2021 HKDSE Chemistry Practice Paper with Answers

Solid \(W\) has a high melting point, does not conduct electricity when solid, but conducts when molten or aqueous; this is characteristic of a giant ionic structure containing mobile ions upon melting/dissolving. Solid \(Z\) has a very high melting point and is a non-conductor in all states, which is typical of a giant covalent (network) structure.

A: 1 mark

The ease of reduction of a metal oxide reflects the reactivity of the metal. Metal \(Y\) is the most reactive because its oxide cannot be reduced by carbon. Metal \(X\) is of intermediate reactivity because its oxide is reduced by carbon. Metal \(Z\) is the least reactive because its oxide decomposes upon heating alone. Therefore, the decreasing order of reactivity is \(Y > X > Z\).

A: 1 mark

Number of moles of ions added:
- \(n(\text{Ca}^{2+}) = 0.0200\times 0.10 = 0.0020\text{ mol}\)
- \(n(\text{Cl}^-) = 2 \times (0.0200\times 0.10) = 0.0040\text{ mol}\)
- \(n(\text{Ag}^+) = 0.0300\times 0.10 = 0.0030\text{ mol}\)
- \(n(\text{NO}_3^-) = 0.0300\times 0.10 = 0.0030\text{ mol}\)

Precipitation reaction: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\).
\(\text{Ag}^+\) is the limiting reactant (\(0.0030\text{ mol}\) reacts with \(0.0030\text{ mol}\) of \(\text{Cl}^-\)). Nearly all \(\text{Ag}^+\) ions are precipitated out, leaving only an extremely small concentration due to the slight solubility of \(\text{AgCl}\). Thus, \(\text{Ag}^+(\text{aq})\) has the lowest concentration.

C: 1 mark

Target reaction: \(\text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g})\).
Using Hess's Law:
\(\Delta H_\text{f}^\theta = \Delta H_1 + \Delta H_2 - \Delta H_3 = -393.5 + (-571.6) - (-890.4) = -74.7\text{ kJ mol}^{-1}\).

A: 1 mark

In this galvanic cell, zinc is more reactive and undergoes oxidation (anode: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\)), so the mass of the zinc electrode decreases and electrons flow from zinc to copper in the external circuit. At the copper cathode, reduction occurs (\(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\)), which depletes positive charge in the cathode compartment. Thus, \(\text{K}^+(\text{aq})\) cations from the salt bridge migrate towards the copper half-cell to maintain electrical neutrality.

B: 1 mark

The carbonyl-containing acyclic isomers of \(\text{C}_4\text{H}_8\text{O}\) are aldehydes and ketones:
1. Butanal: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
2. 2-Methylpropanal: \((\text{CH}_3)_2\text{CHCHO}\)
3. Butanone: \(\text{CH}_3\text{COCH}_2\text{CH}_3\)

Hence, there are 3 structural isomers in total.

B: 1 mark

(1) Decreasing the volume increases the total pressure. The system shifts to the side with fewer moles of gas (from 3 moles on the left to 2 moles on the right), so it shifts to the right. Correct.
(2) The forward reaction is exothermic (\(\Delta H < 0\)). Increasing the temperature shifts the equilibrium to the endothermic side (left). Incorrect.
(3) Increasing the concentration of a reactant (\(\text{O}_2\)) shifts the equilibrium position to the right to consume the added reactant. Correct.

B: 1 mark

(1) \(\text{Na}_2\text{O}\) is a basic oxide that reacts with water to form strongly alkaline \(\text{NaOH}(\text{aq})\). Correct.
(2) \(\text{Al}_2\text{O}_3\) is an amphoteric oxide that reacts with acids (like \(\text{HCl}\)) and alkalis (like \(\text{NaOH}\)). Correct.
(3) \(\text{SO}_2\) is an acidic oxide that dissolves in water to form sulphurous acid (\(\text{H}_2\text{SO}_3\)), giving a solution with \(\text{pH} < 7\). Incorrect.

A: 1 mark

A substance with a high melting point that is soluble in water, does not conduct electricity when solid, but conducts when dissolved in water consists of oppositely charged ions held in a giant ionic lattice. In the molten or aqueous state, ions are mobile to conduct electricity, whereas in the solid state, ions are held in fixed positions.

B (1 mark)
- Giant ionic structure is characteristic of high melting points, non-conduction in solid state, and electrical conductivity in aqueous solution due to mobile ions.

\( \text{X} \) reacts vigorously with cold water, so it is the most reactive metal (e.g., alkali/alkaline earth metal like Na or Ca).
Thermal decomposition of carbonate of \( \text{Y} \) yields the metal directly, which indicates metal \( \text{Y} \) is very unreactive (e.g., Ag).
The carbonate of \( \text{Z} \) decomposes to form metal oxide and \( \text{CO}_2 \), which is typical of moderately reactive metals (e.g., Cu, Zn, Pb).
Therefore, the reactivity order from lowest to highest is \( \text{Y} < \text{Z} < \text{X} \).

A (1 mark)

In \( \text{MnO}_4^- \), the oxidation number of \( \text{Mn} \) is \( +7 \), which decreases to \( +2 \) in \( \text{Mn}^{2+} \); thus \( \text{MnO}_4^- \) acts as an oxidizing agent and is reduced.
In \( \text{H}_2\text{C}_2\text{O}_4 \), the oxidation number of carbon is \( +3 \) (since \( 2(+1) + 2x + 4(-2) = 0 \Rightarrow 2x = +6 \Rightarrow x = +3 \)). In \( \text{CO}_2 \), the oxidation number of carbon is \( +4 \). Therefore, carbon is oxidized and its oxidation number increases by 1 per carbon atom.

C (1 mark)
- The oxidation number of carbon changes from +3 in oxalic acid to +4 in carbon dioxide.

Initial moles of \( \text{Ca}^{2+} = 0.0200 \times 0.15 = 0.0030\,\text{mol} \).
Initial moles of \( \text{Cl}^- = 2 \times 0.0030 = 0.0060\,\text{mol} \).
Initial moles of \( \text{Ag}^+ = 0.0300 \times 0.10 = 0.0030\,\text{mol} \).
Precipitation reaction: \( \text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s) \).
\( \text{Ag}^+ \) is the limiting reactant (\( 0.0030\,\text{mol} \)), consuming \( 0.0030\,\text{mol} \) of \( \text{Cl}^- \).
Remaining moles of \( \text{Cl}^- = 0.0060 - 0.0030 = 0.0030\,\text{mol} \).
Total volume of solution \( = 20.0 + 30.0 = 50.0\,\text{cm}^3 = 0.0500\,\text{dm}^3 \).
Concentration of \( \text{Cl}^- = \frac{0.0030\,\text{mol}}{0.0500\,\text{dm}^3} = 0.060\,\text{M} \).

B (1 mark)
- Correct deduction of total chloride ions, consumed chloride ions, and dividing by total volume (50.0 cm³).

Formation equation of \( \text{CH}_3\text{OH}(l) \):
\[ \text{C}(\text{graphite}) + 2\,\text{H}_2(g) + \frac{1}{2}\,\text{O}_2(g) \rightarrow \text{CH}_3\text{OH}(l) \]
Using enthalpy of combustion:
\[ \Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products}) \]
\[ \Delta H_f^\ominus = [\Delta H_c^\ominus(\text{C}) + 2\Delta H_c^\ominus(\text{H}_2)] - [\Delta H_c^\ominus(\text{CH}_3\text{OH})] \]
\[ \Delta H_f^\ominus = [-393.5 + 2(-285.8)] - [-726.0] \]
\[ \Delta H_f^\ominus = [-393.5 - 571.6] + 726.0 = -965.1 + 726.0 = -239.1\,\text{kJ}\,\text{mol}^{-1} \]

D (1 mark)
- Applying Hess's Law with enthalpies of combustion: ΔHf = (-393.5 + 2(-285.8)) - (-726.0) = -239.1 kJ mol⁻¹.

Let us examine the structure of 3-bromo-2-methylpentan-2-ol:
\( \text{CH}_3-\text{C}(\text{CH}_3)(\text{OH})-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3 \).
- C1: \( -\text{CH}_3 \) (not chiral, 3 identical H)
- C2: bonded to \( -\text{OH} \), two \( -\text{CH}_3 \) groups, and \( -\text{CH}(\text{Br})\text{CH}_2\text{CH}_3 \) (not chiral, two methyl groups are identical)
- C3: bonded to \( -\text{H} \), \( -\text{Br} \), \( -\text{C}(\text{CH}_3)_2\text{OH} \), and \( -\text{CH}_2\text{CH}_3 \) (all four groups are different, so C3 is a chiral centre)
- C4: \( -\text{CH}_2- \) (not chiral, 2 identical H)
- C5: \( -\text{CH}_3 \) (not chiral, 3 identical H)
Hence, there is only 1 chiral carbon atom.

A (1 mark)
- C3 is bonded to 4 different groups: -H, -Br, -CH2CH3, and -C(CH3)2OH. C2 has two identical methyl groups.

The value of the equilibrium constant \( K_c \) depends only on temperature.
Since the forward reaction is exothermic (\( \Delta H < 0 \)), decreasing the temperature shifts the equilibrium position to the right, increasing the ratio of products to reactants at equilibrium, and thus increasing the value of \( K_c \).
Changes in pressure, concentration, or addition of a catalyst do not alter the value of \( K_c \).

B (1 mark)
- Kc is only affected by temperature. For an exothermic reaction, decreasing temperature increases Kc.

The first statement is true: \( \text{AlCl}_3(aq) \) is acidic, so it turns blue litmus red.
The second statement is also true: \( \text{Al}^{3+} \) has a high charge density, forming \( [\text{Al}(\text{H}_2\text{O})_6]^{3+} \), which hydrolyses to release \( \text{H}^+(aq) \) via \( [\text{Al}(\text{H}_2\text{O})_6]^{3+} \rightleftharpoons [\text{Al}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}^+ \).
The second statement correctly explains why the solution is acidic and turns blue litmus red.

A (1 mark)
- Both statements are true and the 2nd statement is a correct explanation of the 1st statement.

Magnesium fluoride (\(\text{MgF}_2\)) possesses a giant ionic lattice structure. In the solid state, its ions are held in fixed lattice positions and cannot move freely, so it cannot conduct electricity. When melted or dissolved in water, the ionic bonds break and mobile ions become free to move and carry electric charge. Silicon dioxide has a giant covalent network and does not conduct electricity even when molten. Graphite conducts electricity in the solid state. Polyethene has a simple molecular structure with a low melting point and does not conduct electricity.

B (1 mark)

Acidified dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\), orange) is a strong oxidising agent that oxidises sulphite ion (\(\text{SO}_3^{2-}\)) to sulphate ion (\(\text{SO}_4^{2-}\)), while being reduced to chromium(III) ion (\(\text{Cr}^{3+}\), green). Hence, the solution changes colour from orange to green, and \(\text{SO}_3^{2-}\) acts as the reducing agent.

B (1 mark)

Metal \(Z\) reacts vigorously with cold water, indicating it is very high in the reactivity series (such as Na or Ca). Metal \(X\) does not react with cold water but displaces \(Y\) from its salt solution, meaning \(X\) is more reactive than \(Y\). Metal \(Y\) is extracted merely by thermal decomposition of its oxide, indicating it is very unreactive (such as Ag or Hg). Therefore, the descending order of reactivity is \(Z > X > Y\).

A (1 mark)

Number of moles of \((\text{COOH})_2 = 0.150\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 3.75 \times 10^{-3}\text{ mol}\).
From the stoichiometric ratio \(1 : 2\), the number of moles of \(\text{NaOH} = 2 \times 3.75 \times 10^{-3}\text{ mol} = 7.50 \times 10^{-3}\text{ mol}\).
Molarity of \(\text{NaOH} = \frac{7.50 \times 10^{-3}\text{ mol}}{\frac{18.75}{1000}\text{ dm}^3} = 0.400\text{ M}\).

C (1 mark)

The equation for the formation of methanol is:
\[\text{C(graphite)} + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH}(\text{l})\]
Using Hess's Law:
\[\Delta H_\text{f}^\ominus = \sum \Delta H_\text{c}^\ominus(\text{reactants}) - \sum \Delta H_\text{c}^\ominus(\text{products})\]
\[\Delta H_\text{f}^\ominus = [\Delta H_\text{c}^\ominus[\text{C}] + 2\Delta H_\text{c}^\ominus[\text{H}_2]] - [\Delta H_\text{c}^\ominus[\text{CH}_3\text{OH}]]\]
\[\Delta H_\text{f}^\ominus = [-393.5 + 2(-285.8)] - (-726.0) = -965.1 + 726.0 = -239.1\text{ kJ mol}^{-1}\]

B (1 mark)

1. Find the longest continuous carbon chain containing both substituents: 5 carbon atoms (pentane).
2. Number the carbon chain from the end that gives the substituents the lowest locants: numbering from left gives locants 2 for chloro and 3 for methyl (sum = 5), whereas from right gives 3, 4 (sum = 7).
3. List the substituent prefixes alphabetically: 'chloro' precedes 'methyl'.
Therefore, the systematic name is 2-chloro-3-methylpentane.

A (1 mark)

In a fuel cell, reduction occurs at the cathode (positive electrode), where oxygen gas is reduced in the presence of water and hydroxide ions:
\[\text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^- \rightarrow 4\text{OH}^-(\text{aq})\]
At the anode (negative electrode), hydrogen gas is oxidised:
\[\text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + 2\text{e}^-\]

A (1 mark)

The structural 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 4 structural isomers in total.

C (1 mark)

Substance \(W\) has a high melting point and conducts electricity only in the molten or aqueous state due to the presence of mobile ions that are free to move. In the solid state, its ions are held rigidly in fixed positions in a giant ionic lattice. Thus, \(W\) has a giant ionic structure.

B (1 mark)

Zinc is higher than silver in the electrochemical series, so \(\text{Zn}\) acts as the anode (undergoes oxidation: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\)) and electrons flow from the \(\text{Zn}\) electrode to the \(\text{Ag}\) electrode through the external circuit. At the cathode, \(\text{Ag}^+\) ions are reduced (\(\text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)}\)), decreasing the concentration of \(\text{Ag}^+\) in the cathodic half-cell. The mass of the zinc electrode decreases as zinc dissolves.

C (1 mark)

Initial moles:
\(n(\text{Ca}^{2+}) = 0.0200 \times 0.30 = 0.0060\,\text{mol}\)
\(n(\text{Cl}^-) = 2 \times 0.0060 = 0.0120\,\text{mol}\)
\(n(\text{Ag}^+) = 0.0300 \times 0.20 = 0.0060\,\text{mol}\)
\(n(\text{NO}_3^-) = 0.0300 \times 0.20 = 0.0060\,\text{mol}\)

Reaction: \(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\)
All \(0.0060\,\text{mol}\) of \(\text{Ag}^+\) reacts with \(0.0060\,\text{mol}\) of \(\text{Cl}^-\).

Remaining amounts in the total volume of \(50.0\,\text{cm}^3\):
- \(n(\text{Cl}^-) = 0.0120 - 0.0060 = 0.0060\,\text{mol}\)
- \(n(\text{Ca}^{2+}) = 0.0060\,\text{mol}\)
- \(n(\text{NO}_3^-) = 0.0060\,\text{mol}\)
- \(n(\text{Ag}^+) \approx 0\,\text{mol}\)

Since \(\text{Ca}^{2+}\), \(\text{Cl}^-\), and \(\text{NO}_3^-\) all have \(0.0060\,\text{mol}\) in the same final solution, \([\text{Ca}^{2+}] = [\text{Cl}^-] = [\text{NO}_3^-] = \frac{0.0060}{0.050} = 0.12\,\text{M}\). Thus, \(\text{NO}_3^-\), \(\text{Ca}^{2+}\), and \(\text{Cl}^-\) share the highest concentration.

A (1 mark)

Formation equation: \(2\text{C(graphite)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}\)
Using combustion data:
\(\Delta H_f^\ominus[\text{C}_2\text{H}_6\text{(g)}] = 2\Delta H_c^\ominus[\text{C}] + 3\Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{C}_2\text{H}_6]\)
\(= 2(-393.5) + 3(-285.8) - (-1560.7)\)
\(= -787.0 - 857.4 + 1560.7 = -83.7\,\text{kJ mol}^{-1}\).

D (1 mark)

1. Identify the principal functional group: aldehyde (\(-\text{CHO}\)), which assigns carbon-1 to the aldehyde carbon.
2. Longest carbon chain containing the principal group: 5 carbons (pentanal).
3. Numbering the chain: \(\text{C1} = \text{CHO}\), \(\text{C2} = \text{CH(CH}_3)\), \(\text{C3} = \text{CH}_2\), \(\text{C4} = \text{CH(Br)}\), \(\text{C5} = \text{CH}_3\).
4. Substituents in alphabetical order: 4-bromo and 2-methyl.
Therefore, the systematic IUPAC name is 4-bromo-2-methylpentanal.

B (1 mark)

Across Period 3 from left to right, the nature of oxides changes progressively from strongly basic (\(\text{Na}_2\text{O}\), \(\text{MgO}\)) to amphoteric (\(\text{Al}_2\text{O}_3\)) and then to acidic (\(\text{SiO}_2\), \(\text{P}_4\text{O}_{10}\), \(\text{SO}_2\), \(\text{Cl}_2\text{O}_7\)).

A (1 mark)

The equilibrium constant \(K_c\) depends ONLY on temperature. Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature shifts the equilibrium position to the right (endothermic direction) according to Le Chatelier's principle, thereby increasing the value of \(K_c\). Changes in pressure or volume shift equilibrium without changing \(K_c\).

C (1 mark)

1st statement is true: butan-2-ol is oxidised by acidified potassium dichromate (orange \(\text{Cr}_2\text{O}_7^{2-}\)), reducing it to chromium(III) ions (green \(\text{Cr}^{3+}\)).
2nd statement is true: butan-2-ol has the \(-\text{OH}\) group attached to a carbon bonded to two other carbon atoms (secondary alcohol), and oxidation of a secondary alcohol yields a ketone (butanone).
The 2nd statement correctly explains why the redox reaction (and corresponding colour change) occurs in the 1st statement.

A (1 mark)

The low sublimation temperature (\(70\ ^\circ\text{C}\)) and lack of electrical conductivity in liquid state indicate that \(EF_3\) has a simple molecular structure. Molecules in simple molecular structures are held together by weak van der Waals' forces, which require relatively little thermal energy to overcome.

B (1 mark)

Calculate the change in oxidation number for each option:
- In A: \(\underline{\text{N}}\text{O}_3^- \rightarrow \underline{\text{N}}\text{H}_4^+\), oxidation number changes from \(+5\) to \(-3\), a decrease of \(8\).
- In B: \(\underline{\text{S}}\text{O}_4^{2-} \rightarrow \underline{\text{S}}\text{O}_2\), oxidation number changes from \(+6\) to \(+4\), a decrease of \(2\).
- In C: \(\underline{\text{Cr}}_2\text{O}_7^{2-} \rightarrow \underline{\text{Cr}}^{3+}\), oxidation number changes from \(+6\) to \(+3\), a decrease of \(3\).
- In D: \(\underline{\text{Mn}}\text{O}_4^- \rightarrow \underline{\text{Mn}}\text{O}_2\), oxidation number changes from \(+7\) to \(+4\), a decrease of \(3\).
Thus, option A represents the greatest decrease.

A (1 mark)

Number of moles of \(\text{Al} = \frac{5.40\text{ g}}{27.0\text{ g mol}^{-1}} = 0.200\text{ mol}\).
From the equation, the reaction of \(2\text{ mol}\) of \(\text{Al}\) releases \(1676\text{ kJ}\) of heat.
Therefore, heat released by \(0.200\text{ mol}\) of \(\text{Al} = 1676\text{ kJ} \times \frac{0.200\text{ mol}}{2\text{ mol}} = 167.6\text{ kJ}\).

B (1 mark)

The structural 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: \(\text{CH}_3\text{CH}_2\text{CH(Cl)}\text{CH}_3\)
3. 1-chloro-2-methylpropane: \((\text{CH}_3)_2\text{CHCH}_2\text{Cl}\)
4. 2-chloro-2-methylpropane: \((\text{CH}_3)_3\text{CCl}\)
There are 4 distinct structural isomers in total.

C (1 mark)

(a) A more reactive metal displaces a less reactive metal from its aqueous salt solution.
- Metal \(\text{Q}\) displaces both \(\text{P}\) and \(\text{R}\), hence \(\text{Q}\) is more reactive than \(\text{P}\) and \(\text{R}\).
- Metal \(\text{P}\) displaces \(\text{R}\), showing \(\text{P}\) is more reactive than \(\text{R}\).
- Metal \(\text{R}\) cannot displace either \(\text{P}\) or \(\text{Q}\), confirming \(\text{R}\) is the least reactive.
Thus, decreasing order of reactivity: \(\text{Q} > \text{P} > \text{R}\).

(b) The ionic equation is:
\[ \text{Q(s)} + \text{P}^{2+}\text{(aq)} \rightarrow \text{Q}^{2+}\text{(aq)} + \text{P(s)} \]

(c) Since \(\text{R}\) is of relatively low reactivity, its oxide can be reduced by heating with carbon (coke/charcoal) or with hydrogen gas / town gas:
\[ \text{RO(s)} + \text{C(s)} \xrightarrow{\Delta} \text{R(s)} + \text{CO(g)} \]

(a) [3 marks]
- Correct order: \(\text{Q} > \text{P} > \text{R}\) (1 mark)
- Justification that \(\text{Q}\) is most reactive as it displaces both \(\text{P}\) and \(\text{R}\) (1 mark)
- Justification that \(\text{P}\) is more reactive than \(\text{R}\) as \(\text{P}\) displaces \(\text{R}\) (1 mark)

(b) [1 mark]
- \(\text{Q(s)} + \text{P}^{2+}\text{(aq)} \rightarrow \text{Q}^{2+}\text{(aq)} + \text{P(s)}\) (1 mark; state symbols not strictly required if formula correct, award 0 if spectator ions are included).

(c) [2 marks]
- Heating with carbon / charcoal / hydrogen (1 mark)
- Appropriate condition: strong heating / high temperature (1 mark)

(a) The analyte is acidic containing unreacted \(\text{HCl}\). Phenolphthalein is colourless in acid and turns pale pink at the endpoint when slightly alkaline.

(b) Number of moles of \(\text{NaOH}\) used in titration:
\[ n(\text{NaOH}) = 0.100\text{ mol dm}^{-3} \times \frac{24.50}{1000}\text{ dm}^3 = 2.45 \times 10^{-3}\text{ mol} \]
Since \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\), the mole ratio is \(1:1\).
Moles of \(\text{HCl}\) in \(25.0\text{ cm}^3 = 2.45 \times 10^{-3}\text{ mol}\).
Moles of \(\text{HCl}\) in the total \(250.0\text{ cm}^3\) volumetric solution:
\[ n(\text{HCl})_{\text{unreacted}} = 2.45 \times 10^{-3} \times \frac{250.0}{25.0} = 0.0245\text{ mol} \]

(c) Initial moles of \(\text{HCl}\) added:
\[ n(\text{HCl})_{\text{initial}} = 1.20\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0600\text{ mol} \]
Moles of \(\text{HCl}\) reacted with \(\text{MgCO}_3\):
\[ n(\text{HCl})_{\text{reacted}} = 0.0600 - 0.0245 = 0.0355\text{ mol} \]
The reaction equation is:
\[ \text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]
Moles of \(\text{MgCO}_3 = \frac{0.0355}{2} = 0.01775\text{ mol}\).
Molar mass of \(\text{MgCO}_3 = 24.3 + 12.0 + 3(16.0) = 84.3\text{ g mol}^{-1}\).
Mass of \(\text{MgCO}_3 = 0.01775\text{ mol} \times 84.3\text{ g mol}^{-1} = 1.4963\text{ g}\).
Percentage by mass:
\[ \% = \frac{1.4963\text{ g}}{1.85\text{ g}} \times 100\% = 80.88\% \approx 80.9\% \text{ (or } 80.8\%\text{)} \]

(a) [1 mark]
- Colourless to (pale / light) pink (1 mark; do not accept red or pink to colourless)

(b) [2 marks]
- Moles in \(25.0\text{ cm}^3 = 0.100 \times 0.02450 = 2.45 \times 10^{-3}\text{ mol}\) (1* mark)
- Total unreacted moles in \(250.0\text{ cm}^3 = 2.45 \times 10^{-3} \times 10 = 0.0245\text{ mol}\) (1 mark)

(c) [4 marks]
- Initial moles of \(\text{HCl} = 1.20 \times 0.050 = 0.0600\text{ mol}\) (1* mark)
- Moles of \(\text{HCl}\) reacted \(= 0.0600 - 0.0245 = 0.0355\text{ mol}\) AND Moles of \(\text{MgCO}_3 = 0.0355 / 2 = 0.01775\text{ mol}\) (1* mark)
- Mass of \(\text{MgCO}_3 = 0.01775 \times 84.3 = 1.496\text{ g}\) (1* mark)
- Percentage mass \(= (1.496 / 1.85) \times 100\% = 80.8\%\) or \(80.9\%\) (accept range: 80.7% to 81.0%) (1 mark)

(a) Silicon dioxide has a giant covalent structure (giant network). To melt it, numerous strong Si-O covalent bonds throughout the lattice must be broken, requiring a very large amount of thermal energy.
In contrast, sulfur dioxide has a simple molecular structure. Molecules are held together only by weak intermolecular forces (van der Waals' forces), which require very little thermal energy to overcome.

(b) In solid silicon dioxide, all valence electrons of silicon and oxygen are fixed/localised in covalent bonds. There are no mobile delocalised electrons or mobile ions to act as charge carriers.

(c) Outermost shell electron diagram for \(\text{SO}_2\):
Sulfur has 6 valence electrons; each oxygen has 6 valence electrons.
Sulfur forms one coordinate bond and one double bond with oxygen, or two double bonds with an expanded octet. Showing standard octet: S has 1 lone pair, a double bond to one O (4 sharing electrons), and a dative bond / single pair shared from S to the other O, with full octets around all atoms.

(a) [3 marks]
- \(\text{SiO}_2\) has a giant covalent structure / network, while \(\text{SO}_2\) has a simple molecular structure. (1 mark)
- Melting \(\text{SiO}_2\) requires breaking extensive / many strong covalent bonds. (1 mark)
- Melting \(\text{SO}_2\) requires overcoming weak intermolecular forces / van der Waals' forces only. (1 mark)

(b) [1 mark]
- All valence electrons are held tightly in covalent bonds / absence of delocalised electrons and mobile ions. (1 mark)

(c) [2 marks]
- Correct sharing between S and both O atoms (1 mark)
- Correct non-bonding pairs on all atoms giving complete outer shells (1 mark)

(a) Equation:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)} + \frac{9}{2}\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \]

(b) Heat absorbed by water:
\[ Q = m c \Delta T = 200.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (46.8 - 22.4)\text{ K} = 20398.4\text{ J} = 20.3984\text{ kJ} \]
Mass of propan-1-ol burned \(= 124.62 - 123.78 = 0.84\text{ g}\).
Molar mass of propan-1-ol \(= 3(12.0) + 8(1.0) + 16.0 = 60.0\text{ g mol}^{-1}\).
Moles of propan-1-ol \(= \frac{0.84\text{ g}}{60.0\text{ g mol}^{-1}} = 0.0140\text{ mol}\).
Enthalpy change of combustion:
\[ \Delta H_c = -\frac{Q}{n} = -\frac{20.3984\text{ kJ}}{0.0140\text{ mol}} = -1457\text{ kJ mol}^{-1} \approx -1.46 \times 10^3\text{ kJ mol}^{-1} \]

(c) Two sources of discrepancy:
1. Significant heat loss to the surroundings and the calorimeter container.
2. Incomplete combustion of propan-1-ol (forming soot / CO).
(Alternatively: loss of fuel by evaporation between weighings).

(d) Add windshields/draught screens around the setup, or cover the copper can with a lid.

(a) [1 mark]
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \frac{9}{2}\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\) (or balanced with integer coefficients: \(2\text{C}_3\text{H}_7\text{OH} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 8\text{H}_2\text{O}\)) (1 mark)

(b) [3 marks]
- Calculation of heat absorbed: \(Q = 200.0 \times 4.18 \times 24.4 = 20398.4\text{ J}\) (1* mark)
- Moles of propan-1-ol: \(0.84 / 60.0 = 0.0140\text{ mol}\) (1* mark)
- \(\Delta H_c = -20.3984 / 0.0140 = -1457\text{ kJ mol}^{-1}\) (correct negative sign, value and units; accept \(-1450\) to \(-1460\text{ kJ mol}^{-1}\)) (1 mark)

(c) [2 marks]
- Heat loss to surrounding air / container not accounted for (1 mark)
- Incomplete combustion occurred / evaporation of propan-1-ol (1 mark)

(d) [1 mark]
- Put a lid on the can / use draught shields around the flame / use a bomb calorimeter (1 mark)

(a) The equilibrium constant expression is:
\[ K_c = \frac{[\text{PCl}_3\text{(g)}][\text{Cl}_2\text{(g)}]}{[\text{PCl}_5\text{(g)}]} \]
Unit: \(\frac{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}{\text{mol dm}^{-3}} = \text{mol dm}^{-3}\).

(b) ICE table (moles in \(2.0\text{ dm}^3\)):
- Initial: \(n(\text{PCl}_5) = 0.60\), \(n(\text{PCl}_3) = 0\), \(n(\text{Cl}_2) = 0\)
- Change: \(-0.24\), \(+0.24\), \(+0.24\)
- Equilibrium: \(n(\text{PCl}_5) = 0.60 - 0.24 = 0.36\text{ mol}\), \(n(\text{PCl}_3) = 0.24\text{ mol}\), \(n(\text{Cl}_2) = 0.24\text{ mol}\)

Equilibrium concentrations:
\[ [\text{PCl}_5] = \frac{0.36}{2.0} = 0.18\text{ mol dm}^{-3} \]
\[ [\text{PCl}_3] = \frac{0.24}{2.0} = 0.12\text{ mol dm}^{-3} \]
\[ [\text{Cl}_2] = \frac{0.24}{2.0} = 0.12\text{ mol dm}^{-3} \]

\[ K_c = \frac{0.12 \times 0.12}{0.18} = 0.080\text{ mol dm}^{-3} \]

(c) When temperature is increased, according to Le Chatelier's principle, the equilibrium position shifts to the right (forward direction) to absorb the heat supplied, because the forward reaction is endothermic (\(\Delta H > 0\)).

(a) [1 mark]
- \(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\) with correct unit \(\text{mol dm}^{-3}\) (1 mark)

(b) [3 marks]
- Equilibrium moles: \(n(\text{PCl}_5) = 0.36\text{ mol}\) and \(n(\text{PCl}_3) = 0.24\text{ mol}\) (1* mark)
- Dividing moles by \(2.0\text{ dm}^3\) to find concentrations (1* mark)
- Correct calculated value: \(K_c = 0.080\text{ mol dm}^{-3}\) (accept \(0.08\)) (1 mark)

(c) [2 marks]
- Equilibrium shifts to the right / forward direction (1 mark)
- Forward reaction is endothermic / absorbs heat (1 mark)

(a) Iron is more reactive / has a greater tendency to lose electrons (higher reducing power) than silver. Hence, oxidation occurs at the iron electrode: \(\text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2e^-\). Therefore, the iron electrode is the negative electrode (anode).

(b) At the silver electrode, silver ions are reduced: \(\text{Ag}^+\text{(aq)} + e^- \rightarrow \text{Ag(s)}\). A silvery/grey crystalline deposit forms on the electrode (or the electrode gains mass / becomes thicker).

(c) The salt bridge completes the circuit and maintains electrical neutrality in the two half-cell solutions by allowing migration of ions (e.g., \(\text{K}^+\) ions into the cathode compartment and \(\text{NO}_3^-\) ions into the anode compartment).

(d) Combining the two half-equations:
\[ \text{Fe(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2\text{Ag(s)} \]

(a) [2 marks]
- Iron (electrode) (1 mark)
- Iron has a greater tendency to release electrons / higher reducing power than silver (1 mark)

(b) [1 mark]
- Shiny grey/silvery solid deposited / electrode increases in mass / thickens (1 mark; do not accept 'turns silver' without deposit/mass mention)

(c) [1 mark]
- Maintains electrical neutrality in both half-cells / allows ions to migrate to complete the circuit (1 mark)

(d) [2 marks]
- Correct species on both sides: \(\text{Fe(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2\text{Ag(s)}\) (1 mark)
- Balanced with correct charges (1 mark)

(a) The cotton wool plug allows \(\text{CO}_2\) gas to escape freely into the atmosphere while preventing liquid droplets/acid spray from escaping during effervescence.

(b) Carbon dioxide gas is generated in the reaction (\(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\)) and escapes into the surrounding air, leading to a decrease in total mass.

(c)(i) Powdered \(\text{CaCO}_3\) has a greater total surface area exposed to the acid. This increases the frequency of collisions (number of collisions per unit time) between \(\text{H}^+\) ions and \(\text{CaCO}_3\), hence increasing the frequency of effective collisions and the initial reaction rate.

(c)(ii) The final mass loss remains the same. Since \(\text{CaCO}_3\) is in excess, \(\text{HCl(aq)}\) is the limiting reactant. Because the number of moles of \(\text{HCl}\) used is identical in both trials, the theoretical yield of \(\text{CO}_2\) gas produced is identical.

(a) [1 mark]
- Prevent loss of acid spray / liquid splashing while allowing \(\text{CO}_2\) gas to escape (1 mark)

(b) [1 mark]
- Escape of carbon dioxide gas (1 mark)

(c)(i) [3 marks]
- Initial rate is higher / increases (1 mark)
- Powder has larger total surface area (1 mark)
- Higher frequency of effective collisions / more collisions per unit time (1 mark)

(c)(ii) [1 mark]
- Final mass loss is the same because \(\text{HCl}\) is the limiting reactant / same amount of \(\text{HCl}\) reacts completely to give the same amount of \(\text{CO}_2\) (1 mark)

(a) Reagent \(\text{P}\): Steam (\(\text{H}_2\text{O(g)}\)) in the presence of concentrated \(\text{H}_3\text{PO}_4\) catalyst at high temperature (\(\approx 300\ ^\circ\text{C}\)) and pressure (\(\approx 60\text{ atm}\)); OR concentrated \(\text{H}_2\text{SO}_4\) followed by water.

(b) According to Markovnikov's rule, addition of water across propene yields propan-2-ol as the major product.

(c) Oxidation of secondary alcohol (propan-2-ol) produces propanone (a ketone), which has the formula \(\text{CH}_3\text{COCH}_3\). Ketones react with 2,4-DNP but do not reduce Tollens' reagent.

(d) Acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) changes from orange (\(\text{Cr}_2\text{O}_7^{2-}\)) to green (\(\text{Cr}^{3+}\)).

(e) Test: Bubble each gas through a solution of bromine in 1,1,1-trichloroethane (or aqueous bromine water) in the dark / without UV light.
Observation: Propene rapidly decolourises the reddish-brown/orange bromine solution, whereas propane causes no observable change.

(a) [2 marks]
- Steam / \(\text{H}_2\text{O(g)}\) (1 mark)
- Concentrated \(\text{H}_3\text{PO}_4\) catalyst / heat and pressure (OR conc. \(\text{H}_2\text{SO}_4\) followed by \(\text{H}_2\text{O}\)) (1 mark)

(b) [1 mark]
- Propan-2-ol (1 mark)

(c) [1 mark]
- Correct structural formula of propanone: \(\text{CH}_3\text{COCH}_3\) (1 mark)

(d) [1 mark]
- (Solution turns from) orange to green (1 mark)

(e) [2 marks]
- Reagent: Bromine in organic solvent / Bromine water in the dark (1 mark)
- Observation: Propene turns bromine from brown/orange to colourless, while propane remains brown/orange (1 mark)

(a) The salt bridge completes the electrical circuit and maintains electrical neutrality in both half-cells through the migration of cations and anions.
(b) \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\): \(\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + \text{e}^-\).
(c) The purple permanganate ions \(\text{MnO}_4^-\text{(aq)}\) are reduced to practically colourless \(\text{Mn}^{2+}\text{(aq)}\) (or pale pink).
(d) Oxidation occurs in Half-cell B (anode, negative electrode) releasing electrons, which flow through the external circuit to Half-cell A (cathode, positive electrode) where reduction occurs.
(e) Acidified \(\text{MnO}_4^-\text{(aq)}\) has a higher standard reduction potential (stronger oxidising power) than acidified \(\text{Cr}_2\text{O}_7^{2-}\text{(aq)}\). Replacing it results in a lower cell electromotive force (e.m.f.), thus decreasing the voltmeter reading.

(a) Complete the electrical circuit / maintain electrical neutrality (1 mark).
(b) \(\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + \text{e}^-\) (1 mark; state symbols optional).
(c) Purple to colourless / very pale pink (1 mark; do not accept just "pink").
(d) From Half-cell B to Half-cell A / from electrode in B to electrode in A (1 mark).
(e) Acidified \(\text{MnO}_4^-\text{(aq)}\) is a stronger oxidising agent than acidified \(\text{Cr}_2\text{O}_7^{2-}\text{(aq)}\) / has higher reduction potential (1 mark);
leading to a smaller potential difference / lower e.m.f. of the cell (1 mark).

(a) The original commercial vinegar is too concentrated; diluting it avoids using excessively large volumes of \(\text{NaOH(aq)}\) or refilling the burette during each run.
(b) The mixture changes from colourless to pale pink.
(c) Discard trial value (\(21.80\text{ cm}^3\)).
Titre 1: \(22.30 - 1.00 = 21.30\text{ cm}^3\)
Titre 2: \(22.25 - 1.05 = 21.20\text{ cm}^3\)
Titre 3: \(22.35 - 1.05 = 21.30\text{ cm}^3\)
Average volume \(= \frac{21.30 + 21.20 + 21.30}{3} = 21.27\text{ cm}^3\) (or \(21.28\text{ cm}^3\) with correct decimal places).
(d) \(\text{Moles of NaOH} = 0.120 \times \frac{21.27}{1000} = 2.552 \times 10^{-3}\text{ mol}\).
\(\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\)
\(\text{Moles of CH}_3\text{COOH in } 25.0\text{ cm}^3 \text{ diluted sample} = 2.552 \times 10^{-3}\text{ mol}\).
\([\text{CH}_3\text{COOH}]\text{ in diluted sample} = \frac{2.552 \times 10^{-3}}{0.0250} = 0.1021\text{ M}\).
\([\text{CH}_3\text{COOH}]\text{ in original vinegar} = 0.1021 \times 10 = 1.021\text{ M}\).
\(\text{Molar mass of CH}_3\text{COOH} = 12.0 \times 2 + 1.0 \times 4 + 16.0 \times 2 = 60.0\text{ g mol}^{-1}\).
\(\text{Concentration in g dm}^{-3} = 1.021 \times 60.0 = 61.3\text{ g dm}^{-3}\).

(a) To avoid using an excessively large volume of standard \(\text{NaOH(aq)}\) / so that titre volume is within readable burette range (1 mark).
(b) From colourless to (persistent) pale pink (1 mark; reject red or purple).
(c) Average titre calculation excluding the trial: \(21.27\text{ cm}^3\) or \(21.28\text{ cm}^3\) or \(21.30\text{ cm}^3\) with correct unit (1 mark).
(d) Moles of \(\text{NaOH}\) / moles of \(\text{CH}_3\text{COOH}\) in diluted aliquot calculated correctly (1 mark);
Concentration of \(\text{CH}_3\text{COOH}\) in original vinegar in \(\text{mol dm}^{-3}\) obtained by multiplying dilution factor of 10 (1 mark);
Final answer converted to \(\text{g dm}^{-3}\): \(61.3\text{ g dm}^{-3}\) (range: 61.2 – 61.4) (1 mark).

(a) Cis-trans isomerism requires: (1) restricted rotation about a double bond (or ring); (2) two different groups attached to each carbon atom of the double bond.
(b) For cis-but-2-en-1-ol, both hydrogen atoms are on one side of the \(\text{C}=\text{C}\) bond, while the \(-\text{CH}_3\) and \(-\text{CH}_2\text{OH}\) groups are on the other side.
(c) (i) \(\text{Cr}_2\text{O}_7^{2-}\) is reduced to \(\text{Cr}^{3+}\), so the solution changes from orange to green.
(ii) Primary alcohol group \(-\text{CH}_2\text{OH}\) is oxidized under reflux to a carboxylic acid group \(-\text{COOH}\), forming but-2-enoic acid, \(\text{CH}_3\text{CH}=\text{CHCOOH}\).
(d) Test for unsaturation (\(\text{C}=\text{C}\)): Add bromine dissolved in an organic solvent (or bromine water) in the dark. But-2-en-1-ol undergoes addition and decolourises the orange/brown solution. Butan-1-ol does not react in the dark and remains orange/brown.

(a) Restricted rotation around \(\text{C}=\text{C}\) bond AND each carbon atom of the double bond is bonded to two different groups (1 mark for both conditions).
(b) Correct structural formula showing cis-geometry (1 mark).
(c) (i) Orange to green (1 mark).
(ii) \(\text{CH}_3\text{CH}=\text{CHCOOH}\) / correct structure of but-2-enoic acid (1 mark).
(d) Reagent: \(\text{Br}_2\text{(in organic solvent)}\) / \(\text{Br}_2\text{(aq)}\) in the dark (1 mark);
Observation: But-2-en-1-ol decolourises bromine solution / turns orange/brown to colourless; butan-1-ol gives no observable change / remains orange/brown (1 mark for correct comparative observations).

(a) Molar mass of \(\text{CaCl}_2 = 40.1 + 2(35.5) = 111.1\text{ g mol}^{-1}\).
\(\text{Moles of }\text{CaCl}_2 = \frac{4.44}{111.1} = 0.0400\text{ mol}\).
\(\text{Heat released, } q_1 = mc\Delta T = 100.0 \times 4.18 \times 7.8 = 3260.4\text{ J} = 3.260\text{ kJ}\).
\(\Delta H_1 = -\frac{3.2604}{0.0400} = -81.5\text{ kJ mol}^{-1}\).

(b) Molar mass of \(\text{CaCl}_2\cdot 2\text{H}_2\text{O} = 111.1 + 2(18.0) = 147.1\text{ g mol}^{-1}\).
\(\text{Moles of }\text{CaCl}_2\cdot 2\text{H}_2\text{O} = \frac{5.88}{147.1} = 0.0400\text{ mol}\).
\(\text{Heat absorbed, } q_2 = mc\Delta T = 100.0 \times 4.18 \times 0.9 = 376.2\text{ J} = 0.3762\text{ kJ}\).
\(\Delta H_2 = +\frac{0.3762}{0.0400} = +9.41\text{ kJ mol}^{-1}\).

(c) According to Hess's Law:
\(\text{CaCl}_2\text{(s)} + 2\text{H}_2\text{O(l)} \xrightarrow{\Delta H_{\text{r}}} \text{CaCl}_2\cdot 2\text{H}_2\text{O(s)}\)
Adding water to both gives \(\text{CaCl}_2\text{(aq)}\).
\(\Delta H_{\text{r}} + \Delta H_2 = \Delta H_1 \implies \Delta H_{\text{r}} = \Delta H_1 - \Delta H_2 = -81.5 - (+9.41) = -90.9\text{ kJ mol}^{-1}\).

(d) Negligible heat loss to surroundings, or negligible heat capacity of polystyrene cup and thermometer.

(a) Calculation of heat released and moles: \(q = 3.26\text{ kJ}\), \(n = 0.0400\text{ mol}\) (1 mark);
\(\Delta H_1 = -81.5\text{ kJ mol}^{-1}\) with correct sign and unit (1 mark).
(b) Calculation of heat absorbed and moles: \(q = 0.376\text{ kJ}\), \(n = 0.0400\text{ mol}\) (1 mark);
\(\Delta H_2 = +9.41\text{ kJ mol}^{-1}\) with correct sign and unit (1 mark).
(c) Applying Hess's Law / enthalpy cycle: \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\) (1 mark);
\(\Delta H_{\text{r}} = -90.9\text{ kJ mol}^{-1}\) (accept \(-90.8\) to \(-91.0\text{ kJ mol}^{-1}\)) (1 mark).
(d) Heat capacity of cup is negligible / no heat loss to the surroundings / specific heat capacity of solution is equal to that of pure water (1 mark).

(a) \(K_{\text{c}} = \frac{[\text{N}_2\text{O}_4\text{(g)}]}{[\text{NO}_2\text{(g)}]^2}\).
Units: \(\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1}\text{ dm}^3\).

(b)
Reaction: \(2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}\)
Initial moles: \(\text{NO}_2 = 0.80\text{ mol}\), \(\text{N}_2\text{O}_4 = 0\text{ mol}\).
Moles of \(\text{N}_2\text{O}_4\) at equilibrium \(= 0.30\text{ mol}\).
Moles of \(\text{NO}_2\) reacted \(= 2 \times 0.30 = 0.60\text{ mol}\).
Moles of \(\text{NO}_2\) at equilibrium \(= 0.80 - 0.60 = 0.20\text{ mol}\).
Equilibrium concentrations in \(2.0\text{ dm}^3\):
\([\text{NO}_2] = \frac{0.20}{2.0} = 0.10\text{ mol dm}^{-3}\)
\([\text{N}_2\text{O}_4] = \frac{0.30}{2.0} = 0.15\text{ mol dm}^{-3}\)
\(K_{\text{c}} = \frac{0.15}{(0.10)^2} = \frac{0.15}{0.010} = 15\text{ mol}^{-1}\text{ dm}^3\).

(c) (i) Since \(\Delta H < 0\) (forward reaction is exothermic), according to Le Chatelier's principle, an increase in temperature favours the endothermic reverse direction. Thus, the equilibrium shifts to the left.
(ii) Shifting to the left decreases \([\text{N}_2\text{O}_4]\) and increases \([\text{NO}_2]\), causing the numerical value of \(K_{\text{c}}\) to decrease.

(a) Expression for \(K_{\text{c}}\) with correct powers (1 mark); correct units \(\text{mol}^{-1}\text{ dm}^3\) (1 mark).
(b) Finding equilibrium moles/concentrations of \(\text{NO}_2\) and \(\text{N}_2\text{O}_4\) (1 mark);
Correct value of \(K_{\text{c}} = 15\) (1 mark).
(c) (i) Equilibrium shifts to the left because the forward reaction is exothermic / reverse reaction is endothermic (1 mark).
(ii) Value of \(K_{\text{c}}\) decreases (1 mark).

(a) (i)
Given:
$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$
$\ln(3.0) = -\frac{52.0 \times 10^3}{8.31}\left(\frac{1}{T_2} - \frac{1}{298}\right)$
$1.0986 = -6257.52 \left(\frac{1}{T_2} - 0.003356\right)$
$\frac{1}{T_2} - 0.003356 = -1.7556 \times 10^{-4}$
$\frac{1}{T_2} = 0.0031804\text{ K}^{-1}$
$T_2 = 314.4\text{ K} \approx 314\text{ K}$ (or $41.4\text{ }^\circ\text{C}$)

(a) (ii)
(1) Overall order $= 2 + 1 = 3$.
(2) $\text{Rate} = k[\text{NO}]^2[\text{O}_2] \implies \text{mol dm}^{-3}\text{ s}^{-1} = k(\text{mol dm}^{-3})^3 \implies k = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}$.
(3) When temperature increases, the average kinetic energy of the molecules increases and the Maxwell-Boltzmann distribution shifts to higher energy levels. Consequently, the fraction / number of molecules with kinetic energy equal to or greater than the activation energy ($E \ge E_a$) increases significantly, leading to a much higher frequency of effective collisions.

(b)
(i) Vanadium(V) oxide / $\text{V}_2\text{O}_5$.
(ii) Higher temperature is not used because the forward reaction is exothermic; increasing temperature shifts the equilibrium position to the left, decreasing the equilibrium yield of $\text{SO}_3$. A lower temperature is not used because the rate of reaction would be too slow to be economically viable.
(iii) The equilibrium yield of $\text{SO}_3$ at 1–2 atm is already sufficiently high (around 98%). Generating and maintaining high pressures requires thick-walled reaction vessels and compressors, leading to high capital, equipment, and maintenance costs without significant yield improvement.
(iv) The direct dissolution of $\text{SO}_3$ in water is extremely exothermic and vaporises water, creating an uncontrollable, corrosive mist of sulphuric acid droplets that is difficult to condense and collect safely.

(c) (i)
- Route 1:
$$\text{Molar mass of } \text{C}_2\text{H}_4\text{O} = 2(12.0) + 4(1.0) + 16.0 = 44.0\text{ g mol}^{-1}$$
$$\text{Total molar mass of all reactants} = [2(12.0)+4(1.0)] + [2(35.5)] + [40.1 + 2(16.0+1.0)] = 28.0 + 71.0 + 74.1 = 173.1\text{ g mol}^{-1}$$
$$\text{Atom economy of Route 1} = \frac{44.0}{173.1} \times 100\% = 25.4\%$$

- Route 2:
$$\text{Total molar mass of all reactants} = 28.0 + 0.5(32.0) = 44.0\text{ g mol}^{-1}$$
$$\text{Atom economy of Route 2} = \frac{44.0}{44.0} \times 100\% = 100\%$$

(c) (ii)
1. Route 2 uses non-toxic $\text{O}_2$ (or air) instead of highly toxic and corrosive chlorine gas ($\text{Cl}_2$), reducing safety hazards.
2. Route 2 produces no by-products or waste, whereas Route 1 produces large amounts of unwanted calcium chloride ($\text{CaCl}_2$) waste that requires disposal.

(c) (iii)
1. Conserves raw materials / reduces the waste of unreacted starting materials.
2. Increases the overall conversion / efficiency of reactants into products, thereby reducing manufacturing costs.

(a) (i) (2 marks)
- Correct substitution into the Arrhenius equation (1 mark)
- Correct calculation of temperature: $314\text{ K}$ or $41.4\text{ }^\circ\text{C}$ (Accept: $314 - 315\text{ K}$ or $41 - 42\text{ }^\circ\text{C}$) (1 mark)

(a) (ii) (5 marks)
- (1) 3 / third order (1 mark)
- (2) $\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}$ (Accept: $\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}$) (1 mark)
- (3) As temperature increases, the Maxwell-Boltzmann distribution shifts to the right / flattens (1 mark);
The fraction / number of molecules having kinetic energy $\ge E_a$ increases significantly (1 mark);
Frequency of effective collisions increases (1 mark).

(b) (6 marks)
- (i) Vanadium(V) oxide / vanadium pentoxide / $\text{V}_2\text{O}_5$ (1 mark)
- (ii) Forward reaction is exothermic, higher temperature shifts equilibrium to the left / reduces yield (1 mark);
Lower temperature results in a rate of reaction that is too slow (1 mark).
- (iii) The yield of $\text{SO}_3$ is already very high (98–99%) at atmospheric pressure (1 mark);
High pressure requires high capital / equipment / energy costs (1 mark).
- (iv) Direct reaction with water is highly exothermic and forms a dense mist of $\text{H}_2\text{SO}_4$ which is hard to condense / hazardous (1 mark).

(c) (7 marks)
- (i) Route 1: $\frac{44.0}{173.1} \times 100\% = 25.4\%$ (Accept: $25.4\% - 25.42\%$) (2 marks: 1 mark for calculating total reactant mass 173.1, 1 mark for answer);
Route 2: $100\%$ (1 mark)
- (ii) Any TWO of the following (1 mark each, max 2 marks):
- Uses safer/less hazardous reagents (uses $\text{O}_2$ instead of toxic $\text{Cl}_2$)
- Avoids generation of solid waste (no $\text{CaCl}_2$ waste produced)
- Uses a catalyst which operates under milder conditions
- (iii) Any TWO of the following (1 mark each, max 2 marks):
- Conserves raw materials / minimizes chemical waste
- Increases the overall conversion / efficiency of the process
- Lowers production costs

(a)
(i) Add Tollens' reagent (ammoniacal silver nitrate solution) and warm in a water bath.
- Propanal: forms a silver mirror / shiny silver deposit on the tube wall (or grey precipitate).
- Propanone: no observable change / solution remains colourless.
(Alternative: Fehling's solution warmed; propanal forms a brick-red precipitate while propanone gives no observable change.)

(ii) Add dilute \(\text{HCl(aq)}\) followed by \(\text{BaCl}_2(\text{aq})\) (or \(\text{Ba(NO}_3)_2(\text{aq})\)).
- \(\text{Na}_2\text{SO}_4(\text{aq})\): forms a white precipitate (\(\text{BaSO}_4\)).
- \(\text{Na}_2\text{SO}_3(\text{aq})\): no precipitate forms (or colourless gas evolved that turns acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) from orange to green).

(iii) The orange solution turns green.

---

(b)
(i)
1. Add aqueous sodium hydrogencarbonate (\(\text{NaHCO}_3\)) or aqueous sodium carbonate (\(\text{Na}_2\text{CO}_3\)) to the crude mixture in a separating funnel.
2. Invert and shake the funnel gently, opening the stopcock repeatedly to release the carbon dioxide gas pressure built up.
3. Allow the two layers to separate, and discard the lower aqueous layer containing the sodium salts of the acids, collecting the upper organic layer.

(ii) (1) Anhydrous magnesium sulfate (\(\text{MgSO}_4\)) / anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) / anhydrous calcium chloride (\(\text{CaCl}_2\)).

(2)
- The IR spectrum of the product shows a strong absorption peak at \(1680\text{ to }1800\text{ cm}^{-1}\) (\(\text{C=O}\) stretching) and at \(1000\text{ to }1300\text{ cm}^{-1}\) (\(\text{C-O}\) stretching), confirming the presence of ethyl ethanoate.
- The absence of a broad absorption band at \(2500\text{ to }3300\text{ cm}^{-1}\) (\(\text{O-H}\) stretching of carboxylic acid) confirms the complete removal of ethanoic acid.
- The absence of a broad absorption band at \(3230\text{ to }3670\text{ cm}^{-1}\) (\(\text{O-H}\) stretching of alcohol) confirms the complete removal of ethanol.

---

(c)
(i)
- Mole ratio of elements in X:
$$\text{C} : \text{H} : \text{O} = \frac{72.0}{12.0} : \frac{6.7}{1.0} : \frac{21.3}{16.0} = 6.0 : 6.7 : 1.33 = 4.5 : 5.0 : 1 = 9 : 10 : 2$$
- Empirical formula is \(\text{C}_9\text{H}_{10}\text{O}_2\) (empirical formula mass \(= 9 \times 12.0 + 10 \times 1.0 + 2 \times 16.0 = 150.0\)).
- Since relative molecular mass \(= m/z \text{ of molecular ion} = 150\), the molecular formula of X is \(\text{C}_9\text{H}_{10}\text{O}_2\).

(ii)
- The base peak at \(m/z = 119\) corresponds to \([\text{CH}_3\text{C}_6\text{H}_4\text{CO}]^+\) (or \([\text{C}_8\text{H}_7\text{O}]^+\)), which results from the cleavage of \(\text{---OCH}_3\) (loss of mass 31).
- Since X is an ester of molecular formula \(\text{C}_9\text{H}_{10}\text{O}_2\), the structure of X is methyl 4-methylbenzoate: \(\text{CH}_3\text{C}_6\text{H}_4\text{COOCH}_3\) (accept methyl 2-methylbenzoate or methyl 3-methylbenzoate).

(iii) (1) \(\text{CH}_3\text{C}_6\text{H}_4\text{COOCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{C}_6\text{H}_4\text{COONa} + \text{CH}_3\text{OH}\)

(2)
- Total moles of \(\text{NaOH}\) added \(= 0.500\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.02500\text{ mol}\)
- Moles of excess \(\text{NaOH}\) \(= \text{moles of HCl} = 0.400\text{ mol dm}^{-3} \times 0.02840\text{ dm}^3 = 0.01136\text{ mol}\)
- Moles of \(\text{NaOH}\) reacted with X \(= 0.02500 - 0.01136 = 0.01364\text{ mol}\)
- Since mole ratio of X : \(\text{NaOH}\) is \(1 : 1\), moles of X \(= 0.01364\text{ mol}\)
- Mass of X \(= 0.01364\text{ mol} \times 150.0\text{ g mol}^{-1} = 2.046\text{ g}\)
- Percentage by mass of X \(= \frac{2.046\text{ g}}{2.25\text{ g}} \times 100\% = 90.9\%\) (accept 90.93% to 91.0%)

(a)
(i)
- Reagent and condition: Tollens' reagent (and warm) / Fehling's solution (and warm) (1 mark)
- Observation: Propanal gives silver mirror (or brick-red ppt with Fehling's) while propanone gives no observable change (1 mark)

(ii)
- Reagent: Add dilute \(\text{HCl(aq)}\) and \(\text{BaCl}_2(\text{aq})\) / \(\text{Ba(NO}_3)_2(\text{aq})\) (1 mark)
- Observation: \(\text{Na}_2\text{SO}_4(\text{aq})\) gives a white precipitate whereas \(\text{Na}_2\text{SO}_3(\text{aq})\) gives no precipitate / dissolves with gas evolution (1 mark)

(iii)
- Solution changes from orange to green (1 mark)

(b)
(i)
- Add \(\text{NaHCO}_3(\text{aq})\) / \(\text{Na}_2\text{CO}_3(\text{aq})\) to the crude sample in the separating funnel (1 mark)
- Shake and invert the separating funnel, releasing gas pressure from time to time via the tap (1 mark)
- Allow layers to settle and discard the aqueous layer / collect the upper organic layer (1 mark)

(ii)(1)
- Anhydrous \(\text{MgSO}_4\) / anhydrous \(\text{Na}_2\text{SO}_4\) / anhydrous \(\text{CaCl}_2\) (1 mark)

(ii)(2)
- The presence of absorption peak at \(1680\text{--}1800\text{ cm}^{-1}\) (\(\text{C=O}\)) and/or \(1000\text{--}1300\text{ cm}^{-1}\) (\(\text{C-O}\)) corresponding to ester (1 mark)
- Absence of absorption peak at \(2500\text{--}3300\text{ cm}^{-1}\) corresponding to carboxylic acid \(\text{O-H}\) (1 mark)
- Absence of absorption peak at \(3230\text{--}3670\text{ cm}^{-1}\) corresponding to alcohol \(\text{O-H}\) (1 mark)

(c)
(i)
- Correct calculation of mole ratio: \(\text{C} : \text{H} : \text{O} = 9 : 10 : 2\) (1 mark)
- Deducing molecular formula \(\text{C}_9\text{H}_{10}\text{O}_2\) with molar mass \(= 150\text{ g mol}^{-1}\) (1 mark)

(ii)
- Stating that peak at \(m/z = 119\) corresponds to \([\text{CH}_3\text{C}_6\text{H}_4\text{CO}]^+\) or \([\text{C}_8\text{H}_7\text{O}]^+\) (1 mark)
- Correct structural formula: \(\text{CH}_3\text{C}_6\text{H}_4\text{COOCH}_3\) (1 mark)

(iii)(1)
- Correct balanced chemical equation (state symbols not required) (1 mark)

(iii)(2)
- Number of moles of excess \(\text{NaOH} = 0.400 \times 0.02840 = 0.01136\text{ mol}\) (1 mark)
- Number of moles of X \(= 0.02500 - 0.01136 = 0.01364\text{ mol}\) (1 mark)
- Percentage by mass \(= \frac{0.01364 \times 150.0}{2.25} \times 100\% = 90.9\%\) (accept 90.9% to 91.0%) (1 mark)