2025 HKDSE Chemistry Answers & Marking Scheme

Pathway (1): Heating propan-1-ol with concentrated sulfuric acid at \(170^\circ\text{C}\) undergoes dehydration to form propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)). Acid-catalyzed hydration of propene with steam follows Markovnikov's rule, yielding propan-2-ol as the major product. This is correct.
Pathway (2): Oxidation of propan-1-ol with acidified dichromate yields propanal or propanoic acid. Reduction of these with \(\text{NaBH}_4\) yields propan-1-ol again, not propan-2-ol. This is incorrect.
Pathway (3): Dehydration of propan-1-ol gives propene. Hydrohalogenation of propene with \(\text{HBr}\) yields 2-bromopropane as the major product. Nucleophilic substitution of 2-bromopropane with aqueous \(\text{NaOH}\) yields propan-2-ol. This is correct.
Therefore, (1) and (3) only are correct.

Award 1 mark for option B. 0 marks for other options.

According to Hess's Law:
\(\Delta H_{\text{rxn}} = \sum \Delta H_c^\theta(\text{Reactants}) - \sum \Delta H_c^\theta(\text{Products})\)
Since liquid water (\(\text{H}_2\text{O(l)}\)) is already a product of combustion, its standard enthalpy of combustion is zero (\(\Delta H_c^\theta = 0\)).
Therefore,
\(\Delta H_{\text{rxn}} = \Delta H_c^\theta(\text{C}_2\text{H}_4\text{(g)}) - \Delta H_c^\theta(\text{C}_2\text{H}_5\text{OH(l)}) = -1411 - (-1367) = -44 \text{ kJ mol}^{-1}\).

Award 1 mark for option A. 0 marks for other options.

The reduction half-equation is:
\(\text{XeF}_2 + 2e^- \rightarrow \text{Xe} + 2\text{F}^-\)
The oxidation half-equation is:
\(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-\)
Multiply the reduction equation by 2 to balance the electrons:
\(2\text{XeF}_2 + 4e^- \rightarrow 2\text{Xe} + 4\text{F}^-\)
Combine both equations to get the overall balanced equation:
\(2\text{XeF}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Xe} + \text{O}_2 + 4\text{HF}\)
Thus, 2 moles of \(\text{XeF}_2\) react to produce 1 mole of \(\text{O}_2\). The ratio is \(2 : 1\).

Award 1 mark for option B. 0 marks for other options.

Statement (1) is incorrect because the rate constant \(k = A e^{-E_a/RT}\) depends on both \(E_a\) and the pre-exponential factor \(A\). A reaction with higher \(E_a\) can have a larger \(k\) if its \(A\) factor is significantly larger than that of the other reaction.
Statement (2) is incorrect because activation energy \(E_a\) is a constant characteristic barrier of a reaction pathway and is independent of temperature.
Statement (3) is correct as it is the linear form of the Arrhenius equation: \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\).

Award 1 mark for option B. 0 marks for other options.

The strong and broad absorption band at around \(3300\text{ cm}^{-1}\) indicates the presence of an \(\text{O}-\text{H}\) group (alcohol). The sharp band at around \(1640\text{ cm}^{-1}\) indicates a \(\text{C}=\text{C}\) group (alkene).
Among the options:
- \(\text{CH}_3\text{CH}_2\text{CHO}\) has a \(\text{C}=\text{O}\) group (absorption around \(1700\text{ cm}^{-1}\)), but no \(\text{O}-\text{H}\) or \(\text{C}=\text{C}\).
- \(\text{CH}_3\text{COCH}_3\) has a \(\text{C}=\text{O}\) group, but no \(\text{O}-\text{H}\) or \(\text{C}=\text{C}\).
- \(\text{CH}_2=\text{CHCH}_2\text{OH}\) (prop-2-en-1-ol) contains both a \(\text{C}=\text{C}\) group and an \(\text{O}-\text{H}\) group.
- \(\text{CH}_3\text{OCH}=\text{CH}_2\) contains a \(\text{C}=\text{C}\) group but no \(\text{O}-\text{H}\) group.
Thus, \(X\) is most likely \(\text{CH}_2=\text{CHCH}_2\text{OH}\).

Award 1 mark for option C. 0 marks for other options.

Set up the ICE table for moles:
- Initial: \(\text{PCl}_5 = 1.00\text{ mol}\); \(\text{PCl}_3 = 0\); \(\text{Cl}_2 = 0\)
- Change: \(\text{PCl}_5 = -0.40\text{ mol}\); \(\text{PCl}_3 = +0.40\text{ mol}\); \(\text{Cl}_2 = +0.40\text{ mol}\)
- Equilibrium: \(\text{PCl}_5 = 0.60\text{ mol}\); \(\text{PCl}_3 = 0.40\text{ mol}\); \(\text{Cl}_2 = 0.40\text{ mol}\)
Divide moles by volume (\(2.00\text{ dm}^3\)) to get concentrations:
\([\text{PCl}_5] = 0.60 / 2.00 = 0.30\text{ mol dm}^{-3}\)
\([\text{PCl}_3] = 0.40 / 2.00 = 0.20\text{ mol dm}^{-3}\)
\([\text{Cl}_2] = 0.40 / 2.00 = 0.20\text{ mol dm}^{-3}\)
Calculate \(K_c\):
\(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.20 \times 0.20}{0.30} = 0.133\text{ mol dm}^{-3} \approx 0.13\text{ mol dm}^{-3}\).

Award 1 mark for option A. 0 marks for other options.

Statement (1) is correct: Transition metals must have at least one stable ion with an incomplete d-subshell. Since zinc's only stable ion \(\text{Zn}^{2+}\) has a filled d-subshell (\([Ar] 3d^{10}\)), it is not classified as a transition metal.
Statement (2) is correct: Ligands split d-orbitals of a metal ion into different energy levels; absorption of visible light promotes d-electrons from lower to higher split d-orbitals (d-d transitions).
Statement (3) is correct: Fe(II) can be oxidized to Fe(III) by \(\text{S}_2\text{O}_8^{2-}\) and then reduced back to Fe(II) by \(\text{I}^-\), bypassing the high activation energy of the direct reaction between two negatively charged ions.

Award 1 mark for option D. 0 marks for other options.

Enantiomerism requires the presence of at least one chiral center (a carbon atom bonded to four different groups).
- Butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)) has no chiral carbon.
- Butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)) has a chiral carbon at C2, which is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\). Thus, it has a chiral center and exhibits enantiomerism.
- Butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) has no chiral carbon.
- 2-Methylpropan-2-ol (\((\text{CH}_3)_3\text{COH}\)) has no chiral carbon.

Award 1 mark for option B. 0 marks for other options.

- \(\text{CO}_2\) is linear (2 bonding pairs, 0 lone pairs on C), while \(\text{SO}_2\) is bent (2 bonding pairs, 1 lone pair on S).
- \(\text{BF}_3\) is trigonal planar (3 bonding pairs, 0 lone pairs on B), while \(\text{NH}_3\) is trigonal pyramidal (3 bonding pairs, 1 lone pair on N).
- Both \(\text{CH}_4\) and \(\text{NH}_4^+\) have 4 bonding pairs and 0 lone pairs around the central atom, giving them both a tetrahedral molecular shape.
- \(\text{H}_2\text{O\) is bent (2 bonding pairs, 2 lone pairs on O), while \(\text{BeCl}_2\) is linear (2 bonding pairs, 0 lone pairs on Be).

Award 1 mark for option C. 0 marks for other options.

All four compounds have similar relative molecular masses (around 72–74), meaning their dispersion forces (Van der Waals' forces) are of similar magnitude.
- Butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)) contains a highly polar \(-\text{OH}\) group that forms strong intermolecular hydrogen bonds.
- Butanal (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)) and ethoxyethane (\(\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3\)) are polar but cannot form hydrogen bonds with themselves, experiencing only dipole-dipole forces.
- Pentane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\)) is nonpolar and experiences only weak dispersion forces.
Therefore, butan-1-ol has the strongest intermolecular forces and the highest boiling point.

Award 1 mark for option B. 0 marks for other options.

According to Hess's Law, the enthalpy change of a reaction can be calculated from the standard enthalpy changes of combustion of reactants and products:
\(\Delta H^\theta = \sum \Delta H_c^\theta[\text{reactants}] - \sum \Delta H_c^\theta[\text{products}]\)

For the reaction: \(3C_2H_2(g) \rightarrow C_6H_6(l)\)
\(\Delta H^\theta = 3 \times \Delta H_c^\theta [C_2H_2(g)] - 1 \times \Delta H_c^\theta [C_6H_6(l)]\)
\(\Delta H^\theta = 3 \times (-1300\text{ kJ mol}^{-1}) - (-3268\text{ kJ mol}^{-1})\)
\(\Delta H^\theta = -3900 + 3268 = -632\text{ kJ mol}^{-1}\).

Therefore, option A is the correct answer.

1 mark for the correct answer (A). No marks for incorrect or multiple answers.

The organic compound \(X\) with formula \(C_4H_8O\) has 1 degree of unsaturation. Since it does not react with Tollens' reagent but can be reduced to an alcohol, it must be a ketone rather than an aldehyde. Therefore, \(X\) is butanone (\(CH_3COCH_2CH_3\)). Reduction of butanone by \(LiAlH_4\) gives butan-2-ol (\(CH_3CH(OH)CH_2CH_3\)), which is alcohol \(Y\).

(1) Correct: Dehydration of butan-2-ol can eliminate hydrogen from C1 or C3, producing a mixture of but-1-ene and but-2-ene.
(2) Correct: The C2 of butan-2-ol is bonded to four different groups (\(-H\), \(-OH\), \(-CH_3\), \(-CH_2CH_3\)), making it a chiral carbon atom.
(3) Incorrect: Butan-2-ol is a secondary alcohol. Its oxidation yields butanone (a ketone), not a carboxylic acid.

Hence, only (1) and (2) are correct.

1 mark for the correct answer (A). No marks for incorrect or multiple answers.

1. When the volume of the container is halved, the concentrations of all gaseous species (including \(NO_2\)) instantly double due to the volume decrease.
2. According to Le Chatelier's Principle, the increase in overall pressure shifts the equilibrium to the side with fewer gas molecules (the reactant side, to the left) to oppose the pressure increase. Thus, some \(NO_2\) reacts to form \(N_2O_4\), causing the concentration of \(NO_2\) to decrease gradually.
3. However, the shift in equilibrium only partially opposes the initial change. Therefore, the new equilibrium concentration of \(NO_2\) remains higher than its initial value before the volume change.

Thus, the concentration increases instantaneously, then decreases gradually to a value higher than its initial value.

1 mark for the correct answer (A). No marks for incorrect or multiple answers.

(1) Sulfur in \(S\) has an oxidation number of 0. In products, it forms \(K_2S\) (oxidation number of S is -2, reduced) and \(K_2SO_3\) (oxidation number of S is +4, oxidized). Thus, sulfur undergoes disproportionation. (Correct)
(2) Sulfur in reactants is in \(SO_2\) (+4) and \(H_2S\) (-2). Both react to form elemental \(S\) (0). This is comproportionation, where different species are oxidized and reduced to form the same product. (Incorrect)
(3) Sulfur in \(S_2O_3^{2-}\) has an average oxidation number of +2. In the products, it forms \(S\) (oxidation number 0, reduced) and \(SO_2\) (oxidation number +4, oxidized). Thus, sulfur undergoes disproportionation. (Correct)

Therefore, (1) and (3) only are correct.

1 mark for the correct answer (B). No marks for incorrect or multiple answers.

According to the Arrhenius equation:
\(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\)

The slope of the straight line obtained by plotting \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\).

\(\text{Slope} = -\frac{E_a}{R} = -1.20 \times 10^4\text{ K}\)
\(E_a = 1.20 \times 10^4 \times R = 1.20 \times 10^4 \times 8.314 = 99768\text{ J mol}^{-1} \approx 99.8\text{ kJ mol}^{-1}\).

Therefore, option B is correct.

1 mark for the correct answer (B). No marks for incorrect or multiple answers.

Let us determine the molecular shapes of each pair:
a) \(CO_2\) is linear (two double bonds, zero lone pairs on C). \(SO_2\) is bent (V-shaped) because sulfur has two bonding domains and one lone pair.
b) \(BF_3\) is trigonal planar (three bonding pairs, zero lone pairs on B). \(NH_3\) is trigonal pyramidal (three bonding pairs, one lone pair on N).
c) Both \(CH_4\) and \(NH_4^+\) have 4 bonding domains and 0 lone pairs around the central atom, so both are tetrahedral.
d) \(H_2O\) is bent (two bonding pairs, two lone pairs). \(BeCl_2\) is linear (two bonding pairs, zero lone pairs).

Therefore, option C is the correct answer.

1 mark for the correct answer (C). No marks for incorrect or multiple answers.

1. The molecular formula \(C_3H_6O_2\) corresponds to acyclic isomers with one double bond equivalent (either an acid or an ester).
2. The strong absorption peak around \(1740\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(C=O\)).
3. The absence of a broad absorption peak in the \(2500 - 3300\text{ cm}^{-1}\) region indicates that there is no carboxylic acid \(O-H\) group. This rules out propanoic acid (1).
4. Both methyl ethanoate (2) and ethyl methanoate (3) are esters. They have the formula \(C_3H_6O_2\), possess a carbonyl group (\(C=O\)), and do not contain an \(O-H\) group. Therefore, they match the spectrum.

Hence, (2) and (3) only are correct.

1 mark for the correct answer (C). No marks for incorrect or multiple answers.

Let's analyze the statements:
a) The concentration of \(Cu^{2+}(aq)\) in the electrolyte remains virtually constant because the amount of copper oxidized at the anode is almost equal to the amount of copper reduced and deposited at the cathode. (Incorrect)
b) Noble metal impurities (like gold and platinum) are less reactive than copper, so they are not oxidized. They detach from the dissolving anode and fall to the bottom of the cell as 'anode slime'. (Incorrect)
c) Zinc and iron impurities are oxidized to \(Zn^{2+}\) and \(Fe^{2+}\) ions at the anode, but they remain in the solution because they are much harder to reduce than \(Cu^{2+}\) ions. (Incorrect)
d) Because insoluble noble metal impurities (e.g. Au, Pt) fall off the anode as sludge, they contribute to the mass lost by the anode but do not deposit on the cathode. Thus, the mass lost by the anode is greater than the mass gained by the cathode. (Correct)

1 mark for the correct answer (D). No marks for incorrect or multiple answers.

(1) Correct: Both HCl and \(CH_3COOH\) are monoprotic acids. Since they have the same concentration (\(0.10\text{ mol dm}^{-3}\)) and volume (\(25.0\text{ cm}^3\)), they contain the same amount of potential hydrogen ions to react with NaOH. During neutralization, the equilibrium of \(CH_3COOH\) dissociation shifts completely to the right. Thus, the volume of NaOH required for complete neutralization is identical.
(2) Incorrect: HCl is a strong acid that is completely ionized in water, while \(CH_3COOH\) is a weak acid that only partially ionizes. The concentration of mobile ions is much higher in HCl, leading to a much higher electrical conductivity.
(3) Incorrect: The initial rate of reaction with magnesium ribbon depends on the concentration of \(H^+(aq)\) ions. Since HCl has a much higher \(H^+(aq)\) concentration, its initial reaction rate is significantly faster.

Therefore, only (1) is correct.

1 mark for the correct answer (A). No marks for incorrect or multiple answers.

For an acyclic compound of formula \(C_4H_7Br\), there is one carbon-carbon double bond (degree of unsaturation = 1). We can find the isomers by considering the carbon skeletons:

1. Skeleton: But-1-ene (\(CH_2=CH-CH_2-CH_3\))
- 1-bromobut-1-ene (\(CHBr=CH-CH_2-CH_3\))
- 2-bromobut-1-ene (\(CH_2=CBr-CH_2-CH_3\))
- 3-bromobut-1-ene (\(CH_2=CH-CHBr-CH_3\))
- 4-bromobut-1-ene (\(CH_2=CH-CH_2-CH_2Br\))
(4 isomers)

2. Skeleton: But-2-ene (\(CH_3-CH=CH-CH_3\))
- 1-bromobut-2-ene (\(CH_2Br-CH=CH-CH_3\))
- 2-bromobut-2-ene (\(CH_3-CBr=CH-CH_3\))
(2 isomers)

3. Skeleton: 2-Methylpropene (\((CH_3)_2C=CH_2\))
- 1-bromo-2-methylpropene (\((CH_3)_2C=CHBr\))
- 3-bromo-2-methylpropene (\(CH_2Br-C(CH_3)=CH_2\))
(2 isomers)

Total structural isomers = 4 + 2 + 2 = 8.

Therefore, option D is correct.

1 mark for the correct answer (D). No marks for incorrect or multiple answers.

According to Hess's Law:
\(\Delta H_r^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\)
\(\Delta H_r^\theta = \Delta H_c^\theta[C_2H_4(g)] + \Delta H_c^\theta[H_2(g)] - \Delta H_c^\theta[C_2H_6(g)]\)
\(\Delta H_r^\theta = (-1411) + (-286) - (-1560) = -137\text{ kJ mol}^{-1}\)

Correct option is A. Award 1 mark for the correct calculation using Hess's Law based on combustion data.

Let's analyze butyl 2-methylpropanoate (Option B). Hydrolysis gives butan-1-ol (Y, 4 carbons) and 2-methylpropanoic acid (Z, 4 carbons). Oxidation of butan-1-ol (Y) gives butanoic acid (Y'), which has the formula \(CH_3CH_2CH_2COOH\). The formula of Z is \(CH_3CH(CH_3)COOH\). Butanoic acid and 2-methylpropanoic acid are structural isomers of each other.

Award 1 mark for identifying the correct ester that hydrolyzes to form an alcohol and a carboxylic acid of equal carbon count which can be oxidized to mutually isomeric acids.

(1) is correct: decreasing volume increases concentration instantly (color intensifies), then equilibrium shifts right to reduce molecules (color fades), but final concentration is still higher than initial. (2) is incorrect: adding inert gas at constant volume does not change partial pressures of reactants/products, so equilibrium does not shift. (3) is incorrect: since the reaction is exothermic, increasing temperature shifts equilibrium left, decreasing \(K_c\).

Award 1 mark for analyzing Le Chatelier's principle and equilibrium constant temperature dependence correctly to find that only statement (1) is correct.

The half-equations are:
Oxidation: \(2I^- \rightarrow I_2 + 2e^-\)
Reduction: \(NO_2^- + 2H^+ + e^- \rightarrow NO + H_2O\)
Multiply the reduction half-equation by 2 to balance electrons:
\(2NO_2^- + 4H^+ + 2e^- \rightarrow 2NO + 2H_2O\)
Combine both:
\(2I^- + 2NO_2^- + 4H^+ \rightarrow I_2 + 2NO + 2H_2O\)
Thus, \(x=2, y=2, z=4, a=1, b=2, c=2\). The value of \(z\) is 4.

Award 1 mark for obtaining the correct balanced equation and identifying that \(z = 4\).

(1) Both \(NH_4^+\) and \(CH_4\) have 4 bond pairs and 0 lone pairs around the central atom, giving them a tetrahedral shape.
(2) Both \(H_3O^+\) and \(NH_3\) have 3 bond pairs and 1 lone pair around the central atom, giving them a trigonal pyramidal shape.
(3) \(BF_3\) has 3 bond pairs and 0 lone pairs (trigonal planar), while \(NF_3\) has 3 bond pairs and 1 lone pair (trigonal pyramidal). They have different shapes.

Award 1 mark for analyzing electron pair repulsion theory correctly to determine that only pairs (1) and (2) share the same shape.

(1) It contains the amide linkage \(-\text{NH}-\text{CO}-\), so it is a polyamide. (2) The linkage splits to form amine groups \(-\text{NH}_2\) and carboxylic acid groups \(-\text{COOH}\). Thus, the monomers are \(\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2\) (ethane-1,2-diamine) and \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) (butanedioic acid). (3) Polyamides are susceptible to alkaline hydrolysis upon heating with aqueous sodium hydroxide.

Award 1 mark if all three statements are determined to be correct.

Since \(1.2\text{ mol}\) of \(R\) is formed, the change in mole of \(P\) and \(Q\) is each \(-0.6\text{ mol}\) because of the 1:1:2 stoichiometry.
At equilibrium:
\(n_P = 1.0 - 0.6 = 0.4\text{ mol}\)
\(n_Q = 1.0 - 0.6 = 0.4\text{ mol}\)
\(n_R = 1.2\text{ mol}\)
Since the total number of moles of gas is the same on both sides, the volume term cancels out:
\(K_c = \frac{[R]^2}{[P][Q]} = \frac{(1.2 / 5.0)^2}{(0.4 / 5.0)(0.4 / 5.0)} = \frac{1.2^2}{0.4 \times 0.4} = 9.00\)

Award 1 mark for calculating the correct value of 9.00.

Using the Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
Given \(\frac{k_2}{k_1} = 2\), \(T_1 = 300\text{ K}\), and \(T_2 = 310\text{ K}\):
\(\ln(2) = \frac{E_a}{8.31} \left(\frac{1}{300} - \frac{1}{310}\right)\)
\(0.693 = \frac{E_a}{8.31} \times 1.075 \times 10^{-4}\)
\(E_a = \frac{0.693 \times 8.31}{1.075 \times 10^{-4}} \approx 53560\text{ J mol}^{-1} \approx 53.6\text{ kJ mol}^{-1}\)

Award 1 mark for correct mathematical manipulation of the Arrhenius equation to find the activation energy.

(1) Correct: Propanoic acid has a broad characteristic \(O-H\) absorption band at \(2500\text{–}3300\text{ cm}^{-1}\), which is absent in methyl ethanoate.
(2) Correct: Propanoic acid is a carboxylic acid and reacts with \(NaHCO_3(aq)\) to release \(CO_2\) gas (bubbles), whereas methyl ethanoate does not react.
(3) Incorrect: Neither compound can be oxidized by acidified potassium dichromate(VI) solution, so no observable change will occur in either.

Award 1 mark for identifying that only methods (1) and (2) can successfully distinguish the two compounds.

(1) is correct: since the same current flows for the same duration, the same amount of charge passes through both. The cathode reaction is \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\) in both setups, depositing the same mass of copper.
(2) is correct: in Setup X, water is oxidized at the anode (\(2H_2O \rightarrow O_2 + 4H^+ + 4e^-\)), producing \(H^+\) ions and lowering the pH. In Setup Y, the copper anode dissolves (\(Cu \rightarrow Cu^{2+} + 2e^-\)), so no \(H^+\) is produced and pH remains constant.
(3) is incorrect: oxygen gas is evolved at the anode in Setup X, while the active copper anode in Setup Y dissolves without producing any gas.

Award 1 mark for analyzing both electrolysis setups correctly and determining that statements (1) and (2) are correct.

Using Hess's Law, \(\Delta H^\theta = \sum \Delta H_f^\theta(\text{products}) - \sum \Delta H_f^\theta(\text{reactants})\). For the products: \(2 \times \Delta H_f^\theta[\text{H}_2\text{O}(l)] = 2 \times (-286) = -572\text{ kJ mol}^{-1}\) (elemental \(\text{N}_2\) has \(\Delta H_f^\theta = 0\)). For the reactants: \(2 \times \Delta H_f^\theta[\text{N}_2\text{O}(g)] + \Delta H_f^\theta[\text{N}_2\text{H}_4(l)] = 2 \times (82) + 51 = +215\text{ kJ mol}^{-1}\). Thus, \(\Delta H^\theta = -572 - (+215) = -787\text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer A. No fractional marks are given.

In iron(II) oxalate, both the \(\text{Fe}^{2+}\) cation and the \(\text{C}_2\text{O}_4^{2-}\) anion are oxidized by acidified \(\text{MnO}_4^-\). The oxidation half-reaction for \(\text{FeC}_2\text{O}_4\) is: \(\text{FeC}_2\text{O}_4 \rightarrow \text{Fe}^{3+} + 2\text{CO}_2 + 3e^-\). The reduction half-reaction of permanganate is: \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). To balance the electrons transferred, 3 moles of \(\text{MnO}_4^-\) are required to react with 5 moles of \(\text{FeC}_2\text{O}_4\). Therefore, 1.0 mole of \(\text{FeC}_2\text{O}_4\) requires \(3/5 = 0.60\) moles of \(\text{KMnO}_4\).

Award 1 mark for the correct answer C. No fractional marks are given.

Heating propan-1-ol with concentrated sulfuric acid at \(180^\circ\text{C}\) causes elimination (dehydration) to yield propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)). Subsequently, heating propene with dilute sulfuric acid (hydration) yields propan-2-ol as the major product because the addition of water follows Markovnikov's rule.

Award 1 mark for the correct answer A. No fractional marks are given.

(1) is correct: Using Arrhenius equation \(\ln(k_2/k_1) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2}) \Rightarrow \ln(2) = \frac{E_a}{8.314} (\frac{1}{300} - \frac{1}{310}) \Rightarrow 0.693 = \frac{E_a}{8.314} (1.075 \times 10^{-4}) \Rightarrow E_a \approx 53.6\text{ kJ mol}^{-1} \approx 54\text{ kJ mol}^{-1}\). (2) is correct: A catalyst provides an alternative pathway with lower activation energy, which increases the value of rate constant \(k\). (3) is incorrect: According to \(\ln k = -\frac{E_a}{RT} + \ln A\), the plot has a negative slope of \(-\frac{E_a}{R}\).

Award 1 mark for the correct answer A. No fractional marks are given.

The very broad absorption band at \(2500 - 3300\text{ cm}^{-1}\) indicates the presence of a carboxylic acid \(\text{O-H}\) group. The strong peak at \(1715\text{ cm}^{-1}\) is due to the \(\text{C=O}\) stretch. Together, they suggest a carboxylic acid group. Among the options, only \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\) (butanoic acid) is a carboxylic acid with the formula \(\text{C}_4\text{H}_8\text{O}_2\). The mass spectrum peak at \(m/z = 45\) corresponds to the \([\text{COOH}]^+\) fragment, which is highly characteristic of carboxylic acids.

Award 1 mark for the correct answer A. No fractional marks are given.

Initial concentrations: \([\text{A}]_0 = 2.0 / 2.0 = 1.0\text{ mol dm}^{-3}\), \([\text{B}]_0 = 3.0 / 2.0 = 1.5\text{ mol dm}^{-3}\). At equilibrium, \([\text{C}]_{\text{eq}} = 1.0\text{ mol dm}^{-3}\). From stoichiometry, change in concentrations: \(\Delta [\text{C}] = +1.0\text{ mol dm}^{-3}\), \(\Delta [\text{B}] = -1.0\text{ mol dm}^{-3}\), \(\Delta [\text{A}] = -0.5\text{ mol dm}^{-3}\). Equilibrium concentrations: \([\text{A}]_{\text{eq}} = 1.0 - 0.5 = 0.5\text{ mol dm}^{-3}\), \([\text{B}]_{\text{eq}} = 1.5 - 1.0 = 0.5\text{ mol dm}^{-3}\). \(K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2} = \frac{(1.0)^2}{(0.5) \times (0.5)^2} = 8.0\text{ dm}^3\text{ mol}^{-1}\).

Award 1 mark for the correct answer C. No fractional marks are given.

(a) \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH(OH)CH}_3\text{(l)}\)\n\n(b) By Hess's Law, the enthalpy change of formation is:\n\(\Delta H_f^\ominus = 3 \times \Delta H_c^\ominus[\text{C(graphite)}] + 4 \times \Delta H_c^\ominus[\text{H}_2\text{(g)}] - \Delta H_c^\ominus[\text{propan-2-ol(l)}]\)\n\(\Delta H_f^\ominus = 3(-393.5) + 4(-285.8) - (-2006.0)\)\n\(\Delta H_f^\ominus = -1180.5 - 1143.2 + 2006.0 = -317.7 \text{ kJ mol}^{-1}\)\n\n(c) Carbon (graphite), hydrogen gas, and oxygen gas do not react directly with each other under standard conditions to form propan-2-ol. Even if they reacted, they would form a complex mixture of other organic compounds (such as other alkanes, alkenes, or carbon monoxide) rather than yielding exclusively propan-2-ol, making direct measurement of the enthalpy change impossible.

(a) 1 mark for the correct balanced equation with state symbols.\n(b) 1 mark for drawing/stating a valid Hess's Law cycle or relationship; 1 mark for correct calculation steps; 1 mark for the final answer with correct units (\(-317.7 \text{ kJ mol}^{-1}\)). (No mark for wrong sign).\n(c) 1 mark for explaining that the elements do not react directly under standard conditions; 1.14 marks for explaining that side reactions occur / other products would form and direct measurement is impossible.

The synthesis can be accomplished in three main steps:\n\nStep 1: Hydration of propene to form propan-2-ol.\nReagent and conditions: \(\text{H}_2\text{O(g)}\), in the presence of phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)) catalyst at high temperature and pressure (e.g., \(300^\circ\text{C}\), \(60\text{ atm}\)).\nEquation/Intermediate: \(\text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH(OH)CH}_3\) (propan-2-ol).\n\nStep 2: Oxidation of ethanol to form ethanoic acid.\nReagent and conditions: Acidified potassium dichromate solution (\(\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+\text{(aq)}\)), heated under reflux.\nEquation/Intermediate: \(\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}\) (ethanoic acid).\n\nStep 3: Esterification to form propan-2-yl ethanoate.\nReagent and conditions: React propan-2-ol (from Step 1) with ethanoic acid (from Step 2) in the presence of a few drops of concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) catalyst, heated under reflux.\nEquation/Product: \(\text{CH}_3\text{COOH} + \text{CH}_3\text{CH(OH)CH}_3 \rightleftharpoons \text{CH}_3\text{COOCH(CH}_3)_2 + \text{H}_2\text{O}\).

Step 1: 1 mark for reagent and conditions (\(\text{H}_2\text{O(g)}\), \(\text{H}_3\text{PO}_4\), heat); 1 mark for identifying the intermediate propan-2-ol.\nStep 2: 1 mark for reagent and conditions (\(\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+\text{(aq)}\), reflux); 1 mark for identifying ethanoic acid.\nStep 3: 1 mark for reaction of the intermediate alcohol and acid in the presence of conc. \(\text{H}_2\text{SO}_4\) and heat under reflux; 1.14 marks for drawing the correct structure of propan-2-yl ethanoate.

(a) \nLet's construct an ICE table:\nInitial moles: \(n(\text{CO}) = 1.00\text{ mol}\), \(n(\text{H}_2) = 2.00\text{ mol}\), \(n(\text{CH}_3\text{OH}) = 0\text{ mol}\).\nAt equilibrium, \(n(\text{CH}_3\text{OH}) = 0.40\text{ mol}\).\nTherefore, change in moles: \(\Delta n(\text{CH}_3\text{OH}) = +0.40\text{ mol}\), \(\Delta n(\text{CO}) = -0.40\text{ mol}\), \(\Delta n(\text{H}_2) = -2(0.40) = -0.80\text{ mol}\).\n\nEquilibrium moles:\n\(n(\text{CO}) = 1.00 - 0.40 = 0.60\text{ mol}\)\n\(n(\text{H}_2) = 2.00 - 0.80 = 1.20\text{ mol}\)\n\(n(\text{CH}_3\text{OH}) = 0.40\text{ mol}\)\n\nEquilibrium concentrations (since volume \(V = 5.0\text{ dm}^3\)):\n\([\text{CO}]_{eq} = \frac{0.60\text{ mol}}{5.0\text{ dm}^3} = 0.12\text{ mol dm}^{-3}\)\n\([\text{H}_2]_{eq} = \frac{1.20\text{ mol}}{5.0\text{ dm}^3} = 0.24\text{ mol dm}^{-3}\)\n\([\text{CH}_3\text{OH}]_{eq} = \frac{0.40\text{ mol}}{5.0\text{ dm}^3} = 0.080\text{ mol dm}^{-3}\)\n\nExpression for \(K_c\):\n\(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\)\n\(K_c = \frac{0.080}{(0.12) \times (0.24)^2} = \frac{0.080}{0.12 \times 0.0576} = \frac{0.080}{0.006912} \approx 11.57\text{ dm}^6\text{ mol}^{-2}\) (accept 11.6)\n\n(b) Decreasing the volume of the container increases the total pressure of the system. According to Le Chatelier's Principle, the system will shift in the direction that opposes this change by reducing the total number of moles of gaseous particles. Since there are 3 moles of gas on the reactant side (\(1\text{ CO} + 2\text{H}_2\)) and only 1 mole of gas on the product side (\(1\text{ CH}_3\text{OH}\)), the equilibrium position will shift to the right (forward direction).

(a) 1 mark for finding correct equilibrium moles of CO and H2; 1 mark for calculating correct equilibrium concentrations of all species; 1 mark for correct substitution into the Kc expression; 1 mark for the correct numerical value (11.57 or 11.6) with correct units (\(\text{dm}^6\text{ mol}^{-2}\)).\n(b) 1 mark for predicting that the equilibrium shifts to the right; 1.14 marks for explaining that decreased volume increases pressure and the system shifts towards the side with fewer gas moles.

(a) The overall order of reaction is the sum of the individual orders with respect to each reactant.\nOverall order = \(1\) (for A) + \(2\) (for B) = \(3\).\n\n(b) From the rate equation:\n\(\text{Rate} = k[\text{A}][\text{B}]^2\)\n\(4.0 \times 10^{-4} \text{ mol dm}^{-3} \text{s}^{-1} = k (0.10 \text{ mol dm}^{-3}) (0.20 \text{ mol dm}^{-3})^2\)\n\(4.0 \times 10^{-4} = k (0.10) (0.040)\)\n\(4.0 \times 10^{-4} = 0.0040 k\)\n\(k = \frac{4.0 \times 10^{-4}}{4.0 \times 10^{-3}} = 0.10\)\n\nUnits of \(k\):\n\(k = \frac{\text{Rate}}{[\text{A}][\text{B}]^2} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\).\n\n(c) The rate constant \(k\) increases with increasing temperature. \nWhen the temperature increases, the average kinetic energy of the reactant molecules increases. Consequently, a much larger fraction of molecules possess kinetic energy equal to or greater than the activation energy (\(E_a\)) of the reaction. This results in a higher frequency of successful collisions, and according to the Arrhenius equation (\(k = A e^{-E_a / RT}\)), \(k\) increases exponentially with temperature.

(a) 1 mark for stating overall order is 3.\n(b) 1 mark for correct calculation of numerical value of \(k\) (0.10); 1.14 marks for the correct unit (\(\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\)).\n(c) 1 mark for stating \(k\) increases; 1 mark for explaining that higher temperature increases average kinetic energy of molecules; 1 mark for stating that a larger fraction of molecules have energy \(\ge E_a\) resulting in more effective collisions.

(a)\n(i) The sharp peak at \(1715\text{ cm}^{-1}\) corresponds to the stretching vibration of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad peak at \(3230 - 3670\text{ cm}^{-1}\) indicates that there is no hydroxyl group (\(-\text{OH}\)).\n(ii) Since \(X\) has the formula \(\text{C}_3\text{H}_6\text{O}\), contains a carbonyl group, and does not react with Tollens' reagent (which means it is not an aldehyde), it must be a ketone. The only ketone with three carbons is propanone.\nStructural formula: \(\text{CH}_3\text{COCH}_3\)\nSystematic name: propanone.\n\n(b)\n(i) The broad peak at \(3230 - 3670\text{ cm}^{-1}\) indicates the presence of a hydroxyl group (\(-\text{OH}\)) of an alcohol. The sharp peak at \(1640\text{ cm}^{-1}\) corresponds to the carbon-carbon double bond (\(\text{C}=\text{C}\)) stretching vibration.\n(ii) Chemical test: Add bromine dissolved in organic solvent (or bromine water) to both compounds in the dark.\nObservation for \(Y\): The orange/brown color of bromine is rapidly decolourised (since \(Y\) contains a \(\text{C}=\text{C}\) double bond which undergoes addition reaction).\nObservation for \(X\): No immediate observable change (or the orange/brown color remains).

(a)(i) 1 mark for identifying the carbonyl group (\(\text{C}=\text{O}\)).\n(a)(ii) 1 mark for the correct structural formula of propanone (\(\text{CH}_3\text{COCH}_3\)); 1 mark for the correct systematic name (propanone).\n(b)(i) 1 mark for carbon-carbon double bond (\(\text{C}=\text{C}\)); 1 mark for hydroxyl group (\(-\text{OH}\)).\n(b)(ii) 1 mark for suggesting a valid reagent (e.g. bromine in tetrachloromethane / bromine water / acidified \(\text{K}_2\text{Cr}_2\text{O}_7\)); 0.14 marks for stating the correct corresponding observations for both \(X\) and \(Y\).

(a) Observation: The copper anode dissolves / decreases in size / becomes thinner.\nHalf-equation: \(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\)\n\n(b) \nFirst, calculate the total quantity of electricity (charge) passed, \(Q\):\n\(Q = I \times t\)\n\(Q = 1.50 \text{ A} \times (40.0 \times 60) \text{ s} = 3600 \text{ C}\)\n\nNext, calculate the number of moles of electrons passed:\n\(n(\text{e}^-) = \frac{Q}{F} = \frac{3600 \text{ C}}{96500 \text{ C mol}^{-1}} = 0.03731 \text{ mol}\)\n\nFrom the anode half-equation, \(1\text{ mol}\) of Cu dissolves for every \(2\text{ mol}\) of electrons lost.\n\(n(\text{Cu dissolved}) = \frac{1}{2} \times n(\text{e}^-) = \frac{0.03731}{2} = 0.01866 \text{ mol}\)\n\nFinally, calculate the change in mass of the anode:\n\(\Delta m = n(\text{Cu}) \times \text{Molar mass of Cu}\)\n\(\Delta m = 0.01866 \text{ mol} \times 63.5 \text{ g mol}^{-1} = 1.18 \text{ g}\)\nTherefore, the mass of the anode decreases by \(1.18 \text{ g}\) (or the change in mass is \(-1.18 \text{ g}\)).\n\n(c) The blue color of the solution is due to the presence of hydrated copper(II) ions, \(\text{Cu}^{2+}\text{(aq)}\). During electrolysis with copper active electrodes, for every mole of \(\text{Cu}^{2+}\) ions reduced and deposited as copper metal at the cathode, an equal mole of \(\text{Cu}^{2+}\) ions is produced by the oxidation of the copper anode. Consequently, the concentration of \(\text{Cu}^{2+}\text{(aq)}\) in the solution remains constant, and the blue color of the solution does not fade.

(a) 1 mark for stating the anode dissolves; 1 mark for the correct half-equation.\n(b) 1 mark for calculating correct charge \(Q = 3600 \text{ C}\); 1 mark for calculating correct moles of copper reacted; 1 mark for the correct mass change (decrease of \(1.18 \text{ g}\)) with appropriate units.\n(c) 1.14 marks for explaining that the rate of oxidation of Cu at the anode equals the rate of reduction of \(\text{Cu}^{2+}\) at the cathode, keeping the concentration of \(\text{Cu}^{2+}\text{(aq)}\) constant.

(a) \n*cis*-isomer (maleic acid):\n```\n HOOC COOH\n \\ /\n C = C\n / \\\n H H\n```\n*trans*-isomer (fumaric acid):\n```\n HOOC H\n \\ /\n C = C\n / \\\n H COOH\n```\n\n(b) The type of stereoisomerism is *cis-trans* isomerism (or geometrical isomerism).\nIt exists because:\n1. There is restricted rotation about the carbon-carbon double bond (\(\text{C}=\text{C}\)) due to the presence of the \(\pi\)-bond.\n2. Each of the carbon atoms in the double bond is attached to two different groups (specifically, \(-\text{H}\) and \(-\text{COOH}\)).\n\n(c) The *cis*-isomer undergoes intramolecular dehydration upon heating.\nIn the *cis*-isomer, the two carboxyl groups (\(-\text{COOH}\)) are on the same side of the \(\text{C}=\text{C}\) double bond and are spatially close to each other. This proximity allows them to easily react with each other to eliminate a water molecule and form a cyclic anhydride. In contrast, in the *trans*-isomer, the two carboxyl groups are on opposite sides of the double bond and are too far apart to react with each other.

(a) 1 mark for correct structure of the *cis*-isomer; 1 mark for correct structure of the *trans*-isomer.\n(b) 1 mark for stating "cis-trans isomerism" (or geometrical isomerism); 1.14 marks for explaining both: restricted rotation around \(\text{C}=\text{C}\) and two different groups on each carbon of the double bond.\n(c) 1 mark for identifying the *cis*-isomer; 1 mark for explaining that the \(-\text{COOH}\) groups in the *cis*-isomer are spatially close to each other, allowing dehydration, whereas they are too far apart in the *trans*-isomer.

(a) Transition metals exhibit several unique properties. Two characteristic properties other than catalysis are:\n1. Variable oxidation states (e.g., Fe can exist as \(\text{Fe}^{2+}\) or \(\text{Fe}^{3+}\)).\n2. Formation of coloured ions/compounds (e.g., aqueous \(\text{Cu}^{2+}\) is blue, while \(\text{Fe}^{3+}\) is yellow/brown).\n(Note: Formation of complexes is also an acceptable answer. General metallic properties such as high melting point are not accepted as chemical characteristics unique to transition metals).\n\n(b)\n(i) The catalyst used in the Contact Process is vanadium(V) oxide (\(\text{V}_2\text{O}_5\)).\n(ii) The catalyst provides an alternative reaction pathway with a lower activation energy (\(E_a\)). As a result, a larger fraction of reactant molecules have kinetic energy equal to or greater than this lower activation energy, leading to a higher frequency of successful collisions and thus a faster reaction rate.\n(iii) It is a heterogeneous catalyst. This is because the catalyst (\(\text{V}_2\text{O}_5\)) is in a different physical state (solid phase) than the reactants (sulfur dioxide and oxygen, which are in the gas phase).

(a) 1 mark for each correct property listed (Max 2 marks). (Accept: variable oxidation states, coloured ions, complex formation).\n(b)(i) 1 mark for Vanadium(V) oxide / \(\text{V}_2\text{O}_5\).\n(b)(ii) 1.14 marks for stating that the catalyst provides an alternative reaction pathway with lower activation energy, increasing the fraction of molecules with energy \(\ge E_a\).\n(b)(iii) 1 mark for identifying it as a heterogeneous catalyst; 1 mark for explaining that the catalyst is in a different phase/state (solid) than the reactants (gas).

(a) \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\)\n\n(b) \nInitial moles of \(\text{HCl}\) added:\n\(n(\text{HCl})_{initial} = 1.00 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.0500 \text{ mol}\)\n\nIn the titration:\n\(n(\text{NaOH}) = 0.100 \text{ mol dm}^{-3} \times 0.02600 \text{ dm}^3 = 0.00260 \text{ mol}\)\n\nSince the neutralisation reaction is \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\):\nMoles of unreacted \(\text{HCl}\) in the \(25.0 \text{ cm}^3\) aliquot = \(0.00260 \text{ mol}\).\n\nTotal moles of unreacted \(\text{HCl}\) in the \(250.0 \text{ cm}^3\) volumetric flask:\n\(n(\text{HCl})_{excess} = 0.00260 \text{ mol} \times \frac{250.0 \text{ cm}^3}{25.0 \text{ cm}^3} = 0.0260 \text{ mol}\)\n\nMoles of \(\text{HCl}\) that reacted with the eggshell sample:\n\(n(\text{HCl})_{reacted} = 0.0500 \text{ mol} - 0.0260 \text{ mol} = 0.0240 \text{ mol}\)\n\n(c) \nFrom the equation in part (a), \(1\text{ mol}\) of \(\text{CaCO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\).\nMoles of \(\text{CaCO}_3\) in the eggshell sample:\n\(n(\text{CaCO}_3) = \frac{1}{2} \times n(\text{HCl})_{reacted} = \frac{0.0240 \text{ mol}}{2} = 0.0120 \text{ mol}\)\n\nMolar mass of \(\text{CaCO}_3 = 40.1 + 12.0 + 3(16.0) = 100.1 \text{ g mol}^{-1}\)\n\nMass of \(\text{CaCO}_3\) in the sample:\n\(m = 0.0120 \text{ mol} \times 100.1 \text{ g mol}^{-1} = 1.2012 \text{ g}\)\n\nPercentage by mass of \(\text{CaCO}_3\) in the eggshell sample:\n\(\% \text{ mass} = \frac{1.2012 \text{ g}}{1.50 \text{ g}} \times 100\% = 80.08\%\) (or \(80.1\%\))

(a) 1 mark for the correct chemical equation with state symbols.\n(b) 1 mark for calculating correct initial moles of HCl; 1 mark for calculating correct excess moles of HCl in the flask; 1.14 marks for subtracting to find the correct moles of HCl reacted with CaCO3 (0.0240 mol).\n(c) 1 mark for converting moles of reacted HCl to mass of CaCO3 (1.2012 g); 1 mark for the correct percentage by mass (80.08% or 80.1%).

(a) \nFor \(\text{BF}_3\): The central boron atom shares its 3 valence electrons with 3 fluorine atoms to form 3 single covalent bonds. There are no lone pairs on the central B atom.\nFor \(\text{NF}_3\): The central nitrogen atom shares 3 of its 5 valence electrons with 3 fluorine atoms to form 3 single covalent bonds, leaving 1 lone pair on the N atom.\n(Fluorine atoms in both structures should have 3 lone pairs each to complete their octets).\n\n(b) \n- \(\text{BF}_3\): Shape is trigonal planar; bond angle is \(120^\circ\).\n- \(\text{NF}_3\): Shape is trigonal pyramidal; bond angle is \(102.5^\circ\) to \(107^\circ\) (due to lone pair-bonding pair repulsion being greater than bonding pair-bonding pair repulsion).\n\n(c) \nIn \(\text{BF}_3\), the molecule has a highly symmetrical trigonal planar geometry. The dipole moments of the three highly polar \(\text{B}-\text{F}\) bonds are equal in magnitude and point in symmetrically opposing directions. Therefore, the dipoles cancel each other out completely, resulting in a net dipole moment of zero (non-polar molecule).\nIn \(\text{NF}_3\), the presence of the lone pair on the nitrogen atom causes the molecule to have an asymmetrical trigonal pyramidal geometry. The dipoles of the three polar \(\text{N}-\text{F}\) bonds do not cancel each other out, resulting in a net molecular dipole moment (polar molecule).

(a) 1 mark for correct electron-dot structure of \(\text{BF}_3\); 1 mark for correct electron-dot structure of \(\text{NF}_3\) (including lone pair on N and octets on F).\n(b) 1.14 marks for correctly stating the shape (trigonal planar) and bond angle (\(120^\circ\)) of \(\text{BF}_3\); 1 mark for correctly stating the shape (trigonal pyramidal) and bond angle (accept \(102^\circ\) to \(107^\circ\)) of \(\text{NF}_3\).\n(c) 1 mark for explaining that \(\text{BF}_3\) is highly symmetrical, so bond dipoles cancel out; 1 mark for explaining that \(\text{NF}_3\) is asymmetrical (due to lone pair), so dipoles do not cancel.

(a) Catalytic cracking (or thermal cracking).\n\n(b)\n(i) Addition polymerization.\n(ii) Structure of repeating unit:\n```\n H H\n | |\n -[C - C]-\n | |\n H H\n```\n(Or \(-\text{[CH}_2\text{-CH}_2\text{]}-\))\n\n(c) \nLow-Density Polyethene (LDPE) has highly branched polymer chains. The branches prevent the polymer chains from packing closely together. This results in weaker intermolecular forces (Van der Waals' forces) between chains, giving LDPE a lower density and making it highly flexible (soft).\nHigh-Density Polyethene (HDPE) consists of straight, unbranched polymer chains that can pack very closely together in a well-ordered, crystalline manner. This results in stronger intermolecular forces between the chains, giving HDPE a higher density and making it more rigid (less flexible).

(a) 1 mark for cracking (catalytic/thermal cracking).\n(b)(i) 1 mark for addition polymerization.\n(b)(ii) 1 mark for drawing correct repeating unit showing open bonds on ends.\n(c) 1 mark for stating LDPE is highly branched and HDPE is unbranched/straight-chain; 1 mark for linking branching to the packing of polymer chains (LDPE cannot pack closely, HDPE packs closely); 1.14 marks for linking packing to density and flexibility (LDPE has weaker intermolecular forces, lower density, and higher flexibility; HDPE has stronger intermolecular forces, higher density, and higher rigidity).

(a) The primary pollutant is sulfur dioxide (\(\text{SO}_2\)).\nIn the atmosphere, sulfur dioxide reacts with water to form sulfurous acid:\n\(\text{SO}_2\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_3\text{(aq)}\)\nAlternatively, it can be oxidized and then dissolve to form sulfuric acid:\n\(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{H}_2\text{SO}_4\text{(aq)}\)\n(Accept nitrogen dioxide \(\text{NO}_2\) and its reaction to form nitric acid and nitrous acid: \(2\text{NO}_2 + \text{H}_2\text{O} \rightarrow \text{HNO}_3 + \text{HNO}_2\)).\n\n(b) Any one of:\n- Acidification of lakes and rivers, killing fish and other aquatic organisms.\n- Damage to leaves and roots of trees, leading to forest decline.\n- Corrosion of metal bridges and weathering of limestone/marble historical buildings.\n\n(c)\n(i) \(2\text{CaCO}_3\text{(s)} + 2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CaSO}_4\text{(s)} + 2\text{CO}_2\text{(g)}\)\n(ii) Calcium sulfate (gypsum) is highly useful in the construction industry as a major component of plasterboard, plaster, and cement. This process turns a harmful pollutant into a commercial product, reducing waste and landfilling.

(a) 1 mark for sulfur dioxide (\(\text{SO}_2\)) or nitrogen dioxide (\(\text{NO}_2\)); 1 mark for a correct balanced equation representing its conversion to acid.\n(b) 1 mark for any correct environmental impact.\n(c)(i) 1.14 marks for correct balanced equation.\n(c)(ii) 1 mark for stating calcium sulfate is gypsum; 1 mark for describing its use in construction / plasterboards / cement.

(a) The standard enthalpy change of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (298 K, 1 atm).

(b) The equation for the formation of ethanoic acid is:
\(2\text{C}(s) + 2\text{H}_2(g) + \text{O}_2(g) \rightarrow \text{CH}_3\text{COOH}(l)\)

According to Hess's Law:
\(\Delta H_f^\theta = \sum \Delta H_c^\theta[\text{reactants}] - \sum \Delta H_c^\theta[\text{products}]\)
\(\Delta H_f^\theta = 2 \times \Delta H_c^\theta[\text{C(graphite)}] + 2 \times \Delta H_c^\theta[\text{H}_2(g)] - \Delta H_c^\theta[\text{CH}_3\text{COOH}(l)]\)
\(\Delta H_f^\theta = 2(-394) + 2(-286) - (-874)\)
\(\Delta H_f^\theta = -788 - 572 + 874 = -486\text{ kJ mol}^{-1}\)

(c) Carbon (graphite), hydrogen, and oxygen do not react directly under standard conditions to form ethanoic acid. If they are allowed to react under other conditions, a mixture of various other compounds (such as carbon monoxide, hydrocarbons, or water) would form instead of pure ethanoic acid.

(a) Enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. [1 mark]
(b) Correct equation or enthalpy cycle setup: \(\Delta H_f^\theta = 2 \times \Delta H_c^\theta[\text{C}] + 2 \times \Delta H_c^\theta[\text{H}_2] - \Delta H_c^\theta[\text{CH}_3\text{COOH}]\) [1 mark]
- Correct substitution of values: \(2(-394) + 2(-286) - (-874)\) [1 mark]
- Correct final value with unit: \(-486\text{ kJ mol}^{-1}\) [1 mark]
(c) Carbon, hydrogen, and oxygen do not react directly to form ethanoic acid under standard conditions / other reactions or side-products occur. [1 mark for stating they do not react directly, 1 mark for mentioning the formation of other side products or incomplete reaction]

(a) The equilibrium constant expression is:
\(K_c = \frac{[\text{NO}_2(g)]^2}{[\text{N}_2\text{O}_4(g)]}\)
Unit: \(\text{mol dm}^{-3}\)

(b) Initial concentration of \(\text{N}_2\text{O}_4(g) = \frac{0.80\text{ mol}}{2.0\text{ dm}^3} = 0.40\text{ mol dm}^{-3}\)
At equilibrium, \([\text{NO}_2(g)] = 0.30\text{ mol dm}^{-3}\).
According to the stoichiometric equation, the change in concentration of \(\text{N}_2\text{O}_4(g)\) is:
\(\Delta [\text{N}_2\text{O}_4(g)] = -\frac{1}{2} \Delta [\text{NO}_2(g)] = -\frac{0.30}{2} = -0.15\text{ mol dm}^{-3}\)

Therefore, the equilibrium concentration of \(\text{N}_2\text{O}_4(g) = 0.40 - 0.15 = 0.25\text{ mol dm}^{-3}\)

\(K_c = \frac{(0.30)^2}{0.25} = 0.36\text{ mol dm}^{-3}\)

(c) When the volume decreases, the total pressure inside the container increases. According to Le Chatelier's Principle, the system will shift the equilibrium position to the left (towards the reactant side with fewer number of moles of gaseous molecules, i.e., 1 mole compared to 2 moles) to counteract the increase in pressure.

(a) Correct expression for \(K_c\) [1 mark]; Correct unit: \(\text{mol dm}^{-3}\) [1 mark]
(b) Correct initial concentration of \(\text{N}_2\text{O}_4\) (\(0.40\text{ mol dm}^{-3}\)) [1 mark]; Correct calculation of equilibrium concentration of \(\text{N}_2\text{O}_4\) (\(0.25\text{ mol dm}^{-3}\)) [1 mark]; Correct calculation of \(K_c\) value (\(0.36\)) [1 mark]
(c) Equilibrium position shifts to the left because decreasing volume increases pressure, and the system favors the side with fewer gas moles to lower the pressure. [1 mark]

(a) In \(\text{CH}_3\text{OH}\), let the oxidation number of carbon be \(x\).
\(x + 4(+1) + (-2) = 0 \Rightarrow x = -2\)
In \(\text{CO}_2\), let the oxidation number of carbon be \(y\).
\(y + 2(-2) = 0 \Rightarrow y = +4\)
The oxidation number of carbon changes from \(-2\) to \(+4\) (an increase of 6).

(b) At the anode, methanol is oxidized to carbon dioxide gas:
\(\text{CH}_3\text{OH}(l) + \text{H}_2\text{O}(l) \rightarrow \text{CO}_2(g) + 6\text{H}^+(aq) + 6e^-\)

(c) Advantage: Methanol is a liquid at room temperature and pressure, which makes it much easier and safer to store, handle, and transport compared to hydrogen gas, which requires high-pressure tanks or cryogenic cooling.
Disadvantage: A direct methanol fuel cell produces carbon dioxide (a greenhouse gas) during operation, whereas a hydrogen-oxygen fuel cell only produces water as a byproduct, making it completely zero-emission.

(a) Correct oxidation number of carbon in \(\text{CH}_3\text{OH}\) (\(-2\)) [1 mark]; Correct oxidation number of carbon in \(\text{CO}_2\) (\(+4\)) and stating the change (increases by 6) [1 mark].
(b) Correct reactants and products: \(\text{CH}_3\text{OH} + \text{H}_2\text{O} \rightarrow \text{CO}_2 + 6\text{H}^+\) [1 mark]; Correctly balanced electrons and states: \(+ 6e^-\), and matching coefficients [1 mark].
(c) One logical advantage (e.g., liquid fuel, easier storage/transport) [1 mark]; One logical disadvantage (e.g., greenhouse gas emission of CO2, toxicity of methanol) [1 mark].

(a) The conversion of but-2-ene to 2-chlorobutane is an addition reaction (specifically, electrophilic addition) with HCl(g).

(b) Step 2 is the nucleophilic substitution of a haloalkane to form an alcohol. Reagent: Sodium hydroxide solution (or NaOH(aq) / KOH(aq)). Condition: Heat under reflux.

(c) Compound X is formed by oxidizing a secondary alcohol (butan-2-ol) to a ketone (butanone).
Its structural formula is:
\(\text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_3\)
The acidified potassium dichromate solution is reduced from Cr2O7^2- (orange) to Cr^3+ (green). Thus, the colour change is orange to green.

(d) Fractional distillation (or simple distillation) can be used to separate butanone from the reaction mixture because they have different boiling points.

(a) Addition (or electrophilic addition) [1 mark]. Reject: substitution / elimination.
(b) Reagent: aqueous sodium hydroxide / NaOH(aq) / aqueous potassium hydroxide / KOH(aq) [1 mark]. (Reject: concentrated NaOH / ethanolic NaOH).
Condition: heat under reflux / heat [1 mark].
(c) Correct structural formula of butanone (clearly showing \(\text{C}=\text{O}\) group) [1 mark]; Color change: from orange to green [1 mark].
(d) Fractional distillation / distillation / simple distillation [1 mark].

Part (a)
(i) Use the Arrhenius Equation:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
Substitute the values:
\(\ln\left(\frac{1.2 \times 10^{-2}}{2.5 \times 10^{-3}}\right) = \frac{E_a}{8.314} \left(\frac{1}{500} - \frac{1}{550}\right)\)
\(\ln(4.8) = \frac{E_a}{8.314} \left(1.818 \times 10^{-4}\right)\)
\(1.5686 = 2.187 \times 10^{-5} \times E_a\)
\(E_a = 71727 \text{ J mol}^{-1} = 71.7 \text{ kJ mol}^{-1}\)

(ii) The unit of the rate constant \(k\) is \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). The general unit for a rate constant is \((\text{mol dm}^{-3})^{1-n} \text{ s}^{-1}\), where \(n\) is the overall reaction order.
Here, \(1-n = -2 \Rightarrow n = 3\).
Therefore, the overall reaction order is 3.

(iii) Maxwell-Boltzmann distribution: The curve at \(550 \text{ K}\) is flatter, has its peak shifted to a higher energy (to the right), and lies below the \(500 \text{ K}\) curve at lower energy levels. An activation energy line (\(E_a\)) is drawn as a vertical line on the right side. The area under the curve to the right of \(E_a\) represents the fraction of molecules with kinetic energy greater than or equal to \(E_a\). At \(550 \text{ K}\), this area is significantly larger than at \(500 \text{ K}\), meaning a greater fraction of molecules has enough energy to react, resulting in a higher rate of effective collisions and thus a faster reaction rate.

Part (b)
(i) Rate: High pressure increases the concentration of gaseous reactants, leading to an increase in collision frequency and thus a faster reaction rate.
Equilibrium yield: In the reaction, 3 moles of gaseous reactants form 1 mole of gaseous product. According to Le Chatelier's Principle, increasing pressure shifts the equilibrium to the right (fewer gas molecules), increasing the equilibrium yield of methanol.

(ii) The forward reaction is exothermic. Lowering the temperature shifts the equilibrium to the right, favoring methanol production. However, at lower temperatures, the rate of reaction is too slow for commercial viability. Thus, a compromise temperature of \(250^\circ\text{C}\) is chosen to balance a high reaction rate and an acceptable equilibrium yield.

(iii) The copper-zinc oxide catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. However, it has no effect on the equilibrium position, as it increases the rates of the forward and reverse reactions equally.

Part (c)
(i) Reaction A:
Formula mass of desired product (\(\text{CH}_3\text{OH}\)) = \(12.0 + 4(1.0) + 16.0 = 32.0 \text{ g mol}^{-1}\)
Total formula mass of reactants (\(\text{CO}_2 + 3\text{H}_2\)) = \((12.0 + 32.0) + 3(2.0) = 50.0 \text{ g mol}^{-1}\)
Atom economy of Reaction A = \(\frac{32.0}{50.0} \times 100\% = 64.0\%\)

Reaction B:
Total formula mass of reactants (\(\text{CO} + 2\text{H}_2\)) = \((12.0 + 16.0) + 2(2.0) = 32.0 \text{ g mol}^{-1}\)
Atom economy of Reaction B = \(\frac{32.0}{32.0} \times 100\% = 100\%\)

(ii) In terms of atom economy, Reaction B is greener because it has a \(100\%\) atom economy, whereas Reaction A has only \(64.0\%\) due to the production of water as a waste product.
However, in terms of environmental sustainability, Reaction A is greener because it uses \(\text{CO}_2\) (a major greenhouse gas) captured from industrial emissions as a feedstock, thereby contributing to carbon mitigation. Furthermore, \(\text{CO}\) used in Reaction B is highly toxic and usually derived from fossil fuels, making Reaction A a more sustainable and safer long-term choice.

Part (a) (Total: 8 marks)
(i)
- Applying Arrhenius equation correctly: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\) (1)
- Correct substitution of given parameters (1)
- Correct calculation of \(E_a\) value in J (\(71727 \text{ J mol}^{-1}\)) (1)
- Expressing final answer with correct unit and sig figs (\(71.7 \text{ kJ mol}^{-1}\) or \(72 \text{ kJ mol}^{-1}\)) (1)

(ii)
- Deduce that overall order is 3 (1)
- Explain that \([\text{mol dm}^{-3}]^{1-n} \text{ s}^{-1} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) leading to \(1-n = -2 \Rightarrow n=3\) (1)

(iii)
- Sketch: showing the curve of \(550 \text{ K}\) having a lower peak, shifted to the right compared to \(500 \text{ K}\), and showing vertical line of \(E_a\) on the energy axis. (1)
- Explanation: stating that at \(550 \text{ K}\), a larger fraction of molecules have energy \(\ge E_a\) (larger area under the curve to the right of \(E_a\)), resulting in a higher frequency of effective collisions and thus increased rate. (1)

Part (b) (Total: 7 marks)
(i)
- Concentration: High pressure increases the concentration / partial pressure of gaseous reactants, leading to more frequent collisions, increasing the reaction rate. (1)
- Position of equilibrium: Since there are fewer moles of gas on the product side (1 mole) than on the reactant side (3 moles), according to Le Chatelier's Principle, high pressure shifts the equilibrium position to the right, increasing the yield of methanol. (1)
- Engineering/Economic limitation: extremely high pressures require very expensive, thick-walled reaction vessels, making 50-100 atm a reasonable practical limit. (1)

(ii)
- Lowering temperature shifts the equilibrium to the right because the forward reaction is exothermic, which increases the equilibrium yield. (1)
- However, lowering temperature decreases the reaction rate. Therefore, \(250^\circ\text{C}\) is a compromise temperature used to obtain a fast enough rate and an acceptable yield in a realistic time frame. (1)

(iii)
- Rate: The catalyst increases the rate of both forward and reverse reactions by providing an alternative pathway with a lower activation energy. (1)
- Equilibrium position: The catalyst has no effect on the equilibrium position because it speeds up the forward and backward reactions to the same extent. (1)

Part (c) (Total: 5 marks)
(i)
- Calculation of Reaction A atom economy = \(64.0\%\) (showing working) (1)
- Calculation of Reaction B atom economy = \(100\%\) (showing working) (1)

(ii)
- Comparison based on atom economy: Reaction B is greener as all reactant atoms end up in the desired product (atom economy = 100%), while Reaction A generates water as a waste product (atom economy = 64%). (1)
- Comparison based on greenhouse gas mitigation: Reaction A utilizes captured \(\text{CO}_2\) (a major greenhouse gas), which is environmentally beneficial as carbon capture and utilization (CCU). (1)
- Comparison based on toxic materials/source: Reaction B uses carbon monoxide (highly toxic gas) which is typically derived from fossil fuels, while Reaction A uses safer greenhouse gases and reduces fossil fuel reliance. Conclude that overall Reaction A may be greener. (1)

(a)(i) The absence of a broad absorption peak in the $3200 - 3600\text{ cm}^{-1}$ region indicates that the compound does not contain an oxygen-hydrogen ($\text{O-H}$) bond. This rules out propan-1-ol (which is an alcohol and contains an $\text{O-H}$ group).
Alternatively, the presence of a strong peak at $1715\text{ cm}^{-1}$ indicates a carbonyl group ($\text{C=O}$), which is present in propanal and propanone but not propan-1-ol.

(ii) Tollens' reagent (or Fehling's solution / acidified potassium dichromate solution):
Add Tollens' reagent to both compounds in separate test tubes and warm the mixtures in a water bath.
- Propanal (aldehyde) will form a silver mirror on the inner wall of the test tube.
- Propanone (ketone) will show no observable change.

(iii) $\text{CH}_3\text{CO}^+$ (or $\text{C}_2\text{H}_3\text{O}^+$)

(b)(i) The separation is based on the difference in the adsorption of different components on the stationary phase and their solubility in the mobile phase. Components with weaker adsorption on the stationary phase and higher solubility in the mobile phase travel faster up the plate.

(ii)(1) Aspirin will have a larger $R_f$ value.
Since aspirin is less polar than salicylic acid, it has weaker intermolecular forces (adsorption) with the polar silica gel stationary phase, and greater relative solubility/affinity for the less polar mobile phase. Thus, aspirin travels faster and further up the plate.

(ii)(2) Use ultraviolet (UV) light to view the plate, or place the plate in a sealed chamber containing iodine crystals to develop the spots.

(c)(i) $\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}$

(ii) $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$

(iii) Indicator: Starch solution.
Color change at the end point: From blue (or blue-black) to colorless.

(iv)
Number of moles of $\text{S}_2\text{O}_3^{2-}$ used = $0.100\text{ mol dm}^{-3} \times 0.02420\text{ dm}^3 = 2.42 \times 10^{-3}\text{ mol}$

According to the equations:
$\text{ClO}^- \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}$

Number of moles of $\text{ClO}^-$ in $25.0\text{ cm}^3$ diluted solution = $\frac{1}{2} \times 2.42 \times 10^{-3}\text{ mol} = 1.21 \times 10^{-3}\text{ mol}$

Number of moles of $\text{ClO}^-$ in $250.0\text{ cm}^3$ diluted solution = $1.21 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 1.21 \times 10^{-2}\text{ mol}$

This amount is present in the original $10.0\text{ cm}^3$ bleach sample.
Concentration of $\text{NaClO}$ in original bleach = $\frac{1.21 \times 10^{-2}\text{ mol}}{0.0100\text{ dm}^3} = 1.21\text{ M}$

Concentration of $\text{NaClO}$ in $\text{g dm}^{-3}$ = $1.21\text{ mol dm}^{-3} \times 74.5\text{ g mol}^{-1} = 90.1\text{ g dm}^{-3}$ (or $90.15\text{ g dm}^{-3}$)

Part (a) [6 marks]
- (i) Absence of absorption peak at 3200-3600 cm^-1 indicates absence of O-H group [1 mark], ruling out propan-1-ol [1 mark].
- (ii) Correct reagent and condition: Tollens' reagent, warm [1 mark]. Observation: propanal forms silver mirror [1 mark], propanone has no observable change [1 mark].
- (iii) [CH3CO]^+ (or [C2H3O]^+) [1 mark] (Note: positive charge must be included).

Part (b) [6 marks]
- (i) Separation depends on the relative adsorption of components on stationary phase and solubility in mobile phase [1 mark]. Components with weaker adsorption/higher solubility move faster [1 mark].
- (ii)(1) Aspirin has a larger Rf value [1 mark]. Aspirin is less polar, leading to weaker intermolecular forces (adsorption) with polar silica gel [1 mark], and greater relative affinity/solubility in the mobile phase [1 mark].
- (ii)(2) UV light / Iodine chamber [1 mark].

Part (c) [8 marks]
- (i) ClO^- + 2I^- + 2H^+ -> Cl^- + I2 + H2O [1 mark]
- (ii) I2 + 2S2O3^2- -> 2I^- + S4O6^2- [1 mark]
- (iii) Starch [1 mark]; Blue to colorless (Reject: clear) [1 mark].
- (iv) Moles of S2O3^2- = 2.42 x 10^-3 mol [1 mark]; Moles of ClO^- in aliquot = 1.21 x 10^-3 mol [1 mark]; Concentration in original bleach = 1.21 M [1 mark]; Mass concentration = 90.1 g dm^-3 (Accept 90.15 or 90.2) [1 mark].