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Part 1:
(a) Moles of \(\text{HCl}\) = concentration \(\times\) volume = \(0.100\text{ mol dm}^{-3} \times (20.00 / 1000)\text{ dm}^3 = 2.00 \times 10^{-3}\text{ mol}\).
(b) Moles of \(M_2CO_3\) in \(25.0\text{ cm}^3\) = Moles of \(\text{HCl} / 2 = 2.00 \times 10^{-3} / 2 = 1.00 \times 10^{-3}\text{ mol}\).
(c) Moles of \(M_2CO_3\) in \(250\text{ cm}^3\) = \(1.00 \times 10^{-3} \times 10 = 1.00 \times 10^{-2}\text{ mol}\).
Molar mass of \(M_2CO_3 \cdot xH_2O\) = mass / moles = \(2.86\text{ g} / 1.00 \times 10^{-2}\text{ mol} = 286\text{ g mol}^{-1}\).

Part 2:
(d) Mass of water lost = \(5.72\text{ g} - 2.12\text{ g} = 3.60\text{ g}\).
Moles of water lost = \(3.60\text{ g} / 18.0\text{ g mol}^{-1} = 0.20\text{ mol}\).
(e) Since 5.72 g is exactly double the mass of the 2.86 g sample used in the titration, the amount of hydrated carbonate heated is \(2.00 \times 10^{-2}\text{ mol}\).
Ratio of \(\text{H}_2\text{O}\) to \(M_2CO_3\) = \(0.20\text{ mol} / 0.020\text{ mol} = 10\). Therefore, \(x = 10\).
Molar mass of anhydrous residue \(M_2CO_3\) = mass / moles = \(2.12\text{ g} / 0.020\text{ mol} = 106\text{ g mol}^{-1}\).
\(M_{\text{r}}\) of \(M_2CO_3 = 2(A_{\text{r}}\text{ of } M) + 12.0 + 3(16.0) = 106\) \(\implies 2(A_{\text{r}}\text{ of } M) + 60.0 = 106 \implies 2(A_{\text{r}}\text{ of } M) = 46.0 \implies A_{\text{r}}\text{ of } M = 23.0\text{ g mol}^{-1}\).
Therefore, metal \(M\) is sodium (\(\text{Na}\)).

[1 mark] Moles of \(\text{HCl}\) = \(0.02000\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.00 \times 10^{-3}\text{ mol}\).
[1 mark] Moles of \(M_2CO_3\) = \(2.00 \times 10^{-3} / 2 = 1.00 \times 10^{-3}\text{ mol}\) (consequential on part a).
[3 marks] Moles of \(M_2CO_3\) in \(250\text{ cm}^3 = 1.00 \times 10^{-2}\text{ mol}\) [1 mark]. Molar mass calculation: \(2.86 / 0.0100\) [1 mark] = \(286\text{ (g mol}^{-1}\text{)}\) [1 mark].
[2 marks] Mass of water lost = \(5.72 - 2.12 = 3.60\text{ g}\) [1 mark]. Moles of water lost = \(3.60 / 18.0 = 0.20\text{ mol}\) [1 mark].
[3 marks] Ratio \(x = 0.20 / 0.020 = 10\) (or by alternative valid method) [1 mark]. Molar mass of \(M_2CO_3 = 106\text{ g mol}^{-1}\) [1 mark]. Identification of \(M\) as Sodium/\(\text{Na}\) (consequential on calculated \(A_{\text{r}} = 23.0\)) [1 mark].

(a) First ionisation energy is the enthalpy change/energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
(b) The third ionisation energy of sodium represents the removal of an electron from a gaseous \(\text{Na}^{2+}\) ion: \(\text{Na}^{2+}(g) \rightarrow \text{Na}^{3+}(g) + e^-\).
(c) Phosphorus has the outer electron configuration \(3s^2 3p^3\), where the 3p subshell consists of three singly occupied orbitals. Sulfur has the configuration \(3s^2 3p^4\), where one of the 3p orbitals contains a pair of electrons. The mutual repulsion between these two paired electrons in the same 3p orbital in sulfur makes it easier to remove the outer electron than from the singly occupied orbital in phosphorus, resulting in a lower first ionisation energy.
(d) The possible combinations for \(\text{Cl}_2^+\) are:
\(^{35}\text{Cl}\)-\(^{35}\text{Cl}^+\) \(\rightarrow m/z = 70\)
\(^{35}\text{Cl}\)-\(^{37}\text{Cl}^+\) and \(^{37}\text{Cl}\)-\(^{35}\text{Cl}^+\) \(\rightarrow m/z = 72\)
\(^{37}\text{Cl}\)-\(^{37}\text{Cl}^+\) \(\rightarrow m/z = 74\)
The relative abundances are calculated by probability:
\(m/z\) 70: \(0.75 \times 0.75 = 0.5625\)
\(m/z\) 72: \(2 \times (0.75 \times 0.25) = 0.375\)
\(m/z\) 74: \(0.25 \times 0.25 = 0.0625\)
Dividing by the smallest value (0.0625) gives a ratio of 9 : 6 : 1 (or 56.25% : 37.5% : 6.25%).

[3 marks] Enthalpy change/energy required to remove one electron from each atom [1 mark] in one mole of gaseous atoms [1 mark] to form gaseous 1+ ions [1 mark].
[2 marks] \(\text{Na}^{2+}(g) \rightarrow \text{Na}^{3+}(g) + e^-\) (Species correct: [1 mark], State symbols correct: [1 mark]).
[3 marks] Sulfur has a paired electron in a 3p orbital / \(3p^4\) configuration [1 mark]; Phosphorus has singly occupied 3p orbitals / \(3p^3\) configuration [1 mark]; Repulsion between paired electrons in the same orbital makes the outer electron easier to remove in sulfur [1 mark].
[2 marks] \(m/z\) values: 70, 72, and 74 [1 mark]. Ratio: 9 : 6 : 1 (or 56.25% : 37.5% : 6.25%) [1 mark].

(a) \(\text{ClF}_3\):
- Chlorine has 7 valence electrons, shares 3 with fluorine, leaving 2 lone pairs (total 5 electron pairs).
- Shape: T-shaped.
- Bond angle: ~\(87.5^\circ\) (accept any angle between \(84^\circ\) and \(89^\circ\)).
\(\text{ClF}_4^-\):
- Chlorine has 7 valence electrons + 1 extra negative charge = 8 valence electrons. Shares 4 with fluorine, leaving 2 lone pairs (total 6 electron pairs).
- Shape: Square planar.
- Bond angle: \(90^\circ\).
(b) Conductivity Comparison:
- Sodium fluoride (\(\text{NaF}\)): Has a giant ionic lattice. In the solid state, ions are fixed in position and cannot move, so it does not conduct. In the molten/liquid state, the ionic bonds are broken and the ions (\(\text{Na}^+\) and \(\text{F}^-\)) are free to move and carry charge, so it conducts electrical current.
- Silicon tetrafluoride (\(\text{SiF}_4\)): Has a simple molecular structure with covalent bonding. It does not conduct electricity in either solid or liquid states because there are no free-moving ions or delocalised electrons to carry charge.

[6 marks] \(\text{ClF}_3\) shape drawn with 2 lone pairs shown on Cl [1 mark], Name of shape: T-shaped [1 mark], Bond angle: ~\(87.5^\circ\) (accept 84-89) [1 mark]. \(\text{ClF}_4^-\) shape drawn with 2 lone pairs (above and below the plane) [1 mark], Name of shape: Square planar [1 mark], Bond angle: \(90^\circ\) [1 mark].
[4 marks] \(\text{NaF}\): does not conduct as solid but conducts as liquid [1 mark] because ions are fixed in solid but free to move in liquid [1 mark]. \(\text{SiF}_4\): does not conduct in solid or liquid [1 mark] because it is simple molecular/has no mobile ions or delocalised electrons [1 mark].

(a) Across Period 3, the atomic radius decreases. This is because the number of protons in the nucleus increases (increasing nuclear charge), while the extra electrons are added to the same main energy level, providing similar shielding. Therefore, the attraction between the nucleus and the outer electrons becomes stronger, pulling the shell closer to the nucleus.
(b) Silicon dioxide has the formula \(\text{SiO}_2\); it is an acidic oxide (insoluble in water but reacts with concentrated bases). Phosphorus(V) oxide has the formula \(\text{P}_4\text{O}_{10}\) (or \(\text{P}_2\text{O}_5\)); it is a strongly acidic oxide that reacts vigorously with water to form phosphoric(V) acid.
Equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) (or \(\text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_4\)).
(c) Both sulfur and phosphorus exist as simple molecules held together by van der Waals forces. \(\text{S}_8\) is a larger molecule with more electrons than \(\text{P}_4\). As a result, the van der Waals forces between \(\text{S}_8\) molecules are stronger and require more thermal energy to overcome, leading to a higher melting point.

[3 marks] Trend: Atomic radius decreases [1 mark]. Reason: Nuclear charge / proton number increases [1 mark]. Shielding remains similar / electrons in same shell, leading to stronger attraction of outer shell [1 mark].
[5 marks] \(\text{SiO}_2\) [1 mark], \(\text{P}_4\text{O}_{10}\) (or \(\text{P}_2\text{O}_5\)) [1 mark]. Acid-base character: \(\text{SiO}_2\) is acidic, \(\text{P}_4\text{O}_{10}\) is strongly acidic [1 mark]. Equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) [2 marks - 1 for reactants and products, 1 for balancing].
[2 marks] \(\text{S}_8\) is larger / has more electrons than \(\text{P}_4\) [1 mark], so van der Waals forces between \(\text{S}_8\) molecules are stronger / require more energy to break [1 mark].

(a)(i) Oxidation state of N in \(\text{NO}_3^-\) is +5. Oxidation state of N in \(\text{NH}_4^+\) is -3.
(a)(ii) Oxidation half-equation: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\). Reduction half-equation: \(\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}\).
(a)(iii) To equalise the electrons transferred, multiply the zinc oxidation half-equation by 4: \(4\text{Zn} \rightarrow 4\text{Zn}^{2+} + 8e^-\). Combining this with the reduction half-equation gives: \(4\text{Zn} + \text{NO}_3^- + 10\text{H}^+ \rightarrow 4\text{Zn}^{2+} + \text{NH}_4^+ + 3\text{H}_2\text{O}\). (so \(a=4\), \(b=1\), \(c=10\), \(d=4\), \(e=1\), \(f=3\)).
(b)(i) Dichromate half-equation: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\).
Iodide half-equation: \(2\text{I}^- \rightarrow \text{I}_2 + 2e^-\). Multiply by 3 to balance electrons: \(6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-\).
Overall ionic equation: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}\).
(b)(ii) The initial solution of acidified potassium dichromate(VI) is orange. Upon reaction, iodine is produced, which turns the solution brown (or green-brown due to the green chromium(III) ions mixing with brown iodine).

[2 marks] (a)(i) N in \(\text{NO}_3^-\) is +5 [1 mark], N in \(\text{NH}_4^+\) is -3 [1 mark].
[2 marks] (a)(ii) \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\) [1 mark], \(\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}\) [1 mark].
[2 marks] (a)(iii) \(4\text{Zn} + \text{NO}_3^- + 10\text{H}^+ \rightarrow 4\text{Zn}^{2+} + \text{NH}_4^+ + 3\text{H}_2\text{O}\) (Reactants and products correct: [1 mark], balancing correct: [1 mark]).
[3 marks] (b)(i) \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}\) (Species correct: [1 mark], balanced \(\text{H}^+/\text{H}_2\text{O}\): [1 mark], fully balanced: [1 mark]).
[1 mark] (b)(ii) Orange to brown / green-brown [1 mark] (accept orange to green/brown).

(a) Trends in solubility:
- Group 2 hydroxides become more soluble down the group.
- Group 2 sulfates become less soluble down the group.
Uses:
- Magnesium hydroxide is used in medicine as an antacid (to neutralise excess stomach acid / treat indigestion) or as a laxative.
- Barium sulfate is used in medicine as a 'barium meal' (contrast medium) for X-ray imaging of the digestive system.
(b) Ionic equation: \(\text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Mg(OH)}_2(s)\).
Observation: A white precipitate is formed.
(c)(i) Equation: \(\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2\).
(c)(ii) Comparison: Barium is more reactive than calcium.
Explanation: Barium has more electron shells and thus a larger atomic radius and more shielding than calcium. This reduces the electrostatic attraction between the positive nucleus and the outer shell electrons, making the outer electrons easier to lose (lower first and second ionisation energies).

[4 marks] Hydroxides solubility increases and sulfates solubility decreases [1 mark for both]. Mg(OH)2 use: antacid / laxative [1 mark]. BaSO4 use: barium meal / X-ray contrast [1 mark].
[2 marks] \(\text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Mg(OH)}_2(s)\) (equation with state symbols: [1 mark]), white precipitate [1 mark].
[1 mark] (c)(i) \(\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2\) [1 mark].
[3 marks] (c)(ii) Barium is more reactive [1 mark]. Barium has a larger atomic radius / more electron shielding [1 mark]. Outer electrons are less strongly attracted / easier to lose [1 mark].

(a)(i) Equation: \(\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HClO}\) (accept single-headed arrow).
(a)(ii) Benefit: Kills disease-causing microorganisms / bacteria / pathogens, preventing water-borne diseases.
Risk: Chlorine is toxic / can react with naturally occurring organic substances in water to form chlorinated organic compounds (such as trihalomethanes) which are carcinogenic.
(b)(i) Equation: \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) (or \(2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl}\)).
Role of sulfuric acid: Proton donor / acid.
(b)(ii) Yellow solid: Sulfur (\(\text{S}\)).
Gas with the smell of rotten eggs: Hydrogen sulfide (\(\text{H}_2\text{S}\)).
Half-equation: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8e^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) (or \(\text{SO}_4^{2-} + 10\text{H}^+ + 8e^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)).
(b)(iii) Explanation: Iodide ions (\(\text{I}^-\)) are larger and have more shielding than chloride ions (\(\text{Cl}^-\)). Therefore, the outer electrons on iodide ions are less strongly attracted to the nucleus, making them easier to lose. This means iodide ions are much stronger reducing agents than chloride ions and can reduce sulfuric acid all the way to sulfur and hydrogen sulfide, whereas chloride ions are not strong enough reducing agents to reduce sulfuric acid (only an acid-base reaction occurs).

[1 mark] (a)(i) \(\text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{HCl} + \text{HClO}\) (accept reversible arrow) [1 mark].
[2 marks] (a)(ii) Benefit: Kills pathogens/bacteria [1 mark]. Risk: Toxic / forms chlorinated organic compounds / carcinogens [1 mark].
[2 marks] (b)(i) \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) [1 mark]. Role: acid / proton donor [1 mark].
[3 marks] (b)(ii) Yellow solid: Sulfur / \(\text{S}\) [1 mark]. Rotten egg gas: Hydrogen sulfide / \(\text{H}_2\text{S}\) [1 mark]. Half-equation: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8e^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) [1 mark].
[2 marks] (b)(iii) Iodide is a stronger reducing agent than chloride [1 mark]. Iodide ion is larger / has more shielding / holds outer electrons less strongly, so loses electrons more easily [1 mark].

For part (a), the heat energy released is calculated using \(q = m c \Delta T\). Here, \(m = 150.0\\ \text{g}\), \(c = 4.18\\ \text{J g}^{-1}\\ \text{K}^{-1}\), and \(\Delta T = 41.8 - 20.2 = 21.6\\ \text{K}\). Thus, \(q = 150.0 \times 4.18 \times 21.6 = 13543.2\\ \text{J} = 13.54\\ \text{kJ}\). For part (b), the molar mass of methanol is \(32.0\\ \text{g mol}^{-1}\). Moles of methanol = \(0.800 / 32.0 = 0.0250\\ \text{mol}\). The enthalpy change is \(\Delta H_c = -q / n = -13.5432 / 0.0250 = -541.7\\ \text{kJ mol}^{-1}\) (rounds to \(-542\\ \text{kJ mol}^{-1}\)). For part (c), incomplete combustion of methanol or evaporation of alcohol from the wick can cause this. For part (d), using a lid or a draft shield reduces heat loss.

(a) Calculation of heat: \(q = 150.0 \times 4.18 \times 21.6 = 13.54\\ \text{kJ}\) (1 mark for correct calculation of temperature rise and heat, 1 mark for correct unit and conversion to kJ). (b) Moles of methanol = 0.0250 mol (1 mark). Division of energy by moles with correct negative sign: \(-13.54 / 0.0250 = -542\\ \text{kJ mol}^{-1}\) (1 mark for calculation, 1 mark for negative sign and final value). (c) Incomplete combustion / evaporation of methanol (1 mark). (d) Use a lid / wind shield / insulate the calorimeter (1.78 marks).

Part (a): The electrophilic addition of HBr to but-1-ene can yield either 1-bromobutane or 2-bromobutane. 2-bromobutane is the major product because it is formed via the secondary carbocation, \(CH_3CH_2CH^+CH_3\), which is more stable than the primary carbocation, \(CH_3CH_2CH_2CH_2^+\). This stability is due to the electron-releasing inductive effect of the two alkyl groups attached to the positively charged carbon, which disperses the positive charge. Part (b): The mechanism begins with the C=C pi bond attacking the \(H^{\delta+}\) of the H-Br molecule. Simultaneously, the H-Br bond breaks heterolytically, yielding a bromide ion. The secondary carbocation intermediate is formed. Finally, the lone pair on the bromide ion attacks the positively charged carbon to yield 2-bromobutane.

(a) IUPAC name: 2-bromobutane (1 mark). Stability explanation: Secondary carbocation is more stable than primary carbocation (1 mark). Inductive effect explanation: Two alkyl groups release electron density towards the carbocation center, stabilizing it more than one alkyl group does (1.78 marks). (b) Curly arrow from C=C double bond to H of HBr (1 mark). Curly arrow from H-Br bond to Br (1 mark). Structure of secondary carbocation shown with positive charge on carbon 2 (1 mark). Curly arrow from lone pair on bromide ion to the carbocation carbon (1 mark).

Part (a): At temperature \(T_2\), which is higher than \(T_1\), the Maxwell-Boltzmann distribution shifts to the right and has a lower peak. The total area under both curves remains constant. The activation energy \(E_a\) is a fixed point on the x-axis. At \(T_2\), the area under the curve to the right of \(E_a\) is significantly larger. This shows that a much larger fraction of molecules have energy greater than or equal to the activation energy, leading to a higher frequency of successful collisions and thus a faster rate. Part (b): A catalyst increases the reaction rate by providing an alternative pathway with lower activation energy. This means that at a given temperature, a greater fraction of reactant molecules have sufficient energy to react, increasing the rate of successful collisions.

(a) Curve for \(T_1\) starting at origin and asymptotic to the energy axis (1 mark). Curve for \(T_2\) showing a lower peak, shifted to the right, crossing the \(T_1\) curve once (1 mark). Activation energy \(E_a\) labeled on the x-axis with shaded areas (1 mark). Explanation that at higher temperature, a larger fraction of molecules have energy \(\ge E_a\) (1 mark). Greater frequency of successful collisions (0.78 marks). (b) Catalyst provides an alternative reaction pathway (1 mark) with lower activation energy (1 mark). Thus more molecules have energy \(\ge\) this lower activation energy (1 mark).

Part (a): UV light provides the quantum of energy necessary to break the covalent bond in the chlorine molecule homolytically, forming chlorine free radicals. Part (b): Initiation: \(Cl_2 \rightarrow 2Cl^{\bullet}\). Propagation Step 1: \(CH_4 + Cl^{\bullet} \rightarrow CH_3^{\bullet} + HCl\). Propagation Step 2: \(CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}\). Termination: Any reaction where two radicals combine to form a stable molecule, such as \(2CH_3^{\bullet} \rightarrow C_2H_6\) or \(CH_3^{\bullet} + Cl^{\bullet} \rightarrow CH_3Cl\) or \(2Cl^{\bullet} \rightarrow Cl_2\). Part (c): A mixture is obtained because the product chloromethane can undergo further propagation steps with chlorine radicals, leading to dichloromethane, trichloromethane, and tetrachloromethane. To maximize chloromethane, a high ratio of methane to chlorine is used so that chlorine radicals are more likely to collide with methane than with chloromethane.

(a) Homolytic fission of the Cl-Cl bond / creation of chlorine free radicals (1 mark). (b) Initiation equation (1 mark). First propagation equation (1 mark). Second propagation equation (1 mark). Correct termination equation (1 mark). (c) Explanation: further substitution occurs as chloromethane reacts with chlorine radicals (1 mark). Condition: use a large excess of methane (0.78 marks).

Part (a): The change in moles of \(SO_2\) is \(1.50 - 0.30 = 1.20\\ \text{mol}\). Based on the stoichiometry of the reaction, \(1.20\\ \text{mol}\) of \(SO_2\) reacts with \(0.60\\ \text{mol}\) of \(O_2\) to produce \(1.20\\ \text{mol}\) of \(SO_3\). Therefore, at equilibrium, \(n(O_2) = 1.00 - 0.60 = 0.40\\ \text{mol}\), and \(n(SO_3) = 1.20\\ \text{mol}\). Part (b): The equilibrium constant expression is \(K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}\). Part (c): First, calculate the equilibrium concentrations by dividing the moles by the volume (2.00 \(\text{dm}^3\)): \([SO_2] = 0.30 / 2.00 = 0.15\\ \text{mol dm}^{-3}\), \([O_2] = 0.40 / 2.00 = 0.20\\ \text{mol dm}^{-3}\), \([SO_3] = 1.20 / 2.00 = 0.60\\ \text{mol dm}^{-3}\\). Substitute these values into the \(K_c\) expression: \(K_c = \frac{0.60^2}{0.15^2 \times 0.20} = \frac{0.36}{0.0225 \times 0.20} = 80\\. The units of \)K_c\) are \(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^3} = \text{mol}^{-1}\\ \text{dm}^3\).

(a) Moles of \(SO_3\) at equilibrium is 1.20 mol (1 mark). Moles of \(O_2\) at equilibrium is 0.40 mol (1 mark). (b) Correct expression for \(K_c\) (1 mark). (c) Division of moles by 2.00 to calculate concentrations (1 mark). Correct substitution of concentrations into the formula (1 mark). Final calculated value of 80 (1 mark). Units: \(mol^{-1}\ dm^3\) (0.78 marks).

Part (a): To oxidize the primary alcohol butan-1-ol to the aldehyde butanal, we use acidified potassium dichromate(VI) (\(H^+ / K_2Cr_2O_7\)) and heat gently under distillation conditions so that the volatile aldehyde is distilled off immediately, preventing further oxidation. The orange dichromate(VI) ions are reduced to green chromium(III) ions. Part (b): Heating under reflux with excess oxidizing agent ensures complete oxidation to the carboxylic acid, which is butanoic acid. Part (c): Dehydration is an elimination reaction that removes water from butan-1-ol. The equation is \(CH_3CH_2CH_2CH_2OH \rightarrow CH_3CH_2CH=CH_2 + H_2O\). The catalyst is concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)), and the major alkene product is but-1-ene.

(a) Reagents: Acidified potassium dichromate(VI) / \(H^+ / K_2Cr_2O_7\) (1 mark). Conditions: Distillation / warm and distill (1 mark). Colour change: Orange to green (1 mark). (b) Product: Butanoic acid (1 mark). (c) Correct structural equation (1 mark). Catalyst: Concentrated sulfuric or phosphoric acid (1 mark). Product: But-1-ene (0.78 marks).

Part (a): In 1-bromobutane, the carbon-bromine bond is polar due to the electronegativity difference, with the carbon being \(\delta+\) and the bromine being \(\delta-\). The hydroxide ion acts as a nucleophile; a curly arrow starts from the lone pair on the oxygen of \(OH^-\) and points to the \(\delta+\) carbon. Another curly arrow starts from the C-Br bond and points to the Br atom. This results in the formation of butan-1-ol and a bromide ion. Part (b): 1-iodobutane reacts faster than 1-bromobutane. This is because the C-I bond is weaker (has a lower bond enthalpy) than the C-Br bond due to the larger atomic radius of iodine, making the bond longer and easier to break. Bond enthalpy is the dominant factor determining the rate of nucleophilic substitution.

(a) Correct dipole representation on the C-Br bond (1 mark). Curly arrow from lone pair on \(OH^-\)\ oxygen to the \(\delta+\) carbon (1 mark). Curly arrow from C-Br bond to Br (1 mark). Correct products: butan-1-ol and bromide ion (1 mark). (b) 1-iodobutane has a faster rate of hydrolysis (1 mark). The C-I bond is weaker / has a lower bond enthalpy than the C-Br bond (1 mark). Bond enthalpy, rather than bond polarity, determines the rate of reaction (0.78 marks).

Part (a): The three alkene isomers are: 1. But-1-ene, \(CH_3CH_2CH=CH_2\); 2. But-2-ene, \(CH_3CH=CHCH_3\); 3. 2-methylpropene, \(CH_2=C(CH_3)_2\). Part (b): But-2-ene exhibits E-Z (stereoisomerism). This arises because of the restricted rotation around the carbon-carbon double bond, which consists of a sigma and a pi bond. Furthermore, each of the two carbon atoms involved in the double bond is bonded to two distinct groups: a hydrogen atom and a methyl group. Part (c): In \(E\)-but-2-ene, the high-priority methyl groups are on opposite sides of the double bond. In \(Z\)-but-2-ene, the methyl groups are on the same side.

(a) Structural formula and name of but-1-ene (1 mark). Structural formula and name of but-2-ene (1 mark). Structural formula and name of 2-methylpropene (1 mark). (b) Identifies but-2-ene (1 mark). Explanation of restricted rotation around C=C (1 mark). Explanation that each double-bonded carbon is attached to two different groups (0.78 marks). (c) Correctly drawn and labeled structure of \(E\)-but-2-ene (1 mark). Correctly drawn and labeled structure of \(Z\)-but-2-ene (1 mark).

(a) The balanced chemical equation using structural formulas:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{NaBr}\)

(b) Mechanism:
1. The hydroxide ion, \(\text{:OH}^-\), acts as a nucleophile. Draw a lone pair on the oxygen of the hydroxide ion and a negative charge.
2. Label the carbon atom bonded to bromine as \(\delta^+\) and the bromine atom as \(\delta^-\).
3. Draw a curly arrow starting from the lone pair on the oxygen of the hydroxide ion to the \(\delta^+\) carbon atom.
4. Draw a second curly arrow starting from the \(\text{C}-\text{Br}\) bond to the \(\delta^-\)\ bromine atom.
5. The products of this step are butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)) and a bromide ion (\(\text{Br}^-\)).

(c) Trend and Explanation:
- Trend: The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane (i.e., 1-iodobutane reacts the fastest and 1-chlorobutane reacts the slowest).
- Reason 1: The C-X bond enthalpy decreases down the group (\(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\)), meaning the C-I bond is the weakest and requires the least energy to break.
- Reason 2: Bond enthalpy (bond strength) is the dominant factor determining the rate of nucleophilic substitution in halogenoalkanes, overriding the effect of bond polarity.

(d) Precipitate and Ionic Equation:
- Color: Yellow
- Ionic Equation: \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{AgI}(\text{s})\)

(a) [1 mark]
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{NaBr}\) (or ionic equivalent: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{Br}^-\)). Do not accept molecular formula \(\text{C}_4\text{H}_{9}\text{Br}\).

(b) [3 marks]
- M1: Curly arrow from lone pair on \(\text{:OH}^-\)'s oxygen to the C atom bonded to Br.
- M2: Correct dipoles: \(\delta^+\) on C atom and \(\delta^-\) on Br atom.
- M3: Curly arrow from \(\text{C}-\text{Br}\) bond to Br atom.

(c) [3 marks]
- M1: Correct trend: rate increases from 1-chlorobutane to 1-iodobutane (or 1-iodobutane is fastest / 1-chlorobutane is slowest).
- M2: C-I bond is weaker than C-Br bond, which is weaker than C-Cl bond / C-X bond strength/enthalpy decreases from Cl to I.
- M3: State that bond enthalpy/strength is the dominant factor (not bond polarity) or that the weaker C-I bond breaks most easily.

(d) [1 mark]
- Yellow precipitate AND \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{AgI}(\text{s})\). (Both required for 1 mark, including state symbols).

Part (a): The expression for \( K_c \) is \( K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \).

Part (b): Using the stoichiometry of the equation:
- Initial moles: \( \text{PCl}_5 = 1.20 \), \( \text{PCl}_3 = 0.00 \), \( \text{Cl}_2 = 0.00 \).
- Change in moles: \( \text{Cl}_2 \) increases by \( +0.40 \), so \( \text{PCl}_3 \) increases by \( +0.40 \) and \( \text{PCl}_5 \) decreases by \( -0.40 \).
- Equilibrium moles: \( \text{PCl}_5 = 1.20 - 0.40 = 0.80 \) mol; \( \text{PCl}_3 = 0.40 \) mol; \( \text{Cl}_2 = 0.40 \) mol.
- Equilibrium concentrations (divided by volume \( 2.00\text{ dm}^3 \)):
\( [\text{PCl}_5] = 0.80 / 2.00 = 0.40\text{ mol dm}^{-3} \)
\( [\text{PCl}_3] = 0.40 / 2.00 = 0.20\text{ mol dm}^{-3} \)
\( [\text{Cl}_2] = 0.40 / 2.00 = 0.20\text{ mol dm}^{-3} \).
- Substituting values into the \( K_c \) expression: \( K_c = \frac{0.20 \times 0.20}{0.40} = 0.10 \).
- Units of \( K_c \): \( \frac{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}{\text{mol dm}^{-3}} = \text{mol dm}^{-3} \).

Part (c): Since the forward reaction is endothermic, increasing the temperature shifts the equilibrium position to the right to absorb the added heat. This increases the concentration of products and decreases the concentration of reactants, so the value of \( K_c \) increases.

Part (a): [1 mark] for the correct expression of \( K_c \) with square brackets.
Part (b): [1 mark] for determining equilibrium moles of all species (\( \text{PCl}_5 = 0.80 \), \( \text{PCl}_3 = 0.40 \), \( \text{Cl}_2 = 0.40 \)). [1 mark] for calculating concentrations of all species by dividing moles by 2.00. [1 mark] for correct substitution of concentrations into \( K_c \) expression. [1 mark] for the final value of 0.10. [1 mark] for the correct units (\( \text{mol dm}^{-3} \)).
Part (c): [1 mark] for stating that \( K_c \) increases. [1 mark] for explaining that the equilibrium shifts to the right (endothermic direction) to oppose the increase in temperature.

Part (a): When concentrated sulfuric acid is added to solid sodium chloride, steamy white fumes of hydrogen chloride gas are observed. The reaction is: \( \text{NaCl(s)} + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl(g)} \) (or forming \( \text{Na}_2\text{SO}_4 \)).

Part (b): Hydrogen sulfide has a characteristic bad egg smell. Iodide ions are larger than chloride ions because they have more electron shells. This results in greater shielding, meaning the outer electron is further from the nucleus and less strongly attracted. Therefore, iodide ions lose electrons more easily and act as stronger reducing agents.

Part (c): In \( \text{H}_2\text{SO}_4 \), sulfur has an oxidation state of +6. In \( \text{H}_2\text{S} \), sulfur has an oxidation state of -2. The half-equation is: \( \text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O} \).

Part (a): [1 mark] for white steamy fumes. [1 mark] for the correct balanced equation (accept state symbols omitted).
Part (b): [1 mark] for bad egg smell / smell of rotten eggs. [1 mark] for explaining that iodide ions are larger / have more electron shells / more shielded. [1 mark] for stating that the outer electron is more easily lost.
Part (c): [1 mark] for identifying reactants and products (\( \text{H}_2\text{SO}_4 \) and \( \text{H}_2\text{S} \)). [1 mark] for correct number of electrons (\( 8\text{e}^- \)) and protons (\( 8\text{H}^+ \)). [1 mark] for overall balancing including \( 4\text{H}_2\text{O} \).

Part (a): Going down Group 2, the solubility of the hydroxides increases, whereas the solubility of the sulfates decreases.

Part (b): Barium sulfate is virtually insoluble in water and acid, meaning it does not dissolve or dissociate in the stomach. As a result, toxic barium ions (\( \text{Ba}^{2+} \)) cannot be absorbed into the bloodstream and it passes safely through the body.

Part (c): Add aqueous sodium hydroxide (or any soluble hydroxide). Magnesium sulfate will react to form a white precipitate of magnesium hydroxide because magnesium hydroxide is sparingly soluble. Barium chloride will show no visible reaction (remains a colourless solution) because barium hydroxide is highly soluble. The ionic equation for the reaction is: \( \text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s}) \).

Part (a): [1 mark] for hydroxide solubility increases down the group. [1 mark] for sulfate solubility decreases down the group.
Part (b): [1 mark] for stating that barium sulfate is insoluble. [1 mark] for explaining that it cannot be absorbed into the body / does not release free toxic \( \text{Ba}^{2+} \) ions.
Part (c): [1 mark] for naming a suitable reagent (aqueous sodium hydroxide / NaOH). [1 mark] for white precipitate with magnesium sulfate. [1 mark] for no reaction / colourless solution with barium chloride. [1 mark] for correct ionic equation with state symbols (\( \text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s}) \)).

Part (a): When temperature increases, the peak of the Maxwell-Boltzmann distribution curve shifts to the right (higher energy) and becomes lower. The entire curve flattens out, representing a higher average energy of the molecules.

Part (b): For a reaction to occur, colliding molecules must have energy greater than or equal to the activation energy (\( E_a \)). At a higher temperature, the distribution curve shifts so that a much larger fraction of molecules have energy \( \ge E_a \). This significantly increases the frequency of successful collisions, leading to a much higher reaction rate.

Part (c): A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy (\( E_c < E_a \)). This means that at the same temperature, a greater proportion of the reacting molecules have kinetic energy greater than or equal to this lower activation energy, increasing the rate of successful collisions.

Part (a): [1 mark] for peak shifting to the right (higher energy). [1 mark] for peak height being lower.
Part (b): [1 mark] for mentioning that molecules must collide with energy \( \ge E_a \) to react. [1 mark] for stating that at a higher temperature, a larger fraction/proportion of molecules have energy \( \ge E_a \). [1 mark] for increased frequency of collisions. [1 mark] for higher frequency of successful/effective collisions.
Part (c): [1 mark] for stating that the catalyst provides an alternative pathway with lower activation energy. [1 mark] for explaining that a greater proportion of molecules have energy \( \ge \) this lower activation energy.

Part (a): In \( \text{Cr}_2\text{O}_7^{2-} \), \( 2x + 7(-2) = -2 \Rightarrow 2x = 12 \Rightarrow x = +6 \). In \( \text{Cr}^{3+} \), the oxidation state is +3.

Part (b): The half-equations are:
Oxidation: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \)
Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)
Multiplying the oxidation half-equation by 6 and combining gives:
\( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \).

Part (c):
1. Calculate the moles of \( \text{Cr}_2\text{O}_7^{2-} \) used:
\( n(\text{Cr}_2\text{O}_7^{2-}) = \text{concentration} \times \text{volume} = 0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\text{ mol} \).
2. Determine moles of \( \text{Fe}^{2+} \) using the 1:6 ratio from the equation:
\( n(\text{Fe}^{2+}) = 6 \times n(\text{Cr}_2\text{O}_7^{2-}) = 6 \times 4.50 \times 10^{-4} = 2.70 \times 10^{-3}\text{ mol} \).
3. Calculate concentration of \( \text{Fe}^{2+} \) in the \( 25.0\text{ cm}^3 \) sample:
\( [\text{Fe}^{2+}] = \frac{\text{moles}}{\text{volume}} = \frac{2.70 \times 10^{-3}}{0.0250} = 0.108\text{ mol dm}^{-3} \).

Part (a): [1 mark] for +6 (or 6) in \( \text{Cr}_2\text{O}_7^{2-} \). [1 mark] for +3 (or 3) in \( \text{Cr}^{3+} \).
Part (b): [1 mark] for writing correct reactants and products. [1 mark] for correctly balancing the charges with \( 14\text{H}^+ \) and \( 7\text{H}_2\text{O} \). [1 mark] for overall correct stoichiometry (1:14:6 to 2:6:7).
Part (c): [1 mark] for calculating moles of \( \text{Cr}_2\text{O}_7^{2-} \) (\( 4.50 \times 10^{-4}\text{ mol} \)). [1 mark] for multiplying by 6 to find moles of \( \text{Fe}^{2+} \) (\( 2.70 \times 10^{-3}\text{ mol} \)). [1 mark] for correct final concentration of \( 0.108\text{ mol dm}^{-3} \) (accept 3 significant figures).

Part (a): Divide the mass percentage of each element by its relative atomic mass (\( A_r \)):
- Carbon: \( 54.5 / 12.0 = 4.542 \)
- Hydrogen: \( 9.1 / 1.0 = 9.100 \)
- Oxygen: \( 36.4 / 16.0 = 2.275 \)
Divide by the smallest value (2.275):
- C: \( 4.542 / 2.275 = 2.00 \)
- H: \( 9.100 / 2.275 = 4.00 \)
- O: \( 2.275 / 2.275 = 1.00 \)
This gives the empirical formula: \( \text{C}_2\text{H}_4\text{O} \).

Part (b): The relative empirical formula mass of \( \text{C}_2\text{H}_4\text{O} \) is \( (2 \times 12.0) + (4 \times 1.0) + 16.0 = 44.0 \).
Ratio of \( M_r \) to empirical mass: \( 88.0 / 44.0 = 2 \).
Therefore, the molecular formula of \( \mathbf{X} \) is \( \text{C}_4\text{H}_8\text{O}_2 \).

Part (c): Use the ideal gas equation: \( PV = nRT \).
- Moles \( n = \frac{\text{mass}}{M_r} = \frac{0.220}{88.0} = 2.50 \times 10^{-3}\text{ mol} \).
- Temperature \( T = 100 + 273 = 373\text{ K} \).
- Pressure \( P = 101\text{ kPa} = 101000\text{ Pa} \).
Rearranging for volume: \( V = \frac{nRT}{P} = \frac{2.50 \times 10^{-3} \times 8.31 \times 373}{101000} = 7.67 \times 10^{-5}\text{ m}^3 \).
Convert volume to \( \text{cm}^3 \) (multiply by \( 10^6 \)):
\( V = 7.67 \times 10^{-5} \times 10^6 = 76.7\text{ cm}^3 \).

Part (a): [1 mark] for dividing percentages by relative atomic masses. [1 mark] for dividing by the smallest number to get ratio 2 : 4 : 1. [1 mark] for concluding empirical formula is \( \text{C}_2\text{H}_4\text{O} \).
Part (b): [1 mark] for the correct molecular formula \( \text{C}_4\text{H}_8\text{O}_2 \).
Part (c): [1 mark] for calculating moles \( n = 2.50 \times 10^{-3}\text{ mol} \). [1 mark] for converting temperature to 373 K and pressure to 101000 Pa. [1 mark] for rearranging the ideal gas equation and substituting values to get volume in \( \text{m}^3 \) (\( 7.67 \times 10^{-5}\text{ m}^3 \)). [1 mark] for correct conversion to \( \text{cm}^3 \) giving final answer \( 76.7\text{ cm}^3 \) (accept 76.6 to 76.8).

Part (a): The atomic radius decreases across Period 3 from sodium to chlorine. This is because the number of protons in the nucleus increases (increasing nuclear charge), while electrons are added to the same main outer shell, so shielding remains approximately constant. Therefore, there is a stronger electrostatic attraction between the nucleus and the outer electrons, pulling them closer to the nucleus.

Part (b): Sodium oxide is a basic oxide that reacts with water to form sodium hydroxide: \( \text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} \). Since sodium hydroxide is a strong soluble base, the resulting solution is highly alkaline with a pH of approximately 13 or 14.

Part (c): Phosphorus(V) oxide is an acidic oxide. It reacts with bases like sodium hydroxide to form a salt (sodium phosphate) and water: \( \text{P}_4\text{O}_{10} + 12\text{NaOH} \rightarrow 4\text{Na}_3\text{PO}_4 + 6\text{H}_2\text{O} \).

Part (a): [1 mark] for stating atomic radius decreases. [1 mark] for stating nuclear charge / proton number increases. [1 mark] for stating outer electrons are in the same shell / shielding is constant, leading to stronger attraction.
Part (b): [1 mark] for the correct balanced equation (accept state symbols omitted). [1 mark] for stating a pH in the range 13–14.
Part (c): [1 mark] for correct formulas of reactants (\( \text{P}_4\text{O}_{10} \) and \( \text{NaOH} \)). [1 mark] for correct products (\( \text{Na}_3\text{PO}_4 \) and \( \text{H}_2\text{O} \)). [1 mark] for correct balancing (coefficients 1, 12, 4, 6).

Part (a): The standard enthalpy of combustion is the enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions (\( 100\text{ kPa} \) and a specified temperature, usually \( 298\text{ K} \)), with all reactants and products in their standard states.

Part (b): Use the formula \( q = m c \Delta T \):
- mass of water \( m = 100.0\text{ g} \)
- temperature change \( \Delta T = 35.7 - 20.2 = 15.5\text{ K} \)
- specific heat capacity \( c = 4.18\text{ J g}^{-1}\text{ K}^{-1} \)
\( q = 100.0 \times 4.18 \times 15.5 = 6479\text{ J} = 6.48\text{ kJ} \).

Part (c):
1. Calculate the moles of propan-1-ol burned:
\( n = \frac{\text{mass}}{M_r} = \frac{0.600}{60.0} = 0.0100\text{ mol} \).
2. Calculate the enthalpy change per mole:
\( \Delta H_c = -\frac{q}{n} = -\frac{6.479\text{ kJ}}{0.0100\text{ mol}} = -648\text{ kJ mol}^{-1} \). (The negative sign must be included as combustion is exothermic).

Part (d): Some heat is lost to the surrounding air or the copper calorimeter itself, rather than heating the water. Alternatively, incomplete combustion of the alcohol can occur, yielding less energy.

Part (a): [1 mark] for enthalpy change when 1 mole of substance is burned completely in oxygen. [1 mark] for specifying standard conditions / standard states.
Part (b): [1 mark] for calculating \( \Delta T = 15.5\text{ K} \). [1 mark] for substituting correctly into the equation to get \( 6.48\text{ kJ} \) (or \( 6.5\text{ kJ} \) / \( 6479\text{ J} \)).
Part (c): [1 mark] for calculating moles of propan-1-ol as \( 0.0100\text{ mol} \). [1 mark] for dividing \( q \) by moles (value \( 648 \)). [1 mark] for writing the final answer with a negative sign (\( -648\text{ kJ mol}^{-1} \) or \( -650\text{ kJ mol}^{-1} \)).
Part (d): [1 mark] for heat loss to the surroundings / incomplete combustion / evaporation of alcohol.

(a)
\(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\)

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

(b)
Using the stoichiometry of the equation:
- Initial moles: \(\text{CO} = 1.20\), \(\text{H}_2 = 2.40\), \(\text{CH}_3\text{OH} = 0\)
- Change in moles to reach equilibrium: \(\text{CH}_3\text{OH} = +0.40\)
- Therefore, change in \(\text{CO} = -0.40\), change in \(\text{H}_2 = -2 \times 0.40 = -0.80\)

Equilibrium amounts:
- \(n(\text{CO}) = 1.20 - 0.40 = 0.80\text{ mol}\)
- \(n(\text{H}_2) = 2.40 - 0.80 = 1.60\text{ mol}\)

Equilibrium concentrations (in \(5.00\text{ dm}^3\)):
- \([\text{CH}_3\text{OH}] = \frac{0.40}{5.00} = 0.080\text{ mol dm}^{-3}\)
- \([\text{CO}] = \frac{0.80}{5.00} = 0.160\text{ mol dm}^{-3}\)
- \([\text{H}_2] = \frac{1.60}{5.00} = 0.320\text{ mol dm}^{-3}\)

Calculating \(K_c\):
\(K_c = \frac{0.080}{0.160 \times (0.320)^2} = \frac{0.080}{0.160 \times 0.1024} = \frac{0.080}{0.016384} = 4.88\text{ mol}^{-2}\text{ dm}^6\) (or \(4.9\text{ mol}^{-2}\text{ dm}^6\))

(c)
- Value of \(K_c\) decreases.
- Explanation: The forward reaction is exothermic. According to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium to the left (the endothermic direction) to absorb the heat, reducing the concentration of products and increasing the concentration of reactants.

(a)
- M1: \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\) (square brackets required, do not accept round brackets) [1 mark]
- M2: \(\text{mol}^{-2}\text{ dm}^6\) (or \(\text{dm}^6\text{ mol}^{-2}\)) [1 mark]

(b)
- M1: Deduces correct equilibrium moles: \(n(\text{CO}) = 0.80\text{ mol}\) AND \(n(\text{H}_2) = 1.60\text{ mol}\) [1 mark]
- M2: Divides all moles by \(5.00\) to find equilibrium concentrations: \([\text{CO}] = 0.160\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.320\text{ mol dm}^{-3}\), \([\text{CH}_3\text{OH}] = 0.080\text{ mol dm}^{-3}\) (allow consequential error on moles from M1) [1 mark]
- M3: Sets up the correct calculation: \(K_c = \frac{0.080}{0.160 \times 0.320^2}\) [1 mark]
- M4: Correct final calculated value of \(K_c = 4.88\) (or \(4.9\)) [1 mark]

(c)
- M1: \(K_c\) decreases [1 mark]
- M2: (Forward) reaction is exothermic / equilibrium shifts to the left / favours reverse reaction [1 mark]

(a)
The half-equation for the reduction of dichromate(VI) ions:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

(b)
The half-equation for the oxidation of iron(II) is:
\(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)

Multiplying this by 6 and combining with the dichromate(VI) half-equation gives the overall ionic equation:
\(6\text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 6\text{Fe}^{3+} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

(c)
Step 1: Calculate moles of dichromate(VI) ions used:
\(n(\text{Cr}_2\text{O}_7^{2-}) = C \times V = 0.0200\text{ mol dm}^{-3} \times \frac{22.50}{1000}\text{ dm}^3 = 4.50 \times 10^{-4}\text{ mol}\)

Step 2: Use the reaction stoichiometry to find the moles of \(\text{Fe}^{2+}\):
From the equation in (b), \(1\text{ mol}\) of \(\text{Cr}_2\text{O}_7^{2-}\) reacts with \(6\text{ mol}\) of \(\text{Fe}^{2+}\).
\(n(\text{Fe}^{2+}) = 6 \times n(\text{Cr}_2\text{O}_7^{2-}) = 6 \times 4.50 \times 10^{-4}\text{ mol} = 2.70 \times 10^{-3}\text{ mol}\)

Step 3: Calculate the concentration of \(\text{Fe}^{2+}\) in the \(25.0\text{ cm}^3\) sample:
\([\text{Fe}^{2+}] = \frac{n}{V} = \frac{2.70 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.108\text{ mol dm}^{-3}\)

(a)
- M1: For showing reactants and products correctly balanced for Cr and O:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) [1 mark]
- M2: For adding \(6\text{e}^-\); overall equation fully balanced: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) [1 mark]

(b)
- M1: \(6\text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 6\text{Fe}^{3+} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) [1 mark]

(c)
- M1: Calculates moles of dichromate(VI): \(n(\text{Cr}_2\text{O}_7^{2-}) = 0.0200 \times 0.0225 = 4.50 \times 10^{-4}\text{ mol}\) [1 mark]
- M2: Identifies the 1:6 ratio between \(\text{Cr}_2\text{O}_7^{2-}\) and \(\text{Fe}^{2+}\) (either state it or use it implicitly; allow consequential error from (b)) [1 mark]
- M3: Calculates moles of \(\text{Fe}^{2+}\):
\(n(\text{Fe}^{2+}) = 6 \times 4.50 \times 10^{-4} = 2.70 \times 10^{-3}\text{ mol}\) [1 mark]
- M4: Correct concentration calculation method: \(\text{Concentration} = \frac{2.70 \times 10^{-3}}{0.0250}\) [1 mark]
- M5: Correct final concentration: \(0.108\text{ mol dm}^{-3}\) (3 sig figs required) [1 mark]