2025 HKDSE Chemistry Practice Paper with Answers

Statement (1) is correct: The compound contains a carbon-carbon double bond (\( \text{C}=\text{C} \)) and a secondary alcohol group (\( -\text{CH(OH)}- \)), both of which are readily oxidised by acidified \( \text{KMnO}_4\text{(aq)} \), resulting in decolourization from purple to colourless/pale pink.

Statement (2) is incorrect: The hydroxyl group is on a secondary carbon atom (C-4). Oxidation of a secondary alcohol by acidified \( \text{K}_2\text{Cr}_2\text{O}_7\text{(aq)} \) yields a ketone (hex-5-en-2-one), not a carboxylic acid.

Statement (3) is correct: Carbon-4 is a chiral carbon (asymmetric carbon) because it is bonded to four different groups: \( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{CH}_2\text{CH}=\text{CH}_2 \). Hence, the molecule exhibits enantiomerism.

C (1 mark) — (1) and (3) are correct.

According to Hess's Law, the enthalpy change of formation can be calculated from enthalpy changes of combustion:

\( \Delta H^\circ_\text{f}[\text{CH}_3\text{OCH}_3\text{(g)}] = 2 \times \Delta H^\circ_\text{c}[\text{C(graphite)}] + 3 \times \Delta H^\circ_\text{c}[\text{H}_2\text{(g)}] - \Delta H^\circ_\text{c}[\text{CH}_3\text{OCH}_3\text{(g)}] \)

\( \Delta H^\circ_\text{f} = 2(-393.5) + 3(-285.8) - (-1460.0) \)
\( \Delta H^\circ_\text{f} = -787.0 - 857.4 + 1460.0 = -184.4\text{ kJ mol}^{-1} \)

A (1 mark) — \( -184.4\text{ kJ mol}^{-1} \).

The value of the equilibrium constant \( K_\text{c} \) is temperature-dependent and is unaffected by changes in concentration, pressure, volume, or the addition of a catalyst.

Since the forward reaction is exothermic (\( \Delta H < 0 \)), decreasing the temperature shifts the equilibrium position to the right (favouring the exothermic reaction) to release heat, thereby increasing the value of \( K_\text{c} = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} \).

C (1 mark) — Decreasing the temperature of the system.

Zinc is more reactive (a stronger reducing agent) than copper, so zinc undergoes oxidation at the anode (\( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \)), causing the zinc electrode mass to decrease.

Electrons travel through the external circuit from the zinc electrode (anode) to the copper electrode (cathode).

As \( \text{Zn}^{2+} \) ions are generated in Half-cell 1, negative ions (\( \text{NO}_3^- \)) from the salt bridge migrate towards Half-cell 1 (the zinc half-cell) to maintain electrical neutrality. In Half-cell 2, \( \text{Cu}^{2+} \) ions are reduced to copper metal (\( \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \)), not oxidised.

B (1 mark) — Nitrate ions (\( \text{NO}_3^- \)) in the salt bridge migrate towards the beaker containing \( \text{ZnSO}_4\text{(aq)} \).

1. Increasing the temperature shifts the entire Maxwell–Boltzmann distribution curve to higher energy values (the curve broadens and its peak lowers and moves to the right), which increases the fraction (proportion) of molecules having kinetic energy \( E \ge E_\text{a} \). The activation energy \( E_\text{a} \) itself remains unchanged.

2. Adding a catalyst provides an alternative pathway with a lower activation energy, thus altering (decreasing) \( E_\text{a} \).

3. Increasing pressure or concentration at constant temperature increases the total number of particles per unit volume, but does not alter the fraction of molecules exceeding \( E_\text{a} \).

B (1 mark) — Increasing the temperature of the gas.

In Step 1, heating propan-1-ol with concentrated sulfuric acid results in acid-catalysed dehydration to form propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)). In Step 2, addition of hydrogen bromide (\(\text{HBr(g)}\)) across the carbon-carbon double bond of propene follows Markovnikov's rule, yielding 2-bromopropane as the major product.

A (1 mark): Step 1 involves dehydration of propan-1-ol to propene using conc. \(\text{H}_2\text{SO}_4\) and heat; Step 2 involves electrophilic addition of \(\text{HBr}\) to give 2-bromopropane.

The equation for the standard enthalpy change of formation of methoxymethane is:
\(2\text{C(graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OCH}_3\text{(g)}\)

Using Hess's Law with standard enthalpy changes of combustion:
\(\Delta H_f^\circ[\text{CH}_3\text{OCH}_3\text{(g)}] = \sum \Delta H_c^\circ(\text{reactants}) - \sum \Delta H_c^\circ(\text{products})\)
\(\Delta H_f^\circ = 2\Delta H_c^\circ[\text{C(graphite)}] + 3\Delta H_c^\circ[\text{H}_2\text{(g)}] - \Delta H_c^\circ[\text{CH}_3\text{OCH}_3\text{(g)}]\)
\(\Delta H_f^\circ = 2(-393.5) + 3(-285.8) - (-1460.0)\)
\(\Delta H_f^\circ = -787.0 - 857.4 + 1460.0 = -184.4\text{ kJ mol}^{-1}\).

A (1 mark): Correct application of Hess's Law to calculate \(\Delta H_f^\circ = 2(-393.5) + 3(-285.8) - (-1460.0) = -184.4\text{ kJ mol}^{-1}\).

(1) is correct: \(\text{Ag}^+\) has a higher standard reduction potential (is a stronger oxidising agent) than \(\text{Cu}^{2+}\), so \(\text{Ag}^+\) is discharged preferentially at the cathode before \(\text{Cu}^{2+}\).
(2) is correct: At the anode, \(\text{OH}^-\text{(aq)}\) ions from water are oxidised preferentially over \(\text{NO}_3^-\text{(aq)}\) ions to form \(\text{O}_2\text{(g)}\) according to \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\).
(3) is correct: The discharge of \(\text{OH}^-\text{(aq)}\) ions at the anode leaves an excess of \(\text{H}^+\text{(aq)}\) ions in the electrolyte, increasing \([\text{H}^+\text{(aq)}]\) and thus decreasing the pH.

D (1 mark): All three statements (1), (2), and (3) are correct based on the electrochemical series, preferential discharge rules, and changes in ion concentration during electrolysis.

According to the stoichiometry of the reaction \(2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}\):
- Moles of \(\text{N}_2\text{O}_4\) formed at equilibrium = \(0.12\text{ mol}\)
- Moles of \(\text{NO}_2\) consumed = \(2 \times 0.12 = 0.24\text{ mol}\)
- Equilibrium moles of \(\text{NO}_2\) = \(0.40 - 0.24 = 0.16\text{ mol}\)

Equilibrium concentrations in the \(2.0\text{ dm}^3\) container:
- \([\text{NO}_2] = \frac{0.16\text{ mol}}{2.0\text{ dm}^3} = 0.080\text{ mol dm}^{-3}\)
- \([\text{N}_2\text{O}_4] = \frac{0.12\text{ mol}}{2.0\text{ dm}^3} = 0.060\text{ mol dm}^{-3}\)

\(K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} = \frac{0.060}{(0.080)^2} = \frac{0.060}{0.0064} = 9.38\text{ dm}^3\text{ mol}^{-1}\).

B (1 mark): Calculation correctly accounts for the 2:1 mole ratio in consumption of \(\text{NO}_2\) and the container volume \(2.0\text{ dm}^3\) to evaluate \(K_c = \frac{0.060}{(0.080)^2} = 9.38\text{ dm}^3\text{ mol}^{-1}\).

1. Formation of an orange precipitate with 2,4-dinitrophenylhydrazine indicates the presence of a carbonyl group (aldehyde or ketone). This eliminates ethyl ethanoate (ester) and 2-methylpropanoic acid (carboxylic acid).
2. No reaction with Tollens' reagent confirms that \(X\) is a ketone rather than an aldehyde. This eliminates 4-hydroxybutanal.
3. The IR absorption at \(1715\text{ cm}^{-1}\) corresponds to a \(\text{C}=\text{O}\) stretch (ketone), and the broad absorption at \(3200\text{--}3600\text{ cm}^{-1}\) corresponds to an \(\text{O}-\text{H}\) stretch (alcohol). Therefore, \(X\) is 3-hydroxybutan-2-one (\(\text{CH}_3\text{COCH(OH)CH}_3\)).

B (1 mark): 3-hydroxybutan-2-one has a ketone group (giving a positive 2,4-DNP test, negative Tollens' test, and \(1715\text{ cm}^{-1}\) peak) and a secondary alcohol group (giving the \(3200\text{--}3600\text{ cm}^{-1}\) peak).

The equation for the standard enthalpy change of formation of \(\text{CH}_3\text{OCH}_3\text{(g)}\) is:
\[2\text{C(graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OCH}_3\text{(g)}\]
Using Hess's Law with standard enthalpy changes of combustion:
\[\Delta H^\circ_\text{f}[\text{CH}_3\text{OCH}_3\text{(g)}] = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})\]
\[\Delta H^\circ_\text{f} = 2\Delta H^\circ_\text{c}[\text{C(graphite)}] + 3\Delta H^\circ_\text{c}[\text{H}_2\text{(g)}] - \Delta H^\circ_\text{c}[\text{CH}_3\text{OCH}_3\text{(g)}]\]
\[\Delta H^\circ_\text{f} = 2(-393.5) + 3(-285.8) - (-1460.0) = -787.0 - 857.4 + 1460.0 = -184.4\text{ kJ mol}^{-1}\]

A (1 mark)
- Award 1 mark for the correct answer A.
- Incorrect sign gives +184.4 kJ mol⁻¹ (B).
- Omitting stoichiometric coefficients gives incorrect value.

A. Pent-3-en-2-ol, \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH(OH)}-\text{CH}_3\):
- The \(\text{C}=\text{C}\) double bond has two different groups attached to each carbon atom (\(-\text{H}\) and \(-\text{CH}_3\) at C4; \(-\text{H}\) and \(-\text{CH(OH)CH}_3\) at C3), so it exhibits cis-trans isomerism.
- Carbon-2 is bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\)), making it a chiral carbon, so it exhibits enantiomerism.

B. Pent-1-en-3-ol has a terminal \(\text{C}=\text{C}\) bond (two H atoms on C1), so it cannot exhibit cis-trans isomerism.
C. 3-Methylpent-2-ene does not contain any chiral carbon.
D. 3,4-Dichlorohexane has no double bond and cannot exhibit cis-trans isomerism.

A (1 mark)
- Option A contains both a stereogenic double bond (restricted rotation with 2 different groups on each carbon) and an asymmetric/chiral carbon centre.

(1) Correct. The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, lowering the temperature favours the exothermic direction (forward), shifting the equilibrium position to the right.
(2) Correct. Reducing the container volume increases the total pressure. The equilibrium shifts toward the side with fewer moles of gas. The forward side has 2 moles of gas while the reactant side has \(1 + 2 = 3\) moles of gas, so the equilibrium shifts to the right.
(3) Incorrect. Adding an inert gas (Ar) at constant volume increases the total pressure, but does not alter the concentrations or partial pressures of the reacting gases. Therefore, the equilibrium position does not shift.

A (1 mark)
- (1) and (2) only are correct.
- (3) is incorrect because partial pressures of reactants/products remain unchanged at constant volume.

Let the rate equation be \(\text{Rate} = k[\text{W}]^m[\text{X}]^n\).
1. Comparing Exp 1 and Exp 2:
\(\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{4.80 \times 10^{-3}}{1.20 \times 10^{-3}} = 4 = \left(\frac{0.20}{0.10}\right)^m = 2^m \implies m = 2\).
2. Comparing Exp 2 and Exp 3:
\(\frac{\text{Rate}_3}{\text{Rate}_2} = \frac{1.44 \times 10^{-2}}{4.80 \times 10^{-3}} = 3 = \left(\frac{0.30}{0.10}\right)^n = 3^n \implies n = 1\).
3. Calculate the rate constant \(k\):
\(k = \frac{\text{Rate}_1}{[\text{W}]^2[\text{X}]} = \frac{1.20 \times 10^{-3}}{(0.10)^2(0.10)} = 1.20\text{ mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).
4. Calculate the rate for the new concentrations:
\(\text{Rate} = 1.20 \times (0.30)^2 \times (0.20) = 1.20 \times 0.090 \times 0.20 = 2.16 \times 10^{-2}\text{ mol dm}^{-3}\text{ s}^{-1}\).

C (1 mark)
- Deducing 2nd order with respect to W and 1st order with respect to X gives the correct rate equation.
- Calculation gives 2.16 × 10⁻² mol dm⁻³ s⁻¹.

(1) Correct. Acidified \(\text{Cr}_2\text{O}_7^{2-}\) undergoes reduction at the cathode (Half-cell 1): \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\). \(\text{Fe}^{2+}\) undergoes oxidation at the anode (Half-cell 2): \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). Electrons flow from anode (Half-cell 2) to cathode (Half-cell 1) in the external circuit.
(2) Correct. In Half-cell 1, \(\text{H}^+\) ions are consumed during the reduction of \(\text{Cr}_2\text{O}_7^{2-}\), so the concentration of \(\text{H}^+\) decreases, causing the pH of the solution to increase.
(3) Correct. In a functioning cell, cations (\(\text{K}^+\)) in the salt bridge migrate towards the cathode compartment (Half-cell 1) to maintain electrical neutrality as positive charges (\(\text{H}^+\)) are consumed.

D (1 mark)
- (1), (2), and (3) are all chemically correct interpretations of the electrochemical cell operating principles.

1. Addition of $\text{HBr}$ to propene follows Markovnikov's rule, yielding 2-bromopropane as the major product $\mathbf{P}$.
2. Alkaline hydrolysis of 2-bromopropane with $\text{NaOH(aq)}$ produces propan-2-ol, a secondary alcohol ($\mathbf{Q}$).
3. Oxidation of propan-2-ol with acidified $\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}$ yields propanone, a ketone ($\mathbf{R}$).

Evaluating the statements:
- (1) is correct: Ketones (including propanone) undergo condensation reactions with 2,4-dinitrophenylhydrazine to form orange/yellow precipitates.
- (2) is correct: Propanone can be reduced by $\text{NaBH}_4$ back to propan-2-ol, which is a secondary alcohol.
- (3) is incorrect: Propanone lacks a oxidisable carbonyl hydrogen and cannot be oxidised further by acidified $\text{KMnO}_4\text{(aq)}$ under normal conditions, so no decolourisation occurs.

Thus, statements (1) and (2) only are correct.

C (1 mark): Both statements (1) and (2) are correct, while statement (3) is incorrect.

Target reaction for the standard enthalpy change of formation of ethene:
$$2\text{C(graphite)} + 2\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_4\text{(g)}$$

Adding reaction (3) and reaction (4) gives the combustion of ethene:
$$\text{C}_2\text{H}_4\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$$
$$\Delta H_{\text{comb}}[\text{C}_2\text{H}_4\text{(g)}] = \Delta H_3 + \Delta H_4 = (-44.2) + (-1367.3) = -1411.5\text{ kJ mol}^{-1}$$

Using the enthalpy of formation definition:
$$\Delta H_{\text{comb}}[\text{C}_2\text{H}_4\text{(g)}] = 2\Delta H_f^\ominus[\text{CO}_2\text{(g)}] + 2\Delta H_f^\ominus[\text{H}_2\text{O(l)}] - \Delta H_f^\ominus[\text{C}_2\text{H}_4\text{(g)}]$$
$$-1411.5 = 2(-393.5) + 2(-285.8) - \Delta H_f^\ominus[\text{C}_2\text{H}_4\text{(g)}]$$
$$-1411.5 = -787.0 - 571.6 - \Delta H_f^\ominus[\text{C}_2\text{H}_4\text{(g)}]$$
$$-1411.5 = -1358.6 - \Delta H_f^\ominus[\text{C}_2\text{H}_4\text{(g)}]$$
$$\Delta H_f^\ominus[\text{C}_2\text{H}_4\text{(g)}] = -1358.6 + 1411.5 = +52.9\text{ kJ mol}^{-1}$$

B (1 mark): Correct application of Hess's Law yielding $+52.9\text{ kJ mol}^{-1}$.

1. Comparing standard reduction potentials: $\text{MnO}_4^-\text{(aq)}$ is a stronger oxidising agent than $\text{Fe}^{3+}\text{(aq)}$.
2. In Half-cell $\mathbf{Y}$ (cathode / positive electrode), reduction occurs:
$$\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5\text{e}^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)}$$
Since $\text{H}^+$ is consumed, $[\text{H}^+]$ decreases and the pH of solution $\mathbf{Y}$ increases (making C incorrect).
3. In Half-cell $\mathbf{X}$ (anode / negative electrode), oxidation occurs:
$$\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + \text{e}^-$$
As $\text{Fe}^{3+}$ ions are produced, positive charge accumulates in beaker $\mathbf{X}$. To maintain electrical neutrality, anions ($\text{NO}_3^-$) from the salt bridge migrate into beaker $\mathbf{X}$ (making B correct).
4. Electrons flow from the negative electrode (Half-cell $\mathbf{X}$) to the positive electrode (Half-cell $\mathbf{Y}$) in the external circuit (making A and D incorrect).

B (1 mark): Correct deduction of ion migration direction towards the anode half-cell.

(1) is correct: As temperature increases from $T_1$ to $T_2$, average kinetic energy increases, shifting the peak to the right (higher energy) and lowering its height.
(2) is correct: The area under the Maxwell-Boltzmann distribution curve represents the total number of gaseous particles. Since the mass (number of particles) of the gas sample is fixed, the total area remains constant at different temperatures.
(3) is correct: A catalyst provides an alternative pathway with lower activation energy ($E_a$), shifting the $E_a$ line to the left on the kinetic energy axis, thereby increasing the fraction of particles with energy $\ge E_a$, without altering the molecular energy distribution curve itself.

Therefore, (1), (2), and (3) are all correct.

D (1 mark): All three statements (1), (2), and (3) are correct.

1. The equilibrium constant $K_c$ is dependent solely on temperature.
2. The forward reaction is endothermic ($\Delta H > 0$). According to Le Chatelier's principle, increasing the temperature favours the forward endothermic reaction, shifting the position of equilibrium to the right.
3. Consequently, increasing temperature increases both the equilibrium yield of $\text{Cl}_2\text{(g)}$ and the value of $K_c$.
4. Adding more $\text{PCl}_5\text{(g)}$ increases the amount of $\text{Cl}_2\text{(g)}$, but $K_c$ remains unchanged.
5. Decreasing volume shifts the equilibrium to the left (side with fewer gas moles) and does not change $K_c$.
6. Adding a catalyst increases the rate of both forward and reverse reactions equally without affecting equilibrium yield or $K_c$.

C (1 mark): Increasing temperature increases both yield and Kc for an endothermic process.

Compound \(\mathbf{W}\) gives a positive test with 2,4-dinitrophenylhydrazine (indicating a carbonyl group, either aldehyde or ketone) and also forms a silver mirror with Tollens' reagent, confirming that \(\mathbf{W}\) is an aldehyde. Since oxidation of \(\mathbf{W}\) gives butanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)), \(\mathbf{W}\) must be butanal (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)).

When butanoic acid reacts with ethanol in the presence of concentrated sulfuric acid under reflux, an esterification reaction takes place to produce ethyl butanoate (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_3\)).

Therefore, Compound \(\mathbf{W}\) is butanal and Compound \(\mathbf{Y}\) is ethyl butanoate.

B (1 mark)
- Identification of \(\mathbf{W}\) as butanal from Tollens' reagent reaction.
- Correct naming of ester \(\mathbf{Y}\) formed from butanoic acid and ethanol as ethyl butanoate.

The equation representing the standard enthalpy change of formation of liquid ethanol is:
$$2\text{C(graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{OH(l)}$$

Using Hess's Law:
$$\Delta H_\text{f}^\ominus [\text{CH}_3\text{CH}_2\text{OH(l)}] = 2\Delta H_\text{c}^\ominus[\text{C(graphite)}] + 3\Delta H_\text{c}^\ominus[\text{H}_2\text{(g)}] - \Delta H_\text{c}^\ominus[\text{CH}_3\text{CH}_2\text{OH(l)}]$$

$$\Delta H_\text{f}^\ominus = 2(-393.5) + 3(-285.8) - (-1367.3)$$
$$\Delta H_\text{f}^\ominus = -787.0 - 857.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}$$

A (1 mark)
- Full calculation: \(2(-393.5) + 3(-285.8) - (-1367.3) = -277.1\text{ kJ mol}^{-1}\).

In the electrolysis of concentrated (\(1.0\text{ M}\)) \(\text{CuCl}_2\text{(aq)}\) with graphite (inert) electrodes:
- At the cathode: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). A reddish-brown layer of copper solid is deposited on the cathode. Hence, (1) is correct.
- At the anode: \(2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\). Chlorine gas is evolved, which bleaches moist litmus paper (turns red then white). Hence, (2) is correct.
- The overall reaction is \(\text{CuCl}_2\text{(aq)} \rightarrow \text{Cu(s)} + \text{Cl}_2\text{(g)}\). Neither \(\text{H}^+\) nor \(\text{OH}^-\) is produced or consumed in significant amounts, so the pH of the solution remains relatively neutral/stable rather than decreasing significantly. Hence, (3) is incorrect.

B (1 mark)
- (1) is correct: \(\text{Cu}^{2+}\) is preferentially discharged at the cathode forming copper metal.
- (2) is correct: \(\text{Cl}^-\) is preferentially discharged due to its relatively high concentration, giving \(\text{Cl}_2\) which bleaches litmus.
- (3) is incorrect: The bulk solution does not generate significant \(\text{H}^+\) ions.

The forward reaction is exothermic (\(\Delta H < 0\)).
According to Le Chatelier's Principle:
- When the temperature is increased, the equilibrium position shifts in the endothermic direction (to the left/backward) to absorb the added thermal energy.
- As the system shifts to the left, \([\text{SO}_3\text{(g)}]\) decreases, while \([\text{SO}_2\text{(g)}]\) and \([\text{O}_2\text{(g)}]\) increase.
- The equilibrium constant \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\) decreases as temperature increases for an exothermic forward reaction.
- The total number of gas moles increases (shifting from 2 moles of gas on the right to 3 moles on the left), so the total pressure in the fixed-volume container increases.

C (1 mark)
- Increasing temperature for an exothermic reaction shifts the equilibrium to the left, decreasing \(K_c\) and decreasing the concentration of \(\text{SO}_3\text{(g)}\).

In Experiment \(\mathbf{P}\):
- \(n(\text{HCl}) = 0.0500 \times 0.40 = 0.020\text{ mol}\).
- Since zinc is in excess, \(\text{HCl}\) is the limiting reactant. \(n(\text{H}_2) = \frac{1}{2} n(\text{HCl}) = 0.010\text{ mol}\).

For Curve \(\mathbf{Q}\):
1. Initial rate is HIGHER \(\rightarrow\) requires higher concentration of \(\text{H}^+\text{(aq)}\) (and/or higher temperature, or zinc powder instead of granulated zinc).
2. Total volume of \(\text{H}_2\) is exactly HALF that of \(\mathbf{P}\) \(\rightarrow\) requires \(n(\text{H}^+) = 0.010\text{ mol}\) (so that \(n(\text{H}_2) = 0.0050\text{ mol}\)).

Let us analyze the options:
- Option A: \(25.0\text{ cm}^3\) of \(0.20\text{ M HCl}\) \(\rightarrow\) \(n(\text{H}^+) = 0.0050\text{ mol}\) (gives 1/4 volume, lower rate).
- Option B: \(50.0\text{ cm}^3\) of \(0.20\text{ M HCl}\) \(\rightarrow\) \(n(\text{H}^+) = 0.010\text{ mol}\) (gives 1/2 volume, but \([\text{H}^+] = 0.20\text{ M} < 0.40\text{ M}\), so lower initial rate).
- Option C: \(100.0\text{ cm}^3\) of \(0.10\text{ M HCl}\) \(\rightarrow\) \(n(\text{H}^+) = 0.010\text{ mol}\) (gives 1/2 volume, but lower initial rate).
- Option D: \(12.5\text{ cm}^3\) of \(0.80\text{ M HCl}\) \(\rightarrow\) \(n(\text{H}^+) = 0.0125 \times 0.80 = 0.010\text{ mol}\) (gives exactly half volume) AND \([\text{HCl}] = 0.80\text{ M} > 0.40\text{ M}\) (gives a higher initial rate).

D (1 mark)
- Calculation of limiting moles of \(\text{H}^+\): \(12.5\text{ cm}^3 \times 0.80\text{ M} = 0.010\text{ mol}\), which is half of the original \(0.020\text{ mol}\).
- Higher concentration (\(0.80\text{ M} > 0.40\text{ M}\)) ensures a faster initial reaction rate.

The formation equation for ethanoic acid is:
\[2\text{C(graphite)} + 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{COOH(l)}\]

Using Hess's Law and standard enthalpy changes of combustion:
\[\Delta H^\circ_\text{f}[\text{CH}_3\text{COOH(l)}] = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})\]
\[\Delta H^\circ_\text{f} = 2\,\Delta H^\circ_\text{c}[\text{C}] + 2\,\Delta H^\circ_\text{c}[\text{H}_2] - \Delta H^\circ_\text{c}[\text{CH}_3\text{COOH}]\]
\[\Delta H^\circ_\text{f} = 2(-394) + 2(-286) - (-874)\]
\[\Delta H^\circ_\text{f} = -788 - 572 + 874 = -486\text{ kJ mol}^{-1}\]

Award [1] for option A.
- Option B incorrectly omits stoichiometric coefficients.
- Option C has incorrect arithmetic sign.
- Option D incorrectly omits the enthalpy of combustion of ethanoic acid.

1. Formation of an orange precipitate with 2,4-dinitrophenylhydrazine confirms that compound \(\text{X}\) contains a carbonyl group (aldehyde or ketone). Statement (1) is correct.
2. Since \(\text{X}\) does not react with (does not reduce) acidified potassium dichromate solution, it cannot be an aldehyde. Therefore, \(\text{X}\) is a ketone. The only four-carbon ketone is butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)). Statement (2) is correct.
3. Reduction of butan-2-one using \(\text{NaBH}_4\) yields butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)). Carbon-2 in butan-2-ol is bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\)), making it a chiral molecule. Statement (3) is correct.

Award [1] for option D (all statements (1), (2), and (3) are correct).

(1) The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's Principle, decreasing the temperature shifts the equilibrium position to the right, increasing the number of moles of \(\text{C(g)}\). (Correct)
(2) Adding an inert gas like argon at constant volume does not alter the partial pressures or concentrations of \(\text{A}\), \(\text{B}\), or \(\text{C}\). Thus, the equilibrium position remains unchanged. (Incorrect)
(3) Decreasing the volume increases the total pressure of the system. The system responds by shifting towards the side with fewer moles of gas (reactants: 3 moles of gas; products: 2 moles of gas). Thus, the equilibrium shifts to the right, increasing the number of moles of \(\text{C(g)}\). (Correct)

Award [1] for option C ((1) and (3) only).

In both cells:
- Cathode reaction: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). Copper metal is deposited on the cathode in both cells, so the mass of the cathode increases in both Cell 1 and Cell 2 (Option C is correct).

In Cell 1 (inert Pt anode):
- Anode reaction: \(2\text{H}_2\text{O(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^-\). \(\text{H}^+\) ions are produced, so the pH decreases (not increases; Option A is incorrect).
- \(\text{Cu}^{2+}\) is consumed without being replenished, so the blue colour fades.

In Cell 2 (active Cu anode):
- Anode reaction: \(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\). The rate of dissolution of copper at the anode equals the rate of copper deposition at the cathode. The concentration of \(\text{Cu}^{2+}\) remains constant, so the blue colour does not fade (Option B is incorrect).
- Oxygen is only produced in Cell 1, not Cell 2 (Option D is incorrect).

Award [1] for option C.

1. The IR spectrum of \(\text{P}\) shows a broad band at \(2500-3300\text{ cm}^{-1}\) (O–H stretch of carboxylic acid) and \(1715\text{ cm}^{-1}\) (C=O stretch). With formula \(\text{C}_3\text{H}_6\text{O}_2\), \(\text{P}\) is propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)). Propanoic acid is acidic and turns moist blue litmus paper red. Statement (1) is correct.
2. Compound \(\text{Q}\) hydrolyses under alkaline conditions to form an alcohol and a carboxylate salt, confirming that \(\text{Q}\) is an ester. The isomeric esters with molecular formula \(\text{C}_3\text{H}_6\text{O}_2\) are ethyl methanoate (\(\text{HCOOCH}_2\text{CH}_3\)) and methyl ethanoate (\(\text{CH}_3\text{COOCH}_3\)). Statement (2) is correct.
3. Carboxylic acids (Compound \(\text{P}\)) react with \(\text{NaHCO}_3\text{(aq)}\) in an acid-carbonate reaction to evolve \(\text{CO}_2\text{(g)}\) bubbles. Esters (Compound \(\text{Q}\)) do not react with \(\text{NaHCO}_3\text{(aq)}\). Statement (3) is correct.

Award [1] for option D (all statements (1), (2), and (3) are correct).

Working backwards from the end product:
1. Hydrolysis of nitrile (Compound Z) by heating with dilute acid gives a carboxylic acid containing the same number of carbons. Since the product is butanoic acid (4 carbons), Compound Z must be butanenitrile ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}$).
2. Nucleophilic substitution of a haloalkane (Compound Y) using ethanolic $\text{KCN}$ introduces a nitrile group ($-\text{CN}$), lengthening the carbon chain by one carbon. Thus, Compound Y must be 1-chloropropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$, 3 carbons).
3. Reaction of an alcohol (Compound X) with $\text{PCl}_3$ replaces the hydroxyl group with a chlorine atom. Therefore, Compound X must be propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$).

A (1 mark): Correctly identifies propan-1-ol through backward step-by-step carbon chain analysis.

(1) is incorrect: Adding an inert gas at constant volume increases total pressure, but does not change the partial pressures or concentrations of the reacting species. Thus, the equilibrium position does not shift.
(2) is correct: Increasing the temperature increases the average kinetic energy of all reacting molecules, increasing the collision frequency and the fraction of molecules having energy equal to or greater than the activation energy. Hence, the rates of both forward and reverse reactions increase.
(3) is correct: The forward reaction is exothermic ($\Delta H < 0$). By Le Chatelier's principle, lowering the temperature shifts the equilibrium to the forward side, increasing $[\text{N}_2\text{O}_4]$ and decreasing $[\text{NO}_2]$. Since $K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}$, the value of $K_c$ increases.

D (1 mark): Recognises that inert gas at constant volume has no effect on equilibrium position, while temperature elevation increases reaction rate and exothermic reactions have larger $K_c$ at lower temperatures.

- Iron half-cell has a more negative standard reduction potential ($E^\circ = -0.44\text{ V}$), acting as the anode (negative terminal) where oxidation occurs: $\text{Fe(s)} \to \text{Fe}^{2+}\text{(aq)} + 2\text{e}^-$.
- Graphite half-cell acts as the cathode (positive terminal) where reduction occurs: $\text{Fe}^{3+}\text{(aq)} + \text{e}^- \to \text{Fe}^{2+}\text{(aq)}$.
- A is incorrect: Electrons flow from the anode (iron electrode) to the cathode (graphite electrode) via the external wire.
- B is incorrect: Reduction in half-cell 2 produces soluble $\text{Fe}^{2+}\text{(aq)}$ ions; no solid metal precipitates onto the graphite electrode, so its mass remains unchanged.
- C is correct: In Half-cell 1, $\text{Fe}^{2+}$ ions are continuously generated (positive charge build-up). To maintain electrical neutrality, anions (such as $\text{NO}_3^-$) from the salt bridge migrate into Half-cell 1.
- D is incorrect: In Half-cell 2, $\text{Fe}^{3+}$ is reduced to $\text{Fe}^{2+}$, so the concentration of $\text{Fe}^{2+}\text{(aq)}$ increases.

C (1 mark): Correctly identifies the anode half-cell and the direction of ion movement in the salt bridge.

The formation equation of propane is:
$$3\text{C(graphite)} + 4\text{H}_2\text{(g)} \to \text{C}_3\text{H}_8\text{(g)}$$
Using Hess's Law via enthalpy of combustion values:
$$\Delta H_\text{f}^\circ[\text{C}_3\text{H}_8\text{(g)}] = \sum \Delta H_\text{c}^\circ(\text{reactants}) - \sum \Delta H_\text{c}^\circ(\text{products})$$
$$\Delta H_\text{f}^\circ = 3\Delta H_\text{c}^\circ[\text{C(graphite)}] + 4\Delta H_\text{c}^\circ[\text{H}_2\text{(g)}] - \Delta H_\text{c}^\circ[\text{C}_3\text{H}_8\text{(g)}]$$
$$\Delta H_\text{f}^\circ = 3(-393.5) + 4(-285.8) - (-2220.0)$$
$$\Delta H_\text{f}^\circ = -1180.5 - 1143.2 + 2220.0 = -2323.7 + 2220.0 = -103.7\text{ kJ mol}^{-1}$$
Thus, the correct answer is A.

A (1 mark): Correctly applies Hess's Law using combustion data: $[3(-393.5) + 4(-285.8) - (-2220.0)] = -103.7\text{ kJ mol}^{-1}$.

Since $\text{CaCO}_3$ is in excess, $\text{HCl}$ is the limiting reactant. The number of moles of $\text{HCl} = 0.050\text{ dm}^3 \times 1.0\text{ M} = 0.050\text{ mol}$, which determines the maximum volume of $\text{CO}_2$ evolved.

- (1) Using powdered $\text{CaCO}_3$ increases the surface area in contact with the acid, thereby increasing collision frequency and initial rate. The amount of limiting reagent ($\text{HCl}$) is unchanged ($0.050\text{ mol}$), yielding the same final volume of gas.
- (2) Using $2.0\text{ M HCl}$ increases acid concentration, resulting in more frequent collisions and a higher initial rate. Moles of $\text{HCl} = 0.025\text{ dm}^3 \times 2.0\text{ M} = 0.050\text{ mol}$, yielding the exact same final volume of gas.
- (3) Raising temperature increases the average kinetic energy of the particles, leading to a higher fraction of collisions with $E \ge E_a$ and a greater collision frequency, hence increasing initial rate. The amount of limiting reagent is identical ($0.050\text{ mol}$), producing the same total volume of gas.

Therefore, (1), (2), and (3) are all correct.

D (1 mark): Identifies that surface area, concentration, and temperature all accelerate initial rate while keeping the limiting reagent quantity constant.

1. Reaction with \( \text{NaHCO}_3\text{(aq)} \): A colourless gas (\( \text{CO}_2 \)) is evolved, indicating that \( X \) contains a carboxyl group (\( -\text{COOH} \)). All four options possess a \( -\text{COOH} \) group.
2. Reaction with acidified \( \text{K}_2\text{Cr}_2\text{O}_7\text{(aq)} \): The solution turns orange to green, indicating that \( X \) can be oxidized (i.e. it possesses a primary or secondary alcohol group). Option C, \( (\text{CH}_3)_2\text{C(OH)COOH} \), contains a tertiary alcohol which cannot be readily oxidized, so C is eliminated.
3. Stereochemistry (no chiral carbon):
- In Option A (\( \text{HOCH}_2\text{CH}_2\text{CH}_2\text{COOH} \)), none of the carbon atoms are attached to four different groups. Thus, it is achiral and has no enantiomers.
- In Option B (\( \text{CH}_3\text{CH(OH)CH}_2\text{COOH} \)), C-3 is bonded to \( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{CH}_2\text{COOH} \) (chiral).
- In Option D (\( \text{HOCH}_2\text{CH(CH}_3)\text{COOH} \)), C-2 is bonded to \( -\text{H} \), \( -\text{CH}_3 \), \( -\text{CH}_2\text{OH} \), and \( -\text{COOH} \) (chiral).

Hence, \( X \) must be \( \text{HOCH}_2\text{CH}_2\text{CH}_2\text{COOH} \).

A (1 mark)
- Option A correctly satisfies all three criteria: carboxylic acid function, oxidizable primary alcohol group, and absence of an asymmetric (chiral) carbon center.

(a) Combustion equation:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l) + \frac{9}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)\) (or \(2\text{C}_3\text{H}_7\text{OH} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 8\text{H}_2\text{O}\))

(b)(i)
1. Temperature change: \(\Delta T = 54.8 - 21.4 = 33.4\ ^\circ\text{C}\)
2. Heat absorbed by water: \(q = mc\Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 33.4\text{ K} = 20941.8\text{ J} = 20.94\text{ kJ}\)
3. Number of moles of propan-1-ol burned: \(n = \frac{0.750\text{ g}}{60.1\text{ g mol}^{-1}} = 0.01248\text{ mol}\)
4. Enthalpy change of combustion: \(\Delta H_c = -\frac{q}{n} = -\frac{20.9418\text{ kJ}}{0.01248\text{ mol}} = -1678\text{ kJ mol}^{-1} \approx -1680\text{ kJ mol}^{-1}\) (to 3 sig. fig.)

(b)(ii)
Any TWO of the following:
- Heat loss to the surroundings / beaker.
- Incomplete combustion of propan-1-ol (forming carbon monoxide / soot).
- Evaporation of propan-1-ol from the wick without burning.
- Heat capacity of the copper beaker was not accounted for.

(a) Correct balanced equation with correct state symbols or without state symbols [1 mark].

(b)(i)
- Correct calculation of heat energy \(q = 20.94\text{ kJ}\) [1* mark].
- Correct calculation of moles of propan-1-ol \(n = 0.01248\text{ mol}\) [1* mark].
- Correct value and negative sign for \(\Delta H_c = -1680\text{ kJ mol}^{-1}\) (accept \(-1678\) to \(-1680\text{ kJ mol}^{-1}\)) [1 mark].

(b)(ii)
- Any TWO valid reasons (1 mark for each point, max 2 marks):
1. Heat lost to the surroundings / beaker / calorimeter [1 mark].
2. Incomplete combustion of propan-1-ol [1 mark].
3. Evaporation of propan-1-ol from the burner [1 mark].

(a)(i) Compound \(\mathbf{X}\) is 2-hydroxypropanenitrile: \(\text{CH}_3\text{CH(OH)CN}\).

(a)(ii) \(\text{NaCN}\) acts as a catalyst / provides cyanide ions (\(\text{CN}^-\)) to increase the concentration of the nucleophile.

(b) Hydrolysis (or acid-catalysed hydrolysis).

(c)(i) The central carbon atom (C-2) is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{COOH}\), making it a chiral centre (asymmetric carbon atom).

(c)(ii) Three-dimensional representations:
Enantiomer 1:
Central C with:
- \(-\text{COOH}\) (in-plane)
- \(-\text{CH}_3\) (in-plane)
- \(-\text{OH}\) (wedged / pointing forward)
- \(-\text{H}\) (dashed / pointing backward)

Enantiomer 2: Non-superimposable mirror image of Enantiomer 1.

(a)(i) Correct structural formula of \(\text{CH}_3\text{CH(OH)CN}\) [1 mark].

(a)(ii) Catalyst / nucleophile source / generates \(\text{CN}^-\) [1 mark].

(b) Hydrolysis / acid hydrolysis [1 mark] (Reject: nucleophilic substitution).

(c)(i) Presence of a chiral carbon / asymmetric carbon bonded to four different groups / atoms [1 mark].

(c)(ii)
- Correct 3D tetrahedral representation with wedge and dash bonds for one enantiomer [1 mark].
- Non-superimposable mirror image of the first enantiomer [1 mark].

(a)
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\)
Units: \(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \text{mol}^{-1}\text{ dm}^3\) (or \(\text{dm}^3\text{ mol}^{-1}\))

(b)
Initial moles:
\(n(\text{SO}_2) = 0.800\text{ mol}\)
\(n(\text{O}_2) = 0.600\text{ mol}\)
\(n(\text{SO}_3) = 0\text{ mol}\)

At equilibrium, \(n(\text{SO}_3) = 0.500\text{ mol}\).
Change in moles:
\(\Delta n(\text{SO}_3) = +0.500\text{ mol}\)
\(\Delta n(\text{SO}_2) = -0.500\text{ mol}\)
\(\Delta n(\text{O}_2) = -\frac{1}{2}(0.500) = -0.250\text{ mol}\)

Equilibrium moles:
\(n(\text{SO}_2) = 0.800 - 0.500 = 0.300\text{ mol}\)
\(n(\text{O}_2) = 0.600 - 0.250 = 0.350\text{ mol}\)
\(n(\text{SO}_3) = 0.500\text{ mol}\)

Equilibrium concentrations (in a \(2.00\text{ dm}^3\) vessel):
\([\text{SO}_2] = \frac{0.300}{2.00} = 0.150\text{ mol dm}^{-3}\)
\([\text{O}_2] = \frac{0.350}{2.00} = 0.175\text{ mol dm}^{-3}\)
\([\text{SO}_3] = \frac{0.500}{2.00} = 0.250\text{ mol dm}^{-3}\)

\(K_c = \frac{(0.250)^2}{(0.150)^2 \times (0.175)} = \frac{0.0625}{0.0225 \times 0.175} = \frac{0.0625}{0.0039375} \approx 15.87...\)
Wait, recalculating:
\(0.0625 / 0.0039375 = 15.87\text{ mol}^{-1}\text{ dm}^3\).

(c) The yield of \(\text{SO}_3(g)\) decreases. The forward reaction is exothermic (\(\Delta H < 0\)). By Le Chatelier's principle, an increase in temperature favours the endothermic backward reaction to absorb the added thermal energy, thus shifting the equilibrium position to the left.

(a) Correct expression for \(K_c\) AND correct units \(\text{mol}^{-1}\text{ dm}^3\) / \(\text{dm}^3\text{ mol}^{-1}\) [1 mark].

(b)
- Deducing correct equilibrium moles: \(n(\text{SO}_2) = 0.300\text{ mol}\) and \(n(\text{O}_2) = 0.350\text{ mol}\) [1* mark].
- Calculating correct equilibrium concentrations by dividing by volume \(2.00\text{ dm}^3\) [1* mark].
- Correct final value of \(K_c = 15.9\text{ mol}^{-1}\text{ dm}^3\) (accept 15.8 to 15.9) [1 mark].

(c)
- Stating that the yield decreases [1 mark].
- Explaining that the forward reaction is exothermic, so increasing temperature shifts equilibrium to the left / endothermic direction [1 mark].

(a)(i) Electrons flow from the zinc electrode (anode/negative pole) to the silver electrode (cathode/positive pole) through the external wire.

(a)(ii) Completes the circuit by allowing migration of ions, and maintains electrical neutrality in both half-cells.

(a)(iii) Overall ionic equation:
\(\text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)\)

(b)(i) Reaction at anode (oxidation of water/hydroxide ions):
\(2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\) (or \(4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^-\))

(b)(ii)
1. Total charge passed: \(Q = I \times t = 2.50\text{ A} \times (40.0 \times 60\text{ s}) = 6000\text{ C}\)
2. Moles of electrons passed: \(n(e^-) = \frac{Q}{F} = \frac{6000\text{ C}}{96500\text{ C mol}^{-1}} = 0.06218\text{ mol}\)
3. Cathode reaction: \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\)
Moles of \(\text{Cu}\) deposited: \(n(\text{Cu}) = \frac{n(e^-)}{2} = \frac{0.06218}{2} = 0.03109\text{ mol}\)
4. Mass of \(\text{Cu}\) deposited: \(m = n \times M = 0.03109\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.974\text{ g} \approx 1.97\text{ g}\) (to 3 sig. fig.)

(a)(i) From zinc (electrode) to silver (electrode) [1 mark].

(a)(ii) Complete the circuit / allow ion movement to maintain electrical neutrality [1 mark].

(a)(iii) \(\text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)\) [1 mark].

(b)(i) \(2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\) (or equivalent) [1 mark].

(b)(ii)
- Calculating total charge \(Q = 6000\text{ C}\) and moles of electrons \(n(e^-) = 0.06218\text{ mol}\) [1* mark].
- Calculating mass of \(\text{Cu} = 1.97\text{ g}\) (accept \(1.97\) to \(1.98\text{ g}\)) [1 mark].

(a) The formation of an orange precipitate with 2,4-dinitrophenylhydrazine indicates a carbonyl group (\(\text{C=O}\)). The negative Tollens' test confirms that it is NOT an aldehyde. Hence, \(\mathbf{Y}\) contains a ketone group.

(b) Yes. The strong peak at \(1715\text{ cm}^{-1}\) corresponds to \(\text{C=O}\) bond absorption in ketones, and the absence of broad absorption in \(2500\text{--}3600\text{ cm}^{-1}\) confirms the absence of \(\text{O-H}\) stretching (which would belong to an alcohol or carboxylic acid).

(c)(i)
Empirical formula mass of \(\text{C}_4\text{H}_8\text{O} = 4(12.0) + 8(1.0) + 16.0 = 72.0\).
Since relative molecular mass is \(72.0\), molecular formula = \(\text{C}_4\text{H}_8\text{O}\).
Since it is a ketone with 4 carbons, its structure is butanone: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) (or \(\text{CH}_3\text{C(=O)CH}_2\text{CH}_3\)).

(c)(ii)
Reduction of butanone with \(\text{NaBH}_4\) gives butan-2-ol: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
Compound \(\mathbf{Z}\) exhibits enantiomerism because C-2 is bonded to 4 different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\)), making it chiral.

(a) Ketone / keto group [1 mark] (Reject: carbonyl alone).

(b) Yes, \(1715\text{ cm}^{-1}\) corresponds to \(\text{C=O}\) stretching, and no absorption at \(2500\text{--}3600\text{ cm}^{-1}\) indicates absence of \(\text{O-H}\) group [1 mark].

(c)(i)
- Molecular formula: \(\text{C}_4\text{H}_8\text{O}\) [1 mark].
- Structural formula: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) / butan-2-one [1 mark].

(c)(ii)
- Structural formula of \(\mathbf{Z}\): \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) [1 mark].
- States that \(\mathbf{Z}\) exhibits enantiomerism AND explains that it has a chiral carbon / asymmetric carbon (bonded to 4 different groups) [1 mark].

(a) \(\mathbf{X}\) (\(\text{C}_4\text{H}_8\text{O}\)) does not react with acidified potassium dichromate, meaning it is not an aldehyde or primary/secondary alcohol; being reduced by \(\text{NaBH}_4\) indicates it is a ketone. The only 4-carbon ketone is butan-2-one (butanone), \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

(b) Reduction of butan-2-one with \(\text{NaBH}_4\) yields the secondary alcohol butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).

(c) Dehydration of butan-2-ol gives but-2-ene (which exists as cis-but-2-ene and trans-but-2-ene) and but-1-ene. \(\mathbf{P}\) and \(\mathbf{Q}\) are cis-trans isomers (geometric isomers).
- cis-but-2-ene: both methyl groups on the same side of the \(\text{C=C}\) double bond.
- trans-but-2-ene: methyl groups on opposite sides of the \(\text{C=C}\) double bond.

(d) \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{H}_2\text{O}\)

(a) Structural formula of \(\text{CH}_3\text{COCH}_2\text{CH}_3\) [1]
Systematic name: butan-2-one / butanone [1]

(b) Structural formula of \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) [1]

(c) Cis-trans isomerism / geometric isomerism [1]
Correct 3D drawings of both cis-but-2-ene and trans-but-2-ene showing correct geometry around \(\text{C=C}\) [1]

(d) \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{H}_2\text{O}\) (or \(\text{C}_4\text{H}_{10}\text{O} \rightarrow \text{C}_4\text{H}_8 + \text{H}_2\text{O}\)) [1]

(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 (\(298\,\text{K}\) and \(1\,\text{atm}\)).

(b) The formation reaction is:
\(\text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \xrightarrow{\Delta H_f^\ominus} \text{CH}_3\text{OH}(\text{l})\)
Combustion products for both reactants and product are \(\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\).
Cycle: \(\Delta H_f^\ominus[\text{CH}_3\text{OH}(\text{l})] = \Delta H_c^\ominus[\text{C}(\text{graphite})] + 2\Delta H_c^\ominus[\text{H}_2(\text{g})] - \Delta H_c^\ominus[\text{CH}_3\text{OH}(\text{l})]\).

(c) Calculation:
\(\Delta H_f^\ominus = (-393.5) + 2(-285.8) - (-726.0)\)
\(\Delta H_f^\ominus = -393.5 - 571.6 + 726.0 = -239.1\,\text{kJ}\,\text{mol}^{-1}\).

(a) Enthalpy change when 1 mole of a substance [1]
is completely burnt in oxygen under standard conditions (\(298\,\text{K}\), \(1\,\text{atm}\)) [1]

(b) Correctly drawn cycle showing reactants \(\text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\), product \(\text{CH}_3\text{OH}(\text{l})\), and combustion products \(\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\) [1]
Correct arrow directions with labels [1]

(c) \(\Delta H_f^\ominus = \Delta H_c^\ominus[\text{C}] + 2\Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{CH}_3\text{OH}]\) (Method mark) [1*]
\(= -393.5 + 2(-285.8) - (-726.0) = -239.1\,\text{kJ}\,\text{mol}^{-1}\) (Unit required, 3-4 sig. figs.) [1]

(a) Aluminium is more reactive (higher in the electrochemical series) than silver, so aluminium has a stronger tendency to lose electrons (undergo oxidation) and acts as the anode (negative electrode).

(b) At the cathode, reduction of silver ions occurs:
\(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})\)

(c) Overall ionic equation:
\(\text{Al}(\text{s}) + 3\text{Ag}^+(\text{aq}) \rightarrow \text{Al}^{3+}(\text{aq}) + 3\text{Ag}(\text{s})\)

(d) Function of salt bridge: It completes the electrical circuit and maintains electrical neutrality in both half-cells by allowing ion migration. \(\text{K}^+(\text{aq})\) cations migrate into the cathode compartment (silver half-cell) to replenish the positive charge consumed as \(\text{Ag}^+\) ions are reduced to \(\text{Ag}\).

(a) Aluminium / \(\text{Al}\) [1]
\(\text{Al}\) is higher in ECS / has greater tendency to lose electrons / oxidised more readily than \(\text{Ag}\) [1]

(b) \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})\) [1]

(c) \(\text{Al}(\text{s}) + 3\text{Ag}^+(\text{aq}) \rightarrow \text{Al}^{3+}(\text{aq}) + 3\text{Ag}(\text{s})\) [1]

(d) Completing the circuit / maintaining electrical neutrality [1]
\(\text{K}^+\) moves towards the \(\text{Ag}\) half-cell / cathode compartment to balance the consumption of \(\text{Ag}^+\) cations [1]

(a) Both \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\) are negatively charged ions. Strong electrostatic repulsion between like charges creates a high activation energy barrier. As a result, very few collisions possess kinetic energy equal to or exceeding the activation energy at room temperature.

(b)(i) \(\text{Fe}^{3+}\) can initiate the reaction by undergoing Step 2 first: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\). The generated \(\text{Fe}^{2+}\) then reacts with \(\text{S}_2\text{O}_8^{2-}\) via Step 1 to regenerate \(\text{Fe}^{3+}\). Since \(\text{Fe}^{3+}\) is regenerated at the end, it functions as a catalyst.

(b)(ii) The catalyst provides an alternative pathway involving oppositely charged ions (cation-anion collisions), which has a lower activation energy (\(E_a\)). Therefore, a greater fraction of colliding particles have kinetic energy equal to or greater than the new activation energy, increasing the frequency of effective collisions.

(a) Both reactants are negatively charged / anions, resulting in electrostatic repulsion between them [1]
High activation energy / small fraction of effective collisions [1]

(b)(i) \(\text{Fe}^{3+}\) reacts with \(\text{I}^-\) in Step 2 to form \(\text{Fe}^{2+}\) and \(\text{I}_2\) [1]
\(\text{Fe}^{2+}\) then reacts with \(\text{S}_2\text{O}_8^{2-}\) in Step 1 to regenerate \(\text{Fe}^{3+}\) [1]

(b)(ii) Provides an alternative reaction pathway with a lower activation energy (\(E_a\)) [1]
Greater fraction of particles possess energy \(\ge E_a\) / higher frequency of effective collisions [1]

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

(b) Initial moles:
\(n(\text{N}_2\text{O}_4) = 0.40\,\text{mol}\), \(n(\text{NO}_2) = 0\,\text{mol}\)
At equilibrium:
\(n(\text{NO}_2) = 0.16\,\text{mol}\)
Moles of \(\text{N}_2\text{O}_4\) reacted \(= \frac{0.16}{2} = 0.08\,\text{mol}\)
\(n(\text{N}_2\text{O}_4)_{\text{eq}} = 0.40 - 0.08 = 0.32\,\text{mol}\)
Equilibrium concentrations in \(2.0\,\text{dm}^3\):
\([\text{NO}_2] = \frac{0.16}{2.0} = 0.080\,\text{mol}\,\text{dm}^{-3}\)
\([\text{N}_2\text{O}_4] = \frac{0.32}{2.0} = 0.16\,\text{mol}\,\text{dm}^{-3}\)
\(K_c = \frac{(0.080)^2}{0.16} = \frac{0.0064}{0.16} = 0.040\,\text{mol}\,\text{dm}^{-3}\).

(c) The forward reaction is endothermic (\(\Delta H > 0\)). According to Le Chatelier's principle, an increase in temperature shifts the equilibrium position to the right (forward direction) to absorb added heat. This increases the concentration of \(\text{NO}_2(\text{g})\), so the mixture turns darker brown.

(a) \(K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}\) [1]
Unit: \(\text{mol}\,\text{dm}^{-3}\) [1]

(b) Equilibrium moles/concentrations calculation: \([\text{N}_2\text{O}_4] = 0.16\,\text{mol}\,\text{dm}^{-3}\) and \([\text{NO}_2] = 0.080\,\text{mol}\,\text{dm}^{-3}\) (Method mark) [1*]
\(K_c = 0.040\) (or \(4.0 \times 10^{-2}\)) (allow missing unit if penalised in (a)) [1]

(c) Colour becomes darker brown / brown colour intensifies [1]
Forward reaction is endothermic / absorbs heat, so increasing temperature shifts equilibrium to the right [1]

(a) Standard enthalpy change of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (\(298\text{ K}\) and \(1\text{ atm}\)).

(b) Equation for the standard enthalpy change of formation of propan-1-ol(l):
$$3\text{C(graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \longrightarrow \text{C}_3\text{H}_7\text{OH(l)}$$

By Hess's Law:
$$\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})$$
$$\Delta H_f^\ominus = [3(-393.5) + 4(-285.8)] - [-2021.0]$$
$$\Delta H_f^\ominus = [-1180.5 - 1143.2] + 2021.0 = -2323.7 + 2021.0 = -302.7\text{ kJ mol}^{-1}$$

(c) Any two valid reasons:
1. Heat loss to the surroundings / calorimeter / thermometer.
2. Incomplete combustion of propan-1-ol (forming soot/carbon monoxide).
3. Evaporation of propan-1-ol from the wick between weighings.

(a) Enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (1)

(b) Formation equation or Hess's cycle correctly set up (1*)
$$\Delta H_f^\ominus = 3(-393.5) + 4(-285.8) - (-2021.0)$$ (1*)
$$\Delta H_f^\ominus = -302.7\text{ kJ mol}^{-1}$$ (1) [Award full marks for correct answer with appropriate working; correct sign and unit required]

(c) Any TWO of the following (1 mark each, max 2):
- Heat lost to the surrounding air / beaker / apparatus.
- Incomplete combustion occurs / carbon or carbon monoxide formed.
- Heat capacity of the calorimeter / beaker was not taken into account.
- Loss of fuel by evaporation from the burner wick.

(a) $$K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}$$

(b) Set up an ICE table:
$$\begin{array}{lcccc}
& \text{CO(g)} & + & \text{H}_2\text{O(g)} & \rightleftharpoons & \text{CO}_2\text{(g)} & + & \text{H}_2\text{(g)} \\
\text{Initial / mol} & 0.40 & & 0.80 & & 0 & & 0 \\
\text{Change / mol} & -0.30 & & -0.30 & & +0.30 & & +0.30 \\
\text{Equilibrium / mol} & 0.10 & & 0.50 & & 0.30 & & 0.30 \\
\text{Equilibrium conc. / M} & 0.050 & & 0.25 & & 0.15 & & 0.15
\end{array}$$

$$K_c = \frac{(0.15)(0.15)}{(0.050)(0.25)} = \frac{0.0225}{0.0125} = 1.8$$
(Note: \(K_c\) is dimensionless for this reaction as the units cancel).

(c) (i) No effect on yield. The total number of moles of gaseous reactants equals the total number of moles of gaseous products (\(1 + 1 = 1 + 1\)), so pressure change does not shift the equilibrium position.
(ii) No effect on yield. A catalyst increases the rates of both the forward and reverse reactions to the same extent.

(a) $$K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}$$ (1)
[Reject if round brackets are used]

(b) Finding equilibrium moles: \(n(\text{CO}) = 0.10\text{ mol}\), \(n(\text{H}_2\text{O}) = 0.50\text{ mol}\), \(n(\text{H}_2) = 0.30\text{ mol}\) (1*)
Correct concentrations divided by \(2.0\text{ dm}^3\) (1*)
$$K_c = 1.8$$ (1)

(c) (i) No change / no effect AND equal number of moles of gas on both sides of the equation (1)
(ii) No change / no effect AND catalyst speeds up forward and reverse reactions equally (1)

(a) The salt bridge completes the electrical circuit by allowing ions to migrate and maintains electrical neutrality in both half-cell solutions.

(b) Cathode: Graphite electrode.
Ionic half-equation:
$$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} + 6\text{e}^- \longrightarrow 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$$

(c) The solution turns from orange (due to \(\text{Cr}_2\text{O}_7^{2-}\)) to green (due to \(\text{Cr}^{3+}\)).

(d) At the zinc anode:
$$\text{Zn(s)} \longrightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-$$
$$\text{Number of moles of Zn reacted} = \frac{0.0150\text{ mol}}{2} = 0.00750\text{ mol}$$
$$\text{Change in mass of Zn rod} = 0.00750\text{ mol} \times 65.4\text{ g mol}^{-1} = 0.4905\text{ g} \approx 0.491\text{ g (decrease)}$$

(a) Complete the circuit / allow migration of ions AND maintain electrical neutrality (1)

(b) Cathode: Graphite / carbon (1)
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$ (1)

(c) From orange to green (1)

(d) $$n(\text{Zn}) = \frac{0.0150}{2} = 0.00750\text{ mol}$$ (1*)
$$\text{Mass change} = 0.00750 \times 65.4 = 0.491\text{ g}$$ (decrease of \(0.491\text{ g}\)) (1)

(a) (i) Addition reaction (or electrophilic addition).
(ii) The minor product is 1-bromopropane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\).

(b) (i) Compound \(\mathbf{X}\) is propan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_3\)), which is oxidized to propanone (\(\mathbf{Y}\)). The acidified dichromate solution changes from orange to green.
(ii) Compound \(\mathbf{Y}\) is propanone:
$$\text{CH}_3-\text{C}(=\text{O})-\text{CH}_3$$

(c) Valid distinguishing test:
Test 1 (using acidified dichromate):
- Reagent: Warm with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}\).
- Observation with \(\mathbf{X}\) (propan-2-ol): Solution turns from orange to green.
- Observation with \(\mathbf{Y}\) (propanone): Solution remains orange / no observable change.

OR Test 2 (using 2,4-dinitrophenylhydrazine):
- Reagent: Add 2,4-dinitrophenylhydrazine solution.
- Observation with \(\mathbf{X}\): No precipitate formed.
- Observation with \(\mathbf{Y}\): Orange/yellow precipitate formed.

(a) (i) Addition / Electrophilic addition (1)
(ii) Correct structural formula of 1-bromopropane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) (1)

(b) (i) From orange to green (1)
(ii) Correct structural formula of propanone: \(\text{CH}_3\text{COCH}_3\) (1)

(c) Reagent: Acidified \(\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}\) and warm / acidified \(\text{KMnO}_4\text{(aq)}\) / \(\text{Na(s)}\) / 2,4-DNP (1)
Expected observations correctly stated for both compounds (1)
[e.g., for acidified dichromate: propan-2-ol turns orange solution green; propanone shows no colour change]

(a) (i)
Comparing Experiments 1 and 2:
$[\text{H}_2]$ remains constant at $1.5 \times 10^{-3}\text{ M}$.
$[\text{NO}]$ increases by a factor of $\frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2$.
Rate increases by a factor of $\frac{7.20 \times 10^{-4}}{1.80 \times 10^{-4}} = 4 = 2^2$.
Therefore, the reaction is second order with respect to $\text{NO}$.

Comparing Experiments 1 and 3:
$[\text{NO}]$ remains constant at $2.0 \times 10^{-3}\text{ M}$.
$[\text{H}_2]$ increases by a factor of $\frac{4.5 \times 10^{-3}}{1.5 \times 10^{-3}} = 3$.
Rate increases by a factor of $\frac{5.40 \times 10^{-4}}{1.80 \times 10^{-4}} = 3 = 3^1$.
Therefore, the reaction is first order with respect to $\text{H}_2$.

Rate equation: $\text{Rate} = k[\text{NO}]^2[\text{H}_2]$

(ii)
Using data from Experiment 1:
$$1.80 \times 10^{-4} = k (2.0 \times 10^{-3})^2 (1.5 \times 10^{-3})$$
$$1.80 \times 10^{-4} = k (6.0 \times 10^{-9})$$
$$k = \frac{1.80 \times 10^{-4}}{6.0 \times 10^{-9}} = 3.00 \times 10^4\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}$$

(iii)
Using the Arrhenius equation in two-point form:
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
$$\ln\left(\frac{1.50 \times 10^5}{3.00 \times 10^4}\right) = \frac{E_a}{8.314}\left(\frac{1}{1050} - \frac{1}{1150}\right)$$
$$\ln(5.00) = \frac{E_a}{8.314}\left(9.5238 \times 10^{-4} - 8.6957 \times 10^{-4}\right)$$
$$1.6094 = \frac{E_a}{8.314}(8.2816 \times 10^{-5})$$
$$E_a = \frac{1.6094 \times 8.314}{8.2816 \times 10^{-5}} = 1.6157 \times 10^5\text{ J mol}^{-1} = 162\text{ kJ mol}^{-1}$$

(b) (i)
(1) Anode: $2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-$
(2) Cathode: $2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$ (or $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$)

(ii)
(1) The cation-exchange membrane is selectively permeable to cations ($\text{Na}^+$ ions) and prevents the migration of chloride ions ($\text{Cl}^-$) into the cathode compartment, preventing the $\text{NaOH}$ product from being contaminated with $\text{NaCl}$. It also blocks $\text{OH}^-$ ions from moving to the anode to react with $\text{Cl}_2$.
(2) The membrane prevents the chlorine gas ($\text{Cl}_2$) generated at the anode and hydrogen gas ($\text{H}_2$) generated at the cathode from coming into contact and forming an explosive mixture.

(iii)
1. It does not use toxic mercury, completely eliminating the risk of mercury pollution in wastewater and food chains.
2. It consumes significantly less electrical energy per unit mass of product compared to the mercury cell.

(c) (i)
Molar mass of $\text{CH}_3\text{CH}_2\text{COOCH}_3 = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 88.0\text{ g mol}^{-1}$

For Route 1:
Total mass of reactants = Molar mass of $\text{CH}_3\text{CH}_2\text{COOH} (74.0) + \text{CH}_3\text{OH} (32.0) = 106.0\text{ g mol}^{-1}$
$$\text{Atom Economy} = \frac{88.0}{106.0} \times 100\% = 83.0\%$$

For Route 2:
Total mass of reactants = Molar mass of $\text{CH}_2=\text{CH}_2 (28.0) + \text{CO} (28.0) + \text{CH}_3\text{OH} (32.0) = 88.0\text{ g mol}^{-1}$
$$\text{Atom Economy} = \frac{88.0}{88.0} \times 100\% = 100\%$$

(ii)
1. Route 2 uses a catalytic addition reaction that avoids the use of strong concentrated mineral acid catalysts (e.g. concentrated $\text{H}_2\text{SO}_4$), reducing corrosive hazards and acid waste neutralization.
2. Route 2 does not produce water as a byproduct, eliminating the need for energy-intensive separation/drying steps to shift the equilibrium or purify the ester.

(iii)
Route 2 uses carbon monoxide ($\text{CO}$), which is an extremely toxic, odourless, and highly flammable compressed gas, posing severe occupational safety risks.

(a)(i)
- Deducing 2nd order wrt [NO] with clear reasoning/working (1)
- Deducing 1st order wrt [H2] with clear reasoning/working (1)
- Correct rate equation: $\text{Rate} = k[\text{NO}]^2[\text{H}_2]$ (1)

(a)(ii)
- Correct value of $k = 3.00 \times 10^4$ (or $3 \times 10^4$) (1)
- Correct unit: $\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}$ (or $\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}$) (1)

(a)(iii)
- Correct substitution into the Arrhenius equation (1*)
- Correct rearrangement and calculation of numerical value (1*)
- Final answer: $162\text{ kJ mol}^{-1}$ (accept $161.5 - 162.0\text{ kJ mol}^{-1}$) with correct unit (1)

(b)(i)
- (1) $2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-$ (1)
- (2) $2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$ OR $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$ (1)

(b)(ii)
- (1) Membrane allows only $\text{Na}^+$ to pass / blocks $\text{Cl}^-$ and $\text{OH}^-$, preventing $\text{NaCl}$ contamination and unwanted side reactions (2)
- (2) Physically separates $\text{Cl}_2(g)$ and $\text{H}_2(g)$ so they cannot mix to form an explosive gas mixture (1)

(b)(iii)
- ANY TWO of the following (1 mark each, max 2 marks):
* Avoids the use/discharge of highly toxic mercury (preventing bioaccumulation / minamata-like pollution) (1)
* Lower electrical energy consumption / higher energy efficiency (1)
* Membrane cells operate with lower overall maintenance hazard (1)

(c)(i)
- Atom economy of Route 1 = 83.0% (1)
- Atom economy of Route 2 = 100% (1)

(c)(ii)
- ANY TWO valid green chemistry principles (1 mark each, max 2 marks):
* Avoids corrosive / hazardous acid catalyst (e.g. conc. $\text{H}_2\text{SO}_4$) used in conventional esterification (1)
* Avoids generating aqueous waste / eliminates energy-intensive separation of water from the ester (1)
* Higher effective conversion as it is not limited by unfavorable esterification equilibrium (1)

(c)(iii)
- Carbon monoxide ($\text{CO}$) is highly toxic / flammable / high pressure required (1)

Part (a)
(i) (1) TLC separation is based on the difference in the extent of adsorption / partition of different components between the stationary phase (polar silica gel) and the mobile phase (developing solvent). Components with higher polarity are adsorbed more strongly onto the stationary phase and move more slowly.
(2) Compound Y (benzoic acid) has a larger \(R_\text{f}\) value. Compound X (4-hydroxybenzoic acid) possesses an additional phenolic \(-\text{OH}\) group, which allows it to form stronger hydrogen bonds / dipole-dipole attractions with the polar silica gel stationary phase, causing it to be retained more strongly and travel a shorter distance.

(ii) According to the partition law, a constant partition coefficient \(K_\text{D}\) governs the distribution of the solute between the two immiscible solvents at equilibrium. Performing multiple extractions with smaller portions repeatedly shifts the equilibrium, extracting a higher total mass of benzoic acid compared to a single extraction with the entire volume.

Part (b)
(i) Ester group (\(-\text{COO}-\)) [strong \(\text{C}=\text{O}\) peak at \(1740\text{ cm}^{-1}\) and absence of broad \(\text{O}-\text{H}\) peak at \(2500 - 3600\text{ cm}^{-1}\)].

(ii) (1) \([\text{CH}_3\text{COO}]^+\) or \([\text{COOCH}_3]^+\)
(2) Any two of: \(\text{CH}_3\text{COOCH(CH}_3)_2\), \(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\), \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3\), \((\text{CH}_3)_2\text{CHCOOCH}_3\).

(iii) (1) Alkaline hydrolysis of compound Z yields an alcohol W and a carboxylate. Alcohol W gives a positive iodoform test (forms yellow precipitate with \(\text{I}_2 / \text{NaOH}\)), meaning W must contain the \(\text{CH}_3\text{CH(OH)}-\) group. The only 3-carbon alcohol with this structure is propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\). Thus, the acid moiety must be ethanoic acid (2 carbons), and compound Z is isopropyl ethanoate (propan-2-yl ethanoate), \(\text{CH}_3\text{COOCH(CH}_3)_2\).
(2) Formula: \(\text{CHI}_3\); Name: Triiodomethane (or iodoform).

Part (c)
(i) \(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 6\text{Fe}^{2+}(\text{aq}) + 14\text{H}^+(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 6\text{Fe}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\)

(ii) Number of moles of \(\text{Cr}_2\text{O}_7^{2-}\) used:
\(n(\text{Cr}_2\text{O}_7^{2-}) = 0.0180\text{ mol dm}^{-3} \times \frac{14.50}{1000}\text{ dm}^3 = 2.610 \times 10^{-4}\text{ mol}\)

From the stoichiometric ratio, \(n(\text{Fe}^{2+}) : n(\text{Cr}_2\text{O}_7^{2-}) = 6 : 1\):
\(n(\text{Fe}^{2+})\text{ in } 25.00\text{ cm}^3 = 6 \times 2.610 \times 10^{-4}\text{ mol} = 1.566 \times 10^{-3}\text{ mol}\)

Molar concentration of \(\text{Fe}^{2+}\):
\([\text{Fe}^{2+}] = \frac{1.566 \times 10^{-3}\text{ mol}}{0.02500\text{ dm}^3} = 0.06264\text{ M}\)

Mass concentration of \(\text{Fe}^{2+}\):
\(\text{Concentration} = 0.06264\text{ mol dm}^{-3} \times 55.8\text{ g mol}^{-1} = 3.495\text{ g dm}^{-3} \approx 3.50\text{ g dm}^{-3}\) (3 sig. figs.)

(iii) \(\text{Cl}^-(\text{aq})\) ions from hydrochloric acid would be oxidized by the strong oxidizing agent \(\text{Cr}_2\text{O}_7^{2-}(\text{aq})\) to chlorine gas (\(\text{Cl}_2\)), consuming extra titrant and leading to an overestimation of the \(\text{Fe}^{2+}\) concentration.

(a)(i)(1) Different components have different relative affinities / extents of adsorption to the stationary phase and solubility in the mobile phase. (1)
More strongly adsorbed components move more slowly up the plate. (1)

(a)(i)(2) Compound Y / benzoic acid. (1)
Compound X contains an extra \(-\text{OH}\) group, forming stronger hydrogen bonding with the polar silica stationary phase, thus being held more strongly (lower \(R_\text{f}\)). (1)

(a)(ii) Successive extractions exploit the equilibrium partition coefficient repeatedly; the fraction of solute remaining unextracted decreases exponentially with the number of extraction cycles. (2)
[Award 1 mark for stating that multiple extractions extract a greater cumulative amount / higher percentage of solute; 1 mark for explanation referencing partition equilibrium / distribution ratio.]

(b)(i) Ester group / \(-\text{COO}-\) (1)

(b)(ii)(1) \([\text{CH}_3\text{COO}]^+\) or \([\text{COOCH}_3]^+\) (1)

(b)(ii)(2) Correct structural formulae of any TWO esters of formula \(\text{C}_5\text{H}_{10}\text{O}_2\) giving \(m/z = 59\) (e.g. \(\text{CH}_3\text{COOCH(CH}_3)_2\), \(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\), \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3\)). (1 + 1)

(b)(iii)(1) Deduction of alcohol W as propan-2-ol from the positive iodoform test (1)
Structural formula of Z: \(\text{CH}_3\text{COOCH(CH}_3)_2\) (1)

(b)(iii)(2) Formula: \(\text{CHI}_3\) (1)
Name: Triiodomethane / iodoform (1)

(c)(i) \(\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\) (1)

(c)(ii) Moles of \(\text{Cr}_2\text{O}_7^{2-} = 2.610 \times 10^{-4}\text{ mol}\) (1*)
Moles of \(\text{Fe}^{2+} = 1.566 \times 10^{-3}\text{ mol}\) (1*)
Concentration \(= 3.50\text{ g dm}^{-3}\) (acceptable range: 3.49 – 3.50; 3 sig. figs. with correct unit) (1)

(c)(iii) \(\text{Cr}_2\text{O}_7^{2-}\) would oxidize \(\text{Cl}^-\) to \(\text{Cl}_2\) (1)
Consumes more \(\text{Cr}_2\text{O}_7^{2-}\) titrant / leads to an inaccurate (overestimated) \(\text{Fe}^{2+}\) value (1)