2024 Edexcel IAL Chemistry (YCH11) Practice Paper with Answers

The successive ionization energies show a very large increase between the 3rd and 4th ionization energies (from 2745 to 11577 \(\text{kJ mol}^{-1}\)). This indicates that the 4th electron is being removed from an inner quantum shell, closer to the nucleus and less shielded. Therefore, element X has three electrons in its outer shell (Valence = +3). It forms a \(\text{X}^{3+}\) ion. The oxide ion is \(\text{O}^{2-}\). To form a neutral ionic compound, the formula must be \(\text{X}_2\text{O}_3\).

[1 mark] C is correct.

- Identify the large jump between the 3rd and 4th ionization energies, indicating three valence electrons.
- Determine the formula of the oxide by combining \(\text{X}^{3+}\) and \(\text{O}^{2-}\) ions.

First, convert all values to SI units:
\(p = 101\text{ kPa} = 101000\text{ Pa}\)
\(V = 252\text{ cm}^3 = 252 \times 10^{-6}\text{ m}^3\)
\(T = 373\text{ K}\)

Using the ideal gas equation, \(pV = nRT\):
\(n = \frac{pV}{RT} = \frac{101000 \times 252 \times 10^{-6}}{8.31 \times 373} = \frac{25.452}{3099.63} \approx 0.00821\text{ mol}\)

Now, calculate the molar mass (\(M\)):
\(M = \frac{m}{n} = \frac{0.460\text{ g}}{0.00821\text{ mol}} \approx 56.0\text{ g mol}^{-1}\)

The empirical formula weight of \(\text{CH}_2\) is:
\(12.0 + (2 \times 1.0) = 14.0\text{ g mol}^{-1}\)

Ratio \(= \frac{56.0}{14.0} = 4\).
Therefore, the molecular formula is \(\text{C}_4\text{H}_8\).

[1 mark] C is the correct answer.

- Convert pressure and volume to SI units correctly.
- Calculate the amount in moles using \(n = \frac{pV}{RT}\).
- Divide mass by moles to obtain the molar mass of 56 g/mol.
- Relate the molar mass to the empirical formula mass (14 g/mol) to find the multiplier of 4.

Let's analyze the valence electron pairs around the central atom for each option:
- \(\text{BF}_3\): Boron has 3 valence electrons, bonded to 3 fluorine atoms. 3 bonding pairs, 0 lone pairs. Shape is trigonal planar.
- \(\text{H}_3\text{O}^+\): Oxygen normally has 6 valence electrons, but the positive charge means it effectively has 5. It forms 3 single bonds with H. This results in 3 bonding pairs and 1 lone pair (total of 4 electron pairs, tetrahedral arrangement). The shape is trigonal pyramidal.
- \(\text{CH}_3^+\): Carbon has 4 valence electrons, minus 1 for the positive charge = 3. Bonded to 3 hydrogens. 3 bonding pairs, 0 lone pairs. Shape is trigonal planar.
- \(\text{ICl}_3\): Iodine has 7 valence electrons, bonded to 3 chlorine atoms. This gives 3 bonding pairs and 2 lone pairs. Shape is T-shaped.

[1 mark] B is correct.

- Determine the number of bonding and lone pairs on the central atom of each species.
- Identify that \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair, which corresponds to a trigonal pyramidal shape.

A termination step in a free-radical mechanism is defined as a step in which two free radicals react together to form a stable, non-radical covalent molecule.
- Option A is a propagation step (radical + molecule \(\rightarrow\) radical + molecule).
- Option B is a propagation step (radical + molecule \(\rightarrow\) radical + molecule).
- Option C is a termination step (radical + radical \(\rightarrow\) molecule).
- Option D is an initiation step (molecule \(\rightarrow\) radical + radical).

[1 mark] C is correct.

- Recognize that a termination step must involve two radical reactants combining to form a non-radical product.

For an alkene to exhibit E/Z isomerism, each carbon of the double bond must be bonded to two different atoms or groups of atoms.
- In 2-methylbut-2-ene, carbon-2 is bonded to two methyl groups. No E/Z isomerism is possible.
- In 2-methylbut-1-ene, carbon-1 is bonded to two hydrogen atoms. No E/Z isomerism is possible.
- In pent-1-ene, carbon-1 is bonded to two hydrogen atoms. No E/Z isomerism is possible.
- In pent-2-ene (\(\text{CH}_3\text{CH=CHCH}_2\text{CH}_3\)), carbon-2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\), and carbon-3 is bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Since both carbons have two different groups attached, E/Z isomerism is possible.

[1 mark] D is correct.

- Identify that E/Z isomerism requires each carbon in the double bond to have two different substituents.
- Test each option by drawing or analyzing its structural formula.

Phosphorus has the outer electronic configuration \(3s^2 3p^3\), with three singly-occupied p-orbitals (half-filled p-subshell). Sulfur has the outer electronic configuration \(3s^2 3p^4\), with one paired set of electrons in a p-orbital. The electrostatic repulsion between these two paired electrons in the same 3p orbital makes the first outer electron easier to remove in sulfur than in phosphorus, resulting in a lower first ionization energy.

[1 mark] C is correct.

- Recall the electronic configurations of phosphorus and sulfur.
- Identify spin-pair repulsion in the 3p orbital of sulfur as the reason for the deviation in the periodic trend of ionization energy.

Atom economy is calculated as:
\(\text{Atom economy} = \frac{\text{Total molar mass of desired products}}{\text{Total molar mass of all reactants}} \times 100\%\)

1. Desired product is copper (2 mol of Cu):
\(\text{Mass of desired product} = 2 \times 63.5 = 127.0\text{ g mol}^{-1}\)

2. Reactants are 2 mol of CuO and 1 mol of C:
\(\text{Mass of reactants} = 2 \times (63.5 + 16.0) + 12.0 = 2 \times 79.5 + 12.0 = 159.0 + 12.0 = 171.0\text{ g mol}^{-1}\)

3. Calculate percentage atom economy:
\(\text{Atom economy} = \frac{127.0}{171.0} \times 100\% \approx 74.27\%\)

Rounding to 3 significant figures gives 74.3%.

[1 mark] B is correct.

- Calculate total mass of desired products (2 moles of Cu).
- Calculate total mass of reactants.
- Calculate the percentage and select the correct option.

Lattice energy is proportional to the electrostatic force of attraction between the ions, which is determined by \(\frac{q_1 q_2}{r^2}\) where \(q_1\) and \(q_2\) are the charges on the ions and \(r\) is the sum of the ionic radii.
- Magnesium oxide (\(\text{MgO}\)) contains \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) ions. The product of charges is \(2 \times 2 = 4\), whereas for \(\text{NaF}\), \(\text{NaCl}\), and \(\text{MgBr}_2\) (considering single cation-anion pairs), the charge product is lower.
- Additionally, \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) are relatively small ions, which decreases \(r\) and increases the electrostatic force. Therefore, \(\text{MgO}\) has the most exothermic (most negative) lattice energy among the options.

[1 mark] C is correct.

- Relate lattice energy to ionic charge and ionic radius.
- Identify that the higher charges on both ions in \(\text{MgO}\) (+2 and -2) lead to a much stronger electrostatic attraction than in the other compounds.

There is a very large increase between the third (\(2745\text{ kJ mol}^{-1}\) and fourth (\(11577\text{ kJ mol}^{-1}\)) ionization energies. This indicates that the fourth electron is removed from an inner shell, which is closer to the nucleus and experiences much stronger electrostatic attraction. Therefore, element \(X\) has three valence electrons and is in Group 3 (Group 13) of the Periodic Table. In its compounds, it most stably exhibits a \(+3\) oxidation state. Thus, the chloride formed is \(X\text{Cl}_3\).

C: Correct option (1 mark). Incorrect options are based on identifying other groups: A (Group 1), B (Group 2), D (Group 4).

First, calculate the molar masses of the desired product and all reactants:\
- Desired product: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\) \
- Reactant 1: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} = (3 \times 12.0) + (7 \times 1.0) + 79.9 = 122.9\text{ g mol}^{-1}\) \
- Reactant 2: \(\text{NaOH} = 23.0 + 16.0 + 1.0 = 40.0\text{ g mol}^{-1}\) \
- Total mass of reactants = \(122.9 + 40.0 = 162.9\text{ g mol}^{-1}\) \
\
Percentage atom economy = \(\frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\) \
Percentage atom economy = \(\frac{60.0}{162.9} \times 100 = 36.83\\%\) which rounds to \(36.8\\%\).

B: Correct option (1 mark). Incorrect option A is calculated by dividing 25 by 162.9. Incorrect option C is calculated by using 95/162.9. Incorrect option D is the atom economy if NaBr was the desired product.

The amide ion, \(\text{NH}_2^-\), has a central nitrogen atom with 2 bonding pairs and 2 lone pairs of electrons (isoelectronic with \(\text{H}_2\text{O}\)). According to VSEPR theory, these four electron pairs arrange themselves tetrahedrally to minimize repulsion, but the presence of two lone pairs repels the bonding pairs more strongly. This reduces the ideal tetrahedral angle of \(109.5^\circ\) by approximately \(2.5^\circ\) per lone pair, resulting in a bent shape with a bond angle of around \(104.5^\circ\).\
- \(\text{NH}_4^+\) is tetrahedral with a bond angle of \(109.5^\circ\).\
- \(\text{H}_3\text{O}^+\) is trigonal pyramidal with a bond angle of approximately \(107^\circ\).\
- \(\text{CO}_2\) is linear with a bond angle of \(180^\circ\).

C: Correct option (1 mark). Incorrect option A corresponds to 109.5 degrees. Incorrect option B corresponds to 107 degrees. Incorrect option D corresponds to 180 degrees.

A termination step in a radical mechanism is a reaction where two free radicals combine to form a stable, non-radical covalent molecule.\
- Option A is a propagation step.\
- Option B is also a propagation step.\
- Option C shows two radicals (\(\text{CH}_3^\bullet\) and \(\text{Cl}^\bullet\)) reacting together to form chloromethane (no radicals remain), which is a termination step.\
- Option D represents the initiation step.

C: Correct option (1 mark). Incorrect option A is propagation. Incorrect option B is propagation. Incorrect option D is initiation.

During electrophilic addition to propene, the hydrogen atom from sulfuric acid acts as the electrophile. The pi electrons of the double bond attack the \(\text{H}^+\) (or \(\text{H}^{\delta+}\) of \(\text{H}_2\text{SO}_4\)), breaking the double bond in the rate-determining step. This step forms a carbocation intermediate. According to Markovnikov's rule, the electrophile adds to the carbon with more hydrogen atoms to yield the more stable carbocation. The secondary carbocation, \(\text{CH}_3\text{CH}^+\text{CH}_3\), is more stable than the primary carbocation, \(\text{CH}_3\text{CH}_2\text{CH}_2^+\), due to the electron-donating inductive effect of two alkyl groups. Therefore, the secondary carbocation is the major intermediate. Options C and D are neutral products, not intermediates.

B: Correct option (1 mark). Incorrect option A is the minor primary carbocation intermediate. Incorrect options C and D are products of the reaction, not intermediates of the rate-determining step.

The reaction of an alkaline earth metal (Group 2) with hydrochloric acid is represented by:\
\(M(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow M\text{Cl}_2(\text{aq}) + \text{H}_2(\text{g})\)\
\
First, calculate the moles of hydrogen gas produced:\
\(n(\text{H}_2) = \frac{\text{Volume of gas}}{\text{Molar volume at rtp}} = \frac{120\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0050\text{ mol}\)\
\
From the 1:1 molar ratio, the moles of metal \(M\) reacted is also \(0.0050\text{ mol}\).\
\
Next, calculate the relative atomic mass of \(M\):\
\
\(A_{\text{r}} = \frac{\text{Mass of metal}}{\text{Moles of metal}} = \frac{0.120\text{ g}}{0.0050\text{ mol}} = 24.0\text{ g mol}^{-1}\)\
\
Comparing this to the relative atomic masses of Group 2 metals (\(\text{Be} = 9.0\), \(\text{Mg} = 24.3\), \(\text{Ca} = 40.1\), \(\text{Sr} = 87.6\)), the metal is magnesium.

B: Correct option (1 mark). Incorrect options correspond to other Group 2 metals if a different calculation or stoichiometry was used.

The atomic number of iron is 26, meaning a neutral iron atom has 26 electrons. Its ground state electron configuration is:\
\
\(\text{Fe}: [\text{Ar}] 4\text{s}^2 3\text{d}^6\)\
\
When d-block transition metals form cations, electrons are lost from the outermost \(4\text{s}\) subshell before any \(3\text{d}\) electrons are removed. To form \(\text{Fe}^{3+}\), three electrons are removed: first, two electrons from the \(4\text{s}\) subshell are lost, leaving \([\text{Ar}] 3\text{d}^6\), and then one electron from the \(3\text{d}\) subshell is lost, leaving \([\text{Ar}] 3\text{d}^5\). Therefore, the ground state configuration of the \(\text{Fe}^{3+}\) ion is \([\text{Ar}] 3\text{d}^5\).

A: Correct option (1 mark). Incorrect option B is based on keeping the 4s electrons and removing three d electrons. Incorrect option C is a miscalculation. Incorrect option D is the electronic configuration of the Fe2+ ion.

The polarity of a covalent bond depends on the difference in electronegativity between the two bonded atoms. The greater the difference in electronegativity, the more polar the bond. Fluorine (electronegativity 4.0) is the most electronegative element in the periodic table, so the electronegativity difference between carbon (2.5) and fluorine (4.0) is 1.5. This is larger than the differences for the other bonds: C-H (0.4), C-Cl (0.5), and C-O (1.0). Thus, the C-F bond is the most polar.

B: Correct option (1 mark). Other options have lower electronegativity differences and are thus less polar.

1. Calculate the number of moles of \( \text{CO}_2 \) produced:
\( \text{Moles of } \text{CO}_2 = \frac{360 \text{ cm}^3}{24000 \text{ cm}^3 \text{ mol}^{-1}} = 0.0150 \text{ mol} \)

2. Determine the molar mass of the carbonate, \( \text{MCO}_3 \):
From the equation: \( \text{MCO}_3(s) \rightarrow \text{MO}(s) + \text{CO}_2(g) \), the molar ratio is 1:1.
\( \text{Moles of } \text{MCO}_3 = 0.0150 \text{ mol} \)
\( \text{Molar mass of } \text{MCO}_3 = \frac{1.50 \text{ g}}{0.0150 \text{ mol}} = 100 \text{ g mol}^{-1} \)

3. Calculate the relative atomic mass of \( \text{M} \):
\( A_r(\text{M}) + A_r(\text{C}) + 3 \times A_r(\text{O}) = 100 \)
\( A_r(\text{M}) + 12.0 + 48.0 = 100 \)
\( A_r(\text{M}) = 40.0 \text{ g mol}^{-1} \)
This corresponds to calcium.

1 mark: Correctly calculates the relative atomic mass of the metal as 40.0 g/mol and identifies it as calcium (B).

1. Identify the group of the element from the ionization energies:
There is a huge increase between the 3rd and 4th ionization energies (from \( 2745 \text{ to } 11577 \text{ kJ mol}^{-1} \)). This indicates that the 4th electron is removed from a shell closer to the nucleus.
Therefore, element \( \text{Y} \) has 3 valence electrons and belongs to Group 3 (Group 13).

2. Determine the formula of its oxide:
Since \( \text{Y} \) has 3 valence electrons, it forms \( \text{Y}^{3+} \) ions.
Oxygen forms \( \text{O}^{2-} \) ions.
To balance the charges, the formula of the oxide is \( \text{Y}_2\text{O}_3 \).

1 mark: Correctly identifies \( \text{Y}_2\text{O}_3 \) as the formula based on the large jump between the 3rd and 4th ionization energies (B).

- \( \text{CO}_2 \) has a linear shape with no permanent dipole due to symmetry.
- \( \text{BCl}_3 \) has a trigonal planar shape with no permanent dipole due to symmetry.
- \( \text{SF}_6 \) has an octahedral shape with no permanent dipole due to symmetry.
- \( \text{SO}_2 \) has a non-linear (bent) shape because the central sulfur atom has two bonding pairs (double bonds) and one lone pair. The asymmetrical shape means the individual polar bond dipoles do not cancel, resulting in a permanent molecular dipole.

1 mark: Correctly identifies \( \text{SO}_2 \) as having a bent shape and a permanent dipole (B).

For a molecule to exhibit geometric (\( E \)/\( Z \)) isomerism, each carbon of the double bond must be bonded to two different groups.
- In 2-methylbut-2-ene, \( (\text{CH}_3)_2\text{C}=\text{CHCH}_3 \), one carbon of the double bond is bonded to two identical methyl groups.
- In 2-methylbut-1-ene, \( \text{CH}_2=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3 \), the terminal carbon is bonded to two identical hydrogen atoms.
- In pent-1-ene, \( \text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}_3 \), the terminal carbon is bonded to two identical hydrogen atoms.
- In pent-2-ene, \( \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3 \), C2 is bonded to \( -\text{H} \) and \( -\text{CH}_3 \) (two different groups), and C3 is bonded to \( -\text{H} \) and \( -\text{CH}_2\text{CH}_3 \) (two different groups). Therefore, it exhibits geometric isomerism.

1 mark: Correctly identifies pent-2-ene as the alkene exhibiting E/Z isomerism (D).

(a) First ionization energy is the energy required per mole to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. (b) There is a very large increase / jump between the 3rd and 4th ionization energies (from 2745 to 11577 kJ mol^{-1}). This indicates that the fourth electron is being removed from an inner electron shell, which experiences significantly less shielding and stronger electrostatic attraction to the nucleus. This shows there are 3 electrons in the outer shell, so the element is in Group 13 / Group 3. Since the element is in Period 3, element Y must be aluminum (Al). (c) \( A_r = \frac{(69 \times 60.11) + (71 \times 39.89)}{100} = \frac{4147.59 + 2832.19}{100} = 69.80 \) (to 2 d.p.). (d) A neutral gallium atom has the configuration \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^1 \). When it forms a \( \text{Ga}^{3+} \) ion, it loses the three highest-energy outer electrons from the 4s and 4p sub-shells. Therefore, the configuration of \( \text{Ga}^{3+} \) is \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} \).

(a) [3 marks total] - Energy required per mole / enthalpy change per mole for the removal of one mole of electrons (1 mark) - from one mole of gaseous atoms (1 mark) - to form one mole of gaseous 1+ ions (1 mark). (b) [4 marks total] - Identifies a large jump between the 3rd and 4th ionization energies (1 mark) - Explains that the 4th electron is removed from an inner shell / a shell closer to the nucleus / a shell with less shielding (1 mark) - Deduces there are 3 valence/outer electrons, placing Y in Group 3 / 13 (1 mark) - Identifies Y as aluminum / Al (1 mark). (c) [3 marks total] - Correct expression for weighted average: \( \frac{(69 \times 60.11) + (71 \times 39.89)}{100} \) (1 mark) - Correct calculation of value to 69.7978 (1 mark) - Evaluated correctly to 2 decimal places: 69.80 (1 mark). (d) [2 marks total] - Correct sub-shells shown up to 3d (1 mark) - Correct number of electrons in each sub-shell: \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} \) (1 mark).

(a) Methyl orange starts as yellow in the alkaline/carbonate solution and turns to orange (peach/pink) at the end-point. (b) \( n(\text{HCl}) = 0.100 \text{ mol dm}^{-3} \times \frac{25.00}{1000} \text{ dm}^3 = 2.50 \times 10^{-3} \text{ mol} \). (c) (i) From the equation, 1 mole of \( \text{Na}_2\text{CO}_3 \) reacts with 2 moles of \( \text{HCl} \), so \( n(\text{Na}_2\text{CO}_3) \text{ in 25.0 cm}^3 = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3} \text{ mol} \). (ii) In the 250.0 cm^{3} volumetric flask, the volume is 10 times greater: \( 1.25 \times 10^{-3} \times 10 = 1.25 \times 10^{-2} \text{ mol} \) (or 0.0125 mol). (d) \( M_r(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}) = \frac{3.58 \text{ g}}{0.0125 \text{ mol}} = 286.4 \text{ g mol}^{-1} \). The molar mass of anhydrous \( \text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1} \). The mass of water of crystallization is \( 286.4 - 106.0 = 180.4 \text{ g mol}^{-1} \). Therefore, \( x = \frac{180.4}{18.0} = 10.02 \approx 10 \). (e) There is no effect on the volume of hydrochloric acid needed because the number of moles of sodium carbonate transferred into the conical flask is fixed by the pipette (25.0 cm^{3}). Extra water in the flask only dilutes the mixture but does not change the amount of carbonate moles. Consequently, there is also no effect on the calculated value of \( x \).

(a) [1 mark total] - Yellow to orange / peach / pink (1 mark) [Reject: yellow to red]. (b) [2 marks total] - Correct volume conversion: \( \frac{25.00}{1000} \) (1 mark) - Correct calculation of moles: \( 2.50 \times 10^{-3} \) mol (1 mark). (c) [2 marks total] - (i) Correct division of (b) by 2: \( 1.25 \times 10^{-3} \) mol (1 mark) - (ii) Correct multiplication of (c)(i) by 10: \( 1.25 \times 10^{-2} \) (or 0.0125) mol (1 mark). (d) [4 marks total] - Calculate molar mass of hydrated salt: \( \frac{3.58}{\text{answer (c)(ii)}} = 286.4 \text{ g mol}^{-1} \) (1 mark) - Calculate molar mass of anhydrous salt: \( 106.0 \text{ g mol}^{-1} \) (1 mark) - Subtract anhydrous mass from hydrated mass: \( 180.4 \text{ g mol}^{-1} \) (1 mark) - Divide by 18.0 to find \( x = 10 \) (must be an integer) (1 mark). (e) [3 marks total] - State that volume of acid needed is unchanged (1 mark) - State that moles of sodium carbonate added remain the same (1 mark) - State that the calculated value of \( x \) is unchanged (1 mark).

(a) (i) Carbon dioxide: Carbon forms double covalent bonds with both oxygen atoms. Carbon shares four of its electrons (two double bonds, total 4 shared pairs), leaving 0 lone pairs on carbon. Each oxygen atom has 2 shared pairs and 2 lone pairs (4 non-bonding electrons). (ii) Sulfur tetrafluoride: Sulfur forms four single covalent bonds with four fluorine atoms, leaving 1 lone pair of electrons on the sulfur atom (expanded octet with 10 electrons in S outer shell). Each fluorine atom has 1 shared pair and 3 lone pairs in its outer shell. (b) (i) \( \text{CO}_2 \): Linear shape, 180° bond angle. Carbon has two bonding regions (two double bonds) and zero lone pairs. The bonding regions repel each other equally to be as far apart as possible to minimize repulsion. (ii) \( \text{SF}_4 \): Seesaw shape, bond angles around 87° / 102° (accept <90° and <120°). Sulfur has five areas of electron density: 4 bonding pairs and 1 lone pair. Lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, which pushes the bonding pairs closer together. (c) In \( \text{CCl}_4 \), the molecular geometry is highly symmetrical (tetrahedral). The individual polar C-Cl bond dipoles are of equal magnitude and act in opposite directions, so they cancel each other out completely, leaving no net molecular dipole. In \( \text{CHCl}_3 \), the geometry is also tetrahedral but asymmetrical because the C-H bond has a different dipole/electronegativity difference compared to the polar C-Cl bonds. Therefore, the dipoles do not cancel, resulting in a net molecular dipole.

(a) [4 marks total] - (i) Correct double bonds between C and O (1 mark) - (i) Correct lone pairs (two pairs) on each O and no lone pairs on C (1 mark) - (ii) Correct 4 single S-F bonds and 1 lone pair on S (1 mark) - (ii) Correct 3 lone pairs on each F atom (1 mark). (b) [5 marks total] - \( \text{CO}_2 \) shape is linear and bond angle is 180° (1 mark) - \( \text{CO}_2 \) explanation: 2 bonding regions (or double bonds) repel each other to minimize repulsion / maximize distance (1 mark) - \( \text{SF}_4 \) shape is seesaw / sawhorse and bond angle(s) are <90° and <120° (or around 87°/102°) (1 mark) - \( \text{SF}_4 \) explanation: 4 bonding pairs and 1 lone pair (total 5 electron pairs) (1 mark) - Repulsion order: lone pair-bonding pair > bonding pair-bonding pair (1 mark). (c) [3 marks total] - \( \text{CCl}_4 \) is symmetrical (tetrahedral) (1 mark) - In \( \text{CCl}_4 \), the dipoles cancel out (resulting in no net dipole) (1 mark) - In \( \text{CHCl}_3 \), the geometry is asymmetrical / dipoles do not cancel (leading to a net dipole) (1 mark).

(a) (i) The ultraviolet light provides the energy to cause homolytic fission of the covalent Br-Br bond, forming bromine free radicals. (ii) **Initiation**: \( \text{Br}_2 \xrightarrow{\text{UV}} 2\text{Br}^\bullet \). **Propagation Step 1**: \( \text{CH}_3\text{CH}_3 + \text{Br}^\bullet \to \text{CH}_3\text{CH}_2^\bullet + \text{HBr} \). **Propagation Step 2**: \( \text{CH}_3\text{CH}_2^\bullet + \text{Br}_2 \to \text{CH}_3\text{CH}_2\text{Br} + \text{Br}^\bullet \). **Termination (forming bromoethane)**: \( \text{CH}_3\text{CH}_2^\bullet + \text{Br}^\bullet \to \text{CH}_3\text{CH}_2\text{Br} \). (iii) The termination step that yields a 4-carbon alkane (butane) occurs when two ethyl radicals combine: \( 2\text{CH}_3\text{CH}_2^\bullet \to \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \). (b) (i) Structural isomerism refers to compounds that share the same molecular formula but have different structural formulae (their atoms are bonded in a different arrangement). (ii) The four structural isomers of \( \text{C}_4\text{H}_9\text{Cl} \) are: 1. **1-chlorobutane**: Skeletal structure is a straight 4-carbon chain with a chlorine atom attached to the first carbon: \( \text{Cl-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \). 2. **2-chlorobutane**: Skeletal structure is a straight 4-carbon chain with a chlorine atom attached to the second carbon: \( \text{CH}_3-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3 \). 3. **1-chloro-2-methylpropane**: Skeletal structure has a 3-carbon main chain, a methyl group on the second carbon, and a chlorine atom on the first carbon: \( \text{Cl-CH}_2-\text{CH(CH}_3)-\text{CH}_3 \). 4. **2-chloro-2-methylpropane**: Skeletal structure has a 3-carbon main chain, with both a methyl group and a chlorine atom attached to the second carbon: \( (\text{CH}_3)_3\text{C-Cl} \).

(a) [6 marks total] - (i) homolytic fission of the Br-Br bond / generates bromine free radicals (1 mark) - (ii) Initiation step equation with radical dot: \( \text{Br}_2 \to 2\text{Br}^\bullet \) (1 mark) - (ii) Propagation Step 1: \( \text{C}_2\text{H}_6 + \text{Br}^\bullet \to \text{C}_2\text{H}_5^\bullet + \text{HBr} \) (1 mark) - (ii) Propagation Step 2: \( \text{C}_2\text{H}_5^\bullet + \text{Br}_2 \to \text{C}_2\text{H}_5\text{Br} + \text{Br}^\bullet \) (1 mark) - (ii) Termination Step (to bromoethane): \( \text{C}_2\text{H}_5^\bullet + \text{Br}^\bullet \to \text{C}_2\text{H}_5\text{Br} \) (1 mark) - (iii) Termination to form butane: \( 2\text{C}_2\text{H}_5^\bullet \to \text{C}_4\text{H}_{10} \) (1 mark). (b) [6 marks total] - (i) Same molecular formula (1 mark) - (i) Different structural formula / different connectivity of atoms (1 mark) - (ii) Correct skeletal structure and IUPAC name for 1-chlorobutane (1 mark) - (ii) Correct skeletal structure and IUPAC name for 2-chlorobutane (1 mark) - (ii) Correct skeletal structure and IUPAC name for 1-chloro-2-methylpropane (1 mark) - (ii) Correct skeletal structure and IUPAC name for 2-chloro-2-methylpropane (1 mark).

(a) (i) Step 1: But-1-ene has a high electron density in the \( \pi \)-bond. A curly arrow goes from the double bond of but-1-ene to the \( \text{H} \) atom of the polar \( \text{H}^{\delta+}-\text{Br}^{\delta-} \) molecule. A concurrent curly arrow goes from the \( \text{H-Br} \) bond to the \( \text{Br} \) atom, forming a bromide ion, \( \text{Br}^- \). Intermediate: A secondary carbocation is formed: \( \text{CH}_3-\text{C}^+\text{H}-\text{CH}_2-\text{CH}_3 \). Step 2: A curly arrow goes from the lone pair of the bromide ion, \( \text{Br}^- \), to the positive carbon-2 atom. Product: 2-bromobutane. (ii) The reaction can proceed through either a secondary carbocation (leading to 2-bromobutane), or a primary carbocation (leading to 1-bromobutane). The secondary carbocation is more stable than the primary carbocation because it has two electron-donating alkyl groups attached to the positively charged carbon, which disperse the positive charge more effectively than the single alkyl group attached to the primary carbocation (the inductive effect). (b) (i) Monomer: \( \text{CH}_2=\text{CH-CH}_3 \). Equation: \( n \text{ CH}_2=\text{CH(CH}_3) \to -[-[\text{CH}_2-\text{CH(CH}_3)]-]_n- \). (ii) Poly(propene) has strong, non-polar C-C and C-H single bonds. It is chemically inert and cannot be easily attacked by nucleophiles, water, or microbial enzymes, preventing biodegradation. (c) Reagent: Bromine water (or aqueous bromine). Observation with but-1-ene: The orange/yellow/brown solution is decolorized / turns colorless. Observation with butane: The solution remains orange/yellow/brown (no reaction).

(a) [7 marks total] - (i) Curly arrow from C=C double bond to H of H-Br AND curly arrow from H-Br bond to Br (with partial charges shown) (1 mark) - (i) Correct structure of secondary carbocation intermediate (1 mark) - (i) Curly arrow from lone pair of Br- to the C+ carbocation (1 mark) - (i) Structure of final product 2-bromobutane (1 mark) - (ii) 2-bromobutane is formed via a secondary carbocation (1 mark) - (ii) Secondary carbocation is more stable than primary carbocation (1 mark) - (ii) Due to the electron-donating inductive effect of two alkyl groups (compared to one) (1 mark). (b) [3 marks total] - (i) Propene monomer drawn correctly (1 mark) - (i) Poly(propene) repeat unit drawn correctly with square brackets and 'n' in an equation (1 mark) - (ii) Polymer is inert / has strong C-C / C-H non-polar bonds that cannot be attacked by enzymes (1 mark). (c) [2 marks total] - Reagent: Bromine water / aqueous bromine (1 mark) - Observations: but-1-ene decolorizes bromine water AND butane does not react / remains orange (1 mark).

The equation for the formation of methanol is:
\(\text{C(s)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\)

Using Hess's Law and standard enthalpies of combustion:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
\(\Delta_f H^\theta = [\Delta_c H^\theta(\text{C(s)}) + 2 \times \Delta_c H^\theta(\text{H}_2\text{(g)})] - \Delta_c H^\theta(\text{CH}_3\text{OH(l)})\)
\(\Delta_f H^\theta = [-394 + 2(-286)] - [-726]\)
\(\Delta_f H^\theta = [-394 - 572] + 726 = -966 + 726 = -240\text{ kJ mol}^{-1}\).

1 mark: Correctly calculates \(-240\text{ kJ mol}^{-1}\).

The thermal stability of Group 2 nitrates increases down the group. The magnesium ion (\(\text{Mg}^{2+}\)) has a smaller ionic radius than the barium ion (\(\text{Ba}^{2+}\)) while carrying the same charge. This gives the \(\text{Mg}^{2+}\) ion a higher charge density, allowing it to polarise the electron cloud of the nitrate anion more strongly, weakening the bonds within the nitrate ion and lowering the temperature required for its decomposition.

1 mark: Correctly identifies that the smaller magnesium ion with higher charge density polarises the nitrate ion.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. The C-I bond is the weakest (lowest bond enthalpy) and breaks most easily, leading to the fastest rate of hydrolysis and appearance of the silver halide precipitate. The C-Cl bond is the strongest (highest bond enthalpy) and breaks most slowly. Therefore, the order from fastest to slowest is 1-iodobutane > 1-bromobutane > 1-chlorobutane.

1 mark: Correctly identifies the rate of hydrolysis order based on carbon-halogen bond strength.

The molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) represents an alcohol or an ether. Since it is oxidised by acidified potassium dichromate(VI) (turning orange to green), it must be a primary or secondary alcohol (tertiary alcohols like 2-methylpropan-2-ol do not oxidise). The product of the oxidation does not react with Tollens' reagent (ruling out aldehydes from primary alcohols like butan-1-ol and 2-methylpropan-1-ol) and cannot be further oxidised to a carboxylic acid. Thus, the product must be a ketone, and the original compound \(X\) must be a secondary alcohol, which is butan-2-ol.

1 mark: Correctly identifies the secondary alcohol, butan-2-ol.

The boiling temperature of HF is exceptionally high due to strong hydrogen bonding between its molecules. From HCl to HI, the molecules cannot form hydrogen bonds. The trend from HCl to HI is dominated by the increasing strength of London forces, which increase because of the larger number of electrons per molecule, requiring more thermal energy to overcome.

1 mark: Correctly relates the boiling temperature trend from HCl to HI to the strength of London forces.

A catalyst provides an alternative pathway with a lower activation energy (\(E_a\)). It does not alter the distribution of kinetic energies of the molecules, meaning the Maxwell-Boltzmann distribution curve itself remains completely unchanged at a constant temperature.

1 mark: Correctly identifies that the curve is unchanged but a different pathway with lower \(E_a\) is introduced.

The oxidation states of sulfur in each species are:
- \(\text{H}_2\text{S}: -2\)
- \(\text{S}: 0\)
- \(\text{SO}_2: +4\)
- \(\text{H}_2\text{SO}_4: +6\)

Therefore, the order from lowest to highest oxidation state is \(\text{H}_2\text{S} < \text{S} < \text{SO}_2 < \text{H}_2\text{SO}_4\).

1 mark: Correctly orders the sulfur species by oxidation state.

Total mass of solution \(m = 50.0 + 50.0 = 100.0\text{ g}\).
Heat energy released, \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.50\text{ K} = 2717\text{ J} = 2.717\text{ kJ}\).
Moles of water formed = \(n(\text{H}^+) = 50.0 \times 10^{-3}\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\).
\(\Delta H = -\frac{q}{n} = -\frac{2.717\text{ kJ}}{0.0500\text{ mol}} = -54.34\text{ kJ mol}^{-1}\).
To 3 significant figures, this is \(-54.3\text{ kJ mol}^{-1}\).

1 mark: Correctly calculates \(-54.3\text{ kJ mol}^{-1}\).

The reaction for the formation of cyclopropane is: \(3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_6(\text{g})\). Using Hess's Law with enthalpies of combustion: \(\Delta H_\text{f}^\ominus = \sum \Delta H_\text{c}^\ominus(\text{reactants}) - \sum \Delta H_\text{c}^\ominus(\text{products})\). Therefore, \(\Delta H_\text{f}^\ominus = [3 \times (-394) + 3 \times (-286)] - (-2091) = [-1182 - 858] + 2091 = -2040 + 2091 = +51\text{ kJ mol}^{-1}\).

Correct answer is B (1 mark).

As Group 2 is descended, the ionic radius of the cation increases while its charge remains \(+2\). Therefore, the charge density and polarizing power of the cation decrease. The cation causes less distortion of the electron cloud of the nitrate anion, making the nitrate more thermally stable and requiring higher temperatures to decompose.

Correct answer is A (1 mark).

Iodoalkanes react faster than chloroalkanes because the C-I bond is weaker and has a lower bond enthalpy than the C-Cl bond. Furthermore, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react much faster than primary halogenoalkanes (such as 1-iodobutane) because they undergo substitution via the \(S_\text{N}1\) mechanism, which involves the formation of a stable tertiary carbocation intermediate.

Correct answer is D (1 mark).

Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) contains a highly polar O-H bond, allowing it to form intermolecular hydrogen bonds. Hydrogen bonds are significantly stronger than the permanent dipole-dipole forces in 1-fluoropropane and methoxyethane, and the London forces in butane. Thus, more energy is required to overcome these intermolecular forces, resulting in the highest boiling temperature.

Correct answer is B (1 mark).

A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_\text{a}\)). It does not affect the temperature or the kinetic energy of the gas molecules. Therefore, the shape of the Maxwell-Boltzmann distribution curve remains exactly the same, but the activation energy threshold line shifts to the left, indicating a greater fraction of molecules have energy greater than or equal to the new activation energy.

Correct answer is C (1 mark).

In the reaction of KI with concentrated sulfuric acid, iodide ions (\(\text{I}^-\)) act as a strong reducing agent and reduce the sulfur in sulfuric acid from an oxidation state of +6 to +4 (in \(\text{SO}_2\)), 0 (in \(\text{S}\)), and -2 (in \(\text{H}_2\text{S}\)). Iodine (\(\text{I}_2\)) is an oxidation product, and hydrogen iodide (\(\text{HI}\)) and potassium hydrogensulfate (\(\text{KHSO}_4\)) are products of an acid-base reaction, not redox products.

Correct answer is A (1 mark).

Secondary alcohols are oxidized to ketones. Since ketones cannot be oxidized further, the mixture is heated under reflux to ensure complete reaction and then separated using fractional distillation. Primary alcohols oxidize to aldehydes (using immediate distillation) or carboxylic acids (under reflux). Tertiary alcohols do not undergo oxidation under these conditions.

Correct answer is B (1 mark).

Total mass of solution \(m = 50.0 + 50.0 = 100.0\text{ g}\). Heat released \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 6.50\ ^\circ\text{C} = 2717\text{ J} = 2.717\text{ kJ}\). Number of moles of water formed \(n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). Enthalpy change of neutralisation \(\Delta H = -\frac{q}{n} = -\frac{2.717\text{ kJ}}{0.0500\text{ mol}} = -54.34\text{ kJ mol}^{-1}\), which rounds to \(-54.3\text{ kJ mol}^{-1}\).

Correct answer is A (1 mark).

The equation for the formation of propan-1-ol from its elements in their standard states is:
\(3\text{C(s)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\)

Using a Hess's cycle based on combustion products:
\(\Delta_{\text{f}}H^{\ominus} = \sum \Delta_{\text{c}}H^{\ominus}(\text{reactants}) - \sum \Delta_{\text{c}}H^{\ominus}(\text{products})\)

\(\Delta_{\text{f}}H^{\ominus} = [3 \times \Delta_{\text{c}}H^{\ominus}(\text{C}) + 4 \times \Delta_{\text{c}}H^{\ominus}(\text{H}_2)] - [\Delta_{\text{c}}H^{\ominus}(\text{C}_3\text{H}_7\text{OH})]\)

\(\Delta_{\text{f}}H^{\ominus} = [3 \times (-394) + 4 \times (-286)] - [-2021]\)

\(\Delta_{\text{f}}H^{\ominus} = [-1182 - 1144] + 2021\)

\(\Delta_{\text{f}}H^{\ominus} = -2326 + 2021 = -305\text{ kJ mol}^{-1}\)

1 mark: Correct calculation of the standard enthalpy of formation with the correct sign (-305 kJ mol^-1).

The rate of hydrolysis of halogenoalkanes depends on bond enthalpy and mechanism:
1. The C-I bond is weaker than the C-Cl bond, so iodoalkanes react faster than chloroalkanes.
2. Tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) undergo rapid hydrolysis via an \(S_N1\) pathway because they form a highly stable tertiary carbocation intermediate, whereas primary halogenoalkanes react much slower via the \(S_N2\) pathway.

Therefore, 2-iodo-2-methylpropane is the most reactive.

1 mark: Correctly identifies 2-iodo-2-methylpropane as the fastest reacting halogenoalkane (D).

Group 2 nitrates undergo thermal decomposition when heated to produce the metal oxide, nitrogen dioxide gas, and oxygen gas.
The balanced chemical equation is:
\(2\text{Ba(NO}_3)_2\text{(s)} \rightarrow 2\text{BaO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\)

Option B represents the decomposition pattern of Group 1 nitrates (except lithium).

1 mark: Correct balanced chemical equation for the thermal decomposition of barium nitrate (A).

All three molecules have similar molar masses (~58–60 g/mol), so the strength of their London forces is comparable, but their other intermolecular forces differ:
1. Butane is non-polar and only experiences weak London forces.
2. Propanal is a polar molecule containing a carbonyl group (\(\text{C}=\text{O}\)), so it experiences permanent dipole-dipole forces in addition to London forces.
3. Propan-1-ol contains a highly polar \(\text{-OH}\) group capable of forming strong intermolecular hydrogen bonds.

Since hydrogen bonds are stronger than permanent dipole-dipole forces, which in turn are stronger than London forces, the boiling temperature increases in the order: butane < propanal < propan-1-ol.

1 mark: Correct order (A) based on the relative strengths of London forces, permanent dipole-dipole forces, and hydrogen bonding.

(a) Heating is required to decompose the solid, so the heat supplied cannot be separated from the heat of the reaction. Also, the decomposition may be incomplete.

(b) Heat absorbed: \(q = m \cdot c \cdot \Delta T = 50.0 \times 4.18 \times 5.8 = 1212.2\text{ J} = 1.2122\text{ kJ}\).
Moles of \(NaHCO_3 = \frac{3.36}{84.0} = 0.0400\text{ mol}\).
\(\Delta H_1 = +\frac{1.2122}{0.0400} = +30.3\text{ kJ mol}^{-1}\).

(c)(i) Hess's Law Cycle:
Top-left: \(2NaHCO_3(s)\)
Top-right: \(Na_2CO_3(s) + H_2O(l) + CO_2(g)\)
Bottom: \(2NaCl(aq) + 2H_2O(l) + 2CO_2(g)\)
Arrow from Top-left to Bottom labelled \(2 \times \Delta H_1\).
Arrow from Top-right to Bottom labelled \(\Delta H_2\).
Arrow from Top-left to Top-right labelled \(\Delta H_r\).

(ii) From the cycle: \(\Delta H_r + \Delta H_2 = 2\Delta H_1\)
\(\Delta H_r = 2\Delta H_1 - \Delta H_2 = 2(+30.3) - (-28.4) = +60.6 + 28.4 = +89.0\text{ kJ mol}^{-1}\).

(d) Any two of: the specific heat capacity of the acid is assumed to be 4.18 but is different; the mass of the solid was ignored in the total mass of the reacting mixture; some carbon dioxide gas escapes carrying some water vapor and heat with it; some heat is absorbed by the calorimeter/cup/thermometer.

(a)
M1: Heating is required to decompose the solid, so direct temperature change of the reaction cannot be measured. (1)
M2: The reaction may be incomplete. (1)

(b)
M1: Calculate heat change: \(q = 50.0 \times 4.18 \times 5.8 = 1212.2\text{ J}\) (1)
M2: Calculate moles of NaHCO3: \(\frac{3.36}{84.0} = 0.0400\text{ mol}\) (1)
M3: Calculate enthalpy change: \(\frac{1.2122}{0.0400} = 30.3\text{ kJ mol}^{-1}\) (1)
M4: Correct sign and 3 significant figures: \(+30.3\text{ kJ mol}^{-1}\) (1)

(c)(i)
M1: Draw cycle with three correct corners containing all correct species. (1)
M2: Correct arrow directions and coefficients (LHS arrow is \(2\Delta H_1\), RHS arrow is \(\Delta H_2\), top arrow is \(\Delta H_r\)). (1)

(c)(ii)
M1: Use Hess's Law relation: \(\Delta H_r = 2\Delta H_1 - \Delta H_2\) (1)
M2: Correct calculation: \(2(+30.3) - (-28.4) = +89.0\text{ kJ mol}^{-1}\) (1)

(d)
M1 & M2: Any two reasons (1 mark each):
- Specific heat capacity of the solution is assumed to be that of pure water (4.18 J g-1 °C-1) but it contains dissolved ions.
- The mass of the solid (3.36 g) added to the cup was neglected in the heat capacity mass calculation.
- Carbon dioxide gas escapes, carrying water vapor or heat.
- Heat absorbed by the polystyrene cup and thermometer was not included.

(a)(i) \(KCl(s) + H_2SO_4(l) \rightarrow KHSO_4(s) + HCl(g)\) (or \(2KCl(s) + H_2SO_4(l) \rightarrow K_2SO_4(s) + 2HCl(g)\))
(ii) Misty/steamy fumes of hydrogen chloride gas, \(HCl(g)\).

(b)(i) Dark grey solid: Iodine, \(I_2(s)\)
Yellow solid: Sulfur, \(S(s)\)
Gas with bad egg smell: Hydrogen sulfide, \(H_2S(g)\)

(ii) Iodide ions (\(I^-\)) are larger and have more shielding than chloride ions (\(Cl^-\)), so the outer electrons are less strongly attracted to the nucleus and are lost more easily, making iodide a much stronger reducing agent than chloride. Chloride ions are not strong enough to reduce sulfuric acid, so only an acid-base displacement occurs.

(iii) The sulfur is reduced from +6 in \(H_2SO_4\) to -2 in \(H_2S\). This is an 8-electron reduction:
\(H_2SO_4 + 8H^+ + 8e^- \rightarrow H_2S + 4H_2O\) (or \(SO_4^{2-} + 10H^+ + 8e^- \rightarrow H_2S + 4H_2O\)).

(a)(i)
M1: Correct formulae: \(KCl + H_2SO_4 \rightarrow KHSO_4 + HCl\) (1)
M2: Correct state symbols: \(KCl(s) + H_2SO_4(l) \rightarrow KHSO_4(s) + HCl(g)\) (1)

(a)(ii)
M1: Misty/steamy white fumes of hydrogen chloride (1)

(b)(i)
M1: Dark grey solid: Iodine / \(I_2\) (1)
M2: Yellow solid: Sulfur / \(S\) (1)
M3: Gas with bad egg smell: Hydrogen sulfide / \(H_2S\) (1)

(b)(ii)
M1: Iodide is a stronger reducing agent than chloride. (1)
M2: Iodide ions are larger and have more shielding, so their outer electrons are lost more easily. (1)
M3: Chloride ions are not strong enough to reduce sulfuric acid (no change in oxidation state of S). (1)

(b)(iii)
M1: Correct species: \(H_2SO_4\) on LHS, \(H_2S\) and \(H_2O\) on RHS. (1)
M2: Correct balancing of H and electrons: \(8H^+\) and \(8e^-\) on LHS. (1)
M3: Fully balanced equation: \(H_2SO_4 + 8H^+ + 8e^- \rightarrow H_2S + 4H_2O\) (1)

(a)(i) \((CH_3)_3CBr + KOH \rightarrow (CH_3)_3COH + KBr\).
Product name: 2-methylpropan-2-ol.

(ii) Mechanism:
Step 1: Draw the structure of 2-bromo-2-methylpropane. Draw a curly arrow starting from the C-Br bond to the Br atom. This forms the tertiary carbocation intermediate, \((CH_3)_3C^+\), and a bromide ion, \(Br^-\).
Step 2: Draw the tertiary carbocation intermediate. Draw a curly arrow from a lone pair on the oxygen of the hydroxide ion (\(:\text{OH}^-\)) to the positive carbon atom.
Product: 2-methylpropan-2-ol.

(b)(i) Base / proton acceptor.
(ii) Structural formula: \((CH_3)_2C=CH_2\). Name: methylpropene (or 2-methylpropene).

(c)(i) O-H bond (alcohol); wavenumber range: 3200 - 3600 cm\(^{-1}\) (or 3200 - 3750 cm\(^{-1}\)).
(ii) The fingerprint region (below 1500 cm\(^{-1}\)) contains a unique, complex pattern of peaks for each compound. Comparing the spectrum to a database of known reference spectra to find an exact match confirms the compound's identity.

(a)(i)
M1: Correct organic structural formulae equation: \((CH_3)_3CBr + KOH \rightarrow (CH_3)_3COH + KBr\) (1)
M2: Product name: 2-methylpropan-2-ol (1)

(a)(ii)
M1: Curly arrow starting from C-Br bond to Br. (1)
M2: Correct carbocation intermediate: \((CH_3)_3C^+\). (1)
M3: Curly arrow from a lone pair on the oxygen of \(OH^-\) to the positive carbon of the carbocation. (1)
M4: Correctly drawn product structure. (1)

(b)(i)
M1: Base / proton acceptor (1)

(b)(ii)
M1: Structural formula of methylpropene: \((CH_3)_2C=CH_2\) (or showing double bond) (1)
M2: Name: methylpropene / 2-methylpropene (1)

(c)(i)
M1: O-H bond (alcohol) (1)
M2: 3200 - 3600 cm\(^{-1}\) (or 3200 - 3750 cm\(^{-1}\)) (1)

(c)(ii)
M1: Compare fingerprint region (below 1500 cm-1) to database of known spectra to find exact match. (1)

(a) Propane has only London forces between its molecules.
Fluoromethane has London forces and permanent dipole-dipole forces.
Methanol has London forces, permanent dipole-dipole forces, and hydrogen bonding.
Hydrogen bonds are significantly stronger than permanent dipole-dipole and London forces, requiring much more energy to overcome, which is why methanol has the highest boiling temperature (+65 \(^\circ\)C).
Propane has a higher boiling temperature than fluoromethane (-42 \(^\circ\)C vs -78 \(^\circ\)C) because propane has more electrons (26 vs 18). This results in stronger London forces in propane that require more energy to break than the combined London forces and permanent dipole-dipole forces in fluoromethane.

(b) Methanol can form hydrogen bonds with water molecules. The energy released when these new hydrogen bonds form is similar to the energy required to break the hydrogen bonds between water molecules and those between methanol molecules. Chloromethane only has London forces and permanent dipole-dipole forces and cannot form hydrogen bonds with water. The weak interactions formed between chloromethane and water cannot overcome the strong hydrogen bonds between water molecules.

(c) Water molecules can form up to two hydrogen bonds per molecule on average because each water molecule has two hydrogen atoms and two lone pairs on oxygen. Methanol can only form one hydrogen bond per molecule on average (due to only having one O-H hydrogen). Thus, water has a more extensive hydrogen-bonded network requiring more energy to break.

(a)
M1: Propane has London forces (only); Fluoromethane has London and permanent dipole-dipole forces; Methanol has London, permanent dipole-dipole, and hydrogen bonds. (1)
M2: Hydrogen bonding is the strongest of these intermolecular forces and requires the most energy to break. (1)
M3: Propane has a higher boiling point than fluoromethane because propane has more electrons (26 vs 18). (1)
M4: This results in stronger/more significant London forces in propane. (1)
M5: These stronger London forces in propane require more energy to overcome than the combined London and dipole-dipole forces in fluoromethane. (1)
M6: Logical explanation connecting all forces to the order of boiling temperatures: Methanol > Propane > Fluoromethane. (1)

(b)
M1: Methanol forms hydrogen bonds with water molecules. (1)
M2: The energy released by forming new hydrogen bonds compensates for breaking existing ones in pure liquids. (1)
M3: Chloromethane cannot form hydrogen bonds with water (only dipole-dipole/London forces). (1)
M4: The interactions formed between chloromethane and water are too weak to overcome the strong hydrogen bonds between water molecules. (1)

(c)
M1: Water can form more hydrogen bonds per molecule on average (up to two). (1)
M2: Methanol can only form one hydrogen bond per molecule on average. (1)

(a)(i) A yellow precipitate of sulfur forms, which gradually makes the solution opaque, obscuring a black cross marked on a piece of paper underneath the reaction flask.
(ii) An increase in temperature increases the average kinetic energy of the particles, so they move faster and collide more frequently. Crucially, a much larger fraction of the colliding particles have energy greater than or equal to the activation energy (\(E_a\)), leading to a significantly higher frequency of successful collisions.

(b) The Maxwell-Boltzmann distribution:
- x-axis: Kinetic energy / Energy
- y-axis: Number of molecules / Fraction of molecules
- Curve \(T_1\): Starts at origin, rises to a peak, and then decreases asymmetrically towards the x-axis, never touching it.
- Curve \(T_2\): Starts at origin, has a lower peak shifted to the right, and crosses the \(T_1\) curve once.
- Activation energy \(E_a\) marked as a vertical line on the right side of the peak.
- Shaded area under the \(T_2\) curve to the right of the \(E_a\) line.

(c)(i) Increasing the temperature shifts the equilibrium to the right (towards \(NO_2\)) because the forward reaction is endothermic (\(\Delta H\) is positive). This shift causes the mixture to become a darker brown.
(ii) Increasing the pressure shifts the equilibrium to the left (towards \(N_2O_4\)) because the reactant side has fewer moles of gas (1 mole on LHS vs 2 moles on RHS). This shift causes the mixture to become lighter/paler brown.

(a)(i)
M1: Formation of a yellow precipitate that obscures a black cross. (1)

(a)(ii)
M1: Particles have more kinetic energy / move faster, leading to more frequent collisions. (1)
M2: A much larger fraction of particles have energy greater than or equal to the activation energy (\(E \ge E_a\)). (1)
M3: Consequently, the frequency of successful collisions increases. (1)

(b)
M1: Correctly labelled axes: x-axis = Energy/Kinetic Energy, y-axis = Number/Fraction of molecules. (1)
M2: Curve \(T_1\) starts at the origin, rises to a peak, and asymptotes to the x-axis, and curve \(T_2\) has a lower peak shifted to the right. (1)
M3: Activation energy \(E_a\) correctly labelled on the x-axis to the right of the peaks. (1)
M4: Area under the \(T_2\) curve to the right of \(E_a\) shaded. (1)

(c)(i)
M1: Equilibrium shifts to the right because the forward reaction is endothermic. (1)
M2: The mixture becomes darker brown. (1)

(c)(ii)
M1: Equilibrium shifts to the left because there are fewer moles of gas on the LHS (1 mole of gas vs 2 moles of gas on the RHS). (1)
M2: The mixture becomes lighter/paler brown. (1)

**(a)** Dilute sulfuric acid is used to prevent the hydrolysis/oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) by dissolved oxygen, and to provide the \(\text{H}^+\) ions required for the reduction of \(\text{MnO}_4^-\).

**(b)**
1. Dissolve the weighed sample completely in a small volume of dilute sulfuric acid/deionised water in a beaker, stirring with a clean glass rod.
2. Pour the solution carefully into a 250 cm\(^3\) volumetric flask using a clean funnel.
3. Rinse the beaker, glass rod, and funnel several times with deionised water, adding all washings to the flask.
4. Add deionised water to the flask until the bottom of the meniscus is exactly on the graduation line, then stopper the flask and invert it several times to mix thoroughly.

**(c)** Colorless to permanent pale pink.

**(d)**
- Moles of \(\text{MnO}_4^-\in 25.00\text{ cm}^3\):
\(n(\text{MnO}_4^-) = 0.0200 \times \frac{25.00}{1000} = 5.00 \times 10^{-4}\text{ mol}\)
- Moles of \(\text{Fe}^{2+}\in 25.00\text{ cm}^3\):
\(n(\text{Fe}^{2+}) = 5 \times (5.00 \times 10^{-4}) = 2.50 \times 10^{-3}\text{ mol}\)
- Moles of \(\text{Fe}^{2+}\in 250.0\text{ cm}^3\):
\(2.50 \times 10^{-3} \times 10 = 0.0250\text{ mol}\)
- Moles of hydrated salt \(\text{(NH}_4\text{)}_2\text{Fe(SO}_4\text{)}_2 \cdot x\text{H}_2\text{O} = 0.0250\text{ mol}\)
- Molar mass of the hydrated salt:
\(M_r = \frac{9.80}{0.0250} = 392.0\text{ g mol}^{-1}\)
- Formula mass of anhydrous \(\text{(NH}_4\text{)}_2\text{Fe(SO}_4\text{)}_2\):
\(M_r(\text{anhydrous}) = 2 \times [14.0 + (4 \times 1.0)] + 55.8 + 2 \times [32.1 + (4 \times 16.0)] = 36.0 + 55.8 + 192.2 = 284.0\text{ g mol}^{-1}\)
- Mass of water of crystallisation:
\(18.0x = 392.0 - 284.0 = 108.0\text{ g mol}^{-1}\)
\(x = \frac{108.0}{18.0} = 6\)

**(e)** If some \(\text{Fe}^{2+}\) was oxidised to \(\text{Fe}^{3+}\), a smaller volume of \(\text{KMnO}_4\) would be needed for the titration. This would result in a lower calculated number of moles of \(\text{Fe}^{2+}\) (and thus salt), leading to a higher calculated molar mass of the salt, and therefore an overestimate (higher value) of \(x\).

**(a)**
- 1 Mark: To prevent the hydrolysis/oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) OR to supply \(\text{H}^+\) ions for the redox reaction.

**(b)**
- 1 Mark: Dissolve the solid in a beaker using a smaller volume of acid/water and stir with a glass rod.
- 1 Mark: Quantitative transfer: rinse beaker, stirring rod, and funnel, transferring all washings to the volumetric flask.
- 1 Mark: Fill with deionised water until the bottom of the meniscus is aligned with the graduation mark.
- 1 Mark: Stopper and invert the flask several times to ensure homogeneity.

**(c)**
- 1 Mark: Colorless to permanent pale pink (reject 'clear' to pink).

**(d)**
- 1 Mark: Calculate moles of \(\text{MnO}_4^-\in 25.00\text{ cm}^3 = 5.00 \times 10^{-4}\text{ mol}\).
- 1 Mark: Calculate moles of \(\text{Fe}^{2+}\) in 25.00 cm\(^3\) using 1:5 ratio = \(2.50 \times 10^{-3}\text{ mol}\).
- 1 Mark: Calculate moles of salt in 250.0 cm\(^3\) = \(0.0250\text{ mol}\).
- 1 Mark: Calculate the experimental molar mass of the hydrated salt = \(392.0\text{ g mol}^{-1}\).
- 1 Mark: Calculate the formula mass of anhydrous salt = \(284.0\text{ g mol}^{-1}\).
- 0.5 Mark: Solve for \(x = 6\).

**(e)**
- 1 Mark: Calculated value of \(x\) would be higher (overestimated) because a smaller titre leads to smaller moles of salt calculated, yielding a larger calculated \(M_r\).

**(a)**
The diagram should show:
- A pear-shaped or round-bottomed flask containing reaction mixture, with a heat source below (electric mantle/water bath preferred due to flammability).
- A Liebig condenser fitted vertically into the flask neck.
- Water flow: water in at the bottom nozzle, water out at the top nozzle.
- The top of the condenser must remain open (no stopper).

**(b)**
- Add the crude product into the separating funnel, then add sodium carbonate/sodium hydrogencarbonate solution.
- Stopper the funnel, invert it, and shake it gently, releasing the built-up pressure by opening the tap intermittently.
- Allow the mixture to separate into two layers; then run off and discard the lower aqueous layer, collecting the upper organic (cyclohexene) layer.

**(c)**
**(i)** Suitable drying agent: Anhydrous calcium chloride / anhydrous magnesium sulfate / anhydrous sodium sulfate.
The liquid is completely dry when it changes from cloudy/turbid to completely clear/transparent.
**(ii)** Remove the drying agent by gravity filtration or decanting.

**(d)**
**(i)** The bulb of the thermometer should be placed at the junction/level of the side-arm opening of the distillation head to ensure it is fully bathed in the vapour distilling over.
**(ii)** Reagent: Bromine water.
Observation: Orange/yellow/brown solution decolourises (becomes colourless).
**(iii)** \(\text{C}_6\text{H}_{10} + \text{Br}_2 \rightarrow \text{C}_6\text{H}_{10}\text{Br}_2\)

**(a)**
- 1 Mark: Flask and vertical condenser with correct joint connection (no gaps, not sealed at top).
- 1 Mark: Correct water jacket labelling showing water entering at the bottom and leaving at the top.
- 1 Mark: Safe heating source labelled (e.g. heating mantle, water bath, or sand bath; reject Bunsen burner directly heating flammable mixture).

**(b)**
- 1 Mark: Add sodium hydrogencarbonate / sodium carbonate solution to neutralise acid.
- 1 Mark: Shake and invert the separating funnel, releasing pressure at intervals.
- 1 Mark: Allow layers to separate, then run off and discard the aqueous layer (retaining the organic layer).

**(c)**
- **(i)** 1 Mark: Anhydrous \(\text{CaCl}_2\) / \(\text{MgSO}_4\) / \(\text{Na}_2\text{SO}_4\) (reject anhydrous copper(II) sulfate or cobalt(II) chloride).
- 1 Mark: Liquid turns from cloudy to clear.
- **(ii)** 1 Mark: Filtration / decantation.

**(d)**
- **(i)** 1 Mark: Level with the side-arm of the T-piece/distillation head.
- **(ii)** 1 Mark: Bromine water / aqueous bromine (accept acidified \(\text{KMnO}_4\)).
- 0.5 Mark: Orange/yellow/brown to colourless (purple to colourless for \(\text{KMnO}_4\)). Reject 'clear'.
- **(iii)** 1 Mark: Correct structural/molecular equation: \(\text{C}_6\text{H}_{10} + \text{Br}_2 \rightarrow \text{C}_6\text{H}_{10}\text{Br}_2\) (accept skeletal or structural formulas).

**(a)** Cation: Calcium ion, \(\text{Ca}^{2+}\) (brick-red flame test).

**(b)**
**(i)** Anion: Bromide ion, \(\text{Br}^-\).
**(ii)** Add dilute aqueous ammonia: the cream precipitate does not dissolve.
Add concentrated aqueous ammonia: the cream precipitate dissolves to form a colourless solution.
**(iii)** \(\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)\)

**(c)**
**(i)** Carbonate ion, \(\text{CO}_3^{2-}\) (or hydrogencarbonate, \(\text{HCO}_3^-\)).
**(ii)** \(\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)\)
**(iii)** Hydrochloric acid contains chloride ions (\(\text{Cl}^-\)), which would react with the silver nitrate solution to form a white precipitate of silver chloride (\(\text{AgCl}\)), masking the test for the halide ion originally present in the sample.

**(d)**
- Carbonate ions react with silver ions to form a white precipitate of silver carbonate, \(\text{Ag}_2\text{CO}_3(s)\) (\(\text{2Ag}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{Ag}_2\text{CO}_3(s)\)), which would interfere with the identification of the bromide precipitate.
- Adding dilute nitric acid decomposes the carbonate ions into carbon dioxide and water before silver nitrate is added, preventing the formation of silver carbonate.

**(a)**
- 1 Mark: Calcium (ion) / \(\text{Ca}^{2+}\).

**(b)**
- **(i)** 1 Mark: \(\text{Br}^-\).
- **(ii)** 1 Mark: Does not dissolve in dilute ammonia.
- 1 Mark: Dissolves in concentrated ammonia.
- **(iii)** 1.5 Marks: \(\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)\) (1 mark for correct species, 0.5 mark for state symbols).

**(c)**
- **(i)** 1 Mark: Carbonate / \(\text{CO}_3^{2-}\) (accept hydrogencarbonate / \(\text{HCO}_3^-\)).
- **(ii)** 1.5 Marks: \(\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)\) (1 mark for correct balanced equation, 0.5 mark for state symbols).
- **(iii)** 1.5 Marks: HCl contains \(\text{Cl}^-\) ions (1 mark) which react with \(\text{Ag}^+\) to form a white precipitate of \(\text{AgCl}\) (0.5 mark).

**(d)**
- 1 Mark: Carbonate reacts with \(\text{Ag}^+\) to form a white precipitate of silver carbonate / \(\text{Ag}_2\text{CO}_3\).
- 1 Mark: Nitric acid reacts with and destroys/removes carbonate ions.
- 1 Mark: This prevents the formation of \(\text{Ag}_2\text{CO}_3\), ensuring any precipitate formed is solely due to the halide.

**(a)**
**(i)** To ensure both solutions are at the same temperature before mixing, or to calculate an accurate mean initial temperature.
**(ii)** Polystyrene is an excellent thermal insulator, which minimises heat transfer to/from the surroundings, and has a negligible heat capacity.
**(iii)** To ensure that all of the hydrochloric acid reacts completely (so that hydrochloric acid is the limiting reagent).

**(b)**
**(i)**
- Total volume of mixture = \(50.0 + 50.0 = 100.0\text{ cm}^3\).
- Mass of mixture, \(m = 100.0\text{ g}\).
- Temperature change, \(\Delta T = 26.9 - 20.2 = 6.7\text{ }^\circ\text{C}\).
- Heat energy released, \(q = m c \Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J}\) (or 2.80 kJ).

**(ii)**
- Moles of \(\text{HCl}\) used = \(1.00 \times \frac{50.0}{1000} = 0.0500\text{ mol}\).
- Moles of \(\text{NaOH}\) used = \(1.05 \times \frac{50.0}{1000} = 0.0525\text{ mol}\).
- \(\text{HCl}\) is the limiting reagent, so moles of water formed = 0.0500 mol.
- \(\Delta H_{neut} = -\frac{q}{n} = -\frac{2800.6\text{ J}}{0.0500\text{ mol}} = -56012\text{ J mol}^{-1} = -56.0\text{ kJ mol}^{-1}\) (to 3 significant figures).

**(c)**
**(i)**
- Heat is still lost to the surroundings (even with a polystyrene cup).
- The specific heat capacity of the solution is assumed to be that of pure water, but it is actually slightly different due to dissolved ions.
- The heat capacity of the polystyrene cup/thermometer is neglected.
**(ii)** The temperature rise would be smaller. Ethanoic acid is a weak acid and is only partially dissociated. Some of the heat energy released during neutralisation is absorbed to dissociate the remaining undissociated ethanoic acid molecules.

**(a)**
- **(i)** 1 Mark: To establish a reliable starting temperature (or to check they are at the same temperature/calculate a mean temperature).
- **(ii)** 1 Mark: Polystyrene is a good thermal insulator / reduces heat loss to surroundings.
- 0.5 Mark: Polystyrene has low heat capacity.
- **(iii)** 1 Mark: To ensure hydrochloric acid is the limiting reactant (or that all acid is completely neutralised).

**(b)**
- **(i)** 1 Mark: Mass of solution = 100.0 g.
- 0.5 Mark: \(\Delta T = 6.7\text{ }^\circ\text{C}\).
- 1 Mark: \(q = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J}\) (accept 2800 J or 2.80 kJ).
- **(ii)** 1 Mark: Moles of \(\text{HCl}\) = 0.0500 mol (identifying it as the limiting reagent).
- 1 Mark: Divide heat by moles of limiting reagent: \(\frac{2.8006}{0.0500}\).
- 1 Mark: Correct calculation of value (56.0) and negative sign.
- 0.5 Mark: Expressed to 3 significant figures with units: \(-56.0\text{ kJ mol}^{-1}\) (accept \(-56000\text{ J mol}^{-1}\)).

**(c)**
- **(i)** 2 Marks: Any two from:
- Heat loss to the surroundings / beaker.
- Specific heat capacity of the solution is less than pure water.
- Heat capacity of the thermometer/cup was ignored.
- **(ii)** 1 Mark: Temperature rise is lower/smaller because ethanoic acid is a weak acid / only partially dissociated.
- 0.5 Mark: Heat/energy is required to dissociate the ethanoic acid.

For a second-order reaction overall, the rate equation is: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). Rearranging for the rate constant gives: \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}\). Substituting the units: \(k = \frac{\text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}}{(\text{mol}\ \text{dm}^{-3})(\text{mol}\ \text{dm}^{-3})} = \text{dm}^3\ \text{mol}^{-1}\ \text{s}^{-1}\).

[1 mark] B is the correct option.

\(\Delta S^\ominus_{\text{system}} = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\). Substituting the values: \(\Delta S^\ominus_{\text{system}} = (39.7 + 213.6) - 92.9 = +160.4\ \text{J}\ \text{mol}^{-1}\ \text{K}^{-1}\).

[1 mark] C is the correct option.

Since the forward reaction is endothermic, increasing the temperature shifts the equilibrium in the direction of the endothermic reaction, which is the forward direction. This increases the yield of \(\text{NO}_2(\text{g})\). Since the ratio of products to reactants at equilibrium increases, the value of \(K_p\) also increases.

[1 mark] D is the correct option.

Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\). First, calculate \(\text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76\). Since the volumes are equal, the ratio of concentrations is equal to the ratio of moles: \(\text{pH} = 4.76 + \log_{10}\left(\frac{0.0500}{0.100}\right) = 4.76 + \log_{10}(0.5) = 4.76 - 0.30 = 4.46\).

[1 mark] A is the correct option.

The carbonyl carbon atom in propanal is planar. The nucleophile \(\text{CN}^-\)\ can attack this planar carbon with equal probability from above or below the plane. This yields an equimolar mixture of both optical isomers (a racemic mixture), which is optically inactive.

[1 mark] B is the correct option.

The iodoform reaction is characteristic of compounds containing a methyl carbonyl group (\(\text{CH}_3\text{CO}-\)) or a methyl hydroxymethyl group (\(\text{CH}_3\text{CH(OH)}-\)). Butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) contains a methyl carbonyl group, so it reacts to form a yellow precipitate of triiodomethane (\(\text{CHI}_3\)).

[1 mark] C is the correct option.

From the Arrhenius equation: \(\text{gradient} = -\frac{E_a}{R}\). Therefore: \(E_a = -\text{gradient} \times R = -(-6.50 \times 10^3\ \text{K}) \times 8.31\ \text{J}\ \text{mol}^{-1}\ \text{K}^{-1} = +54015\ \text{J}\ \text{mol}^{-1} = +54.0\ \text{kJ}\ \text{mol}^{-1}\).

[1 mark] A is the correct option.

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). \(\text{H}_2\text{CO}_3\) is the conjugate acid of the conjugate base \(\text{HCO}_3^-\).

[1 mark] D is the correct option.

Let the original rate be \(\text{Rate}_1 = k[\text{A}][\text{B}]^2\). When \([\text{A}]\) is doubled to \(2[\text{A}]\) and \([\text{B}]\) is halved to \(0.5[\text{B}]\), the new rate is: \(\text{Rate}_2 = k(2[\text{A}])(0.5[\text{B}])^2 = k \cdot 2[\text{A}] \cdot 0.25[\text{B}]^2 = 0.5 k[\text{A}][\text{B}]^2 = 0.5 \text{Rate}_1\). Therefore, the rate decreases by a factor of 2.

1 mark for the correct option (B).

Using the thermodynamic relationship: \(\Delta S_{\text{surroundings}}^{\ominus} = -\frac{\Delta H^{\ominus}}{T} = -\frac{-120000\text{ J mol}^{-1}}{298\text{ K}} = +402.7\text{ J K}^{-1}\text{ mol}^{-1}\). Now, calculate total entropy change: \(\Delta S_{\text{total}}^{\ominus} = \Delta S_{\text{system}}^{\ominus} + \Delta S_{\text{surroundings}}^{\ominus} = -350 + 402.7 = +52.7\text{ J K}^{-1}\text{ mol}^{-1} \approx +53\text{ J K}^{-1}\text{ mol}^{-1}\).

1 mark for the correct option (A).

Calculate moles: \(n(\text{HCOOH}) = 0.0500 \times 0.100 = 0.00500\text{ mol}\); \(n(\text{HCOO}^-) = 0.0500 \times 0.050 = 0.00250\text{ mol}\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HCOO}^-]}{[\text{HCOOH}]}\right) = -\log_{10}(1.6 \times 10^{-4}) + \log_{10}\left(\frac{0.00250}{0.00500}\right) = 3.80 + \log_{10}(0.5) = 3.80 - 0.30 = 3.50\).

1 mark for the correct option (A).

In 2-methylbutanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOH}\)), the carbon at position 2 is bonded to four different groups: a hydrogen atom, a methyl group (\(-\text{CH}_3\)), an ethyl group (\(-\text{CH}_2\text{CH}_3\)), and a carboxylic acid group (\(-\text{COOH}\)). Therefore, it contains an asymmetric carbon atom (chiral centre) and is optically active.

1 mark for the correct option (B).

The partial pressure of each component is given by the product of its mole fraction and the total pressure: \(p(\text{SO}_3) = zP\), \(p(\text{SO}_2) = xP\), and \(p(\text{O}_2) = yP\). Substituting these into the equilibrium expression: \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 p(\text{O}_2)} = \frac{(zP)^2}{(xP)^2 (yP)} = \frac{z^2 P^2}{x^2 y P^3} = \frac{z^2}{x^2 y P}\).

1 mark for the correct option (A).

According to the Arrhenius equation in linear form, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\), the gradient is equal to \(-\frac{E_a}{R}\). Therefore, \(-\frac{E_a}{R} = -9.50 \times 10^3\text{ K}\), which gives \(E_a = 9.50 \times 10^3\text{ K} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 7.89 \times 10^4\text{ J mol}^{-1} \approx +79.0\text{ kJ mol}^{-1}\).

1 mark for the correct option (B).

Compound X is a carboxylic acid because it reacts with \(\text{Na}_2\text{CO}_3\) to release \(\text{CO}_2\). With four carbon atoms, X must be butanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)). Reduction of a carboxylic acid with \(\text{LiAlH}_4\) in dry ether yields a primary alcohol. Thus, reducing butanoic acid yields butan-1-ol.

1 mark for the correct option (A).

A Brønsted-Lowry acid is a proton donor. In equation (a), \(\text{H}_2\text{O}\) donates a proton to \(\text{NH}_3\) to form \(\text{NH}_4^+\), acting as a Brønsted-Lowry acid. In equations (b), (c), and (d), the first reactant accepts a proton (acts as a Brønsted-Lowry base).

1 mark for the correct option (A).

First, calculate the rate constant, \(k\), using the initial rate and concentrations:
\[\text{rate} = k[\text{A}][\text{B}]^2\]
\[1.2 \times 10^{-3} = k(0.10)(0.20)^2\]
\[1.2 \times 10^{-3} = k(0.10)(0.040) = 0.0040 k\]
\[k = \frac{1.2 \times 10^{-3}}{0.0040} = 0.30 \text{ dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\]

Now, calculate the new rate under the second set of conditions:
\[\text{rate} = 0.30 \times [0.20] \times [0.10]^2\]
\[\text{rate} = 0.30 \times 0.20 \times 0.010 = 6.0 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}\]

1 mark: correct option selected (B)

First, calculate the system entropy change, \(\Delta S_{\text{sys}}^\theta\):
\[\Delta S_{\text{sys}}^\theta = \Sigma S^\theta(\text{products}) - \Sigma S^\theta(\text{reactants})\]
\[\Delta S_{\text{sys}}^\theta = (40 + 214) - 93 = +161 \text{ J K}^{-1} \text{ mol}^{-1}\]

For the reaction to be thermodynamically feasible, the Gibbs free energy change must be less than or equal to zero (\(\Delta G^\theta \le 0\)):
\[\Delta G^\theta = \Delta H^\theta - T\Delta S_{\text{sys}}^\theta \le 0\]
\[T \ge \frac{\Delta H^\theta}{\Delta S_{\text{sys}}^\theta}\]

Convert \(\Delta H^\theta\) from \(\text{kJ mol}^{-1}\) to \(\text{J mol}^{-1}\):
\[\Delta H^\theta = +178 \times 10^3 \text{ J mol}^{-1} = +178000 \text{ J mol}^{-1}\]

Calculate the temperature, \(T\):
\[T \ge \frac{178000}{161} = 1105.6 \text{ K} \approx 1106 \text{ K}\]

1 mark: correct option selected (C)

Use the Henderson-Hasselbalch equation (or rearrange the \(K_a\) expression):
\[\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\]

First, calculate \(\text{p}K_a\):
\[\text{p}K_a = -\log_{10}(1.7 \times 10^{-5}) = 4.77\]

Now, calculate pH:
\[\text{pH} = 4.77 + \log_{10}\left(\frac{0.20}{0.10}\right)\]
\[\text{pH} = 4.77 + \log_{10}(2) = 4.77 + 0.30 = 5.07\]

1 mark: correct option selected (C)

- To react with Tollens' reagent, the compound must be an aldehyde. This rules out 3-methylbutan-2-one (C) because it is a ketone.
- Reduction of an aldehyde by \(\text{LiAlH}_4\) forms a primary alcohol.
- Reduction of 2-methylbutanal (\(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CHO}\)) forms 2-methylbutan-1-ol (\(\text{CH}_3\text{CH}_2\text{C}^*\text{H}(\text{CH}_3)\text{CH}_2\text{OH}\)).
- The C2 carbon in 2-methylbutan-1-ol is chiral (asymmetric) because it is bonded to four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{CH}_2\text{OH}\). Therefore, 2-methylbutan-1-ol is a chiral primary alcohol.
- In contrast, reducing 3-methylbutanal forms 3-methylbutan-1-ol (\((\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{OH}\)), and reducing pentanal forms pentan-1-ol, neither of which has a chiral centre.

1 mark: correct option selected (A)

(a)
- Comparing Run 1 and Run 2: \([\text{B}]\) is kept constant, \([\text{A}]\) doubles, and the rate quadruples (\(4.80 \times 10^{-4} / 1.20 \times 10^{-4} = 4\)). Therefore, the reaction is second order with respect to A.
- Comparing Run 1 and Run 3: \([\text{A}]\) is kept constant, \([\text{B}]\) doubles, and the rate remains unchanged. Therefore, the reaction is zero order with respect to B.
- The rate equation is: \(\text{Rate} = k[\text{A}]^2\).

(b) (i)
- Substituting values from Run 1 into the rate equation:
\(1.20 \times 10^{-4} = k(0.10)^2\)
\(k = \frac{1.20 \times 10^{-4}}{0.010} = 1.20 \times 10^{-2}\)
- Units: \(\text{dm}^3 \text{mol}^{-1} \text{s}^{-1}\).

(ii)
- Step 1 is the rate-determining step because it is designated as the slow step.
- This step involves only two molecules of A and zero molecules of B.
- Consequently, the rate equation derived from this mechanism depends only on \([\text{A}]^2\), which matches the experimentally determined rate equation: \(\text{Rate} = k[\text{A}]^2\).

(c)
- Rearranging the Arrhenius equation for two temperatures:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
- Substituting the given values:
\(\ln\left(\frac{4.80 \times 10^{-2}}{1.20 \times 10^{-2}}\right) = \frac{E_a}{8.31} \left(\frac{1}{298} - \frac{1}{318}\right)\)
\(\ln(4.00) = \frac{E_a}{8.31} \left(3.3557 \times 10^{-3} - 3.1447 \times 10^{-3}\right)\)
\(1.3863 = \frac{E_a}{8.31} \left(2.1105 \times 10^{-4}\right)\)
\(11.520 = E_a \left(2.1105 \times 10^{-4}\right)\)
\(E_a = 54584 \text{ J mol}^{-1} = +54.6 \text{ kJ mol}^{-1}\).

(a)
- **M1**: Identifies order with respect to A as 2 with clear explanation linking doubling of concentration to a fourfold increase in rate (1)
- **M2**: Identifies order with respect to B as 0 with clear explanation linking doubling of concentration to no change in rate (1)
- **M3**: Correct rate equation formula corresponding to deduced orders (1)
- **M4**: Uses correct algebraic notation for rate equation with \(k\) (1)

(b)(i)
- **M1**: Calculation of value of \(k = 1.20 \times 10^{-2}\) (or \(0.012\)) (1)
- **M2**: Correct units of \(\text{dm}^3 \text{mol}^{-1} \text{s}^{-1}\) (1)

(b)(ii)
- **M1**: Identifies Step 1 as the rate-determining step as it is the slowest step (1)
- **M2**: Explains that the rate-determining step involves only reactant A molecules (2 molecules) and no B molecules (1)
- **M3**: Links this molecularity directly to the overall rate equation exponents (1)

(c)
- **M1**: Uses correct Arrhenius relationship expression (1)
- **M2**: Calculates the ratio \(\ln(4) = 1.386\) (1)
- **M3**: Calculates \(\left(\frac{1}{298} - \frac{1}{318}\right) = 2.11 \times 10^{-4}\) (1)
- **M4**: Evaluates \(E_a\) in Joules: \(54584 \text{ J mol}^{-1}\) (1)
- **M5**: Converts correctly to \(\text{kJ mol}^{-1}\) to 3 sig figs: \(+54.6 \text{ kJ mol}^{-1}\) (1) [Accept range \(+54.5\) to \(+54.7\)]

(a)
- A gas (\(\text{CO}_2\)) is produced from a solid reactant (\(\text{BaCO}_3\)).
- Gas particles are significantly more disordered than solid particles and have higher positional entropy, resulting in an overall increase in entropy.

(b)
- \(\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\)
- \(\Delta S^{\ominus}_{\text{system}} = (70 + 214) - 112 = +172 \text{ J K}^{-1} \text{mol}^{-1}\)

(c) (i)
- \(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{\Delta H^{\ominus}}{T}\)
- \(\Delta S^{\ominus}_{\text{surroundings}} = -\frac{269000 \text{ J mol}^{-1}}{298 \text{ K}} = -902.68 \text{ J K}^{-1} \text{mol}^{-1} \approx -903 \text{ J K}^{-1} \text{mol}^{-1}\)

(ii)
- \(\Delta S^{\ominus}_{\text{total}} = \Delta S^{\ominus}_{\text{system}} + \Delta S^{\ominus}_{\text{surroundings}}\)
- \(\Delta S^{\ominus}_{\text{total}} = +172 + (-902.68) = -730.68 \text{ J K}^{-1} \text{mol}^{-1} \approx -731 \text{ J K}^{-1} \text{mol}^{-1}\)
- The reaction is **not feasible** at \(298 \text{ K}\) because \(\Delta S^{\ominus}_{\text{total}}\), is negative.

(iii)
- For the reaction to become feasible, \(\Delta S^{\ominus}_{\text{total}} \ge 0\).
- At the transition temperature, \(\Delta S^{\ominus}_{\text{system}} + \Delta S^{\ominus}_{\text{surroundings}} = 0 \implies \Delta S^{\ominus}_{\text{system}} - \frac{\Delta H^{\ominus}}{T} = 0\).
- \(T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}_{\text{system}}} = \frac{269000 \text{ J mol}^{-1}}{172 \text{ J K}^{-1} \text{mol}^{-1}} = 1563.95 \text{ K} \approx 1564 \text{ K}\) (or \(1560 \text{ K}\) to 3 s.f.).

(a)
- **M1**: Mentions that a gas is produced from a solid (1)
- **M2**: Explains that gas molecules are much more disordered / have higher positional entropy than solid particles (1)

(b)
- **M1**: Correct algebraic expression (products - reactants) (1)
- **M2**: Correct substitution: \((70 + 214) - 112\) (1)
- **M3**: Correct calculation of \(+172\) with standard units \(\text{J K}^{-1} \text{mol}^{-1}\) (1)

(c)(i)
- **M1**: Correct equation: \(\Delta S = -\frac{\Delta H}{T}\) (with unit conversion of kJ to J) (1)
- **M2**: Evaluation to \(-903\) (or \(-902.7\)) \(\text{J K}^{-1} \text{mol}^{-1}\) (1)

(c)(ii)
- **M1**: Correct summation of system and surroundings values (1)
- **M2**: Calculation of \(-731\) (or \(-730.7\)) \(\text{J K}^{-1} \text{mol}^{-1}\) (1)
- **M3**: Concludes that the reaction is not feasible because the total entropy change is negative (1)

(c)(iii)
- **M1**: States the threshold condition for feasibility is \(\Delta S^{\ominus}_{\text{total}} = 0\) (1)
- **M2**: Rearranges equation to \(T = \frac{\Delta H}{\Delta S_{\text{system}}}\) (1)
- **M3**: Correct conversion of units (use of \(269000\) or matching units) (1)
- **M4**: Evaluation to \(1564 \text{ K}\) (accept range \(1560 \text{ K}\) to \(1564 \text{ K}\)) (1)

(a)
- A Brønsted-Lowry acid is a proton (\(\text{H}^+\)) donor.

(b) (i)
- \(K_a = \frac{[\text{H}^+(aq)][\text{HCOO}^-(aq)]}{[\text{HCOOH}(aq)]}\)

(ii)
- Assuming \([\text{H}^+(aq)] \approx [\text{HCOO}^-(aq)]\) and that the concentration of methanoic acid at equilibrium is approximately equal to its initial concentration (\(0.250 \text{ mol dm}^{-3}\)):
\(K_a \approx \frac{[\text{H}^+]^2}{[\text{HCOOH}]}\)
\(1.60 \times 10^{-4} = \frac{[\text{H}^+]^2}{0.250}\)
\([\text{H}^+]^2 = 4.00 \times 10^{-5}\)
\([\text{H}^+] = 6.325 \times 10^{-3} \text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(6.325 \times 10^{-3}) = 2.20\)
- **Assumptions**:
1. The concentration of \(\text{H}^+\) from the self-ionization of water is negligible.
2. The degree of dissociation of methanoic acid is so small that the equilibrium concentration of the acid is practically equal to its initial concentration.

(c) (i)
- Initial moles of \(\text{HCOOH} = 0.0500 \text{ dm}^3 \times 0.250 \text{ mol dm}^{-3} = 0.0125 \text{ mol}\)
- Moles of \(\text{NaOH} = 0.0500 \text{ dm}^3 \times 0.150 \text{ mol dm}^{-3} = 0.00750 \text{ mol}\)
- When mixed, the neutralization reaction occurs:
\(\text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O}\)
- Moles of \(\text{HCOO}^-\text{(aq)}\) formed = \(0.00750 \text{ mol}\)
- Moles of \(\text{HCOOH(aq)}\) remaining = \(0.0125 - 0.00750 = 0.00500 \text{ mol}\)
- Total volume of solution is \(100.0 \text{ cm}^3\).
- \([\text{H}^+] = K_a \times \frac{n(\text{HCOOH})}{n(\text{HCOO}^-)} = 1.60 \times 10^{-4} \times \frac{0.00500}{0.00750} = 1.067 \times 10^{-4} \text{ mol dm}^{-3}\)
- \(\text{pH} = -\log_{10}(1.067 \times 10^{-4}) = 3.97\)

(ii)
- When \(\text{H}^+\) ions are added, they react with the reservoir of conjugate base (methanoate ions):
\(\text{HCOO}^-(aq) + \text{H}^+(aq) \rightarrow \text{HCOOH}(aq)\)
- Because the concentration of \(\text{HCOO}^-\) is large relative to the added acid, the ratio of \(\frac{[\text{HCOOH}]}{[\text{HCOO}^-]}\) remains almost constant, meaning the pH changes very little.

(a)
- **M1**: Defines Brønsted-Lowry acid as a proton / \(\text{H}^+\) donor (1)

(b)(i)
- **M1**: Correct expression for \(K_a\) with state symbols optional but must have species correctly written (1)

(b)(ii)
- **M1**: Correct calculation of \([\text{H}^+] = 6.325 \times 10^{-3} \text{ mol dm}^{-3}\) and pH = \(2.20\) (1)
- **M2**: States assumption 1: dissociation of the weak acid is negligible / \([\text{HCOOH}]_{\text{equilibrium}} \approx [\text{HCOOH}]_{\text{initial}}\) (1)
- **M3**: States assumption 2: concentration of hydrogen ions from water is negligible / \([\text{H}^+] \approx [\text{HCOO}^-]\) (1)

(c)(i)
- **M1**: Calculates initial moles of both reactants correctly (0.0125 and 0.0075) (1)
- **M2**: Deduces equilibrium moles of \(\text{HCOO}^-\) (0.0075) and remaining \(\text{HCOOH}\) (0.0050) (1)
- **M3**: Substitutes correct mole values into a valid buffer equation (1)
- **M4**: Calculates \([\text{H}^+] = 1.07 \times 10^{-4} \text{ mol dm}^{-3}\) (1)
- **M5**: Evaluates pH correctly to \(3.97\) (accept 2 decimal places only) (1)

(c)(ii)
- **M1**: Writes correct ionic equation: \(\text{HCOO}^-(aq) + \text{H}^+(aq) \rightarrow \text{HCOOH}(aq)\) (1)
- **M2**: States that added hydrogen ions are removed by reaction with a large reservoir of methanoate ions (1)
- **M3**: Explains that the ratio of weak acid to conjugate base stays virtually constant, minimizing pH change (1)

(a) (i)
- **Reagent**: Potassium cyanide (\(\text{KCN}\)) and dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)) (or a solution buffered to pH 6-8).
- **Conditions**: Warm/room temperature, inside a **fume cupboard** (to prevent exposure to toxic HCN gas).

(ii)
- First step: Nucleophilic attack of cyanide ion on the carbonyl carbon. A curly arrow must go from the lone pair of \(\text{:C}\text{N}^-\)\ to the carbonyl carbon. A second curly arrow must go from the double bond of \(\text{C}=\text{O}\) to the oxygen atom.
- Dipoles \(\text{C}^{\delta+}=\text{O}^{\delta-}\) should be marked on the starting aldehyde.
- Intermediate: An alkoxide intermediate is formed, represented as \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(O}^-\text{)CN}\) with a negative charge on the oxygen atom.
- Second step: Protonation of the intermediate. A curly arrow must go from the lone pair of the negative oxygen to a hydrogen ion (\(\text{H}^+\) or the hydrogen of \(\text{HCN}\)) to form the 2-hydroxypentanenitrile product.

(b) (i)
- The carbonyl carbon in butanal is **planar** (with respect to the trigonal planar geometry around the sp2 carbon).
- The nucleophile (\(\text{CN}^-\)) can attack this planar carbon atom with equal probability from either **above or below the plane**.
- This yields a **50:50 / equimolar mixture** of the two enantiomers (a racemic mixture).
- Since the two enantiomers rotate plane-polarized light in equal magnitudes but in opposite directions, their optical rotations cancel each other out, resulting in no net optical activity.

(ii)
- **Ketone**: Propanone, \(\text{CH}_3\text{COCH}_3\) (accept structural formula).
- **Explanation**: The product formed from the nucleophilic addition of HCN to propanone is 2-hydroxy-2-methylpropanenitrile, \(\text{(CH}_3)_2\text{C(OH)CN}\).
- Since the reactant (propanone) is symmetrical, the central carbon in the product is attached to two identical methyl groups (alongside an \(-\text{OH}\) and a \(-\text{CN}\) group). Thus, it has no chiral center and is inherently optically inactive.

(a)(i)
- **M1**: Identifies reagents: \(\text{KCN}\) and dilute acid (e.g. \(\text{H}_2\text{SO}_4\)) (1)
- **M2**: States safety requirement: fume cupboard (1)

(a)(ii)
- **M1**: Correct partial charges \(\text{C}^{\delta+}=\text{O}^{\delta-}\) and a curly arrow starting from lone pair of carbon in \(\text{CN}^-\)\ to the carbonyl carbon (1)
- **M2**: Curly arrow from \(\text{C}=\text{O}\) double bond to the oxygen atom (1)
- **M3**: Correct structure of the intermediate with a negative charge on oxygen and the nitrile group attached (1)
- **M4**: Curly arrow from lone pair of intermediate oxygen to \(\text{H}^+\) (or \(\text{HCN}\)) leading to correct final product (1)

(b)(i)
- **M1**: States that the carbonyl group in butanal is planar around the carbon atom (1)
- **M2**: Explains that attack by the nucleophile is equally likely from above or below the plane (1)
- **M3**: Identifies that this produces an equimolar (50:50) mixture of enantiomers (or racemic mixture) (1)
- **M4**: States that the optical effects cancel out (1)

(b)(ii)
- **M1**: Identifies Propanone (name or formula \(\text{CH}_3\text{COCH}_3\)) (1)
- **M2**: Draws/states correct structure of product: \(\text{(CH}_3)_2\text{C(OH)CN}\) (1)
- **M3**: Explains that the starting ketone is symmetrical / contains identical alkyl groups on both sides of carbonyl (1)
- **M4**: Explains that the product has two identical groups (methyl groups) attached to the central carbon, hence no chiral center exists (1)

(a)
- \(K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \times p(\text{H}_2)^2}\)
- Units: \(\text{atm}^{-2}\) (since \(\frac{\text{atm}}{\text{atm} \times \text{atm}^2} = \text{atm}^{-2}\))

(b) (i)
- Reaction: \(\text{CO} + 2\text{H}_2 \rightleftharpoons \text{CH}_3\text{OH}\)
- Moles of CO reacting = \(0.40 \text{ mol}\), so moles of CO at equilibrium = \(1.00 - 0.40 = 0.60 \text{ mol}\)
- Moles of \(\text{H}_2\) reacting = \(2 \times 0.40 = 0.80 \text{ mol}\), so moles of \(\text{H}_2\) at equilibrium = \(2.00 - 0.80 = 1.20 \text{ mol}\)

(ii)
- Total moles at equilibrium = \(0.60 + 1.20 + 0.40 = 2.20 \text{ mol}\)
- Mole fraction of CO: \(\chi(\text{CO}) = \frac{0.60}{2.20} = 0.273\)
- Mole fraction of \(\text{H}_2\): \(\chi(\text{H}_2) = \frac{1.20}{2.20} = 0.545\)
- Mole fraction of \(\text{CH}_3\text{OH}\): \(\chi(\text{CH}_3\text{OH}) = \frac{0.40}{2.20} = 0.182\)

(iii)
- Partial pressure = mole fraction \(\times\) total pressure (\(50.0 \text{ atm}\))
- \(p(\text{CO}) = 0.2727 \times 50.0 = 13.64 \text{ atm} \approx 13.6 \text{ atm}\)
- \(p(\text{H}_2) = 0.5455 \times 50.0 = 27.27 \text{ atm} \approx 27.3 \text{ atm}\)
- \(p(\text{CH}_3\text{OH}) = 0.1818 \times 50.0 = 9.09 \text{ atm} \approx 9.1 \text{ atm}\)

(iv)
- \(K_p = \frac{9.09}{13.64 \times (27.27)^2} = \frac{9.09}{10143} = 8.96 \times 10^{-4} \text{ atm}^{-2}\) (allow \(8.9 \times 10^{-4}\) to \(9.0 \times 10^{-4}\))

(c) (i)
- **Value of \(K_p\)**: Decreases.
- **Yield of methanol**: Decreases.
- **Explanation**: The forward reaction is exothermic. According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the heat. Thus, the ratio of product-to-reactant concentrations decreases, lowering \(K_p\).

(ii)
- **Value of \(K_p\)**: Remains unchanged (as only a change in temperature can affect the equilibrium constant).
- **Yield of methanol**: Increases.
- **Explanation**: Increasing pressure shifts the position of equilibrium to the side with fewer gas molecules (the right-hand side: 3 moles of reactant gas vs. 1 mole of product gas) to oppose the pressure increase.

(a)
- **M1**: Correct expression for \(K_p\) (penalize square brackets if used) (1)
- **M2**: Correct units \(\text{atm}^{-2}\) (1)

(b)(i)
- **M1**: Calculates moles of CO at equilibrium = \(0.60 \text{ mol}\) (1)
- **M2**: Calculates moles of \(\text{H}_2\) at equilibrium = \(1.20 \text{ mol}\) (1)

(b)(ii)
- **M1**: Calculates total moles of gas at equilibrium = \(2.20 \text{ mol}\) (1)
- **M2**: Calculates mole fractions for all three components correctly (allow 3 s.f.) (1)

(b)(iii)
- **M1**: Correctly applies formula \(p = \chi \times P_{\text{total}}\) (1)
- **M2**: Evaluation of three partial pressures (\(13.6 \text{ atm}\), \(27.3 \text{ atm}\), and \(9.1 \text{ atm}\)) (1)

(b)(iv)
- **M1**: Substitutes values into expression (1)
- **M2**: Calculates correct final value of \(K_p\) in range \(8.9 \times 10^{-4}\) to \(9.0 \times 10^{-4}\) (1)

(c)(i)
- **M1**: Correctly states both decrease (1)
- **M2**: Explains shift in endothermic back-reaction direction (1)

(c)(ii)
- **M1**: Correctly states \(K_p\) is unchanged, and yield increases (1)
- **M2**: Explains shift to side with fewer gas moles (3 vs 1) (1)

The electronic configuration of a neutral copper atom is \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\). To form a \(\text{Cu}^{2+}\) ion, two electrons are removed, starting with the outer \(4\text{s}\) orbital and then one from the \(3\text{d}\) orbitals, leaving the configuration as \([\text{Ar}] 3\text{d}^9\).

1 mark: Correct identification of \([\text{Ar}] 3\text{d}^9\) as the electronic configuration of the copper(II) ion.

For a reaction to be feasible, \(E^\ominus_{\text{cell}}\) must be positive. Calculating \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\) for each option:

- Option A: \(E^\ominus_{\text{cell}} = 0.15 - 0.77 = -0.62\text{ V}\) (not feasible)
- Option B: \(E^\ominus_{\text{cell}} = 0.34 - 0.77 = -0.43\text{ V}\) (not feasible)
- Option C: \(E^\ominus_{\text{cell}} = 0.77 - 0.15 = +0.62\text{ V}\) (feasible)
- Option D: \(E^\ominus_{\text{cell}} = -0.41 - 0.34 = -0.75\text{ V}\) (not feasible)

1 mark: Correctly identifies option C by calculating standard cell potentials.

Concentrated sulfuric acid is a stronger acid than concentrated nitric acid and protonates it. The subsequent loss of a water molecule produces the nitronium ion (\(\text{NO}_2^+\)), which acts as the electrophile. The balanced equation is \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\).

1 mark: Correctly identifies the balanced equation for the generation of the nitronium ion.

Ethylamine is more basic than ammonia due to the electron-releasing inductive effect of the ethyl group, which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton. Phenylamine is less basic than ammonia because the lone pair on the nitrogen atom is partially delocalised into the benzene ring's \(\pi\)-system, making it less available to accept a proton. Hence, the order is 2 > 1 > 3.

1 mark: Correctly orders the compounds by decreasing basic strength (2 > 1 > 3).

The reaction between \(\text{MnO}_4^-\) and \(\text{C}_2\text{O}_4^{2-}\) produces \(\text{Mn}^{2+}\) ions. Since \(\text{Mn}^{2+}\) acts as a catalyst for this reaction and is a product of it, this is an example of autocatalysis. Because all reactants and the catalyst are in the aqueous phase, it is a homogeneous autocatalytic process.

1 mark: Correctly identifies the reaction as homogeneous autocatalysis with \(\text{Mn}^{2+}\) as the catalyst.

The standard laboratory method for reducing nitrobenzene to phenylamine involves heating with tin (\(\text{Sn}\)) and concentrated hydrochloric acid under reflux. This initially forms the phenylammonium salt. Adding aqueous sodium hydroxide is then required to liberate the free phenylamine base.

1 mark: Correctly identifies the reagents and conditions for the reduction of nitrobenzene to phenylamine.

Addition of excess concentrated hydrochloric acid results in a ligand exchange reaction where water ligands are replaced by larger chloride ligands. Due to steric hindrance of the larger \(\text{Cl}^-\) ligands, only four can coordinate to the cobalt(II) ion, resulting in the tetrahedral complex \([\text{CoCl}_4]^{2-}\), which is blue.

1 mark: Correctly identifies the formula and tetrahedral geometry of the blue complex.

By IUPAC convention, the half-cell where oxidation occurs (the anode, with the more negative standard electrode potential, which is the iron half-cell) is written on the left: \(\text{Fe}(\text{s}) | \text{Fe}^{2+}(\text{aq})\). The half-cell where reduction occurs (the cathode, with the more positive electrode potential, which is the silver half-cell) is written on the right: \(\text{Ag}^+(\text{aq}) | \text{Ag}(\text{s})\). The standard cell potential is calculated as \(E^\ominus_{\text{cell}} = E^\ominus_{\text{right}} - E^\ominus_{\text{left}} = +0.80 - (-0.44) = +1.24\text{ V}\).

1 mark: Correctly identifies the standard IUPAC cell representation and calculates the correct cell potential of +1.24 V.

Addition of excess concentrated ammonia to aqueous copper(II) ions results in partial ligand substitution. Four of the water ligands are replaced by four ammonia ligands, forming the deep-blue tetraamminediacuacopper(II) complex, \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

[1] Correct option B

Methylbenzene contains a methyl group (\(-\text{CH}_3\)), which is electron-donating via a positive inductive effect (+I). This increases the electron density of the benzene ring, making it more susceptible to electrophilic attack by the nitronium ion (\(\text{NO}_2^+\)) compared to benzene. Nitrobenzene (containing \(-\text{NO}_2\)) and benzoic acid (containing \(-\text{COOH}\)) have electron-withdrawing groups, which deactivate the ring and slow down the rate of electrophilic nitration.

[1] Correct option C

Diethylamine is a secondary aliphatic amine. Aliphatic alkyl groups are electron-donating (+I effect), which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton. A secondary amine has two such alkyl groups, making it more basic than a primary aliphatic amine (ethylamine) and ammonia. Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is delocalized into the aromatic \(\pi\)-system, making it much less available for protonation.

[1] Correct option D

For this reaction, the iron(III)/iron(II) system undergoes reduction (the cathode half-cell) and the iodine/iodide system undergoes oxidation (the anode half-cell). Using the equation \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\):

\(E^\ominus_{\text{cell}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\).

[1] Correct option B

Peroxodisulfate and iodide ions are both negatively charged and repel each other, giving the uncatalysed reaction a high activation energy. The homogeneous catalyst \(\text{Fe}^{2+}\) (a positively charged ion) initially reduces the peroxodisulfate ion to sulfate, being itself oxidized to \(\text{Fe}^{3+}\):

\(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)

In the second step, the regenerated \(\text{Fe}^{3+}\) oxidizes iodide to iodine, reforming the catalyst \(\text{Fe}^{2+}\).

[1] Correct option A

Benzene is first nitrated using concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\) at around 50–55 °C to form nitrobenzene. Nitrobenzene is then reduced to phenylamine using tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) under reflux, followed by adding aqueous sodium hydroxide (\(\text{NaOH}\)) to liberate the free phenylamine amine from its protonated salt.

[1] Correct option B

At a low pH (significantly below the isoelectric point), the concentration of \(\text{H}^+\) ions is high. This acidic environment protonates both the basic amine group (\(-\text{NH}_2 \rightarrow -\text{NH}_3^+\)) and keeps the carboxylic acid group in its protonated neutral form (\(-\text{COOH}\)). Consequently, the overall species is a cation: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\).

[1] Correct option B

For a spontaneous cell, the half-cell with the more positive \(E^\ominus\) value behaves as the cathode (reduction on the right-hand side), and the half-cell with the less positive \(E^\ominus\) value behaves as the anode (oxidation on the left-hand side).

Anode reaction (left): \(\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-\), represented as \(\text{Cu}(\text{s}) \mid \text{Cu}^{2+}(\text{aq})\).
Cathode reaction (right): \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\). Since both iron ions are in the aqueous phase, they are separated by a comma, and an inert platinum electrode is required to transfer electrons: \(\text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) \mid \text{Pt}(\text{s})\).

Combining these with a salt bridge (\(\parallel\)) gives the conventional cell diagram: \(\text{Cu}(\text{s}) \mid \text{Cu}^{2+}(\text{aq}) \parallel \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) \mid \text{Pt}(\text{s})\).

[1] Correct option A

When aqueous sodium hydroxide is added to a solution of \([Cr(H_{2}O)_{6}]^{3+}\) ions, a green precipitate of chromium(III) hydroxide, \([Cr(OH)_{3}(H_{2}O)_{3}]\), initially forms because chromium(III) is amphoteric. In excess sodium hydroxide, this precipitate dissolves to form a dark green solution containing the hexahydroxochromate(III) ion, \([Cr(OH)_{6}]^{3-}\). Addition of hydrogen peroxide and boiling would be required to oxidize this to the yellow chromate(VI) ion, \([CrO_{4}]^{2-}\).

1 mark: correct option identified.

Alkyl groups are electron-donating due to the inductive effect. This increases the electron density on the nitrogen atom, making the lone pair more available for donation to a proton. Therefore, secondary aliphatic amines (such as diethylamine) are stronger bases than primary aliphatic amines (such as ethylamine), which are stronger bases than ammonia. Phenylamine is the weakest base because the lone pair on the nitrogen atom is partially delocalised into the benzene ring's \(\pi\)-electron system, making it much less available to accept a proton. Hence, the correct order is Diethylamine > Ethylamine > Ammonia > Phenylamine.

1 mark: correct option identified.

For a reaction to be thermodynamically feasible, the standard cell potential \(E^{\ominus}_{\text{cell}}\) must be positive. Let us calculate the standard cell potential for each option: (a) \(E^{\ominus}_{\text{cell}} = -0.76 - (+0.34) = -1.10\text{ V}\) (Not feasible). (b) \(E^{\ominus}_{\text{cell}} = +0.34 - (+0.80) = -0.46\text{ V}\) (Not feasible). (c) \(E^{\ominus}_{\text{cell}} = -0.76 - (+0.80) = -1.56\text{ V}\) (Not feasible). (d) \(E^{\ominus}_{\text{cell}} = +0.34 - (-0.76) = +1.10\text{ V}\) (Feasible, since \(E^{\ominus}_{\text{cell}}\) is positive).

1 mark: correct option identified.

The standard laboratory method for reducing nitrobenzene to phenylamine involves heating nitrobenzene under reflux with tin (Sn) and concentrated hydrochloric acid (HCl). This first forms the phenylammonium salt, \([C_{6}H_{5}NH_{3}]^{+}Cl^{-}\). Addition of a strong alkali, such as aqueous sodium hydroxide (NaOH), is then required to liberate the free amine, phenylamine (\(C_{6}H_{5}NH_{2}\)).

1 mark: correct option identified.

(a) When ligands approach a transition metal ion, the electric field splits the five degenerate d-orbitals into two groups of different energy levels. Electrons in the lower d-orbitals can absorb energy from the visible light spectrum and be promoted to a higher d-orbital (d-d transition). The energy absorbed corresponds to a specific wavelength of light (\(\Delta E = h\nu\)). The color transmitted or reflected is complementary to the color absorbed. Sc\(^{3+}\) has a \(d^0\) configuration and Zn\(^{2+}\) has a \(d^{10}\) configuration. Since Sc\(^{3+}\) has no d-electrons to promote, and Zn\(^{2+}\) has no vacant d-orbitals to receive promoted electrons, no d-d transitions can occur, rendering these ions colorless.

(b)(i) \([Cu(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CuCl_4]^{2-}(aq) + 6H_2O(l)\)
(ii) The reaction is a ligand substitution (or ligand exchange) reaction. The coordination number changes from 6 to 4. The shape of the complex changes from octahedral to tetrahedral.

(c) The ligand substitution with 1,2-diaminoethane (en) is represented by: \([Cu(H_2O)_6]^{2+} + 3\text{en} \rightarrow [Cu(\text{en})_3]^{2+} + 6H_2O\). In this reaction, 4 reacting particles produce 7 product particles, resulting in a large increase in system entropy (\(\Delta S^{\ominus}\) is positive). For monodentate ammonia: \([Cu(H_2O)_6]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O\), where 5 particles form 5 particles, so there is little change in entropy. Since the coordinate bonds formed and broken are of similar strength, the enthalpy change \(\Delta H^{\ominus} \approx 0\). According to \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\), the highly positive \(\Delta S^{\ominus}\) for the chelate reaction makes \(\Delta G^{\ominus}\) highly negative, driving the reaction to completion and yielding a much more stable complex.

**(a)**
- M1: d-orbitals split into two sets of different energy levels due to presence of ligands (1)
- M2: Electrons absorb energy/light in the visible region (1)
- M3: Electron promoted from lower to higher energy d-orbitals / d-d transition (1)
- M4: Color seen is the complementary color / transmitted light (1)
- M5: Sc3+ has no d-electrons (d0) OR Zn2+ has full d-orbitals (d10), so no d-d transitions are possible (1)

**(b)(i)**
- M1: Correct reactants and products (1)
- M2: Correct balancing (1)

**(b)(ii)**
- M1: Ligand substitution / ligand exchange (1)
- M2: Coordination number changes from 6 to 4 (1)
- M3: Shape changes from octahedral to tetrahedral (1)

**(c)**
- M1: Explains that substituting with 1,2-diaminoethane increases the total number of particles in solution (4 reactant particles to 7 product particles) (1)
- M2: This leads to an increase in entropy / positive \(\Delta S^{\ominus}\) (1)
- M3: Enthalpy change is approximately zero (\(\Delta H^{\ominus} \approx 0\)) because similar coordinate bonds are broken and formed (1)
- M4: Since \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\), a highly positive \(\Delta S^{\ominus}\) leads to a highly negative \(\Delta G^{\ominus}\), making the chelated complex thermodynamically very stable (1)

(a)(i) Reagents: Concentrated nitric acid (\(HNO_3\)) and concentrated sulfuric acid (\(H_2SO_4\)). Conditions: Temperature maintained at \(50^\circ\text{C}\) to \(55^\circ\text{C}\) (under reflux).
(ii) Electrophile generation: \(HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-\).
Mechanism: A curly arrow goes from the delocalized \(\pi\) ring of benzene to the nitrogen atom of the \(NO_2^+\) electrophile. An intermediate is drawn containing a hexadienyl carbocation (horseshoe ring open towards the carbon bonded to the H and \(NO_2\) group, with a positive charge inside the ring). A curly arrow goes from the C-H bond back into the ring, restoring the aromatic system and releasing an \(H^+\) ion.

(b)(i) Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl), heated under reflux, followed by the addition of aqueous sodium hydroxide (NaOH) solution.
(ii) \(C_6H_5NO_2 + 6[H] \rightarrow C_6H_5NH_2 + 2H_2O\)

(c)(i) A Br%nsted-Lowry base is a proton (\(H^+\)) acceptor.
(ii) In phenylamine, the lone pair of electrons on the nitrogen atom is partially delocalized into the \(\pi\)-system of the benzene ring. This decreases the electron density on the nitrogen atom, making its lone pair less available to accept a proton. In contrast, in ethylamine, the alkyl (ethyl) group is electron-releasing due to the positive inductive effect (+I effect). This increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton.

**(a)(i)**
- M1: Concentrated nitric acid and concentrated sulfuric acid (both required) (1)
- M2: Temperature between \(50^\circ\text{C}\) and \(55^\circ\text{C}\) / under reflux (1)

**(a)(ii)**
- M1: Equation showing generation of \(NO_2^+\) electrophile (1)
- M2: Curly arrow from benzene ring to \(NO_2^+\) (1)
- M3: Correct structure of intermediate (positive charge in a horseshoe open towards the sp3 carbon) (1)
- M4: Curly arrow from C-H bond back into the ring to form nitrobenzene and regenerate \(H^+\) (1)

**(b)(i)**
- M1: Tin and concentrated hydrochloric acid (reflux/heated) (1)
- M2: Sodium hydroxide solution (added afterwards to liberate the free amine) (1)

**(b)(ii)**
- M1: \(C_6H_5NO_2 + 6[H] \rightarrow C_6H_5NH_2 + 2H_2O\) (1)

**(c)(i)**
- M1: Proton / \(H^+\) acceptor (1)

**(c)(ii)**
- M1: In phenylamine, the lone pair on nitrogen is delocalized into the benzene ring / \(\pi\) system (1)
- M2: Decreases electron density on nitrogen / lone pair less available to accept proton (1)
- M3: In ethylamine, the alkyl group is electron-donating / positive inductive effect (+I) (1)
- M4: Increases electron density on nitrogen / lone pair more available to accept proton (1)

(a)(i) \(MnO_4^-(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\)

(ii)
1. Moles of \(MnO_4^-\) used: \(n(MnO_4^-) = 0.0200 \text{ mol dm}^{-3} \times \frac{22.40}{1000} \text{ dm}^3 = 4.48 \times 10^{-4}\text{ mol}\)
2. Moles of \(Fe^{2+}\) in \(25.0\text{ cm}^3\): Since the ratio of \(MnO_4^-\colon Fe^{2+}\) is 1:5, \(n(Fe^{2+}) = 5 \times 4.48 \times 10^{-4} = 2.24 \times 10^{-3}\text{ mol}\)
3. Moles of \(Fe^{2+}\) in \(250.0\text{ cm}^3\): \(n_{\text{total}}(Fe^{2+}) = 2.24 \times 10^{-3} \times \frac{250.0}{25.0} = 0.0224\text{ mol}\)
4. Mass of iron in the alloy: \(m(Fe) = 0.0224\text{ mol} \times 55.8\text{ g mol}^{-1} = 1.24992\text{ g}\)
5. Percentage by mass of iron: \(\text{Percentage} = \frac{1.24992}{1.45} \times 100\\% = 86.2\\%\)

(b)(i) \(Cr^{2+}(aq)\) is the stronger reducing agent. This is because its standard electrode potential is more negative (\(E^{\ominus} = -0.41\text{ V}\)) than that of the \(Fe^{3+}/Fe^{2+}\) system (\(E^{\ominus} = +0.77\text{ V}\)), meaning it has a greater tendency to lose electrons and shift its equilibrium to the left.

(ii) The overall cell reaction is: \(4Fe^{2+}(aq) + O_2(g) + 4H^+(aq) \rightarrow 4Fe^{3+}(aq) + 2H_2O(l)\)
\(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{reduction}} - E^{\ominus}_{\text{oxidation}} = 1.23\text{ V} - 0.77\text{ V} = +0.46\text{ V}\)
Since \(E^{\ominus}_{\text{cell}}\) is positive (\(> 0\)), the reaction is thermodynamically feasible.

(c) The reaction may have a high activation energy, making the rate of reaction extremely slow (kinetic inhibition). Additionally, standard electrode potentials only apply to standard conditions (e.g., \(1\text{ mol dm}^{-3}\), \(100\text{ kPa}\)); non-standard concentrations or conditions can alter the actual electrode potential and make a feasible reaction unfeasible (or vice versa).

**(a)(i)**
- M1: Correct formulas for reactants and products (1)
- M2: Correct balancing (1)

**(a)(ii)**
- M1: Calculates moles of \(MnO_4^-\) = \(4.48 \times 10^{-4}\text{ mol}\) (1)
- M2: Calculates moles of \(Fe^{2+}\) in \(25.0\text{ cm}^3\) = \(2.24 \times 10^{-3}\text{ mol}\) (1)
- M3: Scales to \(250.0\text{ cm}^3\) = \(0.0224\text{ mol}\) (1)
- M4: Calculates mass of iron = \(1.25\text{ g}\) (1)
- M5: Calculates percentage by mass = \(86.2\\%\) (Accept 86.2% - 86.5% depending on intermediate rounding) (1)

**(b)(i)**
- M1: Identifies \(Cr^{2+}(aq)\) as the stronger reducing agent (1)
- M2: Explains that it has the more negative / less positive standard electrode potential value (1)

**(b)(ii)**
- M1: Correct balanced overall equation (1)
- M2: Calculates \(E^{\ominus}_{\text{cell}} = +0.46\text{ V}\) (1)
- M3: Concludes that the reaction is feasible because \(E^{\ominus}_{\text{cell}} > 0\) (1)

**(c)**
- M1: High activation energy / reaction is kinetically stable / slow rate (1)
- M2: Non-standard concentrations / conditions are used (which alters the actual electrode potential values) (1)

(a)(i) 2-aminobutanedioic acid.
(ii)
At pH 1: \(^+H_3N-CH(CH_2COOH)-COOH\) (both carboxylic acid groups are fully protonated, amine group is protonated as an ammonium ion).
At pH 13: \(H_2N-CH(CH_2COO^-)-COO^-\) (both carboxylic acid groups are deprotonated, amine group is unprotonated).
(iii) Aspartic acid exists as a zwitterion in the solid state. There are strong electrostatic attractions (ionic bonds) between the oppositely charged groups of neighboring zwitterions. Hexanedioic acid, however, only experiences intermolecular hydrogen bonding and London dispersion forces, which are significantly weaker than ionic attractions and require much less energy to break.

(b)(i) Monomer 1: Ethane-1,2-diamine, \(H_2N-CH_2-CH_2-NH_2\).
Monomer 2: Pentanedioic acid, \(HOOC-CH_2-CH_2-CH_2-COOH\) (or pentanedioyl dichloride, \(ClOC-CH_2-CH_2-CH_2-COCl\)).
(ii) Condensation polymerization.

(c)(i)
Reaction with acid: \(CH_3CH(NH_3^+)COO^- + H^+ \rightarrow CH_3CH(NH_3^+)COOH\)
Reaction with alkali: \(CH_3CH(NH_3^+)COO^- + OH^- \rightarrow CH_3CH(NH_2)COO^- + H_2O\)
(ii) The isoelectric point is the pH value at which the amino acid has no net electrical charge (exists entirely or predominantly as a neutral zwitterion).

(d) \(R_f\) value is calculated as: \(R_f = \frac{\text{Distance travelled by the amino acid}}{\text{Distance travelled by the solvent front}}\).
The colorless spots are visualized by spraying the plate with ninhydrin and heating (which produces purple/blue spots) or by placing the plate under ultraviolet (UV) light.

**(a)(i)**
- M1: 2-aminobutanedioic acid (1)

**(a)(ii)**
- M1: Structure at pH 1 with protonated amine (\(NH_3^+\) / \(H_3N^+\)) and both protonated carboxyls (\(COOH\)) (1)
- M2: Structure at pH 13 with neutral amine (\(NH_2\)) and both carboxylates (\(COO^-\) / \(OOC^-\)) (1)

**(a)(iii)**
- M1: Aspartic acid exists as a zwitterion (1)
- M2: Strong ionic forces / electrostatic attractions between oppositely charged zwitterionic groups (1)
- M3: Hexanedioic acid has only hydrogen bonds / intermolecular forces which are weaker and require less energy to overcome (1)

**(b)(i)**
- M1: \(H_2N-CH_2-CH_2-NH_2\) (1)
- M2: \(HOOC-(CH_2)_3-COOH\) OR \(ClOC-(CH_2)_3-COCl\) (1)

**(b)(ii)**
- M1: Condensation (polymerization) (1)

**(c)(i)**
- M1: Correct equation for reaction with \(H^+\) (1)
- M2: Correct equation for reaction with \(OH^-\) (1)

**(c)(ii)**
- M1: The pH at which the amino acid has no net charge / exists as a neutral zwitterion (1)

**(d)**
- M1: \(R_f\) formula: distance moved by solute/spot divided by distance moved by solvent (1)
- M2: Spray with ninhydrin (and heat) OR use UV lamp (1)

(a)(i)
(A) Yellow
(B) +4
(C) Green
(ii) Zinc (and a dilute acid such as hydrochloric acid or sulfuric acid).

(b)(i) A homogeneous catalyst is in the same physical state (or phase) as the reactants (e.g., both are dissolved in aqueous solution). A heterogeneous catalyst is in a different physical state (or phase) than the reactants (e.g., a solid catalyst with gaseous reactants).
(ii) The mechanism occurs as follows:
1. Gaseous reactants are adsorbed onto active sites on the solid catalyst surface.
2. Vanadium(V) oxide oxidizes sulfur dioxide to sulfur trioxide, being itself reduced to vanadium(IV) oxide:
\(SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4\)
3. Vanadium(IV) oxide is then oxidized back to vanadium(V) oxide by oxygen:
\(2V_2O_4 + O_2 \rightarrow 2V_2O_5\)
4. The product, sulfur trioxide, desorbs from the surface of the catalyst.

(c)(i) Both \(S_2O_8^{2-}\) and \(I^-\) are negatively charged ions. Because they carry the same charge, they strongly repel each other. This results in an exceptionally high activation energy for the uncatalyzed reaction, making the rate very slow.
(ii) Step 1: Iron(II) reduces peroxodisulfate and is oxidized to iron(III):
\(S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}\)
Step 2: Iron(III) then oxidizes iodide and is reduced back to iron(II):
\(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\)

**(a)(i)**
- M1: (A) Yellow (1)
- M2: (B) +4 (1)
- M3: (C) Green (1)

**(a)(ii)**
- M1: Zinc / Zn (in acidic solution / with acid) (1)

**(b)(i)**
- M1: Homogeneous: Catalyst is in the same phase/state as the reactants (1)
- M2: Heterogeneous: Catalyst is in a different phase/state from the reactants (1)

**(b)(ii)**
- M1: Adsorption of reactants onto the catalyst surface / active sites (1)
- M2: Equation: \(SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4\) (or balanced alternative) (1)
- M3: Equation: \(2V_2O_4 + O_2 \rightarrow 2V_2O_5\) (or balanced alternative) (1)
- M4: Desorption of products from the catalyst surface (1)

**(c)(i)**
- M1: Both reactant ions are negatively charged (1)
- M2: Repulsion between like charges leads to a high activation energy (1)

**(c)(ii)**
- M1: \(S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}\) (1)
- M2: \(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\) (1)

(a)
- Dropwise addition of NaOH produces a green precipitate of iron(II) hydroxide: \( \text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)} \).
- The precipitate is insoluble in excess NaOH, so no further change is seen.
- On standing, the green precipitate turns brown on the surface because it is oxidised by oxygen in the air to iron(III) hydroxide: \( 4\text{Fe(OH)}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3 \).

(b)
1. Weigh exactly 9.80 g of the hydrated salt in a weighing bottle or beaker on a balance.
2. Dissolve the salt in about 100 cm³ of deionised water in a beaker, adding a small volume of dilute sulfuric acid to prevent hydrolysis of the iron(II) ions.
3. Transfer the solution quantitatively to a 250 cm³ volumetric flask, rinsing the beaker and glass rod with deionised water and adding all washings to the flask.
4. Add deionised water carefully until the bottom of the meniscus is exactly on the graduation mark at eye level.
5. Stopper and invert the flask multiple times to ensure the solution is homogeneous.

(c)
- Moles of \( \text{MnO}_4^- \) in titration: \( n(\text{MnO}_4^-) = 0.0200 \text{ mol dm}^{-3} \times 0.02500 \text{ dm}^3 = 5.00 \times 10^{-4} \text{ mol} \).
- From the ionic equation, \( 5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \), the mole ratio of \( \text{Fe}^{2+} : \text{MnO}_4^- \) is \( 5:1 \).
- Moles of \( \text{Fe}^{2+} \) in 25.0 cm³: \( n(\text{Fe}^{2+}) = 5 \times (5.00 \times 10^{-4} \text{ mol}) = 2.50 \times 10^{-3} \text{ mol} \).
- Moles of \( \text{Fe}^{2+} \) in the total 250.0 cm³ solution: \( 2.50 \times 10^{-3} \text{ mol} \times 10 = 2.50 \times 10^{-2} \text{ mol} \).
- Molar mass of the hydrated salt: \( M_r = \frac{9.80 \text{ g}}{2.50 \times 10^{-2} \text{ mol}} = 392.0 \text{ g mol}^{-1} \).
- Molar mass of anhydrous \( \text{Fe(NH}_4)_2(\text{SO}_4)_2 \): \( 55.8 + 2 \times (14.0 + 4.0) + 2 \times (32.1 + 64.0) = 55.8 + 36.0 + 192.2 = 284.0 \text{ g mol}^{-1} \).
- Mass of water of crystallisation: \( 18.0x = 392.0 - 284.0 = 108.0 \text{ g mol}^{-1} \).
- Therefore, \( x = \frac{108.0}{18.0} = 6 \).

(a) [3 marks]
- Green precipitate formed upon dropwise addition [1]
- Precipitate remains insoluble in excess sodium hydroxide [1]
- Precipitate turns brown at the liquid surface/on standing in air [1]

(b) [4 marks]
- Dissolves the weighed solid in a beaker using deionised water (and adds dilute acid) [1]
- Transfers the solution to a 250 cm³ volumetric flask with rinsings/washings [1]
- Fills the flask with deionised water until the bottom of the meniscus is aligned with the graduation mark [1]
- Stoppers and inverts several times to mix thoroughly [1]

(c) [6 marks]
- Moles of manganate(VII) = \( 5.00 \times 10^{-4} \text{ mol} \) [1]
- Moles of iron(II) in 25.0 cm³ = \( 2.50 \times 10^{-3} \text{ mol} \) [1]
- Moles of iron(II) in 250.0 cm³ = \( 2.50 \times 10^{-2} \text{ mol} \) [1]
- Molar mass of hydrated salt = \( 392 \text{ g mol}^{-1} \) [1]
- Molar mass of anhydrous salt = \( 284 \text{ g mol}^{-1} \) [1]
- Correct calculation showing \( x = 6 \) [1]

[Accept alternative pathways with correct intermediate steps. Reject rounded atomic mass units if they yield non-integer results.]

(a) The nitration reaction is highly exothermic. Keeping the temperature below \( 10 \ ^\circ\text{C} \) prevents further nitration (e.g., formation of dinitrobenzoate products) and controls the reaction rate.

(b) Buchner filtration is much faster than gravity filtration and draws more liquid out, leaving the crude solid crystals significantly drier.

(c)
1. Dissolve the crude methyl 3-nitrobenzoate in the minimum volume of a hot suitable solvent (such as ethanol).
2. Filter the hot solution quickly (e.g., through a pre-warmed funnel) to remove any insoluble impurities.
3. Allow the filtrate to cool slowly to room temperature, then place it in an ice bath to allow the pure crystals of methyl 3-nitrobenzoate to recrystallise.
4. Filter the crystals using Buchner filtration, wash them with a small volume of ice-cold solvent, and dry them in a desiccator or warm oven.

(d)
- Mass of methyl benzoate used: \( 2.50 \text{ cm}^3 \times 1.09 \text{ g cm}^{-3} = 2.725 \text{ g} \).
- Moles of methyl benzoate: \( n = \frac{2.725 \text{ g}}{136.1 \text{ g mol}^{-1}} = 0.02002 \text{ mol} \).
- Theoretical maximum yield of methyl 3-nitrobenzoate: \( 0.02002 \text{ mol} \times 181.1 \text{ g mol}^{-1} = 3.626 \text{ g} \).
- Percentage yield: \( \% \text{ yield} = \frac{2.45 \text{ g}}{3.626 \text{ g}} \times 100 = 67.57\% \approx 67.6\% \).

(a) [2 marks]
- To prevent further nitration / multiple nitration / formation of dinitro products [1]
- To control the highly exothermic reaction / prevent runaway reaction/decomposition [1]

(b) [2 marks]
- It is much faster/saves time [1]
- It leaves the solid product much drier [1]

(c) [4 marks]
- Dissolve the crude solid in the minimum volume of hot solvent [1]
- Filter hot (to remove insoluble impurities) [1]
- Cool slowly (first to room temperature, then in ice) to crystallise [1]
- Filter under reduced pressure, wash with a small portion of cold solvent, and dry [1]

(d) [4 marks]
- Calculates mass of methyl benzoate = \( 2.725 \text{ g} \) [1]
- Calculates moles of methyl benzoate = \( 0.0200 \text{ mol} \) [1]
- Calculates theoretical mass of product = \( 3.63 \text{ g} \) [1]
- Calculates percentage yield = \( 67.6\% \) [1] (Accept range: \( 67.5\% - 67.7\% \))

(a) Aqueous iodine is brown/yellow and absorbs light. As the reaction proceeds, iodine is consumed and the reaction mixture becomes colorless, causing the absorbance of the solution to decrease. A calibration curve (using known concentrations of iodine) is used to convert absorbance readings directly to iodine concentration over time.

(b)
(i) Concentration of iodine against time: A straight, downward-sloping line (negative gradient) indicating a constant rate of reaction over time.
(ii) Rate of reaction against concentration of iodine: A horizontal line, showing that the rate of reaction is completely independent of the concentration of iodine.

(c) \( \text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+] \)
\( \text{Units of } k = \frac{\text{Rate}}{[\text{CH}_3\text{COCH}_3][\text{H}^+]} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \).

(d) The Arrhenius equation is \( \ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A \).
Thus, \( \text{gradient} = -\frac{E_a}{R} \).
\( -6750 \text{ K} = -\frac{E_a}{8.31 \text{ J mol}^{-1} \text{ K}^{-1}} \)
\( E_a = 6750 \times 8.31 = 56092.5 \text{ J mol}^{-1} \).
To convert to \( \text{kJ mol}^{-1} \): \( E_a = \frac{56092.5}{1000} = 56.09 \text{ kJ mol}^{-1} \approx 56.1 \text{ kJ mol}^{-1} \).

(e) By keeping the concentrations of propanone and acid very high, their concentrations remain effectively constant during the reaction. Consequently, the rate of reaction only changes as a function of iodine concentration (creating pseudo-order conditions relative to iodine).

(a) [2 marks]
- Iodine is colored (brown/yellow) and as it is consumed, absorbance decreases [1]
- A calibration curve is used to convert absorbance readings to concentrations [1]

(b) [3 marks]
- (i) Straight line with negative gradient (pointing down) [1]
- (ii) Horizontal straight line [1]
- Correct linkage of zero order to rate being independent of concentration [1]

(c) [2 marks]
- \( \text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \) (Award [1] for correct units but wrong time unit e.g., \( \text{min}^{-1} \), or [1] for reciprocal units inversion)

(d) [4 marks]
- States relation: \( \text{gradient} = -E_a / R \) [1]
- Calculates energy: \( 56092.5 \text{ J mol}^{-1} \) [1]
- Divides by 1000 to convert to \( \text{kJ mol}^{-1} \) [1]
- Correct final value \( +56.1 \) (must have positive sign or value and 3 sig figs) [1]

(e) [2 marks]
- To ensure the concentrations of propanone and hydrogen ions remain virtually constant during the reaction [1]
- So that the change in rate is only due to the changing concentration of iodine / establishes pseudo-order conditions [1]

(a)
(i) Over time, the electrodes of pH meters drift and the glass membrane undergoes changes, meaning they can yield systematically incorrect readings. Calibration ensures accurate and reliable measurements.
(ii) Rinse the pH electrode with deionised water, then submerge it in a buffer solution of a known pH (e.g., pH 7.00). Adjust the pH meter until the reading matches the buffer's value. Repeat this process with at least one other buffer of a different pH (e.g., pH 4.00 or pH 10.00) to standardise across the range of interest.

(b)
(i) At the half-neutralisation point, \( \text{pH} = \text{p}K_a \).
Therefore, \( \text{p}K_a = 4.76 \).
\( K_a = 10^{-\text{p}K_a} = 10^{-4.76} = 1.74 \times 10^{-5} \text{ mol dm}^{-3} \).

(ii) Dissociation equation: \( \text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)} \).
The acid dissociation constant expression is: \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \).
At half-neutralisation, exactly half of the weak acid \( \text{HA} \) has reacted with the added \( \text{NaOH} \) to produce the conjugate base \( \text{A}^- \). Therefore, \( [\text{HA}] = [\text{A}^-] \).
Substituting this equality into the \( K_a \) expression gives: \( K_a = [\text{H}^+] \).
Taking the negative logarithm of both sides: \( -\log_{10} K_a = -\log_{10}[\text{H}^+] \implies \text{p}K_a = \text{pH} \).

(c) Phenolphthalein is the most suitable indicator.
Reason: The pH at the equivalence point is 8.5, which falls within the color change interval of phenolphthalein (8.3 - 10.0). This range perfectly coincides with the vertical, rapid pH-change portion of the titration curve. Methyl orange and bromothymol blue would change color way before the equivalence point is reached.

(a) [3 marks]
- (i) Calibration corrects for systematic errors/drift in electrode response [1]
- (ii) Rinse the probe with deionised water and place in a buffer of known pH (e.g. pH 7.00), adjusting the scale [1]
- Standardise with a second buffer solution (e.g. pH 4.00 or 10.00) [1]

(b)(i) [2 marks]
- Identifies \( \text{pH} = \text{p}K_a = 4.76 \) [1]
- Calculates \( K_a = 1.74 \times 10^{-5} \text{ mol dm}^{-3} \) (Accept range \( 1.7 \times 10^{-5} \) to \( 1.8 \times 10^{-5} \); ignore missing/incorrect units) [1]

(b)(ii) [3 marks]
- Equation: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \) (state symbols not required but must be reversible arrow) [1]
- States that at half-neutralisation, \( [\text{HA}] = [\text{A}^-] \) [1]
- Shows substitution into \( K_a \) equation to prove \( K_a = [\text{H}^+] \) and hence \( \text{p}K_a = \text{pH} \) [1]

(c) [4 marks]
- Identifies phenolphthalein [1]
- States that the equivalence pH of 8.5 is within the color-change range of phenolphthalein (8.3 - 10.0) [1]
- States that this range lies entirely on the steep vertical section of the pH curve [1]
- Rejects methyl orange/bromothymol blue because they change color in the acidic/neutral region (before the equivalence point is reached) [1]