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1. Calculate the mass of water lost: \(3.57\text{ g} - 1.32\text{ g} = 2.25\text{ g}\). 2. Calculate the amount in moles of anhydrous sodium carbonate: \(n(\text{Na}_2\text{CO}_3) = \frac{1.32\text{ g}}{106.0\text{ g mol}^{-1}} = 0.01245\text{ mol}\). 3. Calculate the amount in moles of water: \(n(\text{H}_2\text{O}) = \frac{2.25\text{ g}}{18.0\text{ g mol}^{-1}} = 0.125\text{ mol}\). 4. Determine the ratio \(x\): \(x = \frac{0.125}{0.01245} \approx 10\).

[1 mark] C is the correct answer. Award 1 mark for the correct choice. Reject all other options.

1. Calculate initial moles: \(n(\text{CH}_3\text{COOH}) = 0.0200\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.00 \times 10^{-3}\text{ mol}\), \(n(\text{NaOH}) = 0.0100\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.00 \times 10^{-3}\text{ mol}\). 2. Determine moles after neutralization: \(n(\text{CH}_3\text{COO}^-) = 1.00 \times 10^{-3}\text{ mol}\), \(n(\text{CH}_3\text{COOH})_{\text{remaining}} = 2.00 \times 10^{-3} - 1.00 \times 10^{-3} = 1.00 \times 10^{-3}\text{ mol}\). 3. Calculate pH: Since \([\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]\), then \(\text{pH} = \text{p}K_a = -\log_{10}(1.75 \times 10^{-5}) = 4.76\).

[1 mark] B is the correct answer. Award 1 mark for the correct choice. Reject all other options.

1. The standard cell potential is: \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = 0.77 - (-0.74) = +1.51\text{ V}\). 2. The standard electrode potential for \(\text{Fe}^{3+}/\text{Fe}^{2+}\) is more positive than that of \(\text{Cr}^{3+}/\text{Cr}\). Thus, \(\text{Fe}^{3+}\) undergoes reduction and \(\text{Cr}\) undergoes oxidation. 3. This means that \(\text{Cr}(\text{s})\) reduces \(\text{Fe}^{3+}(\text{aq})\).

[1 mark] A is the correct answer. Award 1 mark for the correct choice. Reject all other options.

1. The methyl group is an alkyl group that is electron-donating by inductive effects. This increases electron density in the aromatic ring, making it more nucleophilic and reactive, so it reacts more rapidly than benzene. 2. Electron-donating groups like methyl direct incoming electrophiles predominantly to the 2- and 4-positions (ortho/para directing).

[1 mark] C is the correct answer. Award 1 mark for the correct choice. Reject all other options.

Using the Born-Haber cycle: \(\Delta_f H^\theta = \Delta_{at} H^\theta(\text{Na}) + I_1(\text{Na}) + \Delta_{at} H^\theta(\text{Cl}) + EA_1(\text{Cl}) + \Delta_{\text{LE}} H^\theta\). Rearranging to solve for lattice enthalpy: \(-411 = 107 + 496 + 122 - 349 + \Delta_{\text{LE}} H^\theta\), which simplifies to \(-411 = 376 + \Delta_{\text{LE}} H^\theta\). Therefore, \(\Delta_{\text{LE}} H^\theta = -411 - 376 = -787\text{ kJ mol}^{-1}\).

[1 mark] A is the correct answer. Award 1 mark for the correct choice. Reject all other options.

At a highly alkaline pH of 12: 1. The carboxylic acid group is deprotonated: \(-\text{COOH} \rightarrow -\text{COO}^-\). 2. The amine group exists in its basic, unprotonated form: \(-\text{NH}_2\). 3. Therefore, the dominant species is the anion \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).

[1 mark] C is the correct answer. Award 1 mark for the correct choice. Reject all other options.

1. The electronic configuration of a neutral Fe atom is \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). 2. Transition metal atoms lose their \(4\text{s}\) electrons first when forming ions. 3. To form \(\text{Fe}^{3+}\), three electrons are removed: two from \(4\text{s}\) and one from \(3\text{d}\), resulting in \([\text{Ar}] 3\text{d}^5\).

[1 mark] A is the correct answer. Award 1 mark for the correct choice. Reject all other options.

1. Write the new rate expression: \(\text{Rate}_{\text{new}} = k \left( 0.5[\text{A}] \right) \left( 3[\text{B}] \right)^2\). 2. Simplify the terms: \(\text{Rate}_{\text{new}} = k \times 0.5[\text{A}] \times 9[\text{B}]^2 = 4.5 \times k[\text{A}][\text{B}]^2\). 3. Therefore, the rate increases by a factor of 4.5.

[1 mark] B is the correct answer. Award 1 mark for the correct choice. Reject all other options.

First, calculate the number of moles of \( \text{H}_2 \) gas produced using the ideal gas equation, \( pV = nRT \):

Convert units:
\( p = 1.00 \times 10^5\text{ Pa} \)
\( V = 153\text{ cm}^3 = 153 \times 10^{-6}\text{ m}^3 \)
\( T = 298\text{ K} \)

\[ n(\text{H}_2) = \frac{pV}{RT} = \frac{1.00 \times 10^5 \times 153 \times 10^{-6}}{8.314 \times 298} = 6.175 \times 10^{-3}\text{ mol} \]

From the equation, \( 1\text{ mol} \) of \( \text{M} \) produces \( 1\text{ mol} \) of \( \text{H}_2 \). Therefore, \( n(\text{M}) = 6.175 \times 10^{-3}\text{ mol} \).

Now, calculate the molar mass (\( M \)) of \( \text{M} \):

\[ M = \frac{\text{mass}}{n} = \frac{0.248\text{ g}}{6.175 \times 10^{-3}\text{ mol}} = 40.16\text{ g mol}^{-1} \]

Comparing this to the relative atomic masses of Group 2 metals: Calcium (\( \text{Ca} \)) has a relative atomic mass of \( 40.1\text{ g mol}^{-1} \).

[1 mark] award for identifying Calcium (B) as the correct option.
- Method: Correct application of \( pV = nRT \) to find moles of hydrogen, use of stoichiometry to relate to metal moles, and calculation of atomic mass to identify the metal.

According to Hess' Law, the Born-Haber cycle equation for \( \text{MgCl}_2 \) is:

\( \Delta_f H[\text{MgCl}_2\text{(s)}] = \Delta_{\text{at}} H[\text{Mg(s)}] + \text{IE}_1[\text{Mg(g)}] + \text{IE}_2[\text{Mg(g)}] + 2 \times \Delta_{\text{at}} H[\text{Cl}_2\text{(g)}] + 2 \times \text{EA}_1[\text{Cl(g)}] + \Delta_{\text{LE}} H[\text{MgCl}_2\text{(s)}] \)

Substitute the given values into the equation:

\( -642 = 148 + 738 + 1451 + (2 \times 121) + (2 \times -349) + \Delta_{\text{LE}} H \)

\( -642 = 148 + 738 + 1451 + 242 - 698 + \Delta_{\text{LE}} H \)

\( -642 = 1881 + \Delta_{\text{LE}} H \)

\( \Delta_{\text{LE}} H = -642 - 1881 = -2523\text{ kJ mol}^{-1} \)

[1 mark] award for identifying \( -2523\text{ kJ mol}^{-1} \) (A) as the correct value.
- Correctly accounts for two moles of gaseous chlorine atoms and chlorine ions (multiplying both atomisation and electron affinity values of chlorine by 2).

For a reaction to be feasible, Gibbs free energy change must be less than or equal to zero:

\( \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus \le 0 \)

Rearranging for temperature \( T \):

\( T \ge \frac{\Delta H^\ominus}{\Delta S^\ominus} \)

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

\( \Delta H^\ominus = 142 \times 10^3\text{ J mol}^{-1} \)

\( T \ge \frac{142000}{185} = 767.57\text{ K} \approx 768\text{ K} \)

Therefore, the reaction becomes feasible at temperatures above \( 768\text{ K} \).

[1 mark] award for identifying "Above \( 768\text{ K} \)" (B) as the correct range.
- Correct conversion of enthalpy units and rearrangement of the Gibbs equation.

1. Find initial moles:
- \( n(\text{propanoic acid}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol} \)
- \( n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 0.0075\text{ mol} \)

2. Solve reaction stoichiometry (weak acid partially neutralised by strong base):
- \( n(\text{propanoate ions, } \text{A}^-) = n(\text{NaOH}) = 0.0075\text{ mol} \)
- \( n(\text{remaining acid, } \text{HA}) = 0.0125 - 0.0075 = 0.0050\text{ mol} \)

3. Use buffer equation to find \( [\text{H}^+] \):
- \( [\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{0.0050}{0.0075} = 9.00 \times 10^{-6}\text{ mol dm}^{-3} \)

4. Calculate pH:
- \( \text{pH} = -\log_{10}(9.00 \times 10^{-6}) = 5.05 \)

[1 mark] award for identifying 5.05 (C) as the correct pH value.
- Step-by-step method calculation correctly determines acid excess and salt formation moles.

For a spontaneous reaction, the standard cell potential \( E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation}) \) must be positive.

Let's evaluate each mixture:
- **A**: \( \text{Fe}^{3+}\text{(aq)} \) and \( \text{SO}_4^{2-}\text{(aq)} \). The potential cell reaction has \( E^\ominus_{\text{cell}} = 0.77 - 2.01 = -1.24\text{ V} \) (non-spontaneous).
- **B**: \( \text{Fe}^{2+}\text{(aq)} \) and \( \text{I}_2\text{(aq)} \). Here \( \text{I}_2 \) must act as oxidising agent: \( E^\ominus_{\text{cell}} = 0.54 - 0.77 = -0.23\text{ V} \) (non-spontaneous).
- **C**: \( \text{Fe}^{3+}\text{(aq)} \) and \( \text{I}^-\text{(aq)} \). Here \( \text{Fe}^{3+} \) is reduced and \( \text{I}^- \) is oxidised: \( E^\ominus_{\text{cell}} = 0.77 - 0.54 = +0.23\text{ V} \) (spontaneous).
- **D**: \( \text{Cu}^{2+}\text{(aq)} \) and \( \text{I}_2\text{(aq)} \). Here \( \text{Cu}^{2+} \) would need to oxidise \( \text{I}^- \) (not present) or \( \text{I}_2 \) would need to oxidise \( \text{Cu}^+ \) (not present). Even if reacting, \( E^\ominus_{\text{cell}} = 0.15 - 0.54 = -0.39\text{ V} \) (non-spontaneous).

[1 mark] award for identifying C as the spontaneous reaction.
- Under standard conditions, the species with the more positive electrode potential (\( \text{Fe}^{3+} \)) will undergo reduction and oxidise the species with the less positive potential (\( \text{I}^- \)).

- **Ligands**: Ethane-1,2-diamine (en) is a neutral, bidentate ligand. There are two 'en' ligands, forming 4 coordinate bonds. Chloride ions (\( \text{Cl}^- \)) are monodentate, forming 1 coordinate bond each (2 total).
- **Coordination number**: The total number of coordinate bonds is \( 4 + 2 = 6 \).
- **Shape**: A transition metal complex with a coordination number of 6 has an octahedral shape.
- **Oxidation state**: Let \( x \) be the oxidation state of cobalt. Since 'en' is neutral and chloride has a \( -1 \) charge:
\[ x + 2(0) + 2(-1) = +1 \Rightarrow x = +3 \]

[1 mark] award for identifying the correct parameters: Oxidation state = +3, Coordination number = 6, Shape = Octahedral (C).

1. Set up the ICE table (moles):
- Initial: \( n(\text{SO}_2) = 2.00 \), \( n(\text{O}_2) = 2.00 \), \( n(\text{SO}_3) = 0 \)
- Change: \( n(\text{SO}_3) = +1.60 \), therefore \( n(\text{SO}_2) = -1.60 \) and \( n(\text{O}_2) = -0.80 \) due to standard mole ratios.
- Equilibrium moles: \( n(\text{SO}_2) = 0.40 \), \( n(\text{O}_2) = 1.20 \), \( n(\text{SO}_3) = 1.60 \)

2. Convert equilibrium moles to concentration by dividing by the volume of \( 2.00\text{ dm}^3 \):
- \( [\text{SO}_2] = \frac{0.40\text{ mol}}{2.00\text{ dm}^3} = 0.20\text{ mol dm}^{-3} \)
- \( [\text{O}_2] = \frac{1.20\text{ mol}}{2.00\text{ dm}^3} = 0.60\text{ mol dm}^{-3} \)
- \( [\text{SO}_3] = \frac{1.60\text{ mol}}{2.00\text{ dm}^3} = 0.80\text{ mol dm}^{-3} \)

3. Write and calculate the expression for \( K_c \):
- \( K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \)
- \( K_c = \frac{(0.80)^2}{(0.20)^2 \times 0.60} = \frac{0.64}{0.04 \times 0.60} = 26.67\text{ dm}^3\text{ mol}^{-1} \approx 26.7 \)

[1 mark] award for identifying 26.7 (B) as the correct value.
- Requires accurate calculation of equilibrium moles, conversion to concentration using flask volume, and applying the equilibrium expression.

(a) A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton (\(\text{H}^+\)).

(b) Initial moles of propanoic acid (\(\text{HA}\)) = \(0.0500 \times 0.250 = 0.0125\text{ mol}\).
Initial moles of \(\text{NaOH}\) = \(0.0300 \times 0.150 = 0.00450\text{ mol}\).
Reaction: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\).
Remaining moles of propanoic acid = \(0.0125 - 0.00450 = 0.00800\text{ mol}\).
Moles of propanoate ions (\(\text{A}^-\)) formed = \(0.00450\text{ mol}\).
\([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.35 \times 10^{-5} \times \frac{0.00800}{0.00450} = 2.40 \times 10^{-5}\text{ mol dm}^{-3}\).
\(\text{pH} = -\log_{10}(2.40 \times 10^{-5}) = 4.62\).

(c) Added moles of \(\text{H}^+\) = \(0.00100 \times 1.00 = 0.00100\text{ mol}\).
Added \(\text{H}^+\) reacts with propanoate ions: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\).
New moles of propanoate (\(\text{A}^-\)) = \(0.00450 - 0.00100 = 0.00350\text{ mol}\).
New moles of propanoic acid (\(\text{HA}\)) = \(0.00800 + 0.00100 = 0.00900\text{ mol}\).
New \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.00900}{0.00350} = 3.471 \times 10^{-5}\text{ mol dm}^{-3}\).
New \(\text{pH} = -\log_{10}(3.471 \times 10^{-5}) = 4.46\).

(a) [2 marks]
- M1: Two species that differ by an \(\text{H}^+\) / proton.
- M2: Identifies the acid as the proton donor and the conjugate base as the proton acceptor.

(b) [6 marks]
- M1: Calculates initial moles of acid as \(0.0125\text{ mol}\) and alkali as \(0.00450\text{ mol}\).
- M2: Moles of \(\text{CH}_3\text{CH}_2\text{COO}^-\) formed = \(0.00450\text{ mol}\).
- M3: Moles of \(\text{CH}_3\text{CH}_2\text{COOH}\) remaining = \(0.00800\text{ mol}\).
- M4: Correct expression for \(K_a\) or Henderson-Hasselbalch equation used.
- M5: Calculates \([\text{H}^+] = 2.40 \times 10^{-5}\text{ mol dm}^{-3}\).
- M6: Calculates \(\text{pH} = 4.62\) (must be 2 decimal places).

(c) [5 marks]
- M1: Added moles of \(\text{H}^+\) = \(0.00100\text{ mol}\).
- M2: Calculates new moles: \(\text{HA} = 0.00900\text{ mol}\) and \(\text{A}^- = 0.00350\text{ mol}\).
- M3: Calculates new \(\text{pH} = 4.46\) (must be 2 decimal places).
- M4: Gives equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\).
- M5: Explains that the conjugate base reacts with the added \(\text{H}^+\) to minimize the change in pH.

(a) Observation: Pale blue solution turns yellow-green.
Equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\).
Shape: Tetrahedral.

(b) The complex has cis and trans stereoisomers. The cis isomer is chiral and has two enantiomers (optical isomers), whereas the trans isomer is achiral (has a plane of symmetry):
- Trans isomer: Cl ligands are opposite each other (bond angle \(180^\circ\)).
- Cis isomers: Cl ligands are adjacent (bond angle \(90^\circ\)). These exist as a pair of non-superimposable mirror images.

(c) Moles of \(\text{MnO}_4^-\): \(0.01840 \times 0.0200 = 3.68 \times 10^{-4}\text{ mol}\).
Equation: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\).
Moles of \(\text{Fe}^{2+}\) = \(5 \times 3.68 \times 10^{-4} = 1.84 \times 10^{-3}\text{ mol}\).
Mass of \(\text{Fe}\) = \(1.84 \times 10^{-3} \times 55.8 = 0.10267\text{ g}\).
Percentage by mass = \(\frac{0.10267}{0.235} \times 100 = 43.69\% \approx 43.7\%\).

(a) [4 marks]
- M1: Pale blue solution changes to green/yellow.
- M2: Correct reactants and products in equation.
- M3: Balanced equation.
- M4: Shape is tetrahedral.

(b) [5 marks]
- M1: 3D drawing of trans-isomer (Cl-Co-Cl angle of \(180^\circ\)).
- M2: 3D drawing of cis-isomer (Cl-Co-Cl angle of \(90^\circ\)).
- M3: 3D drawing of the mirror image of the cis-isomer.
- M4: Labels cis and trans isomers correctly.
- M5: Labels the two cis isomers as optical isomers.

(c) [4 marks]
- M1: Calculates moles of \(\text{MnO}_4^-\): \(3.68 \times 10^{-4}\text{ mol}\).
- M2: Calculates moles of \(\text{Fe}^{2+}\): \(1.84 \times 10^{-3}\text{ mol}\) (by using 1:5 ratio).
- M3: Mass of \(\text{Fe}\) = \(0.10267\text{ g}\) (accept using \(A_r = 55.8\) or \(56\)).
- M4: Calculates percentage by mass = \(43.7\%\) (accept \(43.69\%\) or \(43.8\%\) if \(56\) is used).

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

(b) The cell diagram must show:
- Two beakers: one with \(\text{I}_2(\text{aq})/\text{I}^-(\text{aq})\) and one with \(\text{Fe}^{3+}(\text{aq})/\text{Fe}^{2+}(\text{aq})\).
- Platinum (Pt) electrodes in both beakers (since both systems consist of aqueous species in different oxidation states).
- A salt bridge connecting the two beakers.
- A voltmeter in the external wire connecting the two Pt electrodes.
- Standard conditions: \(298\text{ K}\) (or \(25^\circ\text{C}\)) and concentration of ions at \(1.0\text{ mol dm}^{-3}\).
- Electron flow: flow of electrons is from the negative electrode (half-cell 2 with \(E^\theta = +0.54\text{ V}\)) to the positive electrode (half-cell 3 with \(E^\theta = +0.77\text{ V}\)).

(c) The uncatalyzed reaction, \(\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2\), involves two negatively charged ions (anions) which repel each other, giving a high activation energy.

In the catalyzed reaction:
Step 1: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\).
This is feasible because \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\) is more positive than \(E^\theta(\text{I}_2/\text{I}^-) = +0.54\text{ V}\).

Step 2: \(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\).
This is feasible because \(E^\theta(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}) = +2.01\text{ V}\) is more positive than \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\).

(a) [3 marks]
- M1: Electromotive force / EMF of a half-cell compared with a standard hydrogen electrode.
- M2: Temperature of \(298\text{ K}\) (or \(25^\circ\text{C}\)) AND pressure of \(100\text{ kPa}\).
- M3: Concentration of all solutions is \(1.00\text{ mol dm}^{-3}\).

(b) [5 marks]
- M1: Complete circuit with two Platinum (Pt) electrodes.
- M2: Salt bridge connecting the two beakers.
- M3: Solution labels (\(\text{I}_2/\text{I}^-\)) and (\(\text{Fe}^{2+}/\text{Fe}^{3+}\)) with standard concentrations.
- M4: Temperature of \(298\text{ K}\) specified.
- M5: Shows direction of electron flow from the iodine half-cell to the iron half-cell on the wire.

(c) [4 marks]
- M1: Explains that the uncatalyzed reaction has high activation energy due to repulsion between two negatively charged ions.
- M2: Writes correct equation for the oxidation of iodide by iron(III) ions.
- M3: Writes correct equation for the reduction of peroxodisulfate by iron(II) ions.
- M4: Explains feasibility using standard electrode potentials (e.g. comparing the standard potentials for each step).

(a) Enthalpy change of atomisation is the enthalpy change that accompanies the formation of one mole of gaseous atoms from the element in its standard state under standard conditions.

(b)(i) The Born-Haber cycle consists of:
- Zero enthalpy level: \(\text{Ca}(\text{s}) + \text{F}_2(\text{g})\).
- Bottom level: \(\text{CaF}_2(\text{s})\).
- Upward steps:
1. Atomisation of calcium: \(\text{Ca}(\text{g}) + \text{F}_2(\text{g})\) (\(+178\text{ kJ mol}^{-1}\))
2. Bond enthalpy of fluorine to atomise \(\text{F}_2\): \(\text{Ca}(\text{g}) + 2\text{F}(\text{g})\) (\(+158\text{ kJ mol}^{-1}\))
3. First ionisation of Ca: \(\text{Ca}^+(\text{g}) + 2\text{F}(\text{g}) + \text{e}^-\) (\(+590\text{ kJ mol}^{-1}\))
4. Second ionisation of Ca: \(\text{Ca}^{2+}(\text{g}) + 2\text{F}(\text{g}) + 2\text{e}^-\) (\(+1145\text{ kJ mol}^{-1}\))
- Downward steps:
5. Electron affinity of 2 moles of F: \(\text{Ca}^{2+}(\text{g}) + 2\text{F}^-(\text{g})\) (\(2 \times (-328) = -656\text{ kJ mol}^{-1}\))
6. Lattice formation enthalpy: \(\text{Ca}^{2+}(\text{g}) + 2\text{F}^-(\text{g}) \rightarrow \text{CaF}_2(\text{s})\).

(ii) Using Hess's law:
\(\Delta H_f = \Delta H_{at}(\text{Ca}) + \text{Bond Enthalpy}(\text{F}_2) + 1\text{st IE}(\text{Ca}) + 2\text{nd IE}(\text{Ca}) + 2\text{EA}(\text{F}) + \Delta H_{LE}\)
\(-1220 = 178 + 158 + 590 + 1145 + 2(-328) + \Delta H_{LE}\)
\(-1220 = 2071 - 656 + \Delta H_{LE}\)
\(-1220 = 1415 + \Delta H_{LE}\)
\(\Delta H_{LE} = -1220 - 1415 = -2635\text{ kJ mol}^{-1}\).

(c) The lattice enthalpy of calcium chloride (\(\text{CaCl}_2\)) is less negative/exothermic than that of calcium fluoride (\(\text{CaF}_2\)).
- This is because the chloride ion (\(\text{Cl}^-\)) has a larger ionic radius than the fluoride ion (\(\text{F}^-\)).
- Both ions have the same charge (\(-1\)), but the larger chloride ion has lower charge density.
- Consequently, there is weaker electrostatic attraction between the calcium cation and chloride anions than in the calcium fluoride lattice.

(a) [2 marks]
- M1: Enthalpy change for the formation of one mole of gaseous atoms.
- M2: From the element in its standard state under standard conditions.

(b)(i) [5 marks]
- M1: Zero line with reactants \(\text{Ca}(\text{s}) + \text{F}_2(\text{g})\) and product line with \(\text{CaF}_2(\text{s})\).
- M2: Atomisation steps with correct species and states.
- M3: Ionisation steps of calcium with correct species and states.
- M4: Electron affinity step with \(2 \times (-328) = -656\text{ kJ mol}^{-1}\).
- M5: Arrow for lattice enthalpy linking gaseous ions to solid lattice.

(b)(ii) [3 marks]
- M1: Sums all endothermic steps to get \(+2071\text{ kJ mol}^{-1}\).
- M2: Accounts for doubling of electron affinity of F (\(-656\text{ kJ mol}^{-1}\)).
- M3: Correctly calculates \(\Delta H_{LE} = -2635\text{ kJ mol}^{-1}\) (must have negative sign and units).

(c) [4 marks]
- M1: States \(\text{CaCl}_2\) has a less negative / less exothermic lattice enthalpy.
- M2: Identifies that the chloride ion is larger than the fluoride ion.
- M3: Mentions that the chloride ion has lower charge density (same charge, larger volume).
- M4: Concludes that electrostatic attraction between ions is weaker.

(a)(i) Order with respect to \(\text{I}^-\):
- Compare Exp 1 and Exp 2: \([\text{S}_2\text{O}_8^{2-}]\) is constant. \([\text{I}^-]\) doubles (from \(0.0150\) to \(0.0300\)), rate doubles (from \(1.50 \times 10^{-5}\) to \(3.00 \times 10^{-5}\)). Thus, first order with respect to \(\text{I}^-\).
Order with respect to \(\text{S}_2\text{O}_8^{2-}\):
- Compare Exp 1 and Exp 3: \([\text{I}^-]\) is constant. \([\text{S}_2\text{O}_8^{2-}]\) doubles (from \(0.0100\) to \(0.0200\)), rate doubles (from \(1.50 \times 10^{-5}\) to \(3.00 \times 10^{-5}\)). Thus, first order with respect to \(\text{S}_2\text{O}_8^{2-}\).
Rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

(a)(ii) Using Experiment 1 data:
\(1.50 \times 10^{-5} = k \times 0.0100 \times 0.0150\)
\(1.50 \times 10^{-5} = k \times 1.50 \times 10^{-4}\)
\(k = 0.100\).
Units of \(k\): \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(b) The Arrhenius equation relationship is: \(\text{gradient} = -\frac{E_a}{R}\).
\(-9620 = -\frac{E_a}{8.314}\)
\(E_a = 9620 \times 8.314 = 79980.68\text{ J mol}^{-1}\).
Converting to \(\text{kJ mol}^{-1}\): \(E_a = 80.0\text{ kJ mol}^{-1}\) (to 3 significant figures).

(c) A catalyst provides an alternative pathway with a lower activation energy. Consequently, a greater fraction of molecules have energy equal to or greater than the activation energy (as represented by the shaded area on a Boltzmann distribution), increasing the rate of successful collisions.

(a)(i) [4 marks]
- M1: Identifies first order for \(\text{I}^-\) and explains using Exp 1 & Exp 2.
- M2: Identifies first order for \(\text{S}_2\text{O}_8^{2-}\) and explains using Exp 1 & Exp 3.
- M3: Writes correct rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (Allow ECF from deduced orders).
- M4: Logic is clear and cohesive.

(a)(ii) [3 marks]
- M1: Rearranges equation to solve for \(k\).
- M2: Calculates correct value: \(0.100\) (or \(0.1\)).
- M3: Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(b) [3 marks]
- M1: Uses expression: \(\text{gradient} = -E_a / R\).
- M2: Calculates \(E_a\) in Joules: \(79981\text{ J mol}^{-1}\).
- M3: Converts to \(\text{kJ mol}^{-1}\) to 3 sig figs: \(80.0\text{ kJ mol}^{-1}\).

(c) [2 marks]
- M1: Alternative pathway with lower activation energy.
- M2: More molecules have energy \(\ge E_a\) (relates to the Boltzmann distribution).

(a)
- Benzene has a delocalised pi-system with 6 pi-electrons spread over 6 carbon atoms, whereas cyclohexene has a localised pi-bond with 2 pi-electrons between 2 carbon atoms.
- The localized pi-bond in cyclohexene has a higher pi-electron density than the delocalised system in benzene.
- As a result, cyclohexene polarises electrophiles more effectively and is more susceptible to electrophilic attack.
- Benzene undergoes electrophilic substitution rather than addition because addition would disrupt the stable delocalised aromatic system (loss of resonance energy).

(b)
- Generation of electrophile:
\(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\) (or \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\)).
- Mechanism:
1. A curly arrow is drawn from the delocalised pi-ring of methylbenzene to the \(\text{NO}_2^+\).
2. An intermediate is drawn showing a horseshoe shape positive charge in the ring, open towards the carbon containing both the methylbenzene hydrogen and the nitro group.
3. A curly arrow is drawn from the \(\text{C}-\text{H}\) bond pointing back into the ring.
4. The final products are 4-nitromethylbenzene (or 2-nitromethylbenzene) and \(\text{H}^+\).

(c)
- Step 1: Bromination of nitrobenzene using \(\text{Br}_2\) and a halogen carrier catalyst (such as \(\text{AlBr}_3\) or \(\text{FeBr}_3\)) to form 3-bromonitrobenzene (the nitro group is meta-directing).
- Step 2: Reduction of 3-bromonitrobenzene using tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) under reflux, followed by addition of aqueous sodium hydroxide (\(\text{NaOH}\)) to form 3-bromophenylamine.

(a) [5 marks]
- M1: Benzene has delocalised pi-electrons, cyclohexene has localized pi-electrons.
- M2: Cyclohexene has a higher pi-electron density than benzene.
- M3: Cyclohexene polarizes electrophiles more easily.
- M4: Addition to benzene would destroy aromaticity / stable delocalised system.
- M5: Substitution allows benzene to retain its aromatic stability.

(b) [5 marks]
- M1: Balanced equation for formation of \(\text{NO}_2^+\).
- M2: Curly arrow from the ring to \(\text{NO}_2^+\).
- M3: Correctly drawn intermediate showing positive charge inside the open ring (horseshoe opening towards sp3 carbon).
- M4: Curly arrow from the \(\text{C}-\text{H}\) bond of that carbon back into the ring.
- M5: Correct products: nitromethylbenzene and \(\text{H}^+\).

(c) [3 marks]
- M1: Step 1: \(\text{Br}_2\) AND halogen carrier catalyst (e.g. \(\text{FeBr}_3\), \(\text{AlBr}_3\)).
- M2: Identifies intermediate structure as 3-bromonitrobenzene.
- M3: Step 2: \(\text{Sn}\) and conc. \(\text{HCl}\) under reflux, followed by \(\text{NaOH}(\text{aq})\).

(a) The strong, broad peak at \(2500 - 3300\text{ cm}^{-1}\) indicates the \(\text{O}-\text{H}\) bond of a carboxylic acid. The sharp peak at \(1715\text{ cm}^{-1}\) indicates a \(\text{C}=\text{O}\) bond. Therefore, the functional group is a carboxylic acid.

(b) Compound X has 4 carbons but only 3 peaks, which indicates that there are two equivalent carbon environments (symmetry), meaning two equivalent methyl groups.

(c)(i) Compound X is 2-methylpropanoic acid, \(\text{(CH}_3)_2\text{CHCOOH}\).
- Peak at \(\delta = 12.0\text{ ppm}\) (1H, singlet) is the carboxylic acid proton (\(\text{-COOH}\)). It is a singlet because there are no hydrogen atoms on the adjacent carbon.
- Peak at \(\delta = 1.20\text{ ppm}\) (6H, doublet) is due to the 6 equivalent protons of the two methyl groups: \(\text{-CH(CH}_3)_2\). According to the \(n+1\) rule, since they are adjacent to 1 proton on the \(\text{-CH-}\) carbon, they split into a doublet (\(1+1 = 2\)).
- Peak at \(\delta = 2.65\text{ ppm}\) (1H, septet) is due to the single proton of the \(\text{-CH-}\) group. It is adjacent to the 6 protons of the two methyl groups, so it is split into a septet (\(6+1 = 7\)).

(c)(ii) The fragment at \(m/z = 45\) is the \(\text{COOH}^+\) ion.
Equation: \([\text{C}_4\text{H}_8\text{O}_2\text{]}^{+\bullet} \rightarrow \text{COOH}^+ + \text{C}_3\text{H}_7^\bullet\) (or \([\text{(CH}_3)_2\text{CHCOOH]}^{+\bullet} \rightarrow \text{COOH}^+ + \text{CH(CH}_3)_2^\bullet\)).

(a) [2 marks]
- M1: Identifies \(\text{O}-\text{H}\) (carboxylic acid) from the \(2500 - 3300\text{ cm}^{-1}\) band.
- M2: Identifies \(\text{C}=\text{O}\) from the \(1715\text{ cm}^{-1}\) peak, concluding the functional group is a carboxylic acid.

(b) [1 mark]
- M1: Explains that there is symmetry resulting in two equivalent carbon environments (equivalent methyl groups).

(c)(i) [7 marks]
- M1: Correct structural formula of 2-methylpropanoic acid shown.
- M2: Assigns \(\delta = 12.0\text{ ppm}\) to the \(\text{-COOH}\) proton.
- M3: Assigns \(\delta = 1.20\text{ ppm}\) to the 6 methyl protons (\(\text{-CH}_3\)).
- M4: Assigns \(\delta = 2.65\text{ ppm}\) to the \(\text{-CH-}\) proton.
- M5: Explains the doublet splitting of methyl protons because they are adjacent to 1 proton.
- M6: Explains the septet splitting of the CH proton because it is adjacent to 6 equivalent protons.
- M7: Structured explanation is complete and logical.

(c)(ii) [3 marks]
- M1: Gives correct formula of the fragment: \(\text{COOH}^+\).
- M2: Gives correct radical formula: \(\text{C}_3\text{H}_7^\bullet\) or \(\text{CH(CH}_3)_2^\bullet\).
- M3: Balanced equation with correct charges and radical notation: \([\text{C}_4\text{H}_8\text{O}_2\text{]}^{+\bullet} \rightarrow \text{COOH}^+ + \text{C}_3\text{H}_7^\bullet\).

First, calculate the initial moles of propanoic acid and sodium hydroxide: \(n(\text{acid}) = 0.0500 \times 0.150 = 7.50 \times 10^{-3}\text{ mol}\); \(n(\text{NaOH}) = 0.0250 \times 0.120 = 3.00 \times 10^{-3}\text{ mol}\). The sodium hydroxide reacts completely with the acid to form sodium propanoate: \(n(\text{propanoate}) = 3.00 \times 10^{-3}\text{ mol}\) and \(n(\text{acid remaining}) = 7.50 \times 10^{-3} - 3.00 \times 10^{-3} = 4.50 \times 10^{-3}\text{ mol}\). Using the buffer equation: \([\text{H}^+] = K_a \times \frac{n(\text{acid})}{n(\text{conjugate base})} = 1.35 \times 10^{-5} \times \frac{4.50 \times 10^{-3}}{3.00 \times 10^{-3}} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\). Finally, \(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.69\).

Award 1 mark for the correct option A.

The reaction of aromatic compounds with concentrated nitric acid and sulfuric acid is an electrophilic aromatic substitution (nitration). The rate of nitration depends on the electron density of the benzene ring. Methyl groups are electron-donating (activating), which increases electron density and increases the rate of reaction. Halogens (chlorine) are electron-withdrawing (deactivating), and nitro groups are strongly electron-withdrawing (deactivating). Therefore, methylbenzene reacts the fastest.

Award 1 mark for the correct option C.

A species with a more positive standard electrode potential will oxidize a species with a less positive standard electrode potential. (1) Since \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\) is more positive than \(E^\theta(\text{I}_2/\text{I}^-) = +0.54\text{ V}\), \(\text{Fe}^{3+}\) can oxidize \(\text{I}^-\). (2) Since \(E^\theta(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}) = +2.01\text{ V}\) is more positive than \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\), \(\text{S}_2\text{O}_8^{2-}\) can oxidize \(\text{Fe}^{2+}\). (3) Since \(E^\theta(\text{I}_2/\text{I}^-) = +0.54\text{ V}\) is less positive than \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\), \(\text{I}_2\) cannot oxidize \(\text{Fe}^{2+}\). Therefore, only statements 1 and 2 are correct.

Award 1 mark for the correct option B.

Acid hydrolysis of a peptide using excess hydrochloric acid breaks the amide (peptide) bonds to yield protonated amino acids. The amino acid residues are alanine (with a \(-\text{CH}_3\) side chain), serine (with a \(-\text{CH}_2\text{OH}\) side chain), and glycine (with a \(-\text{H}\) side chain). In strongly acidic conditions, the amine group becomes protonated to \(-\text{NH}_3^+\) while the carboxylic acid group remains as \(-\text{COOH}\). Thus, alanine becomes \(\text{H}_3\text{N}^+\text{CH}(\text{CH}_3)\text{COOH}\).

Award 1 mark for the correct option A.

Adding excess aqueous ammonia to hexaaquacopper(II) ions leads to partial ligand substitution, where four water molecules are replaced by four ammonia molecules to form the deep-blue complex tetraamminedihydratocopper(II), \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

Award 1 mark for the correct option B.

The splitting pattern with a triplet integrating to \(3\text{H}\) at \(\delta = 1.3\text{ ppm}\) and a quartet integrating to \(2\text{H}\) at \(\delta = 4.1\text{ ppm}\) indicates an ethyl group directly attached to an oxygen atom, \(-\text{OCH}_2\text{CH}_3\). The singlet integrating to \(3\text{H}\) at \(\delta = 2.0\text{ ppm}\) indicates a methyl group adjacent to a carbonyl group, \(\text{CH}_3\text{CO-}\). Joining these fragments gives \(\text{CH}_3\text{COOCH}_2\text{CH}_3\), which is ethyl ethanoate.

Award 1 mark for the correct option B.

Let the initial rate be \(\text{rate}_1 = k[\text{NO}]^2[\text{Cl}_2]\). If \([\text{NO}]\) is tripled and \([\text{Cl}_2]\) is halved, the new rate is \(\text{rate}_2 = k(3[\text{NO}])^2(0.5[\text{Cl}_2]) = 9 \times 0.5 \times k[\text{NO}]^2[\text{Cl}_2] = 4.5 \times \text{rate}_1\). Therefore, the rate increases by a factor of 4.5.

Award 1 mark for the correct option A.

Condensation polymers containing amide linkages are polyamides. Polyamides are formed from the reaction of a dicarboxylic acid and a diamine (or from an amino acid). Hexanedioic acid (dicarboxylic acid) and hexane-1,6-diamine (diamine) react to form Nylon 6,6, which is a polyamide. Ethane-1,2-diol and benzene-1,4-dicarboxylic acid form a polyester. Butane-1,4-diol and butane-1,4-dioic acid also form a polyester. Propene and phenylethene form an addition polymer.

Award 1 mark for the correct option B.

First, calculate the initial moles of propanoic acid and sodium hydroxide: \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\), \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}\). The sodium hydroxide reacts completely with the propanoic acid: \(n(\text{CH}_3\text{CH}_2\text{COO}^-) = n(\text{NaOH}) = 3.00 \times 10^{-3}\text{ mol}\), \(n(\text{CH}_3\text{CH}_2\text{COOH})_{\text{remaining}} = 7.50 \times 10^{-3} - 3.00 \times 10^{-3} = 4.50 \times 10^{-3}\text{ mol}\). Using the acid dissociation constant expression: \(K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\), \([\text{H}^+] = K_{\text{a}} \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{4.50 \times 10^{-3}}{3.00 \times 10^{-3}} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\). Calculate the pH: \(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.69\).

Award 1 mark for the correct answer A. Incorrect choices correspond to: B (assumes pH = pKa = 4.87), C (inverted the acid/conjugate base ratio), and D (omitted the subtraction of reacted acid moles).

For a reaction to be feasible under standard conditions, the overall standard cell potential (\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\)) must be positive. For A: \(E^\theta_{\text{cell}} = +0.77\text{ V} - (+0.80\text{ V}) = -0.03\text{ V}\) (not feasible). For B: \(E^\theta_{\text{cell}} = +0.54\text{ V} - (+0.77\text{ V}) = -0.23\text{ V}\) (not feasible). For C: \(E^\theta_{\text{cell}} = +0.80\text{ V} - (+1.36\text{ V}) = -0.56\text{ V}\) (not feasible). For D: \(E^\theta_{\text{cell}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\) (feasible).

Award 1 mark for the correct answer D. No partial marks.

Optical isomerism occurs in octahedral transition metal complexes that lack a plane of symmetry (chirality). \(\text{cis-}[\text{Co}(\text{en})_2\text{Cl}_2]^+\) has no plane of symmetry because of the asymmetric arrangement of the two bidentate ethylenediamine (en) ligands relative to each other, allowing it to exist as non-superimposable mirror images. The trans isomer, the hexaammine complex, and the cis-tetraammine complex all possess planes of symmetry and are therefore optically inactive.

Award 1 mark for the correct answer C. No partial marks.

The carboxylic acid group (\(-\text{COOH}\)) has an electron-withdrawing carbonyl group adjacent to the benzene ring. This decreases the electron density of the aromatic ring, making it less susceptible to electrophilic attack (deactivated) so it reacts slower than benzene. Additionally, electron-withdrawing groups such as \(-\text{COOH}\) direct incoming electrophiles primarily to the 3-position (meta-directing).

Award 1 mark for the correct answer D. No partial marks.

Acidic hydrolysis of a peptide bond cleaves the amide (\(-\text{CONH}-\)) link to yield amine and carboxylic acid groups. Under strongly acidic conditions (excess \(\text{HCl}\)), the amine groups are fully protonated to form ammonium ions (\(-\text{NH}_3^+\)), while the carboxylic acid groups (\(-\text{COOH}\)) remain un-ionized. Therefore, the products are protonated glycine, \(\text{H}_3\text{N}^+\text{-CH}_2\text{-COOH}\), and protonated alanine, \(\text{H}_3\text{N}^+\text{-CH(CH}_3\text{)-COOH}\).

Award 1 mark for the correct answer B. Incorrect choices: A represents alkaline hydrolysis products; C represents zwitterionic forms which do not dominate in excess strong acid; D represents unprotonated free amino acids.

We determine the number of distinct carbon environments (which equals the number of peaks in a proton-decoupled \(^{13}\text{C}\) NMR spectrum) for each isomer: Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) has 4 distinct carbons. Butanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)) has 4 distinct carbons. Methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)) has 4 distinct carbons. 2-Methylpropanoic acid (\((\text{CH}_3)_2\text{CHCOOH}\)) has a plane of symmetry making the two methyl groups equivalent; hence it has exactly 3 distinct carbon environments.

Award 1 mark for the correct answer D. No partial marks.

Let the initial rate be \(\text{rate}_1 = k[P]^2[Q]\). If the concentration of \(P\) is tripled to \(3[P]\) and \(Q\) is halved to \(0.5[Q]\), the new rate is: \(\text{rate}_2 = k(3[P])^2(0.5[Q]) = 9 \times 0.5 \times k[P]^2[Q] = 4.5 \times \text{rate}_1\). Therefore, the rate increases by a factor of 4.5.

Award 1 mark for the correct answer C. Incorrect choices: A assumes a first-order dependency on P (\(3 \times 0.5 = 1.5\)); B is another incorrect multiplication; D assumes Q is doubled instead of halved (\(9 \times 2 = 18\)).

(a) Equation for electrophile generation:
\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\) (or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\)).

(b) Mechanism:
1. Curly arrow from the benzene ring of methyl benzoate to the \(\text{NO}_2^+\) ion.
2. Correct structure of the intermediate showing a tetrahedrally bonded carbon with a \(\text{H}\) and a \(\text{NO}_2\) group, and a partial ring positive charge spread over the remaining 5 carbon atoms (open towards the \(\text{sp}^3\) carbon).
3. Curly arrow from the \(\text{C}-\text{H}\) bond of the intermediate back into the ring to restore aromaticity, yielding the 3-nitro product and releasing \(\text{H}^+\).
4. Directing effect explanation: The ester carbonyl group is electron-withdrawing by induction/resonance. It deactivates the ring, but specifically destabilises the carbocation intermediates formed by ortho- and para-attack because it places a positive charge adjacent to the partially positive carbonyl carbon. Therefore, meta- (3-) substitution is favoured as it is the least deactivated pathway.

(c) Yield Calculation:
- Moles of methyl benzoate = \(4.08 / 136.0 = 0.0300\text{ mol}\).
- Theoretical yield of methyl 3-nitrobenzoate = \(0.0300\text{ mol}\).
- Theoretical mass of methyl 3-nitrobenzoate = \(0.0300 \times 181.0 = 5.43\text{ g}\).
- Percentage yield = \((3.80 / 5.43) \times 100 = 69.98\%\) (or using moles: \(3.80 / 181.0 = 0.0210\text{ mol}\); \(0.0210 / 0.0300 \times 100 = 70.0\%\) depending on intermediate rounding. If using unrounded values: \(0.020994 / 0.0300 \times 100 = 69.98\% \approx 70.0\%\). Accept 69.9% to 70.0%).

(a) [1 mark]: Correct balanced equation for electrophile generation.

(b) [5 marks total]:
- [1 mark] for curly arrow from the pi-system of benzene to the sulfur/nitrogen-based electrophile, \(\text{NO}_2^+\).
- [1 mark] for correct structure of the meta-substituted intermediate showing the tetrahedral carbon with \(\text{H}\) and \(\text{NO}_2\), and the horseshoe positive charge pointing to that carbon.
- [1 mark] for curly arrow from the \(\text{C}-\text{H}\) bond back into the ring system.
- [1 mark] for explaining that the ester (carbonyl) group is electron-withdrawing.
- [1 mark] for explaining that it deactivates the 2- and 4-positions more than the 3-position (or that meta-attack is less unstable / avoids placing positive charges adjacent to each other).

(c) [3 marks total]:
- [1 mark] for calculating moles of methyl benzoate = 0.0300 mol.
- [1 mark] for calculating theoretical mass of methyl 3-nitrobenzoate = 5.43 g (or moles of product = 0.0210 mol).
- [1 mark] for calculating percentage yield = 69.9% or 70.0% (allow 69.9% to 70.0% based on rounding, must be 3 significant figures).

(a) 3D structures: Draw a central carbon with four different groups tetrahedral-wise: \(-\text{H}\), \(-\text{NH}_2\), \(-\text{COOH}\), and \(-\text{CH}_2\text{CH}_3\). The second isomer must be a non-superimposable mirror image of the first, using wedges and dashes.

(b) Hydrolysis products:
(i) Under acidic conditions (HCl), the amide bond is cleaved and the amine groups are protonated to ammonium ions. The carboxylic acid groups remain uncharged.
- Product 1 (from alanine): \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\) (or 2-aminopropanoic acid with protonated amine group).
- Product 2 (from 2-aminobutanoic acid): \(\text{CH}_3\text{CH}_2\text{CH}(\text{NH}_3^+)\text{COOH}\).
(ii) Under alkaline conditions (NaOH), the amide bond is cleaved and the carboxylic acid groups are deprotonated to carboxylate ions. The amine groups remain uncharged.
- Product 1: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\) (or sodium salt).
- Product 2: \(\text{CH}_3\text{CH}_2\text{CH}(\text{NH}_2)\text{COO}^-\).

(c)(i) Monomers: The amide bonds are cleaved to regenerate the amino acids. These are alanine (2-aminopropanoic acid) and 2-aminobutanoic acid.
Structures: \(\text{H}_2\text{NCH}(\text{CH}_3)\text{COOH}\) and \(\text{H}_2\text{NCH}(\text{CH}_2\text{CH}_3)\text{COOH}\).
(ii) Condensation polymerisation (since a small molecule, water, is eliminated during polymerisation).

(a) [2 marks]:
- [1 mark] for drawing one 3D tetrahedral representation of 2-aminobutanoic acid showing wedges/dashes.
- [1 mark] for drawing the second isomer as a clear, non-superimposable mirror image.

(b) [4 marks total]:
- [1 mark] for correct protonated N-terminal product under acidic hydrolysis: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\) (or \(\text{Cl}^-\) salt).
- [1 mark] for correct protonated C-terminal product under acidic hydrolysis: \(\text{CH}_3\text{CH}_2\text{CH}(\text{NH}_3^+)\text{COOH}\).
- [1 mark] for correct deprotonated N-terminal product under alkaline hydrolysis: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).
- [1 mark] for correct deprotonated C-terminal product under alkaline hydrolysis: \(\text{CH}_3\text{CH}_2\text{CH}(\text{NH}_2)\text{COO}^-\).

(c) [3 marks total]:
- [1 mark] for structure of 2-aminopropanoic acid (alanine) monomer.
- [1 mark] for structure of 2-aminobutanoic acid monomer.
- [1 mark] for identifying the reaction as 'condensation' polymerisation.

(a) Reagent: Tollens' reagent (ammoniacal silver nitrate) [or Fehling's solution].
Observation with 3-methylbutanal: Silver mirror [or red precipitate].
Observation with pentan-3-one: No change / no reaction.

(b)(i) HCN is an extremely toxic, volatile gas (or liquid boiling at 26 °C). Using NaCN and a strong acid in situ is safer and provides a higher concentration of the nucleophile \(\text{CN}^-\).
(ii) Mechanism:
- Step 1: Curly arrow from the lone pair on the carbon of the \(\text{CN}^-\) ion to the carbonyl carbon atom of 3-methylbutanal. Dipole on carbonyl group: \(\text{C}^{\delta+}=\text{O}^{\delta-}\). Curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom.
- Intermediate: \((\text{CH}_3)_2\text{CHCH}_2\text{CH}(\text{O}^-)\text{CN}\).
- Step 2: Curly arrow from the lone pair of the negative oxygen atom of the intermediate to a hydrogen ion (\(\text{H}^+\) from acid or water).
- Product: \((\text{CH}_3)_2\text{CHCH}_2\text{CH}(\text{OH})\text{CN}\).

(c)(i) \(\text{NaBH}_4\) (in aqueous or ethanolic solution) reduces carbonyls but not alkenes.
(ii) The nucleophile \(\text{H}^-\) (hydride ion) is attracted to the carbon atom of the polar \(\text{C}=\text{O}\) bond because it is electron-deficient (\(\text{C}^{\delta+}\)). In contrast, the \(\text{C}=\text{C}\) double bond is a region of high electron density (non-polar) and repels the nucleophilic hydride ion.

(a) [2 marks total]:
- [1 mark] for Tollens' reagent (ammoniacal silver nitrate) or Fehling's/Benedict's solution.
- [1 mark] for correct observation: silver mirror/precipitate (or red precipitate for Fehling's) with 3-methylbutanal AND no change with pentan-3-one.

(b) [5 marks total]:
- [1 mark] for explaining safety: HCN is toxic / volatile / a gas (or NaCN/acid provides higher concentration of \(\text{CN}^-\)).
- [1 mark] for curly arrow from the lone pair of \(\text{CN}^-\) (must be on carbon) to the carbonyl carbon AND correct dipole \(\text{C}^{\delta+}=\text{O}^{\delta-}\) shown.
- [1 mark] for curly arrow from the \(\text{C}=\text{O}\) pi-bond to the oxygen atom.
- [1 mark] for the correct intermediate structure with a negative charge on the oxygen atom.
- [1 mark] for curly arrow from the oxygen lone pair to \(\text{H}^+\) to form the correct hydroxynitrile product.

(c) [2 marks total]:
- [1 mark] for identifying \(\text{NaBH}_4\) as the reagent.
- [1 mark] for explaining that the nucleophile \(\text{H}^-\) attacks the polar \(\text{C}^{\delta+}=\text{O}^{\delta-}\) carbon but is repelled by the high electron density of the non-polar \(\text{C}=\text{C}\) bond.

(a) Comparative analysis:
Method 1: Ethanoic acid
- Reagents/conditions: Concentrated sulfuric acid (catalyst) and heat.
- Advantage: Cheaper / safer reagents / less hazardous than using ethanoyl chloride.
- Disadvantage: Reversible reaction / low yield / requires separation of reactants and products.

Method 2: Ethanoic anhydride
- Reagents/conditions: Warm gently (no catalyst required, though acid can be added).
- Advantage: Reaction is non-reversible (higher yield than acid) / safer and less violent than using ethanoyl chloride.
- Disadvantage: Reaction is slower than with ethanoyl chloride / produces ethanoic acid as a waste product (atom economy is lower than with acyl chloride).

Method 3: Ethanoyl chloride
- Reagents/conditions: Room temperature / no catalyst required.
- Advantage: Reaction is very fast / goes to completion (non-reversible, high yield).
- Disadvantage: Produces toxic, corrosive hydrogen chloride (HCl) gas / very violent and hazardous reaction / ethanoyl chloride is expensive.

(b)(i) Equation: \(\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}\).
(ii) Yield Calculation:
- Moles of propan-1-ol used = \(9.01 / 60.0 = 0.1502\text{ mol}\).
- Theoretical yield of propyl ethanoate = \(0.1502\text{ mol}\).
- Theoretical mass of propyl ethanoate = \(0.1502 \times 102.0 = 15.32\text{ g}\).
- Percentage yield = \((11.02 / 15.32) \times 100 = 71.93\% \approx 72.0\%\) (if using intermediate rounded moles: \(0.150\text{ mol} \times 102.0 = 15.3\text{ g}\); \(11.02 / 15.3 \times 100 = 72.0\%\)). Allow 71.9% to 72.0%.

(a) [6 marks total]:
- [2 marks] for Method 1: reagent (conc. \(\text{H}_2\text{SO}_4\) catalyst) AND one advantage (cheap/safe) AND one disadvantage (reversible/equilibrium reaction).
- [2 marks] for Method 2: reagent (none/warm) AND one advantage (safer than acyl chloride / higher yield than carboxylic acid) AND one disadvantage (slower than acyl chloride / produces carboxylic acid byproduct).
- [2 marks] for Method 3: reagent (none) AND one advantage (fast, irreversible, high yield) AND one disadvantage (produces toxic HCl gas / violent reaction).

(b) [3 marks total]:
- [1 mark] for correct balanced equation: \(\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}\).
- [1 mark] for calculating theoretical yield: \(0.150\text{ mol}\) or \(15.3\text{ g}\).
- [1 mark] for percentage yield = 72.0% (allow 71.9% - 72.0%, must be to 3 significant figures).

(a) Calculation:
- Carbon: \(72.00 / 12.0 = 6.00\text{ mol}\)
- Hydrogen: \(12.00 / 1.0 = 12.00\text{ mol}\)
- Oxygen: \(16.00 / 16.0 = 1.00\text{ mol}\)
Dividing by the smallest value (1.00) gives the ratio \(6 : 12 : 1\).
Empirical formula is \(\text{C}_6\text{H}_{12}\text{O}\).

(b)(i) Since the molecular ion peak is at \(m/z = 100\), and the empirical mass is \(6 \times 12.0 + 12 \times 1.0 + 16.0 = 100\), the molecular formula is the same as the empirical formula: \(\text{C}_6\text{H}_{12}\text{O}\).
(ii) Fragment at \(m/z = 43\) can be \(\text{CH}_3\text{CO}^+\) (acylium ion) or \(\text{C}_3\text{H}_7^+\) (propyl/isopropyl carbocation).

(c) Structure and analysis:
- IR peak at \(1715\text{ cm}^{-1}\) indicates a carbonyl (\(\text{C}=\text{O}\)) group. No peak in the range \(3200\text{-}3600\text{ cm}^{-1}\) rules out an alcohol (\(\text{O}-\text{H}\)) group.
- NMR peak analysis:
1. Singlet at \(\delta = 2.1\text{ ppm}\) (relative peak area 3): indicates a methyl group adjacent to an environment with no adjacent protons, specifically a methyl ketone: \(\text{CH}_3-\text{CO}-\).
2. Doublet at \(\delta = 0.9\text{ ppm}\) (relative peak area 6): indicates two identical methyl groups adjacent to a single proton, i.e., \(-\text{CH}(\text{CH}_3)_2\) (isopropyl group).
3. Multiplet at \(\delta = 2.1\text{ ppm}\) (relative peak area 1): indicates a \(-\text{CH}-\) proton split by many adjacent protons (e.g., the 6 protons of the two methyl groups and the 2 protons of the adjacent methylene group).
4. Doublet at \(\delta = 2.3\text{ ppm}\) (relative peak area 2): indicates a \(-\text{CH}_2-\) group adjacent to one proton (the \(-\text{CH}-\) proton) and also shifted downfield by being adjacent to the carbonyl group.
- Putting the fragments together gives: \(\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}(\text{CH}_3)_2\) (4-methylpentan-2-one).

(a) [2 marks total]:
- [1 mark] for calculating the moles of C, H, and O correctly (6.00, 12.00, 1.00).
- [1 mark] for showing the simplest whole-number ratio is 6:12:1.

(b) [2 marks total]:
- [1 mark] for molecular formula \(\text{C}_6\text{H}_{12}\text{O}\).
- [1 mark] for a correct positive ion formula: \(\text{CH}_3\text{CO}^+\) or \(\text{C}_3\text{H}_7^+\) (must include positive charge).

(c) [5 marks total]:
- [1 mark] for identifying the carbonyl (\(\text{C}=\text{O}\)) from IR and ruling out alcohol/carboxylic acid.
- [1 mark] for attributing the singlet at \(\delta = 2.1\text{ ppm}\) to a \(\text{CH}_3-\text{CO}-\) group.
- [1 mark] for attributing the doublet at \(\delta = 0.9\text{ ppm}\) (area 6) to an isopropyl group \(-\text{CH}(\text{CH}_3)_2\).
- [1 mark] for explaining the doublet at \(\delta = 2.3\text{ ppm}\) as a \(-\text{CH}_2-\) group coupled to a single proton.
- [1 mark] for drawing the correct final structure of 4-methylpentan-2-one.

(a)(i) Stationary phase: the phase that does not move (e.g., silica gel on a plate) where adsorption of compounds occurs. Mobile phase: the phase that moves (solvent) carrying the soluble compounds with it.
(ii) Separation depends on the relative adsorption to the stationary phase vs solubility in the mobile phase. Polar compounds adsorb more strongly to a polar stationary phase (e.g., silica) and travel slower, while less polar compounds are more soluble in a less polar mobile phase and travel faster.

(b)(i) There is a linear relationship between peak area and concentration: \(\text{Peak Area} = 150 \times \text{Concentration}\%\). For a peak area of 375, the concentration is \(375 / 150 = 2.5\%\) vol/vol.
(ii) Dilution factor = \(100.0\text{ cm}^3 / 10.0\text{ cm}^3 = 10\).
Therefore, the concentration in the original mouthwash = \(2.5\% \times 10 = 25.0\%\) vol/vol (or % vol/vol).
(iii) Alternative methods:
- Direct density measurement / hydrometer (since ethanol has a lower density than water).
- Redox titration of ethanol using acidified potassium dichromate(VI) and measuring the concentration of excess dichromate or chromium(III) spectrophotometrically.

(a) [4 marks total]:
- [1 mark] for stationary phase definition (does not move, adsorbs components).
- [1 mark] for mobile phase definition (moves, carries components).
- [1 mark] for mentioning 'adsorption' to stationary phase and 'solubility' in mobile phase.
- [1 mark] for explaining that stronger adsorption leads to slower movement (smaller Rf) or greater solubility leads to faster movement (larger Rf).

(b) [5 marks total]:
- [1 mark] for showing concentration in diluted sample = 2.5% vol/vol (e.g. \(375 / 150 = 2.5\) or showing a linear plot calculation).
- [1 mark] for using the dilution factor of 10.
- [1 mark] for calculating original concentration = 25.0% (must show correct unit or value 25.0).
- [2 marks] for stating a valid alternative method (e.g., density/hydrometer, redox titration with acidified dichromate, refractive index), with 1 mark for the method name and 1 mark for how it works.

(a) Synthetic route:
- Starting material: Methyl 2-hydroxybenzoate, \(\text{C}_6\text{H}_4(\text{OH})\text{COOCH}_3\).
- Step 1: Hydrolysis of the methyl ester. Reagents: Aqueous sodium hydroxide (followed by acidification with dilute hydrochloric acid) and heat under reflux. This produces 2-hydroxybenzoic acid (salicylic acid).
- Intermediate: 2-hydroxybenzoic acid, \(\text{C}_6\text{H}_4(\text{OH})\text{COOH}\).
- Step 2: Esterification to form the ethyl ester. Reagents: Ethanol and concentrated sulfuric acid (catalyst), heat under reflux.
- Product: Ethyl 2-hydroxybenzoate, \(\text{C}_6\text{H}_4(\text{OH})\text{COOCH}_2\text{CH}_3\).

(b) Purification of organic liquid:
1. Transfer the crude reaction mixture to a separating funnel, add aqueous sodium hydrogencarbonate (to neutralise acid catalyst / remaining acids) and release pressure build-up of \(\text{CO}_2\).
2. Separate the organic layer from the aqueous layer (retaining the organic layer).
3. Dry the organic liquid by adding an anhydrous inorganic salt (e.g., anhydrous \(\text{MgSO}_4\) or \(\text{Na}_2\text{SO}_4\)) until the liquid is clear.
4. Decant or filter the liquid into a distillation flask and distill, collecting the fraction boiling at the specific boiling point of ethyl 2-hydroxybenzoate.

(a) [5 marks total]:
- [1 mark] for structure of starting material and intermediate 2-hydroxybenzoic acid.
- [1 mark] for correct reagents/conditions for Step 1: aqueous NaOH (or acid) and heat, followed by acid (H+) if alkaline hydrolysis is used.
- [1 mark] for structure of final product ethyl 2-hydroxybenzoate.
- [1 mark] for correct reagents/conditions for Step 2: ethanol and concentrated sulfuric acid, heat.
- [1 mark] for logic / correct connectivity of the steps.

(b) [4 marks total]:
- [1 mark] for using a separating funnel to remove aqueous acid/impurities, e.g., by adding sodium hydrogencarbonate.
- [1 mark] for running off/separating the organic layer.
- [1 mark] for drying with an anhydrous salt (e.g., \(\text{MgSO}_4\) or \(\text{CaCl}_2\) or \(\text{Na}_2\text{SO}_4\)) until clear.
- [1 mark] for fractional distillation (collecting at correct/constant boiling point).

(a)(i) Procedure:
- Set up three test tubes, each containing ethanol (as a mutual solvent) and a few drops of one of the haloalkanes (1-chlorobutane, 1-bromobutane, 1-iodobutane).
- Add aqueous silver nitrate, \(\text{AgNO}_3\)(aq), to each test tube and place them in a water bath at around 50–60 °C.
- Measure the time taken for a precipitate to appear in each tube.
- Key observation: White precipitate (AgCl) forms slowest; cream precipitate (AgBr) forms at intermediate rate; yellow precipitate (AgI) forms fastest.
(ii) Bond enthalpy explanation:
- The rate depends on the ease of breaking the carbon-halogen bond.
- Bond enthalpies decrease down the group: \(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\).
- The \(\text{C}-\text{I}\) bond is the weakest and requires the least energy to break, so 1-iodobutane hydrolyses fastest. The \(\text{C}-\text{Cl}\) bond is the strongest, so 1-chlorobutane hydrolyses slowest.

(b)(i) Skeletal structures of the three alkenes:
- But-1-ene: \(\text{CH}_2=\text{CHCH}_2\text{CH}_3\)
- (E)-but-2-ene: \(\text{CH}_3\text{CH}=\text{CHCH}_3\) (with trans methyl groups)
- (Z)-but-2-ene: \(\text{CH}_3\text{CH}=\text{CHCH}_3\) (with cis methyl groups)
(ii) Explanation:
- Elimination of water from butan-2-ol involves the removal of the hydroxyl group (\(-\text{OH}\)) and a hydrogen atom from an adjacent carbon atom (either C1 or C3).
- If a hydrogen is removed from C1, but-1-ene is formed.
- If a hydrogen is removed from C3, but-2-ene is formed.
- But-2-ene has stereoisomers because there is restricted rotation about the \(\text{C}=\text{C}\) double bond and both carbons of the double bond have two different groups attached (\(-\text{H}\) and \(-\text{CH}_3\)). This gives rise to (E)-but-2-ene and (Z)-but-2-ene.

(a) [5 marks total]:
- [1 mark] for using aqueous silver nitrate and ethanol (solvent) and heating.
- [1 mark] for measuring the time taken for a precipitate to form.
- [1 mark] for identifying the correct order of precipitation: yellow (fastest), cream (middle), white (slowest).
- [1 mark] for stating that rate of hydrolysis increases down Group 7: \(\text{I} > \text{Br} > \text{Cl}\).
- [1 mark] for explaining that bond enthalpy of \(\text{C}-\text{X}\) decreases down the group (\(\text{C}-\text{I}\) is weakest / breaks easiest).

(b) [4 marks total]:
- [2 marks] for drawing correct skeletal structures of all three alkenes: but-1-ene, (E)-but-2-ene, and (Z)-but-2-ene (1 mark for but-1-ene and 1 mark for the two stereoisomers of but-2-ene correctly drawn).
- [1 mark] for explaining that elimination can remove hydrogen from either carbon 1 or carbon 3.
- [1 mark] for explaining that but-2-ene exhibits E/Z isomerism due to restricted rotation around the \(\text{C}=\text{C}\) double bond and two different groups on each C.

(a)(i) 1. Calculate molar mass of compound \(\textbf{X}\) (\(\text{C}_9\text{H}_{10}\text{O}_3\)): \(M_r = 9 \times 12.0 + 10 \times 1.0 + 3 \times 16.0 = 166.0\text{ g mol}^{-1}\). \( \) 2. Calculate the concentration of the acid, \([\text{HA}]\): \(\text{Moles of } \textbf{X} = \frac{1.245}{166.0} = 0.00750\text{ mol}\). \([\text{HA}] = \frac{0.00750}{0.2500\text{ dm}^3} = 0.0300\text{ mol dm}^{-3}\). \( \) 3. Calculate \([\text{H}^+]\) from pH: \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2.84} = 1.4454 \times 10^{-3}\text{ mol dm}^{-3}\). \( \) 4. Calculate \(K_a\) using approximation \(K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]}\): \(K_a = \frac{(1.4454 \times 10^{-3})^2}{0.0300} = 6.964 \times 10^{-5}\text{ mol dm}^{-3}\) (Accept \(6.96 \times 10^{-5}\text{ mol dm}^{-3}\)). \( \) 5. Calculate \(\text{p}K_a\): \(\text{p}K_a = -\log_{10}(6.964 \times 10^{-5}) = 4.16\). \( \) 6. Assumptions: \([\text{H}^+] \approx [\text{A}^-]\) (water dissociation is negligible) and \([\text{HA}]_{\text{eqm}} \approx [\text{HA}]_{\text{initial}}\) (acid dissociation is negligible). \( \) (a)(ii) 1. Initial moles: \(\text{Moles of HA} = \frac{100.0}{1000} \times 0.0300 = 0.00300\text{ mol}\); \(\text{Moles of NaOH} = \frac{50.0}{1000} \times 0.0150 = 0.000750\text{ mol}\). \( \) 2. Equilibrium moles: \(\text{Moles of HA remaining} = 0.00300 - 0.000750 = 0.00225\text{ mol}\); \(\text{Moles of A}^- \text{ formed} = 0.000750\text{ mol}\). \( \) 3. Calculate \([\text{H}^+]\): \([\text{H}^+] = K_a \times \frac{\text{moles of HA}}{\text{moles of A}^-} = (6.964 \times 10^{-5}) \times \frac{0.00225}{0.000750} = 2.089 \times 10^{-4}\text{ mol dm}^{-3}\). \( \) 4. Calculate pH: \(\text{pH} = -\log_{10}(2.089 \times 10^{-4}) = 3.68\).

(a)(i) M1: [HA] = 0.0300 mol dm-3 (derived from moles = 0.00750). M2: [H+] = 1.45 x 10^-3 mol dm-3 (or 1.445 x 10^-3). M3: Ka = 6.96 x 10^-5 mol dm-3 (accept range 6.95 x 10^-5 to 7.00 x 10^-5). M4: pKa = 4.16 (consequent on Ka). M5: State both assumptions: [H+] = [A-] AND [HA]eqm = [HA]initial. (a)(ii) M1: Initial moles: n(HA) = 0.00300 mol AND n(NaOH) = 0.000750 mol. M2: Equilibrium moles: n(HA) = 0.00225 mol AND n(A-) = 0.000750 mol. M3: [H+] = Ka * (0.00225 / 0.000750) OR equivalent buffer equation. M4: pH = 3.68 (allow 3.67 to 3.69 to 2 decimal places).

(a) The diagram must include: 1. A standard hydrogen electrode (SHE) as one half-cell containing a platinum electrode, \(1.00\text{ mol dm}^{-3}\) \(\text{H}^+(\text{aq})\) solution, and hydrogen gas bubbling at \(100\text{ kPa}\). 2. A vanadium half-cell containing a platinum electrode immersed in a solution containing both \(\text{V}^{3+}(\text{aq})\) and \(\text{V}^{2+}(\text{aq})\), each at a concentration of \(1.00\text{ mol dm}^{-3}\). 3. A high-resistance voltmeter connecting the two electrodes. 4. A salt bridge (e.g., filter paper soaked in aqueous potassium nitrate) connecting the two solutions. 5. Standard conditions: temperature of \(298\text{ K}\) / \(25^\circ\text{C}\) and pressures of all gases at \(100\text{ kPa}\) / \(1\text{ bar}\). (b) Since \(E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\) is more positive than \(E^\theta(\text{VO}^{2+}/\text{V}^{3+}) = +0.34\text{ V}\), \(\text{Fe}^{3+}\) will oxidize \(\text{V}^{3+}\) to \(\text{VO}^{2+}\). Overall equation: \(\text{Fe}^{3+}(\text{aq}) + \text{V}^{3+}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{Fe}^{2+}(\text{aq}) + \)\text{VO}^{2+}(\text{aq}) + 2\text{H}^+(\text{aq})\). Cell potential: \(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = +0.77 - (+0.34) = +0.43\text{ V}\). Since \(E^\theta_{\text{cell}} > 0\), the reaction is feasible. (c) Titration 1: Moles of \(\text{Fe}^{2+}\) used = \(0.0200 \times \frac{37.50}{1000} = 7.50 \times 10^{-4}\text{ mol}\). Since the reduction of \(\text{VO}_2^+\) to \(\text{VO}^{2+}\) is a 1-electron change, the molar ratio of \(\text{VO}_2^+\) to \(\text{Fe}^{2+}\) is 1:1. Therefore, total moles of vanadium in the 25.0 \(\text{cm}^3\) sample = \(7.50 \times 10^{-4}\text{ mol}\). Titration 2: Moles of \(\text{MnO}_4^-\right.\) used = \(0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\text{ mol}\). Since \(\text{MnO}_4^-\right.\) is reduced to \(\text{Mn}^{2+}\), it accepts 5 electrons per mole. Moles of electrons accepted by \(\text{MnO}_4^-\right.\) = \(5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\text{ mol}\). Thus, the moles of electrons lost by \(7.50 \times 10^{-4}\text{ mol}\) of \(\text{V}^{n+}\) to become \(\text{VO}_2^+\) (+5 state) is \(2.25 \times 10^{-3}\text{ mol}\). Change in oxidation state = \(\frac{2.25 \times 10^{-3}}{7.50 \times 10^{-4}} = 3.0\). Therefore, \(5 - n = 3\), which gives \(n = 2\). The original vanadium ion is \(\text{V}^{2+}\).

Part (a): 5 marks. 1 mark: Standard hydrogen electrode drawn with Pt electrode, labeled H2 gas and 1.00 mol dm^-3 H+(aq). 1 mark: Vanadium half-cell containing Pt electrode with a mixture of V3+(aq) and V2+(aq). 1 mark: Salt bridge and high resistance voltmeter shown connecting the half-cells. 1 mark: All aqueous ions labeled as 1.00 mol dm^-3. 1 mark: Standard conditions stated: 298 K, 100 kPa. Part (b): 3 marks. 1 mark: Correct balanced overall equation: Fe3+(aq) + V3+(aq) + H2O(l) -> Fe2+(aq) + VO2+(aq) + 2H+(aq). 1 mark: E_cell = +0.43 V. 1 mark: Reaction is feasible because E_cell is positive. Part (c): 6 marks. 1 mark: Moles of Fe2+ calculated as 7.50 x 10^-4 mol. 1 mark: States total V moles in 25.0 cm^3 = 7.50 x 10^-4 mol. 1 mark: Moles of MnO4- calculated as 4.50 x 10^-4 mol. 1 mark: Calculates moles of electrons accepted by MnO4- as 2.25 x 10^-3 mol. 1 mark: Shows change in oxidation state of vanadium is 3. 1 mark: Deduces n = 2 (concluding V2+).

(a) A strong acid completely dissociates in aqueous solution, whereas a weak acid only partially dissociates. The expression is: \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\). (b) Initial moles of propanoic acid = \(0.0400 \times 0.150 = 6.00 \times 10^{-3}\text{ mol}\). Initial moles of \(\text{NaOH}\) added = \(0.0600 \times 0.0800 = 4.80 \times 10^{-3}\text{ mol}\). Acid-base neutralization reaction: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\). Remaining moles of propanoic acid = \(6.00 \times 10^{-3} - 4.80 \times 10^{-3} = 1.20 \times 10^{-3}\text{ mol}\). Moles of propanoate ions formed = \(4.80 \times 10^{-3}\text{ mol}\). Since both species are in the same volume, we can use the mole ratio: \([\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{CH}_2\text{COOH}]}{[\text{CH}_3\text{CH}_2\text{COO}^-]} = 1.35 \times 10^{-5} \times \frac{1.20 \times 10^{-3}}{4.80 \times 10^{-3}} = 3.375 \times 10^{-6}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(3.375 \times 10^{-6}) = 5.47\). (c) Moles of \(\text{H}^+\) added from \(\text{HCl}\) = \(0.00100 \times 1.00 = 1.00 \times 10^{-3}\text{ mol}\). The added \(\text{H}^+\) reacts with the conjugate base: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\). New moles of propanoate ions = \(4.80 \times 10^{-3} - 1.00 \times 10^{-3} = 3.80 \times 10^{-3}\text{ mol}\). New moles of propanoic acid = \(1.20 \times 10^{-3} + 1.00 \times 10^{-3} = 2.20 \times 10^{-3}\text{ mol}\). New \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{2.20 \times 10^{-3}}{3.80 \times 10^{-3}} = 7.816 \times 10^{-6}\text{ mol dm}^{-3}\). New \(\text{pH} = -\log_{10}(7.816 \times 10^{-6}) = 5.11\).

Part (a): 2 marks. 1 mark: Strong acid dissociates completely and weak acid dissociates partially. 1 mark: Correct Ka expression. Part (b): 6 marks. 1 mark: Moles of propanoic acid = 6.00 x 10^-3 mol. 1 mark: Moles of NaOH = 4.80 x 10^-3 mol. 1 mark: Moles of remaining propanoic acid = 1.20 x 10^-3 mol. 1 mark: Moles of propanoate ions = 4.80 x 10^-3 mol. 1 mark: Correct calculation of [H+] = 3.38 x 10^-6 mol dm^-3. 1 mark: pH = 5.47 (must be 2 dp). Part (c): 6 marks. 1 mark: Moles of H+ added = 1.00 x 10^-3 mol. 1 mark: Calculates new moles of propanoate ions = 3.80 x 10^-3 mol. 1 mark: Calculates new moles of propanoic acid = 2.20 x 10^-3 mol. 1 mark: Correct expression or calculation for new [H+] = 7.82 x 10^-6 mol dm^-3. 2 marks: pH = 5.11 (must be 2 dp). Allow 1 mark if pH calculation is mathematically correct following an arithmetic error in moles.

(a) Compound B is propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)). Compound C is 2-hydroxybutanenitrile (\(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\)). Reagents for Stage 2: \(\text{NaCN}\) and dilute acid (such as \(\text{H}_2\text{SO}_4\)) or \(\text{HCN}\) in alkaline buffer (pH 5-8). Condition: Room temperature. Mechanism: 1. Curly arrow from the lone pair on the carbon of \(^-\text{CN}\) to the carbonyl carbon. 2. Dipole shown on carbonyl group: \(\delta^+\) on carbon, \(\delta^-\) on oxygen. 3. Curly arrow from the C=O double bond to the oxygen atom. 4. Intermediate alkoxide ion formed with negative charge on oxygen and a lone pair. 5. Curly arrow from the lone pair of the intermediate oxygen to an \(\text{H}^+\) ion to form the hydroxyl group. (b) During the oxidation of propan-1-ol to propanal, the broad \(\text{O-H}\) alcohol absorption at \(3200\text{--}3600\text{ cm}^{-1}\) disappears. A sharp \(\text{C=O}\) carbonyl absorption at \(1630\text{--}1820\text{ cm}^{-1}\) appears in the product. (c) The repeating unit of the polyester is: \(\text{-[O-CH(CH}_2\text{CH}_3)\text{-CO]-}\). (d) Molar mass of 2-hydroxybutanoic acid (\(\text{C}_4\text{H}_8\text{O}_3\)) = \(104.0\text{ g mol}^{-1}\). Moles of 2-hydroxybutanoic acid = \(\frac{1.56}{104.0} = 0.0150\text{ mol}\). Since both the \(\text{-OH}\) and \(\text{-COOH}\) groups react with sodium, 1 mole of 2-hydroxybutanoic acid reacts with 2 moles of sodium to produce 1 mole of \(\text{H}_2\) gas. Therefore, moles of \(\text{H}_2 = 0.0150\text{ mol}\). Volume of \(\text{H}_2 = 0.0150 \times 24000 = 360\text{ cm}^3\).

Part (a): 6 marks. 1 mark: Structures of Compound B (propanal) and C (2-hydroxybutanenitrile) correct. 1 mark: Reagents: NaCN and H2SO4 (or HCN) at room temperature. 1 mark: Curly arrow from :CN- to C=O carbon, with correct delta+/- dipoles. 1 mark: Curly arrow from C=O double bond to O. 1 mark: Structure of intermediate alkoxide correct. 1 mark: Curly arrow from O:- to H+. Part (b): 3 marks. 1 mark: Alcohol O-H absorption peak disappears at 3200-3600 cm^-1. 1 mark: Carbonyl C=O absorption peak appears at 1630-1820 cm^-1. 1 mark: Correctly matches change in peaks to the functional group transformation. Part (c): 2 marks. 2 marks: Correct repeating unit of the polyester showing open bonds at both ends. (Allow 1 mark for incorrect carbon chain but correct ester linkages). Part (d): 3 marks. 1 mark: Moles of 2-hydroxybutanoic acid = 0.0150 mol. 1 mark: Identifies 1:1 stoichiometry between organic compound and H2 (both -OH and -COOH react). 1 mark: Volume of H2 = 360 cm^3.

(a) Ionic equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\). Shape of \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) is octahedral. Shape of \([\text{CoCl}_4]^{2-}\) is tetrahedral. The coordination number decreases from 6 to 4 because chloride ligands are larger than water ligands, so fewer chloride ligands can fit around the central cobalt(II) ion. (b) Comparing Experiment 1 and 2: when \([[\text{Fe}(\text{H}_2\text{O})_6]^{3+}]\) is doubled (with \([\text{SCN}^-]\) constant), the rate doubles. Therefore, the reaction is first order with respect to \([[\text{Fe}(\text{H}_2\text{O})_6]^{3+}]\). Comparing Experiment 1 and 3: when \([\text{SCN}^-]\) is tripled (with \([[\text{Fe}(\text{H}_2\text{O})_6]^{3+}]\) constant), the rate triples. Therefore, the reaction is first order with respect to \([\text{SCN}^-]\). Rate equation: \(\text{Rate} = k[[\text{Fe}(\text{H}_2\text{O})_6]^{3+}][\text{SCN}^-]\). Rate constant: \(1.6 \times 10^{-4} = k(4.0 \times 10^{-3})(2.0 \times 10^{-3})\) which gives \(1.6 \times 10^{-4} = k(8.0 \times 10^{-6})\), so \(k = 20\). Units of \(k = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). (c) The presence of thiocyanate and water ligands causes the 3d orbitals of the iron(III) ion to split into two different energy levels. d-electrons absorb specific frequencies/wavelengths of visible light energy to be promoted from a lower d-orbital to a higher d-orbital (d-d transition). The light that is not absorbed is transmitted or reflected, which is seen as the complementary color (red).

Part (a): 4 marks. 1 mark: Correct balanced equilibrium equation. 1 mark: Correct shapes (octahedral and tetrahedral). 1 mark: States coordination number changes from 6 to 4. 1 mark: Reason: Chloride ligands are larger than water ligands. Part (b): 6 marks. 1 mark: Deduces 1st order with respect to [Fe(H2O)6]3+ with clear explanation. 1 mark: Deduces 1st order with respect to SCN- with clear explanation. 1 mark: Correct rate equation. 2 marks: Correct numerical value of k = 20 (1 mark for working, 1 mark for value). 1 mark: Units of k: dm^3 mol^-1 s^-1. Part (c): 4 marks. 1 mark: Ligands cause d-orbitals to split. 1 mark: Electrons absorb visible light energy. 1 mark: Electrons are promoted to higher energy d-orbitals. 1 mark: Transmitted/reflected complementary color is observed.

(a) Equation: \(\text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \rightarrow \text{HOC}_6\text{H}_4\text{NO}_2 + \text{H}_2\text{O}\) (showing 4-nitrophenol structure). Phenol has a lone pair of electrons on the oxygen of the \(\text{-OH}\) group which is partially delocalized into the \(\pi\)-ring system. This increases the electron density of the benzene ring in phenol, making it more attractive to electrophiles and polarizing the electrophile more effectively than benzene. Therefore, no concentrated acid catalyst (like \(\text{H}_2\text{SO}_4\)) is needed. (b) Reagents: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), heated under reflux (followed by addition of aqueous \(\text{NaOH}\)). Equation: \(\text{HOC}_6\text{H}_4\text{NO}_2 + 6[\text{H}] \rightarrow \text{HOC}_6\text{H}_4\text{NH}_2 + 2\text{H}_2\text{O}\). (c) Structure of ethanoic anhydride: \(\text{CH}_3\text{COOCOCH}_3\). Structure of paracetamol: \(\text{HOC}_6\text{H}_4\text{NHCOCH}_3\). Other organic product: Ethanoic acid. (d) Recrystallization steps and purposes: 1. Dissolve the crude product in the minimum volume of hot solvent (to ensure a saturated solution so that maximum recrystallization occurs upon cooling). 2. Filter hot (to remove any insoluble impurities). 3. Cool the filtrate in ice to allow crystals of paracetamol to reform (leaving soluble impurities in the solution), then filter the crystals under reduced pressure (using a Buchner funnel), wash with a small portion of cold solvent, and dry.

Part (a): 5 marks. 2 marks: Correct balanced equation with structural formulas (1 mark for reactants/products, 1 mark for correct 4-nitrophenol isomer). 1 mark: Oxygen lone pair on -OH group is delocalized into the pi-ring. 1 mark: Increases the electron density of the benzene ring. 1 mark: Ring is more susceptible to electrophilic attack / polarizes electrophiles more easily. Part (b): 3 marks. 1 mark: Reagents: Sn and conc. HCl. 1 mark: Conditions: Heat under reflux (accept subsequent addition of NaOH). 1 mark: Equation: HOC6H4NO2 + 6[H] -> HOC6H4NH2 + 2H2O. Part (c): 3 marks. 1 mark: Correct structure of ethanoic anhydride. 1 mark: Correct structure of paracetamol. 1 mark: Ethanoic acid named. Part (d): 3 marks. 1 mark: Dissolve in minimum volume of hot solvent. 1 mark: Hot filtration (to remove insoluble impurities) followed by cooling in ice. 1 mark: Filter cold under reduced pressure, wash with cold solvent and dry.