2023 Edexcel IAL Chemistry (YCH11) Practice Paper with Answers

1. Calculate the amount in moles of \(\text{NaHCO}_3\):
\(n(\text{NaHCO}_3) = \frac{1.68\text{ g}}{84.0\text{ g mol}^{-1}} = 0.0200\text{ mol}\)

2. Use the stoichiometric ratio from the balanced equation:
\(2\text{ mol}\) of \(\text{NaHCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\).
Therefore, \(n(\text{CO}_2) = \frac{0.0200\text{ mol}}{2} = 0.0100\text{ mol}\)

3. Calculate the volume of \(\text{CO}_2\) at r.t.p.:
\(V = 0.0100\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.240\text{ dm}^3\)

4. Convert the volume to \(\text{cm}^3\):
\(0.240\text{ dm}^3 \times 1000 = 240\text{ cm}^3\)

Award 1 mark for the correct option B.
- Incorrect option A (120) arises from failing to divide by 2 for stoichiometry.
- Incorrect option C (480) arises from multiplying by 2 instead of dividing by 2.
- Incorrect option D (960) arises from incorrect stoichiometric calculation.

To determine the group of element \(\text{X}\), look for the largest relative jump between successive ionization energies:
- 1st to 2nd: Increase of \(1239\text{ kJ mol}^{-1}\)
- 2nd to 3rd: Increase of \(928\text{ kJ mol}^{-1}\)
- 3rd to 4th: Increase of \(8833\text{ kJ mol}^{-1}\) (very large jump)
- 4th to 5th: Increase of \(3253\text{ kJ mol}^{-1}\)

The huge jump between the 3rd and 4th ionization energies indicates that the fourth electron is being removed from an inner quantum shell. This means there are three valence electrons in the outermost shell of element \(\text{X}\), so it belongs to Group 3.

Award 1 mark for the correct option C.
- Reject option A (Group 1 would show a large jump between the 1st and 2nd ionization energies).
- Reject option B (Group 2 would show a large jump between the 2nd and 3rd ionization energies).
- Reject option D (Group 4 would show a large jump between the 4th and 5th ionization energies).

Let's analyze the electron pair arrangements around the central atoms:
- A: \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs. This is based on a tetrahedral arrangement of electron pairs.
- B: \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair. This is based on a tetrahedral arrangement of electron pairs (giving a trigonal pyramidal shape).
- C: \(\text{BF}_4^-\) has 4 bonding pairs and 0 lone pairs. This is based on a tetrahedral arrangement of electron pairs.
- D: \(\text{SF}_4\) has 4 bonding pairs and 1 lone pair (total of 5 electron pairs) around the central sulfur atom. This is based on a trigonal bipyramidal arrangement of electron pairs (giving a seesaw shape).

Award 1 mark for the correct option D.
- Reject options A, B, and C as they all have 4 electron pairs around the central atom, which form a tetrahedral spatial arrangement.

2-methylbutane has the structure \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\). There are four distinct chemical environments for the hydrogen atoms:
1. Substitution of H on either of the two equivalent methyl groups at position 1 or 2 (methyl branch) yields: 1-chloro-2-methylbutane.
2. Substitution of H on the tertiary carbon at position 2 yields: 2-chloro-2-methylbutane.
3. Substitution of H on the secondary carbon at position 3 yields: 2-chloro-3-methylbutane.
4. Substitution of H on the primary carbon at position 4 yields: 1-chloro-3-methylbutane.

Therefore, exactly 4 structural isomers are possible.

Award 1 mark for the correct option C.
- Option A (2) is incorrect.
- Option B (3) is incorrect because it misses one of the distinct carbon positions.
- Option D (5) is incorrect as it double-counts equivalent methyl groups.

The reaction proceeds via a carbocation intermediate:
- Addition of \(\text{H}^+\) to the C1 carbon forms a secondary carbocation at C2: \(\text{CH}_3-\text{CH}^+-\text{CH}_3\).
- Addition of \(\text{H}^+\) to the C2 carbon forms a primary carbocation at C1: \(\text{CH}_3-\text{CH}_2-\text{CH}_2^+\).

The secondary carbocation is more stable than the primary carbocation due to the electron-releasing inductive effect of the two adjacent methyl (alkyl) groups, which disperse the positive charge. Therefore, the pathway via the more stable secondary carbocation is favored, making 2-bromopropane the major product.

Award 1 mark for the correct option A.
- Reject option B (describes the primary carbocation as more stable, which is incorrect).
- Reject option C (steric hindrance does not determine the major product here).
- Reject option D (propene is unsymmetrical, but HBr is also an unsymmetrical electrophile, and this does not explain the major product preference).

1. Find the mass of carbon from the \(\text{CO}_2\):
\(m(\text{C}) = 0.440\text{ g} \times \frac{12.0}{44.0} = 0.120\text{ g}\)

2. Find the mass of hydrogen from the \(\text{H}_2\text{O}\):
\(m(\text{H}) = 0.270\text{ g} \times \frac{2.0}{18.0} = 0.030\text{ g}\)

3. Find the mass of oxygen by subtraction:
\(m(\text{O}) = 0.230\text{ g} - 0.120\text{ g} - 0.030\text{ g} = 0.080\text{ g}\)

4. Convert masses to moles:
\(n(\text{C}) = \frac{0.120}{12.0} = 0.010\text{ mol}\)
\(n(\text{H}) = \frac{0.030}{1.0} = 0.030\text{ mol}\)
\(n(\text{O}) = \frac{0.080}{16.0} = 0.005\text{ mol}\)

5. Determine the simplest whole-number ratio by dividing by the smallest value (0.005):
\(\text{C} : \text{H} : \text{O} = \frac{0.010}{0.005} : \frac{0.030}{0.005} : \frac{0.005}{0.005} = 2 : 6 : 1\)

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

Award 1 mark for the correct option C.
- Reject other options as they represent incorrect molar ratios of the elements.

Cobalt has an atomic number of \(Z = 27\).

1. The electronic configuration of a neutral cobalt atom (\(\text{Co}\)) is:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7 4\text{s}^2\)

2. When transition metals form ions, they lose electrons from the outermost \(4\text{s}\) orbital first before losing any from the \(3\text{d}\) orbital.

3. For \(\text{Co}^{2+}\), two electrons are removed from the \(4\text{s}\) subshell, resulting in:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7\)

Award 1 mark for the correct option A.
- Option B is incorrect as it shows electrons incorrectly remaining in the \(4\text{s}\) subshell while being lost from \(3\text{d}\).
- Options C and D represent incorrect numbers of total electrons or incorrect distributions.

- \(\text{NH}_3\) is trigonal pyramidal and asymmetric, so its highly polar \(\text{N}-\text{H}\) bond dipoles do not cancel, making it a polar molecule.
- \(\text{SF}_6\) is octahedral and highly symmetrical. Although each \(\text{S}-\text{F}\) bond is highly polar due to the large electronegativity difference, the symmetrical distribution causes the individual bond dipoles to completely cancel out, resulting in a net dipole moment of zero.
- \(\text{H}_2\text{O}\) is bent and asymmetric, so its polar bond dipoles do not cancel.
- \(\text{CHCl}_3\) is tetrahedral but asymmetric because of the different atoms bonded to carbon, so its dipoles do not cancel.

Award 1 mark for the correct option B.
- Reject options A, C, and D because they are asymmetric molecules that retain a net molecular dipole moment.

Calculate the moles of each element in \(100\text{ g}\) of the compound: \(n(\text{C}) = \frac{62.0}{12.0} = 5.17\text{ mol}\), \(n(\text{H}) = \frac{10.4}{1.0} = 10.4\text{ mol}\), \(n(\text{O}) = \frac{27.6}{16.0} = 1.725\text{ mol}\). Dividing by the smallest value (\(1.725\)) gives: \(\text{C} = 3\), \(\text{H} = 6\), \(\text{O} = 1\). Thus, the empirical formula is \(\text{C}_3\text{H}_6\text{O}\).

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

First ionization energy generally increases across Period 3 due to increasing nuclear charge and constant shielding. However, there is a dip at \(\text{Al}\) because its outer electron is in a higher-energy \(3\text{p}\) orbital, which is more shielded and easier to remove than the \(3\text{s}\) electron of \(\text{Mg}\). Thus, the first ionization energy of \(\text{Al}\) is lower than that of \(\text{Mg}\). The order of increasing first ionization energy is: \(\text{Na} < \text{Al} < \text{Mg} < \text{Si}\).

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

The \(\text{NH}_2^-\) ion has 2 bonding pairs and 2 lone pairs on the central nitrogen atom. This results in a bent shape based on a tetrahedral electron geometry. Due to strong lone pair-lone pair repulsion, the bond angle is reduced from the tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\).

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

A termination step involves the combination of two free radicals to form a stable molecule, eliminating free radicals. The equation \(\text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\) shows two radicals reacting to form a single neutral molecule.

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

The reaction proceeds via a carbocation intermediate. The secondary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\)) is more stable than the primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+\)) due to the electron-releasing inductive effect of the two adjacent alkyl groups, leading to the preferential formation of 2-bromobutane.

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

The equation for thermal decomposition is \(\text{CaCO}_3\text{(s)} \rightarrow \text{CaO\text{(s)}} + \text{CO}_2\text{(g)}\). The amount of \(\text{CaCO}_3\) is \(10.0 / 100.1 = 0.0999\text{ mol}\). This produces \(0.0999\text{ mol}\) of \(\text{CO}_2\). The volume is \(0.0999 \times 24.0 = 2.40\text{ dm}^3\).

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

The largest jump in ionization energy occurs between the 3rd and the 4th ionization energies (from 2745 to 11578 kJ/mol), which indicates that the fourth electron is being removed from an inner, more stable electron shell. This means element \(Y\) has three valence electrons, placing it in Group 13 (Group 3).

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

According to Fajans' Rules, covalent character increases as the polarizing power of the cation increases. Cation polarizing power increases with high positive charge and small ionic radius (high charge density). Among \(\text{Na}^+\), \(\text{Mg}^{2+}\), \(\text{Al}^{3+}\), and \(\text{Ca}^{2+}\), the \(\text{Al}^{3+}\) cation has the highest charge and smallest radius, making it highly polarizing and imparting the greatest covalent character to \(\text{AlCl}_3\).

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

There is a very large increase between the third and fourth ionization energies (\(2745 \text{ to } 11577 \text{ kJ mol}^{-1}\)). This indicates that the fourth electron is removed from an inner quantum shell, which is closer to the nucleus and experiences much less shielding. Therefore, the element has three valence electrons in its outer shell. In Period 3, the element with three outer electrons is aluminium (Group 13/3).

[1 mark] C - Aluminium. Correctly identifies the element based on the position of the largest jump in ionization energies.

1. Calculate the amount in moles of \( \text{CO}_2 \) gas collected:
$$\text{Moles of } \text{CO}_2 = \frac{504\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0210\text{ mol}$$

2. Determine the moles of \( \text{MCO}_3 \):
Since the ratio of \( \text{MCO}_3 : \text{CO}_2 \) is \( 1:1 \), the moles of \( \text{MCO}_3 \) decomposed is also \( 0.0210\text{ mol} \).

3. Calculate the molar mass of \( \text{MCO}_3 \):
$$\text{Molar mass of } \text{MCO}_3 = \frac{2.10\text{ g}}{0.0210\text{ mol}} = 100\text{ g mol}^{-1}$$

4. Calculate the relative atomic mass of M:
$$\text{Molar mass of } \text{CO}_3^{2-} = 12.0 + (3 \times 16.0) = 60.0\text{ g mol}^{-1}$$
$$A_r(\text{M}) = 100 - 60.0 = 40.0\text{ g mol}^{-1}$$

Looking at the periodic table, the Group 2 metal with an atomic mass of \( 40.1\text{ g mol}^{-1} \) is calcium (Ca).

[1 mark] B - Calcium. Correctly identifies the metal through calculating the molar mass of the metal carbonate.

In tetrafluoromethane (\( \text{CF}_4 \)), the C-F bonds are highly polar due to the large difference in electronegativity between carbon and fluorine. However, the molecule is symmetrical with a tetrahedral shape, meaning the dipole moments of the four C-F bonds cancel each other out, resulting in a non-polar molecule overall (zero net dipole moment).
- Ammonia (\( \text{NH}_3 \)) is trigonal pyramidal and asymmetric due to the lone pair, so it is polar.
- Sulfur tetrafluoride (\( \text{SF}_4 \)) has a see-saw shape due to a lone pair on the sulfur atom, which makes it asymmetric and polar.
- Chloromethane (\( \text{CH}_3\text{Cl} \)) is tetrahedral but asymmetric because of the different electronegativities of H and Cl, making it polar.

[1 mark] C - Tetrafluoromethane, \( \text{CF}_4 \). Correctly identifies that a symmetrical tetrahedral shape causes individual bond dipoles to cancel out.

2-methylbut-2-ene has the structure \( \text{(CH}_3\text{)}_2\text{C=CH-CH}_3 \).
During the electrophilic addition of HBr, the electrophile \( \text{H}^+ \) adds to the double bond to form the most stable carbocation intermediate:
- Addition of \( \text{H}^+ \) to C3 forms a tertiary carbocation at C2: \( \text{(CH}_3\text{)}_2\text{C}^+-\text{CH}_2\text{CH}_3 \).
- Addition of \( \text{H}^+ \) to C2 forms a secondary carbocation at C3: \( \text{(CH}_3\text{)}_2\text{CH}-\text{CH}^+-\text{CH}_3 \).

A tertiary carbocation is more stable than a secondary carbocation due to the electron-releasing inductive effect of the three alkyl groups attached to the positively charged carbon atom. The bromide ion (\( \text{Br}^- \)) then attacks the tertiary carbocation to form the major product, 2-bromo-2-methylbutane.

[1 mark] B - 2-bromo-2-methylbutane, formed via the more stable tertiary carbocation. Correct major product name and mechanistic explanation.

Assume a 100 g sample of the compound. Mass of carbon = \(85.7\text{ g}\), mass of hydrogen = \(100 - 85.7 = 14.3\text{ g}\). Moles of carbon = \(85.7 / 12.0 = 7.14\text{ mol}\). Moles of hydrogen = \(14.3 / 1.0 = 14.3\text{ mol}\). Divide by the smallest value to find the ratio: \(7.14 / 7.14 = 1\) for carbon, and \(14.3 / 7.14 = 2\) for hydrogen. Therefore, the empirical formula is \(\text{CH}_2\).

M1: Divides percentage mass by relative atomic mass for both carbon and hydrogen (\(85.7/12.0\) and \(14.3/1.0\)) [1 mark]. M2: Determines the simplest whole-number molar ratio of \(1 : 2\) [1 mark]. A1: Concludes with the correct empirical formula of \(\text{CH}_2\) [0.5 marks].

Across Period 3, from sodium to argon, the number of protons in the nucleus increases, leading to an increased nuclear charge. The outer electrons are added to the same energy level, which means they experience similar shielding. Consequently, there is a stronger electrostatic attraction between the positive nucleus and the outer electrons, requiring more energy to remove an electron.

M1: States that the first ionization energy generally increases across Period 3 [0.5 marks]. M2: Explains that nuclear charge (or number of protons) increases [1 mark]. M3: Explains that shielding remains relatively constant, resulting in stronger attraction on outer electrons [1 mark].

In graphite, each carbon atom is covalently bonded to three other carbon atoms in a hexagonal planar arrangement. This leaves one valence electron per carbon atom delocalised. These delocalised electrons are free to move between layers and carry electrical charge. In contrast, in diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms via strong, localised covalent bonds. All valence electrons are held tightly within these bonds, leaving no mobile charged particles to conduct electricity.

M1: Explains that graphite has three bonds per carbon, leaving one delocalised electron per atom free to move [1 mark]. M2: Explains that diamond has four covalent bonds per carbon, with all electrons localised / no free-moving electrons [1 mark]. A1: Links free-moving electrons in graphite to electrical conductivity and lack thereof in diamond [0.5 marks].

In the first propagation step, a chlorine free radical abstracts a hydrogen atom from ethane to form an ethyl radical and hydrogen chloride: \(\text{C}_2\text{H}_6 + \text{Cl}^\bullet \rightarrow \text{C}_2\text{H}_5^\bullet + \text{HCl}\). In the second propagation step, the ethyl radical reacts with a chlorine molecule to produce chloroethane and regenerate the chlorine free radical: \(\text{C}_2\text{H}_5^\bullet + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^\bullet\).

M1: Correct first propagation step including radical dots: \(\text{C}_2\text{H}_6 + \text{Cl}^\bullet \rightarrow \text{C}_2\text{H}_5^\bullet + \text{HCl}\) [1 mark]. M2: Correct second propagation step including radical dots: \(\text{C}_2\text{H}_5^\bullet + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^\bullet\) [1 mark]. A1: Fully correct chemical formulas and radical dot representation throughout [0.5 marks].

The reaction of propene with \(\text{HBr}\) proceeds via a carbocation intermediate. Addition of a hydrogen ion to carbon-1 forms a secondary carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_3\)), whereas addition to carbon-2 forms a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)). The secondary carbocation is more stable than the primary carbocation because it has two electron-releasing alkyl (methyl) groups that reduce the positive charge on the carbocation carbon via the inductive effect. The major product (2-bromopropane) is formed from the more stable secondary carbocation.

M1: Identifies that the major product is formed via a secondary carbocation intermediate and the minor product via a primary carbocation [1 mark]. M2: States that the secondary carbocation intermediate is more stable than the primary carbocation [1 mark]. A1: Explains stability in terms of the positive inductive effect of two electron-releasing alkyl groups compared to one [0.5 marks].

Atom economy is calculated as: \(\text{Atom economy} = (\text{Molar mass of desired product} / \text{Total molar mass of reactants}) \times 100\%\). Desired product is cyclohexene, mass \(= 82.0\text{ g mol}^{-1}\). The only reactant is cyclohexanol, mass \(= 100.0\text{ g mol}^{-1}\). Therefore, \(\text{Atom economy} = (82.0 / 100.0) \times 100 = 82.0\%\).

M1: Recalls or correctly applies the formula for atom economy [1 mark]. M2: Substitutes values correctly: \((82.0 / 100.0) \times 100\) [1 mark]. A1: Gives final answer as \(82.0\%\) to three significant figures [0.5 marks].

Relative atomic mass (\(A_r\)) is the weighted average of the isotopic masses: \(A_r = \frac{(20 \times 90.48) + (21 \times 0.27) + (22 \times 9.25)}{100} = \frac{1809.6 + 5.67 + 203.5}{100} = \frac{2018.77}{100} = 20.1877\). Rounding to two decimal places gives \(20.19\).

M1: Correct expression for the weighted average calculation [1 mark]. M2: Correct calculation of raw value (\(20.1877\)) [1 mark]. A1: Correct rounding to two decimal places: \(20.19\) [0.5 marks].

The ammonia (\(\text{NH}_3\)) molecule has a trigonal pyramidal shape because the central nitrogen atom has three bonding pairs and one lone pair of electrons. These four electron pairs arrange themselves tetrahedrally to minimize repulsion, but the presence of the lone pair results in a pyramidal geometry. Boron trifluoride (\(\text{BF}_3\)) has a trigonal planar shape because the central boron atom has only three bonding pairs and no lone pairs. These three bonding pairs repel each other equally to achieve maximum separation at \(120^\circ\) bond angles.

M1: Identifies the correct shapes: \(\text{NH}_3\) is trigonal pyramidal and \(\text{BF}_3\) is trigonal planar [1 mark]. M2: Explains that \(\text{NH}_3\) has three bonding pairs and one lone pair, while \(\text{BF}_3\) has three bonding pairs and zero lone pairs [1 mark]. A1: Refers to electron pair repulsion theory (i.e., repulsion to maximize separation / lone pair-bonding pair repulsion being greater than bonding pair-bonding pair repulsion) [0.5 marks].

1. **Calculate the moles of carbon:**
\(n(\text{CO}_2) = \frac{0.660\text{ g}}{44.0\text{ g mol}^{-1}} = 0.0150\text{ mol}\)
Since each mole of \(\text{CO}_2\) contains one mole of carbon, \(n(\text{C}) = 0.0150\text{ mol}\).
Mass of carbon = \(0.0150\text{ mol} \times 12.0\text{ g mol}^{-1} = 0.180\text{ g}\).

2. **Calculate the moles of hydrogen:**
\(n(\text{H}_2\text{O}) = \frac{0.360\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0200\text{ mol}\)
Since each mole of \(\text{H}_2\text{O}\) contains two moles of hydrogen, \(n(\text{H}) = 2 \times 0.0200 = 0.0400\text{ mol}\).
Mass of hydrogen = \(0.0400\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.040\text{ g}\).

3. **Calculate the mass and moles of oxygen:**
Mass of oxygen = \(0.300\text{ g} - (0.180\text{ g} + 0.040\text{ g}) = 0.080\text{ g}\).
\(n(\text{O}) = \frac{0.080\text{ g}}{16.0\text{ g mol}^{-1}} = 0.0050\text{ mol}\).

4. **Determine the simplest ratio:**
\(\text{C} : \text{H} : \text{O} = 0.0150 : 0.0400 : 0.0050\)
Dividing all values by the smallest number (\(0.0050\)):
\(\text{C} = \frac{0.0150}{0.0050} = 3\)
\(\text{H} = \frac{0.0400}{0.0050} = 8\)
\(\text{O} = \frac{0.0050}{0.0050} = 1\)

The empirical formula of **X** is \(\text{C}_3\text{H}_8\text{O}\).

- **[1 Mark]** For calculating \(0.0150\text{ mol}\) of \(\text{C}\) (or \(0.180\text{ g}\)) AND \(0.0400\text{ mol}\) of \(\text{H}\) (or \(0.040\text{ g}\)).
- **[1 Mark]** For calculating the mass of oxygen as \(0.080\text{ g}\) and converting this to \(0.0050\text{ mol}\).
- **[0.5 Marks]** For finding the simplest whole-number molar ratio and writing the correct empirical formula \(\text{C}_3\text{H}_8\text{O}\) (or \(\text{C}_3\text{H}_8\text{O}_1\)).

The first three ionization energies are relatively low, followed by a very large increase (jump) between the 3rd and 4th ionization energies (from \(2745\text{ kJ mol}^{-1}\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the first three electrons are removed from the outer shell, whereas the fourth electron is removed from an inner shell closer to the nucleus, experiencing significantly less shielding and stronger nuclear attraction. Therefore, element **Y** has three valence electrons, which identifies it as Aluminium (\(\text{Al}\)), a Group 3 (Group 13) element in Period 3 with the electronic configuration \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^1\).

- **[1 Mark]** Identifies element **Y** as aluminium / \(\text{Al}\).
- **[1 Mark]** Explains that there is a large increase / jump between the 3rd and 4th ionization energies.
- **[0.5 Marks]** Explains that this jump indicates the 4th electron is removed from an inner shell / a shell closer to the nucleus (or the \(2\text{p}\) subshell instead of the \(3\text{rd}\) shell).

For \(E/Z\) (geometric) isomerism to occur, two conditions must be met:
1. There must be restricted rotation about the carbon-carbon double bond (due to the presence of a \(\pi\)-bond).
2. Each carbon atom of the \(\text{C}=\text{C}\) double bond must be attached to two different groups or atoms.

Applying this to 3-methylpent-2-ene:
- The double bond restricts rotation.
- Carbon-2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups).
- Carbon-3 is bonded to \(-\text{CH}_3\) and \(-\text{CH}_2\text{CH}_3\) (two different groups).

- **[1 Mark]** Explains that there is restricted rotation about the \(\text{C}=\text{C}\) double bond.
- **[1 Mark]** Explains that each carbon of the double bond is bonded to two different groups/atoms.
- **[0.5 Marks]** Explicitly identifies the two different groups attached to each carbon of the double bond in 3-methylpent-2-ene: Carbon-2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\), and Carbon-3 is bonded to \(-\text{CH}_3\) and \(-\text{CH}_2\text{CH}_3\) (or ethyl).

Using Hess's Law and the enthalpy changes of combustion:
\(\Delta H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\)

Substitute the given combustion values into the expression:
\(\Delta H^\ominus = [\Delta_c H^\ominus(\text{C}_3\text{H}_6\text{(g)}) + \Delta_c H^\ominus(\text{H}_2\text{(g)})] - [\Delta_c H^\ominus(\text{C}_3\text{H}_8\text{(g)})]\)
\(\Delta H^\ominus = [-2058 + (-286)] - [-2220]\)
\(\Delta H^\ominus = -2344 - (-2220)\)
\(\Delta H^\ominus = -124\text{ kJ mol}^{-1}\)

- **[1 Mark]** For drawing a correct Hess's cycle or stating the correct expression relating enthalpy of reaction to combustion enthalpies: \(\Delta H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\).
- **[1 Mark]** For correct substitution of values: \(-2058 - 286 - (-2220)\).
- **[0.5 Marks]** For the correct final answer of \(-124\text{ kJ mol}^{-1}\) (must include the negative sign and correct units; accept without units if already specified in the question).

Step 1: Calculate the total percentage of the remaining three isotopes:
\(100\% - 5.80\% = 94.20\%\)

Step 2: Calculate the sum of the ratio parts for the three remaining isotopes:
\(23.4 + 0.560 + 1.00 = 24.96\)

Step 3: Calculate the percentage abundance of each remaining isotope:
- Abundance of \(^{56}\text{Fe} = \frac{23.4}{24.96} \times 94.20\% = 88.3125\%\)
- Abundance of \(^{57}\text{Fe} = \frac{0.560}{24.96} \times 94.20\% = 2.1122\%\)
- Abundance of \(^{58}\text{Fe} = \frac{1.00}{24.96} \times 94.20\% = 3.7753\%\)

Step 4: Calculate the relative atomic mass (\(A_r\)):
\(A_r = \frac{(54 \times 5.80) + (56 \times 88.3125) + (57 \times 2.1122) + (58 \times 3.7753)}{100}\)
\(A_r = \frac{313.20 + 4945.50 + 120.395 + 218.967}{100}\)
\(A_r = \frac{5598.06}{100} = 55.98\)

- Mark 1: Deduce that the total abundance of \(^{56}\text{Fe}\), \(^{57}\text{Fe}\), and \(^{58}\text{Fe}\) is \(94.20\%\).
- Mark 2: Correct calculation of the percentage abundance of \(^{56}\text{Fe}\) as \(88.31\%\) (accept \(88.3\%\)).
- Mark 3: Correct calculation of the percentage abundances of \(^{57}\text{Fe}\) (\(2.11\%\)) and \(^{58}\text{Fe}\) (\(3.78\%\) or \(3.77\%\)).
- Mark 4: Correct setup of the relative atomic mass equation using their calculated abundances.
- Mark 5: Final answer of \(55.98\) (allow \(55.99\) from rounding differences), given to 2 decimal places. (No marks for final answer if not given to 2 decimal places).

Step 1: Calculate the mass and moles of carbon in the product CO2:
\(n(\text{CO}_2) = \frac{2.64\text{ g}}{44.0\text{ g mol}^{-1}} = 0.0600\text{ mol}\)
Since 1 mole of \(\text{CO}_2\) contains 1 mole of carbon:
\(n(\text{C}) = 0.0600\text{ mol}\)
\(m(\text{C}) = 0.0600\text{ mol} \times 12.0\text{ g mol}^{-1} = 0.720\text{ g}\)

Step 2: Calculate the mass and moles of hydrogen in the product H2O:
\(n(\text{H}_2\text{O}) = \frac{1.08\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0600\text{ mol}\)
Since 1 mole of \(\text{H}_2\text{O}\) contains 2 moles of hydrogen:
\(n(\text{H}) = 2 \times 0.0600 = 0.120\text{ mol}\)
\(m(\text{H}) = 0.120\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.120\text{ g}\)

Step 3: Calculate the mass and moles of oxygen in compound Y by subtraction:
\(m(\text{O}) = 1.48\text{ g} - (0.720\text{ g} + 0.120\text{ g}) = 0.640\text{ g}\)
\(n(\text{O}) = \frac{0.640\text{ g}}{16.0\text{ g mol}^{-1}} = 0.0400\text{ mol}\)

Step 4: Find the simplest whole number ratio:
\(\text{C} : \text{H} : \text{O} = 0.0600 : 0.120 : 0.0400\)
Divide by the smallest value (0.0400):
\(\text{C} = 1.50\), \(\text{H} = 3.00\), \(\text{O} = 1.00\)
Multiply by 2 to obtain whole numbers:
\(\text{C} : \text{H} : \text{O} = 3 : 6 : 2\)

Therefore, the empirical formula of Y is \(\text{C}_3\text{H}_6\text{O}_2\).

- Mark 1: Correct calculation of the moles of C (\(0.0600\text{ mol}\)) or mass of C (\(0.720\text{ g}\)).
- Mark 2: Correct calculation of the moles of H (\(0.120\text{ mol}\)) or mass of H (\(0.120\text{ g}\)).
- Mark 3: Correct subtraction to find the mass of O (\(0.640\text{ g}\)) and calculation of its moles (\(0.0400\text{ mol}\)).
- Mark 4: Correct calculation of the mole ratio of C : H : O as \(1.5 : 3 : 1\) (or \(0.0600 : 0.120 : 0.0400\)).
- Mark 5: Correct conversion of the ratio to the simplest whole-number ratio, giving the final empirical formula \(\text{C}_3\text{H}_6\text{O}_2\).

According to electron pair repulsion theory (VSEPR), the shape of a molecule or ion is determined by the mutual repulsion of electron pairs around the central atom, which arrange themselves as far apart as possible to minimize repulsion.

The central nitrogen atom has a different number of lone pairs and bonding pairs in each species:
1. \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs. To minimize repulsion, these four pairs point to the corners of a tetrahedron, giving a tetrahedral shape with a bond angle of \(109.5^\circ\).
2. \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair. Lone pairs repel more strongly than bonding pairs. This squeezes the bond angle down from the tetrahedral angle by about \(2.5^\circ\), resulting in a trigonal pyramidal shape with a bond angle of \(107^\circ\).
3. \(\text{NH}_2^-\) has 2 bonding pairs and 2 lone pairs. The strong repulsion of the two lone pairs reduces the bond angle even further, resulting in a non-linear (or bent / V-shaped) geometry with a bond angle of \(104.5^\circ\).

- Mark 1: State that electron pairs around the central nitrogen atom repel each other and move as far apart as possible to minimize repulsion.
- Mark 2: Identify the correct number of bonding and lone pairs for all three species: \(\text{NH}_4^+\) (4 bonding, 0 lone), \(\text{NH}_3\) (3 bonding, 1 lone), and \(\text{NH}_2^-\) (2 bonding, 2 lone).
- Mark 3: State the principle that lone pair-lone pair repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion.
- Mark 4: State the correct shapes for all three species: \(\text{NH}_4^+\) is tetrahedral, \(\text{NH}_3\) is trigonal pyramidal, and \(\text{NH}_2^-\) is non-linear / bent / V-shaped.
- Mark 5: State the correct bond angles for all three species: \(109.5^\circ\) (for \(\text{NH}_4^+\)), \(107^\circ\) (for \(\text{NH}_3\)), and \(104.5^\circ\) (allow \(104^\circ - 105^\circ\) for \(\text{NH}_2^-\)).

(a) In the presence of UV light, homolytic fission of the Cl-Cl bond occurs to produce chlorine radicals:
Initiation: \(\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet\)

In the propagation steps, these radicals react with the alkane to generate alkyl radicals, which then react with more halogen molecules:
Propagation 1: \(\text{C}_2\text{H}_6 + \text{Cl}^\bullet \rightarrow \text{C}_2\text{H}_5^\bullet + \text{HCl}\)
Propagation 2: \(\text{C}_2\text{H}_5^\bullet + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^\bullet\)

(b) Butane is formed as a minor product in a termination step. When two ethyl radicals, \(\text{C}_2\text{H}_5^\bullet\), collide and combine, they form a covalent bond to yield a butane molecule:
Termination step: \(2\text{C}_2\text{H}_5^\bullet \rightarrow \text{C}_4\text{H}_{10}\)

- Mark 1: Correct equation for initiation showing UV light/condition and radical dots: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\).
- Mark 2: Correct equation for first propagation step: \(\text{C}_2\text{H}_6 + \text{Cl}^\bullet \rightarrow \text{C}_2\text{H}_5^\bullet + \text{HCl}\).
- Mark 3: Correct equation for second propagation step: \(\text{C}_2\text{H}_5^\bullet + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^\bullet\).
- Mark 4: State that butane is formed via a termination step when two ethyl radicals collide/combine.
- Mark 5: Correct chemical equation for this termination reaction: \(2\text{C}_2\text{H}_5^\bullet \rightarrow \text{C}_4\text{H}_{10}\) (accept \(\text{C}_2\text{H}_5^\bullet + \text{C}_2\text{H}_5^\bullet \rightarrow \text{C}_4\text{H}_{10}\)).

The mechanism is electrophilic addition:

Step 1: The \(\pi\)-bond of the propene double bond attacks the partially positive hydrogen atom of \(\text{H}^{\delta+}-\text{Br}^{\delta-}\). This is shown with a curly arrow starting from the double bond to the \(\text{H}\) atom, and another curly arrow from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom. This forms a bromide ion (\(\text{Br}^-\)) and a secondary carbocation intermediate: \(\text{CH}_3\text{CH}^+\text{CH}_3\).

Step 2: The bromide ion attacks the positively charged carbon atom of the secondary carbocation. This is shown with a curly arrow from a lone pair on the \(\text{Br}^-\) ion to the carbon with the positive charge, yielding 2-bromopropane.

Explanation of major product:
Addition of \(\text{H}^+\) to the terminal carbon forms a secondary carbocation, \(\text{CH}_3\text{CH}^+\text{CH}_3\). Addition of \(\text{H}^+\) to the central carbon forms a primary carbocation, \(\text{CH}_3\text{CH}_2\text{CH}_2^+\). The secondary carbocation is more stable than the primary carbocation because it has two electron-releasing methyl groups (alkyl groups) that help disperse the positive charge (+I inductive effect). The reaction thus proceeds preferentially via the more stable secondary carbocation intermediate, making 2-bromopropane the major product.

- Mark 1: Correct first step of mechanism: curly arrow from \(\text{C}=\text{C}\) double bond to the \(\text{H}\) of \(\text{H}-\text{Br}\), and curly arrow from the \(\text{H}-\text{Br}\) bond to \(\text{Br}\), with dipoles on \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) correctly shown.
- Mark 2: Correct structure of the secondary carbocation intermediate (\(\text{CH}_3\text{CH}^+\text{CH}_3\)) and the bromide ion with its lone pair and negative charge (\(:\text{Br}^-\)).
- Mark 3: Correct second step of mechanism: curly arrow from the lone pair of the bromide ion to the positive carbon atom of the carbocation.
- Mark 4: State that the secondary carbocation intermediate is more stable than the primary carbocation.
- Mark 5: Explain that alkyl groups are electron-releasing (or have an inductive effect / +I) which helps disperse the positive charge on the carbocation.

Step 1: Calculate the moles of \(\text{HCl}\) used in the titration:
\(n(\text{HCl}) = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\)

Step 2: Use the reaction stoichiometry to find the moles of \(\text{Na}_2\text{CO}_3\) in the \(25.0\text{ cm}^3\) sample:
According to the equation, 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\).
\(n(\text{Na}_2\text{CO}_3) \text{ in } 25.0\text{ cm}^3 = \frac{2.50 \times 10^{-3}\text{ mol}}{2} = 1.25 \times 10^{-3}\text{ mol}\)

Step 3: Calculate the moles of \(\text{Na}_2\text{CO}_3\) in the full \(250.0\text{ cm}^3\) volumetric flask:
\(n(\text{Na}_2\text{CO}_3) \text{ in } 250.0\text{ cm}^3 = 1.25 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 1.25 \times 10^{-2}\text{ mol}\)

Step 4: Calculate the molar mass (\(M_r\)) of the hydrated salt:
\(M_r(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}) = \frac{\text{mass}}{\text{moles}} = \frac{3.58\text{ g}}{1.25 \times 10^{-2}\text{ mol}} = 286.4\text{ g mol}^{-1}\)

Step 5: Determine the value of \(x\):
\(M_r(\text{Na}_2\text{CO}_3) = 2(22.99) + 12.01 + 3(16.00) = 105.99\text{ g mol}^{-1}\)
Mass of water of crystallisation = \(286.4 - 105.99 = 180.41\text{ g mol}^{-1}\)
\(x = \frac{180.41}{18.02} \approx 10.0\)

Thus, \(x = 10\).

- Mark 1: Calculate the amount of \(\text{HCl}\) used: \(2.50 \times 10^{-3}\text{ mol}\).
- Mark 2: Calculate the amount of \(\text{Na}_2\text{CO}_3\) in the titrated \(25.0\text{ cm}^3\) portion: \(1.25 \times 10^{-3}\text{ mol}\) (by dividing by 2).
- Mark 3: Calculate the amount of \(\text{Na}_2\text{CO}_3\) in the whole \(250.0\text{ cm}^3\) flask: \(1.25 \times 10^{-2}\text{ mol}\) (by multiplying by 10).
- Mark 4: Calculate the experimental molar mass of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\) as \(286.4\text{ g mol}^{-1}\).
- Mark 5: Obtain the final integer value \(x = 10\) using a correct mathematical method (accept 10 from correct calculation).

Butan-1-ol is a straight-chain primary alcohol. It has a larger surface area in contact with neighbouring molecules compared to its branched isomers (2-methylpropan-1-ol and 2-methylpropan-2-ol) and secondary isomer (butan-2-ol). Consequently, it has stronger London forces, requiring more energy to overcome, resulting in the highest boiling temperature.

[1] A - Butan-1-ol

As Group 2 is descended, the cationic radius increases while the ionic charge remains at \(+2\). This means the barium ion has a much lower charge density than the magnesium ion. Consequently, the barium ion polarises the carbonate ion's electron cloud to a lesser extent, making the carbonate ion more stable and requiring more thermal energy to decompose.

[1] B - The barium ion has a lower charge density and polarises the carbonate ion to a lesser extent.

The rate of hydrolysis of halogenoalkanes is determined by the bond enthalpy of the carbon-halogen bond, not by the bond polarity. The \(\text{C–I}\) bond has the lowest bond enthalpy (weakest bond) among the three, so it is broken most easily and therefore reacts the fastest.

[1] B - 1-iodobutane reacts the fastest because the \(\text{C–I}\) bond is the weakest.

A catalyst provides an alternative pathway with a lower activation energy, effectively shifting the activation energy barrier, \(E_{\text{a}}\), to the left (a lower value) on the Maxwell-Boltzmann distribution. Because the temperature remains constant, the actual distribution of molecular kinetic energies (represented by the shape and area of the curve) is completely unaffected.

[1] C - It does not alter the shape of the curve, but shifts the activation energy barrier to a lower value.

To calculate the enthalpy change of reaction from mean bond enthalpies: \(\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})\). Bonds broken: \(1 \times \text{C–H} = 413\text{ kJ mol}^{-1}\) and \(1 \times \text{Br–Br} = 193\text{ kJ mol}^{-1}\). Total energy required to break bonds = \(413 + 193 = 606\text{ kJ mol}^{-1}\). Bonds formed: \(1 \times \text{C–Br} = 290\text{ kJ mol}^{-1}\) and \(1 \times \text{H–Br} = 366\text{ kJ mol}^{-1}\). Total energy released on forming bonds = \(290 + 366 = 656\text{ kJ mol}^{-1}\). \(\Delta H = 606 - 656 = -50\text{ kJ mol}^{-1}\).

[1] A - \(-50\)

Bromide ions are oxidized by concentrated sulfuric acid to bromine, while sulfuric acid is reduced to sulfur dioxide. Some hydrogen bromide gas is also formed as an initial acid-base product. However, bromide is not a strong enough reducing agent to reduce sulfur further to hydrogen sulfide, \(\text{H}_2\text{S}\) (which is only produced in the reaction with iodide ions).

[1] D - Hydrogen sulfide

The original alcohol has molecular formula \(\text{C}_3\text{H}_8\text{O}\), which must be either propan-1-ol or propan-2-ol. Since the product \(\mathbf{Y}\) has a strong peak at \(1715\text{ cm}^{-1}\) (carbonyl group, \(\text{C=O}\)) but no \(\text{O–H}\) stretch for carboxylic acids (\(2500-3300\text{ cm}^{-1}\)) or alcohols (\(3200-3600\text{ cm}^{-1}\)), it must be a ketone (propanone). A ketone is formed from the oxidation of a secondary alcohol. Therefore, the original alcohol must be propan-2-ol.

[1] B - Propan-2-ol

As Group 2 is descended, the solubility of Group 2 hydroxides increases (magnesium hydroxide is sparingly soluble, whereas barium hydroxide is much more soluble). Conversely, the solubility of Group 2 sulfates decreases (magnesium sulfate is highly soluble, whereas barium sulfate is insoluble).

[1] A - The solubility of the hydroxides increases and the solubility of the sulfates decreases.

The chemical equation for the standard enthalpy change of formation of propane is:
\(3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\)

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

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

\(\Delta_f H^\theta = [3 \times (-393.5)] + [4 \times (-285.8)] - (-2219.2)\)

\(\Delta_f H^\theta = -1180.5 - 1143.2 + 2219.2 = -104.5\text{ kJ mol}^{-1}\)

1 mark: Correct calculation showing \(-104.5\text{ kJ mol}^{-1}\) (Option A).

Propan-1-ol is an alcohol and contains an \(\text{O}-\text{H}\) polar bond, enabling it to form strong intermolecular hydrogen bonds. Butane and methylpropane are alkanes and only experience London forces. Methoxyethane is an ether and possesses permanent dipole-dipole forces but cannot form hydrogen bonds with itself. Hydrogen bonds are significantly stronger than London forces and permanent dipole-dipole forces, requiring more energy to overcome, resulting in a higher boiling temperature for propan-1-ol.

1 mark: Correctly identifies propan-1-ol (Option C) as having the highest boiling temperature due to hydrogen bonding.

The solubility of Group 2 hydroxides increases down the group, so barium hydroxide is more soluble than calcium hydroxide. On the other hand, the solubility of Group 2 sulfates decreases down the group, which means magnesium sulfate is highly soluble while barium sulfate is extremely insoluble.

1 mark: Correctly identifies the solubility trend of Group 2 hydroxides (Option C).

A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_a\)). This shifts the minimum energy required for reaction (represented by the activation energy line on the Maxwell-Boltzmann distribution) to the left. The shape of the distribution curve itself (including the peak and total area) does not change because the temperature remains constant.

1 mark: Correctly identifies that only the activation energy line shifts to the left (Option C).

The \(\text{S}_\text{N}1\) mechanism of a tertiary halogenoalkane like 2-bromo-2-methylpropane involves a slow, rate-determining step where the carbon-halogen bond breaks heterolytically to form a stable tertiary carbocation intermediate, \(\text{(CH}_3\text{)}_3\text{C}^+\). The rate equation is \(\text{rate} = k[\text{(CH}_3\text{)}_3\text{CBr}]\), which is first-order overall.

1 mark: Correctly identifies the stable tertiary carbocation intermediate (Option C).

Iodide ions are strong reducing agents and reduce the sulfur in concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\), oxidation state +6) to various lower oxidation states, producing \(\text{SO}_2\) (+4), \(\text{S}\) (0), and \(\text{H}_2\text{S}\) (-2). Sulfur trioxide (\(\text{SO}_3\)) contains sulfur in the +6 oxidation state and is not formed as a reduction product.

1 mark: Identifies sulfur trioxide as the species that is not a product of this reaction (Option D).

The absorption peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl (\(\text{C}=\text{O}\)) group. The absence of a broad absorption in the \(3200-3600\text{ cm}^{-1}\) region indicates that there is no hydroxyl (\(\text{O}-\text{H}\)) group. Propanone (\(\text{C}_3\text{H}_6\text{O}\)) contains a carbonyl group but no hydroxyl group, matching the data perfectly. Propan-1-ol and prop-2-en-1-ol contain \(\text{O}-\text{H}\) groups, and propanoic acid has the wrong molecular formula (\(\text{C}_3\text{H}_6\text{O}_2\)) and contains an \(\text{O}-\text{H}\) group.

1 mark: Correctly identifies propanone (Option B) as matching the spectroscopic details.

According to Le Chatelier's principle, since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat. There are 3 moles of gaseous reactants on the left and 2 moles of gaseous products on the right. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas (to the right). Therefore, decreasing temperature and increasing pressure will both shift the position of equilibrium to the right.

1 mark: Correctly identifies the combination of decreasing temperature and increasing pressure (Option B) to shift the equilibrium to the right.

The boiling temperatures of hydrogen halides decrease from \( \text{HF} \) to \( \text{HCl} \), and then increase from \( \text{HCl} \) to \( \text{HI} \). \( \text{HF} \) has the highest boiling temperature (\( 293\text{ K} \)) because it forms strong hydrogen bonds. For \( \text{HCl} \), \( \text{HBr} \), and \( \text{HI} \), the primary intermolecular forces are London forces, which increase in strength as the number of electrons in the molecule increases, resulting in the overall decreasing order of boiling temperatures: \( \text{HF} > \text{HI} > \text{HBr} > \text{HCl} \).

1 mark for the correct option: A

Sulfuric acid (\( \text{H}_2\text{SO}_4 \)) contains sulfur in the +6 oxidation state. When reduced by iodide ions, one of the products is elemental sulfur (\( \text{S} \)), which is a yellow solid. The oxidation state changes from +6 to 0. Other possible reduction products are sulfur dioxide (\( \text{SO}_2 \), +6 to +4) and hydrogen sulfide (\( \text{H}_2\text{S} \), +6 to -2).

1 mark for the correct option: B

Two factors affect the rate of hydrolysis of halogenoalkanes: the nature of the halogen and the structure of the carbon skeleton. Bromides undergo hydrolysis faster than chlorides because the C-Br bond is weaker (lower bond enthalpy) than the C-Cl bond. Furthermore, tertiary halogenoalkanes (such as 2-bromo-2-methylpropane) react much faster than secondary (2-bromobutane) and primary (1-bromobutane) halogenoalkanes because they undergo nucleophilic substitution via the \( \text{S}_\text{N}1 \) mechanism involving a very stable tertiary carbocation intermediate. Therefore, 2-bromo-2-methylpropane reacts the fastest.

1 mark for the correct option: D

Enthalpy change, \( \Delta H = \sum(\text{bond enthalpies of reactants}) - \sum(\text{bond enthalpies of products}) \)

Bonds broken (reactants):
- \( 4 \times (\text{C}-\text{H}) = 4 \times 413 = 1652 \text{ kJ mol}^{-1} \)
- \( 2 \times (\text{O}-\text{H}) = 2 \times 464 = 928 \text{ kJ mol}^{-1} \)
Total energy to break bonds = \( 1652 + 928 = 2580 \text{ kJ mol}^{-1} \)

Bonds formed (products):
- \( 1 \times (\text{C}\equiv\text{O}) = 1077 \text{ kJ mol}^{-1} \)
- \( 3 \times (\text{H}-\text{H}) = 3 \times 436 = 1308 \text{ kJ mol}^{-1} \)
Total energy released forming bonds = \( 1077 + 1308 = 2385 \text{ kJ mol}^{-1} \)

\( \Delta H = 2580 - 2385 = +195 \text{ kJ mol}^{-1} \).

1 mark for the correct option: B

First, calculate the heat energy transferred to the water using \(q = m c \Delta T\):
\(q = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^{\circ}\text{C}^{-1} \times 24.5\text{ }^{\circ}\text{C} = 15361.5\text{ J} = 15.3615\text{ kJ}\)

Next, calculate the number of moles of methanol burned:
\(n(\text{CH}_3\text{OH}) = \frac{\text{mass}}{\text{Molar Mass}} = \frac{0.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.0250\text{ mol}\)

Finally, calculate the experimental enthalpy of combustion, \(\Delta_c H\):
\(\Delta_c H = -\frac{q}{n} = -\frac{15.3615\text{ kJ}}{0.0250\text{ mol}} = -614.46\text{ kJ mol}^{-1}\)

Rounding to 3 significant figures (appropriate to the experimental data) gives \(-614\text{ kJ mol}^{-1}\) (or \(-610\text{ kJ mol}^{-1}\) if using 2 significant figures).

M1: Calculates the heat energy transferred correctly: \(q = 15.36\text{ kJ}\) (allow 15400 J or similar). [1 mark]
M2: Calculates the amount of methanol burned: \(n = 0.025\text{ mol}\). [1 mark]
M3: Calculates the final enthalpy change with a negative sign and correct units: \(-614\text{ kJ mol}^{-1}\) (allow \(-610\text{ kJ mol}^{-1}\) based on 2 s.f.). [0.5 marks]

Although 1-chlorobutane has a stronger permanent dipole-dipole force due to the higher electronegativity of chlorine compared to iodine, the dominant intermolecular force in these halogenoalkanes is London (dispersion) forces. Iodine has many more electrons than chlorine, which makes the electron cloud of 1-iodobutane far more polarizable. This leads to significantly stronger London forces between 1-iodobutane molecules compared to 1-chlorobutane molecules. Consequently, more thermal energy is required to overcome these stronger London forces, resulting in a higher boiling temperature.

M1: States that both compounds have London forces, but 1-iodobutane has more electrons (or a larger electron cloud). [1 mark]
M2: Explains that the London forces in 1-iodobutane are stronger or more easily polarized. [1 mark]
M3: States that these stronger London forces require more energy to overcome than the intermolecular forces in 1-chlorobutane. [0.5 marks]

The magnesium ion (\(\text{Mg}^{2+}\)) has a much smaller ionic radius than the barium ion (\(\text{Ba}^{2+}\)) while carrying the same charge. Therefore, the \(\text{Mg}^{2+}\) ion has a much higher charge density. This high charge density allows the \(\text{Mg}^{2+}\) ion to strongly polarise and distort the electron cloud of the neighbouring nitrate anion (\(\text{NO}_3^-\)). This distortion weakens the covalent nitrogen-oxygen bonds within the nitrate group, making it much easier to break down thermally at a lower temperature.

M1: Compares ionic size and charge density: \(\text{Mg}^{2+}\) is smaller and has a higher charge density than \(\text{Ba}^{2+}\). [1 mark]
M2: Explains that \(\text{Mg}^{2+}\) polarises / distorts the electron cloud of the nitrate ion (\(\text{NO}_3^-\)) more effectively. [1 mark]
M3: States that this weakens the \(\text{N}-\text{O}\) bond, making the nitrate ion easier to decompose. [0.5 marks]

The presence of two molecular ion peaks of almost equal intensity (\(1:1\) ratio) separated by 2 units at \(m/z = 156\) and \(m/z = 158\) indicates that the compound contains a single bromine atom, which has two naturally occurring isotopes: \({}^{79}\text{Br}\) and \({}^{81}\text{Br}\) in roughly equal abundance.

Subtracting the mass of bromine (79) from the molecular ion peak mass (156) gives:
\(156 - 79 = 77\)

A mass of 77 corresponds to a phenyl group (\(\text{C}_6\text{H}_5\)), since \(6 \times 12 + 5 \times 1 = 77\).

Therefore, the molecular formula of the compound is \(\text{C}_6\text{H}_5\text{Br}\) (bromobenzene).

M1: Identifies the halogen as bromine / \(\text{Br}\) because of the \(1:1\) peak ratio for \(M\) and \(M+2\). [1 mark]
M2: Calculates the mass of the organic part of the molecule (\(156 - 79 = 77\) or \(158 - 81 = 77\)). [1 mark]
M3: Deduces the molecular formula as \(\text{C}_6\text{H}_5\text{Br}\). [0.5 marks]

A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_{cat}\)). On a Maxwell-Boltzmann distribution curve, the activation energy barrier is shifted to the left (to a lower energy value). This means that a significantly larger fraction of the reactant molecules (represented by the area under the curve to the right of the new activation energy line) now possess energy greater than or equal to the lower activation energy. As a result, the frequency of successful collisions increases, which increases the rate of reaction.

M1: States that a catalyst provides an alternative pathway with a lower activation energy. [1 mark]
M2: Explains that this shifts the activation energy to the left on the Maxwell-Boltzmann distribution. [1 mark]
M3: Concludes that a greater fraction/proportion of molecules have energy \(\ge E_{cat}\), leading to more successful collisions per unit time. [0.5 marks]

The strong absorption band at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad absorption band in the region \(3200-3750\text{ cm}^{-1}\) indicates that there is no alcohol (\(\text{O}-\text{H}\)) group present.

Since the molecular formula is \(\text{C}_3\text{H}_6\text{O}\) and it contains a carbonyl group, the compound must be a ketone or an aldehyde. With three carbon atoms, the only possibilities are propanone (a ketone) or propanal (an aldehyde). Propanone has its carbonyl absorption around \(1715\text{ cm}^{-1}\).

Accept either propanone or propanal as both fit the molecular formula and the diagnostic infrared peaks described.

M1: Identifies the functional group as carbonyl / \(\text{C}=\text{O}\) AND states the absence of an alcohol / \(\text{O}-\text{H}\) group. [1 mark]
M2: Uses the molecular formula \(\text{C}_3\text{H}_6\text{O}\) to deduce that it is a ketone or aldehyde. [1 mark]
M3: Names or draws propanone (or propanal). [0.5 marks]

When chlorine gas is bubbled into cold, dilute sodium hydroxide, it undergoes a disproportionation reaction to form chloride ions and chlorate(I) ions:
\(\text{Cl}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cl}^-(\text{aq}) + \text{ClO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

The oxidation numbers of chlorine are:
- In \(\text{Cl}_2\): 0
- In \(\text{Cl}^-\): -1 (reduced)
- In \(\text{ClO}^-\): +1 (oxidised)

Since the same element (chlorine) is simultaneously oxidised and reduced, this is a disproportionation reaction.

M1: Correct ionic equation with state symbols: \(\text{Cl}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cl}^-(\text{aq}) + \text{ClO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\). [1 mark]
M2: Correctly states the three oxidation numbers: 0 in \(\text{Cl}_2\), -1 in \(\text{Cl}^-\), and +1 in \(\text{ClO}^-\). [1 mark]
M3: Mentions that chlorine is both oxidised and reduced in the same reaction. [0.5 marks]

The equation for the formation of ethanol is:
\(2\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{l})\)

Using Hess's Law and the enthalpy of combustion data:
\(\Delta_f H^{\theta} = \sum \Delta_c H^{\theta}(\text{reactants}) - \sum \Delta_c H^{\theta}(\text{products})\)
\(\Delta_f H^{\theta} = [2 \times \Delta_c H^{\theta}(\text{C}) + 3 \times \Delta_c H^{\theta}(\text{H}_2)] - [\Delta_c H^{\theta}(\text{C}_2\text{H}_5\text{OH})]\)
\(\Delta_f H^{\theta} = [2 \times (-393.5) + 3 \times (-285.8)] - [-1367.3]\)
\(\Delta_f H^{\theta} = [-787.0 - 857.4] + 1367.3\)
\(\Delta_f H^{\theta} = -1644.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\)

M1: Correct cycle or expression for Hess's Law using combustion data: \(2 \times \Delta_c H^{\theta}(\text{C}) + 3 \times \Delta_c H^{\theta}(\text{H}_2) - \Delta_c H^{\theta}(\text{ethanol})\). [1 mark]
M2: Correct substitution of values: \([-787.0 - 857.4] - (-1367.3)\). [1 mark]
M3: Final answer calculated with correct sign and unit: \(-277.1\text{ kJ mol}^{-1}\). [0.5 marks]

Ethanol has a highly polar \(\text{O-H}\) bond allowing it to form hydrogen bonds between molecules. Methoxymethane cannot form hydrogen bonds because it lacks hydrogen atoms directly bonded to oxygen, so it only has London forces and permanent dipole-dipole forces. Hydrogen bonds are much stronger than London forces and permanent dipole-dipole forces, requiring significantly more thermal energy to overcome, which results in a higher boiling temperature for ethanol.

M1 (1 mark): Identifies that ethanol forms intermolecular hydrogen bonds due to the polar \(\text{O-H}\) group. M2 (1 mark): Identifies that methoxymethane only has London forces and permanent dipole-dipole forces (or cannot form hydrogen bonds). M3 (0.5 marks): Explains that hydrogen bonds are stronger and require more energy to break than London and permanent dipole-dipole forces.

When chlorine gas is added to hot, concentrated sodium hydroxide, it disproportionates according to the ionic equation: \(\text{3Cl}_2(\text{g}) + \text{6OH}^-(\text{aq}) \rightarrow \text{5Cl}^-(\text{aq}) + \text{ClO}_3^-(\text{aq}) + \text{3H}_2\text{O}(\text{l})\). Chlorine starts with an oxidation number of \(0\) in \(\text{Cl}_2\). In the chloride product (\(\text{Cl}^-\)), the oxidation state of chlorine is \(-1\) (reduction). In the chlorate(V) product (\(\text{ClO}_3^-\)), the oxidation state of chlorine is \(+5\) (oxidation). Since chlorine is both oxidized (\(0 \rightarrow +5\)) and reduced (\(0 \rightarrow -1\)) in the same reaction, it is a disproportionation reaction.

M1 (1 mark): Balanced ionic equation: \(\text{3Cl}_2 + \text{6OH}^- \rightarrow \text{5Cl}^- + \text{ClO}_3^- + \text{3H}_2\text{O}\) (allow state symbols omitted). M2 (1 mark): Identifies the oxidation numbers of chlorine in the products as \(-1\) (in \(\text{Cl}^-\)) and \(+5\) (in \(\text{ClO}_3^-\)). M3 (0.5 marks): Defines disproportionation in this context as chlorine being simultaneously oxidized and reduced.

(a) Ethanol acts as a mutual solvent (cosolvent). Halogenoalkanes are insoluble in water but soluble in ethanol, while aqueous silver nitrate can mix with ethanol, allowing the reactants to meet in a single phase. (b) A cream precipitate is formed. The ionic equation is \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\). (c) The precipitates appear in the order: 1-iodobutane first, then 1-bromobutane, and 1-chlorobutane last. This is because the carbon-halogen bond enthalpy decreases down Group 7 (\(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\)). The weaker the bond, the less energy is required to break it, lowering the activation energy and increasing the rate of hydrolysis.

M1 (a): Suggests ethanol is a mutual solvent / dissolves both the halogenoalkanes and the aqueous silver nitrate. M2 (b): Identifies the precipitate as cream. M3 (b): Balanced ionic equation with correct state symbols: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\). M4 (c): States the order of appearance: 1-iodobutane, then 1-bromobutane, then 1-chlorobutane. M5 (c): Explains that C-X bond strength / bond enthalpy decreases down the group (C-I is the weakest). M6 (c): Connects weaker bond to lower activation energy / faster rate of reaction (and notes bond enthalpy is the dominant factor over bond polarity).

(a) The balanced equation is \(2\text{Mg(NO}_3)_2 \rightarrow 2\text{MgO} + 4\text{NO}_2 + \text{O}_2\). (b) Magnesium and barium are both in Group 2, so the magnesium ion (\(\text{Mg}^{2+}\)) and barium ion (\(\text{Ba}^{2+}\)) both have a \(2+\) charge. However, the magnesium ion has a much smaller ionic radius than the barium ion. Therefore, the magnesium ion has a significantly higher charge density. This high charge density allows the magnesium ion to strongly polarise and distort the electron cloud of the nitrate ion (\(\text{NO}_3^-\)). This polarisation weakens the covalent nitrogen-oxygen bonds within the nitrate ion, making it easier to break and causing it to decompose at a lower temperature.

M1 (a): Correct balanced equation: \(2\text{Mg(NO}_3)_2 \rightarrow 2\text{MgO} + 4\text{NO}_2 + \text{O}_2\) (allow correct fractional coefficients). M2 (b): States that magnesium ion has a smaller ionic radius than barium ion. M3 (b): States that both cations have the same charge / are both \(2+\). M4 (b): Deduces that the magnesium ion has a higher charge density. M5 (b): Explains that \(\text{Mg}^{2+}\) polarises or distorts the electron cloud of the nitrate ion (\(\text{NO}_3^-\)) more than \(\text{Ba}^{2+}\). M6 (b): Concludes that this polarisation weakens the \(\text{N}-\text{O}\) covalent bond within the anion, requiring less thermal energy to break.

(a) Temperature rise \(\Delta T = 41.8 - 19.5 = 22.3\ ^\circ\text{C}\). Heat energy \(q = m c \Delta T = 150.0 \times 4.18 \times 22.3 = 13982.1\text{ J} = 13.98\text{ kJ}\) (or \(14.0\text{ kJ}\)). (b) Moles of methanol burned \(n = \frac{0.68}{32.0} = 0.02125\text{ mol}\). Enthalpy change \(\Delta H_\text{c} = -\frac{q}{n} = -\frac{13.9821}{0.02125} = -657.98\text{ kJ mol}^{-1}\). Expressed to 2 significant figures (consistent with \(0.68\text{ g}\)), \(\Delta H_\text{c} = -660\text{ kJ mol}^{-1}\) (or \(-658\text{ kJ mol}^{-1}\) to 3 significant figures). (c) Incomplete combustion of methanol (producing carbon monoxide or soot) OR evaporation of methanol from the wick of the burner after weighing.

M1 (a): Calculates temperature change of \(22.3\ ^\circ\text{C}\) and substitutes correctly into expression: \(150.0 \times 4.18 \times 22.3\). M2 (a): Correct evaluation of heat energy: \(14.0\text{ kJ}\) or \(13.98\text{ kJ}\) (accept \(13982\text{ J}\)). M3 (b): Calculates moles of methanol correctly as \(0.02125\text{ mol}\). M4 (b): Divides heat energy (in \(\text{kJ}\)) by moles of methanol. M5 (b): Obtains \(-660\text{ kJ mol}^{-1}\) (2 s.f.) or \(-658\text{ kJ mol}^{-1}\) (3 s.f.) with negative sign included. M6 (c): Suggests incomplete combustion of methanol / evaporation of methanol from the wick (Do not accept 'heat loss' or 'heat absorbed by the copper beaker' as these are excluded by the prompt).

1. The rate of nucleophilic substitution depends on the strength of the carbon-halogen bond rather than its polarity. 2. The C-Cl bond is more polar than the C-Br bond, which would suggest 1-chlorobutane reacts faster if polarity were dominant. 3. However, the C-Br bond enthalpy (approx 276 kJ/mol) is significantly lower than the C-Cl bond enthalpy (approx 338 kJ/mol), making the C-Br bond much easier to break. This lower activation energy results in a faster rate of reaction for 1-bromobutane.

1 mark: Stating that bond enthalpy/strength is the dominant factor determining the rate (not bond polarity). 1 mark: Stating that the C-Br bond is weaker / has a lower bond enthalpy than the C-Cl bond. 0.5 marks: Stating that the C-Br bond breaks more easily / requires less energy to break.

1. Magnesium and barium are both in Group 2, so their cations both carry a 2+ charge. 2. However, the ionic radius of Mg2+ is much smaller than that of Ba2+. 3. This gives the Mg2+ ion a much higher charge density. 4. The highly charge-dense Mg2+ ion polarizes the electron cloud of the carbonate (CO32-) ion more strongly. 5. This polarization weakens the C-O covalent bonds within the carbonate group, meaning less thermal energy is required to decompose magnesium carbonate compared to barium carbonate.

1 mark: Identify that Mg2+ is smaller or has a higher charge density than Ba2+. 1 mark: Explain that Mg2+ polarizes the carbonate ion / electron cloud more strongly. 0.5 marks: Conclude that this weakens the C-O bond in the carbonate ion, requiring less thermal energy to break.

1. Calculate heat energy transferred to water: \(Q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 12.0\ ^\circ\text{C} = 2508\text{ J} = 2.508\text{ kJ}\). 2. Calculate the moles of methanol burned: \(n = \text{mass} / M_r = 0.400\text{ g} / 32.0\text{ g mol}^{-1} = 0.0125\text{ mol}\). 3. Calculate the enthalpy change per mole: \(\Delta H_c = -Q / n = -2.508\text{ kJ} / 0.0125\text{ mol} = -200.64\text{ kJ mol}^{-1}\). 4. Rounding to 3 significant figures gives \(-201\text{ kJ mol}^{-1}\).

1 mark: Correct calculation of heat energy transferred, \(Q = 2.508\text{ kJ}\) (or 2508 J). 1 mark: Correct calculation of moles of methanol, \(n = 0.0125\text{ mol}\). 0.5 marks: Correct final value of \(-201\text{ kJ mol}^{-1}\) (must include negative sign and be rounded to 3 significant figures, accept -200.6).

1. Both molecules experience hydrogen bonding (as well as London forces and permanent dipole-dipole forces) due to the presence of -OH groups. 2. Butane-1,4-diol contains two -OH groups per molecule, whereas butane-1-ol has only one -OH group per molecule. 3. This means that butane-1,4-diol can form a more extensive network of hydrogen bonds (more hydrogen bonds per molecule) with neighboring molecules. 4. Consequently, more energy is required to overcome these stronger intermolecular attractions, resulting in a much higher boiling point.

1 mark: State that both compounds form hydrogen bonds. 1 mark: Identify that butane-1,4-diol has two -OH groups whereas butane-1-ol has only one. 0.5 marks: Explain that more hydrogen bonds are formed between molecules, requiring more energy to break.

**(a) Calculation of heat energy, \(q\):**
- Temperature rise, \(\Delta T = 43.2 - 19.5 = 23.7\text{ }^\circ\text{C}\) (or \(23.7\text{ K}\))
- \(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 23.7\text{ K}\)
- \(q = 14859.9\text{ J} = 14.86\text{ kJ}\) (or \(14.9\text{ kJ}\))

**(b) Calculation of enthalpy change of combustion, \(\Delta H_c\):**
- Mass of pentan-1-ol burnt = \(184.35 - 183.73 = 0.62\text{ g}\)
- Moles of pentan-1-ol burnt, \(n = \frac{0.62\text{ g}}{88.0\text{ g mol}^{-1}} = 0.007045\text{ mol}\)
- \(\Delta H_c = -\frac{q}{n} = -\frac{14.8599\text{ kJ}}{0.007045\text{ mol}} = -2109.15\text{ kJ mol}^{-1}\)
- Rounded to 3 significant figures: \(-2110\text{ kJ mol}^{-1}\) (or \(-2.11 \times 10^3\text{ kJ mol}^{-1}\))

**(c) Reasons for experimental value being less exothermic:**
- Heat loss to the surrounding air or draft.
- Incomplete combustion of pentan-1-ol (forming soot/carbon or carbon monoxide instead of carbon dioxide).
- Evaporation of alcohol from the wick after extinguishing but before weighing.
- Heat capacity of the copper calorimeter vessel was not accounted for in the calculation.

**(d) Advantage of copper calorimeter:**
- Copper is a much better thermal conductor than glass, which ensures a faster and more efficient transfer of heat from the flame to the water, thereby reducing heat loss to the surroundings during the heating period.

**(a)** [2 Marks]
- **M1:** Correct calculation of temperature rise: \(\Delta T = 23.7\text{ }^\circ\text{C}\) (1)
- **M2:** Correct calculation of \(q\) in kJ: \(14.86\text{ kJ}\) (or \(14.9\text{ kJ}\)) (1)
*Allow TE for M2 if incorrect temperature change is used.*

**(b)** [3 Marks]
- **M1:** Correct calculation of mass of pentan-1-ol burnt (\(0.62\text{ g}\)) and converting it to moles: \(0.007045\text{ mol}\) (1)
- **M2:** Division of \(q\) by \(n\) (1)
- **M3:** Correct evaluation to 3 s.f. with a negative sign: \(-2110\text{ kJ mol}^{-1}\) (or \(-2.11 \times 10^3\text{ kJ mol}^{-1}\)) (1)
*Allow TE from part (a) and/or M1.*

**(c)** [2 Marks]
- **M1:** Heat lost to surroundings / draft (1)
- **M2:** Incomplete combustion / formation of soot or CO (1)
*Accept: Evaporation of alcohol from the wick after extinguishing but before weighing.*
*Do not accept: Student parallax errors or general volume/mass errors (as these are student measurement errors).*

**(d)** [1 Mark]
- **M1:** Copper is a better thermal conductor (than glass), so it allows faster / more efficient heat transfer from the flame to the water (reducing heat loss to air) (1)

Heating to constant mass is carried out by heating, cooling, weighing, and repeating this sequence until two consecutive mass measurements are identical. This confirms that all of the water of crystallisation has been driven off from the hydrated salt. If any water remained in the crucible, the measured mass loss would be lower than the actual mass of water, resulting in a lower calculated value of \(x\) than the true value.

[1 mark] Heating to constant mass ensures all the water of crystallisation has been completely driven off / evaporated.
[1 mark] If water remained, the mass loss (or mass of water) would be underestimated, leading to an inaccurately low value of \(x\).

Concentrated hydrochloric acid is used to remove any unreacted starting material, butan-1-ol, because the alcohol is highly soluble in concentrated acid. Since the density of 1-bromobutane (\(1.27\text{ g cm}^{-3}\)) is greater than the density of the aqueous acid (\(1.18\text{ g cm}^{-3}\)), the organic product, 1-bromobutane, will form the lower layer.

[1 mark] To remove unreacted butan-1-ol.
[1 mark] The lower layer (as its density is greater than that of the aqueous layer).

To test for the ammonium ion, warm the solid sample with aqueous sodium hydroxide (\(\text{NaOH}\)). This releases ammonia gas (\(\text{NH}_3\)), which can be identified by its characteristic pungent smell or, more reliably, by holding damp red litmus paper at the mouth of the tube. The litmus paper turns blue because ammonia is an alkaline gas.

[1 mark] Add aqueous sodium hydroxide and warm / heat.
[1 mark] Test the gas with damp red litmus paper which turns blue (or damp universal indicator paper turns blue/purple, or dense white fumes with concentrated HCl/glass rod dipped in HCl).

Polystyrene is a good thermal insulator compared to glass, so using a polystyrene cup minimises heat loss to the surroundings. If heat is lost, the maximum temperature rise reached by the reaction mixture will be lower than expected. Consequently, the calculated heat energy \(q = mc\Delta T\) will be lower, resulting in an experimental enthalpy change that is less exothermic (less negative) than the literature value.

[1 mark] Polystyrene is a thermal insulator / poor conductor of heat (which reduces/minimises heat loss to the surroundings).
[1 mark] The calculated enthalpy change of neutralisation will be less exothermic / less negative (or lower magnitude) because the temperature change (\(\Delta T\)) is smaller than it should be.

Fehling's solution is a mild oxidising agent. When heated with propanal (an aldehyde), it oxidises the aldehyde to a carboxylic acid and is itself reduced, forming a red/brick-red precipitate of copper(I) oxide (\(\text{Cu}_2\text{O}\)). Propan-1-ol (a primary alcohol) is not oxidised by Fehling's solution under these conditions, so the solution remains blue.

[1 mark] For propanal, a red precipitate (or orange/brown precipitate) forms.
[1 mark] For propan-1-ol, no reaction / no change / stays blue.

A titre is calculated from the difference between two burette readings (initial and final). Therefore, the total uncertainty is the sum of the uncertainties of both readings: \(\text{Total uncertainty} = 2 \times 0.05\text{ cm}^3 = 0.10\text{ cm}^3\). The percentage uncertainty is calculated as: \(\frac{0.10}{20.00} \times 100\% = 0.5\%\).

[1 mark] Multiplies the uncertainty by 2 to get \(0.10\text{ cm}^3\) (total uncertainty for two readings).
[1 mark] Calculates \(\frac{0.10}{20.00} \times 100\% = 0.5\%\) (allow 0.50%).

Halogenoalkanes are insoluble in water, while aqueous silver nitrate is polar and soluble in water. Ethanol acts as a common solvent (mutual solvent) to allow the reactants to mix and react in a single phase. Ethanol is highly flammable, which is a significant safety hazard. To minimise this risk, heating should be conducted using a water bath, hot plate, or electric heating mantle instead of a naked Bunsen burner flame.

[1 mark] Ethanol acts as a mutual solvent to dissolve both the halogenoalkanes and the aqueous reactants/silver nitrate (or allows them to mix).
[1 mark] Ethanol is flammable, so use a water bath/heating mantle/hot plate instead of an open flame / Bunsen burner.

Copper(II) oxide is added in excess to ensure that all of the sulfuric acid is completely neutralised. This is important because any remaining unreacted sulfuric acid would become concentrated during the evaporation stage, which would contaminate the copper(II) sulfate crystals. The unreacted, insoluble copper(II) oxide solid is then removed from the hot reaction mixture by gravity filtration using a funnel and filter paper.

[1 mark] Excess copper(II) oxide ensures all the sulfuric acid reacts / is completely neutralised.
[1 mark] The excess solid is removed by filtration / filtering.

To confirm the presence of iodide ions, dilute nitric acid followed by aqueous silver nitrate is added to the solution, producing a yellow precipitate of silver iodide. When concentrated ammonia solution is added to this mixture, the yellow precipitate remains completely insoluble, confirming the presence of iodide (unlike silver chloride which dissolves in dilute ammonia, and silver bromide which dissolves in concentrated ammonia).

M1: Addition of aqueous silver nitrate results in a yellow precipitate. (1) M2: Addition of concentrated ammonia results in the yellow precipitate remaining insoluble / not dissolving. (1)

Heating to constant mass ensures that all the water of crystallization has been completely driven off from the hydrated salt. This is practically achieved by heating the crucible and its contents, allowing it to cool in a desiccator, weighing it, and repeating this sequence until consecutive mass readings are identical.

M1: To ensure all water of crystallization is driven off / evaporated. (1) M2: Heat, cool, and weigh repeatedly until consecutive mass measurements are identical. (1)

The residual distilled water in the burette dilutes the hydrochloric acid solution, lowering its concentration. As a result, a larger volume (titre) of this diluted acid is required to react completely with the analyte. In calculations, this larger volume leads to an overestimate of the moles of acid used, which in turn leads to a calculated concentration of the analyte that is higher than the actual concentration.

M1: Distilled water dilutes the acid, so a larger titre/volume of acid is needed to reach the end-point. (1) M2: A larger titre leads to a higher calculated concentration of the analyte. (1)

Using the minimum volume of hot solvent ensures that a hot, saturated solution of the organic compound is formed. This allows the maximum amount of the product to recrystallize out of the solution when it cools. If too much solvent is used, a significant portion of the organic solid remains dissolved in the excess solvent even at lower temperatures, leading to a much lower yield of recovered crystals.

M1: Minimum volume is used to ensure a hot saturated solution is formed so that crystallization occurs on cooling. (1) M2: Too much solvent keeps more product dissolved when cold, reducing the overall yield. (1)

First, calculate the heat energy released: \( q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times (34.3 - 21.5) \text{ K} = 50.0 \times 4.18 \times 12.8 = 2675.2 \text{ J} = 2.6752 \text{ kJ} \). Second, calculate the amount of substance in moles of copper(II) sulfate used: \( n = c \times V = 0.200 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.0100 \text{ mol} \). Third, calculate the molar enthalpy change of reaction: \( \Delta H = -\frac{q}{n} = -\frac{2.6752 \text{ kJ}}{0.0100 \text{ mol}} = -267.52 \text{ kJ mol}^{-1} \approx -268 \text{ kJ mol}^{-1} \) (to 3 significant figures). A key systematic error is heat loss to the surroundings (or heat absorbed by the polystyrene cup itself). Because heat is lost, the measured temperature rise is lower than the theoretical maximum, which results in the calculated value of \( \Delta H \) being less negative (less exothermic) than the actual value.

M1: Calculates heat energy released, \( q = 2.68 \text{ kJ} \) (or \( 2680 \text{ J} \)) [1 mark]. M2: Calculates moles of \( \text{CuSO}_4 = 0.0100 \text{ mol} \) [1 mark]. M3: Calculates molar enthalpy change, \( \Delta H = -268 \text{ kJ mol}^{-1} \) (must have negative sign, 3 significant figures, and correct unit) [1 mark]. M4: Identifies a valid systematic error: heat lost to the surroundings / heat absorbed by the cup/calorimeter [1 mark]. M5: Explains the effect of this error: measured temperature rise is smaller, making the calculated enthalpy change less negative / less exothermic [1 mark].

The brick-red flame color in Test 1 identifies the cation as the calcium ion, \( \text{Ca}^{2+} \). The cream precipitate in Test 2 which is insoluble in dilute ammonia but soluble in concentrated ammonia in Test 3 identifies the halide anion as the bromide ion, \( \text{Br}^- \). The ionic equation for Test 2 is the formation of the silver bromide precipitate: \( \text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)} \). When calcium ions react with aqueous sulfate ions, a white precipitate of calcium sulfate forms according to the ionic equation: \( \text{Ca}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{CaSO}_4\text{(s)} \).

M1: Identifies the cation as \( \text{Ca}^{2+} \) / calcium [1 mark]. M2: Identifies the anion as \( \text{Br}^- \) / bromide [1 mark]. M3: Writes the correct ionic equation for Test 2: \( \text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)} \) including state symbols [1 mark]. M4: Identifies calcium sulfate, \( \text{CaSO}_4 \), as the precipitate [1 mark]. M5: Writes the correct ionic equation for the precipitation of calcium sulfate: \( \text{Ca}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{CaSO}_4\text{(s)} \) including state symbols [1 mark].

To ensure a fair comparison, the student should add equal volumes (or equal moles) of 1-chlorobutane, 1-bromobutane, and 1-iodobutane to separate test tubes. To each tube, the same volume of ethanol (to act as a mutual solvent) and aqueous silver nitrate must be added. The tubes should be placed in a water bath held at a constant temperature. The student then measures the time taken for a precipitate to appear in each test tube. The expected observations are: 1-iodobutane forms a yellow precipitate (silver iodide) fastest; 1-bromobutane forms a cream precipitate (silver bromide) next; 1-chlorobutane forms a white precipitate (silver chloride) slowest. The rate of reaction depends on the strength of the carbon-halogen bond being broken. The C-I bond has the lowest bond enthalpy (is the weakest), meaning it is broken most easily and reacts fastest. The C-Cl bond has the highest bond enthalpy (is the strongest), meaning it requires the most energy to break and reacts slowest.

M1: Describes a fair experimental setup: equal amounts of halogenoalkanes, use of ethanol as a common solvent, same amount of aqueous silver nitrate, and keeping temperature constant in a water bath [1 mark]. M2: Identifies precipitate colors correctly: white for AgCl, cream for AgBr, and yellow for AgI, and states that the rate is determined by measuring the time taken for the precipitate to appear [1 mark]. M3: Identifies correct order of reaction rates: 1-iodobutane is fastest, followed by 1-bromobutane, and 1-chlorobutane is slowest [1 mark]. M4: Explains that reaction rate is determined by the ease of breaking the carbon-halogen (C-X) bond [1 mark]. M5: Connects the trend to bond enthalpy: C-I has the lowest bond enthalpy (weakest bond) so it breaks fastest, while C-Cl has the highest bond enthalpy (strongest bond) so it breaks slowest (reject explanations based on bond polarity) [1 mark].

First, pour the crude distillate into a separating funnel and add aqueous sodium hydrogencarbonate solution. Shake the mixture gently, inverting the funnel and periodically opening the tap to release the build-up of pressure from carbon dioxide gas. This neutralises the acidic impurities. Second, allow the layers to separate. Cyclohexene is less dense than water and is insoluble in it, so it forms the upper organic layer. Run off and discard the lower aqueous layer. Third, wash the organic layer with water in the separating funnel, let the layers separate, and run off the aqueous layer again to remove residual inorganic impurities. Fourth, collect the organic layer in a dry conical flask and add an anhydrous inorganic salt, such as anhydrous calcium chloride, as a drying agent. Swirl the flask and allow it to stand until the liquid changes from cloudy to completely clear, indicating all water has been absorbed. Fifth, decant or filter the dried cyclohexene into a distillation flask and perform a final fractional distillation, collecting only the fraction that distills at the boiling point of cyclohexene, which is approximately \( 83 \text{ }^\circ\text{C} \).

M1: Describes adding sodium hydrogencarbonate solution to the mixture in a separating funnel and shaking while releasing pressure to neutralise acidic impurities [1 mark]. M2: Explains separation of layers: cyclohexene forms the upper organic layer (since it is less dense) and the aqueous layer is run off [1 mark]. M3: Describes adding a suitable anhydrous inorganic drying agent (e.g., anhydrous calcium chloride / calcium sulfate / magnesium sulfate / sodium sulfate) [1 mark]. M4: Explains that the drying agent is added until the liquid becomes clear / is decanted or filtered to separate it from the drying agent [1 mark]. M5: Identifies the final step as fractional distillation, collecting the fraction at the boiling point of cyclohexene (approximately \( 83 \text{ }^\circ\text{C} \)) [1 mark].

For part (a): Since a two-reading difference method was used to weigh the solid, the total uncertainty in the mass is twice the balance uncertainty: \( 2 \times 0.01 \text{ g} = 0.02 \text{ g} \). The percentage uncertainty in mass is: \( \frac{0.02 \text{ g}}{2.86 \text{ g}} \times 100\% = 0.6993\% \approx 0.70\% \). For part (b): The percentage uncertainty in volume is: \( \frac{0.20 \text{ cm}^3}{250.0 \text{ cm}^3} \times 100\% = 0.080\% \). For part (c): Comparing the two calculated percentage uncertainties, we see that the percentage uncertainty of the balance (\( 0.70\% \)) is much larger than that of the volumetric flask (\( 0.080\% \)). Therefore, the balance is the major contributor to the overall uncertainty in the concentration of the standard solution.

M1: Calculates the total absolute uncertainty in mass as \( \pm 0.02 \text{ g} \) (recognising that two readings are taken on the balance) [1 mark]. M2: Calculates the percentage uncertainty in mass as \( \frac{0.02}{2.86} \times 100 = 0.70\% \) (or \( 0.699\% \)) [1 mark]. M3: Calculates the percentage uncertainty in volume as \( \frac{0.20}{250.0} \times 100 = 0.080\% \) [1 mark]. M4: Identifies that the balance contributes more to the overall uncertainty [1 mark]. M5: Justifies the conclusion by comparing the percentage uncertainties explicitly (e.g., stating that \( 0.70\% \) is greater than \( 0.080\% \)) [1 mark].

The rate equation is \(\text{Rate} = k[\text{A}]^2[\text{B}]\). Rearranging for \(k\) gives \(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]}\). Substituting the units: \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark: correct selection of option B. Correctly derived units of the rate constant.

First, calculate the entropy change of the surroundings: \(\Delta S^\theta_{\text{surroundings}} = -\frac{\Delta H^\theta}{T} = -\frac{+178 \times 10^3\text{ J mol}^{-1}}{298\text{ K}} = -597.3\text{ J K}^{-1}\text{ mol}^{-1}\). Then, calculate the total entropy change: \(\Delta S^\theta_{\text{total}} = \Delta S^\theta_{\text{system}} + \Delta S^\theta_{\text{surroundings}} = +160 - 597.3 = -437.3\text{ J K}^{-1}\text{ mol}^{-1}\), which rounds to \(-437\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark: correct selection of option A. Correctly calculated entropy of surroundings and total entropy change.

Total number of moles at equilibrium = \(0.20 + 0.10 + 0.70 = 1.00\text{ mol}\). The partial pressures are: \(p(\text{SO}_2) = \frac{0.20}{1.00}P = 0.20P\), \(p(\text{O}_2) = \frac{0.10}{1.00}P = 0.10P\), and \(p(\text{SO}_3) = \frac{0.70}{1.00}P = 0.70P\). Substituting these into the \(K_p\) expression: \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \cdot p(\text{O}_2)} = \frac{(0.70P)^2}{(0.20P)^2 \cdot (0.10P)} = \frac{0.49 P^2}{0.04 P^2 \cdot 0.10 P} = \frac{0.49}{0.004 P} = \frac{122.5}{P}\).

1 mark: correct selection of option D. Calculated total moles, partial pressures, and correctly simplified the expression for Kp.

Initial moles: propanoic acid = \(0.0500 \times 0.100 = 5.00 \times 10^{-3}\text{ mol}\), \(\text{NaOH} = 0.0500 \times 0.0500 = 2.50 \times 10^{-3}\text{ mol}\). Since hydroxide ions are the limiting reactant, they react completely: moles of remaining propanoic acid = \(5.00 \times 10^{-3} - 2.50 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol}\), and moles of propanoate ions formed = \(2.50 \times 10^{-3}\text{ mol}\). Because the amounts of weak acid and its conjugate base are equal, \([\text{HA}] = [\text{A}^-]\). Thus, \(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

1 mark: correct selection of option B. Determined moles of acid and conjugate base and calculated the pH.

A yellow precipitate with iodine in alkaline solution is given by compounds with a \(\text{CH}_3\text{CO}-\) group or a \(\text{CH}_3\text{CH(OH)}-\) group. A positive Tollens' test (silver mirror) is given by aldehydes. Ethanal (\(\text{CH}_3\text{CHO}\)) is an aldehyde and contains a methyl carbonyl group, so it gives positive results for both tests. Propanone and ethanol do not react with Tollens' reagent, while butanal does not contain the methyl carbonyl group.

1 mark: correct selection of option A. Correctly identified the functional group requirements for both Tollens' and iodoform tests.

An \(\text{S}_\text{N}1\) mechanism proceeds via a flat carbocation intermediate. The nucleophile (\(\text{OH}^-\)) can attack this planar carbocation with equal probability from either the front or the back face. This leads to the formation of an equal mixture of the two optical enantiomers, resulting in a racemic mixture.

1 mark: correct selection of option B. Understood that SN1 mechanism involves a planar carbocation intermediate leading to racemisation.

The gradient of the plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\). Therefore: \(-\frac{E_a}{R} = -9.62 \times 10^3\text{ K}\) which gives \(E_a = 9.62 \times 10^3 \times 8.31 = 79942.2\text{ J mol}^{-1} \approx 79.9\text{ kJ mol}^{-1}\).

1 mark: correct selection of option B. Correctly calculated activation energy using the gradient and R.

Barium hydroxide, \(\text{Ba(OH)}_2\), is a strong base that fully dissociates to yield two moles of hydroxide ions per mole of base: \([\text{OH}^-] = 2 \times 0.040 = 0.080\text{ mol dm}^{-3}\). Using the relation \(K_w = [\text{H}^+][\text{OH}^-]\), the hydrogen ion concentration is \([\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.080} = 1.25 \times 10^{-13}\text{ mol dm}^{-3}\). The pH is then calculated as: \(\text{pH} = -\log_{10}(1.25 \times 10^{-13}) = 12.90\).

1 mark: correct selection of option D. Calculated hydroxide ion concentration, hydrogen ion concentration, and resulting pH.

The initial rate is given by \(\text{rate}_1 = k[\text{A}][\text{B}]^2\). When \([\text{A}]\) is doubled to \(2[\text{A}]\) and \([\text{B}]\) is halved to \(0.5[\text{B}]\), the new rate is: \(\text{rate}_2 = k(2[\text{A}])(0.5[\text{B}])^2 = 2 \times 0.25 \times k[\text{A}][\text{B}]^2 = 0.5 \times \text{rate}_1\). Therefore, the rate of reaction is halved.

1 mark: Correctly identifies that the rate is halved (Option B).

First, calculate the entropy change of the surroundings: \(\Delta S_{\text{surroundings}}^\ominus = -\frac{\Delta H^\ominus}{T} = -\frac{-95.0 \times 10^3\text{ J mol}^{-1}}{298\text{ K}} = +318.8\text{ J K}^{-1}\text{ mol}^{-1}\). Next, calculate the total entropy change: \(\Delta S_{\text{total}}^\ominus = \Delta S_{\text{system}}^\ominus + \Delta S_{\text{surroundings}}^\ominus = -120 + 318.8 = +198.8\text{ J K}^{-1}\text{ mol}^{-1}\), which rounds to \(+199\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark: Correctly calculates surroundings entropy change and adds it to system entropy change to obtain +199 J K-1 mol-1 (Option C).

The expression for the equilibrium constant \(K_p\) is: \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \cdot p(\text{O}_2)}\). Substituting the given values: \(K_p = \frac{(0.80)^2}{(0.20)^2 \cdot (0.40)} = \frac{0.64}{0.040 \cdot (0.40)} = \frac{0.64}{0.016} = 40\).

1 mark: Writes correct Kp expression and evaluates the numerical value to be 40 (Option C).

Using the buffer equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\). First, calculate \\text{p}K_a = -\log_{10}(1.6 \times 10^{-4}) = 3.80\). Then, substitute the concentrations: \(\text{pH} = 3.80 + \log_{10}\left(\frac{0.20}{0.10}\right) = 3.80 + \log_{10}(2) = 3.80 + 0.30 = 4.10\).

1 mark: Correctly calculates the pH of the buffer solution to 2 decimal places as 4.10 (Option C).

The product is 2-hydroxybutanenitrile, which does contain a chiral carbon and can exist as optical isomers. The reactant propanal has a planar carbonyl group (\(\text{C}=\text{O}\)) at the aldehydic carbon. The cyanide nucleophile (\(\text{CN}^-\)) can attack this planar carbon from either side (above or below) with equal probability. This yields a 1:1 mixture (racemic mixture) of the two enantiomers, which is optically inactive because the optical rotation of one enantiomer is cancelled by the equal and opposite rotation of the other.

1 mark: Correctly explains the formation of a racemic mixture due to equal probability of attack on the planar carbonyl group (Option C).

From the Arrhenius equation in logarithmic form, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). The gradient of the line is given by \(\text{gradient} = -\frac{E_a}{R}\). Therefore, \(E_a = -\text{gradient} \times R = -(-6.5 \times 10^3\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 5.40 \times 10^4\text{ J mol}^{-1} = +54\text{ kJ mol}^{-1}\).

1 mark: Correctly calculates Ea using gradient = -Ea/R and converts units to kJ mol-1 (Option C).

The titration involves a weak acid (ethanoic acid) and a strong base (sodium hydroxide). At the equivalence point, the solution contains sodium ethanoate, which undergoes basic hydrolysis, resulting in an alkaline equivalence point (typically \(\text{pH} \approx 8.7\)). The vertical region of this titration curve lies entirely in the alkaline pH range. Therefore, phenolphthalein, with a pH transition range of 8.3 to 10.0, is the only suitable indicator.

1 mark: Correctly identifies phenolphthalein as the only suitable indicator for a weak acid-strong base titration (Option D).

Propanone is a ketone and propanal is an aldehyde. Tollens' reagent is a mild oxidizing agent that oxidizes aldehydes (propanal) to carboxylic acids, reducing the silver complex to form a silver mirror. Ketones cannot be easily oxidized and do not react with Tollens' reagent. 2,4-DNP reacts with both to form orange precipitates, sodium carbonate reacts with neither, and phosphorus pentachloride is not a diagnostic test for distinguishing aldehydes from ketones.

1 mark: Identifies Tollens' reagent as the reagent that distinguishes aldehydes from ketones (Option B).

The rate equation is given as \(\text{rate} = k[A]^2[B]\). Rearranging for \(k\) gives \(k = \frac{\text{rate}}{[A]^2[B]}\). The units of rate are \(\text{mol dm}^{-3}\text{ s}^{-1}\) and the units of concentration are \(\text{mol dm}^{-3}\). Substituting these into the equation: \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). This corresponds to option C.

1 mark: Correctly identifies the units of the third-order rate constant as \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

For a reaction to be thermodynamically feasible, the Gibbs free energy change must be negative (\(\Delta G < 0\)). Using the Gibbs equation: \(\Delta G = \Delta H - T\Delta S_{\text{system}}\). At the point of feasibility, \(\Delta G = 0\), so \(T = \frac{\Delta H}{\Delta S_{\text{system}}}\). Converting \(\Delta H\) to \(\text{J mol}^{-1}\) gives \(135000\text{ J mol}^{-1}\). Therefore, \(T = \frac{135000}{320} = 421.875\text{ K}\). Above this temperature (approximately \(422\text{ K}\)), the reaction is feasible. This corresponds to option B.

1 mark: Correctly calculates the temperature limit as \(422\text{ K}\).

For a weak monoprotic acid, we can use the approximation: \([\text{H}^+] = \sqrt{K_{\text{a}} \times [\text{HA}]}\). Substituting the given values: \([\text{H}^+] = \sqrt{1.80 \times 10^{-5} \times 0.150} = \sqrt{2.70 \times 10^{-6}} = 1.643 \times 10^{-3}\text{ mol dm}^{-3}\). The pH is calculated as: \(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.643 \times 10^{-3}) = 2.78\). This matches option B.

1 mark: Correctly calculates the pH as \(2.78\).

The carbonyl group (\(\text{C=O}\)) in propanal is planar around the carbonyl carbon atom. The cyanide nucleophile (\(\text{CN}^-\)) can attack the planar carbonyl group with equal probability from either above or below the plane. This produces equal amounts of the two enantiomers (a 50:50 racemic mixture). This matches option B.

1 mark: Correctly identifies that equal probability of attack from above or below the planar carbonyl group leads to a racemic mixture.

To calculate the standard entropy change of the system:

\[ \Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants}) \]

Substitute the given values into the equation:

\[ \Delta S^{\ominus}_{\text{system}} = 2 \times S^{\ominus}[\text{NH}_3(\text{g})] - \left( S^{\ominus}[\text{N}_2(\text{g})] + 3 \times S^{\ominus}[\text{H}_2(\text{g})] \right) \]

\[ \Delta S^{\ominus}_{\text{system}} = 2(192.3) - [191.6 + 3(130.6)] \]

\[ \Delta S^{\ominus}_{\text{system}} = 384.6 - [191.6 + 391.8] \]

\[ \Delta S^{\ominus}_{\text{system}} = 384.6 - 583.4 = -198.8\text{ J K}^{-1}\text{ mol}^{-1} \]

- M1 (1 mark): Correct expression or substitution of terms into the entropy equation.
- M2 (1 mark): Correct arithmetic calculation yielding \(-198.8\).
- M3 (0.5 marks): Correct units \(\text{J K}^{-1}\text{ mol}^{-1}\) (or \(\text{J mol}^{-1}\text{ K}^{-1}\)).

Use the logarithmic form of the Arrhenius equation:

\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_{\text{a}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Substitute the given values into the equation:

\[ \ln\left(\frac{4.80 \times 10^{-3}}{1.20 \times 10^{-3}}\right) = -\frac{E_{\text{a}}}{8.31}\left(\frac{1}{318} - \frac{1}{298}\right) \]

\[ \ln(4) = -\frac{E_{\text{a}}}{8.31}\left(0.00314465 - 0.00335570\right) \]

\[ 1.3863 = -\frac{E_{\text{a}}}{8.31}\left(-2.1105 \times 10^{-4}\right) \]

\[ 1.3863 = E_{\text{a}} \times 2.5397 \times 10^{-5} \]

\[ E_{\text{a}} = \frac{1.3863}{2.5397 \times 10^{-5}} = 54585\text{ J mol}^{-1} = 54.6\text{ kJ mol}^{-1} \]

- M1 (1 mark): Correct expression or rearrangement of the Arrhenius equation showing how to calculate \(E_{\text{a}}\).
- M2 (1 mark): Correct intermediate calculations of \(\ln(4)\) and the temperature differences.
- M3 (0.5 marks): Correct final value of \(54.6\text{ kJ mol}^{-1}\) (accept values between \(54.5\) and \(55.0\)). Deduct 0.5 marks for incorrect or missing units.

First, calculate the chemical amount (moles) of the acid and conjugate base in the buffer solution:

\[ n(\text{propanoic acid}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol} \]

\[ n(\text{propanoate ions}) = 0.0500\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol} \]

Now, use the buffer equation:

\[ [\text{H}^+] = K_{\text{a}} \times \frac{[\text{propanoic acid}]}{[\text{propanoate}]} = 1.35 \times 10^{-5} \times \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 2.70 \times 10^{-5}\text{ mol dm}^{-3} \]

Finally, calculate the pH:

\[ \text{pH} = -\log_{10}(2.70 \times 10^{-5}) = 4.57 \]

- M1 (1 mark): Calculates moles of propanoic acid (\(5.00 \times 10^{-3}\text{ mol}\)) and sodium propanoate (\(2.50 \times 10^{-3}\text{ mol}\)).
- M2 (1 mark): Correct substitution of moles/concentrations into the \(K_{\text{a}}\) or Henderson-Hasselbalch equation to find \([\text{H}^+] = 2.70 \times 10^{-5}\text{ mol dm}^{-3}\).
- M3 (0.5 marks): Correct final value of \(4.57\) (must be given to 2 decimal places).

1. Propanal contains an unsymmetrical carbonyl group. The carbonyl carbon is planar, allowing the nucleophile (\(\text{CN}^-\)) to attack from above or below the plane with equal probability. This yields an equimolar (50:50) mixture of the two optical enantiomers, forming a racemic mixture which is optically inactive due to external compensation.
2. The product of this reaction with propanal has a chiral center (carbon bonded to four different groups: \(\text{-H}\), \(\text{-OH}\), \(\text{-CH}_2\text{CH}_3\), and \(\text{-CN}\)).
3. In propanone, the addition of \(\text{CN}^-\) yields 2-hydroxy-2-methylpropanenitrile. Since the central carbon is bonded to two identical methyl groups, the product does not possess a chiral carbon and thus has no optical isomers.

- M1 (1 mark): Explains that the planar carbonyl group in propanal allows equal probability of attack from above or below, forming a 50:50 (racemic) mixture of enantiomers.
- M2 (1 mark): Identifies that the product with propanal has a chiral carbon (chiral center), whereas the product with propanone has two identical methyl groups attached to the central carbon.
- M3 (0.5 marks): Explicitly states that the lack of a chiral carbon in propanone's product is why no optical isomers are produced.

From Dalton's Law of Partial Pressures:

\[ P_{\text{total}} = p(\text{PCl}_5) + p(\text{PCl}_3) + p(\text{Cl}_2) = 0.840\text{ atm} \]

Given \(p(\text{PCl}_5) = 0.240\text{ atm}\):

\[ p(\text{PCl}_3) + p(\text{Cl}_2) = 0.840 - 0.240 = 0.600\text{ atm} \]

Since \(\text{PCl}_5\) dissociates to produce \(\text{PCl}_3\) and \(\text{Cl}_2\) in a 1:1 mole ratio, their equilibrium partial pressures must be equal:

\[ p(\text{PCl}_3) = p(\text{Cl}_2) = \frac{0.600\text{ atm}}{2} = 0.300\text{ atm} \]

Write the expression for \(K_p\):

\[ K_p = \frac{p(\text{PCl}_3) \times p(\text{Cl}_2)}{p(\text{PCl}_5)} \]

Substitute the partial pressures:

\[ K_p = \frac{0.300 \times 0.300}{0.240} = 0.375\text{ atm} \]

- M1 (1 mark): Correctly calculates the equilibrium partial pressures of both \(\text{PCl}_3\) and \(\text{Cl}_2\) as \(0.300\text{ atm}\).
- M2 (1 mark): Correctly writes the expression for \(K_p\) and calculates the numerical value of \(0.375\).
- M3 (0.5 marks): States the correct units: \(\text{atm}\).

Apply the Born-Haber cycle equation for the formation of sodium chloride:

\[ \Delta H^{\ominus}_{\text{f}} = \Delta H^{\ominus}_{\text{at}}[\text{Na}] + 1^{\text{st}}\text{ IE}[\text{Na}] + \Delta H^{\ominus}_{\text{at}}[\text{Cl}_2] + 1^{\text{st}}\text{ EA}[\text{Cl}] + \Delta H^{\ominus}_{\text{latt}} \]

Substitute the known standard values into the energy cycle:

\[ -411 = +107 + 496 + 122 + (-349) + \Delta H^{\ominus}_{\text{latt}} \]

\[ -411 = +376 + \Delta H^{\ominus}_{\text{latt}} \]

Rearrange to solve for the standard lattice energy:

\[ \Delta H^{\ominus}_{\text{latt}} = -411 - 376 = -787\text{ kJ mol}^{-1} \]

- M1 (1 mark): Correctly sets up the Born-Haber cycle equation linking standard formation enthalpy to the individual terms.
- M2 (1 mark): Correct substitution of values with correct signs (giving standard total of reactants as \(+376\)).
- M3 (0.5 marks): Correct final value of \(-787\text{ kJ mol}^{-1}\) (the negative sign and units must be specified).

1. Compare Experiments 1 and 2: \([\text{B}]\) is constant, and \([\text{A}]\) doubles. The initial rate increases by a factor of \(\frac{9.6 \times 10^{-4}}{2.4 \times 10^{-4}} = 4 = 2^2\). Thus, the order of reaction with respect to A is 2.
2. Compare Experiments 1 and 3: \([\text{A}]\) is constant, and \([\text{B}]\) doubles. The initial rate increases by a factor of \(\frac{4.8 \times 10^{-4}}{2.4 \times 10^{-4}} = 2 = 2^1\). Thus, the order of reaction with respect to B is 1.
3. The rate equation is: \(\text{Rate} = k[\text{A}]^2[\text{B}]\).
4. Calculate \(k\) using the values from Experiment 1:

\[ 2.4 \times 10^{-4} = k(0.10)^2(0.15) \]

\[ 2.4 \times 10^{-4} = k(0.0015) \]

\[ k = \frac{2.4 \times 10^{-4}}{0.0015} = 0.16 \]

5. Determine units:

\[ \text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \]

- M1 (1 mark): Deduces the reaction orders with respect to A (second order) and B (first order), stating the rate equation \(\text{Rate} = k[\text{A}]^2[\text{B}]\).
- M2 (1 mark): Calculates the numerical value of \(k\) as \(0.16\).
- M3 (0.5 marks): States the correct rate constant units: \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) (or \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)).

1. Since the alcohol \(\mathbf{Z}\) is oxidized to a ketone, \(\mathbf{Z}\) must be a secondary alcohol.
2. The simplest secondary alcohol is propan-2-ol (having 3 carbons).
3. Since the starting ester \(\mathbf{X}\) has a total of 4 carbons, and alcohol \(\mathbf{Z}\) contains 3 carbons, the carboxylic acid \(\mathbf{Y}\) must have exactly 1 carbon. The 1-carbon carboxylic acid is methanoic acid.
4. Therefore, the ester \(\mathbf{X}\) is formed from methanoic acid and propan-2-ol, which gives the IUPAC name propan-2-yl methanoate (also accepted as isopropyl methanoate).

- \(\mathbf{X}\): propan-2-yl methanoate
- \(\mathbf{Y}\): methanoic acid
- \(\mathbf{Z}\): propan-2-ol

- M1 (1 mark): Correctly identifies alcohol \(\mathbf{Z}\) as propan-2-ol (based on being a secondary alcohol and carbon count).
- M2 (1 mark): Correctly identifies carboxylic acid \(\mathbf{Y}\) as methanoic acid.
- M3 (0.5 marks): Correctly identifies ester \(\mathbf{X}\) as propan-2-yl methanoate (allow isopropyl methanoate).

1. Rearrange the rate equation to make the rate constant, \(k\), the subject:

$$k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}$$

2. Substitute the given values into the equation:

$$k = \frac{1.28 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{(0.0400\text{ mol dm}^{-3}) \times (0.0800\text{ mol dm}^{-3})} = \frac{1.28 \times 10^{-4}}{3.20 \times 10^{-3}} = 0.0400$$

3. Determine the units of \(k\):

$$\text{Units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}$$

Combining the value and units gives \(k = 0.0400\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(4.00 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)).

• M1 (1 mark): Rearranges rate equation correctly and calculates numerical value of \(k = 0.0400\) (or \(4.00 \times 10^{-2}\)).
• M2 (1 mark): Correct units of \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (accept in any order, e.g. \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)).
• M3 (0.5 marks): Final numerical answer given to 3 significant figures.

1. Calculate the entropy change of the surroundings, \(\Delta S_{\text{surroundings}}^{\ominus}\):

$$\Delta S_{\text{surroundings}}^{\ominus} = -\frac{\Delta H^{\ominus}}{T}$$

Convert \(\Delta H^{\ominus}\) from \(\text{kJ mol}^{-1}\) to \(\text{J mol}^{-1}\):

$$\Delta H^{\ominus} = +178.0 \times 10^3\text{ J mol}^{-1} = +178000\text{ J mol}^{-1}$$

$$\Delta S_{\text{surroundings}}^{\ominus} = -\frac{178000\text{ J mol}^{-1}}{298\text{ K}} = -597.315\text{ J K}^{-1}\text{ mol}^{-1}$$

2. Calculate the total entropy change, \(\Delta S_{\text{total}}^{\ominus}\):

$$\Delta S_{\text{total}}^{\ominus} = \Delta S_{\text{system}}^{\ominus} + \Delta S_{\text{surroundings}}^{\ominus}$$

$$\Delta S_{\text{total}}^{\ominus} = +160.4 + (-597.315) = -436.915\text{ J K}^{-1}\text{ mol}^{-1}$$

Rounding to 3 significant figures gives \(-437\text{ J K}^{-1}\text{ mol}^{-1}\) (or \(-436.9\text{ J K}^{-1}\text{ mol}^{-1}\) to 4 significant figures).

• M1 (1 mark): Calculates \ΔS_surroundings correctly as \(-597.3\text{ J K}^{-1}\text{ mol}^{-1}\) (or shows correct conversion of \(\Delta H\) to Joules in expression).
• M2 (1 mark): Correctly adds system and surroundings values to calculate \(\Delta S_{\text{total}}\).
• M3 (0.5 marks): Answer must have a negative sign, units of \(\text{J K}^{-1}\text{ mol}^{-1}\), and be rounded to 3 or 4 significant figures (\(-437\) or \(-436.9\)).

1. Write the weak acid approximation expression for \(K_{\text{a}}\):

$$K_{\text{a}} \approx \frac{[\text{H}^+]^2}{[\text{HA}]}$$

2. Rearrange to solve for \([\text{H}^+]\):

$$[\text{H}^+]^2 = K_{\text{a}} \times [\text{HA}]$$

$$[\text{H}^+]^2 = (1.35 \times 10^{-5}\text{ mol dm}^{-3}) \times (0.150\text{ mol dm}^{-3}) = 2.025 \times 10^{-6}\text{ mol}^2\text{ dm}^{-6}$$

$$[\text{H}^+] = \sqrt{2.025 \times 10^{-6}} = 1.423 \times 10^{-3}\text{ mol dm}^{-3}$$

3. Calculate the pH:

$$\text{pH} = -\log_{10}[\text{H}^+]$$

$$\text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.8468$$

Rounding to 2 decimal places gives \(2.85\).

• M1 (1 mark): Recalls or uses the expression \([\text{H}^+] = \sqrt{K_{\text{a}} \times [\text{HA}]}\) with numbers correctly substituted.
• M2 (1 mark): Calculates \([\text{H}^+]\) as \(1.42 \times 10^{-3}\text{ mol dm}^{-3}\) and correctly converts this concentration to pH.
• M3 (0.5 marks): Gives final pH to 2 decimal places (2.85).

1. The carbonyl group (\(\text{C}=\text{O}\)) is planar about the carbonyl carbon.
2. The hydride ion (\(\text{H}^-\)) nucleophile has an equal probability of attacking this planar carbon atom from either above or below the plane.
3. Therefore, equal amounts (a 50:50 or racemic mixture) of the two enantiomers (\((R)\)-butan-2-ol and \((S)\)-butan-2-ol) are produced.
4. The optical rotations of the two enantiomers are equal in magnitude but opposite in direction, so they cancel each other out, resulting in no net optical activity.

• M1 (1 mark): States that the carbonyl group / \(\text{C}=\text{O}\) bond is planar (accept 'planar about the carbonyl carbon' / 'trigonal planar carbon').
• M2 (1 mark): Explains that nucleophilic attack / attack by hydride ion (\(\text{H}^-\)) is equally likely from either side / from above and below the plane.
• M3 (0.5 marks): Concludes that this results in a racemic mixture (equal amounts of enantiomers) whose optical rotations cancel.

Step 1: Calculate the left-hand side of the equation: \(\ln(k_2/k_1) = \ln(3.10 \times 10^{-3} / 1.25 \times 10^{-4}) = \ln(24.8) = 3.2109\). Step 2: Calculate the reciprocal temperature difference: \(1/T_2 - 1/T_1 = 1/620 - 1/550 = 0.0016129 - 0.0018182 = -2.053 \times 10^{-4}\text{ K}^{-1}\). Step 3: Rearrange to solve for \(E_a\): \(3.2109 = -\frac{E_a}{8.31} \times (-2.053 \times 10^{-4})\) which gives \(E_a = \frac{3.2109 \times 8.31}{2.053 \times 10^{-4}} = 129981\text{ J mol}^{-1} = 130\text{ kJ mol}^{-1}\) (to 3 s.f.). Step 4: A catalyst provides an alternative pathway with a lower activation energy, which increases the proportion of molecules with energy greater than or equal to the activation energy. Consequently, the rate constant \(k\) increases.

M1 (1 mark): Correct calculation of \(\ln(k_2/k_1) = 3.21\). M2 (1 mark): Correct calculation of \((1/T_2 - 1/T_1) = -2.05 \times 10^{-4}\text{ K}^{-1}\). M3 (1.5 marks): Rearrangement and calculation of \(E_a = 130\text{ kJ mol}^{-1}\) (allow 129-131, deduct 0.5 marks for incorrect units or significant figures). M4 (1 mark): Explanation that catalyst provides an alternative pathway with lower activation energy. M5 (1 mark): Correct description that the rate constant, \(k\), increases.

(a) \(\Delta S_{\text{system}}^\ominus = \Sigma S^\ominus\text{(products)} - \Sigma S^\ominus\text{(reactants)} = (210.8 + 213.6) - (240.1 + 197.6) = 424.4 - 437.7 = -13.3\text{ J K}^{-1}\text{ mol}^{-1}\). (b) \(\Delta S_{\text{surroundings}}^\ominus = -\frac{\Delta H^\ominus}{T} = -\frac{-226.4 \times 10^3}{298} = +759.7\text{ J K}^{-1}\text{ mol}^{-1}\). \(\Delta S_{\text{total}}^\ominus = \Delta S_{\text{system}}^\ominus + \Delta S_{\text{surroundings}}^\ominus = -13.3 + 759.7 = +746.4\text{ J K}^{-1}\text{ mol}^{-1}\) (or \(+746\text{ J K}^{-1}\text{ mol}^{-1}\)). (c) As temperature increases, \(\Delta S_{\text{surroundings}}\text{ } (=-\Delta H/T)\) becomes less positive. Since \(\Delta S_{\text{system}}\) is negative, the total entropy change \(\Delta S_{\text{total}}\) becomes less positive, so the reaction becomes less feasible.

M1 (1.5 marks): \(\Delta S_{\text{system}}^\ominus = -13.3\text{ J K}^{-1}\text{ mol}^{-1}\) (1 mark for correct setup, 0.5 mark for correct value and negative sign). M2 (1 mark): \(\Delta S_{\text{surroundings}}^\ominus = +759.7\text{ J K}^{-1}\text{ mol}^{-1}\) (allow +760). M3 (1 mark): \(\Delta S_{\text{total}}^\ominus = +746.4\text{ J K}^{-1}\text{ mol}^{-1}\) (allow +746 or +747). M4 (1 mark): Correct explanation that \(\Delta S_{\text{surroundings}}\) decreases / becomes less positive as temperature increases. M5 (1 mark): Feasibility decreases as temperature increases because \(\Delta S_{\text{total}}\) becomes less positive.

(a) Step 1: Calculate initial moles of acid: \(n(\text{acid}) = 0.0500 \times 0.150 = 7.50 \times 10^{-3}\text{ mol}\). Step 2: Calculate initial moles of \(\text{NaOH}\): \(n(\text{NaOH}) = 0.0300 \times 0.100 = 3.00 \times 10^{-3}\text{ mol}\). Step 3: \(\text{NaOH}\) reacts completely with the acid: Remaining \(n(\text{acid}) = 7.50 \times 10^{-3} - 3.00 \times 10^{-3} = 4.50 \times 10^{-3}\text{ mol}\); Formed \(n(\text{propanoate}^-) = 3.00 \times 10^{-3}\text{ mol}\). Step 4: \([\text{H}^+] = K_a \times \frac{n(\text{acid})}{n(\text{salt})} = 1.35 \times 10^{-5} \times \frac{4.50 \times 10^{-3}}{3.00 \times 10^{-3}} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\). Step 5: \(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6935 \approx 4.69\). (b) Added \(\text{OH}^-\)$ reacts with the excess \(\text{CH}_3\text{CH}_2\text{COOH}\) present in the buffer: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\), maintaining a nearly constant concentration of \([\text{H}^+]\) and pH.

M1 (1 mark): Calculation of initial moles of acid (7.50 x 10^-3) and base (3.00 x 10^-3). M2 (1 mark): Calculation of excess acid moles (4.50 x 10^-3). M3 (1 mark): Calculation of salt moles formed (3.00 x 10^-3). M4 (1 mark): Calculation of hydrogen ion concentration \([\text{H}^+] = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\). M5 (0.5 marks): Correct calculation of pH to 2 decimal places (4.69). M6 (1 mark): Explanation of buffer action with a balanced equation or description of reaction of added \(\text{OH}^-\)

(a) In the nucleophilic addition reaction, the nucleophile is the cyanide ion, \(\text{CN}^-\). It attacks the carbon atom of the polar carbonyl group (which carries a partial positive charge, \(\delta^+\)), represented by a curly arrow from the lone pair of the carbon on the \(\text{CN}^-\) to the carbonyl carbon. Another curly arrow shows the movement of the electron pair from the \(\text{C}=\text{O}\) double bond onto the electronegative oxygen atom (\(\delta^-\)). This forms a tetrahedral intermediate, \(\text{CH}_3\text{CH}_2\text{C}(\text{O}^-)(\text{CN})\text{CH}_3\), which then gets protonated by \(\text{H}^+\) (or \(\text{HCN}\)) via a curly arrow from the lone pair on the oxygen to the hydrogen ion to yield the final hydroxynitrile product, 2-hydroxy-2-methylbutanenitrile. (b) The carbonyl group in butanone is planar around the carbonyl carbon. The nucleophile \(\text{CN}^-\) has an equal probability of attacking the planar carbonyl group from either above or below the plane. This leads to the formation of equal amounts of the two enantiomers (a racemic mixture), which rotate plane-polarized light in equal and opposite directions, making the mixture optically inactive.

M1 (1 mark): Identification of \(\text{CN}^-\) as the nucleophile, and correct description of the polar \(\text{C}^{\delta+}=\text{O}^{\delta-}\) group. M2 (1 mark): Correct description of the first step with curly arrow from \(\text{CN}^-\) lone pair to carbon, and from the double bond to oxygen. M3 (1.5 marks): Identification and structure of the tetrahedral intermediate and the final protonation step to yield 2-hydroxy-2-methylbutanenitrile. M4 (1 mark): Explanation that the carbonyl group / carbon-oxygen double bond is planar. M5 (1 mark): Explanation that attack can occur with equal probability from either side of the plane, resulting in a racemic mixture which is optically inactive.

First, determine the moles of propanoic acid (HA) and propanoate ions (A^-) in the mixture. Moles of HA = \(0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\). Moles of A^- = \(0.0500\text{ dm}^3 \times 0.080\text{ mol dm}^{-3} = 4.00 \times 10^{-3}\text{ mol}\). Using the weak acid equilibrium expression: \([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.35 \times 10^{-5} \times \frac{7.50 \times 10^{-3}}{4.00 \times 10^{-3}} = 2.53125 \times 10^{-5}\text{ mol dm}^{-3}\). Finally, calculate the pH: \(\text{pH} = -\log_{10}(2.53125 \times 10^{-5}) = 4.5966\), which rounds to 4.60.

1 mark: Correct calculation of the amount of substance (in moles) or concentrations of both propanoic acid and propanoate ions in the mixture. 1 mark: Correct substitution of values into the Ka or Henderson-Hasselbalch expression to find [H+] or pH. 0.5 mark: Final pH value of 4.60 (allow range 4.59 to 4.61).

For a reaction to be thermodynamically feasible, \(\Delta G \le 0\). At the transition temperature, \(\Delta G = 0\), which leads to \(\Delta H - T\Delta S = 0\), or \(T = \frac{\Delta H}{\Delta S}\). First, convert \(\Delta H\) to \(\text{J mol}^{-1}\): \(\Delta H = 178 \times 10^3\text{ J mol}^{-1}\). Then, calculate \(T = \frac{178000}{160} = 1112.5\text{ K}\). To three significant figures, the temperature is 1110 K.

1 mark: State or use the condition for feasibility that \(\Delta G = 0\) (or \(\Delta G < 0\)) and correctly convert the units of either enthalpy or entropy so that they are consistent (e.g., \(178\text{ kJ}\) to \(178000\text{ J}\)). 1 mark: Correct rearrangement of the Gibbs free energy equation to make T the subject: \(T = \frac{\Delta H}{\Delta S}\) and substitution of values. 0.5 mark: Correct calculation of the final temperature as 1110 K (allow 1112.5 K or 1113 K).

Use the integrated form of the Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\), which can be written as \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\). Substituting the given values: \(\ln\left(\frac{1.20 \times 10^{-2}}{2.40 \times 10^{-3}}\right) = \frac{E_a}{8.31}\left(\frac{1}{300} - \frac{1}{320}\right)\). This simplifies to \(\ln(5.00) = \frac{E_a}{8.31}\left(3.333 \times 10^{-3} - 3.125 \times 10^{-3}\right)\), which is \(1.6094 = \frac{E_a}{8.31}\left(2.0833 \times 10^{-4}\right)\). Solving for \(E_a\): \(E_a = \frac{1.6094 \times 8.31}{2.0833 \times 10^{-4}} = 64195\text{ J mol}^{-1}\). Converting to \(\text{kJ mol}^{-1}\) gives \(64.2\text{ kJ mol}^{-1}\).

1 mark: Correct substitution of values into the Arrhenius equation. 1 mark: Correct calculation of intermediate terms, \(\ln(5.00) = 1.61\) and \(\Delta(1/T) = 2.08 \times 10^{-4}\text{ K}^{-1}\). 0.5 mark: Correct calculation of Ea in kJ mol^-1, with value 64.2 (accept range 64.0 to 64.3).

Butanone contains a planar carbonyl group (C=O) at the reaction site. When the hydride ion nucleophile (H^-) attacks the carbonyl carbon, it can do so with equal probability from either above or below the plane of the molecule. Since the environment on either side of the flat carbonyl group is equally accessible, equal amounts (a 50:50 ratio) of both the (R)- and (S)-enantiomers of butan-2-ol are formed. This equimolar mixture is a racemic mixture, and because the opposite optical rotations of the two enantiomers cancel each other out, the resulting mixture has no net optical activity.

1 mark: Identify that the carbonyl group (or carbonyl carbon) in butanone is planar. 1 mark: Explain that the nucleophile (hydride ion / H^-) can attack with equal probability from either side (or above and below the plane). 0.5 mark: Conclude that this forms a racemic mixture (or an equimolar mixture of enantiomers) where the optical rotations cancel out.

Part (a): Write the expression for the acid dissociation constant: \(K_a = \frac{[\text{H}^+][\text{Lac}^-]}{[\text{HLac}]}\). Assuming that the concentration of hydrogen ions at equilibrium is equal to the concentration of lactate ions, and that the concentration of undissociated lactic acid is approximately equal to the initial concentration: \(K_a = \frac{[\text{H}^+]^2}{[\text{HLac}]}\). Rearranging gives: \([\text{H}^+]^2 = K_a \times [\text{HLac}] = 1.38 \times 10^{-4} \times 0.150 = 2.07 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\). Taking the square root: \([\text{H}^+] = \sqrt{2.07 \times 10^{-5}} = 4.5497 \times 10^{-3}\text{ mol dm}^{-3}\). Calculate pH: \(\text{pH} = -\log_{10}(4.5497 \times 10^{-3}) = 2.34\). Part (b): Calculate the initial moles of lactic acid (HLac): \(n(\text{HLac}) = \frac{25.0}{1000} \times 0.150 = 3.75 \times 10^{-3}\text{ mol}\). Calculate the moles of sodium hydroxide added: \(n(\text{NaOH}) = \frac{15.0}{1000} \times 0.100 = 1.50 \times 10^{-3}\text{ mol}\). The added sodium hydroxide reacts completely with the lactic acid: \(\text{HLac} + \text{OH}^- \rightarrow \text{Lac}^- + \text{H}_2\text{O}\). Moles of remaining lactic acid: \(n(\text{HLac})_{\text{remaining}} = 3.75 \times 10^{-3} - 1.50 \times 10^{-3} = 2.25 \times 10^{-3}\text{ mol}\). Moles of lactate ions formed: \(n(\text{Lac}^-)_{\text{formed}} = 1.50 \times 10^{-3}\text{ mol}\). The total volume is \(40.0\text{ cm}^3\), but because the volumes cancel in the ratio, we can use moles directly in the expression: \([\text{H}^+] = K_a \times \frac{n(\text{HLac})}{n(\text{Lac}^-)} = 1.38 \times 10^{-4} \times \frac{2.25 \times 10^{-3}}{1.50 \times 10^{-3}} = 2.07 \times 10^{-4}\text{ mol dm}^{-3}\). Calculate the pH: \(\text{pH} = -\log_{10}(2.07 \times 10^{-4}) = 3.68\).

Part (a): [3 Marks] M1: Shows correct expression for Ka or rearranged expression for [H+]^2, and calculates [H+]^2 = 2.07 x 10^-5 mol^2 dm^-6. M2: Calculates [H+] = 4.55 x 10^-3 mol dm^-3 (allow 4.5497 x 10^-3). M3: Calculates pH = 2.34 (must be 2 decimal places). Part (b): [5 Marks] M1: Calculates initial moles of HLac (3.75 x 10^-3 mol) AND NaOH (1.50 x 10^-3 mol). M2: Calculates remaining HLac (2.25 x 10^-3 mol) AND formed Lac^- (1.50 x 10^-3 mol). M3: Shows correct buffer expression or uses Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[weak acid]). M4: Calculates [H+] = 2.07 x 10^-4 mol dm^-3 (or pKa = 3.86). M5: Calculates pH = 3.68 (must be 2 decimal places). (Allow TE throughout for calculation errors, provided the method is chemically sound.)

To obtain a standard cell potential of \(+0.93\text{ V}\), we need to select a reduction and an oxidation half-cell such that \(E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.93\text{ V}\).

Let \(E^\ominus_{\text{reduction}} = E^\ominus(\text{Ce}^{4+}/\text{Ce}^{3+}) = +1.70\text{ V}\).
Let \(E^\ominus_{\text{oxidation}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\).

Then, \(E^\ominus_{\text{cell}} = +1.70 - (+0.77) = +0.93\text{ V}\).

The spontaneous cell reaction involves the reduction of \(\text{Ce}^{4+}\) to \(\text{Ce}^{3+}\) and the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\).
This corresponds to the oxidation of \(\text{Fe}^{2+}\) ions by \(\text{Ce}^{4+}\) ions.

1 mark for the correct option (A).
- Reject other options because they yield different cell potentials:
- B yields \(+0.77 - 0.54 = +0.23\text{ V}\)
- C yields \(+1.70 - 0.54 = +1.16\text{ V}\)
- D yields a non-spontaneous potential of \(-0.93\text{ V}\).

Let's determine the electronic configurations:
- \(\text{Cr}\) has the exceptional ground-state electronic configuration \([\text{Ar}] 4\text{s}^1 3\text{d}^5\). To form \(\text{Cr}^{2+}\), we remove two electrons (one from \(4\text{s}\) and one from \(3\text{d}\)), resulting in \([\text{Ar}] 3\text{d}^4\).
- \(\text{Mn}^{2+}\) is derived from \(\text{Mn}\) (\([\text{Ar}] 4\text{s}^2 3\text{d}^5\)), which loses two \(4\text{s}\) electrons to give \([\text{Ar}] 3\text{d}^5\).
- \(\text{Fe}^{3+}\) is derived from \(\text{Fe}\) (\([\text{Ar}] 4\text{s}^2 3\text{d}^6\)), which loses two \(4\text{s}\) and one \(3\text{d}\) electrons to give \([\text{Ar}] 3\text{d}^5\).
- \(\text{V}^{3+}\) is derived from \(\text{V}\) (\([\text{Ar}] 4\text{s}^2 3\text{d}^3\)), which loses two \(4\text{s}\) and one \(3\text{d}\) electrons to give \([\text{Ar}] 3\text{d}^2\).

1 mark for identifying \(\text{Cr}^{2+}\) as the species with a \(3\text{d}^4\) configuration.

In the nitration of benzene, concentrated sulfuric acid reacts with concentrated nitric acid to generate the active electrophile, \(\text{NO}_2^+\) (the nitronium ion), according to the equation:
\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\).
Here, \(\text{H}_2\text{SO}_4\) acts as a Brønsted-Lowry acid (proton donor) to protonate \(\text{HNO}_3\), which subsequently loses a water molecule. Because the sulfuric acid is regenerated at the end of the mechanism, it acts as a catalyst.

1 mark for the correct role of concentrated sulfuric acid (C).

- Phenylamine is the weakest base because the lone pair on the nitrogen atom is partially delocalized into the aromatic \(\pi\)-system, making it less available to accept a proton.
- Ammonia does not have any delocalization or electron-donating alkyl groups.
- Ethylamine is the strongest base among the three because the ethyl group is electron-releasing (positive inductive effect), increasing the electron density on the nitrogen atom and making the lone pair more available to accept a proton.
Therefore, the correct order is: phenylamine < ammonia < ethylamine.

1 mark for the correct order (A).

- Step 2 involves the reduction of nitrobenzene to phenylamine. This is achieved using tin (Sn) and concentrated hydrochloric acid (HCl), heating under reflux, followed by treatment with sodium hydroxide (NaOH) to release the free amine.
- Step 3 is diazotization, which requires nitrous acid, generated in situ from sodium nitrate(III) (\(\text{NaNO}_2\)) and dilute hydrochloric acid (HCl) at a low temperature (below \(10^\circ\text{C}\), usually \(0\text{--}5^\circ\text{C}\)) to prevent decomposition of the diazonium salt.

1 mark for identifying the correct combination of reagents and conditions for both steps (A).

- The silver half-cell has the more positive standard electrode potential (\(+0.80\text{ V}\)), so reduction occurs here: \(\text{Ag}^+(\text{aq}) + \text{e}^- \to \text{Ag(s)}\).
- The copper half-cell has the less positive potential (\(+0.34\text{ V}\)), so oxidation occurs here: \(\text{Cu(s)} \to \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-\).
- In standard cell notation, the oxidation half-cell (anode) is written on the left, and the reduction half-cell (cathode) is written on the right:
\(\text{Cu(s)} \mid \text{Cu}^{2+}(\text{aq}) \parallel \text{Ag}^+(\text{aq}) \mid \text{Ag(s)}\)
- The standard cell potential is calculated as:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +0.80 - (+0.34) = +0.46\text{ V}\).

1 mark for the correct standard IUPAC cell representation and EMF (A).

The color of a transition metal complex ion is determined by the size of the energy gap (\(\Delta E\)) between split d-orbitals. This energy split is affected by:
1. The identity of the metal ion.
2. The oxidation state of the metal ion.
3. The coordination number (geometry of the complex).
4. The nature of the ligands coordinated to the metal ion.

The concentration of the complex ion in the solution only affects the intensity of the light absorbed (absorbance), not the specific wavelength of maximum absorption (the color itself).

1 mark for selecting option C.

- In phenol, one of the lone pairs of electrons on the oxygen atom of the \(\text{-OH}\) group is partially delocalized into the aromatic ring's \(\pi\)-system.
- This delocalization increases the electron density of the benzene ring compared to benzene itself.
- The higher electron density makes phenol more effective at polarizing incoming electrophiles (such as \(\text{Br}_2\)), allowing it to react without needing a halogen carrier catalyst.

1 mark for the correct explanation of phenol's increased reactivity (B).

Under standard conditions, a species with a more positive standard electrode potential can oxidize the reduced form of a species with a less positive standard electrode potential.
- Since \( E^\ominus(\text{MnO}_4^- / \text{Mn}^{2+}) = +1.51\text{ V} \) is more positive than \( E^\ominus(\text{Cl}_2 / \text{Cl}^-) = +1.36\text{ V} \), \( \text{MnO}_4^-(\text{aq}) \) in acidic conditions is a strong enough oxidizing agent to oxidize \( \text{Cl}^-(\text{aq}) \) to \( \text{Cl}_2(\text{aq}) \).
- Conversely, \( \text{Fe}^{3+}(\text{aq}) \) (\( E^\ominus = +0.77\text{ V} \)) is not a strong enough oxidizing agent to oxidize \( \text{Cl}^- \). Therefore, C is correct.

[1] C

The atomic number of cobalt is 27, so a neutral cobalt atom has the electronic configuration: \( [\text{Ar}] \\, 3\text{d}^7 4\text{s}^2 \). When forming transition metal cations, the electrons from the \( 4\text{s} \) orbital are lost first. To form the \( \text{Co}^{3+} \) ion, three electrons must be removed: two from the \( 4\text{s} \) orbital and one from the \( 3\text{d} \) orbital. This leaves an electronic configuration of \( [\text{Ar}] \\, 3\text{d}^6 \).

[1] A

The complex \( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \) has geometric isomers:
1. The trans isomer, where the two chloride ligands are opposite to each other (at an angle of 180°). This isomer has a plane of symmetry and is therefore achiral (it has no optical isomers).
2. The cis isomer, where the two chloride ligands are adjacent to each other (at an angle of 90°). This isomer lacks a plane of symmetry and is chiral, existing as a pair of non-superimposable mirror images (enantiomers).

Thus, there are 3 stereoisomers in total: one trans isomer and two cis enantiomers.

[1] C

The ester group \( -\text{COOCH}_3 \) is attached to the benzene ring via the carbonyl carbon (\( \text{C}=\text{O} \)). The carbonyl carbon is electron-deficient and acts as an electron-withdrawing group via both inductive and mesomeric effects. This deactivates the aromatic ring, making it less reactive toward electrophiles, so the nitration is slower than that of benzene. Furthermore, electron-withdrawing groups (except halogens) are meta-directing (3-directing), which means the incoming electrophile (\( \text{NO}_2^+ \)) is directed predominantly to the 3-position.

[1] B

The basic strength of nitrogen compounds depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton:
1. Diethylamine is a secondary aliphatic amine. It has two electron-donating ethyl groups (+I effect) which increase the electron density on the nitrogen, making it the most basic of the group.
2. Ethylamine is a primary aliphatic amine. It has one electron-donating ethyl group, making it more basic than ammonia but less basic than diethylamine.
3. Ammonia has no alkyl groups to increase electron density on the nitrogen.
4. Phenylamine is an aromatic amine. The lone pair of electrons on the nitrogen atom is delocalized into the \( \pi \)-system of the benzene ring, making it much less available to accept a proton, thus making it the weakest base.

Therefore, the correct order is: Diethylamine > Ethylamine > Ammonia > Phenylamine.

[1] A

In a highly alkaline solution (high concentration of \( \text{OH}^- \) ions), both the amine and carboxylic acid groups are in their deprotonated forms.
- The carboxylic acid group (\( -\text{COOH} \)) loses a proton to form a carboxylate anion (\( -\text{COO}^- \)).
- The amine group (\( -\text{NH}_3^+ \) or \( -\text{NH}_2 \)) remains in its unprotonated form (\( -\text{NH}_2 \)).
Therefore, the major species is \( \text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^- \).

[1] C

In the laboratory, the reduction of nitrobenzene to phenylamine is standardly achieved by heating nitrobenzene under reflux with tin (\( \text{Sn} \)) and concentrated hydrochloric acid. This initially produces the phenylammonium ion (\( \text{C}_6\text{H}_5\text{NH}_3^+ \)). Aqueous sodium hydroxide (\( \text{NaOH} \)) is then added to liberate the free amine, phenylamine (\( \text{C}_6\text{H}_5\text{NH}_2 \)).

[1] B

The uncatalyzed reaction between peroxodisulfate (\( \text{S}_2\text{O}_8^{2-} \)) and iodide (\( \text{I}^- \)) ions is slow because both reactants are anions. The electrostatic repulsion between ions of the same charge leads to a high activation energy.

With \( \text{Fe}^{2+} \) acting as a homogeneous catalyst, the mechanism proceeds in two steps:
1. \( 2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq}) \)
2. \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \)

Both steps involve collision between oppositely charged ions (positive iron ions colliding with negative reactant ions), which has a much lower activation energy. Option B is incorrect because \( \text{Fe}^{2+} \) reduces (does not oxidize) \( \text{S}_2\text{O}_8^{2-} \).

[1] C

Under standard conditions, a reaction is thermodynamically feasible if the standard cell potential, \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\), is positive.

Zinc acts as the reducing agent (so the half-cell reaction is \(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-\) with \(E^\ominus_{\text{oxidation}} = -0.76\text{ V}\)):

1. Reduction of \(\text{VO}_2^+\) to \(\text{VO}^{2+}\):
\(E^\ominus_{\text{cell}} = +1.00 - (-0.76) = +1.76\text{ V}\) (feasible)

2. Reduction of \(\text{VO}^{2+}\) to \(\text{V}^{3+}\):
\(E^\ominus_{\text{cell}} = +0.34 - (-0.76) = +1.10\text{ V}\) (feasible)

3. Reduction of \(\text{V}^{3+}\) to \(\text{V}^{2+}\):
\(E^\ominus_{\text{cell}} = -0.26 - (-0.76) = +0.50\text{ V}\) (feasible)

4. Reduction of \(\text{V}^{2+}\) to \(\text{V}\) metal:
\(E^\ominus_{\text{cell}} = -1.13 - (-0.76) = -0.37\text{ V}\) (not feasible)

Therefore, the reduction stops at vanadium(II) ions, \(\text{V}^{2+}\), which form a violet solution.

[1 mark] C - Correct identification of the final vanadium species as V2+ and its violet colour based on standard electrode potentials.

The methyl group in methylbenzene is electron-donating due to the positive inductive effect (+I). This increases the electron density of the delocalized \(\pi\)-system of the benzene ring, making it more nucleophilic and therefore more susceptible to attack by electrophiles such as the nitronium ion (\(\text{NO}_2^+\)). Additionally, the electron-donating effect stabilizes the positively charged carbocation intermediate formed during the rate-determining step, lowering the activation energy.

[1 mark] B - Correct explanation of the inductive effect of the methyl group increasing electron density on the benzene ring.

Transition metal complexes are typically coloured due to d-d electronic transitions. When ligand-field splitting occurs in an octahedral environment, electrons in the lower-energy d-orbitals can absorb specific wavelengths of visible light to be promoted to higher-energy d-orbitals.

- \(\text{Sc}^{3+}\) has the electron configuration \([\text{Ar}] 3d^0\). Since it has no d-electrons, d-d transitions are impossible, making the complex ion \([\text{Sc}(\text{H}_2\text{O})_6]^{3+}\) colourless.
- \(\text{Cu}^{2+}\) (\(3d^9\)), \(\text{Ti}^{3+}\) (\(3d^1\)), and \(\text{Fe}^{3+}\) (\(3d^5\)) all have partially filled d-orbitals and thus absorb visible light to form coloured solutions.

[1 mark] B - Correctly identifies Sc3+ as forming a colourless complex due to its 3d0 electronic configuration.

The basic strength of a nitrogen compound depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton:

1. In phenylamine (III), the lone pair of electrons on the nitrogen atom overlaps with, and is partially delocalized into, the \(\pi\)-system of the benzene ring. This significantly decreases the electron density on the nitrogen, making it the weakest base.
2. In ethylamine (II), the ethyl group is electron-donating (+I inductive effect), which increases the electron density on the nitrogen atom compared to ammonia (I). This makes the lone pair more available to accept a proton.
3. Ammonia (I) is intermediate in basic strength between phenylamine and ethylamine.

Thus, the order of increasing basicity is III < I < II.

[1 mark] B - Correct sequence of basicity (III < I < II) based on delocalization and inductive effects.

Calculate standard cell potential: \(E^{\theta}_{\text{cell}} = E^{\theta}_{\text{reduction}} - E^{\theta}_{\text{oxidation}} = +0.77\text{ V} - (-0.41\text{ V}) = +1.18\text{ V}\). Write the overall spontaneous equation by reversing the more negative half-equation and combining them: \(\text{Cr}^{2+}(aq) + \text{Fe}^{3+}(aq) \rightarrow \text{Cr}^{3+}(aq) + \text{Fe}^{2+}(aq)\). The reducing agent is the species that is oxidized, which is \(\text{Cr}^{2+}(aq)\).

M1: \(E^{\theta}_{\text{cell}} = +1.18\text{ V}\) (1 mark) (accept without sign but deduct 0.5 if unit is missing or incorrect).
M2: Spontaneous overall equation: \(\text{Cr}^{2+}(aq) + \text{Fe}^{3+}(aq) \rightarrow \text{Cr}^{3+}(aq) + \text{Fe}^{2+}(aq)\) (1 mark) (allow equilibrium arrow, ignore state symbols).
M3: Reducing agent identified as \(\text{Cr}^{2+}\) or \(\text{Cr}^{2+}(aq)\) (0.5 marks) (reject chromium / \(\text{Cr}\)).

In transition metal complexes, the presence of ligands causes the d-orbitals to split into two groups of different energy levels. \(\text{Cu}^{2+}\) has an electronic configuration of \([\text{Ar}]3\text{d}^9\), meaning it has an incomplete d-subshell. An electron can be promoted from a lower energy d-orbital to a higher energy d-orbital by absorbing energy from the visible spectrum (d-d transition). The remaining unabsorbed frequencies of light are transmitted, giving the blue color. \(\text{Zn}^{2+}\) has a configuration of \([\text{Ar}]3\text{d}^{10}\). Because its d-subshell is completely full, no electrons can be promoted to a higher d-orbital, so no visible light is absorbed and the solution is colorless.

M1: State electronic configuration/state of d-orbitals: \(\text{Cu}^{2+}\) has an incomplete/partially filled d-subshell (or \(3\text{d}^9\)) AND \(\text{Zn}^{2+}\) has a full d-subshell (or \(3\text{d}^{10}\)) (1 mark).
M2: Explain that d-orbitals split in energy (due to ligands) (0.5 marks).
M3: Explain that in \(\text{Cu}^{2+}\) electrons can be excited / undergo d-d transitions by absorbing energy/light from the visible spectrum (causing color), but this is not possible in \(\text{Zn}^{2+}\) because there is no vacant d-orbital (1 mark).

The methyl group is an electron-donating group due to the inductive effect (or hyperconjugation). This pushes electron density into the delocalized \(\pi\)-system of the benzene ring. The increased electron density makes the ring a stronger nucleophile, polarising and attracting the electrophile (such as \(\text{NO}_2^+\)) more strongly and accelerating the rate of substitution.

M1: Identify that the methyl group is electron-donating / has a positive inductive effect (1 mark).
M2: Explain that this increases the electron density of the delocalized \(\pi\) ring system (0.5 marks).
M3: State that this makes the ring more susceptible to attack by electrophiles / attracts the electrophile more readily (1 mark).

The basic strength depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton. Phenylamine is the weakest base because the lone pair on the nitrogen atom is delocalized into the benzene ring's \(\pi\)-electron cloud, reducing its availability. Ammonia is stronger than phenylamine because its lone pair is localized. Ethylamine is the strongest because the ethyl group is electron-donating, which increases the electron density on the nitrogen, making the lone pair even more available.

M1: Correct order: phenylamine < ammonia < ethylamine (1 mark) (accept formulas).
M2: For phenylamine, state that the lone pair on the nitrogen atom is delocalized into the benzene ring (1 mark).
M3: Explain that this delocalization makes the lone pair less available to accept a proton / coordinate to an \(\text{H}^+\) ion (0.5 marks).

At the anode (oxidation): hydrogen gas reacts with hydroxide ions to form water, releasing electrons: \(\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2e^-\). At the cathode (reduction): oxygen gas reacts with water and accepts electrons to form hydroxide ions: \(\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-\). An advantage of fuel cells is that they are more efficient at converting chemical energy to electrical energy than internal combustion engines, and their only product at the point of use is water, meaning no greenhouse gases or harmful pollutants (like \(\text{NO}_x\)) are produced.

M1: Anode half-equation: \(\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2e^-\) (1 mark) (or multiples; allow equilibrium sign).
M2: Cathode half-equation: \(\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-\) (1 mark) (or multiples; allow equilibrium sign).
M3: Any valid advantage: e.g., higher thermodynamic efficiency, quieter operation, or no greenhouse gas/CO2/pollutant emissions at point of use (0.5 marks).

During this ligand substitution reaction, four of the water ligands in the hexaaquacopper(II) ion are replaced by four ammonia ligands to form the tetraamminedihydroxocopper(II) ion: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}\). The transition metal ion has a total of 6 coordinate bonds (4 from ammonia, 2 from water), so the coordination number is 6, and the shape is octahedral (or distorted octahedral due to Jahn-Teller distortion).

M1: Correct balanced equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}\) (1 mark).
M2: Coordination number = 6 (0.5 marks).
M3: Shape = Octahedral / distorted octahedral (1 mark) (reject square planar, as the complex still contains two water ligands).

Alkaline potassium manganate(VII) is a strong oxidizing agent. During reflux, the purple MnO4^- ion is reduced to insoluble manganese(IV) oxide, which forms a brown precipitate. The methyl group is oxidized to a carboxylate group, forming soluble benzoate ions (\(\text{C}_6\text{H}_5\text{COO}^-\)). In the second step, dilute hydrochloric acid is added to protonate the benzoate ions to form benzoic acid (\(\text{C}_6\text{H}_5\text{COOH}\)), which is poorly soluble in cold water and precipitates out.

M1: Color change: Purple (solution) to brown precipitate/solid (1 mark) (accept purple to brown/black, reject purple to colorless).
M2: Role: Oxidizing agent / oxidant (0.5 marks).
M3: Reason for adding acid: To protonate the benzoate ion / convert benzoate to benzoic acid / release benzoic acid from its salt (1 mark) (accept: \(\text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH}\)).

The reaction between a dicarboxylic acid and a diamine is a condensation polymerization. For each amide link formed, a molecule of water (\(\text{H}_2\text{O}\)) is eliminated. The repeating unit consists of the residue of benzene-1,4-dicarboxylic acid linked to the residue of benzene-1,4-diamine via an amide bond: \([-\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-]\).

M1: Type: Condensation (polymerization) (0.5 marks).
M2: Small molecule: Water / \(\text{H}_2\text{O}\) (1 mark).
M3: Drawing or clear structural representation of the repeating unit with open bonds at the ends: e.g., \(-\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-\) (1 mark) (allow structures showing full benzene rings; must show the amide group \(-\text{CONH}-\) correctly linked).

The standard cell potential is calculated using: \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\theta(\text{I}_2/\text{I}^-) = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\). Since \(E^\theta_{\text{cell}}\) is positive, the reaction is thermodynamically feasible under standard conditions because a positive cell potential corresponds to a negative free energy change (\(\Delta G^\theta < 0\)) and a positive total entropy change (\(\Delta S_{\text{total}} > 0\)).

1 mark: Correct calculation of \(E^\theta_{\text{cell}} = +0.23\text{ V}\) (both sign and units required). 1 mark: Stating that the reaction is thermodynamically feasible. 0.5 mark: Explaining that feasibility is due to the positive value of \(E^\theta_{\text{cell}}\) (or linking to \(\Delta G^\theta < 0\) or \(\Delta S_{\text{total}} > 0\)).

The methyl group in methylbenzene is an electron-donating group due to its positive inductive effect. This increases the electron density of the delocalised pi-system in the benzene ring. As a result, the ring in methylbenzene becomes more nucleophilic and is more readily attacked by the electrophile (such as \(\text{NO}_2^+\)) compared to benzene.

1 mark: Stating that the methyl group has a positive inductive effect / is electron-donating. 1 mark: Explaining that this increases the electron density of the benzene ring. 0.5 mark: Concluding that this makes the ring more susceptible to attack by the electrophile.

When concentrated hydrochloric acid is added, chloride ligands replace the water ligands to form tetrachlorocuprate(II) ions: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\). The product is \([\text{CuCl}_4]^{2-}\) which has a tetrahedral geometry. The coordination number changes from 6 to 4 because chloride ions are significantly larger than water molecules, resulting in steric hindrance and mutual repulsion that prevents six chloride ions from fitting around the central \(\text{Cu}^{2+}\) ion.

1 mark: Correct formula of the complex \([\text{CuCl}_4]^{2-}\) and its tetrahedral shape. 1 mark: Explaining that chloride ligands are larger than water molecules. 0.5 mark: Mentioning steric hindrance or repulsion as the reason why only four chloride ligands can coordinate to the copper ion.

The order of increasing basic strength is: phenylamine < ammonia < ethylamine. In ethylamine, the alkyl (ethyl) group is electron-donating due to the positive inductive effect, which increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton. In phenylamine, the lone pair of electrons on the nitrogen atom is partially delocalised into the benzene ring's pi-electron system. This decreases the electron density on the nitrogen atom, making the lone pair less available to accept a proton.

1 mark: Correct order of basic strength (phenylamine < ammonia < ethylamine). 1 mark: For explaining that ethylamine is a stronger base because the ethyl group is electron-donating (positive inductive effect), increasing the availability of the nitrogen lone pair. 0.5 mark: For explaining that phenylamine is a weaker base because the nitrogen lone pair is delocalised into the benzene ring, decreasing its availability to accept a proton.

1. Calculate the moles of sodium thiosulfate used:
\(n(\text{Na}_2\text{S}_2\text{O}_3) = \frac{22.40}{1000} \times 0.150 = 3.36 \times 10^{-3}\text{ mol}\)

2. Deduce the stoichiometry of the reactions:
\(2\text{Cu}^{2+}(aq) + 4\text{I}^-(aq) \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)\)
\(\text{I}_2(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 2\text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq)\)
From these equations, we find that \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\), which simplifies to a 1:1 molar ratio: \(n(\text{Cu}^{2+}) = n(\text{S}_2\text{O}_3^{2-})\).

3. Determine the moles of copper in the sample:
\(n(\text{Cu}) = 3.36 \times 10^{-3}\text{ mol}\)

4. Calculate the mass of copper using its molar mass (\(A_r(\text{Cu}) = 63.5\)):
\(m(\text{Cu}) = 3.36 \times 10^{-3}\text{ mol} \times 63.5\text{ g mol}^{-1} = 0.21336\text{ g}\)

5. Calculate the percentage of copper in the brass alloy:
\(\%\text{ Cu} = \frac{0.21336}{0.320} \times 100\% = 66.675\%\)

Given the starting values, the final answer should be rounded to three significant figures, giving \(66.7\%\).

- **M1**: Correct calculation of moles of thiosulfate ions: \(3.36 \times 10^{-3}\text{ mol}\) (1.00 mark)
- **M2**: Correctly identifies the 1:1 molar ratio between \(\text{Cu}^{2+}\) and \(\text{S}_2\text{O}_3^{2-}\) (1.00 mark)
- **M3**: States that \(n(\text{Cu}^{2+}) = 3.36 \times 10^{-3}\text{ mol}\) (1.00 mark)
- **M4**: Calculates the mass of copper as \(0.21336\text{ g}\) (or \(0.2135\text{ g}\) if using 63.55) (1.00 mark)
- **M5**: Calculates the percentage to an appropriate number of significant figures (3 s.f.): \(66.7\%\) (1.25 marks). (Accept \(66.7\%\) or \(66.8\%\) depending on exact \(A_r\) value chosen).

1. **Generation of the electrophile:**
Concentrated sulfuric acid acts as a stronger acid than nitric acid, protonating it to eventually yield the nitronium ion (\(\text{NO}_2^+\)):
\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\)
(Accept: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_2\text{O} + \text{HSO}_4^-\))

2. **Mechanism:**
- A curly arrow must start from the delocalised \(\pi\) system of the benzene ring and point to the nitrogen atom of the \(\text{NO}_2^+\).
- An intermediate carbocation (cyclohexadienyl cation) is formed. It must contain the \(sp^3\) hybridized carbon bonded to both \(\text{-H}\) and \(\text{-NO}_2\), with a positive charge located inside a partial ring (horseshoe shape spanning the other 5 carbon atoms, open towards the \(sp^3\) carbon).
- A curly arrow starts from the C-H bond of the \(sp^3\) carbon and points back into the ring, restoring the stable delocalised \(\pi\) aromatic system.
- The products are nitrobenzene and an \(\text{H}^+\). The hydrogen ion is accepted by \(\text{HSO}_4^-\), reforming the catalyst \(\text{H}_2\text{SO}_4\).

- **M1**: Correct equation for the generation of the \(\text{NO}_2^+\)$ electrophile (1.00 mark)
- **M2**: States that sulfuric acid acts as an acid/catalyst, or protonates the nitric acid (1.00 mark)
- **M3**: Curly arrow drawn correctly from the benzene ring to the electrophile \(\text{NO}_2^+\) (1.00 mark)
- **M4**: Correct representation of the cationic horseshoe intermediate with a positive charge inside (1.25 marks)
- **M5**: Curly arrow from the C-H bond of the intermediate pointing into the ring to restore aromaticity, yielding nitrobenzene and \(\text{H}^+\) (1.00 mark)

1. **Initial state and d-orbital splitting:**
In aqueous copper(II) solution, copper exists as the hexaaquacopper(II) complex, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\). The presence of water ligands causes the 3d orbitals of copper to split into two different energy levels separated by an energy gap, \(\Delta E\).

2. **Color of aqueous copper(II):**
When visible light passes through the solution, d-electrons absorb a specific frequency of light energy equal to \(\Delta E\) to be promoted from the lower energy level to the higher energy level (d-d transition). The light blue color observed is the complementary color of the absorbed red/orange light.

3. **Ligand Exchange:**
Adding excess concentrated hydrochloric acid introduces a high concentration of chloride ligands, which substitute the water ligands:
\([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\)

4. **Geometry and Splitting Change:**
The complex geometry changes from octahedral to tetrahedral, and the ligand field strength of chloride is weaker than water. This significantly alters the d-orbital splitting energy, \(\Delta E\).

5. **Color change:**
As a result of the different \(\Delta E\), the complex absorbs a different frequency/wavelength of light, and a different complementary color (yellow-green) is transmitted and seen.

- **M1**: Identifies hexaaquacopper(II) as octahedral and explains that d-orbitals split into two energy levels (1.00 mark)
- **M2**: Explains color absorption due to d-d electronic transition where energy \(\Delta E = h\nu\) is absorbed, transmitting the complementary light blue color (1.00 mark)
- **M3**: Provides a balanced equation or clear description of ligand exchange forming the tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\) (1.25 marks)
- **M4**: Identifies that the geometry changes from octahedral to tetrahedral (1.00 mark)
- **M5**: Explains that the change in geometry and ligand alters \(\Delta E\), meaning different frequencies of light are absorbed, causing the observed color change to yellow-green (1.00 mark)

1. **Order of basicity:**
Ethylamine > Ammonia > Phenylamine (most basic to least basic).

2. **Definition of basicity:**
A base acts as a proton acceptor (Brønsted-Lowry base). The strength of a nitrogenous base depends on the availability of the lone pair of electrons on the nitrogen atom to form a dative covalent bond with a proton (\(\text{H}^+\)).

3. **Explanation for Ethylamine:**
The ethyl group is an electron-releasing group (positive inductive effect). This increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton, making ethylamine more basic than ammonia.

4. **Explanation for Phenylamine:**
The lone pair of electrons on the nitrogen atom overlaps with, and becomes delocalised into, the \(\pi\)-system of the benzene ring. This significantly reduces the electron density on the nitrogen, making the lone pair far less available to accept a proton, making phenylamine less basic than ammonia.

5. **Ammonia:**
Ammonia has no alkyl or aryl groups attached to the nitrogen, so it does not experience inductive electron release or ring delocalisation, leaving its basicity intermediate between the two organic amines.

- **M1**: States the correct order of basicity: ethylamine > ammonia > phenylamine (1.00 mark)
- **M2**: States that basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (1.00 mark)
- **M3**: Explains that the ethyl group in ethylamine is electron-donating / has a positive inductive effect, which increases electron density on the nitrogen (1.25 marks)
- **M4**: Explains that in phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring (1.25 marks)
- **M5**: States that this delocalisation makes the lone pair less available to accept a proton compared to ammonia, which has no such groups (0.75 marks)

Mono-nitration of methylbenzene is achieved by reacting it with a mixture of concentrated nitric acid and concentrated sulfuric acid. The concentrated sulfuric acid acts as a catalyst by protonating nitric acid to generate the electrophilic nitronium ion \(\text{NO}_2^+\). The temperature must be maintained between \(50\text{ }^\circ\text{C}\) and \(55\text{ }^\circ\text{C}\) (or below \(60\text{ }^\circ\text{C}\)) to ensure mono-substitution and prevent further nitration to di- or tri-nitromethylbenzene.

1.0 mark for concentrated nitric acid \(\text{HNO}_3\). 0.5 marks for concentrated sulfuric acid \(\text{H}_2\text{SO}_4\) catalyst. 1.0 mark for specifying a temperature of \(50\) to \(55\text{ }^\circ\text{C}\) (or any temperature range below \(60\text{ }^\circ\text{C}\), but above room temperature). Accept conc. nitric acid and conc. sulfuric acid. Reject dilute acids.

The reduction of a nitro group to an amine group requires 6 moles of [H] per mole of nitro compound, forming 2 moles of water: \(\text{O}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + 6[\text{H}] \rightarrow \text{H}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + 2\text{H}_2\text{O}\). Since the reaction is performed in strongly acidic conditions (HCl), the amine group is protonated to form an ammonium salt (\(-\text{NH}_3^+\text{Cl}^-\)). Sodium hydroxide is added to neutralize the remaining acid and deprotonate the salt to liberate the free amine (benzocaine) as a precipitate.

1.0 mark for the balanced equation with 6[H] and 2H2O (accept structural, skeletal, or molecular formulas for reactants and products). 1.0 mark for explaining that sodium hydroxide neutralizes the acid and deprotonates the ammonium salt / phenylammonium ion. 0.5 marks for stating that this action liberates the free amine / benzocaine.

The reduction half-equation for dichromate(VI) in acidic conditions is: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\). The starting solution containing dichromate(VI) ions is orange, and upon reduction to chromium(III) ions, the solution turns green.

1.0 mark for the correct species in the half-equation (Cr2O7^2-, H^+, e^-, Cr^3+, H2O). 0.5 marks for correct balancing of the half-equation (Cr2O7^2- + 14H^+ + 6e^- -> 2Cr^3+ + 7H2O). 1.0 mark for the colour change from orange to green.

To form the diazonium salt, benzocaine is reacted with nitrous acid, which is generated in situ from sodium nitrate(III) (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)). The temperature must be maintained between \(0\text{ }^\circ\text{C}\) and \(10\text{ }^\circ\text{C}\) (typically below \(5\text{ }^\circ\text{C}\)) to prevent the decomposition of the unstable diazonium ion. Coupling this diazonium salt with an alkaline solution of naphthalen-2-ol produces an azo dye, which appears as an orange-red precipitate.

1.0 mark for reagents: sodium nitrate(III) / sodium nitrite (\(\text{NaNO}_2\)) and hydrochloric acid (\(\text{HCl}\)) (or nitrous acid / \(\text{HNO}_2\)). 0.5 marks for specifying a temperature range between \(0\text{ }^\circ\text{C}\) and \(10\text{ }^\circ\text{C}\) (or below \(10\text{ }^\circ\text{C}\), but above freezing). 1.0 mark for observing an orange/red precipitate/dye.

### Part (a)
* **Reagents & Conditions:** Heat under reflux with alkaline potassium manganate(VII) / \(\text{KMnO}_4\), followed by acidification with dilute hydrochloric acid / \(\text{HCl}\) (or dilute sulfuric acid / \(\text{H}_2\text{SO}_4\)).
* **Colour change:** Purple solution to brown precipitate (before acidification), which dissolves on acidification to form a colourless solution.

### Part (b)
* **Equation:** \(\text{O}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + 6[H] \rightarrow \text{H}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + 2\text{H}_2\text{O}\)
* **Reagent:** Aqueous sodium hydroxide / \(\text{NaOH(aq)}\) (or sodium carbonate / \(\text{Na}_2\text{CO}_3\)).

### Part (c)
* **Step 1:** Calculate the number of moles of 4-nitrotoluene used:
\(n(\text{4-nitrotoluene}) = \frac{6.855\text{ g}}{137.1\text{ g mol}^{-1}} = 0.0500\text{ mol}\)
* **Step 2:** The theoretical yield of benzocaine is in a 1:1 ratio, which is \(0.0500\text{ mol}\).
* **Step 3:** Calculate the actual moles of benzocaine produced using the \(55.0\%\) yield:
\(n(\text{actual}) = 0.0500\text{ mol} \times 0.550 = 0.0275\text{ mol}\)
* **Step 4:** Calculate the mass of benzocaine obtained:
\(m(\text{benzocaine}) = 0.0275\text{ mol} \times 165.2\text{ g mol}^{-1} = 4.543\text{ g}\)
* Rounded to 3 significant figures: **\(4.54\text{ g}\)**

**Part (a) [3 Marks]**
* **M1:** Alkaline potassium manganate(VII) / \(\text{KMnO}_4\) and heat under reflux. (1)
* **M2:** Acidification with dilute mineral acid (e.g., \(\text{HCl}\) or \(\text{H}_2\text{SO}_4\)). (1)
* **M3:** Colour change: Purple to brown precipitate / black solid. (1)
* *Accept:* Acidified potassium dichromate(VI) with heat under reflux for 1 mark if alkaline manganate is not given, with colour change orange to green.

**Part (b) [2 Marks]**
* **M4:** Balanced equation: \(\text{O}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + 6[H] \rightarrow \text{H}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + 2\text{H}_2\text{O}\) (1)
* *Accept molecular formula:* \(\text{C}_9\text{H}_9\text{NO}_4 + 6[H] \rightarrow \text{C}_9\text{H}_{11}\text{NO}_2 + 2\text{H}_2\text{O}\)
* **M5:** Sodium hydroxide / \(\text{NaOH}\) (or sodium carbonate / \(\text{Na}_2\text{CO}_3\) / ammonia / \(\text{NH}_3\)). (1)

**Part (c) [4 Marks]**
* **M6:** Calculates moles of 4-nitrotoluene: \(n = \frac{6.855}{137.1} = 0.0500\text{ mol}\) (1)
* **M7:** Identifies 1:1 ratio of reactant to product, yielding theoretical moles of benzocaine as \(0.0500\text{ mol}\) (or calculates theoretical mass: \(0.0500 \times 165.2 = 8.26\text{ g}\)). (1)
* **M8:** Applies \(55.0\%\) yield: \(0.0500 \times 0.550 = 0.0275\text{ mol}\) (or \(8.26 \times 0.550 = 4.543\text{ g}\)). (1)
* **M9:** Final mass calculation to 3 significant figures: \(4.54\text{ g}\) (1)
* *Note:* Allow Error Carried Forward (ECF) from incorrect mole values, provided final answer is rounded to 3 SF.

When sodium hydroxide is added dropwise to aqueous chromium(III) ions, a green or grey-green precipitate of chromium(III) hydroxide, \( [Cr(H_2O)_3(OH)_3] \), is formed. Upon adding excess sodium hydroxide, this precipitate acts amphoterically and dissolves to form a dark green solution containing the hexahydroxochromate(III) complex ion, \( [Cr(OH)_6]^{3-} \).

Mark 1: For stating that a green / grey-green precipitate forms (accept 'green solid'). Mark 2: For stating that the precipitate dissolves / clears in excess sodium hydroxide to give a green / dark green solution.

The chemical purpose of adding saturated sodium hydrogencarbonate solution is to neutralise any remaining acidic reagents or impurities (such as hydrochloric acid or carboxylic acids) present in the organic layer. This reaction produces carbon dioxide gas, which causes a build-up of pressure. To manage this safely, the student must invert the separating funnel periodically and open the tap (away from any other people) to release the gas.

Mark 1: For stating that sodium hydrogencarbonate neutralises / reacts with acidic impurities. Mark 2: For explaining that pressure is released by inverting the funnel and opening the tap periodically.

The absorption at \( 1715\text{ cm}^{-1} \) indicates a carbonyl group, \( C=O \). With the formula \( C_3H_6O \), the compound must be either propanone or propanal. Propanone has six equivalent protons, so it produces only a single singlet peak in its proton NMR spectrum. In contrast, propanal has three distinct proton environments (\( -CH_3 \), \( -CH_2- \), and \( -CHO \)) which would result in three separate signals showing spin-spin splitting (a triplet, a multiplet/quartet, and a triplet).

Mark 1: For identifying the compound as propanone. Mark 2: For explaining that propanone has only one proton environment / singlet, whereas propanal has three proton environments / would show three signals / splitting.

Sodium thiosulfate reacts rapidly with any iodine (\( I_2 \)) as soon as it is formed, reducing it back to iodide ions (\( I^- \)). Once all the sodium thiosulfate is completely consumed, any further iodine produced remains in solution and reacts with the starch indicator to produce a sudden blue-black colour. Because the amount of sodium thiosulfate is small and constant, the time taken for the colour change corresponds to the time required to produce a fixed, known amount of iodine, which allows the initial rate of reaction to be determined.

Mark 1: For explaining that thiosulfate reacts with / removes the iodine as it forms, delaying the starch-iodine color change. Mark 2: For stating that once thiosulfate is exhausted, the blue-black color appears, allowing the time for a fixed amount of iodine to be produced to be measured (to determine initial rate).

In a standard laboratory calculation, only the mass of the solution and its specific heat capacity are used to calculate the heat energy transferred (using \( q = mc\Delta T \)). If the heat capacities of the polystyrene cup, thermometer, and stirrer are ignored, the calculated heat energy absorbed by the system is lower than the actual total heat energy released because some of the heat was absorbed by the apparatus, causing a smaller temperature rise in the water. This results in a less exothermic (less negative) experimental enthalpy value.

Mark 1: For identifying that the heat capacity of the calorimeter apparatus / polystyrene cup / thermometer / stirrer was ignored. Mark 2: For explaining that some heat was absorbed by the apparatus, leading to a smaller recorded temperature rise / underestimate of the total heat released.

To calibrate a pH meter, the student must first rinse the pH electrode thoroughly with distilled or deionised water. The electrode is then immersed in a calibration buffer solution of known pH (often pH 7.00), and the meter is adjusted to read this exact value. The electrode is then rinsed again and immersed in a second buffer solution of a different known pH (typically pH 4.01 for acidic titrations or pH 10.01 for basic titrations) to calibrate the slope of the meter.

Mark 1: For rinsing the electrode with distilled/deionised water and placing it into a buffer solution of known pH to calibrate/adjust the meter. Mark 2: For repeating the process with a second buffer solution of a different known pH (to calibrate the slope).

Adding excess concentrated hydrochloric acid replaces the water ligands in the pink octahedral hexaaquacobalt(II) ion, \( [Co(H_2O)_6]^{2+} \), with chloride ligands. Because chloride ions are larger and carry a negative charge, only four chloride ligands can fit around the cobalt ion, forming the blue tetrachlorocobaltate(II) ion, \( [CoCl_4]^{2-} \). This complex ion has a tetrahedral shape.

Mark 1: For the correct formula of the complex ion: \( [CoCl_4]^{2-} \) (accept cobalt tetrachlorate ion if charge is clear). Mark 2: For stating the shape is tetrahedral.

The student should measure the melting temperature range of the dry benzoic acid crystals. If the sample is pure, it will melt sharply (over a narrow range of 1–2 °C) at the literature melting temperature of benzoic acid (approximately 122 °C), confirming both its identity and high purity. If impurities are present, the melting temperature will be lower than the literature value, and the melting process will occur over a wider, broader temperature range.

Mark 1: For comparing the measured melting temperature range to the literature/database value (to confirm identity). Mark 2: For stating that a sharp melting temperature range indicates purity, whereas impurities lower the melting point and broaden the melting range.

1. Identify the metal ion:
The green solution and green precipitate with NaOH and NH3, which are both insoluble in excess, indicate the presence of iron(II) ions, \( \text{Fe}^{2+} \).
(Note: Chromium(III) forms green solutions but contains \( \text{Cr}^{3+} \), and its precipitate is soluble in excess NaOH. Nickel(II) is green, but its hydroxide precipitate dissolves in excess NH3 to form a blue solution).

2. Write the ionic equation:
\( \text{Fe}^{2+}(\text{aq}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s}) \)

M1: For identifying the metal ion as \( \text{Fe}^{2+} \) or iron(II) (1 mark)
M2: For writing the correct ionic equation with state symbols: \( \text{Fe}^{2+}(\text{aq}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s}) \) (1 mark)
Allow: \( [\text{Fe}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{Fe}(\text{H}_2\text{O})_4(\text{OH})_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \)
Reject: Fe without charge, or incorrect/missing state symbols.

1. Explain pressure build-up:
The reaction between sodium hydrogencarbonate (\( \text{NaHCO}_3 \)) and acid impurities produces carbon dioxide (\( \text{CO}_2 \)) gas. In a closed separating funnel, this gas causes a build-up of pressure.

2. State how to release:
To release the pressure safely, invert the separating funnel (with the stopper secured) and open the tap to vent the gas.

M1: Reaction of hydrogencarbonate with acid produces carbon dioxide gas / \( \text{CO}_2 \) (which builds up pressure) (1 mark).
M2: Invert the separating funnel and open the tap / stopcock (to release the gas) (1 mark).
Reject: Simply opening the stopper.

To ensure all the water of crystallisation has been removed, the student should use the technique of heating to constant mass. This involves:
1. Heating the crucible and contents, allowing it to cool, and weighing it.
2. Repeating the process of heating, cooling, and weighing.
3. Checking that consecutive mass measurements are identical (or within a very small tolerance, e.g., \( \pm 0.01\text{ g} \)).

M1: Heat (and cool) and weigh the crucible and contents, and repeat this process (1 mark).
M2: Until two consecutive mass readings are constant / identical (1 mark).
Alternative: Award 2 marks for 'heat to constant mass'.

1. Identify the bond:
Propan-1-ol is an alcohol and contains an O–H group, whereas propanal is an aldehyde and contains a C=O group but no O–H group. Thus, the O–H (alcohol) absorption band will be present in propan-1-ol but absent from propanal.

2. Wavenumber range:
According to the Edexcel Data Booklet, the characteristic absorption range for an O–H (alcohol) bond is \( 3200\text{--}3600\text{ cm}^{-1} \) (broad peak).

M1: O–H (alcohol / hydroxy) bond (1 mark).
M2: Wavenumber range of \( 3200\text{--}3600\text{ cm}^{-1} \) (1 mark).
Allow: Ranges within \( 3200\text{--}3750\text{ cm}^{-1} \).
Reject: O–H (carboxylic acid) or any range that does not overlap with the correct range.

1. Moles of \(\text{MnO}_4^-\): \(n = 0.0150 \times \frac{22.80}{1000} = 3.42 \times 10^{-4}\text{ mol}\). 2. Moles of \(\text{Fe}^{2+}\) in 25.0 \(\text{cm}^3\) sample: The ratio of \(\text{MnO}_4^-\text{ : Fe}^{2+}\) is 1:5. Therefore, \(n(\text{Fe}^{2+}) = 5 \times 3.42 \times 10^{-4} = 1.71 \times 10^{-3}\text{ mol}\). 3. Scale up to 250.0 \(\text{cm}^3\): \(n(\text{Fe}^{2+})_{\text{total}} = 1.71 \times 10^{-3} \times 10 = 1.71 \times 10^{-2}\text{ mol}\). 4. Mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\): \(\text{Mass} = 1.71 \times 10^{-2} \times 278.0 = 4.7538\text{ g}\). 5. Percentage purity: \(\text{Purity} = \frac{4.7538}{5.80} \times 100 = 81.962\% \approx 82.0\%\).

[1 mark] Correct calculation of moles of manganate(VII). [1 mark] Multiplies moles of manganate(VII) by 5 to find moles of iron(II) in the aliquot. [1 mark] Multiplies by 10 to scale up to the total volumetric flask volume. [1 mark] Calculates mass of pure hydrated iron(II) sulfate by multiplying by 278.0. [1 mark] Calculates the percentage purity to 3 significant figures.

1. Calculate \(\ln(k_2/k_1)\): \(\ln(5.85 \times 10^{-3} / 1.45 \times 10^{-3}) = \ln(4.03448) = 1.3949\). 2. Calculate the temperature reciprocal term: \(\frac{1}{313} - \frac{1}{293} = 0.0031949 - 0.0034130 = -2.181 \times 10^{-4}\text{ K}^{-1}\). 3. Rearrange the equation for \(E_{\text{a}}\): \(E_{\text{a}} = -\frac{R \cdot \ln(k_2/k_1)}{\frac{1}{T_2} - \frac{1}{T_1}}\). 4. Calculate \(E_{\text{a}}\) in \(\text{J mol}^{-1}\): \(E_{\text{a}} = -\frac{8.31 \times 1.3949}{-2.181 \times 10^{-4}} = 53151.7\text{ J mol}^{-1}\). 5. Convert to \(\text{kJ mol}^{-1}\) and round to 3 sig figs: \(53.2\text{ kJ mol}^{-1}\).

[1 mark] Correct calculation of ln(k2/k1). [1 mark] Correct calculation of (1/T2 - 1/T1). [1 mark] Rearrangement of Arrhenius equation to solve for Ea. [1 mark] Correct calculation of Ea in Joules (53100 to 53200 J mol-1). [1 mark] Conversion to kJ mol-1 and rounding to 3 significant figures with units.

(a) Anhydrous calcium chloride (\(\text{CaCl}_2\)), anhydrous magnesium sulfate (\(\text{MgSO}_4\)), or anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) can be used. When dry, the liquid becomes clear (loses its cloudiness) and the drying agent flows freely as a fine powder rather than clumping. (b) Moles of cyclohexanol = \(12.0 / 100.2 = 0.11976\text{ mol}\). Theoretical yield of cyclohexene = \(0.11976\text{ mol}\). Theoretical mass of cyclohexene = \(0.11976 \times 82.1 = 9.8323\text{ g}\). Percentage yield = \(\frac{5.40}{9.8323} \times 100 = 54.919\% \approx 54.9\%\).

[1 mark] Identifies a suitable anhydrous drying agent (e.g., CaCl2, MgSO4, Na2SO4). [1 mark] Explains that the liquid changes from cloudy to clear OR the solid stops clumping/flows freely. [1 mark] Calculates moles of cyclohexanol. [1 mark] Calculates the theoretical mass of cyclohexene. [1 mark] Calculates the percentage yield to 3 significant figures.

(a) The formation of a green precipitate with sodium hydroxide that turns brown in air indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). The formation of a blue precipitate with NaOH and a yellow-green solution with excess concentrated hydrochloric acid (due to tetrachlorocuprate(II) formation) indicates the presence of copper(II) ions, \(\text{Cu}^{2+}\). (b) Precipitate A is iron(II) hydroxide. The formation equation is: \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\). (c) The oxidation of iron(II) hydroxide to iron(III) hydroxide by atmospheric oxygen is represented by: \(4\text{Fe(OH)}_2(\text{s}) + \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{Fe(OH)}_3(\text{s})\).

[1 mark] Identifies iron(II) / Fe2+ as one of the cations. [1 mark] Identifies copper(II) / Cu2+ as the other cation. [1 mark] Explains precipitates A as Fe(OH)2 and B as Cu(OH)2. [1 mark] Writes a balanced ionic equation for the formation of Fe(OH)2 with state symbols. [1 mark] Writes a balanced equation for the oxidation of Fe(OH)2 to Fe(OH)3 by oxygen.

(a) 1. Mass of solution, \(m = 50.0 + 50.0 = 100.0\text{ g}\). Temperature change, \(\Delta T = 29.3 - 21.2 = 8.1\text{ K}\). 2. Heat released: \(q = m \cdot c \cdot \Delta T = 100.0 \times 4.18 \times 8.1 = 3385.8\text{ J} = 3.3858\text{ kJ}\). 3. Moles of acid/alkali reacting: \(n = c \cdot V = 1.20 \times 0.0500 = 0.0600\text{ mol}\). 4. Enthalpy change: \(\Delta H_{\text{neut}} = -\frac{3.3858}{0.0600} = -56.43\text{ kJ mol}^{-1} \approx -56.4\text{ kJ mol}^{-1}\). (b) Some heat is absorbed by the polystyrene cup/thermometer, which was not accounted for in the calculation.

[1 mark] Correct calculation of heat energy released, q = 3.39 kJ (or 3385.8 J). [1 mark] Correct calculation of moles of reactants/water formed = 0.0600 mol. [1 mark] Division of heat by moles to yield 56.4. [1 mark] Correct negative sign and unit of kJ mol-1. [1 mark] Feasible reason given (e.g., heat capacity of the cup/apparatus was neglected, or incomplete mixing).