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(a) Since HX is a weak acid:
\( K_a = \frac{[H^+][X^-]}{[HX]} \approx \frac{[H^+]^2}{[HX]} \)
\( [H^+] = \sqrt{K_a \times [HX]} = \sqrt{1.80 \times 10^{-5} \times 0.150} = \sqrt{2.70 \times 10^{-6}} = 1.643 \times 10^{-3} \text{ mol dm}^{-3} \)
\( \text{pH} = -\log_{10}(1.643 \times 10^{-3}) = 2.78 \)

(b) Moles of HX initially = \( \frac{25.0}{1000} \times 0.150 = 3.75 \times 10^{-3} \text{ mol} \)
Moles of NaOH added = \( \frac{15.0}{1000} \times 0.120 = 1.80 \times 10^{-3} \text{ mol} \)
The reaction is: \( \text{HX} + \text{NaOH} \rightarrow \text{NaX} + \text{H}_2\text{O} \)
Moles of HX remaining = \( 3.75 \times 10^{-3} - 1.80 \times 10^{-3} = 1.95 \times 10^{-3} \text{ mol} \)
Moles of \( X^- \) formed = \( 1.80 \times 10^{-3} \text{ mol} \)
Using the buffer equation:
\( [H^+] = K_a \times \frac{n(\text{HX})}{n(\text{X}^-)} = 1.80 \times 10^{-5} \times \frac{1.95 \times 10^{-3}}{1.80 \times 10^{-3}} = 1.95 \times 10^{-5} \text{ mol dm}^{-3} \)
\( \text{pH} = -\log_{10}(1.95 \times 10^{-5}) = 4.71 \)

(c) Added \( \text{H}^+ \) ions react with the conjugate base \( \text{X}^- \) in the buffer:
\( \text{X}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{HX}(\text{aq}) \)

(a) [Total: 3 marks]
- 1 mark for calculating \( [H^+] = 1.64 \times 10^{-3} \text{ mol dm}^{-3} \)
- 1 mark for correct pH expression \( \text{pH} = -\log_{10}[H^+] \)
- 1 mark for final pH of 2.78 (must be 2 d.p.)

(b) [Total: 5.5 marks]
- 1 mark for calculating moles of HX = \( 3.75 \times 10^{-3} \) AND moles of NaOH = \( 1.80 \times 10^{-3} \)
- 1 mark for calculating remaining moles of HX = \( 1.95 \times 10^{-3} \)
- 1 mark for identifying moles of \( X^- \) = \( 1.80 \times 10^{-3} \)
- 1 mark for substituting values correctly into the \( [H^+] \) expression
- 1 mark for calculating \( [H^+] = 1.95 \times 10^{-5} \text{ mol dm}^{-3} \)
- 0.5 marks for final pH of 4.71

(c) [Total: 2 marks]
- 2 marks for \( \text{X}^- + \text{H}^+ \rightarrow \text{HX} \) (accept state symbols omitted; award 1 mark if NaX is used: \( \text{NaX} + \text{HCl} \rightarrow \text{HX} + \text{NaCl} \))

(a) When dilute aqueous ammonia is added to \( [Fe(H_2O)_6]^{2+} \), a green precipitate is formed. This precipitate is insoluble in excess ammonia. The equation is:
\( [Fe(H_2O)_6]^{2+}(\text{aq}) + 2\text{NH}_3(\text{aq}) \rightarrow \text{Fe}(H_2O)_4(\text{OH})_2(\text{s}) + 2\text{NH}_4^+(\text{aq}) \)

(b) Upon standing in air, the green precipitate on the surface turns brown / red-brown due to oxidation by oxygen in the air. The equation is:
\( 4\text{Fe}(\text{OH})_2(\text{s}) + \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{Fe}(\text{OH})_3(\text{s}) \)
(Accept: \( 4\text{Fe}(H_2O)_4(\text{OH})_2 + \text{O}_2 \rightarrow 4\text{Fe}(H_2O)_3(\text{OH})_3 + 2\text{H}_2\text{O} \))

(c) The overall ionic equation for the titration is:
\( 5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \)
Moles of \( \text{Fe}^{2+} = \frac{25.0}{1000} \times 0.0800 = 2.00 \times 10^{-3} \text{ mol} \)
From the 5:1 stoichiometry, moles of \( \text{MnO}_4^- \) required = \( \frac{2.00 \times 10^{-3}}{5} = 4.00 \times 10^{-4} \text{ mol} \)
Volume of \( \text{KMnO}_4 = \frac{\text{moles}}{\text{concentration}} = \frac{4.00 \times 10^{-4}}{0.0200} = 0.0200 \text{ dm}^3 = 20.0 \text{ cm}^3 \)

(a) [Total: 3.5 marks]
- 1 mark for green precipitate
- 1 mark for stating it is insoluble in excess ammonia
- 1.5 marks for correct equation (deduct 0.5 marks if state symbols are missing or incorrect)

(b) [Total: 3 marks]
- 1 mark for precipitate turning brown / red-brown
- 1 mark for explaining it is due to oxidation by air/oxygen
- 1 mark for correct balanced equation

(c) [Total: 4 marks]
- 1 mark for correct overall titration equation
- 1 mark for moles of \( \text{Fe}^{2+} = 2.00 \times 10^{-3} \text{ mol} \)
- 1 mark for moles of \( \text{MnO}_4^- = 4.00 \times 10^{-4} \text{ mol} \)
- 1 mark for correct volume of 20.0 \( \text{cm}^3 \)

(a) The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (100 kPa and 298 K).

(b) The lattice enthalpy of dissociation is the enthalpy change when 1 mole of an ionic crystal lattice is completely separated into its constituent gaseous ions under standard conditions.

(c) We construct a Born-Haber cycle for \( \text{CaCl}_2(\text{s}) \):
\( \Delta_f H^\ominus(\text{CaCl}_2) = \Delta_{at} H^\ominus(\text{Ca}) + \text{1st IE}(\text{Ca}) + \text{2nd IE}(\text{Ca}) + \Delta_{diss} H^\ominus(\text{Cl}_2) + 2 \times \text{EA}(\text{Cl}) - \Delta_{latt}H^\ominus(\text{diss}) \)
Substituting the values:
\( -796 = +178 + 590 + 1145 + 242 + 2(-349) - \Delta_{latt}H^\ominus(\text{diss}) \)
\( -796 = 2155 - 698 - \Delta_{latt}H^\ominus(\text{diss}) \)
\( -796 = 1457 - \Delta_{latt}H^\ominus(\text{diss}) \)
\( \Delta_{latt}H^\ominus(\text{diss}) = 1457 + 796 = +2253 \text{ kJ mol}^{-1} \)

(a) [Total: 2 marks]
- 1 mark for "enthalpy change when 1 mole of compound is formed from its elements"
- 1 mark for specifying "in their standard states" and "under standard conditions"

(b) [Total: 2 marks]
- 1 mark for "enthalpy change when 1 mole of an ionic compound/solid is converted"
- 1 mark for "into gaseous ions"

(c) [Total: 6.5 marks]
- 1 mark for correctly using 2 x Electron affinity of chlorine \( (2 \times -349 = -698) \)
- 1 mark for using the correct atomisation value of chlorine (using the bond dissociation enthalpy of \( \text{Cl}_2 \) directly as it produces 2 moles of \( \text{Cl}(\text{g}) \))
- 2.5 marks for setting up a correct Born-Haber calculation expression / cycle layout
- 2 marks for final correct value of \( +2253 \text{ kJ mol}^{-1} \) (award 1 mark if magnitude 2253 is correct but sign is negative; award 1 mark for +2132 if only 1 x EA was used)

(a) Rearranging the rate equation for \( k \):
\( k = \frac{\text{rate}}{[\text{A}]^2[\text{B}]} = \frac{3.46 \times 10^{-4}}{(0.120)^2 \times (0.080)} = \frac{3.46 \times 10^{-4}}{0.0144 \times 0.080} = \frac{3.46 \times 10^{-4}}{1.152 \times 10^{-3}} = 0.3003 \approx 0.300 \text{ dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \)

(b) Since the rate equation is \( \text{rate} = k[\text{A}]^2[\text{B}] \):
- Doubling [A] increases the rate by a factor of \( 2^2 = 4 \).
- Halving [B] decreases the rate by a factor of \( 0.5 \).
- Overall rate change = \( 4 \times 0.5 = 2 \) (the rate is doubled).

(c) Conduct the reaction at a series of different temperatures to find \( k \) at each temperature \( T \) (in Kelvin).
Plot a graph of \( \ln k \) on the y-axis against \( 1/T \) on the x-axis.
This gives a straight line with a negative gradient.
Calculate the gradient of the line: \( \text{gradient} = -\frac{E_a}{R} \).
Therefore, the activation energy is calculated as: \( E_a = -\text{gradient} \times R \).

(a) [Total: 4 marks]
- 1 mark for correct rearrangement: \( k = \frac{\text{rate}}{[\text{A}]^2[\text{B}]} \)
- 1 mark for correct value 0.300 (or 0.30)
- 2 marks for correct units: \( \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \) (award 1 mark if almost correct, e.g. \( \text{dm}^6 \text{ mol}^{-2} \text{ s} \))

(b) [Total: 2.5 marks]
- 1 mark for stating rate increases by factor of 4 due to doubling [A]
- 1 mark for stating rate decreases by half due to halving [B]
- 0.5 marks for concluding that the overall rate doubles (increases by a factor of 2)

(c) [Total: 4 marks]
- 1 mark for stating to measure \( k \) at different temperatures (minimum 5) and converting \( T \) to Kelvin
- 1 mark for plotting \( \ln k \) against \( 1/T \)
- 1 mark for identifying that the gradient is equal to \( -E_a / R \)
- 1 mark for showing \( E_a = -\text{gradient} \times R \) (or equivalent explanation)

(a) Let's determine the equilibrium moles using an ICE table:
Initial moles:
\( n(\text{N}_2) = 1.50 \text{ mol} \)
\( n(\text{H}_2) = 3.50 \text{ mol} \)
\( n(\text{NH}_3) = 0.00 \text{ mol} \)

Change in moles:
Since 0.60 mol of \( \text{NH}_3 \) is formed, the change in moles of \( \text{NH}_3 \) is \( +0.60 \text{ mol} \).
By stoichiometry:
Change in \( n(\text{N}_2) = -\frac{0.60}{2} = -0.30 \text{ mol} \)
Change in \( n(\text{H}_2) = -3 \times 0.30 = -0.90 \text{ mol} \)

Equilibrium moles:
\( n(\text{N}_2) = 1.50 - 0.30 = 1.20 \text{ mol} \)
\( n(\text{H}_2) = 3.50 - 0.90 = 2.60 \text{ mol} \)
\( n(\text{NH}_3) = 0.60 \text{ mol} \)

Equilibrium concentrations (divide by \( V = 5.00 \text{ dm}^3 \)):
\( [\text{N}_2] = \frac{1.20}{5.00} = 0.240 \text{ mol dm}^{-3} \)
\( [\text{H}_2] = \frac{2.60}{5.00} = 0.520 \text{ mol dm}^{-3} \)
\( [\text{NH}_3] = \frac{0.60}{5.00} = 0.120 \text{ mol dm}^{-3} \)

Expression for \( K_c \):
\( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \)

Calculation:
\( K_c = \frac{(0.120)^2}{(0.240)(0.520)^3} = \frac{0.0144}{0.240 \times 0.140608} = \frac{0.0144}{0.033746} = 0.4267 \approx 0.427 \)

Units:
\( \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \text{mol}^{-2} \text{ dm}^6 \)

(b) Since \( K_c \) decreases as temperature increases, the equilibrium concentration of products decreases and the equilibrium concentration of reactants increases. This means the equilibrium has shifted to the left (in the reverse direction).
According to Le Chatelier's principle, an increase in temperature shifts the equilibrium in the endothermic direction to absorb the added heat. Since the equilibrium shifted to the left, the reverse reaction must be endothermic. Therefore, the forward reaction is exothermic.

(a) [Total: 6.5 marks]
- 2.5 marks for calculating equilibrium moles: \( N_2 = 1.20 \text{ mol} \) (1 mark), \( H_2 = 2.60 \text{ mol} \) (1 mark), \( NH_3 = 0.60 \text{ mol} \) (0.5 marks)
- 1 mark for dividing by volume to obtain equilibrium concentrations: \( [N_2] = 0.240, [H_2] = 0.520, [NH_3] = 0.120 \)
- 1 mark for correct expression of \( K_c \)
- 1 mark for correct calculation of \( K_c = 0.427 \) (accept range 0.426 to 0.43)
- 1 mark for correct units of \( \text{dm}^6 \text{ mol}^{-2} \) (or \( \text{mol}^{-2} \text{ dm}^6 \))

(b) [Total: 4 marks]
- 1 mark for identifying that a decrease in \( K_c \) indicates the equilibrium has shifted to the left (favouring reactants)
- 1 mark for stating that increasing temperature shifts the equilibrium in the endothermic direction
- 1 mark for connecting the shift to the left to show the reverse reaction is endothermic
- 1 mark for concluding that the forward reaction is exothermic

(a)(i) The half-cell with the lower electrode potential (more negative/less positive) undergoes oxidation and is placed on the left: \( 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \).
The conventional representation is:
\( \text{Pt(s)} \mid \text{I}^-(\text{aq}), \text{I}_2(\text{aq}) \parallel \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) \mid \text{Pt(s)} \)
(Accept comma or phase boundary between ions in same phase, but Pt(s) must be on the outer boundaries).

(a)(ii) \( E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77 \text{ V} - 0.54 \text{ V} = +0.23 \text{ V} \).

(a)(iii) Combining the half-equations:
\( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \)

(b) Feasibility:
\( E^\ominus(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}) \) (\( +2.01 \text{ V} \)) is more positive than \( E^\ominus(\text{I}_2/\text{I}^-) \) (\( +0.54 \text{ V} \)), giving \( E^\ominus_{\text{cell}} = +1.47 \text{ V} \). Since \( E^\ominus_{\text{cell}} > 0 \), the reaction is feasible.
Why it is slow:
Both reactant ions (\( \text{S}_2\text{O}_8^{2-} \) and \( \text{I}^- \)) are negatively charged. They repel each other, which leads to a very high activation energy.
Homogeneous catalysis by \( \text{Fe}^{2+} \):
Step 1: \( \text{Fe}^{2+} \) reduces peroxodisulfate ions:
\( 2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq}) \)
Step 2: The generated \( \text{Fe}^{3+} \) ions then oxidize the iodide ions, regenerating the catalyst:
\( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \)
Each of these individual steps involves reactions between oppositely charged ions (positive iron ions and negative reactant ions), which has a lower activation energy and hence proceeds much faster.

(a)(i) [Total: 3 marks]
- 1 mark for Pt(s) on both outer terminals
- 1 mark for correct phase boundaries (vertical lines) and salt bridge (double vertical line)
- 1 mark for correct left-to-right order (oxidation half-cell on left, reduction on right)

(a)(ii) [Total: 1 mark]
- 1 mark for +0.23 V (unit required)

(a)(iii) [Total: 1.5 marks]
- 1.5 marks for correct balanced equation (allow 1 mark for correct reactants and products but unbalanced)

(b) [Total: 5 marks]
- 1 mark for explaining feasibility using electrode potentials (e.g. \( E^\ominus_{\text{cell}} = +1.47 \text{ V} \) which is positive)
- 1 mark for explaining high activation energy due to repulsion between two negatively charged reactant ions
- 1 mark for Equation 1: \( 2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-} \)
- 1 mark for Equation 2: \( 2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2 \)
- 1 mark for explaining that both steps involve oppositely charged species, avoiding repulsion and lowering activation energy

(a) Silicon dioxide (\( \text{SiO}_2 \)) has a giant covalent macromolecular structure. It contains many strong covalent bonds between silicon and oxygen atoms throughout the structure, which require a large amount of thermal energy to break, resulting in a very high melting point.
Sulfur trioxide (\( \text{SO}_3 \)) has a simple molecular structure. There are only weak intermolecular forces (van der Waals forces and dipole-dipole attractions) between individual molecules, which require very little energy to overcome, resulting in a much lower melting point.

(b)(i) Reaction: \( \text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq}) \)
pH of the resulting phosphoric acid solution is approximately 0 to 2.

(b)(ii) Reaction: \( \text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq}) \)
pH of the resulting sodium hydroxide solution is approximately 12 to 14.

(c) Aluminium oxide is amphoteric and reacts with hot concentrated/excess sodium hydroxide to form a soluble aluminate salt:
\( \text{Al}_2\text{O}_3(\text{s}) + 2\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4](\text{aq}) \)
(Accept ionic equation: \( \text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2[\text{Al}(\text{OH})_4]^-(\text{aq}) \))

(a) [Total: 4.5 marks]
- 1 mark for identifying silicon dioxide has a giant covalent structure
- 1 mark for explaining that melting silicon dioxide requires breaking strong covalent bonds
- 1 mark for identifying sulfur trioxide has a simple molecular structure
- 1 mark for explaining that melting sulfur trioxide only requires overcoming weak intermolecular forces
- 0.5 marks for concluding that more energy is needed to break the bonds in silicon dioxide than to overcome the intermolecular forces in sulfur trioxide

(b)(i) [Total: 2.5 marks]
- 1.5 marks for correct balanced equation
- 1 mark for pH in the range 0 to 2

(b)(ii) [Total: 2.5 marks]
- 1.5 marks for correct balanced equation
- 1 mark for pH in the range 12 to 14

(c) [Total: 1 mark]
- 1 mark for correct balanced equation (either molecular or ionic; ignore state symbols)

(a)(i) \( \text{Cl}_2(\text{g}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{NaClO}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \)

(a)(ii) In \( \text{NaCl} \), the oxidation state of chlorine is \( -1 \).
In \( \text{NaClO} \), the oxidation state of chlorine is \( +1 \).

(a)(iii) This is a disproportionation reaction (chlorine is both oxidized and reduced simultaneously).

(b) Electronegativity decreases down Group 7 from fluorine to iodine because:
- The atomic radius increases as there are more electron shells.
- There is more shielding of the positive nuclear charge by inner shells.
- Consequently, there is a weaker electrostatic attraction between the positive nucleus and the shared pair of electrons in a covalent bond.

(c)(i) \( \text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl}(\text{g}) \)
(Accept: \( 2\text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{Na}_2\text{SO}_4(\text{s}) + 2\text{HCl}(\text{g}) \))

(c)(ii) The dark grey solid is iodine (\( \text{I}_2 \)).
The gas with the smell of bad eggs is hydrogen sulfide (\( \text{H}_2\text{S} \)).
The half-equation for the reduction of sulfuric acid to hydrogen sulfide is:
\( \text{H}_2\text{SO}_4 + 8\text{H}^+ + 8e^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O} \)
(Accept: \( \text{SO}_4^{2-} + 10\text{H}^+ + 8e^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O} \))

(a)(i) [Total: 1.5 marks]
- 1.5 marks for correct balanced equation (1 mark if correct formulas but unbalanced)

(a)(ii) [Total: 2 marks]
- 1 mark for \( -1 \) (for \( \text{NaCl} \))
- 1 mark for \( +1 \) (for \( \text{NaClO} \))

(a)(iii) [Total: 1 mark]
- 1 mark for "disproportionation"

(b) [Total: 3 marks]
- 1 mark for stating electronegativity decreases down the group
- 1 mark for mentioning increased atomic radius / more shells OR increased shielding
- 1 mark for weaker attraction between the nucleus and the bonding/shared pair of electrons

(c)(i) [Total: 1.5 marks]
- 1.5 marks for correct balanced equation (allow 1 mark for correct reactants and products but unbalanced)

(c)(ii) [Total: 1.5 marks]
- 0.5 marks for identifying dark grey solid as iodine (\( \text{I}_2 \))
- 0.5 marks for identifying gas as hydrogen sulfide (\( \text{H}_2\text{S} \))
- 0.5 marks for correct balanced reduction half-equation

Part (a): Moles of propanoic acid = \(0.0500 \times 0.250 = 0.0125\text{ mol}\). Moles of \(NaOH\) = \(0.0250 \times 0.300 = 0.00750\text{ mol}\). The \(NaOH\) reacts with propanoic acid to produce sodium propanoate and water: \(CH_3CH_2COOH + NaOH \rightarrow CH_3CH_2COONa + H_2O\). Moles of propanoate ions formed = \(0.00750\text{ mol}\). Moles of remaining propanoic acid = \(0.0125 - 0.00750 = 0.00500\text{ mol}\). Using \(K_a = \frac{[H^+][A^-]}{[HA]}\), we get \([H^+] = K_a \times \frac{[HA]}{[A^-]} = K_a \times \frac{n(HA)}{n(A^-)} = 1.35 \times 10^{-5} \times \frac{0.00500}{0.00750} = 9.00 \times 10^{-6}\text{ mol dm}^{-3}\). \(pH = -\log_{10}(9.00 \times 10^{-6}) = 5.05\). Part (b): Ionic equation: \(CH_3CH_2COO^- + H^+ \rightarrow CH_3CH_2COOH\). When a small amount of \(H^+\) is added, it reacts with the reservoir of propanoate ions (conjugate base) to form propanoic acid. The ratio of propanoic acid to propanoate ions remains almost constant, so the pH changes very little. Part (c): Halfway through transition, \([HIn] = [In^-]\), so \(K_{in} = [H^+]\) and \(pH = pK_{in}\). \(pH = -\log_{10}(7.94 \times 10^{-5}) = 4.10\).

Part (a) [5.5 marks total]: M1: Moles of \(CH_3CH_2COOH\) = \(0.0125\text{ mol}\) (1 mark). M2: Moles of \(NaOH\) = \(0.00750\text{ mol}\) (1 mark). M3: Moles of \(CH_3CH_2COO^-\ formado = 0.00750\text{ mol}\) and excess \(CH_3CH_2COOH = 0.00500\text{ mol}\) (1 mark). M4: Rearranging \(K_a\) to find \([H^+]\) (1 mark). M5: \([H^+] = 9.00 \times 10^{-6}\text{ mol dm}^{-3}\) (1 mark). M5.5: \(pH = 5.05\) (2 d.p. only) (0.5 mark). Part (b) [3 marks total]: M1: \(CH_3CH_2COO^- + H^+ \rightarrow CH_3CH_2COOH\) (1 mark). M2: Added \(H^+\) reacts with the large reservoir of conjugate base / propanoate ions (1 mark). M3: Ratio of \([HA]/[A^-]\) (or concentrations of acid and conjugate base) remains almost constant (1 mark). Part (c) [2 marks total]: M1: State that \([HIn] = [In^-]\) hence \([H^+] = K_{in}\) (1 mark). M2: \(pH = 4.10\) (1 mark).

Part (a): Test Tube 1: Reaction of \(Fe^{2+}\) with carbonate forms iron(II) carbonate precipitate. Equation: \(Fe^{2+}(aq) + CO_3^{2-}(aq) \rightarrow FeCO_3(s)\). Observation: Green precipitate. Test Tube 2: \(Fe^{3+}\) is strongly acidic and reacts with carbonate to form iron(III) hydroxide precipitate and carbon dioxide gas. Equation: \(2[Fe(H_2O)_6]^{3+}(aq) + 3CO_3^{2-}(aq) \rightarrow 2Fe(H_2O)_3(OH)_3(s) + 3CO_2(g) + 3H_2O(l)\). Observation: Brown precipitate and bubbles/effervescence. Part (b): \(Al^{3+}\) has a higher charge and smaller ionic size than \(Co^{2+}\), giving it a much higher charge density (charge-to-size ratio). The high charge density of \(Al^{3+}\) polarises the O-H bonds in the water ligands more strongly, weakening the O-H bond and allowing a proton (\(H^+\)) to be released more readily into the solution. Part (c): The reaction with excess ammonia is a ligand substitution: \([Cu(H_2O)_6]^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 4H_2O(l)\). The product complex has 4 ammonia ligands and 2 water ligands, giving a total coordination number of 6.

Part (a) [4.5 marks total]: Test Tube 1: M1: \(Fe^{2+}(aq) + CO_3^{2-}(aq) \rightarrow FeCO_3(s)\) (1 mark). M2: Observation: green precipitate / green solid (0.5 mark). Test Tube 2: M3: \(2[Fe(H_2O)_6]^{3+} + 3CO_3^{2-} \rightarrow 2Fe(H_2O)_3(OH)_3 + 3CO_2 + 3H_2O\) (1.5 marks: 1 mark for correct formulas of products, 0.5 mark for correct balancing). M4: Observation: brown/red-brown precipitate AND bubbles/effervescence (1.5 marks: 1 mark for brown precipitate, 0.5 mark for bubbles/effervescence). Part (b) [3 marks total]: M1: \(Al^{3+}\) has higher charge and smaller size / higher charge density than \(Co^{2+}\) (1 mark). M2: \(Al^{3+}\) polarises the O-H bonds in water ligands more strongly (1 mark). M3: This weakens the O-H bond, releasing \(H^+\) ions more easily (1 mark). Part (c) [3 marks total]: M1: Correct formulas for reactants and products in \([Cu(H_2O)_6]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O\) (1 mark). M2: Balanced equation (1 mark). M3: Coordination number is 6 (1 mark).

For (a): The equation is CH3CH2CH2CH2Br + KCN -> CH3CH2CH2CH2CN + KBr. The IUPAC name is pentanenitrile. The mechanism is nucleophilic substitution. For (b): The mechanism involves the nucleophilic attack of the lone pair on the nitrogen atom of pentylamine onto the delta-positive carbon atom of chloromethane, with the simultaneous loss of a chloride ion. This forms an intermediate alkylammonium salt, [CH3(CH2)4NH2CH3]+Cl-. A second amine molecule (or a chloride ion) then removes a proton from the intermediate to yield the neutral secondary amine, N-methylpentylamine. For (c): The product, N-methylpentylamine, is a secondary amine and still contains a nucleophilic lone pair on the nitrogen atom. It can react further with remaining chloromethane to form a tertiary amine and ultimately a quaternary ammonium salt. To maximise the yield of the secondary amine, a large excess of pentylamine relative to chloromethane is used. For (d): Moles of 1-bromobutane = 25.0 / 137.0 = 0.1825 mol. Theoretical moles of pentylamine = 0.1825 mol. Actual moles of pentylamine = 0.1825 * 0.685 = 0.1250 mol. Mass of pentylamine = 0.1250 * 87.0 = 10.87 g, which rounds to 10.9 g.

(a) Equation: CH3CH2CH2CH2Br + CN- -> CH3CH2CH2CH2CN + Br- (1 mark). Name: Pentanenitrile (1 mark). Mechanism: Nucleophilic substitution (1 mark). (b) Curly arrow from N lone pair of pentylamine to C of chloromethane (1 mark). Curly arrow from C-Cl bond to Cl (1 mark). Correct intermediate structure [CH3(CH2)4NH2CH3]+ (1 mark). Curly arrow showing loss of proton from intermediate (1 mark). (c) Secondary amine has lone pair and reacts further with chloromethane (2 marks). Excess pentylamine used to maximise yield (1 mark). (d) Moles of 1-bromobutane = 0.1825 mol (1 mark). Mass of pentylamine = 10.9 g (0.6 marks).

For (a): Moles of propanoic acid (HA) = 0.150 dm3 * 0.220 mol dm-3 = 0.0330 mol. Moles of propanoate ions (A-) = 0.100 dm3 * 0.150 mol dm-3 = 0.0150 mol. Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) or [H+] = Ka * [HA]/[A-]. [H+] = 1.35 x 10^-5 * (0.0330 / 0.0150) = 2.97 x 10^-5 mol dm-3. pH = -log10(2.97 x 10^-5) = 4.53. For (b): Moles of OH- added = 0.00500 dm3 * 0.400 mol dm-3 = 0.00200 mol. The added OH- reacts with the weak acid: HA + OH- -> A- + H2O. New moles of HA = 0.0330 - 0.00200 = 0.0310 mol. New moles of A- = 0.0150 + 0.00200 = 0.0170 mol. New [H+] = 1.35 x 10^-5 * (0.0310 / 0.0170) = 2.462 x 10^-5 mol dm-3. New pH = -log10(2.462 x 10^-5) = 4.61. For (c): When HCl is added, H+ ions react with propanoate ions: C2H5COO-(aq) + H+(aq) -> C2H5COOH(aq). The concentration ratio of the acid to conjugate base changes very little, hence the pH remains almost constant.

(a) Moles of acid and conjugate base calculated (1 mark). Correct [H+] expression and calculation (1 mark). Initial pH = 4.53 (1 mark). (b) Moles of OH- added = 0.00200 mol (1 mark). New moles of HA = 0.0310 and A- = 0.0170 mol (2 marks). New pH = 4.61 (3 marks). (c) Equation: C2H5COO- + H+ -> C2H5COOH (1 mark). Explanation: propanoate ions remove added H+ ions, maintaining a stable ratio of HA/A- (1.6 marks).

For (a): Comparing Exp 1 and Exp 2: [I-] is constant, [S2O8^2-] doubles, and the rate doubles (1.25 x 10^-4 to 2.50 x 10^-4). Thus, the order with respect to [S2O8^2-] is 1. Comparing Exp 2 and Exp 3: [S2O8^2-] is constant, [I-] triples, and the rate triples (2.50 x 10^-4 to 7.50 x 10^-4). Thus, the order with respect to [I-] is 1. For (b): The rate equation is Rate = k[S2O8^2-][I-]. k = Rate / ([S2O8^2-][I-]) = 1.25 x 10^-4 / (0.0400 * 0.0200) = 0.15625, which rounds to 0.156 dm3 mol-1 s-1. For (c): Arrhenius equation: ln(k2 / k1) = -Ea / R * (1/T2 - 1/T1). Here, k1 = 0.156 at 298 K and k2 = 0.525 at 318 K. ln(0.525 / 0.15625) = 1.2119. (1/T1 - 1/T2) = 1/298 - 1/318 = 2.111 x 10^-4 K^-1. Ea = ln(3.36) * R / (2.111 x 10^-4) = 1.2119 * 8.314 / 2.111 x 10^-4 = 47736 J mol-1 = 47.7 kJ mol-1.

(a) First-order wrt S2O8^2- with justification (1 mark). First-order wrt I- with justification (1 mark). Deductions correct (1 mark). (b) Rate = k[S2O8^2-][I-] (1 mark). k = 0.156 (1 mark). Units: dm3 mol-1 s-1 (1 mark). (c) Uses ln(k2/k1) = -Ea/R * (1/T2 - 1/T1) (1 mark). Calculates ln(k2/k1) correctly (1 mark). Calculates (1/T1 - 1/T2) correctly (1 mark). Rearranges expression for Ea (1 mark). Correct final value of Ea in kJ mol-1 = 47.5 - 48.0 kJ mol-1 (1.6 marks).

For (a): The absorption at 1740 cm-1 is caused by a C=O bond. Carboxylic acids have a broad O-H absorption between 2500-3000 cm-1; the absence of this peak confirms X is not a carboxylic acid. For (b): The 0.95 ppm triplet (3H) indicates a methyl group adjacent to a CH2 group (-CH2-CH3). The 1.65 ppm multiplet (2H) indicates a CH2 group adjacent to 5 neighboring protons, i.e., sandwiched in -CH2-CH2-CH3. The 2.01 ppm singlet (3H) indicates a CH3 group with no neighboring protons, placed next to the carbonyl carbon (CH3-CO-). The 4.05 ppm triplet (2H) is downfield, meaning a CH2 group adjacent to an oxygen atom (-O-CH2-CH2-). For (c): Combining these fragments gives the structure CH3COOCH2CH2CH3, which is propyl ethanoate. For (d): There are 5 distinct carbon environments in propyl ethanoate, resulting in 5 peaks in the 13C NMR spectrum.

(a) C=O bond identified (1 mark). No O-H absorption confirms it is not an acid (1 mark). (b) 1H NMR analysis: Triplet at 0.95 ppm is CH3-CH2 (1.5 marks). Multiplet at 1.65 ppm is CH2-CH2-CH3 (1.5 marks). Singlet at 2.01 ppm is CH3-CO (1.5 marks). Triplet at 4.05 ppm is -O-CH2-CH2- (1.5 marks). (c) Correct structural formula (1 mark). Correct name: Propyl ethanoate (1 mark). (d) 5 peaks in 13C NMR (1.6 marks).

For (a): The mechanism is nucleophilic addition. The cyanide ion acts as a nucleophile. A curly arrow is drawn from the lone pair on the carbon of the CN- ion to the delta-positive carbonyl carbon of butanal. Simultaneously, a curly arrow is drawn from the C=O double bond to the oxygen atom. This forms an intermediate with a negative charge on the oxygen: CH3CH2CH2CH(O-)CN. Finally, a curly arrow from the lone pair on the intermediate's oxygen to a hydrogen ion (H+) produces the hydroxynitrile, 2-hydroxypentanenitrile. For (b): The carbonyl carbon in butanal is planar. The CN- nucleophile can attack from either side of this plane with equal probability. This leads to the formation of equal amounts (a 50:50 mixture) of the two enantiomers (a racemic mixture). The optical rotations of the enantiomers cancel each other out, making the mixture optically inactive. For (c): Nucleophilic addition. For (d): HCN gas is highly toxic and volatile. Using KCN with dilute acid generates HCN in situ, minimizing exposure risks and avoiding storage of a lethal gas. Additionally, KCN provides a higher concentration of the CN- nucleophile because it is fully dissociated, unlike weak acid HCN.

(a) Curly arrow from CN- lone pair to carbonyl C (1 mark). Curly arrow from C=O bond to O (1 mark). Correct intermediate structure (1 mark). Curly arrow from intermediate O- lone pair to H+ (1 mark). (b) Carbonyl carbon is planar (1 mark). Equal probability of attack from above/below (1 mark). Leads to 50:50 mixture of enantiomers (1 mark). Optical activities cancel (1 mark). (c) Nucleophilic addition (1 mark). (d) HCN gas is highly toxic (1 mark). In situ generation is safer and provides higher nucleophile concentration (1.6 marks).

For (a): Lactic acid has both an alcohol and a carboxylic acid group. Condensation polymerisation occurs through esterification, yielding the repeating unit: -[O-CH(CH3)-CO]-. For (b): PLA contains ester links (-CO-O-) which are polar and susceptible to hydrolysis by water or action of microorganisms. Poly(ethene) contains only non-polar C-C and C-H bonds in its backbone, which are highly stable and resistant to chemical attack. For (c): (i) Benzene-1,4-dicarboxylic acid is HOOC-C6H4-COOH. Benzene-1,4-diamine is H2N-C6H4-NH2. (ii) The repeating unit of Kevlar is -[CO-C6H4-CONH-C6H4-NH]-. For (d): Kevlar is a polyamide. It is exceptionally strong due to extensive hydrogen bonding between polymer chains, forming between the oxygen of the carbonyl groups (C=O) and the hydrogen of the amide groups (N-H). The flat benzene rings also pack tightly together, maximising these intermolecular forces.

(a) Correct ester link (1 mark). Correct repeating unit of PLA (1 mark). (b) PLA has polar ester links (1 mark). Hydrolysed by microorganisms/water (1 mark). Poly(ethene) has inert, non-polar C-C bonds (1 mark). (c) (i) Correct structure of dicarboxylic acid (1 mark). Correct structure of diamine (1 mark). (ii) Correct amide link shown (1 mark). Full repeating unit of Kevlar (1 mark). (d) Polyamide (1 mark). Hydrogen bonding between chains explained (1.6 marks).

For (a): The reagents are concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4) maintained at a temperature below 15 degrees Celsius. For (b): HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-. For (c): Electrophilic substitution mechanism: An arrow is drawn from the benzene ring (specifically at C-3 relative to the ester group) to the NO2+ electrophile. This forms a resonance-stabilised intermediate containing a partial ring with a positive charge (horseshoe facing C-3) and both H and NO2 attached to C-3. A curly arrow is then drawn from the C-H bond back into the ring, restoring the aromatic system and releasing an H+ ion. For (d): The ester substituent (-COOCH3) is an electron-withdrawing group. It deactivates the ortho (2) and para (4) positions more than the meta (3) position by destabilising adjacent positive charges in the resonance structures, directing the electrophile to the 3-position. For (e): Moles of methyl benzoate = 4.08 / 136.0 = 0.0300 mol. Theoretical mass of methyl 3-nitrobenzoate = 0.0300 mol * 181.0 g mol-1 = 5.43 g. Percentage yield = (3.90 / 5.43) * 100 = 71.8%.

(a) Concentrated HNO3 and concentrated H2SO4 (1 mark). Low temperature (< 15 degrees) (1 mark). (b) Correct equation (1 mark). (c) Arrow from ring to NO2+ (1 mark). Correct intermediate structure (1 mark). Arrow from C-H bond back to ring (1 mark). Restoration of ring shown (1 mark). (d) Ester group is electron-withdrawing, meta-directing (2 marks). (e) Moles of reactant = 0.0300 mol (1 mark). Theoretical mass = 5.43 g (1 mark). Yield = 71.8% (0.6 marks).

For (a): Moles of NaOH = 0.120 mol dm-3 * 0.02500 dm3 = 0.00300 mol. For (b): Reaction stoichiometry is 1 acid : 2 NaOH. Moles of acid in 25.0 cm3 = 0.00300 / 2 = 0.00150 mol. Moles of acid in 250.0 cm3 = 0.00150 * 10 = 0.0150 mol. For (c): Mass of acid dissolved = 1.89 g. Molar mass (Mr) of hydrated acid = 1.89 g / 0.0150 mol = 126.0 g mol-1. Molar mass of anhydrous acid (H2C2O4) = 2.0 + 24.0 + 64.0 = 90.0 g mol-1. Mass of water of crystallisation = 126.0 - 90.0 = 36.0 g mol-1. Since Mr of H2O = 18.0, x = 36.0 / 18.0 = 2. For (d): The titration uses a weak acid and a strong base, so phenolphthalein is the most suitable indicator. Its colour change at the end point is from colourless to pink.

(a) Moles of NaOH = 0.00300 mol (2 marks). (b) Moles of acid in 25.0 cm3 = 0.00150 mol (1 mark). Moles of acid in 250.0 cm3 = 0.0150 mol (2 marks). (c) Molar mass of hydrated acid = 126.0 g mol-1 (2 marks). Molar mass of anhydrous acid = 90.0 g mol-1 (1 mark). Difference = 36.0 (1 mark). x = 2 (1 mark). (d) Phenolphthalein (1 mark). Colourless to pink (0.6 marks).

### Part (a)
- Comparing Experiments 1 and 2: The concentration of \([\text{OH}^-]\) is kept constant. The concentration of \([\text{C}_4\text{H}_9\text{Br}]\) is doubled (from \(0.150\) to \(0.300\ \text{mol dm}^{-3}\)), and the initial rate doubles (from \(1.80 \times 10^{-4}\) to \(3.60 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}\)). Therefore, the order of reaction with respect to \(\text{C}_4\text{H}_9\text{Br}\) is **1**.
- Comparing Experiments 1 and 3: The concentration of \([\text{C}_4\text{H}_9\text{Br}]\) is kept constant. The concentration of \([\text{OH}^-]\) is doubled (from \(0.100\) to \(0.200\ \text{mol dm}^{-3}\)), but the initial rate remains unchanged (\(1.80 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}\)). Therefore, the order of reaction with respect to \(\text{OH}^-\cle\) is **0**.

### Part (b)
- **Rate Equation:** \(\text{Rate} = k[\text{C}_4\text{H}_9\text{Br}]\)
- **Calculation of \(k\):**
Using data from Experiment 1:
\[1.80 \times 10^{-4} = k \times 0.150\]
\[k = \frac{1.80 \times 10^{-4}}{0.150} = 1.20 \times 10^{-3}\]
- **Units:**
\[\text{units of } k = \frac{\text{mol dm}^{-3}\ \text{s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}\]

### Part (c)
Since the rate-determining step depends only on the concentration of the halogenoalkane, this reaction proceeds via an \(\text{S}_\text{N}1\) mechanism. Isomer **Y** is the tertiary halogenoalkane, 2-bromo-2-methylpropane.
- **Step 1:** The heterolytic fission of the \(\text{C}-\text{Br}\) bond (slow step).
- Draw the structure of 2-bromo-2-methylpropane, showing the polar \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) bond.
- Draw a curly arrow originating from the \(\text{C}-\text{Br}\) single bond to the bromine atom.
- Show the formation of a planar tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\).
- **Step 2:** Nucleophilic attack (fast step).
- Draw the hydroxide ion, \(:\text{OH}^-\), with a lone pair on the oxygen atom and a negative charge.
- Draw a curly arrow starting from the lone pair on the oxygen atom of the hydroxide ion to the positively charged carbon atom of the carbocation.
- Show the formation of the product, 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).

### Part (d)
- **IUPAC Name:** 2-bromo-2-methylpropane
- **Explanation:** 2-bromo-2-methylpropane is a tertiary halogenoalkane. It reacts via the \(\text{S}_\text{N}1\) mechanism because the tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\), is highly stable. This stability is due to the positive inductive effect of three electron-releasing alkyl (methyl) groups, which disperse the positive charge. (Additionally, steric hindrance from the bulky methyl groups prevents a direct backside attack by the nucleophile required in the \(\text{S}_\text{N}2\) mechanism).

### Part (a) [3 Marks]
- **M1:** State that the order with respect to \([\text{C}_4\text{H}_9\text{Br}]\) is 1, and the order with respect to \([\text{OH}^-]\) is 0. (1 mark)
- **M2:** Explain that doubling \([\text{C}_4\text{H}_9\text{Br}]\) from Exp 1 to 2 (with \([\text{OH}^-]\) constant) doubles the rate, hence order is 1. (1 mark)
- **M3:** Explain that doubling \([\text{OH}^-]\) from Exp 1 to 3 (with \([\text{C}_4\text{H}_9\text{Br}]\) constant) has no effect on the rate, hence order is 0. (1 mark)

### Part (b) [3 Marks]
- **M1:** Correct rate equation: \(\text{Rate} = k[\text{C}_4\text{H}_9\text{Br}]\) (1 mark) *[Allow ECF from part (a) if consistent]*
- **M2:** Correct calculation of \(k = 1.20 \times 10^{-3}\) (or \(1.2 \times 10^{-3}\)) (1 mark)
- **M3:** Correct unit: \(\text{s}^{-1}\) (1 mark)

### Part (c) [4 Marks]
- **M1:** Correct skeletal or structural drawing of 2-bromo-2-methylpropane with a curly arrow starting from the \(\text{C}-\text{Br}\) bond to the \(\text{Br}\) atom. (1 mark)
- **M2:** Structure of the tertiary carbocation intermediate \((\text{CH}_3)_3\text{C}^+\) with a clear positive charge on the central carbon. (1 mark)
- **M3:** Hydroxide ion drawn with a lone pair and negative charge \((:\text{OH}^-)\), and a curly arrow starting from the lone pair to the positive carbon of the carbocation. (1 mark)
- **M4:** Correct final structure of 2-methylpropan-2-ol. (1 mark)

### Part (d) [2 Marks]
- **M1:** Correctly identifies the IUPAC name as **2-bromo-2-methylpropane**. (1 mark) *[Do not accept 1-bromobutane or 2-bromobutane]*
- **M2:** Mentions that the tertiary carbocation is stabilized by the positive inductive effect of three alkyl (or methyl) groups OR mentions steric hindrance preventing an \(\text{S}_\text{N}2\) attack. (1 mark)

(a) The functional group present is a carboxylic acid, -COOH. The ionic equation is \(\text{H}^+ + \text{HCO}_3^- \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). (b) Compound X is 2-methylpropanoic acid, \((\text{CH}_3)_2\text{CHCOOH}\). The peak at \(\delta = 11.5\text{ ppm}\) (singlet, area 1) corresponds to the acidic proton in the \(-\text{COOH}\) group. The peak at \(\delta = 1.2\text{ ppm}\) (doublet, area 6) corresponds to the six protons of the two equivalent methyl groups \((\text{CH}_3)_2-\), which are split into a doublet by the single adjacent \(-\text{CH}-\) proton. The peak at \(\delta = 2.5\text{ ppm}\) (septet, area 1) corresponds to the single \(-\text{CH}-\) proton, which is split into a septet by the six adjacent methyl protons. (c) To prepare the ester, heat X under reflux with methanol in the presence of a concentrated sulfuric acid catalyst. Pour the reaction mixture into a separating funnel, wash with sodium hydrogencarbonate solution to remove acid impurities, separate the organic layer, dry with an anhydrous drying agent like calcium chloride, and perform a simple distillation, collecting the fraction at the boiling point of the ester.

Part (a): [2 marks] 1 mark for the correct functional group (-COOH). 1 mark for a valid ionic equation. Part (b): [5 marks] 1 mark for identifying X as 2-methylpropanoic acid. 1 mark for assigning the singlet at 11.5 ppm to the -COOH proton. 1 mark for assigning the doublet at 1.2 ppm to the methyl protons. 1 mark for explaining the doublet splitting (split by 1 adjacent CH proton). 1 mark for assigning the septet at 2.5 ppm to the CH proton split by 6 methyl protons. Part (c): [3 marks] 1 mark for heating under reflux with methanol and concentrated sulfuric acid catalyst. 1 mark for separating funnel wash with sodium hydrogencarbonate and separating the organic layer. 1 mark for drying with an anhydrous salt and distilling to collect the ester fraction.

(a) Add sodium hydroxide solution dropwise to solution A. A green precipitate of iron(II) hydroxide forms, and a brown precipitate of iron(III) hydroxide forms. Equations: \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe}(\text{H}_2\text{O})_4(\text{OH})_2(\text{s}) + 2\text{H}_2\text{O}(\text{l})\) and \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe}(\text{H}_2\text{O})_3(\text{OH})_3(\text{s}) + 3\text{H}_2\text{O}(\text{l})\). (b)(i) \(5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}\). (ii) Hydrochloric acid contains chloride ions, which would be oxidized to chlorine gas by the manganate(VII) ions. This would lead to an incorrectly large volume of manganate(VII) being used in the titration. (iii) Moles of \(\text{MnO}_4^- = 0.0200 \times 18.45 / 1000 = 3.69 \times 10^{-4}\text{ mol}\). Moles of \(\text{Fe}^{2+} = 5 \times 3.69 \times 10^{-4} = 1.845 \times 10^{-3}\text{ mol}\). Concentration of \(\text{Fe}^{2+} = 1.845 \times 10^{-3} / 0.0250 = 0.0738\text{ mol dm}^{-3}\).

Part (a): [4 marks] 1 mark for adding NaOH dropwise. 1 mark for green and brown precipitate observations. 1 mark for each of the two correct balanced equations with state symbols. Part (b)(i): [1 mark] 1 mark for the correct balanced redox equation. Part (b)(ii): [2 marks] 1 mark for stating that chloride ions would be oxidized to chlorine. 1 mark for stating this increases the volume of manganate(VII) needed. Part (b)(iii): [3 marks] 1 mark for calculating moles of manganate(VII) as 3.69 x 10^-4. 1 mark for multiplying by 5 to find moles of iron(II) as 1.845 x 10^-3. 1 mark for calculating the final concentration as 0.0738 mol dm^-3.

(a) Before adding NaOH, \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.75 \times 10^{-5} \times 0.100} = 1.323 \times 10^{-3}\text{ mol dm}^{-3}\). pH = \(-\log_{10}(1.323 \times 10^{-3}) = 2.88\). Assumption: \([\text{H}^+] = [\text{A}^-]\) (or the concentration of HA at equilibrium is equal to its initial concentration). (b) Initial moles of HA = \(0.0250 \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\). Moles of NaOH added = \(0.0100 \times 0.150 = 1.50 \times 10^{-3}\text{ mol}\). This reacts to leave \(2.50 \times 10^{-3} - 1.50 \times 10^{-3} = 1.00 \times 10^{-3}\text{ mol}\) of HA, and produces \(1.50 \times 10^{-3}\text{ mol}\) of \(\text{A}^-\). Using the buffer equation: \([\text{H}^+] = K_a \times (n_\text{HA} / n_{\text{A}^-}) = 1.75 \times 10^{-5} \times (1.00 \times 10^{-3} / 1.50 \times 10^{-3}) = 1.167 \times 10^{-5}\text{ mol dm}^{-3}\). pH = \(-\log_{10}(1.167 \times 10^{-5}) = 4.93\). (c) Phenolphthalein is suitable because the titration involves a weak acid and a strong base, which has an equivalence point pH above 7. The pH range of phenolphthalein (8.3 - 10.0) matches the rapid vertical pH change of this titration curve.

Part (a): [3 marks] 1 mark for correct setting up of the expression [H+] = sqrt(Ka x [HA]). 1 mark for pH = 2.88 (must be to 2 decimal places). 1 mark for stating the assumption [H+] = [A-] or negligible ionization of HA. Part (b): [4 marks] 1 mark for initial moles of HA (2.50 x 10^-3) and NaOH (1.50 x 10^-3). 1 mark for calculating remaining moles of HA (1.00 x 10^-3) and formed moles of A- (1.50 x 10^-3). 1 mark for [H+] = 1.17 x 10^-5. 1 mark for pH = 4.93. Part (c): [3 marks] 1 mark for phenolphthalein. 1 mark for stating that equivalence point pH is above 7. 1 mark for explaining that the indicator color change range (8.3-10.0) lies within the vertical section of the pH curve.

(a) \(\text{NH}_4\text{Cl(s)} \rightarrow \text{NH}_4^+(\text{aq}) + \text{Cl}^-(\text{aq})\). (b) Temperature change \(\Delta T = 16.1 - 20.2 = -4.1\ ^\circ\text{C}\). Using \(q = mc\Delta T\) where mass \(m = 100.0\text{ g}\) (assuming the mass of the solute is neglected and density of water is \(1.00\text{ g cm}^{-3}\)): \(q = 100.0 \times 4.18 \times 4.1 = 1713.8\text{ J}\) (accept \(1710\text{ J}\)). (If mass includes solid, \(m = 105.35\text{ g}\), then \(q = 105.35 \times 4.18 \times 4.1 = 1805.6\text{ J}\)). Assumptions: The specific heat capacity of the solution is equal to that of pure water (\(4.18\text{ J g}^{-1}\text{ K}^{-1}\)), and either the mass of the solution is equal to the mass of the water or the solid contributes to the mass but not the heat capacity. (c) Moles of \(\text{NH}_4\text{Cl} = 5.35 / 53.5 = 0.100\text{ mol}\). \(\Delta H_{\text{sol}} = +q / n = +1.7138\text{ kJ} / 0.100\text{ mol} = +17.1\text{ kJ mol}^{-1}\) (or \(+18.1\text{ kJ mol}^{-1}\) if mass of solid is included). (d) Two improvements: 1. Record the temperature at regular intervals (e.g., every minute) before and after mixing, and plot a temperature-time graph to extrapolate the minimum temperature to the time of mixing. 2. Use a high-precision digital thermometer or temperature sensor with smaller uncertainty instead of a standard thermometer. 3. Powder the solid completely to ensure rapid dissolution.

Part (a): [1 mark] 1 mark for the correct balanced equation with state symbols. Part (b): [3 marks] 1 mark for temperature change of 4.1. 1 mark for q calculation (1710 J or 1806 J depending on mass used). 1 mark for stating the correct assumptions (specific heat capacity same as water and choice of mass). Part (c): [3 marks] 1 mark for calculating moles as 0.100. 1 mark for correct division of heat by moles to get value (17.1 or 18.1). 1 mark for a positive sign. Part (d): [3 marks] Any two from: Plot a temperature-time cooling curve and extrapolate to the time of mixing [1]; use a more precise thermometer or probe [1]; ensure the solid is finely ground/powdered [1].

(a)(i) \(\text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_4^-\). (a)(ii) Mechanism: 1. A curly arrow from the double bond of the benzene ring (para-position relative to the methyl group) to the carbon of the \(\text{CH}_3^+\) electrophile. 2. Draw the tetrahedral intermediate showing a partial positive charge inside the ring and both \(-\text{H}\) and \(-\text{CH}_3\) groups attached to the para-carbon. 3. A curly arrow from the C-H bond into the ring to restore the delocalized system. (b) Reagent: Alkaline potassium manganate(VII) (followed by addition of dilute acid to precipitate the dicarboxylic acid). Conditions: Heat under reflux. (c) Dissolve the crude solid in the minimum volume of hot solvent. Filter the hot solution to remove insoluble impurities. Allow the filtrate to cool slowly to room temperature (or in an ice bath) so that crystals of pure acid reform while soluble impurities remain dissolved. Filter the crystals under reduced pressure using a Buchner funnel, wash with a small volume of cold solvent, and dry.

Part (a)(i): [1 mark] 1 mark for correct electrophile generation equation. Part (a)(ii): [3 marks] 1 mark for curly arrow from ring to CH3+. 1 mark for correct intermediate with positive charge inside ring. 1 mark for curly arrow from C-H bond to reform ring. Part (b): [2 marks] 1 mark for alkaline potassium manganate(VII) / acidified potassium manganate(VII). 1 mark for heating under reflux. Part (c): [4 marks] 1 mark for dissolving in minimum volume of hot solvent. 1 mark for hot filtration. 1 mark for cooling to recrystallize. 1 mark for filtration under reduced pressure, washing with cold solvent, and drying.

(a) \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). (b) Moles of HCl initial = \(0.0500\text{ dm}^3 \times 0.500\text{ mol dm}^{-3} = 0.0250\text{ mol}\). (c) Moles of NaOH in titration = \(0.01420\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.42 \times 10^{-3}\text{ mol}\). Since HCl and NaOH react in a 1:1 ratio, moles of HCl in \(25.0\text{ cm}^3\) sample = \(1.42 \times 10^{-3}\text{ mol}\). Moles of HCl in \(250\text{ cm}^3\) volumetric flask = \(1.42 \times 10^{-3} \times 10 = 0.0142\text{ mol}\). (d) Moles of HCl reacted with \(\text{CaCO}_3 = 0.0250\text{ mol} - 0.0142\text{ mol} = 0.0108\text{ mol}\). From the chemical equation, 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of HCl, so moles of \(\text{CaCO}_3 = 0.0108 / 2 = 0.00540\text{ mol}\). Mass of \(\text{CaCO}_3 = 0.00540\text{ mol} \times 100.1\text{ g mol}^{-1} = 0.54054\text{ g}\) (or \(0.541\text{ g}\)). (e) Percentage by mass = \((0.54054 / 1.25) \times 100 = 43.24\%\) (accept \(43.2\%\) or \(43.3\%\)).

Part (a): [1 mark] 1 mark for correct balanced equation. Part (b): [1 mark] 1 mark for 0.0250 mol. Part (c): [3 marks] 1 mark for moles of NaOH = 1.42 x 10^-3 mol. 1 mark for equating to moles of HCl in 25.0 cm3. 1 mark for scaling up by 10 to get 0.0142 mol. Part (d): [3 marks] 1 mark for subtracting to find reacted HCl = 0.0108 mol. 1 mark for dividing by 2 to find moles of CaCO3 = 0.00540 mol. 1 mark for mass of CaCO3 = 0.541 g. Part (e): [2 marks] 1 mark for dividing the calculated mass by 1.25 and multiplying by 100. 1 mark for correct final percentage to 3 significant figures (43.2% or 43.3%).

First, calculate the moles of compound X: \(n = \frac{m}{M_r} = \frac{1.48}{74.0} = 0.0200 \text{ mol}\). Next, convert the temperature to Kelvin: \(T = 127 + 273 = 400 \text{ K}\). Convert the pressure to Pascals: \(P = 100 \times 10^3 \text{ Pa} = 100,000 \text{ Pa}\). Use the ideal gas equation \(PV = nRT\) to find the volume in \(\text{m}^3\): \(V = \frac{nRT}{P} = \frac{0.0200 \times 8.31 \times 400}{100,000} = 6.648 \times 10^{-4} \text{ m}^3\). Finally, convert \(\text{m}^3\) to \(\text{cm}^3\) by multiplying by \(10^6\): \(V = 6.648 \times 10^{-4} \times 10^6 = 665 \text{ cm}^3\) (to 3 significant figures).

Award 1 mark for selecting the correct option (b). Distractors are based on incorrect unit conversions.

The accurate molecular mass of \(C_4H_8O\) is \(4(12.0000) + 8(1.0078) + 15.9949 = 72.0573\), which matches the mass spectrum. An IR absorption at \(1715 \text{ cm}^{-1}\) indicates a carbonyl group (\(C=O\)). The absence of a broad band above \(3200 \text{ cm}^{-1}\) rules out an alcohol or carboxylic acid. Therefore, the compound is a ketone, specifically butanone.

Award 1 mark for the correct option (b). Distractors have incorrect molecular masses or conflicting functional group identification.

First, calculate initial moles of propanoic acid: \(n(\text{HA}) = 0.0250 \times 0.100 = 2.50 \times 10^{-3} \text{ mol}\). Moles of NaOH added: \(n(\text{NaOH}) = 0.0150 \times 0.0800 = 1.20 \times 10^{-3} \text{ mol}\). During the reaction, \(1.20 \times 10^{-3} \text{ mol}\) of acid is neutralised to form \(1.20 \times 10^{-3} \text{ mol}\) of propanoate ions (\(\text{A}^-\)). Remaining propanoic acid: \(n(\text{HA}) = 2.50 \times 10^{-3} - 1.20 \times 10^{-3} = 1.30 \times 10^{-3} \text{ mol}\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = -\log_{10}(1.35 \times 10^{-5}) + \log_{10}\left(\frac{1.20 \times 10^{-3}}{1.30 \times 10^{-3}}\right) = 4.870 - 0.035 = 4.83\).

Award 1 mark for the correct option (a). Distractors are based on incorrect ratio formulations or forgetting to subtract the reacted moles of acid.

A small amount of ammonia acts as a base to form a pale blue precipitate of copper(II) hydroxide: \(Cu(H_2O)_4(OH)_2\). When excess ammonia is added, ligand substitution occurs where four ammonia ligands replace four water ligands, forming a deep blue solution containing \([Cu(NH_3)_4(H_2O)_2]^{2+}\) due to Jahn-Teller distortion and steric reasons.

Award 1 mark for the correct option (b). Distractors incorrectly suggest full substitution to hexaamminecopper(II) or suggest the precipitate is insoluble.

To prepare 3-nitrobenzoic acid (meta-substituted), the meta-directing substituent must be added first. The methyl group (from Friedel-Crafts alkylation in Step 1) is ortho/para-directing, so we cannot nitrate first. Thus, we first synthesize methylbenzene, oxidise it with acidified \(\text{KMnO}_4\) in Step 2 to benzoic acid (which has a meta-directing carboxyl group), and finally nitrate it in Step 3 using conc. \(\text{HNO}_3 / \text{H}_2\text{SO}_4\) to direct the nitro group to the 3-position.

Award 1 mark for the correct option (c). Other options lead to 2- or 4-nitrobenzoic acid, or fail due to the strongly deactivating nature of the nitro group preventing alkylation.

Total mass of the solution \(m = 50.0 + 50.0 = 100.0 \text{ g}\). Temperature change \(\Delta T = 26.2 - 19.5 = 6.7 \text{ K}\). Heat evolved \(q = m c \Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6 \text{ J} = 2.8006 \text{ kJ}\). Moles of water formed: \(n = 0.0500 \text{ dm}^3 \times 1.00 \text{ mol dm}^{-3} = 0.0500 \text{ mol}\). Enthalpy change of neutralisation: \(\Delta H_{\text{neu}} = -\frac{q}{n} = -\frac{2.8006 \text{ kJ}}{0.0500 \text{ mol}} = -56.0 \text{ kJ mol}^{-1}\).

Award 1 mark for the correct option (a). Distractor (b) fails to divide by moles or uses total moles of reactants, distractor (c) forgets the 1:1 ratio, and (d) has the incorrect positive sign.

Tollens' reagent contains \([Ag(NH_3)_2]^+\) ions and is a mild oxidising agent. It reacts only with aldehydes (pentanal) to form a silver mirror. Ketones (pentan-2-one and pentan-3-one) and alcohols (pentan-1-ol) do not react with Tollens' reagent, thereby uniquely identifying pentanal.

Award 1 mark for the correct option (b). Distractor (a) also reacts with the primary alcohol, (c) does not react with any of them, and (d) would identify pentan-2-one instead.

In pure water, \([\text{H}^+] = [\text{OH}^-]\), so the solution is always neutral. At 310 K, \(K_w = [\text{H}^+]^2 = 2.40 \times 10^{-14}\). Taking the square root gives \([\text{H}^+] = 1.549 \times 10^{-7} \text{ mol dm}^{-3}\). Thus, \(\text{pH} = -\log_{10}(1.549 \times 10^{-7}) = 6.81\).

Award 1 mark for the correct option (c). Distractor (b) falsely assumes that pH < 7.00 must be acidic, and (a) assumes pH 7 is neutral at all temperatures.

First, calculate the initial moles of HA and NaOH. Moles of HA = (40.0 / 1000) * 0.150 = 6.00 x 10-3 mol. Moles of NaOH = (10.0 / 1000) * 0.200 = 2.00 x 10-3 mol. The NaOH reacts completely with the HA to form the conjugate base A-. Moles of A- formed = 2.00 x 10-3 mol. Moles of HA remaining = 6.00 x 10-3 - 2.00 x 10-3 = 4.00 x 10-3 mol. Using the buffer equation: [H+] = Ka * (moles of HA / moles of A-) = (2.00 x 10-5) * (4.00 x 10-3 / 2.00 x 10-3) = 4.00 x 10-5 mol dm-3. The pH = -log10[H+] = -log10(4.00 x 10-5) = 4.40.

1 mark for the correct buffer pH calculation resulting in 4.40 (Option B).

The aqueous Cr3+ ion forms a green precipitate of Cr(OH)3(H2O)3 with dropwise NaOH. Since chromium(III) hydroxide is amphoteric, it dissolves in excess NaOH to form the dark green [Cr(OH)6]3- solution. Cr3+ is a 3+ ion, so reacting with sodium carbonate produces carbon dioxide gas and a precipitate of Cr(OH)3(H2O)3 due to acidity of the hexaaqua ion. Fe2+ forms a green precipitate but does not dissolve in excess NaOH. Fe3+ forms a brown precipitate. Cu2+ forms a blue precipitate.

1 mark for identifying Cr3+ as the transition metal ion (Option C).

The combustion reaction is: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g). Bonds broken: 4 x (C-H) + 2 x (O=O) = 4(413) + 2(496) = 1652 + 992 = 2644 kJ mol-1. Bonds made: 2 x (C=O) + 4 x (O-H) = 2(743) + 4(463) = 1486 + 1852 = 3338 kJ mol-1. Enthalpy change = energy for bonds broken - energy for bonds made = 2644 - 3338 = -694 kJ mol-1.

1 mark for calculating the correct enthalpy change of -694 kJ mol-1 (Option A).

The standard synthesis of phenylamine from benzene involves: 1) Nitration of benzene using a nitrating mixture of concentrated HNO3 and concentrated H2SO4 at 50-55 degrees Celsius to form nitrobenzene. 2) Reduction of nitrobenzene using tin (Sn) and concentrated HCl under reflux, followed by addition of NaOH to release the free amine.

1 mark for choosing the correct reaction sequence involving nitration with concentrated acids and reduction with Sn/HCl (Option B).

Calculate the accurate masses of the options: A) C3H6O2: (3 x 12.0000) + (6 x 1.0078) + (2 x 15.9949) = 36.0000 + 6.0468 + 31.9898 = 74.0366. B) C4H10O: (4 x 12.0000) + (10 x 1.0078) + 15.9949 = 48.0000 + 10.0780 + 15.9949 = 74.0729. C) C2H6N2O: (2 x 12.0000) + (6 x 1.0078) + (2 x 14.0031) + 15.9949 = 24.0000 + 6.0468 + 28.0062 + 15.9949 = 74.0479. D) C3H10N2: (3 x 12.0000) + (10 x 1.0078) + (2 x 14.0031) = 36.0000 + 10.0780 + 28.0062 = 74.0842. The accurate mass of 74.0366 corresponds precisely to C3H6O2.

1 mark for identifying the correct molecular formula C3H6O2 (Option A).

Reaction: CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g). Initial moles of CaCO3 = 1.50 / 100.1 = 0.0150 mol. Initial moles of HCl = 0.0200 dm3 * 1.00 mol dm-3 = 0.0200 mol. According to the stoichiometry, 1 mole of CaCO3 requires 2 moles of HCl. Thus, 0.0150 mol of CaCO3 requires 0.0300 mol of HCl. Since we only have 0.0200 mol of HCl, HCl is the limiting reagent. Moles of CO2 produced = moles of HCl reacted / 2 = 0.0200 / 2 = 0.0100 mol. Volume of CO2 at RTP = 0.0100 mol * 24000 cm3 mol-1 = 240 cm3.

1 mark for the correct volume calculation of 240 cm3 based on the limiting reagent (Option B).

Addition of excess concentrated HCl to aqueous copper(II) ions results in the ligand substitution reaction: [Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O. The copper changes coordination number from 6 (in [Cu(H2O)6]2+) to 4 (in [CuCl4]2-). The geometry changes from octahedral to tetrahedral because chloride ligands are larger and repel each other more, so they fit in a tetrahedral arrangement rather than a square planar or octahedral one.

1 mark for identifying that the coordination number decreases from 6 to 4 and the geometry changes from octahedral to tetrahedral (Option A).

Substitute the experimental values into the rate equation: Rate = k[A][B]^2 => 2.0 x 10-4 = k * (0.050) * (0.100)^2 => 2.0 x 10-4 = k * 5.0 x 10-4 => k = 0.40. To find the units: k = Rate / ([A][B]^2) = (mol dm-3 s-1) / ((mol dm-3) * (mol dm-3)^2) = (mol dm-3 s-1) / (mol3 dm-9) = mol-2 dm6 s-1 (or dm6 mol-2 s-1). This corresponds to Option A.

1 mark for calculating k = 0.40 and deriving the correct units dm6 mol-2 s-1 (Option A).

Ethyl ethanoate has the molecular formula \(C_4H_8O_2\). The absorption at \(1740\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad peak at \(2500\text{--}3300\text{ cm}^{-1}\) rules out an O-H bond of a carboxylic acid (ruling out butanoic acid). The absence of a broad peak at \(3230\text{--}3550\text{ cm}^{-1}\) rules out an O-H bond of an alcohol (ruling out 2-hydroxybutanal and butane-1,4-diol).

[1 mark] B - Correctly identifies ethyl ethanoate based on the presence of a C=O absorption and the absence of any O-H absorption.

First calculate the initial moles: \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0250 \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\); \(n(\text{NaOH}) = 0.0250 \times 0.050 = 1.25 \times 10^{-3}\text{ mol}\). Since NaOH is the limiting reagent, it reacts completely with the acid to produce \(1.25 \times 10^{-3}\text{ mol}\) of propanoate ions (\(A^-\)), leaving \(2.50 \times 10^{-3} - 1.25 \times 10^{-3} = 1.25 \times 10^{-3}\text{ mol}\) of unreacted propanoic acid (\(HA\)). Since \([HA] = [A^-]\) in the mixture, the pH expression simplifies to \(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

[1 mark] B - Correctly calculates moles of weak acid and conjugate base, and applies the pH expression for a buffer.

When dropwise ammonia is added to \([Cu(H_2O)_6]^{2+}\), it acts as a Bronsted-Lowry base to form a blue precipitate of copper(II) hydroxide, \(Cu(OH)_2(H_2O)_4(s)\). When excess ammonia is added, ligand substitution occurs to form the soluble deep-blue complex ion \([Cu(NH_3)_4(H_2O)_2]^{2+}(aq)\).

[1 mark] C - Correctly identifies the observations for dropwise and excess addition of ammonia to copper(II) ions.

The equation for the combustion of ethane is: \(\text{C}_2\text{H}_6\text{(g)} + 3.5\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}\). Using Hess's law: \(\Delta H_c^\theta = 2 \times \Delta H_f^\theta(\text{CO}_2\text{(g)}) + 3 \times \Delta H_f^\theta(\text{H}_2\text{O(l)}) - \Delta H_f^\theta(\text{C}_2\text{H}_6\text{(g)}) = 2(-393.5) + 3(-285.8) - (-84.7) = -787.0 - 857.4 + 84.7 = -1559.7\text{ kJ mol}^{-1}\).

[1 mark] A - Correctly balances the combustion equation and applies Hess's law to calculate the enthalpy of combustion.

Using the ideal gas equation: \(pV = nRT\). Convert units: \(p = 102 \times 10^3\text{ Pa}\), \(V = 125 \times 10^{-6}\text{ m}^3\), \(T = 310\text{ K}\). Compute moles: \(n = \frac{pV}{RT} = \frac{102 \times 10^3 \times 125 \times 10^{-6}}{8.31 \times 310} = \frac{12.75}{2576.1} = 4.949 \times 10^{-3}\text{ mol}\). Compute relative molecular mass: \(M_r = \frac{m}{n} = \frac{0.355}{4.949 \times 10^{-3}} = 71.7\).

[1 mark] C - Correctly converts units, calculates moles using the ideal gas equation, and finds the relative molecular mass.

Nitrobenzene is reduced to phenylamine (aniline) using tin (Sn) and concentrated hydrochloric acid under reflux, followed by addition of sodium hydroxide to release phenylamine from its salt. Phenylamine then undergoes nucleophilic addition-elimination with ethanoil chloride to form N-phenylethanamide.

[1 mark] A - Correctly identifies the reagents for both the reduction of nitrobenzene and the subsequent acylation of the amine.

A reaction is feasible when \(\Delta G^\theta \le 0\). Since \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), feasibility is achieved when \(T \ge \frac{\Delta H^\theta}{\Delta S^\theta}\). Converting \(\Delta H^\theta\) to \(\text{J mol}^{-1}\): \(135 \times 10^3\text{ J mol}^{-1}\). Therefore, \(T \ge \frac{135000}{285} \approx 473.68\text{ K}\). Rounding to the nearest whole number gives \(474\text{ K}\).

[1 mark] C - Correctly converts units and uses the Gibbs free energy equation to determine the minimum temperature for feasibility.

Propene is an unsymmetrical alkene. Electrophilic addition of HBr proceeds via the more stable carbocation intermediate. The secondary carbocation, \(\text{CH}_3\text{C}^+\text{HCH}_3\), is more stable than the primary carbocation, \(\text{CH}_3\text{CH}_2\text{C}^+\text{H}_2\), due to the positive inductive effect of two electron-releasing methyl groups.

[1 mark] C - Correctly identifies that the reaction proceeds via the more stable secondary carbocation due to the positive inductive effect of two alkyl groups.

1. The IR spectrum shows a broad peak at \(3350\text{ cm}^{-1}\), indicating the presence of an alcohol (\(\text{O}-\text{H}\)) group, and a strong peak at \(1715\text{ cm}^{-1}\), indicating a carbonyl (\(\text{C}=\text{O}\)) group. 2. This rules out ethyl ethanoate and methyl propanoate, which are esters and do not contain an \(\text{O}-\text{H}\) group. 3. In the \(^1\text{H}\) NMR spectrum, the singlet at \(\delta = 2.1\text{ ppm}\) (3H) corresponds to a \(\text{CH}_3-\text{C}(=\text{O})\) group. 3-hydroxybutanal would have a doublet for its methyl protons. 4. The two triplets (each 2H) correspond to adjacent \(\text{-CH}_2-\text{CH}_2-\) groups. The triplet at \(\delta = 3.8\text{ ppm}\) is shifted downfield due to the adjacent electronegative oxygen atom of the \(\text{-OH}\) group. 5. Therefore, the structure is \(\text{CH}_3\text{COCH}_2\text{CH}_2\text{OH}\), which is 4-hydroxybutan-2-one.

[1 mark] C is correct. (A and D incorrect due to no O-H peak; B incorrect as methyl protons would split to a doublet).

1. Step 1 requires free-radical substitution of an alkane (ethane) to a halogenoalkane (bromoethane), which needs \(\text{Br}_2\) and UV light. 2. Step 2 requires nucleophilic substitution of the halogenoalkane to a nitrile (propanenitrile), using \(\text{KCN}\) in aqueous ethanol under reflux. 3. Step 3 is the acid hydrolysis of a nitrile to a carboxylic acid (propanoic acid), which requires dilute aqueous acid (such as \(\text{HCl}\) or \(\text{H}_2\text{SO}_4\)) and heat under reflux.

[1 mark] A is correct. (B is incorrect as Step 1 does not proceed with HBr; C is incorrect as Step 2 produces an amine; D is incorrect as Step 3 reduces the nitrile to an amine).

1. Moles of \(\text{HCOOH}\) initially = \(2.50 \times 10^{-3}\text{ mol}\). 2. Moles of \(\text{NaOH}\) added = \(1.80 \times 10^{-3}\text{ mol}\). 3. Since they react in a 1:1 ratio, the remaining \(\text{HCOOH} = 2.50 \times 10^{-3} - 1.80 \times 10^{-3} = 0.70 \times 10^{-3}\text{ mol}\), and the produced \(\text{HCOO}^- = 1.80 \times 10^{-3}\text{ mol}\). 4. Use the buffer expression: \([\text{H}^+] = K_{\text{a}} \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.78 \times 10^{-4} \times \frac{0.70 \times 10^{-3}}{1.80 \times 10^{-3}} = 6.92 \times 10^{-5}\text{ mol dm}^{-3}\). 5. \(\text{pH} = -\log_{10}(6.92 \times 10^{-5}) = 4.16\).

[1 mark] C is correct. (A is incorrect because ratio was inverted; B is simply pKa; D is incorrect due to a calculation error using direct ratios without reaction subtraction).

1. Aluminium ions, \(\text{Al}^{3+}(\text{aq})\), form a white precipitate of \(\text{Al(OH)}_3(\text{H}_2\text{O})_3\) with dropwise sodium hydroxide. 2. Because \(\text{Al(OH)}_3\) is amphoteric, it dissolves in excess sodium hydroxide to form the soluble, colourless complex \([\text{Al(OH)}_4]^-\). 3. Due to the high charge-to-size ratio (polarising power) of the 3+ aquated metal ion, it acts as an acid and reacts with carbonate ions to form carbon dioxide gas and a white precipitate of aluminium hydroxide.

[1 mark] C is correct. (A forms a precipitate that is insoluble in excess NaOH and no gas with carbonate; B forms a brown precipitate; D forms a blue precipitate).

1. The heat released in kJ is \(q = \frac{50.0 \times 4.18 \times 7.5}{1000}\). 2. The amount of substance in moles is \(n = \frac{5.00}{159.6}\). 3. The reaction is exothermic (temperature increased), so the enthalpy of solution must be negative: \(\Delta H = -\frac{q}{n} = -\frac{50.0 \times 4.18 \times 7.5 \times 159.6}{5.00 \times 1000}\).

[1 mark] A is correct. (B is incorrect as it has a positive sign; C is inverted; D fails to convert J to kJ).

1. \(n(\text{NH}_4\text{NO}_3) = \frac{2.00}{80.0} = 0.0250\text{ mol}\). 2. Under the conditions (\(400\text{ K}\)), water is a gas. Total gas stoichiometry: \(2\text{ moles of reactant} \rightarrow 7\text{ moles of gas}\). Total moles of gas \(n_{\text{gas}} = 0.0250 \times \frac{7}{2} = 0.0875\text{ mol}\). 3. From ideal gas law: \(V = \frac{nRT}{p} = \frac{0.0875 \times 8.31 \times 400}{100\,000} = 0.0029085\text{ m}^3\). 4. Convert to \(\text{dm}^3\): \(V = 2.91\text{ dm}^3\).

[1 mark] C is correct. (A is incorrect due to incorrect stoichiometry; B is incorrect because water is incorrectly treated as a liquid; D is incorrect because the ratio was multiplied by 7 instead of 7/2).