2024 IB DP Chemistry Practice Paper with Answers

First, write the balanced equation for the combustion of ethane: \(2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)\). The amount of ethane in moles is: \(n(C_2H_6) = \frac{3.0\text{ g}}{30.0\text{ g mol}^{-1}} = 0.10\text{ mol}\). From the stoichiometric ratio, 1 mole of \(C_2H_6\) produces 2 moles of \(CO_2\). Therefore, \(n(CO_2) = 0.10\text{ mol} \times 2 = 0.20\text{ mol}\). The mass of \(CO_2\) produced (using \(M_r(CO_2) = 44.0\text{ g mol}^{-1}\)) is: \(m(CO_2) = 0.20\text{ mol} \times 44.0\text{ g mol}^{-1} = 8.8\text{ g}\).

Award [1] for the correct answer (B). Alternatively, award [1] for showing correct conversion of ethane to moles (0.10 mol), stoichiometric conversion to carbon dioxide (0.20 mol), and final calculated mass of 8.8 g.

\(SF_4\) has 34 valence electrons, resulting in 5 electron domains around the sulfur central atom (4 bonding pairs and 1 lone pair). This gives a trigonal bipyramidal electron domain geometry and a see-saw molecular geometry. \(XeF_4\) has a square planar geometry, while \(SiF_4\) and \(BF_4^-\) both have tetrahedral geometries.

Award [1] for the correct answer (A).

According to the Br%nsted-Lowry theory, a conjugate base is formed when an acid donates a proton (\(H^+\)). Therefore, when \(H_2PO_4^-\) loses a proton, it forms \(HPO_4^{2-}\).

Award [1] for the correct answer (A).

Let's calculate the oxidation states of nitrogen in each option: In \(NO_2\), the oxidation state of N is +4. In \(N_2O\), the oxidation state of N is +1. In \(NH_4Cl\), the nitrogen exists as the ammonium ion \(NH_4^+\), where N has an oxidation state of -3. In \(HNO_3\), the oxidation state of N is +5. The lowest oxidation state is -3.

Award [1] for the correct answer (C).

Enthalpy change can be estimated using the bond enthalpies: \(\Delta H^\ominus = \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed})\). Bonds broken: \(1 \times (H-H) + 1 \times (Cl-Cl) = 436 + 242 = 678\text{ kJ mol}^{-1}\). Bonds formed: \(2 \times (H-Cl) = 2 \times 431 = 862\text{ kJ mol}^{-1}\). \(\Delta H^\ominus = 678 - 862 = -184\text{ kJ mol}^{-1}\).

Award [1] for the correct answer (B).

Comparing Run 1 and Run 2: The concentration of B is constant while the concentration of A is doubled. The rate doubles, so the reaction is first-order with respect to A (\(\text{rate} \propto [A]^1\)). Comparing Run 1 and Run 3: The concentration of A is constant while the concentration of B is doubled. The rate quadruples (\(8.0 \times 10^{-4} / 2.0 \times 10^{-4} = 4\)), so the reaction is second-order with respect to B (\(\text{rate} \propto [B]^2\)). Combining these, the rate expression is \(\text{Rate} = k[A][B]^2\).

Award [1] for the correct answer (C).

Amines are classified as primary, secondary, or tertiary based on the number of alkyl (or aryl) groups bonded directly to the nitrogen atom. \((CH_3)_3N\) has three methyl groups bonded to the nitrogen atom, so it is a tertiary amine. \(CH_3CH_2CH_2NH_2\) and \((CH_3)_2CHNH_2\) are primary amines, and \((CH_3)_2NH\) is a secondary amine.

Award [1] for the correct answer (D).

Spontaneity is determined by the Gibbs free energy change equation: \(\Delta G = \Delta H - T\Delta S\). For a reaction to be spontaneous, \(\Delta G\) must be negative. Since \(\Delta H > 0\) (unfavorable endothermic process) and \(\Delta S > 0\) (favorable entropy increase), the reaction will only become spontaneous (\(\Delta G < 0\)) when the temperature \(T\) is high enough so that the term \(T\Delta S\) is larger in magnitude than \(\Delta H\).

Award [1] for the correct answer (C).

\(\text{XeF}_4\) has 8 valence electrons from xenon and 4 from the fluorines, plus 2 lone pairs on the central atom (total 12 electrons / 6 domains around Xe). This gives an octahedral electron domain geometry and a square planar molecular geometry. \(\text{SF}_4\) has 6 valence electrons from sulfur and 4 from the fluorines, plus 1 lone pair on the central atom (total 10 electrons / 5 domains around S). This gives a trigonal bipyramidal electron domain geometry and a seesaw molecular geometry.

Award 1 mark for the correct option A. No partial marks.

Comparing the first and second experiments, doubling [A] while keeping [B] constant increases the rate by a factor of 4 (\(8.0 \times 10^{-4} / 2.0 \times 10^{-4} = 4\)), so the order with respect to A is 2. Comparing the first and third experiments, doubling [B] while keeping [A] constant has no effect on the rate, so the order with respect to B is 0. The overall order of reaction is 2 + 0 = 2.

Award 1 mark for the correct option B. No partial marks.

The pH of 4.0 means that \([\text{H}^+] = 10^{-4}\text{ mol dm}^{-3}\). For a weak monoprotic acid, we can assume \([\text{H}^+] \approx [\text{X}^-] = 10^{-4}\text{ mol dm}^{-3}\). Using the equilibrium expression \(K_a = \frac{[\text{H}^+][\text{X}^-]}{[\text{HX}]_{\text{eq}}}\), we get \(K_a = \frac{(10^{-4})^2}{0.010 - 10^{-4}} \approx \frac{10^{-8}}{10^{-2}} = 10^{-6}\text{ mol dm}^{-3}\).

Award 1 mark for the correct option B. No partial marks.

Assume 100 g of the compound. Mass of N = 30.4 g, mass of O = 100.0 - 30.4 = 69.6 g. Number of moles of N = \(30.4 / 14.0 \approx 2.17\text{ mol}\) and moles of O = \(69.6 / 16.0 \approx 4.35\text{ mol}\). The mole ratio N : O is \(2.17 : 4.35\), which simplifies to \(1 : 2\). Thus, the empirical formula is \(\text{NO}_2\).

Award 1 mark for the correct option B. No partial marks.

Total mass of the mixture is \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\\. Heat released: \)q = m c \\Delta T = 100.0\\text{ g} \\times 4.18\\text{ J g}^{-1}\\ \\text{K}^{-1} \\times 6.0\\text{ K} = 2508\\text{ J} = 2.508\\text{ kJ}\\. Number of moles of water formed = \(1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\\. Enthalpy change: \)\\Delta H = -\\frac{2.508\\text{ kJ}}{0.0500\\text{ mol}} = -50.16\\text{ kJ mol}^{-1} \\approx -50.2\\text{ kJ mol}^{-1}\\.

Award 1 mark for the correct option A. No partial marks.

The reaction between \(\text{Fe}^{3+}\) and \(\text{I}^-\) is represented by: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\). The cell potential for this is \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = 0.77\text{ V} - 0.54\text{ V} = +0.23\text{ V}\). Since \(E^\theta_{\text{cell}} > 0\), the forward reaction is spontaneous under standard conditions. Thus, \(\text{Fe}^{3+}\) spontaneously oxidizes \(\text{I}^-\).

Award 1 mark for the correct option A. No partial marks.

Aspirin contains a carboxylic acid group (\(-\text{COOH}\)) and an ester group (\(-\text{OCOCH}_3\)). It does not contain a ketone group. Therefore, only I and II are correct.

Award 1 mark for the correct option A. No partial marks.

For spontaneity, \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta < 0\). Converting units: \(-120000\\text{ J mol}^{-1} - T(-400\\text{ J K}^{-1}\text{ mol}^{-1}) < 0 \implies -120000 + 400T < 0 \implies 400T < 120000 \implies T < 300\\text{ K}\). Thus, the reaction is spontaneous only at temperatures below \(300\\text{ K}\).

Award 1 mark for the correct option A. No partial marks.

To find the empirical formula, calculate the amount (in moles) of each element in \(100\text{ g}\) of the compound:

- \(\text{Moles of C} = \frac{60.0\text{ g}}{12.0\text{ g mol}^{-1}} = 5.0\text{ mol}\)
- \(\text{Moles of H} = \frac{13.3\text{ g}}{1.0\text{ g mol}^{-1}} = 13.3\text{ mol}\)
- \(\text{Moles of O} = \frac{26.7\text{ g}}{16.0\text{ g mol}^{-1}} \approx 1.67\text{ mol}\)

Divide each value by the smallest number of moles (\(1.67\)) to find the simplest whole-number ratio:
- \(\text{C} = \frac{5.0}{1.67} \approx 3\)
- \(\text{H} = \frac{13.3}{1.67} \approx 8\)
- \(\text{O} = \frac{1.67}{1.67} = 1\)

Therefore, the empirical formula is \(\text{C}_3\text{H}_8\text{O}\).

[1 mark] for selecting option B.
[0 marks] for other options.

The structural formula of propenenitrile is \(\text{CH}_2=\text{CH}-\text{C}\equiv\text{N}\). We can count the bonds as follows:

- Three \(\text{C}-\text{H}\) single bonds: 3 \(\sigma\) bonds
- One \(\text{C}=\text{C}\) double bond: 1 \(\sigma\) bond and 1 \(\pi\) bond
- One \(\text{C}-\text{C}\) single bond: 1 \(\sigma\) bond
- One \(\text{C}\equiv\text{N}\) triple bond: 1 \(\sigma\) bond and 2 \(\pi\) bonds

Total \(\sigma\) bonds = \(3 + 1 + 1 + 1 = 6\)
Total \(\pi\) bonds = \(1 + 2 = 3\)

[1 mark] for selecting option C.
[0 marks] for other options.

Use the bond enthalpy equation: \(\Delta H = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})\)

**Bonds broken (reactants):**
- 1 x \(\text{C}=\text{C}\) = \(614\text{ kJ mol}^{-1}\)
- 4 x \(\text{C}-\text{H}\) = \(4 \times 414 = 1656\text{ kJ mol}^{-1}\)
- 1 x \(\text{H}-\text{H}\) = \(436\text{ kJ mol}^{-1}\)
Total bonds broken = \(614 + 1656 + 436 = 2706\text{ kJ mol}^{-1}\)

**Bonds formed (products):**
- 1 x \(\text{C}-\text{C}\) = \(346\text{ kJ mol}^{-1}\)
- 6 x \(\text{C}-\text{H}\) = \(6 \times 414 = 2484\text{ kJ mol}^{-1}\)
Total bonds formed = \(346 + 2484 = 2830\text{ kJ mol}^{-1}\)

**Enthalpy Change:**
\(\Delta H = 2706 - 2830 = -124\text{ kJ mol}^{-1}\).

[1 mark] for selecting option A.
[0 marks] for other options.

The rate-determining step is the slow step (Step 2):
\(\text{Rate} = k_2 [\text{NOCl}_2][\text{NO}]\)

Because \(\text{NOCl}_2\) is an intermediate, we express its concentration using the fast pre-equilibrium in Step 1:
\(K_c = \frac{[\text{NOCl}_2]}{[\text{NO}][\text{Cl}_2]} \Rightarrow [\text{NOCl}_2] = K_c [\text{NO}][\text{Cl}_2]\)

Substituting this into the rate equation of the slow step gives:
\(\text{Rate} = k_2 K_c [\text{NO}]^2 [\text{Cl}_2] = k[\text{NO}]^2[\text{Cl}_2]\).

[1 mark] for selecting option C.
[0 marks] for other options.

The relationship is defined by: \(K_w = [\text{H}^+][\text{OH}^-]\).

Solving for \([\text{H}^+]\):
\([\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-4}} = 0.5 \times 10^{-10} = 5.0 \times 10^{-11}\text{ mol dm}^{-3}\).

[1 mark] for selecting option B.
[0 marks] for other options.

A redox reaction is spontaneous under standard conditions if its standard cell potential (\(E^\theta_{\text{cell}}\)) is positive:

- **Reaction I:** \(\text{Fe}\) is oxidized (\(-0.44\text{ V}\)) and \(\text{Cu}^{2+}\) is reduced (\(+0.34\text{ V}\)).
\(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}\) (Spontaneous)

- **Reaction II:** \(\text{Cu}\) is oxidized (\(+0.34\text{ V}\)) and \(\text{Ag}^+\) is reduced (\(+0.80\text{ V}\)).
\(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = +0.80\text{ V} - (+0.34\text{ V}) = +0.46\text{ V}\) (Spontaneous)

- **Reaction III:** \(\text{Ag}\) is oxidized (\(+0.80\text{ V}\)) and \(\text{Fe}^{2+}\) is reduced (\(-0.44\text{ V}\)).
\(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = -0.44\text{ V} - (+0.80\text{ V}) = -1.24\text{ V}\) (Non-spontaneous)

Thus, only reactions I and II are spontaneous.

[1 mark] for selecting option B.
[0 marks] for other options.

Analyzing the structural formula \(\text{CH}_3\text{CH(OH)CH}_2\text{COOCH}_3\):

- The \(\text{-OH}\) group on the second carbon atom represents an **alcohol** functional group.
- The \(\text{-COOCH}_3\) group at the terminus represents an **ester** functional group.

Therefore, the compound contains both alcohol and ester functional groups.

[1 mark] for selecting option B.
[0 marks] for other options.

Spontaneity requires that Gibbs free energy change is negative (\(\Delta G^\theta < 0\)):

\(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)

Convert \(\Delta H^\theta\) to Joules:
\(\Delta H^\theta = -92\,000\text{ J mol}^{-1}\)

Set \(\Delta G^\theta < 0\):
\(-92\,000 - T(-198) < 0\)
\(-92\,000 + 198 T < 0\)
\(198 T < 92\,000\)
\(T < \frac{92\,000}{198} \approx 465\text{ K}\)

Since both \(\Delta H^\theta\) and \(\Delta S^\theta\) are negative, the reaction is spontaneous only at temperatures below \(465\text{ K}\).

[1 mark] for selecting option B.
[0 marks] for other options.

The symmetric resonance structure of the azide ion is \(\ddot{\text{N}}=\text{N}=\ddot{\text{N}}\). This structure has a total of 16 valence electrons. For the central nitrogen atom: formal charge = \(5 - 0 - \frac{1}{2}(8) = +1\). For each of the outer nitrogen atoms: formal charge = \(5 - 4 - \frac{1}{2}(4) = -1\).

Award [1] for the correct choice A.
Award [0] for incorrect choices.

The balanced combustion equations are:

1) \(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\)

Therefore, \(0.10\text{ mol}\) of \(\text{C}_3\text{H}_8\) requires \(0.10 \times 5 = 0.50\text{ mol}\) of \(\text{O}_2\).

2) \(\text{C}_4\text{H}_{10}(\text{g}) + 6.5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\)

Therefore, \(0.20\text{ mol}\) of \(\text{C}_4\text{H}_{10}\) requires \(0.20 \times 6.5 = 1.30\text{ mol}\) of \(\text{O}_2\).

Total \(\text{O}_2\) required = \(0.50 + 1.30 = 1.80\text{ mol}\).

Award [1] for the correct choice C.
Award [0] for incorrect choices.

All three species are amphiprotic because they can both donate a proton (acting as an acid) or accept a proton (acting as a base):

- \(\text{HCO}_3^-\) can donate to form \(\text{CO}_3^{2-}\) or accept to form \(\text{H}_2\text{CO}_3\).

- \(\text{H}_2\text{PO}_4^-\) can donate to form \(\text{HPO}_4^{2-}\) or accept to form \(\text{H}_3\text{PO}_4\).

- \(\text{HS}^-\) can donate to form \(\text{S}^{2-}\) or accept to form \(\text{H}_2\text{S}\).

Award [1] for the correct choice D.
Award [0] for incorrect choices.

Let the initial rate be \(\text{Rate}_1 = k[\text{X}]^2[\text{Y}]\). If the concentration of \(\text{X}\) is doubled to \(2[\text{X}]\) and the concentration of \(\text{Y}\) is halved to \(0.5[\text{Y}]\), the new rate is:

\(\text{Rate}_2 = k(2[\text{X}])^2(0.5[\text{Y}]) = k(4[\text{X}]^2)(0.5[\text{Y}]) = 2k[\text{X}]^2[\text{Y}] = 2\text{Rate}_1\).

Thus, the initial rate increases by a factor of 2.

Award [1] for the correct choice B.
Award [0] for incorrect choices.

For a spontaneous cell reaction, the half-cell with the higher reduction potential undergoes reduction, and the half-cell with the lower reduction potential undergoes oxidation:

Reduction: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{e}^- \rightarrow 2\text{Fe}^{2+}(\text{aq})\)

Oxidation: \(\text{Sn}^{2+}(\text{aq}) \rightarrow \text{Sn}^{4+}(\text{aq}) + 2\text{e}^-\)

Overall reaction: \(2\text{Fe}^{3+}(\text{aq}) + \text{Sn}^{2+}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{Sn}^{4+}(\text{aq})\)

Standard cell potential: \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V}\).

Award [1] for the correct choice A.
Award [0] for incorrect choices.

The molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) has one double bond equivalent (degree of unsaturation). The absence of a broad IR absorption band between \(2500\text{ and }3300\text{ cm}^{-1}\) shows that there are no \(\text{O}-\text{H}\) bonds present (ruling out carboxylic acids and alcohols). Since the compound does not react with sodium metal, it lacks an acidic proton. Therefore, the compound is an ester (for example, ethyl ethanoate).

Award [1] for the correct choice B.
Award [0] for incorrect choices.

In the first experiment, the number of moles reacting is \(0.050\text{ mol}\) in a total volume of \(100.0\text{ cm}^3\). The temperature change is proportional to \(\frac{n}{V_{\text{total}}} = \frac{0.050}{100.0} = 0.0005\text{ mol/cm}^3\).

In the second experiment, both the amount of reactants (\(0.025\text{ mol}\)) and the total volume (\(50.0\text{ cm}^3\)) are halved. The ratio of heat produced to total volume remains identical: \(\frac{0.025}{50.0} = 0.0005\text{ mol/cm}^3\). Thus, the temperature rise is unchanged at \(\Delta T\).

Award [1] for the correct choice C.
Award [0] for incorrect choices.

Spontaneity is determined by the Gibbs free energy change, \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\). For a reaction to be spontaneous, \(\Delta G^\theta\) must be negative. Since both \(\Delta H^\theta\) and \(\Delta S^\theta\) are positive, the entropy term \(-T\Delta S^\theta\) is negative. \(\Delta G^\theta\) only becomes negative when the temperature is high enough that the magnitude of \(T\Delta S^\theta\) is greater than \(\Delta H^\theta\).

Award [1] for the correct choice C.
Award [0] for incorrect choices.

To find the formal charge (FC) on the central oxygen atom in ozone, we use the formula:

\(\text{FC} = V - N - \frac{1}{2}B\)

where:
- \(V\) is the number of valence electrons of the neutral atom (6 for oxygen).
- \(N\) is the number of non-bonding valence electrons.
- \(B\) is the number of bonding electrons.

In the dominant Lewis resonance structures of ozone (\(\text{O}=\text{O}^+-\text{O}^-\)), the central oxygen atom forms one double bond and one single bond (a total of 3 bonding pairs, so 6 bonding electrons) and has 1 lone pair (2 non-bonding electrons).

\(\text{FC} = 6 - 2 - \frac{1}{2}(6) = +1\)

Award 1 mark for the correct option (C).
- Reject all other options.

- \(\text{HCO}_3^-\): can donate a proton to form \(\text{CO}_3^{2-}\) (acid) or accept a proton to form \(\text{H}_2\text{CO}_3\) (base).
- \(\text{H}_2\text{PO}_4^-\): can donate a proton to form \(\text{HPO}_4^{2-}\) (acid) or accept a proton to form \(\text{H}_3\text{PO}_4\) (base).
- \(\text{NH}_4^+\): can only donate a proton to form \(\text{NH}_3\) (acid). It cannot accept another proton to form \(\text{NH}_5^{2+}\) under normal aqueous conditions.

Therefore, only I and II are amphiprotic (can act as both a Brønsted–Lowry acid and base).

Award 1 mark for the correct option (A).
- Reject all other options.

1. Find the limiting reactant:
According to the stoichiometry, \(1\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) requires \(3\text{ mol}\) of \(\text{CO}\).
For \(2.0\text{ mol}\) of \(\text{Fe}_2\text{O}_3\), the required amount of \(\text{CO}\) is \(2.0 \times 3 = 6.0\text{ mol}\).
Since we have \(9.0\text{ mol}\) of \(\text{CO}\) (which is more than \(6.0\text{ mol}\)), \(\text{Fe}_2\text{O}_3\) is the limiting reactant.

2. Find the excess reactant remaining:
Reacted \(\text{CO} = 6.0\text{ mol}\).
Remaining \(\text{CO} = 9.0\text{ mol} - 6.0\text{ mol} = 3.0\text{ mol}\).

Award 1 mark for the correct option (A).
- Reject all other options.

The half-cell with the more negative standard electrode potential undergoes oxidation at the anode.
Therefore, the anode reaction is: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\).

The half-cell with the more positive electrode potential undergoes reduction at the cathode:
\(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\).

The standard cell potential is:
\(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.77\text{ V} - (-0.76\text{ V}) = +1.53\text{ V}\).

Award 1 mark for the correct option (A).
- Reject all other options.

Let the original rate be \(\text{Rate}_1 = k[\text{A}][\text{B}]^2\).

When the concentration of \(\text{A}\) is doubled (\(2[\text{A}]\)) and \(\text{B}\) is halved (\(0.5[\text{B}]\)):
\(\text{Rate}_2 = k(2[\text{A}])(0.5[\text{B}])^2\)
\(\text{Rate}_2 = k \cdot 2[\text{A}] \cdot 0.25[\text{B}]^2\)
\(\text{Rate}_2 = 0.5 \cdot k[\text{A}][\text{B}]^2 = 0.5 \cdot \text{Rate}_1\)

Thus, the rate is halved.

Award 1 mark for the correct option (B).
- Reject all other options.

Analyzing the name 4-hydroxypent-2-enoic acid:
1. "4-hydroxy" indicates the presence of a hydroxyl group (\(-\text{OH}\)).
2. "pent-2-en" indicates the presence of an alkene group, containing a carbon-carbon double bond (alkenyl group).
3. "oic acid" indicates the presence of a carboxylic acid group (carboxyl group, \(-\text{COOH}\)).

Therefore, the functional groups are hydroxyl, alkenyl, and carboxyl.

Award 1 mark for the correct option (A).
- Reject all other options.

1. Determine total mass of solution:
\(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3 = 100.0\text{ g}\) (since density is \(1.0\text{ g cm}^{-3}\)).

2. Calculate heat energy released (\(q\)):
\(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.2\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 6.0\ ^\circ\text{C} = 2520\text{ J} = 2.52\text{ kJ}\).

3. Calculate the number of moles of water formed:
\(n(\text{H}_2\text{O}) = n(\text{H}^+) = 1.0\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.050\text{ mol}\).

4. Calculate the enthalpy of neutralization:
\(\Delta H = -\frac{q}{n} = -\frac{2.52\text{ kJ}}{0.050\text{ mol}} = -50.4\text{ kJ mol}^{-1}\).
(Note: The sign is negative because the temperature rose, indicating an exothermic reaction.)

Award 1 mark for the correct option (A).
- Reject all other options.

According to the combined gas law / ideal gas relationship:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

Given:
- Initial conditions: \(P_1 = P\), \(V_1 = V\), \(T_1 = T\)
- Final conditions: \(P_2 = 0.5P\), \(T_2 = 2T\)

Substitute the values:

\(\frac{P \cdot V}{T} = \frac{0.5P \cdot V_2}{2T}\)

Cancel out \(P\) and \(T\):

\(V = \frac{0.5 \cdot V_2}{2}\)

\(V = 0.25 \cdot V_2\)

\(V_2 = 4V\)

Award 1 mark for the correct option (D).
- Reject all other options.

(a)
- **\(\text{SO}_3\) Lewis Structure:** Total valence electrons = 24. Central sulfur expands its octet to form 3 double bonds with the three oxygen atoms.
*Formal Charge (FC) calculation:*
FC(S) = 6 - 6 - 0 = 0.
FC(O) = 6 - 4 - 2 = 0.
This is the most stable structure because all formal charges are zero.
- **\(\text{SO}_3^{2-}\) Lewis Structure:** Total valence electrons = 26. Central sulfur forms one double bond with one oxygen and two single bonds with the other two oxygens, and retains one lone pair.
*Formal Charge calculation:*
FC(S) = 6 - 5 - 2 = 0.
FC(double-bonded O) = 6 - 4 - 2 = 0.
FC(single-bonded O) = 6 - 6 - 1 = -1 (each).
This minimized formal charge distribution makes it the most stable structure.

(b)
- **\(\text{SO}_3\):** 3 electron domains around S.
*Electron domain geometry:* Trigonal planar.
*Molecular geometry:* Trigonal planar.
*Bond angle:* Exactly \(120^\circ\).
- **\(\text{SO}_3^{2-}\):** 4 electron domains around S (3 bonding, 1 non-bonding).
*Electron domain geometry:* Tetrahedral.
*Molecular geometry:* Trigonal pyramidal.
*Bond angle:* \(< 109.5^\circ\) (typically \(104^\circ - 107^\circ\)) due to lone pair-bonding pair repulsion.

(c)
- Resonance refers to the use of two or more Lewis structures to represent a single molecule or ion that cannot be described accurately by only one structure.
- In the carbonate ion (\(\text{CO}_3^{2-}\)), three resonance structures can be drawn, each containing one double bond and two single bonds.
- In reality, the \(\pi\) electrons are delocalized across all three C-O bonds, giving a bond order of 1.33 for each bond.
- Experimental data shows that all three carbon-oxygen bond lengths are identical (approximately \(129\text{ pm}\)), which is shorter than a standard C-O single bond (\(143\text{ pm}\)) and longer than a standard C=O double bond (\(122\text{ pm}\)).

**(a) [Max 5 marks]**
- Award [1 mark] for correct \(\text{SO}_3\) Lewis structure with 3 double bonds and all valence electrons shown.
- Award [1 mark] for showing correct formal charges for S (0) and O (0) to justify stability.
- Award [1 mark] for correct \(\text{SO}_3^{2-}\) Lewis structure showing 1 double bond, 2 single bonds, and 1 lone pair on S, with overall charge bracketed or individual atoms labeled.
- Award [1 mark] for showing correct formal charges for S (0), double-bonded O (0), and single-bonded O (-1).
- Award [1 mark] for a statement justifying that structures minimizing formal charges are the most stable.

**(b) [Max 4 marks]**
- Award [1 mark] for \(\text{SO}_3\) geometries (trigonal planar for both).
- Award [1 mark] for \(\text{SO}_3\) bond angle (exactly \(120^\circ\)).
- Award [1 mark] for \(\text{SO}_3^{2-}\) geometries (tetrahedral domain and trigonal pyramidal molecular).
- Award [1 mark] for \(\text{SO}_3^{2-}\) bond angle (accept any value in the range \(104^\circ\) to \(107^\circ\) or explicitly stated as \(< 109.5^\circ\)).

**(c) [Max 4 marks]**
- Award [1 mark] for defining resonance as the delocalization of electrons across multiple bonding positions represented by a hybrid of contributing structures.
- Award [1 mark] for stating that carbonate has three equivalent C-O bonds (bond order 1.33).
- Award [1 mark] for stating that experimental bond lengths are all equal.
- Award [1 mark] for explaining that the bond lengths are intermediate between single and double carbon-oxygen bonds.

(a) (i) Dissociation equation:
\(\text{C}_2\text{H}_5\text{COOH}(aq) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(aq) + \text{H}^+(aq)\)
*(Accept \(\text{C}_2\text{H}_5\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(aq) + \text{H}_3\text{O}^+(aq)\))*
(ii) \(K_a\) expression:
\(K_a = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]}\)
*(Accept \(\text{H}_3\text{O}^+\) in place of \(\text{H}^+\))*

(b) (i) \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2.94} = 1.15 \times 10^{-3}\text{ mol dm}^{-3}\)
(ii) Calculation:
- Since \([\text{H}^+] = [\text{C}_2\text{H}_5\text{COO}^-] = 1.15 \times 10^{-3}\text{ mol dm}^{-3}\):
*Using exact equilibrium concentration:* \([\text{C}_2\text{H}_5\text{COOH}]_{\text{eq}} = 0.150 - 0.00115 = 0.149\text{ mol dm}^{-3}\)
\(K_a = \frac{(1.15 \times 10^{-3})^2}{0.149} = 8.86 \times 10^{-6}\text{ mol dm}^{-3}\)
*Using approximation:* \([\text{C}_2\text{H}_5\text{COOH}]_{\text{eq}} \approx 0.150\text{ mol dm}^{-3}\)
\(K_a = \frac{(1.15 \times 10^{-3})^2}{0.150} = 8.82 \times 10^{-6}\text{ mol dm}^{-3}\)
- \(pK_a = -\log_{10}(K_a)\):
If \(K_a = 8.86 \times 10^{-6}\), \(pK_a = 5.05\).
If \(K_a = 8.82 \times 10^{-6}\), \(pK_a = 5.05\) (or \(5.06\)).
- **Assumptions:**
1. \([\text{H}^+] = [\text{C}_2\text{H}_5\text{COO}^-]\) (autoionization of water is negligible).
2. \([\text{C}_2\text{H}_5\text{COOH}]_{\text{equilibrium}} \approx [\text{C}_2\text{H}_5\text{COOH}]_{\text{initial}}\) (dissociation is negligible).

(c) (i)
- The buffer contains a reservoir of weak acid (\(\text{C}_2\text{H}_5\text{COOH}\)) and its conjugate base (\(\text{C}_2\text{H}_5\text{COO}^-\)).
- When a strong acid is added, the added \(\text{H}^+\) is consumed by the conjugate base:
\(\text{C}_2\text{H}_5\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{C}_2\text{H}_5\text{COOH}(aq)\)
- This prevents the concentration of free hydrogen ions from increasing significantly, maintaining a stable pH.
(ii) Calculation:
- Standard buffer formula: \(\text{pH} = pK_a + \log_{10} \left( \frac{[\text{conjugate base}]}{[\text{acid}]} \right)\)
- Since equal volumes (\(50.0\text{ cm}^3\) each) are mixed, final volume is \(100.0\text{ cm}^3\).
- \([\text{C}_2\text{H}_5\text{COOH}]_{\text{final}} = \frac{50.0}{100.0} \times 0.150 = 0.0750\text{ mol dm}^{-3}\)
- \([\text{C}_2\text{H}_5\text{COO}^-]_{\text{final}} = \frac{50.0}{100.0} \times 0.100 = 0.0500\text{ mol dm}^{-3}\)
- \(\text{pH} = 5.05 + \log_{10} \left( \frac{0.0500}{0.0750} \right) = 5.05 - 0.176 = 4.87\)
*(Accept 4.88 if using slightly different rounded values)*

**(a) [Max 2 marks]**
- Award [1 mark] for a correct balanced equation with reversible arrows (state symbols not strictly required but expected).
- Award [1 mark] for a correct \(K_a\) expression matching the equation.

**(b) [Max 5 marks]**
- Award [1 mark] for \([\text{H}^+] = 1.15 \times 10^{-3}\text{ mol dm}^{-3}\).
- Award [1 mark] for a correct calculation of \(K_a\) showing values substituted (accept \(8.82 \times 10^{-6}\) to \(8.9
\times 10^{-6}\)).
- Award [1 mark] for \(pK_a = 5.05\) or \(5.06\) (must be consistent with calculated \(K_a\)).
- Award [2 marks] for two clearly stated assumptions (1 mark each).

**(c) [Max 6 marks]**
- Award [1 mark] for explaining that the conjugate base neutralizes added \(\text{H}^+\).
- Award [1 mark] for correct ionic equation: \(\text{C}_2\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_2\text{H}_5\text{COOH}\).
- Award [1 mark] for noting that this leaves free \([\text{H}^+]\) relatively unchanged.
- Award [1 mark] for calculating correct concentrations of base and acid in the mixture (or identifying the ratio as \(0.0500 / 0.0750 = 2/3\)).
- Award [1 mark] for correct substitution into the Henderson-Hasselbalch equation.
- Award [1 mark] for final pH of \(4.87\) (or \(4.88\)).

(a) (i) Calculations:
- \(M_r(\text{CO}_2) = 44.01\text{ g mol}^{-1}\)
- \(\text{Mass of C} = 3.52\text{ g} \times \frac{12.01}{44.01} = 0.961\text{ g}\)
- \(M_r(\text{H}_2\text{O}) = 18.02\text{ g mol}^{-1}\)
- \(\text{Mass of H} = 1.44\text{ g} \times \frac{2 \times 1.01}{18.02} = 0.161\text{ g}\)
(ii) Calculation:
- \(\text{Mass of O} = 1.76\text{ g} - (0.961\text{ g} + 0.161\text{ g}) = 0.638\text{ g}\)
(iii) Mole ratio:
- \(n(\text{C}) = \frac{0.961}{12.01} = 0.0800\text{ mol}\)
- \(n(\text{H}) = \frac{0.161}{1.01} = 0.159\text{ mol} \approx 0.160\text{ mol}\)
- \(n(\text{O}) = \frac{0.638}{16.00} = 0.0399\text{ mol} \approx 0.0400\text{ mol}\)
- Ratio \(\text{C} : \text{H} : \text{O} = \frac{0.0800}{0.0399} : \frac{0.160}{0.0399} : \frac{0.0399}{0.0399} = 2 : 4 : 1\).
- Empirical formula is \(\text{C}_2\text{H}_4\text{O}\).

(b) Molecular formula:
- Molar mass of empirical unit \(\text{C}_2\text{H}_4\text{O} = 2(12.01) + 4(1.01) + 16.00 = 44.06\text{ g mol}^{-1}\).
- Ratio of molecular mass to empirical formula mass = \(\frac{88.1}{44.06} \approx 2\).
- Molecular formula = \(2 \times (\text{C}_2\text{H}_4\text{O}) = \text{C}_4\text{H}_8\text{O}_2\).

(c) (i) Balanced equation:
\(\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\)
*(Accept single arrow and condensed structural formulas like \(\text{C}_2\text{H}_4\text{O}_2\) and \(\text{C}_2\text{H}_6\text{O}\))*
(ii) Percentage yield:
- \(M_r(\text{ethanol}) = 46.08\text{ g mol}^{-1}\)
- \(n(\text{ethanol}) = \frac{15.0\text{ g}}{46.08\text{ g mol}^{-1}} = 0.3255\text{ mol}\)
- Theoretical yield of ethyl ethanoate = \(0.3255\text{ mol}\)
- \(M_r(\text{ethyl ethanoate}) = 88.12\text{ g mol}^{-1}\)
- Theoretical mass of ethyl ethanoate = \(0.3255\text{ mol} \times 88.12\text{ g mol}^{-1} = 28.68\text{ g}\)
- \(\text{Percentage Yield} = \frac{\text{Actual Mass}}{\text{Theoretical Mass}} \times 100 = \frac{18.2\text{ g}}{28.68\text{ g}} \times 100 = 63.5\%\)

**(a) [Max 6 marks]**
- Award [1 mark] for mass of C (\(0.961\text{ g}\)) and [1 mark] for mass of H (\(0.161\text{ g}\)).
- Award [1 mark] for mass of O (\(0.638\text{ g}\)) based on subtraction.
- Award [1 mark] for calculating the moles of each element correctly.
- Award [1 mark] for finding the simplest integer ratio.
- Award [1 mark] for writing the correct empirical formula (\(\text{C}_2\text{H}_4\text{O}\)).

**(b) [Max 1 mark]**
- Award [1 mark] for \(\text{C}_4\text{H}_8\text{O}_2\) with supporting working showing comparison of molar masses.

**(c) [Max 6 marks]**
- Award [2 marks] for correct reactants, products, and stoichiometry (award [1 mark] if formulas are slightly incorrect but reaction type is clearly represented, or if water is omitted).
- Award [1 mark] for moles of ethanol = \(0.3255\text{ mol}\).
- Award [1 mark] for theoretical yield of ethyl ethanoate = \(28.68\text{ g}\).
- Award [1 mark] for percentage yield formulation.
- Award [1 mark] for final answer of \(63.5\%\) (accept range \(63.4\% - 63.6\%\)).

(a) (i) Calculation of heat released:
- \(\text{Total mass of mixture } (m) = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\)
- \(\Delta T = 28.1^\circ\text{C} - 21.4^\circ\text{C} = 6.7\text{ K}\) (or \(6.7^\circ\text{C}\))
- \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.80\text{ kJ}\)
(ii) Calculation of \(\Delta H_{\text{neut}}\):
- \(n(\text{H}^+) = n(\text{OH}^-) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
- \(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.0\text{ kJ mol}^{-1}\)
*(Negative sign is required for exothermic neutralization)*

(b) (i) Two major systematic errors:
1. Heat loss to the environment (the polystyrene cup is not a perfect insulator).
2. Heat absorbed by the calorimeter container, thermometer, and stirrer is not accounted for in the calculation.
(ii) Explanation:
- Ethanoic acid is a weak acid and exists primarily as undissociated molecules in solution.
- Some energy must be absorbed to break the O-H bonds to fully dissociate the acid during neutralization, which reduces the net amount of heat released.

(c) Reaction Pathway Diagram:
- Y-axis labeled "Enthalpy" (or "Potential Energy") and X-axis labeled "Reaction coordinate" (or "Progress of reaction").
- Horizontal line for Reactants: \(\text{H}^+(aq) + \text{OH}^-(aq)\) (or \(\text{HCl} + \text{NaOH}\)) at higher energy level.
- Horizontal line for Products: \(\text{H}_2\text{O}(l)\) (or \(\text{NaCl} + \text{H}_2\text{O}\)) at lower energy level.
- Curve starts at reactants, goes up to a maximum, then drops to products.
- Double-headed arrow or upward single-headed arrow from Reactants to the peak labeled \(E_a\).
- Downward single-headed arrow from Reactants to Products labeled \(\Delta H\).

**(a) [Max 5 marks]**
- Award [1 mark] for correctly calculating \(\Delta T = 6.7\text{ K}\) and mass = \(100.0\text{ g}\).
- Award [1 mark] for \(q = 2.80\text{ kJ}\) (or \(2801\text{ J}\)).
- Award [1 mark] for calculating moles of reactants = \(0.0500\text{ mol}\).
- Award [1 mark] for the correct magnitude of \(56.0\text{ kJ mol}^{-1}\).
- Award [1 mark] for the negative sign on \(\Delta H\).

**(b) [Max 4 marks]**
- Award [2 marks] for any two valid sources of systematic error (1 mark each). Do not accept human errors like 'misreading the thermometer'.
- Award [1 mark] for stating weak acids are only partially dissociated.
- Award [1 mark] for stating that dissociation is endothermic / requires energy, reducing overall exothermic heat release.

**(c) [Max 4 marks]**
- Award [1 mark] for correct axis labeling.
- Award [1 mark] for showing reactants at a higher level than products.
- Award [1 mark] for showing \(E_a\) correctly labeled.
- Award [1 mark] for showing \(\Delta H\) correctly labeled (with arrow pointing downwards, or stating value is negative).

(a) (i) Oxidation state of Cr:
- Let oxidation state of Cr be \(x\).
- \(2(x) + 7(-2) = -2 \Rightarrow 2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6\)
(ii) Reduction half-equation:
- Balance chromium: \(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}\)
- Balance oxygen with water: \(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
- Balance hydrogen with protons: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
- Balance charge with electrons: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
(iii) Overall balanced redox equation:
- Oxidation half-equation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\)
- Multiply oxidation by 6 to match electrons: \(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-\)
- Add equations: \(\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 6\text{Fe}^{3+}(aq) + 7\text{H}_2\text{O}(l)\)

(b) (i) Standard conditions:
- Concentration of electrolyte solutions = \(1.00\text{ mol dm}^{-3}\)
- Temperature = \(298\text{ K}\) (or \(25^\circ\text{C}\))
- Pressure = \(100\text{ kPa}\) / \(1\text{ bar}\) (if gas is involved)
*(Any two are required for full marks)*
(ii) Anode and Cathode:
- **Anode (oxidation occurs):** Magnesium electrode (it has the lower/more negative standard reduction potential, \(-2.37\text{ V}\)).
- **Cathode (reduction occurs):** Silver electrode (it has the higher/more positive standard reduction potential, \(+0.80\text{ V}\)).
(iii) Cell notation:
- \(\text{Mg}(s) \mid \text{Mg}^{2+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag}(s)\)
*(Accept without state symbols, but order must be anode on left, cathode on right)*
(iv) Standard cell potential:
- \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = (+0.80\text{ V}) - (-2.37\text{ V}) = +3.17\text{ V}\)

**(a) [Max 5 marks]**
- Award [1 mark] for +6 (or VI, accept 6).
- Award [2 marks] for correct reduction half-equation (award [1 mark] if species are correct but coefficients/charges are unbalanced).
- Award [2 marks] for correct overall equation (award [1 mark] if reactants and products are correct but unbalanced).

**(b) [Max 8 marks]**
- Award [2 marks] for any two correct standard conditions (1 mark each).
- Award [1 mark] for correctly identifying Anode as Mg (with brief justification).
- Award [1 mark] for correctly identifying Cathode as Ag (with brief justification).
- Award [2 marks] for correct cell notation (award [1 mark] if order is reversed or double vertical bar is missing).
- Award [1 mark] for standard potential formula and [1 mark] for correct calculation of \(+3.17\text{ V}\) (must include unit V).

(a) (i) Order determination:
- **With respect to \(\text{NO}\):** Compare Run 1 and Run 2 where \([\text{H}_2]\) is constant at \(0.100\text{ mol dm}^{-3}\).
As \([\text{NO}]\) doubles (from 0.100 to 0.200), the rate increases by a factor of \(\frac{4.80 \times 10^{-3}}{1.20 \times 10^{-3}} = 4\).
Since \(2^2 = 4\), the reaction is **second order** with respect to \(\text{NO}\).
- **With respect to \(\text{H}_2\):** Compare Run 1 and Run 3 where \([\text{NO}]\) is constant at \(0.100\text{ mol dm}^{-3}\).
As \([\text{H}_2]\) doubles (from 0.100 to 0.200), the rate increases by a factor of \(\frac{2.40 \times 10^{-3}}{1.20 \times 10^{-3}} = 2\).
Since \(2^1 = 2\), the reaction is **first order** with respect to \(\text{H}_2\).

(ii) Rate expression:
\(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\)

(iii) Calculation of \(k\):
- Using data from Run 1:
\(1.20 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.100\text{ mol dm}^{-3})^2 \times (0.100\text{ mol dm}^{-3})\)
\(1.20 \times 10^{-3} = k \times (1.00 \times 10^{-3})\)
\(k = 1.20\)
- Units:
\(\text{Units} = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\) (or \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\))

(b) (i) Consistency of mechanism:
- The rate-determining step is the slowest elementary step, which is Step 2:
\(\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2]\)
- However, \(\text{N}_2\text{O}_2\) is an intermediate. From the fast pre-equilibrium in Step 1:
\(K_{\text{eq}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \Rightarrow [\text{N}_2\text{O}_2] = K_{\text{eq}}[\text{NO}]^2\)
- Substituting this expression into the rate equation for Step 2:
\(\text{Rate} = k_2 K_{\text{eq}}[\text{NO}]^2[\text{H}_2]\)
Since \(k_2 K_{\text{eq}} = k\) (constant), the rate law becomes \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\), which perfectly matches the experimental rate expression.

(ii) Intermediates:
- Intermediates are species produced in one elementary step and consumed in another, and do not appear in the overall chemical equation.
- The intermediates in this mechanism are **\(\text{N}_2\text{O}_2\)** and **\(\text{N}_2\text{O}\)**.

**(a) [Max 8 marks]**
- Award [2 marks] for order wrt \(\text{NO}\) (1 mark for correct order, 1 mark for clear reasoning showing the calculation).
- Award [2 marks] for order wrt \(\text{H}_2\) (1 mark for correct order, 1 mark for clear reasoning showing the calculation).
- Award [1 mark] for correct rate expression matching deduced orders.
- Award [2 marks] for rate constant value (\(1.20\)).
- Award [1 mark] for correct units (\(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\) or equivalent order of terms).

**(b) [Max 5 marks]**
- Award [1 mark] for identifying Step 2 as the rate-determining step and writing its rate expression.
- Award [1 mark] for expressing intermediate concentration in terms of reactants from the pre-equilibrium.
- Award [1 mark] for showing substitution to yield the final rate expression.
- Award [2 marks] for identifying both intermediates (\(\text{N}_2\text{O}_2\) and \(\text{N}_2\text{O}\)) (1 mark for each; deduct 1 mark if extra substances that are not intermediates are listed).

(a) Full structural formulas:
- **Butan-1-ol:** A linear chain of 4 carbons with the hydroxyl group on the first carbon:
\(\text{H}-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{H})-\text{O}-\text{H}\)
- **Butan-2-ol:** A linear chain of 4 carbons with the hydroxyl group on the second carbon:
\(\text{H}-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{OH})-\text{C}(\text{H})(\text{H})-\text{H}\)
- **2-methylpropan-2-ol:** A central carbon bonded to three methyl groups and one hydroxyl group:
\((\text{CH}_3)_3\text{C-OH}\) fully drawn out showing all C-H and O-H bonds.

(b) Classifications:
- **Butan-1-ol:** Primary alcohol (the carbon bonded to the -OH group is attached to only 1 other carbon atom).
- **Butan-2-ol:** Secondary alcohol (the carbon bonded to the -OH group is attached to 2 other carbon atoms).
- **2-methylpropan-2-ol:** Tertiary alcohol (the carbon bonded to the -OH group is attached to 3 other carbon atoms).

(c) (i) \(^1\text{H}\) NMR signals:
- **Butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)):** 5 chemical environments of H, so **5 signals** are expected.
- **2-methylpropan-2-ol (\((\text{CH}_3)_3\text{C-OH}\)):** 2 chemical environments of H (9 equivalent methyl protons and 1 hydroxyl proton), so **2 signals** are expected.
(ii) Different homologous series (Ethers):
- Isomer: Diethyl ether, \(\text{CH}_3\text{CH}_2\text{O\text{CH}_2\text{CH}_3\). IUPAC name: **ethoxyethane**.
- *Alternatively:* Methylpropyl ether, \(\text{CH}_3\text{O\text{CH}_2\text{CH}_2\text{CH}_3\). IUPAC name: **1-methoxypropane**.

(d) Distinguishing chemical test:
- **Reagent and condition:** Acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\)) under heating/reflux.
- **Observations:**
* **Butan-1-ol (primary alcohol):** Oxidized to butanal / butanoic acid, resulting in a color change from **orange to green**.
* **2-methylpropan-2-ol (tertiary alcohol):** Resistant to oxidation under these conditions, so the solution **remains orange**.

**(a) [Max 3 marks]**
- Award [1 mark] for each correct full structural formula showing all atoms and bonds (including the O-H bond).

**(b) [Max 3 marks]**
- Award [1 mark] for each correct classification (butan-1-ol = primary, butan-2-ol = secondary, 2-methylpropan-2-ol = tertiary).

**(c) [Max 4 marks]**
- Award [1 mark] for 5 signals in butan-1-ol.
- Award [1 mark] for 2 signals in 2-methylpropan-2-ol.
- Award [1 mark] for drawing or identifying a valid ether structure (like diethyl ether).
- Award [1 mark] for its correct corresponding IUPAC name (e.g., ethoxyethane or 1-methoxypropane).

**(d) [Max 3 marks]**
- Award [1 mark] for stating correct reagent and condition (acidified potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7\) with heat/warmth. Accept acidified potassium manganate(VII) / \(\text{KMnO}_4\)).
- Award [1 mark] for correct color change with butan-1-ol (orange to green; or purple to colorless if manganate is used).
- Award [1 mark] for stating no reaction/color change with 2-methylpropan-2-ol (remains orange; or remains purple if manganate is used).

(a) Mass of solution = \(50.0\text{ g} + 3.30\text{ g} = 53.30\text{ g}\). Temperature change, \(\Delta T = 31.8 - 21.2 = 10.6\text{ K}\). Heat evolved, \(q = m \cdot c \cdot \Delta T = 53.30\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.6\text{ K} = 2361.6\text{ J} = 2.3616\text{ kJ}\). (If only the mass of water is used: \(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.6\text{ K} = 2215.4\text{ J} = 2.2154\text{ kJ}\)). Moles of \(\text{CaCl}_2\) = \(3.30\text{ g} / 110.98\text{ g mol}^{-1} = 0.02974\text{ mol}\). \(\Delta H_{\text{sol}} = -q / n\). If using mass of solution: \(\Delta H_{\text{sol}} = -2.3616\text{ kJ} / 0.02974\text{ mol} = -79.4\text{ kJ mol}^{-1}\). If using mass of water: \(\Delta H_{\text{sol}} = -2.2154\text{ kJ} / 0.02974\text{ mol} = -74.5\text{ kJ mol}^{-1}\). (b) Absolute uncertainty in \(\Delta T = 0.1 + 0.1 = 0.2\,^{\circ}\text{C}\). Percentage uncertainty = \((0.2 / 10.6) \times 100 = 1.9\%\) (or \(1.89\%\)). (c) Heat loss to the surroundings is a major systematic error. This causes the measured temperature rise to be smaller than the theoretical value, leading to a calculated enthalpy of solution that is less exothermic (closer to zero / smaller in magnitude).

[a] 2.5 Marks: 1 mark for calculating correct heat evolved, q (accept 2.36 kJ using mass of solution, or 2.22 kJ using mass of water); 1 mark for calculating correct moles of CaCl2 (0.0297 mol); 0.5 marks for dividing -q by n to obtain -79.4 kJ/mol or -74.5 kJ/mol (sign and value must both be correct). [b] 1 Mark: 0.5 marks for absolute uncertainty of 0.2 K, 0.5 marks for percentage uncertainty of 1.9%. [c] 1 Mark: 0.5 marks for identifying heat loss to surroundings (or neglecting heat capacity of calorimeter), 0.5 marks for explaining that this decreases the temperature rise and makes the calculated value less exothermic.

(a) Comparing Exp 1 and Exp 2: \([\text{I}^-]\) remains constant, while \([\text{S}_2\text{O}_8^{2-}]\) doubles. The initial rate doubles (from \(1.20 \times 10^{-4}\) to \(2.40 \times 10^{-4}\)). Thus, the reaction is first-order with respect to \(\text{S}_2\text{O}_8^{2-}\). Comparing Exp 2 and Exp 3: \([\text{S}_2\text{O}_8^{2-}]\) remains constant, while \([\text{I}^-]\) doubles. The initial rate doubles (from \(2.40 \times 10^{-4}\) to \(4.80 \times 10^{-4}\)). Thus, the reaction is first-order with respect to \(\text{I}^-\). The rate equation is: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). (b) Substituting values from Experiment 1 into the rate equation: \(1.20 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.100\text{ mol dm}^{-3}) \times (0.100\text{ mol dm}^{-3})\). Solving for \(k\): \(k = 1.20 \times 10^{-4} / 0.0100 = 0.0120\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). (c) The rate constant increases with temperature. According to collision theory, an increase in temperature increases the average kinetic energy of the reacting particles, meaning a significantly larger fraction of collisions will have energy equal to or greater than the activation energy (\(E_{\text{a}}\)), resulting in more successful collisions per unit time.

[a] 2 Marks: 1 mark for correct deduction of first-order with respect to both reactants with justification, 1 mark for writing the correct rate equation. [b] 1.5 Marks: 1 mark for correct numerical value of k (0.0120 or 1.20 x 10^-2), 0.5 marks for correct units (dm3 mol-1 s-1). [c] 1 Mark: 0.5 marks for stating k increases, 0.5 marks for explaining that a higher fraction of molecules have energy greater than or equal to Ea.

(a) \(T = 99.0\,^{\circ}\text{C} + 273.15 = 372.2\text{ K}\) (or \(372\text{ K}\)). \(V = 250.0\text{ cm}^3 \times (1\text{ m} / 100\text{ cm})^3 = 2.500 \times 10^{-4}\text{ m}^3\). (b) Using \(PV = nRT\), the number of moles is \(n = PV / RT = (101.3 \times 10^3\text{ Pa} \times 2.500 \times 10^{-4}\text{ m}^3) / (8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 372.2\text{ K}) = 25.325 / 3092.98 = 0.008188\text{ mol}\). (Using \(T = 372\text{ K}\) gives \(0.008193\text{ mol}\)). Molar mass \(M = \text{mass} / n = 0.595\text{ g} / 0.008188\text{ mol} = 72.7\text{ g mol}^{-1}\) (or \(72.6\text{ g mol}^{-1}\) if using \(372\text{ K}\)). (c) If the liquid did not completely vaporize, the residual unvaporized liquid remains in the flask alongside the condensed vapor. This increases the measured mass of the liquid, resulting in an artificially high calculated molar mass because molar mass is directly proportional to the measured mass of the liquid.

[a] 1 Mark: 0.5 marks for Kelvin conversion (372.2 K or 372 K), 0.5 marks for volume conversion (2.500 x 10^-4 m3). [b] 2.5 Marks: 1 mark for correct substitution into ideal gas equation, 1 mark for calculating correct moles (approx. 0.0082 mol), 0.5 marks for calculating the final molar mass of 72.6 to 72.7 g/mol with correct units. [c] 1 Mark: 0.5 marks for stating that molar mass would be higher/overestimated, 0.5 marks for explaining that the measured mass of the liquid is artificially increased.

(a) The strong, sharp absorption band at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C=O}\)). The broad absorption band centered at \(3000\text{ cm}^{-1}\) corresponds to the hydrogen-bonded \(\text{O-H}\) stretch of a carboxylic acid (which overlaps the \(\text{C-H}\) stretching region). Therefore, **X** contains a carboxylic acid functional group (\(-\text{COOH}\)). (b) Based on the molecular formula and functional group, **X** is propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\). The \(^1\text{H-NMR}\) assignment is: - Triplet at \(\delta = 1.1\text{ ppm}\) (3H) corresponds to the methyl protons (\(-\text{CH}_3\)) adjacent to a \(-\text{CH}_2-\). The adjacent carbon has 2 hydrogens, so the splitting is \(2 + 1 = 3\) (triplet). - Quartet at \(\delta = 2.3\text{ ppm}\) (2H) corresponds to the methylene protons (\(-\text{CH}_2-\)) adjacent to a \(-\text{CH}_3\). The adjacent carbon has 3 hydrogens, so the splitting is \(3 + 1 = 4\) (quartet). - Singlet at \(\delta = 11.5\text{ ppm}\) (1H) is the acidic carboxylic acid proton (\(-\text{COOH}\)\), which has no neighboring hydrogens on adjacent carbons, thus showing no splitting (singlet). (c) Methyl ethanoate, \(\text{CH}_3\text{COOCH}_3\), is a symmetric ester with respect to neighboring hydrogens. It will display two signals: one for the methyl group adjacent to the carbonyl, and one for the methoxy group. Since neither methyl group has adjacent protons on neighboring carbons, both signals will be singlets.

[a] 1.5 Marks: 0.5 marks for identifying C=O carbonyl, 0.5 marks for identifying O-H of carboxylic acid (with correct wavenumber references), 0.5 marks for deducing the functional group is a carboxylic acid. [b] 2 Marks: 0.5 marks for the correct structure of propanoic acid (or CH3CH2COOH), 0.5 marks for explaining the triplet (3H, adjacent to 2 protons), 0.5 marks for explaining the quartet (2H, adjacent to 3 protons), 0.5 marks for explaining the singlet (1H, no adjacent protons). [c] 1 Mark: 0.5 marks for stating 2 signals, 0.5 marks for stating that both are singlets.

(a) Salicylic acid (2-hydroxybenzoic acid) consists of a benzene ring with an ortho-hydroxyl group (\(-\text{OH}\)) and a carboxylic acid group (\(-\text{COOH}\)). Aspirin (acetylsalicylic acid) consists of a benzene ring with an ortho-acetyloxy group (\(-\text{OCOCH}_3\)) and a carboxylic acid group (\(-\text{COOH}\)). The functional group formed during the reaction is an ester (\(-\text{COO}-\)) via the esterification of the phenolic hydroxyl group. (b) Ice-cold water is used to wash away any remaining soluble impurities (such as unreacted ethanoic acid or catalyst) from the surface of the crystals. Cold water is chosen because aspirin is only minimally soluble in cold water, minimizing the loss of the desired yield due to dissolution. (c) Pure aspirin does not contain a phenolic hydroxyl group (\(-\text{OH}\)), which is present in salicylic acid. Salicylic acid will show a characteristic broad phenol \(\text{O-H}\) absorption band at \(3200-3600\text{ cm}^{-1}\), which is absent in pure aspirin. Furthermore, aspirin will show two distinct carbonyl (\(\text{C=O}\)) stretching peaks (one around \(1750\text{ cm}^{-1}\) for the ester and one around \(1680\text{ cm}^{-1}\) for the carboxylic acid), whereas salicylic acid will only show one carbonyl peak.

[a] 2 Marks: 1 mark for correct structural formulas (or very clear description) of salicylic acid and aspirin, 1 mark for identifying the ester group as the one formed. [b] 1 Mark: 0.5 marks for stating it washes away soluble impurities, 0.5 marks for stating it minimizes product dissolution/loss. [c] 1.5 Marks: 1 mark for identifying the presence of phenol O-H stretch in salicylic acid and its absence in pure aspirin, 0.5 marks for mentioning the two carbonyl stretches in aspirin compared to one in salicylic acid.

(a) Neutralization equations: (1) \(\text{Mg(OH)}_2(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + 2\text{H}_2\text{O}(l)\); (2) \(\text{NaHCO}_3(s) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\). (b) Molar mass of \(\text{Mg(OH)}_2 = 24.31 + 2(16.00 + 1.01) = 58.33\text{ g mol}^{-1}\). Moles of \(\text{Mg(OH)}_2\) in \(1.00\text{ g} = 1.00 / 58.33 = 0.0171\text{ mol}\). Since 1 mol of \(\text{Mg(OH)}_2\) neutralizes 2 mol of \(\text{HCl}\), \(1.00\text{ g}\) neutralizes \(2 \times 0.0171 = 0.0342\text{ mol of }\text{HCl}\). Molar mass of \(\text{NaHCO}_3 = 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01\text{ g mol}^{-1}\). Moles of \(\text{NaHCO}_3\) in \(1.00\text{ g} = 1.00 / 84.01 = 0.0119\text{ mol}\). Since 1 mol of \(\text{NaHCO}_3\) neutralizes 1 mol of \(\text{HCl}\), \(1.00\text{ g}\) neutralizes \(0.0119\text{ mol of }\text{HCl}\). Thus, \(\text{Mg(OH)}_2\) neutralizes more acid per gram. (c) A side effect is bloating, belching, or flatulence. This is caused by the production of carbon dioxide (\(\text{CO}_2\)) gas, which increases the pressure in the stomach.

[a] 2 Marks: 1 mark for the balanced equation of Mg(OH)2, 1 mark for the balanced equation of NaHCO3. [b] 1.5 Marks: 0.5 marks for calculating moles of HCl neutralized by 1g of Mg(OH)2 (0.0342 mol), 0.5 marks for calculating moles of HCl neutralized by 1g of NaHCO3 (0.0119 mol), 0.5 marks for concluding Mg(OH)2 is more effective. [c] 1 Mark: 0.5 marks for stating bloating/belching/flatulence, 0.5 marks for linking it to the release of carbon dioxide gas.

(a) The beta-lactam ring is a cyclic amide consisting of three carbon atoms and one nitrogen atom in a four-membered ring. The bond angles are forced to be approximately \(90^{\circ}\), which is much smaller than the normal sp3 tetrahedral angle (\(109.5^{\circ}\)) and sp2 trigonal planar angle (\(120^{\circ}\)). This creates severe ring strain, making the ring highly reactive as opening it relieves this strain. (b) Penicillin's structure mimics the D-alanyl-D-alanine terminal of the peptidoglycan precursor. The high reactivity of the beta-lactam ring allows it to undergo a nucleophilic attack by a serine residue at the active site of the transpeptidase enzyme. This covalently bonds and permanently deactivates the enzyme, stopping cell wall cross-linking, which causes the bacterial cell to burst and die (lysis). (c) Bacteria develop resistance by producing beta-lactamase (penicillinase) enzymes, which hydrolyze the beta-lactam ring, making the drug inactive. To overcome this, the side-chain of penicillin is chemically modified (producing drugs like methicillin) to introduce bulky groups that create steric hindrance, preventing the beta-lactamase from binding while still allowing the drug to access the bacterial transpeptidase.

[a] 2 Marks: 1 mark for explaining that the four-membered ring forces bond angles of approx 90 degrees (which are smaller than normal tetrahedral/trigonal planar angles), 1 mark for stating that this creates severe angle/ring strain which is relieved upon ring opening. [b] 1.5 Marks: 0.5 marks for stating penicillin mimics the D-alanyl-D-alanine precursor, 1 mark for explaining that the ring undergoes nucleophilic attack by serine at the active site of transpeptidase, forming a covalent bond and inactivating the enzyme. [c] 1 Mark: 0.5 marks for mentioning bacteria produce beta-lactamase to open/hydrolyze the ring, 0.5 marks for stating that modifying the side-chain introduces steric hindrance to prevent beta-lactamase binding.

(a) Using \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\): Convert \(\Delta S^{\ominus}\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\): \(\Delta S^{\ominus} = -85.0 \times 10^{-3}\text{ kJ K}^{-1}\text{ mol}^{-1}\). \(\Delta G^{\ominus} = -45.0\text{ kJ mol}^{-1} - \left(310\text{ K} \times (-0.0850\text{ kJ K}^{-1}\text{ mol}^{-1})\right) = -45.0 - (-26.35) = -18.7\text{ kJ mol}^{-1}\) (or \(-18.65\text{ kJ mol}^{-1}\)). Since \(\Delta G^{\ominus} < 0\), the binding process is spontaneous at \(310\text{ K}\). (b) Although water displacement (the hydrophobic effect) increases entropy, the association of two highly mobile, independent species (the drug and the receptor) into a single, highly constrained complex restricts translational and rotational degrees of freedom. This conformational and vibrational restriction represents a major decrease in entropy, which outweighs the positive entropy from displaced water. (c) The process becomes non-spontaneous when \(\Delta G^{\ominus} \ge 0\). At the temperature boundary where \(\Delta G^{\ominus} = 0\): \(T = \Delta H^{\ominus} / \Delta S^{\ominus} = -45000\text{ J mol}^{-1} / -85.0\text{ J K}^{-1}\text{ mol}^{-1} = 529.4\text{ K}\). Converting to Celsius: \(529.4 - 273.15 = 256.3\,^{\circ}\text{C}\) (accept \(256\,^{\circ}\text{C}\) or \(256.2\,^{\circ}\text{C}\)).

[a] 2 Marks: 1 mark for correct substitution and calculation of Delta G (-18.7 kJ/mol), 1 mark for concluding that the process is spontaneous because Delta G is negative. [b] 1.5 Marks: 1 mark for explaining that two independent molecules combine to form a single restricted complex (loss of rotational/translational freedom), 0.5 marks for noting this conformational constraint outweighing the positive entropy contribution of displaced water. [c] 1 Mark: 0.5 marks for setting Delta G = 0 to calculate temperature in Kelvin (529 K), 0.5 marks for converting to Celsius (256 degrees C).

(a) \(\Delta T = 35.7^\circ\text{C} - 20.2^\circ\text{C} = 15.5\text{ K}\) (or \(15.5^\circ\text{C}\)). Total volume = \(50.0 + 50.0 = 100.0\text{ cm}^3\), so mass \(m = 100.0\text{ g}\). \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 15.5\text{ K} = 6479\text{ J} = 6.48\text{ kJ}\). (b) Equation: \(\text{H}_3\text{PO}_4\text{(aq)} + 3\text{NaOH(aq)} \rightarrow \text{Na}_3\text{PO}_4\text{(aq)} + 3\text{H}_2\text{O(l)}\). Moles of \(\text{H}_3\text{PO}_4 = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). Moles of \(\text{NaOH} = 0.0500\text{ dm}^3 \times 2.50\text{ mol dm}^{-3} = 0.125\text{ mol}\). Since \(3 \times 0.0500 = 0.150\text{ mol}\) of \(\text{NaOH}\) is needed to react completely with the phosphoric acid, \(\text{NaOH}\) is the limiting reactant (as only \(0.125\text{ mol}\) is present). Moles of \(\text{H}_2\text{O}\) formed = moles of \(\text{NaOH}\) reacted (since 1 mole of \(\text{OH}^-\)\ reacts to form 1 mole of \(\text{H}_2\text{O}\)) = \(0.125\text{ mol}\). \(\Delta H_{\text{neut}} = -\frac{6.479\text{ kJ}}{0.125\text{ mol}} = -51.8\text{ kJ mol}^{-1}\). (c) Phosphoric acid is a weak acid and only partially dissociates in aqueous solution. Energy is required (absorbed) to break covalent bonds (specifically the \(\text{O-H}\) bonds) to fully ionize the acid during the neutralization process, making the overall process less exothermic.

(a) Award [0.5] for calculating \(\Delta T = 15.5\text{ K}\). Award [1.0] for correct substitution and final answer of \(6.48\text{ [kJ]}\) (or \(6.5\text{ [kJ]}\)). (b) Award [1.0] for identifying \(\text{NaOH}\) as the limiting reactant with showing calculation/reasoning. Award [1.0] for \(\Delta H_{\text{neut}} = -51.8\text{ [kJ mol}^{-1}]\) (accept range \(-51.8\) to \(-52.0\); negative sign is required for the mark). (c) Award [1.0] for stating that phosphoric acid is weak/partially ionized AND energy is absorbed/required to fully dissociate/ionize the acid (or break O-H bonds).

(a) Recrystallization procedure: Dissolve the crude aspirin in the minimum volume of hot solvent (such as ethanol or water). Allow the solution to cool slowly so that pure aspirin crystals precipitate out of the solution while soluble impurities remain dissolved. Filter the cold mixture under vacuum to isolate the crystals, wash them with a small amount of ice-cold solvent, and dry. (b) Purity determination: Determine the melting point of the crystalline product using a melting point apparatus. Pure aspirin has a sharp melting point at its literature value (around \(136^\circ\text{C}\)). Impure aspirin will melt over a wider temperature range and at a lower temperature than the pure substance. (c) Bioavailability: Converting aspirin to its sodium salt (ionic form) greatly increases its solubility in aqueous solutions (such as gastric juice). This leads to a faster rate of dissolution in the stomach. Consequently, the drug is absorbed more rapidly and efficiently into the bloodstream, increasing its bioavailability.

(a) Award [0.5] for dissolving crude aspirin in the minimum volume of hot solvent. Award [0.5] for cooling the solution to allow pure crystals to form (while impurities stay in solution). Award [0.5] for filtering the crystals and washing with cold solvent. (b) Award [0.5] for stating "determine the melting point". Award [1.0] for explaining that impure aspirin melts over a wide/broad range AND at a lower temperature (than pure aspirin/literature value). (c) Award [0.5] for stating that the sodium salt increases solubility in water/aqueous media. Award [0.5] for stating that this leads to a faster rate of dissolution. Award [0.5] for explaining that faster/greater dissolution leads to quicker absorption into the bloodstream, increasing bioavailability.