2023 Cambridge IAL Chemistry (9701) Practice Paper with Answers

The oxide of element \(X\) is amphoteric because it reacts with both acids and bases. In Period 3, the only element that forms an amphoteric oxide is aluminium (\(\text{Al}_2\text{O}_3\)). The element immediately to the right of aluminium in Period 3 is silicon (\(\text{Si}\)). The oxide of silicon is silicon dioxide, which has the chemical formula \(\text{SiO}_2\).

1 mark: Correctly identifies the element immediately to the right of the amphoteric oxide-forming element as silicon, leading to the formula \(\text{SiO}_2\). Reject other options.

Vanadium exhibits multiple oxidation states, each associated with a characteristic color. The starting yellow solution contains vanadium in the +5 oxidation state as \([\text{VO}_2(\text{H}_2\text{O})_4]^{+}\). Reduction yields the blue \([\text{VO}(\text{H}_2\text{O})_5]^{2+}\) ion (oxidation state +4), followed by the green \([\text{V}(\text{H}_2\text{O})_6]^{3+}\) ion (oxidation state +3), and finally the violet \([\text{V}(\text{H}_2\text{O})_6]^{2+}\) ion (oxidation state +2). Thus, the green color is due to the +3 ion.

1 mark: Identifies the green species as \([\text{V}(\text{H}_2\text{O})_6]^{3+}\). Reject options representing other oxidation states of vanadium.

A stability constant represents the equilibrium constant for the formation of a complex ion. A very large value of \(K_{\text{stab}}\) indicates that the equilibrium lies far to the right, meaning the product complex \(Q\) is highly stable compared to the reactant complex \(P\). It does not provide any information regarding the rate of reaction (kinetics).

1 mark: Selects statement C because a large stability constant signifies high thermodynamic stability of the complex and that the equilibrium position lies far to the right. Reject statements A, B, and D.

The methyl group in methylbenzene is ortho/para-directing. In contrast, the carboxylic acid group (\(\text{-COOH}\)) is meta-directing (which corresponds to position 3). To prepare 3-nitrobenzoic acid, the methyl group must first be oxidized to a carboxylic acid using alkaline potassium manganate(VII) under heat followed by acidification. The resulting benzoic acid is then nitrated using concentrated nitric and sulfuric acids under heat, which directs the nitro group to the 3-position.

1 mark: Recognizes that oxidation must occur before nitration to ensure meta-direction (3-position). Reject option A as it would produce 2- or 4-nitrobenzoic acid.

To increase the carbon chain length by one from a halogenoalkane (2-bromopropane) to form compound \(Z\) (which must be a nitrile, 2-methylpropanenitrile), nucleophilic substitution with cyanide is required. This reaction utilizes potassium cyanide (\(\text{KCN}\)) dissolved in ethanol, heated under reflux.

1 mark: Correctly identifies KCN in ethanol under reflux as the correct reagent and conditions for chain-extending nucleophilic substitution of a halogenoalkane. Reject options with aqueous NaOH, HCN/NaCN, or ammonia.

Lattice energy depends on the electrostatic attraction between oppositely charged ions in a crystal lattice. This is governed by two factors: ionic charge and ionic radius. The product of ionic charges in \(\text{CaO}\) is \(|(+2) \times (-2)| = 4\), whereas in \(\text{KCl}\) it is \(|(+1) \times (-1)| = 1\). Furthermore, the ionic radii of \(\text{Ca}^{2+}\) and \(\text{O}^{2-}\) are smaller than those of \(\text{K}^{+}\) and \(\text{Cl}^{-}\). The combination of higher charges and smaller radii leads to a much stronger attraction and hence a much more exothermic lattice energy for \(\text{CaO}\).

1 mark: Correctly links higher ionic charges and smaller ionic sizes in \(\text{CaO}\) to a more exothermic lattice energy. Reject options involving ionisation energies or electron affinities as key drivers of lattice energy trends.

A reaction becomes thermodynamically feasible when the Gibbs free energy change, \(\Delta G^\ominus\), is less than or equal to zero. Using the Gibbs equation, \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), we can find the transition temperature by setting \(\Delta G^\ominus = 0\): \(T = \frac{\Delta H^\ominus}{\Delta S^\ominus} = \frac{135 \times 10^3 \text{ J mol}^{-1}}{285 \text{ J K}^{-1} \text{ mol}^{-1}} \approx 473.7 \text{ K}\). Since both enthalpy and entropy changes are positive, the reaction becomes feasible (\(\Delta G^\ominus < 0\)) at temperatures above 474 K.

1 mark: Correctly calculates the feasibility temperature threshold by converting kJ to J and evaluating the division. Reject other temperatures.

Silicon is a macromolecular (giant covalent) element. Each silicon atom is tetrahedrally bonded to four other silicon atoms by strong covalent bonds that require a very large amount of thermal energy to break, giving it the highest melting point of all elements in Period 3. Metallic elements (sodium, magnesium, aluminium) and simple molecular elements (phosphorus, sulfur, chlorine, argon) have lower melting points.

1 mark: Correctly identifies Silicon as the element with the highest melting point in Period 3 due to its giant covalent structure. Reject all other elements.

The electronic configurations of the outer shells of these elements are: Phosphorus (\(\text{P}\)): \(3\text{s}^2 3\text{p}^3\) (3 unpaired electrons in the \(3\text{p}\) subshell, following Hund's rule); Sulfur (\(\text{S}\)): \(3\text{s}^2 3\text{p}^4\) (2 unpaired electrons); Chlorine (\(\text{Cl}\)): \(3\text{s}^2 3\text{p}^5\) (1 unpaired electron); Argon (\(\text{Ar}\)): \(3\text{s}^2 3\text{p}^6\) (0 unpaired electrons). Therefore, phosphorus has the highest number of unpaired electrons.

1 mark for identifying phosphorus as having 3 unpaired electrons due to half-filled \(3\text{p}\) subshell, whereas S, Cl, and Ar have more paired electrons.

The oxide of \(X\) is amphoteric because it reacts with both acids and bases; this is aluminium oxide (\(\text{Al}_2\text{O}_3\)), so \(X\) is aluminium (\(\text{Al}\)). The oxide of \(Y\) is insoluble in water and acts as a weak acid, reacting only with hot, concentrated strong bases like \(\text{NaOH}\); this is silicon dioxide (\(\text{SiO}_2\)), so \(Y\) is silicon (\(\text{Si}\)).

1 mark for identifying \(X\) as \(\text{Al}\) due to its amphoteric oxide, and \(Y\) as \(\text{Si}\) due to its giant covalent acidic oxide which only reacts with concentrated alkali.

Aqueous \(\text{Cu}^{2+}\) ions react with excess ammonia to form the deep blue complex \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\). Aqueous \(\text{Fe}^{2+}\) ions react with sodium hydroxide to form a green precipitate of \(\text{Fe}(\text{OH})_2\), which is insoluble in excess sodium hydroxide. Note that \(\text{Cr}^{3+}\) also forms a green precipitate with \(\text{NaOH}\), but it dissolves in excess sodium hydroxide to give a dark green solution.

1 mark for identifying \(X = \text{Cu}^{2+}\) and \(Y = \text{Fe}^{2+}\) based on the observation of the deep blue complex and the insoluble green precipitate.

\(\text{Cu}^{2+}\) has a \(\text{d}^9\) configuration, containing a partially filled \(\text{3d}\) subshell. Splitting of these orbitals by ligands allows \(\text{d-d}\) transitions where red-orange light is absorbed, and the complementary blue light is transmitted. \(\text{Zn}^{2+}\) has a \(\text{d}^{10}\) configuration, so its \(\text{3d}\) subshell is completely filled, meaning no \(\text{d-d}\) transition can occur, making it colourless.

1 mark for the correct explanation of \(\text{d-d}\) transitions in \(\text{Cu}^{2+}\) with a partially filled \(\text{3d}\) subshell and the absence of such transitions in the fully filled \(\text{3d}^{10}\) of \(\text{Zn}^{2+}\).

Step 1 involves the reduction of the nitro group to an amine, which is done using tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\) under reflux, followed by addition of aqueous \(\text{NaOH}\) to liberate the phenylamine from its salt. Step 2 involves diazotisation using nitrous acid (generated from \(\text{NaNO}_2\) and \(\text{HCl}\)) at low temperature (\(0-10\ ^\circ\text{C}\)) to prevent decomposition of the diazonium salt.

1 mark for selecting the correct reagent combination and temperature conditions for both steps.

Step 1 is nucleophilic substitution of a halogenoalkane to form an alcohol, which requires aqueous sodium hydroxide, \(\text{NaOH}(\text{aq})\), heated under reflux. Using \(\text{NaOH}\) in ethanol would lead to elimination, producing propene. Step 2 is the oxidation of a secondary alcohol to a ketone, which requires an oxidizing agent such as acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\), heated under reflux.

1 mark for identifying the correct aqueous conditions for substitution in Step 1 and the oxidizing agent for Step 2.

Using the Born-Haber cycle equation: \(\Delta H_{\text{f}}^{\ominus} = \Delta H_{\text{at}}^{\ominus}[\text{Na}(\text{s})] + \text{IE}_1[\text{Na}(\text{g})] + \Delta H_{\text{at}}^{\ominus}[\text{Cl}_2(\text{g})] + \text{EA}_1[\text{Cl}(\text{g})] + \Delta H_{\text{latt}}^{\ominus}\). Substitute the values: \(-411 = +107 + 496 + 121 - 349 + \Delta H_{\text{latt}}^{\ominus}\) which simplifies to \(-411 = 375 + \Delta H_{\text{latt}}^{\ominus}\). Thus, \(\Delta H_{\text{latt}}^{\ominus} = -411 - 375 = -786\text{ kJ mol}^{-1}\).

1 mark for setting up the Born-Haber equation correctly and calculating the lattice energy as \(-786\text{ kJ mol}^{-1}\).

\(\text{SiCl}_4\) is a simple molecular covalent compound. When added to water, it undergoes complete and rapid hydrolysis to form silicon dioxide (or orthosilicic acid) and hydrogen chloride gas, which dissolves to form a highly acidic solution with a \(\text{pH}\) of approximately \(1-2\): \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})\).

1 mark for identifying covalent bonding in \(\text{SiCl}_4\) and the strongly acidic \(\text{pH}\) (\(1-2\)) resulting from the hydrolysis product \(\text{HCl}\).

The properties of the chloride and oxide described correspond to silicon.

- Silicon tetrachloride (\(\text{SiCl}_4\)) reacts vigorously with water to form white, misty fumes of hydrogen chloride (\(\text{HCl}\)) and a strongly acidic solution:
\(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g})\)

- Silicon dioxide (\(\text{SiO}_2\)) is an acidic oxide that is giant covalent (macromolecular) and therefore insoluble in water. It does not react with acids like nitric acid, but it reacts with hot, concentrated sodium hydroxide to form soluble sodium silicate:
\(\text{SiO}_2(\text{s}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SiO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

Aluminium oxide is amphoteric and would react with nitric acid. Phosphorus and sulfur oxides are soluble in water.

1 mark for identifying silicon as the correct element based on the properties of its chloride and oxide.

The reaction is a ligand substitution reaction:

\([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)

- The hexaaquacopper(II) ion, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), has a coordination number of 6 and an octahedral geometry.
- The tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\), has a coordination number of 4 and a tetrahedral geometry (due to the larger size and mutual repulsion of the chloride ligands).
- The oxidation state of copper remains +2 in both complexes, so this is not a redox reaction.

1 mark for identifying that the geometry changes from octahedral to tetrahedral.

To synthesize 2-hydroxybutanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{COOH}\)) from propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)):

1. Propan-1-ol must be oxidized to propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)). Since propanal is an aldehyde, we must heat the reaction mixture and immediately distil off the product to prevent further oxidation to propanoic acid. Acidified potassium dichromate(VI) is used.

2. Propanal is then reacted with HCN in the presence of NaCN (nucleophilic addition) to form 2-hydroxybutanenitrile, \(\text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CN}\).

3. The nitrile group (\(-\text{CN}\)) is then hydrolyzed to a carboxylic acid group (\(-\text{COOH}\)) by heating under reflux with dilute hydrochloric acid, forming 2-hydroxybutanoic acid.

1 mark for selecting the correct three-step sequence of oxidation (distil), nucleophilic addition, and acid hydrolysis.

For a reaction to change from spontaneous to non-spontaneous, we find the temperature at which the Gibbs free energy change, \(\Delta G^\ominus\), is equal to zero:

\(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus = 0\)

Therefore:

\(T = \frac{\Delta H^\ominus}{\Delta S^\ominus}\)

Ensure that the units of \(\Delta H^\ominus\) and \(\Delta S^\ominus\) are consistent. Converting \(\Delta H^\ominus\) to \(\text{J mol}^{-1}\):

\(\Delta H^\ominus = -114 \times 1000 = -114000\text{ J mol}^{-1}\)

\(T = \frac{-114000\text{ J mol}^{-1}}{-146\text{ J K}^{-1}\text{ mol}^{-1}} \approx 780.82\text{ K}\)

Rounding to three significant figures gives 781 K. The reaction is spontaneous below 781 K and non-spontaneous above 781 K.

1 mark for correct calculation of T = 781 K using consistent units.

Let's analyze each clue:

- Element \(W\) reacts vigorously with cold water to form an alkaline solution and a gas. This is characteristic of sodium (\(\text{Na}\)), which forms \(\text{NaOH}(\text{aq})\) and \(\text{H}_2(\text{g})\).
- Element \(X\) forms an oxide with a macromolecular (giant covalent) structure. This is silicon (\(\text{Si}\)), which forms silicon dioxide (\(\text{SiO}_2\)).
- Element \(Y\) has a lower first ionisation energy than the element immediately preceding it in the Periodic Table, and forms an oxide that is amphoteric. This is aluminium (\(\text{Al}\)). Its first ionisation energy (\(578\text{ kJ mol}^{-1}\)) is lower than that of magnesium (\(738\text{ kJ mol}^{-1}\)) because its outer electron is in a \(3\text{p}\) orbital rather than a \(3\text{s}\) orbital. Its oxide (\(\text{Al}_2\text{O}_3\)) is amphoteric.
- Element \(Z\) has the highest first ionisation energy of the four and forms an oxide which is a gas at room temperature. Sulfur (\(\text{S}\)) has a first ionisation energy of \(1000\text{ kJ mol}^{-1}\), which is higher than \(\text{Na}\), \(\text{Al}\), and \(\text{Si}\), and it forms sulfur dioxide (\(\text{SO}_2\)), which is a gas.

1 mark for identifying the correct set of elements based on Period 3 chemical and physical trends.

A transition metal complex's color is determined by the size of the d-orbital splitting \(\Delta\). This splitting depends on the nature of the ligands, the oxidation state of the metal, and the coordination geometry.

- Option A changes the ligand (water to ammonia), which changes \(\Delta\) (ammonia is a stronger-field ligand), resulting in a color change from pale blue to dark blue.
- Option B changes the oxidation state of the metal from +2 to +3, which changes the size of the d-orbital splitting and the number of d-electrons, resulting in a color change from pale green to yellow/brown.
- Option C replaces \(\text{H}_2\text{O}\) with \(\text{D}_2\text{O}\). Since isotope substitution does not alter the electronic distribution or the chemical characteristics of the ligand, the ligand field strength is virtually unchanged, so the color remains the same.
- Option D changes the geometry from octahedral to tetrahedral, altering the d-orbital splitting scheme and changing the color from pink to blue.

1 mark for identifying that replacing ligands with their isotopes (heavy water) does not alter the visible spectrum/color.

Let's analyze the observations:

1. Compound \(P\) (\(\text{C}_4\text{H}_{10}\text{O}\)) is oxidized by acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) (the orange solution turns green due to \(\text{Cr}^{3+}\) formation), so \(P\) must be a primary or secondary alcohol. Tertiary alcohols cannot be oxidized. This rules out option D, \((\text{CH}_3)_3\text{COH}\).

2. The oxidation product \(Q\) has the formula \(\text{C}_4\text{H}_8\text{O}\) and does not react with Tollens' reagent. This means \(Q\) is a ketone, and therefore \(P\) must be a secondary alcohol. This rules out options A and C, which are primary alcohols.

3. Compound \(P\) reacts with alkaline aqueous iodine to give a yellow precipitate of triiodomethane (\(\text{CHI}_3\)). This positive iodoform test indicates the presence of a \(\text{CH}_3\text{CH}(\text{OH})-\) group in the alcohol. Butan-2-ol, \(\text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_3\), has this group and yields a positive test.

1 mark for deducing that compound P must be butan-2-ol based on oxidation to a ketone and a positive iodoform test.

The equation for the formation of propane is:

\(3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\)

Using Hess's law and enthalpy changes of combustion:

\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)

\(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C(graphite)}) + 4 \times \Delta H_c^\ominus(\text{H}_2\text{(g)})] - \Delta H_c^\ominus(\text{C}_3\text{H}_8\text{(g)})\)

\(\Delta H_f^\ominus = [3 \times (-394) + 4 \times (-286)] - (-2220)\)

\(\Delta H_f^\ominus = [-1182 - 1144] + 2220\)

\(\Delta H_f^\ominus = -2326 + 2220 = -106\text{ kJ mol}^{-1}\)

1 mark for calculating the correct enthalpy change of formation as -106 kJ/mol.

Sodium oxide is a metal oxide that reacts with water to form a strongly alkaline solution of sodium hydroxide (pH approx. 14). Sulfur trioxide is a non-metal oxide that reacts with water to form a strongly acidic solution of sulfuric acid (pH approx. 1). Therefore, \(\text{Na}_2\text{O}\) is a basic oxide and \(\text{SO}_3\) is an acidic oxide, making option A the correct choice.

1 mark: Correctly identifies the pH trends and acid-base classifications of sodium oxide and sulfur trioxide.

In the presence of ligands, the d-orbitals of transition metal ions split into two energy levels. When white light passes through, d-electrons absorb specific frequencies of light (in this case, red light) as they are promoted from the lower to the higher energy d-orbitals. The remaining unabsorbed, complementary light (blue) is transmitted, giving the solution its color. This is an absorption process, not an emission process.

1 mark: Correctly identifies the d-d transition absorption mechanism and the relationship between absorbed and transmitted light.

To obtain a 3-substituted product, the first group introduced must be 3-directing. The nitro group (\(-\text{NO}_2\)) is 3-directing, whereas \(-\text{Br}\) and \(-\text{NH}_2\) are 2,4-directing. Nitrating benzene first gives nitrobenzene. Subsequent bromination yields 3-bromonitrobenzene. Finally, reducing the nitro group with Sn and concentrated HCl (followed by adding NaOH to release the free amine) yields 3-bromophenylamine. If bromination is done first (option B), the nitro group would be directed to the 2- or 4-position. If reduction is done before bromination (option C), the amine group would direct the bromine to the 2,4-positions.

1 mark: Correctly deduces the synthetic sequence based on the directing effects of functional groups on the benzene ring.

A reaction is spontaneous (feasible) when \(\Delta G^\theta \le 0\). Since \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), we find the temperature at which \(\Delta G^\theta = 0\): \(T = \frac{\Delta H^\theta}{\Delta S^\theta}\). Convert \(\Delta H^\theta\) from kJ to J: \(\Delta H^\theta = 178000\text{ J mol}^{-1}\). Therefore, \(T = \frac{178000\text{ J mol}^{-1}}{160\text{ J K}^{-1}\text{ mol}^{-1}} = 1112.5\text{ K}\). Thus, the minimum temperature is approximately 1113 K.

1 mark: Correctly calculates the minimum temperature by applying the Gibbs free energy equation, converting units, and rounding to the nearest integer.

The large increase between the 3rd and 4th ionisation energies (2745 to 11578 \(\text{kJ mol}^{-1}\)) indicates that the 4th electron is removed from a shell closer to the nucleus. Therefore, X has 3 valence electrons and is in Group 13, which means it is aluminium. Aluminium is a metal (making option B incorrect), its oxide is amphoteric (option A is correct), its chloride (\(\text{Al}_2\text{Cl}_6\)) has a simple molecular dimer structure (making option C incorrect), and aluminium does not react vigorously with cold water due to the protective oxide layer (making option D incorrect).

1 mark: Correctly identifies the element from successive ionisation energies and selects the correct property of its oxide.

A higher stability constant (\(K_{\text{stab}}\)) indicates a more stable complex because the equilibrium lies further to the right. Since \(1.2 \times 10^{13}\) is much greater than \(4.2 \times 10^5\), the tetraamminecopper(II) complex is far more stable than the tetrachlorocuprate(II) complex. Option D is incorrect because the tetrachlorocuprate(II) ion is tetrahedral (coordination number 4) whereas the tetraamminecopper(II) ion is octahedral (coordination number 6, containing two water ligands).

1 mark: Correctly identifies that a larger stability constant indicates a more stable complex ion.

Step 1 converts a primary alcohol into a primary halogenoalkane using phosphorus tribromide, \(\text{PBr}_3\). Step 2 is a nucleophilic substitution that increases the carbon chain length using ethanolic potassium cyanide, \(\text{KCN}\), under reflux. Step 3 is an acid hydrolysis of the nitrile functional group to a carboxylic acid using a dilute strong acid, such as \(\text{HCl}\), heated under reflux.

1 mark: Correctly identifies the reagents and reaction conditions for each of the three steps of the organic synthesis.

By constructing a Born-Haber cycle: \(\Delta H_{\text{f}}^\theta = \Delta H_{\text{at}}^\theta(\text{Na}) + \text{IE}_1(\text{Na}) + \Delta H_{\text{at}}^\theta(\text{Cl}) + \text{EA}_1(\text{Cl}) + \Delta H_{\text{latt}}^\theta(\text{NaCl})\). Substituting the given values: \(-411 = +107 + 496 + 121 - 349 + \Delta H_{\text{latt}}^\theta\). \(-411 = 375 + \Delta H_{\text{latt}}^\theta\). Therefore, \(\Delta H_{\text{latt}}^\theta = -411 - 375 = -786\text{ kJ mol}^{-1}\).

1 mark: Correctly applies Hess's Law to set up the Born-Haber cycle calculation and determines the correct negative value for lattice enthalpy.

Silicon(IV) chloride reacts with water to produce silicon dioxide and hydrogen chloride:
\(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\)
Therefore, 1.0 mol of \(\text{SiCl}_4\) produces 4.0 mol of \(\text{HCl}\).

Phosphorus(V) chloride reacts with water to produce phosphoric acid and hydrogen chloride:
\(\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}\)
Therefore, 1.0 mol of \(\text{PCl}_5\) produces 5.0 mol of \(\text{HCl}\).

The total amount of \(\text{HCl}\) produced is \(4.0 + 5.0 = 9.0\text{ mol}\).

1 mark for identifying the correct total moles of HCl (9.0 mol). Award 0 marks for any other option.

The complex \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) can exhibit both geometric (cis-trans) and optical isomerism.
1. The \(\text{trans}\)-isomer has a plane of symmetry, making it achiral. Thus, it only exists as a single stereoisomer.
2. The \(\text{cis}\)-isomer lacks a plane of symmetry, making it chiral. It therefore exists as a pair of non-superimposable mirror images (a pair of enantiomers).

In total, there are 1 (trans) + 2 (cis-enantiomers) = 3 stereoisomers.

1 mark for the correct number of stereoisomers (3). Award 0 marks for incorrect options.

Option B is correct because:
1. Nitration of methylbenzene using \(\text{HNO}_3 / \text{H}_2\text{SO}_4\) yields a mixture of 2- and 4-nitromethylbenzene (as the methyl group is 2,4-directing). The 4-nitromethylbenzene is isolated.
2. The methyl group of 4-nitromethylbenzene is oxidized to a carboxylic acid group using hot alkaline \(\text{KMnO}_4\) followed by acidification, producing 4-nitrobenzoic acid. The nitro group is stable to this oxidation.
3. The nitro group is then reduced to an amino group using Sn and concentrated \(\text{HCl}\), followed by addition of aqueous sodium hydroxide, yielding 4-aminobenzoic acid.

Option A is incorrect because oxidizing methylbenzene first produces benzoic acid. The carboxylic acid group is 3-directing, so subsequent nitration would yield 3-nitrobenzoic acid, eventually leading to 3-aminobenzoic acid.
Option C is incorrect because reducing the nitro group first yields 4-methylphenylamine. Subsequent oxidation of the methyl group with \(\text{KMnO}_4\) would destroy/oxidize the amino group.

1 mark for identifying sequence B as correct. Award 0 marks for incorrect sequences.

Lattice energy is directly proportional to the product of the ionic charges and inversely proportional to the sum of the ionic radii: \(\text{Lattice Energy} \propto \frac{q_+ q_-}{r_+ + r_-}\).
- The fluoride ion (\(\text{F}^-\)) has a smaller ionic radius than the chloride ion (\(\text{Cl}^-\)). Thus, the interionic distance in \(\text{MgF}_2\) is smaller than in \(\text{MgCl}_2\), resulting in stronger electrostatic attractions and a much more exothermic lattice energy.
- As the Group 2 cation group is descended from \(\text{Mg}^{2+}\) to \(\text{Sr}^{2+}\), the ionic radius of the cation increases. This increases the interionic distance, making the lattice energy less exothermic.

1 mark for identifying B as the correct explanation. Award 0 marks for incorrect statements.

Entropy is a measure of the disorder of a system.
- In reaction C, a solid reactant decomposes into two gaseous products: \(\text{NH}_4\text{Cl(s)} \rightarrow \text{NH}_3\text{(g)} + \text{HCl(g)}\). This represents a large increase in the number of moles of gas, leading to a highly positive standard entropy change (\(\Delta S^\theta > 0\)).
- Reactions A and B involve a net decrease in the number of moles of gas (\(\Delta S^\theta < 0\)).
- Reaction D is the freezing of liquid water to solid ice, which decreases disorder (\(\Delta S^\theta < 0\)).

1 mark for identifying reaction C. Award 0 marks for any other option.

- Silicon has a giant covalent structure with localized electrons, making it a semiconductor. At room temperature, it has very low electrical conductivity.
- Sodium, magnesium, and aluminium are metals with delocalised valence electrons. Their electrical conductivity increases with the number of outer shell electrons contributed per atom to the sea of delocalised electrons (Na contributes 1, Mg contributes 2, and Al contributes 3).
- Therefore, the order of increasing electrical conductivity is: \(\text{Si} < \text{Na} < \text{Mg} < \text{Al}\).

1 mark for identifying B as the correct sequence. Award 0 marks for incorrect sequences.

To find the equilibrium constant, \(K\), for the reaction:
\[ [\text{CuCl}_4]^{2-} + 4\text{NH}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{Cl}^- \]

We can combine the reverse of the first equilibrium with the forward of the second equilibrium:

1) \([\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O} \rightleftharpoons [\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^-\quad K_1 = \frac{1}{K_{\text{stab}}([\text{CuCl}_4]^{2-})} = \frac{1}{4.0 \times 10^5}\)

2) \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}\quad K_2 = K_{\text{stab}}([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}) = 1.2 \times 10^{13}\)

Adding these two equations gives the target equation, so the overall equilibrium constant is:
\[ K = K_1 \times K_2 = \frac{1.2 \times 10^{13}}{4.0 \times 10^5} = 3.0 \times 10^7 \]

1 mark for identifying C as the correct statement. Award 0 marks for incorrect options.

1. Compound X is a bromoalkane (either 1-bromopropane or 2-bromopropane). Undergoing elimination with hot ethanolic KOH, both isomers yield the same alkene product Y, which is propene, \(\text{CH}_3\text{CH}=\text{CH}_2\).
2. Propene (Y) reacts with cold dilute acidified \(\text{KMnO}_4\) via mild oxidation (hydroxylation across the carbon-carbon double bond) to form a diol.
3. The addition of two hydroxyl groups across the double bond of propene results in propane-1,2-diol (Z).

1 mark for identifying C as the correct compound Z. Award 0 marks for incorrect structures.

(a) When \(\text{SO}_3\) reacts with water, a vigorous, exothermic reaction occurs to form sulfuric acid: \(\text{SO}_3(\text{l}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_4(\text{aq})\). The pH of this solution is 1-2. When \(\text{P}_4\text{O}_{10}\) reacts with water, a similarly vigorous, exothermic reaction occurs to form phosphoric(V) acid: \(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\). The pH of this solution is also 1-2. (b) Silicon is in Period 3 and has empty, low-lying 3d orbitals in its valence shell. These 3d orbitals can accept a lone pair of electrons from a water molecule, initiating nucleophilic attack and subsequent hydrolysis: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\). Carbon is in Period 2 and has no d-orbitals in its outer shell. It cannot coordinate with water molecules, hence \(\text{CCl}_4\) does not hydrolyse. (c) \(\text{NaCl}\) forms a neutral solution (pH ~7) because the \(\text{Na}^+\) ion has a low charge density and does not polarise the surrounding water molecules. \(\text{MgCl}_2\) forms a slightly acidic solution (pH ~6-6.5) because the smaller \(\text{Mg}^{2+}\) ion has a higher charge density, which polarises the coordinated water molecules, facilitating the release of minor amounts of \(\text{H}^+\) ions into the solution.

(a) 1 mark: Vigorous/exothermic reaction described for both. 1 mark: pH of 1-2 stated for both. 1 mark: Balanced equation for \(\text{SO}_3\) reaction. 1 mark: Balanced equation for \(\text{P}_4\text{O}_{10}\) reaction. 2 marks: Explaining both form strongly acidic solutions due to full dissociation/proton release (1 mark for each explanation). (b) 1 mark: Silicon has vacant 3d orbitals. 1 mark: Carbon does not have d-orbitals in its valence shell. 1 mark: Water can form a dative bond with Si but not with C. 1 mark: Correct balanced equation for \(\text{SiCl}_4\) hydrolysis. (c) 1 mark: pH 7 for \(\text{NaCl}\) and pH 6-6.5 for \(\text{MgCl}_2\). 1 mark: Explanation involving high charge density of \(\text{Mg}^{2+}\) causing polarisation of water molecules.

(a) The lattice energy is the enthalpy change when 1 mole of an ionic crystalline compound is formed from its constituent gaseous ions under standard conditions of 298 K and 101 kPa. (b) The Born-Haber cycle must show the standard states: \(\text{Ca(s)} + \text{F}_2\text{(g)}\) going to \(\text{CaF}_2\text{(s)}\) via: 1. Atomisation of Ca: \(\text{Ca(s)} \rightarrow \text{Ca(g)}\) (\(\Delta H_{at}^{\theta}\)). 2. First and second ionisation of Ca: \(\text{Ca(g)} \rightarrow \text{Ca}^{2+}\text{(g)} + 2\text{e}^-\). 3. Atomisation of fluorine: \(\text{F}_2\text{(g)} \rightarrow 2\text{F(g)}\) (which is \(2 \times \Delta H_{at}^{\theta}\) of F). 4. Electron affinity of F: \(2\text{F(g)} + 2\text{e}^- \rightarrow 2\text{F}^-\text{(g)}\) (which is \(2 \times EA_1\) of F). 5. Lattice formation: \(\text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)} \rightarrow \text{CaF}_2\text{(s)}\) (\(\Delta H_{latt}^{\theta}\)). (c) Using Hess's Law: \(\Delta H_f^{\theta} = \Delta H_{at}^{\theta}(\text{Ca}) + IE_1(\text{Ca}) + IE_2(\text{Ca}) + 2 \times \Delta H_{at}^{\theta}(\text{F}) + 2 \times EA_1(\text{F}) + \Delta H_{latt}^{\theta}\). Substituting the values: \(-1220 = 178 + 590 + 1145 + 2(79) + 2(-328) + \Delta H_{latt}^{\theta}\). \(-1220 = 178 + 590 + 1145 + 158 - 656 + \Delta H_{latt}^{\theta}\). \(-1220 = 1415 + \Delta H_{latt}^{\theta}\). \(\Delta H_{latt}^{\theta} = -1220 - 1415 = -2635\text{ kJ mol}^{-1}\). (d) The \(\text{O}^{2-}\) ion has a greater charge (2-) compared to the \(\text{F}^{-}\) ion (1-). The higher charge leads to stronger electrostatic forces of attraction between the oppositely charged ions in the lattice, resulting in a significantly more exothermic lattice energy.

(a) 1 mark: Enthalpy change when 1 mole of solid ionic compound is formed. 1 mark: Stating 'from its gaseous ions' under standard conditions. (b) 4 marks: 1 mark for correct elements and compound in standard states; 1 mark for atomisation steps with correct state symbols; 1 mark for ionisation steps and electron affinity steps; 1 mark for correct direction of all arrows corresponding to sign. (c) 4 marks: 1 mark for correct expression of Hess's Law; 1 mark for doubling the atomisation and electron affinity values of fluorine; 1 mark for correct arithmetic setup; 1 mark for final calculated value of -2635 kJ mol^-1 (must include units and negative sign). (d) 2 marks: 1 mark for stating oxide ion has a 2- charge whereas fluoride has a 1- charge; 1 mark for relating this to stronger electrostatic attraction between the ions.

(a) (i) Reagent: Hydrogen bromide gas, \(\text{HBr}\). Condition: Room temperature. (ii) Reagent: Aqueous sodium hydroxide, \(\text{NaOH(aq)}\). Condition: Heat under reflux. (iii) Reagent: Acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\). Condition: Heat under reflux. (b) (i) Reaction of propene with \(\text{HBr}\) is an electrophilic addition. Addition of \(\text{H}^+\) to the double bond can form either a secondary carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_3\)) or a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)). The secondary carbocation is more stable than the primary carbocation because it has two electron-donating alkyl groups that reduce the positive charge density on the carbon atom (inductive effect). Therefore, the secondary carbocation is formed preferentially, leading to 2-bromopropane as the major product. (ii) Draw: 1. Propene structure showing the double bond. \(\text{H}-\text{Br}\) shown with \(\delta+\) on \(\text{H}\) and \(\delta-\) on \(\text{Br}\). 2. A curly arrow from the double bond of propene to the \(\text{H}\) of \(\text{H}-\text{Br}\). 3. A curly arrow from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom. 4. Intermediate structure: secondary carbocation \(\text{CH}_3\text{CH}^+\text{CH}_3\) and bromide ion \(\text{:Br}^-\). 5. A curly arrow from a lone pair on the bromide ion to the positive carbon atom of the carbocation.

(a)(i) 1 mark: HBr (gas or in polar solvent) at room temperature. (ii) 2 marks: NaOH(aq) / KOH(aq) [1] and heat under reflux [1]. (iii) 2 marks: Acidified potassium dichromate(VI) (or acidified potassium manganate(VII)) [1] and heat under reflux [1]. (b)(i) 3 marks: 1 mark for stating reaction proceeds via a secondary carbocation intermediate; 1 mark for stating secondary carbocation is more stable than the primary carbocation; 1 mark for attributing this stability to the electron-donating inductive effect of two methyl/alkyl groups. (b)(ii) 4 marks: 1 mark for correct dipoles on H-Br; 1 mark for curly arrow from C=C double bond to H; 1 mark for curly arrow from H-Br bond to Br; 1 mark for showing correct secondary carbocation structure and final curly arrow from lone pair on Br- to carbocation carbon.

(a) A transition element is defined as a d-block element that forms at least one stable ion with an incomplete d-subshell. The electronic configuration of a neutral copper atom (atomic number 29) is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\) (note the extra stability of the full 3d subshell). The copper(II) ion, \(\text{Cu}^{2+}\), is formed by losing the \(4\text{s}\) electron first, followed by one \(3\text{d}\) electron, giving: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\). (b) (i) The initial color of the hexaaquacopper(II) solution is pale blue. Upon adding concentrated hydrochloric acid, the ligand exchange reaction occurs, resulting in a yellow or yellow-green solution. (ii) The balanced equation is: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\). The tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\), has a tetrahedral shape because the large, negatively charged chloride ligands experience steric hindrance and electrostatic repulsion, preventing an octahedral configuration. (c) (i) When ammonia is added dropwise, a pale blue precipitate of copper(II) hydroxide, \(\text{Cu(OH)}_2\text{(s)}\), is formed due to ammonia acting as a weak base. Upon adding excess ammonia, ligand exchange occurs, the precipitate dissolves, and a deep blue solution is formed. (ii) The formula of the deep blue complex ion is \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

(a) 3 marks: 1 mark for correct definition of transition element; 1 mark for configuration of Cu: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\); 1 mark for configuration of \(\text{Cu}^{2+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\). (b)(i) 2 marks: 1 mark for pale blue/blue; 1 mark for yellow / yellow-green. (b)(ii) 3 marks: 2 marks for balanced equation (1 mark for reactants/products, 1 mark for charges and stoichiometry); 1 mark for stating 'tetrahedral'. (c)(i) 2 marks: 1 mark for pale blue precipitate; 1 mark for dissolving in excess to form a deep blue solution. (c)(ii) 2 marks: 1 mark for \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) (accept \([\text{Cu}(\text{NH}_3)_4]^{2+}\)); 1 mark for the overall 2+ charge.

(a) The thermal stability of the hydrogen halides decreases down the group from \(\text{HF}\) to \(\text{HI}\). This is because as you go down Group 17, the halogen atoms increase in atomic radius. This increases the \(\text{H}-\text{X}\) bond length, which weakens the electrostatic attraction between the bonding electrons and the nuclei, thereby decreasing the \(\text{H}-\text{X}\) bond energy. (b) (i) Solid sodium chloride undergoes a displacement reaction: \(\text{NaCl(s)} + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl(g)}\) (or forming \(\text{Na}_2\text{SO}_4\)). The role of the sulfuric acid is to act as an acid (proton donor). (ii) Purple vapor: Iodine gas, \(\text{I}_2\text{(g)}\). Yellow solid: Sulfur, \(\text{S(s)}\). Gas with smell of bad eggs: Hydrogen sulfide, \(\text{H}_2\text{S(g)}\). Iodide ions, \(\text{I}^-\), are much stronger reducing agents than chloride ions, \(\text{Cl}^-\). This is because the ionic radius of \(\text{I}^-\rightleftharpoons\) is larger, so its outer valence electrons are more shielded from the nucleus and are lost more easily. As a result, iodide is capable of reducing the sulfur in concentrated sulfuric acid (from oxidation state +6 to 0 in sulfur and -2 in \(\text{H}_2\text{S}\)), whereas chloride cannot. (c) Cold, dilute NaOH: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). Hot, concentrated NaOH: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\).

(a) 3 marks: 1 mark for stating thermal stability decreases down the group; 1 mark for describing the increasing atomic size/bond length; 1 mark for explaining that bond strength/energy decreases. (b)(i) 2 marks: 1 mark for correct equation; 1 mark for identifying the role of sulfuric acid as an acid / proton donor (reject oxidising agent). (b)(ii) 5 marks: 3 marks for correct identification of the species: purple vapor = \(\text{I}_2\) [1], yellow solid = \(\text{S}\) [1], gas with bad egg smell = \(\text{H}_2\text{S}\) [1]. 2 marks for explanation: 1 mark for stating iodide is a stronger reducing agent than chloride; 1 mark for explaining that valence electrons of iodide are more shielded / lost more easily. (c) 2 marks: 1 mark for correct balanced equation for cold NaOH; 1 mark for correct balanced equation for hot NaOH.

(a)
Moles of anhydrous \(\text{MgSO}_4 = \frac{3.01}{120.4} = 0.0250\text{ mol}\).
Moles of hydrated \(\text{MgSO}_4 \cdot 7\text{H}_2\text{O} = \frac{6.16}{246.4} = 0.0250\text{ mol}\).

(b)
For Experiment 1:
\(\Delta T = 31.1 - 21.5 = +9.6\text{ }^\circ\text{C}\).
\(q_1 = m c \Delta T = 50.0 \times 4.18 \times 9.6 = 2006.4\text{ J}\) (or \(2.01\text{ kJ}\)).

For Experiment 2:
\(\Delta T = 19.6 - 21.8 = -2.2\text{ }^\circ\text{C}\).
\(q_2 = m c \Delta T = 50.0 \times 4.18 \times (-2.2) = -459.8\text{ J}\) (or \(-0.460\text{ kJ}\)).

(c)
\(\Delta H_1 = -\frac{2006.4 \times 10^{-3}}{0.0250} = -80.3\text{ kJ mol}^{-1}\).
\(\Delta H_2 = -\frac{-459.8 \times 10^{-3}}{0.0250} = +18.4\text{ kJ mol}^{-1}\).

(d)
According to Hess's law: \(\Delta H_{\text{hydration}} = \Delta H_1 - \Delta H_2\).
\(\Delta H_{\text{hydration}} = -80.3 - (+18.4) = -98.7\text{ kJ mol}^{-1}\).

(e)
Two temperature readings are taken, so total uncertainty \(= 2 \times 0.1 = 0.2\text{ }^\circ\text{C}\).
Percentage uncertainty \(= \frac{0.2}{9.6} \times 100\% = 2.08\%\) (or \(2.1\%\)).
Improvement: Use a lid on the polystyrene cup, or place the cup inside a larger beaker filled with cotton wool insulation.

- Moles calculation: [1 mark] for both moles calculated as 0.0250 mol with working.
- Heat calculation: [2 marks] for calculating both q values correctly (q1 = 2006.4 J, q2 = -459.8 J / 459.8 J absorbed).
- Enthalpy changes: [2 marks] for correct signs and values of \(\Delta H_1 = -80.3\text{ kJ mol}^{-1}\) and \(\Delta H_2 = +18.4\text{ kJ mol}^{-1}\) (allow ecf from q).
- Hess's law hydration calculation: [2.3 marks] for correct hydration enthalpy of \(-98.7\text{ kJ mol}^{-1}\) with correct units and sign.
- Percentage uncertainty: [2 marks] for calculating \(2.08\%\) (or \(2.1\%\)) using a total uncertainty of \(0.2\text{ }^\circ\text{C}\).
- Improvement: [4 marks] for suggesting a lid or extra insulation and explaining how it reduces heat loss to the surroundings.

(a)
(i) \(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})\)
(ii) \(\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})\)

(b)
Moles of thiosulfate \(= \text{concentration} \times \text{volume} = 0.100 \times \frac{22.50}{1000} = 2.25 \times 10^{-3}\text{ mol}\).

(c)
From the balanced equations, \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\).
Therefore, the mole ratio of \(\text{Cu}^{2+} : \text{S}_2\text{O}_3^{2-} = 1 : 1\).
Moles of \(\text{Cu}^{2+}\) in \(25.0\text{ cm}^3 = 2.25 \times 10^{-3}\text{ mol}\).
Concentration of \(\text{Cu}^{2+}\) in FA 1 \(= \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900\text{ mol dm}^{-3}\).

(d)
Starch forms an insoluble/strongly bound complex with iodine when the iodine concentration is high. If starch is added too early, this complex does not easily release iodine, leading to an inaccurate and delayed end-point.

(e)
The blue-black suspension disappears to leave an off-white/cream suspension (due to the presence of the solid \(\text{CuI}\) precipitate).

- Balanced equations: [2 marks] (1 mark for each correct equation).
- Moles of thiosulfate: [2 marks] for correct value with working (2.25 x 10^-3 mol).
- Molar concentration of copper(II): [3.3 marks] for showing 1:1 ratio and calculating 0.0900 mol dm^-3 with appropriate 3 significant figures.
- Starch indicator explanation: [3 marks] for noting that starch forms an irreversible/insoluble complex with high concentration of iodine.
- End-point appearance: [3 marks] for stating that the blue-black color disappears leaving a cream/off-white precipitate/suspension.

(a)
Mass of \(\text{MCO}_3\) used \(= 17.70 - 15.20 = 2.50\text{ g}\).
Mass of \(\text{CO}_2\) lost \(= 17.70 - 16.60 = 1.10\text{ g}\).

(b)
Moles of \(\text{CO}_2\) evolved \(= \frac{1.10}{44.0} = 0.0250\text{ mol}\).

(c)
Since the reaction stoichiometry is \(\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})\), \(1\text{ mol}\) of \(\text{MCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\).
Moles of \(\text{MCO}_3\) reacted \(= 0.0250\text{ mol}\).
\(M_{\text{r}}\) of \(\text{MCO}_3 = \frac{2.50\text{ g}}{0.0250\text{ mol}} = 100.0\text{ g mol}^{-1}\).
\(A_{\text{r}}\)(M) \(= 100.0 - 12.0 - 3(16.0) = 40.0\text{ g mol}^{-1}\).
Therefore, the metal \(\text{M}\) is Calcium (\(\text{Ca}\)).

(d)
Bubble the gas through limewater (aqueous calcium hydroxide). The limewater turns cloudy/milky due to the formation of a fine white precipitate of calcium carbonate.

(e)
Thermal stability increases down the group. As you descend the group, the ionic radius of the \(\text{M}^{2+}\) cation increases, so its charge density decreases. Consequently, it has a weaker polarising effect on the carbonate ion's electron cloud, distorting the \(\text{C-O}\) bonds less, making the carbonate ion more stable to thermal decomposition.

- Mass calculation: [1 mark] for correct masses of MCO3 (2.50 g) and CO2 (1.10 g).
- Moles of CO2: [2 marks] for calculating 0.0250 mol with clear working.
- Identity of M: [3.3 marks] for calculating Mr = 100.0, Ar = 40.0, and correctly identifying M as Calcium (Ca).
- Limewater test: [2 marks] for the limewater reagent and the 'cloudy/milky' observation.
- Thermal stability trend: [5 marks] (1 mark for stating that stability increases, 1 mark for cation size increasing/charge density decreasing, 1 mark for less polarising power of the cation, 1 mark for less distortion of the carbonate ion/C-O bond).

(a) Lattice energy is the enthalpy change when one mole of an ionic crystalline solid is formed from its separate gaseous ions under standard conditions.
(b) The algebraic expression is: \(\Delta H_f^\ominus(MgF_2) = \Delta H_{at}^\ominus(Mg) + IE_1(Mg) + IE_2(Mg) + E(F-F) + 2 \times EA(F) + \Delta H_{latt}^\ominus(MgF_2)\).
(c) Rearranging the expression: \(\Delta H_{latt}^\ominus(MgF_2) = \Delta H_f^\ominus(MgF_2) - [\Delta H_{at}^\ominus(Mg) + IE_1(Mg) + IE_2(Mg) + E(F-F) + 2 \times EA(F)]\).
Substituting the values: \(\Delta H_{latt}^\ominus(MgF_2) = -1124 - [148 + 738 + 1451 + 158 + 2(-328)]\).
\(\Delta H_{latt}^\ominus(MgF_2) = -1124 - [2495 - 656] = -1124 - 1839 = -2963 \text{ kJ mol}^{-1}\).
(d) The calcium ion, \(Ca^{2+}\), has a larger ionic radius than the magnesium ion, \(Mg^{2+}\). Therefore, there is a weaker electrostatic attraction between \(Ca^{2+}\) and \(F^-\), making the lattice energy less exothermic.

(a) 1 mark for stating it is the enthalpy change for the formation of 1 mole of ionic solid; 1 mark for specifying it is from its gaseous ions under standard conditions.
(b) 1 mark for correct Born-Haber cycle diagram/pathway; 2 marks for the correct algebraic formula linking all components.
(c) 1 mark for multiplying the electron affinity of fluorine by 2; 1 mark for using the bond energy of \(F_2\) as equivalent to the atomisation of 1 mole of \(F_2\) gas; 1 mark for correct rearrangement of the equation; 1.1 marks for the correct value of \(-2963 \text{ kJ mol}^{-1}\) with correct sign and units.
(d) 1 mark for stating that \(Ca^{2+}\) is larger than \(Mg^{2+}\); 1 mark for explaining that this leads to weaker electrostatic attraction between the ions.

(a) The stability constant, \(K_{stab}\), is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.
(b) Ligand exchange occurs because ammonia is a stronger ligand than water, forming a more stable complex. The solution changes color from light blue to deep blue. Equation: \([Cu(H_2O)_6]^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 4H_2O(l)\).
(c)(i) \(en\) is a bidentate ligand, whereas \(NH_3\) is monodentate. Chelating ligands form more stable complexes because the reaction involves an increase in entropy (chelate effect) as more free particles are released into solution.
(c)(ii) \(K_{stab} = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[Cu^{2+}][NH_3]^4}\). Rearranging gives: \([Cu^{2+}] = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{K_{stab} \times [NH_3]^4} = \frac{0.10}{(1.2 \times 10^{13}) \times 1.5^4} = \frac{0.10}{1.2 \times 10^{13} \times 5.0625} = 1.646 \times 10^{-15} \text{ mol dm}^{-3}\), which rounds to \(1.6 \times 10^{-15} \text{ mol dm}^{-3}\).

(a) 1 mark for stating it is an equilibrium constant for the formation of a complex; 1 mark for specifying it is from its constituent ions/molecules in a solvent.
(b) 1 mark for stating ammonia is a stronger ligand / forms a more stable complex; 1 mark for 'deep blue' solution; 1 mark for correct ionic equation.
(c)(i) 1 mark for identifying \(en\) as bidentate and \(NH_3\) as monodentate; 1 mark for explaining that chelation leads to an increase in entropy/disorder (chelate effect).
(c)(ii) 1 mark for the correct expression of \(K_{stab}\); 2 marks for substituting values and algebraic work; 1.1 marks for the correct final answer \(1.6 \times 10^{-15} \text{ mol dm}^{-3}\) (allow 1.6 to 1.65).

(a) \(NaCl\) simply dissolves in water to form a neutral solution (pH = 7) because sodium and chloride ions do not undergo hydrolysis: \(NaCl(s) + aq \rightarrow Na^+(aq) + Cl^-(aq)\).
\(PCl_5\) undergoes vigorous hydrolysis with water to form phosphoric acid and hydrogen chloride gas, yielding a highly acidic solution (pH = 1 or 2): \(PCl_5(s) + 4H_2O(l) \rightarrow H_3PO_4(aq) + 5HCl(aq)\).
(b) Silicon has vacant, low-lying 3d orbitals which can accept a lone pair of electrons from a water molecule to initiate hydrolysis. Carbon does not have d-orbitals in its outer valence shell (\(n=2\)) and is too small to allow water molecules to attack the carbon atom due to steric hindrance.
(c) Thermal stability increases down the group. Down the group, the cationic radius increases (from \(Mg^{2+}\) to \(Ba^{2+}\)) while the charge remains constant (+2). Consequently, the charge density of the cation decreases, and its polarizing power on the large, electron cloud of the carbonate ion (\(CO_3^{2-}\)) decreases. This means less weakening of the C-O bond, requiring more thermal energy to decompose.

(a) 1 mark for stating \(NaCl\) dissolves to give a neutral solution of pH 7; 1 mark for stating \(PCl_5\) hydrolyses to give an acidic solution of pH 1-2; 2 marks for the correct equation for the hydrolysis of \(PCl_5\).
(b) 1 mark for stating silicon has vacant 3d orbitals; 1 mark for stating carbon has no available d-orbitals; 1.1 marks for explaining that water can coordinate/form a dative bond to the silicon atom, but steric hindrance or lack of d-orbitals prevents this in carbon.
(c) 1 mark for stating thermal stability increases down the group; 1 mark for stating cationic radius increases/charge density of cation decreases; 1 mark for stating polarizing power of the cation decreases; 1 mark for stating that there is less distortion/polarization of the carbonate ion.

(a)(i)
Step 1: Oxidize methylbenzene to benzoic acid. Reagents: Alkaline potassium manganate(VII) (\(KMnO_4\)), heat under reflux, followed by acidification with dilute hydrochloric acid (\(HCl\)). Intermediate: benzoic acid, \(C_6H_5COOH\).
Step 2: Nitrate benzoic acid to 3-nitrobenzoic acid. Reagents: Concentrated nitric acid (\(HNO_3\)) and concentrated sulfuric acid (\(H_2SO_4\)), warm at 50-60 °C. Intermediate: 3-nitrobenzoic acid, \(3-NO_2C_6H_4COOH\).
Step 3: Reduce 3-nitrobenzoic acid to 3-aminobenzoic acid. Reagents: Tin (\(Sn\)) and concentrated hydrochloric acid (\(HCl\)), heat under reflux, followed by addition of aqueous sodium hydroxide (\(NaOH\)) to release the free amine. Product: 3-aminobenzoic acid.
(a)(ii) The ester group (\(-COOCH_3\)) in methyl benzoate is electron-withdrawing and deactivates the ring, directing incoming electrophiles to the 3-position (meta-directing). The methyl group (\(-CH_3\)) in methylbenzene is electron-donating, activating the ring and directing electrophiles to the 2- and 4-positions (ortho/para-directing).
(b)(i) Bromine water, \(Br_2(aq)\), at room temperature.
(b)(ii) Benzoyl chloride, \(C_6H_5COCl\), at room temperature in the presence of base (such as aq \(NaOH\)).

(a)(i) 2 marks for Step 1: 1 mark for alkaline \(KMnO_4\) + heat + acidification, 1 mark for drawing the structure of benzoic acid. 2 marks for Step 2: 1 mark for conc. \(HNO_3\) + conc. \(H_2SO_4\) + heat, 1 mark for drawing the structure of 3-nitrobenzoic acid. 2 marks for Step 3: 1 mark for \(Sn\) + conc. \(HCl\) + heat, followed by addition of alkali, 1 mark for drawing the structure of 3-aminobenzoic acid.
(a)(ii) 1 mark for identifying \(-COOCH_3\) as electron-withdrawing and meta-directing; 1.1 marks for identifying \(-CH_3\) as electron-donating and ortho/para-directing.
(b)(i) 1.5 marks for bromine water (or \(Br_2(aq)\)) at room temperature.
(b)(ii) 1.5 marks for benzoyl chloride (\(C_6H_5COCl\)) and base (e.g. NaOH) at room temperature.

(a) Entropy is a measure of the disorder or randomness of a system. For the reaction \(CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\), \(\Delta S^\ominus\) is positive because a solid reactant decomposes to produce a gaseous product, which has much higher disorder/entropy.
(b) Since \(\Delta H^\ominus\) is positive (endothermic) and \(\Delta S^\ominus\) is positive, the term \(-T\Delta S^\ominus\) becomes increasingly negative as temperature \(T\) increases. At low temperatures, \(\Delta H^\ominus > T\Delta S^\ominus\), making \(\Delta G^\ominus > 0\) (not feasible). At high temperatures, \(T\Delta S^\ominus\) exceeds \(\Delta H^\ominus\), so \(\Delta G^\ominus\) becomes negative (feasible).
(c) First, calculate the standard entropy change: \(\Delta S^\ominus = S^\ominus[BaO(s)] + S^\ominus[CO_2(g)] - S^\ominus[BaCO_3(s)] = 70.4 + 213.6 - 112.1 = +171.9 \text{ J K}^{-1} \text{ mol}^{-1} = +0.1719 \text{ kJ K}^{-1} \text{ mol}^{-1}\).
For the reaction to be feasible, \(\Delta G^\ominus \le 0\). Therefore: \(\Delta H^\ominus - T\Delta S^\ominus = 0 \implies T = \frac{\Delta H^\ominus}{\Delta S^\ominus} = \frac{269.3 \text{ kJ mol}^{-1}}{0.1719 \text{ kJ K}^{-1} \text{ mol}^{-1}} = 1566.6 \text{ K}\). This rounds to \(1567 \text{ K}\).

(a) 1 mark for defining entropy as a measure of disorder/randomness; 1 mark for predicting positive \(\Delta S^\ominus\); 1 mark for explaining that a gas is formed from a solid, which increases disorder.
(b) 1 mark for stating that both \(\Delta H^\ominus\) and \(\Delta S^\ominus\) are positive; 1 mark for noting that \(-T\Delta S^\ominus\) becomes more negative as temperature increases; 1 mark for concluding that at high temperatures, \(\Delta G^\ominus\) becomes negative/spontaneous.
(c) 1 mark for calculating \(\Delta S^\ominus = +171.9 \text{ J K}^{-1} \text{ mol}^{-1}\); 1 mark for converting \(\Delta S^\ominus\) to \(\text{kJ K}^{-1} \text{ mol}^{-1}\) (i.e., \(0.1719\)); 1 mark for stating the condition for feasibility is \(\Delta G^\ominus \le 0\); 1 mark for substitution of values; 1.1 marks for the final calculated temperature of \(1567 \text{ K}\) (allow \(1566.6 \text{ K}\) to \(1570 \text{ K}\)).

(a) Standard electrode potential is the electromotive force (e.m.f.) of a half-cell connected to a standard hydrogen electrode under standard conditions of 298 K, 1 atm pressure, and 1.0 mol dm\(-3\) concentration of ions.
(b)(i) Half-equations:
\(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-\)
\(MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l)\)
Overall reaction:
\(MnO_4^-(aq) + 8H^+(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l) + 5Fe^{3+}(aq)\).
(b)(ii) \(E^\ominus_{cell} = E^\ominus_{reduction} - E^\ominus_{oxidation} = 1.51 - 0.77 = +0.74 \text{ V}\).
(c) Using the Nernst equation for \(Fe^{3+} + e^- \rightleftharpoons Fe^{2+}\), where \(z=1\):
\(E = +0.77 + \frac{0.059}{1}\log_{10}\left(\frac{0.050}{0.80}\right)\)
\(E = +0.77 + 0.059\log_{10}(0.0625)\)
\(E = +0.77 + 0.059 \times (-1.204) = 0.77 - 0.071 = +0.699 \text{ V}\), which rounds to \(+0.70 \text{ V}\).

(a) 1 mark for 'e.m.f. of a half-cell compared to standard hydrogen electrode'; 1 mark for specifying standard conditions of 298 K, 1 atm, and 1.0 mol dm\(^{-3}\).
(b)(i) 1 mark for both half-equations correct; 2 marks for the correct overall equation.
(b)(ii) 1.5 marks for \(+0.74 \text{ V}\).
(c) 1 mark for identifying \(z = 1\); 1 mark for setting up the correct ratio \(\frac{0.050}{0.80}\); 1.5 marks for correct log calculation; 1.1 marks for the final value of \(+0.70 \text{ V}\) (allow \(+0.699 \text{ V}\) to \(+0.70 \text{ V}\)).

(a)(i) Comparing Exp 1 and Exp 2: \([I^-]\) is constant, \([S_2O_8^{2-}]\) doubles, and initial rate doubles. Therefore, the reaction is 1st order with respect to \(S_2O_8^{2-}\).
Comparing Exp 1 and Exp 3: \([S_2O_8^{2-}]\) is constant, \([I^-]\) doubles, and initial rate doubles. Therefore, the reaction is 1st order with respect to \(I^-\).
(a)(ii) \(\text{Rate} = k[S_2O_8^{2-}][I^-]\).
(a)(iii) Using Exp 1 data: \(1.5 \times 10^{-5} = k (0.010)(0.020) \implies k = 0.075\).
Units: \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\).
(b) \(Fe^{2+}\) acts as a homogeneous catalyst because it is in the same phase as the reactants. The uncatalysed reaction is slow because it requires two negatively charged anions to collide, which have high electrostatic repulsion. \(Fe^{2+}\) provides an alternative pathway with a lower activation energy:
Step 1: \(S_2O_8^{2-}(aq) + 2Fe^{2+}(aq) \rightarrow 2SO_4^{2-}(aq) + 2Fe^{3+}(aq)\)
Step 2: \(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\).

(a)(i) 1.5 marks for order with respect to \(S_2O_8^{2-}\) being 1st order with explanation; 1.5 marks for order with respect to \(I^-\space\) being 1st order with explanation.
(a)(ii) 1 mark for correct rate equation.
(a)(iii) 1 mark for calculation of value \(k = 0.075\); 1.1 marks for the units \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\); 1 mark for showing working.
(b) 1 mark for stating that direct reaction of two anions is slow due to electrostatic repulsion; 1 mark for stating that the catalyst provides an alternative pathway with lower activation energy; 1 mark for Step 1 equation; 1 mark for Step 2 equation.

(a) Order of basic strength: phenylamine < ammonia < ethylamine.
Explanation: Basic strength depends on the availability of the nitrogen lone pair to accept a proton.
In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the \(\pi\)-system of the benzene ring, making it less available to accept a proton.
In ammonia, there is no such delocalisation or alkyl group effects.
In ethylamine, the ethyl group is electron-donating by a positive inductive effect, which increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton.
(b)(i) Nitrous acid is prepared by reacting sodium nitrite, \(NaNO_2\), with dilute hydrochloric acid, \(HCl\).
(b)(ii) Compound X is benzenediazonium chloride, \(C_6H_5N_2^+ Cl^-\). Reagents: \(NaNO_2\) and dilute \(HCl\); Temperature: 0 to 10 °C.
(b)(iii) The dye formed is 4-hydroxyphenylazobenzene (or 4-(phenylazo)phenol), which has the structure: \(C_6H_5-N=N-C_6H_4-OH\) (where the \(-N=N-\) link is at the 4-position relative to the \(-OH\) group on the second benzene ring). Reaction type: Coupling reaction (or electrophilic substitution).

(a) 1 mark for correct order: phenylamine < ammonia < ethylamine; 1 mark for linking basicity to the availability of the lone pair on the N atom; 1 mark for delocalisation of lone pair in phenylamine; 1 mark for electron-donating inductive effect of the ethyl group in ethylamine; 1 mark for comparing both to ammonia.
(b)(i) 1.5 marks for sodium nitrite (\(NaNO_2\)) and dilute hydrochloric acid (\(HCl\)).
(b)(ii) 1 mark for drawing the structure of benzenediazonium chloride; 1.1 marks for stating reagents and temperature range 0-10 °C.
(b)(iii) 1.5 marks for drawing the correct structure of the azo dye (4-hydroxyphenylazobenzene); 1 mark for stating 'coupling reaction' or 'electrophilic substitution'.

**(a)**
The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion from its constituent ions or simpler complexes in a solvent/aqueous solution.

**(b)**
Water is the solvent, so its concentration remains effectively constant and is omitted from the expression:
\[K_{\text{stab}} = \frac{[[\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{en}]^2}\]
Units:
\[\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6 \text{ mol}^{-2}\]

**(c)**
1. Find the equilibrium concentration of \([\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}\):
Since the initial concentration of \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) is \(0.050\text{ mol dm}^{-3}\) and only \(1.2 \times 10^{-10}\text{ mol dm}^{-3}\) remains:
\[[[\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}]_{\text{eq}} = 0.050 - 1.2 \times 10^{-10} \approx 0.050\text{ mol dm}^{-3}\]

2. Find the equilibrium concentration of \(\text{en}\):
Two moles of \(\text{en}\) react per mole of complex formed:
\[[\text{en}]_{\text{eq}} = 0.250 - 2(0.050) = 0.150\text{ mol dm}^{-3}\]

3. Calculate \(K_{\text{stab}}\):
\[K_{\text{stab}} = \frac{0.050}{(1.2 \times 10^{-10}) \times (0.150)^2}\]
\[K_{\text{stab}} = \frac{0.050}{1.2 \times 10^{-10} \times 0.0225} = \frac{0.050}{2.7 \times 10^{-12}} = 1.85 \times 10^{10}\text{ dm}^6 \text{ mol}^{-2}\]
*(Accept 1.9 \(\times 10^{10}\))*

**(d)**
- In the ligand substitution reaction with 1,2-diaminoethane (en), 3 reactant particles (1 complex ion + 2 bidentate ligands) form 5 product particles (1 complex ion + 4 water molecules). This results in an increase in the number of particles in solution, leading to a significant increase in entropy (\(\Delta S^\ominus\) is highly positive).
- In the reaction with ammonia, 5 reactant particles (1 complex ion + 4 monodentate ligands) react to produce 5 product particles (1 complex ion + 4 water molecules). There is no net change in the number of particles, so the entropy change (\(\Delta S^\ominus\)) is close to zero.
- Since \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\) and \(\Delta G^\ominus = -RT \ln K\), a more positive \(\Delta S^\ominus\) for the en reaction makes \(\Delta G^\ominus\) more negative, resulting in a much larger value of \(K_{\text{stab}}\). This is known as the chelate effect.

**(a)** [2 marks]
- 1 mark: Equilibrium constant for the formation of a complex ion.
- 1 mark: From its constituent ions / ligands / simpler complexes in a solvent.

**(b)** [2 marks]
- 1 mark: Correct expression for \(K_{\text{stab}}\).
- 1 mark: Correct units: \(\text{dm}^6 \text{ mol}^{-2}\).

**(c)** [4 marks]
- 1 mark: \([[\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}]_{\text{eq}} = 0.050\text{ mol dm}^{-3}\) (or showing subtraction of \(1.2 \times 10^{-10}\)).
- 1 mark: \([\text{en}]_{\text{eq}} = 0.150\text{ mol dm}^{-3}\) (correctly subtracting \(2 \times 0.050\)).
- 1 mark: Correct substitution into the expression: \(K_{\text{stab}} = \frac{0.050}{(1.2 \times 10^{-10}) \times (0.150)^2}\).
- 1 mark: Final calculated value of \(1.85 \times 10^{10}\) or \(1.9 \times 10^{10}\).

**(d)** [3 marks]
- 1 mark: Identifies that the reaction with \(\text{en}\) increases the number of particles/species in solution (3 to 5), leading to a positive entropy change (\(\Delta S^\ominus > 0\)), whereas the reaction with \(\text{NH}_3\) shows no change in the number of particles (5 to 5).
- 1 mark: Explains that \(\Delta S^\ominus\) is more positive for the \(\text{en}\) reaction.
- 1 mark: Links the more positive \(\Delta S^\ominus\) to a more negative \(\Delta G^\ominus\), which increases the value of the equilibrium constant (\(K_{\text{stab}}\)).

**(a)** The Hess's Law cycle is constructed as follows:

\[\begin{array}{ccc} \text{CoCl}_2(\text{s}) + 6\text{H}_2\text{O}(\text{l}) & \xrightarrow{\Delta H_{\text{hyd}}} & \text{CoCl}_2\cdot6\text{H}_2\text{O}(\text{s}) \\ \searrow{\Delta H_1} & & \swarrow{\Delta H_2} \\ & \text{Co}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq}) & \end{array}\]

From the cycle, the algebraic expression is:
\[\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2\]

**(b)**
\[\text{Moles of } \text{CoCl}_2 = 0.400\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0200\text{ mol}\]
\[\text{Mass} = 0.0200\text{ mol} \times 129.9\text{ g mol}^{-1} = 2.60\text{ g}\]

**(c)**
* Independent variable: Time (for a single run to record the temperature curve) or the identity of the salt used (when comparing the two experiments).
* Dependent variable: Temperature (for a single run) or temperature change (\(\Delta T\)) (when comparing the two experiments).

**(d)**
1. Place a polystyrene cup inside a \(250\text{ cm}^3\) beaker to provide support and thermal insulation.
2. Use a volumetric pipette to measure and transfer \(50.0\text{ cm}^3\) of distilled water into the polystyrene cup.
3. Support a thermometer (graduated to \(0.1\text{ }^\circ\text{C}\)) in the water and record the temperature every minute for 3 minutes, stirring continuously.
4. At the 4th minute, add the pre-weighed anhydrous cobalt(II) chloride (\(2.60\text{ g}\)) rapidly into the cup, stirring vigorously, but do not record the temperature at this exact minute.
5. Record the temperature of the solution every minute from the 5th minute to the 10th minute.
6. Plot a graph of temperature against time. Extrapolate the temperature readings before addition and the cooling curve back to the 4th minute (the time of mixing) to determine the corrected maximum temperature change, \(\Delta T\).

**(e) (i)**
\[\text{Moles of } \text{CoCl}_2\cdot6\text{H}_2\text{O} = \frac{4.76\text{ g}}{237.9\text{ g mol}^{-1}} = 0.0200\text{ mol}\]
\[\text{Moles of water of crystallization} = 6 \times 0.0200 = 0.120\text{ mol}\]
\[\text{Mass of water of crystallization} = 0.120\text{ mol} \times 18.0\text{ g mol}^{-1} = 2.16\text{ g}\]
Since density of water is \(1.00\text{ g cm}^{-3}\), this represents \(2.16\text{ cm}^3\) of water.
\[\text{Volume of distilled water to be added} = 50.0 - 2.16 = 47.84\text{ cm}^3 \approx 47.8\text{ cm}^3\]

**(e) (ii)**
\[q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 2.4\text{ K} = 501.6\text{ J} = 0.5016\text{ kJ}\]
Since temperature decreased, the process is endothermic:
\[\Delta H_2 = + \frac{0.5016\text{ kJ}}{0.0200\text{ mol}} = +25.08\text{ kJ mol}^{-1} \approx +25.1\text{ kJ mol}^{-1}\]

**(f)**
* Hazard: Cobalt(II) chloride is toxic if swallowed / skin sensitizer / carcinogenic.
* Precaution: Wear protective nitrile gloves / avoid skin contact / wash hands thoroughly after handling / perform in a fume cupboard if handling fine powder.

* **(a)** [2 marks]
* 1 mark for drawing the correct Hess's cycle showing both solid reactant and solid product pointing downwards to the same aqueous ions: \(\text{Co}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq})\).
* 1 mark for stating the correct algebraic relationship: \(\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2\).
* **(b)** [1 mark]
* 1 mark for calculating \(2.60\text{ g}\) (accept \(2.598\text{ g}\)).
* **(c)** [2 marks]
* 1 mark for correctly identifying the independent variable (time or identity of salt).
* 1 mark for correctly identifying the dependent variable (temperature or temperature change).
* **(d)** [5 marks]
* 1 mark for specifying the use of a polystyrene cup placed in a beaker for insulation, and measuring water using a volumetric pipette or burette.
* 1 mark for weighing by difference (weighing container before and after adding the salt to find exact mass transferred).
* 1 mark for measuring the temperature at regular intervals (e.g. every minute) for several minutes prior to salt addition.
* 1 mark for adding the salt rapidly at a defined minute (e.g. 4th minute) without recording the temperature at that exact moment, followed by continuous stirring.
* 1 mark for measuring post-addition temperatures at regular intervals and using extrapolation of the temperature-time lines back to the time of mixing to find the corrected temperature change.
* **(e)(i)** [2 marks]
* 1 mark for calculating the mass of water in the crystal: \(0.0200\text{ mol} \times 6 \times 18.0 = 2.16\text{ g}\).
* 1 mark for showing subtraction of \(2.16\text{ cm}^3\) from \(50.0\text{ cm}^3\) to yield \(47.8\text{ cm}^3\).
* **(e)(ii)** [2 marks]
* 1 mark for calculating the heat change, \(q = 50.0 \times 4.18 \times 2.4 = 501.6\text{ J}\) or \(0.5016\text{ kJ}\).
* 1 mark for calculating \(\Delta H_2 = +25.1\text{ kJ mol}^{-1}\) (must have positive sign, 3 significant figures, and correct units).
* **(f)** [1 mark]
* 1 mark for identifying a valid hazard (toxic / skin sensitizer / carcinogen) matched with a valid safety precaution (wear gloves / safety goggles / use fume cupboard).

**(a)** The completed table with values of \(\ln([R\text{-Br}])\) calculated to 3 decimal places is:

| Time, \(t\text{ / s}\) | \([R\text{-Br}]\text{ / mol dm}^{-3}\) | \(\ln([R\text{-Br}])\) |
| :---: | :---: | :---: |
| 0 | 0.100 | -2.303 |
| 100 | 0.067 | -2.703 |
| 200 | 0.045 | -3.101 |
| 300 | 0.018 | -4.017 |
| 400 | 0.020 | -3.912 |
| 500 | 0.0135 | -4.305 |

**(b)**
* Plot \(\ln([R\text{-Br}])\) on the vertical y-axis (ranging from \(-2.0\) to \(-4.5\)).
* Plot Time, \(t\) on the horizontal x-axis (ranging from \(0\) to \(500\)).
* The points at \(t = 0, 100, 200, 400, 500\) lie very close to a straight line. The point at \(t = 300\) lies significantly below the line.
* Draw a straight line of best fit through the non-anomalous points.

**(c)**
* The anomalous reading occurred at \(t = 300\text{ s}\).
* Reason for lower concentration: The student may have failed to quench the sample immediately when it was withdrawn. As a result, the hydrolysis reaction continued during the delay before titration, leading to a lower remaining concentration of \((CH_3)_3CBr\) than expected.

**(d)**
Select two points on the line of best fit that are widely separated (more than half the length of the drawn line). For example:
* Point 1: \((50, -2.503)\)
* Point 2: \((450, -4.103)\)

\[\text{Gradient} = \frac{-4.103 - (-2.503)}{450 - 50} = \frac{-1.60}{400} = -4.00 \times 10^{-3}\text{ s}^{-1}\]

**(e)**
* **Order:** First-order, because the graph of \(\ln([R\text{-Br}])\) against time is a straight line.
* **Rate constant, \(k\):** For a first-order reaction, \(\ln[A]_t = -kt + \ln[A]_0\). Therefore, \(k = -\text{gradient}\).
\[k = 4.0 \times 10^{-3}\]
* **Units:** \(\text{s}^{-1}\)

**(f)**
* The student should plot a graph of \([R\text{-Br}]\) against time, \(t\).
* Measure at least two successive half-lives (e.g., the time taken for concentration to fall from \(0.100\) to \(0.050\text{ mol dm}^{-3}\), and from \(0.050\) to \(0.025\text{ mol dm}^{-3}\)).
* If the half-lives are constant (approximately \(173\text{ s}\)), this confirms the reaction is first-order.

* **(a)** [2 marks]
* 1 mark for calculating all 6 logarithmic values correctly.
* 1 mark for presenting all values to exactly 3 decimal places (with correct negative signs).
* **(b)** [4 marks]
* 1 mark for clearly labeled axes with appropriate units: y-axis labeled with \(\ln([R\text{-Br}])\) (no units) and x-axis labeled with \(t\text{ / s}\).
* 1 mark for choosing a linear scale where plotted points occupy more than half of the graph grid on both axes.
* 1 mark for plotting all 6 points accurately.
* 1 mark for drawing a single, straight line of best fit, ignoring the anomalous point at \(t = 300\text{ s}\).
* **(c)** [2 marks]
* 1 mark for identifying the anomaly at \(t = 300\text{ s}\).
* 1 mark for suggesting a valid experimental reason for the lower reactant concentration (e.g. delay in quenching/titration, incomplete quenching, or temporary increase in temperature during that time interval).
* **(d)** [2 marks]
* 1 mark for stating the coordinates of two points on the line of best fit which are separated by at least half the length of the line.
* 1 mark for calculating the gradient correctly to 2 or 3 significant figures (expected value is approximately \(-4.0 \times 10^{-3}\)).
* **(e)** [3 marks]
* 1 mark for identifying first-order and stating that the plot of \(\ln([R\text{-Br}])\) vs time is linear.
* 1 mark for calculating \(k = -\text{gradient}\) (must match the absolute value of the calculated gradient, e.g., \(4.0 \times 10^{-3}\)).
* 1 mark for stating the correct units of \(k\) as \(\text{s}^{-1}\).
* **(f)** [2 marks]
* 1 mark for explaining that the student should determine successive half-lives from the concentration-time plot.
* 1 mark for stating that the half-lives must be constant to confirm a first-order reaction.