2024 IB DP Chemistry Practice Paper with Answers

The central xenon atom in \( \text{XeF}_4 \) has 8 valence electrons plus 4 electrons from the fluorine atoms, giving a total of 12 electrons (6 electron pairs) around the central atom. This results in an octahedral electron domain geometry. To minimize electron-pair repulsion, the two lone pairs are positioned opposite to each other, resulting in a square planar molecular geometry.

- \( \text{SF}_4 \) has 5 electron domains around the central sulfur atom, which results in a see-saw molecular geometry.
- \( \text{CF}_4 \) and \( \text{BF}_4^- \) both have 4 electron domains and 0 lone pairs around the central atom, giving a tetrahedral molecular geometry.

[1] Award 1 mark for selecting option B.
[0] Other responses or no response do not receive marks.

- The reaction is exothermic (\( \Delta H < 0 \)). Decreasing the temperature will shift the position of equilibrium to the right to favor the exothermic forward reaction, increasing the yield of \( \text{C} \). Because temperature is the only factor that changes the value of \( K_c \), decreasing the temperature of an exothermic reaction will increase the value of \( K_c \).
- Increasing the pressure shifts the equilibrium to the right because there are fewer moles of gas on the product side (2 moles) than on the reactant side (3 moles). However, pressure changes do not alter the value of \( K_c \).
- Adding a catalyst increases the rate of both forward and reverse reactions equally, so it has no effect on the equilibrium yield or the value of \( K_c \).
- Increasing the concentration of reactant \( \text{A} \) shifts the equilibrium position to the right but does not alter the value of \( K_c \).

[1] Award 1 mark for selecting option B.
[0] Other responses do not receive marks.

The pH of a solution is given by the formula:
\( \text{pH} = -\log_{10}[\text{H}^+] \)

Therefore, the concentration of hydrogen ions is calculated as:
\( [\text{H}^+] = 10^{-\text{pH}} \)

- At pH 4.0: \( [\text{H}^+] = 1.0 \times 10^{-4} \text{ mol dm}^{-3} \)
- At pH 2.0: \( [\text{H}^+] = 1.0 \times 10^{-2} \text{ mol dm}^{-3} \)

The factor of increase in hydrogen ion concentration is:
\( \frac{1.0 \times 10^{-2}}{1.0 \times 10^{-4}} = 10^2 = 100 \)

[1] Award 1 mark for selecting option C.
[0] Other responses do not receive marks.

Let's determine the electron configurations:
- \( \text{V} \) has the configuration \( [\text{Ar}] 4s^2 3d^3 \). When forming the \( \text{V}^{2+} \) ion, it loses the two \( 4s \) electrons, leaving the configuration as \( [\text{Ar}] 3d^3 \). These three electrons occupy separate \( d \) orbitals singly according to Hund's rule, giving exactly 3 unpaired electrons.
- \( \text{Fe}^{3+} \) has the configuration \( [\text{Ar}] 3d^5 \), which has 5 unpaired electrons.
- \( \text{Ni}^{2+} \) has the configuration \( [\text{Ar}] 3d^8 \), which has 2 unpaired electrons.
- \( \text{Ti}^{2+} \) has the configuration \( [\text{Ar}] 3d^2 \), which has 2 unpaired electrons.

[1] Award 1 mark for selecting option A.
[0] Other responses do not receive marks.

First, calculate the heat energy absorbed by the water using the heat equation:
\( q = m c \Delta T \)
\( q = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 10.0 \text{ K} = 4180 \text{ J} = 4.18 \text{ kJ} \)

Since combustion reactions are exothermic, heat is released to the surroundings, meaning the enthalpy change is negative:
\( \Delta H_c = -\frac{q}{n} = -\frac{4.18 \text{ kJ}}{0.010 \text{ mol}} = -418 \text{ kJ mol}^{-1} \)

[1] Award 1 mark for selecting option C.
[0] Other responses do not receive marks.

Let's assign the oxidation state of chlorine (\( x \)) in each species, using \( -2 \) for oxygen and \( -1 \) for the overall charge on the polyatomic ions:

- In \( \text{ClO}_2^- \): \( x + 2(-2) = -1 \implies x = +3 \)
- In \( \text{ClO}_3^- \): \( x + 3(-2) = -1 \implies x = +5 \)
- In \( \text{ClO}^- \): \( x + (-2) = -1 \implies x = +1 \)
- In \( \text{Cl}_2\text{O}_7 \): \( 2x + 7(-2) = 0 \implies 2x = 14 \implies x = +7 \)

Chlorine reaches its highest oxidation state of \( +7 \) in \( \text{Cl}_2\text{O}_7 \).

[1] Award 1 mark for selecting option D.
[0] Other responses do not receive marks.

- A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy, \( E_a \).
- Increasing reactant concentrations and decreasing particle size of a solid both increase the collision frequency of the reacting particles but do not alter the activation energy of the reaction.
- Increasing the temperature increases the average kinetic energy of the particles, leading to a larger fraction of particles having energy greater than or equal to \( E_a \), but the activation energy value itself remains unchanged.

[1] Award 1 mark for selecting option C.
[0] Other responses do not receive marks.

Using the combined gas law relation:
\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)

Let the initial conditions be:
\( P_1 = P \), \( V_1 = V \), \( T_1 = T \)

The new conditions are:
\( P_2 = 0.5P \), \( T_2 = 2T \)

Substituting these values into the equation:
\( \frac{P \cdot V}{T} = \frac{0.5P \cdot V_2}{2T} \)

Simplifying and solving for \( V_2 \):
\( P \cdot V \cdot 2T = 0.5P \cdot V_2 \cdot T \)
\( 2 P V T = 0.5 P T V_2 \)
\( V_2 = \frac{2}{0.5} V = 4V \)

[1] Award 1 mark for selecting option D.
[0] Other responses do not receive marks.

To determine the relative bond lengths, we can determine the bond order of the nitrogen-oxygen bonds in each species:

1. In \(\text{NO}_2^+\), nitrogen has 5 valence electrons, loses 1 (leaving 4 valence electrons), and forms two double bonds with the oxygen atoms: \([\text{O}=\text{N}=\text{O}]^+\). The bond order is 2.0.
2. In \(\text{NO}_2^-\), there is resonance between one single and one double bond, giving a bond order of 1.5.
3. In \(\text{NO}_3^-\), there is resonance across three nitrogen-oxygen bonds, giving a bond order of \(4/3 \approx 1.33\).
4. In \(\text{HNO}_3\), the bond to the hydroxyl group is a single covalent bond (bond order 1.0).

Since bond length is inversely proportional to bond order, the highest bond order (2.0 in \(\text{NO}_2^+\)) corresponds to the shortest bond length.

Award [1] for the correct answer A.
- Reject other options because they have lower bond orders (1.5, 1.33, or 1.0) and therefore longer bond lengths.

1. Calculate the concentration of hydrogen ions:
\([\text{H}^+] = 0.010 \times 0.10\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\).

2. For a weak acid, the dissociation constant \(K_{\text{a}}\):
\(K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \approx \frac{(1.0 \times 10^{-3})^2}{0.10} = 1.0 \times 10^{-5}\).

3. Calculate \(\text{p}K_{\text{a}}\):
\(\text{p}K_{\text{a}} = -\log_{10}(1.0 \times 10^{-5}) = 5.0\).

Award [1] for the correct answer C.
- Award [0] for any other option.

Lattice enthalpy (\(\Delta H^\theta_{\text{lat}}\)) is defined as: \(\text{NaCl(s)} \rightarrow \text{Na}^+(\text{g}) + \text{Cl}^-(\text{g})\).

From the Born-Haber cycle:
\(\Delta H^\theta_{\text{f}} = \Delta H^\theta_{\text{at}}(\text{Na}) + \text{IE}_1(\text{Na}) + \frac{1}{2} E_{\text{b}}(\text{Cl}-\text{Cl}) + \text{EA}(\text{Cl}) - \Delta H^\theta_{\text{lat}}\)

Substituting the values:
\(-411 = 108 + 496 + \frac{1}{2}(242) + (-349) - \Delta H^\theta_{\text{lat}}\)
\(-411 = 108 + 496 + 121 - 349 - \Delta H^\theta_{\text{lat}}\)
\(-411 = 376 - \Delta H^\theta_{\text{lat}}\)
\(\Delta H^\theta_{\text{lat}} = 376 + 411 = +787\text{ kJ mol}^{-1}\).

Award [1] for the correct answer A.
- Award [0] for incorrect calculations (such as failing to halve the bond enthalpy of \(\text{Cl}_2\) or using incorrect signs).

- The overall equation is found by adding Step 1 and Step 2 and canceling common species: \(\text{NO}_2(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{NO}(\text{g}) + \text{CO}_2(\text{g})\). Option A is incorrect.
- The rate equation is determined by the slowest step (Step 1). The reactants in Step 1 are two molecules of \(\text{NO}_2\), so the rate equation is \(\text{rate} = k[\text{NO}_2]^2\). Option B is correct.
- \(\text{CO}\) only participates in the fast step, so changes in its concentration do not affect the reaction rate. Option C is incorrect.
- \(\text{NO}_3\) is produced in the first step and consumed in the second step, making it a reaction intermediate, not a catalyst. Option D is incorrect.

Award [1] for B.
- Award [0] for other choices.

- At the cathode (negative electrode), both \(\text{Cu}^{2+}(\text{aq})\) and \(\text{H}^+(\text{aq})\) are attracted. Since copper has a more positive reduction potential than hydrogen (\(E^\theta = +0.34\text{ V}\)), \(\text{Cu}^{2+}\) is preferentially reduced to form \(\text{Cu}(\text{s})\).
- At the anode (positive electrode), both \(\text{SO}_4^{2-}(\text{aq})\) and \(\text{OH}^-(\text{aq})\) / \(\text{H}_2\text{O(l)}\) are attracted. Water is preferentially oxidized to form \(\text{O}_2(\text{g})\) and \(\text{H}^+(\text{aq})\) because sulfate is extremely stable and very difficult to oxidize.
Therefore, the products are oxygen gas at the anode and solid copper at the cathode.

Award [1] for the correct combination (Anode: \(\text{O}_2(\text{g})\); Cathode: \(\text{Cu}(\text{s})\)).
- Reject other options.

From the ideal gas equation:
\(PV = nRT \implies V = \frac{nRT}{P}\).

Let the new volume be \(V'\):
\(V' = \frac{nR(2T)}{\frac{1}{2}P} = 4 \left(\frac{nRT}{P}\right) = 4V\).

Award [1] for D.
- Reject A, B, and C as they represent incorrect applications of the ideal gas law relationship.

1. **Alcohol classification**: The carbon atom bonded to the hydroxyl (\(-\text{OH}\)) group is bonded to two other carbon atoms (the methyl group \(\text{CH}_3-\) and the methylene group \(-\text{CH}_2-\)). Thus, it is a **secondary alcohol**.
2. **Amine classification**: The nitrogen atom of the amine group is bonded to two carbon atoms (the methylene group \(-\text{CH}_2-\) and the methyl group \(-\text{CH}_3\)). Thus, it is a **secondary amine**.

Award [1] for B.
- Reject other classifications.

1. Calculate the heat released (\(q\)):
\(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\)
\(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.2\text{ J g}^{-1}\text{ K}^{-1} \times 6.0\text{ K} = 2520\text{ J} = 2.52\text{ kJ}\)

2. Calculate the moles of reactant (limiting reactant):
\(n(\text{HCl}) = n(\text{NaOH}) = c \times V = 1.0\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.050\text{ mol}\)

3. Calculate the enthalpy of neutralization (\(\Delta H\)):
\(\Delta H = -\frac{q}{n} = -\frac{2.52\text{ kJ}}{0.050\text{ mol}} = -50.4\text{ kJ mol}^{-1}\).

Award [1] for A.
- Award [0] for other values. Note that the negative sign is required because the reaction is exothermic.

Using the combined gas law:

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

Rearranging to solve for \(P_2\):

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

Substitute the given values:

\(P_2 = \frac{100\text{ kPa} \times 2.0\text{ dm}^3 \times 900\text{ K}}{300\text{ K} \times 4.0\text{ dm}^3}\)

\(P_2 = 100\text{ kPa} \times \left(\frac{2.0}{4.0}\right) \times \left(\frac{900}{300}\right) = 100 \times 0.5 \times 3 = 150\text{ kPa}\)

[1 mark] awarded for selecting option b.
No partial marks.

A dative covalent bond (or coordinate bond) is formed when one atom provides both electrons for a shared pair.

- In \(\text{NH}_3\), \(\text{CH}_4\), and \(\text{CO}_2\), all bonds are normal covalent bonds where each atom contributes one electron to each shared pair.
- In the hydronium ion (\(\text{H}_3\text{O}^+\)), the oxygen atom of water (\(\text{H}_2\text{O}\)) uses one of its lone pairs to form a bond with a hydrogen ion (\(\text{H}^+\)), which has no electrons. This represents a dative covalent bond.

[1 mark] awarded for selecting option b.
No partial marks.

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

When \([\text{A}]\) is doubled to \(2[\text{A}]\) and \([\text{B}]\) is tripled to \(3[\text{B}]\), the new rate is:

\(\text{Rate}_2 = k(2[\text{A}])(3[\text{B}])^2\)

\(\text{Rate}_2 = k \times 2[\text{A}] \times 9[\text{B}]^2 = 18 \times k[\text{A}][\text{B}]^2 = 18 \times \text{Rate}_1\)

Therefore, the rate increases by a factor of 18.

[1 mark] awarded for selecting option d.
No partial marks.

A conjugate acid-base pair consists of two species that differ by only one single proton (\(\text{H}^+\)).

- \(\text{HCO}_3^-\) acts as an acid (proton donor) and forms its conjugate base, \(\text{CO}_3^{2-}\), upon losing a proton.
- Therefore, \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\) constitute a conjugate acid-base pair.

[1 mark] awarded for selecting option b.
No partial marks.

Step 1: Balance the half-reactions:
- Reduction: \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
- Oxidation: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\)

Step 2: Equalize the number of electrons transferred by multiplying the oxidation half-reaction by 5:
- \(5\text{Fe}^{2+}(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + 5\text{e}^-\)

Step 3: Combine the half-reactions:
- \(\text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)

The coefficient of \(\text{H}^+\) is 8.

[1 mark] awarded for selecting option c.
No partial marks.

Phosphorus (\(\text{P}\)) has an atomic number of 15, so a neutral phosphorus atom has 15 electrons. Its ground-state configuration is:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3\)

The phosphide ion, \(\text{P}^{3-}\), is formed by gaining three electrons to achieve a full valence shell. Adding these 3 electrons results in 18 total electrons. The configuration becomes:
\(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\)

[1 mark] awarded for selecting option b.
No partial marks.

1. Calculate the total mass of the mixture:
\(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\)
Assuming density is \(1.00\text{ g cm}^{-3}\), \(m = 100.0\text{ g}\).

2. Calculate the heat released (\(q\)):
\(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.0\text{ K} = 2508\text{ J} = 2.508\text{ kJ}\).

3. Calculate the amount of substance reacted (\(n\)):
\(n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)
\(n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)
Since they react in a 1:1 ratio, \(n(\text{H}_2\text{O}\text{ formed}) = 0.0500\text{ mol}\).

4. Calculate the enthalpy change (\(\Delta H\)) per mole:
\(\Delta H = -\frac{q}{n} = -\frac{2.508\text{ kJ}}{0.0500\text{ mol}} = -50.16\text{ kJ mol}^{-1} \approx -50.2\text{ kJ mol}^{-1}\).

[1 mark] awarded for selecting option b.
No partial marks.

1. Find the number of hydrogen atoms in one molecule of ethanol (\(\text{C}_2\text{H}_5\text{OH}\)):
There are \(5 + 1 = 6\) hydrogen atoms per molecule.

2. Calculate the number of moles of hydrogen atoms in \(0.20\text{ mol}\) of ethanol:
\(n(\text{H}) = 0.20\text{ mol} \times 6 = 1.2\text{ mol}\)

3. Calculate the number of hydrogen atoms using Avogadro's constant:
\(N(\text{H}) = 1.2\text{ mol} \times 6.0 \times 10^{23}\text{ mol}^{-1} = 7.2 \times 10^{23}\).

[1 mark] awarded for selecting option c.
No partial marks.

The ideal gas equation can be written as:
\(PV = nRT\)

Since the number of moles is defined as:
\(n = \frac{m}{M}\)

where \(m\) is the mass and \(M\) is the molar mass, we can substitute this into the ideal gas equation:
\(PV = \frac{m}{M}RT\)

Rearranging for molar mass \(M\):
\(M = \frac{m}{V} \cdot \frac{RT}{P}\)

Since density is defined as \(\rho = \frac{m}{V}\), we can substitute this into the equation:
\(M = \frac{\rho RT}{P}\)

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

1. **Hybridization of \(\text{NH}_4^+\)**:
- The central nitrogen atom has 4 bonding domains (4 single bonds to hydrogen) and 0 lone pairs.
- The steric number is 4, which corresponds to \(sp^3\) hybridization and a tetrahedral electron domain geometry.

2. **Molecular Geometry of \(\text{H}_3\text{O}^+\)**:
- The central oxygen atom has 3 bonding domains (3 single bonds to hydrogen) and 1 lone pair (valence electrons: \(6 \text{ (from O)} + 3 \text{ (from H)} - 1 \text{ (positive charge)} = 8\) electrons, which makes 4 electron domains: 3 bonding and 1 lone pair).
- This results in a tetrahedral electron domain geometry but a **trigonal pyramidal** molecular geometry.

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

The rate constant \(k\) is a constant at a specific temperature. It is independent of reactant concentrations and the volume of the reaction vessel.
According to the Arrhenius equation:
\(k = A e^{-\frac{E_a}{RT}}\)
An increase in absolute temperature, \(T\), increases the value of the rate constant \(k\). Therefore, option D is correct.

[1 mark] for selecting option D.
[0 marks] for any other option.

A conjugate acid-base pair consists of two species that differ by exactly one hydrogen ion (proton, \(\text{H}^+\)).

- \(\text{H}_2\text{PO}_4^-\) acts as a Brønsted–Lowry acid by donating a proton to become \(\text{HPO}_4^{2-}\), which is its conjugate base.
- \(\text{HCO}_3^-\) acts as a Brønsted–Lowry base by accepting a proton to become \(\text{H}_2\text{CO}_3\), which is its conjugate acid.

Thus, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) form a conjugate acid-base pair (Option A).

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

1. The atomic number of iron (\(\text{Fe}\)) is 26.
2. The ground-state electron configuration of neutral iron is: \([\text{Ar}] 4\text{s}^2 3\text{d}^6\).
3. When transition metals form positive ions, they lose electrons from the highest energy shell first (the \(4\text{s}\) orbital is emptied before the \(3\text{d}\) orbitals).
4. To form the \(\text{Fe}^{3+}\) ion, we remove 3 electrons: 2 from the \(4\text{s}\) subshell and 1 from the \(3\text{d}\) subshell.
5. This leaves the electron configuration of \(\text{Fe}^{3+}\) as: \([\text{Ar}] 3\text{d}^5\).

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

1. **Determine anode and cathode**:
- Since the standard reduction potential of zinc is more negative (\(-0.76\text{ V}\) compared to \(+0.34\text{ V}\)), zinc is oxidized at the anode:
\(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\)
- Copper is reduced at the cathode:
\(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\)

2. **Analyze the options**:
- **A is incorrect**: Electrons flow from the negative electrode (anode, Zn) to the positive electrode (cathode, Cu) through the external circuit.
- **B is correct**: Because zinc is oxidized, the concentration of \(\text{Zn}^{2+}(\text{aq})\) in the zinc half-cell increases over time.
- **C is incorrect**: Anions from the salt bridge migrate towards the anode (zinc half-cell) to balance the positive charge of newly formed \(\text{Zn}^{2+}\) ions.
- **D is incorrect**: Reduction occurs at the copper electrode (cathode).

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

(a)
- Comparing Expt 1 and Expt 2: \([\text{S}_2\text{O}_8^{2-}]\) is constant, \([\text{I}^-]\) doubles, and initial rate doubles (\(1.15 \times 10^{-4} \rightarrow 2.30 \times 10^{-4}\)). Therefore, the reaction is first order with respect to \(\text{I}^-\).
- Comparing Expt 1 and Expt 3: \([\text{I}^-]\) is constant, \([\text{S}_2\text{O}_8^{2-}]\) doubles, and initial rate doubles (\(1.15 \times 10^{-4} \rightarrow 2.30 \times 10^{-4}\)). Therefore, the reaction is first order with respect to \(\text{S}_2\text{O}_8^{2-}\).
- Overall rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

(b)
- Rearranging: \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)
- Substituting values from Expt 1: \(k = \frac{1.15 \times 10^{-4}}{(0.0800)(0.0400)} = 0.0359\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(3.59 \times 10^{-2}\)).

(c)
- (i) Intermediate: \(\text{SO}_4\text{I}^-(aq)\) (as it is formed in Step 1 and consumed in Step 2).
- (ii) The rate-determining step is the slow step (Step 1). The rate expression is determined by the reactants of this step: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\), which matches the experimentally determined rate expression.

(d)
- Sketch: Plot of 'number of particles' vs 'kinetic energy'.
- The curve at \(318\text{ K}\) has its peak lower and shifted to the right compared to the curve at \(298\text{ K}\).
- A vertical line representing \(E_a\) is drawn. The area under the curve to the right of \(E_a\) is significantly larger for the \(318\text{ K}\) curve.
- Explanation: At a higher temperature, a larger fraction of reactant particles have kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)), resulting in a higher frequency of successful collisions.

(a) [3.5 marks]:
- [1 mark] for explaining why the reaction is 1st order with respect to \(\text{I}^-\).
- [1 mark] for explaining why the reaction is 1st order with respect to \(\text{S}_2\text{O}_8^{2-}\).
- [1.5 marks] for the correct rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (award [1 mark] if \(k\) is omitted).

(b) [3 marks]:
- [1 mark] for correct substitution of values.
- [1 mark] for \(0.0359\) (accept range \(0.0359 - 0.0360\)).
- [1 mark] for unit: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) [3 marks]:
- [1 mark] for \(\text{SO}_4\text{I}^-\).
- [1 mark] for identifying Step 1 as the rate-determining step.
- [1 mark] for linking molecularity of the rate-determining step to the overall rate law.

(d) [3 marks]:
- [1 mark] for a sketch showing both curves correctly (\(318\text{ K}\) peak lower and shifted to the right, both starting at origin).
- [1 mark] for labeling axes (y: number of particles / fraction of particles; x: kinetic energy) and marking \(E_a\).
- [1 mark] for explaining that more particles have \(E \ge E_a\) at higher temperature, causing more frequent successful collisions.

(a) (i)
- \(\text{XeF}_4\): Central Xe atom has 4 single bonds to F and 2 lone pairs. Each F has 3 lone pairs (octets filled). [1 mark]
- \(\text{H}_3\text{O}^+\): Central O atom has 3 single bonds to H and 1 lone pair. The structure is enclosed in square brackets with a '+' charge outside. [1 mark]
(ii)
- \(\text{XeF}_4\): Square planar geometry; ideal bond angles are \(90^\circ\) and \(180^\circ\). [1 mark]
- \(\text{H}_3\text{O}^+\): Trigonal pyramidal geometry; ideal bond angle is approximately \(107^\circ\) (accept range \(105^\circ - 108^\circ\)). [1 mark]

(b)
- In \(\text{CO}_2\), the carbon-to-oxygen bonds are localized double bonds, meaning a bond order of 2.0. [1 mark]
- In \(\text{CO}_3^{2-}\), the pi electrons are delocalized across three resonance structures, resulting in a bond order of \(4/3 \approx 1.33\). [1 mark]
- A higher bond order represents a greater accumulation of electron density between the nuclei, leading to shorter and stronger bonds. Thus, the bonds in \(\text{CO}_2\) are shorter and stronger than those in \(\text{CO}_3^{2-}\). [1.5 marks]

(c)
- \(\text{C}_2\text{H}_2\) (ethyne): \(sp\) hybridization; formed by the mixing of one \(s\) orbital and one \(p\) orbital, leaving two unhybridized \(p\) orbitals to form two pi bonds. [1 mark]
- \(\text{C}_2\text{H}_4\) (ethene): \(sp^2\) hybridization; formed by the mixing of one \(s\) orbital and two \(p\) orbitals, leaving one unhybridized \(p\) orbital to form one pi bond. [1 mark]

(d) (i)
- Methanol (\(\text{CH}_3\text{OH}\)): Hydrogen bonding. [0.5 marks]
- Fluoromethane (\(\text{CH}_3\text{F}\)): Dipole-dipole forces. [0.5 marks]
- Ethane (\(\text{CH}_3\text{CH}_3\)): London dispersion forces. [0.5 marks]
(ii)
- Trend: Ethane < Fluoromethane < Methanol [0.5 marks]
- Explanation: London dispersion forces are the weakest, followed by dipole-dipole forces. Hydrogen bonding in methanol is the strongest, requiring the most thermal energy to overcome during transition to the gas phase. [1 mark]

(a) [4 marks]:
- [1 mark] for correct Lewis structure of \(\text{XeF}_4\).
- [1 mark] for correct Lewis structure of \(\text{H}_3\text{O}^+\).
- [1 mark] for 'square planar' and \(90^\circ\) (or \(90^\circ\) and \(180^\circ\)).
- [1 mark] for 'trigonal pyramidal' and \(107^\circ\) (accept values between \(104^\circ\) and \(109^\circ\)).

(b) [3.5 marks]:
- [1 mark] for stating the bond order of 2 in \(\text{CO}_2\).
- [1 mark] for stating the resonance in \(\text{CO}_3^{2-}\) results in delocalization and a bond order of 1.33.
- [1.5 marks] for correctly stating and explaining that \(\text{CO}_2\) has shorter and stronger carbon-oxygen bonds due to its higher bond order.

(c) [2 marks]:
- [1 mark] for \(sp\) hybridization and explanation of 1s + 1p mixing in ethyne.
- [1 mark] for \(sp^2\) hybridization and explanation of 1s + 2p mixing in ethene.

(d) [3 marks]:
- [1.5 marks] for correctly identifying dominant IMF for each (0.5 marks each).
- [0.5 marks] for correct order: ethane < fluoromethane < methanol.
- [1 mark] for linking strength of IMF to boiling point.

(a)
- Absolute temperature: \(T = 99.2 + 273.15 = 372.35\text{ K} \approx 372.4\text{ K}\) [1 mark]
- Absolute uncertainty of addition/subtraction remains the same: \(\pm 0.5\text{ K}\). [1 mark]

(b)
- Percentage uncertainty in pressure: \(\frac{0.2\text{ kPa}}{101.3\text{ kPa}} \times 100\% = 0.197\% \approx 0.20\%\). [1 mark]

(c)
- Unit conversions:
- \(P = 101.3\text{ kPa} = 1.013 \times 10^5\text{ Pa}\)
- \(V = 145.0\text{ cm}^3 = 1.450 \times 10^{-4}\text{ m}^3\)
- \(T = 372.4\text{ K}\) [1 mark]
- \(n = \frac{PV}{RT} = \frac{(1.013 \times 10^5\text{ Pa})(1.450 \times 10^{-4}\text{ m}^3)}{(8.31\text{ J K}^{-1}\text{ mol}^{-1})(372.4\text{ K})}\) [1 mark]
- \(n = 4.75 \times 10^{-3}\text{ mol}\) (or \(0.00475\text{ mol}\)). [1 mark]

(d)
- Molar mass: \(M = \frac{m}{n} = \frac{0.452\text{ g}}{4.75 \times 10^{-3}\text{ mol}} = 95.2\text{ g mol}^{-1}\). [1 mark]
- Summing percentage uncertainties:
- \(\% \Delta m = \frac{0.002}{0.452} \times 100\% = 0.442\%\) [0.5 marks]
- \(\% \Delta V = \frac{0.5}{145.0} \times 100\% = 0.345\%\) [0.5 marks]
- \(\% \Delta T = \frac{0.5}{372.4} \times 100\% = 0.134\%\) [0.5 marks]
- \(\% \Delta P = 0.197\%\)
- Total \(\% \Delta M = 0.442\% + 0.345\% + 0.134\% + 0.197\% = 1.118\% \approx 1.1\%\). [1 mark]
- Absolute uncertainty: \(\Delta M = 95.2 \times 0.01118 = 1.06 \approx 1.1\text{ g mol}^{-1}\) (or \(1\text{ g mol}^{-1}\)). [0.5 marks]
- Expressing final answer: \(95.2 \pm 1.1\text{ g mol}^{-1}\) (or \(95 \pm 1\text{ g mol}^{-1}\)). [0.5 marks]

(e)
- Great deviation occurs at low temperatures and high pressures. [1 mark]
- At low temperatures, gas particles move slowly, allowing weak intermolecular forces to pull them together. At high pressures, gas particles are close together, and the volume of the particles themselves is no longer negligible compared to the total volume of the container. [1 mark]

(a) [2 marks]:
- [1 mark] for correct absolute temperature (372.4 K).
- [1 mark] for absolute uncertainty of \(\pm 0.5\text{ K}\).

(b) [1 mark]:
- [1 mark] for pressure uncertainty = 0.20% (or 0.197%).

(c) [3 marks]:
- [1 mark] for converting gas variables to SI units (V in \(\text{m}^3\), P in \(\text{Pa}\)).
- [1 mark] for correct substitution in \(n = \frac{PV}{RT}\).
- [1 mark] for final answer \(4.75 \times 10^{-3}\text{ mol}\).

(d) [4.5 marks]:
- [1 mark] for calculating \(M = 95.2\text{ g mol}^{-1}\).
- [1.5 marks] for calculating individual percentage uncertainties of \(m\), \(V\), and \(T\) (0.5 each).
- [1 mark] for sum of percentage uncertainties = 1.12%.
- [1 mark] for absolute uncertainty (1.1 or 1) and reporting in correct format (e.g., \(95 \pm 1\text{ g mol}^{-1}\) or \(95.2 \pm 1.1\text{ g mol}^{-1}\)).

(e) [2 marks]:
- [1 mark] for identifying "low temperature and high pressure".
- [1 mark] for explaining the effects of intermolecular forces and particle volume under these conditions.

(a)
- Dissociation equation: \(\text{HNO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NO}_2^-(aq) + \text{H}_3\text{O}^+(aq)\) (or \(\text{HNO}_2(aq) \rightleftharpoons \text{H}^+(aq) + \text{NO}_2^-(aq)\)). [1 mark]
- \(K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \approx \frac{[\text{H}^+]^2}{[\text{HNO}_2]_0}\) [0.5 marks]
- \([\text{H}^+] = \sqrt{K_a \cdot [\text{HNO}_2]_0} = \sqrt{(5.62 \times 10^{-4})(0.120)} = 8.21 \times 10^{-3}\text{ mol dm}^{-3}\) [1 mark]
- \(\text{pH} = -\log(8.21 \times 10^{-3}) = 2.09\). [1 mark]

(b)
- Adding \(12.5\text{ cm}^3\) of KOH is exactly half of the volume needed to reach the equivalence point (\(25.0\text{ cm}^3\)), marking the half-equivalence point. [1 mark]
- At this point, the concentrations of weak acid and its conjugate base are equal: \([\text{HNO}_2] = [\text{NO}_2^-]\), forming a buffer. [1 mark]
- Therefore, \(\text{pH} = p K_a = -\log(5.62 \times 10^{-4}) = 3.25\). [1 mark]

(c)
- At the equivalence point, nitrous acid is completely converted to nitrite ions (\(\text{NO}_2^-\)), which act as a weak conjugate base and hydrolyze water:
\(\text{NO}_2^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HNO}_2(aq) + \text{OH}^-(aq)\) [1 mark]
The generation of hydroxide ions (\(\text{OH}^-\)) causes the pH to be greater than 7 at the equivalence point. [1 mark]
- Calculation:
- Final volume at equivalence = \(25.0 + 25.0 = 50.0\text{ cm}^3\).
- \([\text{NO}_2^-] = \frac{0.120\text{ mol dm}^{-3}}{2} = 0.0600\text{ mol dm}^{-3}\). [0.5 marks]
- \(K_b = \frac{K_w}{K_a} = \frac{1.00 \times 10^{-14}}{5.62 \times 10^{-4}} = 1.78 \times 10^{-11}\text{ mol dm}^{-3}\). [0.5 marks]
- \([\text{OH}^-] = \sqrt{K_b \cdot [\text{NO}_2^-]} = \sqrt{(1.78 \times 10^{-11})(0.0600)} = 1.03 \times 10^{-6}\text{ mol dm}^{-3}\). [0.5 marks]
- \(\text{pOH} = -\log(1.03 \times 10^{-6}) = 5.99 \rightarrow \text{pH} = 14.00 - 5.99 = 8.01\). [0.5 marks]

(d)
- Phenolphthalein is more suitable. [1 mark]
- Because its transition range (around its \(p K_a\) of 9.3) corresponds closely with the steep vertical region of the titration curve which spans around pH 7 to 10 for a weak acid-strong base titration. Methyl red (\(p K_a = 5.1\)) would change color prematurely in the acidic buffer region. [1 mark]

(a) [3.5 marks]:
- [1 mark] for correct dissociation equation with reversible arrows.
- [1 mark] for setting up expression and finding \([\text{H}^+] = 8.21 \times 10^{-3}\text{ mol dm}^{-3}\).
- [1.5 marks] for pH = 2.09.

(b) [3 marks]:
- [1 mark] for recognizing this is the half-equivalence point.
- [1 mark] for stating \([\text{HNO}_2] = [\text{NO}_2^-]\) and \(\text{pH} = p K_a\).
- [1 mark] for calculating pH = 3.25.

(c) [4 marks]:
- [1 mark] for correct ionic equation showing hydrolysis of nitrite ion.
- [1 mark] for explaining that hydrolysis yields \(\text{OH}^-\), raising pH.
- [1 mark] for calculating \([\text{NO}_2^-] = 0.0600\text{ mol dm}^{-3}\) and \(K_b = 1.78 \times 10^{-11}\).
- [1 mark] for calculating \([\text{OH}^-]\) and obtaining \(\text{pH} = 8.01\) (accept range 8.00–8.05).

(d) [2 marks]:
- [1 mark] for choosing phenolphthalein.
- [1 mark] for explaining that the transition region of phenolphthalein matches the basic equivalence point (pH ~ 8), whereas methyl red would change color too early in the buffer region.

(a) Dissolving is not instantaneous, and heat is lost to the surroundings as the reaction occurs. Extrapolating back to the time of mixing corrects for this heat loss.

(b) \(\Delta T = 29.4 - 21.2 = 8.2\text{ K}\)
\(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 8.2\text{ K} = 1713.8\text{ J}\).
Rounding to 3 significant figures gives \(1710\text{ J}\) (or \(1.71\text{ kJ}\)).

(c) Absolute uncertainty in \(\Delta T = 0.1\text{ }^\circ\text{C} + 0.1\text{ }^\circ\text{C} = 0.2\text{ }^\circ\text{C}\).
Percentage uncertainty = \(\frac{0.2}{8.2} \times 100\% = 2.44\%\) (or \(2.4\%\)).

(d) \(n(\text{CuSO}_4) = \frac{4.00\text{ g}}{159.61\text{ g mol}^{-1}} = 0.02506\text{ mol}\).
\(\Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{1.7138\text{ kJ}}{0.02506\text{ mol}} = -68.4\text{ kJ mol}^{-1}\).

(e) Polystyrene is a good thermal insulator, while copper is a good thermal conductor. Using a copper calorimeter increases heat loss to the surroundings, resulting in a lower maximum temperature and a smaller temperature change, \(\Delta T\).

(a) Award [1] for mentioning correction for heat loss to the surroundings.
(b) Award [1] for \(1710\text{ J}\) or \(1.71\text{ kJ}\) (accept \(1713.8\text{ J}\)).
(c) Award [1] for calculating absolute uncertainty in temperature change as \(0.2\text{ }^\circ\text{C}\), and [1] for calculating percentage uncertainty as \(2.4\%\) (accept \(2.44\%\)).
(d) Award [1] for calculating moles as \(0.0251\text{ mol}\), and [1] for \(-68.4\text{ kJ mol}^{-1}\) (must include negative sign for mark; accept range \(-68.3\) to \(-68.5\)).
(e) Award [1] for stating that \(\Delta T\) will be smaller AND explaining that copper is a better conductor of heat / has lower insulating capacity than polystyrene.

(a) Mass of condensed liquid = \(84.815\text{ g} - 84.112\text{ g} = 0.703\text{ g}\).

(b) Temperature in Kelvin: \(T = 98.5 + 273.15 = 371.7\text{ K}\) (or \(371.65\text{ K}\)).
Volume in \(\text{m}^3\): \(V = 250.0\text{ cm}^3 \times 10^{-6}\text{ m}^3\text{ cm}^{-3} = 2.500 \times 10^{-4}\text{ m}^3\).

(c) Pressure in Pascals: \(P = 101.3\text{ kPa} = 1.013 \times 10^5\text{ Pa}\).
Using \(PV = nRT\):
\(n = \frac{PV}{RT} = \frac{1.013 \times 10^5\text{ Pa} \times 2.500 \times 10^{-4}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 371.65\text{ K}} = \frac{25.325}{3088.4} = 8.20 \times 10^{-3}\text{ mol}\) (or \(0.00820\text{ mol}\)).

(d) Molar mass \(M = \frac{m}{n} = \frac{0.703\text{ g}}{8.199 \times 10^{-3}\text{ mol}} = 85.7\text{ g mol}^{-1}\).

(e) Water remaining on the outside of the flask adds mass to the final weighing, making the calculated mass of the condensed liquid higher than it actually is. Since \(M = \frac{m}{n}\) and \(n\) remains unchanged, a larger mass value results in an overestimated (higher) calculated molar mass.

(a) Award [1] for \(0.703\text{ g}\).
(b) Award [1] for \(T = 371.7\text{ K}\) (accept \(371.65\text{ K}\) or \(372\text{ K}\)), and [1] for \(V = 2.500 \times 10^{-4}\text{ m}^3\).
(c) Award [1] for correct substitution into rearranged equation (converting \(P\) to Pa is required), and [1] for \(8.20 \times 10^{-3}\text{ mol}\) (accept range \(8.19 \times 10^{-3}\) to \(8.21 \times 10^{-3}\)).
(d) Award [1] for \(85.7\text{ g mol}^{-1}\) (accept \(85.6\) to \(85.8\); allow ECF from (a) and (c)).
(e) Award [1] for stating that the calculated molar mass would be higher/overestimated, and [1] for explaining that water on the outside increases the apparent mass of the condensed liquid (mass of liquid is artificially high).

**(a)**
* **Phenol / phenolic hydroxyl group** [1 mark]
* **Secondary alcohol / aliphatic hydroxyl group** [1 mark]
* **Ethanoic anhydride / acetic anhydride / acetyl chloride** [1 mark]
*(Note: Do not accept just \(\text{-OH}\) or "hydroxyl" for both groups; the distinction between the phenol and the aliphatic alcohol must be clear).*

**(b)**
* Morphine contains highly polar hydroxyl groups (\(\text{-OH}\)) which readily form hydrogen bonds with water, making it hydrophilic and poorly soluble in lipids. [1 mark]
* In diamorphine, these hydroxyl groups are replaced by ester groups (\(\text{-OCOCH}_3\)), which are much less polar and cannot act as hydrogen bond donors. [1 mark]
* This makes diamorphine highly lipophilic (lipid-soluble), allowing it to easily dissolve in and diffuse across the lipid bilayer of the blood-brain barrier. [1 mark]

* **Part (a):**
* Award 1 mark for "phenol" (or "phenolic hydroxyl").
* Award 1 mark for "secondary alcohol" (or "aliphatic hydroxyl" / "alcohol").
* Award 1 mark for "ethanoic anhydride" (accept "acetic anhydride", "acetyl chloride", or "ethanoic acid").
* **Part (b):**
* Award 1 mark for identifying that morphine is more polar due to hydroxyl groups or that diamorphine is less polar due to ester groups.
* Award 1 mark for linking the decreased polarity of diamorphine to increased lipid solubility (lipophilicity) or decreased hydrogen bonding capacity with water.
* Award 1 mark for explaining that the blood-brain barrier consists of a lipid bilayer, which highly lipid-soluble substances (like diamorphine) can easily cross by passive diffusion.

**(a)**
* Both drugs are **neuraminidase inhibitors**. [1 mark]
* They bind to the active site of the neuraminidase enzyme on the viral surface, preventing it from cleaving sialic acid receptors on the host cell membrane, which stops newly replicated virions from being released and spreading to healthy cells. [1 mark]

**(b)**
* **Present in zanamivir but absent in oseltamivir:** Carboxylic acid / hydroxyl group (or alcohol) / guanidino group. [1 mark]
* **Present in oseltamivir but absent in zanamivir:** Ester. [1 mark]

**(c)**
* Oseltamivir is an ester prodrug; its relatively non-polar ester structure makes it lipid-soluble enough to be easily absorbed from the gastrointestinal tract, after which it is hydrolyzed in the liver to its active carboxylate form. [1 mark]
* Zanamivir contains highly polar carboxylic acid, guanidino, and multiple hydroxyl groups, making it highly water-soluble but poorly absorbed across the gastrointestinal tract. Delivering it via inhalation targets it directly to the respiratory tract. [1 mark]

* **Part (a):**
* Award 1 mark for identifying both as neuraminidase inhibitors (or blocking/inhibiting neuraminidase).
* Award 1 mark for explaining that this prevents the release of new viral particles from the host cell.
* **Part (b):**
* Award 1 mark for any correct group unique to zanamivir (e.g., carboxylic acid, hydroxyl/alcohol, or guanidino).
* Award 1 mark for "ester" as unique to oseltamivir (accept "alkene / alkenyl" or "ether" only if contrasting properly, though ester is the primary distinct structural feature).
* **Part (c):**
* Award 1 mark for stating that oseltamivir is less polar / lipid-soluble (due to its ester group) and can be absorbed through the gut / acts as a prodrug.
* Award 1 mark for stating that zanamivir is too polar / water-soluble (due to multiple hydroxyl/carboxyl/guanidino groups) to be absorbed orally, requiring inhalation to go directly to the lungs.

**(a) (i)**
* It emits low-energy gamma (\(\gamma\)) radiation, which easily escapes the patient's body to be detected by external cameras but causes very little ionization damage to cells. [1 mark]
* It has a short half-life of 6 hours, which is long enough for imaging to take place but short enough that patient exposure is minimized. [1 mark]
*(Accept other valid reasons, e.g., it can easily bind to a wide range of carrier molecules/biomolecules).

**(a) (ii)**
\(^{99\text{m}}_{43}\text{Tc} \rightarrow\ ^{99}_{43}\text{Tc} + \gamma\)
*(Accept: \(^{99\text{m}}\text{Tc} \rightarrow\ ^{99}\text{Tc} + \gamma\))* [1 mark]

**(b) (i)**
* Alpha-emitting isotopes are conjugated/bonded to monoclonal antibodies that specifically bind to unique antigen proteins expressed on the target cancer cells. [1 mark]

**(b) (ii)**
* Alpha particles have high mass and charge (\(2+\)), giving them high ionizing power (high linear energy transfer) that causes lethal double-stranded DNA breaks in the target cells. [1 mark]
* They have an extremely short range in tissue (only a few cell diameters / \(50\text{--}100\ \mu\text{m}\)), meaning their destructive energy is entirely deposited within the tumor, leaving surrounding healthy tissues undamaged. [1 mark]

* **Part (a)(i):**
* Award 1 mark for each valid reason, up to a maximum of 2.
* **Part (a)(ii):**
* Award 1 mark for the correct equation. Reject if mass numbers or atomic numbers do not balance (if shown), or if gamma (\(\gamma\)) is missing.
* **Part (b)(i):**
* Award 1 mark for mentioning the conjugation/binding of the isotope to monoclonal antibodies that target cancer-specific receptors.
* **Part (b)(ii):**
* Award 1 mark for mentioning the high ionizing power / high charge/mass / high linear energy transfer (LET) of alpha particles.
* Award 1 mark for mentioning the extremely short range of alpha radiation (limiting damage to the immediate target area / preventing collateral damage).