2024 IB DP Chemistry 模擬試題連答案詳解

Formal charge is calculated as: Valence electrons - Non-bonding electrons - 0.5 * Bonding electrons. For \(\text{H}_3\text{O}^+\), the central oxygen atom has 6 valence electrons, 2 non-bonding electrons (one lone pair), and 6 bonding electrons (three single bonds). This gives a formal charge of: \(6 - 2 - \frac{1}{2}(6) = +1\).

Award [1] for the correct option A. No marks for any other response.

There is a very large relative jump between the third (2745) and fourth (11577) ionization energies. This indicates that the fourth electron is being removed from an inner core shell, meaning the element has three valence electrons (Group 13). Thus, the element forms a \(\text{M}^{3+}\) ion. Since the oxide ion is \(\text{O}^{2-}\), the empirical formula of the oxide is \(\text{M}_2\text{O}_3\).

Award [1] for the correct option B. No marks for any other response.

Convert the mass percentages to moles per 100 g: \(n(\text{C}) = \frac{85.6}{12.01} = 7.13\,\text{mol}\), \(n(\text{H}) = \frac{14.4}{1.01} = 14.26\,\text{mol}\). Dividing each by the smallest value (7.13) yields a ratio of \(1\,\text{C} : 2\,\text{H}\). The empirical formula is therefore \(\text{CH}_2\).

Award [1] for the correct option B. No marks for any other response.

The heat released in the reaction is calculated by \(q = m \cdot c \cdot \Delta T = 100.0 \times 4.18 \times 6.0\) (in J). To convert this heat to kilojoules, it is divided by 1000. The number of moles of water formed is calculated from the limiting reactant: \(n = 0.0500\,\text{dm}^3 \times 1.00\,\text{mol}\,\text{dm}^{-3} = 0.0500\,\text{mol}\). The enthalpy change of neutralization is defined as \(\Delta H^\ominus = -\frac{q}{n}\), which gives \(-\frac{100.0 \times 4.18 \times 6.0}{0.0500 \times 1000}\).

Award [1] for the correct option A. No marks for any other response.

A Brønsted–Lowry base is defined as a proton (\(\text{H}^+\)) acceptor. In reaction C, the \(\text{H}_2\text{PO}_4^-\) ion accepts a proton from the hydronium ion (\(\text{H}_3\text{O}^+\)) to form phosphoric acid (\(\text{H}_3\text{PO}_4\)), acting as a base.

Award [1] for the correct option C. No marks for any other response.

The slow step in a reaction mechanism is the rate-determining step. The rate expression depends on the concentrations of the reactants in the rate-determining step, each raised to the power of its stoichiometric coefficient in that step. For Step 1, the reactants are \(\text{NO}_2\) and \(\text{F}_2\), giving \(\text{Rate} = k[\text{NO}_2][\text{F}_2]\).

Award [1] for the correct option B. No marks for any other response.

Let the oxidation state of sulfur be represented by \(x\). Oxygen typically has an oxidation state of \(-2\). The sum of the oxidation states in the polyatomic ion must equal the overall charge: \(2(x) + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\).

Award [1] for the correct option B. No marks for any other response.

Option A, \(\text{CH}_3\text{COCH}_2\text{COOCH}_3\), contains a ketone carbonyl group (\(\text{-CO-}\) bonded to two carbon atoms) and an ester group (\(\text{-COO-}\) linked to a methyl group). Option B contains an alcohol and an ester. Option C contains a ketone and a carboxylic acid. Option D is an ester only.

Award [1] for the correct option A. No marks for any other response.

\( \text{O}_3 \) has 3 electron domains around the central oxygen atom (one lone pair, one single bond, one double bond). The electron domain geometry is trigonal planar, and repulsion from the lone pair reduces the bond angle from \( 120^\circ \) to approximately \( 117^\circ \), which is closest to \( 120^\circ \). \( \text{H}_2\text{O} \) is bent with \( 104.5^\circ \), \( \text{NH}_4^+ \) is tetrahedral with \( 109.5^\circ \), and \( \text{CO}_2 \) is linear with \( 180^\circ \).

[1 mark] Correct choice is A. Award 1 mark for identifying that ozone has trigonal planar electron domain geometry leading to a bond angle of approximately 117 degrees.

Comparing Trial 1 and Trial 2: \( [\text{B}] \) is constant, \( [\text{A}] \) doubles, and the rate doubles, so the reaction is first-order with respect to A. Comparing Trial 1 and Trial 3: \( [\text{A}] \) is constant, \( [\text{B}] \) doubles, and the rate quadruples, so the reaction is second-order with respect to B. The rate expression is \( \text{Rate} = k[\text{A}][\text{B}]^2 \). The overall order is \( 1 + 2 = 3 \).

[1 mark] Correct choice is C. Award 1 mark for deducing that the order with respect to A is 1 and with respect to B is 2, and summing them to get 3.

Assume 100 g of the compound. Moles of P = \( 43.6 / 30.97 \approx 1.41\text{ mol} \). Moles of O = \( (100 - 43.6) / 16.00 = 56.4 / 16.00 \approx 3.53\text{ mol} \). Divide by the smaller number of moles: \( \text{P} = 1.41 / 1.41 = 1 \), \( \text{O} = 3.53 / 1.41 \approx 2.5 \). Multiplying by 2 to get whole numbers gives a ratio of \( \text{P}_2\text{O}_5 \).

[1 mark] Correct choice is C. Award 1 mark for calculating the mole ratio of P to O as 1 : 2.5 and simplifying to the empirical formula of P2O5.

Heat released \( q = m c \Delta T = 100.0 \times 4.18 \times 10.0\text{ J} = \frac{100.0 \times 4.18 \times 10.0}{1000}\text{ kJ} \). Moles of solute \( n = \frac{5.0}{40.0}\text{ mol} \). Enthalpy change \( \Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{100.0 \times 4.18 \times 10.0}{1000} \times \frac{40.0}{5.0} = -\frac{100.0 \times 4.18 \times 10.0 \times 40.0}{5.0 \times 1000}\text{ kJ mol}^{-1} \).

[1 mark] Correct choice is A. Award 1 mark for correctly formulating heat q, moles n, and combining them with a negative sign indicating an exothermic reaction.

For pure water, \( [\text{H}^+] = [\text{OH}^-] \), so the water is neutral regardless of the temperature. Since \( K_{\text{w}} = 5.5 \times 10^{-14} \), \( [\text{H}^+] = \sqrt{K_{\text{w}}} = \sqrt{5.5 \times 10^{-14}} \approx 2.3 \times 10^{-7}\text{ mol dm}^{-3} \). The \( \text{pH} = -\log[\text{H}^+] \approx 6.6 \), which is less than 7.

[1 mark] Correct choice is B. Award 1 mark for realizing that pure water is always neutral and that a larger Kw value results in pH < 7.

The oxidation state of chlorine in \( \text{Cl}_2 \) is 0. In \( \text{ClO}_3^- \), the oxidation state of chlorine is +5 (oxidized). In \( \text{Cl}^- \), the oxidation state of chlorine is -1 (reduced). Therefore, chlorine undergoes both oxidation and reduction (disproportionation).

[1 mark] Correct choice is C. Award 1 mark for calculating the oxidation states of Cl in reactants and products to identify disproportionation.

The structure contains the \( \text{-CONH-} \) group, which consists of a carbonyl group (\( \text{C=O} \)) bonded directly to a nitrogen atom. This is the characteristic functional group of an amide (specifically, a secondary amide).

[1 mark] Correct choice is B. Award 1 mark for identifying the amide functional group from the molecular formula.

Cobalt (atomic number 27) has the ground-state electron configuration \( [\text{Ar}] 4\text{s}^2 3\text{d}^7 \). When transition metals form cations, they lose the outer s-orbital electrons first. Therefore, the \( \text{Co}^{2+} \) ion loses the two \( 4\text{s} \) electrons, leaving the configuration \( [\text{Ar}] 3\text{d}^7 \).

[1 mark] Correct choice is C. Award 1 mark for correctly identifying that 4s electrons are lost before 3d electrons, resulting in the [Ar] 3d7 configuration.

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 volume \(V_2\): \(V_2 = V_1 \cdot \frac{P_1}{P_2} \cdot \frac{T_2}{T_1}\). Substituting the given values: \(V_2 = V \cdot \frac{P}{4P} \cdot \frac{2T}{T} = V \cdot \frac{1}{4} \cdot 2 = 0.5V\).

Award [1] for the correct answer A.

Sulfur has a first ionization energy slightly lower than phosphorus. This discontinuity occurs because sulfur's outer electron configuration is \(3s^2 3p^4\), which contains one pair of electrons in a 3p orbital. The repulsion between these paired electrons makes it easier to remove one electron compared to the half-filled \(3s^2 3p^3\) configuration of phosphorus.

Award [1] for the correct explanation of spin-pair repulsion (C).

The standard enthalpy of atomization is defined as the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. The standard state of bromine is a liquid, \(\text{Br}_2(l)\). Therefore, the equation for the production of 1 mole of gaseous bromine atoms is \(\frac{1}{2}\text{Br}_2(l) \rightarrow \text{Br}(g)\).

Award [1] for identifying the correct equation representing standard state and 1 mole of gaseous atoms (B).

1. Calculate moles of methanol: \(n = \frac{1.60\text{ g}}{32.0\text{ g mol}^{-1}} = 0.0500\text{ mol}\). 2. Calculate heat absorbed by water: \(q = m c \Delta T = 200.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 30.0\text{ K} = 25080\text{ J} = 25.08\text{ kJ}\). 3. Calculate enthalpy of combustion: \(\Delta H_c = -\frac{q}{n} = -\frac{25.08\text{ kJ}}{0.0500\text{ mol}} = -501.6\text{ kJ mol}^{-1}\), which rounds to -502.

Award [1] for calculating the correct enthalpy change (A).

For a multi-step reaction mechanism, the rate-determining step (slow step) determines the rate law. In mechanism A, the slow step is Step 1: \(\text{A} + \text{B} \rightarrow \text{I}\). Because this is an elementary step, its rate law is \(\text{Rate} = k[\text{A}][\text{B}]\), which matches the experimental rate law. Mechanism C would have a rate law of \(\text{Rate} = k[\text{A}]^2[\text{B}]\).

Award [1] for identifying the mechanism where the reactants in the slow step match the rate expression (A).

The two half-reactions are: Oxidation: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\). Reduction: \(2\text{BrO}_3^- + 12\text{H}^+ + 10e^- \rightarrow \text{Br}_2 + 6\text{H}_2\text{O}\). To balance electrons, multiply the oxidation half-reaction by 5: \(10\text{Br}^- \rightarrow 5\text{Br}_2 + 10e^-\). Add the two half-reactions: \(2\text{BrO}_3^- + 10\text{Br}^- + 12\text{H}^+ \rightarrow 6\text{Br}_2 + 6\text{H}_2\text{O}\). Dividing all coefficients by 2 gives the simplest whole-number ratio: \(\text{BrO}_3^- + 5\text{Br}^- + 6\text{H}^+ \rightarrow 3\text{Br}_2 + 3\text{H}_2\text{O}\). The ratio of \(\text{BrO}_3^-\) to \(\text{Br}^-\) is therefore 1 : 5.

Award [1] for the correct ratio 1:5 (C).

Examine the structure of methyl propenoate: \(\text{H}_2\text{C}=\text{CH}-\text{C}(=\text{O})-\text{O}-\text{CH}_3\). Sigma (\(\sigma\)) bonds: 6 \(\text{C}-\text{H}\) single bonds, 1 \(\text{C}-\text{C}\) single bond, 1 \(\text{C}-\text{O}\) single bond, 1 \(\text{O}-\text{C}\) single bond, 1 from the \(\text{C}=\text{C}\) double bond, and 1 from the \(\text{C}=\text{O}\) double bond. Total = 11 \(\sigma\) bonds. Pi (\(\pi\)) bonds: 1 from the \(\text{C}=\text{C}\) double bond and 1 from the \(\text{C}=\text{O}\) double bond. Total = 2 \(\pi\) bonds.

Award [1] for the correct counts of 11 sigma and 2 pi bonds (C).

Using the approximation for weak acids: \([\text{H}^+] \approx \sqrt{K_a \cdot c} = \sqrt{(1.0 \times 10^{-5}) \times 0.10} = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\). Therefore, \(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.0\).

Award [1] for the correct pH of 3.0 (B).

The formal charge (FC) can be calculated using the formula: \(FC = V - N - \frac{1}{2}B\), where \(V\) is the number of valence electrons, \(N\) is the number of non-bonding valence electrons, and \(B\) is the number of bonding electrons. For the terminal nitrogen atom (which has one lone pair and forms a triple bond): \(FC = 5 - 2 - \frac{1}{2}(6) = 0\). For the central nitrogen atom (which forms a triple bond and a single bond): \(FC = 5 - 0 - \frac{1}{2}(8) = +1\). For the oxygen atom (which has three lone pairs and forms a single bond): \(FC = 6 - 6 - \frac{1}{2}(2) = -1\). Therefore, the formal charges are: Terminal N: 0; Central N: +1; Oxygen: -1.

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

First, determine the equilibrium concentrations of each species: \([\text{A}] = 0.40\text{ mol} / 2.0\text{ dm}^3 = 0.20\text{ mol dm}^{-3}\), \([\text{B}] = 0.80\text{ mol} / 2.0\text{ dm}^3 = 0.40\text{ mol dm}^{-3}\), \([\text{C}] = 0.80\text{ mol} / 2.0\text{ dm}^3 = 0.40\text{ mol dm}^{-3}\). Next, write the expression for the equilibrium constant: \(K_c = \frac{[\text{B}][\text{C}]}{[\text{A}]^2}\). Substitute the equilibrium concentrations: \(K_c = \frac{0.40 \times 0.40}{(0.20)^2} = \frac{0.16}{0.04} = 4.0\). Alternatively, because the total number of moles on both sides is equal, volume cancels out, and using moles directly gives: \(K_c = \frac{0.80 \times 0.80}{0.40^2} = 4.0\).

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

A buffer solution is formed when a weak acid/base is in excess compared to a strong base/acid. In option A: we have \(5.0\text{ mmol}\) of \(\text{NH}_3\) (weak base) reacting with \(2.5\text{ mmol}\) of \(\text{HCl}\) (strong acid). This leaves \(2.5\text{ mmol}\) of unreacted \(\text{NH}_3\) and produces \(2.5\text{ mmol}\) of \(\text{NH}_4^+\) (its conjugate acid), forming a basic buffer with pH greater than 7. In option B: we have \(5.0\text{ mmol}\) of \(\text{CH}_3\text{COOH}\) reacting with \(2.5\text{ mmol}\) of \(\text{NaOH}\), which forms an acidic buffer with pH less than 7. In option C: \(\text{HCl}\) is in excess, so the solution is strongly acidic. In option D: equal stoichiometric amounts react, leaving only the salt \(\text{CH}_3\text{COONa}\) (not a buffer).

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

The second ionization energy is defined as the energy required to remove one mole of electrons from one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions. Thus, the corresponding equation is: \(\text{Ca}^+\text{(g)} \rightarrow \text{Ca}^{2+}\text{(g)} + \text{e}^-\).

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

1. Calculate heat absorbed, \(q\): \(q = m \cdot c \cdot \Delta T\) where \(m = 52.0\text{ g}\) (mass of solution), \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\), and \(\Delta T = 20.0 - 14.0 = 6.0\text{ K}\). \(q = 52.0 \times 4.18 \times 6.0 = 1304.16\text{ J} = 1.30416\text{ kJ}\). 2. Calculate moles of salt, \(n\): \(n = 2.00\text{ g} / 80.0\text{ g mol}^{-1} = 0.0250\text{ mol}\). 3. Calculate \(\Delta H_{\text{sol}}\): Since the temperature decreased, the process is endothermic, so \(\Delta H_{\text{sol}}\) is positive. \(\Delta H_{\text{sol}} = +\frac{1.30416\text{ kJ}}{0.0250\text{ mol}} = +52.2\text{ kJ mol}^{-1}\).

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

The rate-determining step is the slow step (Step 1). The rate equation is derived directly from the reactants of this step: \(\text{rate} = k[\text{NO}_2][\text{F}_2]\). Since the rate equation is first-order with respect to \(\text{NO}_2\) and first-order with respect to \(\text{F}_2\), the overall order of the reaction is \(1 + 1 = 2\), which makes statement C correct. Statement A is incorrect because the coefficient of \(\text{NO}_2\) in the overall reaction does not dictate the rate equation. Statement B is incorrect because \(\text{F}\) is an intermediate, not a catalyst. Statement D is incorrect because changing the concentration of \(\text{NO}_2\) will affect the rate of the reaction.

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

(a) The Lewis structure of \(\text{SF}_4\) has sulfur in the center single-bonded to four fluorine atoms, with one lone pair on the sulfur atom. Each fluorine atom has three lone pairs. S has 10 valence electrons surrounding it (an expanded octet).

(b) The sulfur atom in \(\text{SF}_4\) has 5 electron domains (4 bonding pairs, 1 lone pair). According to VSEPR theory, this corresponds to a see-saw (or sawhorse) molecular geometry. The approximate bond angles are less than 90\(^\circ\) (axial-equatorial) and less than 120\(^\circ\) (equatorial-equatorial) due to the greater repulsion from the lone pair.

(c) \(\text{SF}_6\) is symmetrical (octahedral geometry), so the individual S–F bond dipoles cancel each other out, resulting in a net dipole moment of zero (non-polar). \(\text{SF}_4\) is asymmetrical (see-saw geometry), so the individual S–F bond dipoles do not cancel each other out, resulting in a net dipole moment (polar).

(d) In \(\text{SO}_2\) (which has resonance structures with a double and single bond, or two double bonds), there are 2 sigma (\(\sigma\)) bonds and either 1 or 2 pi (\(\pi\)) bonds. The sulfur atom has 3 electron domains (2 bonding domains, 1 lone pair), so its hybridization is \(\text{sp}^2\).

(a)
- Award [1] for showing four S–F single bonds and one lone pair on the central sulfur atom.
- Award [1] for showing three lone pairs on each of the four fluorine atoms.
[2 marks]

(b)
- Award [1] for see-saw / sawhorse geometry.
- Award [1] for stating bond angles are < 90\(^\circ\) and < 120\(^\circ\) (accept approx. 90\(^\circ\) and 120\(^\circ\)).
[2 marks]

(c)
- Award [1] for stating that \(\text{SF}_6\) is octahedral/symmetrical, so S–F bond dipoles cancel out.
- Award [1] for stating that \(\text{SF}_4\) is see-saw/asymmetrical, so S–F bond dipoles do not cancel out.
[2 marks]

(d)
- Award [1] for stating 2 sigma (\(\sigma\)) and 1 pi (\(\pi\)) bond (or 2 pi (\(\pi\)) bonds).
- Award [1] for stating \(\text{sp}^2\) hybridization of the sulfur atom.
[2 marks]

(a) To find the order with respect to \(\text{NO}\), compare Experiments 1 and 2 where \([\text{H}_2]\) is kept constant at \(0.010\text{ mol dm}^{-3}\). As \([\text{NO}]\) doubles from \(0.010\) to \(0.020\text{ mol dm}^{-3}\), the rate increases by a factor of \(\frac{4.8 \times 10^{-4}}{1.2 \times 10^{-4}} = 4\) (or \(2^2\)). Therefore, the reaction is second order with respect to \(\text{NO}\).
To find the order with respect to \(\text{H}_2\), compare Experiments 2 and 3 where \([\text{NO}]\) is kept constant at \(0.020\text{ mol dm}^{-3}\). As \([\text{H}_2]\) doubles from \(0.010\) to \(0.020\text{ mol dm}^{-3}\), the rate increases by a factor of \(\frac{9.6 \times 10^{-4}}{4.8 \times 10^{-4}} = 2\) (or \(2^1\)). Therefore, the reaction is first order with respect to \(\text{H}_2\).

(b) The rate expression is: \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\).

(c) Substituting data from Experiment 1 into the rate expression:
\(1.2 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.010\text{ mol dm}^{-3})^2 (0.010\text{ mol dm}^{-3})\)
\(1.2 \times 10^{-4} = k (1.0 \times 10^{-6})\)
\(k = 120\)
Units of \(k\): \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

(d) The rate-determining step (slow step) is Step 2:
\(\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2]\)
Since \(\text{N}_2\text{O}_2\) is an intermediate, we express its concentration using the fast equilibrium of Step 1:
\(K_c = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \Rightarrow [\text{N}_2\text{O}_2] = K_c[\text{NO}]^2\)
Substituting this back into the rate equation for Step 2:
\(\text{Rate} = k_2(K_c[\text{NO}]^2)[\text{H}_2] = k[\text{NO}]^2[\text{H}_2]\), where \(k = k_2 K_c\). This is consistent with the experimental rate expression.

(a)
- Award [1] for second order with respect to \(\text{NO}\) and showing comparison of Exp 1 and 2.
- Award [1] for first order with respect to \(\text{H}_2\) and showing comparison of Exp 2 and 3.
- Award [1] for clear explanation logic.
[3 marks]

(b)
- Award [1] for \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\).
[1 mark]

(c)
- Award [1] for the value \(120\).
- Award [1] for correct units: \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\) (or \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)).
[2 marks]

(d)
- Award [1] for identifying that \(\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2]\) from the rate-determining step.
- Award [1] for substituting \([\text{N}_2\text{O}_2] = K_c[\text{NO}]^2\) from the fast equilibrium to obtain the overall rate expression.
[2 marks]

(a) Equation:
\(\text{C}_2\text{H}_5\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^-(aq) + \text{H}_3\text{O}^+(aq)\)
Conjugate pairs:
- \(\text{C}_2\text{H}_5\text{COOH}\) (acid) and \(\text{C}_2\text{H}_5\text{COO}^-\) (conjugate base)
- \(\text{H}_2\text{O}\) (base) and \(\text{H}_3\text{O}^+\) (conjugate acid)

(b) \(K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_5\text{COO}^-]}{[\text{C}_2\text{H}_5\text{COOH}]}\)
Assumption: Since propanoic acid is a weak acid, \([\text{H}^+] \approx [\text{C}_2\text{H}_5\text{COO}^-]\) and \([\text{C}_2\text{H}_5\text{COOH}]_{\text{eq}} \approx [\text{C}_2\text{H}_5\text{COOH}]_{\text{initial}} = 0.100\text{ mol dm}^{-3}\).
\(1.35 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.100}\)
\([\text{H}^+]^2 = 1.35 \times 10^{-6}\)
\([\text{H}^+] = 1.162 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.162 \times 10^{-3}) = 2.93\)

(c) Initial moles of acid: \(n(\text{acid}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\)
Initial moles of \(\text{NaOH}\): \(n(\text{NaOH}) = 0.0500\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\)
Reaction: \(\text{C}_2\text{H}_5\text{COOH} + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{COONa} + \text{H}_2\text{O}\)
Remaining moles of acid: \(5.00 \times 10^{-3} - 2.50 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol}\)
Moles of conjugate base formed: \(2.50 \times 10^{-3}\text{ mol}\)
Since \([\text{acid}] = [\text{conjugate base}]\), \(\text{pH} = \text{p}K_a\):
\(\text{pH} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\)

(a)
- Award [1] for a balanced dissociation equation with equilibrium arrows.
- Award [1] for identifying conjugate acid-base pairs correctly.
[2 marks]

(b)
- Award [1] for stating the assumption \([\text{H}^+] \approx [\text{C}_2\text{H}_5\text{COO}^-]\) OR \([\text{C}_2\text{H}_5\text{COOH}] \approx 0.100\text{ mol dm}^{-3}\).
- Award [1] for calculating \([\text{H}^+] = 1.16 \times 10^{-3}\text{ mol dm}^{-3}\).
- Award [1] for \(\text{pH} = 2.93\) (accept 2.9 or 2.94).
[3 marks]

(c)
- Award [1] for calculating initial moles of both reactants (\(5.00 \times 10^{-3}\text{ mol}\) and \(2.50 \times 10^{-3}\text{ mol}\)).
- Award [1] for establishing that after reaction, \([\text{acid}] = [\text{conjugate base}]\).
- Award [1] for calculating \(\text{pH} = \text{p}K_a = 4.87\).
[3 marks]

(a) The anode is where oxidation occurs. Since nickel has the more negative reduction potential (\(-0.25\text{ V}\) vs \(+0.34\text{ V}\) for copper), it is oxidized and serves as the anode.
Half-equation: \(\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-\)

(b) Standard cell potential:
\(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.34\text{ V} - (-0.25\text{ V}) = +0.59\text{ V}\)
Overall reaction: \(\text{Ni}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Ni}^{2+}(aq) + \text{Cu}(s)\)

(c) Standard IUPAC cell notation (anode on left, cathode on right):
\(\text{Ni}(s) \mid \text{Ni}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s)\)

(d) Electrons flow from the nickel anode to the copper cathode through the external wire. Cations in the salt bridge move towards the copper half-cell / cathode (to balance the decrease in positive charge as \(\text{Cu}^{2+}\) ions are reduced to \(\text{Cu}(s)\)).

(a)
- Award [1] for identifying Nickel (Ni) as the anode.
- Award [1] for the correct oxidation half-equation: \(\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-\).
[2 marks]

(b)
- Award [1] for calculating \(E^\theta_{\text{cell}} = +0.59\text{ V}\).
- Award [1] for correct overall equation: \(\text{Ni}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Ni}^{2+}(aq) + \text{Cu}(s)\).
[2 marks]

(c)
- Award [2] for correct standard cell notation: \(\text{Ni}(s) \mid \text{Ni}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s)\).
- Award [1] if the positions of anode/cathode are reversed but the phases are correct.
[2 marks]

(d)
- Award [1] for stating electrons flow from Ni to Cu through the external circuit.
- Award [1] for stating cations in the salt bridge move towards the copper half-cell / cathode.
[2 marks]

(a) Identify the variables and convert to SI units:
\(m = 0.324\text{ g}\)
\(T = 98.0 + 273.15 = 371.15\text{ K}\) (or 371 K)
\(P = 101.3\text{ kPa} = 1.013 \times 10^5\text{ Pa}\)
\(V = 118.0\text{ cm}^3 = 1.180 \times 10^{-4}\text{ m}^3\)
\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)

Using the ideal gas equation: \(PV = nRT\)
\(n = \frac{PV}{RT} = \frac{1.013 \times 10^5 \times 1.180 \times 10^{-4}}{8.31 \times 371.15}\)
\(n = \frac{11.953}{3084.26} = 3.876 \times 10^{-3}\text{ mol}\)

Molar mass (\(M\)):
\(M = \frac{m}{n} = \frac{0.324\text{ g}}{3.876 \times 10^{-3}\text{ mol}} = 83.6\text{ g mol}^{-1}\) (Accept 83.5 to 84.0 depending on rounding).

(b) Real gases deviate from ideal behavior because:
1. At high pressure, the volume of gas molecules themselves is no longer negligible compared to the total volume of the container.
2. At low temperatures, gas molecules move slower and intermolecular attractive forces become significant.

(c) Volume in \(\text{dm}^3\): \(V = 0.1180\text{ dm}^3\).
Density (\(d\)): \(d = \frac{m}{V} = \frac{0.324\text{ g}}{0.1180\text{ dm}^3} = 2.75\text{ g dm}^{-3}\).

(a)
- Award [1] for correct unit conversions: \(T = 371.15\text{ K}\) (or 371 K) AND \(V = 1.180 \times 10^{-4}\text{ m}^3\).
- Award [1] for correct rearrangement of ideal gas equation: \(n = \frac{PV}{RT}\).
- Award [1] for calculating \(n = 3.88 \times 10^{-3}\text{ mol}\).
- Award [1] for calculating molar mass \(83.6\text{ g mol}^{-1}\) (accept value in the range \(83.5 - 84.0\)).
[4 marks]

(b)
- Award [1] for explaining high pressure deviation: volume of molecules is significant.
- Award [1] for explaining low temperature deviation: intermolecular forces become significant.
[2 marks]

(c)
- Award [1] for volume conversion to \(0.1180\text{ dm}^3\).
- Award [1] for density = \(2.75\text{ g dm}^{-3}\).
[2 marks]

(a) Lattice enthalpy of dissociation is the enthalpy change required to separate one mole of an ionic solid into its gaseous ions under standard conditions.

(b)
(i) \(\text{Mg}(s) + \text{Cl}_2(g) \rightarrow \text{MgCl}_2(s)\)
(ii) \(\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g)\)

(c) The bond enthalpy of chlorine (\(\text{Cl}-\text{Cl}\)) corresponds to breaking 1 mole of \(\text{Cl}_2(g)\) bonds to form 2 moles of \(\text{Cl}(g)\). The standard enthalpy of atomization of chlorine is half of the bond enthalpy: \(\Delta H_{\text{at}}^\theta(\text{Cl}) = \frac{1}{2} \text{E}(\text{Cl}-\text{Cl}) = +121\text{ kJ mol}^{-1}\).

(d) Using the Born-Haber cycle equation:
\(\Delta H_f^\theta = \Delta H_{\text{at}}^\theta(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + \text{E}(\text{Cl}-\text{Cl}) + 2 \times \text{EA}(\text{Cl}) - \Delta H_{\text{lat}}\)

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

Therefore, the standard lattice enthalpy of dissociation is \(+2522\text{ kJ mol}^{-1}\).

(a)
- Award [1] for stating: enthalpy change to form gaseous ions from 1 mole of solid ionic compound.
[1 mark]

(b)
- (i) Award [1] for: \(\text{Mg}(s) + \text{Cl}_2(g) \rightarrow \text{MgCl}_2(s)\) (must include state symbols).
- (ii) Award [1] for: \(\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g)\) (must include state symbols).
[2 marks]

(c)
- Award [1] for stating that the bond enthalpy is twice the enthalpy of atomization (or \(\Delta H_{\text{at}}^\theta = \frac{1}{2} \text{E}\)).
[1 mark]

(d)
- Award [1] for correctly multiplying the electron affinity of chlorine by 2 (\(-698\text{ kJ mol}^{-1}\)).
- Award [1] for recognizing that 2 moles of gaseous Cl atoms require 1 mole of Cl2 bonds to be broken (no division of bond enthalpy by 2 needed).
- Award [1] for setting up the Born-Haber cycle equation correctly: \(\Delta H_{\text{lat}} = 148 + 738 + 1451 + 242 - 698 - (-641)\).
- Award [1] for correct final answer: \(+2522\text{ kJ mol}^{-1}\).
[4 marks]

(a)
Independent variable: Time
Dependent variable: Volume of oxygen gas (or gas evolved)

(b) (i)
Average rate between \(0\) and \(20\ \text{s}\):
\(\text{Rate} = \frac{26.0\ \text{cm}^3 - 0.0\ \text{cm}^3}{20.0\ \text{s} - 0\ \text{s}} = 1.30\ \text{cm}^3\ \text{s}^{-1}\)

Average rate between \(40\) and \(60\ \text{s}\):
\(\text{Rate} = \frac{46.0\ \text{cm}^3 - 40.0\ \text{cm}^3}{60.0\ \text{s} - 40.0\ \text{s}} = \frac{6.0\ \text{cm}^3}{20.0\ \text{s}} = 0.30\ \text{cm}^3\ \text{s}^{-1}\)

(ii)
Over time, the concentration of the reactant (\(\text{H}_2\text{O}_2\)) decreases. This results in a lower frequency of collisions between reactant particles per unit volume, reducing the rate of successful collisions and thus decreasing the rate of reaction.

(c) (i)
The calculated initial rate of reaction will be lower than the actual initial rate, because some oxygen gas escapes before the stopper is sealed, leading to an underestimate of the early volume of gas produced.

(ii)
Use a reaction vessel where the catalyst is suspended above the solution (e.g., in a small vial or on a thread) and can be dropped into the reactant by tilting or releasing without opening the system, OR use a syringe to inject the liquid reactant through a rubber septum into a flask containing the catalyst.

(a) Award [1] for both identified correctly, or [0.5] for only one.
- Independent variable: Time
- Dependent variable: Volume of oxygen gas (or gas evolved)

(b) (i) Award [1] for: \(1.30\ \text{cm}^3\ \text{s}^{-1}\) (or \(\text{cm}^3/\text{s}\))
Award [1] for: \(0.30\ \text{cm}^3\ \text{s}^{-1}\) (or \(\text{cm}^3/\text{s}\))
*Note: Units must be present in at least one of the answers to gain full marks.*

(ii) Award [1.5] maximum:
- Concentration of hydrogen peroxide / reactant decreases [0.5]
- Fewer reactant particles per unit volume, leading to lower frequency of collisions [0.5]
- Fewer successful collisions per unit time [0.5]

(c) (i) Award [1] maximum:
- Initial rate is lower/underestimated [0.5]
- Gas escapes before the system is sealed, so initial volumes are recorded as lower than actual [0.5]

(ii) Award [2] maximum:
- Suspend the catalyst in a small container/vial inside the flask and shake/tilt to start the reaction without opening [1] OR inject hydrogen peroxide via a syringe through a rubber septum into the flask containing the catalyst [1]
- Award [1] for any other logical setup that allows mixing without opening the system to the atmosphere.

(a) (i)
Mass of condensed vapor = \(85.834\ \text{g} - 85.250\ \text{g} = 0.584\ \text{g}\)
Absolute uncertainty of mass = \(\pm(0.001 + 0.001) = \pm 0.002\ \text{g}\)

(ii)
Percentage uncertainty of volume = \(\frac{0.5}{250.0} \times 100 = 0.20\%\) (or \(0.2\%\))

(b)
Convert units to SI units:
\(T = 99.5 + 273.15 = 372.65\ \text{K}\) (accept \(372.7\ \text{K}\) or \(373\ \text{K}\))
\(V = 250.0 \times 10^{-6}\ \text{m}^3 = 2.500 \times 10^{-4}\ \text{m}^3\) (or \(0.2500\ \text{dm}^3\))
\(P = 100.8 \times 10^3\ \text{Pa}\) (or \(100.8\ \text{kPa}\))

Using \(PV = \frac{m}{M}RT\), rearrange for \(M\):
\(M = \frac{mRT}{PV}\)
\(M = \frac{0.584\ \text{g} \times 8.31\ \text{J}\ \text{K}^{-1}\ \text{mol}^{-1} \times 372.65\ \text{K}}{100.8 \times 10^3\ \text{Pa} \times 2.500 \times 10^{-4}\ \text{m}^3}\)
\(M = \frac{1808.49}{25.2} = 71.765\ \text{g}\ \text{mol}^{-1}\)
Rounding to 3 significant figures: \(71.8\ \text{g}\ \text{mol}^{-1}\) (accept \(71.7\ \text{g}\ \text{mol}^{-1}\) if \(373\ \text{K}\) or \(273\ \text{K}\) conversion is used).

(c) (i)
Higher. If the liquid does not completely vaporize, excess liquid remains in the flask, artificially increasing the measured mass of condensed vapor (\(m\)), which leads to a higher calculated molar mass.

(ii)
Higher. Water on the outside of the flask adds to the measured mass of the flask + condensed liquid, resulting in an overestimate of the vapor mass (\(m\)) and a higher calculated molar mass.

(a) (i) Award [1.5]:
- \(0.584\ \text{g}\) [0.5]
- \(\pm 0.002\ \text{g}\) [1]

(ii) Award [1]:
- \(0.20\%\) (accept \(0.2\%\))

(b) Award [3] maximum:
- Temperature conversion to \(372.65\ \text{K}\) / \(373\ \text{K}\) AND Volume conversion to \(2.500 \times 10^{-4}\ \text{m}^3\) (or compatible units for \(P\) and \(V\)) [1]
- Correct rearrangement: \(M = \frac{mRT}{PV}\) [1]
- Final answer to 3 significant figures: \(71.8\ \text{g}\ \text{mol}^{-1}\) (accept \(71.7\ \text{g}\ \text{mol}^{-1}\)) [1]
*Note: Award [2] max if units are missing/incorrect, or if the number of significant figures is incorrect.*

(c) (i) Award [1]:
- Higher / Overestimated AND because excess liquid remains which increases the mass (\(m\)).

(ii) Award [1]:
- Higher / Overestimated AND because the external water increases the final mass reading of the flask, overestimating the vapor mass (\(m\)).

Part (a):
In the preparation of aspirin, salicylic acid is acetylated using ethanoic anhydride. This reaction is catalyzed by a strong acid, typically concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) or concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)). The reaction produces aspirin along with ethanoic acid (\(\text{CH}_3\text{COOH}\)) as a side-product.

Part (b):
Aspirin is relatively insoluble in water due to the presence of the hydrophobic benzene ring and carboxylic acid group. Its solubility (and thus bioavailability) is improved by converting it into an ionic salt, sodium acetylsalicylate, by reacting it with a strong base such as sodium hydroxide (\(\text{NaOH}\)) or sodium hydrogencarbonate (\(\text{NaHCO}_3\)). The ionic nature of the salt allows it to form strong ion-dipole interactions with polar water molecules.
Equation:
\(\text{C}_9\text{H}_8\text{O}_4\text{(s)} + \text{NaOH(aq)} \rightarrow \text{NaC}_9\text{H}_7\text{O}_4\text{(aq)} + \text{H}_2\text{O(l)}\)
(Alternative with sodium hydrogencarbonate is also accepted: \(\text{C}_9\text{H}_8\text{O}_4\text{(s)} + \text{NaHCO}_3\text{(aq)} \rightarrow \text{NaC}_9\text{H}_7\text{O}_4\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\))

Part (a):
- Award [1] for concentrated sulfuric acid / \(\text{H}_2\text{SO}_4\) OR concentrated phosphoric acid / \(\text{H}_3\text{PO}_4\). Reject \(\text{HCl}\) or dilute acids.
- Award [1] for \(\text{CH}_3\text{COOH}\) / ethanoic acid.

Part (b):
- Award [1] for stating that aspirin is reacted with a strong base / \(\text{NaOH}\) / \(\text{NaHCO}_3\) to convert it to an ionic/sodium salt.
- Award [1] for explaining that this increases solubility because the ionic salt forms stronger intermolecular forces (ion-dipole) with water.
- Award [1] for a correct balanced equation: \(\text{C}_9\text{H}_8\text{O}_4 + \text{NaOH} \rightarrow \text{NaC}_9\text{H}_7\text{O}_4 + \text{H}_2\text{O}\) (or alternative with \(\text{NaHCO}_3\)). State symbols are not required.

Part (a):
Morphine contains two hydroxyl (\(-\text{OH}\)) groups (one phenolic and one aliphatic) that undergo conversion to ester groups during the synthesis of diamorphine.

Part (b):
Morphine contains polar hydroxyl groups that participate in hydrogen bonding, making it relatively polar and less soluble in lipids. Diamorphine contains two ester groups, which are much less polar. Consequently, diamorphine is more lipid-soluble. The blood-brain barrier consists of a lipid-rich, non-polar membrane. Because of its lipophilic character, diamorphine crosses this barrier much more rapidly and efficiently than morphine, leading to higher concentration at opioid receptors and a stronger, faster effect.

Part (c):
Diamorphine is synthesized from morphine via an esterification (or condensation/acetylation) reaction using ethanoic anhydride (or acetyl chloride) as the reagent.

Part (a):
- Award [1] for "hydroxyl" or "phenol" or "alcohol". Do NOT accept hydroxide.

Part (b):
- Award [1] for stating that morphine is polar/hydrophilic due to hydroxyl groups AND diamorphine is less polar/hydrophobic/lipophilic due to ester groups.
- Award [1] for stating that the blood-brain barrier is lipid-based / non-polar / lipid-soluble.
- Award [1] for concluding that diamorphine penetrates / crosses the blood-brain barrier more readily/faster.

Part (c):
- Award [1] for esterification / condensation / acetylation AND ethanoic anhydride / acetyl chloride.

Part (a):
The neutralization of hydrochloric acid by magnesium hydroxide produces magnesium chloride and water:
\(\text{Mg(OH)}_2\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + 2\text{H}_2\text{O(l)}\)

Part (b):
1. Calculate the amount in moles of \(\text{HCl}\):
\(n(\text{HCl}) = 0.120\text{ mol dm}^{-3} \times 0.1500\text{ dm}^3 = 0.0180\text{ mol}\)
2. Determine the moles of \(\text{Mg(OH)}_2\) required from the stoichiometric ratio (1:2):
\(n(\text{Mg(OH)}_2) = \frac{0.0180\text{ mol}}{2} = 0.00900\text{ mol}\)
3. Calculate the mass of \(\text{Mg(OH)}_2\):
\(m = 0.00900\text{ mol} \times 58.33\text{ g mol}^{-1} = 0.525\text{ g}\)

Part (c):
- Ranitidine is an \(\text{H}_2\)-receptor antagonist. It acts by blocking histamine receptors on parietal cells, preventing the signal that triggers gastric acid production.
- Omeprazole is a proton pump inhibitor (PPI). It works downstream by directly binding to and inhibiting the \(\text{H}^+/\text{K}^+\)-ATPase enzyme (the proton pump) that releases hydrogen ions into the gastric juice.

Part (a):
- Award [1] for a correct balanced equation: \(\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}\). State symbols are not required.

Part (b):
- Award [1] for calculating \(n(\text{HCl}) = 0.0180\text{ mol}\) AND \(n(\text{Mg(OH)}_2) = 0.00900\text{ mol}\).
- Award [1] for \(0.525\text{ g}\) (accept answers in range \(0.52\) to \(0.53\)). Award [2] for correct final answer.

Part (c):
- Award [1] for stating ranitidine blocks \(\text{H}_2\) / histamine receptors (in parietal cells).
- Award [1] for stating omeprazole inhibits proton pumps / \(\text{H}^+/\text{K}^+\)-ATPase enzyme.

Part (a):
Both oseltamivir and zanamivir are neuraminidase inhibitors. The enzyme neuraminidase is responsible for cleaving sialic acid on the glycoprotein receptors of host cell membranes, allowing newly formed viral particles (virions) to escape the host cell. By inhibiting this enzyme, the virions remain trapped on the host cell surface, halting further viral replication and spreading to healthy cells.

Part (b):
Comparing their structures, zanamivir has a guanidino group (\(-\text{NH}-\text{C}(=\text{NH})-\text{NH}_2\)) and multiple hydroxyl (\(-\text{OH}\)) groups (polyhydroxyl tail) which are absent in oseltamivir (oseltamivir contains an ester and an ether group instead).

Part (c):
Large quantities of oseltamivir and its active metabolite are excreted into waterways through sewage systems. This continuous low-dose exposure in aquatic environments, particularly to wild fowl (natural reservoirs of influenza), can lead to the natural selection and emergence of antiviral-resistant strains of the influenza virus, rendering the medication ineffective.

Part (a):
- Award [1] for naming the enzyme neuraminidase.
- Award [1] for explaining that inhibition prevents newly formed virus particles / virions from escaping/being released from infected cells.

Part (b):
- Award [1] each for identifying any two of: guanidino group (accept guanidine) / hydroxyl group (accept alcohol / polyhydroxyl). Do NOT accept amine, carboxamide, or carboxylic acid (as these are not absent/distinguishing in the same way, or are not in zanamivir).

Part (c):
- Award [1] for any of: leads to development of drug-resistant strains of influenza (in wild bird populations/aquatic environments) OR bioaccumulation in food chains OR toxicity to aquatic ecosystems.