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Using the Born-Haber cycle: \(\Delta H_f^{\ominus}(MgO) = \Delta H_{at}^{\ominus}(Mg) + IE_1(Mg) + IE_2(Mg) + \Delta H_{at}^{\ominus}(O) + EA_1(O) + EA_2(O) + \Delta H_{latt}^{\ominus}(MgO)\). Substituting the values: \(-602 = 148 + 738 + 1450 + 249 - 141 + 798 + \Delta H_{latt}^{\ominus}(MgO)\). Rearranging gives: \(\Delta H_{latt}^{\ominus}(MgO) = -602 - 3242 = -3844\text{ kJ mol}^{-1}\).

1 mark for the correct calculation leading to \(-3844\text{ kJ mol}^{-1}\).

Using the Nernst equation: \(E = E^{\ominus} + \frac{0.059}{z} \log_{10} \left( \frac{[\text{oxidised}]}{[\text{reduced}]} \right)\). Here, \(z = 1\), \([\text{oxidised}] = [Fe^{3+}] = 0.050\text{ mol dm}^{-3}\), and \([\text{reduced}] = [Fe^{2+}] = 1.0\text{ mol dm}^{-3}\). Substituting the values: \(E = 0.77 + 0.059 \log_{10} \left( \frac{0.050}{1.0} \right) = 0.77 + 0.059 \times (-1.301) = 0.77 - 0.077 = +0.69\text{ V}\).

1 mark for the correct calculation of \(+0.69\text{ V}\).

The conversion of \(C_4H_{10}O_2\) to \(C_4H_6O_3\) involves the loss of 4 hydrogen atoms and the gain of 1 oxygen atom. This change corresponds to the oxidation of one primary alcohol group to a carboxylic acid group (loss of 2 H, gain of 1 O) and one secondary alcohol group to a ketone (loss of 2 H). Therefore, compound X must contain exactly one primary alcohol group and one secondary alcohol group. Only \(HOCH_2CH_2CH(OH)CH_3\) fits this structure.

1 mark for identifying the structure with exactly one primary and one secondary alcohol group.

Initial moles of propanoic acid = \(0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\). Moles of NaOH added = \(0.0500\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 0.00250\text{ mol}\). The reaction consumes \(0.00250\text{ mol}\) of propanoic acid to produce \(0.00250\text{ mol}\) of sodium propanoate. Remaining propanoic acid = \(0.00500 - 0.00250 = 0.00250\text{ mol}\). Since moles of weak acid equal moles of conjugate base, the ratio \([\text{acid}] / [\text{conjugate base}] = 1\), meaning \(pH = pK_a\). Therefore, \(pH = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

1 mark for calculating the pH as \(4.87\).

\(k = \frac{\text{Rate}}{[A]^2[B]} = \frac{4.0 \times 10^{-4}}{(0.10)^2 \times 0.20} = \frac{4.0 \times 10^{-4}}{0.010 \times 0.20} = \frac{4.0 \times 10^{-4}}{2.0 \times 10^{-3}} = 0.20\). The unit of \(k\) is obtained by dividing \(\text{mol dm}^{-3}\text{ s}^{-1}\) by \((\text{mol dm}^{-3})^3\), which gives \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for the correct calculation of value 0.20 and units \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

The high concentration of chloride ions replaces water ligands in a ligand substitution reaction. Due to the steric hindrance and electrostatic repulsion between the negative chloride ligands, the coordination number changes from 6 to 4, forming the tetrahedral complex \([CuCl_4]^{2-}\).

1 mark for identifying the tetrahedral complex \([CuCl_4]^{2-}\) and ligand substitution.

Mass of C = \(3.30 \times \frac{12.0}{44.0} = 0.90\text{ g}\). Mass of H = \(1.80 \times \frac{2.0}{18.0} = 0.20\text{ g}\). Mass of O = \(1.50 - 0.90 - 0.20 = 0.40\text{ g}\). Moles of C = \(\frac{0.90}{12.0} = 0.075\text{ mol}\). Moles of H = \(\frac{0.20}{1.0} = 0.20\text{ mol}\). Moles of O = \(\frac{0.40}{16.0} = 0.025\text{ mol}\). Dividing all moles by the smallest value (0.025) gives a ratio of C:H:O = 3:8:1. Therefore, the empirical formula is \(C_3H_8O\).

1 mark for the correct calculation of the empirical formula \(C_3H_8O\).

Hydrolysis of halogenoalkanes is fastest for iodoalkanes due to the low bond enthalpy of the C-I bond. In addition, tertiary halogenoalkanes react much faster than primary or secondary halogenoalkanes because they react via the \(S_N1\) mechanism, which involves a stable tertiary carbocation intermediate. Thus, 2-iodo-2-methylpropane, a tertiary iodoalkane, reacts the most rapidly.

1 mark for identifying 2-iodo-2-methylpropane as the fastest reacting halogenoalkane.

Lattice energy magnitude depends on the charges of the ions and their ionic radii. \(\text{MgO}\) consists of \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) (high charges, small radii), so it has the most exothermic lattice energy. \(\text{CaO}\) also has +2 and -2 charges, but \(\text{Ca}^{2+}\) is larger than \(\text{Mg}^{2+}\), making its lattice energy less exothermic than \(\text{MgO}\). \(\text{NaCl}\) and \(\text{KBr}\) have +1 and -1 charges, so they have much less exothermic lattice energies than \(\text{MgO}\) and \(\text{CaO}\). Within these, \(\text{Na}^+\) and \(\text{Cl}^-\) are smaller than \(\text{K}^+\) and \(\text{Br}^-\), so \(\text{NaCl}\) is more exothermic than \(\text{KBr}\). Thus, the correct sequence is \(\text{MgO} > \text{CaO} > \text{NaCl} > \text{KBr}\).

Award 1 mark for the correct option (A).

Using the Nernst equation: \(E = E^\ominus + \frac{0.059}{z}\log[\text{Cu}^{2+}]\). Here, \(E^\ominus = +0.34\text{ V}\), \(z = 2\) (since 2 electrons are transferred), and \([\text{Cu}^{2+}] = 0.010\text{ mol dm}^{-3}\). Substituting these values: \(E = +0.34 + \frac{0.059}{2}\log(0.010) = +0.34 + 0.0295 \times (-2) = +0.34 - 0.059 = +0.281\text{ V}\).

Award 1 mark for the correct option (B).

Benzoic acid is a carboxylic acid and is acidic enough to react with carbonates or hydrogencarbonates like \(\text{NaHCO}_3\) to release \(\text{CO}_2\) gas. Phenol is a much weaker acid than carboxylic acids and does not react with \(\text{NaHCO}_3\). Both react with sodium metal to produce hydrogen gas. Both react with aqueous sodium hydroxide in an acid-base neutralization (with no gas evolved). Only phenol reacts readily with aqueous bromine to give a white precipitate of 2,4,6-tribromophenol at room temperature.

Award 1 mark for the correct option (A).

First, calculate the \(\text{p}K_\text{a}\) of propanoic acid: \(\text{p}K_\text{a} = -\log(1.35 \times 10^{-5}) = 4.87\). Since equal volumes of the acid and salt are mixed, their relative concentration ratio remains proportional to their original amounts: \([\text{acid}] = 0.0500\text{ mol dm}^{-3}\) and \([\text{salt}] = 0.0250\text{ mol dm}^{-3}\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_\text{a} + \log\frac{[\text{salt}]}{[\text{acid}]} = 4.87 + \log\left(\frac{0.0250}{0.0500}\right) = 4.87 + \log(0.5) = 4.87 - 0.30 = 4.57\).

Award 1 mark for the correct option (A).

Comparing Experiments 1 and 2: [Y] is kept constant at \(0.10\text{ mol dm}^{-3}\) while [X] is doubled from \(0.10\) to \(0.20\text{ mol dm}^{-3}\). This causes the rate to quadruple (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)), which means the reaction is second order with respect to X. Comparing Experiments 2 and 3: [X] is kept constant at \(0.20\text{ mol dm}^{-3}\) while [Y] is doubled from \(0.10\) to \(0.20\text{ mol dm}^{-3}\). This causes the rate to double (from \(8.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)), which means the reaction is first order with respect to Y. Therefore, the rate equation is: \(\text{Rate} = k[\text{X}]^2[\text{Y}]\).

Award 1 mark for the correct option (B).

The electronic configurations are as follows:
- Iron atom, \(\text{Fe}\): \([\text{Ar}] 3\text{d}^6 4\text{s}^2\).
- \(\text{Fe}^{2+}\) (loses 2 electrons from 4s): \([\text{Ar}] 3\text{d}^6\).
- \(\text{Fe}^{3+}\) (loses another electron from 3d): \([\text{Ar}] 3\text{d}^5\).
- Chromium atom, \(\text{Cr}\): \([\text{Ar}] 3\text{d}^5 4\text{s}^1\), so \(\text{Cr}^{2+}\) is \([\text{Ar}] 3\text{d}^4\).
- Cobalt atom, \(\text{Co}\): \([\text{Ar}] 3\text{d}^7 4\text{s}^2\), so \(\text{Co}^{3+}\) is \([\text{Ar}] 3\text{d}^6\).

Award 1 mark for the correct option (C).

In a catalytic converter, nitrogen monoxide (a toxic pollutant) is reduced to nitrogen gas, and carbon monoxide is oxidized to carbon dioxide: \(2\text{CO}(\text{g}) + 2\text{NO}(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + \text{N}_2(\text{g})\). Reactions B and C represent sulfur combustion and its oxidation, which are not catalyzed in a car's catalytic converter. Reaction D occurs inside the engine cylinders under high temperature and pressure, producing the pollutant NO rather than removing it.

Award 1 mark for the correct option (A).

The methyl group (\(-\text{CH}_3\)) is an electron-donating group (+I inductive effect) which activates the benzene ring, making the nitration of methylbenzene faster than that of benzene. The methyl group is 2,4-directing, directing the incoming nitronium electrophile to positions 2 and 4. The electrophile in this electrophilic substitution reaction is the nitronium ion, \(\text{NO}_2^+\). Concentrated sulfuric acid acts as a Brønsted-Lowry acid (proton donor) to help generate the electrophile from nitric acid.

Award 1 mark for the correct option (B).

The relationship is given by: \(\Delta H^\ominus_{\text{sol}}(\text{MgCl}_2) = \Delta H^\ominus_{\text{hyd}}(\text{Mg}^{2+}) + 2 \times \Delta H^\ominus_{\text{hyd}}(\text{Cl}^-) - \Delta H^\ominus_{\text{latt}}(\text{MgCl}_2)\). Substituting the values: \(\Delta H^\ominus_{\text{sol}} = -1920 + 2(-364) - (-2526) = -1920 - 728 + 2526 = -122\text{ kJ mol}^{-1}\).

1 mark for correct calculation and sign.

Using the Nernst equation: \(E = E^\ominus + \frac{0.059}{z} \log [\text{Cu}^{2+}]\). Here, \(E^\ominus = +0.34\text{ V}\), \(z = 2\) (since two electrons are transferred), and \([\text{Cu}^{2+}] = 0.010\text{ mol dm}^{-3}\). Therefore, \(E = 0.34 + \frac{0.059}{2} \log_{10}(0.010) = 0.34 + 0.0295 \times (-2) = 0.34 - 0.059 = +0.281\text{ V}\).

1 mark for correct application of the Nernst equation with z = 2.

First, determine the order with respect to each reactant. Comparing Experiment 1 and 2, doubling \([\text{A}]\) while keeping \([\text{B}]\) constant increases the rate by a factor of 4 (from \(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3}\)), meaning the reaction is second order with respect to A. Comparing Experiment 1 and 3, doubling \([\text{B}]\) while keeping \([\text{A}]\) constant doubles the rate (from \(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\)), meaning the reaction is first order with respect to B. The rate equation is \(\text{Rate} = k[\text{A}]^2[\text{B}]\). Substitute Experiment 1 values: \(1.2 \times 10^{-3} = k(0.10)^2(0.10) \Rightarrow k = 1.2\). The units of \(k\) are calculated as: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

1 mark for determining the correct order of reaction, calculating the value of k, and establishing the correct units.

Transition metal complexes are colored because the d-orbitals split into two non-degenerate energy levels when ligands approach. An electron from a lower d-orbital absorbs light energy of a specific wavelength in the red-orange region of the visible spectrum during promotion to a higher d-orbital (d-d transition). The color observed (blue) is the complementary transmitted color.

1 mark for the correct explanation of split d-orbitals and complementary color absorption/transmission.

The methyl group on methylbenzene is an electron-donating alkyl group that activates the benzene ring and directs electrophilic substitution to the ortho (2-) and para (4-) positions. Under mild conditions (\(30\ ^\circ\text{C}\)), mononitration occurs, producing a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene.

1 mark for identifying the correct directing effect and substitution level.

To synthesize benzenediazonium chloride from phenylamine (Step 1), nitrous acid is required, which is prepared in situ from sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)) at a low temperature (typically between \(0\ ^\circ\text{C}\) and \(10\ ^\circ\text{C}\)). At higher temperatures, the diazonium salt decomposes. In Step 2, the diazonium salt is coupled with phenol in alkaline conditions (such as \(\text{NaOH}(aq)\)) to deprotonate phenol to the highly reactive phenoxide ion, forming the azo dye at room temperature.

1 mark for identifying the correct diazotisation reagents and temperature, followed by the correct coupling conditions.

After mixing equal volumes, the concentrations of both the acid and its salt are halved, but the ratio \(\frac{[\text{salt}]}{[\text{acid}]}\) remains the same: \(\frac{0.050}{0.100} = 0.50\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)\). Here, \(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\). Thus, \(\text{pH} = 4.87 + \log_{10}(0.50) = 4.87 - 0.30 = 4.57\).

1 mark for calculating the correct buffer pH.

This is a polyester. Hydrolysis of the ester linkages reveals that it is synthesized from a diol and a dicarboxylic acid. The diol segment is \(\text{–O–CH}_2\text{–CH}_2\text{–O–}\), corresponding to ethane-1,2-diol. The dicarboxylic acid segment is \(\text{–CO–CH}_2\text{–CH}_2\text{–CO–}\), corresponding to butanedioic acid (with 4 carbons in total).

1 mark for identifying the correct diol and dicarboxylic acid monomers.

The relationship between the enthalpy change of solution, lattice energy (formation), and hydration enthalpies is given by:
\(\Delta H^\ominus_{\text{sol}} = -\Delta H^\ominus_{\text{latt}} + \Delta H^\ominus_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H^\ominus_{\text{hyd}}(\text{Cl}^{-})\)

Substituting the given values:
\(-150 = -(-2524) + (-1920) + 2\Delta H^\ominus_{\text{hyd}}(\text{Cl}^{-})\)
\(-150 = 2524 - 1920 + 2\Delta H^\ominus_{\text{hyd}}(\text{Cl}^{-})\)
\(-150 = 604 + 2\Delta H^\ominus_{\text{hyd}}(\text{Cl}^{-})\)
\(-754 = 2\Delta H^\ominus_{\text{hyd}}(\text{Cl}^{-})\)
\(\Delta H^\ominus_{\text{hyd}}(\text{Cl}^{-}) = -377 \text{ kJ mol}^{-1}\)

Award 1 mark for Option A. Standard Born-Haber cycle calculation: Method mark for setting up the correct equation, accuracy mark for finding the value of -377 kJ mol^-1.

For the half-reaction \(\text{Ag}^{+}(aq) + \text{e}^{-} \rightleftharpoons \text{Ag}(s)\), the number of transferred electrons \(z = 1\).
Using the Nernst equation:
\(E = E^\ominus + 0.059 \log_{10}[\text{Ag}^{+}(aq)]\)
\(E = +0.80 + 0.059 \log_{10}(0.025)\)
\(E = +0.80 + 0.059 \times (-1.602)\)
\(E = +0.80 - 0.095 = +0.705 \text{ V} \approx +0.71 \text{ V}\)

Award 1 mark for Option A. Correct substitution into the Nernst equation with z = 1 to obtain +0.71 V.

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\)

First, calculate \(\text{p}K_a\):
\(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\)

Now, calculate the pH:
\(\text{pH} = 4.87 + \log_{10}\left(\frac{0.100}{0.150}\right)\)
\(\text{pH} = 4.87 + \log_{10}(0.667) = 4.87 - 0.176 = 4.694 \approx 4.69\)

Award 1 mark for Option B. One mark for correctly using the buffer pH expression to arrive at pH = 4.69.

Transition metal complexes are colored because the presence of ligands causes the d-orbitals to split into two non-degenerate energy levels. Electrons in the lower d-orbitals absorb light in the visible spectrum to be promoted to the higher d-orbitals (d-d transition). The light absorbed by \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) is in the red-orange region of the spectrum, and the unabsorbed (complementary) blue light is transmitted, making the solution appear blue.

Award 1 mark for Option A. Understanding that red-orange light is absorbed for promotion and blue light is transmitted.

1. Determine the order with respect to A:
Comparing Exp 1 and Exp 2, [B] is constant. As [A] doubles, the rate increases by a factor of 4 (from 2.0 to 8.0). Therefore, the reaction is second-order with respect to A.

2. Determine the order with respect to B:
Comparing Exp 2 and Exp 3, [A] is constant. As [B] doubles, the rate doubles (from 8.0 to 16.0). Therefore, the reaction is first-order with respect to B.

3. Write the rate equation:
\(\text{Rate} = k[\text{A}]^2[\text{B}]\)

4. Calculate the rate constant \(k\) using Exp 1:
\(2.0 \times 10^{-4} = k(0.10)^2(0.10)\)
\(2.0 \times 10^{-4} = k(1.0 \times 10^{-3})\)
\(k = 0.20\)

5. Determine units:
\(\text{units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^2 (\text{mol dm}^{-3})} = \text{mol}^{-2} \text{ dm}^6 \text{ s}^{-1} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\)

Award 1 mark for Option A. 1 mark for correctly determining the rate expression, calculating k = 0.20, and assigning the correct units.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond being broken, not its polarity. The C–I bond in 1-iodobutane has a lower bond enthalpy (bond energy) than the C–Cl bond in 1-chlorobutane, meaning less energy is required to break the C–I bond. Thus, 1-iodobutane reacts much faster to release iodide ions, which immediately react with \(\text{Ag}^{+}(aq)\) to form a yellow precipitate of silver iodide.

Award 1 mark for Option B. Correctly linking higher reactivity to lower bond enthalpy (strength of C-X bond).

The methyl group is an alkyl group, which is electron-donating due to the inductive effect. This increases the electron density on the aromatic ring, making it more susceptible to electrophilic attack than benzene (activating the ring). The methyl group is an ortho/para director, meaning it directs incoming electrophiles (such as \(\text{NO}_2^{+}\)) to positions 2 (ortho) and 4 (para).

Award 1 mark for Option A. Understanding that the methyl group is electron-donating, activating, and directs to positions 2 and 4.

- In Step 1, propene reacts with \(\text{HBr}(g)\) to form 2-bromopropane (compound X) as the major product according to Markovnikov's rule.
- In Step 2, nucleophilic substitution of the bromoalkane with cyanide ions using \(\text{KCN}\) in ethanol under reflux yields 2-methylpropanenitrile (compound Y).
- In Step 3, acid hydrolysis of the nitrile group using dilute \(\text{HCl}(aq)\) under reflux converts the nitrile (\(-\text{CN}\)) group to a carboxylic acid (\(-\text{COOH}\)) group, forming 2-methylpropanoic acid.

Award 1 mark for Option A (Row A). Understanding that nucleophilic substitution with cyanide is achieved by KCN in ethanol under reflux, and nitrile hydrolysis is achieved by heating with dilute aqueous acid.

According to Hess's law, the standard enthalpy of formation of CaCl2(s) is equal to the sum of the enthalpy changes of the steps to produce gaseous Ca2+ and 2Cl- ions, plus the lattice energy. Ca(s) to Ca(g) is +178. Ca(g) to Ca2+(g) is +590 + 1145 = +1735. Cl2(g) to 2Cl(g) is 2 * (+121) = +242. 2Cl(g) + 2e- to 2Cl-(g) is 2 * (-349) = -698. Total gaseous ion formation steps = 178 + 1735 + 242 - 698 = +1457. Therefore, -796 = +1457 + Lattice Energy, which gives Lattice Energy = -796 - 1457 = -2253 kJ mol-1.

Award 1 mark for correct selection of Option A. Method: correctly summing the steps of the cycle with appropriate stoichiometry and subtracting from standard enthalpy of formation.

Using the Nernst equation: E = E-standard + (0.059 / z) * log10([Cu2+]). Here, z = 2, E-standard = +0.34 V, and [Cu2+] = 0.025 mol dm-3. E = +0.34 + (0.059 / 2) * log10(0.025) = +0.34 + 0.0295 * (-1.602) = +0.34 - 0.047 = +0.29 V.

Award 1 mark for correct selection of Option A. Deduct marks for using z = 1 or for incorrect sign in the log term calculation.

The stability constant is defined as: Kstab = [[Fe(H2O)5(SCN)]2+] / ([[Fe(H2O)6]3+] * [SCN-]). Rearranging to solve for the product concentration: [[Fe(H2O)5(SCN)]2+] = Kstab * [[Fe(H2O)6]3+] * [SCN-] = (2.0 * 10^3) * (1.5 * 10^-4) * (3.0 * 10^-4) = 9.0 * 10^-5 mol dm-3.

Award 1 mark for correct selection of Option A. Verify that the correct stability constant expression is used with water omitted.

Rearranging the rate equation for k: k = Rate / ([A]^2 * [B]). Substituting the given values: k = (2.0 * 10^-3) / ((0.10)^2 * 0.10) = 2.0. The units of k are: (mol dm-3 s-1) / (mol dm-3)^3 = dm6 mol-2 s-1.

Award 1 mark for correct selection of Option A. Checks: both the numerical value of 2.0 and the units of dm6 mol-2 s-1 must be correct.

Stage 1 (diazotisation) requires nitrous acid, which is generated in situ from NaNO2 and HCl at temperatures below 10 degrees C (typically 5 degrees C) to prevent decomposition of the diazonium salt. The coupling of benzenediazonium chloride with phenol forms 4-hydroxyphenylazobenzene, which contains the azo group (-N=N-) linking the two benzene rings, giving the formula C6H5N=NC6H4OH.

Award 1 mark for correct selection of Option A. Correctly identifies reagents (NaNO2 + HCl), temperature (< 10 degrees C), and the correct azo linkage.

Under Condition 1 (UV light, heat), free-radical substitution occurs on the methyl side chain of methylbenzene, yielding (chloromethyl)benzene, C6H5CH2Cl. Under Condition 2 (anhydrous AlCl3 catalyst, dark), electrophilic aromatic substitution occurs on the benzene ring. Since the methyl group is 2,4-directing, the major products are a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.

Award 1 mark for correct selection of Option A. Distinguishes side-chain substitution (Condition 1) from ring substitution (Condition 2).

Acyl chlorides (C6H5COCl) are highly reactive and easily hydrolysed by water at room temperature due to the highly positive carbonyl carbon. Halogenoalkanes like (chloromethyl)benzene (C6H5CH2Cl) undergo nucleophilic substitution but require heating with water or aqueous alkali. Halogenoarenes like chlorobenzene (C6H5Cl) are extremely unreactive and do not undergo hydrolysis under normal conditions because the lone pair on the chlorine atom is delocalised into the benzene ring, strengthening the C-Cl bond. Thus, the order of decreasing ease of hydrolysis is C6H5COCl > C6H5CH2Cl > C6H5Cl.

Award 1 mark for correct selection of Option A. Order must correctly show that acyl chloride is most reactive, followed by alkyl halide, and aryl halide is the least reactive.

A reaction is feasible when Delta G-standard < 0. Since Delta G-standard = Delta H-standard - T * Delta S-standard, feasibility requires Delta H-standard - T * Delta S-standard < 0, which rearranges to T > Delta H-standard / Delta S-standard. Converting Delta H-standard to J mol-1 gives 178 * 10^3 J mol-1. T > (178 * 10^3) / 160 = 1112.5 K. Therefore, the reaction is feasible at temperatures above approximately 1113 K.

Award 1 mark for correct selection of Option A. Requires correct conversion of units and setting up the inequality Delta G-standard < 0.

(a)(i) Moles of initial \(\text{HCl} = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). (a)(ii) Moles of \(\text{NaOH} = 0.0250\text{ dm}^3 \times 0.300\text{ mol dm}^{-3} = 0.00750\text{ mol}\). (a)(iii) In the \(25.0\text{ cm}^3\) sample, the excess \(\text{HCl} = 0.00750\text{ mol}\). In the total \(100\text{ cm}^3\) solution, excess \(\text{HCl} = 0.00750 \times 4 = 0.0300\text{ mol}\). Reacted \(\text{HCl} = 0.0500 - 0.0300 = 0.0200\text{ mol}\). (a)(iv) The reaction is \(M\text{CO}_3 + 2\text{HCl} \rightarrow M\text{Cl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). Moles of \(M\text{CO}_3 = 0.0200 / 2 = 0.0100\text{ mol}\). (a)(v) \(M_r\) of \(M\text{CO}_3 = 1.00\text{ g} / 0.0100\text{ mol} = 100\text{ g mol}^{-1}\). Since \(\text{CO}_3^{2-} = 60.0\text{ g mol}^{-1}\), \(A_r\) of \(M = 100 - 60.0 = 40.0\text{ g mol}^{-1}\). The metal \(M\) is calcium (Ca). (b)(i) Gaseous ion is \(\text{Ca}^{2+}\), with configuration \(1s^2 2s^2 2p^6 3s^2 3p^6\). (b)(ii) Calcium has outer electrons in the fourth shell (4s), whereas magnesium has outer electrons in the third shell (3s). Calcium has greater shielding and a larger atomic radius, resulting in weaker nuclear attraction on outer electrons than in magnesium. (b)(iii) The second ionisation energy of magnesium is represented by: \(\text{Mg}^+\text{(g)} \rightarrow \text{Mg}^{2+}\text{(g)} + \text{e}^-\).

(a)(i) [1 mark] 0.0500 mol. (a)(ii) [1 mark] 0.00750 mol. (a)(iii) [1 mark] Total excess HCl = 0.0300 mol AND reacted HCl = 0.0200 mol. (a)(iv) [1 mark] 0.0100 mol. (a)(v) [3 marks] Mr of MCO3 = 100 [1], Ar of M = 40 [1], metal identified as calcium/Ca [1]. (b)(i) [1 mark] 1s2 2s2 2p6 3s2 3p6. (b)(ii) [2 marks] Outer electrons of Ca in higher energy level/more shielded [1], weaker electrostatic attraction between nucleus and outer electrons in Ca than Mg [1]. (b)(iii) [2 marks] Correct species and balancing [1], state symbols (g) on both Mg species [1].

(a) The standard enthalpy change of combustion is the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions (298 K, 101 kPa). (b)(i) The standard enthalpy change of formation involves forming one mole of compound from its constituent elements in their standard states: \(2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{OH(l)}\). (b)(ii) Using the Hess's Law cycle: \(\Delta H_f^\ominus[\text{CH}_3\text{CH}_2\text{OH(l)}] = 2 \times \Delta H_c^\ominus[\text{C}] + 3 \times \Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{CH}_3\text{CH}_2\text{OH(l)}] = 2(-393.5) + 3(-285.8) - (-1367.3) = -787.0 - 857.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\). (c)(i) Heat energy absorbed, \(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (58.5 - 21.5)\text{ K} = 150.0 \times 4.18 \times 37.0 = 23202\text{ J} = 23.2\text{ kJ}\). (c)(ii) Moles of ethanol = \(0.920\text{ g} / 46.0\text{ g mol}^{-1} = 0.0200\text{ mol}\). (c)(iii) Experimental enthalpy change of combustion, \(\Delta H_c = -q / n = -23.202\text{ kJ} / 0.0200\text{ mol} = -1160\text{ kJ mol}^{-1}\). (c)(iv) Incomplete combustion of ethanol (forming soot or carbon monoxide instead of carbon dioxide) or evaporation of ethanol directly from the wick.

(a) [2 marks] Enthalpy change when 1 mole of substance is burned completely in excess oxygen [1], under standard conditions [1]. (b)(i) [2 marks] Correct equation [1], correct state symbols [1]. (b)(ii) [3 marks] Correct application of Hess's Law [1], correct substitution [1], correct calculation to -277.1 kJ/mol [1]. (c)(i) [1 mark] 23.2 kJ. (c)(ii) [1 mark] 0.0200 mol. (c)(iii) [2 marks] division of energy by moles with negative sign [1], value -1160 kJ/mol [1]. (c)(iv) [1 mark] Incomplete combustion OR evaporation of ethanol from the wick.

(a)(i) Nitrogen and oxygen react under extreme heat: \(\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO(g)}\). (a)(ii) The triple bond between nitrogen atoms in \(\text{N}_2\) is exceptionally strong, requiring a very high activation energy to break. This is only available at high temperatures. (b)(i) Since \(\text{NO}\) gas is in the same phase as the gaseous reactants, it acts as a homogeneous catalyst. (b)(ii) First, \(\text{NO}\) is oxidized by atmospheric oxygen: \(\text{NO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{NO}_2\text{(g)}\). Second, \(\text{NO}_2\) oxidizes sulfur dioxide and is regenerated: \(\text{NO}_2\text{(g)} + \text{SO}_2\text{(g)} \rightarrow \text{SO}_3\text{(g)} + \text{NO(g)}\). (b)(iii) Sulfur trioxide reacts with water vapor in the atmosphere to form strong sulfuric acid: \(\text{SO}_3\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)}\), which dissociates to release \(\text{H}^+\) ions, lowering pH. (c)(i) Transition metals such as Platinum (Pt), Palladium (Pd), or Rhodium (Rh) are used. (c)(ii) Carbon monoxide is oxidized and nitrogen monoxide is reduced: \(2\text{CO(g)} + 2\text{NO(g)} \rightarrow 2\text{CO}_2\text{(g)} + \text{N}_2\text{(g)}\). (c)(iii) This is heterogeneous catalysis. Gaseous reactants are adsorbed onto the solid metal surface, which weakens their bonds and lowers the activation energy. The reaction takes place on the surface, and the products are then desorbed.

(a)(i) [1 mark] N2 + O2 -> 2NO. (a)(ii) [2 marks] N2 has a strong triple bond [1], high temperature is needed to overcome high activation energy [1]. (b)(i) [1 mark] Homogeneous catalyst. (b)(ii) [2 marks] NO + 0.5O2 -> NO2 [1], NO2 + SO2 -> SO3 + NO [1]. (b)(iii) [2 marks] Equation: SO3 + H2O -> H2SO4 [1], explanation of lowering pH / acidifying rainwater [1]. (c)(i) [1 mark] Platinum / palladium / rhodium. (c)(ii) [1 mark] 2CO + 2NO -> 2CO2 + N2. (c)(iii) [2 marks] Heterogeneous catalysis [1], adsorption, bond weakening, reaction, desorption [1].

(a) The skeletal formula consists of a four-carbon zig-zag chain with an -OH group attached to the second carbon. (b)(i) Optical isomerism requires a chiral carbon (a carbon bonded to four different atoms or groups of atoms). In butan-2-ol, Carbon-2 is chiral as it is bonded to -H, -OH, -CH3, and -CH2CH3. (b)(ii) Draw a central carbon with two bonds in the plane of the page, one wedge bond, and one dashed bond, and its mirror image. (c)(i) Reagents are acidified potassium dichromate(VI) (or sodium dichromate(VI)) and the mixture is heated under reflux. (c)(ii) The oxidation of butan-2-ol (a secondary alcohol) yields butanone, which has the structure \(\text{CH}_3\text{COCH}_2\text{CH}_3\). Its IUPAC name is butanone. (c)(iii) The orange dichromate(VI) ions are reduced to green chromium(III) ions, so the solution turns from orange to green. (d)(i) Dehydration is achieved using concentrated sulfuric acid (or phosphoric acid) as a catalyst with heating (or passing the vapor over heated aluminium oxide). (d)(ii) Dehydration can eliminate hydrogen from either Carbon-1 or Carbon-3. Elimination from Carbon-1 yields but-1-ene. Elimination from Carbon-3 yields but-2-ene, which exhibits stereoisomerism as cis-but-2-ene (Z-but-2-ene) and trans-but-2-ene (E-but-2-ene). Thus, the three alkenes are but-1-ene, cis-but-2-ene, and trans-but-2-ene.

(a) [1 mark] Correct skeletal formula. (b)(i) [2 marks] Presence of a chiral carbon [1], list of 4 groups for C2: -H, -OH, -CH3, -CH2CH3 [1]. (b)(ii) [2 marks] Correct 3D representation of one enantiomer with wedge/dash [1], correct mirror image [1]. (c)(i) [2 marks] Acidified potassium dichromate(VI) [1], heat under reflux [1]. (c)(ii) [2 marks] Correct structure of butanone [1], correct name: butanone [1]. (c)(iii) [1 mark] Orange to green color change. (d)(i) [1 mark] Concentrated H2SO4 and heat. (d)(ii) [1 mark] but-1-ene, cis-but-2-ene, and trans-but-2-ene (all three required).

(a) \(K_p = \frac{p(\text{N}_2\text{O}_4)}{(p(\text{NO}_2))^2}\). Units = \(\text{Pa} / \text{Pa}^2 = \text{Pa}^{-1}\) (or \(\text{kPa}^{-1}\)). (b)(i) According to the stoichiometry: \(2\text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4\). Moles of \(\text{N}_2\text{O}_4\) at equilibrium = \(0.30\text{ mol}\). Moles of \(\text{NO}_2\) reacted = \(2 \times 0.30 = 0.60\text{ mol}\). Moles of \(\text{NO}_2\) remaining at equilibrium = \(1.00 - 0.60 = 0.40\text{ mol}\). (b)(ii) Total moles of gas = \(0.40 + 0.30 = 0.70\text{ mol}\). Mole fraction of \(\text{NO}_2 = 0.40 / 0.70 = 0.571\). Mole fraction of \(\text{N}_2\text{O}_4 = 0.30 / 0.70 = 0.429\). (b)(iii) Partial pressure = mole fraction \times total pressure. \(p(\text{NO}_2) = 0.571 \times 2.00 \times 10^5\text{ Pa} = 1.14 \times 10^5\text{ Pa}\) (or \(114.3\text{ kPa}\)). \(p(\text{N}_2\text{O}_4) = 0.429 \times 2.00 \times 10^5\text{ Pa} = 8.57 \times 10^4\text{ Pa}\) (or \(85.7\text{ kPa}\)). (b)(iv) \(K_p = \frac{8.57 \times 10^4\text{ Pa}}{(1.14 \times 10^5\text{ Pa})^2} = 6.56 \times 10^{-6}\text{ Pa}^{-1}\) (or \(6.56 \times 10^{-3}\text{ kPa}^{-1}\)). (c) Since the forward reaction is exothermic (\(\Delta H^\ominus < 0\)), increasing the temperature will cause the equilibrium position to shift to the left (the endothermic direction) to absorb the added heat. This decreases the concentration of products and increases the concentration of reactants, so the value of \(K_p\) decreases.

(a) [2 marks] Correct Kp expression [1], correct units corresponding to the pressure unit used (Pa^-1 or kPa^-1) [1]. (b)(i) [2 marks] Moles of N2O4 = 0.30 mol [1], moles of NO2 = 0.40 mol [1]. (b)(ii) [2 marks] Mole fraction of NO2 = 0.571 [1], mole fraction of N2O4 = 0.429 [1]. (b)(iii) [2 marks] p(NO2) = 1.14 x 10^5 Pa [1], p(N2O4) = 8.57 x 10^4 Pa [1]. (b)(iv) [2 marks] Correct calculation [1], final answer with units: 6.56 x 10^-6 Pa^-1 (or 6.56 x 10^-3 kPa^-1) [1]. (c) [2 marks] Shift to left [1], value of Kp decreases [1].

(a) Concordant titres are those within \( 0.10\text{ cm}^3 \) of each other: Titration 1 (\( 24.20\text{ cm}^3 \)), Titration 2 (\( 24.15\text{ cm}^3 \)), and Titration 3 (\( 24.25\text{ cm}^3 \)). Mean titre = \( (24.20 + 24.15 + 24.25) / 3 = 24.20\text{ cm}^3 \). (b) Moles of \( \text{Na}_2\text{S}_2\text{O}_3 \) = \( (24.20 / 1000) \times 0.100 = 2.42 \times 10^{-3}\text{ mol} \). (c) From the 1:2 stoichiometry of \( \text{I}_2 : \text{S}_2\text{O}_3^{2-} \), moles of \( \text{I}_2 \) = \( 2.42 \times 10^{-3} / 2 = 1.21 \times 10^{-3}\text{ mol} \). From the 1:1 stoichiometry of \( \text{ClO}^- : \text{I}_2 \), moles of \( \text{ClO}^- \) in \( 25.0\text{ cm}^3 \) of FA 1 = \( 1.21 \times 10^{-3}\text{ mol} \). (d) Concentration of \( \text{ClO}^- \) in FA 1 = \( 1.21 \times 10^{-3}\text{ mol} / 0.0250\text{ dm}^3 = 0.0484\text{ mol dm}^{-3} \). (e) Dilution factor = \( 250 / 25 = 10 \). Concentration in original bleach = \( 0.0484 \times 10 = 0.484\text{ mol dm}^{-3} \). (f) Percentage uncertainty = \( (0.06 / 25.0) \times 100\% = 0.24\% \). (g) To ensure all sodium chlorate(I) is completely reduced to liberate equivalent iodine, and to dissolve the liberated iodine as triiodide ions. (h) Starch is the indicator, forming a dark blue complex with triiodide that goes colourless at the endpoint. It is added near the endpoint because starch binds iodine strongly; if added too early, it traps iodine irreversibly and causes inaccurate, premature endpoints.

[1 mark] Concordant titres selected correctly. [1 mark] Mean titre calculated correctly to 2 decimal places with units. [1 mark] Moles of thiosulfate calculated correctly to 3 significant figures. [1 mark] Moles of iodine calculated correctly. [1 mark] Moles of chlorate(I) calculated correctly. [1 mark] Concentration of FA 1 calculated correctly. [1 mark] Dilution factor identified as 10. [1 mark] Concentration of original bleach calculated correctly with units. [1 mark] Percentage uncertainty calculated correctly to 2 significant figures. [1 mark] Explaining that excess iodide ensures complete reaction of the chlorate(I). [1 mark] Starch role as indicator with clear color change (blue-black to colorless). [1 mark] Explaining why starch is added near the endpoint to avoid irreversible binding.

(a) The cooling rate from 5 to 10 min is a constant \( 0.7^{\circ}\text{C/min} \). Extrapolating back to 4 min: \( T_{\text{max}} = 33.5 + 0.7 = 34.2^{\circ}\text{C} \). \( \Delta T = 34.2 - 19.5 = 14.7^{\circ}\text{C} \). (b) \( q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 14.7\text{ K} = 3072.3\text{ J} = 3.07\text{ kJ} \). (c) Moles of \( \text{CuSO}_4 \) = \( 0.0500\text{ dm}^3 \times 0.500\text{ mol dm}^{-3} = 0.0250\text{ mol} \). (d) Moles of \( \text{Zn} \) = \( 3.00\text{ g} / 65.4\text{ g mol}^{-1} = 0.0459\text{ mol} \). Since \( 0.0459 > 0.0250 \), Zn is in excess. (e) \( \Delta H = -q / \text{moles of CuSO}_4 = -3.0723\text{ kJ} / 0.0250\text{ mol} = -122.89\text{ kJ mol}^{-1} \), which is \( -123\text{ kJ mol}^{-1} \) to 3 significant figures. (f) Heat loss to the surroundings is a major systematic error. Improvement: add a lid to the polystyrene cup, or place the cup inside another polystyrene cup to provide extra insulation. (g) Zinc powder has a larger surface area than granules, which increases the rate of reaction, ensuring that the maximum temperature is reached rapidly before significant heat loss occurs.

[1 mark] Finding the cooling rate of 0.7 C/min. [1 mark] Extrapolating the maximum temperature to 34.2 C. [1 mark] Calculating Delta T as 14.7 C. [1 mark] Calculating heat energy q with the correct mass of 50.0 g. [1 mark] Converting q to 3.07 kJ or 3070 J. [1 mark] Calculating moles of CuSO4 as 0.0250 mol. [1 mark] Calculating moles of Zn as 0.0459 mol. [1 mark] Explicitly comparing the moles of Zn and CuSO4 to show excess. [1 mark] Dividing the heat energy by the correct moles. [1 mark] Correctly stating the value of Delta H as -123 kJ mol^-1 with the negative sign. [1 mark] Stating heat loss to surroundings as the major systematic error. [1 mark] Suggesting a lid or double cups as an appropriate improvement. [1 mark] Explaining that powder increases surface area and reaction rate to minimize heat loss.

(a) The gas is carbon dioxide (CO2) because it turns limewater cloudy. The anion is carbonate (CO3^2-) because it effervesces with dilute acid to produce CO2. (b) Cation 1 is iron(II), Fe^2+, confirmed by the green precipitate with NaOH that turns brown on standing. Cation 2 is copper(II), Cu^2+, confirmed by the green precipitate dissolving in excess ammonia to give a deep blue solution, and by forming a red-brown precipitate with potassium hexacyanoferrate(II). (c) (i) CO3^2-(s) + 2H^+(aq) -> CO2(g) + H2O(l). (ii) Fe^2+(aq) + 2OH^-(aq) -> Fe(OH)2(s). (iii) [Cu(H2O)6]^2+(aq) + 4NH3(aq) -> [Cu(NH3)4(H2O)2]^2+(aq) + 4H2O(l) [or Cu^2+(aq) + 4NH3(aq) -> [Cu(NH3)4]^2+(aq)]. (d) The black residue components are copper(II) oxide (CuO) and iron(III) oxide (Fe2O3) or iron(II) oxide (FeO). (e) The green precipitate is iron(II) hydroxide, Fe(OH)2, which is oxidised by oxygen in the air on the surface to brown iron(III) hydroxide, Fe(OH)3.

[1 mark] Identifying gas as carbon dioxide and anion as carbonate. [1 mark] Citing limewater cloudiness and effervescence with acid as evidence. [1 mark] Identifying Cation 1 as iron(II) (Fe^2+). [1 mark] Explaining that the green precipitate turning brown with NaOH is the evidence. [1 mark] Identifying Cation 2 as copper(II) (Cu^2+). [1 mark] Explaining that the deep blue solution with excess ammonia or red-brown precipitate with hexacyanoferrate(II) is the evidence. [1 mark] Correct ionic equation for carbonate with acid. [1 mark] Correct ionic equation for Fe(OH)2 formation. [2 marks] Correct equation for [Cu(NH3)4(H2O)2]^2+ formation (1 mark for reactants and products, 1 mark for correct balancing/charges). [2 marks] Identifying CuO (1 mark) and Fe2O3 or FeO (1 mark) as the components of the black residue. [1 mark] Stating that Fe(OH)2 is oxidised by atmospheric oxygen to Fe(OH)3.

(a) Lattice energy is the standard enthalpy change when 1 mole of an ionic crystalline solid is formed from its separate gaseous ions.

(b) Using the Born-Haber cycle relation:
\(\Delta H_{\text{f}}^{\ominus} = \Delta H_{\text{at}}^{\ominus}(\text{Ba}) + \text{IE}_1(\text{Ba}) + \text{IE}_2(\text{Ba}) + \text{Bond Enthalpy}(\text{F}-\text{F}) + 2 \times \text{EA}_1(\text{F}) + \Delta H_{\text{latt}}^{\ominus}(\text{BaF}_2)\)

Substituting the given values:
\(-1205 = +180 + 503 + 965 + 158 + 2(-328) + \Delta H_{\text{latt}}^{\ominus}(\text{BaF}_2)\)
\(-1205 = 180 + 503 + 965 + 158 - 656 + \Delta H_{\text{latt}}^{\ominus}(\text{BaF}_2)\)
\(-1205 = 1150 + \Delta H_{\text{latt}}^{\ominus}(\text{BaF}_2)\)
\(\Delta H_{\text{latt}}^{\ominus}(\text{BaF}_2) = -1205 - 1150 = -2355 \text{ kJ mol}^{-1}\)

(c) Magnesium ions, \(\text{Mg}^{2+}\), have a much smaller ionic radius than barium ions, \(\text{Ba}^{2+}\). Therefore, the magnesium ion has a much higher charge density, leading to stronger electrostatic forces of attraction between the metal cation and fluoride anions, which releases more energy when the ionic lattice is formed.

Part (a):
- 1 mark: Enthalpy change when 1 mole of ionic solid is formed from its gaseous ions.
- 1 mark: Under standard conditions.

Part (b):
- 1 mark: Correctly represents Born-Haber cycle diagram or algebraic equation.
- 1 mark: Correctly includes 2 x EA of F: 2 x (-328) = -656 kJ mol^-1.
- 1 mark: Correctly identifies that bond enthalpy of F-F (+158 kJ mol^-1) represents the preparation of two moles of atomic fluorine.
- 1 mark: Correctly sums barium's atomisation and ionisation energies: 180 + 503 + 965 = +1648 kJ mol^-1.
- 1 mark: Performs mathematical rearrangement correctly: -1205 = 1150 + Lattice Energy.
- 1 mark: Final correct answer with negative sign and units: -2355 kJ mol^-1.

Part (c):
- 1 mark: Mg^2+ has a smaller ionic radius than Ba^2+.
- 1 mark: Mg^2+ has a higher charge density.
- 1 mark: Stronger electrostatic forces of attraction between Mg^2+ and F^-.

(a) \(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{reduction}} - E^{\ominus}_{\text{oxidation}} = +0.77 - (-0.26) = +1.03 \text{ V}\).
Since \(E^{\ominus}_{\text{cell}} > 0\), the reaction is thermodynamically feasible under standard conditions.

(b) Cell representation:
\(\text{Pt}(\text{s}) | \text{V}^{2+}(\text{aq}), \text{V}^{3+}(\text{aq}) || \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt}(\text{s})\)

(c) For the iron half-cell, \(z = 1\).
\(E = E^{\ominus} + \frac{0.059}{1}\log_{10}\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}\)
\(E = +0.77 + 0.059 \log_{10}\left(\frac{0.050}{0.800}\right)\)
\(E = +0.77 + 0.059 \log_{10}(0.0625)\)
\(E = +0.77 + 0.059(-1.204) = +0.77 - 0.071 = +0.699 \text{ V}\) (or \(+0.70 \text{ V}\)).

(d) The standard hydrogen electrode consists of:
- Hydrogen gas at 100 kPa (or 1 bar pressure).
- An aqueous solution of hydrogen ions, \(\text{H}^+\), at a concentration of \(1.00 \text{ mol dm}^{-3}\).
- A platinum electrode covered in platinum black (finely divided platinum catalyst).
- A temperature maintained at 298 K (25 °C).

Part (a):
- 1 mark: Calculation of cell potential showing working: 0.77 - (-0.26) = +1.03 V.
- 1 mark: States the reaction is feasible because the standard cell potential is positive.

Part (b):
- 1 mark: Correctly identifies platinum, Pt(s), as the inert electrode on both outer ends.
- 1 mark: Correct notation: Pt(s) | V^2+(aq), V^3+(aq) || Fe^3+(aq), Fe^2+(aq) | Pt(s).

Part (c):
- 1 mark: Correct substitution into Nernst equation with z = 1.
- 1 mark: Log term evaluated correctly: log(0.0625) = -1.20.
- 1 mark: Correct final potential with correct sign: +0.70 V (or +0.699 V).

Part (d):
- 1 mark: H2 gas at 100 kPa / 1 bar.
- 1 mark: [H^+] = 1.00 mol dm^-3.
- 1 mark: Platinum electrode / platinum black catalyst.
- 1 mark: Temp = 298 K.

(a) A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added to it.

(b) Equation: \(\text{C}_2\text{H}_5\text{COOH}(\text{aq}) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq})\)
Expression: \(K_{\text{a}} = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]}\)

(c) (i) Total volume = 50.0 + 50.0 = 100.0 cm^3.
\([\text{C}_2\text{H}_5\text{COOH}] = 0.200 \times \frac{50.0}{100.0} = 0.100 \text{ mol dm}^{-3}\)
\([\text{C}_2\text{H}_5\text{COO}^-] = 0.100 \times \frac{50.0}{100.0} = 0.050 \text{ mol dm}^{-3}\)

(ii) \([\text{H}^+] = K_{\text{a}} \times \frac{[\text{C}_2\text{H}_5\text{COOH}]}{[\text{C}_2\text{H}_5\text{COO}^-]}\)
\([\text{H}^+] = (1.35 \times 10^{-5}) \times \frac{0.100}{0.050} = 2.70 \times 10^{-5} \text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(2.70 \times 10^{-5}) = 4.57\)

(d) When H^+ is added, the conjugate base (propanoate ions) reacts with the added H^+ ions to remove them:
\(\text{C}_2\text{H}_5\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{C}_2\text{H}_5\text{COOH}(\text{aq})\)
Since the reservoir of propanoate ions is large, the ratio of acid to salt remains almost unchanged, keeping pH relatively constant.

Part (a):
- 1 mark: Resists/minimises changes in pH.
- 1 mark: When small amounts of acid or base/alkali are added.

Part (b):
- 1 mark: Correct dissociation equation with reversible arrows.
- 1 mark: Correct expression for Ka.

Part (c)(i):
- 1 mark: Correct concentration of propanoic acid (0.100 mol dm^-3).
- 1 mark: Correct concentration of propanoate ions (0.050 mol dm^-3).

Part (c)(ii):
- 1 mark: Correctly relates pH, pKa, and salt/acid ratio or [H+] setup.
- 1 mark: Calculates [H+] = 2.70 x 10^-5 mol dm^-3.
- 1 mark: Final pH answer = 4.57 (must be 2 decimal places).

Part (d):
- 1 mark: Correct ionic equation: C2H5COO^- + H^+ -> C2H5COOH.
- 1 mark: Explanation that propanoate acts as a reservoir to consume added protons.

(a) In a transition metal complex, the presence of ligands causes the degenerate 3d orbitals to split into two sets of non-degenerate energy levels with an energy gap, \(\Delta E\). When visible light passes through the complex, an electron in a lower-energy d-orbital absorbs a photon of specific frequency/wavelength and is promoted to a higher-energy d-orbital (d-d transition). The energy absorbed is given by \(\Delta E = h\nu = \frac{hc}{\lambda}\). The remaining wavelengths of light are transmitted or reflected, which are seen as the complementary color.

(b) Observations:
- On adding a small amount of aqueous ammonia, a pale blue precipitate of copper(II) hydroxide, \(\text{Cu(OH)}_2\), is formed.
- On adding excess ammonia, the precipitate dissolves to form a deep blue/dark blue solution.
- Overall equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)

(c) The scandium(III) ion, \(\text{Sc}^{3+}\), has the electronic configuration \([\text{Ar}]\), which means its 3d subshell is completely empty (\(3\text{d}^0\)). Since there are no d-electrons, no d-d electronic transitions can occur, and hence no visible light is absorbed.

Part (a):
- 1 mark: Ligands cause the five d-orbitals to split into two non-degenerate energy levels.
- 1 mark: Electrons absorb energy/photons from the visible light spectrum.
- 1 mark: An electron is promoted from a lower d-orbital to a higher d-orbital (d-d transition).
- 1 mark: The energy of light absorbed is related to the split gap by E = hf (or E = hc/λ).
- 1 mark: The complementary color (the light not absorbed) is transmitted/seen.

Part (b):
- 1 mark: Pale blue precipitate formed initially.
- 1 mark: Dissolves in excess to form a deep/dark blue solution.
- 2 marks: Correct balanced ionic equation (1 mark for reactants/products, 1 mark for state symbols/balancing).

Part (c):
- 1 mark: Sc^3+ has empty d-orbitals / 3d^0 electronic configuration.
- 1 mark: Therefore, d-d transition cannot occur.

(a)
- To find the order with respect to A, compare Expt 1 and Expt 2: [B] and [C] are held constant, [A] is doubled (from 0.10 to 0.20). The rate increases by a factor of \(\frac{4.80 \times 10^{-3}}{1.20 \times 10^{-3}} = 4\). Therefore, the reaction is second order with respect to A.
- To find the order with respect to B, compare Expt 1 and Expt 3: [A] and [C] are held constant, [B] is doubled (from 0.10 to 0.20). The rate remains unchanged (\(1.20 \times 10^{-3}\)). Therefore, the reaction is zero order with respect to B.
- To find the order with respect to C, compare Expt 1 and Expt 4: [A] and [B] are held constant, [C] is doubled (from 0.10 to 0.20). The rate doubles (from \(1.20 \times 10^{-3}\) to \(2.40 \times 10^{-3}\)). Therefore, the reaction is first order with respect to C.

(b) Rate equation: \(\text{Rate} = k[\text{A}]^2[\text{C}]\)

(c) Using Expt 1:
\(1.20 \times 10^{-3} = k (0.10)^2(0.10)\)
\(1.20 \times 10^{-3} = k (0.0010)\)
\(k = 1.20 \text{ dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\)

(d) The rate constant \(k\) increases with increasing temperature. An increase in temperature increases the average kinetic energy of the reacting molecules, meaning a much higher fraction of reactant molecules have kinetic energy greater than or equal to the activation energy (\(E_{\text{a}}\)). This leads to a higher frequency of successful collisions, increasing the rate, and since concentrations are unchanged, \(k\) must increase.

Part (a):
- 1 mark: Order wrt A = 2, with reasoning comparing Expt 1 & 2.
- 1 mark: Order wrt B = 0, with reasoning comparing Expt 1 & 3.
- 1 mark: Order wrt C = 1, with reasoning comparing Expt 1 & 4.
- 1 mark: Clear logical deduction linking changes in concentration to changes in rate.

Part (b):
- 1 mark: Rate = k[A]^2[C] (must not include [B]).

Part (c):
- 1 mark: Correct substitution of Expt 1 values.
- 1 mark: Calculation of k = 1.20 (or 1.2).
- 1 mark: Correct units: dm^6 mol^-2 s^-1.

Part (d):
- 1 mark: Rate constant k increases with increasing temperature.
- 1 mark: More particles have energy >= activation energy (E_a).
- 1 mark: More successful/effective collisions per unit time.

(a) Reagents: Concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) catalyst.
Conditions: Maintain temperature at \(50 - 55^\circ\text{C}\).
Equation: \(\text{C}_6\text{H}_6 + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}\)

(b) Reagents: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), followed by the addition of aqueous sodium hydroxide (\(\text{NaOH}\)) to liberate phenylamine from its salt.
Conditions: Heat under reflux.

(c) (i) Prepared by reacting sodium nitrite (\(\text{NaNO}_2\)) with dilute hydrochloric acid (\(\text{HCl}\)) at low temperature.
(ii) Temperature range: \(0 - 10^\circ\text{C}\) (or below \(10^\circ\text{C}\), but above \(0^\circ\text{C}\)). If the temperature is below \(0^\circ\text{C}\), the rate of reaction is too slow. If the temperature is above \(10^\circ\text{C}\), the benzenediazonium chloride decomposes to form phenol and nitrogen gas.

(d) The structural formula of the azo dye (4-hydroxyphenylazobenzene) is:
\(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH}\) (with the -OH group at the para-position relative to the azo linkage).

Part (a):
- 1 mark: Reagents: Conc HNO3 and Conc H2SO4.
- 1 mark: Temperature: 50-55 °C.
- 1 mark: Equation: C6H6 + HNO3 -> C6H5NO2 + H2O.

Part (b):
- 1 mark: Reagents: Tin (Sn) and concentrated HCl, followed by aqueous NaOH.
- 1 mark: Condition: Reflux / heat.

Part (c)(i):
- 1 mark: Sodium nitrite (NaNO2) + dilute HCl.

Part (c)(ii):
- 1 mark: Temp between 0 °C and 10 °C.
- 1 mark: Above 10 °C, benzenediazonium ion decomposes / phenol and N2 gas form.
- 1 mark: Below 0 °C, reaction is too slow.

Part (d):
- 2 marks: Correct structure showing the azo bridge (-N=N-) linking two benzene rings, with -OH at the 4-position of one ring. (1 mark for -N=N- bridge; 1 mark for correct para arrangement of -OH).

(a) Order: Chlorobenzene < 1-chlorobutane < Ethanoyl chloride.

Explanation:
- Chlorobenzene is the least reactive because the lone pair of electrons on the chlorine atom overlaps with the pi-delocalised 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 electron-rich pi cloud repels attacking nucleophiles.
- 1-chlorobutane has a simple polar C-Cl bond with no resonance stabilization, allowing typical nucleophilic substitution to occur upon heating.
- Ethanoyl chloride is the most reactive because the carbonyl carbon is bonded to both a highly electronegative oxygen atom and a chlorine atom. This leaves the carbonyl carbon highly electron-deficient (highly delta-positive), making it extremely susceptible to nucleophilic attack even at room temperature.

(b)
(i) \(\text{CH}_3\text{COCl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{HCl}\)
(ii) \(\text{CH}_3\text{COCl} + \text{C}_2\text{H}_5\text{NH}_2 \rightarrow \text{CH}_3\text{CONHC}_2\text{H}_5 + \text{HCl}\)

(c) The mechanism is \(\text{S}_\text{N}2\):
- The nucleophile \(\text{OH}^-\)`s lone pair attacks the electron-deficient carbon atom attached to the chlorine from the opposite side.
- A transition state is formed with partial bonds to both OH and Cl (indicated by dashed lines and square brackets with a negative charge).
- The C-Cl bond breaks, and Cl^- leaves as a chloride ion, yielding butan-1-ol.

Part (a):
- 1 mark: Correct order: chlorobenzene < 1-chlorobutane < ethanoyl chloride.
- 1 mark: Chlorobenzene has lone pair overlap with the pi ring system / C-Cl has partial double bond character.
- 1 mark: Therefore, C-Cl bond in chlorobenzene is very strong / resists nucleophilic attack.
- 1 mark: Ethanoyl chloride has a highly delta-positive carbonyl carbon due to O and Cl.
- 1 mark: This makes it highly vulnerable to nucleophilic attack.
- 1 mark: 1-chlorobutane is moderately reactive with standard polar C-Cl bond.

Part (b):
- 1 mark: Correct hydrolysis equation: CH3COCl + H2O -> CH3COOH + HCl.
- 2 marks: Correct equation for amide formation: CH3COCl + C2H5NH2 -> CH3CONHC2H5 + HCl (1 mark for correct structures; 1 mark for balanced equation).

Part (c):
- 1 mark: States Sn2 mechanism.
- 1 mark: Describes or draws transition state with partial C-OH and C-Cl bonds and correct arrow pointing from lone pair on OH^- to C.

(a) The methyl group is an electron-donating group due to the positive inductive effect (+I effect). This increases the electron density of the delocalised pi system of the benzene ring, making it a stronger nucleophile and more attractive to electrophiles. It also stabilizes the carbocation intermediate formed during the reaction.

(b) Mechanism:
1. Generation of the electrophile:
\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\)
(or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\))

2. Electrophilic attack:
A pair of electrons from the benzene ring attacks the \(\text{NO}_2^+\) electrophile at the para (4-) position, forming a dative covalent bond and creating a positively charged arenium intermediate (Wheland intermediate) with a broken pi system.

3. Deprotonation and regeneration of catalyst:
The C-H bond at position 4 breaks, returning its pair of electrons to the ring to restore the aromatic stability. The hydrogen ion (\(\text{H}^+\)) reacts with \(\text{HSO}_4^-\) to regenerate the \(\text{H}_2\text{SO}_4\) catalyst:
\(\text{H}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4\)

(c) Reagents: Alkaline potassium manganate(VII), \(\text{KMnO}_4(\text{aq})\), followed by acidification with dilute sulfuric acid, \(\text{H}_2\text{SO}_4(\text{aq})\).
Conditions: Heat under reflux.

Part (a):
- 1 mark: Methyl group is electron-donating / has positive inductive (+I) effect.
- 1 mark: Increases the electron density of the benzene ring.
- 1 mark: Attracts electrophiles more strongly / stabilizes the carbocation intermediate.

Part (b):
- 1 mark: Correct equation for electrophile (NO2^+) generation.
- 1 mark: Curly arrow from the benzene ring of methylbenzene to the nitrogen of NO2^+.
- 1 mark: Correct structure of the intermediate showing a horseshoe positive ring open toward C-4.
- 1 mark: Curly arrow from the C-H bond back into the ring.
- 1 mark: Correct final structure of 4-nitromethylbenzene.
- 1 mark: Correct equation for the regeneration of the H2SO4 catalyst.

Part (c):
- 1 mark: Reagents: Alkaline KMnO4 followed by dilute acid (H2SO4 or HCl).
- 1 mark: Condition: Heat under reflux.

(a) The standard electrode potential, \(E^\ominus\), is the electromotive force (emf) or voltage of a half-cell coupled to a standard hydrogen electrode (SHE), measured under standard conditions: a temperature of \(298\text{ K}\), pressure of \(1\text{ bar}\) (or \(1\text{ atm}\) / \(100\text{ kPa}\)) for any gaseous reactants/products, and a concentration of \(1.00\text{ mol dm}^{-3}\) for all solutions.

(b) The standard hydrogen electrode consists of:
- Hydrogen gas at a pressure of \(1\text{ bar}\) (or \(1\text{ atm}\) / \(100\text{ kPa}\))
- An aqueous solution of \(\text{H}^+\) ions with a concentration of \(1.00\text{ mol dm}^{-3}\) (e.g., \(1.0\text{ mol dm}^{-3}\ \text{HCl}\) or \(0.5\text{ mol dm}^{-3}\ \text{H}_2\text{SO}_4\))
- An inert platinum electrode (typically coated in finely divided platinum black)
- Maintained at a temperature of \(298\text{ K}\) (or \(25^\circ\text{C}\))

(c) The silver electrode has a more positive standard reduction potential (\(+0.80\text{ V}\)) than the copper electrode (\(+0.34\text{ V}\)), so \(\text{Ag}^+\) is reduced and \(\text{Cu}\) is oxidised. This yields the overall equation:
\(\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)\)
The standard cell potential is:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = 0.80 - 0.34 = +0.46\text{ V}\)

(d)(i) Since the copper electrode acts as the positive terminal (where reduction occurs), the reduction potential of the copper half-cell is higher than that of the non-standard silver half-cell.
\(E_{\text{cell}} = E_{\text{Cu}^{2+}/\text{Cu}} - E_{\text{Ag}^+/\text{Ag}}\)
Since the copper half-cell is standard, \(E_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}\).
\(0.12\text{ V} = 0.34\text{ V} - E_{\text{Ag}^+/\text{Ag}}\)
\(E_{\text{Ag}^+/\text{Ag}} = 0.34 - 0.12 = +0.22\text{ V}\)

(d)(ii) Using the Nernst equation for the silver electrode (where \(z = 1\) for \(\text{Ag}^+ + e^- \rightleftharpoons \text{Ag}\)):
\(E_{\text{Ag}^+/\text{Ag}} = E^\ominus_{\text{Ag}^+/\text{Ag}} + 0.059 \log [\text{Ag}^+(aq)]\)
\(+0.22 = +0.80 + 0.059 \log [\text{Ag}^+(aq)]\)
\(-0.58 = 0.059 \log [\text{Ag}^+(aq)]\)
\(\log [\text{Ag}^+(aq)] = -9.8305\)
\([\text{Ag}^+(aq)] = 10^{-9.8305} = 1.48 \times 10^{-10}\text{ mol dm}^{-3}\) (or \(1.5 \times 10^{-10}\text{ mol dm}^{-3}\)).

(a)
M1: Electrode potential / voltage of a half-cell connected to a standard hydrogen electrode (SHE). [1]
M2: Measured under standard conditions: temperature of \(298\text{ K}\), solution concentration of \(1.00\text{ mol dm}^{-3}\), and gas pressure of \(1\text{ bar}\) (or \(1\text{ atm}\) / \(100\text{ kPa}\)). [1]

(b)
M1: \(\text{H}_2\) gas at \(1\text{ atm}\) / \(1\text{ bar}\) / \(100\text{ kPa}\) AND \(\text{H}^+(aq)\) ions at \(1.00\text{ mol dm}^{-3}\). [1]
M2: Platinum (Pt) electrode at \(298\text{ K}\) / \(25^\circ\text{C}\). [1]

(c)
M1: Correct overall ionic equation: \(\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)\) (ignore state symbols). [1]
M2: Correct \(E^\ominus_{\text{cell}} = +0.46\text{ V}\). [1]

(d)(i)
M1: Correct calculation showing \(E_{\text{Ag}^+/\text{Ag}} = +0.22\text{ V}\) (allow 1.11 marks for correct value; award 0.5 marks for \(-0.22\text{ V}\) or just \(0.22\)). [1.11]

(d)(ii)
M1: Correct substitution: \(+0.22 = +0.80 + 0.059 \log [\text{Ag}^+]\). [1]
M2: Correct simplification to find \(\log [\text{Ag}^+] = -9.83\) (or \(-9.8\)). [1]
M3: Correct value: \(1.48 \times 10^{-10}\) (or \(1.5 \times 10^{-10}\)). [1]
M4: Correct units: \(\text{mol dm}^{-3}\) (or \(\text{M}\)). [1]

Part (a):
Using \(1/T\) and \(\ln(k)\):
Exp 1: \(1/298.2 = 0.003353 \text{ K}^{-1}\), \(\ln(4.20 \times 10^{-5}) = -10.08\)
Exp 2: \(1/303.2 = 0.003298 \text{ K}^{-1}\), \(\ln(8.11 \times 10^{-5}) = -9.42\)
Exp 3: \(1/308.2 = 0.003245 \text{ K}^{-1}\), \(\ln(2.10 \times 10^{-4}) = -8.47\)
Exp 4: \(1/313.2 = 0.003193 \text{ K}^{-1}\), \(\ln(2.89 \times 10^{-4}) = -8.15\)
Exp 5: \(1/318.2 = 0.003143 \text{ K}^{-1}\), \(\ln(5.26 \times 10^{-4}) = -7.55\)
Exp 6: \(1/323.2 = 0.003094 \text{ K}^{-1}\), \(\ln(9.40 \times 10^{-4}) = -6.97\)

Part (b):
Comparing the trends, Exp 3 has \(\ln(k) = -8.47\) which is more positive than expected (expected is around \(-8.78\)). The anomalous experiment is Experiment 3. A possible experimental error is that the temperature of the water bath was higher than the recorded \(308.2 \text{ K}\), or the flask was not allowed to reach thermal equilibrium before timing began.

Part (c):
According to the Arrhenius equation: \(\ln(k) = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln(A)\).
The gradient \(m\) of the line of best fit is equal to \(-\frac{E_a}{R}\). Therefore, \(E_a = -\text{gradient} \times R\).

Part (d):
Using non-anomalous points Exp 1 and Exp 6:
\(\text{Gradient} = \frac{-6.97 - (-10.08)}{0.003094 - 0.003353} = \frac{3.11}{-0.000259} = -12008 \text{ K}\) (accept \(-11800\) to \(-12200 \text{ K}\)).
\(E_a = -(-12008) \times 8.314 = 99834 \text{ J mol}^{-1} = 99.8 \text{ kJ mol}^{-1}\) (accept \(98.0\) to \(101.4 \text{ kJ mol}^{-1}\)).

Part (e):
\% \text{ Uncertainty} = \frac{0.5}{24.0} \times 100\% = 2.08\%\).

Part (f):
At \(353 \text{ K}\), the reaction rate would be extremely high, meaning the nitrogen gas would be evolved too rapidly to be recorded accurately at specified time intervals, leading to large human/timing errors.

Part (a) [3 marks]:
- 1 mark for all six 1/T values correct (0.003353, 0.003298, 0.003245, 0.003193, 0.003143, 0.003094).
- 1 mark for all six ln(k) values correct (-10.08, -9.42, -8.47, -8.15, -7.55, -6.97).
- 1 mark for correct significant figures (1/T to 4 decimal places, ln(k) to 2 decimal places).

Part (b) [2 marks]:
- 1 mark for identifying Experiment 3 as the anomaly.
- 1 mark for stating that the actual temperature was higher than recorded / inadequate temperature control.

Part (c) [2 marks]:
- 1 mark for stating that gradient = -Ea/R.
- 1 mark for stating Ea = -gradient x R.

Part (d) [5 marks]:
- 1 mark for selecting two points from the line of best fit (excluding the anomalous point) and calculating the change in y and change in x.
- 1 mark for calculating the gradient in the range -11800 to -12200 K.
- 1 mark for multiplying the gradient by -8.314 to obtain Ea in J/mol.
- 1 mark for dividing by 1000 to convert to kJ/mol.
- 1 mark for the final answer to 3 significant figures with units (98.0 to 101.4 kJ/mol).

Part (e) [1 mark]:
- 1 mark for 2.08% (allow 2.1%).

Part (f) [2 marks]:
- 1 mark for stating that the rate is too fast / reaction occurs too quickly.
- 1 mark for linking this to increased error in volume/time measurement.

Part (a):
Calculating \(\log_{10}(C)\):
Exp 1: \(\log_{10}(0.0010) = -3.00\)
Exp 2: \(\log_{10}(0.0050) = -2.30\)
Exp 3: \(\log_{10}(0.0250) = -1.60\)
Exp 4: \(\log_{10}(0.100) = -1.00\)
Exp 5: \(\log_{10}(0.500) = -0.30\)
Exp 6: \(\log_{10}(1.00) = 0.00\)

Part (b):
Independent variable: Concentration of \(M^{n+}\) (or \(\log_{10}(C)\)).
Dependent variable: Cell potential (\(E_{\text{cell}}\)).

Part (c):
Experiment 3 is anomalous. At \(\log_{10}(C) = -1.60\), \(E_{\text{cell}}\) is \(0.320 \text{ V}\), which is higher than the trend predicts (expected is around \(0.293 \text{ V}\)). A practical laboratory error that would cause this is if the electrode or the beaker was not properly rinsed and had residue of a more concentrated solution of \(M^{n+}\) remaining.

Part (d):
- \(E^{\theta}\) is the y-intercept of the graph of \(E_{\text{cell}}\) against \(\log_{10}(C)\) (since \(\log_{10}(C) = 0\) at the y-intercept).
- The gradient of the line is equal to \(\frac{0.059}{n}\). Therefore, \(n = \frac{0.059}{\text{gradient}}\).

Part (e):
Using the non-anomalous points, e.g., Exp 1 \((-3.00, 0.251)\) and Exp 6 \((0.00, 0.340)\):
\(\text{Gradient} = \frac{0.340 - 0.251}{0 - (-3.00)} = \frac{0.089}{3.00} = 0.0297 \text{ V}\) (allow \(0.0290\) to \(0.0305 \text{ V}\)).
\(n = \frac{0.059}{0.0297} = 1.99 \approx 2\).
\(E^{\theta} = 0.340 \text{ V}\) (the y-intercept).

Part (f):
At the SHE, the half-reaction is \(\text{H}_2(\text{g}) \rightleftharpoons 2\text{H}^+(\text{aq}) + 2\text{e}^-\). Lowering the pressure of \(\text{H}_2\) shifts this equilibrium to the left, making the SHE electrode potential slightly more positive (above \(0.00 \text{ V}\)). Since \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\), an increase in the anode potential (SHE) results in a slightly smaller (less positive) cell potential, \(E_{\text{cell}}\).

Part (a) [2 marks]:
- 1 mark for all log10(C) values calculated correctly.
- 1 mark for quoting all values to 2 decimal places (-3.00, -2.30, -1.60, -1.00, -0.30, 0.00).

Part (b) [2 marks]:
- 1 mark for correctly identifying log10(C) or concentration as the independent variable.
- 1 mark for identifying cell potential (E_cell) as the dependent variable.

Part (c) [2 marks]:
- 1 mark for identifying Experiment 3 as the anomalous point.
- 1 mark for suggesting contamination of the beaker/electrode with a more concentrated solution of M^n+.

Part (d) [3 marks]:
- 1 mark for stating that standard electrode potential (E^theta) is the y-intercept / value of E_cell at log10(C) = 0.
- 1 mark for stating that gradient = 0.059 / n.
- 1 mark for rearranging to give n = 0.059 / gradient.

Part (e) [4 marks]:
- 1 mark for calculating the gradient to 3 significant figures using two points on the line (excluding Experiment 3).
- 1 mark for obtaining a gradient in the range 0.0290 to 0.0305 V.
- 1 mark for calculating n = 2 (accept any value from 1.9 to 2.1, rounded to 2).
- 1 mark for identifying E^theta as 0.340 V (allow 0.338 to 0.342 V).

Part (f) [2 marks]:
- 1 mark for stating that a lower pressure of H2 shifts the equilibrium to the left, raising the electrode potential of the SHE (making it > 0 V).
- 1 mark for explaining that this decreases the cell potential (E_cell).