2024 IB DP Chemistry 模擬試題連答案詳解

In a first-order reaction, the rate expression is given by \(\text{Rate} = k[A]\). This means the rate is directly proportional to the concentration of reactant A. Therefore, doubling the concentration of reactant A will double the rate of reaction. Doubling the volume of the vessel would halve the concentration, decreasing the rate. Changing the temperature or adding a catalyst changes the rate constant \(k\) exponentially rather than linearly.

Award 1 mark for the correct option B. Award 0 marks for any other option.

The hydronium ion, \(\text{H}_3\text{O}^+\), has 8 valence electrons (6 from oxygen, 3 from three hydrogen atoms, minus 1 for the positive charge). The central oxygen atom forms three single bonds with the hydrogen atoms and has one lone pair of electrons. This results in a tetrahedral electron-domain geometry and a trigonal pyramidal molecular geometry. \(\text{BF}_3\) is trigonal planar with no lone pairs on the central boron. \(\text{CO}_3^{2-}\) is trigonal planar. \(\text{CH}_4\) is tetrahedral.

Award 1 mark for the correct option B. Award 0 marks for any other option.

Let \(x\) be the oxidation state of nitrogen in each species: In \(\text{NH}_4^+\): \(x + 4(+1) = +1 \Rightarrow x = -3\). In \(\text{NO}_2^-\): \(x + 2(-2) = -1 \Rightarrow x = +3\). In \(\text{N}_2\text{O}\): \(2x + (-2) = 0 \Rightarrow x = +1\). In \(\text{NO}_3^-\): \(x + 3(-2) = -1 \Rightarrow x = +5\). Thus, nitrogen has the highest oxidation state in \(\text{NO}_3^-\) (\(+5\)).

Award 1 mark for the correct option D. Award 0 marks for any other option.

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). In this reaction, \(\text{HSO}_4^-\) acts as a Br\text{\o}nsted-Lowry acid by donating a proton to form its conjugate base, \(\text{SO}_4^{2-}\). therefore, \(\text{HSO}_4^-\) and \(\text{SO}_4^{2-}\) form a conjugate acid-base pair.

Award 1 mark for the correct option C. Award 0 marks for any other option.

Lattice enthalpy is defined as the enthalpy change that occurs when one mole of an ionic compound is broken down into its constituent gaseous ions under standard conditions. This is represented by the endothermic process: \(\text{NaCl}(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g)\). (Some curricula define it as the reverse exothermic process, but only option A correctly involves gaseous ions and the solid lattice in a 1:1 ratio matching this definition).

Award 1 mark for the correct option A. Award 0 marks for any other option.

Across period 3 from sodium to chlorine, the nuclear charge increases while the shielding effect remains relatively constant. This increases the effective nuclear charge felt by the valence electrons, which generally increases the first ionization energy. Atomic radius and metallic character decrease across the period. Ionic radius does not show a simple continuous increase across the entire period (it decreases for the metals, jumps up for the non-metals, and then decreases again).

Award 1 mark for the correct option C. Award 0 marks for any other option.

A homogeneous mixture has a uniform composition and a single phase throughout. An aqueous solution of sodium chloride is a homogeneous mixture because the dissolved ions are distributed uniformly throughout the water. Sand and water, oil and water, and sulfur with iron filings are all heterogeneous mixtures because they do not have a uniform composition and contain distinct phases.

Award 1 mark for the correct option B. Award 0 marks for any other option.

Doubling the volumes of both reactants doubles the number of moles of acid and base that react, which doubles the total heat energy released (\(q_{\text{new}} = 2q\)). However, the total volume of the reaction mixture also doubles (from \(100.0\text{ cm}^3\) to \(200.0\text{ cm}^3\)), which doubles the mass of solution being heated (\(m_{\text{new}} = 2m\)). Since \(\Delta T = \frac{q}{m \cdot c}\), the new temperature change is \(\Delta T_{\text{new}} = \frac{2q}{2m \cdot c} = \Delta T\). Thus, the temperature rise remains the same.

Award 1 mark for the correct option B. Award 0 marks for any other option.

The rate-determining step is the slowest step (Step 2), so the rate equation is: \(\text{Rate} = k_2 [\text{NOCl}_2][\text{NO}]\). Since \(\text{NOCl}_2\) is an intermediate, its concentration is determined from the fast pre-equilibrium in Step 1: \(K_c = \frac{[\text{NOCl}_2]}{[\text{NO}][\text{Cl}_2]}\) which gives \([\text{NOCl}_2] = K_c [\text{NO}][\text{Cl}_2]\). Substituting this into the rate equation of the slow step gives: \(\text{Rate} = k_2 K_c [\text{NO}][\text{Cl}_2][\text{NO}] = k [\text{NO}]^2[\text{Cl}_2]\).

Award 1 mark for the correct option B. Award 0 marks for other options.

The logarithmic form of the Arrhenius equation is: \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). This matches the equation of a straight line, \(y = mx + c\), where the gradient is \(m = -\frac{E_a}{R}\). Rearranging this expression to solve for activation energy gives \(E_a = -m R\).

Award 1 mark for the correct option A. Award 0 marks for other options.

Formal charge (FC) is calculated as: \(FC = V - N - \frac{1}{2}B\), where \(V\) is the number of valence electrons of the free atom, \(N\) is the number of non-bonding valence electrons, and \(B\) is the number of bonding electrons. For the ammonium ion, \(\text{NH}_4^+\), the central nitrogen atom has 5 valence electrons, 0 non-bonding electrons, and 8 bonding electrons (4 single bonds): \(FC = 5 - 0 - \frac{1}{2}(8) = +1\). For \(\text{BF}_3\), B has \(3 - 0 - 3 = 0\). For \(\text{CO}_3^{2-}\), C has \(4 - 0 - 4 = 0\). For \(\text{SF}_6\), S has \(6 - 0 - 6 = 0\).

Award 1 mark for the correct option C. Award 0 marks for other options.

For a spontaneous process, the standard cell potential must be positive. The half-reaction with the more positive standard reduction potential will undergo reduction at the cathode: \(\text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq) \quad E^\ominus = +0.77\text{ V}\). The other half-reaction is reversed as oxidation at the anode: \(\text{Sn}^{2+} (aq) \rightarrow \text{Sn}^{4+} (aq) + 2e^- \quad E^\ominus = +0.15\text{ V}\). The overall spontaneous reaction is \(2\text{Fe}^{3+} (aq) + \text{Sn}^{2+} (aq) \rightarrow 2\text{Fe}^{2+} (aq) + \text{Sn}^{4+} (aq)\), and the cell potential is \(E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +0.77 - (+0.15) = +0.62\text{ V}\).

Award 1 mark for the correct option B. Award 0 marks for other options.

The initial hydrogen ion concentration is \([\text{H}^+] = 10^{-2.0} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}\). Diluting the solution by a factor of 100 decreases the concentration to \([\text{H}^+]_{\text{new}} = \frac{1.0 \times 10^{-2}}{100} = 1.0 \times 10^{-4}\text{ mol dm}^{-3}\), which corresponds to a new pH of 4.0. Using \(K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\) at \(298\text{ K}\), the hydroxide ion concentration is \([\text{OH}^-] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10}\text{ mol dm}^{-3}\).

Award 1 mark for the correct option A. Award 0 marks for other options.

According to the Born-Haber cycle, the enthalpy of formation can be represented by: \(\Delta H_f^\ominus = \Delta H_{\text{at}}^\ominus(\text{Na}) + \Delta H_{\text{IE}}^\ominus(\text{Na}) + \Delta H_{\text{at}}^\ominus(\text{Cl}) + \Delta H_{\text{EA}}^\ominus(\text{Cl}) - \Delta H_{\text{lat}}^\ominus\). Substituting the given values: \(-411 = 107 + 496 + 122 - 349 - \Delta H_{\text{lat}}^\ominus\). This simplifies to: \(-411 = 376 - \Delta H_{\text{lat}}^\ominus\), which gives \(\Delta H_{\text{lat}}^\ominus = 376 - (-411) = +787\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option A. Award 0 marks for other options.

A Lewis acid-base reaction involves the donation of a pair of non-bonding electrons from a Lewis base to a species with an incomplete octet or vacant orbitals acting as a Lewis acid. In option A, the nitrogen atom in ammonia (\(\text{NH}_3\)) acts as a Lewis base and donates its lone pair of electrons to the boron atom in boron trifluoride (\(\text{BF}_3\)) to form an adduct containing a coordinate covalent bond. Options B and C are redox processes, and D is a precipitation reaction.

Award 1 mark for the correct option A. Award 0 marks for other options.

First, find the total mass of the mixture: \(m = (50.0 + 50.0)\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 100.0\text{ g}\). The heat energy released is \(q = m c \Delta T = 100.0 \times c \times \Delta T\text{ J} = 0.100 \times c \times \Delta T\text{ kJ}\). The amount in moles of reactants in the neutralization is: \(n = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). Because the temperature rises, the reaction is exothermic and the enthalpy change is negative: \(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{0.100 \times c \times \Delta T}{0.0500} = -2.00 \times c \times \Delta T\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option A. Award 0 marks for other options.

To find the rate expression, we determine the order of reaction with respect to each reactant. Comparing Experiment 1 and Experiment 2: the concentration of B is constant at \(0.10\text{ mol dm}^{-3}\), while the concentration of A doubles from \(0.10\) to \(0.20\text{ mol dm}^{-3}\). The initial rate increases by a factor of 4 (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)). Since \(2^2 = 4\), the reaction is second order with respect to A. Comparing Experiment 1 and Experiment 3: the concentration of A is constant at \(0.10\text{ mol dm}^{-3}\), while the concentration of B doubles from \(0.10\) to \(0.20\text{ mol dm}^{-3}\). The initial rate remains constant at \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is zero order with respect to B. Combining these results gives the rate expression: \(\text{Rate} = k[\text{A}]^2\).

Award [1] for the correct choice C. No partial marks are awarded.

A catalyst speeds up a chemical reaction by providing an alternative pathway with a lower activation energy (\(E_{\text{a}}\)). In a reversible reaction, it lowers the activation energy for both the forward and reverse reactions by the same amount, thus increasing both rates equally. It does not alter the position of equilibrium, the equilibrium constant, or the average kinetic energy of the particles (which depends only on temperature).

Award [1] for the correct choice C. No partial marks are awarded.

Formal charge (FC) is calculated as: \(\text{FC} = V - N - \frac{1}{2}B\), where \(V\) is the number of valence electrons of the free atom, \(N\) is the number of non-bonding valence electrons, and \(B\) is the number of bonding electrons. For the double-bonded oxygen atom (\(\text{O}_d\)): \(V = 6\), \(N = 4\) (two lone pairs), and \(B = 4\) (four electrons in two shared bonds). \(\text{FC} = 6 - 4 - \frac{1}{2}(4) = 0\). For the single-bonded oxygen atom (\(\text{O}_s\)): \(V = 6\), \(N = 6\) (three lone pairs), and \(B = 2\) (two electrons in one shared bond). \(\text{FC} = 6 - 6 - \frac{1}{2}(2) = -1\).

Award [1] for the correct choice A. No partial marks are awarded.

Both \(\text{H}_2\text{O}\) and \(\text{OF}_2\) have a central oxygen atom bonded to two other atoms and possess two lone pairs of electrons. According to VSEPR theory, this results in a tetrahedral electron domain geometry and a bent (V-shaped) molecular geometry. In contrast: \(\text{NH}_3\) is trigonal pyramidal whereas \(\text{BF}_3\) is trigonal planar; \(\text{CO}_2\) is linear whereas \(\text{SO}_2\) is bent; and \(\text{PCl}_5\) is trigonal bipyramidal whereas \(\text{SF}_6\) is octahedral.

Award [1] for the correct choice C. No partial marks are awarded.

The half-reaction with the more positive reduction potential (\(\text{Fe}^{3+}/\text{Fe}^{2+}\) at \(+0.77\text{ V}\)) occurs as reduction at the cathode. The half-reaction with the less positive potential (\(\text{Sn}^{4+}/\text{Sn}^{2+}\) at \(+0.15\text{ V}\)) must be reversed to occur as oxidation at the anode: \(\text{Sn}^{2+}(\text{aq}) \rightarrow \text{Sn}^{4+}(\text{aq}) + 2\text{e}^-\), with \(E^\theta_{\text{ox}} = -0.15\text{ V}\). Multiplying the reduction half-reaction by 2 to balance electrons and summing them gives the spontaneous cell reaction: \(2\text{Fe}^{3+}(\text{aq}) + \text{Sn}^{2+}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{Sn}^{4+}(\text{aq})\). The standard cell potential is calculated using \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V}\). Note that cell potentials are intensive properties and are not multiplied by stoichiometric coefficients.

Award [1] for the correct choice A. No partial marks are awarded.

First, find the hydrogen ion concentration: \([\text{H}^+] = 10^{-\text{pH}} = 1.0 \times 10^{-4}\text{ mol dm}^{-3}\). For a weak monoprotic acid \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), we assume that \([\text{H}^+] = [\text{A}^-] = 1.0 \times 10^{-4}\text{ mol dm}^{-3}\). Since the acid is weak, the concentration of undissociated acid at equilibrium remains approximately equal to the initial concentration: \([\text{HA}] \approx 0.10\text{ mol dm}^{-3}\). Substituting these into the acid dissociation constant expression: \(K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \approx \frac{(1.0 \times 10^{-4})^2}{0.10} = \frac{1.0 \times 10^{-8}}{1.0 \times 10^{-1}} = 1.0 \times 10^{-7}\text{ mol dm}^{-3}\).

Award [1] for the correct choice B. No partial marks are awarded.

Lattice enthalpy (\(\Delta H^\theta_{\text{lat}}\)) is defined as the enthalpy change when one mole of an crystalline ionic compound is broken down into its constituent gaseous ions under standard conditions: \(\text{NaCl}(s) \rightarrow \text{Na}^+(\text{g}) + \text{Cl}^-(\text{g})\). Option B represents the standard enthalpy of solution. Option C represents the standard enthalpy of formation. Option D represents the standard enthalpy of hydration of gaseous ions (into solution, although written incorrectly here).

Award [1] for the correct choice A. No partial marks are awarded.

Across Period 3 from left to right, the oxides transition from basic to amphoteric to acidic. \(\text{Na}_2\text{O}\) and \(\text{MgO}\) are basic oxides. \(\text{Al}_2\text{O}_3\) is amphoteric (reacting with both acids and bases). \(\text{SiO}_2\), \(\text{P}_4\text{O}_{10}\), and \(\text{SO}_3\) are acidic oxides. Therefore, classification B is correct as \(\text{SO}_3\) is acidic and \(\text{Al}_2\text{O}_3\) is amphoteric.

Award [1] for the correct choice B. No partial marks are awarded.

In sulfur dioxide (\(SO_2\)), the sulfur central atom has 6 valence electrons. It forms one double bond and one coordinate/single covalent bond with two oxygen atoms, leaving one lone pair on the sulfur atom. This gives a steric number of 3 (two bonding domains and one non-bonding domain). Therefore, the hybridization of sulfur is \(sp^2\). The three electron domains adopt a trigonal planar electron-domain geometry, but due to the presence of the lone pair, the molecular geometry is bent (V-shaped).

[1 mark] for B. No partial marks.

Halving the volume of the reaction vessel doubles the concentration of all reactants in the gas phase or in solution: \([A]_{\text{new}} = 2[A]\) and \([B]_{\text{new}} = 2[B]\). Substituting these into the rate expression: \(\text{rate}_{\text{new}} = k(2[A])^2(2[B]) = k \cdot 4[A]^2 \cdot 2[B] = 8 \cdot k[A]^2[B] = 8 \cdot \text{rate}_{\text{old}}\). Therefore, the rate increases by a factor of 8.

[1 mark] for selecting C. No partial credit.

The reduction potential of copper (\(+0.34\text{ V}\)) is more positive than that of iron (\(-0.45\text{ V}\)), so copper ions will be reduced and metallic iron will be oxidized. Reduction: \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\). Oxidation: \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\). Overall spontaneous reaction: \(Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)\). The standard cell potential is: \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.34\text{ V} - (-0.45\text{ V}) = +0.79\text{ V}\).

[1 mark] for selecting B. No partial credit.

A conjugate acid-base pair consists of two species that differ by exactly one hydrogen ion (\(H^+\)), where the acid has one more proton than the base. In option B, \(H_2PO_4^-\) is the acid and \(HPO_4^{2-}\) is the conjugate base, which is correct because: \(H_2PO_4^-(aq) \rightleftharpoons HPO_4^{2-}(aq) + H^+(aq)\). In option A, the acid should be \(H_3O^+\) and the base \(H_2O\). In option C, the acid should be \(NH_4^+\) and the base \(NH_3\). In option D, the conjugate base of \(H_2SO_4\) is \(HSO_4^-\), not \(SO_4^{2-}\).

[1 mark] for B. No partial credit.

The chemical equation for the standard combustion of ethene is: \(C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)\). Using Hess's Law and standard enthalpies of formation: \(\Delta H^\ominus_{\text{comb}} = \sum \Delta H^\ominus_{\text{f}}(\text{products}) - \sum \Delta H^\ominus_{\text{f}}(\text{reactants}) = [2 \cdot \Delta H^\ominus_{\text{f}}(CO_2) + 2 \cdot \Delta H^\ominus_{\text{f}}(H_2O)] - \Delta H^\ominus_{\text{f}}(C_2H_4)\). Substituting the values: \(\Delta H^\ominus_{\text{comb}} = [2(-393.5) + 2(-285.8)] - (+52.5) = [-787.0 - 571.6] - 52.5 = -1358.6 - 52.5 = -1411.1\text{ kJ mol}^{-1}\).

[1 mark] for A. No partial marks.

Across Period 3, the atomic radius decreases because electrons are added to the same outer energy level while the nuclear charge increases, increasing the attraction. Thus statement A is incorrect. The first ionization energy generally increases across the period because of the increasing nuclear charge and decrease in atomic radius, holding the outer electrons more tightly. Thus statement B is correct. Oxides of the elements change from basic (e.g., \(Na_2O\)) to amphoteric (e.g., \(Al_2O_3\)) to acidic (e.g., \(SO_3\)). Thus statement C is incorrect. Electronegativity increases across the period. Thus statement D is incorrect.

[1 mark] for B. No partial marks.

Alkanes are saturated hydrocarbons and undergo substitution reactions with halogens in the presence of UV light. The UV light provides the energy needed to homolytically cleave the \(Cl-Cl\) bond, producing chlorine free radicals which initiate a chain reaction. This process is called free-radical substitution.

[1 mark] for C. No partial marks.

3-hydroxypropanal (\(HO-CH_2-CH_2-CHO\)) contains an aldehyde group (which includes a carbonyl group, \(C=O\)) and a hydroxyl group (\(-OH\)) on a different carbon atom. Thus, it is not a carboxylic acid but contains both functional groups. Propanoic acid contains both groups on the same carbon, making it a carboxylic acid. Methyl ethanoate is an ester and has no hydroxyl group. Propanone is a ketone and has no hydroxyl group.

[1 mark] for C. No partial marks.

In \(\text{O}_3\) (ozone), the central oxygen atom has three electron domains (one single bond, one double bond, and one lone pair in resonance structures), which corresponds to a trigonal planar electron domain geometry and \(sp^2\) hybridization. Repulsion from the lone pair reduces the bond angle slightly from \(120^\circ\) to approximately \(117^\circ\). In contrast, \(\text{H}_3\text{O}^+\), \(\text{NH}_4^+\), and \(\text{OF}_2\) all have four electron domains around their central atoms, corresponding to \(sp^3\) hybridization with bond angles of approximately \(107^\circ\), \(109.5^\circ\), and \(103^\circ\) respectively.

[1 mark] for selecting option A. All other options are incorrect because their central atoms have \(sp^3\) hybridization.

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

[1 mark] for selecting option B. Any other selected option receives 0 marks.

The sum of the oxidation states of all atoms in a polyatomic ion must equal the overall charge of the ion. Oxygen has a standard oxidation state of -2. Let \(x\) be the oxidation state of molybdenum: \(7(x) + 24(-2) = -6\), which simplifies to \(7x - 48 = -6\), hence \(7x = +42\) and \(x = +6\).

[1 mark] for selecting option C.

A Brønsted–Lowry base is defined as a proton (\(\text{H}^+\)) acceptor. In reaction C, \(\text{H}_2\text{PO}_4^-\) accepts a proton from acetic acid (\(\text{CH}_3\text{COOH}\)) to become \(\text{H}_3\text{PO}_4\). In options A, B, and D, \(\text{H}_2\text{PO}_4^-\) donates a proton to form \(\text{HPO}_4^{2-}\), meaning it acts as a Brønsted–Lowry acid.

[1 mark] for selecting option C. Reject other options because they show the species acting as an acid.

The standard enthalpy of atomization, \(\Delta H_{at}^\theta\), is defined as the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions. Under standard conditions (298 K, 100 kPa), bromine exists as a diatomic liquid, \(\text{Br}_2(l)\). Therefore, to form exactly one mole of gaseous bromine atoms, the equation is: \(\frac{1}{2}\text{Br}_2(l) \rightarrow \text{Br}(g)\).

[1 mark] for selecting option A. Other options are incorrect because they either start with a gaseous reactant (not the standard state) or produce two moles of gaseous atoms rather than one.

\(\text{P}_4\text{O}_{10}\) reacts with water to form phosphoric acid, \(\text{H}_3\text{PO}_4\). \(\text{SO}_3\) reacts with water to form sulfuric acid, \(\text{H}_2\text{SO}_4\). Although \(\text{SiO}_2\) is classified as an acidic oxide because it reacts with concentrated alkalis, it is giant covalent and completely insoluble in water, so it does not react with or dissolve in water to form an acidic solution. Therefore, only I and III form an acidic solution.

[1 mark] for selecting option B. Reject any answer containing statement II.

In the first experiment: \(n(\text{H}_2\text{O}\text{ formed}) = 0.0500\text{ dm}^3 \times 1.0\text{ mol dm}^{-3} = 0.050\text{ mol}\). The heat released is \(q_1 = n \times \Delta H_{\text{neut}}\). The total mass of the mixture is \(100.0\text{ g}\) (assuming a density of \(1.0\text{ g cm}^{-3}\)). Thus, \(\Delta T_1 = \frac{q_1}{m_1 c} = \frac{0.050 \times \Delta H_{\text{neut}}}{100.0 \times c} = \Delta T\). In the second experiment: \(n(\text{H}_2\text{O}\text{ formed}) = 0.0250\text{ dm}^3 \times 2.0\text{ mol dm}^{-3} = 0.050\text{ mol}\). The heat released is exactly the same, so \(q_2 = q_1\). However, the total mass of the mixture is now only \(25.0 + 25.0 = 50.0\text{ g}\), which is half of the first experiment's mass. Since \(\Delta T = \frac{q}{m c}\), halving the mass while keeping the heat energy constant doubles the temperature change: \(\Delta T_2 = 2 \Delta T_1 = 2\Delta T\).

[1 mark] for selecting option C.

A tertiary alcohol is characterized by having the hydroxyl (\(-\text{OH}\)) group attached to a carbon atom that is directly bonded to three other carbon atoms. In 2-methylbutan-2-ol, the structural formula is \(\text{CH}_3-\text{C}(\text{CH}_3)(\text{OH})-\text{CH}_2-\text{CH}_3\). The carbon with the \(-\text{OH}\) group is bonded to two methyl groups and one ethyl group (three carbons total), making it a tertiary alcohol. In 3-methylbutan-2-ol and pentan-3-ol, the carbon with the \(-\text{OH}\) is bonded to two other carbon atoms (secondary alcohols). In 2,2-dimethylpropan-1-ol, the carbon with the \(-\text{OH}\) is bonded to only one other carbon atom (primary alcohol).

[1 mark] for selecting option A.

(a) Comparing Experiments 1 and 2: [H2] is kept constant, while [NO] is doubled. The initial rate increases by a factor of 4 (from 2.4 x 10^-6 to 9.6 x 10^-6 mol dm^-3 s^-1). Since \(2^2 = 4\), the reaction is second order with respect to NO.

Comparing Experiments 2 and 3: [NO] is kept constant, while [H2] is tripled. The initial rate increases by a factor of 3 (from 9.6 x 10^-6 to 2.88 x 10^-5 mol dm^-3 s^-1). Since \(3^1 = 3\), the reaction is first order with respect to H2.

(b) Rate expression: Rate = \(k[\text{NO}]^2[\text{H}_2]\).
Using Experiment 1 data to calculate k:
\(2.4 \times 10^{-6}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.010\text{ mol dm}^{-3})^2 \times (0.010\text{ mol dm}^{-3})\)
\(k = \frac{2.4 \times 10^{-6}}{1.0 \times 10^{-6}} = 2.4\)
Units: \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

(c) The rate-determining step is the slow step (Step 2): Rate = \(k_2[\text{N}_2\text{O}_2][\text{H}_2]\).
Since \(\text{N}_2\text{O}_2\) is an intermediate, its concentration is determined by the fast equilibrium in Step 1: \(K_{\text{eq}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2}\), so \([\text{N}_2\text{O}_2] = K_{\text{eq}}[\text{NO}]^2\).
Substituting this into the rate equation: Rate = \(k_2 \times K_{\text{eq}} \times [\text{NO}]^2[\text{H}_2] = k[\text{NO}]^2[\text{H}_2]\), which is consistent with the experimentally determined rate expression.

(d) An increase in temperature shifts the Maxwell-Boltzmann distribution curve to the right (higher average kinetic energy) and flattens it. This means a much greater fraction of reactant particles have kinetic energy greater than or equal to the activation energy (\(E \ge E_{\text{a}}\)), leading to a higher frequency of successful collisions and thus an increased rate of reaction.

(a) [4 marks]
- 1 mark for correctly comparing Exp 1 & 2 to show [NO] doubles and rate quadruples.
- 1 mark for deducing order with respect to NO is 2.
- 1 mark for correctly comparing Exp 2 & 3 to show [H2] triples and rate triples.
- 1 mark for deducing order with respect to H2 is 1.

(b) [3 marks]
- 1 mark for correct rate expression: Rate = \(k[\text{NO}]^2[\text{H}_2]\).
- 1 mark for correct calculation of k = 2.4.
- 1 mark for correct units of \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

(c) [3 marks]
- 1 mark for identifying Step 2 as the rate-determining step and writing its rate equation.
- 1 mark for using Step 1 equilibrium to express [N2O2] in terms of [NO]^2.
- 1 mark for substituting and showing the final rate expression matches the experimental one.

(d) [3 marks]
- 1 mark for stating that increasing temperature increases the average kinetic energy of the particles.
- 1 mark for explaining that a larger fraction of molecules now have energy greater than or equal to the activation energy.
- 1 mark for stating this increases the frequency of successful collisions.

(a) Lewis Structures:
- SF4: Sulfur is the central atom, single-bonded to 4 fluorine atoms. Total valence electrons = 6 + 4(7) = 34. S has 4 bonding pairs and 1 lone pair (expanded octet). Each fluorine has 3 lone pairs.
- XeF4: Xenon is the central atom, single-bonded to 4 fluorine atoms. Total valence electrons = 8 + 4(7) = 36. Xe has 4 bonding pairs and 2 lone pairs. Each fluorine has 3 lone pairs.

(b) Geometry and Bond Angles:
- SF4: 5 electron domains around S (4 bonding, 1 lone pair) gives a trigonal bipyramidal electron domain geometry. The molecular geometry is see-saw. The axial-equatorial bond angles are less than 90 degrees (approx. 87 degrees) and equatorial-equatorial angles are less than 120 degrees (approx. 102 degrees).
- XeF4: 6 electron domains around Xe (4 bonding, 2 lone pairs) gives an octahedral electron domain geometry. The lone pairs position themselves opposite to each other to minimize repulsion, yielding a square planar molecular geometry. The F-Xe-F bond angles are exactly 90 degrees and 180 degrees.

(c) Polarity:
- SF4 is polar. Its see-saw geometry is asymmetrical, so the polar S-F bond dipoles do not cancel each other out, resulting in a net dipole moment.
- XeF4 is non-polar. Its square planar geometry is highly symmetrical, so the individual polar Xe-F bond dipoles cancel each other out, resulting in a net dipole moment of zero.

(d) Intermolecular forces:
- In SF4, because it is polar, the intermolecular forces include both dipole-dipole forces and London (dispersion) forces.
- In XeF4, because it is non-polar, the only intermolecular forces present are London (dispersion) forces.

(a) [4 marks]
- 2 marks for SF4: Correctly showing 4 S-F single bonds, 1 lone pair on S, and 3 lone pairs on each of the 4 F atoms.
- 2 marks for XeF4: Correctly showing 4 Xe-F single bonds, 2 lone pairs on Xe, and 3 lone pairs on each of the 4 F atoms.

(b) [4 marks]
- 1 mark for molecular geometry of SF4: see-saw.
- 1 mark for bond angles of SF4: < 120 degrees and < 90 degrees.
- 1 mark for molecular geometry of XeF4: square planar.
- 1 mark for bond angles of XeF4: 90 degrees.

(c) [2 marks]
- 1 mark for SF4 being polar because its asymmetric structure means dipoles do not cancel.
- 1 mark for XeF4 being non-polar because its symmetric structure means dipoles cancel.

(d) [3 marks]
- 1 mark for stating that both molecules have London (dispersion) forces.
- 1 mark for stating that SF4 also has dipole-dipole forces.
- 1 mark for explicitly stating that XeF4 has only London dispersion forces.

(a) Copper has a more positive standard reduction potential than zinc (+0.34 V > -0.76 V), meaning copper(II) ions are reduced and zinc metal is oxidized.
- Anode (Oxidation): \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\)
- Cathode (Reduction): \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)

(b) The overall cell equation is:
\(\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}\)
The standard cell potential is:
\(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = (+0.34\text{ V}) - (-0.76\text{ V}) = +1.10\text{ V}\)

(c) Electrons flow through the external wire from the zinc anode (negative electrode) to the copper cathode (positive electrode).
In the salt bridge:
- Anions (\(\text{NO}_3^-\)) migrate into the zinc half-cell (anode compartment) to balance the positive charge of produced \(\text{Zn}^{2+}\) ions.
- Cations (\(\text{K}^+\)) migrate into the copper half-cell (cathode compartment) to balance the decrease in positive charge as \(\text{Cu}^{2+}\) ions are reduced.

(d) In the electrolysis of aqueous \(\text{CuSO}_4\):
- At the cathode (negative electrode), \(\text{Cu}^{2+}\) ions are reduced in preference to \(\text{H}^+\) ions because copper has a more positive reduction potential. Product: Copper metal, \(\text{Cu(s)}\).
- At the anode (positive electrode), water is oxidized in preference to sulfate ions (\(\text{SO}_4^{2-}\)). Product: Oxygen gas, \(\text{O}_2\text{(g)}\).
- The half-equation at the anode is:
\(2\text{H}_2\text{O(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^-\)

(a) [3 marks]
- 1 mark for anode oxidation half-equation: Zn(s) -> Zn2+(aq) + 2e-.
- 1 mark for cathode reduction half-equation: Cu2+(aq) + 2e- -> Cu(s).
- 1 mark for correctly identifying anode and cathode.

(b) [3 marks]
- 1 mark for overall equation: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s).
- 1 mark for correct calculation setup: E = +0.34 - (-0.76).
- 1 mark for final value of +1.10 V.

(c) [3 marks]
- 1 mark for electron flow from Zn (anode) to Cu (cathode).
- 1 mark for anions (NO3-) migrating to the anode (Zn) compartment.
- 1 mark for cations (K+) migrating to the cathode (Cu) compartment.

(d) [4 marks]
- 1 mark for copper product at the cathode.
- 1 mark for oxygen product at the anode.
- 2 marks for correct anode half-equation: 2H2O(l) -> O2(g) + 4H+(aq) + 4e-.

(a) Standard lattice enthalpy of dissociation is the enthalpy change when one mole of an ionic compound is broken down into its constituent gaseous ions under standard conditions (298 K, 100 kPa).

(b) Using the Born-Haber cycle, the equation relating these terms is:
\(\Delta H_{\text{f}}^\theta(\text{MgCl}_2) = \Delta H_{\text{at}}^\theta(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + 2 \times \Delta H_{\text{at}}^\theta(\text{Cl}) + 2 \times \text{EA}_1(\text{Cl}) - \Delta H_{\text{lat}}^\theta(\text{MgCl}_2)\)

Substitute the given values:
\(-641 = +148 + 738 + 1451 + 2(121) + 2(-349) - \Delta H_{\text{lat}}^\theta\)
\(-641 = 148 + 738 + 1451 + 242 - 698 - \Delta H_{\text{lat}}^\theta\)
\(-641 = 1881 - \Delta H_{\text{lat}}^\theta\)
\(\Delta H_{\text{lat}}^\theta = 1881 + 641 = +2522\text{ kJ mol}^{-1}\)

(c) Lattice enthalpy depends on the electrostatic attraction between ions:
- Magnesium ions (\(\text{Mg}^{2+}\)) have a higher positive charge (+2) than sodium ions (\(\text{Na}^+\), +1).
- Magnesium ions also have a smaller ionic radius than sodium ions, allowing them to get closer to chloride ions.
- This results in a much stronger electrostatic attraction between the ions in the magnesium chloride lattice, requiring more energy to disrupt it.

(d) If the experimental lattice enthalpy is more exothermic than the theoretical value, it indicates that the bonding has covalent character (due to polarization of the chloride anion by the small, highly charged magnesium cation).

(a) [2 marks]
- 1 mark for 'enthalpy change when one mole of ionic compound is broken into gaseous ions'.
- 1 mark for specifying 'under standard conditions'.

(b) [6 marks]
- 1 mark for representing correct steps for magnesium: atomization (+148) + IE1 (+738) + IE2 (+1451).
- 2 marks for correctly multiplying chlorine atomization and electron affinity by 2: \(2 \times 121 = 242\) and \(2 \times (-349) = -698\).
- 1 mark for correct algebraic setup: \(-641 = 1881 - \Delta H_{\text{lat}}\).
- 2 marks for correct calculation of \(+2522\text{ kJ mol}^{-1}\) (award 1 mark if sign is incorrect, e.g. -2522).

(c) [3 marks]
- 1 mark for stating Mg2+ has a higher charge than Na+ (+2 vs +1).
- 1 mark for stating Mg2+ has a smaller ionic radius than Na+.
- 1 mark for relating this to stronger electrostatic attractions between ions in the MgCl2 lattice.

(d) [2 marks]
- 1 mark for identifying that the bonding has covalent character.
- 1 mark for explaining that the Mg2+ cation polarizes the Cl- anion.

(a) Equation for reaction with water:
\(\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}\)

Since ethanoic acid is a weak acid:
\(K_{\text{a}} = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} \approx \frac{[\text{H}_3\text{O}^+]^2}{C_{\text{acid}}}\)
\([\text{H}_3\text{O}^+] = \sqrt{K_{\text{a}} \times C_{\text{acid}}} = \sqrt{1.8 \times 10^{-5} \times 0.100} = 1.34 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87\)

(b) When 12.5 cm^3 of NaOH is added, half of the ethanoic acid is neutralized:
\(\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}\)
This forms a buffer solution containing equal concentrations of the weak acid (\(\text{CH}_3\text{COOH}\)) and its conjugate base (\(\text{CH}_3\text{COO}^-\)).
Since \([\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]\):
\([\text{H}_3\text{O}^+] = K_{\text{a}}\)
\(\text{pH} = \text{p}K_{\text{a}} = -\log_{10}(1.8 \times 10^{-5}) = 4.74\)

(c) At the equivalence point, 25.0 cm^3 of NaOH is added. The total volume of the solution is \(50.0\text{ cm}^3 = 0.0500\text{ dm}^3\).
\([\text{CH}_3\text{COO}^-] = \frac{2.50 \times 10^{-3}\text{ mol}}{0.0500\text{ dm}^3} = 0.0500\text{ mol dm}^{-3}\).

The conjugate base hydrolyzes water:
\(\text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COOH(aq)} + \text{OH}^-\text{(aq)}\)
\(K_{\text{b}} = \frac{K_{\text{w}}}{K_{\text{a}}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\)
\([\text{OH}^-] = \sqrt{K_{\text{b}} \times [\text{CH}_3\text{COO}^-]} = \sqrt{5.56 \times 10^{-10} \times 0.0500} = 5.27 \times 10^{-6}\text{ mol dm}^{-3}\)
\(\text{pOH} = -\log_{10}(5.27 \times 10^{-6}) = 5.28\)
\(\text{pH} = 14.00 - 5.28 = 8.72\)

The pH is not neutral because the basic salt anion (\(\text{CH}_3\text{COO}^-\)) hydrolyzes to produce hydroxide ions (\(\text{OH}^-\)).

(d) Phenolphthalein. The equivalence point occurs at pH = 8.72, which lies within the transition range of phenolphthalein (8.3 - 10.0), allowing for a sharp color change at the end point.

(a) [4 marks]
- 1 mark for correct reversible equation: CH3COOH + H2O <-> CH3COO- + H3O+.
- 1 mark for showing [H+] = sqrt(Ka * C).
- 1 mark for [H+] = 1.34 x 10^-3 mol dm^-3.
- 1 mark for pH = 2.87.

(b) [3 marks]
- 1 mark for explaining that half-neutralization creates equal amounts of weak acid and conjugate base.
- 1 mark for identifying that this constitutes an acidic buffer system.
- 1 mark for calculating pH = 4.74.

(c) [4 marks]
- 1 mark for calculating [CH3COO-] = 0.0500 mol dm^-3.
- 1 mark for calculating Kb = 5.56 x 10^-10 and [OH-] = 5.27 x 10^-6 mol dm^-3.
- 1 mark for calculating pH = 8.72.
- 1 mark for explaining that the conjugate base CH3COO- undergoes hydrolysis with water to yield OH-.

(d) [2 marks]
- 1 mark for choosing Phenolphthalein.
- 1 mark for justifying that the equivalence point pH (8.72) lies within its color change range.

(a) Electronegativity increases across Period 3 from sodium to chlorine.
- The nuclear charge (number of protons) increases, which increases the attraction between the nucleus and bonding electrons.
- The shielding effect remains relatively constant because electrons are added to the same main energy level (third shell).
- Therefore, the atomic radius decreases, and the nucleus exerts a stronger pull on the shared bonding pair of electrons.

(b) Suitable oxides include \(\text{SO}_2\), \(\text{SO}_3\), and \(\text{P}_4\text{O}_{10}\).
Examples of equations (choose any two):
1. \(\text{SO}_3\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)}\)
2. \(\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\)
3. \(\text{SO}_2\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_3\text{(aq)}\)

(c) Structure and bonding:
- Silicon dioxide, \(\text{SiO}_2\), has a giant covalent macromolecular structure. Each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement.
- Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), has a simple molecular covalent structure. Strong covalent bonds hold the atoms together within each molecule, but only weak London dispersion forces exist between separate molecules.
- Melting \(\text{SiO}_2\) requires breaking many strong covalent bonds throughout the giant lattice, which requires a large amount of energy. Melting \(\text{P}_4\text{O}_{10}\) only requires overcoming weak intermolecular forces, which requires much less energy.

(d) Reactions of aluminum oxide:
(i) With HCl:
\(\text{Al}_2\text{O}_3\text{(s)} + 6\text{HCl(aq)} \rightarrow 2\text{AlCl}_3\text{(aq)} + 3\text{H}_2\text{O(l)}\)
(ii) With NaOH:
\(\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{Na[Al(OH)}_4]\text{(aq)}\) (or \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} \rightarrow 2\text{NaAlO}_2\text{(aq)} + \text{H}_2\text{O(l)}\))

(a) [3 marks]
- 1 mark for stating that electronegativity increases.
- 1 mark for explaining that nuclear charge increases while shielding remains constant.
- 1 mark for relating this to a stronger attraction between the nucleus and the bonding electrons.

(b) [4 marks]
- 2 marks for identifying any two of: P4O10, SO2, SO3, Cl2O7.
- 2 marks for writing two correctly balanced chemical equations.

(c) [4 marks]
- 1 mark for describing SiO2 as giant covalent and P4O10 as simple molecular.
- 1 mark for explaining that melting SiO2 requires breaking strong covalent bonds.
- 1 mark for explaining that melting P4O10 requires overcoming weak intermolecular forces.
- 1 mark for concluding that breaking covalent bonds requires significantly more energy than overcoming intermolecular forces.

(d) [2 marks]
- 1 mark for correct balanced equation for reaction with HCl: Al2O3 + 6HCl -> 2AlCl3 + 3H2O.
- 1 mark for correct balanced equation for reaction with NaOH: Al2O3 + 2NaOH + 3H2O -> 2Na[Al(OH)4].

(a) Calculate heat absorbed by water (q):
\(\Delta T = 38.5 - 21.5 = 17.0\text{ K}\)
\(q = m \cdot c \cdot \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 17.0\text{ K} = 10659\text{ J} = 10.7\text{ kJ}\).

(b) Calculate moles of methanol burned:
Mass of methanol burned = \(245.32 - 244.72 = 0.60\text{ g}\).
\(n(\text{CH}_3\text{OH}) = \frac{0.60\text{ g}}{32.05\text{ g mol}^{-1}} = 0.0187\text{ mol}\).

(c) Calculate enthalpy of combustion (\(\Delta H_{\text{c}}\)):
\(\Delta H_{\text{c}} = -\frac{q}{n} = -\frac{10.659\text{ kJ}}{0.01872\text{ mol}} = -569\text{ kJ mol}^{-1}\) (or \(-570\text{ kJ mol}^{-1}\) to 2 significant figures).

(d) Two major systematic errors:
1. Heat loss to the surrounding air and the copper can. Not all heat released is transferred to the water, making \(\Delta T\) and thus the calculated \(\Delta H_{\text{c}}\) less exothermic (smaller value).
- Improvement: Use draft shields, or use a bomb calorimeter.
2. Incomplete combustion of methanol (evidenced by soot forming on the bottom of the can).
- Improvement: Ensure a plentiful oxygen supply.

(e) Standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions (298 K, 100 kPa), with all reactants and products in their standard states.

(a) [2 marks]
- 1 mark for correct calculation of temperature change: DT = 17.0 K.
- 1 mark for correct calculation of q = 10.7 kJ.

(b) [2 marks]
- 1 mark for correct mass of methanol burned = 0.60 g.
- 1 mark for correct calculation of moles = 0.0187 mol.

(c) [3 marks]
- 1 mark for dividing q by n.
- 1 mark for correct magnitude: 569 or 570 kJ mol^-1.
- 1 mark for the negative sign.

(d) [4 marks]
- 1 mark for identifying heat loss to surroundings.
- 1 mark for identifying incomplete combustion.
- 1 mark for explaining how one of these errors leads to a less exothermic value.
- 1 mark for a valid corresponding improvement.

(e) [2 marks]
- 1 mark for 'enthalpy change when one mole of substance is completely burned in oxygen'.
- 1 mark for specifying 'under standard conditions'.

For (a): The cooling rate after mixing is constant at \((28.4 - 28.0) = 0.4^\circ\text{C min}^{-1}\). Extrapolating back 1.0 minute from \(t = 4.0\text{ min}\) to the mixing time at \(t = 3.0\text{ min}\) gives an extrapolated maximum temperature of \(28.4 + 0.4 = 28.8^\circ\text{C}\). Thus, the maximum temperature change is \(\Delta T = 28.8 - 21.0 = 7.8^\circ\text{C}\) (or \(7.8\text{ K}\)).
For (b): Heat released, \(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 7.8\text{ K} = 1630.2\text{ J} = 1.63\text{ kJ}\). Moles of \(\text{CuSO}_4 = 5.00\text{ g} / 159.62\text{ g mol}^{-1} = 0.03132\text{ mol}\). \(\Delta H_{\text{sol}} = -q / n = -1.6302 / 0.03132 = -52.05\text{ kJ mol}^{-1}\). Rounding to 3 significant figures gives \(-52.0\text{ kJ mol}^{-1}\).
For (c): Glass is a better thermal conductor than polystyrene, leading to greater heat loss to the surroundings (or the glass beaker absorbs more heat). This results in a smaller observed temperature rise (\(\Delta T\)) and a calculated enthalpy of solution that is less exothermic (less negative/more positive) than the actual value.

Part (a) [2 marks]:
- Award [1] for calculating the extrapolated maximum temperature of \(28.8^\circ\text{C}\) (or showing the calculation: \(28.4 + 0.4 = 28.8\)).
- Award [1] for determining \(\Delta T = 7.8^\circ\text{C}\) (or \(7.8\text{ K}\)).

Part (b) [3 marks]:
- Award [1] for calculating heat released: \(q = 1.63\text{ kJ}\) (or \(1630\text{ J}\)).
- Award [1] for calculating moles of \(\text{CuSO}_4 = 0.0313\text{ mol}\).
- Award [1] for final enthalpy change: \(-52.0\text{ kJ mol}^{-1}\) (must include negative sign and correct units; accept range from \(-51.9\) to \(-52.1\)).

Part (c) [2 marks]:
- Award [1] for identifying greater heat loss to the surroundings / glass calorimeter absorbs more heat.
- Award [1] for stating that the calculated enthalpy value becomes less exothermic / less negative / more positive.

For (a): Comparing Exp 1 and Exp 2, \([\text{I}^-]\) is constant while \([\text{S}_2\text{O}_8^{2-}]\) doubles. The time decreases from \(45.0\text{ s}\) to \(22.5\text{ s}\), meaning the rate (which is proportional to \(1/t\)) doubles. Thus, the order with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1.
Comparing Exp 2 and Exp 3, \([\text{S}_2\text{O}_8^{2-}]\) is constant while \([\text{I}^-]\) halves. The time increases from \(22.5\text{ s}\) to \(45.0\text{ s}\), meaning the rate halves. Thus, the order with respect to \(\text{I}^-\) is 1.
For (b): The rate expression is \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). The overall order of reaction is the sum of the individual orders: \(1 + 1 = 2\).
For (c): The rate constant, \(k\), increases as temperature increases. According to collision theory, increasing temperature increases the average kinetic energy of the reactant particles. Consequently, a significantly larger fraction of colliding particles have energy equal to or greater than the activation energy (\(E_a\)), leading to a higher frequency of successful (effective) collisions.

Part (a) [3 marks]:
- Award [1] for deducing first-order for \(\text{S}_2\text{O}_8^{2-}\) and explaining that doubling the concentration doubles the rate (time halves).
- Award [1] for deducing first-order for \(\text{I}^-\) and explaining that halving the concentration halves the rate (time doubles).
- Award [1] for explicitly linking rate as inversely proportional to time (\(\text{rate} \propto 1/t\)) in the explanation.

Part (b) [2 marks]:
- Award [1] for \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (accept equivalent representations; formulas must be chemically recognizable).
- Award [1] for stating the overall order is 2.

Part (c) [3 marks]:
- Award [1] for stating that the rate constant \(k\) increases.
- Award [1] for explaining that a larger fraction of particles have energy \(\ge E_a\) (activation energy).
- Award [1] for stating there is a greater frequency of successful/effective collisions.

1. Calculate theoretical moles of salicylic acid: \(n = \frac{5.00\text{ g}}{138.12\text{ g mol}^{-1}} = 0.0362\text{ mol}\). Since salicylic acid is the limiting reactant and reacts in a 1:1 ratio, the theoretical yield of aspirin is \(0.0362\text{ mol} \times 180.17\text{ g mol}^{-1} = 6.52\text{ g}\). 2. Calculate the percentage yield: \(\text{Percentage yield} = \frac{4.85\text{ g}}{6.52\text{ g}} \times 100 = 74.4\%\). 3. Identify a reason for loss: Product remains dissolved in the cold solvent during recrystallization, incomplete reaction, or loss during filtration.

[1 mark] for calculating the moles of salicylic acid (0.0362 mol). [1 mark] for calculating the theoretical yield of aspirin (6.52 g). [1 mark] for the correct percentage yield (74.4%). [1 mark] for identifying a valid reason for the low yield (e.g., product loss during recrystallization or incomplete reaction).

1. The beta-lactam ring is a four-membered ring containing a carbonyl group and a nitrogen atom. Because of the geometry, the bond angles are forced to be approximately \(90^\circ\) instead of their preferred tetrahedral (\(109.5^\circ\)) or trigonal planar (\(120^\circ\)) angles. This creates significant ring strain. 2. The ring strain makes the amide group highly reactive, causing the ring to open easily in the presence of the bacterial transpeptidase enzyme, covalently bonding to and inactivating it, preventing bacterial cell wall synthesis. 3. Some bacteria develop resistance by secreting beta-lactamase (penicillinase) enzymes, which open the beta-lactam ring through hydrolysis before the drug can reach its target enzyme.

[1 mark] for identifying that the beta-lactam ring has highly strained bond angles of approximately \(90^\circ\). [1 mark] for explaining that this strain makes the amide bond highly reactive/easy to open. [1 mark] for explaining that the open ring binds to and inhibits transpeptidase (disrupting cell wall synthesis). [1 mark] for stating that bacterial resistance occurs via the production of beta-lactamase/penicillinase which hydrolyzes the ring.

1. Morphine contains two hydroxyl (-OH) groups, whereas diamorphine contains two ester (ethanoate, -OCOCH\(_3\)) groups. 2. This chemical modification converts highly polar hydroxyl groups into less polar ester groups. 3. The reduction in polarity makes diamorphine significantly more lipid-soluble. 4. Consequently, diamorphine crosses the non-polar blood-brain barrier much faster and in higher concentrations than morphine, leading to its significantly higher potency.

[1 mark] for identifying the ester (or ethanoate) groups in diamorphine. [1 mark] for noting the hydroxyl groups in morphine. [1 mark] for explaining that esters make diamorphine less polar / more lipid-soluble. [1 mark] for explaining that this increased lipid solubility allows diamorphine to cross the blood-brain barrier more efficiently, increasing its potency.

1. Write the chemical equation: \(Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)\). 2. Find the moles of \(Mg(OH)_2\): \(n = \frac{0.450\text{ g}}{58.32\text{ g mol}^{-1}} = 7.716 \times 10^{-3}\text{ mol}\). 3. Use the stoichiometric ratio (1:2) to find the moles of \(HCl\): \(n_{HCl} = 2 \times 7.716 \times 10^{-3} = 1.543 \times 10^{-2}\text{ mol}\). 4. Calculate the volume of \(HCl\): \(V = \frac{n}{C} = \frac{1.543 \times 10^{-2}\text{ mol}}{0.120\text{ mol dm}^{-3}} = 0.1286\text{ dm}^3 = 129\text{ cm}^3\).

[1 mark] for writing the correct balanced equation: \(Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)\). [1 mark] for calculating the moles of \(Mg(OH)_2\) (\(7.72 \times 10^{-3}\text{ mol}\)). [1 mark] for doubling the moles of \(Mg(OH)_2\) to find the moles of \(HCl\) (\(1.54 \times 10^{-2}\text{ mol}\)). [1 mark] for calculating the correct volume in \(\text{cm}^3\) (129 \(\text{cm}^3\)).

1. Target: Both drugs target the neuraminidase enzyme on the surface of the influenza virus. 2. Mechanism: By inhibiting neuraminidase, these drugs prevent the cleavage of sialic acid residues on the host cell membrane. This prevents newly synthesized viral particles from escaping the host cell to infect other cells, halting viral replication/spread. 3. Common functional group: Both molecules contain an alkenyl group (carbon-carbon double bond, C=C) and an amide (carboxamide) group.

[1 mark] for stating that the target is the neuraminidase enzyme. [2 marks] for explaining that inhibiting this enzyme prevents newly replicated virus particles from detaching/escaping from the host cell membrane, stopping viral spread. [1 mark] for identifying the alkenyl (carbon-carbon double bond) or carboxamide group as common to both structures.

1. Write the nuclear equation: \(^{177}_{71}\text{Lu} \rightarrow ^{177}_{72}\text{Hf} + ^{0}_{-1}e\) (or \(\beta^-\)). The mass number remains 177, while the atomic number increases from 71 to 72 (yielding Hafnium, Hf). 2. Calculate the number of half-lives that have elapsed: \(N = \frac{26.60\text{ days}}{6.65\text{ days}} = 4.0\). 3. Calculate the fraction remaining: \((\frac{1}{2})^4 = \frac{1}{16} = 0.0625\). 4. Convert to a percentage: \(0.0625 \times 100 = 6.25\%\).

[1 mark] for the reactants side and correct symbol for Lutetium-177. [1 mark] for the correct products side including Hafnium-177 (\(^{177}_{72}\text{Hf}\)) and a beta particle (\(^{0}_{-1}e\)). [1 mark] for determining that 4 half-lives have elapsed. [1 mark] for the correct final percentage of 6.25%.

1. Represent the weak acid dissociation of aspirin (HA) in water: \(HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)\). 2. In the stomach, the highly acidic environment (pH ~ 1.5) provides a high concentration of hydrogen ions (\([H^+]\)). According to Le Chatelier's principle, this high concentration of products shifts the dissociation equilibrium to the left, favoring the unionized form (\(HA\)). 3. The unionized form (\(HA\)) is uncharged/neutral and thus highly lipid-soluble, allowing it to easily pass through the phospholipid bilayer of stomach membranes by passive diffusion. 4. In the small intestine, the alkaline environment (pH ~ 8.0) has a low concentration of \([H^+]\), shifting the equilibrium to the right to favor the ionized form (\(A^-\)). The charged carboxylate ion is highly polar and water-soluble, making it unable to easily cross the non-polar cell membranes, resulting in slower absorption.

[1 mark] for presenting the dissociation equilibrium: \(HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)\). [1 mark] for explaining that the high \([H^+]\) in the stomach shifts the equilibrium to the left (unionized form, HA). [1 mark] for noting that the unionized form is neutral, lipid-soluble, and can easily cross cell membranes. [1 mark] for explaining that in the small intestine, the low \([H^+]\) shifts the equilibrium to the right, producing the ionic, polar form (\(A^-\)) which is poorly absorbed.