An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 Cambridge International A Level Chemistry (9620) paper. Not affiliated with or reproduced from Cambridge.
Section CH01
Answer all questions in the spaces provided. Unit 1 assessment focusing on Inorganic 1 and Physical 1 content.
6 Question · 72 marks
Question 1 · Structured/Calculations
12 marks
A student prepared a sample of a hydrated copper(II) ammonium sulfate salt, \(\text{Cu}_x(\text{NH}_4)_y(\text{SO}_4)_z \cdot n\text{H}_2\text{O}\). To determine its formula, the student performed three analyses:
1. Determination of copper(II) ions (\(\text{Cu}^{2+}\)): A \(1.00\text{ g}\) sample of the salt was dissolved in water and treated with an excess of potassium iodide, \(\text{KI}\): \(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})\) The liberated iodine was titrated with \(0.100\text{ mol dm}^{-3}\) sodium thiosulfate solution, \(\text{Na}_2\text{S}_2\text{O}_3\), using starch indicator: \(\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})\) The titration required \(25.00\text{ cm}^3\) of the thiosulfate solution.
2. Determination of sulfate ions (\(\text{SO}_4^{2-}\)): Another \(1.00\text{ g}\) sample of the salt was dissolved in water and an excess of barium chloride solution, \(\text{BaCl}_2\), was added. The precipitate of barium sulfate, \(\text{BaSO}_4\), was filtered, washed, dried, and weighed. The mass of the dry precipitate was \(1.167\text{ g}\).
3. The remaining percentage by mass of the salt is due to ammonium ions (\(\text{NH}_4^+\)) and water of crystallisation (\(\text{H}_2\text{O}\)).
(a) Calculate the moles of \(\text{Cu}^{2+}\) ions in the \(1.00\text{ g}\) sample. (3 marks) (b) Calculate the moles of \(\text{SO}_4^{2-}\) ions in the \(1.00\text{ g}\) sample. (3 marks) (c) Given that the mole ratio of \(\text{Cu}^{2+} : \text{NH}_4^+ : \text{SO}_4^{2-}\) is \(1 : 2 : 2\), determine the percentage by mass of \(\text{H}_2\text{O}\) in the hydrated salt. (4 marks) (d) Hence, determine the value of \(n\) in the formula \(\text{Cu(NH}_4)_2(\text{SO}_4)_2 \cdot n\text{H}_2\text{O}\). (2 marks)
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Worked solution
For (a): Moles of \(\text{S}_2\text{O}_3^{2-}\) used = \(0.02500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\). From the equations, \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\), so the mole ratio is \(1:1\). Moles of \(\text{Cu}^{2+}\) = \(2.50 \times 10^{-3}\text{ mol}\).
For (b): Moles of \(\text{BaSO}_4\) = \(\frac{1.167\text{ g}}{233.4\text{ g mol}^{-1}} = 5.00 \times 10^{-3}\text{ mol}\). Since 1 mole of \(\text{BaSO}_4\) contains 1 mole of sulfate, moles of \(\text{SO}_4^{2-}\) = \(5.00 \times 10^{-3}\text{ mol}\).
For (c): Mass of \(\text{Cu}^{2+}\) in sample = \(2.50 \times 10^{-3}\text{ mol} \times 63.5\text{ g mol}^{-1} = 0.15875\text{ g}\). Mass of \(\text{SO}_4^{2-}\) in sample = \(5.00 \times 10^{-3}\text{ mol} \times 96.1\text{ g mol}^{-1} = 0.4805\text{ g}\). Using the ratio \(\text{Cu}^{2+} : \text{NH}_4^+ = 1 : 2\): Moles of \(\text{NH}_4^+\) = \(2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3}\text{ mol}\). Mass of \(\text{NH}_4^+\) = \(5.00 \times 10^{-3}\text{ mol} \times 18.0\text{ g mol}^{-1} = 0.0900\text{ g}\). Total mass of anhydrous components = \(0.15875 + 0.4805 + 0.0900 = 0.72925\text{ g}\). Mass of water of crystallisation = \(1.00\text{ g} - 0.72925\text{ g} = 0.27075\text{ g}\). Percentage of water = \(\frac{0.27075}{1.00} \times 100\% = 27.1\%\).
For (d): Moles of \(\text{H}_2\text{O}\) = \(\frac{0.27075\text{ g}}{18.0\text{ g mol}^{-1}} = 0.01504\text{ mol}\). Ratio \(n = \frac{\text{moles of }\text{H}_2\text{O}}{\text{moles of }\text{Cu}^{2+}} = \frac{0.01504}{2.50 \times 10^{-3}} = 6.02 \approx 6\). Therefore, \(n = 6\).
Marking scheme
Part (a): 1 mark: Calculation of moles of thiosulfate (\(2.50 \times 10^{-3}\text{ mol}\)). 1 mark: Deducing the 1:1 mole ratio between \(\text{Cu}^{2+}\) and \(\text{S}_2\text{O}_3^{2-}\). 1 mark: Correct calculation of moles of \(\text{Cu}^{2+}\) (\(2.50 \times 10^{-3}\text{ mol}\)).
Part (b): 1 mark: Correct \(M_r\) of barium sulfate (\(233.4\)). 1 mark: Calculation of moles of barium sulfate (\(5.00 \times 10^{-3}\text{ mol}\)). 1 mark: Deducing the 1:1 ratio to find moles of sulfate (\(5.00 \times 10^{-3}\text{ mol}\)).
Part (c): 1 mark: Correct calculation of masses of copper and sulfate in the sample (\(0.15875\text{ g}\) and \(0.4805\text{ g}\)). 1 mark: Deducing the moles (\(5.00 \times 10^{-3}\text{ mol}\)) and mass (\(0.0900\text{ g}\)) of ammonium ions. 1 mark: Determining mass of water (\(0.27075\text{ g}\)) by subtracting anhydrous mass from 1.00 g. 1 mark: Correct percentage of water (\(27.1\%\) or \(27.08\%\)). Accept range \(27.0 - 27.1\%\).
Part (d): 1 mark: Correct calculation of moles of water (\(0.0150\text{ mol}\)). 1 mark: Deducing that \(n = 6\) (must be an integer, allow ecf from previous parts).
Question 2 · Structured/Calculations
12 marks
A sample of selenium contains four isotopes: \(^{76}\text{Se}\), \(^{78}\text{Se}\), \(^{80}\text{Se}\), and \(^{82}\text{Se}\). A time-of-flight (TOF) mass spectrometer is used to analyse the sample.
(a) Describe how ions are formed in the electrospray ionisation process of a TOF mass spectrometer and write an equation for the ionisation of a selenium atom, \(\text{Se}\), by this method. (3 marks) (b) Explain how the ions are accelerated and why they travel at different speeds through the drift region. (3 marks) (c) The relative abundances of the four isotopes in the sample are: - \(^{76}\text{Se}\): \(9.0\%\) - \(^{78}\text{Se}\): \(23.5\%\) - \(^{80}\text{Se}\): \(49.5\%\) - \(^{82}\text{Se}\): \(18.0\%\) Calculate the relative atomic mass (\(A_r\)) of this sample of selenium. Give your answer to 2 decimal places. (3 marks) (d) In the TOF mass spectrometer, an ion of \(^{80}\text{Se}^+\) has a kinetic energy of \(1.35 \times 10^{-13}\text{ J}\). The drift tube has a length of \(1.40\text{ m}\). Calculate the time of flight of this ion in the drift tube. The Avogadro constant, \(L = 6.022 \times 10^{23}\text{ mol}^{-1}\). (Recall: \(KE = \frac{1}{2}mv^2\)) (3 marks)
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Worked solution
For (a): In electrospray ionisation, the sample is dissolved in a volatile solvent and injected through a fine needle attached to a high voltage power supply. The particles gain a proton (\(\text{H}^+\)) from the solvent as they leave the needle, forming positive ions. Equation: \(\text{Se} + \text{H}^+ \rightarrow \text{SeH}^+\) (or \(\text{Se}(\text{g}) + \text{H}^+\text{(aq)} \rightarrow \text{SeH}^+\text{(g)}\)).
For (b): Ions are accelerated by an electric field to have the same kinetic energy. Since \(KE = \frac{1}{2}mv^2\), velocity is inversely proportional to the square root of the mass (\(v = \sqrt{\frac{2KE}{m}}\)). Therefore, lighter ions travel faster and reach the detector first, while heavier ions travel slower.
For (d): Mass of 1 mole of \(^{80}\text{Se}^+\) ions is \(80\text{ g} = 0.080\text{ kg}\). Mass of one ion: \(m = \frac{0.080\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 1.3285 \times 10^{-25}\text{ kg}\). Rearranging \(KE = \frac{1}{2}mv^2\) for velocity \(v\): \(v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1.35 \times 10^{-13}\text{ J}}{1.3285 \times 10^{-25}\text{ kg}}} = 1.4256 \times 10^6\text{ m s}^{-1}\). Time of flight: \(t = \frac{d}{v} = \frac{1.40\text{ m}}{1.4256 \times 10^6\text{ m s}^{-1}} = 9.82 \times 10^{-7}\text{ s}\).
Marking scheme
Part (a): 1 mark: Sample dissolved in volatile solvent and forced through a high voltage nozzle/needle. 1 mark: Gained a proton (\(\text{H}^+\)) to form positive ions. 1 mark: Correct equation: \(\text{Se} + \text{H}^+ \rightarrow \text{SeH}^+\).
Part (b): 1 mark: Accelerated by an electric field. 1 mark: State that all ions are accelerated to the same kinetic energy. 1 mark: Velocity depends on mass / lighter ions travel faster than heavier ions.
Part (c): 1 mark: Correct setup of the weighted average equation. 1 mark: Correct calculation of total mass (7953). 1 mark: Correct division to give 79.53 (must be to 2 decimal places).
Part (d): 1 mark: Correct calculation of the mass of a single selenium-80 ion in kg (\(1.33 \times 10^{-25}\text{ kg}\)). 1 mark: Correct calculation of velocity (\(1.43 \times 10^6\text{ m s}^{-1}\)). 1 mark: Correct calculation of time of flight (\(9.82 \times 10^{-7}\text{ s}\)). Accept \(9.8 \times 10^{-7}\text{ s}\) to \(9.85 \times 10^{-7}\text{ s}\).
Question 3 · Structured/Calculations
12 marks
A student determined the enthalpy change of solution for anhydrous calcium chloride, \(\text{CaCl}_2(\text{s})\), by calorimetry.
(a) The student added \(6.00\text{ g}\) of \(\text{CaCl}_2(\text{s})\) to \(80.0\text{ g}\) of water at \(19.5\text{ }^\circ\text{C}\). The temperature increased to a maximum of \(32.8\text{ }^\circ\text{C}\). Calculate the heat energy released, \(q\), in Joules. Assume that the specific heat capacity of the solution is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\) and its mass is the sum of the water and the dissolved salt. (2 marks) (b) Calculate the enthalpy change of solution, \(\Delta H_{\text{sol}}\), of \(\text{CaCl}_2(\text{s})\) in \(\text{kJ mol}^{-1}\). (3 marks) (c) The student also wants to determine the enthalpy of hydration of anhydrous \(\text{CaCl}_2\) to form hydrated calcium chloride, \(\text{CaCl}_2 \cdot 6\text{H}_2\text{O}(\text{s})\). The enthalpy of solution of \(\text{CaCl}_2 \cdot 6\text{H}_2\text{O}(\text{s})\) is \(+19.0\text{ kJ mol}^{-1}\). The enthalpy of hydration of \(\text{CaCl}_2(\text{s})\) is represented by: \(\text{CaCl}_2(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow \text{CaCl}_2 \cdot 6\text{H}_2\text{O}(\text{s}) \quad \Delta H_{\text{hyd}}\) Construct a Hess's Law cycle and calculate the enthalpy of hydration, \(\Delta H_{\text{hyd}}\), of anhydrous calcium chloride. (3 marks) (d) In another experiment, the enthalpy of combustion of a liquid alcohol, propan-1-ol, was determined using a spirit burner. State why the experimental value obtained is significantly less exothermic than the literature value. Give two reasons. (2 marks) (e) Define the term standard enthalpy of formation. (2 marks)
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Worked solution
For (a): Mass of solution, \(m = 80.0 + 6.00 = 86.0\text{ g}\). Temperature change, \(\Delta T = 32.8 - 19.5 = 13.3\text{ K}\). Heat energy, \(q = m \cdot c \cdot \Delta T = 86.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 13.3\text{ K} = 4781.086\text{ J}\).
For (b): Moles of \(\text{CaCl}_2\) = \(\frac{6.00\text{ g}}{111.1\text{ g mol}^{-1}} = 0.054005\text{ mol}\). Enthalpy change of solution, \(\Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{4.781086\text{ kJ}}{0.054005\text{ mol}} = -88.53\text{ kJ mol}^{-1}\) (or \(-88.5\text{ kJ mol}^{-1}\)).
For (c): The cycle relates \(\text{CaCl}_2(\text{s}) + 6\text{H}_2\text{O}(\text{l})\) to the aqueous solution of \(\text{CaCl}_2\). \(\Delta H_{\text{sol}}(\text{CaCl}_2(\text{s})) = \Delta H_{\text{hyd}} + \Delta H_{\text{sol}}(\text{CaCl}_2 \cdot 6\text{H}_2\text{O}(\text{s}))\). \(-88.5 = \Delta H_{\text{hyd}} + (+19.0)\). \(\Delta H_{\text{hyd}} = -88.5 - 19.0 = -107.5\text{ kJ mol}^{-1}\).
For (d): Any two of: 1. Heat loss to the surroundings / beaker. 2. Incomplete combustion of propan-1-ol. 3. Evaporation of alcohol from the wick. 4. Non-standard conditions used.
For (e): The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (100 kPa and 298 K).
Marking scheme
Part (a): 1 mark: Mass of solution correctly identified as 86.0 g. 1 mark: Calculation of heat energy as 4781 J (accept 4780 J or 4.78 kJ).
Part (b): 1 mark: Correct calculation of moles of calcium chloride (0.0540 mol). 1 mark: Correct conversion of heat energy to kJ and division by moles. 1 mark: Correct answer of -88.5 kJ mol^-1 (must include negative sign and units). Accept range -88.3 to -88.6.
Part (c): 1 mark: Drawing or describing a correct Hess's Law cycle. 1 mark: Showing correct algebraic relationship (\(\Delta H_{\text{hyd}} = \Delta H_{\text{sol}}(\text{anhydrous}) - \Delta H_{\text{sol}}(\text{hydrated})\)). 1 mark: Correct calculation of \(-107.5\text{ kJ mol}^{-1}\) (or value consistent with part b; must be negative).
Part (d): 1 mark each for any two valid reasons (maximum 2 marks): - Heat loss to surroundings/beaker. - Incomplete combustion of alcohol. - Evaporation of alcohol from wick. - Non-standard conditions.
Part (e): 1 mark: Enthalpy change when 1 mole of a compound is formed from its constituent elements. 1 mark: Reference to elements in their standard states under standard conditions.
Question 4 · Structured/Calculations
12 marks
This question is about Group 2 elements and their compounds.
(a) Write an equation for the reaction of magnesium with steam. State one observation for this reaction. (2 marks) (b) Describe the trend in the solubility of Group 2 hydroxides down the group. Hence, write an ionic equation for the reaction that occurs when aqueous sodium hydroxide is added to a solution of magnesium chloride, and state what you would observe. (3 marks) (c) A student is given a solution containing an unknown Group 2 halide. The student adds dilute hydrochloric acid followed by barium chloride solution. No precipitate is observed. The student then adds dilute nitric acid followed by silver nitrate solution. A cream precipitate is formed. (i) Explain why the student adds dilute hydrochloric acid before adding barium chloride solution, and identify the ion that is shown to be absent by the lack of precipitate. (2 marks) (ii) Identify the halide ion present in the solution, and write the simplest ionic equation (including state symbols) for the formation of the cream precipitate. (2 marks) (d) Barium sulfate is highly toxic if ingested, yet it is used safely in medicine as a 'barium meal' to image the digestive system. (i) Explain why barium sulfate can be safely swallowed for this purpose. (1 mark) (ii) Write an equation for the reaction used to remove sulfur dioxide (\(\text{SO}_2\)) from flue gases using calcium oxide. (2 marks)
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Worked solution
For (a): Equation: \(\text{Mg(s)} + \text{H}_2\text{O(g)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)}\) Observation: White light / white solid (powder) formed.
For (b): Trend: Solubility of Group 2 hydroxides increases down the group. Ionic equation: \(\text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s)}\) Observation: White precipitate.
For (c): (i) Dilute hydrochloric acid is added to react with and remove any carbonate (\(\text{CO}_3^{2-}\)) or sulfite (\(\text{SO}_3^{2-}\)) impurities which would otherwise form a white precipitate with barium ions. The absent ion is sulfate (\(\text{SO}_4^{2-}\)). (ii) The halide is bromide (\(\text{Br}^-\)). Equation: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr(s)}\).
For (d): (i) Barium sulfate is extremely insoluble, so it is not absorbed into tissues or the bloodstream (no free toxic barium ions are released). (ii) Equation: \(\text{CaO(s)} + \text{SO}_2(\text{g}) \rightarrow \text{CaSO}_3(\text{s})\).
Marking scheme
Part (a): 1 mark: Correct balanced equation: \(\text{Mg} + \text{H}_2\text{O} \rightarrow \text{MgO} + \text{H}_2\). 1 mark: Bright white light or white powder/solid.
Part (b): 1 mark: Correct trend (solubility increases down the group). 1 mark: Correct ionic equation: \(\text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s})\). 1 mark: White precipitate.
Part (c): (i) 1 mark: To remove carbonate/sulfite ions (which would form a false positive precipitate of barium carbonate/sulfite). 1 mark: Sulfate ion (\(\text{SO}_4^{2-}\)). (ii) 1 mark: Bromide ion (\(\text{Br}^-\)). 1 mark: Correct ionic equation with state symbols: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr(s)}\).
Part (d): (i) 1 mark: Barium sulfate is insoluble. (ii) 2 marks: Correct equation: \(\text{CaO} + \text{SO}_2 \rightarrow \text{CaSO}_3\) (allow 1 mark for correct reactants and products but unbalanced, or wrong state symbols).
Question 5 · Structured/Calculations
12 marks
This question is about the halogens and their halide compounds.
(a) Chlorine reacts with cold, dilute aqueous sodium hydroxide to form a solution used as bleach. Write an equation for this reaction and state the oxidation states of chlorine in the chlorine-containing products. Explain why this is a disproportionation reaction. (3 marks) (b) Write an equation for the reaction of chlorine with water in the presence of sunlight. State one commercial use of chlorine and explain why this use is justified despite chlorine's toxicity. (3 marks) (c) Solid sodium bromide reacts with concentrated sulfuric acid in a redox reaction. (i) Write a balanced equation for this reaction. (2 marks) (ii) State two observations that would be made during this reaction and identify the species responsible for each observation. (2 marks) (d) Contrast the reducing ability of fluoride ions with iodide ions. Refer to their reactions with concentrated sulfuric acid to support your answer. (2 marks)
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Worked solution
For (a): Equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) Oxidation states: \(-1\) in \(\text{NaCl}\) and \(+1\) in \(\text{NaClO}\). Disproportionation: Chlorine is simultaneously oxidised (from 0 to +1 in \(\text{NaClO}\)) and reduced (from 0 to -1 in \(\text{NaCl}\)) in the same reaction.
For (b): Equation: \(2\text{Cl}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HCl} + \text{O}_2\) Use: Water treatment / chlorination of drinking water / sanitation of swimming pools. Justification: The benefits of killing disease-causing pathogens outweigh the health risks associated with toxic chlorine gas.
For (c): (i) \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}\). (ii) Any two observations with correct species: - Orange/brown gas/vapour due to \(\text{Br}_2\). - Pungent/choking gas due to \(\text{SO}_2\). - White misty fumes due to \(\text{HBr}\).
For (d): Iodide ions are much stronger reducing agents than fluoride ions. Fluoride ions are poor reducing agents and do not reduce concentrated sulfuric acid (only acid-base reaction occurs, forming \(\text{HF}\)), whereas iodide ions readily reduce sulfur in sulfuric acid from \(+6\) to \(+4\) (\(\text{SO}_2\)), \(0\) (\(\text{S}\)), and \(-2\) (\(\text{H}_2\text{S}\)).
Marking scheme
Part (a): 1 mark: Correct equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). 1 mark: Correct oxidation states: -1 in NaCl and +1 in NaClO. 1 mark: Simultaneously oxidised and reduced in the same reaction.
Part (b): 1 mark: Correct equation: \(2\text{Cl}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HCl} + \text{O}_2\). 1 mark: Correct use (water treatment / purification). 1 mark: Justification (benefits of killing bacteria outweigh risks of toxicity).
Part (c): (i) 2 marks: Correct balanced equation (1 mark for reactants and products, 1 mark for balancing: \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}\)). (ii) 2 marks: Two observations with correct chemical identities (1 mark each): - Orange/brown fumes/liquid due to \(\text{Br}_2\). - Choking gas/pungent gas due to \(\text{SO}_2\). - White misty fumes due to \(\text{HBr}\).
Part (d): 1 mark: Iodide is a stronger reducing agent than fluoride (or converse). 1 mark: Fluoride does not reduce sulfuric acid (acid-base only), whereas iodide reduces it to \(\text{SO}_2\)/\(\text{S}\)/\(\text{H}_2\text{S}\).
Question 6 · Structured/Calculations
12 marks
This question is about molecular shapes, polarity, and physical properties of compounds.
(a) Phosphorus forms two common chlorides, \(\text{PCl}_3\) and \(\text{PCl}_5\). (i) Predict the shape of a \(\text{PCl}_3\) molecule. Use valence shell electron pair repulsion (VSEPR) theory to explain your answer. (3 marks) (ii) The \(\text{PCl}_4^+\) ion is found in solid phosphorus pentachloride. Predict the shape of the \(\text{PCl}_4^+\) ion and state its bond angle. (2 marks) (b) Explain why phosphorus trichloride, \(\text{PCl}_3\), has a dipole moment (is polar), whereas phosphorus pentachloride, \(\text{PCl}_5\), does not have a dipole moment (is non-polar). (3 marks) (c) Silicon dioxide (\(\text{SiO}_2\)) and sulfur trioxide (\(\text{SO}_3\)) are both oxides of Period 3 elements. Compare the structure and bonding in silicon dioxide and sulfur trioxide. Explain how the differences in structure and bonding lead to the differences in their melting points. (4 marks)
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Worked solution
For (a): (i) Shape of \(\text{PCl}_3\): Trigonal pyramidal. Explanation: Phosphorus has 5 valence electrons. It forms 3 single bonds to chlorine, leaving 1 lone pair. There are 4 electron pairs around the central phosphorus atom. To minimise repulsion, the electron pairs adopt a tetrahedral arrangement. Since lone pairs repel more than bonding pairs, the bond angle is compressed, resulting in a trigonal pyramidal molecular geometry. (ii) Shape of \(\text{PCl}_4^+\): Tetrahedral. Bond angle: \(109.5^\circ\).
For (b): In \(\text{PCl}_3\), the molecule has an unsymmetrical (trigonal pyramidal) geometry due to the lone pair on phosphorus. Thus, the individual polar \(\text{P-Cl}\) bonds do not cancel, resulting in a net dipole moment (polar). In \(\text{PCl}_5\), the molecule has a highly symmetrical (trigonal bipyramidal) geometry with no lone pairs. The individual polar \(\text{P-Cl}\) bond dipoles cancel out completely, giving no net dipole moment (non-polar).
For (c): - Silicon dioxide (\(\text{SiO}_2\)) has a giant covalent macromolecular structure with strong covalent bonds between silicon and oxygen atoms throughout the lattice. - Sulfur trioxide (\(\text{SO}_3\)) has a simple molecular structure with weak intermolecular van der Waals / dipole-dipole forces holding the molecules together. - Melting silicon dioxide requires a large amount of energy to break many strong covalent bonds. - Melting sulfur trioxide requires relatively little energy to overcome the weak intermolecular forces between the molecules, while the strong covalent bonds within the molecules remain intact. Therefore, silicon dioxide has a much higher melting point than sulfur trioxide.
Marking scheme
Part (a): (i) 1 mark: Trigonal pyramidal. 1 mark: P has 3 bonding pairs and 1 lone pair (or 4 electron pairs in total). 1 mark: Lone pairs repel more than bonding pairs (reducing the bond angle). (ii) 1 mark: Tetrahedral. 1 mark: \(109.5^\circ\) (allow \(109^\circ\) to \(109.5^\circ\)).
Part (b): 1 mark: Chlorine is more electronegative than phosphorus, so P-Cl bonds are polar. 1 mark: In \(\text{PCl}_3\), the shape is unsymmetrical, so individual dipoles do not cancel (polar). 1 mark: In \(\text{PCl}_5\), the shape is symmetrical, so the individual dipoles cancel (non-polar).
Part (c): 1 mark: \(\text{SiO}_2\) is giant covalent / macromolecular and \(\text{SO}_3\) is simple molecular. 1 mark: Forces between \(\text{SO}_3\) molecules are weak intermolecular forces (van der Waals / London dispersion / dipole-dipole). 1 mark: Melting \(\text{SiO}_2\) requires breaking strong covalent bonds (which require high energy). 1 mark: Melting \(\text{SO}_3\) only requires overcoming weak intermolecular forces (which require low energy).
Section CH02
Answer all questions in the spaces provided. Unit 2 assessment focusing on Organic 1 and Physical 1 content.
8 Question · 70 marks
Question 1 · Structured
8.75 marks
A sample of gas is placed in a closed vessel at temperature \(T_1\).
(a) Sketch a Maxwell-Boltzmann distribution of molecular energies for this sample. Label the axes, the most probable energy (\(E_{mp}\)), the mean energy (\(E_{mean}\)), and the activation energy (\(E_a\)).
(b) Add a second curve representing the distribution at a higher temperature \(T_2\).
(c) Explain, in terms of collision theory and your diagram, how a catalyst increases the rate of reaction.
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Worked solution
Diagram details: (a) The y-axis is labeled 'Number of molecules' (or 'Fraction of molecules') and the x-axis is labeled 'Energy'. The curve starts at the origin (0,0), rises to a peak, and then tails off to the right, approaching but never touching the x-axis. \(E_{mp}\) is located slightly to the left of the peak, and \(E_{mean}\) is located slightly to the right of the peak. \(E_a\) is marked on the right tail. (b) The curve for \(T_2\) has a lower peak shifted to the right, crosses the \(T_1\) curve once, and is higher than the \(T_1\) curve at high energies. (c) A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_c\)). Because \(E_c\) is lower than \(E_a\), a larger fraction of molecules have energy greater than or equal to the activation energy. This increases the frequency of successful collisions, thereby increasing the rate of reaction.
Marking scheme
1.75 marks: Correctly labeled axes and general shape of the \(T_1\) curve starting at the origin and not touching the x-axis. 2.00 marks: \(E_{mp}\) marked at the peak, and \(E_{mean}\) marked to the right of the peak. 1.00 mark: \(E_a\) marked on the x-axis to the right of the peak. 1.75 marks: \(T_2\) curve peak is lower and shifted to the right, crosses \(T_1\) once, and is higher at high energy. 2.25 marks: Explanation of catalyst: provides alternative pathway with lower \(E_a\) (1.25 marks); more molecules have energy \(\ge E_a\), increasing rate (1.00 mark).
Question 2 · Structured
8.75 marks
2-Bromobutane reacts with aqueous sodium hydroxide to form butan-2-ol.
(a) Name the mechanism for this reaction.
(b) Outline the mechanism for this reaction, using curly arrows to show the movement of electron pairs and showing any partial charges on the relevant bonds.
(c) Explain why 2-iodobutane reacts faster than 2-bromobutane under the same conditions.
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Worked solution
(a) Nucleophilic substitution. (b) 2-bromobutane has a \(\delta+\) carbon bonded to a \(\delta-\) bromine. A curly arrow starts from the lone pair on the oxygen of the hydroxide ion (\(OH^-\)) and points to the \(\delta+\) carbon. A second curly arrow starts from the C-Br bond and points to the bromine atom. This leads to the formation of butan-2-ol and a bromide ion (\(Br^-\)). (c) The carbon-iodine (C-I) bond is weaker than the carbon-bromine (C-Br) bond. This is because iodine has a larger atomic radius, resulting in a longer, weaker bond. Less energy is required to break the C-I bond, so the rate of reaction is faster.
Marking scheme
1.00 mark: Mechanism name: Nucleophilic substitution. 4.75 marks: Mechanism details: 1.25 marks for curly arrow from \(OH^-\). lone pair to C \(\delta+\); 1.00 mark for \(\delta+\) on C and \(\delta-\). on Br; 1.25 marks for curly arrow from C-Br bond to Br; 1.25 marks for correct structures. 3.00 marks: Explanation: C-I bond is weaker than C-Br bond (1.50 marks), due to larger atomic radius/longer bond (0.75 marks), requiring less energy to break (0.75 marks).
Question 3 · Structured
8.75 marks
Propene reacts with hydrogen bromide at room temperature to form a mixture of two isomeric halogenoalkanes.
(a) Outline the mechanism for the reaction that forms the major product, showing curly arrows and partial/formal charges.
(b) Identify the major product and explain why it is formed in preference to the minor product.
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Worked solution
(a) Propene (\(CH_3CH=CH_2\)) reacts via electrophilic addition. The hydrogen bromide molecule has \(\delta+\) on H and \(\delta-\) on Br. A curly arrow goes from the double bond of propene to the H of H-Br. A second curly arrow goes from the H-Br bond to the Br. This forms a secondary carbocation intermediate: \(CH_3CH(+)CH_3\). A curly arrow then goes from the lone pair on the bromide ion (\(Br^-\)) to the positive carbon atom, forming 2-bromopropane. (b) The major product is 2-bromopropane. It is formed via a secondary carbocation intermediate, which is more stable than the primary carbocation intermediate that leads to 1-bromopropane. The secondary carbocation is stabilized by the positive inductive effect of two electron-releasing methyl groups, compared to only one in the primary carbocation.
Marking scheme
4.75 marks: Mechanism: 1.25 marks for curly arrow from C=C to H \(\delta+\) and showing charges; 1.00 mark for curly arrow from H-Br bond to Br; 1.25 marks for structure of secondary carbocation; 1.25 marks for curly arrow from \(Br^-\) lone pair to C+. 4.00 marks: Explanation: 1.00 mark for 2-bromopropane; 1.00 mark for secondary carbocation intermediate; 1.00 mark for stability comparison; 1.00 mark for positive inductive effect of two alkyl groups.
Question 4 · Structured
8.75 marks
In a calorimetry experiment, a student added 5.00 g of anhydrous copper(II) sulfate (\(M_r\) = 159.6) to 50.0 g of water. The temperature of the water increased from 19.5 \(^\circ\)C to 28.2 \(^\circ\)C.
(a) Calculate the heat energy, q, in Joules released. (Assume the specific heat capacity is 4.18 J K\(^{-1}\) g\(^{-1}\) and the density of water is 1.00 g cm\(^{-3}\)).
(b) Calculate the enthalpy change of solution, \(\Delta H\), in kJ mol\(^{-1}\), for anhydrous copper(II) sulfate. Include the sign.
(c) Suggest one major source of experimental error and a method to reduce it.
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Worked solution
(a) \(q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J K}^{-1}\text{ g}^{-1} \times (28.2 - 19.5)\text{ K} = 50.0 \times 4.18 \times 8.7 = 1818.3\text{ J}\). (If using mass of solution = 55.0 g: \(q = 55.0 \times 4.18 \times 8.7 = 2000.1\text{ J}\)). (b) Moles of \(CuSO_4\) = \(5.00\text{ g} / 159.6\text{ g mol}^{-1} = 0.03133\text{ mol}\). \(\Delta H = - (q / 1000) / \text{moles}\). For \(q = 1818.3\text{ J}\): \(\Delta H = -1.8183 / 0.03133 = -58.0\text{ kJ mol}^{-1}\). For \(q = 2000.1\text{ J}\): \(\Delta H = -2.0001 / 0.03133 = -63.8\text{ kJ mol}^{-1}\). (c) Heat loss to the surroundings is the main error. This can be reduced by using a lid on the polystyrene cup or placing the cup inside a beaker with cotton wool insulation.
Marking scheme
2.75 marks for part (a): \(\Delta T\) = 8.7 \(^\circ\)C (0.75 mark), substitution (1.00 mark), correct q (1.00 mark). 3.00 marks for part (b): Moles = 0.03133 mol (1.00 mark), conversion to kJ/mol (1.00 mark), correct value with negative sign (1.00 mark). 3.00 marks for part (c): Identifying heat loss (1.50 marks), suggesting a lid/insulation (1.50 marks).
Question 5 · Structured
8.75 marks
An organic compound X (molecular formula \(C_4H_{10}O\)) is heated under reflux with acidified potassium dichromate(VI) to form compound Y, which turns blue litmus paper red.
(a) Deduce the structures of X and Y.
(b) Describe how the apparatus is modified to obtain an aldehyde instead of a carboxylic acid when oxidizing a primary alcohol.
(c) Describe a chemical test to distinguish between an aldehyde and a ketone, including reagents and observations.
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Worked solution
(a) Since \(C_4H_{10}O\) is oxidized under reflux to a carboxylic acid (Y, which turns blue litmus red), X must be a primary alcohol. Possible structures for X: butan-1-ol (\(CH_3CH_2CH_2CH_2OH\)) or 2-methylpropan-1-ol (\((CH_3)_2CHCH_2OH\)). Corresponding structures for Y: butanoic acid (\(CH_3CH_2CH_2COOH\)) or 2-methylpropanoic acid (\((CH_3)_2CHCOOH\)). (b) To obtain the aldehyde, use distillation apparatus instead of reflux. Gaseous aldehyde is distilled off immediately as it forms, preventing further oxidation. This is possible because the aldehyde has a lower boiling point than the alcohol. (c) Add Tollens' reagent and warm. Aldehyde: silver mirror forms on the walls of the tube. Ketone: no change. Alternatively, use Fehling's solution and warm: Aldehyde: blue solution forms a red precipitate. Ketone: no change.
Marking scheme
3.00 marks for part (a): Structure of X (1.50 marks), structure of Y (1.50 marks). 2.75 marks for part (b): Distillation (1.00 mark), distilling off aldehyde immediately (1.00 mark), based on lower boiling point (0.75 mark). 3.00 marks for part (c): Reagent (1.00 mark), aldehyde result (1.00 mark), ketone result (1.00 mark).
Question 6 · Structured
8.75 marks
A mixture of 2.00 mol of ethyl ethanoate, 2.00 mol of water, 1.00 mol of ethanoic acid, and 1.00 mol of ethanol reached equilibrium in a closed vessel of volume V dm\(^3\) at a constant temperature: \(CH_3COOCH_2CH_3(l) + H_2O(l) \rightleftharpoons CH_3COOH(l) + CH_3CH_2OH(l)\). At equilibrium, 1.45 mol of ethanoic acid was present.
(a) Write the \(K_c\) expression.
(b) Calculate the equilibrium moles of all reactants and products.
(c) Calculate the value of \(K_c\) and explain why volume V is not needed.
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Worked solution
(a) \(K_c = \frac{[CH_3COOH][CH_3CH_2OH]}{[CH_3COOCH_2CH_3][H_2O]}\). (b) Since the initial amount of ethanoic acid was 1.00 mol and it increased to 1.45 mol, the change in moles of ethanoic acid is +0.45 mol. By stoichiometry (1:1:1:1), ethanol increases by +0.45 mol to 1.45 mol. Ethyl ethanoate and water both decrease by 0.45 mol to 1.55 mol each. (c) Substituting into the \(K_c\) expression: \(K_c = \frac{(1.45/V) \times (1.45/V)}{(1.55/V) \times (1.55/V)} = \frac{1.45 \times 1.45}{1.55 \times 1.55} = 0.875\). Volume V is not required because there are two mole terms in both the numerator and denominator, which cancels out the volume terms.
Marking scheme
1.25 marks for part (a): Correct \(K_c\) expression. 3.00 marks for part (b): 1.00 mark for change calculation (+0.45 mol), 1.00 mark for 1.55 mol for reactants, 1.00 mark for 1.45 mol for ethanol. 4.50 marks for part (c): Correct substitutions (1.50 marks), \(K_c\) calculation = 0.875 (1.50 marks), explanation of volume cancellation due to equal moles (1.50 marks).
Question 7 · Structured
8.75 marks
Chlorine reacts with methane in the presence of UV light.
(a) State the role of UV light and write an equation for the initiation step.
(b) Write equations for the two propagation steps to form chloromethane.
(c) Write an equation for a termination step that produces ethane.
(d) Explain why a large excess of methane is used to maximize the yield of chloromethane.
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Worked solution
(a) UV light provides the energy required to break the Cl-Cl bond homolytically (homolytic fission). Initiation step: \(Cl_2 \rightarrow 2Cl^\bullet\). (b) First propagation step: \(CH_4 + Cl^\bullet \rightarrow CH_3^\bullet + HCl\). Second propagation step: \(CH_3^\bullet + Cl_2 \rightarrow CH_3Cl + Cl^\bullet\). (c) Termination step producing ethane: \(2CH_3^\bullet \rightarrow C_2H_6\). (d) An excess of methane ensures that chlorine radicals are far more likely to collide with methane molecules than with the chloromethane product. This minimizes further substitution reactions (which would yield dichloromethane, trichloromethane, etc.) and thus maximizes the yield of chloromethane.
Marking scheme
2.00 marks for part (a): Homolytic fission of Cl-Cl (1.00 mark), initiation equation (1.00 mark). 2.50 marks for part (b): First propagation (1.25 marks), second propagation (1.25 marks). 1.25 marks for part (c): Termination to form ethane (1.25 marks). 3.00 marks for part (d): Collision probability explanation (1.50 marks), prevention of further substitution (1.50 marks).
Question 8 · Structured
8.75 marks
An unknown organic compound Z has the molecular formula \(C_3H_6O\).
(a) Identify the infrared absorption range in the spectrum of Z that would distinguish whether Z is an alcohol or a carbonyl compound, stating the bond responsible for each.
(b) The spectrum of Z has a strong, sharp peak at 1715 cm\(^{-1}\) but no peak at 3230-3550 cm\(^{-1}\). Name the functional group present.
(c) The mass spectrum of Z shows a molecular ion peak at m/z = 58 and a major fragment peak at m/z = 43. Deduce the structural formula of Z and write an equation for the fragmentation that produces the peak at m/z = 43.
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Worked solution
(a) An alcohol is distinguished by a broad O-H absorption in the 3230-3550 cm\(^{-1}\) range. A carbonyl compound is distinguished by a strong, sharp C=O absorption in the 1680-1750 cm\(^{-1}\) range. (b) The presence of a peak at 1715 cm\(^{-1}\) (C=O) and the absence of a peak at 3230-3550 cm\(^{-1}\) (O-H) shows that Z contains a carbonyl group (specifically a ketone or aldehyde). (c) Z has molecular formula \(C_3H_6O\) and a carbonyl group. Since it must have a molecular ion peak at 58, and a major fragment at 43 (loss of 15, which is a \(CH_3\) group), Z is propanone, \(CH_3COCH_3\). The fragmentation equation is: \([CH_3COCH_3]^{+\bullet} \rightarrow [CH_3CO]^+ + ^\bullet CH_3\). The ion responsible for the peak at m/z = 43 is the acylium ion \([CH_3CO]^+\).
Marking scheme
2.75 marks for part (a): O-H absorption range (1.25 marks), C=O absorption range (1.50 marks). 1.50 marks for part (b): Identification of carbonyl group (1.50 marks). 4.50 marks for part (c): Deducing Z is propanone (1.50 marks), writing correct fragmentation equation (1.50 marks), identifying fragment ion \([CH_3CO]^+\) (1.50 marks).
Section CH03
Answer all questions in the spaces provided. Unit 3 assessment focusing on Inorganic 2 and Physical 2 content.
9 Question · 79.92 marks
Question 1 · Structured/Complex Physical
8.88 marks
The Born-Haber cycle can be used to determine the lattice enthalpy of ionic compounds. A student is investigating calcium nitride, \(\text{Ca}_3\text{N}_2\).
Use the following data to calculate the experimental lattice enthalpy of formation of calcium nitride:
- Enthalpy of formation of \(\text{Ca}_3\text{N}_2\text{(s)}\) = \(-457\text{ kJ mol}^{-1}\) - 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 enthalpy of nitrogen, \(\text{N}_2\text{(g)}\) = \(+945\text{ kJ mol}^{-1}\) - First electron affinity of nitrogen = \(-115\text{ kJ mol}^{-1}\) - Second electron affinity of nitrogen = \(+782\text{ kJ mol}^{-1}\) - Third electron affinity of nitrogen = \(+1410\text{ kJ mol}^{-1}\)
Show your workings clearly.
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Worked solution
To calculate the experimental lattice enthalpy of formation of \(\text{Ca}_3\text{N}_2(s)\):
3. Atomisation of nitrogen: \(\text{N}_2\text{(g)} \rightarrow 2\text{N(g)}\) \(\Delta H = +945\text{ kJ mol}^{-1}\)
4. Electron affinity of nitrogen: \(2\text{N(g)} + 6e^- \rightarrow 2\text{N}^{3-}\text{(g)}\) \(\Delta H = 2 \times (-115 + 782 + 1410) = 2 \times 2077 = +4154\text{ kJ mol}^{-1}\)
5. Sum of enthalpy changes to form gaseous ions from elements in standard states: \(\Delta H_{\text{ions}} = +534 + 5205 + 945 + 4154 = +10838\text{ kJ mol}^{-1}\)
- M1: Multiplies atomisation of Ca by 3 (\(+534\)) and dissociation of N2 by 1 (\(+945\)) [1 mark] - M2: Calculates total ionisation energy of 3 moles of Ca correctly (\(+5205\)) [2 marks] - M3: Calculates total electron affinity of 2 moles of N correctly (\(+4154\)) [2 marks] - M4: Sums up these values to obtain total gaseous ion enthalpy (\(+10838\)) [1 mark] - M5: Uses the correct algebraic relationship: \(\Delta H_{\text{Latt}} = \Delta H_f - \Delta H_{\text{ions}}\) [1.88 marks] - M6: Obtains correct answer of \(-11295\text{ kJ mol}^{-1}\) [1 mark] (Accept without units if kJ/mol is assumed, reject positive value).
Question 2 · Structured/Complex Physical
8.88 marks
A buffer solution is prepared by mixing \(150.0\text{ cm}^3\) of \(0.240\text{ mol dm}^{-3}\) propanoic acid (\(\text{C}_2\text{H}_5\text{COOH}\), \(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) and \(100.0\text{ cm}^3\) of \(0.350\text{ mol dm}^{-3}\) sodium propanoate (\(\text{C}_2\text{H}_5\text{COONa}\)).
Calculate the pH of this buffer solution after \(5.00\text{ cm}^3\) of \(1.50\text{ mol dm}^{-3}\) hydrochloric acid is added. Assume that the volumes are additive.
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Worked solution
1. Calculate the initial moles of propanoic acid (HA) and propanoate ions (\(\text{A}^-\)): \(n(\text{HA}) = 0.1500 \times 0.240 = 0.0360\text{ mol}\) \(n(\text{A}^-) = 0.1000 \times 0.350 = 0.0350\text{ mol}\)
2. Calculate the moles of \(\text{H}^+\) added from HCl: \(n(\text{H}^+\text{ added}) = 0.00500 \times 1.50 = 0.00750\text{ mol}\)
3. Set up the reaction of the buffer conjugate base with the added acid: \(\text{A}^- + \text{H}^+ \rightarrow \text{HA}\)
4. Calculate the new moles of acid and conjugate base at equilibrium: \(n(\text{HA})_{\text{new}} = 0.0360 + 0.00750 = 0.0435\text{ mol}\)
- M1: Correct initial moles of HA (\(0.0360\text{ mol}\)) and A- (\(0.0350\text{ mol}\)) [2 marks] - M2: Correct moles of added H+ (\(0.00750\text{ mol}\)) [1 mark] - M3: Correct calculation of new moles of HA (\(0.0435\text{ mol}\)) and A- (\(0.0275\text{ mol}\)) [2 marks] - M4: Correct substitution into the Ka expression to find [H+] [1.88 marks] - M5: Correct pH of 4.67 (Allow 4.7, must be to 2 decimal places in line with typical pH requirements) [2 marks]
Question 3 · Structured/Complex Physical
8.88 marks
The rate constant, \(k\), for the decomposition of dinitrogen pentoxide was determined at two different temperatures.
- At \(298\text{ K}\), \(k_1 = 3.46 \times 10^{-5}\text{ s}^{-1}\) - At \(338\text{ K}\), \(k_2 = 4.87 \times 10^{-3}\text{ s}^{-1}\)
Use the Arrhenius equation to calculate the activation energy, \(E_a\), for this reaction in \(\text{kJ mol}^{-1}\).
The gas constant, \(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\).
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Worked solution
We use the two-point form of the Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
Substitute the given values into the equation: \(\ln\left(\frac{4.87 \times 10^{-3}}{3.46 \times 10^{-5}}\right) = -\frac{E_a}{8.314} \left(\frac{1}{338} - \frac{1}{298}\right)\)
Calculate the natural log of the rate constant ratio: \(\ln(140.751) = 4.9470\)
Calculate the temperature term: \(\left(\frac{1}{338} - \frac{1}{298}\right) = 0.0029586 - 0.0033557 = -3.971 \times 10^{-4}\text{ K}^{-1}\)
Rearrange to solve for \(E_a\): \(4.9470 = -\frac{E_a}{8.314} \times (-3.971 \times 10^{-4})\)
Rounding to 3 significant figures gives \(104\text{ kJ mol}^{-1}\).
Marking scheme
- M1: Selects or rearranges the Arrhenius equation correctly [2 marks] - M2: Calculates the ratio of k values and its natural logarithm correctly (\(4.95\)) [2 marks] - M3: Calculates the correct value for the temperature difference (\(-3.97 \times 10^{-4}\)) [2 marks] - M4: Correct rearrangement to find Ea in J/mol [1.88 marks] - M5: Correct final value of 104 (or 103.6) with units of kJ mol-1 [1 mark]
Question 4 · Structured/Complex Physical
8.88 marks
An electrochemical cell is set up using the following two half-cells under standard conditions:
(a) Write the conventional representation of this cell.
(b) Calculate the standard cell potential (\(E^{\ominus}_{\text{cell}}\)) and state the direction of electron flow in the external circuit.
(c) Under non-standard conditions, the concentration of \(\text{Fe}^{2+}(aq)\) is decreased. Explain, using Le Chatelier's principle, how this change affects the value of the cell potential.
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Worked solution
(a) Conventional representation: The left electrode is the anode (oxidation) and the right electrode is the cathode (reduction). Since both species in the iron half-cell are aqueous ions, an inert platinum electrode is required. \(\text{Pt}(s) \mid \text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag}(s)\)
(b) Standard cell potential: \(E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{cathode}} - E^{\ominus}_{\text{anode}} = +0.80 - (+0.77) = +0.03\text{ V}\) Electrons flow from the negative electrode (Pt anode) to the positive electrode (Ag cathode).
(c) The overall cell reaction is: \(\text{Fe}^{2+}(aq) + \text{Ag}^+(aq) \rightleftharpoons \text{Fe}^{3+}(aq) + \text{Ag}(s)\) Decreasing the concentration of \([\text{Fe}^{2+}]\) shifts the position of the overall cell reaction equilibrium to the left. This reduces the thermodynamic driving force for the forward reaction, which decreases the cell potential.
Marking scheme
- M1: Writes correct cell representation including Pt electrode, phases, and double salt bridge line [3 marks] - M2: Calculates cell potential correctly as +0.03 V [1 mark] - M3: States electron flow is from Pt to Ag electrode [1 mark] - M4: States overall equation of the cell [1 mark] - M5: Explains that lowering [Fe2+] shifts equilibrium left, decreasing cell potential [2.88 marks]
Question 5 · Structured/Complex Physical
8.88 marks
A sample of sulfur trioxide gas, \(\text{SO}_3\), was placed in a sealed vessel at a temperature of \(900\text{ K}\) and an initial pressure of \(2.50 \times 10^5\text{ Pa}\).
The sulfur trioxide partially decomposed according to the equation:
At equilibrium, the total pressure in the vessel was \(3.10 \times 10^5\text{ Pa}\).
Calculate the equilibrium partial pressure of each gas, and hence calculate the value of the equilibrium constant, \(K_p\), at this temperature. Include units with your answer.
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Worked solution
Let \(2p\) be the reduction in the partial pressure of \(\text{SO}_3\) as it reaches equilibrium.
- M1: Sets up the partial pressures in terms of a variable p [2 marks] - M2: Calculates the value of p as \(0.60 \times 10^5\text{ Pa}\) [1 mark] - M3: Obtains correct individual equilibrium pressures for SO3, SO2 and O2 [2 marks] - M4: Writes the correct Kp expression [1 mark] - M5: Calculates the correct numerical value of Kp (\(5.11 \times 10^4\)) [1.88 marks] - M6: Provides correct units of Pa (or N m-2) [1 mark]
Question 6 · Structured/Complex Physical
8.88 marks
A solution contains the transition metal complex ion \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}\).
(a) Write an equation for the reaction of \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}\) with an excess of ethane-1,2-diamine (en). State the type of reaction and explain the sign of the entropy change (\(\Delta S^{\ominus}\)) for this reaction.
(b) Explain why the product complex formed in (a) shows optical isomerism and describe the spatial relationship between the two optical isomers.
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Worked solution
(a) Equation for the reaction: \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}(aq) + 3\text{en}(aq) \rightarrow [\text{Cr}(\text{en})_3]^{3+}(aq) + 6\text{H}_2\text{O}(l)\) Reaction type: Ligand substitution (or chelation / bidentate ligand exchange). Entropy change: \(\Delta S^{\ominus}\) is positive. This is because 4 particles of reactants produce 7 particles of products. This increase in the number of particles increases disorder in the system.
(b) Optical isomerism: The bidentate complex \([\text{Cr}(\text{en})_3]^{3+}\) is chiral because it lacks a plane of symmetry. This results in the formation of two non-superimposable mirror images.
Marking scheme
- M1: Writes the balanced equation with correct formulas and balancing coefficients [2 marks] - M2: Identifies reaction as ligand substitution/exchange [1 mark] - M3: Explains positive entropy change due to increase in number of moles/particles from 4 to 7 [2 marks] - M4: States that optical isomerism occurs because the complex lacks a plane of symmetry [1.88 marks] - M5: Explains that the isomers are non-superimposable mirror images of each other [2 marks]
Question 7 · Structured/Complex Physical
8.88 marks
The oxides of Period 3 elements exhibit a transition from basic to acidic character across the period.
(a) Write a balanced chemical equation for the reaction of phosphorus(V) oxide with water, and state the pH of the resulting solution.
(b) Silicon dioxide, \(\text{SiO}_2\), does not react with water, but it does react with hot concentrated sodium hydroxide. Write an equation for this reaction and explain why silicon dioxide does not dissolve in water despite containing polar covalent bonds.
(c) Write an equation for the reaction of aluminium oxide, \(\text{Al}_2\text{O}_3\), with hydrochloric acid, demonstrating its basic character in this context.
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Worked solution
(a) Phosphorus(V) oxide and water: \(\text{P}_4\text{O}_{10}(s) + 6\text{H}_2\text{O}(l) \rightarrow 4\text{H}_3\text{PO}_4(aq)\) pH = 0 to 2 (highly acidic).
(b) Silicon dioxide and NaOH: \(\text{SiO}_2(s) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SiO}_3(aq) + \text{H}_2\text{O}(l)\) Explanation: Silicon dioxide has a giant covalent (macromolecular) structure. The strong covalent bonds within the giant lattice require high energy to break, which cannot be compensated for by the weaker intermolecular attractions (hydration) formed with water.
- M1: Balanced equation for phosphorus(V) oxide with water [2 marks] - M2: Correct pH (0, 1, or 2) [1 mark] - M3: Balanced equation for SiO2 with NaOH [2 marks] - M4: Explains that SiO2 has giant covalent structure with strong covalent bonds that cannot be broken by solvation [1.88 marks] - M5: Balanced equation for Al2O3 with HCl [2 marks]
Question 8 · Structured/Complex Physical
8.88 marks
A student carries out a series of tests on an aqueous solution containing iron(III) ions, \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\).
(a) Describe the observations when a few drops of aqueous sodium carbonate, \(\text{Na}_2\text{CO}_3\), are added to the solution of iron(III) ions. Write a balanced ionic equation for this reaction.
(b) When excess concentrated hydrochloric acid is added to a solution of iron(III) ions, a yellow tetrahedral complex is formed. Write an equation for this reaction and state the coordination number of the iron ion in the product.
(c) Explain why iron(III) ions are more acidic in aqueous solution than iron(II) ions.
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Worked solution
(a) Observations: Brown precipitate and effervescence (bubbles of gas). Balanced ionic equation: \(2[\text{Fe}(\text{H}_2\text{O})_6]^{3+}(aq) + 3\text{CO}_3^{2-}(aq) \rightarrow 2[\text{Fe}(\text{H}_2\text{O})_3(\text{OH})_3](s) + 3\text{CO}_2(g) + 3\text{H}_2\text{O}(l)\)
(b) Reaction with concentrated HCl: \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}(aq) + 4\text{Cl}^-(aq) \rightarrow [\text{FeCl}_4]^-(aq) + 6\text{H}_2\text{O}(l)\) Coordination number of iron in the product = 4.
(c) The iron(III) ion has a higher charge (+3 vs +2) and a smaller ionic radius than iron(II). This gives \(\text{Fe}^{3+}\) a significantly higher charge density. Consequently, \(\text{Fe}^{3+}\) polarises the O-H bonds of the coordinated water ligands much more strongly, making the release of H+ ions into solution easier.
Marking scheme
- M1: Identifies both observations: brown precipitate and bubbles/effervescence [2 marks] - M2: Writes the correct balanced equation for carbonate addition [2 marks] - M3: Writes the correct equation for HCl addition [1 mark] - M4: Identifies coordination number of 4 [1 mark] - M5: States Fe3+ has a higher charge and smaller size (or greater charge density) than Fe2+ [1.88 marks] - M6: Explains that this causes greater polarization of the O-H bond in coordinated water, facilitating H+ release [1 mark]
Question 9 · structured
8.88 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.250\text{ mol dm}^{-3}\) propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) and \(75.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)). The acid dissociation constant, \(K_a\), of propanoic acid is \(1.35 \times 10^{-5}\text{ mol dm}^{-3}\) at \(298\text{ K}\).
(a) Calculate the pH of this buffer solution at \(298\text{ K}\). Give your answer to 2 decimal places.
(b) A \(2.00\text{ cm}^3\) sample of \(1.00\text{ mol dm}^{-3}\) nitric acid (\(\text{HNO}_3\)) is added to this buffer solution. Calculate the pH of the resulting solution at \(298\text{ K}\). Give your answer to 2 decimal places.
(c) Write an ionic equation to show how the propanoate ions in this buffer solution react with the added hydrogen ions.
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Worked solution
(a) First, calculate the initial moles of propanoic acid (HA) and propanoate ions (A\(^-\)): \(n(\text{HA}) = \frac{50.0}{1000} \times 0.250 = 0.0125\text{ mol}\) \(n(\text{A}^-) = \frac{75.0}{1000} \times 0.150 = 0.01125\text{ mol}\)
Using the expression for the acid dissociation constant: \([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.35 \times 10^{-5} \times \frac{0.0125}{0.01125} = 1.50 \times 10^{-5}\text{ mol dm}^{-3}\)
(b) Calculate the moles of \(\text{H}^+\) added from \(\text{HNO}_3\): \(n(\text{H}^+) = \frac{2.00}{1000} \times 1.00 = 0.00200\text{ mol}\)
These added hydrogen ions react with the propanoate ions to form propanoic acid: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
Calculate the new moles of propanoate ions (A\(^-\)) and propanoic acid (HA): \(n(\text{A}^-)_{\text{new}} = 0.01125 - 0.00200 = 0.00925\text{ mol}\) \(n(\text{HA})_{\text{new}} = 0.0125 + 0.00200 = 0.0145\text{ mol}\)
Calculate the new \([\text{H}^+]\): \([\text{H}^+]_{\text{new}} = 1.35 \times 10^{-5} \times \frac{0.0145}{0.00925} = 2.1162 \times 10^{-5}\text{ mol dm}^{-3}\)
Calculate the new pH: \(\text{pH} = -\log_{10}(2.1162 \times 10^{-5}) = 4.67\) (to 2 decimal places).
(c) The reaction of the buffer conjugate base with added acid is represented by: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
Marking scheme
Part (a) [3 Marks]: - M1: Correct calculation of moles: \(n(\text{propanoic acid}) = 0.0125\text{ mol}\) AND \(n(\text{propanoate}) = 0.01125\text{ mol}\). - M2: Correct calculation of \([\text{H}^+] = 1.50 \times 10^{-5}\text{ mol dm}^{-3}\) or correct substitution into the Henderson-Hasselbalch equation. - M3: Correct calculation of pH to 2 dp = 4.82. (Reject other decimal places).
Part (b) [5 Marks]: - M4: Correct calculation of moles of added \(\text{H}^+ = 0.00200\text{ mol}\). - M5: Correct calculation of new moles of propanoic acid = \(0.0145\text{ mol}\). - M6: Correct calculation of new moles of propanoate ions = \(0.00925\text{ mol}\). - M7: Correct calculation of new \([\text{H}^+] = 2.12 \times 10^{-5}\text{ mol dm}^{-3}\) (allow consequential error from M5/M6). - M8: Correct calculation of final pH to 2 dp = 4.67 (allow consequential error, accept 4.68 depending on intermediate rounding; e.g. if using unrounded figures pH = 4.67).
Part (c) [1 Mark]: - M9: Correct ionic equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\) (accept state symbols, do not penalize lack thereof).
Section CH04
Answer all questions in the spaces provided. Unit 4 assessment focusing on Organic 2 and Physical 2 content.
10 Question · 80 marks
Question 1 · Structured
8 marks
A buffer solution of volume \(500\text{ cm}^3\) is prepared by mixing \(0.200\text{ mol dm}^{-3}\) propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) with \(0.150\text{ mol dm}^{-3}\) sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)). The \(K_a\) of propanoic acid at \(298\text{ K}\) is \(1.35 \times 10^{-5}\text{ mol dm}^{-3}\).
(a) Calculate the pH of this buffer solution. [3 marks]
(b) Calculate the pH of the buffer solution after \(0.0050\text{ mol}\) of solid sodium hydroxide, \(\text{NaOH}\), is dissolved in it. Assume the volume remains unchanged. [5 marks]
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Worked solution
(a) First, write the expression for the acid dissociation constant: \(K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)
Rearrange to solve for \([\text{H}^+]\): \([\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{CH}_2\text{COOH}]}{[\text{CH}_3\text{CH}_2\text{COO}^-]}\)
Substitute the concentrations into the rearranged equation: \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.200}{0.150} = 1.80 \times 10^{-5}\text{ mol dm}^{-3}\)
Calculate the pH: \(\text{pH} = -\log_{10}(1.80 \times 10^{-5}) = 4.74\)
(b) Calculate the initial moles of propanoic acid and propanoate ions in the \(500\text{ cm}^3\) solution: \(\text{Moles of } \text{CH}_3\text{CH}_2\text{COOH} = 0.200 \times 0.500 = 0.100\text{ mol}\) \(\text{Moles of } \text{CH}_3\text{CH}_2\text{COO}^- = 0.150 \times 0.500 = 0.0750\text{ mol}\)
When \(0.0050\text{ mol}\) of \(\text{NaOH}\) is added, it reacts with the propanoic acid: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\)
Calculate the new moles of acid and salt: \(\text{New moles of } \text{CH}_3\text{CH}_2\text{COOH} = 0.100 - 0.0050 = 0.0950\text{ mol}\) \(\text{New moles of } \text{CH}_3\text{CH}_2\text{COO}^- = 0.0750 + 0.0050 = 0.0800\text{ mol}\)
Calculate the new hydrogen ion concentration: \([\text{H}^+] = K_a \times \frac{\text{moles of acid}}{\text{moles of salt}}\) \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0950}{0.0800} = 1.6031 \times 10^{-5}\text{ mol dm}^{-3}\)
Calculate the new pH: \(\text{pH} = -\log_{10}(1.6031 \times 10^{-5}) = 4.795 = 4.80\)
Marking scheme
(a) - M1: Expresses the Henderson-Hasselbalch equation or rearranged \(K_a\) expression correctly: \([\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}\) (1 mark) - M2: Calculates \([\text{H}^+] = 1.80 \times 10^{-5}\text{ mol dm}^{-3}\) (1 mark) - M3: Calculates pH as 4.74 (must be to 2 d.p.) (1 mark)
(b) - M1: Calculates initial moles of propanoic acid (\(0.100\text{ mol}\)) AND propanoate (\(0.0750\text{ mol}\)) (1 mark) - M2: Deducts added \(\text{OH}^-\) from acid moles to find new moles of acid = \(0.0950\text{ mol}\) (1 mark) - M3: Adds added \(\text{OH}^-\) to salt moles to find new moles of salt = \(0.0800\text{ mol}\) (1 mark) - M4: Calculates new \([\text{H}^+] = 1.60 \times 10^{-5}\text{ mol dm}^{-3}\) (1 mark) - M5: Calculates new pH as 4.80 (must be to 2 d.p.; allow consequential error from previous steps) (1 mark)
Question 2 · Structured
8 marks
The reaction between peroxodisulfate(VI) ions and iodide ions is represented by: \(\text{S}_2\text{O}_8^{2-}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{SO}_4^{2-}(aq) + \text{I}_2(aq)\)
Initial rate data obtained at \(298\text{ K}\) are given in the table below:
(a) Deduce the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\). Explain your reasoning. [3 marks]
(b) Write the rate equation and calculate the rate constant, \(k\), at \(298\text{ K}\), stating its units. [3 marks]
(c) In another experiment at \(320\text{ K}\), the rate constant is \(0.450\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). Calculate the activation energy, \(E_a\), for this reaction in \(\text{kJ mol}^{-1}\). (\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)) [2 marks]
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Worked solution
(a) - Comparing Experiments 1 and 2: \([\text{I}^-]\) is constant. Doubling \([\text{S}_2\text{O}_8^{2-}]\) (from 0.040 to 0.080) doubles the rate (from \(1.28 \times 10^{-4}\) to \(2.56 \times 10^{-4}\)). Therefore, the order with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1 (first order). - Comparing Experiments 2 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant. Halving \([\text{I}^-]\) (from 0.080 to 0.040) halves the rate (from \(2.56 \times 10^{-4}\) to \(1.28 \times 10^{-4}\)). Therefore, the order with respect to \(\text{I}^-\) is 1 (first order).
(a) - M1: Deduces order with respect to \(\text{S}_2\text{O}_8^{2-}\) is first order and explains using Exp 1 & 2 (rate doubles as concentration doubles) (1 mark) - M2: Deduces order with respect to \(\text{I}^-\) is first order and explains using Exp 2 & 3 (rate halves as concentration halves) (1 mark) - M3: Standard logical structure of explanation with correct link to experimental rates (1 mark)
(c) - M1: Correct substitution of values into the Arrhenius expression (1 mark) - M2: Calculates correct activation energy value \(87.2\text{ kJ mol}^{-1}\) (accept range \(86.5\) to \(88.0\) depending on intermediate rounding) (1 mark)
Question 3 · Structured
8 marks
Butanone reacts with a mixture of potassium cyanide and dilute sulfuric acid to form a hydroxynitrile.
(a) Outline the mechanism for this reaction, showing all relevant dipole charges, curly arrows, and lone pairs. [4 marks]
(b) Explain why the product of this reaction is a racemic mixture and has no effect on plane-polarized light. [3 marks]
(c) State the reagent and conditions required to convert this hydroxynitrile into a carboxylic acid. [1 mark]
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Worked solution
(a) The mechanism is nucleophilic addition: 1. Show a dipole on the carbonyl group of butanone: \(\text{C}^{\delta+}=\text{O}^{\delta-}\). 2. Draw a curly arrow from the lone pair of the carbon atom of the nucleophile \(:\text{CN}^-\) to the carbonyl carbon (\(\text{C}^{\delta+}\)). 3. Draw a curly arrow from the double bond of \(\text{C}=\text{O}\) to the oxygen atom. 4. Draw the intermediate structure: \(\text{CH}_3\text{CH}_2\text{C(O}^-\text{)(CN)CH}_3\). 5. Draw a curly arrow from the lone pair on the intermediate oxygen ion (\(\text{O}^-\)) to an \(\text{H}^+\) ion to yield the product, 2-hydroxy-2-methylbutanenitrile.
(b) The carbonyl group in butanone is planar at the reacting carbon atom. The nucleophile \(\text{CN}^-\) can attack this carbon atom with equal probability from either above or below the plane. This results in the formation of an equimolar mixture (racemic mixture) of two enantiomers. The rotation of plane-polarized light by each enantiomer is equal in magnitude but opposite in direction, so they cancel each other out, leaving no net effect on plane-polarized light.
(c) Acidic hydrolysis of the nitrile group: Heat under reflux with dilute hydrochloric acid (or dilute sulfuric acid).
Marking scheme
(a) - M1: Curly arrow from the lone pair of \(:\text{CN}^-\) to the carbonyl carbon with partial charges (\(\text{C}^{\delta+}\) and \(\text{O}^{\delta-}\)) correctly shown on the reactant (1 mark) - M2: Curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom (1 mark) - M3: Correct intermediate tetrahedral alkoxide structure including negative charge on oxygen (1 mark) - M4: Curly arrow from the intermediate oxygen lone pair to \(\text{H}^+\) to produce the final product (1 mark)
(b) - M1: States that the carbonyl group is planar (or planar around the carbonyl carbon) (1 mark) - M2: States that attack by \(\text{CN}^-\) can happen from either side (above/below) with equal probability (1 mark) - M3: Explains that this forms a racemic/equimolar mixture of enantiomers whose optical rotations cancel out (1 mark)
(c) - M1: Reagent: dilute hydrochloric acid/dilute sulfuric acid AND Conditions: heat under reflux (1 mark)
Question 4 · Structured
8 marks
Ethyl benzoate (\(\text{C}_6\text{H}_5\text{COOCH}_2\text{CH}_3\)) and benzoyl chloride (\(\text{C}_6\text{H}_5\text{COCl}\)) are both derivatives of benzoic acid.
(a) Write an equation for the base-catalyzed hydrolysis of ethyl benzoate using aqueous sodium hydroxide. [2 marks]
(b) Benzoyl chloride reacts rapidly with water at room temperature. (i) Write an equation for this reaction and name the organic product. [2 marks] (ii) Describe and explain the relative rates of reaction of benzoyl chloride and ethyl benzoate with water at room temperature. [4 marks]
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Worked solution
(a) Under alkaline conditions, hydrolysis of the ester produces the carboxylate salt and an alcohol: \(\text{C}_6\text{H}_5\text{COOCH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{COONa} + \text{CH}_3\text{CH}_2\text{OH}\)
(b) (i) Hydrolysis of acyl chloride with water: \(\text{C}_6\text{H}_5\text{COCl} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{COOH} + \text{HCl}\) Organic product name: benzoic acid
(ii) Benzoyl chloride is far more reactive than ethyl benzoate (reacts vigorously at room temperature with water, while ethyl benzoate is inert to neutral water at room temperature and requires an acid/base catalyst and heat to hydrolyze). This difference in reactivity is explained by: 1. The carbonyl carbon in benzoyl chloride is more electron-deficient (more \(\delta+\)) because of the strongly electronegative chlorine atom drawing electron density away. 2. In contrast, in ethyl benzoate, the oxygen atom of the ethoxy group has lone pairs that can be delocalized into the carbonyl group, reducing its partial positive charge and making it less vulnerable to nucleophilic attack. 3. Chlorine is a much better leaving group (as a stable chloride ion, \(\text{Cl}^-\)) compared to the ethoxide ion (\(\text{CH}_3\text{CH}_2\text{O}^-\)).
(b)(ii) - M1: States that benzoyl chloride reacts much faster/rapidly/vigorously at room temperature compared to ethyl benzoate (1 mark) - M2: Identifies that the carbonyl carbon in benzoyl chloride is more electron-deficient/\(\delta+\)/polar (1 mark) - M3: Explains that chlorine is a better leaving group than the ethoxide group (1 mark) - M4: Explains that resonance/delocalization in the ester group reduces the electrophilicity of its carbonyl carbon (1 mark)
Question 5 · Structured
8 marks
Ethylamine, phenylamine, and ammonia are all weak bases.
(a) Arrange ethylamine, phenylamine, and ammonia in order of increasing base strength. Explain your answer in terms of their structures. [5 marks]
(b) Phenylamine can be prepared in the laboratory from nitrobenzene. (i) State the reagents needed for this two-stage reduction. [2 marks] (ii) Write a balanced equation for the overall conversion of nitrobenzene to phenylamine, representing the reducing agent as \([\text{H}]\). [1 mark]
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Worked solution
(a) Order of increasing base strength: phenylamine < ammonia < ethylamine
Explanation: - A base acts as a proton acceptor; the strength of the base depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (\(\text{H}^+\)). - In ethylamine, the ethyl group is an electron-releasing alkyl group (+I inductive effect). This increases the electron density on the nitrogen atom, making its lone pair more available for protonation compared to ammonia. - In phenylamine, the lone pair of electrons on the nitrogen atom is partially delocalized into the delocalized \(\pi\)-system of the benzene ring. This significantly decreases the electron density on the nitrogen atom, making the lone pair much less available to accept a proton compared to ammonia.
(b) (i) Reagents: Stage 1: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), heated under reflux. Stage 2: Sodium hydroxide (\(\text{NaOH}\)) (to liberate the free amine from its salt).
(a) - M1: Correct order: Phenylamine, ammonia, ethylamine (1 mark) - M2: Relates basic strength to availability of the lone pair of electrons on the nitrogen atom to accept a proton (1 mark) - M3: Explains that ethylamine is a stronger base than ammonia because the ethyl group is electron-releasing (positive inductive effect) (1 mark) - M4: Explains that phenylamine is a weaker base because the nitrogen lone pair is delocalized into the aromatic ring (1 mark) - M5: Clarifies that delocalization reduces the availability of the lone pair on nitrogen (1 mark)
(b)(i) - M1: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) (1 mark) - M2: Aqueous sodium hydroxide (\(\text{NaOH}\)) (1 mark)
Consider the gas-phase equilibrium: \(2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \quad \Delta H = -197\text{ kJ mol}^{-1}\)
In an experiment, \(2.00\text{ mol}\) of \(\text{SO}_2\) and \(1.50\text{ mol}\) of \(\text{O}_2\) were mixed in a sealed vessel at a constant temperature. When equilibrium was reached, \(1.60\text{ mol}\) of \(\text{SO}_3\) had formed, and the total pressure was \(150\text{ kPa}\).
(a) Calculate the equilibrium amounts of \(\text{SO}_2(g)\) and \(\text{O}_2(g)\) in moles. [2 marks]
(b) Calculate the partial pressure of each gas at equilibrium. [2 marks]
(c) Calculate the value of \(K_p\) for this reaction at this temperature and state its units. [3 marks]
(d) State the effect, if any, of increasing the temperature on the value of \(K_p\). Explain your answer. [1 mark]
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Worked solution
(a) Using the reaction stoichiometry: At equilibrium, \(1.60\text{ mol}\) of \(\text{SO}_3\) is produced. Since 2 moles of \(\text{SO}_2\) reacts to form 2 moles of \(\text{SO}_3\), the amount of \(\text{SO}_2\) reacted is \(1.60\text{ mol}\). \(\text{Equilibrium mol of } \text{SO}_2 = 2.00 - 1.60 = 0.40\text{ mol}\)
Since 1 mole of \(\text{O}_2\) reacts to form 2 moles of \(\text{SO}_3\), the amount of \(\text{O}_2\) reacted is \(1.60 / 2 = 0.80\text{ mol}\). \(\text{Equilibrium mol of } \text{O}_2 = 1.50 - 0.80 = 0.70\text{ mol}\)
(b) Calculate the total moles of gas at equilibrium: \(\text{Total moles} = 0.40\text{ (SO}_2\text{)} + 0.70\text{ (O}_2\text{)} + 1.60\text{ (SO}_3\text{)} = 2.70\text{ mol}\)
Calculate the partial pressure of each gas (\(p_i = X_i \times P_{\text{total}}\)): \(p(\text{SO}_2) = \frac{0.40}{2.70} \times 150 = 22.22\text{ kPa}\) \(p(\text{O}_2) = \frac{0.70}{2.70} \times 150 = 38.89\text{ kPa}\) \(p(\text{SO}_3) = \frac{1.60}{2.70} \times 150 = 88.89\text{ kPa}\)
(c) Write the expression for \(K_p\): \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}\)
(d) The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier’s principle, increasing the temperature will shift the equilibrium to the left (in the endothermic direction), reducing the yield of products and increasing reactants. Therefore, the value of \(K_p\) decreases.
(b) - M1: Total moles = \(2.70\text{ mol}\) (0.5 mark) - M2: Calculates correct partial pressures: \(p(\text{SO}_2) = 22.2\text{ kPa}\), \(p(\text{O}_2) = 38.9\text{ kPa}\), and \(p(\text{SO}_3) = 88.9\text{ kPa}\) (1.5 marks; lose 0.5 marks for each error)
(c) - M1: Gives correct \(K_p\) expression: \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}\) (1 mark) - M2: Calculates value of \(K_p = 0.411\) (accept range \(0.409\) to \(0.413\)) (1 mark) - M3: Correct units: \(\text{kPa}^{-1}\) (or \(\text{Pa}^{-1}\) with appropriate value conversion) (1 mark)
(d) - M1: States that \(K_p\) decreases because the forward reaction is exothermic (equilibrium shifts left) (1 mark)
Question 7 · Structured
8 marks
Alanine (\(\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}\)) has an isoelectric point of 6.0.
(a) Draw the structure of alanine at pH 1.0, pH 6.0, and pH 12.0. [3 marks]
(b) A mixture of three amino acids (alanine, aspartic acid, and lysine) was separated using thin-layer chromatography (TLC) on silica gel, using a polar solvent. (i) Describe how the spots of amino acids can be made visible on the chromatogram. [1 mark] (ii) Aspartic acid has a polar carboxyl side-chain (\(-\text{CH}_2\text{COOH}\)), while alanine has a non-polar methyl side-chain. Predict and explain the relative \(R_f\) values of alanine and aspartic acid on a silica gel plate. [2 marks]
(c) Draw the structure of the tripeptide Ala-Ala-Ala, showing all peptide bonds clearly. [2 marks]
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(a) - At pH 1.0 (strongly acidic): Both the basic amino group is protonated, and the acidic carboxyl group is uncharged: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\). - At pH 6.0 (isoelectric point): Exists as a zwitterion: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\). - At pH 12.0 (strongly alkaline): The amino group is uncharged, and the carboxyl group is deprotonated: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).
(b) (i) Amino acids are colorless, so they are visualized by spraying the plate with ninhydrin solution (and heating) to turn them purple, or by viewing the plate under a UV lamp. (ii) Aspartic acid has a highly polar side chain containing a carboxylic acid group, which forms strong hydrogen bonds with the polar stationary phase (silica gel). Therefore, it moves more slowly and has a lower \(R_f\) value. Alanine, with its non-polar methyl side-chain, has weaker interactions with the stationary phase and dissolves more readily in the mobile phase, thus traveling further and having a higher \(R_f\) value.
(c) The tripeptide Ala-Ala-Ala is formed by condensation reactions connecting three alanine units through peptide (amide) bonds: \(\text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{CO}-\text{NH}-\text{CH}(\text{CH}_3)-\text{CO}-\text{NH}-\text{CH}(\text{CH}_3)-\text{COOH}\)
Marking scheme
(a) - M1: Structure at pH 1.0: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\) (1 mark) - M2: Structure at pH 6.0: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\) (1 mark) - M3: Structure at pH 12.0: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\) (1 mark)
(b)(i) - M1: Spray with ninhydrin (and heat) OR use UV light (1 mark)
(b)(ii) - M1: Predicts alanine has a higher \(R_f\) value than aspartic acid (1 mark) - M2: Explains that aspartic acid is more polar, thus has a stronger affinity/adsorption to the polar silica stationary phase (1 mark)
(c) - M1: Shows two peptide (\(-\text{CO}-\text{NH}-\)) links between three carbon-carbon chains correctly (1 mark) - M2: Complete correct structure of tripeptide with correct terminal groups (\(\text{NH}_2\) and \(\text{COOH}\)) and three methyl branch groups (1 mark)
Question 8 · Structured
8 marks
Consider the following synthetic route starting from propanal:
(a) Name and draw the displayed formula of Compound A. Circle the chiral carbon atom in the structure of Compound A. [3 marks]
(b) Compound B is a carboxylic acid. (i) Draw the skeletal structures of the two enantiomers of Compound B, showing their three-dimensional relationship. [2 marks] (ii) Explain how a physical property of these two enantiomers can be used to distinguish between them, and why they cannot be distinguished using simple physical properties like boiling point. [3 marks]
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(a) - Compound A is formed from the nucleophilic addition of HCN to propanal, producing: \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\). - Name: 2-hydroxybutanenitrile. - Displayed formula: Must draw every bond explicitly (including the \(\text{O}-\text{H}\) bond and the \(\text{C}\equiv\text{N}\) triple bond). - The chiral carbon atom is carbon-2, which is bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{C}\equiv\text{N}\)). Circle this carbon atom.
(b) (i) Compound B is 2-hydroxybutanoic acid (\(\text{CH}_3\text{CH}_2\text{CH(OH)COOH}\)). To show the enantiomers, draw skeletal structures as three-dimensional non-superimposable mirror images around the chiral center (C2), using wedge/dash bonds to show stereochemistry: - One enantiomer has \(-\text{OH}\) on a wedge and \(-\text{H}\) on a dash (or vice versa). - The mirror image should show the corresponding spatial arrangement.
(ii) Distinguishing enantiomers: - Pass plane-polarized light through separate solutions of each enantiomer. - One enantiomer will rotate the plane of polarized light clockwise (dextrorotatory), while the other enantiomer will rotate it by the same angle anticlockwise (laevorotatory). - They cannot be distinguished by simple physical properties like boiling point or solubility because the spatial arrangement of atoms relative to each other is identical, resulting in identical intermolecular forces (hydrogen bonding, dipole-dipole, and London dispersion forces).
Marking scheme
(a) - M1: Names Compound A correctly: 2-hydroxybutanenitrile (1 mark) - M2: Correct displayed formula of Compound A, showing all bonds (including \(\text{O}-\text{H}\) and \(\text{C}\equiv\text{N}\)) (1 mark) - M3: Correctly circles/identifies the chiral carbon atom (carbon-2) (1 mark)
(b)(i) - M1: Correct skeletal formula of 2-hydroxybutanoic acid (1 mark) - M2: Draws both enantiomers as non-superimposable mirror images using wedge/dash stereochemical bonds to represent the 3D tetrahedral environment (1 mark)
(b)(ii) - M1: States that they can be distinguished by passing plane-polarized light through their solutions (1 mark) - M2: Explains that they rotate the plane of polarized light in equal and opposite directions (clockwise vs anticlockwise) (1 mark) - M3: Explains that other physical properties (e.g. boiling point, solubility) are identical because they have identical intermolecular forces (1 mark)
Question 9 · Structured
8 marks
A buffer solution is prepared at 298 K 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 \(75.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)). The acid dissociation constant, \(K_a\), of propanoic acid at 298 K is \(1.35 \times 10^{-5}\text{ mol dm}^{-3}\).
(a) Calculate the pH of this buffer solution. Give your answer to 2 decimal places. [3 marks]
(b) Calculate the new pH of this buffer solution after the addition of \(2.00\text{ cm}^3\) of \(0.500\text{ mol dm}^{-3}\) hydrochloric acid. Give your answer to 2 decimal places. [5 marks]
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(b) Calculate pH after addition of \(\text{HCl}\): 1. Find moles of \(\text{H}^+\) added: \(n(\text{H}^+) = \frac{2.00}{1000} \times 0.500 = 0.00100\text{ mol}\)
2. Since the added strong acid reacts with the propanoate conjugate base: \(\text{A}^- + \text{H}^+ \rightarrow \text{HA}\)
3. Calculate the new moles of \(\text{HA}\) and \(\text{A}^-\): \(\text{New } n(\text{HA}) = 0.0125 + 0.00100 = 0.0135\text{ mol}\) \(\text{New } n(\text{A}^-) = 0.01125 - 0.00100 = 0.01025\text{ mol}\)
4. Calculate the new \([\text{H}^+]\): \([\text{H}^+] = (1.35 \times 10^{-5}) \times \frac{0.0135}{0.01025} = 1.778 \times 10^{-5}\text{ mol dm}^{-3}\)
5. Calculate the new pH: \(\text{pH} = -\log_{10}(1.778 \times 10^{-5}) = 4.7501 \approx 4.75\)
Marking scheme
(a) * **M1**: Correct calculation of moles of HA (\(0.0125\text{ mol}\)) AND moles of \(\text{A}^-\)\ (\(0.01125\text{ mol}\)). [1 mark] * **M2**: Correct rearranged expression for \([\text{H}^+]\) or Henderson-Hasselbalch equation showing substitution of these moles. [1 mark] * **M3**: Final pH = \(4.82\) (must be 2 d.p. for full marks). [1 mark]
(b) * **M4**: Correct calculation of moles of \(\text{H}^+\) added (\(1.00 \times 10^{-3}\text{ mol}\)). [1 mark] * **M5**: Correct subtraction of added \(\text{H}^+\) from moles of conjugate base to give new \(n(\text{A}^-) = 0.01025\text{ mol}\). [1 mark] * **M6**: Correct addition of added \(\text{H}^+\) to moles of weak acid to give new \(n(\text{HA}) = 0.0135\text{ mol}\). [1 mark] * **M7**: Correct calculation of new \([\text{H}^+] = 1.778 \times 10^{-5}\text{ mol dm}^{-3}\) (allow ecf from M5 and M6). [1 mark] * **M8**: Final pH = \(4.75\) (must be 2 d.p. for final accuracy mark; allow ecf from M7). [1 mark]
Question 10 · Structured
8 marks
Benzene can be converted into the aromatic amide N-phenylethanamide via a three-step synthetic pathway.
(a) Step 1 involves the nitration of benzene using a mixture of concentrated nitric acid and concentrated sulfuric acid. (i) Write an equation to show how the electrophile, \(\text{NO}_2^+\), is generated in this reaction mixture. [1 mark] (ii) Draw the mechanism for the reaction of benzene with \(\text{NO}_2^+\) to form nitrobenzene. [3 marks]
(b) Identify the reagents and conditions required for the reduction of nitrobenzene to phenylamine in Step 2. [2 marks]
(c) In Step 3, phenylamine is reacted with ethanoyl chloride to produce N-phenylethanamide. Write a balanced chemical equation for this reaction and state the IUPAC name of the organic byproduct if any (or the formula of the inorganic byproduct). [2 marks]
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(ii) Mechanism: - A curly arrow from the delocalised \(\pi\)-system (inner ring) of benzene to the \(\text{N}\) in \(\text{NO}_2^+\). - Correct drawing of the intermediate: A hexagon with a partially broken ring containing a positive charge (the horseshoe shape must face the carbon containing both the \(\text{H}\) and the \(\text{NO}_2\) group, and span across 5 carbons). - A curly arrow from the C–H bond back into the partially broken ring to restore aromaticity, yielding nitrobenzene and \(\text{H}^+\).
(b) Reagents and conditions for Step 2: - Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) [1 mark] - Heat / Reflux [1 mark] - (followed by addition of concentrated aqueous NaOH to liberate the free amine from the phenylammonium salt - implied but first two are the primary marks).
(a)(ii) * **M2**: Curly arrow from the benzene ring to the nitrogen atom of \(\text{NO}_2^+\). [1 mark] * **M3**: Correct intermediate structure (showing \(\text{H}\) and \(\text{NO}_2\) on same carbon, horseshoe positively charged ring pointing towards this carbon). [1 mark] * **M4**: Curly arrow from C–H bond back into the horseshoe ring. [1 mark]
(b) * **M5**: Reagents: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) (accept \(\text{Fe}\) and \(\text{HCl}\)). [1 mark] * **M6**: Conditions: Heat under reflux (or simply 'heat' / 'reflux'). Reject 'room temperature'. [1 mark]
(c) * **M7**: Correct balanced chemical equation: \(\text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{HCl}\) (Structures can be shown as displayed/structural/skeletal formulae). [1 mark] * **M8**: Correctly identifying the byproduct as hydrogen chloride (\(\text{HCl}\)). [1 mark]
CH05 Section A
Answer all questions in the spaces provided. Focus on practical laboratory methods.
3 Question · 30 marks
Question 1 · Structured Practical
10 marks
A student designed an apparatus to determine the enthalpy change of solution of anhydrous copper(II) sulfate. Part 1 (3 marks): Describe the graphical method the student should use to find an accurate value for the temperature change (\(\Delta T\)) at the time of mixing (4th minute), including measurements taken before and after mixing. Part 2 (4 marks): In a specific experiment, 3.99 g of anhydrous copper(II) sulfate (\(M_r = 159.6\)) was added to \(50.0\text{ cm}^3\) of water. The initial temperature of the water was \(20.2\ ^\circ\text{C}\). Using the graphical method, the extrapolated maximum temperature was found to be \(28.4\ ^\circ\text{C}\). Calculate the enthalpy change of solution of anhydrous copper(II) sulfate in \(\text{kJ mol}^{-1}\). Show your working. (Assume the density of the solution is \(1.00\text{ g cm}^{-3}\), the specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\), and the heat capacity of the calorimeter is negligible). Part 3 (3 marks): State three practical modifications to the apparatus (other than using a bomb calorimeter) that would improve the accuracy of this experimental value by reducing heat loss.
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Worked solution
Part 1: Record the temperature of the water every minute for 3 minutes. At the 4th minute, add the anhydrous copper(II) sulfate and do not record the temperature. Record the temperature every minute from 5 to 10 minutes while stirring. Plot temperature against time. Draw two lines of best fit: one through the points before mixing (0 to 3 minutes) and one through the cooling curve points after mixing (5 to 10 minutes). Extrapolate both lines to the 4th minute. Find the difference in temperature between these two lines at the 4th minute to obtain \(\Delta T\). Part 2: Calculate heat energy change, \(q = m c \Delta T\). Here, \(m = 50.0\text{ g}\), \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\), and \(\Delta T = 28.4 - 20.2 = 8.2\text{ K}\). \(q = 50.0 \times 4.18 \times 8.2 = 1713.8\text{ J} = 1.7138\text{ kJ}\). Calculate moles of \(CuSO_4\): \(n = 3.99 / 159.6 = 0.0250\text{ mol}\). Calculate enthalpy change: \(\Delta H = -q / n = -1.7138 / 0.0250 = -68.552\text{ kJ mol}^{-1}\), which rounds to \(-68.6\text{ kJ mol}^{-1}\) (exothermic, so negative sign is required). Part 3: Use a polystyrene lid to cover the cup; place the polystyrene cup inside a beaker to provide air insulation; use double-layered polystyrene cups for extra insulation.
Marking scheme
Part 1: [1 mark] Measure temperature every minute before and after mixing, leaving a gap at the exact minute of mixing. [1 mark] Plot temperature against time and draw two lines of best fit (one before mixing, one after). [1 mark] Extrapolate both lines to the time of mixing (4th minute) and find the difference. Part 2: [1 mark] Calculate temperature rise = 8.2 K. [1 mark] Calculate heat energy change (q) = 1.71 kJ (or 1714 J). [1 mark] Calculate moles of CuSO4 = 0.0250 mol. [1 mark] Calculate \(\Delta H\) = -68.6 \(\text{kJ mol}^{-1}\) (must have negative sign and be to 3 sig figs). Part 3: [3 marks] Any three from: Use a plastic lid on the cup; Place the polystyrene cup inside a beaker; Nest the cup inside another cup (double-layered cup); Wrap the cup in cotton wool or mineral wool insulation; Use a digital thermometer with a higher resolution (e.g. 0.1 or 0.01 °C).
Question 2 · Structured Practical
10 marks
A student investigates a weak acid-strong base titration using a pH probe. They titrate \(25.0\text{ cm}^3\) of a weak monoprotic acid, HA, with \(0.100\text{ mol dm}^{-3}\) sodium hydroxide (NaOH) solution. Part 1 (3 marks): Describe how the student should calibrate the pH probe before carrying out the titration to ensure accurate pH readings. Part 2 (3 marks): Explain how the student can use the resulting titration curve (pH against volume of NaOH added) to find: (i) the equivalence point, and (ii) the acid dissociation constant (\(K_a\)) of HA. Part 3 (4 marks): The equivalence point was reached after adding \(18.80\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) NaOH. Calculate the concentration of the weak acid HA. At the half-neutralisation point, the pH of the mixture was 4.76. Calculate the \(K_a\) value of the weak acid HA, stating its units.
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Worked solution
Part 1: Wash the pH probe with distilled water. Dip the probe into a buffer solution of known pH (such as pH 7.00) and adjust the meter to read this value. Repeat the process with at least one other buffer solution (such as pH 4.00 or pH 10.00) to ensure calibration across a wide pH range. Part 2: (i) The equivalence point is located at the midpoint of the near-vertical section of the titration curve. (ii) Locate the volume of NaOH at the equivalence point, divide this volume by two to find the half-neutralisation volume. Read the pH at this half-neutralisation volume; at this point, \(pH = pK_a\). Thus, \(K_a = 10^{-pH}\). Part 3: Moles of NaOH used = \(0.100 \times (18.80 / 1000) = 1.88 \times 10^{-3}\text{ mol}\). Since HA is a monoprotic acid, moles of HA = \(1.88 \times 10^{-3}\text{ mol}\). Concentration of HA = \(1.88 \times 10^{-3} / 0.0250 = 0.0752\text{ mol dm}^{-3}\). At half-neutralisation, \(pH = pK_a = 4.76\). Therefore, \(K_a = 10^{-4.76} = 1.7378 \times 10^{-5}\text{ mol dm}^{-3}\), which rounds to \(1.74 \times 10^{-5}\text{ mol dm}^{-3}\).
Marking scheme
Part 1: [1 mark] Rinse the pH probe with deionised/distilled water. [1 mark] Place the probe in a buffer solution of known pH (e.g., 7.00) and calibrate/adjust the meter. [1 mark] Use a second buffer of different pH (e.g., 4.00) to calibrate across a range. Part 2: [1 mark] Equivalence point is the midpoint of the vertical/steep portion of the curve. [1 mark] Half-neutralisation volume is half of the equivalence volume. [1 mark] At half-neutralisation \(pH = pK_a\), so \(K_a = 10^{-pH}\). Part 3: [1 mark] Calculate moles of NaOH = \(1.88 \times 10^{-3}\text{ mol}\). [1 mark] Calculate concentration of HA = \(0.0752\text{ mol dm}^{-3}\). [1 mark] Identify that \(pH = pK_a = 4.76\) at half-neutralisation. [1 mark] Calculate \(K_a = 1.74 \times 10^{-5}\text{ mol dm}^{-3}\) (allow \(1.7 \times 10^{-5}\) to \(1.8 \times 10^{-5}\); accept correct units).
Question 3 · Structured Practical
10 marks
A student determines the mass of \(FeSO_4\) in an iron supplement tablet by titration with potassium manganate(VII) solution. Part 1 (4 marks): Describe the steps the student should take to prepare exactly \(250.0\text{ cm}^3\) of solution in a volumetric flask from a single crushed tablet to ensure quantitative transfer and high accuracy. Part 2 (3 marks): Explain why dilute sulfuric acid must be added to the conical flask before starting the titration, and explain why hydrochloric acid is unsuitable for this purpose. Part 3 (3 marks): The student titrated \(25.0\text{ cm}^3\) portions of the prepared tablet solution against \(0.0150\text{ mol dm}^{-3}\) \(KMnO_4\) solution. The mean titre was \(14.30\text{ cm}^3\). Calculate the mass of anhydrous \(FeSO_4\) (\(M_r = 151.9\)) in the original tablet. (The ionic equation is \(MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\)).
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Worked solution
Part 1: Crush the tablet using a pestle and mortar. Dissolve the crushed tablet in a small volume of dilute sulfuric acid (or deionised water) in a beaker, stirring with a glass rod. Transfer the solution to a \(250.0\text{ cm}^3\) volumetric flask through a funnel. Rinse the beaker, glass rod, and funnel with deionised water, adding all washings to the volumetric flask. Make up the solution to the graduation mark with deionised water until the bottom of the meniscus is on the line. Stopper the flask and invert several times to mix. Part 2: Dilute sulfuric acid is added to supply the \(H^+\) ions required for the reduction of \(MnO_4^-\) to \(Mn^{2+}\). Hydrochloric acid is unsuitable because the \(MnO_4^-\) ions are strong enough oxidising agents to oxidise the \(Cl^-\) ions in the acid to \(Cl_2\) gas, leading to an inaccurately large titre of \(KMnO_4\). Part 3: Moles of \(MnO_4^-\) in the titration = \(0.0150 \times (14.30 / 1000) = 2.145 \times 10^{-4}\text{ mol}\). From the stoichiometric equation, 1 mole of \(MnO_4^-\) reacts with 5 moles of \(Fe^{2+}\). Moles of \(Fe^{2+}\) in \(25.0\text{ cm}^3\) = \(5 \times 2.145 \times 10^{-4} = 1.0725 \times 10^{-3}\text{ mol}\). Moles of \(Fe^{2+}\) in the total \(250.0\text{ cm}^3\) volumetric flask = \(1.0725 \times 10^{-3} \times 10 = 0.010725\text{ mol}\). Mass of anhydrous \(FeSO_4\) = \(0.010725 \times 151.9 = 1.6291\text{ g}\), which rounds to \(1.63\text{ g}\) (to 3 significant figures).
Marking scheme
Part 1: [1 mark] Dissolve crushed tablet in beaker and transfer quantitatively (using washes of beaker, rod, and funnel) into the volumetric flask. [1 mark] Fill with deionised water until the bottom of the meniscus is on the graduation mark. [1 mark] Stopper and invert the flask multiple times to ensure the solution is homogeneous. [1 mark] Mention use of dilute acid during dissolution to prevent oxidation of \(Fe^{2+}\) to \(Fe^{3+}\). Part 2: [1 mark] Sulfuric acid provides \(H^+\) ions needed for the reduction of \(MnO_4^-\). [1 mark] \(Cl^-\) from HCl would be oxidised to \(Cl_2\) by \(MnO_4^-\). [1 mark] This would result in an anomalously high titre value. Part 3: [1 mark] Moles of \(MnO_4^-\) = \(2.145 \times 10^{-4}\text{ mol}\). [1 mark] Moles of \(Fe^{2+}\) in \(250\text{ cm}^3\) = \(0.010725\text{ mol}\). [1 mark] Mass of anhydrous \(FeSO_4\) = \(1.63\text{ g}\) (accept \(1.62\text{ g}\) to \(1.64\text{ g}\); must be 3 sig figs).
CH05 Section B
For each question select the best response from options A, B, C, or D.
30 Question · 30 marks
Question 1 · 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 (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) and \(25.0\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\) sodium hydroxide.
What is the pH of this buffer solution at \(298\text{ K}\)?
A.4.69
B.4.87
C.5.05
D.5.23
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Worked solution
First, calculate the initial number of moles of propanoic acid (\(HA\)) and sodium hydroxide (\(NaOH\)):
1 mark for the correct pH value (C). - Reject: other options due to incorrect mole calculations or inverting the buffer ratio.
Question 2 · Multiple Choice
1 marks
The Born-Haber cycle for calcium chloride, \(\text{CaCl}_2(s)\), is constructed using the following data:
- Enthalpy of formation of \(\text{CaCl}_2(s) = -796\text{ kJ mol}^{-1}\) - Enthalpy of atomisation of \(\text{Ca}(s) = +178\text{ kJ mol}^{-1}\) - First ionisation energy of \(\text{Ca}(g) = +590\text{ kJ mol}^{-1}\) - Second ionisation energy of \(\text{Ca}(g) = +1145\text{ kJ mol}^{-1}\) - Bond dissociation enthalpy of \(\text{Cl}_2(g) = +242\text{ kJ mol}^{-1}\) - First electron affinity of \(\text{Cl}(g) = -349\text{ kJ mol}^{-1}\)
What is the lattice enthalpy of dissociation for calcium chloride, \(\text{CaCl}_2(s)\)?
A.+661\text{ kJ mol}^{-1}
B.+2253\text{ kJ mol}^{-1}
C.+2495\text{ kJ mol}^{-1}
D.+2602\text{ kJ mol}^{-1}
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Worked solution
The Born-Haber cycle equation for the lattice dissociation enthalpy (\(\Delta_L H^\ominus\)) of \(\text{CaCl}_2(s)\) is:
1 mark for the correct lattice enthalpy value (B). - Reject: options with incorrect handling of the stoichiometric factor for the electron affinity of chlorine or incorrect signs.
Question 3 · Multiple Choice
1 marks
The reaction between substances \(\text{A}\) and \(\text{B}\) is studied at a constant temperature. The following initial rate data were obtained:
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Worked solution
First, determine the order of reaction with respect to each reactant: - Comparing Experiments 1 and 2: \([\text{B}]\) is kept constant, while \([\text{A}]\) is doubled. The rate increases by a factor of \(\frac{8.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 4 = 2^2\). Therefore, the reaction is second order with respect to \(\text{A}\). - Comparing Experiments 2 and 3: \([\text{A}]\) is kept constant, while \([\text{B}]\) is doubled. The rate increases by a factor of \(\frac{1.6 \times 10^{-3}}{8.0 \times 10^{-4}} = 2 = 2^1\). Therefore, the reaction is first order with respect to \(\text{B}\).
The rate equation is: \(\text{Rate} = k [\text{A}]^2 [\text{B}]\).
Using the data from Experiment 1 to calculate the rate constant \(k\):
What is the standard cell potential (\(E^\ominus_{\text{cell}}\)) and the species that acts as the reducing agent when an electrochemical cell constructed from these half-cells operates under standard conditions?
A.E^\ominus_{\text{cell}} = +0.56\text{ V}; reducing agent is \text{Fe}^{2+}(aq)
B.E^\ominus_{\text{cell}} = +0.56\text{ V}; reducing agent is \text{Cr}^{3+}(aq)
C.E^\ominus_{\text{cell}} = +2.10\text{ V}; reducing agent is \text{Fe}^{2+}(aq)
D.E^\ominus_{\text{cell}} = +2.10\text{ V}; reducing agent is \text{Cr}_2\text{O}_7^{2-}(aq)
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Since \(\text{Fe}^{2+}(aq)\) is oxidized to \(\text{Fe}^{3+}(aq)\), it acts as the electron donor (reducing agent).
Marking scheme
1 mark for the correct cell potential and identification of the reducing agent (A). - Reject: options that add the electrode potentials or identify the wrong species as the reducing agent.
Question 5 · Multiple Choice
1 marks
Which of the following transition metal ions has the ground-state electron configuration \([\text{Ar}] 3d^2\)?
A.\text{Ti}^{3+}
B.\text{V}^{3+}
C.\text{Cr}^{3+}
D.\text{Mn}^{3+}
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Worked solution
Let us determine the electron configuration of each species: - \(\text{Ti}\) (atomic number 22): \([\text{Ar}] 4s^2 3d^2\). The \(\text{Ti}^{3+}\) ion loses three electrons (\(4s^2\) and one \(3d\)), giving \([\text{Ar}] 3d^1\). - \(\text{V}\) (atomic number 23): \([\text{Ar}] 4s^2 3d^3\). The \(\text{V}^{3+}\) ion loses three electrons (\(4s^2\) and one \(3d\)), giving \([\text{Ar}] 3d^2\). - \(\text{Cr}\) (atomic number 24): \([\text{Ar}] 4s^1 3d^5\). The \(\text{Cr}^{3+}\) ion loses three electrons (\(4s^1\) and two \(3d\)), giving \([\text{Ar}] 3d^3\). - \(\text{Mn}\) (atomic number 25): \([\text{Ar}] 4s^2 3d^5\). The \(\text{Mn}^{3+}\) ion loses three electrons (\(4s^2\) and one \(3d\)), giving \([\text{Ar}] 3d^4\).
Marking scheme
1 mark for selecting the correct ion (B). - Reject: other options due to incorrect electron counts or forgetting that \(4s\) electrons are lost first.
Question 6 · Multiple Choice
1 marks
An aqueous solution containing a metal ion, \(\text{M}^{3+}(aq)\), reacts with a limited amount of aqueous sodium hydroxide to form a green precipitate. This precipitate dissolves in excess aqueous sodium hydroxide to form a deep green solution.
What is the identity of the metal ion, \(\text{M}^{3+}(aq)\)?
A.\text{Al}^{3+}(aq)
B.\text{Cr}^{3+}(aq)
C.\text{Fe}^{3+}(aq)
D.\text{Fe}^{2+}(aq)
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Worked solution
- \(\text{Al}^{3+}(aq)\) reacts with \(\text{NaOH}\) to form a white precipitate which dissolves in excess to form a colorless solution. - \(\text{Cr}^{3+}(aq)\) reacts with a limited amount of \(\text{NaOH}\) to form a green precipitate, \(\text{Cr(H}_2\text{O)}_3\text{(OH)}_3(s)\), which dissolves in excess \(\text{NaOH}\) to form a deep green solution of \([\text{Cr(OH)}_6]^{3-}(aq)\). This is due to its amphoteric nature. - \(\text{Fe}^{3+}(aq)\) reacts with \(\text{NaOH}\) to form a brown precipitate that is insoluble in excess \(\text{NaOH}\). - \(\text{Fe}^{2+}(aq)\) is not a \(3+\) ion and its green precipitate does not dissolve in excess \(\text{NaOH}\).
Marking scheme
1 mark for identifying the correct metal ion (B). - Reject: other options based on precipitate colors or solubility in excess alkali.
Question 7 · Multiple Choice
1 marks
A student titrates \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) aqueous ammonia (\(\text{NH}_3\), a weak base) in a conical flask with \(0.100\text{ mol dm}^{-3}\) hydrochloric acid (\(\text{HCl}\), a strong acid) from a burette.
Which of the following indicators is most suitable for detecting the equivalence point of this titration, and what is its color change at the end-point?
A.Phenolphthalein; color change from pink to colorless
B.Phenolphthalein; color change from colorless to pink
C.Methyl orange; color change from yellow to orange/red
D.Methyl orange; color change from red to yellow
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Worked solution
In a titration of a weak base (\(\text{NH}_3\)) with a strong acid (\(\text{HCl}\)), the equivalence point occurs at a pH below 7 (typically around pH 5). - Phenolphthalein changes color in the pH range 8.3 - 10.0, so it is unsuitable as it would change color long before the equivalence point is reached. - Methyl orange changes color in the pH range 3.1 - 4.4, which lies within the region of rapid pH change for this titration. - Since the weak base (\(\text{NH}_3\)) is in the flask initially, the indicator starts in alkaline conditions (where methyl orange is yellow) and changes to acidic conditions (where methyl orange is red/orange). Thus, the correct color change at the end-point is yellow to orange/red.
Marking scheme
1 mark for selecting the correct indicator and its color change (C). - Reject: other options due to incorrect indicator selection or incorrect direction of color change.
1 mark for the correct minimum temperature (C). - Reject: other options due to conversion errors, division errors, or inverting the division.
Question 9 · multiple-choice
1 marks
A mixture is prepared by adding 15.0 cm3 of 0.100 mol dm-3 sodium hydroxide solution to 25.0 cm3 of 0.100 mol dm-3 propanoic acid. The acid dissociation constant, Ka, of propanoic acid is 1.35 x 10-5 mol dm-3. What is the pH of the resulting buffer solution?
A.4.69
B.5.05
C.4.87
D.3.43off-by-one error (uncorrected pH)
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Worked solution
First, calculate the initial moles of propanoic acid: n(HA) = 0.0250 dm3 x 0.100 mol dm-3 = 2.50 x 10-3 mol. Next, calculate the moles of NaOH added: n(OH-) = 0.0150 dm3 x 0.100 mol dm-3 = 1.50 x 10-3 mol. The added OH- reacts completely with the propanoic acid to form propanoate ions: n(A-) = 1.50 x 10-3 mol, leaving unreacted propanoic acid: n(HA) = 2.50 x 10-3 mol - 1.50 x 10-3 mol = 1.00 x 10-3 mol. Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Since pKa = -log(1.35 x 10-5) = 4.87, then pH = 4.87 + log(1.50 x 10-3 / 1.00 x 10-3) = 4.87 + 0.18 = 5.05.
Marking scheme
Award 1 mark for the correct calculation and selection of Option B. Reject other options.
Question 10 · multiple-choice
1 marks
The reaction between peroxodisulfate ions, S2O82-(aq), and iodide ions, I-(aq), is catalyzed by Fe2+(aq) ions. Which of the following statements explains why Fe2+(aq) ions are effective as a homogeneous catalyst in this reaction?
A.They oxidize iodide ions to iodine and are themselves reduced to solid metallic iron.
B.They change the overall reaction from being endothermic to being highly exothermic.
C.They provide an alternative pathway with steps involving collisions between oppositely charged species, reducing the activation energy.
D.They increase the rate of reaction by increasing the equilibrium constant of the process.
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Worked solution
The uncatalyzed reaction requires two negatively charged ions (S2O82- and I-) to collide, which experiences high electrostatic repulsion and results in a high activation energy. The Fe2+ catalyst provides an alternative two-step pathway where each step involves collisions between oppositely charged species: Fe2+ and S2O82-, then Fe3+ and I-. These attractive forces lower the activation energy of the rate-determining step.
Marking scheme
Award 1 mark for the correct explanation that the steps involve collisions between oppositely charged species (Option C).
Question 11 · multiple-choice
1 marks
Consider the following reaction: X(s) -> Y(s) + Z(g) for which the standard enthalpy change is +117 kJ mol-1. The standard entropies of the species involved are: S(X) = 82.4 J K-1 mol-1, S(Y) = 146.5 J K-1 mol-1, and S(Z) = 205.1 J K-1 mol-1. Above which temperature does this reaction become feasible?
A.162 K
B.333 K
C.435 K
D.624 K
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Worked solution
First, calculate the standard entropy change: S = S(Y) + S(Z) - S(X) = 146.5 + 205.1 - 82.4 = 269.2 J K-1 mol-1. The reaction is feasible when G <= 0, which occurs when T >= H/S. Convert H to J mol-1: H = 117000 J mol-1. Thus, T >= 117000 / 269.2 = 434.6 K, which rounds to 435 K.
Marking scheme
Award 1 mark for calculating the correct temperature limit of 435 K (Option C).
Question 12 · multiple-choice
1 marks
The rate equation for a chemical reaction is rate = k[A]2[B]. If the initial concentration of A is tripled and the initial concentration of B is halved, by what factor does the initial rate of the reaction change?
A.1.5
B.3.0
C.4.5
D.9.0
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Worked solution
By substituting the new concentrations into the rate expression: new rate = k x (3[A])2 x (0.5[B]) = k x 9[A]2 x 0.5[B] = 4.5 x k[A]2[B] = 4.5 x original rate. Therefore, the rate increases by a factor of 4.5.
Marking scheme
Award 1 mark for identifying the correct factor of 4.5 (Option C).
Question 13 · multiple-choice
1 marks
Use the standard electrode potentials given below to determine which mixture will undergo a spontaneous redox reaction under standard conditions. [1] Fe3+(aq) + e- <=> Fe2+(aq) (E = +0.77 V); [2] Cr3+(aq) + 3e- <=> Cr(s) (E = -0.74 V); [3] Cl2(g) + 2e- <=> 2Cl-(aq) (E = +1.36 V); [4] Fe2+(aq) + 2e- <=> Fe(s) (E = -0.44 V).
A.Cr3+(aq) and Cl-(aq)
B.Fe3+(aq) and Cl-(aq)
C.Fe2+(aq) and Cr(s)
D.Cl-(aq) and Fe(s)
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Worked solution
A spontaneous redox reaction occurs if the cell potential (E_cell = E_reduction - E_oxidation) is positive. For a mixture of Fe2+(aq) and Cr(s), Fe2+ is reduced to Fe(s) (E = -0.44 V) and Cr(s) is oxidized to Cr3+(aq) (E = -0.74 V). The cell potential is E_cell = -0.44 - (-0.74) = +0.30 V. Since E_cell > 0, the reaction is spontaneous under standard conditions.
Marking scheme
Award 1 mark for identifying the spontaneous mixture of Fe2+(aq) and Cr(s) (Option C).
Question 14 · multiple-choice
1 marks
A sample of a volatile liquid of mass 0.252 g is completely vaporized at 373 K and 101 kPa. The volume of the gas produced is 82.0 cm3. What is the relative molecular mass (Mr) of the liquid? (The gas constant R = 8.314 J K-1 mol-1)
A.25.3
B.94.3
C.114.1
D.253.2
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Worked solution
Use the ideal gas equation: pV = nRT. Convert all given values to SI units: p = 101000 Pa, V = 82.0 x 10-6 m3, T = 373 K. Calculate moles of gas: n = pV / RT = (101000 x 82.0 x 10-6) / (8.314 x 373) = 8.282 / 3101.12 = 0.002671 mol. Calculate molecular mass: Mr = mass / n = 0.252 g / 0.002671 mol = 94.3 g mol-1.
Marking scheme
Award 1 mark for the correct calculation of Mr as 94.3 (Option B).
Question 15 · multiple-choice
1 marks
The successive ionization energies of a Period 3 element X are: 1st = 578 kJ mol-1, 2nd = 1817 kJ mol-1, 3rd = 2745 kJ mol-1, 4th = 11577 kJ mol-1, and 5th = 14842 kJ mol-1. In which group of the Periodic Table is element X located?
A.Group 2
B.Group 13
C.Group 14
D.Group 15
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Worked solution
The largest increase in successive ionization energy occurs between the third and fourth values (2745 to 11577 kJ mol-1). This large jump indicates that the fourth electron is removed from an inner quantum shell. Therefore, element X must have 3 electrons in its outer shell, placing it in Group 13 (Aluminium).
Marking scheme
Award 1 mark for determining that the element belongs to Group 13 (Option B).
Question 16 · multiple-choice
1 marks
Given the standard enthalpy changes of combustion below, what is the standard enthalpy of formation of propane, C3H8(g)? H_combustion[C(s, graphite)] = -393.5 kJ mol-1, H_combustion[H2(g)] = -285.8 kJ mol-1, H_combustion[C3H8(g)] = -2219.2 kJ mol-1.
A.-104.5 kJ mol-1
B.+104.5 kJ mol-1
C.-1539.9 kJ mol-1
D.-4542.9 kJ mol-1
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Worked solution
The chemical equation for the formation of propane is: 3C(s) + 4H2(g) -> C3H8(g). Using a Hess's cycle based on combustion data: H_formation = sum(H_combustion of reactants) - sum(H_combustion of products) = [3 x (-393.5) + 4 x (-285.8)] - [-2219.2] = [-1180.5 - 1143.2] + 2219.2 = -2323.7 + 2219.2 = -104.5 kJ mol-1.
Marking scheme
Award 1 mark for the correct enthalpy of formation value of -104.5 kJ mol-1 (Option A).
Question 17 · Multiple Choice
1 marks
For the ligand substitution reaction shown below: \([Co(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CoCl_4]^{2-}(aq) + 6H_2O(l)\) What is the primary thermodynamic driving force that makes this reaction feasible at higher temperatures?
A.The decrease in coordination number from 6 to 4 releases energy due to decreased steric strain.
B.The coordination geometry changes from octahedral to tetrahedral, which is inherently more stable.
C.The reaction is highly exothermic because the Co-Cl bond is stronger than the Co-O bond.
D.There is a positive entropy change because the number of particles in solution increases from 5 to 7.
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Worked solution
The reaction has a positive entropy change (\(\Delta S^\theta > 0\)) because 5 reactant particles in solution react to produce 7 product particles. Since \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), a positive entropy change makes the free energy change increasingly negative (more feasible) as temperature (\(T\)) increases.
Marking scheme
1 mark for selecting correct option D.
Question 18 · Multiple Choice
1 marks
The following experimental data can be used to construct a Born-Haber cycle for calcium chloride, \(CaCl_2\): 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 = \(+1145 \text{ kJ mol}^{-1}\); Enthalpy of atomisation of chlorine = \(+121 \text{ kJ mol}^{-1}\); Electron affinity of chlorine = \(-349 \text{ kJ mol}^{-1}\); Enthalpy of formation of calcium chloride = \(-796 \text{ kJ mol}^{-1}\). What is the experimental lattice enthalpy of dissociation of calcium chloride?
A.\(+1457 \text{ kJ mol}^{-1}\)
B.\(+1904 \text{ kJ mol}^{-1}\)
C.\(+2253 \text{ kJ mol}^{-1}\)
D.\(+2595 \text{ kJ mol}^{-1}\)
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A buffer solution is prepared at \(298 \text{ K}\) by mixing \(25.0 \text{ cm}^3\) of \(0.100 \text{ mol dm}^{-3}\) methanoic acid (\(HCOOH\), \(K_a = 1.6 \times 10^{-4} \text{ mol dm}^{-3}\)) with \(15.0 \text{ cm}^3\) of \(0.100 \text{ mol dm}^{-3}\) sodium hydroxide (\(NaOH\)). What is the pH of the resulting buffer solution at \(298 \text{ K}\)?
A.3.62
B.3.80
C.3.98
D.4.18
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Worked solution
Moles of \(HCOOH\) initially = \(0.0250 \times 0.100 = 2.50 \times 10^{-3} \text{ mol}\). Moles of \(NaOH\) added = \(0.0150 \times 0.100 = 1.50 \times 10^{-3} \text{ mol}\). After neutralisation: Moles of \(HCOO^-\text{(aq)}\) formed = \(1.50 \times 10^{-3} \text{ mol}\); Moles of \(HCOOH\) remaining = \(2.50 \times 10^{-3} - 1.50 \times 10^{-3} = 1.00 \times 10^{-3} \text{ mol}\). Using the buffer equation: \(pH = pK_a + \log_{10}([HCOO^-]/[HCOOH])\). Since \(pK_a = -\log_{10}(1.6 \times 10^{-4}) = 3.80\), then \(pH = 3.80 + \log_{10}(1.50 \times 10^{-3} / 1.00 \times 10^{-3}) = 3.80 + 0.18 = 3.98\).
Marking scheme
1 mark for selecting correct option C.
Question 20 · Multiple Choice
1 marks
The table shows the initial rates of reaction for different initial concentrations of three reactants, \(A\), \(B\), and \(C\), in the reaction: \(A + B + C \rightarrow \text{products}\). Experiment 1: \([A]=0.10\), \([B]=0.10\), \([C]=0.10\), Initial Rate = \(2.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}\). Experiment 2: \([A]=0.20\), \([B]=0.10\), \([C]=0.10\), Initial Rate = \(4.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}\). Experiment 3: \([A]=0.10\), \([B]=0.20\), \([C]=0.10\), Initial Rate = \(8.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}\). Experiment 4: \([A]=0.10\), \([B]=0.10\), \([C]=0.20\), Initial Rate = \(2.0 \times 10^{-4} \text{ mol dm}^{-3}\text{s}^{-1}\). What is the overall order of the reaction and the units of the rate constant \(k\)?
A.Overall order = 2, units = \(\text{mol}^{-1} \text{dm}^3 \text{s}^{-1}\)
B.Overall order = 3, units = \(\text{mol}^{-2} \text{dm}^6 \text{s}^{-1}\)
C.Overall order = 3, units = \(\text{mol}^2 \text{dm}^{-6} \text{s}^{-1}\)
D.Overall order = 4, units = \(\text{mol}^{-3} \text{dm}^9 \text{s}^{-1}\)
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Worked solution
Compare Exp 1 and Exp 2: doubling \([A]\) doubles the rate, so the reaction is 1st order with respect to \(A\). Compare Exp 1 and Exp 3: doubling \([B]\) quadruples the rate, so the reaction is 2nd order with respect to \(B\). Compare Exp 1 and Exp 4: doubling \([C]\) does not change the rate, so the reaction is 0th order with respect to \(C\). Rate Equation: \(\text{Rate} = k[A][B]^2\). Overall order = \(1 + 2 + 0 = 3\). Units of \(k = \text{Rate}/([A][B]^2) = \text{mol dm}^{-3}\text{s}^{-1} / ((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2) = \text{mol}^{-2}\text{dm}^6\text{s}^{-1}\).
Marking scheme
1 mark for selecting correct option B.
Question 21 · Multiple Choice
1 marks
Some standard electrode potentials are shown below: \(V^{3+}(aq) + e^- \rightleftharpoons V^{2+}(aq) \quad E^\theta = -0.26 \text{ V}\); \(Co^{2+}(aq) + 2e^- \rightleftharpoons Co(s) \quad E^\theta = -0.28 \text{ V}\); \(Cr^{3+}(aq) + e^- \rightleftharpoons Cr^{2+}(aq) \quad E^\theta = -0.41 \text{ V}\); \(Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s) \quad E^\theta = -0.76 \text{ V}\). Which of the following species can reduce \(V^{3+}(aq)\) to \(V^{2+}(aq)\) but cannot reduce \(Cr^{3+}(aq)\) to \(Cr^{2+}(aq)\) under standard conditions?
A.\(Co(s)\)
B.\(Zn(s)\)
C.\(Cr^{2+}(aq)\)
D.\(V^{2+}(aq)\)
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Worked solution
For a reducing agent to reduce a species, the standard electrode potential of the reducing agent's half-cell must be more negative than that of the species being reduced. To reduce \(V^{3+}\) (\(E^\theta = -0.26 \text{ V}\)), the reducing agent's \(E^\theta\) must be less than \(-0.26 \text{ V}\). To fail to reduce \(Cr^{3+}\) (\(E^\theta = -0.41 \text{ V}\)), the reducing agent's \(E^\theta\) must be greater than \(-0.41 \text{ V}\). Only the \(Co^{2+}/Co\) half-cell (\(E^\theta = -0.28 \text{ V}\)) meets this criteria. Therefore, \(Co(s)\) is the correct species.
Marking scheme
1 mark for selecting correct option A.
Question 22 · Multiple Choice
1 marks
Which of the following statements about the oxides of Period 3 elements is correct?
A.Aluminum oxide (\(Al_2O_3\)) is amphoteric and reacts with both hydrochloric acid and aqueous sodium hydroxide.
B.Silicon dioxide (\(SiO_2\)) dissolves readily in water to form a weakly acidic solution of pH 6.
C.Sulfur dioxide (\(SO_2\)) reacts with water to form sulfuric(VI) acid (\(H_2SO_4\)).
D.Tetraphosphorus decaoxide (\(P_4O_{10}\)) has a macromolecular covalent structure and a high melting point.
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Worked solution
Aluminum oxide (\(Al_2O_3\)) is amphoteric, so it reacts with both acids (hydrochloric acid) and bases (sodium hydroxide) to form salts and water. Silicon dioxide is insoluble in water. Sulfur dioxide reacts with water to form sulfurous acid (\(H_2SO_3\)), not sulfuric(VI) acid. Tetraphosphorus decaoxide has a molecular covalent structure, not macromolecular.
Marking scheme
1 mark for selecting correct option A.
Question 23 · Multiple Choice
1 marks
In the reaction between peroxodisulfate ions (\(S_2O_8^{2-}\)) and iodide ions (\(I^-\)) catalysed by iron(II) ions (\(Fe^{2+}\)), which statement correctly describes the pathway of the catalysis?
A.\(Fe^{2+}\) increases the entropy of activation of the rate-determining step.
B.\(Fe^{2+}\) is first oxidized to \(Fe^{3+}\) by peroxodisulfate ions, and then \(Fe^{3+}\) is reduced back to \(Fe^{2+}\) by iodide ions.
C.\(Fe^{2+}\) acts as a heterogeneous catalyst, providing active sites for reactant adsorption.
D.The reaction rate is low without a catalyst because the collision of oppositely charged ions is thermodynamically unfavourable.
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Worked solution
This is homogeneous catalysis. Iron(II) ions are first oxidized to iron(III) ions by peroxodisulfate: \(2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}\). In the second step, the newly formed iron(III) ions are reduced back to iron(II) ions by iodide: \(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\). This sequence avoids the slow direct collision of two negatively charged reactant ions.
Marking scheme
1 mark for selecting correct option B.
Question 24 · Multiple Choice
1 marks
Which of the following isomeric alcohols can exhibit optical isomerism?
A.2-methylbutan-2-ol
B.3-methylpentan-3-ol
C.pentan-3-ol
D.3-methylhexan-3-ol
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Worked solution
To exhibit optical isomerism, a molecule must contain a chiral carbon atom bonded to four different groups. In 3-methylhexan-3-ol, carbon-3 is bonded to a methyl group (\(-CH_3\)), an ethyl group (\(-CH_2CH_3\)), a propyl group (\(-CH_2CH_2CH_3\)), and a hydroxyl group (\(-OH\)). Because all four groups are different, this molecule exists as a pair of optical isomers.
Marking scheme
1 mark for selecting correct option D.
Question 25 · Multiple Choice
1 marks
An ionic compound \( \text{MX}_2 \) has the following standard enthalpy changes:
- Enthalpy of atomisation of \( \text{M(s)} = +150\text{ kJ mol}^{-1} \) - First ionisation energy of \( \text{M(g)} = +590\text{ kJ mol}^{-1} \) - Second ionisation energy of \( \text{M(g)} = +1150\text{ kJ mol}^{-1} \) - Enthalpy of atomisation of \( \text{X}_2\text{(g)} = +120\text{ kJ mol}^{-1} \) (per mole of X atoms formed) - Electron affinity of \( \text{X(g)} = -350\text{ kJ mol}^{-1} \) - Enthalpy of formation of \( \text{MX}_2\text{(s)} = -640\text{ kJ mol}^{-1} \)
What is the lattice enthalpy of formation of \( \text{MX}_2\text{(s)} \)?
A.\( -2070\text{ kJ mol}^{-1} \)
B.\( -1840\text{ kJ mol}^{-1} \)
C.\( -2310\text{ kJ mol}^{-1} \)
D.\( -2530\text{ kJ mol}^{-1} \)
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Worked solution
To find the lattice enthalpy of formation (\( \Delta H^{\ominus}_{L,form} \)), we construct a Born-Haber cycle for the formation of \( \text{MX}_2(\text{s}) \):
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Worked solution
1. Determine the order with respect to A: Compare Exp 1 and Exp 2: \( [B] \) is constant, \( [A] \) doubles (\( 0.10 \to 0.20 \)), and the rate increases by a factor of 4 (\( 2.0 \times 10^{-4} \to 8.0 \times 10^{-4} \)). Therefore, the reaction is second order with respect to A.
2. Determine the order with respect to B: Compare Exp 2 and Exp 3: \( [A] \) is constant, \( [B] \) doubles (\( 0.10 \to 0.20 \)), and the rate increases by a factor of 2 (\( 8.0 \times 10^{-4} \to 1.6 \times 10^{-3} \)). Therefore, the reaction is first order with respect to B.
3. Write the rate equation: \( \text{Rate} = k[A]^2[B] \)
4. Calculate \( k \) using data from Experiment 1: \( k = \frac{\text{Rate}}{[A]^2[B]} = \frac{2.0 \times 10^{-4}}{(0.10)^2(0.10)} = \frac{2.0 \times 10^{-4}}{1.0 \times 10^{-3}} = 0.20 \)
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Worked solution
1. To obtain the maximum standard cell potential, we pair the two half-cells such that the more positive potential undergoes reduction (at the right-hand electrode / cathode) and the more negative potential undergoes oxidation (at the left-hand electrode / anode): - Cathode (reduction): \( \text{Fe}^{3+}(\text{aq}) + \text{e}^- \to \text{Fe}^{2+}(\text{aq}) \) (\( E^{\ominus} = +0.77\text{ V} \)) - Anode (oxidation): \( \text{Cr(s)} \to \text{Cr}^{3+}(\text{aq}) + 3\text{e}^- \) (\( E^{\ominus} = -0.74\text{ V} \))
3. Write the standard cell notation: - Oxidation on the left: \( \text{Cr(s)} | \text{Cr}^{3+}(\text{aq}) \) - Reduction on the right: \( \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt(s)} \) (platinum electrode is used because both Fe ions are in aqueous solution) - Combined: \( \text{Cr(s)} | \text{Cr}^{3+}(\text{aq}) || \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt(s)} \) with \( E^{\ominus}_{cell} = +1.51\text{ V} \).
Marking scheme
1 mark for the correct option A. Give 0 marks for any other option selected.
Question 29 · Multiple Choice
1 marks
Which of the following transition metal complexes can exist as a pair of optical isomers?
A.\( [\text{Co}(\text{en})_3]^{3+} \) (where en is the bidentate ligand ethane-1,2-diamine)
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Worked solution
Optical isomerism occurs in octahedral complexes containing two or more bidentate ligands, where the mirror images are non-superimposable (chiral). - Option A: \( [\text{Co}(\text{en})_3]^{3+} \) contains three bidentate ligands in an octahedral arrangement and has no plane of symmetry, allowing it to exist as non-superimposable mirror images (D and L enantiomers). - Option B: The trans-isomer of \( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \) has a plane of symmetry containing the cobalt and the two chloride ligands, meaning it is achiral. - Option C: \( [\text{Pt}(\text{NH}_3)_2\text{Cl}_2] \) is square planar and has a molecular plane of symmetry. - Option D: \( [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \) is a symmetrical octahedral complex with six identical monodentate ligands.
Marking scheme
1 mark for the correct option A. Give 0 marks for any other option selected.
Question 30 · Multiple Choice
1 marks
When an excess of concentrated aqueous ammonia is added to an aqueous solution containing hexaaquacopper(II) ions, \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \), a deep-blue solution is formed.
What is the formula of the copper(II) complex present in this deep-blue solution?
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Worked solution
When excess aqueous ammonia is added to \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \), a ligand substitution reaction occurs. Four of the water ligands are substituted by four ammonia molecules to form a deep-blue solution containing the complex ion \( [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} \).
The chemical equation for this reaction is: \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \to [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \)
Marking scheme
1 mark for the correct option A. Give 0 marks for any other option selected.
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