2023 Edexcel IAL Chemistry (YCH11) 模擬試題連答案詳解

1. Write the balanced equation: \(\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\). 2. Calculate the number of moles of \(\text{HCl}\): \(n(\text{HCl}) = 0.200\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0100\text{ mol}\). 3. Use the stoichiometric ratio to find moles of \(\text{H}_2\): \(n(\text{H}_2) = \frac{0.0100}{2} = 0.00500\text{ mol}\). 4. Calculate the volume of \(\text{H}_2\): \(V = 0.00500\text{ mol} \times 24000\text{ cm}^3\text{ mol}^{-1} = 120\text{ cm}^3\).

[1 mark] A: Correctly identifies the volume of hydrogen as 120 cm3. Reject other volumes based on incorrect stoichiometry or conversion errors.

1. Find the molar ratio of the elements by dividing each percentage by its relative atomic mass: \(n(\text{C}) = 54.5 / 12.0 = 4.54\), \(n(\text{H}) = 9.1 / 1.0 = 9.1\), \(n(\text{O}) = 36.4 / 16.0 = 2.275\). 2. Divide by the smallest value (2.275): \(\text{C} = 4.54 / 2.275 \approx 2\), \(\text{H} = 9.1 / 2.275 \approx 4\), \(\text{O} = 2.275 / 2.275 = 1\). 3. The empirical formula is \(\text{C}_2\text{H}_4\text{O}\).

[1 mark] B: Correctly calculates the empirical formula as C2H4O.

A neutral copper atom has the electronic configuration \([\text{Ar}] 4\text{s}^1 3\text{d}^{10}\). To form the \(\text{Cu}^{2+}\) ion, two electrons are removed. Electrons are removed from the outer subshell first, so the single \(4\text{s}\) electron is lost, followed by one \(3\text{d}\) electron, giving the electronic configuration \([\text{Ar}] 3\text{d}^9\).

[1 mark] A: Correctly identifies [Ar] 3d9 as the electronic configuration of Cu2+.

The successive ionization energies are: 1st = \(578\), 2nd = \(1817\), 3rd = \(2745\), 4th = \(11577\). There is a very large jump between the third and fourth ionization energies (nearly a four-fold increase). This indicates that the fourth electron is removed from an inner quantum shell. Therefore, there are three electrons in the outer shell, meaning the element belongs to Group 13 (Group 3).

[1 mark] C: Correctly identifies Group 13 (Group 3) based on the large jump between the 3rd and 4th ionization energies.

\(\text{BF}_4^-\)\ has 4 bonding pairs and 0 lone pairs around the central boron atom, resulting in a tetrahedral geometry with a bond angle of \(109.5^\circ\). \(\text{NH}_3\) and \(\text{H}_3\text{O}^+\) have 3 bonding pairs and 1 lone pair (trigonal pyramidal, \(\sim 107^\circ\)). \(\text{SF}_4\) has 4 bonding pairs and 1 lone pair (see-saw).

[1 mark] C: Correctly identifies BF4- as tetrahedral.

Covalent character in ionic compounds is maximized when a small, highly charged cation polarizes a large, easily polarized anion (Fajans' rules). Among the options: \(\text{Li}^+\) is a much smaller cation than \(\text{Cs}^+\), so it has a higher charge density and greater polarizing power. \(\text{I}^-\) is a much larger anion than \(\text{F}^-\), so it is more easily polarized. Therefore, \(\text{LiI}\) has the greatest covalent character.

[1 mark] B: Correctly identifies LiI as having the greatest covalent character.

A termination step involves two free radicals combining to form a stable, non-radical molecule. In this mechanism, the reaction \(\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\) is a termination step because two methyl radicals combine to form ethane.

[1 mark] C: Correctly identifies the termination step involving two radicals.

According to Markovnikov's rule, the electrophilic addition of \(\text{HBr}\) to propene proceeds via the more stable secondary carbocation intermediate, \(\text{CH}_3\text{CH}^+\text{CH}_3\), rather than the less stable primary carbocation intermediate, \(\text{CH}_3\text{CH}_2\text{CH}_2^+\). Thus, the bromide ion attacks the secondary carbocation, making 2-bromopropane the major product.

[1 mark] B: Correctly identifies 2-bromopropane as the major product.

First, calculate the number of moles of carbon dioxide gas produced:
\[n(CO_2) = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.025\text{ mol}\]

Since the molar ratio of \(XCO_3\) to \(CO_2\) is 1:1, the moles of \(XCO_3\) reacted is also \(0.025\text{ mol}\).

Now, calculate the molar mass of \(XCO_3\):
\[M_r(XCO_3) = \frac{3.69\text{ g}}{0.025\text{ mol}} = 147.6\text{ g mol}^{-1}\]

Subtract the molar mass of the carbonate group (\(CO_3^{2-}\)) to find the relative atomic mass of \(X\):
\[A_r(X) = 147.6 - (12.0 + 3 \times 16.0) = 147.6 - 60.0 = 87.6\text{ g mol}^{-1}\]

Looking at the periodic table, the group 2 element with a relative atomic mass of 87.6 is Strontium (\(Sr\)).

- Correct calculation of moles of \(CO_2\) (1 mark)
- Correct calculation of molar mass of \(XCO_3\) and identification of \(X\) as Strontium (1 mark)

Identify the large jump in ionization energy. There is a huge increase between the 3rd and 4th ionization energies (from \(2745\text{ kJ mol}^{-1}\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is removed from an inner shell, which is closer to the nucleus and experiences much less shielding. Therefore, element \(Y\) has 3 valence electrons and forms a stable \(Y^{3+}\) ion.

The oxide ion is \(O^{2-}\). To form a neutral ionic compound, two \(Y^{3+}\) ions combine with three \(O^{2-}\) ions, giving the empirical formula \(Y_2O_3\).

- Identification of the valence of \(Y\) as 3+ (1 mark)
- Correct formula deduction (1 mark)

Let's examine the arrangement of electron pairs (electron geometry) around the central atom for each species:
- \(NH_4^+\) has 4 bonding pairs and 0 lone pairs around the central nitrogen atom. Electron geometry: tetrahedral.
- \(H_3O^+\) has 3 bonding pairs and 1 lone pair around the central oxygen atom (total of 4 electron pairs). Electron geometry: tetrahedral.
- \(BF_4^-\) has 4 bonding pairs and 0 lone pairs around the central boron atom. Electron geometry: tetrahedral.
- \(CO_3^{2-}\) has 3 regions of electron density (two single bonds, one double bond) around the central carbon atom. Electron geometry: trigonal planar.

Therefore, \(CO_3^{2-}\) does not have a tetrahedral electron pair geometry.

- Correct identifying of electron pair geometries (1 mark)

The structure of 2-methylbutane is \(CH_3-CH(CH_3)-CH_2-CH_3\).
Monochlorination can replace a hydrogen atom on any of the carbons:
1. Substitution at C1 or the methyl branch on C2 gives: \(CH_2Cl-CH(CH_3)-CH_2-CH_3\) (1-chloro-2-methylbutane)
2. Substitution at C2 gives: \((CH_3)_2C(Cl)-CH_2-CH_3\) (2-chloro-2-methylbutane)
3. Substitution at C3 gives: \(CH_3-CH(CH_3)-CHCl-CH_3\) (2-chloro-3-methylbutane)
4. Substitution at C4 gives: \(CH_3-CH(CH_3)-CH_2-CH_2Cl\) (1-chloro-3-methylbutane)

Therefore, exactly 4 structural isomers can be formed.

- Drawing or naming the unique structural isomers (1 mark)

For a compound to exhibit E/Z isomerism, each carbon of the \(C=C\) double bond must be attached to two different groups:
- 1-chloropropene (\(CHCl=CHCH_3\)): C1 is bonded to -H and -Cl (different groups); C2 is bonded to -H and -\(CH_3\) (different groups). It exhibits E/Z isomerism.
- 2-chloroprop-1-ene (\(CH_2=C(Cl)CH_3\)): C1 is bonded to two -H atoms (identical). No E/Z isomerism.
- 1,1-dichloropropene (\(CCl_2=CHCH_3\)): C1 is bonded to two -Cl atoms (identical). No E/Z isomerism.
- 2-methylbut-2-ene (\((CH_3)_2C=CHCH_3\)): C2 is bonded to two -\(CH_3\) groups (identical). No E/Z isomerism.

- Application of E/Z isomerism rules to each compound (1 mark)

Atom economy is defined as:
\[\text{Atom economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\\%\]

- Desired product: bromoethane (\(C_2H_5Br\), \(108.9\text{ g mol}^{-1}\))
- Reactants: \(C_2H_5OH\) and \(HBr\)
- Total mass of reactants = \(46.0 + 80.9 = 126.9\text{ g mol}^{-1}\)

\[\text{Atom economy} = \frac{108.9}{126.9} \times 100\\% \approx 85.81\\%\]

Thus, the correct answer is 85.8%.

- Correct calculation of total mass of reactants and fraction for atom economy (1 mark)

Calculate the weighted average of the isotopic masses:
\[A_r = \frac{\sum (\text{Isotopic mass} \times \text{Relative abundance})}{\sum \text{Relative abundance}}\]

\[A_r = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{79.0 + 10.0 + 11.0}\]

\[A_r = \frac{1896 + 250 + 286}{100} = \frac{2432}{100} = 24.32\]

- Correct application of the relative atomic mass formula (1 mark)

According to Fajans' rules, covalent character in ionic compounds is maximized when there is strong polarization of the anion by the cation.
Polarization is maximized by:
1. A small, highly charged cation (which has a high charge density and polarizing power).
2. A large, highly charged anion (which is more easily polarized due to its outer electrons being further from the nucleus and highly shielded).

- Comparing cations: \(Li^+\) is much smaller than \(Cs^+\), so \(Li^+\) has much greater polarizing power.
- Comparing anions: \(I^-\) is much larger than \(F^-\), so \(I^-\) is much more polarizable.

Refining the choices: lithium iodide (\(LiI\)) combines the smallest cation with the largest anion, resulting in the greatest polarization and the greatest degree of covalent character.

- Application of Fajans' rules to deduce that the smallest cation and largest anion yield the greatest covalent character (1 mark)

First, calculate the amount in moles of carbon dioxide gas collected: \(n(\text{CO}_2) = \frac{360\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0150\text{ mol}\). According to the equation: \(\text{MCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{MCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\), the mole ratio of \(\text{MCO}_3\) to \(\text{CO}_2\) is 1:1. Therefore, \(n(\text{MCO}_3) = 0.0150\text{ mol}\). Next, find the molar mass of \(\text{MCO}_3\): \(M_r(\text{MCO}_3) = \frac{1.50\text{ g}}{0.0150\text{ mol}} = 100\text{ g mol}^{-1}\). Subtract the molar mass of the carbonate group (\(12.0 + 3 \times 16.0 = 60.0\text{ g mol}^{-1}\)) to find the relative atomic mass of \(\text{M}\): \(A_r(\text{M}) = 100 - 60.0 = 40.0\text{ g mol}^{-1}\). This value corresponds to Calcium (\(A_r = 40.1\text{ g mol}^{-1}\)).

[1 mark] - Correctly identifies Calcium as the metal.

The successive ionization energies show a very large increase (jump) between the 3rd and the 4th ionization energies (from \(2745\text{ kJ mol}^{-1}\) to \(11577\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner quantum shell, meaning there are three electrons in the outermost shell of element \(X\). Therefore, \(X\) forms a stable \(3+\) ion (\(X^{3+}\)). Oxygen forms a \(2-\)-charged oxide ion (\(\text{O}^{2-}\)). Combining these ions to form a neutral compound gives the formula \(\text{X}_2\text{O}_3\) (which corresponds to Aluminium oxide, \(\text{Al}_2\text{O}_3\)).

[1 mark] - Correctly identifies the formula as X2O3.

The hydronium ion, \(\text{H}_3\text{O}^+\), has three bonding pairs of electrons and one lone pair on the oxygen atom. According to VSEPR theory, these four electron pairs arrange themselves in a tetrahedral geometry to minimize repulsion. The presence of one lone pair results in a trigonal pyramidal molecular geometry, which is non-planar. On the other hand, \(\text{BF}_3\) and \(\text{CO}_3^{2-}\) are trigonal planar, and \(\text{C}_2\text{H}_4\) is planar.

[1 mark] - Correctly selects H3O+ as the non-planar species.

For an alkene to exhibit geometric (\(E\)/\(Z\)) isomerism, each carbon of the \(\text{C}=\text{C}\) double bond must be attached to two different groups. In hex-3-ene (\(\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}_2\text{CH}_3\)), carbon-3 is bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\) (two different groups), and carbon-4 is also bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Thus, it can form both \((E)\)-hex-3-ene and \((Z)\)-hex-3-ene. In all other options, at least one of the double-bonded carbon atoms is bonded to two identical groups (e.g., two methyl groups in 2-methylbut-2-ene, or two hydrogen atoms in propene and 2-methylpent-1-ene).

[1 mark] - Correctly identifies hex-3-ene as capable of geometric isomerism.

(a)(i) Heating to "constant mass" means heating the sample, cooling it, weighing it, and repeating this sequence until two consecutive mass readings are identical. This is necessary to ensure that the thermal decomposition is complete and all water of crystallisation and gaseous products have been entirely driven off.

(ii) The thermal decomposition of basic copper carbonate, \(\text{CuCO}_3\cdot\text{Cu(OH)}_2\), produces copper(II) oxide, carbon dioxide, and water. Therefore, the two gaseous products are carbon dioxide and water vapour (steam).

(b)(i) \(n(\text{HCl}) = c \times V = 0.200\text{ mol dm}^{-3} \times \frac{23.45}{1000}\text{ dm}^3 = 4.69 \times 10^{-3}\text{ mol}\)

(ii) From the balanced equation, \(1\text{ mol}\) of \(\text{M}_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\).
\(n(\text{M}_2\text{CO}_3) \text{ in } 25.0\text{ cm}^3 = \frac{n(\text{HCl})}{2} = \frac{4.69 \times 10^{-3}}{2} = 2.345 \times 10^{-3}\text{ mol}\)

(iii) First, find the amount of \(\text{M}_2\text{CO}_3\) in the total \(250.0\text{ cm}^3\) volumetric flask:
\(n(\text{M}_2\text{CO}_3) \text{ in } 250.0\text{ cm}^3 = 2.345 \times 10^{-3}\text{ mol} \times 10 = 2.345 \times 10^{-2}\text{ mol}\)
Next, calculate the relative formula mass (\(M_r\)):
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{3.24\text{ g}}{2.345 \times 10^{-2}\text{ mol}} = 138.17\text{ g mol}^{-1}\) (Accept \(138\text{ g mol}^{-1}\))

(iv) The formula of the carbonate is \(\text{M}_2\text{CO}_3\).
\(M_r(\text{M}_2\text{CO}_3) = 2 \times A_r(\text{M}) + 12.0 + (3 \times 16.0) = 138.17\)
\(2 \times A_r(\text{M}) + 60.0 = 138.17\)
\(2 \times A_r(\text{M}) = 78.17\)
\(A_r(\text{M}) = 39.09 \approx 39.1\text{ g mol}^{-1}\)
Comparing with the Periodic Table, the metal is Potassium (\(\text{K}\)).

(v) Suitable safety precautions include:
- Use a pipette filler when pipetting the solutions because \(\text{HCl}\) and the carbonate solution are irritants/corrosive and should not be pipetted by mouth.
- Wear gloves to avoid skin contact with the acid/carbonate, which can cause irritation.

(a)(i)
- M1: Heat, weigh, and repeat the heating process (1)
- M2: Until the mass remains identical / constant to ensure the reaction is complete (1)

(a)(ii)
- M1: Carbon dioxide (1)
- M2: Water vapour / steam (1)

(b)(i)
- M1: \(4.69 \times 10^{-3}\text{ mol}\) (1)

(b)(ii)
- M1: \(2.345 \times 10^{-3}\text{ mol}\) (allow consequential marking from (b)(i)) (1)

(b)(iii)
- M1: Multiplies moles by 10 to find moles in \(250.0\text{ cm}^3\) \(= 2.345 \times 10^{-2}\text{ mol}\) (1)
- M2: Uses \(M_r = \frac{\text{mass}}{\text{moles}}\) (1)
- M3: Calculates value of \(138\) or \(138.17\text{ g mol}^{-1}\) (1)

(b)(iv)
- M1: Correct subtraction of carbonate group mass (60) and division by 2 to yield \(A_r = 39.1\) (1)
- M2: Correctly identifies \(\text{M}\) as Potassium / \(\text{K}\) (1)

(b)(v)
- M1: Wear lab gloves because hydrochloric acid is an irritant / use a pipette filler to avoid mouth contact with chemicals (1)

(a)(i) Positive ions are accelerated using an electric field / charged plates with a potential difference, such that all ions are accelerated to have the same / constant kinetic energy.

(ii) \(A_r = \frac{(63 \times 69.17) + (65 \times 30.83)}{100} = \frac{4357.71 + 2003.95}{100} = \frac{6361.66}{100} = 63.6166 \approx 63.62\)
Comparing with the Periodic Table, element X is Copper (\(\text{Cu}\)).

(b)(i) Copper has an anomalous stable configuration with a filled \(3d\) subshell:
\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\)

(ii) To form the \(\text{Cu}^{2+}\) ion, electrons are lost from the \(4s\) subshell first, then from the \(3d\) subshell:
\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\)

(c)(i) Magnesium has its outer electron in the \(3s\) subshell, whereas Aluminium has its outer electron in the \(3p\) subshell. The \(3p\) subshell of Aluminium is at a higher energy level and is shielded by the inner \(3s^2\) electrons, making it easier to remove and requiring less energy.

(ii) Phosphorus has three unpaired electrons in its \(3p\) subshell (\(3p^3\)), while Sulfur has four \(3p\) electrons (\(3p^4\)), meaning one \(3p\) orbital contains a pair of electrons. The mutual repulsion between the two paired electrons in the same orbital of Sulfur makes the outer electron easier to remove compared to Phosphorus.

(a)(i)
- M1: Accelerated by an electric field / plates with a potential difference (1)
- M2: To give all ions the same kinetic energy (1)

(a)(ii)
- M1: Correct set up for relative atomic mass calculation: \(\frac{(63 \times 69.17) + (65 \times 30.83)}{100}\) (1)
- M2: Evaluates to \(63.62\) (must be to 2 d.p.) (1)
- M3: Identifies element X as Copper / \(\text{Cu}\) (1)

(b)(i)
- M1: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (do not accept configuration ending in \(3d^9 4s^2\)) (1)

(b)(ii)
- M1: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\) (1)

(c)(i)
- M1: Al's outer electron is in a \(3p\) subshell/orbital, whereas Mg's outer electron is in a \(3s\) subshell/orbital (1)
- M2: The \(3p\) subshell is higher in energy / further from nucleus / shielded by \(3s\) electrons (1)

(c)(ii)
- M1: Phosphorus has configuration \(3p^3\) / unpaired electrons, and Sulfur has \(3p^4\) / paired electrons in one of its \(3p\) orbitals (1)
- M2: Mentions spin-pairing / electron-electron repulsion in the paired orbital of Sulfur (1)
- M3: This repulsion makes the outer electron easier to remove (hence lower first ionisation energy) (1)

(a)(i) The dot-and-cross diagram of \(\text{PH}_3\) shows a central phosphorus atom with 5 outer electrons. It shares 3 electrons with 3 hydrogen atoms, creating 3 single covalent bonds (3 bonding pairs). It retains 1 lone pair on the phosphorus atom.

(ii) The shape of the \(\text{PH}_3\) molecule is trigonal pyramidal. The predicted bond angle is \(107^\circ\) (accept values in the range of \(93.0^\circ - 107.5^\circ\) as phosphorus's lower electronegativity decreases the repulsion compared to ammonia, but the general electron pair repulsion theory applies).
Explanation:
- There are four areas of electron density / four electron pairs around the central phosphorus atom: 3 bonding pairs and 1 lone pair.
- Electron pairs repel each other to get as far apart as possible to minimise repulsion.
- Lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, which pushes the bonding pairs closer together, reducing the angle from tetrahedral \(109.5^\circ\).

(iii) The shape of the \(\text{BF}_3\) molecule is trigonal planar.
Explanation of non-polarity:
- Fluorine is much more electronegative than boron, so the B–F bonds are polar.
- However, because the molecule is perfectly symmetrical (trigonal planar, bond angles of \(120^\circ\)), the individual dipoles point in opposite directions and cancel each other out, leaving no overall net dipole.

(b)(i) In graphite, each carbon atom is bonded to three others in planar hexagonal sheets, leaving one delocalised electron per carbon atom. These delocalised electrons are free to move between layers and carry electric charge. In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms by strong localised covalent bonds. There are no free-moving/delocalised electrons to carry current.

(ii) Diamond has a giant covalent / macromolecular structure. To melt diamond, a vast number of strong covalent carbon-carbon bonds must be broken. This process requires an extremely high amount of thermal energy, resulting in a very high melting temperature.

(a)(i)
- M1: Correct dot-and-cross diagram showing three shared pairs of electrons between P and H, and one lone pair on the P atom (1)

(a)(ii)
- M1: Shape: Trigonal pyramidal (1)
- M2: Bond angle: \(107^\circ\) (allow any angle from \(93^\circ\) to \(108^\circ\)) (1)
- M3: Explanation: Four electron pairs (3 bonding, 1 lone pair) around the central P atom repel each other to minimise repulsion (1)
- M4: Lone pairs repel more than bonding pairs (reducing the bond angle) (1)

(a)(iii)
- M1: Shape: Trigonal planar (1)
- M2: The molecule is symmetrical (1)
- M3: The individual polar bond dipoles cancel out (resulting in no net dipole) (1)

(b)(i)
- M1: Graphite has delocalised electrons that are free to move (and carry charge) (1)
- M2: Diamond has all outer electrons localised in single covalent bonds (no free-moving charged particles) (1)

(b)(ii)
- M1: Giant covalent lattice / macromolecular structure (1)
- M2: Requires a large amount of energy to break many strong covalent bonds (1)

(a)(i) The skeletal formula of 2,2,4-trimethylpentane consists of a 5-carbon main chain (pentane). On Carbon-2, there are two methyl groups, and on Carbon-4, there is one methyl group:

` \ / `
` \___/\ `
` / \ `
` / \ `
(A standard skeletal structure showing a pentane zigzag with a cross at C2 and a single line at C4).

(ii) Molecular formula: \(\text{C}_8\text{H}_{18}\).
The straight-chain structural isomer of an octane isomer is octane (or n-octane).

(b)(i) Cracking of decane to produce propene and one other alkane:
\(\text{C}_{10}\text{H}_{22}\text{(l)} \rightarrow \text{C}_3\text{H}_6\text{(g)} + \text{C}_7\text{H}_{16}\text{(l)}\)

(ii) Cracking is economically important because:
- It converts long-chain alkanes, which are in low demand and surplus supply from fractional distillation, into shorter-chain alkanes, which are highly demanded as volatile fuels (like petrol).
- It produces alkenes (like propene and ethene), which are essential chemical feedstocks used to manufacture polymers and plastics.

(c)(i) The role of UV radiation is to provide the energy required to break the covalent bond in the chlorine molecule homolytically to create chlorine free radicals (initiation step).

(ii) Propagation steps:
- Step 1: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow {}^\bullet\text{CH}_3 + \text{HCl}\)
- Step 2: \({}^\bullet\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\)

(iii) Termination step forming ethane:
\(2\,{}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\)
This arises when two methyl free radicals, produced during propagation, collide and react (combine) together.

(iv) It is not suitable because further substitution reactions occur (e.g., chloromethane reacting with chlorine to form dichloromethane, trichloromethane, and tetrachloromethane), resulting in a complex mixture of products that must be separated.

(a)(i)
- M1: Correct skeletal formula of 2,2,4-trimethylpentane (1)

(a)(ii)
- M1: \(\text{C}_8\text{H}_{18}\) (1)
- M2: octane / n-octane (1)

(b)(i)
- M1: Balanced equation: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \text{C}_7\text{H}_{16}\) (ignore state symbols) (1)

(b)(ii)
- M1: Converts low-demand long-chain hydrocarbons to higher-demand shorter-chain hydrocarbons (used as fuels) (1)
- M2: Produces alkenes used in polymer/plastic production (1)

(c)(i)
- M1: Breaks the Cl–Cl bond homolytically / initiates the reaction by forming chlorine free radicals (1)

(c)(ii)
- M1: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow {}^\bullet\text{CH}_3 + \text{HCl}\) (1)
- M2: \({}^\bullet\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\) (1)

(c)(iii)
- M1: Correct equation: \(2\,{}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\) (1)
- M2: Explains that two methyl radicals collide and bond together (1)

(c)(iv)
- M1: Further substitution occurs / multiple substitution reactions occur yielding a mixture of products (e.g., \(\text{CH}_2\text{Cl}_2\), etc.) (1)

(a)(i) Stereoisomers are molecules with the same structural formula but have a different arrangement of their atoms in three-dimensional space.

(ii) The displayed formulae show every atom and every bond explicitly:
- \(E\)-but-2-ene (trans-but-2-ene):
H-C-H groups are opposite.
H H
\ /
C=C
/ \
H-C-H H-C-H
(Methyl groups trans to each other)
- \(Z\)-but-2-ene (cis-but-2-ene):
H-C-H groups are on same side.
H-C-H H-C-H
\ /
C=C
/ \
H H
(Methyl groups cis to each other)

(iii) Stereoisomerism occurs because of the restricted rotation about the carbon-to-carbon double bond (\(\text{C=C}\)), which is caused by the presence of the \(\pi\) bond.
In but-2-ene, each carbon of the \(\text{C=C}\) double bond is attached to two different groups (a hydrogen atom and a methyl group). In but-1-ene, one of the carbon atoms in the double bond is attached to two identical groups (two hydrogen atoms), so swapping their positions does not create a new spatial isomer.

(b)(i) Electrophilic addition mechanism:
1. Propene is \(\text{CH}_3-\text{CH}=\text{CH}_2\). Hydrogen bromide is polar: \(\text{H}^{\delta+}-\text{Br}^{\delta-}\).
2. A curly arrow starts from the double bond of propene and points to the hydrogen atom of \(\text{H}-\text{Br}\). Simultaneously, a curly arrow starts from the \(\text{H}-\text{Br}\) bond and points to the bromine atom.
3. This leads to the formation of a secondary carbocation intermediate: \(\text{CH}_3-\text{CH}^+-\text{CH}_3\) and a bromide ion \(\text{Br}^-\).
4. A curly arrow starts from a lone pair on the \(\text{Br}^-\) ion and points to the positively charged carbon atom of the carbocation.
5. The final product is 2-bromopropane.

(ii) The reaction proceeds via a secondary carbocation intermediate, \(\text{CH}_3-\text{CH}^+-\text{CH}_3\), rather than a primary carbocation, \(\text{CH}_3-\text{CH}_2-\text{CH}_2^+\). Secondary carbocations are more stable than primary carbocations because they have two electron-releasing methyl groups (inductive effect) that help disperse the positive charge, compared to only one in a primary carbocation. Thus, the secondary carbocation forms faster and leads to the major product.

(c) The equation for polymerisation:
\[n\,\text{CH}_2\text{=CH(CH}_3\text{)} \rightarrow -[-\text{CH}_2-\text{CH(CH}_3\text{)}-]-_n\]

(a)(i)
- M1: Molecules with the same structural formula but a different arrangement of atoms in space (1)

(a)(ii)
- M1: Correct displayed formula of \(E\)-but-2-ene, with all bonds shown (1)
- M2: Correct displayed formula of \(Z\)-but-2-ene, with all bonds shown (1)

(a)(iii)
- M1: Restricted rotation around the \(\text{C=C}\) double bond (due to the pi bond) (1)
- M2: but-2-ene has two different groups attached to each carbon of the double bond, while but-1-ene has two identical hydrogen atoms on one carbon (1)

(b)(i)
- M1: Correct dipole on H-Br (\(\text{H}^{\delta+}-\text{Br}^{\delta-}\)) and curly arrow from the double bond to the \(\text{H}\) atom (1)
- M2: Curly arrow from the H-Br bond to the \(\text{Br}\) atom (1)
- M3: Correct structure of the secondary carbocation intermediate, \(\text{CH}_3\text{CH}^+\text{CH}_3\) (1)
- M4: Bromide ion shown with a lone pair and negative charge, with a curly arrow pointing from the lone pair to the positively charged carbon atom (1)

(b)(ii)
- M1: The reaction goes via a secondary carbocation (rather than a primary carbocation) (1)
- M2: Secondary carbocations are more stable than primary carbocations due to the inductive electron-releasing effect of two alkyl groups (1)

(c)
- M1: Balanced equation with \(n\) reactants and repeating unit structure showing open-ended bonds crossing brackets with subscript \(n\) (1)

The boiling temperature of isomeric halogenoalkanes depends on the surface contact area between molecules. 1-bromobutane is a straight-chain molecule with no branching, allowing greater surface contact and thus stronger London forces. Branching (as in 2-bromo-2-methylpropane) prevents close packing and reduces the surface area of contact, resulting in weaker London forces and a lower boiling temperature.

[1] Correct option selected.

Magnesium nitrate undergoes thermal decomposition when heated strongly: \( 2\text{Mg(NO}_3)_2(\text{s}) \rightarrow 2\text{MgO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g}) \). Nitrogen dioxide, \( \text{NO}_2 \), is a brown gas. Magnesium oxide, \( \text{MgO} \), is a white solid residue. Oxygen is also evolved but is colourless.

[1] Correct option selected.

The rate of hydrolysis of halogenoalkanes depends on the carbon-halogen bond strength. Since the C-I bond is the weakest (lowest bond enthalpy) among the carbon-halogen bonds, 1-iodobutane reacts the fastest. The iodide ions released react with silver ions from silver nitrate to form a yellow precipitate of silver iodide (\( \text{AgI} \)).

[1] Correct option selected.

The equation for the formation of propane is: \( 3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)} \). Using a Hess's Law cycle based on combustion data: \( \Delta H^\theta_f = 3\Delta H^\theta_c\text{(C)} + 4\Delta H^\theta_c\text{(H}_2) - \Delta H^\theta_c\text{(C}_3\text{H}_8) = 3(-394) + 4(-286) - (-2220) = -1182 - 1144 + 2220 = -106\text{ kJ mol}^{-1} \).

[1] Correct option selected.

A catalyst provides an alternative reaction pathway with a lower activation energy. This increases the rate of both the forward and reverse reactions equally. It does not affect the position of equilibrium, the yield of products, the value of the equilibrium constant \( K_c \), or the overall enthalpy change of the reaction.

[1] Correct option selected.

The peak at \( 1715\text{ cm}^{-1} \) indicates a carbonyl group (\( \text{C}=\text{O} \)). The absence of a broad peak at \( 3200\text{--}3750\text{ cm}^{-1} \) shows there is no \( \text{O-H} \) group, ruling out alcohols such as but-2-en-1-ol. Since the compound has a carbonyl group but does not react with Fehling's solution, it is not an aldehyde. Thus, it must be the ketone butan-2-one, \( \text{CH}_3\text{CH}_2\text{COCH}_3 \).

[1] Correct option selected.

The reaction of chlorine with cold, dilute aqueous sodium hydroxide is: \( \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \). In sodium chloride (\( \text{NaCl} \)), the oxidation state of chlorine is -1. In sodium chlorate(I) (\( \text{NaClO} \)), the oxidation state of chlorine is +1.

[1] Correct option selected.

Because 3-bromo-3-methylhexane is a tertiary halogenoalkane, it undergoes hydrolysis via an \( \text{S}_\text{N}1 \) mechanism. The first step involves the rate-determining loss of the bromide ion to form a planar carbocation intermediate. The hydroxide nucleophile can then attack this planar carbocation with equal probability from either side, resulting in a racemic mixture (equal amounts of both optical isomers) which has no net optical activity.

[1] Correct option selected.

Down Group 2, the ionic radius of the metal cation increases while its charge remains constant at \(+2\). This decrease in charge density reduces the polarizing power of the cation. Consequently, the electron cloud of the carbonate ion is distorted to a lesser extent, which prevents the weakening of the C-O bonds within the carbonate ion. Therefore, more thermal energy is needed to decompose the carbonate, resulting in increased thermal stability down the group.

1 mark for the correct option (A).

The chemical equation for the standard enthalpy of formation of butane is:

\(4\text{C(s, graphite)} + 5\text{H}_2\text{(g)} \rightarrow \text{C}_4\text{H}_{10}\text{(g)}\)

Using a Hess's Law cycle based on standard enthalpy changes of combustion:

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

\(\Delta_f H^\ominus = [4 \times (-393.5) + 5 \times (-285.8)] - (-2876.5)\)

\(\Delta_f H^\ominus = [-1574.0 - 1429.0] + 2876.5\)

\(\Delta_f H^\ominus = -3003.0 + 2876.5 = -126.5\text{ kJ mol}^{-1}\)

1 mark for the correct option (A).

The rate of hydrolysis of halogenoalkanes depends on two factors: the strength of the C-X bond (bond enthalpy) and the structural class of the halogenoalkane (primary, secondary, or tertiary). The C-I bond is much weaker than C-Br and C-Cl bonds, making iodoalkanes hydrolyze more rapidly than bromo- and chloroalkanes. Additionally, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react significantly faster than primary halogenoalkanes (such as 1-iodobutane) because they react via the \(S_{\text{N}}1\) mechanism, which proceeds through a stable tertiary carbocation intermediate.

1 mark for the correct option (D).

Propane (\(M_r = 44\)), methoxymethane (\(M_r = 46\)), and ethanol (\(M_r = 46\)) have very similar molecular masses but different types of intermolecular forces:

1. Propane is non-polar and experiences only weak London forces.
2. Methoxymethane is polar and experiences permanent dipole-dipole forces in addition to London forces.
3. Ethanol contains an -OH group and experiences strong hydrogen bonding in addition to London forces and permanent dipole-dipole forces.

Since hydrogen bonds are the strongest of these intermolecular forces, followed by permanent dipole-dipole forces and London forces, the energy required to separate the molecules increases in the order: propane < methoxymethane < ethanol.

1 mark for the correct option (A).

**(a) Thermal stability of Group 2 nitrates:**

* Both the magnesium ion (\(\text{Mg}^{2+}\)) and the barium ion (\(\text{Ba}^{2+}\)) have the same charge of \(2+\).
* The magnesium ion has a much smaller ionic radius than the barium ion.
* Therefore, the magnesium ion has a much higher charge density than the barium ion.
* This higher charge density allows the magnesium ion to polarise the nitrate anion (specifically, weakening the \(\text{N}-\text{O}\) bond) more strongly, making thermal decomposition occur at a lower temperature.

**(b)(i) Observations indicating redox reactions:**

* Orange-brown fumes/vapour (due to the formation of bromine, \(\text{Br}_2\)).
* A choking gas/pungent smell (due to the formation of sulfur dioxide, \(\text{SO}_2\)).

**(b)(ii) Ionic half-equation for the reduction of sulfuric acid:**

\(\text{H}_2\text{SO}_4 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\)
*(Accept: \(\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\))*

**(b)(iii) Overall equation:**

\(2\text{NaBr} + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{NaHSO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}\)
*(Accept: \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}\))*

**(a) Explain thermal stability trend (Max 4 marks):**
* **M1:** State that \(\text{Mg}^{2+}\) has a smaller ionic radius than \(\text{Ba}^{2+}\) / \(\text{Mg}^{2+}\) is smaller than \(\text{Ba}^{2+}\) (1)
* **M2:** State that both cations have the same charge (\(2+\)) (1)
* **M3:** State that \(\text{Mg}^{2+}\) has a higher charge density (1)
* **M4:** State that \(\text{Mg}^{2+}\) polarises the nitrate ion (or weakens the \(\text{N}-\text{O}\) bond) more effectively (1)

**(b)(i) Observations (Max 2 marks):**
* **M1:** Brown/orange fumes or liquid (for bromine) (1)
* **M2:** Choking gas / pungent smell (for sulfur dioxide) (1)
* *Reject:* Misty fumes (not a redox product, although present as \(\text{HBr}\))

**(b)(ii) Ionic half-equation (1.75 marks):**
* **M1:** Correct reactants and products: \(\text{H}_2\text{SO}_4\) (or \(\text{SO}_4^{2-} + 4\text{H}^+\)) and \(\text{SO}_2\) (1)
* **M2:** Correct balancing of electrons and water: \(2\text{e}^-\), \(2\text{H}^+\) (or \(4\text{H}^+\)), and \(2\text{H}_2\text{O}\) (0.75)

**(b)(iii) Overall equation (Max 2 marks):**
* **M1:** Correct products: \(\text{NaHSO}_4\) (or \(\text{Na}_2\text{SO}_4\)), \(\text{SO}_2\), \(\text{Br}_2\), \(\text{H}_2\text{O}\) (1)
* **M2:** Fully balanced equation (1)

**(a) Enthalpy change calculation:**

* **Step 1: Calculate the temperature change (\(\Delta T\))**
\(\Delta T = 26.9 - 20.2 = 6.7\ ^\circ\text{C}\)
* **Step 2: Calculate the heat energy released (\(q\))**
Total mass of solution \(m = 50.0 + 50.0 = 100.0\text{ g}\) (since density is \(1.00\text{ g cm}^{-3}\))
\(q = m c \Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J} = 2.8006\text{ kJ}\)
* **Step 3: Calculate the moles of water formed**
\(n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)
\(n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)
Since the reaction is \(1:1\), \(0.0500\text{ mol}\) of \(\text{H}_2\text{O}\) is produced.
* **Step 4: Calculate the enthalpy change of neutralisation (\(\Delta H_{neut}\))**
\(\Delta H_{neut} = -\frac{q}{n} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.0\text{ kJ mol}^{-1}\) (to 3 significant figures)

**(b) Source of uncertainty and reduction:**

* **Uncertainty:** Reading the temperature using a thermometer with large graduations (e.g. \(1\ ^\circ\text{C}\) intervals) leading to high percentage uncertainty.
* **Reduction:** Use a thermometer with smaller graduations (e.g., \(0.1\ ^\circ\text{C}\) intervals) or a digital temperature probe.
* *(Accept: volume measurement using a measuring cylinder is imprecise; use a volumetric pipette instead).*

**(c) Comparison with ethanoic acid:**

* Ethanoic acid is a weak acid and is only partially dissociated in aqueous solution.
* Some energy is required to fully dissociate the ethanoic acid molecules into \(\text{CH}_3\text{COO}^-\游离离子\) and \(\text{H}^+\) ions during the reaction.
* Since this dissociation process is endothermic, it absorbs some of the heat energy released, resulting in a less exothermic net enthalpy change and a smaller temperature rise.

**(d) Enthalpy level diagram:**

An enthalpy level diagram showing reactants (\(\text{H}^+ + \text{OH}^-\)) at a higher level than products (\(\text{H}_2\text{O(l)}\)), with an energy hump representing activation energy (\(E_a\)), and a downward-pointing arrow representing \(\Delta H\).

**(a) Calculation (Max 4 marks):**
* **M1:** Calculate temperature difference correctly: \(6.7\ ^\circ\text{C}\) (1)
* **M2:** Correct calculation of \(q\): \(2.8006\text{ kJ}\) (or \(2800.6\text{ J}\)) (1)
* **M3:** Calculate moles of acid/alkali correctly: \(0.0500\text{ mol}\) (1)
* **M4:** Final value with correct sign and 3 sig figs: \(-56.0\text{ kJ mol}^{-1}\) (1)

**(b) Uncertainty (Max 2 marks):**
* **M1:** Identify a valid non-heat-loss source of uncertainty (e.g., temperature reading uncertainty OR volume of acid/alkali measured with measuring cylinder) (1)
* **M2:** Identify a correct way to reduce this uncertainty (e.g., use a thermometer with \(0.1\ ^\circ\text{C}\) divisions / digital probe OR use a volumetric pipette) (1)

**(c) Ethanoic acid explanation (Max 2.75 marks):**
* **M1:** State that ethanoic acid is a weak acid / partially dissociated (1)
* **M2:** Explain that energy is needed to dissociate / ionise the remaining ethanoic acid molecules (1)
* **M3:** Deduce that the net reaction is less exothermic / less heat is released overall (0.75)

**(d) Enthalpy level diagram (1 mark):**
* **M1:** Correct diagram showing reactants higher than products, labeled axes (Enthalpy vs Progress of reaction), and downward arrow for \(\Delta H\) / hump for \(E_a\) (1)

**(a)(i) Reagents and conditions for hydrolysis:**

* **Reagents:** Aqueous silver nitrate (\(\text{AgNO}_3\)) and ethanol.
* **Conditions:** Heat the mixture in a water bath (usually around \(50\text{--}60\ ^\circ\text{C}\)) to ensure a constant temperature.
* **Role of Ethanol:** Acts as a mutual solvent to dissolve both the halogenoalkanes and the aqueous silver nitrate.
* **Monitoring:** Record the time taken for a precipitate of the silver halide (white for chloride, cream for bromide, yellow for iodide) to appear.

**(a)(ii) Order of the rates of hydrolysis:**

* **Order:** 1-iodobutane reacts fastest, followed by 1-bromobutane, and 1-chlorobutane is the slowest.
* **Reasoning:** The rate of reaction depends on the strength of the carbon-halogen bond (bond enthalpy), not bond polarity.
* **Trend:** The bond enthalpy decreases down the group (\(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\)) because the halogen atom gets larger and the orbital overlap is weaker.
* Therefore, the \(\text{C}-\text{I}\) bond is the easiest to break, so 1-iodobutane hydrolyses most rapidly.

**(b)(i) Compound X structure and name:**

* **Skeletal Formula:** A four-carbon chain with a bromine atom on carbon 2:
` /\_/` with `Br` attached to the second carbon.
* **IUPAC Name:** 2-bromobutane

**(b)(ii) Elimination products:**

Three alkenes are formed:
1. **But-1-ene**: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\)
2. **(E)-but-2-ene** (trans-but-2-ene): with methyl groups on opposite sides of the double bond.
3. **(Z)-but-2-ene** (cis-but-2-ene): with methyl groups on the same side of the double bond.

**(a)(i) Experimental details (Max 3.75 marks):**
* **M1:** Add ethanol (as a mutual solvent) (1)
* **M2:** Add aqueous silver nitrate (1)
* **M3:** Heat using a water bath / constant temperature (0.75)
* **M4:** Measure the time taken for a precipitate to form (1)

**(a)(ii) Explanation of rates (Max 3 marks):**
* **M1:** Correct order of rate: 1-iodobutane > 1-bromobutane > 1-chlorobutane (1)
* **M2:** State that the carbon-halogen bond strength / bond enthalpy decreases down Group 7 (\(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\)) (1)
* **M3:** State that the bond enthalpy is the dominant factor / weaker bonds require less energy to break and therefore react faster (1)

**(b)(i) Compound X (Max 2 marks):**
* **M1:** Correct skeletal formula of 2-bromobutane (1)
* **M2:** Correct IUPAC name: 2-bromobutane (1)

**(b)(ii) Elimination products (1 mark):**
* **M1:** Draw all three structures: but-1-ene, (E)-but-2-ene, and (Z)-but-2-ene (1)
*(Allow structural/skeletal/displayed formulae, but stereochemistry must be clearly shown for the E/Z isomers)*

**(a)(i) & (ii) Maxwell-Boltzmann distribution diagram:**

* **Curve at \(T_1\):** Starts at the origin, rises to a peak, and then gradually approaches the x-axis asymptotically.
* **Activation Energy (\(E_a\)):** Marked on the x-axis with a vertical line, and the area to the right of it under the curve at \(T_1\) is shaded.
* **Curve at \(T_2\) (higher temperature):** The peak of the curve at \(T_2\) is shifted to the right (higher energy) and has a lower height than the peak of the curve at \(T_1\). The curve at \(T_2\) must only cross the curve at \(T_1\) once, and must lie above the \(T_1\) curve at high energy values.
* **Explanation of rate increase:**
* At higher temperature \(T_2\), a larger area under the curve lies to the right of the activation energy \(E_a\).
* This means a greater fraction of molecules have energy greater than or equal to the activation energy.
* Therefore, there are more frequent successful collisions per unit time, which increases the rate of reaction.

**(b) Effect of increasing pressure:**

* **Effect on Yield:** The yield of \(\text{NO}_2\text{(g)}\) increases.
* **Explanation:** There are 3 moles of gaseous reactants on the left-hand side and only 2 moles of gaseous product on the right-hand side. Increasing pressure shifts the equilibrium position to the side with fewer gas moles (the right-hand side) to decrease the pressure.
* **Effect on \(K_c\):** The value of \(K_c\) remains unchanged. This is because the equilibrium constant, \(K_c\), is only affected by changes in temperature, not pressure.

**(a)(i) Maxwell-Boltzmann curve at \(T_1\) (Max 3 marks):**
* **M1:** Curve starts at the origin and does not touch the x-axis at high energy (1)
* **M2:** Activation energy \(E_a\) correctly labeled on the x-axis (1)
* **M3:** Shaded area under the curve to the right of \(E_a\) representing reacting molecules (1)

**(a)(ii) Curve at \(T_2\) and explanation (Max 3.75 marks):**
* **M1:** Peak of curve at \(T_2\) is lower than peak at \(T_1\) (1)
* **M2:** Peak of curve at \(T_2\) is to the right of peak at \(T_1\) (1)
* **M3:** Explanation: Area under the curve to the right of \(E_a\) is larger at \(T_2\) / a greater fraction of molecules have energy \(\ge E_a\) (1)
* **M4:** Result: More frequent successful collisions per unit time (0.75)

**(b) Pressure and Equilibrium (Max 3 marks):**
* **M1:** State that the yield of \(\text{NO}_2\) increases (1)
* **M2:** Reason for yield increase: equilibrium shifts to the right because there are fewer moles of gas on the right (2 moles) than on the left (3 moles) (1)
* **M3:** State that the value of \(K_c\) remains unchanged / is constant because it only depends on temperature (1)

**(a)**
1. Calculate heat energy transferred (q):
\(q = m \times c \times \Delta T\)
\(q = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 28.2\ ^\circ\text{C} = 17681.4\text{ J} = 17.6814\text{ kJ}\)

2. Calculate moles of butan-1-ol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{0.578\text{ g}}{74.0\text{ g mol}^{-1}} = 0.0078108\text{ mol}\)

3. Calculate enthalpy change of combustion (\(\Delta H_c\)):
\(\Delta H_c = -\frac{q}{n} = -\frac{17.6814\text{ kJ}}{0.0078108\text{ mol}} = -2263.7\text{ kJ mol}^{-1}\)

4. Appropriate significant figures (3 sig figs based on 0.578 g and 28.2 °C):
\(\Delta H_c = -2260\text{ kJ mol}^{-1}\)

**(b)(i)**
\(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\)
(Alternatively: \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\))

**(b)(ii)**
* Hazard: Corrosive (causes severe burns / skin damage).
* Precaution: Wear protective gloves (e.g., nitrile gloves) or handle in a fume cupboard.

**(b)(iii)**
* Purpose: To neutralise any remaining acid / acid catalyst (e.g., \(\text{H}_2\text{SO}_4\) / \(\text{HBr}\)).
* Precaution: Invert the separating funnel and open the tap periodically to release pressure build-up from carbon dioxide gas.

**(c)**
* Method: Add equal amounts of 1-bromobutane and 1-chlorobutane to separate test tubes. Add aqueous silver nitrate solution (aq) and ethanol (acting as a mutual solvent).
* Conditions: Place both test tubes in a water bath at the same temperature (e.g., \(50\text{--}60\ ^\circ\text{C}\)) to control temperature.
* Observation: A cream precipitate (of silver bromide) will form first/faster with 1-bromobutane, while a white precipitate (of silver chloride) will form slower with 1-chlorobutane.
* Explanation: The C-Br bond is weaker / has a lower bond enthalpy than the C-Cl bond.
* Therefore, the C-Br bond is broken more easily / requires less activation energy, leading to a faster rate of hydrolysis.

**(d)(i)**
Concentrated phosphoric acid is a weaker oxidising agent than concentrated sulfuric acid (which would oxidise/char the alcohol or produce toxic sulfur dioxide gas).

**(d)(ii)**
* Names of the three isomers: but-1-ene, (E)-but-2-ene (or trans-but-2-ene), and (Z)-but-2-ene (or cis-but-2-ene).
* Structural isomerism: Occurs because the hydrogen atom can be eliminated from carbon-1 or carbon-3, resulting in the double bond being located in different positions (between C1 and C2, or between C2 and C3).
* Stereoisomerism (geometric/E-Z isomerism): Occurs in but-2-ene due to restricted rotation about the C=C double bond, where both carbon atoms of the double bond are attached to two different groups (a hydrogen atom and a methyl group).

**(d)(iii)**
* Difference 1: The peak for the O-H (alcohol) bond at \(3200\text{--}3750\text{ cm}^{-1}\) will completely disappear.
* Difference 2: A new peak for the C=C (alkene) bond will appear in the range of \(1620\text{--}1670\text{ cm}^{-1}\).

**(a) [Total: 4 marks]**
* **M1**: Correct calculation of heat energy transferred, \(q = 17.68\text{ kJ}\) (or \(17681.4\text{ J}\)) (1)
* **M2**: Correct calculation of moles of butan-1-ol, \(n = 0.00781\text{ mol}\) (1)
* **M3**: Calculates \(\Delta H_c = \frac{\text{M1}}{\text{M2}} = -2264\text{ kJ mol}^{-1}\) (or value matching their rounded/unrounded numbers) (1)
* **M4**: Correct sign (-) AND final answer given to 3 significant figures: \(-2260\text{ kJ mol}^{-1}\) (1)
* *Note*: If sign is missing, lose M4. Accept \(-2300\text{ kJ mol}^{-1}\) ONLY if mass of 0.578 g was rounded to 2 sig figs in a student's working.

**(b)(i) [Total: 1 mark]**
* **M1**: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (1)
* *Accept*: \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\)

**(b)(ii) [Total: 2 marks]**
* **M1**: State hazard: Corrosive / causes severe skin burns or eye damage (1). *Reject 'toxic' or 'flammable'.*
* **M2**: State precaution: Wear protective gloves / handle in a fume cupboard / use a dropper to transfer (1). *Reject 'wear safety goggles' / 'wear lab coat' as these are standard lab requirements.*

**(b)(iii) [Total: 2 marks]**
* **M1**: Purpose: Neutralise any remaining acid / phosphoric acid / sulfuric acid / HBr (1). *Accept 'remove acid'.*
* **M2**: Precaution: Release pressure build-up periodically by opening the tap with the funnel inverted (1). *Accept 'venting the funnel'.*

**(c) [Total: 5 marks]**
* **M1**: Reagents: Silver nitrate solution AND ethanol (1)
* **M2**: Control variable / Conditions: Place test tubes in a water bath at the same temperature (1)
* **M3**: Observation: Cream precipitate forms faster/sooner than the white precipitate (1)
* **M4**: Bond strength comparison: The C-Br bond is weaker / has lower bond enthalpy than the C-Cl bond (1)
* **M5**: Explanation: C-Br bond breaks more easily / requires less energy to break than the C-Cl bond (1)

**(d)(i) [Total: 1 mark]**
* **M1**: Phosphoric acid is a weaker oxidising agent / does not oxidise or char the organic reactants / does not produce toxic sulfur dioxide gas (1)

**(d)(ii) [Total: 4 marks]**
* **M1**: Identifies all three isomers: but-1-ene, (E)-but-2-ene (or trans-but-2-ene), and (Z)-but-2-ene (or cis-but-2-ene) (1)
* **M2**: Explains structural isomerism: Hydrogen can be removed from C1 or C3, producing double bonds in different positions (1)
* **M3**: Explains stereoisomerism: restricted rotation about C=C bond (1)
* **M4**: Explains condition for E/Z: each carbon in the C=C double bond is attached to two different groups (specifically, \(-\text{H}\) and \(-\text{CH}_3\)) (1)

**(d)(iii) [Total: 2 marks]**
* **M1**: Complete disappearance of the O-H (alcohol) absorption band in the region \(3200\text{--}3750\text{ cm}^{-1}\) (1)
* **M2**: Appearance of a C=C (alkene) absorption band in the region \(1620\text{--}1670\text{ cm}^{-1}\) (1)
* *Note*: Exact ranges must match standard database values (ranges containing these values are acceptable).

(a) The reaction between concentrated sulfuric acid and water/alcohols is highly exothermic. Adding it slowly with cooling prevents the mixture from boiling uncontrollably and splashing.
(b) The diagram must show a round-bottomed or pear-shaped flask with a vertical condenser. Water must enter at the bottom of the condenser jacket and leave at the top. The top of the condenser must remain open to prevent pressure build-up.
(c) Transfer the distillate to a separating funnel. Allow the layers to separate. Since 1-bromobutane is denser than water, it forms the lower layer. Run off and collect the lower layer.
(d) Washing with sodium hydrogencarbonate neutralises any remaining acidic impurities (such as HBr or \( \text{H}_2\text{SO}_4 \)). Equation: \( \text{H}^+(aq) + \text{HCO}_3^-(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) \).
(e) Suitable drying agent is anhydrous calcium chloride (or anhydrous sodium sulfate). The liquid changes from cloudy to clear.
(f) Moles of butan-1-ol = \( 10.0 / 74.0 = 0.1351 \text{ mol} \). Theoretical yield of 1-bromobutane = \( 0.1351 \times 137.0 = 18.51 \text{ g} \). Percentage yield = \( (11.5 / 18.51) \times 100 = 62.135\% \), which rounds to 62.1%.

(a) 1 mark: To prevent a highly exothermic reaction / boiling over / splattering.
(b) 3 marks: 1 mark for correct flask and vertical condenser setup; 1 mark for water entering at bottom and leaving at top; 1 mark for open top (no stopper).
(c) 2 marks: 1 mark for using a separating funnel; 1 mark for identifying the lower layer as 1-bromobutane and running it off.
(d) 2 marks: 1 mark for stating it neutralises excess acid; 1 mark for the correct ionic equation (including state symbols or not, but species must be correct).
(e) 2 marks: 1 mark for identifying anhydrous calcium chloride / anhydrous calcium sulfate / anhydrous sodium sulfate; 1 mark for describing the change from cloudy to clear / colorless.
(f) 3 marks: 1 mark for calculating moles of butan-1-ol (0.1351 mol); 1 mark for calculating theoretical mass of 1-bromobutane (18.51 g); 1 mark for correct percentage yield to 3 significant figures (62.1%).

(a) The ionic equation is \( \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \).
(b) The assumptions are: 1. The density of the combined solution is \( 1.00 \text{ g cm}^{-3} \); 2. The specific heat capacity of the combined solution is \( 4.18 \text{ J g}^{-1} \text{ K}^{-1} \).
(c) Total mass of mixture = \( 50.0 + 50.0 = 100.0 \text{ g} \). Temperature change \( \Delta T = 31.8 - 18.5 = 13.3 \text{ }^\circ\text{C} \). Heat energy \( q = m c \Delta T = 100.0 \times 4.18 \times 13.3 = 5559.4 \text{ J} = 5.56 \text{ kJ} \).
(d) Moles of \( \text{H}^+ = 0.0500 \times 2.00 = 0.100 \text{ mol} \). Moles of \( \text{OH}^- = 0.0500 \times 2.05 = 0.1025 \text{ mol} \). Since \( \text{H}^+ \) is limiting, \( 0.100 \text{ mol} \) of water is formed. \( \Delta H_{\text{neut}} = -5.5594 / 0.100 = -55.594 \text{ kJ mol}^{-1} \), which rounds to \( -55.6 \text{ kJ mol}^{-1} \).
(e) Heat is absorbed by the polystyrene cup and thermometer; incomplete mixing means the maximum temperature is not fully reached; measurement errors in volume/temperature.
(f) Place a lid on the polystyrene cup (or place the cup inside a beaker lined with cotton wool).

(a) 1 mark: \( \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \) (must include state symbols).
(b) 2 marks: 1 mark for density of solution is \( 1.00 \text{ g cm}^{-3} \); 1 mark for specific heat capacity of solution is \( 4.18 \text{ J g}^{-1} \text{ K}^{-1} \) (same as water).
(c) 2 marks: 1 mark for calculating mass and temperature change; 1 mark for correct calculation of heat in kJ (5.56 kJ).
(d) 4 marks: 1 mark for calculating moles of \( \text{HCl} \) (0.100 mol); 1 mark for identifying that \( \text{HCl} \) is the limiting reagent (or stating moles of water formed = 0.100 mol); 1 mark for dividing heat by moles of water; 1 mark for final answer to 3 sf with negative sign (-55.6 kJ mol\(^{-1}\)).
(e) 2 marks: 1 mark each for any two valid reasons (e.g., heat capacity of the thermometer/cup neglected, incomplete mixing, error in reading the thermometer).
(f) 1 mark: Add a lid / use a vacuum flask / wrap the cup in insulating material.

(a) A lilac flame indicates the presence of potassium ions (\( \text{K}^+ \)).
(b)(i) A green precipitate with NaOH that is insoluble in excess indicates iron(II) ions (\( \text{Fe}^{2+} \)). (ii) The ionic equation is \( \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s) \).
(c)(i) Gas turning limewater cloudy is carbon dioxide, indicating the presence of carbonate ions (\( \text{CO}_3^{2-} \)) or hydrogencarbonate (\( \text{HCO}_3^- \)). (ii) The ionic equation is \( \text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) \).
(d)(i) A cream precipitate with silver nitrate indicates bromide ions (\( \text{Br}^- \)). (ii) Silver bromide dissolves in concentrated aqueous ammonia.
(e) To test for sulfate, add dilute hydrochloric acid (to remove any carbonate impurities) followed by barium chloride solution. A white precipitate of barium sulfate forms.
(f) The four ions present are \( \text{K}^+ \), \( \text{Fe}^{2+} \), \( \text{CO}_3^{2-} \), and \( \text{Br}^- \). Since the mixture consists of two soluble salts, they must be potassium carbonate (\( \text{K}_2\text{CO}_3 \)) and iron(II) bromide (\( \text{FeBr}_2 \)). Note: Iron(II) carbonate is insoluble, so the soluble combination must be \( \text{K}_2\text{CO}_3 \) and \( \text{FeBr}_2 \).

(a) 1 mark: Potassium ion / \( \text{K}^+ \).
(b) 3 marks: (i) 1 mark for Iron(II) / \( \text{Fe}^{2+} \). (ii) 2 marks: 1 mark for correct species, 1 mark for correct state symbols.
(c) 3 marks: (i) 1 mark for Carbonate / \( \text{CO}_3^{2-} \) (accept hydrogencarbonate / \( \text{HCO}_3^- \)). (ii) 2 marks: 1 mark for correct species, 1 mark for correct balancing.
(d) 2 marks: (i) 1 mark for Bromide / \( \text{Br}^- \). (ii) 1 mark for stating that the precipitate dissolves.
(e) 2 marks: 1 mark for adding dilute HCl and \( \text{BaCl}_2 \) solution (or nitric acid and barium nitrate); 1 mark for white precipitate.
(f) 2 marks: 1 mark for \( \text{K}_2\text{CO}_3 \); 1 mark for \( \text{FeBr}_2 \).

(a) To prepare the solution: 1. Dissolve the weighed solid completely in the beaker using a portion of deionised water; 2. Transfer the solution quantitatively to the 250.0 cm\(^3\) volumetric flask using a funnel; 3. Rinse the beaker, glass rod, and funnel with deionised water, adding all washings to the flask; 4. Add deionised water until the bottom of the meniscus is exactly aligned with the graduation mark on the neck of the flask, then stopper and invert several times to ensure a homogeneous solution.
(b)(i) \( \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \).
(ii) Moles of \( \text{HCl} = 0.100 \times (25.00 / 1000) = 2.50 \times 10^{-3} \text{ mol} \).
(iii) Moles of \( \text{Na}_2\text{CO}_3 = 0.5 \times 2.50 \times 10^{-3} = 1.25 \times 10^{-3} \text{ mol} \).
(iv) Moles in 250.0 cm\(^3 = 1.25 \times 10^{-3} \times 10 = 0.0125 \text{ mol} \).
(v) Molar mass of \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} = 3.58 / 0.0125 = 286.4 \text{ g mol}^{-1} \).
(vi) \( M_r(\text{Na}_2\text{CO}_3) = (23.0 \times 2) + 12.0 + (16.0 \times 3) = 106.0 \text{ g mol}^{-1} \). Mass of water per mole of hydrate = \( 286.4 - 106.0 = 180.4 \text{ g mol}^{-1} \). \( x = 180.4 / 18.0 = 10.02 \), so \( x = 10 \).

(a) 4 marks: 1 mark for dissolving the solid in a beaker with a reasonable volume of deionised water; 1 mark for quantitative transfer of the solution and washings into the volumetric flask; 1 mark for filling the flask up to the graduation mark with deionised water (meniscus level); 1 mark for stoppering and inverting the flask.
(b) 8 marks: (i) 1 mark for the correct balanced chemical equation. (ii) 1 mark for calculating moles of HCl (\( 2.50 \times 10^{-3} \text{ mol} \)). (iii) 1 mark for halving the moles of HCl to find moles of sodium carbonate (\( 1.25 \times 10^{-3} \text{ mol} \)). (iv) 1 mark for multiplying by 10 (\( 0.0125 \text{ mol} \)). (v) 2 marks: 1 mark for using mass / moles; 1 mark for correct value (286.4 g mol\(^{-1}\)). (vi) 2 marks: 1 mark for subtracting 106.0 from the calculated molar mass (180.4); 1 mark for dividing by 18.0 to get 10.

Using the rate equation \(\text{rate} = k[A]^2[B]\), if \([A]\) is doubled to \(2[A]\) and \([B]\) is halved to \(0.5[B]\), the new rate is \(\text{rate}' = k(2[A])^2(0.5[B]) = k(4[A]^2)(0.5[B]) = 2k[A]^2[B] = 2 \times \text{rate}\). Therefore, the rate increases by a factor of 2.

1 mark for selecting the correct option B.

First calculate the entropy change of the surroundings: \(\Delta S_{\text{surroundings}}^\ominus = -\frac{\Delta H^\ominus}{T} = -\frac{-92200\text{ J mol}^{-1}}{298\text{ K}} = +309.4\text{ J K}^{-1}\text{ mol}^{-1}\). Since \(\Delta S_{\text{total}}^\ominus = \Delta S_{\text{system}}^\ominus + \Delta S_{\text{surroundings}}^\ominus\), we have \(\Delta S_{\text{system}}^\ominus = \Delta S_{\text{total}}^\ominus - \Delta S_{\text{surroundings}}^\ominus = +110 - 309.4 = -199.4\text{ J K}^{-1}\text{ mol}^{-1}\), which rounds to \(-199\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark for selecting the correct option A.

The sum of the partial pressures at equilibrium equals the total pressure: \(p(\text{total}) = p(X) + p(Y) + p(Z) = 3.0\text{ atm}\). Therefore, \(1.2 + 0.6 + p(Z) = 3.0\text{ atm}\), so \(p(Z) = 1.2\text{ atm}\). The expression for \(K_p\) is: \(K_p = \frac{p(Z)^2}{p(X)^2 \times p(Y)}\). Substituting the values: \(K_p = \frac{1.2^2}{1.2^2 \times 0.6} = \frac{1}{0.6} = 1.67\text{ atm}^{-1}\).

1 mark for selecting the correct option C.

For a weak acid buffer, \([H^+] = K_a \times \frac{[HA]}{[A^-]}\). Since both components are diluted to the same final volume of \(100.0\text{ cm}^3\), we can use the moles of each: moles of propanoic acid = \(0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\); moles of propanoate ions = \(0.0500\text{ dm}^3 \times 0.050\text{ mol dm}^{-3} = 0.00250\text{ mol}\). Thus, \([H^+] = 1.35 \times 10^{-5} \times \frac{0.00500}{0.00250} = 2.70 \times 10^{-5}\text{ mol dm}^{-3}\). Finally, \(\text{pH} = -\log_{10}(2.70 \times 10^{-5}) = 4.57\).

1 mark for selecting the correct option A.

Optical isomerism requires a chiral carbon atom (four different groups attached to it). The alkaline iodine test (iodoform reaction) yields a yellow precipitate of triiodomethane (\(\text{CHI}_3\)) in the presence of a methyl carbonyl group (\(\text{CH}_3\text{C=O}\)) or a secondary alcohol with the structure \(\text{CH}_3\text{CH(OH)}-\). Butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)) contains a chiral centre (carbon-2 has H, OH, \(\text{CH}_3\), and \(\text{CH}_2\text{CH}_3\) groups attached) and contains the \(\text{CH}_3\text{CH(OH)}-\) group, giving a positive iodoform test. Propan-2-ol and pentan-2-one give positive iodoform tests but are not chiral. Pentan-3-one is neither chiral nor does it give a positive iodoform test.

1 mark for selecting the correct option A.

Using the Arrhenius equation in two-point form: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\). Substitute the values: \(\ln(2.0) = -\frac{E_a}{8.31}\left(\frac{1}{310} - \frac{1}{300}\right)\). This simplifies to \(0.69315 = -\frac{E_a}{8.31}\left(-1.075 \times 10^{-4}\right)\), which gives \(0.69315 = E_a \times 1.294 \times 10^{-5}\). Thus, \(E_a = \frac{0.69315}{1.294 \times 10^{-5}} = 53566\text{ J mol}^{-1} \approx 53.6\text{ kJ mol}^{-1}\).

1 mark for selecting the correct option B.

Ammonia is a weak base and hydrochloric acid is a strong acid. The equivalence point of this titration lies in the acidic region (typically around pH 5), and the rapid pH change (vertical section) occurs approximately between pH 3 and 7. The indicator used must have a pH transition range that falls entirely within this vertical section. Methyl orange (pH range 3.1 - 4.4) is ideal for a weak base-strong acid titration.

1 mark for selecting the correct option B.

Hydrolysis of an ester yields a carboxylic acid and an alcohol. Since oxidation of alcohol \(Z\) yields a ketone, \(Z\) must be a secondary alcohol. The simplest secondary alcohol is propan-2-ol, which has 3 carbons. Since ester \(X\) has the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\), it contains 5 carbon atoms. Therefore, carboxylic acid \(Y\) must contain 2 carbon atoms (ethanoic acid). The ester formed from ethanoic acid and propan-2-ol is propan-2-yl ethanoate.

1 mark for selecting the correct option D.

Using the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)

Substituting the given values:
\(\ln\left(\frac{9.60 \times 10^{-4}}{1.20 \times 10^{-4}}\right) = \ln(8) = 2.0794\)

The temperature term is:
\(\left(\frac{1}{300} - \frac{1}{320}\right) = 2.0833 \times 10^{-4}\text{ K}^{-1}\)

Rearranging for \(E_a\):
\(E_a = \frac{2.0794 \times 8.31}{2.0833 \times 10^{-4}} = 82944\text{ J mol}^{-1} = 82.9\text{ kJ mol}^{-1}\).

[1 mark] B - Correct answer.
[0 marks] For any other option.

Comparing Experiments 1 and 2: doubling \([\text{A}]\) at constant \([\text{B}]\) doubles the rate, so the reaction is first order with respect to \(\text{A}\).
Comparing Experiments 2 and 3: tripling \([\text{B}]\) at constant \([\text{A}]\) increases the rate by a factor of \(9\) (from \(4.0 \times 10^{-3}\) to \(3.6 \times 10^{-2}\)), so the reaction is second order with respect to \(\text{B}\).

Thus, the rate equation is:
\(\text{Rate} = k[\text{A}][\text{B}]^2\)
- Overall order = \(1 + 2 = 3\).
- Units of \(k\):
\(k = \frac{\text{Rate}}{[\text{A}][\text{B}]^2} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

[1 mark] B - Correct overall order and units.
[0 marks] For any other option.

1. Calculate \(\Delta S_{\text{system}}^\ominus\):
\(\Delta S_{\text{system}}^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S_{\text{system}}^\ominus = [105 + 70] - [214 + 2(193)] = 175 - 600 = -425\text{ J mol}^{-1}\text{ K}^{-1}\)

2. Calculate \(\Delta S_{\text{surrounding}}^\ominus\):
\(\Delta S_{\text{surrounding}}^\ominus = -\frac{\Delta H^\ominus}{T} = -\frac{-134000\text{ J mol}^{-1}}{298\text{ K}} = +449.7\text{ J mol}^{-1}\text{ K}^{-1}\)

3. Calculate \(\Delta S_{\text{total}}^\ominus\):
\(\Delta S_{\text{total}}^\ominus = \Delta S_{\text{system}}^\ominus + \Delta S_{\text{surrounding}}^\ominus = -425 + 449.7 = +24.7\text{ J mol}^{-1}\text{ K}^{-1} \approx +25\text{ J mol}^{-1}\text{ K}^{-1}\).

[1 mark] C - Correct calculation.
[0 marks] For any other option.

Applying Hess's Law through a Born-Haber cycle:
\(\Delta H_f^\ominus = \Delta H_{\text{at}}^\ominus(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + \text{E}_{\text{bond}}(\text{F}_2) + 2 \times \text{EA}(\text{F}) + \Delta H_{\text{latt}}^\ominus\)

Note that the bond enthalpy of \(\text{F}_2\) directly yields \(2\text{ mol}\) of \(\text{F}(g)\) atoms, so we do not divide it by 2.

\(-1124 = 148 + 738 + 1451 + 158 + 2(-328) + \Delta H_{\text{latt}}^\ominus\)
\(-1124 = 1839 + \Delta H_{\text{latt}}^\ominus\)
\(\Delta H_{\text{latt}}^\ominus = -1124 - 1839 = -2963\text{ kJ mol}^{-1}\).

[1 mark] B - Correct lattice energy calculation.
[0 marks] For any other option.

According to Dalton's Law of partial pressures:
\(p_{\text{total}} = p(\text{N}_2\text{O}_4) + p(\text{NO}_2)\)
\(1.20\text{ atm} = 0.40\text{ atm} + p(\text{NO}_2) \Rightarrow p(\text{NO}_2) = 0.80\text{ atm}\)

The expression for \(K_p\) is:
\(K_p = \frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)}\)

Substituting the partial pressures:
\(K_p = \frac{(0.80)^2}{0.40} = \frac{0.64}{0.40} = 1.60\text{ atm}\).

[1 mark] C - Correct partial pressure subtraction and expression execution.
[0 marks] For any other option.

1. Find initial moles:
- Moles of propanoic acid (\(\text{HA}\)) = \(0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\).
- Moles of \(\text{NaOH}\) (\(\text{OH}^-\)) = \(0.0500\text{ dm}^3 \times 0.050\text{ mol dm}^{-3} = 0.00250\text{ mol}\).

2. Solve the stoichiometry:
\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)
- Moles of \(\text{HA}\) remaining = \(0.00500 - 0.00250 = 0.00250\text{ mol}\).
- Moles of propanoate (\(\text{A}^-\)) formed = \(0.00250\text{ mol}\).

3. Calculate the pH:
Since the remaining moles of weak acid equal the moles of conjugate base formed (\([\text{HA}] = [\text{A}^-]\)), the buffer system simplifies to:
\(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

[1 mark] B - Correct buffer stoichiometry and pH calculation.
[0 marks] For any other option.

For a weak monoprotic acid, the approximation for hydrogen ion concentration is:
\([\text{H}^+] \approx \sqrt{K_a \times [\text{HX}]}\)

Substitute the given values:
\([\text{H}^+] = \sqrt{(4.00 \times 10^{-6}\text{ mol dm}^{-3}) \times (0.0250\text{ mol dm}^{-3})} = \sqrt{1.00 \times 10^{-7}} = 3.162 \times 10^{-4}\text{ mol dm}^{-3}\)

Calculate pH:
\(\text{pH} = -\log_{10}(3.162 \times 10^{-4}) = 3.50\).

[1 mark] B - Correct calculation using weak acid formula.
[0 marks] For any other option.

- Reaction A: Butanone reduced with \(\text{NaBH}_4\) yields butan-2-ol, which has a chiral centre (bonded to H, OH, methyl, and ethyl groups). It is chiral and forms a racemic mixture.
- Reaction B: Propanal + \(\text{HCN}\) yields 2-hydroxybutanenitrile, which has a chiral centre. It is chiral.
- Reaction C: Propanone + \(\text{HCN}\) yields 2-hydroxy-2-methylpropanenitrile. The central carbon is bonded to \(\text{-OH}\), \(\text{-CN}\), and two identical \(\text{-CH}_3\) groups, meaning the product molecule is achiral.
- Reaction D: Phenylethanone + \(\text{LiAlH}_4\) yields 1-phenylethanol, which has a chiral centre.

[1 mark] C - Correct identification of the achiral product.
[0 marks] For any other option.

According to the Arrhenius equation, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\).

The gradient of a plot of \(\ln k\) against \(\frac{1}{T}\) is given by:
\(\text{gradient} = -\frac{E_a}{R}\)

Substituting the given values:
\(-1.20 \times 10^4\text{ K} = -\frac{E_a}{8.31\text{ J K}^{-1}\text{ mol}^{-1}}\)

\(E_a = 1.20 \times 10^4 \times 8.31 = 99720\text{ J mol}^{-1}\)

To convert this to \(\text{kJ mol}^{-1}\):
\(E_a = \frac{99720}{1000} = +99.72\text{ kJ mol}^{-1} \approx +99.7\text{ kJ mol}^{-1}\).

1 mark for the correct calculation of \(E_a\) with correct conversion to \(\text{kJ mol}^{-1}\) (corresponding to B).

- Reject option A: derived by dividing the gradient by \(R\).
- Reject option C: incorrect units (omitted division by 1000 but kept the kJ unit).
- Reject option D: incorrect multiplication/expression.

First, calculate the entropy change of the surroundings, \(\Delta S^{\ominus}_{\text{surroundings}}\):

\(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{\Delta H^{\ominus}}{T}\)

\(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{-54.2 \times 10^3\text{ J mol}^{-1}}{298\text{ K}} = +181.9\text{ J K}^{-1}\text{ mol}^{-1}\)

Now, calculate the total entropy change, \(\Delta S^{\ominus}_{\text{total}}\):

\(\Delta S^{\ominus}_{\text{total}} = \Delta S^{\ominus}_{\text{system}} + \Delta S^{\ominus}_{\text{surroundings}}\)

\(\Delta S^{\ominus}_{\text{total}} = -145 + 181.9 = +36.9\text{ J K}^{-1}\text{ mol}^{-1} \approx +37\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark for the correct calculation of total entropy change (corresponding to C).

- Reject option A: incorrect sign in surroundings entropy calculation.
- Reject option B: incorrect calculation.
- Reject option D: this is only the entropy change of the surroundings.

1. Calculate the initial moles of propanoic acid (\(\text{HA}\)) and sodium hydroxide (\(\text{NaOH}\)):

\(\text{moles of HA} = \frac{50.0}{1000} \times 0.100 = 0.00500\text{ mol}\)

\(\text{moles of NaOH} = \frac{50.0}{1000} \times 0.0500 = 0.00250\text{ mol}\)

2. Determine the moles after reaction:

\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)

All of the \(\text{NaOH}\) reacts, so:

\(\text{moles of HA remaining} = 0.00500 - 0.00250 = 0.00250\text{ mol}\)

\(\text{moles of A}^- \text{ formed} = 0.00250\text{ mol}\)

3. Calculate \([\text{H}^+]\):

Since \([\text{HA}] = [\text{A}^-]\) in the buffer solution:

\([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)

\(\text{pH} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

1 mark for correct buffer pH calculation (corresponding to A).

- Reject option B: assumes initial concentrations are used without stoichiometry reaction.
- Reject option C: incorrect ratio inversion.
- Reject option D: pH of propanoic acid before the addition of NaOH.

- Reaction with 2,4-DNPH indicates a carbonyl group (aldehyde or ketone).
- No reaction with Fehling's solution indicates it is a ketone, not an aldehyde.
- Reaction with \(\text{I}_2/\text{NaOH}\) indicates a methyl ketone structure (\(\text{CH}_3\text{C=O}\)).
- This matches all four options since they are all methyl ketones with molecular formula \(\text{C}_6\text{H}_{12}\text{O}\).
- For **X** to exist as a pair of enantiomers, it must have a chiral carbon atom (a carbon bonded to four different groups).

Let's check the structures:

- **3-methylpentan-2-one**: \(\text{CH}_3\text{COCH(CH}_3)\text{CH}_2\text{CH}_3\). The C3 carbon is chiral because it is bonded to: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{COCH}_3\). This is chiral.

- **4-methylpentan-2-one**: \(\text{CH}_3\text{COCH}_2\text{CH(CH}_3)_2\). No chiral carbon.

- **hexan-2-one**: \(\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{CH}_3\). No chiral carbon.

- **3,3-dimethylbutan-2-one**: \(\text{CH}_3\text{COC(CH}_3)_3\). No chiral carbon.

1 mark for identifying 3-methylpentan-2-one based on the chemical tests and chirality requirement (corresponding to A).

- Reject options B, C, D because although they satisfy all chemical tests, they lack a chiral center.

(a)(i)
- Compare Exp 1 and Exp 2: [A] doubles, [B] and [C] are constant. The rate doubles (from \(1.20 \times 10^{-4}\) to \(2.40 \times 10^{-4}\)). Therefore, the reaction is **first order** with respect to A.
- Compare Exp 1 and Exp 3: [B] doubles, [A] and [C] are constant. The rate quadruples (from \(1.20 \times 10^{-4}\) to \(4.80 \times 10^{-4}\)). Since \(2^2 = 4\), the reaction is **second order** with respect to B.
- Compare Exp 1 and Exp 4: [C] doubles, [A] and [B] are constant. The rate remains unchanged (\(1.20 \times 10^{-4}\)). Therefore, the reaction is **zero order** with respect to C.

(a)(ii)
- The rate equation is: \(\text{Rate} = k[\text{A}][\text{B}]^2\)
- Rearranging for \(k\): \(k = \frac{\text{Rate}}{[\text{A}][\text{B}]^2}\)
- Substituting values from Exp 1: \(k = \frac{1.20 \times 10^{-4}}{0.10 \times (0.10)^2} = \frac{1.20 \times 10^{-4}}{1.00 \times 10^{-3}} = 0.120\)
- Units: \(\frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{dm}^{6} \text{ mol}^{-2} \text{ s}^{-1}\)
- Final value: \(0.120 \text{ dm}^{6} \text{ mol}^{-2} \text{ s}^{-1}\)

(b)
- The rate equation depends only on A and B, where Step 1 is the slow step (rate-determining step).
- The reactants of Step 1 consist of one molecule of A and two molecules of B (since one B molecule cancels out as it appears on both sides, the overall stoichiometry of the slow step is \(\text{A} + \text{B} \rightarrow \text{I}\), but here \(\text{A} + 2\text{B}\) means two molecules of B are involved, and one is regenerated. This perfectly matches the order of 1 for A and 2 for B).

(c)
- Using the Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
- Rearranging: \(E_a = \frac{R \ln(k_2 / k_1)}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}\)
- Compute the terms:
- \(\ln\left(\frac{6.45 \times 10^{-2}}{1.20 \times 10^{-2}}\right) = \ln(5.375) = 1.6818\)
- \(\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{298} - \frac{1}{318} = 3.3557 \times 10^{-3} - 3.1447 \times 10^{-3} = 2.1105 \times 10^{-4} \text{ K}^{-1}\)
- Calculate \(E_a\):
- \(E_a = \frac{8.31 \times 1.6818}{2.1105 \times 10^{-4}} = 66221 \text{ J mol}^{-1}\)
- \(E_a = 66.2 \text{ kJ mol}^{-1}\) (to 3 significant figures)

(a)(i) (4 marks):
- 1 mark: Deduce first order with respect to A + explanation linking [A] doubling to rate doubling.
- 1 mark: Deduce second order with respect to B + explanation linking [B] doubling to rate quadrupling.
- 1 mark: Deduce zero order with respect to C + explanation linking [C] doubling to rate remaining constant.
- 1 mark: Correctly states all three orders.

(a)(ii) (3 marks):
- 1 mark: Correct rate equation: \(\text{Rate} = k[\text{A}][\text{B}]^2\).
- 1 mark: Correct calculation of \(k = 0.120\) (allow error carried forward from incorrect orders).
- 1 mark: Correct units: \(\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\).

(b) (2 marks):
- 1 mark: States that Step 1 is the rate-determining step because it is the slow step.
- 1 mark: Explains that the species in the rate-determining step match the reactants in the rate equation, where 1 molecule of A and 2 molecules of B are involved in Step 1.

(c) (5 marks):
- 1 mark: Correctly recalls the Arrhenius equation in logarithmic form.
- 1 mark: Correct calculation of \(\ln(k_2/k_1) = 1.68\).
- 1 mark: Correct calculation of \(\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = 2.11 \times 10^{-4} \text{ K}^{-1}\).
- 1 mark: Correct substitution to get \(E_a\) in Joules (\(66221 \text{ J mol}^{-1}\)).
- 1 mark: Correct conversion to \(\text{kJ mol}^{-1}\) to 3 s.f. and positive sign: \(+66.2\).

(a)(i)
- \(\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\)
- \(\Delta S^{\ominus}_{\text{system}} = [S^{\ominus}(\text{NH}_2\text{CONH}_2(\text{s})) + S^{\ominus}(\text{H}_2\text{O}(\text{l}))] - [2 \times S^{\ominus}(\text{NH}_3(\text{g})) + S^{\ominus}(\text{CO}_2(\text{g}))]\)
- \(\Delta S^{\ominus}_{\text{system}} = (104.6 + 69.9) - (2 \times 192.3 + 213.6)\)
- \(\Delta S^{\ominus}_{\text{system}} = 174.5 - (384.6 + 213.6) = 174.5 - 598.2 = -423.7 \text{ J K}^{-1} \text{ mol}^{-1}\)

(a)(ii)
- The value of \(\Delta S^{\ominus}_{\text{system}}\) is negative (entropy decreases) because 3 moles of gaseous reactants (highly disordered) react to form 1 mole of solid and 1 mole of liquid (much more ordered states).
- A decrease in the number of gaseous molecules leads to a significant decrease in disorder.

(b)(i)
- \(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{\Delta H^{\ominus}}{T}\)
- \(\Delta H^{\ominus} = -134.0 \text{ kJ mol}^{-1} = -134000 \text{ J mol}^{-1}\)
- \(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{-134000}{298} = +449.7 \text{ J K}^{-1} \text{ mol}^{-1}\) (or \(+450 \text{ J K}^{-1} \text{ mol}^{-1}\))

(b)(ii)
- \(\Delta S^{\ominus}_{\text{total}} = \Delta S^{\ominus}_{\text{system}} + \Delta S^{\ominus}_{\text{surroundings}}\)
- \(\Delta S^{\ominus}_{\text{total}} = -423.7 + 449.7 = +26.0 \text{ J K}^{-1} \text{ mol}^{-1}\)
- Since \(\Delta S^{\ominus}_{\text{total}} > 0\) (positive), the reaction is feasible at \(298 \text{ K}\).

(c)
- As temperature increases, \(\Delta S^{\ominus}_{\text{surroundings}}\) (which is equal to \(-\frac{\Delta H^{\ominus}}{T}\)) becomes less positive (decreases in magnitude).
- Since \(\Delta S^{\ominus}_{\text{system}}\) is negative, the overall \(\Delta S^{\ominus}_{\text{total}}\) decreases and will eventually become negative.
- Therefore, the reaction becomes less feasible as temperature increases.

(d)
- When urea dissolves, the highly ordered solid crystal structure breaks down and the urea molecules disperse randomly in water. This causes a significant increase in the entropy of the system (\(\Delta S^{\ominus}_{\text{system}} > 0\)).
- This positive \(\Delta S^{\ominus}_{\text{system}}\) is larger in magnitude than the negative \(\Delta S^{\ominus}_{\text{surroundings}}\) caused by the endothermic process, resulting in a positive \(\Delta S^{\ominus}_{\text{total}}\) (making the process spontaneous).

(a)(i) (3 marks):
- 1 mark: Correct sum of products: \(174.5\).
- 1 mark: Correct sum of reactants: \(598.2\).
- 1 mark: Correct answer with negative sign and units: \(-423.7 \text{ J K}^{-1} \text{ mol}^{-1}\).

(a)(ii) (2 marks):
- 1 mark: Entropy decreases because reactants are gaseous, and products are solid/liquid.
- 1 mark: Linking the loss of 3 moles of gas to a decrease in disorder.

(b)(i) (2 marks):
- 1 mark: Correct formula and conversion of \(\Delta H\) to \(\text{J mol}^{-1}\).
- 1 mark: Correct calculation with positive sign: \(+449.7 \text{ J K}^{-1} \text{ mol}^{-1}\).

(b)(ii) (2 marks):
- 1 mark: Correct total entropy: \(+26.0 \text{ J K}^{-1} \text{ mol}^{-1}\) (or \(+26.3\) if using 450).
- 1 mark: States reaction is feasible because \(\Delta S_{\text{total}}\) is positive.

(c) (3 marks):
- 1 mark: States \(-\Delta H / T\) (or \(\Delta S_{\text{surroundings}}\)) decreases as \(T\) increases.
- 1 mark: States \(\Delta S_{\text{total}}\) becomes less positive / negative.
- 1 mark: Concludes the reaction becomes less feasible.

(d) (2 marks):
- 1 mark: Dissolution results in an increase in system entropy (or \(\Delta S_{\text{system}}\) is highly positive).
- 1 mark: This outweighs the negative \(\Delta S_{\text{surroundings}}\), making \(\Delta S_{\text{total}}\) positive / \(\Delta G\) negative.

(a)(i)
- \(K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}\)

(a)(ii)
- \(K_a \approx \frac{[\text{H}^+]^2}{[\text{C}_6\text{H}_5\text{COOH}]}\)
- \([\text{H}^+] = \sqrt{K_a \times [\text{C}_6\text{H}_5\text{COOH}]} = \sqrt{6.30 \times 10^{-5} \times 0.0800} = \sqrt{5.04 \times 10^{-6}} = 2.245 \times 10^{-3} \text{ mol dm}^{-3}\)
- \(\text{pH} = -\log_{10}(2.245 \times 10^{-3}) = 2.65\)
- Assumptions:
1. The concentration of \(\text{H}^+\) ions from the ionization of water is negligible (so \([\text{H}^+] \approx [\text{C}_6\text{H}_5\text{COO}^-]\)).
2. The ionization of benzoic acid is so small that the equilibrium concentration of benzoic acid is equal to its initial concentration (\([\text{C}_6\text{H}_5\text{COOH}]_{\text{equilibrium}} \approx 0.0800 \text{ mol dm}^{-3}\)).

(b)(i)
- Determine moles of each component in the buffer mixture:
- Moles of benzoic acid, \(n(\text{HA}) = 0.0800 \times \frac{50.0}{1000} = 4.00 \times 10^{-3} \text{ mol}\)
- Moles of benzoate ions, \(n(\text{A}^-) = 0.100 \times \frac{30.0}{1000} = 3.00 \times 10^{-3} \text{ mol}\)
- Total volume of mixture = \(50.0 + 30.0 = 80.0 \text{ cm}^3\) (cancels out in the ratio)
- Using the expression: \([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)}\)
- \([\text{H}^+] = 6.30 \times 10^{-5} \times \frac{4.00 \times 10^{-3}}{3.00 \times 10^{-3}} = 8.40 \times 10^{-5} \text{ mol dm}^{-3}\)
- Calculate pH:
- \(\text{pH} = -\log_{10}(8.40 \times 10^{-5}) = 4.08\)

(b)(ii)
- When \(\text{H}^+\) ions are added, they react with the conjugate base, benzoate ions:
- \(\text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH}\)
- This removes the added acid, preventing a large change in pH.

(c)
- Benzoic acid is a weak acid and sodium hydroxide is a strong base, so the pH at the equivalence point of this titration will be greater than 7 (in the basic region, typically around pH 8.5).
- Methyl red has a pH range of 4.2 – 6.3, which lies entirely in the acidic region. It will change colour long before the equivalence point is reached, making it unsuitable.

(a)(i) (1 mark):
- 1 mark: Correct expression for \(K_a\).

(a)(ii) (4 marks):
- 1 mark: Correct rearrangement/expression \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]}\).
- 1 mark: Correct calculation of \([\text{H}^+] = 2.25 \times 10^{-3} \text{ mol dm}^{-3}\).
- 1 mark: Correct calculation of pH to 2 decimal places: \(2.65\).
- 1 mark: State both correct assumptions.

(b)(i) (5 marks):
- 1 mark: Calculates moles of benzoic acid: \(4.00 \times 10^{-3} \text{ mol}\).
- 1 mark: Calculates moles of benzoate: \(3.00 \times 10^{-3} \text{ mol}\).
- 1 mark: Correct ratio or concentrations of acid and salt.
- 1 mark: Correct calculation of \([\text{H}^+] = 8.40 \times 10^{-5} \text{ mol dm}^{-3}\).
- 1 mark: Correct calculation of pH to 2 decimal places: \(4.08\).

(b)(ii) (2 marks):
- 1 mark: Correct ionic equation: \(\text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH}\).
- 1 mark: Explains that the added hydronium/hydrogen ions are neutralized/removed.

(c) (2 marks):
- 1 mark: Identifies that the equivalence point of a weak acid–strong base titration is in the alkaline/basic region (pH > 7).
- 1 mark: Explains that methyl red's pH range (4.2 – 6.3) is too low and it will change colour prematurely.

(a)(i)
- **Compound X** reacts with 2,4-DNPH, showing it contains a carbonyl group (aldehyde or ketone).
- It does not react with Fehling's solution, which shows it is not an aldehyde. Therefore, it must be a ketone.
- For a 4-carbon ketone, the only possible structure is butanone.
- Structural formula: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) (or \(\text{CH}_3\text{CH}_2\text{COCH}_3\))
- IUPAC name: butanone

(a)(ii)
- Observation: Remains colourless / no change / no silver mirror.
- Reason: Ketones cannot be oxidized by mild oxidizing agents like Tollens' reagent.

(a)(iii)
- Reducing agent: Lithium tetrahydridoaluminate(III) / \(\text{LiAlH}_4\) in dry ether (or sodium tetrahydridoborate(III) / \(\text{NaBH}_4\) in aqueous/alcoholic solution).
- Organic product: butan-2-ol

(b)(i)
- Step 1: Nucleophile attack. The cyanide ion, \(:\text{CN}^-\), attacks the carbonyl carbon which has a \(\delta+\) charge. A curly arrow goes from the lone pair on the carbon of \(:\text{CN}^-\) to the carbonyl carbon. A curly arrow goes from the double bond of \(\text{C}=\text{O}^{\delta-}\) to the oxygen atom.
- Step 2: Intermediate formation. A tetrahedral intermediate \(\text{CH}_3\text{C}(\text{O}^-)(\text{CN})\text{CH}_2\text{CH}_3\) is formed. The oxygen atom carries a negative charge and a lone pair.
- Step 3: Protonation. A curly arrow goes from the lone pair on the negative oxygen to the hydrogen atom of an \(\text{HCN}\) molecule (or \(\text{H}^+\) ion). This forms the product, 2-hydroxy-2-methylbutanenitrile, and regenerates the catalyst \(\text{CN}^-\).

(b)(ii)
- The carbonyl group (\(\text{C}=\text{O}\)) of butanone is planar around the carbonyl carbon.
- The nucleophile (\(:\text{CN}^-\)) can attack this planar carbon atom with equal probability from either above or below the plane.
- This leads to the formation of equal amounts (a 50:50 ratio) of both enantiomers (a racemic mixture).
- The optical rotation of one enantiomer is cancelled by the equal and opposite optical rotation of the other, making the mixture optically inactive.

(c)
- The strong, sharp absorption band at \(1675 - 1750 \text{ cm}^{-1}\) due to the carbonyl group (\(\text{C}=\text{O}\)) of butanone will disappear.
- A broad, strong absorption band at \(3200 - 3675 \text{ cm}^{-1}\) due to the alcohol group (\(\text{O}-\text{H}\)) will appear, and a sharp absorption band at \(2200 - 2250 \text{ cm}^{-1}\) due to the nitrile group (\(\text{C}\equiv\text{N}\)) will appear.

(a)(i) (2 marks):
- 1 mark: Correct structural formula of butanone.
- 1 mark: Correct IUPAC name: butanone.

(a)(ii) (1 mark):
- 1 mark: States no change / no silver mirror because ketones cannot be oxidized.

(a)(iii) (2 marks):
- 1 mark: Identifies \(\text{LiAlH}_4\) (accept \(\text{NaBH}_4\) / hydrogen with Ni catalyst).
- 1 mark: IUPAC name: butan-2-ol.

(b)(i) (4 marks):
- 1 mark: Correct dipole on the \(\text{C}=\text{O}\) bond (\(\text{C}^{\delta+}=\text{O}^{\delta-}\)) and arrow from the lone pair of \(:\text{CN}^-\) to the \(\text{C}\) atom.
- 1 mark: Curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen.
- 1 mark: Correct structure of the tetrahedral intermediate with oxygen carrying a negative charge.
- 1 mark: Curly arrow from the intermediate's \(\text{O}^-\)'s lone pair to the hydrogen atom of \(\text{HCN}\) (or \(\text{H}^+\)) to form the organic product.

(b)(ii) (3 marks):
- 1 mark: Mentions that the carbonyl group / reactant is planar at the reaction site.
- 1 mark: States that the attack of \(:\text{CN}^-\) can occur with equal probability from above or below the plane.
- 1 mark: Concludes that a racemic mixture (50:50 ratio of enantiomers) is formed which cancels out the optical activity.

(c) (2 marks):
- 1 mark: Disappearance of the peak at \(1675 - 1750 \text{ cm}^{-1}\) (\(\text{C}=\text{O}\)).
- 1 mark: Appearance of a broad peak at \(3200 - 3675 \text{ cm}^{-1}\) (\(\text{O}-\text{H}\)) OR a sharp peak at \(2200 - 2250 \text{ cm}^{-1}\) (\(\text{C}\equiv\text{N}\)).

(a)(i)
- \(K_p = \frac{p(\text{C}_2\text{H}_5\text{OH})}{p(\text{C}_2\text{H}_4) \times p(\text{H}_2\text{O})}\)

(a)(ii)
- Total number of moles at equilibrium, \(n_{\text{total}} = 2.00 + 3.00 + 1.50 = 6.50 \text{ mol}\).
- Mole fraction of ethene, \(x(\text{C}_2\text{H}_4) = \frac{2.00}{6.50} = 0.308\) (or \(0.3077\))
- Mole fraction of steam, \(x(\text{H}_2\text{O}) = \frac{3.00}{6.50} = 0.462\) (or \(0.4615\))
- Mole fraction of ethanol, \(x(\text{C}_2\text{H}_5\text{OH}) = \frac{1.50}{6.50} = 0.231\) (or \(0.2308\))

(a)(iii)
- Partial pressure \(p = x \times P_{\text{total}}\)
- \(p(\text{C}_2\text{H}_4) = 0.3077 \times 60.0 = 18.5 \text{ atm}\)
- \(p(\text{H}_2\text{O}) = 0.4615 \times 60.0 = 27.7 \text{ atm}\)
- \(p(\text{C}_2\text{H}_5\text{OH}) = 0.2308 \times 60.0 = 13.8 \text{ atm}\)

(a)(iv)
- Substitute partial pressures into the \(K_p\) expression:
- \(K_p = \frac{13.85}{18.46 \times 27.69} = 0.0271\) (or \(2.71 \times 10^{-2}\))
- Units: \(\frac{\text{atm}}{\text{atm} \times \text{atm}} = \text{atm}^{-1}\)
- Value: \(0.0271 \text{ atm}^{-1}\)

(b)
- The forward reaction is exothermic (\(\Delta H = -45 \text{ kJ mol}^{-1}\)).
- Therefore, increasing the temperature shifts the equilibrium to the left (the endothermic direction) to absorb heat. This decreases the equilibrium concentration/partial pressure of products and increases those of reactants, so **\(K_p\) decreases**.

(c)(i)
- An increase in pressure shifts the equilibrium in the direction of the fewer moles of gas.
- There are 2 moles of gas on the left-hand side and only 1 mole of gas on the right-hand side, so high pressure shifts the equilibrium to the right, **increasing the yield of ethanol**.

(c)(ii)
- High electrical energy costs are required to run the compressors to maintain high pressures.
- Expensive strong-walled reaction vessels and pipes are required to withstand high pressures safely (explosion hazard).

(a)(i) (1 mark):
- 1 mark: Correct expression for \(K_p\) with partial pressure notation \(p\).

(a)(ii) (2 marks):
- 1 mark: Calculates total moles = \(6.50 \text{ mol}\).
- 1 mark: Correctly calculates all three mole fractions to 3 s.f.

(a)(iii) (2 marks):
- 1 mark: Calculates at least two partial pressures correctly.
- 1 mark: Calculates all three partial pressures correctly to 3 s.f.

(a)(iv) (3 marks):
- 1 mark: Correct substitution of values into \(K_p\) expression.
- 1 mark: Correct calculation of value: \(0.0271\) (allow ECF from incorrect partial pressures).
- 1 mark: Correct units: \(\text{atm}^{-1}\).

(b) (2 marks):
- 1 mark: States that \(K_p\) decreases.
- 1 mark: Explains that the reaction is exothermic, so increasing temperature shifts equilibrium in the reverse direction.

(c)(i) (2 marks):
- 1 mark: Identifies there are 2 moles of gas on the left and 1 mole on the right.
- 1 mark: States that increasing pressure shifts the position of equilibrium to the right, increasing ethanol yield.

(c)(ii) (2 marks):
- 1 mark: Explains the high cost of equipment/reactors to safely withstand pressure.
- 1 mark: Explains high energy cost of compression.

To find the standard cell potential:
\( E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = 1.51 \text{ V} - 0.77 \text{ V} = +0.74 \text{ V} \).

The overall spontaneous reaction is:
\( 5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \).

The coefficient of \( \text{Fe}^{2+}(\text{aq}) \) is 5.

1 mark: Correct standard cell potential of +0.74 V and stoichiometric coefficient of 5.

Let us determine the electronic configurations of the gaseous ions in their ground states:
- \( \text{Fe}^{3+} \): The atomic configuration of Fe is \( [\text{Ar}] 3d^6 4s^2 \). For \( \text{Fe}^{3+} \), it is \( [\text{Ar}] 3d^5 \). According to Hund's rule, these five d-electrons occupy separate orbitals with parallel spins, giving 5 unpaired electrons.
- \( \text{Co}^{2+} \): The atomic configuration of Co is \( [\text{Ar}] 3d^7 4s^2 \). For \( \text{Co}^{2+} \), it is \( [\text{Ar}] 3d^7 \), which has 3 unpaired electrons.
- \( \text{Cr}^{3+} \): The atomic configuration of Cr is \( [\text{Ar}] 3d^5 4s^1 \). For \( \text{Cr}^{3+} \), it is \( [\text{Ar}] 3d^3 \), which has 3 unpaired electrons.
- \( \text{Cu}^{2+} \): The atomic configuration of Cu is \( [\text{Ar}] 3d^{10} 4s^1 \). For \( \text{Cu}^{2+} \), it is \( [\text{Ar}] 3d^9 \), which has 1 unpaired electron.

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

1 mark: Correct identification of Fe3+ with 5 unpaired d-electrons.

The nitration reaction proceeds via electrophilic substitution. The active electrophile is the nitronium ion, \( \text{NO}_2^+ \).
Concentrated sulfuric acid acts as a stronger acid than nitric acid, protonating \( \text{HNO}_3 \) to form \( \text{H}_2\text{NO}_3^+ \), which then loses water to produce the electrophile:
\( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \).
Since \( \text{H}_2\text{SO}_4 \) is regenerated in a subsequent step when the hydrogen ion is removed from the benzenium intermediate, it acts as a catalyst as well as an acid.

1 mark: Correct identification of the electrophile as the nitronium ion and the role of concentrated sulfuric acid as acid and catalyst.

The basicity of nitrogenous bases depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton.
- In phenylamine, the lone pair on the nitrogen atom overlaps with the pi-delocalised system of the benzene ring, reducing its electron density and making it much less available to accept a proton.
- In ethylamine and diethylamine, the electron-releasing alkyl groups increase the electron density on the nitrogen atom through the positive inductive effect, making them stronger bases than ammonia.
Therefore, phenylamine is the weakest base.

1 mark: Correct identification of phenylamine as the weakest base.

The reaction that occurs is:
\( [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \).
- The coordination number changes from 6 in \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \) to 4 in \( [\text{CuCl}_4]^{2-} \).
- The shape changes from octahedral to tetrahedral because chloride ligands are relatively large and negatively charged, leading to stronger electrostatic repulsion than water molecules, which results in a tetrahedral structure.
- No precipitation occurs; both reactant and product complexes are soluble in water.
- The oxidation state of copper remains +2.

1 mark: Correct identification that coordination number changes from 6 to 4 and shape changes from octahedral to tetrahedral.

The methyl group (\( -\text{CH}_3 \)) on methylbenzene is a 2,4-directing group (ortho-/para-directing). If we nitrate first, we will produce 2-nitromethylbenzene and 4-nitromethylbenzene.
In contrast, the carboxylic acid group (\( -\text{COOH} \)) is a 3-directing group (meta-directing).
Therefore, to obtain 3-nitrobenzoic acid, we must first oxidise the methyl group to a carboxylic acid group using alkaline potassium manganate(VII) followed by acid (forming benzoic acid), and then perform nitration of benzoic acid using a mixture of concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \).

1 mark: Correct identification of the pathway that oxidises first to generate the meta-directing carboxyl group and then nitrates.

An increase in pH represents a decrease in the concentration of hydrogen ions, \( [\text{H}^+] \).
By Le Chatelier's principle, decreasing the concentration of a reactant (\( \text{H}^+ \)) shifts the equilibrium position of the half-reaction to the left.
This reduces the tendency for reduction to occur, causing the electrode potential, \( E \), to become less positive (decrease).
Since a lower electrode potential indicates a weaker tendency to undergo reduction, the species acts as a less powerful oxidising agent (its oxidising power decreases).

1 mark: Correct deduction that both the electrode potential and the oxidising power decrease.

At a highly acidic pH of 1.0, there is a very high concentration of hydrogen ions (\( \text{H}^+ \)) in solution.
- The amine group (\( -\text{NH}_2 \)) acts as a base and is fully protonated to form the positively charged ammonium group (\( -\text{NH}_3^+ \)).
- The carboxylate group (\( -\text{COO}^- \)) of the zwitterion is also fully protonated to form the neutral carboxylic acid group (\( -\text{COOH} \)).
Therefore, the predominant species is the cation \( \text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH} \).

1 mark: Correctly identifying the structure where both functional groups are protonated.

To reduce \(\text{VO}_2^+\) (\(E^\theta = +1.00\text{ V}\)) and \(\text{VO}^{2+}\) (\(E^\theta = +0.34\text{ V}\)) to \(\text{V}^{3+}\), the reducing agent must have a standard electrode potential less than \(+0.34\text{ V}\) so that \(E^\theta_{\text{cell}} > 0\) for both reduction steps.

To prevent the further reduction of \(\text{V}^{3+}\) to \(\text{V}^{2+}\) (\(E^\theta = -0.26\text{ V}\)), the reducing agent must have a standard electrode potential greater than \(-0.26\text{ V}\) so that \(E^\theta_{\text{cell}} < 0\) for that step.

Therefore, we require a reducing agent from a redox couple with a standard electrode potential in the range:
\(-0.26\text{ V} < E^\theta < +0.34\text{ V}\).

- The \(\text{Sn}^{4+}/\text{Sn}^{2+}\) system has \(E^\theta = +0.15\text{ V}\), which lies within this range. Thus, \(\text{Sn}^{2+}(\text{aq})\) is the correct reducing agent.
- \(\text{Zn}(\text{s})\) has \(E^\theta = -0.76\text{ V}\), which is below \(-0.26\text{ V}\), meaning it would reduce the vanadium species all the way to \(\text{V}^{2+}\).
- \(\text{Fe}^{2+}(\text{aq})\) and \(\text{I}^-(\text{aq})\) have \(E^\theta\) values of \(+0.77\text{ V}\) and \(+0.54\text{ V}\) respectively. These can only reduce \(\text{VO}_2^+\) to \(\text{VO}^{2+}\) and cannot reduce \(\text{VO}^{2+}\) to \(\text{V}^{3+}\).

Correct answer is C.
- 1 mark for selecting C.
- Reject all other options.

Optical isomerism (enantiomerism) occurs in octahedral complexes that lack a plane of symmetry.

- \(\text{trans}-[\text{Co}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2\text{Cl}_2]^+\) has a plane of symmetry containing the cobalt ion and the two chloride ligands, so it is achiral.
- \(\text{cis}-[\text{Co}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2\text{Cl}_2]^+\) lacks a plane of symmetry due to the relative arrangement of the two bidentate 1,2-diaminoethane (ethylenediamine) ligands. It exists as two non-superimposable mirror images (enantiomers).
- \(\text{trans}-[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\) is symmetrical and achiral.
- \(\text{cis}-[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]\) is a square-planar complex. Square-planar complexes have a plane of symmetry (the molecular plane itself) and therefore cannot exhibit optical isomerism.

Correct answer is B.
- 1 mark for selecting B.
- Reject all other options.

- Option A is incorrect because the electrophile in aromatic nitration is the nitronium ion, which is \(\text{NO}_2^+\) (a cation), not \(\text{NO}_2^-\) (an anion).
- Option B is incorrect because the carboxylic acid group (\(-\text{COOH}\)) is a 3-directing (meta-directing) group, so the primary organic product is 3-nitrobenzoic acid, not 4-nitrobenzoic acid.
- Option C is incorrect because the \(-\text{COOH}\) group is electron-withdrawing by both inductive and mesomeric effects, which decreases the electron density of the benzene ring. This deactivates the ring, meaning benzoic acid reacts more slowly than benzene.
- Option D is correct because the electron-withdrawing nature of the \(-\text{COOH}\) group deactivates the ring toward electrophilic attack and directs the incoming electrophile to the 3-position.

Correct answer is D.
- 1 mark for selecting D.
- Reject all other options.

The basicity of a nitrogen compound depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (\(\text{H}^+\)).

- In ethylamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)), the ethyl group is electron-donating due to the positive inductive effect. This increases the electron density on the nitrogen atom, making its lone pair highly available to accept a proton. Thus, ethylamine is the strongest base listed.
- In ammonia (\(\text{NH}_3\)), there are no alkyl groups to donate electron density, so it is a weaker base than ethylamine.
- In phenylamine (\(\text{C}_6\text{H}_5\text{NH}_2\)), the lone pair of electrons on the nitrogen atom is partially delocalized into the \(\pi\)-system of the benzene ring, making it much less available.
- In ethanamide (\(\text{CH}_3\text{CONH}_2\)), the lone pair of electrons on the nitrogen atom is strongly delocalized onto the highly electronegative oxygen atom of the carbonyl group, making it practically non-basic in water.

Correct answer is C.
- 1 mark for selecting C.
- Reject all other options.

(a)(i) For Zinc as a reducing agent (\( E^\theta = -0.76\text{ V} \)):
- Step 1: \( \text{VO}_2^+ \rightarrow \text{VO}^{2+} \): \( E^\theta_{\text{cell}} = +1.00 - (-0.76) = +1.76\text{ V} \) (feasible since \( >0 \)).
- Step 2: \( \text{VO}^{2+} \rightarrow \text{V}^{3+} \): \( E^\theta_{\text{cell}} = +0.34 - (-0.76) = +1.10\text{ V} \) (feasible since \( >0 \)).
- Step 3: \( \text{V}^{3+} \rightarrow \text{V}^{2+} \): \( E^\theta_{\text{cell}} = -0.26 - (-0.76) = +0.50\text{ V} \) (feasible since \( >0 \)).
Thus, zinc can reduce vanadium through all three stages to \( \text{V}^{2+} \).

For \( \text{Fe}^{2+} \) as a reducing agent (\( E^\theta = +0.77\text{ V} \)):
- Step 1: \( \text{VO}_2^+ \rightarrow \text{VO}^{2+} \): \( E^\theta_{\text{cell}} = +1.00 - (+0.77) = +0.23\text{ V} \) (feasible since \( >0 \)).
- Step 2: \( \text{VO}^{2+} \rightarrow \text{V}^{3+} \): \( E^\theta_{\text{cell}} = +0.34 - (+0.77) = -0.43\text{ V} \) (not feasible since \( <0 \)).
Therefore, \( \text{Fe}^{2+} \) cannot reduce \( \text{VO}^{2+} \) to \( \text{V}^{3+} \) and the reaction stops at \( \text{VO}^{2+} \).

(a)(ii) Violet (accept lavender).

(b)(i) \( \text{Pt}(s) \mid \text{V}^{3+}(aq), \text{VO}^{2+}(aq), \text{H}^+(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s) \)
(Accept the reverse order if standard reduction potentials are equal, but platinum must remain at the vanadium electrode phase boundary: \( \text{Cu}(s) \mid \text{Cu}^{2+}(aq) \parallel \text{VO}^{2+}(aq), \text{V}^{3+}(aq), \text{H}^+(aq) \mid \text{Pt}(s) \))

(b)(ii) The platinum electrode provides an inert conductive surface for transfer of electrons between \( \text{VO}^{2+} \) and \( \text{V}^{3+} \) ions. The standard conditions required for the vanadium half-cell are: Temperature of \( 298\text{ K} \) (or \( 25^\circ\text{C} \)), and solutions of \( [\text{VO}^{2+}] = 1.0\text{ mol dm}^{-3} \), \( [\text{V}^{3+}] = 1.0\text{ mol dm}^{-3} \), and \( [\text{H}^+] = 1.0\text{ mol dm}^{-3} \).

(c)(i) Moles of V-salt = \( \frac{0.150}{249.0} = 6.024 \times 10^{-4}\text{ mol} \).
Since 1 formula unit contains 1 V atom, moles of \( \text{V}^{n+} = 6.024 \times 10^{-4}\text{ mol} \).

(c)(ii) Moles of \( \text{MnO}_4^- = 0.0200 \times \frac{18.00}{1000} = 3.60 \times 10^{-4}\text{ mol} \).

(c)(iii) \( \text{MnO}_4^- \) gains 5 electrons per mole: \( 5 \times (3.60 \times 10^{-4}\text{ mol}) = 1.80 \times 10^{-3}\text{ mol} \) of electrons gained by \( \text{MnO}_4^- \).
By conservation of charge/electrons, the vanadium ions must lose \( 1.80 \times 10^{-3}\text{ mol} \) of electrons.
Change in oxidation state of vanadium = \( \frac{\text{moles of electrons lost}}{\text{moles of vanadium}} = \frac{1.80 \times 10^{-3}}{6.024 \times 10^{-4}} = 2.99 \approx 3 \).
Since the final vanadium species is \( \text{VO}_2^+ \) (where V is in the +5 oxidation state), the initial oxidation state is:
\( n = 5 - 3 = +2 \).
Thus, the original oxidation state was +2.

(a)(i) 4 Marks:
- M1: For calculating feasible cell potentials for Zn reducing \( \text{VO}_2^+ \rightarrow \text{VO}^{2+} \) (\( +1.76\text{ V} \)) AND \( \text{VO}^{2+} \rightarrow \text{V}^{3+} \) (\( +1.10\text{ V} \)).
- M2: For calculating \( E^\theta_{\text{cell}} = +0.50\text{ V} \) for Zn reducing \( \text{V}^{3+} \rightarrow \text{V}^{2+} \) and stating it is feasible.
- M3: For calculating \( E^\theta_{\text{cell}} = +0.23\text{ V} \) for \( \text{Fe}^{2+} \) reducing \( \text{VO}_2^+ \rightarrow \text{VO}^{2+} \) and stating it is feasible.
- M4: For calculating \( E^\theta_{\text{cell}} = -0.43\text{ V} \) for \( \text{Fe}^{2+} \) reducing \( \text{VO}^{2+} \rightarrow \text{V}^{3+} \) and stating it is not feasible (hence reaction stops at \( \text{VO}^{2+} \)).

(a)(ii) 1 Mark:
- Violet / lavender. Reject: purple, blue, green.

(b)(i) 3 Marks:
- M1: Pt electrode shown on vanadium side at the end of the chain, separated by a single vertical line: \( \text{Pt}(s) \mid \dots \)
- M2: Comma used to separate aqueous vanadium species in different oxidation states (\( \text{V}^{3+} \), \( \text{VO}^{2+} \), and optionally \( \text{H}^+ \)), and single vertical lines for phase boundaries with a double vertical line for the salt bridge: \( \dots \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s) \)
- M3: Complete correct representation: \( \text{Pt}(s) \mid \text{V}^{3+}(aq), \text{VO}^{2+}(aq), \text{H}^+(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s) \) (or reverse layout).

(b)(ii) 2 Marks:
- M1: Platinum provides an inert surface for electron transfer / conducts electricity.
- M2: Standard conditions: Temp \( 298\text{ K} \) / \( 25^\circ\text{C} \) AND concentration of all ions (\( \text{VO}^{2+} \), \( \text{V}^{3+} \), \( \text{H}^+ \)) is \( 1.0\text{ mol dm}^{-3} \).

(c)(i) 1 Mark:
- \( 6.024 \times 10^{-4}\text{ mol} \) (accept \( 6.02 \times 10^{-4} \) to \( 6.0 \times 10^{-4} \)).

(c)(ii) 1 Mark:
- \( 3.60 \times 10^{-4}\text{ mol} \) (accept \( 3.6 \times 10^{-4} \)).

(c)(iii) 5.5 Marks:
- M1: Recognises 5 electrons are transferred per mole of \( \text{MnO}_4^- \) (0.5 marks).
- M2: Calculation of moles of electrons transferred: \( 3.60 \times 10^{-4} \times 5 = 1.80 \times 10^{-3}\text{ mol} \) (1 mark).
- M3: Calculates ratio of \( \frac{\text{mol } e^-}{\text{mol V}} = \frac{1.80 \times 10^{-3}}{6.024 \times 10^{-4}} = 2.99 \approx 3 \) (2 marks).
- M4: Deduces original oxidation state: Final state is +5, so initial state = \( 5 - 3 = +2 \) (2 marks). (No marks for +2 without working, except 1 mark max for bald correct answer).

(a)
- In benzene, the \( \pi \)-electrons are delocalized uniformly throughout the carbon ring.
- In phenol, the oxygen atom of the hydroxyl (\( -\text{OH} \)) group has a lone pair of electrons in a p-orbital that overlaps with and is delocalized into the \( \pi \)-system of the ring.
- This significantly increases the electron density of the benzene ring in phenol compared to benzene.
- The higher electron density in phenol is sufficient to polarize the non-polar bromine molecule (\( \text{Br}_2 \)) to form a dipole (\( \delta^+\text{Br}-\text{Br}\delta^- \)) and trigger electrophilic attack without a catalyst.
- Benzene, having lower electron density, cannot sufficiently polarize \( \text{Br}_2 \), hence requiring a halogen carrier catalyst (such as \( \text{FeBr}_3 \)) to generate the stronger electrophile \( \text{Br}^+ \).

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

(b)(ii) The mechanism comprises:
- A curly arrow starting from the delocalized \( \pi \)-ring of methylbenzene pointing to the \( \text{N} \) of the \( \text{NO}_2^+ \) ion.
- Formation of the cationic intermediate (a horseshoe-shaped delocalized positive charge covering 5 carbons, with the open end facing the carbon holding both the \( -\text{H} \) and the new \( -\text{NO}_2 \) group at position 4).
- A curly arrow pointing from the C-H bond at position 4 back into the ring to restore aromaticity.
- Formation of the products: 4-nitromethylbenzene and \( \text{H}^+ \).

(c)(i) The methyl group (\( -\text{CH}_3 \)) is an ortho/para director (2,4-director). If nitrated first, the nitro group will substitute onto positions 2 and 4, forming 2-nitromethylbenzene and 4-nitromethylbenzene. Subsequent oxidation would therefore yield 2-nitrobenzoic acid and 4-nitrobenzoic acid rather than the desired 3-nitrobenzoic acid.

(c)(ii)
- Step 1: React methylbenzene with alkaline potassium manganate(VII) (\( \text{KMnO}_4 \)), heat under reflux, followed by acidification with dilute acid (e.g. \( \text{HCl} \)), to produce benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \)).
- Step 2: React benzoic acid with concentrated nitric acid (\( \text{HNO}_3 \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) at temperatures above 50 °C (e.g., 55–60 °C). Because the \( -\text{COOH} \) group is a meta director (3-director), the nitro group is directed to position 3, yielding 3-nitrobenzoic acid.

(a) 5 Marks:
- M1: Delocalization of \( \pi \)-electrons in benzene explained.
- M2: Delocalization of the lone pair of electrons on the oxygen atom of the \( -\text{OH} \) group of phenol into the ring.
- M3: Phenol has a higher ring electron density than benzene.
- M4: Higher electron density in phenol polarizes bromine molecule / induces a dipole.
- M5: Benzene cannot polarize the bromine molecule, so requires a halogen carrier catalyst.

(b)(i) 1 Mark:
- Balanced equation showing generation of \( \text{NO}_2^+ \).

(b)(ii) 5 Marks:
- M1: Curly arrow from ring to \( \text{NO}_2^+ \).
- M2: Correct structure of intermediate with a positive charge within a horseshoe that covers 5 carbon atoms.
- M3: Horseshoe open end faces the C4 carbon (carbon with \( -\text{H} \) and \( -\text{NO}_2 \)).
- M4: Curly arrow from C-H bond to the interior of the ring to restore aromatic system.
- M5: Structures of reactants (methylbenzene) and products (4-nitromethylbenzene and \( \text{H}^+ \)) are correct.

(c)(i) 2.5 Marks:
- M1: Identifies \( -\text{CH}_3 \) as a 2,4-directing / ortho, para-directing group (1 mark).
- M2: Predicts that 2-nitrobenzoic acid and 4-nitrobenzoic acid would be formed (1.5 marks).

(c)(ii) 4 Marks:
- M1: Reagents and conditions for Step 1: alkaline \( \text{KMnO}_4 \), heat/reflux followed by dilute acid (1.5 marks).
- M2: Correct structure of intermediate (benzoic acid) (0.5 marks).
- M3: Reagents and conditions for Step 2: conc. \( \text{HNO}_3 \) and conc. \( \text{H}_2\text{SO}_4 \), heat (> 50 °C) (1.5 marks).
- M4: Correct structure of final product (3-nitrobenzoic acid) (0.5 marks).

(a)
- Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton.
- In ethylamine, the ethyl group (\( -\text{C}_2\text{H}_5 \)) is electron-donating due to the inductive effect. This increases the electron density on the nitrogen atom, making the lone pair more available to coordinate with a proton (strongest base, lowest \( pK_b \)).
- In phenylamine, the lone pair of electrons on the nitrogen atom overlaps with and is delocalized into the \( \pi \)-system of the benzene ring. This significantly decreases the electron density on the nitrogen atom, making the lone pair much less available to accept a proton (weakest base, highest \( pK_b \)).
- In ammonia, there is no electron-donating alkyl group and no delocalizing benzene ring, so its basicity is intermediate between ethylamine and phenylamine.

(b)(i)
- Reagents: Concentrated nitric acid (\( \text{HNO}_3 \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)).
- Conditions: Maintain temperature between 50 °C and 55 °C.

(b)(ii)
- Reagents: Tin (\( \text{Sn} \)) and concentrated hydrochloric acid (\( \text{HCl} \)), followed by addition of aqueous sodium hydroxide (\( \text{NaOH} \)).
- Equation: \( \text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O} \)

(c)(i)
- Zwitterion of alanine: \( \text{H}_3\text{N}^+-\text{CH(CH}_3)-\text{COO}^- \)
- Structure at pH 1: \( \text{H}_3\text{N}^+-\text{CH(CH}_3)-\text{COOH} \)

(c)(ii)
- Dipeptide 1 (Glycyl-alanine): \( \text{H}_2\text{N}-\text{CH}_2-\text{CO}-\text{NH}-\text{CH(CH}_3)-\text{COOH} \)
- Dipeptide 2 (Alanyl-glycine): \( \text{H}_2\text{N}-\text{CH(CH}_3)-\text{CO}-\text{NH}-\text{CH}_2-\text{COOH} \)
- (Both must show the amide link, \( -\text{CONH}- \), clearly).

(a) 6 Marks:
- M1: Stating that basicity increases with the availability of the lone pair of electrons on the nitrogen atom.
- M2: Ethyl group is electron-donating (positive inductive effect).
- M3: This increases electron density on the nitrogen in ethylamine (making it the strongest base).
- M4: Lone pair of nitrogen in phenylamine is delocalized into the ring.
- M5: This decreases electron density on the nitrogen in phenylamine (making it the weakest base).
- M6: Ammonia has no inductive or delocalization effects, resulting in intermediate basicity.

(b)(i) 2 Marks:
- M1: Concentrated nitric acid AND concentrated sulfuric acid.
- M2: Temperature of 50–55 °C (Accept 'under 60 °C' but reject 'room temperature' or 'above 60 °C').
- Note: Both acids must be specified as 'concentrated' for M1.

(b)(ii) 3.5 Marks:
- M1: Tin (\( \text{Sn} \)) and conc. hydrochloric acid (\( \text{HCl} \)) (1 mark).
- M2: Followed by addition of aqueous sodium hydroxide (\( \text{NaOH} \)) (0.5 marks).
- M3: Correct balanced equation with \( 6[\text{H}] \) and \( 2\text{H}_2\text{O} \) (2 marks; 1 mark for unbalanced but showing correct species).

(c)(i) 2 Marks:
- M1: Correct zwitterion structure: \( \text{H}_3\text{N}^+-\text{CH(CH}_3)-\text{COO}^- \).
- M2: Correct structure at pH 1: \( \text{H}_3\text{N}^+-\text{CH(CH}_3)-\text{COOH} \).

(c)(ii) 4 Marks:
- M1: Correct structure for Gly-Ala dipeptide (1.5 marks).
- M2: Correct structure for Ala-Gly dipeptide (1.5 marks).
- M3: Amide/peptide link \( -\text{CONH}- \) is clearly shown/drawn (1 mark).

(a)(i)
- Electronic configurations: \( \text{Cu}^{2+} \) is \( [\text{Ar}] 3d^9 \) (partially filled d-subshell) while \( \text{Cu}^+ \) is \( [\text{Ar}] 3d^{10} \) (completely filled d-subshell).
- Ligands cause the five 3d orbitals to split into two different energy levels.
- In \( \text{Cu}^{2+} \), because the 3d orbitals are partially filled, electrons can absorb energy from visible light to undergo a d-d transition (promotion from a lower to a higher split d-orbital).
- The non-absorbed wavelengths are transmitted/reflected, producing the complementary blue color.
- In \( \text{Cu}^+ \), the 3d orbitals are completely filled. No d-d transitions are possible because there is no empty/partially filled d-orbital to receive a promoted electron, making it colourless.

(b)(i)
- When added dropwise, a pale blue precipitate of copper(II) hydroxide, \( \text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4 \), is formed:
\( [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 2\text{NH}_3(aq) \rightarrow \text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4(s) + 2\text{NH}_4^+(aq) \)
- When ammonia is added in excess, the precipitate dissolves to form a deep blue solution of \( [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \):
\( \text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4(s) + 4\text{NH}_3(aq) \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(aq) + 2\text{OH}^-(aq) + 2\text{H}_2\text{O}(l) \)
(or overall: \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{NH}_3(aq) \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(aq) + 4\text{H}_2\text{O}(l) \))

(b)(ii)
- Formula: \( [\text{CuCl}_4]^{2-} \)
- Shape: Tetrahedral
- Explanation: Chloride ligands (\( \text{Cl}^- \)) are larger than water molecules (\( \text{H}_2\text{O} \)). Therefore, only four chloride ligands can fit around the central copper(II) ion due to steric hindrance / ligand-ligand repulsion, resulting in a change in coordination number from 6 to 4.

(c)(i) A ligand that donates two lone pairs of electrons from two different atoms to a central metal ion to form two dative covalent bonds.

(c)(ii) Cis-trans (geometric) isomerism and optical isomerism.

(c)(iii) Only the *cis* isomer is chiral and has a non-superimposable mirror image. The drawings must show the octahedral complex containing cobalt in the centre, two Cl ligands in cis positions (90° apart), and two curved lines with 'N' atoms representing the 'en' ligands. The second structure must be the non-superimposable mirror image of the first, showing 3D wedges and dashes.

(a)(i) 5 Marks:
- M1: Identifies \( \text{Cu}^{2+} \) as \( 3d^9 \) (partially filled) and \( \text{Cu}^+ \) as \( 3d^{10} \) (completely filled).
- M2: Stating that 3d orbitals split in energy in the presence of ligands.
- M3: In \( \text{Cu}^{2+} \), electrons absorb energy from visible light to undergo a d-d transition.
- M4: The remaining wavelengths are transmitted/reflected to give the complementary blue colour.
- M5: In \( \text{Cu}^+ \), no d-d transitions are possible because the d-subshell is full, so it is colourless.

(b)(i) 4.5 Marks:
- M1: Observes blue precipitate with dropwise addition (0.5 marks).
- M2: Observes precipitate dissolving to give a deep blue solution in excess (1 mark).
- M3: Correct equation for the precipitation reaction (1.5 marks).
- M4: Correct equation for the excess substitution reaction (1.5 marks).

(b)(ii) 2 Marks:
- M1: Correct formula (\( [\text{CuCl}_4]^{2-} \)) AND shape (tetrahedral) (1 mark).
- M2: Explanation that chloride ions are larger, leading to steric hindrance / fewer ligands fitting (1 mark).

(c)(i) 1 Mark:
- Defines 'bidentate ligand' as forming two dative covalent bonds to the metal ion using two lone pairs of electrons.

(c)(ii) 2 Marks:
- M1: Geometric / cis-trans isomerism (1 mark).
- M2: Optical isomerism (1 mark).

(c)(iii) 3 Marks:
- M1: Recognises that only the *cis* isomer is optically active / has optical isomers (1 mark).
- M2: Draw first 3D octahedral structure of the cis isomer with correct stereochemical notation (wedges/dashes) (1 mark).
- M3: Draw the second structure as a correct non-superimposable mirror image of the first (1 mark).

(a) (i) Concentrated nitric acid reacts with copper to produce nitrogen dioxide, \(NO_2\), which is a highly toxic, brown gas.
(ii) Sodium carbonate is added to neutralize the excess nitric acid (which would otherwise oxidize iodide ions to iodine). Ethanoic acid is then added to dissolve the precipitate of copper(II) carbonate, ensuring the medium is slightly acidic, which is the optimum pH for the reaction with iodide ions without forming unwanted precipitates.

(b) (i) A brown mixture consisting of a white/off-white precipitate of copper(I) iodide in a brown solution of triiodide/iodine is formed.
(ii) \(2Cu^{2+}(aq) + 4I^-(aq) \rightarrow 2CuI(s) + I_2(aq)\)

(c) (i) Starch indicator is used. It is added near the end-point when the solution turns a pale straw/yellow color.
(ii) The blue-black color disappears leaving an off-white/cream suspension of copper(I) iodide.

(d) Moles of \(S_2O_3^{2-}\) in titre: \(n(S_2O_3^{2-}) = 0.100 \times 0.01850 = 1.85 \times 10^{-3} \text{ mol}\).
Since \(2S_2O_3^{2-}(aq) + I_2(aq) \rightarrow S_4O_6^{2-}(aq) + 2I^-(aq)\), and \(2Cu^{2+} \equiv I_2 \equiv 2S_2O_3^{2-}\):
Moles of \(Cu^{2+}\) in 25.0 cm³ = \(1.85 \times 10^{-3} \text{ mol}\).
Moles of \(Cu^{2+}\) in 250.0 cm³ = \(1.85 \times 10^{-2} \text{ mol}\).
Mass of copper: \(m(Cu) = 1.85 \times 10^{-2} \times 63.5 = 1.17475 \text{ g}\).
Percentage of copper in brass: \(\frac{1.17475}{1.50} \times 100\% = 78.317\% \approx 78.3\%\).

(a)(i) 1 Mark: Toxic brown gas (NO2) is produced.
(a)(ii) 1 Mark: Sodium carbonate neutralizes excess mineral acid / nitric acid. 1 Mark: Ethanoic acid provides a mildly acidic buffer / dissolves the copper carbonate precipitate so copper ions are free to react.
(b)(i) 1 Mark: Brown mixture / brown solution with an off-white/white/buff precipitate.
(b)(ii) 1 Mark: 2Cu2+(aq) + 4I-(aq) -> 2CuI(s) + I2(aq) (including state symbols; accept correct ionic equations forming I3-).
(c)(i) 1 Mark: Starch. 1 Mark: Added when the solution is pale yellow / straw-colored (near the end-point).
(c)(ii) 1 Mark: Blue-black to off-white / white suspension (colorless is acceptable only if white solid is mentioned, reject blue-black to colorless).
(d) 4.5 Marks:
- Moles of thiosulfate = 1.85 x 10^-3 mol (1 Mark)
- Moles of copper in 25.0 cm³ = 1.85 x 10^-3 mol (1 Mark)
- Moles of copper in 250.0 cm³ = 1.85 x 10^-2 mol (0.5 Mark)
- Mass of copper = 1.17 g / 1.17475 g (1 Mark)
- Percentage mass of copper = 78.3% (1 Mark) (allow 78% or 78.32%, correct rounding to 2 or 3 sig figs)

(a) Sodium thiosulfate reacts instantaneously with the iodine product as it forms, reducing it back to iodide: \(2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-\). Once all thiosulfate is consumed, free iodine remains in solution and reacts with starch to turn the solution blue-black. This represents a fixed, known amount of peroxodisulfate consumed over time \(t\), making \(1/t\) proportional to the initial rate.

(b) This ensures that the thiosulfate is consumed when only a tiny fraction (less than 5-10%) of the reactants has reacted. Therefore, the reactant concentrations remain approximately constant during the timed interval, and the rate measured is truly the *initial* rate.

(c) (i) Since concentrations are constant, the rate constant \(k\) is directly proportional to \(1/t\) (i.e., \(k \propto 1/t\)).
(ii) From Arrhenius equation, \(\text{gradient} = -\frac{E_a}{R}\).
\(-6500 = -\frac{E_a}{8.31}\)
\(E_a = 6500 \times 8.31 = 54015 \text{ J mol}^{-1} = 54.0 \text{ kJ mol}^{-1}\).

(d) Time halved means the initial rate is doubled (since \(\text{rate} \propto 1/t\)). Because doubling the concentration of iodide ions doubles the rate, the reaction is first order with respect to iodide.

(e) Burette or graduated pipette.

(f) Human reaction time in starting/stopping the stopwatch, or subjectiveness in judging the exact moment the blue-black color appears. This can be minimized by using a colorimeter connected to a data logger.

(a) 3 Marks:
- Thiosulfate reacts with / removes the iodine as soon as it forms (1 Mark)
- Color change occurs when all thiosulfate is consumed, allowing free iodine to react with starch (1 Mark)
- Rate is proportional to 1/t because thiosulfate concentration represents a fixed/constant amount of reaction (1 Mark)
(b) 1.5 Marks:
- Only a small fraction of reactants reacts before the thiosulfate is depleted (1 Mark)
- Ensures reactant concentrations are virtually constant, so the measured rate is the initial rate (0.5 Mark)
(c)(i) 1 Mark: Direct proportionality / \(k \propto 1/t\) (or rate constant is proportional to the rate/inverse of time).
(c)(ii) 3 Marks:
- States gradient = -E_a / R (1 Mark)
- Substitution: E_a = 6500 x 8.31 = 54015 J mol^-1 (1 Mark)
- Final value 54.0 (or 54) kJ mol^-1 with correct units (1 Mark) (accept 54 to 54.02)
(d) 2 Marks:
- Halving time means doubling the rate (1 Mark)
- Since doubling reactant concentration doubles the rate, the order is 1 / first order (1 Mark)
(e) 1 Mark: Burette / volumetric pipette / graduated pipette (reject measuring cylinder).
(f) 1 Mark: Human reaction time / judging the color end-point subjective (0.5 Mark) AND use colorimeter / data logger OR use a cross/background reference (0.5 Mark).

(a) The diagram must show:
- Pear-shaped/round-bottomed flask with a vertical condenser fitted into the neck.
- Condenser open at the top with water inlet at the bottom and outlet at the top.
- Heat source (e.g., water bath or electric heating mantle, not a direct Bunsen burner flame because the organic compounds are highly flammable).

(b) (i) Vacuum filtration / suction filtration (using a Büchner funnel).
(ii) It is much faster than gravity filtration and leaves the solid sample drier.

(c) Recrystallization steps:
1. Dissolve the crude aspirin in the minimum volume of hot solvent (ethanol or water) (1 Mark). Purpose: To obtain a saturated solution so that crystallization occurs upon cooling, while maximizing yield (0.5 Mark).
2. Filter the hot solution quickly (using gravity filtration with a fluted filter paper) (0.5 Mark). Purpose: To remove insoluble impurities (0.5 Mark).
3. Allow the filtrate to cool slowly to room temperature, then place it in an ice bath (0.5 Mark). Purpose: To allow pure crystals to form while soluble impurities remain dissolved in the cold solvent (0.5 Mark).
4. Filter under reduced pressure, wash with a small volume of ice-cold solvent (0.5 Mark). Purpose: To rinse away any remaining soluble impurities from the surface of the crystals without redissolving them (0.5 Mark).

(d) Calculation:
- Theoretical moles of salicylic acid = \(\frac{2.50}{138.0} = 0.01812 \text{ mol}\)
- Theoretical mass of aspirin = \(0.01812 \times 180.0 = 3.2616 \text{ g}\)
- Percentage yield = \(\frac{2.15}{3.2616} \times 100\% = 65.92\% \approx 65.9\%\).

(a) 3 Marks:
- Pear-shaped/round-bottom flask with a vertical condenser (condenser must not be sealed at the top) (1 Mark)
- Water jacket with flow in at the bottom, out at the top (1 Mark)
- Suitable heat source indicated (water bath/heating mantle) (1 Mark) (Reject Bunsen burner directly heating the flask)
(b)(i) 1 Mark: Vacuum/suction filtration (or Büchner filtration).
(b)(ii) 2 Marks: 1 Mark for faster filtration; 1 Mark for drier product.
(c) 4.5 Marks:
- Dissolve in minimum volume (0.5 Mark) of hot solvent (0.5 Mark) + purpose: to ensure a saturated solution / high yield (0.5 Mark)
- Filter hot (0.5 Mark) + purpose: to remove insoluble impurities (0.5 Mark)
- Cool in ice (0.5 Mark) + purpose: to allow crystals to recrystallize / soluble impurities stay dissolved (0.5 Mark)
- Wash crystals with ice-cold solvent (0.5 Mark) + purpose: to wash off soluble surface impurities without dissolving product (0.5 Mark)
(d) 2 Marks:
- Calculates theoretical yield of aspirin = 3.26 g / 0.0181 mol (1 Mark)
- Calculates percentage yield = 65.9% (1 Mark) (Accept range 65.9% - 66.0%)

(a) The yellow-orange color of the solution is due to the presence of the hexaaquairon(III) ion: \([Fe(H_2O)_6]^{3+}\).

(b) Observation: A red-brown / brown precipitate is formed, which is insoluble in excess sodium hydroxide.
Ionic Equation: \([Fe(H_2O)_6]^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(H_2O)_3(OH)_3(s) + 3H_2O(l)\) or \(Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)\).

(c) (i) A white precipitate (of silver chloride, \(AgCl\)) forms.
(ii) The white precipitate dissolves completely in dilute aqueous ammonia to form a colorless solution.
Equation: \(AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)\).

(d) (i) Hydrochloric acid is added to react with and remove any carbonate (\(CO_3^{2-}\)) or sulfite (\(SO_3^{2-}\)) ions present. This prevents the false positive formation of insoluble barium carbonate or barium sulfite precipitates.
(ii) A white precipitate (of barium sulfate, \(BaSO_4\)) is formed.
(iii) \(Ba^{2+}(aq) + SO_4^{2+}(aq) \rightarrow BaSO_4(s)\).

(e) Add potassium thiocyanate solution (\(KSCN(aq)\)) / thiocyanate ions to the solution of **Y**. A blood-red solution / deep-red solution is observed (due to the formation of the \([Fe(H_2O)_5(SCN)]^{2+}\) complex).

(a) 1 Mark: [Fe(H2O)6]3+ (Accept Fe3+)
(b) 3 Marks:
- Brown / red-brown precipitate (1 Mark)
- Insoluble in excess (1 Mark)
- [Fe(H2O)6]3+(aq) + 3OH-(aq) -> Fe(H2O)3(OH)3(s) + 3H2O(l) (or Fe3+ + 3OH- -> Fe(OH)3) (1 Mark) (must have correct state symbols)
(c)(i) 1 Mark: White precipitate.
(c)(ii) 2 Marks:
- Precipitate dissolves / forms a colorless solution (1 Mark)
- AgCl(s) + 2NH3(aq) -> [Ag(NH3)2]+(aq) + Cl-(aq) (1 Mark)
(d)(i) 1.5 Marks:
- To react with / remove carbonate or sulfite ions (1 Mark)
- Which would otherwise form white precipitates of barium carbonate / barium sulfite (0.5 Mark)
(d)(ii) 1 Mark: White precipitate.
(d)(iii) 1 Mark: Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ignore state symbols)
(e) 2 Marks:
- Add potassium thiocyanate solution / KSCN (or K4[Fe(CN)6]) (1 Mark)
- Blood-red / deep-red solution (or dark blue / Prussian blue precipitate) (1 Mark)