2024 IB DP Chemistry 模擬試題連答案詳解

Using the combined gas law equation:

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

Rearranging to solve for the final volume, \(V_2\):

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

Substitute the given values:

\(V_2 = 2.0\text{ dm}^3 \times \frac{100\text{ kPa}}{400\text{ kPa}} \times \frac{150\text{ K}}{300\text{ K}}\)

\(V_2 = 2.0 \times \frac{1}{4} \times \frac{1}{2} = 2.0 \times \frac{1}{8} = 0.25\text{ dm}^3\)

[1 mark] for the correct answer A.
Award 0 marks for any other choice.

The units of rate are \(\text{mol dm}^{-3}\text{ s}^{-1}\).

The overall reaction order is \(2 + 1 = 3\).

The concentration terms have units of \((\text{mol dm}^{-3})^3 = \text{mol}^3\text{ dm}^{-9}\).

Rearranging for the rate constant \(k\):

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

Units of \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\), which is equivalent to \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

[1 mark] for the correct answer B.
Award 0 marks for any other choice.

The heat released in the reaction, \(q\), is directly proportional to the amount of reactants used. Halving the volume of the solution halves the number of moles of reactants, which halves the total heat energy released (\(q_2 = 0.5 q_1\)). However, the total mass of the solution being heated is also halved (\(m_2 = 0.5 m_1\)). Since \(\Delta T = \frac{q}{m \cdot c}\), halving both \(q\) and \(m\) leaves the temperature change \(\Delta T\) unchanged.

[1 mark] for the correct answer C.
Award 0 marks for any other choice.

A conjugate base is formed when a species donates/loses a proton (\(\text{H}^+\)).

\(\text{H}_2\text{PO}_4^- \rightarrow \text{HPO}_4^{2-} + \text{H}^+\)

Therefore, \(\text{HPO}_4^{2-}\) is the conjugate base of \(\text{H}_2\text{PO}_4^-\).

[1 mark] for the correct answer B.
Award 0 marks for any other choice.

We determine the oxidation states of sulfur in each option:

- In \(\text{SO}_3^{2-}\): \(x + 3(-2) = -2 \Rightarrow x = +4\)
- In \(\text{S}_2\text{O}_3^{2-}\): \(2x + 3(-2) = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\)
- In \(\text{SO}_4^{2-}\): \(x + 4(-2) = -2 \Rightarrow x = +6\)
- In \(\text{S}_8\): elemental form, so the oxidation state is \(0\)

The highest oxidation state is \(+6\) in \(\text{SO}_4^{2-}\).

[1 mark] for the correct answer C.
Award 0 marks for any other choice.

Ethyl propanoate is an ester with the molecular structure \(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3\). It contains the ester (carboxylate) functional group.

[1 mark] for the correct answer B.
Award 0 marks for any other choice.

Let's analyze the shapes using VSEPR theory:

- \(\text{BF}_3\) has 3 bonding domains and 0 lone pairs on the central B atom, resulting in a trigonal planar geometry.
- \(\text{NH}_3\) has 3 bonding domains and 1 lone pair on the central N atom, resulting in a trigonal pyramidal geometry.
- \(\text{CH}_4\) has 4 bonding domains and 0 lone pairs on the central C atom, resulting in a tetrahedral geometry.
- \(\text{H}_2\text{O}\) has 2 bonding domains and 2 lone pairs on the central O atom, resulting in a bent geometry.

[1 mark] for the correct answer B.
Award 0 marks for any other choice.

For the ion \(^{37}\text{Cl}^-\):
- Atomic number (Z) of chlorine is 17, which means there are 17 protons.
- Mass number (A) is 37, so the number of neutrons is \(A - Z = 37 - 17 = 20\).
- The charge is \(-1\), indicating that the ion has one more electron than protons: \(17 + 1 = 18\) electrons.

[1 mark] for the correct answer A.
Award 0 marks for any other choice.

According to the ideal gas equation, \(PV = nRT\). For a fixed mass of gas, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Substituting the given changes: \(P_1 = P\), \(V_1 = V\), \(T_1 = T\), \(P_2 = 0.5P\), and \(T_2 = 2T\). Therefore, \(\frac{P \times V}{T} = \frac{0.5P \times V_2}{2T}\). Solving for \(V_2\) gives \(V_2 = \frac{2 \times P \times V}{0.5 \times P} = 4V\).

Award [1] for the correct answer (d). Award [0] for incorrect answers.

\(\text{Ba(OH)}_2\) is a strong dibasic base and dissociates completely: \(\text{Ba(OH)}_2(\text{aq}) \rightarrow \text{Ba}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})\). Thus, \([\text{OH}^-] = 2 \times (5.0 \times 10^{-4} \text{ mol dm}^{-3}) = 1.0 \times 10^{-3} \text{ mol dm}^{-3}\). Since \(\text{pOH} = -\log_{10}[\text{OH}^-]\), \(\text{pOH} = -\log_{10}(1.0 \times 10^{-3}) = 3.0\). At \(298\text{ K}\), \(\text{pH} + \text{pOH} = 14.0\), so \(\text{pH} = 14.0 - 3.0 = 11.0\).

Award [1] for the correct answer (d). Award [0] for incorrect answers.

The electron configuration of a neutral cobalt atom (Z = 27) is \([\text{Ar}] 4\text{s}^2 3\text{d}^7\). When forming the \(\text{Co}^{2+}\) ion, the two valence electrons are lost from the outermost \(4\text{s}\) orbital first, giving the configuration \([\text{Ar}] 3\text{d}^7\). According to Hund's rule, the seven electrons populate the five d-orbitals as follows: two orbitals are doubly occupied, and three orbitals are singly occupied. Therefore, there are 3 unpaired electrons.

Award [1] for the correct answer (c). Award [0] for incorrect answers.

Assuming a \(100\text{ g}\) sample, the mass of nitrogen is \(30.4\text{ g}\) and the mass of oxygen is \(100 - 30.4 = 69.6\text{ g}\). The number of moles of each element is: \(n(\text{N}) = 30.4\text{ g} / 14.0\text{ g mol}^{-1} \approx 2.17\text{ mol}\) and \(n(\text{O}) = 69.6\text{ g} / 16.0\text{ g mol}^{-1} \approx 4.35\text{ mol}\). Dividing both values by the smaller number of moles (2.17) yields a ratio of \(\text{N} : \text{O} \approx 1 : 2\). The empirical formula is therefore \(\text{NO}_2\).

Award [1] for the correct answer (b). Award [0] for incorrect answers.

In \(\text{H}_2\text{C}_2\text{O}_4\), hydrogen has an oxidation state of \(+1\) and oxygen has \(-2\). Setting up the equation for carbon: \(2(+1) + 2(x) + 4(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow x = +3\). In \(\text{CO}_2\), oxygen is \(-2\), so carbon has an oxidation state of \(+4\). Since the oxidation state of carbon increases from \(+3\) to \(+4\), it is oxidized. Manganese decreases from \(+7\) in \(\text{MnO}_4^-\) to \(+2\) in \(\text{Mn}^{2+}\), so it is reduced.

Award [1] for the correct answer (c). Award [0] for incorrect answers.

The rate expression is \(\text{Rate} = k[\text{A}]^1[\text{B}]^0 = k[\text{A}]\). Rearranging for the rate constant gives \(k = \text{Rate} / [\text{A}]\). The unit of reaction rate is \(\text{mol dm}^{-3}\text{ s}^{-1}\) and the unit of concentration \([\text{A}]\) is \(\text{mol dm}^{-3}\). Substituting these units gives \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}\).

Award [1] for the correct answer (a). Award [0] for incorrect answers.

The temperature rise \(\Delta T\) is given by \(\Delta T = \frac{q}{m \cdot c}\). By halving the volumes of both solutions, the number of moles of reactants reacting is halved, which halves the total heat energy released (\(q' = 0.5q\)). However, the total mass of the mixture (which is proportional to the total volume) is also halved (\(m' = 0.5m\)). Therefore, the temperature change is \(\Delta T' = \frac{0.5q}{0.5m \cdot c} = \Delta T\), which is identical to the initial temperature rise.

Award [1] for the correct answer (c). Award [0] for incorrect answers.

Phosphorus is in group 15 and has 5 valence electrons. In \(\text{PCl}_3\), it shares 3 electrons with three chlorine atoms, leaving one lone pair on the phosphorus atom. With 3 bonding domains and 1 non-bonding domain, the electron domain geometry is tetrahedral, and the molecular geometry is trigonal pyramidal. \(\text{BF}_3\), \(\text{SO}_3\), and \(\text{CO}_3^{2-}\) all have 3 electron domains with no lone pairs on the central atom, resulting in trigonal planar molecular geometries.

Award [1] for the correct answer (c). Award [0] for incorrect answers.

The number of moles of \(\text{CH}_4\) is calculated as: \(n = \frac{8.0\text{ g}}{16.0\text{ g mol}^{-1}} = 0.50\text{ mol}\). Each molecule of \(\text{CH}_4\) contains 5 atoms (1 carbon atom and 4 hydrogen atoms). Therefore, the total number of moles of atoms is \(0.50\text{ mol} \times 5 = 2.5\text{ mol}\). Thus, the total number of atoms is \(2.5 L\).

Award [1] for correct answer (B). Award [0] for incorrect choices.

According to the ideal gas equation relationship, \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Rearranging to solve for the final volume \(V_2\) gives: \(V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}\). Substituting the changes: \(V_2 = V \times \frac{P}{0.5P} \times \frac{2T}{T} = V \times 2 \times 2 = 4V\).

Award [1] for correct answer (C). Award [0] for incorrect choices.

The atomic number of iron (Fe) is 26, which gives the ground-state electron configuration of neutral iron as \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). To form the \(\text{Fe}^{3+}\) ion, three electrons must be removed. Electrons are lost first from the outer \(4\text{s}\) orbital (2 electrons) and then from the \(3\text{d}\) orbitals (1 electron). This yields a configuration of \([\text{Ar}] 3\text{d}^5\).

Award [1] for correct answer (A). Award [0] for incorrect choices.

The nitrogen atom has 5 valence electrons. The negative charge adds 1 electron, for a total of 6 valence electrons on the central atom. The nitrogen forms two single N-H covalent bonds (using 2 electrons), which leaves 4 non-bonding valence electrons (or 2 lone pairs). With 2 bonding domains and 2 non-bonding domains, the electron domain geometry is tetrahedral, but the molecular geometry is bent / V-shaped.

Award [1] for correct answer (B). Award [0] for incorrect choices.

An increase in temperature means molecules have a higher average kinetic energy, shifting the distribution peak to the right (higher energy). To maintain the same total area under the curve (representing the total number of molecules), the peak must also flatten out, resulting in a lower fraction of molecules having the most probable energy.

Award [1] for correct answer (A). Award [0] for incorrect choices.

First, calculate the heat absorbed by water: \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.0\text{ K} = 2090\text{ J} = 2.09\text{ kJ}\). Since the water temperature increases, the reaction is exothermic, making the enthalpy change negative. Enthalpy change per mole of reactant: \(\Delta H = -\frac{2.09\text{ kJ}}{0.0100\text{ mol}} = -209\text{ kJ mol}^{-1}\).

Award [1] for correct answer (A). Award [0] for incorrect choices.

A reducing agent is oxidized during a redox reaction. The oxidation state of Fe in \(\text{Fe}^{2+}\) increases from +2 to +3 in \(\text{Fe}^{3+}\), meaning it loses an electron (oxidation). Therefore, \(\text{Fe}^{2+}\) is the reducing agent.

Award [1] for correct answer (B). Award [0] for incorrect choices.

The structure is an ester of the general form R-COO-R'. The R' group attached to the oxygen is a methyl group (\(-\text{CH}_3\)). The R-COO portion is derived from propanoic acid and contains 3 carbons (\(\text{CH}_3\text{CH}_2\text{COO}-\)), making it a propanoate group. Combining these gives the name methyl propanoate.

Award [1] for correct answer (A). Award [0] for incorrect choices.

1. Calculate the amount in moles of \( \text{C}_2\text{H}_6 \):
\( n(\text{C}_2\text{H}_6) = \frac{\text{mass}}{\text{molar mass}} = \frac{3.0 \text{ g}}{30 \text{ g mol}^{-1}} = 0.10 \text{ mol} \).

2. Determine the moles of hydrogen atoms:
Each mole of \( \text{C}_2\text{H}_6 \) contains \( 6 \) moles of hydrogen atoms.
\( n(\text{H}) = 0.10 \text{ mol} \times 6 = 0.60 \text{ mol} \).

3. Calculate the number of hydrogen atoms:
\( N(\text{H}) = 0.60 \text{ mol} \times 6.0 \times 10^{23} \text{ mol}^{-1} = 3.6 \times 10^{23} \).

Award [1] for the correct answer B.
Reject incorrect calculations.

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

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

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

Rearranging to solve for the new volume \( V_2 \):
\( V_2 = \frac{P \cdot V \cdot 2 T}{T \cdot 0.5 P} = \frac{2}{0.5} V = 4 V \)

Award [1] for the correct answer D.

The atomic number of Cobalt (\( \text{Co} \)) is 27, so a neutral cobalt atom has 27 electrons.
Its ground-state electron configuration is:
\( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 4\text{s}^2 3\text{d}^7 \) or \( [\text{Ar}] 4\text{s}^2 3\text{d}^7 \).

When transition metal atoms form positive ions, they lose electrons from the outermost shell (highest main energy level, \( n = 4 \)) first.
Therefore, the two \( 4\text{s} \) electrons are removed to form the \( \text{Co}^{2+} \) ion, leaving the configuration as:
\( [\text{Ar}] 3\text{d}^7 \).

Award [1] for the correct answer C.

- \( \text{BF}_3 \): Boron has 3 valence electrons and forms 3 bonding pairs. There are no lone pairs, so the geometry is trigonal planar.
- \( \text{SO}_3 \): Sulfur has 6 valence electrons and forms double bonds with 3 oxygen atoms. It has 3 electron domains and no lone pairs, leading to a trigonal planar geometry.
- \( \text{H}_3\text{O}^+ \): Oxygen has 6 valence electrons. Losing 1 electron for the overall positive charge leaves 5. Bonding with 3 hydrogen atoms uses 3 electrons, leaving 1 lone pair. With 3 bonding pairs and 1 lone pair, the shape is trigonal pyramidal.
- \( \text{CO}_3^{2-} \): Carbon has 4 valence electrons. Adding 2 from the negative charge gives 6. It forms bonds with 3 oxygen atoms (3 electron domains, 0 lone pairs), yielding a trigonal planar geometry.

Award [1] for the correct answer C.

For a second-order reaction, the rate equation can be written as:
\( \text{Rate} = k [A]^2 \)

Rearranging for the rate constant \( k \):
\( k = \frac{\text{Rate}}{[A]^2} \)

Substitute the standard units:
- Unit of Rate = \( \text{mol dm}^{-3} \text{s}^{-1} \)
- Unit of Concentration \( [A] = \text{mol dm}^{-3} \)

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

Award [1] for the correct answer B.

According to the Brønsted-Lowry theory, a conjugate base is formed when an acid donates a proton (\( \text{H}^+ \)).

The reaction of \( \text{HPO}_4^{2-} \) acting as an acid is:
\( \text{HPO}_4^{2-} \rightleftharpoons \text{PO}_4^{3-} + \text{H}^+ \)

Therefore, \( \text{PO}_4^{3-} \) is the conjugate base of \( \text{HPO}_4^{2-} \).

Award [1] for the correct answer C.

The sum of the oxidation states of all atoms in a polyatomic ion must equal the overall charge of the ion.

Let the oxidation state of sulfur be \( x \).
The oxidation state of oxygen is typically \( -2 \).

Setting up the equation:
\( 2x + 3(-2) = -2 \)
\( 2x - 6 = -2 \)
\( 2x = +4 \)
\( x = +2 \)

Therefore, the oxidation state of sulfur is \( +2 \).

Award [1] for the correct answer B.

1. Identify the functional group: The compound contains the ester linkage (\( -\text{COO}- \)).
2. Determine the two components of the ester name:
- The alkyl group attached directly to the single-bonded oxygen: \( -\text{CH}_2\text{CH}_3 \) is an ethyl group.
- The carboxylic acid chain containing the carbonyl group has 4 carbons in its main chain (butanoate), numbered from the carbonyl carbon as C1:
- C1: Carbonyl carbon (\( \text{C}=\text{O} \))
- C2: \( -\text{CH}_2- \)
- C3: \( -\text{CH(CH}_3)- \) (contains a methyl substituent)
- C4: \( -\text{CH}_3 \)
This component is "3-methylbutanoate".
3. Combine these parts: "ethyl 3-methylbutanoate".

Award [1] for the correct answer B.

The decomposition reaction is: \(MCO_3(s) \rightarrow MO(s) + CO_2(g)\). Based on the stoichiometry, \(1\text{ mol}\) of \(MCO_3\) decomposes to produce \(1\text{ mol}\) of \(MO\). Therefore, \(0.10\text{ mol}\) of \(MCO_3\) yields \(0.10\text{ mol}\) of \(MO\). The molar mass of \(MO\) is calculated as: \(\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{4.0\text{ g}}{0.10\text{ mol}} = 40\text{ g mol}^{-1}\). Since the molar mass of \(MO = A_r(M) + A_r(O) = A_r(M) + 16 = 40\text{ g mol}^{-1}\), the relative atomic mass of the metal \(M\) is \(40 - 16 = 24\).

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

Using the combined gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Rearranging to solve for the final pressure gives: \(P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)\). Since the volume is halved, the volume ratio \(\frac{V_1}{V_2} = 2\). Since the temperature is doubled, the temperature ratio \(\frac{T_2}{T_1} = 2\). Substituting the values: \(P_2 = 100\text{ kPa} \times 2 \times 2 = 400\text{ kPa}\).

[1] Award 1 mark for the correct option D. Reject all other options.

The ground-state electron configuration of a neutral iron atom is \([Ar]\,3d^6\,4s^2\). When forming transition metal cations, electrons are lost from the outermost \(s\)-subshell (\(4s\)) before the \(d\)-subshell (\(3d\)). To form the \(\text{Fe}^{3+}\) ion, three electrons are removed: two from the \(4s\) subshell and one from the \(3d\) subshell, resulting in the configuration \([Ar]\,3d^5\).

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

In the hydronium ion, \(H_3O^+\), the central oxygen atom has 6 valence electrons, plus 3 from the hydrogen atoms, minus 1 for the positive charge, giving 8 valence electrons (4 pairs). These 4 electron domains adopt a tetrahedral arrangement. With 3 bonding pairs and 1 lone pair, the molecular geometry is trigonal pyramidal. In contrast, \(BF_3\) is trigonal planar, \(ClF_3\) is T-shaped, and \(SF_4\) is seesaw (distorted tetrahedral).

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

The overall rate of a reaction is determined by its slowest step, also known as the rate-determining step. For this mechanism, Step 1 is the slow step. The rate equation for this elementary step is directly derived from its reactant stoichiometry: \(\text{Rate} = k[NO_2][NO_2] = k[NO_2]^2\). Therefore, the overall rate expression is \(\text{Rate} = k[NO_2]^2\).

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

An acidic buffer solution contains a weak acid and its conjugate base in comparable amounts. When equal volumes of the reactants in option B are mixed, the limiting reactant is the strong base \(NaOH\). It reacts completely with half of the weak acid \(CH_3COOH\) to produce \(CH_3COONa\) (the conjugate base). This leaves an equimolar mixture of unreacted weak acid \(CH_3COOH\) and its conjugate base \(CH_3COO^-\), establishing a buffer system. Option A produces a neutral salt, option C completely neutralizes the acid to its salt (not a buffer), and option D leaves an excess of strong base.

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

Reduction involves a decrease in the oxidation state of an element. In the reactant \(KMnO_4\), the oxidation state of manganese is \(+7\) (calculated as: \(+1 + x + 4(-2) = 0 \Rightarrow x = +7\)). In the product \(MnSO_4\), manganese is present as the \(\text{Mn}^{2+}\) ion with an oxidation state of \(+2\). Since its oxidation state decreases from \(+7\) to \(+2\), manganese is reduced. Carbon undergoes oxidation, increasing from \(+3\) in \(H_2C_2O_4\) to \(+4\) in \(CO_2\).

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

An ester is named as an alkyl alkanoate. The alkyl group (derived from the parent alcohol) is attached to the oxygen atom, which is a single-carbon methyl group (\(-CH_3\)). The alkanoate group (derived from the parent carboxylic acid) has a carbon chain of three carbons (\(CH_3CH_2COO-\)), corresponding to propanoate. Therefore, the systematic IUPAC name is methyl propanoate.

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

(a) Assumptions of an ideal gas:
1. The volume occupied by the gas molecules themselves is negligible compared to the total volume of the container.
2. There are no intermolecular forces of attraction or repulsion between the gas particles.

(b) Using the ideal gas equation: \(PV = nRT\)
Convert units to SI:
\(P = 100.8\text{ kPa} = 100.8 \times 10^3\text{ Pa}\)
\(V = 250.0\text{ cm}^3 = 2.500 \times 10^{-4}\text{ m}^3\)
\(T = 98.5 + 273.15 = 371.65\text{ K}\)
\(n = \frac{PV}{RT} = \frac{(100.8 \times 10^3\text{ Pa}) \times (2.500 \times 10^{-4}\text{ m}^3)}{8.314\text{ J K}^{-1}\text{ mol}^{-1} \times 371.65\text{ K}} = 0.008154\text{ mol} \approx 0.00815\text{ mol}\)

(c) \(M = \frac{m}{n} = \frac{0.842\text{ g}}{0.008154\text{ mol}} = 103.3\text{ g mol}^{-1} \approx 103\text{ g mol}^{-1}\)

(d) Calculated molar mass: \(M_{\text{calc}} = \frac{m R T}{P V_{\text{calc}}}\)
Actual molar mass: \(M_{\text{actual}} = \frac{m R T}{P V_{\text{actual}}}\)
Percentage error = \(\left| \frac{M_{\text{calc}} - M_{\text{actual}}}{M_{\text{actual}}} \right| \times 100\%
= \left| \frac{\frac{1}{V_{\text{calc}}} - \frac{1}{V_{\text{actual}}}}{\frac{1}{V_{\text{actual}}}} \right| \times 100\%
= \left| \frac{V_{\text{actual}} - V_{\text{calc}}}{V_{\text{calc}}} \right| \times 100\%
= \frac{|245.0 - 250.0|}{250.0} \times 100\% = 2.00\%\).
Since a larger volume was used in the denominator, the calculated molar mass is underestimated.

(e) At high pressures, gas molecules are forced close together. Intermolecular forces become significant, attracting particles to each other, which reduces the force and frequency of collisions with the container walls.

(a) Award [1] for each correct assumption up to [2 max]:
- Gas molecules have negligible volume / are point masses.
- No intermolecular forces between particles.
- Collisions are perfectly elastic.

(b) Award [3] total:
- [1] for converting \(T\) to \(371.65\text{ K}\) or \(372\text{ K}\).
- [1] for converting \(V\) to \(2.500 \times 10^{-4}\text{ m}^3\).
- [1] for correct final answer of \(0.00815\text{ mol}\) (allow \(0.0081\) to \(0.0082\)).

(c) Award [2] total:
- [1] for correct division formula of mass by calculated moles.
- [1] for correct final answer of \(103\text{ g mol}^{-1}\) (allow ECF from b).

(d) Award [3.25] total:
- [1] for stating "underestimated".
- [2.25] for calculating the correct percentage error of \(2.00\%\) (award [1.25] for partially correct working or using the individual masses to get \(2.0\%\)).

(e) Award [1] for stating that close proximity of molecules at high pressure makes intermolecular forces significant, pulling particles together and reducing collision pressure.

(a) First ionization energy is the energy required to remove one mole of electrons [1] from one mole of gaseous atoms to form one mole of gaseous singly charged positive ions [1].

(b) (i) Copper atom: \(\text{Cu}: 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\)
(ii) Copper(II) ion: \(\text{Cu}^{2+}: 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\) (electrons are lost from the \(4\text{s}\) subshell first).

(c) Copper has an electron configuration of \([Ar] 3\text{d}^{10} 4\text{s}^1\) instead of \([Ar] 3\text{d}^9 4\text{s}^2\). This is because the completely filled \(3\text{d}\) subshell is structurally more symmetrical and lower in energy, which provides greater electronic stability.

(d) (i) Phosphorus has the valence configuration \(3\text{s}^2 3\text{p}^3\) (three unpaired electrons in the \(3\text{p}\) orbitals). Sulfur has the configuration \(3\text{s}^2 3\text{p}^4\) (one paired set of electrons in a \(3\text{p}\) orbital). The repulsion between the two paired electrons in the same orbital in sulfur makes it easier to remove one electron compared to the singly-occupied orbitals of phosphorus.

(ii) Across Period 3 (from Na to Ar), the number of protons (nuclear charge) increases, while the shielding effect remains approximately constant because electrons are added to the same main energy level. As a result, the atomic radius decreases, and the effective nuclear charge increases, holding the outermost valence electrons more strongly.

(a) Award [2] total:
- [1] for "energy to remove one mole of electrons from one mole of atoms".
- [1] for "gaseous state" (atoms and ions).

(b) (i) Award [1] for \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\) or \([Ar] 3\text{d}^{10} 4\text{s}^1\).
(ii) Award [1] for \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\) or \([Ar] 3\text{d}^9\).

(c) Award [1.25] for explaining that a full \(3\text{d}^{10}\) subshell offers enhanced stability/lower energy state compared to \(3\text{d}^9 4\text{s}^2\).

(d) (i) Award [3] total:
- [1] for stating that phosphorus has singly occupied \(3\text{p}\) orbitals while sulfur has one paired \(3\text{p}\) orbital.
- [1] for identifying "inter-electron repulsion" or "spin-pairing repulsion" in sulfur.
- [1] for concluding that less energy is required to remove the electron from sulfur.

(ii) Award [3] total:
- [1] for increasing nuclear charge / more protons.
- [1] for constant shielding / same number of inner shells.
- [1] for stronger electrostatic attraction on outer electrons / smaller atomic radius.

(a) A Brønsted-Lowry acid is a proton (\(\text{H}^+\)) donor [1]. The conjugate base of lactic acid is the lactate ion, \(\text{CH}_3\text{CH(OH)COO}^-\) [1].

(b) (i) \(K_{\text{a}} = \frac{[\text{CH}_3\text{CH(OH)COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH(OH)COOH}]}\) [1]
(ii) Let \([\text{H}^+] = [\text{CH}_3\text{CH(OH)COO}^-] = x\).
Assumption: \([\text{CH}_3\text{CH(OH)COOH}]_{\text{eq}} \approx 0.150\text{ mol dm}^{-3}\) (dissociation is negligible) [1].
\(1.38 \times 10^{-4} = \frac{x^2}{0.150} \implies x^2 = 2.07 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}\) [1].
\(x = [\text{H}^+] = \sqrt{2.07 \times 10^{-5}} = 4.55 \times 10^{-3}\text{ mol dm}^{-3}\) [1].
\(\text{pH} = -\log_{10}(4.55 \times 10^{-3}) = 2.34\) [1].

(c) (i) Initial amount of lactic acid (HA) = \(0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 0.00750\text{ mol}\).
Initial amount of \(\text{NaOH}\) added = \(0.0500\text{ dm}^3 \times 0.0800\text{ mol dm}^{-3} = 0.00400\text{ mol}\).
Reaction: \(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\).
Remaining acid = \(0.00750 - 0.00400 = 0.00350\text{ mol}\) [1].
Produced lactate ion (\(\text{A}^-\)) = \(0.00400\text{ mol}\) [1].
Total volume of mixture = \(100.0\text{ cm}^3 = 0.100\text{ dm}^3\).
\([\text{HA}] = \frac{0.00350\text{ mol}}{0.100\text{ dm}^3} = 0.0350\text{ mol dm}^{-3}\).
\([\text{A}^-] = \frac{0.00400\text{ mol}}{0.100\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}\) [0.25 for both concentration calculations].

(ii) \(\text{p}K_{\text{a}} = -\log_{10}(1.38 \times 10^{-4}) = 3.86\) [1].
Using Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_{\text{a}} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 3.86 + \log_{10}\left(\frac{0.0400}{0.0350}\right) = 3.86 + 0.058 = 3.92\) [1].

(a) Award [2] total:
- [1] for "proton donor" or "\(\text{H}^+\)-donor".
- [1] for "\(\text{CH}_3\text{CH(OH)COO}^-\) / lactate ion".

(b) (i) Award [1] for correct expression.
(ii) Award [4] total:
- [1] for stating correct assumption.
- [1] for setting up correct equation \(x^2 = K_{\text{a}} \times [\text{HA}]\).
- [1] for \([\text{H}^+] = 4.55 \times 10^{-3}\text{ mol dm}^{-3}\).
- [1] for \(\text{pH} = 2.34\).

(c) (i) Award [2.25] total:
- [1] for calculating remaining moles of acid \(0.00350\text{ mol}\).
- [1] for calculating moles of conjugate base formed \(0.00400\text{ mol}\).
- [0.25] for dividing both values by \(0.100\text{ dm}^3\) to get \([\text{HA}] = 0.0350\text{ M}\) and \([\text{A}^-] = 0.0400\text{ M}\).

(ii) Award [2] total:
- [1] for calculating \(\text{p}K_{\text{a}} = 3.86\).
- [1] for final \(\text{pH} = 3.92\) (allow ECF from (c)(i)).

(a) (i) Comparing Exp 1 and Exp 2: \([\text{H}_2]\) is kept constant. As \([\text{NO}]\) doubles, the rate increases by a factor of \(\frac{4.80 \times 10^{-4}}{1.20 \times 10^{-4}} = 4 = 2^2\). Thus, the order with respect to \(\text{NO}\) is 2 [1.5].
(ii) Comparing Exp 2 and Exp 3: \([\text{NO}]\) is kept constant. As \([\text{H}_2]\) doubles, the rate increases by a factor of \(\frac{9.60 \times 10^{-4}}{4.80 \times 10^{-4}} = 2 = 2^1\). Thus, the order with respect to \(\text{H}_2\) is 1 [1.5].

(b) Rate Law: \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\) [1].
Value of \(k\): \(k = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]} = \frac{1.20 \times 10^{-4}}{(0.010)^2 \times (0.010)} = 120\) [1].
Units of \(k\): \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) [1.25].

(c) A suitable mechanism must yield the rate law \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\).
Step 1: \(2\text{NO}(g) \rightleftharpoons \text{N}_2\text{O}_2(g)\) (fast equilibrium)
Step 2: \(\text{N}_2\text{O}_2(g) + \text{H}_2(g) \rightarrow \text{N}_2\text{O}(g) + \text{H}_2\text{O}(g)\) (slow / Rate-Determining Step) [1]
Step 3: \(\text{N}_2\text{O}(g) + \text{H}_2(g) \rightarrow \text{N}_2(g) + \text{H}_2\text{O}(g)\) (fast) [1]
Overall reaction matches: \(2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}\). Since the RDS involves \(\text{N}_2\text{O}_2\) (derived from \(2\text{NO}\)) and \(\text{H}_2\), the rate law is correctly predicted [1].

(d) The rate constant \(k\) increases [1]. A catalyst provides an alternative pathway with a lower activation energy (\(E_{\text{a}}\)). According to collision theory, a larger fraction of reactant collisions will possess kinetic energy greater than or equal to this lower activation energy [1].

(a) (i) Award [1.5] for showing the calculation/logic leading to order of 2.
(ii) Award [1.5] for showing the calculation/logic leading to order of 1.

(b) Award [3.25] total:
- [1] for the correct rate law expression.
- [1] for the correct value \(120\).
- [1.25] for the correct units \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

(c) Award [3] total:
- [1] for writing a consistent series of elementary steps.
- [1] for identifying the slow step (RDS).
- [1] for demonstrating that the steps correctly add up to the overall chemical equation.

(d) Award [2] total:
- [1] for stating "rate constant increases".
- [1] for explaining that a catalyst lowers activation energy, leading to a higher fraction of successful/effective collisions.

(a) Since copper has a more positive standard reduction potential (\(+0.34\text{ V}\)), it is reduced at the cathode [1]. Chromium has a more negative potential (\(-0.74\text{ V}\)), so it is oxidized at the anode [1].
To balance the electron transfer (6 electrons), the half-equations are multiplied and added:
\(2\text{Cr}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cu}(s)\) [1].

(b) \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +0.34\text{ V} - (-0.74\text{ V}) = +1.08\text{ V}\) [1].

(c) Using \(\Delta G^\theta = -nFE^\theta_{\text{cell}}\):
\(n = 6\) (moles of electrons transferred in the balanced equation) [1].
\(\Delta G^\theta = -6 \times 96485\text{ C mol}^{-1} \times 1.08\text{ V} = -625123\text{ J mol}^{-1} = -625\text{ kJ mol}^{-1}\) [2.25].

(d) (i) The salt bridge completes the circuit [0.75] and allows the migration of ions to maintain overall charge neutrality in both half-cells [0.75].

(ii) In the anode half-cell, oxidation occurs producing \(\text{Cr}^{3+}\) ions, causing an accumulation of positive charge. Anions from the salt bridge migrate into this half-cell to balance the charge [1.25]. In the cathode half-cell, reduction of \(\text{Cu}^{2+}\) occurs, causing a deficit of positive charge. Cations from the salt bridge migrate into this half-cell to balance the charge [1.25].

(a) Award [3] total:
- [1] for identifying copper as the cathode.
- [1] for identifying chromium as the anode.
- [1] for the correctly balanced overall chemical equation.

(b) Award [1] for \(+1.08\text{ V}\) (must include the correct sign).

(c) Award [3.25] total:
- [1] for correctly identifying \(n = 6\).
- [2.25] for the correct final answer of \(-625\text{ kJ mol}^{-1}\) (or \(-6.25 \times 10^5\text{ J mol}^{-1}\)). Award [1.25] if the only error is using \(n = 1, 2\), or \(3\) (producing ECF values).

(d) (i) Award [1.5] total:
- [0.75] for completing the electrical circuit.
- [0.75] for maintaining charge/ionic neutrality.
(ii) Award [2.5] total:
- [1.25] for explaining anion migration towards the anode (chromium half-cell) to counter positive charge build-up.
- [1.25] for explaining cation migration towards the cathode (copper half-cell) to counter the loss of copper ions.

(a) Full structural formula:
\(\text{H}-\text{O}-\text{C}(\text{H}_2)-\text{C}(\text{H}_2)-\text{C}(\text{H}_2)-\text{C}(=\text{O})-\text{O}-\text{H}\) [1].
The two functional groups are:
1. Carboxyl (or carboxylic acid group) [1].
2. Hydroxyl (or alcohol group) [1].

(b) (i) The cyclic ester (lactone) formed is \(\gamma\)-butyrolactone:
It consists of a 5-membered ring containing 4 carbon atoms and 1 oxygen atom, with one of the ring carbons adjacent to the oxygen having a double-bonded oxygen (carbonyl group) [2].

(ii) This reaction is an intramolecular condensation (or esterification / nucleophilic substitution) [1]. It proceeds readily because a 5-membered ring has low ring strain (bond angles close to the ideal tetrahedral angle of \(109.5^\circ\)), making it thermodynamically and kinetically favorable [1.25].

(c) To distinguish the two isomers:
Add aqueous sodium hydrogencarbonate (\(\text{NaHCO}_3(aq)\)) or sodium carbonate (\(\text{Na}_2\text{CO}_3(aq)\)) to separate samples of each compound [1].
- 4-hydroxybutanoic acid (which contains a carboxylic acid group) will react to produce carbon dioxide gas, observed as effervescence/bubbles [1.5].
- Diethyl carbonate (which does not contain an acidic carboxyl group) will show no visible reaction [1.5].
(Alternative: Use a pH indicator/probe; 4-hydroxybutanoic acid will be acidic while diethyl carbonate is neutral).

(a) Award [3] total:
- [1] for drawing correct full structural formula showing all atoms and bonds.
- [1] for identifying carboxyl.
- [1] for identifying hydroxyl.

(b) (i) Award [2] for drawing a correct 5-membered cyclic ester ring.
(ii) Award [2.25] total:
- [1] for identifying intramolecular condensation or esterification.
- [1.25] for explaining that a 5-membered ring minimizes ring strain / has favorable bond angles.

(c) Award [4] total:
- [1] for specifying a suitable reagent (e.g., \(\text{NaHCO}_3(aq)\), \(\text{Na}_2\text{CO}_3(aq)\), or reactive metal like \(\text{Na}\)).
- [1.5] for the positive observation with 4-hydroxybutanoic acid (effervescence / gas evolution).
- [1.5] for the correct negative observation with diethyl carbonate (no visible reaction).

(a) (i) \(\text{SO}_2\) Lewis structure:
Central \(\text{S}\) has 1 lone pair, and double bonds to both \(\text{O}\) atoms (expanding sulfur's octet to 10 electrons): \(\text{O}=\ddot{\text{S}}=\text{O}\). (Alternatively, a structure with formal charges showing one single and one double bond is acceptable as a contributing resonance structure) [2].

(ii) \(\text{CO}_3^{2-}\) resonance structures:
Three equivalent Lewis structures must be drawn. Each structure shows a central carbon with no lone pairs, double-bonded to one oxygen and single-bonded to two oxygens (which carry negative charges). The structures should be enclosed in brackets with a \(2-\)\ charge [3].

(b) (i) \(\text{SO}_2\) has 3 electron domains around the sulfur atom (2 bonding domains, 1 lone pair domain).
- Electron domain geometry: Trigonal planar [1].
- Molecular geometry: Bent (or V-shaped) [1].

(ii) The expected bond angle is less than \(120^\circ\) (typically \(119^\circ\)) [1].
Explanation: According to VSEPR theory, lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion. The lone pair on the sulfur atom repels the double bonds, forcing them closer together [1.25].

(c) The carbonate ion exhibits resonance, meaning that rather than having one discrete double bond and two single bonds, the \(\pi\) electrons are delocalized equally over all three carbon-oxygen bonds [1]. Consequently, all three \(\text{C-O}\) bonds are identical in length and strength, and are intermediate in character between a single and a double bond (bond order of \(1.33\)) [1].

(a) (i) Award [2] for a correct Lewis structure of \(\text{SO}_2\) showing all valence electrons.
(ii) Award [3] for drawing all three correct resonance structures with brackets and negative charges.

(b) (i) Award [1] for trigonal planar, and [1] for bent.
(ii) Award [2.25] total:
- [1] for predicting a bond angle < \(120^\circ\) (accept \(115^\circ\) to \(119^\circ\)).
- [1.25] for explaining that lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion.

(c) Award [2] total:
- [1] for stating that electrons are delocalized across all three bonds.
- [1] for explaining that all bonds are equal in length and strength (intermediate between single and double / bond order of 1.33).

(a) \(\Delta S^\theta = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants})\)
\(\Delta S^\theta = [S^\theta(\text{CaO}(s)) + S^\theta(\text{CO}_2(g))] - S^\theta(\text{CaCO}_3(s))\)
\(\Delta S^\theta = [39.7 + 213.7] - 92.9 = +160.5\text{ J K}^{-1}\text{ mol}^{-1}\) [2].

(b) The positive sign of \(\Delta S^\theta\) indicates an increase in disorder [0.5]. This is because a solid reactant is converted into a solid and a gas. Gaseous molecules have much higher vibrational, translational, and rotational freedom (positional entropy) than solid molecules [1].

(c) \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)
Convert \(\Delta S^\theta\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\): \(\Delta S^\theta = +0.1605\text{ kJ K}^{-1}\text{ mol}^{-1}\) [1].
\(\Delta G^\theta = 178.3\text{ kJ mol}^{-1} - (298\text{ K} \times 0.1605\text{ kJ K}^{-1}\text{ mol}^{-1}) = 178.3 - 47.83 = +130.5\text{ kJ mol}^{-1}\) [1].
Since \(\Delta G^\theta > 0\), the reaction is non-spontaneous at \(298\text{ K}\) [1].

(d) For the reaction to become spontaneous, \(\Delta G^\theta < 0\), which occurs when \(T\Delta S^\theta > \Delta H^\theta\) [1].
\(T_{\text{min}} = \frac{\Delta H^\theta}{\Delta S^\theta} = \frac{178.3\text{ kJ mol}^{-1}}{0.1605\text{ kJ K}^{-1}\text{ mol}^{-1}}\) [1].
\(T_{\text{min}} = 1110.9\text{ K} \approx 1111\text{ K}\) [1.25].

(e) The relationship is \(\Delta G^\theta = -RT \ln K\) [0.5].
At \(298\text{ K}\), \(\Delta G^\theta\) is highly positive, meaning \(K \ll 1\). Thus, the equilibrium lies far to the left (reactants are heavily favored) [1].

(a) Award [2] for correct calculation of \(+160.5\text{ J K}^{-1}\text{ mol}^{-1}\) with correct units.

(b) Award [1.5] total:
- [0.5] for identifying that entropy increases.
- [1] for explaining that gases are much more disordered than solids.

(c) Award [3] total:
- [1] for correct conversion of units (either \(\Delta H\) to J or \(\Delta S\) to kJ).
- [1] for correct numerical calculation of \(+130.5\text{ kJ mol}^{-1}\).
- [1] for concluding that the reaction is non-spontaneous because \(\Delta G^\theta > 0\).

(d) Award [3.25] total:
- [1] for setting \(\Delta G^\theta = 0\) to find the threshold temperature.
- [1] for substituting correct values into the expression \(T = \frac{\Delta H}{\Delta S}\).
- [1.25] for the correct final temperature of \(1111\text{ K}\) (or \(1110.9\text{ K}\)).

(e) Award [1.5] total:
- [0.5] for the formula \(\Delta G^\theta = -RT \ln K\).
- [1] for stating that the equilibrium lies far to the left (reactants favored).

(a) \(\Delta T = 37.8\text{ }^\circ\text{C} - 21.2\text{ }^\circ\text{C} = 16.6\text{ }^\circ\text{C}\).
Since subtraction was performed, the absolute uncertainties are added: \(0.1 + 0.1 = \pm 0.2\text{ }^\circ\text{C}\).

(b) Percentage uncertainty in mass: \(\frac{0.5}{50.0} \times 100\% = 1.0\%\).
Percentage uncertainty in \(\Delta T\): \(\frac{0.2}{16.6} \times 100\% = 1.2\%\).
Percentage uncertainty in \(q\): \(1.0\% + 1.2\% = 2.2\%\).

(c) Heat energy released, \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 16.6\text{ K} = 3469.4\text{ J} = 3.47\text{ kJ}\).
Moles of \(\text{CaCl}_2\): \(n = \frac{5.00\text{ g}}{110.98\text{ g mol}^{-1}} = 0.04505\text{ mol}\).
Enthalpy change of solution, \(\Delta H_{\text{sol}} = -\frac{3.4694\text{ kJ}}{0.04505\text{ mol}} = -77.0\text{ kJ mol}^{-1}\) (exothermic, hence negative sign).

(d) Systematic error: Heat loss to surroundings (resulting in a lower maximum temperature and a less exothermic value).
Modification: Place a well-fitting lid on the polystyrene cup, double-cup the polystyrene calorimeter for better insulation, or plot a temperature-time cooling curve and extrapolate to the mixing time.

(a) [1.5 marks]
- \(16.6\text{ }^\circ\text{C}\) [0.5 marks]
- \(\pm 0.2\text{ }^\circ\text{C}\) [1 mark]

(b) [2 marks]
- Correct calculation of percentage uncertainty for \(m\) and \(\Delta T\) (\(1.0\%\) and \(1.2\%\)) [1 mark]
- Overall percentage uncertainty is \(2.2\%\) [1 mark]

(c) [3 marks]
- \(q = 3.47\text{ kJ}\) (or \(3469.4\text{ J}\)) [1 mark]
- \(n(\text{CaCl}_2) = 0.04505\text{ mol}\) [1 mark]
- \(\Delta H_{\text{sol}} = -77.0\text{ kJ mol}^{-1}\) (accept range \(-76.9\) to \(-77.1\)) [1 mark] (award mark only if negative sign is included)

(d) [1 mark]
- Identification of heat loss AND addition of lid/insulation OR extrapolation of temperature from cooling curve [1 mark]. Accept 'heat absorbed by calorimeter' and 'determine calorimeter heat capacity'.

(a) Convert parameters to SI units:
\(V = 48.0\text{ cm}^3 = 48.0 \times 10^{-6}\text{ m}^3\)
\(P = 100\text{ kPa} = 1.00 \times 10^5\text{ Pa}\)
\(T = 298\text{ K}\)
\(n = \frac{PV}{RT} = \frac{1.00 \times 10^5 \times 48.0 \times 10^{-6}}{8.31 \times 298} = \frac{4.80}{2476.38} = 1.94 \times 10^{-3}\text{ mol}\).

(b) The molar volume of an ideal gas at \(298\text{ K}\) and \(100\text{ kPa}\) is:
\(V_{\text{m}} = \frac{RT}{P} = \frac{8.31 \times 298}{1.00 \times 10^5} = 0.02476\text{ m}^3\text{ mol}^{-1} = 24.8\text{ dm}^3\text{ mol}^{-1} = 24800\text{ cm}^3\text{ mol}^{-1}\).
Rate of \(\text{O}_2\) production in \(\text{mol s}^{-1}\):
\(\text{Rate} = \frac{0.80\text{ cm}^3\text{ s}^{-1}}{24800\text{ cm}^3\text{ mol}^{-1}} = 3.23 \times 10^{-5}\text{ mol s}^{-1}\).
According to the stoichiometry, \(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\), the rate of consumption of \(\text{H}_2\text{O}_2\) is twice the rate of production of \(\text{O}_2\):
\(\text{Rate of } \text{H}_2\text{O}_2 \text{ consumption} = 2 \times 3.23 \times 10^{-5}\text{ mol s}^{-1} = 6.45 \times 10^{-5}\text{ mol s}^{-1}\) (or \(6.5 \times 10^{-5}\text{ mol s}^{-1}\)).

(c) Carbon dioxide is moderately soluble in water because it undergoes a chemical reaction to establish an equilibrium with carbonic acid (\(\text{CO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{CO}_3(aq)\)), causing a significant loss in the volume of gas collected. Oxygen gas is non-polar and has extremely low solubility in water, meaning negligible gas is dissolved and the downward displacement of water provides highly accurate results.

(a) [2 marks]
- Correctly converting volume to \(\text{m}^3\) or using standard molar volume [1 mark]
- Correct calculation of moles: \(1.94 \times 10^{-3}\text{ mol}\) (accept range \(1.93 \times 10^{-3}\) to \(1.94 \times 10^{-3}\)) [1 mark]

(b) [2.5 marks]
- Finding molar volume at experimental conditions (\(24.8\text{ dm}^3\text{ mol}^{-1}\) or equivalent) [0.5 marks]
- Determining molar rate of oxygen production: \(3.23 \times 10^{-5}\text{ mol s}^{-1}\) [1 mark]
- Multiplying by 2 to get consumption rate of hydrogen peroxide: \(6.45 \times 10^{-5}\text{ mol s}^{-1}\) (accept \(6.5 \times 10^{-5}\)) [1 mark]

(c) [3 marks]
- Identifies that \(\text{CO}_2\) is soluble in water / reacts with water [1 mark]
- States that this solubility decreases the volume of \(\text{CO}_2\) collected over water, making the displacement method inaccurate [1 mark]
- Explains that \(\text{O}_2\) is non-polar / has very low solubility, so minimal gas is lost to the water, making both methods reliable [1 mark]

**(a)**
* Reagent: Ethanoic anhydride (or acetyl chloride). [1 mark]
* Catalyst: Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) or concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)). [1 mark]

**(b)**
* Dissolve the crude aspirin in the minimum volume of a hot solvent (e.g., ethanol or a mixture of water and ethanol). [1 mark]
* Allow the solution to cool slowly so that pure aspirin recrystallizes while impurities remain dissolved, then filter the crystals and wash with cold solvent. [1 mark]

**(c)**
* Purified aspirin will melt at a higher temperature (closer to the literature value of 136 °C). [1 mark]
* It will have a much sharper/narrower melting point range (typically 1–2 °C), whereas impure aspirin melts over a lower and wider temperature range. [1 mark]

**(a)**
* Award [1] for identifying ethanoic anhydride / acetyl chloride (accept ethanoic acid).
* Award [1] for identifying concentrated sulfuric acid / phosphoric acid.

**(b)**
* Award [1] for dissolving in the **minimum volume** of **hot solvent**.
* Award [1] for cooling to crystallize and filtering the crystals (and washing with cold solvent).

**(c)**
* Award [1] for stating that the purified product has a **higher** melting point.
* Award [1] for stating that the purified product has a **narrower** / **shorter** melting point range.

**(a)**
* Name of ring: **Beta-lactam ring** (or 2-azetidinone). [1 mark]
* Explanation of reactivity: The ring consists of one nitrogen and three carbon atoms with bond angles of approximately \(90^\circ\) (instead of the preferred \(109.5^\circ\) or \(120^\circ\) for \(\text{sp}^3\) and \(\text{sp}^2\) hybridized atoms). This causes severe **ring strain**, making the amide bond highly reactive. [1 mark]
* Mechanism of action: The ring easily opens and covalently binds to (and deactivates) the enzyme **transpeptidase**, which is responsible for cross-linking the peptidoglycan cell wall, leading to cell lysis. [1 mark]

**(b)**
* Resistance mechanism: Resistant bacteria produce the enzyme **beta-lactamase** (or penicillinase). [1 mark]
* Enzyme action: This enzyme hydrolyzes/opens the beta-lactam ring before it can reach the target, rendering the antibiotic inactive. [1 mark]
* Chemical modification: Chemists modify the **R side-chain** with bulky groups (creating steric hindrance, e.g., in methicillin) to prevent the beta-lactamase enzyme from binding to the drug, while still allowing the beta-lactam ring to access the bacterial cell wall enzymes. [1 mark]

**(a)**
* Award [1] for identifying the **beta-lactam ring**.
* Award [1] for explaining that the \(90^\circ\) bond angles create **high ring strain** / make the ring highly unstable/reactive.
* Award [1] for stating that the ring opens and inhibits the enzyme (transpeptidase) responsible for bacterial cell wall synthesis / cross-linking.

**(b)**
* Award [1] for stating that resistant bacteria produce **beta-lactamase** / **penicillinase**.
* Award [1] for explaining that this enzyme hydrolyzes/breaks the beta-lactam ring (rendering it inactive).
* Award [1] for stating that chemists modify the **R side-chain** / add bulky groups to cause steric hindrance (preventing enzyme access).

**(a)**
* Morphine contains two **hydroxyl** (\(-\text{OH}\)) groups (one phenolic hydroxyl and one secondary alcohol hydroxyl). [1 mark]
* In diamorphine, these are converted into **ester / ethanoate / acetate** groups. [10 mark]

**(b)**
* Solubility: Diamorphine is much more **lipid-soluble** (lipophilic / hydrophobic) than morphine, because the polar hydroxyl groups in morphine are replaced by less polar, more hydrophobic ester groups. [1 mark]
* Blood-brain barrier: Due to its high lipid solubility, diamorphine crosses the non-polar **blood-brain barrier** much more rapidly and efficiently than morphine. [1 mark]

**(c)**
* Medical advantage: It is highly potent, has a **faster onset of action**, and can be administered in smaller doses for severe pain management (such as terminal cancer pain). [1 mark]
* Risk/disadvantage: It has a much **higher potential for addiction / physical dependence** and a higher risk of fatal respiratory depression (overdose). [1 mark]

**(a)**
* Award [1] for **hydroxyl** (accept alcohol).
* Award [1] for **phenol** (accept phenolic hydroxyl / phenolic OH).

**(b)**
* Award [1] for stating that diamorphine is **more lipid-soluble / lipophilic** (due to having ester groups instead of polar hydroxyl groups).
* Award [1] for explaining that this allows diamorphine to **penetrate/cross the blood-brain barrier** more easily/faster.

**(c)**
* Award [1] for a valid medical advantage (e.g., faster pain relief, high potency, smaller doses required, effective for extreme terminal pain).
* Award [1] for a valid risk (e.g., high addiction liability, severe withdrawal symptoms, higher risk of overdose/respiratory depression).

**(a)**
* Equation: \(\text{Mg(OH)}_2\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + 2\text{H}_2\text{O(l)}\) [1 mark]

**(b)**
* **Step 1:** Calculate the concentration of \(\text{H}^+\) ions in the stomach acid:
\([\text{H}^+] = 10^{-\text{pH}} = 10^{-1.30} = 0.0501\text{ mol dm}^{-3}\) [1 mark]
* **Step 2:** Calculate the number of moles of \(\text{HCl}\) in \(150\text{ cm}^3\) (\(0.150\text{ dm}^3\)):
\(n(\text{HCl}) = 0.0501\text{ mol dm}^{-3} \times 0.150\text{ dm}^3 = 7.52 \times 10^{-3}\text{ mol}\)
* **Step 3:** Calculate the moles and mass of \(\text{Mg(OH)}_2\) needed:
\(n(\text{Mg(OH)}_2) = \frac{7.52 \times 10^{-3}}{2} = 3.76 \times 10^{-3}\text{ mol}\) [1 mark]
\(\text{mass} = 3.76 \times 10^{-3}\text{ mol} \times 58.33\text{ g mol}^{-1} = 0.219\text{ g}\) (or \(0.22\text{ g}\)) [1 mark]

**(c)**
* Antacids neutralize stomach acid **chemically** after it has already been produced/secreted into the stomach. [1 mark]
* Ranitidine acts as an **\(\text{H}_2\)-receptor antagonist** to **prevent/reduce the secretion/production** of stomach acid at the source. [1 mark]

**(a)**
* Award [1] for a fully balanced equation: \(\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}\).

**(b)**
* Award [1] for calculating \([\text{H}^+] = 0.0501\text{ mol dm}^{-3}\).
* Award [1] for determining \(n(\text{Mg(OH)}_2) = 3.76 \times 10^{-3}\text{ mol}\) (using the 1:2 molar ratio).
* Award [1] for the final mass of \(0.219\text{ g}\) (accept \(0.22\text{ g}\)).

**(c)**
* Award [1] for stating that antacids work by **neutralizing** existing acid.
* Award [1] for stating that ranitidine works by **preventing/reducing the secretion** of stomach acid (by blocking \(\text{H}_2\)/histamine receptors).

**(a)**
* Replication and release: The virus infects a host cell and uses its machinery to replicate. When new viral particles bud out, they are anchored to the host cell surface by glycoproteins binding to sialic acid. [1 mark]
* Role of Neuraminidase: The viral enzyme **neuraminidase** cleaves this sialic acid link, releasing the new viral particles so they can infect other cells. [1 mark]
* Action of Inhibitors: Neuraminidase inhibitors (like oseltamivir and zanamivir) mimic sialic acid and bind to the active site of neuraminidase, blocking its activity. This prevents the release of new viruses, limiting the spread of infection. [1 mark]

**(b)**
* **(i)** Zanamivir contains multiple **hydroxyl** (\(-\text{OH}\)) groups / a **guanidino** group (not present in oseltamivir). Oseltamivir has an **ester** group, whereas zanamivir has a **carboxylic acid** group. [1 mark]
* **(ii)**
* Zanamivir is highly polar (due to the many hydroxyl and guanidino groups), which leads to **low oral bioavailability** because it cannot easily pass through polar/non-polar cell membranes in the digestive tract. Thus, it must be administered by **inhalation** (directly to the lungs). [1 mark]
* Oseltamivir is administered as an **ester prodrug**, making it less polar and more easily absorbed by the digestive system, allowing for **oral administration** (as a pill/liquid). [1 mark]

**(a)**
* Award [1] for describing that new viral particles bud off but remain attached to the host cell membrane (via sialic acid/glycoproteins).
* Award [1] for explaining that **neuraminidase** is the enzyme that cleaves this link to release the virus.
* Award [1] for explaining that inhibitors bind to the active site of neuraminidase, blocking release/spread of the virus.

**(b)**
* **(i)** Award [1] for identifying a correct structural difference (e.g., zanamivir has **guanidino** group / multiple **hydroxyl** groups, while oseltamivir has ether/ester group).
* **(ii)**
* Award [1] for explaining that zanamivir's high polarity / many polar groups results in poor oral absorption, requiring **inhalation**.
* Award [1] for explaining that oseltamivir is less polar (or a prodrug) and can be taken **orally**.