2 marks)

c

In an acidic LPG fuel cell, O₂ enters at Electrode P (leaving as H₂O) and C₃H₈ + H₂O enters at Electrode Q (leaving as CO₂). State the polarity of Electrode P. (1 mark)

d

Write the half-equation at Electrode Q. (2 marks)

e

Identify one difference between this fuel cell and a secondary cell during discharge. (1 mark)

Study design: all parts are in the current course.

Show worked solution

a. [2 marks].

$$\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \qquad \Delta H = -2220 \text{ kJ mol}^{-1}$$

b. [2 marks]. $m = 0.510 \times 33.7 = 17.19$ kg $= 1.719 \times 10^4$ g. Using the Data Book value of 50.5 kJ g⁻¹:

$$E = 1.719 \times 10^4 \times 50.5 = 8.68 \times 10^5 \text{ kJ} \;(868 \text{ MJ}).$$

(Using moles, $\tfrac{17\,187}{44.0} \times 2220 = 8.67 \times 10^5$ kJ, which rounds the same way.)

c. [1 mark]. Positive (+). O₂ is reduced at P, so P is the cathode, which is positive in a fuel cell.

d. [2 marks]. Propane is oxidised at Q:

$$\text{C}_3\text{H}_8(g) + 6\text{H}_2\text{O}(l) \rightarrow 3\text{CO}_2(g) + 20\text{H}^+(aq) + 20e^-$$

e. [1 mark]. For example: the fuel cell has its reactants (C₃H₈ and O₂) supplied continuously, whereas a secondary cell contains a fixed amount of reactants that is used up during discharge. (Also accepted: products leave the fuel cell but stay in contact with the electrodes in a secondary cell.) Both cells must be mentioned.

From the report. In a (32% full marks), states had to be correct and $\Delta H$ had to have its sign and unit. In d (38% full marks), the second mark was for balancing. In e, answers that described only one of the two cells did not get the mark.

Question 8 (11 marks)

A student investigates how ethanol concentration affects the oxidation rate of ethanol in white wine. Four beakers hold 250.00 g in total: wine with 0, 5.00, 10.00 or 15.00 g of pure ethanol added. After two weeks, 25.00 mL aliquots are titrated with 0.0100 M NaOH (phenolphthalein) until three concordant titres are obtained. A technician found that the fresh wine was 8.12% m/m ethanol, with $9.45 \times 10^{-4}$ M acid.

a

How does the titration method affect the precision of the data? (1 mark)

b

Calculate the initial ethanol concentration in Beaker 2 in % m/m. (3 marks)

c

What is the independent variable? (1 mark)

d

The pipette was rinsed with distilled water instead of the beaker contents. How does this affect the calculated acid concentration? (2 marks)

e

Why does the acidity relate to the oxidation of ethanol? (1 mark)

f

Mean titres fall from about 24.6 mL (Beaker 1) and 21.3 mL (Beaker 2) to under 3 mL (Beakers 3 and 4). i. Write a conclusion about the oxidation rate. (1 mark) ii. Discuss your conclusion: make two points about its limitations or an explanation for the results. (2 marks)

Study design: all parts are in the current course (investigation skills; acid-base titration is assumed from Units 1 and 2).

Show worked solution

a. [1 mark]. Step 11 requires three concordant titres (within a narrow range), so the results used will be close together: the data will be precise.

b. [3 marks].

$$m(\text{ethanol in } 245.00 \text{ g wine}) = 245.00 \times 0.0812 = 19.89 \text{ g}$$

$$m(\text{total ethanol}) = 19.89 + 5.00 = 24.89 \text{ g}, \qquad \% \text{ m/m} = \frac{24.89}{250.00} \times 100 = 9.96\%.$$

c. [1 mark]

The concentration of ethanol in the beaker (the mass of ethanol added).

d. [2 marks]

The water left in the pipette dilutes the aliquot, so it contains less acid. Less NaOH is needed (a smaller titre), and the calculated acid concentration is lower than the true value.

e. [1 mark]

Oxidation of ethanol produces ethanoic acid (a carboxylic acid), so more acid means more ethanol has been oxidised.

f. i. [1 mark]

As the concentration of ethanol increases, the rate of oxidation of ethanol decreases (less acid forms in two weeks).

ii. [2 marks]

Any two points, for example:

• this goes against collision theory, which predicts a faster rate at higher concentration, so something else is limiting the reaction;
• high ethanol concentrations may inactivate (denature the enzymes of) the bacteria that oxidise the ethanol;
• the ethanol may only be oxidised as far as ethanal, which is not detected by an acid-base titration;
• the beakers started with different masses of wine, so different amounts of acid, and the large jump between Beakers 2 and 3 is out of proportion to the steady change in ethanol, which limits the conclusion.

From the report. Question 8 was hard because the data showed a rate falling as concentration rose: part a (20% correct), b (19% full marks, 67% scored zero), e (22%), f.i (19%) and f.ii (0% full marks) were all weak. Many fell back on rote-learned answers instead of reasoning from the method and data given.

Question 9 (8 marks)

a. Describe and explain the variation in flashpoint between organic molecules: define flashpoint and refer to structure and bonding. (4 marks)

b. Ethene and hydrogen react very slowly at SLC but quickly with a platinum catalyst. Use the Maxwell-Boltzmann distribution to explain why, referring to activation energy. (4 marks)

Study design: a (flashpoint) is outside the current course. In b, catalysts, alternative pathways and activation energy are in, but Maxwell-Boltzmann curves are no longer named.

Show worked solution

a. [4 marks].

• Flashpoint is the lowest temperature at which a liquid gives off enough vapour to ignite momentarily when a flame (ignition source) is applied.
• The stronger the intermolecular forces, the less volatile the substance, so the higher its flashpoint.
• Structure sets the strength of those forces: chain length, branching and functional groups (polarity).
• For example: flashpoint rises with chain length in the alkanes (more dispersion forces between larger molecules); a branched isomer such as 2-methylheptane has a lower flashpoint than octane (less surface contact); and propanoic acid (hydrogen bonding) has a higher flashpoint than its isomer methyl ethanoate (only dipole-dipole and dispersion forces).

b. [4 marks].

• The Maxwell-Boltzmann curve shows the number of particles with each kinetic energy; the total area under it is the total number of particles.
• Only particles with energy at least equal to the activation energy can collide successfully; that proportion is the area under the curve to the right of $E_a$.
• Without a catalyst, $E_a$ is high, so only a small proportion of H₂ and C₂H₄ particles have enough energy: the reaction is slow.
• Platinum provides an alternative pathway with a lower activation energy. The line moves left, so a much larger proportion of particles have enough energy, there are more successful collisions per unit time, and the rate increases.

From the report. In a (7% full marks), the definition needed "lowest temperature", "vapour" and "ignition source", and the top mark went to answers that compared two related molecules. In b (10% full marks), many wrote everything they knew about the curve instead of linking the lower activation energy to a larger area under the curve (a larger proportion of successful collisions).

General advice from the 2022 report

• Read the command term. "Name" needs a name, not a formula (Q5b); "compare" needs a direct comparison (Q3b.ii); answers must go in the box when the question says so.
• Use precise collision-theory language: successful collisions per unit time (frequency), and the proportion of collisions with energy at least equal to the activation energy.
• Be consistent in redox: the anode is where oxidation occurs, and its sign depends on the cell type; contradicting yourself (Q2b.i) loses marks.
• Use what a spectrum does not show. No broad acid O-H band rules out a carboxylic acid; no splitting rules out neighbouring hydrogens (Q5).
• Reason from unfamiliar data instead of reciting rote answers, especially in experimental design questions (Q6c and Q8).