An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 13 (Multiple Choice)
Answer all 40 questions by choosing the correct option A, B, C or D on the separate answer sheet.
40 Question · 40 marks
Question 1 · MCQ
1 marks
10 cm\(^3\) of a gaseous hydrocarbon was reacted with 50 cm\(^3\) of oxygen (an excess) by combustion. After cooling to room temperature, the total volume of gas remaining was 40 cm\(^3\). Passing this gas mixture through aqueous sodium hydroxide caused the volume of the gas to decrease to 10 cm\(^3\). What is the molecular formula of the hydrocarbon?
A.C3H4
B.C3H6
C.C3H8
D.C4H8
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Worked solution
Let the formula of the hydrocarbon be \(C_xH_y\). 1. The volume of \(CO_2\) produced is equal to the decrease in volume when the gas mixture is passed through aqueous NaOH. Thus, volume of \(CO_2 = 40 \text{ cm}^3 - 10 \text{ cm}^3 = 30 \text{ cm}^3\). 2. Since 10 cm\(^3\) of \(C_xH_y\) produced 30 cm\(^3\) of \(CO_2\), the ratio is \(1 : 3\), which means \(x = 3\). 3. The remaining 10 cm\(^3\) of gas after NaOH treatment is the unreacted excess oxygen. Since the initial volume of oxygen was 50 cm\(^3\), the volume of oxygen reacted was \(50 \text{ cm}^3 - 10 \text{ cm}^3 = 40 \text{ cm}^3\). 4. The combustion equation is: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). Since 10 cm\(^3\) of \(C_xH_y\) reacts with 40 cm\(^3\) of \(O_2\), \(x + \frac{y}{4} = 4\). Since \(x = 3\), we have \(3 + \frac{y}{4} = 4 \Rightarrow \frac{y}{4} = 1 \Rightarrow y = 4\). Therefore, the molecular formula of the hydrocarbon is \(C_3H_4\).
Marking scheme
Award 1 mark for the correct option A. Reject all other options.
Question 2 · MCQ
1 marks
A sample of 4.34 g of a hydrated metal sulfate, \(M_x(SO_4)_y \cdot nH_2O\), is heated until all the water of crystallisation is lost, leaving a residue of anhydrous metal sulfate of mass 2.54 g. In a separate experiment, 0.010 mol of this anhydrous metal sulfate is found to contain 0.020 mol of metal ions, \(M^{z+}\). What is the formula of the hydrated metal sulfate? (The relative formula mass of the anhydrous metal sulfate is 254.)
A.MSO4 . 5H2O
B.M2SO4 . 5H2O
C.MSO4 . 10H2O
D.M2SO4 . 10H2O
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Worked solution
1. Find the mass of water lost: \(4.34 \text{ g} - 2.54 \text{ g} = 1.80 \text{ g}\). 2. Calculate the amount in moles of water lost: \(n(H_2O) = \frac{1.80 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.10 \text{ mol}\). 3. Calculate the amount in moles of anhydrous metal sulfate in the residue: \(n(M_x(SO_4)_y) = \frac{2.54 \text{ g}}{254 \text{ g mol}^{-1}} = 0.010 \text{ mol}\). 4. Determine the ratio of anhydrous salt to water: \(0.010 : 0.10 = 1 : 10\). Thus, \(n = 10\). 5. From the separate experiment, 0.010 mol of anhydrous metal sulfate contains 0.020 mol of \(M^{z+}\) ions, which means 1 mol of formula unit contains 2 mol of \(M^{z+}\) ions. Therefore, \(x = 2\). This corresponds to \(M_2SO_4\). Thus, the formula is \(M_2SO_4 \cdot 10H_2O\).
Marking scheme
Award 1 mark for the correct option D. Reject all other options.
Question 3 · MCQ
1 marks
The standard enthalpy changes of combustion of carbon, hydrogen, and methanoic acid are given in the table:
The standard enthalpy change of formation of liquid methanoic acid is represented by the equation:
\(C(s) + H_2(g) + O_2(g) \rightarrow HCOOH(l)\)
What is the standard enthalpy change of formation, \(\Delta H^\theta_f\), of liquid methanoic acid?
A.-942.3 kJ mol^-1
B.-416.3 kJ mol^-1
C.-155.3 kJ mol^-1
D.+416.3 kJ mol^-1
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Worked solution
By Hess's Law, the enthalpy change of a reaction can be calculated from enthalpy changes of combustion using: \(\Delta H^\theta_r = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\). For the formation reaction: \(\Delta H^\theta_f = [\Delta H^\theta_c(C(s)) + \Delta H^\theta_c(H_2(g))] - [\Delta H^\theta_c(HCOOH(l))]\). Note that the enthalpy change of combustion of \(O_2(g)\) is zero. \(\Delta H^\theta_f = [-393.5 + (-285.8)] - [-263.0]\) \(\Delta H^\theta_f = -679.3 + 263.0 = -416.3 \text{ kJ mol}^{-1}\).
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 4 · MCQ
1 marks
The enthalpy change of solution of anhydrous magnesium sulfate, \(MgSO_4(s)\), is \(-91.2 \text{ kJ mol}^{-1}\). The enthalpy change of solution of hydrated magnesium sulfate, \(MgSO_4 \cdot 7H_2O(s)\), is \(+13.8 \text{ kJ mol}^{-1}\).
The hydration of anhydrous magnesium sulfate is represented by:
What is the standard enthalpy change of hydration, \(\Delta H^\theta_{\text{hyd}}\), of anhydrous magnesium sulfate?
A.-105.0 kJ mol^-1
B.-77.4 kJ mol^-1
C.+77.4 kJ mol^-1
D.+105.0 kJ mol^-1
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Worked solution
Using a Hess's law cycle, we can express the dissolution of anhydrous magnesium sulfate with water as a two-step process: hydration to the heptahydrate followed by dissolution of the heptahydrate. \(\Delta H^\theta_{\text{sol}}[MgSO_4(s)] = \Delta H^\theta_{\text{hyd}} + \Delta H^\theta_{\text{sol}}[MgSO_4 \cdot 7H_2O(s)]\) Substituting the values: \(-91.2 = \Delta H^\theta_{\text{hyd}} + (+13.8)\) \(\Delta H^\theta_{\text{hyd}} = -91.2 - 13.8 = -105.0 \text{ kJ mol}^{-1}\).
Marking scheme
Award 1 mark for the correct option A. Reject all other options.
Question 5 · MCQ
1 marks
An unsymmetrical alkene \(X\) has the molecular formula \(C_5H_{10}\). When \(X\) reacts with cold, concentrated sulfuric acid, followed by warming with water, a mixture of two isomeric alcohols, \(Y\) and \(Z\), is formed. Alcohol \(Y\) is a secondary alcohol that can be oxidised to a ketone. Alcohol \(Z\) is a tertiary alcohol that is resistant to oxidation by acidified potassium dichromate(VI). What is the IUPAC name of the alkene \(X\)?
A.2-methylbut-1-ene
B.2-methylbut-2-ene
C.3-methylbut-1-ene
D.pent-2-ene
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Worked solution
Hydration of 2-methylbut-2-ene, \(CH_3-C(CH_3)=CH-CH_3\), can occur at two different positions across the double bond: 1. Addition of the hydroxyl group to C2 yields 2-methylbutan-2-ol, which is a tertiary alcohol (resistant to oxidation). 2. Addition of the hydroxyl group to C3 yields 3-methylbutan-2-ol, which is a secondary alcohol (can be oxidised to 3-methylbutan-2-one, a ketone). None of the other options yield both a secondary and a tertiary alcohol as isomers.
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 6 · MCQ
1 marks
A hydrocarbon \(W\) with molecular formula \(C_6H_{10}\) is reacted with hot, concentrated, acidified potassium manganate(VII). The only organic product formed is a single compound \(V\) with molecular formula \(C_6H_{10}O_2\). Compound \(V\) reacts with 2,4-dinitrophenylhydrazine (2,4-DNPH) reagent to form an orange precipitate, but does not react with Tollens' reagent. What is the IUPAC name of the hydrocarbon \(W\)?
A.cyclohexene
B.1-methylcyclopentene
C.1,2-dimethylcyclobutene
D.3-methylcyclopentene
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Worked solution
1. \(W\) has 6 carbons and the product \(V\) also has 6 carbons, meaning the ring is cleaved without losing carbon atoms as \(CO_2\). 2. \(V\) reacts with 2,4-DNPH but not Tollens' reagent, indicating it is a diketone (no aldehyde or carboxylic acid groups). 3. For cleavage to yield a diketone, both carbon atoms involved in the alkene double bond must be substituted with alkyl groups (i.e., they must be tertiary carbons with no attached hydrogen atoms). 4. 1,2-dimethylcyclobutene consists of a four-membered ring where the double bond carbons are both bonded to methyl groups. Cleavage of this double bond yields hexane-2,5-dione, which is a diketone.
Marking scheme
Award 1 mark for the correct option C. Reject all other options.
Question 7 · MCQ
1 marks
The successive ionisation energies, in kJ mol\(^{-1}\), for an element \(T\) in Period 3 of the Periodic Table are shown below:
Which statement about element \(T\) or its compounds is correct?
A.Its oxide is basic only.
B.Its chloride exists as a dimer, T2Cl6, in the vapour phase.
C.The element reacts vigorously with cold water to release hydrogen gas.
D.Its oxide has a simple molecular structure with a low melting point.
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Worked solution
The massive jump between the 3rd and 4th ionisation energies (from 2745 to 11578 kJ mol\(^{-1}\)) indicates that there are 3 valence electrons, so element \(T\) is in Group 13. Since it is in Period 3, \(T\) is aluminium (Al). - Aluminium chloride dimerises in the gas phase to form \(Al_2Cl_6\) via coordinate covalent bonding (Option B). - Aluminium oxide is amphoteric, not basic only (Option A). - Aluminium does not react vigorously with cold water due to its passivating oxide layer (Option C). - Aluminium oxide has a giant ionic lattice structure with a very high melting point (Option D).
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 8 · MCQ
1 marks
Which statement correctly explains why the first ionisation energy of sulfur is lower than that of phosphorus?
A.The 3p electrons in a sulfur atom are more shielded from the nucleus than those in a phosphorus atom.
B.In a sulfur atom, there is spin-pair repulsion between two electrons in one of the 3p orbitals.
C.The outer electron of a sulfur atom is in a higher energy subshell than that of a phosphorus atom.
D.The nuclear charge of a sulfur atom is less than that of a phosphorus atom.
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Worked solution
Phosphorus has the outer electronic configuration \(3s^2 3p^3\) where the 3p subshell is half-filled with three unpaired electrons. Sulfur has the configuration \(3s^2 3p^4\). In sulfur, one of the 3p orbitals is occupied by two electrons. The mutual repulsion between these paired electrons (spin-pair repulsion) makes it easier to remove the outer electron from sulfur than from phosphorus, resulting in a lower first ionisation energy.
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 9 · MCQ
1 marks
A mixture of 0.10 mol of anhydrous magnesium nitrate, \( \text{Mg(NO}_3\text{)}_2 \), and 0.10 mol of anhydrous magnesium carbonate, \( \text{MgCO}_3 \), is heated strongly until both compounds have completely decomposed.
The gaseous products are collected and allowed to cool to room temperature and pressure (r.t.p.).
What is the total volume of gas collected at r.t.p.?
(Assume all gaseous products behave ideally. \( \text{NO}_2 \) remains gaseous at r.t.p.)
A.2.4 dm3
B.6.0 dm3
C.8.4 dm3
D.12.0 dm3
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Worked solution
First, write the balanced chemical equations for the thermal decomposition of both magnesium compounds:
From 0.10 mol of \( \text{MgCO}_3 \): - Moles of \( \text{CO}_2(g) = 0.10 \text{ mol} \)
Total moles of gas collected at r.t.p. = \( 0.25 \text{ mol} + 0.10 \text{ mol} = 0.35 \text{ mol} \)
Using the molar volume of a gas at r.t.p. (\( 24.0 \text{ dm}^3\text{ mol}^{-1} \)): \( \text{Volume of gas} = 0.35 \text{ mol} \times 24.0 \text{ dm}^3\text{ mol}^{-1} = 8.4 \text{ dm}^3 \).
Marking scheme
Award 1 mark for the correct option (C).
Breakdown: - Identify correct stoichiometry of both decompositions (0.25 mol of gas from nitrate and 0.10 mol of gas from carbonate). - Calculate total moles of gas as 0.35 mol. - Convert to volume at r.t.p. by multiplying by 24.0 to yield 8.4 dm3.
Question 10 · MCQ
1 marks
An excess of solid zinc powder is added to \( 50.0 \text{ cm}^3 \) of \( 0.200 \text{ mol dm}^{-3} \) silver nitrate solution.
2. Determine the theoretical yield of silver in moles: Based on the 1:1 stoichiometry between \( \text{Ag}^+ \) and \( \text{Ag} \), the theoretical moles of \( \text{Ag} \) produced is \( 0.0100 \text{ mol} \).
3. Calculate the theoretical mass of silver: \( \text{Theoretical mass} = 0.0100 \text{ mol} \times 107.9 \text{ g mol}^{-1} = 1.079 \text{ g} \)
What is the standard enthalpy of formation of liquid methanol in \( \text{kJ mol}^{-1} \)?
A.-1405.3 kJ mol-1
B.-239.1 kJ mol-1
C.+46.7 kJ mol-1
D.+239.1 kJ mol-1
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Worked solution
The equation for the formation of liquid methanol is: \( \text{C}(s) + 2\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CH}_3\text{OH}(l) \)
Using Hess's Law and standard enthalpies of combustion: \( \Delta H_f^\ominus = \sum \Delta H_c^\ominus (\text{reactants}) - \sum \Delta H_c^\ominus (\text{products}) \)
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Worked solution
Using Hess's Law based on bond energies: \( \Delta H^\ominus = \sum \text{Bond energies of bonds broken} - \sum \text{Bond energies of bonds formed} \)
Let the bond energy of \( \text{C=C} \) be \( x \).
Bonds broken: - 1 \( \text{C=C} \) bond = \( x \) - 1 \( \text{C-C} \) bond = 347 - 6 \( \text{C-H} \) bonds = \( 6 \times 413 = 2478 \) - 1 \( \text{H-H} \) bond = 436 Total bonds broken = \( x + 347 + 2478 + 436 = x + 3261 \text{ kJ mol}^{-1} \)
Set up the equation: \( -124 = (x + 3261) - 3998 \) \( -124 = x - 737 \) \( x = 737 - 124 = 613 \text{ kJ mol}^{-1} \).
Marking scheme
Award 1 mark for the correct option (B).
Breakdown: - Correctly count all chemical bonds broken and formed. - Solve the equation: bonds broken minus bonds formed = -124 kJ mol-1. - Obtain 613 kJ mol-1.
Question 13 · MCQ
1 marks
When 2-methylbut-2-ene reacts with cold, concentrated sulfuric acid, followed by warming with water, a major organic product is formed.
Which intermediate carbocation is formed in the rate-determining step to yield this major product?
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Worked solution
2-methylbut-2-ene has the structure \( \text{(CH}_3\text{)}_2\text{C=CHCH}_3 \).
During the electrophilic addition of \( \text{H}^+ \) (from the acid) in the first, rate-determining step, two carbocations can be formed:
1) If the hydrogen adds to C3, a tertiary carbocation is formed at C2: \( \text{(CH}_3\text{)}_2\text{C}^+\text{CH}_2\text{CH}_3 \). 2) If the hydrogen adds to C2, a secondary carbocation is formed at C3: \( \text{(CH}_3\text{)}_2\text{CH}\text{C}^+\text{HCH}_3 \).
A tertiary carbocation is much more stable than a secondary carbocation due to the electron-donating inductive effect of three alkyl groups, which disperses the positive charge on the carbon. Therefore, the reaction proceeds preferentially through the tertiary carbocation intermediate, yielding the major product.
Marking scheme
Award 1 mark for the correct option (A).
Breakdown: - Identify the correct structure of 2-methylbut-2-ene. - Explain carbocation stability (tertiary carbocation is more stable than secondary carbocation). - Identify the correct tertiary intermediate formula.
Question 14 · MCQ
1 marks
An unknown alkene, X, is reacted with hot, concentrated, acidified potassium manganate(VII), \( \text{KMnO}_4 \).
The only organic products obtained are propanone, \( \text{CH}_3\text{COCH}_3 \), and propanoic acid, \( \text{CH}_3\text{CH}_2\text{CO}_2\text{H} \).
What is the IUPAC name of alkene X?
A.2-methylpent-2-ene
B.4-methylpent-2-ene
C.3-methylpent-2-ene
D.2-methylhex-2-ene
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Worked solution
Oxidative cleavage of an alkene using hot, concentrated, acidified \( \text{KMnO}_4 \) cleaves the \( \text{C=C} \) double bond: - A ketone product (propanone, \( \text{(CH}_3\text{)}_2\text{C=O} \)) indicates that one of the double-bonded carbons was attached to two methyl groups: \( \text{(CH}_3\text{)}_2\text{C}= \). - A carboxylic acid product (propanoic acid, \( \text{CH}_3\text{CH}_2\text{CO}_2\text{H} \)) indicates that the other double-bonded carbon was attached to a hydrogen atom and an ethyl group: \( =\text{CHCH}_2\text{CH}_3 \).
Joining these two pieces together gives alkene X: \( \text{(CH}_3\text{)}_2\text{C=CHCH}_2\text{CH}_3 \)
The longest continuous carbon chain containing the double bond has 5 carbons (pentene). Numbering from the left gives the double bond at carbon 2, and a methyl branch at carbon 2. Thus, the IUPAC name is 2-methylpent-2-ene.
Marking scheme
Award 1 mark for the correct option (A).
Breakdown: - Deduce the structural formula of alkene X from cleavage products. - Apply IUPAC rules correctly to name the alkene as 2-methylpent-2-ene.
Question 15 · MCQ
1 marks
The table below shows the first six successive ionisation energies, in \( \text{kJ mol}^{-1} \), of an element Y in Period 3 of the Periodic Table.
A.Its first ionisation energy is higher than that of magnesium.
B.Its oxide is amphoteric and reacts with both acids and alkalis.
C.Its chloride has a giant ionic structure with a very high melting point.
D.It reacts with cold water to produce a solution with a pH of 13.
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Worked solution
1. Identify the group of Y: There is a very large jump between the 3rd and 4th ionisation energies (from 2745 to 11578 kJ mol-1). This indicates that the 4th electron is removed from an inner core shell, meaning element Y has 3 valence electrons and is in Group 13.
2. Identify the element: Since Y is in Period 3 and Group 13, element Y is aluminium (Al).
3. Evaluate the statements: - A: Aluminium's first ionisation energy is lower than that of magnesium due to its 3p electron being in a higher energy subshell with greater shielding. - B: Aluminium oxide (\( \text{Al}_2\text{O}_3 \)) is amphoteric, reacting with both acids and alkalis. This is correct. - C: Aluminium chloride (\( \text{AlCl}_3 \)) has a covalent dimer structure at moderate temperatures, and is not giant ionic. - D: Aluminium does not react readily with cold water due to its passivating oxide layer, and does not form a solution of pH 13.
Marking scheme
Award 1 mark for the correct option (B).
Breakdown: - Identify element Y as Aluminium (Al) from the large jump between the 3rd and 4th ionisation energies. - Recall that aluminium oxide is amphoteric.
Question 16 · MCQ
1 marks
The first ionisation energies of the elements in Period 3 generally increase from left to right. However, there are exceptions (dips) in this trend.
Between which two elements in Period 3 is there a decrease in first ionisation energy due to spin-pair repulsion in a p-orbital?
A.Magnesium and aluminium
B.Phosphorus and sulfur
C.Silicon and phosphorus
D.Sulfur and chlorine
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Worked solution
Let's analyze the electronic configurations of Period 3 elements where dips occur:
1) Between Magnesium (\( 1s^2 2s^2 2p^6 3s^2 \)) and Aluminium (\( 1s^2 2s^2 2p^6 3s^2 3p^1 \)): The dip occurs because the 3p electron of Al is in a higher energy subshell and is shielded by the 3s electrons.
2) Between Phosphorus (\( [\text{Ne}] 3s^2 3p_x^1 3p_y^1 3p_z^1 \)) and Sulfur (\( [\text{Ne}] 3s^2 3p_x^2 3p_y^1 3p_z^1 \)): In phosphorus, the three 3p electrons occupy separate orbitals singly. In sulfur, the fourth 3p electron must pair up in one of these 3p orbitals. The repulsion between the two spin-paired electrons in the same orbital makes it easier to remove the outer electron from sulfur than phosphorus, leading to a decrease in first ionisation energy.
Thus, spin-pair repulsion causes the dip between phosphorus and sulfur.
Marking scheme
Award 1 mark for the correct option (B).
Breakdown: - Correctly distinguish the two anomalous dips in the trend of first ionisation energies across Period 3. - Identify that the dip between P and S is due to spin-pair repulsion.
Question 17 · MCQ
1 marks
A \(10\text{ cm}^3\) sample of gaseous propane, \(\text{C}_3\text{H}_8\), was mixed with \(70\text{ cm}^3\) of oxygen (an excess) and exploded in a sealed vessel. The vessel was cooled back to room temperature and pressure, and the remaining gas mixture was then shaken with excess aqueous sodium hydroxide. What is the total reduction in the volume of gas from the start of the experiment to the end of the procedure? (All gas volumes are measured at room temperature and pressure.)
A.\(30\text{ cm}^3\)
B.\(40\text{ cm}^3\)
C.\(50\text{ cm}^3\)
D.\(60\text{ cm}^3\)
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Worked solution
The balanced chemical equation for the complete combustion of propane is: \(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\). Initially, we have \(10\text{ cm}^3\) of propane and \(70\text{ cm}^3\) of oxygen, giving a total starting volume of \(80\text{ cm}^3\). Propane is the limiting reactant, so \(10\text{ cm}^3\) of propane reacts with \(5 \times 10 = 50\text{ cm}^3\) of oxygen. This leaves \(70 - 50 = 20\text{ cm}^3\) of unreacted oxygen. The reaction produces \(3 \times 10 = 30\text{ cm}^3\) of carbon dioxide gas. Water is a liquid at room temperature and pressure, so its volume is negligible. Thus, after the reaction and cooling, the gas mixture contains \(20\text{ cm}^3\) of oxygen and \(30\text{ cm}^3\) of carbon dioxide. Shaking this mixture with aqueous sodium hydroxide completely absorbs the acidic carbon dioxide gas, leaving only \(20\text{ cm}^3\) of oxygen. The final gas volume is therefore \(20\text{ cm}^3\). The total reduction in gas volume from the start is \(80\text{ cm}^3 - 20\text{ cm}^3 = 60\text{ cm}^3\).
Marking scheme
1 mark for the correct option D. Method: Identify initial total volume (80 cm3). Use reaction stoichiometry to find volume of unreacted oxygen (20 cm3) and produced carbon dioxide (30 cm3). Determine final volume after absorption of carbon dioxide (20 cm3). Calculate the overall change in volume.
Question 18 · MCQ
1 marks
A sample of \(5.72\text{ g}\) of hydrated sodium carbonate, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\), is dissolved in distilled water and made up to \(250.0\text{ cm}^3\) of solution in a volumetric flask. A \(25.0\text{ cm}^3\) portion of this solution is titrated against \(0.200\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\). It requires \(20.0\text{ cm}^3\) of the acid for complete neutralisation: \(\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\). What is the value of \(x\)? (Relative atomic masses: \(\text{H}=1.0\), \(\text{C}=12.0\), \(\text{O}=16.0\), \(\text{Na}=23.0\))
A.5
B.7
C.10
D.12
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Worked solution
First, calculate the moles of hydrochloric acid used in the titration: \(n(\text{HCl}) = 0.0200\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.00400\text{ mol}\). According to the equation, \(2\text{ moles}\) of \(\text{HCl}\) react with \(1\text{ mole}\) of \(\text{Na}_2\text{CO}_3\). Therefore, the number of moles of \(\text{Na}_2\text{CO}_3\) in the \(25.0\text{ cm}^3\) aliquot is: \(n(\text{Na}_2\text{CO}_3) = \frac{0.00400}{2} = 0.00200\text{ mol}\). Since the total volume of the solution is \(250.0\text{ cm}^3\), the total number of moles of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\) in the original sample is: \(0.00200 \times 10 = 0.0200\text{ mol}\). The molar mass of the hydrated salt is: \(M_r = \frac{5.72\text{ g}}{0.0200\text{ mol}} = 286\text{ g mol}^{-1}\). The formula mass of anhydrous sodium carbonate is: \(2 \times 23.0 + 12.0 + 3 \times 16.0 = 106.0\text{ g mol}^{-1}\). Therefore, the mass of the water of crystallisation in one mole of the hydrated salt is: \(286 - 106 = 180\text{ g mol}^{-1}\). Since the molar mass of water is \(18.0\text{ g mol}^{-1}\), the value of \(x\) is: \(x = \frac{180}{18.0} = 10\).
Marking scheme
1 mark for the correct option C. Method: Calculate moles of HCl, relate to moles of sodium carbonate in the aliquot, scale up to the full 250 cm3 volume, determine the molar mass of the hydrated compound, subtract the mass of anhydrous sodium carbonate, and divide by the molar mass of water.
Question 19 · MCQ
1 marks
The standard enthalpy change of combustion of iron, \(\text{Fe(s)}\), to form iron(III) oxide, \(\text{Fe}_2\text{O}_3(\text{s})\), is \(-411\text{ kJ mol}^{-1}\). The standard enthalpy change of combustion of carbon (graphite) to form carbon dioxide, \(\text{CO}_2(\text{g})\), is \(-394\text{ kJ mol}^{-1}\). What is the enthalpy change for the reduction of iron(III) oxide by carbon according to the equation shown below? \(2\text{Fe}_2\text{O}_3(\text{s}) + 3\text{C(s)} \rightarrow 4\text{Fe(s)} + 3\text{CO}_2(\text{g})\)
A.\(-462\text{ kJ mol}^{-1}\)
B.\(-360\text{ kJ mol}^{-1}\)
C.\(+360\text{ kJ mol}^{-1}\)
D.\(+462\text{ kJ mol}^{-1}\)
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Worked solution
The standard enthalpy of formation, \(\Delta H_f^\ominus\), of a substance can be related to the enthalpy of combustion. The enthalpy of combustion of iron is for the reaction: \(\text{Fe(s)} + \frac{3}{4}\text{O}_2(\text{g}) \rightarrow \frac{1}{2}\text{Fe}_2\text{O}_3(\text{s})\) with \(\Delta H = -411\text{ kJ mol}^{-1}\). Thus, the standard enthalpy of formation of iron(III) oxide is: \(\Delta H_f^\ominus[\text{Fe}_2\text{O}_3(\text{s})] = 2 \times (-411) = -822\text{ kJ mol}^{-1}\). The standard enthalpy of formation of carbon dioxide is equal to the combustion enthalpy of carbon: \(\Delta H_f^\ominus[\text{CO}_2(\text{g})] = -394\text{ kJ mol}^{-1}\). For the reaction: \(2\text{Fe}_2\text{O}_3(\text{s}) + 3\text{C(s)} \rightarrow 4\text{Fe(s)} + 3\text{CO}_2(\text{g})\), the enthalpy change is calculated using Hess's Law: \(\Delta H_r^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants}) = [4 \times 0 + 3 \times (-394)] - [2 \times (-822) + 3 \times 0] = -1182 - (-1644) = +462\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct option D. Method: Convert standard enthalpy of combustion of iron to standard enthalpy of formation of iron(III) oxide (multiply by 2). Use Hess's Law formula (products minus reactants) to calculate the final enthalpy change of reaction.
Question 20 · MCQ
1 marks
The standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), at 298 K, for three substances are given in the table below:
What is the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH(l)}\)?
A.\(-513\text{ kJ mol}^{-1}\)
B.\(-394\text{ kJ mol}^{-1}\)
C.\(+513\text{ kJ mol}^{-1}\)
D.\(+847\text{ kJ mol}^{-1}\)
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Worked solution
The equation representing the formation of propanoic acid from its elements in their standard states is: \(3\text{C(s)} + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{CH}_2\text{COOH(l)}\). Using Hess's Law and the enthalpies of combustion: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). Therefore: \(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C}) + 3 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{propanoic acid}) = [3(-394) + 3(-286)] - (-1527) = [-1182 - 858] + 1527 = -2040 + 1527 = -513\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct option A. Method: Write the correct formation equation for propanoic acid, construct the Hess cycle using combustion data, apply the formula (sum of combustion of reactants minus combustion of products), and perform the arithmetic calculation.
Question 21 · MCQ
1 marks
An alkene, \(X\), reacts with hot, concentrated, acidified \(\text{KMnO}_4\) to give propanone as the only organic product. Which statement about \(X\) and its reactions is correct?
A.\(X\) has the molecular formula \(\text{C}_5\text{H}_{10}\).
B.\(X\) displays cis-trans (geometric) isomerism.
C.The reaction of \(X\) with cold, dilute, acidified \(\text{KMnO}_4\) produces 2,3-dimethylbutane-2,3-diol.
D.The reaction of \(X\) with steam in the presence of an acid catalyst produces a secondary alcohol.
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Worked solution
When an alkene is cleaved by hot, concentrated, acidified \(\text{KMnO}_4\) and gives propanone as the only organic product, the alkene must be symmetrical and contain two identical \((\text{CH}_3)_2\text{C}=\) groups. This identifies alkene \(X\) as 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\). Let us evaluate the options: - A: The molecular formula of 2,3-dimethylbut-2-ene is \(\text{C}_6\text{H}_{12}\), not \(\text{C}_5\text{H}_{10}\). - B: Each carbon atom in the double bond is bonded to two identical methyl groups, so \(X\) does not exhibit cis-trans isomerism. - C: The reaction of alkenes with cold, dilute, acidified \(\text{KMnO}_4\) results in dihydroxylation across the double bond, converting 2,3-dimethylbut-2-ene to 2,3-dimethylbutane-2,3-diol. This is correct. - D: Reaction with steam in the presence of an acid catalyst (hydration) forms 2,3-dimethylbutan-2-ol, which is a tertiary alcohol, not a secondary alcohol.
Marking scheme
1 mark for the correct option C. Method: Identify the structure of the alkene X from the oxidation product, and analyze each statement concerning its structural formula, stereoisomerism, dihydroxylation reaction, and hydration reaction.
Question 22 · MCQ
1 marks
When 2-methylbut-2-ene reacts with hydrogen bromide, \(\text{HBr}\), two bromoalkanes are formed. One is the major product and the other is the minor product. Which option correctly describes the first step of the electrophilic addition mechanism and identifies the major organic product?
A.The \(\pi\) bond of the alkene attacks the \(\delta+\) hydrogen atom of \(\text{HBr}\) to form a tertiary carbocation; the major product is 2-bromo-2-methylbutane.
B.The \(\pi\) bond of the alkene attacks the \(\delta+\) hydrogen atom of \(\text{HBr}\) to form a secondary carbocation; the major product is 2-bromo-3-methylbutane.
C.A bromide ion attacks a carbon atom of the \(\text{C}=\text{C}\) bond to form a carbanion; the major product is 2-bromo-2-methylbutane.
D.A bromine radical attacks a carbon atom of the \(\text{C}=\text{C}\) bond to form a radical intermediate; the major product is 2-bromo-3-methylbutane.
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Worked solution
The mechanism of electrophilic addition of \(\text{HBr}\) to 2-methylbut-2-ene involves the \(\pi\) electrons of the double bond acting as a nucleophile and attacking the electrophilic, electron-deficient \(\text{H}^{\delta+}\) of the polar \(\text{H}-\text{Br}\) molecule. In the first step, this results in the formation of a carbocation intermediate. The major product is formed via the more stable carbocation. The two possible carbocations are: a tertiary carbocation, \(\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3\), and a secondary carbocation, \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}^+-\text{CH}_3\). The tertiary carbocation is more stable because of the greater electron-releasing inductive effect of the three alkyl groups. Consequently, the bromide ion attacks this tertiary carbocation in the second step to form the major product, 2-bromo-2-methylbutane.
Marking scheme
1 mark for the correct option A. Method: Identify the electrophile and nucleophile in the first step of the electrophilic addition, understand carbocation stability rules (tertiary vs secondary), and determine the structure of the major addition product.
Question 23 · MCQ
1 marks
An element, \(Y\), in Period 3 of the Periodic Table has the following first six successive ionisation energies (in \(\text{kJ mol}^{-1}\)):
A.\(Y\) forms an ionic chloride with the formula \(Y\text{Cl}_3\).
B.The oxide of \(Y\) reacts with water to form an alkaline solution.
C.\(Y\) is in Group 15 of the Periodic Table.
D.The first ionisation energy of the element immediately to the left of \(Y\) in the Periodic Table is higher than that of \(Y\).
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Worked solution
The successive ionisation energies of element \(Y\) show a relatively steady increase for the first five electrons, but a very large jump between the 5th and the 6th ionisation energies (from 6274 to 21269 \(\text{kJ mol}^{-1}\)). This indicates that the 6th electron is removed from a shell closer to the nucleus (an inner quantum shell). Therefore, \(Y\) has 5 valence electrons and belongs to Group 15 of the Periodic Table. In Period 3, this element is Phosphorus (\(\text{P}\)). Let's evaluate the statements: - A: Phosphorus reacts with chlorine to form covalent chlorides like \(\text{PCl}_3\) and \(\text{PCl}_5\), not ionic ones. - B: Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) reacts with water to form phosphoric(V) acid, which is a strongly acidic solution, not alkaline. - C: \(Y\) is in Group 15. This is correct. - D: The element immediately to the left of Phosphorus in Period 3 is Silicon (\(\text{Si}\)), which has a lower first ionisation energy (786 \(\text{kJ mol}^{-1}\)) than Phosphorus (1012 \(\text{kJ mol}^{-1}\)) because Phosphorus has a larger nuclear charge and a stable, half-filled 3p subshell.
Marking scheme
1 mark for the correct option C. Method: Identify the group of the element from the location of the large jump in successive ionisation energies, determine its identity as phosphorus, and assess periodic properties including bonding, acidity/basicity of oxides, and first ionisation energy trends.
Question 24 · MCQ
1 marks
The first ionisation energies of five consecutive elements in Period 3, in order of increasing atomic number, are given below:
- Element 1: 738 \(\text{kJ mol}^{-1}\) - Element 2: 578 \(\text{kJ mol}^{-1}\) - Element 3: 786 \(\text{kJ mol}^{-1}\) - Element 4: 1012 \(\text{kJ mol}^{-1}\) - Element 5: 1000 \(\text{kJ mol}^{-1}\)
Which statement about these elements is correct?
A.Element 2 has a greater nuclear charge than Element 1, and its outer electron is in a 3s orbital.
B.The decrease in first ionisation energy from Element 4 to Element 5 is due to spin-pair repulsion in a 3p orbital.
C.Element 3 has a lower first ionisation energy than Element 4 because its outer electron experiences significantly more shielding.
D.Element 1 is a non-metal that forms a covalent chloride.
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Worked solution
By matching the first ionisation energies of the consecutive Period 3 elements, we can identify them: - Element 1: Magnesium (Mg, \(738\text{ kJ mol}^{-1}\)) - Element 2: Aluminium (Al, \(578\text{ kJ mol}^{-1}\)) - Element 3: Silicon (Si, \(786\text{ kJ mol}^{-1}\)) - Element 4: Phosphorus (P, \(1012\text{ kJ mol}^{-1}\)) - Element 5: Sulfur (S, \(1000\text{ kJ mol}^{-1}\))
Let us analyse the options: - A: Element 2 (Aluminium) has a greater nuclear charge than Element 1 (Magnesium), but its outer electron is in a 3p orbital, which is higher in energy than the 3s orbital of Magnesium. - B: The decrease in first ionisation energy from Element 4 (P, \(3\text{p}^3\)) to Element 5 (S, \(3\text{p}^4\)) occurs because sulfur has paired electrons in one of its 3p orbitals. The mutual repulsion between these two paired electrons (spin-pair repulsion) makes it easier to remove one of them compared to the unpaired 3p electrons in phosphorus. This is correct. - C: Element 3 (Si) has a lower first ionisation energy than Element 4 (P) because Phosphorus has a greater nuclear charge, while the shielding of the outer electrons remains relatively constant across the period. - D: Element 1 (Magnesium) is a metal that forms an ionic chloride, \(\text{MgCl}_2\).
Marking scheme
1 mark for the correct option B. Method: Identify the elements from the list of first ionisation energies, use knowledge of electronic configurations of Period 3 elements, and apply the concept of spin-pair repulsion to explain the drop in ionisation energy between Group 15 and Group 16.
Question 25 · MCQ
1 marks
A sample of 0.200 mol of an anhydrous metal nitrate, \( \text{M(NO}_3)_2 \), is heated strongly until it decomposes completely.\ \ \( 2\text{M(NO}_3)_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g}) \)\ \ What is the total volume of gas, measured at room temperature and pressure (r.t.p.), produced during this decomposition?\ \ [Molar volume of gas at r.t.p. = \( 24.0\text{ dm}^3\text{ mol}^{-1} \)]
A.\( 2.40\text{ dm}^3 \)
B.\( 4.80\text{ dm}^3 \)
C.\( 9.60\text{ dm}^3 \)
D.\( 12.0\text{ dm}^3 \)
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Worked solution
1. Identify the molar ratio between \( \text{M(NO}_3)_2 \) and the gaseous products.\ According to the equation, \( 2\text{ mol} \) of \( \text{M(NO}_3)_2 \) produces \( 4\text{ mol} \) of \( \text{NO}_2(\text{g}) \) and \( 1\text{ mol} \) of \( \text{O}_2(\text{g}) \), giving a total of \( 5\text{ mol} \) of gas.\ \ 2. Calculate the moles of gas produced from \( 0.200\text{ mol} \) of \( \text{M(NO}_3)_2 \):\ \( \text{Moles of gas} = 0.200\text{ mol} \times \frac{5}{2} = 0.500\text{ mol} \)\ \ 3. Calculate the volume of gas at r.t.p.:\ \( \text{Volume} = 0.500\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 12.0\text{ dm}^3 \).
Marking scheme
1 mark for the correct answer D.\ - Award 1 mark for correct calculation of total gas volume (12.0 dm³).\ - Reject other options: A (only O2 volume), C (only NO2 volume), B (incorrect mole ratio).
Question 26 · MCQ
1 marks
A \( 25.0\text{ cm}^3 \) sample of \( 0.120\text{ mol dm}^{-3} \) sodium thiosulfate solution, \( \text{Na}_2\text{S}_2\text{O}_3 \), reacts completely with \( 20.0\text{ cm}^3 \) of an aqueous iodine solution, \( \text{I}_2 \), according to the equation:\ \ \( 2\text{S}_2\text{O}_3^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \)\ \ What is the concentration of the iodine solution?
A.\( 0.0375\text{ mol dm}^{-3} \)
B.\( 0.0750\text{ mol dm}^{-3} \)
C.\( 0.150\text{ mol dm}^{-3} \)
D.\( 0.300\text{ mol dm}^{-3} \)
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Worked solution
1. Calculate the moles of thiosulfate ions used:\ \( n(\text{S}_2\text{O}_3^{2-}) = 0.120\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.00300\text{ mol} \)\ \ 2. Determine the moles of iodine that react:\ According to the stoichiometric equation, \( 2\text{ mol} \) of \( \text{S}_2\text{O}_3^{2-} \) reacts with \( 1\text{ mol} \) of \( \text{I}_2 \).\ \( n(\text{I}_2) = \frac{0.00300\text{ mol}}{2} = 0.00150\text{ mol} \)\ \ 3. Calculate the concentration of the iodine solution:\ \( [\text{I}_2] = \frac{0.00150\text{ mol}}{0.0200\text{ dm}^3} = 0.0750\text{ mol dm}^{-3} \).
Marking scheme
1 mark for the correct answer B.\ - Award 1 mark for calculating the correct concentration of 0.0750 mol dm⁻³.\ - Reject A (if divided by 2 twice), C (if multiplied by 2 instead of dividing), D (other scaling error).
Question 27 · MCQ
1 marks
The standard enthalpy changes of combustion, \( \Delta H^\ominus_\text{c} \), for carbon, hydrogen, and propanoic acid, \( \text{CH}_3\text{CH}_2\text{CO}_2\text{H}(\text{l}) \), are shown in the table.\ \ | Substance | \( \Delta H^\ominus_\text{c} / \text{kJ mol}^{-1} \) |\ | :--- | :---: |\ | \( \text{C(s)} \) | \( -393.5 \) |\ | \( \text{H}_2(\text{g}) \) | \( -285.8 \) |\ | \( \text{CH}_3\text{CH}_2\text{CO}_2\text{H(l)} \) | \( -1527.0 \) |\ \ What is the standard enthalpy change of formation, \( \Delta H^\ominus_\text{f} \), of propanoic acid?\ \ \( 3\text{C(s)} + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{CH}_2\text{CO}_2\text{H(l)} \)
A.\( -510.9\text{ kJ mol}^{-1} \)
B.\( -847.7\text{ kJ mol}^{-1} \)
C.\( +510.9\text{ kJ mol}^{-1} \)
D.\( -2037.9\text{ kJ mol}^{-1} \)
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1 mark for the correct answer A.\ - Award 1 mark for the correct value with correct sign (-510.9 kJ mol⁻¹).\ - Reject C (+510.9 kJ mol⁻¹ due to sign inversion), D (-2037.9 kJ mol⁻¹ from forgetting the product term), B (-847.7 kJ mol⁻¹ from stoich error).
Question 28 · MCQ
1 marks
The equation for the hydration of ethene to ethanol is shown:\ \ \( \text{C}_2\text{H}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{g}) \)\ \ The standard enthalpy changes of formation, \( \Delta H^\ominus_\text{f} \), are given:\ - \( \Delta H^\ominus_\text{f}[\text{C}_2\text{H}_4(\text{g})] = +52.3\text{ kJ mol}^{-1} \)\ - \( \Delta H^\ominus_\text{f}[\text{H}_2\text{O}(\text{g})] = -241.8\text{ kJ mol}^{-1} \)\ - \( \Delta H^\ominus_\text{f}[\text{C}_2\text{H}_5\text{OH}(\text{g})] = -235.1\text{ kJ mol}^{-1} \)\ \ What is the standard enthalpy change of this hydration reaction?
A.\( -45.6\text{ kJ mol}^{-1} \)
B.\( -324.6\text{ kJ mol}^{-1} \)
C.\( +45.6\text{ kJ mol}^{-1} \)
D.\( -529.2\text{ kJ mol}^{-1} \)
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1 mark for the correct answer A.\ - Award 1 mark for correct value and sign (-45.6 kJ mol⁻¹).\ - Reject C (+45.6 kJ mol⁻¹ due to inverted subtraction), B or D (due to incorrect handling of signs or summation of values).
Question 29 · MCQ
1 marks
When 2-methylbut-1-ene, \( \text{CH}_2\text{=C(CH}_3)\text{CH}_2\text{CH}_3 \), reacts with hydrogen bromide, \( \text{HBr} \), the major organic product formed is 2-bromo-2-methylbutane.\ \ Which statement correctly explains why this is the major product?
A.The tertiary carbocation intermediate formed is more stable than the alternative primary carbocation intermediate due to the electron-donating inductive effect of three alkyl groups.
B.The secondary carbocation intermediate formed is more stable than the alternative primary carbocation intermediate due to the electron-withdrawing inductive effect of the alkyl groups.
C.The tertiary carbocation intermediate is more stable because the bromine atom is highly electronegative and stabilizes the positive charge prior to attack.
D.The primary carbocation intermediate is more stable because it is less sterically hindered, allowing faster reaction with the bromide ion.
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Worked solution
During electrophilic addition of \( \text{HBr} \) to \( \text{CH}_2\text{=C(CH}_3)\text{CH}_2\text{CH}_3 \), the electrophile \( \text{H}^+ \) adds first. \ - Addition of \( \text{H}^+ \) to the \( \text{C}_1 \) carbon atom forms a tertiary carbocation: \( \text{CH}_3\text{-C}^+(\text{CH}_3)\text{CH}_2\text{CH}_3 \).\ - Addition of \( \text{H}^+ \) to the \( \text{C}_2 \) carbon atom forms a primary carbocation: \( ^+\text{CH}_2\text{-CH(CH}_3)\text{CH}_2\text{CH}_3 \).\ \ The tertiary carbocation intermediate is significantly more stable than the primary carbocation because the three electron-donating alkyl groups disperse the positive charge on the carbon. Therefore, the pathway via the tertiary carbocation is favored, and subsequent attack by \( \text{Br}^- \) yields 2-bromo-2-methylbutane as the major product.
Marking scheme
1 mark for the correct answer A.\ - Award 1 mark for identifying that the tertiary carbocation is stabilized by the electron-donating inductive effect of the alkyl groups.\ - Reject B (incorrect carbocation class/inductive effect description), C (incorrect reasoning about bromine stabilizing the cation), D (incorrect relative stability of carbocations).
Question 30 · MCQ
1 marks
An organic compound, \( \text{X} \), has the molecular formula \( \text{C}_6\text{H}_8 \).\ \ When \( \text{X} \) is reacted with hot, concentrated, acidified potassium manganate(VII), the only organic product obtained is propanedioic acid, \( \text{HOOC-CH}_2\text{-COOH} \).\ \ Which compound is \( \text{X} \)?
A.cyclohexa-1,3-diene
B.cyclohexa-1,4-diene
C.hexa-1,3-diene
D.hexa-1,4-diene
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Worked solution
1. Analyze the molecular formula \( \text{C}_6\text{H}_8 \). This corresponds to a hydrocarbon with a total of three degrees of unsaturation (e.g., a ring with two double bonds).\ 2. Hot, concentrated, acidified \( \text{KMnO}_4 \) cleaves carbon-carbon double bonds (\( \text{C=C} \)). If the double bond has a hydrogen atom attached to each carbon (\( \text{-CH=CH-} \)), it is oxidized to two carboxylic acid groups (\( \text{-COOH} \)).\ 3. Let's test cyclohexa-1,4-diene:\ - Structure: a cyclic 6-carbon ring with double bonds at 1 and 4.\ - The two \( \text{-CH}_2- \) groups at positions 3 and 6 are separated by the double bonds at 1-2 and 4-5.\ - Cleaving both \( \text{C=C} \) double bonds splits the ring into two identical fragments: each containing one \( \text{-CH}_2- \) group bonded to two \( \text{-CH=} \) groups that oxidize to \( \text{-COOH} \) groups.\ - This yields two molecules of \( \text{HOOC-CH}_2\text{-COOH} \) (propanedioic acid) and no other organic products.\ \ 4. For cyclohexa-1,3-diene:\ - Cleavage of \( \text{C1=C2} \) and \( \text{C3=C4} \) yields ethanedioic acid (\( \text{HOOC-COOH} \)) and butanedioic acid (\( \text{HOOC-CH}_2\text{-CH}_2\text{-COOH} \)).\ 5. For acyclic dienes like hexa-1,3-diene or hexa-1,4-diene, cleavage would produce different mixtures of smaller carboxylic acids and potentially carbon dioxide.
Marking scheme
1 mark for the correct answer B.\ - Award 1 mark for identifying cyclohexa-1,4-diene as the precursor that symmetrically cleaves to give propanedioic acid.\ - Reject A (yields ethanedioic and butanedioic acids), C and D (yield different acyclic products/carbon dioxide).
Question 31 · MCQ
1 marks
The table shows the successive ionisation energies, in \( \text{kJ mol}^{-1} \), of an element \( \text{T} \) in Period 3 of the Periodic Table.\ \ | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |\ | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\ | 1000 | 2250 | 3360 | 4560 | 7010 | 8500 | 27100 |\ \ In which group of the Periodic Table is element \( \text{T} \)?
A.Group 14
B.Group 15
C.Group 16
D.Group 17
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Worked solution
1. Look at the ratios or differences between successive ionisation energies to identify the major jump, which corresponds to removing an electron from a lower principal quantum shell:\ - \( \text{IE}_2 / \text{IE}_1 = 2.25 \)\ - \( \text{IE}_3 / \text{IE}_2 = 1.49 \)\ - \( \text{IE}_4 / \text{IE}_3 = 1.36 \)\ - \( \text{IE}_5 / \text{IE}_4 = 1.54 \)\ - \( \text{IE}_6 / \text{IE}_5 = 1.21 \)\ - \( \text{IE}_7 / \text{IE}_6 = 3.19 \) (A very large jump: \( 27100 - 8500 = 18600\text{ kJ mol}^{-1} \)).\ \ 2. The sudden jump between the 6th and 7th ionisation energies indicates that the 7th electron is removed from an inner shell.\ 3. Therefore, element \( \text{T} \) has 6 valence electrons in its outer shell.\ 4. An element in Period 3 with 6 outer electrons belongs to Group 16 (sulfur).
Marking scheme
1 mark for the correct answer C.\ - Award 1 mark for correctly identifying the jump between the 6th and 7th ionisation energies and assigning it to Group 16.\ - Reject A, B, and D because they correspond to 4, 5, or 7 valence electrons respectively.
Question 32 · MCQ
1 marks
Which statement correctly explains why the first ionisation energy of sulfur is lower than that of phosphorus?
A.The outer electron in a sulfur atom is in a \( 3\text{d} \) orbital, which is higher in energy and more shielded than the \( 3\text{p} \) orbital of phosphorus.
B.In sulfur, the outer electron is removed from a paired \( 3\text{p} \) orbital, and the spin-pair repulsion makes it easier to remove than an unpaired electron from a phosphorus atom.
C.Sulfur has a smaller nuclear charge than phosphorus, resulting in a weaker attraction between the nucleus and the outer electrons.
D.The outer electrons of phosphorus experience significantly less shielding from the inner-shell electrons than those of sulfur.
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Worked solution
1. Write the electronic configurations of phosphorus and sulfur:\ - Phosphorus (atomic number 15): \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3 \) (three unpaired electrons in the three \( 3\text{p} \) orbitals).\ - Sulfur (atomic number 16): \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^4 \) (one of the \( 3\text{p} \) orbitals contains a pair of electrons with opposite spins).\ \ 2. State the effect of electron pairing:\ In the paired \( 3\text{p} \) orbital of sulfur, there is mutual repulsion between the two electrons (spin-pair repulsion). This repulsion makes it easier to remove one of these paired electrons compared to removing one of the unpaired electrons from the half-filled \( 3\text{p} \) subshell in phosphorus, despite sulfur having a higher nuclear charge.
Marking scheme
1 mark for the correct answer B.\ - Award 1 mark for correctly identifying spin-pair repulsion as the reason for the lower first ionisation energy of sulfur.\ - Reject A (sulfur's outer electron is in a 3p orbital, not 3d), C (sulfur has a larger nuclear charge of +16 than phosphorus +15), D (shielding is approximately the same for elements in the same period).
Question 33 · MCQ
1 marks
A mixture of 15 cm³ of a gaseous hydrocarbon X and 100 cm³ of oxygen (an excess) was exploded in a sealed vessel. After cooling to room temperature and pressure, the total volume of gas remaining was 70 cm³. When this remaining gas was shaken with concentrated aqueous potassium hydroxide (which absorbs carbon dioxide), the volume decreased to 25 cm³. What is the molecular formula of hydrocarbon X?
A.C2H6
B.C3H6
C.C3H8
D.C4H10
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Worked solution
The decrease in gas volume after shaking with KOH is \(70 - 25 = 45\text{ cm}^3\), which represents the volume of \(\text{CO}_2\) gas produced. Since \(15\text{ cm}^3\) of X produces \(45\text{ cm}^3\) of \(\text{CO}_2\), the ratio of X to \(\text{CO}_2\) is 1:3, meaning X contains 3 carbon atoms. The remaining gas after KOH absorption is \(25\text{ cm}^3\), which is the unreacted oxygen. The volume of oxygen reacted is therefore \(100 - 25 = 75\text{ cm}^3\). The volume ratio of X to reacted oxygen is \(15:75 = 1:5\). The combustion equation is \(\text{C}_3\text{H}_y + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\). Comparing oxygen atoms, \(5 \times 2 = 10\) on the left, and \(3 \times 2 = 6\) in \(\text{CO}_2\), so there must be 4 oxygen atoms in water, indicating \(4\text{H}_2\text{O}\) is formed, which gives 8 hydrogen atoms. Thus, X is \(\text{C}_3\text{H}_8\).
Marking scheme
Award 1 mark for selecting correct option C.
Question 34 · MCQ
1 marks
A student added 25.0 cm³ of 0.200 mol dm⁻³ aqueous copper(II) sulfate to 30.0 cm³ of 0.400 mol dm⁻³ aqueous sodium hydroxide. A blue precipitate of copper(II) hydroxide was formed: \(\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{Cu(OH)}_2(\text{s})\). After filtering the mixture, what is the concentration of \(\text{OH}^{-}(\text{aq})\) remaining in the filtrate?
A.0.036 mol dm⁻³
B.0.040 mol dm⁻³
C.0.073 mol dm⁻³
D.0.218 mol dm⁻³
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Worked solution
Initial moles of \(\text{Cu}^{2+} = 0.0250\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\). Initial moles of \(\text{OH}^{-} = 0.0300\text{ dm}^3 \times 0.400\text{ mol dm}^{-3} = 1.20 \times 10^{-2}\text{ mol}\). From the equation, 1 mole of \(\text{Cu}^{2+}\) reacts with 2 moles of \(\text{OH}^{-}\). Moles of \(\text{OH}^{-}\) reacted = \(2 \times 5.00 \times 10^{-3} = 1.00 \times 10^{-2}\text{ mol}\). Moles of \(\text{OH}^{-}\) remaining = \(1.20 \times 10^{-2} - 1.00 \times 10^{-2} = 2.00 \times 10^{-3}\text{ mol}\). The total volume of the mixture is \(25.0 + 30.0 = 55.0\text{ cm}^3 = 0.055\text{ dm}^3\). Concentration of remaining \(\text{OH}^{-} = \frac{2.00 \times 10^{-3}\text{ mol}}{0.055\text{ dm}^3} = 0.0364\text{ mol dm}^{-3}\).
Marking scheme
Award 1 mark for selecting correct option A.
Question 35 · MCQ
1 marks
The standard enthalpy changes of combustion of graphite, hydrogen and butane are given in the table: graphite, \(\text{C(s)} = -394\text{ kJ mol}^{-1}\); hydrogen, \(\text{H}_2\text{(g)} = -286\text{ kJ mol}^{-1}\); butane, \(\text{C}_4\text{H}_{10}\text{(g)} = -2877\text{ kJ mol}^{-1}\). What is the standard enthalpy change of formation of butane, \(\text{C}_4\text{H}_{10}\text{(g)}\)?
A.-3557 kJ mol⁻¹
B.-129 kJ mol⁻¹
C.+129 kJ mol⁻¹
D.+2197 kJ mol⁻¹
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Worked solution
The formation equation for butane is: \(4\text{C(s)} + 5\text{H}_2\text{(g)} \rightarrow \text{C}_4\text{H}_{10}\text{(g)}\). According to Hess's Law using combustion enthalpies: \(\Delta H_f^\ominus = 4 \times \Delta H_c^\ominus[\text{C(s)}] + 5 \times \Delta H_c^\ominus[\text{H}_2\text{(g)}] - \Delta H_c^\ominus[\text{C}_4\text{H}_{10}\text{(g)}]\). Substituting the values: \(\Delta H_f^\ominus = 4(-394) + 5(-286) - (-2877) = -1576 - 1430 + 2877 = -129\text{ kJ mol}^{-1}\).
Marking scheme
Award 1 mark for selecting correct option B.
Question 36 · MCQ
1 marks
Consider the reaction: \(\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(s)} + 3\text{CO}_2\text{(g)}\), for which the standard enthalpy change is \(-25\text{ kJ mol}^{-1}\). The standard enthalpy change of formation of \(\text{CO(g)}\) is \(-111\text{ kJ mol}^{-1}\) and that of \(\text{CO}_2\text{(g)}\) is \(-394\text{ kJ mol}^{-1}\). What is the standard enthalpy change of formation of \(\text{Fe}_2\text{O}_3\text{(s)}\)?
A.-1540 kJ mol⁻¹
B.-874 kJ mol⁻¹
C.-824 kJ mol⁻¹
D.-258 kJ mol⁻¹
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Worked solution
Let \(\Delta H_f^\ominus[\text{Fe}_2\text{O}_3\text{(s)}]\) be \(x\). For the reaction, \(\Delta H_{\text{reaction}} = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})\). Since \(\text{Fe(s)}\) is an element in its standard state, \(\Delta H_f^\ominus[\text{Fe(s)}] = 0\). Substituting the given values: \(-25 = [2(0) + 3(-394)] - [x + 3(-111)]\). \(-25 = -1182 - x + 333\). \(-25 = -849 - x\). \(x = -849 + 25 = -824\text{ kJ mol}^{-1}\).
Marking scheme
Award 1 mark for selecting correct option C.
Question 37 · MCQ
1 marks
Which statement about the reaction between propene and hydrogen bromide, HBr, is correct?
A.The major product is 1-bromopropane because the secondary carbocation intermediate is more stable than the primary carbocation intermediate.
B.The major product is 2-bromopropane because the secondary carbocation intermediate is more stable than the primary carbocation intermediate.
C.The major product is 1-bromopropane because the tertiary carbocation intermediate is more stable than the secondary carbocation intermediate.
D.The major product is 2-bromopropane because the tertiary carbocation intermediate is more stable than the secondary carbocation intermediate.
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Worked solution
In the electrophilic addition of HBr to propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)), the major product is formed via the more stable carbocation intermediate. The secondary carbocation (\(\text{CH}_3\text{CH}^{+}\text{CH}_3\)) is more stable than the primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^{+}\)) due to the positive inductive effect of two electron-releasing methyl groups. Attack of bromide on this secondary carbocation yields 2-bromopropane as the major product.
Marking scheme
Award 1 mark for selecting correct option B.
Question 38 · MCQ
1 marks
An organic compound Y is treated with hot, concentrated, acidified potassium manganate(VII). The only organic products formed are propanone and carbon dioxide. What is the identity of compound Y?
A.2-methylbut-2-ene
B.2-methylpropene
C.but-2-ene
D.propene
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Worked solution
Hot, concentrated, acidified \(\text{KMnO}_4\) cleaves the \(\text{C}=\text{C}\) double bond. A double-bonded carbon with two alkyl substituents, \((\text{CH}_3)_2\text{C}=\), is oxidised to a ketone (propanone, \((\text{CH}_3)_2\text{C}=\text{O}\)). A double-bonded carbon with two hydrogen atoms, \(=\text{CH}_2\), is oxidised to carbon dioxide (\(\text{CO}_2\)). Therefore, compound Y must contain both \((\text{CH}_3)_2\text{C}=\) and \(=\text{CH}_2\) groups, which corresponds to 2-methylpropene, \((\text{CH}_3)_2\text{C}=\text{CH}_2\).
Marking scheme
Award 1 mark for selecting correct option B.
Question 39 · MCQ
1 marks
The successive ionisation energies, in kJ mol⁻¹, of an element Z in Period 3 are shown: 1st: 578, 2nd: 1817, 3rd: 2745, 4th: 11578, 5th: 14831. In which group of the Periodic Table is element Z?
A.Group 2
B.Group 13
C.Group 14
D.Group 15
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Worked solution
Successive ionisation energies increase as electrons are removed. A very large increase (a jump) occurs between the 3rd and 4th ionisation energies (from 2745 to 11578 kJ mol⁻¹). This indicates that the fourth electron is being removed from an inner core shell which experiences much less shielding and stronger nuclear attraction. Thus, element Z has 3 valence electrons in its outer shell, which places it in Group 13.
Marking scheme
Award 1 mark for selecting correct option B.
Question 40 · MCQ
1 marks
Which statement best explains why the first ionisation energy of sulfur is less than the first ionisation energy of phosphorus?
A.The outer electron in a sulfur atom is in a 3p orbital, whereas in a phosphorus atom it is in a 3s orbital.
B.Sulfur has a greater nuclear charge than phosphorus, so there is greater attraction between the nucleus and the outer electrons.
C.The outer electron in a sulfur atom experiences greater shielding than the outer electron in a phosphorus atom.
D.In a sulfur atom, there is spin-pair repulsion between two electrons in the same 3p orbital, which does not occur in a phosphorus atom.
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Worked solution
Phosphorus has the outer electronic configuration \(3\text{s}^2 3\text{p}^3\), with three singly-occupied \(3\text{p}\) orbitals. Sulfur has the outer electronic configuration \(3\text{s}^2 3\text{p}^4\), with one doubly-occupied (paired) \(3\text{p}\) orbital and two singly-occupied \(3\text{p}\) orbitals. The mutual repulsion between the two paired electrons in the same \(3\text{p}\) orbital of sulfur makes it easier to remove one of these electrons, resulting in a lower first ionisation energy than that of phosphorus despite sulfur's higher nuclear charge.
Marking scheme
Award 1 mark for selecting correct option D.
Paper 23 (AS Level Structured)
Answer all questions in the spaces provided on the question paper.
4 Question · 60 marks
Question 1 · Structured
15 marks
An alloy of magnesium and aluminium of mass 1.026 g was fully reacted with an excess of dilute hydrochloric acid. The chemical equations for the reactions are: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g) and 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g). The total volume of hydrogen gas collected was 1.20 dm3 measured at room temperature and pressure (r.t.p.). [Assume Ar: Mg = 24.3, Al = 27.0; molar volume of gas at r.t.p. = 24.0 dm3 mol-1] (a) State the oxidation numbers of magnesium and aluminium before and after the reaction with dilute hydrochloric acid. Identify which species has been oxidized in this reaction. (3 marks) (b) Write the ionic equation, including state symbols, for the reaction of aluminium with dilute hydrochloric acid. (2 marks) (c) Using the experimental data, calculate the percentage by mass of magnesium in the alloy. Show all your working. (6 marks) (d) In a separate experiment, 3.00 g of magnesium is reacted with 1.12 dm3 of oxygen gas (measured at r.t.p.) to form magnesium oxide: 2Mg(s) + O2(g) -> 2MgO(s). Determine the limiting reactant and calculate the mass of magnesium oxide formed. (4 marks)
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Worked solution
(a) Before the reaction, both magnesium and aluminium are in their elemental states, so their oxidation numbers are 0. After the reaction, magnesium exists as Mg2+ and aluminium as Al3+, so their oxidation numbers are +2 and +3 respectively. Both magnesium and aluminium have been oxidized because their oxidation numbers increased. (b) The ionic equation is: 2Al(s) + 6H+(aq) -> 2Al3+(aq) + 3H2(g). (c) Total moles of H2 collected = 1.20 / 24.0 = 0.050 mol. Let the mass of Mg in the alloy be x g, so the mass of Al is (1.026 - x) g. Moles of Mg = x / 24.3. Moles of H2 produced from Mg = x / 24.3. Moles of Al = (1.026 - x) / 27.0. Moles of H2 produced from Al = 1.5 * (1.026 - x) / 27.0 = (1.026 - x) / 18.0. The total moles of H2 is: (x / 24.3) + ((1.026 - x) / 18.0) = 0.050. Multiplying both sides by 437.4 (24.3 * 18.0): 18.0x + 24.3(1.026 - x) = 21.87. 18.0x + 24.9318 - 24.3x = 21.87. -6.3x = -3.0618. x = 0.486 g. Mass of Mg = 0.486 g. Percentage by mass of Mg = (0.486 / 1.026) * 100% = 47.37% (or 47.4%). (d) Moles of Mg = 3.00 / 24.3 = 0.1235 mol. Moles of O2 = 1.12 / 24.0 = 0.0467 mol. Stoichiometric ratio of Mg : O2 is 2 : 1. Thus, 0.0467 mol of O2 requires 2 * 0.0467 = 0.0934 mol of Mg. Since we have 0.1235 mol of Mg (which is more than 0.0934 mol), Mg is in excess, and O2 is the limiting reactant. Moles of MgO formed = 2 * 0.0467 = 0.0934 mol. Mass of MgO = 0.0934 * (24.3 + 16.0) = 0.0934 * 40.3 = 3.76 g.
Marking scheme
(a) [1 mark] for correct oxidation numbers of Mg (0 to +2) and Al (0 to +3). [1 mark] for stating that both metals are oxidized. [1 mark] for linking oxidation to the increase in oxidation numbers. (b) [1 mark] for balanced ionic species: 2Al + 6H+ -> 2Al3+ + 3H2. [1 mark] for correct state symbols: s, aq, aq, g. (c) [1 mark] for calculating total moles of H2 = 0.050 mol. [1 mark] for expressing moles of H2 from Mg as x / 24.3. [1 mark] for expressing moles of H2 from Al as 1.5 * (1.026 - x) / 27.0. [1 mark] for setting up the algebraic equation. [1 mark] for solving the mass of Mg as 0.486 g. [1 mark] for calculating the final percentage as 47.4% (accept 47.3% to 47.5%). (d) [1 mark] for calculating moles of Mg (0.123) and O2 (0.0467). [1 mark] for identifying O2 as the limiting reactant with a clear comparison. [1 mark] for calculating moles of MgO formed (0.0934 mol). [1 mark] for the final mass of MgO = 3.76 g (accept 3.75 g to 3.80 g).
Question 2 · Structured
15 marks
(a) Define the following terms: (i) Standard enthalpy change of formation. (2 marks) (ii) Standard enthalpy change of combustion. (2 marks) (b) Propan-2-ol, C3H8O(l), is a widely used solvent. (i) Write the balanced chemical equation, including state symbols, for the reaction representing the standard enthalpy change of formation of liquid propan-2-ol. (2 marks) (ii) Construct a Hess's Law cycle that can be used to determine the standard enthalpy change of formation of liquid propan-2-ol from the standard enthalpy changes of combustion of carbon, hydrogen, and propan-2-ol. (3 marks) (iii) Calculate the standard enthalpy change of formation of liquid propan-2-ol, in kJ/mol, using the following data: standard enthalpy change of combustion of C(s) = -393.5 kJ/mol, standard enthalpy change of combustion of H2(g) = -285.8 kJ/mol, and standard enthalpy change of combustion of C3H8O(l) = -2005.8 kJ/mol. (3 marks) (c) The standard enthalpy change of vaporisation of propan-2-ol is +47.5 kJ/mol. Determine the standard enthalpy change of formation of gaseous propan-2-ol, C3H8O(g). (3 marks)
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Worked solution
(a)(i) Standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K, 100 kPa). (a)(ii) Standard enthalpy change of combustion is the enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions. (b)(i) Equation: 3C(s) + 4H2(g) + 0.5O2(g) -> C3H8O(l). (b)(ii) Cycle: Reactants: 3C(s) + 4H2(g) + 0.5O2(g) are at the top-left; Product: C3H8O(l) is at the top-right. The arrow connecting them directly is standard enthalpy of formation. Combustion products: 3CO2(g) + 4H2O(l) are at the bottom. Arrow from reactants to combustion products has value 3*combustion of C + 4*combustion of H2. Arrow from product to combustion products has value combustion of C3H8O(l). (b)(iii) Calculation: standard enthalpy of formation = 3*(-393.5) + 4*(-285.8) - (-2005.8) = -1180.5 - 1143.2 + 2005.8 = -317.9 kJ/mol. (c) Formation of gaseous propan-2-ol is represented by: 3C(s) + 4H2(g) + 0.5O2(g) -> C3H8O(g). This is equivalent to forming the liquid first and then vaporising it. Thus, standard enthalpy of formation of gaseous propan-2-ol = standard enthalpy of formation of liquid propan-2-ol + standard enthalpy of vaporisation = -317.9 + 47.5 = -270.4 kJ/mol.
Marking scheme
(a)(i) [1 mark] for formation of 1 mole of compound from elements. [1 mark] for standard states and standard conditions. (a)(ii) [1 mark] for complete combustion of 1 mole of substance in excess oxygen. [1 mark] for standard conditions. (b)(i) [1 mark] for correct balanced species. [1 mark] for correct state symbols. (b)(ii) [1 mark] for showing correct cycle species. [1 mark] for correct arrows with correct direction. [1 mark] for correct labeling of enthalpy terms. (b)(iii) [1 mark] for 3*(-393.5) + 4*(-285.8). [1 mark] for subtracting (-2005.8). [1 mark] for correct final value of -317.9 kJ/mol with sign and units. (c) [1 mark] for identifying relation: formation of gas = formation of liquid + vaporisation. [1 mark] for correct substitution. [1 mark] for correct final value of -270.4 kJ/mol with sign and units.
Question 3 · Structured
15 marks
(a) Pent-2-ene is a hydrocarbon with the molecular formula C5H10. It exhibits stereoisomerism. Draw and label the cis and trans (or E and Z) isomers of pent-2-ene, and explain why pent-2-ene can exhibit stereoisomerism while pent-1-ene cannot. (4 marks) (b) 2-methylbut-2-ene is another structural isomer of C5H10. (i) Complete the chemical equation below for the reaction of 2-methylbut-2-ene with cold, dilute, alkaline potassium manganate(VII), KMnO4. Draw the structural formula of the organic product and state the observed colour change: (CH3)2C=CHCH3 + [O] + H2O -> ... (3 marks) (ii) When 2-methylbut-2-ene reacts with hydrogen bromide, HBr, a mixture of two structural isomers is formed: a major product and a minor product. Draw the skeletal structures of the major and minor products. Explain, with reference to the stability of the carbocation intermediates, why one product is formed as the major product. (5 marks) (iii) 2-methylbut-2-ene is reacted with hot, concentrated, acidified potassium manganate(VII). The double bond is completely cleaved. State the structural formulas and names of the two organic products formed in this reaction. (3 marks)
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Worked solution
(a) Cis-pent-2-ene has the two alkyl groups (or the two hydrogen atoms) on the same side of the double bond. Trans-pent-2-ene has the groups on opposite sides. Pent-2-ene exhibits stereoisomerism because there is restricted rotation about the C=C double bond, and each carbon of the C=C bond is attached to two different groups (one has -H and -CH3, and the other has -H and -CH2CH3). Pent-1-ene cannot show stereoisomerism because one of the double-bonded carbons is attached to two identical groups (two hydrogen atoms). (b)(i) Under cold, dilute, alkaline conditions, alkenes are oxidised to diols. The organic product is 2-methylbutane-2,3-diol, which has the structural formula (CH3)2C(OH)CH(OH)CH3. The purple potassium manganate(VII) solution is decolourised (or a brown precipitate of MnO2 is formed). (b)(ii) The major product is 2-bromo-2-methylbutane. The minor product is 2-bromo-3-methylbutane. In the first step of electrophilic addition, H+ adds to the double bond. Addition of H+ to C3 forms a tertiary carbocation: (CH3)2C+-CH2CH3. Addition of H+ to C2 forms a secondary carbocation: (CH3)2CH-C+H-CH3. The tertiary carbocation is more stable than the secondary carbocation because of the greater positive inductive effect of three electron-releasing alkyl groups compared to two, which stabilizes the positive charge more effectively. (b)(iii) Cleavage of the double bond in 2-methylbut-2-ene yields propanone, CH3COCH3, and ethanoic acid, CH3COOH.
Marking scheme
(a) [1 mark] for correctly drawn structures of cis- and trans-pent-2-ene with labels. [1 mark] for stating restricted rotation about the C=C bond. [1 mark] for explaining that pent-2-ene has two different groups on each carbon of the double bond. [1 mark] for explaining that pent-1-ene has two identical hydrogen atoms on one C of the double bond. (b)(i) [1 mark] for the correct formula of the diol product: (CH3)2C(OH)CH(OH)CH3. [1 mark] for balancing the equation. [1 mark] for stating correct colour change: purple to colourless (or purple to brown precipitate). (b)(ii) [1 mark] for drawing the correct skeletal structure of the major product (2-bromo-2-methylbutane). [1 mark] for drawing the correct skeletal structure of the minor product (2-bromo-3-methylbutane). [1 mark] for identifying the tertiary and secondary carbocations. [1 mark] for stating that the tertiary carbocation is more stable. [1 mark] for explaining stability via the positive inductive effect of alkyl groups. (b)(iii) [1 mark] for naming both products: propanone and ethanoic acid. [2 marks] for drawing both correct structural formulas (1 mark each).
Question 4 · Structured
15 marks
(a) Define the term first ionisation energy. (3 marks) (b) Write an equation, including state symbols, that represents the third ionisation energy of sodium. (2 marks) (c) The successive ionisation energies, in kJ/mol, of a Period 3 element, Q, are shown below: 1st: 578, 2nd: 1817, 3rd: 2745, 4th: 11578, 5th: 14831, 6th: 18378. (i) Identify element Q. Justify your answer using the data provided. (3 marks) (ii) Explain why successive ionisation energies of element Q show a general increase. (2 marks) (d) Explain the following observations: (i) The first ionisation energy of sulfur is lower than that of phosphorus, even though sulfur has a greater nuclear charge. (3 marks) (ii) The first ionisation energy of magnesium is greater than that of aluminium. (2 marks)
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Worked solution
(a) First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. (b) The third ionisation energy of sodium is represented by: Na2+(g) -> Na3+(g) + e-. (c)(i) Element Q is aluminium (Al). There is a very large increase (or jump) between the third (2745 kJ/mol) and fourth (11578 kJ/mol) ionisation energies. This indicates that the fourth electron is being removed from an inner quantum shell which is closer to the nucleus, meaning there are three electrons in the outer shell. (c)(ii) Successive ionisation energies increase because after each electron is removed, the remaining electrons experience a stronger electrostatic attraction from the nucleus (same nuclear charge attracting fewer electrons) and there is also less electron-electron repulsion. (d)(i) Phosphorus has the valence electronic configuration 3s2 3p3, where each of the three 3p orbitals is singly occupied. Sulfur has the valence configuration 3s2 3p4, which contains one doubly occupied 3p orbital. The repulsion between the two paired electrons in the same 3p orbital of sulfur makes it easier to remove one of these electrons compared to the singly occupied orbitals of phosphorus. (d)(ii) Magnesium has the configuration 1s2 2s2 2p6 3s2. Aluminium has the configuration 1s2 2s2 2p6 3s2 3p1. The electron being removed from magnesium is in a 3s orbital, while the electron being removed from aluminium is in a 3p orbital. The 3p orbital of aluminium is at a higher energy level and is more shielded from the nucleus by the 3s electrons, making the outer electron in aluminium easier to remove.
Marking scheme
(a) [1 mark] for 'energy required to remove one mole of electrons'. [1 mark] for 'from one mole of gaseous atoms'. [1 mark] for 'to form one mole of gaseous 1+ ions'. (b) [1 mark] for correct species: Na2+ -> Na3+ + e-. [1 mark] for state symbols (g) on both ions. (c)(i) [1 mark] for identifying Aluminium / Al. [1 mark] for highlighting the large jump between 3rd and 4th ionisation energies. [1 mark] for concluding that the 4th electron is removed from an inner shell / there are 3 valence electrons. (c)(ii) [1 mark] for stating that the same number of protons attract fewer electrons. [1 mark] for pointing out the reduction in electron-electron shielding/repulsion. (d)(i) [1 mark] for writing correct configurations of P (3p3) and S (3p4). [1 mark] for mentioning that S has a paired electron in a 3p orbital. [1 mark] for explaining that spin-pair repulsion in S makes it easier to remove the electron. (d)(ii) [1 mark] for stating that Mg's electron is in 3s and Al's is in 3p. [1 mark] for explaining that the 3p orbital is higher in energy and experiences greater shielding.
Paper 33 (Advanced Practical Skills)
Answer all questions. Show all your working and use appropriate units.
3 Question · 39.99 marks
Question 1 · Practical
13.33 marks
A student performs a volumetric titration to determine the exact concentration of a monobasic acid, \( \text{HX} \).
The student is provided with: - **FA 1**: An aqueous solution of the monobasic acid \( \text{HX} \) of unknown concentration. - **FA 2**: \( 0.0800 \text{ mol dm}^{-3} \) standard sodium hydroxide, \( \text{NaOH} \) solution.
The student uses phenolphthalein as an indicator and records the following burette readings: - **Rough**: Final reading = \( 24.50 \text{ cm}^3 \), Initial reading = \( 0.00 \text{ cm}^3 \) - **Titration 1**: Final reading = \( 23.85 \text{ cm}^3 \), Initial reading = \( 0.20 \text{ cm}^3 \) - **Titration 2**: Final reading = \( 47.70 \text{ cm}^3 \), Initial reading = \( 24.00 \text{ cm}^3 \) - **Titration 3**: Final reading = \( 24.35 \text{ cm}^3 \), Initial reading = \( 0.65 \text{ cm}^3 \)
(a) Complete the calculation of the volume of **FA 2** added for each titration (Rough, Titration 1, Titration 2, and Titration 3).
(b) Select the concordant titres and calculate the mean titre of **FA 2** that should be used for subsequent calculations. Show your working.
(c) Calculate the number of moles of \( \text{NaOH} \) present in the mean titre volume of **FA 2** calculated in (b).
(d) Write the balanced chemical equation for the neutralisation reaction and calculate the concentration, in \( \text{mol dm}^{-3} \), of the acid **FA 1**, assuming that a \( 25.0 \text{ cm}^3 \) pipette was used to measure **FA 1** for each titration.
(e) Calculate the percentage uncertainty in the volume of **FA 2** added during Titration 1, given that each individual burette reading has an uncertainty of \( \pm 0.05 \text{ cm}^3 \).
(f) Suggest two procedural practices during a titration that would increase the precision and accuracy of the end-point determination.
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(b) Concordant titres are within \( 0.10 \text{ cm}^3 \) of each other: Titrations 1, 2, and 3. Mean titre = \( \frac{23.65 + 23.70 + 23.70}{3} = 23.68 \text{ cm}^3 \) (or if only Titrations 2 & 3 are used, Mean titre = \( 23.70 \text{ cm}^3 \)).
Since the mole ratio is 1:1, moles of \( \text{HX} \) in \( 25.0 \text{ cm}^3 \) = moles of \( \text{NaOH} \) in the mean titre. Using \( 1.894 \times 10^{-3} \text{ mol} \): \( [\text{HX}] = \frac{1.894 \times 10^{-3} \text{ mol}}{0.0250 \text{ dm}^3} = 0.0758 \text{ mol dm}^{-3} \) Using \( 1.896 \times 10^{-3} \text{ mol} \): \( [\text{HX}] = \frac{1.896 \times 10^{-3} \text{ mol}}{0.0250 \text{ dm}^3} = 0.0758 \text{ mol dm}^{-3} \) (to 3 s.f.)
(e) Percentage uncertainty: Total uncertainty for a titre = \( 2 \times (\pm 0.05 \text{ cm}^3) = \pm 0.10 \text{ cm}^3 \) Percentage uncertainty = \( \frac{0.10}{23.65} \times 100\% = 0.423\% \)
(f) Improvements: 1. Place a white tile under the conical flask to observe the colour change clearly. 2. Wash down the inside walls of the conical flask with distilled water near the end-point to ensure all reactants are mixed in the solution.
Marking scheme
Total: 13 marks - (a) [2 marks]: 1 mark for calculating all 4 volumes correctly; 1 mark for recording all values to 2 decimal places. - (b) [2 marks]: 1 mark for selecting Titrations 1, 2, and 3 (or 2 and 3); 1 mark for calculating the correct mean to 2 decimal places with units. - (c) [2 marks]: 1 mark for the correct formula for moles; 1 mark for the correct calculation to 3 significant figures. - (d) [2 marks]: 1 mark for the balanced equation with state symbols; 1 mark for calculating the concentration of HX to 3 significant figures with units. - (e) [2 marks]: 1 mark for doubling the reading uncertainty to get 0.10 cm3; 1 mark for calculating the correct percentage uncertainty (0.423% or 0.42%). - (f) [3 marks]: 1 mark each for two valid procedural improvements (up to 3 marks max).
Question 2 · Practical
13.33 marks
A student carries out a thermochemical investigation using a simple calorimeter to determine the enthalpy change of hydration of anhydrous copper(II) sulfate using Hess's Law.
\( 3.99 \text{ g} \) of anhydrous \( \text{CuSO}_4 \) (\( M_{\text{r}} = 159.6 \)) was dissolved in \( 50.0 \text{ cm}^3 \) of water in a polystyrene cup. The temperature of the water increased from \( 21.5\ ^\circ\text{C} \) to \( 30.1\ ^\circ\text{C} \).
\( 6.24 \text{ g} \) of hydrated \( \text{CuSO}_4\cdot5\text{H}_2\text{O} \) (\( M_{\text{r}} = 249.6 \)) was dissolved in \( 45.0 \text{ cm}^3 \) of water in a polystyrene cup. The temperature of the water decreased from \( 22.0\ ^\circ\text{C} \) to \( 20.8\ ^\circ\text{C} \).
[Assume the specific heat capacity of all solutions is \( 4.18 \text{ J g}^{-1}\ ^\circ\text{C}^{-1} \) and the density is \( 1.00 \text{ g cm}^{-3} \). Assume the mass of the solution is equal to the mass of the water added.]
(a) For **Experiment 1**: (i) Calculate the heat energy released, \( q_1 \), in joules. (ii) Calculate the molar enthalpy change of solution of anhydrous copper(II) sulfate, \( \Delta H_1 \), in \( \text{kJ mol}^{-1} \). Include the appropriate sign.
(b) For **Experiment 2**: (i) Calculate the heat energy absorbed, \( q_2 \), in joules. (ii) Calculate the molar enthalpy change of solution of hydrated copper(II) sulfate, \( \Delta H_2 \), in \( \text{kJ mol}^{-1} \). Include the appropriate sign.
(c) Construct a Hess's Law energy cycle to relate \( \Delta H_{\text{hyd}} \), \( \Delta H_1 \), and \( \Delta H_2 \) for the following reaction: \( \text{CuSO}_4(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow \text{CuSO}_4\cdot5\text{H}_2\text{O}(\text{s}) \) Calculate the enthalpy change of hydration, \( \Delta H_{\text{hyd}} \), in \( \text{kJ mol}^{-1} \).
(d) State two major sources of heat-related errors in this practical setup and how each could be minimised.
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(c) Hess's Law Cycle: \( \text{CuSO}_4(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \xrightarrow{\Delta H_{\text{hyd}}} \text{CuSO}_4\cdot5\text{H}_2\text{O}(\text{s}) \) Both react with excess water to form \( \text{CuSO}_4(\text{aq}) \) via \( \Delta H_1 \) and \( \Delta H_2 \) respectively. \( \Delta H_1 = \Delta H_{\text{hyd}} + \Delta H_2 \) \( \Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2 = -71.9 - (+9.03) = -80.93 \text{ kJ mol}^{-1} \)
(d) Errors and improvements: 1. Heat loss to surroundings: Place a lid on the polystyrene cup or place the cup inside a larger beaker packed with cotton wool for insulation. 2. Heat absorbed by the thermometer / cup itself: Correct for cooling by taking temperature readings at regular intervals over time and plotting a temperature-time cooling curve to extrapolate back to the time of mixing.
Marking scheme
Total: 13 marks - (a)(i) [1 mark]: Correct calculation of q1 in J (1797 J or 1.80 kJ). - (a)(ii) [2 marks]: 1 mark for correct mole calculation (0.0250 mol); 1 mark for correct calculation of ΔH1 with negative sign and units. - (b)(i) [1 mark]: Correct calculation of q2 in J (226 J or 0.226 kJ). - (b)(ii) [2 marks]: 1 mark for correct mole calculation (0.0250 mol); 1 mark for correct calculation of ΔH2 with positive sign and units. - (c) [3 marks]: 1 mark for constructing a clear Hess's cycle diagram; 1 mark for correct mathematical equation connecting the steps; 1 mark for the final enthalpy value (-80.9 kJ mol^-1 or -80.93 kJ mol^-1) with the correct sign. - (d) [4 marks]: 1 mark each for identifying two valid errors; 1 mark each for the corresponding realistic improvements.
Question 3 · Practical
13.33 marks
You are provided with three solutions: **FA 3**, **FA 4**, and **FA 5**, each containing one of the following compounds: - Aqueous sodium but-2-enedioate (an alkene salt) - Potassium chloride, \( \text{KCl} \) - Potassium iodide, \( \text{KI} \)
You are also provided with solid **FA 6**, which is an anhydrous carbonate of a Group 2 metal.
(a) Describe the expected observations when **FA 4** and **FA 5** are tested to identify their halide ions. Tabulate your answer to show observations for: - **Test 1**: Addition of dilute nitric acid followed by aqueous silver nitrate. - **Test 2**: Addition of dilute aqueous ammonia to the mixture from Test 1. - **Test 3**: Addition of concentrated aqueous ammonia to the mixture from Test 1.
(b) Write the ionic equation, including state symbols, for the precipitation reaction occurring in **Test 1** with **FA 5**.
(c) Describe a chemical test that can be carried out to confirm that **FA 3** contains a carbon-carbon double bond. State the reagent, the conditions, and the expected positive observation.
(d) A student carries out the thermal decomposition of solid **FA 6** in a test-tube and passes the gas evolved through limewater. (i) State the observation in the limewater and write the equation for the thermal decomposition of **FA 6**, representing the metal as \( \text{M} \). (ii) Describe the trend in the thermal stability of Group 2 carbonates down the group and explain this trend in terms of ionic radius, charge density, and polarising power of the metal cation.
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Worked solution
(a) Expected Observations Table:
| Test | Observation with FA 4 (contains \( \text{Cl}^- \)) | Observation with FA 5 (contains \( \text{I}^- \)) | | :--- | :--- | :--- | | **Test 1** (aq \( \text{AgNO}_3 \)) | White precipitate forms | Yellow precipitate forms | | **Test 2** (dilute \( \text{NH}_3 \)) | Precipitate dissolves to form a colourless solution | Precipitate remains / insoluble | | **Test 3** (conc \( \text{NH}_3 \)) | Precipitate dissolves to form a colourless solution | Precipitate remains / insoluble |
(c) Test for alkene (\( \text{C=C} \) double bond): - **Reagent**: Bromine water / \( \text{Br}_2(\text{aq}) \) - **Conditions**: Room temperature - **Observation**: The orange/yellow-brown solution becomes colourless. [Alternative: Cold, alkaline dilute \( \text{KMnO}_4(\text{aq}) \) which turns from purple to a brown precipitate or colourless solution.]
(ii) Trend and explanation: - **Trend**: Thermal stability of Group 2 carbonates increases down the group. - **Explanation**: 1. Going down the group, the ionic radius of the \( \text{M}^{2+} \) cation increases, while keeping the same \( 2+ \) charge. 2. This causes the charge density of the cation to decrease. 3. Consequently, the larger cation has lower polarising power and distorts the electron cloud of the carbonate ion (\( \text{CO}_3^{2-} \)) to a lesser extent. 4. This leads to less weakening of the carbon-oxygen (C-O) covalent bonds within the carbonate ion, requiring a higher temperature to decompose.
Marking scheme
Total: 14 marks - (a) [4 marks]: 1 mark for correct table structure with columns/rows; 1 mark for white vs yellow precipitate; 1 mark for correct solubility behaviour of FA 4 precipitate in ammonia; 1 mark for correct insolubility of FA 5 precipitate in both dilute and concentrated ammonia. - (b) [2 marks]: 1 mark for Ag+ + I- -> AgI; 1 mark for correct state symbols (aq, aq -> s). - (c) [2 marks]: 1 mark for stating bromine water (or cold alkaline KMnO4); 1 mark for colour change from orange/brown to colourless (or purple to colourless/brown precipitate). - (d)(i) [2 marks]: 1 mark for limewater turning cloudy; 1 mark for the balanced equation with correct states of symbols. - (d)(ii) [4 marks]: 1 mark for stating that thermal stability increases down the group; 1 mark for stating that ionic radius increases / charge density decreases; 1 mark for explaining that polarising power of the cation decreases; 1 mark for relating this to less distortion of the carbonate ion and less weakening of the C-O bond.
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