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There is a very large increase in ionisation energy between the 3rd and 4th ionisation energies (from 2745 to 11577 \(\text{kJ mol}^{-1}\)). This indicates that the 4th electron is removed from an inner electron shell closer to the nucleus, meaning the element has 3 valence electrons and belongs to Group 13 (Group 3). The Period 3 element in Group 13 is aluminium.

1 mark for identifying Aluminium (C) based on the location of the largest jump in successive ionisation energies.

The balanced equation for the complete combustion of butane is:
\(\text{C}_4\text{H}_{10}\text{(g)} + 6.5\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 5\text{H}_2\text{O}\text{(l)}\)

From the stoichiometry, 1 mole of butane reacts with 6.5 moles of oxygen.
Number of moles of oxygen required = \(0.20 \text{ mol} \times 6.5 = 1.30\text{ mol}\).
Volume of oxygen at RTP = \(1.30\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 31.2\text{ dm}^3\).

1 mark for the correct calculation leading to 31.2 \(\text{dm}^3\) (C).

The hydronium ion, \(\text{H}_3\text{O}^+\), has 3 bonding pairs and 1 lone pair of electrons on the oxygen atom, giving it a trigonal pyramidal shape. The repulsion from the lone pair reduces the bond angle from the tetrahedral angle of \(109.5^\circ\) to approximately \(107^\circ\), which is identical to the geometry of ammonia (\(\text{NH}_3\)).

1 mark for selecting \(\text{H}_3\text{O}^+\).

Let the abundance of \(^{65}X\) be \(y\) (as a fraction) and the abundance of \(^{63}X\) be \(1 - y\).

Using the formula for relative atomic mass:
\(63(1 - y) + 65y = 63.6\)
\(63 - 63y + 65y = 63.6\)
\(2y = 0.6\)
\(y = 0.3\)

Therefore, the percentage abundance of \(^{65}X\) is 30%.

1 mark for the correct calculation of 30% (A).

In a propagation step, a free radical reacts with a stable molecule to produce a new free radical and a new stable molecule. Option B shows a chlorine radical reacting with methane to form a methyl radical and hydrogen chloride, which is a key propagation step.

1 mark for selecting option B.

For a molecule to exhibit E/Z isomerism, both carbon atoms involved in the double bond must have two different groups attached to them. In pent-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3\)), Carbon-2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\), while Carbon-3 is bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Thus, it has distinct E and Z isomers.

1 mark for selecting pent-2-ene (D).

Find the molar ratio of the elements:

\(\text{Moles of C} = \frac{52.2}{12.0} = 4.35\)
\(\text{Moles of H} = \frac{13.0}{1.0} = 13.0\)
\(\text{Moles of O} = \frac{34.8}{16.0} = 2.175\)

Divide by the smallest value (2.175):

\(\text{C}: \frac{4.35}{2.175} = 2\)
\(\text{H}: \frac{13.0}{2.175} = 5.98 \approx 6\)
\(\text{O}: \frac{2.175}{2.175} = 1\)

The empirical formula is \(\text{C}_2\text{H}_6\text{O}\).

1 mark for the correct empirical formula derivation (B).

Boron trichloride, \(\text{BCl}_3\), has three polar \(\text{B}-\text{Cl}\) bonds due to the difference in electronegativity between boron and chlorine. However, because the molecule is trigonal planar and symmetrical, the individual dipoles cancel each other out completely, resulting in a net dipole moment of zero.

1 mark for selecting \(\text{BCl}_3\) (C).

First, calculate the amount of zinc in moles:
\(n(\text{Zn}) = \frac{3.27\text{ g}}{65.4\text{ g mol}^{-1}} = 0.0500\text{ mol}\)

From the balanced chemical equation, the molar ratio of \(\text{Zn}\) to \(\text{H}_2\) gas is \(1:1\):
\(n(\text{H}_2) = 0.0500\text{ mol}\)

Next, calculate the volume of hydrogen gas produced:
\(V(\text{H}_2) = 0.0500\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3\)

1 mark: Correct option selected (B).

The largest relative jump in ionization energy occurs between the third (2745) and fourth (11577) ionization energies. This indicates that the fourth electron is being removed from a lower, fully-filled electron shell, meaning element \(\text{X}\) has three valence electrons (Group 13 / Group 3). The stable ion formed by this element is \(\text{X}^{3+}\). The oxide ion is \(\text{O}^{2-}\). Therefore, the formula of the oxide is \(\text{X}_2\text{O}_3\).

1 mark: Correct option selected (C).

The central phosphorus atom has 5 valence electrons. The positive charge means we subtract 1 electron, leaving 4 valence electrons. These 4 electrons are used to form 4 single covalent bonds with 4 chlorine atoms. There are 4 bonding pairs and 0 lone pairs of electrons around the central phosphorus atom. According to VSEPR theory, 4 bonding pairs and 0 lone pairs adopt a tetrahedral geometry to minimize electron-pair repulsion.

1 mark: Correct option selected (A).

1. Find the longest continuous carbon chain: The longest chain containing the maximum number of carbons has 5 carbons, so the parent name is pentane.
2. Number the chain from the end that gives the substituents the lowest possible numbers: Numbering from the left gives substituents at positions 2 (methyl) and 3 (ethyl). Numbering from the right would give substituents at positions 3 (ethyl) and 4 (methyl). Thus, we number from left to right.
3. List the substituents in alphabetical order: 'ethyl' comes before 'methyl'. Therefore, the correct name is 3-ethyl-2-methylpentane.

1 mark: Correct option selected (B).

During the electrophilic addition of \(\text{HBr}\) to propene, the electrophile (\(\text{H}^+\)) adds to the double bond to form a carbocation intermediate. Addition of \(\text{H}^+\clef\) to the terminal carbon forms a secondary carbocation, \(\text{CH}_3\text{C}^+\text{HCH}_3\). Addition of \(\text{H}^+\clef\) to the middle carbon forms a primary carbocation, \(\text{CH}_3\text{CH}_2\text{C}^+\text{H}_2\). The secondary carbocation is more stable than the primary carbocation due to the electron-donating inductive effect of the two alkyl groups, so the reaction proceeds via the more stable secondary carbocation to yield 2-bromopropane as the major product.

1 mark: Correct option selected (A).

Percentage yield is a measure of the efficiency of converting reactants to the desired products in practice. A high percentage yield indicates that most of the limiting reactant was successfully converted into products. Atom economy is a theoretical measure of how much of the mass of the starting reactants ends up in the desired final product. A low atom economy indicates that a significant proportion of the reactants forms unwanted side products (waste).

1 mark: Correct option selected (A).

A neutral copper atom (\(\text{Cu}\)) has the electronic configuration: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\) (or \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\)). When forming transition metal cations, electrons are always lost from the \(4\text{s}\) orbital before the \(3\text{d}\) orbital. Therefore, losing one electron to form the \(\text{Cu}^+\) ion results in the loss of the single \(4\text{s}\) electron, giving the configuration: \([\text{Ar}] 3\text{d}^{10}\).

1 mark: Correct option selected (B).

Boron trifluoride (\(\text{BF}_3\)) contains polar \(\text{B-F}\) bonds due to the high electronegativity difference between boron and fluorine. However, \(\text{BF}_3\) has a trigonal planar geometry which is highly symmetrical. The individual dipole moments of the three polar bonds cancel each other out completely, leaving the molecule with no net dipole moment (overall non-polar).
- \(\text{NH}_3\) is polar (trigonal pyramidal, dipoles do not cancel).
- \(\text{CHCl}_3\) is polar (tetrahedral but asymmetrical, dipoles do not cancel).
- \(\text{H}_2\text{O}\) is polar (non-linear/bent, dipoles do not cancel).

1 mark: Correct option selected (C).

The successive ionisation energies show a very large increase (a 'jump') between the 5th and the 6th ionisation energies (from \(6274\text{ kJ mol}^{-1}\) to \(21267\text{ kJ mol}^{-1}\)). This indicates that the 6th electron is being removed from an inner quantum shell closer to the nucleus, which experiences significantly less shielding. Therefore, element \(X\) has 5 electrons in its outermost shell and belongs to Group 15 (Group 5). The Period 3 element in Group 15 is phosphorus (\(\text{P}\)).

[1 mark] for selecting B (Phosphorus).
Award 0 marks for incorrect identification.

First, find the number of moles of carbon dioxide gas produced:
\(n(\text{CO}_2) = \frac{288\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0120\text{ mol}\)

From the 1:1 stoichiometry of the chemical equation:
\(n(\text{CaCO}_3) = n(\text{CO}_2) = 0.0120\text{ mol}\)

Calculate the mass of pure calcium carbonate:
\(m(\text{CaCO}_3) = 0.0120\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.2012\text{ g}\)

Determine the percentage purity:
\(\text{Purity} = \frac{1.2012\text{ g}}{1.50\text{ g}} \times 100\% = 80.08\% \approx 80.1\%\)

[1 mark] for selecting C (80.1%).
Award 0 marks for any other response.

1. \(\text{NH}_2^-\): The central nitrogen atom has 2 bonding pairs of electrons and 2 lone pairs of electrons. Repulsion between the two lone pairs is greater than between lone pair-bond pair and bond pair-bond pair, reducing the bond angle from the tetrahedral angle to approximately \(104.5^\circ\).

2. \(\text{NH}_3\): The central nitrogen atom has 3 bonding pairs and 1 lone pair of electrons. The lone pair repels the bonding pairs, reducing the tetrahedral angle to approximately \(107^\circ\).

3. \(\text{NH}_4^+\): The central nitrogen atom has 4 bonding pairs and 0 lone pairs of electrons. The 4 bonding pairs repel each other equally, resulting in a perfect tetrahedral geometry with a bond angle of \(109.5^\circ\).

Therefore, the order of increasing bond angle is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).

[1 mark] for selecting B.
Award 0 marks for incorrect choices.

The electrophilic addition of \(\text{HBr}\) to but-1-ene proceeds via carbocation intermediates. Addition of the hydrogen ion to carbon-1 forms a secondary carbocation (\(\text{CH}_3\text{CH}_2\text{C}^+\text{HCH}_3\)), whereas addition to carbon-2 forms a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{C}^+\text{H}_2\)). The secondary carbocation intermediate is more stable than the primary carbocation because the positive charge on the carbon atom is stabilised by the electron-releasing inductive effect of two alkyl groups (an ethyl group and a methyl group), compared to only one alkyl group (a propyl group) in the primary carbocation.

[1 mark] for selecting A.
Award 0 marks for incorrect explanations.

(a) \(M_r((NH_4)_2Fe(SO_4)_2) = (14.0 \times 2) + (1.0 \times 8) + 55.8 + (32.1 \times 2) + (16.0 \times 8) = 284.0 \text{ g mol}^{-1}\).

(b)(i) \(MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\).

(b)(ii) Moles of \(MnO_4^-\) in titration = \(0.0200 \times (10.00 / 1000) = 2.00 \times 10^{-4} \text{ mol}\). Moles of \(Fe^{2+}\) in \(25.0 \text{ cm}^3 = 5 \times (2.00 \times 10^{-4}) = 1.00 \times 10^{-3} \text{ mol}\). Moles of \(Fe^{2+}\) in \(250.0 \text{ cm}^3 = 1.00 \times 10^{-3} \times 10 = 1.00 \times 10^{-2} \text{ mol}\).

(b)(iii) Moles of hydrated salt in 3.92 g = \(1.00 \times 10^{-2} \text{ mol}\). \(M_r\) of hydrated salt = \(3.92 / (1.00 \times 10^{-2}) = 392 \text{ g mol}^{-1}\). Mass of water of crystallization per mole = \(392 - 284 = 108 \text{ g}\). \(x = 108 / 18.0 = 6\).

(b)(iv) Colorless / pale green to pale pink.

(a) M1: Correctly summing individual atomic masses (1). M2: Correct calculation of 284.0 g/mol (1).
(b)(i) M1: All species correct (1). M2: Correct balancing (1).
(b)(ii) M1: Calculates moles of MnO4- as 2.00 x 10^-4 mol (1). M2: Calculates moles of Fe2+ in 25.0 cm3 as 1.00 x 10^-3 mol (1). M3: Scales up to 250.0 cm3 to get 1.00 x 10^-2 mol (1).
(b)(iii) M1: Calculates Mr of hydrated salt as 392 (1). M2: Calculates mass of water as 108 g (1). M3: Divides by 18.0 to find x (1). M4: Correct integer value of x = 6 (1).
(b)(iv) M1: Colorless (or very pale green) to (permanent) pale pink (1).

(a) The first ionization energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.

(b)(i) Element X is Aluminium / Al. There is a very large increase (jump) between the third and fourth ionization energies. This indicates that the fourth electron is being removed from a lower energy shell closer to the nucleus (inner shell), meaning there are three electrons in the outer shell.

(b)(ii) \(Al^{2+}(g) \rightarrow Al^{3+}(g) + e^-\) (Accept \(X^{2+}(g) \rightarrow X^{3+}(g) + e^-\)).

(c) Phosphorus has the outer shell electron configuration of \(3s^2 3p^3\) with singly occupied p-orbitals, whereas sulfur has the configuration \(3s^2 3p^4\). In sulfur, one of the 3p orbitals contains a pair of electrons. The repulsion between these paired electrons in the same orbital makes it easier to remove the outer electron from sulfur, requiring less energy despite the greater nuclear charge of sulfur.

(a) M1: removal of one mole of electrons from one mole of atoms (1). M2: gaseous state specified for both reactants and products (1). M3: formation of one mole of 1+ gaseous ions (1).
(b)(i) M1: Identifies Element X as Aluminium / Al (1). M2: States there is a large jump between the 3rd and 4th ionization energies (1). M3: Explains that the 4th electron is removed from a shell closer to the nucleus / shell with less shielding / inner shell (1).
(b)(ii) M1: Correct chemical species (Al2+ and Al3+) (1). M2: Correct gaseous state symbols on both sides (1).
(c) M1: Mentions electron configuration of P as ending in 3p3 and S as 3p4 (1). M2: Identifies that P has singly occupied 3p orbitals (1). M3: Identifies that S has a paired set of electrons in a 3p orbital (1). M4: Explains that the mutual repulsion of paired electrons in S makes it easier to remove (1).

(a) Boron trifluoride, \(BF_3\), has three bonding pairs of electrons and zero lone pairs around the central boron atom. According to electron pair repulsion theory, these pairs repel each other equally to minimize repulsion, resulting in a trigonal planar shape with a bond angle of \(120^\circ\). In contrast, nitrogen trifluoride, \(NF_3\), has three bonding pairs and one lone pair of electrons around the central nitrogen atom. The four electron pairs arrange themselves tetrahedrally, but the shape of the molecule is trigonal pyramidal. Because lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, the bond angle is reduced from the tetrahedral angle to approximately \(107^\circ\).

(b) Magnesium fluoride has ionic bonding. It is held together by the strong electrostatic forces of attraction between oppositely charged magnesium ions, \(Mg^{2+}\), and fluoride ions, \(F^-\), arranged in a giant ionic lattice. A large amount of thermal energy is needed to overcome these strong electrostatic forces throughout the giant lattice, which explains its high melting point.

(c) In \(CO_3^{2-}\), the carbon atom forms one double covalent bond with one oxygen atom (sharing 4 electrons) and single covalent bonds with two other oxygen atoms (sharing 2 electrons each). The two single-bonded oxygen atoms each carry a negative charge (possessing an extra electron). The carbon has no lone pairs, the double-bonded oxygen has 2 lone pairs (4 electrons), and each single-bonded oxygen has 3 lone pairs (6 electrons). The entire structure is enclosed in square brackets with a 2- charge outside.

(a) M1: BF3 has 3 bonding pairs (BP) and 0 lone pairs (LP) (1). M2: BF3 shape is trigonal planar with a bond angle of 120 degrees (1). M3: NF3 has 3 BP and 1 LP (1). M4: NF3 shape is trigonal pyramidal with a bond angle of 107 degrees (accept 106-108 degrees) (1). M5: Electron pairs arrange themselves to minimize repulsion (1). M6: Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion (1).
(b) M1: Identifies bonding as ionic (1). M2: Describes strong electrostatic attraction between Mg2+ and F- ions (1). M3: States it has a giant ionic lattice structure (1). M4: Explains that high energy is required to break these strong bonds (1).
(c) M1: Correct double bond (C=O) and two single bonds (C-O) shown with dots and crosses (1). M2: Correct number of non-bonding outer electrons on all oxygen atoms and overall 2- charge indicated (1).

(a) But-2-ene exhibits stereoisomerism because there is restricted rotation around the carbon-carbon double bond (\(C=C\)) due to the presence of a \(\pi\) bond, and each carbon atom in the double bond is attached to two different groups: a hydrogen atom and a methyl group. In but-1-ene, one of the double-bonded carbon atoms is attached to two identical groups (hydrogen atoms), so stereoisomerism is not possible. The skeletal structures are: cis-but-2-ene (or (Z)-but-2-ene) with both methyl groups on the same side, and trans-but-2-ene (or (E)-but-2-ene) with methyl groups on opposite sides.

(b)(i) Mechanism:
1. Electrophilic attack: A curly arrow starts from the C=C double bond pointing to the partially positive hydrogen atom in \(H^{\delta+}-Br^{\delta-}\).
2. H-Br bond cleavage: A curly arrow starts from the H-Br covalent bond pointing to the bromine atom.
3. Carbocation intermediate: The secondary carbocation \(CH_3-CH^+-CH_2-CH_3\) is formed alongside a bromide ion, \(:Br^-\).
4. Nucleophilic attack: A curly arrow starts from the lone pair on the bromide ion pointing to the positively charged carbon atom of the carbocation.
5. Product: 2-bromobutane is formed.

(b)(ii) The major product is 2-bromobutane. The reaction of but-1-ene with \(HBr\) can proceed via two different carbocation intermediates: a secondary carbocation (\(CH_3-CH^+-CH_2-CH_3\)) or a primary carbocation (\(CH_3-CH_2-CH_2-CH_2^+\)). The secondary carbocation is more stable than the primary carbocation because it has two electron-donating alkyl groups (inductive effect) stabilizing the positive charge compared to only one in the primary carbocation. Therefore, the secondary carbocation forms more rapidly, leading to a higher yield of 2-bromobutane.

(a) M1: Restricted rotation about the C=C bond (1). M2: Each carbon of the double bond must be attached to two different groups (1). M3: Correctly drawn skeletal structures for both isomers (1). M4: Correct IUPAC names: cis-but-2-ene / (Z)-but-2-ene and trans-but-2-ene / (E)-but-2-ene (1).
(b)(i) M1: Curly arrow from C=C bond to H of H-Br (1). M2: Dipoles on H-Br and curly arrow from H-Br bond to Br (1). M3: Correct structure of the secondary carbocation intermediate (1). M4: Curly arrow from the lone pair of Br- to the C+ of the carbocation (1). M5: Correct final structure of 2-bromobutane (1).
(b)(ii) M1: Identifies 2-bromobutane as the major product (1). M2: States that the reaction goes via a secondary carbocation which is more stable than a primary carbocation (1). M3: Explains that the alkyl groups have an electron-donating inductive effect that stabilizes the positive charge (1).

(a) Crude oil is a mixture of hydrocarbons. As the carbon chain length increases, the strength of the intermolecular London (dispersion) forces increases because there are more electrons and a larger surface area of contact. Consequently, longer-chain hydrocarbons have higher boiling points and condense lower down the fractionating column where temperatures are higher, while shorter-chain hydrocarbons have lower boiling points and condense near the cooler top.

(b) \(C_3H_8 \rightarrow CH_4 + C_2H_4\).

(c)(i) UV light provides the energy needed to undergo homolytic fission of the chlorine-chlorine covalent bond, generating chlorine free radicals.

(c)(ii)
Initiation: \(Cl_2 \xrightarrow{UV} 2Cl^\bullet\)
Propagation 1: \(CH_4 + Cl^\bullet \rightarrow \bullet CH_3 + HCl\)
Propagation 2: \(\bullet CH_3 + Cl_2 \rightarrow CH_3Cl + Cl^\bullet\)
Termination: \(\bullet CH_3 + Cl^\bullet \rightarrow CH_3Cl\)

(c)(iii) During the reaction, methyl radicals (\(\bullet CH_3\)) are produced in the first propagation step. In a termination step, two of these methyl free radicals can collide and combine to form a covalent bond, resulting in the formation of ethane. Equation: \(2\bullet CH_3 \rightarrow C_2H_6\).

(a) M1: Mentions London forces (dispersion forces) (1). M2: Explains that London forces increase with carbon chain length / molecular size / contact area (1). M3: Relates this to boiling point differences and separation along the column (1).
(b) M1: Correct formula of reactants and products: C3H8 -> CH4 + C2H4 (1).
(c)(i) M1: Energy to break Cl-Cl bond / homolytic fission (1).
(c)(ii) M1: Correct initiation equation (1). M2: Correct propagation step 1 (1). M3: Correct propagation step 2 (1). M4: Correct termination step producing chloromethane (1). (Radical dots must be clearly shown on Cl and CH3).
(c)(iii) M1: Identifies collision/combination of two methyl radicals (1). M2: Identifies this as a termination reaction (1). M3: Correct balanced equation: 2.CH3 -> C2H6 (1).

First, calculate the energy absorbed by the water (\(q\)): \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 38.5\text{ K} = 16093\text{ J} = 16.093\text{ kJ}\). Next, calculate the amount of methanol burned in moles (\(n\)): \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.025\text{ mol}\). Finally, calculate the molar enthalpy change of combustion: \(\Delta_c H = -\frac{q}{n} = -\frac{16.093\text{ kJ}}{0.025\text{ mol}} = -643.72\text{ kJ mol}^{-1} \approx -644\text{ kJ mol}^{-1}\).

1 mark for Option C. Incorrect options result from wrong calculations (A is just the value of q in kJ, B represents an incorrect scaling, and D is incorrect by a factor of 2).

Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) contains an \(\text{O}-\text{H}\) group and is capable of forming intermolecular hydrogen bonds, which are the strongest type of intermolecular force. Propanal and methoxyethane have permanent dipole-dipole forces, and butane has only London forces. Therefore, propan-1-ol has the highest boiling temperature.

1 mark for Option A.

The \(\text{Mg}^{2+}\) ion is smaller and has a higher charge density than the \(\text{Ba}^{2+}\) ion. Thus, the \(\text{Mg}^{2+}\) ion polarises the large carbonate (\(\text{CO}_3^{2-}\)) ion more strongly, weakening the carbon-oxygen bonds within the carbonate group and decreasing its thermal stability.

1 mark for Option A.

2-bromo-2-methylpropane is a tertiary halogenoalkane. Because of the steric hindrance of the methyl groups and the high stability of the tertiary carbocation intermediate formed, tertiary halogenoalkanes undergo nucleophilic substitution via the two-step \(\text{S}_\text{N}1\) mechanism.

1 mark for Option B.

Increasing the temperature of a reaction mixture shifts the peak of the Maxwell-Boltzmann distribution to the right (higher energy) and downwards (lower fraction of molecules). The activation energy (\(E_a\)) is a constant value for a given reaction pathway and does not change with temperature.

1 mark for Option B.

To find \(\Delta H\), calculate \(\sum (\text{bond enthalpies of reactants}) - \sum (\text{bond enthalpies of products})\). Focusing on only the bonds that change: Bonds broken: 1 \(\text{C}=\text{C}\) (\(+612\)) and 1 \(\text{H}-\text{H}\) (\(+436\)) = \(+1048\text{ kJ mol}^{-1}\). Bonds formed: 1 \(\text{C}-\text{C}\) (\(-347\)) and 2 \(\text{C}-\text{H}\) (\(-2 \times 413 = -826\)) = \(-1173\text{ kJ mol}^{-1}\). \(\Delta H = 1048 - 1173 = -125\text{ kJ mol}^{-1}\).

1 mark for Option A.

Iodide ions are very strong reducing agents and reduce concentrated sulfuric acid to a mixture of products, including sulfur dioxide (\(\text{SO}_2\)), elemental sulfur (\(\text{S}\)), and hydrogen sulfide (\(\text{H}_2\text{S}\)). Elemental sulfur is a yellow solid.

1 mark for Option B.

The strong absorption peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad peak between \(3200-3600\text{ cm}^{-1}\) shows there is no hydroxyl (\(\text{O}-\text{H}\)) group. Propanone has the formula \(\text{C}_3\text{H}_6\text{O}\) and a carbonyl group but no hydroxyl group, which matches this data perfectly.

1 mark for Option B.

To find the standard enthalpy of formation of propane:

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

Using Hess's Law and standard enthalpies of combustion:

\(\Delta_f H^\theta = [3 \times \Delta_c H^\theta(\text{C}) + 4 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{C}_3\text{H}_8)\)

\(\Delta_f H^\theta = [3 \times (-394) + 4 \times (-286)] - (-2219)\)

\(\Delta_f H^\theta = [-1182 - 1144] + 2219\)

\(\Delta_f H^\theta = -2326 + 2219 = -107\text{ kJ mol}^{-1}\)

Award 1 mark for the correct answer (C).
Method: award 1 mark for the correct application of Hess's Law using combustion data.

HF has the highest boiling temperature due to strong intermolecular hydrogen bonding, which is absent in the other hydrogen halides. For HCl, HBr, and HI, the boiling temperature increases as the number of electrons increases because the strength of London (dispersion) forces increases. Thus, the order of increasing boiling temperature is \(\text{HCl} < \text{HBr} < \text{HI} < \text{HF}\).

Award 1 mark for the correct answer (B).

The barium ion (\(\text{Ba}^{2+}\)) has a much larger ionic radius than the magnesium ion (\(\text{Mg}^{2+}\)), resulting in a lower charge density. Therefore, the \(\text{Ba}^{2+}\) ion polarizes the carbonate (\(\text{CO}_3^{2-}\) ) electron cloud less than \(\text{Mg}^{2+}\) does. Since polarization weakens the carbon-oxygen bond within the carbonate ion, barium carbonate is thermally more stable and requires a higher temperature to decompose.

Award 1 mark for the correct answer (D).

The rate of hydrolysis of halogenoalkanes depends on the strength of the C–X bond (bond enthalpy), not the bond polarity. The C–I bond is the weakest bond (lowest bond enthalpy) because iodine has the largest atomic radius, resulting in the poorest orbital overlap. Hence, the C–I bond is broken most easily, and 1-iodobutane hydrolyzes fastest.

Award 1 mark for the correct answer (A).

A catalyst provides an alternative reaction pathway with a lower activation energy, so the activation energy (\(E_a\)) decreases. However, because the temperature is held constant, the kinetic energy of the particles remains unchanged. Thus, the shape and the peak of the Maxwell-Boltzmann energy distribution curve do not change.

Award 1 mark for the correct answer (B).

Iodide ions (\(\text{I}^-\)) are strong reducing agents and reduce sulfuric acid (\(\text{H}_2\text{SO}_4\), sulfur oxidation state +6) to sulfur dioxide (\(\text{SO}_2\), +4), sulfur (\(\text{S}\), 0), and hydrogen sulfide (\(\text{H}_2\text{S}\), -2). Iodine (\(\text{I}_2\)) is the oxidation product of iodide. Hydrogen iodide (\(\text{HI}\)) is formed via a non-redox acid-base reaction.

Award 1 mark for the correct answer (C).

The molecular formula \(\text{C}_3\text{H}_8\text{O}\) represents propan-1-ol or propan-2-ol.
- Heating propan-2-ol (a secondary alcohol) under reflux with acidified dichromate(VI) oxidizes it to propanone. The ketone propanone has a \(\text{C}=\text{O}\) bond (strong absorption at \(1715\text{ cm}^{-1}\)) but no \(\text{O}-\text{H}\) bond, so there is no absorption above \(3000\text{ cm}^{-1}\).
- Heating propan-1-ol (a primary alcohol) under reflux with excess oxidising agent produces propanoic acid. Propanoic acid contains both a \(\text{C}=\text{O}\) bond and a very broad carboxylic acid \(\text{O}-\text{H}\) absorption in the range \(2500\text{–}3300\text{ cm}^{-1}\).
Therefore, \(\mathbf{Y}\) must be propan-2-ol.

Award 1 mark for the correct answer (A).

1. Mass of solution, \(m = 50.0 + 50.0 = 100.0\text{ g}\).
2. Temperature change, \(\Delta T = 26.1 - 19.5 = 6.6\\ ^\circ\text{C}\).
3. Heat energy released, \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\\ ^\circ\text{C}^{-1} \times 6.6\\ ^\circ\text{C} = 2758.8\text{ J} = 2.7588\text{ kJ}\).
4. Moles of water formed: \(n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\).
5. Enthalpy of neutralization: \(\Delta_{neut} H = -\frac{q}{n} = -\frac{2.7588\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer (D).

The correct option is A. Both propan-1-ol and propan-2-ol can form hydrogen bonds, which are stronger than the permanent dipole-dipole forces in propanone and the London forces in propane. Propan-1-ol has a straight-chain structure, whereas propan-2-ol is more branched (the hydroxyl group is on the central carbon, making the molecule more spherical). This reduces the surface area contact between propan-2-ol molecules, leading to weaker London forces than in propan-1-ol. Therefore, propan-1-ol has the highest boiling temperature.

1 mark: Correct option selected (A).

The correct option is C. The reaction of concentrated sulfuric acid with potassium iodide (a powerful reducing agent) reduces sulfur from \(+6\) in \(\text{H}_2\text{SO}_4\) to various products, including sulfur dioxide (\(\text{SO}_2\), oxidation state \(+4\)), sulfur (\(\text{S}\), oxidation state \(0\)), and hydrogen sulfide (\(\text{H}_2\text{S}\), oxidation state \(-2\pack\)). The gas with a distinct rotten egg smell is hydrogen sulfide (\(\text{H}_2\text{S}\)), where sulfur has an oxidation state of \(-2\).

1 mark: Correct option selected (C).

The correct option is D. The rate of hydrolysis of halogenoalkanes is determined by both the C-X bond strength (C-I > C-Br > C-Cl in terms of reactivity because bond enthalpy decreases down the group) and the structure of the halogenoalkane (tertiary > secondary > primary). 2-bromo-2-methylpropane is a tertiary bromoalkane, which undergoes rapid nucleophilic substitution via an \(\text{S}_\text{N}1\) mechanism because the tertiary carbocation intermediate formed is very stable. It is faster than primary and secondary bromoalkanes, and much faster than chloroalkanes due to the weaker C-Br bond compared to the C-Cl bond.

1 mark: Correct option selected (D).

The correct option is A. When temperature increases, the Maxwell-Boltzmann distribution curve flattens and shifts to the right. The peak of the curve corresponds to the most probable molecular energy, which therefore increases (shifts to a higher energy value). Additionally, the entire distribution curve shifts towards higher energies, meaning a significantly greater fraction of molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)).

1 mark: Correct option selected (A).

(a)
1. Place \(50.0\text{ cm}^3\) of hydrochloric acid into a polystyrene cup supported in a beaker.
2. Measure the temperature of the acid every minute for 3 minutes.
3. At the 4th minute, add the weighed \(2.50\text{ g}\) of \(\text{KHCO}_3\), stir continuously but do not record the temperature.
4. Record the temperature every minute from the 5th to the 10th minute.
5. Plot a graph of temperature against time, draw lines of best fit before and after mixing, and extrapolate both lines to the 4th minute to find the theoretical temperature change.

(b) (i)
- Mass of solution \(m = 50.0\text{ g}\) (assuming the mass of solid is neglected as standard, or including it is sometimes acceptable, but standard is using solution volume as mass).
- Heat energy absorbed:
\(q = m \times c \times \Delta T = 50.0 \times 4.18 \times 3.3 = 689.7\text{ J} = 0.6897\text{ kJ}\).
- Moles of \(\text{KHCO}_3\):
\(n = \frac{2.50}{100.1} = 0.024975\text{ mol}\).
- Molar enthalpy change \(\Delta H_1\):
\(\Delta H_1 = +\frac{0.6897}{0.024975} = +27.616\text{ kJ mol}^{-1}\).
- Rounded to 3 significant figures: \(\Delta H_1 = +27.6\text{ kJ mol}^{-1}\).

(ii)
Using the Hess cycle:
\[2\text{KHCO}_3(\text{s}) \xrightarrow{\Delta H_{\text{r}}} \text{K}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\]
Both react with \(\text{HCl}\) to form \(2\text{KCl}(\text{aq}) + 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\).
- Direct route: \(2 \times \Delta H_1 = 2 \times (+27.6) = +55.2\text{ kJ}\).
- Indirect route: \(\Delta H_{\text{r}} + \Delta H_2\).
- Therefore, \(\Delta H_{\text{r}} = 2\Delta H_1 - \Delta H_2 = +55.2 - (-34.8) = +90.0\text{ kJ mol}^{-1}\).

(c) Direct heating of \(\text{KHCO}_3\) requires high temperatures to decompose, making it impossible to measure the heat change directly because heat is continuously supplied, and the temperature cannot be kept constant.

(d)
1. Assuming the specific heat capacity of the solution is the same as water (\(4.18\)), whereas the actual ionic solution has a lower specific heat capacity, leading to an overestimation of heat transfer.
2. Incomplete reaction/dissolution of the solid potassium hydrogencarbonate, resulting in a smaller measured temperature change.

(a) (Max 4 marks):
- M1: Measure the initial temperature of the acid for a few minutes to establish a stable baseline (1)
- M2: Add the solid at a specific minute (e.g., 4th minute) without recording, then record temperature at regular intervals (1)
- M3: Plot a graph of temperature against time (1)
- M4: Extrapolate both lines of best fit to the time of mixing to determine the accurate temperature change (1)

(b)(i) (Max 3 marks):
- M1: Calculation of energy: \(q = 50.0 \times 4.18 \times 3.3 = 689.7\text{ J}\) or \(0.6897\text{ kJ}\) (1)
- M2: Calculation of moles of \(\text{KHCO}_3 = 2.50 / 100.1 = 0.02498\text{ mol}\) (1)
- M3: Correct division and positive sign (endothermic): \(+27.6\text{ kJ mol}^{-1}\) (3 s.f.) (1). (Accept \(+27.5\) to \(+27.7\))

(b)(ii) (Max 2 marks):
- M1: Application of Hess's Law expression: \(\Delta H_{\text{r}} = 2\Delta H_1 - \Delta H_2\) (1)
- M2: Correct calculation: \(2(27.6) - (-34.8) = +90.0\text{ kJ mol}^{-1}\) (1). (Allow consequential error from (b)(i))

(c) (Max 1 mark):
- M1: Decomposition requires strong heating / cannot be done at room temperature (1)

(d) (Max 2 marks):
- M1: Specific heat capacity of solution is different from water / heat capacity of polystyrene cup is neglected (1)
- M2: Incomplete reaction / solid does not completely dissolve (1)

(a) (i) Ethanol acts as a mutual solvent that allows the halogenoalkane (insoluble in water) and the aqueous silver nitrate to mix and react in a single phase.

(ii)
- 1-iodobutane forms a yellow precipitate.
- 1-bromobutane forms a cream precipitate.
- 1-chlorobutane forms a white precipitate.
- Rate order (fastest to slowest): 1-iodobutane > 1-bromobutane > 1-chlorobutane.

(iii) The reaction rate depends on the strength of the C-X bond that is broken during hydrolysis. Down Group 7, the C-X bond enthalpy decreases because the halogen atoms get larger, making the shared pair of electrons further from the nucleus, weakening the covalent bond. Therefore, the C-I bond is the easiest/fastest to break, while the C-Cl bond is the hardest/slowest.

(b) (i)
- Step 1: Heterolytic fission of the C-Br bond. A curly arrow starts from the C-Br bond and points to the Br atom, forming a tertiary carbocation \(((\text{CH}_3)_3\text{C}^+)\) and a bromide ion \(\text{Br}^-\).
- Step 2: Nucleophilic attack. A curly arrow starts from a lone pair on the oxygen of a water molecule (the nucleophile in hydrolysis) and points to the positive carbon of the carbocation, forming a protonated alcohol intermediate: \(((\text{CH}_3)_3\text{C}-\text{OH}_2)^+\).
- Step 3: Deprotonation. A curly arrow from the O-H bond to the oxygen atom, yielding 2-methylpropan-2-ol and releasing \(\text{H}^+\).

(ii) 2-bromo-2-methylpropane is a tertiary halogenoalkane. It can form a highly stable tertiary carbocation intermediate because of the electron-donating inductive effect of the three methyl groups. 1-bromobutane is a primary halogenoalkane, and the primary carbocation that would form is highly unstable; instead, it reacts via a single transition state in an \(S_{\text{N}}2\) mechanism where steric hindrance is low.

(a)(i) (Max 1 mark):
- M1: To act as a common solvent / dissolve both the organic halogenoalkane and the aqueous silver nitrate (1)

(a)(ii) (Max 3 marks):
- M1: Correctly identifies colors: 1-chlorobutane is white, 1-bromobutane is cream, 1-iodobutane is yellow (1)
- M2: Order: 1-iodobutane is fastest, 1-chlorobutane is slowest (1)
- M3: Precipitate forms quickest with iodide / slowest with chloride (1)

(a)(iii) (Max 3 marks):
- M1: The reaction rate depends on the C-X bond enthalpy / bond strength, not bond polarity (1)
- M2: C-I bond is the weakest / has the lowest bond enthalpy (1)
- M3: C-Cl bond is the strongest / has the highest bond enthalpy (1)

(b)(i) (Max 3 marks):
- M1: Curly arrow from C-Br bond to Br (1)
- M2: Structure of tertiary carbocation intermediate showing positive charge on central carbon (1)
- M3: Curly arrow from O of \(\text{H}_2\text{O}\) (or \(\text{OH}^-\)) to the carbocation carbon to form product (1)

(b)(ii) (Max 2 marks):
- M1: Tertiary carbocation is stabilized by the inductive effect of three alkyl groups (1)
- M2: Primary carbocation is unstable / primary carbon has less steric hindrance allowing direct nucleophilic attack (S_N2) (1)

(a)
- Butane: London forces (dispersion forces / instantaneous dipole-induced dipole).
- Propanal: Permanent dipole-permanent dipole forces (due to polar C=O bond).
- Propan-1-ol: Hydrogen bonding (due to highly polar O-H bond).

(b)
- Butane: \(-0.5\ ^\circ\text{C}\)
- Propanal: \(49\ ^\circ\text{C}\)
- Propan-1-ol: \(97\ ^\circ\text{C}\)

Justification:
- All three have similar molar masses, so their London forces are of a similar magnitude.
- Butane has only weak London forces, which require the least thermal energy to overcome, resulting in the lowest boiling point.
- Propanal also has permanent dipole-dipole forces, which are stronger than London forces and require more energy to overcome.
- Propan-1-ol has hydrogen bonds (the strongest type of intermolecular force), requiring the most energy to overcome, resulting in the highest boiling point.

(c) (i) Propan-1-ol can form hydrogen bonds with water because both molecules contain highly polar O-H bonds.
Diagram requirements:
- A water molecule showing \(\text{H}-\text{O}-\text{H}\) with \(\delta^-\)\ on oxygen and \(\delta^+\)\ on hydrogen.
- A propan-1-ol molecule showing \(\text{R}-\text{O}-\text{H}\) with \(\delta^-\)\ on oxygen and \(\delta^+\)\ on hydrogen.
- Lone pairs shown on the oxygen atoms (two pairs each).
- A dashed or dotted line representing the hydrogen bond, drawn directly from a lone pair of one oxygen atom to the hydrogen atom of the other molecule.
- The angle \(\text{O}\cdots\text{H}-\text{O}\) should be approximately linear (\(180^\circ\)).

(ii) Butane is non-polar and cannot form hydrogen bonds with water. The weak London forces between butane and water are not strong enough to overcome the hydrogen bonds between the water molecules.

(a) (Max 3 marks):
- M1: Butane: London forces (1)
- M2: Propanal: Permanent dipole-dipole forces (1)
- M3: Propan-1-ol: Hydrogen bonding (1)

(b) (Max 5 marks):
- M1: Correct association of boiling temperatures to all three compounds (1)
- M2: Identifies that all three have similar molar masses / similar strength of London forces (1)
- M3: Butane has only London forces which are weak / easily broken (1)
- M4: Propanal has permanent dipole-dipole forces which are stronger than London forces (1)
- M5: Propan-1-ol has hydrogen bonding which is the strongest intermolecular force (1)

(c)(i) (Max 3 marks):
- M1: Explanation: Propan-1-ol can form hydrogen bonds with water molecules (1)
- M2: Diagram: Showing correct dipole charges (\(\delta^+\) on H and \(\delta^-\) on O) and lone pairs on O (1)
- M3: Diagram: Showing dotted line for hydrogen bond from O lone pair of one molecule to H of the other (1)

(c)(ii) (Max 1 mark):
- M1: Butane is non-polar / cannot form hydrogen bonds with water (1)

(a) (i)
- Equation: \(\text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{l}) \rightarrow \text{NaHSO}_4(\text{s}) + \text{HCl}(\text{g})\) (or \(2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl}\)).
- Observation: Misty fumes / white fumes.
- Gas evolved: Hydrogen chloride.

(ii)
- Yellow solid: Sulfur (\(\text{S}\)).
- Gas with bad-egg smell: Hydrogen sulfide (\(\text{H}_2\text{S}\)).
- Role of sulfuric acid: Oxidizing agent (it oxidizes the iodide ions to iodine, while being reduced itself).

(iii)
- Half-equation:
\[\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\]

(b)
- An iodide ion (\(\text{I}^-\)) has more electron shells / a larger ionic radius than a chloride ion (\(\text{Cl}^-\)).
- There is greater shielding of the outermost electrons from the nucleus in the iodide ion.
- Therefore, the attraction between the nucleus and the outermost electrons is weaker in the iodide ion.
- It is much easier for an iodide ion to lose an electron to act as a reducing agent.

(a)(i) (Max 3 marks):
- M1: Correct balanced equation (allow either acid salt or normal salt product) (1)
- M2: State observation: misty fumes / white fumes (1)
- M3: Identify gas: hydrogen chloride / \(\text{HCl}\) (1)

(a)(ii) (Max 3 marks):
- M1: Yellow solid: sulfur / \(\text{S}\) (1)
- M2: Bad-egg gas: hydrogen sulfide / \(\text{H}_2\text{S}\) (1)
- M3: Role: oxidizing agent / oxidizes iodide (1)

(a)(iii) (Max 2 marks):
- M1: Correct species: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) (1)
- M2: Correct charge balance with \(8\text{e}^-\)\ on the reactant side (1)

(b) (Max 4 marks):
- M1: Iodide ion is larger / has more shells than chloride ion (1)
- M2: Iodide has greater shielding (1)
- M3: Weaker attraction between the outer electrons and the nucleus (1)
- M4: Easier for the iodide ion to lose an electron (1)

(a) (i) & (ii)
- y-axis labeled 'Number of molecules' (or 'Fraction of molecules' / 'Probability density').
- x-axis labeled 'Energy' (or 'Kinetic energy').
- The curve for \(T_1\) starts at the origin, rises to a peak, and then asymptotically approaches the x-axis but never touches it.
- \(E_{\text{mp}}\) is labeled at the peak of the curve.
- \(E_{\text{a}}\) is labeled as a vertical line to the right of the peak, with the area under the curve to the right of this line representing the active molecules.
- The curve for \(T_2\) must have its peak lower and shifted to the right compared to \(T_1\), with the tail of the curve at higher energies being above the \(T_1\) curve, and starting at the origin.

(iii)
- Only molecules with energy greater than or equal to the activation energy (\(E \ge E_{\text{a}}\)) can react successfully upon collision.
- When temperature increases from \(T_1\) to \(T_2\), the distribution shifts to higher average energies.
- This causes a very large increase in the fraction (or area under the curve) of molecules with energy \(E \ge E_{\text{a}}\).
- Consequently, there is a much higher frequency of successful collisions, leading to a much faster rate of reaction.

(b) (i)
- Initial effect: The gas mixture immediately becomes darker brown because the volume is reduced, which increases the concentration of the brown \(\text{NO}_2\) molecules.
- Subsequent effect: The mixture then gradually becomes paler/lighter brown. According to Le Chatelier's Principle, the increase in pressure causes the equilibrium to shift to the side with fewer gas molecules (the right-hand side, with 1 mole of gas instead of 2) to reduce the pressure. This produces more colorless \(\text{N}_2\text{O}_4\).

(ii)
- Since the forward reaction is exothermic (\(\Delta H < 0\)), an increase in temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the added heat.
- This decreases the concentration of the product (\(\text{N}_2\text{O}_4\)) and increases the concentration of the reactant (\(\text{NO}_2\)).
- Therefore, the value of \(K_{\text{c}} = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}\) decreases.

(a)(i) (Max 3 marks):
- M1: Correctly labeled axes (y: number of molecules; x: energy) and curve starting at origin (1)
- M2: Correct positioning of \(E_{\text{mp}}\) at the peak and \(E_{\text{a}}\) to the right of the peak (1)
- M3: Curve asymptotically approaches the x-axis but does not touch or cross it (1)

(a)(ii) (Max 2 marks):
- M1: \(T_2\) curve has a lower peak shifted to the right of \(T_1\) (1)
- M2: \(T_2\) curve is above \(T_1\) at high energies (to the right of \(E_{\text{a}}\)) and starts at origin (1)

(a)(iii) (Max 3 marks):
- M1: Reaction requires collisions to have energy \(\ge E_{\text{a}}\) (1)
- M2: Shaded area to the right of \(E_{\text{a}}\) is significantly larger at \(T_2\) than at \(T_1\) (1)
- M3: A much larger fraction of molecules have energy \(\ge E_{\text{a}}\), resulting in a much higher frequency of successful collisions (1)

(b)(i) (Max 2 marks):
- M1: Mixture initially darkens because concentration of \(\text{NO}_2\) increases instantly (1)
- M2: Then it becomes paler because equilibrium shifts to the right (fewer gas moles) to decrease pressure (1)

(b)(ii) (Max 2 marks):
- M1: Increasing temperature shifts the equilibrium to the left / in the endothermic direction (1)
- M2: The concentration of product decreases (or reactant increases), so \(K_{\text{c}}\) decreases (1)

(a) Heating to constant mass ensures that all the water of crystallisation has been completely driven off from the salt.
(b) The lid prevents any solid from spitting or spattering out of the crucible during heating, which would cause a loss of solid. It is kept slightly ajar to allow the water vapour to escape freely from the crucible.
(c) Mass of anhydrous zinc sulfate = 20.67 g - 18.25 g = 2.42 g. Mass of water lost = 22.56 g - 20.67 g = 1.89 g.
(d) Moles of \( \text{ZnSO}_4 = 2.42 / 161.5 = 0.01498 \text{ mol} \). Moles of \( \text{H}_2\text{O} = 1.89 / 18.0 = 0.1050 \text{ mol} \). Ratio of moles of \( \text{H}_2\text{O} \) to \( \text{ZnSO}_4 = 0.1050 / 0.01498 = 7.01 \). Therefore, x = 7.
(e) If the anhydrous salt absorbs water vapour, the recorded mass of the anhydrous salt will be higher than it should be. This will make the calculated mass of water lost (initial mass minus final mass) smaller than the actual amount of water. Consequently, the calculated ratio of moles of water to anhydrous salt (x) will be lower than the true value.
(f) Two mass measurements are required to find the mass of the hydrated salt. Total uncertainty = \( 2 \times 0.005 \text{ g} = 0.01 \text{ g} \). Mass of hydrated salt = 22.56 g - 18.25 g = 4.31 g. Percentage uncertainty = \( (0.01 / 4.31) \times 100\% = 0.23\% \).

(a) 1 mark: To ensure all water of crystallisation has been removed / reaction is complete.
(b) 1 mark: Lid prevents loss of solid by spitting/spattering. 1 mark: Ajar to allow water vapour to escape.
(c) 1 mark: Mass of anhydrous salt = 2.42 g. 1 mark: Mass of water = 1.89 g.
(d) 1 mark: Moles of \( \text{ZnSO}_4 \) = 0.01498 mol (or 0.0150 mol). 1 mark: Moles of \( \text{H}_2\text{O} \) = 0.1050 mol. 1 mark: Correct mole ratio calculation (approx. 7.01). 1 mark: x = 7 (must be nearest integer).
(e) 1 mark: Measured mass of anhydrous salt is too high. 1 mark: Calculated mass of water lost is too low. 0.5 marks: Calculated value of x is lower.
(f) 0.5 marks: Total uncertainty of 0.01 g. 0.5 marks: Percentage uncertainty of 0.23% (accept 0.2% or 0.232%).

(a) Plot a graph of temperature against time. Draw a line of best fit through the cooling points (from 5 to 10 minutes) and extrapolate this line back to the 4th minute (the time of mixing). Also, extrapolate the initial temperature line to the 4th minute. The difference between these two temperature values at the 4th minute represents \( \Delta T \). This extrapolation is necessary because heat is lost to the surroundings from the moment the reaction starts; this method corrects for heat loss.
(b) \( \Delta T = 32.6 - 19.5 = 13.1 \text{ ^\circ C} \).
(c) \( q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ ^\circ C}^{-1} \times 13.1 \text{ ^\circ C} = 2737.9 \text{ J} \).
(d) Moles of \( \text{CuSO}_4 = 0.0500 \text{ dm}^3 \times 0.200 \text{ mol dm}^{-3} = 0.0100 \text{ mol} \).
(e) \( \Delta H = -q / \text{moles} = -2.7379 \text{ kJ} / 0.0100 \text{ mol} = -273.79 \text{ kJ mol}^{-1} \). Rounding to 3 significant figures gives \( -274 \text{ kJ mol}^{-1} \).
(f) One source: Heat loss to the surroundings. This can be minimized by using a lid on the polystyrene cup and placing the cup inside a glass beaker for insulation. Second source: Heat absorbed by the polystyrene cup/thermometer. This can be minimized by calibrating the calorimeter to find its heat capacity and including it in the calculation.

(a) 1 mark: Plot temperature against time and extrapolate the cooling curve back to 4 minutes. 1 mark: Find difference between extrapolated temperature and initial temperature at 4 minutes. 1 mark: Explain that it corrects for heat lost to surroundings during the reaction.
(b) 1 mark: \( \Delta T = 13.1 \text{ ^\circ C} \).
(c) 1 mark: Substitution into \( q = mc\Delta T \). 1 mark: \( q = 2737.9 \text{ J} \) (accept 2738 J or 2.74 kJ).
(d) 1 mark: Moles = 0.0100 mol.
(e) 1 mark: Correct division of q by moles of CuSO4. 1 mark: Negative sign included. 1 mark: Value in kJ mol-1 (-274). 0.5 marks: Expressed to 3 significant figures.
(f) 1 mark: Heat loss to surroundings AND use of a lid/insulating beaker. 1 mark: Heat capacity of cup ignored AND calibrate cup/determine calorimeter constant.

(a) \( \text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr} \) (or \( 2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr} \)).
(b)(i) Reflux prevents the loss of volatile organic reactants and products (such as butan-1-ol and 1-bromobutane) by vaporisation, as any vapours condense and drip back into the reaction flask. This allows the reaction to be heated over a prolonged period to increase yield.
(b)(ii) A round-bottomed or pear-shaped flask is fitted with a vertical Liebig condenser. Water must enter the condenser jacket at the bottom and leave from the top to ensure the jacket is completely filled with cold water. The top of the condenser must remain open (unsealed) to prevent pressure build-up, which could cause the apparatus to explode.
(c)(i) The mixture is poured into a separating funnel. Sodium hydrogencarbonate solution is added to neutralise any acidic impurities. The funnel is stoppered, shaken, and inverted periodically. The tap is opened while inverted to vent and release pressure from carbon dioxide gas. The funnel is placed in a stand, the layers are allowed to separate, and the lower organic layer (which is denser than water) is run off into a clean flask.
(c)(ii) Suitable drying agent: Anhydrous calcium chloride (\( \text{CaCl}_2 \)) or anhydrous magnesium sulfate (\( \text{MgSO}_4 \)) or anhydrous sodium sulfate (\( \text{Na}_2\text{SO}_4 \)). The appearance of the organic liquid before adding the drying agent is cloudy or turbid, and it becomes clear after the drying process is complete.

(a) 1 mark: \( \text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr} \) (allow \( 2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr} \)).
(b)(i) 1 mark: Prevents loss of volatile organic components / reactants / products. 1 mark: Allows heating to speed up reaction without losing substance.
(b)(ii) 1 mark: Condenser is fitted vertically. 1 mark: Cold water enters at the bottom and exits at the top. 1 mark: The top must be open to prevent pressure build-up.
(c)(i) 1 mark: Add sodium hydrogencarbonate solution to neutralise acid. 1 mark: Shake and invert the funnel, venting regularly by opening the tap to release \( \text{CO}_2 \). 1 mark: Allow layers to separate. 1 mark: Run off the lower layer (since 1-bromobutane is denser than water).
(c)(ii) 1 mark: Anhydrous calcium chloride / anhydrous magnesium sulfate / anhydrous sodium sulfate. 1 mark: Turbid/cloudy before drying. 0.5 marks: Clear after drying.

(a)(i) Sodium ion, \( \text{Na}^+ \) (or \( \text{Na} \)).
(a)(ii) Dip the platinum or nichrome wire into concentrated hydrochloric acid, then hold it in the hot (blue) Bunsen burner flame. Repeat this process until the wire no longer imparts any color to the flame.
(b)(i) Iron(II) ion, \( \text{Fe}^{2+} \).
(b)(ii) \( \text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)} \).
(c)(i) Bromide ion, \( \text{Br}^- \).
(c)(ii) Add dilute aqueous ammonia to the precipitate; the cream precipitate does not dissolve. Then add concentrated aqueous ammonia; the cream precipitate dissolves to form a clear, colorless solution.
(c)(iii) Nitric acid is added to react with and remove any interfering anions, such as carbonate (\( \text{CO}_3^{2-} \)) or sulfite (\( \text{SO}_3^{2-} \)) ions, which would otherwise form precipitates with silver ions and give a false positive.
(d) \( \text{NaBr} \) and \( \text{FeBr}_2 \).

(a)(i) 1 mark: Sodium / \( \text{Na}^+ \).
(a)(ii) 1 mark: Use concentrated hydrochloric acid / \( \text{HCl} \). 1 mark: Heat in Bunsen flame until no color is produced.
(b)(i) 1 mark: Iron(II) / \( \text{Fe}^{2+} \) (reject \( \text{Fe}^{3+} \) or chromium(III)).
(b)(ii) 1 mark: Correct formulae (\( \text{Fe}^{2+} + 2\text{OH}^- \rightarrow \text{Fe(OH)}_2 \)). 1 mark: Correct state symbols (aq, aq -> s).
(c)(i) 1 mark: Bromide / \( \text{Br}^- \).
(c)(ii) 1 mark: Addition of dilute ammonia results in no reaction/precipitate remains. 1.5 marks: Addition of concentrated ammonia results in the precipitate dissolving (accept 1 mark if concentrated ammonia is mentioned without specifying dilute, but full 2.5 marks requires both stages of the test).
(c)(iii) 1 mark: To remove carbonate / sulfite ions / impurities. 1 mark: To prevent the formation of other silver precipitates which would give a false positive.
(d) 1 mark: \( \text{NaBr} \) AND \( \text{FeBr}_2 \) (both must be correct).

According to the Arrhenius equation, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). Therefore, the gradient of the plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\). Thus, \(E_a = -\text{gradient} \times R = -(-1.20 \times 10^4\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 9.972 \times 10^4\text{ J mol}^{-1} = +99.7\text{ kJ mol}^{-1}\).

[1] A is correct. Award 1 mark for the correct calculation and selection.

From the Born-Haber cycle: \(\Delta H_f^{\ominus} = \Delta H_{at}^{\ominus}(\text{Ca}) + 1st\,IE(\text{Ca}) + 2nd\,IE(\text{Ca}) + \Delta H_{at}^{\ominus}(\text{O}) + 1st\,EA(\text{O}) + 2nd\,EA(\text{O}) + \Delta H_{latt}^{\ominus}\). Rearranging gives: \(\Delta H_{latt}^{\ominus} = -635 - [178 + 590 + 1145 + 249 - 141 + 798] = -635 - 2819 = -3454\text{ kJ mol}^{-1}\).

[1] A is correct. Award 1 mark for the correct application of the Born-Haber cycle and calculation.

For a reaction to be feasible, \(\Delta G^{\ominus} \le 0\). At the temperature of feasibility limit, \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S_{\text{system}}^{\ominus} = 0\). Thus, \(T = \frac{\Delta H^{\ominus}}{\Delta S_{\text{system}}^{\ominus}} = \frac{57.2 \times 10^3\text{ J mol}^{-1}}{175.8\text{ J K}^{-1}\text{ mol}^{-1}} = 325.4\text{ K}\). Converting to degrees Celsius: \(325.4 - 273.15 = 52.25^{\circ}\text{C} \approx 52.3^{\circ}\text{C}\).

[1] B is correct. Award 1 mark for calculating the correct temperature limit and converting to Celsius.

The value of the equilibrium constant \(K_p\) depends solely on temperature. Since the forward reaction is exothermic, decreasing the temperature shifts the position of equilibrium to the right, which increases the partial pressure of \(\text{SO}_3\) and decreases those of \(\text{SO}_2\) and \(\text{O}_2\), leading to an increase in the value of \(K_p\).

[1] B is correct. Award 1 mark for identifying that decreasing temperature increases the equilibrium constant for an exothermic reaction.

The hydrogen ion concentration can be calculated using \([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = (1.8 \times 10^{-5}) \times \frac{0.20}{0.10} = 3.6 \times 10^{-5}\text{ mol dm}^{-3}\). The pH is given by \(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(3.6 \times 10^{-5}) = 4.44\).

[1] B is correct. Award 1 mark for the correct calculation of buffer pH.

A Brønsted-Lowry base is a proton acceptor. In the forward reaction, \(\text{HNO}_3\) accepts a proton from \(\text{H}_2\text{SO}_4\) to form \(\text{H}_2\text{NO}_3^+\), so \(\text{HNO}_3\) is a base. In the backward reaction, \(\text{HSO}_4^-\) accepts a proton from \(\text{H}_2\text{NO}_3^+\) to reform \(\text{H}_2\text{SO}_4\), so \(\text{HSO}_4^-\) acts as a base.

[1] C is correct. Award 1 mark for identifying the two Brønsted-Lowry bases.

Optical activity is exhibited by molecules containing an asymmetric carbon atom (chiral centre). 2-hydroxypropanenitrile has the structure \(\text{CH}_3\text{CH(OH)CN}\), where the central carbon atom is bonded to four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{OH}\), and \(-\text{CN}\), making it a chiral molecule.

[1] C is correct. Award 1 mark for identifying the compound with a chiral carbon atom.

The reaction with 2,4-dinitrophenylhydrazine indicates a carbonyl group (aldehyde or ketone). The lack of reaction with Fehling's solution rules out aldehydes (such as propanal). The lack of a yellow precipitate with alkaline iodine rules out methyl ketones (such as propanone and pentan-2-one). Therefore, the compound must be pentan-3-one, which is a ketone but does not contain a methyl carbonyl group.

[1] C is correct. Award 1 mark for identifying pentan-3-one based on the chemical tests described.

The rate-determining step (the slowest step) must involve only the reactants that appear in the rate equation, with stoichiometry matching their order in the rate equation. Since the rate equation is \(\text{rate} = k[X]^2\), the rate-determining step must involve two molecules of \(X\) and zero molecules of \(Y\). In Mechanism B, Step 1 is the slow step and has two \(X\) reactants: \(2X \rightarrow X_2\). This matches the rate equation.

1 mark for the correct option B.

\(\Delta S_{\text{system}}^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\) \(\Delta S_{\text{system}}^{\ominus} = S^{\ominus}(CH_3OH(l)) - [S^{\ominus}(CO(g)) + 2 \times S^{\ominus}(H_2(g))]\) \(\Delta S_{\text{system}}^{\ominus} = 126.8 - [197.6 + 2 \times 130.6]\) \(\Delta S_{\text{system}}^{\ominus} = 126.8 - [197.6 + 261.2]\) \(\Delta S_{\text{system}}^{\ominus} = 126.8 - 458.8 = -332.0\text{ J K}^{-1}\text{ mol}^{-1}\)

1 mark for the correct option A.

The equilibrium constant \(K_p\) is only affected by changes in temperature. Therefore, if the temperature is kept constant, \(K_p\) remains constant. According to Le Chatelier's principle, an increase in total pressure shifts the position of equilibrium to the side with fewer gas moles (the right-hand side, where there are 2 moles of gas compared to 3 moles on the left). This increases the yield of \(SO_3(g)\) and therefore its mole fraction.

1 mark for the correct option B.

For a weak monoprotic acid: \([H^+] \approx \sqrt{K_a \times [HA]}\) \([H^+] = \sqrt{(2.5 \times 10^{-5}) \times 0.040} = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\) \(\text{pH} = -\log_{10}[H^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.00\)

1 mark for the correct option B.

A chiral carbon (or asymmetric carbon) is bonded to four different groups. In 2-hydroxypropanenitrile (\(CH_3CH(OH)CN\)), the second carbon is bonded to: a methyl group (\(-CH_3\)), a hydrogen atom (\(-H\)), a hydroxyl group (\(-OH\)), and a nitrile group (\(-CN\)). Thus, it is chiral and can exist as enantiomers which rotate the plane of plane-polarized light. Propan-2-ol has two identical methyl groups attached to the central carbon. Butanone is a ketone with a flat carbonyl carbon and no other carbon with four different groups. 2,2-dimethylpropanoic acid has three identical methyl groups on the central carbon.

1 mark for the correct option A.

According to the logarithmic form of the Arrhenius equation: \(\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A\) This is in the form \(y = mx + c\), where \(y = \ln k\) and \(x = \frac{1}{T}\). Thus, a graph of \(\ln k\) against \(\frac{1}{T}\) is a straight line with a gradient of \(-\frac{E_a}{R}\).

1 mark for the correct option B.

Pentan-2-one is a methyl ketone (\(CH_3COCH_2CH_2CH_3\)) and will give a positive result with the triiodomethane (iodoform) test (yellow precipitate of \(CHI_3\)) when reacted with iodine in alkaline solution. Pentan-3-one (\(CH_3CH_2COCH_2CH_3\)) is not a methyl ketone and will not react. Tollens' and Fehling's reagents only oxidize aldehydes, not ketones. Acidified potassium dichromate(VI) is an oxidizing agent that does not oxidize ketones.

1 mark for the correct option C.

An acidic buffer solution (pH < 7) contains a weak acid and its conjugate base. When \(0.10\text{ mol}\) of \(CH_3COOH\) is mixed with \(0.05\text{ mol}\) of \(NaOH\), they react: \(CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O\) This uses up all \(0.05\text{ mol}\) of \(NaOH\), producing \(0.05\text{ mol}\) of \(CH_3COONa\) (conjugate base) and leaving \(0.05\text{ mol}\) of unreacted \(CH_3COOH\) (weak acid). This mixture containing comparable amounts of weak acid and its conjugate base is an acidic buffer. Option A is a mixture of two acids. Option C contains excess strong base. Option D forms a basic buffer with pH > 7.

1 mark for the correct option B.

The standard entropy change of the system is calculated using the formula: \(\Delta S_{\text{system}}^{\ominus} = \sum S_{\text{products}}^{\ominus} - \sum S_{\text{reactants}}^{\ominus}\). Substituting the values: \(\Delta S_{\text{system}}^{\ominus} = [2 \times S^{\ominus}(\text{NO}_2(g))] - [2 \times S^{\ominus}(\text{NO}(g)) + S^{\ominus}(\text{O}_2(g))] = [2 \times 240.0] - [2 \times 210.8 + 205.0] = 480.0 - 626.6 = -146.6\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark for correct selection of option B. Accept: \(-146.6\text{ J K}^{-1}\text{ mol}^{-1}\). Reject: positive values or other calculations.

Comparing Exp 1 and Exp 2: doubling \([X]\) quadruples the rate, so the reaction is second order with respect to X. Comparing Exp 2 and Exp 3: doubling \([Y]\) doubles the rate, so the reaction is first order with respect to Y. Thus, \(\text{Rate} = k[X]^2[Y]\). Rearranging for \(k\): \(k = \frac{\text{Rate}}{[X]^2[Y]} = \frac{2.0 \times 10^{-4}}{(0.10)^2 \times 0.10} = 0.20\). Units of \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for correct selection of option C. Accept: \(0.20\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

First calculate the moles of propanoic acid and sodium propanoate: \(n(\text{acid}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\); \(n(\text{salt}) = 0.0500\text{ dm}^3 \times 0.050\text{ mol dm}^{-3} = 0.00250\text{ mol}\). Using the buffer equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)\). \(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\). Therefore, \(\text{pH} = 4.87 + \log_{10}\left(\frac{0.00250}{0.00500}\right) = 4.87 - 0.30 = 4.57\).

1 mark for correct selection of option D. Accept: \(4.57\).

When propanone reacts with hydrogen cyanide, nucleophilic addition occurs to form 2-hydroxy-2-methylpropanenitrile. In this molecule, the central carbon is bonded to two identical methyl groups, an \(-OH\) group, and a \(-CN\) group, which means it does not contain a chiral carbon atom (it is achiral). The other options all react to form products with at least one carbon atom bonded to four different groups (a chiral carbon).

1 mark for correct selection of option A. Accept: Propanone.

**(a)**
- Comparing Experiments 1 and 2: $[\text{Br}^-]$ and $[\text{H}^+]$ are constant. $[\text{BrO}_3^-]$ doubles (from 0.10 to 0.20), and the initial rate doubles (from $1.20 \times 10^{-3}$ to $2.40 \times 10^{-3}$). Therefore, the reaction is **first order** with respect to $\text{BrO}_3^-$.
- Comparing Experiments 1 and 3: $[\text{BrO}_3^-]$ and $[\text{H}^+]$ are constant. $[\text{Br}^-]$ doubles (from 0.10 to 0.20), and the initial rate doubles (from $1.20 \times 10^{-3}$ to $2.40 \times 10^{-3}$). Therefore, the reaction is **first order** with respect to $\text{Br}^-$.
- Comparing Experiments 1 and 4: $[\text{BrO}_3^-]$ and $[\text{Br}^-]$ are constant. $[\text{H}^+]$ doubles (from 0.10 to 0.20), and the initial rate quadruples (from $1.20 \times 10^{-3}$ to $4.80 \times 10^{-3}$ ($2^2 = 4$)). Therefore, the reaction is **second order** with respect to $\text{H}^+$.

**(b)**
- Rate equation: $\text{Rate} = k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2$
- To find $k$, substitute values from Experiment 1:
$$1.20 \times 10^{-3} = k (0.10)(0.10)(0.10)^2$$
$$1.20 \times 10^{-3} = k (1.0 \times 10^{-4})$$
$$k = \frac{1.20 \times 10^{-3}}{1.0 \times 10^{-4}} = 12.0$$
- Units: $[\text{mol dm}^{-3} \text{ s}^{-1}] / [\text{mol dm}^{-3}]^4 = \text{dm}^9 \text{ mol}^{-3} \text{ s}^{-1}$
- Value: $12.0\text{ dm}^9 \text{ mol}^{-3} \text{ s}^{-1}$

**(c)**
- The overall reaction involves the collision of 12 reactant particles ($1\text{BrO}_3^- + 5\text{Br}^- + 6\text{H}^+$).
- The simultaneous collision of 12 particles with sufficient energy and correct orientation is extremely unlikely / statistically impossible.
- This implies that the reaction must proceed via a multi-step mechanism involving simpler molecular steps (with a slow rate-determining step).

**(d)**
- Use the Arrhenius equation in logarithmic form:
$$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
- Substitute the known values:
$$k_1 = 12.0, \quad T_1 = 298\text{ K}$$
$$k_2 = 48.0, \quad T_2 = 318\text{ K}$$
$$\ln\left(\frac{48.0}{12.0}\right) = -\frac{E_a}{8.31} \left(\frac{1}{318} - \frac{1}{298}\right)$$
$$\ln(4.0) = -\frac{E_a}{8.31} (0.00314465 - 0.00335570)$$
$$1.3863 = -\frac{E_a}{8.31} (-0.00021105)$$
$$1.3863 = 2.5397 \times 10^{-5} \times E_a$$
$$E_a = \frac{1.3863}{2.5397 \times 10^{-5}} = 54585\text{ J mol}^{-1}$$
$$E_a = +54.6\text{ kJ mol}^{-1}$$

**(a) [Total: 5 marks]**
- M1: Deduces 1st order for $[\text{BrO}_3^-]$ with reference to Exp 1 & 2 showing concentration doubles and rate doubles (1)
- M2: Deduces 1st order for $[\text{Br}^-]$ with reference to Exp 1 & 3 showing concentration doubles and rate doubles (1)
- M3: Deduces 2nd order for $[\text{H}^+]$ with reference to Exp 1 & 4 showing concentration doubles and rate quadruples (1)
- M4: Clear mathematical or logical reasoning shown for all three deductions (2) (Allow 1 mark if reasoning is partially complete/clear).

**(b) [Total: 3 marks]**
- M1: Correct rate equation matching part (a) (1) (e.g., $\text{Rate} = k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2$)
- M2: Correct value of $k = 12.0$ (or consequential on wrong orders) (1)
- M3: Correct units $\text{dm}^9 \text{ mol}^{-3} \text{ s}^{-1}$ (consequential on rate equation) (1)

**(c) [Total: 3.5 marks]**
- M1: Identifies that a single-step reaction requires a simultaneous collision of 12 particles (1)
- M2: Explains that such a collision is highly improbable/unlikely (1)
- M3: States that the reaction must proceed via a series of simpler steps / multi-step mechanism (1)
- M4: Recognises that there is a slow step / rate-determining step in this pathway (0.5)

**(d) [Total: 6 marks]**
- M1: Recalls or correctly selects Arrhenius formula in logarithmic form (1)
- M2: Correctly calculates temperature terms: $\frac{1}{318} - \frac{1}{298} = -0.00021105\text{ K}^{-1}$ (1)
- M3: Calculates $\ln(k_2/k_1) = 1.3863$ (1)
- M4: Correct substitution and rearrangement of equation to find $E_a$ (1)
- M5: Obtains $E_a \approx 54600\text{ J mol}^{-1}$ (1)
- M6: Converts correctly to $+54.6\text{ kJ mol}^{-1}$ with positive sign and to 3 sig figs (1)

**(a)**
The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions ($298\text{ K}$ and $1\text{ atm}$).

**(b)**
- By Hess's Law, the enthalpy of formation of $\text{CaCl}_2(\text{s})$ is equal to the sum of the enthalpy changes of atomisation, ionisation, electron affinity, and lattice enthalpy:
$$\Delta H_f^\ominus = \Delta H_{at}^\ominus(\text{Ca}) + \text{1st IE}(\text{Ca}) + \text{2nd IE}(\text{Ca}) + 2 \times \Delta H_{at}^\ominus(\text{Cl}) + 2 \times \text{1st EA}(\text{Cl}) + \Delta H_{latt}^\ominus$$
- Substitute the values:
$$-796 = +178 + 590 + 1145 + 2(121) + 2(-349) + \Delta H_{latt}^\ominus$$
$$-796 = 178 + 590 + 1145 + 242 - 698 + \Delta H_{latt}^\ominus$$
$$-796 = 1457 + \Delta H_{latt}^\ominus$$
$$\Delta H_{latt}^\ominus = -796 - 1457 = -2253\text{ kJ mol}^{-1}$$

**(c)**
- The experimental value ($-2253\text{ kJ mol}^{-1}$) is very close to the theoretical value ($-2223\text{ kJ mol}^{-1}$), differing by only $30\text{ kJ mol}^{-1}$ ($1.3\%$).
- This indicates that the bonding in calcium chloride is almost entirely ionic / has highly dominant ionic character (closely matching the purely ionic model of point charges in a lattice).
- The slight discrepancy indicates a tiny degree of polarization / minor covalent character (due to the $\text{Ca}^{2+}$ ion polarizing the $\text{Cl}^-$ electron cloud).

**(d)**
- First, calculate the standard entropy change of the system, $\Delta S^\ominus$:
$$\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})$$
$$\Delta S^\ominus = (S^\ominus[\text{CaO}(\text{s})] + S^\ominus[\text{CO}_2(\text{g})]) - S^\ominus[\text{CaCO}_3(\text{s})]$$
$$\Delta S^\ominus = (39.7 + 213.6) - 92.9 = 253.3 - 92.9 = +160.4\text{ J K}^{-1}\text{ mol}^{-1}$$
- Convert $\Delta S^\ominus$ to $\text{kJ K}^{-1}\text{ mol}^{-1}$:
$$\Delta S^\ominus = 160.4 / 1000 = 0.1604\text{ kJ K}^{-1}\text{ mol}^{-1}$$
- For the reaction to be thermodynamically feasible, $\Delta G^\ominus \le 0$.
$$\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus \le 0$$
$$T\Delta S^\ominus \ge \Delta H^\ominus$$
$$T \ge \frac{\Delta H^\ominus}{\Delta S^\ominus} = \frac{178.2}{0.1604} = 1110.97\text{ K}$$
- The minimum temperature is $1111\text{ K}$ (or $1110.97\text{ K}$).

**(a) [Total: 2 marks]**
- M1: Enthalpy change for the formation of 1 mole of gaseous atoms (1)
- M2: From the element in its standard state under standard conditions (1)

**(b) [Total: 6 marks]**
- M1: Standard Born-Haber cycle drawn or represented mathematically using a Hess's law equation (1)
- M2: Multiplies the enthalpy of atomisation of chlorine by 2 ($+242$) (1)
- M3: Multiplies the first electron affinity of chlorine by 2 ($-698$) (1)
- M4: Includes both first and second ionisation energies of calcium ($590 + 1145 = 1735$) (1)
- M5: Correctly rearranges equation to solve for lattice enthalpy (1)
- M6: Obtains $-2253\text{ kJ mol}^{-1}$ with correct unit and sign (1)

**(c) [Total: 3.5 marks]**
- M1: States that the calculated and theoretical values are close to each other (1)
- M2: Deduces that the bonding is predominantly/highly ionic (1)
- M3: Explains that any small difference is due to some polarization of the anion by the cation / slight covalent character (1)
- M4: Refers specifically to the high charge/charge density of $\text{Ca}^{2+}$ or the polarization of $\text{Cl}^-$ (0.5)

**(d) [Total: 6 marks]**
- M1: Calculates entropy change: $\Delta S = (39.7 + 213.6) - 92.9 = +160.4\text{ J K}^{-1}\text{ mol}^{-1}$ (1)
- M2: Converts $\Delta S$ to $\text{kJ}$ or $\Delta H$ to $\text{J}$ correctly (1)
- M3: Identifies condition for feasibility as $\Delta G \le 0$ or uses $T = \Delta H / \Delta S$ (1)
- M4: Correct substitution of values into formula (1)
- M5: Evaluates $T = 1111\text{ K}$ (accept 1110.97 to 1111) (1)
- M6: Specifies unit as Kelvin (K) and matches standard calculation conventions (1)

**(a)**
- A conjugate acid-base pair consists of two species that differ only by one proton ($\text{H}^+$).
- Pair 1: Acid = $\text{CH}_3\text{CH}_2\text{COOH}$, Conjugate base = $\text{CH}_3\text{CH}_2\text{COO}^-$
- Pair 2: Base = $\text{H}_2\text{O}$, Conjugate acid = $\text{H}_3\text{O}^+$

**(b)**
- Weak acid dissociation: $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$
- Using the approximation $[\text{H}^+] \approx [\text{A}^-]$:
$$K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]}$$
- Using the approximation $[\text{HA}]_{equilibrium} \approx [\text{HA}]_{initial}$:
$$[\text{H}^+]^2 = K_a \times [\text{HA}]_{initial}$$
$$[\text{H}^+]^2 = (1.35 \times 10^{-5}) \times 0.150 = 2.025 \times 10^{-6}\text{ mol}^2\text{ dm}^{-6}$$
$$[\text{H}^+] = \sqrt{2.025 \times 10^{-6}} = 1.423 \times 10^{-3}\text{ mol dm}^{-3}$$
- Calculating pH:
$$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.423 \times 10^{-3}) = 2.8468 \approx 2.85$$
- Approximations:
1. The concentration of $\text{H}^+$ ions from the auto-ionisation of water is negligible (hence $[\text{H}^+] = [\text{A}^-]$).
2. The ionization of the weak acid is negligible / so small that the equilibrium concentration of the acid is equal to the initial concentration ($[\text{HA}]_{eq} = [\text{HA}]_{initial}$).

**(c)**
- Moles of propanoic acid ($\text{HA}$) initially:
$$n(\text{HA}) = \text{Volume} \times \text{Concentration} = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}$$
- Moles of hydroxide ions ($\text{OH}^-$) added:
$$n(\text{OH}^-) = 0.0300\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}$$
- The reaction occurring is:
$$\text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})$$
- Since $\text{OH}^-$ is the limiting reagent, all of it reacts:
- Moles of $\text{HA}$ remaining $= 7.50 \times 10^{-3} - 3.00 \times 10^{-3} = 4.50 \times 10^{-3}\text{ mol}$
- Moles of $\text{A}^-$ (propanoate ions) formed $= 3.00 \times 10^{-3}\text{ mol}$
- Total volume of the mixture $= 50.0 + 30.0 = 80.0\text{ cm}^3 = 0.0800\text{ dm}^3$
- $[\text{HA}] = \frac{4.50 \times 10^{-3}}{0.0800} = 0.05625\text{ mol dm}^{-3}$
- $[\text{A}^-] = \frac{3.00 \times 10^{-3}}{0.0800} = 0.0375\text{ mol dm}^{-3}$
- Substitute into the buffer equation $[\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}$:
$$[\text{H}^+] = 1.35 \times 10^{-5} \times \left(\frac{0.05625}{0.0375}\right) = 1.35 \times 10^{-5} \times 1.5 = 2.025 \times 10^{-5}\text{ mol dm}^{-3}$$
- Calculate the final pH:
$$\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6935 \approx 4.69$$

**(a) [Total: 3.5 marks]**
- M1: Defines conjugate pair as two species that differ by one proton ($\text{H}^+$) (1)
- M2: Identifies Pair 1: $\text{CH}_3\text{CH}_2\text{COOH}$ (acid) and $\text{CH}_3\text{CH}_2\text{COO}^-$ (base) (1)
- M3: Identifies Pair 2: $\text{H}_3\text{O}^+$ (acid) and $\text{H}_2\text{O}$ (base) (1)
- M4: Correctly identifies the relative roles (acid vs base) in both pairs (0.5)

**(b) [Total: 5 marks]**
- M1: Writes the correct expression for $K_a$ or $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$ (1)
- M2: Correct calculation of $[\text{H}^+] = 1.42 \times 10^{-3}\text{ mol dm}^{-3}$ (1)
- M3: Calculates pH as $2.85$ (allow 2.8) (1)
- M4: State approximation 1: $[\text{H}^+] = [\text{A}^-]$ (or ionization of water is negligible) (1)
- M5: State approximation 2: $[\text{HA}]_{eq} = [\text{HA}]_{initial}$ (or ionization of acid is negligible) (1)

**(c) [Total: 9 marks]**
- M1: Calculates initial moles of propanoic acid $= 7.50 \times 10^{-3}\text{ mol}$ (1)
- M2: Calculates moles of $\text{NaOH}$ added $= 3.00 \times 10^{-3}\text{ mol}$ (1)
- M3: Shows subtraction to find remaining moles of acid $= 4.50 \times 10^{-3}\text{ mol}$ (1)
- M4: Identifies moles of sodium propanoate formed $= 3.00 \times 10^{-3}\text{ mol}$ (1)
- M5: Uses total volume to calculate concentrations of acid and salt (or shows volume cancels) (1)
- M6: Substitutes values correctly into $K_a$ expression or Henderson-Hasselbalch equation (1)
- M7: Calculates $[\text{H}^+] = 2.025 \times 10^{-5}\text{ mol dm}^{-3}$ (2)
- M8: Correctly calculates pH $= 4.69$ (to 2 decimal places) (1)

**(a)**
(i) The three structural isomers containing a carbonyl group are:
- Butanal: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$
- 2-methylpropanal: $(\text{CH}_3)_2\text{CHCHO}$
- Butanone: $\text{CH}_3\text{COCH}_2\text{CH}_3$

(ii) Tollens' reagent oxidized aldehydes but not ketones. Thus:
- Butanal and 2-methylpropanal will react to form a silver mirror.
- Butanone does not react.

(iii) Reduction of butanone yields butan-2-ol:
$$\text{CH}_3\text{COCH}_2\text{CH}_3 \xrightarrow{[\text{H}]} \text{CH}_3\text{C}^*\text{H(OH)CH}_2\text{CH}_3$$
- Butan-2-ol contains a chiral carbon at position 2 (indicated with an asterisk *), because it is bonded to four different groups: $-\text{H}$, $-\text{OH}$, $-\text{CH}_3$, and $-\text{CH}_2\text{CH}_3$.
- This alcohol is optically active because it lacks a plane of symmetry, exists as two non-superimposable mirror image enantiomers, and rotates the plane of plane-polarized light.

**(b)**
(i) Mechanism steps:
1. The nucleophile is the cyanide ion, $^-\!:\text{CN}$, which has a lone pair of electrons on the carbon atom.
2. The carbonyl carbon of butanone is polar with a dipole: $\text{C}^{\delta+}=\text{O}^{\delta-}$.
3. Draw a curly arrow from the lone pair on the carbon of $^-\!:\text{CN}$ to the $\text{C}^{\delta+}$ atom.
4. Draw a curly arrow from the $\text{C}=\text{O}$ double bond to the oxygen atom, forming an intermediate alkoxide ion with a negative charge: $\text{CH}_3\text{C}(\text{O}^-)(\text{CN})\text{CH}_2\text{CH}_3$.
5. The negative oxygen atom has a lone pair. Draw a curly arrow from this lone pair to the $\text{H}^+$ ion (or the hydrogen atom of an $\text{HCN}$ molecule with subsequent transfer of electrons to the carbon of $\text{CN}^-$).
6. This forms the final product: 2-hydroxy-2-methylbutanenitrile.

(ii) Reasons for forming a racemic mixture:
- The carbonyl group, $\text{C}=\text{O}$, in butanone is planar around the carbonyl carbon.
- The nucleophile ($\text{CN}^-$) has an equal probability of attacking the planar carbonyl group from either side (above or below the molecular plane).
- This results in the formation of equal amounts of the two enantiomers, which rotate plane-polarized light in equal and opposite directions, so the optical activity cancels out, making the mixture optically inactive.

**(a)(i) [Total: 3 marks]**
- M1: Draws/names butanal correctly (1)
- M2: Draws/names 2-methylpropanal correctly (1)
- M3: Draws/names butanone correctly (1)

**(a)(ii) [Total: 2 marks]**
- M1: Identifies both butanal and 2-methylpropanal as reacting positive with Tollens' (1)
- M2: Identifies butanone as not reacting (1)

**(a)(iii) [Total: 4 marks]**
- M1: Identifies butanone as the isomer (1)
- M2: Draws the correct skeletal structure of butan-2-ol with an asterisk on C2 (1)
- M3: Explains that C2 is bonded to 4 different groups (named or shown) (1)
- M4: Explains that optical activity occurs because it rotates the plane of plane-polarized light (1)

**(b)(i) [Total: 5.5 marks]**
- M1: Shows correct dipole $\text{C}^{\delta+}=\text{O}^{\delta-}$ on butanone (1)
- M2: Draws curly arrow from the lone pair on $^-\!:\text{CN}$ to the carbonyl carbon (1)
- M3: Draws curly arrow from $\text{C}=\text{O}$ double bond to oxygen (1)
- M4: Correct structure of intermediate with a negative charge on oxygen (1)
- M5: Draws curly arrow from lone pair on $\text{O}^-$ to $\text{H}^+$ (or $\text{H}$ in $\text{HCN}$) (1)
- M6: Draws final structure of the hydroxynitrile correctly (0.5)

**(b)(ii) [Total: 3 marks]**
- M1: States that the carbonyl group (or carbon atom) is planar (1)
- M2: Explains that attack by the nucleophile is equally likely from above or below the plane (1)
- M3: Concludes that enantiomers are formed in equal quantities / a 50:50 / racemic mixture is obtained (1)

The overall cell reaction is \(2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(aq)\). The standard cell potential is calculated as: \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{I}_2/\text{I}^-) = +0.77 - (+0.54) = +0.23\text{ V}\). Since the standard cell potential is positive, the reaction is thermodynamically feasible.

1 mark for the correct option (A). Correct calculation of cell potential (+0.23 V) and linking the positive potential to thermodynamic feasibility.

Cobalt has an atomic number of 27. The ground state electronic configuration of a neutral Co atom is \([\text{Ar}] 3d^7 4s^2\). When forming the \(\text{Co}^{3+}\) ion, three electrons are removed. The two \(4s\) electrons are removed first, followed by one \(3d\) electron, leaving \([\text{Ar}] 3d^6\).

1 mark for the correct option (A). Awarded for identifying that electrons are lost from the 4s orbital first, then 3d, leading to the [Ar] 3d6 configuration.

The nitration of methylbenzene is an electrophilic aromatic substitution reaction. The active electrophile is the nitronium ion, \(\text{NO}_2^+\). The methyl group is electron-donating by induction, which activates the ring and directs substitution primarily to the ortho (2-) and para (4-) positions.

1 mark for the correct option (A). Identifying the electrophile as NO2+ and understanding the directing effect of the methyl group.

Ethylamine is the strongest base among the options. The ethyl group is an electron-donating group (via inductive effect), which increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton compared to ammonia. Phenylamine has its nitrogen lone pair delocalised into the benzene ring, and benzamide has its lone pair delocalised into the carbonyl group, making them much weaker bases.

1 mark for the correct option (A). Understanding how inductive effects and delocalisation affect the availability of the nitrogen lone pair for protonation.

The ligand \(\text{1,2-diaminoethane}\) (\(\text{en}\)) is bidentate, meaning each molecule forms two coordinate bonds. Since there are two \(\text{en}\) ligands, they form a total of four coordinate bonds. The two chloride ions are monodentate, forming one coordinate bond each. Thus, the total coordination number is \(4 + 2 = 6\). The charge of the complex ion is explicitly shown as \(+1\) in the chemical formula.

1 mark for the correct option (A). Correct identification of the coordination number as 6 (due to bidentate nature of en) and the overall charge of the complex as +1.

Hexan-3-ol is a secondary alcohol: \(\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3\). Reaction of a Grignard reagent with an aldehyde yields a secondary alcohol. The propyl group (\(-\text{CH}_2\text{CH}_2\text{CH}_3\)) is supplied by the propylmagnesium bromide. The remaining part of the chain, containing three carbons and the carbonyl group, must come from propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)).

1 mark for the correct option (A). Identifying that an aldehyde is needed to produce a secondary alcohol, and matching the carbon chain lengths of the reactants to the product.

The relationship between standard Gibbs energy and standard cell potential is given by \(\Delta G^\ominus = -n F E^\ominus_{\text{cell}}\). Since \(\Delta G^\ominus = -T \Delta S_{\text{total}}^\ominus\), we can equate the two expressions: \(-T \Delta S_{\text{total}}^\ominus = -n F E^\ominus_{\text{cell}}\). Dividing both sides by \(-T\) yields \(\Delta S_{\text{total}}^\ominus = \frac{n F E^\ominus_{\text{cell}}}{T}\).

1 mark for the correct option (A). Derived from equating standard free energy expressions.

In Friedel-Crafts acylation, anhydrous aluminium chloride (\(\text{AlCl}_3\)) acts as a Lewis acid (electron-pair acceptor) that reacts with ethanoyl chloride (\(\text{CH}_3\text{COCl}\)) to form the electrophilic acylium ion (\(\text{CH}_3\text{CO}^+\) and \(\text{AlCl}_4^-\). It therefore catalyzes the reaction by generating the required electrophile.

1 mark for the correct option (A). Identifying AlCl3 as a Lewis acid catalyst involved in electrophile generation.

The reaction involves the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) and the reduction of \(\text{Ag}^+\) to \(\text{Ag}\). Therefore, \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = E^\theta(\text{Ag}^+/\text{Ag}) - E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.80\text{ V} - (+0.77\text{ V}) = +0.03\text{ V}\). Since the cell potential is positive, the reaction is thermodynamically feasible under standard conditions.

1 mark: Correctly calculates cell potential as +0.03 V and identifies it as thermodynamically feasible.

Manganese has atomic number 25, with electronic configuration \([Ar] 3d^5 4s^2\). When forming the \(\text{Mn}^{2+}\) ion, the two 4s electrons are lost, resulting in \([Ar] 3d^5\).

1 mark: Correctly identifies Mn2+ as having the [Ar] 3d5 electronic configuration.

When excess concentrated hydrochloric acid is added to aqueous copper(II) ions, a ligand exchange reaction occurs where water ligands are replaced by chloride ligands. Due to the large size and negative charge of chloride ions, only four can fit around the copper central ion, forming a tetrahedral \([\text{CuCl}_4]^{2-}\) complex.

1 mark: Correctly identifies the formula and tetrahedral shape of the tetrachlorocuprate(II) ion.

The active electrophile is the nitronium ion, \(\text{NO}_2^+\). It is generated when concentrated nitric acid reacts with concentrated sulfuric acid, which acts as a catalyst and a stronger acid to protonate nitric acid, followed by the loss of a water molecule.

1 mark: Identifies the nitronium ion (NO2+) and its generation from nitric and sulfuric acids.

Phenylamine is the weakest base because the lone pair on the nitrogen atom is partially delocalised into the benzene ring, making it less available to accept a proton. Ethylamine is the strongest base because the alkyl group is electron-releasing, which increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton than in ammonia.

1 mark: Correctly orders the three compounds by increasing basicity: Phenylamine < Ammonia < Ethylamine.

At a high pH of 12, hydroxide ions remove a proton from the protonated amino group of the zwitterion, leaving an uncharged amino group (\(-\text{NH}_2\)) and a negatively charged carboxylate group (\(-\text{COO}^-\)).

1 mark: Identifies the correct anionic structure of alanine under alkaline conditions.

Step 2 is the reduction of nitrobenzene to phenylamine using tin (Sn) and concentrated hydrochloric acid, followed by alkaline workup. Step 3 is diazotisation using nitrous acid, prepared in-situ from sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid, maintained below \(10\ ^\circ\text{C}\) to prevent decomposition.

1 mark: Identifies correct reagents and temperature control for steps 2 and 3.

At the anode (where oxidation occurs) of an alkaline hydrogen-oxygen fuel cell, hydrogen gas reacts with hydroxide ions to form water and release electrons: \(\text{H}_2(g) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + 2e^-\).

1 mark: Identifies the correct oxidation half-equation under alkaline conditions.

The relationship between the standard cell potential and the total entropy change of the system and surroundings at equilibrium is given by:

\(\Delta S^{\circ}_{\text{total}} = \frac{n F E^{\circ}_{\text{cell}}}{T}\)

Substitute the given values into the equation:
- \(n = 2\)
- \(F = 96500\text{ C mol}^{-1}\)
- \(E^{\circ}_{\text{cell}} = +0.34\text{ V}\)
- \(T = 298\text{ K}\)

\(\Delta S^{\circ}_{\text{total}} = \frac{2 \times 96500 \times 0.34}{298} = \frac{65620}{298} \approx +220.2\text{ J K}^{-1}\text{ mol}^{-1}\)

This rounds to \(+220\text{ J K}^{-1}\text{ mol}^{-1}\).

- Correct answer is B (1 mark).
- Incorrect options:
- A: Calculates with \(n = 1\) instead of \(2\).
- C: Incorrect sign (entropy change must be positive since \(E^{\circ}_{\text{cell}} > 0\)).
- D: Uses the numerator \(nFE^{\circ}\) without dividing by temperature \(T\).

The reaction that takes place is:
\([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)

1. **Coordination number**: The starting complex has 6 water ligands, so the coordination number is 6. The product complex has 4 chloride ligands, so the coordination number is 4. The coordination number decreases from 6 to 4.
2. **Geometry**: Tetrachloro complexes of cobalt(II) are tetrahedral in shape due to the steric hindrance of the large chloride ligands.
3. **Oxidation state**: In \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), Co has an oxidation state of +2 (since water is neutral). In \([\text{CoCl}_4]^{2-}\), cobalt has an oxidation state of \(x + 4(-1) = -2 \implies x = +2\). There is no change in the oxidation state of cobalt.

- Correct answer is A (1 mark).
- Incorrect options:
- B: Incorrect geometry (it is tetrahedral, not square planar).
- C: Incorrect coordination number and oxidation state.
- D: Incorrect oxidation state change.

In the first step of the reaction to generate the electrophile:
\(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-\)

Here, concentrated sulfuric acid acts as a Brønsted-Lowry acid (proton donor) because it is a stronger acid than nitric acid. Concentrated nitric acid accepts a proton (acting as a Brønsted-Lowry base) to form the protonated intermediate \(\text{H}_2\text{NO}_3^+\), which then splits into water and the nitronium ion (\(\text{NO}_2^+\)). Therefore, the reactant acting as the base is \(\text{HNO}_3\).

- Correct answer is A (1 mark).
- Incorrect options:
- B: Concentrated sulfuric acid is the proton donor (acid).
- C: This is the product electrophile, not a reactant.
- D: This is a product / conjugate base, not a starting reactant acting as a base in the initial proton transfer.

1. **Ethanamide (I)**: The lone pair of electrons on the nitrogen atom is heavily delocalised into the adjacent \(\text{C}=\text{O}\) pi system, making it virtually unavailable to accept a proton. This is the weakest base.
2. **Phenylamine (II)**: The lone pair on the nitrogen atom is partially delocalised into the benzene ring's pi system, which decreases its availability to accept protons compared to ammonia.
3. **Ammonia (III)**: Standard weak base. The lone pair is localised on the nitrogen atom.
4. **Ethylamine (IV)**: The ethyl group is an electron-releasing alkyl group (positive inductive effect), which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton. This is the strongest base.

Thus, the increasing order of basic strength is: I < II < III < IV.

- Correct answer is A (1 mark).
- Incorrect options:
- B: Incorrectly places phenylamine as a weaker base than ethanamide.
- C: Reverses the sequence entirely (strongest to weakest).
- D: Swaps the position of ammonia and phenylamine.

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

(a)(ii) The coordination number changes from 6 to 4 because chloride ligands (\(\text{Cl}^-\)) are significantly larger than water molecules (\(\text{H}_2\text{O}\)). Therefore, fewer chloride ligands can fit around the central cobalt(II) ion due to steric hindrance and mutual electrostatic repulsion between the negatively charged chloride ligands.

(a)(iii) Ligand substitution (or ligand exchange).

(b)(i)
Oxidation half-equation: \(\text{VO}^{2+} + \text{H}_2\text{O} \rightarrow \text{VO}_2^+ + 2\text{H}^+ + \text{e}^-\)
Reduction half-equation: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
Overall ionic equation: \(5\text{VO}^{2+} + \text{MnO}_4^- + \text{H}_2\text{O} \rightarrow 5\text{VO}_2^+ + \text{Mn}^{2+} + 2\text{H}^+\)

(b)(ii)
Moles of \(\text{MnO}_4^-\text{(aq)}\) used:
\(n(\text{MnO}_4^-) = 0.0200 \times \frac{18.50}{1000} = 3.70 \times 10^{-4}\text{ mol}\)

Using the stoichiometry of the overall reaction (\(5\text{VO}^{2+} : 1\text{MnO}_4^-\)):
\(n(\text{VO}^{2+}) = 5 \times 3.70 \times 10^{-4} = 1.85 \times 10^{-3}\text{ mol}\)

Since 1 mole of original \(\text{VO}_2^+\) was reduced to form 1 mole of \(\text{VO}^{2+}\):
\(n(\text{VO}_2^+) = 1.85 \times 10^{-3}\text{ mol}\)

Concentration of original \(\text{VO}_2^+\) solution in 25.0 cm\(^3\):
\([\text{VO}_2^+] = \frac{1.85 \times 10^{-3}}{0.0250} = 0.0740\text{ mol dm}^{-3}\)

(c)
- The presence of ligands causes the degenerate 3d orbitals of the transition metal ion to split into two non-degenerate energy levels with an energy gap, \(\Delta E\).
- d-electrons in the lower energy level can absorb a specific frequency of visible light and undergo promotion to the higher energy level (d-d transition).
- The energy absorbed is related to the frequency of light by the equation \(\Delta E = h\nu\).
- The remaining unabsorbed wavelengths of light are transmitted or reflected, producing the complementary color observed.
- Different ligands alter the size of the d-orbital splitting energy gap (\(\Delta E\)), which changes the wavelength/frequency of light absorbed and therefore changes the color of the complex.

(a)(i) 1 Mark: Correct balanced equation with state symbols not strictly required but formulas must be correct.
(a)(ii) 3 Marks:
- 1 Mark for identifying that coordination number changes from 6 to 4.
- 1 Mark for stating that chloride ions are larger than water molecules.
- 1 Mark for explaining that fewer chloride ligands can fit due to steric hindrance / repulsion.
(a)(iii) 1 Mark: Ligand substitution / ligand exchange.

(b)(i) 3 Marks:
- 1 Mark for correct oxidation half-equation.
- 1 Mark for correct reduction half-equation.
- 1 Mark for correct overall balanced equation.
(b)(ii) 4 Marks:
- 1 Mark for calculating moles of MnO4-.
- 1 Mark for multiplying moles by 5 to find moles of VO2+.
- 1 Mark for linking moles of VO2+ to original VO2+ (1:1 ratio).
- 1 Mark for correct final concentration of 0.0740 mol dm^-3 with appropriate units.

(c) 5.5 Marks:
- 1 Mark: Ligands cause 3d orbitals to split into different energy levels.
- 1 Mark: Electrons are promoted from lower to higher d-orbital level by absorbing light.
- 1 Mark: Reference to energy equation \(\Delta E = h\nu\) or \(\Delta E = hc/\lambda\).
- 1 Mark: The color observed is the complementary color to that absorbed.
- 1.5 Marks: Different ligands cause different degrees of splitting (different \(\Delta E\)), leading to different frequencies of light absorbed.

(a)(i) \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\) (or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\))
Sulfuric acid acts as a Brønsted-Lowry acid / proton donor, protonating the nitric acid.

(a)(ii) Mechanism steps:
- Curly arrow from the delocalized \(\pi\)-ring of benzene to the nitrogen of the \(\text{NO}_2^+\) ion.
- Drawing of the intermediate cation with a partial ring (horseshoe shape covering at least 5 carbons, open towards the \(\text{sp}^3\) carbon) and a positive charge inside the ring.
- Curly arrow from the C-H bond of the \(\text{sp}^3\) carbon back into the ring to restore the aromatic system, generating nitrobenzene and \(\text{H}^+\).

(b)
- The oxygen atom of the hydroxyl (\(-\text{OH}\)) group has a lone pair of electrons.
- This lone pair is partially delocalized into the \(\pi\)-system of the benzene ring.
- This increases the electron density of the aromatic ring.
- Consequently, the ring in phenol polarizes and attracts electrophiles much more effectively than benzene.

(c)(i) Reagents: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)). Conditions: Heat under reflux, followed by the addition of aqueous sodium hydroxide (\(\text{NaOH}\)) to liberate the free amine.

(c)(ii) Reagent: Ethanoic anhydride (or acetyl chloride).
Structure: \(\text{HO}-\text{C}_6\text{H}_4-\text{NHCOCH}_3\) (where the substituent groups are para to each other on the benzene ring).
Type of reaction: Acylation / nucleophilic addition-elimination / condensation.

(c)(iii)
- High atom economy reduces the quantity of unwanted waste products, making the reaction more environmentally sustainable and cost-effective (less cost for waste disposal).
- Fewer steps minimize product loss at each purification stage, resulting in a higher overall percentage yield and lower production costs.

(a)(i) 2 Marks:
- 1 Mark for the correct equation for electrophile generation.
- 1 Mark for identifying that H2SO4 acts as a proton donor/acid.
(a)(ii) 3 Marks:
- 1 Mark for curly arrow from ring to NO2+.
- 1 Mark for correctly drawn intermediate including charge.
- 1 Mark for curly arrow restoring aromaticity to form products.

(b) 4 Marks:
- 1 Mark for pointing out the lone pair on the oxygen of the -OH group.
- 1 Mark for stating this lone pair delocalizes into the ring.
- 1 Mark for stating this increases electron density of the ring.
- 1 Mark for explaining it polarizes/attracts electrophiles more easily.

(c)(i) 2 Marks:
- 1 Mark for Sn + conc. HCl.
- 1 Mark for reflux, followed by NaOH(aq).
(c)(ii) 3.5 Marks:
- 1 Mark for ethanoic anhydride (accept acetyl chloride).
- 1.5 Marks for correct structural formula of paracetamol.
- 1 Mark for acylation / condensation / nucleophilic addition-elimination.
(c)(iii) 3 Marks:
- 1 Mark for explaining high atom economy decreases waste / increases sustainability.
- 1 Mark for explaining fewer steps reduce loss of yield at each step.
- 1 Mark for general economic/process efficiency benefit (e.g. lower cost).

(a)(i) Standard Hydrogen Electrode (SHE) requirements:
- Hydrogen gas at \(100\text{ kPa}\) (or \(1\text{ bar}\)) pressure.
- Solution containing \(\text{H}^+\text{(aq)}\) ions at a concentration of \(1.00\text{ mol dm}^{-3}\).
- Temperature of \(298\text{ K}\) (or \(25^\circ\text{C}\)).
- Platinum electrode coated in platinum black.

(a)(ii) Labeled diagram must show:
- The SHE with \(\text{H}_2\) gas entering and \(\text{H}^+\text{(aq)}\) solution.
- The iron half-cell containing a solution with equal concentrations (\(1.00\text{ mol dm}^{-3}\)) of both \(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\) ions.
- High-resistance voltmeter connecting the electrodes.
- A salt bridge filled with a suitable inert electrolyte (e.g., KNO3) dipping into both solutions.
- Platinum (Pt) is used as the electrode in the iron half-cell because it is inert (does not react with the solution) but provides a catalytic surface to allow electron transfer.

(b)(i) \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +2.01 - (+0.54) = +1.47\text{ V}\).
The reaction is slow because both reacting species, \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\), are negatively charged anions. Electrostatic repulsion between the similarly charged ions results in a high activation energy.

(b)(ii)
First, \(\text{Fe}^{2+}\) ions are oxidized by \(\text{S}_2\text{O}_8^{2-}\):
\(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)

Then, the generated \(\text{Fe}^{3+}\) ions oxidize the \(\text{I}^-\) ions, regenerating the catalyst:
\(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

Both steps involve collision between oppositely charged ions, which lowers the activation energy of each individual step compared to the uncatalyzed route.

(c) First, determine the \(E^\ominus_{\text{cell}}\) for the reaction between \(\text{Fe}^{3+}\) and \(\text{I}^-\):
\(E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{I}_2/\text{I}^-) = +0.77 - (+0.54) = +0.23\text{ V}\)

For the balanced equation: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightleftharpoons 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\), the number of electrons transferred is \(n = 2\).

Calculate \(\Delta G^\ominus\):
\[\Delta G^\ominus = -nFE^\ominus_{\text{cell}} = -2 \times 96500 \times 0.23 = -44390\text{ J mol}^{-1}\ (\text{or } -44.39\text{ kJ mol}^{-1})\\]

Calculate \(\ln K\):
\[\ln K = -\frac{\Delta G^\ominus}{RT} = -\frac{-44390}{8.31 \times 298} = \frac{44390}{2476.38} = 17.925 \approx 17.9\\]

*(Note: If \(n = 1\) is used for the single-electron stoichiometric equation, \(\Delta G^\ominus = -22195\text{ J mol}^{-1}\) and \(\ln K = 8.96\). Both are accepted if working is consistent with the chosen equation)*

(a)(i) 3 Marks:
- 1 Mark: [H+] = 1.00 mol dm^-3 and H2(g) at 100 kPa / 1 bar.
- 1 Mark: Temperature at 298 K.
- 1 Mark: Platinum electrode coated in platinum black.
(a)(ii) 4 Marks:
- 1 Mark: Correct diagram showing SHE connected to the Fe3+/Fe2+ half-cell with salt bridge.
- 1 Mark: Both solutions and gases labeled with standard concentrations.
- 1 Mark: Voltmeter in circuit.
- 1 Mark: Platinum electrode specified for iron half-cell because it is inert.

(b)(i) 3 Marks:
- 1 Mark: E_cell = +1.47 V.
- 1 Mark: Reacting species are both negatively charged / anions.
- 1 Mark: High activation energy due to electrostatic repulsion.
(b)(ii) 4 Marks:
- 1 Mark: Correct equation for reaction of Fe2+ with S2O8^2-.
- 1 Mark: Correct equation for reaction of Fe3+ with I-.
- 2 Marks: Explanation that steps involve collision of oppositely charged ions, reducing Ea.

(c) 3.5 Marks:
- 1 Mark: E_cell calculation of +0.23 V.
- 1 Mark: Correct calculation of Delta G = -44.39 kJ mol^-1 (or -22.20 kJ mol^-1 if n=1).
- 1 Mark: Correct rearrangement to find ln K.
- 0.5 Marks: Correct final answer to 3 significant figures (17.9 for n=2, or 8.96 for n=1).

(a)
Order of increasing base strength: Phenylamine < Ammonia < Ethylamine

Explanation:
- Basicity depends on the ability of the lone pair of electrons on the nitrogen atom to accept a proton (\(\text{H}^+\)).
- In phenylamine, the lone pair of electrons on the nitrogen atom is partially delocalized into the \(\pi\)-system of the benzene ring. This significantly decreases the electron density on the nitrogen, making the lone pair less available to accept a proton.
- In ethylamine, the alkyl (ethyl) group is electron-releasing due to the inductive effect (\(+\text{I}\) effect). This increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton than in ammonia.

(b)(i)
- At \(\text{pH} = 1\) (highly acidic): Both amine groups are protonated to form ammonium cations, while the carboxylic acid remains un-ionized:
\([\text{H}_3\text{N}^+-(\text{CH}_2)_4-\text{CH}(\text{NH}_3^+)-\text{COOH}]\)
- At \(\text{pH} = 13\) (highly alkaline): Both amine groups are neutral (unprotonated), and the carboxylic acid group is deprotonated to form a carboxylate anion:
\([\text{H}_2\text{N}-(\text{CH}_2)_4-\text{CH}(\text{NH}_2)-\text{COO}^-]\)

(b)(ii)
- Isoelectric point is the pH at which an amino acid exists as a zwitterion with no net electrical charge.
- Structure of lysine at its isoelectric point (\(\text{pH} = 9.7\)) has one protonated amine group and one deprotonated carboxylate group:
\(\text{H}_3\text{N}^+-(\text{CH}_2)_4-\text{CH}(\text{NH}_2)-\text{COO}^-\)
*(or with the terminal amine group protonated and the alpha-amine group neutral, e.g. \(\text{H}_2\text{N}-(\text{CH}_2)_4-\text{CH}(\text{NH}_3^+)-\text{COO}^-\))*

(c)(i)
- Benzene-1,4-dicarboxylic acid: \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) (para-substituted benzene)
- Benzene-1,4-diamine: \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2\) (para-substituted benzene)

(c)(ii) Repeating unit of Kevlar:
\(-[\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\text{NH}-\text{C}_6\text{H}_4-\text{NH}]_n-\)

(c)(iii)
- The polymer chains are planar and linear, allowing them to align and pack closely together.
- Extensive hydrogen bonds form between the \(\text{C}=\text{O}\) group of one chain and the \(\text{N}-\text{H}\) group of an adjacent chain.
- The cumulative strength of these many hydrogen bonds along the aligned chains yields very high tensile strength.

(a) 5 Marks:
- 1 Mark: Correct order of basicity: phenylamine < ammonia < ethylamine.
- 1 Mark: Basicity defined by proton-donating/accepting availability of N's lone pair.
- 1.5 Marks: Phenylamine explanation (delocalization of lone pair into ring reduces availability).
- 1.5 Marks: Ethylamine explanation (+I inductive effect of alkyl group increases availability).

(b)(i) 3 Marks:
- 1.5 Marks: Correct structure at pH 1 showing both nitrogens protonated.
- 1.5 Marks: Correct structure at pH 13 showing carboxylate anion and neutral amines.
(b)(ii) 2.5 Marks:
- 1 Mark: Definition of isoelectric point as pH with net zero charge.
- 1.5 Marks: Correct structure of zwitterionic lysine (one amine protonated, carboxylate deprotonated).

(c)(i) 2 Marks:
- 1 Mark: Structure of benzene-1,4-dicarboxylic acid.
- 1 Mark: Structure of benzene-1,4-diamine.
(c)(ii) 2 Marks:
- 2 Marks for correct repeating unit with free bonds at ends and amide linkage visible.
(c)(iii) 3 Marks:
- 1 Mark: Linear polymer chains pack closely together.
- 1 Mark: Hydrogen bonding between C=O and N-H of adjacent chains.
- 1 Mark: High cumulative strength from many hydrogen bonds.

(a) Under acidic conditions, the half-equations are:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
\(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-\)
Multiplying the first by 2 and the second by 5 gives the overall equation:
\(2\text{MnO}_4^-(\text{aq}) + 5\text{C}_2\text{O}_4^{2-}(\text{aq}) + 16\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 10\text{CO}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})\)

(b) Potassium manganate(VII) is self-indicating. The purple \(\text{MnO}_4^-\)\ ions are reduced to almost colourless \(\text{Mn}^{2+}\)\ ions during the reaction. At the end-point, the first excess drop of \(\text{MnO}_4^-\)\ turns the solution from colourless to a permanent pale pink.

(c) Step 1: Moles of \(\text{MnO}_4^-\)
\(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{24.15}{1000}\text{ dm}^3 = 4.83 \times 10^{-4}\text{ mol}\)

Step 2: Moles of oxalate in \(25.0\text{ cm}^3\)
From stoichiometry, \(5\text{ mol of } \text{C}_2\text{O}_4^{2-} \equiv 2\text{ mol of } \text{MnO}_4^-\)
\(n(\text{C}_2\text{O}_4^{2-}) = 4.83 \times 10^{-4} \times \frac{5}{2} = 1.2075 \times 10^{-3}\text{ mol}\)

Step 3: Moles of complex in \(25.0\text{ cm}^3\)
Since 1 mole of \(\text{K}_3[\text{Fe}(\text{C}_2\text{O}_4)_3] \cdot 3\text{H}_2\text{O}\) contains 3 moles of oxalate ligands:
\(n(\text{complex}) = \frac{1.2075 \times 10^{-3}}{3} = 4.025 \times 10^{-4}\text{ mol}\)

Step 4: Moles of complex in \(250.0\text{ cm}^3\)
\(n(\text{total}) = 4.025 \times 10^{-4} \times 10 = 4.025 \times 10^{-3}\text{ mol}\)

Step 5: Mass of pure complex
\(m = 4.025 \times 10^{-3}\text{ mol} \times 491.2\text{ g mol}^{-1} = 1.9771\text{ g}\)

Step 6: Percentage purity
\(\text{Purity} = \frac{1.9771}{2.350} \times 100\% = 84.13\% \approx 84.1\%\)

(d) Potassium manganate(VII) is an oxidising agent and stains skin/clothing; wear protective gloves.

(e) The reaction between \(\text{MnO}_4^-\)\ and \(\text{C}_2\text{O}_4^{2-}\)\ has a high activation energy due to repulsion between two negative ions. If conducted at room temperature, the rate is too slow; the pink colour does not fade quickly, leading to a premature (too low) titre reading because the student stops adding titrant early.

(a) [2 marks]
- 1 mark for correct species on reactant and product sides.
- 1 mark for correct balancing of all species including charge and state symbols.

(b) [2 marks]
- 1 mark for stating that it is self-indicating (purple to pale pink transition).
- 1 mark for specifying the end-point colour change: colourless to pale pink.

(c) [6 marks]
- 1 mark for calculating moles of manganate(VII) (\(4.83 \times 10^{-4}\)).
- 1 mark for using the correct 5:2 ratio to get moles of oxalate (\(1.2075 \times 10^{-3}\)).
- 1 mark for dividing by 3 to find moles of complex in the aliquot (\(4.025 \times 10^{-4}\)).
- 1 mark for scaling up by 10 for the volumetric flask (\(4.025 \times 10^{-3}\)).
- 1 mark for calculating mass of pure complex (\(1.977\text{ g}\)).
- 1 mark for calculating the correct final percentage purity (84.1%).

(d) [0.5 marks]
- 0.5 marks for stating: wear gloves (as it stains skin / is an irritant/oxidant).

(e) [2 marks]
- 1 mark for mentioning the reaction has high activation energy / repulsion between anions slows down the rate.
- 1 mark for noting that at room temperature, the pink color persists too long, leading to a false endpoint (lower titre).

(a) Ethanoic anhydride is cheaper, less corrosive, less vulnerable to rapid hydrolysis, and does not produce toxic, hazardous hydrogen chloride (HCl) gas fumes as a byproduct.

(b) Recrystallisation steps:
1. Dissolve the crude aspirin in the minimum volume of hot solvent (ethanol/water mixture).
2. Filter the hot solution through fluted filter paper to remove insoluble impurities.
3. Allow the filtrate to cool slowly to room temperature (and then in ice) to allow pure aspirin crystals to reform.
4. Filter under reduced pressure (using a Büchner funnel) to isolate the crystals, wash with a small volume of ice-cold solvent, and dry.

(c) Calculation:
\(n(2\text{-hydroxybenzoic acid}) = \frac{3.00}{138.0} = 0.02174\text{ mol}\)
Since stoichiometry is 1:1, theoretical moles of aspirin = \(0.02174\text{ mol}\).
Theoretical mass of aspirin = \(0.02174 \times 180.0 = 3.913\text{ g}\).
\(\text{Percentage yield} = \frac{2.45}{3.913} \times 100\% = 62.61\% \approx 62.6\%\).

(d) Place a dry sample of aspirin in a capillary tube and heat slowly in a melting point apparatus. A pure sample will have a sharp melting range (within \(1-2\text{ }^\circ\text{C}\)) that is close to the literature value of \(135\text{ }^\circ\text{C}\). An impure sample will melt over a wider range and at a lower temperature than the literature value.

(e) Visualisation: Place the plate under a UV lamp (or expose to iodine chamber). Observation: The purified product lane will only show a single spot corresponding to aspirin and will show no spot at the same \(R_f\) value as the reference spot of 2-hydroxybenzoic acid.

(a) [2 marks]
- 1 mark for stating that it does not produce toxic hydrogen chloride gas.
- 1 mark for stating that it is less corrosive / cheaper / reacts less violently.

(b) [4 marks]
- 1 mark for dissolving in the minimum volume of hot solvent.
- 1 mark for hot filtration to remove insoluble impurities.
- 1 mark for cooling slowly to crystallise (and cooling in ice).
- 1 mark for filtration under reduced pressure (using Büchner funnel), washing with cold solvent, and drying.

(c) [3 marks]
- 1 mark for calculating moles of reactant (0.0217 mol).
- 1 mark for calculating theoretical mass of aspirin (3.91 g).
- 1 mark for final percentage yield to 3 sig figs (62.6%).

(d) [2 marks]
- 1 mark for noting that a pure sample melts sharply at a temperature close to the literature value.
- 1 mark for noting that impurities lower the melting point and broaden the melting temperature range.

(e) [1.5 marks]
- 0.5 marks for specifying UV light or iodine stain.
- 1 mark for stating that there should be only a single spot and no spot corresponding to 2-hydroxybenzoic acid.

(a) Quenching is the process of stopping the reaction at a specific moment. This is achieved by adding sodium hydrogencarbonate (\(\text{NaHCO}_3\)) which neutralises the acid catalyst (\(\text{H}^+\)). Without the acid catalyst, the rate of iodination becomes negligible (effectively zero), allowing the exact concentration of iodine at that time to be accurately determined.

(b) Indicator: Starch solution. Colour change: From blue-black to colourless at the end-point.

(c) (i) The order of reaction with respect to iodine is zero. Reasoning: A graph of concentration of a reactant against time that is a straight line indicates a constant rate of reaction, which is independent of the reactant concentration.
(ii) To find the order with respect to propanone, repeat the experiment several times by varying the initial concentration of propanone, while keeping the concentrations of iodine and the acid catalyst constant. Determine the initial rate (or slope of the straight-line plot) for each run. Plot rate against \([\text{propanone}]\) to deduce the order.

(d) Because iodine is zero order, it does not participate in the rate-determining step (the slowest step) of the reaction mechanism. It must react in a subsequent, fast step. The slow, rate-determining step must involve only propanone and the proton catalyst (forming an enol intermediate).

(a) [3 marks]
- 1 mark for stating that quenching stops/freezes the reaction.
- 1 mark for stating that sodium hydrogencarbonate is added to neutralise the acid catalyst.
- 1 mark for explaining that stopping the reaction ensures the measured concentration reflects the exact time of sampling.

(b) [2 marks]
- 1 mark for identifying starch indicator.
- 1 mark for correct colour change: blue-black to colourless (reject 'clear').

(c) [5 marks]
- (i) 1 mark for identifying zero order.
- (ii) 1 mark for explaining that a constant gradient of the concentration-time graph indicates rate is independent of concentration.
- (iii) 1 mark for stating that experiments must be repeated with varying concentrations of propanone.
- (iv) 1 mark for keeping other factors constant (temperature, [H+]).
- (v) 1 mark for measuring the rate from each gradient and plotting/comparing rates.

(d) [2.5 marks]
- 1 mark for stating that iodine is not involved in the rate-determining (slow) step.
- 1 mark for stating that it reacts in a subsequent, fast step.
- 0.5 marks for linking to the rate-determining step involving only propanone and acid catalyst.

(a) (i) The anion is sulfate, \(\text{SO}_4^{2-}\).
(ii) The white precipitate is barium sulfate. The ionic equation is:
\(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)

(b) (i) Precipitate **Y** is iron(II) hydroxide, \(\text{Fe}(\text{OH})_2\). The transition metal cation present in **X** is \(\text{Fe}^{2+}\) (or \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\)).
(ii) The green iron(II) hydroxide is oxidised by oxygen in the air to form brown iron(III) hydroxide, \(\text{Fe}(\text{OH})_3\). This is an oxidation/redox reaction.
(iii) The gas is ammonia, \(\text{NH}_3\). The second cation present in **X** is the ammonium ion, \(\text{NH}_4^+\).

(c) Step 1: Calculate the mass of water of crystallisation lost:
\(m(\text{H}_2\text{O}) = 3.92\text{ g} - 2.84\text{ g} = 1.08\text{ g}\)

Step 2: Calculate moles of anhydrous residue:
\(n(\text{anhydrous}) = \frac{2.84\text{ g}}{284.1\text{ g mol}^{-1}} = 0.0100\text{ mol}\)

Step 3: Calculate moles of water lost:
\(n(\text{H}_2\text{O}) = \frac{1.08\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0600\text{ mol}\)

Step 4: Find the ratio of moles of water to anhydrous salt:
\(x = \frac{0.0600}{0.0100} = 6\)
Thus, the formula is \((\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \cdot 6\text{H}_2\text{O}\).

(a) [3 marks]
- (i) 1 mark for identifying Sulfate / \(\text{SO}_4^{2-}\).
- (ii) 1 mark for correct formulas of reactants and products.
- 1 mark for correct state symbols (\(\text{aq}\) and \(\text{s}\)).

(b) [6 marks]
- (i) 1 mark for stating Iron(II) hydroxide (or \(\text{Fe}(\text{OH})_2\)) and 1 mark for transition metal cation formula \(\text{Fe}^{2+}\).
- (ii) 1 mark for identifying oxidation by oxygen in air, and 1 mark for mentioning formation of iron(III) hydroxide (or \(\text{Fe}(\text{OH})_3\)).
- (iii) 1 mark for identifying ammonia (\(\text{NH}_3\)) gas, and 1 mark for the formula of the ammonium ion \(\text{NH}_4^+\).

(c) [3.5 marks]
- 1 mark for calculating mass of water lost (\(1.08\text{ g}\)).
- 1 mark for calculating moles of anhydrous salt (\(0.0100\text{ mol}\)).
- 1 mark for calculating moles of water (\(0.0600\text{ mol}\)).
- 0.5 marks for calculating the correct mole ratio of 6.