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In deuterium oxide (\(D_2O\)), labile protons such as those in the hydroxyl (\(-\text{OH}\) and carboxylic acid (\(-\text{COOH}\)) groups undergo rapid exchange with deuterium. Since deuterium does not absorb in the standard proton NMR frequency range, these signals disappear. The remaining protons on the carbon chain are: the three protons of the methyl group (\(-\text{CH}_3\)), which are adjacent to one proton and thus split into a doublet; and the single proton of the CH group (\(-\text{CH}-\)), which is adjacent to three methyl protons and thus split into a quartet. Therefore, only two peaks are observed: a doublet and a quartet.

[1] Mark is awarded for selecting option A. Correctly identifying that labile protons exchange with deuterium and identifying the doublet and quartet splitting patterns for the remaining non-exchanging protons.

An oxidizing agent can oxidize a species if its reduction potential \(E^\theta\) is more positive than that of the species being oxidized. To oxidize \(\text{I}^-\rightleftharpoons \frac{1}{2}\text{I}_2 + \text{e}^-\), the oxidising agent must have \(E^\theta > +0.54\text{ V}\). To not oxidize \(\text{Fe}^{2+}\rightleftharpoons \text{Fe}^{3+} + \text{e}^-\), the oxidising agent must have \(E^\theta < +0.77\text{ V}\). Thus, we look for an oxidizing agent with \(E^\theta\) between \(+0.54\text{ V}\) and \(+0.77\text{ V}\). \(\text{H}_3\text{AsO}_4\) is an oxidizing agent with \(E^\theta = +0.56\text{ V}\), which satisfies this condition.

[1] Mark is awarded for selecting option C. Correctly identifying that the oxidizing agent must have an electrode potential between \(+0.54\text{ V}\) and \(+0.77\text{ V}\) and matching this to \(\text{H}_3\text{AsO}_4\).

Using the Arrhenius equation in two-point form: \(\ln(k_2/k_1) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})\). Substituting the values: \(E_a = 50000\text{ J mol}^{-1}\), \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\), \(T_1 = 300\text{ K}\), and \(T_2 = 310\text{ K}\). This gives \(\ln(k_2/k_1) = \frac{50000}{8.31} (\frac{1}{300} - \frac{1}{310}) = 6016.85 \times 0.0001075 = 0.647\). Therefore, \(k_2/k_1 = e^{0.647} \approx 1.91\). This represents an increase by a factor of approximately 1.9.

[1] Mark is awarded for selecting option B. Correct substitution of values into the Arrhenius equation and calculation of the ratio of rate constants.

The reaction equations are: \(\text{Cl}_2 + 2\text{I}^- \to \text{I}_2 + 2\text{Cl}^-\), and \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \to 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\). Thus, \(1\text{ mol of Cl}_2 \equiv 2\text{ mol of S}_2\text{O}_3^{2-}\). The moles of thiosulfate used are: \(n(\text{S}_2\text{O}_3^{2-}) = 0.100\text{ mol dm}^{-3} \times 0.01850\text{ dm}^3 = 1.85 \times 10^{-3}\text{ mol}\). Moles of \(\text{Cl}_2 = \frac{1.85 \times 10^{-3}}{2} = 9.25 \times 10^{-4}\text{ mol}\). The concentration of \(\text{Cl}_2\) is \(\frac{9.25 \times 10^{-4}\text{ mol}}{0.0250\text{ dm}^3} = 0.0370\text{ mol dm}^{-3}\).

[1] Mark is awarded for selecting option B. Correct determination of the 1:2 stoichiometric ratio between chlorine and thiosulfate, calculation of moles of thiosulfate, and conversion to concentration.

Let the change in moles be \(x\). At equilibrium: \(n(\text{Cl}_2) = x = 0.40\text{ mol}\), \(n(\text{PCl}_3) = x = 0.40\text{ mol}\), and \(n(\text{PCl}_5) = 1.00 - x = 0.60\text{ mol}\). Since the volume is \(2.00\text{ dm}^3\), the concentrations are: \([\text{Cl}_2] = 0.20\text{ mol dm}^{-3}\), \([\text{PCl}_3] = 0.20\text{ mol dm}^{-3}\), and \([\text{PCl}_5] = 0.30\text{ mol dm}^{-3}\). Substituting these into the equilibrium expression: \(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.20 \times 0.20}{0.30} = 0.133\text{ mol dm}^{-3}\).

[1] Mark is awarded for selecting option A. Correct calculation of equilibrium concentrations and substituting into the \(K_c\) expression to find the correct value.

Acidic hydrolysis of a peptide bond yields the constituent amino acids (glycine and alanine). Under strongly acidic conditions (excess dilute HCl), any basic amine groups (\(-\text{NH}_2\)) are fully protonated to form ammonium ions (\(-\text{NH}_3^+\)), while the carboxylic acid groups (\(-\text{COOH}\)) remain un-ionized. Therefore, the dominant species in solution are the fully protonated cationic forms: \(\text{H}_3\text{N}^+\text{CH}_2\text{COOH}\) and \(\text{H}_3\text{N}^+\text{CH}(\text{CH}_3)\text{COOH}\).

[1] Mark is awarded for selecting option C. Correctly identifying that acidic hydrolysis splits the peptide bond and protonates the amine groups to form cationic species.

First, find the moles of each species: \(n(\text{H}_2\text{CO}_3) = 0.0500 \times 0.100 = 5.00 \times 10^{-3}\text{ mol}\) and \(n(\text{HCO}_3^-) = 0.0500 \times 0.0500 = 2.50 \times 10^{-3}\text{ mol}\). In the buffer solution, \([\text{H}^+] = K_{a1} \times \frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} = 4.50 \times 10^{-7} \times \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 9.00 \times 10^{-7}\text{ mol dm}^{-3}\). The pH is \(-\log_{10}(9.00 \times 10^{-7}) = 6.05\).

[1] Mark is awarded for selecting option A. Correctly calculating the moles of acid and conjugate base, and applying the buffer equation to find the pH.

Down Group 2, ionic radius increases while the charge remains \(+2\). This causes the charge density of the cation to decrease. A lower charge density means the cation has weaker polarizing power, distorting the electron cloud of the large carbonate anion less effectively. Consequently, the carbonate ion is more stable and requires higher temperatures to undergo thermal decomposition.

[1] Mark is awarded for selecting option B. Correctly explaining that larger cation size down Group 2 decreases charge density and reduces polarization of the carbonate anion, increasing thermal stability.

1. **Singlet at \(\delta = 2.0\text{ ppm}\) (3H)** indicates a methyl group attached directly to a carbonyl carbon: \(\text{CH}_3\text{C}=\text{O}\).
2. **Quartet at \(\delta = 4.1\text{ ppm}\) (2H)** indicates a \(-\text{CH}_2-\) group adjacent to a \(-\text{CH}_3\) group (due to the \(n+1\) rule, \(3+1=4\) lines). The downfield shift to \(4.1\text{ ppm}\) shows it is directly bonded to an oxygen atom of an ester: \(-\text{O}-\text{CH}_2-\).
3. **Triplet at \(\delta = 1.2\text{ ppm}\) (3H)** indicates a \(-\text{CH}_3\) group adjacent to a \(-\text{CH}_2-\) group (due to the \(n+1\) rule, \(2+1=3\) lines).
Combining these fragments gives ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

Award 1 mark for selecting option A (Ethyl ethanoate).

1. Identify the oxidation states of chlorine in the reactants and products:
* Reactant: \(\text{Cl}_2\) (oxidation state = 0)
* Products: \(\text{Cl}^-\) (oxidation state = -1, reduction) and \(\text{ClO}_3^-\) (oxidation state = +5, oxidation).
2. Determine the number of chlorine atoms undergoing each change:
* 5 chlorine atoms are reduced from 0 to -1 (forming \(5\text{Cl}^-\)).
* 1 chlorine atom is oxidized from 0 to +5 (forming \(1\text{ClO}_3^-\)).
3. Therefore, the ratio of reduced Cl to oxidized Cl is 5 : 1.

Award 1 mark for selecting option C (5 : 1).

Use the Arrhenius equation in logarithmic form:

\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_{\text{a}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)

Substitute the given values:

* \(E_{\text{a}} = 50.0\text{ kJ mol}^{-1} = 50000\text{ J mol}^{-1}\)
* \(T_1 = 300\text{ K}\)
* \(T_2 = 310\text{ K}\)

\(\ln\left(\frac{k_2}{k_1}\right) = \frac{50000}{8.314} \left(\frac{1}{300} - \frac{1}{310}\right)\)

\(\ln\left(\frac{k_2}{k_1}\right) = 6013.95 \times 0.00010753 = 0.6467\)

\(\frac{k_2}{k_1} = e^{0.6467} \approx 1.91\)

The rate constant increases by a factor of approximately 1.9.

Award 1 mark for selecting option B (1.9).

1. Calculate moles of \(\text{S}_2\text{O}_3^{2-}\) reacted:
\(n(\text{S}_2\text{O}_3^{2-}) = c \times V = 0.100\text{ mol dm}^{-3} \times 0.01500\text{ dm}^3 = 1.50 \times 10^{-3}\text{ mol}\).
2. Find moles of \(\text{I}_2\) in the titration:
From the equation, 2 moles of \(\text{S}_2\text{O}_3^{2-}\) react with 1 mole of \(\text{I}_2\).
\(n(\text{I}_2) = 0.5 \times 1.50 \times 10^{-3} = 7.50 \times 10^{-4}\text{ mol}\).
3. Find moles of \(\text{ClO}^-\) in the \(25.0\text{ cm}^3\) aliquot:
Since 1 mole of \(\text{ClO}^-\) produces 1 mole of \(\text{I}_2\), \(n(\text{ClO}^-) = 7.50 \times 10^{-4}\text{ mol}\).
4. Scale up to find moles of \(\text{ClO}^-\) in the \(250.0\text{ cm}^3\) volumetric flask:
\(n(\text{ClO}^-)_{\text{total}} = 7.50 \times 10^{-4} \times \frac{250.0}{25.0} = 7.50 \times 10^{-3}\text{ mol}\).
5. Calculate original concentration in the \(10.0\text{ cm}^3\) bleach sample:
\(c = \frac{n}{V} = \frac{7.50 \times 10^{-3}\text{ mol}}{0.0100\text{ dm}^3} = 0.750\text{ mol dm}^{-3}\).

Award 1 mark for selecting option D (0.750 mol dm\(^{-3}\)).

1. Write the expression for the equilibrium constant, \(K_{\text{c}}\):
\(K_{\text{c}} = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}\)
2. Convert moles to concentration using the volume (\(2.0\text{ dm}^3\)):
* \([\text{NO}_2] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}\)
* \([\text{N}_2\text{O}_4] = \frac{0.80\text{ mol}}{2.0\text{ dm}^3} = 0.40\text{ mol dm}^{-3}\)
3. Substitute the concentration values into the \(K_{\text{c}}\) expression:
\(K_{\text{c}} = \frac{0.40}{(0.20)^2} = \frac{0.40}{0.040} = 10\text{ dm}^3\text{ mol}^{-1}\).

Award 1 mark for selecting option C (10 dm\(^3\) mol\(^{-1}\)).

First, calculate the \(M_{\text{r}}\) of possible fragments and their corresponding protonated ions (\([M+\text{H}]^+\) values, where \(m/z = M_{\text{r}} + 1\)):
1. **Gly-Ala dipeptide**: \(M_{\text{r}} = 75 + 89 - 18 = 146\) (loss of one \(\text{H}_2\text{O}\)). \([M+\text{H}]^+ = 147\). (Matches Option A)
2. **Gly-Ser dipeptide**: \(M_{\text{r}} = 75 + 105 - 18 = 162\). \([M+\text{H}]^+ = 163\). (Matches Option B)
3. **Ala-Ser dipeptide**: \(M_{\text{r}} = 89 + 105 - 18 = 176\). \([M+\text{H}]^+ = 177\).
4. **Tripeptide (Gly-Ala-Ser)**: \(M_{\text{r}} = 75 + 89 + 105 - (2 \times 18) = 233\) (loss of two \(\text{H}_2\text{O}\)). \([M+\text{H}]^+ = 234\). (Matches Option D)

Since \(m/z = 191\) does not correspond to any valid combination of the amino acid components, it is not a possible protonated fragment ion.

Award 1 mark for selecting option C (191).

A redox reaction is thermodynamically feasible if \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} > 0\).

* **Option A**: \(\text{Fe}^{3+}\) is reduced (\(+0.77\text{ V}\)) and \(\text{I}^-\) is oxidized (\(+0.54\text{ V}\)). \(E^\theta_{\text{cell}} = +0.77 - (+0.54) = +0.23\text{ V}\) (Feasible).
* **Option B**: \(\text{I}_2\) is reduced (\(+0.54\text{ V}\)) and \(\text{S}_2\text{O}_3^{2-}\) is oxidized (\(+0.09\text{ V}\)). \(E^\theta_{\text{cell}} = +0.54 - (+0.09) = +0.45\text{ V}\) (Feasible).
* **Option C**: \(\text{Fe}^{3+}\) is reduced (\(+0.77\text{ V}\)) and \(\text{Ni}\) is oxidized (\(-0.25\text{ V}\)). \(E^\theta_{\text{cell}} = +0.77 - (-0.25) = +1.02\text{ V}\) (Feasible).
* **Option D**: \(\text{S}_4\text{O}_6^{2-}\) is reduced (\(+0.09\text{ V}\)) and \(\text{I}^-\) is oxidized (\(+0.54\text{ V}\)). \(E^\theta_{\text{cell}} = +0.09 - (+0.54) = -0.45\text{ V}\) (NOT Feasible, as \(E^\theta_{\text{cell}} < 0\)).

Award 1 mark for selecting option D.

1. **Determine orders of reaction**:
* Comparing Exp 1 and 2: \([\text{B}]\) and \([\text{C}]\) are constant. \([\text{A}]\) doubles, rate increases by a factor of 4 (\(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). Thus, the order with respect to \(\text{A}\) is 2.
* Comparing Exp 2 and 3: \([\text{A}]\) and \([\text{C}]\) are constant. \([\text{B}]\) doubles, rate is unchanged. Thus, the order with respect to \(\text{B}\) is 0.
* Comparing Exp 2 and 4: \([\text{A}]\) and \([\text{B}]\) are constant. \([\text{C}]\) doubles, rate doubles (\(8.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\)). Thus, the order with respect to \(\text{C}\) is 1.
2. **Write the rate equation**:
\(\text{Rate} = k[\text{A}]^2[\text{C}]\)
3. **Calculate the units of \(k\)**:
\(k = \frac{\text{Rate}}{[\text{A}]^2[\text{C}]}\)
\(\text{Units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

Award 1 mark for selecting option B (mol\(^{-2}\) dm\(^6\) s\(^{-1}\)).

The cell potential is calculated as \(E^{\theta}_{\text{cell}} = E^{\theta}_{\text{reduction}} - E^{\theta}_{\text{oxidation}} = 1.33\text{ V} - 0.77\text{ V} = +0.56\text{ V}\). The balanced half-equations show that dichromate(VI) accepts 6 electrons, whilst iron(II) loses 1 electron. Therefore, 6 moles of \(Fe^{2+}\) are oxidised by 1 mole of \(Cr_2O_7^{2-}\).

1 mark for the correct option A (correct calculation of +0.56 V and 1:6 stoichiometry).

When temperature increases, the Maxwell-Boltzmann distribution curve flattens and shifts to the right. The peak of the curve, representing the most probable molecular energy, moves to a higher energy. The total area under the curve represents the total number of particles, which remains constant.

1 mark for identifying that the peak moves to a higher energy and the total area stays constant (Option B).

The singlet at \(\delta = 2.0\text{ ppm}\) (3H) corresponds to a \(\text{CH}_3\text{CO-}\) group next to a carbonyl. The quartet at \(\delta = 4.1\text{ ppm}\) (2H) and triplet at \(\delta = 1.2\text{ ppm}\) (3H) form an ethyl ester group (\(-\text{OCH}_2\text{CH}_3\)) since the quartet is shifted downfield by the adjacent oxygen atom. Putting these pieces together gives ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

1 mark for correct identification of ethyl ethanoate (Option B).

Atom economy is calculated as (Molar mass of desired product / Total molar mass of all reactants) x 100%. Desired product is hydrazine (\(N_2H_4\)): \(M_r = 2(14.0) + 4(1.0) = 32.0\). Total molar mass of reactants: \(M_r(NaClO) + 2 \times M_r(NH_3) = (23.0 + 35.5 + 16.0) + 2(14.0 + 3.0) = 74.5 + 34.0 = 108.5\). Atom economy = \((32.0 / 108.5) \times 100\% = 29.49\% \approx 29.5\%\).

1 mark for calculating the correct percentage atom economy as 29.5% (Option A).

The equilibrium expression is \(K_c = [N_2O_4] / [NO_2]^2\). Substituting the equilibrium values: \(K_c = 0.80 / (0.40)^2 = 0.80 / 0.16 = 5.0\). The units are \((\text{mol dm}^{-3}) / (\text{mol dm}^{-3})^2 = \text{dm}^3\text{ mol}^{-1}\).

1 mark for finding both the correct value (5.0) and unit (dm3 mol-1) corresponding to Option B.

The 3:1 ratio of the \(M\) and \(M+2\) peaks indicates the presence of a single chlorine atom, which consists of the isotopes \(^{35}Cl\) (approx. 75%) and \(^{37}Cl\) (approx. 25%). The compound must also have a molecular mass corresponding to the \(M\) peak at \(m/z = 156\) (using \(^{35}Cl\)). (3-chloropropyl)benzene has the formula \(C_9H_{13}Cl\). The molecular mass with \(^{35}Cl\) is \(9(12) + 13(1) + 35 = 156\). Bromobenzene has peaks of equal intensity (1:1 ratio) at \(m/z = 156\) and \(158\).

1 mark for selecting (3-chloropropyl)benzene based on isotopic ratio of chlorine and mass calculation (Option B).

Iodide ions are strong reducing agents and reduce the sulfur in concentrated sulfuric acid (oxidation state +6) to various lower oxidation states: sulfur dioxide (\(SO_2\), +4), sulfur (\(S\), 0), and hydrogen sulfide (\(H_2S\), -2). Hydrogen gas (\(H_2\)) is not produced as a reduction product of sulfuric acid in this reaction; the hydrogen atoms remain in the oxidation state of +1 in substances like water, hydrogen iodide, and hydrogen sulfide.

1 mark for identifying Hydrogen as not being a reduction product of this reaction (Option D).

Comparing Exp 1 and Exp 2: when \([B]\) is kept constant, doubling \([A]\) quadruples the rate (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)), indicating a second-order reaction with respect to A. Comparing Exp 2 and Exp 3: when \([A]\) is kept constant, doubling \([B]\) doubles the rate (from \(8.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\)), indicating a first-order reaction with respect to B. The rate equation is \(\text{Rate} = k[A]^2[B]\).

1 mark for deducing the correct rate equation of Rate = k[A]2[B] (Option B).

Methyl propanoate has the formula \(\text{CH}_3\text{CH}_2\text{COOCH}_3\). The protons in the methyl group of the ethyl chain (adjacent to a \(\text{-CH}_2-\) group) produce a triplet integrating to 3H at \(\delta = 1.2\text{ ppm}\). The protons in the \(\text{-CH}_2-\) group (adjacent to \(\text{-CH}_3\) and the carbonyl group) produce a quartet integrating to 2H at \(\delta = 2.3\text{ ppm}\). The protons on the ester methyl group (\(\text{-OCH}_3\)) are next to no adjacent protons and are shielded by oxygen, giving a singlet integrating to 3H at \(\delta = 3.7\text{ ppm}\).

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

Iodide (\(\text{I}^-\)) is a stronger reducing agent (more easily oxidized) than bromide (\(\text{Br}^-\)) because the standard electrode potential for iodine/iodide is less positive than that of bromine/bromide. Therefore, iodide ions are oxidized first to iodine (\(\text{I}_2\)). Chlorine gas (\(\text{Cl}_2\)) acts as the oxidizing agent and is reduced to chloride ions (\(\text{Cl}^-\)), where its oxidation state changes from 0 in \(\text{Cl}_2\) to -1 in \(\text{Cl}^-\).

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

Let the initial rate be \(\text{Rate}_1 = k[\text{A}][\text{B}]^2\). When \([\text{A}]\) is halved and \([\text{B}]\) is doubled: \(\text{Rate}_2 = k (0.5[\text{A}])(2[\text{B}])^2 = k \cdot 0.5[\text{A}] \cdot 4[\text{B}]^2 = 2k[\text{A}][\text{B}]^2 = 2 \times \text{Rate}_1\). Therefore, the rate doubles. To find the units of \(k\): \(k = \text{Rate} / ([\text{A}][\text{B}]^2)\). Units of \(k = \text{mol dm}^{-3}\text{ s}^{-1} / ((\text{mol dm}^{-3})(\text{mol dm}^{-3})^2) = \text{dm}^6\text{mol}^{-2}\text{s}^{-1}\).

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

First, calculate the amount in moles of iodine: \(\text{Moles of I}_2 = 0.0200\text{ mol dm}^{-3} \times (18.20 / 1000)\text{ dm}^3 = 3.64 \times 10^{-4}\text{ mol}\). Next, determine the moles of thiosulfate using the 2:1 ratio: \(\text{Moles of S}_2\text{O}_3^{2-} = 2 \times 3.64 \times 10^{-4}\text{ mol} = 7.28 \times 10^{-4}\text{ mol}\). Finally, calculate the concentration: \(\text{Concentration} = (7.28 \times 10^{-4}\text{ mol}) / (0.0250\text{ dm}^3) = 0.02912\text{ mol dm}^{-3} \approx 0.0291\text{ mol dm}^{-3}\).

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

The equilibrium constant, \(K_c\), is only affected by temperature. Because the forward reaction is exothermic, decreasing the temperature shifts the equilibrium position to the right to oppose the change, which increases the concentration of the products relative to the reactants at equilibrium. This leads to both an increase in the equilibrium constant, \(K_c\), and an increased equilibrium yield of \(\text{SO}_3\). While increasing the pressure or removing \(\text{SO}_3\) increases the yield of \(\text{SO}_3\), they do not alter the value of \(K_c\). Adding a catalyst increases the rate of reaction but does not affect the yield or the value of \(K_c\).

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

For a first-order reaction, the half-life is constant and independent of the starting concentration. Let us determine how many half-lives have elapsed as the concentration falls from \(0.80\text{ mol dm}^{-3}\) to \(0.10\text{ mol dm}^{-3}\): After 1 half-life: \(0.80 \rightarrow 0.40\text{ mol dm}^{-3}\); After 2 half-lives: \(0.40 \rightarrow 0.20\text{ mol dm}^{-3}\); After 3 half-lives: \(0.20 \rightarrow 0.10\text{ mol dm}^{-3}\). Since 3 half-lives are equal to 90 minutes, we have \(3 \times t_{1/2} = 90\text{ minutes}\), which gives \(t_{1/2} = 30\text{ minutes}\).

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

(a)(i) Anode (oxidation): \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\). Cathode (reduction): \(2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq})\) (or \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g})\)). (a)(ii) Calculate total charge passed: \(Q = I \times t = 5.00\text{ A} \times (2.00 \times 3600\text{ s}) = 36000\text{ C}\). Moles of electrons: \(n(\text{e}^-) = 36000 / 96500 = 0.3731\text{ mol}\). Moles of \(\text{Cl}_2\) produced: \(n(\text{Cl}_2) = 0.3731 / 2 = 0.1865\text{ mol}\). Volume of \(\text{Cl}_2\) gas: \(V = 0.1865 \times 24.0 = 4.48\text{ dm}^3\). (b) Moles of \(\text{S}_2\text{O}_3^{2-}\) used: \(n(\text{S}_2\text{O}_3^{2-}) = (22.40 / 1000) \times 0.100 = 2.24 \times 10^{-3}\text{ mol}\). Moles of \(\text{I}_2\) reacted: \(n(\text{I}_2) = 2.24 \times 10^{-3} / 2 = 1.12 \times 10^{-3}\text{ mol}\). Moles of \(\text{ClO}^-\) in the titrated \(25.0\text{ cm}^3\) sample: \(n(\text{ClO}^-) = n(\text{I}_2) = 1.12 \times 10^{-3}\text{ mol}\). Moles of \(\text{ClO}^-\) in the \(250.0\text{ cm}^3\) volumetric flask: \(1.12 \times 10^{-3} \times 10 = 1.12 \times 10^{-2}\text{ mol}\). Concentration of \(\text{ClO}^-\) in original \(10.0\text{ cm}^3\) bleach: \(c = 1.12 \times 10^{-2}\text{ mol} / 0.0100\text{ dm}^3 = 1.12\text{ mol dm}^{-3}\). (c) In the reaction \(2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-\), iodine atoms change oxidation state from 0 in \(\text{I}_2\) to -1 in \(\text{I}^-\), which is reduction because the oxidation state decreases. Sulfur atoms change oxidation state from +2 in \(\text{S}_2\text{O}_3^{2-}\) to +2.5 in \(\text{S}_4\text{O}_6^{2-}\), which is oxidation because the oxidation state increases. (d) Starch forms a strong, stable, and irreversible complex with high concentrations of iodine. If added at the beginning, the complex does not easily release iodine, resulting in an inaccurate or sluggish end-point. Adding it near the end-point (when the solution is pale straw yellow) ensures all remaining iodine is quickly and reversibly released, giving a sharp and distinct color change from blue-black to colourless.

(a)(i) 2 marks: 1 mark for correct anode reaction and identification, 1 mark for correct cathode reaction and identification. (a)(ii) 3 marks: 1 mark for calculating charge \(Q = 36000\text{ C}\), 1 mark for calculating moles of electrons and dividing by 2 to get \(0.1865\text{ mol}\) of \(\text{Cl}_2\), 1 mark for final volume of \(4.48\text{ dm}^3\). (b) 6 marks: 1 mark for moles of thiosulfate, 1 mark for moles of iodine, 1 mark for moles of chlorate(I) in the sample, 1 mark for scaling up to 250 cm3, 1 mark for dividing by original volume (10.0 cm3), 1 mark for correct final value of \(1.12\text{ mol dm}^{-3}\) with units. (c) 3 marks: 1 mark for iodine oxidation states (0 to -1) and reduction, 1 mark for sulfur oxidation states (+2 to +2.5) and oxidation, 1 mark for linking oxidation/reduction definitions to change in oxidation states. (d) 2 marks: 1 mark for stating starch-iodine complex is irreversible/insoluble at high concentrations, 1 mark for stating that adding it near the end-point yields a sharp, reversible color change.

(a)(i) Order with respect to \(\text{NO}\): Comparing Experiments 1 and 2, \([\text{O}_2]\) is constant while \([\text{NO}]\) doubles. The initial rate increases by a factor of \(1.12 \times 10^{-4} / 2.80 \times 10^{-5} = 4\). Since \(2^2 = 4\), the order with respect to \(\text{NO}\) is 2. Order with respect to \(\text{O}_2\): Comparing Experiments 1 and 3, \([\text{NO}]\) is constant while \([\text{O}_2]\) triples. The initial rate increases by a factor of \(8.40 \times 10^{-5} / 2.80 \times 10^{-5} = 3\). Since \(3^1 = 3\), the order with respect to \(\text{O}_2\) is 1. (a)(ii) \(\text{Rate} = k[\text{NO}]^2[\text{O}_2]\). (a)(iii) Substitution using Experiment 1: \(2.80 \times 10^{-5} = k \times (0.0100)^2 \times (0.0100) \Rightarrow k = 2.80 \times 10^{-5} / 1.00 \times 10^{-6} = 28.0\). Units: \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\). (b) Step 2 is the slow (rate-determining) step: \(\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{O}_2]\). From Step 1, the fast pre-equilibrium has \(K_c = [\text{N}_2\text{O}_2]/[\text{NO}]^2\), so \([\text{N}_2\text{O}_2] = K_c[\text{NO}]^2\). Substituting this expression into the rate equation gives: \(\text{Rate} = k_2 K_c [\text{NO}]^2[\text{O}_2] = k[\text{NO}]^2[\text{O}_2]\), which is consistent with the experimental rate equation. (c) From the Arrhenius equation, the gradient of \(\ln k\) vs \(1/T\) is equal to \(-E_a/R\). \(-E_a/R = -1450 \Rightarrow E_a = 1450 \times 8.31 = 12049.5\text{ J mol}^{-1}\). Dividing by 1000 to convert to \(\text{kJ mol}^{-1}\) gives \(E_a = +12.0\text{ kJ mol}^{-1}\) (to 3 sig figs). (d) An increase in temperature shifts the Maxwell-Boltzmann distribution curve to the right, lowering the peak and spreading it out. This means a significantly higher fraction of the gaseous molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)), leading to a higher frequency of successful collisions. In a catalytic converter, nitrogen oxides (\(\text{NO}_x\)) are adsorbed onto the transition metal (Pt/Pd/Rh) catalyst surface, weakening internal bonds and allowing them to react with lower activation energy to yield harmless gases (\(\text{N}_2\) and \(\text{O}_2\)).

(a)(i) 2 marks: 1 mark for order of NO with correct comparison of Exp 1 and 2, 1 mark for order of O2 with correct comparison of Exp 1 and 3. (a)(ii) 1 mark for correct rate equation: \(\text{Rate} = k[\text{NO}]^2[\text{O}_2]\). (a)(iii) 3 marks: 1 mark for correct substitution of values, 1 mark for correct numerical value of \(28.0\), 1 mark for units \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\). (b) 3 marks: 1 mark for setting up rate equation of rate-determining step, 1 mark for substituting \([\text{N}_2\text{O}_2]\) from Step 1 equilibrium, 1 mark for showing consistency with experimental rate equation. (c) 3 marks: 1 mark for linking gradient to \(-E_a / R\), 1 mark for correct calculations in Joules, 1 mark for final value of \(+12.0\text{ kJ mol}^{-1}\). (d) 4 marks: 1 mark for explaining temperature effect on Maxwell-Boltzmann curve, 1 mark for linking this to the fraction of successful collisions, 1 mark for explaining transition metal catalyst adsorption, 1 mark for explaining lowering of activation energy.

(a)(i) Mass of water lost: \(3.570\text{ g} - 1.950\text{ g} = 1.620\text{ g}\). Moles of \(\text{H}_2\text{O} = 1.620 / 18.0 = 0.0900\text{ mol}\). (a)(ii) The chemist should heat the sample, weigh it, and repeat the heating and weighing process until the mass remains completely constant (heating to constant mass). (b)(i) Moles of \(\text{AgCl} = 4.302 / 143.4 = 0.0300\text{ mol}\). Since there is 1 mole of chloride ions in 1 mole of silver chloride, moles of \(\text{Cl}^- = 0.0300\text{ mol}\). (b)(ii) \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\). (c)(i) Moles of \(\text{MCl}_2\) in the sample is half the moles of chloride ions, because the formula contains two chloride ions per metal ion. \(n(\text{MCl}_2) = 0.0300 / 2 = 0.0150\text{ mol}\). Ratio of \(\text{H}_2\text{O} : \text{MCl}_2 = 0.0900 / 0.0150 = 6\). Thus, \(x = 6\). (c)(ii) Molar mass of \(\text{MCl}_2 = 1.950\text{ g} / 0.0150\text{ mol} = 130.0\text{ g mol}^{-1}\). Molar mass of \(\text{M} = 130.0 - 2(35.5) = 130.0 - 71.0 = 59.0\text{ g mol}^{-1}\). Looking at the periodic table, this corresponds to Cobalt (Co) (actual molar mass \(58.9\text{ g mol}^{-1}\)), which matches the blue (anhydrous) to pink (hydrated) transition metal chloride colour change. (d)(i) A coordinate bond is a covalent bond in which both of the shared electrons are provided by the same atom or ligand. (d)(ii) When ligands coordinate to a transition metal ion, they cause the d-orbitals to split into two different energy levels. Electrons in lower energy d-orbitals absorb a specific frequency of visible light to be promoted (excited) to a higher energy d-orbital. The remaining, non-absorbed frequencies are transmitted or reflected and are seen as the complementary colour.

(a)(i) 2 marks: 1 mark for mass of water lost (1.620 g), 1 mark for moles of water (0.0900 mol). (a)(ii) 1 mark for heating to constant mass. (b)(i) 2 marks: 1 mark for moles of AgCl (0.0300 mol), 1 mark for equating this to moles of chloride. (b)(ii) 1 mark for correct ionic equation with state symbols. (c)(i) 3 marks: 1 mark for moles of \(\text{MCl}_2\) (0.0150 mol), 1 mark for setting up the 1:6 ratio, 1 mark for concluding \(x = 6\). (c)(ii) 4 marks: 1 mark for calculating molar mass of anhydrous salt (130.0), 1 mark for calculating molar mass of metal (59.0), 1 mark for identifying cobalt / Co, 1 mark for referencing the blue-to-pink color change as characteristic of cobalt(II) chloride. (d)(i) 1 mark for definition of coordinate bond. (d)(ii) 2 marks: 1 mark for stating d-orbitals split, 1 mark for explaining promotion of electrons by absorption of visible light.

(a) \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\). Units: \(\frac{\text{mol dm}^{-3}}{\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6\) (or \(\text{dm}^6\text{ mol}^{-2}\)). (b)(i) Let the change in moles of \(\text{CH}_3\text{OH}\) be \(+0.60\text{ mol}\). According to stoichiometry, change in \(\text{CO}\) is \(-0.60\text{ mol}\) and change in \(\text{H}_2\) is \(-1.20\text{ mol}\). Equilibrium moles: \(n(\text{CO}) = 1.50 - 0.60 = 0.90\text{ mol}\), \(n(\text{H}_2) = 3.00 - 1.20 = 1.80\text{ mol}\), \(n(\text{CH}_3\text{OH}) = 0.60\text{ mol}\). Dividing by volume of \(5.00\text{ dm}^3\): \([\text{CO}] = 0.90 / 5.00 = 0.180\text{ mol dm}^{-3}\), \([\text{H}_2] = 1.80 / 5.00 = 0.360\text{ mol dm}^{-3}\), \([\text{CH}_3\text{OH}] = 0.60 / 5.00 = 0.120\text{ mol dm}^{-3}\). (b)(ii) Substituting into the equilibrium constant expression: \(K_c = \frac{0.120}{0.180 \times (0.360)^2} = \frac{0.120}{0.023328} = 5.14\text{ dm}^6\text{ mol}^{-2}\) (to 3 sig figs). (c)(i) An increase in temperature causes the position of equilibrium to shift to the left (in the endothermic direction) to absorb the added heat energy. Since the forward reaction is exothermic, the concentrations of reactants increase while that of methanol decreases. Therefore, \(K_c\) decreases. (c)(ii) Halving the volume of the vessel increases the overall gas pressure. The position of equilibrium shifts to the right (the side with fewer gas molecules: 1 mole on the right vs 3 moles on the left) to reduce pressure. However, the value of \(K_c\) remains completely unchanged because temperature is constant. (d) High pressures are chosen because they increase both the rate of reaction (particles closer together) and the yield of methanol (shifts equilibrium to the right), but is limited by the safety and capital cost of high-pressure equipment. A moderate temperature (500 K) is a compromise: low temperatures would shift the equilibrium right to increase yield but result in an extremely slow rate. A copper-based catalyst allows a commercially viable rate at this moderate temperature without reducing the yield.

(a) 3 marks: 2 marks for correct formula, 1 mark for units \(\text{dm}^6\text{ mol}^{-2}\). (b)(i) 3 marks: 1 mark for correct equilibrium moles, 1 mark for dividing all moles by volume, 1 mark for correct final concentrations. (b)(ii) 3 marks: 1 mark for correct substitution of values, 1 mark for correct arithmetic, 1 mark for final answer of \(5.14\). (c)(i) 3 marks: 1 mark for shifting left, 1 mark for explaining in terms of the endothermic direction, 1 mark for stating \(K_c\) decreases. (c)(ii) 2 marks: 1 mark for shifting right, 1 mark for stating \(K_c\) remains constant. (d) 2 marks: 1 mark for explaining high pressure benefits vs cost/safety, 1 mark for explaining moderate temperature + catalyst as a rate/yield compromise.

(a) Calculate empirical formula from mass percentage: \(n(\text{C}) = 46.15 / 12.0 = 3.846\text{ mol}\); \(n(\text{H}) = 7.69 / 1.0 = 7.69\text{ mol}\); \(n(\text{O}) = 46.15 / 16.0 = 2.884\text{ mol}\). Divide by the smallest (2.884): \(\text{C} = 1.33\), \(\text{H} = 2.67\), \(\text{O} = 1.00\). Multiply by 3 to obtain whole numbers: \(\text{C} = 4\), \(\text{H} = 8\), \(\text{O} = 3\). Empirical formula is \(\text{C}_4\text{H}_8\text{O}_3\). The empirical formula mass is \((4 \times 12.0) + (8 \times 1.0) + (3 \times 16.0) = 104\text{ g mol}^{-1}\). Because the molecular ion peak is at \(m/z = 104\), the molecular formula is also \(\text{C}_4\text{H}_8\text{O}_3\). (b) Functional groups: \(2500 - 3300\text{ cm}^{-1}\): O-H stretch of a carboxylic acid; \(3200 - 3600\text{ cm}^{-1}\): O-H stretch of an alcohol; \(1715\text{ cm}^{-1}\): C=O stretch of a carboxylic acid (carbonyl group). (c) Based on the molecular formula and IR evidence of carboxylic acid and alcohol functional groups, \(Y\) is 3-hydroxybutanoic acid, \(\text{CH}_3\text{CH(OH)CH}_2\text{COOH}\). Carbon-13 NMR Assignments: \(\delta = 178\text{ ppm}\) corresponds to the highly deshielded carbonyl carbon of \(\text{-COOH}\); \(\delta = 64\text{ ppm}\) is the carbon attached to oxygen (\(\text{-CH(OH)-}\)); \(\delta = 43\text{ ppm}\) is the \(\text{-CH}_2-\) carbon adjacent to the carbonyl; \(\delta = 22\text{ ppm}\) is the methyl carbon (\(\text{-CH}_3\)). Proton NMR Assignments and Splitting: \(\delta = 1.2\text{ ppm}\) (doublet, 3H) is the methyl group (\(\text{-CH}_3\)) split by the adjacent single \(\text{-CH-}\) proton (\(n+1=2\)); \(\delta = 2.5\text{ ppm}\) (doublet, 2H) is the \(\text{-CH}_2-\) group split by the adjacent \(\text{-CH-}\) proton (\(n+1=2\)); \(\delta = 4.2\text{ ppm}\) (multiplet, 1H) is the \(\text{-CH-}\) proton coupled to the 3 methyl protons and 2 methylene protons (giving a complex multiplet); \(\delta = 4.8\text{ ppm}\) is the alcohol \(\text{-OH}\) proton, and \(\delta = 12.1\text{ ppm}\) is the acidic \(\text{-COOH}\) proton. Added \(\text{D}_2\text{O}\) undergoes rapid chemical exchange with the labile protons: \(\text{R-OH} + \text{D}_2\text{O} \rightleftharpoons \text{R-OD} + \text{HOD}\) and \(\text{R-COOH} + \text{D}_2\text{O} \rightleftharpoons \text{R-COOD} + \text{HOD}\). Since deuterium does not absorb in the standard \(^1\text{H}\) NMR frequency range, these signals disappear. (d) Condensation of \(\text{HO-CH(CH}_3\text{)-CH}_2\text{-COOH}\) yields a polyester repeating unit: \(\text{[-O-CH(CH}_3\text{)-CH}_2\text{-CO-]}\).

(a) 3 marks: 1 mark for correct molar ratio calculation, 1 mark for correct empirical formula \(\text{C}_4\text{H}_8\text{O}_3\), 1 mark for linking empirical mass to \(m/z = 104\) to deduce molecular formula. (b) 3 marks: 1 mark for O-H carboxylic acid, 1 mark for O-H alcohol, 1 mark for C=O carboxylic acid. (c) 8 marks (Level of Response style): 7-8 marks for clear deduction of 3-hydroxybutanoic acid structure, correct assignment of all 13C and 1H NMR peaks with explanation of splitting, and description of deuterium exchange. 4-6 marks for mostly correct structure/isomer, major NMR peaks assigned, partial explanation of splitting and D2O exchange. 1-3 marks for partial structure or key functional groups identified with some correct NMR assignments. (d) 2 marks: 1 mark for correct ester backbone linkage, 1 mark for correct methyl and chain substituents shown as part of the repeating unit.

(a) Peak at \(3300\text{ cm}^{-1}\) indicates an alcohol hydroxyl group (\(-OH\)). Peak at \(1735\text{ cm}^{-1}\) indicates a carbonyl group (\(C=O\)) of an ester (or carboxylic acid).

(b) \(M_r\) of \(C_7H_{12}O_5 = 7 \times 12.0 + 12 \times 1.0 + 5 \times 16.0 = 176.0\).
Percentage of carbon = \(\frac{7 \times 12.0}{176.0} \times 100\% = 47.7\%\).

(c) Level of Response (LOR) marking:
- IR data: Shows presence of alcohol \(-OH\) and ester/acid \(C=O\).
- MS data: Molecular ion peak at \(m/z = 176\) matches the molecular mass of \(C_7H_{12}O_5\).
- \(^1\text{H}\) NMR assignments:
- \(\delta = 11.00\text{ ppm}\) (1H, broad singlet, disappears with \(D_2O\)) is the carboxylic acid \(-COOH\) proton.
- \(\delta = 3.50\text{ ppm}\) (1H, broad singlet, disappears with \(D_2O\)) is the alcohol \(-OH\) proton.
- \(\delta = 1.25\text{ ppm}\) (3H, doublet) is the \(CH_3\) attached to the \(CH(OH)\) carbon, split by 1 adjacent H.
- \(\delta = 1.45\text{ ppm}\) (3H, doublet) is the \(CH_3\) attached to the \(CH(O-CO)\) carbon, split by 1 adjacent H.
- \(\delta = 2.50\text{ ppm}\) (2H, doublet) is the \(-CH_2-\) group adjacent to the \(CH(OH)\) proton.
- \(\delta = 4.20\text{ ppm}\) (1H, sextet) is the \(-CH(OH)-\) proton, coupled to 3 protons from \(CH_3\) and 2 from \(CH_2\) (total of 5 adjacent protons, \(n+1=6\)).
- \(\delta = 5.10\text{ ppm}\) (1H, quartet) is the ester-linked \(-CH(O-CO)-\) proton, coupled to 3 protons of \(CH_3\) (\(n+1=4\)).

(d) Coupling explanations:
- For the doublet at \(\delta = 1.25\text{ ppm}\), the \(CH_3\) protons are adjacent to exactly one proton (\(n = 1\)) on the neighboring carbon. According to the \(n+1\) rule, this splits the signal into \(1 + 1 = 2\) peaks (a doublet).
- For the sextet at \(\delta = 4.20\text{ ppm}\), the single proton of the \(-CH(OH)-\) group is adjacent to five protons in total (three from the adjacent \(-CH_3\) group and two from the adjacent \(-CH_2-\) group). According to the \(n+1\) rule, this splits the signal into \(5 + 1 = 6\) peaks (a sextet).

(e) There are 2 chiral centres (the two \(CH\) carbons bonded to oxygen atoms). Structures must show 3D tetrahedral geometry (wedges and dashes) around at least one of these chiral centres to show enantiomeric/diastereomeric relationship.

(f) Equation:
\(CH_3-CH(OH)-CH_2-CO-O-CH(CH_3)-COOH + 2NaOH \rightarrow CH_3-CH(OH)-CH_2-COONa + CH_3-CH(OH)-COONa + H_2O\)
(or ionic equivalent: \(CH_3CH(OH)CH_2COOCH(CH_3)COOH + 2OH^- \rightarrow CH_3CH(OH)CH_2COO^- + CH_3CH(OH)COO^- + H_2O\))

(a) [2 marks]
- 1 mark for identifying alcohol hydroxyl group (\(-OH\))
- 1 mark for identifying carbonyl group (\(C=O\)) / ester / carboxylic acid

(b) [2 marks]
- 1 mark for showing \(M_r = 176.0\)
- 1 mark for final percentage of \(47.7\%\) (accept 47.72% to 47.73%)

(c) [6 marks] Level of Response:
- Level 3 (5-6 marks): Compiles comprehensive evidence from IR, MS, and NMR. Fully assigns all 7 proton environments, detailing chemical shifts, integrations, and splitting patterns, demonstrating a cohesive proof.
- Level 2 (3-4 marks): Partially assigns NMR peaks (at least 4) and links them to structure. Mentions IR and MS.
- Level 1 (1-2 marks): Basic identification of some NMR peaks or functional groups from IR.

(d) [3 marks]
- 1 mark for stating the \(n+1\) rule
- 1 mark for explanation of the doublet (coupled to 1 adjacent H)
- 1 mark for explanation of the sextet (coupled to 5 adjacent H: 3 on \(CH_3\) and 2 on \(CH_2\))

(e) [4 marks]
- 1 mark for stating '2' chiral centres
- 3 marks for drawing two distinct stereoisomers with 3D tetrahedral representation (using wedges/dashes) around the chiral centres

(f) [3 marks]
- 1 mark for reactant formula and correct stoichiometry (reacts with \(2NaOH\) or \(2OH^-\))
- 1 mark for correct structures of the two sodium carboxylate salts (sodium 3-hydroxybutanoate and sodium 2-hydroxypropanoate/lactate)
- 1 mark for water as a product

(a) Equation: \(ClO^-(aq) + 2I^-(aq) + 2H^+(aq) \rightarrow Cl^-(aq) + I_2(aq) + H_2O(l)\)

(b) Iodide ions act as a reducing agent (or are oxidized); they lose electrons (one electron per iodide ion: \(2I^- \rightarrow I_2 + 2e^-\)).

(c)(i) The colour change at the end-point is blue-black to colourless. Starch should be added near the end-point when the titration mixture becomes a pale straw-yellow colour.
(ii) Calculation:
- Moles of \(S_2O_3^{2-}\) used: \(n = C \times V = 0.100\text{ mol dm}^{-3} \times 0.02240\text{ dm}^3 = 2.24 \times 10^{-3}\text{ mol}\).
- From titration equation: \(n(I_2) = 0.5 \times n(S_2O_3^{2-}) = 1.12 \times 10^{-3}\text{ mol}\).
- From equation in (a): \(n(ClO^-)\) in \(25.0\text{ cm}^3\) aliquot \(= n(I_2) = 1.12 \times 10^{-3}\text{ mol}\).
- Moles of \(ClO^-\) in \(250.0\text{ cm}^3\) volumetric flask \(= 1.12 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 1.12 \times 10^{-2}\text{ mol}\).
- This came from \(10.0\text{ cm}^3\) of original bleach. Concentration in original bleach: \(C = \frac{1.12 \times 10^{-2}\text{ mol}}{0.0100\text{ dm}^3} = 1.12\text{ mol dm}^{-3}\).

(d)(i) Oxidation numbers:
- \(NaClO_3\): \(+5\)
- \(HCl\): \(-1\)
- \(ClO_2\): \(+4\)
- \(Cl_2\): \(0\)
- \(NaCl\): \(-1\)
- Species oxidized: \(HCl\) (chlorine changes from \(-1\) to \(0\) in \(Cl_2\)).
- Species reduced: \(NaClO_3\) (chlorine changes from \(+5\) to \(+4\) in \(ClO_2\)).

(ii) Formula: \(\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100\%\)
- Desired product mass: \(2 \times M_r(ClO_2) = 2 \times [35.5 + 2(16.0)] = 2 \times 67.5 = 135.0\text{ g mol}^{-1}\)
- Reactant mass: \(2 \times M_r(NaClO_3) + 4 \times M_r(HCl) = 2(23.0 + 35.5 + 48.0) + 4(36.5) = 2(106.5) + 146.0 = 359.0\text{ g mol}^{-1}\)
- \(\text{Atom economy} = \frac{135.0}{359.0} \times 100\% = 37.6\%\) (or 38%).

(a) [3 marks]
- 1 mark for correct reactants and products
- 1 mark for correct charge balance
- 1 mark for correct atom balance and state symbols

(b) [1 mark]
- 1 mark for stating that iodide ions act as a reducing agent / lose electrons / are oxidized

(c)(i) [3 marks]
- 1 mark for colour change: blue-black to colourless
- 1 mark for adding starch near/just before the end-point
- 1 mark for stating when the solution is pale straw-yellow
(ii) [5 marks]
- 1 mark for calculating moles of \(S_2O_3^{2-} = 2.24 \times 10^{-3}\text{ mol}\)
- 1 mark for calculating moles of \(I_2 = 1.12 \times 10^{-3}\text{ mol}\)
- 1 mark for equating moles of \(ClO^- = 1.12 \times 10^{-3}\text{ mol}\) in aliquot
- 1 mark for scaling up to \(250.0\text{ cm}^3\) volumetric flask \(= 1.12 \times 10^{-2}\text{ mol}\)
- 1 mark for final concentration \(= 1.12\text{ mol dm}^{-3}\) (3 sig figs required)

(d)(i) [4 marks]
- 1 mark for all correct oxidation numbers of chlorine (\(+5\), \(-1\), \(+4\), \(0\), \(-1\))
- 1 mark for identifying oxidized species (\(HCl\) / \(Cl^-\)) with explanation (O.N. increases from \(-1\) to \(0\))
- 2 marks for identifying reduced species (\(NaClO_3\) / \(ClO_3^-\)) with explanation (O.N. decreases from \(+5\) to \(+4\))
(ii) [4 marks]
- 1 mark for correct mass of desired products (\(135.0\))
- 1 mark for correct mass of all reactants (\(359.0\))
- 2 marks for correct calculation of atom economy (\(37.6\%\))

(a) Both \(S_2O_8^{2-}\) and \(I^-\) are negatively charged ions. Due to electrostatic repulsion between like charges, there is a high energy barrier (high activation energy) to bring them close enough to react, making the rate very slow.

(b)(i) Order of reaction:
- Comparing Exp 1 & Exp 2: \([I^-]\) is constant. \([S_2O_8^{2-}]\) doubles (0.050 to 0.100), and the initial rate doubles (\(2.20 \times 10^{-4}\) to \(4.40 \times 10^{-4}\)). Thus, the order with respect to \(S_2O_8^{2-}\) is 1.
- Comparing Exp 1 & Exp 3: \([S_2O_8^{2-}]\) is constant. \([I^-]\) doubles (0.050 to 0.100), and the initial rate doubles (\(2.20 \times 10^{-4}\) to \(4.40 \times 10^{-4}\)). Thus, the order with respect to \(I^-\) is 1.
- Rate equation: \(\text{Rate} = k[S_2O_8^{2-}][I^-]\).
(ii) Calculation of \(k\):
Using Experiment 1 values:
\(2.20 \times 10^{-4} = k(0.050)(0.050)\)
\(2.20 \times 10^{-4} = k(0.0025)\)
\(k = 0.088\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
Units: \(\text{mol dm}^{-3}\text{ s}^{-1} / (\text{mol dm}^{-3})^2 = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) \(\text{Gradient} = -\frac{E_a}{R} = -6450\text{ K}\).
\(E_a = 6450 \times 8.314 = 53625.3\text{ J mol}^{-1} = 53.6\text{ kJ mol}^{-1}\).

(d)(i) Homogeneous catalysis provides an alternative pathway with a lower activation energy. Instead of two negative ions colliding directly, the catalyst reacts sequentially:
Step 1: \(2Fe^{2+}(aq) + S_2O_8^{2-}(aq) \rightarrow 2Fe^{3+}(aq) + 2SO_4^{2-}(aq)\)
Step 2: \(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\)
Because each of these steps involves collisions between oppositely charged ions (positive and negative), the electrostatic repulsion is absent, resulting in a much lower activation energy and a faster reaction rate.
(ii) \(Fe^{3+}(aq)\) works because the sequence of reactions can proceed in the reverse order of starting materials, starting with Step 2 (oxidation of iodide by \(Fe^{3+}\)) followed by Step 1 (reduction of peroxodisulfate by the generated \(Fe^{2+}\)). Since the iron species is regenerated in a catalytic cycle, either oxidation state works as an effective catalyst.

(a) [2 marks]
- 1 mark for mentioning electrostatic repulsion between two negative ions
- 1 mark for linking repulsion to a high activation energy

(b)(i) [4 marks]
- 1 mark for working showing order with respect to \(S_2O_8^{2-}\) is 1
- 1 mark for working showing order with respect to \(I^-\) is 1
- 2 marks for correct rate equation: \(\text{Rate} = k[S_2O_8^{2-}][I^-]\)
(ii) [3 marks]
- 1 mark for correct value \(0.088\) (accept \(8.8 \times 10^{-2}\))
- 2 marks for correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(c) [3 marks]
- 1 mark for equating \(\text{gradient} = -E_a/R\)
- 1 mark for calculating \(E_a\) in Joules \(53625\text{ J mol}^{-1}\)
- 1 mark for final answer in \(\text{kJ mol}^{-1} = 53.6\) (accept 53.6 to 54.0)

(d)(i) [4 marks]
- 1 mark for stating that catalysis provides an alternative pathway with a lower activation energy
- 1 mark for Equation 1: \(2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}\)
- 1 mark for Equation 2: \(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\)
- 1 mark for explaining that steps involve collisions of opposite charges reducing electrostatic barrier
(ii) [4 marks]
- 2 marks for explaining that \(Fe^{3+}\) reacts first with iodide to form \(Fe^{2+}\) and iodine
- 2 marks for explaining that the generated \(Fe^{2+}\) then reacts with peroxodisulfate to regenerate \(Fe^{3+}\), completing the catalytic cycle

(a)(i) \(1.0\text{ m}^3 = 1000\text{ dm}^3\).
Mass of bromide = \(65\text{ mg dm}^{-3} \times 1000\text{ dm}^3 = 65000\text{ mg} = 65\text{ g}\).
(ii) Density of seawater = \(1.025\text{ g cm}^{-3} = 1025\text{ g dm}^{-3}\).
Mass of \(1.0\text{ dm}^3\) of seawater = \(1025\text{ g}\).
Concentration in ppm = \(\frac{\text{mass of bromide in g}}{\text{mass of seawater in g}} \times 10^6 = \frac{0.065\text{ g}}{1025\text{ g}} \times 10^6 = 63.4\text{ ppm}\).

(b)(i) Seawater processed in 24 hours = \(1500\text{ m}^3\text{ h}^{-1} \times 24\text{ h} = 36000\text{ m}^3 = 3.6 \times 10^7\text{ dm}^3\).
Mass of \(Br^-\) in this volume = \(3.6 \times 10^7\text{ dm}^3 \times 65\text{ mg dm}^{-3} = 2.34 \times 10^9\text{ mg} = 2340\text{ kg}\).
Moles of \(Br^- = \frac{2.34 \times 10^6\text{ g}}{79.9\text{ g mol}^{-1}} = 29286.6\text{ mol}\).
Moles of \(Br_2 = 14643.3\text{ mol}\).
Mass of \(Br_2 = 14643.3\text{ mol} \times 159.8\text{ g mol}^{-1} = 2.34 \times 10^6\text{ g} = 2340\text{ kg}\).

(ii) Recovered mass of \(Br_2 = 2340\text{ kg} \times 0.85 = 1989\text{ kg} = 1.989 \times 10^6\text{ g}\).
Moles of recovered \(Br_2 = \frac{1.989 \times 10^6\text{ g}}{159.8\text{ g mol}^{-1}} = 12446.8\text{ mol}\).
From equation, 1 mole of \(SO_2\) reacts with 1 mole of \(Br_2\).
Moles of \(SO_2\) required = \(12446.8\text{ mol}\).
Volume of \(SO_2\) at RTP = \(12446.8\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 2.99 \times 10^5\text{ dm}^3\).

(c)(i) Mass of \(Br_2\) added = \(21.6\text{ g} - 5.60\text{ g} = 16.0\text{ g}\).
Moles of \(Br_2\) reacted = \(\frac{16.0\text{ g}}{159.8\text{ g mol}^{-1}} = 0.100\text{ mol}\).
Moles of alkene = \(0.100\text{ mol}\) (since addition is 1:1).
\(M_r\) of alkene = \(\frac{5.60\text{ g}}{0.100\text{ mol}} = 56.0\text{ g mol}^{-1}\).
General formula of alkene = \(C_nH_{2n} \Rightarrow 14n = 56 \Rightarrow n = 4\).
Thus, the molecular formula of alkene \(Y\) is \(C_4H_{8}\).

(ii) Isomers can be any two of:
- But-1-ene: \(CH_3-CH_2-CH=CH_2\)
- But-2-ene: \(CH_3-CH=CH-CH_3\)
- Methylpropene: \((CH_3)_2C=CH_2\)

(a)(i) [1 mark]
- 1 mark for correct mass of \(65\text{ g}\)
(ii) [3 marks]
- 1 mark for calculating mass of \(1\text{ dm}^3\) of seawater = \(1025\text{ g}\)
- 1 mark for setting up ratio \(0.065 / 1025\)
- 1 mark for final value of \(63.4\text{ ppm}\) (accept 63.4 to 63.5)

(b)(i) [4 marks]
- 1 mark for volume of water per day = \(3.6 \times 10^7\text{ dm}^3\)
- 1 mark for mass of bromide = \(2340\text{ kg}\) or \(2.34 \times 10^6\text{ g}\)
- 1 mark for dividing moles of bromide by 2 to find moles of \(Br_2 = 14643\text{ mol}\)
- 1 mark for converting to mass of \(Br_2 = 2340\text{ kg}\)
(ii) [5 marks]
- 1 mark for calculating recovered mass of bromine (\(1989\text{ kg}\) or \(12447\text{ mol}\))
- 1 mark for correct mole ratio from equation (\(SO_2 : Br_2 = 1:1\))
- 1 mark for stating moles of \(SO_2 = 12447\text{ mol}\)
- 1 mark for multiplying by 24.0
- 1 mark for final volume \(2.99 \times 10^5\text{ dm}^3\) (accept 2.98 x 10^5 to 3.00 x 10^5)

(c)(i) [5 marks]
- 1 mark for finding mass of added bromine = \(16.0\text{ g}\)
- 1 mark for moles of bromine = \(0.100\text{ mol}\)
- 1 mark for stating moles of alkene = \(0.100\text{ mol}\)
- 1 mark for finding \(M_r\) of alkene = \(56.0\text{ g mol}^{-1}\)
- 1 mark for molecular formula \(C_4H_8\)
(ii) [2 marks]
- 1 mark for each correct drawn structural or skeletal formula of two structural isomers of \(C_4H_8\) (but-1-ene, but-2-ene, or methylpropene)

(a) \(K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\).
Units: \(\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6\).

(b) ICE table calculation:
- Initial moles: \(CO = 1.00\), \(H_2 = 2.00\), \(CH_3OH = 0.00\).
- Change in moles: Since \(0.40\text{ mol}\) of \(CH_3OH\) is formed:
- \(CO\) used \(= 0.40\text{ mol}\), remaining \(CO = 1.00 - 0.40 = 0.60\text{ mol}\).
- \(H_2\) used \(= 2 \times 0.40 = 0.80\text{ mol}\), remaining \(H_2 = 2.00 - 0.80 = 1.20\text{ mol}\).
- Equilibrium concentrations (Volume = \(5.00\text{ dm}^3\)):
- \([CO] = \frac{0.60}{5.00} = 0.120\text{ mol dm}^{-3}\)
- \([H_2] = \frac{1.20}{5.00} = 0.240\text{ mol dm}^{-3}\)
- \([CH_3OH] = \frac{0.40}{5.00} = 0.080\text{ mol dm}^{-3}\)
- \(K_c = \frac{0.080}{(0.120) \times (0.240)^2} = \frac{0.080}{0.120 \times 0.0576} = \frac{0.080}{0.006912} = 11.57\text{ dm}^6\text{ mol}^{-2}\).
- To 3 sig figs: \(11.6\text{ dm}^6\text{ mol}^{-2}\).

(c)(i) Increasing temperature: Since the forward reaction is exothermic, according to Le Chatelier's principle, increasing the temperature shifts the equilibrium to the left (the endothermic direction) to absorb the heat. Consequently, concentration of reactants increases and product decreases, so the value of \(K_c\) decreases.
(ii) Increasing pressure: \(K_c\) is temperature-dependent only, so its value is unaffected. Decreasing the volume shifts the equilibrium to the right (towards the side with fewer gas moles: 3 moles on LHS vs 1 mole on RHS) to reduce the pressure.

(d) Industrially balanced conditions:
- Temperature (\(250^\circ\text{C}\)): A lower temperature would increase the equilibrium yield because the reaction is exothermic. However, the rate of reaction would be too slow. \(250^\circ\text{C}\) is a compromise temperature that ensures a sufficiently high rate of reaction.
- Pressure (\(50-100\text{ atm}\)): Higher pressure increases both the rate of reaction (higher collision frequency) and the equilibrium yield (shifts right). However, generating and containing extremely high pressures is expensive (cost of thick-walled reactors and compressors) and poses safety risks. \(50-100\text{ atm}\) is an economic compromise.
- Catalyst: A copper-based catalyst increases the rate of reaction by providing an alternative pathway with a lower activation energy, allowing the reaction to proceed rapidly at a lower compromise temperature than would otherwise be possible.

(a) [2 marks]
- 1 mark for correct expression \(K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\)
- 1 mark for units \(\text{mol}^{-2}\text{ dm}^6\) (accept \(\text{dm}^6\text{ mol}^{-2}\))

(b) [6 marks]
- 1 mark for calculating equilibrium moles of \(CO = 0.60\text{ mol}\)
- 1 mark for calculating equilibrium moles of \(H_2 = 1.20\text{ mol}\)
- 1 mark for dividing moles by \(5.00\text{ dm}^3\) to get all concentrations (\([CO] = 0.12\), \([H_2] = 0.24\), \([CH_3OH] = 0.08\))
- 1 mark for substituting concentrations into their \(K_c\) expression
- 1 mark for calculating \(11.57\)
- 1 mark for giving final answer to 3 significant figures: \(11.6\)

(c) [6 marks]
- (i) 1 mark for stating \(K_c\) decreases
- 1 mark for stating equilibrium shifts left
- 1 mark for linking to exothermic forward reaction
- (ii) 1 mark for stating \(K_c\) remains unchanged
- 1 mark for stating equilibrium shifts right
- 1 mark for linking to fewer moles of gas on the right-hand side (3 vs 1)

(d) [6 marks]
- 2 marks for temperature discussion (compromise between rate and yield with explanation)
- 2 marks for pressure discussion (benefit to rate and yield balanced against safety/energy costs)
- 2 marks for catalyst discussion (increases rate, allows lower temperature to be used without affecting yield)

**(a)**
- Sodium thiosulfate reacts immediately with any iodine (\(\text{I}_2\)) produced: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\).
- This prevents iodine from reacting with starch, keeping the solution colourless as long as thiosulfate is present.
- Once all thiosulfate is completely consumed, the next trace of iodine produced remains in solution and reacts instantly with the starch indicator to form a blue-black complex.
- Since the amount of thiosulfate is small and constant, the time taken (\(t\)) for the colour to appear represents the time taken for a fixed, small amount of reactants to react. The initial rate is therefore proportional to \(1/t\) (or can be calculated as \(\Delta [\text{S}_2\text{O}_8^{2-}] / t\)).

**(b)**
- **Order with respect to \(\text{S}_2\text{O}_8^{2-}\)**: Compare Run 1 and Run 2. The concentration of \(\text{S}_2\text{O}_8^{2-}\) doubles (from 0.040 to 0.080 \(\text{mol dm}^{-3}\)) while \([\text{I}^-]\) remains constant (0.080 \(\text{mol dm}^{-3}\)). The time taken halves (from 44.0 s to 22.0 s), which means the rate doubles. Therefore, the reaction is first-order with respect to \(\text{S}_2\text{O}_8^{2-}\).
- **Order with respect to \(\text{I}^-\)**: Compare Run 1 and Run 3. The concentration of \(\text{I}^-\) halves (from 0.080 to 0.040 \(\text{mol dm}^{-3}\)) while \([\text{S}_2\text{O}_8^{2-}]\) remains constant (0.040 \(\text{mol dm}^{-3}\)). The time taken doubles (from 44.0 s to 88.0 s), which means the rate halves. Therefore, the reaction is first-order with respect to \(\text{I}^-\).
- **Rate Equation**: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)
- **Change in concentration of peroxodisulfate**:
Based on reaction stoichiometry: \(1\text{ mol of }\text{S}_2\text{O}_8^{2-} \equiv 1\text{ mol of }\text{I}_2 \equiv 2\text{ mol of }\text{S}_2\text{O}_3^{2-}\).
Therefore, \(\Delta [\text{S}_2\text{O}_8^{2-}] = \frac{1}{2} [\text{S}_2\text{O}_3^{2-}]_{\text{initial}} = 0.5 \times 1.00 \times 10^{-3} = 5.00 \times 10^{-4}\text{ mol dm}^{-3}\).
- **Rate for Run 1**: \(\text{Rate} = \frac{5.00 \times 10^{-4}\text{ mol dm}^{-3}}{44.0\text{ s}} = 1.136 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\).
- **Calculation of \(k\)**:
\(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.136 \times 10^{-5}}{0.040 \times 0.080} = 3.55 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(3.56 \times 10^{-3}\) with intermediate rounding).

**(c)**
- **Experiment**: Carry out the clock reaction under identical conditions (reactant concentrations, volumes, temperature) without catalyst, and then repeat with a small volume of \(\text{Fe}^{2+}(\text{aq})\) added. Show that the time for the blue-black colour to appear is significantly shorter (higher rate) when \(\text{Fe}^{2+}\) is present.
- **Mechanism Equations**:
1) \(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)
2) \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

**(a)** [4 marks]
- Thiosulfate reacts immediately with iodine: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\) (1)
- Keeps iodine concentration at zero as long as thiosulfate remains (1)
- Once thiosulfate is completely consumed, free iodine reacts with starch to turn blue-black (1)
- Constant, small amount of thiosulfate ensures initial rate is proportional to \(1/t\) (1)

**(b)** [7 marks]
- Order w.r.t. \(\text{S}_2\text{O}_8^{2-}\) is 1 with justification from Runs 1 & 2 (1)
- Order w.r.t. \(\text{I}^-\)// is 1 with justification from Runs 1 & 3 (1)
- Correct rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (1)
- \(\Delta [\text{S}_2\text{O}_8^{2-}] = 5.00 \times 10^{-4}\text{ mol dm}^{-3}\) (based on 1:2 stoichiometry with thiosulfate) (1)
- Correct calculation of Rate for Run 1 = \(1.14 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\) (1)
- Correct calculation of \(k = 3.55 \times 10^{-3}\) to \(3.56 \times 10^{-3}\) (1)
- Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (1)

**(c)** [4 marks]
- Measure time for colour change with and without \(\text{Fe}^{2+}\) under identical conditions, showing shorter time with \(\text{Fe}^{2+}\) (2)
- Correct equation showing reduction of peroxodisulfate by \(\text{Fe}^{2+}\) (1)
- Correct equation showing oxidation of iodide by \(\text{Fe}^{3+}\) (1)

**(a)**
(i) \(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})\)
(ii) \(\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})\)

**(b)**
Mean titre = \(\frac{21.80 + 21.75 + 21.85}{3} = 21.80\text{ cm}^3\).
Justification: Only runs 1, 2, and 3 are concordant (within \(0.10\text{ cm}^3\) of each other). The rough titre is excluded.

**(c)**
- \(\text{Moles of } \text{S}_2\text{O}_3^{2-} \text{ in mean titre} = 0.100\text{ mol dm}^{-3} \times \frac{21.80}{1000}\text{ dm}^3 = 2.18 \times 10^{-3}\text{ mol}\)
- From the equations, \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\). Thus, the ratio of \(\text{Cu}^{2+}\) to \(\text{S}_2\text{O}_3^{2-}\) is 1:1.
- \(\text{Moles of } \text{Cu}^{2+} \text{ in } 25.0\text{ cm}^3 \text{ aliquot} = 2.18 \times 10^{-3}\text{ mol}\)
- \(\text{Moles of } \text{Cu}^{2+} \text{ in } 250.0\text{ cm}^3 \text{ total solution} = 2.18 \times 10^{-3} \times 10 = 2.18 \times 10^{-2}\text{ mol}\)
- \(\text{Mass of copper in brass sample} = 2.18 \times 10^{-2}\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.3843\text{ g}\)
- \(\text{Percentage of copper in brass} = \frac{1.3843\text{ g}}{2.82\text{ g}} \times 100\% = 49.1\%\) (or \(49.09\%\)).

**(d)**
- **Colour change**: Blue-black to cream / off-white (due to the presence of the white \(\text{CuI}\) precipitate slurry).
- **Starch addition explanation**: If added at the start when iodine concentration is high, starch binds strongly and irreversibly to iodine. This prevents some iodine from being reduced near the end-point, leading to an inaccurate and lower titre value.

**(e)**
- **Hazard**: Toxic nitrogen dioxide (\(\text{NO}_2\)) gas is produced when brass (copper) reacts with concentrated nitric acid.
- **Minimisation**: Conduct the dissolving step inside a functioning fume cupboard.

**(f)**
- Weighing by difference involves two readings (empty container and container + sample).
- \(\text{Total uncertainty} = 2 \times 0.005\text{ g} = 0.010\text{ g}\).
- \(\text{Percentage uncertainty} = \frac{0.010}{2.82} \times 100 = 0.355\%\) (accept \(0.35\%\) or \(0.36\%\); allow \(0.18\%\) if only a single reading is assumed).

**(a)** [2 marks]
- (i) Correct equation with state symbols (1)
- (ii) Correct equation with state symbols (1)

**(b)** [1 mark]
- Mean titre = \(21.80\text{ cm}^3\) with justification of concordant titres (1)

**(c)** [5 marks]
- Moles of thiosulfate = \(2.18 \times 10^{-3}\text{ mol}\) (1)
- Correct 1:1 molar ratio between copper and thiosulfate shown/stated (1)
- Moles of copper in \(250.0\text{ cm}^3 = 2.18 \times 10^{-2}\text{ mol}\) (1)
- Mass of copper = \(1.38\text{ g}\) (1)
- Percentage of copper = \(49.1\%\) (allow \(49.09\%\)) (1)

**(d)** [3 marks]
- Blue-black to cream / off-white (1)
- Iodine binds irreversibly/strongly to starch at high concentrations (1)
- Leads to an inaccurate/underestimated end-point (1)

**(e)** [2 marks]
- Toxic / brown gas / nitrogen dioxide (\(\text{NO}_2\)) (1)
- Use of fume cupboard (1)

**(f)** [2 marks]
- Total uncertainty = \(0.010\text{ g}\) (1)
- Percentage uncertainty = \(0.35\%\) or \(0.355\%\) (1) (allow \(0.18\%\) for single reading calculation with 1 mark max)

**(a)**
Key components of the distillation setup:
- A round-bottom or pear-shaped flask containing the mixture, heated using a heating mantle or water bath (no naked flames as alcohol products are flammable).
- A thermometer positioned with its bulb exactly at the T-junction of the distillation head to accurately measure the boiling point of the vapour being condensed.
- A condenser with water entering through the lower connector (bottom) and exiting from the upper connector (top). This ensures the condenser is fully filled with cold water, maximising cooling efficiency.
- Anti-bumping granules added to the flask before heating to promote smooth, controlled boiling and prevent liquid from splashing into the condenser.
- A receiver adapter and collection vessel; the apparatus must not be sealed airtight at the end to prevent pressure build-up upon heating.

**(b)**
**Level of Response Analysis**:
- **Mass Spectrum**: The molecular ion peak at \(m/z = 74\) indicates that the relative molecular mass (\(M_{\text{r}}\)) of alcohol **Z** is 74.
- **Infrared Spectrum**: The broad peak at \(3200-3600\text{ cm}^{-1}\) indicates the presence of an alcohol group (\(\text{O-H}\) stretch). The absorption at \(2850-2960\text{ cm}^{-1}\) is due to aliphatic \(\text{C-H}\) bonds. The absence of a peak at \(1700\text{ cm}^{-1}\) confirms there is no carbonyl group (\(\text{C=O}\)), which is consistent with **Z** being an alcohol.
- **Molecular Formula**: For a saturated aliphatic alcohol (\(\text{C}_n\text{H}_{2n+2}\text{O}\)): \(12n + 2n + 2 + 16 = 74 \Rightarrow 14n + 18 = 74 \Rightarrow n = 4\). Thus, the formula is \(\text{C}_4\text{H}_{10}\text{O}\) (a butanol isomer).
- **NMR - \(\text{D}_2\text{O}\) Shake**: Peak 5 (singlet, 1H) disappears on adding deuterium oxide. This is because the labile \(\text{O-H}\) proton undergoes rapid deuterium exchange to form \(\text{-OD}\), which does not absorb in \({}^1\text{H}\) NMR. This confirms Peak 5 represents the alcohol hydroxyl proton.
- **NMR - Splitting and Structure**:
- Peak 1 (triplet, 3H) at \(\delta = 0.9\text{ ppm}\) represents a terminal methyl group (\(\text{-CH}_3\)) adjacent to a \(\text{-CH}_2-\). Under the \(n+1\) rule, 2 adjacent protons split the signal into a triplet.
- Peak 4 (triplet, 2H) at \(\delta = 3.6\text{ ppm}\) is shifted downfield due to the adjacent electronegative oxygen atom (\(\text{-CH}_2\text{-O-}\)). Its triplet splitting indicates that this \(\text{-CH}_2-\) group is adjacent to another \(\text{-CH}_2-\) group (2 adjacent protons).
- The presence of two multiplets (Peak 2 and 3), each integrating to 2H, represents the internal \(\text{-CH}_2-\) groups of a straight chain.
- **Conclusion**: Compound **Z** is **butan-1-ol** (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)).

**(c)**
- **TLC Setup**: Draw a pencil line about 1 cm from the bottom of a silica TLC plate. Spot reference samples of reactant **X**, product **Z**, and samples of the reaction mixture taken at different times on this line.
- **Development**: Place the plate in a beaker/tank containing a small volume of a suitable mobile phase (solvent), ensuring the solvent level is below the pencil line. Cover with a lid to saturate the atmosphere.
- **Measurement**: Allow the solvent to ascend near the top. Remove the plate, quickly draw a pencil line along the solvent front, and let it dry.
- **Calculation**: Locate spots (using UV light or iodine stain). Calculate \(R_{\text{f}}\) values for each spot: \(R_{\text{f}} = \frac{\text{distance moved by spot}}{\text{distance moved by solvent front}}\).
- **Monitoring Completion**: The reaction is complete when the spot corresponding to reactant **X** has completely disappeared from the reaction mixture lane on the plate, and only spots corresponding to the products are visible.

**(a)** [4 marks]
- Diagram/description showing round-bottom flask heated with water bath/heating mantle (no naked flame) (1)
- Thermometer correctly positioned at T-junction/condenser entrance (1)
- Condenser with cooling water entering at the bottom and exiting at the top (1)
- Addition of anti-bumping granules before heating AND system not sealed airtight (1)

**(b)** [6 marks]
- **Level 3 (5-6 marks)**: Complete and detailed interpretation of all three analytical datasets. Correctly identifies relative molecular mass, alcohol functional group, molecular formula (C4H10O), and explains both the D2O shake and the NMR splitting patterns to deduce the structure as butan-1-ol.
- **Level 2 (3-4 marks)**: Interprets mass spec and IR spectra successfully. Identifies that Z is a butanol isomer. Explains the D2O shake or NMR splitting patterns with some success.
- **Level 1 (1-2 marks)**: Identifies some individual spectroscopic features (e.g. O-H bond from IR or molecular weight of 74) without integrating them to find the correct structure.

**(c)** [5 marks]
- Draw pencil line on plate, spot reactant X, product Z, and reaction mixture (1)
- Run in solvent tank with level below line, covered with a lid (1)
- Mark solvent front, dry, and visualise spots using UV light/iodine (1)
- Show calculation: \(R_{\text{f}} = \text{distance of spot} / \text{distance of solvent front}\) (1)
- Reaction complete when the spot corresponding to reactant **X** is completely absent from the reaction lane (1)

**(a)**
- The catalyst increases the rates of both the forward and reverse reactions equally, enabling the system to reach equilibrium much faster.
- The esterification reaction is slow at room temperature; leaving the sealed flask for a week ensures that chemical equilibrium has definitely been reached before analysis.

**(b)**
- \(K_{\text{c}} = \frac{[\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{CH}_2\text{COOH}][\text{CH}_3\text{CH}_2\text{OH}]}\)
- Units: None (or unitless) because the concentration terms cancel out: \(\frac{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})} = 1\).

**(c)**
(i) \(\text{Moles of } \text{H}^+ \text{ from catalyst} = 0.010\text{ mol}\) (stated directly in the prompt).
(ii) \(\text{Moles of NaOH used in titration} = \text{volume} \times \text{concentration} = \frac{38.00}{1000}\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0380\text{ mol}\).
Since both acids react with \(\text{NaOH}\) in a 1:1 molar ratio, the total moles of acid neutralised is \(0.0380\text{ mol}\).
(iii) \(\text{Moles of propanoic acid at equilibrium} = \text{total moles of acid neutralised} - \text{moles of catalyst } \text{H}^+\)
\(= 0.0380 - 0.010 = 0.028\text{ mol}\).
(iv)
- \(\text{Moles of propanoic acid reacted} = 0.100\text{ mol (initial)} - 0.028\text{ mol (equilibrium)} = 0.072\text{ mol}\).
- Since the stoichiometry is 1:1 for all reactants and products:
- \(\text{Equilibrium moles of ethanol} = 0.120\text{ mol (initial)} - 0.072\text{ mol (reacted)} = 0.048\text{ mol}\).
- \(\text{Equilibrium moles of ethyl propanoate} = 0.072\text{ mol}\).
- \(\text{Equilibrium moles of water} = 0.072\text{ mol}\) (since initial water content was negligible).
(v) Since the total volume \(V\) of the mixture is the same for all species, the volume terms in the concentration expressions cancel out:
\(K_{\text{c}} = \frac{(0.072 / V) \times (0.072 / V)}{(0.028 / V) \times (0.048 / V)} = \frac{0.072 \times 0.072}{0.028 \times 0.048} = \frac{0.005184}{0.001344} = 3.86\) (accept \(3.9\)).

**(d)**
- The titration is carried out rapidly at room temperature. Without heating and in a diluted state, the rate of the esterification/hydrolysis reactions is extremely slow, so no significant shift in the position of equilibrium occurs during the short time of the titration.

**(a)** [2 marks]
- Catalyst increases rate of both forward and reverse reactions / system reaches equilibrium faster (1)
- Left for a week to ensure equilibrium has actually been established at room temperature (1)

**(b)** [2 marks]
- Correct expression for \(K_{\text{c}}\) (1)
- Units specified as none / unitless (1)

**(c)** [10 marks]
- (i) \(0.010\text{ mol}\) (1)
- (ii) Moles of \(\text{NaOH} = 0.0380\text{ mol}\) (1) AND total moles of acid = \(0.0380\text{ mol}\) (1)
- (iii) \(\text{Equilibrium propanoic acid} = 0.0380 - 0.010 = 0.028\text{ mol}\) (1)
- (iv) Moles of propanoic acid reacted = \(0.072\text{ mol}\) (1); \(\text{equilibrium ethanol} = 0.048\text{ mol}\) (1); \(\text{equilibrium ester and water} = 0.072\text{ mol each}\) (1)
- (v) Correct substitution into the Kc expression (1); correct calculation: \(K_{\text{c}} = 3.86\) (allow \(3.9\)) (1); correct justification that volume terms cancel out (1)

**(d)** [1 mark]
- Titration is quick/at room temperature, so the rate of reaction is negligible and the equilibrium position does not change significantly during titration (1)