2025 AQA IAL Chemistry (9620) Practice Paper with Answers

(a) First ionization energy is the enthalpy change or energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.

(b) Phosphorus has the outer electronic configuration \(3\text{s}^2 3\text{p}^3\) where the \(3\text{p}\) subshell contains three singly-occupied orbitals. Sulfur has the configuration \(3\text{s}^2 3\text{p}^4\) where one of the \(3\text{p}\) orbitals contains a pair of electrons. The mutual repulsion between the paired electrons in the sulfur \(3\text{p}\) orbital makes it easier to remove one electron compared to the singly-occupied orbitals in phosphorus.

(c) Mass of one \(^{69}\text{Ga}^+\) ion in kg:
\(m = \frac{69 \times 10^{-3} \text{ kg mol}^{-1}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 1.1458 \times 10^{-25} \text{ kg}\)
Using \(KE = \frac{1}{2}mv^2\):
\(v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1.15 \times 10^{-16}}{1.1458 \times 10^{-25}}} = 4.480 \times 10^4 \text{ m s}^{-1}\)
Using \(t = \frac{d}{v}\):
\(t = \frac{1.50}{4.480 \times 10^4} = 3.35 \times 10^{-5} \text{ s}\)

(a)
M1: Enthalpy change/energy required for 1 mole of gaseous atoms (1)
M2: to lose one electron per atom (1)
M3: to form 1 mole of gaseous 1+ ions (1)

(b)
M1: Phosphorus has configuration ending in \(3\text{p}^3\) and sulfur in \(3\text{p}^4\) (or orbital diagram) (1)
M2: In sulfur, there is a paired electron in a \(3\text{p}\) orbital (1)
M3: Repulsion between paired electrons makes it easier to remove (1)

(c)
M1: Calculates mass of one ion: \(1.146 \times 10^{-25} \text{ kg}\) (1)
M2: Rearranges formula: \(v = \sqrt{\frac{2 \times KE}{m}}\) (1)
M3: Calculates velocity: \(4.48 \times 10^4 \text{ m s}^{-1}\) (1)
M4: Calculates time: \(3.35 \times 10^{-5} \text{ s}\) (accept range \(3.34 \times 10^{-5}\) to \(3.36 \times 10^{-5}\)) (1)

(a)(i) Moles of \(\text{HCl} = 0.100 \times \frac{22.00}{1000} = 2.20 \times 10^{-3} \text{ mol}\).

(a)(ii) Reacting ratio of \(\text{Na}_2\text{CO}_3 : \text{HCl}\) is \(1 : 2\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0 \text{ cm}^3 = \frac{2.20 \times 10^{-3}}{2} = 1.10 \times 10^{-3} \text{ mol}\).
Moles in \(250.0 \text{ cm}^3 = 1.10 \times 10^{-3} \times 10 = 0.0110 \text{ mol}\).

(a)(iii) \(M_r(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}) = \frac{\text{mass}}{\text{moles}} = \frac{3.15 \text{ g}}{0.0110 \text{ mol}} = 286.4 \text{ g mol}^{-1}\).
\(M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0 \text{ g mol}^{-1}\).
Mass of water of crystallisation \(= 286.4 - 106.0 = 180.4 \text{ g mol}^{-1}\).
\(x = \frac{180.4}{18.0} = 10.02 \approx 10\).

(b)(i) \(PV = nRT \implies PV = \frac{m}{M}RT \implies M = \frac{mRT}{PV}\).
Since density is \(1.80 \text{ g dm}^{-3}\), we can assume \(1.00 \text{ dm}^3\) of gas (\(1.00 \times 10^{-3} \text{ m}^3\)) has a mass of \(1.80 \text{ g}\).
\(M = \frac{1.80 \times 8.314 \times 298}{101300 \times 1.00 \times 10^{-3}} = 44.0 \text{ g mol}^{-1}\).

(b)(ii) A hydrocarbon with \(M_r = 44.0\) is propane, so its molecular formula is \(\text{C}_3\text{H}_8\).

(a)(i)
M1: \(2.20 \times 10^{-3} \text{ mol}\) (1)

(a)(ii)
M1: Moles in \(25.0 \text{ cm}^3 = 1.10 \times 10^{-3} \text{ mol}\) (1)
M2: Moles in \(250.0 \text{ cm}^3 = 0.0110 \text{ mol}\) (1)

(a)(iii)
M1: Calculates \(M_r = 286.4\) (or 286) (1)
M2: Calculates mass of water \(= 180.4\) (1)
M3: Finds \(x = 10\) (must be an integer) (1)

(b)(i)
M1: Correct conversions: \(V = 1.00 \times 10^{-3} \text{ m}^3\) and rearranges to make \(M\) the subject (1)
M2: Correct substitution of values into the rearranged ideal gas equation (1)
M3: Calculates \(M_r = 44.0 \text{ g mol}^{-1}\) (1)

(b)(ii)
M1: \(\text{C}_3\text{H}_8\) (1)

(a)(i) Shape: Trigonal pyramidal. There are 3 bonding pairs and 1 lone pair of electrons around the central phosphorus atom. The drawing should show a central P with one lone pair and three Cl atoms bonded using wedge/dash representation. The bond angle is around \(100^\circ\) to \(108^\circ\) (actual value is \(100^\circ\)).

(a)(ii) Shape: Octahedral.
Bond angle: \(90^\circ\) (and \(180^\circ\)).
Explanation: The phosphorus atom in \(\text{PCl}_6^-\)
has 6 bonding pairs of electrons and 0 lone pairs. According to VSEPR theory, these electron pairs repel each other equally and move as far apart as possible to minimise repulsion, resulting in an octahedral arrangement.

(b) Chlorine is a larger atom than fluorine, so \(\text{PCl}_3\) has more electrons than \(\text{PF}_3\). Therefore, the temporary dipoles are larger, making the van der Waals' (or London/induced dipole-dipole) forces stronger between \(\text{PCl}_3\) molecules. These require more energy to overcome than the combined permanent dipole-dipole and van der Waals' forces in \(\text{PF}_3\).

(a)(i)
M1: Correct 3D drawing of \(\text{PCl}_3\) with a lone pair (1)
M2: Trigonal pyramidal (1)
M3: Any angle in the range \(100^\circ\) to \(108^\circ\) (1)

(a)(ii)
M1: Octahedral (1)
M2: \(90^\circ\) (or \(90^\circ\) and \(180^\circ\)) (1)
M3: Phosphorus has 6 bonding pairs and 0 lone pairs (1)
M4: Electron pairs repel to minimise repulsion / get as far apart as possible (1)

(b)
M1: \(\text{PCl}_3\) has more electrons than \(\text{PF}_3\) (1)
M2: Van der Waals' / London dispersion / temporary dipole-induced dipole forces are stronger (1)
M3: Require more energy to overcome (1)

(a) 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) \(\Delta H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})\)
\(\Delta_f H^\ominus(\text{products}) = [0 + 4 \times (-241.8)] = -967.2 \text{ kJ mol}^{-1}\)
\(\Delta_f H^\ominus(\text{reactants}) = [50.6 + 2 \times (-187.8)] = -325.0 \text{ kJ mol}^{-1}\)
\(\Delta H^\ominus = -967.2 - (-325.0) = -642.2 \text{ kJ mol}^{-1}\).

(c) Heat energy transferred, \(q = m c \Delta T\)
\(m = 50.0 \text{ g}\) (mass of water)
\(\Delta T = 31.2 - 19.5 = 11.7 \text{ K}\)
\(q = 50.0 \times 4.18 \times 11.7 = 2445.3 \text{ J} = 2.4453 \text{ kJ}\)
Moles of \(\text{CuSO}_4 = \frac{5.00}{159.6} = 0.03133 \text{ mol}\)
\(\Delta H_{\text{soln}} = - \frac{2.4453}{0.03133} = -78.05 \text{ kJ mol}^{-1}\) (negative sign because temperature increased/exothermic reaction).

(a)
M1: Enthalpy change when 1 mole of a compound is formed (1)
M2: from its constituent elements in their standard states (1)
M3: under standard conditions / at 100 kPa and 298 K (1)

(b)
M1: Correct expression used: \(\sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})\) (1)
M2: Correct substituted numbers (e.g., \(-967.2 - (-325.0)\)) (1)
M3: Calculates \(-642.2 \text{ kJ mol}^{-1}\) (1)

(c)
M1: Calculates heat energy: \(q = 50.0 \times 4.18 \times 11.7 = 2.445 \text{ kJ}\) (1)
M2: Calculates moles of \(\text{CuSO}_4 = 0.03133 \text{ mol}\) (1)
M3: Divides energy by moles (1)
M4: Calculates \(-78.1 \text{ kJ mol}^{-1}\) (accept \(-78.0\) to \(-78.1\); must have negative sign) (1)
(Note: if mass of 55.0 g is used, \(q = 2.690 \text{ kJ}\) leading to \(-85.9 \text{ kJ mol}^{-1}\), award 4 marks)

(a)(i) \(\text{I}_2\) moles decreased by \(1.50 - 0.44 = 1.06 \text{ mol}\).
Therefore, \(\text{H}_2\) moles also decrease by \(1.06 \text{ mol}\):
Equilibrium \(\text{H}_2 = 1.50 - 1.06 = 0.44 \text{ mol}\).
\(\text{HI}\) moles increase by \(2 \times 1.06 = 2.12 \text{ mol}\).
Equilibrium \(\text{HI} = 2.12 \text{ mol}\).

(a)(ii) \(K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\)

(a)(iii) \(K_c = \frac{(2.12 / V)^2}{(0.44 / V) \times (0.44 / V)} = \frac{2.12^2}{0.44^2} = 23.2\).
The volume term \(V\) is not needed because there is an equal number of moles of gas on both sides of the equation, meaning the volume terms in the numerator and denominator cancel out.

(b) Exothermic. Since \(K_c\) decreases with an increase in temperature, the equilibrium has shifted to the left (reactants side) to oppose the increase in temperature. An increase in temperature shifts the equilibrium in the endothermic direction, meaning the reverse reaction is endothermic, so the forward reaction must be exothermic.

(c)(i) No effect.
(c)(ii) Increases.

(a)(i)
M1: \(\text{H}_2 = 0.44 \text{ mol}\) (1)
M2: \(\text{HI} = 2.12 \text{ mol}\) (1)

(a)(ii)
M1: \(K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\) (1)

(a)(iii)
M1: Substitution of values: \(\frac{2.12^2}{0.44 \times 0.44}\) (1)
M2: Calculates \(K_c = 23.2\) (accept range 23 to 23.3) (1)
M3: Explains that volume cancels out because the number of moles of reactants equals the moles of products (1)

(b)
M1: Exothermic (1)
M2: Since \(K_c\) decreases, the equilibrium has shifted to the left/reactants side; temperature increase shifts in the endothermic direction, so the reverse reaction is endothermic (1)

(c)(i)
M1: No effect (1)

(c)(ii)
M1: Increases (1)

(a)(i) The equation is:
\(\text{KCl} + \text{H}_2\text{SO}_4 \to \text{KHSO}_4 + \text{HCl}\) (or \(2\text{KCl} + \text{H}_2\text{SO}_4 \to \text{K}_2\text{SO}_4 + 2\text{HCl}\))
The role of sulfuric acid is an acid (or proton donor).

(a)(ii) Yellow solid: Sulfur (\(\text{S}\)).
Gas with bad egg smell: Hydrogen sulfide (\(\text{H}_2\text{S}\)).
Explanation: Iodide ions (\(\text{I}^-\)) are stronger reducing agents than chloride ions (\(\text{Cl}^-\)), because the outer electrons of iodide are further from the nucleus and more shielded, so they are lost more easily. Thus, \(\text{I}^-\) can reduce the sulfur in sulfuric acid further (from +6 to 0 in \(\text{S}\) and -2 in \(\text{H}_2\text{S}\)), whereas \(\text{Cl}^-\) cannot reduce sulfuric acid.

(b)(i) Observation: The colourless solution turns orange.
Ionic equation: \(\text{Cl}_2 + 2\text{Br}^- \to 2\text{Cl}^- + \text{Br}_2\)

(b)(ii) Chlorine is a stronger oxidizing agent than bromine, so it oxidizes bromide ions to bromine.

(c) The equation is:
\(\text{Cl}_2 + 2\text{NaOH} \to \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\)
The oxidation states are \(-1\) in \(\text{NaCl}\) and \(+1\) in \(\text{NaClO}\).

(a)(i)
M1: \(\text{KCl} + \text{H}_2\text{SO}_4 \to \text{KHSO}_4 + \text{HCl}\) (or \(2\text{KCl} + \text{H}_2\text{SO}_4 \to \text{K}_2\text{SO}_4 + 2\text{HCl}\)) (1)
M2: Acid / proton donor (reject oxidizing agent) (1)

(a)(ii)
M1: Yellow solid = sulfur (\(\text{S}\) or \(\text{S}_8\)) (1)
M2: Bad egg gas = hydrogen sulfide (\(\text{H}_2\text{S}\)) (1)
M3: Iodide ions are stronger reducing agents than chloride ions (1)

(b)(i)
M1: Solution turns orange / yellow-orange (1)
M2: \(\text{Cl}_2 + 2\text{Br}^- \to 2\text{Cl}^- + \text{Br}_2\) (1)

(b)(ii)
M1: Chlorine is a stronger oxidizing agent than bromine (1)

(c)
M1: \(\text{Cl}_2 + 2\text{NaOH} \to \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (1)
M2: \(-1\) (for \(\text{NaCl}\)) and \(+1\) (for \(\text{NaClO}\)) (1)

(a) Solubility of the hydroxides increases down the group.

(b) Reagent: Aqueous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) or sulfuric acid (\(\text{H}_2\text{SO}_4\)).
Observation with \(\text{Ba}^{2+}\): A thick white precipitate forms (of \(\text{BaSO}_4\)).
Observation with \(\text{Mg}^{2+}\): Solution remains colourless / no precipitate forms.
(Alternative reagent: \(\text{NaOH}\). Observation with \(\text{Mg}^{2+}\): white precipitate of \(\text{Mg(OH)}_2\); observation with \(\text{Ba}^{2+}\): no precipitate / remains colourless).

(c)(i) \(\text{TiCl}_4 + 2\text{Mg} \to \text{Ti} + 2\text{MgCl}_2\)

(c)(ii) Magnesium acts as a reducing agent / it is an electron donor / loses electrons.

(d)(i) It is used to neutralise acidic soils and raise the pH.

(d)(ii) \(\text{Ba(OH)}_2 + 2\text{HCl} \to \text{BaCl}_2 + 2\text{H}_2\text{O}\)

(e) It is used as an antacid (to treat indigestion by neutralising excess stomach acid). It is safe to use because it is very weakly alkaline and has very low solubility, so it does not irritate or damage the stomach lining or esophagus.

(a)
M1: Solubility increases down the group (1)

(b)
M1: Reagent: Aqueous sodium sulfate / sulfuric acid (or sodium hydroxide) (1)
M2: \(\text{Ba}^{2+}\) observation: white precipitate (or no precipitate with NaOH) (1)
M3: \(\text{Mg}^{2+}\) observation: no precipitate / remains colourless (or white precipitate with NaOH) (1)

(c)(i)
M1: \(\text{TiCl}_4 + 2\text{Mg} \to \text{Ti} + 2\text{MgCl}_2\) (1)

(c)(ii)
M1: Reducing agent / electron donor / loses electrons (1)

(d)(i)
M1: To neutralise acidic soil / adjust soil pH (1)

(d)(ii)
M1: \(\text{Ba(OH)}_2 + 2\text{HCl} \to \text{BaCl}_2 + 2\text{H}_2\text{O}\) (1)

(e)
M1: Antacid / neutralise stomach acid / indigestion remedy (1)
M2: It is weakly alkaline / relatively insoluble so does not cause harm/tissue damage (1)

(a) Temperature change, \(\Delta T = 71.8 - 21.3 = 50.5\text{ K}\) (or \(^{\circ}\text{C}\))
Using \(q = mc\Delta T\):
\(q = 250.0 \times 4.18 \times 50.5 = 52772.5\text{ J} = 52.8\text{ kJ}\) (accept 52.77 kJ to 52.8 kJ).

(b) Moles of X burned: \(n = \frac{52.7725\text{ kJ}}{3984\text{ kJ mol}^{-1}} = 0.013246\text{ mol}\).
Relative molecular mass: \(M_r = \frac{\text{mass}}{n} = \frac{1.35\text{ g}}{0.013246\text{ mol}} = 101.9\text{ g mol}^{-1}\) (accept 101.8 - 102.0).

(c) General formula for a saturated monohydric alcohol is \(C_n H_{2n+1}OH\) or \(C_n H_{2n+2}O\).
\(12.0n + 1.0(2n + 2) + 16.0 = 101.9\)
\(14n + 18 = 101.9 \implies 14n = 83.9 \implies n = 6.0\).
Molecular formula: \(C_6H_{14}O\) (or \(C_6H_{13}OH\)).
Since it is a straight-chain alcohol (and primary is implied by standard combustion questions), the IUPAC name is hexan-1-ol (accept hexanol).

(a) [2 marks]
1 mark for correct calculation of temperature change (50.5 K).
1 mark for correct calculation of heat energy (52.8 kJ or 52.77 kJ). (Consequent on temperature change).

(b) [3 marks]
1 mark for calculating moles of X (0.0132 mol).
1 mark for the correct method to calculate Mr (mass / moles).
1 mark for the final Mr value of 101.9 (range 101.8 - 102.0) given to 1 d.p.

(c) [2.77 marks]
1 mark for showing how n is derived (e.g., 14n + 18 = 101.9 or showing 101.9 - 18 = 83.9).
1 mark for the correct molecular formula (C6H14O or C6H13OH).
0.77 marks for the correct IUPAC name (hexan-1-ol / hexanol).

(a) A graph of fraction (or number) of molecules with energy \(E\) (y-axis) against Energy, \(E\) (x-axis).
The curve must start at the origin \((0,0)\), rise to a single peak, and decay gradually to the right, approaching but never reaching the x-axis (asymptote).
- \(E_{mp}\) (most probable energy) is the energy value directly below the peak.
- \(E_{mean}\) (mean energy) is slightly to the right of the peak.
- \(E_a\) (activation energy) is a point on the far right of the curve.

(b) At a higher temperature, \(T_2\):
- The peak of the curve is lower and shifts to the right.
- The curve starts at the origin and crosses the \(T_1\) curve once, remaining higher than the \(T_1\) curve at higher energies.
- The area under the curve to the right of the activation energy (\(E_a\)) is significantly larger, meaning a much larger fraction of molecules have energy greater than or equal to the activation energy.
- Consequently, there are more successful/effective collisions per unit time, which increases the rate of reaction.

(a) [3.77 marks]
1 mark for correctly labeled axes (y: fraction/number of molecules; x: energy/E).
1 mark for correct shape starting at origin and not touching x-axis at high energy.
1.77 marks for correct relative positions of Emp (under peak), Emean (right of peak), and Ea (further right).

(b) [4 marks]
1 mark for describing the shift of the T2 peak (lower and to the right).
1 mark for stating that the area under the curve representing molecules with energy \(\ge E_a\) is larger.
1 mark for explaining that a greater fraction of molecules now have energy \(\ge E_a\).
1 mark for linking this to more successful collisions per unit time (or higher frequency of successful collisions).

(a) \(K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}\)
Since there are equal numbers of moles on both sides of the equation (2 moles of reactants and 2 moles of products), the volume terms (\(V\)) in the concentration expressions cancel out in the fraction.

(b) Using ICE method to find equilibrium moles:
Reactants/Products: \(CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O\)
Initial moles: 1.50, 1.80, 0, 0
Change in moles: -0.95, -0.95, +0.95, +0.95
Equilibrium moles: 0.55, 0.85, 0.95, 0.95

Substitute these moles directly into the \(K_c\) expression:
\(K_c = \frac{0.95 \times 0.95}{0.55 \times 0.85} = \frac{0.9025}{0.4675} = 1.93\) (accept 1.9).
Units: None (they cancel).

(c) Catalyst has no effect on the value of \(K_c\) (it only increases the rate at which equilibrium is reached).

(a) [2.77 marks]
1 mark for correct expression of Kc.
1.77 marks for explaining that there are equal moles on both sides, hence the volume terms (V) cancel out.

(b) [4 marks]
1 mark for equilibrium moles of ester and water both equal to 0.95.
1 mark for equilibrium moles of ethanoic acid = 0.55 and ethanol = 0.85.
1 mark for correct substitution into the Kc expression.
1 mark for correct Kc value of 1.93 (or 1.9). Reject any units.

(c) [1 mark]
1 mark for stating no effect / unchanged.

(a)
- Initiation step: \(Cl_2 \xrightarrow{UV} 2Cl\cdot\)
- Propagation step 1: \(CH_3CH_2CH_2CH_3 + Cl\cdot \rightarrow CH_3CH_2\dot{C}HCH_3 + HCl\) (major secondary butyl radical is formed)
- Propagation step 2: \(CH_3CH_2\dot{C}HCH_3 + Cl_2 \rightarrow CH_3CH_2CH(Cl)CH_3 + Cl\cdot\)

(b)
- Termination step: \(2CH_3CH_2\dot{C}HCH_3 \rightarrow CH_3CH_2CH(CH_3)CH(CH_3)CH_2CH_3\) (or written as \(2C_4H_9\cdot \rightarrow C_8H_{18}\))
- Name of hydrocarbon: 3,4-dimethylhexane.

(c) Further substitution reactions can occur on the product (e.g. producing di-, tri- substituted halogenoalkanes, such as dichlorobutanes), and isomer mixtures are produced (e.g. 1-chlorobutane). This results in low yields of the desired monochloroalkane and requires separation.

(a) [3 marks]
1 mark for correct initiation equation.
1 mark for correct propagation step 1 (must show radical on carbon-2).
1 mark for correct propagation step 2.
Accept structural, skeletal, or molecular formulas for radicals, but radicals must be clearly shown with dot.

(b) [2.77 marks]
1.77 marks for correct termination equation showing dimerization of secondary butyl radicals to form 3,4-dimethylhexane.
1 mark for correct IUPAC name: 3,4-dimethylhexane.

(c) [2 marks]
1 mark for mentioning further substitution/multi-substitution can occur (forming di/tri-halo compounds).
1 mark for noting that a mixture of isomers/products is formed, making separation necessary.

(a) Major product: 2-bromopropane.

Mechanism (Electrophilic Addition):
1. \(HBr\) has a permanent dipole shown as \(H^{\delta+} - Br^{\delta-}\).
2. A curly arrow is drawn from the double bond (\(C=C\)) of propene to the \(H\) atom of \(H-Br\).
3. Simultaneously, a curly arrow is drawn from the \(H-Br\) bond to the \(Br\) atom.
4. This forms a secondary carbocation intermediate: \(CH_3\overset{+}{C}HCH_3\) and a bromide ion, \(Br^-\), which has a lone pair.
5. A curly arrow is drawn from the lone pair on the \(Br^-\)-ion to the positively charged carbon atom of the carbocation to form 2-bromopropane.

(b) The major product is formed via the secondary carbocation intermediate, while the minor product (1-bromopropane) is formed via a primary carbocation (\(CH_3CH_2CH_2^+\)).
- The secondary carbocation is more stable than the primary carbocation.
- This is because the secondary carbocation has two electron-releasing methyl groups (alkyl groups) that reduce the positive charge on the carbon atom via the positive inductive effect, whereas the primary carbocation has only one alkyl group.

(a) [4.77 marks]
1 mark for identifying 2-bromopropane as the major product.
1 mark for correct curly arrow from C=C bond to H of HBr and showing dipole \(H^{\delta+} - Br^{\delta-}\).
1 mark for correct curly arrow from H-Br bond to Br.
1 mark for the correct structure of the secondary carbocation intermediate (\(CH_3\overset{+}{C}HCH_3\)).
0.77 marks for correct curly arrow from lone pair of \(Br^-\)
to \(C^+\) of carbocation.

(b) [3 marks]
1 mark for identifying that the major product goes via a secondary carbocation and the minor via a primary carbocation.
1 mark for stating that the secondary carbocation is more stable.
1 mark for explaining stability in terms of the positive inductive effect / electron-releasing nature of two alkyl groups versus one.

(a) Under ethanolic conditions, elimination occurs:
- Organic product structure: \((CH_3)_2C=CH_2\) (or skeletal equivalent).
- IUPAC name: 2-methylpropene (accept methylpropene).
- Role of hydroxide ion: Base (accepts a proton / proton acceptor).

(b) Mechanism of \(S_N1\) reaction:
- Step 1: The C-Br bond breaks heterolytically. A curly arrow is drawn from the C-Br bond to the Br atom. This forms a tertiary carbocation intermediate, \((CH_3)_3C^+\), and a bromide ion (\(Br^-\)).
- Step 2: The hydroxide ion attacks the carbocation. A curly arrow is drawn from the lone pair of the oxygen in the \(OH^-\)
ion to the positively charged carbon atom of the carbocation, forming 2-methylpropan-2-ol.

Reason for \(S_N1\) pathway:
- 2-Bromo-2-methylpropane is a tertiary halogenoalkane. The tertiary carbocation intermediate is highly stable due to the positive inductive effect of three electron-donating methyl groups.
- Steric hindrance around the tertiary carbon prevents the nucleophile from attacking directly from the backside (which is required for an \(S_N2\) mechanism).

(a) [3.77 marks]
1 mark for drawing the correct structural formula of 2-methylpropene.
1 mark for the correct name (2-methylpropene or methylpropene).
1.77 marks for stating that the role of hydroxide is a base.

(b) [4 marks]
1 mark for correct curly arrow from C-Br bond to Br in 2-bromo-2-methylpropane.
1 mark for correct structure of the tertiary carbocation intermediate.
1 mark for correct curly arrow from lone pair on \(OH^-\)
to \(C^+\) of the carbocation.
1 mark for explaining that the tertiary carbocation is stabilized by the positive inductive effect of three alkyl groups and steric hindrance prevents the SN2 backside attack.

(a) Oxidation of pentan-2-ol (a secondary alcohol) yields a ketone:
Equation: \(CH_3CH(OH)CH_2CH_2CH_3 + [O] \rightarrow CH_3COCH_2CH_2CH_3 + H_2O\)
- Organic product: pentan-2-one
- Colour change: Orange to Green (due to reduction of \(Cr_2O_7^{2-}\) to \(Cr^{3+}\))

(b) (i) The dehydration of pentan-2-ol involves elimination of water, forming a C=C double bond. This can occur between C1 and C2, or between C2 and C3.
Three isomers formed:
1. Pent-1-ene: \(CH_2=CHCH_2CH_2CH_3\)
2. (E)-pent-2-ene (trans-pent-2-ene)
3. (Z)-pent-2-ene (cis-pent-2-ene)

(ii) Explanation: Elimination of H can occur from either C1 or C3 adjacent to the C2 hydroxyl carbon. This produces pent-1-ene and pent-2-ene. Pent-2-ene exhibits stereoisomerism (E/Z isomerism) because there is restricted rotation about the C=C double bond, and each carbon atom of the C=C double bond is attached to two different groups (\(-H\) and \(-CH_3\) on C2; \(-H\) and \(-CH_2CH_3\) on C3).

(a) [3.77 marks]
1.77 marks for correct balanced equation (including [O] and H2O, formulas can be structural or molecular).
1 mark for identifying the product as pentan-2-one.
1 mark for stating the colour change is orange to green.

(b) (i) [3 marks]
1 mark for drawing/identifying pent-1-ene.
1 mark for drawing/identifying (E)-pent-2-ene (showing correct geometry or distinct label).
1 mark for drawing/identifying (Z)-pent-2-ene (showing correct geometry or distinct label).

(b) (ii) [1 mark]
1 mark for explaining that H can be lost from C1 or C3, and pent-2-ene exists as E/Z stereoisomers due to restricted C=C rotation and different groups on double-bonded carbons.

(a) Identification of Y: Propan-1-ol.
Justification:
- The broad absorption band at \(3350\text{ cm}^{-1}\) is characteristic of an \(O-H\) alcohol bond (standard range: \(3230 - 3550\text{ cm}^{-1}\)).
- The lack of a peak between \(1680\text{ cm}^{-1}\) and \(1750\text{ cm}^{-1}\) indicates the absence of a \(C=O\) (carbonyl) group. This rules out propanal, propanone, and propanoic acid, leaving propan-1-ol as the only possible compound.

(b) Formula of propan-1-ol: \(C_3H_8O\).
- Molecular mass (\(M_r\)) of \(C_3H_8O\) = \(3 \times 12.0 + 8 \times 1.0 + 16.0 = 60.0\).
- The m/z of the molecular ion (\(M^+\)) is therefore 60.
- The fragment ion responsible for the peak at m/z = 31 is \([CH_2OH]^+\) (or \(CH_2OH^+\)). This is formed by cleavage of the C-C bond adjacent to the oxygen atom: \(CH_3CH_2-CH_2OH\).

(a) [3.77 marks]
1 mark for identifying Y as propan-1-ol.
1 mark for assigning 3350 cm^-1 to the O-H alcohol bond.
1.77 marks for stating that the lack of peak at 1680-1750 cm^-1 means no C=O bond is present, which rules out the other three compounds.

(b) [4 marks]
2 marks for correct m/z value of the molecular ion (60).
2 marks for the correct formula of the fragment ion ([CH2OH]+ or CH2OH+). Must include the positive charge; lose 1 mark if charge is missing.

(a)(i) Nucleophilic substitution.

(a)(ii)
- M1: Curly arrow from the lone pair on the oxygen atom of the hydroxide ion \(\text{(:OH}^-\text{)}\) to the carbon atom attached to the bromine atom.
- M2: Correct dipole on the \(\text{C}-\text{Br}\) bond showing \(\text{C}^{\delta+}\) and \(\text{Br}^{\delta-}\).
- M3: Curly arrow from the \(\text{C}-\text{Br}\) bond to the bromine atom.
- M4: Correct structures of the organic product (butan-1-ol) and the leaving bromide ion (\(\text{Br}^-\)).

(b)(i) Base (or proton acceptor).

(b)(ii)
- M1: Correct skeletal drawings of (E)-but-2-ene (trans) and (Z)-but-2-ene (cis):
- (E)-but-2-ene shows the two methyl groups on opposite sides of the double bond.
- (Z)-but-2-ene shows the two methyl groups on the same side of the double bond.
- M2: State the type of stereoisomerism as E-Z isomerism / geometric isomerism.

(a)(i)
- 1 mark: Nucleophilic substitution (reject electrophilic substitution / nucleophilic elimination).

(a)(ii)
- 1 mark: Curly arrow from lone pair on \(\text{OH}^-\)\ to the \(\text{C}^{\delta+}\) atom.
- 1 mark: Correct \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) dipole.
- 1 mark: Curly arrow from the \(\text{C}-\text{Br}\) bond to the \(\text{Br}\) atom.
- 1 mark: Correct products: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) and \(\text{Br}^-\).

(b)(i)
- 1 mark: Base / proton acceptor / H⁺ acceptor (reject nucleophile).

(b)(ii)
- 1 mark: Correct skeletal structures of both E-but-2-ene and Z-but-2-ene.
- 1 mark: E-Z isomerism / geometric isomerism (accept cis-trans isomerism).

(a) \( \Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants}) \)
\( \Delta S^\ominus = (26.9 + 213.6) - 65.7 = 174.8\text{ J K}^{-1}\text{ mol}^{-1} \).

(b) Feasibility is achieved when \( \Delta G \le 0 \). At the threshold temperature, \( \Delta G = 0 \), so \( T = \frac{\Delta H^\ominus}{\Delta S^\ominus} \).
Converting \( \Delta H^\ominus \) to \( \text{J mol}^{-1} \): \( 117 \times 10^3\text{ J mol}^{-1} \).
\( T = \frac{117000}{174.8} = 669.3\text{ K} \) (or 669 K).

(c) Increasing the temperature increases the feasibility of this reaction (the reaction becomes more feasible / \( \Delta G \) becomes more negative).
This is because the entropy change (\( \Delta S^\ominus \)) is positive. Therefore, the \( -T\Delta S \) term becomes more negative as \( T \) increases, eventually outweighing the positive \( \Delta H \) term.

(a) M1: Correct mathematical expression shown (1 mark)
M2: Correct calculated value of 174.8 (accept 175) with correct units (1 mark)

(b) M1: State or use \( \Delta G = 0 \) or rearrange equation to \( T = \frac{\Delta H}{\Delta S} \) (1 mark)
M2: Conversion of \( \Delta H \) to \( 117000\text{ J mol}^{-1} \) (1 mark)
M3: Correct temperature of 669 K or 669.3 K (1 mark)

(c) M1: Feasibility increases / \( \Delta G \) becomes more negative as temperature increases (1 mark)
M2: Reference to the positive sign of \( \Delta S \) (1 mark)
M3: Explanation that the negative \( -T\Delta S \) term outweighs the positive \( \Delta H \) term at higher temperatures (1 mark)

(a) \( \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \) or equivalent, e.g., \( \ln k = -\frac{E_a}{RT} + \ln A \).

(b) Let's use \( \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \):
\( \ln\left(\frac{4.98 \times 10^{-4}}{3.46 \times 10^{-5}}\right) = \ln(14.393) = 2.667 \)
\( \left(\frac{1}{298} - \frac{1}{318}\right) = 0.003356 - 0.003145 = 2.110 \times 10^{-4}\text{ K}^{-1} \)
\( 2.667 = \frac{E_a}{8.31} \times 2.110 \times 10^{-4} \)
\( E_a = \frac{2.667 \times 8.31}{2.110 \times 10^{-4}} = 105025\text{ J mol}^{-1} \)
Converting to \( \text{kJ mol}^{-1} \): \( E_a = 105\text{ kJ mol}^{-1} \) (accept range 104 - 106 due to rounding).

(c) A small increase in temperature leads to a significantly larger proportion of reactant molecules having kinetic energy greater than or equal to the activation energy (\( E \ge E_a \)). This greatly increases the frequency of successful collisions.

(a) M1: Correct Arrhenius equation written in logarithmic form (1 mark)

(b) M1: Correct calculation of \( \ln(k_2/k_1) = 2.67 \) (1 mark)
M2: Correct calculation of \( (1/T_1 - 1/T_2) = 2.11 \times 10^{-4}\text{ K}^{-1} \) (1 mark)
M3: Correct algebraic rearrangement showing \( E_a \) as the subject (1 mark)
M4: Calculated value of \( E_a \) in \( \text{J mol}^{-1} \) (105000) (1 mark)
M5: Division by 1000 to express final answer as 105 (or 104-106) \( \text{kJ mol}^{-1} \) (1 mark)

(c) M1: Greater proportion/fraction of molecules have energy \( \ge E_a \) (1 mark)
M2: Leading to a significantly greater frequency of successful/effective collisions (1 mark)

(a) \( K_p = \frac{p(CH_3OH)}{p(CO) \times p(H_2)^2} \)

(b) From the stoichiometry:
Change in \( CO = -0.60\text{ mol} \), so equilibrium \( CO = 1.50 - 0.60 = 0.90\text{ mol} \).
Change in \( H_2 = -2 \times 0.60 = -1.20\text{ mol} \), so equilibrium \( H_2 = 3.00 - 1.20 = 1.80\text{ mol} \).

(c) Total moles at equilibrium \( = 0.90 + 1.80 + 0.60 = 3.30\text{ mol} \).
Partial pressures:
\( p(CO) = \frac{0.90}{3.30} \times 450 = 122.7\text{ kPa} \) (or 123 kPa)
\( p(H_2) = \frac{1.80}{3.30} \times 450 = 245.5\text{ kPa} \) (or 245 kPa)
\( p(CH_3OH) = \frac{0.60}{3.30} \times 450 = 81.8\text{ kPa} \) (or 82 kPa)

(d) \( K_p = \frac{81.8}{122.7 \times (245.5)^2} = \frac{81.8}{7.395 \times 10^6} = 1.11 \times 10^{-5}\text{ kPa}^{-2} \).
Units: \( \text{kPa}^{-2} \).

(a) M1: Correct expression for \( K_p \) using partial pressures (use of square brackets is penalised) (1 mark)

(b) M1: Equilibrium amount of \( CO = 0.90\text{ mol} \) (1 mark)
M2: Equilibrium amount of \( H_2 = 1.80\text{ mol} \) (1 mark)

(c) M1: Total moles calculated as \( 3.30\text{ mol} \) (1 mark)
M2: Correct mole fractions calculated for each gas (1 mark)
M3: Correct calculation of all three partial pressures: \( p(CO) = 123\text{ kPa} \), \( p(H_2) = 245\text{ kPa} \), \( p(CH_3OH) = 82\text{ kPa} \) (allow ecf from b) (1 mark)

(d) M1: Correct calculation of \( K_p = 1.11 \times 10^{-5} \) (allow ecf from c; accept range 1.09 - 1.12 x 10^-5) (1 mark)
M2: Correct units of \( \text{kPa}^{-2} \) (1 mark)

(a) A Brønsted-Lowry acid is a proton (or \( H^+ \)) donor.

(b) \( HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq) \) (Accept with \( H_2O \) forming \( H_3O^+ \)).

(c) Calculate moles of \( HCOOH \):
\( \text{moles} = 0.0400\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 6.00 \times 10^{-3}\text{ mol} \).
Calculate moles of \( HCOO^- \):
\( \text{moles} = 0.0600\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 6.00 \times 10^{-3}\text{ mol} \).
Total volume of the buffer is \( 100.0\text{ cm}^3 \).
Since both species are in the same volume, we can use their moles directly in the buffer equation:
\( [H^+] = K_a \times \frac{[HCOOH]}{[HCOO^-]} = 1.78 \times 10^{-4} \times \left(\frac{6.00 \times 10^{-3}}{6.00 \times 10^{-3}}\right) = 1.78 \times 10^{-4}\text{ mol dm}^{-3} \).
Calculate pH:
\( pH = -\log_{10}[H^+] = -\log_{10}(1.78 \times 10^{-4}) = 3.75 \).

(d) The pH will remain almost constant.

(a) M1: Proton / \( H^+ \) donor (1 mark)

(b) M1: Correct reversible dissociation equation with appropriate state symbols (1 mark)

(c) M1: Calculate moles of \( HCOOH = 6.00 \times 10^{-3}\text{ mol} \) (1 mark)
M2: Calculate moles of \( HCOO^- = 6.00 \times 10^{-3}\text{ mol} \) (1 mark)
M3: State or use buffer expression rearranged for \( [H^+] \) (1 mark)
M4: Determine that \( [H^+] = 1.78 \times 10^{-4}\text{ mol dm}^{-3} \) (1 mark)
M5: Calculate pH = 3.75 (accept 3.8) (1 mark)

(d) M1: Remain almost constant / negligibly change (1 mark)

(a) The oxidation half-cell (lower \( E^\ominus \)) is on the left:
\( Pt(s) \mid Fe^{2+}(aq), Fe^{3+}(aq) \parallel MnO_4^-(aq), H^+(aq), Mn^{2+}(aq) \mid Pt(s) \).

(b) \( E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +1.51 - (+0.77) = +0.74\text{ V} \).

(c) The spontaneous reaction is:
\( 5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O \).

(d) Yes, chloride ions can be oxidized by acidified manganate(VII).
- \( E^\ominus(MnO_4^-/Mn^{2+}) = +1.51\text{ V} \) is more positive than \( E^\ominus(Cl_2/Cl^-) = +1.36\text{ V} \).
- This means \( MnO_4^- \) is a stronger oxidizing agent than \( Cl_2 \).
- The cell EMF for this oxidation reaction is positive (\( +1.51 - 1.36 = +0.15\text{ V} \)), making the reaction thermodynamically feasible.

(a) M1: Correctly show left-hand electrode as Pt and correct order of iron species (1 mark)
M2: Correctly show right-hand electrode as Pt, include acid/species, and use salt bridge symbol \( \parallel \) (1 mark)

(b) M1: Correct calculation of cell EMF = +0.74 V (or 0.74 V) (1 mark)

(c) M1: Correct reactants and products (1 mark)
M2: Equation fully balanced (1 mark)

(d) M1: Yes, oxidation is feasible, AND state that \( E^\ominus(MnO_4^-) \) is more positive than \( E^\ominus(Cl_2/Cl^-) \) (1 mark)
M2: Identify \( MnO_4^- \) as the stronger oxidizing agent / identify that the cell potential is positive (1 mark)
M3: Conclude that the oxidation reaction is thermodynamically feasible (1 mark)

(a) Shape: octahedral; bond angle: \( 90^\circ \) (accept \( 180^\circ \)); color: blue.

(b) (i) \( [Cu(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CuCl_4]^{2-} + 6H_2O \).

(ii) Chloride ligands (\( Cl^- \)) are larger than water molecules (\( H_2O \)). Therefore, only four chloride ligands can fit around the central \( Cu^{2+} \) ion due to steric hindrance / mutual repulsion.

(iii) Tetrahedral.

(a) M1: Shape = octahedral (1 mark)
M2: Bond angle = \( 90^\circ \) (1 mark)
M3: Color = blue (1 mark)

(b) (i) M1: Correct formulas for reactants and products (1 mark)
M2: Correctly balanced equation (1 mark)

(ii) M1: Chloride ligands are larger than water ligands (1 mark)
M2: Less space around the central copper ion / steric hindrance prevents six chloride ligands fitting (1 mark)

(iii) M1: Tetrahedral (1 mark)

(a) Sodium oxide (\( Na_2O \)): ionic bonding and giant ionic lattice/structure. Sulfur dioxide (\( SO_2 \)): covalent bonding and simple molecular structure.

(b) \( Na_2O(s) + H_2O(l) \rightarrow 2NaOH(aq) \) (or ionic: \( Na_2O + H_2O \rightarrow 2Na^+ + 2OH^- \)). Approximate pH: 13 or 14.

(c) (i) \( Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O \) (or ionic: \( Al_2O_3 + 6H^+ \rightarrow 2Al^{3+} + 3H_2O \)).

(ii) \( Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4] \) (or ionic: \( Al_2O_3 + 2OH^- + 3H_2O \rightarrow 2[Al(OH)_4]^- \)).

(iii) It has a giant ionic lattice with very strong electrostatic attractions between the highly charged ions (\( Al^{3+} \) and \( O^{2-} \)) which require a massive amount of energy to break.

(a) M1: \( Na_2O \) is giant ionic (1 mark)
M2: \( SO_2 \) is covalent / simple molecular (1 mark)

(b) M1: Correct balanced equation for \( Na_2O \) with water (1 mark)
M2: pH in range 12-14 (1 mark)

(c) (i) M1: Correct balanced equation with \( HCl \) or \( H^+ \) (1 mark)

(ii) M1: Correct formulas for reactants and products (1 mark)
M2: Balanced equation (1 mark)

(iii) M1: Strong electrostatic attractions between ions (\( Al^{3+} \) and \( O^{2-} \)) in giant ionic structure require significant thermal energy to overcome (1 mark)

(a) The \( Fe^{3+} \) ion has a higher charge and a smaller ionic radius than the \( Fe^{2+} \) ion, which gives it a higher charge density.
Therefore, the \( Fe^{3+} \) ion is more polarizing and attracts the electrons in the coordinate-bonded water ligands more strongly.
This weakens the \( O-H \) bonds in the water ligands, facilitating the release of hydrogen ions (\( H^+ \)) into the solution, making it more acidic.

(b) Observation: A green precipitate forms (which slowly turns brown at the surface upon contact with air).
Ionic Equation: \( [Fe(H_2O)_6]^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(H_2O)_4(OH)_2(s) + 2H_2O(l) \).

(c) (i) Gas: Carbon dioxide / \( CO_2 \).
(ii) Equation: \( 2[Fe(H_2O)_6]^{3+}(aq) + 3CO_3^{2-}(aq) \rightarrow 2Fe(H_2O)_3(OH)_3(s) + 3CO_2(g) + 3H_2O(l) \).

(a) M1: \( Fe^{3+} \) has a higher charge density than \( Fe^{2+} \) (or higher charge and smaller size) (1 mark)
M2: \( Fe^{3+} \) is more polarizing (1 mark)
M3: Weakens the \( O-H \) bonds in the ligand, releasing \( H^+ \) ions more easily (1 mark)

(b) M1: Green precipitate (1 mark)
M2: Correct reactants and products in equation (1 mark)
M3: Correct balancing and state symbols: (s) for product and (aq)/(l) (1 mark)

(c) (i) M1: Carbon dioxide / \( CO_2 \) (1 mark)

(ii) M1: Correct balanced ionic equation (1 mark)

(a) First, calculate the initial moles of propanoic acid: \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\). Next, calculate the moles of sodium hydroxide added: \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\). The sodium hydroxide reacts completely with the propanoic acid: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\). Moles of propanoic acid remaining: \(7.50 \times 10^{-3}\text{ mol} - 2.50 \times 10^{-3}\text{ mol} = 5.00 \times 10^{-3}\text{ mol}\). Moles of propanoate ions formed: \(2.50 \times 10^{-3}\text{ mol}\). Using the buffer expression: \([\text{H}^+] = K_a \times \frac{n(\text{acid})}{n(\text{base})} = 1.35 \times 10^{-5} \times \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 2.70 \times 10^{-5}\text{ mol dm}^{-3}\). Finally, calculate the pH: \(\text{pH} = -\log_{10}(2.70 \times 10^{-5}) = 4.57\). (b) When hydrochloric acid is added, the extra \(\text{H}^+\) ions react with the conjugate base (propanoate ions) in the buffer: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\). Because the added \(\text{H}^+\) ions are removed, the ratio of \([\text{acid}]/[\text{conjugate base}]\) changes very little, and the pH remains almost constant.

Part (a): 5 marks total. M1: Moles of propanoic acid = \(7.50 \times 10^{-3}\text{ mol}\) AND moles of NaOH = \(2.50 \times 10^{-3}\text{ mol}\). M2: Moles of remaining propanoic acid = \(5.00 \times 10^{-3}\text{ mol}\). M3: Moles of propanoate ions = \(2.50 \times 10^{-3}\text{ mol}\). M4: Correct substitution into \([\text{H}^+]\) expression yielding \([\text{H}^+] = 2.70 \times 10^{-5}\text{ mol dm}^{-3}\). M5: Correct calculation of pH = 4.57 (accept range 4.56 to 4.57). Part (b): 3 marks total. M1: Added \(\text{H}^+\) ions react with the propanoate ions (conjugate base). M2: Balanced ionic equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\) (accept state symbols, equilibrium sign). M3: Ratio of \([\text{acid}]/[\text{conjugate base}]\) (or concentrations of both species) remains relatively constant.

(a)(i) The \(\text{Fe}^{3+}\) ion has a high charge density (high charge and small ionic radius). This high charge density strongly polarises the electron density in the O-H bonds of the coordinated water ligands. This weakens the O-H bonds, allowing a hydrogen ion (proton) to be released easily to the solvent water molecule. Equation: \([\text{Fe}(\text{H}_2\text{O})_6]^{3+} + \text{H}_2\text{O} \rightleftharpoons [\text{Fe}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}_3\text{O}^+\) (or with \(\text{H}^+\) on the right). (a)(ii) When sodium carbonate is added, a brown (or red-brown) precipitate forms, and there is effervescence (bubbles of gas evolved). Equation: \(2[\text{Fe}(\text{H}_2\text{O})_6]^{3+} + 3\text{CO}_3^{2-} \rightarrow 2\text{Fe}(\text{H}_2\text{O})_3(\text{OH})_3 + 3\text{CO}_2 + 3\text{H}_2\text{O}\). (b) When excess concentrated HCl is added, ligand substitution occurs to form the tetrachloroferrate(III) ion, \([\text{FeCl}_4]^-\). The coordination number changes from 6 to 4, and the shape changes from octahedral to tetrahedral.

Part (a)(i): 3 marks. M1: \(\text{Fe}^{3+}\) has high charge density / high charge-to-size ratio. M2: Polarises / weakens the O-H bond (in coordinated water). M3: Correct equation: \([\text{Fe}(\text{H}_2\text{O})_6]^{3+} + \text{H}_2\text{O} \rightleftharpoons [\text{Fe}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}_3\text{O}^+\) (or \([\text{Fe}(\text{H}_2\text{O})_6]^{3+} \rightleftharpoons [\text{Fe}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}^+\)). Part (a)(ii): 3 marks. M1: Observation 1: Brown precipitate / red-brown solid. M2: Observation 2: Effervescence / bubbles of gas / fizzing. M3: Balanced ionic equation: \(2[\text{Fe}(\text{H}_2\text{O})_6]^{3+} + 3\text{CO}_3^{2-} \rightarrow 2\text{Fe}(\text{H}_2\text{O})_3(\text{OH})_3 + 3\text{CO}_2 + 3\text{H}_2\text{O}\) (accept \(\text{Fe(OH)}_3\) instead of the hydrated precipitate). Part (b): 2 marks. M1: Correct formula: \([\text{FeCl}_4]^-\) (accept \(\text{FeCl}_4^-\)). M2: Coordination number changes from 6 to 4 AND shape changes from octahedral to tetrahedral (both required).

(a) (i)
- Order with respect to A: Compare Experiment 1 and 2. [B] and [C] are kept constant. When [A] is doubled (from 0.10 to 0.20), the rate doubles (from \(1.20 \times 10^{-3}\) to \(2.40 \times 10^{-3}\)). Therefore, the reaction is 1st order with respect to A.
- Order with respect to B: Compare Experiment 1 and 3. [A] and [C] are kept constant. When [B] is doubled (from 0.10 to 0.20), the rate quadruples (from \(1.20 \times 10^{-3}\) to \(4.80 \times 10^{-3}\)). Therefore, the reaction is 2nd order with respect to B.
- Order with respect to C: Compare Experiment 1 and 4. [A] and [B] are kept constant. When [C] is tripled (from 0.10 to 0.30), the rate remains unchanged (\(1.20 \times 10^{-3}\)). Therefore, the reaction is 0th order with respect to C.

(a) (ii)
- Rate equation: \( \text{Rate} = k[A][B]^2 \)
- Using Experiment 1 data to calculate \(k\):
\( 1.20 \times 10^{-3} = k [0.10][0.10]^2 \)
\( 1.20 \times 10^{-3} = k [0.10][0.010] \)
\( 1.20 \times 10^{-3} = k \times 1.00 \times 10^{-3} \)
\( k = 1.20 \)
- Units: \( \text{Rate} / ([\text{A}][\text{B}]^2) = \text{mol dm}^{-3} \text{s}^{-1} / (\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2) = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1} \)

(b)
- The rate equation depends on [X] and [Y] (each first order), which means 1 molecule of X and 1 molecule of Y are involved in the rate-determining step.
- Step 1 (slow/RDS): \( X + Y \rightarrow I \) (where I is an intermediate)
- Step 2 (fast): \( I + X \rightarrow Z \)
- Sum of steps: \( X + Y + I + X \rightarrow I + Z \), which simplifies to \( 2X + Y \rightarrow Z \).

(a) (i)
- 1 mark for order 1 with respect to A and reasoning.
- 1 mark for order 2 with respect to B and reasoning.
- 1 mark for order 0 with respect to C and reasoning.
- 1 mark for linking logic clearly to the specific experiments in each case.

(a) (ii)
- 1 mark for correct rate equation: \( \text{Rate} = k[A][B]^2 \) (allow consequential error from (i)).
- 1 mark for correct value: \( k = 1.20 \) (allow 1.2).
- 1 mark for correct units: \( \text{mol}^{-2} \text{dm}^6 \text{s}^{-1} \).

(b)
- 1 mark for Step 1 showing \( X + Y \rightarrow I \) and labelled as slow / rate-determining step.
- 1 mark for Step 2 showing \( I + X \rightarrow Z \) and labelled as fast.
- 1 mark for showing both equations sum up to the overall equation \( 2X + Y \rightarrow Z \) (or stating this clearly).

(a) (i)
\( K_a = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \)

(a) (ii)
Using the weak acid approximation \( [\text{H}^+] \approx [\text{CH}_3\text{CH}_2\text{COO}^-] \):
\( K_a = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]} \)
\( [\text{H}^+] = \sqrt{K_a \times [\text{CH}_3\text{CH}_2\text{COOH}]} = \sqrt{1.35 \times 10^{-5} \times 0.150} \)
\( [\text{H}^+] = \sqrt{2.025 \times 10^{-6}} = 1.423 \times 10^{-3} \text{ mol dm}^{-3} \)
\( \text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.8468 \approx 2.85 \)

(b) (i)
When a small amount of acid (\(\text{H}^+\)) is added, the large reservoir of conjugate base (propanoate ions) reacts with it to form weak, undissociated propanoic acid:
\( \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH} \)
Because the added \(\text{H}^+\)-ions are removed, the concentration of free \(\text{H}^+\) remains nearly constant, and the pH change is resisted.

(b) (ii)
From the expression of \(K_a\):
\( K_a = [\text{H}^+] \times \frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \)
Rearranging for the ratio:
\( \frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} = \frac{K_a}{[\text{H}^+]} \)
Calculate \([\text{H}^+]\) from pH:
\( [\text{H}^+] = 10^{-\text{pH}} = 10^{-4.80} = 1.585 \times 10^{-5} \text{ mol dm}^{-3} \)
Calculate ratio:
\( \text{Ratio} = \frac{1.35 \times 10^{-5}}{1.585 \times 10^{-5}} = 0.8517 \approx 0.85 \) (or \( 0.852 \))

(a) (i)
- 1 mark for correct expression: \( K_a = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \) (square brackets required, state symbols not needed).

(a) (ii)
- 1 mark for showing \( [\text{H}^+]^2 = K_a \times [\text{HA}] \) or \( [\text{H}^+] = \sqrt{1.35 \times 10^{-5} \times 0.150} \).
- 1 mark for calculating \( [\text{H}^+] = 1.42 \times 10^{-3} \text{ mol dm}^{-3} \).
- 1 mark for pH = 2.85 (must be exactly 2 decimal places).

(b) (i)
- 1 mark for equation: \( \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH} \).
- 1 mark for explaining that the added \(\text{H}^+\) is consumed by the large reservoir of propanoate, maintaining a stable pH.

(b) (ii)
- 1 mark for rearranging to \( \frac{[\text{A}^-]}{[\text{HA}]} = \frac{K_a}{[\text{H}^+]} \).
- 1 mark for calculating \( [\text{H}^+] = 1.585 \times 10^{-5} \text{ mol dm}^{-3} \).
- 1 mark for substituting numbers correctly.
- 1 mark for final ratio of 0.85 or 0.852 (accept answers that round to 0.85).

(a)
- \( K_p = \frac{(p\text{NO}_2)^2}{p\text{N}_2\text{O}_4} \)
- Units: \( \text{kPa}^2 / \text{kPa} = \text{kPa} \)

(b) (i)
- Initial moles: \( n(\text{N}_2\text{O}_4) = 1.00 \text{ mol} \), \( n(\text{NO}_2) = 0.00 \text{ mol} \).
- Since \(40.0\%\) has dissociated, change in \( n(\text{N}_2\text{O}_4) = -0.400 \text{ mol} \).
- Moles of \( \text{N}_2\text{O}_4 \) at equilibrium = \( 1.00 - 0.400 = 0.600 \text{ mol} \).
- Moles of \( \text{NO}_2 \) formed at equilibrium = \( 2 \times 0.400 = 0.800 \text{ mol} \).

(b) (ii)
- Total moles at equilibrium = \( 0.600 + 0.800 = 1.400 \text{ mol} \).
- Mole fraction of \( \text{N}_2\text{O}_4 \): \( x(\text{N}_2\text{O}_4) = \frac{0.600}{1.400} = 0.429 \) (or \( 3/7 \)).
- Mole fraction of \( \text{NO}_2 \): \( x(\text{NO}_2) = \frac{0.800}{1.400} = 0.571 \) (or \( 4/7 \)).

(b) (iii)
- Partial pressure = mole fraction \( \times \) total pressure
- \( p(\text{N}_2\text{O}_4) = 0.4286 \times 150 \text{ kPa} = 64.3 \text{ kPa} \) (or \( 64.29 \text{ kPa} \)).
- \( p(\text{NO}_2) = 0.5714 \times 150 \text{ kPa} = 85.7 \text{ kPa} \) (or \( 85.71 \text{ kPa} \)).

(b) (iv)
- \( K_p = \frac{(85.71)^2}{64.29} = \frac{7346.2}{64.29} = 114.27 \approx 114 \text{ kPa} \) (accept range 114 - 114.3).

(a)
- 1 mark for correct \(K_p\) expression: \( K_p = \frac{(p\text{NO}_2)^2}{p\text{N}_2\text{O}_4} \) (must use lowercase p / partial pressure notation; square brackets not accepted).
- 1 mark for correct units: \( \text{kPa} \).

(b) (i)
- 1 mark for \( n(\text{N}_2\text{O}_4) = 0.600 \text{ mol} \).
- 1 mark for \( n(\text{NO}_2) = 0.800 \text{ mol} \).

(b) (ii)
- 1 mark for calculating total moles = \(1.400 \text{ mol}\).
- 1 mark for both mole fractions correct (0.429 and 0.571, or fractions 3/7 and 4/7).

(b) (iii)
- 1 mark for \( p(\text{N}_2\text{O}_4) = 64.3 \text{ kPa} \) (or consequential to (ii)).
- 1 mark for \( p(\text{NO}_2) = 85.7 \text{ kPa} \) (or consequential to (ii)).

(b) (iv)
- 1 mark for correct substitution into the \(K_p\) formula.
- 1 mark for the final answer of 114 (accept range 114 to 114.3).

(a)
- Reagents: Potassium cyanide (\(\text{KCN}\)) and dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)) (or hydrochloric acid, \(\text{HCl}\)).
- Conditions: Room temperature / aqueous-alcoholic solution.

(b)
Mechanism details:
- The carbonyl group in butanal has a dipole, shown as \(\text{C}^{\delta+}=\text{O}^{\delta-}\).
- A curly arrow starts from the lone pair on the carbon atom of the cyanide ion (\(:\text{CN}^-\)) and points to the carbonyl carbon (\(\text{C}^{\delta+}\)).
- A curly arrow starts from the \(\text{C}=\text{O}\) double bond and points to the oxygen atom.
- This produces an intermediate alkoxide ion: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(O}^-\text{)CN}\).
- A curly arrow starts from a lone pair on the negative oxygen atom (\(\text{O}^-\)) and points to a hydrogen ion (\(\text{H}^+\)) from the acid.
- This forms the product: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)CN}\).

(c) (i)
- Draw the central carbon atom (C2) with tetrahedral bonds (two in the plane of the page, one wedge pointing forward, and one dashed wedge pointing away).
- The groups attached to C2 are: \(\text{-H}\), \(\text{-OH}\), \(\text{-CN}\), and \(\text{-CH}_2\text{CH}_2\text{CH}_3\).
- Draw the second molecule as a mirror image of the first, showing the reflection across a vertical mirror line.

(c) (ii)
- The carbonyl group (or the carbon atom in the carbonyl group) in butanal is planar.
- The cyanide nucleophile (\(\text{CN}^-\)) has an equal probability of attacking this planar group from either side (above or below the plane).
- This results in a racemic mixture (equal amounts of the two enantiomers).
- The optical rotation caused by each enantiomer cancels out, making the mixture optically inactive.

(a)
- 1 mark for reagents: \(\text{KCN}\) and dilute acid (accept \(\text{H}_2\text{SO}_4\) or \(\text{HCl}\)). (Do not accept HCN alone as it is not commonly used due to safety).
- 1 mark for conditions: Room temperature / aqueous / pH around 6 to 8.

(b)
- 1 mark for dipole on \(\text{C}=\text{O}\) and curly arrow from lone pair on carbon of \(:\text{CN}^-\) to the carbonyl carbon.
- 1 mark for curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom.
- 1 mark for the correct structure of the intermediate with a negative charge on the oxygen.
- 1 mark for curly arrow from lone pair on \(\text{O}^-\)\ to \(\text{H}^+\) (or to the H of water/HCN).

(c) (i)
- 1 mark for drawing one tetrahedral 3D structure showing wedges and dashes with correct groups: \(\text{-H}\), \(\text{-OH}\), \(\text{-CN}\), and \(\text{-C}_3\text{H}_7\).
- 1 mark for drawing the mirror image correctly.

(c) (ii)
- 1 mark for stating that the carbonyl group is planar (or flat) around the carbonyl carbon.
- 1 mark for stating that the nucleophile can attack from either side with equal probability, forming a racemic mixture.

(a) (i)
\( \text{C}_6\text{H}_5\text{COOH} + \text{CH}_3\text{OH} \rightleftharpoons \text{C}_6\text{H}_5\text{COOCH}_3 + \text{H}_2\text{O} \)
(reversible reaction)

(a) (ii)
Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) (accept concentrated phosphoric acid).

(b) (i)
\( \text{C}_6\text{H}_5\text{COCl} + \text{CH}_3\text{OH} \rightarrow \text{C}_6\text{H}_5\text{COOCH}_3 + \text{HCl} \)

(b) (ii)
Mechanism:
1. The nucleophile methanol attacks the carbonyl carbon. A curly arrow is drawn from the lone pair on the oxygen of \(\text{CH}_3\text{OH}\) to the carbon of the \(\text{C}=\text{O}\) group. A second curly arrow is drawn from the double bond of \(\text{C}=\text{O}\) to the oxygen atom.
2. This forms a tetrahedral intermediate: \(\text{C}_6\text{H}_5\text{-C(O}^-\text{)(Cl)-O}^+\text{H(CH}_3\text{)}\).
3. Eliminate the chloride ion and proton to reform the carbonyl group. A curly arrow is drawn from the lone pair on \(\text{O}^-\)\ to reform the \(\text{C}=\text{O}\) bond, and a curly arrow is drawn from the \(\text{C-Cl}\) bond to the chlorine atom (releasing \(\text{Cl}^-\)).
4. Deprotonate the oxygen. A curly arrow is drawn from the \(\text{O-H}\) bond to the positive oxygen atom.

(c)
- Advantage of Route 2: The reaction goes to completion (is irreversible) / occurs rapidly at room temperature / gives a higher yield / does not require heating or catalyst.
- Disadvantage of Route 2: It produces toxic/corrosive hydrogen chloride (HCl) gas / benzoyl chloride is more expensive / reacts violently with water.

(a) (i)
- 1 mark for correct equation (reversible arrow is preferred but not strictly required for this mark).

(a) (ii)
- 1 mark for concentrated sulfuric acid (reject dilute sulfuric acid; accept concentrated phosphoric acid).

(b) (i)
- 1 mark for correct equation: \( \text{C}_6\text{H}_5\text{COCl} + \text{CH}_3\text{OH} \rightarrow \text{C}_6\text{H}_5\text{COOCH}_3 + \text{HCl} \).

(b) (ii)
- 1 mark for arrow from lone pair on oxygen of \(\text{CH}_3\text{OH}\) to the carbonyl carbon.
- 1 mark for arrow from \(\text{C}=\text{O}\) bond to oxygen.
- 1 mark for correct structure of the tetrahedral intermediate (must show negative charge on carbonyl oxygen and positive charge on methanol oxygen).
- 1 mark for three simultaneous or sequential arrows: reforming \(\text{C}=\text{O}\), breaking \(\text{C-Cl}\) bond, and breaking \(\text{O-H}\) bond.

(c)
- 1 mark for contrasting the nature of the two reactions (Route 1 is reversible/slow, Route 2 is irreversible/fast).
- 1 mark for any correct advantage of Route 2 (e.g. goes to completion, higher yield, no heating required).
- 1 mark for any correct disadvantage of Route 2 (e.g. toxic/corrosive HCl fumes produced, reactant is moisture sensitive).

(a) (i)
- Temperature between \(50\,^\circ\text{C}\) and \(55\,^\circ\text{C}\) (accept below \(60\,^\circ\text{C}\)).

(a) (ii)
- Equation: \( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+ \)
(accept \( \text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O} \))

(a) (iii)
- Mechanism:
1. Draw a curly arrow from the ring of benzene to the nitrogen of the \(\text{NO}_2^+\) ion.
2. Draw the intermediate structure. This must show a horseshoe shape inside the ring, with the open end facing the carbon bonded to the \(\text{-H}\) and \(\text{-NO}_2\) groups, and a positive charge (+) inside the partial ring.
3. Draw a curly arrow from the C-H bond back into the ring to restore the aromatic system.
4. Show the products: Nitrobenzene and \(\text{H}^+\).

(b) (i)
- Stage 1: Tin (Sn) and concentrated hydrochloric acid (HCl) (refluxed).
- Stage 2: Sodium hydroxide (NaOH) (to liberate the free amine from its salt).

(b) (ii)
- Manufacture of polyurethane/aniline dyes/pharmaceuticals.

(a) (i)
- 1 mark for temperature between \(50\,^\circ\text{C}\) and \(55\,^\circ\text{C}\) (accept any value or range within \(50 - 60\,^\circ\text{C}\)).

(a) (ii)
- 1 mark for correct reactants and products.
- 1 mark for correct balancing and charges.

(a) (iii)
- 1 mark for curly arrow from the circle of benzene to the nitrogen atom of \(\text{NO}_2^+\).
- 1 mark for drawing the intermediate correctly (horseshoe must open towards sp3 carbon, positive charge must be inside the horseshoe, both H and \(\text{NO}_2\) must be drawn on C1).
- 1 mark for arrow from C-H bond to the ring.
- 1 mark for the final products (nitrobenzene and \(\text{H}^+\)).

(b) (i)
- 1 mark for Tin (Sn) and concentrated hydrochloric acid (HCl) (accept Fe and HCl).
- 1 mark for addition of sodium hydroxide / alkali (NaOH).

(b) (ii)
- 1 mark for making azo-dyes / making polyurethanes / making pharmaceuticals / making antioxidants.

(a) (i)
- Kevlar is made by reacting the carboxylic acid groups of benzene-1,4-dicarboxylic acid with the amine groups of benzene-1,4-diamine.
- Structure of repeating unit: \( \text{[-CO-C}_6\text{H}_4\text{-CO-NH-C}_6\text{H}_4\text{-NH-]} \) showing open bonds on each end.

(a) (ii)
- Classification: Polyamide.
- Small molecule: Water (\(\text{H}_2\text{O}\)).

(b)
- Basicity depends on the ability of the lone pair of electrons on the nitrogen atom to accept a proton (\(\text{H}^+\)).
- In methylamine, the methyl group is electron-releasing due to the positive inductive effect (+I effect). This increases electron density on the nitrogen, making the lone pair more available to accept a proton, so it is more basic than ammonia.
- In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring's \(\pi\)-system. This decreases electron density on the nitrogen, making the lone pair less available to accept a proton, so it is less basic than ammonia.
- Basicity order: methylamine > ammonia > phenylamine.

(c) (i)
- Zwitterion of glycine (\(\text{H}_2\text{NCH}_2\text{COOH}\)): \( \text{H}_3\text{N}^+\text{-CH}_2\text{-COO}^- \).

(c) (ii)
- The zwitterion has ionic charges which allow strong electrostatic attractions (ionic bonding) between different molecules in the solid lattice. These require much more energy to overcome than the intermolecular forces (such as hydrogen bonding) in typical neutral organic molecules of similar mass.

(a) (i)
- 1 mark for correct amide linkage (\(\text{-CO-NH-}\)).
- 1 mark for the correct aromatic rings linked at position 1,4 (para-positions) with open bonds at both ends of the unit.

(a) (ii)
- 1 mark for polyamide.
- 1 mark for water / \(\text{H}_2\text{O}\).

(b)
- 1 mark for stating that basicity is determined by the availability of the lone pair of electrons on the nitrogen atom.
- 1 mark for explaining that the methyl group has a positive inductive effect which increases the lone pair availability (making methylamine more basic than ammonia).
- 1 mark for explaining that the lone pair in phenylamine is delocalised into the benzene ring, decreasing its availability (making phenylamine less basic than ammonia).
- 1 mark for giving the correct order: methylamine > ammonia > phenylamine.

(c) (i)
- 1 mark for correct structure: \( \text{H}_3\text{N}^+\text{-CH}_2\text{-COO}^- \) (must show charges on correct atoms).

(c) (ii)
- 1 mark for explaining that there are strong electrostatic attractions / ionic bonds between zwitterions in the lattice.

(a)
- The sharp peak at \(1740\text{ cm}^{-1}\) is due to a carbonyl group (\(\text{C}=\text{O}\) stretch), which is found in esters, aldehydes, ketones, and carboxylic acids.
- Since there is no broad absorption in \(2500-3000\text{ cm}^{-1}\), the carboxylic acid \(\text{O-H}\) group is absent.
- Since there is no broad absorption in \(3200-3600\text{ cm}^{-1}\), the alcohol \(\text{O-H}\) group is absent.
- Since the formula is \(\text{C}_4\text{H}_8\text{O}_2\) and it has two oxygen atoms but no \(\text{-OH}\) groups, it must be an ester.

(b) (i)
- Integration values (3 : 2 : 3) indicate the relative ratio of hydrogen atoms in each chemical environment (representing \(3\text{H}\), \(2\text{H}\), and \(3\text{H}\) respectively).

(b) (ii)
- The triplet (3H) and quartet (2H) pairing indicates the presence of an ethyl group (\(\text{-CH}_2\text{CH}_3\)).
- According to the \(n+1\) rule, the triplet at \(\delta = 1.15\text{ ppm}\) (\(\text{-CH}_3\)) arises because it has \(n=2\) adjacent protons (the \(\text{-CH}_2\text{-}\) group).
- The quartet at \(\delta = 2.30\text{ ppm}\) (\(\text{-CH}_2\text{-}\)) arises because it has \(n=3\) adjacent protons (the \(\text{-CH}_3\) group).

(b) (iii)
- The singlet at \(\delta = 3.65\text{ ppm}\) (3H) represents a methyl group attached to an oxygen (\(\text{-OCH}_3\)) because of its high downfield chemical shift.
- The quartet at \(\delta = 2.30\text{ ppm}\) represents a \(\text{-CH}_2\text{-}\) group adjacent to the carbonyl carbon (\(\text{-CH}_2\text{CO-}\)).
- Combining these pieces: \(\text{CH}_3\text{CH}_2\text{COOCH}_3\).
- IUPAC name: Methyl propanoate.

(b) (iv)
- Number of peaks in \({}^{13}\text{C}\) NMR: 4 peaks.
- Justification: Methyl propanoate has 4 carbon atoms, and each is in a unique chemical environment, so each carbon produces a separate peak.

(a)
- 1 mark for identifying the ester group from the \(\text{C}=\text{O}\) stretch at \(1740\text{ cm}^{-1}\) and absence of \(\text{O-H}\).
- 1 mark for stating that carboxylic acids are absent due to the lack of \(\text{O-H}\) stretch at \(2500-3000\text{ cm}^{-1}\).
- 1 mark for stating that alcohols are absent due to the lack of \(\text{O-H}\) stretch at \(3200-3600\text{ cm}^{-1}\).

(b) (i)
- 1 mark for explaining that integration shows the relative number of protons in each environment (3H, 2H, and 3H).

(b) (ii)
- 1 mark for identifying the ethyl group (\(\text{-CH}_2\text{CH}_3\)).
- 1 mark for explaining the splitting using the \(n+1\) rule (triplet split by 2 neighbouring protons, quartet split by 3 neighbouring protons).

(b) (iii)
- 1 mark for drawing/writing correct structure: \(\text{CH}_3\text{CH}_2\text{COOCH}_3\).
- 1 mark for correct IUPAC name: Methyl propanoate.

(b) (iv)
- 1 mark for 4 peaks.
- 1 mark for stating that there are 4 different carbon environments.

(a) Dilute sulfuric acid is used to prevent the hydrolysis of \(\text{Fe}^{2+}\) ions / to prevent the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) by atmospheric oxygen.

(b) Steps to prepare the solution accurately:
1. Dissolve the solid in a beaker using a volume of acid/water that is less than \(250\text{ cm}^3\), stirring with a glass rod.
2. Transfer the solution quantitatively into the volumetric flask, rinsing the beaker and glass rod with distilled water and adding the washings to the flask.
3. Fill the flask with distilled water until the bottom of the meniscus is on the graduation mark, then stopper and invert the flask several times to ensure thorough mixing.

(c) The ionic equation is:
\(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)

(d) Calculations:
Moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = 0.0150 \times \frac{14.50}{1000} = 2.175 \times 10^{-4}\text{ mol}\)
Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\):
\(n(\text{Fe}^{2+}) = 5 \times 2.175 \times 10^{-4} = 1.0875 \times 10^{-3}\text{ mol}\)
Moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\) (and hence in \(3.02\text{ g}\)):
\(1.0875 \times 10^{-2}\text{ mol}\)
Mass of anhydrous \(\text{FeSO}_4\):
\(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (16.0 \times 4) = 151.9\text{ g mol}^{-1}\)
\(m(\text{FeSO}_4) = 1.0875 \times 10^{-2} \times 151.9 = 1.652\text{ g}\)
Mass of water of crystallisation:
\(3.02 - 1.652 = 1.368\text{ g}\)
Moles of water:
\(n(\text{H}_2\text{O}) = \frac{1.368}{18.0} = 0.0760\text{ mol}\)
Ratio \(x = \frac{0.0760}{0.010875} = 6.9885 \approx 6.99\)

(e) Since weighing by difference involves two separate readings, the total uncertainty is \(2 \times 0.01 = 0.02\text{ g}\).
Percentage uncertainty = \(\frac{0.02}{3.02} \times 100 = 0.662\%\) (or \(0.66\%\)).

(a) [1 mark] - For stating that it prevents oxidation of iron(II) to iron(III) OR prevents hydrolysis/precipitation of iron hydroxides.
(b) [3 marks] - 1 mark for dissolving the solid in a beaker with a volume less than 250 cm3 and transferring washings/rinsings; 1 mark for making up to the mark so the bottom of the meniscus is on the line; 1 mark for inverting/shaking to mix.
(c) [1 mark] - Correct balanced ionic equation: \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\).
(d) [4 marks] - 1 mark for calculating moles of manganate(VII) (\(2.175 \times 10^{-4}\text{ mol}\)); 1 mark for scaling up to 250 cm3 to get moles of iron(II) (\(1.0875 \times 10^{-2}\text{ mol}\)); 1 mark for finding mass of water (\(1.368\text{ g}\)) and moles of water (\(0.0760\text{ mol}\)); 1 mark for calculating \(x = 6.99\) (accept 7 or 7.00 if correct working shown).
(e) [1 mark] - For \(0.66\%\) (accept \(0.33\%\) only if explicitly justified by a single-weighing assumption).

(a) Sodium thiosulfate reacts rapidly with any iodine produced, converting it back to iodide. Once all the thiosulfate is consumed, the iodine remaining reacts with the starch indicator to turn the solution blue-black. This ensures the time measured is for a fixed, small amount of reaction to occur, representing initial rate.

(b) Water is added to keep the total volume constant so that the volume of each reactant added is directly proportional to its initial concentration in the reaction mixture.

(c) Deduction of orders:
- Comparing Run 1 and Run 2: Volume of \(\text{KI}\) doubles while volume of \(\text{K}_2\text{S}_2\text{O}_8\) and total volume remain constant. The time taken is halved, meaning the rate (which is proportional to \(1/t\)) doubles. Therefore, the reaction is 1st order with respect to \(\text{I}^-\).
- Comparing Run 1 and Run 3: Volume of \(\text{K}_2\text{S}_2\text{O}_8\) doubles while volume of \(\text{KI}\) and total volume remain constant. The time taken is halved, meaning the rate doubles. Therefore, the reaction is 1st order with respect to \(\text{S}_2\text{O}_8^{2-}\).

(d) Rate equation: \(\text{Rate} = k[\text{I}^-][\text{S}_2\text{O}_8^{2-}]\)
In Run 1:
\([\text{I}^-] = 0.100 \times \frac{10.0}{40.0} = 0.0250\text{ mol dm}^{-3}\)
\([\text{S}_2\text{O}_8^{2-}] = 0.050 \times \frac{10.0}{40.0} = 0.0125\text{ mol dm}^{-3}\)
\(\text{Rate} = \frac{1}{44} = 0.02273\text{ s}^{-1}\)
\(k = \frac{0.02273}{0.0250 \times 0.0125} = 72.73 \approx 72.7\)
Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(a) [2 marks] - 1 mark for stating that sodium thiosulfate reacts with the iodine as it is formed; 1 mark for explaining that when thiosulfate runs out, the starch turns blue-black to show a fixed amount of reaction has occurred.
(b) [1 mark] - To keep the total volume constant so concentration is proportional to volume.
(c) [4 marks] - 1 mark for Run 1 vs Run 2 comparison showing concentration of iodide doubles and rate doubles; 1 mark for concluding 1st order for \(\text{I}^-\); 1 mark for Run 1 vs Run 3 comparison showing concentration of peroxodisulfate doubles and rate doubles; 1 mark for concluding 1st order for \(\text{S}_2\text{O}_8^{2-}\).
(d) [3 marks] - 1 mark for correct rate equation; 1 mark for correct calculation of \(k = 72.7\) (accept range 72.6 - 72.8); 1 mark for correct units \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(a) The diagram must feature:
1. A pear-shaped or round-bottomed flask containing the reaction mixture, heated safely (e.g., using a water bath, oil bath, or heating mantle, NOT a bare Bunsen flame since the organic substances are highly flammable).
2. A vertical Liebig condenser fitted directly to the flask neck.
3. Water entering the condenser jacket at the bottom and leaving at the top, with the top of the condenser left open to the atmosphere (not stoppered).

(b) Reflux prevents the loss of volatile reactants, products, and solvents by vaporisation, allowing prolonged heating to increase the yield/rate.

(c) Pour the crude mixture into the separating funnel and add aqueous sodium hydrogencarbonate (or sodium carbonate) solution. Shake the mixture and invert the funnel, opening the tap periodically to release pressure. Observations: Effervescence / bubbling / fizzing. The unreacted ethanoic acid reacts to form water-soluble sodium ethanoate, which enters the lower aqueous layer, which can then be run off and discarded.

(d) Use an anhydrous inorganic salt such as anhydrous calcium chloride (\(\text{CaCl}_2\)), magnesium sulfate (\(\text{MgSO}_4\)), or sodium sulfate (\(\text{Na}_2\text{SO}_4\)). The liquid is dry when it turns from cloudy to completely clear/transparent.

(e) Perform simple distillation (or fractional distillation) and collect the fraction that distils at the boiling point of ethyl ethanoate (approximately \(77^\circ\text{C}\)).

(a) [3 marks] - 1 mark for vertical condenser with correct water flow (in at bottom, out at top); 1 mark for pear-shaped/round-bottomed flask with open top (no stopper); 1 mark for indicating a safe heat source (heating mantle/water bath/oil bath) and labellings.
(b) [1 mark] - To prevent loss of volatile components / reactants / products by evaporation.
(c) [3 marks] - 1 mark for adding aqueous sodium carbonate/hydrogencarbonate; 1 mark for stating that pressure must be released (venting) periodically; 1 mark for noting the observation of effervescence/fizzing.
(d) [2 marks] - 1 mark for identifying a suitable anhydrous drying agent (e.g. anhydrous \(\text{CaCl}_2\), \(\text{MgSO}_4\), or \(\text{Na}_2\text{SO}_4\)); 1 mark for stating that the mixture becomes clear / no longer cloudy.
(e) [1 mark] - Redistillation collecting the fraction boiling at \(77^\circ\text{C}\) (accept range \(74 - 79^\circ\text{C}\)).

1. Calculate the percentage uncertainty in the volume of \(\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}\):
\% \text{ uncertainty in volume} = \frac{0.1\text{ cm}^3}{10.0\text{ cm}^3} \times 100 = 1.0\%

2. Calculate the percentage uncertainty in the time, \(t\):
\% \text{ uncertainty in time} = \frac{0.5\text{ s}}{50.0\text{ s}} \times 100 = 1.0\%

3. Combine these percentage uncertainties:
\text{Combined percentage uncertainty} = 1.0\% + 1.0\% = 2.0\%

1 mark: Correct calculation of the individual uncertainties (1.0% and 1.0%) and summing them to give 2.0% (C).

- Reject: 1.0% (A)
- Reject: 1.5% (B)
- Reject: 3.0% (D)

- \(X\) has the formula \(\text{C}_3\text{H}_6\text{O}\) and does not react with Tollens' reagent, meaning it is a ketone, so \(X\) must be propanone.
- Reduction of propanone with \(\text{NaBH}_4\) gives the secondary alcohol, propan-2-ol (\(Y\)).
- Dehydration of propan-2-ol using concentrated \(\text{H}_2\text{SO}_4\) yields propene (\(Z\)).
- Propan-2-ol (\(Y\)), being a secondary alcohol, can be oxidized by acidified potassium dichromate(VI) to regenerate propanone (\(X\)). This makes option D correct.
- A is incorrect because ketones cannot be easily oxidized by acidified potassium dichromate(VI).
- B is incorrect because propan-2-ol is a secondary alcohol, not a primary alcohol.
- C is incorrect because propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)) cannot exhibit cis-trans isomerism.

1 mark: Correctly identifies that \(X\) is propanone, \(Y\) is propan-2-ol, and \(Z\) is propene, and deduces that the secondary alcohol \(Y\) is oxidized back to the ketone \(X\) (D).

- Reject: A, B, C

1. Write the oxidation half-equations for the species in \(\text{FeC}_2\text{O}_4\):
\(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\)
\(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-\)
Each mole of \(\text{FeC}_2\text{O}_4\) loses \(1 + 2 = 3\) moles of electrons.

2. Write the reduction half-equation for manganate(VII):
\(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
Each mole of \(\text{MnO}_4^-\) gains 5 moles of electrons.

3. Equate electron gain and loss:
\(3 \times n(\text{FeC}_2\text{O}_4) = 5 \times n(\text{MnO}_4^-)\)
\(n(\text{FeC}_2\text{O}_4) = \frac{5}{3} n(\text{MnO}_4^-)\)

4. Calculate moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = 0.0300\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 6.00 \times 10^{-4}\text{ mol}\)

5. Calculate moles of \(\text{FeC}_2\text{O}_4\):
\(n(\text{FeC}_2\text{O}_4) = \frac{5}{3} \times (6.00 \times 10^{-4}\text{ mol}) = 1.00 \times 10^{-3}\text{ mol}\)

6. Calculate concentration of \(\text{FeC}_2\text{O}_4\):
\(C = \frac{1.00 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}\)

1 mark: Correctly identifies that both \(\text{Fe}^{2+}\) and \(\text{C}_2\text{O}_4^{2-}\) are oxidized, establishes the 3:5 stoichiometry, and calculates the concentration as \(0.0400\text{ mol dm}^{-3}\) (B).

- Reject: \(0.0144\text{ mol dm}^{-3}\) (A)
- Reject: \(0.0600\text{ mol dm}^{-3}\) (C)
- Reject: \(0.120\text{ mol dm}^{-3}\) (D)

1. Calculate \(\text{p}K_a\):
\(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.870\)

2. Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)
\(4.80 = 4.870 + \log_{10}\left(\frac{n(\text{A}^-)}{n(\text{HA})}\right)\)
\(-0.070 = \log_{10}\left(\frac{n(\text{A}^-)}{n(\text{HA})}\right)\)
\(\frac{n(\text{A}^-)}{n(\text{HA})} = 10^{-0.070} = 0.8511\)

3. Calculate \(n(\text{HA})\):
\(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\)

4. Calculate required \(n(\text{A}^-)\):
\(n(\text{A}^-) = 0.8511 \times (5.00 \times 10^{-3}\text{ mol}) = 4.255 \times 10^{-3}\text{ mol}\)

5. Calculate the volume \(V\) of \(0.200\text{ mol dm}^{-3}\) sodium propanoate:
\(V = \frac{4.255 \times 10^{-3}\text{ mol}}{0.200\text{ mol dm}^{-3}} = 0.02128\text{ dm}^3 = 21.3\text{ cm}^3\)

1 mark: Correctly calculates \(\text{p}K_a\), sets up the ratio of acid to conjugate base, and determines the correct volume to be \(21.3\text{ cm}^3\) (B).

- Reject: A, C, D

- A is incorrect: a salt bridge contains a dissolved salt, not a metal wire, as it must allow ion migration to maintain charge balance.
- B is incorrect: zinc ions do not migrate across the salt bridge to the copper half-cell; rather, ions from the salt bridge migrate to maintain neutrality.
- C is correct: potassium nitrate is the standard choice for a salt bridge because neither potassium ions nor nitrate ions form precipitates with the standard electrolytes used (such as sulfates or chlorides of zinc and copper).
- D is incorrect: electrons flow from the anode (oxidation) to the cathode (reduction) in the external circuit.

1 mark: Correctly identifies that potassium nitrate is used because its constituent ions do not react with the electrolytes in the half-cells (C).

- Reject: A, B, D

- A is correct: sodium hydrogencarbonate neutralizes any acid carry-over.
- B is correct: anhydrous calcium chloride is a suitable drying agent for removing traces of water from the organic layer.
- C is correct: cyclohexene has a boiling point of \(83\text{ }^\circ\text{C}\), so collecting the fraction between \(80\text{ }^\circ\text{C}\) and \(85\text{ }^\circ\text{C}\) isolates the pure product.
- D is incorrect: washing with concentrated phosphoric acid would add more acid catalyst back into the mixture, contaminate the cyclohexene, and cause side-reactions rather than purify it. Acid is removed by washing with water and neutralizing with a base.

1 mark: Identifies that washing with concentrated phosphoric acid is not a valid purification step for cyclohexene (D).

- Reject: A, B, C

The standard enthalpy of atomisation is defined as the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.
For chlorine, the standard state is gaseous diatomic molecules, \(\text{Cl}_2\text{(g)}\).
To form 1 mole of gaseous chlorine atoms, the equation must be:
\(\frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{Cl(g)}\).
Option A represents the standard bond dissociation enthalpy of chlorine, which yields 2 moles of gaseous chlorine atoms.

1 mark: Recognizes the definition of standard enthalpy of atomisation as forming 1 mole of gaseous atoms from the element in its standard state (B).

- Reject: A (bond dissociation enthalpy)
- Reject: C, D

1. Calculate mass of anhydrous \(\text{MgSO}_4\):
\(20.60\text{ g} - 18.20\text{ g} = 2.40\text{ g}\)

2. Calculate mass of water lost:
\(23.12\text{ g} - 20.60\text{ g} = 2.52\text{ g}\)

3. Calculate moles of anhydrous \(\text{MgSO}_4\) (\(M_r = 120.4\text{ g mol}^{-1}\)):
\(n(\text{MgSO}_4) = \frac{2.40}{120.4} = 0.01993\text{ mol}\)

4. Calculate moles of water lost (\(M_r = 18.0\text{ g mol}^{-1}\)):
\(n(\text{H}_2\text{O}) = \frac{2.52}{18.0} = 0.1400\text{ mol}\)

5. Calculate the molar ratio \(x\):
\(x = \frac{0.1400}{0.01993} = 7.02\text{, which rounds to } 7\)

1 mark: Correctly calculates moles of anhydrous salt and water to find the mole ratio of 7 (C).

- Reject: 5 (A)
- Reject: 6 (B)
- Reject: 10 (D)

Percentage uncertainty is calculated as (uncertainty / measured value) * 100. The volume of sodium hydroxide added is measured using the burette. Therefore, the percentage uncertainty is (0.10 / 24.50) * 100 = 0.408%, which rounds to 0.41%.

1 mark for correct calculation of the percentage uncertainty for the burette reading. Reject other combinations of uncertainty calculations.

The hot filtration removes insoluble impurities. Cooling the filtrate in ice allows the desired organic product to recrystallise while leaving soluble impurities in solution. Cold filtration under reduced pressure separates the crystals from the solvent and soluble impurities. Washing with a small amount of cold solvent removes any adhering impurities without dissolving the crystals. Drying removes the remaining solvent.

1 mark for identifying the correct, standard sequence of recrystallisation steps.

The white precipitate with NaOH that dissolves in excess to a colourless solution is characteristic of amphoteric metal hydroxides, specifically Al(OH)3 forming [Al(OH)4]-. The reaction with carbonate producing carbon dioxide gas and a white precipitate of Al(OH)3(H2O)3 confirms the presence of the 3+ ion, Al3+, which has a high enough charge density to polarise water ligands and act as an acid.

1 mark for the correct identification of Al3+.

Sodium hydrogencarbonate is added to neutralise any acid carried over during distillation. Cyclohexene is less dense than water, so it forms the upper layer; the lower aqueous layer is discarded. Anhydrous calcium chloride is used to dry the organic layer. Decanting and redistillation at the boiling point of cyclohexene (83 degrees Celsius) yields the pure product.

1 mark for the correct chemical process sequence for cyclohexene purification.

The reaction equation is MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O. First, find the moles of MnO4-: n(MnO4-) = C * V = 0.0200 * 0.02250 = 4.50 * 10^-4 mol. Since the ratio of MnO4- to Fe2+ is 1:5, the moles of Fe2+ = 5 * (4.50 * 10^-4) = 2.25 * 10^-3 mol. The concentration of Fe2+ is n / V = (2.25 * 10^-3) / 0.0250 = 0.0900 mol dm^-3.

1 mark for the correct calculation of the concentration of Fe2+.

The Rf value is 4.8 / 12.0 = 0.40. A lower Rf value indicates that the compound has travelled less, which means it has a stronger affinity for the stationary phase relative to the mobile phase compared to a compound with a higher Rf value (such as 0.75).

1 mark for identifying the correct Rf value and its chemical significance in terms of chromatographic separation dynamics.

The rate of reaction is proportional to 1/t. In Experiment 1, the rate is proportional to 1/80 = 0.0125. In Experiment 2, the concentration is doubled and the rate is proportional to 1/40 = 0.0250 (which is double the rate of Experiment 1). In Experiment 3, the concentration is doubled again and the rate is proportional to 1/20 = 0.0500 (which is double the rate of Experiment 2). Since the rate doubles when the concentration of iodide ions doubles, the reaction is first order with respect to iodide.

1 mark for selecting First order with correct rate relationships.

The total volume is 100.0 cm3, which corresponds to a mass of 100.0 g. The temperature change is 26.9 - 20.2 = 6.7 K. The heat released is q = m * c * dT = 100.0 * 4.18 * 6.7 = 2800.6 J = 2.80 kJ. The number of moles of water formed (equal to the moles of acid/alkali reacting) is n = C * V = 1.00 * 0.0500 = 0.0500 mol. The enthalpy change is -q / n = -2.8006 / 0.0500 = -56.0 kJ mol-1.

1 mark for the correct calculation of the enthalpy change of neutralisation, including the correct negative sign.

First, calculate the moles of \(MnO_4^-\):
\(n(MnO_4^-) = C \times V = 0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\) mol.

Using the redox equation:
\(5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O\)

The reacting ratio is 5 moles of \(Fe^{2+}\) to 1 mole of \(MnO_4^-\).
\(n(Fe^{2+}) = 5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\) mol.

Since 1 mole of \(FeSO_4\) produces 1 mole of \(Fe^{2+}\):
\(m(FeSO_4) = n \times M_r = 2.25 \times 10^{-3} \times 151.9 = 0.3418\) g.

Calculate the percentage by mass:
\% \text{ purity} = \frac{0.3418}{1.25} \times 100 = 27.3\%.

1 mark for the correct calculation leading to 27.3%.
- Allow 1 mark for correct answer (C).
- Reject other choices: A uses 1:1 ratio; B uses 1:2.5 ratio; D uses 1:10 ratio.

The initial rate of reaction is inversely proportional to the time taken for the colour change to occur (\(\text{Rate} \propto \frac{1}{t}\)). If the time taken doubles, the rate is halved.

When the concentration of \(S_2O_8^{2-}\) is halved, the rate is also halved. Since the rate change is directly proportional to the concentration change, the reaction is first order with respect to \(S_2O_8^{2-}\).

1 mark for identifying the correct order (B).
- Allow 1 mark for correct option.
- Reject zero, second, and third order options.

The reactions described are characteristic of copper(II) ions, \(Cu^{2+}\).
- Addition of dropwise NaOH(aq) forms a pale blue precipitate of copper(II) hydroxide, \(Cu(OH)_2(H_2O)_4\), which is insoluble in excess sodium hydroxide.
- Addition of dropwise ammonia forms the same pale blue precipitate. In excess ammonia, ligand exchange occurs to form the deep blue soluble complex \([Cu(NH_3)_4(H_2O)_2]^{2+}\).

1 mark for selecting the correct metal ion (C).
- Reject cobalt, iron(II), or chromium(III) based on precipitate colour and solubility characteristics.

To account for heat loss to the surroundings before the maximum temperature is achieved, the student must extrapolate the initial temperature line (before mixing) forward to the time of mixing (minute 4) and the cooling curve (after the reaction has finished, from minutes 5 to 10) back to the time of mixing (minute 4). The difference between these two temperature values at the 4th minute gives the theoretical maximum temperature rise, \(\Delta T\).

1 mark for the correct method of extrapolation to the time of mixing (minute 4).

In recrystallisation, the desired compound should be very soluble in the chosen solvent at high temperatures (near the boiling point) but virtually insoluble or only sparingly soluble in it at low temperatures (room temperature or below). This ensures that a high yield of crystals is recovered upon cooling while impurities remain dissolved.

1 mark for identifying the correct solubility characteristics of a recrystallisation solvent.

A more polar stationary phase will have stronger intermolecular attractions (such as hydrogen bonding or dipole-dipole interactions) with the polar amino acid. As a result, the polar amino acid will spend more time adsorbed to the stationary phase and travel a shorter distance relative to the solvent front. Therefore, its \(R_f\) value will decrease.

1 mark for identifying that the retention/interaction with the stationary phase increases, reducing the \(R_f\) value.

Fehling's solution is a mild oxidising agent that oxidises aldehydes (like propanal) but does not react with ketones (like propanone).
- Propanal reduces the blue copper(II) ions in Fehling's solution to form a red-orange precipitate of copper(I) oxide (\(Cu_2O\)).
- Propanone yields no reaction.
- Acidified potassium dichromate(VI) would also react with propan-1-ol, so it is not a direct differentiator without further steps, and Tollens' reagent reacts with propanal to form a silver mirror (option D incorrectly states it forms a mirror with propanone).

1 mark for selecting Fehling's solution and its correct matching observation.

The titration of a weak acid (propanoic acid) with a strong base (sodium hydroxide) produces an equivalence point in the alkaline region (pH \(\approx\) 8.5–9.0).
- Phenolphthalein has a pH transition range of 8.3–10.0, which perfectly overlaps with the rapid pH change at the equivalence point of this titration.
- When adding alkali to acid, the colour change of phenolphthalein at the end-point is from colourless to pale pink.

1 mark for identifying the correct indicator (phenolphthalein) and its corresponding colour change.

First, calculate the mass of anhydrous \(\text{BaCl}_2\):
\(20.75\text{ g} - 18.25\text{ g} = 2.50\text{ g}\)

Calculate the mass of \(\text{H}_2\text{O}\) lost:
\(21.18\text{ g} - 20.75\text{ g} = 0.43\text{ g}\)

Calculate the moles of anhydrous \(\text{BaCl}_2\) (\(M_{\text{r}} = 208.3\text{ g mol}^{-1}\)):
\(n = \frac{2.50}{208.3} = 0.0120\text{ mol}\)

Calculate the moles of \(\text{H}_2\text{O}\) (\(M_{\text{r}} = 18.0\text{ g mol}^{-1}\)):
\(n = \frac{0.43}{18.0} = 0.0239\text{ mol}\)

Determine the simplest molar ratio:
\(\frac{0.0239}{0.0120} \approx 2\)

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

1 mark: correct calculation of masses, moles, and ratio leading to B.

Sodium hydrogencarbonate reacts with residual acid from the reaction mixture to produce carbon dioxide gas (\(\text{CO}_2\)). This gas causes a build-up of pressure inside the closed separating funnel. Inverting the funnel and opening the tap periodically releases this pressure safely to prevent the stopper from blowing out.

1 mark: correct selection of option C.

The balanced ionic equation for the titration is:
\(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)

Calculate the moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = 0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\text{ mol}\)

Using the 1:5 stoichiometry, calculate the moles of \(\text{Fe}^{2+}\):
\(n(\text{Fe}^{2+}) = 5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\text{ mol}\)

Calculate the concentration of \(\text{Fe}^{2+}\) in the \(25.0\text{ cm}^3\) sample:
\(c = \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900\text{ mol dm}^{-3}\)

1 mark: correct calculation of concentration from stoichiometry, yielding 0.0900 mol dm^-3.

Calculate heat energy released (\(q\)):
\(q = m c \Delta T = 55.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 7.5\text{ K} = 1724.25\text{ J} = 1.72425\text{ kJ}\)

Calculate moles of anhydrous \(\text{CuSO}_4\):
\(n = \frac{5.00}{159.6} = 0.03133\text{ mol}\)

Calculate enthalpy change of solution (\(\Delta H_{\text{sol}}\)):
\(\Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{1.72425}{0.03133} = -55.0\text{ kJ mol}^{-1}\) (the reaction is exothermic as temperature increased).

1 mark: correct calculation of the enthalpy change of solution to 3 sig figs, including sign.

The rate of reaction is proportional to \([\text{H}_2\text{O}_2][\text{I}^-]\).

For Experiment 1:
\(\text{Relative rate}_1 \propto 0.040 \times 0.050 = 0.0020\)

For Experiment 2:
\(\text{Relative rate}_2 \propto 0.080 \times 0.100 = 0.0080\)

The rate in Experiment 2 is \(\frac{0.0080}{0.0020} = 4\) times faster than in Experiment 1.

Since the time taken (\(t\)) is inversely proportional to the rate:
\(t_2 = \frac{t_1}{4} = \frac{50\text{ s}}{4} = 12.5\text{ s}\).

1 mark: correct calculation showing a four-fold increase in rate and corresponding quartering of the time taken.

Reaction 1: Dropwise ammonia deprotonates water ligands to form the neutral, insoluble pale blue precipitate \([\text{Cu}(\text{H}_2\text{O})_4(\text{OH})_2]\).
Reaction 2: Excess ammonia results in ligand substitution, forming the deep blue solution containing \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).
Reaction 3: Concentrated \(\text{HCl}\) provides \(\text{Cl}^-\) ligands, substituting the water ligands to form the tetrahedral yellow-green complex \([\text{CuCl}_4]^{2-}\).

1 mark: correct identification of all three chemical species.