An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 AQA A Level Chemistry 7405 paper. Not affiliated with or reproduced from AQA.
Paper 1 Section A
Answer all questions. Show all working for calculations.
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PastPaper.question 1 · Structured
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The reaction between peroxodisulfate(VI) ions and iodide ions is represented by the following equation:
a) A series of experiments was carried out at a constant temperature to determine the rate equation for this reaction. The results are shown in the table below:
Deduce the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\), explaining your reasoning.
b) Write the rate equation for this reaction and calculate the value of the rate constant, \(k\), stating its units.
c) Explain why this reaction is slow in the absence of a catalyst, and write the formula of a transition metal ion that can act as a catalyst for this reaction.
d) In a separate experiment, the rate constant for this reaction was determined at two different temperatures: - At \(298 \text{ K}\), \(k_1 = 0.0375 \text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}\) - At \(318 \text{ K}\), \(k_2 = 0.1500 \text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)
Calculate the activation energy, \(E_a\), for this reaction in \(\text{kJ mol}^{-1}\). The gas constant, \(R = 8.31 \text{ J K}^{-1}\text{ mol}^{-1}\).
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PastPaper.workedSolution
a) Comparing Experiments 1 and 2: \([\text{I}^-]\) is constant, \([\text{S}_2\text{O}_8^{2-}]\) is doubled (from 0.0400 to 0.0800), and the initial rate is doubled (from \(1.20 \times 10^{-4}\) to \(2.40 \times 10^{-4}\)). Therefore, the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1. Comparing Experiments 2 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant, \([\text{I}^-]\) is doubled (from 0.0800 to 0.1600), and the initial rate is doubled (from \(2.40 \times 10^{-4}\) to \(4.80 \times 10^{-4}\)). Therefore, the order of reaction with respect to \(\text{I}^-\) is 1.
b) Rate equation: \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) Using Experiment 1 data: \(1.20 \times 10^{-4} = k \times 0.0400 \times 0.0800\) \(k = \frac{1.20 \times 10^{-4}}{0.0032} = 0.0375\) Units of \(k\): \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\).
c) The reaction involves two negatively charged ions (anions) repelling each other, leading to high activation energy. The transition metal ion that acts as a catalyst is \(\text{Fe}^{2+}\) (or \(\text{Fe}^{3+}\)).
a) 2 marks: - 1 mark for order 1 with respect to peroxodisulfate with explanation (e.g., doubling concentration doubles rate). - 1 mark for order 1 with respect to iodide with explanation (e.g., doubling concentration doubles rate).
b) 3 marks: - 1 mark for correct rate equation: \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (Allow consequential error from part a). - 1 mark for calculating \(k = 0.0375\) (Accept 0.038). - 1 mark for units: \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\).
c) 3 marks: - 1 mark for identifying that both reactants are negatively charged / ions of the same charge. - 1 mark for stating that they repel each other (leading to high activation energy). - 1 mark for identifying \(\text{Fe}^{2+}\) or \(\text{Fe}^{3+}\) as the catalyst.
d) 5.125 marks: - 1 mark for rearranging the Arrhenius equation or substituting values correctly: \(\ln(4) = \frac{E_a}{8.31} \left(\frac{1}{298} - \frac{1}{318}\right)\). - 1 mark for calculating the temperature difference term correctly: \(2.11 \times 10^{-4} \text{ K}^{-1}\). - 1 mark for calculating \(\ln(4) = 1.386\). - 1.125 marks for evaluating \(E_a\) in J/mol: \(54598 \text{ J mol}^{-1}\). - 1 mark for converting to kJ/mol to 3 significant figures: \(54.6 \text{ kJ mol}^{-1}\).
PastPaper.question 2 · Structured
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The acid-catalysed iodination of propanone is represented by the equation:
a) Describe how a colorimeter can be used to monitor the progress of this reaction. Explain why the absorbance changes.
b) Explain why a large excess of propanone and acid is used in this experiment to find the order of reaction with respect to iodine.
c) In a different rate study, a 'clock reaction' is used. Explain how the time taken, \(t\), for a visible change to occur is related to the initial rate of reaction in a clock reaction.
d) In a study of the rate of a reaction, the rate constant \(k\) was found to be \(4.50 \times 10^{-3} \text{ s}^{-1}\) at \(25.0 ^\circ\text{C}\) with an activation energy, \(E_a\), of \(62.0 \text{ kJ mol}^{-1}\).
Calculate the value of the Arrhenius constant, \(A\), for this reaction. State its units. The gas constant, \(R = 8.31 \text{ J K}^{-1}\text{ mol}^{-1}\).
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PastPaper.workedSolution
a) Iodine is colored (brown/yellow/orange) whereas the other reactants and products are colorless. As the reaction proceeds, iodine is consumed, so the intensity of the color decreases. A colorimeter can measure the light absorbed by the solution at regular time intervals. The absorbance is directly proportional to the concentration of iodine, so monitoring absorbance over time allows us to monitor the change in iodine concentration.
b) By using a large excess of propanone and hydrochloric acid, their concentrations remain effectively constant throughout the reaction. Therefore, any change in rate is solely due to the change in the concentration of iodine, allowing the order with respect to iodine to be determined.
c) In a clock reaction, the time taken, \(t\), is measured for a fixed, small amount of reactant to be consumed (or product to be formed). Since this represents a constant change in concentration, \(\Delta [\text{I}_2]\), the initial rate of reaction is inversely proportional to the time taken (i.e., \(\text{Rate} \propto \frac{1}{t}\)).
d) Convert temperature to Kelvin: \(T = 25.0 + 273.15 = 298.15 \text{ K}\) (or \(298 \text{ K}\)). Convert \(E_a\) to J/mol: \(E_a = 62.0 \times 1000 = 62000 \text{ J mol}^{-1}\). From the Arrhenius equation: \(k = A e^{-\frac{E_a}{RT}}\) \(A = k e^{\frac{E_a}{RT}}\) \(\frac{E_a}{RT} = \frac{62000}{8.31 \times 298} = 25.0365\) \(e^{25.0365} = 7.468 \times 10^{10}\) \(A = (4.50 \times 10^{-3}) \times (7.468 \times 10^{10}) = 3.36 \times 10^8\) Units of \(A\): \(\text{s}^{-1}\).
PastPaper.markingScheme
a) 3 marks: - 1 mark for stating iodine is colored and products/other reactants are colorless. - 1 mark for stating that absorbance decreases as iodine is consumed. - 1 mark for linking absorbance to concentration (directly proportional).
b) 2 marks: - 1 mark for stating that their concentrations remain effectively constant. - 1 mark for stating that the rate then only depends on the concentration of iodine.
c) 3 marks: - 1 mark for noting that a fixed, small amount of reactant is consumed / product formed before the color change occurs. - 1 mark for stating that rate is change in concentration over time. - 1 mark for concluding that \(\text{Rate} \propto \frac{1}{t}\) (or rate is proportional to \(1/t\)).
d) 5.125 marks: - 1 mark for converting temperature to Kelvin (\(298 \text{ K}\)) and \(E_a\) to \(62000 \text{ J mol}^{-1}\). - 1 mark for rearranging: \(A = k e^{\frac{E_a}{RT}}\). - 1 mark for calculating exponent term: \(\frac{E_a}{RT} = 25.04\) or \(e^{25.04} = 7.47 \times 10^{10}\). - 1.125 marks for evaluating \(A\) correctly as \(3.36 \times 10^8\) (accept range \(3.30 \times 10^8\) to \(3.42 \times 10^8\) depending on rounding of \(T\) to 298 or 298.15). - 1 mark for units: \(\text{s}^{-1}\).
PastPaper.question 3 · Structured
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A student carries out a titration by adding \(0.120 \text{ mol dm}^{-3}\) sodium hydroxide solution to \(25.0 \text{ cm}^3\) of \(0.150 \text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5} \text{ mol dm}^{-3}\)).
a) Calculate the pH of the initial propanoic acid solution before any sodium hydroxide is added. Show your working.
b) Calculate the volume, in \(\text{cm}^3\), of the sodium hydroxide solution required to reach the equivalence point of this titration.
c) Calculate the pH of the titration mixture after the addition of \(15.0 \text{ cm}^3\) of the \(0.120 \text{ mol dm}^{-3}\) sodium hydroxide solution.
d) Suggest a suitable indicator for this titration. Justify your choice by referring to the pH range of the indicator and the nature of the equivalence point.
d) Phenolphthalein. The equivalence point for a weak acid-strong base titration is greater than pH 7 (at around 8–9). Phenolphthalein has a pH range of 8.3–10.0, which perfectly matches the vertical region of rapid pH change during this titration.
PastPaper.markingScheme
a) 3 marks: - 1 mark for weak acid expression or substitution: \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]}\). - 1 mark for \([\text{H}^+] = 1.42 \times 10^{-3} \text{ mol dm}^{-3}\). - 1 mark for pH = 2.85 (must be 2 decimal places).
b) 2 marks: - 1 mark for calculating moles of propanoic acid: \(3.75 \times 10^{-3} \text{ mol}\). - 1 mark for volume of NaOH: \(31.25 \text{ cm}^3\) (accept 31.3).
c) 6 marks: - 1 mark for calculating moles of \(\text{OH}^-\right.\) added = \(1.80 \times 10^{-3} \text{ mol}\). - 1 mark for calculating remaining moles of \(\text{HA}\) = \(1.95 \times 10^{-3} \text{ mol}\). - 1 mark for calculating moles of \(\text{A}^-\right.\) formed = \(1.80 \times 10^{-3} \text{ mol}\). - 1 mark for using the buffer equation correctly (concentrations or moles ratio). - 1 mark for \([\text{H}^+] = 1.46 \times 10^{-5} \text{ mol dm}^{-3}\). - 1 mark for pH = 4.84 (must be 2 decimal places).
d) 2.125 marks: - 1 mark for identifying phenolphthalein. - 1.125 marks for justifying: the equivalence point occurs at a pH > 7 (alkaline) which falls within the indicator's range of 8.3–10.0 / color change range.
a) The value of the ionic product of water, \(K_w\), is \(5.48 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}\) at \(50.0 ^\circ\text{C}\).
Calculate the pH of pure water at this temperature. Explain whether this water is acidic, alkaline, or neutral.
b) A solution is prepared by mixing \(35.0 \text{ cm}^3\) of \(0.150 \text{ mol dm}^{-3}\) hydrochloric acid with \(25.0 \text{ cm}^3\) of \(0.200 \text{ mol dm}^{-3}\) barium hydroxide solution at \(25.0 ^\circ\text{C}\) (where \(K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}\)).
Calculate the pH of the resulting mixture. Show all of your working.
c) Explain, with the aid of an ionic equation, how an acidic buffer solution containing ethanoic acid and sodium ethanoate resists change in pH when a small amount of sodium hydroxide is added.
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PastPaper.workedSolution
a) In pure water, \([\text{H}^+] = [\text{OH}^-]\). Therefore, \(K_w = [\text{H}^+]^2\). \([\text{H}^+] = \sqrt{K_w} = \sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7} \text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(2.34 \times 10^{-7}) = 6.63\). The water is neutral because the concentration of hydrogen ions still equals the concentration of hydroxide ions (\([\text{H}^+] = [\text{OH}^-]\)).
b) Moles of \(\text{H}^+\right.\) from \(\text{HCl}\): \(n(\text{H}^+) = 0.0350 \times 0.150 = 5.25 \times 10^{-3} \text{ mol}\). Barium hydroxide, \(\text{Ba(OH)}_2\), dissociates to release 2 moles of \(\text{OH}^-\right.\) per mole: Moles of \(\text{Ba(OH)}_2\) = \(0.0250 \times 0.200 = 5.00 \times 10^{-3} \text{ mol}\). Moles of \(\text{OH}^-\right.\) = \(2 \times 5.00 \times 10^{-3} = 1.00 \times 10^{-2} \text{ mol}\). Excess moles of \(\text{OH}^-\right.\): \(n(\text{OH}^-)_{\text{excess}} = 1.00 \times 10^{-2} - 5.25 \times 10^{-3} = 4.75 \times 10^{-3} \text{ mol}\). Total volume of solution = \(35.0 + 25.0 = 60.0 \text{ cm}^3 = 0.0600 \text{ dm}^3\). Concentration of excess \(\text{OH}^-\right.\): \([\text{OH}^-] = \frac{4.75 \times 10^{-3}}{0.0600} = 0.07917 \text{ mol dm}^{-3}\). Using \(K_w\): \([\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.07917} = 1.263 \times 10^{-13} \text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(1.263 \times 10^{-13}) = 12.90\).
c) When sodium hydroxide is added, it provides hydroxide ions (\(\text{OH}^-\right.\)). These react with the weak acid, ethanoic acid (\(\text{CH}_3\text{COOH}\)), to form ethanoate ions and water: \(\text{CH}_3\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\) Since the added hydroxide ions are consumed by reacting with the weak acid, the ratio of weak acid to its conjugate base changes very little, and the pH remains almost constant.
PastPaper.markingScheme
a) 4 marks: - 1 mark for relationship \([\text{H}^+] = \sqrt{K_w}\). - 1 mark for calculating \([\text{H}^+] = 2.34 \times 10^{-7} \text{ mol dm}^{-3}\). - 1 mark for pH = 6.63. - 1 mark for explanation: neutral because \([\text{H}^+] = [\text{OH}^-]\).
b) 7.125 marks: - 1 mark for moles of \(\text{H}^+\right.\) = \(5.25 \times 10^{-3} \text{ mol}\). - 1 mark for realizing \(\text{Ba(OH)}_2\) produces 2 moles of \(\text{OH}^-\right.\). - 1 mark for moles of \(\text{OH}^-\right.\) = \(1.00 \times 10^{-2} \text{ mol}\). - 1 mark for excess moles of \(\text{OH}^-\right.\) = \(4.75 \times 10^{-3} \text{ mol}\). - 1 mark for total volume = \(0.0600 \text{ dm}^3\) and calculating \([\text{OH}^-] = 0.0792 \text{ mol dm}^{-3}\). - 1.125 marks for calculating \([\text{H}^+] = 1.26 \times 10^{-13} \text{ mol dm}^{-3}\). - 1 mark for pH = 12.90 (accept 12.9).
c) 2 marks: - 1 mark for correct ionic equation: \(\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}\). - 1 mark for explanation: the added \(\text{OH}^-\right.\) is removed/consumed by the large reservoir of ethanoic acid, keeping the ratio of acid to salt/base relatively constant.
PastPaper.question 5 · Structured
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Chromium is a transition metal that exhibits a range of oxidation states and coordinates with various ligands to form complex ions.
a) State the full electronic configuration of a chromium atom and a \(\text{Cr}^{3+}\) ion. Explain why the electronic configuration of a chromium atom is considered unusual.
b) Describe the observations made and write equations for the reactions that occur when excess aqueous ammonia is added dropwise to a solution containing \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}\) ions.
c) Green chromium(III) chloride can exist as different isomers in solution. One isomer has the formula \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}\).
Calculate the mass of silver chloride (\(\text{AgCl}\), \(M_r = 143.4\)) precipitate formed when \(50.0 \text{ cm}^3\) of a \(0.120 \text{ mol dm}^{-3}\) solution of this isomer is reacted with an excess of silver nitrate solution.
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PastPaper.workedSolution
a) Electronic configuration of Cr atom: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\) (or \([\text{Ar}]3d^5 4s^1\)). Electronic configuration of \(\text{Cr}^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\) (or \([\text{Ar}]3d^3\)). It is unusual because an electron from the \(4s\) subshell is promoted to the \(3d\) subshell to give a half-filled \(3d^5\) subshell, which is exceptionally stable due to minimized electron repulsion.
b) When ammonia is added dropwise, it acts as a weak base, accepting protons from water ligands: \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}(\text{aq}) + 3\text{NH}_3(\text{aq}) \rightarrow [\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3](\text{s}) + 3\text{NH}_4^+(\text{aq})\) Observation: A green solution forms a green precipitate. When excess ammonia is added, ligand substitution occurs because ammonia acts as a ligand: \([\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3](\text{s}) + 6\text{NH}_3(\text{aq}) \rightarrow [\text{Cr}(\text{NH}_3)_6]^{3+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) + 3\text{OH}^-(\text{aq})\) Observation: The green precipitate dissolves to form a purple solution.
c) Moles of isomer in solution = \(C \times V = 0.120 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 6.00 \times 10^{-3} \text{ mol}\). In the complex \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}\), only the two chloride ions outside the coordination sphere dissociate in solution to form free \(\text{Cl}^-\right.\) ions: \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}(\text{aq}) \rightarrow [\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\). Thus, 1 mole of the isomer produces 2 moles of free chloride ions. Moles of \(\text{Cl}^-\right.\) = \(2 \times 6.00 \times 10^{-3} = 1.20 \times 10^{-2} \text{ mol}\). Reaction with silver ions: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\). Moles of \(\text{AgCl}\) precipitate = \(1.20 \times 10^{-2} \text{ mol}\). Mass of \(\text{AgCl}\) = \(1.20 \times 10^{-2} \text{ mol} \times 143.4 \text{ g mol}^{-1} = 1.72 \text{ g}\).
PastPaper.markingScheme
a) 3 marks: - 1 mark for correct Cr atom configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\). - 1 mark for correct \(\text{Cr}^{3+}\) configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\). - 1 mark for explanation of stability/half-filled 3d subshell.
b) 5 marks: - 1 mark for green precipitate formula/equation 1: \([\text{Cr}(\text{H}_2\text{O})_6]^{3+} + 3\text{NH}_3 \rightarrow [\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3] + 3\text{NH}_4^+\). - 1 mark for observation of green precipitate from green solution. - 1 mark for purple solution formula/equation 2: \([\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3] + 6\text{NH}_3 \rightarrow [\text{Cr}(\text{NH}_3)_6]^{3+} + 3\text{H}_2\text{O} + 3\text{OH}^-\). - 1 mark for observation of precipitate dissolving to form a purple solution. - 1 mark for correctly balancing both equations.
c) 5.125 marks: - 1 mark for calculating moles of isomer: \(6.00 \times 10^{-3} \text{ mol}\). - 1 mark for realizing only 2 chloride ions are outside the coordination sphere/dissociate. - 1 mark for calculating moles of free chloride ions = \(1.20 \times 10^{-2} \text{ mol}\). - 1.125 marks for calculating mass of \(\text{AgCl}\) = \(1.72 \text{ g}\) (allow 1.721 g). - 1 mark for using appropriate 3 significant figures.
PastPaper.question 6 · Structured
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Ethane-1,2-diamine (often abbreviated as 'en') is a bidentate ligand that reacts with transition metal ions.
a) Define the term bidentate ligand. Suggest the formula and structure of one other common bidentate ligand.
b) A solution containing hexaaquairon(III) ions, \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\), is reacted with a solution of ethane-1,2-diamine to form a tris-complex.
Write an equation for this complete ligand substitution reaction. Explain, in terms of entropy change (\(\Delta S^\theta\)) and enthalpy change (\(\Delta H^\theta\)), why this reaction is highly feasible. This is known as the chelate effect.
c) Complex ions are often colored. State the shape and coordination number of the \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\) ion. Explain how the d-orbitals of a transition metal ion are split in energy when ligands approach to form this complex.
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PastPaper.workedSolution
a) A bidentate ligand is a species that can donate two lone pairs of electrons from two different atoms to a central metal ion to form two coordinate (dative covalent) bonds. Another common bidentate ligand is the ethanedioate (oxalate) ion, \(\text{C}_2\text{O}_4^{2-}\) (or its structure: \(^{-}OOC-COO^{-}\)).
b) Equation: \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}(\text{aq}) + 3\text{en}(\text{aq}) \rightarrow [\text{Fe}(\text{en})_3]^{3+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\) Explanation: - In this reaction, 4 particles on the reactant side react to form 7 particles on the product side. - This increase in the number of species in solution results in a significant increase in disorder, meaning the system has a large positive entropy change (\(\Delta S^\theta > 0\)). - The enthalpy change (\(\Delta H^\theta\)) for this reaction is very small/near zero because the coordination number remains 6, and similar strength Fe-N bonds are formed as Fe-O bonds broken. - Since Gibbs free energy change is given by \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), the large positive \(\Delta S^\theta\) and negligible \(\Delta H^\theta\) make \(\Delta G^\theta\) highly negative. Thus, the reaction is highly feasible and thermodynamically stable.
c) Shape: Octahedral. Coordination number: 6. Splitting of d-orbitals: As the ligands approach the metal ion, their lone pairs repel the electrons in the metal's d-orbitals. Because the ligands approach along the Cartesian axes, the d-orbitals pointing directly along the axes (\(d_{x^2-y^2}\) and \(d_{z^2}\)) experience greater repulsion and rise to a higher energy level, while the other three d-orbitals (\(d_{xy}\), \(d_{yz}\), \(d_{xz}\)) pointing between the axes experience less repulsion and remain at a lower energy level. This splits the five degenerate d-orbitals into two distinct energy groups.
PastPaper.markingScheme
a) 3 marks: - 1 mark for definition: donates two lone pairs from different atoms to form two coordinate bonds. - 1 mark for naming another bidentate ligand: e.g. ethanedioate / oxalate / \(\text{C}_2\text{O}_4^{2-}\). - 1 mark for drawing/showing structure of this ligand with correct donor atoms.
b) 7 marks: - 2 marks for equation: 1 mark for correct formulas, 1 mark for correct balancing. - 1 mark for explaining that number of particles increases (from 4 to 7). - 1 mark for stating entropy change (\(\Delta S^\theta\)) is highly positive. - 1 mark for stating enthalpy change (\(\Delta H^\theta\)) is approximately zero due to breaking/making similar bonds. - 1 mark for linking to \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\). - 1 mark for concluding that \(\Delta G^\theta\) is highly negative, making the reaction highly feasible.
c) 3.125 marks: - 1 mark for octahedral shape and coordination number 6. - 1.125 marks for explaining that ligand lone pairs repel metal d-orbital electrons differently depending on orbital orientation. - 1 mark for stating that the 5 d-orbitals split into two sets of different energy levels.
PastPaper.question 7 · Structured
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Enthalpy changes of solution and hydration are closely linked to the lattice energy of ionic solids.
a) Define the term enthalpy of lattice formation.
b) Use the data below to calculate the lattice enthalpy of formation of magnesium chloride, \(\text{MgCl}_2(\text{s})\).
- Enthalpy of atomisation of Mg(s) = \(+148 \text{ kJ mol}^{-1}\) - First ionisation energy of Mg(g) = \(+738 \text{ kJ mol}^{-1}\) - Second ionisation energy of Mg(g) = \(+1451 \text{ kJ mol}^{-1}\) - Enthalpy of atomisation of chlorine, \(\text{Cl}_2\)(g) = \(+121 \text{ kJ mol}^{-1}\) - Electron affinity of chlorine, Cl(g) = \(-349 \text{ kJ mol}^{-1}\) - Enthalpy of formation of magnesium chloride, \(\text{MgCl}_2(\text{s})\) = \(-642 \text{ kJ mol}^{-1}\)
c) The enthalpy of solution of magnesium chloride is \(-155 \text{ kJ mol}^{-1}\) and the hydration enthalpy of magnesium ions, \(\text{Mg}^{2+}(\text{g})\), is \(-1920 \text{ kJ mol}^{-1}\).
Calculate the hydration enthalpy of chloride ions, \(\text{Cl}^-(\text{g})\), using your answer to part (b). (If you were unable to calculate an answer for part (b), you may use \(-2500 \text{ kJ mol}^{-1}\) as a representative value, though this is not the correct answer).
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PastPaper.workedSolution
a) Enthalpy of lattice formation is the enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent gaseous ions under standard conditions.
a) 2 marks: - 1 mark for enthalpy change when 1 mole of ionic crystal lattice is formed. - 1 mark for from its constituent gaseous ions (under standard conditions).
b) 6 marks: - 1 mark for showing \(2 \times \Delta H_{at}^\theta(\text{Cl}) = 242\). - 1 mark for showing \(2 \times \text{EA}(\text{Cl}) = -698\). - 2 marks for setting up the calculation correctly: \(\Delta H_{L,form}^\theta = \Delta H_f^\theta - (\text{all other terms})\). - 2 marks for calculating the correct value: \(-2523 \text{ kJ mol}^{-1}\) (allow 1 mark for correct magnitude but wrong sign, i.e., \(+2523\)).
c) 5.125 marks: - 1 mark for writing the correct algebraic relationship: \(\Delta H_{sol}^\theta = -\Delta H_{L,form}^\theta + \Delta H_{hyd}^\theta(\text{Mg}^{2+}) + 2\Delta H_{hyd}^\theta(\text{Cl}^-)\). - 1 mark for correctly substituting values (using their answer to part b or the representative value). - 1 mark for resolving the negative signs of lattice formation (i.e. using \(+2523\) as lattice dissociation energy). - 1.125 marks for calculating \(2 \times \Delta H_{hyd}^\theta(\text{Cl}^-) = -758\) (or \(-735\)). - 1 mark for final correct value with correct units and sign: \(-379 \text{ kJ mol}^{-1}\) (or \(-367.5 \text{ kJ mol}^{-1}\) / \(-368 \text{ kJ mol}^{-1}\) if using representative value).
PastPaper.question 8 · Structured
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A student wanted to determine the percentage purity of a solid sample of calcium carbonate, \(\text{CaCO}_3\), using a back titration.
A \(1.25 \text{ g}\) sample of the impure calcium carbonate was completely dissolved in \(50.0 \text{ cm}^3\) of \(1.00 \text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\) (which was an excess).
The resulting solution was transferred to a volumetric flask and made up to exactly \(250.0 \text{ cm}^3\) with distilled water.
A \(25.0 \text{ cm}^3\) portion of this solution was titrated against \(0.150 \text{ mol dm}^{-3}\) sodium hydroxide, \(\text{NaOH}\), requiring \(18.40 \text{ cm}^3\) for complete neutralisation.
a) Write balanced equations for: (i) The reaction between calcium carbonate and hydrochloric acid. (ii) The reaction between sodium hydroxide and hydrochloric acid.
b) Calculate the moles of sodium hydroxide used in the titration.
c) Calculate the moles of hydrochloric acid in excess in the \(250.0 \text{ cm}^3\) volumetric flask.
d) Calculate the moles of hydrochloric acid that reacted with the calcium carbonate.
e) Calculate the percentage purity of the calcium carbonate sample. Give your answer to 3 significant figures. (\(M_r(\text{CaCO}_3) = 100.1\))
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a) (i) \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\) (ii) \(\text{HCl}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})\)
b) Moles of \(\text{NaOH}\) = \(C \times V = 0.150 \text{ mol dm}^{-3} \times 0.01840 \text{ dm}^3 = 2.76 \times 10^{-3} \text{ mol}\).
c) From the 1:1 reaction stoichiometry of \(\text{NaOH}\) and \(\text{HCl}\), the moles of excess \(\text{HCl}\) in the \(25.0 \text{ cm}^3\) portion = \(2.76 \times 10^{-3} \text{ mol}\). The total volume of the volumetric flask is \(250.0 \text{ cm}^3\). Therefore, the total moles of excess \(\text{HCl}\) in the volumetric flask = \(10 \times 2.76 \times 10^{-3} = 2.76 \times 10^{-2} \text{ mol}\).
d) Initial moles of \(\text{HCl}\) added = \(C \times V = 1.00 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 5.00 \times 10^{-2} \text{ mol}\). Moles of \(\text{HCl}\) that reacted with calcium carbonate = \(5.00 \times 10^{-2} - 2.76 \times 10^{-2} = 2.24 \times 10^{-2} \text{ mol}\).
e) From the stoichiometry of reaction (i): 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\). Moles of \(\text{CaCO}_3\) in the sample = \(\frac{2.24 \times 10^{-2}}{2} = 1.12 \times 10^{-2} \text{ mol}\). Mass of pure \(\text{CaCO}_3\) = \(1.12 \times 10^{-2} \text{ mol} \times 100.1 \text{ g mol}^{-1} = 1.1211 \text{ g}\). Percentage purity = \(\frac{1.1211 \text{ g}}{1.25 \text{ g}} \times 100\% = 89.688\% = 89.7\%\).
PastPaper.markingScheme
a) 2 marks: - 1 mark for correct balanced equation of \(\text{CaCO}_3\) and \(\text{HCl}\). - 1 mark for correct balanced equation of \(\text{HCl}\) and \(\text{NaOH}\).
b) 2 marks: - 1 mark for setting up formula \(C \times V\). - 1 mark for calculating \(2.76 \times 10^{-3} \text{ mol}\).
c) 3 marks: - 1 mark for matching moles of \(\text{HCl}\) in portion to moles of \(\text{NaOH}\). - 1 mark for using scaling factor of 10 (\(250.0 / 25.0\)). - 1 mark for calculating \(2.76 \times 10^{-2} \text{ mol}\).
d) 3 marks: - 1 mark for calculating initial moles of \(\text{HCl}\) = \(5.00 \times 10^{-2} \text{ mol}\). - 1 mark for subtracting excess moles from initial moles. - 1 mark for obtaining \(2.24 \times 10^{-2} \text{ mol}\).
e) 3.125 marks: - 1 mark for dividing moles of \(\text{HCl}\) by 2 to get moles of \(\text{CaCO}_3\) (\(1.12 \times 10^{-2} \text{ mol}\)). - 1.125 marks for calculating mass of \(\text{CaCO}_3\) (\(1.121 \text{ g}\)). - 1 mark for calculating percentage purity to 3 significant figures (\(89.7\%\)).
Paper 2 Section A
Answer all questions. Drawings of organic mechanisms must show precise curly arrow origins.
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PastPaper.question 1 · structured
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Butanone reacts with potassium cyanide followed by dilute sulfuric acid to form a hydroxynitrile.
(a) Outline the mechanism for this reaction, showing the formation of the intermediate. Use curly arrows to show the movement of electron pairs. Name the mechanism. [5 marks]
(b) The product of this reaction contains an asymmetric carbon atom but is optically inactive. Explain why this is the case, referencing the shape of the organic reactant and the probability of attack. [3.5 marks]
(c) State the IUPAC name of the product. Identify the type of stereoisomerism shown by the product and state how two such isomers can be distinguished. [2 marks]
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**(a) Mechanism and Name:** * **Name:** Nucleophilic addition * **Mechanism:** 1. Curly arrow from the lone pair of electrons on the carbon atom of \(\text{CN}^-\) to the carbonyl carbon atom (\(\text{C}=\text{O}\)). 2. Curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom. 3. Intermediate structure: \(\text{CH}_3\text{CH}_2\text{C}(\text{O}^-)(\text{CN})\text{CH}_3\) (with a negative charge and lone pair on oxygen). 4. Curly arrow from the lone pair on the oxygen atom of the intermediate to a proton (\(\text{H}^+\)).
**(b) Explanation of Optical Inactivity:** * The carbonyl carbon and its three attached groups in butanone lie in a flat, planar arrangement (planar shape around the \(\text{C}=\text{O}\) group). * The nucleophilic cyanide ion (\(\text{CN}^-\)) is equally likely to attack this planar carbon atom from either above or below the plane (equal probability of attack). * This produces an equimolar (50:50) mixture of the two optical isomers (enantiomers), known as a racemic mixture. * Each enantiomer rotates the plane of polarized light by equal angles but in opposite directions; therefore, the optical rotations cancel each other out, making the mixture optically inactive.
**(c) Product Name and Isomerism:** * **IUPAC Name:** 2-hydroxy-2-methylbutanenitrile * **Stereoisomerism:** Optical isomerism * **Distinction:** Pass plane-polarized light through separate samples of the isomers. One enantiomer will rotate the plane of polarization clockwise, and the other will rotate it anticlockwise by an equal angle.
PastPaper.markingScheme
**(a) [Total: 5 marks]** * Nucleophilic addition (1) * Curly arrow from lone pair on carbon of \(\text{CN}^-\) to the \(\text{C}=\text{O}\) carbon (1) * Curly arrow from \(\text{C}=\text{O}\) double bond to the oxygen (1) * Correct structure of the intermediate showing negative charge and lone pair on oxygen (1) * Curly arrow from the lone pair on the \(\text{O}^-\) of the intermediate to \(\text{H}^+\) (1)
**(b) [Total: 3.5 marks]** * Butanone (or carbonyl group) is planar around the carbon-oxygen double bond (1) * Nucleophile / \(\text{CN}^-\) has an equal probability of attack from above or below the plane (1) * Forms a 50:50 / equimolar mixture of enantiomers / a racemic mixture (1) * The opposite rotations of plane-polarized light cancel out (0.5)
**(c) [Total: 2 marks]** * IUPAC Name: 2-hydroxy-2-methylbutanenitrile (1) * Optical isomerism AND rotate plane-polarized light in opposite directions / equal and opposite angles (1)
PastPaper.question 2 · structured
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Methylbenzene can be converted into aromatic derivatives via electrophilic substitution.
(a) Write an equation to show how the electrophile, \(\text{NO}_2^+\), is generated from concentrated nitric acid and concentrated sulfuric acid. [1.5 marks]
(b) Outline the mechanism for the nitration of methylbenzene to form 1-methyl-4-nitrobenzene (4-nitrotoluene). Show all necessary curly arrows and intermediate charges. [4 marks]
(c) 1-methyl-4-nitrobenzene can be reduced to 4-methylphenylamine. (i) State the reagents and conditions required for this reduction. [2 marks] (ii) Write an equation for this reduction using [H] to represent the reducing agent. [1 mark]
(d) Give one major commercial use of aromatic amines like 4-methylphenylamine. [2 marks]
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**(a) Generation of Electrophile:** The reaction between concentrated nitric and sulfuric acids generates the nitronium ion (\(\text{NO}_2^+\)): \[ \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \] *(Alternatively, \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_2\text{O} + \text{HSO}_4^-\) is acceptable.)*
**(b) Electrophilic Substitution Mechanism:** 1. A curly arrow starts from the delocalised pi ring of the methylbenzene ring (specifically at position 4) and points to the nitrogen atom of the \(\text{NO}_2^+\) ion. 2. This forms a cyclohexadienyl carbocation intermediate. The ring must have a broken delocalised system represented by a horseshoe spanning from carbon 2 to carbon 6, open towards carbon 4, with a positive charge located inside the horseshoe. Carbon 4 must be shown explicitly bonded to both a hydrogen atom and the \(\text{NO}_2\) group. 3. A curly arrow starts from the C-H bond on carbon 4 and points back into the center of the ring to restore the delocalised system. 4. The products are 1-methyl-4-nitrobenzene and an \(\text{H}^+\) ion.
**(c) Reduction of 1-methyl-4-nitrobenzene:** *(i) Reagents and Conditions:* * Reagent: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) (Accept Iron / \(\text{HCl}\)). * Conditions: Heating under reflux, followed by addition of sodium hydroxide solution (\(\text{NaOH(aq)}\)) to liberate the free amine. *(ii) Equation:* \[ \text{CH}_3\text{C}_6\text{H}_4\text{NO}_2 + 6[\text{H}] \rightarrow \text{CH}_3\text{C}_6\text{H}_4\text{NH}_2 + 2\text{H}_2\text{O} \]
**(d) Commercial Use:** * Used in the manufacturing of organic dyestuffs, specifically azo dyes, or pharmaceuticals.
**(b) [Total: 4 marks]** * Curly arrow from benzene ring to \(\text{NO}_2^+\) (1) * Correct structure of the cyclohexadienyl intermediate (horseshoe open towards carbon 4 with positive charge inside) (1) * Both H and \(\text{NO}_2\) attached to the same carbon (carbon 4) in the intermediate (1) * Curly arrow from C-H bond back into the ring to restore aromaticity (1)
**(c) [Total: 3 marks]** * (i) Tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\) (1) * (i) Heating under reflux followed by \(\text{NaOH}\) (1) * (ii) Balanced equation: \(\text{CH}_3\text{C}_6\text{H}_4\text{NO}_2 + 6[\text{H}] \rightarrow \text{CH}_3\text{C}_6\text{H}_4\text{NH}_2 + 2\text{H}_2\text{O}\) (1)
**(d) [Total: 2 marks]** * Making dyes / dyestuffs / azo dyes (2) (Accept production of pharmaceuticals / polyurethane polymers)
PastPaper.question 3 · structured
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An organic compound **X** has the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\). Chemical tests reveal that **X** is an ester.
The proton-decoupled \({}^{13}\text{C}\) NMR spectrum of **X** shows exactly 4 peaks.
The \({}^1\text{H}\) NMR spectrum of **X** shows the following signals: * A singlet at \(\delta = 2.0\text{ ppm}\) (integration value = 3H) * A doublet at \(\delta = 1.2\text{ ppm}\) (integration value = 6H) * A septet (multiplet) at \(\delta = 4.9\text{ ppm}\) (integration value = 1H)
Explain how this spectroscopic data can be used to determine the structure of **X**. In your answer: (a) Explain the splitting pattern and integration value of each of the three \({}^1\text{H}\) NMR signals. [4.5 marks] (b) Explain how the number of peaks in the \({}^{13}\text{C}\) NMR spectrum is consistent with your structure. [1.5 marks] (c) Draw the structural formula of **X** and state its IUPAC name. [4.5 marks]
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**(a) Analysis of the \({}^1\text{H}\) NMR Signals:** * **Signal at \(\delta = 2.0\text{ ppm}\) (3H, singlet):** - The integration of 3H indicates a methyl group (\(\text{-CH}_3\)). - The singlet splitting pattern (singlet) indicates that there are zero protons on the adjacent carbon atom (\(n=0\) adjacent protons, so \(n+1 = 1\) peak). - This methyl group is adjacent to the carbonyl group of the ester (\(\text{CH}_3\text{CO-}\)).
* **Signal at \(\delta = 1.2\text{ ppm}\) (6H, doublet):** - The integration of 6H indicates two identical, equivalent methyl groups (e.g., in an isopropyl group, \(\text{-(CH}_3)_2\)). - The doublet splitting pattern indicates that these methyl groups are adjacent to a carbon atom with exactly one proton (\(n=1\) adjacent proton, so \(n+1 = 2\) peaks).
* **Signal at \(\delta = 4.9\text{ ppm}\) (1H, septet):** - The integration of 1H indicates a single CH proton (\(\text{-CH-}\)). - The septet splitting pattern indicates that this proton is adjacent to six equivalent protons (\(n=6\) adjacent protons, so \(n+1 = 7\) peaks). - The high chemical shift (\(\delta = 4.9\text{ ppm}\)) indicates that this CH carbon is directly attached to the highly electronegative oxygen atom of the ester group (\(\text{-CO-O-CH-}\)).
**(b) Consistency with \({}^{13}\text{C}\) NMR Spectrum:** * The structure contains 5 carbon atoms in total. * However, the \({}^{13}\text{C}\) NMR spectrum shows only 4 peaks. This indicates that there are 4 different carbon environments. * This is consistent with our proposed structure, because the two methyl carbons of the isopropyl group are equivalent (experience the same electronic environment) and thus give a single combined peak. * The other three carbon environments are: the carbonyl carbon (\(\text{C}=\text{O}\)), the acetate methyl carbon (\(\text{CH}_3\text{CO-}\)), and the isopropyl CH carbon (\(\text{-CH-}\)).
**(c) Structural Formula and IUPAC Name:** * **Structural Formula:** \(\text{CH}_3\text{COOCH}(\text{CH}_3)_2\) (or drawn out fully showing all bonds) * **IUPAC Name:** propan-2-yl ethanoate (accept isopropyl ethanoate)
PastPaper.markingScheme
**(a) [Total: 4.5 marks]** * Singlet at \(\delta = 2.0\text{ ppm}\) (3H): 3H means \(\text{CH}_3\) group (0.5), singlet means no adjacent protons (0.5), next to carbonyl/ester group (0.5). * Doublet at \(\delta = 1.2\text{ ppm}\) (6H): 6H means two identical/equivalent \(\text{CH}_3\) groups (0.5), doublet means adjacent to 1 CH proton (1.0). * Septet at \(\delta = 4.9\text{ ppm}\) (1H): 1H means CH group (0.5), septet means adjacent to 6 protons (0.5), high shift because bonded to oxygen (0.5).
**(b) [Total: 1.5 marks]** * 4 peaks indicates 4 distinct carbon environments (0.5). * The two methyl groups in the isopropyl group are equivalent/symmetrical, giving one peak (1.0).
The halogenoalkane 2-bromobutane can undergo different reactions depending on the conditions used.
(a) When 2-bromobutane is heated with hot, aqueous potassium hydroxide, it forms butan-2-ol. (i) Name the mechanism for this reaction. [1 mark] (ii) Outline the mechanism for this reaction, using curly arrows. [3.5 marks]
(b) When 2-bromobutane is heated with hot, ethanolic potassium hydroxide, an elimination reaction occurs to form a mixture of three isomeric alkenes. (i) Draw the skeletal structures of all three alkenes formed. [3 marks] (ii) State which of these three alkenes is the major product and explain why it is formed in a larger amount. [3 marks]
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**(a) Nucleophilic Substitution of 2-bromobutane:** *(i) Mechanism Name:* * Nucleophilic substitution *(ii) Mechanism Outline:* 1. Curly arrow starts from a lone pair of electrons on the oxygen of the hydroxide ion (\(\text{OH}^-\)) and points to the \(\delta^+\) carbon atom bonded to the bromine atom. 2. Curly arrow starts from the polar \(\text{C}-\text{Br}\) bond and points to the bromine atom, showing the departure of the bromide leaving group. 3. The intermediate/product state is represented with the correct formation of butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)) and a bromide ion (\(\text{Br}^-\)).
**(b) Elimination of 2-bromobutane:** *(i) Skeletal Structures:* * The three isomeric alkenes formed are: 1. But-1-ene: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\) 2. (E)-but-2-ene (trans-but-2-ene): \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) with methyl groups trans to each other. 3. (Z)-but-2-ene (cis-but-2-ene): \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) with methyl groups cis to each other.
*(ii) Major Product and Explanation:* * **Major Product:** (E)-but-2-ene (or trans-but-2-ene). * **Explanation:** - But-2-ene is more stable than but-1-ene because highly substituted alkenes are more stable. - Between the stereoisomers, (E)-but-2-ene is more stable than (Z)-but-2-ene because the bulky methyl groups are on opposite sides of the planar double bond. - This arrangement minimises steric hindrance (steric strain / repulsion) between the methyl groups, lowering the potential energy of the molecule.
PastPaper.markingScheme
**(a) [Total: 4.5 marks]** * (i) Nucleophilic substitution (1) * (ii) Curly arrow from lone pair on \(\text{OH}^-\) to the C bonded to Br (1.5) * (ii) Curly arrow from \(\text{C}-\text{Br}\) bond to Br (1.5) * (ii) Show correct products butan-2-ol and \(\text{Br}^-\) (0.5)
**(b) [Total: 6 marks]** * (i) Drawing of but-1-ene (1) * (i) Drawing of (E)-but-2-ene (1) * (i) Drawing of (Z)-but-2-ene (1) * (ii) Identifies (E)-but-2-ene / trans-but-2-ene as the major product (1) * (ii) States that it is more stable than (Z)-but-2-ene (0.5) and but-1-ene (0.5) * (ii) Explains stability in terms of minimised steric hindrance/repulsion between methyl groups (1)
PastPaper.question 5 · structured
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N-phenylethanamide is an important organic compound used as an intermediate in the synthesis of pharmaceuticals.
(a) Phenylamine (\(\text{C}_6\text{H}_5\text{NH}_2\)) reacts with ethanoyl chloride (\(\text{CH}_3\text{COCl}\)) to form N-phenylethanamide. (i) Write a balanced chemical equation for this reaction. [1.5 marks] (ii) Name the mechanism of this reaction. [1 mark]
(b) Outline the mechanism for this reaction. Show all necessary curly arrows, lone pairs, and charges. [5 marks]
(c) Ethanoic anhydride (\(\text{(CH}_3\text{CO)}_2\text{O}\)) is frequently used instead of ethanoyl chloride in the industrial synthesis of N-phenylethanamide, even though the atom economy of the reaction with ethanoic anhydride is lower. State two practical or safety reasons why ethanoic anhydride is preferred over ethanoyl chloride. [3 marks]
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**(a) Synthesis of N-phenylethanamide:** *(i) Equation:* \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{HCl} \] *(Note: \(2\text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-\) is also acceptable.)*
**(b) Mechanism Outline:** 1. Curly arrow starts from the lone pair on the nitrogen atom of phenylamine (\(\text{C}_6\text{H}_5\ddot{\text{N}}\text{H}_2\)) and points to the carbonyl carbon atom of ethanoyl chloride (\(\text{CH}_3\text{COCl}\)). 2. Curly arrow starts from one of the bonds in the carbonyl \(\text{C}=\text{O}\) double bond and points to the oxygen atom. 3. This forms a tetrahedral intermediate which has a negative charge on the oxygen (\(\text{O}^-\)), a positive charge on the nitrogen (\(\text{N}^+\)), and the chlorine atom still attached. 4. Curly arrow starts from a lone pair on the negative oxygen (\(\text{O}^-\)) and points back to form the \(\text{C}=\text{O}\) double bond. 5. Curly arrow starts from the \(\text{C}-\text{Cl}\) bond and points to the chlorine atom to show elimination of \(\text{Cl}^-\). 6. Curly arrow starts from one of the \(\text{N}-\text{H}\) bonds in the intermediate and points to the positive nitrogen atom to remove a proton (\(\text{H}^+\)).
**(c) Comparison of Reagents:** Any two reasons from the following (1.5 marks each): * Ethanoic anhydride does not produce toxic and highly corrosive hydrogen chloride (\(\text{HCl}\)) gas (it produces relatively harmless ethanoic acid instead). * The reaction with ethanoic anhydride is less violent / less exothermic, making it safer and easier to control on an industrial scale. * Ethanoic anhydride is cheaper to purchase than ethanoyl chloride. * Ethanoic anhydride is less volatile / less easily hydrolysed by moisture in the air than ethanoyl chloride, making it easier to store and handle.
PastPaper.markingScheme
**(a) [Total: 2.5 marks]** * (i) Correct structures and balanced equation (1.5) (Deduct 0.5 for minor balancing error) * (ii) Nucleophilic addition-elimination (1)
**(b) [Total: 5 marks]** * Curly arrow from lone pair on \(\text{N}\) of phenylamine to \(\text{C}=\text{O}\) carbon (1) * Curly arrow from \(\text{C}=\text{O}\) double bond to \(\text{O}\) (1) * Correct structure of intermediate with charges (\(\text{O}^-\) and \(\text{N}^+\)) (1) * Curly arrow from lone pair on \(\text{O}^-\) to form \(\text{C}=\text{O}\) and arrow from \(\text{C}-\text{Cl}\) to \(\text{Cl}\) (1) * Curly arrow from \(\text{N}-\text{H}\) bond to \(\text{N}\) (1)
**(c) [Total: 3 marks]** * Reason 1: No toxic/corrosive \(\text{HCl}\) fumes produced (1.5) * Reason 2: Reaction is less violent/exothermic / easier to control OR cheaper / less vulnerable to moisture (1.5)
PastPaper.question 6 · structured
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The polyester commonly known as PET (polyethylene terephthalate) is widely used to make synthetic fibers and plastic bottles. It is manufactured from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.
(a) Draw the repeating unit of PET. Show all atoms and bonds in the ester linkage. [2 marks]
(b) Write an equation for the formation of PET from \(n\) molecules of benzene-1,4-dicarboxylic acid and \(n\) molecules of ethane-1,2-diol. [2 marks]
(c) Unlike addition polymers such as poly(ethene), polyesters like PET are biodegradable. Explain, in terms of their bonding, why polyesters are biodegradable. [2.5 marks]
(d) High-resolution \({}^1\text{H}\) NMR spectroscopy is used to verify the purity of the ethane-1,2-diol monomer. Describe the expected \({}^1\text{H}\) NMR spectrum of pure ethane-1,2-diol. In your answer, state the number of peaks, their integration ratio, and their splitting patterns. [4 marks]
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PastPaper.workedSolution
**(a) Repeating Unit of PET:** The repeating unit is formed by a condensation reaction, eliminating water. The structure is: \[ \text{[-O-CH}_2\text{-CH}_2\text{-O-CO-C}_6\text{H}_4\text{-CO-]} \] (Ensure that both the \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\) and \(\text{-CO-C}_6\text{H}_4\text{-CO-}\) parts are linked via an ester bond \(\text{-O-CO-}\) and the open bonds are shown at each end of the repeating unit.)
**(b) Equation for Polymerisation:** \[ n\,\text{HOOC-C}_6\text{H}_4\text{-COOH} + n\,\text{HO-CH}_2\text{-CH}_2\text{-OH} \rightarrow \text{[-CO-C}_6\text{H}_4\text{-CO-O-CH}_2\text{-CH}_2\text{-O-]}_n + 2n\,\text{H}_2\text{O} \] (Accept \((2n-1)\text{H}_2\text{O}\) as well, but \(2n\text{H}_2\text{O}\) is standard in AQA specifications for polymer equations starting from \(n\) units of each monomer.)
**(c) Explanation of Biodegradability:** * Polyesters contain polar ester linkages (\(\text{C}=\text{O}\) and \(\text{C}-\text{O}\) bonds) due to the difference in electronegativity between carbon and oxygen. * This makes them susceptible to nucleophilic attack by water (hydrolysis). * Consequently, they can be broken down naturally by microorganisms or enzymes. * In contrast, addition polymers consist of a non-polar, unreactive carbon-carbon single bond backbone (\(\text{C}-\text{C}\)), which cannot be hydrolysed.
**(d) \({}^1\text{H}\) NMR Spectrum of Ethane-1,2-diol (\(\text{HO-CH}_2\text{-CH}_2\text{-OH}\)):** * **Number of peaks:** 2 peaks (because there are two distinct proton environments: the \(\text{-CH}_2-\) protons and the \(\text{-OH}\) protons). * **Integration ratio:** 4 : 2 (or simplified 2 : 1) (corresponding to 4 \(\text{-CH}_2-\) protons and 2 \(\text{-OH}\) protons). * **Splitting patterns:** Both peaks appear as singlets. - The \(\text{-OH}\) proton peak is a singlet (proton exchange with traces of moisture/solvent prevents coupling under normal conditions). - The \(\text{-CH}_2-\) protons are equivalent to each other and thus do not split each other, appearing as a singlet.
PastPaper.markingScheme
**(a) [Total: 2 marks]** * Correct repeating unit structure with open bonds at ends (2) (Deduct 1 mark if ester linkage is drawn incorrectly or if water is not fully eliminated)
**(b) [Total: 2 marks]** * Correct structural formula of polymer product (1) * Correct balancing with \(n\) monomers and \(2n\,\text{H}_2\text{O}\) (1)
**(c) [Total: 2.5 marks]** * Polyesters contain polar ester links / polar \(\text{C}=\text{O}\) or \(\text{C}-\text{O}\) bonds (1) * Can be hydrolysed / broken down by water/nucleophiles (1) * Catalysed by enzymes / microorganisms in nature (0.5)
**(d) [Total: 4 marks]** * Explains two proton environments: \(\text{-CH}_2-\) and \(\text{-OH}\) (1) * Number of peaks = 2 (1) * Integration ratio = 4:2 or 2:1 (1) * Splitting = both are singlets (1)
PastPaper.question 7 · structured
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An unknown carbonyl compound **Y** has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).
(a) When **Y** is warmed with Fehling's solution, a red precipitate is formed. (i) Identify the functional group present in **Y**. [1 mark] (ii) Assuming **Y** has a straight carbon chain, write the structural formula of **Y**. [1 mark]
(b) Another isomer **Z** of \(\text{C}_4\text{H}_8\text{O}\) is a ketone that gives a positive result in the triiodomethane (iodoform) test. (i) Draw the skeletal structure of **Z**. [1 mark] (ii) State the reagents used for the triiodomethane test and describe the observation for a positive result. [2 marks] (iii) Write a balanced chemical equation for the reaction of **Z** with the reagents of the triiodomethane test to form a yellow precipitate of triiodomethane (\(\text{CHI}_3\)). [2.5 marks]
(c) **Z** can be reduced back to an alcohol using sodium tetrahydridoborate(III) (\(\text{NaBH}_4\)). Outline the mechanism for this reduction, using \(\text{H}^-\) to represent the hydride ion. [3 marks]
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**(a) Identification of Y:** *(i) Functional Group:* * Aldehyde group (or \(\text{-CHO}\) group). *(ii) Structural Formula of Y:* * Since **Y** is a straight-chain aldehyde with 4 carbons, it is butanal: \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \]
**(b) Ketone Isomer Z and Triiodomethane Test:** *(i) Skeletal Structure of Z:* * **Z** is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\). *(ii) Reagents and Observation:* * Reagents: Iodine (\(\text{I}_2\)) and sodium hydroxide (\(\text{NaOH}\)) solution (or aqueous alkaline iodine). * Observation: Yellow precipitate (with an antiseptic smell). *(iii) Balanced Equation:* \[ \text{CH}_3\text{COCH}_2\text{CH}_3 + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3 + \text{CH}_3\text{CH}_2\text{COONa} + 3\text{NaI} + 3\text{H}_2\text{O} \] *(Alternatively, the ionic form is acceptable: \(\text{CH}_3\text{COCH}_2\text{CH}_3 + 3\text{I}_2 + 4\text{OH}^- \rightarrow \text{CHI}_3 + \text{CH}_3\text{CH}_2\text{COO}^- + 3\text{I}^- + 3\text{H}_2\text{O}\).)*
**(c) Reduction Mechanism:** 1. Curly arrow from the lone pair of electrons on the hydride ion (\(\text{H}^-\)) to the carbonyl carbon atom of butanone. 2. Curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom. 3. This forms an intermediate alkoxide ion: \(\text{CH}_3\text{CH(O}^-)\text{CH}_2\text{CH}_3\). 4. Curly arrow from the lone pair on the negative oxygen atom (\(\text{O}^-\)) of the intermediate to a proton (\(\text{H}^+\)) (from water or acid) to yield butan-2-ol.
PastPaper.markingScheme
**(a) [Total: 2 marks]** * (i) Aldehyde / \(\text{-CHO}\) group (1) * (ii) Correct structural formula: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\) (1)
**(b) [Total: 5.5 marks]** * (i) Skeletal structure of butanone (1) * (ii) Iodine (\(\text{I}_2\)) AND sodium hydroxide (\(\text{NaOH}\)) (1) * (ii) Yellow precipitate (1) * (iii) Correct formula of products: \(\text{CHI}_3\) and \(\text{CH}_3\text{CH}_2\text{COONa}\) (or \(\text{CH}_3\text{CH}_2\text{COO}^-\)) (1) * (iii) Balanced equation (1.5)
**(c) [Total: 3 marks]** * Curly arrow from lone pair on \(\text{H}^-\) to \(\text{C}=\text{O}\) carbon (1) * Curly arrow from \(\text{C}=\text{O}\) double bond to oxygen (1) * Curly arrow from lone pair on \(\text{O}^-\) of intermediate to \(\text{H}^+\) (or \(\text{H}_2\text{O}\)) (0.5) * Correct final organic product (butan-2-ol) drawn (0.5)
PastPaper.question 8 · structured
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The following three-step synthesis is carried out starting with propene:
(a) Name and outline the mechanism for Step 1, showing the major product, Compound **A**. Use curly arrows to show the movement of electron pairs. [4.5 marks]
(b) Explain why Compound **A** is the major product in Step 1 rather than its isomer, referring to carbocation stability. [2.5 marks]
(c) State the IUPAC names of Compound **B** and Compound **C**. [1.5 marks]
(d) State the role of the concentrated sulfuric acid in Step 3. [1 mark]
(e) Predict the number of peaks in the proton-decoupled \({}^{13}\text{C}\) NMR spectrum of Compound **C**. [1 mark]
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**(a) Name and Mechanism for Step 1:** * **Mechanism Name:** Electrophilic addition * **Mechanism Outline:** 1. The dipole in \(\text{H}-\text{Br}\) is represented with \(\delta^+\) on \(\text{H}\) and \(\delta^-\) on \(\text{Br}\). 2. A curly arrow starts from the \(\text{C}=\text{C}\) double bond of propene and points to the \(\delta^+\) \(\text{H}\) atom. 3. A curly arrow starts from the \(\text{H}-\text{Br}\) bond and points to the \(\text{Br}\) atom. 4. This forms a secondary carbocation intermediate: \(\text{CH}_3\text{CH}^+\text{CH}_3\) (with a positive charge on the central carbon atom) and a bromide ion (\(\text{Br}^-\)). 5. A curly arrow starts from a lone pair on the bromide ion (\(\text{Br}^-\)) and points to the positive carbon atom of the carbocation, yielding the major product, 2-bromopropane (Compound **A**).
**(b) Explanation of Major Product Formation:** * The reaction proceeds via the secondary carbocation intermediate (\(\text{CH}_3\text{CH}^+\text{CH}_3\)) rather than the primary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)). * The secondary carbocation is more stable than the primary carbocation. * This is because the secondary carbocation has two electron-releasing alkyl (methyl) groups adjacent to the positive carbon, whereas the primary carbocation has only one alkyl group. * These alkyl groups push electron density towards the positive carbon (via the inductive effect), which stabilises the positive charge more effectively.
**(c) IUPAC Names:** * **Compound B:** propan-2-ol (from nucleophilic substitution of 2-bromopropane with \(\text{NaOH(aq)}\)). * **Compound C:** propan-2-yl ethanoate (accept isopropyl ethanoate) (from esterification of propan-2-ol with ethanoic acid).
**(d) Role of Concentrated Sulfuric Acid:** * Catalyst (Accept dehydrating agent).
**(e) Number of Peaks in \({}^{13}\text{C}\) NMR:** * Compound **C** is propan-2-yl ethanoate (\(\text{CH}_3\text{COOCH(CH}_3)_2\)). * There are 4 distinct carbon environments: 1. The carbonyl carbon (\(\text{C}=\text{O}\)) 2. The methyl carbon of the ethanoate group (\(\text{CH}_3\text{CO-}\)) 3. The CH carbon of the isopropyl group (\(\text{-CH-}\)) 4. The two equivalent methyl carbons of the isopropyl group (\(\text{-(CH}_3)_2\)) * Therefore, there are **4 peaks**.
PastPaper.markingScheme
**(a) [Total: 4.5 marks]** * Electrophilic addition (1) * Curly arrow from \(\text{C}=\text{C}\) double bond to \(\text{H}\) of \(\text{H}-\text{Br}\) (and dipole shown on \(\text{H}-\text{Br}\)) (1) * Curly arrow from \(\text{H}-\text{Br}\) bond to \(\text{Br}\) (1) * Correct structure of secondary carbocation intermediate (0.5) * Curly arrow from lone pair on \(\text{Br}^-\) to the positive carbon of carbocation (1)
**(b) [Total: 2.5 marks]** * Major product goes via secondary carbocation intermediate (1) * Secondary carbocation is more stable than primary carbocation (1) * Due to more electron-releasing alkyl groups (or inductive effect) stabilising the positive charge (0.5)
An ester, A (\(\text{C}_5\text{H}_9\text{O}_2\text{Cl}\)), is hydrolysed by heating under reflux with dilute hydrochloric acid. This reaction produces two organic compounds: a carboxylic acid, B, and a chlorinated alcohol, C (\(\text{C}_3\text{H}_7\text{OCl}\)).
The \(^1\text{H}\) NMR spectrum of B contains only two signals: - A singlet at \(\delta = 2.1\text{ ppm}\) (integration value of 3) - A singlet at \(\delta = 11.5\text{ ppm}\) (integration value of 1)
The \(^1\text{H}\) NMR spectrum of C contains four signals: - A doublet at \(\delta = 1.48\text{ ppm}\) (3H) - A singlet at \(\delta = 2.45\text{ ppm}\) (1H) - A doublet at \(\delta = 3.65\text{ ppm}\) (2H) - A multiplet at \(\delta = 4.12\text{ ppm}\) (1H)
(a) Deduce the structure of the carboxylic acid, B. Use the given \(^1\text{H}\) NMR data to justify your answer. [2.5 marks]
(b) Deduce the structure of the chlorinated alcohol, C. Explain how the integration values and splitting patterns of the peaks support your structure. [3.5 marks]
(c) Ester A can be synthesised by reacting alcohol C with an acyl chloride, D, in the presence of pyridine. Identify acyl chloride D and draw the structure of ester A. [1.5 marks]
(d) Outline the mechanism for the reaction between acyl chloride D and alcohol C to form ester A. You may use the abbreviation \(\text{R-OH}\) for alcohol C in your mechanism, where \(\text{R}\) represents the alkyl group. Show all relevant lone pairs and curly arrows. [3.0 marks]
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(a) Structure of B: Ethanoic acid, \(\text{CH}_3\text{COOH}\). Justification: The singlet at \(\delta = 11.5\text{ ppm}\) (1H) is characteristic of an acidic proton in a carboxylic acid group (\(-\text{COOH}\)). The singlet at \(\delta = 2.1\text{ ppm}\) (3H) corresponds to a methyl group adjacent to a carbonyl group (\(\text{CH}_3\text{CO}-\)) with no adjacent protons to cause coupling.
(b) Structure of C: 2-chloropropan-1-ol, \(\text{CH}_3\text{CH(Cl)CH}_2\text{OH}\). Justification: - The doublet at \(\delta = 1.48\text{ ppm}\) (3H) indicates a methyl group (\(-\text{CH}_3\)) adjacent to a single proton (\(-\text{CH}-\)), causing a split into a doublet (\(n+1 = 2\)). - The doublet at \(\delta = 3.65\text{ ppm}\) (2H) represents a \(-\text{CH}_2-\) group adjacent to a single proton (\(-\text{CH}-\)), causing a split into a doublet. Its chemical shift is relatively high due to the neighbouring oxygen atom. - The multiplet at \(\delta = 4.12\text{ ppm}\) (1H) represents the proton on the central carbon (\(-\text{CH(Cl)}-\)). It is adjacent to 5 protons (3 from \(-\text{CH}_3\) and 2 from \(-\text{CH}_2-\)), causing coupling into a multiplet (sextet). It is highly deshielded due to the strongly electronegative chlorine atom attached to the same carbon. - The singlet at \(\delta = 2.45\text{ ppm}\) (1H) is characteristic of the alcohol group (\(-\text{OH}\)) proton, which does not show splitting under standard conditions.
(d) Mechanism of nucleophilic addition-elimination: 1. First step (Nucleophilic attack): A curly arrow starts from the lone pair on the oxygen of \(\text{R-OH}\) to the carbonyl carbon of \(\text{CH}_3\text{COCl}\). Simultaneously, a curly arrow starts from the \(\text{C}=\text{O}\) double bond to the carbonyl oxygen atom. 2. Intermediate structure: Draw the correct structure of the tetrahedral intermediate: \(\text{CH}_3-\text{C}(\text{O}^-)(\text{Cl})-\text{O}^+(\text{H})-\text{R}\). Both formal charges (\(\text{O}^-\) and \(\text{O}^+\)) must be explicitly shown. 3. Second step (Elimination and deprotonation): A curly arrow starts from a lone pair on the \(\text{O}^-\) of the intermediate to reform the \(\text{C}=\text{O}\) double bond. A curly arrow starts from the \(\text{C}-\text{Cl}\) bond to the chlorine atom to eliminate a chloride ion (\(\text{Cl}^-\)). A curly arrow starts from the \(\text{O}-\text{H}\) bond to the positive oxygen atom to regenerate neutral oxygen, yielding the ester and \(\text{HCl}\).
PastPaper.markingScheme
(a) Deduce B & Justify [2.5 marks]: - M1: Correct structure of ethanoic acid (displayed, structural, or skeletal) [1 mark]. Accept \(\text{CH}_3\text{COOH}\). - M2: Identifies \(\delta = 11.5\text{ ppm}\) peak as the carboxylic acid \(\text{O}-\text{H}\) proton [0.5 marks]. - M3: Identifies \(\delta = 2.1\text{ ppm}\) peak as a \(\text{CH}_3-\) adjacent to \(\text{C}=\text{O}\) (or acetyl group) [1 mark].
(b) Deduce C & Justify [3.5 marks]: - M1: Correct structure of 2-chloropropan-1-ol, \(\text{CH}_3\text{CH(Cl)CH}_2\text{OH}\) [1 mark]. (Reject 3-chloropropan-1-ol or 1-chloropropan-2-ol). - M2: Explains \(\delta = 1.48\text{ ppm}\) (3H doublet) as \(\text{CH}_3\) coupled with 1 neighboring \(\text{CH}\) proton [0.5 marks]. - M3: Explains \(\delta = 3.65\text{ ppm}\) (2H doublet) as \(\text{CH}_2\) adjacent to \(\text{CH}\) proton (deshielded by oxygen) [0.5 marks]. - M4: Explains \(\delta = 4.12\text{ ppm}\) (1H multiplet) as \(\text{CH}\) coupled to both \(\text{CH}_3\) and \(\text{CH}_2\) (total 5 adjacent protons) and heavily deshielded by the electronegative \(\text{Cl}\) atom [1 mark]. - M5: Explains \(\delta = 2.45\text{ ppm}\) (1H singlet) as the \(\text{O}-\text{H}\) proton [0.5 marks].
(c) Acyl chloride D & Ester A [1.5 marks]: - M1: Correctly names or draws ethanoyl chloride (\(\text{CH}_3\text{COCl}\)) [0.5 marks]. - M2: Correct structure of ester A: \(\text{CH}_3\text{COOCH}_2\text{CH(Cl)CH}_3\) [1 mark].
(d) Mechanism [3.0 marks]: - M1: Curly arrow from oxygen lone pair of \(\text{ROH}\) to carbonyl carbon AND curly arrow from \(\text{C}=\text{O}\) double bond to oxygen [1 mark]. (Both required for M1. Must show precise arrow origins from lone pairs/bonds). - M2: Correct tetrahedral intermediate structure including correct formal charges (\(\text{O}^-\) and \(\text{O}^+\)) [1 mark]. - M3: Curly arrow from \(\text{O}^-\) lone pair to reform \(\text{C}=\text{O}\) AND curly arrow from \(\text{C}-\text{Cl}\) bond to \(\text{Cl}\) AND curly arrow from \(\text{O}-\text{H}\) bond to \(\text{O}\) [1 mark]. (All three required for M3).
Compound P, with the molecular formula \(\text{C}_9\text{H}_{10}\text{O}\), is an aromatic ketone that contains a disubstituted benzene ring. The \(^{13}\text{C}\) NMR spectrum of P shows exactly 7 peaks. The \(^1\text{H}\) NMR spectrum of P shows: - A singlet at \(\delta = 2.4\text{ ppm}\) (3H) - A singlet at \(\delta = 2.6\text{ ppm}\) (3H) - A doublet at \(\delta = 7.3\text{ ppm}\) (2H) - A doublet at \(\delta = 7.8\text{ ppm}\) (2H)
P reacts with \(\text{NaBH}_4\) in aqueous methanol to form Compound Q.
(a) Deduce the structure of Compound P. Explain how the \(^{13}\text{C}\) and \(^1\text{H}\) NMR spectra support this structure. [3.0 marks]
(b) Identify the organic product Q and name the mechanism for the reduction of P to Q. [1.5 marks]
(c) Outline the mechanism for this reduction reaction. In your mechanism, use \(\text{H}^-\) to represent the nucleophile from \(\text{NaBH}_4\). Show all relevant curly arrows, lone pairs, and the structure of the intermediate and organic product. [3.5 marks]
(d) Explain why the sample of Q produced in this reaction is optically inactive, despite the presence of a chiral centre in its structure. [2.5 marks]
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(a) Structure of P: 1-(4-methylphenyl)ethanone (also accepted: 4-methylacetophenone or \(\text{CH}_3-\text{C}_6\text{H}_4-\text{COCH}_3\)). Justification: - The \(^{13}\text{C}\) NMR spectrum showing 7 peaks indicates a symmetrical 1,4-disubstituted benzene ring. There are 4 unique ring carbon environments due to the plane of symmetry. This, plus the carbonyl carbon (\(\text{C}=\text{O}\)) and two different methyl carbons (\(-\text{CH}_3\)), gives exactly 7 environments. - The \(^1\text{H}\) NMR spectrum contains two doublets at \(\delta = 7.3\text{ ppm}\) and \(7.8\text{ ppm}\) (each integrating to 2H), which is a characteristic pattern for a 1,4-disubstituted (para-substituted) benzene ring. - The singlet at \(\delta = 2.4\text{ ppm}\) (3H) is a methyl group attached directly to the benzene ring (\(\text{Ar}-\text{CH}_3\)). - The singlet at \(\delta = 2.6\text{ ppm}\) (3H) is a methyl group adjacent to the carbonyl (\(\text{CH}_3\text{CO}-\)).
(c) Mechanism of Nucleophilic Addition: 1. First step: A curly arrow starts from the lone pair on the hydride ion (\(\text{H}^-\)) to the carbonyl carbon (\(\text{C}=\text{O}\)) of 1-(4-methylphenyl)ethanone. Simultaneously, a curly arrow starts from the \(\text{C}=\text{O}\) double bond to the oxygen atom. 2. Intermediate structure: Draw the alkoxide intermediate with a single \(\text{C}-\text{O}\) bond and a negative charge on the oxygen atom: \(\text{CH}_3-\text{C}_6\text{H}_4-\text{C}(\text{O}^-)(\text{H})-\text{CH}_3\). 3. Protonation: A curly arrow starts from a lone pair on the \(\text{O}^-\) of the intermediate to a proton source (such as \(\text{H}^+\) or the \(\text{H}\) of a water molecule, \(\text{H}-\text{OH}\)). If water is used, show a curly arrow from the \(\text{O}-\text{H}\) bond of water to the oxygen of water. 4. Final Product: Correct structure of the alcohol Q.
(d) Optical Inactivity Explanation: - The carbonyl group (\(\text{C}=\text{O}\)) on the reactant P is planar around the carbonyl carbon. - The nucleophile (\(\text{H}^-\)) can attack this planar carbon atom with equal probability from either side (above or below the plane). - This leads to the formation of a racemic mixture (enantiomeric mixture in a 50:50 ratio). - Since both enantiomers are present in equal amounts, their rotation of plane-polarised light is equal and opposite, and thus cancels out, leaving the product optically inactive.
PastPaper.markingScheme
(a) Structure & Justification [3.0 marks]: - M1: Correct structure of 1-(4-methylphenyl)ethanone [1 mark]. - M2: Explains that 7 peaks in \(^{13}\text{C}\) NMR indicate symmetry of 1,4-disubstitution (4 aromatic C environments + 1 carbonyl C + 2 methyl Cs) [1 mark]. - M3: Explains that two 2H doublets in the aromatic region of \(^1\text{H}\) NMR confirm 1,4-disubstitution on the benzene ring, and identifies the two 3H singlets as the ring-bound methyl and acetyl methyl groups [1 mark].
(b) Product Q & Mechanism name [1.5 marks]: - M1: Correct structure of 1-(4-methylphenyl)ethanol [1 mark]. - M2: Nucleophilic addition [0.5 marks]. (Reject nucleophilic substitution or addition-elimination).
(c) Mechanism [3.5 marks]: - M1: Curly arrow from lone pair on \(\text{H}^-\) to carbonyl carbon [1 mark]. - M2: Curly arrow from \(\text{C}=\text{O}\) double bond to oxygen [1 mark]. - M3: Correct intermediate structure with a negative charge on the oxygen atom [1 mark]. - M4: Curly arrow from lone pair on intermediate oxygen to \(\text{H}^+\) (or to \(\text{H}\) of \(\text{H}_2\text{O}\) or \(\text{CH}_3\text{OH}\) with the \(\text{O}-\text{H}\) bond breaking arrow) to form the final product Q [0.5 marks].
(d) Optical Inactivity Explanation [2.5 marks]: - M1: Carbonyl group (or carbonyl carbon) in P is planar [1 mark]. (Do not accept "the molecule is planar"). - M2: Equal probability of attack by hydride (\(\text{H}^-\)) from either side / above or below the plane [1 mark]. - M3: Form a racemic mixture (or equimolar mixture of enantiomers) which cancels out optical activity / has no net rotation of plane-polarised light [0.5 marks].
Paper 3 Section A
Answer all questions in this section. Questions focus on practical details and synoptic chemistry.
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PastPaper.question 1 · practical
12 PastPaper.marks
A student investigates the rate of reaction between propanone and iodine in acidic conditions: \(\text{CH}_3\text{COCH}_3 + \text{I}_2 \xrightarrow{\text{H}^+} \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^-\). The student uses a colorimeter to measure the concentration of iodine over time. (a) Explain why a colorimeter can be used to monitor this reaction and describe how the student would use a calibration curve to convert colorimeter readings into concentration of iodine. (3 marks) (b) Explain why sodium hydrogencarbonate (\(\text{NaHCO}_3\)) is added to samples quenched from the reaction mixture if a titrimetric method was used instead. (2 marks) (c) The student obtains a graph of concentration of \(\text{I}_2\) against time, which is a straight line. State the order of reaction with respect to iodine and explain how the graph shows this. (2 marks) (d) In another experiment, the student changes the initial concentrations of propanone and hydrochloric acid. The table of initial rates is given: [Propanone] / mol dm\(^{-3}\), [HCl] / mol dm\(^{-3}\), [\(\text{I}_2\)] / mol dm\(^{-3}\), Initial rate / mol dm\(^{-3}\) s\(^{-1}\): (1) 0.40, 0.20, 0.010, \(1.20 \times 10^{-5}\); (2) 0.80, 0.20, 0.010, \(2.40 \times 10^{-5}\); (3) 0.80, 0.40, 0.020, \(4.80 \times 10^{-5}\). Deduce the rate equation and calculate the rate constant, \(k\), including units. (5 marks)
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(a) Iodine is coloured (brown/yellow/orange) while all other reactants and products are colourless. To use a calibration curve, prepare several standard solutions of iodine of known concentrations. Measure the absorbance of each solution and plot a graph of absorbance against concentration. Read the concentration of the unknown reaction mixture from the calibration curve using its measured absorbance. (b) Sodium hydrogencarbonate neutralises the acid catalyst (H+), which stops (quenches) the reaction at that specific time. (c) The order with respect to iodine is 0 (zero order) because the gradient of the graph of concentration against time is constant (straight line), meaning the rate of reaction is constant and independent of the concentration of iodine. (d) From Expt 1 to Expt 2, [Propanone] doubles, [HCl] is constant, [I2] is constant, and the rate doubles. Therefore, the reaction is first order with respect to propanone. Since the order with respect to [I2] is zero, comparing Expt 2 and Expt 3: [Propanone] is constant, [HCl] doubles, [I2] doubles (no effect), and the rate doubles. Therefore, the reaction is first order with respect to H+ (or HCl). Rate equation: Rate = k[CH3COCH3][H+]. Using Expt 1: 1.20 x 10^-5 = k(0.40)(0.20) => k = 1.50 x 10^-4. Units: mol^-1 dm^3 s^-1.
PastPaper.markingScheme
(a) 1 mark for mentioning iodine is coloured and other species are colourless. 1 mark for plotting a graph of absorbance against known concentrations of iodine. 1 mark for explaining how to read unknown concentration from absorbance. (b) 1 mark for neutralising the acid catalyst. 1 mark for stopping the reaction (quenching). (c) 1 mark for zero order. 1 mark for explanation (constant gradient / rate is independent of concentration). (d) 1 mark for showing first order with respect to propanone. 1 mark for showing first order with respect to H+. 1 mark for correct rate equation. 1 mark for correct value of k (1.50 x 10^-4). 1 mark for correct units (mol^-1 dm^3 s^-1).
PastPaper.question 2 · practical
12 PastPaper.marks
This question is about halogenoalkanes. (a) Write a mechanism for the reaction of 2-bromo-2-methylpropane with hydroxide ions to form 2-methylpropan-2-ol. Show any curly arrows and relevant lone pairs. (4 marks) (b) A student wants to measure the rate of hydrolysis of various halogenoalkanes. Describe a practical method to compare the rate of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Include the reagents used, what is observed, and how the relative rate is determined. Explain the trend in rates observed. (6 marks) (c) Explain why ethanol is used as a solvent in this practical. (2 marks)
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(a) The mechanism is SN1. Step 1: Curly arrow starts from the C-Br bond to the Br atom. Step 2: Draw the tertiary carbocation intermediate (CH3)3C+ and a Br- ion. Step 3: Draw a curly arrow from a lone pair on the oxygen of the OH- ion to the positively charged carbon atom of the carbocation. Step 4: Draw the correct structure of 2-methylpropan-2-ol. (b) Add equal volumes of ethanol and the respective halogenoalkane into three separate test tubes. Add aqueous silver nitrate solution (AgNO3) to each tube. Place the tubes in a water bath at a constant temperature (e.g. 50-60 degrees C) and start a timer. Record the time taken for a precipitate to appear in each test tube. The faster the precipitate forms, the faster the rate of hydrolysis. The order of rates is 1-iodobutane (fastest, yellow precipitate) > 1-bromobutane (cream precipitate) > 1-chlorobutane (slowest, white precipitate). This is because bond enthalpy decreases in the order C-Cl > C-Br > C-I, making the C-I bond the easiest to break. (c) Halogenoalkanes are insoluble in water, but dissolve in ethanol. Ethanol acts as a mutual solvent to allow the halogenoalkanes and aqueous silver nitrate to mix in a single phase, ensuring they can react.
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(a) 1 mark for arrow from C-Br bond to Br. 1 mark for correct structure of tertiary carbocation. 1 mark for arrow from lone pair of OH- to carbocation carbon. 1 mark for correct product. (b) 1 mark for adding silver nitrate. 1 mark for maintaining constant temperature (water bath) / equal volumes. 1 mark for measuring time to see precipitate. 1 mark for identifying the precipitate colours / trend in rate (I > Br > Cl). 2 marks for explaining the trend in terms of bond enthalpies (C-Cl is stronger than C-I, so C-I breaks most easily). (c) 1 mark for stating halogenoalkanes are insoluble in water. 1 mark for stating ethanol acts as a mutual solvent / miscible with both.
PastPaper.question 3 · practical
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A student is given an impure sample of iron(II) sulfate crystals (\(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\)) of mass 8.25 g. The sample is dissolved in dilute sulfuric acid and made up to \(250.0\text{ cm}^3\) of solution in a volumetric flask. A \(25.0\text{ cm}^3\) sample of this solution is pipetted into a conical flask, acidified with extra dilute sulfuric acid, and titrated against \(0.0200\text{ mol dm}^{-3}\) \(\text{KMnO}_4\). The mean titre is \(23.40\text{ cm}^3\). (a) State the role of the sulfuric acid added to the conical flask and explain why hydrochloric acid or nitric acid cannot be used instead. (3 marks) (b) Write the overall ionic equation for the reaction between \(\text{Fe}^{2+}\) and \(\text{MnO}_4^-\). (2 marks) (c) State the color change at the end-point of this titration. (1 mark) (d) Calculate the percentage purity of the \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) sample. (Molar mass of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O} = 278.0\text{ g mol}^{-1}\)). (5 marks) (e) State one precaution the student should take when filling the burette with potassium manganate(VII) solution. (1 mark)
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(a) Sulfuric acid provides the H+ ions required for the reduction of MnO4-. Hydrochloric acid cannot be used because Cl- ions would be oxidized to Cl2 by MnO4-, leading to an artificially high titre. Nitric acid cannot be used because nitrate ions (NO3-) are oxidizing agents and would oxidize Fe2+ to Fe3+ before the titration begins. (b) MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+ (c) Colourless (or pale green) to permanent pale pink. (d) Moles of MnO4- in titre = 0.0200 x 0.02340 = 4.68 x 10^-4 mol. Moles of Fe2+ in 25.0 cm3 = 5 x 4.68 x 10^-4 = 2.34 x 10^-3 mol. Moles of Fe2+ in 250.0 cm3 = 2.34 x 10^-2 mol. Mass of FeSO4.7H2O in 250 cm3 = 2.34 x 10^-2 x 278.0 = 6.5052 g. Percentage purity = (6.5052 / 8.25) x 100 = 78.9%. (e) Fill the burette below eye level (e.g., on the floor or a low stool) / use a funnel and ensure it is removed before starting.
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(a) 1 mark for H+ source. 1 mark for explaining HCl oxidation. 1 mark for explaining HNO3 oxidation. (b) 2 marks for balanced ionic equation (1 mark for species, 1 mark for balancing). (c) 1 mark for colourless/pale green to pale pink. (d) 1 mark for calculating moles of MnO4-. 1 mark for calculating moles of Fe2+ in 25 cm3. 1 mark for scaling up to 250 cm3. 1 mark for mass of FeSO4.7H2O. 1 mark for final percentage (78.9%). (e) 1 mark for filling below eye level or removing funnel.
PastPaper.question 4 · practical
12 PastPaper.marks
A student performs a pH titration to determine the \(K_a\) of a weak monoprotic acid, propanoic acid (HA), by titrating a \(25.0\text{ cm}^3\) sample of the acid with \(0.100\text{ mol dm}^{-3}\) sodium hydroxide (NaOH). (a) The student adds \(15.0\text{ cm}^3\) of NaOH and notices that a buffer solution is formed. Define the term 'buffer solution' and write an equation to show how this buffer system resists change in pH when a small amount of acid (\(\text{H}^+\)) is added. (2 marks) (b) Calculate the pH of a \(0.120\text{ mol dm}^{-3}\) solution of propanoic acid given that its \(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\) at 298 K. Show your working. (3 marks) (c) Calculate the pH of the buffer solution formed when \(15.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) NaOH is added to \(25.0\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\) propanoic acid. (5 marks) (d) Explain why a student should calibrate a pH meter before carrying out the titration, and briefly describe how this calibration is done. (2 marks)
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(a) A buffer solution is one that minimizes pH changes when small amounts of acid or base are added. Equation: A- + H+ -> HA. (b) Ka = [H+]^2 / [HA] => [H+] = sqrt(Ka * [HA]) = sqrt(1.35 x 10^-5 * 0.120) = 1.2728 x 10^-3 mol dm^-3. pH = -log10(1.2728 x 10^-3) = 2.90. (c) Initial moles of HA = 0.0250 * 0.120 = 3.00 x 10^-3 mol. Moles of NaOH added = 0.0150 * 0.100 = 1.50 x 10^-3 mol. Moles of HA remaining = 3.00 x 10^-3 - 1.50 x 10^-3 = 1.50 x 10^-3 mol. Moles of A- formed = 1.50 x 10^-3 mol. Since moles of HA = moles of A-, [H+] = Ka = 1.35 x 10^-5 mol dm^-3. pH = pKa = -log10(1.35 x 10^-5) = 4.87. (d) pH meters can drift over time and give inaccurate readings. To calibrate, rinse the probe with deionised water, dip it into a buffer solution of known pH (e.g., pH 7.00), adjust the reading, and repeat with at least one other buffer solution (e.g., pH 4.00).
PastPaper.markingScheme
(a) 1 mark for buffer definition. 1 mark for equation (A- + H+ -> HA). (b) 1 mark for [H+] calculation. 1 mark for pH = 2.90. 1 mark for pH to 2 dp. (c) 1 mark for initial moles of HA. 1 mark for moles of NaOH. 1 mark for moles of HA remaining. 1 mark for moles of A- formed. 1 mark for calculating pH = 4.87. (d) 1 mark for explaining calibration is needed to correct for drift/accuracy loss. 1 mark for using buffer solutions of known pH.
PastPaper.question 5 · practical
12 PastPaper.marks
This question is about the determination of enthalpy of solution of anhydrous calcium chloride (\(\text{CaCl}_2\)). A student dissolves \(5.00\text{ g}\) of anhydrous \(\text{CaCl}_2\) in \(50.0\text{ cm}^3\) of water in a polystyrene cup. The temperature rises from \(20.2\text{ ^oC}\) to \(37.8\text{ ^oC}\). Assume the density of water is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). (a) Calculate the heat energy released, \(q\), in Joules. (2 marks) (b) Calculate the enthalpy change of solution, \(\Delta H_{\text{sol}}\), of anhydrous calcium chloride in \(\text{kJ mol}^{-1}\). (Molar mass of \(\text{CaCl}_2 = 111.0\text{ g mol}^{-1}\)). (3 marks) (c) Draw a Hess's Law cycle (Born-Haber cycle) for the dissolution of \(\text{CaCl}_2\) in water, showing the relationship between lattice enthalpy of dissociation (\(\Delta H_{\text{L,diss}}\)), enthalpy of hydration of ions (\(\Delta H_{\text{hyd}}\)), and enthalpy of solution (\(\Delta H_{\text{sol}}\)). Write an equation relating these terms. (4 marks) (d) Explain why the experimental value of enthalpy of solution using a polystyrene cup is less exothermic than the theoretical value. Suggest one practical improvement to the experiment to obtain a more accurate value. (3 marks)
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(a) Delta T = 37.8 - 20.2 = 17.6 K. If using mass of water, m = 50.0 g: q = m * c * Delta T = 50.0 * 4.18 * 17.6 = 3678.4 J. If using total mass of water + solid, m = 55.0 g: q = 55.0 * 4.18 * 17.6 = 4046.2 J. (b) Moles of CaCl2 = 5.00 / 111.0 = 0.04505 mol. Enthalpy of solution = -q / (moles * 1000). For q = 3.6784 kJ: Delta H_sol = -3.6784 / 0.04505 = -81.7 kJ mol^-1. For q = 4.0462 kJ: Delta H_sol = -4.0462 / 0.04505 = -89.8 kJ mol^-1. (c) The cycle has CaCl2(s) at the top left. An arrow goes downwards or across to Ca2+(g) + 2Cl-(g) representing Delta H_L,diss. From Ca2+(g) + 2Cl-(g), arrows go to Ca2+(aq) + 2Cl-(aq) representing hydration enthalpies: Delta H_hyd(Ca2+) + 2 * Delta H_hyd(Cl-). From CaCl2(s) directly to Ca2+(aq) + 2Cl-(aq) is Delta H_sol. Equation: Delta H_sol = Delta H_L,diss + Delta H_hyd(Ca2+) + 2 * Delta H_hyd(Cl-). (d) The experimental value is less exothermic because some heat is lost to the surroundings (air, thermometer, cup). To improve accuracy, the student can use a lid to reduce heat loss, use extra insulation (e.g. nested cups), or record temperature at regular intervals to plot a cooling curve and extrapolate back to the time of mixing.
PastPaper.markingScheme
(a) 1 mark for calculating temperature change (17.6 K). 1 mark for correct calculation of q (3678 J using m=50 g or 4046 J using m=55 g). (b) 1 mark for moles of CaCl2 (0.04505 mol). 1 mark for dividing kJ by mol. 1 mark for negative sign and value (-81.7 or -89.8 kJ mol^-1). (c) 1 mark for correctly showing gaseous ions. 1 mark for correctly showing aqueous ions. 1 mark for correct stoichiometry (2Cl-). 1 mark for correct equation. (d) 1 mark for heat loss to surroundings. 2 marks for practical improvements: 1 mark for adding a lid / extra insulation, 1 mark for temperature-time cooling curve extrapolation.
Paper 3 Section B
Answer all multiple-choice questions. Select the single best answer for each.
30 PastPaper.question · 30 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
The rate equation for the reaction between peroxodisulfate ions and iodide ions is:
In an experiment, the initial rate of reaction was found to be \(1.20 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\) when the concentration of \(\text{I}^-\) was \(0.100\text{ mol dm}^{-3}\) and \(\text{S}_2\text{O}_8^{2-}\) was \(0.0500\text{ mol dm}^{-3}\).
What is the value and unit of the rate constant, \(k\)?
1 mark for the correct selection of option A. Correctly rearranges the rate equation, evaluates the value as 0.024, and determines the unit as dm3 mol-1 s-1.
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
The rate constant \(k\) for a reaction was measured at several temperatures. A plot of \(\ln k\) against \(1/T\) (where \(T\) is in Kelvin) gave a straight line with a gradient of \(-1.20 \times 10^4\text{ K}\).
What is the activation energy, \(E_a\), of this reaction? (Gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\))
A.\(+1.44\text{ kJ mol}^{-1}\)
B.\(+99.7\text{ kJ mol}^{-1}\)
C.\(-99.7\text{ kJ mol}^{-1}\)
D.\(+1.20 \times 10^4\text{ kJ mol}^{-1}\)
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According to the Arrhenius equation in its logarithmic form:
\[\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\]
A plot of \(\ln k\) against \(1/T\) yields a straight line with gradient \(m = -\frac{E_a}{R}\).
1 mark for the correct selection of option C. Correctly determines the remaining moles of propanoic acid, moles of propanoate produced, and applies the acid dissociation constant expression to calculate a pH of 4.83.
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
At \(298\text{ K}\), \(50.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}(\text{aq)\), is mixed with \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) barium hydroxide, \(\text{Ba(OH)}_2(\text{aq})\).
What is the pH of the resulting solution at \(298\text{ K}\)? (Take \(K_w = 1.00 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\) at \(298\text{ K}\))
A.\(1.60\)
B.\(12.40\)
C.\(12.10\)
D.\(1.90\)
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PastPaper.workedSolution
1. Moles of \(\text{H}^+\) from HCl: \[n(\text{H}^+) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\]
2. Moles of \(\text{OH}^-\) from \(\text{Ba(OH)}_2\) (note there are 2 moles of \(\text{OH}^-\) per mole of barium hydroxide): \[n(\text{OH}^-) = 2 \times (0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3}) = 1.00 \times 10^{-2}\text{ mol} = 10.0 \times 10^{-3}\text{ mol}\]
3. Determine the excess reactant: \(\text{OH}^-\) is in excess by: \[10.0 \times 10^{-3} - 7.50 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol}\]
4. Calculate the concentration of excess \(\text{OH}^-\): \[\text{Total volume} = 50.0 + 50.0 = 100.0\text{ cm}^3 = 0.100\text{ dm}^3\]
1 mark for the correct selection of option B. Correctly accounts for the stoichiometry of barium hydroxide (yielding 2 OH- ions), determines the excess concentration of hydroxide in the total mixture volume, and correctly finds the pH as 12.40.
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
When an excess of aqueous ammonia is added to a solution containing hexaaquacopper(II) ions, a deep-blue solution is formed.
What is the formula of the copper(II) complex present in this deep-blue solution?
When excess ammonia is added to hexaaquacopper(II) ions, a ligand substitution reaction occurs. Four water molecules are substituted by four ammonia molecules to form a deep-blue solution containing the octahedral complex \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).
The overall equation is: \[[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}\]
PastPaper.markingScheme
1 mark for the correct selection of option B. Recognises that only 4 of the water molecules are substituted by ammonia, leaving a mixed ligand complex.
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
A \(25.0\text{ cm}^3\) sample of a solution containing both iron(II) ions and ethanedioate ions, \(\text{FeC}_2\text{O}_4\), was acidified and titrated with \(0.0200\text{ mol dm}^{-3}\) potassium manganate(VII) solution. Both the \(\text{Fe}^{2+}\) and \(\text{C}_2\text{O}_4^{2-}\) ions are oxidised in this titration.
The titration required \(30.0\text{ cm}^3\) of the potassium manganate(VII) solution to reach the end point.
What is the concentration of the \(\text{FeC}_2\text{O}_4\) solution?
A.\(0.0080\text{ mol dm}^{-3}\)
B.\(0.0120\text{ mol dm}^{-3}\)
C.\(0.0400\text{ mol dm}^{-3}\)
D.\(0.0240\text{ mol dm}^{-3}\)
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PastPaper.workedSolution
1. Write down oxidation half-equations: - \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\) (1 mole of electrons per mole of \(\text{Fe}^{2+}\)) - \(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-\) (2 moles of electrons per mole of \(\text{C}_2\text{O}_4^{2-}\)) - Combined oxidation for \(\text{FeC}_2\text{O}_4\): \[\text{FeC}_2\text{O}_4 \rightarrow \text{Fe}^{3+} + 2\text{CO}_2 + 3\text{e}^-\] Thus, 1 mole of \(\text{FeC}_2\text{O}_4\) releases 3 moles of electrons.
2. Write down reduction half-equation for manganate(VII): \[\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\] Thus, 1 mole of \(\text{MnO}_4^-\) accepts 5 moles of electrons.
3. Equate electron gain and loss: \[3 \times \text{MnO}_4^- \equiv 5 \times \text{FeC}_2\text{O}_4\] \[\text{Mole ratio: } \frac{n(\text{FeC}_2\text{O}_4)}{n(\text{MnO}_4^-)} = \frac{5}{3}\]
1 mark for the correct selection of option C. Correctly determines the combined electron output of the iron(II) ethanedioate and balances it with the reduction of manganate(VII) to yield the 5:3 reacting mole ratio.
The standard enthalpy change, \(\Delta H^{\theta}\), is \(+180\text{ kJ mol}^{-1}\) and the standard entropy change, \(\Delta S^{\theta}\), is \(+160\text{ J K}^{-1}\text{ mol}^{-1}\).
Assuming these values do not change with temperature, above what temperature does this reaction become feasible?
A.\(1125\text{ }^{\circ}\text{C}\)
B.\(1125\text{ K}\)
C.\(852\text{ K}\)
D.\(1.13\text{ K}\)
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PastPaper.workedSolution
A reaction becomes feasible when the Gibbs free energy change, \(\Delta G\), is less than or equal to zero:
\[\Delta G = \Delta H - T\Delta S \le 0\]
To find the boundary temperature, set \(\Delta G = 0\):
\[T = \frac{\Delta H}{\Delta S}\]
Convert \(\Delta H\) to \(\text{J mol}^{-1}\):
\[\Delta H = +180 \times 10^3\text{ J mol}^{-1}\]
Calculate \(T\):
\[T = \frac{180,000}{160} = 1125\text{ K}\]
PastPaper.markingScheme
1 mark for the correct selection of option B. Uses the equation Delta G = Delta H - T(Delta S), sets Delta G = 0, converts units correctly, and calculates T = 1125 K.
PastPaper.question 8 · multiple-choice
1 PastPaper.marks
Four different halogenoalkanes are heated separately with aqueous silver nitrate in ethanol.
Which halogenoalkane reacts fastest to form a precipitate of silver halide?
A.1-fluorobutane
B.1-chlorobutane
C.1-bromobutane
D.1-iodobutane
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PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond.
The C-I bond is the longest and weakest (lowest bond enthalpy) of the carbon-halogen bonds listed, meaning it is broken most easily. Consequently, 1-iodobutane reacts the fastest to release iodide ions, which immediately react with silver ions to form a yellow precipitate of silver iodide (\(\text{AgI}\)).
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1 mark for the correct selection of option D. Recognises that C-I has the lowest bond enthalpy and therefore undergoes nucleophilic substitution at the fastest rate.
PastPaper.question 9 · multiple-choice
1 PastPaper.marks
For a reaction \(2\text{A} + \text{B} \rightarrow \text{C}\), the rate equation is determined to be \(\text{Rate} = k[\text{A}]^2[\text{B}]\). At a constant temperature, a mixture with initial concentrations \([\text{A}] = 0.20\text{ mol dm}^{-3}\) and \([\text{B}] = 0.40\text{ mol dm}^{-3}\) has an initial rate of reaction of \(1.60 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the value and units of the rate constant, \(k\), at this temperature?
Rearranging the rate equation to solve for the rate constant: \(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]}\). Substituting the given values: \(k = \frac{1.60 \times 10^{-3}}{(0.20)^2 \times 0.40} = \frac{1.60 \times 10^{-3}}{0.040 \times 0.40} = \frac{1.60 \times 10^{-3}}{0.016} = 0.10\). To determine the units: \(\text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). Therefore, \(k = 0.10\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
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1 mark for the correct calculation of the value (0.10) and correct rate constant units (dm^6 mol^-2 s^-1).
PastPaper.question 10 · multiple-choice
1 PastPaper.marks
At \(298\text{ K}\), \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) is mixed with \(15.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting buffer solution?
1 mark for calculating the stoichiometry of the buffer system correctly to determine a pH value of 5.05.
PastPaper.question 11 · multiple-choice
1 PastPaper.marks
When excess concentrated hydrochloric acid is added to an aqueous solution containing \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) ions, a chemical reaction occurs. Which row correctly describes the changes in coordination number, molecular shape, and colour of the cobalt complex?
A.Coordination number decreases from 6 to 4; shape changes from octahedral to tetrahedral; colour changes from pink to blue.
B.Coordination number increases from 4 to 6; shape changes from tetrahedral to octahedral; colour changes from blue to pink.
C.Coordination number remains 6; shape remains octahedral; colour changes from pink to yellow.
D.Coordination number decreases from 6 to 4; shape changes from octahedral to square planar; colour changes from pink to blue.
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PastPaper.workedSolution
Adding concentrated hydrochloric acid results in ligand substitution: \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightleftharpoons [\text{CoCl}_4]^{2-}\text{(aq)} + 6\text{H}_2\text{O}\text{(l)}\). The initial hexaaquacobalt(II) complex is pink, octahedral, and has a coordination number of 6. The product tetrachlorocobaltate(II) complex is blue, tetrahedral, and has a coordination number of 4 because of the larger size of the chloride ligand which prevents six-coordination.
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1 mark for selecting the option showing a decrease in coordination number (6 to 4), a change in shape from octahedral to tetrahedral, and a colour transition from pink to blue.
PastPaper.question 12 · multiple-choice
1 PastPaper.marks
An industrial chemical process has a standard enthalpy change \(\Delta H^\ominus = +135\text{ kJ mol}^{-1}\) and standard entropy change \(\Delta S^\ominus = +285\text{ J K}^{-1}\text{ mol}^{-1}\). Assuming these values do not change with temperature, what is the minimum temperature above which this reaction becomes feasible?
A.\(2.11\text{ K}\)
B.\(474\text{ K}\)
C.\(211\text{ K}\)
D.\(474^\circ\text{C}\)
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PastPaper.workedSolution
A reaction is feasible when \(\Delta G \le 0\). Using \(\Delta G = \Delta H - T\Delta S\), at the boundary of feasibility we set \(\Delta G = 0\), which gives \(T = \frac{\Delta H}{\Delta S}\). Converting \(\Delta H\) to Joules gives \(135000\text{ J mol}^{-1}\). Thus, \(T = \frac{135000}{285} \approx 473.68\text{ K}\). This means the reaction is feasible above \(474\text{ K}\).
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1 mark for the correct calculation of temperature in Kelvin using the Gibbs free energy relationship.
PastPaper.question 13 · multiple-choice
1 PastPaper.marks
Which of the following halogenoalkanes is hydrolysed most rapidly when warmed with aqueous silver nitrate in ethanol?
A.1-chlorobutane
B.1-iodobutane
C.2-chloro-2-methylpropane
D.2-iodo-2-methylpropane
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PastPaper.workedSolution
The rate of hydrolysis is dictated by two factors: (1) bond strength, where C-I is weaker and breaks more easily than C-Cl; and (2) substitution mechanism, where tertiary halogenoalkanes hydrolyse much faster than primary halogenoalkanes via the \(\text{S}_\text{N}1\) path, which is driven by the formation of a highly stable tertiary carbocation. Therefore, 2-iodo-2-methylpropane (tertiary iodoalkane) is hydrolysed most rapidly.
PastPaper.markingScheme
1 mark for identifying that 2-iodo-2-methylpropane undergoes the most rapid hydrolysis due to its weak C-I bond and tertiary carbocation stability.
PastPaper.question 14 · multiple-choice
1 PastPaper.marks
A sample of an unknown anhydrous Group 2 metal carbonate, \(\text{MCO}_3\), with a mass of \(0.369\text{ g}\) was reacted with excess dilute hydrochloric acid. The carbon dioxide gas evolved was collected and measured to be \(61.3\text{ cm}^3\) at \(298\text{ K}\) and \(101\text{ kPa}\). Identify the metal \(\text{M}\). (The gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\))
A.Magnesium
B.Calcium
C.Strontium
D.Barium
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PastPaper.workedSolution
1. Find moles of \(\text{CO}_2\) using \(pV = nRT\): \(n = \frac{pV}{RT} = \frac{(101 \times 10^3\text{ Pa}) \times (61.3 \times 10^{-6}\text{ m}^3)}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 298\text{ K}} = \frac{6.1913}{2476.38} \approx 0.00250\text{ mol}\). 2. Since 1 mole of \(\text{MCO}_3\) yields 1 mole of \(\text{CO}_2\), moles of \(\text{MCO}_3 = 0.00250\text{ mol}\). 3. Molar mass of \(\text{MCO}_3 = \frac{0.369\text{ g}}{0.00250\text{ mol}} = 147.6\text{ g mol}^{-1}\). 4. Molar mass of \(\text{M} = 147.6 - 12.0 - (3 \times 16.0) = 87.6\text{ g mol}^{-1}\). Comparing with the periodic table, the Group 2 element with \(A_r \approx 87.6\) is Strontium (Sr).
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1 mark for correctly applying the ideal gas equation and stoichiometry to find the identity of the metal as Strontium.
PastPaper.question 15 · multiple-choice
1 PastPaper.marks
The rate constant, \(k\), for a reaction was measured at several different temperatures. A graph of \(\ln k\) against \(\frac{1}{T}\) (where \(T\) is the temperature in Kelvin) was plotted and gave a straight-line graph with a gradient of \(-1.20 \times 10^4\text{ K}\). What is the activation energy, \(E_a\), of this reaction? (The gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\))
A.\(1.44\text{ kJ mol}^{-1}\)
B.\(99.7\text{ kJ mol}^{-1}\)
C.\(120\text{ kJ mol}^{-1}\)
D.\(9.97 \times 10^4\text{ kJ mol}^{-1}\)
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PastPaper.workedSolution
From the Arrhenius equation: \(\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A\). This shows that the gradient of a plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\). Therefore: \(-\frac{E_a}{R} = -1.20 \times 10^4\), which leads to \(E_a = 1.20 \times 10^4 \times 8.31 = 99720\text{ J mol}^{-1}\). Converting to \(\text{kJ mol}^{-1}\) gives \(99.7\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark for correctly linking the gradient of the Arrhenius plot to activation energy and performing the unit conversion to obtain 99.7 kJ mol^-1.
PastPaper.question 16 · multiple-choice
1 PastPaper.marks
The standard electrode potentials for two half-equations are shown below: \(\text{Fe}^{3+}\text{(aq)} + \text{e}^- \rightleftharpoons \text{Fe}^{2+}\text{(aq)} \quad E^\ominus = +0.77\text{ V}\) \(\text{S}_2\text{O}_8^{2-}\text{(aq)} + 2\text{e}^- \rightleftharpoons 2\text{SO}_4^{2-}\text{(aq)} \quad E^\ominus = +2.01\text{ V}\)
Which of the following statements is correct under standard conditions?
A.\(\text{Fe}^{2+}\text{(aq)}\) can reduce \(\text{S}_2\text{O}_8^{2-}\text{(aq)}\), and the EMF of the cell is \(+1.24\text{ V}\).
B.\(\text{Fe}^{3+}\text{(aq)}\) can oxidise \(\text{SO}_4^{2-}\text{(aq)}\), and the EMF of the cell is \(+1.24\text{ V}\).
C.\(\text{Fe}^{2+}\text{(aq)}\) can reduce \(\text{S}_2\text{O}_8^{2-}\text{(aq)}\), and the EMF of the cell is \(+2.78\text{ V}\).
D.\(\text{SO}_4^{2-}\text{(aq)}\) can reduce \(\text{Fe}^{3+}\text{(aq)}\), and the EMF of the cell is \(+1.24\text{ V}\).
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The half-cell with the more positive potential is the reduction half-cell (cathode): \(\text{S}_2\text{O}_8^{2-}\text{(aq)} + 2\text{e}^- \rightarrow 2\text{SO}_4^{2-}\text{(aq)}\). The half-cell with the less positive potential is the oxidation half-cell (anode), meaning the reaction runs in reverse: \(\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + \text{e}^-\). Combining these reactions shows that \(\text{Fe}^{2+}\text{(aq)}\) is oxidized and reduces \(\text{S}_2\text{O}_8^{2-}\text{(aq)}\). The EMF of the cell is calculated as \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +2.01\text{ V} - (+0.77\text{ V}) = +1.24\text{ V}\).
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1 mark for the correct identification of the feasible redox process and standard cell EMF calculation of +1.24 V.
PastPaper.question 17 · multiple choice
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The rate constant, \(k\), for a first-order reaction is determined at two temperatures: At \(300\text{ K}\), \(k = 2.4 \times 10^{-3}\text{ s}^{-1}\) and at \(320\text{ K}\), \(k = 1.2 \times 10^{-2}\text{ s}^{-1}\). What is the activation energy, \(E_a\), for this reaction in \(\text{kJ mol}^{-1}\)? (The gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\))
A.13.4
B.64.2
C.134
D.6.42
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We use the Arrhenius equation in the form: \(\ln(k_2/k_1) = -\frac{E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})\). Substituting the given values: \(\ln(1.2 \times 10^{-2} / 2.4 \times 10^{-3}) = -\frac{E_a}{8.31} (\frac{1}{320} - \frac{1}{300})\). This simplifies to: \(\ln(5) = -\frac{E_a}{8.31} (-2.0833 \times 10^{-4})\). Thus, \(1.6094 = E_a \times 2.507 \times 10^{-5}\), giving \(E_a = 64197\text{ J mol}^{-1}\), which is \(64.2\text{ kJ mol}^{-1}\).
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1 mark for the correct option selection (B). Reject other values due to calculation errors or incorrect temperature unit handling.
PastPaper.question 18 · multiple choice
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A buffer solution is prepared by mixing \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(15.0\text{ cm}^3\) of \(0.080\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting buffer solution at \(298\text{ K}\)?
1 mark for the correct option selection (B). Distractors account for failing to subtract reacted acid moles or inverting the buffer ratio.
PastPaper.question 19 · multiple choice
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When excess concentrated hydrochloric acid is added to an aqueous solution containing \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) ions, a chemical reaction occurs. Which of the following correctly describes the color change of the solution and the change in coordination number of the cobalt ion?
A.Color change: Pink to blue; Coordination number: Decreases from 6 to 4
B.Color change: Blue to pink; Coordination number: Increases from 4 to 6
C.Color change: Pink to green; Coordination number: Remains 6
D.Color change: Green to blue; Coordination number: Decreases from 6 to 4
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The aqueous hexaaquacobalt(II) ion, \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), is pink and has a coordination number of 6. Addition of excess concentrated hydrochloric acid leads to ligand substitution forming the tetrachlorocobaltate(II) ion, \([\text{CoCl}_4]^{2-}\), which is blue and has a coordination number of 4 because chloride ligands are larger and experience more steric hindrance than water.
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1 mark for the correct option selection (A). Reject other combinations because of incorrect color changes or coordination numbers.
PastPaper.question 20 · multiple choice
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For a particular reaction, \(\Delta H^\theta = +135\text{ kJ mol}^{-1}\) and \(\Delta S^\theta = +285\text{ J K}^{-1}\text{ mol}^{-1}\). Under standard conditions, at what temperature does this reaction become feasible?
A.Above \(474\text{ }^\circ\text{C}\)
B.Above \(201\text{ }^\circ\text{C}\)
C.Below \(201\text{ }^\circ\text{C}\)
D.Below \(474\text{ }^\circ\text{C}\)
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A reaction is feasible when \(\Delta G \le 0\). Setting \(\Delta G = \Delta H - T\Delta S = 0\), we find the transition temperature \(T = \frac{\Delta H}{\Delta S} = \frac{135000\text{ J mol}^{-1}}{285\text{ J K}^{-1}\text{ mol}^{-1}} = 473.68\text{ K}\). Converting to Celsius: \(473.68 - 273.15 = 200.53\text{ }^\circ\text{C}\). Since both enthalpy and entropy changes are positive, the reaction is feasible at temperatures above this value (i.e., above \(201\text{ }^\circ\text{C}\)).
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1 mark for the correct option selection (B). Distractors account for forgetting to convert Kelvin to Celsius or misapplying the inequality direction.
PastPaper.question 21 · multiple choice
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Which of the following halogenoalkanes reacts most rapidly when heated with aqueous silver nitrate in ethanol?
A.1-chlorobutane
B.2-chloro-2-methylpropane
C.1-iodobutane
D.2-iodo-2-methylpropane
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The rate of nucleophilic substitution (hydrolysis) depends on two factors: the strength of the carbon-halogen bond and the mechanism pathway. The C-I bond is weaker and more easily broken than the C-Cl bond, meaning iodoalkanes react faster than chloroalkanes. Furthermore, tertiary halogenoalkanes react much faster than primary halogenoalkanes because they react via the \(S_N1\) mechanism, which involves the formation of a highly stable tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.
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1 mark for the correct option selection (D). Reject other options because of either stronger C-X bonds or primary halogenoalkane structures.
PastPaper.question 22 · multiple choice
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A sample of \(1.43\text{ g}\) of a Group 2 metal carbonate, \(\text{MCO}_3\), was reacted completely with excess hydrochloric acid: \(\text{MCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\). The volume of \(\text{CO}_2\) gas collected at \(298\text{ K}\) and \(100\text{ kPa}\) was \(3.54 \times 10^{-4}\text{ m}^3\). What is the identity of the metal, \(\text{M}\)? (The gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\))
A.Magnesium
B.Calcium
C.Strontium
D.Barium
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First, calculate the moles of carbon dioxide produced using \(pV = nRT\): \(n = \frac{pV}{RT} = \frac{100 \times 10^3\text{ Pa} \times 3.54 \times 10^{-4}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 298\text{ K}} = \frac{35.4}{2476.38} = 0.014295\text{ mol}\). Since the stoichiometry of the reaction is 1:1, there are \(0.014295\text{ mol}\) of \(\text{MCO}_3\). Thus, \(M_r(\text{MCO}_3) = \frac{1.43\text{ g}}{0.014295\text{ mol}} = 100.03\text{ g mol}^{-1}\). Subtracting the mass of carbonate (\(\text{CO}_3^{2-}\) = \(12.0 + 3 \times 16.0 = 60.0\text{ g mol}^{-1}\)) gives the relative atomic mass of the metal: \(100.03 - 60.0 = 40.03\text{ g mol}^{-1}\). This corresponds to Calcium (\(A_r = 40.1\)).
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1 mark for the correct option selection (B). Reject other Group 2 metals due to mismatched molecular masses.
PastPaper.question 23 · multiple choice
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For the reaction \(\text{P} + \text{Q} \rightarrow \text{R}\), the following initial rate data were obtained at a constant temperature: Experiment 1: \([\text{P}] = 0.10\text{ mol dm}^{-3}\), \([\text{Q}] = 0.10\text{ mol dm}^{-3}\), rate = \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Experiment 2: \([\text{P}] = 0.20\text{ mol dm}^{-3}\), \([\text{Q}] = 0.10\text{ mol dm}^{-3}\), rate = \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Experiment 3: \([\text{P}] = 0.20\text{ mol dm}^{-3}\), \([\text{Q}] = 0.20\text{ mol dm}^{-3}\), rate = \(1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the value and units of the rate constant, \(k\), for this reaction?
Comparing Experiments 1 and 2: when \([\text{P}]\) doubles, the rate increases by a factor of 4 (\(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)), so the order with respect to \(\text{P}\) is 2. Comparing Experiments 2 and 3: when \([\text{Q}]\) doubles, the rate increases by a factor of 2 (\(8.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\)), so the order with respect to \(\text{Q}\) is 1. Thus, the rate equation is: \(\text{rate} = k[\text{P}]^2[\text{Q}]\). Using Experiment 1: \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3})^2(0.10\text{ mol dm}^{-3})\), yielding \(k = 0.20\). Units: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).
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1 mark for the correct option selection (A). Other options are incorrect in either value or units.
PastPaper.question 24 · multiple choice
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At \(310\text{ K}\) (normal human body temperature), the ionic product of water, \(K_w\), is \(2.4 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\). Which of the following statements correctly describes pure water at \(310\text{ K}\)?
A.The pH is 6.81 and the water is acidic.
B.The pH is 6.81 and the water is neutral.
C.The pH is 7.00 and the water is neutral.
D.The pH is 7.19 and the water is alkaline.
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By definition, pure water is always neutral because the concentration of hydrogen ions, \([\text{H}^+]\), is exactly equal to the concentration of hydroxide ions, \([\text{OH}^-]\). At \(310\text{ K}\), \(K_w = [\text{H}^+]^2 = 2.4 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\). Taking the square root gives \([\text{H}^+] = 1.55 \times 10^{-7}\text{ mol dm}^{-3}\). The pH is then \(-\log_{10}(1.55 \times 10^{-7}) = 6.81\). Because the water is pure, it remains neutral.
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1 mark for the correct option selection (B). Distractors test the common misconception that neutrality is defined strictly by pH 7.00 rather than \([\text{H}^+] = [\text{OH}^-]\).
PastPaper.question 25 · Multiple Choice
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For the reaction \(2A + B + 2C \rightarrow D\), the following initial rate data were obtained at a constant temperature:
What are the units of the rate constant, \(k\), for this reaction?
A.\(\text{mol}^{-1} \text{dm}^3 \text{s}^{-1}\)
B.\(\text{mol}^{-2} \text{dm}^6 \text{s}^{-1}\)
C.\(\text{mol}^{-3} \text{dm}^9 \text{s}^{-1}\)
D.\(\text{s}^{-1}\)
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1. Deduce the order with respect to each reactant: - Comparing Experiments 1 and 2: \([A]\) doubles while \([B]\) and \([C]\) remain constant; the rate doubles (from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\)). Therefore, the order with respect to \(A\) is 1. - Comparing Experiments 1 and 3: \([B]\) doubles while \([A]\) and \([C]\) remain constant; the rate quadruples (from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). Therefore, the order with respect to \(B\) is 2. - Comparing Experiments 1 and 4: \([C]\) doubles while \([A]\) and \([B]\) remain constant; the rate remains unchanged. Therefore, the order with respect to \(C\) is 0.
2. Write the rate equation: $$\text{Rate} = k[A][B]^2$$
3. Determine the units of \(k\): $$k = \frac{\text{Rate}}{[A][B]^2}$$ $$\text{Units} = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{\text{mol}^3 \text{dm}^{-9}} = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1}$$
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1 mark for the correct units of the rate constant based on the deduced rate equation (Option B).
PastPaper.question 26 · Multiple Choice
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A buffer solution is prepared by mixing \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(15.0\text{ cm}^3\) of \(0.0800\text{ mol dm}^{-3}\) sodium hydroxide at \(298\text{ K}\).
1 mark for calculating the correct pH value (Option B).
PastPaper.question 27 · Multiple Choice
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Which of the following statements about the complex ion \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) (where \(\text{en} = \text{1,2-diaminoethane}\)) is correct?
A.It has a coordination number of 4 and cannot exhibit optical isomerism.
B.It has a coordination number of 6 and the trans-isomer is optically active.
C.It has a coordination number of 6 and only the cis-isomer exhibits optical isomerism.
D.It has a coordination number of 6 and neither isomer is optically active.
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1. Identify coordination number: - 1,2-diaminoethane (\(\text{en}\)) is a bidentate ligand, so each \(\text{en}\) forms two coordinate bonds with the cobalt ion. - Since there are two \(\text{en}\) ligands and two monodentate chloride (\(\text{Cl}^-\)) ligands, the coordination number is \(2 \times 2 + 2 = 6\) (octahedral geometry).
2. Analyse isomerism: - Octahedral complexes of the type \([\text{M}(\text{en})_2\text{X}_2]\) exist as cis and trans stereoisomers. - The trans-isomer has a plane of symmetry and is achiral, meaning it does not show optical isomerism. - The cis-isomer lacks a plane of symmetry and has non-superimposable mirror images (enantiomers), hence it is optically active.
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1 mark for identifying the correct coordination number and stereochemical property (Option C).
PastPaper.question 28 · Multiple Choice
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Using the following data, calculate the lattice dissociation enthalpy of \(\text{CaCl}_2(s)\):
- Enthalpy of formation of \(\text{CaCl}_2(s) = -796\text{ kJ mol}^{-1}\) - 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 dissociation enthalpy of chlorine \(= +242\text{ kJ mol}^{-1}\) - Electron affinity of chlorine \(= -349\text{ kJ mol}^{-1}\)
A.\(+2253\text{ kJ mol}^{-1}\)
B.\(+2602\text{ kJ mol}^{-1}\)
C.\(+2495\text{ kJ mol}^{-1}\)
D.\(+661\text{ kJ mol}^{-1}\)
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By applying Hess's Law to a Born-Haber cycle for \(\text{CaCl}_2(s)\):
1 mark for the correct calculation of lattice dissociation enthalpy (Option A).
PastPaper.question 29 · Multiple Choice
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Which of the following halogenoalkanes reacts most rapidly when heated with aqueous silver nitrate?
A.1-chlorobutane
B.1-iodobutane
C.2-chloro-2-methylpropane
D.2-iodo-2-methylpropane
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1. The rate of nucleophilic substitution of halogenoalkanes depends on the strength of the carbon-halogen bond. The C-I bond has the lowest bond enthalpy and is broken most easily, so iodoalkanes react faster than chloroalkanes. 2. The rate also depends on the structure of the halogenoalkane. Tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react extremely rapidly via the \(S_N1\) mechanism, which involves the formation of a stable tertiary carbocation intermediate. This mechanism has a much lower activation energy than the \(S_N2\) mechanism undergone by primary halogenoalkanes (such as 1-iodobutane).
Therefore, 2-iodo-2-methylpropane reacts the fastest.
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1 mark for identifying the tertiary iodoalkane as the fastest-reacting species (Option D).
PastPaper.question 30 · Multiple Choice
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A sample of an unknown volatile liquid with a mass of \(0.352\text{ g}\) was vaporised in a gas syringe at a temperature of \(98\text{ ^\circ C}\) and a pressure of \(101\text{ kPa}\). The volume of gas collected was \(118\text{ cm}^3\).
What is the relative molecular mass (\(M_r\)) of the volatile liquid? (The gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\))