2024 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

1. The volume of CO₂ produced corresponds to the decrease in volume when shaken with KOH: Volume of CO₂ = 55 cm³ - 25 cm³ = 30 cm³. 2. The remaining volume of 25 cm³ is the unreacted excess oxygen. 3. The volume of oxygen reacted is: Volume of O₂ reacted = 70 cm³ - 25 cm³ = 45 cm³. 4. 10 cm³ of the hydrocarbon reacts with 45 cm³ of O₂ to produce 30 cm³ of CO₂. The mole ratio of hydrocarbon to oxygen to carbon dioxide is 1 : 4.5 : 3. 5. From the carbon balance, the number of carbon atoms per molecule of hydrocarbon is 3. 6. From the oxygen balance for C₃H_y + 4.5 O₂ -> 3 CO₂ + 0.5y H₂O: 4.5 * 2 = (3 * 2) + 0.5y, which gives 9 = 6 + 0.5y, so y = 6. Thus, the molecular formula is C₃H₆.

1 mark for the correct option B. Award 1 mark for correct deduction of the volume of carbon dioxide and reacted oxygen, and using the mole ratios to find the empirical / molecular formula.

1. Compare Exp 1 and Exp 2: [Q] and [R] are constant. Doubling [P] doubles the rate, so the reaction is first-order with respect to P. 2. Compare Exp 1 and Exp 3: [P] and [R] are constant. Doubling [Q] quadruples the rate, so the reaction is second-order with respect to Q. 3. Compare Exp 1 and Exp 4: [P] and [Q] are constant. Doubling [R] does not affect the rate, so the reaction is zero-order with respect to R. 4. The rate equation is Rate = k[P][Q]². The overall order is 1 + 2 + 0 = 3. 5. Units of k = Rate / ([P][Q]²) = (mol dm⁻³ s⁻¹) / (mol dm⁻³ * (mol dm⁻³)²) = dm⁶ mol⁻² s⁻¹.

1 mark for the correct option A. Award 1 mark for deducing individual orders, overall order, and deriving units of the rate constant.

When excess concentrated hydrochloric acid is added to aqueous copper(II) ions, a ligand exchange reaction occurs where water ligands are replaced by chloride ligands to form the tetrachlorocuprate(II) ion, [CuCl₄]²⁻. This complex has a coordination number of 4 and a tetrahedral geometry.

1 mark for the correct option A. Award 1 mark for identifying the correct chemical formula and geometry of the chloro complex of copper(II).

1. Diethylamine is a secondary aliphatic amine. Its two electron-donating ethyl groups increase the electron density on the nitrogen atom, making its lone pair more available to accept a proton than in ethylamine. 2. Ethylamine is a primary aliphatic amine, more basic than ammonia due to one electron-donating ethyl group. 3. Ammonia is more basic than phenylamine. 4. Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is delocalised into the aromatic pi system, making it much less available to accept a proton.

1 mark for the correct option A. Award 1 mark for correctly comparing the availability of the nitrogen lone pairs based on inductive and delocalisation effects.

Applying Hess's Law to the Born-Haber cycle for CaCl₂(s): ΔH_fᶿ = ΔH_atᶿ(Ca) + IE₁(Ca) + IE₂(Ca) + Bond Enthalpy(Cl₂) + 2 * EA₁(Cl) + ΔH_lattᶿ. Note that the atomisation of chlorine to 2 moles of gas atoms is equivalent to the bond dissociation energy of Cl₂ (+242 kJ mol⁻¹), and the electron affinity is multiplied by 2 because two moles of chloride ions are formed: 2 * (-349) = -698 kJ mol⁻¹. Substituting these values: -796 = 178 + 590 + 1145 + 242 - 698 + ΔH_lattᶿ. This gives -796 = 1457 + ΔH_lattᶿ. Therefore, ΔH_lattᶿ = -796 - 1457 = -2253 kJ mol⁻¹.

1 mark for the correct option A. Award 1 mark for using the correct coefficients for Cl atomisation and electron affinity, and correctly calculating the lattice energy.

For a reaction to be feasible, ΔGᶿ <= 0. Using ΔGᶿ = ΔHᶿ - TΔSᶿ, at the boundary of feasibility ΔGᶿ = 0, so T = ΔHᶿ / ΔSᶿ. Converting ΔHᶿ to J mol⁻¹: ΔHᶿ = 176 * 10³ J mol⁻¹. Substituting the values: T = 176000 / 285 = 617.5 K, which rounds to approximately 618 K.

1 mark for the correct option C. Award 1 mark for correctly using the feasibility condition and converting temperature units properly.

1. Reaction with 2,4-DNPH indicates a carbonyl group (aldehyde or ketone). 2. Lack of reaction with Tollens' reagent shows that W is a ketone, not an aldehyde. This eliminates butanal and methylpropanal. 3. The reaction with alkaline aqueous iodine (the iodoform test) to yield a yellow precipitate indicates the presence of a methyl carbonyl (CH₃CO-) group. 4. A 4-carbon ketone with a methyl carbonyl group must be butanone (CH₃COCH₂CH₃).

1 mark for the correct option B. Award 1 mark for deducing that the functional groups correspond to a methyl ketone of formula C₄H₈O.

Diazotisation of phenylamine is carried out using nitrous acid (HNO₂), which is generated in situ from sodium nitrite (NaNO₂) and a dilute acid such as HCl. The reaction must be kept cold (0-10 °C) because the resulting benzenediazonium salt is unstable and decomposes to phenol and nitrogen gas at higher temperatures.

1 mark for the correct option A. Award 1 mark for identifying the correct diazotising mixture and the required low temperature condition.

Let the formula of the hydrocarbon be \(C_x H_y\). The combustion equation is: \(C_x H_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O (l)\). Since water is a liquid at room temperature and pressure, its volume is negligible. Shaking the remaining gases with excess aqueous \(NaOH\) removes \(CO_2\). The decrease in volume is the volume of \(CO_2\) produced: \(\text{Volume of } CO_2 = 110\text{ cm}^3 - 50\text{ cm}^3 = 60\text{ cm}^3\). Since \(20\text{ cm}^3\) of \(C_x H_y\) produces \(60\text{ cm}^3\) of \(CO_2\), we have: \(20x = 60 \implies x = 3\). The remaining \(50\text{ cm}^3\) of gas is the unreacted oxygen. The volume of oxygen reacted is: \(150\text{ cm}^3 - 50\text{ cm}^3 = 100\text{ cm}^3\). Using the stoichiometry of \(O_2\): \(20\left(x + \frac{y}{4}\right) = 100\). Substituting \(x = 3\): \(3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8\). Thus, the molecular formula of \(X\) is \(C_3H_8\).

Award 1 mark for the correct option D.
- Method: Relate the volume decrease of 60 cm³ to the volume of CO2 produced to find x = 3.
- Method: Find the volume of reacted oxygen (100 cm³) and use the mole ratio to find y = 8.
- Accuracy: Correctly identify the formula as C3H8.

The reaction that occurs is a ligand exchange: \([Cu(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CuCl_4]^{2-}(aq) + 6H_2O(l)\). The coordination number decreases from 6 (in the octahedral aquo-complex) to 4 (in the tetrahedral tetrachlorocuprate(II) complex). A change in geometry from octahedral to tetrahedral alters the d-orbital splitting pattern and the magnitude of the d-orbital splitting energy (\(\Delta E\)). Because \(\Delta E\) changes, the wavelengths of light absorbed during d-d electron transitions change, leading to a change in the complementary colour observed (from blue to greenish-yellow).

Award 1 mark for the correct option C.
- Reject A as the coordination number decreases, not increases.
- Reject B as Cl⁻ is a weaker field ligand than H₂O.
- Reject D as copper remains in the +2 oxidation state and is not reduced.

Comparing Experiments 1 and 2, \([Q]\) is constant while \([P]\) is doubled. The initial rate increases by a factor of: \(\frac{9.6 \times 10^{-4}}{2.4 \times 10^{-4}} = 4\). Since \(2^2 = 4\), the reaction is second order with respect to \(P\). Comparing Experiments 2 and 3, \([P]\) is constant while \([Q]\) is doubled. The initial rate increases by a factor of: \(\frac{1.92 \times 10^{-3}}{9.6 \times 10^{-4}} = 2\). Since \(2^1 = 2\), the reaction is first order with respect to \(Q\). Thus, the rate equation is \(\text{Rate} = k[P]^2[Q]\). To calculate the rate constant, \(k\), using data from Experiment 1: \(2.4 \times 10^{-4} = k \times (0.10)^2 \times (0.10) \implies 2.4 \times 10^{-4} = k \times 1.0 \times 10^{-3} \implies k = 0.24\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Award 1 mark for the correct option B.
- Method: Correctly deduce the orders of reaction for P (2nd order) and Q (1st order).
- Method: Use the rate equation and data from any experiment to calculate the rate constant k = 0.24.

In Stage 1, nitrous acid is prepared in situ by reacting sodium nitrite (not sodium nitrate) with dilute acid, so Option A is incorrect. If the temperature of Stage 1 exceeds 10 degrees Celsius (such as 50 degrees Celsius), the diazonium salt decomposes to form phenol (not nitrobenzene), so Option B is incorrect. In Stage 2, the coupling reaction involves electrophilic aromatic substitution, where the benzenediazonium ion acts as an electrophile (not a nucleophile), so Option C is incorrect. Option D is correct because the coupling of the benzenediazonium ion with alkaline phenol produces 4-hydroxyphenylazobenzene, which contains a stable azo group (\(-\text{N}=\text{N}-\)) linking two benzene rings.

Award 1 mark for the correct option D.
- Reject A because sodium nitrate is used instead of sodium nitrite.
- Reject B because phenol is formed at 50 °C.
- Reject C because the diazonium ion acts as an electrophile, not a nucleophile.

Step 1: Reacting (bromomethyl)benzene with potassium cyanide (\(KCN\)) in aqueous ethanol under reflux introduces an extra carbon atom via nucleophilic substitution, yielding phenylacetonitrile, \(C_6H_5CH_2CN\). Step 2: Acid hydrolysis of the nitrile group using dilute hydrochloric acid under reflux yields phenylacetic acid, \(C_6H_5CH_2COOH\). Step 3: Reduction of the carboxylic acid group using lithium tetrahydridoaluminate (\(LiAlH_4\)) in dry ether yields the primary alcohol, 2-phenylethanol, \(C_6H_5CH_2CH_2OH\).

Award 1 mark for the correct option B.
- Method: Understand the need to add one carbon atom using cyanide, followed by hydrolysis to a carboxylic acid and subsequent reduction.
- Accuracy: Correctly identify that reducing a nitrile directly (D) yields an amine, not an alcohol.

1. Reacting with 2,4-DNPH means \(Z\) is a carbonyl compound (aldehyde or ketone). 2. Not reacting with Tollens' reagent means \(Z\) is a ketone. 3. Reacting with alkaline aqueous iodine (iodoform test) means \(Z\) is a methyl ketone containing the \(CH_3C(=O)-\) group. Ketones with molecular formula \(C_5H_{10}O\) that are methyl ketones have the general structure \(CH_3 - CO - R\), where \(R\) is a 3-carbon alkyl group (\(-C_3H_7\)). There are exactly two possible isomers for \(R\): the propyl group, giving pentan-2-one, and the isopropyl group, giving 3-methylbutan-2-one. Pentan-3-one is not a methyl ketone and is negative in the iodoform test. Thus, exactly 2 structural isomers fit the description.

Award 1 mark for the correct option B.
- Method: Identify that the compound must be a methyl ketone.
- Method: Draw the possible methyl ketones with 5 carbons: pentan-2-one and 3-methylbutan-2-one.
- Accuracy: Determine that there are exactly 2 isomers.

Option A is incorrect because primary halogenoalkanes react mainly via the \(S_N2\) mechanism, which involves a single transition state and no carbocation intermediate. Option B is incorrect because tertiary halogenoalkanes react mainly via the \(S_N1\) mechanism, which has a two-step mechanism with a carbocation intermediate. Option C is incorrect because primary halogenoalkanes undergo \(S_N2\) reactions which lead to inversion of stereochemistry rather than racemisation (and the specific compound has no chiral centre anyway). Option D is correct: tertiary halogenoalkanes react much faster than primary halogenoalkanes because the tertiary carbocation intermediate, \((CH_3)_3C^+\), is highly stabilized by the positive inductive effects of the three electron-releasing methyl groups, lowering the activation energy for the rate-determining first step of the \(S_N1\) pathway.

Award 1 mark for the correct option D.
- Reject A because primary halogenoalkanes react via S_N2, which has no carbocation intermediate.
- Reject B because tertiary halogenoalkanes react via S_N1.
- Reject C because S_N2 does not involve racemisation.

1. Thermal stability: Going down Group 17, the size of the halogen atom increases, so the \(H-X\) bond length increases and the \(H-X\) bond energy decreases. Since the bonds are weaker, the hydrogen halides decompose more easily upon heating. Thus, thermal decomposition becomes easier. 2. Reducing ability: Iodide ions (\(I^-\)) are the strongest reducing agents among the stable halide ions. Only solid potassium iodide is strong enough to reduce the sulfur in concentrated sulfuric acid from the +6 oxidation state to the -2 oxidation state in hydrogen sulfide (\(H_2S\)). Potassium bromide can only reduce sulfuric acid to sulfur dioxide (\(SO_2\)), and potassium chloride cannot reduce it at all. Therefore, row B is correct.

Award 1 mark for the correct option B.
- Method: Identify that the ease of thermal decomposition increases down the group as the H-X bond weakens.
- Method: Identify iodide as the only halide capable of reducing concentrated sulfuric acid to hydrogen sulfide.

Calculate the moles of carbon atoms in the CO2 produced: \(n(\text{C}) = 3.77 / 44.0 = 0.0857\text{ mol}\). Calculate the moles of hydrogen atoms in the H2O produced: \(n(\text{H}) = 2 \times (1.54 / 18.0) = 0.1711\text{ mol}\). The mole ratio of carbon to hydrogen is \(0.0857 : 0.1711 \approx 1 : 2\). Therefore, the empirical formula of the hydrocarbon is \(\text{CH}_2\). Among the choices, only \(\text{C}_5\text{H}_{10}\) has the empirical formula \(\text{CH}_2\).

1 mark for the correct option B. [1]

A higher value of the stability constant \(K_{\text{stab}}\) indicates a more stable complex. The stability constant for the \([\text{Co}(\text{en})_3]^{2+}\) complex is the largest by several orders of magnitude (\(3.2 \times 10^{13}\)), which is primarily driven by the chelate effect. Therefore, this species is the most stable and will be present in the highest concentration.

1 mark for the correct option C. [1]

Comparing Exp 1 and Exp 2: doubling \([\text{A}]\) doubles the rate, so the order with respect to A is 1. Comparing Exp 1 and Exp 3: doubling \([\text{B}]\) quadruples the rate, so the order with respect to B is 2. Comparing Exp 1 and Exp 4: doubling \([\text{C}]\) has no effect on the rate, so the order with respect to C is 0. The rate equation is Rate = \(k[\text{A}][\text{B}]^2\). The units of \(k\) are given by: \(\text{mol dm}^{-3}\text{ s}^{-1} / (\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2) = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for the correct option B. [1]

Diazotisation of an aromatic amine to form a diazonium salt requires nitrous acid, which is generated in situ from sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)). The reaction must be performed at low temperatures (between \(0\ ^\circ\text{C}\) and \(10\ ^\circ\text{C}\)) to prevent decomposition of the unstable diazonium ion into a phenol.

1 mark for the correct option A. [1]

A positive test with 2,4-DNPH (orange precipitate) indicates that \(Y\) is a carbonyl compound (either an aldehyde or a ketone). The negative test with Fehling's solution (no red precipitate) indicates that \(Y\) is not an aldehyde. Therefore, \(Y\) must be a ketone. The only ketone with the molecular formula \(\text{C}_4\text{H}_8\text{O}\) is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

1 mark for the correct option B. [1]

Determine the amount (in moles) of weak acid and conjugate base: \(n(\text{acid}) = 0.0500 \times 0.100 = 0.0050\text{ mol}\), \(n(\text{base}) = 0.0500 \times 0.050 = 0.0025\text{ mol}\). Use the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_{\text{a}} + \log_{10}([\text{base}]/[\text{acid}])\). \(\text{p}K_{\text{a}} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\). \(\text{pH} = 4.87 + \log_{10}(0.0025 / 0.0050) = 4.87 + \log_{10}(0.5) = 4.87 - 0.30 = 4.57\).

1 mark for the correct option B. [1]

A reaction is feasible when \(\Delta G^\ominus < 0\). The transition temperature is when \(\Delta G^\ominus = 0\), meaning \(\Delta H^\ominus - T\Delta S^\ominus = 0\), or \(T = \Delta H^\ominus / \Delta S^\ominus\). Convert \(\Delta H^\ominus\) to J mol-1: \(-92.0 \times 10^3\text{ J mol}^{-1}\). Calculate \(T = (-92.0 \times 10^3) / -198 = 464.6\text{ K} \approx 465\text{ K}\).

1 mark for the correct option C. [1]

The absorption at \(3200\text{--}3600\text{ cm}^{-1}\) represents an alcohol \(\text{O--H}\) stretch (carboxylic acid \(\text{O--H}\) is broader and typically around \(2500\text{--}3000\text{ cm}^{-1}\)). The peak at \(1715\text{ cm}^{-1}\) represents a carbonyl \(\text{C=O}\) stretch. Thus, compound \(Z\) must possess both an alcohol group and a carbonyl group. 4-hydroxypentan-2-one is the only option containing both functions.

1 mark for the correct option B. [1]

1. Calculate the mass and moles of carbon from \(\text{CO}_2\):
\(\text{Mass of C} = 2.64\text{ g} \times \frac{12.0}{44.0} = 0.72\text{ g}\)
\(\text{Moles of C} = \frac{0.72\text{ g}}{12.0\text{ g mol}^{-1}} = 0.06\text{ mol}\)

2. Calculate the mass and moles of hydrogen from \(\text{H}_2\text{O}\):
\(\text{Mass of H} = 1.62\text{ g} \times \frac{2.0}{18.0} = 0.18\text{ g}\)
\(\text{Moles of H} = \frac{0.18\text{ g}}{1.0\text{ g mol}^{-1}} = 0.18\text{ mol}\)

3. Calculate the mass and moles of oxygen by subtraction:
\(\text{Mass of O} = 1.38\text{ g} - (0.72\text{ g} + 0.18\text{ g}) = 0.48\text{ g}\)
\(\text{Moles of O} = \frac{0.48\text{ g}}{16.0\text{ g mol}^{-1}} = 0.03\text{ mol}\)

4. Determine the simplest whole-number ratio of atoms:
\(\text{C} : \text{H} : \text{O} = \frac{0.06}{0.03} : \frac{0.18}{0.03} : \frac{0.03}{0.03} = 2 : 6 : 1\)

Therefore, the empirical formula of Y is \(\text{C}_2\text{H}_6\text{O}\).

1 mark for the correct option (C).
- Accept: Option C.
- Reject: Any other option.

When ligands approach a transition metal ion, the degenerate d-orbitals split into two groups of different energy levels. When visible light shines on the complex, an electron in a lower energy d-orbital absorbs a photon of a specific frequency/wavelength and is promoted to a higher energy d-orbital (known as a d-d transition). The colour observed is the complementary colour to the light absorbed. Thus, absorption (not emission) of light causes the colour.

1 mark for the correct option (A).
- Accept: Option A.
- Reject: Option B (incorrectly attributes colour to emission), Option C (incorrectly attributes colour to coordinate bond formation energy), Option D (incorrectly mentions 3d to 4s transitions).

1. Since doubling \([\text{P}]\) quadruples the rate (increases by a factor of \(2^2 = 4\)), the reaction is second-order with respect to \(\text{P}\).
2. When \(\text{P}\) is in a large excess, its concentration remains effectively constant. Under these conditions, the rate is proportional only to \([\text{Q}]^y\), where \(y\) is the order with respect to \(\text{Q}\). Since the half-life of this reaction is constant and independent of the initial concentration of \(\text{Q}\), the reaction must be first-order with respect to \(\text{Q}\) (\(y = 1\)).

Therefore, the rate equation is \(\text{rate} = k[\text{P}]^2[\text{Q}]\).

1 mark for the correct option (B).
- Accept: Option B.
- Reject: All other options.

The basic strength of an amine depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton:
1. In phenylamine (\(\text{C}_6\text{H}_5\text{NH}_2\)), the lone pair of electrons on the nitrogen atom is delocalised into the benzene \(\pi\) ring system, making it the least available and therefore the weakest base.
2. In ammonia (\(\text{NH}_3\)), there are no alkyl groups to release electron density.
3. In ethylamine (\(\text{C}_2\text{H}_5\text{NH}_2\)), the electron-donating ethyl group (+I inductive effect) increases the electron density on the nitrogen atom.
4. In diethylamine (\((\text{C}_2\text{H}_5)_2\text{NH}\)), there are two electron-donating ethyl groups, which increase the electron density on the nitrogen atom further and stabilise the conjugate cation in aqueous solution.

Hence, diethylamine is the strongest base.

1 mark for the correct option (C).
- Accept: Option C.
- Reject: Any other option.

To convert propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), 3 carbons) to butanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\), 4 carbons), we need to extend the carbon chain by one carbon:
- Step 1: React propan-1-ol with \(\text{SOCl}_2\) to form 1-chloropropane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}\)).
- Step 2: Nucleophilic substitution of the chlorine atom with a cyanide group using \(\text{KCN}\) in ethanol under reflux to form butanenitrile (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\)).
- Step 3: Acid hydrolysis of the nitrile group using dilute hydrochloric acid under reflux to form butanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)).

Therefore, option A is correct.

1 mark for the correct option (A).
- Accept: Option A.
- Reject: Option B (oxidation produces propanoic acid which does not react with KCN), Option C (HCN is a weak nucleophile and does not substitute well; NaOH hydrolysis would yield the sodium salt of butanoic acid, not the free acid directly), Option D (elimination produces propene, which does not react with HCN).

1. Tollens' reagent is used to distinguish aldehydes from ketones. A positive result (silver mirror) indicates the presence of an aldehyde. Therefore, the compound must be an aldehyde (eliminates propanone and ethanol).
2. Alkaline aqueous iodine (the iodoform reaction) gives a yellow precipitate of triiodomethane (\(\text{CHI}_3\)) with compounds containing a \(\text{CH}_3\text{CO}-\) group or a \(\text{CH}_3\text{CH}(\text{OH})-\).
3. Ethanal (\(\text{CH}_3\text{CHO}\)) contains both the aldehyde group (positive Tollens' test) and the methyl carbonyl group, \(\text{CH}_3\text{CO}-\) (positive iodoform test).
4. Propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) is an aldehyde but does not contain a \(\text{CH}_3\text{CO}-\) group, so it gives a negative iodoform test.

Therefore, the compound is ethanal.

1 mark for the correct option (A).
- Accept: Option A.
- Reject: Option B (ketone, negative Tollens'), Option C (no methyl carbonyl group, negative iodoform), Option D (alcohol, negative Tollens').

The conversion of 2-bromo-2-methylpropane (a tertiary halogenoalkane) to 2-methylpropene (an alkene) is an elimination reaction. Elimination is favored by using a strong base like \(\text{NaOH}\) dissolved in hot ethanol (ethanolic solution) under reflux. Aqueous \(\text{NaOH}\) under reflux would favor nucleophilic substitution instead, yielding 2-methylpropan-2-ol. Therefore, option B is correct.

1 mark for the correct option (B).
- Accept: Option B.
- Reject: Option A (leads to nucleophilic substitution), Option C and D (lead to nucleophilic substitution products, nitriles and amines respectively).

Let's evaluate each statement:
- Option A: Incorrect. HF has the highest boiling point among the hydrogen halides because of strong intermolecular hydrogen bonding, whereas others experience weaker permanent dipole-dipole and London dispersion forces.
- Option B: Incorrect. Thermal stability decreases down the group because the H–X bond length increases, leading to weaker bonds with lower bond enthalpies, which are more easily broken on heating.
- Option C: Correct. HF is a weak acid due to the high strength of the H–F bond, preventing complete dissociation. Down the group, the H–X bond enthalpy decreases dramatically (H–Cl > H–Br > H–I), which makes it much easier for the molecules to dissociate in water and release \(\text{H}^+\) ions. Therefore, acid strength increases down the group.
- Option D: Incorrect. Iodide is a strong reducing agent and reduces concentrated sulfuric acid to hydrogen sulfide (\(\text{H}_2\text{S}\)), sulfur dioxide (\(\text{SO}_2\)), and sulfur (\(\text{S}\)). Thus, HI is not the only gaseous product.

1 mark for the correct option (C).
- Accept: Option C.
- Reject: Options A, B, and D.

1. Calculate the number of moles of \(\text{CO}_2\) gas evolved:
\(\text{Moles of } \text{CO}_2 = \frac{540\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0225\text{ mol}\)

2. From the stoichiometry of the reaction:
\(\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)
\(1\text{ mole}\) of \(\text{CaCO}_3\) produces \(1\text{ mole}\) of \(\text{CO}_2\). Therefore, there are \(0.0225\text{ mol}\) of \(\text{CaCO}_3\) in the mixture.

3. Calculate the mass of \(\text{CaCO}_3\):
\(\text{Mass of } \text{CaCO}_3 = 0.0225\text{ mol} \times 100.1\text{ g mol}^{-1} = 2.25\text{ g}\)

4. Calculate the mass of \(\text{CaCl}_2\) in the mixture:
\(\text{Mass of } \text{CaCl}_2 = 5.00\text{ g} - 2.25\text{ g} = 2.75\text{ g}\)

5. Calculate the percentage by mass of \(\text{CaCl}_2\):
\(\text{Percentage of } \text{CaCl}_2 = \frac{2.75\text{ g}}{5.00\text{ g}} \times 100\% = 55.0\%\)

Award 1 mark for the correct answer B.
- Method: Calculate moles of CO2, relate to moles of CaCO3, find mass of CaCl2 and convert to %.
- Reject other choices as they represent wrong stoichiometry or calculating the % of CaCO3 instead of CaCl2.

Titanium has the outer electronic configuration \([\text{Ar}] 3\text{d}^2 4\text{s}^2\). When it forms the \(\text{Ti}^{4+}\) ion, it loses all four outer electrons, resulting in the electronic configuration \([\text{Ar}]\) (or \(3\text{d}^0\)). Since there are no electrons in the d-orbitals of the \(\text{Ti}^{4+}\) ion, d-d electronic transitions cannot take place, and the complex does not absorb light in the visible spectrum. Thus, it appears colourless.

Award 1 mark for selecting B.
- A is incorrect because Ti(IV) has 3d0, not 3d10.
- C is incorrect because d-d transitions are impossible regardless of energy without d-electrons.
- D is incorrect because water does cause splitting, but there are no electrons to split.

Let the original rate be \(\text{Rate}_1 = k[X]^2[Y]\).
When the concentration of X is halved to \(0.5[X]\) and the concentration of Y is tripled to \(3[Y]\), the new rate, \(\text{Rate}_2\), is given by:
\(\text{Rate}_2 = k(0.5[X])^2(3[Y]) = k(0.25[X]^2)(3[Y]) = 0.75 k[X]^2[Y] = 0.75 \times \text{Rate}_1\).

Therefore, the ratio of the new rate to the original rate is:
\(\frac{\text{Rate}_2}{\text{Rate}_1} = 0.75\).

Award 1 mark for B.
- Correctly square the change in [X] (0.5^2 = 0.25) and multiply by the change in [Y] (3) to get 0.75.
- Incorrect choices represent failure to square the change in [X] (leading to 1.50) or other mathematical errors.

The hydroxyl group (-OH) on the benzene ring of 2-methylphenol is a highly activating, ortho/para-directing group. The methyl group (-CH3) is also activating and ortho/para-directing, but the -OH group is a much stronger activating group and dominates the regiochemistry. The positions relative to -OH (C1) are:
- C2: occupied by -CH3
- C4: para position (vacant)
- C6: ortho position (vacant)

Electrophilic aromatic substitution (coupling) preferentially occurs at the less sterically hindered para position (C4) relative to the highly activating -OH group. Thus, substitution occurs at Position 4.

Award 1 mark for B.
- Identify -OH as the dominant directing group (ortho/para).
- Identify that position 4 is para to -OH and is less sterically hindered than position 6 (ortho).
- Distinguish from position 3 or 5, which are meta to the directing -OH group.

- Step 1: Converting a primary alcohol (propan-1-ol) to a primary halogenoalkane (1-bromopropane) requires a halogenating agent such as \(\text{PBr}_3\) with heat.
- Step 2: Nucleophilic substitution of the bromide with a nitrile group requires potassium cyanide (\(\text{KCN}\)) in ethanol, heated under reflux. Using \(\text{HCN}\) is incorrect as it is a weak nucleophile and highly toxic gas.
- Step 3: Hydrolysis of the nitrile to a carboxylic acid (butanoic acid) requires heating under reflux with a dilute mineral acid, such as dilute \(\text{HCl}(aq)\).

Award 1 mark for B.
- Step 1: recognize PBr3/heat or NaBr/H2SO4 as appropriate for alcohols.
- Step 2: recognize KCN in ethanol/reflux (CN- is the nucleophile, HCN is incorrect).
- Step 3: recognize acidic hydrolysis of nitrile to carboxylic acid.

1. W reacts with 2,4-DNPH, so it is a carbonyl compound (aldehyde or ketone).
2. W does not react with Tollens' reagent, so W is a ketone. Since W has 4 carbon atoms (\(\text{C}_4\text{H}_8\text{O}\)), W must be butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)).
3. Reduction of butanone (W) with \(\text{NaBH}_4\) yields butan-2-ol (X), which is a secondary alcohol: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
4. Evaluating the statements for X (butan-2-ol):
- A: Dehydration of butan-2-ol yields but-1-ene and but-2-ene (which shows cis-trans isomerism), so more than one alkene is formed. (Incorrect)
- B: Alcohols do not react with Tollens' reagent. (Incorrect)
- C: Butan-2-ol is a secondary alcohol. (Incorrect)
- D: Butan-2-ol contains the methyl secondary alcohol structure, \(\text{CH}_3\text{CH(OH)}-\), so it undergoes the triiodomethane (iodoform) reaction to give a yellow precipitate. (Correct)

Award 1 mark for D.
- Correctly deduce that W is butanone and X is butan-2-ol.
- Correctly identify that butan-2-ol contains the CH3CH(OH)- group and thus gives a positive iodoform test (yellow precipitate of CHI3).

Two main factors determine the rate of hydrolysis of halogenoalkanes:
1. **Mechanism type (tertiary vs. primary):** Tertiary halogenoalkanes (3 and 4) react via the \(\text{S}_N1\) mechanism, which is much faster than the \(\text{S}_N2\) mechanism of primary halogenoalkanes (1 and 2), because of the stable tertiary carbocation intermediate.
2. **Carbon-halogen bond strength:** C-Br bonds are weaker than C-Cl bonds, so bromoalkanes hydrolyse faster than their chloroalkane analogues.

Combining these factors:
- Tertiary bromoalkane (4) is the fastest.
- Tertiary chloroalkane (3) is the second fastest.
- Primary bromoalkane (2) is the third fastest.
- Primary chloroalkane (1) is the slowest.

Thus, the correct order from fastest to slowest is 4 \(\rightarrow\) 3 \(\rightarrow\) 2 \(\rightarrow\) 1.

Award 1 mark for A.
- Deduce that tertiary halogenoalkanes (3, 4) hydrolyse faster than primary halogenoalkanes (1, 2) due to SN1 mechanism superiority.
- Deduce that bromoalkanes hydrolyse faster than chloroalkanes due to lower bond dissociation energy of C-Br compared to C-Cl.

When concentrated sulfuric acid is added to solid potassium iodide (containing \(\text{I}^-\) ions):
1. Highly acidic hydrogen iodide gas (\(\text{HI}\)) is formed, which fumes in moist air.
2. Since \(\text{I}^-\) is a very strong reducing agent, it reduces the concentrated sulfuric acid to various products, including sulfur dioxide (\(\text{SO}_2\)), sulfur (\(\text{S}\)), and hydrogen sulfide (\(\text{H}_2\text{S}\)).
3. The yellow solid observed is elemental sulfur (\(\text{S}\)).
4. The iodide ions are oxidized to iodine (\(\text{I}_2\)), which is seen as purple vapours in the reaction vessel.

In contrast, bromide ions only reduce sulfuric acid to sulfur dioxide (no yellow sulfur is produced), and chloride ions cannot reduce concentrated sulfuric acid at all.

Award 1 mark for B.
- Identify that only iodide (I-) is strong enough to reduce sulfuric acid to yellow elemental sulfur (S) and produce purple vapours of iodine (I2).
- Reject A because bromide does not reduce sulfuric acid to elemental sulfur.
- Reject C because iodine is not a yellow solid (it is dark grey/purple).

(a)(i) Calculate the molar mass of anhydrous cobalt(II) sulfate: \(M_r(\text{CoSO}_4) = 58.9 + 32.1 + (4 \times 16.0) = 155.0\text{ g mol}^{-1}\). Number of moles of anhydrous \(\text{CoSO}_4 = 2.325 / 155.0 = 0.0150\text{ mol}\). (a)(ii) Mass of water lost = \(4.215 - 2.325 = 1.890\text{ g}\). Moles of water lost = \(1.890 / 18.0 = 0.105\text{ mol}\). Ratio of \(\text{H}_2\text{O}\) to \(\text{CoSO}_4 = 0.105 / 0.0150 = 7\). Thus, \(x = 7\). (b)(i) The ionic equation is \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). (b)(ii) Since 1 mole of \(\text{CoSO}_4\) yields 1 mole of \(\text{BaSO}_4\), moles of \(\text{BaSO}_4 = 0.0150\text{ mol}\). Molar mass of \(\text{BaSO}_4 = 137.3 + 32.1 + (4 \times 16.0) = 233.4\text{ g mol}^{-1}\). Mass of \(\text{BaSO}_4 = 0.0150 \times 233.4 = 3.50\text{ g}\). (c) Convert temperature and volume to SI units: \(T = 373\text{ K}\), \(V = 85.3 \times 10^{-6}\text{ m}^3\). Apply the ideal gas equation \(pV = nRT\): \(n = pV / RT = (1.01 \times 10^5 \times 85.3 \times 10^{-6}) / (8.31 \times 373) = 8.6153 / 3099.63 = 2.78 \times 10^{-3}\text{ mol}\). Calculate relative molecular mass: \(M_r = \text{mass} / n = 0.200 / (2.78 \times 10^{-3}) = 72.0\). (d)(i) Divide mass percentages by respective atomic masses: C: \(83.3 / 12.0 = 6.94\), H: \(16.7 / 1.0 = 16.7\). Divide by the smallest value: C: \(6.94 / 6.94 = 1.0\), H: \(16.7 / 6.94 = 2.4\). Multiply by 5 to obtain whole numbers: C = 5, H = 12. Therefore, the empirical formula is \(\text{C}_5\text{H}_{12}\). (d)(ii) The empirical formula mass of \(\text{C}_5\text{H}_{12}\) is \((5 \times 12.0) + (12 \times 1.0) = 72.0\). Since this matches the calculated \(M_r\) of 72.0, the molecular formula is also \(\text{C}_5\text{H}_{12}\).

(a)(i) M1: Calculation of Mr of CoSO4 as 155.0. M2: Correct calculation of moles to three significant figures (0.0150 mol). (a)(ii) M1: Calculation of mass of water lost (1.890 g) and moles of water (0.105 mol). M2: Correct ratio deduction of x = 7. (b)(i) M1: Correct species: Ba2+ and SO42- reacting to form BaSO4. M2: Correct state symbols: (aq) for reactants and (s) for product. (b)(ii) M1: Calculation of Mr of BaSO4 as 233.4. M2: Final mass calculation of 3.50 g (accept 3.5 g). (c) M1: Temperature conversion to 373 K and volume conversion to 85.3 x 10^-6 m3. M2: Rearrangement of pV=nRT to solve for n. M3: Calculation of n as 2.78 x 10^-3 mol. M4: Relative molecular mass calculated as 72.0 (allow range 71.8 to 72.2). (d)(i) M1: Correct division of percentages by Ar values. M2: Obtaining the simplified ratio of 1 to 2.4 and scaling up to C5H12. (d)(ii) M1: Deduction of C5H12 by comparing empirical mass to molecular mass.

(a)(i) The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane. This is because the rate depends on the bond strength (bond enthalpy) of the carbon-halogen bond. The C-I bond has the lowest bond enthalpy and is the easiest to break, while the C-Cl bond has the highest bond enthalpy and is the hardest to break. (a)(ii) The precipitate formed with 1-bromobutane is cream-colored silver bromide, AgBr. (b)(i) Step 1: Heterolytic fission of the C-Br bond occurs. A curly arrow starts from the C-Br bond and points to the Br atom. This produces a tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\). Step 2: The hydroxide ion, \(\text{OH}^-\), acts as a nucleophile. A curly arrow starts from a lone pair on the oxygen of the \(\text{OH}^-\)- ion and points to the positively charged carbon atom of the carbocation, forming 2-methylpropan-2-ol. (b)(ii) The reaction proceeds via \(\text{S}_\text{N}1\) because 2-bromo-2-methylpropane is a tertiary halogenoalkane. The tertiary carbocation intermediate is highly stable due to the electron-donating inductive effect of the three methyl groups. Additionally, the bulky methyl groups around the central carbon cause steric hindrance, preventing a nucleophile from attacking via the back-side route required for an \(\text{S}_\text{N}2\) mechanism. (c)(i) The elimination of HBr from 2-bromobutane can occur in two directions. Elimination involving hydrogen from C1 yields but-1-ene. Elimination involving hydrogen from C3 yields but-2-ene, which exists as a pair of stereoisomers: cis-but-2-ene (or (Z)-but-2-ene) and trans-but-2-ene (or (E)-but-2-ene). (c)(ii) Cis-but-2-ene and trans-but-2-ene exhibit geometrical (or cis-trans / stereoisomerism) isomerism. This is because there is restricted rotation around the carbon-carbon double bond (C=C), and each carbon atom of the double bond is attached to two different groups (a hydrogen atom and a methyl group).

(a)(i) M1: State that the rate of hydrolysis increases from chlorine to iodine. M2: Relate rate to the bond strength/enthalpy of the C-X bond. M3: Explain that C-I is the weakest bond and therefore breaks most easily. (a)(ii) M1: Cream (accept off-white; reject yellow or white). (b)(i) M1: Step 1 shows heterolytic fission of C-Br bond to form a carbocation intermediate. M2: Curly arrow from C-Br bond to Br. M3: Curly arrow from lone pair on OH- to carbocation carbon. M4: Structure of tertiary carbocation shown with positive charge on the central carbon. (b)(ii) M1: Explain carbocation stability due to positive inductive effect of three alkyl groups. M2: Mention steric hindrance preventing back-side nucleophilic attack. (c)(i) M1: But-1-ene. M2: cis-but-2-ene (or (Z)-but-2-ene). M3: trans-but-2-ene (or (E)-but-2-ene). (c)(ii) M1: Geometrical / cis-trans isomerism. M2: Restricted rotation about C=C bond (and two different groups attached to each C).

(a)(i) Chlorine reacts with cold, dilute aqueous sodium hydroxide according to the equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). (a)(ii) The oxidation number of chlorine in the reactant \(\text{Cl}_2\) is 0. In the products: in \(\text{NaCl}\) it is -1, and in \(\text{NaClO}\) it is +1. (a)(iii) This is a disproportionation reaction because chlorine is simultaneously oxidized and reduced. (a)(iv) In hot, concentrated NaOH, chlorine undergoes disproportionation to form sodium chlorate(V), \(\text{NaClO}_3\), instead of sodium chlorate(I). The oxidation state of chlorine in \(\text{NaClO}_3\) is +5. (b)(i) Observation: Steamy / misty / white fumes are observed. Equation: \(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\) (or \(2\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Na}_2\text{SO}_4\text{(s)} + 2\text{HCl(g)}\)). (b)(ii) The dark grey solid is iodine, \(\text{I}_2\). The yellow solid is sulfur, \(\text{S}\). The gas with the smell of rotten eggs is hydrogen sulfide, \(\text{H}_2\text{S}\). (b)(iii) Iodide ions, \(\text{I}^-\), are stronger reducing agents than chloride ions, \(\text{Cl}^-\). This is because the iodide ion has a larger ionic radius and has more electron shielding. Consequently, the attraction between the nucleus and the outermost electrons is weaker, making it easier for iodide ions to lose electrons and oxidize back to iodine.

(a)(i) M1: Reactants and products correct (Cl2, NaOH, NaCl, NaClO, H2O). M2: Correct balancing. (a)(ii) M1: Reactant Cl2 = 0. M2: Products NaCl = -1 and NaClO = +1. (a)(iii) M1: Disproportionation. (a)(iv) M1: Identifies NaClO3 (sodium chlorate(V)). M2: States oxidation state of chlorine is +5. (b)(i) M1: Observation of misty/steamy/white fumes. M2: Correct species in equation. M3: Correct balancing of the acid-base equation. (b)(ii) M1: Dark grey solid is iodine (I2). M2: Yellow solid is sulfur (S). M3: Rotten egg gas is hydrogen sulfide (H2S). (b)(iii) M1: States that iodide is a stronger reducing agent than chloride. M2: Explains in terms of larger ionic radius / more shielding / valence electrons more easily lost.

(a)(i) Isomer 1: Propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\). Isomer 2: Propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\). (a)(ii) Compound X is propan-1-ol. Propanoic acid (W, \(\text{C}_3\text{H}_6\text{O}_2\)) is a carboxylic acid, which is formed by the complete oxidation of the primary alcohol, propan-1-ol. If X were propan-2-ol (a secondary alcohol), it would be oxidized to propanone (\(\text{C}_3\text{H}_6\text{O}\)), which cannot be oxidized further to a carboxylic acid with three carbon atoms. (a)(iii) The color change is from orange (due to \(\text{Cr}_2\text{O}_7^{2-}\)) to green (due to \(\text{Cr}^{3+}\)). (b)(i) The two characteristic IR absorptions for propanoic acid (W) are: 1. A broad absorption due to the O-H (carboxylic acid) bond in the range \(2500 - 3000\text{ cm}^{-1}\). 2. A strong, sharp absorption due to the C=O bond in the range \(1670 - 1740\text{ cm}^{-1}\). (b)(ii) To obtain the aldehyde V (propanal) as the main product, the reaction mixture should be distilled immediately (distillation with addition) so that the volatile aldehyde is distilled off as soon as it forms, preventing further oxidation to propanoic acid. Additionally, a limiting amount of the oxidizing agent (potassium dichromate(VI)) should be used. (c)(i) An orange, yellow, or red precipitate is formed when 2,4-DNPH is added to a carbonyl compound. (c)(ii) This test identifies the carbonyl functional group (aldehydes and ketones). (c)(iii) Test: Heat gently with Tollens' reagent (ammoniacal silver nitrate) or Fehling's solution. With V (propanal, an aldehyde), a silver mirror is observed on the walls of the tube (or a red precipitate with Fehling's). With Z (propanone, a ketone), no visible change occurs / the solution remains colorless.

(a)(i) M1: Correct structural formula of propan-1-ol. M2: Correct structural formula of propan-2-ol. (a)(ii) M1: Identifies X as propan-1-ol. M2: Explains that primary alcohols oxidize to carboxylic acids whereas secondary alcohols oxidize to ketones. (a)(iii) M1: Orange to green. (b)(i) M1: O-H (carboxylic acid) bond absorption at 2500-3000 cm-1. M2: C=O bond absorption at 1670-1740 cm-1. (b)(ii) M1: Use of distillation (to remove product as it forms). M2: Use of excess alcohol / limiting oxidizing agent. (c)(i) M1: Orange / yellow / red precipitate (or solid). (c)(ii) M1: Carbonyl group (accept aldehyde or ketone). (c)(iii) M1: Named reagent: Tollens' reagent (or Fehling's solution / acidified potassium dichromate(VI)). M2: Heat / warm. M3: Observation with V: silver mirror / black precipitate (or red precipitate for Fehling's, or orange to green for dichromate). M4: Observation with Z: no change / remains colorless / no reaction.

First, we calculate the average initial temperature \(T_{\text{avg}}\) and temperature rise \(\Delta T\) for each experiment: Experiment 1: \(T_{\text{avg}} = (21.0+21.2)/2 = 21.1\text{ }^{\circ}\text{C}\), \(\Delta T = 26.3 - 21.1 = 5.2\text{ }^{\circ}\text{C}\). Experiment 2: \(T_{\text{avg}} = 21.1\text{ }^{\circ}\text{C}\), \(\Delta T = 31.4 - 21.1 = 10.3\text{ }^{\circ}\text{C}\). Experiment 3: \(T_{\text{avg}} = 21.2\text{ }^{\circ}\text{C}\), \(\Delta T = 33.9 - 21.2 = 12.7\text{ }^{\circ}\text{C}\). Experiment 4: \(T_{\text{avg}} = 21.2\text{ }^{\circ}\text{C}\), \(\Delta T = 31.3 - 21.2 = 10.1\text{ }^{\circ}\text{C}\). Experiment 5: \(T_{\text{avg}} = 21.3\text{ }^{\circ}\text{C}\), \(\Delta T = 26.2 - 21.3 = 4.9\text{ }^{\circ}\text{C}\). Plotting \(\Delta T\) against the volume of FB 1 reveals two straight lines of best fit: an increasing line passing through (10.0, 5.2) and (20.0, 10.3), and a decreasing line passing through (30.0, 10.1) and (40.0, 4.9). Solving the equations of these lines: Line 1: \(\Delta T = 0.51 \times V\). Line 2: \(\Delta T = -0.52 \times V + 25.7\). Their intersection occurs at \(V_{\text{neutral}} = 25.0\text{ cm}^3\) of FB 1, which corresponds to \(25.0\text{ cm}^3\) of FB 2. Since \(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\), the moles of \(NaOH\) reacted are \(0.0250\text{ dm}^3 \times 2.50\text{ mol dm}^{-3} = 0.0625\text{ mol}\). Thus, the moles of \(H_2SO_4\) in \(25.0\text{ cm}^3\) are \(0.0625 / 2 = 0.03125\text{ mol}\). The concentration of FB 1 is \(0.03125\text{ mol} / 0.0250\text{ dm}^3 = 1.25\text{ mol dm}^{-3}\).

- [1 mark]: Table with appropriate headers, units (e.g. Volume of FB 1 / \(cm^3\), Temperature / °C), and consistent precision.
- [1 mark]: Correct calculation of average initial temperatures (\(T_{\text{avg}}\)) to 1 decimal place.
- [1 mark]: Correct calculation of \(\Delta T\) for all 5 experiments.
- [2 marks]: Graph plotting with correctly labeled axes, linear scales, and points accurately plotted.
- [1 mark]: Two straight, intersecting lines of best fit correctly drawn.
- [1 mark]: Value of \(V_{\text{neutral}}\) read correctly from the intersection as 25.0 \(cm^3\).
- [1 mark]: Calculation of moles of NaOH reacting = 0.0625 mol.
- [1 mark]: Use of the correct 1:2 stoichiometric ratio (0.03125 mol of \(H_2SO_4\)).
- [2 marks]: Correct calculation of concentration of FB 1 as 1.25 \(\text{mol dm}^{-3}\) with 3 significant figures.
- [1 mark]: Explanation of the source of error in using a measuring cylinder (less precise than a burette).
- [1 mark]: Correct calculation of percentage error of measuring cylinder: \((0.5 / 10.0) \times 100\% = 5.0\%\).

First, we calculate the concentration of iodide ions \([I^-]\) using the dilution formula \(C_1 V_1 = C_2 V_2\) where \(V_2 = 42.00\text{ cm}^3\) and \(C_1 = 0.200\text{ mol dm}^{-3}\). Experiment 1: \([I^-] = 0.200 \times (20.00 / 42.00) = 0.0952\text{ mol dm}^{-3}\), \(1/t = 1 / 34.0 = 0.0294\text{ s}^{-1}\). Experiment 2: \([I^-] = 0.200 \times (15.00 / 42.00) = 0.0714\text{ mol dm}^{-3}\), \(1/t = 1 / 45.0 = 0.0222\text{ s}^{-1}\). Experiment 3: \([I^-] = 0.200 \times (10.00 / 42.00) = 0.0476\text{ mol dm}^{-3}\), \(1/t = 1 / 68.0 = 0.0147\text{ s}^{-1}\). Experiment 4: \([I^-] = 0.200 \times (5.00 / 42.00) = 0.0238\text{ mol dm}^{-3}\), \(1/t = 1 / 136.0 = 0.00735\text{ s}^{-1}\). Comparing Experiment 1 and 3: the concentration of iodide ions is halved (0.0952 to 0.0476), and the rate is also halved (0.0294 to 0.0147). This direct proportionality means the reaction is first-order with respect to \(I^-\).

- [2 marks]: Table presenting all raw and calculated data with appropriate headings and units.
- [1 mark]: Correctly calculated \([I^-]\) values to 3 significant figures.
- [2 marks]: Correctly calculated relative rate (\(1/t\)) values to 3 significant figures with units (\(s^{-1}\)).
- [2 marks]: Explanation showing that doubling the concentration of \(I^-\) (e.g., from Exp 4 to Exp 3) doubles the rate of reaction.
- [1 mark]: Deducing the order is 1.
- [2 marks]: Identifying a major source of error: difficulty in judging the exact start or end-point color change, or temperature variations.
- [3 marks]: Suggesting a valid improvement: using a colorimeter with a data logger to detect the exact onset of color change, or using a thermostatic water bath to keep temperature constant.

The concordant titres are Titre 1 (19.75 \(cm^3\)), Titre 2 (19.65 \(cm^3\)), and Titre 3 (19.70 \(cm^3\)), as they are all within 0.10 \(cm^3\). (i) Mean titre = \((19.75 + 19.65 + 19.70) / 3 = 19.70\text{ cm}^3\). (ii) Moles of \(HCl\) used in the titration = \(19.70 \times 10^{-3}\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 3.94 \times 10^{-3}\text{ mol}\). Since 1 mole of \(HCl\) reacts with 1 mole of \(NH_3\), the moles of \(NH_3\) in the \(25.0\text{ cm}^3\) sample = \(3.94 \times 10^{-3}\text{ mol}\). The moles of \(NH_3\) in the \(250.0\text{ cm}^3\) volumetric flask = \(3.94 \times 10^{-3} \times 10 = 3.94 \times 10^{-2}\text{ mol}\). (iii) Mass of \(NH_3\) in FB 1 = \(3.94 \times 10^{-2}\text{ mol} \times 17.03\text{ g mol}^{-1} = 0.67098\text{ g}\). Percentage by mass of ammonia in FB 1 = \((0.67098 / 2.42) \times 100 = 27.72\%\), which is 27.7% to 3 significant figures.

- [2 marks]: Table of titration results showing initial and final burette readings to 2 decimal places with units, and clearly indicating the concordant titres.
- [1 mark]: Correct calculation of mean titre = 19.70 \(cm^3\).
- [2 marks]: Correct calculation of moles of HCl = \(3.94 \times 10^{-3}\text{ mol}\).
- [2 marks]: Correct calculation of moles of \(NH_3\) in \(250.0\text{ cm}^3\) = \(3.94 \times 10^{-2}\text{ mol}\) (10x multiplier applied).
- [2 marks]: Correct calculation of mass of \(NH_3\) = 0.671 g.
- [2 marks]: Correct calculation of percentage by mass of ammonia = 27.7%.
- [2 marks]: Appropriate and consistent use of significant figures (3 or 4) throughout the entire calculation.

(a) Electronic configuration of \( \text{Co} \) is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7 4\text{s}^2 \). The configuration of \( \text{Co}^{2+} \) is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7 \). In the presence of ligands, the five degenerate d-orbitals split into two sets of different energy levels. Electrons absorb a photon of visible light and are promoted from the lower energy d-orbitals to the higher energy d-orbitals (d-d transition). The remaining non-absorbed wavelengths of light are transmitted, resulting in a colored solution.

(b) (i) \( K_{\text{stab}} = \frac{[[\text{CoCl}_4]^{2-}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4} \). The units are \( \text{dm}^{12} \text{mol}^{-4} \).
(ii) The color changes from pink (due to \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \)) to blue (due to \( [\text{CoCl}_4]^{2-} \)).

(c) (i) \( [\text{Co}(\text{en})_3]^{2+} \) is much more thermodynamically stable than \( [\text{Co}(\text{NH}_3)_6]^{2+} \) as indicated by its significantly higher \( K_{\text{stab}} \) value (\( 10^{13.8} \gg 10^{4.6} \)).
(ii) The reaction \( [\text{Co}(\text{NH}_3)_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightleftharpoons [\text{Co}(\text{en})_3]^{2+}(\text{aq}) + 6\text{NH}_3(\text{aq}) \) proceeds with an increase in the number of free particles in solution (from 4 reactant species to 7 product species). This results in a large positive entropy change, \( \Delta S^\ominus > 0 \), which drives the reaction to the right (the chelate effect).

(d) (i) Cis-trans (geometrical) isomerism and optical isomerism.
(ii) The cis-isomer is chiral (optically active). The student should draw a central cobalt ion octahedrally coordinated by two bidentate 'en' ligands (each spanning adjacent/cis positions) and two chloride ligands in cis positions, demonstrating a non-superimposable mirror image.

(a)
- M1: Correct electronic configuration of \( \text{Co}^{2+} \): \( [\text{Ar}]3\text{d}^7 \) or fully written out. [1]
- M2: Explanation of d-orbital splitting by ligands into two levels. [1]
- M3: Absorption of visible light causing d-d electron promotion, transmitting the complementary color. [1]

(b)
- M1: Correct expression for \( K_{\text{stab}} \) (water omitted). [1]
- M2: Correct units: \( \text{dm}^{12} \text{mol}^{-4} \). [1]
- M3: Color change: Pink to blue. [0.5]

(c)
- M1: States that \( [\text{Co}(\text{en})_3]^{2+} \) is more stable because its \( K_{\text{stab}} \) is larger. [1]
- M2: Mentions ligand exchange equation or particle count change (from 4 particles on LHS to 7 on RHS). [1]
- M3: Relates this to a positive entropy change (\( \Delta S^\ominus > 0 \)). [1]
- M4: Correctly identifies the phenomenon as the chelate effect. [1]

(d)
- M1: Identifies both cis-trans (geometrical) and optical isomerism. [1]
- M2: Draws the cis-isomer showing the 3D octahedral geometry correctly. [1]
- M3: Shows the bidentate ligands connected to adjacent coordination sites. [1]

(a) Comparing Run 1 and Run 2: \( [\text{B}] \) is constant. \( [\text{A}] \) is doubled, and the initial rate doubles (from \( 2.0 \times 10^{-4} \) to \( 4.0 \times 10^{-4} \)). Therefore, the reaction is first order with respect to A.
Comparing Run 1 and Run 3: \( [\text{A}] \) is constant. \( [\text{B}] \) is tripled, and the rate increases by a factor of \( 9 \) (from \( 2.0 \times 10^{-4} \) to \( 1.8 \times 10^{-3} \)). Since \( 3^2 = 9 \), the reaction is second order with respect to B.
Rate equation: \( \text{Rate} = k[\text{A}][\text{B}]^2 \).

(b) Using data from Run 1:
\( 2.0 \times 10^{-4} = k(0.10)(0.10)^2 \)
\( 2.0 \times 10^{-4} = k(1.0 \times 10^{-3}) \)
\( k = 0.20 \)
Units of \( k \): \( \text{mol dm}^{-3}\text{ s}^{-1} / (\text{mol dm}^{-3} \times \text{mol}^2\text{ dm}^{-6}) = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

(c) The rate-determining step (slow step) must involve one molecule of A and two molecules of B overall up to this step. A proposed mechanism is:
Step 1 (Slow): \( \text{A} + \text{B} \rightarrow \text{I} \) (where I is an intermediate)
Step 2 (Fast): \( \text{I} + \text{B} \rightarrow \text{C} + \text{D} \)
(Alternative: \( 2\text{B} \rightleftharpoons \text{B}_2 \) fast, followed by \( \text{B}_2 + \text{A} \rightarrow \text{C} + \text{D} \) slow).

(d) When temperature increases, the value of the rate constant \( k \) increases. This is because the average kinetic energy of the molecules increases, shifting the Boltzmann distribution to the right. Consequently, a much larger fraction of colliding molecules possess kinetic energy greater than or equal to the activation energy (\( E_a \)), leading to a higher frequency of successful collisions.

(a)
- M1: Deduces first order for A with clear working comparing Run 1 and 2. [1]
- M2: Deduces second order for B with clear working comparing Run 1 and 3. [1]
- M3: Correctly states the rate equation: \( \text{Rate} = k[\text{A}][\text{B}]^2 \). [2]

(b)
- M1: Correct numerical calculation of \( k = 0.20 \). [1.5]
- M2: Correct units: \( \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \). [1]

(c)
- M1: Proposes a logical multi-step mechanism where reactants sum to \( \text{A} + 2\text{B} \rightarrow \text{C} + \text{D} \). [1]
- M2: Rate-determining step includes the correct stoichiometry matching the rate equation. [1]
- M3: Clearly identifies which step is slow/fast. [1]

(d)
- M1: States that \( k \) increases with temperature. [1]
- M2: Explains using Maxwell-Boltzmann distribution that more particles have energy \( \ge E_a \). [1]
- M3: Explains that there is a higher frequency of successful collisions. [1]

(a) Step 1: Nitration of methylbenzene.
- Reagents & conditions: Concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \), temperature maintained around \( 50-55^\circ\text{C} \).
- Intermediate product: 4-nitrotoluene (separated from the 2-isomer).

Step 2: Oxidation of the methyl group.
- Reagents & conditions: Acidified potassium manganate(VII), \( \text{KMnO}_4 / \text{H}^+ \), heat under reflux.
- Intermediate product: 4-nitrobenzoic acid.

Step 3: Reduction of the nitro group.
- Reagents & conditions: Tin (Sn) and concentrated hydrochloric acid, heated under reflux, followed by addition of aqueous sodium hydroxide to release the free amine.
- Product: 4-aminobenzoic acid.

(b) (i) \( \text{Cl}^-\text{H}_3\text{N}^+-\text{C}_6\text{H}_4-\text{COOH} \) (protonation of the basic amine group to form an ammonium salt).
(ii) \( \text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{COOCH}_2\text{CH}_3 \) (esterification of the carboxylic acid group).
(iii) \( \text{H}_2\text{N}-\text{C}_6\text{H}_2\text{Br}_2-\text{COOH} \) where bromine atoms substitute at positions 3 and 5 relative to the carboxylic acid group (directing effect of the strongly activating \( -\text{NH}_2 \) group at position 4 directs incoming electrophiles to positions 3 and 5, orthoto the amine).
(iv) An azo dye: \( \text{HO}-\text{C}_6\text{H}_4-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{COOH} \).

(a)
- M1: Step 1 reagents and conditions: conc. \( \text{HNO}_3 \) + conc. \( \text{H}_2\text{SO}_4 \), \( 50-60^\circ\text{C} \). [1]
- M2: Structure of 4-nitrotoluene. [1]
- M3: Step 2 reagents and conditions: \( \text{KMnO}_4 \) (alkaline or acidic), heat/reflux. [1]
- M4: Structure of 4-nitrobenzoic acid. [1]
- M5: Step 3 reagents and conditions: Sn + conc. \( \text{HCl} \), heat, followed by NaOH(aq). [1]
- M6: Clearly links the sequence in a viable logical order. [1]

(b)
- (i) M1: Correct ammonium salt structure showing protonation of amine only. [1.5]
- (ii) M1: Correct ester structure. [1.5]
- (iii) M1: Shows substitution of two bromine atoms ortho to the amine group. [2.5]
- (iv) M1: Correct structures of the azo dye coupling product. [1]

(a) Step 1: Benzene to methylbenzene
- Reagents and conditions: Chloromethane, \( \text{CH}_3\text{Cl} \), in the presence of anhydrous aluminium chloride, \( \text{AlCl}_3 \), catalyst. Room temperature.
- Product: Methylbenzene, \( \text{C}_6\text{H}_5\text{CH}_3 \).

Step 2: Methylbenzene to (bromomethyl)benzene
- Reagents and conditions: Bromine, \( \text{Br}_2 \), with ultraviolet (UV) light / heat under reflux.
- Product: (Bromomethyl)benzene, \( \text{C}_6\text{H}_5\text{CH}_2\text{Br} \).

Step 3: (Bromomethyl)benzene to phenylethanenitrile
- Reagents and conditions: Potassium cyanide, \( \text{KCN} \) in aqueous ethanol, heat under reflux.
- Product: Phenylethanenitrile, \( \text{C}_6\text{H}_5\text{CH}_2\text{CN} \).

Step 4: Phenylethanenitrile to phenylethanoic acid
- Reagents and conditions: Dilute hydrochloric acid (or dilute sulfuric acid), heat under reflux.
- Product: Phenylethanoic acid, \( \text{C}_6\text{H}_5\text{CH}_2\text{COOH} \).

(b) Relative ease of hydrolysis (highest to lowest reactivity):
(bromomethyl)benzene > 1-bromobutane > bromobenzene.

Explanation:
- Bromobenzene: Extremely resistant to nucleophilic attack because the p-orbital on the bromine atom overlaps with the pi-delocalized electron system of the benzene ring. This imparts partial double bond character to the C-Br bond, making it stronger and harder to break. Furthermore, the high electron density of the pi ring repels incoming nucleophiles.
- 1-Bromobutane: Standard primary halogenoalkane. Hydrolysis occurs via nucleophilic substitution (predominantly \( \text{S}_\text{N}2 \)), where the polar C-Br bond is attacked by the nucleophile.
- (Bromomethyl)benzene: Highly reactive toward hydrolysis because the C-Br bond cleavage (via \( \text{S}_\text{N}1 \)) produces a benzyl carbocation, \( \text{C}_6\text{H}_5\text{CH}_2^+ \), which is highly stabilized by the delocalization of the positive charge over the pi system of the benzene ring.

(a)
- M1: Step 1 reagents (\( \text{CH}_3\text{Cl} \) + anhydrous \( \text{AlCl}_3 \)) and product methylbenzene. [1.5]
- M2: Step 2 reagents (\( \text{Br}_2 \) + UV light/heat) and product (bromomethyl)benzene. [1.5]
- M3: Step 3 reagents (KCN in ethanol, reflux) and product phenylethanenitrile. [2]
- M4: Step 4 reagents (dilute strong acid, reflux) and product phenylethanoic acid. [2]

(b)
- M1: Correctly orders reactivity: (bromomethyl)benzene > 1-bromobutane > bromobenzene. [1.5]
- M2: Explains that bromobenzene has partial double bond character due to p-orbital overlap with the pi ring, making C-Br exceptionally strong. [1.5]
- M3: Explains that the benzene ring repels nucleophiles in bromobenzene. [0.5]
- M4: Explains that (bromomethyl)benzene undergoes rapid hydrolysis due to the highly stable benzyl carbocation/transition state stabilized by the aromatic ring. [2]

(a) Calculate the mass of carbon in the sample:
\( n(\text{CO}_2) = \frac{3.52}{44.0} = 0.0800\text{ mol} \)
\( m(\text{C}) = 0.0800 \times 12.0 = 0.960\text{ g} \)

Calculate the mass of hydrogen in the sample:
\( n(\text{H}_2\text{O}) = \frac{1.44}{18.0} = 0.0800\text{ mol} \)
\( m(\text{H}) = 2 \times 0.0800 \times 1.0 = 0.160\text{ g} \)

Calculate the mass of oxygen in the sample:
\( m(\text{O}) = 1.44 - (0.960 + 0.160) = 0.320\text{ g} \)

Find the mole ratios:
\( n(\text{C}) = \frac{0.960}{12.0} = 0.080\text{ mol} \)
\( n(\text{H}) = \frac{0.160}{1.0} = 0.160\text{ mol} \)
\( n(\text{O}) = \frac{0.320}{16.0} = 0.020\text{ mol} \)

Divide by the smallest value (0.020):
\( \text{C} = \frac{0.080}{0.020} = 4 \)
\( \text{H} = \frac{0.160}{0.020} = 8 \)
\( \text{O} = \frac{0.020}{0.020} = 1 \n
Empirical formula is \) \text{C}_4\text{H}_8\text{O} \).

(b) (i) Ideal gas equation: \( pV = nRT \)
(ii) Rearranging to solve for molecular mass, \( M_r = \frac{mRT}{pV} \)
Convert units to SI:
\( T = 350\text{ K} \)
\( p = 101 \times 10^3\text{ Pa} \)
\( V = 173 \times 10^{-6}\text{ m}^3 \)
\( R = 8.31\text{ J K}^{-1}\text{ mol}^{-1} \)

\( M_r = \frac{0.432 \times 8.31 \times 350}{101 \times 10^3 \times 173 \times 10^{-6}} = \frac{1256.47}{17.473} = 71.9 \approx 72.0 \)

Since the empirical formula mass of \( \text{C}_4\text{H}_8\text{O} \) is \( 4(12) + 8(1) + 16 = 72 \), the molecular formula is also \( \text{C}_4\text{H}_8\text{O} \).

(c) (i) It forms a precipitate with 2,4-DNPH (carbonyl compound) but does not react with Fehling's solution (not an aldehyde). Therefore, Y is a ketone.
(ii) Structural formula: \( \text{CH}_3\text{COCH}_2\text{CH}_3 \). IUPAC name: Butanone.

(a)
- M1: Calculates moles of C (0.080 mol) and H (0.160 mol). [1.5]
- M2: Calculates mass of oxygen correctly (0.320 g). [1.5]
- M3: Calculates moles of oxygen (0.020 mol). [1]
- M4: Correct empirical formula \( \text{C}_4\text{H}_8\text{O} \). [1]

(b)
- M1: States \( pV = nRT \) or \( pV = \frac{m}{M_r}RT \). [1]
- M2: Correct conversion of volume to \( \text{m}^3 \) and pressure to \( \text{Pa} \). [1]
- M3: Computes \( M_r \approx 72 \). [1.5]
- M4: Deduce molecular formula is \( \text{C}_4\text{H}_8\text{O} \). [1]

(c)
- M1: Identifies Y as a ketone. [1]
- M2: Draws correct structure of butanone. [1]
- M3: Names the compound as butanone. [1]

(a) (i) Enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
(ii) Enthalpy change when one mole of an ionic crystalline solid is formed from its gaseous ions at infinite separation under standard conditions.

(b) By Hess's Law, the sum of enthalpies around the cycle equals the enthalpy of formation:
\( \Delta H_{\text{f}}^\ominus = \Delta H_{\text{at}}^\ominus(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + 2 \times \Delta H_{\text{at}}^\ominus(\text{F}) + 2 \times \text{EA}_1(\text{F}) + \Delta H_{\text{latt}}^\ominus \)

Note that the atomisation of fluorine molecular gas \( \text{F}_2(\text{g}) \rightarrow 2\text{F}(\text{g}) \) is equivalent to the bond energy of F-F: \( 2 \times \Delta H_{\text{at}}^\ominus(\text{F}) = E(\text{F-F}) = +158\text{ kJ mol}^{-1} \).

Substitute the values:
\( -1220 = +178 + 590 + 1145 + 158 + 2(-328) + \Delta H_{\text{latt}}^\ominus \)
\( -1220 = 2071 - 656 + \Delta H_{\text{latt}}^\ominus \)
\( -1220 = 1415 + \Delta H_{\text{latt}}^\ominus \)
\( \Delta H_{\text{latt}}^\ominus = -1220 - 1415 = -2635\text{ kJ mol}^{-1} \).

(c) Calcium fluoride is a highly ionic compound because calcium has a low charge density and fluorine is small and highly electronegative, leading to negligible polarization.
For zinc iodide, the zinc ion (\( \text{Zn}^{2+} \)) has a small ionic radius and high charge, giving it a high charge density and making it strongly polarizing. The iodide ion (\( \text{I}^- \)) is large and has its outer electrons far from the nucleus, making it highly polarizable. This causes significant polarization of the iodide electron cloud, introducing a significant degree of covalent character into the bonding, which makes the experimental lattice energy more exothermic than the theoretical purely ionic value.

(a)
- M1: Defines \( \Delta H_{\text{f}}^\ominus \) correctly (1 mole, from elements in standard states). [1.5]
- M2: Defines \( \Delta H_{\text{latt}}^\ominus \) correctly (1 mole of solid, from gaseous ions). [1.5]

(b)
- M1: Correctly constructs Born-Haber cycle diagram or algebraic equation. [2]
- M2: Uses standard F-F bond energy correctly as equivalent to 2 moles of atomic fluorine. [1]
- M3: Multiplies electron affinity of fluorine by 2. [1]
- M4: Correctly calculates \( \Delta H_{\text{latt}}^\ominus = -2635\text{ kJ mol}^{-1} \). [1]

(c)
- M1: Mentions calcium fluoride is almost completely ionic with negligible polarization. [1]
- M2: Explains zinc ion has a high charge density / strongly polarizing power. [1.5]
- M3: Explains iodide ion is large and highly polarizable. [1]
- M4: Concludes that polarization leads to significant covalent character in \( \text{ZnI}_2 \). [1]

(a) Under standard conditions:
\( E^\ominus_{\text{cell}} = E^\ominus(\text{cathode}) - E^\ominus(\text{anode}) = +0.80 - (+0.77) = +0.03\text{ V} \).
The positive potential indicates the forward reaction is feasible:
\( \text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s}) \).

(b) (i) For Half-cell 1:
\( E_1 = +0.77 + \frac{0.059}{1}\log\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]} \)
\( E_1 = +0.77 + 0.059 \log\frac{0.050}{1.0} = +0.77 + 0.059(-1.301) = +0.77 - 0.077 = +0.693\text{ V} \).

(ii) For Half-cell 2:
\( E_2 = +0.80 + \frac{0.059}{1}\log[\text{Ag}^+] \)
\( E_2 = +0.80 + 0.059 \log(0.10) = +0.80 + 0.059(-1.0) = +0.741\text{ V} \).

(iii) The cell potential is:
\( E_{\text{cell}} = E_2 - E_1 = +0.741 - (+0.693) = +0.048\text{ V} \).

(c) As the cell operates, the forward reaction proceeds. The concentration of reactants (\( \text{Fe}^{2+} \) and \( \text{Ag}^+ \)) decreases, while the concentration of products (\( \text{Fe}^{3+} \)) increases.
This shifts the positions of both equilibria: the potential of Half-cell 1 (\( E_1 \)) increases and the potential of Half-cell 2 (\( E_2 \)) decreases. Consequently, \( E_{\text{cell}} \) decreases, and the reaction becomes less feasible. When \( E_{\text{cell}} \) reaches \( 0\text{ V} \), the system is at equilibrium, and the reaction is no longer feasible in either direction.

(a)
- M1: Correct calculation of standard cell potential: \( E^\ominus_{\text{cell}} = +0.03\text{ V} \). [1]
- M2: Correct overall cell equation with balanced states. [2]

(b)
- M1: Correct application of Nernst equation for Half-cell 1 showing substituted values. [1.5]
- M2: Correct calculation of \( E_1 = +0.693\text{ V} \). [1]
- M3: Correct calculation of \( E_2 = +0.741\text{ V} \). [1.5]
- M4: Correct calculation of \( E_{\text{cell}} = +0.048\text{ V} \). [1.5]

(c)
- M1: Mentions concentration changes as the reaction progresses (reactants decrease, products increase). [1]
- M2: Relates concentration changes to shifting half-cell potentials. [1]
- M3: Identifies that \( E_{\text{cell}} \) decreases over time. [1]
- M4: States that feasibility ceases when \( E_{\text{cell}} = 0\text{ V} \) (at equilibrium). [1]

(a) (i) When \( \text{H}^+ \) ions are added, they react with the conjugate base (propanoate ions) in the buffer:
\( \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) \).
This keeps the concentration of hydrogen ions relatively constant.
(ii) When \( \text{OH}^- \) ions are added, they are neutralized by the weak acid (propanoic acid) present in the buffer:
\( \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \).
This prevents a significant increase in pH.

(b) First, calculate the moles of each component in the final mixture:
\( n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0400 \times 0.150 = 6.00 \times 10^{-3}\text{ mol} \)
\( n(\text{CH}_3\text{CH}_2\text{COO}^-) = 0.0600 \times 0.100 = 6.00 \times 10^{-3}\text{ mol} \)

Using the Henderson-Hasselbalch equation:
\( \text{pH} = \text{p}K_a + \log\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right) \)
Since the moles of acid and base are equal in the same final volume of \( 100.0\text{ cm}^3 \), their concentrations are equal:
\( \text{pH} = \text{p}K_a + \log(1) = \text{p}K_a \)
\( \text{p}K_a = -\log(1.35 \times 10^{-5}) = 4.87 \)
\( \text{pH} = 4.87 \).

(c) (i) A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a single proton (\( \text{H}^+ \)).
(ii) For the equilibrium:
\( \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq) \n- Pair 1: \) \text{CH}_3\text{CH}_2\text{COOH} \) (acid) and \( \text{CH}_3\text{CH}_2\text{COO}^- \) (base).
- Pair 2: \( \text{H}_3\text{O}^+ \) (acid) and \( \text{H}_2\text{O} \) (base).

(a)
- M1: Explains reaction with added \( \text{H}^+ \) with a correct ionic equation. [2]
- M2: Explains reaction with added \( \text{OH}^- \) with a correct ionic equation. [2]

(b)
- M1: Calculates moles of propanoic acid (\( 6.00 \times 10^{-3}\text{ mol} \)). [1]
- M2: Calculates moles of propanoate (\( 6.00 \times 10^{-3}\text{ mol} \)). [1]
- M3: Shows correct substitution into the buffer formula. [1.5]
- M4: Calculates pH = 4.87 (allow 4.9). [1]

(c)
- M1: Defines conjugate acid-base pair as species differing by one proton. [1.5]
- M2: Identifies propanoic acid / propanoate pair. [1.25]
- M3: Identifies hydronium / water pair. [1.25]

(a) The independent variable is the temperature, \(T\), which is controlled and varied by the student. The dependent variable is the time, \(t\) (or the rate of reaction, which is proportional to \(1/t\)), which is measured.

(b) Calculations:
For Experiment 3:
\(1/T = 1 / 310 = 3.2258 \times 10^{-3} \approx 3.23 \times 10^{-3} \text{ K}^{-1}\)
\(\ln(1/t) = \ln(1/21) = -3.0445 \approx -3.04\)

For Experiment 5:
\(1/T = 1 / 330 = 3.0303 \times 10^{-3} \approx 3.03 \times 10^{-3} \text{ K}^{-1}\)
\(\ln(1/t) = \ln(1/6) = -1.7917 \approx -1.79\)

(c) Arrhenius equation in log form: \(\ln(k) = -\frac{E_a}{RT} + \ln(A)\). Since rate is proportional to \(1/t\), the plot of \(\ln(1/t)\) against \(1/T\) has gradient \(m = -\frac{E_a}{R}\). Since the y-axis is dimensionless and the x-axis has unit \(\text{K}^{-1}\), the gradient \(m\) has the unit \(\text{K}\).

(d) Experiment 4 is anomalous. For Experiment 4, the measured time of 18 seconds is significantly longer than the expected trend (which should be around 11-12 seconds based on other points). A longer time indicates a slower reaction. A likely practical cause is that the temperature of the reaction mixture dropped below the intended \(320 \text{ K}\) during the run, or there was an accidental dilution of one of the reactant solutions, lowering their initial concentrations.

(e) Gradient calculation using Experiment 1 and Experiment 5:
\(\Delta y = -1.79 - (-4.43) = 2.64\)
\(\Delta x = 3.03 \times 10^{-3} - 3.45 \times 10^{-3} = -0.42 \times 10^{-3} \text{ K}^{-1}\)
\(m = \frac{2.64}{-0.42 \times 10^{-3}} = -6285.7 \text{ K}\) (accept range: \(-6280\) to \(-6300\))
\(E_a = -m \times R = -(-6285.7) \times 8.31 = 52234 \text{ J mol}^{-1} = 52.2 \text{ kJ mol}^{-1}\) (accept range: \(52.1\) to \(52.4\)).

(f) To maintain temperature stability:
1. Place both reactant vessels in a thermostatically controlled water bath.
2. Allow sufficient time for the separate solutions to reach thermal equilibrium with the water bath before mixing them.
3. Measure and record the temperature of the mixture both at the start and at the end of the reaction, using the average temperature for calculations to account for any small heat loss.

Part (a):
- 1 Mark: Correctly identifies the independent variable as Temperature (T).
- 1 Mark: Correctly identifies the dependent variable as Time (t) (or rate of reaction / \(1/t\)).

Part (b):
- 1 Mark: Correctly calculates \(1/T\) and \(\ln(1/t)\) for Expt 3: \(3.23 \times 10^{-3} \text{ K}^{-1}\) and \(-3.04\).
- 1 Mark: Correctly calculates \(1/T\) and \(\ln(1/t)\) for Expt 5: \(3.03 \times 10^{-3} \text{ K}^{-1}\) and \(-1.79\).

Part (c):
- 1 Mark: Correct equation relating gradient to activation energy: \(m = -\frac{E_a}{R}\) (or \(E_a = -mR\)).
- 1 Mark: Correct unit for the gradient: \(\text{K}\) (Kelvin).

Part (d):
- 1 Mark: Identifies Experiment 4 as the anomalous point (reaction is too slow / time is too long).
- 1 Mark: Suggests a logical reason (e.g. temperature fell during the run / reactant concentration was too low / dilution error). Reject: human timing delay of a few milliseconds (this is too small to account for the large discrepancy).

Part (e):
- 1 Mark: Correctly sets up the gradient expression using the coordinates: \(\frac{-1.79 - (-4.43)}{3.03 \times 10^{-3} - 3.45 \times 10^{-3}}\).
- 1 Mark: Calculates the gradient as \(-6286\text{ K}\) (accept range \(-6280\) to \(-6300\)).
- 1 Mark: Multiplies gradient by \(8.31\) and converts from \(\text{J}\) to \(\text{kJ}\).
- 1 Mark: Final answer for \(E_a\) of \(52.2\text{ kJ mol}^{-1}\) (accept range \(52.1\) to \(52.4\)) given to 3 significant figures with correct units.

Part (f):
- 1 Mark: Mentions using a thermostatically controlled water bath.
- 1 Mark: Explains that reactant solutions must be allowed to equilibrate in the water bath before mixing.
- 1 Mark: Suggests measuring the temperature before and after mixing and calculating an average temperature.

(a) The balanced ionic equation for the reaction of the complex cation with acid is:
\[[\text{Cu}(\text{NH}_3)_x]^{2+}(aq) + x\text{H}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + x\text{NH}_4^+(aq)\]

(b) Titration procedure:
- Place the \(25.0 \text{ cm}^3\) aliquot of the acidic mixture into a clean conical flask.
- Add 2-3 drops of methyl orange (or bromophenol blue) indicator. (Note: Phenolphthalein is not suitable here because \(\text{NH}_4^+\) is present, which is a weak acid and would buffer the pH, causing an early or gradual endpoint if a high-pH indicator is used; methyl orange is ideal as its transition pH is in the acidic region).
- Fill a clean, rinsed burette with \(0.0250 \text{ mol dm}^{-3}\) \(\text{NaOH}(aq)\).
- Titrate until the indicator permanently changes color. With methyl orange, the color changes from red (acidic) to yellow/orange at the endpoint.

(c) Titration Calculations:
- Moles of \(\text{NaOH}\) used in the titration:
\(n(\text{NaOH}) = 20.00 \times 10^{-3} \text{ dm}^3 \times 0.0250 \text{ mol dm}^{-3} = 5.00 \times 10^{-4} \text{ mol}\).
- Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the moles of excess \(\text{HCl}\) in the \(25.0 \text{ cm}^3\) aliquot = \(5.00 \times 10^{-4} \text{ mol}\).
- The total excess \(\text{HCl}\) in the \(250.0 \text{ cm}^3\) volumetric flask:
\(n(\text{HCl})_{\text{excess, total}} = 5.00 \times 10^{-4} \text{ mol} \times \frac{250.0}{25.0} = 5.00 \times 10^{-3} \text{ mol}\).
- The initial moles of \(\text{HCl}\) added:
\(n(\text{HCl})_{\text{initial}} = 50.0 \times 10^{-3} \text{ dm}^3 \times 0.500 \text{ mol dm}^{-3} = 2.50 \times 10^{-2} \text{ mol}\).
- Moles of \(\text{HCl}\) that reacted with ammonia ligands:
\(n(\text{HCl})_{\text{reacted}} = 2.50 \times 10^{-2} - 5.00 \times 10^{-3} = 2.00 \times 10^{-2} \text{ mol}\).
- Since \(1 \text{ mol}\) of \(\text{NH}_3\) reacts with \(1 \text{ mol}\) of \(\text{H}^+\), the moles of \(\text{NH}_3\) in the original sample = \(2.00 \times 10^{-2} \text{ mol}\) (or \(0.0200 \text{ mol}\)).

(d) Gravimetric analysis of sulfate:
- Moles of \(\text{BaSO}_4 = \frac{\text{mass of } \text{BaSO}_4}{M_r(\text{BaSO}_4)} = \frac{1.167 \text{ g}}{233.4 \text{ g mol}^{-1}} = 0.00500 \text{ mol}\).
- Since \(1 \text{ mol}\) of \(\text{BaSO}_4\) contains \(1 \text{ mol}\) of \(\text{SO}_4^{2-}\), moles of sulfate in the \(1.228 \text{ g}\) sample = \(0.00500 \text{ mol}\).

(e) Formula determination:
(i) Ratio of \(\text{NH}_3 : \text{SO}_4^{2-} = \frac{0.0200 \text{ mol}}{0.00500 \text{ mol}} = 4\).
Therefore, \(x = 4\).

(ii) Since \(1 \text{ mol}\) of the complex contains \(1 \text{ mol}\) of \(\text{SO}_4^{2-}\), the total moles of the complex in the \(1.228 \text{ g}\) sample is equal to the moles of sulfate = \(0.00500 \text{ mol}\).
- Molar mass (\(M_r\)) of the complex:
\(M_r = \frac{\text{mass of sample}}{\text{moles of complex}} = \frac{1.228 \text{ g}}{0.00500 \text{ mol}} = 245.6 \text{ g mol}^{-1}\).
- The formula is \([\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \cdot y\text{H}_2\text{O}\).
Calculate the molar mass of the anhydrous part, \([\text{Cu}(\text{NH}_3)_4]\text{SO}_4\):
\(M_r([\text{Cu}(\text{NH}_3)_4]\text{SO}_4) = 63.5 + 4(17.0) + 96.1 = 63.5 + 68.0 + 96.1 = 227.6 \text{ g mol}^{-1}\).
- The mass of water of crystallization per mole of complex is:
\(18.0y = 245.6 - 227.6 = 18.0\).
- Therefore, \(y = 1\).
- The complete formula is \([\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \cdot \text{H}_2\text{O}\).

Part (a):
- 1 Mark: Correct reactant and product formulas: \([\text{Cu}(\text{NH}_3)_x]^{2+}\), \(\text{H}^+\), \(\text{Cu}^{2+}\), and \(\text{NH}_4^+\).
- 1 Mark: Fully balanced equation with correct stoichiometry: \(x\text{H}^+\) and \(x\text{NH}_4^+\).

Part (b):
- 1 Mark: Mentions filling burette with NaOH and titrating into the aliquot in the conical flask.
- 1 Mark: Selects methyl orange or bromophenol blue as indicator. (Reject: phenolphthalein).
- 1 Mark: Describes correct color change for the chosen indicator (e.g., methyl orange changes from red to yellow/orange).

Part (c):
- 1 Mark: Calculates moles of NaOH used: \(5.00 \times 10^{-4} \text{ mol}\).
- 1 Mark: Scaled up to find total unreacted HCl: \(5.00 \times 10^{-3} \text{ mol}\).
- 1 Mark: Calculates initial moles of HCl added: \(2.50 \times 10^{-2} \text{ mol}\).
- 1 Mark: Correct final value of moles of \(\text{NH}_3\): \(0.0200 \text{ mol}\).

Part (d):
- 1 Mark: Uses the formula \(n = m / M_r\) with \(1.167\) and \(233.4\).
- 1 Mark: Correctly calculates moles of \(\text{SO}_4^{2-}\) as \(0.00500 \text{ mol}\).

Part (e):
- 1 Mark: Deduces \(x = 4\) by dividing moles of \(\text{NH}_3\) by moles of \(\text{SO}_4^{2-}\).
- 1 Mark: Correctly calculates the molar mass of the complex as \(245.6 \text{ g mol}^{-1}\).
- 1 Mark: Subtracts the molar mass of the anhydrous salt \(227.6 \text{ g mol}^{-1}\) to find the mass of water as \(18.0 \text{ g mol}^{-1}\).
- 1 Mark: Deduces \(y = 1\) and states the full formula as \([\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \cdot \text{H}_2\text{O}\).