2024 AQA IAL Chemistry (9620) Practice Paper with Answers

For part (a), the electric field accelerates the positive ions to give them all the same kinetic energy. The sample must be ionized because only positive ions can be accelerated by the electric field and detected by the ion detector. For part (b), the positive ions hit the detector (a negative plate) and gain electrons, which generates an electric current. The size of the current is directly proportional to the abundance of that isotope. For part (c), the mass of one \(^{81}\text{Br}^+\) ion is calculated as follows: Mass of 1 mole of \(^{81}\text{Br} = 0.081\text{ kg mol}^{-1}\). Mass of one ion = \(0.081 / (6.022 \times 10^{23}) = 1.345 \times 10^{-25}\text{ kg}\). Since \(KE = \frac{1}{2} m v^2\), the velocity \(v\) of the \(^{81}\text{Br}^+\) ion is \(v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 2.15 \times 10^{-15}}{1.345 \times 10^{-25}}} = \sqrt{3.197 \times 10^{10}} = 1.788 \times 10^5\text{ m s}^{-1}\). The time taken \(t\) is given by \(t = \frac{d}{v} = \frac{1.35}{1.788 \times 10^5} = 7.55 \times 10^{-6}\text{ s}\).

Part (a): [1 mark] for stating that the electric field accelerates the positive ions (or gives them all the same kinetic energy). [1 mark] for explaining that only ions can be accelerated/detected. Part (b): [1 mark] for stating that ions gain electrons at the detector, generating a current. [1 mark] for stating that the current size is proportional to abundance. Part (c): [1 mark] for mass of 1 mole of 81Br in kg (0.081 kg). [1 mark] for mass of one ion = 1.35 x 10^-25 kg. [1 mark] for the rearranged formula v = sqrt(2KE/m). [1 mark] for the velocity of 1.79 x 10^5 m s^-1. [1 mark] for the formula t = d/v. [1 mark] for the correct final time of 7.55 x 10^-6 s (accept 7.5 x 10^-6 to 7.6 x 10^-6 s).

For part (a): Mass of water lost = \(4.722 - 3.282 = 1.440\text{ g}\). \(M_r(\text{Ca(NO}_3\text{)}_2) = 164.1\). Moles of \(\text{Ca(NO}_3\text{)}_2 = 3.282 / 164.1 = 0.0200\text{ mol}\). Moles of \(\text{H}_2\text{O} = 1.440 / 18.0 = 0.0800\text{ mol}\). Ratio of moles = \(0.0800 / 0.0200 = 4\), so \(y = 4\). For part (b): Moles of anhydrous calcium nitrate decomposed = \(0.0200\text{ mol}\). According to the reaction equation, \(2\text{ mol}\) of \(\text{Ca(NO}_3\text{)}_2\) yields \(5\text{ mol}\) of gas molecules (\(4\text{NO}_2 + \text{O}_2\)). Total moles of gas, \(n = 0.0200 \times 2.5 = 0.0500\text{ mol}\). Using the ideal gas equation: \(pV = nRT \implies V = nRT / p\). \(V = (0.0500 \times 8.314 \times 425) / (1.02 \times 10^5) = 1.732 \times 10^{-3}\text{ m}^3\). Converting to \(\text{dm}^3\): \(V = 1.73\text{ dm}^3\).

Part (a): [1 mark] for calculating mass of water lost = 1.440 g. [1 mark] for moles of calcium nitrate = 0.0200 mol. [1 mark] for moles of water = 0.0800 mol. [1 mark] for finding ratio y = 4. Part (b): [1 mark] for stating ideal gas equation pV = nRT. [1 mark] for calculating total moles of gas = 0.0500 mol (from 0.0200 x 2.5). [1 mark] for rearranging V = nRT / p. [1 mark] for correct substitution of values. [1 mark] for volume in m3 = 1.732 x 10^-3. [1 mark] for correct final volume to 3 sig figs = 1.73 dm3.

For part (a)(i), Cl has 7 valence electrons + 3 from F = 10 electrons around Cl (5 pairs: 3 bonding pairs, 2 lone pairs). The shape is T-shaped. For part (a)(ii), the bond angle is slightly less than \(90^\circ\) (accept \(85-89^\circ\) or less than \(90^\circ\)). Electron pairs repel to a position of minimum repulsion. Lone pairs repel more strongly than bonding pairs, pushing the bonding pairs closer together. For part (b)(i), in \(\text{ClF}_2^-\), Cl has 7 valence electrons + 1 from negative charge + 2 from F = 10 electrons (5 pairs: 2 bonding pairs, 3 lone pairs). For part (b)(ii), the shape is linear as the 3 lone pairs occupy the equatorial positions to minimize repulsion. For part (c), \(\text{CCl}_4\) is a highly symmetrical tetrahedral molecule, so individual polar C-Cl bond dipoles cancel each other out perfectly. In \(\text{CHCl}_3\), the molecule is asymmetrical and the C-Cl and C-H dipoles do not cancel, leading to a net molecular dipole.

Part (a)(i): [1 mark] for drawing showing Cl with three single bonds to F and two lone pairs. [1 mark] for naming the shape 'T-shaped'. Part (a)(ii): [1 mark] for bond angle of 85-89 degrees (or less than 90 degrees). [1 mark] for stating that electron pairs repel to get as far apart as possible (minimize repulsion). [1 mark] for stating that lone pairs repel more than bonding pairs. Part (b)(i): [1 mark] for 2 bonding pairs. [1 mark] for 3 lone pairs. Part (b)(ii): [1 mark] for linear shape. Part (c): [1 mark] for stating CCl4 is symmetrical and dipoles cancel. [1 mark] for stating CHCl3 is asymmetrical and dipoles do not cancel.

For part (a): \(q = m c \Delta T = 55.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 9.2\text{ K} = 2115\text{ J} = 2.115\text{ kJ}\) (accept \(2.12\text{ kJ}\)). For part (b): \(n(\text{CuSO}_4) = 5.00 / 159.6 = 0.03133\text{ mol}\) (accept \(0.0313\text{ mol}\)). For part (c): \(\Delta H_{\text{sol}} = -q / n = -2.115 / 0.03133 = -67.5\text{ kJ mol}^{-1}\) (accept range \(-67.5\) to \(-67.6\)). If a student used \(m = 50.0\text{ g}\) in (a), then \(q = 1.923\text{ kJ}\) and \(\Delta H_{\text{sol}} = -61.4\text{ kJ mol}^{-1}\), which is awarded full marks for error-carried-forward. For part (d): Using Hess's law, \(\Delta H_{\text{rxn}} = \Delta H_{\text{sol1}} - \Delta H_{\text{sol2}} = -67.5 - (+11.5) = -79.0\text{ kJ mol}^{-1}\) (or \(-65.0 - 11.5 = -76.5\text{ kJ mol}^{-1}\) if using the alternative value). For part (e): Heat loss to the surroundings/beaker; heat capacity of the cup/thermometer not considered; incomplete dissolving of the anhydrous copper(II) sulfate.

Part (a): [1 mark] for temperature change = 9.2 K. [1 mark] for q = 2.115 kJ or 2.12 kJ (allow 1.92 kJ if mass 50.0g was used). Part (b): [1 mark] for n = 0.0313 mol. Part (c): [1 mark] for dividing q by n with negative sign. [1 mark] for -67.5 kJ mol^-1 (allow -61.4 kJ mol^-1 if 50.0g mass was used). Part (d): [1 mark] for correct Hess's cycle or algebraic formula. [1 mark] for correct substitution of values. [1 mark] for correct final answer of -79.0 kJ mol^-1 (or -76.5 kJ mol^-1 using alternative value). Part (e): [2 marks] for any two of: heat loss to surroundings; incomplete dissolution of solid; neglecting heat capacity of calorimeter/thermometer.

For part (a)(i): \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). For part (a)(ii): It is a disproportionation reaction because chlorine is simultaneously oxidized and reduced. The oxidation state of Cl in \(\text{Cl}_2\) is 0; it changes to -1 in \(\text{NaCl}\) (reduction) and to +1 in \(\text{NaClO}\) (oxidation). For part (b)(i): Red-brown/orange fumes or vapor (of bromines) and a choking gas / acidic gas (of sulfur dioxide / hydrogen bromide). For part (b)(ii): \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\) (or ionic version is acceptable). For part (c): Add acidified silver nitrate solution. Sodium chloride gives a white precipitate, while sodium iodide gives a yellow precipitate. Equation: \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{AgI}(\text{s})\) (or for Cl-).

Part (a)(i): [1 mark] for Cl2 + 2NaOH -> NaCl + NaClO + H2O (or ionic equivalent). Part (a)(ii): [1 mark] for disproportionation. [1 mark] for stating Cl is both oxidized and reduced. [1 mark] for stating oxidation state changes from 0 to -1 and +1. Part (b)(i): [2 marks] for any two observations: brown/orange fumes (or liquid); choking gas; steamy/misty fumes. Part (b)(ii): [2 marks] for 2NaBr + 2H2SO4 -> Na2SO4 + Br2 + SO2 + 2H2O (1 mark for correct species, 1 mark for balancing). Part (c): [1 mark] for adding silver nitrate solution (acidified with HNO3). [1 mark] for white precipitate with NaCl and yellow precipitate with NaI (including equation: Ag+(aq) + I-(aq) -> AgI(s) or chloride equivalent).

For part (a): \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). For part (b): \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\). For part (c): The color change is from colorless (or pale green) to pale pink. For part (d): Moles of \(\text{MnO}_4^-\)\ used in the titration = \(0.0220 \times (23.15 / 1000) = 5.093 \times 10^{-4}\text{ mol}\). According to the equation, \(1\text{ mol}\) of \(\text{MnO}_4^-\)\ reacts with \(5\text{ mol}\) of \(\text{Fe}^{2+}\). Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\) aliquot = \(5 \times 5.093 \times 10^{-4} = 2.5465 \times 10^{-3}\text{ mol}\). Moles of \(\text{Fe}^{2+}\) in the original \(250.0\text{ cm}^3\) solution = \(2.5465 \times 10^{-3} \times 10 = 0.025465\text{ mol}\). Mass of iron in the original sample = \(0.025465\text{ mol} \times 55.8\text{ g mol}^{-1} = 1.421\text{ g}\) (accept \(1.42\text{ g}\)). For part (e): Percentage by mass = \((1.421 / 1.65) \times 100 = 86.1\%\) (accept range \(86.1 - 86.2\%\)).

Part (a): [1 mark] for correct species. [1 mark] for correct balancing with H+ and electrons. Part (b): [1 mark] for overall balanced ionic equation. Part (c): [1 mark] for colorless to pale/permanent pink. Part (d): [1 mark] for calculating moles of MnO4- = 5.09 x 10^-4 mol. [1 mark] for moles of Fe2+ in 25cm3 = 2.55 x 10^-3 mol. [1 mark] for multiplying by 10 to get moles of Fe2+ in 250cm3 = 2.55 x 10^-2 mol. [2 marks] for calculating the mass of Fe = 1.42 g (1 mark for multiplying by 55.8, 1 mark for correct accuracy). Part (e): [1 mark] for percentage by mass = 86.1% (allow 86.1 to 86.2%).

For part (a): Solubility of Group 2 hydroxides increases down the group. For part (b): Solubility of Group 2 sulfates decreases down the group. For part (c): Barium sulfate is highly insoluble, so it does not dissolve in the digestive system and cannot be absorbed into the bloodstream. For part (d): Magnesium hydroxide is used as an antacid (or to treat indigestion / neutralize stomach acid). Equation: \(\text{Mg(OH)}_2 + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + 2\text{H}_2\text{O}\) (or \(\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}\)). For part (e): \(\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2\). Observation: Fizzing / effervescence, the mixture gets warm, or a white precipitate forms. For part (f)(i): \(\text{TiCl}_4 + 2\text{Mg} \rightarrow \text{Ti} + 2\text{MgCl}_2\). For part (f)(ii): Magnesium acts as a reducing agent.

Part (a): [1 mark] for solubility increases down the group. Part (b): [1 mark] for solubility decreases down the group. Part (c): [1 mark] for stating BaSO4 is highly insoluble. [1 mark] for explaining that it is not absorbed into the body/bloodstream. Part (d): [1 mark] for use as an antacid / indigestion remedy. [1 mark] for equation: H+ + OH- -> H2O or Mg(OH)2 + 2H+ -> Mg2+ + 2H2O. Part (e): [1 mark] for Ca + 2H2O -> Ca(OH)2 + H2. [1 mark] for observation (e.g., fizzing/effervescence, white precipitate). Part (f)(i): [1 mark] for TiCl4 + 2Mg -> Ti + 2MgCl2. Part (f)(ii): [1 mark] for stating reducing agent.

(a)
\(\Delta T = 26.8 - 20.2 = 6.6\text{ K}\)
Total mass of solution, \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\)
\(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.6\text{ K} = 2758.8\text{ J} = 2.76\text{ kJ}\) (to 3 significant figures).

(b)
\(n(\text{HCl}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
\(n(\text{NaOH}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
Moles of \(\text{H}_2\text{O}\) formed = \(0.0500\text{ mol}\)
\(\Delta H = -\frac{q}{n} = -\frac{2.7588\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\).

(c)
1. Heat loss from the calorimeter/cup to the surroundings.
2. The heat capacity of the calorimeter/polystyrene cup/thermometer is not included in the calculation.

(d)
Record the temperature of the acid every minute for 3 minutes before mixing. At the 4th minute, add the alkali but do not record the temperature. Continue recording the temperature every minute from the 5th minute to the 10th minute. Plot a graph of temperature against time, draw lines of best fit for both sections, and extrapolate the cooling curve back to the 4th minute to find the theoretical maximum temperature change.

(a) [3 marks]
- 1 mark for calculating \(\Delta T = 6.6\text{ K}\) and solution mass \(m = 100.0\text{ g}\).
- 1 mark for calculating \(q = 2758.8\text{ J}\).
- 1 mark for converting to \(2.76\text{ kJ}\) (accept 2.759 or 2.8 kJ if rounded, but must match sig figs of temperature).

(b) [3 marks]
- 1 mark for finding moles of reaction/water formed = 0.0500 mol.
- 1 mark for dividing \(q\) by moles.
- 1 mark for final answer of \(-55.2\text{ kJ mol}^{-1}\) (must include the negative sign and correct units; accept range -55.0 to -55.2 based on rounding in (a)).

(c) [2 marks]
- 1 mark for heat loss to the surroundings / beaker / air.
- 1 mark for heat absorbed by the cup / thermometer (or heat capacity of the cup neglected).

(d) [2 marks]
- 1 mark for recording temperature at regular intervals before and after mixing.
- 1 mark for plotting a graph of temperature vs. time and extrapolating back to the point/time of mixing to find the maximum temperature rise.

(a)
- First step: The double bond in but-1-ene (\(\text{CH}_3\text{CH}_2\text{CH=CH}_2\)) attacks the electropositive hydrogen in \(\text{H}^{\delta+}\text{-Br}^{\delta-}\).
- A curly arrow starts from the middle of the \(\text{C=C}\) double bond and points to the \(\text{H}\) of \(\text{H-Br}\).
- A concurrent curly arrow starts from the \(\text{H-Br}\) bond and points to the \(\text{Br}\) atom.
- Second step: This forms a secondary carbocation intermediate: \(\text{CH}_3\text{CH}_2\text{C}^+\text{H-CH}_3\) and a bromide ion (\(\text{Br}^-\)) with at least one lone pair shown.
- A curly arrow starts from the lone pair on \(\text{Br}^-\) and points to the positively charged carbon (\(\text{C}^+\)) of the carbocation.

(b) 2-bromobutane

(c)
- The major product is formed via a secondary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{C}^+\text{H-CH}_3\)).
- The minor product would be formed via a primary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{C}^+\text{H}_2\)).
- Secondary carbocations are more stable than primary carbocations.
- This is because secondary carbocations have two electron-donating alkyl groups (an ethyl group and a methyl group) that stabilize the positive charge by the inductive effect, whereas primary carbocations have only one (a propyl group).

(a) [5 marks]
- 1 mark for correct \(\text{H}^{\delta+}\text{-Br}^{\delta-}\) dipoles.
- 1 mark for curly arrow from \(\text{C=C}\) double bond to \(\text{H}\) of \(\text{H-Br}\).
- 1 mark for curly arrow from \(\text{H-Br}\) bond to \(\text{Br}\).
- 1 mark for drawing the correct secondary carbocation structure (\(\text{CH}_3\text{CH}_2\text{C}^+\text{H-CH}_3\)).
- 1 mark for curly arrow from lone pair on \(\text{Br}^-\) to the \(\text{C}^+\).

(b) [1 mark]
- 1 mark for 2-bromobutane (do not accept 2-bromobut-1-ene or other incorrect isomers).

(c) [4 marks]
- 1 mark for identifying that the major product goes via a secondary carbocation intermediate.
- 1 mark for identifying that the minor product goes via a primary carbocation intermediate.
- 1 mark for stating that secondary carbocations are more stable than primary carbocations.
- 1 mark for explaining stability in terms of the inductive / electron-donating effect of alkyl groups.

(a)
- Draw 2-bromopropane showing the polar \(\text{C}^{\delta+}\text{-Br}^{\delta-}\) bond.
- Draw a curly arrow starting from a lone pair on the oxygen of the hydroxide ion (\(\text{OH}^-\)) pointing directly to the \(\text{C}^{\delta+}\) atom.
- Draw a curly arrow starting from the center of the \(\text{C-Br}\) bond pointing to the bromine atom.
- This yields propan-2-ol and a bromide ion.

(b)
- Show a hydrogen atom on one of the adjacent methyl groups of 2-bromopropane (\(\text{CH}_3\text{CH(Br)CH}_3\)).
- Draw a curly arrow from the lone pair on the oxygen of the hydroxide ion (\(\text{OH}^-\)) pointing to one of these adjacent hydrogen atoms.
- Draw a curly arrow from the \(\text{C-H}\) bond of that hydrogen pointing to the single bond between the carbon atoms to form the \(\text{C=C}\) double bond.
- Draw a curly arrow from the \(\text{C-Br}\) bond pointing to the bromine atom.
- The organic product is propene (\(\text{CH}_3\text{CH=CH}_2\)).

(c)
- (i) Nucleophile (or electron pair donor).
- (ii) Base (or proton acceptor).

(a) [4 marks]
- 1 mark for showing \(\text{C}^{\delta+}\text{-Br}^{\delta-}\) dipoles correctly on the halogenoalkane.
- 1 mark for curly arrow from lone pair on oxygen of \(\text{OH}^-\) to the central carbon.
- 1 mark for curly arrow from the \(\text{C-Br}\) bond to the bromine atom.
- 1 mark for correct structures of products (propan-2-ol and \(\text{Br}^-\)).

(b) [4 marks]
- 1 mark for curly arrow from lone pair on oxygen of \(\text{OH}^-\) to H on adjacent carbon.
- 1 mark for curly arrow from the adjacent \(\text{C-H}\) bond to the adjacent \(\text{C-C}\) bond.
- 1 mark for curly arrow from the \(\text{C-Br}\) bond to the bromine atom.
- 1 mark for stating IUPAC name of the product is 'propene' (accept displayed formula of propene).

(c) [2 marks]
- 1 mark for (i) nucleophile / electron pair donor.
- 1 mark for (ii) base / proton acceptor.

(a) Le Chatelier's principle states that when a system at equilibrium is subjected to a change in conditions (e.g., concentration, temperature, or pressure), the system shifts its position of equilibrium in a direction that opposes/minimizes the effect of that change.

(b)
Set up an ICE (Initial, Change, Equilibrium) table:
- Initial moles: \(\text{A} = 1.50\), \(\text{B} = 2.00\), \(\text{C} = 0\)
- Equilibrium moles of \(\text{C} = 0.60\text{ mol}\). Thus, change in moles of \(\text{C} = +0.60\text{ mol}\).
- From the stoichiometry, change in B = \(-0.60\text{ mol}\), change in A = \(-0.30\text{ mol}\).
- Equilibrium moles:
- \(\text{A} = 1.50 - 0.30 = 1.20\text{ mol}\)
- \(\text{B} = 2.00 - 0.60 = 1.40\text{ mol}\)
- \(\text{C} = 0.60\text{ mol}\)

Convert equilibrium moles to concentrations using \(V = 5.00\text{ dm}^3\):
- \([\text{A}] = \frac{1.20\text{ mol}}{5.00\text{ dm}^3} = 0.240\text{ mol dm}^{-3}\)
- \([\text{B}] = \frac{1.40\text{ mol}}{5.00\text{ dm}^3} = 0.280\text{ mol dm}^{-3}\)
- \([\text{C}] = \frac{0.60\text{ mol}}{5.00\text{ dm}^3} = 0.120\text{ mol dm}^{-3}\)

Write expression for \(K_c\):
\[K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}\]

Substitute the equilibrium concentrations:
\[K_c = \frac{(0.120)^2}{(0.240) \times (0.280)^2} = \frac{0.0144}{0.240 \times 0.0784} = \frac{0.0144}{0.018816} \approx 0.765\text{ dm}^3\text{ mol}^{-1}\]

Units:
\[\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{mol}^{-1}\text{ dm}^3\]

(c)
- The value of \(K_c\) will decrease.
- Since the forward reaction is exothermic, an increase in temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the heat. This decreases the concentration of products and increases the concentration of reactants, resulting in a smaller value for \(K_c\).

(a) [2 marks]
- 1 mark for referring to a system at equilibrium subjected to change/disturbance.
- 1 mark for stating the system shifts/opposes/minimizes that change.

(b) [6 marks]
- 1 mark for calculating correct equilibrium moles of A (1.20 mol) and B (1.40 mol).
- 1 mark for dividing moles by volume (5.00) to find equilibrium concentrations.
- 1 mark for the correct expression of \(K_c\).
- 1 mark for calculating the numerical value: 0.765 (accept 0.76 to 0.77).
- 1 mark for correct units: \(\text{dm}^3\text{ mol}^{-1}\) (or \(\text{mol}^{-1}\text{ dm}^3\)).
- 1 mark for working shown clearly.

(c) [2 marks]
- 1 mark for stating that \(K_c\) decreases.
- 1 mark for explaining that equilibrium shifts to the left / in the endothermic direction to oppose the temperature rise.

(a)
- Structural formula: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (or \(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OH}\)).
- IUPAC Name: Butan-1-ol (or 2-methylpropan-1-ol).

(b)
- Reagent: Acidified potassium dichromate(VI) (or \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\)).
- Conditions: Use distillation apparatus (or immediate distillation) and a limited amount of the oxidizing agent (or excess alcohol) to prevent further oxidation to a carboxylic acid.

(c)
- The blue solution forms a brick-red precipitate (of copper(I) oxide).

(d)
- Equation: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_2\text{CH}_3 + \text{H}_2\text{O}\)
- Product is butanone.

(e)
- 2-methylpropan-2-ol is a tertiary alcohol.
- Tertiary alcohols cannot be easily oxidized because the carbon atom holding the hydroxyl (\(\text{-OH}\)) group is bonded to three other carbons and does not have any hydrogen atoms attached to it. Oxidation requires the breaking of a \(\text{C-H}\) bond on this carbon, which is not present in tertiary alcohols.

(a) [2 marks]
- 1 mark for correct structural formula (either butan-1-ol or 2-methylpropan-1-ol).
- 1 mark for correct corresponding IUPAC name.

(b) [3 marks]
- 1 mark for naming acidified potassium dichromate(VI) (accept \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\); reject dichromate without acid).
- 1 mark for mentioning immediate distillation / distillation setup.
- 1 mark for using a limited amount of the oxidizing agent / excess alcohol.

(c) [1 mark]
- 1 mark for blue solution to red/brick-red precipitate.

(d) [2 marks]
- 1 mark for correct reactant and products in the equation.
- 1 mark for balancing with \([\text{O}]\) and \(\text{H}_2\text{O}\).

(e) [2 marks]
- 1 mark for identifying it as a tertiary alcohol.
- 1 mark for explaining that there is no hydrogen atom on the carbon bonded to the hydroxyl group (or that a C-C bond would have to break, which is too strong).

(a)
- Initiation:
\[\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet\]
- Propagation 1:
\[\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\]
- Propagation 2:
\[\text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\]
- Termination (any of the following that leads to a stable product):
\[\text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\]
\[\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\]
\[\text{Cl}^\bullet + \text{Cl}^\bullet \rightarrow \text{Cl}_2\]

(b)
- UV light provides the activation energy needed to break the strong \(\text{Cl-Cl}\) covalent bond homolytically to form chlorine free radicals.

(c)
(i)
- Step 1:
\[\text{CH}_3\text{Cl} + \text{Cl}^\bullet \rightarrow \text{CH}_2\text{Cl}^\bullet + \text{HCl}\]
- Step 2:
\[\text{CH}_2\text{Cl}^\bullet + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}_2 + \text{Cl}^\bullet\]
(ii)
- Use a large excess of methane (\(\text{CH}_4\)).
- This increases the probability of a chlorine radical (\(\text{Cl}^\bullet\)) colliding with a methane molecule rather than a chloromethane (\(\text{CH}_3\text{Cl}\)) molecule, thus reducing further substitution.

(a) [5 marks]
- 1 mark for correct initiation equation with UV indicated (above the arrow or in text).
- 1 mark for propagation step 1 (including radical dots on \(\text{Cl}\) and \(\text{CH}_3\)).
- 1 mark for propagation step 2 (including radical dots on \(\text{CH}_3\) and \(\text{Cl}\)).
- 1 mark for a valid termination equation.
- 1 mark for all radicals having correctly positioned radical dots (e.g., \(\text{C}^\bullet\) on the methyl radical, not on the hydrogen).

(b) [1 mark]
- 1 mark for stating that UV light provides energy to break the \(\text{Cl-Cl}\) bond / for homolytic fission.

(c) [4 marks]
- 1 mark for (i) first propagation equation forming the chloromethyl radical.
- 1 mark for (i) second propagation equation forming dichloromethane.
- 1 mark for (ii) excess of methane.
- 1 mark for (ii) explaining that it makes collisions between chlorine radicals and methane far more likely than with chloromethane.

(a)
- Axes: The y-axis must be labeled 'Number of molecules' (or 'Fraction of molecules') and the x-axis must be labeled 'Energy' (or 'Kinetic energy').
- Curve \(T_1\): Must start at the origin \((0,0)\), rise to a single peak, and decay asymptotically towards the x-axis (it must not touch the x-axis at high energy).
- The activation energy, \(E_a\), should be marked as a vertical line on the x-axis towards the right-hand side.

(b)
- Curve \(T_2\): Must start at the origin, have its peak shifted to the right (higher energy) and be lower in height than the peak of \(T_1\). The tail of the curve must lie above the tail of \(T_1\) at high energy.
- Differences:
1. The peak is lower for the curve at higher temperature (\(T_2\)).
2. The peak is shifted to the right / to a higher energy.

(c)
- The activation energy, \(E_a\), remains constant.
- A small increase in temperature from \(T_1\) to \(T_2\) results in a much larger area under the curve to the right of \(E_a\). This means there is a significantly larger fraction of molecules with energy greater than or equal to the activation energy, leading to a much higher frequency of successful collisions.

(d)
- A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_{cat}\)).
- This shifts the minimum energy required to react to the left on the Maxwell-Boltzmann distribution curve. Consequently, a much larger fraction of molecules have energy greater than or equal to this new, lower activation energy, increasing the rate of reaction.

(a) [3 marks]
- 1 mark for correct axes labels: y-axis = Number / fraction of molecules; x-axis = Kinetic Energy.
- 1 mark for correct shape of curve starting at origin and asymptotic to the x-axis at high energy.
- 1 mark for showing \(E_a\) clearly marked on the x-axis.

(b) [3 marks]
- 1 mark for drawing the \(T_2\) curve correctly: peak shifted right, lower maximum, and tail crossing over \(T_1\).
- 1 mark for stating peak is lower at \(T_2\).
- 1 mark for stating peak shifts to higher energy / to the right.

(c) [2 marks]
- 1 mark for stating that a significantly larger fraction/number of molecules have energy \(\ge E_a\) at \(T_2\) (or referring to the area under the curve to the right of \(E_a\)).
- 1 mark for stating this results in a significantly higher frequency of successful collisions.

(d) [2 marks]
- 1 mark for stating that the catalyst provides an alternative pathway with a lower activation energy.
- 1 mark for explaining that a greater fraction of molecules now have energy \(\ge\) this lower activation energy.

(a) Comparing Experiment 1 and Experiment 2:
- \([\text{I}^-]\) is constant.
- \([\text{S}_2\text{O}_8^{2-}]\) doubles (from 0.0100 to 0.0200).
- The rate doubles (from \(1.50 \times 10^{-5}\) to \(3.00 \times 10^{-5}\)).
- Therefore, the reaction is 1st order with respect to \(\text{S}_2\text{O}_8^{2-}\).

Comparing Experiment 2 and Experiment 3:
- \([\text{S}_2\text{O}_8^{2-}]\) is constant.
- \([\text{I}^-]\) doubles (from 0.0200 to 0.0400).
- The rate doubles (from \(3.00 \times 10^{-5}\) to \(6.00 \times 10^{-5}\)).
- Therefore, the reaction is 1st order with respect to \(\text{I}^-\).

(b) Rate equation:
\[\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\]

To find \(k\), substitute values from Experiment 1:
\[1.50 \times 10^{-5} = k (0.0100) (0.0200)\]
\[k = \frac{1.50 \times 10^{-5}}{2.00 \times 10^{-4}} = 0.0750\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\]

(c) The reaction between two negatively charged ions has high activation energy due to electrostatic repulsion. \(\text{Fe}^{2+}\) ions act as a homogeneous catalyst because they are positive and can react with negative reactant ions without repulsion, providing an alternative route with lower activation energy.

Catalytic cycle:
- Step 1: Oxidation of \(\text{Fe}^{2+}\) by \(\text{S}_2\text{O}_8^{2-}\):
\[2\text{Fe}^{2+}(aq) + \text{S}_2\text{O}_8^{2-}(aq) \rightarrow 2\text{Fe}^{3+}(aq) + 2\text{SO}_4^{2-}(aq)\]

- Step 2: Reduction of \(\text{Fe}^{3+}\) back to \(\text{Fe}^{2+}\) by \(\text{I}^-\):
\[2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(aq)\]

(a) [3 marks]
- 1 mark for order wrt \(\text{S}_2\text{O}_8^{2-}\) is 1.
- 1 mark for order wrt \(\text{I}^-\) is 1.
- 1 mark for showing valid working (e.g. comparing Exp 1/2 and 2/3).

(b) [3 marks]
- 1 mark for correct Rate Equation: \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).
- 1 mark for \(k = 0.0750\) (accept \(0.075\)).
- 1 mark for units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) [4 marks]
- 1 mark for explaining that electrostatic repulsion between negatively charged ions results in a high activation energy.
- 1 mark for stating that \(\text{Fe}^{2+}\) provides an alternative reaction pathway of lower activation energy.
- 1 mark for correct first equation: \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\) (or ionic equivalent).
- 1 mark for correct second equation: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) (or ionic equivalent).

(a)(i) Formula of \(X\): \([\text{Co}(\text{en})_3]^{2+}\)
Coordination number: 6 (since 1,2-diaminoethane is a bidentate ligand, and there are three of them, forming six coordinate bonds).

(a)(ii) Equation:
\[[\text{Co}(\text{H}_2\text{O})_6]^{2+}(aq) + 3\text{en}(aq) \rightarrow [\text{Co}(\text{en})_3]^{2+}(aq) + 6\text{H}_2\text{O}(l)\]

(a)(iii)
- **Entropy change (\(\Delta S^{\theta}\)):** The reaction goes from 4 reactant species to 7 product species. This increase in the number of particles increases disorder, resulting in a large positive entropy change (\(\Delta S^{\theta} > 0\)).
- **Enthalpy change (\(\Delta H^{\theta}\)):** The enthalpy change is approximately zero (\(\Delta H^{\theta} \approx 0\)) because six Co-O coordinate bonds are broken and six similar Co-N coordinate bonds are formed.
- **Gibbs Free Energy (\(\Delta G^{\theta}\)):** Since \(\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}\), a very positive \(T\Delta S^{\theta}\) and near-zero \(\Delta H^{\theta}\) makes \(\Delta G^{\theta}\) highly negative, making the reaction thermodynamically highly feasible.

(b)
- Formula of \(Y\): \([\text{CoCl}_4]^{2-}\)
- Shape: Tetrahedral
- Color: Blue

(a)(i) [2 marks]
- 1 mark for correct formula: \([\text{Co}(\text{en})_3]^{2+}\).
- 1 mark for coordination number: 6.

(a)(ii) [2 marks]
- 1 mark for correct reactants and products (allow missing state symbols).
- 1 mark for correct balancing: 3 \(\text{en}\) and 6 \(\text{H}_2\text{O}\).

(a)(iii) [3 marks]
- 1 mark for stating that the number of species increases from 4 to 7, resulting in an increase in disorder / positive entropy change (\(\Delta S^{\theta} > 0\)).
- 1 mark for stating that \(\Delta H^{\theta}\) is close to zero because similar coordinate bonds are broken and made (Co-O vs Co-N).
- 1 mark for linking this to a negative Gibbs free energy change (\(\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta} < 0\)).

(b) [3 marks]
- 1 mark for formula: \([\text{CoCl}_4]^{2-}\).
- 1 mark for shape: Tetrahedral.
- 1 mark for color: Blue.

(a) First, calculate the initial moles of propanoic acid (\(\text{HA}\)) and propanoate ions (\(\text{A}^-\)) in the buffer:
\[n(\text{HA}) = \text{Volume (dm}^3) \times \text{Concentration} = 0.250 \times 0.150 = 0.0375\text{ mol}\]
\[n(\text{A}^-) = \text{Volume (dm}^3) \times \text{Concentration} = 0.150 \times 0.200 = 0.0300\text{ mol}\]

Using the acid dissociation expression:
\[K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]

Since the volumes cancel in the concentration ratio:
\[[\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)}\]
\[[\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0375}{0.0300} = 1.6875 \times 10^{-5}\text{ mol dm}^{-3}\]
\[\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.6875 \times 10^{-5}) = 4.7726 \approx 4.77\]

(b) Moles of \(\text{OH}^-\Change\) added:
\[n(\text{OH}^-) = \frac{2.00}{1000} \times 1.00 = 0.00200\text{ mol}\]

When \(\text{NaOH}\) is added, \(\text{OH}^-\Change\) reacts with \(\text{C}_2\text{H}_5\text{COOH}\):
\[\text{C}_2\text{H}_5\text{COOH} + \text{OH}^- \rightarrow \text{C}_2\text{H}_5\text{COO}^- + \text{H}_2\text{O}\]

New moles of propanoic acid (\(\text{HA}\)):
\[n(\text{HA})_{\text{new}} = 0.0375 - 0.00200 = 0.0355\text{ mol}\]

New moles of propanoate ions (\(\text{A}^-\)):
\[n(\text{A}^-)_{\text{new}} = 0.0300 + 0.00200 = 0.0320\text{ mol}\]

Calculate the new \([\text{H}^+]\):
\[[\text{H}^+] = K_a \times \frac{n(\text{HA})_{\text{new}}}{n(\text{A}^-)_{\text{new}}}\]
\[[\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0355}{0.0320} = 1.4977 \times 10^{-5}\text{ mol dm}^{-3}\]
\[\text{pH} = -\log_{10}(1.4977 \times 10^{-5}) = 4.8246 \approx 4.82\]

(a) [4 marks]
- 1 mark for \(n(\text{propanoic acid}) = 0.0375\text{ mol}\).
- 1 mark for \(n(\text{propanoate}) = 0.0300\text{ mol}\).
- 1 mark for calculating \([\text{H}^+] = 1.69 \times 10^{-5}\text{ mol dm}^{-3}\).
- 1 mark for pH = 4.77 (accept 4.77 to 4.78).

(b) [6 marks]
- 1 mark for \(n(\text{OH}^-\text{ added}) = 0.00200\text{ mol}\).
- 1 mark for calculating new \(n(\text{propanoic acid}) = 0.0355\text{ mol}\).
- 1 mark for calculating new \(n(\text{propanoate}) = 0.0320\text{ mol}\).
- 1 mark for showing correct use of the ratio \(\frac{0.0355}{0.0320}\) in the \(K_a\) expression.
- 1 mark for calculating new \([\text{H}^+] = 1.50 \times 10^{-5}\text{ mol dm}^{-3}\).
- 1 mark for pH = 4.82 (accept 4.82 to 4.83).

(a) The standard enthalpy of atomisation is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.

(b) A standard Born-Haber cycle should include:
- Lower line containing: \(\text{MgCl}_2(s)\)
- Element line (zero/reference line) containing: \(\text{Mg}(s) + \text{Cl}_2(g)\)
- Arrow down from elements to compound for enthalpy of formation (\(\Delta_f H^{\theta} = -642\text{ kJ mol}^{-1}\)).
- Upward steps:
1. Atomisation of Mg: \(\text{Mg}(g) + \text{Cl}_2(g)\) (\(\Delta H = +148\))
2. First ionisation of Mg: \(\text{Mg}^+(g) + \text{Cl}_2(g) + e^-\Change\) (\(\Delta H = +738\))
3. Second ionisation of Mg: \(\text{Mg}^{2+}(g) + \text{Cl}_2(g) + 2e^-\Change\) (\(\Delta H = +1451\))
4. Atomisation of chlorine: \(\text{Mg}^{2+}(g) + 2\text{Cl}(g) + 2e^-\Change\) (\(\Delta H = 2 \times 121 = +242\))
- Downward steps:
5. Electron affinity of chlorine: \(\text{Mg}^{2+}(g) + 2\text{Cl}^-(g)\) (\(\Delta H = 2 \times (-349) = -698\))
6. Lattice formation enthalpy (\(\Delta_{\text{L}} H^{\theta}\)) pointing down from gaseous ions to \(\text{MgCl}_2(s)\).

(c) Using Hess's law to solve for the lattice enthalpy of formation (\(\Delta_{\text{latt}}H_{\text{form}}\)):
\[\Delta_f H^{\theta} = \Delta_{\text{at}}H^{\theta}(\text{Mg}) + IE_1(\text{Mg}) + IE_2(\text{Mg}) + 2\Delta_{\text{at}}H^{\theta}(\text{Cl}) + 2EA(\text{Cl}) + \Delta_{\text{latt}}H_{\text{form}}\]
\[-642 = 148 + 738 + 1451 + 2(121) + 2(-349) + \Delta_{\text{latt}}H_{\text{form}}\]
\[-642 = 1881 + \Delta_{\text{latt}}H_{\text{form}}\]
\[\Delta_{\text{latt}}H_{\text{form}} = -642 - 1881 = -2523\text{ kJ mol}^{-1}\]

(Note: If lattice dissociation energy is calculated, \(\Delta_{\text{latt}}H_{\text{diss}} = +2523\text{ kJ mol}^{-1}\)).

(a) [2 marks]
- 1 mark for 'enthalpy change when one mole of gaseous atoms is formed'.
- 1 mark for 'from the element in its standard state'.

(b) [4 marks]
- 1 mark for correct elements at the standard state baseline level and correct compound at the lower level.
- 1 mark for all correct upward arrows (representing atomisations and ionisations of Mg and Cl).
- 1 mark for correct downward arrow for electron affinity (with 2 moles of electrons / 2 chlorine atoms).
- 1 mark for all states (s, g, l) correctly represented on the diagram.

(c) [4 marks]
- 1 mark for doubling the chlorine atomisation enthalpy (\(2 \times 121 = 242\)) AND doubling the electron affinity of chlorine (\(2 \times -349 = -698\)).
- 1 mark for setting up a correct mathematical equation representing the Born-Haber cycle.
- 1 mark for correct calculation of the sum of the cycle steps (\(1881\text{ kJ mol}^{-1}\)).
- 1 mark for the correct final value with units: \(-2523\text{ kJ mol}^{-1}\) (allow \(+2523\text{ kJ mol}^{-1}\) if explicitly defined as lattice dissociation enthalpy).

(a) The negative electrode corresponds to the half-cell with the less positive (more negative) potential, which is the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell.

Conventional representation (negative on left):
\[\text{Pt}(s) \mid \text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq) \parallel \text{Cr}_2\text{O}_7^{2-}(aq), \text{H}^+(aq), \text{Cr}^{3+}(aq) \mid \text{Pt}(s)\]

(b) Standard EMF:
\[E^{\theta}_{\text{cell}} = E^{\theta}_{\text{right}} - E^{\theta}_{\text{left}} = 1.33 - 0.77 = +0.56\text{ V}\]

(c) To balance electrons, multiply the oxidation half-equation by 6:
\[6\text{Fe}^{2+}(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 6e^-\]

Add this to the reduction half-equation:
\[\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 6\text{Fe}^{3+}(aq)\]

(d) The salt bridge completes the electrical circuit without mixing the solutions. It contains a solution of an unreactive electrolyte (e.g., potassium nitrate) allowing ions to move between half-cells to maintain electrical neutrality / charge balance.

(e) Decreasing \([\text{H}^+]\) shifts the position of the \(\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}\) equilibrium to the left. This causes the electrode potential of Half-cell 2 to become less positive (decrease), which in turn decreases the overall EMF of the cell.

(a) [3 marks]
- 1 mark for \(\text{Pt}(s)\) at both outer ends.
- 1 mark for correct half-cell species on correct sides separated by a salt bridge (\(\parallel\)).
- 1 mark for correct phase boundaries (\(\mid\)) and comma separators for species in the same phase.

(b) [1 mark]
- 1 mark for \(+0.56\text{ V}\) (must include sign).

(c) [2 marks]
- 1 mark for correct reactants and products.
- 1 mark for correct balancing.

(d) [2 marks]
- 1 mark for stating that it completes the circuit.
- 1 mark for stating that it allows ions to flow to maintain charge balance/neutrality.

(e) [2 marks]
- 1 mark for predicting that the EMF decreases.
- 1 mark for explaining that equilibrium shifts to the left / electrode potential of Half-cell 2 decreases.

(a)(i)
- **Observation:** Blue precipitate.
- **Equation:**
\[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Cu}(\text{H}_2\text{O})_4(\text{OH})_2(s) + 2\text{H}_2\text{O}(l)\]
*(Accept: \(\text{Cu}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Cu}(\text{OH})_2(s)\))*

(a)(ii)
- **Observation:** Initial blue precipitate dissolves to form a deep blue solution.
- **Equation:**
\[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{NH}_3(aq) \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(aq) + 4\text{H}_2\text{O}(l)\]

(b)(i)
- **Observation:** Brown (or red-brown) precipitate and bubbles of gas/effervescence.
- **Equation:**
\[2[\text{Fe}(\text{H}_2\text{O})_6]^{3+}(aq) + 3\text{CO}_3^{2-}(aq) \rightarrow 2\text{Fe}(\text{H}_2\text{O})_3(\text{OH})_3(s) + 3\text{CO}_2(g) + 3\text{H}_2\text{O}(l)\]

(b)(ii)
- \(\text{Fe}^{3+}\) has a high charge density (large 3+ charge and small ionic radius) compared to \(\text{Fe}^{2+}\).
- The high charge density of \(\text{Fe}^{3+}\) strongly polarizes the O-H bonds of the coordinated water ligands, making the solution acidic enough (releasing \(\text{H}^+\) ions) to react with carbonate ions and produce \(\text{CO}_2\) gas.
- \(\text{Fe}^{2+}\) has a lower charge density, is less polarizing, and is not acidic enough to react with carbonate ions in this way, so it simply precipitates as \(\text{FeCO}_3\).

(a)(i) [2 marks]
- 1 mark for observation: Blue precipitate.
- 1 mark for correct equation (molecular or ionic).

(a)(ii) [3 marks]
- 1 mark for observation: Deep blue solution (or blue precipitate dissolving to deep blue solution).
- 2 marks for equation: 1 mark for correct product \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\), 1 mark for correct balancing.

(b)(i) [3 marks]
- 1 mark for observation: Brown / red-brown precipitate AND effervescence/bubbles.
- 2 marks for equation: 1 mark for correct products \(\text{Fe}(\text{H}_2\text{O})_3(\text{OH})_3\) and \(\text{CO}_2\), 1 mark for correct stoichiometry.

(b)(ii) [2 marks]
- 1 mark for stating that \(\text{Fe}^{3+}\) has a higher charge density (or is more polarizing) than \(\text{Fe}^{2+}\).
- 1 mark for explaining that \(\text{Fe}^{3+}\) polarizes coordinated water ligands strongly, making the complex acidic enough to release \(\text{H}^+\) which reacts with carbonate to form \(\text{CO}_2\).

(a) Expression for \(K_p\):
\[K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}\]
*(Must use "p" for partial pressures, square brackets represent concentration and are not accepted)*

(b) Let's set up the ICE (Initial, Change, Equilibrium) table:
- **Initial moles:** \(n(\text{SO}_2) = 2.00\), \(n(\text{O}_2) = 1.50\), \(n(\text{SO}_3) = 0\)
- **At equilibrium:** \(n(\text{SO}_3) = 0.80\text{ mol}\)
- Since \(2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3\):
- Moles of \(\text{SO}_2\) reacted = \(0.80\text{ mol}\)
- Moles of \(\text{O}_2\) reacted = \(0.40\text{ mol}\)
- **Equilibrium moles:**
- \(n(\text{SO}_2) = 2.00 - 0.80 = 1.20\text{ mol}\)
- \(n(\text{O}_2) = 1.50 - 0.40 = 1.10\text{ mol}\)

(c) Total equilibrium moles:
\[n_{\text{total}} = 1.20 + 1.10 + 0.80 = 3.10\text{ mol}\]

Partial pressures (\(p = \frac{n_{\text{gas}}}{n_{\text{total}}} \times P_{\text{total}}\)):
\[p(\text{SO}_2) = \frac{1.20}{3.10} \times 150 = 58.06\text{ kPa} \approx 58.1\text{ kPa}\]
\[p(\text{O}_2) = \frac{1.10}{3.10} \times 150 = 53.23\text{ kPa} \approx 53.2\text{ kPa}\]
\[p(\text{SO}_3) = \frac{0.80}{3.10} \times 150 = 38.71\text{ kPa} \approx 38.7\text{ kPa}\]

(d) Substitute partial pressures into \(K_p\):
\[K_p = \frac{(38.71)^2}{(58.06)^2 \times (53.23)} = \frac{1498.46}{3370.96 \times 53.23} = \frac{1498.46}{179436.4} = 8.35 \times 10^{-3}\text{ kPa}^{-1}\]
*(Or \(8.35 \times 10^{-6}\text{ Pa}^{-1}\) if converted to Pa)*

(e) There is no effect on \(K_p\) (only temperature changes the value of \(K_p\)).

(a) [1 mark]
- 1 mark for the correct \(K_p\) expression using partial pressure notation (\(p\)).

(b) [2 marks]
- 1 mark for \(n(\text{SO}_2) = 1.20\text{ mol}\).
- 1 mark for \(n(\text{O}_2) = 1.10\text{ mol}\).

(c) [3 marks]
- 1 mark for calculating total equilibrium moles = \(3.10\text{ mol}\).
- 1 mark for calculating mole fractions.
- 1 mark for all three partial pressures correct: \(p(\text{SO}_2) = 58.1\text{ kPa}\), \(p(\text{O}_2) = 53.2\text{ kPa}\), \(p(\text{SO}_3) = 38.7\text{ kPa}\).

(d) [3 marks]
- 1 mark for correct substitution into \(K_p\).
- 1 mark for the value: \(8.35 \times 10^{-3}\) (allow \(8.3 \times 10^{-3}\) to \(8.4 \times 10^{-3}\)).
- 1 mark for units: \(\text{kPa}^{-1}\) (or \(\text{Pa}^{-1}\) if pressure was converted to Pa, giving \(8.35 \times 10^{-6}\)).

(e) [1 mark]
- 1 mark for stating "no effect" / "remains constant".

(a) For 1 mole of \(\text{FeC}_2\text{O}_4\), both \(\text{Fe}^{2+}\) and \(\text{C}_2\text{O}_4^{2-}\) are oxidized:
- \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\Change\) (loses \(1e^-\Change\))
- \(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-\Change\) (loses \(2e^-\Change\))
Thus, the total oxidation half-equation is:
\[\text{FeC}_2\text{O}_4 \rightarrow \text{Fe}^{3+} + 2\text{CO}_2 + 3e^-\]

To combine with \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\), we multiply the oxidation half-equation by 5 and the reduction half-equation by 3 to balance the electron change (to 15 electrons):
\[5\text{FeC}_2\text{O}_4 \rightarrow 5\text{Fe}^{3+} + 10\text{CO}_2 + 15e^-\]
\[3\text{MnO}_4^- + 24\text{H}^+ + 15e^- \rightarrow 3\text{Mn}^{2+} + 12\text{H}_2\text{O}\]

Adding these equations gives:
\[3\text{MnO}_4^- + 24\text{H}^+ + 5\text{FeC}_2\text{O}_4 \rightarrow 3\text{Mn}^{2+} + 12\text{H}_2\text{O} + 5\text{Fe}^{3+} + 10\text{CO}_2\]

(b) Moles of \(\text{MnO}_4^-\Change\) in \(24.00\text{ cm}^3\):
\[n = 0.02400\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 4.80 \times 10^{-4}\text{ mol}\]

(c) The stoichiometric ratio of \(\text{MnO}_4^- : \text{FeC}_2\text{O}_4\) is \(3:5\).
- Moles of \(\text{FeC}_2\text{O}_4\) in \(25.0\text{ cm}^3\):
\[n(\text{FeC}_2\text{O}_4) = 4.80 \times 10^{-4} \times \frac{5}{3} = 8.00 \times 10^{-4}\text{ mol}\]
- Moles of \(\text{FeC}_2\text{O}_4\) in the original \(250.0\text{ cm}^3\) solution:
\[n(\text{FeC}_2\text{O}_4)_{\text{orig}} = 8.00 \times 10^{-4} \times 10 = 8.00 \times 10^{-3}\text{ mol}\]

(d) Relative formula mass (\(M_r\)) of \(\text{FeC}_2\text{O}_4 \cdot x\text{H}_2\text{O}\):
\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{1.436\text{ g}}{8.00 \times 10^{-3}\text{ mol}} = 179.5\]

(e) Calculate the formula mass of anhydrous \(\text{FeC}_2\text{O}_4\):
\[M_r(\text{FeC}_2\text{O}_4) = 55.8 + 2(12.0) + 4(16.0) = 55.8 + 24.0 + 64.0 = 143.8\]

Mass contribution of water (\(x\text{H}_2\text{O}\)):
\[M_r(x\text{H}_2\text{O}) = 179.5 - 143.8 = 35.7\]
\[x = \frac{35.7}{18.0} = 1.983 \approx 2\]

Therefore, the value of \(x\) is 2.

(a) [2 marks]
- 1 mark for constructing the correct combined oxidation half-equation for \(\text{FeC}_2\text{O}_4\) showing the loss of 3 electrons.
- 1 mark for multiplying the two half-equations by 5 and 3 respectively, canceling the electrons, and obtaining the correct overall equation.

(b) [1 mark]
- 1 mark for \(n = 4.80 \times 10^{-4}\text{ mol}\).

(c) [2 marks]
- 1 mark for multiplying moles of manganate by \(5/3\) to get \(8.00 \times 10^{-4}\text{ mol}\) in the titration sample.
- 1 mark for scaling up by a factor of 10 to get \(8.00 \times 10^{-3}\text{ mol}\) in \(250.0\text{ cm}^3\).

(d) [2 marks]
- 1 mark for setting up formula: \(M_r = \text{mass} / \text{moles}\).
- 1 mark for correct calculation of \(M_r = 179.5\) (accept 180).

(e) [3 marks]
- 1 mark for calculation of anhydrous \(M_r(\text{FeC}_2\text{O}_4) = 143.8\).
- 1 mark for subtracting to find the mass of water of crystallization (\(35.7\) or \(36.2\)).
- 1 mark for dividing by 18.0 to get \(x = 2\) (must be an integer).

Part (a):
- Comparing Experiment 1 and 2: \([F_2]\) is kept constant at \(0.100\text{ mol dm}^{-3}\) while \([NO_2]\) is doubled from \(0.100\) to \(0.200\text{ mol dm}^{-3}\). The initial rate doubles from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Thus, the order of reaction with respect to \(NO_2\) is 1 (first order).
- Comparing Experiment 2 and 3: \([NO_2]\) is kept constant at \(0.200\text{ mol dm}^{-3}\) while \([F_2]\) is doubled from \(0.100\) to \(0.200\text{ mol dm}^{-3}\). The initial rate doubles from \(4.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Thus, the order of reaction with respect to \(F_2\) is 1 (first order).

Part (b):
- The rate equation is: \(\text{Rate} = k[NO_2][F_2]\)
- To find \(k\), substitute the data from Experiment 1:
\(2.0 \times 10^{-4} = k(0.100)(0.100)\)
\(k = \frac{2.0 \times 10^{-4}}{0.0100} = 0.020\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
- Units: \(\text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

Part (c):
- Step 1 is the rate-determining step because it is the slow step of the mechanism.
- The rate-determining step determines the overall rate of reaction.
- The reactants in Step 1 are one molecule of \(NO_2\) and one molecule of \(F_2\), which matches the rate equation \(\text{Rate} = k[NO_2]^1[F_2]^1\), making the mechanism consistent.

Part (a) [4 marks]:
- 1 mark for stating first order (or 1) with respect to \(NO_2\).
- 1 mark for explaining that doubling \([NO_2]\) doubles the rate when \([F_2]\) is constant.
- 1 mark for stating first order (or 1) with respect to \(F_2\).
- 1 mark for explaining that doubling \([F_2]\) doubles the rate when \([NO_2]\) is constant.

Part (b) [3 marks]:
- 1 mark for rate equation: \(\text{Rate} = k[NO_2][F_2]\) (consequential on part a).
- 1 mark for calculation: \(k = 0.020\) (or \(2.0 \times 10^{-2}\)).
- 1 mark for units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

Part (c) [3 marks]:
- 1 mark for stating that Step 1 is the rate-determining step because it is the slow step.
- 1 mark for stating that the rate-determining step determines the rate equation.
- 1 mark for linking the reactants in Step 1 to the species in the rate equation (one molecule of each).

Part (a):
\(K_a = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]}\)

Part (b):
1. Calculate initial moles of reactants:
- \(n(\text{C}_2\text{H}_5\text{COOH}) = \frac{50.0}{1000} \times 0.200 = 0.0100\text{ mol}\)
- \(n(\text{OH}^-) = \frac{25.0}{1000} \times 0.150 = 0.00375\text{ mol}\)
2. Write the reaction equation and calculate moles after reaction:
\(\text{C}_2\text{H}_5\text{COOH} + \text{OH}^- \rightarrow \text{C}_2\text{H}_5\text{COO}^- + \text{H}_2\text{O}\)
- Moles of remaining \(\text{C}_2\text{H}_5\text{COOH} = 0.0100 - 0.00375 = 0.00625\text{ mol}\)
- Moles of formed \(\text{C}_2\text{H}_5\text{COO}^- = 0.00375\text{ mol}\)
3. Calculate \([\text{H}^+]\):
Since volume terms cancel, we can substitute moles directly into the rearranged \(K_a\) expression:
\([\text{H}^+] = K_a \times \frac{n(\text{C}_2\text{H}_5\text{COOH})}{n(\text{C}_2\text{H}_5\text{COO}^-)} = 1.35 \times 10^{-5} \times \frac{0.00625}{0.00375} = 2.25 \times 10^{-5}\text{ mol dm}^{-3}\)
4. Calculate pH:
\(\text{pH} = -\log_{10}(2.25 \times 10^{-5}) = 4.65\) (to 2 decimal places).

Part (c):
- Ionic equation: \(\text{C}_2\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_2\text{H}_5\text{COOH}\)
- When a small amount of acid is added, the large reservoir of conjugate base propanoate ions (\(\text{C}_2\text{H}_5\text{COO}^-\)) reacts with the added \(\text{H}^+\) ions to form weak, mostly undissociated propanoic acid molecules. This keeps the concentration of free hydrogen ions, and hence the pH, virtually constant.

Part (a) [1 mark]:
- 1 mark for correct \(K_a = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]}\) (charges must be shown).

Part (b) [6 marks]:
- 1 mark for calculating initial moles of acid: \(0.0100\text{ mol}\).
- 1 mark for calculating initial moles of hydroxide: \(0.00375\text{ mol}\).
- 1 mark for calculating equilibrium moles of propanoic acid: \(0.00625\text{ mol}\).
- 1 mark for calculating equilibrium moles of propanoate ion: \(0.00375\text{ mol}\).
- 1 mark for correct rearranged expression for \([\text{H}^+]\) or correct use of Henderson-Hasselbalch equation.
- 1 mark for \(\text{pH} = 4.65\) (must be given to 2 decimal places. Allow range 4.64 - 4.66).

Part (c) [3 marks]:
- 1 mark for correct equation: \(\text{C}_2\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_2\text{H}_5\text{COOH}\).
- 1 mark for identifying that added \(\text{H}^+\) reacts with propanoate ions.
- 1 mark for explaining that the free hydrogen ion concentration remains nearly unchanged.

Part (a):
- The peak at \(1740\text{ cm}^{-1}\) is characteristic of a carbonyl carbon-oxygen double bond (\(\text{C=O}\)). The lack of a broad peak at \(3200 - 3600\text{ cm}^{-1}\) indicates the absence of an \(\text{O-H}\) group. This, combined with the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\), indicates the functional group is an ester.
- Proton NMR environment assignments:
- \(\delta = 0.95\text{ ppm}\) (3H): Methyl group protons (\(\text{-CH}_3\)) of an alkyl chain.
- \(\delta = 1.65\text{ ppm}\) (2H): Protons of a methylene group (\(\text{-CH}_2-\)) in an alkyl chain.
- \(\delta = 2.01\text{ ppm}\) (3H): Methyl group protons next to an electron-withdrawing group, specifically the carbonyl group (\(\text{CH}_3\text{CO-}\)).
- \(\delta = 4.05\text{ ppm}\) (2H): Methylene group protons directly bonded to the electronegative ester oxygen atom (\(\text{-OCH}_2-\)).

Part (b):
- The singlet at \(\delta = 2.01\text{ ppm}\) has no adjacent hydrogen atoms on neighboring carbons (\(n+1 = 1 \rightarrow n=0\)). This confirms a isolated methyl group, \(\text{CH}_3\text{CO-}\).
- The triplet at \(\delta = 4.05\text{ ppm}\) (2H) is split by 2 neighboring protons on an adjacent carbon (\(n+1 = 3 \rightarrow n=2\)), meaning it is adjacent to a \(\text{-CH}_2-\).
- The sextet at \(\delta = 1.65\text{ ppm}\) (2H) is split by 5 neighboring protons (\(n+1=6 \rightarrow n=5\)), meaning it is adjacent to both a \(\text{-CH}_3\) group (3H) and a \(\text{-CH}_2-\) group (2H).
- The triplet at \(\delta = 0.95\text{ ppm}\) (3H) is adjacent to a \(\text{-CH}_2-\) group.
- Combining these pieces reveals a propyl ester chain: \(\text{-O-CH}_2\text{-CH}_2\text{-CH}_3\).

Part (c):
- Combining \(\text{CH}_3\text{CO-}\) and \(\text{-OCH}_2\text{-CH}_2\text{-CH}_3\) yields propyl ethanoate.
- Systematic name: propyl ethanoate.
- Skeletal formula: shows a 2-carbon acyl side (including the carbonyl) connected via oxygen to a 3-carbon chain.

Part (a) [5 marks]:
- 1 mark for identifying the functional group as an ester from \(\text{C=O}\) stretch and absence of \(\text{O-H}\).
- 1 mark for assigning \(\delta = 0.95\text{ ppm}\) to a \(\text{-CH}_3\) group.
- 1 mark for assigning \(\delta = 1.65\text{ ppm}\) to a \(\text{-CH}_2-\) group.
- 1 mark for assigning \(\delta = 2.01\text{ ppm}\) to a methyl group next to a carbonyl (\(\text{CH}_3\text{CO-}\)).
- 1 mark for assigning \(\delta = 4.05\text{ ppm}\) to a \(\text{-OCH}_2-\) group.

Part (b) [3 marks]:
- 1 mark for explaining that the singlet at \(2.01\text{ ppm}\) indicates no adjacent hydrogen atoms (confirming \(\text{CH}_3\text{CO-}\)).
- 1 mark for explaining that the triplets at \(0.95\text{ ppm}\) and \(4.05\text{ ppm}\) and the sextet at \(1.65\text{ ppm}\) establish the propyl group sequence (\(\text{-CH}_2\text{-CH}_2\text{-CH}_3\)).
- 1 mark for explicit reference to the \(n+1\) rule in splitting assignments.

Part (c) [2 marks]:
- 1 mark for correct systematic name: propyl ethanoate.
- 1 mark for correct skeletal structure.

Part (a):
- Equation: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\)
(Alternatively: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_2\text{O} + \text{HSO}_4^-\))
- Role of sulfuric acid: Catalyst / Brønsted-Lowry acid / proton donor.

Part (b):
- The mechanism is electrophilic aromatic substitution:
1. Draw a curly arrow from the benzene ring of methyl benzoate (from the double bond at carbon 3) pointing to the nitrogen atom of the \(\text{NO}_2^+\) ion.
2. Draw the carbocation intermediate. This intermediate must show carbon 3 bonded to both the hydrogen atom (\(\text{-H}\)) and the nitro group (\(\text{-NO}_2\)). The ring must contain a positive charge and a horseshoe shape representing the delocalized system over 5 carbons, open towards carbon 3.
3. Draw a curly arrow from the \(\text{C-H}\) bond at carbon 3 back into the ring to restore the aromatic system.
4. Show the products: methyl 3-nitrobenzoate and \(\text{H}^+\).

Part (c):
- The carbonyl group (\(\text{-C=O}\)) in the ester is highly polar and strongly electron-withdrawing.
- It withdraws electron density from the benzene ring by inductive and resonance effects, deactivating the ring overall.
- Specifically, it withdraws electron density most strongly from the 2- (ortho) and 4- (para) positions. This places a partial positive charge on those carbons, so electrophilic attack at these positions would produce a very unstable carbocation intermediate with positive charges on adjacent carbons. Meta (3-position) attack is favored as it avoids this unstable configuration.

Part (a) [2 marks]:
- 1 mark for correct equation showing generation of \(\text{NO}_2^+\).
- 1 mark for identifying sulfuric acid as a catalyst or proton donor/acid.

Part (b) [5 marks]:
- 1 mark for curly arrow from ring to \(\text{NO}_2^+\).
- 1 mark for showing both \(\text{-H}\) and \(\text{-NO}_2\) bonded to carbon 3 in the intermediate.
- 1 mark for correct intermediate structure with a positive charge and a horseshoe shape spanning 5 carbons, open towards the \(\text{sp}^3\) carbon.
- 1 mark for curly arrow from the \(\text{C-H}\) bond back into the ring.
- 1 mark for correct structure of methyl 3-nitrobenzoate.

Part (c) [3 marks]:
- 1 mark for stating that the ester group is electron-withdrawing / deactivating.
- 1 mark for explaining that it withdraws electron density primarily from ortho/para (2 and 4) positions.
- 1 mark for explaining that attack at the meta (3) position is preferred because it avoids a highly unstable carbocation intermediate where positive charges are adjacent.

Part (a)(i):
- The monomers are \(\text{H}_2\text{N-C}_6\text{H}_4\text{-NH}_2\) and \(\text{HOOC-C}_6\text{H}_4\text{-COOH}\).
- The repeating unit is formed by eliminating water molecules:
\(\text{-HN-C}_6\text{H}_4\text{-NH-CO-C}_6\text{H}_4\text{-CO-}\) with trailing extension bonds at both ends.

Part (a)(ii):
- Reaction type: Condensation polymerization.
- Small molecule: Water (\(\text{H}_2\text{O}\)).

Part (b):
- Polyamides contain amide/peptide linkages (\(\text{-CO-NH-}\)). The carbonyl group (\(\text{C=O}\)) and \(\text{C-N}\) bonds are polar, rendering the carbonyl carbon electron-deficient (\(\delta+\)) and susceptible to nucleophilic attack by water (hydrolysis).
- Consequently, polyamides can be broken down by micro-organisms/enzymes via hydrolysis.
- In contrast, poly(ethene) is a polyalkene consisting of a non-polar carbon-carbon backbone (\(\text{C-C}\) single bonds) and non-polar \(\text{C-H}\) bonds.
- These bonds are highly stable and unreactive, resisting chemical or biological attack, hence making polyalkenes non-biodegradable.

Part (c):
- (i) At pH 1 (strongly acidic), the amine group is protonated to form an ammonium ion, while the carboxyl group remains protonated:
\(\text{H}_3\text{N}^+\text{-CH(CH}_3\text{)-COOH}\)
- (ii) At pH 12 (strongly alkaline), the carboxyl group is deprotonated to form a carboxylate ion, while the amine group remains unprotonated:
\(\text{H}_2\text{N-CH(CH}_3\text{)-COO}^-\)
- (iii) The species containing both positive and negative charges with an overall neutral charge is called a **zwitterion**.

Part (a) [3 marks]:
- 2 marks for drawing the correct repeating unit of Kevlar with extension bonds. (Allow 1 mark if there is a minor error in drawing the benzene rings but the amide linkage is correct).
- 1 mark for 'condensation' and 'water' (both required).

Part (b) [4 marks]:
- 1 mark for identifying the presence of polar amide linkages (\(\text{-CO-NH-}\)) in polyamides.
- 1 mark for stating that polyamides can be hydrolyzed (by water/microorganisms).
- 1 mark for stating that polyalkenes have non-polar \(\text{C-C}\) and \(\text{C-H}\) bonds.
- 1 mark for explaining that the non-polar nature of these bonds makes polyalkenes inert / resistant to chemical or biological attack.

Part (c) [3 marks]:
- 1 mark for correct protonated structure at pH 1: \(\text{H}_3\text{N}^+\text{-CH(CH}_3\text{)-COOH}\).
- 1 mark for correct deprotonated structure at pH 12: \(\text{H}_2\text{N-CH(CH}_3\text{)-COO}^-\).
- 1 mark for 'zwitterion'.

Part (a):
\(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \cdot p(\text{O}_2)}\) (where \(p\) represents partial pressure. Do not use square brackets).

Part (b):
- Initial moles: \(\text{SO}_2 = 2.00\text{ mol}\), \(\text{O}_2 = 1.00\text{ mol}\), \(\text{SO}_3 = 0.00\text{ mol}\).
- Moles of \(\text{SO}_3\) at equilibrium = \(1.60\text{ mol}\).
- Reacted moles: \(1.60\text{ mol}\) of \(\text{SO}_2\) reacted, \(\frac{1}{2} \times 1.60 = 0.80\text{ mol}\) of \(\text{O}_2\) reacted.
- Equilibrium moles:
- \(n(\text{SO}_2) = 2.00 - 1.60 = 0.40\text{ mol}\)
- \(n(\text{O}_2) = 1.00 - 0.80 = 0.20\text{ mol}\)

Part (c):
1. Total moles of gas at equilibrium:
\(n_{\text{total}} = 0.40 + 0.20 + 1.60 = 2.20\text{ mol}\)
2. Calculate the mole fraction of each gas:
- \(x(\text{SO}_2) = \frac{0.40}{2.20} = 0.1818\)
- \(x(\text{O}_2) = \frac{0.20}{2.20} = 0.0909\)
- \(x(\text{SO}_3) = \frac{1.60}{2.20} = 0.7273\)
3. Calculate the partial pressure of each gas (\(p = x \times P_{\text{total}}\)) using \(P_{\text{total}} = 250\text{ kPa}\):
- \(p(\text{SO}_2) = 0.1818 \times 250 = 45.45\text{ kPa}\)
- \(p(\text{O}_2) = 0.0909 \times 250 = 22.73\text{ kPa}\)
- \(p(\text{SO}_3) = 0.7273 \times 250 = 181.82\text{ kPa}\)
4. Calculate \(K_p\):
\(K_p = \frac{(181.82)^2}{(45.45)^2 \times 22.73} = \frac{33058.5}{2065.7 \times 22.73} = \frac{33058.5}{46953.4} = 0.704\text{ kPa}^{-1}\)
- (If calculated using Pascals, \(P_{\text{total}} = 2.50 \times 10^5\text{ Pa}\):
\(p(\text{SO}_2) = 4.545 \times 10^4\text{ Pa}\), \(p(\text{O}_2) = 2.273 \times 10^4\text{ Pa}\), \(p(\text{SO}_3) = 1.818 \times 10^5\text{ Pa}\).
\(K_p = 7.04 \times 10^{-4}\text{ Pa}^{-1}\)).
- Units of \(K_p\):
\(\text{units} = \frac{\text{kPa}^2}{\text{kPa}^2 \cdot \text{kPa}} = \text{kPa}^{-1}\) (or \(\text{Pa}^{-1}\)).

Part (a) [1 mark]:
- 1 mark for correct expression: \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \cdot p(\text{O}_2)}\) (reject square brackets []).

Part (b) [2 marks]:
- 1 mark for equilibrium \(n(\text{SO}_2) = 0.40\text{ mol}\).
- 1 mark for equilibrium \(n(\text{O}_2) = 0.20\text{ mol}\).

Part (c) [7 marks]:
- 1 mark for calculating total equilibrium moles = \(2.20\text{ mol}\).
- 2 marks for calculating the partial pressures of all three gases (allow 1 mark for two correct).
- 2 marks for substitution and calculation of \(K_p = 0.704\) (allow range \(0.700 - 0.710\)).
- 1 mark for correct unit: \(\text{kPa}^{-1}\) (or \(7.04 \times 10^{-4}\text{ Pa}^{-1}\)).
- 1 mark for showing logical layout of calculations with working.

Part (a):
Mechanism steps:
1. Draw butanone, \(\text{CH}_3\text{-CO-CH}_2\text{CH}_3\), with a partial positive charge (\(\delta+\)) on the carbonyl carbon and a partial negative charge (\(\delta-\)) on the oxygen.
2. Show the \(\text{CN}^-\)_ nucleophile with a lone pair on the carbon atom.
3. Draw a curly arrow from the lone pair of the carbon in \(\text{CN}^-\)_ to the carbonyl carbon.
4. Draw a curly arrow from the double bond of \(\text{C=O}\) to the oxygen atom.
5. Draw the intermediate species: \(\text{CH}_3\text{-C(O}^-\text{)(CN)-CH}_2\text{CH}_3\) showing a lone pair and negative charge on oxygen.
6. Draw a curly arrow from the negative oxygen's lone pair to a hydrogen ion (\(\text{H}^+\)) to complete the addition.
7. Draw the final product: 2-hydroxy-2-methylbutanenitrile.

Part (b):
- The carbonyl group (\(\text{C=O}\)) is planar around the carbonyl carbon atom.
- The nucleophile \(\text{CN}^-\) can attack this planar carbon atom with equal probability from either side (above or below the plane).
- This results in the formation of equal amounts (a 50:50 mixture) of the two optical isomers (enantiomers), which is a racemic mixture.
- Because the enantiomers rotate plane-polarized light in equal and opposite directions, their effects cancel out, resulting in no overall optical activity.

Part (c):
- Reagent: Tollens' reagent (ammoniacal silver nitrate) [or Fehling's solution].
- Observation with butanal: A silver mirror forms [or red precipitate with Fehling's solution].

Part (a) [5 marks]:
- 1 mark for correct partial charges (\(\delta+\)/\(\delta-\)) on \(\text{C=O}\) in butanone.
- 1 mark for curly arrow from carbon lone pair on \(\text{CN}^-\) to the carbonyl carbon.
- 1 mark for curly arrow from the double bond of \(\text{C=O}\) to the oxygen atom.
- 1 mark for the correct structure of the intermediate anion with negative charge on oxygen.
- 1 mark for curly arrow from oxygen lone pair to \(\text{H}^+\) (or to H of H-Cl with bond breaking).

Part (a):
- Reagents: Concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)).
- Conditions: Maintain a temperature of \(50 - 55^\circ\text{C}\) (to prevent multiple nitrations).
- Mechanism name: Electrophilic substitution.

Part (b):
- Reagents and conditions: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), heated under reflux, followed by treatment with aqueous sodium hydroxide (\(\text{NaOH}\)) to liberate the free amine from its salt.
- Equation:
\(\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}\)

Part (c):
- (i) Acyl chloride: Ethanoyl chloride (\(\text{CH}_3\text{COCl}\)).
- (ii) Balanced equation:
\(\text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{HCl}\)
- (iii) Functional group: Amide (or secondary / N-substituted amide).

Part (a) [3 marks]:
- 1 mark for concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\) (both required, 'concentrated' must be stated).
- 1 mark for warm / temperature of \(50 - 55^\circ\text{C}\) (accept a range between \(45 - 60^\circ\text{C}\)).
- 1 mark for 'electrophilic substitution'.

Part (b) [3 marks]:
- 1 mark for Tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\) (accept iron, \(\text{Fe}\), and \(\text{HCl}\)).
- 1 mark for heating/reflux, followed by treatment with alkali (\(\text{NaOH}\)).
- 1 mark for correct equation: \(\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}\).

Part (c) [4 marks]:
- 1 mark for 'ethanoyl chloride'.
- 2 marks for balanced equation (1 mark for correct reactants and products, 1 mark for balancing; accept ionic equivalents).
- 1 mark for naming the functional group as 'amide' or 'N-substituted amide'.

(a) Use the Arrhenius equation in its logarithmic form:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)

Substitute the values:
\(\ln\left(\frac{9.85 \times 10^{-3}}{1.45 \times 10^{-3}}\right) = \ln(6.7931) = 1.9159\)

\(\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{313} - \frac{1}{293} = 3.1949 \times 10^{-3} - 3.4130 \times 10^{-3} = -2.181 \times 10^{-4}\text{ K}^{-1}\)

Substitute back into the equation:
\(1.9159 = -\frac{E_a}{8.31} \times (-2.181 \times 10^{-4})\)

\(\frac{E_a}{8.31} = \frac{1.9159}{2.181 \times 10^{-4}} = 8784.5\)

\(E_a = 8784.5 \times 8.31 = 72999\text{ J mol}^{-1} = 73.0\text{ kJ mol}^{-1}\)

(b) Separate reactants are placed in a water bath to allow them to reach thermal equilibrium before mixing. A thermometer is used to measure the temperature throughout the reaction to ensure it remains stable. Precise temperature control is essential because the rate constant \(k\) varies exponentially with temperature, meaning even small temperature variations lead to significant errors in calculated \(k\) values.

(c) A catalyst decreases \(E_a\) by providing an alternative reaction pathway with lower activation energy. The pre-exponential factor, \(A\), remains approximately unchanged, as it is primarily a function of collision frequency and molecular orientation geometry, which are not altered by the catalyst.

(a)
M1: Calculates \(\ln(k_2/k_1) = 1.92\) (or separate \(\ln k\) values: -4.62 and -6.54) [1 mark]
M2: Calculates \(\Delta(1/T) = -2.18 \times 10^{-4}\text{ K}^{-1}\) (or individual values: \(3.19 \times 10^{-3}\) and \(3.41 \times 10^{-3}\)) [1 mark]
M3: Rearranges equation to find gradient or ratio: \(\frac{E_a}{R} = 8785\) [1 mark]
M4: Correctly calculates \(E_a = 73000\text{ J mol}^{-1}\) (allow range 72.8 to 73.2) [1 mark]
M5: Converts value to \(\text{kJ mol}^{-1}\) to give 73.0 (accept 73) [1 mark]

(b)
M1: Reactants placed in a water bath to reach thermal equilibrium before mixing, monitored with a thermometer [1 mark]
M2: Because \(k\) is exponentially dependent on temperature, so small deviations produce massive errors in the Arrhenius gradient [1 mark]

(c)
M1: Catalyst decreases \(E_a\) [1 mark]
M2: By providing an alternative pathway with lower activation energy [1 mark]
M3: The pre-exponential factor \(A\) remains unchanged/constant [1 mark]

(a) Add a few drops of sodium hydroxide solution to aqueous silver nitrate to form a brown precipitate of silver(I) oxide, \( \text{Ag}_2\text{O} \). Then, add dilute aqueous ammonia dropwise until the brown precipitate just dissolves to form a colorless solution containing the diamminesilver(I) complex ion, \( [\text{Ag}(\text{NH}_3)_2]^+ \).

(b) (i) Under alkaline conditions, ethanal is oxidized to the ethanoate ion:
\(\text{CH}_3\text{CHO} + 3\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + 2\text{H}_2\text{O} + 2\text{e}^-\)
(Accept the neutral/acidic alternative if balanced: \(\text{CH}_3\text{CHO} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + 2\text{H}^+ + 2\text{e}^-\))

(ii) The diamminesilver(I) complex ion is reduced to metallic silver:
\([\text{Ag}(\text{NH}_3)_2]^+ + \text{e}^- \rightarrow \text{Ag} + 2\text{NH}_3\)
Observation: A silver mirror forms on the inner wall of the test tube.

(c) Aldehydes are easily oxidized because they have a hydrogen atom attached to the carbonyl carbon. Ketones do not possess this hydrogen atom; oxidation of a ketone would require the cleavage of a strong carbon-carbon (C-C) bond, which cannot be achieved by a mild oxidizing agent like Tollens' reagent.

(a)
M1: Add NaOH(aq) to silver nitrate to produce a brown precipitate [1 mark]
M2: Correct formula of precipitate: \(\text{Ag}_2\text{O}\) [1 mark]
M3: Add dilute ammonia dropwise until the precipitate dissolves to give a colorless solution [1 mark]
M4: Correct formula of complex: \([\text{Ag}(\text{NH}_3)_2]^+\) [1 mark]

(b)(i)
M1: Correct organic reactant and product: \(\text{CH}_3\text{CHO}\) and \(\text{CH}_3\text{COO}^-\) (or \(\text{CH}_3\text{COOH}\)) [1 mark]
M2: Balanced equation including electrons: \(\text{CH}_3\text{CHO} + 3\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + 2\text{H}_2\text{O} + 2\text{e}^-\) [1 mark]

(b)(ii)
M1: Balanced reduction half-equation: \([\text{Ag}(\text{NH}_3)_2]^+ + \text{e}^- \rightarrow \text{Ag} + 2\text{NH}_3\) [1 mark]
M2: Observation: Silver mirror / grey precipitate [1 mark]

(c)
M1: Aldehydes contain a C-H bond on the carbonyl carbon which allows easy oxidation [1 mark]
M2: Ketones lack this C-H bond, and oxidizing them requires breaking a strong C-C bond, which Tollens' reagent is not strong enough to do [1 mark]

(a) Sodium thiosulfate reacts rapidly with any iodine formed in the main reaction, converting it back to iodide:
\(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\)
This prevents the iodine from reacting with the starch. Once all the sodium thiosulfate has been completely consumed, any further iodine produced remains in solution and reacts with the starch indicator, causing a sudden, sharp blue-black color change that marks a fixed amount of progress in the reaction.

(b) Since the initial rate is proportional to \(\frac{1}{t}\):
- Rate 1 \(\propto 1/120 = 0.00833\text{ s}^{-1}\)
- Rate 2 \(\propto 1/60 = 0.0167\text{ s}^{-1}\)
- Rate 3 \(\propto 1/60 = 0.0167\text{ s}^{-1}\)

Comparing Experiments 1 and 2: \([\text{H}_2\text{O}_2]\) is kept constant, \([\text{I}^-]\) doubles (from 0.020 to 0.040), and the rate doubles (from 0.00833 to 0.0167). Therefore, the reaction is first order with respect to \(\text{I}^-\).

Comparing Experiments 1 and 3: \([\text{I}^-]\) is kept constant, \([\text{H}_2\text{O}_2]\) doubles (from 0.050 to 0.100), and the rate doubles (from 0.00833 to 0.0167). Therefore, the reaction is first order with respect to \(\text{H}_2\text{O}_2\).

(c) The initial rate is approximated as \(\frac{1}{t}\) because the blue-black color occurs after only a very small, negligible fraction of the reactants has been consumed, meaning the reactant concentrations remain essentially constant during this brief interval. If the concentration of sodium thiosulfate is too high, it will take a long time to be consumed. Consequently, a significant portion of the reactants will be depleted before the endpoint is reached, causing the actual rate to slow down significantly. Under these conditions, the average rate over time \(t\) is no longer a valid approximation of the initial rate.

(a)
M1: States that thiosulfate reacts with iodine to reduce it back to iodide, preventing starch from turning blue-black [1 mark]
M2: Provides the balanced equation: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\) [1 mark]
M3: Explains that the color change occurs suddenly once all thiosulfate is exhausted, indicating a fixed amount of reaction [1 mark]

(b)
M1: Connects rate to \(1/t\) (calculates numerical rates or states inverse relationship) [1 mark]
M2: Compares Exp 1 & 2 to show doubling \([\text{I}^-]\) doubles the rate, concluding 1st order with respect to \(\text{I}^-\). [1 mark]
M3: Compares Exp 1 & 3 to show doubling \([\text{H}_2\text{O}_2]\) doubles the rate, concluding 1st order with respect to \(\text{H}_2\text{O}_2\). [1 mark]
M4: Clearly links deduction to calculated rates/working [1 mark]

(c)
M1: States that only a very small/negligible fraction of reactants is consumed during the timed interval [1 mark]
M2: Too high thiosulfate concentration means the reaction is monitored for too long / reactants are significantly depleted [1 mark]
M3: The rate decreases during the run, making \(1/t\) a poor approximation of the initial rate [1 mark]

Platinum(II) typically forms square planar complexes when bonded to four ligands, such as in cisplatin, \([Pt(NH_3)_2Cl_2]\). The tetrachlorocuprate(II) ion, \([CuCl_4]^{2-}\), has a tetrahedral geometry due to the larger size of the chloride ligands. Complexes with coordination number 6, such as \([Co(NH_3)_6]^{3+}\) and \([Fe(H_2O)_6]^{2+}\), are octahedral.

1 mark: B is selected. 0 marks for any other option.

To calculate the pH of an acidic buffer solution, use the equation: \([H^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}\). Substituting the given values: \([H^+] = 1.74 \times 10^{-5} \times \frac{0.150}{0.100} = 2.61 \times 10^{-5}\text{ mol dm}^{-3}\). Then calculate pH: \(\text{pH} = -\log_{10}(2.61 \times 10^{-5}) = 4.58\).

1 mark: A is selected. 0 marks for any other option.

The standard cell potential is calculated as \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\). The half-cell with the more positive potential undergoes reduction (\(Fe^{3+} + e^- \rightarrow Fe^{2+}\)) and the other undergoes oxidation (\(2I^- \rightarrow I_2 + 2e^-\)). Combining these gives the feasible overall reaction: \(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\). The cell potential is \(E^\theta_{\text{cell}} = +0.77 - (+0.54) = +0.23\text{ V}\).

1 mark: B is selected. 0 marks for any other option.

A reaction is thermodynamically feasible when the Gibbs free energy change is negative (\(\Delta G^\theta < 0\)). Using \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), we find feasibility when \(T > \frac{\Delta H^\theta}{\Delta S^\theta}\). Converting \(\Delta H^\theta\) to Joules gives \(135000\text{ J mol}^{-1}\). Therefore, \(T > \frac{135000}{310} = 435.5\text{ K}\).

1 mark: C is selected. 0 marks for any other option.

Letting the initial rate be \(\text{rate}_1 = k[X]^2[Y]\). When \([X]\) is doubled, the rate increases by a factor of \(2^2 = 4\). When \([Y]\) is halved, the rate decreases by a factor of \(0.5\). The combined effect is \(4 \times 0.5 = 2\), meaning the rate is doubled.

1 mark: B is selected. 0 marks for any other option.

At a high pH of 12 (strongly alkaline), both functional groups are deprotonated. The carboxylic acid group (\(-\text{COOH}\)) loses a proton to become a carboxylate ion (\(-\text{COO}^-\)), and the protonated amino group exists in its neutral, deprotonated form (\(-\text{NH}_2\)). Thus, the structure is \(\text{H}_2\text{NCH(CH}_3\text{)COO}^-\).

1 mark: C is selected. 0 marks for any other option.

In the preparation of the electrophile for nitration, concentrated sulfuric acid protonates concentrated nitric acid, which then dehydrates to yield the nitronium ion (\(\text{NO}_2^+\)). This is represented by the stoichiometry: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\).

1 mark: B is selected. 0 marks for any other option.

Aqueous \(Fe^{3+}\) ions have a high charge density and are highly polarizing, forming strongly acidic solutions containing \([Fe(H_2O)_6]^{3+}\). When carbonate ions are added, they act as a base to deprotonate the hexaaquairon(III) complex rather than participating in simple precipitation. This results in the formation of a brown precipitate of iron(III) hydroxide, \(Fe(OH)_3(s)\), and the release of carbon dioxide gas, \(CO_2(g)\).

1 mark: B is selected. 0 marks for any other option.

Using the Gibbs free energy equation, \(\Delta G = \Delta H - T\Delta S\). A reaction is feasible when \(\Delta G < 0\). At the temperature boundary where the reaction ceases to be feasible, \(\Delta G = 0\). Thus, \(T = \frac{\Delta H}{\Delta S} = \frac{-95 \times 10^3\text{ J mol}^{-1}}{-210\text{ J K}^{-1}\text{ mol}^{-1}} = 452.38\text{ K}\). Since both enthalpy and entropy changes are negative, the reaction is feasible at lower temperatures and becomes non-feasible above 452 K.

1 mark for correct calculation of the transition temperature (452 K) and selecting option A.

The reaction is third order overall (first order in X, second order in Y). The units of the rate constant are: \(k = \frac{\text{rate}}{[\text{X}][\text{Y}]^2} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\). If \([\text{X}]\) is tripled and \([\text{Y}]\) is halved, the new rate is \(k (3[\text{X}]) (0.5[\text{Y}])^2 = 3 \times 0.25 \times k[\text{X}][\text{Y}]^2 = 0.75 \times \text{rate}\). Therefore, the rate is multiplied by 0.75.

1 mark for correct deduction of both the rate constant units and the rate multiplier factor, corresponding to option A.

EMF = \(E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 1.51 - 0.77 = +0.74\text{ V}\). The manganese half-cell has the more positive potential and undergoes reduction. The iron half-cell has the less positive potential and undergoes oxidation. Electrons flow from the oxidation half-cell to the reduction half-cell, which is from the Fe(III)/Fe(II) electrode to the Mn(VII)/Mn(II) electrode.

1 mark for calculating the EMF as +0.74 V and identifying that electron flow is from the iron electrode to the manganese electrode.

Initial moles of propanoic acid = \(25.0 \times 10^{-3} \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\). Moles of \(\text{NaOH}\) added = \(12.5 \times 10^{-3} \times 0.100 = 1.25 \times 10^{-3}\text{ mol}\). Since NaOH is the limiting reagent, it reacts completely to convert \(1.25 \times 10^{-3}\text{ mol}\) of propanoic acid into propanoate ions. Remaining propanoic acid = \(2.50 \times 10^{-3} - 1.25 \times 10^{-3} = 1.25 \times 10^{-3}\text{ mol}\). Propanoate ions formed = \(1.25 \times 10^{-3}\text{ mol}\). This represents the half-neutralisation point, where \([\text{HA}] = [\text{A}^-]\). Thus, \(\text{pH} = \text{p}K_{\text{a}} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

1 mark for recognizing the half-neutralisation point and calculating pH = pKa = 4.87, selecting option A.

\(\text{Fe}^{3+}\) has the outer electron configuration \([\text{Ar}]3d^5\). By Hund's rule, these 5 electrons occupy the five d-orbitals singly, resulting in 5 unpaired d-electrons. \(\text{Cr}^{3+}\) has the configuration \([\text{Ar}]3d^3\) with 3 unpaired electrons. \(\text{Cu}^{2+}\) has the configuration \([\text{Ar}]3d^9\) with 1 unpaired electron. \(\text{Co}^{2+}\) has the configuration \([\text{Ar}]3d^7\) with 3 unpaired electrons. Therefore, \(\text{Fe}^{3+}\) has the greatest number of unpaired d-electrons.

1 mark for identifying Fe3+ as having 5 unpaired d-electrons.

The transition metal ion is \(\text{Cu}^{2+}\). When a small amount of aqueous ammonia is added, it acts as a weak base, forming the blue precipitate of copper(II) hydroxide. Upon the addition of excess ammonia, ligand substitution occurs, forming the soluble deep blue tetraamminedifluorocopper(II) complex, which is represented as \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

1 mark for identifying the copper(II) ion and the correct tetraamminedifluorocopper(II) formula.

The methyl group in methylbenzene is electron-releasing via the inductive effect. This increases the overall electron density of the benzene ring, making it more attractive to the incoming electrophile. Additionally, this increased electron density helps to stabilise the positive charge of the intermediate carbocation formed during the electrophilic attack, lowering the activation energy.

1 mark for identifying that the electron-releasing inductive effect of the methyl group increases ring electron density and stabilises the carbocation intermediate.

The repeating unit represents a polyester. Cleaving the ester linkage (\(\text{–O–CO–}\)) shows that the diol component is \(\text{HO–CH}_2\text{–CH}_2\text{–OH}\) (ethane-1,2-diol) and the dicarboxylic acid component is \(\text{HOOC–(CH}_2)_4\text{–COOH}\) (hexanedioic acid).

1 mark for correctly identifying both monomers as ethane-1,2-diol and hexanedioic acid.

A reaction becomes feasible when \(\Delta G^\theta \le 0\).
Using the Gibbs free energy equation:
\(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)

Setting \(\Delta G^\theta = 0\):
\(0 = 135 \times 10^3\text{ J mol}^{-1} - T(180\text{ J K}^{-1}\text{ mol}^{-1})\)

Solving for \(T\):
\(T = \frac{135000}{180} = 750\text{ K}\)

Therefore, the reaction becomes feasible at temperatures above 750 K.

Award 1 mark for the correct option (B).

- Correct identification of the feasibility condition \(\Delta G = 0\) to find the threshold temperature.
- Conversion of \(\Delta H^\theta\) from \(\text{kJ mol}^{-1}\) to \(\text{J mol}^{-1}\) is required.

First, determine the order with respect to each reactant:
1. Comparing Exp 1 and Exp 2: doubling \([\text{P}]\) increases the rate by a factor of 4 (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). Therefore, the reaction is second order with respect to P.
2. Comparing Exp 1 and Exp 3: doubling \([\text{Q}]\) doubles the rate (from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\)). Therefore, the reaction is first order with respect to Q.

The rate equation is:
\(\text{Rate} = k[\text{P}]^2[\text{Q}]\)

To find the units of the rate constant \(k\):
\(k = \frac{\text{Rate}}{[\text{P}]^2[\text{Q}]}\)

Substituting the units:
\(\text{Units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)

Award 1 mark for the correct option (B).

- M1: Deduce that the rate equation is rate = \(k[\text{P}]^2[\text{Q}]\).
- M2: Correctly simplify the unit expression to get \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

For a feasible electrochemical cell, the half-cell with the more positive (less negative) standard electrode potential undergoes reduction, and the one with the more negative standard electrode potential undergoes oxidation.

Here, \(E^\theta(X^{2+}/X) = -0.40\text{ V}\) is more positive than \(E^\theta(Y^{3+}/Y) = -0.76\text{ V}\).

Therefore:
- Reduction (cathode): \(X^{2+}(aq) + 2e^- \rightarrow X(s)\)
- Oxidation (anode): \(Y(s) \rightarrow Y^{3+}(aq) + 3e^-\)

The cell potential is calculated as:
\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\)
\(E^\theta_{\text{cell}} = -0.40\text{ V} - (-0.76\text{ V}) = +0.36\text{ V}\)

Award 1 mark for the correct option (C).

- Correct identification of the reduction and oxidation half-cells.
- Correct arithmetic to obtain +0.36 V.

First, calculate the concentration of weak acid \([\text{HA}]\) and conjugate base \([\text{A}^-]\) in the final mixture. Because equal volumes (\(50.0\text{ cm}^3\) each) are mixed, the total volume is doubled, so the concentrations are halved:
\([\text{HA}] = \frac{0.100}{2} = 0.0500\text{ mol dm}^{-3}\)
\([\text{A}^-] = \frac{0.0500}{2} = 0.0250\text{ mol dm}^{-3}\)

Alternatively, use the moles directly:
\(n(\text{HA}) = 0.0500 \times 0.100 = 0.00500\text{ mol}\)
\(n(\text{A}^-) = 0.0500 \times 0.0500 = 0.00250\text{ mol}\)

Using the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)

Calculate \(\text{p}K_a\):
\(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\)

Calculate pH:
\(\text{pH} = 4.87 + \log_{10}\left(\frac{0.0250}{0.0500}\right) = 4.87 + \log_{10}(0.5) = 4.87 - 0.30 = 4.57\)

Award 1 mark for the correct option (B).

- Correct application of the Henderson-Hasselbalch equation or equilibrium expression.
- Correct subtraction of the log term.

The atomic number of iron (Fe) is 26.
The electronic configuration of a neutral Fe atom in its ground state is:
\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\) or \([\text{Ar}] 3d^6 4s^2\).

When a transition metal forms a positive ion, electrons are lost first from the outer \(4s\) sub-shell before the \(3d\) sub-shell.
To form \(\text{Fe}^{3+}\), three electrons are removed:
- Two from the \(4s\) sub-shell
- One from the \(3d\) sub-shell

This results in the configuration:
\([\text{Ar}] 3d^5\).

Award 1 mark for the correct option (A).

- Recognises that 4s electrons are lost before 3d electrons when forming transition metal cations.

When excess aqueous ammonia is added to \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), a ligand substitution reaction occurs where four water molecules are replaced by four ammonia ligands. This reaction is incomplete due to steric and electronic factors, resulting in the deep-blue mixed-ligand complex:
\([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\)

Award 1 mark for the correct option (B).

- Knowledge that four ammonia molecules substitute into the coordination sphere of copper(II) to form a deep-blue complex.

\(\text{Al}_2\text{O}_3\) is ionic with covalent character and is insoluble in water (no change to water's pH).
\(\text{SiO}_2\) has a giant covalent structure and is insoluble in water.
\(\text{P}_4\text{O}_{10}\) reacts with water to form phosphoric(V) acid (\(\text{H}_3\text{P}\text{O}_4\)), which is a weak/medium acid.
\(\text{SO}_3\) reacts vigorously with water to form sulfuric(VI) acid (\(\text{H}_2\text{S}\text{O}_4\)), which is a very strong acid and completely dissociates, yielding the lowest pH (near 0).

Award 1 mark for the correct option (D).

- Correctly links the chemical nature of the acid formed by Period 3 oxides in water to the resulting pH value.

According to the logarithmic form of the Arrhenius equation:
\(\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A\)

A plot of \(\ln k\) against \(\frac{1}{T}\) gives a straight line with:
\(\text{Gradient} = -\frac{E_a}{R}\)

Given that the gradient is \(-1.20 \times 10^4\text{ K}\):
\(-\frac{E_a}{R} = -1.20 \times 10^4\)

Therefore:
\(E_a = 1.20 \times 10^4 \times 8.31 = 9.972 \times 10^4\text{ J mol}^{-1}\)

Converting to \(\text{kJ mol}^{-1}\):
\(E_a = \frac{9.972 \times 10^4}{1000} \approx 99.7\text{ kJ mol}^{-1}\)

Award 1 mark for the correct option (B).

- Correct recognition of the Arrhenius equation gradient relation.
- Correct units conversion from J to kJ.

Using the rate equation \(\text{rate} = k[\text{A}]^2[\text{B}]\), if the new concentration of \(\text{A}\) is \(0.5[\text{A}]\) and the new concentration of \(\text{B}\) is \(3[\text{B}]\), the new rate is \(k(0.5[\text{A}])^2(3[\text{B}]) = 0.25 \times 3 \times k[\text{A}]^2[\text{B}] = 0.75 \times \text{rate}\). Thus, the rate is multiplied by 0.75.

1 mark for selecting correct option A. Reject all other options.

A reaction is feasible when \(\Delta G^{\ominus} \le 0\). Since \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\), feasibility requires \(T \ge \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\). Converting enthalpy to J: \(T \ge \frac{58000\text{ J mol}^{-1}}{176.0\text{ J K}^{-1}\text{ mol}^{-1}} = 329.5\text{ K}\), which rounds to 330 K.

1 mark for selecting correct option C. Reject all other options.

Since equal volumes are mixed, the concentrations of both components are halved in the final mixture: \([\text{HCOOH}] = 0.10\text{ mol dm}^{-3}\) and \([\text{HCOONa}] = 0.05\text{ mol dm}^{-3}\). Using the buffer expression: \([\text{H}^+] = K_{\text{a}} \times \frac{[\text{acid}]}{[\text{salt}]} = (1.8 \times 10^{-4}) \times \frac{0.10}{0.05} = 3.6 \times 10^{-4}\text{ mol dm}^{-3}\). Therefore, \(\text{pH} = -\log_{10}(3.6 \times 10^{-4}) = 3.44\).

1 mark for selecting correct option A. Reject all other options.

\(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{reduction}} - E^{\ominus}_{\text{oxidation}} = 0.77\text{ V} - 0.54\text{ V} = +0.23\text{ V}\). The half-cell with the more positive potential (iron) undergoes reduction, while the other (iodine) undergoes oxidation. Electrons are released during oxidation and travel through the external circuit from the negative electrode (iodine/iodide) to the positive electrode (iron(III)/iron(II)).

1 mark for selecting correct option A. Reject all other options.

\(\text{Fe}^{3+}\) in \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\) is a \(3\text{d}^5\) system. Since water is a weak-field ligand, it forms a high-spin complex with 5 unpaired electrons. \(\text{Fe}^{2+}\) is \(3\text{d}^6\) (4 unpaired electrons), \(\text{Cu}^{2+}\) is \(3\text{d}^9\) (1 unpaired electron), and \(\text{Co}^{2+}\) is \(3\text{d}^7\) (3 unpaired electrons).

1 mark for selecting correct option B. Reject all other options.

At pH 2.0, the solution is highly acidic (well below the isoelectric point of alanine, which is 6.0). In acidic conditions, both basic groups are protonated. The carboxylate group accepts a proton to become \(\text{-COOH}\), and the amino group remains protonated as \(\text{-NH}_3^+\). This results in a net positive charge on the molecule: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\).

1 mark for selecting correct option C. Reject all other options.