2024 Edexcel A-Level Chemistry (9CH0) Practice Paper with Answers

There is a massive increase between the fifth and sixth ionization energies (from 6274 to 21269 \(\text{kJ mol}^{-1}\)). This indicates that the sixth electron is being removed from a lower principal quantum shell closer to the nucleus, meaning the element has five valence electrons. In Period 3, the element with five valence electrons is Phosphorus (Group 15).

1 mark for identifying that the large jump in ionization energy occurs between the 5th and 6th ionization energies, indicating 5 valence electrons, and correctly identifying the element as Phosphorus.

To determine the electron pair arrangement around the central atom:

- \(\text{NH}_2^-\): Nitrogen has 5 valence electrons + 1 (from negative charge) = 6. Forming 2 single bonds with H leaves 2 lone pairs. Total of 4 electron pairs, which arrange tetrahedrally.

- \(\text{XeF}_4\): Xenon has 8 valence electrons. Forming 4 single bonds with F leaves 2 lone pairs. Total of 6 electron pairs, which arrange octahedrally (yielding a square planar molecular shape).

- \(\text{PCl}_4^+\): Phosphorus has 5 valence electrons - 1 (from positive charge) = 4. Forming 4 single bonds with Cl leaves 0 lone pairs. Total of 4 electron pairs, which arrange tetrahedrally.

- \(\text{BF}_4^-\): Boron has 3 valence electrons + 1 (from negative charge) = 4. Forming 4 single bonds with F leaves 0 lone pairs. Total of 4 electron pairs, which arrange tetrahedrally.

Therefore, only \(\text{XeF}_4\) is not based on a tetrahedral arrangement.

1 mark for correctly identifying that \(\text{XeF}_4\) has 6 electron pairs (octahedral arrangement, square planar geometry) whereas the others all have 4 electron pairs (tetrahedral arrangement).

Using the Born-Haber cycle:

\(\Delta H_f^{\theta}[\text{MgO(s)}] = \Delta H_{at}^{\theta}[\text{Mg(s)}] + 1\text{st IE}[\text{Mg}] + 2\text{nd IE}[\text{Mg}] + \Delta H_{at}^{\theta}[\text{O}] + 1\text{st EA}[\text{O}] + 2\text{nd EA}[\text{O}] + \Delta H_{latt}^{\theta}[\text{MgO}]\)

Substituting the values:

\(-602 = 148 + 738 + 1451 + 249 + (-141) + 798 + \Delta H_{latt}^{\theta}\)

\(-602 = 3243 + \Delta H_{latt}^{\theta}\)

\(\Delta H_{latt}^{\theta} = -602 - 3243 = -3845 \text{ kJ mol}^{-1}\).

1 mark for correctly setting up the Born-Haber cycle equation and calculating the lattice energy as \(-3845 \text{ kJ mol}^{-1}\).

1. Calculate initial moles of reactants:

\(\text{Moles of propanoic acid (HA)} = 0.0500 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.00500 \text{ mol}\)

\(\text{Moles of NaOH} = 0.0500 \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 0.00250 \text{ mol}\)

2. Determine moles at equilibrium after reaction:

\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)

Since \(\text{OH}^-\rightleftharpoons\) is the limiting reactant, it is fully consumed:

\(\text{Moles of HA remaining} = 0.00500 - 0.00250 = 0.00250 \text{ mol}\)

\(\text{Moles of conjugate base } \text{A}^- \text{ formed} = 0.00250 \text{ mol}\)

3. Use the buffer equation:

\([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}\)

Since the moles of \(\text{HA}\) and \(\text{A}^-\rightleftharpoons\) are equal in the mixture:

\([\text{H}^+] = K_a = 1.35 \times 10^{-5} \text{ mol dm}^{-3}\)

\(\text{pH} = -\log_{10}(1.35 \times 10^{-5}) \approx 4.87\).

1 mark for calculating the correct pH of \(4.87\) by finding equal amounts of weak acid and conjugate base and equating \(\text{pH}\) to \(\text{p}K_a\).

Let's find the electronic configuration of each option:

- \(\text{Cr}\) is \([\text{Ar}] 3\text{d}^5 4\text{s}^1\). Removing 3 electrons gives \(\text{Cr}^{3+}\): \([\text{Ar}] 3\text{d}^3\).

- \(\text{Mn}\) is \([\text{Ar}] 3\text{d}^5 4\text{s}^2\). Removing 2 electrons gives \(\text{Mn}^{2+}\): \([\text{Ar}] 3\text{d}^5\).

- \(\text{Fe}\) is \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). Removing 2 electrons gives \(\text{Fe}^{2+}\): \([\text{Ar}] 3\text{d}^6\).

- \(\text{Co}\) is \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). Removing 3 electrons gives \(\text{Co}^{3+}\): \([\text{Ar}] 3\text{d}^6\).

Therefore, \(\text{Mn}^{2+}\) has the \([\text{Ar}] 3\text{d}^5\) configuration.

1 mark for identifying \(\text{Mn}^{2+}\) as the correct ion with the ground state configuration \([\text{Ar}] 3\text{d}^5\).

As you go down Group 2, the size of the cation (\(\text{M}^{2+}\)) increases while its charge remains \(+2\). This causes the charge density of the cation to decrease, which in turn decreases its polarising power. A less polarising cation causes less distortion (polarisation) of the carbonate ion's electron cloud. Consequently, the carbon-oxygen bonds within the carbonate ion are weakened less, making the carbonate compound more thermally stable (i.e., its decomposition temperature increases).

1 mark for identifying that the thermal decomposition temperature increases down the group and choosing the correct explanation involving larger cation size, lower polarising power, and less distortion of the carbonate ion.

(a) The standard enthalpy of atomisation is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.

(b) Using the Born-Haber cycle relationship:
\(\Delta H_f^{\theta} = \Delta H_{at}^{\theta}(\text{Ca}) + 1st\text{ IE}(\text{Ca}) + 2nd\text{ IE}(\text{Ca}) + 2 \times \Delta H_{at}^{\theta}(\text{Cl}) + 2 \times EA(\text{Cl}) + \Delta H_{lattice}^{\theta}\)

Substituting the values:
\(-796 = 178 + 590 + 1145 + 2(121) + 2(-349) + \Delta H_{lattice}^{\theta}\)
\(-796 = 178 + 590 + 1145 + 242 - 698 + \Delta H_{lattice}^{\theta}\)
\(-796 = 1457 + \Delta H_{lattice}^{\theta}\)
\(\Delta H_{lattice}^{\theta} = -796 - 1457 = -2253\text{ kJ mol}^{-1}\).

(c) The experimental lattice energy (\(-2253\text{ kJ mol}^{-1}\)) is more exothermic than the theoretical value (\(-2223\text{ kJ mol}^{-1}\)). This indicates that the bonding in calcium chloride is not purely ionic and has some covalent character. The calcium cation (\(\text{Ca}^{2+}\)) polarises the electron cloud of the chloride anions.

(a)
- M1: Enthalpy change when one mole of gaseous atoms is formed (1)
- M2: From the element in its standard state under standard conditions (1)

(b)
- M1: Correctly doubling the enthalpy of atomisation of chlorine (\(+242\)) AND the electron affinity of chlorine (\(-698\)) (1)
- M2: Correct algebraic setup of the Born-Haber cycle equation (1)
- M3: Correct evaluation of the intermediate sum of the cycle steps (\(+1457\)) (1)
- M4: Final calculated value: \(-2253\text{ kJ mol}^{-1}\) (must have the negative sign and correct units) (1)

(c)
- M1: Mentions that the experimental value is more exothermic/negative, indicating the compound has covalent character (1)
- M2: Explains that the \(\text{Ca}^{2+}\) cation polarises the electron cloud of the chloride anion (1)

(a) Equation:
\(\text{HCOOH(aq)} + \text{NaOH(aq)} \rightarrow \text{HCOONa(aq)} + \text{H}_2\text{O(l)}\)

(b)
- Moles of methanoic acid: \(n(\text{HCOOH}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol}\)
- Moles of sodium hydroxide: \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.300\text{ mol dm}^{-3} = 0.0075\text{ mol}\)

(c)
The reaction consumes all the NaOH (limiting reactant) and converts an equal amount of methanoic acid to methanoate ions:
- Moles of \(\text{HCOO}^-\text{(aq)}\) formed = \(0.0075\text{ mol}\)
- Moles of \(\text{HCOOH(aq)}\) remaining = \(0.0125\text{ mol} - 0.0075\text{ mol} = 0.0050\text{ mol}\)

Total volume of solution = \(50.0 + 25.0 = 75.0\text{ cm}^3 = 0.075\text{ dm}^3\).
- \([\text{HCOOH}] = \frac{0.0050\text{ mol}}{0.075\text{ dm}^3} = 0.0667\text{ mol dm}^{-3}\)
- \([\text{HCOO}^-] = \frac{0.0075\text{ mol}}{0.075\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\)

Using the acid dissociation expression:
\(K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\)

Rearranging to solve for \([\text{H}^+]\):
\([\text{H}^+] = K_a \times \frac{[\text{HCOOH}]}{[\text{HCOO}^-]} = 1.60 \times 10^{-4} \times \frac{0.0667}{0.100} = 1.067 \times 10^{-4}\text{ mol dm}^{-3}\)

Calculating pH:
\(\text{pH} = -\log_{10}(1.067 \times 10^{-4}) = 3.97\)

(a)
- M1: Correct balanced chemical or ionic equation (1)

(b)
- M1: Correct calculation of moles of \(\text{HCOOH}\) (\(0.0125\text{ mol}\)) (1)
- M2: Correct calculation of moles of \(\text{NaOH}\) (\(0.0075\text{ mol}\)) (1)

(c)
- M1: Deduces moles of methanoic acid remaining (\(0.0050\text{ mol}\)) (1)
- M2: Deduces moles of methanoate ion produced (\(0.0075\text{ mol}\)) (1)
- M3: Calculates concentrations of both components or uses mole ratio directly in the buffer equation (1)
- M4: Correct calculation of hydrogen ion concentration, \([\text{H}^+] = 1.067 \times 10^{-4}\text{ mol dm}^{-3}\) (1)
- M5: Correct calculation of pH to 2 decimal places = \(3.97\) (1)

(a) The coordination number of cobalt is 6. A bidentate ligand is a molecule or ion that donates two lone pairs of electrons to a central metal ion to form two coordinate (dative covalent) bonds.

(b) The three stereoisomers consist of:
1. The *trans*-isomer, where the two chloride ligands are opposite each other (180 degrees apart). This isomer has a plane of symmetry and is optically inactive.
2. A pair of *cis*-isomers (enantiomers), where the two chloride ligands are adjacent to each other (90 degrees apart). These are non-superimposable mirror images of each other and are optically active.

(c) The presence of ligands causes the d-orbitals of the transition metal ion to split into two groups of different energy levels. Electrons absorb a specific frequency of visible light to be promoted from a lower d-orbital to a higher d-orbital (\(\Delta E = h\nu\)). The remaining wavelengths of light that are not absorbed are transmitted or reflected, which we perceive as the complementary color.

(a)
- M1: States coordination number is 6 (1)
- M2: Defines bidentate ligand as donating two lone pairs of electrons to form two coordinate bonds (1)

(b)
- M1: Draws the *trans*-isomer correctly with 3D representation (1)
- M2: Draws both enantiomers of the *cis*-isomer as mirror images (2)
- M3: Clearly identifies that the *cis*-isomers are optically active (or the *trans* is optically inactive) (1)

(c)
- M1: Explains d-orbital splitting in the presence of ligands (1)
- M2: Mentions electron promotion/excitation absorbing visible light, with the remaining light being transmitted/reflected (1)

1. Calculate the initial amount of \(\text{HCl}\) added:
\(n(\text{HCl})_{\text{initial}} = \frac{50.00}{1000} \times 0.500 = 0.0250\text{ mol}\)

2. Calculate the amount of \(\text{NaOH}\) used in the titration:
\(n(\text{NaOH}) = \frac{21.45}{1000} \times 0.100 = 0.002145\text{ mol}\)

3. Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the moles of excess \(\text{HCl}\) in the \(25.00\text{ cm}^3\) aliquot is equal to \(0.002145\text{ mol}\).

4. Scale up to the total \(250.0\text{ cm}^3\) volumetric flask:
\(n(\text{HCl})_{\text{excess total}} = 0.002145 \times \frac{250.0}{25.00} = 0.02145\text{ mol}\)

5. Calculate the amount of \(\text{HCl}\) that reacted with \(\text{BaCO}_3\):
\(n(\text{HCl})_{\text{reacted}} = 0.0250\text{ mol} - 0.02145\text{ mol} = 0.00355\text{ mol}\)

6. The reaction equation is:
\(\text{BaCO}_3 + 2\text{HCl} \rightarrow \text{BaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\)
Therefore, the amount of \(\text{BaCO}_3\) is:
\(n(\text{BaCO}_3) = \frac{0.00355\text{ mol}}{2} = 0.001775\text{ mol}\)

7. Calculate the mass of pure \(\text{BaCO}_3\):
\(M_r(\text{BaCO}_3) = 137.3 + 12.0 + (3 \times 16.0) = 197.3\text{ g mol}^{-1}\)
\(\text{Mass} = 0.001775\text{ mol} \times 197.3\text{ g mol}^{-1} = 0.3502\text{ g}\)

8. Calculate the percentage purity:
\(\text{Percentage Purity} = \frac{0.3502\text{ g}}{1.350\text{ g}} \times 100\% = 25.94\% \approx 25.9\%\)

- M1: Calculates initial moles of \(\text{HCl}\) = \(0.0250\text{ mol}\) (1)
- M2: Calculates moles of \(\text{NaOH}\) in titration = \(0.002145\text{ mol}\) (1)
- M3: Scales up to find total excess moles of \(\text{HCl}\) in \(250.0\text{ cm}^3\) = \(0.02145\text{ mol}\) (1)
- M4: Calculates moles of \(\text{HCl}\) reacted = \(0.00355\text{ mol}\) (1)
- M5: Uses the 1:2 stoichiometry to find moles of \(\text{BaCO}_3\) = \(0.001775\text{ mol}\) (1)
- M6: Calculates molar mass of \(\text{BaCO}_3\) = \(197.3\text{ g mol}^{-1}\) (1)
- M7: Calculates the mass of pure \(\text{BaCO}_3\) = \(0.3502\text{ g}\) (1)
- M8: Expresses purity to three significant figures = \(25.9\%\) (1)

(a) (i) Element X is aluminium (\(\text{Al}\)). There is a large jump between the 3rd and 4th ionisation energies (from \(2745\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is removed from a shell closer to the nucleus (inner shell) experiencing a much stronger electrostatic attraction. Therefore, the element has 3 outer electrons, which corresponds to Group 3 (Group 13), so the element is aluminium.
(ii) \(\text{Al}^{2+}\text{(g)} \rightarrow \text{Al}^{3+}\text{(g)} + \text{e}^-\)

(b) Across Period 3:
1. The nuclear charge (number of protons) increases.
2. The outer electrons are in the same main energy level, so they experience similar shielding.
3. This leads to a stronger electrostatic attraction between the nucleus and outer shell electrons, drawing them closer (reducing atomic radius) and requiring more energy to remove an electron.

(c) Phosphorus has the electronic configuration \([\text{Ne}] 3s^2 3p^3\) (each \(3p\) orbital is singly occupied), whereas sulfur has \([\text{Ne}] 3s^2 3p^4\). In sulfur, one of the \(3p\) orbitals contains a pair of electrons. The mutual repulsion between these two paired electrons in the same orbital makes it easier to remove one, resulting in a lower first ionisation energy.

(a) (i)
- M1: Identifies element as aluminium (or Al) (1)
- M2: Mentions the large jump between the 3rd and 4th ionisation energies (1)
- M3: Connects this to the removal of the 4th electron from an inner shell / indicates 3 valence electrons (1)
(ii)
- M1: Correct equation: \(\text{Al}^{2+}\text{(g)} \rightarrow \text{Al}^{3+}\text{(g)} + \text{e}^-\). (State symbols must be gas) (1)

(b)
- M1: Number of protons / nuclear charge increases (1)
- M2: Shielding remains constant / similar (1)
- M3: Stronger attraction between nucleus and outer electrons / atomic radius decreases (1)

(c)
- M1: Identifies that sulfur has a pair of electrons in a \(3p\) orbital which repel each other (1)

(a) Reduction half-equation:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)

(b) Overall ionic equation:
\(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)

(c) (i)
- Moles of \(\text{MnO}_4^-\) in average titre:
\(n(\text{MnO}_4^-) = 0.01610\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 3.22 \times 10^{-4}\text{ mol}\)

- From the 1:5 stoichiometry, moles of \(\text{Fe}^{2+}\) in the \(25.00\text{ cm}^3\) sample:
\(n(\text{Fe}^{2+}) = 5 \times 3.22 \times 10^{-4} = 1.61 \times 10^{-3}\text{ mol}\)

- Moles of \(\text{Fe}^{2+}\) in the total \(250.0\text{ cm}^3\) solution:
\(n(\text{Fe}^{2+})_{\text{total}} = 1.61 \times 10^{-3} \times 10 = 1.61 \times 10^{-2}\text{ mol}\)

- Since \(1\text{ mol of Fe}^{2+}\) corresponds to \(1\text{ mol of FeSO}_4\), the mass of \(\text{FeSO}_4\) is:
\(\text{Mass} = 1.61 \times 10^{-2}\text{ mol} \times 151.9\text{ g mol}^{-1} = 2.44559\text{ g}\)

- Rounded to 3 significant figures: \(2.45\text{ g}\).

(ii) The end point color change is from colorless (or very pale green) to permanent pale pink.

(a)
- M1: Correctly balanced ionic half-equation (1)

(b)
- M1: Correct overall ionic equation (1)

(c) (i)
- M1: Calculates moles of \(\text{MnO}_4^-\) = \(3.22 \times 10^{-4}\text{ mol}\) (1)
- M2: Applies stoichiometry (1:5 ratio) to find moles of \(\text{Fe}^{2+}\) in \(25.00\text{ cm}^3\) = \(1.61 \times 10^{-3}\text{ mol}\) (1)
- M3: Scales up to the full volume of \(250.0\text{ cm}^3\) (multiplies by 10) = \(1.61 \times 10^{-2}\text{ mol}\) (1)
- M4: Calculates mass of \(\text{FeSO}_4\) = \(2.45\text{ g}\) (must be 3 s.f.) (1)

(ii)
- M1: States color change from colorless (or pale green) to pale pink (1)

(a) (i) Sulfur hexafluoride (\(\text{SF}_6\)) has an **octahedral** shape with a bond angle of **\(90^\circ\)**. There are 6 bonding pairs of electrons and 0 lone pairs around the central sulfur atom. According to VSEPR theory, these electron pairs repel each other equally and arrange themselves as far apart as possible to minimise repulsion.

(ii) The hydronium ion (\(\text{H}_3\text{O}^+\)) has a **trigonal pyramidal** shape with a bond angle of approximately **\(107^\circ\)**. The central oxygen atom has 3 bonding pairs and 1 lone pair of electrons. The lone pair-bonding pair repulsion is stronger than the bonding pair-bonding pair repulsion, which pushes the O-H bonds closer together, compressing the angle from the tetrahedral angle of \(109.5^\circ\).

(b) Water molecules form **hydrogen bonds** between the highly electronegative oxygen atom of one molecule and the hydrogen atom of another. These intermolecular forces are much stronger than the weaker London forces and permanent dipole-dipole interactions found between hydrogen sulfide (\(\text{H}_2\text{S}\)) molecules. Consequently, significantly more thermal energy is needed to overcome the hydrogen bonds in water, leading to a much higher boiling temperature.

(a) (i)
- M1: Octahedral shape (1)
- M2: \(90^\circ\) bond angle (1)
- M3: Explanation based on 6 bonding pairs and 0 lone pairs repelling to minimise repulsion (1)

(a) (ii)
- M1: Trigonal pyramidal shape (1)
- M2: Bond angle of \(107^\circ\) (accept \(106^\circ - 108^\circ\)) (1)
- M3: Explanation of 3 bonding pairs and 1 lone pair, where lone pair repulsion is greater than bonding pair repulsion (1)

(b)
- M1: Mentions water forms hydrogen bonds, whereas \(\text{H}_2\text{S}\) has weaker London/dipole-dipole forces (1)
- M2: States that more energy is required to break the hydrogen bonds in water (1)

(a) As you go down Group 2, the thermal stability of the nitrates increases (they require more heat/higher temperatures to decompose).
Explanation:
1. The ionic radius/size of the Group 2 cation increases down the group while keeping the same charge (\(2+\)).
2. Consequently, the charge density of the cation decreases down the group.
3. The cation with lower charge density has less polarizing power, meaning it distorts the electron cloud of the nitrate ion less.
4. This causes less weakening of the covalent N-O bonds within the nitrate anion, making it harder to break and more thermally stable.

(b) The balanced equation is:
\(2\text{Mg(NO}_3)_2\text{(s)} \rightarrow 2\text{MgO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\)

(c) Observations include: the evolution of a brown gas (nitrogen dioxide, \(\text{NO}_2\)), or a white solid/residue remains (magnesium oxide, \(\text{MgO}\)), or a glowing splint relights (oxygen, \(\text{O}_2\)).

(a)
- M1: Thermal stability increases down the group (1)
- M2: Cation size / ionic radius increases (or charge density decreases) (1)
- M3: Polarising power of the cation decreases (1)
- M4: Polarisation / distortion of the nitrate ion's electron cloud decreases (1)
- M5: Weakening of the \(\text{N}-\text{O}\) covalent bond decreases (1)

(b)
- M1: Correct formulas and balancing: \(2\text{Mg(NO}_3)_2 \rightarrow 2\text{MgO} + 4\text{NO}_2 + \text{O}_2\) (1)
- M2: Correct state symbols: \(\text{(s)}\) for reactants and \(\text{MgO}\), \(\text{(g)}\) for product gases (1)

(c)
- M1: Brown gas produced / white solid formed / glowing splint relights (1)

### Part (a)
Standard enthalpy of atomisation is the enthalpy change when **one mole of gaseous atoms** is formed from its **element in its standard state** under standard conditions.

### Part (b)
To calculate the experimental lattice energy (\(\Delta H_{\text{latt}}^{\ominus}\)), we set up a Born-Haber cycle where:

\(\Delta H_{\text{f}}^{\ominus}[\text{CaI}_2(\text{s})] = \Delta H_{\text{at}}^{\ominus}[\text{Ca}(\text{s})] + 1^{\text{st}}\text{IE}[\text{Ca}(\text{g})] + 2^{\text{nd}}\text{IE}[\text{Ca}(\text{g})] + 2 \times \Delta H_{\text{at}}^{\ominus}[\text{I}_2(\text{s})] + 2 \times 1^{\text{st}}\text{EA}[\text{I}(\text{g})] + \Delta H_{\text{latt}}^{\ominus}[\text{CaI}_2(\text{s})]\)

Substitute the given values:
\(-534 = +178 + 590 + 1145 + 2(107) + 2(-295) + \Delta H_{\text{latt}}^{\ominus}\)

\(-534 = 178 + 590 + 1145 + 214 - 590 + \Delta H_{\text{latt}}^{\ominus}\)

\(-534 = 1537 + \Delta H_{\text{latt}}^{\ominus}\)

\(\Delta H_{\text{latt}}^{\ominus} = -534 - 1537 = -2071 \text{ kJ mol}^{-1}\)

### Part (c)
* The calcium ion (\(\text{Ca}^{2+}\)) polarises the iodide (\(\text{I}^{-}\)) ion (due to its high charge density and the large, polarisable electron cloud of the iodide ion).
* This leads to a degree of covalent character in the bonding in addition to the ionic bonding. This covalent character makes the bond stronger, releasing more energy when the lattice forms.

### Part (a) [2 Marks]
* **M1**: Enthalpy change when 1 mole of gaseous atoms is formed [1]
* **M2**: from its element in its standard state under standard conditions [1]

### Part (b) [4 Marks]
* **M3**: Multiplying atomisation enthalpy of iodine by 2 (\(2 \times 107 = 214\)) AND electron affinity of iodine by 2 (\(2 \times -295 = -590\)) [1]
* **M4**: Setting up a correct algebraic equation/cycle expressing \(\Delta H_{\text{latt}}^{\ominus}\) as the subject:
\(\Delta H_{\text{latt}}^{\ominus} = \Delta H_{\text{f}}^{\ominus} - (\Delta H_{\text{at}}^{\ominus}[\text{Ca}] + 1^{\text{st}}\text{IE}[\text{Ca}] + 2^{\text{nd}}\text{IE}[\text{Ca}] + 2\Delta H_{\text{at}}^{\ominus}[\text{I}] + 2\text{EA}[\text{I}])\) [1]
* **M5**: Correct substitution of values:
\(-534 - (178 + 590 + 1145 + 214 - 590)\) [1]
* **M6**: Correct evaluation to give \(-2071 \text{ (kJ mol}^{-1}\text{)}\) [1]
*(Allow full marks for correct answer with no working. Allow 3 marks for positive value +2071)*

### Part (c) [2 Marks]
* **M7**: Identifies polarization of the iodide ion (by the calcium ion) / iodide ion is large and easily distorted [1]
* **M8**: State that this introduces covalent character (which makes the bonds stronger/more exothermic) [1]

### Part (a)
\(K_{\text{a}} = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)

### Part (b)
1. Calculate initial moles of propanoic acid (\(\text{HA}\)):
\(n(\text{HA})_{\text{initial}} = \frac{50.0}{1000} \times 0.250 = 0.0125 \text{ mol}\)

2. Calculate moles of \(\text{NaOH}\) added (equal to moles of \(\text{A}^-\)\/propanoate ions formed):
\(n(\text{NaOH}) = n(\text{A}^-)_{\text{formed}} = \frac{25.0}{1000} \times 0.150 = 0.00375 \text{ mol}\)

3. Calculate the moles of remaining unreacted propanoic acid:
\(n(\text{HA})_{\text{remaining}} = 0.0125 - 0.00375 = 0.00875 \text{ mol}\)

4. Use the \(K_{\text{a}}\) expression rearranged for \([\text{H}^+]\):
\([\text{H}^+] = K_{\text{a}} \times \frac{[\text{HA}]}{[\text{A}^-]} = K_{\text{a}} \times \frac{n(\text{HA})}{n(\text{A}^-)}\)
\([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.00875}{0.00375}\)
\([\text{H}^+] = 3.15 \times 10^{-5} \text{ mol dm}^{-3}\)

5. Calculate the pH:
\(\text{pH} = -\log_{10}(3.15 \times 10^{-5}) = 4.5017 \approx 4.50\)

### Part (c)
* When \(\text{H}^+\) ions are added, they react with the conjugate base (propanoate ions) in the buffer solution to form propanoic acid:
\(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
* Because there is a large reservoir of propanoate ions, the ratio of \(\frac{[\text{HA}]}{[\text{A}^-]}\) remains almost constant, so the pH changes very little.

### Part (a) [1 Mark]
* **M1**: Correctly written equation: \(K_{\text{a}} = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\) (State symbols not needed, but must have square brackets) [1]

### Part (b) [5 Marks]
* **M2**: Calculates initial moles of propanoic acid = \(0.0125 \text{ mol}\) [1]
* **M3**: Calculates moles of sodium hydroxide added = \(0.00375 \text{ mol}\) (which is also the moles of conjugate base formed) [1]
* **M4**: Calculates moles of excess/remaining propanoic acid = \(0.00875 \text{ mol}\) [1]
* **M5**: Re-arranges and substitutes values into \(K_{\text{a}}\) to find \([\text{H}^+] = 3.15 \times 10^{-5} \text{ mol dm}^{-3}\) [1]
* **M6**: Correct pH calculation to 2 decimal places = \(4.50\) [1]
*(Note: consecutive error propagation (TE) is allowed for calculation steps)*

### Part (c) [2 Marks]
* **M7**: Identifies that added \(\text{H}^+\) ions react with propanoate ions (or writes equation) [1]
* **M8**: States that the ratio of \(\frac{[\text{acid}]}{[\text{conjugate base}]}\) (or \([\text{H}^+]\)) remains virtually constant [1]

According to the Arrhenius equation:
\(\ln(k) = -\frac{E_a}{RT} + \ln(A)\)

Thus, a plot of \(\ln(k)\) against \(1/T\) gives a straight line with a gradient of \(-\frac{E_a}{R}\).

\(\text{Gradient} = -\frac{E_a}{R} = -1.20 \times 10^4\text{ K}\)

\(E_a = 1.20 \times 10^4\text{ K} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 99\,720\text{ J mol}^{-1} = +99.7\text{ kJ mol}^{-1}\).

1 mark for the correct calculation of \(+99.7\text{ kJ mol}^{-1}\).

Propanal is an aldehyde with a planar carbonyl (\(C=O\)) group. The cyanide nucleophile (\(CN^-\)) can attack this planar carbon atom with equal probability from either above or below the plane. This produces an equimolar mixture of two enantiomers (a racemic mixture), which is optically inactive overall.

1 mark for identifying that the product is a racemic mixture due to equal probability of attack on either side of the planar carbonyl group.

The structural formula of ethyl 2-methylpropanoate is \((CH_3)_2CHCOOCH_2CH_3\).

Let's analyze the carbon environments:
1. The two equivalent methyl carbons on the isopropyl group: \((CH_3)_2CH-\) (1 peak)
2. The \(-CH-\) carbon of the isopropyl group (1 peak)
3. The ester carbonyl carbon: \(-CO-\) (1 peak)
4. The methylene carbon of the ethyl group: \(-OCH_2-\) (1 peak)
5. The methyl carbon of the ethyl group: \(-CH_3\) (1 peak)

This gives a total of 5 distinct environments, so there are 5 peaks in the carbon-13 NMR spectrum.

1 mark for identifying the 5 unique carbon environments.

The rate of nucleophilic substitution of halogenoalkanes depends on two factors:
1. Carbon-halogen bond strength: C-I is the weakest bond and breaks easiest, so iodoalkanes react much faster than bromoalkanes or chloroalkanes.
2. Carbocation stability (mechanism): Tertiary halogenoalkanes react via the \(S_N1\) mechanism, which involves a stable tertiary carbocation intermediate, making it very fast.

Hence, 2-iodo-2-methylpropane is a tertiary iodoalkane and will react the fastest.

1 mark for selecting the correct tertiary iodoalkane.

1. The infrared absorption at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl group (\(C=O\)), which rules out propan-1-ol and prop-2-en-1-ol.
2. The remaining choices are the carbonyl isomers propanal (\(CH_3CH_2CHO\)) and propanone (\(CH_3COCH_3\)).
3. In the mass spectrum, the fragment at \(m/z = 43\) corresponds to the stable acylium ion, \([CH_3CO]^+\) (mass \(15 + 28 = 43\)), which is formed by the loss of a methyl radical from propanone:
\([CH_3COCH_3]^{+\bullet} \rightarrow [CH_3CO]^+ + \bullet CH_3\)
Propanal would preferentially fragment to give a peak at \(m/z = 29\) due to \([CH_3CH_2]^+\) or \([CHO]^+\). Thus, the compound must be propanone.

1 mark for identifying the correct carbonyl compound using the infrared and mass spectrometry data.

1. Calculate the moles of ethanol reactant:
\(n(\text{ethanol}) = \frac{11.5\text{ g}}{46.0\text{ g mol}^{-1}} = 0.250\text{ mol}\)

2. The reaction has a 1:1 stoichiometry, so the theoretical yield of ethyl ethanoate is \(0.250\text{ mol}\).

3. Calculate the theoretical mass of ethyl ethanoate:
\(m(\text{theoretical}) = 0.250\text{ mol} \times 88.0\text{ g mol}^{-1} = 22.0\text{ g}\)

4. Calculate the percentage yield:
\(\text{Percentage Yield} = \frac{\text{Actual Mass}}{\text{Theoretical Mass}} \times 100\% = \frac{13.2\text{ g}}{22.0\text{ g}} \times 100\% = 60.0\%\)

1 mark for the correct calculation showing a \(60.0\%\) yield.

Part (a): The mechanism is a nucleophilic addition. 1. The nucleophile is \(\text{CN}^-\). Draw a curly arrow from the lone pair on the carbon of \(\text{CN}^-\) to the carbonyl carbon (\(\text{C}^{\delta+}\)) of propanal. 2. Draw a curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen (\(\text{O}^{\delta-}\)) atom. 3. Draw the structure of the intermediate: \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\). 4. Draw a curly arrow from the lone pair on the negative oxygen atom to a hydrogen ion (\(\text{H}^+\) from the acid). 5. The final product is 2-hydroxybutanenitrile. Part (b): The carbonyl carbon in propanal is planar (trigonal planar geometry). The nucleophile (\(\text{CN}^-\)) has an equal probability of attacking the carbonyl carbon from either above or below the plane. This produces an equimolar (50:50) mixture of the two enantiomers (a racemic mixture). Since the two enantiomers rotate plane-polarized light in opposite directions by equal amounts, the optical activity cancels out, and the product is optically inactive. Part (c): Compound X is 2-hydroxybutanoic acid. The reaction type is acid-catalyzed hydrolysis (or simply hydrolysis).

Part (a): 5.5 marks total
- 1 mark for correct dipoles (\(\text{C}^{\delta+}=\text{O}^{\delta-}\)) on the propanal molecule.
- 1 mark for curly arrow from the lone pair of the carbon of \(\text{CN}^-\) to the carbonyl carbon.
- 1 mark for curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen.
- 1 mark for correct intermediate structure with negative charge on oxygen.
- 1 mark for curly arrow from the lone pair on the \(\text{O}^-\)` of the intermediate to \(\text{H}^+\) (or water).
- 0.5 mark for the correct structure of 2-hydroxybutanenitrile.

Part (b): 3 marks total
- 1 mark for stating that the carbonyl carbon/group of propanal is planar.
- 1 mark for stating that the nucleophile can attack equally from above or below the plane.
- 1 mark for stating that this leads to an equimolar / 50:50 mixture of enantiomers (racemic mixture) which cancel out optical activity.

Part (c): 2 marks total
- 1 mark for correct IUPAC name: 2-hydroxybutanoic acid (Reject: hydroxybutanoic acid without number).
- 1 mark for identifying the reaction type as (acid) hydrolysis.

Part (a): Comparing Exp 1 and Exp 2: \([\text{H}_2]\) is constant. \([\text{NO}]\) is doubled (from 0.10 to 0.20). The rate increases by a factor of \(\frac{4.8 \times 10^{-3}}{1.2 \times 10^{-3}} = 4\). Since \(2^2 = 4\), the order with respect to \(\text{NO}\) is 2. Comparing Exp 1 and Exp 3: \([\text{NO}]\) is constant. \([\text{H}_2]\) is doubled (from 0.10 to 0.20). The rate increases by a factor of \(\frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = 2\). Since \(2^1 = 2\), the order with respect to \(\text{H}_2\) is 1. Part (b): The rate equation is: \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\). Part (c): Rearranging the rate equation: \(k = \frac{\text{Rate}}{[\text{NO}]^2[\text{H}_2]}\). Substituting values from Experiment 1: \(k = \frac{1.2 \times 10^{-3}}{(0.10)^2 \times 0.10} = \frac{1.2 \times 10^{-3}}{1.0 \times 10^{-3}} = 1.2\). Units: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\). Part (d): The rate-determining step must involve the reactants in the rate equation in their stoichiometric ratios: \(2\text{NO} + \text{H}_2\). Step 1 (Slow): \(2\text{NO} + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}\). Step 2 (Fast): \(N_2O + H_2 \rightarrow N_2 + H_2O\). The sum of these steps is: \(2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}\), which matches the overall equation.

Part (a): 4 marks total
- 1 mark for comparing Exp 1 & 2 and showing that doubling [NO] quadruples the rate.
- 1 mark for deducing order with respect to [NO] is 2.
- 1 mark for comparing Exp 1 & 3 and showing that doubling [H2] doubles the rate.
- 1 mark for deducing order with respect to [H2] is 1.

Part (b): 1 mark total
- 1 mark for: Rate = k[NO]^2[H2] (Accept capital K, accept lowercase rate, must have square brackets).

Part (c): 3.5 marks total
- 1 mark for correct rearrangement of the equation.
- 1 mark for correct calculation of value (1.2).
- 1.5 marks for correct units: mol^-2 dm^6 s^-1 (or dm^6 mol^-2 s^-1) (Allow 0.5 mark for mol^-2 dm^6 or s^-1 if partially correct).

Part (d): 2 marks total
- 1 mark for Step 1 showing 2NO + H2 reactant side with products that balance the next step (e.g., N2O + H2O or N2O2 + H2) and labeled as 'slow' or 'rate-determining'.
- 1 mark for Step 2 showing remaining intermediates and reactants reacting to give the correct overall balanced equation (labeled as 'fast').

Part (a)(i): The repeating unit of Terylene consists of an ester linkage. The acid monomer is benzene-1,4-dicarboxylic acid (forming \(-\text{C}(=\text{O})-\text{C}_6\text{H}_4-\text{C}(=\text{O})-\)) and the alcohol monomer is ethane-1,2-diol (forming \(-\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\)). Connecting them gives the repeating unit: \(-[\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\text{C}(=\text{O})-\text{C}_6\text{H}_4-\text{C}(=\text{O})]-\). Brackets must show trailing bonds passing through them. Part (a)(ii): The type of reaction is condensation polymerisation. The small molecule released is water (\(\text{H}_2\text{O}\)). Part (b)(i): The diacyl chloride is benzene-1,4-dicarbonyl chloride. Its skeletal structure features a central benzene ring with \(-\text{COCl}\) groups at positions 1 and 4: \(\text{Cl-C}(=\text{O})-\text{C}_6\text{H}_4-\text{C}(=\text{O})-\text{Cl}\). Part (b)(ii): Advantages: 1. The reaction is much faster at room temperature (does not require heating/catalyst). 2. The reaction goes to completion and is not reversible, giving a higher yield. Disadvantage: The byproduct is hydrogen chloride (\(\text{HCl}\)) gas, which is toxic and highly corrosive. Part (b)(iii): The small molecule released is hydrogen chloride (or \(\text{HCl}\)).

Part (a)(i): 2.5 marks total
- 1 mark for correct ester linkage \(-\text{O}-\text{C}(=\text{O})-\).
- 1 mark for the correct diol and dicarboxylate carbon backbones.
- 0.5 mark for correct trailing bonds at both ends passing through brackets.

Part (a)(ii): 2 marks total
- 1 mark for: Condensation (polymerisation).
- 1 mark for: Water (or \(\text{H}_2\text{O}\)).

Part (b)(i): 1.5 marks total
- 1.5 marks for correct skeletal structure of benzene-1,4-dicarbonyl chloride (1 mark if formula is correct but not skeletal).

Part (b)(ii): 3.5 marks total
- 1 mark for first advantage: Faster reaction / occurs at room temperature / does not require a catalyst.
- 1 mark for second advantage: Not reversible / goes to completion / higher yield.
- 1.5 marks for the disadvantage: Corrosive / toxic / hazardous gaseous byproduct (\(\text{HCl}\)) is formed.

Part (b)(iii): 1 mark total
- 1 mark for: Hydrogen chloride (Accept \(\text{HCl}\)).

Part (a): The strong, broad absorption at \(3300\text{ cm}^{-1}\) is characteristic of an alcohol \(\text{O}-\text{H}\) bond. The strong, sharp absorption at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl \(\text{C}=\text{O}\) bond (ketone). Part (b): Deuterium (\(\text{D}\) or \(^2\text{H}\)) exchange occurs. The deuterium in \(\text{D}_2\text{O}\) replaces the labile proton on the hydroxyl (\(-\text{OH}\)) group: \(\text{R-OH} + \text{D}_2\text{O} \rightleftharpoons \text{R-OD} + \text{HOD}\). Since deuterium does not produce a peak in standard \(^1\text{H}\) NMR spectroscopy, the signal at \(\delta = 3.5\text{ ppm}\) disappears, confirming it was due to an \(-\text{OH}\) proton. Part (c): Let's analyze the peaks: 1. Singlet at \(\delta = 2.1\text{ ppm}\) (3H) indicates a methyl group attached directly to a carbonyl group: \(\text{CH}_3-\text{C}(=\text{O})-\). It is a singlet because there are no hydrogen atoms on the adjacent carbonyl carbon. 2. Doublet at \(\delta = 1.3\text{ ppm}\) (3H) indicates a methyl group adjacent to a carbon with exactly 1 hydrogen atom: \(\text{CH}_3-\text{CH}-\). 3. Singlet at \(\delta = 3.5\text{ ppm}\) (1H) is the hydroxyl proton (\(-\text{OH}\)), which does not couple with adjacent protons. 4. Quartet at \(\delta = 4.2\text{ ppm}\) (1H) is the CH proton adjacent to the methyl group: \(-\text{CH}(\text{OH})-\text{CH}_3\). It is split into a quartet by the three neighboring protons on the methyl group. Combining these fragments: \(\text{CH}_3-\text{CO}-\text{CH}(\text{OH})-\text{CH}_3\). The name is 3-hydroxybutan-2-one.

Part (a): 2 marks total
- 1 mark for: Alcohol (or hydroxyl) group / O-H bond (at 3300 cm^-1).
- 1 mark for: Carbonyl group / C=O bond / ketone (at 1715 cm^-1).

Part (b): 2.5 marks total
- 1 mark for explaining that deuterium replaces the hydrogen of the hydroxyl group.
- 1 mark for explaining that deuterium does not absorb at the proton NMR frequency, causing the peak to disappear.
- 0.5 mark for equation: ROH + D2O -> ROD + HOD (or similar).

Part (c): 6 marks total
- 1.5 marks for deducing that the singlet at 2.1 ppm represents a CH3 group adjacent to a C=O carbonyl.
- 1.5 marks for deducing that the doublet at 1.3 ppm represents a CH3 group adjacent to a CH carbon.
- 1.5 marks for deducing that the quartet at 4.2 ppm represents a CH proton adjacent to a CH3 group, heavily deshielded by the adjacent oxygen and carbonyl.
- 1.5 marks for drawing the correct structural formula of 3-hydroxybutan-2-one with all hydrogens assigned correctly to the corresponding NMR signals.

Part (a): \(K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\). Part (b): Using the approximation: \(K_a \approx \frac{[\text{H}^+]^2}{[\text{HCOOH}]_{\text{initial}}}\). So, \([\text{H}^+] = \sqrt{K_a \times [\text{HCOOH}]} = \sqrt{1.6 \times 10^{-4} \times 0.150} = \sqrt{2.4 \times 10^{-5}} = 4.90 \times 10^{-3}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(4.90 \times 10^{-3}) = 2.31\). Assumptions: 1. \([\text{H}^+] = [\text{HCOO}^-]\) (dissociation of water is negligible). 2. \([\text{HCOOH}]_{\text{equilibrium}} \approx [\text{HCOOH}]_{\text{initial}}\) (dissociation of the weak acid is negligible). Part (c)(i): Reaction: \(\text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O}\). Initial moles: \(\text{Moles of HCOOH} = 0.0250 \times 0.150 = 3.75 \times 10^{-3}\text{ mol}\). \(\text{Moles of NaOH} = 0.0150 \times 0.100 = 1.50 \times 10^{-3}\text{ mol}\). After neutralisation: \(\text{Moles of HCOOH remaining} = 3.75 \times 10^{-3} - 1.50 \times 10^{-3} = 2.25 \times 10^{-3}\text{ mol}\). \(\text{Moles of HCOO}^-\text{ formed} = 1.50 \times 10^{-3}\text{ mol}\). Using Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCOO}^-]}{[\text{HCOOH}]}\right)\). \(\text{p}K_a = -\log_{10}(1.6 \times 10^{-4}) = 3.80\). \(\text{pH} = 3.80 + \log\left(\frac{1.50 \times 10^{-3}}{2.25 \times 10^{-3}}\right) = 3.80 + (-0.176) = 3.62\). Part (c)(ii): When hydrochloric acid (\(\text{H}^+\)) is added, it reacts with the conjugate base, methanoate ions: \(\text{HCOO}^- + \text{H}^+ \rightarrow \text{HCOOH}\). Because the reservoir of \(\text{HCOO}^-\) is large, the ratio \([\text{HCOO}^-]/[\text{HCOOH}]\) changes very little, resisting any significant change in pH.

Part (a): 1 mark total
- 1 mark for correct expression: Ka = [H+][HCOO-] / [HCOOH] (state symbols not required, square brackets required).

Part (b): 4 marks total
- 1 mark for calculating [H+] = 4.90 x 10^-3 mol dm^-3.
- 1 mark for correct pH = 2.31 (must be 2 d.p.).
- 1 mark for assumption 1: [H+] = [HCOO-] (or dissociation of water is negligible).
- 1 mark for assumption 2: [HCOOH]_eqm = [HCOOH]_initial (or dissociation of acid is negligible).

Part (c)(i): 4.5 marks total
- 1 mark for calculating initial moles of HCOOH (3.75 x 10^-3) and NaOH (1.50 x 10^-3).
- 1 mark for calculating remaining HCOOH (2.25 x 10^-3 mol) and HCOO^- formed (1.50 x 10^-3 mol).
- 1 mark for pKa = 3.80.
- 1.5 marks for correct final pH of 3.62 (allow 3.6). (Accept alternative correct methods using Ka expression directly).

Part (c)(ii): 1 mark total
- 1 mark for stating that added H+ reacts with HCOO^- to form HCOOH (or providing the equation HCOO^- + H^+ -> HCOOH).

Part (a)(i): Reagents: concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)). Conditions: Temperature maintained between \(50^\circ\text{C}\) and \(55^\circ\text{C}\) (to prevent dinitration). Part (a)(ii): \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\). (Alternatively: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_2\text{O} + \text{HSO}_4^-\)). Part (a)(iii): 1. A curly arrow from the benzene ring delocalised system to the electrophile \(\text{NO}_2^+\) at position 4. 2. Formation of the intermediate carbocation showing a broken circle (horseshoe) with the opening towards the tetrahedral carbon (bonded to both \(\text{H}\) and \(\text{NO}_2\)), and a positive charge inside the ring. 3. A curly arrow from the \(\text{C}-\text{H}\) bond at position 4 back into the ring to restore the aromatic system. 4. Product 4-methylnitrobenzene formed with \(\text{H}^+\) regenerated. Part (b): Reagent: alkaline potassium manganate(VII) (\(\text{KMnO}_4\)) solution, heated under reflux, followed by acidification with dilute hydrochloric acid or sulfuric acid. Part (c): The methyl group is electron-donating by inductive effect, which increases the electron density on the benzene ring (activating it) specifically directing substitution to positions 2 and 4 (ortho and para).

Part (a)(i): 2 marks total
- 1 mark for concentrated HNO3 and concentrated H2SO4.
- 1 mark for temperature of 50-55 °C (or heated under reflux with temperature control).

Part (a)(ii): 1.5 marks total
- 1.5 marks for correct equation showing generation of NO2^+ (1 mark for reactants and products but unbalanced).

Part (a)(iii): 4 marks total
- 1 mark for curly arrow from ring to carbon of NO2^+.
- 1.5 marks for intermediate structure (0.5 for correct substituent attachments, 1 for correct horseshoe spanning 5 carbons with positive charge inside).
- 1 mark for curly arrow from C-H bond back into the ring.
- 0.5 mark for final structure of 4-methylnitrobenzene.

Part (b): 2 marks total
- 1 mark for alkaline potassium manganate(VII) / KMnO4 heated under reflux.
- 1 mark for acidification (with dilute HCl / H2SO4).

Part (c): 1 mark total
- 1 mark for stating that the methyl group is electron-donating (inductive effect) and stabilizes the intermediate at positions 2 and 4.

Part (a): \(E_a\) is the activation energy (SI units: \(\text{J mol}^{-1}\)); \(R\) is the gas constant (SI units: \(\text{J K}^{-1}\text{ mol}^{-1}\)); \(T\) is the temperature (SI units: \(\text{K}\)). Part (b)(i): At \(T_1 = 298\text{ K}\): \(\frac{1}{T_1} = \frac{1}{298} = 3.36 \times 10^{-3}\text{ K}^{-1}\) and \(\ln k_1 = \ln(3.46 \times 10^{-5}) = -10.27\). At \(T_2 = 328\text{ K}\): \(\frac{1}{T_2} = \frac{1}{328} = 3.05 \times 10^{-3}\text{ K}^{-1}\) and \(\ln k_2 = \ln(1.50 \times 10^{-3}) = -6.50\). Part (b)(ii): Using the equation: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\). Substituting values: \(\ln\left(\frac{1.50 \times 10^{-3}}{3.46 \times 10^{-5}}\right) = -\frac{E_a}{8.31}(3.05 \times 10^{-3} - 3.36 \times 10^{-3})\). \(3.77 = -\frac{E_a}{8.31}(-3.1 \times 10^{-4})\). \(3.77 = E_a \times 3.73 \times 10^{-5}\). \(E_a = \frac{3.77}{3.73 \times 10^{-5}} = 101,000\text{ J mol}^{-1} \approx 101-102\text{ kJ mol}^{-1}\). (Using unrounded values: \(\ln(43.35) = 3.769\); \(\Delta(1/T) = -3.07 \times 10^{-4}\); \(E_a = \frac{3.769 \times 8.31}{3.07 \times 10^{-4}} = 102,000\text{ J mol}^{-1} = 102\text{ kJ mol}^{-1}\)). Part (c): A catalyst provides an alternative pathway with a lower activation energy (\(E_a\)). In the Arrhenius equation, a lower \(E_a\) value decreases the magnitude of the negative exponent \(-\frac{E_a}{RT}\), making \(e^{-\frac{E_a}{RT}}\) larger, which increases the rate constant \(k\) and hence the overall reaction rate.

Part (a): 2.5 marks total
- 1.5 marks for definitions of all three terms.
- 1 mark for correct standard SI units for all three: Ea (J mol^-1), R (J K^-1 mol^-1), T (K).

Part (b)(i): 2 marks total
- 0.5 mark for each correctly calculated value: 1/T1 = 3.36 x 10^-3 K^-1, ln k1 = -10.27, 1/T2 = 3.05 x 10^-3 K^-1, ln k2 = -6.50.

Part (b)(ii): 4 marks total
- 1 mark for correct substitute of terms into the two-point Arrhenius equation.
- 1 mark for calculating the gradient / change: Delta(ln k) = 3.77 and Delta(1/T) = -3.07 x 10^-4.
- 1 mark for calculating Ea in Joules (approx. 102,000 J mol^-1).
- 1 mark for converting to kJ mol^-1 to give 101-102 kJ mol^-1.

Part (c): 2 marks total
- 1 mark for stating that a catalyst lowers Ea.
- 1 mark for explaining that a lower Ea makes -Ea/RT less negative, resulting in a higher rate constant (k).

Part (a)(i): The carbon atom at position 2 is bonded to 4 different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{Br}\). Draw two 3D tetrahedral structures that are mirror images of each other, using wedges for bonds pointing out of the page, dashes for bonds pointing into the page, and normal lines for bonds in the plane of the page. Part (a)(ii): 'Chiral' refers to a molecule (or carbon atom) that is non-superimposable on its mirror image. A 'racemic mixture' is a 50:50 (equimolar) mixture of two enantiomers. Because each enantiomer rotates plane-polarized light by equal and opposite amounts, the net optical activity is zero. Part (b)(i): The mechanism is \(\text{S}_\text{N}1\). 1. The rate-determining step involves the slow loss of the bromide leaving group, forming a flat/planar carbocation intermediate. 2. The hydroxide nucleophile can attack the planar carbocation from either side with equal probability. 3. This results in equal amounts (a racemic mixture) of both enantiomers of butan-2-ol, causing a loss of optical activity. Part (b)(ii): The alternative mechanism is \(\text{S}_\text{N}2\). This is a single concerted step where the nucleophile attacks the carbon atom from the opposite side of the leaving group (backside attack) as the \(\text{C}-\text{Br}\) bond breaks. This results in the complete inversion of configuration (Walden inversion), producing only a single enantiomer, which remains optically active.

Part (a)(i): 2 marks total
- 2 marks for drawing two clear 3D tetrahedral structures that are non-superimposable mirror images of each other (1 mark if 3D representation is incomplete or bonds are drawn incorrectly).

Part (a)(ii): 2 marks total
- 1 mark for defining chiral: non-superimposable on its mirror image (or containing an asymmetric carbon with 4 different groups).
- 1 mark for defining racemic mixture: equimolar / 50:50 mixture of two enantiomers, causing zero overall optical rotation.

Part (b)(i): 3.5 marks total
- 0.5 mark for identifying the mechanism as SN1.
- 1 mark for stating that a planar carbocation intermediate is formed.
- 1 mark for stating that the nucleophile can attack from either side with equal probability.
- 1 mark for concluding that this results in equal amounts of both enantiomers / racemic mixture.

Part (b)(ii): 3 marks total
- 1 mark for identifying the SN2 mechanism.
- 1 mark for stating that nucleophilic attack occurs from the backside / opposite side to the leaving group in a single step.
- 1 mark for stating that this leads to an inversion of configuration (producing a single enantiomer / maintaining optical activity).

(a) Heating to constant mass ensures that all of the water of crystallisation has been completely driven off from the hydrated salt.

(b)
- Mass of hydrated salt = \( 22.17 - 18.42 = 3.75\text{ g} \)
- Mass of anhydrous salt = \( 20.82 - 18.42 = 2.40\text{ g} \)
- Mass of water lost = \( 3.75 - 2.40 = 1.35\text{ g} \)
- Moles of anhydrous \( \text{CuSO}_4 = \frac{2.40}{159.6} = 0.01504\text{ mol} \) (using \( M_r(\text{CuSO}_4) = 63.5 + 32.1 + 64.0 = 159.6\text{ g mol}^{-1} \))
- Moles of \( \text{H}_2\text{O} = \frac{1.35}{18.0} = 0.0750\text{ mol} \)
- Ratio of \( \text{H}_2\text{O} : \text{CuSO}_4 = \frac{0.0750}{0.01504} = 4.99 \approx 5 \)
- Hence, \( x = 5 \).

(c)
- The mass of water lost is calculated from two balance readings (the mass of the crucible + hydrated salt and the mass of the crucible + anhydrous salt after the final heating).
- Absolute uncertainty for 2 readings = \( 2 \times 0.005 = 0.01\text{ g} \)
- Percentage uncertainty = \( \frac{0.01}{1.35} \times 100 = 0.741\% \approx 0.74\% \)

(d)
- The black solid is copper(II) oxide.
- Equation: \( \text{CuSO}_4(\text{s}) \rightarrow \text{CuO}(\text{s}) + \text{SO}_3(\text{g}) \) (or producing \( \text{SO}_2 \) and \( \text{O}_2 \))
- Because sulfur trioxide gas escapes, more mass is lost than just water.
- This makes the mass of water lost appear larger than it actually is, and the mass of the anhydrous residue smaller.
- Therefore, the calculated value of \( x \) would be higher than the actual value.

(a)
- [1 mark] For stating it ensures all water of crystallisation is lost.

(b)
- [1 mark] Correctly calculating mass of water (1.35 g) and anhydrous salt (2.40 g).
- [1 mark] Calculating moles of anhydrous salt (0.0150 mol) and water (0.0750 mol).
- [1 mark] Dividing moles of water by moles of salt (4.99).
- [1 mark] Final integer value of x = 5.

(c)
- [1 mark] Stating that 2 balance readings are involved.
- [1 mark] Calculating total absolute uncertainty as 0.01 g.
- [1 mark] Correctly calculating percentage uncertainty to 2 significant figures (0.74%).

(d)
- [1 mark] Identifying black solid as copper(II) oxide.
- [1 mark] Balanced equation with correct state symbols: \( \text{CuSO}_4(\text{s}) \rightarrow \text{CuO}(\text{s}) + \text{SO}_3(\text{g}) \).
- [1 mark] Explanation that mass loss appears greater due to gas escape.
- [1 mark] Concluding that x would increase/be higher.

(a)
- A weak acid is an acid that only partially dissociates/ionises in aqueous solution.
- Expression: \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)

(b)
- At the half-equivalence point, exactly half of the weak acid HA has reacted with the base to form its conjugate base \( \text{A}^- \).
- Therefore, \( [\text{HA}] = [\text{A}^-] \).
- Substituting this into the \( K_a \) expression: \( K_a = [\text{H}^+] \frac{[\text{A}^-]}{[\text{HA}]} \Rightarrow K_a = [\text{H}^+] \).
- Taking the negative logarithm of both sides gives \( \text{p}K_a = \text{pH} \).
- Therefore, \( \text{p}K_a = 4.76 \).
- \( K_a = 10^{-4.76} = 1.74 \times 10^{-5}\text{ mol dm}^{-3} \).

(c)
- Suitable indicator: Phenolphthalein.
- Reason: The titration is between a weak acid and a strong base, so the pH at the equivalence point is basic (greater than 7) and will fall within phenolphthalein's transition range (8.2 to 10.0).
- Why methyl orange is unsuitable: Its transition range (3.2 – 4.4) is in the acidic region. Since the titration starts with a weak acid (pH around 3), the indicator would change colour almost immediately / long before the equivalence point is reached.

(d)
- Calibration: Rinse the probe with distilled water, place it in a buffer solution of known pH (e.g., pH 4.00), adjust the meter to read that pH, and repeat with at least one other buffer solution (e.g., pH 7.00 or 10.00).
- Necessity: pH meters/probes drift over time and must be calibrated to ensure accuracy of the pH readings.

(a)
- [1 mark] Weak acid definition: partially/incomplete dissociation in water.
- [1 mark] Correct \( K_a \) expression with state symbols or concentration brackets.

(b)
- [1 mark] Explaining that at half-equivalence, half of HA is converted to \( \text{A}^- \), so \( [\text{HA}] = [\text{A}^-] \).
- [1 mark] Showing that \( K_a = [\text{H}^+] \) or \( \text{pH} = \text{p}K_a \).
- [1 mark] Correctly identifying \( \text{p}K_a = 4.76 \).
- [1 mark] Final calculation of \( K_a = 1.74 \times 10^{-5}\text{ mol dm}^{-3} \) (accept \( 1.7 \times 10^{-5} \)).

(c)
- [1 mark] Selecting phenolphthalein.
- [1 mark] Reason: equivalence point pH is basic / >7 and falls within its transition range.
- [1 mark] Explaining that methyl orange changes color in the acidic range.
- [1 mark] Stating that methyl orange would change color too early / long before equivalence.

(d)
- [1 mark] Description of using buffer solutions of known pH (at least two different pH values).
- [1 mark] Stating that calibration corrects for system drift / ensures accurate readings.

(a) A complete reflux diagram must include:
- Pear-shaped or round-bottomed flask with a heat source (e.g. water bath or heating mantle, NOT a bare Bunsen flame as organic reactants are highly flammable).
- Vertical condenser fitted into the flask.
- Water jacket correctly labelled with water entering at the bottom and exiting at the top.
- The top of the condenser must remain open to the atmosphere (no stopper).

(b)
- Molar mass of ethanoic acid, \( M_r(\text{CH}_3\text{COOH}) = 2(12.0) + 4(1.0) + 2(16.0) = 60.0\text{ g mol}^{-1} \)
- Moles of ethanoic acid = \( \frac{12.0}{60.0} = 0.200\text{ mol} \)
- Molar mass of ethanol, \( M_r(\text{CH}_3\text{CH}_2\text{OH}) = 2(12.0) + 6(1.0) + 16.0 = 46.0\text{ g mol}^{-1} \)
- Moles of ethanol = \( \frac{11.5}{46.0} = 0.250\text{ mol} \)
- Since the stoichiometry is 1:1, ethanoic acid is the limiting reactant (0.200 mol < 0.250 mol).
- Theoretical yield of ethyl ethanoate = \( 0.200\text{ mol} \)
- Molar mass of ethyl ethanoate, \( M_r(\text{CH}_3\text{COOCH}_2\text{CH}_3) = 4(12.0) + 8(1.0) + 2(16.0) = 88.0\text{ g mol}^{-1} \)
- Theoretical mass of ethyl ethanoate = \( 0.200 \times 88.0 = 17.6\text{ g} \)
- Percentage yield = \( \frac{9.24}{17.6} \times 100 = 52.5\% \)

(c)
- Purpose: To react with and remove any unreacted ethanoic acid or sulfuric acid catalyst remaining in the distillate.
- Ionic equation: \( \text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O} \) (or \( \text{CO}_3^{2-} + 2\text{CH}_3\text{COOH} \rightarrow 2\text{CH}_3\text{COO}^- + \text{CO}_2 + \text{H}_2\text{O} \))

(d)
- Anhydrous calcium chloride is used as a drying agent to remove residual water from the organic ester layer.
- The student knows enough has been added when the powder no longer clumps together and remains free-flowing / disperses like a fine dust when the flask is swirled.

(a)
- [1 mark] Correctly drawn flask with vertical condenser, open at the top (no gaps at joint).
- [1 mark] Water flow directions correctly indicated (in at bottom, out at top).
- [1 mark] Safe heating source shown (e.g. heating mantle, water bath, or electrical heater; reject Bunsen burner directly heating flask).

(b)
- [1 mark] Calculating moles of ethanoic acid (0.200 mol) and ethanol (0.250 mol).
- [1 mark] Identifying ethanoic acid as the limiting reactant.
- [1 mark] Calculating the theoretical mass of ethyl ethanoate (17.6 g).
- [1 mark] Calculating the percentage yield (52.5%).
- [1 mark] Awarded for correct significant figures (3 s.f.) and clear working throughout.

(c)
- [1 mark] For identifying that sodium carbonate removes acidic impurities / neutralises acids.
- [1 mark] Correct balanced ionic equation (e.g., \( \text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)).

(d)
- [1 mark] Stating that calcium chloride removes remaining water / acts as drying agent.
- [1 mark] Explaining that clumping ceases / powder becomes free-flowing when dry.

(a)
- The reaction produces a fixed amount of iodine (equivalent to the amount of thiosulfate added) before the blue starch-iodine complex can persist.
- Since the change in concentration of iodine (\( \Delta [\text{I}_2] \)) is constant, rate (\( \frac{\Delta [\text{I}_2]}{\Delta t} \)) is inversely proportional to the time taken (\( t \)).

(b)
- **Order with respect to \( \text{S}_2\text{O}_8^{2-} \):** Compare Experiment 1 and 2. \( [\text{I}^-] \) is kept constant at \( 0.0800\text{ mol dm}^{-3} \). \( [\text{S}_2\text{O}_8^{2-}] \) is doubled (from 0.0400 to 0.0800). The time is halved (from 45.0 s to 22.5 s), which means the rate is doubled. Therefore, the reaction is first order with respect to \( \text{S}_2\text{O}_8^{2-} \).
- **Order with respect to \( \text{I}^- \):** Compare Experiment 2 and 3. \( [\text{S}_2\text{O}_8^{2-}] \) is kept constant at \( 0.0800\text{ mol dm}^{-3} \). \( [\text{I}^-] \) is halved (from 0.0800 to 0.0400). The time is doubled (from 22.5 s to 45.0 s), which means the rate is halved. Therefore, the reaction is first order with respect to \( \text{I}^- \).

(c)
- Rate equation: \( \text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-] \)
- Using Experiment 1:
- \( \text{Rate} = \frac{2.00 \times 10^{-4}}{45.0} = 4.444 \times 10^{-6}\text{ mol dm}^{-3}\text{ s}^{-1} \)
- \( k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{4.444 \times 10^{-6}}{(0.0400)(0.0800)} = \frac{4.444 \times 10^{-6}}{0.0032} = 1.389 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \)
- Rounded to 3 significant figures: \( k = 1.39 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \).

(d)
- Yes, it is consistent.
- The rate-determining step involves one molecule/ion of \( \text{S}_2\text{O}_8^{2-} \) reacting with one molecule/ion of \( \text{I}^- \).
- The stoichiometry of the reactants in the rate-determining step matches the orders in the rate equation (first order for both reactants).

(a)
- [1 mark] Stating that a fixed/constant amount of product is formed before the color change.
- [1 mark] Explaining that Rate = change in concentration / time, so when change in concentration is constant, Rate is proportional to 1/time.

(b)
- [1 mark] Comparing Exp 1 and 2 to show rate doubles when peroxodisulfate concentration doubles, concluding 1st order.
- [1 mark] Comparing Exp 2 and 3 to show rate halves when iodide concentration halves, concluding 1st order.
- [2 marks] Clear and logical written reasoning for both deductions.

(c)
- [1 mark] Correct rate equation: \( \text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-] \).
- [1 mark] Calculating the initial rate for Exp 1 (\( 4.44 \times 10^{-6}\text{ mol dm}^{-3}\text{ s}^{-1} \)).
- [1 mark] Correct calculation of \( k = 1.39 \times 10^{-3} \) (or \( 1.4 \times 10^{-3} \)).
- [1 mark] Correct units: \( \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1} \).

(d)
- [1 mark] Stating that the rate-determining step is consistent with the rate equation.
- [1 mark] Explaining that the reactants in the rate-determining step match the species in the rate equation (1 of each species).

(a)
- \( 5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

(b)
- Color change: colourless to pale pink (accept first permanent pink color).
- Why indicator is not required: Potassium manganate(VII) acts as its own indicator (it is self-indicating) because intense purple \( \text{MnO}_4^- \) is reduced to virtually colourless \( \text{Mn}^{2+} \). The first excess drop of \( \text{MnO}_4^- \) turns the solution pale pink.

(c)
- Moles of \( \text{MnO}_4^- \) in \( 25.00\text{ cm}^3 \) titre = \( 0.0200 \times \frac{25.00}{1000} = 5.00 \times 10^{-4}\text{ mol} \)
- Stoichiometric ratio of \( \text{Fe}^{2+} : \text{MnO}_4^- = 5 : 1 \)
- Moles of \( \text{Fe}^{2+} \) in \( 25.0\text{ cm}^3 \) aliquot = \( 5 \times 5.00 \times 10^{-4} = 2.50 \times 10^{-3}\text{ mol} \)
- Moles of \( \text{Fe}^{2+} \) in the original \( 250.0\text{ cm}^3 \) solution = \( 2.50 \times 10^{-3} \times 10 = 0.0250\text{ mol} \)
- Molar mass of hydrated salt \( (\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \cdot y\text{H}_2\text{O} = \frac{\text{mass}}{\text{moles}} = \frac{9.80}{0.0250} = 392.0\text{ g mol}^{-1} \)
- Molar mass of anhydrous salt \( (\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \):
- \( M_r = 2(14.0) + 8(1.0) + 55.8 + 2(32.1) + 8(16.0) = 28.0 + 8.0 + 55.8 + 64.2 + 128.0 = 284.0\text{ g mol}^{-1} \)
- Mass of water in one mole of salt = \( 392.0 - 284.0 = 108.0\text{ g} \)
- Number of water molecules, \( y = \frac{108.0}{18.0} = 6 \)

(d)
- Acidic conditions (specifically \( \text{H}^+ \) ions) are required for the reduction reaction of \( \text{MnO}_4^- \) to occur correctly.
- Dissolving in acid also prevents the air-oxidation of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \) and prevents hydrolysis to form insoluble iron hydroxides (which would make the solution cloudy).

(a)
- [1 mark] Correct reactants and products: \( \text{Fe}^{2+} + \text{MnO}_4^- + \text{H}^+ \rightarrow \text{Fe}^{3+} + \text{Mn}^{2+} + \text{H}_2\text{O} \).
- [1 mark] Correct balancing: \( 5 : 1 : 8 \rightarrow 5 : 1 : 4 \).

(b)
- [1 mark] End point colour change: colourless (or pale green) to pale/permanent pink.
- [1 mark] Explaining that manganate(VII) is self-indicating.

(c)
- [1 mark] Moles of manganate(VII) in titration = \( 5.00 \times 10^{-4}\text{ mol} \).
- [1 mark] Moles of \( \text{Fe}^{2+} \) in titration = \( 2.50 \times 10^{-3}\text{ mol} \).
- [1 mark] Moles of \( \text{Fe}^{2+} \) in \( 250\text{ cm}^3 \) flask = \( 0.0250\text{ mol} \).
- [1 mark] Molar mass of hydrated salt = \( 392.0\text{ g mol}^{-1} \).
- [1 mark] Calculating anhydrous mass as \( 284.0\text{ g mol}^{-1} \).
- [1 mark] Deducting \( y = 6 \).

(d)
- [1 mark] Acid provides hydrogen ions (\( \text{H}^+ \)) for the reduction reaction.
- [1 mark] Prevents the oxidation of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \) / prevents precipitation of iron hydroxides.

(a)
- Plot temperature (y-axis) against time (x-axis).
- Draw a line of best fit through the cooling points (from minutes 5 to 10) and extrapolate this line back to the 4th minute (the time of mixing).
- Extrapolating the linear cooling trend back by 1 minute: the temperature at minute 5 is \( 32.4^\circ\text{C} \) and it decreases by \( 0.5^\circ\text{C} \) per minute. Thus, the extrapolated temperature at minute 4 is \( 32.4 + 0.5 = 32.9^\circ\text{C} \).
- Corrected temperature rise, \( \Delta T = 32.9 - 20.2 = 12.7^\circ\text{C} \).

(b)
- Total volume of reaction mixture = \( 50.0 + 50.0 = 100.0\text{ cm}^3 \).
- Mass of reaction mixture, \( m = 100.0\text{ g} \).
- Heat energy released, \( q = m c \Delta T = 100.0 \times 4.18 \times 12.7 = 5308.6\text{ J} = 5.3086\text{ kJ} \).
- Moles of acid used = \( 2.00 \times 0.0500 = 0.100\text{ mol} \).
- Moles of base used = \( 2.00 \times 0.0500 = 0.100\text{ mol} \).
- Moles of water formed = \( 0.100\text{ mol} \).
- \( \Delta H_{\text{neut}} = -\frac{q}{\text{moles of water}} = -\frac{5.3086}{0.100} = -53.086\text{ kJ mol}^{-1} \).
- To 3 significant figures: \( \Delta H_{\text{neut}} = -53.1\text{ kJ mol}^{-1} \).

(c)
- Source 1: Heat loss to the surroundings / air. This would mean the measured temperature rise is lower than theoretical, making the calculated \( \Delta H_{\text{neut}} \) less exothermic / less negative.
- Source 2: The specific heat capacity of the polystyrene cup and thermometer are neglected (heat is absorbed by the calorimeter itself). This means less heat is recorded as raising the temperature of the water, making \( \Delta H_{\text{neut}} \) less exothermic / less negative.

(a)
- [1 mark] Draw a best-fit line through cooling points (minutes 5-10).
- [1 mark] Extrapolate back to the mixing time (minute 4).
- [1 mark] State the extrapolated maximum temperature is 32.9 °C.
- [1 mark] Correctly calculate \( \Delta T = 12.7^\circ\text{C} \).

(b)
- [1 mark] Calculate mass of solution = 100 g.
- [1 mark] Calculate energy change, \( q = 5308.6\text{ J} \) or \( 5.31\text{ kJ} \).
- [1 mark] Calculate moles of water formed = 0.100 mol.
- [1 mark] Calculate \( \Delta H = -53.1\text{ kJ mol}^{-1} \) (accept -53.0 to -53.1).
- [1 mark] Include a negative sign and correct units (\( \text{kJ mol}^{-1} \)).

(c)
- [1 mark] For identifying heat loss to surroundings / heat absorbed by calorimeter.
- [1 mark] For identifying another valid source of systematic error (e.g., density/SHC of solution is not exactly equal to pure water).
- [1 mark] Explaining that these errors make the experimental value less negative / less exothermic than the literature value.

(a)
- Cation: Calcium ion, \( \text{Ca}^{2+} \) (indicated by brick-red flame test).
- Anion: Chloride ion, \( \text{Cl}^- \) (indicated by white precipitate with silver nitrate that dissolves in dilute ammonia).
- Formula of X: \( \text{CaCl}_2 \).

(b)
- (i) \( \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s}) \)
- (ii) \( \text{AgCl}(\text{s}) + 2\text{NH}_3(\text{aq}) \rightarrow [\text{Ag}(\text{NH}_3)_2]^+(\text{aq}) + \text{Cl}^-(\text{aq}) \)

(c)
- Solubility of Group 2 hydroxides **increases** down the group (from magnesium to barium).
- Solubility of Group 2 sulfates **decreases** down the group.
- Relation to Test 3: Since calcium is near the top of the group, calcium hydroxide, \( \text{Ca}(\text{OH})_2 \), is only sparingly soluble.
- When sodium hydroxide is added, the concentration of hydroxide ions is high enough to exceed the solubility product, causing a white precipitate of \( \text{Ca}(\text{OH})_2 \) to form.
- (If a Group 2 metal further down the group, like barium, was tested, no precipitate would form with NaOH because barium hydroxide is highly soluble).

(a)
- [1 mark] Identify cation as calcium / \( \text{Ca}^{2+} \).
- [1 mark] Identify anion as chloride / \( \text{Cl}^- \).
- [1 mark] Provide formula \( \text{CaCl}_2 \).

(b)
- [1 mark] Correct equation for (i): \( \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s}) \).
- [1 mark] Correct state symbols for (i).
- [1 mark] Correct equation for (ii): \( \text{AgCl}(\text{s}) + 2\text{NH}_3(\text{aq}) \rightarrow [\text{Ag}(\text{NH}_3)_2]^+\text{(aq)} + \text{Cl}^-\text{(aq)} \).
- [1 mark] Correct state symbols for (ii).

(c)
- [1 mark] Stating Group 2 hydroxides become more soluble down the group.
- [1 mark] Stating Group 2 sulfates become less soluble down the group.
- [1 mark] Identifying the white precipitate in Test 3 as calcium hydroxide, \( \text{Ca}(\text{OH})_2 \).
- [1 mark] Explaining that calcium hydroxide is only sparingly/slightly soluble, so it precipitates.
- [1 mark] Contrasting with lower elements (e.g., Ba) which would not precipitate because their hydroxides are too soluble.

(a)
- Functional group: Ester.
- Justification: The sharp absorption at \( 1740\text{ cm}^{-1} \) corresponds to a carbonyl group (\( \text{C}=\text{O} \)). The lack of a broad absorption in the \( 2500 - 3300\text{ cm}^{-1} \) range indicates that there is no carboxylic acid \( \text{O}-\text{H} \) group present.

(b)
- Structure of Y: Ethyl ethanoate, \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \).
- Signal assignments:
- **Triplet at \( \delta = 1.25\text{ ppm} \) (3H):** Belongs to the methyl protons of the ethyl group, \( -\text{CH}_2\text{CH}_3 \). The triplet splitting indicates there are 2 adjacent protons (on the neighboring carbon, using the \( N+1 \) rule, \( N=2 \)).
- **Singlet at \( \delta = 2.05\text{ ppm} \) (3H):** Belongs to the methyl protons directly attached to the carbonyl group, \( \text{CH}_3\text{CO}- \). The singlet indicates there are 0 adjacent protons (\( N=0 \)).
- **Quartet at \( \delta = 4.12\text{ ppm} \) (2H):** Belongs to the \( -\text{O}-\text{CH}_2- \) protons of the ethyl group. The quartet splitting indicates there are 3 adjacent protons (on the neighboring methyl group, \( N=3 \)). The high chemical shift (4.12 ppm) is due to deshielding by the adjacent highly electronegative oxygen atom.

(c)
- Fragment ion: \( [\text{CH}_3\text{COO}]^+ \) or \( [\text{M} - \text{C}_2\text{H}_5]^+ \).
- Formation equation: \( [\text{CH}_3\text{COOCH}_2\text{CH}_3]^{+\bullet} \rightarrow [\text{CH}_3\text{COO}]^+ + \cdot\text{CH}_2\text{CH}_3 \) (loss of ethyl radical).

(d)
- Dissolve Y in a deuterated solvent (such as deuterated chloroform, \( \text{CDCl}_3 \)), which does not contain protons that would interfere with the \( ^1\text{H} \) spectrum.
- Add a small amount of tetramethylsilane (TMS) as an internal reference standard (to set \( \delta = 0\text{ ppm} \)), filter the solution, and transfer it into a clean, dry NMR tube.

(a)
- [1 mark] Identifies functional group as an ester.
- [1 mark] Explains that the peak at 1740 cm^-1 is C=O and the absence of O-H peak rules out a carboxylic acid.

(b)
- [1 mark] Correct structure/name: ethyl ethanoate / \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \).
- [1 mark] Assigns triplet at 1.25 ppm to \( \text{CH}_3 \) adjacent to \( \text{CH}_2 \) (showing \( N+1=3 \)).
- [1 mark] Assigns singlet at 2.05 ppm to \( \text{CH}_3 \) next to carbonyl (showing \( N+1=1 \)).
- [1 mark] Assigns quartet at 4.12 ppm to \( \text{CH}_2 \) adjacent to \( \text{CH}_3 \) (showing \( N+1=4 \)).
- [1 mark] Relates high shift (4.12 ppm) to the deshielding effect of oxygen.
- [1 mark] Logical structure assignment consistent with integration ratios (3H, 3H, 2H).

(c)
- [1 mark] Correctly identifies formula of fragment ion: \( [\text{CH}_3\text{COO}]^+ \) (including charge).
- [1 mark] Correctly balanced fragmentation equation showing molecular ion radical cation reactant and neutral radical product.

(d)
- [1 mark] Choice of deuterated solvent (e.g., \( \text{CDCl}_3 \)) and reason (no protons to interfere).
- [1 mark] Adding TMS as reference standard / transferring to NMR tube.

(a) Moles of \(\text{NaOH}\) reacted = \(0.100\text{ mol dm}^{-3} \times \frac{20.00}{1000}\text{ dm}^3 = 2.00 \times 10^{-3}\text{ mol}\). Since \(HA\) is monoprotic, moles of \(HA\) in the \(25.0\text{ cm}^3\) aliquot = \(2.00 \times 10^{-3}\text{ mol}\). Thus, total moles of \(HA\) in the \(250.0\text{ cm}^3\) flask = \(2.00 \times 10^{-3} \times 10 = 2.00 \times 10^{-2}\text{ mol}\). Molar mass of \(HA = \frac{\text{mass}}{\text{moles}} = \frac{1.22\text{ g}}{2.00 \times 10^{-2}\text{ mol}} = 61.0\text{ g mol}^{-1}\).

(b) (i) \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\).
(ii) At the half-equivalence point, exactly half of the weak acid \(HA\) has been neutralised to form the conjugate base \(A^-\). Therefore, \([\text{HA}] = [\text{A}^-]\). Substituting this equality into the dissociation expression: \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = [\text{H}^+]\). Taking negative logarithms of both sides: \(-\log K_a = -\log[\text{H}^+]\), which yields \(\text{p}K_a = \text{pH}\).
(iii) Since \(\text{p}K_a = \text{pH} = 4.82\), then \(K_a = 10^{-4.82} = 1.51 \times 10^{-5}\text{ mol dm}^{-3}\).

(c) Percentage uncertainty = \(\frac{0.05}{4.82} \times 100\% = 1.04\%\).

(d) The distilled water in the pipette dilutes the weak acid solution, meaning fewer moles of \(HA\) are transferred to the conical flask. This requires a smaller volume (titre) of \(\text{NaOH}\) to reach the endpoint. Consequently, the calculated number of moles of \(HA\) will be smaller than the true value. Because \(\text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}}\), dividing the mass by a smaller number of moles results in a higher (larger) calculated molar mass.

(a) [3 marks total]:
- M1: Moles of NaOH = \(2.00 \times 10^{-3}\text{ mol}\) (1)
- M2: Moles of HA in 250 cm3 = \(2.00 \times 10^{-2}\text{ mol}\) (1)
- M3: Molar mass = 61.0 g mol^-1 (must show dividing 1.22 by 2.00 x 10^-2) (1)

(b) (i) [1 mark total]:
- M1: Correct expression for Ka (charges must be correct) (1)
(ii) [3 marks total]:
- M1: States that \([\text{HA}] = [\text{A}^-]\) at half-equivalence (1)
- M2: Cancels the terms to show \(K_a = [\text{H}^+]\) (1)
- M3: Links this to \(\text{p}K_a = \text{pH}\) via negative logarithms (1)
(iii) [2 marks total]:
- M1: \(K_a = 1.51 \times 10^{-5}\) (allow 1.5 x 10^-5) (1)
- M2: \(\text{mol dm}^{-3}\) (1)

(c) [1 mark total]:
- M1: 1.04% (accept 1% or 1.0%) (1)

(d) [2 marks total]:
- M1: Explains that fewer moles of HA are transferred / smaller titre of NaOH is recorded (1)
- M2: Explains that smaller calculated moles leads to a higher calculated molar mass (1)

(a) The colour change at the endpoint is from colourless (or very pale green) to a permanent pale pink colour.

(b) Dilute sulfuric acid is added to provide the necessary \(\text{H}^+\) ions for the reduction of the manganate(VII) ions to manganese(II) ions, and to prevent the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) by oxygen in the air, or the precipitation of brown manganese dioxide (\(\text{MnO}_2\)).

(c) First, calculate the moles of \(\text{MnO}_4^-\) used in the titration: \(\text{Moles of } \text{MnO}_4^- = 0.0200\text{ mol dm}^{-3} \times \frac{20.00}{1000}\text{ dm}^3 = 4.00 \times 10^{-4}\text{ mol}\).
According to the equation, the reacting ratio is \(1\text{ mol of } \text{MnO}_4^- : 5\text{ mol of } M^{2+}\).
Therefore, moles of \(M^{2+}\) in the \(25.0\text{ cm}^3\) sample = \(5 \times 4.00 \times 10^{-4}\text{ mol} = 2.00 \times 10^{-3}\text{ mol}\).
Since the solution was made up to \(250.0\text{ cm}^3\), the total moles of \(M^{2+}\) (and thus \(M\text{SO}_4 \cdot x\text{H}_2\text{O}\)) in the 5.56 g sample is: \(2.00 \times 10^{-3} \times \frac{250.0}{25.0} = 2.00 \times 10^{-2}\text{ mol}\).
Now calculate the molar mass: \(\text{Molar mass} = \frac{\text{Mass}}{\text{Moles}} = \frac{5.56\text{ g}}{2.00 \times 10^{-2}\text{ mol}} = 278\text{ g mol}^{-1}\).

(d) The molar mass of anhydrous \(\text{FeSO}_4 = 55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1}\).
The mass of the water of crystallisation in one mole of the hydrated salt is: \(278 - 151.9 = 126.1\text{ g mol}^{-1}\).
Since the molar mass of water, \(\text{H}_2\text{O}\), is \(18.0\text{ g mol}^{-1}\):
\(x = \frac{126.1}{18.0} = 7.01\).
Therefore, the value of \(x\) is 7 (giving the formula \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\)).

(e) To test for sulfate ions, add barium chloride solution (\(\text{BaCl}_2\)) acidified with dilute hydrochloric acid (\(\text{HCl}\)) or nitric acid (\(\text{HNO}_3\)) to the salt solution. A white precipitate of barium sulfate (\(\text{BaSO}_4\)) will form.

(a) [1 mark total]:
- M1: Colourless / very pale green to permanent pale pink (1)

(b) [2 marks total]:
- M1: To provide H+ ions / prevent manganese dioxide (brown precipitate) forming (1)
- M2: To prevent premature oxidation of iron(II) ions by dissolved oxygen (1)

(c) [4 marks total]:
- M1: Moles of MnO4- = \(4.00 \times 10^{-4}\text{ mol}\) (1)
- M2: Moles of M2+ in 25 cm3 = \(2.00 \times 10^{-3}\text{ mol}\) (1)
- M3: Moles of M2+ in 250 cm3 = \(2.00 \times 10^{-2}\text{ mol}\) (1)
- M4: Molar mass of salt = \(278\text{ g mol}^{-1}\) (1)

(d) [3 marks total]:
- M1: Molar mass of anhydrous FeSO4 = 151.9 g mol^-1 (1)
- M2: Subtracts to find mass of H2O = 126.1 g mol^-1 (1)
- M3: Division by 18 to give x = 7 (must be a whole number) (1)

(e) [2 marks total]:
- M1: Reagent: Barium chloride solution acidified with HCl or HNO3 (do not accept acidified with H2SO4) (1)
- M2: Observation: White precipitate (1)