An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 Cambridge International A Level Chemistry (9620) paper. Not affiliated with or reproduced from Cambridge.
Section Unit 1: Inorganic 1 and Physical 1
Answer all questions in the spaces provided. Show all working for calculations.
6 Question · 72 marks
Question 1 · Structured
12 marks
This question is about atomic structure and time-of-flight (TOF) mass spectrometry.
(a) Describe how ions are formed in a TOF mass spectrometer using electrospray ionization. Write an equation for this ionization process using a molecule, \(M\). Compare this process with electron impact ionization, writing a corresponding equation for the ionization of \(M\) under electron impact.
(b) In a TOF mass spectrometer, a sample of gallium is analyzed. A single \(^{69}\text{Ga}^+\) ion is accelerated to a kinetic energy of \(1.15 \times 10^{-16}\text{ J}\) over an acceleration region and then enters a drift tube of length \(1.20\text{ m}\).
Calculate the time of flight, in seconds, for this \(^{69}\text{Ga}^+\) ion.
The Avogadro constant, \(L = 6.022 \times 10^{23}\text{ mol}^{-1}\). Kinetic energy, \(KE = \frac{1}{2}mv^2\). Give your answer to 3 significant figures.
(c) Explain why the first ionization energy of sulfur is lower than that of phosphorus, despite sulfur having a higher nuclear charge.
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Worked solution
(a) - Electrospray ionization: The sample is dissolved in a volatile solvent and forced through a fine hypodermic needle connected to a positive high-voltage power supply. The particles gain a proton (\(\text{H}^+\)) from the solvent as they emerge. Equation: \(M + \text{H}^+ \rightarrow MH^+\) - Electron impact ionization: The sample is vaporized and high-energy electrons from an electron gun are fired at it, knocking off an electron from each particle. Equation: \(M \rightarrow M^+ + e^-\) (or \(M(\text{g}) \rightarrow M^+(\text{g}) + e^-\))
(b) 1. Calculate the mass of one \(^{69}\text{Ga}^+\) ion: \(m = \frac{69.0 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 1.1458 \times 10^{-25}\text{ kg}\)
2. Use kinetic energy to find velocity: \(KE = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2KE}{m}}\) \(v = \sqrt{\frac{2 \times 1.15 \times 10^{-16}}{1.1458 \times 10^{-25}}} = \sqrt{2.0073 \times 10^9} = 4.480 \times 10^4\text{ m s}^{-1}\)
3. Calculate time of flight: \(t = \frac{d}{v} = \frac{1.20}{4.480 \times 10^4} = 2.68 \times 10^{-5}\text{ s}\)
(c) - The electron configuration of phosphorus's outer shell is \(3s^2 3p^3\) with three singly occupied 3p orbitals, whereas sulfur's configuration is \(3s^2 3p^4\). - In sulfur, two electrons are paired in one of the 3p orbitals. - Mutual repulsion between these paired electrons in the same orbital makes the outer electron in sulfur easier to remove than an electron from the half-filled 3p subshell of phosphorus.
(b) [Total: 5 marks] - 1 mark: Convert mass of 1 mol of Ga into kg (\(0.069\text{ kg}\)). - 1 mark: Calculate mass of one ion (\(1.15 \times 10^{-25}\text{ kg}\), accept \(1.146 \times 10^{-25}\)). - 1 mark: Correct rearrangement of \(KE = \frac{1}{2}mv^2\) for \(v\). - 1 mark: Correct calculation of velocity (\(4.48 \times 10^4\text{ m s}^{-1}\)). - 1 mark: Correct final time to 3 significant figures (\(2.68 \times 10^{-5}\text{ s}\)).
(c) [Total: 3 marks] - 1 mark: Identify that P has single electrons in 3p orbitals, while S has a paired set of electrons in a 3p orbital. - 1 mark: Reference to electron-electron pairing repulsion in S. - 1 mark: State that repulsion makes the electron easier to remove (requires less energy).
Question 2 · Structured
12 marks
This question is about reacting masses, empirical formulas, and gas calculations.
(a) Compound X is a hydrated metal sulfate with the formula \(\text{M}_2(\text{SO}_4)_3 \cdot x\text{H}_2\text{O}\). A sample of \(4.50\text{ g}\) of Compound X was heated to constant mass to remove all water of crystallization. The mass of the anhydrous residue was \(3.05\text{ g}\).
Show that the value of \(x\) is 9, given that the relative formula mass of the anhydrous metal sulfate, \(\text{M}_2(\text{SO}_4)_3\), is \(342.3\).
(b) Identify metal \(\text{M}\) using your answer from part (a) and a periodic table.
(c) A student prepares a sample of nitrobenzene (\(\text{C}_6\text{H}_5\text{NO}_2\), \(M_r = 123.1\)) by reacting benzene (\(\text{C}_6\text{H}_6\), \(M_r = 78.1\)) with excess concentrated nitric acid:
The student starts with \(15.6\text{ g}\) of benzene and obtains \(18.2\text{ g}\) of pure nitrobenzene. Calculate the percentage yield of nitrobenzene.
(d) Sodium azide (\(\text{NaN}_3\), \(M_r = 65.0\)) decomposes rapidly upon heating to produce sodium metal and nitrogen gas:
Calculate the volume of nitrogen gas, in \(\text{dm}^3\), produced at \(298\text{ K}\) and \(101.3\text{ kPa}\) from the decomposition of \(75.0\text{ g}\) of sodium azide.
The gas constant, \(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\). Give your answer to 3 significant figures.
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Worked solution
(a) - Mass of water lost = \(4.50 - 3.05 = 1.45\text{ g}\) - Moles of water = \(\frac{1.45}{18.0} = 0.08056\text{ mol}\) - Moles of anhydrous residue \(\text{M}_2(\text{SO}_4)_3\) = \(\frac{3.05}{342.3} = 0.00891\text{ mol}\) - Ratio of \(\text{H}_2\text{O} : \text{M}_2(\text{SO}_4)_3 = \frac{0.08056}{0.00891} = 9.04 \approx 9\). Thus, \(x = 9\).
(b) - \(M_r\) of \(\text{M}_2(\text{SO}_4)_3\) = \(342.3\) - \(2 \times A_r(\text{M}) + 3 \times [32.1 + (4 \times 16.0)] = 342.3\) - \(2 \times A_r(\text{M}) + 3 \times 96.1 = 342.3 \Rightarrow 2 \times A_r(\text{M}) + 288.3 = 342.3\) - \(2 \times A_r(\text{M}) = 54.0 \Rightarrow A_r(\text{M}) = 27.0\) - From the periodic table, the metal with \(A_r = 27.0\) is Aluminium (\(\text{Al}\)).
(c) - Moles of benzene used = \(\frac{15.6}{78.1} = 0.1997\text{ mol}\) - Theoretical moles of nitrobenzene = \(0.1997\text{ mol}\) - Theoretical mass of nitrobenzene = \(0.1997 \times 123.1 = 24.59\text{ g}\) - Percentage yield = \(\frac{18.2}{24.59} \times 100\% = 74.01\% \approx 74.0\%\)
(d) - Moles of \(\text{NaN}_3\) = \(\frac{75.0}{65.0} = 1.1538\text{ mol}\) - Moles of \(\text{N}_2\) produced = \(1.1538 \times 1.5 = 1.7308\text{ mol}\) - Convert pressure to Pa: \(P = 101.3 \times 10^3\text{ Pa} = 101300\text{ Pa}\) - Rearrange ideal gas equation: \(V = \frac{nRT}{P}\) - \(V = \frac{1.7308 \times 8.314 \times 298}{101300} = 0.04234\text{ m}^3\) - Convert to \(\text{dm}^3\): \(0.04234 \times 1000 = 42.34\text{ dm}^3\), which rounds to \(42.3\text{ dm}^3\).
Marking scheme
(a) [Total: 3 marks] - 1 mark: Correct calculation of mass of water (\(1.45\text{ g}\)) and moles of water (\(0.0806\text{ mol}\)). - 1 mark: Correct calculation of moles of anhydrous residue (\(0.00891\text{ mol}\)). - 1 mark: Clear calculation showing ratio is approximately 9.
(b) [Total: 1 mark] - 1 mark: Identify Aluminium / Al (must show working or reference to \(A_r = 27.0\)).
(c) [Total: 3 marks] - 1 mark: Calculate moles of benzene (\(0.200\text{ mol}\)). - 1 mark: Calculate theoretical mass of nitrobenzene (\(24.6\text{ g}\)). - 1 mark: Correct percentage yield (\(74.0\%\), accept range \(74.0\% - 74.1\%\)).
(d) [Total: 5 marks] - 1 mark: Calculate moles of sodium azide (\(1.15\text{ mol}\)). - 1 mark: Use stoichiometry to find moles of nitrogen gas (\(1.73\text{ mol}\)). - 1 mark: Convert pressure to Pa (\(101300\text{ Pa}\)). - 1 mark: Correct rearrangement of ideal gas equation and substitution. - 1 mark: Correct final volume in \(\text{dm}^3\) to 3 sig figs (\(42.3\text{ dm}^3\)).
Question 3 · Structured
12 marks
This question is about molecular shapes, bonding, and intermolecular forces.
(a) Predict the shape of the \(\text{ClF}_3\) molecule. Use valence shell electron pair repulsion (VSEPR) theory to explain your answer and state the approximate bond angle.
(b) Compare the shapes and bond angles of ammonia (\(\text{NH}_3\)) and boron trifluoride (\(\text{BF}_3\)). Explain why they have different shapes and angles.
(c) Explain why water (\(\text{H}_2\text{O}\)) has a significantly higher boiling point than hydrogen sulfide (\(\text{H}_2\text{S}\)), despite hydrogen sulfide having a larger relative molecular mass.
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Worked solution
(a) - Chlorine is the central atom and has 7 valence electrons, plus 3 electrons from the 3 fluorine bonds, making 10 electrons (5 pairs) in its outer shell. - There are 3 bonding pairs and 2 lone pairs. - The electron pair geometry is trigonal bipyramidal, but to minimize repulsion, the 2 lone pairs occupy the equatorial positions. - Therefore, the shape is T-shaped. - The lone pairs repel bonding pairs more than bonding pairs repel each other, pushing the equatorial-axial bonds closer, resulting in a bond angle of slightly less than \(90^\circ\) (around \(87.5^\circ\)).
(b) - Ammonia (\(\text{NH}_3\)) has 4 electron pairs around the central nitrogen atom (3 bonding pairs and 1 lone pair). Its shape is trigonal pyramidal and its bond angle is \(107^\circ\) (reduced from the tetrahedral angle of \(109.5^\circ\) by the extra repulsion of the lone pair). - Boron trifluoride (\(\text{BF}_3\)) has only 3 electron pairs around the central boron atom (3 bonding pairs and 0 lone pairs). Its shape is trigonal planar and its bond angle is exactly \(120^\circ\).
(c) - Oxygen is a very small, highly electronegative atom. The polar \(\text{O}-\text{H}\) bonds in \(\text{H}_2\text{O}\) allow strong hydrogen bonds to form between water molecules. - Sulfur is less electronegative than oxygen, so \(\text{H}_2\text{S}\) cannot form hydrogen bonds. It only experiences weaker permanent dipole-dipole interactions and London/van der Waals forces. - Hydrogen bonds are significantly stronger than dipole-dipole and van der Waals forces and require much more energy to overcome, resulting in a higher boiling point for water.
Marking scheme
(a) [Total: 4 marks] - 1 mark: Identify chlorine has 5 electron pairs (3 bonding, 2 lone pairs). - 1 mark: State the shape is T-shaped. - 1 mark: State bond angle is slightly less than \(90^\circ\) (accept range \(85^\circ - 89^\circ\)). - 1 mark: Explain that lone pair-lone pair / lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion.
(b) [Total: 4 marks] - 1 mark: Ammonia shape (trigonal pyramidal) and bond angle (\(107^\circ\), accept \(107-108^\circ\)). - 1 mark: Boron trifluoride shape (trigonal planar) and bond angle (\(120^\circ\)). - 1 mark: Explain that ammonia has a lone pair which repels the bonding pairs more strongly. - 1 mark: Explain that boron trifluoride has no lone pairs, so bonding pairs repel equally.
(c) [Total: 4 marks] - 1 mark: Identify that hydrogen bonding occurs between \(\text{H}_2\text{O}\) molecules. - 1 mark: State that only permanent dipole-dipole and/or van der Waals forces occur between \(\text{H}_2\text{S}\) molecules. - 1 mark: Mention that oxygen is highly electronegative (or more electronegative than sulfur) enabling hydrogen bonding. - 1 mark: Conclude that hydrogen bonds require significantly more thermal energy to break.
Question 4 · Structured
12 marks
This question is about energetics and determination of enthalpy changes.
(a) A student used a simple calorimeter to determine the enthalpy change of combustion of methanol (\(\text{CH}_3\text{OH}\)).
Burning \(0.960\text{ g}\) of methanol raised the temperature of \(150.0\text{ g}\) of water from \(19.5\ ^\circ\text{C}\) to \(45.5\ ^\circ\text{C}\).
The specific heat capacity of water is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).
(i) Calculate the heat energy released, in \(\text{kJ}\), during this combustion. Show your working.
(ii) Calculate the experimental enthalpy change of combustion of methanol, \(\Delta_c H\), in \(\text{kJ mol}^{-1}\).
(b) The student's experimental value is much less exothermic than the accepted data book value (\(-726\text{ kJ mol}^{-1}\)). Suggest two reasons for this discrepancy, other than direct heat loss to the surroundings.
(c) Calculate the theoretical enthalpy change of combustion of gaseous methanol using the mean bond enthalpies given in the table below.
The equation for the reaction is: \[\text{CH}_3\text{OH}(\text{g}) + 1.5\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g})\]
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Worked solution
(a)(i) - \(q = m c \Delta T\) - \(m = 150.0\text{ g}\) - \(\Delta T = 45.5 - 19.5 = 26.0\ ^\circ\text{C}\) (or \(26.0\text{ K}\)) - \(q = 150.0 \times 4.18 \times 26.0 = 16302\text{ J} = 16.302\text{ kJ}\) - Rounded to 3 sig figs: \(16.3\text{ kJ}\).
(a)(ii) - Moles of methanol, \(n = \frac{\text{mass}}{M_r} = \frac{0.960}{32.0} = 0.0300\text{ mol}\) - \(\Delta_c H = -\frac{q}{n} = -\frac{16.302}{0.0300} = -543.4\text{ kJ mol}^{-1}\) - Rounded to 3 sig figs: \(-543\text{ kJ mol}^{-1}\).
(b) - Incomplete combustion of methanol (forming carbon monoxide or carbon/soot instead of carbon dioxide). - Evaporation of methanol from the wick of the burner when not being combusted. - (Also accept: non-standard conditions, heat capacity of the calorimeter beaker is ignored).
(a)(ii) [Total: 3 marks] - 1 mark: Calculate moles of methanol (\(0.0300\text{ mol}\)). - 1 mark: Divide heat by moles (\(16.302 / 0.0300 = 543.4\)). - 1 mark: Negative sign and correct value (\(-543\text{ kJ mol}^{-1}\)).
(b) [Total: 2 marks] - 1 mark: Mention incomplete combustion. - 1 mark: Mention evaporation of fuel from the wick.
(c) [Total: 5 marks] - 1 mark: Correctly list and multiply the bonds broken in reactants. - 1 mark: Calculate total energy of bonds broken as \(2803\text{ kJ}\). - 1 mark: Correctly list and multiply the bonds formed in products. - 1 mark: Calculate total energy of bonds formed as \(3462\text{ kJ}\). - 1 mark: Calculate correct final \(\Delta H\) with negative sign (\(-659\text{ kJ mol}^{-1}\)).
Question 5 · Structured
12 marks
This question is about reaction kinetics, Maxwell-Boltzmann distributions, and catalysts.
(a) Sketch a Maxwell-Boltzmann distribution curve of molecular energies for a sample of gas at a temperature \(T_1\). On the same axes, sketch the distribution for the same sample of gas at a higher temperature \(T_2\). Label the axes and show the activation energy, \(E_a\).
Use your diagram to explain why the rate of reaction increases significantly with a small increase in temperature.
(b) Define the term *activation energy*.
(c) Describe how a catalyst increases the rate of a chemical reaction.
(d) Explain why the presence of a catalyst does not affect the position of an equilibrium or the value of the equilibrium constant, \(K_c\).
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Worked solution
(a) - Axis labeling: y-axis is 'Number of molecules' or 'Fraction of molecules', x-axis is 'Energy' or 'Kinetic energy'. - The curve for \(T_1\) starts at the origin, rises to a peak, and asymptotes to the x-axis. - The curve for \(T_2\) starts at the origin, has a lower peak shifted to the right, and lies above the \(T_1\) curve at higher energies. - Activation energy \(E_a\) is marked on the energy axis. The area under the curves to the right of \(E_a\) represents the molecules with energy greater than or equal to the activation energy. - Explanation: At a higher temperature \(T_2\), the average kinetic energy of the molecules is higher. This causes the distribution curve to flatten and shift to the right. Consequently, there is a significantly larger fraction of molecules with energy \(\ge E_a\) (the shaded area under the \(T_2\) curve is much greater than that under the \(T_1\) curve), leading to a higher rate of successful collisions.
(b) - Activation energy is the minimum energy that colliding particles must possess in order to react successfully.
(c) - A catalyst provides an alternative reaction pathway. - This alternative pathway has a lower activation energy than the uncatalyzed pathway.
(d) - A catalyst increases the rate of both the forward and reverse reactions. - It increases these rates by the exact same factor because it lowers the activation energy for both directions by the same amount. - As a result, the equilibrium state is reached faster, but the final position of equilibrium and the value of \(K_c\) (which depends only on temperature) remain unchanged.
Marking scheme
(a) [Total: 6 marks] - 1 mark: Correctly labeled axes (y: number/fraction of molecules, x: energy). - 1 mark: Correct shape for \(T_1\) (starts at origin, does not touch x-axis at high energy). - 1 mark: Correct shape for \(T_2\) (lower peak, shifted to the right, cross-over once). - 1 mark: Activation energy (\(E_a\)) correctly placed on x-axis. - 1 mark: Mention that at higher temperature, average kinetic energy increases, shifting distribution to higher values. - 1 mark: Explain that a much larger fraction of molecules have energy \(\ge E_a\), leading to more frequent successful collisions.
(d) [Total: 3 marks] - 1 mark: Catalyst increases rate of both forward and reverse reactions. - 1 mark: Increases them by the same factor. - 1 mark: Concludes that the position of equilibrium and \(K_c\) value are unchanged.
Question 6 · Structured
12 marks
This question is about Group 7 (the halogens) and halide ions.
(a) Chlorine reacts with cold, dilute aqueous sodium hydroxide.
(i) Write an equation for this reaction.
(ii) State the oxidation state of chlorine in each of the chlorine-containing products formed in this reaction.
(iii) Give one commercial use of the solution mixture formed in this reaction.
(b) Sodium halides react with concentrated sulfuric acid.
(i) Describe the observations made when solid sodium iodide reacts with concentrated sulfuric acid. Explain these observations by writing balanced chemical equations for the formation of two different reduction products of sulfuric acid.
(ii) Explain why sodium chloride only undergoes an acid-base reaction, and not a redox reaction, with concentrated sulfuric acid.
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(a)(ii) - In \(\text{NaCl}\) (or chloride ion), the oxidation state of chlorine is \(-1\). - In \(\text{NaClO}\) (or chlorate(I) ion), the oxidation state of chlorine is \(+1\).
(a)(iii) - Used as bleach / disinfectant / water treatment / killing bacteria.
(b)(i) - Observations (any two): - Purple vapor or black/grey solid (due to iodine, \(\text{I}_2\)) - Rotten egg smell (due to hydrogen sulfide, \(\text{H}_2\text{S}\)) - Yellow solid (due to sulfur, \(\text{S}\)) - Misty fumes (due to hydrogen iodide, \(\text{HI}\)) - Equations showing reduction of sulfuric acid: - To sulfur dioxide (\(\text{SO}_2\)) (S has gone from \(+6\) to \(+4\)): \(2\text{NaI} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\) (or ionic: \(2\text{I}^- + \text{H}_2\text{SO}_4 + 2\text{H}^+ \rightarrow \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\)) - To hydrogen sulfide (\(\text{H}_2\text{S}\)) (S has gone from \(+6\) to \(-2\)): \(8\text{NaI} + 5\text{H}_2\text{SO}_4 \rightarrow 4\text{Na}_2\text{SO}_4 + 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}\) (or ionic: \(8\text{I}^- + \text{H}_2\text{SO}_4 + 8\text{H}^+ \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}\))
(b)(ii) - Chloride ions (\(\text{Cl}^-\)) are weaker reducing agents than iodide ions (\(\text{I}^-\)) because the outer electrons of chloride are closer to the nucleus and less shielded, so they are held more strongly. - Consequently, sulfuric acid is not a strong enough oxidizing agent to oxidize chloride ions, so only an acid-base proton transfer occurs (producing \(\text{HCl}\) misty fumes, but no change in oxidation state of sulfur).
(a)(ii) [Total: 2 marks] - 1 mark: Correct state of \(-1\) for \(\text{NaCl}\). - 1 mark: Correct state of \(+1\) for \(\text{NaClO}\).
(a)(iii) [Total: 1 mark] - 1 mark: State commercial use (e.g. bleach, disinfectant).
(b)(i) [Total: 6 marks] - 1 mark: Identify purple vapor or black solid as iodine. - 1 mark: Identify yellow solid (sulfur) or rotten egg smell (\(\text{H}_2\text{S}\)). - 2 marks: Balanced equation for production of sulfur dioxide (1 mark for correct reactants/products, 1 mark for balancing). - 2 marks: Balanced equation for production of hydrogen sulfide (1 mark for correct reactants/products, 1 mark for balancing).
(b)(ii) [Total: 2 marks] - 1 mark: State that chloride is a weaker reducing agent (or less easily oxidized than iodide). - 1 mark: Explain that chloride cannot reduce sulfur / sulfuric acid is not a strong enough oxidizing agent to react with chloride.
Section Unit 2: Organic 1 and Physical 1
Answer all questions in the spaces provided. Show all working for calculations.
9 Question · 69.60000000000001 marks
Question 1 · Structured Questions
7.7 marks
A student carries out an experiment to determine the enthalpy of combustion of propan-1-ol (\(C_3H_7OH\)). The student burns a sample of propan-1-ol and uses the heat released to heat 150.0 g of water in a copper calorimeter.
The initial temperature of the water is \(21.2\ ^\circ\text{C}\) and the maximum temperature reached is \(53.7\ ^\circ\text{C}\).
The mass of the spirit burner containing propan-1-ol is \(124.35\text{ g}\) before combustion and \(123.53\text{ g}\) after combustion.
(The specific heat capacity of water is \(4.18\text{ J K}^{-1}\text{ g}^{-1}\)).
(a) Calculate the heat energy, \(q\), in kJ, transferred to the water. [2 marks]
(b) Calculate the amount, in moles, of propan-1-ol burned. [2 marks]
(c) Calculate the experimental enthalpy of combustion, \(\Delta_c H\), in \(\text{kJ mol}^{-1}\), of propan-1-ol. Give your answer to an appropriate number of significant figures. [2 marks]
(d) State two reasons why the experimental value of \(\Delta_c H\) is significantly less exothermic than the accepted database value. [1.7 marks]
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Worked solution
(a) Temperature rise, \(\Delta T = 53.7 - 21.2 = 32.5\text{ K}\). \(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J K}^{-1}\text{ g}^{-1} \times 32.5\text{ K} = 20377.5\text{ J} = 20.4\text{ kJ}\) (to 3 sig figs).
(b) Mass of propan-1-ol burned = \(124.35 - 123.53 = 0.82\text{ g}\). \(M_r\) of propan-1-ol (\(C_3H_8O\)) = \(60.0\text{ g mol}^{-1}\). Amount of propan-1-ol = \(\frac{0.82}{60.0} = 0.0137\text{ mol}\) (to 3 sig figs).
(c) \(\Delta_c H = -\frac{q}{\text{moles}} = -\frac{20.3775\text{ kJ}}{0.01367\text{ mol}} = -1490\text{ kJ mol}^{-1}\) (to 3 sig figs, accept range -1480 to -1500 depending on intermediate rounding).
(d) Heat loss to the surroundings/calorimeter; incomplete combustion of propan-1-ol (forming carbon monoxide or soot); evaporation of alcohol from the wick after extinguishing.
Marking scheme
(a) [2 marks] - 1 mark for correct temperature change (\(32.5\text{ K}\)) - 1 mark for correct calculation of heat energy in kJ (\(20.4\text{ kJ}\))
(b) [2 marks] - 1 mark for mass of propan-1-ol (\(0.82\text{ g}\)) - 1 mark for correct amount of moles (\(0.0137\text{ mol}\))
(c) [2 marks] - 1 mark for dividing \(q\) by moles - 1 mark for correct final value with negative sign and units (accept \(-1480\) to \(-1500\text{ kJ mol}^{-1}\))
(d) [1.7 marks] - 1 mark for first valid reason (e.g. heat loss) - 0.7 marks for second valid reason (e.g. incomplete combustion)
Question 2 · Structured Questions
7.7 marks
The rate of chemical reactions can be highly sensitive to changes in temperature.
(a) Sketch a Maxwell-Boltzmann distribution of molecular energies for a gas at temperature \(T_1\). On the same axes, sketch a second curve to show the distribution at a higher temperature, \(T_2\). Label the activation energy, \(E_a\), on your sketch. [4 marks]
(b) With reference to your sketch, explain in terms of collision theory why the rate of reaction increases when the temperature is increased. [3.7 marks]
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Worked solution
(a) The sketch should show: 1. Both curves starting at the origin (0,0) and asymptoting towards the x-axis at high energy (never touching the axis). 2. The curve for \(T_2\) has a lower peak (maximum) which is shifted to the right of the peak for \(T_1\). 3. The curves only cross once. 4. The activation energy \(E_a\) marked on the energy axis, with the area under the curve to the right of \(E_a\) larger for \(T_2\) than \(T_1\).
(b) When temperature is increased from \(T_1\) to \(T_2\), the average kinetic energy of the molecules increases, shifting the molecular energy distribution to the right. Consequently, a much larger fraction of molecules possess energy greater than or equal to the activation energy (\(E \ge E_a\)). This results in a significantly higher frequency of successful collisions (more successful collisions per unit time), thereby increasing the rate of reaction.
Marking scheme
(a) [4 marks] - 1 mark for labeled axes (y-axis: Fraction/number of molecules, x-axis: Energy). - 1 mark for curve \(T_1\) starting at origin and asymptoting to x-axis. - 1 mark for curve \(T_2\) having a peak that is lower and shifted to the right, crossing \(T_1\) only once. - 1 mark for correctly locating and labeling activation energy \(E_a\) to the right of the peaks.
(b) [3.7 marks] - 1 mark for stating that at higher temperature, average kinetic energy of molecules increases. - 1 mark for explaining that a much greater proportion of molecules have energy \(\ge E_a\). - 1.7 marks for stating this leads to a higher frequency of successful collisions (or more successful collisions per second).
Question 3 · Structured Questions
7.7 marks
Ethanol reacts with ethanoic acid in the presence of an acid catalyst to form ethyl ethanoate and water:
(a) Write an expression for the equilibrium constant, \(K_c\), for this reaction. [1 mark]
(b) In an experiment, 1.50 mol of ethanoic acid and 1.50 mol of ethanol are mixed in a flask and left to reach equilibrium at a temperature \(T\). The equilibrium mixture is found to contain 0.50 mol of ethanoic acid. Calculate the amounts, in mol, of ethanol, ethyl ethanoate, and water present in this equilibrium mixture. [3 marks]
(c) Calculate the value of \(K_c\) at temperature \(T\). Show your working. State the units of \(K_c\), if any. [2.7 marks]
(d) The forward reaction is exothermic. Explain what happens to the value of \(K_c\) when the temperature is increased. [1 mark]
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(b) Let \(x\) be the change in moles of reactants at equilibrium. \text{Moles of ethanoic acid at equilibrium} = 1.50 - x = 0.50\text{ mol} \implies x = 1.00\text{ mol}. \text{Equilibrium amount of ethanol} = 1.50 - 1.00 = 0.50\text{ mol}. \text{Equilibrium amount of ethyl ethanoate} = 1.00\text{ mol}. \text{Equilibrium amount of water} = 1.00\text{ mol}.
(c) Substituting the equilibrium concentrations (using volume \(V\)) into the \(K_c\) expression: \(K_c = \frac{(1.00/V) \times (1.00/V)}{(0.50/V) \times (0.50/V)} = \frac{1.00 \times 1.00}{0.50 \times 0.50} = 4.0\). Since the total number of moles is the same on both sides, the units cancel out. Therefore, \(K_c\) has no units.
(d) Since the forward reaction is exothermic, an increase in temperature shifts the position of equilibrium to the left (in the endothermic direction) to absorb the added heat. As a result, the concentrations of reactants increase and products decrease, so the value of \(K_c\) decreases.
Marking scheme
(a) [1 mark] - 1 mark for correct expression with square brackets.
(b) [3 marks] - 1 mark for equilibrium amount of ethanol = \(0.50\text{ mol}\). - 1 mark for equilibrium amount of ethyl ethanoate = \(1.00\text{ mol}\). - 1 mark for equilibrium amount of water = \(1.00\text{ mol}\).
(c) [2.7 marks] - 1 mark for substituting equilibrium amounts into the expression (showing volume cancels or is omitted correctly). - 1 mark for calculating \(K_c = 4.0\). - 0.7 marks for stating there are no units.
(d) [1 mark] - 1 mark for stating that \(K_c\) decreases because the equilibrium shifts to the left in the endothermic direction.
Question 4 · Structured Questions
7.7 marks
Halogenoalkanes undergo nucleophilic substitution reactions when treated with suitable nucleophiles.
(a) Write a balanced chemical equation for the reaction of 1-bromobutane with aqueous sodium hydroxide to form butan-1-ol. Identify the type of reaction mechanism that occurs. [2 marks]
(b) Draw the detailed mechanism for this reaction. Use curly arrows to show the movement of electron pairs, and include any relevant partial charges (\(\delta+\) and \(\delta-\)). [4 marks]
(c) If 2-bromo-2-methylpropane is used instead of 1-bromobutane under identical conditions, a different mechanism dominates. Name this mechanism and explain why 2-bromo-2-methylpropane reacts via this pathway. [1.7 marks]
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(b) Mechanism steps: 1. Draw 1-bromobutane showing partial positive charge on the C attached to Br (\(\text{C}^{\delta+}\)) and partial negative charge on the Br (\(\text{Br}^{\delta-}\)). 2. Show a curly arrow starting from a lone pair on the oxygen of the hydroxide ion (\(\text{OH}^-\)) pointing directly to the \(\text{C}^{\delta+}\). 3. Show a curly arrow starting from the center of the \(\text{C}-\text{Br}\) bond pointing to the \(\text{Br}\) atom. 4. Draw the products: butan-1-ol and a bromide ion (\(\text{Br}^-\)).
(c) Mechanism: \(S_N1\) (Nucleophilic substitution unimolecular). Reason: 2-bromo-2-methylpropane is a tertiary halogenoalkane. It forms a tertiary carbocation intermediate, which is highly stable due to the positive inductive electron-releasing effect of the three methyl groups. Steric hindrance also prevents direct access of the nucleophile to the carbon atom in an \(S_N2\) mechanism.
Marking scheme
(a) [2 marks] - 1 mark for correct balanced equation. - 1 mark for naming nucleophilic substitution.
(b) [4 marks] - 1 mark for showing correct dipoles \(\text{C}^{\delta+}\) and \(\text{Br}^{\delta-}\) on the C-Br bond. - 1 mark for curly arrow from the lone pair of \(\text{OH}^-\)_ to the carbon atom. - 1 mark for curly arrow from the C-Br bond to the Br atom. - 1 mark for correct structures of products (butan-1-ol and \(\text{Br}^-\ Triton\)).
(c) [1.7 marks] - 0.7 marks for naming \(S_N1\). - 1 mark for explaining that the tertiary carbocation intermediate is stable due to the positive inductive effect of three methyl groups (or steric hindrance prevents \(S_N2\)).
Question 5 · Structured Questions
7.7 marks
But-1-ene reacts with hydrogen bromide (\(\text{HBr}\)) at room temperature to form two isomeric halogenoalkanes.
(a) State the names of the major and minor products of this reaction, and explain why the major product is formed in a much greater yield. [3 marks]
(b) Draw the mechanism for the formation of the major product. Include curly arrows, relevant dipoles (\(\delta+\) and \(\delta-\)), and the structure of the carbocation intermediate. [4.7 marks]
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Worked solution
(a) Major product: 2-bromobutane. Minor product: 1-bromobutane. Explanation: The reaction proceeds via carbocation intermediates. The major product (2-bromobutane) is formed via a secondary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\)), whereas the minor product is formed via a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+\)). The secondary carbocation is more stable than the primary carbocation due to the positive inductive effect of two electron-releasing alkyl groups compared to only one in the primary carbocation.
(b) Mechanism details: 1. Show HBr with dipoles: \(\text{H}^{\delta+}-\text{Br}^{\delta-}\). 2. Draw a curly arrow starting from the \(\text{C}=\text{C}\) double bond of but-1-ene to the \(\text{H}^{\delta+}\). 3. Draw a curly arrow from the H-Br bond to the Br atom. 4. Draw the secondary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\)) and a bromide ion (\(\text{Br}^-\)) with a lone pair and negative charge. 5. Draw a curly arrow from the lone pair on \(\text{Br}^-\) to the positive carbon of the carbocation. 6. Draw the final product: 2-bromobutane.
Marking scheme
(a) [3 marks] - 1 mark for naming both products correctly (major = 2-bromobutane, minor = 1-bromobutane). - 1 mark for identifying that the major pathway goes via the secondary carbocation, which is more stable than the primary carbocation. - 1 mark for explaining stability in terms of the positive inductive effect of two alkyl groups.
(b) [4.7 marks] - 1 mark for correct dipoles on \(\text{H}-\text{Br}\) and curly arrow from the double bond to the \(\text{H}\). - 1 mark for curly arrow from the \(\text{H}-\text{Br}\) bond to \(\text{Br}\). - 1 mark for correct structure of the secondary carbocation intermediate. - 1 mark for drawing \(\text{Br}^-\)_ with a lone pair and a curly arrow to the positive carbon atom. - 0.7 marks for the correct final structure of 2-bromobutane.
Question 6 · Structured Questions
7.7 marks
Alcohols can be oxidized under different conditions to yield different carbonyl compounds.
(a) Propan-1-ol can be oxidized to form either propanal or propanoic acid. (i) State the reagents and conditions required to oxidize propan-1-ol to propanal. Explain how the experimental setup ensures that propanal is obtained as the main product. [3.7 marks] (ii) Write a balanced chemical equation for the oxidation of propan-1-ol to propanal. Use \([\text{O}]\) to represent the oxidizing agent. [1 mark]
(b) State a chemical test that can distinguish between propanal and propan-2-ol. Describe the expected observation for each compound. [3 marks]
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Worked solution
(a)(i) Reagents: Acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7\) in dilute \(\text{H}_2\text{SO}_4\)). Conditions: Heat and distil immediately. Explanation: Propanal is an aldehyde and cannot form hydrogen bonds with other propanal molecules, so it has a lower boiling point than propan-1-ol and propanoic acid. By using distillation apparatus, propanal vaporizes as soon as it is formed and is distilled off out of the oxidizing mixture before it can undergo further oxidation to propanoic acid.
(b) Test: Tollens' reagent (or Fehling's solution). Observation with propanal: Formation of a silver mirror on the walls of the tube (or red precipitate with Fehling's solution). Observation with propan-2-ol: No visible reaction / solution remains colourless (or solution remains blue with Fehling's solution).
Marking scheme
(a)(i) [3.7 marks] - 1 mark for specifying acidified potassium dichromate(VI) (accept \(\text{H}^+/\text{Cr}_2\text{O}_7^{2-}\)). - 1 mark for stating distillation/distil immediately. - 1.7 marks for explaining that propanal has a lower boiling point (due to lack of hydrogen bonding) and is removed before further oxidation occurs.
(a)(ii) [1 mark] - 1 mark for correct balanced equation.
(b) [3 marks] - 1 mark for choosing a suitable test (Tollens' reagent or Fehling's solution; reject simple acidified potassium dichromate as both would react/oxidize eventually). - 1 mark for correct positive observation with propanal (silver mirror / red precipitate). - 1 mark for correct observation with propan-2-ol (no reaction/no change).
Question 7 · Structured Questions
7.7 marks
An unknown organic compound \(Y\) contains carbon, hydrogen, and oxygen only. It was analyzed using mass spectrometry, infrared (IR) spectroscopy, and chemical tests.
- Mass spectrometry of \(Y\) shows a molecular ion peak at \(m/z = 74\). - The infrared spectrum of \(Y\) shows a broad, strong absorption band at \(3330\text{ cm}^{-1}\) and no absorption peak in the region \(1680 - 1750\text{ cm}^{-1}\). - When \(Y\) is warmed with acidified potassium dichromate(VI), the orange solution turns green, but \(Y\) cannot be oxidized to form a carboxylic acid.
(a) Use the information from the infrared spectrum to identify the functional group present in \(Y\) and rule out another common oxygen-containing functional group. [2 marks]
(b) Deduce the molecular formula of \(Y\), showing your working. [2 marks]
(c) Deduce the structure and the IUPAC name of \(Y\). Explain your reasoning using all the experimental evidence provided. [3.7 marks]
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Worked solution
(a) The broad, strong absorption band at \(3330\text{ cm}^{-1}\) indicates the presence of an alcohol (\(\text{O}-\text{H}\) stretch) group. The absence of any peak in the region \(1680-1750\text{ cm}^{-1}\) rules out the presence of a carbonyl group (\(\text{C}=\text{O}\) stretch), which means \(Y\) is not an aldehyde, ketone, or carboxylic acid.
(b) Since \(Y\) is an alcohol and contains no other functional groups, its general molecular formula is \(\text{C}_n\text{H}_{2n+2}\text{O}\). The molecular ion peak at \(m/z = 74\) indicates \(M_r = 74\). \(12n + (2n + 2) + 16.0 = 74\) \(14n + 18 = 74 \implies 14n = 56 \implies n = 4\). Therefore, the molecular formula of \(Y\) is \(\text{C}_4\text{H}_{10}\text{O}\).
(c) Since \(Y\) is oxidized by acidified potassium dichromate(VI) (turning orange to green), it must be a primary or secondary alcohol (as tertiary alcohols do not undergo oxidation). However, because \(Y\) cannot be oxidized to a carboxylic acid, it cannot be a primary alcohol. Therefore, \(Y\) must be a secondary alcohol. The only secondary alcohol with molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) is butan-2-ol. Structure: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\). IUPAC name: Butan-2-ol.
Marking scheme
(a) [2 marks] - 1 mark for identifying alcohol (\(\text{O}-\text{H}\) ) group from the \(3330\text{ cm}^{-1}\) band. - 1 mark for ruling out carbonyl (\(\text{C}=\text{O}\) ) group due to lack of absorption in the \(1680-1750\text{ cm}^{-1}\) region.
(b) [2 marks] - 1 mark for using the general formula of saturated monohydric alcohols (\(\text{C}_n\text{H}_{2n+2}\text{O}\) ) or setting up the algebraic equation. - 1 mark for correct deduction of the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
(c) [3.7 marks] - 1 mark for concluding that \(Y\) must be a secondary alcohol because it is oxidized but not to a carboxylic acid. - 1 mark for identifying butan-2-ol as the unique secondary alcohol. - 1 mark for correct structural drawing. - 0.7 marks for correct IUPAC name.
Question 8 · Structured Questions
7.7 marks
Methane reacts with chlorine in the presence of ultraviolet (UV) light to form chloromethane and hydrogen chloride.
(a) Explain the purpose of the ultraviolet light in this reaction. [1 mark]
(b) Write equations for the initiation, propagation, and termination steps involved in this reaction to form chloromethane. [4 marks]
(c) Apart from chloromethane, other halogenated organic compounds are formed as by-products in this reaction. (i) Write a propagation step that leads to the formation of dichloromethane. [1 mark] (ii) Explain how the experimental conditions can be modified to minimize the formation of these by-products and maximize the yield of chloromethane. [1.7 marks]
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Worked solution
(a) UV light provides the energy required to break the covalent chlorine-chlorine bond homolytically to form highly reactive chlorine free radicals (\(\text{Cl}^\bullet\)).
Termination (any one of the following): \({}^\bullet\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\) (or \(2{}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\) or \(2\text{Cl}^\bullet \rightarrow \text{Cl}_2\)).
(c)(ii) To maximize the yield of chloromethane and minimize further substitution products, a large excess of methane should be used. This significantly increases the probability of a chlorine radical colliding with a methane molecule rather than a chloromethane molecule.
Marking scheme
(a) [1 mark] - 1 mark for explaining that UV light provides energy for homolytic fission of the \(\text{Cl}-\text{Cl}\) bond.
(b) [4 marks] - 1 mark for correct initiation equation. - 1 mark for first propagation equation. - 1 mark for second propagation equation. - 1 mark for any valid termination equation. Note: Radical dots must be shown clearly on the correct atoms.
(c)(i) [1 mark] - 1 mark for either of the correct propagation steps leading to dichloromethane.
(c)(ii) [1.7 marks] - 0.7 marks for stating \"excess methane\" is used. - 1 mark for explaining that this increases the likelihood of \(\text{Cl}^\bullet\)_ colliding with \(\text{CH}_4\)_ rather than \(\text{CH}_3\text{Cl}\".
Question 9 · structured
8 marks
A student carried out a laboratory experiment to determine the enthalpy of combustion of propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) using a spirit burner and a copper calorimeter.
The student's recorded data are shown below:
Mass of water in calorimeter = \(150.0\text{ g}\)
Initial temperature of water = \(19.5\text{ }^\circ\text{C}\)
Final temperature of water = \(38.1\text{ }^\circ\text{C}\)
Initial mass of spirit burner + propan-1-ol = \(122.45\text{ g}\)
Final mass of spirit burner + propan-1-ol = \(121.93\text{ g}\)
Specific heat capacity of water, \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)
(a) Calculate the heat energy, \(q\), released in kJ during this combustion experiment. (2 marks)
(b) Calculate the enthalpy change of combustion of propan-1-ol, \(\Delta H_c\), in \(\text{kJ mol}^{-1}\) from this experiment. Give your answer to 3 significant figures and include a sign. (3 marks)
(c) State two reasons, other than heat loss to the surroundings, why the experimental value is significantly less exothermic than the accepted literature value. (2 marks)
(d) Suggest one modification to the experimental apparatus (excluding changing the calorimeter type or replacing the burner) that would decrease the heat lost to the surroundings. (1 mark)
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Worked solution
(a) First, calculate the temperature change: \(\Delta T = 38.1 - 19.5 = 18.6\text{ K}\) (or \(^\circ\text{C}\))
Now use the heat energy equation: \(q = m c \Delta T\) \(q = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 18.6\text{ K} = 11662.2\text{ J}\)
Convert Joules to kilojoules: \(q = \frac{11662.2}{1000} = 11.6622\text{ kJ}\) (rounds to \(11.7\text{ kJ}\))
(b) First, calculate the mass of propan-1-ol burned: \(\text{mass burned} = 122.45 - 121.93 = 0.52\text{ g}\)
Calculate the molar mass (\(M_r\)) of propan-1-ol (\(\text{C}_3\text{H}_8\text{O}\)): \(M_r = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\)
Calculate the moles of propan-1-ol burned: \(n = \frac{0.52}{60.0} = 0.008667\text{ mol}\)
Calculate the enthalpy change of combustion (\(\Delta H_c\)): \(\Delta H_c = -\frac{q}{n} = -\frac{11.6622\text{ kJ}}{0.008667\text{ mol}} = -1345.6\text{ kJ mol}^{-1}\)
Rounding to 3 significant figures gives \(-1350\text{ kJ mol}^{-1}\) (or \(-1340\text{ kJ mol}^{-1}\) if intermediate rounding of moles to \(0.0087\) was used).
(c) Other reasons for the lower experimental value include: 1. Incomplete combustion of the fuel (forming carbon monoxide or carbon soot instead of carbon dioxide, which releases less energy). 2. Loss of fuel by evaporation from the wick between weighing and lighting/extinguishing. 3. The heat capacity of the copper calorimeter itself and the thermometer was not accounted for (some heat is absorbed by the copper container rather than the water).
(d) To minimize heat loss to the surrounding air, the student could: - Add a lid to the copper calorimeter. - Place draught shields (or wind shields) around the apparatus.
Marking scheme
(a) [2 marks] • M1: Correct calculation of \(\Delta T = 18.6\text{ K}\) and substitution into \(q = mc\Delta T\) \(q = 150.0 \times 4.18 \times 18.6\) [1 mark] • M2: Correct calculation of energy in kJ \(11.7\text{ kJ}\) (or \(11.66\text{ kJ}\)) [1 mark] Note: Allow recovery of M2 if final value is correct in kJ. Award 1 mark maximum for 11662 J (correct value but in J).
(b) [3 marks] • M1: Correct calculation of moles of propan-1-ol: \(\text{moles} = \frac{0.52}{60.0} = 0.00867\text{ mol}\) (allow \(0.0087\text{ mol}\)) [1 mark] • M2: Expression or calculation of \(\frac{\text{energy}}{\text{moles}}\): \(\frac{11.6622}{0.008667}\) [1 mark] • M3: Final value to 3 significant figures with negative sign: \(-1350\text{ kJ mol}^{-1}\) (allow range \(-1340\) to \(-1350\text{ kJ mol}^{-1}\) depending on intermediate rounding) [1 mark] Note: Consequential marking (CE) allowed from part (a). If positive sign or no sign is given, penalise M3.
(c) [2 marks] Any two from: • Incomplete combustion / formation of soot or CO [1 mark] • Evaporation of alcohol from the wick [1 mark] • Heat absorbed by the copper calorimeter / thermometer (not accounted for in the calculation) [1 mark] Do not accept: "Heat loss to surroundings" (excluded by question).
(d) [1 mark] • Add a lid (to the calorimeter) OR use a wind shield / draught screens [1 mark] Do not accept: "Use a bomb calorimeter" (as this replaces the setup).
Section Unit 3: Inorganic 2 and Physical 2
Answer all questions in the spaces provided. Show all working for calculations.
10 Question · 80 marks
Question 1 · Structured
8 marks
(a) Define the term lattice enthalpy of dissociation. (2)
(b) Use the following data to calculate the lattice enthalpy of dissociation for magnesium chloride (\(\text{MgCl}_2\)). - Enthalpy of formation of \(\text{MgCl}_2(\text{s})\) = \(-642 \text{ kJ mol}^{-1}\) - Enthalpy of atomisation of \(\text{Mg}(\text{s})\) = \(+148 \text{ kJ mol}^{-1}\) - First ionisation energy of \(\text{Mg}(\text{g})\) = \(+738 \text{ kJ mol}^{-1}\) - Second ionisation energy of \(\text{Mg}(\text{g})\) = \(+1451 \text{ kJ mol}^{-1}\) - Bond dissociation enthalpy of \(\text{Cl}_2(\text{g})\) = \(+242 \text{ kJ mol}^{-1}\) - First electron affinity of \(\text{Cl}(\text{g})\) = \(-349 \text{ kJ mol}^{-1}\) (4)
(c) Explain why the experimental lattice dissociation enthalpy of magnesium iodide (\(\text{MgI}_2\)) is significantly larger than the value calculated using a perfect ionic model. (2)
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Worked solution
Part (a): Lattice enthalpy of dissociation is the enthalpy change when one mole of an ionic compound is completely separated into its constituent gaseous ions under standard conditions.
Part (c): Magnesium iodide exhibits significant covalent character. The small, highly charged \(\text{Mg}^{2+}\) cation polarises the large, easily polarisable iodide anion (\(\text{I}^-\)), causing sharing of electron density. The perfect ionic model assumes spheres with only electrostatic attraction, but the additional covalent character strengthens the bonding, leading to a larger experimental lattice dissociation enthalpy than that predicted by the model.
Marking scheme
Part (a): M1: Enthalpy change when 1 mole of an ionic substance/crystal is separated into its constituent gaseous ions (1) M2: Under standard conditions (1)
Part (c): M1: Mention that the iodide ion is polarised by the magnesium ion, introducing covalent character (1) M2: Covalent character makes the bonding stronger than predicted by the pure electrostatic/ionic model (1)
Question 2 · Structured
8 marks
The reaction between peroxodisulfate(VI) ions and iodide ions is: \(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq})\). The following initial rates data were obtained at constant temperature: - Run 1: \([\text{S}_2\text{O}_8^{2-}] = 0.0400 \text{ mol dm}^{-3}\), \([\text{I}^-] = 0.0800 \text{ mol dm}^{-3}\), Initial Rate = \(2.20 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}\) - Run 2: \([\text{S}_2\text{O}_8^{2-}] = 0.0800 \text{ mol dm}^{-3}\), \([\text{I}^-] = 0.0800 \text{ mol dm}^{-3}\), Initial Rate = \(4.40 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}\) - Run 3: \([\text{S}_2\text{O}_8^{2-}] = 0.0800 \text{ mol dm}^{-3}\), \([\text{I}^-] = 0.1600 \text{ mol dm}^{-3}\), Initial Rate = \(8.80 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}\)
(a) Deduce the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\). Explain your reasoning. (3)
(b) Write the rate equation for this reaction. (1)
(c) Calculate the rate constant, \(k\), for this reaction and state its units. Give your answer to 3 significant figures. (3)
(d) Suggest why this reaction has a high activation energy even though it is thermodynamically feasible. (1)
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Worked solution
Part (a): Comparing Run 1 and Run 2, the concentration of \(\text{S}_2\text{O}_8^{2-}\) is doubled while the concentration of \(\text{I}^-\right.\) is kept constant. The rate doubles (from \(2.20 \times 10^{-4}\) to \(4.40 \times 10^{-4}\)), so the order with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1. Comparing Run 2 and Run 3, the concentration of \(\text{I}^-\right.\) is doubled while the concentration of \(\text{S}_2\text{O}_8^{2-}\) is kept constant. The rate doubles (from \(4.40 \times 10^{-4}\) to \(8.80 \times 10^{-4}\)), so the order with respect to \(\text{I}^-\right.\) is 1.
Part (b): \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).
Part (c): Using Run 1 data: \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{2.20 \times 10^{-4}}{0.0400 \times 0.0800} = 0.06875 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\). To 3 significant figures, \(k = 0.0688 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\).
Part (d): The reacting species are both negatively charged ions (anions), which repel each other, requiring significant kinetic energy to overcome this electrostatic repulsion.
Marking scheme
Part (a): M1: Order wrt \(\text{S}_2\text{O}_8^{2-}\) is 1 because doubling concentration doubles rate (from Run 1 to 2) (1) M2: Order wrt \(\text{I}^-\right.\) is 1 because doubling concentration doubles rate (from Run 2 to 3) (1) M3: Clear logic of keeping other variable constant in comparisons (1)
Part (b): M1: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (1)
Part (c): M1: Correct rearrangement: \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}\) (1) M2: Value of 0.0688 (accept 0.06875) (1) M3: Units of \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\) (1)
Part (d): M1: Repulsion between similarly/negatively charged ions (1)
Question 3 · Structured
8 marks
An electrochemical cell is set up using the following two standard half-cells: 1. \(\text{Fe}^{3+}(\text{aq}) + e^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\) \(E^\theta = +0.77 \text{ V}\) 2. \(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6e^- \rightleftharpoons 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\) \(E^\theta = +1.33 \text{ V}\)
(a) Write the conventional representation for this electrochemical cell under standard conditions. (3)
(b) Calculate the standard electromotive force (EMF) of this cell. (1)
(c) Write the overall ionic equation for the spontaneous reaction that occurs when the cell discharges. (2)
(d) Explain the effect, if any, on the EMF of this cell if the concentration of \(\text{H}^+(\text{aq})\) in the chromium half-cell is decreased. (2)
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Worked solution
Part (a): The iron half-cell has the less positive standard electrode potential, so it undergoes oxidation and forms the left-hand electrode (anode). The chromium half-cell undergoes reduction and forms the right-hand electrode (cathode). Both half-cells require inert platinum electrodes. The conventional representation is: \(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) \parallel \text{Cr}_2\text{O}_7^{2-}(\text{aq}), \text{H}^+(\text{aq}), \text{Cr}^{3+}(\text{aq}) \mid \text{Pt}(\text{s})\).
Part (d): If \([\text{H}^+(\text{aq})]\) is decreased, the position of the equilibrium \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightleftharpoons 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) shifts to the left to oppose the change. This decreases the electrode potential of the chromium half-cell, making it less positive. Therefore, the cell EMF decreases.
Marking scheme
Part (a): M1: Correctly place iron on the left, chromium on the right (1) M2: Include Pt(s) on both ends with phase boundaries (1) M3: Correct representation of species: \(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) \parallel \text{Cr}_2\text{O}_7^{2-}(\text{aq}), \text{H}^+(\text{aq}), \text{Cr}^{3+}(\text{aq}) \mid \text{Pt}(\text{s})\) (commas between ions in the same phase, vertical lines for phase changes) (1)
Part (b): M1: \(+0.56 \text{ V}\) (1)
Part (c): M1: Correct species on both sides of equation (1) M2: Balanced equation: \(6\text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 6\text{Fe}^{3+} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) (1)
Part (d): M1: Equilibrium shifts to the left / less reduction occurs in the chromium half-cell (1) M2: Electrode potential of the chromium half-cell decreases, so EMF decreases (1)
Question 4 · Structured
8 marks
Propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) has an acid dissociation constant, \(K_a\), of \(1.35 \times 10^{-5} \text{ mol dm}^{-3}\) at \(298 \text{ K}\).
(a) Calculate the pH of a \(0.150 \text{ mol dm}^{-3}\) solution of propanoic acid at this temperature. Show your working. (3)
(b) A buffer solution is prepared by mixing \(50.0 \text{ cm}^3\) of this \(0.150 \text{ mol dm}^{-3}\) propanoic acid with \(30.0 \text{ cm}^3\) of \(0.200 \text{ mol dm}^{-3}\) sodium propanoate solution. Calculate the pH of this buffer solution. (3)
(c) Explain, using an ionic equation, how this buffer solution resists a change in pH when a small amount of hydrochloric acid is added. (2)
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Part (b): Moles of propanoic acid (\(\text{HA}\)) = \(0.0500 \times 0.150 = 7.50 \times 10^{-3} \text{ mol}\). Moles of propanoate ions (\(\text{A}^-\)) = \(0.0300 \times 0.200 = 6.00 \times 10^{-3} \text{ mol}\). Since both are in the same total volume, we can use the ratio of moles in the \(K_a\) expression: \([\text{H}^+] = K_a \times \frac{\text{moles of HA}}{\text{moles of A}^-} = 1.35 \times 10^{-5} \times \frac{7.50 \times 10^{-3}}{6.00 \times 10^{-3}} = 1.6875 \times 10^{-5} \text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(1.6875 \times 10^{-5}) = 4.77\).
Part (c): When \(\text{H}^+\) from HCl is added, it reacts with the propanoate conjugate base to form propanoic acid: \(\text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(\text{aq})\). Since the added \(\text{H}^+\) ions are removed, the pH remains almost constant.
Part (b): M1: Calculate moles of HA (\(7.50 \times 10^{-3}\)) AND moles of \(\text{A}^-\right.\) (\(6.00 \times 10^{-3}\)) (1) M2: Correctly calculate \([\text{H}^+] = 1.688 \times 10^{-5} \text{ mol dm}^{-3}\) (1) M3: pH = 4.77 (1)
Part (c): M1: Ionic equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\) (1) M2: Explain that the conjugate base removes the added \(\text{H}^+\) ions (1)
Question 5 · Structured
8 marks
This question is about Period 3 oxides.
(a) Describe the structure and bonding in silicon dioxide (\(\text{SiO}_2\)) and sulfur trioxide (\(\text{SO}_3\)), and explain how this accounts for the difference in their melting points. (4)
(b) Write balanced chemical equations for the reactions of: - (i) Phosphorus(V) oxide with excess water. (1) - (ii) Aluminum oxide with hot aqueous sodium hydroxide. (2)
(c) State the approximate pH of the solution formed when sulfur dioxide gas dissolves in water. (1)
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Worked solution
Part (a): Silicon dioxide (\(\text{SiO}_2\)) is a giant covalent/macromolecular structure with strong, covalent bonds extending throughout a giant 3D lattice. Sulfur trioxide (\(\text{SO}_3\)) is a simple molecular structure with weak van der Waals forces between the individual molecules. Melting silicon dioxide requires breaking many strong covalent bonds, which requires a large amount of energy, whereas melting sulfur trioxide only requires overcoming the weak intermolecular van der Waals forces, which requires much less energy. Consequently, \(\text{SiO}_2\) has a much higher melting point than \(\text{SO}_3\).
Part (b)(i): \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\).
Part (c): The pH is approximately 2 (accept range 1-3).
Marking scheme
Part (a): M1: \(\text{SiO}_2\) has giant covalent structure with strong covalent bonds, \(\text{SO}_3\) has simple molecular structure with weak intermolecular/van der Waals forces (1) M2: Breaking strong covalent bonds in \(\text{SiO}_2\) requires high energy (1) M3: Overcoming weak van der Waals forces in \(\text{SO}_3\) requires low energy (1) M4: Conclusion linking energy required to high vs low melting point (1)
(a) Aqueous cobalt(II) chloride contains the hexaaquacobalt(II) ion, \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\). State the color and shape of this complex ion. (2)
(b) When excess concentrated hydrochloric acid is added to aqueous cobalt(II) chloride, a reaction occurs. - (i) Write a balanced equation for this ligand substitution reaction. (2) - (ii) State the color and shape of the new complex ion formed, and explain why the coordination number changes. (3)
(c) State the type of isomerism shown by the complex ion \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\). (1)
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Worked solution
Part (a): The hexaaquacobalt(II) ion is pink in color and has an octahedral shape.
Part (b)(i): \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\).
Part (b)(ii): The new complex ion \([\text{CoCl}_4]^{2-}\) is blue in color and has a tetrahedral shape. The coordination number changes from 6 to 4 because chloride ligands are larger than water ligands (and carry a negative charge, causing greater electrostatic repulsion), so fewer chloride ligands can fit around the central cobalt(II) ion.
Part (c): It shows stereoisomerism (specifically cis-trans / geometric isomerism).
Marking scheme
Part (a): M1: Pink (accept pink-red) (1) M2: Octahedral (1)
Part (b): M1: Correct formulas: \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) and \([\text{CoCl}_4]^{2-}\) (1) M2: Correct balancing: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\) (1) M3: Blue and Tetrahedral (both correct for 1 mark) (1) M4: Chloride ligands are larger / have negative charge (1) M5: Fewer ligands can fit around the metal ion / steric hindrance / repulsion (1)
Part (c): M1: cis-trans / geometric / stereoisomerism (1)
Question 7 · Structured
8 marks
Describe the chemical tests and observations used to distinguish between solutions containing transition metal and metalloid ions.
(a) Explain the observations made when aqueous sodium hydroxide is added dropwise and then in excess to separate solutions of iron(III) chloride and aluminium chloride. Include an ionic equation for any reaction where the precipitate dissolves. (4)
(b) Explain the difference in observations when aqueous sodium carbonate is added to separate solutions of iron(II) sulfate and iron(III) chloride. (4)
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Worked solution
Part (a): With iron(III) chloride, dropwise addition of NaOH forms a red-brown/brown precipitate of \(\text{Fe(OH)}_3(\text{H}_2\text{O})_3\) which is insoluble in excess NaOH. With aluminium chloride, dropwise addition forms a white precipitate of \(\text{Al(OH)}_3(\text{H}_2\text{O})_3\) which dissolves in excess NaOH to form a colorless solution containing the tetrahydroxoaluminate ion, \([\text{Al(OH)}_4]^-\). The ionic equation is: \(\text{Al(OH)}_3(\text{H}_2\text{O})_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Al(OH)}_4]^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\).
Part (b): With \(\text{Fe}^{2+}(\text{aq})\), sodium carbonate reacts directly to form a green precipitate of iron(II) carbonate, \(\text{FeCO}_3\), with no gas produced. With \(\text{Fe}^{3+}(\text{aq})\), the high charge-to-size ratio of the \(\text{Fe}^{3+}\) ion polarises the coordinated water ligands, making the hexaaqua ion strongly acidic. The carbonate ions act as a base, reacting with \(\text{H}^+\) to produce carbon dioxide gas (bubbles/effervescence) and a brown precipitate of iron(III) hydroxide rather than iron(III) carbonate.
Marking scheme
Part (a): M1: Iron(III) forms brown precipitate, insoluble in excess (1) M2: Aluminium forms white precipitate, dissolving in excess to form colorless solution (1) M3: Balanced ionic equation: \(\text{Al(OH)}_3(\text{H}_2\text{O})_3 + \text{OH}^- \rightarrow [\text{Al(OH)}_4]^- + 3\text{H}_2\text{O}\) (or equivalent) (2)
Part (b): M1: \(\text{Fe}^{2+}\) forms a green precipitate (\(\text{FeCO}_3\)) and no gas (1) M2: \(\text{Fe}^{3+}\) forms a brown precipitate and bubbles/effervescence of \(\text{CO}_2\) (1) M3: \(\text{Fe}^{3+}\) is more highly charged and polarizing than \(\text{Fe}^{2+}\) / makes hexaaqua ion more acidic (1) M4: Carbonate reacts as a base to deprotonate the iron(III) complex to form the hydroxide (1)
Question 8 · Structured
8 marks
Methanol is synthesised industrially by the homogeneous gas-phase reaction: \(\text{CO}(\text{g}) + 2\text{H}_2(\text{g}) \rightleftharpoons \text{CH}_3\text{OH}(\text{g})\)
A mixture of \(1.50 \text{ mol}\) of \(\text{CO}\) and \(3.00 \text{ mol}\) of \(\text{H}_2\) was sealed in a container and allowed to reach equilibrium at temperature \(T\) under a constant total pressure of \(24.0 \text{ MPa}\). At equilibrium, the mixture contained \(0.60 \text{ mol}\) of \(\text{CH}_3\text{OH}\).
(a) Calculate the equilibrium amounts, in moles, of \(\text{CO}\) and \(\text{H}_2\) present in the mixture. (2)
(b) Calculate the partial pressure of each gas at equilibrium. (3)
(c) Write the expression for the equilibrium constant, \(K_p\), and calculate its value at temperature \(T\), including its units. (3)
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Worked solution
Part (a): Moles of \(\text{CO}\) at equilibrium = \(1.50 - 0.60 = 0.90 \text{ mol}\). Moles of \(\text{H}_2\) at equilibrium = \(3.00 - 2(0.60) = 1.80 \text{ mol}\).
Chromium is extracted from chromium(III) oxide by reduction with carbon according to the following equation: \(\text{Cr}_2\text{O}_3(\text{s}) + 3\text{C}(\text{s}) \rightarrow 2\text{Cr}(\text{s}) + 3\text{CO}(\text{g})\)
(a) Explain why the standard entropy change (\(\Delta S^\circ\)) for this reaction is positive. [2 marks]
(b) Calculate the standard enthalpy change (\(\Delta H^\circ\)), in \(\text{kJ mol}^{-1}\), for this reaction using the standard enthalpies of formation provided: - \(\Delta_f H^\circ[\text{Cr}_2\text{O}_3(\text{s})] = -1140\text{ kJ mol}^{-1}\) - \(\Delta_f H^\circ[\text{CO}(\text{g})] = -111\text{ kJ mol}^{-1}\) [2 marks]
(c) Calculate the standard entropy change (\(\Delta S^\circ\)), in \(\text{J K}^{-1}\text{ mol}^{-1}\), for this reaction using the following standard entropies: - \(S^\circ[\text{Cr}_2\text{O}_3(\text{s})] = 81.2\text{ J K}^{-1}\text{ mol}^{-1}\) - \(S^\circ[\text{C}(\text{s})] = 5.7\text{ J K}^{-1}\text{ mol}^{-1}\) - \(S^\circ[\text{Cr}(\text{s})] = 23.8\text{ J K}^{-1}\text{ mol}^{-1}\) - \(S^\circ[\text{CO}(\text{g})] = 197.6\text{ J K}^{-1}\text{ mol}^{-1}\) [2 marks]
(d) Calculate the minimum temperature, in Kelvin, at which this reaction becomes feasible. [2 marks]
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Worked solution
(a) The reactants are both solids, which are highly ordered (low entropy). The reaction produces 3 moles of gas (carbon monoxide) per mole of chromium(III) oxide. Gases have a high degree of disorder and much larger standard entropies than solids, leading to a large positive entropy change.
(d) For a reaction to be feasible, \(\Delta G \le 0\). At the transition point of feasibility, \(\Delta G = 0\), so: \(T = \frac{\Delta H^\circ}{\Delta S^\circ}\) Convert \(\Delta H^\circ\) to \(\text{J mol}^{-1}\): \(807 \times 1000 = 807,000\text{ J mol}^{-1}\) \(T = \frac{807,000}{542.1} = 1488.65\text{ K}\) Rounding to 3 or 4 significant figures yields \(1489\text{ K}\) or \(1490\text{ K}\).
Marking scheme
(a) - 1 mark: Stating that 3 moles of gas are produced from solid reactants. - 1 mark: Stating that gases are much more disordered / have higher entropy than solids.
(b) - 1 mark: Correct expression or working, e.g., \(3 \times (-111) - (-1140)\). - 1 mark: Correct value of \(+807\text{ kJ mol}^{-1}\) (sign must be correct).
(d) - 1 mark: Rearranging \(\Delta G = \Delta H - T\Delta S\) to make \(T\) the subject, and converting \(\Delta H\) to J (or \(\Delta S\) to kJ), i.e., \(T = \frac{807,000}{542.1}\). - 1 mark: Correct final temperature of \(1489\text{ K}\) (accept range \(1488\text{ K}\) to \(1490\text{ K}\) depending on rounding).
Question 10 · Structured
8 marks
When cobalt(II) chloride is dissolved in water, a pink solution containing the octahedral complex ion \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) is formed.
(a) Concentrated hydrochloric acid is added dropwise to this pink solution until it turns completely blue. Write an equation for this reaction and state both the change in coordination number and the change in shape of the cobalt complex. [3 marks]
(b) Explain why transition metal complex ions, such as those of cobalt, are coloured. [3 marks]
(c) The blue cobalt complex has a different colour from the pink cobalt complex. State two factors, other than coordination number and shape, that affect the colour of a transition metal complex. [2 marks]
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Worked solution
(a) The reaction represents a ligand substitution where water ligands are replaced by chloride ions: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\) - The coordination number of cobalt changes from 6 to 4. - The shape of the complex changes from octahedral (in \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\)) to tetrahedral (in \([\text{CoCl}_4]^{2-}\)).
(b) Transition metal ions have partially filled d-orbitals. When ligands coordinate to the metal ion, their presence causes the five d-orbitals to split into two groups of different energy levels (non-degenerate). Electrons in lower-energy d-orbitals absorb specific frequencies/wavelengths of visible light to be promoted to a higher-energy d-orbital (d-d transition). The remaining wavelengths of light are not absorbed; they are transmitted or reflected, which is perceived by the eye as the complementary colour.
(c) Other factors that affect the d-orbital splitting energy (and hence the colour) are: - The oxidation state of the transition metal ion (e.g., \(\text{Co}^{2+}\) vs. \(\text{Co}^{3+}\)). - The identity/nature of the ligands (e.g., \(\text{H}_2\text{O}\) vs. \(\text{NH}_3\)). - The identity of the metal ion itself.
Marking scheme
(a) - 1 mark: Correct balanced equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\) (charges and states must be balanced; accept \(\text{HCl}\) on reactants side with balancing \(\text{H}^+\)). - 1 mark: Coordination number changes from 6 to 4. - 1 mark: Shape changes from octahedral to tetrahedral.
(b) - 1 mark: d-orbitals are split in energy by ligands (or exist in different energy levels). - 1 mark: d-electrons absorb energy/light/photons of specific frequencies in the visible spectrum to be promoted (d-d transition). - 1 mark: Remaining light/frequencies are transmitted/reflected to show the complementary colour.
(c) - 1 mark: Oxidation state of the transition metal. - 1 mark: Identity / nature of the ligand. (Accept: Identity of the central transition metal ion)
Section Unit 4: Organic 2 and Physical 2
Answer all questions in the spaces provided. Show all working for calculations.
9 Question · 79.39999999999999 marks
Question 1 · Structured
8.8 marks
The reaction between peroxodisulfate(VI) ions and iodide ions is represented by the following equation: \(S_2O_8^{2-}(aq) + 2I^-(aq) \rightarrow 2SO_4^{2-}(aq) + I_2(aq)\)
A series of experiments was carried out to investigate the kinetics of this reaction at constant temperature. - In Experiment 1, with \([S_2O_8^{2-}] = 0.0100\ mol\ dm^{-3}\) and \([I^-] = 0.0150\ mol\ dm^{-3}\), the initial rate was \(2.20 \times 10^{-5}\ mol\ dm^{-3}\ s^{-1}\). - In Experiment 2, doubling the concentration of \(S_2O_8^{2-}\) while keeping \([I^-]\) constant doubled the initial rate. - In Experiment 3, doubling the concentration of \(I^-\post\) while keeping \([S_2O_8^{2-}]\) constant also doubled the initial rate.
(a) Deduce the order of reaction with respect to each reactant, giving a reason for your answer. (b) Write the rate equation for the reaction. (c) Calculate the rate constant, \(k\), and state its units.
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Worked solution
(a) The order with respect to \(S_2O_8^{2-}\) is 1 because doubling its concentration (while holding \([I^-]\) constant) doubles the rate of reaction. The order with respect to \(I^-\post\) is also 1 because doubling its concentration (while holding \([S_2O_8^{2-}]\) constant) doubles the rate of reaction. (b) Rate equation: \(Rate = k[S_2O_8^{2-}][I^-]\) (c) Using Experiment 1 data: \(k = \frac{Rate}{[S_2O_8^{2-}][I^-]} = \frac{2.20 \times 10^{-5}}{0.0100 \times 0.0150} = 0.147\ dm^3\ mol^{-1}\ s^{-1}\) (or \(0.15\))
Marking scheme
- Deduce order for S2O8^2- is 1 with explanation: 2 marks - Deduce order for I- is 1 with explanation: 2 marks - Correct rate equation: 1 mark - Calculation of k showing working (0.147 or 0.15): 2.8 marks - Correct units (dm3 mol-1 s-1): 1 mark
Question 2 · Structured
8.8 marks
The experimental Born-Haber cycle values for magnesium sulfide, \(MgS(s)\), are given below: - Enthalpy of formation of \(MgS(s) = -346\ kJ\ mol^{-1}\) - Enthalpy of atomisation of \(Mg(s) = +148\ kJ\ mol^{-1}\) - First ionisation energy of \(Mg(g) = +738\ kJ\ mol^{-1}\) - Second ionisation energy of \(Mg(g) = +1451\ kJ\ mol^{-1}\) - Enthalpy of atomisation of \(S(s) = +279\ kJ\ mol^{-1}\) - First electron affinity of \(S(g) = -200\ kJ\ mol^{-1}\) - Second electron affinity of \(S(g) = +640\ kJ\ mol^{-1}\)
(a) Define the term 'lattice enthalpy of dissociation'. (b) Calculate the lattice enthalpy of dissociation for \(MgS(s)\). (c) Explain why the experimental lattice enthalpy of dissociation for \(MgS(s)\) is greater than the theoretical value calculated from a purely ionic model.
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Worked solution
(a) Lattice enthalpy of dissociation is the enthalpy change when 1 mole of an ionic compound is completely separated into its constituent gaseous ions under standard conditions. (b) Applying Hess's Law: \(\Delta H_f^{\theta} = \Delta H_{at}^{\theta}(Mg) + 1st\ IE(Mg) + 2nd\ IE(Mg) + \Delta H_{at}^{\theta}(S) + 1st\ EA(S) + 2nd\ EA(S) - \Delta H_{L,diss}^{\theta}\) \(-346 = 148 + 738 + 1451 + 279 - 200 + 640 - \Delta H_{L,diss}^{\theta}\) \(-346 = 3056 - \Delta H_{L,diss}^{\theta}\) \(\Delta H_{L,diss}^{\theta} = 3056 + 346 = +3402\ kJ\ mol^{-1}\) (c) The theoretical model assumes purely electrostatic attraction between perfectly spherical ions (pure ionic model). In reality, \(MgS\) has significant covalent character because the small, highly charged \(Mg^{2+}\) cation polarises the large electron cloud of the \(S^{2-}\) anion, drawing the electron density between the nuclei and creating stronger bonds than predicted.
Marking scheme
- Definition of lattice dissociation enthalpy: 2 marks - Correct Hess's Law expression / cycle setup: 2 marks - Correct calculation of +3402 kJ mol-1: 2.8 marks - Explanation of covalent character / polarisation of sulfur anion by magnesium cation: 2 marks
Question 3 · Structured
8.8 marks
A buffer solution is prepared at \(298\ K\) by mixing \(50.0\ cm^3\) of \(0.250\ mol\ dm^{-3}\) propanoic acid (\(CH_3CH_2COOH\)) with \(30.0\ cm^3\) of \(0.150\ mol\ dm^{-3}\) sodium hydroxide (\(NaOH\)). The acid dissociation constant, \(K_a\), of propanoic acid is \(1.35 \times 10^{-5}\ mol\ dm^{-3}\) at \(298\ K\).
(a) Write an equation for the reaction that occurs when these two solutions are mixed. (b) Calculate the pH of the resulting buffer solution. Give your answer to 2 decimal places.
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Worked solution
(a) Equation: \(CH_3CH_2COOH(aq) + NaOH(aq) \rightarrow CH_3CH_2COONa(aq) + H_2O(l)\) (b) Calculations: Initial moles of \(CH_3CH_2COOH = 0.0500 \times 0.250 = 0.0125\ mol\) Initial moles of \(NaOH = 0.0300 \times 0.150 = 0.00450\ mol\) All \(NaOH\) reacts, neutralizing \(0.00450\ mol\) of the acid and forming \(0.00450\ mol\) of \(CH_3CH_2COO^-\). Moles of acid remaining = \(0.0125 - 0.00450 = 0.00800\ mol\) Moles of propanoate ions formed = \(0.00450\ mol\) Using the Henderson-Hasselbalch equation or \(K_a\) rearrangement: \([H^+] = K_a \times \frac{n(acid)}{n(conjugate\ base)} = 1.35 \times 10^{-5} \times \frac{0.00800}{0.00450} = 2.40 \times 10^{-5}\ mol\ dm^{-3}\) \(pH = -\log_{10}(2.40 \times 10^{-5}) = 4.62\)
Marking scheme
- Balanced chemical equation: 1 mark - Calculation of initial moles of acid and alkali: 2 marks - Calculation of remaining moles of acid and formed moles of salt: 2 marks - Correct expression/working for [H+]: 1.8 marks - pH calculation to 2 d.p. (4.62): 2 marks
Question 4 · Structured
8.8 marks
Use the following standard electrode potential data to answer the questions: - Half-equation 1: \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq)\) \(E^{\theta} = +0.77\ V\) - Half-equation 2: \(MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightleftharpoons Mn^{2+}(aq) + 4H_2O(l)\) \(E^{\theta} = +1.51\ V\) - Half-equation 3: \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\) \(E^{\theta} = +0.80\ V\)
(a) Write the conventional cell representation for a standard electrochemical cell constructed from the \(Fe^{3+}/Fe^{2+}\) half-cell and the \(Ag^+/Ag\) half-cell. (b) Calculate the EMF of this cell under standard conditions, and identify the direction of electron flow in the external circuit. (c) Deduce, with the aid of an overall ionic equation and a cell potential calculation, whether acidified potassium manganate(VII) can oxidize iron(II) ions to iron(III) ions under standard conditions.
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Worked solution
(a) Conventional representation: \(Pt(s) | Fe^{2+}(aq), Fe^{3+}(aq) || Ag^+(aq) | Ag(s)\) (b) \(E^{\theta}_{cell} = E^{\theta}_{right} - E^{\theta}_{left} = +0.80 - (+0.77) = +0.03\ V\). Electrons flow from the more negative half-cell to the more positive half-cell (from the Platinum electrode of the iron half-cell to the Silver electrode). (c) Manganate(VII) has a more positive potential (\(+1.51\ V\)) than iron(III)/iron(II) (\(+0.77\ V\)). Overall potential: \(E^{\theta}_{cell} = +1.51 - (+0.77) = +0.74\ V\). Since \(E^{\theta}_{cell} > 0\), the reaction is feasible. Overall equation: \(MnO_4^-(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\)
Marking scheme
- Cell representation (correct phase boundaries, salt bridge, state symbols, Pt electrode): 2 marks - EMF calculation (+0.03 V): 1 mark - Direction of electron flow (from Pt to Ag): 1 mark - Overall cell potential (+0.74 V) and feasibility deduction: 1.8 marks - Overall balanced equation: 3 marks
Question 5 · Structured
8.8 marks
Aqueous copper(II) ions exist as the pale blue hexaaquacopper(II) complex, \([Cu(H_2O)_6]^{2+}\).
(a) When concentrated hydrochloric acid is added to aqueous copper(II) sulfate, a yellow-green solution containing the complex ion \([CuCl_4]^{2-}\) is formed. (i) Write an equation for this reaction. (ii) State the coordination number and shape of both the reactant and product complex ions. (b) Explain, in terms of entropy change and the chelate effect, why substituting the water ligands in \([Cu(H_2O)_6]^{2+}\) with the bidentate ligand ethane-1,2-diamine (abbreviated as 'en') is a highly thermodynamically favorable process: \([Cu(H_2O)_6]^{2+}(aq) + 3en(aq) \rightleftharpoons [Cu(en)_3]^{2+}(aq) + 6H_2O(l)\)
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Worked solution
(a) (i) \([Cu(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CuCl_4]^{2-}(aq) + 6H_2O(l)\) (ii) Reactant complex: Coordination number = 6, Shape = Octahedral. Product complex: Coordination number = 4, Shape = Tetrahedral. (b) In the forward reaction, 4 reactant particles (one complex ion and three bidentate ligands) react to form 7 product particles (one complex ion and six water molecules). This increase in the total number of species in solution creates a highly positive entropy change (\(\Delta S^{\theta} > 0\)). Since the enthalpy change is minimal (as similar metal-nitrogen and metal-oxygen bond strengths are broken/formed), the free energy change \(\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}\) becomes highly negative, driving the reaction far to the right.
Marking scheme
- Equation with correct formulas and balancing: 2 marks - Correct coordination numbers (6 for reactant, 4 for product): 1 mark - Correct shapes (octahedral and tetrahedral): 1.8 marks - Reference to the increase in number of particles (from 4 to 7): 2 marks - Explanation linking entropy change (positive) and free energy change (negative): 2 marks
Question 6 · Structured
8.8 marks
Synthetic polymers have diverse applications depending on their physical and chemical stability.
(a) Terylene (PET) is a common polyester synthesized from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. (i) Draw the repeating unit of Terylene, showing all bonds in the ester linkage. (ii) State the type of polymerisation involved and name the small molecule eliminated. (b) Poly(lactic acid) (PLA) is a biodegradable polymer made from 2-hydroxypropanoic acid. (i) Draw the structural formula of 2-hydroxypropanoic acid. (ii) Explain why PLA is biodegradable whereas poly(ethene) is not, with reference to the chemical bonds present in their backbones.
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Worked solution
(a) (i) Repeating unit of Terylene: \(-O-CH_2-CH_2-O-CO-C_6H_4-CO-\) (ii) Condensation polymerisation; water (\(H_2O\)) is eliminated. (b) (i) Structural formula of 2-hydroxypropanoic acid: \(CH_3CH(OH)COOH\) (ii) PLA contains polar ester linkages (\(C-O\) carbonyl ester bonds) in its polymer backbone, which can be attacked by nucleophiles like water in the environment, leading to hydrolysis. Poly(ethene) has only non-polar carbon-carbon (\(C-C\)) single bonds in its backbone, which are highly stable and lack polar sites for nucleophilic attack, making it resistant to biodegradation.
Marking scheme
- Correct repeating unit of Terylene: 2 marks - Correct polymerisation type (condensation) and molecule (water): 1.8 marks - Correct structural formula of 2-hydroxypropanoic acid: 2 marks - Explanation of polar ester linkages in PLA and hydrolysis: 2 marks - Explanation of non-polar C-C bonds in poly(ethene): 1 mark
Question 7 · Structured
8.8 marks
Phenylamine (\(C_6H_5NH_2\)) is synthesized from benzene in a two-stage process. Stage 1: Benzene is converted into nitrobenzene. Stage 2: Nitrobenzene is reduced to phenylamine.
(a) Identify the reagents and conditions required for the nitration of benzene in Stage 1, and write an equation showing the formation of the electrophile. (b) State the reagents used to reduce nitrobenzene to phenylamine in Stage 2. (c) Explain why phenylamine is a weaker base than ethylamine (\(CH_3CH_2NH_2\)).
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Worked solution
(a) Reagents: Concentrated nitric acid (\(HNO_3\)) and concentrated sulfuric acid (\(H_2SO_4\) catalyst). Conditions: Warm at \(50\ ^\circ C\) to \(55\ ^\circ C\). Electrophile formation: \(HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-\) (b) Tin (\(Sn\)) and concentrated hydrochloric acid (\(HCl\)), followed by the addition of sodium hydroxide (\(NaOH\)) to liberate the free amine. (c) Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (\(H^+\)). In phenylamine, the lone pair on the nitrogen atom is partially delocalised into the \(\pi\)-electron system of the benzene ring, reducing its availability. In ethylamine, the alkyl group (ethyl) is electron-releasing due to the positive inductive effect, which increases the electron density on the nitrogen atom and makes the lone pair more available.
Marking scheme
- Reagents and conditions for Stage 1: 2 marks - Equation for electrophile formation: 1.8 marks - Reagents for Stage 2 (Sn + HCl then NaOH): 2 marks - Explanation of delocalisation in phenylamine: 1.5 marks - Explanation of positive inductive effect in ethylamine: 1.5 marks
Question 8 · Structured
8.8 marks
Butanal (\(CH_3CH_2CH_2CHO\)) reacts with acidified potassium cyanide (\(KCN/H^+\)) to form 2-hydroxypentanenitrile.
(a) Describe the nucleophilic addition mechanism for this reaction. Outline key features including relevant curly arrows, dipoles, and lone pairs. (b) The product, 2-hydroxypentanenitrile, contains a chiral carbon atom. Explain why the product mixture formed in this reaction is optically inactive.
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Worked solution
(a) Mechanism description: The cyanide ion (\(:CN^-\)) acts as a nucleophile. Its lone pair on the carbon attacks the carbon of the polar carbonyl group (\(C^{\delta+}=O^{\delta-}\)). The \(C=O\) double bond breaks, transferring a pair of electrons to the oxygen to form an intermediate with a negatively charged oxygen (\(O^-\)). The oxygen intermediate then uses its lone pair to capture a proton (\(H^+\)) to form the hydroxyl (\(-OH\)) group of 2-hydroxypentanenitrile. (b) The carbonyl carbon in the starting material, butanal, is planar. The cyanide nucleophile can attack this planar carbon atom with equal probability from either above or below the plane. This yields an equimolar (50:50) mixture of the two optical isomers (enantiomers), known as a racemic mixture. Since the two enantiomers rotate plane-polarised light in equal but opposite directions, their effects cancel out, resulting in no overall optical activity.
Marking scheme
- Description of nucleophilic attack (lone pair on C of CN- to carbonyl C, dipole on C=O): 2 marks - Description of intermediate and protonation: 1.8 marks - Identification of planar carbonyl group: 2 marks - Explanation of equal probability of attack from above or below: 2 marks - Explanation of equal amounts of enantiomers (racemate) and cancellation of optical rotation: 1 mark
Question 9 · Structured Questions
9 marks
Propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) is a weak acid that dissociates in water according to the following equation:
(a) A buffer solution is prepared by mixing \(150\text{ cm}^3\) of \(0.350\text{ mol dm}^{-3}\) propanoic acid with \(250\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\) sodium hydroxide solution. The acid dissociation constant, \(K_{\text{a}}\), of propanoic acid is \(1.35 \times 10^{-5}\text{ mol dm}^{-3}\) at \(298\text{ K}\). Calculate the pH of this buffer solution. Give your answer to 2 decimal places. [5 marks]
(b) Calculate the new pH of the buffer solution after the addition of \(0.010\text{ mol}\) of nitric acid (\(\text{HNO}_3\)). Assume that there is no change in volume. Give your answer to 2 decimal places. [4 marks]
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Worked solution
**Part (a)**
1. Calculate the initial amount, in moles, of propanoic acid and sodium hydroxide: \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.150\text{ dm}^3 \times 0.350\text{ mol dm}^{-3} = 0.0525\text{ mol}\) \(n(\text{NaOH}) = 0.250\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 0.0300\text{ mol}\)
2. Determine the remaining amount of propanoic acid and the amount of sodium propanoate formed after reaction: Propanoic acid reacts with hydroxide ions in a 1:1 molar ratio, and is in excess: \(n(\text{CH}_3\text{CH}_2\text{COOH})_{\text{remaining}} = 0.0525 - 0.0300 = 0.0225\text{ mol}\) \(n(\text{CH}_3\text{CH}_2\text{COO}^-)_{\text{formed}} = 0.0300\text{ mol}\)
3. Use the acid dissociation constant expression to find \([\text{H}^+]\): \(K_{\text{a}} = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)
Since the total volume is the same for both species, the volumes cancel out, allowing us to use moles directly: \([\text{H}^+] = K_{\text{a}} \times \frac{n(\text{CH}_3\text{CH}_2\text{COOH})}{n(\text{CH}_3\text{CH}_2\text{COO}^-)}\) \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0225}{0.0300} = 1.0125 \times 10^{-5}\text{ mol dm}^{-3}\)
1. Identify the reaction of the buffer with added \(\text{H}^+\) ions: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
2. Calculate the new moles of weak acid and conjugate base: \(n(\text{CH}_3\text{CH}_2\text{COOH})_{\text{new}} = 0.0225\text{ mol} + 0.010\text{ mol} = 0.0325\text{ mol}\) \(n(\text{CH}_3\text{CH}_2\text{COO}^-)_{\text{new}} = 0.0300\text{ mol} - 0.010\text{ mol} = 0.0200\text{ mol}\)
3. Calculate the new \([\text{H}^+]\): \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0325}{0.0200} = 2.19375 \times 10^{-5}\text{ mol dm}^{-3}\)
4. Calculate the new pH: \(\text{pH} = -\log_{10}(2.19375 \times 10^{-5}) = 4.6588 \approx 4.66\)
Marking scheme
**Part (a) [5 Marks]** * **M1**: Calculates initial moles of propanoic acid (0.0525 mol) AND NaOH (0.0300 mol). * **M2**: Correctly calculates remaining moles of propanoic acid as \(0.0525 - 0.0300 = 0.0225\text{ mol}\). * **M3**: Correctly states that moles of propanoate ions formed is 0.0300 mol. * **M4**: Uses the \(K_{\text{a}}\) expression correctly to find \([\text{H}^+]\) (e.g., \([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0225}{0.0300}\) OR \(1.0125 \times 10^{-5}\text{ mol dm}^{-3}\)). * **M5**: Gives the correct pH value to 2 decimal places (4.99). *(Accept ECF throughout for subsequent steps if an early arithmetic error occurs.)*
**Part (b) [4 Marks]** * **M1**: Recognises that added \(\text{H}^+\) reacts with propanoate to increase propanoic acid moles and decrease propanoate moles. * **M2**: Correctly calculates both new moles: propanoic acid = 0.0325 mol AND propanoate = 0.0200 mol. * **M3**: Calculates new \([\text{H}^+] = 2.19 \times 10^{-5}\text{ mol dm}^{-3}\). * **M4**: Calculates new pH to 2 decimal places (4.66). *(Accept ECF from student's values in part (a).)*
Unit 5: Section A (Practical)
Answer all questions in the spaces provided.
3 Question · 30 marks
Question 1 · Structured Practical Questions
10 marks
A student carries out a titration to determine the value of \(x\) in hydrated ammonium iron(II) sulfate, \((NH_4)_2Fe(SO_4)_2 \cdot xH_2O\).
**Part (a)** Describe how the student should prepare exactly \(250\text{ cm}^3\) of a standard solution of hydrated ammonium iron(II) sulfate starting from a known mass of the solid.
**Part (b)** Write the ionic equation for the reaction that occurs during the titration between \(Fe^{2+}(aq)\) and acidified \(MnO_4^-(aq)\) ions.
**Part (c)** The student prepares the standard solution by dissolving \(9.80\text{ g}\) of the hydrated salt to make \(250\text{ cm}^3\) of solution. A \(25.0\text{ cm}^3\) portion of this solution is titrated with \(0.0200\text{ mol dm}^{-3}\) potassium manganate(VII) solution. The average titre is \(25.00\text{ cm}^3\).
Use these data to calculate the value of \(x\) in \((NH_4)_2Fe(SO_4)_2 \cdot xH_2O\). Show your working.
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**Part (a)** 1. Weigh the mass of the solid in a weighing bottle or beaker on a balance. Dissolve the solid in a beaker using a volume of deionised water that is less than \(250\text{ cm}^3\). 2. Quantitatively transfer the solution to a \(250\text{ cm}^3\) volumetric flask. 3. Rinse the beaker, glass rod, and funnel with deionised water, adding all the washings into the volumetric flask. 4. Add deionised water to the volumetric flask until the bottom of the meniscus is exactly on the graduation line. Invert the flask several times to ensure a homogeneous solution.
**Part (b)** The balanced ionic equation is: \(MnO_4^-(aq) + 8H^+(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\)
**Part (c)** 1. Calculate the amount of \(MnO_4^-\right) used in the titration: \)n(MnO_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{25.00}{1000}\text{ dm}^3 = 5.00 \times 10^{-4}\text{ mol}\)
2. Determine the amount of \(Fe^{2+}\) in the \(25.0\text{ cm}^3\) portion: From the stoichiometry, \(n(Fe^{2+}) = 5 \times n(MnO_4^-) = 5 \times 5.00 \times 10^{-4} = 2.50 \times 10^{-3}\text{ mol}\)
3. Calculate the amount of \(Fe^{2+}\) in the \(250\text{ cm}^3\) standard solution: \(n(Fe^{2+})\text{ total} = 2.50 \times 10^{-3}\text{ mol} \times 10 = 2.50 \times 10^{-2}\text{ mol}\)
4. Calculate the \(M_r\) of the hydrated salt: \(M_r = \frac{\text{mass}}{\text{moles}} = \frac{9.80\text{ g}}{2.50 \times 10^{-2}\text{ mol}} = 392.0\text{ g mol}^{-1}\)
5. Determine the value of \(x\): \(M_r((NH_4)_2Fe(SO_4)_2) = 2 \times [14.0 + 4(1.0)] + 55.8 + 2 \times [32.1 + 4(16.0)] = 284.0\text{ g mol}^{-1}\) \(M_r\text{ of } xH_2O = 392.0 - 284.0 = 108.0\text{ g mol}^{-1}\) \(x = \frac{108.0}{18.0} = 6\)
Marking scheme
**Part (a) [4 marks]** - **M1:** Weigh sample in a weighing bottle/beaker and dissolve in deionised water in a beaker. (1) - **M2:** Transfer quantitatively to a \(250\text{ cm}^3\) volumetric flask. (1) - **M3:** Rinse beaker and glass rod and transfer the washings to the flask. (1) - **M4:** Make up to the mark/graduation line with deionised water so the bottom of the meniscus is on the line, then invert to mix. (1)
**Part (c) [4 marks]** - **M1:** Calculates moles of \(MnO_4^-\right) = \)5.00 \times 10^{-4}\text{ mol}\). (1) - **M2:** Calculates moles of \(Fe^{2+}\) in \(250\text{ cm}^3\) = \(2.50 \times 10^{-2}\text{ mol}\). (1) - **M3:** Calculates \(M_r\) of hydrated salt = \(392.0\text{ g mol}^{-1}\). (1) - **M4:** Calculates \(x = 6\) (working must be shown). (1) *Allow 1 mark for calculating \(M_r\) of anhydrous salt as \(284.0\text{ g mol}^{-1}\) if value of \(x\) is incorrect due to a minor arithmetic error.*
Question 2 · Structured Practical Questions
10 marks
A student investigates the kinetics of the reaction between propanone and iodine in acidic conditions:
The progress of the reaction is monitored using a colorimeter.
**Part (a)** Explain why a colorimeter is a suitable instrument for monitoring the progress of this reaction. (1 mark)
**Part (b)** Suggest a suitable color filter to use with the colorimeter in this experiment. Justify your suggestion. (2 marks)
**Part (c)** Outline how the student could use a series of standard solutions of iodine to construct a calibration curve. (3 marks)
**Part (d)** The student obtains a concentration-time graph for iodine which is a straight line with a constant negative gradient. Explain what this graph shows about the order of reaction with respect to iodine. (2 marks)
**Part (e)** In another set of experiments, the student doubles the concentration of propanone while keeping the concentration of acid and iodine constant. The initial rate of the reaction doubles. Determine the order of reaction with respect to propanone and justify your answer. (2 marks)
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Worked solution
**Part (a)** Iodine is a colored species (yellow/brown) while all other reactants and products in the reaction mixture are colorless. As the reaction proceeds, the color intensity decreases, which can be measured quantitatively as a decrease in light absorbance.
**Part (b)** A blue or violet filter (wavelength between 400–490 nm) is suitable. This is because blue is the complementary color to yellow/brown, so yellow/brown iodine solutions absorb light in the blue-violet region most strongly, giving the maximum sensitivity of absorbance changes.
**Part (c)** 1. Prepare a series of iodine solutions of known, different concentrations (using dilution of a standard stock solution). 2. Measure the absorbance of each solution in the colorimeter using the chosen filter. 3. Plot a graph of absorbance on the y-axis against concentration of iodine on the x-axis.
**Part (d)** A straight-line graph with a constant negative gradient means that the rate of reaction (\(\text{rate} = -\frac{\Delta[I_2]}{\Delta t}\)) is constant and does not change as the concentration of iodine decreases. This indicates that the reaction is zero order with respect to iodine.
**Part (e)** When the concentration of propanone doubles (\(\times 2\)), the rate also doubles (\(\times 2^1\)). Since the rate is directly proportional to the concentration of propanone, the order of reaction with respect to propanone is 1 (first order).
Marking scheme
**Part (a) [1 mark]** - **M1:** Iodine is colored (yellow/brown/orange) and other reactants/products are colorless, so absorbance changes can be monitored. (1)
**Part (b) [2 marks]** - **M1:** Suggests blue / violet filter (or wavelength range 400-490 nm). (1) - **M2:** Justification: Blue is the complementary color to yellow/brown OR iodine absorbs blue light most strongly / maximum absorbance is achieved. (1)
**Part (c) [3 marks]** - **M1:** Prepare several/at least 5 standard iodine solutions of known concentrations. (1) - **M2:** Measure the absorbance of each solution in the colorimeter. (1) - **M3:** Plot a graph of absorbance against concentration. (1)
**Part (d) [2 marks]** - **M1:** The rate / gradient is constant (independent of concentration). (1) - **M2:** Order with respect to iodine is zero. (1)
**Part (e) [2 marks]** - **M1:** Order with respect to propanone is 1 / first order. (1) - **M2:** Because the rate is directly proportional to concentration (or doubling concentration doubles the rate). (1)
Question 3 · Structured Practical Questions
10 marks
A student synthesized a sample of aspirin (acetylsalicylic acid, \(M_r = 180.0\)) from salicylic acid (\(M_r = 138.0\)). The student purified the crude product by recrystallization.
**Part (a)** Outline the experimental steps the student should take to recrystallize the crude aspirin to obtain a pure sample. For each step, explain the chemical reasoning behind it. (5 marks)
**Part (b)** Describe how the student could determine if the recrystallized sample of aspirin is pure, and describe the expected observations for both a pure sample and an impure sample. (3 marks)
**Part (c)** The student began with \(5.00\text{ g}\) of salicylic acid. After recrystallization, the student obtained a \(65.0\%\) yield of pure aspirin. Calculate the mass of pure aspirin obtained. Show your working. (2 marks)
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Worked solution
**Part (a)** 1. Dissolve the crude aspirin in the minimum volume of hot solvent (such as ethanol or water). *Reason:* Minimum volume is used to ensure a saturated solution is formed so that the maximum amount of crystals will precipitate out on cooling. 2. Filter the hot solution (optional but recommended to remove insoluble impurities). 3. Allow the hot solution to cool slowly to room temperature and then place in an ice bath. *Reason:* Cooling decreases the solubility of aspirin, causing pure crystals to reform while soluble impurities remain dissolved in the solvent. 4. Filter the crystals under reduced pressure using a Buchner flask and funnel. *Reason:* This separates the crystals from the filtrate containing soluble impurities and dries the crystals faster. 5. Wash the crystals with a small volume of ice-cold solvent, and dry them. *Reason:* Cold solvent removes any soluble impurities remaining on the surface of the crystals without dissolving the aspirin.
**Part (b)** 1. Use a melting point apparatus to determine the melting point of the crystals. 2. A pure sample will melt sharply at the literature melting point of aspirin (approx. \(135\text{–}136\text{ }^\circ\text{C}\)). 3. An impure sample will melt over a wider temperature range and at a lower temperature than the literature value.
**Part (c)** 1. Calculate moles of salicylic acid used: \(n(\text{salicylic acid}) = \frac{5.00\text{ g}}{138.0\text{ g mol}^{-1}} = 0.03623\text{ mol}\)
2. Since the stoichiometry of the reaction is \(1:1\), the theoretical yield of aspirin is: \(m(\text{aspirin, theoretical}) = 0.03623\text{ mol} \times 180.0\text{ g mol}^{-1} = 6.521\text{ g}\)
3. Calculate the actual mass obtained using the \(65.0\%\) yield: \(m(\text{aspirin, actual}) = 6.521\text{ g} \times 0.650 = 4.24\text{ g}\) (to 3 significant figures)
Marking scheme
**Part (a) [5 marks]** - **M1:** Dissolve crude solid in the minimum volume of hot solvent. (1) - **M2:** Explanation: Minimum volume ensures a saturated solution (to maximize crystal yield on cooling). (1) - **M3:** Cool the solution (in ice). (1) - **M4:** Explanation: Decreases solubility of aspirin so crystals reform (while soluble impurities stay in solution). (1) - **M5:** Filter under reduced pressure (using Buchner funnel/flask), wash with cold solvent and dry. (1)
**Part (b) [3 marks]** - **M1:** Determine the melting point (using melting point apparatus / capillary tube in oil bath). (1) - **M2:** Pure sample: Melts sharply at the literature value / \(135\text{–}136\text{ }^\circ\text{C}\). (1) - **M3:** Impure sample: Melts over a wide range / melts at a lower temperature than literature. (1)
**Part (c) [2 marks]** - **M1:** Calculates theoretical yield: \(\frac{5.00}{138.0} \times 180.0 = 6.52\text{ g}\) (or moles of salicylic acid = \(0.0362\text{ mol}\)). (1) - **M2:** Calculates actual mass of aspirin: \(6.521 \times 0.650 = 4.24\text{ g}\) (accept \(4.23\) to \(4.25\text{ g}\)). (1)
Unit 5: Section B (Synoptic Multiple Choice)
Answer all questions. For each question select the best response.
30 Question · 30 marks
Question 1 · Multiple Choice
1 marks
When ethylenediamine (\(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\), abbreviated as \(\text{en}\)) reacts with aqueous copper(II) ions, a stable complex \([\text{Cu(en)}_3]^{2+}\) is formed according to the following equation: \([\text{Cu(H}_2\text{O)}_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightarrow [\text{Cu(en)}_3]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\). What is the primary thermodynamic reason for the high stability of this complex compared to the hexaaquacopper(II) complex?
A.There is a large decrease in enthalpy because \(\text{Cu}-\text{N}\) coordinate bonds are much stronger than \(\text{Cu}-\text{O}\) coordinate bonds.
B.There is a large increase in entropy because seven moles of species are formed from four moles of reactants.
C.There is a large decrease in activation energy because the reaction proceeds via an alternative low-energy mechanism.
D.There is a large increase in entropy because the geometry of the complex changes from octahedral to tetrahedral.
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Worked solution
The reaction of one hexaaquacopper(II) ion with three bidentate ethylenediamine ligands produces one \([\text{Cu(en)}_3]^{2+}\) complex and six monodentate water molecules. The total number of particles in solution increases from 4 to 7. This leads to a significant increase in disorder, resulting in a positive change in entropy (\(\Delta S^\theta > 0\)), which drives the reaction forward and makes the complex highly stable.
Marking scheme
1 mark: Correctly identifies that the increase in entropy due to the increase in the number of particles (from 4 to 7 moles) in solution is the primary driving force.
Question 2 · Multiple Choice
1 marks
A chemist prepares ethyl ethanoate by reacting ethanoic acid (\(M_r = 60.0\)) and excess ethanol (\(M_r = 46.0\)): \(\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\). When \(12.0\text{ g}\) of ethanoic acid is reacted, \(11.0\text{ g}\) of ethyl ethanoate (\(M_r = 88.0\)) is obtained. What is the percentage yield of ethyl ethanoate?
A.56.8%
B.62.5%
C.83.0%
D.91.7%
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Worked solution
First, calculate the moles of ethanoic acid: \(n(\text{CH}_3\text{COOH}) = \frac{12.0}{60.0} = 0.200\text{ mol}\). Since the stoichiometry is 1:1, the theoretical moles of ethyl ethanoate is \(0.200\text{ mol}\). The theoretical mass of ethyl ethanoate is \(0.200\text{ mol} \times 88.0\text{ g mol}^{-1} = 17.6\text{ g}\). The percentage yield is \(\frac{11.0}{17.6} \times 100\% = 62.5\%\).
Marking scheme
1 mark: Correct calculation of moles of ethanoic acid, theoretical mass of ethyl ethanoate, and subsequent percentage yield.
Question 3 · Multiple Choice
1 marks
Which of the following halogenoalkanes reacts most rapidly when heated with aqueous silver nitrate in ethanol at \(50\text{ }^\circ\text{C}\)?
A.1-chlorobutane
B.1-iodobutane
C.2-chloro-2-methylpropane
D.2-iodo-2-methylpropane
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Worked solution
The rate of hydrolysis of halogenoalkanes depends on two main factors: the strength of the C-X bond and the mechanism. C-I bonds are significantly weaker than C-Cl bonds, meaning iodobutanes react faster than chlorobutanes. Tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react extremely rapidly via the \(\text{S}_\text{N}1\) mechanism because they form a stable tertiary carbocation intermediate, whereas primary halogenoalkanes react much slower via the \(\text{S}_\text{N}2\) mechanism. Therefore, 2-iodo-2-methylpropane reacts most rapidly.
Marking scheme
1 mark: Correctly identifies the tertiary iodoalkane as the most reactive due to the weakest carbon-halogen bond and stable carbocation formation.
Question 4 · Multiple Choice
1 marks
The amino acid alanine has an isoelectric point (pI) of \(6.0\). Which structure represents the predominant species of alanine in an aqueous solution at \(\text{pH } 2.0\)?
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Worked solution
At a pH of 2.0 (which is well below the isoelectric point of 6.0), the solution is strongly acidic with a high concentration of \(\text{H}^+\) ions. Both the basic amine group and the weakly basic carboxylate group are fully protonated. The amine group exists as \(-\text{NH}_3^+\) and the carboxylic acid group exists as \(-\text{COOH}\). Thus, the predominant species is the cation \(\text{H}_3\text{N}^+\text{-CH(CH}_3\text{)-COOH}\).
Marking scheme
1 mark: Correctly identifies that at pH < pI, both groups are protonated to form a positive cation.
Question 5 · Multiple Choice
1 marks
Which of the following elements has the highest first ionization energy?
A.Nitrogen
B.Oxygen
C.Phosphorus
D.Sulfur
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Worked solution
First, compare Period 2 elements (N and O) with Period 3 elements (P and S). Period 2 elements have their outer electrons in the second shell, which is closer to the nucleus and experiences less shielding, giving them higher ionization energies than Period 3 elements. Next, compare Nitrogen and Oxygen: Nitrogen has the electron configuration \(1s^2 2s^2 2p^3\) with a stable, half-filled p-subshell. Oxygen has the configuration \(1s^2 2s^2 2p^4\). The paired electron in one of the \(2p\) orbitals of oxygen experiences mutual repulsion, making it easier to remove despite the higher nuclear charge. Thus, Nitrogen has a higher first ionization energy than Oxygen, and the highest overall.
Marking scheme
1 mark: Correctly identifies Nitrogen as having the highest first ionization energy due to period trends and the half-filled p-subshell repulsion exception in Oxygen.
Question 6 · Multiple Choice
1 marks
Consider the thermal decomposition of calcium carbonate: \(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \quad \Delta H^\theta = +178\text{ kJ mol}^{-1}\). At what minimum temperature (in \(\text{K}\)) does this reaction become feasible, given that the standard entropy change (\(\Delta S^\theta\)) is \(+161\text{ J K}^{-1}\text{ mol}^{-1}\)?
A.1.11 K
B.288 K
C.904 K
D.1106 K
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Worked solution
A reaction is feasible when \(\Delta G^\theta \le 0\). Using the relation \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), we find the transition temperature where \(\Delta G^\theta = 0\): \(T = \frac{\Delta H^\theta}{\Delta S^\theta}\). Convert \(\Delta H^\theta\) to \(\text{J mol}^{-1}\): \(\Delta H^\theta = 178 \times 10^3\text{ J mol}^{-1}\). Thus, \(T = \frac{178000}{161} \approx 1105.59\text{ K}\). Rounding to the nearest whole Kelvin gives \(1106\text{ K}\).
Marking scheme
1 mark: Correct conversion of enthalpy units to J and calculation of the threshold temperature.
Question 7 · Multiple Choice
1 marks
Which of the following is the sulfur-containing reduction product formed when solid sodium bromide reacts with concentrated sulfuric acid?
A.\(\text{SO}_2\)
B.\(\text{S}\)
C.\(\text{H}_2\text{S}\)
D.\(\text{SO}_3\)
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Worked solution
Concentrated sulfuric acid acts as an oxidizing agent. Bromide ions are oxidized to bromine gas (\(\text{Br}_2\)), and they are strong enough reducing agents to reduce sulfuric acid (where sulfur is in the +6 state) to sulfur dioxide gas (\(\text{SO}_2\)), where sulfur is in the +4 state. Bromide is not strong enough to reduce it further to elemental sulfur or hydrogen sulfide (unlike iodide ions, which can reduce it to \(\text{S}\) and \(\text{H}_2\text{S}\)).
Marking scheme
1 mark: Correctly identifies sulfur dioxide as the unique sulfur-containing reduction product for the reaction with bromide.
Question 8 · Multiple Choice
1 marks
The following initial rate data was obtained for the reaction \(2\text{A} + \text{B} + \text{C} \rightarrow \text{D} + \text{E}\): [Experiment 1: \([\text{A}] = 0.10\), \([\text{B}] = 0.10\), \([\text{C}] = 0.10\), rate = \(2.0 \times 10^{-4}\)]; [Experiment 2: \([\text{A}] = 0.20\), \([\text{B}] = 0.10\), \([\text{C}] = 0.10\), rate = \(4.0 \times 10^{-4}\)]; [Experiment 3: \([\text{A}] = 0.10\), \([\text{B}] = 0.20\), \([\text{C}] = 0.10\), rate = \(8.0 \times 10^{-4}\)]; [Experiment 4: \([\text{A}] = 0.10\), \([\text{B}] = 0.10\), \([\text{C}] = 0.20\), rate = \(2.0 \times 10^{-4}\)] (all concentrations in \(\text{mol dm}^{-3}\), rates in \(\text{mol dm}^{-3}\text{ s}^{-1}\)). What is the overall order of this reaction?
A.1
B.2
C.3
D.4
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Worked solution
Let rate = \(k[\text{A}]^x[\text{B}]^y[\text{C}]^z\). Comparing Exp 1 and Exp 2: \([\text{A}]\) doubles while \([\text{B}]\) and \([\text{C}]\) remain constant; the rate doubles, so \(x = 1\). Comparing Exp 1 and Exp 3: \([\text{B}]\) doubles while \([\text{A}]\) and \([\text{C}]\) remain constant; the rate quadruples, so \(y = 2\). Comparing Exp 1 and Exp 4: \([\text{C}]\) doubles while \([\text{A}]\) and \([\text{B}]\) remain constant; the rate remains unchanged, so \(z = 0\). The rate equation is rate = \(k[\text{A}][\text{B}]^2\). The overall order is \(1 + 2 + 0 = 3\).
Marking scheme
1 mark: Deduces the individual orders of A, B, and C correctly, and sums them to obtain the overall order of 3.
Question 9 · multiple_choice
1 marks
Which of the following statements about ligand substitution reactions and the colours of transition metal complexes is correct?
A.In the substitution of \([Cu(H_2O)_6]^{2+}\) with excess concentrated hydrochloric acid, the coordination number remains 6.
B.When excess aqueous ammonia is added to \([Cu(H_2O)_6]^{2+}\), the complex \([Cu(NH_3)_6]^{2+}\) is formed.
C.The colour of transition metal complexes is due to the absorption of light corresponding to the energy gap \(\Delta E\) between split d-orbitals.
D.Complete substitution of \(H_2O\) ligands in \([Fe(H_2O)_6]^{3+}\) by \(C_2O_4^{2-}\) ligands occurs with no change in the entropy of the system.
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Worked solution
In transition metal complexes, the d-orbitals split into two sets of non-degenerate energy levels. When light is absorbed, an electron is promoted from a lower d-orbital to a higher d-orbital. The energy of the absorbed light (\(\Delta E = h\nu\)) corresponds to this energy gap. Option A is incorrect because substitution of \([Cu(H_2O)_6]^{2+}\) with excess concentrated hydrochloric acid forms the tetrahedral \([CuCl_4]^{2-}\) complex (coordination number changes from 6 to 4). Option B is incorrect because reaction with excess aqueous ammonia yields \([Cu(NH_3)_4(H_2O)_2]^{2+}\). Option D is incorrect because replacing six monodentate ligands with three bidentate ligands increases entropy (positive \(\Delta S\)) due to an increase in the number of particles in solution.
Marking scheme
1 mark for the correct option (C). No partial marks.
Question 10 · multiple_choice
1 marks
A 1.25 g sample of an impure metal carbonate, \(MCO_3\) (where \(M\) is a Group 2 metal with a molar mass of 84.3 g mol\(^{-1}\) for \(MCO_3\)), was reacted completely with 50.0 cm\(^3\) of 0.500 mol dm\(^{-3}\) hydrochloric acid.
The excess acid required 28.4 cm\(^3\) of 0.100 mol dm\(^{-3}\) sodium hydroxide solution for complete neutralisation. What is the percentage purity by mass of \(MCO_3\) in the sample?
A.36.4%
B.74.7%
C.91.0%
D.37.3%
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3. Calculate moles of \(HCl\) reacted with \(MCO_3\): \(n(HCl)_{\text{reacted}} = 0.0250 - 0.00284 = 0.02216 \text{ mol}\)
4. Determine moles of \(MCO_3\): The reaction is \(MCO_3 + 2HCl \rightarrow MCl_2 + CO_2 + H_2O\). \(n(MCO_3) = \frac{0.02216}{2} = 0.01108 \text{ mol}\)
5. Calculate pure mass and percentage purity: \(\text{Mass of } MCO_3 = 0.01108 \text{ mol} \times 84.3 \text{ g mol}^{-1} = 0.934 \text{ g}\) \(\text{Purity} = \left(\frac{0.934 \text{ g}}{1.25 \text{ g}}\right) \times 100\% \approx 74.7\%\)
Marking scheme
1 mark for the correct calculation and option (B). No partial marks.
Question 11 · multiple_choice
1 marks
Under certain conditions, 2-bromo-2-methylpropane reacts with aqueous sodium hydroxide. Which of the following statements about this reaction is correct?
A.The reaction mechanism is primary nucleophilic substitution (\(S_N2\)).
B.The rate of the reaction depends on the concentration of both 2-bromo-2-methylpropane and hydroxide ions.
C.The rate-determining step involves the heterolytic fission of the C–Br bond to form a tertiary carbocation.
D.Changing the solvent from water to ethanol has no effect on the rate of reaction.
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Worked solution
2-bromo-2-methylpropane is a tertiary halogenoalkane. It undergoes nucleophilic substitution primarily via the \(S_N1\) mechanism. The first, slow (rate-determining) step of the \(S_N1\) mechanism involves the heterolytic cleavage of the C–Br bond to form a stable tertiary carbocation intermediate. Therefore, Option C is correct. Option A is incorrect as the mechanism is \(S_N1\), not \(S_N2\). Option B is incorrect because the rate depends only on the concentration of the halogenoalkane. Option D is incorrect because solvent polarity strongly affects the rate of carbocation formation in an \(S_N1\) pathway.
Marking scheme
1 mark for selecting the correct statement (C). No partial marks.
Question 12 · multiple_choice
1 marks
Glutamic acid is a dicarboxylic amino acid with the structure \(HOOC-CH_2-CH_2-CH(NH_2)-COOH\). Which of the following represents the major organic species present in an aqueous solution of glutamic acid at a pH of 12?
A.\(^-OOC-CH_2-CH_2-CH(NH_2)-COO^-\)
B.\(^-OOC-CH_2-CH_2-CH(NH_3^+)-COO^-\)
C.\(HOOC-CH_2-CH_2-CH(NH_3^+)-COOH\)
D.\(^-OOC-CH_2-CH_2-CH(NH_2)-COOH\)
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Worked solution
At highly alkaline pH values (such as pH = 12), the concentration of \(OH^-\) ions is very high, causing deprotonation of all acidic sites. Both carboxylic acid groups (\(-COOH\)) lose protons to become carboxylate anions (\(-COO^-\)), and the amino group remains in its basic, unprotonated form (\(-NH_2\)). This results in the species \(^-OOC-CH_2-CH_2-CH(NH_2)-COO^-\).
Marking scheme
1 mark for the correct deprotonated structure (A). No partial marks.
Question 13 · multiple_choice
1 marks
The successive ionisation energies of a Period 3 element \(X\) are \(786\), \(1577\), \(3232\), \(4356\), \(16091\), and \(19785\) kJ mol\(^{-1}\). Which formula represents a stable compound formed between element \(X\) and oxygen?
A.\(X_2O\)
B.\(XO\)
C.\(XO_2\)
D.\(X_2O_3\)
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Worked solution
By looking at the successive ionisation energies: 1st (786) \(\rightarrow\) 2nd (1577) \(\rightarrow\) 3rd (3232) \(\rightarrow\) 4th (4356) are relatively gradual increases. There is a huge jump between the 4th and 5th ionisation energies (from 4356 to 16091 kJ mol\(^{-1}\)). This indicates that the 5th electron is removed from an inner quantum shell, meaning element \(X\) has 4 valence electrons and belongs to Group 14 (Group 4). In Period 3, this element is Silicon (\(Si\)). Silicon forms a stable oxide with the formula \(SiO_2\), represented as \(XO_2\).
Marking scheme
1 mark for the correct formula (C) based on the valence electronic structure of the element. No partial marks.
Question 14 · multiple_choice
1 marks
Consider the following thermodynamic data for the formation of magnesium chloride, \(MgCl_2\)(s):
- Enthalpy of formation of \(MgCl_2\)(s) = \(-642\text{ kJ mol}^{-1}\) - 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, \(\Delta H_{at}[\frac{1}{2}Cl_2]\) = \(+122\text{ kJ mol}^{-1}\) - First electron affinity of \(Cl\)(g) = \(-349\text{ kJ mol}^{-1}\)
What is the lattice enthalpy of formation (\(\Delta H_{L}^{\ominus}\)) of \(MgCl_2\)(s)?
A.\(-2525\text{ kJ mol}^{-1}\)
B.\(-2176\text{ kJ mol}^{-1}\)
C.\(-1826\text{ kJ mol}^{-1}\)
D.\(-2874\text{ kJ mol}^{-1}\)
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1 mark for the correct calculation and option (A). No partial marks.
Question 15 · multiple_choice
1 marks
The rate equation for the reaction between two substances, \(P\) and \(Q\), is:
\(Rate = k [P]^2 [Q]\)
When the initial concentrations of both \(P\) and \(Q\) are \(0.20\text{ mol dm}^{-3}\), the initial rate of reaction is \(1.6 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the value and unit of the rate constant, \(k\), for this reaction?
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Worked solution
1. Substitute the experimental values into the rate equation: \(1.6 \times 10^{-4} = k \times (0.20)^2 \times (0.20)\) \(1.6 \times 10^{-4} = k \times 0.040 \times 0.20 = k \times 0.0080\) \(k = \frac{1.6 \times 10^{-4}}{8.0 \times 10^{-3}} = 2.0 \times 10^{-2}\)
2. Determine the units of \(k\): \(k = \frac{Rate}{[P]^2 [Q]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\) which is written as \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
Marking scheme
1 mark for the correct value and units of the rate constant (B). No partial marks.
Question 16 · multiple_choice
1 marks
Which of the following statements/equations regarding the reactions of Period 3 oxides is incorrect?
D.\(MgO(s) + H_2O(l) \rightarrow Mg(OH)_2(aq)\) (readily forming a highly alkaline solution of pH 14)
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Worked solution
Magnesium oxide is a basic oxide that is only sparingly soluble in water, reacting slowly to form a saturated solution of magnesium hydroxide, \(Mg(OH)_2\), which has a pH of approximately 9 to 10. It is not highly soluble enough to form a highly alkaline solution of pH 14. Therefore, option D is incorrect. Options A, B, and C are correct descriptions of the chemical reactions of \(P_4O_{10}\), \(Al_2O_3\), and \(SiO_2\) respectively.
Marking scheme
1 mark for identifying the incorrect statement (D). No partial marks.
Question 17 · Multiple Choice
1 marks
What is the oxidation state and coordination number of chromium in the complex ion \([\text{Cr}(\text{C}_2\text{O}_4)_2(\text{H}_2\text{O})_2]^-\)?
A.Oxidation state: +3, Coordination number: 4
B.Oxidation state: +3, Coordination number: 6
C.Oxidation state: +5, Coordination number: 4
D.Oxidation state: +5, Coordination number: 6
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Worked solution
The oxalate ligand, \(\text{C}_2\text{O}_4^{2-}\), is a bidentate ligand carrying a 2- charge. The water ligand, \(\text{H}_2\text{O}\), is a neutral monodentate ligand. Let \(x\) represent the oxidation state of chromium: \(x + 2(-2) + 2(0) = -1\), which solves to \(x = +3\). The coordination number is calculated by adding the number of dative covalent bonds formed: 2 bidentate ligands and 2 monodentate ligands give a total coordination number of \((2 \times 2) + 2 = 6\).
Marking scheme
1 mark for the correct option (B) based on correct determination of oxidation state (+3) and coordination number (6).
Question 18 · Multiple Choice
1 marks
A \(0.310\text{ g}\) sample of an organic compound containing only carbon, hydrogen, and oxygen was completely combusted in excess oxygen. It produced \(0.440\text{ g}\) of \(\text{CO}_2\) and \(0.270\text{ g}\) of \(\text{H}_2\text{O}\). What is the empirical formula of the compound?
A.\(\text{CH}_2\text{O}\)
B.\(\text{CH}_3\text{O}\)
C.\(\text{C}_2\text{H}_6\text{O}\)
D.\(\text{C}_2\text{H}_4\text{O}\)
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Worked solution
First, calculate the moles and mass of carbon: \(n(\text{CO}_2) = \frac{0.440}{44.0} = 0.010\text{ mol}\), so \(n(\text{C}) = 0.010\text{ mol}\). Mass of \(\text{C} = 0.010 \times 12.0 = 0.120\text{ g}\). Next, calculate the moles and mass of hydrogen: \(n(\text{H}_2\text{O}) = \frac{0.270}{18.0} = 0.015\text{ mol}\), so \(n(\text{H}) = 2 \times 0.015 = 0.030\text{ mol}\). Mass of \(\text{H} = 0.030 \times 1.0 = 0.030\text{ g}\). Then, find the mass and moles of oxygen by subtraction: Mass of \(\text{O} = 0.310 - 0.120 - 0.030 = 0.160\text{ g}\). \(n(\text{O}) = \frac{0.160}{16.0} = 0.010\text{ mol}\). The mole ratio is \(\text{C} : \text{H} : \text{O} = 0.010 : 0.030 : 0.010 = 1 : 3 : 1\). Therefore, the empirical formula is \(\text{CH}_3\text{O}\).
Marking scheme
1 mark for the correct empirical formula option (B).
Question 19 · Multiple Choice
1 marks
Which of the following statements correctly explains the relative rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane?
A.1-chlorobutane reacts fastest because the \(\text{C}-\text{Cl}\) bond is the most polar.
B.1-iodobutane reacts fastest because the \(\text{C}-\text{I}\) bond has the lowest bond enthalpy.
C.1-chlorobutane reacts fastest because chlorine is the most electronegative halogen.
D.1-iodobutane reacts slowest because iodine has the largest atomic radius, creating steric hindrance.
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Worked solution
The rate of hydrolysis depends on the carbon-halogen bond strength (enthalpy) rather than bond polarity. The \(\text{C}-\text{I}\) bond is the longest and weakest (lowest bond enthalpy) among the three, so it is broken most easily, resulting in 1-iodobutane undergoing hydrolysis at the fastest rate.
Marking scheme
1 mark for selecting the correct explanation based on bond enthalpy (B).
Question 20 · Multiple Choice
1 marks
What is the predominant structure of alanine (2-aminopropanoic acid) in an aqueous solution of \(\text{pH } 12\)?
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Worked solution
At high pH (alkaline conditions), basic species are deprotonated. The carboxylic acid group loses its proton to form the carboxylate ion (\(-\text{COO}^-\)), and the amino group exists in its unprotonated form (\(-\text{NH}_2\)). Thus, the predominant species is \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).
Marking scheme
1 mark for identifying the correct anionic form of the amino acid at high pH (C).
Question 21 · Multiple Choice
1 marks
The first five successive ionisation energies (\(\text{kJ mol}^{-1}\)) of a Period 3 element, \(X\), are shown below:
Which description fits the oxide of element \(X\)?
A.An acidic oxide with the formula \(X_2\text{O}_3\)
B.An amphoteric oxide with the formula \(X_2\text{O}_3\)
C.A basic oxide with the formula \(X\text{O}\)
D.An amphoteric oxide with the formula \(X\text{O}_2\)
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Worked solution
The large jump between the third and fourth ionisation energies indicates that the element has three valence electrons, which puts it in Group 13. Because it is in Period 3, element \(X\) must be aluminium (\(\text{Al}\)). Aluminium forms \(\text{Al}_2\text{O}_3\), which is an amphoteric oxide with the general formula \(X_2\text{O}_3\).
Marking scheme
1 mark for identifying the element as aluminium and choosing the correct amphoteric oxide option (B).
Question 22 · Multiple Choice
1 marks
For a particular chemical reaction, \(\Delta H^\ominus = +135\text{ kJ mol}^{-1}\) and \(\Delta S^\ominus = +285\text{ J K}^{-1}\text{ mol}^{-1}\). Under standard pressure, at which temperature range does this reaction become feasible?
A.Above \(2.11\text{ K}\)
B.Above \(47.4\text{ K}\)
C.Above \(474\text{ K}\)
D.Above \(2110\text{ K}\)
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Worked solution
A reaction is feasible when \(\Delta G^\ominus \le 0\). Given \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), we find feasibility occurs when \(T \ge \frac{\Delta H^\ominus}{\Delta S^\ominus}\). Converting \(\Delta H^\ominus\) to \(\text{J mol}^{-1}\) gives \(135 \times 10^3\text{ J mol}^{-1}\). Dividing by \(285\text{ J K}^{-1}\text{ mol}^{-1}\) yields \(T \ge 473.68\text{ K}\), which rounds to \(474\text{ K}\).
Marking scheme
1 mark for the correct temperature calculation and inequality selection (C).
Question 23 · Multiple Choice
1 marks
A reaction has the rate equation: \(\text{Rate} = k[P][Q]^2\). In an experiment, the initial rate of reaction was \(4.80 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\) when \([P] = 0.200\text{ mol dm}^{-3}\) and \([Q] = 0.400\text{ mol dm}^{-3}\). What is the value and unit of the rate constant, \(k\), for this reaction?
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Worked solution
Rearrange the rate equation for \(k\): \(k = \frac{\text{Rate}}{[P][Q]^2}\). Substitute the values: \(k = \frac{4.80 \times 10^{-3}}{0.200 \times (0.400)^2} = \frac{4.80 \times 10^{-3}}{0.0320} = 0.150\). To find the unit: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\). Thus, \(k = 0.150\text{ mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).
Marking scheme
1 mark for calculating the correct value and determining the correct unit (B).
Question 24 · Multiple Choice
1 marks
Which row correctly describes the observations and the approximate pH of the resulting solution when silicon tetrachloride, \(\text{SiCl}_4\), is added to excess water?
A.White precipitate and misty fumes; pH = 1–2
B.No precipitate, solution remains clear; pH = 7
C.White precipitate and effervescence of a colorless, odorless gas; pH = 5–6
D.Yellow precipitate and misty fumes; pH = 1–2
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Worked solution
Silicon tetrachloride reacts vigorously with water in a hydrolysis reaction: \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g})\). The silicon dioxide formed is a white precipitate (insoluble solid), and the hydrogen chloride gas evolved is seen as misty white fumes. When the \(\text{HCl}\) gas dissolves in the excess water, it forms highly acidic hydrochloric acid, which has a pH of 1–2.
Marking scheme
1 mark for the correct combination of observations and pH (A).
Question 25 · multiple-choice
1 marks
An investigation into the rate of reaction between peroxodisulfate ions and iodide ions was carried out. The rate equation was determined to be: \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). In an experiment at a constant temperature, the initial concentrations were \([\text{S}_2\text{O}_8^{2-}] = 0.050\text{ mol dm}^{-3}\) and \([\text{I}^-] = 0.080\text{ mol dm}^{-3}\). Under these conditions, the initial rate of reaction was \(1.2 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the value of the rate constant, \(k\), and its units under these conditions?
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Worked solution
The rate equation is given by \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). Rearranging for the rate constant gives \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}\). Substituting the given values: \(k = \frac{1.2 \times 10^{-4}}{0.050 \times 0.080} = \frac{1.2 \times 10^{-4}}{0.0040} = 0.030\). The units of \(k\) are found by substituting the units into the expression: \(\text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). Hence, the correct rate constant is \(0.030\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
Marking scheme
1 mark for the correct calculation of rate constant value and its units. Award 1 mark for choosing option A.
Question 26 · multiple-choice
1 marks
Which of the following statements about the reactions of aqueous transition metal ions is correct?
A.Addition of excess aqueous ammonia to aqueous copper(II) ions produces a blue precipitate of \(\text{Cu(OH)}_2\) that does not redissolve.
B.Addition of aqueous sodium carbonate to aqueous iron(III) ions produces a green precipitate of iron(II) carbonate and carbon dioxide gas.
C.Addition of concentrated hydrochloric acid to aqueous cobalt(II) ions results in a colour change from pink to blue due to the formation of a tetrahedral complex.
D.Addition of sodium hydroxide to aqueous aluminium(III) ions produces a white precipitate of aluminium hydroxide which is insoluble in excess sodium hydroxide.
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Worked solution
Option A is incorrect because the blue precipitate of \(\text{Cu(OH)}_2\) redissolves in excess aqueous ammonia to form a deep blue solution of \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\). Option B is incorrect because aqueous iron(III) reacts with carbonate ions to form a brown precipitate of iron(III) hydroxide and carbon dioxide gas. Option C is correct because the pink octahedral complex \([\text{Co(H}_2\text{O)}_6]^{2+}\) reacts with concentrated \(\text{HCl}\) to form the blue tetrahedral complex \([\text{CoCl}_4]^{2-}\). Option D is incorrect because aluminium hydroxide is amphoteric and dissolves in excess sodium hydroxide to form a colourless solution containing \([\text{Al(OH)}_4]^-\).
Marking scheme
1 mark for identifying the correct chemical reaction and complex ion transition. Award 1 mark for choosing option C.
Question 27 · multiple-choice
1 marks
An organic compound with the molecular formula \(\text{C}_3\text{H}_7\text{NO}_2\) exists as a zwitterion in the solid state. Which of the following represents the predominant structure of this compound in an aqueous solution of \(\text{pH} = 1\)?
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Worked solution
The compound is alanine. In a highly acidic environment (\(\text{pH} = 1\)), the high concentration of \(\text{H}^+\) ions causes protonation of both basic sites. The carboxylate group \(\text{-COO}^-\) is protonated to \(\text{-COOH}\), and the amine group \(\text{-NH}_2\) is protonated to \(\text{-NH}_3^+\) to form the cation. Therefore, the overall species has a positive charge, as represented by \(\text{H}_3\text{N}^+\text{-CH(CH}_3\text{)-COOH}\).
Marking scheme
1 mark for identifying the protonated cationic species at low pH. Award 1 mark for choosing option C.
Question 28 · multiple-choice
1 marks
Consider the following thermodynamic data for magnesium chloride, \(\text{MgCl}_2\): Enthalpy of formation of \(\text{MgCl}_2(s)\): \(-642\text{ kJ mol}^{-1}\), Atomisation enthalpy of magnesium, \(\Delta H_{at}^\theta[\text{Mg}(s)]\): \(+148\text{ kJ mol}^{-1}\), First ionisation energy of magnesium, \(I_1[\text{Mg}]\): \(+738\text{ kJ mol}^{-1}\), Second ionisation energy of magnesium, \(I_2[\text{Mg}]\): \(+1451\text{ kJ mol}^{-1}\), Atomisation enthalpy of chlorine, \(\Delta H_{at}^\theta[\text{Cl}_2(g)]\): \(+121\text{ kJ mol}^{-1}\), Electron affinity of chlorine, \(EA_1[\text{Cl}]\): \(-349\text{ kJ mol}^{-1}\). What is the lattice enthalpy of formation of magnesium chloride?
A.\(-2523\text{ kJ mol}^{-1}\)
B.\(-2402\text{ kJ mol}^{-1}\)
C.\(-2872\text{ kJ mol}^{-1}\)
D.\(+2523\text{ kJ mol}^{-1}\)
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Worked solution
Using a Born-Haber cycle, the enthalpy of formation of magnesium chloride is related to the other enthalpy changes by: \(\Delta H_f^\theta[\text{MgCl}_2(s)] = \Delta H_{at}^\theta[\text{Mg}(s)] + I_1[\text{Mg}] + I_2[\text{Mg}] + 2 \times \Delta H_{at}^\theta[\text{Cl}_2(g)] + 2 \times EA_1[\text{Cl}] + \Delta H_{L}^\theta[\text{MgCl}_2(s)]\). Note that 2 moles of gaseous chlorine atoms are formed, which requires \(2 \times (+121)\text{ kJ mol}^{-1}\), and 2 moles of chlorine atoms are converted to chloride ions, which releases \(2 \times (-349)\text{ kJ mol}^{-1}\). Substituting the values: \(-642 = 148 + 738 + 1451 + 2(121) + 2(-349) + \Delta H_{L}^\theta\). This simplifies to \(-642 = 2337 + 242 - 698 + \Delta H_{L}^\theta\), which is \(-642 = 1881 + \Delta H_{L}^\theta\). Thus, \(\Delta H_{L}^\theta = -642 - 1881 = -2523\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for setting up the Born-Haber cycle and calculating the correct lattice enthalpy of formation. Award 1 mark for choosing option A.
Question 29 · multiple-choice
1 marks
An organic compound, \(\text{X}\), has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\). The proton NMR spectrum of \(\text{X}\) shows three peaks: a triplet at \(\delta = 1.3\text{ ppm}\) (integrating to \(3\text{H}\)), a singlet at \(\delta = 2.0\text{ ppm}\) (integrating to \(3\text{H}\)), and a quartet at \(\delta = 4.1\text{ ppm}\) (integrating to \(2\text{H}\)). Which of the following is the correct IUPAC name for compound \(\text{X}\)?
A.Methyl propanoate
B.Ethyl ethanoate
C.Propyl methanoate
D.Butanoic acid
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Worked solution
The singlet at \(\delta = 2.0\text{ ppm}\) integrating to \(3\text{H}\) indicates a \(\text{CH}_3\) group next to a carbonyl group (\(\text{CH}_3\text{-CO-}\)). The triplet at \(\delta = 1.3\text{ ppm}\) (\(3\text{H}\)) and the quartet at \(\delta = 4.1\text{ ppm}\) (\(2\text{H}\)) indicate an ethyl group adjacent to an oxygen atom (\(\text{-O-CH}_2\text{-CH}_3\)). Combining these fragments gives \(\text{CH}_3\text{-CO-O-CH}_2\text{-CH}_3\), which is ethyl ethanoate.
Marking scheme
1 mark for decoding the splitting patterns and chemical shifts to identify the ester. Award 1 mark for choosing option B.
Question 30 · multiple-choice
1 marks
Consider the following standard electrode potentials: 1. \(\text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq) \quad E^\theta = +0.77\text{ V}\), 2. \(\text{I}_2(aq) + 2e^- \rightleftharpoons 2\text{I}^-(aq) \quad E^\theta = +0.54\text{ V}\), 3. \(\text{S}_4\text{O}_6^{2-}(aq) + 2e^- \rightleftharpoons 2\text{S}_2\text{O}_3^{2-}(aq) \quad E^\theta = +0.08\text{ V}\), 4. \(\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s) \quad E^\theta = -0.76\text{ V}\). Which of the following reactions is NOT thermodynamically feasible under standard conditions?
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Worked solution
A reaction is thermodynamically feasible under standard conditions if the standard cell potential (\(E^\theta_{\text{cell}}\)) is positive. For reaction D, the reduction half-reaction is \(\text{S}_4\text{O}_6^{2-} + 2e^- \rightarrow 2\text{S}_2\text{O}_3^{2-}\) (\(E^\theta = +0.08\text{ V}\)) and the oxidation half-reaction is \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\rightleftharpoons \) (\(E^\theta = +0.77\text{ V}\)). Therefore, \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +0.08 - 0.77 = -0.69\text{ V}\). Since \(E^\theta_{\text{cell}}\) is negative, this reaction is not thermodynamically feasible.
Marking scheme
1 mark for calculating standard cell potentials and identifying the non-feasible reaction with negative cell potential. Award 1 mark for choosing option D.
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