AQA A-Level · Thinka-original Practice Paper

2024 AQA A-Level Chemistry 7405 Practice Paper with Answers

Thinka Jun 2024 AQA A Level-Style Mock — Chemistry 7405

300 marks360 mins2024
AnalysisSample paper
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 AQA A Level Chemistry 7405 paper. Not affiliated with or reproduced from AQA.

Paper 1 (Inorganic and Physical Chemistry)

Answer all questions in the spaces provided.
9 Question · 105.03 marks
Question 1 · structured
11.67 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.250\text{ mol dm}^{-3}\) propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) with \(37.5\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) sodium hydroxide (\(\text{NaOH}\)). The \(K_a\) of propanoic acid is \(1.35 \times 10^{-5}\text{ mol dm}^{-3}\) at \(298\text{ K}\).

(a) Write an equation for the reaction of propanoic acid with sodium hydroxide. [1 mark]

(b) Calculate the pH of this buffer solution at \(298\text{ K}\). Give your answer to 2 decimal places. [5 marks]

(c) A small amount of hydrochloric acid is added to this buffer solution. Explain, with the help of an ionic equation, how the buffer solution resists a change in pH. [3 marks]

(d) Calculate the pH of a \(0.0450\text{ mol dm}^{-3}\) solution of barium hydroxide, \(\text{Ba(OH)}_2\), at \(298\text{ K}\) where \(K_w = 1.00 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\). Give your answer to 2 decimal places. [3 marks]
Show answer & marking scheme

Worked solution

(a) \(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\) (or ionic: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\))

(b)
- Initial moles of \(\text{CH}_3\text{CH}_2\text{COOH}\):
\(n(\text{HA}) = \frac{50.0}{1000} \times 0.250 = 0.0125\text{ mol}\)
- Initial moles of \(\text{NaOH}\):
\(n(\text{OH}^-) = \frac{37.5}{1000} \times 0.150 = 0.005625\text{ mol}\)
- After reaction:
- Moles of propanoate ion formed, \(n(\text{A}^-) = 0.005625\text{ mol}\)
- Moles of propanoic acid remaining, \(n(\text{HA}) = 0.0125 - 0.005625 = 0.006875\text{ mol}\)
- Rearranging \(K_a\) to find \([\text{H}^+]\):
\([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{0.006875}{0.005625} = 1.65 \times 10^{-5}\text{ mol dm}^{-3}\)
- Calculating pH:
\(\text{pH} = -\log_{10}(1.65 \times 10^{-5}) = 4.78\)

(c)
- The added \(\text{H}^+\) ions react with the conjugate base (propanoate ions) in the buffer:
\(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
- The ratio of \(\frac{[\text{HA}]}{[\text{A}^-]}\) changes only slightly, so the pH remains relatively constant.

(d)
- Barium hydroxide is a strong base: \(\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-\)
- \([\text{OH}^-] = 2 \times 0.0450 = 0.0900\text{ mol dm}^{-3}\)
- \([\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.0900} = 1.111 \times 10^{-13}\text{ mol dm}^{-3}\)
- \(\text{pH} = -\log_{10}(1.111 \times 10^{-13}) = 12.95\)

Marking scheme

(a) [1 mark] Correct equation.
(b) [5 marks total]:
- [1 mark] Correct initial moles of propanoic acid and sodium hydroxide.
- [1 mark] Correct moles of propanoate ions formed.
- [1 mark] Correct moles of propanoic acid remaining.
- [1 mark] Correct rearrangement of Ka and calculation of [H+].
- [1 mark] Correct pH to 2 decimal places (4.78).
(c) [3 marks total]:
- [1 mark] Correct ionic equation.
- [1 mark] Statement that added H+ reacts with propanoate ions.
- [1 mark] Explanation that the concentration ratio of HA to A- remains almost unchanged.
(d) [3 marks total]:
- [1 mark] Identifies [OH-] as 0.0900.
- [1 mark] Correct expression or calculation for [H+].
- [1 mark] Correct pH to 2 decimal places (12.95).
Question 2 · structured
11.67 marks
This question is about cobalt and its complexes.

(a) When excess concentrated hydrochloric acid is added to an aqueous solution containing pink \([\text{Co(H}_2\text{O)}_6]^{2+}\) ions, a deep blue solution is formed.

(i) Write an ionic equation for this reaction. [1 mark]

(ii) State the shape of the blue complex ion and explain why the coordination number changes from 6 to 4 in this reaction. [3 marks]

(b) Aqueous cobalt(II) ions react with excess ammonia solution to form a hexaminecobalt(II) complex, which is rapidly oxidized by oxygen in the air to a cobalt(III) complex.

(i) Write the formula of the hexaminecobalt(II) complex and state its shape. [2 marks]

(ii) Explain why transition metal complexes, such as those of cobalt, are colored. [4 marks]

(c) Draw the 3D structures of the two optical isomers of the octahedral complex ion \([\text{Co(en)}_3]^{3+}\), where 'en' represents the bidentate ligand ethane-1,2-diamine. [2 marks]
Show answer & marking scheme

Worked solution

(a)(i) \([\text{Co(H}_2\text{O)}_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)
(a)(ii) Shape: Tetrahedral. Reason: Chloride ligands are larger than water ligands, so fewer chloride ligands can fit around the central cobalt ion due to steric hindrance / electrostatic repulsion.

(b)(i) Formula: \([\text{Co(NH}_3)_6]^{2+}\). Shape: Octahedral.
(b)(ii)
- Ligands split the d-orbitals into two different energy levels.
- Electrons absorb specific wavelengths of visible light to transition from lower to higher energy d-orbitals (d-d transition).
- The remaining unabsorbed wavelengths of light are transmitted or reflected, which forms the complementary color observed.

(c)
- Correct drawings of octahedral Co center showing 3 bidentate 'en' loops.
- The second drawing must be the exact, non-superimposable mirror image of the first.

Marking scheme

(a)(i) [1 mark] Correct ionic equation.
(a)(ii) [3 marks total]:
- [1 mark] Shape: Tetrahedral.
- [1 mark] Chloride ions are larger / have a larger ionic radius than water.
- [1 mark] Only 4 Cl- can fit around the cobalt ion due to steric hindrance / ligand repulsion.
(b)(i) [2 marks total]:
- [1 mark] Formula [Co(NH3)6]2+ (charge must be correct).
- [1 mark] Shape: Octahedral.
(b)(ii) [4 marks total]:
- [1 mark] d-orbitals split in energy by ligands.
- [1 mark] Light/photons absorbed to promote an electron from lower to higher energy level.
- [1 mark] Energy gap corresponds to a frequency in the visible spectrum (E = hv).
- [1 mark] Color seen is the complementary color transmitted/reflected.
(c) [2 marks total]:
- [1 mark] Two octahedral structures drawn.
- [1 mark] Correct mirror-image representation with bidentate rings clearly shown.
Question 3 · structured
11.67 marks
This question is about calcium chloride, \(\text{CaCl}_2\).

(a) Define the term *enthalpy of lattice formation*. [2 marks]

(b) Use the following data to calculate the lattice enthalpy of formation of solid calcium chloride, \(\text{CaCl}_2(\text{s})\). [5 marks]

- Enthalpy of atomisation of calcium = \(+178\text{ kJ mol}^{-1}\)
- First ionisation energy of calcium = \(+590\text{ kJ mol}^{-1}\)
- Second ionisation energy of calcium = \(+1145\text{ kJ mol}^{-1}\)
- Bond dissociation enthalpy of chlorine, \(\text{Cl}_2\) = \(+242\text{ kJ mol}^{-1}\)
- Electron affinity of chlorine = \(-348\text{ kJ mol}^{-1}\)
- Standard enthalpy of formation of \(\text{CaCl}_2(\text{s})\) = \(-796\text{ kJ mol}^{-1}\)

(c) Calcium carbonate decomposes at high temperatures:
\(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)
Given that \(\Delta H^\ominus = +178\text{ kJ mol}^{-1}\) and \(\Delta S^\ominus = +160\text{ J K}^{-1}\text{ mol}^{-1}\), calculate the minimum temperature, in Kelvin, at which this reaction becomes feasible. Assume that enthalpy and entropy changes do not vary with temperature. [3 marks]

(d) Explain why the experimental lattice enthalpy of calcium iodide, \(\text{CaI}_2\), is different from the theoretical value calculated using a purely ionic model. [2 marks]
Show answer & marking scheme

Worked solution

(a) Enthalpy change when 1 mole of solid ionic compound is formed from its constituent gaseous ions under standard conditions.

(b)
- By Hess's Law / Born-Haber cycle:
\(\Delta H_f^\ominus(\text{CaCl}_2) = \Delta H_{at}^\ominus(\text{Ca}) + 1st\text{ IE}(\text{Ca}) + 2nd\text{ IE}(\text{Ca}) + \Delta H_{diss}^\ominus(\text{Cl}_2) + 2 \times EA(\text{Cl}) + \Delta H_{L,f}^\ominus\)
- Substitute the values:
\(-796 = +178 + 590 + 1145 + 242 + 2(-348) + \Delta H_{L,f}^\ominus\)
\(-796 = 2155 - 696 + \Delta H_{L,f}^\ominus\)
\(-796 = 1459 + \Delta H_{L,f}^\ominus\)
- Rearrange:
\(\Delta H_{L,f}^\ominus = -796 - 1459 = -2255\text{ kJ mol}^{-1}\)

(c)
- Reaction is feasible when \(\Delta G \le 0\).
- \(\Delta G = \Delta H - T\Delta S\)
- Set \(\Delta G = 0\):
\(T = \frac{\Delta H}{\Delta S} = \frac{178 \times 10^3\text{ J mol}^{-1}}{160\text{ J K}^{-1}\text{ mol}^{-1}} = 1112.5\text{ K}\)
- Rounding to 3 sig figs gives \(1113\text{ K}\) (accept \(1112.5\text{ K}\)).

(d)
- Theoretical calculations assume a purely ionic model with perfectly spherical ions.
- The calcium ion (\(\text{Ca}^{2+}\)) polarizes the large, highly polarizable iodide ion (\(\text{I}^-\)), giving the bonding some covalent character. This extra covalent character makes the experimental lattice enthalpy more exothermic.

Marking scheme

(a) [2 marks total]:
- [1 mark] Enthalpy change for formation of 1 mole of solid ionic lattice/compound.
- [1 mark] From its constituent gaseous ions.
(b) [5 marks total]:
- [1 mark] Account for doubling electron affinity: \(2 \times (-348) = -696\).
- [1 mark] Account for dissociation of \(\text{Cl}_2\) (enthalpy of atomisation of chlorine is half of dissociation, but since we have 2 Cl, we use the full bond enthalpy \(+242\)).
- [1 mark] Correctly link standard enthalpy of formation with other cycle components.
- [1 mark] Correct rearrangement of equation.
- [1 mark] Correct calculation of \(-2255\text{ kJ mol}^{-1}\).
(c) [3 marks total]:
- [1 mark] Conversion of \(\Delta S\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\) (0.160) or \(\Delta H\) to J.
- [1 mark] Setting up equation \(T = \frac{\Delta H}{\Delta S}\).
- [1 mark] \(1113\text{ K}\) or \(1112.5\text{ K}\).
(d) [2 marks total]:
- [1 mark] Identification of polarization of iodide ion by calcium ion.
- [1 mark] Leads to covalent character in bonding.
Question 4 · structured
11.67 marks
The reaction between peroxodisulfate(VIII) ions (\(\text{S}_2\text{O}_8^{2-}\)) and iodide ions (\(\text{I}^-\)) is catalyzed by iron(II) ions.

(a) Explain why this reaction is very slow in the absence of a catalyst. Use two equations to show how \(\text{Fe}^{2+}\) ions catalyze this reaction. [4 marks]

(b) In a series of kinetic experiments for a different reaction, \(P + Q \rightarrow R\), the following results were obtained at a constant temperature:

- Experiment 1: \([P] = 0.150\text{ mol dm}^{-3}\), \([Q] = 0.200\text{ mol dm}^{-3}\), Rate = \(1.80 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
- Experiment 2: \([P] = 0.300\text{ mol dm}^{-3}\), \([Q] = 0.200\text{ mol dm}^{-3}\), Rate = \(7.20 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
- Experiment 3: \([P] = 0.150\text{ mol dm}^{-3}\), \([Q] = 0.400\text{ mol dm}^{-3}\), Rate = \(3.60 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)

(i) Deduce the order of reaction with respect to \(P\) and \(Q\). Explain your reasoning. [2 marks]

(ii) Write the rate equation and calculate the rate constant, \(k\), including units. [3 marks]

(c) The Arrhenius equation can be written as \(\ln k = -\frac{E_a}{RT} + \ln A\).
A plot of \(\ln k\) against \(\frac{1}{T}\) for this reaction has a gradient of \(-7220\text{ K}\). Calculate the activation energy, \(E_a\), in \(\text{kJ mol}^{-1}\). (\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)). [3 marks]
Show answer & marking scheme

Worked solution

(a)
- Peroxodisulfate and iodide are both negatively charged ions. They repel each other, resulting in a very high activation energy.
- Fe2+ ions reduce S2O8^2- ions:
\(\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}\)
- Fe3+ ions then oxidize I^- ions, regenerating the Fe2+ catalyst:
\(2\text{I}^- + 2\text{Fe}^{3+} \rightarrow \text{I}_2 + 2\text{Fe}^{2+}\)

(b)(i)
- Comparing Experiment 1 and 2: [Q] is constant, [P] doubles, and the rate increases by a factor of 4 (\(7.20 \times 10^{-3} / 1.80 \times 10^{-3}\)). Therefore, the reaction is second order with respect to P.
- Comparing Experiment 1 and 3: [P] is constant, [Q] doubles, and the rate increases by a factor of 2 (\(3.60 \times 10^{-3} / 1.80 \times 10^{-3}\)). Therefore, the reaction is first order with respect to Q.

(b)(ii)
- Rate equation: \(\text{Rate} = k[P]^2[Q]\)
- Rearranging for \(k\):
\(k = \frac{\text{Rate}}{[P]^2[Q]} = \frac{1.80 \times 10^{-3}}{(0.150)^2 \times (0.200)} = \frac{1.80 \times 10^{-3}}{0.0045} = 0.400\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)

(c)
- \(\text{Gradient} = -\frac{E_a}{R} = -7220\text{ K}\)
- \(E_a = 7220 \times 8.31 = 59998.2\text{ J mol}^{-1}\)
- Converting to kJ mol-1: \(E_a = 60.0\text{ kJ mol}^{-1}\)

Marking scheme

(a) [4 marks total]:
- [1 mark] State that both reactant ions have a negative charge, so they repel.
- [1 mark] Equation 1: \(\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}\)
- [1 mark] Equation 2: \(2\text{I}^- + 2\text{Fe}^{3+} \rightarrow \text{I}_2 + 2\text{Fe}^{2+}\)
- [1 mark] Mentioning why catalyzed pathway is faster (replaces mutually repelling steps with opposite charge steps).
(b)(i) [2 marks total]:
- [1 mark] Deduce second order for P with explanation.
- [1 mark] Deduce first order for Q with explanation.
(b)(ii) [3 marks total]:
- [1 mark] Correct rate equation matching order.
- [1 mark] Correct calculation of 0.400.
- [1 mark] Correct units: dm6 mol-2 s-1.
(c) [3 marks total]:
- [1 mark] Identifies \(\text{gradient} = -\frac{E_a}{R}\).
- [1 mark] Calculation of Ea in J mol-1 (60000 J mol-1).
- [1 mark] Correct conversion and rounding to 3 sig figs: 60.0 kJ mol-1.
Question 5 · structured
11.67 marks
This question concerns the chemistry of the Group 2 alkaline earth metals and their compounds.

(a) State the trend in solubility of the Group 2 hydroxides down the group from \(\text{Mg(OH)}_2\) to \(\text{Ba(OH)}_2\). Describe and explain how a student could use magnesium hydroxide in medicine and how its solubility relates to this safety. [3 marks]

(b) State the trend in solubility of the Group 2 sulfates down the group. Describe how a student could carry out a chemical test on an aqueous solution to confirm the presence of sulfate ions, including any reagents used, the observations, and the simplest ionic equation for the reaction that occurs. [4 marks]

(c) Barium sulfate is used in medicine as a 'barium meal'. Explain why this use is safe despite barium ions being highly toxic. [2 marks]

(d) Write an equation for the extraction of titanium from titanium(IV) chloride using magnesium. State the role of magnesium in this reaction in terms of electrons. [2 marks]
Show answer & marking scheme

Worked solution

(a)
- Solubility of Group 2 hydroxides increases down the group.
- Magnesium hydroxide is used as an antacid (indigestion remedy) to neutralize excess stomach acid.
- It is safe because it is only sparingly soluble, meaning the concentration of corrosive hydroxide ions in the body remains very low.

(b)
- Solubility of Group 2 sulfates decreases down the group.
- Test: Add dilute hydrochloric acid (to remove any carbonate ions that would give a false positive precipitate), then add barium chloride solution.
- Observation: A white precipitate forms.
- Ionic equation: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)

(c)
- Barium sulfate is highly insoluble in water.
- Therefore, it is not absorbed into the bloodstream/tissues and simply passes through the digestive tract safely.

(d)
- Equation: \(\text{TiCl}_4 + 2\text{Mg} \rightarrow \text{Ti} + 2\text{MgCl}_2\)
- Role of magnesium: Reducing agent (magnesium atoms lose electrons to titanium ions).

Marking scheme

(a) [3 marks total]:
- [1 mark] Solubility of hydroxides increases down the group.
- [1 mark] Used as an antacid / to treat indigestion / neutralizes stomach acid.
- [1 mark] Sparingly soluble, so concentration of OH- is low and not harmful.
(b) [4 marks total]:
- [1 mark] Solubility of sulfates decreases down the group.
- [1 mark] Add HCl followed by BaCl2 solution.
- [1 mark] White precipitate forms.
- [1 mark] \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\) (including state symbols).
(c) [2 marks total]:
- [1 mark] State that barium sulfate is insoluble.
- [1 mark] So it is not absorbed by the body / cannot enter bloodstream.
(d) [2 marks total]:
- [1 mark] Correct equation: \(\text{TiCl}_4 + 2\text{Mg} \rightarrow \text{Ti} + 2\text{MgCl}_2\).
- [1 mark] Reducing agent / donor of electrons.
Question 6 · structured
11.67 marks
An electrochemical cell is set up under standard conditions using the following two half-cells:

- Half-cell A: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\quad E^\ominus = +0.77\text{ V}\)
- Half-cell B: \(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{e}^- \rightleftharpoons 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\quad E^\ominus = +1.33\text{ V}\)

(a) Write the conventional representation for this electrochemical cell, including state symbols and the platinum electrodes. [3 marks]

(b) Calculate the standard electromotive force (EMF) of this cell. [1 mark]

(c) Write an overall ionic equation for the reaction that occurs when this cell is allowed to discharge. [2 marks]

(d) State the function of a salt bridge in an electrochemical cell. Name a suitable chemical that could be used in the salt bridge and explain why it is chosen. [3 marks]

(e) Standard electrode potentials are measured relative to the Standard Hydrogen Electrode (SHE). Describe the conditions required for the SHE and write the half-equation for the reaction occurring at its surface. [3 marks]
Show answer & marking scheme

Worked solution

(a)
Since Half-cell B has the more positive potential, it is the cathode (reduction occurs on the right side) and Half-cell A is the anode (oxidation occurs on the left side).
Conventional cell representation:
\(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) \parallel \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}), 2\text{Cr}^{3+}(\text{aq}) \mid \text{Pt}(\text{s})\)
(Allow commas or semicolons between species in the same phase: \(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) \parallel \text{Cr}_2\text{O}_7^{2-}(\text{aq}), \text{H}^+(\text{aq}), \text{Cr}^{3+}(\text{aq}) \mid \text{Pt}(\text{s})\))

(b)
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 1.33\text{ V} - 0.77\text{ V} = +0.56\text{ V}\)

(c)
Multiply Half-cell A by 6 to balance electrons and reverse it (oxidation):
\(6\text{Fe}^{2+}(\text{aq}) \rightarrow 6\text{Fe}^{3+}(\text{aq}) + 6\text{e}^-\)
Combine with Half-cell B:
\(6\text{Fe}^{2+}(\text{aq}) + \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) \rightarrow 6\text{Fe}^{3+}(\text{aq}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\)

(d)
- Function: Allows the transfer/flow of ions between half-cells to complete the electrical circuit and maintain charge neutrality.
- Substance: Potassium nitrate solution (\(\text{KNO}_3\)) or potassium chloride solution (\(\text{KCl}\)).
- Explanation: These ions are inert and will not react or form precipitates with any of the ions in either half-cell.

(e)
- Conditions: Temperature of \(298\text{ K}\) (\(25^\circ\text{C}\)), gas pressure of \(100\text{ kPa}\) (\(1\text{ bar}\)), and solution concentration of \(1.00\text{ mol dm}^{-3}\) with respect to \(\text{H}^+\).
- Half-equation: \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g})\)

Marking scheme

(a) [3 marks total]:
- [1 mark] Pt(s) shown on both extreme left and right.
- [1 mark] Phase boundaries correct (vertical solid lines) and salt bridge represented by double vertical lines.
- [1 mark] Left side (Fe2+, Fe3+) and right side (Cr2O7^2-, H+, Cr3+) in correct positions with state symbols.
(b) [1 mark] +0.56 V (must include positive sign).
(c) [2 marks total]:
- [1 mark] Left side reactants correct.
- [1 mark] Balanced equation with no electrons shown.
(d) [3 marks total]:
- [1 mark] Purpose: complete circuit / maintain ion/charge balance.
- [1 mark] Potassium nitrate (or alternative inert electrolyte).
- [1 mark] Reason: does not form precipitates / react with the species in either half-cell.
(e) [3 marks total]:
- [1 mark] Temperature (298 K) and Pressure (100 kPa / 1 bar).
- [1 mark] Concentration [H+] = 1.00 mol dm-3.
- [1 mark] Half-equation: \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g})\).
Question 7 · structured
11.67 marks
Sulfur trioxide decomposes at high temperatures according to the following equation:
\(2\text{SO}_3(\text{g}) \rightleftharpoons 2\text{SO}_2(\text{g}) + \text{O}_2(\text{g})\quad \Delta H = +196\text{ kJ mol}^{-1}\)

A sample of \(4.00\text{ mol}\) of sulfur trioxide, \(\text{SO}_3\), was heated in a sealed vessel. At equilibrium, \(1.20\text{ mol}\) of oxygen, \(\text{O}_2\), had been formed and the total pressure in the vessel was \(210\text{ kPa}\).

(a) Calculate the amounts, in moles, of \(\text{SO}_3\) and \(\text{SO}_2\) in the equilibrium mixture. [2 marks]

(b) Calculate the partial pressure of each gas in the equilibrium mixture at this temperature. [2 marks]

(c) Write an expression for the equilibrium constant, \(K_p\), for this reaction, including its units. [2 marks]

(d) Calculate the value of \(K_p\) under these conditions. [3 marks]

(e) State and explain the effect on the value of \(K_p\) if the temperature is increased. [3 marks]
Show answer & marking scheme

Worked solution

(a)
- At start: \(n(\text{SO}_3) = 4.00\text{ mol}\), \(n(\text{SO}_2) = 0\), \(n(\text{O}_2) = 0\)
- Since \(1.20\text{ mol}\) of \(\text{O}_2\) is present at equilibrium, the change in moles of \(\text{SO}_2\) is \(+2 \times 1.20 = +2.40\text{ mol}\)
- The change in moles of \(\text{SO}_3\) is \(-2 \times 1.20 = -2.40\text{ mol}\)
- Equilibrium moles:
- \(n(\text{SO}_3) = 4.00 - 2.40 = 1.60\text{ mol}\)
- \(n(\text{SO}_2) = 2.40\text{ mol}\)
- \(n(\text{O}_2) = 1.20\text{ mol}\)

(b)
- Total equilibrium moles = \(1.60 + 2.40 + 1.20 = 5.20\text{ mol}\)
- Partial pressures:
- \(p(\text{SO}_3) = \frac{1.60}{5.20} \times 210\text{ kPa} = 64.62\text{ kPa}\)
- \(p(\text{SO}_2) = \frac{2.40}{5.20} \times 210\text{ kPa} = 96.92\text{ kPa}\)
- \(p(\text{O}_2) = \frac{1.20}{5.20} \times 210\text{ kPa} = 48.46\text{ kPa}\)

(c)
- \(K_p = \frac{p(\text{SO}_2)^2 \times p(\text{O}_2)}{p(\text{SO}_3)^2}\)
- Units: \(\frac{\text{kPa}^2 \times \text{kPa}}{\text{kPa}^2} = \text{kPa}\)

(d)
- \(K_p = \frac{(96.92)^2 \times 48.46}{(64.62)^2}\)
- \(K_p = \frac{9393.5 \times 48.46}{4175.7} = 109\text{ kPa}\) (or \(1.09 \times 10^5\text{ Pa}\) if converted to standard units).

(e)
- Effect on \(K_p\): \(K_p\) increases.
- Explanation: The forward reaction is endothermic (\(\Delta H = +196\text{ kJ mol}^{-1}\)). According to Le Chatelier's principle, an increase in temperature shifts the equilibrium position to the right (the endothermic direction) to absorb heat, increasing the concentration/partial pressure of products relative to reactants.

Marking scheme

(a) [2 marks total]:
- [1 mark] \(n(\text{SO}_2) = 2.40\text{ mol}\).
- [1 mark] \(n(\text{SO}_3) = 1.60\text{ mol}\).
(b) [2 marks total]:
- [1 mark] Divides each mol by total mol of 5.20 to get mole fractions.
- [1 mark] Multiplies by 210 to obtain correct partial pressures (allow rounding errors: \(p(\text{SO}_3) \approx 64.6\), \(p(\text{SO}_2) \approx 96.9\), \(p(\text{O}_2) \approx 48.5\)).
(c) [2 marks total]:
- [1 mark] Correct expression \(K_p = \frac{p(\text{SO}_2)^2 \cdot p(\text{O}_2)}{p(\text{SO}_3)^2}\) (round brackets required, no square brackets allowed).
- [1 mark] Units: kPa (or Pa if specified in Pa).
(d) [3 marks total]:
- [1 mark] Insertion of correct values into expression.
- [1 mark] Correct calculations.
- [1 mark] Final value: 109 (or 110) kPa (or \(1.09 \times 10^5\text{ Pa}\)).
(e) [3 marks total]:
- [1 mark] Kp increases.
- [1 mark] Forward reaction is endothermic.
- [1 mark] Equilibrium shifts to the right to favor the endothermic reaction.
Question 8 · structured
11.67 marks
This question concerns atomic structure and time-of-flight (TOF) mass spectrometry.

(a) A sample of copper was analyzed in a TOF mass spectrometer.

(i) Describe how copper ions are produced using the electrospray ionization method. Write an equation to represent this process. [3 marks]

(ii) Explain why it is necessary to ionize the copper atoms before they can be analyzed in the mass spectrometer. [1 mark]

(iii) In the drift tube of the mass spectrometer, a \({}^{63}\text{Cu}^+\) ion travels down a \(1.40\text{ m}\) path in \(1.18 \times 10^{-5}\text{ s}\). Calculate the time taken, in seconds, for a \({}^{65}\text{Cu}^+\) ion to travel down the same path under the same acceleration conditions. Give your answer to 3 significant figures.
(Kinetic energy of an ion \(KE = \frac{1}{2}mv^2\)). [4 marks]

(b) Write the full electronic configuration of:
(i) A ground-state copper atom. [1 mark]
(ii) A copper(II) ion, \(\text{Cu}^{2+}\). [1 mark]

(c) Explain why the first ionization energy of sulfur is lower than that of phosphorus. [2 marks]
Show answer & marking scheme

Worked solution

(a)(i)
- The sample is dissolved in a volatile solvent.
- It is injected through a fine hypodermic needle connected to a high voltage supply.
- The particles gain a proton (\(\text{H}^+\)) from the solvent as they exit the needle:
\(\text{Cu} + \text{H}^+ \rightarrow [\text{CuH}]^+\)

(a)(ii)
- Neutral atoms cannot be accelerated by the electric field or detected by the ion detector.

(a)(iii)
- Since all ions are accelerated to have the same kinetic energy:
\(KE = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2\)
- Substitute \(v = \frac{d}{t}\):
\(m_1 \left(\frac{d}{t_1}\right)^2 = m_2 \left(\frac{d}{t_2}\right)^2\)
- Since \(d\) is constant:
\(\frac{m_1}{t_1^2} = \frac{m_2}{t_2^2} \Rightarrow t_2 = t_1 \sqrt{\frac{m_2}{m_1}}\)
- Substitute the values:
\(t_2 = 1.18 \times 10^{-5} \times
\sqrt{\frac{65.0}{63.0}} = 1.18 \times 10^{-5} \times 1.01575 = 1.20 \times 10^{-5}\text{ s}\)

(b)(i)
- \(1
s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (Note: transition metal anomaly where 3d is filled and 4s is half-filled).\s

(b)(ii)
- \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\) (Electrons are lost from 4s first).

(c)
- Phosphorus has a outer electronic configuration of \(3s^2 3p^3\) with three singly occupied 3p orbitals (half-filled subshell).
- Sulfur has a configuration of \(3s^2 3p^4\) with one doubly occupied 3p orbital.
- The repulsion between the paired electrons in the same 3p orbital of sulfur makes it easier to remove the electron.

Marking scheme

(a)(i) [3 marks total]:
- [1 mark] Dissolved in volatile solvent and passed through high-voltage capillary needle.
- [1 mark] Gain a proton to form MH+.
- [1 mark] Equation: \(\text{Cu} + \text{H}^+ \rightarrow [\text{CuH}]^+\).
(a)(ii) [1 mark] Only charged particles/ions can be accelerated by an electric field / detected.
(a)(iii) [4 marks total]:
- [1 mark] Equates kinetic energy: \(m_1 v_1^2 = m_2 v_2^2\) or \(t \propto \sqrt{m}\).
- [1 mark] Correct rearrangement: \(t_2 = t_1 \sqrt{\frac{m_2}{m_1}}\).
- [1 mark] Correct substitution of numbers.
- [1 mark] Answer: \(1.20 \times 10^{-5}\text{ s}\) (must be 3 sig figs).
(b)(i) [1 mark] \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (do not accept 3d9 4s2).
(b)(ii) [1 mark] \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\).
(c) [2 marks total]:
- [1 mark] In sulfur, the outer electron is in a paired 3p orbital.
- [1 mark] Mutual repulsion between these paired electrons makes it easier to remove.
Question 9 · structured
11.67 marks

This question is about the thermodynamics of the dissolution of magnesium chloride, \(MgCl_2\).



(a) Define the term enthalpy of hydration of an ion. [2 marks]



(b) Use the data in Table 1 below to calculate the enthalpy of solution, in \(\text{kJ mol}^{-1}\), of magnesium chloride, \(MgCl_2(s)\). [3 marks]



Table 1





Enthalpy term
Value / \(\text{kJ mol}^{-1}\)




Enthalpy of lattice formation of \(MgCl_2(s)\)
\(-2526\)


Enthalpy of hydration of \(Mg^{2+}(g)\)
\(-1920\)


Enthalpy of hydration of \(Cl^{-}(g)\)
\(-364\)




(c) The entropy change, \(\Delta S^\theta\), for the dissolution of \(MgCl_2(s)\) in water is \(+11.5\text{ J K}^{-1}\text{ mol}^{-1}\).

Calculate the Gibbs free energy change, \(\Delta G^\theta\), in \(\text{kJ mol}^{-1}\), for the dissolution of magnesium chloride at \(298\text{ K}\). State and explain whether the dissolution is feasible at this temperature. [4 marks]



(d) Explain why the enthalpy of hydration of the magnesium ion, \(Mg^{2+}\), is significantly more exothermic than that of the calcium ion, \(Ca^{2+}\). [3 marks]

Show answer & marking scheme

Worked solution

(a) The enthalpy of hydration is defined as the enthalpy change that occurs when 1 mole of gaseous ions is dissolved in water to form an infinitely dilute aqueous solution (or simply to form aqueous ions).



(b) The enthalpy of solution can be calculated using the thermodynamic cycle:

\(\Delta H_{\text{sol}} = \Delta H_{\text{lattice dissociation}} + \sum \Delta H_{\text{hyd}}\)

Since \(\Delta H_{\text{lattice formation}} = -2526\text{ kJ mol}^{-1}\), then \(\Delta H_{\text{lattice dissociation}} = +2526\text{ kJ mol}^{-1}\).

\(\Delta H_{\text{sol}} = +2526 + [\Delta H_{\text{hyd}}(Mg^{2+}) + 2 \times \Delta H_{\text{hyd}}(Cl^{-})]\)

\(\Delta H_{\text{sol}} = +2526 + [-1920 + 2 \times (-364)]\)

\(\Delta H_{\text{sol}} = +2526 + [-1920 - 728]\)

\(\Delta H_{\text{sol}} = +2526 - 2648 = -122\text{ kJ mol}^{-1}\)



(c) Convert entropy change to \(\text{kJ K}^{-1}\text{ mol}^{-1}\):

\(\Delta S^\theta = +11.5\text{ J K}^{-1}\text{ mol}^{-1} = +0.0115\text{ kJ K}^{-1}\text{ mol}^{-1}\)

Use the Gibbs free energy equation:

\(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)

\(\Delta G^\theta = -122 - (298 \times 0.0115)\)

\(\Delta G^\theta = -122 - 3.427 = -125.427\text{ kJ mol}^{-1}\) (or \(-125\text{ kJ mol}^{-1}\) to 3 significant figures).

Since \(\Delta G^\theta\) is negative (\(< 0\)), the dissolution process is feasible at \(298\text{ K}\).



(d) The magnesium ion (\(Mg^{2+}\)) has a smaller ionic radius than the calcium ion (\(Ca^{2+}\)).

Both ions have the same charge (\(2+\)), which means the magnesium ion has a higher charge density.

Therefore, there is a stronger electrostatic attraction between the magnesium ion and the lone pairs of electrons on the oxygen atoms of polar water molecules.

Marking scheme

Part (a) [2 marks]

• M1: Enthalpy change when 1 mole of gaseous ions... (1)

• M2: ...is dissolved in water / becomes aqueous. (1)



Part (b) [3 marks]

• M1: Correctly identifies that lattice dissociation is endothermic (\(+2526\)) or correctly sets up the Hessian cycle equation. (1)

• M2: Multiplies the hydration enthalpy of chloride by 2: \(2 \times (-364) = -728\). (1)

• M3: Correct calculation: \(-122\text{ kJ mol}^{-1}\). (1)

Allow 1 mark overall for an answer of \(+246\text{ kJ mol}^{-1}\) (which forgets to change the sign of the lattice formation enthalpy but correctly doubles the chloride hydration).



Part (c) [4 marks]

• M1: Converts \(\Delta S^\theta\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\) by dividing by 1000: \(0.0115\text{ kJ K}^{-1}\text{ mol}^{-1}\). (1)

• M2: Correct substitution into Gibbs equation: \(\Delta G = -122 - (298 \times 0.0115)\). (1)

• M3: Correct value of \(-125\) (or \(-125.4\)) \(\text{kJ mol}^{-1}\). (1)

• M4: Concludes that the reaction is feasible because \(\Delta G\) is negative / \(< 0\). (1)

Note: Carry forward error (consequential marking) from Part (b).



Part (d) [3 marks]

• M1: Magnesium ion is smaller / has a smaller ionic radius than calcium ion. (1)

• M2: Magnesium ion has a higher charge density. (1)

• M3: Stronger attraction between \(Mg^{2+}\) and the polar water molecules (or oxygen atoms/lone pairs in water). (1)

Paper 2 (Organic and Physical Chemistry)

Answer all questions in the spaces provided.
11 Question · 105.04999999999998 marks
Question 1 · structured
9.55 marks
This question is about rate equations and determining kinetic parameters.

The reaction between hydrogen peroxide and iodide ions in acidic solution is shown below:

\(\text{H}_2\text{O}_2(\text{aq}) + 2\text{I}^-(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\)

A series of experiments was carried out at 298 K to investigate the rate of this reaction. The initial concentrations of the reactants and the initial rates are given in the table below:

\begin{array}{|c|c|c|c|c|}
\hline
\text{Experiment} & [\text{H}_2\text{O}_2] / \text{mol dm}^{-3} & [\text{I}^-] / \text{mol dm}^{-3} & [\text{H}^+] / \text{mol dm}^{-3} & \text{Initial Rate} / \text{mol dm}^{-3}\text{s}^{-1} \\
\hline
1 & 0.010 & 0.010 & 0.10 & 1.15 \times 10^{-6} \\
\hline
2 & 0.020 & 0.010 & 0.10 & 2.30 \times 10^{-6} \\
\hline
3 & 0.010 & 0.020 & 0.10 & 2.30 \times 10^{-6} \\
\hline
4 & 0.010 & 0.010 & 0.20 & 1.15 \times 10^{-6} \\
\hline
\end{array}

(a) Deduce the order of reaction with respect to \(\text{H}_2\text{O}_2\), \(\text{I}^-\), and \(\text{H}^+\). Explain your reasoning. [3.00 marks]

(b) Write the overall rate equation for the reaction, and calculate the value of the rate constant, \(k\), at this temperature. Include the units of \(k\). [3.55 marks]

(c) A student proposes that the reaction proceeds via the following rate-determining step:

\(\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{IO}^- + \text{H}_2\text{O}\)

Explain whether this proposed step is consistent with the rate equation deduced in part (b). [3.00 marks]
Show answer & marking scheme

Worked solution

(a) Comparing Expt 1 and Expt 2: \([\text{I}^-]\) and \([\text{H}^+]\) are constant. \([\text{H}_2\text{O}_2]\) doubles, and the initial rate doubles (from \(1.15 \times 10^{-6}\) to \(2.30 \times 10^{-6}\)). Therefore, the order with respect to \(\text{H}_2\text{O}_2\) is 1.

Comparing Expt 1 and Expt 3: \([\text{H}_2\text{O}_2]\) and \([\text{H}^+]\) are constant. \([\text{I}^-]\) doubles, and the initial rate doubles (from \(1.15 \times 10^{-6}\) to \(2.30 \times 10^{-6}\)). Therefore, the order with respect to \(\text{I}^-\)$ is 1.

Comparing Expt 1 and Expt 4: \([\text{H}_2\text{O}_2]\) and \([\text{I}^-]\) are constant. \([\text{H}^+]\) doubles, but the initial rate remains unchanged (at \(1.15 \times 10^{-6}\)). Therefore, the order with respect to \(\text{H}^+\) is 0.

(b) The rate equation is: \(\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]\)

Using Experiment 1 data to calculate \(k\):
\(k = \frac{\text{Rate}}{[\text{H}_2\text{O}_2][\text{I}^-]} = \frac{1.15 \times 10^{-6}}{0.010 \times 0.010} = 0.0115\) (or \(1.15 \times 10^{-2}\))

Units: \(\frac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3\text{mol}^{-1}\text{s}^{-1}\)

(c) The proposed step is consistent. In a multi-step mechanism, the reactants in the rate-determining step match the species (and their stoichiometric coefficients match the orders) present in the rate equation. Since the rate-determining step involves one molecule of \(\text{H}_2\text{O}_2\) and one ion of \(\text{I}^-\), the derived rate law from this step is indeed \(\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]\).

Marking scheme

(a) [3.00 marks]:
- 1 mark for deducing order of H2O2 is 1 with explanation (doubling conc. doubles rate).
- 1 mark for deducing order of I- is 1 with explanation (doubling conc. doubles rate).
- 1 mark for deducing order of H+ is 0 with explanation (doubling conc. has no effect on rate).

(b) [3.55 marks]:
- 1.00 mark for correct rate equation: Rate = k[H2O2][I-]. (Allow ecf from (a))
- 1.00 mark for rearrangement and calculation of k: k = 0.0115 (or 1.15 x 10^-2).
- 1.00 mark for correct units of k: dm3 mol-1 s-1.
- 0.55 marks for stating the calculation or substitution clearly.

(c) [3.00 marks]:
- 1 mark for stating that it is consistent.
- 1 mark for explaining that the rate-determining step determines the rate of the overall reaction.
- 1 mark for linking the reactants of the RDS (1 H2O2 and 1 I-) to the first-order dependencies in the rate equation.
Question 2 · structured
9.55 marks
This question is about buffer solutions.

A buffer solution was prepared by adding \(25.0 \text{ cm}^3\) of \(0.120 \text{ mol dm}^{-3}\) sodium hydroxide solution to \(50.0 \text{ cm}^3\) of \(0.150 \text{ mol dm}^{-3}\) propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)).

The acid dissociation constant, \(K_a\), of propanoic acid is \(1.35 \times 10^{-5} \text{ mol dm}^{-3}\) at 298 K.

(a) Define the term 'buffer solution'. [2.00 marks]

(b) Calculate the pH of the buffer solution formed at 298 K. Show all your working. [5.55 marks]

(c) State and explain how the pH of this buffer would change, if at all, when a small amount of dilute nitric acid is added. [2.00 marks]
Show answer & marking scheme

Worked solution

(a) A buffer solution is a solution that resists changes in pH when small amounts of an acid or an alkali are added to it.

(b) Step 1: Calculate initial moles.
\(n(\text{CH}_3\text{CH}_2\text{COOH}) = \frac{50.0}{1000} \times 0.150 = 7.50 \times 10^{-3} \text{ mol}\)
\(n(\text{NaOH}) = \frac{25.0}{1000} \times 0.120 = 3.00 \times 10^{-3} \text{ mol}\)

Step 2: Determine moles of weak acid and conjugate base after partial neutralisation.
\(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\)

Moles of \(\text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+\) formed = \(3.00 \times 10^{-3} \text{ mol}\)
Moles of \(\text{CH}_3\text{CH}_2\text{COOH}\) remaining = \(7.50 \times 10^{-3} - 3.00 \times 10^{-3} = 4.50 \times 10^{-3} \text{ mol}\)

Step 3: Calculate \([\text{H}^+]\).
Since both species are in the same volume (\(75.0 \text{ cm}^3\)), the volumes cancel out in the expression:
\([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)}
\)[\text{H}^+] = 1.35 \times 10^{-5} \times \frac{4.50 \times 10^{-3}}{3.00 \times 10^{-3}} = 2.025 \times 10^{-5} \text{ mol dm}^{-3}\)

Step 4: Calculate pH.
\(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6935 \approx 4.69\)

(c) The pH would decrease only very slightly (almost unchanged).
When \(\text{H}^+\) ions from the nitric acid are added, they react with the propanoate conjugate base to form propanoic acid:
\(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
This keeps the concentration of free \(\text{H}^+\) ions virtually constant.

Marking scheme

(a) [2.00 marks]:
- 1 mark for 'resists change in pH' / 'minimises pH change'.
- 1 mark for specifying 'when small amounts of acid or base/alkali are added'.

(b) [5.55 marks]:
- 1.00 mark for calculating initial moles of acid (7.50 x 10^-3) and NaOH (3.00 x 10^-3).
- 1.00 mark for deducing excess moles of propanoic acid = 4.50 x 10^-3 mol.
- 1.00 mark for deducing moles of propanoate formed = 3.00 x 10^-3 mol.
- 1.00 mark for setting up the Ka/pH expression or using Henderson-Hasselbalch equation.
- 1.00 mark for [H+] = 2.025 x 10^-5 mol dm^-3.
- 0.55 marks for final pH of 4.69 (must be to 2 decimal places).

(c) [2.00 marks]:
- 1 mark for stating pH remains virtually constant / decreases only very slightly.
- 1 mark for writing the ionic equation showing consumption of added H+ by propanoate ions.
Question 3 · structured
9.55 marks
This question is about organic synthetic pathways.

A student planned a two-step synthesis to prepare butylamine from 1-chloropropane as shown below:

\(\text{1-chloropropane} \xrightarrow{\text{Step 1}} \text{Butanenitrile} \xrightarrow{\text{Step 2}} \text{Butylamine}\)

(a) Draw the displayed formula of 1-chloropropane and state the reagent and conditions required for Step 1. [3.00 marks]

(b) Outline the mechanism for Step 1, using curly arrows. [3.55 marks]

(c) Identify a suitable reducing agent for Step 2 and write an equation for this reduction using [H] to represent the reducing agent. [3.00 marks]
Show answer & marking scheme

Worked solution

(a) Displayed formula of 1-chloropropane:
An image of 3 carbons in a chain, with single bonds, fully showing all C-H and C-Cl bonds (no structural abbreviations like CH3 allowed for a fully displayed formula):
H H H
| | |
H-C - C - C-Cl
| | |
H H H
Reagent: Potassium cyanide (KCN) or sodium cyanide (NaCN).
Conditions: Aqueous ethanol, heat under reflux.

(b) The mechanism is nucleophilic substitution:
1. A curly arrow starts from the lone pair on the carbon atom of the cyanide ion \(:\text{CN}^-\) pointing to the carbon atom bonded to the chlorine atom in 1-chloropropane.
2. A curly arrow starts from the C-Cl bond pointing to the chlorine atom, showing the heterolytic fission of the C-Cl bond.
3. The products are butanenitrile (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\)) and a chloride ion (\(\text{Cl}^-\)).

(c) Suitable reducing agent: Lithium tetrahydridoaluminate (\(\text{LiAlH}_4\)) in dry ether (accept \(\text{H}_2\) with nickel catalyst).
Equation:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\)

Marking scheme

(a) [3.00 marks]:
- 1 mark for correct displayed formula of 1-chloropropane showing all bonds including C-H and C-Cl.
- 1 mark for identifying KCN/NaCN reagent.
- 1 mark for stating 'aqueous ethanol' and 'reflux/heat'.

(b) [3.55 marks]:
- 1.00 mark for nucleophile (:CN-) showing the lone pair on the carbon and a curly arrow from the lone pair/minus charge to the C-1 atom of 1-chloropropane.
- 1.00 mark for correct dipole showing delta+ on C and delta- on Cl (or implicit in correct arrow).
- 1.00 mark for curly arrow from the C-Cl bond to the Cl atom.
- 0.55 marks for drawing the organic product (butanenitrile) and leaving group (Cl-) correctly.

(c) [3.00 marks]:
- 1 mark for identifying LiAlH4 (or H2 with Ni/Pt/Pd catalyst).
- 1 mark for specifying solvent/condition (dry ether for LiAlH4 or high temp/pressure for H2/Ni).
- 1 mark for correct balanced equation using 4[H].
Question 4 · structured
9.55 marks
This question is about aromatic chemistry and electrophilic substitution.

Benzene undergoes Friedel-Crafts acylation with ethanoyl chloride in the presence of an anhydrous aluminium chloride catalyst to produce phenylethanone.

(a) Write an equation for the reaction of ethanoyl chloride with aluminium chloride to generate the electrophile. [2.00 marks]

(b) Outline the mechanism for the reaction of benzene with this electrophile to form phenylethanone. [4.55 marks]

(c) Explain why methylbenzene reacts faster than benzene in this electrophilic substitution reaction. [3.00 marks]
Show answer & marking scheme

Worked solution

(a) Equation:
\(\text{CH}_3\text{COCl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{CO}^+ + \text{AlCl}_4^-\)

(b) Mechanism:
1. A curly arrow starts from inside the delocalised \(\pi\) ring of benzene pointing to the carbon atom of the \(\text{CH}_3\text{CO}^+\) electrophile.
2. Draw the intermediate structure (a cyclohexadienyl cation):
- The ring should have a partial circle open towards the carbon that has been substituted, with a '+' charge inside the open horseshoe.
- The substituted carbon must be shown bonded to both a hydrogen atom (\(\text{H}\)) and the ethanoyl group (\(-\text{COCH}_3\)).
3. A curly arrow starts from the C-H bond on the substituted carbon pointing back into the open ring to restore the delocalised system.
4. The products are phenylethanone (\(\text{C}_6\text{H}_5\text{COCH}_3\)) and \(\text{H}^+\).

(c) Explanation:
- The methyl group (\(-\text{CH}_3\)) is an electron-donating group (via positive inductive effect).
- This increases the electron density of the benzene ring in methylbenzene compared to benzene.
- As a result, the ring in methylbenzene is a better nucleophile and attracts the electrophile more strongly/reacts faster.

Marking scheme

(a) [2.00 marks]:
- 1 mark for CH3COCl + AlCl3 reacting.
- 1 mark for correct products: CH3CO+ (must show charge on carbon/oxygen, usually carbon) and AlCl4-.

(b) [4.55 marks]:
- 1.00 mark for curly arrow from the benzene ring to the carbon of the CH3CO+ electrophile.
- 1.00 mark for drawing the intermediate correctly (open horseshoe pointing to sp3 carbon, positive charge inside the ring).
- 1.00 mark for showing both H and COCH3 attached to the sp3 carbon in the intermediate.
- 1.00 mark for curly arrow from the C-H bond back into the ring to restore the delocalised system.
- 0.55 marks for overall neatness and correct organic product shown.

(c) [3.00 marks]:
- 1 mark for stating that the methyl group is electron-donating / has a positive inductive effect.
- 1 mark for stating this increases the electron density on the benzene ring.
- 1 mark for stating it is more susceptible to electrophilic attack / attracts the electrophile more readily.
Question 5 · structured
9.55 marks
This question is about structural elucidation using nuclear magnetic resonance (NMR) and infrared (IR) spectroscopy.

An ester, compound **X**, has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\).

The proton NMR spectrum of **X** contains three peaks:
- A singlet at \(\delta = 2.0 \text{ ppm}\), integrating to 3H
- A quartet at \(\delta = 4.1 \text{ ppm}\), integrating to 2H
- A triplet at \(\delta = 1.2 \text{ ppm}\), integrating to 3H

(a) Deduce the structure of compound **X**. Explain how the splitting patterns and integration values in the \(^1\text{H}\) NMR spectrum support your structure. [5.55 marks]

(b) Describe how you could distinguish between compound **X** and its structural isomer, methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)), using both \(^{13}\text{C}\) NMR and infrared spectroscopy. [4.00 marks]
Show answer & marking scheme

Worked solution

(a) Compound **X** is ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

Explanation of NMR data:
- The triplet integrating to 3H at \(\delta = 1.2 \text{ ppm}\) and the quartet integrating to 2H at \(\delta = 4.1 \text{ ppm}\) represent an ethyl group (\(-\text{CH}_2\text{CH}_3\)). The triplet split (\(n+1 = 3\)) indicates the \(-\text{CH}_3\) is adjacent to a \(-\text{CH}_2-\) group. The quartet split (\(n+1 = 4\)) indicates the \(-\text{CH}_2-\) is adjacent to a \(-\text{CH}_3\) group.
- The chemical shift of the quartet at \(\delta = 4.1 \text{ ppm}\) is characteristic of protons on a carbon directly bonded to an ester oxygen (\(-\text{O}-\text{CH}_2-\)).
- The singlet integrating to 3H at \(\delta = 2.0 \text{ ppm}\) represents a isolated methyl group (no adjacent protons, hence singlet) bonded directly to a carbonyl group (\(-\text{CO}-\text{CH}_3\)).

(b) Distinguishing between ethyl ethanoate and methyl propanoate:
- **\(^{13}\text{C}\) NMR spectroscopy**:
- Both isomers have 4 carbon environments and thus will show 4 peaks in their \(^{13}\text{C}\) NMR spectra.
- However, the chemical shifts of the carbons will differ. For instance, in ethyl ethanoate, the carbon adjacent to oxygen (\(-\text{O}-\text{CH}_2-\)) will appear at \(\delta \approx 50\text{--}80 \text{ ppm}\), whereas in methyl propanoate, the carbon adjacent to oxygen is a methyl group (\(-\text{O}-\text{CH}_3\)) which will have a different characteristic chemical shift.
- **Infrared (IR) spectroscopy**:
- Both are esters, so both will exhibit a strong \(\text{C}=\text{O}\) stretch at \(1720\text{--}1750 \text{ cm}^{-1}\) and a \(\text{C}-\text{O}\) stretch at \(1000\text{--}1300 \text{ cm}^{-1}\).
- They can be distinguished by comparing their IR spectra with database spectra in the **fingerprint region** (below \(1500 \text{ cm}^{-1}\)), which will be unique for each compound.

Marking scheme

(a) [5.55 marks]:
- 1.00 mark for deducing the correct structure/name of X as ethyl ethanoate.
- 1.00 mark for identifying the triplet-quartet pair as an ethyl group (-CH2CH3).
- 1.00 mark for linking the triplet splitting to the adjacent CH2 and quartet splitting to the adjacent CH3 (using the n+1 rule).
- 1.00 mark for explaining that the chemical shift of the quartet (4.1 ppm) shows it is adjacent to the ester oxygen.
- 1.00 mark for explaining the singlet at 2.0 ppm is a CH3 group next to a carbonyl (C=O) with no neighbouring hydrogens.
- 0.55 marks for clear logical presentation linking all data points to the structure.

(b) [4.00 marks]:
- 1 mark for stating that both show 4 peaks in 13C NMR (or that 13C NMR cannot be distinguished simply by peak number).
- 1 mark for stating they will have different chemical shifts in 13C NMR (e.g., -O-CH2- vs -O-CH3 environments).
- 1 mark for stating that IR spectra will look very similar because they contain the same functional groups (C=O and C-O).
- 1 mark for explaining that they can be distinguished using the fingerprint region, which is unique for each compound.
Question 6 · structured
9.55 marks
This question is about optical isomerism and carbonyl compounds.

Butanal reacts with hydrogen cyanide (HCN) to form a hydroxynitrile, 2-hydroxypentanenitrile.

(a) Explain, by drawing the structures of the optical isomers, why the product mixture formed in this reaction is optically inactive. [4.55 marks]

(b) Outline the nucleophilic addition mechanism for the reaction of butanal with a mixture of KCN and dilute acid to form 2-hydroxypentanenitrile. [5.00 marks]
Show answer & marking scheme

Worked solution

(a) Explanation:
- Butanal has a planar carbonyl group (\(\text{C}=\text{O}\)) at the reaction site.
- The nucleophile (\(\text{CN}^-\)) has an equal probability of attacking the planar carbonyl carbon from either above or below the plane.
- This results in the formation of equal amounts (a 50:50 / racemic mixture) of two enantiomers (optical isomers).
- The optical rotation of one enantiomer is exactly cancelled out by the opposite rotation of the other enantiomer, making the mixture optically inactive.

Structures of optical isomers (mirror images of 2-hydroxypentanenitrile):
Draw a central chiral carbon (C) bonded tetrahedrally to:
- \(-\text{H}\)
- \(-\text{OH}\)
- \(-\text{CN}\)
- \(-\text{CH}_2\text{CH}_2\text{CH}_3\)
Show both 3D tetrahedral structures (using wedge/dash notation) reflecting across a mirror plane.

(b) Mechanism:
1. Butanal formula is \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\).
2. The \(\text{CN}^-\) ion acts as a nucleophile. A curly arrow starts from the lone pair on the carbon of \(:\text{CN}^-\) to the carbonyl carbon (\(\text{C}=\text{O}\)).
3. A curly arrow starts from the double bond of \(\text{C}=\text{O}\) pointing to the oxygen atom.
4. This forms an intermediate anion: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\).
5. A curly arrow starts from the lone pair on the negatively charged oxygen (\(:\text{O}^-\)) pointing to an \(\text{H}^+\) ion (from the acid).
6. The product is 2-hydroxypentanenitrile: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}(\text{OH})\text{CN}\).

Marking scheme

(a) [4.55 marks]:
- 1.00 mark for stating that the carbonyl group/bond in butanal is planar.
- 1.00 mark for stating that attack by CN- can occur with equal probability from above or below the plane.
- 1.00 mark for explaining this leads to a racemic mixture / 50:50 mixture of enantiomers.
- 1.00 mark for drawing both 3D enantiomers of 2-hydroxypentanenitrile as non-superimposable mirror images using wedge/dash bonds.
- 0.55 marks for stating that the optical rotations cancel out.

(b) [5.00 marks]:
- 1 mark for showing correct partial charges delta+ on C and delta- on O of the C=O bond in butanal.
- 1 mark for curly arrow from lone pair on carbon of :CN- to the carbonyl carbon.
- 1 mark for curly arrow from C=O double bond to the oxygen atom.
- 1 mark for correct structure of the intermediate with a negative charge on the oxygen.
- 1 mark for curly arrow from the O- lone pair to H+ to form the final neutral product.
Question 7 · structured
9.55 marks
This question is about reaction kinetics and the Arrhenius equation.

The rate constant, \(k\), of a reaction was determined at two different temperatures, giving the following results:
- At \(T_1 = 300 \text{ K}\), \(k_1 = 2.45 \times 10^{-4} \text{ s}^{-1}\)
- At \(T_2 = 320 \text{ K}\), \(k_2 = 1.85 \times 10^{-3} \text{ s}^{-1}\)

(a) Use the Arrhenius equation to calculate the activation energy, \(E_a\), for this reaction in \(\text{kJ mol}^{-1}\). The gas constant, \(R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}\). [6.55 marks]

(b) Explain, in terms of the Maxwell-Boltzmann distribution, why a relatively small increase in temperature leads to a large increase in the rate of reaction. [3.00 marks]
Show answer & marking scheme

Worked solution

(a) The logarithmic form of the Arrhenius equation is:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)

Step 1: Calculate the left-hand side (LHS):
\(\ln\left(\frac{1.85 \times 10^{-3}}{2.45 \times 10^{-4}}\right) = \ln(7.5510) = 2.0217\)

Step 2: Calculate the term in brackets:
\(\left(\frac{1}{300} - \frac{1}{320}\right) = 0.0033333 - 0.003125 = 2.0833 \times 10^{-4} \text{ K}^{-1}\)

Step 3: Solve for \(E_a\):
\(2.0217 = \frac{E_a}{8.31} \times 2.0833 \times 10^{-4}\)
\(E_a = \frac{2.0217 \times 8.31}{2.0833 \times 10^{-4}}\)
\(E_a = \frac{16.800}{2.0833 \times 10^{-4}} = 80641 \text{ J mol}^{-1}\)

Step 4: Convert to \(\text{kJ mol}^{-1}\):
\(E_a = 80.64 \text{ kJ mol}^{-1} \approx 80.6 \text{ kJ mol}^{-1}\) (3 s.f.)

(b) According to the Maxwell-Boltzmann distribution:
- An increase in temperature shifts the peak of the curve to a higher energy and lowers its height (broadening the distribution).
- This results in a much larger fraction of molecules having energy equal to or greater than the activation energy (\(E \ge E_a\)), represented by the area under the curve to the right of \(E_a\).
- Consequently, there is a much higher frequency of successful/effective collisions per unit time, vastly increasing the reaction rate.

Marking scheme

(a) [6.55 marks]:
- 1.00 mark for using or stating the correct Arrhenius equation formula: ln(k2/k1) = -Ea/R * (1/T2 - 1/T1) or equivalent.
- 1.00 mark for correctly evaluating ln(k2/k1) = 2.02.
- 1.00 mark for calculating (1/T1 - 1/T2) = 2.08 x 10^-4.
- 1.00 mark for rearranging the equation to solve for Ea.
- 1.00 mark for calculating Ea in J mol-1 (80641 J mol-1).
- 1.00 mark for converting Ea to kJ mol-1 (80.6 kJ mol-1).
- 0.55 marks for expressing the final answer to 3 significant figures.

(b) [3.00 marks]:
- 1 mark for stating that a temperature rise shifts the distribution curve to the right / peak is lower and shifted to higher energy.
- 1 mark for explaining that a significantly larger fraction of molecules now have energy >= Ea (area under curve beyond Ea increases significantly).
- 1 mark for linking this to a much higher frequency of successful collisions.
Question 8 · structured
9.55 marks
This question is about condensation polymers and their environmental impact.

Poly(ethylene terephthalate), commonly known as PET, is a polyester widely used in plastic bottles. It is synthesised from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

(a) Draw the repeating unit of PET and classify the type of polymerisation. [3.00 marks]

(b) Write an equation for the reaction of benzene-1,4-dicarboxylic acid with ethane-1,2-diol to form PET, showing the ester linkage and the small molecule eliminated. [3.55 marks]

(c) Explain why polyesters like PET are biodegradable whereas addition polymers like poly(ethene) are not. [3.00 marks]
Show answer & marking scheme

Worked solution

(a) Classification: Condensation polymerisation.

Repeating unit of PET:
\(-\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\) with open bonds at each end.
(Alternatively, draw out the benzene ring fully in place of \(-\text{C}_6\text{H}_4-\)).

(b) Equation:
\(n \text{ HOOC}-\text{C}_6\text{H}_4-\text{COOH} + n \text{ HO}-\text{CH}_2-\text{CH}_2-\text{OH} \rightarrow -[\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}]_n- + 2n \text{ H}_2\text{O}\)
Note: \(2n - 1\) water molecules is technically exact for a chain of length \(n\), but \(2n\) or \(2n-1\) are both fully accepted at A Level.

(c) Explanation:
- Polyesters contain polar ester linkages (\(\text{C}=\text{O}\) and \(\text{C}-\text{O}\) bonds).
- These polar bonds can be attacked by nucleophiles such as water (hydrolysis) or enzymes in nature, causing the polymer chain to break down.
- In contrast, addition polymers like poly(ethene) have a backbone consisting entirely of non-polar \(\text{C}-\text{C}\) single bonds, which cannot be easily attacked by water or biological systems, making them non-biodegradable.

Marking scheme

(a) [3.00 marks]:
- 1 mark for identifying the reaction as condensation polymerisation.
- 2 marks for drawing the correct repeating unit of PET with open bonds on both ends (deduct 1 mark if ester linkage is drawn incorrectly).

(b) [3.55 marks]:
- 1.00 mark for correct reactant structures: benzene-1,4-dicarboxylic acid and ethane-1,2-diol.
- 1.00 mark for showing the correct polymer product repeating unit in brackets with subscript 'n'.
- 1.00 mark for showing the elimination of water (2n H2O or 2n-1 H2O).
- 0.55 marks for balancing the overall equation with respect to 'n'.

(c) [3.00 marks]:
- 1 mark for stating polyesters have polar ester links (C=O or C-O bonds).
- 1 mark for explaining that these polar links are susceptible to attack by nucleophiles/water (hydrolysis).
- 1 mark for explaining addition polymers have non-polar C-C bonds in the backbone, which cannot be hydrolysed.
Question 9 · structured
9.55 marks
This question is about the kinetics of the hydrolysis of 2-bromo-2-methylpropane.

\((CH_3)_3CBr(aq) + OH^-(aq) \rightarrow (CH_3)_3COH(aq) + Br^-(aq)\)

A series of experiments was carried out at 298 K to determine the rate equation for this reaction. The results are shown in the table below:

| Experiment | Initial \([(CH_3)_3CBr]\) / \(\text{mol dm}^{-3}\) | Initial \([OH^-]\) / \(\text{mol dm}^{-3}\) | Initial rate / \(\text{mol dm}^{-3}\text{ s}^{-1}\) |
| :---: | :---: | :---: | :---: |
| 1 | 0.050 | 0.100 | \(1.25 \times 10^{-4}\) |
| 2 | 0.100 | 0.100 | \(2.50 \times 10^{-4}\) |
| 3 | 0.050 | 0.200 | \(1.25 \times 10^{-4}\) |

**(a)** Deduce the order of reaction with respect to \((CH_3)_3CBr\) and \(OH^-\). Explain your reasoning. (3 marks)

**(b)** Write the rate equation and calculate the rate constant, \(k\), at this temperature. Give the units of \(k\). (2 marks)

**(c)** Write a two-step mechanism for this reaction that is consistent with the rate equation. Identify the rate-determining step. (2 marks)

**(d)** In a different temperature-dependence study, a graph of \(\ln k\) against \(\frac{1}{T}\) was plotted. The gradient of the line was found to be \(-9.62 \times 10^3\text{ K}\). Calculate the activation energy (\(E_a\)) for the reaction in \(\text{kJ mol}^{-1}\). (Gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)). (2.55 marks)
Show answer & marking scheme

Worked solution

**(a)**
- Comparing Exp 1 and Exp 2: \([(CH_3)_3CBr]\) doubles while \([OH^-]\) remains constant. The initial rate doubles from \(1.25 \times 10^{-4}\) to \(2.50 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is **first order** with respect to \((CH_3)_3CBr\).
- Comparing Exp 1 and Exp 3: \([OH^-]\) doubles while \([(CH_3)_3CBr]\) remains constant. The initial rate remains unchanged at \(1.25 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is **zero order** with respect to \(OH^-\).

**(b)**
- Rate equation: \(\text{Rate} = k[(CH_3)_3CBr]\)
- To calculate \(k\) using Experiment 1: \(k = \frac{\text{Rate}}{[(CH_3)_3CBr]} = \frac{1.25 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{0.050\text{ mol dm}^{-3}} = 2.50 \times 10^{-3}\text{ s}^{-1}\)
- Units: \(\text{s}^{-1}\)

**(c)**
Since the reaction is first order with respect to the halogenoalkane and zero order with respect to hydroxide, only the halogenoalkane is involved in the rate-determining step (an \(S_N1\) mechanism):
- **Step 1 (Slow / Rate-determining step):**
\((CH_3)_3CBr \rightarrow (CH_3)_3C^+ + Br^-\)
- **Step 2 (Fast):**
\((CH_3)_3C^+ + OH^- \rightarrow (CH_3)_3COH\)

**(d)**
From the Arrhenius equation, \(\ln k = -\frac{E_a}{RT} + \ln A\), the gradient of a plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\).
- \(\text{Gradient} = -\frac{E_a}{R} = -9.62 \times 10^3\text{ K}\)
- \(E_a = 9.62 \times 10^3 \times 8.31 = 79942.2\text{ J mol}^{-1}\)
- Converting to \(\text{kJ mol}^{-1}\): \(E_a = 79.9\text{ kJ mol}^{-1}\) (to 3 significant figures).

Marking scheme

**(a)**
- **1 mark** for stating first order with respect to \((CH_3)_3CBr\) and linking to the rate doubling when concentration doubles (Exp 1 vs 2).
- **1 mark** for stating zero order with respect to \(OH^-\).
- **1 mark** for explanation linking constant rate to doubling of hydroxide concentration (Exp 1 vs 3).

**(b)**
- **1 mark** for correct rate equation: \(\text{Rate} = k[(CH_3)_3CBr]\) (allow omission of concentration signs if clear, or including \([OH^-]^0\)).
- **1 mark** for correct value of \(k = 2.50 \times 10^{-3}\) (or \(2.5 \times 10^{-3}\)) AND correct unit of \(\text{s}^{-1}\).

**(c)**
- **1 mark** for correct Step 1 equation showing carbocation formation and labeling as slow/RDS.
- **1 mark** for correct Step 2 equation showing reaction of carbocation with hydroxide ion.

**(d)**
- **1 mark** for identifying that \(\text{Gradient} = -\frac{E_a}{R}\).
- **1 mark** for correct calculation of energy in Joules: \(79942\text{ J mol}^{-1}\).
- **0.55 marks** for converting to kJ and giving final answer as \(79.9\text{ kJ mol}^{-1}\) (accept range \(79.9\) to \(80.0\)).
Question 10 · structured
9.55 marks
The synthetic scheme below outlines the multi-step preparation of Compound **X** (an industrial chemical used to manufacture dyes) starting from phenylethane.

\(\text{Phenylethane} \xrightarrow{\text{Step 1: conc. HNO}_3 \text{ + conc. H}_2\text{SO}_4} \text{Compound } \mathbf{Y} \text{ (major para isomer)} \xrightarrow{\text{Step 2: Sn / conc. HCl}} \text{Compound } \mathbf{Z} \xrightarrow{\text{Step 3: CH}_3\text{COCl}} \text{Compound } \mathbf{X}\)

**(a)** Name and outline the mechanism for Step 1. You should include an equation showing how the active electrophile is generated. (4 marks)

**(b)** Identify the essential subsequent step required immediately after heating with \(\text{Sn / conc. HCl}\) in Step 2. Write an overall equation for the reduction of Compound **Y** to Compound **Z**, using \([H]\) to represent the reducing agent. (2.55 marks)

**(c)** Draw the structure of Compound **X** and name the functional group formed in Step 3. (3 marks)
Show answer & marking scheme

Worked solution

**(a)**
- **Mechanism Name:** Electrophilic substitution (or nitration).
- **Electrophile Generation:** \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\)
- **Mechanism Steps:**
1. Curly arrow starts from the delocalised \(\pi\)-system of the phenylethane ring to the nitrogen atom of the \(\text{NO}_2^+\) electrophile.
2. Draw the unstable intermediate with a broken ring (horseshoe shape, open towards the carbon bearing the \(\text{NO}_2\) group and containing a positive charge delocalised over the remaining carbon atoms).
3. Curly arrow from the C-H bond (on the carbon where nitration occurred) back into the ring to restore the aromatic system.

**(b)**
- The subsequent step is the addition of an alkali (such as aqueous sodium hydroxide, \(\text{NaOH(aq)}\)) to liberate the free amine from its protonated salt form (since the acidic conditions of Step 2 leave the amine protonated as a phenylammonium ion).
- **Reduction Equation:**
\(\text{CH}_3\text{CH}_2\text{C}_6\text{H}_4\text{NO}_2 + 6[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{C}_6\text{H}_4\text{NH}_2 + 2\text{H}_2\text{O}\)

**(c)**
- **Structure of Compound X:** \(\text{CH}_3\text{CH}_2\text{-C}_6\text{H}_4\text{-NHCOCH}_3\) (specifically the 1,4-disubstituted isomer: N-(4-ethylphenyl)acetamide, showing the ethyl group and the acetamido group on opposite sides of the benzene ring).
- **Functional Group:** Amide (accept N-substituted amide).

Marking scheme

**(a)**
- **1 mark** for naming the mechanism as **Electrophilic substitution**.
- **1 mark** for the correct equation generating \(\text{NO}_2^+\).
- **1 mark** for curly arrow from ring to \(\text{NO}_2^+\) AND correct horseshoe intermediate with positive charge.
- **1 mark** for curly arrow from C-H bond back into the ring.

**(b)**
- **1 mark** for identifying the addition of \(\text{NaOH}\) / alkali / base.
- **1.55 marks** for the balanced equation with correct structures (or molecular formulas showing 6[H] and producing 2 \(\text{H}_2\text{O}\)).

**(c)**
- **2 marks** for drawing the correct skeletal, displayed, or structural formula of \(\text{CH}_3\text{CH}_2\text{-C}_6\text{H}_4\text{-NHCOCH}_3\) (1,4-isomer). Deduct 1 mark if ortho or meta isomer is drawn.
- **1 mark** for naming the functional group as **amide** (accept peptide link/peptide group).
Question 11 · structured
9.55 marks
This question is about buffer solutions containing weak acids.

A buffer solution is prepared at 298 K by mixing \(50.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\), \(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(30.0\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\) sodium hydroxide solution.

**(a)** Write an equation for the chemical reaction that occurs when these two solutions are mixed. (1 mark)

**(b)** Calculate the pH of the resulting buffer solution at 298 K. Give your answer to 2 decimal places. Show all your working. (5.55 marks)

**(c)** A student adds \(1.00\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) hydrochloric acid to this buffer solution. Write an ionic equation to show how the buffer solution reacts with the added acid, and explain why the pH remains virtually constant. (3 marks)
Show answer & marking scheme

Worked solution

**(a)**
\(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\)

**(b)**
1. **Calculate initial moles of propanoic acid (HA):**
\(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\)

2. **Calculate initial moles of hydroxide ions (\(OH^-\)):**
\(n(\text{OH}^-) = 0.0300\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 3.60 \times 10^{-3}\text{ mol}\)

3. **Calculate moles of substances after neutralisation:**
Since \(\text{HA}\) reacts with \(\text{OH}^-\) in a 1:1 ratio to form \(\text{A}^-\):
- \(n(\text{HA})_\text{remaining} = 7.50 \times 10^{-3} - 3.60 \times 10^{-3} = 3.90 \times 10^{-3}\text{ mol}\)
- \(n(\text{A}^-)_\text{formed} = 3.60 \times 10^{-3}\text{ mol}\)

4. **Calculate \([H^+]\) concentration using the \(K_a\) expression:**
\(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \Rightarrow [\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}\)
Since the volume terms cancel out, we can use moles directly:
\([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{3.90 \times 10^{-3}}{3.60 \times 10^{-3}} = 1.4625 \times 10^{-5}\text{ mol dm}^{-3}\)

5. **Calculate pH:**
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.4625 \times 10^{-5}) = 4.83488 \approx 4.83\)

**(c)**
- When a strong acid is added, the extra \(H^+\) ions react with the conjugate base (propanoate ions) in the buffer solution.
- **Ionic Equation:** \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
- **Explanation:** Because there is a large reservoir of propanoate ions (\(\text{A}^-\)) and propanoic acid (\(\text{HA}\)) compared to the added acid, the ratio of \(\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\) changes only marginally, meaning the pH stays nearly the same.

Marking scheme

**(a)**
- **1 mark** for correct balanced equation (can be molecular or ionic, accept state symbols but not required).

**(b)**
- **1 mark** for calculating initial moles of HA (\(7.50 \times 10^{-3}\)).
- **1 mark** for calculating initial moles of \(\text{OH}^-\)/NaOH (\(3.60 \times 10^{-3}\)).
- **1.55 marks** for calculating remaining moles of HA (\(3.90 \times 10^{-3}\)) and formed moles of \(\text{A}^-\)/propanoate (\(3.60 \times 10^{-3}\)).
- **1 mark** for setting up \(K_a\) expression correctly to find \([H^+] = 1.46 \times 10^{-5}\text{ mol dm}^{-3}\).
- **1 mark** for calculating pH = \(4.83\) (must be strictly 2 decimal places; if rounded incorrectly to 4.8 or 4.84, penalise this mark).

**(c)**
- **1 mark** for correct ionic equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\) (accept \(\text{A}^- + \text{H}^+ \rightarrow \text{HA}\)).
- **1 mark** for identifying that the added \(H^+\) ions are removed by reaction with propanoate ions.
- **1 mark** for explaining that the ratio of \([\text{A}^-]/[\text{HA}]\) remains relatively unchanged, so \([H^+]\) and pH remain stable.

Paper 3 (Synoptic)

Section A contains structured questions. Section B contains multiple choice questions.
35 Question · 90 marks
Question 1 · structured
12 marks
The acid-catalyzed hydrolysis of methyl ethanoate is investigated: \(CH_3COOCH_3(aq) + H_2O(l) \xrightarrow{H^+} CH_3COOH(aq) + CH_3OH(aq)\). A series of experiments is carried out at constant temperature to determine the rate equation. The results are shown below:

Experiment 1: \([CH_3COOCH_3]\) = \(0.200\text{ mol dm}^{-3}\), \([H^+]\) = \(0.100\text{ mol dm}^{-3}\), Initial Rate = \(4.00 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)
Experiment 2: \([CH_3COOCH_3]\) = \(0.200\text{ mol dm}^{-3}\), \([H^+]\) = \(0.200\text{ mol dm}^{-3}\), Initial Rate = \(8.00 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)
Experiment 3: \([CH_3COOCH_3]\) = \(0.400\text{ mol dm}^{-3}\), \([H^+]\) = \(0.100\text{ mol dm}^{-3}\), Initial Rate = \(8.00 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)

(a) Deduce the order of reaction with respect to both methyl ethanoate and hydrogen ions, and write the overall rate equation. [4 marks]

(b) Calculate the value of the rate constant, \(k\), using the data from Experiment 1, and state its units. [3 marks]

(c) In a separate experiment (Experiment 4) at the same temperature, a weak acid \(HX\) of concentration \(0.500\text{ mol dm}^{-3}\) is used as the catalyst instead of the strong acid. The initial concentration of methyl ethanoate is \(0.200\text{ mol dm}^{-3}\). The initial rate of reaction is measured to be \(3.00 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). Calculate the concentration of hydrogen ions in this solution, and hence calculate the acid dissociation constant, \(K_a\), of \(HX\) at this temperature. [5 marks]
Show answer & marking scheme

Worked solution

(a) Comparing Expt 1 and Expt 2: \([H^+]\) doubles while \([CH_3COOCH_3]\) is constant. The initial rate doubles (from \(4.00 \times 10^{-4}\) to \(8.00 \times 10^{-4}\)). Therefore, the reaction is first order with respect to \([H^+]\). Comparing Expt 1 and Expt 3: \([CH_3COOCH_3]\) doubles while \([H^+]\) is constant. The initial rate doubles (from \(4.00 \times 10^{-4}\) to \(8.00 \times 10^{-4}\)). Therefore, the reaction is first order with respect to \([CH_3COOCH_3]\). Rate equation: \(Rate = k[CH_3COOCH_3][H^+]\).

(b) Using Experiment 1 data: \(k = \frac{Rate}{[CH_3COOCH_3][H^+]} = \frac{4.00 \times 10^{-4}}{0.200 \times 0.100} = 0.0200\). Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(c) Using the rate equation and the rate of Experiment 4: \(3.00 \times 10^{-5} = 0.0200 \times 0.200 \times [H^+]\). This gives \([H^+] = \frac{3.00 \times 10^{-5}}{0.00400} = 7.50 \times 10^{-3}\text{ mol dm}^{-3}\). Since \(HX \rightleftharpoons H^+ + X^-\), we assume \([H^+] = [X^-]\). Therefore, \(K_a = \frac{[H^+]^2}{[HX]}\). If dissociation is assumed to be negligible: \(K_a = \frac{(7.50 \times 10^{-3})^2}{0.500} = 1.125 \times 10^{-4}\text{ mol dm}^{-3}\) (rounds to \(1.13 \times 10^{-4}\)). If dissociation is taken into account: \([HX]_{eq} = 0.500 - 7.50 \times 10^{-3} = 0.4925\text{ mol dm}^{-3}\), so \(K_a = \frac{(7.50 \times 10^{-3})^2}{0.4925} = 1.14 \times 10^{-4}\text{ mol dm}^{-3}\).

Marking scheme

Part (a) [4 marks]:
- M1: Order wrt H+ is 1 with justification from Expt 1 and 2 (1)
- M2: Order wrt ester is 1 with justification from Expt 1 and 3 (1)
- M3: Correct rate equation matching deductions: Rate = k[CH3COOCH3][H+] (1)
- M4: Correct justification linking doubling concentration to doubling rate (1)

Part (b) [3 marks]:
- M5: Correct rearrangement of rate equation and calculation: k = 0.0200 (1)
- M6: Correct units: dm3 mol-1 s-1 (1)
- M7: Value matches units and uses 3 sig figs (1)

Part (c) [5 marks]:
- M8: Correct substitution into rate equation to find [H+]: 3.00 x 10^-5 = 0.0200 x 0.200 x [H+] (1)
- M9: Correct calculation of [H+] = 7.50 x 10^-3 mol dm-3 (1)
- M10: Correct expression for Ka: Ka = [H+]^2 / [HX] (1)
- M11: Correct substitution of values into Ka expression (1)
- M12: Correct final value of Ka: 1.13 x 10^-4 OR 1.14 x 10^-4 mol dm-3 with 3 sig figs (1)
Question 2 · structured
12 marks
A student analyzed a green hydrated complex transition metal salt, \(K_xFe(C_2O_4)_y \cdot zH_2O\). A \(2.456\text{ g}\) sample of this salt was dissolved in dilute sulfuric acid and made up to \(250.0\text{ cm}^3\) in a volumetric flask.

(a) A \(25.0\text{ cm}^3\) portion of this solution was warmed to \(60^\circ\text{C}\) and titrated against \(0.0200\text{ mol dm}^{-3}\) \(KMnO_4(aq)\). This step oxidizes only the oxalate (\(C_2O_4^{2-}\)) ions. The average titre was \(30.00\text{ cm}^3\). Write the ionic equation for the reaction of \(MnO_4^-\) with \(C_2O_4^{2-}\) in acid, and calculate the amount, in moles, of \(C_2O_4^{2-}\) in the original \(250.0\text{ cm}^3\) solution. [4 marks]

(b) Another \(25.0\text{ cm}^3\) portion of the original solution was reacted with excess zinc powder to reduce all \(Fe^{3+}\) ions to \(Fe^{2+}\) ions. After filtering off the unreacted zinc, the solution was titrated against \(0.0200\text{ mol dm}^{-3}\) \(KMnO_4(aq)\). In this second titration, both the oxalate and \(Fe^{2+}\) ions are oxidized. The average titre was \(35.00\text{ cm}^3\). Given that \(MnO_4^-\) reacts with \(Fe^{2+}\) in a \(1:5\) ratio, calculate the amount, in moles, of \(Fe^{3+}\) in the original \(250.0\text{ cm}^3\) solution. [3 marks]

(c) Deduce the values of \(x\) and \(y\) in the formula of the salt, assuming the complex is neutral and potassium is present as \(K^+\). [2 marks]

(d) Calculate the value of \(z\) to the nearest integer. [3 marks]
Show answer & marking scheme

Worked solution

(a) The balanced ionic equation for the oxidation of oxalate ions by manganate(VII) is: \(2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O\).
Moles of \(MnO_4^-\) used in the first titration = \(0.0200 \times 0.03000 = 6.00 \times 10^{-4}\text{ mol}\).
Moles of \(C_2O_4^{2-}\) in the \(25.0\text{ cm}^3\) portion = \(6.00 \times 10^{-4} \times 2.5 = 1.50 \times 10^{-3}\text{ mol}\).
Moles of \(C_2O_4^{2-}\) in the original \(250.0\text{ cm}^3\) solution = \(1.50 \times 10^{-3} \times 10 = 1.50 \times 10^{-2}\text{ mol}\).

(b) In the second titration, both \(C_2O_4^{2-}\) and \(Fe^{2+}\) are titrated. Since the amount of oxalate in \(25.0\text{ cm}^3\) is unchanged, it still accounts for \(30.00\text{ cm}^3\) of the \(KMnO_4\) titre. The volume of \(KMnO_4\) required to oxidize only \(Fe^{2+}\) is \(35.00 - 30.00 = 5.00\text{ cm}^3\).
Moles of \(MnO_4^-\) used for \(Fe^{2+}\) = \(0.0200 \times 0.00500 = 1.00 \times 10^{-4}\text{ mol}\).
Since \(1\text{ mol}\) of \(MnO_4^-\) reacts with \(5\text{ mol}\) of \(Fe^{2+}\):
Moles of \(Fe^{2+}\) in the \(25.0\text{ cm}^3\) portion = \(1.00 \times 10^{-4} \times 5 = 5.00 \times 10^{-4}\text{ mol}\).
Moles of \(Fe^{3+}\) in the original \(250.0\text{ cm}^3\) solution = \(5.00 \times 10^{-4} \times 10 = 5.00 \times 10^{-3}\text{ mol}\).

(c) The mole ratio of \(Fe : C_2O_4^{2-}\) in the compound is \(5.00 \times 10^{-3} : 1.50 \times 10^{-2} = 1 : 3\). Thus, \(y = 3\).
For a neutral compound \(K_xFe(C_2O_4)_3\), since \(Fe\) is \(+3\) and \(C_2O_4^{2-}\) is \(-2\): \(x(+1) + 3 + 3(-2) = 0 \Rightarrow x - 3 = 0 \Rightarrow x = 3\).

(d) The formula is \(K_3Fe(C_2O_4)_3 \cdot zH_2O\).
Moles of salt in the \(2.456\text{ g}\) sample is equal to the moles of \(Fe^{3+}\) = \(5.00 \times 10^{-3}\text{ mol}\).
Molar mass (\(M_r\)) of \(K_3Fe(C_2O_4)_3 \cdot zH_2O\) = \\frac{2.456}{5.00 \times 10^{-3}} = 491.2\text{ g mol}^{-1}\).
\(M_r\) of anhydrous \(K_3Fe(C_2O_4)_3\) = \(3(39.1) + 55.8 + 3(88.0) = 117.3 + 55.8 + 264.0 = 437.1\text{ g mol}^{-1}\).
Mass of water of crystallization per mole = \(491.2 - 437.1 = 54.1\text{ g mol}^{-1}\).
\(z = \frac{54.1}{18.0} = 3.01 \approx 3\).

Marking scheme

Part (a) [4 marks]:
- M1: Correct equation: 2MnO4- + 5C2O4^2- + 16H+ -> 2Mn2+ + 10CO2 + 8H2O (1)
- M2: Moles of MnO4- = 6.00 x 10^-4 mol (1)
- M3: Moles of C2O4^2- in 25.0 cm3 = 1.50 x 10^-3 mol (1)
- M4: Moles of C2O4^2- in 250.0 cm3 = 1.50 x 10^-2 mol (1)

Part (b) [3 marks]:
- M5: Deduce volume of KMnO4 for Fe2+ is 5.00 cm3 (1)
- M6: Calculate moles of MnO4- for Fe2+ = 1.00 x 10^-4 mol (1)
- M7: Calculate moles of Fe3+ in original 250 cm3 = 5.00 x 10^-3 mol (1)

Part (c) [2 marks]:
- M8: Find mole ratio Fe3+ : C2O4^2- = 1 : 3 to get y = 3 (1)
- M9: Use charge balance to deduce x = 3 (1)

Part (d) [3 marks]:
- M10: Calculate Mr of hydrated salt = 491.2 g mol-1 (1)
- M11: Calculate Mr of anhydrous salt = 437.1 g mol-1 (1)
- M12: Calculate z = 3 (must be integer) (1)
Question 3 · structured
12 marks
The hydrogen-oxygen fuel cell is an electrochemical cell that produces electricity from the reaction of hydrogen and oxygen.

(a) Write the half-equations for the reactions occurring at the anode (negative electrode) and cathode (positive electrode) in an alkaline hydrogen-oxygen fuel cell. Write an overall equation for the cell reaction. [3 marks]

(b) Standard entropy values (\(S^\theta\)) at \(298\text{ K}\) are given below:
- \(S^\theta(H_2(g)) = 131\text{ J K}^{-1}\text{ mol}^{-1}\)
- \(S^\theta(O_2(g)) = 205\text{ J K}^{-1}\text{ mol}^{-1}\)
- \(S^\theta(H_2O(l)) = 70.0\text{ J K}^{-1}\text{ mol}^{-1}\)

Calculate the standard entropy change (\(\Delta S^\theta\)) for the overall cell reaction: \(2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\). [3 marks]

(c) State and explain the sign of \(\Delta S^\theta\) by referring to the state of the reactants and products. [2 marks]

(d) The standard enthalpy change of combustion of hydrogen is \(-286\text{ kJ mol}^{-1}\) (per mole of \(H_2\)). Calculate the standard Gibbs free energy change (\(\Delta G^\theta\)) for the overall cell reaction at \(298\text{ K}\), and explain whether this reaction becomes more or less feasible at higher temperatures. [4 marks]
Show answer & marking scheme

Worked solution

(a) Anode (oxidation): \(H_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^-\)
Cathode (reduction): \(O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)\)
Overall: \(2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\)

(b) \(\Delta S^\theta = \sum S^\theta\text{(products)} - \sum S^\theta\text{(reactants)}\)
\(\Delta S^\theta = [2 \times S^\theta(H_2O(l))] - [2 \times S^\theta(H_2(g)) + S^\theta(O_2(g))]\)
\(\Delta S^\theta = [2 \times 70.0] - [(2 \times 131) + 205] = 140 - [262 + 205]\)
\(\Delta S^\theta = 140 - 467 = -327\text{ J K}^{-1}\text{ mol}^{-1}\).

(c) The sign is negative (entropy decreases) because \(3\text{ moles}\) of gas reactants (\(2\text{ mol } H_2\) and \(1\text{ mol } O_2\)) react to form \(2\text{ moles}\) of liquid product (\(2\text{ mol } H_2O\)). Liquids have much lower entropy than gases as the molecules are more ordered.

(d) \(\Delta H^\theta\) for the overall reaction of \(2\text{ moles}\) of \(H_2\) is \(2 \times (-286) = -572\text{ kJ mol}^{-1} = -572,000\text{ J mol}^{-1}\).
\(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)
\(\Delta G^\theta = -572,000 - (298 \times -327) = -572,000 - (-97,446) = -474,554\text{ J mol}^{-1} = -475\text{ kJ mol}^{-1}\) (or \(-474.6\text{ kJ mol}^{-1}\)).
As temperature increases, \(-T\Delta S^\theta\) becomes more positive because \(\Delta S^\theta\) is negative. This makes \(\Delta G^\theta\) less negative (closer to zero or positive), so the reaction becomes less feasible at higher temperatures.

Marking scheme

Part (a) [3 marks]:
- M1: Correct anode half-equation (1)
- M2: Correct cathode half-equation (1)
- M3: Correct overall equation (1)

Part (b) [3 marks]:
- M4: Correct expression: Delta S = Sum(S products) - Sum(S reactants) (1)
- M5: Correctly substituting values: 2(70) - (2(131) + 205) (1)
- M6: Correct final value of -327 J K-1 mol-1 (1)

Part (c) [2 marks]:
- M7: Identifies the sign is negative / entropy decreases (1)
- M8: Explains that gas molecules are converting into liquid molecules, representing a decrease in disorder / fewer moles of gas (1)

Part (d) [4 marks]:
- M9: Calculates Delta H = -572 kJ mol-1 (1)
- M10: Correct calculation of Delta G = -475 kJ mol-1 (accept -474.6) (1)
- M11: Links increasing temperature to -T Delta S becoming more positive (1)
- M12: Correctly concludes that feasibility decreases at higher temperatures (1)
Question 4 · structured
12 marks
Benzocaine is a local anesthetic synthesized from methylbenzene. The synthetic route is shown below:

Methylbenzene \(\xrightarrow{\text{Step 1}}\) 4-nitromethylbenzene \(\xrightarrow{\text{Step 2}}\) 4-nitrobenzoic acid \(\xrightarrow{\text{Step 3}}\) ethyl 4-nitrobenzoate \(\xrightarrow{\text{Step 4}}\) Benzocaine (ethyl 4-aminobenzoate)

(a) Identify the inorganic reagents and concentrated acid required for Step 1. Write an equation to show how the active electrophile is generated in this reaction mixture. [3 marks]

(b) Draw the mechanism for the electrophilic substitution reaction in Step 1 to form 4-nitromethylbenzene from methylbenzene. [3 marks]

(c) Step 3 is an esterification. Identify the alcohol and the catalyst required for Step 3, and state the reaction conditions. [2 marks]

(d) Write an equation for the reduction reaction in Step 4, using \([H]\) to represent the reducing agent. [2 marks]

(e) Benzocaine contains an ethyl group. Describe the splitting pattern and relative integration of the peaks expected in the \(^1\text{H}\) NMR spectrum for the protons in this ethyl group. [2 marks]
Show answer & marking scheme

Worked solution

(a) Reagents: Concentrated nitric acid (\(HNO_3\)) and concentrated sulfuric acid (\(H_2SO_4\)).
Equation: \(HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-\)
(Accept \(HNO_3 + H_2SO_4 \rightarrow NO_2^+ + H_2O + HSO_4^-\)).

(b) Mechanism of electrophilic substitution:
- Curly arrow from the benzene ring to the \(NO_2^+\) ion.
- Intermediate with a horseshoe pointing to position 4, with a positive charge inside the ring, and showing both \(H\) and \(NO_2\) attached to the carbon at position 4.
- Curly arrow from the C-H bond back to the ring to restore the delocalized system.

(c) Alcohol: Ethanol (\(CH_3CH_2OH\)). Catalyst: Concentrated sulfuric acid (\(H_2SO_4\)). Conditions: Heat under reflux.

(d) Balanced equation for Step 4 reduction:
\(O_2N-C_6H_4-COOCH_2CH_3 + 6[H] \rightarrow H_2N-C_6H_4-COOCH_2CH_3 + 2H_2O\)

(e) Splitting pattern and integration:
- A triplet (due to \(-CH_3\) adjacent to \(-CH_2-\)), relative integration = 3.
- A quartet (due to \(-CH_2-\) adjacent to \(-CH_3\)), relative integration = 2.

Marking scheme

Part (a) [3 marks]:
- M1: Reagents: Concentrated HNO3 and Concentrated H2SO4 (Both required, reject dilute) (1)
- M2: Correct generation of NO2+ electrophile equation (1)

Part (b) [3 marks]:
- M3: Curly arrow from ring to NO2+ electrophile (1)
- M4: Correct intermediate showing positive charge and horseshoe opening to sp3 carbon (1)
- M5: Curly arrow from C-H bond back into ring to restore delocalization (1)

Part (c) [2 marks]:
- M6: Reagents: Ethanol and Concentrated H2SO4 (1)
- M7: Conditions: Heat under reflux (1)

Part (d) [2 marks]:
- M8: Correct structures of ethyl 4-nitrobenzoate and Benzocaine in equation (1)
- M9: Balanced with 6[H] on left and 2H2O on right (1)

Part (e) [2 marks]:
- M10: Triplet with integration of 3 (1)
- M11: Quartet with integration of 2 (1)
Question 5 · structured
12 marks
An organic compound \(Y\) has the molecular formula \(C_6H_{12}O_2\).

The infrared spectrum of \(Y\) shows a strong absorption peak at \(1735\text{ cm}^{-1}\), but no broad absorption peak in the range \(3200-3600\text{ cm}^{-1}\).
The \(^{13}\text{C}\) NMR spectrum of \(Y\) has exactly 4 peaks.
The \(^1\text{H}\) NMR spectrum of \(Y\) consists of only two singlets: one at \(\delta = 1.4\text{ ppm}\) (relative integration 9) and one at \(\delta = 2.0\text{ ppm}\) (relative integration 3).

(a) Explain how the infrared spectrum helps to deduce the functional group of \(Y\), ruling out other possibilities. [2 marks]

(b) Deduce the structure of \(Y\). Fully explain how all the NMR data (both \(^{13}\text{C}\) and \(^1\text{H}\)) support your deduced structure. [6 marks]

(c) Write the chemical equation for the hydrolysis of \(Y\) when heated with aqueous sodium hydroxide. [2 marks]

(d) After the alkaline hydrolysis of \(Y\) is complete, the mixture is acidified. Suggest a simple chemical test, including reagents and observations, to distinguish between the two organic products formed. [2 marks]
Show answer & marking scheme

Worked solution

(a) The absorption at \(1735\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(C=O\)). The absence of a broad absorption in the \(3200-3600\text{ cm}^{-1}\) region indicates there is no hydroxyl (\(O-H\)) group, which rules out carboxylic acids and alcohols. Therefore, \(Y\) must be an ester.

(b) The structure of \(Y\) is t-butyl ethanoate (or tert-butyl acetate), \(CH_3COOC(CH_3)_3\).
- The singlet at \(\delta = 1.4\text{ ppm}\) with integration 9 corresponds to the 9 equivalent protons of a tert-butyl group, \(-C(CH_3)_3\), which has no adjacent protons to cause splitting.
- The singlet at \(\delta = 2.0\text{ ppm}\) with integration 3 corresponds to the 3 protons of a methyl group adjacent to a carbonyl, \(CH_3CO-\), with no adjacent protons to cause splitting.
- \(^{13}\text{C}\) NMR shows 4 peaks, which matches the 4 carbon environments: \(CH_3-CO\) (1), \(C=O\) (2), the quaternary carbon \(-C(CH_3)_3\) (3), and the 3 equivalent methyl carbons in \(-C(CH_3)_3\) (4).

(c) Hydrolysis equation: \(CH_3COOC(CH_3)_3 + NaOH \rightarrow CH_3COONa + (CH_3)_3COH\).

(d) Upon acidification, the products are ethanoic acid (\(CH_3COOH\)) and 2-methylpropan-2-ol (\((CH_3)_3COH\)).
- Test: Add sodium hydrogencarbonate (\(NaHCO_3(aq)\)) or sodium carbonate (\(Na_2CO_3(aq)\)).
- Observation: Ethanoic acid will produce effervescence/fizzing (carbon dioxide gas). 2-methylpropan-2-ol will show no visible reaction/no bubbles.

Marking scheme

Part (a) [2 marks]:
- M1: Peak at 1735 cm-1 shows C=O is present (1)
- M2: Absence of 3200-3600 cm-1 peak shows O-H is absent, ruling out carboxylic acid/alcohol, hence it is an ester (1)

Part (b) [6 marks]:
- M3: Identifies structure of Y as CH3COOC(CH3)3 (1)
- M4: Links singlet at 1.4 ppm to -C(CH3)3 group and explains singlet splitting (no adjacent protons) (1)
- M5: Links integration of 9 to the 9 equivalent protons (1)
- M6: Links singlet at 2.0 ppm to CH3-CO group with integration 3 (1)
- M7: Explains 13C NMR showing 4 peaks by identifying 4 unique carbon environments (1)
- M8: Explains why the methyl groups in the t-butyl group are equivalent (1)

Part (c) [2 marks]:
- M9: Correct structures of reactants in equation: CH3COOC(CH3)3 + NaOH (1)
- M10: Correct products: CH3COONa + (CH3)3COH (1)

Part (d) [2 marks]:
- M11: Reagent: NaHCO3 / Na2CO3 (or any metal carbonate/reactive metal like Mg) (1)
- M12: Observation: Effervescence with ethanoic acid and no change with the alcohol (1)
Question 6 · multiple-choice
1 marks
Under certain conditions, the rate equation for the reaction \(2A + B \rightarrow C\) is: \(\text{rate} = k[A]^2[B]\). If the concentration of \(A\) is halved and the concentration of \(B\) is tripled, by what factor does the rate change?
  1. A.Decreases by a factor of \(\frac{4}{3}\) (multiplied by 0.75)
  2. B.Decreases by a factor of \(\frac{2}{3}\)
  3. C.Increases by a factor of \(\frac{9}{4}\)
  4. D.Increases by a factor of 3 post-substitutionially-scaled factor of 3.00 if it were first order in A and first order in B only.
Show answer & marking scheme

Worked solution

By the rate equation, if \([A]\) is halved, the rate is multiplied by \((0.5)^2 = 0.25\). If \([B]\) is tripled, the rate is multiplied by \(3\). The overall change factor is \(0.25 \times 3 = 0.75\), which is equivalent to dividing by \(\frac{4}{3}\) (or decreasing by a factor of \(\frac{4}{3}\)).

Marking scheme

1 mark for the correct option (A).
Question 7 · multiple-choice
1 marks
Which of the following statements about ligand substitution and the chelate effect is correct?
  1. A.The substitution of monodentate ligands by multidentate ligands is driven primarily by a highly negative enthalpy change (\(\Delta H\)).
  2. B.When \([Co(H_2O)_6]^{2+}\) reacts with EDTA\(^{4-}\), the coordination number of the cobalt ion decreases.
  3. C.The chelate effect occurs because there is a significant increase in entropy (\(\Delta S\)) when multidentate ligands replace monodentate ligands.
  4. D.The reaction of \([Cu(H_2O)_6]^{2+}\) with excess concentrated hydrochloric acid results in a change in coordination number from 6 to 5.
Show answer & marking scheme

Worked solution

Multidentate ligands replace monodentate ligands because this increases the number of species in solution, leading to a positive entropy change (\(\Delta S > 0\)). Since \(\Delta H\) is typically very small, the free energy change \(\Delta G = \Delta H - T\Delta S\) becomes more negative, making the reaction highly feasible.

Marking scheme

1 mark for the correct option (C).
Question 8 · multiple-choice
1 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting buffer solution at \(298\text{ K}\)?
  1. A.3.87
  2. B.4.87
  3. C.5.17
  4. D.7.00
Show answer & marking scheme

Worked solution

The moles of propanoic acid initially = \(0.0500 \times 0.100 = 5.00 \times 10^{-3}\text{ mol}\). The moles of NaOH added = \(0.0250 \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\). This neutralizes exactly half of the propanoic acid, creating equal amounts of propanoic acid and propanoate ions (\(2.50 \times 10^{-3}\text{ mol}\) of each). At half-neutralisation, \([\text{HA}] = [\text{A}^-]\), so \([\text{H}^+] = K_a\). Therefore, \(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

Marking scheme

1 mark for the correct option (B).
Question 9 · multiple-choice
1 marks
A student wants to synthesize ethyl benzoate from benzonitrile in a multi-step pathway. Which of the following sequences represents the most suitable and feasible set of reagents and conditions?
  1. A.1. Heat under reflux with dilute aqueous \(\text{HCl}\); 2. Heat under reflux with ethanol and concentrated \(\text{H}_2\text{SO}_4\).
  2. B.1. Heat under reflux with ethanolic \(\text{KCN}\); 2. Heat under reflux with ethyl alcohol and concentrated \(\text{HCl}\).
  3. C.1. \(\text{LiAlH}_4\) in dry ether; 2. React with ethyl acid chloride.
  4. D.1. React with steam at high temperature and pressure; 2. React with concentrated \(\text{H}_2\text{SO}_4\) and benzene.
Show answer & marking scheme

Worked solution

Benzonitrile is hydrolyzed to benzoic acid by heating under reflux with a dilute acid like \(\text{HCl}\). The benzoic acid product is then heated under reflux with ethanol in the presence of concentrated \(\text{H}_2\text{SO}_4\) catalyst to undergo esterification and yield ethyl benzoate.

Marking scheme

1 mark for the correct option (A).
Question 10 · multiple-choice
1 marks
Which of the following trends is correct for the Group 2 elements from magnesium to barium?
  1. A.The solubility of their hydroxides decreases, and the solubility of their sulfates increases.
  2. B.The solubility of their hydroxides increases, and the solubility of their sulfates decreases.
  3. C.The first ionisation energy increases because of increased nuclear charge.
  4. D.The atomic radius decreases due to increased nuclear attraction on outer electrons.
Show answer & marking scheme

Worked solution

Down Group 2 from magnesium to barium, the solubility of hydroxides increases (magnesium hydroxide is sparingly soluble, barium hydroxide is soluble), and the solubility of sulfates decreases (magnesium sulfate is soluble, barium sulfate is insoluble).

Marking scheme

1 mark for the correct option (B).
Question 11 · multiple-choice
1 marks
For the reaction: \(2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{SO}_3(\text{g})\), the standard enthalpy change, \(\Delta H^\ominus\), is \(-196\text{ kJ mol}^{-1}\) and the standard entropy change, \(\Delta S^\ominus\), is \(-188\text{ J K}^{-1}\text{ mol}^{-1}\). At what temperature does this reaction cease to be feasible?
  1. A.1.04 K
  2. B.104 K
  3. C.1043 K
  4. D.3715 K
Show answer & marking scheme

Worked solution

A reaction ceases to be feasible when \(\Delta G = 0\). Using \(\Delta G = \Delta H - T\Delta S\), we set \(\Delta H - T\Delta S = 0\), which gives \(T = \frac{\Delta H}{\Delta S}\). Converting \(\Delta H\) to J: \(T = \frac{-196000\text{ J mol}^{-1}}{-188\text{ J K}^{-1}\text{ mol}^{-1}} \approx 1042.6\text{ K}\). Above this temperature, \(\Delta G\) becomes positive, so the reaction ceases to be feasible.

Marking scheme

1 mark for the correct option (C).
Question 12 · multiple-choice
1 marks
A \(0.250\text{ g}\) sample of pure anhydrous metal carbonate, \(\text{M}_2\text{CO}_3\), was dissolved in water and neutralized exactly by \(24.4\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) hydrochloric acid. What is the molar mass of this metal carbonate?
  1. A.68.3\text{ g mol}^{-1}
  2. B.136.6\text{ g mol}^{-1}
  3. C.273.2\text{ g mol}^{-1}
  4. D.546.4\text{ g mol}^{-1}
Show answer & marking scheme

Worked solution

The neutralization reaction is: \(\text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{CO}_2 + \text{H}_2\text{O}\). The moles of \(\text{HCl}\) used = \(0.0244\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.66 \times 10^{-3}\text{ mol}\). Since 1 mole of carbonate reacts with 2 moles of acid, the moles of \(\text{M}_2\text{CO}_3\) = \(3.66 \times 10^{-3} / 2 = 1.83 \times 10^{-3}\text{ mol}\). The molar mass is \(0.250\text{ g} / (1.83 \times 10^{-3}\text{ mol}) = 136.61\text{ g mol}^{-1}\).

Marking scheme

1 mark for the correct option (B).
Question 13 · multiple-choice
1 marks
Standard electrode potentials for two half-cells are given below: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\ominus = +0.77\text{ V}\), \(\text{Cr}^{3+}(\text{aq}) + 3\text{e}^- \rightleftharpoons \text{Cr}(\text{s}) \quad E^\ominus = -0.74\text{ V}\). Which of the following represents the standard cell diagram and the standard EMF (\(E^\ominus_{\text{cell}}\)) of the feasible electrochemical cell made from these half-cells?
  1. A.\(\text{Pt}(\text{s}) | \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) || \text{Cr}^{3+}(\text{aq}) | \text{Cr}(\text{s}) \quad E^\ominus_{\text{cell}} = +1.51\text{ V}\)
  2. B.\(\text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq}) || \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt}(\text{s}) \quad E^\ominus_{\text{cell}} = +1.51\text{ V}\)
  3. C.\(\text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq}) || \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt}(\text{s}) \quad E^\ominus_{\text{cell}} = +0.03\text{ V}\)
  4. D.\(\text{Pt}(\text{s}) | \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) || \text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq}) \quad E^\ominus_{\text{cell}} = -1.51\text{ V}\)
Show answer & marking scheme

Worked solution

The half-cell with the more positive potential (+0.77 V) undergoes reduction, which is written on the right of the cell diagram. Because both ions are in solution, an inert Pt electrode is used: \(\text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt}(\text{s})\). The half-cell with the more negative potential (-0.74 V) undergoes oxidation, which is written on the left of the cell diagram: \(\text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq})\). The standard EMF is calculated as \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77 - (-0.74) = +1.51\text{ V}\).

Marking scheme

1 mark for the correct option (B).
Question 14 · Multiple Choice
1 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_{\text{a}} = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(25.0\text{ cm}^3\) of \(0.080\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting solution at \(298\text{ K}\)?
  1. A.4.52
  2. B.4.69
  3. C.4.87
  4. D.5.05 Limiting Reactant Error (Inverse Ratio)
Show answer & marking scheme

Worked solution

First, calculate the initial moles of propanoic acid (\(\text{HA}\)) and sodium hydroxide (\(\text{NaOH}\)): \(\text{initial moles of HA} = 0.0500 \times 0.100 = 5.00 \times 10^{-3}\text{ mol}\); \(\text{moles of OH}^- \text{ added} = 0.0250 \times 0.080 = 2.00 \times 10^{-3}\text{ mol}\). The added \(\text{OH}^-\right\) reacts with \(\text{HA}\) to produce propanoate ions (\(\text{A}^-\)): \(\text{moles of HA remaining} = 5.00 \times 10^{-3} - 2.00 \times 10^{-3} = 3.00 \times 10^{-3}\text{ mol}\); \(\text{moles of A}^- \text{ formed} = 2.00 \times 10^{-3}\text{ mol}\). Using the buffer equation: \([\text{H}^+] = K_{\text{a}} \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{3.00 \times 10^{-3}}{2.00 \times 10^{-3}} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\). Therefore, \(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.69\).

Marking scheme

1 mark for the correct calculation of pH, yielding 4.69 (B).
Question 15 · Multiple Choice
1 marks
Which of the following complexes can exist as optical isomers?
  1. A.\([\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}\)
  2. B.\(\textit{trans}\text{-}[\text{Co}(\text{en})_2\text{Cl}_2]^+\)
  3. C.\([\text{Cu}(\text{NH}_3)_4]^{2+}\)
  4. D.\(\textit{cis}\text{-}[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]\)
Show answer & marking scheme

Worked solution

\([\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}\) is an octahedral complex with three bidentate oxalate ligands. Because of the spatial arrangement of the chelating ligands, it lacks a plane of symmetry and forms non-superimposable mirror images (enantiomers). The \(\textit{trans}\) isomer in option B has a plane of symmetry, while options C and D are square planar complexes and possess a plane of symmetry.

Marking scheme

1 mark for identifying \([\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}\) as the optically active complex (A).
Question 16 · Multiple Choice
1 marks
The table below contains thermodynamic data for the formation of magnesium chloride, \(\text{MgCl}_2(\text{s})\): \(\Delta H_{\text{at}}^{\ominus}[\text{Mg}] = +148\text{ kJ mol}^{-1}\); \(\text{IE}_1[\text{Mg}] = +738\text{ kJ mol}^{-1}\); \(\text{IE}_2[\text{Mg}] = +1451\text{ kJ mol}^{-1}\); \(\Delta H_{\text{at}}^{\ominus}[\text{Cl}] = +122\text{ kJ mol}^{-1}\); \(\text{EA}[\text{Cl}] = -349\text{ kJ mol}^{-1}\); \(\Delta H_{\text{f}}^{\ominus}[\text{MgCl}_2(\text{s})] = -642\text{ kJ mol}^{-1}\). What is the lattice enthalpy of formation of \(\text{MgCl}_2(\text{s})\)?
  1. A.\(-2525\text{ kJ mol}^{-1}\)
  2. B.\(-2403\text{ kJ mol}^{-1}\)
  3. C.\(-2752\text{ kJ mol}^{-1}\)
  4. D.\(-1883\text{ kJ mol}^{-1}\)
Show answer & marking scheme

Worked solution

Using the Born-Haber cycle: \(\Delta H_{\text{f}}^{\ominus}[\text{MgCl}_2(\text{s})] = \Delta H_{\text{at}}^{\ominus}[\text{Mg}] + \text{IE}_1[\text{Mg}] + \text{IE}_2[\text{Mg}] + 2\Delta H_{\text{at}}^{\ominus}[\text{Cl}] + 2\text{EA}[\text{Cl}] + \Delta H_{\text{L,form}}^{\ominus}\). Substituting the values: \(-642 = 148 + 738 + 1451 + 2(122) + 2(-349) + \Delta H_{\text{L,form}}^{\ominus} \Rightarrow -642 = 1883 + \Delta H_{\text{L,form}}^{\ominus} \Rightarrow \Delta H_{\text{L,form}}^{\ominus} = -2525\text{ kJ mol}^{-1}\).

Marking scheme

1 mark for the correct calculation of lattice enthalpy of formation, yielding -2525 kJ/mol (A).
Question 17 · Multiple Choice
1 marks
A student heats a \(1.38\text{ g}\) sample of hydrated magnesium carbonate, \(\text{MgCO}_3 \cdot x\text{H}_2\text{O}\), until it is completely decomposed to form anhydrous magnesium oxide, carbon dioxide and water vapour: \(\text{MgCO}_3 \cdot x\text{H}_2\text{O}(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g}) + x\text{H}_2\text{O}(\text{g})\). The total volume of gas collected at a temperature of \(500\text{ K}\) and a pressure of \(100\text{ kPa}\) was \(1.66\text{ dm}^3\). What is the value of \(x\) in the formula? (\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\); \(M_{\text{r}}(\text{MgCO}_3) = 84.3\))
  1. A.1
  2. B.2
  3. C.3
  4. D.5
Show answer & marking scheme

Worked solution

First, find the total moles of gas produced using the ideal gas equation: \(n_{\text{gas}} = \frac{pV}{RT} = \frac{100 \times 10^3\text{ Pa} \times 1.66 \times 10^{-3}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 500\text{ K}} = 0.0400\text{ mol}\). From the stoichiometry, 1 mole of \(\text{MgCO}_3 \cdot x\text{H}_2\text{O}\) produces \((1+x)\) moles of gas. Let the moles of reactant be \(n = \frac{1.38}{84.3 + 18.0x}\). Since \(n_{\text{gas}} = n(1+x)\), we have \(\frac{1.38(1+x)}{84.3 + 18.0x} = 0.0400 \Rightarrow 1.38 + 1.38x = 3.372 + 0.72x \Rightarrow 0.66x = 1.992 \Rightarrow x = 3.02 \approx 3\).

Marking scheme

1 mark for the correct determination of x as 3 (C).
Question 18 · Multiple Choice
1 marks
Consider the standard electrode potentials: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \ (E^{\ominus} = +0.77\text{ V})\); \(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{e}^- \rightleftharpoons 2\text{SO}_4^{2-}(\text{aq}) \ (E^{\ominus} = +2.01\text{ V})\); \(\text{I}_2(\text{aq}) + 2\text{e}^- \rightleftharpoons 2\text{I}^-(\text{aq}) \ (E^{\ominus} = +0.54\text{ V})\). Which statement is correct under standard conditions?
  1. A.\(\text{Fe}^{2+}(\text{aq})\) ions can reduce \(\text{I}_2(\text{aq})\) to \(\text{I}^-(\text{aq})\).
  2. B.\(\text{Fe}^{3+}(\text{aq})\) ions can oxidise \(\text{I}^-(\text{aq})\) to \(\text{I}_2(\text{aq})\).
  3. C.\(\text{I}_2(\text{aq})\) can oxidise \(\text{SO}_4^{2-}(\text{aq})\) to \(\text{S}_2\text{O}_8^{2-}(\text{aq})\).
  4. D.\(\text{SO}_4^{2-}(\text{aq})\) is a stronger oxidising agent than \(\text{S}_2\text{O}_8^{2-}(\text{aq})\).
Show answer & marking scheme

Worked solution

For statement B, the oxidation of \(\text{I}^-\right\) by \(\text{Fe}^{3+}\) has a cell potential of \(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{red}} - E^{\ominus}_{\text{ox}} = 0.77\text{ V} - 0.54\text{ V} = +0.23\text{ V}\). Since \(E^{\ominus}_{\text{cell}} > 0\), the reaction is thermodynamically feasible. Statements A and C are not feasible because their cell potentials are negative, and statement D is incorrect because \(\text{S}_2\text{O}_8^{2-}\) has the more positive potential, meaning it is the stronger oxidising agent.

Marking scheme

1 mark for identifying the correct thermodynamic feasibility statement (B).
Question 19 · Multiple Choice
1 marks
Which of the following describes the first step in the mechanism for the reaction of methylbenzene with a mixture of concentrated nitric acid and concentrated sulfuric acid?
  1. A.Electrophilic attack by \(\text{NO}_2^+\) on the methyl group of methylbenzene.
  2. B.Electrophilic attack by \(\text{NO}_2^+\) on the benzene ring of methylbenzene.
  3. C.Nucleophilic attack by \(\text{NO}_3^-\) on the benzene ring of methylbenzene.
  4. D.Free radical substitution of the methyl group by \(\text{NO}_2^{\bullet}\) radicals.
Show answer & marking scheme

Worked solution

The reaction of methylbenzene with concentrated \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\) is electrophilic aromatic substitution (nitration). The electrophile \(\text{NO}_2^+\) (nitronium ion) is generated first, which then attacks the high electron density delocalised pi system of the benzene ring to form a carbocation intermediate.

Marking scheme

1 mark for identifying electrophilic attack by nitronium ion on the benzene ring (B).
Question 20 · Multiple Choice
1 marks
Which sequence of reagents can be used to convert 1-bromobutane into butanone?
  1. A.Heat with \(\text{KCN}\) in aqueous ethanol, followed by reduction with \(\text{LiAlH}_4\), then treatment with \(\text{HNO}_2\).
  2. B.Heat with \(\text{KOH(aq)}\), followed by oxidation with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\).
  3. C.Heat with \(\text{KOH}\) in ethanol, followed by reaction with steam and an acid catalyst, then oxidation with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\).
  4. D.Heat with \(\text{NH}_3\) under pressure, followed by reaction with nitrous acid, then oxidation with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\).
Show answer & marking scheme

Worked solution

To convert 1-bromobutane (primary) into butanone (secondary ketone), we must relocate the functional group to the 2-position. This is achieved by: 1) Elimination using hot ethanolic \(\text{KOH}\) to form but-1-ene; 2) Acid-catalysed hydration of but-1-ene, which follows Markovnikov's rule to yield butan-2-ol as the major product; 3) Oxidation of the secondary alcohol butan-2-ol with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) under reflux to produce butanone.

Marking scheme

1 mark for identifying the correct multi-step synthetic pathway (C).
Question 21 · Multiple Choice
1 marks
An organic compound with the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) has four peaks in its \(^{13}\text{C}\) NMR spectrum and a singlet, a triplet and a quartet in its \(^1\text{H}\) NMR spectrum with an integration ratio of \(3:3:2\) respectively. Which of the following compounds is consistent with this data?
  1. A.Ethyl ethanoate
  2. B.Propyl methanoate
  3. C.Butanoic acid
  4. D.2-Methylpropanoic acid
Show answer & marking scheme

Worked solution

Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) has four distinct carbon environments (hence four peaks in \(^{13}\text{C}\) NMR). In \(^1\text{H}\) NMR, the methyl protons on the carbonyl (\(\text{CH}_3\text{CO}-\)) appear as a singlet (3H), the ester ethyl group gives a triplet (3H, \(\text{CH}_3\)) and a quartet (2H, \(\text{CH}_2\)). The integration ratio of singlet : triplet : quartet is therefore \(3:3:2\). This is perfectly consistent with the given data.

Marking scheme

1 mark for identifying Ethyl ethanoate as the matching compound (A).
Question 22 · multiple_choice
1 marks
A buffer solution is prepared at 298 K by mixing \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(15.0\text{ cm}^3\) of \(0.080\text{ mol dm}^{-3}\) sodium hydroxide.

What is the pH of the resulting buffer solution at 298 K?
  1. A.4.70
  2. B.4.84
  3. C.4.97
  4. D.5.11
Show answer & marking scheme

Worked solution

Step 1: Calculate initial moles of reactants.
Moles of propanoic acid, \(n(\text{HA}) = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\).
Moles of \(\text{NaOH}\), \(n(\text{OH}^-) = 0.0150\text{ dm}^3 \times 0.080\text{ mol dm}^{-3} = 1.20 \times 10^{-3}\text{ mol}\).

Step 2: Determine moles of species after reaction.
\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\).
Moles of \(\text{HA}\) remaining = \(2.50 \times 10^{-3} - 1.20 \times 10^{-3} = 1.30 \times 10^{-3}\text{ mol}\).
Moles of \(\text{A}^-\)[propanoate ions] formed = \(1.20 \times 10^{-3}\text{ mol}\).

Step 3: Calculate \([\text{H}^+]\).
\([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{1.30 \times 10^{-3}}{1.20 \times 10^{-3}} = 1.4625 \times 10^{-5}\text{ mol dm}^{-3}\).

Step 4: Calculate pH.
\(\text{pH} = -\log_{10}(1.4625 \times 10^{-5}) = 4.835 \approx 4.84\).

Marking scheme

1 mark for correct answer (B).
Question 23 · multiple_choice
1 marks
For the reaction \(A + B \rightarrow C\), the following rate data were obtained at a constant temperature:

| Experiment | \([A]\) / \(\text{mol dm}^{-3}\) | \([B]\) / \(\text{mol dm}^{-3}\) | Initial Rate / \(\text{mol dm}^{-3}\text{ s}^{-1}\) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | \(2.0 \times 10^{-4}\) |
| 2 | 0.20 | 0.10 | \(8.0 \times 10^{-4}\) |
| 3 | 0.20 | 0.20 | \(1.6 \times 10^{-3}\) |

What is the value and units of the rate constant, \(k\), for this reaction?
  1. A.\(0.20\text{ mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)
  2. B.\(0.20\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)
  3. C.\(2.0 \times 10^{-2}\text{ mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)
  4. D.\(2.0 \times 10^{-2}\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)
Show answer & marking scheme

Worked solution

Step 1: Determine reaction orders.
Comparing Experiment 1 and 2: \([B]\) is constant, \([A]\) doubles, initial rate increases by a factor of 4 (\(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). Therefore, order with respect to \(A\) is 2.
Comparing Experiment 2 and 3: \([A]\) is constant, \([B]\) doubles, initial rate doubles (\(8.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\)). Therefore, order with respect to \(B\) is 1.

Step 2: Write the rate equation.
\(\text{Rate} = k [A]^2 [B]\).

Step 3: Calculate the value of \(k\) using Experiment 1 data.
\(2.0 \times 10^{-4} = k (0.10)^2 (0.10) \Rightarrow 2.0 \times 10^{-4} = k (1.0 \times 10^{-3}) \Rightarrow k = 0.20\).

Step 4: Determine the units of \(k\).
\(\text{Units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

Marking scheme

1 mark for correct answer (A).
Question 24 · multiple_choice
1 marks
Which of the following statements about the reactions of aqueous transition metal ions is correct?
  1. A.Addition of excess concentrated hydrochloric acid to aqueous cobalt(II) ions produces a blue solution containing \([CoCl_4]^{2-}\).
  2. B.Addition of excess aqueous ammonia to aqueous copper(II) ions produces a pale blue precipitate of \([Cu(NH_3)_4(H_2O)_2]^{2+}\) which does not redissolve.
  3. C.Aqueous iron(III) ions are reduced to iron(II) ions upon addition of sodium carbonate solution, with the evolution of carbon dioxide gas.
  4. D.Addition of sodium hydroxide solution to aqueous chromium(III) ions produces a green precipitate of \(Cr(OH)_3\) which is insoluble in excess sodium hydroxide.
Show answer & marking scheme

Worked solution

a) Correct. Addition of concentrated hydrochloric acid leads to ligand substitution forming tetrahedral \([CoCl_4]^{2-}\), which is blue.
b) Incorrect. Aqueous copper(II) ions form a deep blue solution containing \([Cu(NH_3)_4(H_2O)_2]^{2+}\) when excess ammonia is added because the pale blue copper(II) hydroxide precipitate redissolves.
c) Incorrect. Iron(III) ions are not reduced; they undergo a hydrolysis reaction with carbonate ions to form a brown precipitate of \(Fe(OH)_3(H_2O)_3\) and carbon dioxide gas.
d) Incorrect. Chromium(III) hydroxide is amphoteric and dissolves in excess sodium hydroxide to form a dark green solution of \([Cr(OH)_6]^{3-}\).

Marking scheme

1 mark for correct answer (A).
Question 25 · multiple_choice
1 marks
For a particular chemical reaction, \(\Delta H^\ominus = -115\text{ kJ mol}^{-1}\) and \(\Delta S^\ominus = -145\text{ J K}^{-1}\text{ mol}^{-1}\).

At which temperatures is this reaction thermodynamically feasible?
  1. A.At temperatures below \(793\text{ K}\)
  2. B.At temperatures above \(793\text{ K}\)
  3. C.At temperatures below \(0.793\text{ K}\)
  4. D.The reaction is feasible at all temperatures
Show answer & marking scheme

Worked solution

A reaction is thermodynamically feasible when \(\Delta G^\ominus \le 0\).
\(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\).
Setting \(\Delta G^\ominus = 0\):
\(T = \frac{\Delta H^\ominus}{\Delta S^\ominus} = \frac{-115 \times 10^3\text{ J mol}^{-1}}{-145\text{ J K}^{-1}\text{ mol}^{-1}} = 793.1\text{ K}\).
Because \(\Delta S^\ominus\) is negative, the term \(-T\Delta S^\ominus\) becomes increasingly positive as temperature increases, making \(\Delta G^\ominus\) positive. Therefore, the reaction is only feasible at temperatures below \(793\text{ K}\).

Marking scheme

1 mark for correct answer (A).
Question 26 · multiple_choice
1 marks
In the mononitration of benzene using a mixture of concentrated nitric acid and concentrated sulfuric acid, what is the reactive electrophile and the primary role of the sulfuric acid?
  1. A.The electrophile is \(NO_2^+\) and sulfuric acid acts as a catalyst by protonating nitric acid.
  2. B.The electrophile is \(NO_2^-\) and sulfuric acid acts as an oxidizing agent.
  3. C.The electrophile is \(NO_3^-\) and sulfuric acid acts as a catalyst by deprotonating benzene.
  4. D.The electrophile is \(NO_2^+\) and sulfuric acid acts as a reducing agent.
Show answer & marking scheme

Worked solution

Concentrated sulfuric acid is a stronger acid than nitric acid and protonates it: \(HNO_3 + 2H_2SO_4 \rightleftharpoons NO_2^+ + H_3O^+ + 2HSO_4^-\). The reactive electrophile generated is the nitronium ion (\(NO_2^+\)). Sulfuric acid is regenerated at the end, acting as a catalyst.

Marking scheme

1 mark for correct answer (A).
Question 27 · multiple_choice
1 marks
Which of the following statements correctly describes a chemical test or trend involving Group 2 elements or their compounds?
  1. A.Magnesium hydroxide is highly soluble in water and forms a strongly alkaline solution.
  2. B.Sulfate solubility increases down the group, so barium sulfate is highly soluble.
  3. C.Barium hydroxide is the most soluble Group 2 hydroxide and is used to test for sulfate ions because it forms a precipitate.
  4. D.Acidified barium chloride solution is used to test for sulfate ions, and hydrochloric acid is added to prevent the formation of other insoluble barium salts such as barium carbonate.
Show answer & marking scheme

Worked solution

a) Incorrect; Magnesium hydroxide is sparingly soluble in water and forms a weakly alkaline suspension.
b) Incorrect; Sulfate solubility decreases down the group (barium sulfate is insoluble).
c) Incorrect; Barium chloride or barium nitrate is used to test for sulfate, not barium hydroxide (since we do not want to introduce alkaline hydroxide ions which could complicate precipitation).
d) Correct; Hydrochloric acid is added to react with and remove other ions like carbonate (\(\text{CO}_3^{2-}\)) or sulfite (\(\text{SO}_3^{2-}\)) which would also form a white precipitate with barium ions and give a false-positive result.

Marking scheme

1 mark for correct answer (D).
Question 28 · multiple_choice
1 marks
An organic compound has the molecular formula \(C_4H_8O_2\). Its proton (\(^1H\)) NMR spectrum shows four signals:

- A triplet at \(\delta = 1.0\text{ ppm}\) (integrating to 3 protons)
- A multiplet (sextet) at \(\delta = 1.7\text{ ppm}\) (integrating to 2 protons)
- A triplet at \(\delta = 4.1\text{ ppm}\) (integrating to 2 protons)
- A singlet at \(\delta = 8.0\text{ ppm}\) (integrating to 1 proton)

Which of the following is the correct IUPAC name for this compound?
  1. A.Propyl methanoate
  2. B.Methyl propanoate
  3. C.Ethyl ethanoate
  4. D.Butanoic acid
Show answer & marking scheme

Worked solution

The singlet at \(\delta = 8.0\text{ ppm}\) is characteristic of a formyl proton (\(H-\text{C}=\text{O}\)) of a methanoate ester.
The triplet at \(\delta = 4.1\text{ ppm}\) (2H) is consistent with a \(-\text{O}-\text{CH}_2-\text{R}\) group, split by an adjacent \(-\text{CH}_2-\).
The multiplet at \(\delta = 1.7\text{ ppm}\) (2H) is the middle \(-\text{CH}_2-\), split by both the adjacent \(-\text{O}-\text{CH}_2-\) and terminal \(-\text{CH}_3\).
The triplet at \(\delta = 1.0\text{ ppm}\) (3H) is the terminal methyl group \(-\text{CH}_3\).
Therefore, the compound is propyl methanoate (\(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)).

Marking scheme

1 mark for correct answer (A).
Question 29 · multiple_choice
1 marks
Consider the following multi-step organic synthesis:

\(\text{Propan-1-ol} \xrightarrow{\text{Step 1}} \text{Propanoic acid} \xrightarrow{\text{Step 2}} \text{Propanoyl chloride} \xrightarrow{\text{Step 3}} \text{N-methylpropanamide}\)

Which of the following sets of reagents is correct for each step?
  1. A.Step 1: Acidified \(K_2Cr_2O_7\) under reflux; Step 2: \(PCl_5\); Step 3: Excess \(CH_3NH_2\)
  2. B.Step 1: Acidified \(K_2Cr_2O_7\) under distillation; Step 2: \(HCl(aq)\); Step 3: Excess \(NH_3\)
  3. C.Step 1: Acidified \(KMnO_4\) under distillation; Step 2: \(NaCl\); Step 3: Excess \(CH_3NH_2\)
  4. D.Step 1: Concentrated \(H_2SO_4\); Step 2: \(PCl_5\); Step 3: Excess \((CH_3)_2NH\)
Show answer & marking scheme

Worked solution

Step 1 requires the oxidation of a primary alcohol (propan-1-ol) to a carboxylic acid (propanoic acid), which uses acidified potassium dichromate(VI) under reflux.
Step 2 requires converting a carboxylic acid to an acyl chloride, which is achieved using phosphorus(V) chloride (\(PCl_5\)) or thionyl chloride (\(SOCl_2\)).
Step 3 requires reacting the acyl chloride with a primary amine (methylamine, \(CH_3NH_2\)) to form an N-substituted amide (N-methylpropanamide).

Marking scheme

1 mark for correct answer (A).
Question 30 · Multiple Choice
1 marks
Which of the following transition metal complexes is chiral and can exist as a pair of optical isomers?
  1. A.cis-\([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\)
  2. B.trans-\([\text{Co}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2\text{Cl}_2]^+\)
  3. C.cis-\([\text{Co}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2\text{Cl}_2]^+\)
  4. D.fac-\([\text{Co}(\text{NH}_3)_3\text{Cl}_3]\)
Show answer & marking scheme

Worked solution

Optical isomerism occurs in octahedral complexes containing two or more bidentate ligands in a cis-arrangement. cis-\([\text{Co}(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2)_2\text{Cl}_2]^+\) lacks a plane of symmetry, making its mirror image non-superimposable (chiral). The other options contain either monodentate ligands with a plane of symmetry or a trans-arrangement of bidentate ligands, which introduces a plane of symmetry, making them achiral.

Marking scheme

[1 mark] C is selected as the only chiral complex showing optical isomerism.
Question 31 · Multiple Choice
1 marks
For the reaction \(2\text{P} + \text{Q} \rightarrow \text{R} + \text{S}\), the rate equation is: \(\text{Rate} = k[\text{P}][\text{Q}]^2\). In a first experiment, the initial concentrations of \(\text{P}\) and \(\text{Q}\) are both \(0.10\text{ mol dm}^{-3}\), and the initial rate is \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). The experiment is repeated under the same temperature with initial concentrations \([\text{P}] = 0.20\text{ mol dm}^{-3}\) and \([\text{Q}] = 0.05\text{ mol dm}^{-3}\). What is the new initial rate of reaction?
  1. A.\(3.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)
  2. B.\(6.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)
  3. C.\(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
  4. D.\(2.4 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
Show answer & marking scheme

Worked solution

Let the initial rate in the first experiment be \(\text{Rate}_1 = k[\text{P}]_1[\text{Q}]_1^2\) and in the second experiment be \(\text{Rate}_2 = k[\text{P}]_2[\text{Q}]_2^2\). Dividing the two expressions: \(\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{[\text{P}]_2}{[\text{P}]_1} \times \left(\frac{[\text{Q}]_2}{[\text{Q}]_1}\right)^2 = \left(\frac{0.20}{0.10}\right) \times \left(\frac{0.05}{0.10}\right)^2 = 2 \times (0.5)^2 = 0.5\). Therefore, \(\text{Rate}_2 = 1.2 \times 10^{-3} \times 0.5 = 6.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\).

Marking scheme

[1 mark] B is selected based on the correct application of the orders of reaction.
Question 32 · Multiple Choice
1 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(25.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting buffer solution?
  1. A.3.52
  2. B.4.57
  3. C.4.87
  4. D.5.17
Show answer & marking scheme

Worked solution

Initial moles: \(n(\text{HA}) = 0.0500 \times 0.200 = 0.0100\text{ mol}\); \(n(\text{OH}^-) = 0.0250 \times 0.200 = 0.0050\text{ mol}\). After neutralisation: \(n(\text{HA})_{\text{remaining}} = 0.0100 - 0.0050 = 0.0050\text{ mol}\) and \(n(\text{A}^-)_{\text{formed}} = 0.0050\text{ mol}\). Since \([\text{HA}] = [\text{A}^-]\), the buffer equation \([\text{H}^+] = K_a \frac{[\text{HA}]}{[\text{A}^-]}\) simplifies to \([\text{H}^+] = K_a\). Therefore, \(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

Marking scheme

[1 mark] C is selected by calculating the stoichiometric amounts of weak acid and conjugate base and converting the resulting [H+] to pH.
Question 33 · Multiple Choice
1 marks
Which of the following statements about the trends in properties of the Group 2 elements and their compounds is correct?
  1. A.The first ionisation energy of the elements increases down the group.
  2. B.The solubility of the Group 2 sulfates increases down the group.
  3. C.Magnesium hydroxide is more soluble in water than barium hydroxide.
  4. D.Barium metal reacts more vigorously with water than calcium metal.
Show answer & marking scheme

Worked solution

Reactivity of Group 2 metals with water increases down the group, so barium reacts more vigorously than calcium. Ionisation energy decreases down the group due to increased atomic radius and shielding. Sulfate solubility decreases down the group, while hydroxide solubility increases down the group.

Marking scheme

[1 mark] D is selected based on correct knowledge of Group 2 chemical trends.
Question 34 · Multiple Choice
1 marks
Consider the following three-step synthesis: 1. Propan-1-ol is heated with concentrated sulfuric acid to produce compound X. 2. Compound X is reacted with hydrogen bromide to form the major product, compound Y. 3. Compound Y is heated under reflux with aqueous potassium hydroxide to produce compound Z. What is the IUPAC name of compound Z?
  1. A.Propan-1-ol
  2. B.Propan-2-ol
  3. C.Propene
  4. D.2-bromopropane
Show answer & marking scheme

Worked solution

Step 1: Acid-catalysed dehydration of propan-1-ol yields propene (X). Step 2: Electrophilic addition of HBr to propene follows Markovnikov's rule to give 2-bromopropane (Y) as the major product. Step 3: Nucleophilic substitution of 2-bromopropane with aqueous KOH replaces the bromine atom to form propan-2-ol (Z).

Marking scheme

[1 mark] B is selected as the correct final product of this multi-step pathway.
Question 35 · Multiple Choice
1 marks
Consider the following standard electrode potentials: \(\text{Fe}^{3+}(\text{aq}) + e^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\theta = +0.77\text{ V}\); \(\text{I}_2(\text{aq}) + 2e^- \rightleftharpoons 2\text{I}^-(\text{aq}) \quad E^\theta = +0.54\text{ V}\); \(\text{Fe}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Fe}(\text{s}) \quad E^\theta = -0.44\text{ V}\). Which of the following redox reactions is thermodynamically feasible under standard conditions?
  1. A.\(2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq})\)
  2. B.\(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)
  3. C.\(\text{Fe}^{2+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow \text{Fe}(\text{s}) + \text{I}_2(\text{aq})\)
  4. D.The reaction of \(\text{Fe}^{2+}(\text{aq})\) with \(\text{I}_2(\text{aq})\) to produce \(\text{Fe}(\text{s})\) and \(\text{I}^-(\text{aq})\).
Show answer & marking scheme

Worked solution

A reaction is feasible if the overall cell potential \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} > 0\). For option B, \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\), we have \(E^\theta_{\text{cell}} = E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\theta(\text{I}_2/\text{I}^-) = +0.77 - (+0.54) = +0.23\text{ V}\). Since \(E^\theta_{\text{cell}}\) is positive, this reaction is feasible.

Marking scheme

[1 mark] B is selected as the only reaction with a positive overall cell potential.

Wondering how well you actually know this?

Thinka is an AI practice app for DSE students — unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Start Practising FreeSee pricing

Want more questions like this? Practice unlimited on Thinka — instant answers included.

Start Practising Free