2023 Edexcel A-Level Chemistry (9CH0) Practice Paper with Answers

To identify the element, look for the largest jump between successive ionisation energies, which indicates the removal of an electron from a shell closer to the nucleus (a new inner shell).

- \(1^{\text{st}}\) to \(2^{\text{nd}}\): \(1450 - 738 = 712\text{ kJ mol}^{-1}\)
- \(2^{\text{nd}}\) to \(3^{\text{rd}}\): \(7730 - 1450 = 6280\text{ kJ mol}^{-1}\) (a very large increase)
- \(3^{\text{rd}}\) to \(4^{\text{th}}\): \(10500 - 7730 = 2770\text{ kJ mol}^{-1}\)

The significant increase between the second and third ionisation energies indicates that the third electron is removed from a shell closer to the nucleus. Therefore, the element has two valence electrons and belongs to Group 2. The Group 2 element in Period 3 is magnesium (Mg).

Award 1 mark for the correct answer (B).
- Incorrect answers: A (Group 1), C (Group 13/3), D (Group 14/4).

When sodium sulfate is added to a solution containing \(\text{Ba}^{2+}\) and \(\text{Mg}^{2+}\) ions, precipitation reactions can theoretically occur to form \(\text{BaSO}_4\text{(s)}\) and \(\text{MgSO}_4\text{(s)}\).

According to the solubility trends of Group 2 sulfates, solubility decreases down the group. Magnesium sulfate is highly soluble in water, whereas barium sulfate is extremely insoluble. Therefore, only barium sulfate, \(\text{BaSO}_4\text{(s)}\), will precipitate out of solution, while magnesium ions and sulfate ions remain dissolved.

Award 1 mark for identifying B as the correct option.
- A is incorrect because magnesium sulfate does not precipitate.
- C is incorrect because magnesium sulfate is highly soluble.
- D is incorrect because barium chloride is a soluble reactant, not a precipitate.

Let's determine the electronic configuration of each species starting from their neutral atomic states:
- Cr has atomic number 24: \([Ar] 3d^5 4s^1\). Therefore, \(\text{Cr}^{3+}\) (losing 3 electrons) is \([Ar] 3d^3\).
- Mn has atomic number 25: \([Ar] 3d^5 4s^2\). Therefore, \(\text{Mn}^{2+}\) (losing 2 electrons from the 4s subshell first) is \([Ar] 3d^5\), which matches the target configuration.
- Fe has atomic number 26: \([Ar] 3d^6 4s^2\). Therefore, \(\text{Fe}^{2+}\) is \([Ar] 3d^6\).
- Co has atomic number 27: \([Ar] 3d^7 4s^2\). Therefore, \(\text{Co}^{3+}\) (losing 3 electrons) is \([Ar] 3d^6\).

Award 1 mark for the correct answer (B).
- A is incorrect (3d3).
- C is incorrect (3d6).
- D is incorrect (3d6).

For a weak acid, \(HA \rightleftharpoons H^+ + A^-\), we use the approximation:

\([H^+] \approx \sqrt{K_a \times [HA]}\)

Substitute the given values:

\([H^+] = \sqrt{1.60 \times 10^{-4} \times 0.150} = \sqrt{2.40 \times 10^{-5}} = 4.899 \times 10^{-3}\text{ mol dm}^{-3}\)

Now calculate the pH:

\(\text{pH} = -\log_{10}[H^+] = -\log_{10}(4.899 \times 10^{-3}) = 2.31\)

Award 1 mark for the correct answer (A).
- B is incorrect (corresponds to an incorrect formula or square root error).
- C is incorrect (corresponds to \(pK_a = 3.80\)).
- D is incorrect (corresponds to not taking the square root before calculating pH).

By applying Hess's Law to a Born-Haber cycle:

\(\Delta H_f^{\ominus}(\text{CaCl}_2\text{(s)}) = \Delta H_{at}^{\ominus}(\text{Ca}) + 1^{\text{st}}\text{ IE}(\text{Ca}) + 2^{\text{nd}}\text{ IE}(\text{Ca}) + 2 \times \Delta H_{at}^{\ominus}(\text{Cl}) + 2 \times \text{EA}(\text{Cl}) + \Delta H_{lattice}^{\ominus}\)

Note that the bond enthalpy of \(\text{Cl}-\text{Cl}\) represents the energy needed to produce two moles of Cl gaseous atoms from one mole of \(\text{Cl}_2\text{(g)}\). Therefore, \(2 \times \Delta H_{at}^{\ominus}(\text{Cl}) = \text{bond enthalpy of } \text{Cl}_2 = +242\text{ kJ mol}^{-1}\).

Substitute the values:

\(-796 = 178 + 590 + 1145 + 242 + 2(-349) + \Delta H_{lattice}^{\ominus}\)

\(-796 = 2155 - 698 + \Delta H_{lattice}^{\ominus}\)

\(-796 = 1457 + \Delta H_{lattice}^{\ominus}\)

\[\Delta H_{lattice}^{\ominus} = -796 - 1457 = -2253\text{ kJ mol}^{-1}\\]

Award 1 mark for the correct answer (B).
- A is incorrect (corresponds to sum of endothermic steps only).
- C is incorrect (missed multiplying the electron affinity of chlorine by 2).
- D is incorrect (added the electron affinity instead of subtracting/multiplying by 2 correctly).

To find the oxidation states of iodine in the polyatomic ions:
1. For \(\text{IO}_4^-\): Let the oxidation state of iodine be \(x\). Oxygen has an oxidation state of \(-2\).
\(x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7\)

2. For \(\text{IO}_3^-\): Let the oxidation state of iodine be \(y\).
\(y + 3(-2) = -1 \implies y - 6 = -1 \implies y = +5\)

Thus, the oxidation states are \(+7\) and \(+5\).

Award 1 mark for the correct answer (A).
- B, C, and D are incorrect calculations of the oxidation states of iodine.

Phosphorus has the outer electron configuration \(3s^2 3p^3\) with three singly occupied 3p orbitals. Sulfur has the configuration \(3s^2 3p^4\), meaning one of its 3p orbitals contains a pair of electrons. The mutual repulsion between the two paired electrons in the same 3p orbital of sulfur makes it easier to remove one of them, resulting in a lower first ionization energy compared to phosphorus.

1 Mark: Identify that sulfur has a paired electron in a 3p orbital / phosphorus has only singly occupied 3p orbitals.
1 Mark: Explain that repulsion between the paired electrons in sulfur's orbital makes it easier to remove an electron (requires less energy).

Xenon has 8 valence electrons. Each of the 4 fluorine atoms contributes 1 electron for sharing, making 12 electrons (6 electron pairs) around the central Xe atom. There are 4 bonding pairs and 2 lone pairs. To minimize repulsion, the 2 lone pairs position themselves opposite each other, leading to a square planar geometry.

1 Mark: Square planar.
1 Mark: 2 lone pairs (on xenon).

To balance the half-equation:
1. Balance the oxygen atoms by adding water: \(\text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH}\).
2. Balance the hydrogen atoms by adding hydrogen ions: \(\text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + 4\text{H}^+\).
3. Balance the charge by adding electrons: \(\text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + 4\text{H}^+ + 4\text{e}^-\).

1 Mark: Correct formulas for all species with balanced atoms: \(\text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + 4\text{H}^+\).
1 Mark: Correctly balanced charge with \(4\text{e}^-\) on the right-hand side.

Down Group 2, the ionic radius of the \(M^{2+}\) cation increases while keeping the same \(2+\) charge. This leads to a lower charge density of the cation. A cation with a lower charge density has a weaker polarizing effect on the carbonate ion (\(\text{CO}_3^{2-}\)), resulting in less distortion of the carbonate's carbon-oxygen bonds, thereby making the compound thermally more stable.

1 Mark: State that cation size increases down the group, so charge density decreases.
1 Mark: Relate this to a weaker polarizing effect / less distortion of the carbonate ion (or weakening of the \(\text{C}-\text{O}\) bond).

Using the relationship:
\(\Delta H^\ominus_{\text{sol}} = -\Delta H^\ominus_{\text{latt}} + \Delta H^\ominus_{\text{hyd}}(\text{Na}^+) + \Delta H^\ominus_{\text{hyd}}(\text{Cl}^-)\)
Substituting the values:
\(+4 = -(-787) + (-406) + \Delta H^\ominus_{\text{hyd}}(\text{Cl}^-)\)
\(+4 = 787 - 406 + \Delta H^\ominus_{\text{hyd}}(\text{Cl}^-)\)
\(+4 = 381 + \Delta H^\ominus_{\text{hyd}}(\text{Cl}^-)\)
\(\Delta H^\ominus_{\text{hyd}}(\text{Cl}^-) = 4 - 381 = -377\text{ kJ mol}^{-1}\).

1 Mark: Correct cycle setup / expression: \(\Delta H^\ominus_{\text{hyd}}(\text{Cl}^-) = \Delta H^\ominus_{\text{sol}} + \Delta H^\ominus_{\text{latt}} - \Delta H^\ominus_{\text{hyd}}(\text{Na}^+)\) (or equivalent numerical expression).
1 Mark: Correct answer with negative sign and units: \(-377\text{ kJ mol}^{-1}\).

The expression for \(K_p\) is:
\(K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2) \times p(\text{H}_2)^3}\)
Substitute the values:
\(K_p = \frac{30.0^2}{15.0 \times 45.0^3} = \frac{900}{15.0 \times 91125} = \frac{900}{1366875} = 6.584 \times 10^{-4}\)
Units: \(\frac{\text{kPa}^2}{\text{kPa} \times \text{kPa}^3} = \text{kPa}^{-2}\).

1 Mark: Correct calculation of value: \(6.58 \times 10^{-4}\) (accept \(6.6 \times 10^{-4}\)).
1 Mark: Correct unit: \(\text{kPa}^{-2}\) (independent of value, but dependent on attempting the calculation).

Since equal volumes are mixed, the concentrations are halved, but their ratio remains the same as the initial concentrations.
Using the buffer equation:
\([\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}\)
\([\text{H}^+] = 1.74 \times 10^{-5} \times \frac{0.100}{0.0800} = 2.175 \times 10^{-5}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(2.175 \times 10^{-5}) = 4.662 \approx 4.66\).

1 Mark: Correct calculation of \([\text{H}^+]\) as \(2.175 \times 10^{-5}\text{ mol dm}^{-3}\) or correct substitution into pH equation.
1 Mark: Correct value for pH of \(4.66\) (accept \(4.7\)).

When excess ammonia is added to aqueous copper(II) ions, a ligand exchange reaction occurs where four water ligands are replaced by four ammonia molecules, forming the complex \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\). The color of this complex is deep blue (or dark blue).

1 Mark: Correct formula \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) (must include the \(2+\) charge).
1 Mark: Deep blue / dark blue (reject blue / light blue).

1. Identify the largest jump in ionization energy: the fourth ionization energy (\(11577\text{ kJ mol}^{-1}\)) is much higher than the third (\(2745\text{ kJ mol}^{-1}\)), indicating that the fourth electron is removed from a shell closer to the nucleus (inner shell). This means element \(X\) has 3 valence electrons and belongs to Group 3 (Group 13).
2. Since \(X\) is in Period 3, the element is Aluminium (\(\text{Al}\)).
3. The atomic number of Aluminium is 13, so its full electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^1\).

• M1: Identifies the element as aluminium / Al (based on the large jump between the 3rd and 4th ionization energies indicating 3 outer electrons) (1 mark)
• M2: Gives the correct full electronic configuration: \(1s^2 2s^2 2p^6 3s^2 3p^1\) (allow sub-shell notation like \(3s^2 3p_x^1\)) (1 mark)
[Do not accept noble gas core shorthand such as \([Ne] 3s^2 3p^1\)]

1. Compare the size and charge of the cations: Both \(\text{Mg}^{2+}\) and \(\text{Ba}^{2+}\) have a \(2+\) charge, but \(\text{Mg}^{2+}\) is smaller in size than \(\text{Ba}^{2+}\) (ionic radius increases down the group).
2. Therefore, the magnesium ion has a higher charge density.
3. The \(\text{Mg}^{2+}\) ion polarizes the nitrate (\(\text{NO}_3^-\)) electron cloud more strongly, which weakens the covalent \(\text{N}-\text{O}\) bond within the nitrate ion.
4. Consequently, less thermal energy is required to break the \(\text{N}-\text{O}\) bond and decompose magnesium nitrate.

• M1: States that the \(\text{Mg}^{2+}\) ion has a smaller ionic radius / higher charge density than the \(\text{Ba}^{2+}\) ion (1 mark)
• M2: States that \(\text{Mg}^{2+}\) polarises the nitrate ion (or its electron cloud) more strongly, weakening the \(\text{N}-\text{O}\) covalent bond (so less heat energy is needed) (1 mark)
[Reject: magnesium atom / barium atom]

Using Hess's Law for the Born-Haber cycle:
\(\Delta H_{\text{f}} = \Delta H_{\text{at}}(\text{Na}) + IE_1(\text{Na}) + \Delta H_{\text{at}}(\text{F}) + \Delta H_{\text{EA}}(\text{F}) + \Delta H_{\text{latt}}(\text{NaF})\)

Substitute the values into the equation:
\(-574 = 107 + 496 + 79 + \Delta H_{\text{EA}}(\text{F}) - 923\)

Simplify the right-hand side:
\(-574 = -241 + \Delta H_{\text{EA}}(\text{F})\)

Rearrange to solve for \(\Delta H_{\text{EA}}(\text{F})\):
\(\Delta H_{\text{EA}}(\text{F}) = -574 + 241 = -333\text{ kJ mol}^{-1}\)

• M1: Correctly sets up the equation or shows a partially completed calculation, e.g., \(-574 = 107 + 496 + 79 + \Delta H_{\text{EA}}(\text{F}) - 923\) (or simplifies to \(-574 = -241 + \Delta H_{\text{EA}}(\text{F})\)) (1 mark)
• M2: Correct calculation of \(-333\text{ kJ mol}^{-1}\) (must include negative sign and correct unit) (1 mark)
[Accept: \(-333\) without units if the method is correct, but penalize incorrect units]

1. The complex \([\text{Pt}(\text{NH}_3)_2\text{Cl}_2]\) is square planar and has two different ligands arranged in pairs. It can exist as cis-\([\text{Pt}(\text{NH}_3)_2\text{Cl}_2]\) (cisplatin) and trans-\([\text{Pt}(\text{NH}_3)_2\text{Cl}_2]\) (transplatin). This is a type of stereoisomerism known as cis-trans isomerism (or geometric isomerism).
2. Optical isomerism requires a molecule to be chiral, meaning it lacks a plane of symmetry and its mirror images are non-superimposable. Because square planar complexes are flat (planar), they possess a plane of symmetry (the molecular plane). Thus, their mirror images are superimposable, meaning they cannot show optical isomerism.

• M1: Identifies the stereoisomerism as cis-trans isomerism / geometric isomerism (1 mark)
• M2: Explains that it does not show optical isomerism because it is planar / has a plane of symmetry, making its mirror image superimposable (1 mark)
[Do not accept just 'it has no chiral center' without reference to symmetry or planar geometry]

1. Write the expression for the weak acid dissociation constant:
\(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)

Using the standard approximation that \([\text{H}^+] \approx [\text{A}^-]\) and \([\text{HA}] \approx 0.150\text{ mol dm}^{-3}\):
\(K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]}\)

2. Solve for \([\text{H}^+]\):
\([\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.35 \times 10^{-5} \times 0.150}\)
\([\text{H}^+] = \sqrt{2.025 \times 10^{-6}} = 1.423 \times 10^{-3}\text{ mol dm}^{-3}\)

3. Calculate the pH:
\(\text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.8468\)

Rounded to two decimal places, \(\text{pH} = 2.85\).

• M1: Calculates \([\text{H}^+] = 1.42 \times 10^{-3}\text{ mol dm}^{-3}\) (or shows correct substitution: \([\text{H}^+] = \sqrt{1.35 \times 10^{-5} \times 0.150}\)) (1 mark)
• M2: Correct pH value of 2.85 (must be to exactly two decimal places) (1 mark)
[Alternative: award full marks for correct answer with no working shown]

1. Write the two half-equations:
Reduction: \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
Oxidation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\)

2. Multiply the oxidation half-reaction by 5 to balance the electrons:
\(5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-\)

3. Combine the two half-equations:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}\)

• M1: Identifies the 1:5 reacting ratio of \(\text{MnO}_4^-\) to \(\text{Fe}^{2+}\) (or has correct species on both sides) (1 mark)
• M2: Correctly balanced overall equation with correct species and charges: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}\) (1 mark)
[Ignore state symbols]

1. Write the expression for \(K_p\):
\(K_p = \frac{p(\text{PCl}_3) \times p(\text{Cl}_2)}{p(\text{PCl}_5)}\)

2. Substitute the equilibrium partial pressures:
\(K_p = \frac{0.280 \times 0.280}{0.420}\)
\(K_p = \frac{0.0784}{0.420} \approx 0.1867\text{ atm}\)

3. Round to 3 significant figures: \(0.187\text{ atm}\) (or 2 significant figures: \(0.19\text{ atm}\)).

• M1: Correct numerical calculation resulting in \(0.187\) / \(0.19\) (1 mark)
• M2: Correct units of \(\text{atm}\) (or \(\text{atm}^1\)) (1 mark)
[Accept Pa / kPa if converted correctly, but standard is atm here. If units are omitted or incorrect, do not award M2]

1. Calculate the initial moles of propanoic acid and sodium hydroxide:
- Moles of \(\text{CH}_3\text{CH}_2\text{COOH}\) = \(0.0600\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0150\text{ mol}\)
- Moles of \(\text{NaOH}\) = \(0.0400\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 0.00600\text{ mol}\)

2. Calculate the moles of propanoic acid and propanoate ions in the buffer:
- The reaction is: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\)
- Moles of \(\text{CH}_3\text{CH}_2\text{COOH}\) remaining = \(0.0150 - 0.00600 = 0.00900\text{ mol}\)
- Moles of \(\text{CH}_3\text{CH}_2\text{COO}^-\) formed = \(0.00600\text{ mol}\)

3. Use the acid dissociation constant expression to find \([\text{H}^+]\):
\(K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)
\([\text{H}^+] = K_a \times \frac{n(\text{CH}_3\text{CH}_2\text{COOH})}{n(\text{CH}_3\text{CH}_2\text{COO}^-)}
\)[\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.00900}{0.00600} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\)

4. Calculate pH:
\(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6936 \approx 4.69\)

- M1: Calculates initial moles of propanoic acid (\(0.0150\text{ mol}\)) and sodium hydroxide (\(0.00600\text{ mol}\)).
- M2: Calculates moles of propanoic acid remaining (\(0.00900\text{ mol}\)) and moles of propanoate ions formed (\(0.00600\text{ mol}\)).
- M3: Sets up the correct expression for \([\text{H}^+]\) and calculates its value (\(2.025 \times 10^{-5}\text{ mol dm}^{-3}\)).
- M4: Correctly calculates the pH to 2 decimal places (\(4.69\)). Allow FT from incorrect moles.

1. Calculate the moles of manganate(VII) ions used:
\(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{22.40}{1000}\text{ dm}^3 = 4.48 \times 10^{-4}\text{ mol}\)

2. Determine the moles of \(\text{Fe}^{2+}\) using the 1:5 ratio from the redox equation:
\(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)
\(n(\text{Fe}^{2+}) = 5 \times n(\text{MnO}_4^-) = 5 \times 4.48 \times 10^{-4} = 2.24 \times 10^{-3}\text{ mol}\)

3. Calculate the mass of iron in the alloy:
\(\text{Mass of Fe} = 2.24 \times 10^{-3}\text{ mol} \times 55.8\text{ g mol}^{-1} = 0.12499\text{ g}\)

4. Calculate the percentage by mass of iron:
\(\%\text{ Fe} = \frac{0.12499\text{ g}}{0.280\text{ g}} \times 100\% = 44.64\% \approx 44.6\%\)

- M1: Calculates moles of manganate(VII) ions used (\(4.48 \times 10^{-4}\text{ mol}\)).
- M2: Multiplies moles of manganate(VII) by 5 to obtain moles of iron (\(2.24 \times 10^{-3}\text{ mol}\)).
- M3: Calculates the mass of iron (\(0.125\text{ g}\)).
- M4: Correctly calculates percentage by mass of iron to 3 significant figures (\(44.6\%\)). Accept 44.6 or 45.

1. Determine the equilibrium moles of all species using the stoichiometry:
- Initial: \(\text{SO}_3 = 1.50\text{ mol}\), \(\text{SO}_2 = 0\text{ mol}\), \(\text{O}_2 = 0\text{ mol}\)
- Change: \(\text{SO}_2\) increased by \(0.60\text{ mol}\), so \(\text{SO}_3\) decreased by \(0.60\text{ mol}\) and \(\text{O}_2\) increased by \(0.30\text{ mol}\).
- Equilibrium: \(\text{SO}_3 = 1.50 - 0.60 = 0.90\text{ mol}\); \(\text{SO}_2 = 0.60\text{ mol}\); \(\text{O}_2 = 0.30\text{ mol}\)
- Total moles of gas at equilibrium = \(0.90 + 0.60 + 0.30 = 1.80\text{ mol}\)

2. Calculate partial pressures (\(p = \frac{\text{moles of gas}}{\text{total moles}} \times P_{total}\)):
- \(p(\text{SO}_3) = \frac{0.90}{1.80} \times 2.50\text{ bar} = 1.25\text{ bar}\)
- \(p(\text{SO}_2) = \frac{0.60}{1.80} \times 2.50\text{ bar} = 0.8333\text{ bar}\)
- \(p(\text{O}_2) = \frac{0.30}{1.80} \times 2.50\text{ bar} = 0.4167\text{ bar}\)

3. Set up the \(K_p\) expression:
\(K_p = \frac{p(\text{SO}_2)^2 \cdot p(\text{O}_2)}{p(\text{SO}_3)^2}\)

4. Calculate \(K_p\) and determine units:
\(K_p = \frac{(0.8333)^2 \times 0.4167}{(1.25)^2} = \frac{0.28935}{1.5625} = 0.185\text{ bar}\)

- M1: Calculates equilibrium moles of all species (\(\text{SO}_3 = 0.90\text{ mol}\), \(\text{O}_2 = 0.30\text{ mol}\)) and the total moles at equilibrium (\(1.80\text{ mol}\)).
- M2: Calculates the partial pressure of each species (\(p(\text{SO}_3) = 1.25\text{ bar}\), \(p(\text{SO}_2) = 0.833\text{ bar}\), \(p(\text{O}_2) = 0.417\text{ bar}\)).
- M3: Shows the correct algebraic expression for \(K_p\).
- M4: Correctly calculates \(K_p = 0.185\) (accept \(0.185\) to \(0.19\)) and provides correct units (\(bar\) or \(b\)).

1. State the relationship between the enthalpy of formation and the Born-Haber cycle stages:
\(\Delta H_f^{\ominus} = \Delta H_{at}^{\ominus}[\text{Ca}] + \Delta H_{i1}^{\ominus}[\text{Ca}] + \Delta H_{i2}^{\ominus}[\text{Ca}] + \Delta H_{at}^{\ominus}[\text{O}] + \Delta H_{ea1}^{\ominus}[\text{O}] + \Delta H_{ea2}^{\ominus}[\text{O}] + \Delta H_{latt}^{\ominus}[\text{CaO}]\)

2. Substitute the values into the equation:
\(-635 = (+178) + (+590) + (+1145) + (+249) + (-141) + (+798) + \Delta H_{latt}^{\ominus}\)

3. Sum the other energy terms:
\(\text{Sum} = 178 + 590 + 1145 + 249 - 141 + 798 = +2819 \text{ kJ mol}^{-1}\)

4. Rearrange to find \(\Delta H_{latt}^{\ominus}\):
\(\Delta H_{latt}^{\ominus} = -635 - 2819 = -3454 \text{ kJ mol}^{-1}\)

- M1: Shows an understanding of the Born-Haber cycle by writing a correct expression or drawing a diagram linking the terms.
- M2: Substitutes all numbers correctly into the cycle with correct signs.
- M3: Obtains the correct sum of all non-lattice enthalpy terms (\(+2819\text{ kJ mol}^{-1}\)).
- M4: Correctly calculates the final value as \(-3454 \text{ kJ mol}^{-1}\) (must have negative sign and units).

1. Identify the oxidation and reduction half-cells. The half-cell with the more positive \(E^{\ominus}\) is reduced (chlorine/chloride) and the more negative is oxidized (chromium/chromium(III)).
- Oxidation at anode (left): \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\)
- Reduction at cathode (right): \(\text{Cl}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^-(\text{aq})\)

2. Write the standard cell diagram:
\(\text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq}) \parallel \text{Cl}^-(\text{aq}) | \text{Cl}_2(\text{g}) | \text{Pt}(\text{s})\)

3. Calculate the standard cell potential:
\(E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} = +1.36\text{ V} - (-0.74\text{ V}) = +2.10\text{ V}\)

4. Combine and balance the half-equations to find the overall equation:
\(2\text{Cr}(\text{s}) + 3\text{Cl}_2(\text{g}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 6\text{Cl}^-(\text{aq})\)

- M1: Writes the left side of the cell diagram correctly as \(\text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq})\) (accept with or without state symbols).
- M2: Writes the right side of the cell diagram correctly including \(\text{Pt}(\text{s})\) as \(\text{Cl}^-(\text{aq}) | \text{Cl}_2(\text{g}) | \text{Pt}(\text{s})\).
- M3: Calculates \(E^{\ominus}_{cell} = +2.10\text{ V}\) (must have sign).
- M4: Writes the correct overall balanced equation: \(2\text{Cr}(\text{s}) + 3\text{Cl}_2(\text{g}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 6\text{Cl}^-(\text{aq})\) (accept with or without state symbols).

1. Find the mass of the anhydrous salt (\(\text{CoCl}_2\)) and water lost:
- \(\text{Mass of anhydrous } \text{CoCl}_2 = 22.97 - 21.35 = 1.62\text{ g}\)
- \(\text{Mass of } \text{H}_2\text{O} \text{ lost} = 24.33 - 22.97 = 1.36\text{ g}\)

2. Calculate the molar mass of \(\text{CoCl}_2\) and water:
- \(M_r(\text{CoCl}_2) = 58.9 + (2 \times 35.5) = 129.9\text{ g mol}^{-1}\)
- \(M_r(\text{H}_2\text{O}) = 18.0\text{ g mol}^{-1}\)

3. Calculate the moles of each:
- \(\text{Moles of } \text{CoCl}_2 = \frac{1.62}{129.9} = 0.01247\text{ mol}\)
- \(\text{Moles of } \text{H}_2\text{O} = \frac{1.36}{18.0} = 0.07556\text{ mol}\)

4. Determine the ratio \(x\):
- \(x = \frac{0.07556}{0.01247} = 6.059 \approx 6\)

- M1: Calculates the mass of anhydrous \(\text{CoCl}_2\) (\(1.62\text{ g}\)) and water lost (\(1.36\text{ g}\)).
- M2: Calculates the moles of anhydrous salt (\(0.0125\text{ mol}\)).
- M3: Calculates the moles of water (\(0.0756\text{ mol}\)).
- M4: Correctly determines the ratio \(x = 6\) from their working.

1. Let the percentage abundance of \(^{69}\text{Ga}\) be \(y\%\).
Then, the percentage abundance of \(^{71}\text{Ga}\) is \((100 - y)\%\).

2. Set up the equation for the relative atomic mass:
\(\frac{69y + 71(100 - y)}{100} = 69.72\)

3. Expand and solve the equation for \(y\):
\(69y + 7100 - 71y = 6972\)
\(-2y = 6972 - 7100\)
\(-2y = -128\)
\(y = 64.0\)

4. State the percentage abundances of both isotopes:
- Abundance of \(^{69}\text{Ga} = 64.0\%\)
- Abundance of \(^{71}\text{Ga} = 100 - 64.0 = 36.0\%\)

- M1: Sets up the correct algebraic expression for relative atomic mass, e.g., \(\frac{69y + 71(100 - y)}{100} = 69.72\).
- M2: Successfully expands and simplifies the equation to \(-2y = -128\) or equivalent.
- M3: Finds the percentage abundance of \(^{69}\text{Ga}\) is \(64.0\%\).
- M4: Calculates the percentage abundance of \(^{71}\text{Ga}\) as \(36.0\%\) and clearly states both values.

1. Convert temperature, pressure and volume to SI units:
- \(T = 298\text{ K}\)
- \(p = 101 \times 10^3\text{ Pa}\)
- \(V = 360 \times 10^{-6}\text{ m}^3 = 3.60 \times 10^{-4}\text{ m}^3\)

2. Calculate the number of moles of \(\text{CO}_2\) produced using the ideal gas equation:
\(n = \frac{pV}{RT} = \frac{(101 \times 10^3) \times (3.60 \times 10^{-4})}{8.31 \times 298} = \frac{36.36}{2476.38} = 0.01468\text{ mol}\)

3. Relate moles of \(\text{CO}_2\) to moles of \(\text{BaCO}_3\) and calculate the mass of \(\text{BaCO}_3\):
- Since the ratio is 1:1, \(\text{moles of } \text{BaCO}_3 = 0.01468\text{ mol}\)
- \(M_r(\text{BaCO}_3) = 137.3 + 12.0 + (3 \times 16.0) = 197.3\text{ g mol}^{-1}\)
- \(\text{Mass of } \text{BaCO}_3 = 0.01468\text{ mol} \times 197.3\text{ g mol}^{-1} = 2.896\text{ g}\)

4. Calculate the percentage by mass of barium carbonate:
- \(\%\text{ } \text{BaCO}_3 = \frac{2.896}{4.50} \times 100\% = 64.36\% \approx 64.4\%\)

- M1: Converts volume to \(\text{m}^3\) and pressure to \(\text{Pa}\) and substitutes correctly into \(pV=nRT\).
- M2: Calculates the correct number of moles of \(\text{CO}_2\) (\(0.0147\text{ mol}\)).
- M3: Multiplies moles by \(197.3\text{ g mol}^{-1}\) to find the mass of \(\text{BaCO}_3\) (\(2.90\text{ g}\)).
- M4: Correctly calculates percentage by mass as \(64.4\%\) (accept \(64.3\% - 64.5\%\)).

Step 1: Calculate the initial moles of propanoic acid and sodium hydroxide. Moles of CH\(_3\)CH\(_2\)COOH = 0.0500 dm\(^3\) \(\times\) 0.150 mol dm\(^{-3}\) = 7.50 \(\times\) 10\(^{-3}\) mol. Moles of NaOH = 0.0250 dm\(^3\) \(\times\) 0.100 mol dm\(^{-3}\) = 2.50 \(\times\) 10\(^{-3}\) mol. Step 2: Determine the moles of species present after neutralization. The reaction is CH\(_3\)CH\(_2\)COOH + NaOH \(\rightarrow\) CH\(_3\)CH\(_2\)COONa + H\(_2\)O. Moles of CH\(_3\)CH\(_2\)COO\(^-\) formed = 2.50 \(\times\) 10\(^{-3}\) mol. Moles of remaining CH\(_3\)CH\(_2\)COOH = 7.50 \(\times\) 10\(^{-3}\) - 2.50 \(\times\) 10\(^{-3}\) = 5.00 \(\times\) 10\(^{-3}\) mol. Step 3: Calculate the hydrogen ion concentration. [H\(^+\)] = K\(_a\) \(\times\) (moles of acid / moles of conjugate base) = 1.35 \(\times\) 10\(^{-5}\) \(\times\) (5.00 \(\times\) 10\(^{-3}\) / 2.50 \(\times\) 10\(^{-3}\)) = 2.70 \(\times\) 10\(^{-5}\) mol dm\(^{-3}\). Step 4: Calculate the pH. pH = -log\(_{10}\)(2.70 \(\times\) 10\(^{-5}\)) = 4.5686. To two decimal places, pH = 4.57.

M1: Calculates initial moles of propanoic acid (7.50 \(\times\) 10\(^{-3}\) mol) and sodium hydroxide (2.50 \(\times\) 10\(^{-3}\) mol). M2: Calculates moles of remaining propanoic acid (5.00 \(\times\) 10\(^{-3}\) mol) and propanoate ions formed (2.50 \(\times\) 10\(^{-3}\) mol). M3: Calculates [H\(^+\)] = 2.70 \(\times\) 10\(^{-5}\) mol dm\(^{-3}\) (or shows correct substitution into Henderson-Hasselbalch equation). M4: Calculates pH = 4.57 (must be to 2 decimal places). Correct answer with no working scores 4 marks.

The Born-Haber cycle equation for calcium chloride is: \(\Delta\)H\(^\ominus\)_f(CaCl\(_2\)) = \(\Delta\)H\(^\ominus\)_{at}(Ca) + 1st IE(Ca) + 2nd IE(Ca) + 2 \(\times\) \(\Delta\)H\(^\ominus\)_{at}(Cl) + 2 \(\times\) 1st EA(Cl) + \(\Delta\)H\(^\ominus\)_{latt}(CaCl\(_2\)). Substituting the given values: -796 = +178 + 590 + 1145 + 2(121) + 2(-349) + \(\Delta\)H\(^\ominus\)_{latt}(CaCl\(_2\)). -796 = 178 + 590 + 1145 + 242 - 698 + \(\Delta\)H\(^\ominus\)_{latt}(CaCl\(_2\)). -796 = 1457 + \(\Delta\)H\(^\ominus\)_{latt}(CaCl\(_2\)). Therefore, \(\Delta\)H\(^\ominus\)_{latt}(CaCl\(_2\)) = -796 - 1457 = -2253 kJ mol\(^{-1}\).

M1: Multiplies the enthalpy of atomisation of chlorine by 2 (+242) AND the electron affinity of chlorine by 2 (-698). M2: Constructs a correct algebraic expression or Born-Haber cycle relating these terms. M3: Correctly substitutes values into the expression: -796 = 178 + 590 + 1145 + 242 - 698 + Lattice Energy. M4: Evaluates to obtain -2253 (kJ mol\(^{-1}\)). Correct answer with units or without units scores 4 marks. Accept -2253. Reject +2253.

Using the Arrhenius equation:
\( k = A \exp\left(-\frac{E_a}{RT}\right) \)
Convert \( E_a \) to \( \text{J mol}^{-1} \):
\( E_a = 75.0 \times 1000 = 75000\text{ J mol}^{-1} \)
\( \frac{E_a}{RT} = \frac{75000}{8.31 \times 323} = \frac{75000}{2684.13} \approx 27.942 \)
\( k = 2.50 \times 10^{11} \times e^{-27.942} \approx 2.50 \times 10^{11} \times 7.345 \times 10^{-13} = 0.1836\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \approx 1.84 \times 10^{-1}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \)

1 mark for correct calculation and selection of option A. Correctly converts Ea to J/mol and applies Arrhenius equation.

Reaction A yields ethylamine (a primary amine) because excess ammonia is used. Reaction B involves the nucleophilic substitution of bromoethane by the primary amine ethylamine, which produces diethylamine (a secondary amine). Reaction C reduces the nitrile to propan-1-amine (a primary amine). Reaction D hydrolyses N-phenylethanamide to form phenylamine (a primary amine) and ethanoic acid.

1 mark for selecting option B. Correctly identifies that reaction of a primary amine with a halogenoalkane yields a secondary amine.

Let us look at the carbon environments in each option:
- Methylpropan-2-ol, \( (\text{CH}_3)_3\text{COH} \), has two carbon environments (the three equivalent methyl carbons and the central quaternary carbon). Thus, it has 2 peaks.
- Butan-2-ol, \( \text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}_3 \), has four different carbon environments and therefore has 4 peaks.
- Butanone, \( \text{CH}_3\text{COCH}_2\text{CH}_3 \), has four different carbon environments and therefore has 4 peaks.
- Methylpropan-1-ol, \( (\text{CH}_3)_2\text{CHCH}_2\text{OH} \), has three different carbon environments: the two equivalent methyl carbons (1), the CH carbon (2), and the \( \text{CH}_2 \) carbon (3). Thus, it has exactly 3 peaks.

1 mark for selecting option D. Correctly identifies the number of carbon environments in the isomeric compounds.

Reaction via the \( \text{S}_{\text{N}}1 \) mechanism is predominant in tertiary halogenoalkanes. This is because tertiary carbocations are highly stable due to the positive inductive effect of three electron-releasing alkyl groups. 2-chloro-2-methylpropane is a tertiary halogenoalkane and thus reacts via the \( \text{S}_{\text{N}}1 \) pathway.

1 mark for selecting option C. Correctly identifies 2-chloro-2-methylpropane as a tertiary halogenoalkane reacting via SN1.

Tollens' reagent contains the diamminesilver(I) complex ion, \( [\text{Ag}(\text{NH}_3)_2]^+ \). It oxidises aldehydes (like propanal) to carboxylate ions while being reduced to metallic silver, forming a silver mirror. Ketones (like propanone) are not easily oxidised and do not react with Tollens' reagent. 2,4-DNPH reacts with both to give orange precipitates. \( \text{PCl}_5 \) and \( \text{NaBH}_4 \) react with both compounds but do not provide a visual distinction.

1 mark for selecting option A. Correctly identifies Tollens' reagent as a chemical test to distinguish aldehydes from ketones.

Since the first step has a higher activation energy than the second step, it represents the slower process with the larger energy barrier. Hence, the first step is the rate-determining step. The reaction is exothermic because the reactants are at a higher energy than the products. Transition states are energy maxima and are always less stable (higher in energy) than intermediates (which lie in energy minima). A catalyst has no effect on the equilibrium constant.

1 mark for selecting option B. Correctly identifies the step with higher activation energy as the rate-determining step.

Let us determine the average oxidation state of carbon in both compounds:
- Ethanol, \( \text{C}_2\text{H}_6\text{O} \): Let \( x \) be the oxidation state of carbon. Since H is \( +1 \) and O is \( -2 \), \( 2x + 6(+1) + (-2) = 0 \implies 2x + 4 = 0 \implies x = -2 \).
- Ethanal, \( \text{C}_2\text{H}_4\text{O} \): Let \( y \) be the oxidation state of carbon. \( 2y + 4(+1) + (-2) = 0 \implies 2y + 2 = 0 \implies y = -1 \).
Therefore, the average oxidation state of the carbon atoms changes from \( -2 \) to \( -1 \).

1 mark for selecting option A. Correctly calculates average carbon oxidation states in ethanol and ethanal.

First, find the moles of butan-1-ol reacted:
\( n(\text{butan-1-ol}) = \frac{7.41\text{ g}}{74.1\text{ g mol}^{-1}} = 0.100\text{ mol} \)
Since the reaction stoichiometry is 1:1, the theoretical yield of butyl ethanoate is \( 0.100\text{ mol} \).
Theoretical mass of butyl ethanoate = \( 0.100\text{ mol} \times 116.2\text{ g mol}^{-1} = 11.62\text{ g} \).
Percentage yield = \( \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100\\% = \frac{5.81}{11.62} \times 100\\% = 50.0\\% \).

1 mark for selecting option A. Correctly calculates percentage yield by converting mass of reactant to moles and comparing with theoretical yield.

Using the Arrhenius equation: \(\ln(k_2 / k_1) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\). Substituting the values: \(\ln(1.2 \times 10^{-2} / 3.4 \times 10^{-4}) = \frac{E_a}{8.31} \left(\frac{1}{300} - \frac{1}{350}\right)\). This gives: \(3.564 = \frac{E_a}{8.31} (4.762 \times 10^{-4})\). Solving for \(E_a\) gives \(E_a = \frac{3.564 \times 8.31}{4.762 \times 10^{-4}} = 62191 \text{ J mol}^{-1} = 62.2 \text{ kJ mol}^{-1}\).

1 mark for correct calculation of temperature and rate constant terms. 1 mark for correct calculation of final activation energy value of 62.2 (accept values between 62.0 and 62.4) with correct units.

Benzoyl chloride reacts with excess concentrated ammonia in a nucleophilic addition-elimination reaction. Two moles of ammonia are required: one mole to form the benzamide and one mole to react with the released hydrogen chloride to form ammonium chloride: \(\text{C}_6\text{H}_5\text{COCl} + 2\text{NH}_3 \rightarrow \text{C}_6\text{H}_5\text{CONH}_2 + \text{NH}_4\text{Cl}\).

1 mark for correct chemical formulae of reactants (benzoyl chloride and ammonia) and products (benzamide and ammonium chloride). 1 mark for correct balancing of the equation with 2 moles of ammonia.

Methylbenzene contains 5 unique carbon environments. These environments are: the methyl carbon (1), the aromatic C1 carbon bonded to the methyl group (2), the equivalent ortho-carbons C2/C6 (3), the equivalent meta-carbons C3/C5 (4), and the para-carbon C4 (5). This symmetry results in 5 peaks in the carbon-13 NMR spectrum.

1 mark for identifying 5 peaks. 1 mark for explaining that the molecule has 5 distinct carbon environments due to symmetry in the benzene ring.

Hydrogen cyanide (\(\text{HCN}\)) is a weak acid and dissociates poorly in solution, resulting in an extremely low concentration of the nucleophilic cyanide (\(\text{CN}^-\)) ions. Potassium cyanide is ionic and dissociates fully in water, providing a high concentration of \(\text{CN}^-\) nucleophiles to initiate the attack on the carbonyl carbon.

1 mark for explaining that HCN is a weak acid and dissociates poorly (low concentration of \(\text{CN}^-\)). 1 mark for stating that KCN provides a high concentration of nucleophilic \(\text{CN}^-\) ions.

The overall order of a reaction is defined as the sum of the powers of the concentration terms in the rate equation. For the given rate equation, the order with respect to A is 1 and with respect to B is 2, hence the overall order is \(1 + 2 = 3\).

1 mark for the definition (sum of powers of concentration terms in the rate equation). 1 mark for identifying the overall order as 3.

The molar mass of the desired product, epoxypropane (\(\text{C}_3\text{H}_6\text{O}\)), is \(3 \times 12.0 + 6 \times 1.0 + 16.0 = 58.0 \text{ g mol}^{-1}\). The total mass of all products is \(58.0\) (epoxypropane) + \(18.0\) (water) = \(76.0 \text{ g mol}^{-1}\). The percentage atom economy is calculated as \(\frac{58.0}{76.0} \times 100 = 76.315\%\) which rounds to 76.3%.

1 mark for calculating the correct molar masses of the desired product (58.0) and total products/reactants (76.0). 1 mark for the correct final percentage atom economy of 76.3% (accept 76% or 76.3).

Butan-1-ol contains a highly polar \(\text{O-H}\) bond, allowing it to form strong intermolecular hydrogen bonds between its molecules. Diethyl ether does not contain an \(\text{O-H}\) bond and can only form much weaker permanent dipole-dipole and London forces. More thermal energy is required to break the stronger hydrogen bonds in butan-1-ol.

1 mark for identifying that butan-1-ol forms strong intermolecular hydrogen bonds. 1 mark for explaining that diethyl ether has only weaker intermolecular forces (London and/or dipole-dipole forces) which require less energy to overcome.

Stereoisomerism (E-Z) in alkenes requires restricted rotation about the carbon-carbon double bond (\(\text{C=C}\)) and each carbon of the double bond to be bonded to two different groups. In but-2-ene, both double-bonded carbons are bonded to a methyl group and a hydrogen atom (two different groups). In but-1-ene, one of the double-bonded carbons is attached to two identical hydrogen atoms, preventing geometric isomerism.

1 mark for stating that there is restricted rotation around the \(\text{C=C}\) double bond. 1 mark for explaining that but-2-ene has two different groups on each carbon of the double bond, whereas but-1-ene has two identical groups (hydrogen atoms) on one carbon of the double bond.

1. Rearrange the rate equation to solve for the rate constant, \(k\): \(k = \frac{\text{rate}}{[\text{X}][\text{Y}]^2}\). 2. Substitute the given values into the equation: \(k = \frac{3.10 \times 10^{-4}}{0.400 \times (0.250)^2} = \frac{3.10 \times 10^{-4}}{0.0250} = 0.0124\). 3. Determine the units for \(k\): \(\text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

MP1: Correct numerical calculation of the rate constant, \(k = 0.0124\) (or \(1.24 \times 10^{-2}\)). Allow 1 mark. MP2: Correct units for the rate constant, \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) (or \(\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)). Allow 1 mark. Ignore minor formatting issues with units but must have correct powers.

Ethylamine contains an alkyl (ethyl) group which is electron-releasing due to the positive inductive effect. This increases the electron density on the nitrogen atom, making its lone pair of electrons more available to accept a proton (\(\text{H}^+\)). In contrast, phenylamine has a phenyl group where the lone pair of electrons on the nitrogen atom overlaps with the delocalised \(\pi\)-electron system of the benzene ring. This delocalisation reduces the electron density on the nitrogen atom, making the lone pair less available to accept a proton.

MP1: Explain that the ethyl group in ethylamine is electron-releasing / has a positive inductive effect which increases the electron density on the nitrogen atom / makes the lone pair more available (to accept a proton). (1 mark) MP2: Explain that in phenylamine, the lone pair on the nitrogen atom is delocalised into the benzene ring (making it less available to accept a proton). (1 mark)

The carbonyl group (\(\text{C}=\text{O}\)) in butanal is planar. During the nucleophilic addition reaction, the cyanide ion (\(\text{CN}^-\)) nucleophile has an equal probability of attacking the planar carbonyl carbon from either above or below the plane. This equal-rate attack results in the formation of an equimolar (50:50) mixture of the two optical isomers (enantiomers), known as a racemic mixture. Since the two enantiomers rotate the plane of polarised light by equal angles in opposite directions, the optical activities cancel out, rendering the final product optically inactive.

MP1: State that the carbonyl group / carbon (\(\text{C}=\text{O}\)) is planar. (1 mark) MP2: State that attack (by the \(\text{CN}^-\) nucleophile) is equally likely from above or below (the plane), resulting in an equimolar mixture of enantiomers / a racemic mixture (which is optically inactive). (1 mark)

The structural formula of the ester is \(\text{HCOOCH}(\text{CH}_3)_2\) (isopropyl methanoate). This structure is consistent with the spectrum because it contains exactly three distinct carbon environments: 1. The carbonyl carbon (\(\text{H}\underline{\text{C}}\text{OO}-\)); 2. The CH carbon attached directly to the oxygen (\(-\text{O}\underline{\text{C}}\text{H}(\text{CH}_3)_2\)); 3. The two equivalent methyl carbons (\(-\text{CH}(\underline{\text{C}}\text{H}_3)_2\)) which are in the same environment and therefore produce a single peak.

MP1: Correct structural formula of the ester: \(\text{HCOOCH}(\text{CH}_3)_2\) or displayed/semi-structural equivalent. (1 mark) MP2: Correctly identify the three carbon environments: carbonyl carbon, the \(\text{CH}\) carbon, and the two equivalent methyl (\(\text{CH}_3\)) carbons. (1 mark)

Use the Arrhenius equation in the form: \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
Substitute the values into the equation:
\ln\left(\frac{4.80 \times 10^{-2}}{1.50 \times 10^{-3}}\right) = \frac{E_a}{8.31} \left(\frac{1}{300} - \frac{1}{350}\right)
\ln(32) = \frac{E_a}{8.31} \left(3.333 \times 10^{-3} - 2.857 \times 10^{-3}\right)
3.4657 = \frac{E_a}{8.31} \left(4.762 \times 10^{-4}\right)
E_a = \frac{3.4657 \times 8.31}{4.762 \times 10^{-4}} = 60464\text{ J mol}^{-1}
Convert to \text{kJ mol}^{-1}: \frac{60464}{1000} = 60.5\text{ kJ mol}^{-1} (to 3 significant figures).

1 Mark: Correct substitution into Arrhenius equation, or calculation of \ln(k_2/k_1) = 3.47.
1 Mark: Correct calculation of the temperature term \left(\frac{1}{T_1} - \frac{1}{T_2}\right) = 4.76 \times 10^{-4}\text{ K}^{-1}.
1 Mark: Evaluation of \(E_a\) in \text{J mol}^{-1} as 60500 (or 60464).
1 Mark: Final answer correct to 3 significant figures in \text{kJ mol}^{-1} (60.5).

Calculate moles of glycine used:
moles of glycine = \frac{3.75}{75.0} = 0.0500\text{ mol}
Since the reaction ratio of glycine to Gly-Ala is 1:1, the theoretical moles of Gly-Ala is also \(0.0500\text{ mol}\).
Calculate the theoretical mass of Gly-Ala:
theoretical mass = 0.0500 \times 146.0 = 7.30\text{ g}
Calculate the percentage yield:
\text{Yield} = \frac{2.92}{7.30} \times 100\% = 40.0\%.

1 Mark: Calculate moles of glycine as 0.0500 mol.
1 Mark: State that theoretical moles of dipeptide is 0.0500 mol (from 1:1 ratio).
1 Mark: Calculate theoretical mass of dipeptide as 7.30 g.
1 Mark: Correct calculation of percentage yield to 3 significant figures (40.0%).

1. Calculate mass of Carbon:
Moles of \text{CO}_2 = \frac{4.40}{44.0} = 0.100\text{ mol}
Since each mole of \text{CO}_2 contains 1 mole of carbon, moles of \text{C} = 0.100\text{ mol}. Mass of \text{C} = 0.100 \times 12.0 = 1.20\text{ g}.
2. Calculate mass of Hydrogen:
Moles of \text{H}_2\text{O} = \frac{1.80}{18.0} = 0.100\text{ mol}. Since each mole of \text{H}_2\text{O} contains 2 moles of hydrogen, moles of \text{H} = 0.200\text{ mol}. Mass of \text{H} = 0.200 \times 1.0 = 0.20\text{ g}.
3. Calculate mass and moles of Oxygen:
Mass of \text{O} = 2.20\text{ g} - (1.20 + 0.20) = 0.80\text{ g}. Moles of \text{O} = \frac{0.80}{16.0} = 0.050\text{ mol}.
4. Find simplest whole-number ratio:
\text{Ratio C : H : O} = 0.100 : 0.200 : 0.050 = 2 : 4 : 1. Therefore, the empirical formula is \text{C}_2\text{H}_4\text{O}.

1 Mark: Determine moles of C (0.100 mol) and mass of C (1.20 g).
1 Mark: Determine moles of H (0.200 mol) and mass of H (0.20 g).
1 Mark: Deduce mass of O by subtraction (0.80 g) and convert to moles (0.050 mol).
1 Mark: Determine correct empirical formula (C2H4O) by finding the simplest ratio.

(i) For a first-order reaction, the integrated rate law is \ln[A]_t = -kt + \ln[A]_0. Thus, the gradient of the plot of \ln[A] against time is \(-k\). Therefore, \(k = -(\text{gradient}) = 3.50 \times 10^{-3}\text{ s}^{-1}\).
(ii) Using the first-order rate equation: \text{Rate} = k[\text{halogenoalkane}]
\text{Initial Rate} = (3.50 \times 10^{-3}\text{ s}^{-1}) \times (0.250\text{ mol dm}^{-3}) = 8.75 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}.

1 Mark: Identify rate constant \(k = 3.50 \times 10^{-3}\).
1 Mark: Provide correct units of \text{s}^{-1}.
1 Mark: State or show substitution into the first-order rate expression: \text{Rate} = k[\text{halogenoalkane}].
1 Mark: Correct calculation of initial rate as \(8.75 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\) (accept correct units).

(i) The product of the reaction between propanone and ethylmagnesium bromide is 2-methylbutan-2-ol. The skeletal structure features a 5-carbon continuous chain with a branch, specifically a 4-carbon chain (butane) with a methyl group and a hydroxyl group both on carbon-2.
(ii) Grignard reagents are strong bases and nucleophiles. They react rapidly with water (or moisture) to form alkanes (ethane in this case) and magnesium hydroxide bromide, destroying the reagent.
(iii) The Grignard carbon (nucleophile) attacks the carbonyl carbon (electrophile), which is a nucleophilic addition reaction.
(iv) The intermediate formed prior to hydrolysis has the formula \text{(CH}_3\text{)}_2\text{C(OMgBr)CH}_2\text{CH}_3.

1 Mark: Correctly identify/draw skeletal structure of 2-methylbutan-2-ol.
1 Mark: Explain that moisture/water reacts with the Grignard reagent (to form ethane/alkane) / hydrolyses it.
1 Mark: Correctly name the mechanism as nucleophilic addition.
1 Mark: Draw or write the correct chemical formula of the intermediate: \text{(CH}_3\text{)}_2\text{C(OMgBr)CH}_2\text{CH}_3.

(i) The molecule must contain a level of symmetry such that two carbon atoms are in the exact same chemical environment (equivalent carbons) and therefore produce only one combined peak instead of two separate peaks.
(ii) The two esters of formula \text{C}_5\text{H}_{10}\text{O}_2 with exactly four carbon environments are:
1. Isopropyl ethanoate (or isopropyl acetate), \text{CH}_3\text{COOCH(CH}_3\text{)}_2, where the two methyl carbons on the isopropyl group are equivalent.
2. Methyl 2-methylpropanoate (or methyl isobutyrate), \text{(CH}_3\text{)}_2\text{CHCOOCH}_3, where the two methyl carbons on the 2-methylpropyl group are equivalent.

1 Mark: Explain that four peaks mean two carbon atoms are equivalent / there is symmetry in the molecule.
1 Mark: Specify that this symmetry comes from a dimethyl-substituted carbon (such as an isopropyl group).
1 Mark: Identify Isopropyl ethanoate (systematic name or correct structural formula).
1 Mark: Identify Methyl 2-methylpropanoate (systematic name or correct structural formula).

(i) 2-bromo-2-methylpropane is a tertiary halogenoalkane and reacts via the \text{S}_{\text{N}}1 mechanism. This is because the tertiary carbocation intermediate is highly stable due to the electron-donating inductive effect of three methyl groups. Also, the steric hindrance around the tertiary carbon prevents back-side attack.
1-bromobutane is a primary halogenoalkane and reacts via the \text{S}_{\text{N}}2 mechanism because the primary carbocation is extremely unstable and there is minimal steric hindrance, allowing back-side nucleophilic attack.
(ii) 2-bromo-2-methylpropane reacts faster. The tertiary carbocation intermediate forms rapidly because of its high stability, giving a lower activation energy for the rate-determining step compared to the transition state of the \text{S}_{\text{N}}2 mechanism.

1 Mark: Correctly identify 2-bromo-2-methylpropane as SN1 and 1-bromobutane as SN2.
1 Mark: Explain SN1 for the tertiary halogenoalkane based on carbocation stability / electron-donating inductive effect of alkyl groups.
1 Mark: Explain SN2 for the primary halogenoalkane based on low steric hindrance / instability of primary carbocations.
1 Mark: Identify 2-bromo-2-methylpropane as faster because the tertiary carbocation is more stable / has lower activation energy for carbocation formation.

1. Moles of \text{Cr}_2\text{O}_7^{2-}: \text{moles} = 0.0150 \times \frac{22.40}{1000} = 3.36 \times 10^{-4}\text{ mol}.
2. Moles of \text{Fe}^{2+} in \(25.0\text{ cm}^3\) aliquot:
Stoichiometric ratio is \(1\text{ mol of } \text{Cr}_2\text{O}_7^{2-} : 6\text{ mol of } \text{Fe}^{2+}\).
moles = 6 \times 3.36 \times 10^{-4} = 2.016 \times 10^{-3}\text{ mol}.
3. Moles of \text{Fe}^{2+} in the original \(250.0\text{ cm}^3\) solution:
moles = 2.016 \times 10^{-3} \times 10 = 2.016 \times 10^{-2}\text{ mol}.
4. Mass of iron in the original sample:
mass = 2.016 \times 10^{-2}\text{ mol} \times 55.8\text{ g mol}^{-1} = 1.1249\text{ g}.
5. Percentage by mass of iron in the wire:
\%\text{ Fe} = \frac{1.1249}{2.50} \times 100\% = 45.0\%.

1 Mark: Calculate moles of dichromate: \(3.36 \times 10^{-4}\text{ mol}\).
1 Mark: Multiply by 6 to find moles of iron in aliquot: \(2.016 \times 10^{-3}\text{ mol}\).
1 Mark: Multiply by 10 to scale to 250 cm3: \(2.016 \times 10^{-2}\text{ mol}\).
1 Mark: Convert to mass and calculate percentage yield: 45.0% (accept range 44.9 - 45.1%).

1. Calculate the natural log of the ratio of the rate constants:
\(\ln\left(\frac{1.85 \times 10^{-3}}{1.50 \times 10^{-4}}\right) = \ln(12.333) = 2.5123\)

2. Calculate the difference in reciprocal temperatures:
\(\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{1}{298} - \frac{1}{328} = 0.0033557 - 0.0030488 = 3.069 \times 10^{-4} \text{ K}^{-1}\)

3. Rearrange the Arrhenius equation to solve for \(E_a\):
\(E_a = \frac{R \ln\left(\frac{k_2}{k_1}\right)}{\frac{1}{T_1} - \frac{1}{T_2}}\)

\(E_a = \frac{8.31 \times 2.5123}{3.069 \times 10^{-4}} = 68026 \text{ J mol}^{-1}\)

4. Convert the answer into \(\text{kJ mol}^{-1}\):
\(E_a = \frac{68026}{1000} = 68.026 \text{ kJ mol}^{-1}\)

Rounding to 3 significant figures gives \(68.0 \text{ kJ mol}^{-1}\).

• M1: Correct calculation of the ratio of rate constants or its natural log: \(\ln(k_2/k_1) = 2.51\) (1)
• M2: Correct calculation of \(\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = 3.07 \times 10^{-4} \text{ K}^{-1}\) (1)
• M3: Correct rearrangement of Arrhenius expression and calculation of \(E_a\) in \(\text{J mol}^{-1}\) (values in range \(68000\) to \(68100\)) (1)
• M4: Correct final value of \(68.0 \text{ kJ mol}^{-1}\) to 3 sig figs (accept \(68\)), including the correct division by 1000 and the unit (1)

1. Calculate the amount in moles of salicylic acid used:
\(\text{Moles} = \frac{\text{Mass}}{M_r} = \frac{6.21 \text{ g}}{138.0 \text{ g mol}^{-1}} = 0.0450 \text{ mol}\)

2. Determine the theoretical moles of aspirin expected:
Since the stoichiometry of salicylic acid to aspirin is 1:1, the theoretical yield of aspirin is \(0.0450 \text{ mol}\).

3. Calculate the theoretical mass of aspirin:
\(\text{Theoretical mass} = \text{Moles} \times M_r = 0.0450 \text{ mol} \times 180.0 \text{ g mol}^{-1} = 8.10 \text{ g}\)

4. Calculate the percentage yield:
\(\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{5.12 \text{ g}}{8.10 \text{ g}} \times 100 = 63.2098...\%\)

Rounding to 3 significant figures gives \(63.2\%\).

• M1: Correct calculation of moles of salicylic acid: \(\frac{6.21}{138} = 0.0450 \text{ mol}\) (1)
• M2: Relates 1:1 molar ratio to state that theoretical moles of aspirin = \(0.0450 \text{ mol}\) (1)
• M3: Correct calculation of theoretical mass of aspirin: \(0.0450 \times 180.0 = 8.10 \text{ g}\) (1)
• M4: Correct calculation of percentage yield to 3 significant figures: \(63.2\%\) (1). Allow ECF from minor mathematical slips in previous stages.

1. Calculate the mass of carbon in the sample from the carbon dioxide produced:
\(\text{Moles of CO}_2 = \frac{2.64 \text{ g}}{44.0 \text{ g mol}^{-1}} = 0.060 \text{ mol}\)
Since there is 1 carbon atom in \(\text{CO}_2\), \(\text{Moles of C} = 0.060 \text{ mol}\).
\(\text{Mass of C} = 0.060 \text{ mol} \times 12.0 \text{ g mol}^{-1} = 0.72 \text{ g}\).

2. Calculate the mass of hydrogen in the sample from the water produced:
\(\text{Moles of H}_2\text{O} = \frac{1.62 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.090 \text{ mol}\)
Since there are 2 hydrogen atoms in \(\text{H}_2\text{O}\), \(\text{Moles of H} = 0.180 \text{ mol}\).
\(\text{Mass of H} = 0.180 \text{ mol} \times 1.0 \text{ g mol}^{-1} = 0.18 \text{ g}\).

3. Calculate the mass and moles of oxygen by subtraction:
\(\text{Mass of O} = 1.38 \text{ g} - (0.72 \text{ g} + 0.18 \text{ g}) = 0.48 \text{ g}\).
\(\text{Moles of O} = \frac{0.48 \text{ g}}{16.0 \text{ g mol}^{-1}} = 0.030 \text{ mol}\).

4. Determine the simplest whole-number ratio of the elements:
\(\text{C} : \text{H} : \text{O} = 0.060 : 0.180 : 0.030\)
Dividing all values by 0.030 gives:
\(\text{C} : \text{H} : \text{O} = 2 : 6 : 1\)
So, the empirical formula of **X** is \(\text{C}_2\text{H}_6\text{O}\).

5. Deduce the molecular formula:
The mass of the empirical formula unit \(\text{C}_2\text{H}_6\text{O}\) is:
\((2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\).
Since this matches the given relative molecular mass of 46.0, the molecular formula of **X** is also \(\text{C}_2\text{H}_6\text{O}\).

• M1: Calculates the masses of carbon (\(0.72 \text{ g}\)) and hydrogen (\(0.18 \text{ g}\)) present in the sample (1)
• M2: Subtracts from total mass to find mass of oxygen (\(0.48 \text{ g}\)) and converts all masses to moles: \(\text{C} = 0.060\), \(\text{H} = 0.180\), \(\text{O} = 0.030\) (1)
• M3: Determines the simplest whole-number ratio to arrive at empirical formula: \(\text{C}_2\text{H}_6\text{O}\) (1)
• M4: Deduces that because the empirical formula mass is 46.0, which matches the relative molecular mass of 46.0, the molecular formula is also \(\text{C}_2\text{H}_6\text{O}\) (1)

1. Ninhydrin can be sprayed onto the dry TLC plate to react with the amino acids and produce visible purple/brown spots (alternatively, UV light can be used if the silica has a fluorescent indicator). 2. A lid must be placed on the tank to prevent evaporation of the solvent. This ensures that the atmosphere inside the tank is saturated with solvent vapor, which is necessary for the solvent front to rise evenly up the plate without drying out.

[1 mark] Identifies ninhydrin (accept UV light or iodine vapor). [1 mark] Explains that the lid prevents evaporation of the solvent / keeps the atmosphere saturated with solvent vapor.

1. A pre-warmed funnel prevents premature crystallisation of the organic product on the filter paper or funnel neck, which would block the filter and reduce the yield. 2. Using a minimum volume of hot solvent ensures the solution is saturated at high temperature. Upon cooling, the maximum amount of dissolved organic solid will recrystallise out rather than remaining dissolved in the excess solvent.

[1 mark] To prevent premature crystallisation / precipitation of the product on the filter paper or funnel. [1 mark] To ensure the solution is saturated / to maximize the yield of crystals obtained upon cooling.

1. Sodium thiosulfate reacts rapidly and selectively with any iodine (\(\text{I}_2\)) produced, converting it back to iodide (\(\text{I}^-\)). This prevents the starch indicator from turning blue-black immediately. 2. Once the fixed amount of sodium thiosulfate is completely exhausted, the iodine produced remains in solution and reacts with the starch, causing a sudden blue-black color change. The time taken (\(t\)) to reach this end-point allows the rate to be calculated because rate is proportional to \(1/t\).

[1 mark] Explains that sodium thiosulfate reacts with the iodine produced until it is completely consumed. [1 mark] Explains that the sudden appearance of the blue-black color at a fixed point in the reaction allows the rate to be calculated as proportional to \(1/t\).

1. The student should select a filter of the complementary color to the blue copper(II) solution (which would be a red or orange filter, or the filter that gives the maximum absorbance value when tested with a standard solution). 2. This choice is important because it ensures maximum sensitivity of the absorbance measurements to changes in concentration, resulting in a steeper calibration curve and more accurate determination of unknown concentrations.

[1 mark] Choose a filter of the complementary color (accept red/orange filter) / filter that gives maximum absorbance. [1 mark] To maximize the sensitivity of the readings / ensure maximum absorbance change per unit change in concentration.

1. Rinse the pH probe/electrode with deionised water, place it in a buffer solution of a known pH (typically pH 7.00), and calibrate/adjust the meter to read this value. 2. Rinse the electrode again and place it in a second buffer solution of a different known pH (such as pH 4.00 for acidic measurements or pH 10.00 for alkaline measurements), and adjust the meter to read this second value, ensuring the expected pH range of the test solutions lies between these two calibration points.

[1 mark] Rinse electrode with deionised water and place in a buffer of known pH (e.g., pH 7) and adjust the meter to read that pH. [1 mark] Repeat the process (rinsing first) using a second buffer of a different known pH (e.g., pH 4 or 10) that brackets the expected range of measurements.

To make a fair comparison of the rate at which calcium carbonate and strontium carbonate decompose, the heating conditions and physical state of the reactants must be controlled: 1. The heat source must be identical: use the same Bunsen burner setting (e.g., fully open blue flame) and maintain a constant distance between the flame and the test tube. 2. The surface area/particle size of the solid carbonates must be the same (e.g., both as fine powders), as larger lumps would react more slowly due to reduced surface area.

[1 mark per valid control variable, max 2 marks] Same distance between the flame and the test tube / same intensity of heating (Bunsen flame setting). Same surface area / particle size of the carbonate solids. Same volume/concentration of limewater used (if measuring time for limewater to go cloudy). Reject: 'same temperature' (as temperature is changing during heating).

1. Heating to constant mass ensures that all the water of crystallisation has been completely driven off from the hydrated salt, leaving only the anhydrous compound. 2. The student achieves and confirms this by heating the crucible, allowing it to cool (ideally in a desiccator), weighing it, and then repeating this cycle of heating, cooling, and weighing until two consecutive mass measurements are identical (within experimental error).

[1 mark] To ensure that all of the water of crystallisation has been removed / evaporated. [1 mark] By heating, cooling, and weighing the sample repeatedly until consecutive mass measurements are constant / identical.

1. The phenomenon is autocatalysis (where one of the products of the reaction acts as a catalyst for the reaction). 2. The species responsible is the \(\text{Mn}^{2+}\) ion, which is formed as a product when \(\text{MnO}_4^-\) is reduced, and then catalyzes the further reaction between manganate(VII) and ethanedioate ions.

[1 mark] Identifies the phenomenon as autocatalysis (or the reaction is autocatalytic). [1 mark] Identifies the catalyst as \(\text{Mn}^{2+}\) ions (or manganese(II) ions).

Rinsing the burette with distilled water but not the acid leaves a thin film of water on the inner walls. This water dilutes the hydrochloric acid when it is added, decreasing its concentration. Consequently, a larger volume of this diluted acid is required to react completely with the alkali. When calculating the concentration of the alkali, the nominal (undiluted) concentration of the acid is used alongside this falsely high volume, leading to an overestimation of the moles of acid and hence a higher calculated concentration of the alkali.

M1: State that the acid in the burette is diluted, so a larger volume (titre) of acid is required to reach the end-point. (1 mark) M2: State that using this larger volume in calculations results in a calculated concentration of the alkali that is higher than the actual value. (1 mark)

Cyclohexene and water are immiscible, forming two distinct layers. They can be separated by pouring the mixture into a separating funnel, allowing the layers to settle, and running off the lower aqueous layer. Remaining water dissolved in the organic layer is removed by adding an anhydrous inorganic salt like anhydrous calcium chloride, which acts as a drying agent by absorbing water to form a hydrated solid.

M1: Use a separating funnel to run off the aqueous layer (or collect the organic layer). (1 mark) M2: Add a suitable anhydrous drying agent, such as anhydrous calcium chloride / anhydrous sodium sulfate / anhydrous magnesium sulfate. (1 mark) (Do not accept hydrated salts or indicators like copper(II) sulfate).

The reaction produces iodine: \(\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + 2\text{H}_2\text{O}\). Added sodium thiosulfate instantly reacts with the generated iodine: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\). This keeps the concentration of free iodine at zero. Once all the sodium thiosulfate has been consumed, the free iodine concentration rises and reacts with the starch indicator, turning the solution blue-black. This provides a precise time point at which a fixed, known amount of reactant has been consumed, allowing the initial rate to be determined.

M1: Sodium thiosulfate reacts with the iodine as soon as it is formed, converting it back to iodide. (1 mark) M2: This delays the blue-black starch-iodine color until all the thiosulfate is consumed, providing a fixed point to measure the initial rate of reaction. (1 mark)

To ensure complete thermal decomposition (dehydration), the sample must be heated to constant mass. The student must heat the crucible and its contents, allow them to cool inside a desiccator (to prevent re-absorption of moisture from the air), and weigh the crucible. This sequence (heating, cooling, weighing) must be repeated until consecutive mass readings are identical, indicating that no more water is being lost.

M1: Heat the crucible, allow it to cool in a desiccator, and record the mass. (1 mark) M2: Repeat the cycle of heating, cooling, and weighing until a constant mass / two consecutive identical mass measurements are obtained. (1 mark)

Copper(II) ions absorb light primarily in the red region of the visible spectrum, which is why they appear blue (the complementary colour transmitted). By selecting a red filter, the light passing through the solution corresponds to the wavelength that is most strongly absorbed by the copper(II) ions. This maximises the absorbance readings and provides the greatest sensitivity for detecting changes in concentration. If a blue filter were used, the blue light would be mostly transmitted, resulting in very low absorbance and poor sensitivity.

M1: Red is the complementary colour to blue, so light of this wavelength is absorbed most strongly by the copper(II) ions. (1 mark) M2: This maximises absorbance / provides the highest sensitivity for the calibration curve (whereas a blue filter would transmit blue light, resulting in minimal absorption). (1 mark)

The presence of impurities disrupts the regular crystalline lattice of the solid organic compound, weakening the intermolecular forces. Consequently, less energy is required to melt the solid, which lowers the melting temperature. Furthermore, because different regions of the impure mixture may have varying concentrations of impurities, the sample does not melt sharply at one temperature; instead, it melts gradually over a wider range of temperatures.

M1: The melting temperature of the impure sample will be lower than the literature/accepted value for pure aspirin. (1 mark) M2: The impure sample will melt over a broad/wide range of temperatures (instead of melting sharply at a single temperature). (1 mark)

The salt bridge serves two main purposes. First, it completes the electrical circuit by providing an electrical pathway between the two half-cells without allowing the electrolyte solutions to mix. Second, it maintains charge balance (electrical neutrality) in both half-cells. As electrons flow through the external wire from the anode to the cathode, cations from the salt bridge (\(\text{K}^+\)) migrate into the cathode compartment to balance the buildup of negative charge, while anions (\(\text{NO}_3^-\)) migrate into the anode compartment to balance the buildup of positive charge.

M1: Completes the electrical circuit (by connecting the two half-cells). (1 mark) M2: Maintains electrical neutrality / charge balance in both half-cells by allowing ions to migrate / flow. (1 mark)

To calibrate a pH probe, the student must use standard buffer solutions of known pH. First, the probe is rinsed with distilled or deionised water to remove any contaminants. It is then immersed in the first buffer solution (commonly pH 7.00 or pH 4.00), allowed to stabilise, and the meter is adjusted to match the known value. The probe is rinsed again with deionised water to prevent carryover, placed into a second buffer solution of a different pH (commonly pH 10.00 or pH 4.00), allowed to stabilise, and the second point is calibrated on the meter.

M1: Rinse the pH probe with distilled/deionised water, place it into a buffer solution of known pH (e.g., pH 7.00/4.00), and calibrate the meter. (1 mark) M2: Rinse the probe again and place it into a second buffer solution of a different, known pH (e.g., pH 4.00/9.20/10.00), and calibrate again. (1 mark)

First, determine the overall absolute uncertainty in the titre volume. Since a titre is calculated from two burette readings (initial and final), the absolute uncertainty is: \(2 \times 0.05\text{ cm}^3 = 0.10\text{ cm}^3\). Next, calculate the percentage uncertainty: \(\text{Percentage uncertainty} = \frac{0.10}{24.30} \times 100 = 0.412\%\) (or \(0.41\%\)).

M1: Recognises that the absolute uncertainty of a titre involves two readings and is \(2 \times 0.05 = 0.10\text{ cm}^3\) (1 mark). M2: Calculates percentage uncertainty as \(0.41\%\) or \(0.412\%\) (1 mark).

Using the minimum volume of hot water ensures that the solution is saturated with benzoic acid at high temperature. If too much water is used, a significant amount of the benzoic acid will remain dissolved in the water even when it is cooled to room temperature or ice-cold, which reduces the mass of benzoic acid that recrystallises, lowering the overall yield.

M1: Minimum volume is used to ensure a saturated solution so that benzoic acid crystallises out upon cooling (1 mark). M2: Too much water means more benzoic acid remains dissolved in the cold solvent, reducing the mass of solid recovered or lowering the yield (1 mark).

For \(\text{Cr}^{3+}(\text{aq})\), the initial green precipitate of chromium(III) hydroxide, \(\text{Cr(OH)}_3\), dissolves in excess sodium hydroxide to form a dark green solution containing the \([\text{Cr(OH)}_6]^{3-}\) ion. For \(\text{Fe}^{3+}(\text{aq})\), the brown precipitate of iron(III) hydroxide, \(\text{Fe(OH)}_3\), does not dissolve in excess sodium hydroxide and remains as a brown precipitate.

M1: Identifies that the green precipitate formed with \(\text{Cr}^{3+}\) dissolves in excess \(\text{NaOH}\) to form a dark green solution (1 mark). M2: Identifies that the brown precipitate formed with \(\text{Fe}^{3+}\) does not dissolve or remains insoluble in excess (1 mark).

The sodium thiosulfate reacts instantly with any iodine produced. The amount of sodium thiosulfate must be small to ensure that only a very small, negligible fraction of the main reactants has reacted before the colour change occurs. This means the concentrations of the reactants remain virtually constant during this brief period, so the measured average rate (proportional to \(1/t\)) is a highly accurate approximation of the initial rate.

M1: To ensure only a tiny fraction of the reactants are consumed or reactant concentrations remain virtually constant (1 mark). M2: So that the average rate measured is a close approximation of the initial rate (1 mark).

1. Calculate the amount in moles of manganate(VII) ions used: \(n(\text{MnO}_4^-) = 0.02340\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 4.68 \times 10^{-4}\text{ mol}\). 2. Use the stoichiometric ratio of the reaction, \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\), to determine the moles of iron(II) ions: \(n(\text{Fe}^{2+}) = 5 \times 4.68 \times 10^{-4}\text{ mol} = 2.34 \times 10^{-3}\text{ mol}\). 3. Calculate the mass of iron: \(m(\text{Fe}) = 2.34 \times 10^{-3}\text{ mol} \times 55.8\text{ g mol}^{-1} = 0.130572\text{ g}\). 4. Calculate the percentage by mass of iron in the tablet: \(\%\text{ by mass} = \frac{0.130572\text{ g}}{0.350\text{ g}} \times 100 = 37.306\% \approx 37.3\%\).

Mark 1: For calculating the moles of manganate(VII) ions (\(4.68 \times 10^{-4}\text{ mol}\)). Mark 2: For multiplying the moles of manganate(VII) by 5 to obtain the moles of iron(II) ions (\(2.34 \times 10^{-3}\text{ mol}\)). Mark 3: For calculating the mass of iron (\(0.131\text{ g}\)). Mark 4: For calculating the final percentage by mass to three significant figures (\(37.3\%\)).

1. Calculate initial moles: \(n(HA) = 0.0250\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.75 \times 10^{-3}\text{ mol}\); \(n(\text{NaOH}) = 0.0150\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}\). 2. Find equilibrium moles: Since \(HA\) and \(\text{NaOH}\) react in a 1:1 ratio, moles of \(HA\) remaining = \(3.75 \times 10^{-3} - 3.00 \times 10^{-3} = 7.50 \times 10^{-4}\text{ mol}\); moles of conjugate base \(A^-\)\ produced = \(3.00 \times 10^{-3}\text{ mol}\). 3. Use the buffer equation to find hydrogen ion concentration: \([\text{H}^+] = K_a \times \frac{[HA]}{[A^-]} = K_a \times \frac{n(HA)}{n(A^-)} = 1.80 \times 10^{-5} \times \frac{7.50 \times 10^{-4}}{3.00 \times 10^{-3}} = 4.50 \times 10^{-6}\text{ mol dm}^{-3}\). 4. Calculate pH: \(\text{pH} = -\log_{10}(4.50 \times 10^{-6}) = 5.3468 \approx 5.35\).

Mark 1: For calculating the initial moles of both \(HA\) and \(\text{NaOH}\). Mark 2: For calculating the equilibrium moles of \(HA\) (\(7.50 \times 10^{-4}\text{ mol}\)) and \(A^-\)\ (\(3.00 \times 10^{-3}\text{ mol}\)). Mark 3: For calculating the concentration of \([\text{H}^+]\) (\(4.50 \times 10^{-6}\text{ mol dm}^{-3}\)). Mark 4: For calculating the final pH value to two decimal places (\(5.35\)).

1. Express the Arrhenius equation in logarithmic form for two temperatures: \(\ln(k_2 / k_1) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\). 2. Substitute values: \(\ln(6.00 \times 10^{-3} / 1.50 \times 10^{-3}) = \ln(4.00) = 1.3863\). 3. Calculate reciprocal temperature difference: \(\frac{1}{313\text{ K}} - \frac{1}{293\text{ K}} = 0.0031949 - 0.0034130 = -2.181 \times 10^{-4}\text{ K}^{-1}\). 4. Rearrange and solve for \(E_a\): \(E_a = -\frac{1.3863 \times 8.31}{-2.181 \times 10^{-4}} = 52820\text{ J mol}^{-1} = 52.8\text{ kJ mol}^{-1}\).

Mark 1: For evaluating the left-hand side \(\ln(k_2/k_1) = 1.39\) or \(1.386\). Mark 2: For evaluating the change in reciprocal temperature as \(-2.18 \times 10^{-4}\text{ K}^{-1}\). Mark 3: For rearranging the expression to calculate \(E_a\) in J (\(52800\text{ J mol}^{-1}\)). Mark 4: For converting to \(\text{kJ mol}^{-1}\) and giving the correct answer to three significant figures with units (\(52.8\text{ kJ mol}^{-1}\)).

1. Determine the concentration of copper(II) ions in the brass solution: Since \(A \propto C\), \(C_{\text{brass}} = 0.0500\text{ mol dm}^{-3} \times \frac{0.240}{0.400} = 0.0300\text{ mol dm}^{-3}\). 2. Calculate the moles of copper in the volumetric flask: \(n(\text{Cu}) = 0.0300\text{ mol dm}^{-3} \times 0.250\text{ dm}^3 = 7.50 \times 10^{-3}\text{ mol}\). 3. Calculate the mass of copper in the brass: \(m(\text{Cu}) = 7.50 \times 10^{-3}\text{ mol} \times 63.5\text{ g mol}^{-1} = 0.47625\text{ g}\). 4. Calculate the percentage by mass: \(\%\text{ by mass} = \frac{0.47625\text{ g}}{0.950\text{ g}} \times 100 = 50.13\% \approx 50.1\%\).

Mark 1: For determining the concentration of copper in the brass solution (\(0.0300\text{ mol dm}^{-3}\)). Mark 2: For calculating the moles of copper in the flask (\(7.50 \times 10^{-3}\text{ mol}\)). Mark 3: For calculating the mass of copper in the sample (\(0.476\text{ g}\)). Mark 4: For calculating the final percentage by mass of copper to three significant figures (\(50.1\%\)).

1. Calculate mass of cyclohexanol used: \(m = 12.0\text{ cm}^3 \times 0.962\text{ g cm}^{-3} = 11.544\text{ g}\). 2. Calculate moles of cyclohexanol used: \(n = \frac{11.544\text{ g}}{100.2\text{ g mol}^{-1}} = 0.11521\text{ mol}\). 3. Determine theoretical yield of cyclohexene: Moles of cyclohexene = \(0.11521\text{ mol}\). Theoretical mass = \(0.11521\text{ mol} \times 82.1\text{ g mol}^{-1} = 9.4587\text{ g}\). 4. Calculate percentage yield: \(\%\text{ Yield} = \frac{5.25\text{ g}}{9.4587\text{ g}} \times 100 = 55.504\% \approx 55.5\%\).

Mark 1: For calculating the mass of cyclohexanol (\(11.5\text{ g}\)). Mark 2: For calculating the moles of cyclohexanol (\(0.115\text{ mol}\)). Mark 3: For calculating the theoretical mass of cyclohexene (\(9.46\text{ g}\)). Mark 4: For calculating the percentage yield to three significant figures (\(55.5\%\)).

1. Calculate temperature change: \(\Delta T = 36.8 - 20.2 = 16.6\text{ }^\circ\text{C}\). 2. Calculate heat energy transferred: \(q = m c \Delta T = 55.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 16.6\text{ K} = 3816.34\text{ J} = 3.81634\text{ kJ}\). 3. Calculate moles of calcium chloride dissolved: \(n = \frac{5.00\text{ g}}{111.0\text{ g mol}^{-1}} = 0.045045\text{ mol}\). 4. Calculate molar enthalpy of solution (accounting for temperature rise indicating an exothermic reaction): \(\Delta H_{\text{soln}} = -\frac{q}{n} = -\frac{3.81634\text{ kJ}}{0.045045\text{ mol}} = -84.723\text{ kJ mol}^{-1} \approx -84.7\text{ kJ mol}^{-1}\).

Mark 1: For calculating temperature change (\(16.6\text{ K}\)) and using the correct mass in heat calculation (\(55.0\text{ g}\)). Mark 2: For calculating heat energy released (\(3.82\text{ kJ}\)). Mark 3: For calculating the moles of calcium chloride dissolved (\(0.0450\text{ mol}\)). Mark 4: For the final enthalpy change with a negative sign and correct units to three significant figures (\(-84.7\text{ kJ mol}^{-1}\)).

1. Moles of remaining ethanoic acid at equilibrium: \(n(\text{CH}_3\text{COOH})_{\text{eq}} = 0.0450\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0450\text{ mol}\). 2. Calculate the moles of acid that reacted: \(0.200 - 0.0450 = 0.155\text{ mol}\). 3. Determine equilibrium moles of all components based on 1:1 stoichiometry: \(n(\text{CH}_3\text{COOH})_{\text{eq}} = 0.0450\text{ mol}\); \(n(\text{C}_2\text{H}_5\text{OH})_{\text{eq}} = 0.200 - 0.155 = 0.0450\text{ mol}\); \(n(\text{CH}_3\text{COOC}_2\text{H}_5)_{\text{eq}} = 0.155\text{ mol}\); \(n(\text{H}_2\text{O})_{\text{eq}} = 0.155\text{ mol}\). 4. Write expression and calculate \(K_c\): \(K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} = \frac{0.155 \times 0.155}{0.0450 \times 0.0450} = 11.864 \approx 11.9\). (Note: volumes cancel out in \(K_c\) expression).

Mark 1: For calculating the equilibrium moles of ethanoic acid (\(0.0450\text{ mol}\)). Mark 2: For calculating the equilibrium moles of ester and water (\(0.155\text{ mol}\)). Mark 3: For establishing that the equilibrium moles of ethanol is also \(0.0450\text{ mol}\). Mark 4: For substituting into the \(K_c\) expression correctly to obtain \(11.9\) (no units).

1. Find concentration of the diluted sample: \(C_{\text{dilute}} = \frac{13500}{4500} = 3.00\text{ mg dm}^{-3}\). 2. Determine the dilution factor: \(\text{Dilution factor} = \frac{100.0\text{ cm}^3}{2.00\text{ cm}^3} = 50.0\). 3. Calculate original concentration in \(\text{mg dm}^{-3}\): \(C_{\text{original}} = 3.00\text{ mg dm}^{-3} \times 50.0 = 150\text{ mg dm}^{-3}\). 4. Convert units to \(\text{g dm}^{-3}\): \(150\text{ mg dm}^{-3} \times \frac{1\text{ g}}{1000\text{ mg}} = 0.150\text{ g dm}^{-3}\).

Mark 1: For calculating the concentration of the diluted sample as \(3.00\text{ mg dm}^{-3}\). Mark 2: For determining the dilution factor of 50. Mark 3: For calculating the original concentration in \(\text{mg dm}^{-3}\) (\(150\text{ mg dm}^{-3}\)). Mark 4: For converting to \(\text{g dm}^{-3}\) to give the final answer of \(0.150\text{ g dm}^{-3}\).

1. Moles of \(MnO_4^-\)
\(n(MnO_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{10.00}{1000}\text{ dm}^3 = 2.00 \times 10^{-4}\text{ mol}\)

2. Moles of \(Fe^{2+}\) in \(25.0\text{ cm}^3\)
The reaction ratio is \(5Fe^{2+} : 1MnO_4^-\)
\(n(Fe^{2+}) = 5 \times 2.00 \times 10^{-4}\text{ mol} = 1.00 \times 10^{-3}\text{ mol}\)

3. Moles of \(Fe^{2+}\) in the original \(250.0\text{ cm}^3\) sample
\(n(Fe^{2+})_{\text{total}} = 1.00 \times 10^{-3}\text{ mol} \times 10 = 0.0100\text{ mol}\)

4. Molar mass of the hydrated salt
\(M_r = \frac{3.92\text{ g}}{0.0100\text{ mol}} = 392\text{ g mol}^{-1}\)

5. Value of \(x\)
\(M_r((NH_4)_2Fe(SO_4)_2) = (2 \times 18.0) + 55.8 + (2 \times 96.1) = 284.0\text{ g mol}^{-1}\)
Mass of water of crystallisation = \(392 - 284.0 = 108\text{ g mol}^{-1}\)
\(x = \frac{108}{18.0} = 6\)

• Mark 1: Calculates moles of manganate(VII) ion used (\(2.00 \times 10^{-4}\text{ mol}\)) [1M]
• Mark 2: Correctly applies stoichiometry to find moles of \(Fe^{2+}\) in \(250.0\text{ cm}^3\) (\(0.0100\text{ mol}\)) [1M]
• Mark 3: Calculates molar mass of the hydrated salt as \(392\text{ g mol}^{-1}\) [1M]
• Mark 4: Deduces \(x = 6\) by subtracting anhydrous mass (\(284.0\text{ g mol}^{-1}\)) and dividing by 18.0 [1M]

1. Moles of \(NaOH\) used in titration:
\(n(NaOH) = 0.150\text{ mol dm}^{-3} \times \frac{25.20}{1000}\text{ dm}^3 = 3.78 \times 10^{-3}\text{ mol}\)
This reacts with excess \(HCl\) in the \(10.0\text{ cm}^3\) sample in a 1:1 ratio.

2. Excess moles of \(HCl\) in the total \(100.0\text{ cm}^3\) solution:
\(n(HCl)_{\text{excess}} = 3.78 \times 10^{-3}\text{ mol} \times 10 = 0.0378\text{ mol}\)

3. Initial moles of \(HCl\) added:
\(n(HCl)_{\text{initial}} = 1.00\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0500\text{ mol}\)

4. Moles of \(HCl\) that reacted with \(CaCO_3\):
\(n(HCl)_{\text{reacted}} = 0.0500 - 0.0378 = 0.0122\text{ mol}\)

5. Moles and mass of \(CaCO_3\):
Since \(CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O\):
\(n(CaCO_3) = \frac{0.0122}{2} = 0.00610\text{ mol}\)
\(M_r(CaCO_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1\text{ g mol}^{-1}\)
\(m(CaCO_3) = 0.00610 \times 100.1 = 0.61061\text{ g}\)

6. Percentage by mass:
\(\text{Percentage} = \frac{0.61061}{1.20} \times 100 = 50.88\% \approx 50.9\%\)

• Mark 1: Calculates excess \(HCl\) in \(100.0\text{ cm}^3\) (\(0.0378\text{ mol}\)) [1M]
• Mark 2: Calculates moles of \(HCl\) reacted with the carbonate (\(0.0122\text{ mol}\)) [1M]
• Mark 3: Applies 1:2 stoichiometry to find moles of \(CaCO_3\) (\(0.00610\text{ mol}\)) [1M]
• Mark 4: Calculates mass of \(CaCO_3\) and converts to percentage by mass (\(50.9\%\)) [1M] (Allow 50.8% - 51.0% depending on rounding)

1. Calculate temperature change (\(\Delta T\)):
\(\Delta T = 25.8 - 19.5 = 6.3\text{ K}\)

2. Calculate heat absorbed by the solution:
Total mass of solution = \(50.0 + 50.0 = 100.0\text{ g}\)
\(q_{\text{soln}} = m c \Delta T = 100.0 \times 4.18 \times 6.3 = 2633.4\text{ J}\)

3. Calculate heat absorbed by the calorimeter:
\(q_{\text{cal}} = C_{\text{cal}} \Delta T = 15.0 \times 6.3 = 94.5\text{ J}\)

4. Total heat released by the reaction:
\(q_{\text{total}} = 2633.4 + 94.5 = 2727.9\text{ J} = 2.7279\text{ kJ}\)

5. Moles of water formed:
\(n(H_2O) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)

6. Enthalpy change of neutralisation:
\(\Delta H_{\text{neut}} = -\frac{2.7279\text{ kJ}}{0.0500\text{ mol}} = -54.558\text{ kJ mol}^{-1} \approx -54.6\text{ kJ mol}^{-1}\)

• Mark 1: Calculates the heat energy absorbed by the solution (\(2633.4\text{ J}\) or \(2.63\text{ kJ}\)) [1M]
• Mark 2: Calculates the heat energy absorbed by the calorimeter (\(94.5\text{ J}\)) and sums to find total heat released (\(2727.9\text{ J}\) or \(2.73\text{ kJ}\)) [1M]
• Mark 3: Calculates moles of reaction / water formed (\(0.0500\text{ mol}\)) [1M]
• Mark 4: Final value of \(-54.6\text{ kJ mol}^{-1}\) (must include negative sign and 3 sig figs) [1M]

1. Moles of methyl benzoate starting material:
\(n(\text{starting}) = \frac{6.80\text{ g}}{136.0\text{ g mol}^{-1}} = 0.0500\text{ mol}\)

2. Theoretical yield of methyl 3-nitrobenzoate:
Theoretical mass = \(0.0500\text{ mol} \times 181.0\text{ g mol}^{-1} = 9.05\text{ g}\)

3. Actual mass of pure methyl 3-nitrobenzoate obtained:
\(\text{Actual pure mass} = 6.35\text{ g} \times 0.950 = 6.0325\text{ g}\)

4. Percentage yield:
\(\text{Percentage Yield} = \frac{6.0325\text{ g}}{9.05\text{ g}} \times 100 = 66.657\% \approx 66.7\%\)

• Mark 1: Calculates moles of methyl benzoate starting material (\(0.0500\text{ mol}\)) [1M]
• Mark 2: Calculates theoretical yield in grams (\(9.05\text{ g}\)) [1M]
• Mark 3: Calculates actual mass of pure product (\(6.0325\text{ g}\)) [1M]
• Mark 4: Evaluates percentage yield correctly as \(66.7\%\) (Accept range \(66.6\% - 66.7\%\)) [1M]

Stage 1: Sample preparation. Weigh a known mass of the brass sample accurately. Dissolve the brass in concentrated nitric acid (carrying this out in a fume cupboard due to toxic nitrogen dioxide emissions). Neutralise the excess nitric acid using sodium carbonate solution until a faint precipitate of copper(II) carbonate forms, then add a minimum volume of ethanoic acid to redissolve the precipitate. Transfer the mixture to a volumetric flask and make up to the calibration mark with deionised water, mixing thoroughly. Stage 2: Reaction and Titration. Pipette a known volume of this copper(II) solution into a conical flask. Add an excess of potassium iodide solution (which turns the solution brown due to iodine formation and produces an off-white precipitate of copper(I) iodide). Titrate the mixture against a standard solution of sodium thiosulfate. When the brown colour fades to a pale straw/yellow, add a few drops of starch indicator (turning the mixture blue-black). Continue titration dropwise until the blue-black color completely disappears, leaving an off-white/cream precipitate of copper(I) iodide. Stage 3: Chemistry and Calculations. The chemical reactions are: \(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^{-}(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_{2}(\text{aq})\) and \(\text{I}_{2}(\text{aq}) + 2\text{S}_{2}\text{O}_{3}^{2-}(\text{aq}) \rightarrow 2\text{I}^{-}(\text{aq}) + \text{S}_{4}\text{O}_{6}^{2-}(\text{aq})\). From the titration, the moles of sodium thiosulfate used can be calculated (concentration \(\times\) volume). The stoichiometry shows that 1 mole of thiosulfate reacts with 0.5 moles of iodine, which corresponds to 1 mole of copper(II) ions. Calculate the moles of copper in the pipette volume, scale up to find the total moles in the volumetric flask, and multiply by the molar mass of copper (63.5 \(\text{g mol}^{-1}\)) to find the total mass of copper. Percentage of copper is (mass of copper / mass of brass sample) \(\times 100\)%.

For 5-6 marks (Level 3): The candidate covers all three stages fully with logical structure and correct chemical equations. Clear and accurate description of the endpoints, reagent additions, and calculation sequences. For 3-4 marks (Level 2): The candidate covers two stages fully, or all three stages partially. Some omissions in equations or key practical steps (e.g. forgetting to neutralise acid or missing starch indicator timing). For 1-2 marks (Level 1): The candidate describes elements of at least one stage, such as the basic iodometric equation or the use of starch indicator, but lacks detailed procedure or calculation steps. For 0 marks: No rewardable content.

Stage 1: Dissolution and Hot Filtration. Dissolve the crude benzoic acid in the minimum volume of hot solvent (water). Using the minimum volume ensures the solution is saturated so that crystallization occurs upon cooling. Filter the hot solution through fluted filter paper in a pre-heated funnel. This step removes any insoluble impurities while preventing benzoic acid from crystallising out on the cold funnel or paper. Stage 2: Crystallisation and Cold Filtration. Allow the filtrate to cool slowly to room temperature, and then place it in an ice bath to maximise the yield of crystals. Slow cooling encourages larger, purer crystals to form, leaving soluble impurities dissolved in the cold solvent. Filter the crystals under reduced pressure using a Buchner funnel and flask. Wash the crystals with a small portion of ice-cold solvent to remove any remaining soluble impurities adhering to the surface. Stage 3: Drying and Purity Analysis. Dry the purified crystals in a desiccator or low-temperature oven. Determine the melting point of the dry product using a melting point apparatus. Compare the experimental melting point range with the known literature value (122 \(^{\circ}\text{C}\) for benzoic acid). A pure sample will have a sharp melting point range (within 1-2 degrees) that matches the literature value. If impurities are present, the melting point will be lower and the melting range will be wider.

For 5-6 marks (Level 3): All three stages are fully and logically described with clear justifications for each practical step (minimum volume, hot filtration, slow cooling, cold wash, melting point range analysis). For 3-4 marks (Level 2): Two stages are fully described, or all three are partially described. May miss specific details like the reason for using minimum volume of solvent or the cooling in an ice bath. For 1-2 marks (Level 1): Shows basic understanding of recrystallisation (e.g. dissolving in hot solvent and cooling) but lacks detail on purification steps or melting point determination. For 0 marks: No rewardable content.

Stage 1: Experimental Design. Prepare a series of mixtures where the concentration of iodide ions, \([\text{I}^{-}]\), is varied while keeping the concentration of peroxodisulfate(VIII) ions, \([\text{S}_2\text{O}_8^{2-}]\), constant. Vary the volume of potassium iodide solution used and add appropriate volumes of water to ensure the total volume (and thus the concentration of other components) remains constant. Control the temperature of all solutions using a water bath. Stage 2: The Clock Chemistry. To each mixture, add a constant, small volume of sodium thiosulfate solution (\(\text{Na}_2\text{S}_2\text{O}_3\)) and a few drops of starch indicator. The thiosulfate ions react immediately with any iodine produced: \(\text{I}_{2}(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^{-}(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})\). Once all of the thiosulfate ions are consumed, any further iodine produced reacts with the starch, causing a sudden and distinct colour change to blue-black. Stage 3: Measurements and Analysis. Start a stopwatch as soon as the reactants are mixed, and stop it when the blue-black colour appears. Record this time, \(t\). Because the amount of thiosulfate is constant and small, the initial rate of reaction is proportional to \(1/t\). Plot a graph of \(\log(1/t)\) against \(\log([\text{I}^{-}])\). The gradient of the resulting straight-line graph represents the order of reaction with respect to iodide ions (gradient = \(\text{order}\)). Alternatively, compare pairs of runs directly to deduce the relationship between the relative concentration and the rate (e.g. if concentration doubles and \(1/t\) doubles, the order is first-order).

For 5-6 marks (Level 3): The candidate covers all three stages fully with excellent chemical reasoning. Explains the purpose of varying volume and adding water, details the clock reaction equations with starch indicator, and outlines the correct graphical analysis (rate proportional to \(1/t\) and log-log analysis). For 3-4 marks (Level 2): The candidate addresses two stages fully or all three stages with minor omissions. May miss the log-log graph details or have incomplete equations for the thiosulfate reaction. For 1-2 marks (Level 1): The candidate identifies that rate is related to \(1/t\) or mentions the use of thiosulfate and starch, but lacks cohesive experimental design or analysis steps. For 0 marks: No rewardable content.