2024 HKDSE Chemistry Practice Paper with Answers

Butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)) contains an \(-\text{OH}\) group which allows it to form extensive intermolecular hydrogen bonds. Butane has only weak van der Waals' forces, while methoxyethane and propanal have dipole-dipole attractions. Hydrogen bonds are significantly stronger than permanent dipole-dipole interactions and dispersion forces between molecules of comparable molecular mass, leading to the highest boiling point for butan-1-ol.

D (1 mark)

Mass of oxygen in the oxide \(= 4.80\text{ g} - 3.36\text{ g} = 1.44\text{ g}\).
Number of moles of \(\text{O} = \frac{1.44}{16.0} = 0.090\text{ mol}\).
Since the empirical formula is \(M_2\text{O}_3\), the mole ratio of \(M\) to \(\text{O}\) is \(2 : 3\).
Number of moles of \(M = 0.090 \times \frac{2}{3} = 0.060\text{ mol}\).
Relative atomic mass of \(M = \frac{3.36\text{ g}}{0.060\text{ mol}} = 56.0\).

C (1 mark)

Initial moles of \(\text{H}^+(\text{aq}) = 0.20 \times \frac{25.0}{1000} \times 2 = 0.0100\text{ mol}\).
Initial moles of \(\text{OH}^-(\text{aq}) = 0.30 \times \frac{25.0}{1000} = 0.0075\text{ mol}\).
Remaining moles of \(\text{H}^+(\text{aq}) = 0.0100 - 0.0075 = 0.0025\text{ mol}\).
Total volume of solution \(= 25.0 + 25.0 = 50.0\text{ cm}^3 = 0.050\text{ dm}^3\).
\([\text{H}^+(\text{aq})] = \frac{0.0025\text{ mol}}{0.050\text{ dm}^3} = 0.050\text{ M}\).

B (1 mark)

At the cathode, cations present are \(\text{Cu}^{2+}(\text{aq})\), \(\text{Na}^+(\text{aq})\), and \(\text{H}^+(\text{aq})\). Since \(\text{Cu}^{2+}\) is the strongest oxidising agent among the three cations (it lies lowest in the electrochemical series), it is preferentially discharged: \(\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu}(\text{s})\). Thus, a reddish-brown solid of copper metal is deposited on the cathode.

B (1 mark)

The formation reaction is:
\(\text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH}(\text{l})\)
Using Hess's Law with enthalpies of combustion:
\(\Delta H^\circ_{\text{f}}[\text{CH}_3\text{OH}(\text{l})] = \Delta H^\circ_{\text{c}}[\text{C}(\text{graphite})] + 2\Delta H^\circ_{\text{c}}[\text{H}_2(\text{g})] - \Delta H^\circ_{\text{c}}[\text{CH}_3\text{OH}(\text{l})]\)
\(\Delta H^\circ_{\text{f}} = -393.5 + 2(-285.8) - (-726.0) = -393.5 - 571.6 + 726.0 = -239.1\text{ kJ mol}^{-1}\).

B (1 mark)

The equilibrium constant \(K_c\) depends exclusively on temperature for a given reaction. Changes in concentration, volume, or total pressure affect the position of equilibrium but do not alter \(K_c\). Because the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium to the product side, thereby increasing the equilibrium constant \(K_c\).

C (1 mark)

The structural (constitutional) isomers of \(\text{C}_4\text{H}_9\text{Cl}\) are:
1. 1-chlorobutane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}\))
2. 2-chlorobutane (\(\text{CH}_3\text{CH}_2\text{CH}(\text{Cl})\text{CH}_3\))
3. 1-chloro-2-methylpropane (\((\text{CH}_3)_2\text{CHCH}_2\text{Cl}\))
4. 2-chloro-2-methylpropane (\((\text{CH}_3)_3\text{CCl}\))
Thus, there are exactly 4 structural isomers.

C (1 mark)

The 1st statement is true: \(\text{SiO}_2\) possesses a giant covalent network structure requiring substantial energy to break numerous strong covalent bonds during melting, whereas \(\text{SO}_2\) has a simple molecular structure held by weak intermolecular forces. The 2nd statement is false: the difference in melting points arises from the type of structure and bonding (giant network vs simple molecular with van der Waals forces), not because the covalent bonds within \(\text{SiO}_2\) are inherently stronger than the covalent bonds within \(\text{SO}_2\) molecules.

C (1 mark)

Ammonium chloride (\(\text{NH}_4\text{Cl}\)) consists of ammonium cations (\(\text{NH}_4^+\)) and chloride anions (\(\text{Cl}^-\)) held together by ionic bonds. Within each ammonium ion, the nitrogen atom is bonded to the hydrogen atoms by covalent bonds (including a dative covalent bond). \(\text{SiO}_2\) and \(\text{CCl}_4\) contain only covalent bonds, whereas \(\text{MgCl}_2\) contains only ionic bonds.

A (1 mark): Correctly identifies \(\text{NH}_4\text{Cl}\) as containing both ionic bonding (between \(\text{NH}_4^+\) and \(\text{Cl}^-\)) and covalent bonding (within \(\text{NH}_4^+\)).

The balanced chemical equation is:
\[\text{H}_2\text{Z(aq)} + 2\text{KOH(aq)} \rightarrow \text{K}_2\text{Z(aq)} + 2\text{H}_2\text{O(l)}\]
Number of moles of \(\text{H}_2\text{Z} = 0.050\text{ mol dm}^{-3} \times \frac{20.0}{1000}\text{ dm}^3 = 0.0010\text{ mol}\).
According to the mole ratio, number of moles of \(\text{KOH} = 2 \times 0.0010\text{ mol} = 0.0020\text{ mol}\).
Therefore, the molarity of \(\text{KOH} = \frac{0.0020\text{ mol}}{16.0 / 1000\text{ dm}^3} = 0.125\text{ M}\).

C (1 mark): Correct stoichiometric calculation accounting for the 1:2 mole ratio of dibasic acid to KOH.

Let us determine the decrease in the oxidation number of chlorine in each option:
- In A: \(\text{Cl}_2\) (oxidation number \(0\)) to \(\text{Cl}^-\) (\(-1\)), decrease of \(1\).
- In B: \(\text{ClO}_3^-\) (oxidation number \(+5\)) to \(\text{Cl}^-\) (\(-1\)), decrease of \(6\).
- In C: \(\text{ClO}^-\) (oxidation number \(+1\)) to \(\text{Cl}^-\) (\(-1\)), decrease of \(2\).
- In D: \(\text{Cl}_2\) (oxidation number \(0\)) to \(\text{Cl}^-\) (\(-1\)), decrease of \(1\).
Thus, reaction B exhibits the largest decrease.

B (1 mark): Correct calculation of changes in oxidation states (from +5 in \(\text{ClO}_3^-\) to -1 in \(\text{Cl}^-\), a decrease of 6).

\(\text{CH}_3\text{CH(OH)COOH}\) (lactic acid / 2-hydroxypropanoic acid) contains a secondary alcohol group (\(-\text{CH(OH)}-\)) which can be oxidised by acidified \(\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}\), and a carboxyl group (\(-\text{COOH}\)) which reacts with \(\text{NaHCO}_3\text{(aq)}\) to produce \(\text{CO}_2\text{(g)}\).

A (1 mark): Recognising that the presence of both an oxidisable alcohol group and a carboxylic acid group allows reaction with both reagents.

The formation reaction is:
\[\text{C(graphite)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\]
Using Hess's Law:
\[\Delta H^\circ_{\text{f}}[\text{CH}_3\text{OH(l)}] = \Delta H^\circ_{\text{c}}[\text{C(graphite)}] + 2\Delta H^\circ_{\text{c}}[\text{H}_2\text{(g)}] - \Delta H^\circ_{\text{c}}[\text{CH}_3\text{OH(l)}]\]
\[\Delta H^\circ_{\text{f}} = -394 + 2(-286) - (-726) = -394 - 572 + 726 = -240\text{ kJ mol}^{-1}\]

B (1 mark): Correct application of Hess's Law using combustion enthalpies.

(1) is correct: Since the forward reaction is exothermic (\(\Delta H < 0\)), increasing temperature shifts the equilibrium in the endothermic reverse direction, converting \(1\text{ mol}\) of \(\text{N}_2\text{O}_4\) into \(2\text{ mol}\) of \(\text{NO}_2\), thus increasing the total number of moles of gas.
(2) is correct: Decreasing pressure by expanding volume shifts the equilibrium to the side with more gaseous moles (left side), increasing the total moles of gas.
(3) is incorrect: A catalyst increases the rates of both forward and reverse reactions equally and does not alter the equilibrium position or the total number of moles.

A (1 mark): Correctly identifies (1) and (2) as driving the equilibrium to the left, increasing total moles of gas.

(1) is correct: Each \(\text{H}_2\text{O}\) molecule can form, on average, four hydrogen bonds per molecule in a extensive three-dimensional network, whereas \(\text{HF}\) forms on average only two hydrogen bonds per molecule, giving \(\text{H}_2\text{O}\) a higher boiling point.
(2) is incorrect: The \(\text{C}-\text{H}\) bonds possess bond dipoles due to differences in electronegativity; however, the symmetrical tetrahedral arrangement causes the bond dipoles to cancel out, resulting in a non-polar molecule.
(3) is correct: \(\text{NH}_3\) has a trigonal pyramidal shape with one lone pair that exerts greater electrostatic repulsion than bond pairs, compressing the \(\text{H}-\text{N}-\text{H}\) bond angle to approximately \(107^\circ\), which is smaller than the tetrahedral bond angle of \(109.5^\circ\) in \(\text{CH}_4\).

C (1 mark): Correctly identifies statements (1) and (3) as correct while rejecting statement (2).

Both statements are true. Magnesium is higher than iron in the electrochemical / reactivity series, meaning magnesium has a higher tendency to lose electrons (it is a stronger reducing agent) than iron. This higher reducing power explains why the displacement reaction of copper by magnesium is much more vigorous than that by iron. Therefore, the 2nd statement is a correct explanation of the 1st statement.

A (1 mark): Evaluates both statements as true and confirms that statement 2 correctly explains statement 1.

Silicon dioxide (\(\text{SiO}_2\)) has a giant covalent network structure, and magnesium oxide (\(\text{MgO}\)) has a giant ionic lattice structure. Both possess strong bonding throughout their lattices, leading to very high melting points. In contrast, dry ice (\(\text{CO}_2\)) and sulfur (\(\text{S}_8\)) have simple molecular structures held by weak van der Waals' forces.

B (1 mark) - Correctly identifying the pair containing giant covalent and giant ionic structures.

Number of moles of \(\text{Na} = \frac{4.60\text{ g}}{23.0\text{ g mol}^{-1}} = 0.20\text{ mol}\).
From the stoichiometric ratio, \(n(\text{H}_2) = \frac{1}{2} \times n(\text{Na}) = \frac{1}{2} \times 0.20\text{ mol} = 0.10\text{ mol}\).
Volume of \(\text{H}_2 = 0.10\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 2.40\text{ dm}^3\).

C (1 mark) - Correct calculation of mole ratio and volume of gas produced.

For \(\text{pH} = 2.0\), \([\text{H}^+] = 10^{-2.0} = 0.010\text{ M}\).
Dilution factor is \(\frac{1000.0}{10.0} = 100\).
The new concentration \([\text{H}^+] = \frac{0.010}{100} = 1.0 \times 10^{-4}\text{ M}\).
Thus, \(\text{pH} = -\log(1.0 \times 10^{-4}) = 4.0\).

C (1 mark) - Dilution by a factor of 100 increases the pH of a strong monoprotic acid by 2 units.

(1) is correct: Zn is more reactive than Cu, so Zn acts as the anode where oxidation (\(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\)) occurs.
(2) is incorrect: Electrons flow from anode (Zn) to cathode (Cu) through the external wire.
(3) is correct: To balance the buildup of positive charge (\(\text{Zn}^{2+}\)) in the anode compartment, anions migrate from the salt bridge into the zinc half-cell.
Therefore, (1) and (3) only are correct.

B (1 mark) - Statements (1) and (3) are correct, statement (2) is false.

The formation reaction is: \(\text{C(graphite)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}\).
Using \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\):
\(\Delta H_f^\ominus = [\Delta H_c^\ominus(\text{C}) + 2 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{CH}_4)\)
\(\Delta H_f^\ominus = [-393.5 + 2(-285.8)] - (-890.3) = -965.1 + 890.3 = -74.8\text{ kJ mol}^{-1}\).

A (1 mark) - Correct application of Hess's Law using standard enthalpies of combustion.

The equilibrium constant \(K_c\) depends strictly on temperature. Changes in pressure, volume, concentration, or the addition of a catalyst do not alter the value of \(K_c\). Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature shifts the equilibrium to the right and increases \(K_c\).

B (1 mark) - Only a change in temperature affects \(K_c\); for an endothermic process, increasing temperature increases \(K_c\).

The isomers of \(\text{C}_4\text{H}_9\text{Br}\) are:
1. 1-bromobutane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\))
2. 2-bromobutane (\(\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3\))
3. 1-bromo-2-methylpropane (\((\text{CH}_3)_2\text{CHCH}_2\text{Br}\))
4. 2-bromo-2-methylpropane (\((\text{CH}_3)_3\text{CBr}\))
There are altogether 4 structural isomers.

C (1 mark) - Identifying all 4 structural isomers derived from butane and 2-methylpropane carbon skeletons.

Across Period 3 from left to right:
- Atomic radius decreases due to increasing nuclear charge while electron shielding remains approximately constant.
- Electronegativity increases.
- First ionisation energy generally increases.
- Acidity of the highest oxides increases (from basic \(\text{Na}_2\text{O}\) to acidic \(\text{Cl}_2\text{O}_7\)).

A (1 mark) - Atomic radius decreases across Period 3.

Graphite conducts electricity in the solid state due to delocalised electrons between its hexagonal layers. When heated to high temperatures in a molten/liquid state (or during sublimation/melting under pressure), it retains mobile electrons. More straightforwardly, among common crystalline solids: sodium chloride only conducts in molten/aqueous states; silicon dioxide and sucrose are non-conductors in both solid and molten states; graphite is an electrical conductor in solid form (and retains conductivity).

Award 1 mark for the correct option B.

Number of moles of \( \text{Mg} = \frac{4.86\text{ g}}{24.3\text{ g mol}^{-1}} = 0.200\text{ mol} \).
According to the stoichiometric ratio, \( 1\text{ mol Mg} \equiv 1\text{ mol H}_2 \).
Therefore, moles of \( \text{H}_2\text{ produced} = 0.200\text{ mol} \).
Volume of \( \text{H}_2 = 0.200\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 4.80\text{ dm}^3 \).

Award 1 mark for the correct option C.

In \( \text{H}_2\text{O}_2 \), the oxidation number of oxygen is \( -1 \). In \( \text{O}_2 \), it is \( 0 \). Thus, oxygen is oxidised and \( \text{H}_2\text{O}_2 \) acts as the reducing agent. In \( \text{MnO}_4^- \), manganese decreases its oxidation state from \( +7 \) to \( +2 \) in \( \text{Mn}^{2+} \), so \( \text{MnO}_4^- \) is reduced.

Award 1 mark for the correct option B.

Both \( \text{CH}_4 \) and \( \text{NH}_4^+ \) have 4 bonding pairs and 0 lone pairs on the central atom, giving a tetrahedral shape. \( \text{BF}_3 \) is trigonal planar while \( \text{NH}_3 \) is trigonal pyramidal. \( \text{CO}_2 \) is linear while \( \text{SO}_2 \) is bent/V-shaped. \( \text{H}_2\text{O} \) is bent while \( \text{BeCl}_2 \) is linear.

Award 1 mark for the correct option A.

(1) is correct: The forward reaction is exothermic (\( \Delta H < 0 \)). A decrease in temperature shifts the equilibrium position to the right, increasing moles of \( \text{N}_2\text{O}_4 \).
(2) is correct: Decreasing volume increases the pressure. The system shifts to the side with fewer gas moles (the right side, from 2 moles of gas to 1 mole of gas), thus increasing moles of \( \text{N}_2\text{O}_4 \).
(3) is incorrect: Adding a catalyst only increases the rates of forward and backward reactions equally; it does not change the position of equilibrium or the equilibrium amounts/moles of reactants and products.

Award 1 mark for the correct option A.

\( \Delta H^{\ominus} = \sum \Delta H_f^{\ominus}(\text{products}) - \sum \Delta H_f^{\ominus}(\text{reactants}) \)
\( \Delta H_f^{\ominus}[\text{Fe(s)}] = 0 \)
\( \Delta H^{\ominus} = [2(0) + 3(-393.5)] - [-824.2 + 3(-110.5)] \)
\( \Delta H^{\ominus} = -1180.5 - [-824.2 - 331.5] = -1180.5 - (-1155.7) = -24.8\text{ kJ mol}^{-1} \).

Award 1 mark for the correct option B.

An enantiomer requires a chiral carbon (a carbon atom bonded to four different groups).
- Propan-2-ol: \( \text{CH}_3\text{CH(OH)CH}_3 \) has two identical \( -\text{CH}_3 \) groups attached to C2 (achiral).
- Butan-2-ol: \( \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \) has C2 bonded to \( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{CH}_2\text{CH}_3 \) (4 distinct groups, so chiral).
- Pentan-3-ol: \( \text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3 \) has two identical ethyl groups attached to C3 (achiral).
- 2-Methylpropan-2-ol: \( (\text{CH}_3)_3\text{COH} \) has three identical methyl groups (achiral).

Award 1 mark for the correct option B.

Magnesium is more reactive than iron because it loses electrons more readily and is higher in the reactivity series. Because it is higher in the electrochemical / reactivity series, its activation energy for reaction with aqueous \( \text{H}^+ \) ions is lower, leading to a faster rate of reaction under identical conditions. Thus, both statements are true, and the 2nd statement correctly explains the 1st statement.

Award 1 mark for the correct option A.

1. Calculate total charge passed:
\[ Q = I \times t = 2.50 \text{ A} \times 3860 \text{ s} = 9650 \text{ C} \]
2. Calculate the number of moles of electrons transferred:
\[ n(\text{e}^-) = \frac{Q}{F} = \frac{9650 \text{ C}}{96500 \text{ C mol}^{-1}} = 0.10 \text{ mol} \]
3. Determine \( n \) from the cathode reduction equation \( \text{M}^{n+} + n\text{e}^- \rightarrow \text{M} \):
\[ n = \frac{n(\text{e}^-)}{n(\text{M})} = \frac{0.10 \text{ mol}}{0.050 \text{ mol}} = 2 \]
Therefore, the charge on the metal ion is \( 2+ \).

B (1 mark)
- Correct calculation of charge passed and moles of electrons to deduce \( n = 2 \).

Statement (1) is correct: \( \text{CH}_3\text{COOH} \) is a weak acid that is only slightly ionised in water, whereas the resulting \( \text{CH}_3\text{COONa(aq)} \) is a strong electrolyte that completely dissociates into \( \text{Na}^+ \) and \( \text{CH}_3\text{COO}^- \) ions, giving a significantly higher concentration of mobile ions and thus higher electrical conductivity.
Statement (2) is incorrect: \( \text{CH}_3\text{COO}^- \) undergoes hydrolysis with water (\( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \)), creating an alkaline solution with \( \text{pH} > 7.0 \) at \( 25^\circ\text{C} \).
Statement (3) is correct: Neutralisation reactions between acids and bases are always exothermic (\( \Delta H < 0 \)).
Hence, only (1) and (3) are correct.

C (1 mark)
- (1) is correct due to complete dissociation of sodium ethanoate.
- (2) is incorrect due to salt hydrolysis producing alkaline pH.
- (3) is correct because acid-base neutralisation is exothermic.

(1) can be done in a single step by acid-catalysed hydration of propene (reaction with steam/water in the presence of concentrated \( \text{H}_2\text{SO}_4 \) or \( \text{H}_3\text{PO}_4 \)), following Markovnikov's rule to give propan-2-ol.
(2) can be done in a single step via esterification by heating ethanoic acid under reflux with ethanol in the presence of concentrated \( \text{H}_2\text{SO}_4 \).
(3) cannot be done in a single step because alkanes lack a functional group for direct nucleophilic attack or oxidation to an alcohol; propane must first undergo free-radical halogenation before substitution to form an alcohol.
Thus, only (1) and (2) are correct.

A (1 mark)
- (1) is correct (hydration of alkene).
- (2) is correct (esterification of carboxylic acid with alcohol).
- (3) is incorrect (alkane cannot directly convert to primary alcohol in one step).

- 1st statement is true: Because the forward reaction is endothermic (\( \Delta H > 0 \)), decreasing the temperature shifts the equilibrium to the left, which decreases the equilibrium concentrations of products and increases those of reactants, thus decreasing the numerical value of \( K_c \).
- 2nd statement is true: According to Le Chatelier's principle, lowering the temperature shifts the equilibrium in the direction that releases heat, which is the exothermic direction.
- The 2nd statement is the correct thermodynamic explanation of why the backward (exothermic) reaction is favoured upon cooling, leading to the reduction in the value of \( K_c \).

A (1 mark)
- Both statements are true and the 2nd statement correctly explains the 1st statement based on Le Chatelier's principle and equilibrium constant temperature dependence.

(a) \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\)

(b) Methyl orange is yellow in alkaline/neutral solution and turns orange / red-orange at the end point in this titration.

(c) Number of moles of \(\text{HCl}\) used in titration \(= 0.120 \times \frac{28.40}{1000} = 3.408 \times 10^{-3}\text{ mol}\)

According to the equation, \(\text{Na}_2\text{CO}_3 : \text{HCl} = 1 : 2\)
Number of moles of \(\text{Na}_2\text{CO}_3\) in \(25.00\text{ cm}^3 = \frac{3.408 \times 10^{-3}}{2} = 1.704 \times 10^{-3}\text{ mol}\)

Number of moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3\) solution \(= 1.704 \times 10^{-3} \times \frac{250.0}{25.00} = 1.704 \times 10^{-2}\text{ mol}\)

Molar mass of \(\text{Na}_2\text{CO}_3 = 23.0 \times 2 + 12.0 + 16.0 \times 3 = 106.0\text{ g mol}^{-1}\)
Mass of \(\text{Na}_2\text{CO}_3 = 1.704 \times 10^{-2} \times 106.0 = 1.80624\text{ g}\)

Percentage by mass of \(\text{Na}_2\text{CO}_3 = \frac{1.80624}{2.18} \times 100\% = 82.85\% \approx 82.9\%\) (or \(82.6\% - 82.9\%\) depending on intermediate rounding).

(a) Correct balanced equation with correct formulae: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\) (1 mark)
(b) From yellow to orange / pink / red (1 mark)
(c) Moles of \(\text{HCl} = 3.408 \times 10^{-3}\text{ mol}\) (1 mark)
Moles of \(\text{Na}_2\text{CO}_3\) in total \(250\text{ cm}^3 = 1.704 \times 10^{-2}\text{ mol}\) (1 mark)
Correct mass and percentage: \(82.9\%\) (1 mark)

(a) Mass of reaction mixture \(= 50.0 + 6.02 = 56.02\text{ g}\)
Heat released \(q_1 = m c \Delta T = 56.02 \times 4.2 \times 10.5 = 2470.5\text{ J} = 2.471\text{ kJ}\)
Moles of \(\text{MgSO}_4 = \frac{6.02}{120.4} = 0.0500\text{ mol}\)
\(\Delta H_1 = -\frac{2.471}{0.0500} = -49.4\text{ kJ mol}^{-1}\)

(b) Mass of reaction mixture \(= 43.7 + 12.33 = 56.03\text{ g}\)
Heat absorbed \(q_2 = 56.03 \times 4.2 \times 3.8 = 894.2\text{ J} = 0.8942\text{ kJ}\)
Moles of \(\text{MgSO}_4\cdot 7\text{H}_2\text{O} = \frac{12.33}{246.5} = 0.0500\text{ mol}\)
\(\Delta H_2 = +\frac{0.8942}{0.0500} = +17.9\text{ kJ mol}^{-1}\)

(c) According to Hess's Law:
\(\Delta H_{\text{hydr}} + \Delta H_2 = \Delta H_1\)
\(\Delta H_{\text{hydr}} = \Delta H_1 - \Delta H_2 = -49.4 - (+17.9) = -67.3\text{ kJ mol}^{-1}\)

(d) The rate of reaction of solid with a small quantity of water is slow, and hydration does not go to completion without excess water, forming a wet lump or solution instead of pure hydrated crystals.

(a) Correct calculation of \(q_1\) and \(\Delta H_1\) with negative sign: \(-49.4\text{ kJ mol}^{-1}\) (allow \(-42.0\text{ kJ mol}^{-1}\) if only water mass 50.0 g used) (2 marks)
(b) Correct calculation of \(q_2\) and \(\Delta H_2\) with positive sign: \(+17.9\text{ kJ mol}^{-1}\) (allow \(+15.2\text{ kJ mol}^{-1}\) if only water mass used) (1 mark)
(c) Correct Hess's Law cycle or algebraic expression (1 mark) and correct evaluated \(\Delta H_{\text{hydr}}\) value: \(-67.3\text{ kJ mol}^{-1}\) (1 mark)
(d) Mentioning that the reaction is incomplete / difficult to measure temperature change accurately / hydrated salt dissolves partially in water (1 mark)

(a) \(K_c = \frac{[\text{HI}(\text{g})]^2}{[\text{H}_2(\text{g})][\text{I}_2(\text{g})]} (b) Reaction stoichiometry: \)\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\)
Amount of \(\text{HI}\) formed \(= 1.28\text{ mol}\)
Amount of \(\text{H}_2\) consumed \(= \frac{1.28}{2} = 0.64\text{ mol}\)
Amount of \(\text{I}_2\) consumed \(= 0.64\text{ mol}\)

At equilibrium:
Moles of \(\text{H}_2 = 0.80 - 0.64 = 0.16\text{ mol}\)
Moles of \(\text{I}_2 = 0.80 - 0.64 = 0.16\text{ mol}\)
Moles of \(\text{HI} = 1.28\text{ mol}\)

Equilibrium concentrations (in \(2.0\text{ dm}^3\)):
\([\text{H}_2] = \frac{0.16}{2.0} = 0.080\text{ mol dm}^{-3}\)
\([\text{I}_2] = \frac{0.16}{2.0} = 0.080\text{ mol dm}^{-3}\)
\([\text{HI}] = \frac{1.28}{2.0} = 0.64\text{ mol dm}^{-3}\)

(c) \(K_c = \frac{(0.64)^2}{(0.080)(0.080)} = \frac{0.4096}{0.0064} = 64\) (no units)

(d) Compressing the container increases the pressure. Since the number of moles of gaseous reactants (\(1 + 1 = 2\)) equals the number of moles of gaseous products (\(2\)), a change in pressure has no effect on the position of equilibrium. Therefore, the yield of \(\text{HI}(\text{g})\) remains unchanged.

(a) Correct expression of \(K_c\) (1 mark)
(b) Correct deduction of equilibrium moles and division by volume: \([\text{H}_2] = 0.080\text{ M}\), \([\text{I}_2] = 0.080\text{ M}\), \([\text{HI}] = 0.64\text{ M}\) (2 marks)
(c) Correct calculation of \(K_c = 64\) (1 mark)
(d) States that yield remains unchanged (1 mark); explains that the number of moles of gas on both sides of the equation is equal (1 mark)

(a) Reagent: \(\text{HBr}(\text{g})\) or concentrated \(\text{HBr}(\text{aq})\) at room temperature.
(b) \(\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3\) (2-bromobutane).
(c)(i) Addition / nucleophilic addition of \(\text{HCN}\) to the carbonyl group of butanal.
(c)(ii) Heating / refluxing with dilute acid (e.g. \(\text{HCl}(\text{aq})\) or \(\text{H}_2\text{SO}_4(\text{aq})\)) (acid hydrolysis).
(d)(i) The C-2 carbon atom is asymmetric / chiral because it is attached to four distinct groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{COOH}\), and \(-\text{CH}_2\text{CH}_3\).
(d)(ii) Rotation of the plane of plane-polarised light (optical activity).

(a) \(\text{HBr}\) / hydrogen bromide (1 mark)
(b) Correct structural formula of 2-bromobutane: \(\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3\) (1 mark)
(c)(i) Addition / nucleophilic addition (1 mark)
(c)(ii) Dilute acid (e.g., \(\text{HCl}\) / \(\text{H}_2\text{SO}_4\)) and heat / reflux (1 mark)
(d)(i) Presence of a chiral carbon / asymmetric carbon atom bonded to 4 different groups (1 mark)
(d)(ii) Direction of rotation of plane-polarised light (1 mark)

(a) Silicon has a giant covalent network structure. Melting silicon requires breaking many strong covalent bonds between silicon atoms, which requires a large amount of energy.
In contrast, phosphorus consists of discrete simple molecules (\(\text{P}_4\)) held together by weak van der Waals' forces, which require much less energy to overcome.

(b) Both phosphorus and sulfur have simple molecular structures. An \(\text{S}_8\) molecule has a larger molecular size / higher relative molecular mass / more electrons than a \(\text{P}_4\) molecule. Hence, the van der Waals' forces between \(\text{S}_8\) molecules are stronger than those between \(\text{P}_4\) molecules, requiring more thermal energy to separate them.

(c) Sodium has a giant metallic structure consisting of sodium cations surrounded by a sea of delocalised electrons. The delocalised electrons are mobile and act as charge carriers under an electric potential.

(a) Mentions giant covalent structure of Si with strong covalent bonds (1 mark); mentions simple molecular structure of phosphorus with weak van der Waals' forces (1 mark)
(b) Mentions \(\text{S}_8\) has a larger molecular size / more electrons than \(\text{P}_4\) (1 mark); stronger van der Waals' forces between \(\text{S}_8\) molecules (1 mark)
(c) Mentions giant metallic structure (1 mark) and mobile / delocalised electrons (1 mark)

(a)(i) Anode (oxidation): \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\)
(a)(ii) Cathode (reduction): \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\)

(b) Electrons flow from the zinc electrode (anode, negative terminal) to the copper electrode (cathode, positive terminal) via the external circuit.

(c) \(\text{K}^+\) cations migrate into the cathode compartment (copper half-cell) to balance the excess negative charge left by the reduction of \(\text{Cu}^{2+}\), while \(\text{NO}_3^-\) anions migrate into the anode compartment (zinc half-cell) to balance the excess positive charge from \(\text{Zn}^{2+}\) ions formed.

(d) The cell voltage decreases. Ammonia reacts with \(\text{Cu}^{2+}(\text{aq})\) ions to form the tetraamminecopper(II) complex ion, \([\text{Cu}(\text{NH}_3)_4]^{2+}(\text{aq})\), significantly decreasing the concentration of free \(\text{Cu}^{2+}(\text{aq})\) ions. According to Le Chatelier's principle, the reduction tendency of \(\text{Cu}^{2+}\) decreases (equilibrium shifts to the left), leading to a lower reduction potential and decreased cell voltage.

(a)(i) Correct half-equation: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-\) (1 mark)
(a)(ii) Correct half-equation: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\) (1 mark)
(b) From zinc to copper through the external circuit (1 mark)
(c) \(\text{K}^+\) to \(\text{Cu}^{2+}\) compartment AND \(\text{NO}_3^-\) to \(\text{Zn}^{2+}\) compartment (1 mark)
(d) States that voltage decreases (1 mark); explains that \(\text{NH}_3\) forms complex with \(\text{Cu}^{2+}\) / reduces \([\text{Cu}^{2+}]\) (1 mark)

(a) Moles of \(\text{HCl} = 0.0500 \times 0.40 = 0.0200\text{ mol}\)
\(\text{CaCO}_3\) is in excess, so \(\text{HCl}\) is the limiting reactant.
Moles of \(\text{CO}_2 = \frac{0.0200}{2} = 0.0100\text{ mol}\)
Volume of \(\text{CO}_2 = 0.0100 \times 24.0 = 0.240\text{ dm}^3 = 240\text{ cm}^3\)

(b) Method 1: Collect \(\text{CO}_2(\text{g})\) using a gas syringe and record the gas volume at regular time intervals.
Method 2: Place the flask on an electronic balance (with a cotton wool plug) and record the decrease in mass over time.

(c) At the beginning, the concentration of \(\text{H}^+(\text{aq})\) ions is at its highest, meaning the number of \(\text{H}^+\) ions per unit volume is highest, resulting in the highest frequency of collisions between \(\text{H}^+\) ions and \(\text{CaCO}_3\) surface. As the reaction proceeds, \(\text{H}^+\) ions are consumed, decreasing their concentration and resulting in a lower collision frequency and lower reaction rate.

(d) An increase in temperature increases the average kinetic energy of the reacting particles. A much larger fraction of particles possess energy equal to or greater than the activation energy (\(E_a\)). Therefore, the frequency of effective collisions increases, resulting in a faster rate.

(a) Correct calculation of moles of \(\text{CO}_2\) and volume \(= 240\text{ cm}^3\) (1 mark)
(b) Two correct methods: gas syringe measuring volume / electronic balance measuring mass loss / pH change (1 mark for any two valid methods)
(c) Explains that concentration of \(\text{HCl}\) / \(\text{H}^+\) is highest initially leading to highest collision frequency (1 mark); concentration decreases over time, lowering effective collision frequency (1 mark)
(d) Higher kinetic energy / larger fraction of particles with \(E \ge E_a\) (1 mark); higher frequency of effective collisions (1 mark)

(a) Rusting is an electrochemical redox process. Iron acts as the anode and undergoes oxidation: \(\text{Fe}(\text{s}) \rightarrow \text{Fe}^{2+}(\text{aq}) + 2\text{e}^-\). Oxygen dissolved in water at the surface acts as the cathode and is reduced: \(\text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^- \rightarrow 4\text{OH}^-(\text{aq})\).

(b)(i) Sacrificial protection (or cathodic protection).
(b)(ii) Zinc is higher than iron in the electrochemical / reactivity series. Zinc is oxidised preferentially by losing electrons: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\). These electrons flow to the iron pipe, preventing iron from losing electrons / oxidising. Disadvantage: The zinc blocks corrode away and need regular monitoring and replacement.

(c) Iron is higher in the reactivity series (more reactive) than tin. Once scratched, both metals are in electrical contact in the presence of moisture and oxygen, forming a chemical cell. Iron acts as the anode (negative terminal) and loses electrons more readily to tin (the cathode), which accelerates the oxidation and rusting of iron compared to unprotected iron.

(a) Identifies iron oxidation with half-equation \(\text{Fe} \rightarrow \text{Fe}^{2+} + 2\text{e}^-\) and reduction of \(\text{O}_2\) in water (1 mark)
(b)(i) Sacrificial protection (1 mark)
(b)(ii) Explains that zinc is more reactive and loses electrons preferentially to protect iron (1 mark); states requirement for periodic replacement of zinc (1 mark)
(c) Explains that iron is more reactive than tin (1 mark); iron acts as the anode / loses electrons more readily when in contact with tin, thereby speeding up rusting (1 mark)

(a) \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

(b) (i) Moles of \(\text{HCl}\) added \(= 0.500 \times \frac{50.0}{1000} = 0.0250\text{ mol}\)

(ii) Moles of \(\text{NaOH}\) used \(= 0.400 \times \frac{22.40}{1000} = 8.96 \times 10^{-3}\text{ mol}\)
Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), mole ratio is \(1:1\).
Therefore, unreacted \(\text{HCl} = 8.96 \times 10^{-3}\text{ mol}\).

(iii) Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3 = 0.0250 - 8.96 \times 10^{-3} = 0.01604\text{ mol}\)
From the equation, \(\text{mol of CaCO}_3 = \frac{0.01604}{2} = 8.02 \times 10^{-3}\text{ mol}\)
Molar mass of \(\text{CaCO}_3 = 40.1 + 12.0 + (3 \times 16.0) = 100.1\text{ g mol}^{-1}\)
Mass of \(\text{CaCO}_3 = 8.02 \times 10^{-3} \times 100.1 = 0.8028\text{ g}\)
Percentage by mass \(= \frac{0.8028\text{ g}}{1.25\text{ g}} \times 100\% = 64.2\%\)

(c) Red to orange (or red to yellow).

(a) Correct balanced equation: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) [1 mark]
(b) (i) \(\text{Moles of HCl} = 0.0250\text{ mol}\) [1 mark]
(ii) \(\text{Moles of unreacted HCl} = 8.96 \times 10^{-3}\text{ mol}\) [1 mark]
(iii) Moles of \(\text{CaCO}_3 = 8.02 \times 10^{-3}\text{ mol}\) and mass \(= 0.803\text{ g}\) [1 mark]; Percentage by mass \(= 64.2\%\) (accept 64.2% - 64.3%) [1 mark]
(c) Red to orange / yellow [1 mark]

(a) Zinc is higher than copper in the electrochemical series / reactivity series, so it loses electrons more readily.
Negative electrode (anode): Zinc (Zn)
Half-equation: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\)

(b) Electrons flow from the zinc electrode (anode) to the copper electrode (cathode) through the external circuit.

(c) (i) To complete the electrical circuit by allowing the migration/flow of ions / to maintain electrical neutrality in both half-cells.
(ii) With silver half-cell, \(\text{Cl}^-\text{(aq)}\) ions from \(\text{KCl}\) would react with \(\text{Ag}^+\text{(aq)}\) to form an insoluble precipitate of silver chloride (\(\text{AgCl(s)}\)), which depletes silver ions and blocks ion conduction.

(d) Observation: The blue colour of the \(\text{CuSO}_4\text{(aq)}\) solution becomes paler (or fades).
Explanation: \(\text{Cu}^{2+}\text{(aq)}\) ions are reduced to \(\text{Cu(s)}\) at the cathode (\(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)), decreasing the concentration of \(\text{Cu}^{2+}\text{(aq)}\) ions in the solution.

(a) Zinc / Zn AND \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\) [1 mark]
(b) From zinc to copper through the external circuit [1 mark]
(c) (i) Complete the circuit / balance charges / allow ion migration to maintain electrical neutrality [1 mark]
(ii) \(\text{Cl}^-\text{(aq)}\) reacts with \(\text{Ag}^+\text{(aq)}\) to form an insoluble precipitate of \(\text{AgCl(s)}\) [1 mark]
(d) Blue colour becomes paler / fades [1 mark]; \(\text{Cu}^{2+}\text{(aq)}\) ions are reduced to copper metal / concentration of \(\text{Cu}^{2+}\text{(aq)}\) decreases [1 mark]

(a) Temperature rise \(\Delta T = 46.6 - 21.4 = 25.2\ ^\circ\text{C} = 25.2\text{ K}\)
Heat absorbed by water \(q = m \times c \times \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 25.2\text{ K} = 15800.4\text{ J} = 15.80\text{ kJ}\)

(b) Mass of propan-1-ol burned \(= 188.42 - 187.82 = 0.600\text{ g}\)
Molar mass of \(\text{C}_3\text{H}_7\text{OH} = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\)
Moles of propan-1-ol burned \(= \frac{0.600}{60.0} = 0.0100\text{ mol}\)

(c) Experimental \(\Delta H_c = -\frac{q}{n} = -\frac{15.8004\text{ kJ}}{0.0100\text{ mol}} = -1580\text{ kJ mol}^{-1}\) (or \(-1.58 \times 10^3\text{ kJ mol}^{-1}\))

(d) Any two valid reasons:
1. Heat loss to the surroundings (air, copper can, tripod, thermometer).
2. Incomplete combustion of propan-1-ol (forming soot / CO).
3. Evaporation of propan-1-ol from the wick before or after combustion.
4. Heat capacity of the copper calorimeter was neglected.

(a) \(q = 150.0 \times 4.18 \times 25.2 = 15.8\text{ kJ}\) [1 mark]
(b) \(\text{Mass} = 0.600\text{ g}\), \(\text{Moles} = 0.0100\text{ mol}\) [1 mark]
(c) Correct division: \(\frac{15.8}{0.0100}\) [1 mark]; Correct value with negative sign and units: \(-1580\text{ kJ mol}^{-1}\) [1 mark]
(d) Any two points (1 mark each, max 2 marks):
- Heat loss to the surroundings
- Incomplete combustion of propan-1-ol
- Evaporation of alcohol from the wick
- Heat absorbed by the copper calorimeter / thermometer was not taken into account

(a) \(K_c = \frac{[\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{CH}_2\text{COOH}][\text{CH}_3\text{CH}_2\text{OH}]}\)

(b) ICE table:
Initial moles:
\(\text{CH}_3\text{CH}_2\text{COOH} = 2.00\text{ mol}\)
\(\text{CH}_3\text{CH}_2\text{OH} = 3.00\text{ mol}\)
\(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 = 0\text{ mol}\)
\(\text{H}_2\text{O} = 0\text{ mol}\)

Change in moles:
Ester formed \(= +1.50\text{ mol}\)
\(\text{H}_2\text{O}\) formed \(= +1.50\text{ mol}\)
Acid reacted \(= -1.50\text{ mol}\)
Alcohol reacted \(= -1.50\text{ mol}\)

Equilibrium moles:
Acid \(= 2.00 - 1.50 = 0.50\text{ mol}\)
Alcohol \(= 3.00 - 1.50 = 1.50\text{ mol}\)
Ester \(= 1.50\text{ mol}\)
\(\text{H}_2\text{O} = 1.50\text{ mol}\)

\(K_c = \frac{(1.50 / V)(1.50 / V)}{(0.50 / V)(1.50 / V)} = \frac{1.50 \times 1.50}{0.50 \times 1.50} = \frac{1.50}{0.50} = 3.00\)

(c) The total number of moles of reactants equals the total number of moles of products (or the power of concentration in numerator equals the power in denominator), so the volume terms \(V\) cancel out completely.

(d) \(K_c\) decreases. The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, an increase in temperature favours the endothermic backward reaction, shifting the equilibrium to the left, which reduces product concentrations and increases reactant concentrations, thereby decreasing \(K_c\).

(a) Correct \(K_c\) expression [1 mark]
(b) Correct equilibrium moles of all species (Acid: 0.50, Alcohol: 1.50, Water: 1.50) [1 mark]; Correct calculation of \(K_c = 3.00\) (or 3) [1 mark]
(c) Total number of moles of reactants equals total number of moles of products / volume terms cancel out in the numerator and denominator [1 mark]
(d) \(K_c\) decreases [1 mark]; Since the forward reaction is exothermic, increasing temperature shifts equilibrium to the left / favours backward reaction [1 mark]

(a) Concentrated \(\text{H}_2\text{SO}_4\) with heating (approx. \(170\ ^\circ\text{C}\)) OR \(\text{Al}_2\text{O}_3\text{(s)}\) with strong heating.

(b) Propene undergoes electrophilic addition of water across the double bond following Markovnikov's rule. The \(-\text{OH}\) group attaches to the secondary carbon atom.
Structural formula: \(\text{CH}_3\text{CH(OH)CH}_3\)
Systematic name: propan-2-ol

(c) Propan-2-ol is a secondary alcohol, which is oxidized to a ketone (propanone).
Structural formula of \(\mathbf{Z}\): \(\text{CH}_3\text{COCH}_3\)

(d) The dichromate(VI) ion (\(\text{Cr}_2\text{O}_7^{2-}\)) is reduced to chromium(III) ion (\(\text{Cr}^{3+}\)).
Colour change: Orange to green.

(e) Method 1: Add acidified potassium dichromate(VI) (or acidified potassium permanganate) and warm gently.
- \(\mathbf{W}\) (propan-1-ol): The orange solution turns green (or purple turns colourless) because propan-1-ol can be oxidized further.
- \(\mathbf{Z}\) (propanone): The solution remains orange (or purple) as ketones are resistant to mild oxidation.
OR Method 2: Add 2,4-dinitrophenylhydrazine (2,4-DNP).
- \(\mathbf{Z}\): Yellow / orange precipitate forms.
- \(\mathbf{W}\): No precipitate / no observable change.

(a) \(\text{Conc. H}_2\text{SO}_4\) with heat / \(\text{Al}_2\text{O}_3\) with heat [1 mark]
(b) Structure of propan-2-ol: \(\text{CH}_3\text{CH(OH)CH}_3\) [1 mark]; Systematic name: propan-2-ol [1 mark]
(c) Structure of propanone: \(\text{CH}_3\text{COCH}_3\) [1 mark]
(d) Orange to green [1 mark]
(e) Suitable test with distinct reagent and correct observation for both \(\mathbf{W}\) and \(\mathbf{Z}\) (e.g. 2,4-DNP gives yellow/orange ppt with Z but not W, or acidified \(\text{KMnO}_4\) decolorised by W but not Z) [1 mark]

(a) (i) Sodium oxide has a giant ionic structure with strong ionic bonds between \(\text{Na}^+\) and \(\text{O}^{2-}\) ions.
(ii) Sulfur dioxide has a simple molecular structure with covalent bonds between sulfur and oxygen atoms within the molecule.

(b) Melting \(\text{Na}_2\text{O}\) requires overcoming strong ionic bonds / strong electrostatic forces between oppositely charged ions (\(\text{Na}^+\) and \(\text{O}^{2-}\)), which requires a large amount of energy.
Melting \(\text{SO}_2\) only requires overcoming weak van der Waals' forces (intermolecular forces) between discrete molecules, which requires much less energy.

(c) (i) \(\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}\) (or \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{Na}^+ + 2\text{OH}^-\)).
Sodium oxide is a basic oxide.
(ii) \(\text{SO}_2\text{(g)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{SO}_3\text{(aq)}\).
Sulfur dioxide is an acidic oxide.

(a) (i) Giant ionic structure / ionic bonding [1 mark]
(ii) Simple molecular structure / covalent bonding with intermolecular forces [1 mark]
(b) Giant ionic structure with strong ionic bonds requiring large energy to break vs simple molecular structure with weak van der Waals' forces between molecules requiring little energy [1 mark]
(c) (i) Balanced equation: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\) AND basic oxide [1 mark + 1 mark = 2 marks]
(ii) Balanced equation: \(\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3\) AND acidic oxide [1 mark]

(a) (i) Catalyst: Vanadium(V) oxide ($\text{V}_2\text{O}_5$).
The forward reaction is exothermic ($\Delta H < 0$). By Le Chatelier's principle, lowering temperature shifts equilibrium to the right, giving a higher equilibrium yield of $\text{SO}_3$. However, at low temperature, the reaction rate is too slow to be economically viable. At high temperature, the reaction proceeds faster, but the equilibrium yield decreases significantly. Therefore, $450\text{ }^\circ\text{C}$ is chosen as an optimum compromise temperature between rate and yield.
(ii) The equilibrium yield of $\text{SO}_3$ is already very high (about 98%) at 1 to 2 atm; operating at very high pressures requires expensive high-pressure equipment and high maintenance costs without significant yield improvement.
(iii) Direct reaction of $\text{SO}_3$ with water is extremely exothermic and produces a dense, corrosive mist/fog of sulphuric acid droplets that is difficult to condense and separate. Dissolving in concentrated $\text{H}_2\text{SO}_4$ safely produces oleum ($\text{H}_2\text{S}_2\text{O}_7$), which can then be safely diluted with water.

(b) (i) Molar mass of $\text{C}_2\text{H}_4\text{O} = 2(12.0) + 4(1.0) + 16.0 = 44.0\text{ g mol}^{-1}$.
Total mass of reactants = Molar mass of $\text{C}_2\text{H}_4$ ($28.0$) + $\text{Cl}_2$ ($71.0$) + $\text{Ca(OH)}_2$ ($40.1 + 2(17.0) = 74.1$) = $181.1\text{ g mol}^{-1}$.
$$\text{Atom economy} = \frac{44.0}{181.1} \times 100\% = 24.3\%$$
(ii) 1. Route 2 has $100\%$ atom economy whereas Route 1 has only $24.3\%$.
2. Route 1 uses toxic and hazardous chlorine gas and generates significant amounts of calcium chloride byproduct waste, while Route 2 produces no toxic byproducts.

(c) (i) Comparing Exp 1 and Exp 2: when $[\text{NO}_2]$ doubles (from $0.050$ to $0.100\text{ M}$) with $[\text{F}_2]$ constant, initial rate doubles (from $1.20 \times 10^{-3}$ to $2.40 \times 10^{-3}\text{ M s}^{-1}$). Hence, order with respect to $\text{NO}_2$ is 1.
Comparing Exp 2 and Exp 3: when $[\text{F}_2]$ doubles (from $0.100$ to $0.200\text{ M}$) with $[\text{NO}_2]$ constant, initial rate doubles (from $2.40 \times 10^{-3}$ to $4.80 \times 10^{-3}\text{ M s}^{-1}$). Hence, order with respect to $\text{F}_2$ is 1.
Rate equation: $\text{Rate} = k[\text{NO}_2][\text{F}_2]$
(ii) Using Exp 1:
$$k = \frac{\text{Rate}}{[\text{NO}_2][\text{F}_2]} = \frac{1.20 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}}{(0.050\text{ mol dm}^{-3})(0.100\text{ mol dm}^{-3})} = 0.240\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}$$
(iii) Using $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ with $T_1 = 300\text{ K}$, $T_2 = 350\text{ K}$, $E_a = 45000\text{ J mol}^{-1}$:
$$\ln\left(\frac{k_2}{0.240}\right) = -\frac{45000}{8.31}\left(\frac{1}{350} - \frac{1}{300}\right) = -5415.16 \times (-0.00047619) = 2.5786$$
$$\frac{k_2}{0.240} = e^{2.5786} = 13.18$$
$$k_2 = 0.240 \times 13.18 = 3.16\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}$$
(iv) A catalyst does not change the shape of the Maxwell-Boltzmann distribution curve, but lowers the activation energy ($E_a$). As a result, a larger fraction of molecules have kinetic energy equal to or greater than the lower activation energy, increasing the frequency of effective collisions and thus the reaction rate.

(a) (i) Vanadium(V) oxide / V2O5 (1 mark)
Compromise temperature explanation: forward reaction is exothermic, lower T increases equilibrium yield (1 mark); higher T increases rate of reaction (1 mark). (Total: 3 marks)
(ii) High equilibrium conversion is achieved at atmospheric pressure / building high pressure plant is uneconomical (1 mark)
(iii) Direct dissolution in water produces acid mist / violent exothermic reaction (1 mark); dissolving in H2SO4 forms oleum safely (1 mark). (Total: 2 marks)

(b) (i) Correct formula and molar masses: 44.0 / 181.1 * 100% (1 mark); 24.3% (1 mark). (Total: 2 marks)
(ii) Any two green principles: Route 2 has 100% atom economy (1 mark); Route 2 avoids the use of toxic Cl2 / avoids CaCl2 waste production (1 mark). (Total: 2 marks)

(c) (i) Order with respect to NO2 = 1 (with deduction) (1 mark); Order with respect to F2 = 1 (with deduction) (1 mark); Correct rate equation: Rate = k[NO2][F2] (1 mark). (Total: 3 marks)
(ii) Value of k = 0.240 (1 mark); correct units dm3 mol-1 s-1 (1 mark). (Total: 2 marks)
(iii) Correct substitution into Arrhenius equation (1 mark); calculation of ln(k2/k1) or ratio (1 mark); correct final value of k2 = 3.16 dm3 mol-1 s-1 (accept 3.14 to 3.18) (1 mark). (Total: 3 marks)
(iv) Distribution curve unchanged, but Ea shifted to lower energy (1 mark); fraction of molecules with energy >= Ea increases, leading to more frequent effective collisions (1 mark). (Total: 2 marks)

(a) (i) Add dilute hydrochloric acid / dilute nitric acid, followed by barium chloride / barium nitrate solution. A white precipitate of barium sulphate ($\text{BaSO}_4$) is observed.
(ii) Add excess $\text{NaOH(aq)}$:
- $\text{Fe}^{2+}(\text{aq})$ reacts to form a dirty green precipitate of $\text{Fe(OH)}_2(\text{s})$, which is insoluble in excess $\text{NaOH(aq)}$.
- $\text{Al}^{3+}(\text{aq})$ first forms a white gelatinous precipitate of $\text{Al(OH)}_3(\text{s})$, which dissolves in excess $\text{NaOH(aq)}$ to form a colourless solution containing $[\text{Al(OH)}_4]^-(\text{aq})$ (or $\text{AlO}_2^-$).
Filtration separates the insoluble $\text{Fe(OH)}_2(\text{s})$ as residue from the soluble aluminium complex in the filtrate.
Ionic equation: $\text{Al(OH)}_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Al(OH)}_4]^-(\text{aq})$

(b) (i) Solvent: Hexane (or tetrachloromethane / dichloromethane).
Reason: Hexane is immiscible with water, forms a distinct layer, and non-polar iodine is significantly more soluble in hexane than in water.
(ii) Let $x$ be the mass (in g) of $\text{I}_2$ remaining in the aqueous layer.
Mass of $\text{I}_2$ extracted into hexane layer $= (0.500 - x)\text{ g}$.
Volume of aqueous layer $V_{\text{aq}} = 100.0\text{ cm}^3 = 0.100\text{ dm}^3$.
Volume of hexane layer $V_{\text{org}} = 50.0\text{ cm}^3 = 0.050\text{ dm}^3$.
$$K_D = \frac{[\text{I}_2]_{\text{org}}}{[\text{I}_2]_{\text{aq}}} = \frac{(0.500 - x) / 0.050}{x / 0.100} = 85.0$$
$$\frac{0.500 - x}{x} \times \frac{0.100}{0.050} = 85.0$$
$$\frac{0.500 - x}{x} \times 2 = 85.0 \implies \frac{0.500 - x}{x} = 42.5$$
$$0.500 - x = 42.5x \implies 43.5x = 0.500 \implies x = 0.01149\text{ g} \approx 0.0115\text{ g}$$
(iii) Yes, extracting with multiple smaller portions is more efficient because each successive extraction equilibrates with a fresh portion of solvent, leaving a much smaller total mass of solute in the aqueous phase compared to a single extraction with the same total volume.

(c) (i) - $3200 - 3600\text{ cm}^{-1}$: $\text{O}-\text{H}$ stretching mode (hydroxyl group in alcohol or carboxylic acid).
- $1715\text{ cm}^{-1}$: $\text{C}=\text{O}$ stretching mode (carbonyl group in carboxylic acid/ester/ketone/aldehyde).
(ii) (1) With $m/z = 90$ and presence of $\text{O}-\text{H}$ and $\text{C}=\text{O}$, molecular formula is $\text{C}_3\text{H}_6\text{O}_3$ ($3 \times 12 + 6 \times 1 + 3 \times 16 = 90$).
(2) The fragment at $m/z = 45$ corresponds to $\text{COOH}^+$ (or $\text{CH}_3\text{CHOH}^+$ / $\text{CH}_2\text{CH}_2\text{OH}^+$).
(3) Lactic acid (2-hydroxypropanoic acid): $\text{CH}_3\text{CH(OH)COOH}$ (or 3-hydroxypropanoic acid: $\text{HOCH}_2\text{CH}_2\text{COOH}$, or 2-hydroxyethyl formate).

(d) (i) $[\text{Fe(SCN)}]^{2+}$ is deeply coloured (absorbs visible light), and by Beer-Lambert law, its absorbance is directly proportional to concentration.
(ii) Purpose: To set the instrument absorbance to zero and calibrate for any absorbance caused by the cuvette/solvent.
Substance: Deionised water (or deionised water containing the reagents without $\text{Fe}^{3+}$).
(iii) Concentration in diluted solution:
$$\text{Concentration}_{\text{diluted}} = \frac{\text{Absorbance}}{0.0450} = \frac{0.360}{0.0450} = 8.00\text{ mg dm}^{-3}$$
Dilution factor $= \frac{100.0\text{ cm}^3}{25.0\text{ cm}^3} = 4$.
$$\text{Concentration in original sample} = 8.00\text{ mg dm}^{-3} \times 4 = 32.0\text{ mg dm}^{-3}$$

(a) (i) Add dilute HCl / HNO3 (1 mark); add BaCl2(aq) / Ba(NO3)2(aq) to give a white precipitate (1 mark). (Total: 2 marks)
(ii) Dirty green precipitate forms with Fe2+ and insoluble in excess (1 mark); white precipitate forms with Al3+ and dissolves in excess to give colourless solution (1 mark); Al(OH)3 + OH- -> [Al(OH)4]- (1 mark). (Total: 3 marks)

(b) (i) Hexane / dichloromethane / trichloromethane (1 mark); immiscible with water / high solubility for iodine (1 mark). (Total: 2 marks)
(ii) Setting up partition equilibrium expression with volumes (1 mark); correct algebraic manipulation (1 mark); mass = 0.0115 g (accept 0.0114 to 0.0115 g) (1 mark). (Total: 3 marks)
(iii) Multiple extractions are more efficient because partition equilibrium is established repeatedly with fresh solvent (1 mark).

(c) (i) O-H (hydroxyl / alcohol / carboxylic acid) (1 mark); C=O (carbonyl / carboxylic acid / ester / ketone / aldehyde) (1 mark). (Total: 2 marks)
(ii) (1) C3H6O3 (1 mark)
(2) COOH+ or CH3CHOH+ or CH2CH2OH+ (1 mark)
(3) Correct structural formula (e.g. CH3CH(OH)COOH) (1 mark)

(d) (i) The complex has visible colour / absorbance is proportional to concentration (1 mark)
(ii) To zero the colorimeter / compensate for solvent absorbance (1 mark); deionised water / solvent blank (1 mark). (Total: 2 marks)
(iii) Diluted concentration = 0.360 / 0.0450 = 8.00 mg dm-3 (1 mark); original concentration = 8.00 * 4 = 32.0 mg dm-3 (1 mark). (Total: 2 marks)