MetadataPastPaper.sampleTitle

First, calculate the amount (in moles) of \(Ba^{2+}\) ions used in the titration:
\(\text{Moles of } Ba^{2+} = 0.0500 \text{ mol dm}^{-3} \times \frac{20.0}{1000} \text{ dm}^3 = 1.00 \times 10^{-3} \text{ mol}\).

Since \(Ba^{2+}(aq)\) reacts with \(SO_4^{2-}(aq)\) in a 1:1 ratio to form \(BaSO_4(s)\):
\(\text{Moles of } SO_4^{2-} \text{ in } 25.0 \text{ cm}^3 = 1.00 \times 10^{-3} \text{ mol}\).

Therefore, the number of moles of sulfate (and hence \(MSO_4 \cdot 7H_2O\)) in the entire 250 \(\text{cm}^3\) solution is:
\(\text{Moles of } MSO_4 \cdot 7H_2O = 1.00 \times 10^{-3} \text{ mol} \times 10 = 1.00 \times 10^{-2} \text{ mol}\).

The molar mass (\(M_r\)) of the hydrated salt is calculated as:
\(M_r = \frac{\text{Mass}}{\text{Moles}} = \frac{2.88 \text{ g}}{1.00 \times 10^{-2} \text{ mol}} = 288 \text{ g mol}^{-1}\).

Using the formula of the hydrated salt:
\(M_r(M) + M_r(SO_4) + 7 \times M_r(H_2O) = 288\)
\(M_r(M) + [32.1 + (16.0 \times 4)] + [7 \times (1.0 \times 2 + 16.0)] = 288\)
\(M_r(M) + 96.1 + 126.0 = 288\)
\(M_r(M) + 222.1 = 288\)
\(M_r(M) = 65.9 \text{ g mol}^{-1}\).

Comparing this to the relative atomic masses (\(A_r\)) of the options:
- Cobalt: 58.9
- Iron: 55.8
- Magnesium: 24.3 (Note: Magnesium is not a transition metal, but is also ruled out by mass)
- Zinc: 65.4

The calculated value of 65.9 is closest to Zinc (65.4). Therefore, M is Zinc.

1 mark for the correct answer D.
- Step 1: Calculate moles of barium ions = 1.00 x 10^-3 mol.
- Step 2: Scale up to find moles of salt in 250 cm^3 = 1.00 x 10^-2 mol.
- Step 3: Calculate Mr of hydrated salt = 288 g/mol.
- Step 4: Deduce Ar(M) = 65.9 g/mol and identify Zinc (D).

Optical isomerism (enantiomerism) occurs in molecules or ions that lack a plane of symmetry.

- \([Co(\text{en})_3]^{3+}\) is an octahedral complex with three bidentate ligands. The arrangement of the chelating rings creates a chiral, propeller-like structure that has two non-superimposable mirror images (the \(\Lambda\) and \(\Delta\) isomers). Therefore, it exists as a pair of enantiomers.
- \([Co(NH_3)_4Cl_2]^+\) is an octahedral complex with monodentate ligands. It exhibits cis-trans isomerism, but both the cis and trans isomers contain planes of symmetry and are thus achiral.
- \(\text{trans}-[Pt(NH_3)_2Cl_2]\) and \(\text{cis}-[Pt(NH_3)_2Cl_2]\) are square planar complexes. Because the complex is planar, the plane containing all the atoms is a plane of symmetry, making them achiral.

1 mark for the correct answer B.
- Identification of [Co(en)3]3+ as a chiral complex with no plane of symmetry.
- Understanding that square planar complexes are achiral due to molecular plane of symmetry.
- Understanding that octahedral complexes of the form [Ma4b2]+ do not show enantiomerism.

First, determine the order of reaction with respect to each reactant:
1. Comparing Exp 1 and Exp 2: \([Y]\) is constant while \([X]\) is doubled. The rate increases by a factor of \(\frac{8.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 4\). Since \(2^2 = 4\), the reaction is second order with respect to \(X\).
2. Comparing Exp 2 and Exp 3: \([X]\) is constant while \([Y]\) is doubled. The rate increases by a factor of \(\frac{1.6 \times 10^{-3}}{8.0 \times 10^{-4}} = 2\). Since \(2^1 = 2\), the reaction is first order with respect to \(Y\).

The rate equation is:
\(\text{Rate} = k[X]^2[Y]\)

Now, substitute the values from Experiment 1 to find the rate constant \(k\):
\(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3})^2(0.10\text{ mol dm}^{-3})\)
\(2.0 \times 10^{-4} = k \times 1.0 \times 10^{-3}\)
\(k = \frac{2.0 \times 10^{-4}}{1.0 \times 10^{-3}} = 0.20\)

To find the units of \(k\):
\(\text{Units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \frac{1}{\text{mol}^2\text{ dm}^{-6}\text{ s}} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Thus, \(k = 0.20\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for the correct answer A.
- Step 1: Deduce second order for X and first order for Y.
- Step 2: Calculate numerical value of k = 0.20.
- Step 3: Determine units as dm^6 mol^-2 s^-1.

A reaction is spontaneous under standard conditions if the standard cell potential, \(E^\ominus_{\text{cell}}\), is positive.

Let's calculate \(E^\ominus_{\text{cell}}\) for each reaction:
- A: \(E^\ominus_{\text{cell}} = E^\ominus(Ce^{4+}/Ce^{3+}) - E^\ominus(Fe^{3+}/Fe^{2+}) = +1.61\text{ V} - 0.77\text{ V} = +0.84\text{ V}\) (Spontaneous)
- B: \(E^\ominus_{\text{cell}} = E^\ominus(Fe^{3+}/Fe^{2+}) - E^\ominus(I_2/I^-) = +0.77\text{ V} - 0.54\text{ V} = +0.23\text{ V}\) (Spontaneous)
- C: \(E^\ominus_{\text{cell}} = E^\ominus(I_2/I^-) - E^\ominus(V^{3+}/V^{2+}) = +0.54\text{ V} - (-0.26\text{ V}) = +0.80\text{ V}\) (Spontaneous)
- D: \(E^\ominus_{\text{cell}} = E^\ominus(V^{3+}/V^{2+}) - E^\ominus(Fe^{3+}/Fe^{2+}) = -0.26\text{ V} - 0.77\text{ V} = -1.03\text{ V}\) (Not spontaneous)

Therefore, reaction D has a negative standard cell potential and is not spontaneous.

1 mark for the correct answer D.
- Correct application of E_cell = E_reduction - E_oxidation.
- Identifying that only option D results in a negative overall cell potential (-1.03 V).

Entropy (\(S\)) is a measure of the disorder of a system. The physical state of reactants and products is the major factor determining entropy changes: gases have much higher entropy than liquids, which have higher entropy than solids. An increase in the number of moles of gas during a reaction leads to a positive entropy change (\(\Delta S > 0\)).

- In A, 4 moles of gas react to form 2 moles of gas. There is a decrease in entropy (\(\Delta S < 0\)).
- In B, a solid and an aqueous solution react to form an aqueous solution, a liquid, and a gas. Since a gas (\(CO_2\)) is produced where there were no gaseous reactants, there is a highly significant increase in entropy (\(\Delta S \gg 0\)).
- In C, 3 moles of gas react to form 2 moles of gas. There is a decrease in entropy (\(\Delta S < 0\)).
- In D, ions in solution react to form a highly ordered solid precipitate. There is a decrease in entropy (\(\Delta S < 0\)).

1 mark for the correct answer B.
- Recognizing that the production of gas from solid and aqueous reactants results in a substantial increase in system disorder and entropy.

In the nitration of benzene, the electrophile is the nitronium ion, \(NO_2^+\).

This electrophile is generated in situ by the reaction between concentrated nitric acid and concentrated sulfuric acid:
\(HNO_3 + 2H_2SO_4 \rightleftharpoons NO_2^+ + H_3O^+ + 2HSO_4^-\)

In this reaction, \(HNO_3\) accepts a proton from \(H_2SO_4\), acting as a base. Therefore, \(H_2SO_4\) donates a proton, acting as a Brønsted–Lowry acid. Furthermore, the sulfuric acid is regenerated at the end of the substitution process when a hydrogen ion is lost from the benzene ring to restore aromaticity, so it acts as a catalyst. Thus, option A is correct.

1 mark for the correct answer A.
- Correct identification of NO2+ as the electrophile.
- Correct identification of H2SO4 acting as a Brønsted-Lowry proton donor (acid) and as a catalyst in regenerating the overall process.

According to the official syllabus, the standard laboratory preparation of phenylamine from nitrobenzene involves reducing the nitro group using tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), heated under reflux.

Initially, this reaction forms the phenylammonium ion (\(C_6H_5NH_3^+\)) in the acidic medium. In the second step, aqueous sodium hydroxide (\(\text{NaOH}\)) is added to deprotonate the phenylammonium ion and release the free phenylamine base (\(C_6H_5NH_2\)). Therefore, option B is correct.

1 mark for the correct answer B.
- Identifying Sn and concentrated HCl under reflux as the reducing agent.
- Recognizing the necessity of adding aqueous NaOH to release the free amine from the phenylammonium salt.

The ease of hydrolysis depends on the susceptibility of the carbon atom bonded to the chlorine atom to nucleophilic attack, and the strength of the C-Cl bond:

1. **Propanoyl chloride (acyl chloride):** The carbonyl carbon is highly \(\delta^+\) because it is bonded to two highly electronegative atoms (oxygen and chlorine), and the planar geometry allows easy access for nucleophiles. It hydrolyses rapidly at room temperature with water.
2. **1-chloropropane (halogenoalkane):** The carbon bonded to chlorine is moderately \(\delta^+\). It is not easily attacked by weak nucleophiles like water at room temperature, requiring heating under reflux with a stronger nucleophile like aqueous sodium hydroxide (\(OH^-\)) to undergo hydrolysis.
3. **Chlorobenzene (halogenoarene):** One of the lone pairs of electrons on the chlorine atom overlaps with the delocalised \(\pi\) electron system of the benzene ring. This gives the C-Cl bond partial double bond character, making it exceptionally strong. Furthermore, the high electron density of the \(\pi\) cloud repels incoming nucleophiles. It does not undergo hydrolysis under normal conditions.

Thus, the order of decreasing ease of hydrolysis is: propanoyl chloride > 1-chloropropane > chlorobenzene.

1 mark for the correct answer A.
- Correctly ordering the three classes of organochlorine compounds by reactivity: acyl chloride (most reactive) > halogenoalkane > halogenoarene (least reactive).

Phosphorus is a simple molecular non-metal and does not conduct electricity (it acts as an insulator). Silicon is a giant covalent metalloid and behaves as a semiconductor. Sodium and aluminium are both metallic structures with delocalised electrons that conduct electricity well. Aluminium has more outer-shell delocalised electrons per atom (three) compared to sodium (one), leading to a higher electrical conductivity. Therefore, the order of increasing conductivity is phosphorus < silicon < sodium < aluminium.

1 mark for identifying the correct trend based on the bonding and structure of the elements in Period 3 (insulator < semiconductor < metallic conductor with 1 valence electron < metallic conductor with 3 valence electrons).

Using the Nernst equation: \(E = E^{\theta} + \frac{0.059}{z} \log \left( \frac{[\text{oxidised}]}{[\text{reduced}]} \right)\). Here, the half-cell reaction is \(Fe^{3+}(aq) + e^{-} \rightleftharpoons Fe^{2+}(aq)\), so the number of electrons transferred, \(z\), is 1. The oxidised species is \(Fe^{3+}\) and the reduced species is \(Fe^{2+}\). Substituting the values: \(E = +0.77 + \frac{0.059}{1} \log \left( \frac{1.0}{0.010} \right) = +0.77 + 0.059 \log(100) = +0.77 + 0.059 \times 2 = +0.77 + 0.118 = +0.888 \text{ V}\), which rounds to \(+0.89 \text{ V}\).

1 mark for the correct application of the Nernst equation with z = 1 to obtain a calculated value of +0.89 V.

When solid sodium bromide reacts with concentrated sulfuric acid, the initial acid-base reaction produces hydrogen bromide gas, which is observed as misty fumes. Because bromide ions are moderately strong reducing agents, some of the hydrogen bromide is subsequently oxidised by the sulfuric acid to bromine (observed as a brown vapor), while the sulfuric acid is reduced to sulfur dioxide (a choking gas). Therefore, misty fumes, a brown vapor, and a choking gas are all observed.

1 mark for identifying the correct combination of gaseous products (HBr, Br2, and SO2) and their corresponding physical observations.

The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion in solution from its parent complex. In aqueous systems, the concentration of water is extremely high and remains virtually constant, so it is omitted from the equilibrium expression. The expression is therefore the concentration of the product complex divided by the product of the concentration of the reactant complex and the concentration of the ammonia ligand raised to the power of 4.

1 mark for identifying the correct equilibrium expression that omits water and includes the correct power of 4 for the ammonia concentration.

Nitration of methylbenzene at a moderate temperature (30 degrees Celsius) yields mono-nitration products. Because the methyl group is an electron-donating group (by inductive effect), it activates the ring and directs electrophilic substitution to the ortho (2-) and para (4-) positions. Thus, the major products are 2-nitromethylbenzene and 4-nitromethylbenzene (also named (4-nitrophenyl)methane). 3-nitromethylbenzene is only a minor product because the methyl group is not meta-directing. Side-chain nitration to form (nitromethyl)benzene does not occur under these conditions. Trinitration to form 2,4,6-trinitromethylbenzene requires much more drastic conditions and higher temperatures.

1 mark for identifying that the methyl group directs substitution to the 4-position, making (4-nitrophenyl)methane the major product.

Rearranging the rate equation to solve for the rate constant: \(k = \frac{\text{Rate}}{[A]^2[B]}\). Substituting the given values: \(k = \frac{4.0 \times 10^{-4}}{(0.10)^2 \times 0.20} = \frac{4.0 \times 10^{-4}}{0.01 \times 0.20} = \frac{4.0 \times 10^{-4}}{2.0 \times 10^{-3}} = 0.20\). To determine the units: \(\text{units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{\text{mol}^3 \text{ dm}^{-9}} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\).

1 mark for calculating the correct numerical value of 0.20 and determining the third-order rate constant units as dm^6 mol^-2 s^-1.

Acyl chlorides (Compound X) are highly reactive towards nucleophilic substitution because the carbonyl carbon is strongly electrophilic, being bonded to two highly electronegative atoms (O and Cl), and the reaction proceeds via an addition-elimination mechanism that occurs very rapidly at room temperature. Alkyl halides such as (chloromethyl)benzene (Compound Y) undergo hydrolysis at a moderate rate at room temperature. Aryl halides such as chlorobenzene (Compound Z) do not undergo hydrolysis at room temperature because the lone pair of electrons on the chlorine atom is delocalised into the benzene pi-system, giving the C-Cl bond partial double-bond character and strengthening it significantly. Thus, the order of speed of precipitate formation is X > Y > Z.

1 mark for correctly ordering the compounds by ease of hydrolysis (acyl chloride > alkyl halide > aryl halide).

The balanced chemical equation for the combustion of ethanol is: \(C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)\). According to Hess's law, the standard enthalpy change of combustion is calculated by: \(\Delta H^{\theta}_c = \sum \Delta H^{\theta}_f(\text{products}) - \sum \Delta H^{\theta}_f(\text{reactants})\). Note that the standard enthalpy of formation of \(O_2(g)\) is zero. Therefore: \(\Delta H^{\theta}_c = [2 \times (-394) + 3 \times (-286)] - [-277] = [-788 - 858] + 277 = -1646 + 277 = -1369 \text{ kJ mol}^{-1}\).

1 mark for correctly balancing the combustion equation, applying Hess's law, and arriving at the value of -1369 kJ mol^-1.

Silicon (Si) reacts with chlorine to form silicon tetrachloride, \( \text{SiCl}_4 \), which is a liquid at room temperature. It undergoes rapid hydrolysis with water: \( \text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq}) \). This produces a white precipitate of silicon dioxide and an acidic solution of hydrochloric acid. Aluminium chloride is a solid under standard conditions, phosphorus pentachloride is a solid, and sulfur monochloride/dichloride does not yield a white precipitate of an oxide upon simple hydrolysis.

1 mark for identifying B as the correct answer based on the physical state of the chloride and its hydrolysis products.

According to the Nernst equation, \( E = E^{\ominus} + \frac{0.059}{z} \log \frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]} \). Increasing the concentration of the reactant, \( \text{Fe}^{3+}(\text{aq}) \), increases the value of the ratio and therefore makes the electrode potential, \( E \), more positive (increases it). According to Le Chatelier's principle, increasing the concentration of a reactant shifts the position of equilibrium to the right to oppose the change.

1 mark for selecting A as the correct combination of increased E and a shift to the right.

Using the ideal gas equation: \( pV = nRT = \frac{m}{M_r}RT \), we rearrange to find \( M_r = \frac{mRT}{pV} \). Convert the given quantities to SI units: \( T = 90 + 273.15 = 363.15 \text{ K} \) (or 363 K), \( V = 80.7 \times 10^{-6} \text{ m}^{3} \), \( p = 1.01 \times 10^{5} \text{ Pa} \), \( m = 0.200 \text{ g} \). Calculation: \( M_r = \frac{0.200 \times 8.31 \times 363}{1.01 \times 10^{5} \times 8.07 \times 10^{-5}} = \frac{603.3}{8.15} \approx 74.0 \text{ g mol}^{-1} \).

1 mark for the correct calculation showing \( M_r = 74 \).

Stage 1 is diazotisation, which requires nitrous acid (prepared in situ from \( \text{NaNO}_2 \) and \( \text{HCl} \)) maintained at a low temperature, typically around \( 5 \text{ }^{\circ}\text{C} \), to prevent the decomposition of the benzenediazonium salt. Stage 2 is the coupling reaction, where the cold diazonium salt is reacted with an alkaline solution of phenol, also kept cold (around \( 5 \text{ }^{\circ}\text{C} \)), to produce the azo dye.

1 mark for selecting option A, identifying the correct temperature and reagents for both stages.

The formation reaction of dimethyl ether is: \( 2\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OCH}_3(\text{g}) \). Using Hess's law with combustion data: \( \Delta H_f^{\ominus} = \sum \Delta H_c^{\ominus}(\text{reactants}) - \sum \Delta H_c^{\ominus}(\text{products}) \). Thus, \( \Delta H_f^{\ominus} = [2 \times (-393.5) + 3 \times (-285.8)] - [-1460.0] = [-787.0 - 857.4] + 1460.0 = -1644.4 + 1460.0 = -184.4 \text{ kJ mol}^{-1} \).

1 mark for calculating the correct enthalpy change using combustion values in Hess's cycle.

From Exp 1 and Exp 2: doubling \( [P] \) quadruples the rate, so the reaction is second order with respect to P. From Exp 2 and Exp 3: doubling \( [Q] \) doubles the rate, so the reaction is first order with respect to Q. The rate equation is: \( \text{Rate} = k[P]^2[Q] \). Rearranging for \( k \): \( k = \frac{\text{Rate}}{[P]^2[Q]} \). The units are: \( \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \text{mol}^{-2} \text{ dm}^{6} \text{ s}^{-1} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \).

1 mark for determining the correct rate equation and deriving the units as B.

Benzoyl chloride is an acyl chloride and undergoes extremely rapid nucleophilic acyl substitution (hydrolysis) at room temperature, releasing chloride ions instantly. (Chloromethyl)benzene is a primary halogenoalkane; it undergoes nucleophilic substitution slower than acyl chlorides but will yield a precipitate within a few minutes. Chlorobenzene is an aryl halide; due to the overlap of the chlorine lone pair with the pi electron cloud of the benzene ring, the C-Cl bond has partial double bond character and is highly resistant to hydrolysis, producing no precipitate at room temperature. Thus, the order is: benzoyl chloride > (chloromethyl)benzene > chlorobenzene.

1 mark for ordering the speed of hydrolysis correctly based on structural features.

A larger value of the stability constant, \( K_{\text{stab}} \), indicates a more stable complex ion. Since \( 1.2 \times 10^{13} \gg 4.0 \times 10^5 \), the tetraamminediaphoracopper(II) complex, \( [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \), is significantly more stable than the tetrachlorocuprate(II) complex, \( [\text{CuCl}_4]^{2-} \). Therefore, statement A is correct.

1 mark for selecting A based on comparing the magnitudes of the stability constants.

Silicon exists as a giant covalent macromolecule with strong covalent bonds, which require a large amount of energy to break, resulting in a very high melting point. Phosphorus (P4) and sulfur (S8) exist as simple molecules; sulfur has a higher melting point than phosphorus because S8 molecules have more electrons and stronger London dispersion forces than P4 molecules. Magnesium has a higher melting point than sodium because Mg2+ has a higher charge density and contributes more delocalised electrons.

Award 1 mark for the correct choice of option A.

Benzenediazonium chloride (compound X) is unstable above \(10\ ^\circ\text{C}\). At higher temperatures such as \(50\ ^\circ\text{C}\), it reacts with water (hydrolysis) to form phenol and nitrogen gas. The coupling reaction in Step 2 is an electrophilic substitution reaction. Compound X contains a diazo group containing two nitrogen atoms. Nitrous acid is unstable and must be prepared in situ using sodium nitrite and dilute hydrochloric acid.

Award 1 mark for the correct choice of option A.

Apply the Nernst equation: \(E = E^\theta + \frac{0.059}{z}\log_{10}[Ag^+]\). Here, \(E^\theta = +0.80\text{ V}\), \(z = 1\), and \([Ag^+] = 0.020\text{ mol dm}^{-3}\). Substituting the values: \(E = +0.80 + \frac{0.059}{1}\log_{10}(0.020)\). Since \(\log_{10}(0.020) \approx -1.70\), we get \(E = +0.80 + 0.059 \times (-1.70) \approx +0.80 - 0.100 = +0.70\text{ V}\).

Award 1 mark for the correct choice of option A.

Doubling \([P]\) increases the rate by a factor of 4 (\(2^2\)), so the reaction is second order with respect to \(P\). When both are doubled, the rate increases by 8. Since doubling \([P]\) alone increases the rate by 4, doubling \([Q]\) must increase the rate by 2 (\(4 \times 2 = 8\)), meaning the reaction is first order with respect to \(Q\). Thus, \(\text{Rate} = k[P]^2[Q]\). The units of the rate constant \(k\) are given by: \(k = \frac{\text{Rate}}{[P]^2[Q]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Award 1 mark for the correct choice of option A.

Ethanoyl chloride is an aliphatic acyl chloride and undergoes extremely rapid hydrolysis with cold water at room temperature to form ethanoic acid and HCl gas. Benzoyl chloride reacts much more slowly because the aromatic ring reduces the electrophilicity of the carbonyl carbon and adds steric hindrance. Chloroethane requires heating with an alkali to undergo nucleophilic substitution. Chlorobenzene does not react with water because the C-Cl bond is strengthened by the delocalisation of the chlorine lone pair into the pi aromatic system.

Award 1 mark for the correct choice of option A.

The electronic configuration of \(\text{Cr}^{3+}\) is \([Ar] 3d^3\), which has 3 unpaired d-electrons. The configuration of \(\text{Co}^{2+}\) is \([Ar] 3d^7\). By Hund's rule, a \(3d^7\) subshell has two paired d-orbitals and three unpaired d-orbitals (total of 3 unpaired electrons). \(\text{Fe}^{3+}\) is \([Ar] 3d^5\) (5 unpaired), \(\text{V}^{3+}\) is \([Ar] 3d^2\) (2 unpaired), and \(\text{Ni}^{2+}\) is \([Ar] 3d^8\) (2 unpaired).

Award 1 mark for the correct choice of option A.

The total mass of the solution is \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). The heat energy released is calculated using \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 2842.4\text{ J} = 2.8424\text{ kJ}\). The number of moles of water formed in this neutralisation is \(n = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). Therefore, \(\Delta H_n = -\frac{q}{n} = -\frac{2.8424\text{ kJ}}{0.0500\text{ mol}} = -56.8\text{ kJ mol}^{-1}\).

Award 1 mark for the correct choice of option A.

The methyl group is an alkyl group, which is electron-donating by inductive effect. It activates the benzene ring and directs electrophilic substitution to the 2- (ortho) and 4- (para) positions, making 2-nitromethylbenzene and 4-nitromethylbenzene the major products. The reaction is an electrophilic aromatic substitution, not nucleophilic. Sulfuric acid acts as a catalyst and an acid (proton source) to help generate the nitronium ion (\(NO_2^+\)) electrophile, not as an oxidizing agent.

Award 1 mark for the correct choice of option A.

The reduction in volume upon shaking with KOH is due to the absorption of \(CO_2\). Therefore, the volume of \(CO_2\) produced is \(70 - 30 = 40\text{ cm}^3\). Since 10 cm\(^3\) of \(C_xH_y\) yields 40 cm\(^3\) of \(CO_2\), the carbon subscript \(x\) must be 4. The remaining gas after KOH absorption is the excess \(O_2\), which is 30 cm\(^3\). The volume of \(O_2\) reacted is therefore \(80 - 30 = 50\text{ cm}^3\). The general combustion equation is \(C_4H_y + (4 + y/4)O_2 \rightarrow 4CO_2 + (y/2)H_2O\). The ratio of hydrocarbon to reacted oxygen is 10 : 50, which simplifies to 1 : 5. Thus, \(4 + y/4 = 5\), solving to give \(y = 4\). The molecular formula is \(C_4H_4\).

1 mark for the correct option A. Method: Determine carbon count from the volume of carbon dioxide absorbed by KOH, then determine hydrogen count using the volume of oxygen reacted.

In the presence of water ligands, the 3d subshell of \(Cu^{2+}\) splits into two different energy levels. Electronic transitions occur when d-electrons absorb light of a specific frequency/wavelength in the red-orange region of the visible spectrum and are promoted to the higher energy d-orbitals. The remaining non-absorbed light (the complementary colour, which is blue) is transmitted, making the solution appear blue.

1 mark for the correct option C. Reject theories involving promotion to s-orbitals or emissions of visible light as the cause of the solution's colour.

In the redox reaction between \(Fe^{3+}\) and \(I^-\), \(Fe^{3+}\) acts as the oxidising agent (undergoing reduction) and \(I^-\) acts as the reducing agent (undergoing oxidation). The standard cell potential is calculated using: \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = E^\ominus(Fe^{3+}/Fe^{2+}) - E^\ominus(I_2/I^-) = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\). Since \(E^\ominus_{\text{cell}}\) is positive, the reaction is thermodynamically feasible.

1 mark for the correct option A. Method: Correctly identify which species is reduced and which is oxidised, apply the cell potential equation, and relate the positive potential to thermodynamic feasibility.

Comparing Experiments 1 and 2, doubling \([A]\) while keeping \([B]\) constant increases the rate by a factor of 4, indicating second-order kinetics with respect to A. Comparing Experiments 2 and 3, doubling \([B]\) while keeping \([A]\) constant doubles the rate, indicating first-order kinetics with respect to B. The rate equation is \(Rate = k[A]^2[B]\). The units of \(k\) are calculated as: \(\text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for the correct option B. Method: Determine the order of reaction for both reactants, deduce the rate equation, and calculate the corresponding units for the third-order rate constant.

According to the Born-Haber cycle: \(\Delta H^\ominus_{\text{f}} = \Delta H^\ominus_{\text{at}}[Mg] + \text{1st IE}[Mg] + \text{2nd IE}[Mg] + \Delta H^\ominus_{\text{at}}[O] + \text{1st EA}[O] + \text{2nd EA}[O] + \Delta H^\ominus_{\text{latt}}\). Substituting the given values: \(-602 = 148 + 738 + 1450 + 248 - 141 + 798 + \Delta H^\ominus_{\text{latt}}\). \(-602 = 3241 + \Delta H^\ominus_{\text{latt}}\). Thus, \(\Delta H^\ominus_{\text{latt}} = -602 - 3241 = -3843\text{ kJ mol}^{-1}\).

1 mark for the correct option A. Method: Construct the correct Born-Haber cycle algebraic expression and solve for the unknown lattice energy, keeping track of signs.

The methyl group on methylbenzene is an electron-donating activator that is 2,4-directing. At 50 \(^\circ\)C, mononitration is the predominant process, yielding 4-nitromethylbenzene (and 2-nitromethylbenzene) as major products. The concentrated sulfuric acid acts as a catalyst by protonating the nitric acid to generate the electrophile, \(NO_2^+\).

1 mark for the correct option B. Method: Identify the directing effect of the methyl substituent to select 4-nitromethylbenzene and recognise that sulfuric acid functions as a catalyst.

Phenylamine reacts with nitrous acid at 5 \(^\circ\)C to form benzenediazonium chloride. When this diazonium salt is mixed with phenol in alkaline conditions, an electrophilic substitution reaction occurs. The strongly activating hydroxyl group on phenol directs substitution to the 2- and 4-positions. Due to steric hindrance, coupling occurs almost exclusively at the less-hindered 4-position (para-position), yielding 4-(phenyldiazenyl)phenol.

1 mark for the correct option A. Method: Identify the structure of the azo dye resulting from the coupling of benzenediazonium chloride and phenol at the para-position.

Using the ideal gas equation: \(pV = nRT = \frac{m}{M_r}RT\), which rearranges to \(M_r = \frac{mRT}{pV}\). Converting all values to SI units: \(m = 0.130\text{ g}\), \(T = 80 + 273 = 353\text{ K}\), \(p = 1.01 \times 10^5\text{ Pa}\), and \(V = 42.0 \times 10^{-6}\text{ m}^3\). Substituting these into the formula: \(M_r = \frac{0.130 \times 8.31 \times 353}{1.01 \times 10^5 \times 42.0 \times 10^{-6}} = \frac{381.35}{4.242} \approx 89.9\text{ g mol}^{-1}\). The closest value is 90.0.

1 mark for the correct option A. Method: Correctly convert Celsius to Kelvin and cubic centimetres to cubic metres, and correctly substitute values into the rearranged ideal gas equation.

(a) Moles of C = 3.52 / 44.0 = 0.080 mol. Mass of C = 0.080 * 12.0 = 0.96 g. Moles of H = 2 * (1.80 / 18.0) = 0.20 mol. Mass of H = 0.20 * 1.0 = 0.20 g. Mass of O = 1.48 - 0.96 - 0.20 = 0.32 g. Moles of O = 0.32 / 16.0 = 0.020 mol. Ratio of C : H : O = 0.080/0.020 : 0.20/0.020 : 0.020/0.020 = 4 : 10 : 1. Empirical formula is \(\text{C}_4\text{H}_{10}\text{O}\). (b)(i) \(pV = nRT \implies pV = (m/M_r)RT \implies M_r = mRT / pV\). Convert values: \(T = 100 + 273.15 = 373\text{ K}\), \(V = 96.3 \times 10^{-6}\text{ m}^3\). \(M_r = (0.230 \times 8.31 \times 373) / (1.00 \times 10^5 \times 96.3 \times 10^{-6}) = 712.92 / 9.63 = 74.0\). (ii) Empirical formula mass of \(\text{C}_4\text{H}_{10}\text{O} = (4 \times 12) + (10 \times 1) + 16 = 74\). Since empirical mass = \(M_r\), molecular formula is \(\text{C}_4\text{H}_{10}\text{O}\). (c)(i) Functional group: Tertiary alcohol. It does not react with acidified potassium dichromate(VI) because there is no hydrogen atom attached to the carbon atom bearing the hydroxyl group, preventing oxidation. (ii) Structural formula: \((\text{CH}_3)_3\text{COH}\). (d) Hydrolysis of \(\text{CH}_3\text{COOC}(\text{CH}_3)_3\) (tert-butyl ethanoate): \(\text{CH}_3\text{COOC}(\text{CH}_3)_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + (\text{CH}_3)_3\text{COH}\) (or any other appropriate ester). (e) Dehydration of 2-methylpropan-2-ol gives 2-methylpropene. Skeletal formula: Y-shaped carbon skeleton with a double bond on one branch. Name: 2-methylpropene.

(a) M1: Calculate moles/mass of C and H (1 mark). M2: Calculate mass and moles of O (1 mark). M3: Find the simplest whole-number ratio (1 mark). M4: Correct empirical formula \(\text{C}_4\text{H}_{10}\text{O}\) (1 mark). (b)(i) M1: Convert temperature to K (373 K) and volume to \(\text{m}^3\) (1 mark). M2: Rearrange \(pV = nRT\) and substitute values correctly (1 mark). M3: Calculate \(M_r = 74.0\) to 3 s.f. (1 mark). (ii) M1: Deduce molecular formula is \(\text{C}_4\text{H}_{10}\text{O}\) (1 mark). (c)(i) M1: Identify tertiary alcohol (1 mark). M2: Explain absence of C-H bond on the hydroxyl-bearing carbon (1 mark). (ii) M1: Draw \((\text{CH}_3)_3\text{COH}\) structural formula (1 mark). (d) M1: Identify reactant ester (1 mark). M2: Write correct balanced equation showing hydrolysis products (1 mark). (e) M1: Draw skeletal structure of 2-methylpropene (1 mark). M2: Name 2-methylpropene (1 mark).

(a) The electromotive force (emf) of a half-cell coupled with a standard hydrogen electrode (SHE) measured under standard conditions of 298 K, 1 atm (100 kPa) pressure, and solution concentrations of 1.00 \(\text{mol dm}^{-3}\). (b)(i) \(\text{Fe}^{3+}(\text{aq}) + e^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\), \(E^\ominus = +0.77\text{ V}\) and \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \rightleftharpoons \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\), \(E^\ominus = +1.52\text{ V}\). (ii) \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +1.52 - (+0.77) = +0.75\text{ V}\). Overall reaction: \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) + 5\text{Fe}^{3+}(\text{aq})\). (iii) Platinum (Pt) is used because there are no solid metals involved in the redox reactions of either half-cell to serve as electrodes, and platinum is highly inert and conducting. (c) Using Nernst equation: \(E = +0.77 + \frac{0.059}{1} \log\left(\frac{1.00}{0.0150}\right) = +0.77 + 0.059 \log(66.67) = +0.77 + 0.059(1.824) = +0.77 + 0.108 = +0.88\text{ V}\). (d) In the MnO4-/Mn2+ half-cell: decreasing \([\text{H}^+]\) shifts the position of equilibrium to the left, which decreases the reduction electrode potential of this half-cell (making it less positive). Since \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\), decreasing the potential of the cathode (the reduction half-cell) results in a smaller (decreased) overall cell potential.

(a) M1: Measured against standard hydrogen electrode (1 mark). M2: Standard conditions specified: 298 K, 1.00 \(\text{mol dm}^{-3}\), and 1 bar / 100 kPa (1 mark). (b)(i) M1: Both equations and correct standard potentials given (1 mark). (ii) M1: Calculated cell potential of +0.75 V (1 mark). M2: Balanced overall ionic equation (2 marks - deduct 1 mark for unbalanced charge or state symbols if missing, allow 1 mark for correct formulas). (iii) M1: Identify Platinum electrode (1 mark). M2: State that it is inert and used because no solid metal reactants are present (1 mark). (c) M1: Use z = 1 in the formula (1 mark). M2: Correct calculation of log term (1.824) (1 mark). M3: Answer +0.88 V (1 mark). (d) M1: Shift in equilibrium to LHS due to lower \([\text{H}^+]\) (1 mark). M2: Half-cell potential becomes less positive / decreases (1 mark). M3: Overall cell potential decreases (1 mark).

(a) A transition element is a d-block element that forms at least one stable ion with an incomplete d-subshell. (b)(i) A coordinate bond is formed when the oxygen atom of a water molecule (donor atom) donates a lone pair of electrons into an empty orbital of the copper(II) ion. (ii) Equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightarrow [\text{CuCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\). Coordination number of \([\text{CuCl}_4]^{2-}\) is 4; geometry is tetrahedral. The coordination number decreases from 6 to 4 because chloride ligands are larger than water molecules, creating steric hindrance and allowing only four chloride ligands to coordinate around the central copper ion. (c) In transition metal complexes, the presence of ligands causes the degenerate d-orbitals of the central metal ion to split into two groups of different energy levels. When visible light is shone on the complex, an electron absorbs a specific frequency of light energy and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition). The remaining unabsorbed frequencies of visible light are transmitted or reflected and observed as the complementary color of the absorbed light. (d)(i) A bidentate ligand is a species that donates two lone pairs of electrons to a central metal ion to form two coordinate bonds. (ii) Cis-isomer: the two Cl- ligands are adjacent (90-degree angle to each other). Trans-isomer: the two Cl- ligands are opposite (180-degree angle to each other).

(a) M1: Mentions stable ion with an incomplete d-subshell (1 mark). (b)(i) M1: Oxygen atom donates lone pair to copper(II) (1 mark). M2: Explains coordinate/dative covalent bond (1 mark). (ii) M1: Correct balanced equation (1 mark). M2: Coordination number = 4 and tetrahedral geometry (2 marks - 1 mark for each). M3: Explain size difference / steric hindrance of Cl- (1 mark). (c) M1: Ligands split d-orbitals into two different energy levels (1 mark). M2: Electron absorbed energy from visible light to undergo d-d transition / promotion (1 mark). M3: Reference to equation \(\Delta E = hf\) or transition of electron (1 mark). M4: Explains that unabsorbed / complementary light is transmitted (1 mark). (d)(i) M1: Donates two lone pairs of electrons to form two coordinate bonds (1 mark). (ii) M1: Drawing of cis-isomer (1 mark). M2: Drawing of trans-isomer (1 mark). M3: Correctly labeled cis and trans (1 mark).

(a)(i) Comparing Experiment 1 and 2: \([\text{I}^-]\) is held constant, \([\text{S}_2\text{O}_8^{2-}]\) is doubled (from 0.010 to 0.020 \(\text{mol dm}^{-3}\)), and the initial rate doubles (from \(2.2 \times 10^{-5}\) to \(4.4 \times 10^{-5}\)). Therefore, the reaction is first order with respect to \(\text{S}_2\text{O}_8^{2-}\). Comparing Experiment 2 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is held constant, \([\text{I}^-]\) is doubled (from 0.015 to 0.030 \(\text{mol dm}^{-3}\)), and the initial rate doubles (from \(4.4 \times 10^{-5}\) to \(8.8 \times 10^{-5}\)). Therefore, the reaction is first order with respect to \(\text{I}^-\). (ii) Rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). (iii) Using Experiment 1: \(2.2 \times 10^{-5} = k(0.010)(0.015) \implies k = 2.2 \times 10^{-5} / 1.5 \times 10^{-4} = 0.147\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or 0.15). Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). (b)(i) Both reacting ions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) have negative charges. They repel each other, requiring high energy to collide and react. (ii) Step 1: \(\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}\). Step 2: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\). (iii) Homogeneous catalysis because the catalyst (\(\text{Fe}^{2+}\)) and the reactants are in the same phase (aqueous). (c) Proposed mechanism: Step 1 (Slow / Rate-determining step): \(\text{S}_2\text{O}_8^{2-} + \text{I}^- \rightarrow \text{SO}_4^{2-} + \text{SO}_4\text{I}^-\). Step 2 (Fast): \(\text{SO}_4\text{I}^- + \text{I}^- \rightarrow \text{SO}_4^{2-} + \text{I}_2\).

(a)(i) M1: Compares Exp 1 and 2 to find 1st order wrt \(\text{S}_2\text{O}_8^{2-}\) with explanation (1 mark). M2: Compares Exp 2 and 3 to find 1st order wrt \(\text{I}^-\) with explanation (1 mark). M3: Explicit deduction of both orders based on math/ratios (2 marks). (ii) M1: Correct rate equation matching orders: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (1 mark). (iii) M1: Calculation: \(k = 0.147\) or \(0.15\) (1 mark). M2: Appropriate significant figures (2 or 3 s.f.) (1 mark). M3: Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (1 mark). (b)(i) M1: Identifies electrostatic repulsion between negative ions (1 mark). (ii) M1: Write Equation 1 correctly (1 mark). M2: Write Equation 2 correctly (1 mark). (iii) M1: State homogeneous catalysis + define (same physical state) (1 mark). (c) M1: Rate-determining step contains one \(\text{S}_2\text{O}_8^{2-}\) and one \(\text{I}^-\) (1 mark). M2: Equations of both steps sum up to the overall equation (1 mark). M3: Correctly identifies Step 1 as the slow/rate-determining step (1 mark).

Assuming a standard sample mean titre of \( 31.25\text{ cm}^3 \) is obtained from concordant results:

**(a) & (b)** Mean volume of FA 2 = \( 31.25\text{ cm}^3 \).

**(c)** Moles of \( \text{S}_2\text{O}_3^{2-} \) used = \( \frac{31.25}{1000} \times 0.100 = 3.125 \times 10^{-3}\text{ mol} \).

**(d)** Since 2 moles of \( \text{Cu}^{2+} \) yield 1 mole of \( \text{I}_2 \), which reacts with 2 moles of \( \text{S}_2\text{O}_3^{2-} \), the mole ratio is \( 1:1 \).
Moles of \( \text{Cu}^{2+} \) in \( 25.0\text{ cm}^3 \) of FA 1 = \( 3.125 \times 10^{-3}\text{ mol} \).

**(e)** Concentration of \( \text{Cu}^{2+} \) = \( 3.125 \times 10^{-3} \times \frac{1000}{25.0} = 0.125\text{ mol dm}^{-3} \).

**(f)** Mass concentration of the hydrated salt = \( 31.2\text{ g dm}^{-3} \).
\( M_r \) of \( \text{CuSO}_4 \cdot x\text{H}_2\text{O} \) = \( \frac{31.2}{0.125} = 249.6\text{ g mol}^{-1} \).

**(g)** \( M_r(\text{CuSO}_4) = 63.5 + 32.1 + 4(16.0) = 159.6 \).
Mass of water of crystallisation = \( 249.6 - 159.6 = 90.0 \).
\( x = \frac{90.0}{18.0} = 5.0 \).

- **Table** (4 marks): 1 mark for correct table structure with headers and units; 1 mark for all readings recorded to 2 decimal places (ending in .00 or .05); 2 marks for obtaining two concordant titres within 0.10 cm³.
- **Mean titre** (1 mark): correctly calculated from concordant titres.
- **Moles of thiosulfate** (1 mark): correctly calculated with working shown.
- **Moles of copper** (1 mark): correctly stated as equal to thiosulfate moles based on 1:1 stoichiometry.
- **Concentration** (1.33 marks): correctly calculated by multiplying moles by 40, to 3 or 4 significant figures.
- **Molar Mass** (2 marks): correctly calculated as mass / concentration, with appropriate working.
- **Value of x** (2 marks): subtraction of 159.6, division by 18.0 to give an integer value of 5.

Assuming a standard sample temperature data set is obtained:
Initial temperatures: 0 min = 21.5, 1 min = 21.5, 2 min = 21.5, 3 min = 21.5 °C.
Addition at 4 min.
Post-addition temperatures: 5 min = 34.1, 6 min = 33.5, 7 min = 32.9, 8 min = 32.3, 9 min = 31.7, 10 min = 31.1 °C.

**(a)** The student records these values clearly in a table, specifying units (°C and minutes).

**(b)** Plotting these points reveals constant temperature up to 3 minutes, and a linear cooling curve from 5 to 10 minutes. Extrapolating the cooling curve back to 4 minutes gives a maximum theoretical temperature of \( 34.7^{\circ}\text{C} \).
\( \Delta T = 34.7 - 21.5 = 13.2^{\circ}\text{C} \).

**(c)** Total mass of solution = \( 25.0\text{ cm}^3 + 25.0\text{ cm}^3 = 50.0\text{ g} \).
\( q = mc\Delta T = 50.0 \times 4.18 \times 13.2 = 2758.8\text{ J} = 2.76\text{ kJ} \).

**(d)** Moles of \( \text{CH}_3\text{COOH} \) = \( 0.0250 \times 2.00 = 0.050\text{ mol} \).
Moles of \( \text{NaOH} \) = \( 0.0250 \times 2.00 = 0.050\text{ mol} \).
Moles of \( \text{H}_2\text{O} \) formed = \( 0.050\text{ mol} \).

**(e)** \( \Delta H_{\text{neut}} = -\frac{2.7588\text{ kJ}}{0.050\text{ mol}} = -55.2\text{ kJ mol}^{-1} \).

**(f)** The student's suggestion is incorrect. Glass is a better conductor of heat than polystyrene, which means more heat would be lost to the surroundings during the reaction. This would result in a lower maximum temperature, a smaller calculated \( \Delta T \), and a less accurate enthalpy value. An improvement to further reduce heat loss is to place a lid on the polystyrene cup or wrap it in insulating cotton wool.

- **Table** (2 marks): 1 mark for appropriate columns, headers, and units; 1 mark for all temperature readings recorded to 1 decimal place (e.g. .0 or .5).
- **Graph & Extrapolation** (4 marks): 1 mark for sensible scale and correct axes (Time on x-axis, Temperature on y-axis); 1 mark for accurate plotting; 1 mark for two distinct best-fit straight lines; 1 mark for correct extrapolation at 4 minutes to find \( \Delta T \).
- **Heat Calculation** (2 marks): 1 mark for using the correct mass of 50.0 g; 1 mark for correct calculation of \( q \) in J or kJ.
- **Moles of water** (1 mark): correctly calculated as 0.050 mol.
- **Enthalpy change** (1.33 marks): 1 mark for dividing \( q \) by moles with a negative sign; 0.33 marks for correct conversion to kJ/mol to 3 significant figures.
- **Error analysis** (3 marks): 1 mark for explaining that glass increases heat loss (as it is a better conductor); 1 mark for concluding the student's suggestion is wrong; 1 mark for suggesting a valid improvement (e.g., using a lid or double cups).

**(a)** Expected Observations Table:
- **FA 6 + NaOH (aq)**: Blue precipitate formed, insoluble in excess. On standing, the precipitate turns pinkish-brown.
- **FA 6 + NH3 (aq)**: Blue precipitate formed, soluble in excess to give a yellow-brown/brown solution.
- **FA 6 + AgNO3 (aq)**: White precipitate formed, which is soluble in dilute ammonia.
- **FA 7 + NaOH (aq)**: Red-brown precipitate formed, insoluble in excess.
- **FA 7 + NH3 (aq)**: Red-brown precipitate formed, insoluble in excess.
- **FA 7 + BaCl2 (aq)**: White precipitate formed.
- **FA 7 + KSCN (aq)**: A blood-red solution is formed.

**(b)** **FA 6 Identification**:
- Cation: \( \text{Co}^{2+} \). Reason: Forms a blue precipitate with both NaOH and ammonia, with the ammonia precipitate dissolving in excess to give a brown solution.
- Anion: \( \text{Cl}^- \). Reason: White precipitate formed with silver nitrate which is soluble in dilute ammonia.

**(c)** **FA 7 Identification**:
- Cation: \( \text{Fe}^{3+} \). Reason: Red-brown precipitate with both NaOH and ammonia, and a characteristic blood-red color with thiocyanate ions.
- Anion: \( \text{SO}_4^{2-} \). Reason: White precipitate formed with barium chloride that is insoluble in hydrochloric acid.

**(d) Ionic Equations**:
(i) \( \text{Co}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Co(OH)}_2(\text{s}) \)
(ii) \( \text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) \)

- **Table of Observations** (6 marks):
- 1 mark for FA 6 with NaOH (blue precipitate, insoluble in excess, turns pink/brown on standing).
- 1 mark for FA 6 with NH3 (blue precipitate, dissolves in excess to give brown solution).
- 1 mark for FA 6 with AgNO3 (white ppt soluble in ammonia).
- 1 mark for FA 7 with NaOH and NH3 (red-brown ppt in both, insoluble in excess).
- 1 mark for FA 7 with BaCl2 (white ppt).
- 1 mark for FA 7 with KSCN (blood-red solution).
- **FA 6 Identification** (2 marks): 1 mark for identifying \( \text{Co}^{2+} \) with reasoning; 1 mark for identifying \( \text{Cl}^- \) with reasoning.
- **FA 7 Identification** (2 marks): 1 mark for identifying \( \text{Fe}^{3+} \) with reasoning; 1 mark for identifying \( \text{SO}_4^{2-} \) with reasoning.
- **Ionic Equations** (3.33 marks):
- 1.33 marks for \( \text{Co}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Co(OH)}_2(\text{s}) \) (1 mark for correct formulas, 0.33 marks for all correct state symbols).
- 2 marks for \( \text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) \) (1 mark for correct formulas, 1 mark for all correct state symbols).

(a) The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion from its constituent ions or molecules in a solvent.

(b)(i) Equilibrium equation:
\([Cu(H_2O)_6]^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 4H_2O(l)\)
Expression:
\(K_{\text{stab}} = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][NH_3]^4}\)

(b)(ii) Observation: The deep-blue solution turns yellow-green.
Explanation: Addition of concentrated HCl introduces a very high concentration of \(Cl^-\) ions, which shifts the equilibrium towards the formation of the tetrachlorocuprate(II) complex, \([CuCl_4]^{2-}\), despite its lower stability constant, by Le Chatelier's principle.

(c)(i) Structure of ethylenediamine:
\(\text{H}_2\ddot{\text{N}}-\text{CH}_2-\text{CH}_2-\ddot{\text{N}}\text{H}_2\) with a lone pair on each of the two nitrogen atoms.

(c)(ii) The ligand exchange reaction:
\([Cu(NH_3)_6]^{2+}(aq) + 3en(aq) \rightleftharpoons [Cu(en)_3]^{2+}(aq) + 6NH_3(aq)\)
There are 4 particles on the left-hand side and 7 particles on the right-hand side. This represents an increase in the number of free particles in solution, leading to a significant increase in disorder, so the entropy change \(\Delta S^\ominus\) is positive. Since \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), a positive \(\Delta S^\ominus\) makes \(\Delta G^\ominus\) more negative, resulting in a much larger equilibrium constant (stability constant).

(a)
- M1: Define as the equilibrium constant for the formation of a complex ion from its constituent ions/molecules in a solvent. (1)
- M2: Mention that the concentration of water is omitted / solvent is in large excess. (1)

(b)(i)
- M1: Correctly balanced equation: \([Cu(H_2O)_6]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O\) (1)
- M2: Correct \(K_{\text{stab}}\) expression: \(K_{\text{stab}} = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][NH_3]^4}\) (1)

(b)(ii)
- M1: State that the solution turns yellow-green / green. (1)
- M2: Explain using Le Chatelier's principle (high concentration of chloride ions overrides the lower stability constant to shift equilibrium). (1)

(c)(i)
- M1: Correctly drawn structure showing all atoms in ethylenediamine. (1)
- M2: Show lone pairs on both nitrogen atoms. (1)

(c)(ii)
- M1: Identify that replacing monodentate ligands with bidentate ligands increases the number of particles in solution (from 4 to 7). (1)
- M2: State that this leads to an increase in entropy / positive \(\Delta S^\ominus\). (1)
- M3: Relate positive \(\Delta S^\ominus\) to more negative \(\Delta G^\ominus\), which increases the stability constant \(K_{\text{stab}}\). (1)

(a)(i) Cell diagram:
\(Pt(s) | Fe^{2+}(aq), Fe^{3+}(aq) || Ag^+(aq) | Ag(s)\)

(a)(ii) \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.80 - 0.77 = +0.03\text{ V}\)
Overall reaction:
\(Fe^{2+}(aq) + Ag^+(aq) \rightleftharpoons Fe^{3+}(aq) + Ag(s)\)

(b)(i) Using the Nernst equation:
\(E = +0.80 + \frac{0.059}{1}\log(0.020)\)
\(E = +0.80 + 0.059 \times (-1.699)\)
\(E = +0.80 - 0.10\text{ V} = +0.70\text{ V}\) (or \(+0.699\text{ V}\))

(b)(ii) Since \(E(Ag^+/Ag) = +0.70\text{ V}\) and \(E(Fe^{3+}/Fe^{2+}) = +0.77\text{ V}\), the silver half-cell now acts as the anode (oxidation) and the iron half-cell acts as the cathode (reduction).
\(E_{\text{cell}} = +0.77 - 0.70 = +0.07\text{ V}\).
Because the sign of \(E_{\text{cell}}\) is still positive, the reaction is feasible, but the direction has reversed: \(Ag(s) + Fe^{3+}(aq) \rightarrow Ag^+(aq) + Fe^{2+}(aq)\). The original reaction is no longer feasible.

(c) Standard electrode potentials only give information about thermodynamic feasibility (equilibrium position). A reaction may have a very high activation energy, \(E_a\). At standard conditions, very few reactant particles have sufficient kinetic energy to overcome this activation energy, making the rate of reaction extremely slow (kinetically inert).

(a)(i)
- M1: Correct representation of \(Pt(s) | Fe^{2+}(aq), Fe^{3+}(aq)\) including state symbols. (1)
- M2: Correct salt bridge and silver half-cell \(|| Ag^+(aq) | Ag(s)\). (1)

(a)(ii)
- M1: Correct calculation: \(E^\ominus_{\text{cell}} = +0.03\text{ V}\). (1)
- M2: Balanced equation: \(Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)\) (1)

(b)(i)
- M1: Correct substitution into Nernst equation: \(E = 0.80 + 0.059 \log(0.020)\). (1)
- M2: Accurate calculation of \(E = +0.70\text{ V}\) (or \(+0.699\text{ V}\)). (1)

(b)(ii)
- M1: Correct calculation of new cell potential: \(E_{\text{cell}} = |0.77 - 0.70| = 0.07\text{ V}\). (1)
- M2: Correctly deduce that the reaction is not feasible in the original direction (the feasible direction has reversed). (1)

(c)
- M1: Mention that \(E^\ominus\) only relates to thermodynamic feasibility / stability, not reaction rate / kinetics. (1)
- M2: State that the reaction may have a high activation energy (\(E_a\)). (1)
- M3: Conclude that the rate of reaction is extremely slow / kinetically inert. (1)

(a)(i) Iodine is zero order (as it does not appear in the rate equation).
(a)(ii) Hydrogen ions are first order (power of 1 in the rate equation).

(b)(i) Rate constant calculation:
\(\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\)
\(1.20 \times 10^{-5} = k \times 0.80 \times 0.20\)
\(k = \frac{1.20 \times 10^{-5}}{0.16} = 7.50 \times 10^{-5}\)
Units: \(\text{rate} / ([\text{conc}][\text{conc}]) = \text{mol dm}^{-3}\text{ s}^{-1} / (\text{mol dm}^{-3})^2 = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
Value of \(k = 7.50 \times 10^{-5}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(b)(ii) \(\text{H}^+\) acts as a catalyst, meaning it is used in one step but regenerated in a subsequent step, so its concentration does not change overall. Since the reaction is zero order with respect to iodine, the rate of the reaction remains completely constant (does not change) as iodine is consumed.

(c) A consistent mechanism:
Step 1 (Slow / Rate-determining step):
\(\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow [\text{CH}_3\text{C(OH)CH}_3]^+ \) (protonation) followed by slow enolisation: \(\text{CH}_3\text{C(OH)}=\text{CH}_2 + \text{H}^+\) (or just write: \(\text{CH}_3\text{COCH}_3 + \text{H}^+ \xrightarrow{\text{slow}} [\text{intermediate}]\)).
Step 2 (Fast):
\(\text{CH}_3\text{C(OH)}=\text{CH}_2 + \text{I}_2 \xrightarrow{\text{fast}} \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^-\).
This mechanism is consistent because only propanone and \(\text{H}^+\) are involved in or before the rate-determining step.

(a)(i)
- M1: Zero order. (1)

(a)(ii)
- M1: First order. (1)

(b)(i)
- M1: Correct substitution of values into the rate equation: \(1.20 \times 10^{-5} = k \times 0.80 \times 0.20\). (1)
- M2: Correct calculation of \(k = 7.50 \times 10^{-5}\). (1)
- M3: Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). (1)

(b)(ii)
- M1: State that \(\text{H}^+\) is a catalyst, so it is regenerated and concentration remains constant. (1)
- M2: State that the rate remains constant because the reaction is zero order with respect to \(\text{I}_2\). (1)

(c)
- M1: Show a slow step (rate-determining step) involving only propanone and \(\text{H}^+\). (1)
- M2: Show a fast step involving the intermediate from Step 1 reacting with \(\text{I}_2\). (1)
- M3: Show the regeneration of the catalyst \(\text{H}^+\) in the fast step. (1)
- M4: Explain that this is consistent because the species in the rate equation match the reactants in the rate-determining step (and prior fast equilibria). (1)

(a) \(\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S^\ominus = (39.7 + 213.6) - 92.9\)
\(\Delta S^\ominus = 253.3 - 92.9 = +160.4\text{ J K}^{-1}\text{ mol}^{-1}\)

(b)(i) The reactant is a highly ordered solid (\(\text{CaCO}_3\)), whereas the products include a gas (\(\text{CO}_2\)) and a solid (\(\text{CaO}\)). Gaseous particles have a much greater degree of disorder (more ways of arranging particles/microstates) than solids, resulting in a net increase in disorder and a positive entropy change.

(b)(ii) A reaction is thermodynamically feasible when \(\Delta G^\ominus \le 0\).
Using \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\):
Set \(\Delta G^\ominus = 0\):
\(T = \frac{\Delta H^\ominus}{\Delta S^\ominus}\)
\(\Delta H^\ominus = +178\text{ kJ mol}^{-1} = +178000\text{ J mol}^{-1}\)
\(T = \frac{178000}{160.4} = 1109.7\text{ K}\) (or \(1110\text{ K}\))

(c) At \(T = 1100\text{ K}\):
\(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\)
\(\Delta G^\ominus = 178 - \left(1100 \times \frac{160.4}{1000}\right)\)
\(\Delta G^\ominus = 178 - 176.44 = +1.56\text{ kJ mol}^{-1}\)
Since \(\Delta G^\ominus > 0\) (is positive), the reaction is not thermodynamically feasible at \(1100\text{ K}\) (though it is very close to becoming feasible).

(a)
- M1: Correct formula or substitution: \((39.7 + 213.6) - 92.9\). (1)
- M2: Value: \(+160.4\) (or \(160\)). (1)
- M3: Correct units: \(\text{J K}^{-1}\text{ mol}^{-1}\) (or \(\text{kJ K}^{-1}\text{ mol}^{-1}\) if value is \(+0.1604\)). (1)

(b)(i)
- M1: Identify that a gas is produced from a solid reactant. (1)
- M2: Explain that gas molecules are much more disordered / have many more microstates than solids. (1)

(b)(ii)
- M1: State that feasibility requires \(\Delta G^\ominus \le 0\) / \(\Delta G^\ominus = 0\) for the threshold. (1)
- M2: Convert \(\Delta H^\ominus\) to \(\text{J}\) (\(178000\text{ J}\)) or \(\Delta S^\ominus\) to \(\text{kJ}\) (\(0.1604\text{ kJ}\)). (1)
- M3: Correct calculation of temperature: \(1110\text{ K}\) (accept range \(1109-1110\text{ K}\)). (1)

(c)
- M1: Correct substitution of values into \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\). (1)
- M2: Calculation of \(\Delta G^\ominus = +1.56\text{ kJ mol}^{-1}\) (or \(+1560\text{ J mol}^{-1}\)). (1)
- M3: Deduce that the reaction is not feasible because \(\Delta G^\ominus > 0\). (1)

(a)(i) Reagents: Conc. \(\text{HNO}_3\) and conc. \(\text{H}_2\text{SO}_4\).
Conditions: Temperature at \(50 - 55^\circ\text{C}\).

(a)(ii) Reagents: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), followed by addition of aqueous sodium hydroxide (\(\text{NaOH}\)).
Conditions: Heat under reflux.

(b)(i) Reagents: Sodium nitrite (\(\text{NaNO}_2\)) and concentrated hydrochloric acid (\(\text{HCl}\)) (which react to form nitrous acid, \(\text{HNO}_2\)).
Temperature: \(0 - 10^\circ\text{C}\) (or below \(10^\circ\text{C}\)).

(b)(ii) Equation:
\(\text{C}_6\text{H}_5\text{N}_2^+ + \text{C}_6\text{H}_5\text{OH} + \text{OH}^- \rightarrow \text{C}_6\text{H}_5\text{N=NC}_6\text{H}_4\text{OH} + \text{H}_2\text{O}\)
Structure: \(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH}\) (specifically 4-hydroxyphenylazobenzene, with the azo link para to the OH group).

(c) Azo dyes contain an extensive delocalised \(\pi\) electron system (conjugation) spanning across both benzene rings and the \(-\text{N}=\text{N}-\) azo group. This large conjugated system decreases the energy gap (\(\Delta E\)) between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). Consequently, the molecule absorbs light in the visible region of the electromagnetic spectrum, and the complementary color is observed.

(a)(i)
- M1: Reagents: Conc. \(\text{HNO}_3\) and conc. \(\text{H}_2\text{SO}_4\). (1)
- M2: Conditions: Temperature at \(50-55^\circ\text{C}\) (or range within this). (1)

(a)(ii)
- M1: Reagents: Tin (\(\text{Sn}\)) and conc. \(\text{HCl}\), followed by aqueous \(\text{NaOH}\). (1)
- M2: Conditions: Heat under reflux. (1)

(b)(i)
- M1: Reagents: Nitrous acid (or \(\text{NaNO}_2 + \text{HCl}\)). (1)
- M2: Temperature: \(0 - 10^\circ\text{C}\). (1)

(b)(ii)
- M1: Correct formula for reactants (benzenediazonium ion and phenol/phenoxide). (1)
- M2: Correct structure of the azo dye showing \(-\text{N}=\text{N}-\) linked para to the \(-\text{OH}\) group. (1)
- M3: Balanced equation with \(\text{H}_2\text{O}\) / \(\text{H}^+\) as appropriate. (1)

(c)
- M1: Mention extensive delocalisation / highly conjugated system involving the benzene rings and the azo group. (1)
- M2: Explain that this lowers the energy gap (\(\Delta E\)) so that light in the visible region is absorbed. (1)

(a) Order: Ethanoyl chloride > chloroethane > chlorobenzene.

(b)(i) In chlorobenzene, a lone pair of electrons on the chlorine atom overlaps with the delocalised \(\pi\) system of the benzene ring. This gives the C–Cl bond partial double-bond character, making it much stronger and harder to break. Additionally, the high electron density of the benzene ring repels incoming nucleophiles like water.

(b)(ii) In ethanoyl chloride, the carbonyl carbon is bonded to both a highly electronegative oxygen atom and a chlorine atom. This makes the carbonyl carbon highly electron-deficient (strongly \(\delta+\)), making it very susceptible to nucleophilic attack. The elimination of the chloride ion is also rapid as the carbonyl group stabilizes the transition state.

(c)(i) Equation:
\(\text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{HCl}\)
Organic product: Ethyl ethanoate.

(c)(ii) Equation:
\(\text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{NH}_2 \rightarrow \text{CH}_3\text{CONHCH}_2\text{CH}_3 + \text{HCl}\)
Organic product: N-ethylethanamide.

(a)
- M1: Correct order: Ethanoyl chloride > chloroethane > chlorobenzene. (1)

(b)(i)
- M1: Mention overlap of chlorine's lone pair with the \(\pi\) ring system. (1)
- M2: State that this gives the C–Cl bond partial double bond character / makes it stronger. (1)
- M3: State that the benzene ring repels nucleophilic attack. (1)

(b)(ii)
- M1: State that the carbonyl carbon is highly \(\delta+\) due to being bonded to both O and Cl. (1)
- M2: Explain that this makes it highly susceptible to nucleophilic attack. (1)
- M3: Mention that the carbon is \(sp^2\) hybridized / open to attack / C-Cl bond is easily broken in the elimination step. (1)

(c)(i)
- M1: Correctly balanced equation: \(\text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{HCl}\). (1)
- M2: Name of product: Ethyl ethanoate. (1)

(c)(ii)
- M1: Correctly balanced equation: \(\text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{NH}_2 \rightarrow \text{CH}_3\text{CONHCH}_2\text{CH}_3 + \text{HCl}\) (or balanced with salt). (1)
- M2: Name of product: N-ethylethanamide. (1)

(a)(i) Skeletal structure of 2-aminopropanoic acid: Draw a 3-carbon chain with a carbonyl oxygen and OH at C1, and an \(\text{NH}_2\) group at C2. The C2 carbon is labeled with an asterisk (*).

(a)(ii) 3D structures: Draw a central carbon with tetrahedral geometry (two bonds in the plane, one wedge, one dash). The four groups are \(-\text{H}\), \(-\text{CH}_3\), \(-\text{NH}_2\), and \(-\text{COOH}\). The second structure must be the mirror image of the first.

(b)(i) Zwitterion structure: \(^+H_3N-CH(CH_3)-COO^-\).

(b)(ii) The zwitterion contains permanent positive (\(-\text{NH}_3^+\)) and negative (\(-\text{COO}^-\)) charges. This leads to strong electrostatic forces of attraction (ionic bonds) between neighbouring zwitterions in the lattice. These require a large amount of energy to break, resulting in a very high melting point.

(c)(i) Reaction with HCl: \(^+H_3N-CH(CH_3)-COOH\).

(c)(ii) Reaction with NaOH: \(H_2N-CH(CH_3)-COO^-\).

(c)(iii) Dipeptide structure:
\(\text{H}_2\text{N}-\text{CH(CH}_3)-\text{CONH}-\text{CH(CH}_3)-\text{COOH}\) with the amide group (\(-\text{CONH}-\) circled as the peptide link.

(a)(i)
- M1: Correct skeletal structure of alanine. (1)
- M2: Asterisk (*) correctly placed on C2. (1)

(a)(ii)
- M1: One 3D tetrahedral structure drawn correctly with four different groups. (1)
- M2: Second 3D structure is a non-superimposable mirror image of the first. (1)

(b)(i)
- M1: Correct zwitterionic structure showing \(^+H_3N-CH(CH_3)-COO^-\). (1)

(b)(ii)
- M1: Identify strong electrostatic attractions / ionic bonds between zwitterions. (1)
- M2: State that significant thermal energy is required to overcome these forces. (1)

(c)(i)
- M1: Correct structure with protonated amine group: \(^+H_3N-CH(CH_3)-COOH\). (1)

(c)(ii)
- M1: Correct structure with deprotonated carboxylic group: \(H_2N-CH(CH_3)-COO^-\). (1)

(c)(iii)
- M1: Correct structure of the dipeptide. (1)
- M2: Correctly circle the amide link (\(-CO-NH-\)). (1)

(a)(i) Dissociation equation:
\(\text{CH}_3\text{CH}_2\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-(aq) + \text{H}_3\text{O}^+(aq)\)
Expression:
\(K_a = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)

(a)(ii) Since it is a weak acid:
\([\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]}\)
\([\text{H}^+] = \sqrt{1.35 \times 10^{-5} \times 0.150} = \sqrt{2.025 \times 10^{-6}} = 1.423 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.85\)

(b)(i) Moles of propanoic acid:
\(n(\text{HA}) = \frac{50.0}{1000} \times 0.150 = 7.50 \times 10^{-3}\text{ mol}\)
Moles of NaOH:
\(n(\text{NaOH}) = \frac{25.0}{1000} \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\)

(b)(ii) The reaction is:
\(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\)
Since NaOH is the limiting reagent:
- Moles of propanoate ions formed, \(n(\text{A}^-) = 2.50 \times 10^{-3}\text{ mol}\).
- Moles of propanoic acid remaining, \(n(\text{HA}) = 7.50 \times 10^{-3} - 2.50 \times 10^{-3} = 5.00 \times 10^{-3}\text{ mol}\).
Total volume of solution = \(50.0 + 25.0 = 75.0\text{ cm}^3 = 0.075\text{ dm}^3\).

Concentrations:
- \([\text{CH}_3\text{CH}_2\text{COO}^-] = \frac{2.50 \times 10^{-3}}{0.075} = 0.0333\text{ mol dm}^{-3}\)
- \([\text{CH}_3\text{CH}_2\text{COOH}] = \frac{5.00 \times 10^{-3}}{0.075} = 0.0667\text{ mol dm}^{-3}\)

(b)(iii) Using the buffer equation:
\(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)
\([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.35 \times 10^{-5} \times \frac{0.0667}{0.0333} = 2.70 \times 10^{-5}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(2.70 \times 10^{-5}) = 4.57\)

(a)(i)
- M1: Correct dissociation equation. (1)
- M2: Correct \(K_a\) expression. (1)

(a)(ii)
- M1: Use of correct approximation \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]}\). (1)
- M2: Correct calculation of \([\text{H}^+] = 1.42 \times 10^{-3}\text{ mol dm}^{-3}\). (1)
- M3: Correct pH of \(2.85\) (must be 2 decimal places). (1)

(b)(i)
- M1: Moles of propanoic acid = \(7.50 \times 10^{-3}\text{ mol}\). (1)
- M2: Moles of NaOH = \(2.50 \times 10^{-3}\text{ mol}\). (1)

(b)(ii)
- M1: Correct remaining moles of propanoic acid (\(5.00 \times 10^{-3}\text{ mol}\)) and formed moles of propanoate (\(2.50 \times 10^{-3}\text{ mol}\)). (1)
- M2: Correct calculation of concentrations: \([\text{CH}_3\text{CH}_2\text{COOH}] = 0.0667\text{ mol dm}^{-3}\) and \([\text{CH}_3\text{CH}_2\text{COO}^-] = 0.0333\text{ mol dm}^{-3}\). (1)

(b)(iii)
- M1: Correct substitution of concentrations/moles into the \(K_a\) expression or Henderson-Hasselbalch equation. (1)
- M2: Correct calculation of pH = \(4.57\) (accept \(4.6\)). (1)

**(a)**
- The presence of water ligands causes the 3d orbitals of the cobalt(II) ion to split into two groups of different energy levels.
- An electron absorbs light energy/a photon of a frequency corresponding to the energy gap (\(\Delta E = hf\)) and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition).
- The complementary color (remaining wavelengths of light not absorbed) is transmitted or reflected, which is seen as pink.

**(b)**
(i) \([Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\n(ii) Pink to blue.\n\n**(c)**\n(i) \)K_{\text{stab}} = \frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}][Cl^-]^4}\)
(ii) \(\text{dm}^{12} \text{mol}^{-4}\)

**(d)**
- Calculate \(K_{\text{stab}}\):
\(K_{\text{stab}} = 10^{1.60} = 39.81 \text{ dm}^{12}\text{mol}^{-4}\)
- Set up the ratio from the percentage given:
\(\frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}]} = \frac{90.0}{10.0} = 9.00\)
- Substitute into the expression:
\(39.81 = \frac{9.00}{[Cl^-]^4}\)
- Solve for \([Cl^-]\):
\([Cl^-]^4 = \frac{9.00}{39.81} = 0.2261\)
\([Cl^-] = (0.2261)^{0.25} = 0.689 \text{ mol dm}^{-3}\) (or \(0.69 \text{ mol dm}^{-3}\))

**(a)** [3 marks total]
- M1: Splitting of d-orbitals into two different energy levels by ligands. (1)
- M2: Promotion of an electron from lower to higher d-orbital / d-d transition by absorption of light/photon of specific frequency/wavelength. (1)
- M3: Transmitted / reflected / complementary color is seen. (1)

**(b)** [2 marks total]
- M4: Correctly balanced equation: \([Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\). (1)
- M5: Pink to blue. (1)

**(c)** [2 marks total]
- M6: Correct expression: \(K_{\text{stab}} = \frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}][Cl^-]^4}\). (1)
- M7: Correct units: \(\text{dm}^{12}\text{mol}^{-4}\) or \(\text{mol}^{-4}\text{dm}^{12}\). (1)

**(d)** [4 marks total]
- M8: \(K_{\text{stab}} = 10^{1.60} = 39.8\) (or 39.81). (1)
- M9: Correct ratio of complexes \(= 9.0\) (or \(0.90 / 0.10\)). (1)
- M10: Rearrangement to find \([Cl^-]^4 = 0.226\). (1)
- M11: Correct calculation of \([Cl^-] = 0.69\) or \(0.689\) (allow ECF from M8, M9, M10). (1)

**(a)**
- The presence of water ligands causes the 3d orbitals of the cobalt(II) ion to split into two groups of different energy levels.
- An electron absorbs light energy/a photon of a frequency corresponding to the energy gap (\(\Delta E = hf\)) and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition).
- The complementary color (remaining wavelengths of light not absorbed) is transmitted or reflected, which is seen as pink.

**(b)**
(i) \([Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\n(ii) Pink to blue.\n\n**(c)**\n(i) \)K_{\text{stab}} = \frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}][Cl^-]^4}\)
(ii) \(\text{dm}^{12} \text{mol}^{-4}\)

**(d)**
- Calculate \(K_{\text{stab}}\):
\(K_{\text{stab}} = 10^{1.60} = 39.81 \text{ dm}^{12}\text{mol}^{-4}\)
- Set up the ratio from the percentage given:
\(\frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}]} = \frac{90.0}{10.0} = 9.00\)
- Substitute into the expression:
\(39.81 = \frac{9.00}{[Cl^-]^4}\)
- Solve for \([Cl^-]\):
\([Cl^-]^4 = \frac{9.00}{39.81} = 0.2261\)
\([Cl^-] = (0.2261)^{0.25} = 0.689 \text{ mol dm}^{-3}\) (or \(0.69 \text{ mol dm}^{-3}\))

**(a)** [3 marks total]
- M1: Splitting of d-orbitals into two different energy levels by ligands. (1)
- M2: Promotion of an electron from lower to higher d-orbital / d-d transition by absorption of light/photon of specific frequency/wavelength. (1)
- M3: Transmitted / reflected / complementary color is seen. (1)

**(b)** [2 marks total]
- M4: Correctly balanced equation: \([Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\). (1)
- M5: Pink to blue. (1)

**(c)** [2 marks total]
- M6: Correct expression: \(K_{\text{stab}} = \frac{[[CoCl_4]^{2-}]}{[[Co(H_2O)_6]^{2+}][Cl^-]^4}\). (1)
- M7: Correct units: \(\text{dm}^{12}\text{mol}^{-4}\) or \(\text{mol}^{-4}\text{dm}^{12}\). (1)

**(d)** [4 marks total]
- M8: \(K_{\text{stab}} = 10^{1.60} = 39.8\) (or 39.81). (1)
- M9: Correct ratio of complexes \(= 9.0\) (or \(0.90 / 0.10\)). (1)
- M10: Rearrangement to find \([Cl^-]^4 = 0.226\). (1)
- M11: Correct calculation of \([Cl^-] = 0.69\) or \(0.689\) (allow ECF from M8, M9, M10). (1)

(a)
- Independent variable: The mole fraction of salicylate, \(w\) (or the volume ratio of salicylate to iron(III) ions).
- Dependent variable: Absorbance (measured at 530 nm).

(b)
- Calculation:
\(n(\text{Fe}(\text{NO}_3)_3\cdot 9\text{H}_2\text{O}) = 2.00 \times 10^{-3}\text{ mol dm}^{-3} \times 0.250\text{ dm}^3 = 5.00 \times 10^{-4}\text{ mol}\)
\(m = 5.00 \times 10^{-4}\text{ mol} \times 404.0\text{ g mol}^{-1} = 0.202\text{ g}\)
- Procedure:
1. Weigh exactly \(0.202\text{ g}\) of solid iron(III) nitrate nonahydrate using a 3-decimal place balance and a weighing bottle (weighing by difference).
2. Transfer the solid to a beaker and dissolve it completely in about \(50\text{--}100\text{ cm}^3\) of deionised water.
3. Quantitatively transfer the solution to a \(250\text{ cm}^3\) volumetric flask, rinsing the beaker and glass rod several times with deionised water and adding the washings to the flask.
4. Add deionised water until the bottom of the meniscus is exactly on the graduation mark, insert the stopper, and invert the flask several times to ensure a homogeneous solution.

(c)
- Prepare a \(2.00 \times 10^{-3}\text{ mol dm}^{-3}\) solution of sodium salicylate in the same manner.
- Set up 9 separate test tubes/boiling tubes.
- Fill two separate burettes: one with the \(\text{Fe}^{3+}\) solution and one with the salicylate (\(\text{Sal}^-\)) solution.
- Deliver the volumes to each tube systematically as follows:
- Tube 1: \(1.0\text{ cm}^3\text{ Sal}^- + 9.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.1\))
- Tube 2: \(2.0\text{ cm}^3\text{ Sal}^- + 8.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.2\))
- Tube 3: \(3.0\text{ cm}^3\text{ Sal}^- + 7.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.3\))
- Tube 4: \(4.0\text{ cm}^3\text{ Sal}^- + 6.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.4\))
- Tube 5: \(5.0\text{ cm}^3\text{ Sal}^- + 5.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.5\))
- Tube 6: \(6.0\text{ cm}^3\text{ Sal}^- + 4.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.6\))
- Tube 7: \(7.0\text{ cm}^3\text{ Sal}^- + 3.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.7\))
- Tube 8: \(8.0\text{ cm}^3\text{ Sal}^- + 2.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.8\))
- Tube 9: \(9.0\text{ cm}^3\text{ Sal}^- + 1.0\text{ cm}^3\text{ Fe}^{3+}\) (\(w = 0.9\))
- Mix each tube thoroughly by swirling before placing samples into cuvettes.

(d)
- (i) Graph: The graph should have absorbance on the y-axis (from 0 to 1) and mole fraction of salicylate, \(w\), on the x-axis (from 0 to 1). The shape should show two straight lines ascending from \(w=0\) and descending to \(w=1\), intersecting at a peak which represents the maximum absorbance.
- (ii) To determine \(w_{\text{max}}\), draw two straight lines of best fit through the rising and falling data points. The x-coordinate of the intersection point of these two lines corresponds to \(w_{\text{max}}\).
Since \(w_{\text{max}} = \frac{y}{x+y}\), the ratio of salicylate to iron(III) ions is given by \(\frac{y}{x} = \frac{w_{\text{max}}}{1 - w_{\text{max}}}\). Expressing this as the simplest whole-number ratio determines the values of \(x\) and \(y\).

(e)
- Wear chemically resistant gloves to prevent skin irritation; OR keep the solid away from combustible/flammable organic chemicals.

Part (a) [2 marks]
- M1: Correctly identifies the independent variable as the mole fraction of salicylate, \(w\) (or the volume of salicylate/iron(III) solutions mixed). [1]
- M2: Correctly identifies the dependent variable as absorbance. [1]

Part (b) [4 marks]
- M3: Correct calculation of the mass of \(\text{Fe}(\text{NO}_3)_3\cdot 9\text{H}_2\text{O}\) required = \(0.202\text{ g}\) (must show working and be rounded to 3 significant figures). [1]
- M4: Describes weighing out the solid accurately on a 3-decimal place balance (by difference) and dissolving in a beaker using deionised water. [1]
- M5: Quantitative transfer of the mixture to a \(250\text{ cm}^3\) volumetric flask (including rinsing the beaker and glass rod and adding washings to the flask). [1]
- M6: Making up to the graduation mark with deionised water (bottom of meniscus on the mark) and inverting to mix. [1]

Part (c) [4 marks]
- M7: Explicitly states that a \(2.00 \times 10^{-3}\text{ mol dm}^{-3}\) solution of sodium salicylate must also be prepared/used. [1]
- M8: Explains using two separate burettes (or graduated pipettes) to measure the volumes. [1]
- M9: Specifies a systematic range of 9 mixtures with total volume \(10.0\text{ cm}^3\) (e.g., \(1.0, 2.0, \dots, 9.0\text{ cm}^3\) of salicylate and corresponding \(9.0, 8.0, \dots, 1.0\text{ cm}^3\) of iron(III) ions). [1]
- M10: Mentions thorough mixing of each mixture (e.g. swirling/shaking) before measuring absorbance. [1]

Part (d) [3 marks]
- M11: Sketch shows absorbance on the y-axis, mole fraction of salicylate (\(w\)) on the x-axis, with two intersecting straight lines forming a peak (with absorbance starting and ending near 0 at \(w=0\) and \(w=1\)). [1]
- M12: Explains that \(w_{\text{max}}\) is found from the intersection of two straight lines of best fit drawn through the ascending and descending data points. [1]
- M13: States the formula \(\frac{y}{x} = \frac{w_{\text{max}}}{1-w_{\text{max}}}\) (or \(w_{\text{max}} = \frac{y}{x+y}\)) and explains that the values of \(x\) and \(y\) are determined from the simplest whole-number ratio. [1]

Part (e) [1 mark]
- M14: Suggests wearing protective gloves to avoid contact with the skin irritant, OR keeping the substance away from combustible/flammable materials (due to oxidising properties). [1]

(a)
(i) Independent variable: Concentration of hydrogen peroxide, \([\text{H}_2\text{O}_2]\) (or volume of \(\text{H}_2\text{O}_2\) used).
(ii) Dependent variable: Time, \(t\), for the blue-black color to appear (or rate of reaction represented by \(1/t\)).
(iii) Controlled variables: Temperature; concentrations and volumes of potassium iodide, sulfuric acid, sodium thiosulfate, and starch; total volume of the reaction mixture (\(50.0\text{ cm}^3\)).

(b)
The total volume of \(\text{KI}\), \(\text{H}_2\text{SO}_4\), \(\text{Na}_2\text{S}_2\text{O}_3\), and starch is constant at:
\(10.0 + 10.0 + 5.0 + 1.0 = 26.0\text{ cm}^3\).
Since the total volume of each mixture is \(50.0\text{ cm}^3\), the combined volume of \(\text{H}_2\text{O}_2\) and deionised water must be:
\(50.0 - 26.0 = 24.0\text{ cm}^3\).
Therefore: \(V_{\text{water}} = 24.0 - V_{\text{H}_2\text{O}_2}\).

| Mixture | Vol. of \(\text{H}_2\text{O}_2\) / \(\text{cm}^3\) | Vol. of \(\text{KI}\) / \(\text{cm}^3\) | Vol. of \(\text{H}_2\text{SO}_4\) / \(\text{cm}^3\) | Vol. of \(\text{Na}_2\text{S}_2\text{O}_3\) / \(\text{cm}^3\) | Vol. of starch / \(\text{cm}^3\) | Vol. of water / \(\text{cm}^3\) | Total Vol. / \(\text{cm}^3\) |
|---|---|---|---|---|---|---|---|
| 1 | 4.0 | 10.0 | 10.0 | 5.0 | 1.0 | 20.0 | 50.0 |
| 2 | 8.0 | 10.0 | 10.0 | 5.0 | 1.0 | 16.0 | 50.0 |
| 3 | 12.0 | 10.0 | 10.0 | 5.0 | 1.0 | 12.0 | 50.0 |
| 4 | 16.0 | 10.0 | 10.0 | 5.0 | 1.0 | 8.0 | 50.0 |
| 5 | 20.0 | 10.0 | 10.0 | 5.0 | 1.0 | 4.0 | 50.0 |

(c)
- Use separate burettes to accurately deliver the volumes of \(\text{KI}\), \(\text{H}_2\text{SO}_4\), \(\text{Na}_2\text{S}_2\text{O}_3\), starch, and deionised water into a clean \(100\text{ cm}^3\) beaker (Mixture A).
- Measure the volume of \(\text{H}_2\text{O}_2\) into a separate small beaker or boiling tube (Mixture B).
- Pour Mixture B rapidly into Mixture A, swirl immediately to mix, and start the stopwatch at the exact moment of mixing.
- Stop the stopwatch the instant the solution turns blue-black and record the time, \(t\).
- Repeat this procedure for the other four mixtures, ensuring the temperature remains constant.

(d)
- The initial rate of reaction is represented as \(\frac{1}{t}\), because the amount of thiosulfate reacts with a constant, small amount of iodine produced, meaning the average rate during this short time is a good measure of the initial rate.
- Plot a graph of \(\frac{1}{t}\) (y-axis) against the volume of \(\text{H}_2\text{O}_2\) (x-axis).
- If the graph is a straight line passing through the origin, the rate is directly proportional to the concentration of \(\text{H}_2\text{O}_2\), and therefore the reaction is first-order with respect to \(\text{H}_2\text{O}_2\). Alternatively, a log-log plot of \(\log(1/t)\) against \(\log(V_{\text{H}_2\text{O}_2})\) can be constructed, where the gradient of the straight line equals the order.

(e)
- The concentration of \(\text{Na}_2\text{S}_2\text{O}_3\) must be very low to ensure that only a negligible percentage (e.g. less than 5%) of the reactants is consumed before the end-point is reached. This keeps the concentrations of the reactants effectively constant during the timed interval, ensuring that the measured rate represents the true *initial* rate.

Part (a) [3 marks]
- M1: Identifies the independent variable: concentration or volume of \(\text{H}_2\text{O}_2\). [1]
- M2: Identifies the dependent variable: time taken (\(t\)) or rate (\(1/t\)). [1]
- M3: Identifies any two controlled variables: temperature, and concentrations/volumes of \(\text{KI}\), \(\text{H}_2\text{SO}_4\), \(\text{Na}_2\text{S}_2\text{O}_3\), and starch (or total volume of the mixture). [1]

Part (b) [4 marks]
- M4: Design of table columns includes units in headings (e.g. \(/ \text{cm}^3\)) for all six liquid components and shows a systematic variation of \(\text{H}_2\text{O}_2\) volume from 4.0 to 20.0 \(\text{cm}^3\). [1]
- M5: Shows volumes of \(\text{KI}\) (10.0), \(\text{H}_2\text{SO}_4\) (10.0), \(\text{Na}_2\text{S}_2\text{O}_3\) (5.0), and starch (1.0) constant in all mixtures. [1]
- M6: Correctly calculates water volumes (20.0, 16.0, 12.0, 8.0, 4.0 \(\text{cm}^3\) respectively) ensuring the total volume is consistently \(50.0\text{ cm}^3\). [1]
- M7: Formats all values in the table to 1 decimal place. [1]

Part (c) [4 marks]
- M8: Explains measuring \(\text{KI}\), \(\text{H}_2\text{SO}_4\), \(\text{Na}_2\text{S}_2\text{O}_3\), starch, and water into one reaction vessel, and \(\text{H}_2\text{O}_2\) in a separate container. [1]
- M9: Specifies using burettes or graduated pipettes for precise volume measurements. [1]
- M10: States that the stopwatch must be started at the instant of mixing. [1]
- M11: States that the stopwatch is stopped at the first permanent appearance of the blue-black color. [1]

Part (d) [3 marks]
- M12: Explains that \(\text{rate} \propto 1/t\) because the amount of iodine produced to reach the blue-black end-point is constant. [1]
- M13: Proposes plotting a graph of \(1/t\) vs volume of \(\text{H}_2\text{O}_2\) or \(\log(1/t)\) vs \(\log(V_{\text{H}_2\text{O}_2})\). [1]
- M14: Explains how the order is deduced (a straight line through the origin for the linear plot indicates first-order, OR the gradient of the log-log plot equals the order). [1]

Part (e) [1 mark]
- M15: Explains that a low concentration of thiosulfate is required so that only a tiny fraction of the reactants is consumed before the clock color change, keeping concentrations constant and ensuring the measured rate represents the initial rate. [1]