2024 Cambridge IAS-Level Chemistry (9701) 模擬試題連答案詳解

First, write the balanced chemical equation for the reaction: \( M(s) + 2\text{HCl}(aq) \rightarrow M\text{Cl}_2(aq) + \text{H}_2(g) \). Find the moles of hydrogen gas produced: \( n(\text{H}_2) = \frac{720 \text{ cm}^3}{24000 \text{ cm}^3\text{ mol}^{-1}} = 0.030 \text{ mol} \). Since the mole ratio of \( M \) to \( \text{H}_2 \) is 1:1, the moles of metal reacted is also 0.030 mol. Calculate the relative atomic mass of \( M \): \( A_r(M) = \frac{1.20 \text{ g}}{0.030 \text{ mol}} = 40.0 \text{ g mol}^{-1} \). Looking at the Periodic Table, this corresponds to Calcium (Ca).

1 mark for the correct option B. Award marks for calculating moles of gas (0.030 mol), using stoichiometry to find moles of metal, and determining the molar mass as 40 g/mol to identify calcium.

The successive ionisation energies show a very large jump between the 3rd and 4th ionisation energies (2745 to 11577 \( \text{kJ mol}^{-1} \)). This indicates that the element has three valence electrons and belongs to Group 13. Since it is in Period 3, the element is aluminium (\( \text{Al} \)). Aluminium forms the amphoteric oxide \( \text{Al}_2\text{O}_3 \) (formula \( X_2\text{O}_3 \)).

1 mark for the correct option B. Award marks for identifying that the large jump between the 3rd and 4th ionisation energies implies 3 valence electrons, identifying element X as aluminium, and determining its oxide's formula and amphoteric nature.

The carbonate ion, \( \text{CO}_3^{2-} \), has a central carbon atom with three bonding areas and no lone pairs. According to VSEPR theory, this results in a trigonal planar geometry with bond angles of exactly \( 120^\circ \). In contrast, \( \text{H}_3\text{O}^+ \) is trigonal pyramidal (~\( 107^\circ \)), \( \text{NH}_2^- \) is bent (~\( 104.5^\circ \)), and \( \text{NH}_4^+ \) is tetrahedral (\( 109.5^\circ \)).

1 mark for the correct option C. Award marks for analyzing the electron pair geometry and molecular shapes to identify that the trigonal planar species has a bond angle of 120 degrees.

Silicon is a metalloid with a giant covalent macromolecular structure, meaning melting it requires breaking many strong covalent bonds. Phosphorus (\( \text{P}_4 \)) is a simple molecular non-metal, meaning melting it only requires overcoming weak intermolecular forces (London dispersion forces) between discrete molecules.

1 mark for the correct option B. Award marks for identifying the different structures (giant covalent vs simple molecular) and the relative strengths of the forces holding them together.

In option D, iron(II) hydroxide, \( \text{Fe}(\text{OH})_2 \), is oxidized by oxygen to iron(III) hydroxide, \( \text{Fe}(\text{OH})_3 \). The oxidation state of iron increases from +2 to +3. In the other options, the oxidation state of the transition metal decreases: Cr goes from +6 to +3 in option A; Mn goes from +7 to +2 in option B; Fe goes from +3 to +2 in option C.

1 mark for the correct option D. Award marks for calculating and comparing the oxidation states of the transition metal reactant and product in each equation.

The electrophilic addition of \( \text{HBr} \) to 2-methylbut-2-ene, \( (\text{CH}_3)_2\text{C}=\text{CHCH}_3 \), proceeds via the most stable carbocation intermediate. The hydrogen ion (electrophile) adds to the carbon with more hydrogens (C3), producing a tertiary carbocation at C2 (\( (\text{CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3 \)), which is more stable than the alternative secondary carbocation. The bromide nucleophile then attacks C2 to yield 2-bromo-2-methylbutane as the major product.

1 mark for the correct option B. Award marks for using Markovnikov's rule, determining carbocation stability (tertiary vs secondary), and correctly naming the major product.

Since compound \( Z \) reacts with sodium carbonate to produce carbon dioxide (a gas), it must be a carboxylic acid. Fehling's solution only oxidizes aldehydes, so it does not react with carboxylic acids, which is consistent with the details given. For \( Y \) to be oxidized under reflux to a carboxylic acid, it must be a primary alcohol. Among the choices, butan-1-ol is the only primary alcohol. Butan-2-ol is a secondary alcohol (oxidizes to a ketone, which does not react with sodium carbonate), 2-methylpropan-2-ol is a tertiary alcohol (does not oxidize), and methylpropyl ether is an ether (does not oxidize under these conditions).

1 mark for the correct option C. Award marks for deducing that Z is a carboxylic acid, identifying that Y must therefore be a primary alcohol, and selecting the correct primary alcohol from the options.

The reactants (\( \text{SO}_2, \text{O}_2 \)) and the catalyst (\( \text{NO}_2 \)) are all in the gas phase. Because the catalyst and reactants are in the same phase, this is an example of homogeneous catalysis. Nitrogen dioxide is reduced to nitrogen monoxide (not nitrogen gas) in step 1, and sulfur dioxide is oxidized (not reduced) to sulfur trioxide.

1 mark for the correct option C. Award marks for identifying that the reactants and catalysts are in the same phase, confirming the catalyst type as homogeneous, and checking the correct oxidation state changes.

Silicon is a metalloid in Period 3. It reacts with chlorine to form silicon tetrachloride, \( \text{SiCl}_4 \), which is a liquid at room temperature and hydrolyses violently in water to release white fumes of \( \text{HCl} \). The oxide of silicon is silicon dioxide, \( \text{SiO}_2 \), which has a giant covalent structure and a very high melting point. Aluminium forms a chloride that is solid at room temperature. Phosphorus forms a molecular oxide with a relatively low sublimation point.

Award 1 mark for the correct option (B). Reject all other options.

4-chloropent-2-ene has two centers of stereoisomerism: 1. A carbon-carbon double bond between C2 and C3, which can exist as cis (Z) or trans (E) stereoisomers. 2. A chiral carbon at C4, which can exist as (R) or (S) enantiomers. Since the molecule is unsymmetrical, the total number of stereoisomers is \( 2^n = 2^2 = 4 \) (cis-R, cis-S, trans-R, trans-S).

Award 1 mark for identifying the correct total number of stereoisomers (C).

1. Calculate moles of \( \text{CuO} \): \( n(\text{CuO}) = \frac{15.9\text{ g}}{79.5\text{ g mol}^{-1}} = 0.200\text{ mol} \). 2. According to the stoichiometry, \( 3\text{ mol} \) of \( \text{CuO} \) produces \( 3\text{ mol} \) of \( \text{Cu} \). Therefore, the theoretical yield of \( \text{Cu} \) is \( 0.200\text{ mol} \). 3. Calculate theoretical mass of \( \text{Cu} \): \( \text{mass} = 0.200\text{ mol} \times 63.5\text{ g mol}^{-1} = 12.7\text{ g} \). 4. Calculate percentage yield: \( \text{Yield} = \frac{3.81\text{ g}}{12.7\text{ g}} \times 100\% = 30.0\% \).

Award 1 mark for the correct calculation showing 30.0% yield (B).

1. The formation of an orange precipitate with 2,4-DNPH shows that Y is a carbonyl compound (either an aldehyde or ketone). 2. The lack of reaction with Tollens' reagent indicates Y is a ketone, not an aldehyde. With 4 carbons, the only possible ketone is butanone. 3. Reduction of butanone (a ketone) with \( \text{NaBH}_4 \) yields a secondary alcohol, butan-2-ol.

Award 1 mark for identifying Z as butan-2-ol (B).

The carbonate ion, \( \text{CO}_3^{2-} \), has a central carbon atom with three bonding regions (two single bonds and one double bond in resonance) and zero lone pairs. This results in a symmetrical trigonal planar shape with bond angles of exactly 120°. In contrast: \( \text{SO}_2 \) has a lone pair on S, compressing the angle to less than 120°; \( \text{NF}_3 \) is trigonal pyramidal (around 102°); and \( \text{H}_3\text{O}^+ \) is trigonal pyramidal (around 107°).

Award 1 mark for the correct selection of the carbonate ion (B).

The reaction in a catalytic converter between nitrogen monoxide and carbon monoxide is represented by: \( 2\text{NO(g)} + 2\text{CO(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{CO}_2\text{(g)} \). In \( \text{NO} \), the oxidation state of nitrogen is +2. In \( \text{N}_2 \), it is 0. Therefore, the oxidation state of nitrogen changes from +2 to 0.

Award 1 mark for the correct balanced equation and correct oxidation state change (A).

Dehydration of pentan-2-ol (\( \text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3 \)) can proceed in two directions: 1. Elimination towards C1 yields pent-1-ene (no stereoisomers). 2. Elimination towards C3 yields pent-2-ene, which has cis-trans stereoisomerism, giving two distinct isomers: (E)-pent-2-ene and (Z)-pent-2-ene. Thus, there are 3 distinct alkene products in total.

Award 1 mark for the correct number of alkene products (B).

1. Determine the oxidation state change for sulfur: In \( \text{SO}_3^{2-} \), sulfur has oxidation state +4. In \( \text{SO}_4^{2-} \), sulfur has oxidation state +6. This is a loss of 2 electrons per sulfite ion. 2. Since \( 1.0\text{ mol} \) of \( \text{VO}_2^+ \) reacts with \( 1.0\text{ mol} \) of \( \text{SO}_3^{2-} \), \( 1.0\text{ mol} \) of \( \text{VO}_2^+ \) must gain the 2 electrons released. 3. In \( \text{VO}_2^+ \), the oxidation state of vanadium is +5 (\( V + 2(-2) = +1 \Rightarrow V = +5 \)). 4. Gaining 2 electrons reduces the oxidation state of vanadium to: \( +5 - 2 = +3 \).

Award 1 mark for calculating the correct oxidation state of +3 (B).

1. Element \(X\) must be sodium (\(\text{Na}\)). Its oxide, \(\text{Na}_2\text{O}\), is ionic and reacts with water to form the strong base sodium hydroxide, \(\text{NaOH}\), yielding a highly alkaline solution (pH 13).
2. Element \(Y\) must be silicon (\(\text{Si}\)). Its chloride, \(\text{SiCl}_4\), is covalent and undergoes rapid and violent hydrolysis with water to produce silicon dioxide, \(\text{SiO}_2\), and acidic hydrogen chloride gas, \(\text{HCl}\) (misty/white fumes, pH 2): \(\text{SiCl}_4(l) + 2\text{H}_2\text{O}(l) \rightarrow \text{SiO}_2(s) + 4\text{HCl}(g)\).

Award 1 mark for the correct option (A).
- Reject B because magnesium oxide is only sparingly soluble and yields a solution of pH ~9.
- Reject C because magnesium chloride dissolves to give a nearly neutral solution (pH ~6.5) with no white fumes.
- Reject D because aluminium oxide is insoluble in water.

1. Compound \(W\) reacts with 2,4-DNPH, which indicates it contains a carbonyl group (aldehyde or ketone).
2. Compound \(W\) does not react with Tollens' reagent, indicating it is a ketone, not an aldehyde. Given \(\text{C}_4\text{H}_8\text{O}\), \(W\) must be butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)).
3. Reduction of butanone with \(\text{NaBH}_4\) yields butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)), which is a secondary alcohol (compound \(Z\)).
4. Butan-2-ol contains the \(\text{CH}_3\text{CH(OH)}-\) group, so it reacts with alkaline aqueous iodine (the triiodomethane reaction) to form a yellow precipitate of triiodomethane (\(\text{CHI}_3\)).

Award 1 mark for the correct option (B).
- Reject A because secondary alcohols are oxidized to ketones, not carboxylic acids.
- Reject C because carbon-2 in butan-2-ol is bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\)) and is therefore chiral.
- Reject D because alcohols react with sodium metal to produce hydrogen gas, not carbon dioxide.

Dehydration of pentan-2-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)CH}_3\)) can occur in two directions:
- Elimination of a hydrogen atom from carbon-1 yields pent-1-ene (1 structural isomer, no stereoisomers).
- Elimination of a hydrogen atom from carbon-3 yields pent-2-ene. Pent-2-ene exhibits cis-trans (E/Z) isomerism, existing as cis-pent-2-ene and trans-pent-2-ene (2 stereoisomers).
- Therefore, a total of exactly three isomeric alkenes are formed: pent-1-ene, cis-pent-2-ene, and trans-pent-2-ene.

Award 1 mark for the correct option (B).
- Reject A because pentan-1-ol can only yield pent-1-ene (1 isomer).
- Reject C because pentan-3-ol can only yield pent-2-ene, which exists as 2 isomers (cis and trans).
- Reject D because 2-methylbutan-2-ol yields 2-methylbut-1-ene and 2-methylbut-2-ene, neither of which exhibits stereoisomerism (total of 2 isomers).

In the atmosphere, gaseous sulfur dioxide (\(\text{SO}_2\)) is oxidized to sulfur trioxide (\(\text{SO}_3\)) by gaseous nitrogen dioxide (\(\text{NO}_2\)). The equations for this cycle are:
1) \(\text{SO}_2(g) + \text{NO}_2(g) \rightarrow \text{SO}_3(g) + \text{NO}(g)\)
2) \(\text{NO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{NO}_2(g)\)
Since the reactants and the catalyst (\(\text{NO}_2\)) are in the same physical state (gas), \(\text{NO}_2\) acts as a homogeneous catalyst.

Award 1 mark for the correct option (C).
- Reject A because the catalyst and reactants are in the same phase, making it a homogeneous, not heterogeneous, catalyst.
- Reject B because the reaction between \(\text{NO}\) and \(\text{O}_3\) is extremely rapid and does not require UV catalysis.
- Reject D because in a catalytic converter, \(\text{NO}\) is reduced to nitrogen gas while carbon monoxide is oxidized to carbon dioxide: \(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\).

1. Volume of hydrocarbon = \(10.0\text{ cm}^3\).
2. The remaining gas consists of unreacted \(\text{O}_2\) and product \(\text{CO}_2\) (water is liquid at room temperature and has negligible volume).
3. Aqueous NaOH absorbs \(\text{CO}_2\). The decrease in volume is \(85.0 - 55.0 = 30.0\text{ cm}^3\), which represents the volume of \(\text{CO}_2\) produced.
4. Therefore, 1 mole of hydrocarbon produces 3 moles of \(\text{CO}_2\) (since \(10.0\text{ cm}^3 \rightarrow 30.0\text{ cm}^3\)), so \(x = 3\).
5. The volume of unreacted \(\text{O}_2\) is \(55.0\text{ cm}^3\), which means the volume of \(\text{O}_2\) consumed is \(100.0 - 55.0 = 45.0\text{ cm}^3\).
6. The combustion equation is \(\text{C}_3\text{H}_y + (3 + \frac{y}{4})\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\).
7. The volume ratio of \(\text{O}_2\) to hydrocarbon is \(45.0 / 10.0 = 4.5\). Thus, \(3 + \frac{y}{4} = 4.5 \rightarrow \frac{y}{4} = 1.5 \rightarrow y = 6\).
8. The molecular formula is \(\text{C}_3\text{H}_6\).

Award 1 mark for the correct option (B).
- Reject A because \(\text{C}_3\text{H}_4\) would consume only \(40.0\text{ cm}^3\) of \(\text{O}_2\).
- Reject C because \(\text{C}_3\text{H}_8\) would consume \(50.0\text{ cm}^3\) of \(\text{O}_2\).
- Reject D because \(\text{C}_4\text{H}_8\) would yield \(40.0\text{ cm}^3\) of \(\text{CO}_2\).

1. In \(\text{NH}_2^-\), nitrogen has 2 bonding pairs and 2 lone pairs. The arrangement is based on a tetrahedral structure, but lone pair-lone pair and lone pair-bond pair repulsions reduce the bond angle to approximately \(104.5^\circ\).
2. In \(\text{NH}_3\), nitrogen has 3 bonding pairs and 1 lone pair. The lone pair-bond pair repulsion reduces the tetrahedral angle to approximately \(107^\circ\).
3. In \(\text{NH}_4^+\), nitrogen has 4 bonding pairs and 0 lone pairs. It is a regular tetrahedral shape with a bond angle of \(109.5^\circ\).
4. Therefore, the order of increasing bond angle is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).

Award 1 mark for the correct option (A).
- Reject options B, C, and D as they represent incorrect rankings of the bond angles based on VSEPR theory.

An ester must contain the \(-\text{COO}-\) functional group. We can construct the possible esters by splitting the remaining 3 carbons between the carboxylic acid part and the alcohol part:
1. Methanoates (1-carbon acid): The alcohol part has 3 carbons, which can be propyl or isopropyl.
- Propyl methanoate: \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)
- Isopropyl methanoate: \(\text{HCOOCH(CH}_3)_2\)
2. Ethanoates (2-carbon acid): The alcohol part has 2 carbons.
- Ethyl ethanoate: \(\text{CH}_3\text{COOCH}_2\text{CH}_3\)
3. Propanoates (3-carbon acid): The alcohol part has 1 carbon.
- Methyl propanoate: \(\text{CH}_3\text{CH}_2\text{COOCH}_3\)
There are exactly 4 distinct ester isomers.

Award 1 mark for the correct option (B).
- Reject A, C, and D because the total number of structural isomers that are esters is exactly 4 (carboxylic acid isomers are excluded as the question specifies 'ester isomers').

1. The reaction is an electrophilic addition. The electrophile \(\text{H}^+\) attacks the double bond.
2. In 2-methylbut-1-ene (\(\text{CH}_2=\text{C(CH}_3)\text{CH}_2\text{CH}_3\)), adding the hydrogen to carbon-1 generates a tertiary carbocation at carbon-2: \(\text{CH}_3-\text{C}^+\text{(CH}_3)\text{CH}_2\text{CH}_3\).
3. Adding the hydrogen to carbon-2 would generate a primary carbocation at carbon-1: \(^{+}\text{CH}_2-\text{CH(CH}_3)\text{CH}_2\text{CH}_3\).
4. The tertiary carbocation is much more stable than the primary carbocation due to the positive inductive effect of three electron-releasing alkyl groups (two methyls and one ethyl). Thus, the reaction proceeds predominantly via the tertiary carbocation, forming 2-bromo-2-methylbutane as the major product.

Award 1 mark for the correct option (B).
- Reject A because the carbocation intermediate at C2 is tertiary, not secondary.
- Reject C because alkyl groups are electron-donating (or electron-releasing), not electron-withdrawing.
- Reject D because although bromide is a nucleophile, its nucleophilic strength does not determine the regioselectivity (major vs minor product) of the addition; that is determined by carbocation stability.

Let the formula of the hydrocarbon be \(\text{C}_x\text{H}_y\). The combustion equation is:
\(\text{C}_x\text{H}_y(\text{g}) + (x + \frac{y}{4})\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l})\)

Since the reaction is carried out at room temperature and pressure, the water produced condenses into a liquid with negligible volume.

Initial volume of gas = \(V(\text{C}_x\text{H}_y) + V(\text{O}_2) = 10 + 60 = 70\text{ cm}^3\).

Volume of \(\text{O}_2\) reacted = \(10(x + \frac{y}{4})\text{ cm}^3\).
Volume of remaining \(\text{O}_2\) = \(60 - 10(x + \frac{y}{4})\text{ cm}^3\).
Volume of \(\text{CO}_2\) produced = \(10x\text{ cm}^3\).

Total final gas volume = \(V(\text{remaining O}_2) + V(\text{CO}_2) = [60 - 10(x + \frac{y}{4})] + 10x = 60 - 2.5y\text{ cm}^3\).

We are given that the final gas volume is \(45\text{ cm}^3\):
\(60 - 2.5y = 45\)
\(2.5y = 15\)
\(y = 6\)

Therefore, the hydrocarbon must contain 6 hydrogen atoms. Looking at the options:
A) \(\text{C}_2\text{H}_4\) (y = 4)
B) \(\text{C}_3\text{H}_8\) (y = 8)
C) \(\text{C}_3\text{H}_6\) (y = 6)
D) \(\text{C}_4\text{H}_{10}\) (y = 10)

Thus, \(X\) must be \(\text{C}_3\text{H}_6\).

Award 1 mark for the correct answer (C). Method: Deduce that the change in gas volume depends only on the number of hydrogen atoms in the hydrocarbon, calculate the number of hydrogen atoms to be 6, and identify C₃H₆ as the matching option.

1. Analyze successive ionisation energies of element \(Y\):
- There is a very large increase (jump) between the 3rd and 4th ionisation energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the 4th electron is removed from an inner, lower-energy shell. Therefore, \(Y\) has three valence electrons and belongs to Group 13.

2. Identify \(Y\):
In Period 3, the Group 13 element is aluminium (\(\text{Al}\)). Thus, the oxide of \(Y\) is \(\text{Al}_2\text{O}_3\).

3. Evaluate the options for \(\text{Al}_2\text{O}_3\):
A) It does not react with water to form an alkaline solution (it is insoluble in water).
B) It has a giant ionic structure with significant covalent character, not a giant covalent structure.
C) \(\text{Al}_2\text{O}_3\) is amphoteric, so it reacts with both dilute acids (like \(\text{HCl}\)) and strong bases (like \(\text{NaOH}\)).
D) It is a solid with a very high melting point at room temperature.

Award 1 mark for identifying the element Y as aluminium from the jump between the 3rd and 4th ionisation energies, and correctly stating that aluminium oxide is amphoteric (reacts with both acids and bases).

To determine the bond angles, we use Valence Shell Electron Pair Repulsion (VSEPR) theory:

1. \(\text{NH}_2^-\):
- Nitrogen has 5 valence electrons + 1 from negative charge = 6 electrons. It forms 2 single covalent bonds with H atoms, leaving 2 lone pairs. Electron pair geometry is tetrahedral. Shape is non-linear (bent).
- Due to strong lone pair-lone pair repulsion, the bond angle is compressed significantly to approximately \(104.5^\circ\).

2. \(\text{NH}_3\):
- Nitrogen has 5 valence electrons. It forms 3 single covalent bonds with H atoms, leaving 1 lone pair. Shape is trigonal pyramidal.
- The single lone pair compresses the bond angle from \(109.5^\circ\) to approximately \(107^\circ\).

3. \(\text{NH}_4^+\):
- Nitrogen has 5 valence electrons - 1 from positive charge = 4 electrons. It forms 4 single covalent bonds with H atoms, leaving 0 lone pairs. Shape is tetrahedral.
- With no lone pairs, the bond angle is a perfect tetrahedral angle of \(109.5^\circ\).

Comparing the bond angles: \(104.5^\circ\) (\(\text{NH}_2^-\)) < \(107^\circ\) (\(\text{NH}_3\)) < \(109.5^\circ\) (\(\text{NH}_4^+\)).
Therefore, the correct order is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).

Award 1 mark for the correct choice (A). Method: Determine the number of bonding pairs and lone pairs for each species, apply VSEPR theory to deduce the relative compression of bond angles, and arrange them in the correct ascending order.

We can use the formula:
\(\Delta H^\theta_{\text{rxn}} = \sum \Delta H^\theta_{\text{f}}(\text{products}) - \sum \Delta H^\theta_{\text{f}}(\text{reactants})\)

For the reaction:
\(\text{N}_2\text{H}_4(\text{l}) + \text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\)

We know:
- \(\Delta H^\theta_{\text{rxn}} = -622\text{ kJ mol}^{-1}\)
- \(\Delta H^\theta_{\text{f}}[\text{N}_2(\text{g})] = 0\text{ kJ mol}^{-1}\) (element in its standard state)
- \(\Delta H^\theta_{\text{f}}[\text{O}_2(\text{g})] = 0\text{ kJ mol}^{-1}\) (element in its standard state)
- \(\Delta H^\theta_{\text{f}}[\text{H}_2\text{O}(\text{l})] = -286\text{ kJ mol}^{-1}\)

Substituting these values into the equation:
\(-622 = [ \Delta H^\theta_{\text{f}}(\text{N}_2(\text{g})) + 2 \times \Delta H^\theta_{\text{f}}(\text{H}_2\text{O}(\text{l})) ] - [ \Delta H^\theta_{\text{f}}(\text{N}_2\text{H}_4(\text{l})) + \Delta H^\theta_{\text{f}}(\text{O}_2(\text{g})) ]\)
\(-622 = [ 0 + 2(-286) ] - [ \Delta H^\theta_{\text{f}}(\text{N}_2\text{H}_4(\text{l})) + 0 ]\)
\(-622 = -572 - \Delta H^\theta_{\text{f}}(\text{N}_2\text{H}_4(\text{l}))\)

Rearranging to solve for \(\Delta H^\theta_{\text{f}}(\text{N}_2\text{H}_4(\text{l}))\):
\(\Delta H^\theta_{\text{f}}(\text{N}_2\text{H}_4(\text{l})) = -572 - (-622) = +50\text{ kJ mol}^{-1}\).

Award 1 mark for the correct calculation leading to +50 kJ mol⁻¹ (C). Correct use of stoichiometry (multiplying ΔH_f[H₂O] by 2) and signs is required.

1. Write the structure of 2-methylbut-2-ene:
\((\text{CH}_3)_2\text{C}=\text{CHCH}_3\)

2. When \(\text{HBr}\) reacts with this unsymmetrical alkene, an \(\text{H}^+\) ion adds to one of the double-bonded carbons, generating a carbocation.
- If \(\text{H}^+\) adds to C3: the positive charge ends up on C2, forming the tertiary carbocation, \((\text{CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\).
- If \(\text{H}^+\) adds to C2: the positive charge ends up on C3, forming the secondary carbocation, \((\text{CH}_3)_2\text{CH}-\text{CH}^+-\text{CH}_3\).

3. Determine stability:
Tertiary carbocations are more stable than secondary carbocations due to the positive inductive effect of three electron-releasing alkyl groups (two methyl groups and one ethyl group) which disperse the positive charge.

4. Therefore, the major product is formed via the tertiary carbocation intermediate, which has the formula \((\text{CH}_3)_2\text{C}^+\text{CH}_2\text{CH}_3\).

Award 1 mark for identifying the correct tertiary carbocation formula (A) based on the relative stability of carbocation intermediates during electrophilic addition.

1. Identify compound \(W\):
- \(W\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\), indicating it is an alcohol (or ether).
- Since it is resistant to oxidation by acidified potassium dichromate(VI), \(W\) must be a tertiary alcohol.
- The only tertiary alcohol with four carbon atoms is 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).

2. Identify the reaction:
- Reacting an alcohol with concentrated sulfuric acid at \(170\text{ }^\circ\text{C}\) is an acid-catalysed dehydration (elimination of water) to form an alkene.

3. Determine the product:
- Dehydration of 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\), yields 2-methylpropene:
\((\text{CH}_3)_3\text{COH} \rightarrow (\text{CH}_3)_2\text{C}=\text{CH}_2 + \text{H}_2\text{O}\)

Therefore, the organic product is 2-methylpropene.

Award 1 mark for identifying W as a tertiary alcohol (2-methylpropan-2-ol) and correctly determining its dehydration product under concentrated acid and heat as 2-methylpropene (B).

Let's review the mechanism of nucleophilic addition of \(\text{HCN}\) to ethanal:

- The reaction requires a catalytic amount of \(\text{NaCN}\) or alkali to generate \(\text{CN}^-\) ions, because \(\text{HCN}\) is a very weak acid and does not dissociate sufficiently on its own. Thus, \(\text{CN}^-\) acts as the nucleophile.
- Step 1: The nucleophilic \(\text{CN}^-\) ion attacks the electron-deficient carbonyl carbon (which is partially positive, \(\delta+\)), forming an intermediate alkoxide ion. This confirms statement B is correct.
- Step 2: The intermediate alkoxide ion is protonated by an \(\text{HCN}\) molecule to form the hydroxynitrile (2-hydroxypropanenitrile) and regenerate the \(\text{CN}^-\) catalyst.
- A racemic mixture is formed because the carbonyl group of ethanal is planar, allowing the \(\text{CN}^-\) nucleophile to attack with equal probability from either above or below the plane. However, the intermediate is an alkoxide ion, not a carbocation.
- Ethanal is more reactive than propanone towards nucleophiles because it has less steric hindrance and only one electron-donating methyl group reducing the positive charge on the carbonyl carbon.

Award 1 mark for selecting B. Method: Recall the nucleophilic addition mechanism of carbonyls, identify the nucleophile (CN⁻) and its attack on the carbonyl carbon in the first step, and rule out the incorrect statements.

In this atmospheric process, \(\text{NO}_2\) acts as a catalyst because it speeds up the oxidation of \(\text{SO}_2\) to \(\text{SO}_3\) and is regenerated at the end of the process (as shown by Reaction 2, where \(\text{NO}\) is re-oxidized back to \(\text{NO}_2\) by atmospheric oxygen).

Since all the reactants (\(\text{SO}_2\), \(\text{O}_2\)) and the catalyst (\(\text{NO}_2\)) are in the gas phase, they are in the same physical state. Therefore, \(\text{NO}_2\) behaves as a homogeneous catalyst.

Thus, option B is correct.

Award 1 mark for recognizing that NO₂ behaves as a homogeneous catalyst because it is in the same gaseous phase as the reactants and is regenerated during the reaction cycle (B).

First, analyze the successive ionisation energies of element \(X\).
There is a relatively steady increase from the 1st to the 3rd ionisation energy (\(578 \rightarrow 1817 \rightarrow 2745\text{ kJ\,mol}^{-1}\)).
There is a very large jump between the 3rd and the 4th ionisation energy (from \(2745\) to \(11578\text{ kJ\,mol}^{-1}\)). This indicates that the 4th electron is being removed from a inner shell, closer to the nucleus and less shielded. Therefore, \(X\) has 3 valence electrons and belongs to Group 13.

Let us evaluate the options:
- **A is incorrect**: The oxide of a Group 13 element like aluminium (\(\text{Al}_2\text{O}_3\)) is amphoteric, meaning it reacts with both acids and bases.
- **B is incorrect**: The outermost electron of a Group 13 element resides in a p-orbital (valence configuration is \(ns^2 np^1\)), not a d-orbital.
- **C is incorrect**: Aluminium does not react vigorously with cold water due to the presence of a protective, passivating oxide layer.
- **D is correct**: Since there are 3 electrons in the outer shell, \(X\) is in Group 13.

[1] Award 1 mark for identifying option D as the correct statement based on identifying the large jump between the 3rd and 4th ionisation energies.

Let us identify each element from the given chemical properties of their Period 3 oxides:
1. The oxide of \(P\) is amphoteric (reacts with both acids and bases). This is aluminium oxide, \(\text{Al}_2\text{O}_3\). Thus, \(P\) is Aluminium (\(Z = 13\)).
2. The oxide of \(Q\) dissolves in water to form a strongly alkaline solution. This is sodium oxide, \(\text{Na}_2\text{O}\), which forms \(\text{NaOH}\). Thus, \(Q\) is Sodium (\(Z = 11\)).
3. The oxide of \(R\) has a giant molecular (covalent) structure and is insoluble in water. This is silicon dioxide, \(\text{SiO}_2\). Thus, \(R\) is Silicon (\(Z = 14\)).
4. The oxide of \(S\) reacts with water to form a strong diprotic acid. This is sulfur trioxide, \(\text{SO}_3\), which forms \(\text{H}_2\text{SO}_4\). Thus, \(S\) is Sulfur (\(Z = 16\)).

Arranging them in order of increasing atomic number:
\(Q\) (11) \(< P\) (13) \(< R\) (14) \(< S\) (16).
Therefore, the correct order is \(Q \rightarrow P \rightarrow R \rightarrow S\).

[1] Award 1 mark for identifying the elements from their Period 3 oxide properties and ordering them correctly by atomic number (Option B).

A carbonyl compound contains a \(\text{C}=\text{O}\) double bond. This means we are looking for the structural isomers of aldehydes and ketones with the formula \(\text{C}_5\text{H}_{10}\text{O}\).

Let us list all possible structural isomers:

**Aldehydes (containing a \(-\text{CHO}\) group):**
1. Pentanal: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}\)
2. 2-Methylbutanal: \(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CHO}\)
3. 3-Methylbutanal: \(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CHO}\)
4. 2,2-Dimethylpropanal: \(\text{CH}_3\text{C}(\text{CH}_3)_2\text{CHO}\)

**Ketones (containing a non-terminal \(\text{C}=\text{O}\) group):**
5. Pentan-2-one: \(\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\)
6. Pentan-3-one: \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\)
7. 3-Methylbutan-2-one: \(\text{CH}_3\text{COCH}(\text{CH}_3)_2\)

There are no other aldehydes or ketones. This gives a total of 7 structural isomers.

[1] Award 1 mark for systematically determining the 4 aldehydes and 3 ketones to find the total of 7 structural isomers (Option C).

Let us evaluate the shapes of all the options:
- **\(\text{BF}_3\)**: The boron atom has 3 valence electrons and forms 3 single bonds with fluorine atoms, with zero lone pairs. According to VSEPR theory, this results in a **trigonal planar** geometry with bond angles of \(120^\circ\). (Planar)
- **\(\text{NH}_3\)**: The nitrogen atom has 5 valence electrons, forms 3 single bonds with hydrogen atoms, and has 1 lone pair. This results in a **trigonal pyramidal** geometry with bond angles of approximately \(107^\circ\). Because the nitrogen atom sits above the plane of the three hydrogen atoms, this shape is **not planar**.
- **\(\text{CO}_3^{2-}\)**: The carbon atom has 4 valence electrons (plus 2 from the negative charge), forming three bonding regions (resonance hybrid). It has zero lone pairs, giving a **trigonal planar** shape. (Planar)
- **\(\text{C}_2\text{H}_4\)**: Each carbon atom is \(\text{sp}^2\) hybridised, forming a planar structure around each double-bonded carbon. The entire molecule is coplanar. (Planar)

[1] Award 1 mark for selecting option B based on the presence of a lone pair on nitrogen causing a non-planar trigonal pyramidal shape.

1. Write the balanced chemical equation for the reaction:
$$\text{Mg}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2(\text{g})$$

2. Calculate the number of moles of \(\text{HCl}\) reacted:
$$\text{Moles of HCl} = \text{volume in dm}^3 \times \text{concentration} = 0.0500\text{ dm}^3 \times 2.00\text{ mol\,dm}^{-3} = 0.100\text{ mol}$$

3. Determine the moles of \(\text{H}_2\) produced:
Since magnesium is in excess, \(\text{HCl}\) is the limiting reactant. From the stoichiometry:
$$\text{Moles of H}_2 = \frac{1}{2} \times \text{moles of HCl} = \frac{1}{2} \times 0.100\text{ mol} = 0.0500\text{ mol}$$

4. Calculate the volume of \(\text{H}_2\) gas evolved in \(\text{cm}^3\):
$$\text{Volume of H}_2 = 0.0500\text{ mol} \times 24.0\text{ dm}^3\text{\,mol}^{-1} = 1.20\text{ dm}^3$$
$$\text{Volume in cm}^3 = 1.20 \times 1000 = 1200\text{ cm}^3$$

[1] Award 1 mark for calculating the correct volume of 1200 cm³ using the stoichiometry and gas volume conversion (Option B).

Write down the two half-equations first:

1. **Reduction half-equation**:
$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

2. **Oxidation half-equation**:
$$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$$

To balance the electron transfer, we must multiply the reduction half-equation by 2 (10 electrons total transferred) and the oxidation half-equation by 5 (10 electrons total transferred):
$$2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$$
$$5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10\text{e}^-$$

Combining the equations gives the overall ionic equation:
$$2\text{MnO}_4^- + 16\text{H}^+ + 5\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2$$

The molar ratio of \(\text{MnO}_4^-\)-ions to \(\text{C}_2\text{O}_4^{2-}\)-ions is \(2 : 5\).

[1] Award 1 mark for balancing the redox half-equations using electron transfer and finding the correct mole ratio of 2 : 5 (Option B).

Let us analyze the clues step by step:

1. **Reaction with sodium metal**: Alcohols (containing \(-\text{OH}\) groups) react with sodium metal to evolve hydrogen gas. Since there are 2 oxygen atoms in the formula \(\text{C}_4\text{H}_{10}\text{O}_2\) and the reaction occurs with excess sodium, this compound contains two alcohol groups (a diol).

2. **Oxidation with excess acidified \(\text{K}_2\text{Cr}_2\text{O}_7\)**:
- Primary alcohols are oxidised to carboxylic acids (\(-\text{COOH}\)), which react with sodium carbonate (\(\text{Na}_2\text{CO}_3\)) to form carbon dioxide gas.
- Secondary alcohols are oxidised to ketones (\(\text{C}=\text{O}\)), which do not react with sodium carbonate.
- Tertiary alcohols are resistant to oxidation and remain unchanged (alcohols do not react with sodium carbonate).

Since the final oxidised product does **not** react with sodium carbonate, there must be **no** carboxylic acid groups in the product. This means the original compound \(Y\) cannot contain any primary alcohol groups (which would contain a \(-\text{CH}_2\text{OH}\) group).

Let us evaluate the choices:
- **A**: \(\text{HOCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) contains two primary alcohol groups. It would be oxidised to a dicarboxylic acid, which reacts with sodium carbonate. (Incorrect)
- **B**: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{OH}\) contains one secondary and one primary alcohol group. The primary group oxidises to a carboxylic acid, which reacts with sodium carbonate. (Incorrect)
- **C**: \(\text{CH}_3\text{CH(OH)CH(OH)CH}_3\) contains two secondary alcohol groups. Both are oxidised to ketone groups, forming butane-2,3-dione, which does not react with sodium carbonate. (Correct)
- **D**: \(\text{(CH}_3)_2\text{C(OH)CH}_2\text{OH}\) contains one tertiary and one primary alcohol group. The primary group oxidises to a carboxylic acid, which reacts with sodium carbonate. (Incorrect)

[1] Award 1 mark for deduced elimination of options containing primary alcohol groups to conclude that butane-2,3-diol (Option C) is the correct structure.

Let us deduce the identity of \(Z\) step-by-step:
1. **Reaction with 2,4-DNPH**: The formation of an orange precipitate indicates the presence of a carbonyl group (either an aldehyde or a ketone). This rules out alcohols like but-3-en-1-ol (Option C).
2. **Tollens' reagent**: The absence of a silver mirror indicates that the compound is **not** an aldehyde. Therefore, \(Z\) must be a ketone. This rules out butanal (Option A) and 2-methylpropanal (Option D).
3. **Alkaline aqueous iodine**: The formation of a yellow precipitate (tri-iodomethane/iodoform test) indicates the presence of either a methyl ketone group (\(\text{CH}_3\text{CO-}\)) or a \(\text{CH}_3\text{CH(OH)-}\) group. Since \(Z\) is a ketone with 4 carbon atoms, it must be butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)), which contains the methyl ketone group and thus gives a positive result.

[1] Award 1 mark for analyzing the organic chemical tests to uniquely identify \(Z\) as butanone (Option B).

(a) (i) Bonding: Ionic. Structure: Giant ionic lattice.
(ii) Bonding: Covalent. Structure: Simple molecular.

(b) (i) Sodium oxide: Dissolves to form a colorless solution (or temperature increases).
Equation: \(Na_2O(s) + H_2O(l) \rightarrow 2NaOH(aq)\)
pH: 12-14.

Sulfur trioxide: Reacts violently with a hissing sound/white fumes.
Equation: \(SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\)
pH: 1-2.

(ii) Term: Amphoteric.
Reaction with acid: \(Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l)\)
Reaction with alkali: \(Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)\) (or \(Al_2O_3(s) + 2OH^-(aq) \rightarrow 2AlO_2^-(aq) + H_2O(l)\))

(c)
1. Calculate moles of \(H_3PO_4\) required:
\(n(H_3PO_4) = c \times V = 0.120\text{ mol dm}^{-3} \times 0.500\text{ dm}^3 = 0.0600\text{ mol}\)

2. Write the balanced equation for the reaction:
\(P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4\)

3. Determine the moles of \(P_4O_{10}\) needed:
From the stoichiometry, 1 mole of \(P_4O_{10}\) produces 4 moles of \(H_3PO_4\).
\(n(P_4O_{10}) = \frac{0.0600}{4} = 0.0150\text{ mol}\)

4. Calculate the molar mass of \(P_4O_{10}\):
\(M_r(P_4O_{10}) = (31.0 \times 4) + (16.0 \times 10) = 284.0\text{ g mol}^{-1}\)

5. Calculate the mass of \(P_4O_{10}\):
\(\text{Mass} = 0.0150\text{ mol} \times 284.0\text{ g mol}^{-1} = 4.26\text{ g}\)

(a) (i) [1] for ionic bonding; [1] for giant ionic structure/lattice.
(ii) [1] for covalent bonding; [1] for simple molecular structure.

(b) (i) [1] for sodium oxide: description (dissolves to colorless solution) + equation \(Na_2O + H_2O \rightarrow 2NaOH\) + pH 12-14.
[1] for sulfur trioxide: description (violent reaction / mist / hiss) + equation \(SO_3 + H_2O \rightarrow H_2SO_4\) + pH 1-2.

(ii) [1] for "amphoteric".
[1] for \(Al_2O_3 + 6H^+ \rightarrow 2Al^{3+} + 3H_2O\).
[1] for \(Al_2O_3 + 2OH^- + 3H_2O \rightarrow 2[Al(OH)_4]^-\) (accept sensible alternatives like \(Al_2O_3 + 2OH^- \rightarrow 2AlO_2^- + H_2O\)).

(c)
[1] for moles of \(H_3PO_4 = 0.0600\text{ mol}\).
[1] for moles of \(P_4O_{10} = 0.0150\text{ mol}\) (correctly applying 1:4 ratio).
[1] for \(M_r(P_4O_{10}) = 284.0\).
[1] for final mass of \(4.26\text{ g}\) (must show correct 3 significant figures).

(a) Functional group isomerism.

(b) Reagent: Tollens' reagent (or Fehling's solution / acidified potassium dichromate(VI)).
Observation with propanal: Silver mirror (or grey precipitate) is formed (with Fehling's: red precipitate; with dichromate: color changes from orange to green).
Observation with propanone: No change (remains colorless/blue/orange).

(c) (i) Reagents: \(HCN\) and \(NaCN\) (or \(NaCN\) and dilute \(H_2SO_4\)).
Conditions: Warm / heat (approx. \(10-20^\circ\text{C}\)).

(ii) Mechanism of nucleophilic addition:
1. The nucleophile \(:CN^-\) attacks the carbonyl carbon atom. A curly arrow is drawn from the lone pair of the carbon in \(:CN^-\) to the carbonyl carbon \(C\).
2. The carbonyl group is polar: \(C^{\delta+} = O^{\delta-}\). A curly arrow starts from the double bond of \(C=O\) to the oxygen atom.
3. This forms an intermediate: \(CH_3CH_2CH(O^-)(CN)\), which is tetrahedral. The oxygen atom has a negative charge and a lone pair.
4. A curly arrow is drawn from the lone pair on \(O^-\) to the hydrogen atom of \(HCN\) (or \(H^+\)), and the \(H-CN\) bond breaks (curly arrow from the bond to carbon).

(iii) Chiral centre: A carbon atom bonded to four different groups of atoms.
Explanation of optical inactivity:
The carbonyl carbon in propanal is planar. The \(CN^-\) nucleophile has an equal probability of attacking this planar group from either above or below the plane. This produces an equimolar mixture of the two enantiomers (a racemic mixture), which has no net effect on plane-polarized light.

(iv) The product is 2-hydroxybutanenitrile, \(CH_3CH_2-CH(OH)-CN\). The chiral carbon is C2, which is bonded to: \(-H\), \(-OH\), \(-CN\), and \(-CH_2CH_3\).
The drawings must show a central carbon with tetrahedral geometry (two in-plane bonds, one wedge, one dash) for both enantiomers, demonstrating they are non-superimposable mirror images of each other.

(a) [1] for "functional group isomerism" (reject "structural" as it is too general).

(b) [1] for Tollens' reagent (or Fehling's / acidified \(K_2Cr_2O_7\)).
[1] for correct positive observation with propanal (silver mirror / red ppt / green solution).
[1] for correct negative observation with propanone (no change/no reaction).

(c) (i) [1] for \(HCN\) and \(NaCN\) (or \(NaCN + H_2SO_4\)).
[1] for warm/room temperature (between \(10\) and \(20^\circ\text{C}\)).

(ii) [1] for dipole on \(C=O\) (\(C^{\delta+}\), \(O^{\delta-}\)) and curly arrow from the lone pair on carbon in \(:CN^-\) to the carbonyl carbon.
[1] for curly arrow from the \(C=O\) bond to oxygen.
[1] for correct structure of the tetrahedral intermediate with oxygen carrying a negative charge (\(CH_3CH_2C(H)(O^-)(CN)\)).
[1] for curly arrow from the lone pair of the intermediate oxygen to \(H^+\) or \(HCN\).

(iii) [1] for defining chiral centre as a carbon bonded to four different groups.
[1] for describing the carbonyl group/carbon as planar.
[1] for stating that attack by the nucleophile is equally likely from either side, resulting in a racemic mixture.

(iv) [1] for drawing tetrahedral 3D shapes around the chiral carbon (using wedges and dashes).
[1] for drawing them as non-superimposable mirror images of each other, containing correct groups: \(-H\), \(-OH\), \(-CN\), \(-CH_2CH_3\).

(a) (i) Equation: \(N_2(g) + O_2(g) \rightarrow 2NO(g)\)
Explanation: The \(N \equiv N\) triple bond is extremely strong and has a very high bond energy. A high temperature is needed to provide the high activation energy required to break this bond.

(ii) Equations:
1) \(NO(g) + \frac{1}{2}O_2(g) \rightarrow NO_2(g)\) (or \(2NO + O_2 \rightarrow 2NO_2\))
2) \(NO_2(g) + SO_2(g) \rightarrow NO(g) + SO_3(g)\)

(iii) Environmental consequence: Acid rain (or acidification of lakes / damage to marine life / corrosion of limestone buildings).
Equation: \(SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\)

(b) (i) \(K_p = \frac{p(SO_3)^2}{p(SO_2)^2 \cdot p(O_2)}\)
Units: \(\text{Pa}^{-1}\) (or \(\text{kPa}^{-1}\))

(ii) Moles of reactants and products at equilibrium:
- Initial: \(n(SO_2) = 2.00\text{ mol}\), \(n(O_2) = 1.00\text{ mol}\), \(n(SO_3) = 0\text{ mol}\)
- Moles of \(SO_2\) reacted: \(2.00 \times 0.800 = 1.60\text{ mol}\)
- Equilibrium moles:
\(n(SO_2) = 2.00 - 1.60 = 0.40\text{ mol}\)
\(n(O_2) = 1.00 - 0.80 = 0.20\text{ mol}\)
\(n(SO_3) = 1.60\text{ mol}\)
- Total equilibrium moles: \(n_{total} = 0.40 + 0.20 + 1.60 = 2.20\text{ mol}\)
- Partial pressures: \(p = \frac{n}{n_{total}} \times P_{total}\)
\(p(SO_2) = \frac{0.40}{2.20} \times 1.50 \times 10^5 = 2.73 \times 10^4\text{ Pa}\) (or \(27.3\text{ kPa}\))
\(p(O_2) = \frac{0.20}{2.20} \times 1.50 \times 10^5 = 1.36 \times 10^4\text{ Pa}\) (or \(13.6\text{ kPa}\))
\(p(SO_3) = \frac{1.60}{2.20} \times 1.50 \times 10^5 = 1.09 \times 10^5\text{ Pa}\) (or \(109\text{ kPa}\))

(iii) Calculation of \(K_p\):
\(K_p = \frac{(1.091 \times 10^5)^2}{(2.727 \times 10^4)^2 \times (1.364 \times 10^4)} = 1.17 \times 10^{-3}\text{ Pa}^{-1}\) (or \(1.17\text{ kPa}^{-1}\))

(iv) Yield of \(SO_3\): Decreases. Since the forward reaction is exothermic, according to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic direction (to the left).
Value of \(K_p\): Decreases. Since the equilibrium position shifts to the left, the ratio of product concentration to reactant concentration decreases, which lowers the value of \(K_p\).

(a) (i) [1] for balanced equation: \(N_2(g) + O_2(g) \rightarrow 2NO(g)\).
[1] for explaining that nitrogen has a strong triple bond which requires high activation energy to break.

(ii) [1] for \(2NO + O_2 \rightarrow 2NO_2\) (or \(NO + \frac{1}{2}O_2 \rightarrow NO_2\)).
[1] for \(NO_2 + SO_2 \rightarrow NO + SO_3\).

(iii) [1] for identifying acid rain (or environmental effect of acid rain).
[1] for balanced equation: \(SO_3 + H_2O \rightarrow H_2SO_4\).

(b) (i) [1] for expression: \(K_p = \frac{p(SO_3)^2}{p(SO_2)^2 \cdot p(O_2)}\) (allow square brackets only if labeled as partial pressure, e.g. \(p[SO_3]\)).
[1] for units: \(\text{Pa}^{-1}\) or \(\text{kPa}^{-1}\) (must match expression, allow \(\text{atm}^{-1}\)).

(ii) [1] for calculating equilibrium moles: \(n(SO_2) = 0.40\text{ mol}\), \(n(O_2) = 0.20\text{ mol}\), \(n(SO_3) = 1.60\text{ mol}\), total \(= 2.20\text{ mol}\).
[2] for calculating all three partial pressures correctly:
\(p(SO_2) = 2.73 \times 10^4\text{ Pa}\),
\(p(O_2) = 1.36 \times 10^4\text{ Pa}\),
\(p(SO_3) = 1.09 \times 10^5\text{ Pa}\).
(Deduct 1 mark if one calculation error is made, or if total pressure was not used correctly).

(iii) [1] for correct substitution of partial pressures from (ii) into expression.
[1] for correct numerical answer: \(1.17 \times 10^{-3}\text{ Pa}^{-1}\) (or \(1.17\text{ kPa}^{-1}\)).

(iv) [1] for yield decreases because the forward reaction is exothermic (equilibrium shifts left).
[1] for \(K_p\) decreases because \(K_p\) is only affected by temperature and shifts towards the reactants.

(a) (i) Reagent: Concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)).
Conditions: Heat (approx. \(170-180^\circ\text{C}\)).
(Or reagent: Alumina catalyst, \(Al_2O_3\); conditions: heat, pass vapor over catalyst).

(ii) The three alkenes are:
1) But-1-ene: Skeletal structure is a 4-carbon chain with a double bond between C1 and C2.
2) cis-But-2-ene (or (Z)-but-2-ene): Skeletal structure with the double bond in the middle, and both methyl groups on the same side of the double bond.
3) trans-But-2-ene (or (E)-but-2-ene): Skeletal structure with the double bond in the middle, and the methyl groups on opposite sides of the double bond.

(iii) Isomers: cis-But-2-ene and trans-but-2-ene (or (Z)-but-2-ene and (E)-but-2-ene) exhibit stereoisomerism.
Explanation:
1. There is restricted rotation about the \(C=C\) double bond due to the presence of a \(\pi\) bond.
2. Each of the carbon atoms in the double bond is bonded to two different atoms or groups (a \(-H\) atom and a \(-CH_3\) group).

(b) (i) Mechanism of addition of \(HBr\) to but-2-ene:
1. Polarization of \(H-Br\): \(H^{\delta+} - Br^{\delta-}\).
2. Curly arrow starts from the \(C=C\) double bond of but-2-ene pointing to the \(H^{\delta+}\) of \(H-Br\).
3. Curly arrow starts from the \(H-Br\) bond pointing to the \(Br\) atom, forming a bromide ion, \(Br^-\).
4. This forms a secondary carbocation intermediate: \(CH_3-CH_2-\overset{+}{C}H-CH_3\).
5. A curly arrow starts from the lone pair of the \(Br^-\) ion pointing to the carbon atom carrying the positive charge, yielding 2-bromobutane.

(ii) Major product: 2-bromobutane.
Explanation:
- The addition of \(H^+\) to but-1-ene can form either a secondary carbocation (\(CH_3CH_2\overset{+}{C}HCH_3\)) or a primary carbocation (\(CH_3CH_2CH_2\overset{+}{C}H_2\)).
- The secondary carbocation is more stable than the primary carbocation.
- This is because there are two electron-donating alkyl groups attached to the positively charged carbon atom in the secondary carbocation, which stabilizes the positive charge via the inductive effect more than the single alkyl group in the primary carbocation.

(a) (i) [1] for concentrated \(H_2SO_4\) or concentrated \(H_3PO_4\) (accept \(Al_2O_3\)).
[1] for heat (accept \(170-180^\circ\text{C}\), or high temperature for \(Al_2O_3\)).

(ii) [1] for drawing and naming but-1-ene.
[1] for drawing and naming cis-but-2-ene.
[1] for drawing and naming trans-but-2-ene.

(iii) [1] for identifying cis-but-2-ene and trans-but-2-ene.
[1] for explaining that there is restricted rotation around the \(C=C\) double bond (presence of \(\pi\) bond).
[1] for explaining that each carbon of the double bond is attached to two different groups (\(-H\) and \(-CH_3\)).

(b) (i) [1] for dipole \(H^{\delta+} - Br^{\delta-}\) and curly arrow from the double bond to \(H^{\delta+}\).
[1] for curly arrow from the \(H-Br\) bond to the bromine atom.
[1] for drawing the correct secondary carbocation intermediate with a positive charge on the carbon.
[1] for curly arrow from the lone pair on the \(Br^-\) ion to the positive carbon.

(ii) [1] for stating 2-bromobutane is the major product.
[1] for explaining that the secondary carbocation is more stable than the primary carbocation.
[1] for explaining that the stability is due to the greater inductive (electron-donating) effect of two alkyl groups compared to one.