2024 HKDSE Chemistry Answers & Marking Scheme

The molecular formula \(C_3H_6O_2\) represents a compound with one degree of unsaturation. The IR absorption at \(1740 \text{ cm}^{-1}\) indicates a carbonyl (\(C=O\)) group, while the absence of O-H absorption (neither \(2500-3300 \text{ cm}^{-1}\) nor \(3230-3670 \text{ cm}^{-1}\)) rules out carboxylic acids and alcohols. Thus, X must be an ester, specifically methyl ethanoate (\(CH_3COOCH_3\)).
(1) Esters undergo alkaline hydrolysis to form a carboxylate salt and an alcohol. (Correct)
(2) In mass spectrometry, methyl ethanoate (M = 74) cleaves at the C-O single bond to yield a prominent fragment peak at m/z = 59 (\([COOCH_3]^+\) or \([CH_3COO]^+\)). (Correct)
(3) Only carboxylic acids (not esters) react with \(NaHCO_3\) to release \(CO_2\) gas. (Incorrect)

Select B. Correctly identifying X as an ester (1 mark).

According to the Arrhenius equation, \(\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A\).
Thus, the slope of the plot is equal to \(-\frac{E_a}{R}\).
\(-\frac{E_a}{R} = -6015 \text{ K}\)
\(E_a = 6015 \times 8.31 = 49984.65 \text{ J mol}^{-1} \approx 50.0 \text{ kJ mol}^{-1}\).

Select A. Correct calculation using Arrhenius equation slope formula (1 mark).

(1) But-2-ene reacts with gaseous hydrogen chloride (\(HCl(g)\)) via electrophilic addition to form 2-chlorobutane. (Correct)
(2) 2-chlorobutane undergoes nucleophilic substitution when heated with aqueous sodium hydroxide (\(NaOH(aq)\)) to yield butan-2-ol. (Correct)
(3) Butan-2-ol is a secondary alcohol, which can be oxidized to butanone (a ketone) using acidified potassium dichromate solution under reflux. (Correct)

Select D. Correctly evaluating all three organic conversion reactions (1 mark).

Initial concentrations:
\([X]_0 = 2.0 / 2.0 = 1.0 \text{ mol dm}^{-3}\)
\([Y]_0 = 1.0 / 2.0 = 0.5 \text{ mol dm}^{-3}\)
\([Z]_0 = 0.0 \text{ mol dm}^{-3}\)
At equilibrium, \([Z]_{eq} = 0.40 \text{ mol dm}^{-3}\).
Change in concentrations:
\(\Delta [Z] = +0.40 \text{ mol dm}^{-3}\)
\(\Delta [X] = -0.40 \text{ mol dm}^{-3}\)
\(\Delta [Y] = -0.20 \text{ mol dm}^{-3}\)
Equilibrium concentrations:
\([X]_{eq} = 1.0 - 0.40 = 0.60 \text{ mol dm}^{-3}\)
\([Y]_{eq} = 0.50 - 0.20 = 0.30 \text{ mol dm}^{-3}\)
\(K_c = \frac{[Z]^2}{[X]^2 [Y]} = \frac{(0.40)^2}{(0.60)^2 \times 0.30} = \frac{0.16}{0.36 \times 0.30} \approx 1.48 \text{ dm}^3 \text{ mol}^{-1}\).

Select B. Correct calculation of initial concentrations, equilibrium values, and Kc formula (1 mark).

An oxide that is insoluble in water but dissolves in both dilute acids and dilute alkalis is an amphoteric oxide. Aluminum oxide (\(Al_2O_3\)) is amphoteric and behaves this way:
\(Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l)\)
\(Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)\).

Select B. Correctly identifying aluminium as having an amphoteric oxide (1 mark).

The overall ionic equation for the reaction is:
\(Cr_2O_7^{2-}(aq) + 3SO_2(g) + 2H^+(aq) \rightarrow 2Cr^{3+}(aq) + 3SO_4^{2-}(aq) + H_2O(l)\)
(1) Sulfur in \(SO_2\) (+4) is oxidized to \(SO_4^{2-}\) (+6), so \(SO_2\) is a reducing agent. (Correct)
(2) Chromium in \(Cr_2O_7^{2-}\) (+6) is reduced to \(Cr^{3+}\) (+3). (Correct)
(3) Since \(H^+(aq)\) is consumed, the concentration of hydrogen ions decreases, leading to an increase in pH. (Correct)

Select D. Correctly evaluating all redox change features and pH trend (1 mark).

The chemical equation for the combustion of propane is:
\(C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\)
Using Hess's Law:
\(\Delta H_c^\theta = 3 \times \Delta H_f^\theta[CO_2(g)] + 4 \times \Delta H_f^\theta[H_2O(l)] - \Delta H_f^\theta[C_3H_8(g)]\)
\(\Delta H_c^\theta = 3 \times (-394) + 4 \times (-286) - (-104)\)
\(\Delta H_c^\theta = -1182 - 1144 + 104 = -2222 \text{ kJ mol}^{-1}\).

Select A. Correct application of Hess's Law with correct stoichiometric coefficients (1 mark).

In \(BF_3\), the electronegativity difference between B and F creates polar B-F bonds. However, because of the trigonal planar shape (symmetrical geometry), the dipoles cancel each other out, resulting in a non-polar molecule with a net dipole moment of zero. In contrast, \(NH_3\) (trigonal pyramidal), \(H_2O\) (bent), and \(CH_2Cl_2\) (asymmetric tetrahedral) have geometries that do not allow dipoles to cancel completely, making them polar molecules.

Select C. Correctly relating molecular shape and symmetry to molecular polarity (1 mark).

(1) Gas volume measurement: Since \(CO_2(g)\) is generated, measuring gas volume over time is a standard direct method. (Correct)
(2) Mass measurement: As \(CO_2(g)\) escapes from an open system, the overall mass of the mixture decreases. (Correct)
(3) pH measurement: Hydrochloric acid (\(H^+(aq)\)) is consumed, causing the pH of the solution to increase. A pH meter can continuously track this change. (Correct)

Select D. Correctly identifying all three monitoring methods for this specific reaction (1 mark).

The reaction equation is:
\(H_2C_2O_4(aq) + 2NaOH(aq) \rightarrow Na_2C_2O_4(aq) + 2H_2O(l)\)
Number of moles of \(H_2C_2O_4 = 0.050 \text{ M} \times 0.0250 \text{ dm}^3 = 0.00125 \text{ mol}\).
Since \(H_2C_2O_4\) is dibasic, the number of moles of \(NaOH\) needed is:
\(n(NaOH) = 2 \times 0.00125 = 0.00250 \text{ mol}\).
Concentration of \(NaOH = \frac{0.00250 \text{ mol}}{0.0200 \text{ dm}^3} = 0.125 \text{ M}\).

Select C. Correctly using the 1:2 reaction ratio to calculate NaOH molarity (1 mark).

An absorption peak around \(1735\text{ cm}^{-1}\) indicates the presence of a \(\text{C}=\text{O}\) group. The absence of broad absorption peaks around \(2500-3300\text{ cm}^{-1}\) (carboxylic acid \(\text{O}-\text{H}\)) and \(3230-3670\text{ cm}^{-1}\) (alcohol \(\text{O}-\text{H}\)) indicates that \(X\) does not contain an \(-\text{OH}\) group, which rules out butanoic acid (3). Both ethyl ethanoate (1) and methyl propanoate (2) are esters with the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\). Esters contain a \(\text{C}=\text{O}\) group but no \(-\text{OH}\) group, and they do not react with acidified potassium dichromate solution. Thus, (1) and (2) are possible structures.

Correct Answer: C. 1 mark for identifying that (1) and (2) match the molecular formula and IR spectroscopic data, while (3) is excluded due to the absence of the carboxylic acid O-H absorption band.

According to the Arrhenius equation, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). Therefore, the slope of the plot of \(\ln k\) against \(1/T\) is equal to \(-\frac{E_a}{R}\).

\(\text{Slope} = -1.20 \times 10^4\text{ K} = -\frac{E_a}{8.31\text{ J K}^{-1}\text{ mol}^{-1}}\)

\(E_a = 1.20 \times 10^4 \times 8.31 = 9.972 \times 10^4\text{ J mol}^{-1} = 99.7\text{ kJ mol}^{-1}\).

Correct Answer: B. 1 mark for relating the slope of the Arrhenius plot to activation energy using the gas constant R and converting units from J to kJ.

1. \(P\) reacts with 2,4-DNP to form a yellow precipitate, indicating the presence of a carbonyl group (\(\text{C}=\text{O}\)).
2. \(P\) does not react with Tollens' reagent, indicating it is a ketone and not an aldehyde. This rules out pentanal (C).
3. Reduction of ketones with \(\text{NaBH}_4\) yields secondary alcohols:
- Pentan-3-one is reduced to pentan-3-ol, \(\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3\), which has a symmetric carbon (bonded to two identical ethyl groups) and is therefore achiral.
- Pentan-2-one is reduced to pentan-2-ol (chiral).
- 3-Methylbutan-2-one is reduced to 3-methylbutan-2-ol (chiral).
Thus, \(P\) is pentan-3-one.

Correct Answer: B. 1 mark for deducing that P is a ketone from the carbonyl tests, and verifying that pentan-3-ol is the only achiral product among the possible reduction options.

The reaction quotient \(Q_c\) is calculated as:
\(Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{(2.0/2.0) \times (2.0/2.0)}{(1.0/2.0) \times (1.0/2.0)} = \frac{1.0 \times 1.0}{0.5 \times 0.5} = 4.0\).
Since \(Q_c = K_c = 4.0\), the reaction is already at equilibrium.

Correct Answer: A. 1 mark for correctly calculating the reaction quotient Qc and matching it with the equilibrium constant Kc.

1. Moles of \(\text{NaOH}\) used in the titration:
\(n(\text{NaOH}) = 0.100\text{ mol dm}^{-3} \times \frac{18.50}{1000}\text{ dm}^3 = 1.85 \times 10^{-3}\text{ mol}\)

2. Moles of \(\text{CH}_3\text{COOH}\) in the \(25.00\text{ cm}^3\) diluted sample:
\(n(\text{CH}_3\text{COOH}) = 1.85 \times 10^{-3}\text{ mol}\)

3. Concentration of diluted vinegar:
\(\text{Molarity} = \frac{1.85 \times 10^{-3}\text{ mol}}{0.02500\text{ dm}^3} = 0.0740\text{ M}\)

4. Concentration of the original vinegar (due to a 10-fold dilution):
\(\text{Molarity}_{\text{original}} = 0.0740 \times 10 = 0.740\text{ M}\)

5. Mass concentration of original vinegar:
\(\text{Mass concentration} = 0.740\text{ mol dm}^{-3} \times 60.05\text{ g mol}^{-1} = 44.4\text{ g dm}^{-3}\).

Correct Answer: C. 1 mark for step-by-step volumetric stoichiometry calculations, accounting for the dilution factor of 10 and conversion to mass concentration in g/dm³.

(1) By IUPAC cell notation convention, the anode (where oxidation occurs) is on the left. Thus, statement (1) is correct.
(2) Oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) occurs at the left electrode: \(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\). Statement (2) is correct.
(3) Reduction occurs at the cathode (right side): \(\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)\). As \(\text{H}^+\) is consumed, the concentration of \(\text{H}^+\) decreases, causing the pH to increase. Statement (3) is correct.

Correct Answer: D. 1 mark for successfully analyzing all three points: IUPAC cell notation, the anode oxidation reaction, and the consumption of hydrogen ions during permanganate reduction.

Benzene-1,4-dicarboxylic acid is \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) and benzene-1,4-diamine is \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2\). During condensation polymerisation, dicarboxylic acid monomer units lose \(-\text{OH}\) and diamine monomer units lose \(-\text{H}\), releasing water molecules.
This links the groups with amide bonds \(-\text{CO}-\text{NH}-\).
The repeating unit of the resulting polyamide is \(-\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-\), which is Option A.

Correct Answer: A. 1 mark for deriving the correct repeating unit structure based on amide linkage formation with elimination of water molecules.

A catalyst increases the rate of reaction by providing an alternative pathway with a lower activation energy, which accelerates both the forward and reverse reactions equally. However, it does not alter the position of chemical equilibrium or change the enthalpy change (\(\Delta H\)) of the reaction. Therefore, the catalyst does not affect the equilibrium yield of ammonia in the Haber Process. Option C is incorrect.

Correct Answer: C. 1 mark for recognizing that a catalyst has no effect on equilibrium yields, only on the rate at which equilibrium is reached.

(1) Powdered \(\text{CaCO}_3\) provides a higher surface area, increasing contact and collision frequency between solid particles and acid, thus increasing the reaction rate.
(2) Increasing \(\text{HCl}\) concentration increases the number of acid particles per unit volume, which increases collision frequency and increases the reaction rate.
(3) Raising the temperature increases the kinetic energy of particles, so a larger fraction of particles will have energy exceeding the activation energy, increasing the frequency of successful/effective collisions.
Therefore, all three modifications increase the rate.

Correct Answer: D. 1 mark for correctly determining that surface area (1), concentration (2), and temperature (3) all increase the rate of this heterogeneous reaction.

1. When the volume is halved, the concentration of all gases immediately doubles. Since the concentration of the brown gas \(\text{NO}_2\) instantly doubles, the mixture immediately becomes darker brown.
2. According to Le Chatelier's Principle, reducing the volume (increasing total pressure) shifts the equilibrium to the side with fewer gas moles (the right side, forming colourless \(\text{N}_2\text{O}_4\)).
3. This shift consumes some \(\text{NO}_2\), causing the brown color to gradually fade (become lighter brown).
4. However, the shift only partially opposes the concentration increase. The new equilibrium concentration of \(\text{NO}_2\) is still higher than its original concentration. Thus, the final color is lighter than the peak immediately after compression, but remains darker than its initial color.

Correct Answer: B. 1 mark for explaining both the instantaneous effect of volume reduction (darkening) and the subsequent Le Chatelier's shift (fading to a net darker state).

When the volume is halved, the concentrations of all species initially double, making the instantaneous concentration of \(\text{Cl}_2(g)\) equal to \(0.80\text{ mol dm}^{-3}\). Since the system has more gas moles on the product side (2 moles) than on the reactant side (1 mole), increasing the pressure shifts the equilibrium to the left to reduce the pressure. This shift decreases the concentration of \(\text{Cl}_2(g)\) from the instantaneous value of \(0.80\text{ mol dm}^{-3}\). According to Le Chatelier's Principle, the shift cannot fully counteract the change, so the final equilibrium concentration must remain greater than the initial value of \(0.40\text{ mol dm}^{-3}\). Thus, the concentration is between \(0.40\text{ mol dm}^{-3}\) and \(0.80\text{ mol dm}^{-3}\).

Award 1 mark for selecting option C. No partial marks are awarded for other choices.

Option A, \(\text{(CH}_3)_3\text{COH}\) (2-methylpropan-2-ol), is a tertiary alcohol. It cannot be oxidized by acidified potassium dichromate solution (hence no color change). However, because it contains a hydroxyl group (\(\text{-OH}\)), it reacts with sodium metal to release hydrogen gas (a colorless gas). Option B is a primary alcohol and is easily oxidized. Options C and D do not contain hydroxyl groups and do not react with sodium metal.

Award 1 mark for selecting option A. Other choices score 0.

The desired product is phenol (\(\text{C}_6\text{H}_6\text{O}\)).
Formula mass of phenol = \(12.0 \times 6 + 1.0 \times 6 + 16.0 = 94.0\).
Formula masses of all reactants:
Benzene (\(\text{C}_6\text{H}_6\)) = \(12.0 \times 6 + 1.0 \times 6 = 78.0\).
Propene (\(\text{C}_3\text{H}_6\)) = \(12.0 \times 3 + 1.0 \times 6 = 42.0\).
Oxygen (\(\text{O}_2\)) = \(16.0 \times 2 = 32.0\).
Total formula mass of reactants = \(78.0 + 42.0 + 32.0 = 152.0\).
Atom economy = \(\frac{94.0}{152.0} \times 100\% \approx 61.8\%\).

Award 1 mark for selecting B. Option A represents the atom economy if propanone was the desired product.

The strong absorption peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C=O}\)). The very broad absorption band in the region \(2500 - 3300\text{ cm}^{-1}\) is characteristic of the \(\text{O-H}\) bond stretching of a carboxylic acid. Thus, \(Y\) is propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)). Propanoic acid reacts with sodium hydrogencarbonate solution to produce carbon dioxide (Option A). Propanoic acid cannot reduce Tollens' reagent (Option B). Its \({}^{13}\text{C}\) NMR spectrum has 3 peaks corresponding to its 3 different carbon environments (Option C). It lacks a chiral carbon and thus does not show enantiomerism (Option D).

Award 1 mark for selecting option A.

In reaction (1), the strongly oxidizing acidified \(\text{KMnO}_4(aq)\) oxidizes \(\text{H}_2\text{O}_2\) to oxygen gas (\(\text{O}_2\)). Thus, \(\text{H}_2\text{O}_2\) acts as a reducing agent (oxidation state of O increases from -1 to 0). In reactions (2) and (3), \(\text{H}_2\text{O}_2\) oxidizes \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) and \(\text{I}^-\) to \(\text{I}_2\) respectively, acting as an oxidizing agent (oxidation state of O decreases from -1 to -2).

Award 1 mark for selecting option A.

Let \(\text{M} = \text{mol dm}^{-3}\).
For option A: \(K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}\), unit is \(\frac{\text{M}^2}{\text{M}^2} = \text{dimensionless}\).
For option B: \(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\), unit is \(\frac{\text{M}^2}{\text{M}} = \text{mol dm}^{-3}\).
For option C: \(K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}\), unit is \(\frac{\text{M}}{\text{M}^2} = \text{M}^{-1} = \text{mol}^{-1}\text{ dm}^3\).
For option D: \(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\), unit is \(\frac{\text{M}^2}{\text{M}^4} = \text{M}^{-2} = \text{mol}^{-2}\text{ dm}^6\).

Award 1 mark for selecting option C.

Ethyl propanoate is an ester with the formula \(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3\). Esters are prepared by the condensation of a carboxylic acid and an alcohol. The acyl group (\(\text{CH}_3\text{CH}_2\text{CO-}\)) is derived from propanoic acid, and the alkyl group (\(\text{-CH}_2\text{CH}_3\)) is derived from ethanol. Thus, heating propanoic acid and ethanol under reflux in the presence of concentrated sulfuric acid yields ethyl propanoate.

Award 1 mark for selecting option A.

According to the Arrhenius equation, \(\ln k = -\frac{E_a}{RT} + \ln A\). A plot of \(\ln k\) against \(\frac{1}{T}\) gives a straight line with gradient \(m = -\frac{E_a}{R}\).
Therefore, \(E_a = -\text{gradient} \times R = -(-1.20 \times 10^4\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 9.972 \times 10^4\text{ J mol}^{-1} \approx 99.7\text{ kJ mol}^{-1}\).

Award 1 mark for selecting option C. Option D is incorrect due to incorrect unit conversion (using J instead of kJ).

The peak at \(m/z = 57\) corresponds to the propionyl cation \(\text{CH}_3\text{CH}_2\text{CO}^+\). Cleavage of the ester bond in methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)) generates \(\text{CH}_3\text{CH}_2\text{CO}^+\) (\(m/z = 57\)) and a methoxy radical. Esters like ethyl ethanoate (Option A) or propyl methanoate (Option C) contain parts that would easily form fragments at \(m/z = 43\) (e.g. \(\text{CH}_3\text{CO}^+\) or \(\text{C}_3\text{H}_7^+\)), which are absent here. Thus, methyl propanoate is the most likely structure.

Award 1 mark for selecting option B.

Butane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\)) is non-polar and contains only weak dispersion (Van der Waals') forces. Propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) is polar and experiences stronger dipole-dipole forces in addition to dispersion forces. Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) contains molecules associated via strong intermolecular hydrogen bonds, requiring the highest amount of thermal energy to overcome. Therefore, the boiling point increases in the order: butane < propanal < propan-1-ol.

Award 1 mark for selecting option A.

Since there is a strong absorption at \(1740\text{ cm}^{-1}\) (\(\text{C}=\text{O}\) stretch) but no O-H absorption (carboxylic acid or alcohol), X must be an ester. Among the options, only A, B, and C are esters. The mass spectrum shows a strong signal at \(m/z = 57\), which corresponds to the propanoyl ion ([\(\text{CH}_3\text{CH}_2\text{CO}\)]+). This is formed by the cleavage of the alkoxy group from methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)). Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) would show a strong peak at \(m/z = 43\) ([\(\text{CH}_3\text{CO}\)]+), which is very weak here. Thus, X is methyl propanoate.

Award 1 mark for the correct answer (A). No marks are given for incorrect choices.

Use the Arrhenius equation: \(\ln (k_2 / k_1) = (E_a / R) \cdot ((T_2 - T_1) / (T_1 T_2))\). Here, \(k_2 / k_1 = 4.00\), \(T_1 = 300\text{ K}\), and \(T_2 = 320\text{ K}\). Substituting the values: \(\ln(4.00) = (E_a / 8.31) \cdot ((320 - 300) / (300 \cdot 320))\) which simplifies to \(1.3863 = (E_a / 8.31) \cdot (20 / 96000)\). Solving for \(E_a\) yields \(E_a = (1.3863 \cdot 8.31 \cdot 96000) / 20 = 55297\text{ J mol}^{-1} = 55.3\text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer (B). No marks are given for incorrect choices.

(1) Incorrect: Since compound A is a straight-chain alcohol and oxidizes to a carboxylic acid B (\(\text{C}_4\text{H}_8\text{O}_2\)), it must be a primary alcohol (butan-1-ol). A secondary alcohol would oxidize to a ketone. (2) Correct: Dehydration of butan-1-ol with concentrated acid involves a carbocation intermediate that rearranges to yield but-2-ene as the major product C, which has cis-trans isomers. (3) Correct: The reaction of but-2-ene with HBr yields 2-bromobutane as the major product D, which has a chiral carbon at position 2. Therefore, statements (2) and (3) are correct.

Award 1 mark for the correct answer (D). No marks are given for incorrect choices.

Calculate concentrations: \([\text{SO}_2] = 0.40\text{ mol} / 2.0\text{ dm}^3 = 0.20\text{ mol dm}^{-3}\) and \([\text{SO}_3] = 0.80\text{ mol} / 2.0\text{ dm}^3 = 0.40\text{ mol dm}^{-3}\). Using \(K_c = [\text{SO}_3]^2 / ([\text{SO}_2]^2 [\text{O}_2])\), we have \(0.25 = (0.40)^2 / ((0.20)^2 [\text{O}_2]) = 0.16 / (0.04 [\text{O}_2]) = 4 / [\text{O}_2]\). Thus \([\text{O}_2] = 16.0\text{ mol dm}^{-3}\). Number of moles of \(\text{O}_2 = 16.0\text{ mol dm}^{-3} \cdot 2.0\text{ dm}^3 = 32.0\text{ mol}\).

Award 1 mark for the correct answer (D). No marks are given for incorrect choices.

A is incorrect: \(\text{BF}_3\) has 3 single bonds and 0 lone pairs around boron, so its molecular shape is trigonal planar. B is correct: oxygen in \(\text{H}_3\text{O}^+\) has 3 single bonds and 1 lone pair, giving it a trigonal pyramidal shape. C is incorrect: \(\text{NH}_4^+\) has 4 single bonds and 0 lone pairs around nitrogen, making its shape tetrahedral. D is incorrect: \(\text{CO}_2\) is linear.

Award 1 mark for the correct answer (B). No marks are given for incorrect choices.

(1) Correct: The oxidation state of sulfur in \(\text{SO}_2\) is +4, which is oxidized to sulfate (\(\text{SO}_4^{2-}\)) where its oxidation state is +6. (2) Correct: Orange dichromate ion containing chromium(VI) is reduced to green chromium(III) ion (\(\text{Cr}^{3+}\)). (3) Incorrect: Since \(\text{SO}_2\) undergoes oxidation, it acts as a reducing agent, not an oxidizing agent. Therefore, only statements (1) and (2) are correct.

Award 1 mark for the correct answer (B). No marks are given for incorrect choices.

(a) Arrhenius equation: \(k = A e^{-\frac{E_a}{RT}}\) (or \(\ln k = -\frac{E_a}{RT} + \ln A\)), where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, and \(T\) is the absolute temperature (in Kelvin).
(b) Using the formula: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
\(\ln\left(\frac{1.05 \times 10^{-2}}{2.10 \times 10^{-3}}\right) = \frac{E_a}{8.31}\left(\frac{1}{300} - \frac{1}{320}\right)\)
\(\ln(5) = \frac{E_a}{8.31} \times 2.0833 \times 10^{-4}\)
\(1.6094 = 2.507 \times 10^{-5} E_a\)
\(E_a = 64197 \text{ J mol}^{-1} = 64.2 \text{ kJ mol}^{-1}\).
(c) Adding a catalyst decreases the activation energy \(E_a\).

(a) 1 mark for the correct equation. 1 mark for defining all terms correctly (k, A, Ea, T).
(b) 1 mark for correct substitution into the Arrhenius-type ratio equation. 1 mark for the correct calculated value (64.2). 1 mark for correct unit and 3 significant figures.
(c) 1 mark for stating that Ea decreases.

(a)
Initially: \(n(\text{PCl}_5) = 1.00 \text{ mol}\), \(n(\text{PCl}_3) = 0\), \(n(\text{Cl}_2) = 0\)
Change: \(n(\text{PCl}_5) = -0.60 \text{ mol}\), \(n(\text{PCl}_3) = +0.60 \text{ mol}\), \(n(\text{Cl}_2) = +0.60 \text{ mol}\)
At equilibrium: \(n(\text{PCl}_5) = 0.40 \text{ mol}\), \(n(\text{PCl}_3) = 0.60 \text{ mol}\), \(n(\text{Cl}_2) = 0.60 \text{ mol}\)
Equilibrium concentrations in a \(2.0 \text{ dm}^3\) container:
\([\text{PCl}_5]_{eq} = \frac{0.40}{2.0} = 0.20 \text{ M}\)
\([\text{PCl}_3]_{eq} = \frac{0.60}{2.0} = 0.30 \text{ M}\)
\([\text{Cl}_2]_{eq} = \frac{0.60}{2.0} = 0.30 \text{ M}\)
\(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.30 \times 0.30}{0.20} = 0.45 \text{ mol dm}^{-3}\).

(b) Decreasing the volume increases the pressure of the system. According to Le Chatelier's Principle, the system shifts to oppose this change by moving in the direction with fewer moles of gas. Since the reactants side has 1 mole of gas and products side has 2 moles of gas, the equilibrium position shifts to the left.

(a) 1 mark for finding equilibrium moles of PCl3 and Cl2 (0.60 mol). 1 mark for calculating correct equilibrium concentrations of all species. 1 mark for correct Kc expression. 1 mark for correct value and unit (0.45 mol dm^-3).
(b) 1 mark for stating that decreasing volume increases pressure / concentration, causing shift to side with fewer gas moles. 1 mark for concluding that equilibrium shifts to the left.

(a)
Reactants total molar mass = Molar mass of butan-1-ol + Molar mass of \(\text{O}_2\)
\(\text{Total mass} = 74.0 + 32.0 = 106.0 \text{ g mol}^{-1}\)
Desired product = Butanoic acid (\(88.0 \text{ g mol}^{-1}\))
\(\text{Atom economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\% = \frac{88.0}{106.0} \times 100\% = 83.0\%\).

(b) Any two of the following:
1. Oxygen is a green, non-toxic reagent, whereas potassium dichromate is highly toxic, carcinogenic, and generates toxic heavy metal chromium waste.
2. Oxygen can be sourced easily from air, while dichromate is expensive.
3. The use of a catalyst often allows the reaction to run at lower temperatures or pressures, saving energy, and the catalyst can be reused.

(c) High atom economy minimizes the amount of unwanted waste materials generated, reducing waste treatment/disposal costs and raw material costs.

(a) 1 mark for correct formula of atom economy. 1 mark for correct total reactant mass (106.0 g/mol). 1 mark for final percentage (83.0%).
(b) 1 mark for each valid green chemistry advantage (max 2 marks).
(c) 1 mark for stating reduction of waste / environmental protection / raw material cost efficiency.

(a) \(\text{CH}_3\text{CH}_2^+\) or \(\text{CHO}^+\) (must include the '+' sign for the ion).
(b) (i) Carbonyl group / \(\text{C}=\text{O}\) (or aldehyde group).
(ii) Structural formula of Y: \(\text{CH}_3\text{CH}_2\text{CHO}\) (propanal).
Reasoning:
- The absorption peak at \(1715\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C}=\text{O}\)), and the absence of a broad peak at \(3200-3600\text{ cm}^{-1}\) indicates no \(\text{O}-\text{H}\) group, ruling out unsaturated alcohols.
- Since Y can either be propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) or propanone (\(\text{CH}_3\text{COCH}_3\)), the peak at \(m/z = 29\) matches the fragment \(\text{CH}_3\text{CH}_2^+\) or \(\text{CHO}^+\), which can easily fragment from propanal but not easily from propanone (which fragments to give \(\text{CH}_3\text{CO}^+\) at \(m/z = 43\)).
(c) Handle the compound inside a fume cupboard (or keep away from open flames since it is highly flammable).

(a) 1 mark for CH3CH2^+ or CHO^+. Reject formulas without chemical charge.
(b) (i) 1 mark for carbonyl group / C=O.
(ii) 1 mark for drawing/naming propanal correctly. 1 mark for linking IR data (presence of C=O, absence of O-H). 1 mark for linking m/z = 29 to CH3CH2^+ / CHO^+ to rule out propanone.
(c) 1 mark for fume cupboard / keep away from flames.

(a)
*cis*-but-2-ene:
\( \begin{array}{cc} H_3C & CH_3 \\ \backslash & / \\ C & = C \\ / & \backslash \\ H & H \end{array} \)
*trans*-but-2-ene:
\( \begin{array}{cc} H_3C & H \\ \backslash & / \\ C & = C \\ / & \backslash \\ H & CH_3 \end{array} \)
(b) There is restricted rotation about the \(\text{C}=\text{C}\) double bond. In but-2-ene, each carbon atom of the double bond is bonded to two different groups (\(-\text{H}\) and \(-\text{CH}_3\)), allowing two non-interconvertible spatial arrangements. In but-1-ene, one of the carbon atoms of the double bond is bonded to two identical groups (two \(-\text{H}\) atoms), so no such stereoisomers exist.
(c) (i) Cold, dilute, alkaline \(\text{KMnO}_4\text{(aq)}\) solution.
(ii) Yes, because it contains chiral carbon atoms (carbon-2 and carbon-3 are bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH(OH)CH}_3\)), which can form non-superimposable mirror images.

(a) 1 mark for cis-but-2-ene structure with correct label. 1 mark for trans-but-2-ene structure with correct label.
(b) 1 mark for mentioning restricted rotation around the C=C bond. 1 mark for stating that each double-bonded carbon in but-2-ene has two different groups attached, unlike but-1-ene.
(c) (i) 1 mark for cold, dilute, alkaline KMnO4 (reject acidified KMnO4 as it cleaves C=C).
(ii) 1 mark for saying yes and mentioning the presence of chiral carbon(s) / four different groups attached.

(a) \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)
(b)
- Initial moles of \(\text{HCl} = 1.00 \times 0.0500 = 0.0500 \text{ mol}\)
- Moles of \(\text{NaOH}\) used in titration = \(0.200 \times 0.02880 = 5.76 \times 10^{-3} \text{ mol}\)
- Moles of excess \(\text{HCl}\) in \(25.0 \text{ cm}^3\) aliquot = \(5.76 \times 10^{-3} \text{ mol}\) (since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\))
- Moles of excess \(\text{HCl}\) in \(100.0 \text{ cm}^3\) total diluted solution = \(5.76 \times 10^{-3} \times \frac{100.0}{25.0} = 0.02304 \text{ mol}\)
- Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3 = 0.0500 - 0.02304 = 0.02696 \text{ mol}\) (or \(0.0270 \text{ mol}\)).
(c)
- From the equation, 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\).
- Moles of \(\text{CaCO}_3 = \frac{0.02696}{2} = 0.01348 \text{ mol}\)
- Mass of \(\text{CaCO}_3 = 0.01348 \times 100.1 = 1.349 \text{ g}\)
- Percentage of \(\text{CaCO}_3\) in eggshell = \(\frac{1.349}{1.50} \times 100\% = 89.9\%\) (or \(90.0\%\)).

(a) 1 mark for correct equation with state symbols (optional but formulas must be correct).
(b) 1 mark for initial HCl moles (0.0500 mol). 1 mark for NaOH moles (5.76 x 10^-3 mol) and stating it equals HCl in 25 mL. 1 mark for scaling up to 100 mL (0.02304 mol). 1 mark for calculating reacted moles of HCl (0.02696 mol).
(c) 1 mark for calculating mass of CaCO3 and getting final percentage 89.9% or 90.0% (accept correct range depending on rounding).

(a)
Anode (oxidation): \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\)
Cathode (reduction): \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)
(b)
Function: To complete the electrical circuit and maintain electrical neutrality in both half-cells by allowing migration of ions.
Suitable electrolyte: Potassium nitrate solution (\(\text{KNO}_3\text{(aq)}\)) or ammonium nitrate solution (\(\text{NH}_4\text{NO}_3\text{(aq)}\)).
(c) The cell voltage would increase. Magnesium is more reactive / has a higher tendency to lose electrons than zinc (is higher up in the electrochemical series). The difference in reactivity between Mg and Cu is larger than that between Zn and Cu.

(a) 1 mark for correct anode half-equation. 1 mark for correct cathode half-equation.
(b) 1 mark for correct function (maintain electrical neutrality / complete the circuit). 1 mark for suggesting KNO3(aq) / NH4NO3(aq).
(c) 1 mark for stating that cell voltage increases. 1 mark for explaining in terms of larger difference in position in electrochemical series / reactivity between Mg and Cu.

(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 and 1 atm).
(b) Target equation:
\(2\text{C(graphite)} + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\)
Using Hess's Law:
\(\Delta H_f^\theta[\text{C}_2\text{H}_5\text{OH(l)}] = \sum \Delta H_c^\theta[\text{Reactants}] - \sum \Delta H_c^\theta[\text{Products}]\)
\(\Delta H_f^\theta = 2 \times \Delta H_c^\theta[\text{C(graphite)}] + 3 \times \Delta H_c^\theta[\text{H}_2(g)] - \Delta H_c^\theta[\text{C}_2\text{H}_5\text{OH(l)}]\)
\(\Delta H_f^\theta = 2(-393.5) + 3(-285.8) - (-1367.3)\)
\(\Delta H_f^\theta = -787.0 - 857.4 + 1367.3 = -277.1 \text{ kJ mol}^{-1}\).

(a) 1 mark for 'one mole of substance'. 1 mark for 'completely burned in oxygen under standard conditions'.
(b) 1 mark for constructing a correct Hess's Law cycle or algebraic expression. 1 mark for correct substitution. 1 mark for correct calculated value (-277.1). 1 mark for correct sign and unit (kJ mol^-1).

(a)
Since P is a primary haloalkane:
P: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) (1-bromopropane)
Q: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (propan-1-ol)
R: \(\text{CH}_3\text{CH}_2\text{COOH}\) (propanoic acid)
(b) Reagent: \(\text{NaOH(aq)}\) (or \(\text{KOH(aq)}\)). Condition: Heating under reflux.
(c) Equation: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{OH} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O}\)
Type of reaction: Esterification (or Condensation).

(a) 1 mark for each correct structure of P, Q, and R (accept structural formula or condensed formula).
(b) 1 mark for NaOH(aq)/KOH(aq) and heating.
(c) 1 mark for correct equation showing structural formulae of reactants and products. 1 mark for identifying the reaction as esterification / condensation.

(a) propane < dimethyl ether < ethanol (or \(\text{CH}_3\text{CH}_2\text{CH}_3 < \text{CH}_3\text{OCH}_3 < \text{CH}_3\text{CH}_2\text{OH}\))
(b)
- Ethanol molecules can form intermolecular hydrogen bonds because of the highly polar \(\text{O}-\text{H}\) bond.
- Dimethyl ether molecules cannot form hydrogen bonds because they do not have an \(\text{O}-\text{H}\) bond; they can only form weaker dipole-dipole forces (and Van der Waals' forces).
- Since hydrogen bonds are much stronger than dipole-dipole forces, more energy is required to overcome the intermolecular forces in ethanol than in dimethyl ether, leading to a much higher boiling point.
(c)
- Ethanol molecules can form hydrogen bonds with water molecules, making it highly soluble in water.
- Propane is non-polar and can only form weak Van der Waals' forces with water, which are not strong enough to overcome the strong hydrogen bonds between water molecules.

(a) 1 mark for correct order.
(b) 1 mark for noting hydrogen bonds between ethanol molecules. 1 mark for noting only dipole-dipole forces in dimethyl ether. 1 mark for comparing the strength of forces and energy required.
(c) 1 mark for stating that ethanol forms hydrogen bonds with water. 1 mark for explaining that propane is non-polar / forms only weak Van der Waals' forces which cannot break the hydrogen bonds of water.

(a) Initial concentrations:
\[[\text{CO}]_{\text{initial}} = \frac{1.0\text{ mol}}{2.0\text{ dm}^3} = 0.50\text{ mol dm}^{-3}\]
\[[\text{H}_2]_{\text{initial}} = \frac{2.0\text{ mol}}{2.0\text{ dm}^3} = 1.0\text{ mol dm}^{-3}\]
At equilibrium, \([\text{CH}_3\text{OH}]_{\text{eq}} = \frac{0.4\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}\).

Let the change in concentration of \(\text{CH}_3\text{OH}\) be \(+0.20\text{ mol dm}^{-3}\).
By stoichiometry:
\[[\text{CO}]_{\text{eq}} = 0.50 - 0.20 = 0.30\text{ mol dm}^{-3}\]
\[[\text{H}_2]_{\text{eq}} = 1.0 - 2(0.20) = 0.60\text{ mol dm}^{-3}\]

Calculating \(K_c\):
\[K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} = \frac{0.20}{(0.30)(0.60)^2} = 1.85\text{ dm}^6\text{ mol}^{-2}\]

(b) When the volume is halved, the total pressure increases. According to Le Chatelier's Principle, the equilibrium system shifts to the side with fewer moles of gas molecules (the right side, where 3 moles of gaseous reactants become 1 mole of gaseous product) to reduce the pressure. Therefore, the equilibrium yield of \(\text{CH}_3\text{OH}(g)\) increases.

(a)
- Correct calculation of equilibrium concentrations of all three species: \([\text{CO}] = 0.30\text{ M}\), \([\text{H}_2] = 0.60\text{ M}\), \([\text{CH}_3\text{OH}] = 0.20\text{ M}\) (1 mark)
- Correct expression of \(K_c\) (1 mark)
- Correct numerical value of \(K_c = 1.85\) (accept 1.8 or 1.9) (1 mark)
- Correct unit: \(\text{dm}^6\text{ mol}^{-2}\) (1 mark)

(b)
- State that equilibrium yield of \(\text{CH}_3\text{OH}(g)\) increases (1 mark)
- Explain that decreasing volume increases pressure, causing equilibrium to shift to the side with fewer gas moles (1 mark)

(a) Using the Arrhenius equation in two-temperature form:
\[\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]

Substitute the given values:
\[\ln\left(\frac{5.0 \times 10^{-4}}{3.4 \times 10^{-5}}\right) = -\frac{E_a}{8.31} \left(\frac{1}{318} - \frac{1}{298}\right)\]
\[\ln(14.71) = -\frac{E_a}{8.31} (0.003145 - 0.003356)\]
\[2.688 = -\frac{E_a}{8.31} (-2.11 \times 10^{-4})\]
\[2.688 = 2.539 \times 10^{-5} E_a\]
\[E_a = \frac{2.688}{2.539 \times 10^{-5}} = 105868\text{ J mol}^{-1} \approx 106\text{ kJ mol}^{-1}\]

(b) The sketch should include:
1. Y-axis: "Fraction of molecules" (or "Number of molecules with energy E"); X-axis: "Kinetic energy".
2. Curve for \(298\text{ K}\) should have a higher peak positioned more to the left.
3. Curve for \(318\text{ K}\) should have a lower peak shifted to the right and be flatter/broader.
4. A vertical line on the right hand side labeled as \(E_a\). The area under the curve to the right of \(E_a\) represents the fraction of molecules with energy greater than or equal to \(E_a\) (showing a larger area for the \(318\text{ K}\) curve).

(a)
- Correct formula or substitution of values (1 mark)
- Correct calculation of LHS: \(\ln(14.71) = 2.69\) (1 mark)
- Correct calculation of the temperature change term: \(-2.11 \times 10^{-4}\text{ K}^{-1}\) (1 mark)
- Correct value of \(E_a = 106\text{ kJ mol}^{-1}\) (accept 105 to 107, deduction of 1 mark if unit is missing or wrong) (1 mark)

(b)
- Both axes correctly labeled (Y: Fraction/number of molecules, X: Kinetic energy) AND the vertical line of \(E_a\) is correctly placed (1 mark)
- Correct shapes of the two curves showing \(318\text{ K}\) curve shifted to the right, having a lower peak, and crossing the \(298\text{ K}\) curve once (1 mark)

(a) Reagent X is concentrated sulfuric acid (\(\text{conc. H}_2\text{SO}_4\)) (or phosphoric acid). The condition is heating (around \(170^\circ\text{C}\)). [Alternative: passing propan-1-ol vapor over hot aluminium oxide (\(\text{Al}_2\text{O}_3\)) catalyst].

(b) Compound Y is 2-bromopropane (addition of \(\text{HBr}\) to propene follows Markovnikov's rule). Its structural formula is:
\(\text{CH}_3\text{CH(Br)CH}_3\) (or showing all bonds).

(c) Positional isomerism.

(d) Use the iodoform test. Add iodine in sodium hydroxide solution (\(\text{I}_2 / \text{NaOH}(aq)\)) and warm gently.
- Propan-2-ol (a secondary alcohol containing the \(\text{CH}_3\text{CH(OH)}-\) group) will form a yellow precipitate (triiodomethane).
- Propan-1-ol will show no observable change (or no yellow precipitate).

(a)
- Concentrated sulfuric acid / \(\text{conc. H}_2\text{SO}_4\) [Accept: \(\text{conc. H}_3\text{PO}_4\) / \(\text{Al}_2\text{O}_3\)] (1 mark)
- Heat / \(170^\circ\text{C}\) [Accept: hot catalyst] (1 mark)

(b)
- Correct structural formula of 2-bromopropane (1 mark)

(c)
- Positional isomerism (1 mark)

(d)
- Reagents: Iodine and aqueous sodium hydroxide (\(\text{I}_2\) and \(\text{NaOH}(aq)\)) and warm (1 mark)
- Correct observations: Propan-2-ol gives a yellow precipitate; Propan-1-ol gives no precipitate / no observable change (1 mark)

(a) \(\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\)

(b) Moles of \(\text{NaOH}\) used in the titration:
\[n(\text{NaOH}) = 0.02240\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.24 \times 10^{-3}\text{ mol}\]

Since \(\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)\),
the mole ratio of \(\text{HCl}\) to \(\text{NaOH}\) is 1:1.

Therefore, moles of excess \(\text{HCl}\) in the \(25.00\text{ cm}^3\) aliquot:
\[n(\text{HCl})_{\text{aliquot}} = 2.24 \times 10^{-3}\text{ mol}\]

Moles of excess \(\text{HCl}\) in the entire \(250.0\text{ cm}^3\) volumetric flask:
\[n(\text{HCl})_{\text{excess}} = 2.24 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.00} = 2.24 \times 10^{-2}\text{ mol}\ (0.0224\text{ mol})\]

(c) Initial moles of \(\text{HCl}\) added to the eggshell:
\[n(\text{HCl})_{\text{initial}} = 0.05000\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\]

Moles of \(\text{HCl}\) that reacted with \(\text{CaCO}_3\):
\[n(\text{HCl})_{\text{reacted}} = 0.0500 - 0.0224 = 0.0276\text{ mol}\]

From the equation in (a), 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\).
Moles of \(\text{CaCO}_3\) in the sample:
\[n(\text{CaCO}_3) = \frac{0.0276}{2} = 0.0138\text{ mol}\]

Mass of \(\text{CaCO}_3\):
\[\text{Mass} = 0.0138\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.381\text{ g}\]

Percentage by mass of \(\text{CaCO}_3\) in the eggshell:
\[\text{Percentage} = \frac{1.381\text{ g}}{2.00\text{ g}} \times 100\% = 69.1\%\ (\text{or } 69.07\%)\]

(a)
- Correct chemical equation with balanced stoichiometry and state symbols (1 mark)

(b)
- Calculation of moles of \(\text{NaOH}\) and equating to \(\text{HCl}\) in \(25.00\text{ cm}^3\) (1 mark)
- Correct scale up to \(250.0\text{ cm}^3\) to get \(0.0224\text{ mol}\) (1 mark)

(c)
- Correct calculation of reacted \(\text{HCl} = 0.0276\text{ mol}\) (1 mark)
- Correct determination of moles of \(\text{CaCO}_3 = 0.0138\text{ mol}\) (1 mark)
- Correct mass percentage calculation to obtain \(69.1\%\) (accept 69.0% - 69.2%, 3 significant figures preferred) (1 mark)

Part (a)
(i) Using the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
\(\ln\left(\frac{3.82 \times 10^{-3}}{2.45 \times 10^{-4}}\right) = \frac{E_a}{8.31} \left(\frac{1}{500} - \frac{1}{600}\right)\)
\(\ln(15.592) = \frac{E_a}{8.31} \left(3.333 \times 10^{-4}\right)\)
\(2.747 = \frac{E_a}{8.31} \left(3.333 \times 10^{-4}\right)\)
\(E_a = \frac{2.747 \times 8.31}{3.333 \times 10^{-4}} = 68477\text{ J mol}^{-1} = 68.5\text{ kJ mol}^{-1}\)

(ii) In a Maxwell-Boltzmann distribution curve, the x-axis represents kinetic energy and the y-axis represents the fraction of molecules. A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_{a,\text{cat}}\)). This shifts the activation energy threshold to the left on the energy axis. Consequently, a much larger fraction of molecules possess kinetic energy greater than or equal to the lower activation energy, which is represented by a larger shaded area under the curve. The frequency of effective collisions increases, resulting in a higher reaction rate.

Part (b)
(i) The atom economy is 100%. This is an addition reaction where all reactant atoms are incorporated into the single desired product (ethylene carbonate), resulting in zero waste by-products.

(ii) Formula mass of ethylene glycol \(\text{C}_2\text{H}_6\text{O}_2 = 2(12.0) + 6(1.0) + 2(16.0) = 62.0\text{ g/mol}\).
Method A:
Reactants: \(\text{C}_2\text{H}_4\text{O}\) (44.0 g/mol) + \(\text{H}_2\text{O}\) (18.0 g/mol) = 62.0 g/mol.
\(\text{Atom economy} = \frac{62.0}{62.0} \times 100\% = 100\%\).
Method B:
Reactants: \(\text{CH}_2\text{ClCH}_2\text{Cl}\) (99.0 g/mol) + \(2\text{NaOH}\) (80.0 g/mol) = 179.0 g/mol.
\(\text{Atom economy} = \frac{62.0}{179.0} \times 100\% = 34.6\%\).

(iii) Method A has 100% atom economy and produces no waste, avoiding the disposal of large amounts of inorganic waste (\(\text{NaCl}\)) generated in Method B. Furthermore, Method B uses hazardous chlorinated hydrocarbons (1,2-dichloroethane) and corrosive sodium hydroxide, which pose greater environmental and health risks than the reactants used in Method A (water and ethylene oxide, though ethylene oxide is hazardous, the overall process in B is more hazard-intensive).

Part (c)
(i) Increasing the pressure shifts the equilibrium to the side with fewer gas molecules (the right side, where 3 moles of gaseous reactants form 1 mole of gaseous product). Thus, the equilibrium yield of methanol increases.

(ii) The forward reaction is exothermic. According to Le Chatelier's principle, a lower temperature favors a higher equilibrium yield of methanol, but results in a very slow reaction rate. A higher temperature increases the reaction rate but decreases the yield. Therefore, \(250^\circ\text{C}\) is a compromise temperature that achieves a reasonable reaction rate and an acceptable equilibrium yield in industrial production.

Part (a)
(i) Correct expression of Arrhenius equation (1)
Substitution of values (1)
Correct calculation of \(E_a\) in Joules (1)
Correct conversion to \(\text{kJ mol}^{-1}\) (68.4 to 68.6) with unit (1)
(ii) Mentioning catalyst provides alternative pathway with lower \(E_a\) (1)
Explaining the shift of \(E_a\) to the left on the Maxwell-Boltzmann curve (1)
Explaining that the fraction of molecules with \(E \ge E_{a,\text{cat}}\) increases (larger shaded area) (1)
Concluding that the frequency of effective collisions increases (1)

Part (b)
(i) State 100% (1)
Explain that it is an addition reaction / all reactant atoms end up in the desired single product (1)
(ii) Method A: correct calculation showing 100% (2)
Method B: correct total reactant mass calculation (179.0) (1), correct AE calculation showing 34.6% (1)
(iii) Waste management: Method A produces zero waste while Method B produces \(\text{NaCl}\) waste (1)
Hazard prevention: Method B involves highly corrosive strong alkali/toxic chlorinated organics, while Method A avoids these (1)

Part (c)
(i) State equilibrium yield increases (1) and explain by Le Chatelier's principle (fewer gas molecules on RHS) (1)
(ii) Explain from kinetic perspective (higher temperature increases rate) (1) and thermodynamic perspective (exothermic reaction, lower temperature favors yield) (1)

Part (a)
(i) The strong absorption peak at \(1735\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad peak at \(3200 - 3600\text{ cm}^{-1}\) rules out alcohols (\(-\text{OH}\)), and the absence of a broad peak at \(2500 - 3300\text{ cm}^{-1}\) rules out carboxylic acids (\(-\text{COOH}\)). Therefore, the oxygen-containing functional group in X must be an ester.
(ii) The fragment peak at \(m/z = 29\) corresponds to the ethyl cation, \([\text{CH}_3\text{CH}_2]^+\) (or \([\text{C}_2\text{H}_5]^+\)). The fragment peak at \(m/z = 59\) corresponds to the \([\text{CH}_3\text{COO}]^+\) (or \([\text{C}_2\text{H}_3\text{O}_2]^+\)) cation. The combination of these fragments indicates that X is ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)).
(iii) Ethyl ethanoate.
(iv) Add sodium hydrogencarbonate solution (\(\text{NaHCO}_3(aq)\)) to both compounds separately. Butanoic acid will produce effervescence (colorless gas bubbles), while ethyl ethanoate (X) will show no observable change (and forms two immiscible liquid layers).

Part (b)
(i) The mobile phase (an inert carrier gas) carries the vaporized sample through a column containing a stationary phase. Components in the mixture partition differently between the mobile phase and stationary phase based on their volatility and intermolecular forces, resulting in different migration speeds and separation.
(ii) Ethyl ethanoate has a shorter retention time than butanoic acid. Since the stationary phase is non-polar, the separation depends mainly on the boiling points (volatilities) of the compounds. Ethyl ethanoate has a lower boiling point (\(77^\circ\text{C}\)) and is more volatile than butanoic acid (\(164^\circ\text{C}\)), so it passes through the column faster.
(iii) Gas Chromatography can be coupled with Mass Spectrometry. The combined technique is Gas Chromatography-Mass Spectrometry (GC-MS).

Part (c)
(i) \(\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\)
(ii) Moles of \(\text{HCl}\) used in titration = \(0.0950\text{ mol dm}^{-3} \times 0.02240\text{ dm}^3 = 2.128 \times 10^{-3}\text{ mol}\).
From the equation, mole ratio of \(\text{Na}_2\text{CO}_3 : \text{HCl} = 1 : 2\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(25.00\text{ cm}^3\) aliquot = \(\frac{1}{2} \times 2.128 \times 10^{-3}\text{ mol} = 1.064 \times 10^{-3}\text{ mol}\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3\) original solution = \(1.064 \times 10^{-3}\text{ mol} \times 10 = 0.01064\text{ mol}\).
Molar mass of \(\text{Na}_2\text{CO}_3 = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}\).
Mass of \(\text{Na}_2\text{CO}_3\) in the sample = \(0.01064\text{ mol} \times 106.0\text{ g mol}^{-1} = 1.128\text{ g}\).
Percentage by mass of \(\text{Na}_2\text{CO}_3\) = \(\frac{1.128\text{ g}}{2.50\text{ g}} \times 100\% = 45.1\%\).

Part (a)
(i) Identify carbonyl group (\(\text{C}=\text{O}\)) from \(1735\text{ cm}^{-1}\) (1); rule out alcohol by absence of peak at \(3200-3600\text{ cm}^{-1}\) (1); rule out carboxylic acid by absence of peak at \(2500-3300\text{ cm}^{-1}\) (1)
(ii) Identify \(m/z = 29\) as \([\text{CH}_3\text{CH}_2]^+\) / \([\text{C}_2\text{H}_5]^+\) (with positive charge) (1); identify \(m/z = 59\) as \([\text{CH}_3\text{COO}]^+\) (with positive charge) (1); deduce X as ethyl ethanoate (1)
(iii) Ethyl ethanoate (1)
(iv) State reagent (e.g. \(\text{NaHCO}_3(aq)\) or \(\text{Na}_2\text{CO}_3(aq)\)) (1); state correct observations (effervescence for butanoic acid, no bubble/immiscible layers for X) (1)

Part (b)
(i) State partition of components between mobile gas phase and stationary phase (1); explain that separation is due to different partition coefficients / migration rates (1)
(ii) State ethyl ethanoate has shorter retention time (1); explain based on lower boiling point / higher volatility of ethyl ethanoate in non-polar column (1)
(iii) Mass spectrometry (1); Gas Chromatography-Mass Spectrometry / GC-MS (1)

Part (c)
(i) Correct chemical equation with state symbols (not mandatory unless specified, but balanced) (1)
(ii) Calculate moles of \(\text{HCl}\) (1); calculate moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3\) (1); calculate mass of \(\text{Na}_2\text{CO}_3\) (1); calculate correct percentage (45.1%) with 3 significant figures (1)