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In \(SO_2\), the central sulfur atom has 6 valence electrons. It forms double bonds with two oxygen atoms and retains one lone pair. This gives a total of 3 electron domains, which corresponds to a trigonal planar electron domain geometry and \(sp^2\) hybridization. Because of the lone pair, the molecular geometry (shape) is bent. In contrast:
- \(OF_2\) and \(H_2S\) have 4 electron domains (2 bonding, 2 non-bonding), giving them \(sp^3\) hybridization and a bent shape.
- \(CO_2\) has 2 electron domains (2 bonding, 0 non-bonding), giving it \(sp\) hybridization and a linear shape.

Award 1 mark for the correct choice C. Reject other choices as they do not meet both criteria of being \(sp^2\) hybridized and having a bent geometry.

To determine the bond length, we look at the bond order:
1. In \(CO\) (carbon monoxide), there is a triple bond between carbon and oxygen, giving a bond order of 3. This is the shortest and strongest bond.
2. In \(CO_2\) (carbon dioxide), there are two double bonds, giving a bond order of 2.
3. In \(CO_3^{2-}\) (carbonate ion), the negative charge is delocalized over three carbon-oxygen bonds due to resonance, resulting in a bond order of \(1.33\) (\(\frac{4}{3}\)). This is the longest and weakest bond.

Therefore, the order of increasing bond length (shortest first) is I < II < III.

Award 1 mark for the correct choice A. Confirm that triple bonds are shorter than double bonds, which are shorter than resonance-stabilized partial double bonds.

Butanoic acid is a carboxylic acid with the molecular formula \(C_4H_8O_2\). Structural isomers must have the same molecular formula but different structural arrangements.
- Methyl propanoate is an ester with the formula \(CH_3CH_2COOCH_3\), which also has the molecular formula \(C_4H_8O_2\). Therefore, it is a structural isomer (specifically a functional group isomer) of butanoic acid.
- Butan-2-one is a ketone with the formula \(C_4H_8O\).
- Butan-1-ol is an alcohol with the formula \(C_4H_{10}O\).
- Diethyl ether is an ether with the formula \(C_4H_{10}O\).

Award 1 mark for the correct choice A. Identify that carboxylic acids and esters with the same number of carbons are functional group isomers.

Amines are classified as primary, secondary, or tertiary based on the number of alkyl (or aryl) groups directly attached to the nitrogen atom:
- \((CH_3)_3CNH_2\) has only one carbon-nitrogen bond (the nitrogen is attached to a tertiary butyl group). This is a primary amine.
- \((CH_3)_2CHNHCH_3\) has two carbon-nitrogen bonds (the nitrogen is bonded to an isopropyl group and a methyl group). This is a secondary amine.
- \((CH_3)_3N\) (trimethylamine) has three carbon-nitrogen bonds (the nitrogen is bonded to three methyl groups). This is a tertiary amine.
- \((CH_3CH_2)_2NH\) (diethylamine) has two carbon-nitrogen bonds. This is a secondary amine.

Award 1 mark for the correct choice C. Distinguish between primary, secondary, and tertiary amine classifications by focusing on the substituents on the nitrogen atom.

Let us calculate the oxidation state of sulfur (S) in each species:
- In \(S_2O_3^{2-}\) (thiosulfate): \(2(\text{S}) + 3(-2) = -2 \Rightarrow 2(\text{S}) - 6 = -2 \Rightarrow 2(\text{S}) = +4 \Rightarrow \text{S} = +2\).
- In \(SO_3^{2-}\) (sulfite): \(\text{S} + 3(-2) = -2 \Rightarrow \text{S} - 6 = -2 \Rightarrow \text{S} = +4\).
- In \(SF_6\) (sulfur hexafluoride): Fluorine is more electronegative and always has an oxidation state of \(-1\) in compounds. Therefore, \(\text{S} + 6(-1) = 0 \Rightarrow \text{S} = +6\).
- In \(S_8\) (octasulfur): Since it is an element in its standard state, the oxidation state of sulfur is \(0\).

Therefore, the highest oxidation state of sulfur is \(+6\) in \(SF_6\).

Award 1 mark for the correct choice C. Correctly assign oxidation states for sulfur in all four options.

In a voltaic cell, the half-cell with the more positive standard reduction potential (\(E^\theta\)) undergoes reduction. Since \(E^\theta(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}\) is more positive than \(E^\theta(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}\), copper ions are reduced to copper metal.
- Reduction occurs at the cathode, so the copper electrode is the cathode: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\).
- Oxidation occurs at the anode, so the zinc electrode is the anode: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\).
- Electrons always flow through the external circuit from the anode (oxidation) to the cathode (reduction). Therefore, electrons flow from the zinc electrode to the copper electrode.

Thus, copper is the cathode and electrons flow from zinc to copper.

Award 1 mark for the correct choice B. Correctly identify the cathode and anode based on \(E^\theta\) values, and direct electron flow from anode to cathode.

According to the Brønsted-Lowry theory, a conjugate acid-base pair consists of two species that differ by exactly one hydrogen ion (\(H^+\)).
- \(H_2PO_4^-\) acts as an acid (donates a proton) to become its conjugate base, \(HPO_4^{2-}\). These two species differ by exactly one proton, so they constitute a conjugate acid-base pair.
- None of the other options consist of species that differ by a single proton within the same acid-base system.

Award 1 mark for the correct choice A. Apply Brønsted-Lowry definitions to identify a conjugate acid-base pair differing by exactly one proton.

When the temperature of a gas mixture is increased (Vessel Y):
1. The average kinetic energy of the molecules increases, which causes the Maxwell–Boltzmann distribution curve to flatten and spread to the right. Consequently, the peak of the curve (representing the most probable energy) is lower and shifted to a higher kinetic energy value. Therefore, Vessel Y has a lower peak (maximum) on its curve than Vessel X.
2. At a higher temperature, the frequency of total collisions increases, and a significantly higher fraction of molecules possess kinetic energy equal to or greater than the activation energy (\(E \ge E_a\)). This results in a much higher frequency of successful (fruitful) collisions.

Thus, Vessel Y has a lower peak (maximum) on its curve and a higher frequency of successful collisions.

Award 1 mark for the correct choice C. Understand the effect of temperature on the shape of the Maxwell–Boltzmann distribution curve (lower peak at higher T). Understand the relationship between temperature and successful collision frequency.

To determine the molecular geometry and bond angles, we look at the number of electron domains around the central atom. In \(\text{SO}_2\), the central sulfur atom has 6 valence electrons, forming two bonds and carrying one lone pair, giving 3 electron domains (trigonal planar parent geometry). The molecular geometry is bent, and the presence of the lone pair reduces the bond angle slightly below \(120^\circ\) (to approximately \(119^\circ\)). In \(\text{NO}_2^-\), the central nitrogen atom also has 3 electron domains, giving a bent molecular geometry with a bond angle of approximately \(115^\circ\). In \(\text{OF}_2\), the central oxygen atom has 4 electron domains (tetrahedral parent geometry), giving a bent geometry with a bond angle of approximately \(103^\circ\). Therefore, only I and II have bond angles close to \(120^\circ\).

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

In methylpropene, there are two \(\text{C}-\text{C}\) single bonds (each is 1 \(\sigma\) bond) and one \(\text{C}=\text{C}\) double bond (consisting of 1 \(\sigma\) and 1 \(\pi\) bond). There are also eight \(\text{C}-\text{H}\) single bonds (each is 1 \(\sigma\) bond). Summing these up gives: Total \(\sigma\) bonds = 2 + 1 + 8 = 11. Total \(\pi\) bonds = 1.

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

An alcohol is classified as tertiary (3°) if the carbon atom bonded to the hydroxyl (\(-\text{OH}\)) group is itself directly bonded to three other carbon atoms. In 2-methylbutan-2-ol, the structure is \(\text{CH}_3-\text{C}(\text{OH})(\text{CH}_3)-\text{CH}_2-\text{CH}_3\), where the carbon with the hydroxyl group is bonded to three other carbons (two methyls and one ethyl). In 3-methylbutan-2-ol, the carbon with the hydroxyl group is bonded to two other carbons (secondary). In 2,2-dimethylpropan-1-ol, it is bonded to only one other carbon (primary). In pentan-3-ol, it is bonded to two other carbons (secondary).

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

Boiling points depend on the strength of intermolecular forces. Butan-1-ol, butan-2-ol, and 2-methylpropan-2-ol are isomeric alcohols containing highly polar \(-\text{O}-\text{H}\) bonds, enabling them to form strong intermolecular hydrogen bonds. Ethoxyethane is an ether; since it does not have a hydrogen atom directly bonded to an oxygen atom, it cannot form intermolecular hydrogen bonds with itself. It only experiences dipole-dipole and London dispersion forces, which are much weaker than hydrogen bonding. Thus, ethoxyethane has the lowest boiling point.

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

Let's analyze the oxidation states. In \(\text{MnO}_4^-\), the oxidation state of Mn is +7, and in \(\text{Mn}^{2+}\) it is +2. Since the oxidation state of Mn decreases, it is reduced, making \(\text{MnO}_4^-\) the oxidizing agent. In \(\text{H}_2\text{C}_2\text{O}_4\), the oxidation state of carbon is +3 (since \(2(+1) + 2(y) + 4(-2) = 0 \Rightarrow y = +3\)). In \(\text{CO}_2\), carbon has an oxidation state of +4. Since the oxidation state of carbon increases from +3 to +4, it is oxidized, making \(\text{H}_2\text{C}_2\text{O}_4\) the reducing agent. Hydrogen does not change its oxidation state of +1.

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

Because zinc has a more negative standard reduction potential (\(-0.76\text{ V}\)) than copper (\(+0.34\text{ V}\)), zinc is oxidized at the anode: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-\). Therefore, the mass of the zinc electrode decreases. Copper ions are reduced at the cathode: \(\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu}(\text{s})\), depositing solid copper and increasing the cathode mass. Electrons flow from the anode (Zn) to the cathode (Cu) through the external circuit. Anions in the salt bridge migrate towards the anode (zinc half-cell) to neutralize excess positive charge. The standard cell potential is \(+0.34 - (-0.76) = +1.10\text{ V}\).

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

A Brønsted-Lowry conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). In this reaction, \(\text{H}_2\text{PO}_4^-\) acts as an acid by donating a proton to form its conjugate base, \(\text{HPO}_4^{2-}\). These two species differ by exactly one proton, so they constitute a conjugate pair.

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

A catalyst provides an alternative pathway with a lower activation energy (\(E_a\)). According to the Arrhenius equation, \(k = A e^{-E_a/RT}\), a lower activation energy increases the rate constant, \(k\), resulting in a faster rate. A catalyst does not affect the average kinetic energy of the particles (which is determined solely by temperature), does not shift the position of the equilibrium (it speeds up both the forward and reverse reactions equally), and does not change the enthalpy change (\(\Delta H\)) of the reaction.

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

\(\text{H}_3\text{O}^+\) has 3 bonding pairs of electrons and 1 lone pair of electrons around the central oxygen atom (trigonal pyramidal molecular geometry). The repulsion from the lone pair reduces the ideal tetrahedral bond angle from \(109.5^\circ\) to approximately \(107^\circ\). \(\text{CO}_2\) is linear (\(180^\circ\)), \(\text{BF}_3\) is trigonal planar (\(120^\circ\)), and \(\text{NH}_4^+\) is tetrahedral (\(109.5^\circ\)).

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

By expanding the structure of methyl propenoate to \(\text{CH}_2=\text{CH}-\text{C}(=\text{O})-\text{O}-\text{CH}_3\), we count all individual single bonds (each is 1 \(\sigma\) bond) and double bonds (each has 1 \(\sigma\) and 1 \(\pi\) bond). There are 6 C-H bonds, 1 C-C bond, 1 C-O bond, and 1 O-C bond, which are all single bonds (9 \(\sigma\) bonds). Additionally, there is 1 C=C double bond and 1 C=O double bond, contributing 2 \(\sigma\) and 2 \(\pi\) bonds. The total is 11 \(\sigma\) and 2 \(\pi\) bonds.

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

The structural formula of 2-methylbutan-2-ol is \(\text{CH}_3\text{C(OH)(CH}_3)\text{CH}_2\text{CH}_3\). The carbon atom carrying the hydroxyl (\(-\text{OH}\)) group is directly bonded to three other carbon atoms (two methyl groups and one ethyl group). Therefore, it is classified as a tertiary carbon atom, and the compound is a tertiary alcohol.

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

The longest continuous carbon chain containing the hydroxyl group has 5 carbon atoms (pentan-). To give the hydroxyl group the lowest locant, we number the chain. Since the hydroxyl group is on carbon-3 from both ends, we choose the direction of numbering that gives the methyl substituent the lowest locant. Numbering from left-to-right gives the methyl group position 2, and the hydroxyl group position 3, leading to the name 2-methylpentan-3-ol.

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

Let \(x\) be the oxidation state of molybdenum. Oxygen has a standard oxidation state of \(-2\). The sum of the oxidation states must equal the overall charge of the ion: \(7(x) + 24(-2) = -6\), which simplifies to \(7x - 48 = -6\). Solving for \(x\) gives \(7x = 42\), which means \(x = +6\).

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

Zinc has the more negative standard reduction potential, meaning it is more easily oxidized and acts as the anode: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\). The \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell has the more positive potential, meaning reduction occurs at the cathode: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\), which utilizes an inert platinum (Pt) electrode. Electrons flow in the external circuit from the anode where they are released (the zinc electrode) to the cathode where they are accepted (the platinum electrode). The standard cell potential is \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.77\text{ V} - (-0.76\text{ V}) = +1.53\text{ V}\).

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

A conjugate acid-base pair consists of two chemical species that differ by exactly one proton (\(\text{H}^+\)). In this reaction, \(\text{H}_2\text{PO}_4^-\), acting as a Brønsted-Lowry acid, donates a proton to form its conjugate base \(\text{HPO}_4^{2-}\). Thus, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) form a conjugate acid-base pair.

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

When the concentration of reactant A is doubled while the concentration of B is kept constant, the rate of the reaction doubles. This indicates that the reaction is first-order with respect to A (order = 1). When the concentration of reactant B is doubled while the concentration of A is kept constant, the rate of the reaction increases by a factor of 4 (\(2^2\)). This indicates that the reaction is second-order with respect to B (order = 2). Combining these gives the rate equation: \(\text{rate} = k[\text{A}][\text{B}]^2\).

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

To determine the bond angles, we analyze the Lewis structures and the number of electron domains around the central atom: 1. For \(\text{NO}_2^+\), the nitrogen atom has 2 bonding domains and no lone pairs, leading to a linear molecular geometry with a bond angle of \(180^\circ\). 2. For \(\text{NO}_2^-\), the nitrogen atom has 3 electron domains (one single bond, one double bond, and one lone pair). This results in a trigonal planar electron domain geometry and a bent molecular geometry with a bond angle of slightly less than \(120^\circ\) (approximately \(115^\circ\)). 3. For \(\text{NH}_4^+\), the nitrogen atom has 4 bonding domains and no lone pairs, leading to a tetrahedral molecular geometry with a bond angle of \(109.5^\circ\). 4. For \(\text{H}_3\text{O}^+\), the oxygen atom has 4 electron domains (3 bonding domains and 1 lone pair), leading to a trigonal pyramidal molecular geometry with a bond angle of slightly less than \(109.5^\circ\) (approximately \(107^\circ\)). Therefore, \(\text{NO}_2^-\), with its bent geometry derived from a trigonal planar arrangement, has a bond angle closest to \(120^\circ\).

- [1 mark] Award for selecting option B. - [0 marks] For any other selection.

Let's draw out the Lewis structure of methyl propenoate to identify all single and multiple bonds: The structural formula is \(\text{H}_2\text{C}=\text{CH}-\text{C}(=\text{O})-\text{O}-\text{CH}_3\). Every single covalent bond is a \(\sigma\) bond. Each double bond consists of one \(\sigma\) bond and one \(\pi\) bond. Let's count them: - Six C-H single bonds: 6 \(\sigma\) bonds. - One C=C double bond: 1 \(\sigma\) bond and 1 \(\pi\) bond. - One C-C single bond: 1 \(\sigma\) bond. - One C=O double bond: 1 \(\sigma\) bond and 1 \(\pi\) bond. - Two C-O single bonds: 2 \(\sigma\) bonds. Summing these up: Total \(\sigma\) bonds = \(6 + 1 + 1 + 1 + 2 = 11\). Total \(\pi\) bonds = \(1 + 1 = 2\). Therefore, there are 11 \(\sigma\) bonds and 2 \(\pi\) bonds.

- [1 mark] Award for selecting option B. - [0 marks] For any other selection.

An alcohol is classified as primary (\(1^\circ\)), secondary (\(2^\circ\)), or tertiary (\(3^\circ\)) depending on the number of carbon atoms directly bonded to the carbon carrying the hydroxyl (\(-\text{OH}\)) group: - In 2-methylbutan-2-ol, the carbon carrying the \(-\text{OH}\) group (carbon-2) is bonded to carbon-1 (\(-\text{CH}_3\)), carbon-3 (\(-\text{CH}_2-\)), and a methyl branch (\(-\text{CH}_3\)). Since this carbon is bonded to three other carbons, it is a tertiary alcohol. - In 3-methylbutan-2-ol, the carbon carrying the \(-\text{OH}\) group (carbon-2) is bonded to carbon-1 (\(-\text{CH}_3\)) and carbon-3 (\(-\text{CH}-\)). Since it is bonded to two other carbons, it is a secondary alcohol. - In 2-methylbutan-1-ol, the carbon carrying the \(-\text{OH}\) group (carbon-1) is bonded to carbon-2. Since it is bonded to only one other carbon, it is a primary alcohol. - In 2,2-dimethylpropan-1-ol, the carbon carrying the \(-\text{OH}\) group (carbon-1) is bonded to carbon-2 (which is quaternary). However, since carbon-1 is directly bonded to only one carbon atom, it is still classified as a primary alcohol.

- [1 mark] Award for selecting option A. - [0 marks] For any other selection.

To name the compound using IUPAC rules: 1. Identify the principal functional group: The compound contains an aldehyde group (\(-\text{CHO}\)), which takes highest priority in this molecule. The aldehyde carbon is designated as Carbon-1 (\(\text{C}-1\)). 2. Find the longest continuous carbon chain containing the principal functional group: The chain contains 5 carbons, starting from the aldehyde carbon: \(\text{C}-1\): \(-\text{CHO}\), \(\text{C}-2\): \(-\text{CH}(\text{CH}_3)-\), \(\text{C}-3\): \(-\text{CH}_2-\), \(\text{C}-4\): \(-\text{CH}(\text{Cl})-\), \(\text{C}-5\): \(-\text{CH}_3\). Therefore, the parent name is pentanal. 3. Identify and number the substituents: A methyl group is attached to \(\text{C}-2\). A chlorine atom (chloro) is attached to \(\text{C}-4\). 4. Assemble the name alphabetically: 'chloro' comes before 'methyl' alphabetically. Thus, the correct IUPAC name is 4-chloro-2-methylpentanal.

- [1 mark] Award for selecting option B. - [0 marks] For any other selection.

Let's analyze the oxidation states of chlorine in the reactants and products for each reaction: - Option A: Chlorine changes from \(\text{Cl}_2\) (oxidation state 0) to \(\text{Cl}^-\). The decrease is 1. - Option B: Chlorine changes from \(\text{ClO}_3^-\). In \(\text{ClO}_3^-\): \(x + 3(-2) = -1 \implies x = +5\). It changes to \(\text{Cl}^-\) (oxidation state -1). The decrease is from +5 to -1, which is a decrease of 6. - Option C: Chlorine in \(\text{ClO}_3^-\) changes to \(\text{Cl}_2\). In \(\text{ClO}_3^-\), the oxidation state is +5. In \(\text{Cl}_2\), the oxidation state is 0. The decrease is from +5 to 0, which is a decrease of exactly 5. - Option D: Chlorine changes from \(\text{ClO}_4^-\) (oxidation state +7) to \(\text{Cl}^-\) (oxidation state -1). The decrease is 8. Therefore, the correct reaction is Option C.

- [1 mark] Award for selecting option C. - [0 marks] For any other selection.

Let's evaluate each option: 1. Determine anode and cathode: The zinc half-cell has a more negative standard electrode potential (\(-0.76\ \text{V}\)) than the silver half-cell (\(+0.80\ \text{V}\)). Thus, zinc is more easily oxidized and acts as the anode: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\). Silver ions are more easily reduced and the silver electrode acts as the cathode: \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s}\)). 2. Electron flow: Electrons flow through the external wire from the anode (zinc) to the cathode (silver). Ions, not electrons, flow through the salt bridge. Thus, option A is incorrect and option B is incorrect. 3. Changes in concentration and mass: At the cathode, \(\text{Ag}^+(\text{aq})\) ions are reduced to solid silver, which deposits on the silver electrode. Therefore, the concentration of \(\text{Ag}^+(\text{aq})\) decreases (Option C is correct). At the anode, solid zinc is oxidized to \(\text{Zn}^{2+}(\text{aq})\), meaning the mass of the zinc electrode decreases. Thus, option D is incorrect.

- [1 mark] Award for selecting option C. - [0 marks] For any other selection.

According to the Brønsted-Lowry theory, a conjugate acid-base pair consists of two species that differ from each other by a single proton (\(\text{H}^+\)). Let's trace the proton transfer in the reaction: - \(\text{H}_2\text{PO}_4^-\)(aq) donates a proton to become \(\text{HPO}_4^{2-}\)(aq). Therefore, \(\text{H}_2\text{PO}_4^-\)(aq) is a Brønsted-Lowry acid and \(\text{HPO}_4^{2-}\)(aq) is its conjugate base. This constitutes a conjugate acid-base pair (differing by exactly one proton). - \(\text{HCO}_3^-\)(aq) accepts a proton to become \(\text{H}_2\text{CO}_3\)(aq). Thus, \(\text{HCO}_3^-\)(aq) is a Brønsted-Lowry base and \(\text{H}_2\text{CO}_3\)(aq) is its conjugate acid. Looking at the options, only Option C correctly identifies a pair that differs by a single proton and represents a conjugate acid-base relationship.

- [1 mark] Award for selecting option C. - [0 marks] For any other selection.

Let's analyze the physical and chemical effects of adding a catalyst: - Option A is incorrect: The average kinetic energy of the reactant molecules depends only on the temperature. Since the temperature is constant, the average kinetic energy does not change. - Option B is incorrect: The Maxwell-Boltzmann distribution curve represents the distribution of kinetic energies of molecules at a specific temperature. Because the temperature does not change, the shape and position of the curve remain entirely unchanged. - Option C is correct: A catalyst works by providing an alternative reaction pathway with a lower activation energy (\(E_a\)). This decreases the activation energy for both the forward and the reverse reactions by the same amount. - Option D is incorrect: The enthalpy change (\(\Delta H\)) of the reaction is the difference in potential energy between the products and the reactants. Since a catalyst does not alter the energy levels of the reactants or products, \(\Delta H\) remains unchanged.

- [1 mark] Award for selecting option C. - [0 marks] For any other selection.

To determine the molecular geometry, we look at the number of bonding domains and lone pairs on the central atom using VSEPR theory:

* \(BF_3\) has 3 bonding domains and 0 lone pairs on the boron atom, giving a trigonal planar geometry.
* \(PCl_3\) has 3 bonding domains and 1 lone pair on the phosphorus atom, giving a trigonal pyramidal geometry.
* \(SO_3\) has 3 bonding domains and 0 lone pairs on the sulfur atom, giving a trigonal planar geometry.
* \(ClF_3\) has 3 bonding domains and 2 lone pairs on the chlorine atom, giving a T-shaped geometry.

Therefore, only \(PCl_3\) has a trigonal pyramidal geometry.

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

To determine the bond length, we compare the nitrogen-nitrogen bond orders in each species:

* In \(N_2H_4\), there is a single bond between the nitrogen atoms (bond order = 1).
* In \(N_2H_2\), there is a double bond between the nitrogen atoms (bond order = 2).
* In \(N_2\), there is a triple bond between the nitrogen atoms (bond order = 3).
* In \(N_2F_4\), there is a single bond between the nitrogen atoms (bond order = 1).

Since bond length decreases as bond order increases, the triple bond in \(N_2\) is the shortest.

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

A tertiary alcohol contains a hydroxyl group (\(-OH\)) attached to a carbon atom that is directly bonded to three other carbon atoms:

* Butan-2-ol is a secondary alcohol, as the carbon bonded to the \(-OH\) group is attached to two other carbon atoms.
* 2-Methylpropan-2-ol is a tertiary alcohol, as the central carbon bonded to the \(-OH\) group is attached to three methyl groups: \((CH_3)_3C-OH\).
* 2-Methylpropan-1-ol is a primary alcohol, as the carbon bonded to the \(-OH\) group is attached to only one other carbon atom.
* 3-Methylbutan-2-ol is a secondary alcohol, as the carbon bonded to the \(-OH\) group is attached to two other carbon atoms.

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

The possible structural isomers for the molecular formula \(C_4H_9Cl\) are:

1. 1-chlorobutane: \(CH_3CH_2CH_2CH_2Cl\)
2. 2-chlorobutane: \(CH_3CH_2CH(Cl)CH_3\)
3. 1-chloro-2-methylpropane: \((CH_3)_2CHCH_2Cl\)
4. 2-chloro-2-methylpropane: \((CH_3)_3CCl\)

There are exactly 4 structural isomers.

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

To balance the redox reaction in acidic solution, split it into half-reactions:

1. **Reduction half-reaction:**
* \(\text{MnO}_4^- \rightarrow \text{Mn}^{2+}\)
* Balance O with water: \(\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
* Balance H with \(H^+\): \(\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
* Balance charge with electrons: \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\) (Eq 1)

2. **Oxidation half-reaction:**
* \(\text{H}_2\text{SO}_3 \rightarrow \text{HSO}_4^-\)
* Balance O with water: \(\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^-\)
* Balance H with \(H^+\): \(\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+\)
* Balance charge with electrons: \(\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^-\) (Eq 2)

3. **Equalize and combine electrons (lowest common multiple of 5 and 2 is 10):**
* Multiply Eq 1 by 2:
\(2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}\)
* Multiply Eq 2 by 5:
\(5\text{H}_2\text{SO}_3 + 5\text{H}_2\text{O} \rightarrow 5\text{HSO}_4^- + 15\text{H}^+ + 10e^-\)
* Combine the half-reactions:
\(2\text{MnO}_4^- + 16\text{H}^+ + 5\text{H}_2\text{SO}_3 + 5\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{HSO}_4^- + 15\text{H}^+\)
* Simplify \(H^+\) and \(H_2O\):
\(2\text{MnO}_4^-(aq) + 5\text{H}_2\text{SO}_3(aq) + \text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 5\text{HSO}_4^-(aq) + 3\text{H}_2\text{O}(l)\)

Thus, the coefficient of \(H^+\) is 1.

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

Let's analyze each option:

* Iron has a more negative standard reduction potential (\(-0.44\text{ V}\)) than copper (\(+0.34\text{ V}\)), meaning iron undergoes oxidation (acts as the anode) and copper undergoes reduction (acts as the cathode). This rules out option B.
* The overall cell reaction is \(\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s)\). This rules out option C.
* Electrons flow through the external wire from anode (iron) to cathode (copper). Ions (not electrons) flow through the salt bridge to maintain electrical neutrality. This rules out option A.
* The standard cell potential is:
$$E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.34 - (-0.44) = +0.78\text{ V}$$
This confirms that option D is correct.

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

A conjugate acid-base pair consists of two chemical species that differ by exactly one proton (\(H^+\)):

* \(\text{HPO}_4^{2-}\) donates a proton to form \(\text{PO}_4^{3-}\). Therefore, \(\text{HPO}_4^{2-}\) is the acid and \(\text{PO}_4^{3-}\) is its conjugate base, making them a conjugate acid-base pair.
* \(\text{HCO}_3^-\) accepts a proton to form \(\text{H}_2\text{CO}_3\). Therefore, \(\text{H}_2\text{CO}_3\) is the conjugate acid of \(\text{HCO}_3^-\).

Among the options given, only \(\text{HPO}_4^{2-}\) and \(\text{PO}_4^{3-}\) represent a valid conjugate acid-base pair.

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

Let the initial rate be \(R_1 = k[A]^2[B]\).

If the concentration of A is tripled, the new concentration is \(3[A]\).
If the concentration of B is halved, the new concentration is \(0.5[B]\).

Substitute these changes into the rate law to find the new rate \(R_2\):
$$R_2 = k(3[A])^2(0.5[B])$$
$$R_2 = k(9[A]^2)(0.5[B])$$
$$R_2 = 4.5 k[A]^2[B]$$
$$R_2 = 4.5 R_1$$

Therefore, the rate of reaction increases by a factor of 4.5.

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

(a) An increase in temperature increases the average kinetic energy of the reacting particles. Therefore, a larger fraction of reactant particles have energy greater than or equal to the activation energy (\(E \ge E_a\)). Additionally, the particles move faster, which increases the frequency of collisions. Both effects result in a higher frequency of successful collisions, thus increasing the reaction rate.

(b) Compare Run 1 and Run 2: \([\text{S}_2\text{O}_8^{2-}]\) is kept constant, while \([\text{I}^-]\) is tripled (from 0.020 to 0.060 \(\text{mol dm}^{-3}\)). The rate also triples (from \(1.2 \times 10^{-4}\) to \(3.6 \times 10^{-4}\ \text{mol dm}^{-3}\text{ s}^{-1}\)). This indicates that the reaction is first-order with respect to \([\text{I}^-]\).
Compare Run 1 and Run 3: \([\text{I}^-]\) is kept constant, while \([\text{S}_2\text{O}_8^{2-}]\) is tripled (from 0.050 to 0.150 \(\text{mol dm}^{-3}\)). The rate also triples (from \(1.2 \times 10^{-4}\) to \(3.6 \times 10^{-4}\ \text{mol dm}^{-3}\text{ s}^{-1}\)). This indicates that the reaction is first-order with respect to \([\text{S}_2\text{O}_8^{2-}]\).
Overall Rate Expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

(c) Using the data from Run 1:
\(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.2 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{(0.050\text{ mol dm}^{-3})(0.020\text{ mol dm}^{-3})} = 0.12\)
Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(d) (i) Yes, the mechanism is consistent. The slow step is the rate-determining step, so the rate law is derived from the stoichiometry of the reactants in this step. Since Step 1 involves one mole of \(\text{S}_2\text{O}_8^{2-}\) and one mole of \(\text{I}^-\), the expected rate expression from this mechanism is \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\), which matches the experimental rate expression.
(ii) The intermediate is \(\text{SO}_4\text{I}^-\).

(e) At the higher temperature \(T_2\), the peak of the Maxwell-Boltzmann distribution curve is lower and shifted to the right (higher energy) compared to \(T_1\). The activation energy \(E_a\) remains constant, but the area under the curve to the right of \(E_a\) is significantly larger at \(T_2\), representing a greater proportion of molecules with sufficient energy to react.

(a) Max 3 marks:
- Award 1 mark for stating that increasing temperature increases average kinetic energy.
- Award 1 mark for stating that a greater fraction of collisions have energy \(E \ge E_a\).
- Award 1 mark for stating that the collision frequency increases.

(b) Max 4 marks:
- Award 1 mark for comparing Run 1 and Run 2 to show first-order dependence on \(\text{I}^-\).
- Award 1 mark for comparing Run 1 and Run 3 to show first-order dependence on \(\text{S}_2\text{O}_8^{2-}\).
- Award 1 mark for clear mathematical justification (e.g., rate triples when concentration triples).
- Award 1 mark for the correct rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).

(c) Max 3 marks:
- Award 1 mark for correct setup/substitution: \(1.2 \times 10^{-4} / (0.050 \times 0.020)\).
- Award 1 mark for the correct numerical value: 0.12 (accept 0.120).
- Award 1 mark for correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(\text{M}^{-1}\text{ s}^{-1}\)).

(d) Max 3 marks:
- (i) Award 1 mark for identifying Step 1 as the rate-determining step.
- Award 1 mark for explaining that the reactants in Step 1 match the species and powers in the experimental rate expression.
- (ii) Award 1 mark for identifying \(\text{SO}_4\text{I}^-\).

(e) Max 2 marks:
- Award 1 mark for describing that the curve for higher temperature \(T_2\) has a lower peak shifted to the right.
- Award 1 mark for stating that there is a larger area under the curve to the right of \(E_a\).

(a) (i) The total valence electrons in \(\text{NO}_2^-\): \(5 \text{ (N)} + 2 \times 6 \text{ (O)} + 1 \text{ (charge)} = 18\) electrons (9 pairs).
Structure 1: A central Nitrogen double-bonded to one Oxygen (with two lone pairs) and single-bonded to another Oxygen (with three lone pairs). Nitrogen has one lone pair.
Structure 2: The mirror image, where the positions of the single and double bonds are reversed.

(ii) Formal charge formula: \(\text{FC} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{1}{2}(\text{Bonding electrons})\).
- Double-bonded Oxygen: \(6 - 4 - 2 = 0\)
- Nitrogen: \(5 - 2 - 3 = 0\)
- Single-bonded Oxygen: \(6 - 6 - 1 = -1\)

(b) Nitrogen has three electron domains: a single bond, a double bond, and one lone pair. According to VSEPR theory, three electron domains adopt a trigonal planar electron-domain geometry. The presence of one lone pair results in a bent (or V-shaped) molecular geometry. The bond angle is expected to be less than \(120^\circ\) (specifically around \(115^\circ\)) because the lone pair exerts greater repulsion than the bonding pairs.

(c) Resonance occurs when more than one valid Lewis structure can be drawn for a molecule or ion. In \(\text{NO}_2^-\), the actual structure is a hybrid of the two drawn resonance contributors. Instead of one discrete single bond and one double bond, the electrons are delocalized. Thus, both nitrogen-oxygen bonds are of equal length and strength, intermediate between a single and a double bond (with a bond order of 1.5).

(d) (i) In \(\text{CO}_2\), carbon has two electron domains, giving it a linear molecular geometry. The polar \(\text{C}=\text{O}\) bonds are equal and opposite, so their bond dipoles cancel each other, resulting in a non-polar molecule. In \(\text{SO}_2\), sulfur has three electron domains (two bonding domains and one lone pair), resulting in a bent geometry. The polar \(\text{S}=\text{O}\) bond dipoles do not cancel, giving a net molecular dipole moment, making it polar.

(ii) In liquid \(\text{CO}_2\), only weak London (dispersion) forces are present because it is non-polar. In liquid \(\text{SO}_2\), stronger dipole-dipole forces are present in addition to London (dispersion) forces due to its molecular polarity.

(a) Max 4 marks:
- (i) Award 1 mark for each correct Lewis structure including all non-bonding and bonding valence electrons (total 18 electrons) and brackets with a negative charge [2].
- (ii) Award 1 mark for showing correct calculation/formal charge of 0 on double-bonded O and central N [1].
- Award 1 mark for showing formal charge of -1 on single-bonded O [1].

(b) Max 3 marks:
- Award 1 mark for identifying 3 electron domains around the nitrogen atom.
- Award 1 mark for stating that the molecular geometry is bent / V-shaped.
- Award 1 mark for predicting the bond angle is less than \(120^\circ\) with correct justification (lone pair repulsion).

(c) Max 3 marks:
- Award 1 mark for explaining that resonance involves electron delocalization / hybrid of structures.
- Award 1 mark for stating that both N-O bonds are identical.
- Award 1 mark for stating that the bond lengths are intermediate between single and double bonds (or bond order of 1.5).

(d) Max 5 marks:
- (i) Award 1 mark for stating \(\text{CO}_2\) is linear and \(\text{SO}_2\) is bent.
- Award 1 mark for stating bond dipoles in \(\text{CO}_2\) cancel each other.
- Award 1 mark for stating bond dipoles in \(\text{SO}_2\) do not cancel because of its asymmetric shape.
- (ii) Award 1 mark for stating liquid \(\text{CO}_2\) has only London dispersion forces.
- Award 1 mark for stating liquid \(\text{SO}_2\) has dipole-dipole attractions (and London forces).

(a) (i) Carboxylic acid isomer: Butanoic acid (or 2-methylpropanoic acid).
Full structural formula of butanoic acid: \(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C}(=\text{O})-\text{OH}\) with all bonds shown explicitly.
IUPAC name: Butanoic acid.

(ii) Ester isomer: e.g., Ethyl ethanoate.
Full structural formula of ethyl ethanoate: \(\text{CH}_3-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_3\) with all bonds shown explicitly.
IUPAC name: Ethyl ethanoate (accept methyl propanoate, propyl methanoate, etc.).

(b) The carboxylic acid contains a highly polar \(\text{O}-\text{H}\) bond, allowing molecules to form strong intermolecular hydrogen bonds with each other. In contrast, the ester does not contain any hydrogen atoms directly bonded to an electronegative oxygen atom. Therefore, the ester cannot form intermolecular hydrogen bonds. It is held together only by weaker dipole-dipole and London (dispersion) forces, which require less thermal energy to overcome, resulting in a lower boiling point.

(c) (i) Amine (accept primary amine).
(ii) Ketone / carbonyl.
(iii) Amide / carboxamide.
(iv) Alcohol / hydroxyl (accept secondary alcohol).

(d) (i)
- Primary alcohol: The carbon atom bonded to the hydroxyl (\(\text{-OH}\)) group is attached to only one other carbon atom (or no carbon atoms in the case of methanol).
- Secondary alcohol: The carbon atom bonded to the hydroxyl (\(\text{-OH}\)) group is attached directly to two other carbon atoms.
- Tertiary alcohol: The carbon atom bonded to the hydroxyl (\(\text{-OH}\)) group is attached directly to three other carbon atoms.

(ii) The only tertiary alcohol with molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) is 2-methylpropan-2-ol:
\((\text{CH}_3)_3\text{C}-\text{OH}\) or showing all bonds explicitly with a central carbon bonded to three methyl groups and one hydroxyl group.

(a) Max 4 marks:
- (i) Award 1 mark for correct full structural formula of butanoic acid or 2-methylpropanoic acid (all bonds shown).
- Award 1 mark for correct corresponding IUPAC name.
- (ii) Award 1 mark for correct full structural formula of any C4 ester (e.g. ethyl ethanoate, methyl propanoate).
- Award 1 mark for correct corresponding IUPAC name.

(b) Max 3 marks:
- Award 1 mark for stating that carboxylic acid molecules form intermolecular hydrogen bonds.
- Award 1 mark for stating that ester molecules cannot form hydrogen bonds with each other.
- Award 1 mark for stating that the dipole-dipole/London forces in esters are weaker than hydrogen bonds, needing less energy to break.

(c) Max 4 marks:
- Award 1 mark for each correctly identified functional group:
- (i) amine
- (ii) ketone (accept carbonyl)
- (iii) amide (accept carboxamide)
- (iv) alcohol (accept hydroxyl / secondary alcohol)

(d) Max 4 marks:
- (i) Award 1 mark for each correct definition of primary, secondary, and tertiary environments [3].
- (ii) Award 1 mark for drawing the correct structure of 2-methylpropan-2-ol [1].

(a) (i)
- Oxidation: Loss of electrons.
- Reduction: Gain of electrons.
(ii)
- Oxidation: Increase in oxidation state.
- Reduction: Decrease in oxidation state.

(b)
- Oxidation half-equation: \(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\)
- Reduction half-equation: \(\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)\)
To balance the electrons, multiply the oxidation half-equation by 5:
\(5\text{Fe}^{2+}(aq) \rightarrow 5\text{Fe}^{3+}(aq) + 5e^-\)
Adding the two half-equations gives the overall balanced ionic equation:
\(\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)\)

(c) (i) Zinc is more active/reactive than copper (positioned higher in the activity series) and has a greater tendency to lose electrons. Therefore, zinc undergoes oxidation and acts as the anode, while copper undergoes reduction and acts as the cathode.
(ii)
- Anode (Oxidation): \(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-\)
- Cathode (Reduction): \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\)
(iii) Electrons flow through the external circuit from the zinc electrode (anode) to the copper electrode (cathode).

(d) The salt bridge:
- Completes the electrical circuit by allowing the flow of ions.
- Maintains electrical neutrality in both half-cells by permitting anions to migrate towards the anode and cations to migrate towards the cathode.

(a) Max 4 marks:
- (i) Award 1 mark for defining oxidation as loss of electrons and 1 mark for defining reduction as gain of electrons [2].
- (ii) Award 1 mark for defining oxidation as increase in oxidation state and 1 mark for defining reduction as decrease in oxidation state [2].

(b) Max 4 marks:
- Award 1 mark for correct oxidation half-equation.
- Award 2 marks for correct reduction half-equation (1 mark for correct species, 1 mark for balancing with \(\text{H}^+\) and \(\text{H}_2\text{O}\)).
- Award 1 mark for correct overall balanced chemical equation.

(c) Max 5 marks:
- (i) Award 1 mark for identifying Zinc as the anode and Copper as the cathode.
- Award 1 mark for justifying this based on Zn being more reactive / higher in the activity series.
- (ii) Award 1 mark for correct anode half-equation.
- Award 1 mark for correct cathode half-equation.
- (iii) Award 1 mark for stating that electrons flow from Zn (anode) to Cu (cathode).

(d) Max 2 marks:
- Award 1 mark for stating that it completes the electrical circuit.
- Award 1 mark for stating that it maintains electrical neutrality in the half-cells.

(a) A Brønsted-Lowry acid is a proton (\(\text{H}^+\)) donor, and a Brønsted-Lowry base is a proton (\(\text{H}^+\)) acceptor.
Conjugate acid-base pairs:
- Pair 1: \(\text{NH}_3\) (base) and \(\text{NH}_4^+\) (conjugate acid)
- Pair 2: \(\text{H}_2\text{O}\) (acid) and \(\text{OH}^-\) (conjugate base)

(b) A strong acid dissociates/ionizes completely in aqueous solution, whereas a weak acid only partially dissociates/ionizes in aqueous solution.
Experimental methods to distinguish them at the same concentration:
- Electrical conductivity: A strong acid will have a higher electrical conductivity than a weak acid because it has a higher concentration of mobile ions.
- Reactivity / Rate of reaction: A strong acid will react more rapidly with a metal (e.g., Magnesium) or a carbonate (e.g., Calcium carbonate), producing gas bubbles faster than a weak acid.

(c) (i) Nitric acid is a strong monoprotic acid, so \([\text{H}^+] = [\text{HNO}_3] = 0.050\text{ mol dm}^{-3}\).
\(\text{pH} = -\log[\text{H}^+] = -\log(0.050) = 1.30\).

(ii) Barium hydroxide is a strong base that yields two moles of hydroxide ions per mole:
\(\text{Ba(OH)}_2(aq) \rightarrow \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq)\)
\([\text{OH}^-] = 2 \times 0.10\text{ mol dm}^{-3} = 0.20\text{ mol dm}^{-3}\)
\(\text{pOH} = -\log[\text{OH}^-] = -\log(0.20) = 0.70\)
\(\text{pH} = 14.00 - \text{pOH} = 14.00 - 0.70 = 13.30\).

(d) (i) \(K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}\)

(ii) Assumptions:
- The concentration of \(\text{H}^+\) originating from the autoionization of water is negligible.
- The ionization of the weak acid is so small that the equilibrium concentration of ethanoic acid is approximately equal to its initial concentration (i.e., \([\text{CH}_3\text{COOH}]_{\text{eq}} \approx [\text{CH}_3\text{COOH}]_{\text{initial}} = 0.20\text{ mol dm}^{-3}\)).

(iii) Using the approximation: \(K_a = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{COOH}]}\)
\(1.8 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.20}\)
\([\text{H}^+]^2 = 3.6 \times 10^{-6}\)
\([\text{H}^+] = 1.9 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log(1.9 \times 10^{-3}) = 2.72\).

(a) Max 3 marks:
- Award 1 mark for defining Brønsted-Lowry acid (proton donor) and base (proton acceptor).
- Award 1 mark for identifying \(\text{NH}_3\) / \(\text{NH}_4^+\) as a conjugate pair.
- Award 1 mark for identifying \(\text{H}_2\text{O}\) / \(\text{OH}^-\) as a conjugate pair.

(b) Max 3 marks:
- Award 1 mark for stating that strong acids fully dissociate/ionize while weak acids only partially dissociate/ionize.
- Award 1 mark for outlining a valid method (electrical conductivity or reaction rate with metal/carbonate).
- Award 1 mark for describing the expected result (e.g., strong acid has higher conductivity / faster gas production).

(c) Max 4 marks:
- (i) Award 1 mark for identifying \([\text{H}^+] = 0.050\text{ mol dm}^{-3}\).
- Award 1 mark for calculating \(\text{pH} = 1.30\).
- (ii) Award 1 mark for calculating \([\text{OH}^-] = 0.20\text{ mol dm}^{-3}\).
- Award 1 mark for calculating \(\text{pH} = 13.30\).

(d) Max 5 marks:
- (i) Award 1 mark for the correct expression of \(K_a\).
- (ii) Award 1 mark for each stated assumption (up to 2 marks).
- (iii) Award 1 mark for calculating \([\text{H}^+] = 1.9 \times 10^{-3}\text{ mol dm}^{-3}\).
- Award 1 mark for calculating \(\text{pH} = 2.72\) (accept 2.7).

(a) (i)
- Moles of \(\text{N}_2 = \frac{42.0\text{ g}}{28.02\text{ g mol}^{-1}} = 1.50\text{ mol}\).
- Moles of \(\text{H}_2 = \frac{12.0\text{ g}}{2.02\text{ g mol}^{-1}} = 5.94\text{ mol}\).
According to the stoichiometric ratio, 1 mole of \(\text{N}_2\) requires 3 moles of \(\text{H}_2\).
For \(1.50\text{ mol}\) of \(\text{N}_2\), we require \(3 \times 1.50 = 4.50\text{ mol}\) of \(\text{H}_2\).
Since we have \(5.94\text{ mol}\) of \(\text{H}_2\), hydrogen is in excess, and \(\text{N}_2\) is the limiting reactant.

(ii) From the balanced equation, \(1.50\text{ mol}\) of \(\text{N}_2\) produces \(2 \times 1.50 = 3.00\text{ mol}\) of \(\text{NH}_3\).
Mass of \(\text{NH}_3 = 3.00\text{ mol} \times 17.04\text{ g mol}^{-1} = 51.1\text{ g}\).

(b) Percentage Yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = \frac{38.5\text{ g}}{51.1\text{ g}} \times 100\% = 75.3\%\).
(Accept values around \(75.3\%\) to \(75.4\%\) depending on rounding intermediate steps).

(c) (i)
- Moles of \(\text{Na}_2\text{CO}_3 = \frac{10.6\text{ g}}{105.99\text{ g mol}^{-1}} = 0.100\text{ mol}\).
- The ratio of \(\text{Na}_2\text{CO}_3\) to \(\text{CO}_2\) is 1:1, so \(0.100\text{ mol}\) of \(\text{CO}_2\) is produced.
- Volume of \(\text{CO}_2 = 0.100\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 2.27\text{ dm}^3\).

(ii) The volume would increase because the volume of a gas is directly proportional to its absolute temperature at constant pressure (Charles's Law).

(d) Atom economy is the molecular mass of the desired product divided by the sum of the molecular masses of all reactants, expressed as a percentage.
Reactants: \(\text{Na}_2\text{CO}_3 + 2\text{HCl}\)
Sum of molar masses of reactants = \(105.99 + 2 \times (36.46) = 178.91\text{ g mol}^{-1}\).
Desired product: \(2\text{NaCl}\)
Sum of molar masses of desired product = \(2 \times 58.44 = 116.88\text{ g mol}^{-1}\).
\(\text{Atom Economy} = \frac{116.88}{178.91} \times 100\% = 65.3\%\).

(a) Max 4 marks:
- (i) Award 1 mark for calculating correct moles of both reactants.
- Award 1 mark for identifying \(\text{N}_2\) as the limiting reactant with a valid explanation.
- (ii) Award 1 mark for determining the correct molar amount of \(\text{NH}_3\) (3.00 mol).
- Award 1 mark for calculating the mass of \(\text{NH}_3 = 51.1\text{ g}\) (accept 51.0 g - 51.2 g).

(b) Max 3 marks:
- Award 1 mark for correct percentage yield formula.
- Award 2 marks for correct calculation and answer: \(75.3\%\) (or \(75.4\%\)).

(c) Max 4 marks:
- (i) Award 1 mark for calculating moles of \(\text{Na}_2\text{CO}_3 = 0.100\text{ mol}\).
- Award 1 mark for correct mole ratio and identifying moles of \(\text{CO}_2\) as \(0.100\text{ mol}\).
- Award 1 mark for volume = \(2.27\text{ dm}^3\).
- (ii) Award 1 mark for stating that the volume would increase.

(d) Max 4 marks:
- Award 1 mark for a clear definition of atom economy.
- Award 1 mark for calculating the total mass of reactants (\(178.91\text{ g mol}^{-1}\)).
- Award 1 mark for calculating the total mass of the desired product (\(116.88\text{ g mol}^{-1}\)).
- Award 1 mark for correct atom economy calculation: \(65.3\%\).

(a)
Independent variables: Initial concentration of \(\text{S}_2\text{O}_8^{2-}\) and initial concentration of \(\text{I}^-\).
Dependent variable: Time taken for the color change (or initial rate of reaction).

(b)
Rate of reaction = \(\frac{\Delta[\text{I}_2]}{\Delta t}\)
For Run 1: Rate = \(\frac{2.00 \times 10^{-4} \text{ mol dm}^{-3}}{50.0 \text{ s}} = 4.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1}\)
For Run 2: Rate = \(\frac{2.00 \times 10^{-4} \text{ mol dm}^{-3}}{25.0 \text{ s}} = 8.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1}\)

(c)
(i) Comparing Run 1 and Run 2: \([\text{I}^-]\) is kept constant at \(0.100 \text{ mol dm}^{-3}\). When \([\text{S}_2\text{O}_8^{2-}]\) is doubled (from \(0.100\) to \(0.200 \text{ mol dm}^{-3}\)), the initial rate doubles from \(4.00 \times 10^{-6}\) to \(8.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is first-order with respect to \(\text{S}_2\text{O}_8^{2-}\).
(ii) Comparing Run 1 and Run 3: \([\text{S}_2\text{O}_8^{2-}]\) is kept constant at \(0.100 \text{ mol dm}^{-3}\). When \([\text{I}^-]\) is doubled (from \(0.100\) to \(0.200 \text{ mol dm}^{-3}\)), the rate doubles from \(4.00 \times 10^{-6}\) to \(8.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is first-order with respect to \(\text{I}^-\).

(d)
Rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)
Using Run 1 data to calculate \(k\):
\(4.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1} = k(0.100 \text{ mol dm}^{-3})(0.100 \text{ mol dm}^{-3})\)
\(k = \frac{4.00 \times 10^{-6}}{0.0100} = 4.00 \times 10^{-4}\)
Units: \(\text{dm}^3\text{ mol}^{-1}\text{s}^{-1}\)

(e)
The rate constant, \(k\), will increase with temperature. According to collision theory, an increase in temperature increases the average kinetic energy of the particles, so a significantly larger fraction of reactant particles have kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)), leading to a higher rate of successful collisions per unit time.

(f)
Potential systematic error: Temperature fluctuations during the runs (since the reaction is highly temperature-dependent, a changing temperature shifts the rate constant).
Improvement: Use a thermostatically controlled water bath to keep all solutions at a constant, uniform temperature throughout the experiment.
[Alternative valid systematic errors and improvements are acceptable, e.g., delay in manual timing -> use a colorimeter linked to a light sensor and data logger to record time automatically.]

(a)
Award [1] for both independent variables correct (initial concentration of peroxodisulfate and iodide ions).
Award [1] for the dependent variable correct (time taken for color change / rate).

(b)
Award [1] for calculating Run 1 rate as \(4.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1}\) (or equivalent form).
Award [1] for calculating Run 2 rate as \(8.00 \times 10^{-6} \text{ mol dm}^{-3}\text{ s}^{-1}\) (or equivalent form).
Accept unit-less values if correct units are stated at least once, or vice-versa.

(c)
(i) Award [1] for showing that doubling concentration doubles the rate (using Run 1 and 2). Award [1] for concluding it is first-order.
(ii) Award [1] for showing that doubling concentration doubles the rate (using Run 1 and 3). Award [1] for concluding it is first-order.

(d)
Award [1] for correct rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).
Award [1] for correct calculation of \(k = 4.00 \times 10^{-4}\).
Award [1] for correct units: \(\text{dm}^3\text{ mol}^{-1}\text{s}^{-1}\) (accept \(\text{L mol}^{-1}\text{ s}^{-1}\)).

(e)
Award [1] for stating that \(k\) increases.
Award [1] for explaining that more particles have \(E \ge E_a\) / greater fraction of successful collisions per unit time.

(f)
Award [1] for identifying a valid systematic error (e.g., human reaction time delay, temperature variation).
Award [1] for a corresponding realistic improvement (e.g., colorimeter with data logger, thermostatically controlled water bath).

(a)
To find the oxidation state of carbon in methanol (\( \text{CH}_3\text{OH} \)):
\( \text{C} + 4(+1) + (-2) = 0 \Rightarrow \text{C} + 2 = 0 \Rightarrow \text{C} = -2 \).
For carbon dioxide (\( \text{CO}_2 \)):
\( \text{C} + 2(-2) = 0 \Rightarrow \text{C} - 4 = 0 \Rightarrow \text{C} = +4 \).
Since the oxidation state of carbon increases from -2 to +4, methanol (\( \text{CH}_3\text{OH} \)) is the species oxidized.

(b)
To balance the electrons between the two half-reactions:
Multiply the anode half-reaction by 2 (to yield 12 electrons):
\( 2\text{CH}_3\text{OH}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{CO}_2(g) + 12\text{H}^+(aq) + 12\text{e}^- \)
Multiply the cathode half-reaction by 3 (to consume 12 electrons):
\( 3\text{O}_2(g) + 12\text{H}^+(aq) + 12\text{e}^- \rightarrow 6\text{H}_2\text{O}(l) \)

Adding these two equations together and simplifying water and hydrogen ions gives the overall balanced equation:
\( 2\text{CH}_3\text{OH}(aq) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \)

The standard cell potential is:
\( E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +1.23\text{ V} - (-0.02\text{ V}) = +1.25\text{ V} \).

(c)
Advantage: Methanol is liquid at room temperature, making it much easier and safer to store, handle, and transport than gaseous hydrogen, which requires high pressure or cryogenic storage. Methanol also has a higher volumetric energy density.
Disadvantage: The cell produces carbon dioxide, which is a greenhouse gas, unlike hydrogen fuel cells which only produce water. Additionally, methanol is highly toxic and flammable.

Part (a) [2 marks]:
- Correct oxidation states: C in methanol is -2 AND C in CO2 is +4 [1 mark].
- Correctly identifies methanol/CH3OH as the species oxidized [1 mark].

Part (b) [2.5 marks]:
- Correct overall balanced equation including state symbols [1.5 marks]. Deduct [0.5 marks] if state symbols are missing or incorrect.
- Correct cell potential: E_cell = +1.25 V [1 mark]. Award [0.5 marks] if sign is missing or incorrect.

Part (c) [3 marks]:
- One valid advantage (e.g., methanol is a liquid, easier to store/transport, higher volumetric energy density) [1.5 marks].
- One valid disadvantage (e.g., produces carbon dioxide/greenhouse gas, lower thermodynamic efficiency, methanol is toxic) [1.5 marks].

(a)
A heterogeneous catalyst works by adsorbing reactant molecules onto its solid active surface, which weakens the covalent bonds within the reactants. This provides an alternative reaction pathway with a lower activation energy (\(E_a\)). According to collision theory, at any given temperature, a lower activation energy means a significantly higher fraction of reactant molecules possess energy equal to or greater than the activation energy, resulting in a higher frequency of successful/effective collisions per unit of time, thus increasing the reaction rate.

(b)
The diagram should display:
1. Y-axis labeled as 'Potential Energy' (or 'Enthalpy' / 'Energy') and X-axis labeled as 'Reaction Coordinate' (or 'Progress of Reaction').
2. Reactants level drawn higher than the products level (as the reaction is exothermic).
3. Two energy barrier curves: a higher curve representing the uncatalyzed reaction path, and a lower curve representing the catalyzed reaction path.
4. Labeling of the uncatalyzed activation energy (\(E_{a,\text{uncat}}\)) from reactants to the top of the higher peak.
5. Labeling of the catalyzed activation energy (\(E_{a,\text{cat}}\)) from reactants to the top of the lower peak.
6. Labeling of the enthalpy change (\(\Delta H\)) as the energy difference between reactants and products, showing a downward arrow or indicating a negative value.

(c)
- The overall order is the sum of the individual orders: \( 2 + 0 = 2 \) (second order overall).
- Since the order with respect to CO is 0, doubling its concentration has no effect on the rate.
- Since the order with respect to NO is 2, tripling its concentration increases the rate by a factor of \( 3^2 = 9 \).
- Therefore, the overall rate of reaction increases by a factor of 9.

Part (a) [2.5 marks]:
- States that the catalyst provides an alternative pathway with lower activation energy/Ea [1 mark].
- Explains that a larger fraction/proportion of reactant molecules have energy greater than or equal to Ea [1 mark].
- Mentions that this increases the frequency of successful/effective collisions [0.5 marks].

Part (b) [3 marks]:
- Correctly labeled axes (Y = Potential Energy/Enthalpy/Energy, X = Reaction Coordinate/Progress) [0.5 marks].
- Reactants drawn higher than products to represent an exothermic reaction [0.5 marks].
- Two distinct reaction curves showing a lower peak for the catalyzed reaction [0.5 marks].
- Correctly labeled activation energies: Ea for uncatalyzed and Ea for catalyzed pathways [1 mark].
- Correctly labeled enthalpy change (\(\Delta H\)) [0.5 marks].

Part (c) [2 marks]:
- Identifies the overall order as 2 [1 mark].
- Deduces that the rate increases by a factor of 9 [1 mark].

(a)
The chemical structure of acetylsalicylic acid (aspirin) contains:
1. A carboxylic acid group (\( -\text{COOH} \))
2. An ester group (\( -\text{COO}- \))
3. A phenyl group (arene or benzene ring, \( -\text{C}_6\text{H}_5 \))

(b)
- Chemical Test: Add a few drops of aqueous iron(III) chloride (\( \text{FeCl}_3(aq) \)) to samples of both salicylic acid and the product.
- Observation for Salicylic Acid: The solution turns a deep purple/violet color. This is because salicylic acid contains a phenol group (hydroxyl attached directly to the benzene ring), which forms a colored complex with \( \text{Fe}^{3+} \) ions.
- Observation for Pure Aspirin: No color change is observed (the solution remains its original yellow-brown color). Since the phenol group of salicylic acid was converted into an ester group in aspirin, it no longer forms the purple iron complex.

(c)
- Reaction Type: Esterification (or condensation / nucleophilic acyl substitution).
- Role of acid: Phosphoric acid (or sulfuric acid) acts as an acid catalyst. It protonates the carbonyl oxygen of ethanoic anhydride, making the carbonyl carbon more electrophilic and thus accelerating the nucleophilic attack by the phenolic hydroxyl group of salicylic acid.

Part (a) [3 marks]:
- Identifies three correct functional groups: Carboxylic acid (or carboxyl), Ester, and Phenyl (or arene/benzene ring) [1 mark each, up to 3 marks]. Accept 'ether' or 'carbonyl' only if clearly specified in context, but carboxylic acid, ester, and phenyl are the primary groups.

Part (b) [2.5 marks]:
- Identifies the reagent as iron(III) chloride / FeCl3 [0.5 marks].
- Explains the mechanism: FeCl3 forms a purple complex with the phenol group [1 mark].
- States observation for salicylic acid (turns purple/violet) AND observation for aspirin (no color change/stays yellow-brown) [1 mark].

Part (c) [2 marks]:
- Identifies the reaction type as esterification, condensation, or nucleophilic acyl substitution [1 mark].
- Explains that the acid acts as a catalyst [1 mark].

(a)
- Definition: A buffer solution is a solution that resists changes in pH when small amounts of an acid or a base are added to it.
- Response to added acid: The hydrogen carbonate ions act as a conjugate base and react with the incoming hydronium ions:
\( \text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{CO}_3(aq) \)
- Response to added base: The weak acid (carbonic acid) reacts with the incoming hydroxide ions:
\( \text{H}_2\text{CO}_3(aq) + \text{OH}^-(aq) \rightarrow \text{HCO}_3^-(aq) + \text{H}_2\text{O}(l) \)

(b)
Using the Henderson-Hasselbalch equation:
\( \text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) \)

Substitute the given values:
\( 7.40 = 6.35 + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) \)
\( \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = 7.40 - 6.35 = 1.05 \)

Taking the antilog of both sides:
\( \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 10^{1.05} \approx 11.2 \) (or 11 to two significant figures).

(c)
Hyperventilation causes excessive loss of \( \text{CO}_2(g) \) from the lungs, which lowers the concentration of dissolved \( \text{CO}_2(aq) \) in the blood.
According to Le Chatelier's principle, the system will shift the equilibrium to the left to replace the lost \( \text{CO}_2 \):
\( \text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{CO}_3(aq) \rightarrow \text{CO}_2(aq) + \text{H}_2\text{O}(l) \)
This shift consumes \( \text{H}^+ \) ions, thereby decreasing the concentration of free hydrogen ions in the blood. Consequently, the pH of the blood increases (a condition known as respiratory alkalosis).

Part (a) [3 marks]:
- Correct definition of a buffer solution [1 mark].
- Balanced equation for added acid (\( \text{HCO}_3^- + \text{H}^+ \rightarrow \text{H}_2\text{CO}_3 \)) [1 mark].
- Balanced equation for added base (\( \text{H}_2\text{CO}_3 + \text{OH}^- \rightarrow \text{HCO}_3^- + \text{H}_2\text{O} \)) [1 mark].

Part (b) [2.5 marks]:
- Correct formulation of the Henderson-Hasselbalch equation [1 mark].
- Correct calculation of the difference: \( \text{pH} - \text{p}K_a = 1.05 \) [0.5 marks].
- Final calculated ratio of \( 11.2 \) (or \( 11 \)) [1 mark].

Part (c) [2 marks]:
- Explains that the loss of CO2 shifts the equilibrium to the left [1 mark].
- States that this shift reduces \( [\text{H}^+] \), causing the blood pH to increase [1 mark].