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(a) Identification & Reasoning:
1. The reaction with aqueous silver nitrate in ethanol produces a pale cream precipitate slowly, which indicates the presence of bromide ions (formation of \(AgBr(s)\)). Thus, compound X is a bromoalkane.
2. When Y is oxidized under reflux with excess acidified potassium dichromate(VI) to form a carboxylic acid Z, it means Y must be a primary alcohol. Secondary alcohols would oxidize to ketones, and tertiary alcohols do not oxidize.
3. Since Y is a primary alcohol, the starting bromoalkane X must be a primary bromoalkane.
4. There are two primary bromoalkane isomers with formula \(C_4H_9Br\): 1-bromobutane or 1-bromo-2-methylpropane.
- If X is 1-bromobutane: \(CH_3CH_2CH_2CH_2Br\). Then Y is butan-1-ol: \(CH_3CH_2CH_2CH_2OH\), and Z is butanoic acid: \(CH_3CH_2CH_2COOH\).
- If X is 1-bromo-2-methylpropane: \((CH_3)_2CHCH_2Br\). Then Y is 2-methylpropan-1-ol: \((CH_3)_2CHCH_2OH\), and Z is 2-methylpropanoic acid: \((CH_3)_2CHCOOH\).

(b) (i) Procedure:
1. Add equal volumes/amounts (e.g., \(1\text{ cm}^3\)) of 1-chlorobutane, 1-bromobutane, and 1-iodobutane to separate test tubes.
2. Add a mutual solvent such as ethanol (to dissolve the halogenoalkanes) and aqueous silver nitrate solution to each test tube.
3. Place all test tubes in a water bath held at a constant temperature (e.g., \(50^\circ\text{C}\)) and record the time taken for a precipitate to appear in each test tube.

(b) (ii) Order & Explanation:
- Order: 1-iodobutane (fastest) > 1-bromobutane > 1-chlorobutane (slowest).
- The rate of hydrolysis depends on the strength of the carbon-halogen bond (bond enthalpy).
- Going down Group 7 from Cl to I, the atomic size increases, which weakens the orbital overlap in the C-X bond, increasing bond length and decreasing the C-X bond enthalpy (C-I is the weakest bond, C-Cl is the strongest).
- Although the C-Cl bond is the most polar due to the higher electronegativity of chlorine (which would attract nucleophiles faster), bond enthalpy is the dominant factor. The C-I bond requires the least energy to break, so it reacts the fastest.

(c) (i) Elimination Conditions:
Concentrated potassium hydroxide (or sodium hydroxide) dissolved in ethanol (ethanolic) and heating under reflux.

(c) (ii) Product:
The skeletal structure of the product is a three-carbon chain with a double bond and a methyl group on the middle carbon. The IUPAC name is methylpropene (or 2-methylpropene).

(a) [5 marks total]
- Deduce X is a bromoalkane due to cream precipitate of \(AgBr\) with silver nitrate. (1)
- Deduce Y is a primary alcohol because it oxidizes to a carboxylic acid. (1)
- Identify that X must be a primary bromoalkane. (1)
- State correct names/structures for X, Y, and Z for one isomer set: either 1-bromobutane, butan-1-ol, and butanoic acid OR 1-bromo-2-methylpropane, 2-methylpropan-1-ol, and 2-methylpropanoic acid. (2 marks for all three correct; 1 mark if only one or two are correct).

(b)(i) [3 marks total]
- Use equal volumes/amounts of each halogenoalkane and add ethanol and silver nitrate. (1)
- Control temperature using a water bath. (1)
- Measure the time taken for the precipitate to appear to determine rate. (1)

(b)(ii) [4 marks total]
- State correct order: 1-iodobutane > 1-bromobutane > 1-chlorobutane. (1)
- Identify that C-X bond enthalpy decreases down the group / C-I bond is the weakest. (1)
- Identify that C-Cl bond is the most polar. (1)
- State that bond enthalpy is the deciding factor / C-I bond is broken most easily. (1)

(c)(i) [1 mark total]
- Potassium hydroxide in ethanol / ethanolic KOH AND heating under reflux. (1)

(c)(ii) [2 marks total]
- Correct skeletal structure of methylpropene showing the double bond and branched methyl group. (1)
- Correct IUPAC name: methylpropene / 2-methylpropene. (1)

(a) (i) Heat energy calculation:
Total volume of solution = \(50.0 + 50.0 = 100.0\text{ cm}^3\).
Assume the density of the solution is the same as that of water, \(1.00\text{ g cm}^{-3}\), so mass \(m = 100.0\text{ g}\).
Assume the specific heat capacity of the solution is the same as water, \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\).
Temperature rise, \(\Delta T = 25.2 - 18.5 = 6.7^\circ\text{C} = 6.7\text{ K}\).
\(q = m c \Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J}\) (or \(2.80\text{ kJ}\)).

(a) (ii) Enthalpy of neutralisation:
Moles of \(HCl\) used = \(\frac{50.0}{1000} \times 1.00 = 0.0500\text{ mol}\).
Moles of \(NaOH\) used = \(\frac{50.0}{1000} \times 1.05 = 0.0525\text{ mol}\).
Since \(HCl\) is the limiting reactant, moles of water formed = \(0.0500\text{ mol}\).
\(\Delta H_{neut} = - \frac{q}{\text{moles of water}} = - \frac{2800.6\text{ J}}{0.0500\text{ mol}} = -56012\text{ J mol}^{-1} = -56.0\text{ kJ mol}^{-1}\) (to 3 sig figs).

(b) (i) Reasons for difference:
1. Some heat was absorbed by the polystyrene cup and the thermometer (heat capacity of calorimeter not being negligible).
2. Inaccurate measurements of solution volumes or concentrations (experimental uncertainty).
3. The density and specific heat capacity of the actual aqueous mixtures are slightly different from those of pure water.

(b) (ii) Weak acid neutralisation:
Ethanoic acid is a weak acid and is only partially dissociated in aqueous solution.
Some energy is absorbed to fully dissociate the weak acid molecules into \(H^+\) and \(CH_3COO^-\) ions (as dissociation is an endothermic process). This reduces the net heat released to the surroundings, making the overall process less exothermic.

(c) (i) Formation Equation:
\(2C(s, graphite) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)\)

(c) (ii) Hess's Law calculation:
From Hess's Cycle using combustion values:
\(\Delta H_f^\ominus[C_2H_5OH(l)] = \sum \Delta H_c^\ominus(\text{Reactants}) - \sum \Delta H_c^\ominus(\text{Products})\)
\(\Delta H_f^\ominus = [2 \times (-393.5) + 3 \times (-285.8)] - [-1367.3]\)
\(\Delta H_f^\ominus = [-787.0 - 857.4] + 1367.3 = -1644.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\).

(a)(i) [3 marks total]
- Calculates temperature change correctly: \(\Delta T = 6.7^\circ\text{C}\). (1)
- Calculates heat energy correctly: \(q = 2800.6\text{ J}\) (or \(2.80\text{ kJ}\)). (1)
- States both assumptions: density of solution is \(1.00\text{ g cm}^{-3}\) and specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). (1)

(a)(ii) [4 marks total]
- Calculates moles of \(HCl = 0.0500\text{ mol}\) and moles of \(NaOH = 0.0525\text{ mol}\). (1)
- Identifies \(HCl\) is the limiting reactant, so moles of water formed = \(0.0500\text{ mol}\). (1)
- Calculates value of enthalpy change: \(\frac{2.8006}{0.0500} = 56.0\text{ kJ mol}^{-1}\). (1)
- Correctly applies negative sign and 3 significant figures: \(-56.0\text{ kJ mol}^{-1}\). (1)

(b)(i) [2 marks total]
- Accept any two reasonable points: heat absorbed by the calorimeter / thermometer (1); density/heat capacity of solutions is not exactly that of pure water (1); uncertainty in temperature reading (1).

(b)(ii) [2 marks total]
- Weak acids are only partially dissociated in solution. (1)
- Energy is required/absorbed to complete the dissociation of the acid molecules, making the overall process less exothermic. (1)

(c)(i) [2 marks total]
- Balanced chemical equation for formation: \(2C + 3H_2 + 0.5O_2 \rightarrow C_2H_5OH\). (1)
- All state symbols correct: \(C(s)\) (or graphite), \(H_2(g)\), \(O_2(g)\), \(C_2H_5OH(l)\). (1)

(c)(ii) [2 marks total]
- Shows correct Hess's cycle or algebraic formulation: \(2 \times \Delta H_c[C] + 3 \times \Delta H_c[H_2] - \Delta H_c[C_2H_5OH]\). (1)
- Correct answer with units: \(-277.1\text{ kJ mol}^{-1}\). (1)

(a) (i) Equation:
\(2Mg(NO_3)_2(s) \rightarrow 2MgO(s) + 4NO_2(g) + O_2(g)\)

(a) (ii) Explanation:
1. The magnesium ion (\(Mg^{2+}\)) has a smaller ionic radius than the barium ion (\(Ba^{2+}\)).
2. Both ions have the same charge (+2).
3. Therefore, the magnesium ion has a higher charge density than the barium ion.
4. This higher charge density polarizes and distorts the electron cloud of the nitrate ion (\(NO_3^-\)) more strongly, weakening the covalent oxygen-nitrogen bonds within the nitrate ion, which allows it to decompose at a lower temperature.

(b) (i) Disproportionation:
Equation: \(Cl_2(aq) + H_2O(l) \rightleftharpoons HCl(aq) + HClO(aq)\)
- The oxidation number of chlorine in elemental chlorine (\(Cl_2\)) is 0.
- In hydrochloric acid (\(HCl\)), the oxidation number of chlorine is -1.
- In chloric(I) acid (\(HClO\)), the oxidation number of chlorine is +1.
- Since chlorine has been simultaneously oxidized (from 0 to +1) and reduced (from 0 to -1) in the same chemical reaction, it is a disproportionation reaction.

(b) (ii) Hazard and Benefit:
- Hazard: Chlorine is highly toxic/poisonous and can react with organic compounds in water to form chlorinated organic compounds, which are carcinogenic.
- Benefit: The benefits of killing pathogenic microorganisms (such as bacteria causing cholera and typhoid) outweigh the extremely low risks associated with the concentrations of chlorine used.

(c) (i) Sodium Iodide and Sulfuric Acid:
- The gas with a smell of bad eggs is hydrogen sulfide, \(H_2S\).
- The purple vapour is iodine, \(I_2\).
- Sulfuric acid acts as an oxidizing agent (it oxidizes iodide ions, \(I^-\), to free iodine, while being reduced to hydrogen sulfide).

(a)(i) [2 marks total]
- Correct formulae and balancing: \(2Mg(NO_3)_2 \rightarrow 2MgO + 4NO_2 + O_2\). (1)
- Correct state symbols: \((s)\) for reactants/solid products and \((g)\) for gases. (1)

(a)(ii) [4 marks total]
- State that \(Mg^{2+}\) has a smaller ionic radius / ionic size than \(Ba^{2+}\). (1)
- State both cations have the same (+2) charge. (1)
- Conclude \(Mg^{2+}\) has a higher charge density. (1)
- Explain that \(Mg^{2+}\) polarizes / distorts the nitrate ion electron cloud more, weakening the N-O bond. (1)

(b)(i) [4 marks total]
- Correct equation: \(Cl_2 + H_2O \rightleftharpoons HCl + HClO\). (1)
- State that chlorine in \(Cl_2\) is at oxidation state 0. (1)
- State that chlorine in \(HCl\) is at -1 AND in \(HClO\) is at +1. (1)
- Define disproportionation as the simultaneous oxidation and reduction of the same element in a single reaction. (1)

(b)(ii) [2 marks total]
- State hazard: chlorine is toxic / poisonous / forms toxic chlorinated hydrocarbons. (1)
- State benefit: kills bacteria / sterilises water, and the benefits outweigh the risks. (1)

(c)(i) [3 marks total]
- Identify hydrogen sulfide (\(H_2S\)). (1)
- Identify iodine (\(I_2\)). (1)
- Identify the role of sulfuric acid as an oxidizing agent. (1)

(a) (i) Maxwell-Boltzmann Distribution Features:
1. Axes: The y-axis represents the number of molecules (or fraction of molecules) with a given energy, and the x-axis represents the molecular kinetic energy.
2. Shape at \(T_1\): The curve starts at the origin \((0,0)\), rises to a peak (most probable energy), and then tails off gradually towards the right at higher energies, but never touches the x-axis.
3. Shift at higher temperature \(T_2\): The curve shifts to the right; the peak is lower and broader. The total area under both curves remains identical (representing the same number of molecules).
4. Activation Energy, \(E_a\): Marked as a vertical line on the high-energy tail of the x-axis. The area under the curve to the right of this line represents the number of molecules with energy equal to or greater than \(E_a\).

(a) (ii) Temperature and Rate:
1. At temperature \(T_2\), a much larger fraction of molecules have kinetic energy equal to or greater than the activation energy (\(E \ge E_a\)), which is shown by a larger area under the curve to the right of \(E_a\).
2. Therefore, upon collision, a significantly greater proportion of collisions are successful.
3. Additionally, the molecules move faster, leading to a higher frequency of collisions (though the increase in successful proportion of collisions is the dominant factor).

(b) (i) Effect of Pressure:
- Yield of methanol increases.
- According to Le Chatelier's principle, an increase in pressure shifts the equilibrium in the direction that decreases the pressure, which is towards the side with fewer gas molecules (there are 3 moles of gas on the left and only 1 mole on the right).

(b) (ii) Effect of Temperature:
- Yield of methanol decreases.
- The forward reaction is exothermic (\(\Delta H = -91\text{ kJ mol}^{-1}\)). An increase in temperature shifts the equilibrium to the left (the endothermic direction) to absorb the added thermal energy.

(b) (iii) Compromise Conditions in Industry:
1. Low Temperature: Although low temperature favors a high yield of methanol, the rate of reaction would be extremely slow, making the process economically unviable as it takes too long to reach equilibrium.
2. High Pressure: Although high pressure increases both the rate and the equilibrium yield, generating and maintaining very high pressures requires very expensive thick-walled reactors and compressors, and presents a high safety risk.
3. Compromise: A compromise temperature of \(250^\circ\text{C}\) and pressure of \(50-100\text{ atm}\) provide a sufficient reaction rate and reasonable yield while ensuring cost-effectiveness and safety.
4. Catalyst: The use of a copper catalyst speeds up the rate of reaction, allowing equilibrium to be reached much faster at this moderate compromise temperature.

(a)(i) [4 marks total]
- Labels axes correctly (y-axis: number/fraction of molecules; x-axis: energy). (1)
- Describes curve starting at the origin, having an asymmetrical peak, and tailing off without touching the x-axis. (1)
- Describes high-temperature curve (\(T_2\)) as having a lower peak shifted to the right. (1)
- Explains that activation energy \(E_a\) is a vertical line and the area to its right represents molecules with \(E \ge E_a\). (1)

(a)(ii) [3 marks total]
- State that only molecules with energy \(E \ge E_a\) can react upon collision. (1)
- State that at higher temperature, a much larger proportion / fraction of molecules have energy \(E \ge E_a\). (1)
- Conclude that this results in a higher frequency of successful collisions. (1)

(b)(i) [2 marks total]
- State that equilibrium yield increases. (1)
- Explain that increasing pressure shifts the system to the side with fewer gas moles (3 moles reactant vs 1 mole product). (1)

(b)(ii) [2 marks total]
- State that equilibrium yield decreases. (1)
- Explain that the forward reaction is exothermic, so the equilibrium shifts to the left to absorb heat. (1)

(b)(iii) [4 marks total]
- Explains why low temp is not used: rate is too slow. (1)
- Explains why high pressure is not used: high costs of equipment/energy and safety hazards. (1)
- States that compromise conditions balance rate, yield, and safety/cost. (1)
- States the catalyst increases the rate, allowing a lower, safer, and cheaper compromise temperature to be used. (1)