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According to the Arrhenius equation: \(\ln(k) = -\frac{E_a}{RT} + \ln(A)\).

A plot of \(\ln(k)\) against \(1/T\) yields a straight line with a gradient equal to \(-\frac{E_a}{R}\).

Therefore:
\(\text{Gradient} = -\frac{E_a}{R} = -1.20 \times 10^{4}\text{ K}\)

\(E_a = 1.20 \times 10^{4} \times R = 1.20 \times 10^{4} \times 8.31 = 9.97 \times 10^{4}\text{ J mol}^{-1}\)

To convert this to \(\text{kJ mol}^{-1}\), divide by 1000:
\(E_a = +99.7\text{ kJ mol}^{-1}\).

1 mark for selecting the correct option B. Reject options with negative values, as activation energy is always positive.

In transition metal complexes, ligands split the d-orbitals into lower and higher energy levels. Transition metal complexes are colored because they absorb specific wavelengths of visible light to promote an electron from a lower energy d-orbital to a higher energy d-orbital (d-to-d transition). For a \(d^{10}\) complex, all d-orbitals are fully occupied, so there is no vacancy in the higher energy d-orbitals to accommodate a promoted electron. Consequently, no d-to-d transitions can occur, and the complex is colorless.

1 mark for selecting the correct option A.

First, determine the moles of reactants:
- \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\)
- \(n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 0.0400\text{ mol dm}^{-3} = 0.00200\text{ mol}\)

Sodium hydroxide reacts completely with propanoic acid:
\(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\)

After reaction:
- \(n(\text{CH}_3\text{CH}_2\text{COO}^-) = 0.00200\text{ mol}\)
- \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.00500 - 0.00200 = 0.00300\text{ mol}\)

Using the buffer equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)

\(\text{p}K_a = -\log_{10}(1.30 \times 10^{-5}) = 4.89\)

\(\text{pH} = 4.89 + \log_{10}\left(\frac{0.00200}{0.00300}\right) = 4.89 - 0.18 = 4.71\).

1 mark for selecting the correct option A.

A redox reaction is thermodynamically feasible if the standard cell potential, \(E^\theta_{\text{cell}}\), is positive (\(E^\theta_{\text{cell}} > 0\)).

For option B: Bromine oxidizing iron(II) ions to iron(III) ions.
- Reduction: \(\text{Br}_2(\text{aq}) + 2\text{e}^- \rightarrow 2\text{Br}^-(\text{aq})\) with \(E^\theta = +1.09\text{ V}\)
- Oxidation: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\) with \(E^\theta = -0.77\text{ V}\)

\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +1.09\text{ V} - (+0.77\text{ V}) = +0.32\text{ V}\).

Since \(E^\theta_{\text{cell}} = +0.32\text{ V} > 0\), this reaction is thermodynamically feasible.

1 mark for selecting the correct option B.

Calculate the mass of water lost:
\(m(\text{H}_2\text{O}) = 4.83\text{ g} - 3.75\text{ g} = 1.08\text{ g}\)

Calculate the amount (in moles) of anhydrous copper(II) nitrate and water:
- \(n(\text{Cu(NO}_3)_2) = \frac{3.75\text{ g}}{187.5\text{ g mol}^{-1}} = 0.0200\text{ mol}\)
- \(n(\text{H}_2\text{O}) = \frac{1.08\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0600\text{ mol}\)

Determine the molar ratio:
\(\text{Ratio } (x) = \frac{n(\text{H}_2\text{O})}{n(\text{Cu(NO}_3)_2)} = \frac{0.0600}{0.0200} = 3\).

Refining the formula, we find that \(x = 3\).

1 mark for selecting the correct option B.

Calculate the energy required to break bonds in the reactants (propene and hydrogen):
- Bonds broken:
- 1 \(\text{C}=\text{C}\) bond = \(1 \times 612 = 612\text{ kJ mol}^{-1}\)
- 1 \(\text{H}-\text{H}\) bond = \(1 \times 436 = 436\text{ kJ mol}^{-1}\)
- (Note: We can ignore the other bonds as they remain unchanged, or include them to get the same difference.)
- Total energy input = \(612 + 436 = 1048\text{ kJ mol}^{-1}\)

Calculate the energy released when bonds are formed in the product (propane):
- Bonds formed:
- 1 \(\text{C}-\text{C}\) bond = \(1 \times 347 = 347\text{ kJ mol}^{-1}\)
- 2 \(\text{C}-\text{H}\) bonds = \(2 \times 413 = 826\text{ kJ mol}^{-1}\)
- Total energy output = \(347 + 826 = 1173\text{ kJ mol}^{-1}\)

Calculate the overall enthalpy change:
\(\Delta_r H^\theta = \text{Energy input} - \text{Energy output} = 1048 - 1173 = -125\text{ kJ mol}^{-1}\).

1 mark for selecting the correct option A.

2-chloro-2-methylpropane is a tertiary haloalkane. Tertiary haloalkanes undergo nucleophilic substitution reactions predominantly via the \(\text{S}_\text{N}1\) (substitution nucleophilic unimolecular) mechanism. The first step involves the slow loss of the chloride leaving group to form a stable tertiary carbocation intermediate, which is stabilized by the electron-donating inductive effect of three methyl groups. In the second step, the hydroxide nucleophile rapidly attacks this carbocation to form the alcohol, 2-methylpropan-2-ol.

1 mark for selecting the correct option A.

Atom economy is calculated as:
\(\text{Atom Economy} = \frac{\text{Molecular mass of desired product}}{\text{Total molecular mass of all reactants}} \times 100\)

Desired product: \(\text{C}_6\text{H}_5\text{CH}=\text{CHCHO}\) (\(M_r = 132.0\))
Reactants: \(\text{C}_6\text{H}_5\text{CHO}\) and \(\text{CH}_3\text{CHO}\) (\(\text{Total } M_r = 106.0 + 44.0 = 150.0\))

\(\text{Atom Economy} = \frac{132.0}{150.0} \times 100 = 88.0\%\).

1 mark for selecting the correct option C.

From the first experimental observation, the reaction is first-order with respect to A: \(\text{rate} \propto [\text{A}]^1\).

From the second experimental observation, the reaction is second-order with respect to B: \(\text{rate} \propto [\text{B}]^2\).

Therefore, the rate equation is: \(\text{rate} = k[\text{A}][\text{B}]^2\).

Rearranging for the rate constant \(k\):

\(k = \frac{\text{rate}}{[\text{A}][\text{B}]^2}\)

Substituting the units:

\(k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{(\text{mol}\ \text{dm}^{-3}) \times (\text{mol}\ \text{dm}^{-3})^2} = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{\text{mol}^3\ \text{dm}^{-9}} = \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}\).

1 mark for the correct option (B).

Reject all other options.

The relationship between energy change and wavelength is given by:

\(\Delta E = \frac{hc}{\lambda}\)

Given:

\(\lambda = 650\ \text{nm} = 650 \times 10^{-9}\ \text{m}\)

\(h = 6.63 \times 10^{-34}\ \text{J s}\)

\(c = 3.00 \times 10^8\ \text{m s}^{-1}\)

Substituting these values:

\(\Delta E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{650 \times 10^{-9}} = 3.06 \times 10^{-19}\ \text{J}\).

1 mark for the correct calculation and selection of option C.

Reject B (due to incorrect conversion of nanometres) and other options.

For an acidic buffer solution, the hydrogen ion concentration can be calculated using:

\([\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{conjugate base}]}\)

\([\text{H}^+] = 1.7 \times 10^{-5} \times \frac{0.15}{0.10} = 2.55 \times 10^{-5}\ \text{mol dm}^{-3}\)

Now, calculate the \(\text{pH}\):

\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2.55 \times 10^{-5}) = 4.59\).

1 mark for calculating \(\text{pH} = 4.59\) and selecting option B.

Reject D (which is the pH obtained if the acid/salt ratio is inverted).

A redox reaction is thermodynamically feasible under standard conditions if the cell potential, \(E^\theta_{\text{cell}}\), is positive.

Let us calculate \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\) for each option:

- Option A: \(\text{Cl}_2\) is reduced (\(+1.36\ \text{V}\)) and \(\text{Fe}^{2+}\) is oxidised (\(+0.77\ \text{V}\)). \(E^\theta_{\text{cell}} = +1.36 - (+0.77) = +0.59\ \text{V}\) (Feasible).

- Option B: \(\text{Br}_2\) is reduced (\(+1.09\ \text{V}\)) and \(\text{I}^-\) is oxidised (\(+0.54\ \text{V}\)). \(E^\theta_{\text{cell}} = +1.09 - (+0.54) = +0.55\ \text{V}\) (Feasible).

- Option C: \(\text{Fe}^{3+}\) is reduced (\(+0.77\ \text{V}\)) and \(\text{I}^-\) is oxidised (\(+0.54\ \text{V}\)). \(E^\theta_{\text{cell}} = +0.77 - (+0.54) = +0.23\ \text{V}\) (Feasible).

- Option D: \(\text{I}_2\) is reduced (\(+0.54\ \text{V}\)) and \(\text{Fe}^{2+}\) is oxidised (\(+0.77\ \text{V}\)). \(E^\theta_{\text{cell}} = +0.54 - (+0.77) = -0.23\ \text{V}\) (NOT feasible).

1 mark for identifying the correct reaction (D) which has a negative standard cell potential.

Reject all other options.

First, find the mass of water lost:

\(\text{Mass of water} = 3.15\ \text{g} - 2.68\ \text{g} = 0.47\ \text{g}\)

Calculate the amount in moles of anhydrous \(\text{BaCl}_2\) and \(\text{H}_2\text{O}\):

\(M_r(\text{BaCl}_2) = 137.3 + (2 \times 35.5) = 208.3\ \text{g mol}^{-1}\)

\(n(\text{BaCl}_2) = \frac{2.68\ \text{g}}{208.3\ \text{g mol}^{-1}} \approx 0.01287\ \text{mol}\)

\(M_r(\text{H}_2\text{O}) = (2 \times 1.0) + 16.0 = 18.0\ \text{g mol}^{-1}\)

\(n(\text{H}_2\text{O}) = \frac{0.47\ \text{g}}{18.0\ \text{g mol}^{-1}} \approx 0.02611\ \text{mol}\)

Find the molar ratio \(x\):

\(x = \frac{n(\text{H}_2\text{O})}{n(\text{BaCl}_2)} = \frac{0.02611}{0.01287} \approx 2\)

1 mark for the correct determination of \(x = 2\) and selecting option B.

Reject all other options.

The enthalpy change of reaction can be calculated using bond enthalpies of reactants (bonds broken) and products (bonds formed):

\(\Delta H = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}\)

Bonds broken:
- 1 \(\text{C}=\text{C}\) bond = \(612\ \text{kJ mol}^{-1}\)
- 1 \(\text{H}-\text{H}\) bond = \(436\ \text{kJ mol}^{-1}\)
Total energy to break bonds = \(612 + 436 = +1048\ \text{kJ mol}^{-1}\)
(We can ignore other bonds that do not change during the reaction).

Bonds formed:
- 1 \(\text{C}-\text{C}\) bond = \(347\ \text{kJ mol}^{-1}\)
- 2 \(\text{C}-\text{H}\) bonds = \(2 \times 413 = 826\ \text{kJ mol}^{-1}\)
Total energy released in forming bonds = \(347 + 826 = 1173\ \text{kJ mol}^{-1}\)

\(\Delta H = 1048 - 1173 = -125\ \text{kJ mol}^{-1}\).

1 mark for the correct calculation yielding \(-125\ \text{kJ mol}^{-1}\) and selecting option A.

Reject C (sign error), B (forgetting the C-C bond formation), and D.

2-bromo-2-methylpropane is a tertiary halogenoalkane. Tertiary halogenoalkanes undergo alkaline hydrolysis via an \(\text{S}_{\text{N}}1\) mechanism. This is a two-step process where the first, slow (rate-determining) step is the heterolytic fission of the \(\text{C}-\text{Br}\) bond to form a stable tertiary carbocation intermediate, \(\text{(CH}_3\text{)}_3\text{C}^+\).

Therefore, option C is correct. Options A, B, and D describe an \(\text{S}_{\text{N}}2\) mechanism, which is characteristic of primary halogenoalkanes.

1 mark for selecting option C.

Reject options A, B, and D which describe \(\text{S}_{\text{N}}2\) pathways.

The atom economy is calculated as:

\(\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100\%\)

From the balanced equation:

Desired product = \(2 \times \text{C}_4\text{H}_2\text{O}_3\)
Mass of desired product = \(2 \times 98.0 = 196.0\ \text{g}\)

Reactants = \(2 \times \text{C}_4\text{H}_{10} + 7 \times \text{O}_2\)
Total mass of reactants = \((2 \times 58.0) + (7 \times 32.0) = 116.0 + 224.0 = 340.0\ \text{g}\)

Calculate atom economy:

\(\text{Atom economy} = \frac{196.0}{340.0} \times 100\% = 57.6\%\).

1 mark for the correct calculation of 57.6% and selecting option B.

Reject A (which is the atom economy of the alternative benzene-based route).

The rate equation can be rearranged to solve for the rate constant: \(k = \text{Rate} / ([\text{NO}]^2[\text{H}_2])\). Substituting the corresponding units gives: \(k = (\text{mol dm}^{-3}\text{ s}^{-1}) / ((\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}) = (\text{mol dm}^{-3}\text{ s}^{-1}) / (\text{mol}^3\text{ dm}^{-9}) = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for correct option B. Reject all other options.

Conjugated systems feature alternating single and double bonds, which allow the delocalisation of \(\pi\) electrons. As the length of the conjugated system increases, the energy gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) decreases. This smaller energy gap means the molecule absorbs light at a longer wavelength (lower energy). Octa-1,3,5,7-tetraene has four alternating double bonds, making it the most highly conjugated molecule listed.

1 mark for correct option D. Reject all other options.

Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}([\text{A}^-] / [\text{HA}])\). First, calculate \(\text{p}K_a = -\log_{10}(1.30 \times 10^{-5}) = 4.886\). Then substitute the ratio into the equation: \(\text{pH} = 4.886 + \log_{10}(3.00) = 4.886 + 0.477 = 5.36\) (to 3 significant figures).

1 mark for correct option C. Reject all other options.

For a redox reaction to be thermodynamically feasible under standard conditions, the cell potential must be positive (the species with the more positive \(E^\ominus\) undergoes reduction). Comparing \(\text{Fe}^{3+}/\text{Fe}^{2+}\) (\(+0.77\text{ V}\)) and \(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}\) (\(+2.01\text{ V}\)), the peroxodisulfate half-cell has the more positive potential and will undergo reduction, while the iron half-cell will run in reverse (oxidation). Therefore, \(\text{Fe}^{2+}(\text{aq})\) acts as a reducing agent and reduces \(\text{S}_2\text{O}_8^{2-}(\text{aq})\) to \(\text{SO}_4^{2-}(\text{aq})\).

1 mark for correct option B. Reject all other options.

First, calculate the amount of the precipitate: \(n(\text{precipitate}) = 1.44\text{ g} / 288.7\text{ g mol}^{-1} = 4.988 \times 10^{-3}\text{ mol}\). Since each mole of the complex contains one mole of nickel ions, the amount of nickel is \(4.988 \times 10^{-3}\text{ mol}\). Next, find the mass of nickel: \(m(\text{Ni}) = 4.988 \times 10^{-3}\text{ mol} \times 58.7\text{ g mol}^{-1} = 0.2928\text{ g}\). Finally, calculate the percentage by mass: \(\%\text{Ni} = (0.2928\text{ g} / 2.50\text{ g}) \times 100 = 11.7\%\).

1 mark for correct option B. Reject all other options.

Energy input (bonds broken in reactants): Propan-1-ol has 2 \(\text{C-C}\) bonds, 7 \(\text{C-H}\) bonds, 1 \(\text{C-O}\) bond, and 1 \(\text{O-H}\) bond. Energy = \((2 \times 347) + (7 \times 413) + 358 + 463 = 4406\text{ kJ mol}^{-1}\). For oxygen: \(4.5 \times 498 = 2241\text{ kJ mol}^{-1}\). Total input = \(4406 + 2241 = 6647\text{ kJ mol}^{-1}\). Energy output (bonds formed in products): 3 \(\text{CO}_2\) contains 6 \(\text{C=O}\) bonds: \(6 \times 805 = 4830\text{ kJ mol}^{-1}\). 4 \(\text{H}_2\text{O}\) contains 8 \(\text{O-H}\) bonds: \(8 \times 463 = 3704\text{ kJ mol}^{-1}\). Total output = \(4830 + 3704 = 8534\text{ kJ mol}^{-1}\). Enthalpy change: \(\Delta H = \text{input} - \text{output} = 6647 - 8534 = -1887\text{ kJ mol}^{-1}\).

1 mark for correct option B. Reject all other options.

In the preparation of aspirin, salicylic acid undergoes esterification of its phenolic hydroxyl (\(-\text{OH}\)) group by reacting with ethanoic anhydride. The lone pair of electrons on the phenolic oxygen acts as a nucleophile and attacks the carbonyl carbon of the ethanoic anhydride. This follows a nucleophilic addition-elimination mechanism. The carboxylic acid group on the salicylic acid remains unreacted.

1 mark for correct option C. Reject all other options.

Atom economy is calculated as: \((\text{Molar mass of desired product} / \text{Total molar mass of reactants}) \times 100\). Desired product: \(1 \times \text{N}_2\text{H}_4 = 32.0\text{ g mol}^{-1}\). Reactants: \(2 \times \text{NH}_3 + 1 \times \text{H}_2\text{O}_2 = (2 \times 17.0) + 34.0 = 68.0\text{ g mol}^{-1}\). \(\text{Atom economy} = (32.0 / 68.0) \times 100 = 47.1\%\).

1 mark for correct option B. Reject all other options.

Let the initial rate be \(\text{rate}_1 = k[NO]^2[O_2]\). When \([NO]\) is tripled, the new concentration is \(3[NO]\). When \([O_2]\) is halved, the new concentration is \(0.5[O_2]\). Substituting these into the rate equation gives: \(\text{rate}_2 = k(3[NO])^2(0.5[O_2]) = k(9[NO]^2)(0.5[O_2]) = 4.5 \times k[NO]^2[O_2] = 4.5 \times \text{rate}_1\). Therefore, the rate increases by a factor of 4.5.

1 mark for the correct choice B.

The sulfonate groups on the acid dye are negatively charged (\(-SO_3^-\)). To form strong ionic interactions (salt linkages), the wool fiber must have positive charges. Under highly acidic conditions, the amino groups (\(-NH_2\)) of the polypeptide chains in wool are protonated to form positively charged ammonium groups (\(-NH_3^+\)). This creates a strong electrostatic attraction (ionic bond) between the dye and the fiber.

1 mark for the correct choice B.

Using the Henderson-Hasselbalch equation: \(pH = pK_a + \log_{10}([A^-] / [HA])\). First, calculate \(pK_a\): \(pK_a = -\log_{10}(1.8 \times 10^{-5}) \approx 4.74\). Next, calculate the ratio term: \(\log_{10}(0.15 / 0.05) = \log_{10}(3.0) \approx 0.48\). Now add them together: \(pH = 4.74 + 0.48 = 5.22\).

1 mark for the correct choice C.

For an oxidizing agent to oxidize \(I^-\). to \(I_2\) under standard conditions, the standard reduction potential of the oxidizing agent must be greater than \(+0.54\text{ V}\). To fail to oxidize \(Br^-\). to \(Br_2\), its standard reduction potential must be less than \(+1.09\text{ V}\). The species must also be on the reactant side of a reduction equation. \(Fe^{3+}(aq)\) meets these criteria because \(+0.54\text{ V} < +0.77\text{ V} < +1.09\text{ V}\).

1 mark for the correct choice B.

1. Calculate the moles of \(MnO_4^-\). used in the titration: \(n(MnO_4^-) = 0.0200 \text{ mol dm}^{-3} \times (22.50 / 1000) \text{ dm}^3 = 4.50 \times 10^{-4} \text{ mol}\). 2. The redox equation is: \(MnO_4^-(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\). 3. Find the moles of \(Fe^{2+}\). in the 25.0 cm\(^3\) aliquot: \(n(Fe^{2+}) = 5 \times 4.50 \times 10^{-4} \text{ mol} = 2.25 \times 10^{-3} \text{ mol}\). 4. Scale up to the 250 cm\(^3\) volumetric flask: \(n_{\text{total}}(Fe^{2+}) = 2.25 \times 10^{-3} \text{ mol} \times 10 = 2.25 \times 10^{-2} \text{ mol}\). 5. Calculate the mass of iron: \(m(Fe) = 2.25 \times 10^{-2} \text{ mol} \times 55.8 \text{ g mol}^{-1} = 1.2555 \text{ g}\). 6. Calculate percentage by mass: \(\% \text{ by mass} = (1.2555 \text{ g} / 5.00 \text{ g}) \times 100\% = 25.11\%\), which is approximately \(25.1\%\).

1 mark for the correct choice C.

2-bromo-2-methylpropane is a tertiary halogenoalkane, which reacts via the \(S_N1\) mechanism. The rate-determining step is the heterolytic fission of the C-Br bond to form a stable tertiary carbocation intermediate: \((CH_3)_3CBr \rightarrow (CH_3)_3C^+ + Br^-\). Since only the halogenoalkane molecule is involved in this slow step, the reaction is first-order overall (rate = \(k[(CH_3)_3CBr]\)). Hydroxide ions act as nucleophiles, not electrophiles, and react in the subsequent fast step.

1 mark for the correct choice B.

(a) A correct Maxwell-Boltzmann distribution shows:
- The curve for \(T_2\) having a peak that is lower and shifted to the right compared to \(T_1\).
- Both curves starting at the origin and asymptotic to the x-axis at high energies.
- Activation energy, \(E_a\), marked on the energy axis.
- An increase in temperature from \(T_1\) to \(T_2\) leads to a much larger fraction of molecules having energy greater than or equal to \(E_a\) (represented by the area under the curve to the right of \(E_a\)).
- This greatly increases the frequency of successful collisions per unit time, hence increasing the rate.

(b)
- Comparing Exp 1 and Exp 2: \([\text{NO}]\) doubles, \([\text{H}_2]\) is constant. Rate increases by \(\frac{1.44 \times 10^{-4}}{3.60 \times 10^{-5}} = 4\). Thus, the order with respect to \(\text{NO}\) is 2.
- Comparing Exp 1 and Exp 3: \([\text{H}_2]\) triples, \([\text{NO}]\) is constant. Rate increases by \(\frac{1.08 \times 10^{-4}}{3.60 \times 10^{-5}} = 3\). Thus, the order with respect to \([\text{H}_2]\) is 1.
- Rate Equation: \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\)
- Calculation of \(k\) using Exp 1 data:
\(k = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]} = \frac{3.60 \times 10^{-5}}{(1.20 \times 10^{-3})^2 \times (2.00 \times 10^{-3})} = 12500\)
- Units: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

(c) Proposed multi-step mechanism:
- Step 1 (Fast equilibrium): \(2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2\)
- Step 2 (Slow / Rate-determining step): \(\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}\)
- Step 3 (Fast): \(\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}\)
Overall equation: \(2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}\).
Since the reactants up to and including the slow step are \(2\text{NO}\) and \(1\text{H}_2\), this is consistent with the rate equation.

(d)
- Adsorption: Reactant molecules (\(\text{NO}\) and \(\text{H}_2\)) bond chemically to the active sites on the catalyst's solid surface.
- Bond weakening: The reactant bonds are weakened/broken, facilitating transition states and lowering the activation energy.
- Desorption: The product molecules (\(\text{N}_2\) and \(\text{H}_2\text{O}\)) break bonds with the catalyst surface and escape, freeing up the active sites for further reactant molecules.

(a) [5 marks total]
* 1 mark: Correct shape of Maxwell-Boltzmann curve at \(T_1\) (starts at origin, skewed right, does not touch x-axis at high energy).
* 1 mark: Curve for \(T_2\) has its peak lower and shifted to the right.
* 1 mark: \(E_a\) correctly located on the horizontal axis with shading of the area representing molecules with energy \(\ge E_a\).
* 1 mark: Explains that area under \(T_2\) curve to the right of \(E_a\) is significantly larger than under \(T_1\).
* 1 mark: States that a much larger proportion of collisions have energy \(\ge E_a\), leading to a higher frequency of successful collisions.

(b) [5 marks total]
* 1 mark: Deduce order with respect to \(\text{NO}\) is 2, with clear reasoning from data.
* 1 mark: Deduce order with respect to \(\text{H}_2\) is 1, with clear reasoning from data.
* 1 mark: Correct rate equation matching deductions.
* 1 mark: Correct value of \(k = 12500\) (allow ecf from incorrect rate equation).
* 1 mark: Correct units: \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) (allow ecf from incorrect rate equation).

(c) [3 marks total]
* 2 marks: Three reasonable steps that add up to the overall equation (1 mark for correct intermediate species; 1 mark for correct balancing).
* 1 mark: Correctly identifies the slow step / rate-determining step as the one involving \(2\text{NO}\) and \(1\text{H}_2\) (or intermediates formed from them).

(d) [3 marks total]
* 1 mark: Reactant gases are adsorbed onto active sites of the catalyst surface.
* 1 mark: Bonds in the reactant molecules are weakened (lowering the activation energy).
* 1 mark: Product molecules are desorbed from the surface.

(a) A conjugate acid-base pair consists of two species that differ by one proton (\(\text{H}^+\)).
In the reaction:
\(\text{CH}_3\text{CH}_2\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-(aq) + \text{H}_3\text{O}^+(aq)\)
- Pair 1: \(\text{CH}_3\text{CH}_2\text{COOH}\) (acid) and \(\text{CH}_3\text{CH}_2\text{COO}^-\) (conjugate base).
- Pair 2: \(\text{H}_2\text{O}\) (base) and \(\text{H}_3\text{O}^+\) (conjugate acid).

(b) \(K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)
Using assumptions:
1) \([\text{H}^+] \approx [\text{CH}_3\text{CH}_2\text{COO}^-]\) (negligible dissociation of water).
2) \([\text{CH}_3\text{CH}_2\text{COOH}]_{\text{eq}} \approx [\text{CH}_3\text{CH}_2\text{COOH}]_{\text{initial}}\) (negligible dissociation of the weak acid).

Therefore, \(K_a \approx \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]_{\text{initial}}}\)
\(1.35 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.150}\)
\([\text{H}^+]^2 = 2.025 \times 10^{-6}\)
\([\text{H}^+] = 1.423 \times 10^{-3}\ \text{mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.85\)

(c) (i)
\(\text{pH} = \text{p}K_a + \log_{10}\frac{[\text{A}^-]}{[\text{HA}]}\)
\(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\)
\(4.80 = 4.87 + \log_{10}\frac{[\text{propanoate}]}{[\text{propanoic acid}]}\)
\(\log_{10}\frac{[\text{propanoate}]}{[\text{propanoic acid}]} = -0.07\)
\(\frac{[\text{propanoate}]}{[\text{propanoic acid}]} = 10^{-0.07} = 0.851\) (accept \(0.85\))

(ii) When \(\text{H}^+\) is added, it is consumed by the conjugate base:
\(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
When \(\text{OH}^-\) is added, it is consumed by the weak acid:
\(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\)
Thus, the concentrations of \(\text{H}^+\) and \(\text{OH}^-\) remain almost unchanged, and the pH is minimised from changing.

(d) A sketch with:
- Initial pH at ~2.85.
- A flat buffer region centered around pH 4.87 (the \(\text{p}K_a\)).
- An equivalence point volume of exactly \(25.0\ \text{cm}^3\), occurring at a pH greater than 7 (around pH 8.8).
- A final pH leveling off around 12-13.

(a) [2 marks total]
* 1 mark: Define conjugate acid-base pair as species that differ by one \(\text{H}^+\).
* 1 mark: Correctly identify \(\text{CH}_3\text{CH}_2\text{COOH}/\text{CH}_3\text{CH}_2\text{COO}^-\) AND \(\text{H}_2\text{O}/\text{H}_3\text{O}^+\).

(b) [5 marks total]
* 1 mark: Formula for \(K_a\) or expression \([\text{H}^+] = \sqrt{K_a \times c}\).
* 1 mark: State assumption 1 (negligible ionization of water/all hydrogen ions come from acid).
* 1 mark: State assumption 2 (degree of dissociation of weak acid is very small / initial conc is same as equilibrium conc).
* 1 mark: Correct calculation of \([\text{H}^+] = 1.42 \times 10^{-3}\ \text{mol dm}^{-3}\).
* 1 mark: Correct pH value of \(2.85\) (must be 2 decimal places).

(c) [7 marks total]
- Part (i) [4 marks]:
* 1 mark: Correct calculation of \(\text{p}K_a = 4.87\).
* 1 mark: Rearranging the Henderson-Hasselbalch equation or using \(K_a\) expression correctly.
* 1 mark: Correct deduction that \(\log_{10}(\text{ratio}) = -0.07\).
* 1 mark: Correct ratio of \(0.85\) (or \(0.851\)).
- Part (ii) [3 marks]:
* 1 mark: Equation and description of reaction with added acid (propanoate reacting with \(\text{H}^+\)).
* 1 mark: Equation and description of reaction with added alkali (propanoic acid reacting with \(\text{OH}^-\)).
* 1 mark: Explaining that since added ions are consumed, the ratio changes very little and pH is maintained.

(d) [2 marks total]
* 1 mark: Curve shows weak acid-strong base characteristics (initial pH around 2.5-3.0, buffer region, vertical region at \(25.0\ \text{cm}^3\)).
* 1 mark: Correctly labels equivalence point in the basic region (pH > 7) and maximum buffer capacity area labeled in the flat section.

(a) Mechanism: Nucleophilic Addition-Elimination
1. Curly arrow from the lone pair on the nitrogen of \(\text{CH}_3\text{NH}_2\) to the carbonyl carbon of \(\text{CH}_3\text{CH}_2\text{COCl}\).
2. Curly arrow from the \(\text{C}=\text{O}\) double bond onto the oxygen atom.
3. Drawing of the tetrahedral intermediate with a single bond to oxygen (bearing a negative charge, \(\text{O}^-\)), a positive charge on nitrogen (\(\text{N}^+\)), and the chlorine atom still attached.
4. Curly arrow from the lone pair on the oxygen reforming the \(\text{C}=\text{O}\) double bond, and a curly arrow breaking the \(\text{C}-\text{Cl}\) bond to release \(\text{Cl}^-\).
5. Deprotonation step: A base (such as \(\text{Cl}^-\) or another \(\text{CH}_3\text{NH}_2\) molecule) removes a proton from the nitrogen, with a curly arrow from the \(\text{N}-\text{H}\) bond to the positively charged nitrogen atom to yield the final amide.

(b) Nucleophile (or electron pair donor).

(c) (i) Key IR absorption bands:
- \(\text{C}=\text{O}\) stretch (amide): \(1630 - 1700\ \text{cm}^{-1}\).
- \(\text{N}-\text{H}\) stretch (secondary amide): \(3140 - 3500\ \text{cm}^{-1}\).

(ii) Proton (\(^{1}\text{H}\)) NMR spectrum analysis of \(\text{CH}_3\text{CH}_2\text{CONHCH}_3\):
- There are 4 distinct proton environments.
- Peak 1 (\(\text{CH}_3\) of ethyl group): Relative integration = 3H, split into a triplet (adjacent to \(\text{CH}_2\)).
- Peak 2 (\(\text{CH}_2\) of ethyl group): Relative integration = 2H, split into a quartet (adjacent to \(\text{CH}_3\)).
- Peak 3 (\(\text{N}-\text{CH}_3\) methyl group): Relative integration = 3H, singlet (since adjacent to N and N-H does not split).
- Peak 4 (\(\text{N}-\text{H}\) proton): Relative integration = 1H, singlet (broad).

(a) [6 marks total]
* 1 mark: Curly arrow from the lone pair on the nitrogen of \(\text{CH}_3\text{NH}_2\) to the carbonyl carbon.
* 1 mark: Curly arrow from \(\text{C}=\text{O}\) double bond to oxygen.
* 2 marks: Correct structure of the tetrahedral intermediate (showing \(\text{O}^-\), \(\text{N}^+\), \(\text{Cl}\), and the correct alkyl chains). (1 mark if intermediate has correct species but misses charges).
* 1 mark: Curly arrow from the oxygen lone pair back to reform \(\text{C}=\text{O}\) and curly arrow from the \(\text{C}-\text{Cl}\) bond to chlorine.
* 1 mark: Final step showing deprotonation of the intermediate to form \(\text{CH}_3\text{CH}_2\text{CONHCH}_3\).

(b) [1 mark total]
* 1 mark: Nucleophile (accept electron pair donor; reject base for first step).

(c) [9 marks total]
- Part (i) [3 marks]:
* 1 mark: Identifies \(\text{C}=\text{O}\) stretch in amide and gives range \(1630 - 1700\ \text{cm}^{-1}\).
* 1 mark: Identifies \(\text{N}-\text{H}\) stretch and gives range \(3140 - 3500\ \text{cm}^{-1}\).
* 1 mark: Explicitly states the absence of a broad carboxylic acid \(\text{O}-\text{H}\) stretch or \(\text{C}-\text{Cl}\) bond.
- Part (ii) [6 marks]:
* 1 mark: Identifies 4 proton environments.
* 1 mark: Triplet for \(\text{CH}_3\) of ethyl group with 3H integration.
* 1 mark: Quartet for \(\text{CH}_2\) with 2H integration.
* 1 mark: Singlet for \(\text{N}-\text{CH}_3\) with 3H integration.
* 1 mark: Singlet for \(\text{N}-\text{H}\) with 1H integration.
* 1 mark: Logical reasoning linking splitting to adjacent protons (using \(n+1\) rule).

(a) (i)
- \(\text{Fe}\) atom: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\)
- \(\text{Fe}^{2+}\) ion: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\)
- \(\text{Fe}^{3+}\) ion: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\)
- Stability explanation: The \(\text{Fe}^{3+}\) ion has a half-filled d-subshell (\(3d^5\)), which is particularly stable due to the symmetrical distribution of electron density and reduced electron-electron repulsion.

(ii)
- Precipitation equation:
\(\text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s)\)
- Identity of brown precipitate: Iron(III) hydroxide, \(\text{Fe(OH)}_3\) (or hydrated iron(III) oxide).
- Equation for reaction on standing:
\(4\text{Fe(OH)}_2(s) + \text{O}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow 4\text{Fe(OH)}_3(s)\)

(b) (i) Titration equation:
\(5\text{Fe}^{2+}(aq) + \text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow 5\text{Fe}^{3+}(aq) + \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)\)

(ii) Colour change at endpoint: Colourless to permanent pale pink (reject purple).

(iii) Calculation steps:
1. Moles of \(\text{MnO}_4^-\) in titre:
\(n(\text{MnO}_4^-) = C \times V = 0.0150\ \text{mol dm}^{-3} \times \frac{22.40}{1000}\ \text{dm}^3 = 3.36 \times 10^{-4}\ \text{mol}\)

2. Moles of \(\text{Fe}^{2+}\) in the \(25.0\ \text{cm}^3\) sample:
Since 1 mole of \(\text{MnO}_4^-\) reacts with 5 moles of \(\text{Fe}^{2+}\):
\(n(\text{Fe}^{2+}) = 5 \times 3.36 \times 10^{-4}\ \text{mol} = 1.68 \times 10^{-3}\ \text{mol}\)

3. Moles of \(\text{Fe}^{2+}\) in the original \(250.0\ \text{cm}^3\) flask:
\(n(\text{Fe}^{2+})_{\text{total}} = 1.68 \times 10^{-3}\ \text{mol} \times \frac{250.0}{25.0} = 1.68 \times 10^{-2}\ \text{mol}\)

4. Mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\):
\(\text{Mass} = n \times M_r = 1.68 \times 10^{-2}\ \text{mol} \times 278.0\ \text{g mol}^{-1} = 4.6704\ \text{g}\)

5. Percentage Purity:
\(\text{Percentage Purity} = \frac{4.6704\ \text{g}}{6.50\ \text{g}} \times 100\% = 71.85\% \approx 71.9\%\) (to 3 significant figures).

(a) [7 marks total]
- Part (i) [3 marks]:
* 1 mark: Correct full electron configuration for \(\text{Fe}^{2+}\).
* 1 mark: Correct full electron configuration for \(\text{Fe}^{3+}\).
* 1 mark: Correctly identifies the half-filled d-subshell of \(\text{Fe}^{3+}\) (\(3d^5\)) as providing extra stability.
- Part (ii) [4 marks]:
* 1 mark: Correct ionic precipitation equation with state symbols: \(\text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s)\).
* 1 mark: Correct name/formula of brown precipitate (Iron(III) hydroxide / \(\text{Fe(OH)}_3\)).
* 2 marks: Correctly balanced oxidation equation with oxygen (1 mark for reactants/products, 1 mark for correct balancing: \(4\text{Fe(OH)}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3\)).

(b) [9 marks total]
- Part (i) [2 marks]:
* 2 marks: Correct overall ionic equation. (1 mark for correct species, 1 mark for balancing: 5:1:8 ratio).
- Part (ii) [1 mark]:
* 1 mark: Colourless to pale pink.
- Part (iii) [6 marks]:
* 1 mark: Moles of \(\text{MnO}_4^-\) calculated correctly (\(3.36 \times 10^{-4}\)).
* 1 mark: Moles of \(\text{Fe}^{2+}\) in \(25\ \text{cm}^3\) calculated correctly using stoichiometry (\(1.68 \times 10^{-3}\)).
* 1 mark: Scaling factor of 10 applied correctly to find total moles in \(250\ \text{cm}^3\) (\(1.68 \times 10^{-2}\)).
* 1 mark: Mass of pure salt calculated using \(M_r = 278.0\) (\(4.67\text{ g}\)).
* 1 mark: Percentage calculation set up correctly and evaluated (\(71.9\%\)).
* 1 mark: Final answer rounded to 3 significant figures.

(a) Standard enthalpy of combustion is the enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions (298 K, 100 kPa), with all reactants and products in their standard states.

(b) (i) Balanced equation:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l) + 4.5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)\)
(Note: fractional coefficients are standard as combustion is defined per mole of fuel, but balanced integer values e.g. \(2\text{C}_3\text{H}_7\text{OH} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 8\text{H}_2\text{O}\) are accepted if enthalpy is defined per mole later).

(ii) Enthalpy of combustion calculation:
\(\Delta_c H^\theta = \sum \Delta_f H^\theta(\text{products}) - \sum \Delta_f H^\theta(\text{reactants})\)
\(\Delta_c H^\theta = [3 \times (-394) + 4 \times (-286)] - [-303 + 0]\)
\(\Delta_c H^\theta = [-1182 - 1144] + 303\)
\(\Delta_c H^\theta = -2326 + 303 = -2023\ \text{kJ mol}^{-1}\).

(c) (i) Experimental calculation:
- Heat energy change: \(q = m c \Delta T\)
\(q = 150.0\text{ g} \times 4.18\ \text{J g}^{-1}\ \text{K}^{-1} \times (48.5 - 20.2)\text{ K} = 150.0 \times 4.18 \times 28.3 = 17744.1\text{ J} = 17.7441\text{ kJ}\)
- Moles of propan-1-ol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{0.820}{60.0} = 0.01367\ \text{mol}\)
- Enthalpy of combustion:
\(\Delta H_c = -\frac{q}{n} = -\frac{17.7441\text{ kJ}}{0.01367\ \text{mol}} = -1298\ \text{kJ mol}^{-1} \approx -1300\ \text{kJ mol}^{-1}\) (to 3 sig figs).

(ii) Two reasons:
1. Heat loss to the surroundings / the copper calorimeter.
2. Incomplete combustion of the fuel (forming carbon monoxide or soot instead of carbon dioxide).

(d) For the reaction:
\(\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)\)
- The reaction is highly exothermic, so \(\Delta H^\theta\) is negative.
- There is an increase in the number of gaseous molecules (from 6 to 7 moles of gas), so disorder increases, and \(\Delta S^\theta\) is positive.
- In the equation \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), because \(\Delta H^\theta\) is negative and \(-T\Delta S^\theta\) is negative at all temperatures (as both \(T\) and \(\Delta S^\theta\) are positive), \(\Delta G^\theta\) will always be negative, meaning the reaction is spontaneous at all temperatures.

(a) [2 marks total]
* 1 mark: Enthalpy change when 1 mole of substance is burned completely in oxygen.
* 1 mark: Mentions standard conditions (298 K, 100 kPa/1 atm) and standard states.

(b) [5 marks total]
- Part (i) [2 marks]:
* 1 mark: Correctly balanced equation (fractional coefficients allowed).
* 1 mark: Correct state symbols: \(l\) for propanol and water, \(g\) for oxygen and carbon dioxide.
- Part (ii) [3 marks]:
* 1 mark: Correct Hess's Law cycle setup or algebraic formula expression.
* 1 mark: Calculation of total products value (\(-2326\)).
* 1 mark: Correct calculation of \(-2023\ \text{kJ mol}^{-1}\) (must include negative sign and correct units).

(c) [6 marks total]
- Part (i) [4 marks]:
* 1 mark: Correct calculation of heat energy change \(q = 17.74\text{ kJ}\) (or \(17744\text{ J}\)).
* 1 mark: Correct calculation of moles of propan-1-ol (\(0.0137\ \text{mol}\)).
* 1 mark: Dividing \(q\) by \(n\) with negative sign applied.
* 1 mark: Final answer rounded to 3 significant figures: \(-1300\ \text{kJ mol}^{-1}\) (or \(-1298\ \text{kJ mol}^{-1}\)).
- Part (ii) [2 marks]:
* 1 mark: Heat lost to surroundings (or calorimeter).
* 1 mark: Incomplete combustion (or evaporation of propan-1-ol).

(d) [3 marks total]
* 1 mark: Identifies that combustion is exothermic, so \(\Delta H^\theta < 0\).
* 1 mark: Identifies that gas moles increase (from 6 to 7), so entropy increases, \(\Delta S^\theta > 0\).
* 1 mark: Links both signs to \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\) to show \(\Delta G^\theta\) is always negative regardless of \(T\).

(a) \(\text{CH}_3\text{OCH}_3(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{g})\) (or \(\text{l}\))

(b)(i) When the initial pressure of dimethyl ether doubles (from 40.0 to 80.0 kPa, or 80.0 to 160.0 kPa), the initial rate also doubles (from \(1.20 \times 10^{-4}\) to \(2.40 \times 10^{-4}\text{ kPa s}^{-1}\)). Since the rate is directly proportional to the concentration (pressure) of the reactant, the reaction is first-order with respect to dimethyl ether.

(ii) \(\text{Rate} = k [\text{CH}_3\text{OCH}_3]\) or \(\text{Rate} = k P(\text{CH}_3\text{OCH}_3)\)

(iii) Using Run 1:
\(1.20 \times 10^{-4}\text{ kPa s}^{-1} = k \times 40.0\text{ kPa}\)
\(k = \frac{1.20 \times 10^{-4}}{40.0} = 3.00 \times 10^{-6}\text{ s}^{-1}\)

(c)(i) Step 1
(ii) Steps 2 and 3
(iii) Step 4

(d) On a Maxwell-Boltzmann distribution:
- The y-axis represents the fraction or number of molecules, and the x-axis represents molecular energy.
- At a higher temperature, the peak shifts to the right (higher energy) and is flatter (lower peak height).
- The activation energy, \(E_{\text{a}}\), is a fixed value on the energy axis.
- At the higher temperature, a significantly larger area under the curve lies to the right of \(E_{\text{a}}\), meaning a much higher fraction of molecules possess kinetic energy greater than or equal to \(E_{\text{a}}\).
- According to collision theory, this results in a vastly increased frequency of successful collisions, thus significantly increasing the rate of reaction.

(e) Using the Arrhenius equation ratio:
\(\frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{-\frac{\Delta E_{\text{a}}}{RT}}\)
\(\Delta E_{\text{a}} = E_{\text{a,cat}} - E_{\text{a,uncat}} = 95.0 - 150 = -55.0\text{ kJ mol}^{-1} = -55000\text{ J mol}^{-1}\)
\(\frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{-\frac{-55000}{8.314 \times 600}} = e^{11.026} \approx 6.14 \times 10^4\)

(a) [2 marks]
- Correct balanced equation [1]
- Correct state symbols: gaseous reactants and products (allow liquid or gas for water) [1]

(b)(i) [2 marks]
- Statement that doubling pressure doubles rate [1]
- Conclusion that the reaction is first order [1]

(b)(ii) [1 mark]
- Correct rate equation (can use pressure or concentration) [1]

(b)(iii) [2 marks]
- Correct calculation of numerical value: \(3.00 \times 10^{-6}\) [1]
- Correct units: \(\text{s}^{-1}\) [1]

(c) [4 marks]
- (i) Step 1 is initiation (homolytic fission to form radicals) [1]
- (ii) Steps 2 and 3 are propagation (radicals react and produce new radicals) [2]
- (iii) Step 4 is termination (two radicals combine to form a neutral molecule) [1]

(d) [6 marks]
- Correctly labeled axes on a sketched or described plot (y: fraction of molecules, x: energy) [1]
- Curve starts at origin, asymmetrical with long tail at high energy [1]
- Curve for higher temperature has a lower peak shifted to the right [1]
- Activation energy, \(E_{\text{a}}\), indicated on the x-axis [1]
- Indication that shaded area under the curve representing molecules with energy \(\ge E_{\text{a}}\) is larger at the higher temperature [1]
- Explaining that a higher fraction of molecules have energy \(\ge E_{\text{a}}\), leading to a higher frequency of successful collisions [1]

(e) [3 marks]
- Correctly identifying \(\Delta E_{\text{a}} = -55\text{ kJ mol}^{-1}\) or converting to Joules [1]
- Correct substitution into the Arrhenius ratio formula [1]
- Final answer: \(6.14 \times 10^4\) (accept range \(6.13 \times 10^4\) to \(6.15 \times 10^4\)) [1]

(a)(i) A chromophore is an atom or group of atoms whose presence is responsible for the colour of a compound.

(ii) The nitrogen atoms of the azo group have \(\pi\) bonds and lone pairs that conjugate with the adjacent aromatic rings. This creates a larger, extended delocalised \(\pi\) system. The extensive delocalisation reduces the energy gap between the occupied and unoccupied molecular orbitals, bringing the absorption energy down into the visible spectrum.

(b) Electronic transitions occur when a molecule absorbs a photon of light, promoting an electron from its HOMO to its LUMO. If the compound has a small energy gap (e.g., due to highly conjugated systems), it absorbs light in the visible range (400-700 nm) and the unabsorbed complementary wavelengths are seen as colour. Colourless compounds have a larger HOMO-LUMO energy gap and absorb only higher-energy UV light.

(c) Upon protonation, the hydrogen ion bonds to one of the azo nitrogen atoms. This results in structural rearrangement to a quinonoid form. This increases the extent of conjugation across the molecule, which decreases the energy gap between the HOMO and LUMO. Consequently, the molecule absorbs lower-energy, longer-wavelength light (yellow/green is absorbed), meaning the compound appears red (complementary colour).

(d)(i) Using the Beer-Lambert Law: \(A = \varepsilon c l\)
\(\varepsilon = \frac{A}{c \cdot l} = \frac{0.450}{2.50 \times 10^{-5}\text{ mol dm}^{-3} \times 1.00\text{ cm}} = 18000\text{ dm}^3\text{ mol}^{-1}\text{ cm}^{-1}\)

(ii) Diluted concentration, \(c_{\text{dil}} = \frac{A}{\varepsilon \cdot l} = \frac{0.360}{18000 \times 1.00} = 2.00 \times 10^{-5}\text{ mol dm}^{-3}\)
Original concentration = \(c_{\text{dil}} \times 10 = 2.00 \times 10^{-4}\text{ mol dm}^{-3}\)

(iii) 1. High concentrations of dye can cause molecular interactions, leading to a non-linear relationship (deviation from the law).
2. Suspended solids or other particulate matter in untreated effluent can scatter light, giving false absorbance readings.

(a)(i) [1 mark]
- Correct definition (part of molecule responsible for colour / absorbing visible light) [1]

(a)(ii) [3 marks]
- Mention of lone pairs and/or \(\pi\) electrons on azo nitrogens [1]
- Conjugation with the adjacent benzene/aromatic rings [1]
- Leading to a large, extended delocalised \(\pi\) electron cloud [1]

(b) [4 marks]
- Absorption of light promotes an electron from HOMO to LUMO [1]
- The energy of the absorbed light corresponds to the HOMO-LUMO energy gap (\(E = h\nu\)) [1]
- Colourless compounds have a large gap, absorbing in the UV region [1]
- Highly conjugated compounds have a smaller gap, absorbing in the visible region, showing the complementary colour [1]

(c) [4 marks]
- Protonation occurs at the azo nitrogen (or amino group in resonance) [1]
- Structure changes to a quinonoid system / extends the conjugation [1]
- This decreases the HOMO-LUMO energy gap [1]
- Shifts absorption to longer wavelengths / lower frequencies (resulting in red colour) [1]

(d)(i) [3 marks]
- Correct calculation: \(18000\) or \(1.80 \times 10^4\) [1]
- Correct calculation method showing rearrangement of Beer-Lambert law [1]
- Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ cm}^{-1}\) (or \(\text{L mol}^{-1}\text{ cm}^{-1}\)) [1]

(d)(ii) [3 marks]
- Correct diluted concentration calculation: \(2.00 \times 10^{-5}\text{ mol dm}^{-3}\) [1]
- Correct dilution factor multiplication (\(\times 10\)) [1]
- Final answer: \(2.00 \times 10^{-4}\text{ mol dm}^{-3}\) [1]

(d)(iii) [2 marks]
- Any two valid reasons: suspended solids scattering light, presence of other absorbing contaminants/species at the same wavelength, high concentrations causing intermolecular interactions/non-linearity [2]

(a)(i) \(\text{CO}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{H}_2\text{CO}_3(\text{aq})\)
(ii) \(\text{H}_2\text{CO}_3(\text{aq}) \rightleftharpoons \text{HCO}_3^-(\text{aq}) + \text{H}^+(\text{aq})\)

(b) \(K_{\text{a}} = \frac{[\text{H}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\)
Rearranging to find \([\text{H}^+]\):
\([\text{H}^+] = K_{\text{a}} \times \frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]}\)
Given ratio \([\text{HCO}_3^-]/[\text{H}_2\text{CO}_3] = 20/1\), so \([\text{H}_2\text{CO}_3]/[\text{HCO}_3^-] = 1/20\).
\([\text{H}^+] = 7.9 \times 10^{-7} \times \frac{1}{20} = 3.95 \times 10^{-8}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(3.95 \times 10^{-8}) = 7.403 \approx 7.40\).

(c)(i) New ratio \([\text{HCO}_3^-]/[\text{H}_2\text{CO}_3] = \frac{2.4 \times 10^{-2}}{6.0 \times 10^{-4}} = 40\).
\([\text{H}^+] = 7.9 \times 10^{-7} \times \frac{1}{40} = 1.975 \times 10^{-8}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.975 \times 10^{-8}) = 7.70\).

(ii) The condition is respiratory alkalosis. Exhaling excess \(\text{CO}_2\) shifts the first equilibrium: \(\text{CO}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{H}_2\text{CO}_3(\text{aq})\) to the left. This causes the dissociation equilibrium: \(\text{H}_2\text{CO}_3(\text{aq}) \rightleftharpoons \text{HCO}_3^-(\text{aq}) + \text{H}^+(\text{aq})\) to also shift to the left to replace the lost carbonic acid, thereby consuming \(\text{H}^+\) ions and raising the pH.

(d)(i) In pure water, \([\text{H}^+] = [\text{OH}^-]\).
\(K_{\text{w}} = [\text{H}^+]^2 = 2.4 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\)
\([\text{H}^+] = \sqrt{2.4 \times 10^{-14}} = 1.55 \times 10^{-7}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.55 \times 10^{-7}) = 6.81\).

(ii) Pure water is neutral because the concentration of hydrogen ions equals the concentration of hydroxide ions (\([\text{H}^+] = [\text{OH}^-]\)). Neutrality is defined by this equality, not by pH being exactly 7.00 (which is only true at \(25^\circ\text{C}\)).

(e)(i) Since \(\text{pH} = 7.00\), \([\text{H}^+] = 1.0 \times 10^{-7}\text{ mol dm}^{-3}\).
Using \(K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\):
\(1.0 \times 10^{-7} = \frac{(1.0 \times 10^{-7}) \times [\text{A}^-]}{0.100}\)
\([\text{A}^-] = 0.100\text{ mol dm}^{-3}\).
In \(500\text{ cm}^3\) (\(0.500\text{ dm}^3\)), moles of \(\text{NaA}\) required = \(0.100\text{ mol dm}^{-3} \times 0.500\text{ dm}^3 = 0.0500\text{ mol}\).
Mass of \(\text{NaA} = 0.0500\text{ mol} \times 96.0\text{ g mol}^{-1} = 4.80\text{ g}\).

(ii) When \(\text{OH}^-\) ions are added, they react with the weak acid reservoirs:
\(\text{HA}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{A}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\)
Since there is a large amount of \(\text{HA}\) available, the added hydroxide ions are neutralized with minimal change to the \([\text{A}^-]/[\text{HA}]\) ratio, keeping the pH steady.

(a)(i) [1 mark]
- Balanced equation with state symbols: \(\text{CO}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{H}_2\text{CO}_3(\text{aq})\) [1]

(a)(ii) [1 mark]
- Balanced equation with state symbols: \(\text{H}_2\text{CO}_3(\text{aq}) \rightleftharpoons \text{HCO}_3^-(\text{aq}) + \text{H}^+(\text{aq})\) [1]

(b) [4 marks]
- Expression for \(K_{\text{a}}\) [1]
- Substituting the 20:1 ratio (or showing \([\text{H}^+] = 7.9 \times 10^{-7} / 20\)) [1]
- Calculating \([\text{H}^+] = 3.95 \times 10^{-8}\text{ mol dm}^{-3}\) [1]
- Calculating pH = 7.40 [1]

(c)(i) [2 marks]
- Determination of new ratio or substituting concentrations into buffer equation [1]
- Final pH = 7.70 [1]

(c)(ii) [3 marks]
- Naming condition: respiratory alkalosis [1]
- Stating that equilibrium shifts left to replace lost carbon dioxide [1]
- Stating that this consumes \(\text{H}^+\), leading to pH rise [1]

(d)(i) [2 marks]
- Taking square root of \(K_{\text{w}}\) to find \([\text{H}^+]\) [1]
- Calculating pH = 6.81 [1]

(d)(ii) [2 marks]
- Stating that water is neutral [1]
- Explaining that neutrality means \([\text{H}^+] = [\text{OH}^-]\) [1]

(e)(i) [3 marks]
- Deducing that \([\text{A}^-] = 0.100\text{ mol dm}^{-3}\) (since \(\text{pH} = \text{p}K_{\text{a}}\)) [1]
- Calculating moles of \(\text{NaA} = 0.050\text{ mol}\) [1]
- Calculating mass of \(\text{NaA} = 4.80\text{ g}\) [1]

(e)(ii) [2 marks]
- Writing ionic equation of buffer reaction with \(\text{OH}^-\): \(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\) [1]
- Explaining that the ratio \([\text{A}^-]/[\text{HA}]\) changes only minimally [1]

(a)(i) \(\text{Cl}_2(\text{aq}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq})\) (accept state symbol (g) for chlorine)

(ii) Chlorine is a stronger oxidising agent than bromine because a chlorine atom has a smaller atomic radius and less electron shielding than a bromine atom. Thus, incoming electrons feel a stronger electrostatic attraction to the chlorine nucleus, meaning chlorine gains electrons more readily.

(b)(i) To construct the standard cell:
- Use two separate beakers: one containing a mixture of \(1.0\text{ mol dm}^{-3}\) \(\text{NaCl}\) (or other soluble chloride) and \(1.0\text{ mol dm}^{-3}\) \(\text{Cl}_2\) (or chlorine gas at \(100\text{ kPa}\)); the other containing a mixture of \(1.0\text{ mol dm}^{-3}\) \(\text{NaBr}\) and \(1.0\text{ mol dm}^{-3}\) \(\text{Br}_2\).
- Insert an inert platinum electrode into both solutions.
- Connect the two solutions with a salt bridge filled with an inert electrolyte solution (e.g., saturated potassium nitrate, \(\text{KNO}_3\)).
- Connect the platinum electrodes to a high-resistance voltmeter using copper wires.
- Keep the cell at standard conditions of \(298\text{ K}\) (\(25^\circ\text{C}\)) and pressure of \(100\text{ kPa}\).

(ii) \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = 1.36\text{ V} - 1.09\text{ V} = +0.27\text{ V}\).
Electrons flow from the negative electrode (bromine half-cell, where oxidation occurs) to the positive electrode (chlorine half-cell, where reduction occurs) through the external wire.

(iii) \(\text{Pt(s)} \mid \text{Br}^-(\text{aq}), \text{Br}_2(\text{aq}) \parallel \text{Cl}_2(\text{aq}), \text{Cl}^-(\text{aq}) \mid \text{Pt(s)}\)

(c)(i) \(E = 1.09 - \frac{0.0591}{2} \log \frac{(0.010)^2}{0.20}\)
\(E = 1.09 - 0.02955 \log \left(5.0 \times 10^{-4}\right)\)
\(E = 1.09 - 0.02955 \times (-3.301) = 1.09 + 0.0975 = +1.1875\text{ V} \approx +1.19\text{ V}\).

(ii) If \([\text{Br}^-]\) is decreased, the value of the term \(\frac{[\text{Br}^-]^2}{[\text{Br}_2]}\) decreases. Consequently, the log of this term becomes more negative. Subtracting a more negative value increases \(E\) (makes it more positive).
Qualitatively, from Le Chatelier's principle, reducing the product concentration (\(\text{Br}^-\)) shifts the reduction half-reaction equilibrium \(\text{Br}_2 + 2\text{e}^- \rightleftharpoons 2\text{Br}^-\) to the right, increasing the tendency to accept electrons, making the electrode potential more positive.

(a)(i) [1 mark]
- Correct ionic equation with no spectator ions [1]

(a)(ii) [3 marks]
- Chlorine is a stronger oxidising agent than bromine / is more easily reduced [1]
- Chlorine has smaller atomic radius / less shielding [1]
- Stronger attraction between chlorine nucleus and incoming electrons [1]

(b)(i) [6 marks]
- Platinum used as electrode in both half-cells [1]
- Both solutions are \(1.0\text{ mol dm}^{-3}\) in terms of halide and halogen [1]
- Salt bridge connecting both solutions [1]
- Voltmeter connected across electrodes [1]
- Standard temperature (298 K) and pressure (100 kPa) [1]
- Saturated potassium nitrate or other inert electrolyte for salt bridge [1]

(b)(ii) [2 marks]
- \(E^\theta_{\text{cell}} = +0.27\text{ V}\) [1]
- Electron flow from Br half-cell (anode/negative) to Cl half-cell (cathode/positive) [1]

(b)(iii) [2 marks]
- Correct species and phases shown: Pt(s) on both ends, single/double vertical lines correct [1]
- Correct order of species (oxidation on left, reduction on right) [1]

(c)(i) [3 marks]
- Correct substitution of values into the formula [1]
- Correct evaluation of the logarithmic term: \(-3.30\) [1]
- Final potential: \(+1.19\text{ V}\) (accept \(+1.18\text{ V}\) to \(+1.19\text{ V}\)) [1]

(c)(ii) [3 marks]
- Statement that \(E\) increases / becomes more positive [1]
- Reference to Le Chatelier's principle or reduction equilibrium shifting right [1]
- Explaining that lower product concentration increases driving force for reduction [1]

(a)(i) \(\text{HOC}_6\text{H}_4\text{COOH} + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{CH}_3\text{COOC}_6\text{H}_4\text{COOH} + \text{CH}_3\text{COOH}\)
(Salicylic acid + ethanoic anhydride \(\rightarrow\) acetylsalicylic acid + ethanoic acid)

(ii) The mechanism is classified as nucleophilic addition-elimination (or nucleophilic substitution). The acid catalyst (e.g., \(\text{H}_3\text{PO}_4\)) protonates the carbonyl oxygen of ethanoic anhydride, which increases the positive charge on the carbonyl carbon, making it a stronger electrophile and highly susceptible to nucleophilic attack by the phenolic hydroxyl group.

(b) - Step 1: A lone pair of electrons on the carbonyl oxygen atom of the ester attacks a hydrogen ion (\(\text{H}^+\)) from the catalyst to form a protonated intermediate (bond making between O and H). The \(\pi\) bond of the C=O polarises towards the oxygen.
- Step 2: The nucleophile (water, \(\text{H}_2\text{O}\), with a lone pair on oxygen) attacks the electrophilic carbonyl carbon atom of the protonated ester. Simultaneously, the \(\pi\) bond of the C=O breaks completely, moving its electrons to the carbonyl oxygen.
- Hybridisation and geometry change: The carbonyl carbon starts as \(sp^2\) hybridized with planar geometry. Upon nucleophilic attack, it becomes \(sp^3\) hybridized with tetrahedral geometry in the intermediate.

(c)(i) \(\text{Rate} = k [\text{aspirin}] [\text{OH}^-]\)

(ii) - Step 1 (Slow / Rate-determining step): Hydroxide ion (\(\text{OH}^-\)) attacks the ester carbonyl carbon of aspirin, forming a tetrahedral intermediate.
\(\text{Aspirin} + \text{OH}^- \rightarrow [\text{Intermediate}]^-\)
- Step 2 (Fast): The tetrahedral intermediate collapses, expelling the phenoxide/salicylate leaving group and forming the hydrolysed products (salicylate and acetate).
\([\text{Intermediate}]^- \rightarrow \text{Products}\)
This matches the rate equation because the reactants in the rate equation (aspirin and \(\text{OH}^-\)) are the only reactants in the slow step.

(d)(i) Practical steps and chemical principles:
- Dissolve the impure aspirin in the minimum volume of hot solvent (ethanol/water mixture). (This ensures maximum recovery upon cooling).
- Filter the hot solution to remove any insoluble impurities.
- Allow the filtrate to cool slowly to room temperature and then in an ice bath. Aspirin recrystallises out as solubility drops at low temperatures, while soluble impurities remain dissolved.
- Filter the crystals under reduced pressure (using a Buchner funnel), wash with a small volume of ice-cold solvent, and leave to dry.

(ii) Measure the melting point of the dry product using a melting point apparatus. Compare the experimental melting point range with the literature value (\(135\text{ }^\circ\text{C}\)). Pure aspirin will have a sharp melting point range (typically 1-2 degrees wide) at the literature value. Impure samples will melt over a wider range and at a lower temperature.

(a)(i) [2 marks]
- Correct formulae of reactants [1]
- Correct products shown [1]

(a)(ii) [3 marks]
- Classifying mechanism as nucleophilic addition-elimination or nucleophilic substitution [1]
- Explaining protonation of carbonyl oxygen [1]
- Mention of making carbonyl carbon more electrophilic [1]

(b) [5 marks]
- Step 1 description: curly arrow from carbonyl O to \(\text{H}^+\) [1]
- Step 2 description: curly arrow from lone pair on water O to carbonyl carbon, and opening of C=O bond [1]
- Identifying water as the nucleophile and protonated carbonyl carbon as the electrophile [1]
- Stating hybridisation changes from \(sp^2\) to \(sp^3\) [1]
- Stating geometry changes from planar (or trigonal planar) to tetrahedral [1]

(c)(i) [1 mark]
- Correct rate equation corresponding to the given orders [1]

(c)(ii) [3 marks]
- Step 1 involves reaction of aspirin and \(\text{OH}^-\) [1]
- Step 1 designated as the slow/rate-determining step [1]
- Step 2 showing intermediate breakdown to products, designated as fast [1]

(d)(i) [4 marks]
- Dissolving in minimum volume of hot solvent [1]
- Hot filtration to remove insoluble impurities [1]
- Cool to allow crystals to form while soluble impurities remain in solution [1]
- Filter under vacuum, wash with cold solvent, and dry [1]

(d)(ii) [2 marks]
- Stating that a pure sample has a sharp melting point that matches literature [1]
- Stating that impurities lower the melting point and widen the melting range [1]

(a) Sodium thiosulfate reacts rapidly with any produced iodine: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\). When all thiosulfate is completely consumed, free iodine remains and instantly turns the starch indicator blue-black. Because only a tiny fraction of the reactants is consumed during this time, the average rate of iodine formation (\(\Delta[\text{I}_2] / \Delta t\)) is a reliable estimate of the initial rate. (b) Comparing Run 1 and Run 2: \([\text{I}^-]\) is held constant. \([\text{S}_2\text{O}_8^{2-}]\) is doubled from \(0.0100\) to \(0.0200 \text{ mol dm}^{-3}\). The time halves from \(50.0 \text{ s}\) to \(25.0 \text{ s}\), meaning the rate (which is proportional to \(1/t\)) doubles. Therefore, the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1. Comparing Run 1 and Run 3: \([\text{S}_2\text{O}_8^{2-}]\) is held constant. \([\text{I}^-]\) is doubled from \(0.0200\) to \(0.0400 \text{ mol dm}^{-3}\). The time halves from \(50.0 \text{ s}\) to \(25.0 \text{ s}\), meaning the rate doubles. Therefore, the order of reaction with respect to \(\text{I}^-\) is 1. (c) The concentration of thiosulfate consumed is \(4.00 \times 10^{-4} \text{ mol dm}^{-3}\). According to the stoichiometry, \(2 \text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\) reacts with \(1 \text{ mol}\) of \(\text{I}_2\), which is formed from \(1 \text{ mol}\) of \(\text{S}_2\text{O}_8^{2-}\). Thus, the change in concentration of \(\text{S}_2\text{O}_8^{2-}\) at the point of color change is: \(\Delta [\text{S}_2\text{O}_8^{2-}] = 0.5 \times 4.00 \times 10^{-4} = 2.00 \times 10^{-4} \text{ mol dm}^{-3}\). Initial rate for Run 1 = \(\frac{2.00 \times 10^{-4} \text{ mol dm}^{-3}}{50.0 \text{ s}} = 4.00 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1}\). Using the rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) \(\Rightarrow 4.00 \times 10^{-6} = k(0.0100)(0.0200) \Rightarrow k = 0.0200 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\). (d) Run the same experiment at several different temperatures (e.g., five temperatures between \(20\ ^\circ\text{C}\) and \(60\ ^\circ\text{C}\)) using a water bath. Measure \(t\) and calculate the rate constant \(k\) at each temperature. Plot \(\ln(k)\) against \(1/T\) (where \(T\) is in Kelvin). The gradient of this Arrhenius plot is equal to \(-E_a / R\). Calculate \(E_a\) as \(-\text{gradient} \times R\). (e) A major source of uncertainty is human reaction time in starting and stopping the stopwatch when the color change occurs. This can be minimized by repeating the experiment and calculating a mean, or by using a colorimeter connected to a data logger to accurately detect the onset of absorbance.

Part (a) [2 marks]: 1 mark for stating that thiosulfate reacts with iodine as soon as it is formed. 1 mark for explaining that when thiosulfate is completely reacted, free iodine turns starch blue-black, and since reactant concentrations change very little, this represents the initial rate. Part (b) [4 marks]: 1 mark for calculating or stating that rate is proportional to 1/t. 1 mark for comparing Run 1 and Run 2 to show that doubling [S2O8^2-] doubles the rate, hence order is 1. 1 mark for comparing Run 1 and Run 3 to show that doubling [I^-] doubles the rate, hence order is 1. 1 mark for stating the final correct rate equation: Rate = k[S2O8^2-][I^-]. Part (c) [3 marks]: 1 mark for calculating the initial rate for Run 1 = 4.00 x 10^-6 mol dm^-3 s^-1 (allow ecf from incorrect mole ratio). 1 mark for calculating k = 0.0200. 1 mark for correct units: dm^3 mol^-1 s^-1. Part (d) [4 marks]: 1 mark for stating that the reaction must be run at multiple temperatures (at least 5). 1 mark for stating that k (or 1/t) must be determined at each temperature. 1 mark for plotting ln(k) against 1/T (T in Kelvin). 1 mark for stating that the gradient = -E_a/R and showing how to get E_a. Part (e) [2 marks]: 1 mark for identifying human reaction time as the source of uncertainty. 1 mark for suggesting repeating and averaging OR using a colorimeter/data logger.

(a) As the volume of ethylenediamine increases up to 7.5 cm^3, more ligand is available to coordinate with the copper(II) ions, producing a higher concentration of the colored complex and increasing absorbance. The maximum absorbance is reached at the exact stoichiometric ratio where the maximum amount of complex is formed. Beyond this point, copper(II) becomes the limiting reactant, and since its volume decreases, the concentration of the complex formed decreases, leading to a drop in absorbance. (b) The maximum absorbance occurs at Tube 7, where the volume of copper(II) solution is \(2.5 \text{ cm}^3\) and the volume of ethylenediamine is \(7.5 \text{ cm}^3\). Because the concentrations of both solutions are identical (\(0.10 \text{ mol dm}^{-3}\)), the mole ratio is directly proportional to the volume ratio: \(\text{Ratio} = \frac{\text{Volume of en}}{\text{Volume of Cu}^{2+}} = \frac{7.5}{2.5} = 3\). Thus, the ratio of ligand to metal ion is 3:1, giving the formula \([\text{Cu}(\text{en})_3]^{2+}\). (c) A colorimeter passes a selected wavelength of light through a sample solution contained in a cuvette and measures the intensity of light transmitted relative to a blank. The filter/wavelength chosen is the complementary color of the solution (or the wavelength of maximum absorbance, \(\lambda_{\text{max}}\)). Since the copper-ethylenediamine complex is dark blue-purple, a yellow/green filter (absorbance around \(550 \text{ nm}\)) is used to ensure maximum sensitivity and minimal absorption by unreacted reactants. (d) The complex has a coordination number of 6 because each of the 3 bidentate ethylenediamine ligands forms 2 coordinate bonds. The geometry is octahedral. Draw a central Cu ion with six bonds in octahedral arrangement (four in-plane, one up, one down) using wedge-and-dash representation for 3D depth, with ethylenediamine molecules represented as bidentate rings bridging adjacent sites, and an overall charge of 2+. (e) Each nitrogen atom in ethylenediamine has a lone pair of electrons. A dative covalent (coordinate) bond is formed when these lone pairs are donated into empty 3d or 4s/4p orbitals of the transition metal ion \(\text{Cu}^{2+}\). Since each ligand contains two nitrogen atoms with lone pairs, it forms two dative covalent bonds per ligand.

Part (a) [3 marks]: 1 mark for stating that increasing ligand volume increases the concentration of the colored complex. 1 mark for stating that the maximum absorbance corresponds to the exact stoichiometric ratio of the complex. 1 mark for explaining that after the maximum, copper(II) becomes the limiting reactant, so the amount of complex decreases. Part (b) [3 marks]: 1 mark for identifying Tube 7 as the peak absorbance. 1 mark for calculating the mole ratio of en to Cu2+ as 3:1. 1 mark for identifying the formula as [Cu(en)3]^2+. Part (c) [3 marks]: 1 mark for explaining that a colorimeter measures light absorption through a solution. 1 mark for stating that the filter selected is the complementary color of the complex (green/yellow, around 550 nm). 1 mark for explaining that this provides maximum absorbance and minimizes interference. Part (d) [3 marks]: 1 mark for identifying the coordination number as 6 and geometry as octahedral. 1 mark for drawing an octahedral skeletal structure with 3 bidentate chelating rings. 1 mark for showing correct 3-dimensional bonds and overall 2+ charge. Part (e) [3 marks]: 1 mark for stating that nitrogen in ethylenediamine has a lone pair of electrons. 1 mark for defining a dative covalent bond as the donation of a lone pair from the ligand to the empty orbitals of the Cu2+ ion. 1 mark for explaining that because there are two donor atoms per ethylenediamine molecule, it forms two dative covalent bonds.

(a) First, rinse the pH probe with distilled water. Immerse the probe in a buffer solution of known neutral pH (such as pH 7.00), allow the reading to stabilize, and calibrate the meter to this value. Rinse the probe again and repeat this process with an acidic buffer (such as pH 4.00) to calibrate across the working range. Calibration is necessary because the electrode's potential response drifts over time and temperature shifts can lead to systematic measurement errors. (b) Moles of \(\text{NaOH}\) used at the equivalence point = \(\text{volume} \times \text{concentration} = 0.02000 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 2.00 \times 10^{-3} \text{ mol}\). According to the equation: \(\text{HA}(aq) + \text{NaOH}(aq) \rightarrow \text{NaA}(aq) + \text{H}_2\text{O}(l)\), the mole ratio is 1:1. Therefore, there are \(2.00 \times 10^{-3} \text{ mol}\) of \(\text{HA}\) in the \(25.0 \text{ cm}^3\) sample. Concentration of \(\text{HA}\) = \(\frac{2.00 \times 10^{-3} \text{ mol}}{0.0250 \text{ dm}^3} = 0.0800 \text{ mol dm}^{-3}\). (c) The total volume of the weak acid solution prepared was \(250.0 \text{ cm}^3\). The total moles of \(\text{HA}\) in \(250.0 \text{ cm}^3\) = \(0.0800 \text{ mol dm}^{-3} \times 0.250 \text{ dm}^3 = 0.0200 \text{ mol}\). Molar mass of \(\text{HA}\) = \(\frac{\text{mass}}{\text{moles}} = \frac{1.48 \text{ g}}{0.0200 \text{ mol}} = 74.0 \text{ g mol}^{-1}\). (d) At the half-equivalence point, exactly half of the weak acid has been neutralized. This means \([\text{HA}] = [\text{A}^-]\). Substituting this into the acid dissociation constant expression: \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\), we get \(K_a = [\text{H}^+]\), and therefore \(\text{p}K_a = \text{pH}\). Since the pH at the half-equivalence point is \(4.85\), \(\text{p}K_a = 4.85\). Therefore, \(K_a = 10^{-4.85} = 1.41 \times 10^{-5} \text{ mol dm}^{-3}\). (e) The sketch of the titration curve should feature: 1. A starting pH around 3.0 (calculated as \(\text{pH} = -\log_{10}(\sqrt{1.41 \times 10^{-5} \times 0.0800}) \approx 2.97\); accept between 2.5 and 3.5). 2. A shallow, flat buffer region centered around 10.0 cm^3 of NaOH added with pH around 4.85. 3. A steep, near-vertical rise at the equivalence point of 20.0 cm^3 NaOH, with the center of this vertical rise (the equivalence point) at an alkaline pH (around pH 8.5). 4. A curve that plateaus off at around pH 12 to 13 as excess strong base is added.

Part (a) [3 marks]: 1 mark for rinsing the probe and immersing in buffer of pH 7.00 to adjust/calibrate. 1 mark for repeating with a second buffer (pH 4.00) to calibrate the slope. 1 mark for stating that calibration corrects for electrode drift/aging or temperature changes to ensure accuracy. Part (b) [3 marks]: 1 mark for calculating moles of NaOH = 2.00 x 10^-3 mol. 1 mark for recognizing 1:1 stoichiometry between HA and NaOH. 1 mark for calculating concentration of HA = 0.0800 mol dm^-3. Part (c) [2 marks]: 1 mark for calculating total moles of HA in 250 cm^3 = 0.0200 mol. 1 mark for calculating molar mass = 74.0 g mol^-1. Part (d) [3 marks]: 1 mark for stating that at half-equivalence, [HA] = [A^-]. 1 mark for showing that under this condition, Ka = [H^+] and pKa = pH. 1 mark for calculating Ka = 1.41 x 10^-5 mol dm^-3 (allow 1.4 x 10^-5). Part (e) [4 marks]: 1 mark for starting pH between 2.5 and 3.5. 1 mark for showing a buffer region at 10 cm^3 with a pH of 4.85. 1 mark for a vertical rise occurring at exactly 20 cm^3 with equivalence point pH > 7. 1 mark for curve flattening out at high pH (~12-13).

(a) Starch is used to detect the exact endpoint of the iodometric titration by forming a deep blue-black complex with free iodine. The color change at the endpoint is from blue-black to colorless. Starch must not be added at the beginning of the titration because a high concentration of iodine forms an irreversible complex with starch, which does not release iodine readily, leading to an overestimation of the end-point. It should be added when the solution in the conical flask becomes a pale straw-yellow, indicating that the iodine concentration is low. (b) From Equation 1, \(1 \text{ mol}\) of chlorate(I) ions (\(\text{ClO}^-\)) produces \(1 \text{ mol}\) of iodine (\(\text{I}_2\network\)). From Equation 2, \(1 \text{ mol}\) of iodine reacts with \(2 \text{ mol}\) of thiosulfate ions (\(\text{S}_2\text{O}_3^{2-}\)). Therefore, \(1 \text{ mol}\) of \(\text{ClO}^- \equiv 2 \text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\), showing the overall mole ratio is 1:2. (c) Moles of \(\text{S}_2\text{O}_3^{2-}\) in titration = \(\text{volume} \times \text{concentration} = 0.01620 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 1.62 \times 10^{-3} \text{ mol}\). Since \(\text{ClO}^- : \text{S}_2\text{O}_3^{2-} = 1:2\), moles of \(\text{ClO}^-\)
in the \(25.0 \text{ cm}^3\) aliquot = \(\frac{1.62 \times 10^{-3}}{2} = 8.10 \times 10^{-4} \text{ mol}\). Concentration of \(\text{ClO}^-\) in the diluted bleach = \(\frac{8.10 \times 10^{-4} \text{ mol}}{0.0250 \text{ dm}^3} = 0.0324 \text{ mol dm}^{-3}\). Since the original bleach was diluted from \(10.0 \text{ cm}^3\) to \(250.0 \text{ cm}^3\), the dilution factor is \(\frac{250.0}{10.0} = 25\). Concentration of \(\text{NaClO}\) in the original bleach = \(0.0324 \text{ mol dm}^{-3} \times 25 = 0.810 \text{ mol dm}^{-3}\). (d) Mass concentration of \(\text{NaClO}\) = \(\text{concentration in mol dm}^{-3} \times \text{molar mass} = 0.810 \text{ mol dm}^{-3} \times 74.5 \text{ g mol}^{-1} = 60.3 \text{ g dm}^{-3}\). (e) The hazard is the production of highly toxic chlorine gas (\(\text{Cl}_2\)). When acid is added to bleach (which contains both \(\text{ClO}^-\)
and \(\text{Cl}^-\)), they react to form chlorine gas according to the equation: \(\text{ClO}^-(aq) + \text{Cl}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{Cl}_2(g) + \text{H}_2\text{O}(l)\). This gas is toxic and irritating to the respiratory tract, so this step must be performed in a fume cupboard.

Part (a) [4 marks]: 1 mark for stating that starch forms a blue-black complex with iodine. 1 mark for stating that the endpoint color change is blue-black to colorless. 1 mark for explaining that starch must be added when the solution is pale yellow (not at the start) because high iodine concentration binds starch irreversibly. 1 mark for stating that adding starch late ensures a sharp and accurate endpoint. Part (b) [2 marks]: 1 mark for using Equation 1 to show 1 mol ClO^- produces 1 mol I2. 1 mark for using Equation 2 to show 1 mol I2 reacts with 2 mol S2O3^2- to establish the overall 1:2 ratio. Part (c) [4 marks]: 1 mark for calculating moles of S2O3^2- = 1.62 x 10^-3 mol. 1 mark for calculating moles of ClO^- in 25.0 cm^3 aliquot = 8.10 x 10^-4 mol. 1 mark for calculating the concentration of diluted bleach = 0.0324 mol dm^-3. 1 mark for multiplying by 25 to get the original concentration = 0.810 mol dm^-3. Part (d) [2 marks]: 1 mark for multiplying original concentration by 74.5. 1 mark for the correct final answer of 60.3 g dm^-3 with appropriate units. Part (e) [3 marks]: 1 mark for identifying the hazard as toxic/poisonous chlorine gas. 1 mark for the balanced ionic equation: ClO^- + Cl^- + 2H^+ -> Cl2 + H2O. 1 mark for stating that this should be carried out in a fume cupboard.