2024 Edexcel IAL Chemistry (YCH11) 模擬試題連答案詳解

The successive ionisation energies show a very large jump between the third and fourth ionisation energies (from \(2745 \text{ kJ mol}^{-1}\) to \(11577 \text{ kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from a lower principal quantum shell, which is closer to the nucleus and less shielded. Therefore, the element has three valence electrons in its outer shell and forms a \(3+\) ion, \(X^{3+}\), to achieve a stable noble gas configuration.

1 mark: Correctly identifies the large jump between the 3rd and 4th ionisation energies and deduces the formation of \(X^{3+}\) (option C).

1. Calculate the number of moles of \(\text{CaCO}_3\):
\(n(\text{CaCO}_3) = \frac{10.0\text{ g}}{100.1\text{ g mol}^{-1}} \approx 0.0999\text{ mol}\)

2. From the balanced chemical equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\). Therefore:
\(n(\text{CO}_2) = 0.0999\text{ mol}\)

3. Calculate the volume of \(\text{CO}_2\) at rtp:
\(V = n \times V_m = 0.0999\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} \approx 2.40\text{ dm}^3\).

1 mark: Correct calculation of the volume of carbon dioxide to give \(2.40\text{ dm}^3\) (option B).

Boron trifluoride, \(\text{BF}_3\), has three bonding pairs of electrons and no lone pairs around the central boron atom. According to VSEPR theory, these electron pairs arrange themselves as far apart as possible to minimise repulsion, resulting in a trigonal planar geometry with a bond angle of exactly \(120^\circ\).

Other molecules:
- \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair, giving a trigonal pyramidal shape with a bond angle of approximately \(107^\circ\).
- \(\text{H}_2\text{O}\) has 2 bonding pairs and 2 lone pairs, giving a non-linear (bent) shape with a bond angle of approximately \(104.5^\circ\).
- \(\text{CH}_4\) has 4 bonding pairs and no lone pairs, giving a tetrahedral shape with a bond angle of \(109.5^\circ\).

1 mark: Correctly identifies \(\text{BF}_3\) (option B) as having a trigonal planar shape with a \(120^\circ\) bond angle.

A propagation step must involve a free radical reacting with a stable molecule to produce a new free radical and a new stable molecule.

- \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\) is an initiation step.
- \(\text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\) is a termination step.
- \(\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\) is a termination step.
- \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) is a propagation step, where a chlorine radical abstracts a hydrogen atom from methane, producing a methyl radical and hydrogen chloride.

1 mark: Identifies option C as the correct propagation step equation.

For an alkene to exhibit geometric (E/Z) isomerism, each carbon atom of the double bond must be bonded to two different groups.

- In but-1-ene (\(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\)), the C1 carbon is bonded to two identical hydrogen atoms. No E/Z isomerism.
- In propene (\(\text{CH}_2=\text{CH}-\text{CH}_3\)), the C1 carbon is bonded to two identical hydrogen atoms. No E/Z isomerism.
- In 2-methylbut-2-ene (\((\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_3\)), the C2 carbon is bonded to two identical methyl groups. No E/Z isomerism.
- In but-2-ene (\(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\)), C2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups), and C3 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups). Therefore, it exhibits geometric isomerism.

1 mark: Correctly identifies but-2-ene (option D) as exhibiting geometric isomerism.

First ionisation energy generally increases across Period 3 (from sodium to argon) because the nuclear charge increases (more protons) while shielding remains approximately constant (electrons are added to the same main shell). This results in a stronger electrostatic attraction between the nucleus and the outermost electron, requiring more energy to remove it. Argon, as the noble gas at the end of the period, has the highest first ionisation energy.

1 mark: Correctly identifies Argon (option D) as having the highest first ionisation energy in Period 3.

1. Assume \(100\text{ g}\) of the compound. Thus, mass of \(\text{C} = 85.7\text{ g}\) and mass of \(\text{H} = 100 - 85.7 = 14.3\text{ g}\).

2. Find the number of moles of each element:
- \(n(\text{C}) = \frac{85.7}{12.0} = 7.14\text{ mol}\)
- \(n(\text{H}) = \frac{14.3}{1.0} = 14.3\text{ mol}\)

3. Divide by the smallest number of moles (7.14):
- \(\text{C} = \frac{7.14}{7.14} = 1.0\)
- \(\text{H} = \frac{14.3}{7.14} = 2.0\)

Therefore, the empirical formula is \(\text{CH}_2\).

1 mark: Correct calculation of empirical formula as \(\text{CH}_2\) (option B).

Bond polarity is determined by the difference in electronegativity between the two bonded atoms. Fluorine (F) is the most electronegative element in the periodic table, so the difference in electronegativity between carbon (C) and fluorine (F) is greater than that between carbon and hydrogen (H), chlorine (Cl), or oxygen (O). Therefore, the \(\text{C}-\text{F}\) bond is the most polar.

1 mark: Correctly identifies the \(\text{C}-\text{F}\) bond (option B) as the most polar bond.

There is a very large jump between the third and fourth ionisation energies (from \(2745\text{ kJ mol}^{-1}\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner, more stable shell. Thus, element \(X\) has three valence electrons and belongs to Group 3 (Group 13). Its common stable oxidation state is \(+3\), forming an oxide with the formula \(X_2O_3\).

1 mark: Correctly identifies the oxide formula as \(X_2O_3\) (Option C) by locating the largest jump in ionisation energy between the 3rd and 4th values.

Using the ideal gas equation, \(PV = nRT\). First convert temperature to Kelvin: \(T = 127 + 273 = 400\text{ K}\). Convert pressure to Pascals: \(P = 150 \times 10^3\text{ Pa}\). Rearrange to solve for volume: \(V = \frac{nRT}{P} = \frac{0.200 \times 8.31 \times 400}{150 \times 10^3} = 4.432 \times 10^{-3}\text{ m}^3\). To convert to \(\text{dm}^3\), multiply by \(1000\), giving \(4.43\text{ dm}^3\).

1 mark: Correctly calculates \(4.43\text{ dm}^3\) (Option A) by converting units and applying the ideal gas equation.

Ammonia (\(\text{NH}_3\)) has a trigonal pyramidal shape with three bonding pairs and one lone pair of electrons. The lone pair-bonding pair repulsion reduces the bond angle from the tetrahedral angle of \(109.5^\circ\) to approximately \(107^\circ\). \(\text{BF}_3\) is trigonal planar (\(120^\circ\)), \(\text{NH}_4^+\) is tetrahedral (\(109.5^\circ\)), and \(\text{H}_2\text{O}\) is non-linear/bent (\(104.5^\circ\)).

1 mark: Correctly identifies \(\text{NH}_3\) (Option C) as having a bond angle of \(107^\circ\).

A propagation step begins with a free radical reactant and ends with a different free radical product. The step \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) involves a chlorine radical reacting with methane to form a methyl radical and hydrogen chloride. Option A is an initiation step, and Options C and D are termination steps.

1 mark: Correctly identifies the propagation step (Option B) which produces a new free radical.

The reaction proceeds via an electrophilic addition mechanism. Propene can form either a primary or a secondary carbocation intermediate. The secondary carbocation is more stable than the primary carbocation due to the positive inductive effect of the two electron-releasing methyl groups, which disperses the positive charge. This leads to the preferential formation of 2-bromopropane.

1 mark: Selects Option A, recognising that carbocation stability dictates product distribution in electrophilic additions to unsymmetrical alkenes.

Find the molar ratio of elements: Carbon: \(40.0 / 12 = 3.33\); Hydrogen: \(6.7 / 1 = 6.7\); Oxygen: \(53.3 / 16 = 3.33\). Dividing by the smallest value (3.33) yields an empirical formula of \(\text{CH}_2\text{O}\) (empirical formula mass = \(12 + 2(1) + 16 = 30\)). Since the relative molecular mass is \(180\), the scale factor is \(180 / 30 = 6\). Thus, the molecular formula is \(\text{C}_6\text{H}_{12}\text{O}_6\).

1 mark: Correctly determines the molecular formula is \(\text{C}_6\text{H}_{12}\text{O}_6\) (Option C).

Phosphorus has the outer electronic configuration \(3\text{s}^2 3\text{p}^3\) with three singly occupied \(3\text{p}\) orbitals. Sulfur has the configuration \(3\text{s}^2 3\text{p}^4\) with one fully paired \(3\text{p}\) orbital. The spin-pair repulsion between the two electrons sharing the same orbital in sulfur makes it easier to remove one of these electrons, resulting in a lower first ionisation energy.

1 mark: Selects Option B, attributing the decrease in first ionisation energy from P to S to electron pair repulsion in sulfur's \(3\text{p}\) orbital.

According to Fajan's rules, covalent character in ionic compounds increases with high charge density of the cation (high positive charge and small radius), which leads to greater polarisation of the anion. \(\text{Al}^{3+}\) has a higher positive charge (\(+3\)) and a smaller ionic radius compared to \(\text{Na}^+\), \(\text{Mg}^{2+}\), and \(\text{Ca}^{2+}\). Therefore, the aluminium ion is highly polarising, giving \(\text{AlCl}_3\) the greatest covalent character.

1 mark: Identifies \(\text{AlCl}_3\) (Option C) as having the greatest covalent character due to high polarising power of the \(\text{Al}^{3+}\) cation.

There is a very large increase between the third and fourth ionization energies (from 2745 to 11577 \(\text{kJ mol}^{-1}\)). This indicates that the fourth electron is removed from an inner quantum shell, meaning the element has three valence electrons and belongs to Group 13 (Group 3). The Period 3 element in this group is aluminium, which forms an oxide with the formula \(\text{X}_2\text{O}_3\).

1 mark for correct option C. Reject other options as they correspond to incorrect group oxidation states.

First, calculate the moles of magnesium: \(\text{moles of Mg} = 0.1215 \text{ g} / 24.3 \text{ g mol}^{-1} = 0.00500 \text{ mol}\). According to the equation, 1 mole of Mg produces 1 mole of hydrogen gas, so \(0.00500 \text{ mol}\) of \(\text{H}_2\) is produced. The volume of hydrogen gas is \(0.00500 \text{ mol} \times 24.0 \text{ dm}^3 \text{ mol}^{-1} = 0.120 \text{ dm}^3 = 120 \text{ cm}^3\).

1 mark for correct option B. Reject options with calculation errors or incorrect conversions.

In the amide ion (\(\text{NH}_2^-\)), the nitrogen atom has 2 bonding pairs of electrons and 2 lone pairs of electrons, making 4 electron pairs in total. This gives it a tetrahedral arrangement of electron pairs, but a bent shape. Each lone pair repels more strongly than bonding pairs, reducing the bond angle from the tetrahedral angle of \(109.5^\circ\) by approximately \(2.5^\circ\) per lone pair, resulting in a bond angle of \(104.5^\circ\).

1 mark for correct option C. Reject other options: \(\text{NH}_4^+\) is tetrahedral with a bond angle of \(109.5^\circ\); \(\text{NH}_3\) is trigonal pyramidal with a bond angle of \(107^\circ\); \(\text{CO}_2\) is linear with a bond angle of \(180^\circ\).

Electrophilic addition of \(\text{HBr}\) to propene involves the formation of a carbocation intermediate. The addition of the proton can form either a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)) or a secondary carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_3\)). The secondary carbocation is more stable because it has two electron-releasing methyl groups that disperse the positive charge. The major product, 2-bromopropane, is formed via this more stable secondary carbocation intermediate.

1 mark for correct option A.

(a)(i) The molecular ions are: m/z 158: \( [^{79}\text{Br}^{79}\text{Br}]^+ \), m/z 160: \( [^{79}\text{Br}^{81}\text{Br}]^+ \), m/z 162: \( [^{81}\text{Br}^{81}\text{Br}]^+ \). (ii) The probability of each combination is: m/z 158: \( 0.507 \times 0.507 = 0.257 \); m/z 160: \( 2 \times 0.507 \times 0.493 = 0.500 \); m/z 162: \( 0.493 \times 0.493 = 0.243 \). The ratio is 0.257 : 0.500 : 0.243, which is approximately 1 : 1.95 : 0.95 (or 1 : 2 : 1). (b) \( \text{Br}^+(g) \to \text{Br}^{2+}(g) + e^- \). (c) Arsenic has the outer electron configuration \( 4s^2 4p^3 \) (a stable half-filled 4p subshell). Selenium has \( 4s^2 4p^4 \). The fourth 4p electron in selenium is paired in an orbital. Repulsion between these two paired electrons in the same orbital makes it easier to remove an electron from selenium than from the half-filled p subshell of arsenic, despite selenium having a higher nuclear charge. (d) Moles of \( \text{Br}_2 = 1.20 / (2 \times 79.9) = 1.20 / 159.8 = 0.007509 \text{ mol} \). Moles of Br atoms = \( 2 \times 0.007509 = 0.01502 \text{ mol} \). Number of atoms = \( 0.01502 \times 6.02 \times 10^{23} = 9.04 \times 10^{21} \).

(a)(i) 1 mark for any two correct species, 2 marks for all three correct (including + charge). (ii) 1 mark for correct calculation of probabilities, 1 mark for correct ratio (e.g., 1:1.95:0.95 or 1:2:1). (b) 1 mark for correct chemical species, 1 mark for correct state symbols (gas). (c) 1 mark for stating electronic configurations or noting half-filled subshell in As vs paired electrons in Se. 1 mark for identifying electron pairing in the 4p orbital of Se. 1 mark for identifying that electron-electron repulsion makes the paired electron easier to remove. (d) 1 mark for calculating moles of \( \text{Br}_2 \). 1 mark for multiplying moles of molecules by 2 to get moles of atoms. 1 mark for final correct value to 3 significant figures.

(a) Moles of C = 85.7 / 12.0 = 7.14; Moles of H = 14.3 / 1.0 = 14.3. Ratio is 1 : 2, so the empirical formula is \( \text{CH}_2 \). (b) Rearranging \( pV = nRT \) to find \( n = pV / RT \). Volume in \( \text{m}^3 = 126 \times 10^{-6} \text{ m}^3 \). \( n = (1.01 \times 10^5 \times 126 \times 10^{-6}) / (8.31 \times 373) = 12.726 / 3099.63 = 0.004106 \text{ mol} \). Molar mass \( M = 0.285 / 0.004106 = 69.4 \text{ g mol}^{-1} \). Empirical mass of \( \text{CH}_2 = 14.0 \text{ g mol}^{-1} \). \( 69.4 / 14.0 \approx 5 \), so the molecular formula is \( \text{C}_5\text{H}_{10} \). (c)(i) \( \text{C}_5\text{H}_{10} + \text{HBr} \to \text{C}_5\text{H}_{11}\text{Br} \). (ii) Atom economy is 100%. This is an addition reaction, so there is only one product, meaning all reactant atoms are converted into the desired product. (d) Mass of water lost = 4.35 - 2.51 = 1.84 g. Moles of anhydrous salt = 2.51 / 151.9 = 0.01652 mol. Moles of water lost = 1.84 / 18.0 = 0.1022 mol. Ratio \( x = 0.1022 / 0.01652 = 6.19 \approx 6 \).

(a) 1 mark for correct mole calculations of C and H, 1 mark for correct empirical formula. (b) 1 mark for volume conversion to \( \text{m}^3 \), 1 mark for calculating correct moles of gas, 1 mark for finding molar mass (69.4 g/mol), 1 mark for deducing \( \text{C}_5\text{H}_{10} \). (c)(i) 1 mark for correct balanced equation. (ii) 1 mark for stating 100%, 1 mark for explaining that it is an addition reaction / only one product is formed. (d) 1 mark for finding mass of water lost (1.84 g), 1 mark for calculating moles of anhydrous salt and moles of water, 1 mark for deducing \( x = 6 \).

(a)(i) \( E \)-pent-2-ene has the methyl and ethyl groups on opposite sides of the double bond; \( Z \)-pent-2-ene has them on the same side. (ii) Pent-2-ene has restricted rotation around the carbon-carbon double bond, and each carbon of the double bond is attached to two different groups (H and methyl; H and ethyl). Pent-1-ene does not show stereoisomerism because one of the C=C carbons is attached to two identical groups (two hydrogen atoms). (b)(i) Mechanism: 1. Draw pent-1-ene and H-Br with dipoles \( \text{H}^{\delta+}-\text{Br}^{\delta-} \). 2. Curly arrow from the C=C double bond to the electrophilic H atom. 3. Curly arrow from the H-Br bond to the Br atom. 4. Draw the secondary carbocation intermediate: \( \text{CH}_3-\text{CH}^+-\text{CH}_2-\text{CH}_2-\text{CH}_3 \) and bromide ion with a lone pair. 5. Curly arrow from the lone pair of \( \text{Br}^- \) to the positive carbon. 6. Draw 2-bromopentane. (b)(ii) The reaction proceeds via a secondary carbocation rather than a primary carbocation. The secondary carbocation has two electron-releasing alkyl groups which stabilize the positive charge through the positive inductive effect, making it more stable than the primary carbocation, which only has one alkyl group. Thus, the secondary carbocation forms faster, leading to the major product.

(a)(i) 1 mark for each correctly drawn and labeled skeletal structure. (ii) 1 mark for stating restricted rotation about double bond and two different groups on both carbons of pent-2-ene, 1 mark for stating that pent-1-ene has two identical groups (hydrogen atoms) on one of the double-bonded carbons. (b)(i) 1 mark for dipole on H-Br and curly arrow from C=C to H, 1 mark for curly arrow from H-Br bond to Br, 1 mark for correct structure of secondary carbocation, 1 mark for curly arrow from lone pair on \( \text{Br}^- \) to carbon with positive charge, 1 mark for correct structure of 2-bromopentane. (b)(ii) 1 mark for identifying the secondary carbocation intermediate as more stable than the primary carbocation, 1 mark for explaining that alkyl groups are electron-releasing (positive inductive effect), 1 mark for linking stability to the faster rate of major product formation.

(a)(i) Ammonia has a trigonal pyramidal shape with a bond angle of \( 107^\circ \). Nitrogen has 3 bonding pairs and 1 lone pair of electrons (total 4 pairs). These pairs repel to maximize separation. Lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, squeezing the bond angle from the tetrahedral angle of \( 109.5^\circ \) to \( 107^\circ \). (ii) Boron trifluoride has a trigonal planar shape with a bond angle of \( 120^\circ \). Boron has 3 bonding pairs and 0 lone pairs of electrons. These three pairs repel equally to minimize repulsion, resulting in a planar shape with equal bond angles of \( 120^\circ \). (b)(i) Dative covalent (coordinate) bond. It is formed when the nitrogen atom donates both electrons from its lone pair to the electron-deficient boron atom (which has a vacant orbital). (b)(ii) Around nitrogen, the shape changes from trigonal pyramidal to tetrahedral, and the bond angle increases from \( 107^\circ \) to \( 109.5^\circ \) because the lone pair becomes a bonding pair, eliminating the extra repulsion. Around boron, the shape changes from trigonal planar to tetrahedral, and the bond angle decreases from \( 120^\circ \) to \( 109.5^\circ \) because boron now has 4 bonding pairs instead of 3, resulting in tetrahedral geometry.

(a)(i) 1 mark for trigonal pyramidal shape and \( 107^\circ \) angle. 1 mark for stating there are 3 bonding pairs and 1 lone pair. 1 mark for explaining that lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion. (ii) 1 mark for trigonal planar shape and \( 120^\circ \) angle. 1 mark for stating 3 bonding pairs and 0 lone pairs. 1 mark for stating that the 3 bonding pairs repel equally. (b)(i) 1 mark for naming dative covalent / coordinate bond. 1 mark for explaining that both electrons in the shared pair come from nitrogen's lone pair. (b)(ii) 1 mark for stating N shape becomes tetrahedral and angle increases to \( 109.5^\circ \). 1 mark for explaining this is due to lone pair becoming a bonding pair. 1 mark for stating B shape becomes tetrahedral and angle decreases to \( 109.5^\circ \). 1 mark for explaining this is due to increasing from 3 to 4 bonding pairs.

(a)(i) UV light provides the energy to cause homolytic fission of the chlorine molecule into chlorine radicals. (ii) Initiation: \( \text{Cl}_2 \xrightarrow{UV} 2\text{Cl}^\bullet \). Propagation 1: \( \text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \to \text{CH}_3\text{CH}^\bullet\text{CH}_3 + \text{HCl} \). Propagation 2: \( \text{CH}_3\text{CH}^\bullet\text{CH}_3 + \text{Cl}_2 \to \text{CH}_3\text{CHClCH}_3 + \text{Cl}^\bullet \). Termination to make \( \text{C}_6\text{H}_{14} \): \( 2\text{C}_3\text{H}_7^\bullet \to \text{C}_6\text{H}_{14} \) (accept structural representation like isopropyl or propyl radicals combining). (b) The trace product \( \text{C}_6\text{H}_{14} \) (which is a larger alkane) can only be formed when two propyl / isopropyl free radicals collide and combine (dimerization). This confirms that free radicals are generated as intermediates in the reaction. (c)(i) \( \text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O} \). (ii) Moles of propane = 4.40 / 44.0 = 0.100 mol. Moles of oxygen required = \( 5 \times 0.100 = 0.500 \text{ mol} \). Volume of oxygen = \( 0.500 \times 24.0 = 12.0 \text{ dm}^3 \).

(a)(i) 1 mark for homolytic fission of Cl-Cl bond / creation of chlorine radicals. (ii) 1 mark for initiation step. 1 mark for first propagation step (showing radical on the carbon). 1 mark for second propagation step. 1 mark for termination step showing combination of two propyl radicals. 1 mark for correct radical dots on all radicals. (b) 1 mark for stating that it is formed by combination of two propyl radicals, 1 mark for explaining that this dimerisation confirms the presence of radical intermediates. (c)(i) 1 mark for correct balanced equation. (ii) 1 mark for calculating moles of propane (0.100 mol), 1 mark for using stoichiometry to get moles of oxygen (0.500 mol), 1 mark for correct final volume of oxygen with units (12.0 dm3).

The standard enthalpy of formation of ethanoic acid corresponds to the equation: \(2\text{C}(s) + 2\text{H}_2(g) + \text{O}_2(g) \rightarrow \text{CH}_3\text{COOH}(l)\). Using Hess's Law and enthalpies of combustion: \(\Delta_f H^{\theta} = \sum \Delta_c H^{\theta}(\text{reactants}) - \sum \Delta_c H^{\theta}(\text{products})\). Therefore: \(\Delta_f H^{\theta} = [2 \times (-394) + 2 \times (-286)] - [-874] = [-788 - 572] + 874 = -1360 + 874 = -486\text{ kJ mol}^{-1}\).

[1] Correct calculation of the standard enthalpy of formation, including correct sign and units: -486 kJ mol^-1.

Going down Group 2, the cationic radius increases while the charge remains 2+. This decreases the charge density and therefore the polarizing power of the cation. Consequently, the cation polarizes the nitrate anion less, making the N-O bonds in the anion harder to break, which increases the thermal stability of the nitrate.

[1] Correctly states that thermal stability increases down the group because the increased cationic radius reduces its polarizing power, leading to less polarization of the nitrate anion.

The very broad absorption band between \(2500\text{--}3300\text{ cm}^{-1}\) is characteristic of the O-H stretching in carboxylic acids. The sharp absorption at \(1715\text{ cm}^{-1}\) corresponds to the C=O stretching in a carbonyl group. Together with the molecular formula \(\text{C}_3\text{H}_6\text{O}_2\), this indicates that compound \(X\) is propanoic acid.

[1] Correctly identifies propanoic acid based on the presence of carboxylic acid O-H and C=O functional groups from the infrared data.

\(\text{CH}_3\text{F}\) is polar due to the highly polar C-F bond, so it has permanent dipole-dipole forces as well as London forces. However, it does not form hydrogen bonds because the hydrogen atoms are bonded to carbon, not directly to the highly electronegative fluorine atom. \(\text{HF}\) and \(\text{CH}_3\text{OH}\) form hydrogen bonds, whereas \(\text{CF}_4\) is symmetrical and non-polar, possessing only London forces.

[1] Correctly identifies CH3F as a polar molecule without hydrogen-bonding capability between its molecules.

As temperature increases, the Maxwell-Boltzmann distribution curve flattens, shifting its peak to the right (higher energy) and downwards (lower peak height). Due to this shift, the area under the curve to the right of the activation energy line (\(E_a\)) increases, meaning the fraction of molecules with \(E \ge E_a\) increases.

[1] Correctly identifies that the peak shifts to the right and is lower, and that the fraction of molecules with energy greater than or equal to Ea increases.

The rate of hydrolysis of halogenoalkanes is determined by the strength of the C-halogen bond and the mechanism of the reaction. Iodoalkanes are more reactive than chloroalkanes because the C-I bond is weaker (lower bond enthalpy) than the C-Cl bond. Furthermore, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react much faster than primary halogenoalkanes because they hydrolyze via the stable tertiary carbocation in the \(S_N1\) mechanism. Thus, 2-iodo-2-methylpropane is the most reactive.

[1] Correctly identifies 2-iodo-2-methylpropane as the fastest reacting halogenoalkane based on carbon-halogen bond strength and the tertiary carbocation stabilization.

In this reaction, iodide is a strong reducing agent that reduces sulfur in sulfuric acid from an oxidation state of +6 to +4 in \(\text{SO}_2\), 0 in \(\text{S}\), and -2 in \(\text{H}_2\text{S}\). \(\text{I}_2\) is an oxidation product of iodide, whereas \(\text{KHSO}_4\) and \(\text{HI}\) are products of acid-base reactions where the oxidation state of sulfur remains unchanged (+6).

[1] Correctly identifies the list of only the reduction products of sulfur: SO2, S, and H2S.

The total volume of the mixture is \(100.0\text{ cm}^3\), so its mass \(m = 100.0\text{ g}\). The temperature change \(\Delta T = 26.8 - 20.0 = 6.8\text{ K}\). The heat released is \(q = m \times c \times \Delta T = 100.0 \times 4.18 \times 6.8 = 2842.4\text{ J} = 2.8424\text{ kJ}\). The number of moles of water formed is \(n = 1.00 \times 0.0500 = 0.0500\text{ mol}\). Thus, the enthalpy change of neutralization is \(\Delta_{neut}H = -\frac{q}{n} = -\frac{2.8424}{0.0500} = -56.8\text{ kJ mol}^{-1}\).

[1] Correct calculation of the enthalpy of neutralization to 3 significant figures, including the negative sign: -56.8 kJ mol^-1.

Hydrogen bonding is present between \(\text{HF}\) molecules due to the highly electronegative fluorine atom bonded to hydrogen. \(\text{HCl}\) does not form hydrogen bonds; it only has permanent dipole-dipole interactions and London forces. Since hydrogen bonds are much stronger than dipole-dipole interactions, \(\text{HF}\) has a higher boiling point than \(\text{HCl}\). Moving from \(\text{HCl}\) to \(\text{HI}\), the number of electrons per molecule increases, which increases the strength of the London (instantaneous dipole-induced dipole) forces, making the boiling point of \(\text{HI}\) higher than that of \(\text{HCl}\).

1 mark: Correct option B selected. Reject all other options.

The formation reaction is \(\text{C(graphite)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}\). Using Hess's cycle: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products}) = [\Delta H_c^\ominus(\text{C}) + 2 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{CH}_4) = [-393.5 + 2(-285.8)] - [-890.3] = -965.1 + 890.3 = -74.8\text{ kJ mol}^{-1}\).

1 mark: Correct option C selected. Reject all other options.

Down Group 2, the solubility of the hydroxides increases, and the thermal stability of the nitrates increases. The thermal stability increases because the cationic radius increases, reducing the charge density and polarising power of the cation, which distorts the nitrate anion less.

1 mark: Correct option D selected. Reject all other options.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond (bond enthalpy) rather than bond polarity. The \(\text{C}-\text{Cl}\) bond is the strongest and hardest to break, leading to the slowest reaction. The \(\text{C}-\text{I}\) bond is the weakest and easiest to break, leading to the fastest reaction. Thus, the order of reaction rate is 1-chlorobutane < 1-bromobutane < 1-iodobutane.

1 mark: Correct option A selected. Reject all other options.

A catalyst does not change the temperature or the number of particles, so the Maxwell-Boltzmann distribution curve of molecular energies remains completely unchanged. Instead, the catalyst provides an alternative pathway with a lower activation energy, effectively moving the minimum energy required for reaction to the left.

1 mark: Correct option B selected. Reject all other options.

The peak at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad absorption in the \(3200-3750\text{ cm}^{-1}\) region indicates that there is no hydroxyl (\(\text{O}-\text{H}\)) group present, ruling out alcohols and carboxylic acids. Propanone, a ketone, has the molecular formula \(\text{C}_3\text{H}_6\text{O}\), contains a \(\text{C}=\text{O}\) group, and contains no \(\text{O}-\text{H}\) group, fitting the data perfectly.

1 mark: Correct option D selected. Reject all other options.

The reaction of chlorine with cold dilute \(\text{NaOH}\) is: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). In \(\text{NaCl}\), the oxidation state of chlorine is \(-1\). In \(\text{NaClO}\), sodium is \(+1\) and oxygen is \(-2\), so chlorine must be \(+1\). This is a disproportionation reaction where chlorine is simultaneously oxidized to \(+1\) and reduced to \(-1\).

1 mark: Correct option B selected. Reject all other options.

Since the forward reaction is exothermic (\(\Delta H < 0\)), according to Le Chatelier's principle, decreasing the temperature shifts the equilibrium position in the exothermic direction (to the right) to release heat, thereby increasing the equilibrium yield of \(\text{SO}_3(\text{g})\). Decreasing the pressure shifts the equilibrium to the left (towards more moles of gas). Removing a reactant also shifts the equilibrium to the left. A catalyst only increases the rate of reaction but does not change the equilibrium yield.

1 mark: Correct option A selected. Reject all other options.

Pentan-1-ol has a longer straight carbon chain with five carbon atoms, allowing for a larger surface area of contact between molecules and therefore stronger London forces compared to its branched isomers (2-methylbutan-2-ol and 2,2-dimethylpropan-1-ol). It also contains a hydroxyl group, allowing for hydrogen bonding, which is much stronger than London forces. Butan-1-ol only has four carbon atoms, so it has fewer electrons and weaker London forces than pentan-1-ol. Thus, pentan-1-ol has the highest boiling temperature.

A is correct (1 mark). B, C and D are incorrect because they have either fewer carbon atoms or a more branched carbon chain, resulting in weaker intermolecular forces.

Down Group 2, the ionic radius of the metal cation increases while the charge remains 2+. Consequently, the charge density of the cation decreases, reducing its ability to polarize (distort) the carbonate ion's electron cloud. Since less polarization occurs, the C-O bonds within the carbonate ion are weakened to a lesser extent, requiring more thermal energy to decompose. Thus, thermal stability increases down the group.

A is correct (1 mark). B, C, and D are incorrect because they describe incorrect trends in ionic radius or make incorrect claims about the effect of polarization on thermal stability.

The equation for the formation of propane is: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\). Using the combustion data: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products}) = [3 \times (-393.5) + 4 \times (-285.8)] - (-2219.2) = [-1180.5 - 1143.2] + 2219.2 = -2323.7 + 2219.2 = -104.5\text{ kJ mol}^{-1}\).

A is correct (1 mark). B is incorrect due to a sign error. C is the sum of the reactant combustions without subtracting the product. D is the sum of all values with negative signs.

Distillation of a primary alcohol with acidified potassium dichromate(VI) oxidizes it to an aldehyde. Aldehydes contain a carbonyl group (\(\text{C}=\text{O}\)), which shows a strong, sharp absorption band in the range \(1740 - 1720\text{ cm}^{-1}\). The starting alcohol contains only single bonds (\(\text{C}-\text{O}\), \(\text{O}-\text{H}\), \(\text{C}-\text{H}\)) and has no absorption in this region.

A is correct (1 mark). B is incorrect as this represents the alcohol O-H stretch, which is present in the starting material. C represents a carboxylic acid O-H stretch. D represents an alkene C=C stretch.

(a) First, calculate the temperature rise: \(\Delta T = 48.5 - 19.5 = 29.0\text{ K}\). Using the heat energy equation: \(q = m c \Delta T = 150 \times 4.18 \times 29.0 = 18183\text{ J} = 18.2\text{ kJ}\). (b) Calculate the molar mass of pentan-1-ol: \(M_r(\text{C}_5\text{H}_{12}\text{O}) = (5 \times 12.0) + (12 \times 1.0) + 16.0 = 88.0\text{ g mol}^{-1}\). Moles of pentan-1-ol burned: \(n = 0.620 / 88.0 = 0.007045\text{ mol}\). The enthalpy of combustion: \(\Delta H_c = -q / n = -18.183 / 0.007045 = -2581\text{ kJ mol}^{-1}\) (or \(-2580\text{ kJ mol}^{-1}\) depending on intermediate rounding). (c) Two reasons other than heat loss: 1. Incomplete combustion of pentan-1-ol, leading to the formation of carbon monoxide or carbon soot instead of carbon dioxide. 2. Evaporation of pentan-1-ol from the wick after extinguishing the burner and before weighing. (d) The setup can be improved by adding a draft shield around the burner to prevent drafts, or by using a lid on the copper calorimeter. (e) A bomb calorimeter ensures complete combustion by using excess pressurized pure oxygen, and it is heavily insulated/calibrated to virtually eliminate heat loss to the surroundings.

(a) [3 marks] M1: Calculate temperature rise of 29.0 K (1). M2: Recall and substitute correctly into the expression \(q = m c \Delta T\) (1). M3: Correct evaluation of heat energy as 18.2 kJ (accept 18.18 kJ) (1). (b) [3 marks] M1: Calculate moles of pentan-1-ol as 0.00705 mol (1). M2: Divide kJ by moles to find enthalpy change (1). M3: Give final answer as -2580 or -2581 kJ mol\(^{-1}\), including the negative sign and to 3 significant figures (1). (c) [2 marks] M1: Incomplete combustion (1). M2: Evaporation of alcohol from the wick (1). (d) [2 marks] M1: Use a draft shield (1). M2: Use a lid on the calorimeter (1). (e) [2 marks] M1: Excess oxygen ensures complete combustion (1). M2: Highly insulated setup minimizes heat loss (1).

(a) Add equal volumes of ethanol to three separate test tubes containing 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Add equal volumes of aqueous silver nitrate to each test tube. Place the tubes in a water bath at 50 °C. Measure the time taken for a precipitate to appear. Silver chloride forms a white precipitate, silver bromide forms a cream precipitate, and silver iodide forms a yellow precipitate. (b) The ionic equation representing the precipitate formation is: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\). (c) The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane. This is because the C-X bond enthalpy decreases down Group 7 (C-Cl is strongest, C-I is weakest). The weaker the bond, the less energy is required to break it, making the C-I bond break fastest. Bond strength is the dominant factor here, not bond polarity. (d) The hydrolysis of 1-bromobutane proceeds via an SN2 mechanism. The hydroxide ion acts as a nucleophile. A lone pair of electrons on the oxygen of the hydroxide ion attacks the electron-deficient carbon atom bonded to the bromine. Simultaneously, the pair of electrons in the C-Br bond is transferred entirely to the bromine atom, which departs as a bromide leaving group, yielding butan-1-ol.

(a) [5 marks] M1: Use of ethanol as a common solvent (1). M2: Addition of aqueous silver nitrate (1). M3: Use of a water bath for temperature control (1). M4: Timing the appearance of the precipitate (1). M5: Correctly identifying colors of precipitate (white for Cl, cream for Br, yellow for I) (1). (b) [1 mark] M1: Correct ionic equation with state symbols: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\) (1). (c) [4 marks] M1: Identify that 1-iodobutane reacts fastest and 1-chlorobutane slowest (1). M2: State that C-X bond enthalpy/strength decreases down the group (C-Cl > C-Br > C-I) (1). M3: Explain that weaker bonds require less energy to break (1). M4: State that bond enthalpy is the dominant factor over bond polarity (1). (d) [2 marks] M1: Describe curly arrow from the lone pair of hydroxide oxygen to the carbon (1). M2: Describe curly arrow from the C-Br bond to the bromine atom (1).

(a) Magnesium nitrate decomposes on heating to form magnesium oxide, nitrogen dioxide, and oxygen: \(2\text{Mg(NO}_3)_2(\text{s}) \rightarrow 2\text{MgO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\). (b) Two observations: 1. Formation of brown fumes/gas (nitrogen dioxide). 2. A white solid residue remains (magnesium oxide), or a gas is produced that relights a glowing splint (oxygen). (c) Thermal stability increases down Group 2. This is because the ionic radius of the Group 2 cation increases down the group, while the 2+ charge remains the same. This leads to a decrease in charge density of the cation. The larger cation with lower charge density has a weaker polarizing effect on the nitrate anion, causing less distortion of the N-O bonds, meaning more thermal energy is required to decompose the nitrate. (d) The solubility of Group 2 hydroxides increases down the group. Both the lattice energy and the hydration energy of the hydroxides decrease as the cation size increases. However, the lattice energy decreases much more rapidly than the hydration energy of the cation. Consequently, the enthalpy change of solution becomes more exothermic (or less endothermic) down the group, making the hydroxides more soluble.

(a) [2 marks] M1: Correct balanced chemical formulae (1). M2: Correct state symbols: (s) for reactant and MgO, (g) for \(\text{NO}_2\) and \(\text{O}_2\) (1). (b) [2 marks] M1: Brown gas/fumes (1). M2: White solid remaining / gas that relights a glowing splint (1). (c) [4 marks] M1: State that thermal stability increases down the group (1). M2: Cation size increases / charge density decreases (1). M3: Cation has a weaker polarizing effect on the nitrate anion (1). M4: Less distortion/polarization of the N-O covalent bonds, requiring more energy to break (1). (d) [4 marks] M1: State that solubility of hydroxides increases down the group (1). M2: Both lattice energy and hydration energy decrease down the group (1). M3: Lattice energy decreases more rapidly than hydration energy (1). M4: Enthalpy change of solution becomes more exothermic / less endothermic down the group (1).

(a) Propan-1-ol: Hydrogen bonding. Propanal: Permanent dipole-permanent dipole forces. Butane: London forces (instantaneous dipole-induced dipole forces). (b) Boiling points follow the order: propan-1-ol > propanal > butane. Hydrogen bonds in propan-1-ol are the strongest type of intermolecular force and require the most thermal energy to overcome. Permanent dipole-dipole forces in propanal are stronger than London forces but weaker than hydrogen bonds. London forces in butane are the weakest intermolecular forces and require the least energy to overcome. Since all three molecules have similar molar masses (and thus similar numbers of electrons), their London forces are of similar baseline strength, meaning the differences in boiling points are due to the additional dipole and hydrogen bonding interactions. (c) Infrared spectroscopy can distinguish them because propan-1-ol contains an O-H (alcohol) bond which shows a broad and intense absorption peak in the range \(3200 - 3750\text{ cm}^{-1}\), and lacks a \(\text{C=O}\) stretch. In contrast, propanal contains a carbonyl group with a \(\text{C=O}\) bond, displaying a strong, sharp absorption peak in the range \(1675 - 1750\text{ cm}^{-1}\), and does not show an O-H alcohol stretch.

(a) [3 marks] M1: Propan-1-ol has hydrogen bonding (1). M2: Propanal has permanent dipole-dipole forces (1). M3: Butane has London forces (1). (b) [5 marks] M1: Correct order of boiling points: propan-1-ol > propanal > butane (1). M2: Hydrogen bonding is the strongest IMF, requiring the most thermal energy to overcome (1). M3: Dipole-dipole forces are intermediate in strength (1). M4: London forces are the weakest (1). M5: Reference to similar molar masses / similar number of electrons resulting in comparable baseline London forces (1). (c) [4 marks] M1: Propan-1-ol has an O-H stretch in the range \(3200-3750\text{ cm}^{-1}\) (1). M2: Mention that this O-H peak is broad (1). M3: Propanal has a \(\text{C=O}\) stretch in the range \(1675-1750\text{ cm}^{-1}\) (1). M4: Mention that this \(\text{C=O}\) peak is sharp/strong and no O-H peak is present in propanal (1).

(a) A Maxwell-Boltzmann curve plots number of molecules (y-axis) against kinetic energy (x-axis). The curve starts at the origin, rises to a peak, and then decreases, approaching but never reaching the horizontal axis. When temperature increases from \(T_1\) to \(T_2\), the peak shifts to the right (higher energy) and becomes lower. The activation energy \(E_a\) is a fixed point on the energy axis, and the area to the right represents molecules with energy greater than or equal to \(E_a\). (b) At higher temperature \(T_2\), the distribution is flattened and shifted to the right, meaning a significantly larger fraction of molecules have kinetic energy greater than or equal to the activation energy. This results in a much higher frequency of successful collisions per unit time, thereby increasing the rate of reaction. (c) An increase in temperature decreases the yield of \(\text{SO}_3\) at equilibrium. The forward reaction is exothermic. According to Le Chatelier's Principle, increasing temperature causes the equilibrium to shift in the endothermic (reverse) direction to absorb the added heat, thus reducing the yield of \(\text{SO}_3\). (d) A catalyst has no effect on the position of equilibrium or the yield of \(\text{SO}_3\). It speeds up both the forward and reverse reactions by the same rate by providing an alternative pathway with a lower activation energy, allowing the system to reach equilibrium faster but without altering the composition of the equilibrium mixture.

(a) [4 marks] M1: Axes correctly described: Y-axis as number of molecules, X-axis as kinetic energy (1). M2: Sketch of \(T_1\) starting at origin, rising to a peak, and tailing off (1). M3: Curve for \(T_2\) showing a lower peak shifted to the right (1). M4: Activation energy \(E_a\) correctly located on the X-axis (1). (b) [3 marks] M1: State that a larger proportion of molecules have energy \(\ge E_a\) at higher temperature (1). M2: Reference to the area under the curve to the right of \(E_a\) being larger (1). M3: More successful collisions per unit time (1). (c) [3 marks] M1: Yield of \(\text{SO}_3\) decreases (1). M2: Forward reaction is exothermic (1). M3: Equilibrium shifts to the left / in endothermic direction to absorb heat (1). (d) [2 marks] M1: No effect on position of equilibrium / yield (1). M2: Increases rate of both forward and reverse reactions equally (1).

(a) A slight excess of sodium hydroxide ensures that all of the hydrochloric acid (the limiting reactant) is completely neutralised so that the calculated heat is based on the full amount of acid.

(b) 1. Polystyrene cup: It acts as an excellent thermal insulator to reduce heat conduction. 2. Lid: Placing a lid on top of the cup minimizes heat loss via convection and evaporation.

(c) Temperature change, \( \Delta T = 27.8 - 21.3 = 6.5\text{ }^\circ\text{C} \). Total mass of solution, \( m = 50.0 + 50.0 = 100.0\text{ g} \). Heat energy, \( q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 6.5\text{ }^\circ\text{C} = 2717\text{ J} \). Assumptions: The density of the solution is exactly \( 1.00\text{ g cm}^{-3} \) and the specific heat capacity is \( 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \) (the same as pure water); no heat is absorbed by the polystyrene cup or lost to the air.

(d) Number of moles of \( \text{HCl} \) used: \( n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol} \). Number of moles of \( \text{NaOH} \) used: \( n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 1.05\text{ mol dm}^{-3} = 0.0525\text{ mol} \). Since \( \text{HCl} \) is the limiting reactant, \( 0.0500\text{ mol} \) of water is formed. Enthalpy of neutralisation: \( \Delta H = -\frac{2717\text{ J}}{0.0500\text{ mol}} = -54340\text{ J mol}^{-1} = -54.3\text{ kJ mol}^{-1} \).

(e) Because two temperature readings are taken (initial and final), the absolute uncertainty in the temperature change is \( 2 \times 0.1 = 0.2\text{ }^\circ\text{C} \). Percentage uncertainty = \( \frac{0.2}{6.5} \times 100\% = 3.08\% \) (or \( 3.1\% \)).

Part (a) [1 mark]
- M1: To ensure that all of the acid reacts/is neutralised [1].

Part (b) [2 marks]
- M1: Polystyrene cup acts as a thermal insulator / reduces conduction [1].
- M2: A lid reduces heat loss by convection / evaporation [1].

Part (c) [3 marks]
- M1: Calculates temperature change \( \Delta T = 6.5\text{ }^\circ\text{C} \) and total mass \( 100.0\text{ g} \) [1].
- M2: Calculates heat energy: \( q = 100.0 \times 4.18 \times 6.5 = 2717\text{ J} \) (or \( 2.717\text{ kJ} \)) [1].
- M3: States that the solution is assumed to have the same density/specific heat capacity as water OR no heat is lost to the surroundings [1].

Part (d) [4.5 marks]
- M1: Calculates moles of HCl = \( 0.0500\text{ mol} \) [1].
- M2: Divides heat energy by the moles of HCl: \( \frac{2.717}{0.0500} = 54.34\text{ kJ mol}^{-1} \) [2].
- M3: States the correct value with negative sign and to 3 significant figures: \( -54.3\text{ kJ mol}^{-1} \) [1.5].

Part (e) [2 marks]
- M1: Identifies total absolute uncertainty in temperature difference as \( 0.2\text{ }^\circ\text{C} \) [1].
- M2: Calculates percentage uncertainty: \( \frac{0.2}{6.5} \times 100\% = 3.08\% \) (accept \( 3.1\% \)) [1].

(a) The reaction between concentrated sulfuric acid and water/reaction mixture is highly exothermic. Adding it slowly with cooling prevents the mixture from boiling over or splattering, and minimizes the formation of toxic sulfur dioxide gas or oxidation of bromide ions to bromine.

(b) The diagram should show a pear-shaped or round-bottomed flask with a vertical Liebig condenser attached to its neck. There should be no stopper at the top of the condenser. The water inlet must be shown at the bottom of the condenser jacket and the water outlet at the top of the condenser jacket. A heat source (e.g., heating mantle) should be under the flask.

(c)(i) Pour the mixture into a separating funnel and allow the layers to separate into two distinct phases. Since 1-bromobutane has a density of \( 1.27\text{ g cm}^{-3} \) which is greater than that of water (\( 1.00\text{ g cm}^{-3} \)), the bottom layer is the organic layer. Run off the bottom organic layer into a clean flask.

(c)(ii) Purpose: To neutralise any remaining acidic impurities (such as sulfuric acid or hydrogen bromide). Ionic equation: \( \text{H}^+ + \text{HCO}_3^- \rightarrow \text{H}_2\text{O} + \text{CO}_2 \).

(d) Suitable drying agent: Anhydrous calcium chloride (\( \text{CaCl}_2 \)), anhydrous magnesium sulfate (\( \text{MgSO}_4 \)), or anhydrous sodium sulfate (\( \text{Na}_2\text{SO}_4 \)). When completely dry, the cloudy liquid becomes completely clear and transparent.

Part (a) [2 marks]
- M1: The reaction is highly exothermic / releases a lot of heat [1].
- M2: Slow addition/cooling prevents boiling over / splattering / minimizes toxic fumes (\( \text{SO}_2 \)) / prevents oxidation of bromide to bromine [1].

Part (b) [3 marks]
- M1: Vertical condenser attached to a flask with no stopper at the top [1].
- M2: Water flow in condenser jacket in at the bottom and out at the top [1].
- M3: Heat source shown at the bottom [1].

Part (c)(i) [2.5 marks]
- M1: Use a separating funnel [1].
- M2: Allow layers to separate and collect the lower layer [1].
- M3: States that 1-bromobutane is the lower layer because its density is higher than water's density [0.5].

Part (c)(ii) [3 marks]
- M1: To neutralise remaining acid / acidic impurities / H2SO4 / HBr [1].
- M2: Correct reactants in equation: \( \text{H}^+ \) and \( \text{HCO}_3^- \) [1].
- M3: Correct products: \( \text{H}_2\text{O} \) and \( \text{CO}_2 \) [1].

Part (d) [2 marks]
- M1: Names a correct anhydrous salt (e.g., anhydrous calcium chloride / magnesium sulfate / sodium sulfate) [1]. (Do not accept copper sulfate or cobalt chloride).
- M2: Describes the appearance of the dry liquid as clear / transparent / no longer cloudy [1].

(a) To carry out a flame test: Clean a nichrome or platinum wire by dipping it into concentrated hydrochloric acid and heating it in a blue Bunsen flame until no colour is observed. Redip the wire in concentrated hydrochloric acid, touch it to a small sample of solid **X**, and then place the wire back into the non-luminous (blue) Bunsen flame. If barium is present, an apple-green flame will be observed.

(b)(i) The cation is \( \text{Ba}^{2+} \). Reacting with sulfate ions from sulfuric acid produces insoluble barium sulfate: \( \text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) \).

(b)(ii) Test for halide ions: Add dilute nitric acid followed by silver nitrate solution to solution **Y**. Chloride gives a white precipitate, bromide gives a cream precipitate, and iodide gives a yellow precipitate. To distinguish them: add dilute aqueous ammonia; silver chloride will dissolve. If the precipitate remains, add concentrated aqueous ammonia; silver bromide will dissolve, whereas silver iodide remains insoluble in both dilute and concentrated ammonia.

(c) Observation: The colourless solution turns orange/yellow/brown. Ionic equation: \( \text{Cl}_2(\text{aq}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq}) \).

Part (a) [4 marks]
- M1: Cleans nichrome/platinum wire with concentrated hydrochloric acid and heats in a Bunsen burner flame [1].
- M2: Dips clean wire in concentrated HCl, picks up solid, and places in non-luminous (blue) flame [1].
- M3: Flame test observation for barium: apple-green flame [2].

Part (b)(i) [2 marks]
- M1: Correct chemical species: \( \text{Ba}^{2+} \), \( \text{SO}_4^{2-} \), and \( \text{BaSO}_4 \) [1].
- M2: Correct state symbols: \( (\text{aq}) \) for reactants, \( (\text{s}) \) for product [1].

Part (b)(ii) [4.5 marks]
- M1: Reagents: Add dilute nitric acid followed by silver nitrate solution [1].
- M2: Identifies precipitates: white (chloride), cream (bromide), yellow (iodide) [1.5].
- M3: Silver chloride dissolves in dilute ammonia, silver bromide dissolves in concentrated ammonia (but not dilute), and silver iodide is insoluble in both [2].

Part (c) [2 marks]
- M1: Observation: solution turns orange / yellow / brown [1].
- M2: Equation: \( \text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2 \) [1].

(a) To prepare the standard solution:
1. Weigh exactly \( 1.325\text{ g} \) of anhydrous sodium carbonate on a balance and record the mass.
2. Transfer the solid to a beaker and dissolve it completely in approximately \( 100\text{ cm}^3 \) of deionised water, stirring with a clean glass rod.
3. Transfer the solution quantitatively into a \( 250.0\text{ cm}^3 \) volumetric flask. Rinse the beaker and glass rod with deionised water and add all washings to the flask.
4. Add deionised water dropwise until the bottom of the meniscus lies exactly on the graduation line. Stopper the flask and invert it several times to mix thoroughly.

(b) Moles of \( \text{Na}_2\text{CO}_3 = \frac{1.325\text{ g}}{106.0\text{ g mol}^{-1}} = 0.0125\text{ mol} \).
Concentration = \( \frac{0.0125\text{ mol}}{0.2500\text{ dm}^3} = 0.0500\text{ mol dm}^{-3} \).

(c)(i) At the end-point of adding acid to carbonate with methyl orange, the colour changes from yellow to orange (or peach/pink).

(c)(ii) A pipette is used because it is designed to measure and deliver a highly accurate, single, fixed volume (exactly \( 25.0\text{ cm}^3 \)). A burette is used because it allows a variable volume of solution to be delivered and measured dropwise to determine the exact end-point.

(d)(i) The rough titre is non-concordant and must be excluded. Concordant titres are \( 23.10\text{ cm}^3 \) and \( 23.20\text{ cm}^3 \).
Mean titre = \( \frac{23.10 + 23.20}{2} = 23.15\text{ cm}^3 \).

(d)(ii) Moles of \( \text{Na}_2\text{CO}_3 \) in \( 25.0\text{ cm}^3 \) = \( 0.0500\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 1.25 \times 10^{-3}\text{ mol} \).
Moles of \( \text{HCl} \) reacted = \( 2 \times 1.25 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol} \).
Concentration of \( \text{HCl} \) = \( \frac{2.50 \times 10^{-3}\text{ mol}}{0.02315\text{ dm}^3} = 0.108\text{ mol dm}^{-3} \) (to 3 s.f.).

Part (a) [4 marks]
- M1: Dissolves solid in a beaker with deionised water (less than \( 250\text{ cm}^3 \)) [1].
- M2: Quantitative transfer of solution and washings of beaker/rod into a \( 250.0\text{ cm}^3 \) volumetric flask [1].
- M3: Makes up to the mark with deionised water (bottom of meniscus on the line) [1].
- M4: Stoppers and inverts flask several times to mix [1].

Part (b) [2 marks]
- M1: Calculates moles of sodium carbonate: \( \frac{1.325}{106} = 0.0125\text{ mol} \) [1].
- M2: Calculates concentration: \( \frac{0.0125}{0.2500} = 0.0500\text{ mol dm}^{-3} \) [1].

Part (c) [3.5 marks]
- (i) M1: Yellow to orange / peach / pink [1].
- (ii) M2: Pipette measures a single fixed volume with high precision/accuracy [1].
- M3: Burette measures variable volumes / allows dropwise delivery [1.5].

Part (d) [3 marks]
- (i) M1: Mean titre = \( 23.15\text{ cm}^3 \) (concordant titres only) [1].
- (ii) M2: Moles of \( \text{HCl} = 2 \times (0.0500 \times 0.0250) = 2.50 \times 10^{-3}\text{ mol} \) [1].
- M3: Concentration of \( \text{HCl} = \frac{2.50 \times 10^{-3}}{0.02315} = 0.108\text{ mol dm}^{-3} \) (3 s.f.) [1].

The rate equation shows that the order of reaction with respect to iodine is zero, as iodine does not appear in the rate equation. Therefore, the rate of the reaction is independent of the concentration of iodine. Option B is incorrect because the overall order of the reaction is 2 (1 + 1 = 2). Option C is incorrect because doubling [CH3COCH3] and halving [H+] results in a net multiplier of 2 * 0.5 = 1, meaning the rate remains unchanged. Option D is incorrect because the unit of the rate constant k for a second-order reaction is \(\text{dm}^3 \text{mol}^{-1} \text{s}^{-1}\).

[1] Correct option is A. Reject B, C, and D.

For a reaction to be spontaneous, the total entropy change must be greater than zero: \(\Delta S_{\text{total}} = \Delta S_{\text{system}} - \frac{\Delta H}{T} > 0\). Substituting the given values (with \(\Delta H\) in J/mol): \(-120 - \frac{-45000}{T} > 0\) which simplifies to \(\frac{45000}{T} > 120\) or \(T < 375\text{ K}\). Therefore, the reaction is spontaneous only at temperatures below \(375\text{ K}\), and it becomes non-spontaneous at temperatures above \(375\text{ K}\).

[1] Correct option is A. Reject B, C, and D.

The value of the equilibrium constant \(K_p\) is only affected by temperature. Since the forward reaction is exothermic, decreasing the temperature shifts the equilibrium position to the right to oppose the change, thereby increasing the value of \(K_p\). Changes in pressure, concentrations of reactants, or the addition of a catalyst do not alter the value of \(K_p\).

[1] Correct option is B. Reject A, C, and D.

Using the weak acid approximation: \([\text{H}^+] \approx \sqrt{K_{\text{a}} \times [\text{HA}]} = \sqrt{1.80 \times 10^{-5} \times 0.050} = \sqrt{9.00 \times 10^{-7}} = 9.49 \times 10^{-4}\text{ mol dm}^{-3}\). The pH is given by \(-\log_{10}[\text{H}^+] = -\log_{10}(9.49 \times 10^{-4}) = 3.02\).

[1] Correct option is B. Reject A, C, and D.

Fehling's solution reacts only with aldehydes to form a red precipitate of \(\text{Cu}_2\text{O}\), ruling out propanone and ethanol. Alkaline aqueous iodine reacts with methyl carbonyls (or compounds oxidised to them) to form a yellow precipitate of triiodomethane. Ethanal (\(\text{CH}_3\text{CHO}\)) contains the methyl carbonyl group and gives a positive test, whereas propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) does not contain this group.

[1] Correct option is B. Reject A, C, and D.

For a first-order reaction, \(\text{Rate} = k[\text{N}_2\text{O}_5]\). When \(60\%\) of the reactant has decomposed, \(40\%\) remains, so the concentration of \(\text{N}_2\text{O}_5\) is \(0.040\text{ mol dm}^{-3}\). This gives \(\text{Rate} = (4.30 \times 10^{-4}\text{ s}^{-1}) \times (0.040\text{ mol dm}^{-3}) = 1.72 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\).

[1] Correct option is A. Reject B, C, and D.

At the boiling point of \(373\text{ K}\), the system is at equilibrium, so \(\Delta S_{\text{total}} = 0\). Therefore, \(\Delta S^{\ominus}_{\text{system}} = -\Delta S_{\text{surroundings}} = \frac{\Delta H^{\ominus}_{\text{vap}}}{T} = \frac{+40.7 \times 10^3\text{ J mol}^{-1}}{373\text{ K}} = +109\text{ J K}^{-1}\text{ mol}^{-1}\).

[1] Correct option is B. Reject A, C, and D.

In the mixture, the amount of propanoic acid is \(0.00500\text{ mol}\) and the amount of propanoate ions is \(0.00250\text{ mol}\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_{\text{a}} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 4.87 + \log_{10}\left(\frac{0.00250}{0.00500}\right) = 4.87 - 0.30 = 4.57\).

[1] Correct option is A. Reject B, C, and D.

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

1 mark: Correct calculation of activation energy including sign and units conversion to kJ/mol.

The standard entropy change of the system is calculated as: \(\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants}) = (312 + 223) - 364 = 535 - 364 = +171\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark: Correct calculation of system entropy change with correct sign.

The partial pressure of each gas is: \(p_{\text{N}_2\text{O}_4} = 0.40 P\) and \(p_{\text{NO}_2} = 0.60 P\). The expression for \(K_p\) is: \(K_p = \frac{(p_{\text{NO}_2})^2}{p_{\text{N}_2\text{O}_4}} = \frac{(0.60 P)^2}{0.40 P} = \frac{0.36 P^2}{0.40 P} = 0.90 P\).

1 mark: Correct substitution of mole fractions and total pressure into the Kp expression.

For a weak acid: \([\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]} = \sqrt{1.80 \times 10^{-5} \times 0.150} = \sqrt{2.70 \times 10^{-6}} = 1.643 \times 10^{-3}\text{ mol dm}^{-3}\). Therefore, \(\text{pH} = -\log_{10}(1.643 \times 10^{-3}) = 2.78\).

1 mark: Correct calculation of pH to 2 decimal places.

Reduction of butanone with \(\text{NaBH}_4\) yields butan-2-ol, which has a chiral carbon atom (carbon-2) bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\)). This leads to optical isomers. The other options (propanone, pentan-3-one, and propanal) yield achiral alcohols upon reduction.

1 mark: Correct identification of the ketone that reduces to a chiral alcohol.

Carboxylic acids react with carbonates to produce carbon dioxide gas. The molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) belongs to a carboxylic acid with 4 carbon atoms, which is butanoic acid. Esters (like ethyl ethanoate and methyl propanoate) and aldehydes (like 3-hydroxybutanal) do not react with sodium carbonate.

1 mark: Correct identification of butanoic acid as the carboxylic acid.

Because iodine (\(\text{I}_2\)) does not appear in the rate equation, the reaction is zero-order with respect to iodine. Therefore, changing the concentration of iodine has no effect on the overall rate of reaction. The rate constant \(k\) increases with temperature, and \(\text{H}^+\) is a homogeneous catalyst because it is dissolved in the same aqueous phase.

1 mark: Correct identification that iodine is zero-order and has no effect on rate.

The moles of propanoic acid (\(\text{HA}\)) and propanoate ions (\(\text{A}^-\)) are: \(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\), and \(n(\text{A}^-) = 0.0500\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\). Using the buffer equation: \([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = (1.35 \times 10^{-5}) \times \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 2.70 \times 10^{-5}\text{ mol dm}^{-3}\). Thus, \(\text{pH} = -\log_{10}(2.70 \times 10^{-5}) = 4.57\).

1 mark: Correct calculation of pH.

In mechanism C, Step 2 is the rate-determining step, so \(\text{Rate} = k_2[X_2][Y]\). Since Step 1 is a fast equilibrium, the equilibrium constant can be written as \(K_c = \frac{[X_2]}{[X]^2}\), which gives \([X_2] = K_c[X]^2\). Substituting this into the rate equation for the slow step gives \(\text{Rate} = k_2 K_c [X]^2 [Y] = k[X]^2[Y]\). This is consistent with the experimental rate equation.

[1 mark] C is correct. Award 1 mark for identifying the mechanism where the pre-equilibrium intermediate substitution yields the experimental second-order dependence on X and first-order dependence on Y.

A reaction becomes non-spontaneous when \(\Delta G^\ominus > 0\). The transition occurs at \(\Delta G^\ominus = 0\). Using \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus = 0\), we rearrange to find \(T = \frac{\Delta H^\ominus}{\Delta S^\ominus}\). Converting the enthalpy change to joules gives \(\Delta H^\ominus = -115000\text{ J mol}^{-1}\). Thus, \(T = \frac{-115000}{-145} = 793.1\text{ K}\). Above this temperature, the entropy term dominates, making \(\Delta G^\ominus\) positive (non-spontaneous).

[1 mark] C is correct. Award 1 mark for correct division of standard enthalpy by standard entropy with appropriate unit conversion to yield approximately 793 K.

First calculate the initial moles of the reactants: moles of propanoic acid = \(0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\); moles of sodium hydroxide added = \(0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00250\text{ mol}\). The acid reacts with the hydroxide ions: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\). After reaction, remaining propanoic acid = \(0.00500 - 0.00250 = 0.00250\text{ mol}\), and propanoate ions formed = \(0.00250\text{ mol}\). Since the concentration of the weak acid equals the concentration of its conjugate base, \([\text{H}^+] = K_a\), and therefore \(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

[1 mark] C is correct. Award 1 mark for recognising that the system is at the half-neutralisation point and calculating pH = pKa.

Nucleophilic addition of cyanide to a carbonyl group forms a hydroxynitrile. For the product to be chiral, the central carbon must be bonded to four different groups. Propanone forms 2-hydroxy-2-methylpropanenitrile (two identical methyl groups - achiral). Butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) forms 2-hydroxy-2-methylbutanenitrile, where the central carbon is bonded to \(-\text{OH}\), \(-\text{CN}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\) (four different groups - chiral). Pentan-3-one forms a product with two ethyl groups (achiral). Methanal forms a product with two hydrogens (achiral).

[1 mark] B is correct. Award 1 mark for identifying butanone as the only carbonyl starting material that yields an asymmetric chiral carbon center upon nucleophilic addition of cyanide.

(a) Comparing Exp 1 and 2: \([\text{I}^-]\) is constant, \([\text{S}_2\text{O}_8^{2-}]\) doubles, and rate doubles. Therefore, the order with respect to \([\text{S}_2\text{O}_8^{2-}]\) is 1. Comparing Exp 1 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant, \([\text{I}^-]\) doubles, and rate doubles. Therefore, the order with respect to \([\text{I}^-]\) is 1. The overall rate equation is: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). (b) Rearranging the rate equation: \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.2 \times 10^{-4}}{0.050 \times 0.040} = 0.060\). Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). (c) Step 1 (slow, rate-determining step): \(\text{S}_2\text{O}_8^{2-} + \text{I}^- \rightarrow \text{SO}_4^{2-} + \text{SO}_4\text{I}^-\). Step 2 (fast): \(\text{SO}_4\text{I}^- + \text{I}^- \rightarrow \text{SO}_4^{2-} + \text{I}_2\). (d) Using the Arrhenius relation: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\). \(\ln\left(\frac{1.2 \times 10^{-1}}{3.0 \times 10^{-2}}\right) = \ln(4) = 1.3863\). \(\left(\frac{1}{298} - \frac{1}{318}\right) = 0.0033557 - 0.0031447 = 0.0002110\text{ K}^{-1}\). \(E_a = -\frac{1.3863 \times 8.31}{-0.0002110} = 54599\text{ J mol}^{-1} = 54.6\text{ kJ mol}^{-1}\).

(a) 1 mark for order 1 wrt peroxodisulfate with explanation. 1 mark for order 1 wrt iodide with explanation. 1 mark for correct deduction process. 1 mark for correct rate equation. (b) 1 mark for rearranging formula. 1 mark for correct value 0.060. 1 mark for correct units. (c) 1 mark for Step 1 reacting one peroxodisulfate and one iodide. 1 mark for identifying Step 1 as slow/RDS. 1 mark for a consistent Step 2 that yields the overall stoichiometry. (d) 1 mark for correct calculation of ln(k2/k1). 1 mark for correct calculation of (1/T1 - 1/T2). 1 mark for rearrangement of Arrhenius formula. 1 mark for final activation energy of 54.6 (accept 54.5 to 55.0) with unit kJ/mol.

(a) \(K_a = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]}\). Assuming \([\text{H}^+] = [\text{C}_2\text{H}_5\text{COO}^-]\) and dissociation is negligible, \([\text{H}^+] = \sqrt{K_a \times [\text{acid}]} = \sqrt{1.35 \times 10^{-5} \times 0.150} = 1.423 \times 10^{-3}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.85\). (b)(i) Initial moles of acid = \(0.0500 \times 0.150 = 7.50 \times 10^{-3}\text{ mol}\). Moles of NaOH added = \(0.0250 \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\). Moles of propanoate ions formed = \(2.50 \times 10^{-3}\text{ mol}\). Moles of remaining propanoic acid = \(7.50 \times 10^{-3} - 2.50 \times 10^{-3} = 5.00 \times 10^{-3}\text{ mol}\). (b)(ii) \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)\). \(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\). \(\text{pH} = 4.87 + \log_{10}\left(\frac{2.50 \times 10^{-3}}{5.00 \times 10^{-3}}\right) = 4.87 - 0.30 = 4.57\). (c) The equivalence point occurs when moles of NaOH equal moles of propanoic acid: \(V_{\text{NaOH}} = \frac{0.150 \times 25.0}{0.100} = 37.5\text{ cm}^3\). Phenolphthalein is selected because the equivalence point pH is basic (>7) and its transition range (8.2 - 10.0) lies entirely within the vertical section of the titration curve.

(a) 1 mark for correct Ka expression. 1 mark for assumption statement or working. 1 mark for correct hydrogen ion concentration. 1 mark for pH = 2.85. (b)(i) 1 mark for calculating initial moles of acid. 1 mark for moles of NaOH. 1 mark for correct equilibrium buffer moles of both acid and salt. (b)(ii) 1 mark for calculating pKa. 1 mark for buffer formula substitution. 1 mark for final pH of 4.57. (c) 1 mark for sketching curve starting at pH ~2.9 with a vertical buffer region. 1 mark for vertical section centered at 37.5 cm3. 1 mark for selecting phenolphthalein. 1 mark for explaining that the vertical pH jump matches phenolphthalein's transition range.

(a) The entropy of one mole of a substance under standard conditions of 100 kPa pressure and a specified temperature (normally 298 K). (b)(i) \(\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants}) = (39.7 + 213.6) - 92.9 = +160.4\text{ J K}^{-1}\text{ mol}^{-1}\). The sign is positive because there is an increase in the disorder of the system due to the production of a gas molecule from a solid reactant. (b)(ii) \(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{\Delta H^{\ominus}}{T} = -\frac{+178 \times 10^3\text{ J mol}^{-1}}{298\text{ K}} = -597.3\text{ J K}^{-1}\text{ mol}^{-1}\). (b)(iii) \(\Delta S^{\ominus}_{\text{total}} = \Delta S^{\ominus}_{\text{system}} + \Delta S^{\ominus}_{\text{surroundings}} = +160.4 - 597.3 = -436.9\text{ J K}^{-1}\text{ mol}^{-1}\). The reaction is not feasible at 298 K because the total entropy change is negative. (c) For the reaction to be feasible, \(\Delta S^{\ominus}_{\text{total}} \ge 0\). Therefore, \(\Delta S^{\ominus}_{\text{system}} - \frac{\Delta H^{\ominus}}{T} \ge 0 \Rightarrow T \ge \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}_{\text{system}}}\). \(T = \frac{178000\text{ J mol}^{-1}}{160.4\text{ J K}^{-1}\text{ mol}^{-1}} = 1109.7\text{ K}\) (or \(1110\text{ K}\)).

(a) 1 mark for 'entropy of one mole'. 1 mark for specifying 'under standard conditions of 100 kPa and specified temperature'. (b)(i) 1 mark for working. 1 mark for correct value (+160.4 J K-1 mol-1). 1 mark for explanation involving solid to gas conversion. (b)(ii) 1 mark for formula. 1 mark for converting kJ to J. 1 mark for correct calculation (-597.3 J K-1 mol-1). (b)(iii) 1 mark for correct addition. 1 mark for value (-436.9 J K-1 mol-1). 1 mark for stating not feasible because total entropy is negative. (c) 1 mark for setting up inequality or equation T = dH/dS. 1 mark for correct division with values. 1 mark for correct temperature of 1110 K (or 1109.7 K).

(a)(i) Reagent: Tollens' reagent (or Fehling's solution). Observation: Propanal forms a silver mirror (or red precipitate), while propanone shows no change. (a)(ii) Propanone has only one hydrogen environment. Its spectrum shows: 1 peak (singlet) due to zero adjacent protons on neighboring carbons; Chemical shift range of 2.0-3.0 ppm (protons adjacent to a carbonyl group); Relative peak area (integration) of 6. (b)(i) \(\text{CH}_3\text{CH}_2\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH(OH)CN}\). Catalyst: \(\text{KCN}\) (or trace alkali, \(\text{OH}^-\)). (b)(ii) Mechanism: 1. Curly arrow from lone pair on carbon of \(:\text{CN}^-\) to the carbonyl carbon. 2. Dipole shown on carbonyl group (C is \(\delta^+\), O is \(\delta^-\)) and curly arrow from \(\text{C}=\text{O}\) double bond to the oxygen atom. 3. Structure of intermediate \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\) drawn correctly with charge on oxygen. 4. Curly arrow from lone pair on intermediate \(\text{O}^-\) to a hydrogen ion (or to H of HCN) to form the final product. (b)(iii) The carbonyl carbon of propanal has a planar geometry. The nucleophile (cyanide ion) is equally likely to attack the planar carbonyl carbon from either above or below the plane. This produces equal amounts (a 50:50 ratio) of both enantiomers, whose optical rotations cancel each other out, resulting in no overall optical activity.

(a)(i) 1 mark for reagent (Tollens' or Fehling's). 1 mark for correct observation with propanal vs no reaction for propanone. (a)(ii) 1 mark for stating singlet. 1 mark for correct chemical shift range (2.0-3.0 ppm). 1 mark for relative area of 6. (b)(i) 1 mark for correct equation. 1 mark for catalyst (KCN/NaCN/alkali). (b)(ii) 1 mark for nucleophilic attack curly arrow from CN- to C. 1 mark for dipole and carbonyl pi-bond arrow. 1 mark for intermediate structure. 1 mark for protonation of intermediate to form OH. (b)(iii) 1 mark for stating that the carbonyl carbon is planar. 1 mark for equal probability of attack from either side. 1 mark for stating equal amounts of enantiomers formed which cancel out optical activity.

(a) \(K_p = \frac{p(\text{PCl}_3) \times p(\text{Cl}_2)}{p(\text{PCl}_5)}\). (b)(i) Dissociation percentage = 40.0%. Moles of PCl5 reacted = \(1.50 \times 0.400 = 0.600\text{ mol}\). Moles at equilibrium: \(\text{PCl}_5 = 1.50 - 0.600 = 0.900\text{ mol}\); \(\text{PCl}_3 = 0.600\text{ mol}\); \(\text{Cl}_2 = 0.600\text{ mol}\). (b)(ii) Total equilibrium moles = \(0.900 + 0.600 + 0.600 = 2.100\text{ mol}\). Mole fractions (\(\chi\)): \(\chi(\text{PCl}_5) = \frac{0.900}{2.100} = 0.429\); \(\chi(\text{PCl}_3) = \frac{0.600}{2.100} = 0.286\); \(\chi(\text{Cl}_2) = \frac{0.600}{2.100} = 0.286\). (b)(iii) Partial pressure (\(p = \chi \times P_{\text{total}}\)): \(p(\text{PCl}_5) = 0.4286 \times 1.80 = 0.771\text{ atm}\); \(p(\text{PCl}_3) = 0.2857 \times 1.80 = 0.514\text{ atm}\); \(p(\text{Cl}_2) = 0.2857 \times 1.80 = 0.514\text{ atm}\). (b)(iv) \(K_p = \frac{0.514 \times 0.514}{0.771} = 0.343\). Units: \(\text{atm}\). (c) No change because \(K_p\) is only affected by changes in temperature.

(a) 1 mark for correct Kp expression using partial pressures (p). (b)(i) 1 mark for moles of PCl5 reacted. 1 mark for equilibrium moles of PCl5 (0.90). 1 mark for equilibrium moles of PCl3 and Cl2 (0.60 each). (b)(ii) 1 mark for total moles calculation. 1 mark for mole fraction of PCl5. 1 mark for mole fractions of products. (b)(iii) 1 mark for correct partial pressure of PCl5. 1 mark for partial pressure of PCl3. 1 mark for partial pressure of Cl2. (b)(iv) 1 mark for correct Kp substitution. 1 mark for value (accept 0.340-0.345 depending on rounding). 1 mark for correct unit (atm). (c) 1 mark for stating no change because only temperature changes Kp.

The relationship between standard cell potential and standard total entropy change is given by the equation:

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

Substitute the given values into the equation:

\(\Delta S^\ominus_{\text{total}} = \frac{2 \times 96500 \text{ C mol}^{-1} \times 0.45 \text{ V}}{298 \text{ K}}\)

\(\Delta S^\ominus_{\text{total}} = \frac{86850}{298} = +291.44 \text{ J K}^{-1} \text{ mol}^{-1}\)

To 3 significant figures, this is \(+291 \text{ J K}^{-1} \text{ mol}^{-1}\).

- [1] for calculating \(\Delta S^\ominus_{\text{total}} = +291 \text{ J K}^{-1} \text{ mol}^{-1}\) using the correct formula and values.
- Reject other values due to incorrect use of \(n\) or incorrect signs.

The electronic configurations of the transition metal ions are:
- \(\text{Cr}^{3+}\): \([\text{Ar}] 3\text{d}^3\) (3 unpaired electrons in d-orbitals)
- \(\text{Fe}^{3+}\): \([\text{Ar}] 3\text{d}^5\) (5 unpaired electrons in d-orbitals)
- \(\text{Co}^{2+}\): \([\text{Ar}] 3\text{d}^7\) (3 unpaired electrons in d-orbitals)
- \(\text{Cu}^{2+}\): \([\text{Ar}] 3\text{d}^9\) (1 unpaired electron in d-orbital)

Therefore, \(\text{Fe}^{3+}\) has the highest number of unpaired d-electrons (5).

- [1] for correctly identifying \(\text{Fe}^{3+}\) as having 5 unpaired d-electrons.

The methyl group (\(-\text{CH}_3\)) on the benzene ring is electron-donating via a positive inductive effect. This activates the ring and directs substitution to the 2- and 4-positions (ortho and para positions). Under mild nitration conditions (around \(50\ ^\circ\text{C}\)), mono-nitration occurs, producing a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene.

- [1] for identifying that methylbenzene undergoes mono-nitration at \(50\ ^\circ\text{C}\) to yield mainly a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene.

The pH of the solution is determined by the concentration of hydroxide ions produced. The strength of the base depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton:
- Ethylamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)) is an aliphatic amine. The ethyl group is electron-donating, which increases the electron density on the nitrogen atom, making its lone pair more available. Hence, it is the strongest base and has the highest pH.
- Ammonia (\(\text{NH}_3\)) has no electron-donating alkyl groups and is a weaker base than ethylamine.
- Phenylamine (\(\text{C}_6\text{H}_5\text{NH}_2\)) is a weaker base because the lone pair on the nitrogen atom is delocalized into the \(\pi\)-system of the benzene ring.
- Ethanamide (\(\text{CH}_3\text{CONH}_2\)) is neutral/very weakly basic due to the strongly electron-withdrawing carbonyl group adjacent to the nitrogen.

- [1] for selecting \(\text{CH}_3\text{CH}_2\text{NH}_2\) as the strongest base / highest pH.

In the reactant complex, \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), there are six monodentate water ligands, so the coordination number is 6 (geometry is octahedral).
In the product complex, \([\text{CoCl}_4]^{2-}\), there are four chloride ligands. Since chloride ligands are larger and negatively charged, only four can fit around the cobalt ion. The coordination number is 4 and the geometry is tetrahedral.
Therefore, the correct description is coordination number 6 for the reactant, coordination number 4 for the product, and tetrahedral geometry for the product.

- [1] for selecting the option with reactant coordination number 6, product coordination number 4, and tetrahedral product geometry.

Primary aliphatic amines react with nitrous acid (\(\text{HNO}_2\)) at room temperature to form unstable diazonium salts, which rapidly decompose to release nitrogen gas (\(\text{N}_2\)) and form alcohols.
Let's find the structural isomers of amines with molecular formula \(\text{C}_3\text{H}_9\text{N}\):
1. Propan-1-amine: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2\) (Primary amine) - reacts to evolve \(\text{N}_2\) gas.
2. Propan-2-amine: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{CH}_3\) (Primary amine) - reacts to evolve \(\text{N}_2\) gas.
3. N-methylethanamine: \(\text{CH}_3\text{NHCH}_2\text{CH}_3\) (Secondary amine) - reacts to form a yellow oily nitrosamine, no gas evolved.
4. Trimethylamine: \((\text{CH}_3)_3\text{N}\) (Tertiary amine) - forms a soluble salt, no gas evolved.

Thus, exactly 2 isomers of formula \(\text{C}_3\text{H}_9\text{N}\) react with nitrous acid to evolve gas.

- [1] for identifying that only primary amines react with nitrous acid to produce gas, and that there are exactly 2 primary amine structural isomers of formula \(\text{C}_3\text{H}_9\text{N}\).

Let's look at the standard electrode potentials:
- Left half-cell: \(\text{Fe}^{3+}(\text{aq}) + e^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\ominus = +0.77\text{ V}\)
- Right half-cell: \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+ + 5e^- \rightleftharpoons \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O(l)} \quad E^\ominus = +1.51\text{ V}\)

Since the right-hand electrode has a more positive standard potential, reduction occurs in the right-hand half-cell (cathode):
\(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \to \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O(l)}\)

This process consumes \(\text{H}^+\) ions, reducing their concentration, which increases the pH.
- Left half-cell undergoes oxidation (anode): \(\text{Fe}^{2+}(\text{aq}) \to \text{Fe}^{3+}(\text{aq}) + e^-\). Thus, iron(II) is oxidized, not reduced.
- The platinum electrodes are inert and their mass does not change.
- Electrons flow in the external circuit from the negative electrode (anode, left) to the positive electrode (cathode, right).

- [1] for identifying that the right-hand half-cell consumes \(\text{H}^+\) ions, leading to an increase in pH.

A transition metal species is colored because of d-d transitions. This requires a partially filled d-subshell.
- \(\text{Sc}^{3+}\) has the electronic configuration \([\text{Ar}] 3\text{d}^0\). Because there are no d-electrons, d-d transition cannot occur. Therefore, \([\text{Sc}(\text{H}_2\text{O})_6]^{3+}\) is colorless.
- \(\text{V}^{3+}\) (\(3\text{d}^2\)), \(\text{Cr}^{3+}\) (\(3\text{d}^3\)), and \(\text{Fe}^{3+}\) (\(3\text{d}^5\)) all have partially filled d-orbitals and form colored aqueous complexes.

- [1] for identifying \([\text{Sc}(\text{H}_2\text{O})_6]^{3+}\) as the colorless complex due to its \(3\text{d}^0\) configuration.

In this reaction, \( \text{Ce}^{4+} \) is reduced to \( \text{Ce}^{3+} \) (reduction half-equation has \( E^\ominus = +1.70\text{ V} \)), and \( \text{AsO}_3^{3-} \) is oxidized to \( \text{AsO}_4^{3-} \) (oxidation half-equation has \( E^\ominus = +0.56\text{ V} \)). The standard cell potential is calculated using: \( E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +1.70 - (+0.56) = +1.14\text{ V} \). Note that multiplying the stoichiometry of the half-equations to balance the electrons does not affect the standard electrode potential values.

1 mark for the correct calculation: \( E^\ominus_{\text{cell}} = +1.14\text{ V} \).

Aqueous cobalt(II) sulfate contains the pink, octahedral hexaaquacobalt(II) ion, \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \). When excess concentrated hydrochloric acid is added, a ligand substitution reaction occurs to form the blue, tetrahedral tetrachlorocobaltate(II) ion, \( [\text{CoCl}_4]^{2-} \).

1 mark for selecting the correct option showing 'pink to blue' and 'octahedral to tetrahedral'.

Concentrated sulfuric acid acts as a catalyst in this reaction. It protonates nitric acid, which subsequently loses water to generate the electrophile, the nitronium ion (\( \text{NO}_2^+ \)). It is regenerated at the end of the mechanism.

1 mark for identifying sulfuric acid as an acid catalyst to help generate the electrophile.

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

1 mark for identifying the correct order: phenylamine < ammonia < ethylamine.

The reaction is autocatalyzed by the manganese(II) ions, \( \text{Mn}^{2+} \), which are produced as a product. Initially, the reaction is very slow because it involves the collision of two negatively charged ions. Once some \( \text{Mn}^{2+} \) forms, it acts as an intermediate catalyst via temporary changes in its oxidation state.

1 mark for identifying \( \text{Mn}^{2+}(aq) \) as the autocatalyst.

For recrystallization to succeed, the desired solid product must be highly soluble in the chosen hot solvent but very poorly soluble in the cold solvent. This ensures the maximum yield of crystals upon cooling while leaving soluble impurities behind in the cold filtrate.

1 mark for identifying the solubility temperature-dependency requirement of the ideal recrystallization solvent.

An oxidation reaction is thermodynamically feasible if \( E^\ominus_{\text{cell}} > 0 \). For \( \text{Fe}^{3+} \) oxidizing \( \text{I}^- \): \( E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation}) = 0.77 - 0.54 = +0.23\text{ V} \) (feasible). For \( \text{Fe}^{3+} \) oxidizing \( \text{Br}^- \): \( E^\ominus_{\text{cell}} = 0.77 - 1.09 = -0.32\text{ V} \) (not feasible). Thus, \( \text{Fe}^{3+}(aq) \) can oxidize \( \text{I}^-(aq) \) but cannot oxidize \( \text{Br}^-(aq) \).

1 mark for identifying that \( \text{Fe}^{3+} \) can oxidize \( \text{I}^- \) but not \( \text{Br}^- \).

The lone pair of electrons in a p-orbital on the oxygen atom of the hydroxyl (-OH) group is partially delocalized into the benzene ring's pi-system. This increases the electron density of the ring, making it much more effective at polarizing and attracting electrophiles like bromine without needing a halogen carrier catalyst.

1 mark for identifying that the oxygen lone pair delocalizes into the ring, increasing its electron density.

To find the standard cell potential, subtract the electrode potential of the anode (oxidation) from the cathode (reduction): \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +1.51\text{ V} - (+1.36\text{ V}) = +0.15\text{ V}\). Since the \(\text{MnO}_4^- / \text{Mn}^{2+}\) half-cell has the more positive electrode potential, \(\text{MnO}_4^-\) is reduced (acting as the oxidizing agent), and \(\text{Cl}^-\) is oxidized to \(\text{Cl}_2\) (acting as the reducing agent).

1 mark for correct selection of option A. Correctly calculated cell potential of +0.15 V and identified Cl-(aq) as the species that undergoes oxidation.

The addition of concentrated hydrochloric acid to hexaaquacobalt(II) ions leads to the formation of the tetrachlorocobaltate(II) complex: \([\text{Co(H}_2\text{O)}_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CoCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)\). Because the chloride ligands are large and charged, only four can coordinate around the central cobalt(II) ion, resulting in a tetrahedral geometry. This complex has a distinctive blue color.

1 mark for selecting A. Correctly identified both the tetrahedral shape and the blue color of the tetrachlorocobaltate(II) ion.

Phenol reacts rapidly with bromine water at room temperature without requiring a catalyst (like \(\text{FeBr}_3\)) to form a white precipitate of 2,4,6-tribromophenol. This high reactivity is due to the lone pair of electrons on the oxygen atom of the \(\text{-OH}\) group being partially delocalized into the aromatic \(\pi\)-system, which increases the electron density of the benzene ring and makes it highly susceptible to electrophilic attack.

1 mark for selecting option D. Benzene, chlorobenzene, and methylbenzene do not react rapidly with bromine water without a catalyst and do not form a white precipitate.

Basic strength depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton. In phenylamine, the lone pair is delocalized into the benzene \(\pi\)-system, making it the least available (weakest base). In ethylamine, the alkyl group has an electron-releasing (+I inductive) effect, which increases the electron density on the nitrogen, making its lone pair more available than in ammonia. Therefore, the order is: phenylamine < ammonia < ethylamine.

1 mark for selecting option A. Identifying the correct order based on lone pair availability and inductive/delocalization effects.

(a) Standard electrode potential is the electromotive force (emf) of a half-cell compared with a standard hydrogen electrode under standard conditions of \(298\text{ K}\), \(100\text{ kPa}\) gas pressure, and ion concentrations of \(1.00\text{ mol dm}^{-3}\).

(b) The diagram must show:
1. Standard hydrogen electrode containing a platinum electrode, hydrogen gas entering at \(100\text{ kPa}\), and \(1.00\text{ mol dm}^{-3}\ \text{H}^+(\text{aq})\) solution.
2. \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell containing a platinum electrode immersed in a solution containing both \(\text{Fe}^{3+}(\text{aq})\) and \(\text{Fe}^{2+}(\text{aq})\) ions, both at \(1.00\text{ mol dm}^{-3}\).
3. High-resistance voltmeter connecting the two platinum electrodes.
4. Salt bridge (e.g., filter paper soaked in saturated \(\text{KNO}_3\)) connecting the two solutions.

(c)(i) \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +1.51 - (+0.77) = +0.74\text{ V}\).

(c)(ii) \(\text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\).

(d) Moles of \(\text{MnO}_4^-\text{ used} = \frac{24.25}{1000} \times 0.0200 = 4.85 \times 10^{-4}\text{ mol}\).
Using the balanced equation, \(\text{moles of Fe}^{2+}\text{ in } 25.0\text{ cm}^3 = 5 \times 4.85 \times 10^{-4} = 2.425 \times 10^{-3}\text{ mol}\).
\(\text{Moles of Fe}^{2+}\text{ in } 250\text{ cm}^3 = 10 \times 2.425 \times 10^{-3} = 2.425 \times 10^{-2}\text{ mol}\).
\(\text{Mass of iron in wire} = 2.425 \times 10^{-2} \times 55.8 = 1.35315\text{ g}\).
\(\text{Percentage by mass of iron} = \left(\frac{1.35315}{1.45}\right) \times 100 = 93.32\% \approx 93.3\%\).

(a) 1 mark for emf of a cell containing the half-cell of interest connected to a standard hydrogen electrode. 1 mark for specifying standard conditions: 298 K, 100 kPa / 1 atm pressure, and 1.00 mol dm^-3 concentration of ions.

(b) 1 mark for a standard hydrogen electrode (labeled Pt, H2 gas at 100 kPa / 1 atm, 1 mol dm^-3 H+(aq)). 1 mark for the Fe3+/Fe2+ half-cell (labeled Pt, containing both Fe2+ and Fe3+ at 1 mol dm^-3). 1 mark for high-resistance voltmeter connecting the circuit. 1 mark for a salt bridge connecting the solutions.

(c)(i) 1 mark for +0.74 V.

(c)(ii) 1 mark for correct species, 1 mark for correct balancing.

(d) 1 mark for calculating moles of MnO4- (4.85 x 10^-4 mol). 1 mark for multiplying by 5 to find moles of Fe2+ in 25.0 cm3 (2.425 x 10^-3 mol). 1 mark for multiplying by 10 to find total moles in 250 cm3 (2.425 x 10^-2 mol). 1 mark for calculating mass of iron (1.353 g). 1 mark for calculating percentage to 3 sig figs (93.3%).

(a) Cyclohexene has one double bond and an enthalpy of hydrogenation of \(-120\text{ kJ mol}^{-1}\). If benzene were cyclohexa-1,3,5-triene (with three independent double bonds), the expected enthalpy of hydrogenation would be \(3 \times (-120) = -360\text{ kJ mol}^{-1}\). The experimental enthalpy of hydrogenation of benzene is \(-208\text{ kJ mol}^{-1}\), which is \(152\text{ kJ mol}^{-1}\) less exothermic than expected. This indicates that benzene is much more stable than cyclohexa-1,3,5-triene. This extra stability is due to the delocalisation of the \(\pi\)-electrons across the ring.

(b)(i) \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\) (or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\)).

(b)(ii)
1. A curly arrow from the benzene ring of methylbenzene (at C-4) to the nitrogen atom of \(\text{NO}_2^+\).
2. Correct structure of the intermediate with a positive charge in the partially broken ring (horseshoe open towards C-4, covering 5 carbon atoms) and C-4 showing both H and \(\text{NO}_2\).
3. A curly arrow from the C-H bond on C-4 back into the ring to restore aromaticity.
4. Correct products: 4-nitromethylbenzene and \(\text{H}^+\).

(c)
Step 1: Oxidation of the methyl group to a carboxylic acid group.
- Reagents: Acidified potassium dichromate(VI) or alkaline potassium manganate(VII) followed by acid.
- Conditions: Heat under reflux.
Step 2: Reduction of the nitro group to an amine group.
- Reagents: Tin (Sn) and concentrated hydrochloric acid.
- Conditions: Heat under reflux, followed by the addition of aqueous sodium hydroxide to free the amine.

(a) 1 mark for calculating expected enthalpy of -360 kJ mol^-1. 1 mark for stating benzene is 152 kJ mol^-1 more stable (less exothermic). 1 mark for attributing this stability to delocalisation of pi electrons. 1 mark for referencing the relative values clearly.

(b)(i) 1 mark for correct equation showing electrophile generation.

(b)(ii) 1 mark for curly arrow from ring to NO2+ (attacking C-4). 1 mark for correct intermediate with positive charge inside horseshoe. 1 mark for curly arrow from C-H bond into ring. 1 mark for correct final products.

(c) 1 mark for oxidising agent (e.g., K2Cr2O7/H+ or KMnO4). 1 mark for reflux condition. 1 mark for Sn and conc. HCl. 1 mark for heating/refluxing. 1 mark for adding NaOH(aq) to release the free amine.

(a) Transition metals have partially filled d-orbitals. When ligands approach, the d-orbitals split into two sets of different energy levels. Electrons absorb a specific frequency of visible light and are promoted from the lower energy d-orbital to the higher energy d-orbital (d-d transition). The colour observed is the complementary colour to the absorbed light. Zinc ions (\(\text{Zn}^{2+}\)) have a completely full d-subshell (\(3\text{d}^{10}\)), so there are no empty d-orbitals available for promotion, hence no d-d transition can occur, and zinc complexes are colourless.

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

(b)(ii) The shape changes from octahedral (coordination number 6) to tetrahedral (coordination number 4). The coordination number decreases because chloride ligands are larger than water ligands, so fewer chloride ligands can fit around the central copper(II) ion due to steric hindrance and increased electrostatic repulsion.

(c) Use the calibration equation:
\(0.420 = 15.6 \times [\text{Cu}^{2+}(\text{aq})] + 0.015\)
\(15.6 \times [\text{Cu}^{2+}(\text{aq})] = 0.405\)
\([\text{Cu}^{2+}(\text{aq})] = \frac{0.405}{15.6} = 0.02596\text{ mol dm}^{-3}\)
\(\text{Moles of Cu}^{2+}\text{ in } 250\text{ cm}^3 = 0.02596 \times 0.250 = 6.490 \times 10^{-3}\text{ mol}\).
Since \(1\text{ mol of CuSO}_4 \cdot 5\text{H}_2\text{O}\) yields \(1\text{ mol of Cu}^{2+}\):
\(\text{Mass of CuSO}_4 \cdot 5\text{H}_2\text{O} = 6.490 \times 10^{-3} \times 249.6 = 1.62\text{ g}\) (to 3 significant figures).

(a) 1 mark for d-orbitals splitting. 1 mark for electron promotion from lower to higher d-orbitals (d-d transition). 1 mark for absorption of visible light. 1 mark for observed colour being complementary. 1 mark for zinc having full d-subshell (3d^10) preventing promotion.

(b)(i) 1 mark for correct species. 1 mark for correct balancing.

(b)(ii) 1 mark for octahedral to tetrahedral. 1 mark for noting chloride ligand is larger than water. 1 mark for steric hindrance / repulsion limiting coordination to 4.

(c) 1 mark for finding concentration of Cu2+ (0.02596 mol dm^-3). 1 mark for finding moles in 250 cm3 (6.490 x 10^-3 mol). 1 mark for multiplying by molar mass of 249.6. 1 mark for final answer of 1.62 g.

(a) A base is a proton acceptor; its strength depends on the availability of the lone pair of electrons on the nitrogen atom. In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the \(\pi\)-system of the benzene ring. This reduces the electron density on the nitrogen, making the lone pair less available to accept a proton. In ethylamine, the alkyl (ethyl) group is electron-releasing (+I inductive effect), which increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton.

(b)(i) Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl), followed by the addition of aqueous sodium hydroxide.
Conditions: Heat under reflux.

(b)(ii) \(\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}\).

(c)(i) A zwitterion is an overall neutral molecule containing both positively charged and negatively charged functional groups.

(c)(ii) The structure of the zwitterion of alanine is \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\).

(c)(iii)
In excess acid: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\).
In excess alkali: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).

(c)(iv)
First dipeptide (Ala-Gly): \(\text{H}_2\text{N-CH}(\text{CH}_3)\text{-CONH-CH}_2\text{-COOH}\).
Second dipeptide (Gly-Ala): \(\text{H}_2\text{N-CH}_2\text{-CONH-CH}(\text{CH}_3)\text{-COOH}\).

(a) 1 mark for comparing nitrogen lone pair availability to accept a proton. 1 mark for delocalisation of nitrogen lone pair in phenylamine. 1 mark for ethyl group having +I inductive effect. 1 mark for resulting higher base strength of ethylamine.

(b)(i) 1 mark for Sn and concentrated HCl (reflux). 1 mark for adding NaOH.

(b)(ii) 1 mark for balanced equation with 6[H].

(c)(i) 1 mark for dipolar species / positive and negative charge. 1 mark for net neutral overall charge.

(c)(ii) 1 mark for structure with correct charge positions.

(c)(iii) 1 mark for fully protonated cation in acid. 1 mark for fully deprotonated anion in base.

(c)(iv) 1 mark for each correct dipeptide structure showing the -CONH- link clearly.

(a)
\(\text{Molar mass of AgCl} = 107.9 + 35.5 = 143.4\text{ g mol}^{-1}\).
\(\text{Moles of AgCl} = \frac{2.87}{143.4} = 0.02001\text{ mol}\).
Since \(1\text{ mol of NiCl}_2 \cdot x\text{H}_2\text{O}\) contains \(2\text{ moles of Cl}^-\):
\(\text{Moles of NiCl}_2 \cdot x\text{H}_2\text{O} = \frac{0.02001}{2} = 0.01000\text{ mol}\).
\(\text{Molar mass of NiCl}_2 \cdot x\text{H}_2\text{O} = \frac{2.38}{0.01000} = 238\text{ g mol}^{-1}\).
\(\text{Molar mass of anhydrous NiCl}_2 = 58.7 + 2(35.5) = 129.7\text{ g mol}^{-1}\).
\(\text{Mass of } x\text{H}_2\text{O} = 238 - 129.7 = 108.3\text{ g mol}^{-1}\).
\(x = \frac{108.3}{18.0} = 6.02 \approx 6\).

(b)(i) A bidentate ligand is a molecule or ion that donates two lone pairs of electrons to a central metal ion to form two dative covalent bonds.

(b)(ii) The coordination number of nickel is 6. This complex ion exhibits optical isomerism (enantiomerism) because it has non-superimposable mirror images. The structures of the two optical isomers should show three 'en' rings forming bidentate loops around an octahedral nickel center.

(c)(i) \(\text{Ni}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Ni}(\text{OH})_2(\text{s})\).

(c)(ii) \(\text{Ni}(\text{OH})_2(\text{s}) + 6\text{NH}_3(\text{aq}) \rightarrow [\text{Ni}(\text{NH}_3)_6]^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})\).

(a) 1 mark for molar mass of AgCl and calculating moles of AgCl (0.0200 mol). 1 mark for dividing by 2 to find moles of hydrated nickel salt (0.0100 mol). 1 mark for finding total molar mass (238 g mol^-1). 1 mark for subtracting molar mass of anhydrous NiCl2 (129.7 g mol^-1) to find mass of water (108.3 g mol^-1). 1 mark for dividing by 18 to find x = 6.

(b)(i) 1 mark for donating two lone pairs. 1 mark for forming two dative covalent bonds to the same metal ion.

(b)(ii) 1 mark for coordination number 6. 1 mark for stating it exhibits optical isomerism / non-superimposable mirror images. 2 marks for drawing the correct mirror images (octahedral layout with 3 bidentate loops).

(c)(i) 1 mark for correct balanced equation with state symbols.

(c)(ii) 1 mark for correct reactants and products (forming hexamminenickel(II) complex). 1 mark for correct balancing.

In (a)(i), propanoic acid reacts with sodium hydrogencarbonate to produce carbon dioxide gas, observed as effervescence. In (a)(ii), the ionic equation is \(\text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{HCO}_3^-\text{(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\). In (b)(i), 2,4-DNPH reacts with propanone to form an orange precipitate. In (b)(ii), the precipitate can be filtered, recrystallized, its melting temperature determined, and compared to literature values. In (c), propan-2-ol can be oxidized by warming with acidified potassium dichromate(VI), which turns from orange to green. In (d)(i), W is propanone. In (d)(ii), the yellow precipitate is triiodomethane, \(\text{CHI}_3\).

(a)(i) Effervescence / fizzing / bubbles of gas (1 mark). (a)(ii) \(\text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{HCO}_3^-\text{(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) (2 marks: 1 mark for correct species, 1 mark for state symbols). (b)(i) Orange/yellow/red precipitate/solid (1 mark). (b)(ii) Recrystallize/purify the solid (1 mark) and measure its melting point/temperature and compare with literature values (1 mark). (c) Acidified potassium dichromate(VI) (1 mark) and orange to green observation (1 mark). (d)(i) Propanone (1 mark). (d)(ii) \(\text{CHI}_3\) (1 mark).

In (a), the titration endpoint is colorless to permanent pale pink. In (b), sulfuric acid is used because hydrochloric acid contains chloride ions which would be oxidized to chlorine by manganate(VII). In (c), \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\). In (d), moles of \(\text{MnO}_4^-\) = \(0.0200 \times (20.00 / 1000) = 4.00 \times 10^{-4}\text{ mol}\). Moles of \(\text{Fe}^{2+}\) in \(25.00\text{ cm}^3\) = \(5 \times 4.00 \times 10^{-4} = 2.00 \times 10^{-3}\text{ mol}\). Moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\) = \(2.00 \times 10^{-2}\text{ mol}\). Molar mass of the hydrated salt = \(7.84 / (2.00 \times 10^{-2}) = 392\text{ g mol}^{-1}\). Molar mass of anhydrous \((\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 = 284\text{ g mol}^{-1}\). Mass of water = \(392 - 284 = 108\text{ g mol}^{-1}\). \(x = 108 / 18 = 6\).

(a) Colorless to permanent pale pink (1 mark). (b) Chloride ions from HCl would be oxidized to chlorine gas (1 mark) by the manganate(VII) ions, giving an inaccurately high titre (1 mark). (c) \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\) (2 marks, 1 mark for reactants and products, 1 mark for balancing). (d) Moles of \(\text{MnO}_4^-\) = \(4.00 \times 10^{-4}\text{ mol}\) (1 mark). Moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\) = \(2.00 \times 10^{-2}\text{ mol}\) (1 mark). Molar mass of hydrated salt = \(392\text{ g mol}^{-1}\) (1 mark). Molar mass of anhydrous salt = \(284.0\text{ g mol}^{-1}\) (1 mark). Calculation of \(x = 6\) (1 mark).

In (a), cyclohexanol and cyclohexene can both vaporize; the fractionating column allows cyclohexanol (higher b.p.) to condense and return to the flask, while cyclohexene (lower b.p.) passes to the condenser. In (b)(i), transfer the mixture to a separating funnel, allow layers to separate, and run off the lower layer. In (b)(ii), cyclohexene is on top because its density is lower than water. In (c)(i), anhydrous calcium chloride or magnesium sulfate can be used. In (c)(ii), the liquid turns from cloudy to clear. In (d), moles of cyclohexanol = \(10.0 / 100.2 = 0.0998\text{ mol}\). Theoretical mass of cyclohexene = \(0.0998 \times 82.1 = 8.19\text{ g}\). Percentage yield = \((5.10 / 8.19) \times 100\% = 62.3\%\).

(a) Cyclohexanol and cyclohexene both vaporize (1 mark); the column allows cyclohexanol to condense and return to the flask while cyclohexene passes into the condenser (1 mark). (b)(i) Use a separating funnel (1 mark) and run off the lower layer / decant upper layer (1 mark). (b)(ii) Cyclohexene on top because it has a lower density than water (1 mark). (c)(i) Anhydrous calcium chloride / anhydrous sodium sulfate / anhydrous magnesium sulfate (1 mark). (c)(ii) Liquid turns from cloudy to clear (1 mark). (d) Moles of cyclohexanol = \(0.0998\text{ mol}\) (1 mark). Theoretical mass of cyclohexene = \(8.19\text{ g}\) (1 mark). Percentage yield = \(62.3\%\) (or \(62\%\)) (1 mark).

In (a)(i), the green precipitate insoluble in excess NaOH indicates the presence of \(\text{Fe}^{2+}\) ions. In (a)(ii), the green iron(II) hydroxide is oxidized by oxygen in the air to brown iron(III) hydroxide. In (b)(i), the gas is ammonia, \(\text{NH}_3\). In (b)(ii), the second cation is ammonium, \(\text{NH}_4^+\). In (c)(i), the anion is sulfate, \(\text{SO}_4^{2-}\). In (c)(ii), \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). In (d)(i), a green precipitate forms and then dissolves to give a dark green solution. In (d)(ii), the species is \([\text{Cr(OH)}_6]^{3-}\).

(a)(i) Iron(II) / \(\text{Fe}^{2+}\) (1 mark). (a)(ii) Iron(II) hydroxide is oxidized by atmospheric oxygen to iron(III) hydroxide (1 mark). (b)(i) Ammonia / \(\text{NH}_3\) (1 mark). (b)(ii) \(\text{NH}_4^+\) (1 mark). (c)(i) Sulfate / \(\text{SO}_4^{2-}\) (1 mark). (c)(ii) \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\) (2 marks: 1 mark for species, 1 mark for state symbols). (d)(i) Green precipitate dissolves to form a green solution (1 mark). (d)(ii) \([\text{Cr(OH)}_6]^{3-}\) (or \([\text{Cr(H}_2\text{O)}_2\text{(OH)}_4]^-\)) (2 marks: 1 mark for correct formula, 1 mark for charge).

In (a), extrapolation accounts for heat loss to the surroundings during mixing and ensures the theoretical maximum temperature rise if the reaction had occurred instantaneously is determined. In (b), excess sodium hydroxide ensures that all of the hydrochloric acid is fully neutralized. In (c), mass of solution = \(100.0\text{ g}\). Heat energy \(q = 100.0 \times 4.18 \times 13.5 = 5643\text{ J}\). In (d), moles of \(\text{HCl}\) = \(2.00 \times 0.0500 = 0.100\text{ mol}\). Moles of \(\text{NaOH}\) = \(2.05 \times 0.0500 = 0.1025\text{ mol}\). \(\text{HCl}\) is the limiting reactant, so moles of water formed = \(0.100\text{ mol}\). \(\Delta_n H = - (5.643\text{ kJ}) / 0.100\text{ mol} = -56.4\text{ kJ mol}^{-1}\). In (e), other sources of uncertainty include temperature measurements with the thermometer or measuring solution volumes with a measuring cylinder.

(a) To account for heat loss to the surroundings (1 mark) and find the temperature change if the reaction had been instantaneous (1 mark). (b) To ensure all hydrochloric acid is fully neutralized / acid is the limiting reagent (1 mark). (c) Mass of solution = \(100.0\text{ g}\) (1 mark). \(q = 100.0 \times 4.18 \times 13.5 = 5643\text{ J}\) (accept \(5640\text{ J}\)) (1 mark). (d) Moles of \(\text{HCl}\) reacting = \(0.100\text{ mol}\) (1 mark). Division of heat (in kJ) by moles (1 mark). Negative sign (1 mark). Enthalpy value of \(-56.4\text{ kJ mol}^{-1}\) (1 mark). (e) Uncertainty in volume from measuring cylinder / uncertainty in temperature from thermometer / assumption that the specific heat capacity is that of pure water (1 mark).