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The successive ionisation energies show a very large jump between the third and fourth ionisation energies (from 2745 to 11578 \(kJ\text{ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner quantum shell, which is closer to the nucleus and experiences significantly less shielding. Therefore, element X has three valence electrons and belongs to Group 13 (Group 3).

1 mark: Correctly identify Group 13 (Group 3) based on the location of the largest jump in successive ionisation energies.

The reduction in volume when passed through NaOH is due to the absorption of \(CO_2\), so the volume of \(CO_2\) produced is \(60 - 30 = 30\text{ cm}^3\). Since 10 \(cm^3\) of hydrocarbon produced 30 \(cm^3\) of \(CO_2\), the ratio of C to hydrocarbon is 3:1, so \(x = 3\). The volume of unreacted oxygen is 30 \(cm^3\), which means the volume of oxygen reacted is \(80 - 30 = 50\text{ cm}^3\). The general equation for combustion is \(C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O\). The ratio of reacted oxygen to hydrocarbon is \(50 / 10 = 5\), so \(x + y/4 = 5\). Substituting \(x = 3\) gives \(3 + y/4 = 5\), which yields \(y = 8\). The molecular formula is therefore \(C_3H_8\).

1 mark: Correctly determine the molecular formula by mole calculations based on gas volumes.

The amide ion, \(NH_2^-\), has a central nitrogen atom with 5 valence electrons, plus 1 extra electron from the negative charge, making 6 valence electrons. It forms 2 single covalent bonds with hydrogen atoms, leaving 2 lone pairs of electrons. Based on electron pair repulsion theory, these 4 electron pairs arrange themselves tetrahedrally to minimise repulsion. The presence of two lone pairs compresses the bond angle from the tetrahedral angle of \(109.5^\circ\) by approximately \(2 \times 2.5^\circ = 5^\circ\) to \(104.5^\circ\), resulting in a non-linear (bent) shape.

1 mark: Correctly identify NH2- as having a bent shape with two lone pairs and a bond angle of approximately 104.5 degrees.

There are four structural isomers with the molecular formula \(C_4H_9Cl\). These are: 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane, and 2-chloro-2-methylpropane.

1 mark: Correctly identify that there are exactly 4 structural isomers.

The electrophilic addition of hydrogen bromide to 2-methylbut-2-ene can form either a secondary carbocation or a tertiary carbocation intermediate. The tertiary carbocation is more stable than the secondary carbocation because the three electron-releasing alkyl groups reduce the positive charge density on the positive carbon atom more effectively than two alkyl groups. Therefore, the reaction proceeds predominantly via the more stable tertiary carbocation, forming 2-bromo-2-methylbutane as the major product.

1 mark: Correct explanation of the major product formation based on the higher stability of a tertiary carbocation compared to a secondary carbocation.

Phosphorus has the electron configuration \([Ne] 3s^2 3p^3\) with one electron in each of the three 3p orbitals. Sulfur has the configuration \([Ne] 3s^2 3p^4\), meaning one of the 3p orbitals contains a pair of electrons. The mutual spin-pair repulsion between these two electrons in the same orbital of sulfur makes it easier to remove one of them compared to removing an electron from a singly occupied orbital in phosphorus, leading to a lower first ionisation energy for sulfur.

1 mark: Correct explanation of the ionization energy anomaly between phosphorus and sulfur due to orbital electron-pair repulsion.

To find the empirical formula, calculate the molar ratio of nitrogen to oxygen. Moles of N = \(30.4 / 14.0 = 2.17\). Moles of O = \((100 - 30.4) / 16.0 = 69.6 / 16.0 = 4.35\). The ratio of N to O is \(2.17 : 4.35 \approx 1 : 2\). Therefore, the empirical formula is \(NO_2\).

1 mark: Correctly calculate the empirical formula from mass percentage.

Polarizing power of a cation depends on its charge density, which is proportional to its charge and inversely proportional to its ionic radius. Among the given options, the cations are \(Na^+\), \(Mg^{2+}\), \(Al^{3+}\), and \(Ca^{2+}\). The aluminium cation, \(Al^{3+}\), has the highest positive charge and the smallest ionic radius, giving it the highest charge density and thus the greatest polarizing power.

1 mark: Correctly identify that Al3+ in AlCl3 has the highest charge density and greatest polarizing power.

The largest jump in ionization energy occurs between the third and fourth ionization energies (from \( 2745 \) to \( 11578 \text{ kJ mol}^{-1} \)). This shows that the fourth electron is being removed from an inner quantum shell, indicating that the element has three valence electrons and belongs to Group 13 (Group 3). The element is aluminium, which forms a \( 3+ \) ion. Therefore, the formula of its chloride is \( X\text{Cl}_3 \).

1 mark: Identifies the correct option C based on the sudden increase between the 3rd and 4th ionization energies.

Shaking with NaOH removes \( \text{CO}_2 \), causing a contraction of \( 55.0 \text{ cm}^3 - 25.0 \text{ cm}^3 = 30.0 \text{ cm}^3 \). This is the volume of \( \text{CO}_2 \) produced. Since \( 10.0 \text{ cm}^3 \) of hydrocarbon produced \( 30.0 \text{ cm}^3 \) of \( \text{CO}_2 \), the hydrocarbon has 3 carbon atoms per molecule. The remaining gas is unreacted oxygen, so \( 25.0 \text{ cm}^3 \) of \( \text{O}_2 \) was left over, meaning \( 70.0 \text{ cm}^3 - 25.0 \text{ cm}^3 = 45.0 \text{ cm}^3 \) of \( \text{O}_2 \) reacted. In the combustion of \( \text{C}_3\text{H}_y \), the ratio of \( \text{O}_2 \) to hydrocarbon is \( 45.0 / 10.0 = 4.5 \). The equation is \( \text{C}_3\text{H}_y + 4.5\text{O}_2 \rightarrow 3\text{CO}_2 + (y/2)\text{H}_2\text{O} \). Since \( 3 + y/4 = 4.5 \), we find \( y = 6 \). Thus, the formula is \( \text{C}_3\text{H}_6 \).

1 mark: Correctly calculates the volume of carbon dioxide and oxygen reacted to deduce the molecular formula as \( \text{C}_3\text{H}_6 \) (Option B).

In the amide ion, \( \text{NH}_2^- \), the nitrogen atom has 5 valence electrons plus 1 from the negative charge, giving 6 valence electrons. It has 2 bonding pairs and 2 lone pairs. According to electron-pair repulsion theory (VSEPR), the presence of 2 lone pairs repels the bonding pairs more strongly than bonding pair-bonding pair repulsion, reducing the bond angle from the tetrahedral angle of \( 109.5^\circ \) to approximately \( 104.5^\circ \). This is isoelectronic and isostructural with water.

1 mark: Correctly identifies \( \text{NH}_2^- \) as having a bent shape with two lone pairs and a bond angle of approximately \( 104.5^\circ \) (Option C).

2-methylbutane, \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3 \), has four different hydrogen environments where substitution by chlorine can take place: (1) on either of the two equivalent methyl groups at position 1, yielding 1-chloro-2-methylbutane; (2) on the CH carbon at position 2, yielding 2-chloro-2-methylbutane; (3) on the CH2 carbon at position 3, yielding 2-chloro-3-methylbutane; (4) on the methyl group at position 4, yielding 1-chloro-3-methylbutane. This results in exactly 4 structural isomers.

1 mark: Correctly identifies that there are 4 unique structural isomers formed (Option B).

The reaction of propene with \( \text{HBr} \) is an electrophilic addition. The major product is 2-bromopropane because the reaction proceeds via the more stable secondary carbocation intermediate \( \text{CH}_3\text{CH}^+\text{CH}_3 \) rather than the less stable primary carbocation intermediate \( \text{CH}_3\text{CH}_2\text{CH}_2^+ \). The greater stability of the secondary carbocation is due to the electron-releasing inductive effect of the two adjacent alkyl groups.

1 mark: Identifies that the major product is 2-bromopropane because it is formed via the more stable secondary carbocation intermediate (Option B).

(a) Relative isotopic mass is the mass of an atom of an isotope compared to \(1/12\)th of the mass of an atom of carbon-12.

(b) Let \(x\) be the fractional abundance of \(^{71}\text{Ga}\). The fractional abundance of \(^{69}\text{Ga}\) is \(1 - x\).
\(69.0(1 - x) + 71.0x = 69.723\)
\(69.0 - 69.0x + 71.0x = 69.723\)
\(2.0x = 0.723\)
\(x = 0.3615\)
Percentage abundance of \(^{71}\text{Ga} = 36.2\%\) (to 3 s.f.).

(c) (i) Element \(X\) belongs to Group 15 (or Group 5) of the Periodic Table. This is because there is a very large increase (jump) between the fifth and sixth ionisation energies, which indicates that the sixth electron is being removed from an inner electron shell. Therefore, there are five electrons in the outer shell.
(ii) The fourth ionisation energy is represented by:
\(X^{3+}(g) \to X^{4+}(g) + e^-\) (or \(X^{3+}(g) - e^- \to X^{4+}(g)\))

(d) (i) Phosphorus has the outer electron configuration \(3s^2 3p^3\), where the three \(3p\) orbitals are each singly occupied (half-filled). Sulfur has the configuration \(3s^2 3p^4\), where one of the \(3p\) orbitals contains a pair of electrons. The mutual repulsion between the paired electrons in the \(3p\) orbital of sulfur makes it easier to remove an electron, so less energy is required compared to phosphorus.
(ii) Chlorine has a higher nuclear charge (more protons) than sulfur, but the outer electrons in both atoms experience similar shielding because they are in the same quantum shell. Therefore, the outer electron in chlorine experiences a stronger electrostatic attraction to the nucleus, requiring more energy to remove.

(a)
- M1: Mass of one atom of an isotope (1)
- M2: relative to 1/12th of the mass of an atom of carbon-12 (1)

(b)
- M1: Setting up the algebraic equation: \(69(1-x) + 71x = 69.723\) (or equivalent) (1)
- M2: Solving to find \(x = 0.3615\) (1)
- M3: Converting to percentage to 3 s.f.: \(36.2\%\) (1) [Allow TE for minor arithmetic slips]

(c)(i)
- M1: Group 15 / Group 5 (1)
- M2: Explanation: Sharp/large increase between 5th and 6th ionisation energies (indicating 5 electrons in the outer shell) (1)

(c)(ii)
- M1: Correct species and charges: \(X^{3+} \to X^{4+} + e^-\) (1)
- M2: State symbols: \((g)\) on both \(X^{3+}\) and \(X^{4+}\) (1)

(d)(i)
- M1: P is \(3p^3\) and S is \(3p^4\) (or electronic configuration drawn/written) (1)
- M2: In S there is a paired electron in a \(3p\) orbital / in P all \(3p\) orbitals are singly occupied (1)
- M3: (Mutual/spin) repulsion between the paired electrons in S makes it easier to remove (1)

(d)(ii)
- M1: Chlorine has more protons / greater nuclear charge (1)
- M2: Same shielding / similar shielding (1)
- M3: Stronger electrostatic attraction between nucleus and outer electron (1)

(a) (i) The ionic equation is:
\(\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \to \text{BaSO}_4(s)\)

(ii) Moles of \(\text{BaSO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.70}{233.4} = 0.01157\text{ mol}\) (or \(1.16 \times 10^{-2}\text{ mol}\))

(iii) Since 1 mole of hydrated salt contains 1 mole of sulfate ions, the moles of the hydrated salt in the sample is equal to the moles of barium sulfate formed.
Moles of \(M\text{SO}_4 \cdot x\text{H}_2\text{O} = 0.01157\text{ mol}\)
Molar mass of \(M\text{SO}_4 \cdot x\text{H}_2\text{O} = \frac{3.21\text{ g}}{0.01157\text{ mol}} = 277.4\text{ g mol}^{-1}\)

(b) (i) Mass of water lost = \(4.50\text{ g} - 2.46\text{ g} = 2.04\text{ g}\)
Moles of water lost = \frac{2.04}{18.0} = 0.1133\text{ mol}\)

(ii) Moles of hydrated salt in \(4.50\text{ g}\) of the sample:
\(\text{Moles} = \frac{4.50}{277.4} = 0.01622\text{ mol}\)
Therefore, the ratio of moles of water to moles of hydrated salt is:
\(x = \frac{0.1133}{0.01622} = 6.99 \approx 7\)
Thus, \(x = 7\).

(iii) The molar mass of the hydrated salt is \(277.4\text{ g mol}^{-1}\).
Mass of water of crystallisation = \(7 \times 18.0 = 126.0\text{ g mol}^{-1}\)
Molar mass of anhydrous \(M\text{SO}_4 = 277.4 - 126.0 = 151.4\text{ g mol}^{-1}\)
\(A_r(M) + A_r(\text{S}) + 4 \times A_r(\text{O}) = 151.4\)
\(A_r(M) + 32.1 + 64.0 = 151.4\)
\(A_r(M) = 151.4 - 96.1 = 55.3\)
The transition metal with a relative atomic mass close to 55.3 is Iron (\(\text{Fe}\), actual \(A_r = 55.8\)).

(c) From the reaction equation:
\(2M\text{SO}_4(s) \to 2MO(s) + 2\text{SO}_2(g) + \text{O}_2(g)\)
For every 2 moles of \(M\text{SO}_4\) decomposed, \(3\text{ moles}\) of gas are produced (\(2\text{ mol of } \text{SO}_2\) and \(1\text{ mol of } \text{O}_2\)).
Therefore, the total moles of gas produced from \(0.125\text{ mol}\) of \(M\text{SO}_4\) is:
\(\text{Moles of gas} = 0.125 \times \frac{3}{2} = 0.1875\text{ mol}\)
Total volume of gas = \(0.1875\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 4.50\text{ dm}^3\).

(a)(i)
- M1: Correct species: \(\text{Ba}^{2+} + \text{SO}_4^{2-} \to \text{BaSO}_4\) (1)
- M2: State symbols: \((aq)\) for reactants and \((s)\) for product (1)

(a)(ii)
- M1: Formula: \(\text{moles} = 2.70 / 233.4\) (1)
- M2: Answer: \(0.01157\text{ mol}\) (accept \(0.0115\) to \(0.0116\)) (1)

(a)(iii)
- M1: Calculation of molar mass: \(3.21 / 0.01157 = 277.4\text{ g mol}^{-1}\) (accept range \(276 - 279\)) (1)

(b)(i)
- M1: Mass of water = \(4.50 - 2.46 = 2.04\text{ g}\) (1)
- M2: Moles of water = \(2.04 / 18.0 = 0.113\text{ mol}\) (accept \(0.1133\)) (1)

(b)(ii)
- M1: Moles of hydrated salt = \(4.50 / 277.4 = 0.0162\text{ mol}\) (or alternative valid method) (1)
- M2: Ratio of water to salt = \(0.113 / 0.0162 = 6.98\) (1)
- M3: Deduce integer value: \(x = 7\) (1)

(b)(iii)
- M1: Calculation of \(A_r(M)\): \(277.4 - 126.0 - 96.1 = 55.3\) (accept range \(54.0 - 57.0\) depending on rounding) (1)
- M2: Identification: Iron / \(\text{Fe}\) (1)

(c)
- M1: Calculate total moles of gas produced = \(0.125 \times 1.5 = 0.1875\text{ mol}\) (1)
- M2: Multiply by molar gas volume (24.0) (1)
- M3: Final answer: \(4.50\text{ dm}^3\) (1)

(a) (i) The dot-and-cross diagram for \(\text{PCl}_3\) shows a central phosphorus atom bonded to three chlorine atoms by single covalent bonds (each consisting of one dot and one cross). The phosphorus atom has a lone pair of electrons (two dots or two crosses) remaining in its outer shell. Each chlorine atom has three lone pairs of electrons (total of 8 electrons in each Cl outer shell).

(ii) Shape: Trigonal pyramidal.
Bond angle: \(107^\circ\) (accept any value from \(100^\circ\) to \(108^\circ\)).
Explanation:
- The central phosphorus atom has four electron pairs in its valence shell: three bonding pairs and one lone pair.
- Electron pairs repel each other to be as far apart as possible to minimise repulsion.
- Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, which pushes the bonding pairs closer together, reducing the bond angle from the tetrahedral angle of \(109.5^\circ\).

(iii) Shape: Trigonal bipyramidal.
Bond angles: \(90^\circ\) and \(120^\circ\) (also accept \(180^\circ\)).

(b) (i) \([\text{PCl}_4]^+\): Tetrahedral.
\([\text{PCl}_6]^-\): Octahedral.

(ii)
- \([\text{PCl}_4]^+\) has four bonding pairs of electrons and zero lone pairs around the central phosphorus atom, whereas \(\text{PCl}_3\) has three bonding pairs and one lone pair.
- In \([\text{PCl}_4]^+\), all four electron pairs repel equally, resulting in a perfect tetrahedral shape with a bond angle of \(109.5^\circ\).
- In \(\text{PCl}_3\), the presence of the lone pair causes greater repulsion, squeezing the bond angle to a smaller value (approx. \(107^\circ\)).

(c)
- The \(\text{P-Cl}\) bonds are polar because chlorine is more electronegative than phosphorus, leading to an unequal sharing of electrons and a dipole along each bond.
- The molecule as a whole is polar because its trigonal pyramidal shape is asymmetrical, meaning the individual dipoles do not cancel out, resulting in a net dipole moment.

(a)(i)
- M1: Three shared pairs of electrons (each containing one P and one Cl electron) (1)
- M2: One lone pair on P and three lone pairs on each of the three Cl atoms (1)

(a)(ii)
- M1: Shape: Trigonal pyramidal AND Bond angle: \(107^\circ\) (allow \(100^\circ\) to \(108^\circ\)) (1)
- M2: Phosphorus has 3 bonding pairs and 1 lone pair (1)
- M3: Electron pairs repel to get as far apart as possible (1)
- M4: Lone pairs repel more than bonding pairs (reducing the bond angle) (1)

(a)(iii)
- M1: Shape: Trigonal bipyramidal (1)
- M2: Bond angles: \(90^\circ\) and \(120^\circ\) (both needed) (1)

(b)(i)
- M1: \([\text{PCl}_4]^+\) is tetrahedral (1)
- M2: \([\text{PCl}_6]^-\) is octahedral (1)

(b)(ii)
- M1: \([\text{PCl}_4]^+\) has 4 bonding pairs and 0 lone pairs; \(\text{PCl}_3\) has 3 bonding pairs and 1 lone pair (1)
- M2: In \([\text{PCl}_4]^+\), repulsion between the four bonding pairs is equal, giving a \(109.5^\circ\) bond angle (1)
- M3: In \(\text{PCl}_3\), the lone pair repels the bonding pairs more, reducing the bond angle (1)

(c)
- M1: \(\text{P-Cl}\) bonds are polar because Cl is more electronegative than P (1)
- M2: The molecule is polar because it is asymmetrical / trigonal pyramidal, so bond dipoles do not cancel out (1)

(a) (i) Stereoisomers are molecules that have the same structural formula (and molecular formula) but have a different arrangement of their atoms in space.

(ii)
- (Z)-but-2-ene has the two methyl groups (\(-\text{CH}_3\)) on the same side of the \(\text{C=C}\) double bond (both pointing 'up' or 'down', with \(-\text{H}\) atoms on the opposite side).
- (E)-but-2-ene has the two methyl groups on opposite sides of the \(\text{C=C}\) double bond (diagonal to each other).

(iii)
- But-2-ene has restricted rotation around the \(\text{C=C}\) double bond due to the presence of the \(\pi\)-bond. Additionally, each carbon atom of the double bond is attached to two different groups (a \(-\text{CH}_3\) group and a \(-\text{H}\) atom).
- In but-1-ene, one of the carbon atoms in the double bond is attached to two identical hydrogen atoms, meaning that switching their positions does not result in a new spatial arrangement.

(b) (i) Mechanism:
- The double bond (\(\text{C=C}\)) is an area of high electron density. A curly arrow is drawn starting from the double bond to the hydrogen atom of the \(\text{H-Br}\) molecule.
- The \(\text{H-Br}\) bond is polar, showing \(\delta+\) on the \(\text{H}\) atom and \(\delta-\Default\) on the \(\text{Br}\) atom. A curly arrow is drawn from the \(\text{H-Br}\) covalent bond to the \(\text{Br}\) atom.
- This forms a carbocation intermediate: \(\text{CH}_3-\text{CH}^+-\text{CH}_2-\text{CH}_3\) (secondary carbocation) and a bromide ion, \(\text{Br}^-\), which has a lone pair of electrons.
- A curly arrow is drawn from the lone pair of the \(\text{Br}^-\) ion to the positively charged carbon atom of the carbocation.
- The final product is 2-bromobutane.
- Name of mechanism: Electrophilic addition.

(ii)
- Major product: 2-bromobutane.
- Explanation: The reaction proceeds via two possible carbocation intermediates. The addition of \(\text{H}^+\) to carbon-1 forms a secondary carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\)), whereas the addition of \(\text{H}^+\) to carbon-2 forms a primary carbocation (\(\text{CH}_2^+\text{CH}_2\text{CH}_2\text{CH}_3\)).
- The secondary carbocation is more stable than the primary carbocation because it has two electron-donating alkyl groups attached to the positive carbon, which help disperse the positive charge (known as the positive inductive effect).
- Therefore, the secondary carbocation intermediate is formed preferentially, leading to 2-bromobutane as the major product.

(a)(i)
- M1: Same structural formula / same connectivity (1)
- M2: Different spatial arrangement of atoms / different 3D arrangement (1)

(a)(ii)
- M1: Correctly drawn structure of (Z)-but-2-ene with methyl groups on same side (1)
- M2: Correctly drawn structure of (E)-but-2-ene with methyl groups on opposite sides (1)

(a)(iii)
- M1: Restricted rotation about the \(\text{C=C}\) double bond (1)
- M2: Each carbon of the double bond is attached to two different groups in but-2-ene, but in but-1-ene one carbon is attached to two identical groups (H atoms) (1)

(b)(i)
- M1: Curly arrow from \(\text{C=C}\) double bond to \(\text{H}\) of \(\text{H-Br}\) AND correct dipoles on \(\text{H-Br}\) (\(\delta+\) on H, \(\delta-\) on Br) (1)
- M2: Curly arrow from \(\text{H-Br}\) bond to \(\text{Br}\) (1)
- M3: Correct carbocation intermediate AND \(\text{Br}^-\) with lone pair shown (1)
- M4: Curly arrow from lone pair on \(\text{Br}^-\) to \(\text{C}^+\) of the carbocation (1)
- M5: Name of mechanism: Electrophilic addition (1)

(b)(ii)
- M1: Major product: 2-bromobutane (1)
- M2: Reaction goes via a secondary carbocation rather than a primary carbocation (1)
- M3: Secondary carbocations are more stable than primary carbocations (1)
- M4: Due to the positive inductive effect of two alkyl groups (compared to only one in primary) (1)

Using Hess's Law, the enthalpy change of formation can be calculated from the enthalpy changes of combustion of the reactants minus those of the products:

\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \Delta_c H^\theta(\text{products})\)

\(\Delta_f H^\theta = [\Delta_c H^\theta(\text{C}) + 2 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{CH}_3\text{OH})\)

\(\Delta_f H^\theta = [-394 + 2(-286)] - (-726)\)

\(\Delta_f H^\theta = [-394 - 572] + 726 = -966 + 726 = -240\text{ kJ mol}^{-1}\)

Thus, the correct option is A.

1 mark: Correct calculation of \(-240\text{ kJ mol}^{-1}\) leading to option A.

Propan-1-ol (option C) contains an \(\text{O-H}\) bond and can form hydrogen bonds between its molecules. Hydrogen bonding is the strongest type of intermolecular force among those present in these molecules (compared to London forces only in butane, and permanent dipole-dipole forces in fluoropropane and methoxyethane). Therefore, propan-1-ol requires the most thermal energy to overcome these intermolecular forces, resulting in the highest boiling temperature.

1 mark: Correctly identifies propan-1-ol (option C) as having the highest boiling temperature.

Magnesium nitrate undergoes thermal decomposition to form magnesium oxide (solid), nitrogen dioxide (brown gas), and oxygen (colorless gas):

\(2\text{Mg(NO}_3)_2\text{(s)} \rightarrow 2\text{MgO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\)

Magnesium nitrate decomposes more easily than barium nitrate because the magnesium ion, \(\text{Mg}^{2+}\), has a smaller ionic radius than the barium ion, \(\text{Ba}^{2+}\). Therefore, \(\text{Mg}^{2+}\) has a higher charge density and polarises the nitrate ion more strongly, weakening the \(\text{N-O}\) bond and decreasing the thermal stability. Thus, option B is correct.

1 mark: Correctly identifies the observation of brown gas and the explanation based on the higher charge density of the magnesium ion (option B).

Iodide ions are very strong reducing agents and reduce sulfuric acid (where sulfur has an oxidation state of +6) to sulfur dioxide (+4), elemental sulfur (0), and hydrogen sulfide (-2). In contrast, chloride ions are weaker reducing agents and cannot reduce concentrated sulfuric acid at all (they only undergo an acid-base reaction to produce \(\text{HCl}\) gas). Potassium bromide reduces sulfuric acid only as far as sulfur dioxide (+4). Therefore, potassium iodide reduces sulfur in sulfuric acid to a wider range of lower oxidation states than potassium chloride.

1 mark: Correctly identifies option D.

At a higher temperature, the average kinetic energy of the molecules increases, meaning the most probable energy increases. Thus, the peak of the Maxwell-Boltzmann distribution curve shifts to the right. Since the total number of molecules remains constant, the area under the curve must stay the same, which requires the curve to flatten and the peak to become lower. Hence, option A is correct.

1 mark: Correctly identifies that the peak shifts to the right and becomes lower (option A).

The rate of hydrolysis of halogenoalkanes is determined by:
1. The strength of the carbon-halogen (C-X) bond: Since C-I is the weakest bond, iodoalkanes hydrolyse much faster than bromoalkanes and chloroalkanes.
2. The class of the halogenoalkane: Tertiary halogenoalkanes undergo nucleophilic substitution via the \(S_N1\) mechanism, which is significantly faster than the \(S_N2\) mechanism of primary halogenoalkanes due to the stability of the tertiary carbocation intermediate.

Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) will react the fastest and form a yellow precipitate of silver iodide first.

1 mark: Correctly identifies 2-iodo-2-methylpropane (option B) as reacting the fastest.

The color change from orange to green shows that oxidation has taken place, ruling out the tertiary alcohol 2-methylpropan-2-ol, \(\text{(CH}_3)_3\text{COH}\) (option B), which resists oxidation.
Under reflux, primary alcohols, such as butan-1-ol (option A) and 2-methylpropan-1-ol (option D), are oxidized to carboxylic acids, which would react with sodium carbonate to release carbon dioxide gas.
A secondary alcohol, such as butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) (option C), is oxidized to a ketone (butanone), which does not react with sodium carbonate. Thus, the original compound must be butan-2-ol.

1 mark: Correctly identifies the secondary alcohol (option C) as the correct structure.

The broad infrared peak in the range \(3200-3750\text{ cm}^{-1}\) indicates the presence of an \(\text{O-H}\) group (alcohol), which rules out methoxyethane (option C) and propanal (option D).
The fragment peak at \(m/z = 31\) corresponds to the stable cation \([\text{CH}_2\text{OH}]^+\), which is formed by the fragmentation (cleavage of the \(\text{C-C}\) bond alpha to the oxygen) of primary alcohols like propan-1-ol:

\(\text{CH}_3\text{CH}_2-\text{CH}_2\text{OH}^+ \rightarrow \text{CH}_3\text{CH}_2^\bullet + [\text{CH}_2\text{OH}]^+\)

Propan-2-ol (option B) fragments to produce a major peak at \(m/z = 45\) corresponding to the \([\text{CH(OH)CH}_3]^+\) fragment. Therefore, the compound is propan-1-ol.

1 mark: Correctly identifies propan-1-ol (option A) based on the IR and mass spectrometry evidence.

The equation for the formation of ethane is: 2C(s) + 3H2(g) -> C2H6(g). Using enthalpies of combustion: Enthalpy of formation = [2 * (-394) + 3 * (-286)] - (-1560) = [-788 - 858] + 1560 = -1646 + 1560 = -86 kJ mol^{-1}. Thus, the correct option is A.

1 mark for the correct application of Hess's Law and calculation of the enthalpy of formation.

Down Group 2, cationic radius increases, so charge density and polarizing power decrease. The cation polarizes the nitrate anion less, making it more thermally stable. Thus, thermal stability increases, matching option A.

1 mark for identifying that stability increases down the group and explaining it via decreasing polarizing power of the larger cation.

Butane (CH3CH2CH2CH3) only has weak London forces. 1-chloropropane (CH3CH2CH2Cl) has permanent dipole-dipole forces in addition to London forces. Propan-1-ol (CH3CH2CH2OH) has hydrogen bonding in addition to London forces and permanent dipole-dipole forces. Therefore, boiling point increases in the order: butane < 1-chloropropane < propan-1-ol, which is option A.

1 mark for correctly ranking the compounds based on the strength of their intermolecular forces.

A yellow precipitate indicates silver iodide, so the halogenoalkane must be an iodoalkane. The C-I bond is the weakest C-halogen bond, so it hydrolyses the fastest. Tertiary halogenoalkanes react faster than primary or secondary ones due to the highly stable tertiary carbocation intermediate in SN1. Thus, 2-iodo-2-methylpropane is correct (option D).

1 mark for identifying the correct halogenoalkane based on precipitate color and rate of nucleophilic substitution.

The Maxwell-Boltzmann distribution curve depends only on temperature, so adding a catalyst does not change the shape or position of the curve. A catalyst lowers the activation energy, which shifts the activation energy line (Ea) to the left on the energy axis. This matches option C.

1 mark for identifying that a catalyst lowers the activation energy, effectively shifting its position on the distribution to the left.

2-methylpropan-2-ol, (CH3)3COH, has a molecular mass of 74. Loss of a methyl group (15) gives a highly stable tertiary carbocation/oxonium ion [(CH3)2C=OH]+ at m/z = 59. Because it lacks an ethyl group on the central carbon, it cannot lose an ethyl group to produce the fragment [CH3CH=OH]+ at m/z = 45 (which is common for butan-2-ol). Therefore, the correct alcohol is 2-methylpropan-2-ol, option D.

1 mark for deducing the identity of the alcohol from its fragmentation pattern.

Reaction of Cl2 with hot concentrated NaOH produces NaCl and NaClO3: 3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O. The oxidation state of Cl in NaCl is -1, and in NaClO3 is +5. Thus, the correct option is C.

1 mark for identifying the correct oxidation states of chlorine in hot concentrated sodium hydroxide reaction products.

Heat released, q = m * c * delta T = 50.0 g * 4.18 J g^{-1} K^{-1} * 10.5 K = 2194.5 J = 2.1945 kJ. Moles of CuSO4 = 0.200 * 0.0500 = 0.0100 mol. Since zinc is in excess, CuSO4 is limiting. Delta H = -q / n = -2.1945 kJ / 0.0100 mol = -219.45 kJ mol^{-1}. To 3 significant figures, this is -219 kJ mol^{-1}. This matches option A.

1 mark for calculating the correct molar enthalpy change with the correct sign and to 3 significant figures.

The rate of hydrolysis of halogenoalkanes is determined by the bond enthalpy of the carbon-halogen bond rather than its bond polarity. As we descend Group 7, the carbon-halogen bond enthalpy decreases because the halogen atoms become larger, resulting in longer and weaker bonds. The C-I bond is the weakest and requires the least activation energy to break, so 1-iodobutane undergoes hydrolysis the fastest. The C-Cl bond is the strongest, so 1-chlorobutane reacts the slowest.

[1 Mark] Correct option is b. Option a is incorrect because bond enthalpy, not polarity, is the dominant factor. Option c is incorrect because 1-chlorobutane reacts slowest. Option d is incorrect because 1-iodobutane reacts fastest.

As Group 2 is descended, the ionic radius of the metal cation increases while the 2+ charge remains constant, which decreases its charge density. Consequently, the larger cation causes less polarization (distortion) of the electron cloud of the nitrate anion, making the N-O covalent bonds within the anion harder to break. Therefore, more thermal energy is required to decompose the nitrate, meaning thermal stability increases down the group.

[1 Mark] Correct option is b. Option a is incorrect because thermal stability increases down the group. Option c is incorrect because stability increases and the explanation is faulty. Option d is incorrect because electronegativity decreases down the group and does not explain the trend in polarization.

a) i) \( q = m c \Delta T = 150.0 \times 4.18 \times (54.0 - 18.5) = 22258.5 \text{ J} = 22.3 \text{ kJ} \). ii) \( n = \frac{1.32}{88.0} = 0.0150 \text{ mol} \). \( \Delta_c H = -\frac{22.2585 \text{ kJ}}{0.0150 \text{ mol}} = -1483.9 \text{ kJ mol}^{-1} \). To 3 significant figures, this is \( -1480 \text{ kJ mol}^{-1} \). iii) Heat loss to the surroundings / calorimeter; incomplete combustion of the alcohol; loss of alcohol due to evaporation from the wick. b) \( \text{C}_5\text{H}_{11}\text{OH}(l) + 7.5\text{O}_2(g) \rightarrow 5\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \). \( \Delta_c H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants}) = [5(-394) + 6(-286)] - [-354] = [-1970 - 1716] + 354 = -3332 \text{ kJ mol}^{-1} \). c) \( 5\text{C}(\text{s, graphite}) + 6\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_5\text{H}_{11}\text{OH}(l) \).

a) i) M1: correct expression for heat transferred: \( q = 150.0 \times 4.18 \times 35.5 \) (1) M2: evaluation to \( 22.3 \text{ kJ} \) (accept \( 22.26 \text{ kJ} \) or \( 22 \text{ kJ} \)) (1). ii) M1: calculates moles of pentan-1-ol = \( 0.0150 \text{ mol} \) (1) M2: divides heat energy by moles to give \( 1480 \) (1) M3: negative sign and 3 significant figures: \( -1480 \text{ kJ mol}^{-1} \) (1). iii) Any two from: Heat loss to surroundings (1) Incomplete combustion of alcohol (1) Evaporation of alcohol from the wick (1). b) M1: correct balanced equation for combustion OR correct Hess cycle representation (1) M2: correct calculation of sum of products: \( 5(-394) + 6(-286) = -3686 \text{ kJ mol}^{-1} \) (1) M3: final answer of \( -3332 \text{ kJ mol}^{-1} \) (1). c) M1: left-hand side formula and state symbols: \( 5\text{C}(\text{s}) + 6\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \) (accept graphite for s) (1) M2: right-hand side formula and state symbols: \( \text{C}_5\text{H}_{11}\text{OH}(l) \) (1).

a) i) Thermal stability increases down Group 2 from magnesium carbonate to barium carbonate. As you go down the group, the cationic radius increases while the charge remains 2+. Therefore, the charge density of the cation decreases. The polarization of the carbonate ion by the cation decreases, which makes the C-O bonds in the carbonate ion stronger and harder to break, requiring more heat energy to decompose. ii) \( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \). b) i) Observation: The colourless solution turns brown (or a dark grey precipitate forms). Ionic equation: \( \text{Cl}_2(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{I}_2(aq) \) (or \( \text{I}_2(s) \)). ii) Sulfur-containing products: 1. Sulfur dioxide, \( \text{SO}_2 \): observation is a choking gas/effervescence. 2. Sulfur, \( \text{S} \): observation is a yellow solid. 3. Hydrogen sulfide, \( \text{H}_2\text{S} \): observation is a gas with a smell of rotten eggs.

a) i) M1: thermal stability increases down the group (1) M2: cationic radius increases / charge density decreases (1) M3: polarizing power of the cation decreases (1) M4: less distortion/polarization of the carbonate ion (1). ii) M1: correct balanced equation with all state symbols: \( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \) (1). b) i) M1: observation: solution turns brown / brown solution forms / grey precipitate forms (1) M2: correct ionic equation: \( \text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2 \) (1) M3: all state symbols correct: \( (aq) \) for reactants and \( (aq) \) or \( (s) \) for products (1). ii) M1 & M2: any two correct sulfur-containing products (sulfur dioxide, sulfur, hydrogen sulfide) (2) M3 & M4: correct corresponding observations for each chosen product (choking gas, yellow solid, rotten egg smell) (2).

a) i) Nucleophilic substitution (specifically SN2). ii) Mechanism: The hydroxide ion (\( \text{OH}^- \)) acts as the nucleophile. There is a delta-positive (\( \delta^+ \)) charge on the carbon atom bonded to the chlorine, and a delta-negative (\( \delta^- \)) charge on the chlorine atom. A curly arrow must be drawn from a lone pair on the oxygen of the hydroxide ion to the carbon atom. A second curly arrow must be drawn from the C-Cl bond to the chlorine atom. The products are butan-1-ol and a chloride ion (\( \text{Cl}^- \)). b) i) Ethanol acts as a common solvent / cosolvent to dissolve both the water-insoluble halogenoalkane and the aqueous silver nitrate solution. ii) 1-iodobutane reacts the fastest and forms a yellow precipitate. 1-bromobutane reacts at an intermediate rate and forms a cream precipitate. 1-chlorobutane reacts the slowest and forms a white precipitate. iii) The rate of hydrolysis depends on the carbon-halogen (C-X) bond strength/enthalpy. The C-Cl bond is the strongest because chlorine is the smallest halogen and has the shortest bond length, while the C-I bond is the weakest. The weaker the bond, the lower the activation energy, and the faster the bond is broken, leading to a faster rate of reaction.

a) i) M1: nucleophilic substitution (1). ii) M1: correct dipole \( \delta^+ \) on carbon and \( \delta^- \) on chlorine, and lone pair on hydroxide oxygen (1) M2: curly arrow from lone pair on hydroxide oxygen to the carbon atom (1) M3: curly arrow from the C-Cl bond to the chlorine atom (1) M4: correct products: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \) and \( \text{Cl}^- \) (1). b) i) M1: solvent / cosolvent to allow reactants to mix (1). ii) M1: 1-iodobutane is fastest / forms yellow precipitate (1) M2: 1-bromobutane is intermediate / forms cream precipitate (1) M3: 1-chlorobutane is slowest / forms white precipitate (1). iii) M1: rate depends on C-X bond strength / bond enthalpy (not bond polarity) (1) M2: C-Cl bond is strongest / C-I bond is weakest (1) M3: weaker bonds break more easily / require less energy to break, leading to a faster rate down the group (1).

a) i) Compound **X** is oxidized by acidified potassium dichromate(VI) because the solution turns green, meaning **X** is either a primary or a secondary alcohol. Since compound **Y** does not react with Fehling's solution, **Y** must be a ketone (as aldehydes react with Fehling's solution, but ketones do not). This means **X** must be a secondary alcohol. The only secondary alcohol with molecular formula \( \text{C}_4\text{H}_{10}\text{O} \) is butan-2-ol. Structural formula of **X**: \( \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \). ii) Equation: \( \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_2\text{CH}_3 + \text{H}_2\text{O} \). b) i) Compound **Z** is a tertiary alcohol, 2-methylpropan-2-ol. Tertiary alcohols are resistant to oxidation because the carbon atom bonded to the -OH group does not have a hydrogen atom attached to it, so a C-C bond would have to be broken to oxidize it, which requires too much energy. ii) The displayed formula of 2-methylpropan-2-ol shows three methyl groups attached to a central carbon atom, which is also bonded to an oxygen, which is bonded to a hydrogen. All single bonds (C-H, C-C, C-O, O-H) must be explicitly drawn. c) The IR spectrum of **X** (butan-2-ol) will show a broad absorption band for the O-H (alcohol) bond in the range \( 3200-3750 \text{ cm}^{-1} \), which is absent in **Y**. The IR spectrum of **Y** (butanone) will show a strong absorption band for the C=O (carbonyl) bond in the range \( 1675-1750 \text{ cm}^{-1} \), which is absent in **X**.

a) i) M1: deduction that **X** is a secondary alcohol (1) M2: explanation that **Y** is a ketone because it does not react with Fehling's solution (1) M3: name of **X**: butan-2-ol (1) M4: structural formula of **X**: \( \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \) (1). ii) M1: correct reactants and products (1) M2: balanced equation with water as a product (1). b) i) M1: identification of **Z** as 2-methylpropan-2-ol / tertiary alcohol (1) M2: explanation that tertiary alcohols lack a hydrogen atom on the carbon carrying the -OH group (1). ii) M1: correct displayed formula showing all atoms and all bonds (C-C, C-H, C-O, O-H) (1). c) M1: O-H bond in **X** at \( 3200-3750 \text{ cm}^{-1} \) (1) M2: C=O bond in **Y** at \( 1675-1750 \text{ cm}^{-1} \) (1) M3: stating that O-H is in **X** but not **Y**, and C=O is in **Y** but not **X** (1).

a) i) The y-axis should be labelled 'number of molecules' or 'fraction of molecules'. The x-axis should be labelled 'energy' or 'kinetic energy'. The curve starts at the origin (0,0), rises to a single peak, and slopes down towards the x-axis, becoming asymptotic to it (it must not touch the x-axis or cross it). The activation energy, \( E_a \), should be represented by a vertical line on the right side of the peak. ii) The curve for \( T_2 \) must have a lower peak that is shifted to the right of the peak for \( T_1 \). It must start at the origin, cross the \( T_1 \) curve only once, and lie above the \( T_1 \) curve at all energies higher than the intersection point. iii) A catalyst provides an alternative reaction pathway with a lower activation energy, which can be represented as \( E_{cat} \) to the left of \( E_a \) on the x-axis. As a result, a larger fraction of molecules have kinetic energy greater than or equal to this lower activation energy (the area under the curve to the right of \( E_{cat} \) is larger), leading to more frequent successful collisions and a faster rate of reaction. b) i) Increasing the pressure shifts the position of equilibrium to the right (towards the product, \( \text{SO}_3 \)). This is because there are fewer moles of gas on the right-hand side (2 moles) than on the left-hand side (3 moles). The system shifts to reduce the pressure. ii) The value of the equilibrium constant, \( K_c \), decreases. Since the forward reaction is exothermic (\( \Delta H = -197 \text{ kJ mol}^{-1} \)), increasing the temperature shifts the equilibrium in the endothermic (reverse) direction to absorb the added heat. This decreases the concentration of the products (\( \text{SO}_3 \)) and increases the concentration of the reactants (\( \text{SO}_2 \) and \( \text{O}_2 \)), thus reducing the value of \( K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} \).

a) i) M1: y-axis labelled 'number of molecules' / 'fraction of molecules' AND x-axis labelled 'energy' / 'kinetic energy' (1) M2: correct shape of curve starting at origin and not touching x-axis at high energy (1) M3: \( E_a \) marked with a vertical line on the x-axis (1). ii) M1: peak of \( T_2 \) is lower than peak of \( T_1 \) and shifted to the right (1) M2: \( T_2 \) curve only crosses \( T_1 \) curve once and is higher than \( T_1 \) at high energies (1). iii) M1: catalyst lowers the activation energy / provides alternative pathway with lower activation energy (1) M2: a greater fraction of molecules have energy \( \ge E_{cat} \) (or more frequent successful collisions) (1). b) i) M1: position of equilibrium shifts to the right (1) M2: because there are fewer moles of gas on the product side (2 vs 3 moles on reactant side) (1). ii) M1: \( K_c \) decreases (1) M2: forward reaction is exothermic (1) M3: equilibrium shifts to the left / in the endothermic direction to oppose the temperature rise, decreasing product concentration relative to reactant concentration (1).

(a) Zinc is added in excess to ensure that all the copper(II) ions in the CuSO4 solution react completely. This makes CuSO4 the limiting reagent, allowing for accurate stoichiometric calculations based on CuSO4. (b) A polystyrene cup is a much better thermal insulator than a glass beaker, which significantly reduces the rate of heat loss to the surroundings during the reaction. (c) Extrapolation accounts for the heat lost to the surroundings during the time it takes for the reaction to reach its maximum temperature, providing a more accurate estimate of the theoretical maximum temperature change. (d) Temperature change: dT = 33.5 - 19.5 = 14.0 degrees C. Mass of solution, m = 50.0 g. Heat energy released: q = m * c * dT = 50.0 * 4.18 * 14.0 = 2926 J = 2.926 kJ. Moles of CuSO4: n = conc * vol = 0.200 mol dm-3 * 0.0500 dm3 = 0.0100 mol. Enthalpy change: dH = -q / n = -2.926 kJ / 0.0100 mol = -293 kJ mol-1. (e) Two temperature readings are taken (initial and maximum), so the total uncertainty in dT is 2 * (+/- 0.1 degrees C) = +/- 0.2 degrees C. Percentage uncertainty = (0.2 / 14.0) * 100 = 1.43%.

(a) [1 mark] To ensure all CuSO4 / copper(II) ions react completely / CuSO4 is the limiting reagent. (b) [1 mark] Polystyrene is a good thermal insulator / reduces heat transfer to surroundings. (c) [1 mark] To compensate/correct for heat loss during the reaction. (d) [2 marks] Calculating q = 50.0 * 4.18 * 14.0 = 2926 J (1 mark for correct q formula and values; 1 mark for correct value 2.93 kJ). [1 mark] Moles of CuSO4 = 0.0100 mol. [2 marks] dH = -293 (kJ mol-1) (1 mark for dividing q by moles; 1 mark for correct sign and 3 significant figures). (e) [2 marks] Total uncertainty = 0.2 degrees C (1 mark). Percentage uncertainty = (0.2 / 14.0) * 100 = 1.43% (1 mark; accept 1.4%).

(a) The addition of concentrated sulfuric acid to water/alcohols is highly exothermic. Adding it slowly and cooling in an ice bath prevents the mixture from boiling violently, splashing, or evaporating the volatile organic reactants. (b) The reflux apparatus must have: a round-bottomed or pear-shaped flask heated from below; a vertical condenser securely fitted; water flowing into the condenser jacket at the bottom and out at the top; and the top of the condenser must remain open to the atmosphere (not stoppered) to prevent pressure build-up. (c) (i) The bottom layer is the organic layer because 1-bromobutane has a higher density (1.27 g cm-3) than water (1.00 g cm-3). (ii) Transfer the mixture to a separating funnel, allow the layers to separate cleanly, open the tap to run off the lower organic layer into a clean container, and close the tap before the upper aqueous layer reaches the outlet. (d) (i) Washing with aqueous sodium hydrogencarbonate neutralizes any remaining acidic impurities (such as sulfuric acid or hydrobromic acid). (ii) Neutralization produces carbon dioxide gas, which increases pressure inside the funnel; opening the stopper vents this gas and prevents the stopper from blowing out. (e) A suitable drying agent is anhydrous calcium chloride (or anhydrous sodium sulfate / magnesium sulfate).

(a) [1 mark] Highly exothermic reaction / prevents violent boiling / splashing / evaporation of reactants. (b) [3 marks] Reflux features: flask heated from below with vertical condenser (1 mark); water in at bottom, out at top of condenser (1 mark); top of condenser open/not stoppered (1 mark). (c)(i) [1 mark] Bottom layer is organic as 1-bromobutane is more dense than water. (c)(ii) [2 marks] Use separating funnel (1 mark); let layers separate and run off bottom layer through tap (1 mark). (d)(i) [1 mark] To neutralize acidic impurities / H2SO4 / HBr. (d)(ii) [1 mark] To release/vent CO2 gas / relieve pressure. (e) [1 mark] Anhydrous calcium chloride / anhydrous magnesium sulfate / anhydrous sodium sulfate.

(a) Salt X: The lilac flame test indicates the presence of potassium ions, K+. The cream precipitate with silver nitrate that dissolves only in concentrated ammonia indicates the presence of bromide ions, Br-. Thus, X is potassium bromide (KBr). (b) Salt Y: The apple-green flame test indicates the presence of barium ions, Ba2+. The white precipitate with silver nitrate that dissolves in dilute ammonia indicates the presence of chloride ions, Cl-. Thus, Y is barium chloride (BaCl2). (c) Salt Z: The gas evolved in Test 4 that turns damp red litmus paper blue is ammonia, indicating the presence of ammonium ions, NH4+. The dense white precipitate formed with acidified barium chloride in Test 3 indicates the presence of sulfate ions, SO42-. The chemical formula of Z is (NH4)2SO4. (d) For Test 2 with Y (barium chloride), the precipitation reaction is between silver ions and chloride ions to form silver chloride: Ag+(aq) + Cl-(aq) -> AgCl(s).

(a) [3 marks] Cation = potassium / K+ (1 mark); Anion = bromide / Br- (1 mark); lilac flame shows K+ and cream ppt dissolving in conc NH3 shows Br- (1 mark). (b) [2 marks] Cation = barium / Ba2+ (1 mark); Anion = chloride / Cl- (1 mark). Accept barium chloride. (c) [3 marks] Cation = ammonium / NH4+ (1 mark); Anion = sulfate / SO42- (1 mark); Formula = (NH4)2SO4 (1 mark). (d) [2 marks] Ag+(aq) + Cl-(aq) -> AgCl(s). Correct formulas (1 mark); correct state symbols (1 mark).

(a) Methyl orange in basic solution (calcium hydroxide) starts yellow and turns orange (or peach) at the end-point. (b) The concordant titres are Titrations 1, 2, and 3 (all within 0.10 cm3 of each other). Mean titre = (20.45 + 20.35 + 20.40) / 3 = 20.40 cm3. (c) Ca(OH)2 + 2HCl -> CaCl2 + 2H2O. (d) (i) Moles of HCl = conc * vol = 0.0500 * (20.40 / 1000) = 1.02 * 10^-3 mol. From the equation, 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Moles of Ca(OH)2 = (1.02 * 10^-3) / 2 = 5.10 * 10^-4 mol. Concentration of Ca(OH)2 = moles / volume = (5.10 * 10^-4) / 0.0250 = 0.0204 mol dm-3. (ii) Molar mass of Ca(OH)2 = 40.1 + 2 * (16.0 + 1.0) = 74.1 g mol-1. Concentration in g dm-3 = 0.0204 mol dm-3 * 74.1 g mol-1 = 1.51 g dm-3. (e) Percentage uncertainty = (0.06 / 25.0) * 100 = 0.24%. (f) Adding distilled water does not change the number of moles of calcium hydroxide present in the flask; it only dilutes the solution, so the amount of acid required remains unchanged.

(a) [1 mark] Yellow to orange/peach. (b) [2 marks] Selecting Titrations 1, 2, and 3 (1 mark); mean titre = 20.40 cm3 (1 mark). (c) [1 mark] Ca(OH)2 + 2HCl -> CaCl2 + 2H2O. (d)(i) [3 marks] Moles of HCl = 1.02 * 10^-3 (1 mark); Moles of Ca(OH)2 = 5.10 * 10^-4 (1 mark); Concentration = 0.0204 mol dm-3 (1 mark). (d)(ii) [1 mark] 1.51 g dm-3 (accept answers based on calculated concentration multiplied by 74.1). (e) [1 mark] (0.06 / 25.0) * 100 = 0.24%. (f) [1 mark] It does not change the number of moles of calcium hydroxide / reactants in the flask.

(a) The functional group is a hydroxyl group / alcohol (-OH). Reasoning: Effervescence with sodium indicates an acidic proton (found in both alcohols and carboxylic acids) releasing hydrogen gas. No reaction with sodium hydrogencarbonate rules out carboxylic acids (which would react to produce carbon dioxide gas). Thus, L is an alcohol. (b) L is a secondary alcohol. Reasoning: Test 3 shows L is oxidized by acidified potassium dichromate(VI) (indicated by the orange to green color change), ruling out tertiary alcohols (which cannot be oxidized). Test 4 shows that the oxidation product of L does not react with Fehling's solution, which indicates the product is a ketone rather than an aldehyde. Since primary alcohols oxidize to aldehydes, L must be a secondary alcohol. (c) The simplest straight-chain secondary alcohol with a chiral center is butan-2-ol (or 2-butanol). The chiral carbon is Carbon-2, which is bonded to four different groups: -H, -OH, -CH3, and -CH2CH3. (d) The oxidation product of butan-2-ol is butanone. Its skeletal formula consists of a four-carbon chain with a carbonyl group (C=O) on the second carbon.

(a) [3 marks] Hydroxyl / alcohol / -OH group (1 mark); sodium reaction shows acidic H / presence of alcohol or carboxylic acid (1 mark); no reaction with NaHCO3 rules out carboxylic acid (1 mark). (b) [3 marks] Secondary alcohol (1 mark); Test 3 shows L can be oxidized, ruling out tertiary alcohols (1 mark); Test 4 shows the oxidized product is a ketone / not an aldehyde, ruling out primary alcohols (1 mark). (c) [2 marks] Butan-2-ol / 2-butanol (1 mark); Carbon-2 is the chiral carbon because it is bonded to four different groups / -H, -OH, -CH3, -C2H5 (1 mark). (d) [2 marks] Butanone skeletal structure: four-carbon chain (1 mark); C=O double bond on C2 (1 mark).

From the experimental evidence, the reaction is first order with respect to both \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\). Therefore, the overall rate equation is: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). Solving for \(k\): \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}\). Substituting units: \(\text{Units of } k = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3 \text{mol}^{-1} \text{s}^{-1}\).

[1 mark] for correct calculation of the second-order rate constant units.

The logarithmic form of the Arrhenius equation is \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). The gradient \(m\) is equal to \(-\frac{E_a}{R}\). Therefore: \(-1.20 \times 10^4 = -\frac{E_a}{8.31}\), which gives \(E_a = 1.20 \times 10^4 \times 8.31 = 99720 \text{ J mol}^{-1} = +99.7 \text{ kJ mol}^{-1}\).

[1 mark] for correct calculation of the activation energy with correct sign and units.

First, calculate the standard entropy change of the surroundings: \(\Delta S_{\text{surrounding}}^{\ominus} = -\frac{\Delta H^{\ominus}}{T} = -\frac{-115000 \text{ J mol}^{-1}}{298 \text{ K}} = +385.9 \text{ J K}^{-1} \text{mol}^{-1}\). Next, calculate the total standard entropy change: \(\Delta S_{\text{total}}^{\ominus} = \Delta S_{\text{system}}^{\ominus} + \Delta S_{\text{surrounding}}^{\ominus} = -145 + 385.9 = +240.9 \text{ J K}^{-1} \text{mol}^{-1}\), which rounds to \(+241 \text{ J K}^{-1} \text{mol}^{-1}\).

[1 mark] for correct application of total entropy formula to obtain +241 J K-1 mol-1.

The expression for \(K_p\) is \(K_p = \frac{p(\text{PCl}_3) \times p(\text{Cl}_2)}{p(\text{PCl}_5)}\). Substituting the given equilibrium partial pressures: \(K_p = \frac{0.350 \times 0.350}{0.240} = 0.5104 \text{ atm}\). Rounding to three significant figures yields \(0.510 \text{ atm}\).

[1 mark] for writing the correct Kp expression and calculating its value.

Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\). First, \(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.870\). Because equal volumes of both solution are mixed, the ratio of concentrations in the mixture is the same as the ratio of their initial concentrations: \(\frac{[\text{propanoate}]}{[\text{propanoic acid}]} = \frac{0.050}{0.100} = 0.500\). Thus, \(\text{pH} = 4.870 + \log_{10}(0.500) = 4.870 - 0.301 = 4.569 \approx 4.57\).

[1 mark] for calculating the correct pH of the buffer solution.

Nucleophilic addition of \(\text{HCN}\) to propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) produces 2-hydroxybutanenitrile, \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\). Carbon-2 of this product is a chiral centre because it is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{CN}\). Propanone, methanal, and pentan-3-one do not yield products with a chiral centre because the central carbon ends up bonded to at least two identical groups.

[1 mark] for identifying propanal as the correct precursor.

The positive reaction with 2,4-dinitrophenylhydrazine indicates the presence of a carbonyl group (aldehyde or ketone), ruling out alcohols. The negative reaction with Tollens' reagent rules out aldehydes. Thus, compound X must be a ketone. Since it contains four carbon atoms, its name is butanone.

[1 mark] for identifying butanone from the chemical test results.

A titration of a weak base with a strong acid has an equivalence point in the acidic region (typically around pH 5). The vertical portion of the titration curve lies approximately in the pH 4 to 7 range. Methyl orange, with a pH transition range of 3.1 - 4.4, is the most appropriate indicator because its colour change occurs entirely within the steep section of this titration curve.

[1 mark] for selecting methyl orange as the appropriate indicator for a weak base-strong acid titration.

(a) Compare Experiment 1 and 2: \([\text{S}_2\text{O}_8^{2-}]\) doubles while \([\text{I}^-]\) remains constant, and the initial rate doubles (from \(1.25 \times 10^{-5}\) to \(2.50 \times 10^{-5}\)). Therefore, the reaction is first-order with respect to \(\text{S}_2\text{O}_8^{2-}\).

Compare Experiment 1 and 3: \([\text{I}^-]\) doubles while \([\text{S}_2\text{O}_8^{2-}]\) remains constant, and the initial rate doubles (from \(1.25 \times 10^{-5}\) to \(2.50 \times 10^{-5}\)). Therefore, the reaction is first-order with respect to \(\text{I}^-\).

Rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

(b) From Experiment 1:
\(1.25 \times 10^{-5} = k \times 0.0100 \times 0.0100\)
\(k = \frac{1.25 \times 10^{-5}}{1.00 \times 10^{-4}} = 0.125\)
Units of \(k\) for an overall second-order reaction:
\(\text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(c) The slow step in a reaction mechanism is the rate-determining step. In the proposed mechanism, Step 1 is the slow step. This step involves one molecule of \(\text{S}_2\text{O}_8^{2-}\) and one molecule of \(\text{I}^-\). Therefore, the rate equation depends only on the concentrations of these two reactants to the power of 1, which matches the experimental rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

(d) From the Arrhenius equation: \(\ln k = -\frac{E_a}{RT} + \ln A\)
Comparing with the equation of a straight line, \(y = mx + c\), where \(y = \ln k\) and \(x = 1/T\):
\(\text{gradient } (m) = -\frac{E_a}{R}\)
\(-6540 = -\frac{E_a}{8.31}\)
\(E_a = 6540 \times 8.31 = 54347.4\text{ J mol}^{-1}\)
In \(\text{kJ mol}^{-1}\):
\(E_a = +54.3\text{ kJ mol}^{-1}\) (to 3 significant figures)

(e) Peroxodisulfate and iodide ions are both negatively charged. They repel each other, resulting in a high activation energy. \(\text{Fe}^{2+}\) ions have a positive charge and can catalyse the reaction by reacting with negative peroxodisulfate ions first, then the resulting \(\text{Fe}^{3+}\) ions react with iodide ions, regenerating \(\text{Fe}^{2+}\). This provides an alternative pathway with a lower activation energy.
Equation 1: \(\text{S}_2\text{O}_8^{2-}(aq) + 2\text{Fe}^{2+}(aq) \rightarrow 2\text{SO}_4^{2-}(aq) + 2\text{Fe}^{3+}(aq)\)
Equation 2: \(2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(aq)\)

**Part (a) [3 Marks]**
- 1 Mark: Deduces order with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1, with reference to Exp 1 & 2.
- 1 Mark: Deduces order with respect to \(\text{I}^-\) is 1, with reference to Exp 1 & 3.
- 1 Mark: Correct rate equation matching deduced orders: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

**Part (b) [3 Marks]**
- 1 Mark: Correct rearrangement of rate equation and substitution.
- 1 Mark: Value of \(k = 0.125\) (or \(1.25 \times 10^{-1}\)).
- 1 Mark: Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (allow in any order, e.g. \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)).

**Part (c) [2 Marks]**
- 1 Mark: States that Step 1 is the rate-determining / slow step.
- 1 Mark: Connects the reactants in Step 1 (one of each species) to the first-order dependence of both reactants in the rate equation.

**Part (d) [3 Marks]**
- 1 Mark: Use of \(\text{gradient} = -E_a / R\).
- 1 Mark: Calculates \(E_a\) in Joules: \(54347.4\text{ J mol}^{-1}\).
- 1 Mark: Converts to \(+54.3\text{ kJ mol}^{-1}\) (must have + sign and 3 sig figs).

**Part (e) [3 Marks]**
- 1 Mark: States that \(\text{Fe}^{2+}\) provides an alternative pathway with lower activation energy / avoids repulsion between two negative ions.
- 1 Mark: Correct Equation 1 (balanced with correct states).
- 1 Mark: Correct Equation 2 (balanced with correct states).

(a) Entropy is a measure of the degree of disorder (or randomness) of a system. The sign of \(\Delta S_{\text{system}}^{\ominus}\) is positive. This is because 2 moles of solid reactants decompose to produce 1 mole of solid and 2 moles of gas. Gases have much higher disorder (and therefore entropy) than solids, leading to a significant increase in the system's entropy.

(b) Using \(\Delta S_{\text{system}}^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\):
\(\sum S^{\ominus}(\text{products}) = S^{\ominus}[\text{Na}_2\text{CO}_3(s)] + S^{\ominus}[\text{CO}_2(g)] + S^{\ominus}[\text{H}_2\text{O}(g)] = 136.0 + 213.6 + 188.7 = 538.3\text{ J K}^{-1}\text{ mol}^{-1}\)
\(\sum S^{\ominus}(\text{reactants}) = 2 \times S^{\ominus}[\text{NaHCO}_3(s)] = 2 \times 102.1 = 204.2\text{ J K}^{-1}\text{ mol}^{-1}\)
\(\Delta S_{\text{system}}^{\ominus} = 538.3 - 204.2 = +334.1\text{ J K}^{-1}\text{ mol}^{-1}\)

(c) Using \(\Delta S_{\text{surroundings}}^{\ominus} = -\frac{\Delta H^{\ominus}}{T}\):
Convert \(\Delta H^{\ominus}\) to \(\text{J mol}^{-1}\):
\(\Delta H^{\ominus} = +135.6 \times 1000 = +135600\text{ J mol}^{-1}\)
\(\Delta S_{\text{surroundings}}^{\ominus} = -\frac{135600}{298} = -455.03\text{ J K}^{-1}\text{ mol}^{-1} \approx -455.0\text{ J K}^{-1}\text{ mol}^{-1}\)

(d) \(\Delta S_{\text{total}}^{\ominus} = \Delta S_{\text{system}}^{\ominus} + \Delta S_{\text{surroundings}}^{\ominus} = +334.1 + (-455.03) = -120.9\text{ J K}^{-1}\text{ mol}^{-1}\)
Since the total entropy change is negative (\(\Delta S_{\text{total}}^{\ominus} < 0\)), the reaction is not feasible at \(298\text{ K}\).

(e) For the reaction to be feasible, \(\Delta S_{\text{total}}^{\ominus}\) must be at least zero:
\(\Delta S_{\text{system}}^{\ominus} + \Delta S_{\text{surroundings}}^{\ominus} \ge 0\)
\(\Delta S_{\text{system}}^{\ominus} - \frac{\Delta H^{\ominus}}{T} \ge 0\)
\(T \ge \frac{\Delta H^{\ominus}}{\Delta S_{\text{system}}^{\ominus}}\)
\(T \ge \frac{135600}{334.1} = 405.87\text{ K}\)
So the minimum temperature is \(405.9\text{ K}\) (or \(406\text{ K}\)).

**Part (a) [3 Marks]**
- 1 Mark: Defines entropy as a measure of disorder / randomness.
- 1 Mark: Predicts positive sign.
- 1 Mark: Explains that gases are produced from solid reactants, which increases disorder.

**Part (b) [3 Marks]**
- 1 Mark: Correctly multiplies \(S^{\ominus}[\text{NaHCO}_3(s)]\) by 2 (value: 204.2).
- 1 Mark: Correct expression: \(\sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\).
- 1 Mark: Correct answer with positive sign and units: \(+334.1\text{ J K}^{-1}\text{ mol}^{-1}\).

**Part (c) [3 Marks]**
- 1 Mark: Correct formula \(\Delta S_{\text{surroundings}}^{\ominus} = -\Delta H^{\ominus} / T\).
- 1 Mark: Converts \(\Delta H^{\ominus}\) from \(\text{kJ}\) to \(\text{J}\) (by multiplying by 1000).
- 1 Mark: Correct answer with negative sign and units: \(-455.0\text{ J K}^{-1}\text{ mol}^{-1}\) (allow \(-455\)).

**Part (d) [2 Marks]**
- 1 Mark: Correct calculation: \(-120.9\text{ J K}^{-1}\text{ mol}^{-1}\) (allow \(-121\)).
- 1 Mark: Correct conclusion that the reaction is not feasible because \(\Delta S_{\text{total}}^{\ominus}\) is negative.

**Part (e) [3 Marks]**
- 1 Mark: States condition for feasibility (\(\Delta S_{\text{total}}^{\ominus} \ge 0\)).
- 1 Mark: Correct substitution of values with matched units (i.e. \(135600 / 334.1\)).
- 1 Mark: Correct final temperature: \(405.9\text{ K}\) (allow \(406\text{ K}\)).

(a) \(K_p = \frac{p(\text{COCl}_2)}{p(\text{CO}) \times p(\text{Cl}_2)}\)
Units: \(\frac{\text{atm}}{\text{atm} \times \text{atm}} = \text{atm}^{-1}\)

(b)(i) Since \(1\text{ mol}\) of \(\text{COCl}_2\) is formed from \(1\text{ mol}\) of \(\text{CO}\) and \(1\text{ mol}\) of \(\text{Cl}_2\):
- Equilibrium moles of \(\text{CO} = 0.500 - 0.150 = 0.350\text{ mol}\)
- Equilibrium moles of \(\text{Cl}_2 = 0.400 - 0.150 = 0.250\text{ mol}\)

(b)(ii) Total moles of gas at equilibrium = \(0.350 + 0.250 + 0.150 = 0.750\text{ mol}\)
- Mole fraction of \(\text{CO}\), \(\chi(\text{CO}) = \frac{0.350}{0.750} = 0.467\) (or \(7/15\))
- Mole fraction of \(\text{Cl}_2\), \(\chi(\text{Cl}_2) = \frac{0.250}{0.750} = 0.333\) (or \(1/3\))
- Mole fraction of \(\text{COCl}_2\), \(\chi(\text{COCl}_2) = \frac{0.150}{0.750} = 0.200\) (or \(1/5\))

(b)(iii) Partial pressure = Mole fraction \(\times\) Total pressure (\(2.50\text{ atm}\)):
- \(p(\text{CO}) = 0.4667 \times 2.50 = 1.17\text{ atm}\) (or \(1.167\text{ atm}\))
- \(p(\text{Cl}_2) = 0.3333 \times 2.50 = 0.833\text{ atm}\) (or \(0.8333\text{ atm}\))
- \(p(\text{COCl}_2) = 0.200 \times 2.50 = 0.500\text{ atm}\)

(b)(iv) \(K_p = \frac{0.500}{1.167 \times 0.8333} = 0.514\text{ atm}^{-1}\) (allow \(0.511\) to \(0.515\) depending on rounding)

(c)(i) The yield of phosgene increases. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium in the direction that decreases the total number of gas moles. In this reaction, there are 2 moles of gas on the left-hand side and 1 mole of gas on the right-hand side, so the equilibrium shifts to the right.
(c)(ii) The value of \(K_p\) remains unchanged, as it is only affected by temperature.

**Part (a) [2 Marks]**
- 1 Mark: Correct expression for \(K_p\) (square brackets \([ ]\) not allowed).
- 1 Mark: Correct units: \(\text{atm}^{-1}\) (or equivalent based on expression).

**Part (b)(i) [2 Marks]**
- 1 Mark: Moles of \(\text{CO} = 0.350\).
- 1 Mark: Moles of \(\text{Cl}_2 = 0.250\).

**Part (b)(ii) [2 Marks]**
- 1 Mark: Correctly calculates total moles as \(0.750\).
- 1 Mark: Calculates all three mole fractions correctly (to 3 sig figs).

**Part (b)(iii) [3 Marks]**
- 1 Mark: \(p(\text{CO}) = 1.17\text{ atm}\).
- 1 Mark: \(p(\text{Cl}_2) = 0.833\text{ atm}\).
- 1 Mark: \(p(\text{COCl}_2) = 0.500\text{ atm}\).
(Allow error carried forward from (b)(ii))

**Part (b)(iv) [2 Marks]**
- 1 Mark: Substitution of partial pressures into \(K_p\) expression.
- 1 Mark: Evaluation of \(K_p = 0.514\text{ atm}^{-1}\) (allow \(0.511 - 0.515\)).

**Part (c)(i) [2 Marks]**
- 1 Mark: State that yield increases.
- 1 Mark: Explains that there are fewer moles of gas on the product side (or reactant side has more moles of gas).

**Part (c)(ii) [1 Mark]**
- 1 Mark: State that \(K_p\) is unchanged because temperature is constant.

(a) Dissociation equation:
\(\text{CH}_3\text{CH}_2\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-(aq) + \text{H}_3\text{O}^+(aq)\)
(Accept: \(\text{CH}_3\text{CH}_2\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-(aq) + \text{H}^+(aq)\))

Conjugate acid-base pairs:
- Pair 1: \(\text{CH}_3\text{CH}_2\text{COOH}\) (acid) and \(\text{CH}_3\text{CH}_2\text{COO}^-\) (conjugate base)
- Pair 2: \(\text{H}_2\text{O}\) (base) and \(\text{H}_3\text{O}^+\) (conjugate acid)

(b) For a weak acid, we use the approximation:
\(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \approx \frac{[\text{H}^+]^2}{[\text{HA}]}\)
\([\text{H}^+]^2 = K_a \times [\text{HA}] = 1.35 \times 10^{-5} \times 0.120 = 1.62 \times 10^{-6}\text{ mol}^2\text{ dm}^{-6}\)
\([\text{H}^+] = \sqrt{1.62 \times 10^{-6}} = 1.273 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.273 \times 10^{-3}) = 2.90\)

Assumptions (any one):
- The dissociation of the weak acid is negligible, so \([\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}}\).
- The concentration of \([\text{H}^+]\) from the self-ionisation of water is negligible.

(c)(i)
First, calculate the moles of acid and salt:
- \(\text{Moles of propanoic acid } = \frac{50.0 \times 0.120}{1000} = 6.00 \times 10^{-3}\text{ mol}\)
- \(\text{Moles of sodium propanoate } = \frac{30.0 \times 0.150}{1000} = 4.50 \times 10^{-3}\text{ mol}\)

Since they are in the same total volume, we can substitute moles directly into the expression:
\([\text{H}^+] = K_a \times \frac{\text{moles of acid}}{\text{moles of salt}}\)
\([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{6.00 \times 10^{-3}}{4.50 \times 10^{-3}} = 1.80 \times 10^{-5}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.80 \times 10^{-5}) = 4.74\)

(c)(ii) When hydrochloric acid is added, the extra \(\text{H}^+\) ions react with the propanoate ions (the conjugate base of the buffer) to form weak propanoic acid molecules:
\(\text{CH}_3\text{CH}_2\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(aq)\)
Because there is a large reservoir of propanoate ions, the added \(\text{H}^+\) ions are mostly removed, meaning the ratio of acid to salt changes very little, keeping the pH stable.

**Part (a) [3 Marks]**
- 1 Mark: Correctly balanced reversible equation.
- 1 Mark: Correctly identifies first acid-base pair.
- 1 Mark: Correctly identifies second acid-base pair.

**Part (b) [4 Marks]**
- 1 Mark: Rearranges \(K_a\) to write \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]}\).
- 1 Mark: Calculates \([\text{H}^+] = 1.27 \times 10^{-3}\text{ mol dm}^{-3}\).
- 1 Mark: Correct pH = 2.90 (must be written to 2 decimal places).
- 1 Mark: States a valid assumption (e.g. dissociation is negligible).

**Part (c)(i) [4 Marks]**
- 1 Mark: Calculates moles of propanoic acid = \(6.00 \times 10^{-3}\text{ mol}\).
- 1 Mark: Calculates moles of propanoate ions = \(4.50 \times 10^{-3}\text{ mol}\).
- 1 Mark: Correct calculation of \([\text{H}^+] = 1.80 \times 10^{-5}\text{ mol dm}^{-3}\).
- 1 Mark: Correct pH = 4.74 (must be written to 2 decimal places).

**Part (c)(ii) [3 Marks]**
- 1 Mark: Correct ionic equation for the reaction of propanoate with \(\text{H}^+\).
- 1 Mark: Explains that added \(\text{H}^+\) is removed by reacting with the large reservoir of conjugate base (propanoate ions).
- 1 Mark: States that the ratio of \([\text{HA}]/[\text{A}^-]\) is largely unchanged, so pH remains constant.

(a) Since compound A reacts with Tollens' reagent, it must be an aldehyde. It has the formula \(\text{C}_4\text{H}_8\text{O}\) and is a straight-chain compound, so it is butanal.
- Structural formula: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
- IUPAC Name: butanal

(b)(i) Reaction: Triiodomethane reaction (or iodoform reaction).
- Formula of the yellow precipitate: \(\text{CHI}_3\) (triiodomethane).

(b)(ii) Since compound B is an isomer of \(\text{C}_4\text{H}_8\text{O}\) that does not react with Tollens' reagent, it is a ketone. It gives a positive iodoform test, meaning it contains a \(\text{CH}_3\text{CO}-\) group. Therefore, it is butanone.
- Structural formula: \(\text{CH}_3\text{COCH}_2\text{CH}_3\)
- IUPAC Name: butanone (or butan-2-one)

(c)(i) Class: Secondary alcohol.
- Equation: \(\text{CH}_3\text{COCH}_2\text{CH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)

(c)(ii) Compound C is butan-2-ol. It is optically active because it contains a chiral carbon atom (asymmetric carbon) bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\).

The sample obtained is optically inactive (racemic) because the carbonyl carbon in butanone is planar (trigonal planar geometry, bond angle \(120^{\circ}\)). The nucleophile (hydride ion, \(\text{H}^-\)) has an equal probability of attacking the planar carbonyl group from either side (above or below the plane). This results in the formation of equal amounts (50:50 mixture) of both enantiomers, which rotate plane-polarized light in opposite directions by equal amounts, cancelling each other out.

**Part (a) [3 Marks]**
- 1 Mark: Identifies A as an aldehyde / butanal.
- 1 Mark: Correct structural formula of butanal.
- 1 Mark: Correct IUPAC name: butanal.

**Part (b)(i) [2 Marks]**
- 1 Mark: Names the iodoform / triiodomethane reaction.
- 1 Mark: Correct formula of precipitate: \(\text{CHI}_3\).

**Part (b)(ii) [3 Marks]**
- 1 Mark: Identifies B as a ketone / butanone.
- 1 Mark: Correct structural formula of butanone.
- 1 Mark: Correct IUPAC name: butanone (or butan-2-one).

**Part (c)(i) [2 Marks]**
- 1 Mark: Classifies alcohol C as secondary.
- 1 Mark: Correctly balanced equation using \(2[\text{H}]\).

**Part (c)(ii) [4 Marks]**
- 1 Mark: States that C is optically active because of the chiral / asymmetric carbon atom bonded to four different groups.
- 1 Mark: States that the carbonyl group in butanone is planar.
- 1 Mark: States that attack by \(\text{H}^-\)\ / nucleophile is equally likely from above or below the plane.
- 1 Mark: Concludes that equal amounts (a racemic mixture) of both enantiomers are produced, so the optical rotations cancel out.

(a) The standard enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state, under standard conditions.

(b) Born-Haber cycle diagram layout:
- Bottom level: \(\text{MgCl}_2(s)\)
- Elements level: \(\text{Mg}(s) + \text{Cl}_2(g)\)
- Atomisation: \(\text{Mg}(g) + \text{Cl}_2(g)\) then \(\text{Mg}(g) + 2\text{Cl}(g)\)
- Ionisation: \(\text{Mg}^+(g) + 2\text{Cl}(g) + \text{e}^-\), then \(\text{Mg}^{2+}(g) + 2\text{Cl}(g) + 2\text{e}^-\)
- Electron affinity: \(\text{Mg}^{2+}(g) + 2\text{Cl}^-(g)\)
- Arrows and labels correctly showing \(\Delta H_f^{\ominus}\), \(\Delta H_{\text{at}}^{\ominus}\), \(\text{IE}\), \(\text{EA}\), and \(\Delta H_{\text{latt}}^{\ominus}\).

(c) Using Hess's Law:
\(\Delta H_f^{\ominus} = \Delta H_{\text{at}}^{\ominus}(\text{Mg}) + \text{1st IE}(\text{Mg}) + \text{2nd IE}(\text{Mg}) + 2 \times \Delta H_{\text{at}}^{\ominus}(\text{Cl}) + 2 \times \text{EA}(\text{Cl}) + \Delta H_{\text{latt}}^{\ominus}\)

Substitute the values:
\(-641 = +150 + 738 + 1451 + 2(+122) + 2(-349) + \Delta H_{\text{latt}}^{\ominus}\)
\(-641 = 150 + 738 + 1451 + 244 - 698 + \Delta H_{\text{latt}}^{\ominus}\)
\(-641 = 1885 + \Delta H_{\text{latt}}^{\ominus}\)
\(\Delta H_{\text{latt}}^{\ominus} = -641 - 1885 = -2526\text{ kJ mol}^{-1}\)

(d) The theoretical model assumes a purely ionic model with perfectly spherical ions. The experimental value is more exothermic because the bonding in magnesium chloride has some covalent character. The \(\text{Mg}^{2+}\) ion is small and has a high charge density, which allows it to polarise the electron cloud of the \(\text{Cl}^-\)(chloride) ion, causing a degree of covalent bonding that increases the strength of the lattice.

(e) The enthalpy of solution is given by:
\(\Delta H_{\text{sol}}^{\ominus} = -\Delta H_{\text{latt}}^{\ominus} + \Delta H_{\text{hyd}}^{\ominus}(\text{Mg}^{2+}) + 2 \times \Delta H_{\text{hyd}}^{\ominus}(\text{Cl}^-)\)
\(\Delta H_{\text{sol}}^{\ominus} = -(-2526) + (-1920) + 2(-364)\)
\(\Delta H_{\text{sol}}^{\ominus} = +2526 - 1920 - 728 = -122\text{ kJ mol}^{-1}\)

**Part (a) [2 Marks]**
- 1 Mark: Formation of 1 mole of gaseous atoms.
- 1 Mark: From the element in its standard state, under standard conditions.

**Part (b) [4 Marks]**
- 1 Mark: Correct levels and state symbols for \(\text{Mg}(s) + \text{Cl}_2(g)\) and \(\text{MgCl}_2(s)\) with arrow for \(\Delta H_f\).
- 1 Mark: Correct levels for atomisation steps showing \(2\text{Cl}(g)\) (not \(\text{Cl}\)).
- 1 Mark: Correct levels for both ionisation steps and electron affinity (showing \(2\text{Cl}^-\)).
- 1 Mark: Correct direction of all arrows and labels for each step.

**Part (c) [3 Marks]**
- 1 Mark: Correctly multiplies both atomisation of chlorine (value: 244) and electron affinity (value: -698) by 2.
- 1 Mark: Correct Hess's Law expression / equation set-up.
- 1 Mark: Value of \(\Delta H_{\text{latt}}^{\ominus} = -2526\text{ kJ mol}^{-1}\) (must have negative sign and correct units).

**Part (d) [3 Marks]**
- 1 Mark: States that the theoretical model assumes purely ionic / spherical ions.
- 1 Mark: Explains that there is polarization of the anion / covalent character present.
- 1 Mark: Explains that \(\text{Mg}^{2+}\) has high charge density / polarising power, making the bonds stronger than purely ionic.

**Part (e) [2 Marks]**
- 1 Mark: Correct formula \(\Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \sum \Delta H_{\text{hyd}}\), correctly doubling hydration enthalpy of chloride.
- 1 Mark: Correct calculation with sign and units: \(-122\text{ kJ mol}^{-1}\).

The pink hexaaquacobalt(II) ion, \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), has a coordination number of 6. Upon addition of excess concentrated hydrochloric acid, a ligand substitution reaction occurs to form the blue tetrahedral tetrachlorocobaltate(II) ion, \([\text{Co}\text{Cl}_4]^{2-}\), which has a coordination number of 4. Therefore, the coordination number decreases from 6 to 4, and the final color is blue.

1 mark for the correct option (a). Reject other options.

To oxidize \(\text{Fe}^{2+}(\text{aq})\) to \(\text{Fe}^{3+}(\text{aq})\), the oxidizing agent must belong to a half-cell with a standard electrode potential more positive than \(+0.77\text{ V}\). The bromine half-cell has \(E^\ominus = +1.09\text{ V}\), which is more positive, meaning \(\text{Br}_2(\text{aq})\) can oxidize \(\text{Fe}^{2+}\). The iodine half-cell has \(E^\ominus = +0.54\text{ V}\), which is less positive, so \(\text{I}_2(\text{aq})\) is not strong enough. The actual species undergoing reduction is \(\text{Br}_2(\text{aq})\).

1 mark for the correct option (c). Reject other options.

In the generation of the electrophile, concentrated sulfuric acid donates a proton to concentrated nitric acid: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\). Since it donates a proton, it acts as a Brønsted-Lowry acid. It is also regenerated at the end of the electrophilic substitution mechanism, meaning it acts as a catalyst.

1 mark for the correct option (a). Reject other options.

Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is partially delocalized into the benzene pi-system, making it less available to accept a proton. Ammonia is a stronger base than phenylamine. Ethylamine is the strongest base of the three because the electron-releasing ethyl group (positive inductive effect) increases electron density on the nitrogen atom, making its lone pair more available to accept a proton.

1 mark for the correct option (a). Reject other options.

The reaction involves 4 reactant particles forming 7 product particles in solution. This significant increase in the number of free particles in solution leads to a large increase in entropy (\(\Delta S_{\text{system}}^\ominus > 0\)). Since the number and type of coordination bonds (nitrogen-metal vs. oxygen-metal bonds) are similar, the enthalpy change \(\Delta H^\ominus\) is close to zero. The positive system entropy change drives the reaction to feasibility (\(\Delta S_{\text{total}}^\ominus > 0\)). This is known as the chelate effect.

1 mark for the correct option (a). Reject other options.

Step 1: Methylbenzene is oxidized to benzoic acid by heating under reflux with alkaline potassium manganate(VII), \(\text{KMnO}_4\), followed by acidification with a dilute acid like sulfuric acid. Step 2: Benzoic acid is converted to its acyl chloride (benzoyl chloride) using phosphorus(V) chloride, \(\text{PCl}_5\). Step 3: Benzoyl chloride reacts vigorously with concentrated ammonia, \(\text{NH}_3\), to form benzamide.

1 mark for the correct option (a). Reject other options.

The reaction between two negatively charged ions has a high activation energy due to electrostatic repulsion, making it very slow. \(\text{Fe}^{2+}(\text{aq})\) acts as a homogeneous catalyst because of its variable oxidation states. It first reduces \(\text{S}_2\text{O}_8^{2-}\) to \(\text{SO}_4^{2-}\), becoming oxidized to \(\text{Fe}^{3+}\): \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\). The \(\text{Fe}^{3+}\) then oxidizes \(\text{I}^-\), regenerating the \(\text{Fe}^{2+}\) catalyst: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\). Each individual step involves oppositely charged ions, so the activation energy is much lower and the rate increases.

1 mark for the correct option (a). Reject other options.

At a pH below its isoelectric point (such as pH 2.0, which is highly acidic), the concentration of \(\text{H}^+\) ions is high. Therefore, the carboxylate group is protonated to form \(-\text{COOH}\) (neutral), and the amino group is also protonated to form \(-\text{NH}_3^+\) (positively charged). Thus, the predominant species is the cation \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\).

1 mark for the correct option (a). Reject other options.

To calculate the energy gap per mole of ions:

1. Calculate the energy of a single photon absorbed:
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{530 \times 10^{-9}} = 3.75 \times 10^{-19}\text{ J} \)

2. Multiply by Avogadro's constant to find the energy per mole:
\( E_{\text{mole}} = 3.75 \times 10^{-19} \times 6.02 \times 10^{23} = 225920\text{ J mol}^{-1} \)

3. Convert to \( \text{kJ mol}^{-1} \):
\( \Delta E = 226\text{ kJ mol}^{-1} \)

1 mark for correct answer B.
- A: Incorrect conversion without incorporating Avogadro's constant.
- C: Represents the value of the energy of a single photon in Joules but given the unit \( \text{kJ mol}^{-1} \).
- D: Incorrect conversion showing a factor of 1000 error.

Ethylamine is the most basic because the electron-donating ethyl group increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton than in ammonia. Ammonia is more basic than phenylamine because in phenylamine, the nitrogen lone pair is delocalised into the benzene pi-ring system, reducing its availability. Ethanamide is the least basic because the strongly electron-withdrawing carbonyl group adjacent to the nitrogen severely delocalises the lone pair, making amides virtually neutral in water. Thus, the order is: ethylamine > ammonia > phenylamine > ethanamide.

1 mark for correct answer A.
- B: Incorrect because ammonia is more basic than phenylamine.
- C: Incorrect as it lists the species in order of increasing basicity instead of decreasing basicity.
- D: Incorrect because ethylamine is more basic than ammonia.

(a)(i) Pink species is \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \). Blue species is \( [\text{CoCl}_4]^{2-} \). (a)(ii) Reaction type: Ligand substitution. Coordination number changes from 6 to 4. (a)(iii) The pink complex is octahedral and the blue complex is tetrahedral. The change in geometry changes the d-orbital splitting energy (\( \Delta E \)). d-orbitals split into two energy levels. Electrons absorb a photon of light in the visible region of the electromagnetic spectrum to be promoted to a higher d-orbital (d-d transition). The remaining wavelengths of light are transmitted/reflected to give the complementary color. Since \( \Delta E \) is different for the tetrahedral chloro complex compared to the octahedral aqua complex, a different wavelength of light is absorbed, resulting in a change from pink to blue. (b)(i) \( \text{Cr}(\text{OH})_3(\text{s}) + 3\text{OH}^-(\text{aq}) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(\text{aq}) \) (or with 1 OH- to form tetrahydroxochromate(III)). (b)(ii) \( \text{Cr}(\text{OH})_3(\text{s}) + 3\text{H}^+(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \). (b)(iii) The green precipitate dissolves to form a green solution. (c)(i) Since 0.010 mol of the complex yields 0.010 mol of AgCl, there must be only 1 mol of chloride ions outside the coordination sphere per mole of complex. Therefore, two chloride ions must be inside the coordination sphere. The formula of the compound is \( [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O} \). The complex cation is \( [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ \). (c)(ii) The complex shows cis-trans (geometric) isomerism. In the cis isomer, the two chloride ligands are adjacent to each other (at 90 degrees). In the trans isomer, the two chloride ligands are opposite each other (at 180 degrees).

(a)(i) M1: Formula of \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \) (1 mark). M2: Formula of \( [\text{CoCl}_4]^{2-} \) (1 mark). (a)(ii) M1: Ligand substitution / exchange (1 mark). M2: Coordination number changes from 6 to 4 (1 mark). (a)(iii) M1: Different ligands/geometry cause a change in the d-orbital splitting energy (\( \Delta E \)) (1 mark). M2: Electrons are promoted to higher d-orbitals by absorbing visible light energy (1/2 transitions) (1 mark). M3: \( \Delta E = h\nu \) showing energy absorbed changes (1 mark). M4: The remaining color (complementary color) is transmitted (1 mark). (b)(i) M1: Left hand side and right hand side species correct (1 mark). M2: Correct state symbols (1 mark). (b)(ii) M1: Left hand side and right hand side species correct (1 mark). M2: Correct state symbols (1 mark). (b)(iii) Green precipitate dissolves to form a green solution (1 mark). (c)(i) M1: Ratio of compound to Cl- precipitated is 1:1, so 1 Cl- is outside the sphere (1 mark). M2: Remaining 2 Cl- and 4 H2O are inside the coordination sphere (1 mark). M3: Formula of complex cation is \( [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ \) (1 mark). (c)(ii) M1: Identifies cis-trans / geometric isomerism (1 mark). M2: Cis-isomer description (two chloride ligands adjacent / at 90 degrees) (1 mark). M3: Trans-isomer description (two chloride ligands opposite / at 180 degrees) (1 mark). M4: High-quality drawing or clear 3D descriptions of both (1 mark).

(a)(i) A base is a proton acceptor. In phenylamine, the lone pair of electrons on the nitrogen atom overlaps with the pi-delocalized ring system of benzene. This delocalizes the lone pair, making it less available to accept a proton (H+). In ethylamine, the ethyl group is electron-releasing (positive inductive effect), which increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton. (a)(ii) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^- \). (b)(i) Step 1: Benzene to nitrobenzene. Reagents: Concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4) catalyst. Conditions: Heat at 50-60 °C. Step 2: Nitrobenzene to phenylamine. Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl), heated under reflux, followed by addition of aqueous sodium hydroxide. (b)(ii) Electrophilic substitution. (c)(i) Reagents: Sodium nitrate(III) (or sodium nitrite, NaNO2) and dilute hydrochloric acid (HCl). Temperature: Between 0 °C and 10 °C. (c)(ii) The azo dye is 4-hydroxyphenylazobenzene (or 4-(phenyldiazenyl)phenol). It consists of two benzene rings linked by an azo (-\text{N}=\text{N}-) group, with the phenol ring having the -OH group para to the azo linkage. (d)(i) At pH 1: \( \text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH} \). At pH 12: \( \text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^- \). (d)(ii) Zwitterion structure: \( \text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^- \). Alanine has a high melting temperature because it exists as a zwitterion with strong electrostatic forces of attraction (ionic bonds) between the oppositely charged groups of neighboring molecules, requiring significant energy to break. (d)(iii) Dipeptide 1 (alanyl-glycine): \( \text{CH}_3\text{CH}(\text{NH}_2)\text{CONHCH}_2\text{COOH} \). Dipeptide 2 (glycyl-alanine): \( \text{H}_2\text{NCH}_2\text{CONHCH}(\text{CH}_3)\text{COOH} \).

(a)(i) M1: Lone pair on N in phenylamine overlaps with the pi-delocalized ring system (1 mark). M2: Lone pair is less available to accept H+ (1 mark). M3: Alkyl group in ethylamine is electron-donating, making lone pair more available (1 mark). (a)(ii) M1: Correct equation showing salt formation (1 mark). (b)(i) M1: Step 1 reagents (conc. HNO3 + conc. H2SO4) (1 mark). M2: Step 1 temperature (50-60 °C) (1 mark). M3: Step 2 reagents (Sn + conc. HCl) (1 mark). M4: Step 2 condition (heating under reflux, then NaOH(aq)) (1 mark). (b)(ii) M1: Electrophilic substitution (1 mark). (c)(i) M1: NaNO2 and HCl (1 mark). M2: Temperature 0-10 °C (1 mark). (c)(ii) M1: Correct azo group linking two rings (1 mark). M2: Correct connectivity at para-position of phenol (1 mark). (d)(i) M1: Correct structure at pH 1 (1 mark). M2: Correct structure at pH 12 (1 mark). (d)(ii) M1: Correct zwitterion structure (1 mark). M2: Strong electrostatic forces / ionic bonding (1 mark). M3: Between neighboring zwitterions (1 mark). (d)(iii) M1: Correct structure of alanyl-glycine (1 mark). M2: Correct structure of glycyl-alanine (1 mark).

(a)(i) IUPAC Cell Diagram: \( \text{Pt}(\text{s}) | \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) || \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}), \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) | \text{Pt}(\text{s}) \) (or similar). (a)(ii) \( E^\ominus_{\text{cell}} = 1.51 - 0.77 = +0.74\text{ V} \). Overall equation: \( 5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \). (a)(iii) \( E^\ominus_{\text{cell}} = E^\ominus(\text{Cl}_2/\text{Cl}^-) - E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = 1.36 - 0.77 = +0.59\text{ V} \). Since \( E^\ominus_{\text{cell}} > 0 \), the reaction is feasible and chlorine gas can oxidize \( \text{Fe}^{2+}(\text{aq}) \). (a)(iv) \( E^\ominus(\text{MnO}_4^-/\text{Mn}^{2+}) = +1.51\text{ V} \) is greater than \( E^\ominus(\text{Cl}_2/\text{Cl}^-) = +1.36\text{ V} \). Therefore, manganate(VII) ions will oxidize chloride ions from HCl to chlorine gas, leading to an inaccurate titration value. (b)(i) Colorless / pale green to permanent pale pink. (b)(ii) Portion 1: \( n(\text{MnO}_4^-) = 0.0200 \times 0.01200 = 2.40 \times 10^{-4}\text{ mol} \). \( n(\text{Fe}^{2+}) = 5 \times 2.40 \times 10^{-4} = 1.20 \times 10^{-3}\text{ mol} \) in Portion 1. Total \( n(\text{Fe}^{2+}) \) in 1.50 g sample = \( 2 \times 1.20 \times 10^{-3} = 2.40 \times 10^{-3}\text{ mol} \). Mass of \( \text{Fe}^{2+} \) = \( 2.40 \times 10^{-3} \times 55.8 = 0.13392\text{ g} \). % of \( \text{Fe}^{2+} \) = \( (0.13392 / 1.50) \times 100 = 8.93\% \). Portion 2: \( n(\text{MnO}_4^-) = 0.0200 \times 0.02800 = 5.60 \times 10^{-4}\text{ mol} \). Total moles of iron in Portion 2 = \( 5 \times 5.60 \times 10^{-4} = 2.80 \times 10^{-3}\text{ mol} \). Moles of \( \text{Fe}^{3+} \) in Portion 2 = \( 2.80 \times 10^{-3} - 1.20 \times 10^{-3} = 1.60 \times 10^{-3}\text{ mol} \). Total moles of \( \text{Fe}^{3+} \) in sample = \( 2 \times 1.60 \times 10^{-3} = 3.20 \times 10^{-3}\text{ mol} \). Mass of \( \text{Fe}^{3+} \) = \( 3.20 \times 10^{-3} \times 55.8 = 0.17856\text{ g} \). % of \( \text{Fe}^{3+} \) = \( (0.17856 / 1.50) \times 100 = 11.90\% \).

(a)(i) M1: Pt electrode shown on both sides (1 mark). M2: Correct species and state symbols on left (Fe2+(aq), Fe3+(aq)) (1 mark). M3: Correct species and state symbols on right (MnO4-(aq), H+(aq), Mn2+(aq)) (1 mark). (a)(ii) M1: Correct calculation of cell potential (+0.74 V) (1 mark). M2: Balanced equation (1 mark). M3: Correct state symbols in equation (1/2 error carried forward) (1 mark). (a)(iii) M1: Cell potential calculation: 1.36 - 0.77 = +0.59 V (1 mark). M2: Feasible because E_cell is positive (1 mark). (a)(iv) M1: Manganate(VII) ions will oxidize Cl- ions to Cl2 (1 mark). M2: Because E_theta of MnO4-/Mn2+ is more positive than Cl2/Cl- (1 mark). (b)(i) Colorless / pale green to permanent pale pink (1 mark). (b)(ii) M1: Moles of MnO4- in portion 1 = 2.40 x 10^-4 mol (1 mark). M2: Moles of Fe2+ in portion 1 = 1.20 x 10^-3 mol (1 mark). M3: Total mass of Fe2+ in sample = 0.134 g (1 mark). M4: Percentage of Fe2+ = 8.93% (1 mark). M5: Moles of MnO4- in portion 2 = 5.60 x 10^-4 mol (1 mark). M6: Total moles of Fe in portion 2 = 2.80 x 10^-3 mol (1 mark). M7: Moles of Fe3+ in portion 2 = 1.60 x 10^-3 mol (1 mark). M8: Total mass of Fe3+ in sample = 0.179 g (1 mark). M9: Percentage of Fe3+ = 11.90% (1 mark).

(a)(i) In benzene, each carbon atom is sp2 hybridized and forms three sigma bonds. The remaining p-orbital on each carbon atom overlaps sideways with neighboring p-orbitals to form a ring of delocalized pi electrons above and below the plane of the carbon ring. Cyclohexene has an enthalpy of hydrogenation of -120 kJ mol^-1. If benzene were cyclohexatriene (three localized C=C double bonds), the expected enthalpy of hydrogenation would be 3 x (-120) = -360 kJ mol^-1. However, the actual enthalpy of hydrogenation of benzene is -208 kJ mol^-1. This makes benzene 152 kJ mol^-1 more stable than expected due to the delocalization of its pi electrons. (a)(ii) Addition reactions would involve breaking the stable delocalized pi system, which is energetically unfavorable. Substitution reactions preserve the delocalized ring structure and its thermodynamic stability. (b)(i) Reagent: Chloromethane (CH3Cl). Catalyst: Aluminum chloride (AlCl3). Equation: \( \text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_4^- \). (b)(ii) Step 1: Curly arrow from the benzene ring to the electrophile \( \text{CH}_3^+ \). Step 2: Draw the intermediate carbocation (hexagon with a broken circle/horseshoe containing a positive charge, with both -H and -CH3 bonded to the sp3 carbon). Step 3: Curly arrow from the C-H bond back into the ring to restore the delocalized pi system. Step 4: Products are methylbenzene and H+. (c) Synthesis of 4-aminobenzoic acid from methylbenzene: Step 1: Nitration. Reagents: conc. HNO3 and conc. H2SO4 at 50-60 °C. Intermediate 1: 4-nitromethylbenzene (isolated from 2-nitromethylbenzene). Step 2: Oxidation of the methyl group. Reagents: alkaline KMnO4, heat under reflux, followed by acidification with dilute HCl/H2SO4. Intermediate 2: 4-nitrobenzoic acid. Step 3: Reduction of the nitro group. Reagents: Tin (Sn) and concentrated HCl, heat under reflux, followed by addition of NaOH(aq). Product: 4-aminobenzoic acid.

(a)(i) M1: Sideways overlap of p-orbitals forming a delocalized pi ring system (1 mark). M2: Expected enthalpy of hydrogenation of Kekulé structure = 3 x (-120) = -360 kJ mol^-1 (1 mark). M3: Actual enthalpy of hydrogenation is -208 kJ mol^-1 (1 mark). M4: Benzene is 152 kJ mol^-1 more stable due to delocalization energy (1 mark). (a)(ii) M1: Addition would disrupt the stable delocalized pi system (1 mark). M2: Substitution retains the delocalization and stability of the ring (1 mark). (b)(i) M1: Reactants: CH3Cl + AlCl3 (1 mark). M2: Products: CH3+ + AlCl4- (1 mark). (b)(ii) M1: Curly arrow from ring to CH3+ (1 mark). M2: Correct structure of intermediate carbocation (1 mark). M3: Curly arrow from C-H bond to reform the ring (1 mark). M4: Correct products (1 mark). (c) M1: Step 1 Reagents & Conditions: conc. HNO3 + conc. H2SO4 at 50-60 °C (1 mark). M2: Step 1 Product: 4-nitromethylbenzene (1 mark). M3: Mentions separating the 4-nitro isomer from the 2-nitro isomer (1 mark). M4: Step 2 Reagents & Conditions: alkaline KMnO4, heat under reflux (1 mark). M5: Step 2 acidification: dilute strong acid (1 mark). M6: Step 2 Product: 4-nitrobenzoic acid (1 mark). M7: Step 3 Reagents & Conditions: Sn + conc. HCl, heated, then NaOH(aq) (1 mark). M8: Final Product: Correct structure / name of 4-aminobenzoic acid (1 mark).

(a) Addition of sodium hydroxide solution, \(NaOH(aq)\), to the test tube containing the complex. This forms a red-brown precipitate of iron(III) hydroxide, \(Fe(OH)_3\).
(b) Add aqueous calcium chloride, \(CaCl_2(aq)\). A white precipitate of calcium oxalate, \(CaC_2O_4\), forms, which is insoluble in dilute ethanoic acid but soluble in hydrochloric acid.
(c)(i) Manganate(VII) reduction half-equation: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)
Oxalate oxidation half-equation: \(C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-\)
Combining them to balance electrons: \(2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O\).
(ii) Moles of \(MnO_4^-\): \(n(MnO_4^-) = 0.0200 \times \frac{15.25}{1000} = 3.05 \times 10^{-4}\text{ mol}\).
Using the 2:5 mole ratio, moles of \(C_2O_4^{2-}\) in \(25.0\text{ cm}^3\): \(n(C_2O_4^{2-}) = 3.05 \times 10^{-4} \times \frac{5}{2} = 7.625 \times 10^{-4}\text{ mol}\).
In the original \(250.0\text{ cm}^3\) volumetric flask: \(7.625 \times 10^{-4} \times 10 = 7.625 \times 10^{-3}\text{ mol}\).
Mass of oxalate ions (\(M_r = 88.0\)): \(m = 7.625 \times 10^{-3} \times 88.0 = 0.671\text{ g}\).
Percentage by mass: \(\% = \frac{0.671}{1.25} \times 100\% = 53.68\% \approx 53.7\%\).
(iii) The reaction between \(MnO_4^-\) and \(C_2O_4^{2-}\) involves two negatively charged ions that repel each other, resulting in a very high activation energy and a slow rate at room temperature. Heating to \(60\text{ }^\circ\text{C}\) provides molecules with kinetic energy exceeding the activation energy. The rate increases during the titration because \(Mn^{2+}\) ions, which act as an autocatalyst, are produced as the reaction proceeds.

(a) [2 marks total]:
- Add \(NaOH(aq)\) / \(NH_3(aq)\) (1 mark)
- Red-brown precipitate forms (1 mark)
(b) [2 marks total]:
- Add \(CaCl_2(aq)\) / \(Ca(NO_3)_2(aq)\) (1 mark)
- White precipitate forms (1 mark)
(c)(i) [2 marks total]:
- Correct reactants and products (1 mark)
- Fully balanced equation (1 mark)
(c)(ii) [4 marks total]:
- Correct moles of \(MnO_4^-\) (1 mark)
- Correct moles of oxalate in \(250\text{ cm}^3\) (1 mark)
- Mass of oxalate calculated using \(M_r = 88.0\) (1 mark)
- % calculated to 3 s.f. (1 mark) [Accept \(53.7\%\); allow TE for math errors]
(c)(iii) [2.5 marks total]:
- Repulsion between negative ions / high activation energy (1 mark)
- \(Mn^{2+}\) acts as an autocatalyst (1 mark)
- Acceleration is due to the generation of this catalyst (0.5 mark)

(a) The diagram should depict standard distillation apparatus:
- Round-bottomed or pear-shaped flask containing the reaction mixture, sitting in a heating mantle or water bath.
- Still head with a thermometer placed at the junction to measure vapour temperature.
- Liebig condenser sloped downwards with water entering at the lower end and leaving at the upper end.
- Receiver adapter directing liquid into a collection flask/beaker (open to air to prevent pressure build-up).
(b)(i) Saturated \(NaHCO_3\) neutralises acidic impurities: \(H^+ + HCO_3^- \rightarrow H_2O + CO_2\). Venting is required to release the built-up pressure from \(CO_2\) gas.
(ii) Anhydrous calcium chloride is a drying agent that binds to remaining water molecules. The organic layer changes from cloudy to clear when dry.
(iii) Cyclohexene has a boiling point of \(83\text{ }^\circ\text{C}\), while cyclohexanol has a boiling point of \(161\text{ }^\circ\text{C}\). Distilling and collecting in this narrow temperature window ensures only pure cyclohexene is isolated.
(c)(i) Moles of cyclohexanol starting material: \(n = \frac{12.0}{100.2} = 0.1198\text{ mol}\).
Since the reaction stoichiometry is 1:1, theoretical moles of cyclohexene = \(0.1198\text{ mol}\).
Theoretical mass of cyclohexene: \(m = 0.1198 \times 82.1 = 9.836\text{ g}\).
Percentage yield: \(\% \text{ yield} = \frac{5.40}{9.836} \times 100\% = 54.91\% \approx 54.9\%\).
(ii) Add a small piece of sodium metal to a sample. If no cyclohexanol is present, there will be no bubbling/effervescence. Alternatively, add \(PCl_5\); the absence of cyclohexanol is confirmed if no misty white fumes (of \(HCl\)) are produced.

(a) [3 marks total]:
- Workable distillation setup with flask, heat, condenser, receiver, and thermometer bulb correctly placed at the T-junction (1 mark)
- Correct direction of water flow in condenser (in at bottom, out at top) (1 mark)
- System is open/not sealed, and all components are appropriately labelled (1 mark)
(b)(i) [2 marks total]:
- Neutralise the acid catalyst/phosphoric acid (1 mark)
- Releases pressure of CO2 formed / venting (1 mark)
(b)(ii) [2 marks total]:
- Drying agent / removes trace water (1 mark)
- Liquid turns from cloudy to clear/transparent (1 mark)
(b)(iii) [1.5 marks total]:
- Separates substances by boiling point (0.5 mark)
- Cyclohexene boils at \(83\text{ }^\circ\text{C}\) / cyclohexanol has a higher boiling point due to hydrogen bonding (1 mark)
(c)(i) [3 marks total]:
- Calculates moles of cyclohexanol (1 mark)
- Calculates theoretical mass of cyclohexene (1 mark)
- Evaluates % yield to 3 s.f. (1 mark) [Accept \(54.9\%\)]
(c)(ii) [1 mark total]:
- Add \(PCl_5\) / sodium metal AND state negative result (no misty fumes / no bubbling) (1 mark) [Do not accept bromine water as it tests for the alkene product, not the absence of the alcohol]

(a) Sodium thiosulfate reacts instantly with the iodine generated: \(2S_2O_3^{2-}(aq) + I_2(aq) \rightarrow S_4O_6^{2-}(aq) + 2I^-(aq)\). This prevents iodine from reacting with starch to form the blue-black complex. Once all the thiosulfate is depleted, the next trace of iodine produced immediately reacts with the starch indicator, creating a sharp, sudden color transition to blue-black.
(b)(i) Potassium chloride is added to keep the total concentration of ions (ionic strength) constant across all mixtures. Changing the ion concentration can affect rate constants independently of the reactant concentration.
(ii) Rate is proportional to \(1/t\).
Comparing Experiment 1 and 2: \([S_2O_8^{2-}]\) is kept constant. \([I^-]\) is doubled (volume of \(KI\) increases from \(10.0\text{ cm}^3\) to \(20.0\text{ cm}^3\)). The time decreases from \(45\text{ s}\) to \(22\text{ s}\), which means the rate approximately doubles (increase by factor of \(45/22 = 2.05\)). Therefore, the reaction is first order with respect to \(I^-\).
Comparing Experiment 1 and 3: \([I^-]\) is kept constant. \([S_2O_8^{2-}]\) is doubled (volume of \(K_2S_2O_8\) increases from \(10.0\text{ cm}^3\) to \(20.0\text{ cm}^3\)). The time decreases from \(45\text{ s}\) to \(23\text{ s}\), which means the rate approximately doubles (increase by factor of \(45/23 = 1.96\)). Therefore, the reaction is first order with respect to \(S_2O_8^{2-}\).
(iii) The Arrhenius relationship is \(\ln(k) = -\frac{E_a}{RT} + \text{constant}\). Since \(1/t\) is proportional to the rate constant \(k\), we use \(\ln(1/t) = -\frac{E_a}{RT} + \text{constant}\).
Thus, \(\text{gradient} = -\frac{E_a}{R}\).
\(-6150 = -\frac{E_a}{8.31}\)
\(E_a = 6150 \times 8.31 = 51106.5\text{ J mol}^{-1}\).
Converting to \(\text{kJ mol}^{-1}\): \(E_a = +51.1\text{ kJ mol}^{-1}\) (to 3 s.f.).

(a) [2.5 marks total]:
- Thiosulfate reacts with iodine as soon as it is formed (1 mark)
- Starch turns blue-black once all thiosulfate is consumed (1 mark)
- Allows a constant / fixed amount of iodine to be produced before the endpoint (0.5 mark)
(b)(i) [2 marks total]:
- Keep ionic strength constant / concentration of ions constant (2 marks) [Award 1 mark if they say 'to keep total concentration of potassium ions constant']
(b)(ii) [4 marks total]:
- Compares Exp 1 & 2: doubling [I-] doubles rate, so 1st order (2 marks)
- Compares Exp 1 & 3: doubling [S2O8 2-] doubles rate, so 1st order (2 marks)
(b)(iii) [4 marks total]:
- Recalls / uses relationship: \(gradient = -E_a / R\) (1 mark)
- Calculates numerical value of \(E_a\) as \(51106.5\text{ J}\) (1 mark)
- Converts value to \(\text{kJ mol}^{-1}\) (1 mark)
- Provides positive sign and rounds correctly to 3 s.f. (1 mark) [Accept \(+51.1\text{ kJ mol}^{-1}\)]

(a) Heat loss occurs continuously to the surroundings while the reaction takes place and before the maximum temperature is recorded. Extrapolating the cooling curve back to the time of mixing corrects for this heat loss, giving a more accurate theoretical maximum temperature rise.
(b) Total volume = \(50.0 + 50.0 = 100.0\text{ cm}^3\).
Mass of mixture = \(100.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 100.0\text{ g}\).
\(q = mc\Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 13.4\text{ }^\circ\text{C} = 5601.2\text{ J}\).
(c) Moles of \(HCl\) used: \(n(HCl) = 2.00 \times 0.0500 = 0.100\text{ mol}\).
Moles of \(NaOH\) used: \(n(NaOH) = 2.05 \times 0.0500 = 0.1025\text{ mol}\).
Since \(HCl\) is the limiting reactant, moles of \(H_2O\) formed = \(0.100\text{ mol}\).
\(\Delta H_{neut} = -\frac{q}{1000 \times n(H_2O)} = -\frac{5601.2}{1000 \times 0.100} = -56.012\text{ kJ mol}^{-1}\).
To 3 s.f.: \(-56.0\text{ kJ mol}^{-1}\).
(d) Improvements to reduce heat loss: add a lid to the cup; place the cup inside a secondary insulated beaker (with bubble wrap or cotton wool filling the gap).
(e)(i) \(\Delta T\) will be smaller/lower. (ii) Ethanoic acid is weak and only partially dissociated. Energy must be supplied (an endothermic process) to dissociate the remaining undissociated ethanoic acid molecules, which decreases the overall net enthalpy release of the neutralisation.

(a) [2 marks total]:
- Heat is lost during the reaction (1 mark)
- Extrapolation estimates temperature rise if reaction were instantaneous / correction for heat loss (1 mark)
(b) [2 marks total]:
- Correctly identifies total mass as \(100.0\text{ g}\) (1 mark)
- Calculates \(q = 5601.2\text{ J}\) [accept \(5600\text{ J}\)] (1 mark)
(c) [4 marks total]:
- Calculates moles of \(HCl\) and \(NaOH\) correctly (1 mark)
- Identifies \(HCl\) as limiting and thus \(0.100\text{ mol}\) of water is formed (1 mark)
- Calculates \(\Delta H_{neut} = -56.012\text{ kJ mol}^{-1}\) (1 mark)
- Final answer given to 3 s.f. with negative sign (1 mark) [Accept \(-56.0\text{ kJ mol}^{-1}\)]
(d) [2 marks total]:
- Use a lid (1 mark)
- Double cup / place in a beaker packed with insulation (1 mark)
(e)(i) [1.5 marks total]:
- Value of \(\Delta T\) is lower/smaller (1 mark)
- Due to fewer hydrogen ions immediately available / less heat overall released (0.5 mark)
(e)(ii) [1 mark total]:
- Some energy is absorbed to completely dissociate/ionise the weak ethanoic acid (1 mark)