VCE Chemistry 2025

Every question from the 2025 VCE Chemistry examination. 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 2025 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 2025 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): 7 questions (13, 13, 12, 17, 13, 14 and 8 marks). Show working, give states in equations and use sensible significant figures.

Section A: Multiple choice

Q1

Renewability of biomethane and natural gas. Answer: B - biomethane is renewable (made quickly by anaerobic digestion of plant and animal material); natural gas is a fossil fuel, so it is not. (92%)

Q2

Energy per 100 g of a 65.0 g bar with 45.0 g carbohydrate, 10.0 g protein and 10.0 g fat. Answer: D - using the Data Book (16, 17 and 37 kJ g⁻¹): $45(16) + 10(17) + 10(37) = 1260$ kJ per 65 g, so $1260 \times \tfrac{100}{65} \approx 1940$ kJ $= 1.9 \times 10^6$ J. (34%; unit errors were the most common mistake.)

Q3

Which statement is true of all exothermic reactions? Answer: C - bond breaking absorbs less energy than bond making releases. (71%)

Q4

What would improve the circular economy of bioethanol production? Answer: C - turning the waste CO₂ into a useful product is the only option that deals with waste. (83%)

Q5

Calibration factor for 2.50 A at 12.0 V for 150 s, with a rise from 22.3 °C to 28.4 °C. Answer: D - $\text{CF} = \tfrac{VIt}{\Delta T} = \tfrac{12.0 \times 2.50 \times 150}{6.1} \approx 738$ J °C⁻¹. (93%)

Q6

Energy change per mole of CO₂ in photosynthesis, given $\Delta H_c(\text{glucose}) = -2840$ kJ mol⁻¹. Answer: A - photosynthesis is the reverse of combustion, $+2840$ kJ per 6 mol CO₂, so $\tfrac{2840}{6} = 473.3$ kJ is absorbed per mole. (56%)

Q7

What migrates through the polymer/gel electrolyte of a discharging metal-air cell? Answer: B - metal ions move from the anode towards the cathode. Electrons travel through the external wire, not the electrolyte. (24%; 60% chose A.)

Q8

Which metal-air cell has the highest potential difference? Answer: A - Mg with an acidic electrolyte: $1.23 - (-2.37) = 3.60$ V. The others give Al (alkaline) $0.40 + 1.66 = 2.06$ V, Na (alkaline) $0.40 + 2.71 = 3.11$ V and Zn (acidic) $1.23 + 0.76 = 1.99$ V. (29%; 53% chose C.)

Q9

In which cell are side reactions with moisture less likely? Answer: D - Zn²⁺/Zn ($-0.76$ V) lies above water's reduction ($-0.83$ V), so water does not spontaneously oxidise zinc. Al, Mg and Na all lie below $-0.83$ V and can react with water. (57%)

Q10

Which statements about bioethanol production are correct? All four options were accepted. Statements II (only hydrolysis and fermentation use enzymes) and III (glucose forms in hydrolysis) are correct. Statement I is also true, since fermentation involves both oxidation and reduction of carbon, but no option offered I, II and III together.

Q11

Which design feature significantly improves fuel cell efficiency? Answer: C - porous electrodes give a large surface area for the catalysed reactions. (80%)

Q12

Mass of O₂ consumed when a H₂/O₂ fuel cell delivers 36 000 C. Answer: B - $n(e^-) = \tfrac{36\,000}{96\,500} = 0.373$ mol; O₂ needs 4 e⁻, so $n(\text{O}_2) = 0.0933$ mol and $m = 0.0933 \times 32.0 \approx 2.98$ g. (62%)

Q13

Reduction half-equation during recharge of a Ni-Cd cell. Answer: D - recharge reverses discharge, so Cd(OH)₂ is reduced: Cd(OH)₂ + 2e⁻ → Cd + 2OH⁻. (61%)

Q14

Primary role of the catalyst. Answer: C - it lowers the activation energy, so a greater proportion of collisions are successful. (62%)

Q15

$K_c$ for $4\text{NH}_3 \rightleftharpoons 2\text{N}_2 + 6\text{H}_2$, given $K_c = 0.286$ M⁻² for N₂ + 3H₂ ⇌ 2NH₃. Answer: D - reverse (take the reciprocal) and double (square): $\left(\tfrac{1}{0.286}\right)^2 \approx 12.2$. (56%)

Q16

Concentration-time graph after the temperature of 2NO + O₂ ⇌ 2NO₂ ($\Delta H < 0$) is raised. Answer: C - heating favours the endothermic reverse reaction, so [NO₂] falls gradually while [NO] and [O₂] rise gradually, with no sudden jumps at $t = 10$. (67%)

Q17

Which claimed benefits of the electrolyte-free fuel cell fit "design for energy efficiency"? Answer: C - both: better conduction and fewer interfaces each mean less energy lost as heat. (49%)

Q18

In artificial photosynthesis. Answer: A - sunlight drives the splitting of water: water is oxidised to O₂ and hydrogen gas is produced. Reduction to H₂ happens at the cathode, not the anode, and no external power supply is needed. (41%)

Q19

Order the equipment by resolution, lowest to highest. Answer: C - graduations of 10 mL (beaker), 1 mL (50 mL cylinder), 0.5 mL (pipette) and 0.2 mL (10 mL cylinder). (34%)

Q20

Most likely cause of the scatter about the trend line. Answer: A - points fall both above and below the line, which signals random error, and judging the moment of colour change is the main random source. B and C would cause systematic error. (64%)

Q21

Key advantage of renewable feedstocks. Answer: B - they reduce reliance on finite resources such as fossil fuels. The other options are possible, not guaranteed. (84%)

Q22

Major product of propan-1-ol with butanoic acid (H₂SO₄ catalyst). Answer: D - the ester propyl butanoate, CH₃CH₂CH₂COOCH₂CH₂CH₃. (75%)

Q23

Protein hydrolysis. Answer: C - the amide (peptide) links break and water is a reactant. (63%)

Q24

Step 1 has a 45% yield and the overall yield is 16%. Yield of step 2? Answer: D - $0.45 \times y = 0.16$, so $y \approx 0.356$, about 36%. (64%)

Q25

Separating volatile polar compounds with boiling points similar to ethanol. Answer: C - fractional distillation separates liquids with close boiling points. (59%)

Q26

Molecular formulas of geranial and linalool from their skeletal structures. Answer: C - C₁₀H₁₆O and C₁₀H₁₈O. (60%)

Q27

A test that distinguishes geranial (an aldehyde) from linalool (a tertiary alcohol). Answer: A - acidified dichromate oxidises the aldehyde, turning orange to green; tertiary alcohols are not oxidised. Permanganate would go purple to colourless, not pink to purple; both compounds decolourise bromine (C=C); neither has a carboxyl group to fizz with NaHCO₃. (41%)

Q28

Melting point range of vanillin (82 to 83 °C) blended with significant impurities. Answer: A - impurities lower the melting point and broaden the range: 68 to 79 °C. (45%)

Q29

One mole of a single-fatty-acid triglyceride reacts with 6 mol I₂. Which fatty acid? Answer: B - 6 C=C in three chains means 2 per chain: linoleic acid. (50%)

Q30

¹³C NMR spectrum of pentan-3-one. Answer: A - the symmetrical ketone has three carbon environments, one of them the C=O near 210 ppm. (67%)

Section B

Question 1 (13 marks)

Kim heats 166.0 g of water from 24.1 °C to 90.1 °C by burning 9.0 g of propane, C₃H₈, on a camp stove.

a

Write a balanced thermochemical equation for the complete combustion of C₃H₈. (2 marks)

b

Using the Data Book, calculate the energy released by burning 9.0 g of C₃H₈ ($M = 44.0$ g mol⁻¹). (2 marks)

c. i

Calculate the energy absorbed by the water. (2 marks)

ii

Find the overall energy efficiency. (1 mark)

d

The experiment is repeated in a sealed container holding 0.125 m³ of air (21.0% v/v O₂) at SLC. Find the mass of the reactant in excess. (4 marks)

e

15.3 g of ethanol ($M = 46.0$ g mol⁻¹) gives the same temperature change. Explain why different masses of the fuels are needed. (2 marks)

Show worked solution

a. [2 marks]. Using the molar enthalpy of combustion from the Data Book (which assumes SLC, so water is a liquid):

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

b. [2 marks]. $n(\text{C}_3\text{H}_8) = \dfrac{9.0}{44.0} = 0.2045$ mol, so

$$E = 0.2045 \times 2220 = 454 \text{ kJ} \approx 4.5 \times 10^2 \text{ kJ}.$$

c. i. [2 marks]. $\Delta T = 90.1 - 24.1 = 66.0$ °C.

$$q = mc\Delta T = 166.0 \times 4.18 \times 66.0 = 45\,796 \text{ J} \approx 45.8 \text{ kJ}.$$

ii. [1 mark]. Efficiency $= \dfrac{45.8}{454} \times 100\% \approx 10\%$.

d. [4 marks]. Convert the volume first: 0.125 m³ $= 125$ L.

• $V(\text{O}_2) = 0.210 \times 125 = 26.25$ L, so $n(\text{O}_2) = \dfrac{26.25}{24.8} = 1.058$ mol.
• Oxygen needed $= 5 \times 0.2045 = 1.023$ mol, so O₂ is in excess (all the propane burns).
• Excess $n(\text{O}_2) = 1.058 - 1.023 = 0.036$ mol.
• $m(\text{O}_2) = 0.036 \times 32.0 \approx 1.1$ g of oxygen is left over.

e. [2 marks]. Ethanol releases less energy per gram than propane (about $\tfrac{1370}{46.0} = 29.8$ kJ g⁻¹ against $\tfrac{2220}{44.0} = 50.5$ kJ g⁻¹), so a larger mass is needed for the same energy. This is because ethanol is already partly oxidised: its hydroxy group means one carbon is bonded to oxygen, so less energy is released when it is fully oxidised to CO₂ and H₂O. (Check: $\tfrac{15.3}{46.0} \times 1370 \approx 456$ kJ, about the same as the propane's 454 kJ.)

From the report. In a (42% full marks) many left out $\Delta H$, gave states that did not match $-2220$ kJ, or unbalanced the equation. In b, units were missing or the answer was given as $-454$ kJ "released". Part d was demanding (12% full marks): many could not convert m³ to L (Data Book item 4) or stopped before finding the excess amount. In e (8% full marks), bond-enthalpy arguments were not accepted without full calculations.

Question 2 (13 marks)

Objects are electroplated with chromium from a Cr₂(SO₄)₃ electrolyte containing a little H₂SO₄, with a chromium anode and the object as the cathode: Cr³⁺(aq) + 3e⁻ ⇌ Cr(s), $E^0 = -0.74$ V.

a

Using oxidation numbers, explain why this is reduction. (2 marks)

b

Explain why effective electroplating needs the same ion in both electrode reactions. (2 marks)

c

Use the electrochemical series to justify why the amount of H₂SO₄ must be carefully managed. (3 marks)

d

7.80 g of Cr ($A_r = 52.0$) is plated in 2.50 hours. Calculate the current. (4 marks)

e

Rh³⁺(aq) + 3e⁻ ⇌ Rh(s), $E^0 = +0.76$ V ($A_r = 102.9$). Explain why less current is needed to plate the same mass of Rh as Cr. (2 marks)

Show worked solution

a. [2 marks]

The oxidation number of chromium falls from +3 in Cr³⁺ to 0 in Cr. A decrease in oxidation number is reduction.

b. [2 marks]

For consistent plating, the Cr³⁺ concentration in the electrolyte must stay constant. That happens when Cr³⁺ is produced at the anode, Cr(s) → Cr³⁺(aq) + 3e⁻, at the same rate as it is used up at the cathode, Cr³⁺(aq) + 3e⁻ → Cr(s).

c. [3 marks]

In the electrochemical series, 2H⁺ + 2e⁻ → H₂ (0.00 V) is above Cr³⁺ + 3e⁻ → Cr ($-0.74$ V), so H⁺ is the stronger oxidant. If there is too much acid:

• H⁺ is reduced at the cathode in preference to Cr³⁺, producing H₂ gas (a flammable hazard) and little or no chromium plating;
• H⁺ can also react spontaneously with chromium already deposited, dissolving the plating.

Keeping the H⁺ concentration low (well below 1 M) makes its reduction less favourable.

d. [4 marks].

$$n(\text{Cr}) = \frac{7.80}{52.0} = 0.150 \text{ mol}, \qquad n(e^-) = 3 \times 0.150 = 0.450 \text{ mol}$$

$$Q = n(e^-)F = 0.450 \times 96\,500 = 43\,425 \text{ C}$$

$$I = \frac{Q}{t} = \frac{43\,425}{2.50 \times 3600} \approx 4.83 \text{ A}.$$

e. [2 marks]. Both ions have a 3+ charge, so each atom needs 3 electrons. Rhodium atoms are about twice as heavy as chromium atoms, so the same mass contains fewer moles of Rh, needing fewer moles of electrons: for 7.80 g, $n(\text{Rh}) = 0.0758$ mol and $n(e^-) = 0.227$ mol, against 0.450 mol for Cr. Less charge over the same time means a smaller current (about 2.44 A instead of 4.83 A). The $E^0$ values do not decide this.

From the report. In a, "3+" is a charge, not an oxidation number. Part b (6% full marks) was weak: many did not see that commercial plating needs a constant electrolyte concentration. In c (10% full marks), many called Cr an oxidant, treated H₂SO₄ as a direct source of H₂, or claimed the added water would react. In e (23%), common errors were comparing half-cell voltages and missing that the ions have equal charge.

Question 3 (12 marks)

Methanol is made by CO(g) + 2H₂(g) ⇌ CH₃OH(g), $\Delta H < 0$, at 500 K and 10 000 kPa with a catalyst. At 500 K, $K_c = \dfrac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} = 1.21$ M⁻².

a

Define "reversible reaction". (1 mark)

b. i

Raising the temperature increases the rate. Explain its effect on the yield. (2 marks)

ii

Using Le Châtelier's principle, state and justify one way to increase the yield at constant temperature. (2 marks)

c

After a change, [CO] = 2.00 M, [H₂] = 1.25 M and [CH₃OH] = 5.60 M. Determine what must happen for the system to return to equilibrium. (3 marks)

d

5.00 mol CO and some H₂ are injected into an evacuated 1.00 L container at 500 K. At equilibrium [CH₃OH] = 1.30 M. Calculate the amount of H₂ initially added. (4 marks)

Show worked solution

a. [1 mark]

A reaction that can proceed in both the forward and reverse directions.

b. i. [2 marks]

The forward reaction is exothermic, so increasing the temperature shifts the equilibrium to the left (the endothermic direction), which decreases the yield of CH₃OH.

ii. [2 marks]

Any one, with its reason:

• Decrease the volume (increasing the pressure): the system shifts forward to partially oppose the change, because the forward reaction reduces the number of gas particles (3 to 1).
• Add CO or H₂: the system shifts forward to use up some of the added reactant.
• Remove CH₃OH as it forms: the system shifts forward to replace some of it.

c. [3 marks].

$$Q = \frac{5.60}{2.00 \times 1.25^2} = 1.79 \text{ M}^{-2}.$$

$Q > K_c$ (1.79 > 1.21), so the system must shift backwards (to the left): [CH₃OH] decreases and [CO] and [H₂] increase until $Q = K_c$.

d. [4 marks]. In 1.00 L, amounts in mol equal concentrations in M. Let $x$ mol of CO react:

	CO	H₂	CH₃OH
Initial	5.00	$a$	0
Change	$-x$	$-2x$	$+x$
Equilibrium	$5.00 - x$	$a - 2x$	$x$

$x = 1.30$, so [CO] $= 3.70$ M. Then

$$[\text{H}_2]^2 = \frac{[\text{CH}_3\text{OH}]}{K_c[\text{CO}]} = \frac{1.30}{1.21 \times 3.70} = 0.290, \qquad [\text{H}_2] = 0.539 \text{ M}.$$

$$a = 0.539 + 2 \times 1.30 = 3.14 \text{ M},$$

so 3.14 mol of H₂ was added.

From the report. In a, an equilibrium description ("forward rate equals reverse rate") earned no mark. In b.ii, "increase the pressure" alone was not enough; the method had to be stated (reducing the volume), since adding an inert gas raises pressure without a shift. In d (28% full marks), many did not combine an ICE table with the equilibrium expression, and a common slip was giving a concentration instead of the amount in mol.

Question 4 (17 marks)

A company makes organic compounds from simple hydrocarbons and separates its waste.

a. i

Write a balanced equation, with conditions, for making C₃H₇Cl from C₃H₈. (2 marks)

ii

Name two reactants that form propan-1-amine, C₃H₉N, in one substitution reaction. (2 marks)

b

Boiling points: HCOOH 101 °C, CH₃COOH 118 °C, CH₃CH₂COOH 141 °C; CH₃NH₂ $-6$ °C, CH₃CH₂NH₂ 17 °C, CH₃CH₂CH₂NH₂ 48 °C. i. Referring to structure, explain why the carboxylic acids boil higher than the corresponding amines. (2 marks)

ii

An IR spectrum of one separated sample shows weak peaks near 3300 to 3400 cm⁻¹ and C-H absorption near 2900 cm⁻¹, with no strong peak near 1700 cm⁻¹. Amine or carboxylic acid? (1 mark)

c

Compound X contains oxygen and chlorine (³⁵Cl and ³⁷Cl). Its mass spectrum has peaks at m/z 108 and 110, 93 and 95 (about 3 : 1) and a base peak at 59. i. Identify the m/z of the base peak. (1 mark)

ii

Why would every isomer of X have a peak at m/z = 108? (1 mark)

iii

Identify the species responsible for m/z = 95. (1 mark)

d

Its ¹H NMR spectrum shows three singlets, at 2.8, 2.5 and 1.3 ppm, with relative areas 16.7 : 33.3 : 100. Deduce three things about the structure. (3 marks)

e

Its ¹³C NMR spectrum shows three peaks. What does this indicate? (2 marks)

f. i

Draw a skeletal structure for X consistent with parts c to e. (1 mark)

ii

Name it. (1 mark)

Show worked solution

a. i. [2 marks]. Substitution with chlorine in UV light:

$$\text{C}_3\text{H}_8(g) + \text{Cl}_2(g) \xrightarrow{\text{UV light}} \text{C}_3\text{H}_7\text{Cl}(g) + \text{HCl}(g)$$

ii. [2 marks]

1-chloropropane and ammonia (any primary 3-carbon haloalkane with ammonia was accepted, as was propan-1-ol with ammonia).

b. i. [2 marks]

The carboxyl group (-COOH) has a C=O and an O-H, so carboxylic acid molecules form more, and stronger, hydrogen bonds than amines (they can even pair up as dimers), and they are also heavier than the amine with the same number of carbons. The stronger intermolecular forces need more energy to overcome, so the boiling points are higher.

ii. [1 mark]

Amine. There is no C=O absorption near 1700 cm⁻¹ and no very broad O-H (acid) band from 2500 to 3300 cm⁻¹; the peaks near 3300 to 3400 cm⁻¹ fit N-H.

c. i. [1 mark]

m/z = 59 (the tallest peak, relative intensity 100).

ii. [1 mark]

All isomers have the same molecular formula, C₄H₉ClO. The molecular ion containing ³⁵Cl always gives m/z = 108, whatever the arrangement of atoms (the molar mass itself is 108.5 because of the isotope mix).

iii. [1 mark]

[C₃H₆³⁷ClO]⁺: the ³⁷Cl molecular ion (m/z 110) after losing a CH₃ group (15). The peak at 93 is the matching ³⁵Cl fragment, in the 3 : 1 isotope ratio.

d. [3 marks]

Any three of:

• Three peaks, so three hydrogen environments.
• All peaks are singlets, so no hydrogen environment has hydrogens on the neighbouring carbon.
• The area ratio 16.7 : 33.3 : 100 simplifies to 1 : 2 : 6, fitting 9 hydrogens in total (C₄H₉ClO).
• The 6H signal indicates two equivalent CH₃ groups, so the molecule has symmetry.

e. [2 marks]. Four carbons but only three carbon environments, so two carbons must be equivalent: the molecule is symmetrical (two identical CH₃ groups).

f. i. [1 mark].

ii. [1 mark]. 1-chloro-2-methylpropan-2-ol, (CH₃)₂C(OH)CH₂Cl. Its isomer 2-chloro-2-methylpropan-1-ol, (CH₃)₂CClCH₂OH, fits the data equally well and was also accepted.

From the report. In a.ii (17% full marks) many gave formulas when names were asked for. In b.i, discussing the strength of C=O or O-H bonds, or saying amines have only dispersion forces, was wrong. In c.ii, "same molecular mass" was not accepted, since the molar mass is 108.5. Part c.iii (10%) needed the ³⁷Cl shown and a single positive charge. In d and e, arguments from chemical shift values earned no marks; the evidence had to be the number of peaks, splitting and areas. Only 15% could draw the skeletal structure.

Question 5 (13 marks)

Quinine, C₂₀H₂₄N₂O₂, is measured in tonic water by HPLC; the quinine peak has a retention time of 1.415 minutes. A calibration curve of peak area against concentration uses standards from 15 to 120 mg L⁻¹.

a. i

Explain how retention time can confirm quinine is present. (2 marks)

ii

Explain why the standard solutions must be run under controlled conditions. (2 marks)

iii

Why should the curve not be used for a sample with a mean concentration of 130 mg L⁻¹? (1 mark)

b. i

Quinine's circled groups are a hydroxy group, a tertiary amine nitrogen and an alkene. Using Data Book item 19, name the highest-priority group. (1 mark)

ii

Structure Q has both nitrogen atoms protonated (N-H⁺); structure R is neutral. Which is more likely at low pH? Justify. (2 marks)

c. i

Quinine has four chiral centres. Explain why only one isomer is an effective medicine. (3 marks)

ii

Explain why quinine is a more effective competitive inhibitor at pH 4.7 than at higher pH. (2 marks)

Show worked solution

a. i. [2 marks]

Run a pure quinine standard through the same instrument under identical conditions (same column, mobile phase, flow rate and temperature) and record its retention time. If the tonic water gives a peak at the same retention time (1.415 min), quinine is likely present. (Alternatively, "spike" the sample with pure quinine: if the 1.415 min peak grows, that peak is quinine.)

ii. [2 marks]

Retention time and peak area depend on the conditions. If they vary between runs, peaks shift and areas change for reasons other than concentration, so the calibration curve would be inaccurate and the sample's concentration read from it would be invalid.

iii. [1 mark]

130 mg L⁻¹ is above the highest standard (120 mg L⁻¹), outside the range of the calibration curve, so reading it would mean extrapolating.

b. i. [1 mark]

Hydroxy (hydroxyl).

ii. [2 marks]

Structure Q. At low pH there is a high concentration of H⁺ ions, and the basic nitrogen atoms (amine groups) accept protons, becoming positively charged N-H⁺ groups as in Q.

c. i. [3 marks]

A chiral centre is a carbon atom bonded to four different groups, so it can have two mirror-image arrangements. With four chiral centres, quinine has many stereoisomers, each with a different three-dimensional shape. Only one of these has the exact shape that is complementary to the target site (the enzyme's active site), so only that isomer binds and works as a medicine.

ii. [2 marks]

At pH 4.7, quinine is protonated (structure Q). The positively charged N-H⁺ groups can form ionic attractions with negatively charged groups in the enzyme's active site (and the protonated shape fits better), so quinine binds more strongly and blocks the substrate more effectively than the neutral form found at higher pH.

From the report. Part a.ii was answered well by only 7%. In c.i (14% full marks) answers needed all three ideas: four different groups, uniquely shaped isomers, and only one shape complementary to the active site. Part c.ii (8% full marks) needed the link between the protonated form and stronger binding.

Question 6 (14 marks)

Karolina hypothesises that boiling removes 75% of the oxalate in spinach. She extracts oxalate with 1 M HCl, makes the extract up to 500 mL, holds it at 50 °C, and titrates 25.0 mL aliquots with 0.0020 M acidified KMnO₄: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.

a

Which methodology did she most likely use to develop her hypothesis: case study, fieldwork, literature review or modelling? (1 mark)

b

Using Data Book item 7, what observation marks the end point? (1 mark)

c. i

Using collision theory, why titrate at 50 °C? (1 mark)

ii

Why might the experiment be invalid at room temperature? (1 mark)

d

Titres were 16.10, 15.45, 15.40 and 15.40 mL. Calculate the average titre. (1 mark)

e

The uncooked sample had mass 15.15 g. Calculate the mass of C₂O₄²⁻ ($M = 88.0$ g mol⁻¹) in it. (4 marks)

f

A boiled, rinsed and dried 15.15 g sample contained 0.0545 g of C₂O₄²⁻. Find the percentage change. (1 mark)

g

Explain why insufficient rinsing of the cooked sample would give a result higher than the true value. (2 marks)

h

Discuss whether the findings support the hypothesis. (2 marks)

Show worked solution

a. [1 mark]

Literature review (she researched background information).

b. [1 mark]

The first permanent pale purple colour in the flask: once all the oxalate has reacted, excess MnO₄⁻ (purple) is no longer decolourised. Mn²⁺ is very pale pink and forms throughout, so "pink" is not the end point.

c. i. [1 mark]

At higher temperature the particles have more kinetic energy, so more collisions exceed the activation energy and the reaction is faster (the colour change is immediate at the end point).

ii. [1 mark]

At room temperature the reaction is too slow, so the permanganate would not decolourise straight away and the end point (and titre) would be unreliable. (The report also accepted that it would no longer follow the stated method.)

d. [1 mark]

Use the concordant titres (within 0.10 mL) and leave out the rough first trial:

$$\frac{15.45 + 15.40 + 15.40}{3} = 15.42 \text{ mL}.$$

e. [4 marks].

$$n(\text{MnO}_4^-) = 0.0020 \times 0.01542 = 3.084 \times 10^{-5} \text{ mol}$$

$$n(\text{C}_2\text{O}_4^{2-}) \text{ in } 25.0 \text{ mL} = \tfrac{5}{2} \times 3.084 \times 10^{-5} = 7.71 \times 10^{-5} \text{ mol}$$

$$n \text{ in } 500 \text{ mL} = \tfrac{500}{25.0} \times 7.71 \times 10^{-5} = 1.542 \times 10^{-3} \text{ mol}$$

$$m = 1.542 \times 10^{-3} \times 88.0 = 0.136 \text{ g} \approx 0.14 \text{ g}.$$

f. [1 mark].

$$\frac{0.0545 - 0.136}{0.136} \times 100\% \approx -60\%,$$

a decrease of about 60%.

g. [2 marks]. Boiling water left on the leaves contains dissolved (soluble) oxalate that has already been removed from the spinach. If it is not rinsed away, it is carried into the extract, so more permanganate is needed (a larger titre), and the calculated oxalate content is higher than what actually remains in the leaves.

h. [2 marks]. The hypothesis predicted a 75% reduction. The experiment found about 60%, less than predicted, so the findings do not support the hypothesis (as stated).

From the report. Part b (22%) was weak: "pink" was not accepted because Mn²⁺ is pink and forms throughout. In e (13% full marks) the most common issues were ignoring the 5 : 2 ratio or the dilution from 500 mL. Part f was answered correctly by only 17%.

Question 7 (8 marks)

Traditional blast furnaces reduce Fe₂O₃ with coke at 2200 to 2300 °C, producing CO₂: 2Fe₂O₃ + 3C → 4Fe + 3CO₂ (1600 kg CO₂ and 850 MJ of electricity per tonne of steel). Molten oxide electrolysis (MOE) dissolves Fe₂O₃ in molten CaO at about 1600 °C and electrolyses it with an inert anode (Pt, Ir or ceramic-metal oxide): 2Fe₂O₃ → 4Fe + 3O₂ (about 0 kg CO₂ but 14 400 MJ of electricity per tonne).

a. Describe a requirement for MOE to have a positive impact on UN Sustainable Development Goal 12 (responsible consumption and production) while meeting society's need for steel. (2 marks)

b. Evaluate MOE and the blast furnace against two green chemistry principles, describe one ethical factor, and give a conclusion. (6 marks)

Show worked solution

a. [2 marks]. For example: MOE uses about 17 times as much electricity as the blast furnace (14 400 MJ against 850 MJ per tonne), so that electricity must come from renewable sources. Otherwise burning fossil fuels for power would simply move the CO₂ emissions elsewhere. (Other accepted ideas: using the ceramic-metal oxide anode instead of scarce Pt or Ir, or noting that avoiding coal mining protects finite resources and land.)

b. [6 marks]. A strong response covers, for example:

• Prevention of waste. The blast furnace produces 1600 kg of CO₂ per tonne of steel, a greenhouse gas with little use. MOE's by-product is O₂, which is harmless and useful, with essentially no direct CO₂. MOE is better on this principle.
• Design for energy efficiency. The blast furnace runs at 2200 to 2300 °C, MOE at about 1600 °C, but MOE needs far more electricity. MOE is only the more energy-responsible choice if the electricity (and heat) come from renewables; on a fossil-fuel grid it could be worse. (Atom economy is another valid principle: with Fe as the product, about 70% for MOE against about 63% for the blast furnace.)
• Ethical factor. For example, access to large amounts of renewable electricity varies between countries, so poorer steel-producing regions may be unable to switch, or it raises the question of who gets priority for renewable power. Job losses and retraining in steel communities, or increased mining of rare Pt and Ir, are also valid.
• Conclusion. For example: MOE is the greener process because it eliminates direct CO₂ and has better atom economy, provided it is powered by renewable electricity and uses an abundant anode material; until then, the blast furnace will still be needed to meet global demand.

From the report. Part a was answered fully by only 15%. In b, each mark was for a specific element: two relevant principles, each supported by evidence comparing both processes, one ethical factor, and a conclusion consistent with the evidence. Energy-efficiency evidence could support either process, depending on the electricity source.

General advice from the 2025 report

• Units and signs. "Energy released" is a positive quantity with a unit; watch kJ against J (Section A Q2, Section B Q1b).
• Show every step in multi-step calculations: limiting and excess reagent (Q1d), electrolysis (Q2d), ICE tables with $K_c$ (Q3d) and titration stoichiometry with dilution (Q6e).
• Use precise language: oxidation numbers are written +3, not 3+; say how a pressure change is made; name the observation that marks an end point.
• Use spectra correctly: the number of peaks, splitting and peak areas give structural information; chemical shift values alone were not accepted as evidence in Q4d and Q4e.