2024 OCR AS Level Chemistry B (Salters) - H033 模擬試題連答案詳解

The successive ionisation energies show a very large increase (jump) between the third and fourth ionisation energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from a lower energy level, closer to the nucleus. Therefore, element \(X\) has three valence electrons and forms an \(X^{3+}\) ion. Oxygen forms the oxide ion, \(\text{O}^{2-}\). To achieve electrical neutrality, the formula of the oxide must be \(\text{X}_2\text{O}_3\).

[1 mark] B is the correct answer. Give 1 mark for identifying the correct formula of the oxide.

The cream precipitate is silver bromide (\(\text{AgBr}\)). Silver chloride is white and dissolves in dilute aqueous ammonia. Silver bromide is cream and is insoluble in dilute ammonia but dissolves in concentrated aqueous ammonia. Silver iodide is yellow and is insoluble in both dilute and concentrated ammonia.

[1 mark] C is the correct answer. Give 1 mark for identifying that silver bromide is cream and dissolves in concentrated but not dilute ammonia.

The value of the equilibrium constant \(K_c\) is affected only by temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), a decrease in temperature shifts the equilibrium position to the right to produce more heat, resulting in an increase in the concentrations of products and a decrease in reactants. Thus, decreasing the temperature increases \(K_c\). Changes in concentration, pressure, or the addition of a catalyst do not alter \(K_c\).

[1 mark] B is the correct answer. Give 1 mark for explaining that only a decrease in temperature will increase the equilibrium constant for an exothermic reaction.

The equation for the complete combustion of pentane is: \(\text{C}_5\text{H}_{12}\text{(l)} + 8\text{O}_2\text{(g)} \rightarrow 5\text{CO}_2\text{(g)} + 6\text{H}_2\text{O}\text{(l)}\). Since 1 mole of pentane produces 5 moles of \(\text{CO}_2\) gas, 0.100 mol of pentane will produce \(0.100 \times 5 = 0.500\text{ mol}\) of \(\text{CO}_2\). Using the ideal gas equation: \(V = \frac{nRT}{p}\), we get \(V = \frac{0.500 \times 8.31 \times 298}{100000} = 0.01238\text{ m}^3 = 12.4\text{ dm}^3\).

[1 mark] B is the correct answer. Give 1 mark for the correct calculation of gas volume using stoichiometric ratios and the ideal gas equation.

The thermal stability of Group 2 carbonates increases down the group. As we go down the group, the ionic radius of the metal cation increases, but the charge remains \(2+\). Therefore, the charge density of the cation decreases. As a result, the larger cation has less polarizing power on the carbonate ion (it distorts the electron cloud of the carbonate ion less), making the carbon-oxygen bonds within the carbonate ion harder to break, which increases its thermal stability.

[1 mark] B is the correct answer. Give 1 mark for identifying that the increased thermal stability is due to the decreased polarizing power of the larger metal cation.

The formation reaction is: \(3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\). Using a Hess cycle based on combustion: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\). Therefore, \(\Delta_f H^\theta = 3\Delta_c H^\theta[\text{C(s)}] + 4\Delta_c H^\theta[\text{H}_2\text{(g)}] - \Delta_c H^\theta[\text{C}_3\text{H}_8\text{(g)}] = 3(-394) + 4(-286) - (-2220) = -1182 - 1144 + 2220 = -106\text{ kJ mol}^{-1}\).

[1 mark] A is the correct answer. Give 1 mark for the correct Hess's Law calculation resulting in -106 kJ mol\(^{-1}\).

In a propagation step, a free radical reacts with a neutral molecule to produce a new free radical and a new molecule. Reaction C shows the chlorine radical reacting with a methane molecule to produce a methyl radical and hydrogen chloride, which is a key propagation step. Equation A represents initiation, equation B is incorrect as hydrogen radicals do not form, and equation D represents termination.

[1 mark] C is the correct answer. Give 1 mark for identifying the correct propagation step in the radical mechanism.

Disproportionation is a reaction in which an element is simultaneously oxidised and reduced. In reaction B: \(\text{Cl}_2\text{(g)} + 2\text{NaOH}\text{(aq)} \rightarrow \text{NaCl}\text{(aq)} + \text{NaClO}\text{(aq)} + \text{H}_2\text{O}\text{(l)}\), chlorine starts with an oxidation state of 0 in \(\text{Cl}_2\). In \(\text{NaCl}\), chlorine has an oxidation state of -1 (reduced), and in \(\text{NaClO}\), it has an oxidation state of +1 (oxidised).

[1 mark] B is the correct answer. Give 1 mark for recognizing the disproportionation reaction of chlorine.

Chlorine (\(\text{Cl}_2\)) is a stronger oxidizing agent than iodine (\(\text{I}_2\)). It oxidizes iodide ions (\(\text{I}^-\)) to iodine: \(2\text{I}^-(\text{aq}) + \text{Cl}_2(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{Cl}^-(\text{aq})\). Iodine is non-polar and dissolves preferentially in the organic solvent (cyclohexane) to form a purple upper layer. Because iodide ions lose electrons to form iodine, they are oxidized.

1 mark for identifying the correct salt (sodium iodide) and the correct redox process (oxidation of iodide ions).

The equilibrium constant \(K_c\) is temperature-dependent. Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium to the right to favor the exothermic direction. This increases the concentration of the products relative to the reactants at equilibrium, thus increasing the value of \(K_c\).

1 mark for identifying that decreasing the temperature increases the equilibrium constant \(K_c\) for an exothermic reaction.

1. Calculate the moles of gas:
\(n = \frac{48.0\text{ cm}^3}{24,000\text{ cm}^3\text{ mol}^{-1}} = 0.00200\text{ mol}\)

2. Calculate the molar mass of the hydrocarbon:
\(M = \frac{0.116\text{ g}}{0.00200\text{ mol}} = 58.0\text{ g mol}^{-1}\)

3. Determine which hydrocarbon has a molar mass of \(58.0\text{ g mol}^{-1}\):
- \(\text{C}_3\text{H}_8 = 3(12.0) + 8(1.0) = 44.0\text{ g mol}^{-1}\)
- \(\text{C}_4\text{H}_{10} = 4(12.0) + 10(1.0) = 58.0\text{ g mol}^{-1}\)
- \(\text{C}_4\text{H}_8 = 4(12.0) + 8(1.0) = 56.0\text{ g mol}^{-1}\)
- \(\text{C}_5\text{H}_{12} = 5(12.0) + 12(1.0) = 72.0\text{ g mol}^{-1}\)

Therefore, the molecular formula is \(\text{C}_4\text{H}_{10}\).

1 mark for the correct calculation of molar mass and identification of \(\text{C}_4\text{H}_{10}\).

Down Group 2 (from magnesium to barium):
1. The solubility of the hydroxides increases (magnesium hydroxide is sparingly soluble, whereas barium hydroxide is much more soluble).
2. The thermal stability of the carbonates increases. As the cation size increases down the group, its charge density decreases, so it exerts less polarization on the carbonate anion, making the compound more thermally stable.

1 mark for identifying that both the solubility of Group 2 hydroxides and the thermal stability of Group 2 carbonates increase down the group.

The massive jump between the 3rd and the 4th ionisation energies (from \(2745\) to \(11577\text{ kJ mol}^{-1}\)) indicates that the 4th electron is removed from a shell closer to the nucleus (an inner core shell). Thus, element \(Y\) has 3 valence electrons and forms a stable \(Y^{3+}\) ion. The stable oxide of a \(Y^{3+}\) ion is \(Y_2\text{O}_3\) (as seen in aluminium oxide, \(\text{Al}_2\text{O}_3\)).

1 mark for identifying the correct chemical formula based on successive ionisation energy analysis.

To find the ratio of \(a : b\), we can balance the half-equations:

Oxidation half-equation:
\(\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^-\)

Reduction half-equation:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

Multiplying the oxidation half-equation by 3 to balance the electron transfer (6 electrons total) gives:
\(3\text{SO}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 12\text{H}^+ + 6\text{e}^-\)

Combining the half-equations, we obtain a balanced ratio of 3 moles of \(\text{SO}_2\) to 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\), meaning \(a : b = 3 : 1\).

1 mark for determining the correct balanced mole ratio of reactants as 3 : 1.

The equation for the formation of propene is:
\(3\text{C(graphite)} + 3\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_6(\text{g})\)

Using Hess's law:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)

\(\Delta_f H^\theta = [3 \times (-394) + 3 \times (-286)] - [-2058]\)

\(\Delta_f H^\theta = [-1182 - 858] + 2058 = -2040 + 2058 = +18\text{ kJ mol}^{-1}\).

1 mark for calculating the correct enthalpy change value of +18 kJ mol^-1.

3-methylbut-1-ene has the molecular formula \(\text{C}_5\text{H}_{10}\) and structure \(\text{CH}_2=\text{CH}-\text{CH}(\text{CH}_3)_2\). As an alkene, it undergoes addition polymerization. It does not exhibit stereoisomerism because: 1. It does not show geometric (E/Z) isomerism since the double-bonded C1 is attached to two identical hydrogen atoms. 2. It does not show optical isomerism since there is no chiral carbon atom (the C3 atom is bonded to two identical methyl groups). Thus, it has no stereoisomers. Incorrect options explained: a is incorrect because it has no chiral center; b is incorrect because its molecular formula is \(\text{C}_5\text{H}_{10}\) but its empirical formula is \(\text{CH}_2\); c is incorrect because it reacts with bromine water to form a dibromoalkane, not a diol.

1 mark for identifying that 3-methylbut-1-ene undergoes addition polymerization and does not exhibit stereoisomerism.

Silver chloride (\(AgCl\)) is a white precipitate that dissolves in dilute aqueous ammonia. Silver iodide (\(AgI\)) is a pale yellow precipitate that is insoluble in both dilute and concentrated aqueous ammonia. Therefore, the portion of the precipitate that dissolved was \(AgCl\) (indicating chloride ions were present) and the remaining pale yellow precipitate is \(AgI\) (indicating iodide ions were present).

1 mark for identifying both chloride and iodide correctly (Option B).

The expression for the equilibrium constant is \(K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}\). Substituting the equilibrium values: \(K_c = \frac{(0.80)^2}{(0.20)^2 \times 0.40} = \frac{0.64}{0.04 \times 0.40} = \frac{0.64}{0.016} = 40\). The units are calculated as: \(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \text{dm}^3\text{ mol}^{-1}\). Therefore, \(K_c = 40\text{ dm}^3\text{ mol}^{-1}\).

1 mark for the correct calculation of value and unit (Option A).

First, convert the volume to \(dm^3\): \(60.0\text{ cm}^3 = 0.0600\text{ dm}^3\). Next, calculate the amount in moles of the gas: \(n = \frac{0.0600}{24.0} = 0.00250\text{ mol}\). Find the molar mass: \(M_r = \frac{0.110}{0.00250} = 44.0\text{ g mol}^{-1}\). The alkane with \(M_r = 44.0\) is propane, \(C_3H_8\).

1 mark for calculating the molar mass of 44.0 and matching it to propane (Option C).

There is a very large increase between the third and fourth ionisation energies (from 2745 to 11578 \(kJ\text{ mol}^{-1}\)), which indicates that the fourth electron is removed from a shell closer to the nucleus (an inner shell). This means element \(X\) has three valence electrons in its outer shell. Since the element is in Period 3, its outer electron configuration is \(3s^2 3p^1\).

1 mark for identifying the large increase between the 3rd and 4th ionisation energies and deducing the correct Period 3 outer electron configuration (Option C).

(a) London forces (instantaneous dipole-induced dipole forces) exist between halogen molecules. Down the group, the total number of electrons in each molecule increases. This results in stronger London forces that require more thermal energy to overcome, leading to higher boiling points. (b)(i) \(\text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}\) (ii) The upper organic layer becomes purple/pink/violet. The lower aqueous layer becomes pale yellow/brown/colourless. (c)(i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (ii) Disproportionation. Chlorine is both oxidised and reduced in the same reaction. The oxidation state of chlorine changes from 0 in \(\text{Cl}_2\) to -1 in \(\text{NaCl}\) (reduction) and to +1 in \(\text{NaClO}\) (oxidation). (d) Add concentrated sulfuric acid to both solids. Sodium chloride produces misty fumes of hydrogen chloride, whereas sodium bromide produces orange-brown fumes of bromine gas.

(a) London forces exist between molecules (1) Number of electrons increases down the group (1) Stronger London forces require more energy to break (1). (b)(i) Correct species and balancing: \(\text{Cl}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)}\) (1) Correct state symbols (1). (ii) Organic layer purple/pink/violet (1) Aqueous layer pale yellow/brown/colourless (1). (c)(i) Correct formulas and balanced: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (2). (ii) Disproportionation (1) Chlorine is both oxidised and reduced (0.5) Oxidation state in \(\text{Cl}_2\) is 0, in \(\text{NaCl}\) is -1, and in \(\text{NaClO}\) is +1 (1). (d) Add conc. H2SO4 (1) Correct observations for both (misty fumes vs orange-brown fumes) (1).

(a) \(K_c = \frac{[\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}{[\text{HCl}]^4 [\text{O}_2]}\). Units: \(\text{dm}^3\text{ mol}^{-1}\). (b)(i) The yield of chlorine decreases. The forward reaction is exothermic, so increasing temperature shifts the equilibrium in the endothermic (reverse) direction to absorb heat. (ii) \(K_c\) decreases. (c)(i) Let change in moles of \(\text{Cl}_2\) be +0.80 mol. Initial moles: HCl = 2.40, O2 = 1.20, Cl2 = 0, H2O = 0. Change: HCl = -1.60, O2 = -0.40, Cl2 = +0.80, H2O = +0.80. Equilibrium moles: HCl = 0.80 mol, O2 = 0.80 mol, Cl2 = 0.80 mol, H2O = 0.80 mol. (ii) Volume = 4.00 dm3. Equilibrium concentrations: [HCl] = 0.20, [O2] = 0.20, [Cl2] = 0.20, [H2O] = 0.20 mol dm-3. \(K_c = \frac{(0.20)^2 \times (0.20)^2}{(0.20)^4 \times (0.20)} = \frac{1}{0.20} = 5.0\text{ dm}^3\text{ mol}^{-1}\). (d) A catalyst increases the rate of both the forward and reverse reactions equally. Therefore, the composition of the equilibrium mixture is unchanged, so \(K_c\) remains constant.

(a) Expression: \(K_c = [\text{Cl}_2]^2 [\text{H}_2\text{O}]^2 / ([\text{HCl}]^4 [\text{O}_2])\) (1) Units: \(\text{dm}^3\text{ mol}^{-1}\) (1). (b)(i) Yield decreases (1) shift in endothermic/reverse direction to oppose temperature increase (1). (ii) \(K_c\) decreases (1). (c)(i) HCl equilibrium moles = 0.80 (1) O2 equilibrium moles = 0.80 (1) H2O equilibrium moles = 0.80 (0.5). (ii) Division of all moles by 4.00 to find concentration = 0.20 mol dm-3 (1) Substitution into expression (1) Correct calculation to give 5.0 (1). (d) Catalyst speeds up forward and reverse reactions equally (1) No change to position of equilibrium / concentrations (1).

(a)(i) Mass of C = \(12.0 / 44.0 \times 4.40 = 1.20\text{ g}\). Mass of H = \(2.0 / 18.0 \times 2.70 = 0.30\text{ g}\). Mass of O = \(2.30 - (1.20 + 0.30) = 0.80\text{ g}\). (ii) Moles of C = \(1.20 / 12.0 = 0.10\text{ mol}\). Moles of H = \(0.30 / 1.0 = 0.30\text{ mol}\). Moles of O = \(0.80 / 16.0 = 0.05\text{ mol}\). Ratio C:H:O = \(0.10/0.05 : 0.30/0.05 : 0.05/0.05 = 2 : 6 : 1\). Empirical formula is \(\text{C}_2\text{H}_6\text{O}\). (b)(i) Empirical formula mass of \(\text{C}_2\text{H}_6\text{O} = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\text{ g mol}^{-1}\). Since this equals the molecular ion mass, the molecular formula is \(\text{C}_2\text{H}_6\text{O}\). (ii) Isomer 1: Ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\). Isomer 2: Methoxymethane, \(\text{CH}_3\text{OCH}_3\). (c) \(\text{C}_2\text{H}_6\text{O(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}\).

(a)(i) Mass of C = 1.20 g (1) Mass of H = 0.30 g (1) Mass of O = 0.80 g (1). (ii) Moles of C, H, O all calculated correctly (1) Ratio of elements shown as 2:6:1 (1) Stating empirical formula is \(\text{C}_2\text{H}_6\text{O}\) (1). (b)(i) Shows molecular mass matches empirical mass so molecular formula is \(\text{C}_2\text{H}_6\text{O}\) (1). (ii) Structure of ethanol (1) Name: ethanol (0.5) Structure of methoxymethane (1) Name: methoxymethane / dimethyl ether (1). (c) Correct reactants and products: \(\text{C}_2\text{H}_6\text{O} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\) (1) Correct state symbols (l, g, g, l) (1).

(a)(i) \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3\). (ii) \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\). (b)(i) Across Period 3, the number of protons (nuclear charge) increases. Shielding remains approximately constant as electrons are added to the same main energy shell. Thus, outer electrons experience a stronger attraction to the nucleus and require more energy to remove. (ii) Phosphorus has the configuration \([\text{Ne}] 3\text{s}^2 3\text{p}^3\) with singly occupied 3p orbitals. Sulfur has the configuration \([\text{Ne}] 3\text{s}^2 3\text{p}^4\), with one paired 3p orbital. The repulsion between the two electrons in the paired 3p orbital of sulfur makes it easier to remove the outer electron compared to phosphorus. (c)(i) Electrons are excited to higher energy levels by absorbing energy. When they fall back down to lower energy levels, they emit a photon of electromagnetic radiation. (ii) Each line corresponds to a specific frequency and energy of light emitted during a transition between two defined, fixed energy states.

(a)(i) \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3\) (1). (ii) \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\) (1). (b)(i) Nuclear charge increases (1) shielding is constant / same outer shell (1) stronger attraction to outer electrons (1). (ii) P has singly occupied 3p orbitals (1) S has one paired 3p orbital (1) spin-pair repulsion in S (0.5) makes it easier to remove an electron (1). (c)(i) Electrons excited to higher energy levels (1) fall back to lower levels (1) emitting photons / energy as light (1). (ii) Lines represent transition between fixed energy levels / energy of emission is quantized (1).

(a) Chlorine gas is a stronger oxidizing agent than bromine, so it displaces bromide ions: \(Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)\). The formation of aqueous bromine causes the solution to change from colourless to orange/yellow/brown.

(b) In this reaction, each chlorine atom in \(Cl_2\) gains an electron to form chloride ions: \(Cl_2 + 2e^- \rightarrow 2Cl^-\). The oxidation number of chlorine decreases from 0 to -1 (reduction). Because it accepts electrons from bromide ions (which are oxidized from -1 to 0), chlorine acts as the oxidizing agent.

(c)(i) To test for halide ions, first acidify the solution with dilute nitric acid (to remove any interfering carbonate or hydroxide ions), then add aqueous silver nitrate. White precipitate indicates chloride (\(AgCl\)), cream precipitate indicates bromide (\(AgBr\)), and yellow precipitate indicates iodide (\(AgI\)). To confirm: add dilute aqueous ammonia. Only the white precipitate (\(AgCl\)) dissolves. Add concentrated aqueous ammonia to the remaining precipitates; the cream precipitate (\(AgBr\)) dissolves, but the yellow precipitate (\(AgI\)) remains insoluble.
(ii) The precipitation reaction is represented by: \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\).

(d)(i) Chlorine reacts with cold, dilute sodium hydroxide to form sodium chloride, sodium chlorate(I), and water: \(Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O\).
(ii) Disproportionation is a reaction in which the same element is simultaneously oxidized and reduced. Here, chlorine in \(Cl_2\) has an oxidation state of 0. In \(NaCl\), its oxidation state is -1 (reduced), and in \(NaClO\), its oxidation state is +1 (oxidized).

(a)
- One mark for correct ionic equation: \(Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)\) (ignore state symbols). [1]
- One mark for observation: Solution turns orange / yellow / brown. [1]

(b)
- One mark for stating chlorine gains electrons / bromide loses electrons. [1]
- One mark for stating the oxidation number of chlorine decreases from 0 to -1 (reduced). [1]
- One mark for explaining that because chlorine is reduced / gains electrons, it oxidizes the bromide (acting as an oxidizing agent). [1]

(c)(i)
- One mark for adding nitric acid followed by silver nitrate solution. (Reject hydrochloric acid or sulfuric acid). [1]
- 1.5 marks for correct initial precipitate colours: white (for \(Cl^-\)), cream (for \(Br^-\)), yellow (for \(I^-\)) (0.5 marks each). [1.5]
- One mark for stating dilute ammonia dissolves silver chloride (white ppt). [1]
- One mark for stating concentrated ammonia dissolves silver bromide (cream ppt) but silver iodide (yellow ppt) remains insoluble. [1]

(c)(ii)
- One mark for ionic equation with correct state symbols: \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\). [1]

(d)(i)
- One mark for correct reactants and products: \(Cl_2 + NaOH \rightarrow NaCl + NaClO + H_2O\). [1]
- One mark for correct balancing: \(Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O\). [1]

(d)(ii)
- One mark for defining disproportionation as the simultaneous oxidation and reduction of the same element in a single reaction. [1]
- One mark for state identification: Chlorine goes from 0 (in \(Cl_2\)) to -1 (in \(NaCl\), reduced) and +1 (in \(NaClO\), oxidized). [1]

(a) Dynamic equilibrium is defined by two key criteria: 1) the rate of the forward reaction is equal to the rate of the reverse reaction, and 2) the concentrations of reactants and products remain constant in a closed system.

(b)(i) The equilibrium constant expression is: \(K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}\).
Units calculation: \(\frac{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}{\text{mol dm}^{-3}} = \text{mol dm}^{-3}\).

(b)(ii) Construct an ICE (Initial, Change, Equilibrium) table:
- Initial moles: \(PCl_5 = 2.00\text{ mol}\), \(PCl_3 = 0.00\text{ mol}\), \(Cl_2 = 0.00\text{ mol}\).
- Change in moles: Since 1.20 mol of \(Cl_2\) is present at equilibrium, change for \(Cl_2 = +1.20\text{ mol}\). By stoichiometry (1:1:1 ratio), change for \(PCl_3 = +1.20\text{ mol}\) and change for \(PCl_5 = -1.20\text{ mol}\).
- Equilibrium moles: \(PCl_5 = 2.00 - 1.20 = 0.80\text{ mol}\); \(PCl_3 = 1.20\text{ mol}\); \(Cl_2 = 1.20\text{ mol}\).
- Equilibrium concentrations (divide by \(V = 5.00\text{ dm}^3\)):
- \([PCl_5] = \frac{0.80\text{ mol}}{5.00\text{ dm}^3} = 0.16\text{ mol dm}^{-3}\)
- \([PCl_3] = \frac{1.20\text{ mol}}{5.00\text{ dm}^3} = 0.24\text{ mol dm}^{-3}\)
- \([Cl_2] = \frac{1.20\text{ mol}}{5.00\text{ dm}^3} = 0.24\text{ mol dm}^{-3}\).

(b)(iii) Substitute the equilibrium concentrations into the \(K_c\) expression:
\(K_c = \frac{0.24 \times 0.24}{0.16} = \frac{0.0576}{0.16} = 0.36\text{ mol dm}^{-3}\).

(c)(i) The forward reaction is endothermic. An increase in temperature shifts the equilibrium position to favor the endothermic direction in order to absorb the added heat. Since \(K_c\) increases, the ratio of products to reactants must have increased, indicating that the forward reaction was favored.

(c)(ii) Increasing the total pressure shifts the position of equilibrium to the side with the fewer moles of gas. In this equation, there is 1 mole of gas on the reactant side (left) and 2 moles of gas on the product side (right). Thus, the equilibrium shifts to the left, decreasing the yield of products.

(a)
- One mark for stating that the rate of the forward reaction equals the rate of the reverse reaction. [1]
- One mark for stating that concentrations of reactants and products remain constant / reaction must be in a closed system. [1]

(b)(i)
- One mark for the correct expression: \(K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}\) (brackets must be square to denote concentrations). [1]
- One mark for the correct units: \(\text{mol dm}^{-3}\). [1]

(b)(ii)
- One mark for calculating equilibrium moles of reactants and products: \(PCl_5 = 0.80\text{ mol}\), \(PCl_3 = 1.20\text{ mol}\), \(Cl_2 = 1.20\text{ mol}\). [1]
- Two marks for calculating correct equilibrium concentrations by dividing by 5.00: \([PCl_5] = 0.16\text{ mol dm}^{-3}\), \([PCl_3] = 0.24\text{ mol dm}^{-3}\), \([Cl_2] = 0.24\text{ mol dm}^{-3}\). (Deduct 1 mark for any arithmetic error or failure to divide by volume). [2]

(b)(iii)
- One mark for substituting values correctly into their \(K_c\) expression. [1]
- One mark for the correct final value: 0.36. [1]
- 0.5 marks for correct units matching expression (carried forward if incorrect). [0.5]

(c)(i)
- One mark for identifying the reaction as endothermic. [1]
- One mark for stating that increasing temperature shifts the position of equilibrium in the endothermic direction (to absorb heat). [1]
- One mark for linking the increase in \(K_c\) to an increased yield of products (meaning the forward reaction is favoured). [1]

(c)(ii)
- One mark for predicting that the equilibrium shifts to the left / towards the reactants. [1]
- One mark for explaining that the left side has fewer moles of gas (1 mole) than the right side (2 moles). [1]

(a)(i)
- Moles of \(CO_2 = \frac{4.40\text{ g}}{44.0\text{ g mol}^{-1}} = 0.100\text{ mol}\). Since each mole of \(CO_2\) contains 1 mole of carbon, moles of C in the sample = 0.100 mol. Mass of C = \(0.100\text{ mol} \times 12.0\text{ g mol}^{-1} = 1.20\text{ g}\).
- Moles of \(H_2O = \frac{2.70\text{ g}}{18.0\text{ g mol}^{-1}} = 0.150\text{ mol}\). Since each mole of \(H_2O\) contains 2 moles of hydrogen atoms, moles of H in the sample = \(0.150 \times 2 = 0.300\text{ mol}\). Mass of H = \(0.300\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.300\text{ g}\).
- The remaining mass in the 2.30 g sample is oxygen. Mass of O = \(2.30\text{ g} - (1.20\text{ g} + 0.30\text{ g}) = 0.80\text{ g}\).
- Moles of O = \frac{0.80\text{ g}}{16.0\text{ g mol}^{-1}} = 0.050\text{ mol}.
- Determine the simplest whole-number ratio of the moles of atoms:
- C: \(\frac{0.100}{0.050} = 2\)
- H: \(\frac{0.300}{0.050} = 6\)
- O: \(\frac{0.050}{0.050} = 1\)
- Therefore, the empirical formula of biofuel X is \(C_2H_6O\).

(a)(ii) The empirical formula mass of \(C_2H_6O = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\text{ g mol}^{-1}\). Since this matches the given relative molecular mass of 46.0, the molecular formula is also \(C_2H_6O\).

(b) The complete combustion of ethanol yields carbon dioxide and water: \(C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)\).

(c)(i) Use the formula \(q = m c \Delta T\):
- \(m = 100.0\text{ g}\) (mass of water)
- \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)
- \(\Delta T = 24.5\text{ K}\) (or \(24.5^\circ\text{C}\))
- \(q = 100.0 \times 4.18 \times 24.5 = 10241\text{ J} = 10.241\text{ kJ}\). To three significant figures, \(q = 10.2\text{ kJ}\).

(c)(ii) First, calculate the moles of ethanol burned:
- Moles of ethanol = \(\frac{0.920\text{ g}}{46.0\text{ g mol}^{-1}} = 0.0200\text{ mol}\).
- Enthalpy change of combustion, \(\Delta_c H = -\frac{q}{n} = -\frac{10.241\text{ kJ}}{0.0200\text{ mol}} = -512.05\text{ kJ mol}^{-1}\).
- Rounded to 3 significant figures, \(\Delta_c H = -512\text{ kJ mol}^{-1}\) (must include the negative sign for an exothermic combustion reaction).

(c)(iii) The primary reason for the much lower experimental value is heat loss to the surroundings (air, copper can, draft) rather than being absorbed by the water. Incomplete combustion of ethanol is also an acceptable answer.

(a)(i)
- One mark for finding moles of carbon: \(0.100\text{ mol}\) (or mass = \(1.20\text{ g}\)). [1]
- One mark for finding moles of hydrogen: \(0.300\text{ mol}\) (or mass = \(0.30\text{ g}\)). [1]
- One mark for calculating the mass of oxygen by subtraction: \(2.30 - (1.20 + 0.30) = 0.80\text{ g}\). [1]
- One mark for calculating the moles of oxygen: \(0.050\text{ mol}\). [1]
- One mark for finding the simplest whole-number ratio of \(C:H:O = 2:6:1\) and stating empirical formula is \(C_2H_6O\). [1]

(a)(ii)
- One mark for concluding molecular formula is \(C_2H_6O\) because empirical formula mass matches \(M_r\). [1]

(b)
- One mark for correct reactant and product formulas: \(C_2H_5OH + O_2 \rightarrow CO_2 + H_2O\). [1]
- One mark for correct balancing and state symbols: \(C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)\). [1]

(c)(i)
- One mark for correct substitution into \(q = m c \Delta T\) equation. [1]
- One mark for correct calculation of energy in kJ: \(10.2\text{ kJ}\) (accept 10 or 10.24 kJ). [1]

(c)(ii)
- One mark for calculating moles of ethanol: \(0.0200\text{ mol}\). [1]
- Two marks for calculating \(\Delta H = -512\text{ kJ mol}^{-1}\) (one mark for the numerical value 512, one mark for the negative sign). [2]
- 0.5 marks for quoting the answer to exactly 3 significant figures. [0.5]

(c)(iii)
- One mark for identifying heat loss to surroundings / incomplete combustion. [1]

(a)(i) The thermal stability of Group 2 carbonates increases down the group. Decomposing the carbonate requires distorting the carbonate ion (\(CO_3^{2-}\)) by the cation (\(M^{2+}\)). Down the group, the ionic radius of the \(M^{2+}\) cation increases while its charge remains \(+2\), leading to a lower charge density. Therefore, the cation has a weaker polarizing effect on the carbonate ion, causing less distortion/weakening of the C-O bond. Consequently, more thermal energy is needed to decompose the carbonate down the group.

(a)(ii) The solubility of Group 2 hydroxides increases down the group. As the cation size increases, both the lattice enthalpy of the solid hydroxide and the hydration enthalpy of the cation decrease. However, because the hydroxide ion (\(OH^-\)) is relatively small, the lattice enthalpy decreases more rapidly down the group than the hydration enthalpy of the cation, making the process of dissolving thermodynamically more favorable down the group.

(b)(i) To perform a flame test:
1. Use a clean nichrome or platinum wire (materials that do not produce their own flame color).
2. Dip the wire into concentrated hydrochloric acid to clean it and to convert the metal carbonate into a volatile chloride salt, which vaporizes more easily.
3. Dip the moist wire into the solid sample.
4. Place the wire in the hot, blue (roaring) Bunsen burner flame and observe the color.
- Calcium (\(Ca^{2+}\)) produces a brick-red (or orange-red) flame.
- Strontium (\(Sr^{2+}\)) produces a crimson (or red) flame.
- Barium (\(Ba^{2+}\)) produces an apple-green flame.

(b)(ii) The thermal energy of the Bunsen flame absorbs into the metal ions, promoting/exciting electrons from their ground state to higher energy levels (excited state). Since the excited state is unstable, the electrons drop back down to lower energy levels. The energy difference (\(\Delta E = h\nu\)) is released as electromagnetic radiation (photons). For Group 2 elements, this energy corresponds to wavelengths within the visible light region, producing the characteristic colors observed.

(a)(i)
- One mark for stating that thermal stability increases down the group. [1]
- One mark for stating that cationic radius increases down the group / charge density of the metal cation decreases. [1]
- One mark for stating that the larger cation causes less polarization / distortion of the carbonate ion. [1]
- One mark for linking weaker polarization to less weakening of the C-O bond (requiring higher temperature to break). [1]

(a)(ii)
- One mark for stating that solubility of hydroxides increases down the group. [1]
- 1.5 marks for explaining that both lattice enthalpy and hydration enthalpy decrease, but the lattice enthalpy decreases more rapidly down the group. [1.5]

(b)(i)
- One mark for mentioning the use of a clean nichrome or platinum wire. [1]
- 0.5 marks for mentioning concentrated hydrochloric acid (HCl). [0.5]
- One mark for stating the wire is placed in a hot/blue/roaring Bunsen flame. [1]
- 2 marks for all three correct flame colors: Calcium = brick-red/orange-red, Strontium = crimson/red, Barium = apple-green (deduct 0.5 marks for each missing/incorrect color). [2]

(b)(ii)
- One mark for stating that electrons are excited / promoted to higher energy levels by absorbing heat energy from the flame. [1]
- One mark for stating that excited electrons fall back down to lower / ground state energy levels. [1]
- One mark for stating that energy is emitted as light / electromagnetic radiation. [1]
- 0.5 marks for stating that the color observed corresponds to the specific wavelength / energy difference (\(\Delta E = h\nu\)) of the transitions. [0.5]

To identify the three aqueous halides, the chemist can use the following systematic approach:

1. **Precipitation Test with Silver Nitrate**:
- Add dilute nitric acid, \(HNO_3\), to each solution to remove any interfering carbonate ions.
- Add aqueous silver nitrate, \(AgNO_3(aq)\).
- **Observations**:
- Chloride ions (\(Cl^-\)) produce a **white precipitate** of silver chloride, \(AgCl(s)\).
- Bromide ions (\(Br^-\)) produce a **cream precipitate** of silver bromide, \(AgBr(s)\).
- Iodide ions (\(I^-\)) produce a **yellow precipitate** of silver iodide, \(AgI(s)\).
- **Ionic Equations**:
- \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\)
- \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\)
- \(Ag^+(aq) + I^-(aq) \rightarrow AgI(s)\)

2. **Confirmatory Tests with Ammonia**:
- Because white, cream, and yellow can sometimes be difficult to distinguish, add aqueous ammonia, \(NH_3(aq)\), to the precipitates:
- The white precipitate (\(AgCl\)) **dissolves in dilute ammonia** to form a colourless solution.
- The cream precipitate (\(AgBr\)) **does not dissolve in dilute ammonia**, but **dissolves in concentrated ammonia**.
- The yellow precipitate (\(AgI\)) is **insoluble in both** dilute and concentrated ammonia.

3. **Trend in Ease of Halide Formation**:
- The ease with which halogens form halide ions (gaining an electron to form a \(1-\)\ charge, which represents their oxidising power) **decreases down Group 17**.
- **Explanation in terms of atomic structure**:
- Down the group, the atomic radius increases and the number of inner electron shells (shielding) increases.
- This outweighs the increase in nuclear charge.
- Therefore, the electrostatic attraction between the positive nucleus and the incoming electron in the outer shell becomes weaker, meaning an electron is gained less readily.

**Level 3 (5–6 marks)**
Provides a complete, logically sequenced plan for identifying all three halide ions using both acidified silver nitrate and confirmatory ammonia tests. Correctly identifies all observations and provides all three ionic equations with correct state symbols. Offers a comprehensive and accurate explanation of the trend in the ease of halide formation down Group 17 using appropriate terms (atomic radius, shielding, nuclear attraction).

**Level 2 (3–4 marks)**
Provides a logical plan to identify the three halide ions using acidified silver nitrate, with most observations correct. Includes at least one correct ionic equation with state symbols. Gives a partially correct explanation of the trend down Group 17, missing one key concept (e.g., shielding or atomic radius).

**Level 1 (1–2 marks)**
Identifies at least one reagent (silver nitrate) and some expected observations, but the plan is incomplete or disorganized. Ionic equations may be missing or incorrect. The explanation of the trend down Group 17 is missing or contains major misconceptions.

**Indicative Chemistry Content**
- Reagent: Acidified silver nitrate, \(AgNO_3\).
- Cl- : White precipitate, dissolves in dilute ammonia.
- Br- : Cream precipitate, dissolves only in concentrated ammonia.
- I- : Yellow precipitate, insoluble in concentrated ammonia.
- State symbols: Must be \((aq)\) for ions, \((s)\) for precipitates.
- Trend: Ease of gaining an electron decreases down Group 17.
- Cause: Outer shell is further from the nucleus, has more shielding, and thus experiences weaker electrostatic attraction.

To process the experimental data and calculate \(\Delta_c H\):

1. **Calculate the heat energy transferred (\(q\)) to the water**:
- Use the formula: \(q = m \times c \times \Delta T\)
- \(m\) = mass of water heated (in \(g\)), which is equivalent to its volume in \(cm^3\) (assuming density is \(1.0\,g\,cm^{-3}\)).
- \(c\) = specific heat capacity of water (\(4.18\,J\,g^{-1}\,K^{-1}\)).
- \(\Delta T\) = temperature rise of the water (in \(K\) or \(^{\circ}C\)).

2. **Calculate the amount, in moles, of methanol burned (\(n\))**:
- Calculate the mass change of the spirit burner during combustion: \(m_{burned} = m_{initial} - m_{final}\).
- Calculate the molar mass (\(M_r\)) of methanol (\(32.0\,g\,mol^{-1}\)).
- Use: \(n = \frac{m_{burned}}{M_r}\).

3. **Calculate the enthalpy change of combustion (\(\Delta_c H\))**:
- Convert the heat energy \(q\) from Joules to kilojoules: \(q_{\,kJ} = \frac{q}{1000}\).
- Divide by the number of moles: \(\Delta_c H = -\frac{q_{\,kJ}}{n}\).
- The negative sign must be included because the combustion reaction is exothermic.

**Sources of Error and Practical Improvements**:

- **Error 1: Heat loss to the surrounding air and apparatus**
- *Why it makes it less exothermic*: Heat is dissipated to the surrounding atmosphere or absorbed by the copper beaker itself, so the measured temperature rise \(\Delta T\) is lower than it should be.
- *Improvement*: Use draught shields to protect the flame from air currents, add a lid to the beaker, or use a bomb calorimeter to enclose the system.

- **Error 2: Incomplete combustion of methanol**
- *Why it makes it less exothermic*: Limited oxygen supply leads to the formation of soot (carbon) or carbon monoxide rather than carbon dioxide, releasing less energy per mole of fuel.
- *Improvement*: Ensure a continuous and generous supply of oxygen to the flame, or use a specialized burner with an oxygen feed.

**Level 3 (5–6 marks)**
Detailed and logical description of the quantitative processing steps, including all relevant equations (\(q=mc\Delta T\), \(n=m/M_r\), \(\Delta_c H = -q/n\)) and correct definition of terms (especially that \(m\) is the mass of water, not methanol). Clearly identifies two distinct sources of experimental error and proposes a realistic, effective practical improvement for each.

**Level 2 (3–4 marks)**
Provides a mostly correct method for calculating the enthalpy change of combustion, but may have minor omissions (e.g., forgetting to convert \(J\) to \(kJ\) or omitting the negative sign for exothermic change). Identifies at least one source of error and one corresponding improvement.

**Level 1 (1–2 marks)**
Mention of the equation \(q=mc\Delta T\) and calculating moles, but with significant gaps in the calculation steps (e.g., confusing mass of water and mass of methanol). Identifies heat loss or incomplete combustion as an error, but does not provide suitable or realistic improvements.

**Indicative Chemistry Content**
- Equation: \(q = mc\Delta T\) where \(m\) is mass of water heated, \(c\) is \(4.18\,J\,g^{-1}\,K^{-1}\).
- Moles of methanol: \(n = \text{mass loss of burner} / M_r\) of methanol (\(32.0\,g\,mol^{-1}\)).
- \(\Delta_c H\) calculation: \(-q / n\) (with conversion of \(q\) to \(kJ\)).
- Error 1: Heat lost to the air / copper can. Improvement: Add draught shields / insulate the container / add a lid.
- Error 2: Incomplete combustion (soot/CO formed). Improvement: Burn in pure oxygen / bomb calorimeter.