2023 IB DP Chemistry 模擬試題連答案詳解

The central iodine atom in \( \text{ICl}_4^- \) has 7 valence electrons plus 1 additional electron from the negative charge, giving 8 valence electrons in total. It forms 4 single covalent bonds with chlorine atoms, using 4 electrons. The remaining 4 electrons form 2 non-bonding pairs. The steric number (coordination number) is 6, indicating an octahedral electron domain geometry. The molecular geometry is square planar with 2 non-bonding pairs on the central iodine atom.

Award [1] for the correct combination of square planar geometry and 2 non-bonding pairs. Reject all other options.

Prop-2-enal has the structure \( \text{CH}_2=\text{CH}-\text{CH}=\text{O} \). The single bonds are all \( \sigma \) bonds, and each double bond consists of one \( \sigma \) and one \( \pi \) bond. Specifically, there are four C-H \( \sigma \) bonds, one C-C \( \sigma \) bond, one C=C \( \sigma \) bond, and one C=O \( \sigma \) bond, totaling 7 \( \sigma \) bonds. There is one \( \pi \) bond in the C=C double bond and one \( \pi \) bond in the C=O double bond, totaling 2 \( \pi \) bonds.

Award [1] for the correct identification of 7 \( \sigma \) and 2 \( \pi \) bonds. Reject all other choices.

The most stable Lewis structure of the cyanate ion is \( [:\ddot{\text{O}}-\text{C}\equiv\text{N}:]^- \). In this structure, the formal charges are calculated as follows: FC(O) = 6 valence - 6 non-bonding - 1 bond = -1; FC(C) = 4 valence - 0 non-bonding - 4 bonds = 0; FC(N) = 5 valence - 2 non-bonding - 3 bonds = 0. This structure is more stable than \( [\ddot{\text{O}}=\text{C}=\ddot{\text{N}}]^- \) (where O is 0 and N is -1) because the negative formal charge is located on the more electronegative oxygen atom.

Award [1] for the correct formal charges of -1, 0, and 0 for O, C, and N respectively. Reject all other options.

Adding a catalyst does not change the temperature or the total number of molecules, so the shape and the total area of the Maxwell-Boltzmann distribution curve remain unchanged. Instead, the catalyst provides an alternative reaction pathway with a lower activation energy, which effectively moves the activation energy line to the left, increasing the fraction of molecules with energy greater than or equal to the activation energy.

Award [1] for the correct choice that states the curve shape is unaltered but a lower activation energy pathway is provided.

Comparing Experiments 1 and 2: doubling [A] while keeping [B] constant doubles the rate, indicating the reaction is first-order with respect to A. Comparing Experiments 1 and 3: doubling [B] while keeping [A] constant quadruples the rate, indicating the reaction is second-order with respect to B. The rate expression is therefore Rate = k[A][B]\( ^2 \), giving an overall order of 3. The units for the rate constant are k = Rate / ([A][B]\( ^2 \)) = (mol dm\( ^{-3} \) s\( ^{-1} \)) / (mol dm\( ^{-3} \))\( ^3 \) = dm\( ^6 \) mol\( ^{-2} \) s\( ^{-1} \).

Award [1] for the correct unit of dm\( ^6 \) mol\( ^{-2} \) s\( ^{-1} \). Reject other units.

According to the Arrhenius equation: \( \ln k = -\frac{E_a}{R}\frac{1}{T} + \ln A \). This has the form of a straight-line equation \( y = mx + c \), where \( y = \ln k \) and \( x = \frac{1}{T} \). Thus, the slope of the plot is equal to \( -\frac{E_a}{R} \). Rearranging this relationship yields \( E_a = -\text{slope} \times R \).

Award [1] for the correct expression \( E_a = -\text{slope} \times R \).

The strength of a bond depends on its bond order. In \( \text{CO} \), there is a triple bond (bond order = 3). In \( \text{CO}_2 \), there are double bonds (bond order = 2). In the carbonate ion, \( \text{CO}_3^{2-} \), the negative charge is delocalized over three carbon-oxygen bonds, resulting in a bond order of 1.33. In methanol, \( \text{CH}_3\text{OH} \), there is a single carbon-oxygen bond (bond order = 1). Since bond strength decreases as the bond order decreases, the correct order is CO > CO\( _2 \) > CO\( _3^{2-} \) > CH\( _3 \)OH.

Award [1] for the correct order from strongest (triple bond) to weakest (single bond). Reject all other orders.

When the volumes of both reactants are doubled while keeping the concentrations the same, the moles of reactants that react also double. Consequently, the total amount of heat energy released, \( q \), doubles. However, the total mass (or volume) of the reaction mixture also doubles, meaning there is twice as much solution to be heated. Since \( q = mc\Delta T \), both \( q \) and \( m \) are multiplied by 2, leaving the temperature change \( \Delta T \) unchanged. Therefore, \( \Delta T_2 = \Delta T_1 \).

Award [1] for identifying that doubling both the heat produced and the mass of the solution results in the same temperature change.

To determine the most stable Lewis structure of \( \text{N}_2\text{O} \), we calculate the total number of valence electrons: \( 5 + 5 + 6 = 16 \) electrons (8 pairs).

Let's consider the possible resonance structures:
1. Structure I: \( \text{N}\equiv\text{N}-\text{O} \) with three lone pairs on O and one lone pair on terminal N.
- Formal charge on terminal \( \text{N} = 5 - 2 - 3 = 0 \)
- Formal charge on central \( \text{N} = 5 - 0 - 4 = +1 \)
- Formal charge on terminal \( \text{O} = 6 - 6 - 1 = -1 \)

2. Structure II: \( \text{N}=\text{N}=\text{O} \) with two lone pairs on both terminal atoms.
- Formal charge on terminal \( \text{N} = 5 - 4 - 2 = -1 \)
- Formal charge on central \( \text{N} = 5 - 0 - 4 = +1 \)
- Formal charge on terminal \( \text{O} = 6 - 4 - 2 = 0 \)

Both structures minimize formal charges, but Structure I is more stable because the negative formal charge is on the oxygen atom, which is more electronegative than nitrogen (electronegativities: \( \text{O} = 3.4 \), \( \text{N} = 3.0 \)). Therefore, in the most stable structure, the formal charge on the terminal N is 0 and on O is -1.

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

Let's analyze the bonding in \( \text{CH}_2=\text{CH}-\text{C}\equiv\text{N} \):
1. **Sigma (\(\sigma\)) bonds**:
- Two \( \text{C}-\text{H} \) single bonds in the \( \text{CH}_2 \) group: 2 \(\sigma\) bonds.
- One \( \text{C}=\text{C} \) double bond: 1 \(\sigma\) bond (and 1 \(\pi\) bond).
- One \( \text{C}-\text{H} \) single bond in the middle group: 1 \(\sigma\) bond.
- One \( \text{C}-\text{C} \) single bond connecting to the nitrile group: 1 \(\sigma\) bond.
- One \( \text{C}\equiv\text{N} \) triple bond: 1 \(\sigma\) bond (and 2 \(\pi\) bonds).
- Total \(\sigma\) bonds = \( 2 + 1 + 1 + 1 + 1 = 6 \).

2. **Pi (\(\pi\)) bonds**:
- One \(\pi\) bond from the \( \text{C}=\text{C} \) double bond.
- Two \(\pi\) bonds from the \( \text{C}\equiv\text{N} \) triple bond.
- Total \(\pi\) bonds = \( 1 + 2 = 3 \).

3. **Hybridization**:
- Carbon 1 (in \( \text{CH}_2 \)) has 3 electron domains (bonded to 2 H atoms and 1 C atom): \(sp^2\) hybridized.
- Carbon 2 (middle CH) has 3 electron domains (bonded to 1 H, 1 C, and 1 C): \(sp^2\) hybridized.
- Carbon 3 (in \( \text{C}\equiv\text{N} \)) has 2 electron domains (bonded to 1 C and 1 N): \(sp\) hybridized.
- Total \(sp^2\) hybridized carbon atoms = 2.

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

Let's look at the Lewis structures and geometries of each option:
- **\( \text{SF}_4 \)**: Sulfur has 6 valence electrons, and each F contributes 1, totaling 34 valence electrons. The central S has 4 bonding pairs and 1 lone pair (5 electron domains). The electron domain geometry is trigonal bipyramidal, and the molecular geometry is see-saw. Because of the asymmetrical lone pair, the dipoles do not cancel out, making the molecule polar.
- **\( \text{XeF}_4 \)**: Xenon has 8 valence electrons, totaling 36 valence electrons. The central Xe has 4 bonding pairs and 2 lone pairs (6 electron domains). The molecular geometry is square planar, which is symmetrical and non-polar.
- **\( \text{CF}_4 \)**: Carbon has 4 valence electrons, totaling 32 valence electrons. Carbon has 4 bonding pairs and 0 lone pairs (4 electron domains). The molecular geometry is tetrahedral, which is symmetrical and non-polar.
- **\( \text{BF}_4^- \)**: This is an ion rather than a neutral polar molecule, and it has tetrahedral geometry.

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

Bond length is inversely related to bond order. Higher bond orders result in stronger, shorter bonds.
- **\( \text{CO} \)**: Carbon monoxide has a triple bond (bond order = 3).
- **\( \text{CO}_2 \)**: Carbon dioxide has double bonds (bond order = 2).
- **\( \text{CO}_3^{2-} \)**: The carbonate ion has three resonance structures, giving a bond order of \( 4/3 \approx 1.33 \).
- **\( \text{CH}_3\text{OH} \)**: Methanol has a single \( \text{C}-\text{O} \) bond (bond order = 1).

Comparing bond orders: \( 3 > 2 > 1.33 > 1 \).
Therefore, the increasing order of bond length is: \( \text{CO} < \text{CO}_2 < \text{CO}_3^{2-} < \text{CH}_3\text{OH} \).

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

The units of reaction rate are always \( \text{mol}\,\text{dm}^{-3}\,\text{s}^{-1} \), and the units of concentration are \( \text{mol}\,\text{dm}^{-3} \).

Rearrange the rate expression for \( k \):
\( k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]^{-1}} \)

Substitute the units into the expression:
\( [k] = \frac{\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}}{(\text{mol}\,\text{dm}^{-3})^2 \times (\text{mol}\,\text{dm}^{-3})^{-1}} \)
\( [k] = \frac{\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}}{\text{mol}\,\text{dm}^{-3}} \)
\( [k] = \text{s}^{-1} \)

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

The total area under a Maxwell-Boltzmann curve represents the total number of particles/molecules in the sample.
- Increasing the temperature of the mixture shifts the peak to the right and flattens it, but the total area under the curve remains constant.
- Adding a catalyst lowers the activation energy (represented by shifting the \( E_a \) line to the left), but it does not change the distribution curve itself or the total area.
- Increasing the total number of reactant molecules increases the number of particles, which directly increases the total area under the curve.
- Decreasing the volume at constant temperature increases the frequency of collisions but does not change the total number of particles, so the curve remains unchanged.

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

Let's analyze the statements:
- **Statement I is correct**: Step 1 is the slow step (rate-determining step), so the rate expression is determined solely by the reactants of this step: \( \text{Rate} = k[\text{NO}_2][\text{F}_2] \).
- **Statement II is incorrect**: \( \text{F}(g) \) is produced in Step 1 and consumed in Step 2. Thus, it is a reaction intermediate, not a catalyst (a catalyst is consumed first and regenerated later).
- **Statement III is correct**: The rate law is first order with respect to \( \text{NO}_2 \) and first order with respect to \( \text{F}_2 \). The overall reaction order is \( 1 + 1 = 2 \).

Therefore, only statements I and III are correct.

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

According to the linear form of the Arrhenius equation:
\( \ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A \)

The gradient \( m \) of the plot is given by:
\( m = -\frac{E_a}{R} \)

Given \( m = -1.20 \times 10^4 \text{ K} \):
\( -1.20 \times 10^4 = -\frac{E_a}{8.31} \)
\( E_a = 1.20 \times 10^4 \times 8.31 = 9.972 \times 10^4 \text{ J}\,\text{mol}^{-1} \)

Converting to \( \text{kJ}\,\text{mol}^{-1} \):
\( E_a = \frac{9.972 \times 10^4}{1000} \approx 100 \text{ kJ}\,\text{mol}^{-1} \)

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

To determine the molecular geometry, we find the number of electron domains on the central atom:
- In \( \text{SF}_4 \), sulfur has 6 valence electrons, forms 4 bonds, and has 1 lone pair (5 electron domains, seesaw geometry).
- In \( \text{XeF}_4 \), xenon has 8 valence electrons, forms 4 bonds, and has 2 lone pairs (6 electron domains, square planar geometry).
- In \( \text{CF}_4 \), carbon has 4 valence electrons, forms 4 bonds, and has 0 lone pairs (4 electron domains, tetrahedral geometry).
- In \( \text{BF}_4^- \), boron has 3 valence electrons + 1 extra charge, forms 4 bonds, and has 0 lone pairs (4 electron domains, tetrahedral geometry).

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

- Statement I is correct: As temperature increases, the average kinetic energy of the particles increases, shifting the peak (the most probable kinetic energy) to the right (higher energy).
- Statement II is incorrect: Since the total number of molecules remains constant but more particles occupy higher energy states, the distribution becomes broader, causing the peak height to decrease.
- Statement III is correct: The area under the Maxwell-Boltzmann curve represents the total number of particles in the system, which does not change when the temperature is varied.

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

Bond length is inversely proportional to bond order (higher bond order results in shorter bond length):
- In \( \text{CO} \), there is a triple bond between C and O (bond order = 3).
- In \( \text{CO}_2 \), there are double bonds between C and O (bond order = 2).
- In \( \text{CO}_3^{2-} \), there is resonance, leading to equivalent bonds with a bond order of \( 1.33 \).
- In \( \text{CH}_3\text{OH} \), there is a C-O single bond (bond order = 1).

Therefore, \( \text{CO} \) has the shortest carbon-oxygen bond.

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

Rearranging the rate expression for the rate constant \( k \):
\( k = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]} \)

Substituting the units:
\( k = \frac{\text{mol}\cdot\text{dm}^{-3}\cdot\text{s}^{-1}}{(\text{mol}\cdot\text{dm}^{-3})^3} = \text{mol}^{-2}\cdot\text{dm}^6\cdot\text{s}^{-1} \)

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

Let us map out the chemical structure of methyl propenoate: \( \text{CH}_2=\text{CH}-\text{C}(=\text{O})-\text{O}-\text{CH}_3 \).
1. Count of \( \pi \) bonds: There are two double bonds in the molecule (one C=C bond and one C=O bond). Each double bond contains one \( \sigma \) bond and one \( \pi \) bond. Thus, there are 2 \( \pi \) bonds.
2. Count of \( \sigma \) bonds:
An acyclic molecule containing \( N \) atoms always contains \( N - 1 \) \( \sigma \) bonds.
- Carbon atoms: 4
- Hydrogen atoms: 6
- Oxygen atoms: 2
Total atoms (\( N \)) = 12.
Therefore, the number of \( \sigma \) bonds is \( 12 - 1 = 11 \).

This gives a total of 11 \( \sigma \) and 2 \( \pi \) bonds.

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

The Arrhenius equation in linear form is:
\( \ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A \)

Thus, the slope (gradient) \( m \) is equal to \( -\frac{E_a}{R} \):
\( -\frac{E_a}{R} = -6.0 \times 10^3 \text{ K} \)

\( E_a = 6.0 \times 10^3 \text{ K} \times 8.31 \text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} = 49.86 \times 10^3 \text{ J}\cdot\text{mol}^{-1} \)

Converting to \( \text{kJ}\cdot\text{mol}^{-1} \):
\( E_a \approx 50 \text{ kJ}\cdot\text{mol}^{-1} \)

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

Let us analyze the resonance structures of \( \text{N}_2\text{O} \):
1. \( :\ddot{\text{N}}=\text{N}=\ddot{\text{O}}: \)
- Terminal N: \( 5 - 4 - 2 = -1 \)
- Central N: \( 5 - 0 - 4 = +1 \)
- Terminal O: \( 6 - 4 - 2 = 0 \)

2. \( :\text{N}\equiv\text{N}-\ddot{\text{O}}:^{3-} \) (where oxygen has 3 lone pairs):
- Terminal N: \( 5 - 2 - 3 = 0 \)
- Central N: \( 5 - 0 - 4 = +1 \)
- Terminal O: \( 6 - 6 - 1 = -1 \)

Since oxygen is more electronegative than nitrogen, structure 2 is the most stable and dominant resonance structure. In both structures, the central nitrogen atom has 4 bonds and 0 lone pairs, giving it a formal charge of \( +1 \).

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

The rate-determining step is the slow step (Step 2):
\( \text{Rate} = k_2 [\text{C}][\text{A}] \)

Since \( \text{C} \) is a reaction intermediate, we use the fast pre-equilibrium step (Step 1) to express its concentration in terms of the reactants:
\( K_{eq} = \frac{[\text{C}]}{[\text{A}][\text{B}]} \implies [\text{C}] = K_{eq}[\text{A}][\text{B}] \)

Substituting this back into the rate law for Step 2:
\( \text{Rate} = k_2 (K_{eq}[\text{A}][\text{B}])[\text{A}] = k[\text{A}]^2[\text{B}] \) (where \( k = k_2 K_{eq} \)).

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

In propadiene, the central carbon is involved in two double bonds (two \(\sigma\) bonds and two \(\pi\) bonds), which requires two hybrid orbitals, giving it an \(sp\) hybridization. The terminal carbons are each bonded to two hydrogen atoms via single bonds and to the central carbon via a double bond (three \(\sigma\) bonds and one \(\pi\) bond), requiring three hybrid orbitals, resulting in \(sp^2\) hybridization. Therefore, the central carbon is \(sp\) hybridized, and the terminal carbons are \(sp^2\) hybridized.

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

Comparing Exp 1 and Exp 2: doubling \([\text{A}]\) doubles the rate, so the reaction is 1st order with respect to \(\text{A}\). Comparing Exp 1 and Exp 3: doubling \([\text{B}]\) quadruples the rate, so the reaction is 2nd order with respect to \(\text{B}\). The overall order is \(1 + 2 = 3\). The rate equation is: \(\text{Rate} = k[\text{A}][\text{B}]^2\). Rearranging for \(k\): \(k = \text{Rate} / ([\text{A}][\text{B}]^2)\). Substituting units: \(\text{mol\ dm}^{-3}\ \text{s}^{-1} / (\text{mol\ dm}^{-3} \times (\text{mol\ dm}^{-3})^2) = \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}\).

Award [1] for correct determination of overall order (3) and unit derivation. Select option A.

Carbon monoxide, \(\text{CO}\), has a triple bond (bond order 3.0), which is the shortest. Carbon dioxide, \(\text{CO}_2\), has double bonds (bond order 2.0). The carbonate ion, \(\text{CO}_3^{2-}\), exhibits resonance, giving each C-O bond a bond order of 1.33. Methanol, \(\text{CH}_3\text{OH}\), has a single C-O covalent bond (bond order 1.0), which is the longest. Thus, the order of increasing bond length is \(\text{CO} < \text{CO}_2 < \text{CO}_3^{2-} < \text{CH}_3\text{OH}\).

Award [1] for the correct identification of the trend in bond lengths based on bond order.

The Maxwell-Boltzmann distribution curve represents the distribution of kinetic energies among molecules, which depends solely on temperature. Therefore, at a constant temperature, the curve remains completely unchanged. A catalyst works by providing an alternative reaction pathway with a lower activation energy, so the activation energy barrier effectively decreases (moves to the left on the energy axis of the distribution).

Award [1] for identifying that the curve is unchanged at constant temperature and the activation energy decreases.

All three species have 4 electron domains around the central nitrogen atom. In \(\text{NH}_4^+\), there are 4 bonding pairs and 0 lone pairs, giving a regular tetrahedral geometry with a bond angle of \(109.5^\circ\). In \(\text{NH}_3\), there are 3 bonding pairs and 1 lone pair, causing lone-pair/bonding-pair repulsion to squeeze the angle to approximately \(107^\circ\). In \(\text{NH}_2^-\), there are 2 bonding pairs and 2 lone pairs, causing even stronger repulsion and reducing the angle to approximately \(104.5^\circ\). Thus, the correct increasing order of bond angles is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).

Award [1] for the correct option A. The order must show the species with the most lone pairs having the smallest angle.

The slow step is the rate-determining step: \(\text{Rate} = k_2[\text{NOCl}_2][\text{NO}]\). Because \(\text{NOCl}_2\) is an intermediate, we use the fast equilibrium step to express its concentration in terms of the reactants: \(K_c = [\text{NOCl}_2] / ([\text{NO}][\text{Cl}_2])\), which gives \([\text{NOCl}_2] = K_c[\text{NO}][\text{Cl}_2]\). Substituting this into the rate-determining step rate expression gives: \(\text{Rate} = k_2 K_c [\text{NO}][\text{Cl}_2][\text{NO}] = k[\text{NO}]^2[\text{Cl}_2]\).

Award [1] for correct substitution of the intermediate to yield option B.

\(\text{XeF}_4\) has 6 electron domains (4 bonding pairs, 2 lone pairs) around the central xenon atom, resulting in a square planar molecular geometry. The highly polar Xe-F bonds point towards the corners of a square, and the individual dipole moments cancel each other out due to symmetry. Meanwhile, \(\text{SF}_4\) (seesaw), \(\text{PCl}_3\) (trigonal pyramidal), and \(\text{CH}_2\text{Cl}_2\) (tetrahedral but asymmetric) are all polar molecules with net dipole moments.

Award [1] for selecting the correct symmetrical square-planar molecule XeF4.

According to the Arrhenius equation in its linear form, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\), the slope of the graph of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\). Therefore, \(-\frac{E_a}{R} = -1.20 \times 10^4\), which gives \(E_a = 1.20 \times 10^4 \times 8.31\ \text{J\ mol}^{-1}\). To convert this activation energy into \(\text{kJ\ mol}^{-1}\), we divide the value by 1000, giving \(\frac{1.20 \times 10^4 \times 8.31}{1000}\).

Award [1] for correctly identifying that Slope = -Ea/R and converting the units of the activation energy from J to kJ.

\(SCl_2\) has a central sulfur atom with 6 valence electrons. It forms two single covalent bonds with two chlorine atoms and has two lone pairs remaining. According to VSEPR theory, a tetrahedral electron-domain geometry with two lone pairs results in a bent (V-shaped) molecular geometry with a bond angle slightly less than the ideal tetrahedral angle (approximately \(103^\circ\)). \(CO_2\) is linear (\(180^\circ\)), \(NO_2^-\)

Award [1] for the correct option (C). No marks are awarded for incorrect choices.

Bond strength and length are determined by the bond order. In \(CH_3OH\), the C-O bond is a single bond (bond order = 1). In \(HCHO\), the C-O bond is a double bond (bond order = 2). In \(CO_2\), each C-O bond is a double bond (bond order = 2). In \(CO\), the bond is a triple bond (bond order = 3). A triple bond is shorter and stronger than a double or single bond.

Award [1] for the correct option (C). No marks are awarded for incorrect choices.

Silicon dioxide (\(SiO_2\)) has a giant covalent (macromolecular) structure where every silicon atom is tetrahedrally bonded to four oxygen atoms, requiring a huge amount of energy to break these strong covalent bonds, resulting in a very high melting point. Carbon dioxide (\(CO_2\)) is a simple covalent molecular substance. Its molecules are held together only by weak intermolecular forces (specifically, London dispersion forces), which require very little energy to overcome, making it a gas at room temperature.

Award [1] for the correct option (B). No marks are awarded for incorrect choices.

In boron trifluoride (\(BF_3\)), the boron-fluorine bonds are polar due to the electronegativity difference between B and F. However, the molecule has a highly symmetric trigonal planar geometry. The individual bond dipoles cancel each other out completely, resulting in a net dipole moment of zero (non-polar molecule). In contrast, \(NH_3\) (trigonal pyramidal), \(H_2O\) (bent), and \(CHF_3\) (asymmetric tetrahedral) have geometries where the individual dipoles do not cancel, making them polar molecules.

Award [1] for the correct option (C). No marks are awarded for incorrect choices.

Let the initial rate be \(\text{Rate}_1 = k[A][B]^2\). When \([A]\) is halved to \(0.5[A]\) and \([B]\) is doubled to \(2[B]\), the new rate is: \(\text{Rate}_2 = k(0.5[A])(2[B])^2 = k(0.5[A])(4[B]^2) = 2 \times k[A][B]^2 = 2 \times \text{Rate}_1\). Therefore, the initial rate increases by a factor of 2.

Award [1] for the correct option (C). No marks are awarded for incorrect choices.

A catalyst provides an alternative pathway with a lower activation energy (\(E_a\)). This means a larger fraction of reactant particles have kinetic energy equal to or greater than the new lower activation energy, leading to a higher rate of successful collisions. It does not change the average kinetic energy of the particles (which is only changed by temperature) nor does it alter the enthalpy change (\(\Delta H\)) of the reaction.

Award [1] for the correct option (C). No marks are awarded for incorrect choices.

As temperature increases, the average kinetic energy of the particles increases, shifting the entire distribution to the right. Since the total number of particles (and thus the area under the curve) remains constant, the peak must flatten out and become lower to compensate for the broader distribution at higher energy. Thus, the peak of the curve at \(T_2\) is lower and shifted to the right.

Award [1] for the correct option (D). No marks are awarded for incorrect choices.

The reaction produces oxygen gas, so measuring the volume of gas (A), the mass loss due to the escaping gas in an open system (B), or the gas pressure in a closed system (D) are all highly effective methods to track the reaction rate. However, neither the reactants nor the products are strongly acidic or basic, and there is no significant change in hydrogen ion concentration, making pH measurement (C) unsuitable.

Award [1] for the correct option (C). No marks are awarded for incorrect choices.

(a) Lewis structures:
- For \(\text{SF}_4\): Sulfur is the central atom with 4 single bonds to Fluorine atoms and 1 lone pair on Sulfur (total of 10 valence electrons around Sulfur). Each Fluorine atom has 3 lone pairs (8 valence electrons total per F).
- For \(\text{XeF}_2\): Xenon is the central atom with 2 single bonds to Fluorine atoms and 3 lone pairs on Xenon (total of 10 valence electrons around Xenon). Each Fluorine atom has 3 lone pairs.

(b) Geometries:
- \(\text{SF}_4\): Has 5 electron domains (4 bonding domains, 1 non-bonding domain). Electron domain geometry: trigonal bipyramidal. Molecular geometry: see-saw.
- \(\text{XeF}_2\): Has 5 electron domains (2 bonding domains, 3 non-bonding domains). Electron domain geometry: trigonal bipyramidal. Molecular geometry: linear.

(c) Formal charge calculation:
- Structure A (four single bonds): Central phosphorus has 4 single bonds and 0 lone pairs. Formal charge of \(\text{P} = 5 - (4 + 0) = +1\). Each oxygen has 1 single bond and 3 lone pairs. Formal charge of each \(\text{O} = 6 - (1 + 6) = -1\).
- Structure B (one double bond, three single bonds): Central phosphorus has 5 bonds and 0 lone pairs. Formal charge of \(\text{P} = 5 - (5 + 0) = 0\). Double-bonded oxygen has 2 bonds and 2 lone pairs. Formal charge of double-bonded \(\text{O} = 6 - (2 + 4) = 0\). Three single-bonded oxygens each have 1 bond and 3 lone pairs. Formal charge of each single-bonded \(\text{O} = 6 - (1 + 6) = -1\).
- Evaluation: Structure B is more stable because the formal charges are minimized (the central atom and one oxygen have a formal charge of 0, whereas in Structure A all atoms have non-zero formal charges).

(d) Resonance in ozone (\(\text{O}_3\)):
- Ozone has two main resonance structures where the double bond is alternating between the two outer oxygen atoms.
- In reality, the actual structure is a resonance hybrid where the electrons are delocalized across the molecule.
- The two oxygen-oxygen bonds are identical in length and strength. The bond order is 1.5, which is intermediate between a single bond (bond order 1) and a double bond (bond order 2).

(a) Award [2] for \(\text{SF}_4\) structure showing 1 lone pair on central S and 3 lone pairs on all F atoms. Award [2] for \(\text{XeF}_2\) structure showing 3 lone pairs on central Xe and 3 lone pairs on all F atoms.

(b) Award [1] each for correct electron domain geometry and molecular geometry of \(\text{SF}_4\) (trigonal bipyramidal and see-saw). Award [1] each for correct electron domain geometry and molecular geometry of \(\text{XeF}_2\) (trigonal bipyramidal and linear).

(c) Award [1] for calculating formal charge of P (+1) and O (-1) in structure with four single bonds. Award [1] for calculating formal charge of P (0), double-bonded O (0), and single-bonded O (-1) in structure with one double bond. Award [1] for stating that the sum of formal charges equals the overall charge of the ion (-3) in both cases. Award [1] for explaining that structure B is preferred because it has the lowest formal charges / central atom charge is zero.

(d) Award [1] for stating that resonance involves the delocalization of pi electrons / alternating double and single bonds. Award [1] for stating that the two O-O bonds are equal in length and intermediate between single and double bonds. Award [1] for stating the bond order is 1.5.

(a)
- Comparing Run 1 and Run 2: [B] is held constant at \(0.100\text{ mol dm}^{-3}\), while [A] is doubled from \(0.100\) to \(0.200\text{ mol dm}^{-3}\). The rate increases by a factor of 4 (from \(2.00 \times 10^{-4}\) to \(8.00 \times 10^{-4}\)). Since \(2^2 = 4\), the order with respect to A is 2.
- Comparing Run 2 and Run 3: [A] is held constant at \(0.200\text{ mol dm}^{-3}\), while [B] is doubled from \(0.100\) to \(0.200\text{ mol dm}^{-3}\). The rate does not change (remains \(8.00 \times 10^{-4}\)). Therefore, the order with respect to B is 0.
- Overall rate expression: \(\text{Rate} = k[\text{A}]^2\).

(b) Using data from Run 1:
\(2.00 \times 10^{-4} = k(0.100)^2\)
\(k = \frac{2.00 \times 10^{-4}}{0.0100} = 2.00 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) Suggested mechanism:
- Step 1 (Slow, Rate-determining step): \(2\text{A} \rightarrow \text{I}\) (where \(\text{I}\) is an intermediate)
- Step 2 (Fast): \(\text{I} + \text{B} \rightarrow \text{C}\)
- Net Reaction: \(2\text{A} + \text{B} \rightarrow \text{C}\). Since the slow step contains only 2 molecules of A, the rate law is \(\text{Rate} = k[\text{A}]^2\), which matches the experimental rate law.

(d) Using the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
Given:
\(k_1 = 2.50 \times 10^{-3}\text{ s}^{-1}\)
\(T_1 = 300\text{ K}\)
\(T_2 = 350\text{ K}\)
\(E_a = 55.0\text{ kJ mol}^{-1} = 55000\text{ J mol}^{-1}\)
\(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)

Calculation:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{55000}{8.314} \left(\frac{1}{350} - \frac{1}{300}\right)\)
\(\frac{1}{350} - \frac{1}{300} = -4.762 \times 10^{-4}\text{ K}^{-1}\)
\(\ln\left(\frac{k_2}{k_1}\right) = -6615.35 \times (-4.762 \times 10^{-4}) = 3.150\)
\(\frac{k_2}{k_1} = e^{3.150} = 23.34\)
\(k_2 = 23.34 \times (2.50 \times 10^{-3}\text{ s}^{-1}) = 5.83 \times 10^{-2}\text{ s}^{-1}\).

(a) Award [1] for identifying second order for A with explanation. Award [1] for identifying zero order for B with explanation. Award [2] for correct rate expression: \(\text{Rate} = k[\text{A}]^2\) (deduct 1 mark if k is missing or formatted incorrectly).

(b) Award [1] for correct numerical calculation of \(k = 2.00 \times 10^{-2}\). Award [1] for correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) Award [1] for Step 1 (Slow) equation containing 2 reactant molecules of A. Award [1] for Step 2 (Fast) equation combining intermediate and reactant B to form C. Award [1] for showing that the sum of the steps equals the net stoichiometric equation.

(d) Award [1] for converting \(E_a\) to J/mol (\(55000\text{ J mol}^{-1}\)). Award [1] for correct substitution of values into the Arrhenius equation. Award [2] for calculating the intermediate logarithmic ratio/exponent (value around \(3.15\)). Award [2] for final answer \(5.83 \times 10^{-2}\text{ s}^{-1}\) (accept range \(5.80 \times 10^{-2}\) to \(5.90 \times 10^{-2}\)).

(a) (i) \(\text{C}_7\text{H}_6\text{O}_3\text{ (salicylic acid)} + \text{C}_4\text{H}_6\text{O}_3\text{ (ethanoic anhydride)} \rightarrow \text{C}_9\text{H}_8\text{O}_4\text{ (aspirin)} + \text{C}_2\text{H}_4\text{O}_2\text{ (ethanoic acid)}\)
(ii) Concentrated sulfuric acid / phosphoric acid acts as a catalyst (and dehydrating agent).
(iii) Recrystallization (dissolving crude product in minimum volume of hot solvent, cooling, filtering crystals, washing, and drying).

(b)
- Salicylic acid contains a phenol (hydroxyl) group which has a strong, broad absorption due to the \(\text{O-H}\) bond stretch at \(3200-3600\text{ cm}^{-1}\). This peak is absent in pure aspirin (as the phenol is converted to an ester).
- Aspirin will show two distinct carbonyl (\(\text{C=O}\)) stretching peaks: one for the carboxylic acid (around \(1680-1730\text{ cm}^{-1}\)) and one for the ester group (around \(1700-1750\text{ cm}^{-1}\)). Salicylic acid only has one carbonyl peak (carboxylic acid).

(c) (i)
- The beta-lactam ring is a 4-membered ring containing a nitrogen and a carbonyl group. The bond angles are approximately \(90^\circ\).
- This is highly distorted from the normal \(109.5^\circ\) (\(sp^3\) carbon/nitrogen) and \(120^\circ\) (\(sp^2\) carbonyl carbon) bond angles, leading to severe ring strain.
- This strain weakens the amide bond, making the ring highly reactive.
- Mechanism of action: The ring opens up and covalently/irreversibly binds to the active site of the transpeptidase enzyme (responsible for cross-linking peptidoglycan cell walls of bacteria). Cell wall synthesis is inhibited, causing the cell to burst due to osmotic pressure.

(ii)
- Overuse/misuse of penicillin allows bacteria to adapt and develop resistance by producing penicillinase (beta-lactamase) enzymes.
- These enzymes hydrolyze the amide bond in the beta-lactam ring before the drug can reach the target transpeptidase enzyme, rendering the drug inactive.
- Semi-synthetic penicillins are designed by chemically modifying the side-chain (R-group) of natural penicillin.
- This modification introduces bulky groups (steric hindrance) that prevent the penicillinase enzyme from binding to the beta-lactam ring, while still allowing the drug to target bacterial cell walls.

(a) (i) Award [2] for the fully balanced equation with correct molecular formulas (or structures). Deduct [1] for minor balancing/formula errors.
(ii) Award [1] for stating "catalyst" / "acid catalyst".
(iii) Award [1] for "recrystallization".

(b) Award [1] for noting that salicylic acid has a phenol \(\text{O-H}\) stretch at \(3200-3600\text{ cm}^{-1}\) which is absent in aspirin. Award [1] for noting that aspirin has two carbonyl (\(\text{C=O}\)) peaks (due to the ester and carboxylic acid) whereas salicylic acid has only one. Award [1] for citing correct wavenumbers for these functional groups (e.g. \(\text{O-H}\) broad at \(3200-3600\text{ cm}^{-1}\) or \(\text{C=O}\) at \(1700-1750\text{ cm}^{-1}\)).

(c) (i) Award [1] for mentioning the four-membered ring structure with \(90^\circ\) bond angles. Award [1] for connecting this to ring strain / weakened carbon-nitrogen amide bond. Award [1] for stating that it inhibits the transpeptidase enzyme / bacterial cell wall synthesis. Award [1] for explaining that cell wall rupture leads to osmotic lysis / bacterial death.
(ii) Award [1] for stating that bacteria produce beta-lactamase (penicillinase) which breaks the drug ring. Award [1] for stating that overuse/exposure selects for these resistant strains. Award [1] for stating that semi-synthetic penicillins modify the side chain. Award [1] for explaining that bulky side chains prevent the enzyme from binding due to steric hindrance.

(a) (i)
- Morphine contains two hydroxyl (\(\text{-OH}\)) groups (one phenolic hydroxyl and one secondary alcohol group).
- Diamorphine contains two ester (ethanoate/acetate, \(\text{-OCOCH}_3\)) groups instead of these hydroxyl groups.

(ii)
- The hydroxyl groups in morphine make it relatively polar and capable of forming hydrogen bonds with water.
- In diamorphine, these polar groups are converted to non-polar ester groups, making diamorphine much less polar and highly lipid-soluble.
- The blood-brain barrier is composed of a non-polar lipid bilayer.
- Due to its high lipid solubility, diamorphine crosses the blood-brain barrier much more rapidly and in higher concentrations than morphine, leading to greater potency in the central nervous system.

(iii)
- Advantage: Highly effective in relieving severe/acute pain (e.g., cancer pain, post-surgical pain) by acting directly on the central nervous system (intercepting pain signals).
- Disadvantage: High risk of physiological/psychological addiction, physical dependence, tolerance build-up, or side effects like respiratory depression.

(b) (i)
- Both drugs are neuraminidase inhibitors.
- They bind to the active site of the neuraminidase enzyme on the surface of the influenza virus.
- This prevents the enzyme from cleaving sialic acid, thereby trapping the newly formed viral particles inside the host cell and preventing them from budding off to infect other healthy cells.

(ii) Any two of: carboxamide/secondary amide group (\(\text{-NHCOCH}_3\)), ether group (\(\text{-C-O-C-\/\/}\)), amine/primary amine group (\(\text{-NH}_2\)).

(iii)
- Mass production relies heavily on precursors like shikimic acid (historically extracted from star anise), which can cause global shortages and requires energy-intensive chemical processing.
- Improper disposal or excretion of the active drugs into wastewater can lead to bioaccumulation and the development of drug-resistant viral strains in wild populations (e.g., waterfowl).

(a) (i) Award [1] for identifying hydroxyl / phenol / alcohol in morphine. Award [1] for identifying ester / ethanoate groups in diamorphine.
(ii) Award [1] for stating that diamorphine is less polar than morphine. Award [1] for stating that diamorphine is more lipid-soluble. Award [1] for stating that the blood-brain barrier is a lipid barrier / non-polar membrane. Award [1] for concluding that diamorphine crosses this barrier more rapidly / easily.
(iii) Award [1] for a valid advantage (e.g. effective against severe pain / works directly on CNS). Award [1] for a valid disadvantage (e.g. risk of addiction / tolerance / respiratory depression).

(b) (i) Award [1] for stating they are neuraminidase inhibitors. Award [1] for stating they prevent the cleavage of sialic acid / bond to the active site of the enzyme. Award [1] for explaining that this prevents viral particles from escaping the host cell / spreading.
(ii) Award [2] for identifying any two correct functional groups shared by both (amide / ether / amine). Award [1] if only one is correct.
(iii) Award [1] for identifying issues with precursor shortage (shikimic acid) or solvent/energy waste. Award [1] for identifying contamination of water supply / development of drug-resistant strains.

(a) A Brønsted-Lowry acid is a proton (\(\text{H}^+\)) donor.

(b) (i)
\(K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\)
Assuming \([\text{H}^+] \approx [\text{HCOO}^-]\) and \([\text{HCOOH}]_{\text{eq}} \approx 0.150\text{ mol dm}^{-3}\):
\(1.77 \times 10^{-4} = \frac{[\text{H}^+]^2}{0.150}\)
\([\text{H}^+]^2 = 2.655 \times 10^{-5}\)
\([\text{H}^+] = \sqrt{2.655 \times 10^{-5}} = 5.153 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(5.153 \times 10^{-3}) = 2.29\).

(ii)
1. The dissociation of methanoic acid is very small / negligible compared to its initial concentration (i.e. \([\text{HCOOH}]_{\text{initial}} - [\text{H}^+] \approx [\text{HCOOH}]_{\text{initial}}\)).
2. The concentration of \(\text{H}^+\) produced from the autoionization of water is negligible.

(c) (i)
- The buffer consists of a weak acid (\(\text{HCOOH}\)) and its conjugate base (\(\text{HCOO}^-\)).
- When \(\text{HCl}\) is added, the added \(\text{H}^+\) ions react with the conjugate base \(\text{HCOO}^-\):
\(\text{HCOO}^-(aq) + \text{H}^+(aq) \rightarrow \text{HCOOH}(aq)\)
- This prevents a significant increase in \([\text{H}^+]\), thereby keeping the pH relatively stable.

(ii)
- Initial moles of \(\text{HCOOH}\):
\(n(\text{HCOOH}) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\)
- Moles of \(\text{NaOH}\) added:
\(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\)
- Reaction:
\(\text{HCOOH} + \text{OH}^- \rightarrow \text{HCOO}^- + \text{H}_2\text{O}\)
- Since \(\text{NaOH}\) is the limiting reactant, it reacts completely:
- Remaining moles of \(\text{HCOOH}\) = \(7.50 \times 10^{-3} - 5.00 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol}\)
- Moles of \(\text{HCOO}^-\) formed = \(5.00 \times 10^{-3}\text{ mol}\)
- Using the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HCOO}^-]}{[\text{HCOOH}]}\right)\)
\(\text{p}K_a = -\log_{10}(1.77 \times 10^{-4}) = 3.75\)
\(\text{pH} = 3.75 + \log_{10}\left(\frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}}\right)\)
\(\text{pH} = 3.75 + \log_{10}(2) = 3.75 + 0.30 = 4.05\).

(a) Award [1] for "proton donor" / "\(\text{H}^+\) donor".

(b) (i) Award [1] for writing the equilibrium expression or substituting values. Award [1] for calculating \([\text{H}^+] = 5.15 \times 10^{-3}\text{ mol dm}^{-3}\). Award [1] for \(\text{pH} = 2.29\).
(ii) Award [1] each for any two correct assumptions: dissociation of weak acid is negligible; autoionization of water is negligible; temperature is constant at \(298\text{ K}\).

(c) (i) Award [1] for identifying the buffer components (\(\text{HCOOH}\) and \(\text{HCOO}^-\)). Award [2] for correct equation: \(\text{HCOO}^- + \text{H}^+ \rightarrow \text{HCOOH}\) (state symbols not required). Award [1] for explaining that \([\text{H}^+]\) remains relatively constant because \(\text{H}^+\) is converted into weak acid molecules.
(ii) Award [1] for calculating initial moles of \(\text{HCOOH}\) (\(7.50 \times 10^{-3}\)) and \(\text{NaOH}\) (\(5.00 \times 10^{-3}\)). Award [1] for calculating remaining moles of \(\text{HCOOH}\) (\(2.50 \times 10^{-3}\)) and formed moles of conjugate base (\(5.00 \times 10^{-3}\)). Award [1] for calculating \(\text{p}K_a = 3.75\). Award [1] for correct substitution into Henderson-Hasselbalch equation. Award [1] for final pH value of \(4.05\) (accept range \(4.04\) to \(4.06\)).

(a) The standard enthalpy of atomization is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.

(b) Born-Haber cycle stages:
- Elements: \(\text{Mg}(s) + \text{Cl}_2(g)\)
- Enthalpy of formation path down to \(\text{MgCl}_2(s)\)
- Atomization of Mg: \(\text{Mg}(g) + \text{Cl}_2(g)\)
- Dissociation of \(\text{Cl}_2\): \(\text{Mg}(g) + 2\text{Cl}(g)\)
- First ionization of Mg: \(\text{Mg}^+(g) + e^- + 2\text{Cl}(g)\)
- Second ionization of Mg: \(\text{Mg}^{2+}(g) + 2e^- + 2\text{Cl}(g)\)
- Electron affinity of Cl: \(\text{Mg}^{2+}(g) + 2\text{Cl}^-(g)\)
- Lattice enthalpy path down to \(\text{MgCl}_2(s)\)

(c) Let lattice enthalpy \(\Delta H_{\text{latt}}\text{ (defined as breaking the lattice: endothermic)}\):
\(\Delta H_f^\theta = \Delta H_{\text{at}}^\theta(\text{Mg}) + IE_1(\text{Mg}) + IE_2(\text{Mg}) + E_{\text{diss}}(\text{Cl}_2) + 2 \times EA(\text{Cl}) - \Delta H_{\text{latt}}\)
Substitute the values:
\(-642 = 148 + 738 + 1451 + 242 + 2(-349) - \Delta H_{\text{latt}}\)
\(-642 = 148 + 738 + 1451 + 242 - 698 - \Delta H_{\text{latt}}\)
\(-642 = 1881 - \Delta H_{\text{latt}}\)
\(\Delta H_{\text{latt}} = 1881 + 642 = 2523\text{ kJ mol}^{-1}\).

(d)
- Comparison: The experimental value (\(2523\text{ kJ mol}^{-1}\)) is greater (more endothermic) than the theoretical value (\(2326\text{ kJ mol}^{-1}\)).
- Explanation: The theoretical ionic model assumes purely electrostatic attraction between spherical, non-polarizable ions (perfectly ionic bonding).
- However, magnesium chloride has significant covalent character.
- The small, highly charged magnesium ion (\(\text{Mg}^{2+}\)) polarizes the electron cloud of the larger chloride ion (\(\text{Cl}^-\)), leading to additional covalent orbital overlap. This extra covalent character makes the bonding stronger than predicted by a purely ionic model, requiring more energy to break the lattice.

(a) Award [1] for stating "formation of 1 mole of gaseous atoms from the element in its standard state".

(b) Award [4] for a fully drawn cycle: [1] for correct reactant and product states; [1] for atomization of Mg and dissociation of Cl2 (showing \(2\text{Cl}\)); [1] for both ionization steps of Mg; [1] for electron affinity step showing \(2 \times EA\) of Cl.

(c) Award [1] for standard equation relating all enthalpy changes. Award [1] for multiplying electron affinity of chlorine by 2 (\(-698\text{ kJ}\)). Award [1] for using the correct dissociation enthalpy (\(242\text{ kJ}\)). Award [1] for correct signs for all terms. Award [2] for correct final calculated value of \(+2523\text{ kJ mol}^{-1}\) (accept \(-2523\text{ kJ mol}^{-1}\) if lattice enthalpy is defined as exothermic, but must state definition). Deduct [1] for missing/incorrect units.

(d) Award [1] for stating experimental value is larger than theoretical value. Award [1] for identifying that theoretical model assumes purely ionic bonding. Award [1] for stating that the real bonding has covalent character. Award [1] for explaining that \(\text{Mg}^{2+}\) polarizes the electron cloud of \(\text{Cl}^-\).

**(a)**
(i) A known, small concentration of sodium thiosulfate (\(\text{S}_2\text{O}_3^{2-}\)) and starch indicator are added to the reaction mixture. The thiosulfate reacts rapidly with any produced iodine (\(\text{I}_2\)), reducing it back to iodide (\(\text{I}^-\)). Once all thiosulfate is completely consumed, any further produced \(\text{I}_2\) reacts with starch to form a dark blue-black complex. Since the amount of thiosulfate is constant and small, the time (\(t\)) taken for the color change to occur is inversely proportional to the initial rate of reaction (\(\text{Initial Rate} \propto 1/t\)).
(ii) The dependent variable is the time taken for the blue-black color to appear (or the reaction rate).

**(b)**
(i)
* Comparing Experiments 1 and 2: \([\text{I}^-]\) is constant, while \([\text{S}_2\text{O}_8^{2-}]\) doubles (from 0.050 to 0.100). The initial rate also doubles (from \(1.10 \times 10^{-4}\) to \(2.20 \times 10^{-4}\)). Thus, the reaction is **first order** with respect to \(\text{S}_2\text{O}_8^{2-}\).
* Comparing Experiments 1 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant, while \([\text{I}^-]\) doubles (from 0.050 to 0.100). The initial rate also doubles (from \(1.10 \times 10^{-4}\) to \(2.20 \times 10^{-4}\)). Thus, the reaction is **first order** with respect to \(\text{I}^-\).
(ii) Rate expression: \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)
(iii) The reaction proceeds via a multi-step mechanism, and the reaction orders are determined solely by the species involved in the rate-determining step, not the overall balanced equation stoichiometry.

**(c)**
(i) Using Experiment 1:
\(1.10 \times 10^{-4} = k \times 0.050 \times 0.050\)
\(k = \frac{1.10 \times 10^{-4}}{0.0025} = 0.044\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(\text{L mol}^{-1}\text{ s}^{-1}\))

**(d)**
(i) Logarithmic form of the Arrhenius equation:
\(\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A\)
Plotting \(\ln k\) (y-axis) against \(1/T\) (x-axis) yields a straight line with a slope (gradient) \(m = -\frac{E_a}{R}\). Therefore, \(E_a = -\text{slope} \times R\).

(ii) Calculations:
* At \(T_1 = 298\text{ K}\):
\(1/T_1 = \frac{1}{298} = 3.36 \times 10^{-3}\text{ K}^{-1}\) (or \(0.00336\text{ K}^{-1}\))
\(\ln k_1 = \ln(0.044) = -3.12\)
* At \(T_2 = 313\text{ K}\):
\(1/T_2 = \frac{1}{313} = 3.19 \times 10^{-3}\text{ K}^{-1}\) (or \(0.00319\text{ K}^{-1}\))
\(\ln k_2 = \ln(0.134) = -2.01\)

(iii) Calculate slope, \(m\):
\(m = \frac{\ln k_2 - \ln k_1}{\frac{1}{T_2} - \frac{1}{T_1}}
= \frac{-2.01 - (-3.12)}{3.19 \times 10^{-3} - 3.36 \times 10^{-3}} = \frac{1.11}{-1.7 \times 10^{-4}} \approx -6529\text{ K}\)

(If using unrounded calculator values: \(m = \frac{-2.0100 - (-3.1235)}{0.0031949 - 0.0033557} = \frac{1.1135}{-0.0001608} \approx -6925\text{ K}\))

Using rounded slope (\(-6529\text{ K}\)):
\(E_a = -(-6529\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 54256\text{ J mol}^{-1} \approx 54.3\text{ kJ mol}^{-1}\)

Using unrounded slope (\(-6925\text{ K}\)):
\(E_a = -(-6925\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 57547\text{ J mol}^{-1} \approx 57.5\text{ kJ mol}^{-1}\)

**(a)**
(i) [2] Max:
- Starch and a fixed, limiting amount of sodium thiosulfate (\(\text{S}_2\text{O}_3^{2-}\)) are added which instantly reduces produced iodine back to iodide [1].
- Once all thiosulfate is consumed, free iodine forms a blue-black complex with starch. The time taken (\(t\)) for this sudden color change is measured, making \(1/t\) proportional to initial rate [1].

(ii) [1] Max:
- Time taken (for the blue-black color to appear) / Rate of reaction [1].

**(b)**
(i) [2] Max:
- Order with respect to \([\text{S}_2\text{O}_8^{2-}] = 1\) AND justification (e.g., comparing Exp 1 and 2, doubling concentration doubles rate) [1].
- Order with respect to \([\text{I}^-] = 1\) AND justification (e.g., comparing Exp 1 and 3, doubling concentration doubles rate) [1].

(ii) [1] Max:
- \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) [1].
*Allow error carried forward (ECF) from (b)(i).*

(iii) [1] Max:
- The overall reaction occurs in a series of steps (mechanism) / the stoichiometric coefficients represent overall reaction, but only reactant species in the slow/rate-determining step define the order of reaction [1].

**(c)**
(i) [2] Max:
- Correct numerical value: \(0.044\) (or \(4.4 \times 10^{-2}\)) [1].
- Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(\text{L mol}^{-1}\text{ s}^{-1}\)) [1].

**(d)**
(i) [2] Max:
- \(\ln k = -\frac{E_a}{RT} + \ln A\) (or equivalent form) [1].
- Plotting \(\ln k\) against \(1/T\) gives a straight line where \(\text{slope} = -\frac{E_a}{R}\) (or \(E_a = -\text{slope} \times R\)) [1].

(ii) [2] Max:
- Correct \(1/T\) and \(\ln k\) for \(T = 298\text{ K}\): \(1/T = 3.36 \times 10^{-3}\text{ K}^{-1}\) and \(\ln k = -3.12\) [1].
- Correct \(1/T\) and \(\ln k\) for \(T = 313\text{ K}\): \(1/T = 3.19 \times 10^{-3}\text{ K}^{-1}\) and \(\ln k = -2.01\) [1].
*Accept slight variations in rounding (e.g., \(0.0034\) or \(0.0032\)).*

(iii) [2] Max:
- Correct calculation of slope, e.g., range \(-6500\) to \(-7000\text{ K}\) [1].
- Correct calculation of \(E_a\) in \(\text{kJ mol}^{-1}\), e.g., range \(54.0\) to \(58.5\text{ kJ mol}^{-1}\) [1].
*Award full marks for a correct final answer with working. Correct unit conversion to kJ is required for the accuracy mark.*

(a) The balanced molecular chemical equation is:
\(\text{C}_7\text{H}_6\text{O}_3 + \text{C}_4\text{H}_6\text{O}_3 \rightarrow \text{C}_9\text{H}_8\text{O}_4 + \text{CH}_3\text{COOH}\) (or \(\text{C}_2\text{H}_4\text{O}_2\))

(b) Recrystallization steps:
1. Dissolve the crude aspirin in the minimum volume of hot solvent (such as ethanol or water).
2. Filter the hot solution (using gravity filtration) to remove any insoluble impurities.
3. Allow the filtrate to cool slowly to room temperature (and then in an ice bath) to recrystallize the pure aspirin while keeping soluble impurities dissolved in the mother liquor.
4. Filter the cold mixture under reduced pressure (vacuum filtration) to collect the pure crystals, wash them with a small volume of cold solvent, and dry them.

(c) The crude aspirin sample will:
1. Melt at a lower temperature than the pure sample (which has a literature melting point of approximately \(136\ ^\circ\text{C}\)).
2. Melt over a wider/broader temperature range (rather than the sharp melting point of the pure substance).

(d) Pure salicylic acid contains a phenol group, which exhibits a broad \(\text{O}-\text{H}\) absorption band at \(3200-3600\text{ cm}^{-1}\). This is absent in pure aspirin (where the phenol has been converted to an ester).
*Alternative:* Aspirin has two distinct \(\text{C}=\text{O}\) absorption bands (one for the ester at \(\approx 1750\text{ cm}^{-1}\) and one for the carboxylic acid at \(\approx 1680\text{ cm}^{-1}\)), whereas salicylic acid has only one \(\text{C}=\text{O}\) band (at \(\approx 1660\text{ cm}^{-1}\)).

(a) [2 marks]
- \(\text{C}_7\text{H}_6\text{O}_3 + \text{C}_4\text{H}_6\text{O}_3\) on reactant side [1]
- \(\text{C}_9\text{H}_8\text{O}_4 + \text{C}_2\text{H}_4\text{O}_2\) (or \(\text{CH}_3\text{COOH}\)) on product side [1]

(b) [3 marks]
- Dissolve in the minimum volume of hot solvent [1]
- Filter hot to remove insoluble impurities [1]
- Cool filtrate to crystallize aspirin AND filter crystals/wash with cold solvent [1]

(c) [1.5 marks]
- Lower melting point than the pure sample [1]
- Melts over a wider/broader range [0.5]

(d) [1 mark]
- Mentions phenol \(\text{O}-\text{H}\) absorption in salicylic acid (\(3200-3600\text{ cm}^{-1}\)) which is absent in aspirin OR mentions two distinct \(\text{C}=\text{O}\) peaks in aspirin versus one in salicylic acid [1].

(a) Diamorphine contains two ester (or ethanoate / acetate) groups, which replace the two hydroxyl groups (one phenolic and one aliphatic) found in morphine.

(b)
1. Diamorphine has ester groups which make the molecule less polar (and therefore more lipid-soluble / hydrophobic) compared to morphine, which has polar hydroxyl groups.
2. The blood-brain barrier is composed of a lipid membrane, which restricts the passage of polar molecules.
3. Because diamorphine is highly lipid-soluble, it easily and rapidly penetrates/crosses the blood-brain barrier.
4. Once inside the brain, it is rapidly metabolized (hydrolyzed) back to morphine to bind to opioid receptors, yielding a much higher localized concentration and thus a higher potency.

(c)
- Reagent: Ethanoic anhydride (or ethanoyl chloride).
- Reaction type: Esterification / condensation / acylation (or nucleophilic substitution).

(a) [2 marks]
- Ester / ethanoate / acetate groups [1]
- Identifying that there are two of these groups in diamorphine replacing the two hydroxyl groups in morphine [1]

(b) [3.5 marks]
- Diamorphine is less polar / more lipid-soluble / more hydrophobic than morphine (due to lack of hydroxyl groups) [1]
- Blood-brain barrier is made of lipids / is non-polar [1]
- Diamorphine crosses/penetrates the blood-brain barrier much more easily/rapidly [1]
- Results in higher potency / faster onset [0.5]

(c) [2 marks]
- Reagent: Ethanoic anhydride / acetic anhydride / ethanoyl chloride [1]
- Reaction type: Esterification / condensation / nucleophilic substitution / acylation [1]
*Do not accept 'addition' or 'elimination' on its own.*

(a) Mechanism of action:
1. Neuraminidase is a viral enzyme that cleaves sialic acid residues on the host cell surface, allowing newly replicated viral particles to escape and infect neighboring cells.
2. Antivirals like oseltamivir and zanamivir mimic sialic acid and bind competitively to the active site of neuraminidase.
3. This inhibits the enzyme, trapping the viral particles inside the host cell and preventing them from spreading.

(b)
(i) Common functional groups:
- Amide (carboxamide)
- Alkene (carbon-carbon double bond)
- Ether
- Amine (oseltamivir has a primary amine, zanamivir has a secondary amine, both have amine functionalities)

(ii) Functional groups in zanamivir but not oseltamivir:
- Hydroxyl / alcohol group
- Carboxylic acid (oseltamivir has an ester group instead)
- Guanidino group

(c)
1. Large amounts of antivirals are excreted by patients and enter sewage and municipal water systems because they are not completely degraded in wastewater treatment plants.
2. Exposure of wild viral populations to low concentrations of these active ingredients in the environment promotes natural selection of drug-resistant strains, rendering the medications ineffective in the future.

(a) [2.5 marks]
- Neuraminidase is an enzyme that allows new viral particles to escape the host cell [1]
- Inhibitor binds to the active site of the neuraminidase enzyme [1]
- Prevents release/spread of the virus to other cells [0.5]

(b) [3 marks]
(i) Any two of: Amide/carboxamide, Alkene (C=C), Ether, Amine [2] (1 mark per correct group)
(ii) Any one of: Hydroxyl / alcohol, Carboxylic acid, Guanidino group [1]

(c) [2 marks]
- Excreted drugs enter wastewater/environment intact [1]
- Leads to development of drug-resistant viral strains through natural selection [1]

(a) (i) Nuclear equation:
\(^{90}_{39}\text{Y} \rightarrow \ ^{90}_{40}\text{Zr} + \ ^{\ \ 0}_{-1}\text{e}\) (or \(^{\ \ 0}_{-1}\beta\))

(ii) Calculations:
1. Determine the number of half-lives:
\(n = \frac{192.0\text{ hours}}{64.0\text{ hours}} = 3.0\) half-lives
2. Calculate the fraction remaining:
\(\text{Fraction} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125\)
3. Convert to percentage:
\(\text{Percentage} = 0.125 \times 100\% = 12.5\%\)

(b) (i)
1. Alpha particles have a very short range/penetration depth in human tissue (typically 50–100 \(\mu\text{m}\), or a few cell diameters), which minimizes damage to adjacent healthy cells.
2. Alpha particles have a very high linear energy transfer (high ionizing density), which causes severe, irreparable double-strand DNA breaks in the targeted cancer cells, making them highly cytotoxic.

(ii) Monoclonal antibodies (or specific antibodies / peptides) that target cancer-associated antigens.

(a) [4 marks]
(i) [2 marks]
- \(^{90}_{40}\text{Zr}\) [1]
- \(^{\ \ 0}_{-1}\text{e}\) (or \(^{\ \ 0}_{-1}\beta\)) AND balanced equation [1]
(ii) [2 marks]
- Finding \(n = 3\) half-lives [1]
- Correct final answer of \(12.5\%\) [1]

(b) [3.5 marks]
(i) [2 marks]
- Short range/penetration depth of alpha particles (minimizing damage to healthy tissue) [1]
- High ionizing density / linear energy transfer / causes double-stranded DNA breaks [1]
(ii) [1.5 marks]
- Monoclonal antibodies (accept antibodies or peptides) [1.5]