2022 AQA AS-Level Chemistry 7404 Practice Paper with Answers

Part (a): Sulfur has its outer electron in a 3p orbital that contains a paired set of electrons (3p\(^4\)), whereas phosphorus has three unpaired 3p electrons (3p\(^3\)). The mutual repulsion between the paired electrons in the 3p orbital of sulfur makes it easier to remove one of these electrons compared to phosphorus.

Part (b): \(\text{Ni(g)} \rightarrow \text{Ni}^{+}\text{(g)} + \text{e}^{-}\) (or \(\text{Ni(g)} + \text{e}^{-} \rightarrow \text{Ni}^{+}\text{(g)} + 2\text{e}^{-}\))

Part (c): Mass of one \(^{58}\text{Ni}^{+}\) ion: \(m = \frac{58 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 9.631 \times 10^{-26}\text{ kg}\). Using the kinetic energy equation \(KE = \frac{1}{2}mv^2\): \(v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1.15 \times 10^{-16}\text{ J}}{9.631 \times 10^{-26}\text{ kg}}} = 4.887 \times 10^4\text{ m s}^{-1}\). Using \(t = \frac{d}{v}\): \(t = \frac{1.40\text{ m}}{4.887 \times 10^4\text{ m s}^{-1}} = 2.86 \times 10^{-5}\text{ s}\).

Part (a) [3 marks]:
1 mark: For identifying that sulfur's outer electron is in a paired 3p orbital (3p\(^4\)) and phosphorus has three unpaired electrons (3p\(^3\)).
1 mark: For stating that paired electrons repel each other.
1 mark: For explaining that this repulsion makes it easier to remove the outer electron in sulfur.

Part (b) [2 marks]:
1 mark: For correct chemical symbols and charge on the ion (\(\text{Ni}^{+}\)).
1 mark: For correct state symbols (must be gaseous for both nickel atom and ion).

Part (c) [3.125 marks]:
1 mark: For calculating the mass of one nickel ion in kg (\(9.631 \times 10^{-26}\text{ kg}\)).
1 mark: For calculating the velocity (\(4.887 \times 10^4\text{ m s}^{-1}\)).
1.125 marks: For calculating the correct time of flight to 3 significant figures (\(2.86 \times 10^{-5}\text{ s}\)).

Part (a): \(n(\text{CuO}) = \frac{2.56\text{ g}}{79.5\text{ g mol}^{-1}} = 0.0322\text{ mol}\).

Part (b): From the equation, 2 moles of \(\text{CuO}\) are produced from 1 mole of reactant. Therefore, \(n(\text{reactant}) = \frac{0.0322}{2} = 0.0161\text{ mol}\). Molar mass of reactant \(M_r = \frac{\text{mass}}{\text{moles}} = \frac{3.85\text{ g}}{0.0161\text{ mol}} = 239.1\text{ g mol}^{-1}\). The formula mass of anhydrous basic copper carbonate, \(\text{CuCO}_3 \cdot \text{Cu(OH)}_2\), is \((2 \times 63.5) + 12.0 + (5 \times 16.0) + (2 \times 1.0) = 221.0\text{ g mol}^{-1}\). Mass of water of crystallisation per mole = \(239.1 - 221.0 = 18.1\text{ g mol}^{-1}\). Thus, \(x = \frac{18.1}{18.0} \approx 1\).

Part (c): During rapid or vigorous heating, the rapid evolution of gases may blow some of the solid product out of the crucible (spitting).

Part (d): From the stoichiometry, \(n(\text{CO}_2) = n(\text{reactant}) = 0.0161\text{ mol}\). Using \(pV = nRT\): \(V = \frac{nRT}{p} = \frac{0.0161 \times 8.31 \times 298}{101000} = 3.95 \times 10^{-4}\text{ m}^3 = 0.395\text{ dm}^3\).

Part (a) [1 mark]:
1 mark: For correct calculation of moles of \(\text{CuO}\) (0.0322).

Part (b) [4.125 marks]:
1 mark: For calculating moles of hydrated reactant (0.0161).
1 mark: For finding the total molar mass of the reactant (\(239\text{ to }240\text{ g mol}^{-1}\)).
1 mark: For calculating the anhydrous molar mass (221.0).
1.125 marks: For correctly determining \(x = 1\) (accept working showing integer value).

Part (c) [1 mark]:
1 mark: For suggesting spitting/loss of solid residue during vigorous heating.

Part (d) [2 marks]:
1 mark: For calculating correct moles of \(\text{CO}_2\) (same as reactant moles) and converting pressure to Pa (101000 Pa).
1 mark: For correct volume of \(\text{CO}_2\) in \(\text{dm}^3\) (0.395).

Part (a): Chlorine has 7 outer electrons, plus 3 from fluorine ligands, giving 10 electrons in the outer shell (5 electron pairs). There are 3 bonding pairs and 2 lone pairs. The parent geometry is trigonal bipyramidal. To minimise repulsion, lone pairs occupy equatorial positions, leading to a T-shaped geometry. The repulsion order is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair.

Part (b): In \(\text{ClF}_3\), the shape is asymmetric (T-shaped), so the individual polar C-F bonds do not cancel out, resulting in a net dipole moment. In \(\text{BF}_3\), the shape is trigonal planar (symmetric), so the polar bond dipoles cancel out completely, resulting in a non-polar molecule.

Part (c): Permanent dipole-dipole forces.

Part (a) [4 marks]:
1 mark: For stating there are 3 bonding pairs and 2 lone pairs (total of 5 electron pairs).
1 mark: For stating that electron pairs repel to get as far apart as possible.
1 mark: For stating that lone pairs repel more than bonding pairs.
1 mark: For predicting the T-shaped geometry.

Part (b) [3.125 marks]:
1 mark: For identifying that both molecules contain polar bonds due to electronegativity differences between the central atom and fluorine.
1 mark: For explaining that \(\text{ClF}_3\) is asymmetric, so the dipoles do not cancel.
1.125 marks: For explaining that \(\text{BF}_3\) is symmetric (trigonal planar), so the dipoles cancel out completely.

Part (c) [1 mark]:
1 mark: For specifying permanent dipole-dipole forces.

Part (a): Solubility of Group 2 hydroxides increases down the group. This is because the lattice dissociation enthalpy decreases more rapidly than the hydration enthalpy of the metal ions as the size of the cation increases.

Part (b)(i): Sodium sulfate (or sulfuric acid / ammonium sulfate / any soluble sulfate).

Part (b)(ii): \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\).

Part (b)(iii): First, filter the mixture to obtain the precipitate (barium sulfate) as a residue. Next, wash the residue with cold distilled/deionised water to remove any remaining soluble sodium and chloride ions. Finally, dry the precipitate in a warm oven (or press between filter papers) to remove moisture.

Part (a) [2 marks]:
1 mark: For stating that solubility increases down the group.
1 mark: For explaining this in terms of the relative changes in lattice and hydration enthalpies (or saying the metal ions get larger, making the lattice weaker).

Part (b)(i) [1 mark]:
1 mark: For specifying a soluble sulfate (e.g., sodium sulfate, sulfuric acid). Reject 'sulfate' alone.

Part (b)(ii) [2 marks]:
1 mark: For correct formula of reactants and products.
1 mark: For correct state symbols (must include aq and s).

Part (b)(iii) [3.125 marks]:
1 mark: For filtration.
1 mark: For washing with distilled/deionised water.
1.125 marks: For drying in an oven or with filter paper.

Part (a): \(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\) (accept \(2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl}\)). Role: Acid (proton donor).

Part (b)(i): \(\text{H}_2\text{SO}_4 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\) (or \(\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\)).

Part (b)(ii): \(2\text{H}^+ + \text{H}_2\text{SO}_4 + 2\text{Br}^- \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\) (or \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\)).

Part (b)(iii): Role of sulfuric acid: Oxidising agent. Chloride ions are weaker reducing agents than bromide ions because the chloride ion is smaller, has less electron shielding, and holds its outer electrons more strongly than a bromide ion, making it harder to oxidise.

Part (a) [2 marks]:
1 mark: For correct balanced equation.
1 mark: For stating the role is an acid (or proton donor).

Part (b)(i) [1 mark]:
1 mark: For correct reduction half-equation.

Part (b)(ii) [2 marks]:
2 marks: For correct overall balanced equation (1 mark for correct species, 1 mark for balancing).

Part (b)(iii) [3.125 marks]:
1 mark: For stating the role is an oxidising agent.
1 mark: For stating chloride is a weaker reducing agent than bromide.
1.125 marks: For explaining this in terms of chloride's smaller ionic radius / less shielding / stronger attraction to outer electrons.

Part (a): \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\). Units: \(\text{dm}^6\text{ mol}^{-2}\).

Part (b): Moles at equilibrium: \(\text{CO} = 0.400 - 0.250 = 0.150\text{ mol}\); \(\text{H}_2 = 0.800 - (2 \times 0.250) = 0.300\text{ mol}\); \(\text{CH}_3\text{OH} = 0.250\text{ mol}\). Concentrations at equilibrium (divided by 2.50): \([\text{CO}] = 0.0600\text{ mol dm}^{-3}\); \([\text{H}_2] = 0.1200\text{ mol dm}^{-3}\); \([\text{CH}_3\text{OH}] = 0.1000\text{ mol dm}^{-3}\). \(K_c = \frac{0.1000}{0.0600 \times (0.1200)^2} = 115.74 \approx 116\text{ dm}^6\text{ mol}^{-2}\).

Part (c): The value of \(K_c\) decreases. Since the forward reaction is exothermic, increasing the temperature causes the equilibrium to shift in the endothermic (reverse) direction to absorb heat. This decreases product concentration and increases reactant concentrations, lowering \(K_c\).

Part (a) [2 marks]:
1 mark: For correct expression of \(K_c\).
1 mark: For correct units (\(\text{dm}^6\text{ mol}^{-2}\)).

Part (b) [4.125 marks]:
1 mark: For calculating correct equilibrium moles of \(\text{CO}\) (0.150) and \(\text{H}_2\) (0.300).
1 mark: For dividing moles by volume to get concentrations.
1 mark: For substituting values into \(K_c\) expression.
1.125 marks: For final answer of 116 (or 115.7) to 3 significant figures.

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

Part (a): \(n(\text{Zn}) = \frac{1.50}{65.4} = 0.0229\text{ mol}\). \(n(\text{CuSO}_4) = 0.0500\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0100\text{ mol}\). Since the reaction stoichiometry is 1:1, only 0.0100 mol of Zn is needed. Thus, zinc is in excess.

Part (b): \(q = m c \Delta T\) where \(m = 50.0\text{ g}\), \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\), \(\Delta T = 31.8 - 19.5 = 12.3\text{ K}\). \(q = 50.0 \times 4.18 \times 12.3 = 2570.7\text{ J} = 2.571\text{ kJ}\). The reaction moles (based on the limiting reagent, \(\text{CuSO}_4\)) is \(0.0100\text{ mol}\). \(\Delta H = -\frac{2.571}{0.0100} = -257\text{ kJ mol}^{-1}\) (3 sig figs, negative sign required).

Part (c): 1. Heat loss to the surroundings (air, cup, thermometer). 2. Incomplete reaction of the solid zinc (due to oxide layer or lack of stirring).

Part (a) [2 marks]:
1 mark: For correct calculation of moles of both reactants.
1 mark: For logical comparison showing zinc is in excess.

Part (b) [4.125 marks]:
1 mark: For calculating \(q = 2570.7\text{ J}\) (or \(2.571\text{ kJ}\)).
1 mark: For identifying \(0.0100\text{ mol}\) of limiting reactant is used for division.
1 mark: For division of energy by moles.
1.125 marks: For correct sign (-/negative) and final value of \(-257\text{ kJ mol}^{-1}\) to 3 significant figures.

Part (c) [2 marks]:
1 mark: For each sensible suggestion (e.g., heat loss to surroundings / heat capacity of cup ignored / incomplete reaction).

Part (a): The graph should have 'Number of molecules' on the y-axis and 'Energy' on the x-axis. The curve must start at the origin, rise to a peak, and then tail off asymptotically towards the x-axis. \(E_a\) must be labeled as a vertical line on the right side of the curve.

Part (b): The curve for \(T_2\) must have a peak that is lower and shifted to the right compared to the \(T_1\) curve. The curves must only cross once, and the tail of the \(T_2\) curve must lie above the \(T_1\) curve. At temperature \(T_2\), a significantly greater fraction of molecules (represented by the larger shaded area under the curve to the right of \(E_a\)) have kinetic energy greater than or equal to the activation energy. This results in more frequent successful collisions, increasing the reaction rate.

Part (c): A catalyst provides an alternative pathway with a lower activation energy, \(E_c\). This shifts the activation energy line to the left on the x-axis. Consequently, a much larger fraction of molecules have energy greater than or equal to this new activation energy (\(E \ge E_c\)), leading to more successful collisions per unit time.

Part (a) [3 marks]:
1 mark: For correct axis labels (x-axis: energy, y-axis: number of molecules/particles).
1 mark: For a curve starting at the origin, with a single peak, and approaching but never touching the x-axis at high energy.
1 mark: For showing \(E_a\) correctly labeled.

Part (b) [3.125 marks]:
1 mark: For drawing the \(T_2\) curve correctly (lower peak, shifted to the right, crossing once, tail above \(T_1\)).
1 mark: For stating that at \(T_2\), a larger fraction/number of molecules have energy \(\ge E_a\).
1.125 marks: For relating this to more successful collisions per unit time.

Part (c) [2 marks]:
1 mark: For explaining that a catalyst lowers the activation energy / provides an alternative pathway.
1 mark: For stating that more molecules now have energy \(\ge\) this lower activation energy.

The first ionisation energy of sulfur is lower than that of phosphorus. Phosphorus has a outer electron configuration of \(3s^2 3p^3\) with singly occupied p-orbitals. Sulfur has a configuration of \(3s^2 3p^4\), where one of the 3p orbitals contains a pair of electrons. The mutual repulsion between these paired electrons makes it easier to remove one of them, resulting in a lower first ionisation energy.

1 mark for the correct option (C). No partial marks.

Phosphorus trifluoride (\(PF_3\)) has a central phosphorus atom with 5 valence electrons. It forms three single bonds with fluorine atoms, leaving one lone pair on the phosphorus. Three bonding pairs and one lone pair arrange themselves in a tetrahedral geometry to minimise repulsion, giving the molecule a trigonal pyramidal shape.

1 mark for the correct option (B). No partial marks.

First, find the moles of \(Na_2CO_3\) in the \(25.0\text{ cm}^3\) portion: since \(25.0\text{ cm}^3\) is exactly one-tenth of the \(250\text{ cm}^3\) total volume, it contains \(0.0200\text{ mol}\) of \(Na_2CO_3\). From the equation, 1 mole of \(Na_2CO_3\) reacts with 2 moles of \(HCl\), so \(0.0400\text{ mol}\) of \(HCl\) is required. The volume of \(HCl = \text{moles} / \text{concentration} = 0.0400\text{ mol} / 0.100\text{ mol dm}^{-3} = 0.400\text{ dm}^3\), which is \(400\text{ cm}^3\).

1 mark for the correct option (D). No partial marks.

Barium sulfate is highly insoluble in water and is not absorbed by the body. Because of this, it can be safely swallowed as a 'barium meal' to outline the digestive tract during X-ray imaging, despite barium ions being toxic. Other statements are incorrect: barium sulfate is insoluble, solubility of Group 2 hydroxides increases down the group so magnesium hydroxide is less soluble than calcium hydroxide, and magnesium hydroxide (not calcium hydroxide) is used as an indigestion remedy.

1 mark for the correct option (C). No partial marks.

The iodide ion (\(I^-\)) is the strongest reducing agent among the halide ions. It is capable of reducing the sulfur in concentrated sulfuric acid (oxidation state +6) to hydrogen sulfide, \(H_2S\), in which sulfur has an oxidation state of \(-2\). Bromide can only reduce sulfur to \(SO_2\) (+4), while fluoride and chloride are not strong enough reducing agents to reduce concentrated sulfuric acid.

1 mark for the correct option (D). No partial marks.

In sodium nitrite (\(NaNO_2\)), sodium has an oxidation state of \(+1\) and oxygen has an oxidation state of \(-2\). Let \(x\) be the oxidation state of nitrogen: \(+1 + x + 2(-2) = 0 \Rightarrow x = +3\). In \(HNO_3\) it is \(+5\), in \(NH_4Cl\) it is \(-3\), and in \(N_2O\) it is \(+1\).

1 mark for the correct option (C). No partial marks.

Silicon has a giant covalent macromolecular structure. To melt silicon, many strong covalent bonds between silicon atoms must be broken, which requires a very large amount of thermal energy. Therefore, silicon has the highest melting point of all elements in Period 3.

1 mark for the correct option (B). No partial marks.

The value of the equilibrium constant, \(K_c\), is only affected by temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature will shift the position of the equilibrium to the right to oppose the change, increasing the concentrations of the products and decreasing those of the reactants. This results in a higher value of \(K_c\). Changes in concentration, pressure, or the addition of a catalyst have no effect on the value of \(K_c\).

1 mark for the correct option (B). No partial marks.

The reaction of hydrocarbon combustion is \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). The volume of \(CO_2\) is determined by absorption with NaOH, which is \(45.0\text{ cm}^3\). Since \(15.0\text{ cm}^3\) of hydrocarbon was used, \(x = \frac{45.0}{15.0} = 3\). The total contraction in volume during combustion is \(\Delta V = V(\text{reactants}) - V(\text{products}) = 15.0 + 15.0(x + \frac{y}{4}) - 45.0 = 30.0\text{ cm}^3\). Substituting \(x = 3\) gives \(15.0(3 + \frac{y}{4}) - 30.0 = 30.0\), which simplifies to \(3 + \frac{y}{4} = 4\), so \(y = 4\). The molecular formula is therefore \(C_3H_4\).

1 mark for the correct option (C). Correct working shows determination of carbon count (3) and hydrogen count (4) through the stoichiometric volume relations.

\(NH_2^-\)\ has 2 bonding pairs and 2 lone pairs of electrons around the central Nitrogen atom (giving a tetrahedral arrangement of electron pairs). Due to the greater repulsion from the two lone pairs, the bond angle is reduced from the ideal \(109.5^\circ\) to approximately \(104.5^\circ\), and its molecular shape is non-linear.

1 mark for selecting option A.

Concentrated sulfuric acid reacts with sodium iodide to produce hydrogen iodide, which is a strong reducing agent. Iodide ions reduce the sulfur in sulfuric acid from an oxidation state of +6 to 0, forming yellow solid sulfur. In this process, sulfuric acid acts as an oxidizing agent.

1 mark for selecting option C.

The combustion equation is: \(C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\). Using Hess's law: \(\Delta H_c^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants}) = [3 \times (-394) + 4 \times (-286)] - [-104] = [-1182 - 1144] + 104 = -2326 + 104 = -2222\text{ kJ mol}^{-1}\).

1 mark for selecting option B.

For the reaction \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\): Initial moles of \(N_2 = 2.0\), \(H_2 = 6.0\). At equilibrium, the moles present are: \(N_2 = 2.0 - x\), \(H_2 = 6.0 - 3x = 3(2.0 - x)\), and \(NH_3 = 2x\). The equilibrium concentrations in volume \(V\) are \([N_2] = \frac{2.0 - x}{V}\), \([H_2] = \frac{3(2.0 - x)}{V}\), and \([NH_3] = \frac{2x}{V}\). Substituting these into the equilibrium expression gives \(K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2x/V)^2}{\frac{2.0-x}{V} \times [\frac{3(2.0-x)}{V}]^3} = \frac{4x^2 V^2}{27(2.0-x)^4}\).

1 mark for selecting option A.

Group 2 sulfates become less soluble down the group, so \(MgSO_4\) is highly soluble (no precipitate forms when \(Na_2SO_4\) is added), while \(CaSO_4\), \(SrSO_4\), and \(BaSO_4\) form precipitates. Group 2 hydroxides become more soluble down the group, so \(Mg(OH)_2\) is insoluble and forms a thick white precipitate when \(NaOH\) is added, whereas \(Ba(OH)_2\) is highly soluble. Thus, \(Mg^{2+}\) matches both observation results.

1 mark for selecting option A.

Adding a catalyst does not alter the temperature of the gas, so the distribution of molecular kinetic energies remains completely unchanged (the curve does not shift). However, the catalyst provides an alternative pathway with a lower activation energy, effectively shifting the activation energy threshold to a lower value.

1 mark for selecting option C.

(a) To write the ionic equation, we remove the spectator ions (\(\text{Na}^+\) and \(\text{Cl}^-\)), giving:
\(\text{S}_2\text{O}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{S}(\text{s}) + \text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\).
(b) According to collision theory and the Maxwell-Boltzmann distribution:
- At higher temperatures, the average kinetic energy of the particles increases, causing the distribution curve to flatten and shift to the right.
- The activation energy (\(E_a\)) remains unchanged.
- A significantly larger area under the curve lies to the right of \(E_a\), meaning a much higher fraction of molecules have energy \(E \ge E_a\).
- This results in a much greater frequency of successful/effective collisions.
(c) The independent variable is temperature. Variables to control include concentration of the reactants, the volumes used, the depth of liquid in the beaker (affected by container size), and using the exact same cross.
(d) The reaction rate is defined as change in concentration over time. Since the cross disappears when a fixed amount of sulfur precipitate has formed, \(\Delta[\text{S}]\) is constant. Hence, \(\text{Rate} \propto \frac{1}{t}\).

M1: Correct species in the ionic equation: \(\text{S}_2\text{O}_3^{2-} + 2\text{H}^+ \rightarrow \text{S} + \text{SO}_2 + \text{H}_2\text{O}\) [1 mark].
M2: Correct state symbols on all species: \(\text{S}_2\text{O}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{S}(\text{s}) + \text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\) [1 mark].
M3: At higher temperature, the distribution curve shifts to the right and has a lower peak [1 mark].
M4: Significantly more molecules/particles have energy greater than or equal to the activation energy (\(E \ge E_a\)) [1 mark].
M5: Much higher frequency of successful collisions / more successful collisions per unit time [2 marks].
M6: Any two control variables from: reactant concentration, volume of reactants, flask size/liquid depth, identical cross [2 marks].
M7: Recognises that a constant mass/amount of sulfur is produced when the cross is obscured [1.28 marks].

(a) Butan-2-ol is a secondary alcohol, which oxidizes to a ketone (butanone): \(\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COCH}_3 + \text{H}_2\text{O}\).
(b) Dehydration is an elimination reaction that removes water from an alcohol. It requires a concentrated acid catalyst like concentrated \(\text{H}_2\text{SO}_4\) or concentrated \(\text{H}_3\text{PO}_4\) and heat.
(c)(i) But-1-ene has two hydrogen atoms attached to one of the \(\text{C}=\text{C}\) carbon atoms, so it does not show stereoisomerism.
(ii) But-2-ene shows stereoisomerism. (E)-but-2-ene has the methyl groups on opposite sides, while (Z)-but-2-ene has them on the same side.
(iii) Stereoisomerism occurs because the \(\pi\)-bond prevents rotation around the double bond (restricted rotation) and both carbons of the double bond are attached to two distinct groups: a methyl group and a hydrogen atom.

M1: Correct formulas of reactants and products: \(\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3\) and \(\text{CH}_3\text{CH}_2\text{COCH}_3\) [1 mark].
M2: Correctly balanced equation with \([\text{O}]\) and \(\text{H}_2\text{O}\) [1 mark].
M3: Reagent: Concentrated \(\text{H}_2\text{SO}_4\) or concentrated \(\text{H}_3\text{PO}_4\) [1 mark] (Reject: dilute acid).
M4: Condition: Heat / high temperature (or catalyst of \(\text{Al}_2\text{O}_3\) and heat) [1 mark].
M5: Correct structure of but-1-ene showing all bonds or skeletal [1 mark].
M6: Correct structures of (E)-but-2-ene and (Z)-but-2-ene showing spatial arrangement clearly [1.28 marks].
M7: Correctly named (E)-but-2-ene / trans-but-2-ene and (Z)-but-2-ene / cis-but-2-ene [1 mark].
M8: Restricted/no rotation around the \(\text{C}=\text{C}\) double bond [1 mark].
M9: Two different groups/atoms attached to each of the \(\text{C}=\text{C}\) carbons [1 mark].

(a) Standard nucleophilic substitution mechanism:
1. Show \(\delta+\) on the central carbon atom and \(\delta-\) on the bromine atom of 2-bromopropane.
2. Curly arrow from a lone pair on the oxygen of \(\text{OH}^-\) to the central carbon.
3. Curly arrow from the \(\text{C}-\text{Br}\) bond to the bromine atom.
4. Show the products: \(\text{CH}_3\text{CH(OH)CH}_3\) and \(\text{Br}^-\).
(b) The rate of reaction depends on the strength of the carbon-halogen bond. Because bond enthalpy decreases down Group 7 (\(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\)), the \(\text{C}-\text{I}\) bond is weaker and requires less energy to break than the \(\text{C}-\text{Cl}\) bond.
(c) In hot, ethanolic conditions, elimination occurs where \(\text{OH}^-\) acts as a base by accepting a proton (\(\text{H}^+\)) from a carbon adjacent to the halogen-bearing carbon.
(d) Halogenoalkanes undergo hydrolysis with water (acting as a nucleophile) to release halide ions, which react with silver ions to form precipitates. The rate of precipitate formation is timed:
- 1-iodobutane forms a yellow precipitate (\(\text{AgI}\)) rapidly.
- 1-bromobutane forms a cream precipitate (\(\text{AgBr}\)) moderately fast.
- 1-chlorobutane forms a white precipitate (\(\text{AgCl}\)) very slowly.

M1: Correct dipole \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) [1 mark].
M2: Curly arrow from lone pair of \(\text{OH}^-\) to the central carbon [1 mark].
M3: Curly arrow from the \(\text{C}-\text{Br}\) bond to the bromine atom [1 mark].
M4: Correct products: \(\text{CH}_3\text{CH(OH)CH}_3\) and \(\text{Br}^-\)[1 mark].
M5: \(\text{C}-\text{I}\) bond is weaker / has lower bond enthalpy than \(\text{C}-\text{Cl}\) bond [1 mark].
M6: \(\text{C}-\text{I}\) bond breaks more easily [1 mark] (Reject: references to bond polarity determining the rate).
M7: State that \(\text{OH}^-\) acts as a base / proton acceptor [1 mark].
M8: Method: add \(\text{AgNO}_3(\text{aq})\) in ethanol/water and measure time taken for precipitate to appear [1 mark].
M9: Correct colors: white (\(\text{AgCl}\)), cream (\(\text{AgBr}\)), yellow (\(\text{AgI}\)) [1 mark].
M10: Order of rate: yellow/iodide is fastest, white/chloride is slowest [0.28 marks].

(a) Detailed electrophilic addition mechanism:
1. Hydrogen bromide is polar: \(\text{H}^{\delta+}-\text{Br}^{\delta-}\).
2. Curly arrow starts from the electron-rich \(\text{C}=\text{C}\) double bond to the partially positive \(\text{H}\) atom.
3. Curly arrow from the \(\text{H}-\text{Br}\) single bond to the electronegative \(\text{Br}\) atom.
4. Intermediate is a secondary carbocation, \(\text{CH}_3\text{C}^+\text{HCH}_3\), and a bromide ion, \(\text{Br}^-\).
5. Curly arrow from the lone pair on \(\text{Br}^-\) to the positive carbon atom.
6. The product is 2-bromopropane.
(b) The reaction can form either a primary carbocation (leading to 1-bromopropane) or a secondary carbocation (leading to 2-bromopropane). Secondary carbocations are more stable than primary carbocations because there are two alkyl groups that push electrons towards the positive carbon (the inductive effect), reducing its charge density.
(c) Bromine water is a standard test for unsaturation. Alkenes decolourise bromine water from orange/brown to colourless. The bromine atoms add across the double bond to form 1,2-dibromopropane: \(\text{CH}_3\text{CHBrCH}_2\text{Br}\).
(d) Poly(propene) is an addition polymer. Its repeating unit has a two-carbon main chain with single bonds and extended bonds passing through brackets, with one hydrogen and one methyl group on one carbon, and two hydrogens on the other.

M1: Curly arrow from the \(\text{C}=\text{C}\) double bond to the \(\text{H}\) of \(\text{H}-\text{Br}\) [1 mark].
M2: Curly arrow from the \(\text{H}-\text{Br}\) bond to the bromine atom showing correct partial charges [1 mark].
M3: Structure of the secondary carbocation intermediate \(\text{CH}_3\text{C}^+\text{HCH}_3\) [1 mark].
M4: Curly arrow from lone pair on \(\text{Br}^-\) to the positive carbon of the carbocation [1 mark].
M5: Identify that 2-bromopropane goes via a secondary carbocation and 1-bromopropane via a primary carbocation [1 mark].
M6: State that the secondary carbocation is more stable than the primary carbocation [1 mark].
M7: State that this is due to the electron-donating inductive effect of two alkyl groups [1 mark].
M8: Color change: orange/brown/yellow to colourless [1 mark] (Reject: 'clear').
M9: Correct structure of 1,2-dibromopropane [1 mark].
M10: Correct repeating unit of poly(propene) showing correct connectivity [0.28 marks].

(a) Propanoic acid is a weak acid. Adding a hydrogen carbonate (like \(\text{NaHCO}_3\)) or carbonate (like \(\text{Na}_2\text{CO}_3\)) results in an acid-carbonate reaction that produces carbon dioxide gas, observed as effervescence. Propan-1-ol and propanal will show no visible reaction.
(b) Propanal is an aldehyde. Adding Tollens' reagent and heating in a hot water bath will produce a silver mirror because Tollens' reagent is reduced to metallic silver. Fehling's solution could also be used, forming a brick-red precipitate. Propan-1-ol does not react with Tollens' or Fehling's.
(c)(i) Propanoic acid contains both a carbonyl group (\(\text{C}=\text{O}\)) and an acid hydroxyl group (\(\text{O}-\text{H}\)). Thus, it displays a broad peak at \(2500 - 3000\text{ cm}^{-1}\) (acid O-H) and a sharp, strong peak at \(1680 - 1750\text{ cm}^{-1}\) (C=O). Propan-1-ol only has an alcohol hydroxyl group showing a peak at \(3230 - 3550\text{ cm}^{-1}\) and lacks a C=O peak.
(ii) Propanal contains a carbonyl group (\(\text{C}=\text{O}\)) showing a strong absorption at \(1680 - 1750\text{ cm}^{-1}\) but lacks any O-H absorption. Propan-1-ol contains a broad O-H absorption peak at \(3230 - 3550\text{ cm}^{-1}\) but lacks the C=O absorption peak.
(d) The fingerprint region lies below \(1500\text{ cm}^{-1}\). It contains complex, unique bending vibrations that are characteristic of a specific molecule. An unknown spectrum can be matched directly against library standards to identify the exact compound.

M1: Reagent: Aqueous sodium hydrogencarbonate (or sodium carbonate) [1 mark].
M2: Observation: Effervescence/fizzing with propanoic acid, no visible change with the other two [1 mark].
M3: Reagent: Tollens' reagent / Fehling's solution / Benedict's solution [1 mark].
M4: Observation: Silver mirror / brick-red precipitate with propanal, no change with propan-1-ol [1 mark].
M5: Propanoic acid has a \(\text{C}=\text{O}\) absorption at \(1680 - 1750\text{ cm}^{-1}\) and \(\text{O}-\text{H}\) (acid) absorption at \(2500 - 3000\text{ cm}^{-1}\) [1 mark].
M6: Propan-1-ol has \(\text{O}-\text{H}\) (alcohol) absorption at \(3230 - 3550\text{ cm}^{-1}\) but no \(\text{C}=\text{O}\) absorption [1 mark].
M7: Propanal has \(\text{C}=\text{O}\) absorption at \(1680 - 1750\text{ cm}^{-1}\) but no \(\text{O}-\text{H}\) absorption [1 mark].
M8: Propan-1-ol has \(\text{O}-\text{H}\) (alcohol) absorption at \(3230 - 3550\text{ cm}^{-1}\) but no \(\text{C}=\text{O}\) absorption [1 mark].
M9: Fingerprint region is unique to each molecule and can be compared with a database of known spectra to confirm identity [1.28 marks].

(a) In fractional distillation:
- Crude oil is heated and vaporized before entering the fractionating column.
- The column has a temperature gradient: hotter at the bottom and cooler at the top.
- The vaporized hydrocarbons rise through the column. They cool and condense back to liquids at different levels depending on their boiling points.
- Larger molecules with larger surface areas have stronger van der Waals forces, hence higher boiling points, and condense at the bottom. Shorter-chain hydrocarbons condense near the top.
(b) Structural isomers are defined as compounds sharing the identical molecular formula but possessing different arrangements of atoms, resulting in different structural formulas.
(c)(i) The three skeletal structures of \(\text{C}_5\text{H}_{12}\) are:
1. A simple zig-zag line representing pentane.
2. A four-carbon chain with a methyl branch on carbon-2 representing 2-methylbutane.
3. A central carbon bonded to four methyl groups resembling a cross, representing 2,2-dimethylpropane.
(ii) Their IUPAC names are pentane, 2-methylbutane, and 2,2-dimethylpropane.
(d) Incomplete combustion of pentane: \(\text{C}_5\text{H}_{12} + 5.5\text{O}_2 \rightarrow 5\text{CO} + 6\text{H}_2\text{O}\).

M1: Crude oil is vaporized/heated and introduced into a fractionating column with a temperature gradient (cooler at the top, hotter at the bottom) [1 mark].
M2: Vaporized hydrocarbons rise, cool, and condense at their boiling points [1 mark].
M3: Larger/longer-chain hydrocarbons have higher boiling points and condense lower down, while smaller/shorter-chain hydrocarbons condense near the top [1 mark].
M4: Structural isomerism: same molecular formula, different structural formula [1 mark].
M5: Draw all three correct skeletal structures [2 marks; 1 mark for 2 correct].
M6: Correctly name pentane, 2-methylbutane, and 2,2-dimethylpropane [2 marks; 1 mark for 2 correct].
M7: Correctly balanced equation for incomplete combustion: \(\text{C}_5\text{H}_{12} + 5.5\text{O}_2 \rightarrow 5\text{CO} + 6\text{H}_2\text{O}\) (or doubled/fractional equivalents) [1.28 marks].

(a) Ultraviolet (UV) light provides the energy needed to break the \(\text{Cl}-\text{Cl}\) single covalent bond homolytically. The initiation step is: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\).
(b) In propagation, a chlorine radical abstracts a hydrogen atom from methane to form a methyl radical: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\). Then, the methyl radical reacts with a chlorine molecule to produce chloromethane and regenerate the chlorine radical: \(\text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\).
(c) Ethane is formed when two methyl radicals collide and combine in a termination step: \(2\text{CH}_3^\bullet \rightarrow \text{CH}_3\text{CH}_3\).
(d) The radical reaction is non-selective. Once chloromethane (\(\text{CH}_3\text{Cl}\)) is formed, it can collide with chlorine radicals, leading to further substitution reactions that produce dichloromethane (\(\text{CH}_2\text{Cl}_2\)), trichloromethane (\(\text{CHCl}_3\)), and tetrachloromethane (\(\text{CCl}_4\)). To increase the yield of chloromethane, a massive excess of methane is used. This ensures chlorine radicals are much more likely to collide with unreacted methane rather than chloromethane.

M1: Condition: Ultraviolet / UV light [1 mark].
M2: Equation: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\) [1 mark] (radical dot must be clearly shown on the chlorine atom).
M3: Propagation 1: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) [1 mark].
M4: Propagation 2: \(\text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\) [1 mark].
M5: Termination: \(2\text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\) (or \(\text{CH}_3\text{CH}_3\)) [1 mark].
M6: State that chloromethane product can undergo further propagation / substitution reactions [1 mark].
M7: Write equations showing further substitution, e.g., \(\text{CH}_3\text{Cl} + \text{Cl}^\bullet \rightarrow \text{CH}_2\text{Cl}^\bullet + \text{HCl}\) [1 mark].
M8: Identify that this leads to a mixture of products (dichloromethane, trichloromethane, etc.) [1 mark].
M9: Suggest using a large excess of methane to maximize mono-substituted chloromethane yield [1.28 marks].

The rate of hydrolysis of halogenoalkanes is determined by bond enthalpy (the strength of the carbon-halogen bond) rather than bond polarity. Since the C–I bond is the longest and weakest of the three carbon-halogen bonds (having the lowest bond enthalpy), it requires the least energy to break. Therefore, 1-iodobutane reacts the fastest.

1 mark for the correct option (B).

The reaction of but-1-ene with HBr proceeds via electrophilic addition. The electrophile \(H^+\) adds to but-1-ene to form either a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+\)) or a secondary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\)). The secondary carbocation is more stable due to the inductive effect of the two adjacent alkyl groups, which release electron density toward the positively charged carbon. Thus, the secondary carbocation is formed preferentially, leading to the major product 2-bromobutane.

1 mark for the correct option (B).

Dehydration of pentan-2-ol involves the elimination of water. The hydrogen can be eliminated from either carbon-1 or carbon-3 adjacent to the C-OH group. Elimination of hydrogen from C-1 yields pent-1-ene. Elimination of hydrogen from C-3 yields pent-2-ene. Although pent-2-ene exists as (E)- and (Z)- stereoisomers, the question specifically asks for the number of structural isomers. Therefore, there are exactly 2 structural isomers: pent-1-ene and pent-2-ene.

1 mark for the correct option (B).

In free-radical substitution:
- Option A represents initiation (homolytic fission of Cl–Cl by UV light).
- Option C represents a propagation step where a chlorine radical abstracts a hydrogen atom from methane to produce a methyl radical and hydrogen chloride.
- Options B and D represent termination steps where two radicals combine to form a stable molecule.

1 mark for the correct option (C).

Ethyl propanoate has the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\). Let us examine the molecular formulas of the options:
- Methyl butanoate: \(\text{C}_5\text{H}_{10}\text{O}_2\) (an isomer)
- Pentan-2-one: \(\text{C}_5\text{H}_{10}\text{O}\) (a ketone)
- Propyl propanoate: \(\text{C}_6\text{H}_{12}\text{O}_2\) (6 carbons)
- Butanoic acid: \(\text{C}_4\text{H}_8\text{O}_2\) (4 carbons)
Therefore, methyl butanoate is a structural isomer.

1 mark for the correct option (A).

1. Tollens' reagent: Propanal is an aldehyde and is oxidized by Tollens' reagent to form a silver mirror, whereas propan-1-ol is a primary alcohol and does not react with Tollens' reagent.
2. Infrared spectra: Outside the fingerprint region (above \(1500\text{ cm}^{-1}\)), propanal has a sharp, strong absorption peak for the C=O bond at \(1680\text{--}1750\text{ cm}^{-1}\) and lacks an O-H peak. Propan-1-ol has a broad, strong O-H absorption peak at \(3230\text{--}3550\text{ cm}^{-1}\) and lacks a C=O peak. Thus, they can be easily distinguished by both methods.

1 mark for the correct option (B).

When the temperature of a gas sample is increased, the average kinetic energy of the molecules increases. This causes the Maxwell–Boltzmann distribution curve to flatten and spread out:
1. The peak shifts to the right (higher energy) and downwards (lower fraction of molecules at the peak energy).
2. The area under the curve remains constant because the total number of molecules remains unchanged. Therefore, option C is correct.

1 mark for the correct option (C).

Using Hess's Law:
\(\Delta_c H^\theta = \sum \Delta_f H^\theta(\text{products}) - \sum \Delta_f H^\theta(\text{reactants})\)
Note that the enthalpy of formation of an element in its standard state, \(\text{O}_2(\text{g})\), is zero.

\(\Delta_c H^\theta = [4 \times \Delta_f H^\theta(\text{CO}_2) + 5 \times \Delta_f H^\theta(\text{H}_2\text{O})] - [\Delta_f H^\theta(\text{C}_4\text{H}_{10})]\)

\(\Delta_c H^\theta = [4 \times (-394) + 5 \times (-286)] - [-126]\)

\(\Delta_c H^\theta = [-1576 - 1430] + 126\)

\(\Delta_c H^\theta = -3006 + 126 = -2880\text{ kJ mol}^{-1}\)

Therefore, the correct answer is A.

1 mark for the correct option (A).

The area under a Maxwell-Boltzmann distribution curve represents the total number of molecules in the sample, which remains constant at a given quantity of gas. The peak represents the most probable energy, not the mean energy (which is slightly to the right of the peak). A catalyst lowers the activation energy threshold but does not shift the curve itself. At higher temperatures, the peak shifts to the right and is lower, not to the left and higher.

1 mark for the correct option B.

The rate of nucleophilic substitution (hydrolysis) in halogenoalkanes is determined by the strength of the carbon-halogen bond. The C-I bond has the lowest bond enthalpy (is the weakest bond) and is broken most easily, leading to the fastest rate of reaction compared to C-Br, C-Cl, and C-F bonds.

1 mark for the correct option D.

The reaction proceeds via the more stable carbocation intermediate. Electrophilic addition of H+ to C3 of 2-methylbut-2-ene forms a stable tertiary carbocation at C2, whereas addition of H+ to C2 forms a less stable secondary carbocation at C3. The tertiary carbocation is more stable due to the electron-releasing inductive effect of the three attached alkyl groups. Subsequent attack by the bromide ion at C2 yields 2-bromo-2-methylbutane as the major product.

1 mark for the correct option B.

Butan-2-ol is a secondary alcohol, so it can be oxidized to a ketone (butanone). When dehydrated, the elimination of water can involve a hydrogen atom from C1 (forming but-1-ene) or from C3 (forming but-2-ene). But-2-ene exhibits stereoisomerism as (E)-but-2-ene and (Z)-but-2-ene. Primary alcohols like butan-1-ol and 2-methylpropan-1-ol oxidize to aldehydes and carboxylic acids, not ketones. Tertiary alcohols like 2-methylpropan-2-ol cannot be oxidized easily.

1 mark for the correct option B.

The broad peak in the range 3230–3550 cm^-1 corresponds to the O-H stretch of an alcohol. The lack of an absorption peak in the range 1680–1750 cm^-1 shows there is no C=O carbonyl bond present in the compound. Propanoic acid, propanal, and methyl ethanoate all contain C=O bonds and would display a strong peak in the 1680–1750 cm^-1 range. Therefore, the compound is propan-1-ol.

1 mark for the correct option B.

Enthalpy change = Sum of bonds broken (reactants) - Sum of bonds formed (products). Reactants: 1 x C=C (612) + 4 x C-H (4 x 412 = 1648) + 2 x O-H (2 x 463 = 926) = 3186 kJ mol^-1. Products: 1 x C-C (348) + 5 x C-H (5 x 412 = 2060) + 1 x C-O (360) + 1 x O-H (463) = 3231 kJ mol^-1. Enthalpy change = 3186 - 3231 = -45 kJ mol^-1.

1 mark for the correct option A.

Propagation steps must involve a free radical reacting with a stable molecule to generate a different free radical and a different stable molecule. Equation C shows a methane molecule reacting with a chlorine radical to form a methyl radical and hydrogen chloride, which is a key propagation step.

1 mark for the correct option C.