2023 IB DP Chemistry 模擬試題連答案詳解

In \(\text{SO}_2\), there are 3 electron domains around the central sulfur atom (one lone pair and two bonding domains). This gives a trigonal planar electron domain geometry, leading to a bent molecular geometry with a bond angle of slightly less than \(120^\circ\) (specifically \(119^\circ\)). In contrast, \(\text{H}_2\text{O}\) has 4 electron domains (bent, \(104.5^\circ\)), \(\text{NH}_3\) has 4 electron domains (trigonal pyramidal, \(107^\circ\)), and \(\text{CO}_2\) has 2 electron domains (linear, \(180^\circ\)).

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

The equilibrium constant, \(K_c\), is temperature-dependent only. Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering/decreasing the temperature shifts the position of equilibrium to the right (favouring the exothermic direction to produce heat). This increases the concentration of products and decreases the concentration of reactants, thereby increasing the value of \(K_c\). Changes in pressure, concentration, or the addition of a catalyst do not affect the value of \(K_c\).

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

According to the Brønsted-Lowry theory, a base is a proton (\(\text{H}^+\)) acceptor. In option D, \(\text{H}_2\text{PO}_4^-(aq)\) accepts a proton from hydrochloric acid (\(\text{HCl}\)) to form phosphoric acid (\(\text{H}_3\text{PO}_4\)). In the other reactions, \(\text{H}_2\text{PO}_4^-\rightleftharpoons \text{HPO}_4^{2-}\), showing that it donates a proton and thus acts as a Brønsted-Lowry acid.

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

Assume a \(100\text{ g}\) sample of the oxide. Mass of \(\text{S} = 50.0\text{ g}\) and mass of \(\text{O} = 50.0\text{ g}\). Moles of \(\text{S} = \frac{50.0}{32.07} \approx 1.56\text{ mol}\). Moles of \(\text{O} = \frac{50.0}{16.00} = 3.125\text{ mol}\). Divide by the smallest value to find the ratio: \(\text{S}: \frac{1.56}{1.56} = 1\), \(\text{O}: \frac{3.125}{1.56} \approx 2\). The empirical formula is \(\text{SO}_2\).

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

We can calculate the oxidation states of chlorine in each species: A) In \(\text{ClO}_2^-\), \(\text{Cl} + 2(-2) = -1 \implies \text{Cl} = +3\). B) In \(\text{ClO}_3^-\), \(\text{Cl} + 3(-2) = -1 \implies \text{Cl} = +5\). C) In \(\text{Cl}_2\text{O}_7\), \(2(\text{Cl}) + 7(-2) = 0 \implies 2\text{Cl} = 14 \implies \text{Cl} = +7\). D) In \(\text{HClO}\), \(+1 + \text{Cl} + (-2) = 0 \implies \text{Cl} = +1\). Therefore, chlorine has the highest oxidation state of \(+7\) in \(\text{Cl}_2\text{O}_7\).

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

A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_a\)). This increases the fraction of collisions with energy equal to or greater than the activation energy, thereby increasing the rate of reaction. It does not affect the average kinetic energy (which depends on temperature), nor does it change the overall enthalpy change (\(\Delta H\)) or significantly alter the total number of collisions per second.

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

The structure contains an \(-\text{OH}\) group attached to a saturated carbon atom, which represents an alcohol (hydroxyl) group. It also contains a \(-\text{COO}-\) group bonded to a methyl group (\(-\text{COOCH}_3\)), which is an ester group. Therefore, the compound contains both alcohol and ester functional groups.

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

The atomic number of Cobalt (Co) is 27, which represents the number of protons. The mass number is 59, which is the sum of protons and neutrons, meaning the number of neutrons is \(59 - 27 = 32\). The charge of \(2+\) indicates the species has lost 2 electrons compared to its neutral state, so the number of electrons is \(27 - 2 = 25\). This corresponds to option A.

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

An alcohol is classified as tertiary when the carbon atom bonded to the hydroxyl (-OH) group is directly attached to three other carbon atoms. Let us analyze the structures of the options: a) 2-methylbutan-2-ol has the formula CH3-C(OH)(CH3)-CH2-CH3. The carbon with the -OH group is bonded to three other carbons (two methyl groups and one ethyl group), making it a tertiary alcohol. b) 3-methylbutan-2-ol is CH3-CH(CH3)-CH(OH)-CH3, where the carbon with the -OH group is bonded to two other carbons, making it a secondary alcohol. c) pentan-3-ol is CH3-CH2-CH(OH)-CH2-CH3, which is a secondary alcohol. d) 2,2-dimethylpropan-1-ol is (CH3)3C-CH2-OH, where the carbon with the -OH group is bonded to only one other carbon, making it a primary alcohol.

Award [1] for the correct option A. No marks are given for incorrect options.

To find the molecular geometry, we determine the number of bonding and lone pairs on the central atom: a) In BF3, boron has 3 valence electrons and forms 3 single bonds with no lone pairs, resulting in a trigonal planar geometry. b) In PCl3, phosphorus has 5 valence electrons, forms 3 single bonds, and has 1 lone pair on the central atom. This results in a trigonal pyramidal molecular geometry. c) In SO3, sulfur has 6 valence electrons and forms bonds with 3 oxygen atoms (with no lone pairs on the sulfur), leading to a trigonal planar geometry. d) In CO3^{2-}, the carbonate ion has a trigonal planar geometry due to its 3 electron domains and no lone pairs on the central carbon atom.

Award [1] for the correct option B. No marks are given for incorrect options.

Let us calculate the oxidation state of sulfur in each species: a) In H2SO3: 2(+1) + S + 3(-2) = 0, which gives S = +4. b) In S2O3^{2-}: 2(S) + 3(-2) = -2, which gives 2S = +4, so S = +2. c) In SF6: S + 6(-1) = 0, which gives S = +6. d) In S8: elemental sulfur has an oxidation state of 0. Therefore, sulfur has the highest oxidation state of +6 in SF6.

Award [1] for the correct option C. No marks are given for incorrect options.

A homogeneous mixture has a uniform composition throughout its mass. Brass is an alloy, which is a solid solution of copper and zinc, and is uniform throughout, making it a homogeneous mixture. Suspensions of chalk in water, mixtures of sand and salt, and mixtures of sulfur and iron filings are non-uniform and are classified as heterogeneous mixtures.

Award [1] for the correct option A. No marks are given for incorrect options.

One formula unit of aluminum sulfate, \(\text{Al}_2(\text{SO}_4)_3\), dissociates into 5 ions in total: 2 \(\text{Al}^{3+}\) ions and 3 \(\text{SO}_4^{2-}\) ions. Therefore, the total amount of ions in \(0.20\text{ mol}\) of \(\text{Al}_2(\text{SO}_4)_3\) is \(0.20\text{ mol} \times 5 = 1.0\text{ mol}\). The total number of ions is \(1.0 \times L = L\).

Award [1] for the correct option C. No marks are given for incorrect options.

First, write the balanced chemical equation for the complete combustion of methane: \(\text{CH}_4\text{ (g)} + 2\text{O}_2\text{ (g)} \rightarrow \text{CO}_2\text{ (g)} + 2\text{H}_2\text{O}\text{ (l)}\). Next, calculate the moles of methane: \(n(\text{CH}_4) = \frac{4.0\text{ g}}{16.0\text{ g mol}^{-1}} = 0.25\text{ mol}\). From the stoichiometry of the equation, 1 mole of \(\text{CH}_4\) requires 2 moles of \(\text{O}_2\), so: \(n(\text{O}_2) = 2 \times 0.25\text{ mol} = 0.50\text{ mol}\). Finally, calculate the volume of oxygen at STP: \(V = 0.50\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 11.35\text{ dm}^3\), which rounds to \(11.4\text{ dm}^3\).

Award [1] for the correct option B. No marks are given for incorrect options.

For the ion \(^{79}_{34}\text{X}^{2-}\): The atomic number (subscript 34) represents the number of protons, so Protons = 34. The mass number (superscript 79) is the sum of protons and neutrons, so Neutrons = 79 - 34 = 45. The negative charge of 2- indicates that the ion has two more electrons than protons, so Electrons = 34 + 2 = 36.

Award [1] for the correct option B. No marks are given for incorrect options.

The atomic number of iron (Fe) is 26, so a neutral iron atom has the ground-state electron configuration \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). When transition metals form positive ions, they lose electrons from the outer valence shell (4s subshell) first before losing electrons from the 3d subshell. Therefore, to form \(\text{Fe}^{2+}\), the two electrons in the 4s subshell are removed, leaving the configuration as \([\text{Ar}] 3\text{d}^6\).

Award [1] for the correct option C. No marks are given for incorrect options.

One mole of sulfuric acid, \( \text{H}_2\text{SO}_4 \), contains 7 moles of atoms (2 moles of hydrogen, 1 mole of sulfur, and 4 moles of oxygen). Therefore, 0.50 mol of \( \text{H}_2\text{SO}_4 \) contains \( 0.50 \times 7 = 3.5 \text{ mol} \) of atoms. The number of atoms is equal to the amount in moles multiplied by Avogadro's constant, which is \( 3.5 L \).

Award [1] for the correct choice C. No partial credit.

The carbonate ion, \( \text{CO}_3^{2-} \), has a central carbon atom with three bonding electron domains and no lone pairs, resulting in a trigonal planar geometry. This gives a bond angle of \( 120^\circ \). Ammonia, \( \text{NH}_3 \), and hydronium, \( \text{H}_3\text{O}^+ \), are both trigonal pyramidal with bond angles of around \( 107^\circ \). Ammonium, \( \text{NH}_4^+ \), is tetrahedral with a bond angle of \( 109.5^\circ \).

Award [1] for the correct choice A. No partial credit.

Let's calculate the oxidation states of sulfur: In \( \text{S}_2\text{O}_3^{2-} \), \( 2x + 3(-2) = -2 \Rightarrow x = +2 \). In \( \text{SO}_3^{2-} \), \( x + 3(-2) = -2 \Rightarrow x = +4 \). In \( \text{SO}_4^{2-} \), \( x + 4(-2) = -2 \Rightarrow x = +6 \). In \( \text{S}_8 \), sulfur is in its elemental form, so its oxidation state is 0. Therefore, sulfur has the highest oxidation state in \( \text{SO}_4^{2-} \).

Award [1] for the correct choice C. No partial credit.

Using the equation \( q = mc\Delta T \), where \( m = 50.0 \text{ g} \), \( c = 4.18 \text{ J g}^{-1} \text{ K}^{-1} \), and \( \Delta T = 40.0 - 20.0 = 20.0 \text{ K} \): \( q = 50.0 \times 4.18 \times 20.0 = 1000 \times 4.18 = 4180 \text{ J} = 4.18 \text{ kJ} \).

Award [1] for the correct choice A. No partial credit.

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This allows more reactant particles to have kinetic energy greater than or equal to the activation energy. It does not alter the average kinetic energy of the particles (which depends on temperature) or the enthalpy change of the reaction.

Award [1] for the correct choice C. No partial credit.

The molecular formula shows a carbonyl group (\( \text{C=O} \)) bonded to two carbon chains, which defines the ketone functional group class (specifically, this compound is butanone).

Award [1] for the correct choice B. No partial credit.

A homogeneous mixture has a uniform composition throughout. Aqueous sodium chloride (salt water) is a solution and is uniform throughout, making it homogeneous. Sand and water, and sulfur powder and iron filings are heterogeneous mixtures. Solid sodium chloride is a pure compound, not a mixture.

Award [1] for the correct choice B. No partial credit.

A conjugate acid-base pair consists of two species that differ by only a single proton (\( \text{H}^+ \)). Here, \( \text{HPO}_4^{2-} \) is a Brønsted-Lowry acid that donates a proton to form its conjugate base, \( \text{PO}_4^{3-} \).

Award [1] for the correct choice A. No partial credit.

During a phase change (represented by the plateaus on a heating curve), the temperature of the substance remains constant. Since temperature is a measure of the average kinetic energy of the particles, the average kinetic energy does not change during a plateau. Instead, the heat energy absorbed is used to overcome intermolecular forces, which increases the potential energy of the particles. Therefore, at the plateau at \(T_2\) (which represents boiling), the potential energy of the particles increases. Between \(T_1\) (melting point) and \(T_2\) (boiling point), the substance is entirely in the liquid phase. Phase changes of molecular substances are physical changes and do not involve breaking intramolecular covalent bonds.

Award [1] for the correct answer (B). Reject other options: A is incorrect because temperature (and thus kinetic energy) is constant during plateaus. C is incorrect because between the two plateaus the substance is a liquid only. D is incorrect because boiling is a physical change, not a chemical change breaking intramolecular bonds.

The given molecule is \(\text{CH}_2=\text{CHCH}_2\text{COOCH}_3\). It contains: 1. A carbon-carbon double bond (\(\text{C}=\text{C}\)), which belongs to the alkene homologous series. 2. An ester functional group (\(\text{—COO—}\)), which belongs to the ester class of compounds. Therefore, the molecule belongs to the alkene and ester classes.

Award [1] for selecting A. Reject all other options as they contain incorrect classes (ether, alkyne, ketone) not present in the molecule.

Let's analyze the electron domain geometry and molecular geometry of each species: - \(\text{CO}_3^{2-}\) has 3 electron domains around the central carbon atom (no lone pairs). Its geometry is trigonal planar, which has bond angles of exactly \(120^\circ\) due to resonance. - \(\text{PCl}_3\) has 4 electron domains (3 bonding, 1 lone pair) around the central phosphorus atom. Its molecular geometry is trigonal pyramidal, with bond angles of approximately \(107^\circ\). - \(\text{NH}_4^+\) has 4 bonding electron domains (no lone pairs) around nitrogen. Its geometry is tetrahedral, with bond angles of \(109.5^\circ\). - \(\text{H}_2\text{O}\) has 4 electron domains (2 bonding, 2 lone pairs) around oxygen. Its molecular geometry is bent, with a bond angle of \(104.5^\circ\).

Award [1] for the correct choice (A). Reject other choices as their bond angles are significantly different from \(120^\circ\) (\(107^\circ\), \(109.5^\circ\), and \(104.5^\circ\) respectively).

1. For \(\text{K}_2\text{Cr}_2\text{O}_7\): The sum of oxidation states is zero. Potassium (\(\text{K}\)) has an oxidation state of \(+1\) and oxygen (\(\text{O}\)) has \(-2\). Let \(x\) be the oxidation state of chromium: \(2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6\). 2. For \(\text{Na}_2\text{S}_2\text{O}_3\): Sodium (\(\text{Na}\)) has an oxidation state of \(+1\) and oxygen (\(\text{O}\)) has \(-2\). Let \(y\) be the oxidation state of sulfur: \(2(+1) + 2(y) + 3(-2) = 0 \implies 2 + 2y - 6 = 0 \implies 2y = 4 \implies y = +2\). Thus, \(\text{Cr} = +6\) and \(\text{S} = +2\).

Award [1] for A. Reject other combinations of oxidation states.

Each formula unit of magnesium chloride, \(\text{MgCl}_2\), dissociates into one magnesium ion (\(\text{Mg}^{2+}\)) and two chloride ions (\(\text{Cl}^-\)). Therefore, the number of moles of \(\text{Cl}^-\) ions in \(0.10\text{ mol}\) of \(\text{MgCl}_2\) is: \(n(\text{Cl}^-) = 2 \times 0.10\text{ mol} = 0.20\text{ mol}\). The number of chloride ions is: \(N = n \times L = 0.20\text{ mol} \times 6.0 \times 10^{23}\text{ mol}^{-1} = 1.2 \times 10^{23}\).

Award [1] for correct answer (B). Award [0] for A (which represents moles of \(\text{MgCl}_2\) or \(\text{Mg}^{2+}\) ions converted to particles: \(0.10 \times 6.0 \times 10^{23}\)), or other incorrect calculations.

To determine the limiting reactant, compare the ratio of available moles to the stoichiometric coefficients: For \(\text{Al}\): \(0.40 / 2 = 0.20\). For \(\text{Cl}_2\): \(0.45 / 3 = 0.15\). Since \(0.15 < 0.20\), \(\text{Cl}_2\) is the limiting reactant. Use the limiting reactant to calculate the moles of \(\text{AlCl}_3\) produced: \(n(\text{AlCl}_3) = n(\text{Cl}_2) \times (2 / 3) = 0.45\text{ mol} \times (2 / 3) = 0.30\text{ mol}\).

Award [1] for B. Award [0] for C (which assumes Al is the limiting reactant).

(a)
- In \(\text{Cr}_2\text{O}_7^{2-}\), each oxygen is \(-2\). Let \(x\) be the oxidation state of Cr:
\(2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = +12 \implies x = +6\).
- In \(\text{Cr}^{3+}\), the oxidation state is +3.

(b)
- At the anode (oxidation): \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\)
- At the cathode (reduction): \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\)

(c)
- Overall reaction: To balance electrons, multiply oxidation by 2 and reduction by 3:
\(2\text{Cr}(\text{s}) + 3\text{Cu}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{Cu}(\text{s})\)
- Standard cell potential:
\(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.34\text{ V} - (-0.74\text{ V}) = +1.08\text{ V}\)

(d)
- Electrons flow from the chromium electrode (anode / more negative potential) to the copper electrode (cathode / more positive potential) through the external wire.
- The salt bridge completes the electrical circuit and maintains electrical neutrality by allowing the migration of anions to the anode half-cell and cations to the cathode half-cell.

(e)
- Decreasing the concentration of \(\text{Cu}^{2+}(\text{aq})\) decreases the concentration of a reactant.
- According to Le Chatelier's principle, this shifts the equilibrium position to the left (towards the reactants), reducing the driving force for the forward reaction and lowering the cell potential.

(a)
- +6 (or VI) [1 mark]
- +3 (or III) [1 mark]

(b)
- \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\) [1 mark]
- \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\) [1 mark]
*Note: Accept reversible arrows but penalize missing state symbols or incorrect electron balance.*

(c)
- \(2\text{Cr}(\text{s}) + 3\text{Cu}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{Cu}(\text{s})\) [1 mark] (must be balanced)
- \(E^\theta_{\text{cell}} = +1.08\text{ V}\) [1 mark]

(d)
- Electron flow: Chromium (anode) to copper (cathode) [1 mark]
- Salt bridge: Completes the circuit / maintains electrical neutrality by ion migration [1 mark]

(e)
- Equilibrium shifts to the left / reactants side [1 mark]
- Cell potential decreases [1 mark]

(a)
- Carboxylic acid isomer: Butanoic acid. Structure: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\) (or \(\text{(CH}_3)_2\text{CHCOOH}\), methylpropanoic acid).
- Ester isomer: Ethyl ethanoate. Structure: \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (or methyl propanoate, \(\text{CH}_3\text{CH}_2\text{COOCH}_3\); or propyl methanoate, \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)).

(b)
- Carboxylic acid (butanoic acid): London dispersion forces, dipole-dipole forces, and hydrogen bonding.
- Ester (ethyl ethanoate): London dispersion forces and dipole-dipole forces only (no hydrogen bonding between ester molecules).

(c)
- Butanoic acid has a higher boiling point.
- It contains a polar \(\text{-O-H}\) bond which allows molecules to form strong intermolecular hydrogen bonds (often forming stable dimers). These require more thermal energy to overcome than the weaker dipole-dipole interactions between ester molecules.

(d)
- A structural isomer containing a ketone and a hydroxyl group is 3-hydroxybutan-2-one: \(\text{CH}_3\text{-C(=O)-CH(OH)-CH}_3\).
- The chiral carbon is carbon-3: \(\text{CH}_3\text{-C(=O)-C*H(OH)-CH}_3\) because it is bonded to four different groups: \(\text{-H}\), \(\text{-OH}\), \(\text{-CH}_3\), and \(\text{-COCH}_3\).

(a)
- Correct structure of butanoic acid (or methylpropanoic acid) [1 mark]
- Correct corresponding name: butanoic acid (or methylpropanoic acid) [1 mark]
- Correct structure of ethyl ethanoate (or other valid ester) [1 mark]
- Correct corresponding name: ethyl ethanoate (or other ester name) [1 mark]

(b)
- Carboxylic acid has hydrogen bonding (and London/dipole-dipole forces) [1 mark]
- Ester has dipole-dipole forces and London forces (no hydrogen bonding) [1 mark]

(c)
- Correct prediction: Butanoic acid (carboxylic acid) has the higher boiling point [1 mark]
- Explanation: Carboxylic acid molecules can form strong intermolecular hydrogen bonds (dimers) which require more energy to break than the dipole-dipole attractions in the ester [1 mark]

(d)
- Correct structural formula of 3-hydroxybutan-2-one (or other valid ketone-alcohol) [1 mark]
- Chiral carbon clearly circled/identified on carbon-3 [1 mark]

(a)
- Balanced equation with state symbols:
\(\text{CH}_3\text{OH}(\text{l}) + 1.5\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\)
or:
\(2\text{CH}_3\text{OH}(\text{l}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\)

(b)
- Temperature change: \(\Delta T = 41.1 - 19.5 = 21.6\text{ K}\) (or \(^\circ\text{C}\))
- \(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 21.6\text{ K}\)
- \(q = 13543.2\text{ J} = 13.5\text{ kJ}\) (to 3 significant figures)

(c)
- Moles of methanol: \(M(\text{CH}_3\text{OH}) = 12.01 + 4(1.01) + 16.00 = 32.05\text{ g mol}^{-1}\)
- \(n = \frac{0.800\text{ g}}{32.05\text{ g mol}^{-1}} = 0.02496\text{ mol}\)
- \(\Delta H_{\text{c}} = -\frac{q}{n} = -\frac{13.5432\text{ kJ}}{0.02496\text{ mol}} = -543\text{ kJ mol}^{-1}\) (or \(-542\text{ kJ mol}^{-1}\) depending on rounding)

(d)
- Reasons for error (any two):
1. Heat loss to the surroundings / calorimeter beaker.
2. Incomplete combustion of methanol (producing CO/C instead of \(\text{CO}_2\)).
3. Evaporation of methanol from the wick before/after burning.
4. Specific heat capacity of the copper calorimeter was ignored.
- Minimizing errors:
- Heat loss can be minimized by insulating the beaker or using a draft shield / bomb calorimeter.

(a)
- Correct species and balancing [1 mark]
- Correct state symbols (l, g, g, l) [1 mark]

(b)
- Correct calculation of \(\Delta T = 21.6\text{ K}\) [1 mark]
- \(q = 13.5\text{ kJ}\) (accept \(13.54\text{ kJ}\)) [1 mark]

(c)
- Moles of methanol = \(0.0250\text{ mol}\) [1 mark]
- Division of heat by moles [1 mark]
- \(\Delta H_{\text{c}} = -543\text{ kJ mol}^{-1}\) (must have negative sign; accept range \(-542\) to \(-543\)) [1 mark]

(d)
- Two reasons for discrepancy (e.g. heat loss, incomplete combustion) [2 marks]
- Suggestion to minimize one error (e.g. use draft shield, insulate beaker, use bomb calorimeter) [1 mark]

(a)
- For \(\text{SF}_4\): The central sulfur atom has 6 valence electrons, and each fluorine has 7. Total valence electrons = 34. The Lewis structure shows sulfur forming 4 single bonds with the 4 F atoms, with 1 lone pair on the sulfur atom. Each F atom has 3 lone pairs (8 electrons in total around each F, 10 around S).
- For \(\text{XeF}_4\): The central xenon atom has 8 valence electrons, and each fluorine has 7. Total valence electrons = 36. The Lewis structure shows xenon forming 4 single bonds with the 4 F atoms, with 2 lone pairs on the xenon atom. Each F atom has 3 lone pairs (8 electrons around each F, 12 around Xe).

(b)
- \(\text{SF}_4\):
- Shape: Seesaw (or sawhorse)
- Bond angles: less than \(90^\circ\) and less than \(120^\circ\) (specifically axial-axial is \(<180^\circ\) or \(\approx 173^\circ\), and equatorial-equatorial is \(<120^\circ\) or \(\approx 102^\circ\))
- \(\text{XeF}_4\):
- Shape: Square planar
- Bond angles: \(90^\circ\) (and \(180^\circ\))

(c)
- In \(\text{XeF}_4\), the square planar geometry is highly symmetrical. The four polar Xe-F bond dipoles point in opposite directions and cancel each other out, and the two lone pairs lie directly opposite each other, meaning there is no net dipole moment.
- In \(\text{SF}_4\), the seesaw geometry is unsymmetrical. The dipoles of the polar S-F bonds do not cancel out, resulting in a net molecular dipole moment.

(a)
- Correct Lewis structure for \(\text{SF}_4\) (4 single bonds, 1 lone pair on S, 3 lone pairs on each F) [2 marks]
- Correct Lewis structure for \(\text{XeF}_4\) (4 single bonds, 2 lone pairs on Xe, 3 lone pairs on each F) [2 marks]
*Deduct 1 mark overall if fluorine lone pairs are completely omitted.*

(b)
- \(\text{SF}_4\) shape: Seesaw [1 mark]
- \(\text{SF}_4\) angles: \(<90^\circ\) and \(<120^\circ\) (accept values around \(102^\circ\) and \(173^\circ\)) [1 mark]
- \(\text{XeF}_4\) shape: Square planar [1 mark]
- \(\text{XeF}_4\) angles: \(90^\circ\) (or \(180^\circ\)) [1 mark]

(c)
- \(\text{XeF}_4\) is symmetrical and individual bond dipoles cancel out [1 mark]
- \(\text{SF}_4\) is unsymmetrical (due to the single lone pair) so dipoles do not cancel out [1 mark]

(a)
- **With respect to \(\text{S}_2\text{O}_8^{2-}\)**: Comparing Experiments 1 and 2, \([\text{I}^-]\) is constant, and \([\text{S}_2\text{O}_8^{2-}]\) is doubled (from \(0.010\) to \(0.020\text{ mol dm}^{-3}\)). The rate also doubles (from \(1.5 \times 10^{-5}\) to \(3.0 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\)). Therefore, the reaction is **first order** with respect to \(\text{S}_2\text{O}_8^{2-}\).
- **With respect to \(\text{I}^-\)**: Comparing Experiments 1 and 3, \([\text{S}_2\text{O}_8^{2-}]\) is constant, and \([\text{I}^-]\) is doubled (from \(0.010\) to \(0.020\text{ mol dm}^{-3}\)). The rate also doubles (from \(1.5 \times 10^{-5}\) to \(3.0 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\)). Therefore, the reaction is **first order** with respect to \(\text{I}^-\).

(b)
- \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

(c)
- Using Experiment 1:
\(1.5 \times 10^{-5} = k (0.010)(0.010)\)
\(1.5 \times 10^{-5} = k (1.0 \times 10^{-4})\)
\(k = 0.15\)
- Units: \(\text{mol dm}^{-3}\text{ s}^{-1} / (\text{mol dm}^{-3})^2 = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(d)
- **Sketch details**:
- y-axis labeled as "Number of particles" / "Fraction of particles" and x-axis labeled as "Kinetic energy".
- Both curves start at the origin and asymptotically approach the x-axis at high energy.
- The curve for \(T_2\) (higher temperature) has a lower peak which is shifted to the right compared to \(T_1\).
- Activation energy, \(E_{\text{a}}\), labeled as a vertical line on the x-axis.
- **Explanation**: At temperature \(T_2\), a greater fraction of molecules have kinetic energy equal to or greater than the activation energy (area under the curve to the right of \(E_{\text{a}}\)). This leads to a higher frequency of successful (fruitful) collisions, increasing the reaction rate.

(a)
- Order with respect to \(\text{S}_2\text{O}_8^{2-}\) is first order + justification (comparing Exp 1 & 2) [1.5 marks]
- Order with respect to \(\text{I}^-\) is first order + justification (comparing Exp 1 & 3) [1.5 marks]

(b)
- \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) [1 mark]

(c)
- \(k = 0.15\) [1 mark]
- Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) [1 mark]

(d)
- Correctly sketched Maxwell-Boltzmann curves at two temperatures (correct shifting) [1 mark]
- Correctly labeled axes and \(E_{\text{a}}\) [1 mark]
- More particles have \(E \geq E_{\text{a}}\) at higher temperature [1 mark]
- Higher frequency of successful collisions [1 mark]

**(a)**
- Independent variable: Time (\(t\)) [1]
- Dependent variable: Temperature (\(T\)) [1]

**(b)**
- Heat is lost to the surroundings and calorimeter as the reaction takes place, meaning the maximum recorded temperature is lower than the true theoretical maximum. [1]
- By plotting a temperature-time cooling curve and extrapolating back to the time of mixing (\(t=0\)), we estimate the maximum temperature that would have been reached if the reaction was instantaneous. [1]
- This corrects for the heat lost during the time taken for mixing and thermal equilibration. [1]

**(c)**
- At \(t = 0\), extrapolated temperature: \(T_{\text{extrapolated}} = -0.300(0) + 27.7 = 27.7\ ^\circ\text{C}\) [1]
- Initial temperature, \(T_{\text{initial}} = 21.3\ ^\circ\text{C}\) [1]
- Corrected temperature change: \(\Delta T = 27.7\ ^\circ\text{C} - 21.3\ ^\circ\text{C} = 6.4\ ^\circ\text{C}\) (or \(6.4\text{ K}\)) [1]

**(d)**
- Mass of solution \(m = 100.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 100.0\text{ g}\) [1]
- \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\ \text{K}^{-1} \times 6.4\text{ K} = 2675.2\text{ J}\) (accept \(2680\text{ J}\) or \(2.68 \times 10^3\text{ J}\)) [1]

**(e)**
- Number of moles of \(\text{HCl}\) reacting: \(n = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\) (which reacts completely with \(0.0500\text{ mol}\) of \(\text{NaOH}\) to form \(0.0500\text{ mol}\) of water) [1]
- \(\Delta H_{\text{neut}} = -\frac{q}{1000 \times n} = -\frac{2675.2}{1000 \times 0.0500} = -53.5\text{ kJ mol}^{-1}\) (accept range between \(-53.5\) and \(-53.6\text{ kJ mol}^{-1}\) depending on rounding of \(q\). Negative sign is required for the mark) [1]

**(f)**
- Percentage uncertainty in the volume of \(\text{HCl}\): \(\frac{0.5}{50.0} \times 100\% = 1.0\%\) [1]
- Systematic error: Heat loss to surroundings / heat absorbed by the polystyrene cup itself [1]
- Effect: Calculated \(\Delta T\) and \(q\) values will be smaller than theoretical values, resulting in a less exothermic (less negative/smaller magnitude) \(\Delta H_{\text{neut}}\). [1]

**(a)**
- Award [1] for identifying Time (or \(t\)) as the independent variable.
- Award [1] for identifying Temperature (or \(T\)) as the dependent variable.

**(b)**
- Award [1] for referencing heat loss to surroundings/calorimeter during the process.
- Award [1] for referencing extrapolation back to the time of mixing (\(t = 0\)).
- Award [1] for explaining that extrapolation models instantaneous heat transfer / zero-heat-loss temperature.

**(c)**
- Award [1] for extrapolated maximum temperature of \(27.7\ ^\circ\text{C}\).
- Award [1] for showing initial temperature of \(21.3\ ^\circ\text{C}\).
- Award [1] for corrected temperature change of \(6.4\ ^\circ\text{C}\) (or \(6.4\text{ K}\)).

**(d)**
- Award [1] for correct substitution of values into \(q = mc\Delta T\) (e.g., \(100.0 \times 4.18 \times 6.4\)).
- Award [1] for final calculated energy value \(2675.2\text{ J}\) (accept \(2.68 \times 10^3\text{ J}\) or \(2680\text{ J}\)).

**(e)**
- Award [1] for correctly determining moles of reaction/water formed \(= 0.0500\text{ mol}\).
- Award [1] for \(\Delta H_{\text{neut}} = -53.5\text{ kJ mol}^{-1}\) (or value consistent with part (d), must include the negative sign).

**(f)**
- Award [1] for calculating percentage uncertainty in volume \(= 1.0\%\).
- Award [1] for identifying a valid systematic error (e.g., heat lost to surroundings, heat capacity of cup ignored).
- Award [1] for stating that this error makes the calculated \(\Delta H_{\text{neut}}\) less exothermic / smaller in magnitude.

(a) Equation: \(C_7H_6O_3 + C_4H_6O_3 \rightarrow C_9H_8O_4 + C_2H_4O_2\) (or written using names: salicylic acid + ethanoic anhydride \(\rightarrow\) acetylsalicylic acid + ethanoic acid). Catalyst: Concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)). (b) Dissolve the crude aspirin in the minimum volume of a hot solvent (such as an ethanol/water mixture) in which aspirin is highly soluble when hot but poorly soluble when cold. Filter the hot solution to remove insoluble impurities, then cool the filtrate slowly to allow pure aspirin crystals to form. Filter the crystals under vacuum and dry them. (c) Determine the melting point range of the sample. Pure aspirin has a sharp melting point at approximately 135-136 \(^\circ\)C. An impure sample will melt over a wider temperature range and at a lower temperature than the pure substance.

(a) Award [1] for correct reactants: salicylic acid/\(C_7H_6O_3\) AND ethanoic anhydride/\(C_4H_6O_3\). Award [1] for correct products: acetylsalicylic acid/\(C_9H_8O_4\) AND ethanoic acid/\(C_2H_4O_2\). Award [1] for concentrated \(H_2SO_4\) or concentrated \(H_3PO_4\) catalyst. (b) Award [1] for dissolving the crude product in the minimum volume of hot solvent. Award [1] for cooling the solution to form crystals and filtering to isolate the pure product. (c) Award [1] for stating that the melting point is compared to the literature/pure value (135-136 \(^\circ\)C). Award [0.67] for stating that impurities lower the melting point and broaden the melting temperature range.

(a) The beta-lactam ring is a highly strained four-membered ring containing bond angles of approximately 90\(^\circ\) instead of the preferred tetrahedral or trigonal planar angles. This strain makes the amide bond inside the ring highly reactive. The ring opens and covalently binds to the active site of the transpeptidase enzyme, irreversibly inhibiting cell-wall cross-linking in bacteria. This weakens the cell wall, causing water to enter by osmosis until the bacterium undergoes osmotic lysis (bursts). (b) Resistant bacteria produce an enzyme called beta-lactamase (or penicillinase). This enzyme hydrolyzes the amide link in the beta-lactam ring, opening the ring and rendering the penicillin inactive. Modern penicillins are modified by replacing the side-chain (R group) with a bulkier substituent. This bulkier group introduces steric hindrance, preventing the beta-lactamase enzyme from accessing and binding to the beta-lactam ring while still allowing the drug to bind to its target transpeptidase enzyme. Alternatively, the penicillin can be administered alongside a beta-lactamase inhibitor like clavulanic acid.

(a) Award [1] for noting that the beta-lactam ring has high ring strain / 90\(^\circ\) bond angles. Award [1] for stating this makes the amide bond highly reactive / prone to opening. Award [1] for stating it inhibits the transpeptidase enzyme, preventing cell-wall cross-linking and causing osmotic lysis. (b) Award [1] for stating that resistant bacteria produce the enzyme beta-lactamase / penicillinase. Award [1] for stating that this enzyme hydrolyzes / cleaves the beta-lactam ring. Award [1] for explaining that modifying the side-chain (R group) to be bulkier introduces steric hindrance that blocks beta-lactamase. Award [0.67] for suggesting co-administration with a beta-lactamase inhibitor / clavulanic acid.

(a) Bacteria are living cellular organisms with their own unique metabolic pathways and structures (such as cell walls and 70S ribosomes) that can be targeted specifically by drugs without affecting human cells. In contrast, viruses lack cell structures and metabolic machinery of their own; they reproduce inside host cells by hijacking the host's cellular pathways. Designing drugs that block viral replication without damaging the host cells is extremely difficult. Furthermore, viruses mutate rapidly, changing their surface proteins and rendering treatments ineffective quickly. (b) (i) Both oseltamivir and zanamivir molecules contain an amide (or carboxamide) group and an alkene (carbon-carbon double bond, C=C). (ii) Both drugs act as neuraminidase inhibitors. They bind to the active site of the neuraminidase enzyme on the surface of the influenza virus, preventing it from cleaving sialic acid on the host cell membrane. This prevents newly formed viral particles (virions) from being released from the infected cell, limiting the spread of the virus to healthy cells.

(a) Award [1] for stating bacteria have distinct structures/metabolic pathways that can be targeted. Award [1] for stating viruses rely on host cell machinery for replication. Award [1] for explaining that targeting viral replication risks damaging host cells OR that viruses mutate rapidly. (b) (i) Award [0.83] for identifying amide / carboxamide. Award [0.84] for identifying alkene / C=C double bond (Accept ether / cyclic ether or amine). (ii) Award [1] for identifying them as neuraminidase inhibitors. Award [1] for explaining that they prevent the release of new viral particles (virions) from the infected host cell.