An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V3) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 13 Multiple Choice
There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
The reaction between acidified aqueous sodium thiosulfate and hydrochloric acid produces a cloudy precipitate of sulfur.
If the temperature of the reaction mixture is decreased from \(50\ ^\circ\text{C}\) to \(20\ ^\circ\text{C}\), how do the activation energy, \(E_\text{a}\), and the proportion of colliding molecules with energy greater than or equal to \(E_\text{a}\) change?
Activation energy, \(E_\text{a}\), is a constant characteristic of a given reaction pathway and does not change with temperature. However, decreasing the temperature from \(50\ ^\circ\text{C}\) to \(20\ ^\circ\text{C}\) shifts the Maxwell-Boltzmann distribution of molecular energies to the left, which significantly reduces the proportion of colliding molecules that possess energy greater than or equal to the activation energy.
PastPaper.markingScheme
1 mark: Identify that activation energy is unaffected by temperature change, and that decreasing the temperature decreases the fraction of molecules with \(E \ge E_\text{a}\).
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
An excess of water is added to two separate test-tubes, one containing liquid silicon(IV) chloride, \(\text{SiCl}_4\), and the other containing solid phosphorus(V) chloride, \(\text{PCl}_5\).
Which statement about the observations and the pH of the resulting solutions is correct?
A.Both compounds dissolve without chemical reaction to form neutral solutions of pH 7.
B.Both compounds undergo hydrolysis to form strongly acidic solutions with a pH of 2 or less.
C.Silicon(IV) chloride does not react with water, whereas phosphorus(V) chloride undergoes rapid hydrolysis.
D.Silicon(IV) chloride forms a strongly acidic solution, whereas phosphorus(V) chloride forms an alkaline solution.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Both \(\text{SiCl}_4\) and \(\text{PCl}_5\) undergo complete hydrolysis when reacted with water.
Since both reactions release a large amount of \(\text{HCl}\) which fully dissociates in water, both resulting mixtures are strongly acidic (pH less than 3).
PastPaper.markingScheme
1 mark: Correctly identify that both Period 3 covalent chlorides hydrolyse in water to produce highly acidic solutions.
PastPaper.question 3 · multiple-choice
1 PastPaper.marks
A student added \(2.02\text{ g}\) of potassium nitrate (\(\text{KNO}_3\), \(M_\text{r} = 101.1\)) to \(50.0\text{ cm}^3\) of water in a polystyrene cup. The temperature of the water fell from \(21.4\ ^\circ\text{C}\) to \(18.2\ ^\circ\text{C}\).
Assume the mass of the resulting solution is \(50.0\text{ g}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1}\).
What is the standard enthalpy change of solution, \(\Delta H_\text{sol}^\ominus\), of potassium nitrate?
2. Calculate moles of \(\text{KNO}_3\): \(n = \frac{2.02\text{ g}}{101.1\text{ g mol}^{-1}} = 0.0200\text{ mol}\).
3. Calculate \(\Delta H_\text{sol}\): Because the temperature fell, the reaction is endothermic (\(\Delta H > 0\)). \(\Delta H_\text{sol} = + \frac{0.6688\text{ kJ}}{0.0200\text{ mol}} = +33.44\text{ kJ mol}^{-1} \approx +33.4\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark: Correctly calculate the heat absorbed in kJ, determine the mole quantity, and divide to get the positive enthalpy change of solution value.
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
How many stereoisomers (including both optical and cis-trans isomers) exist for the compound 4-methylhept-5-en-2-ol?
A.2
B.4
C.6
D.8
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Let's analyse the structure of 4-methylhept-5-en-2-ol: $$\text{CH}_3-\text{C}^6\text{H}=\text{C}^5\text{H}-\text{C}^4\text{H}(\text{CH}_3)-\text{C}^3\text{H}_2-\text{C}^2\text{H}(\text{OH})-\text{C}^1\text{H}_3$$
1. Chiral centres: - C2 is a chiral carbon bonded to: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}=\text{CH}-\text{CH}_3\). - C4 is a chiral carbon bonded to: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}=\text{CH}-\text{CH}_3\), and \(-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_3\). Thus, there are 2 chiral centres.
2. Cis-trans isomerism: - The double bond between C5 and C6 can exhibit cis-trans (E/Z) stereoisomerism because C5 has two different groups attached (\(-\text{H}\) and the C1-C4 group) and C6 also has two different groups attached (\(-\text{H}\) and \(-\text{CH}_3\)).
3. Total stereoisomers: With \(n = 2\) chiral centres and \(m = 1\) stereogenic double bond, the total number of stereoisomers is \(2^{n+m} = 2^{2+1} = 2^3 = 8\).
PastPaper.markingScheme
1 mark: Identify 2 chiral carbons and 1 cis-trans double bond, and calculate total stereoisomers as 2^3 = 8.
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
In the reaction between potassium dichromate(VI) and hydrogen sulfide in acidic solution, the chemical equation is:
Which row correctly identifies the changes in the oxidation number of chromium and sulfur?
A.Chromium: from \(+6\) to \(+3\); Sulfur: from \(-2\) to \(0\)
B.Chromium: from \(+7\) to \(+3\); Sulfur: from \(-2\) to \(0\)
C.Chromium: from \(+6\) to \(+3\); Sulfur: from \(-1\) to \(0\)
D.Chromium: from \(+12\) to \(+3\); Sulfur: from \(-2\) to \(+2\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
1. In \(\text{Cr}_2\text{O}_7^{2-}\), oxygen has an oxidation number of \(-2\). \(2x + 7(-2) = -2 \Rightarrow 2x = +12 \Rightarrow x = +6\). In \(\text{Cr}^{3+}\), the oxidation number of chromium is \(+3\). So, Chromium changes from \(+6\) to \(+3\).
2. In \(\text{H}_2\text{S}\), hydrogen is \(+1\). \(2(+1) + y = 0 \Rightarrow y = -2\). In elemental sulfur, \(\text{S}(\text{s})\), the oxidation number is \(0\). So, Sulfur changes from \(-2\) to \(0\).
PastPaper.markingScheme
1 mark: Correctly determine the oxidation numbers of Cr and S in all reactants and products, matching option A.
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
A mixture of \(1.00\text{ mol}\) of nitrogen gas and \(3.00\text{ mol}\) of hydrogen gas is allowed to reach equilibrium in a sealed vessel of volume \(2.00\text{ dm}^3\) at a constant temperature.
1 mark: Calculate correct equilibrium concentrations by factoring in the 2.00 dm3 volume, construct the Kc expression, and calculate the final numerical value.
PastPaper.question 7 · multiple-choice
1 PastPaper.marks
The standard enthalpy changes of formation, \(\Delta H_\text{f}^\ominus\), for some substances are given below: - \(\text{CO}_2(\text{g}) = -393.5\text{ kJ mol}^{-1}\) - \(\text{H}_2\text{O}(\text{l}) = -285.8\text{ kJ mol}^{-1}\) - \(\text{CH}_3\text{OCH}_3(\text{g}) = -184.1\text{ kJ mol}^{-1}\)
What is the standard enthalpy change of combustion, \(\Delta H_\text{c}^\ominus\), of dimethyl ether, \(\text{CH}_3\text{OCH}_3(\text{g})\)?
A.\(-1828.5\text{ kJ mol}^{-1}\)
B.\(-1460.3\text{ kJ mol}^{-1}\)
C.\(-1174.5\text{ kJ mol}^{-1}\)
D.\(+1460.3\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
1. Write the balanced equation for the combustion of dimethyl ether: $$\text{CH}_3\text{OCH}_3(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})$$
2. Apply Hess's law: \(\Delta H_\text{c}^\ominus = \sum \Delta H_\text{f}^\ominus(\text{products}) - \sum \Delta H_\text{f}^\ominus(\text{reactants})\) \(\Delta H_\text{c}^\ominus = [2 \times \Delta H_\text{f}^\ominus(\text{CO}_2) + 3 \times \Delta H_\text{f}^\ominus(\text{H}_2\text{O})] - \Delta H_\text{f}^\ominus(\text{CH}_3\text{OCH}_3)\) (Note: \(\Delta H_\text{f}^\ominus\) for elements like \(\text{O}_2\) is zero)
1 mark: Correctly write the balanced equation, and perform the subtraction sum (products - reactants) including correct stoichiometric coefficients.
PastPaper.question 8 · multiple-choice
1 PastPaper.marks
Four different halogenoalkanes are heated separately with aqueous silver nitrate in ethanol at \(50\ ^\circ\text{C}\).
Which halogenoalkane reacts fastest to form a precipitate of silver halide?
A.1-chlorobutane
B.1-iodobutane
C.2-chloro-2-methylpropane
D.2-iodo-2-methylpropane
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes is determined by two main factors: 1. The strength of the carbon-halogen bond: C-I is weaker and has a lower bond enthalpy than C-Cl, meaning iodoalkanes react faster than chloroalkanes. 2. The class of halogenoalkane: tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) undergo rapid hydrolysis via the \(\text{S}_\text{N}1\) mechanism because they form highly stable tertiary carbocations. Primary halogenoalkanes react much slower via the \(\text{S}_\text{N}2\) mechanism.
Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.
PastPaper.markingScheme
1 mark: Identify that tertiary halogenoalkanes react faster than primary ones, and iodoalkanes react faster than chloroalkanes, concluding that 2-iodo-2-methylpropane is the fastest.
PastPaper.question 9 · multiple-choice
1 PastPaper.marks
The temperature of a reaction mixture is increased by \(10\,^\circ\text{C}\). What happens to the activation energy, \(E_a\), the total frequency of collisions, and the fraction of collisions with energy greater than or equal to \(E_a\)?
A.\(E_a\) remains constant; total frequency of collisions increases; fraction of collisions with \(E \ge E_a\) increases.
B.\(E_a\) decreases; total frequency of collisions increases; fraction of collisions with \(E \ge E_a\) increases.
C.\(E_a\) remains constant; total frequency of collisions remains constant; fraction of collisions with \(E \ge E_a\) increases.
D.\(E_a\) decreases; total frequency of collisions remains constant; fraction of collisions with \(E \ge E_a\) remains constant.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The activation energy, \(E_a\), is a characteristic property of a chemical reaction pathway and is independent of temperature; it does not change when temperature is increased. Increasing the temperature increases the average kinetic energy of the molecules, meaning they move faster and collide more frequently, so the total frequency of collisions increases. Most importantly, the Maxwell-Boltzmann distribution curve flattens and shifts to the right, significantly increasing the fraction of molecules (and thus the fraction of collisions) with energy greater than or equal to \(E_a\).
PastPaper.markingScheme
1 mark for selecting option A. Correctly identifying that \(E_a\) is independent of temperature, while collision frequency and the fraction of high-energy collisions both increase with temperature.
PastPaper.question 10 · multiple-choice
1 PastPaper.marks
A white solid oxide of a Period 3 element, \(X\), reacts violently with water to form a strongly acidic solution. A different liquid chloride of a Period 3 element, \(Y\), also reacts violently with water to form a strongly acidic solution and white fumes.
Phosphorus(V) oxide, \(P_4O_{10}\), is a white solid that reacts violently with water to form phosphoric(V) acid, a strong acid: \(P_4O_{10}(s) + 6H_2O(l) \rightarrow 4H_3PO_4(aq)\). Silicon(IV) chloride, \(SiCl_4\), is a volatile liquid that hydrolyses violently in water to form silicon(IV) oxide precipitate and white fumes of toxic hydrogen chloride gas: \(SiCl_4(l) + 2H_2O(l) \rightarrow SiO_2(s) + 4HCl(g)\). Other options are incorrect because \(SiO_2\) and \(Al_2O_3\) do not react with water, and \(NaCl\) dissolves neutrally in water without reacting violently.
PastPaper.markingScheme
1 mark for option A. Correctly matching the violent reactions of Period 3 oxides and chlorides with water to identify \(P_4O_{10}\) and \(SiCl_4\).
PastPaper.question 11 · multiple-choice
1 PastPaper.marks
In a calorimetry experiment, \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(NaOH(aq)\) is mixed with \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(HCl(aq)\) in a polystyrene cup. The temperature of both solutions is initially \(22.0\,^\circ\text{C}\). After mixing, the maximum temperature reached is \(28.7\,^\circ\text{C}\).
Assuming the density of the final mixture is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\), what is the enthalpy change of neutralisation, \(\Delta H_n\), in \(\text{kJ mol}^{-1}\)?
A.\(-56.0\text{ kJ mol}^{-1}\)
B.\(-28.0\text{ kJ mol}^{-1}\)
C.\(-112\text{ kJ mol}^{-1}\)
D.\(+56.0\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
1. Calculate total volume and mass: \(V = 50.0 + 50.0 = 100.0\text{ cm}^3\), so mass \(m = 100.0\text{ g}\). 2. Calculate temperature change: \(\Delta T = 28.7 - 22.0 = 6.7\text{ K}\). 3. Calculate heat released: \(q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J} = 2.8006\text{ kJ}\). 4. Calculate chemical amount of water formed: \(n = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). 5. Calculate standard enthalpy change of neutralisation: \(\Delta H_n = -q / n = -2.8006 / 0.0500 = -56.0\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark for option A. Correctly calculates heat energy released (\(q\)), chemical amount of reactant, and the final enthalpy change with correct negative sign for exothermic neutralisation.
PastPaper.question 12 · multiple-choice
1 PastPaper.marks
How many isomeric esters have the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\)?
A.2
B.3
C.4
D.5
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
We construct ester structures of the form \(\text{R}^1-\text{COO}-\text{R}^2\), where \(\text{R}^2\) must be an alkyl group (at least 1 carbon) and \(\text{R}^1\) can be hydrogen or an alkyl group. The total number of carbons in \(\text{R}^1\) and \(\text{R}^2\) must be 3. 1. If \(\text{R}^1\) has 0 carbons (H-), \(\text{R}^2\) has 3 carbons: propyl methanoate (\(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)) and isopropyl methanoate (\(\text{HCOOCH}(\text{CH}_3)_2\)). 2. If \(\text{R}^1\) has 1 carbon (\(\text{CH}_3\)-), \(\text{R}^2\) has 2 carbons: ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)). 3. If \(\text{R}^1\) has 2 carbons (\(\text{CH}_3\text{CH}_2\)-), \(\text{R}^2\) has 1 carbon: methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)). There are no other ester combinations, and none exhibit stereoisomerism. Thus, there are exactly 4 isomeric esters.
PastPaper.markingScheme
1 mark for option C. Correct identification of the 4 structural isomers that are esters.
PastPaper.question 13 · multiple-choice
1 PastPaper.marks
In acidic solution, manganate(VII) ions oxidize tin(II) ions to tin(IV) ions, and are themselves reduced to manganese(II) ions.
The unbalanced ionic equation is: \(a\,\text{MnO}_4^-(aq) + b\,\text{Sn}^{2+}(aq) + c\,\text{H}^+(aq) \rightarrow d\,\text{Mn}^{2+}(aq) + e\,\text{Sn}^{4+}(aq) + f\,\text{H}_2\text{O}(l)\)
What is the ratio \(b : a\) in the balanced equation?
A.2 : 1
B.2 : 5
C.5 : 2
D.1 : 2
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Using oxidation numbers: Manganese is reduced from \(+7\) in \(MnO_4^-\) to \(+2\) in \(Mn^{2+}\) (a decrease of 5). Tin is oxidized from \(+2\) in \(Sn^{2+}\) to \(+4\) in \(Sn^{4+}\) (an increase of 2). To balance the electronic charges / change in oxidation states, the ratio of tin oxidized to manganese reduced must be \(5 : 2\). Therefore, \(a = 2\) and \(b = 5\), making the ratio \(b : a = 5 : 2\).
PastPaper.markingScheme
1 mark for option C. Correctly determines the stoichiometric ratio of tin(II) to manganate(VII) by balancing electron transfer or oxidation number changes.
PastPaper.question 14 · multiple-choice
1 PastPaper.marks
The following exothermic reaction is in a state of dynamic equilibrium in a closed vessel:
Which of the following changes will result in an increase in both the equilibrium constant, \(K_p\), and the equilibrium yield of \(\text{SO}_3(g)\)?
A.decreasing the temperature
B.increasing the temperature
C.increasing the total pressure
D.adding a platinum catalyst
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The equilibrium constant, \(K_p\), is only affected by temperature. Since the forward reaction is exothermic, decreasing the temperature shifts the equilibrium position to the right to release heat (according to Le Chatelier's principle), which increases both the equilibrium yield of \(SO_3(g)\) and the equilibrium constant, \(K_p\). Increasing pressure shifts the equilibrium to the right but does not alter the value of \(K_p\).
PastPaper.markingScheme
1 mark for option A. Correctly identifying that only a temperature change alters the equilibrium constant, and a temperature decrease shifts an exothermic reaction towards the products.
PastPaper.question 15 · multiple-choice
1 PastPaper.marks
Using the standard enthalpy changes of formation provided in the table, what is the standard enthalpy change of combustion, \(\Delta H_c^\ominus\), of methane, \(\text{CH}_4(g)\), in \(\text{kJ mol}^{-1}\)?
1 mark for option A. Correct setup of Hess's Law cycle, accounting for stoichiometric coefficients (especially \(2 \times H_2O\)) and carrying out calculation with correct arithmetic signs.
PastPaper.question 16 · multiple-choice
1 PastPaper.marks
The rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane are compared by reacting them with aqueous silver nitrate in ethanol at \(50\,^\circ\text{C\)}.
Which of the following correctly describes the order of rate of reaction (from fastest to slowest) and identifies the correct reason for this trend?
A.Order: 1-iodobutane > 1-bromobutane > 1-chlorobutane; Reason: the \(\text{C}-\text{I}\) bond is the weakest bond and requires the least energy to break.
B.Order: 1-chlorobutane > 1-bromobutane > 1-iodobutane; Reason: chlorine is the most electronegative halogen, making the adjacent carbon atom highly electron-deficient and vulnerable to nucleophilic attack.
C.Order: 1-iodobutane > 1-bromobutane > 1-chlorobutane; Reason: iodine is the most electronegative halogen, which stabilizes the carbocation intermediate.
D.Order: 1-chlorobutane > 1-bromobutane > 1-iodobutane; Reason: the \(\text{C}-\text{Cl}\) bond is the weakest bond and requires the least energy to break.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes depends primarily on the carbon-halogen bond strength (bond enthalpy) and not on bond polarity. Down Group 17, the carbon-halogen bonds become longer and weaker: \(C-Cl\) (\(338\text{ kJ mol}^{-1}\)) > \(C-Br\) (\(276\text{ kJ mol}^{-1}\)) > \(C-I\) (\(238\text{ kJ mol}^{-1}\)). Because the \(C-I\) bond has the lowest bond enthalpy, it is the easiest to break, allowing 1-iodobutane to react the fastest. Conversely, the \(C-Cl\) bond is the strongest, making 1-chlorobutane react the slowest.
PastPaper.markingScheme
1 mark for option A. Correctly identifies 1-iodobutane as hydrolysing the fastest, and explains the trend based on carbon-halogen bond strength/enthalpy rather than polarity.
PastPaper.question 17 · multiple-choice
1 PastPaper.marks
The diagram shows a Maxwell-Boltzmann distribution of molecular energies for a gas-phase reaction. Which statement is correct when the temperature of the gas is increased at constant volume?
A.The peak of the curve shifts to the right and is higher.
B.The activation energy, \(E_a\), decreases.
C.The total area under the curve increases.
D.The fraction of molecules with energy greater than or equal to \(E_a\) increases.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
An increase in temperature causes the Maxwell-Boltzmann distribution curve to flatten and shift to the right. The peak of the curve becomes lower and shifts to a higher energy. The total area under the curve represents the total number of molecules, which remains constant. The activation energy, \(E_a\), is unaffected by temperature. However, the proportion of molecules with energy greater than or equal to \(E_a\) (the area under the curve to the right of \(E_a\)) increases, leading to a higher rate of reaction.
PastPaper.markingScheme
1 mark for identifying that increased temperature increases the fraction of molecules with energy greater than or equal to \(E_a\).
PastPaper.question 18 · multiple-choice
1 PastPaper.marks
Two Period 3 elements, \(X\) and \(Y\), react separately with oxygen to form oxides. The oxide of \(X\) dissolves in water to form a solution with a pH of approximately 2. The oxide of \(Y\) is insoluble in water but reacts with both aqueous sodium hydroxide and dilute hydrochloric acid. Which elements are \(X\) and \(Y\)?
A.\(X = \text{Si}\), \(Y = \text{Al}\)
B.\(X = \text{P}\), \(Y = \text{Al}\)
C.\(X = \text{S}\), \(Y = \text{Si}\)
D.\(X = \text{Na}\), \(Y = \text{Al}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Phosphorus reacts with oxygen to form phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)), which dissolves in water to form phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)), producing a strongly acidic solution with a pH of approximately 2. Aluminium reacts with oxygen to form aluminium oxide (\(\text{Al}_2\text{O}_3\)), which is amphoteric; it is insoluble in water but reacts with both acids and bases. Silicon dioxide (\(\text{SiO}_2\)) is acidic but insoluble in water and only reacts with concentrated alkalis, not dilute HCl.
PastPaper.markingScheme
1 mark for identifying phosphorus oxide as acidic and soluble (pH ~2) and aluminium oxide as amphoteric.
PastPaper.question 19 · multiple-choice
1 PastPaper.marks
In a calorimetry experiment, \(100\text{ cm}^3\) of \(1.0\text{ mol dm}^{-3}\) \(\text{HCl(aq)}\) is mixed with \(100\text{ cm}^3\) of \(1.0\text{ mol dm}^{-3}\) \(\text{NaOH(aq)}\) in a polystyrene cup. The temperature rises by \(5.8\text{ }^{\circ}\text{C}\). Assume the density of the solution is \(1.0\text{ g cm}^{-3}\) and its specific heat capacity is \(4.2\text{ J g}^{-1}\text{ K}^{-1}\). What is the standard enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), for this reaction?
A.\(-97.4\text{ kJ mol}^{-1}\)
B.\(-48.7\text{ kJ mol}^{-1}\)
C.\(-24.4\text{ kJ mol}^{-1}\)
D.\(-4.9\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, calculate the heat released, \(q = m c \Delta T\). Total volume of solution = \(100 + 100 = 200\text{ cm}^3\). Thus, mass \(m = 200\text{ g}\). \(q = 200\text{ g} \times 4.2\text{ J g}^{-1}\text{ K}^{-1} \times 5.8\text{ K} = 4872\text{ J} = 4.872\text{ kJ}\). Number of moles of water formed = \(0.100\text{ dm}^3 \times 1.0\text{ mol dm}^{-3} = 0.10\text{ mol}\). Enthalpy change of neutralisation, \(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{4.872\text{ kJ}}{0.10\text{ mol}} = -48.72\text{ kJ mol}^{-1} \approx -48.7\text{ kJ mol}^{-1}\). The reaction is exothermic, so the value is negative.
PastPaper.markingScheme
1 mark for calculating correct enthalpy of neutralisation with negative sign.
PastPaper.question 20 · multiple-choice
1 PastPaper.marks
How many isomers with the molecular formula \(\text{C}_4\text{H}_8\) can decolourise a solution of bromine in an organic solvent?
A.3
B.4
C.5
D.6
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
To decolourise bromine, the compound must be an alkene. There are four alkene isomers with the molecular formula \(\text{C}_4\text{H}_8\): (1) but-1-ene, (2) *cis*-but-2-ene, (3) *trans*-but-2-ene, and (4) 2-methylpropene. Cyclobutane and methylcyclopropane do not have a \(\text{C}=\text{C}\) double bond and will not readily decolourise bromine under standard conditions.
PastPaper.markingScheme
1 mark for correctly identifying that there are exactly 4 alkene isomers of formula \(\text{C}_4\text{H}_8\).
PastPaper.question 21 · multiple-choice
1 PastPaper.marks
When hot concentrated sulfuric acid reacts with solid potassium iodide, several products are formed including \(\text{I}_2\), \(\text{H}_2\text{S}\), \(\text{SO}_2\), and \(\text{S}\). In which product does sulfur have the lowest oxidation state?
A.\(\text{SO}_2\)
B.\(\text{S}\)
C.\(\text{H}_2\text{S}\)
D.\(\text{H}_2\text{SO}_4\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Let's determine the oxidation state of sulfur in each species: - In \(\text{H}_2\text{SO}_4\), sulfur is in the \(+6\) oxidation state. - In \(\text{SO}_2\), sulfur is in the \(+4\) oxidation state. - In \(\text{S}\), sulfur is in the \(0\) oxidation state. - In \(\text{H}_2\text{S}\), sulfur is in the \(-2\) oxidation state. Therefore, sulfur has the lowest oxidation state (\(-2\)) in \(\text{H}_2\text{S}\).
PastPaper.markingScheme
1 mark for correctly assigning oxidation states and identifying \(\text{H}_2\text{S}\).
PastPaper.question 22 · multiple-choice
1 PastPaper.marks
The following equilibrium is established in a closed container at temperature \(T\): \[\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)} \quad \Delta H = +88\text{ kJ mol}^{-1}\] Which change will result in an increase in the value of the equilibrium constant, \(K_c\)?
A.Increasing the temperature of the container.
B.Decreasing the volume of the container at constant temperature.
C.Adding a catalyst at constant temperature.
D.Adding more \(\text{PCl}_5\text{(g)}\) at constant temperature.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The value of the equilibrium constant, \(K_c\), is solely temperature-dependent. Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature shifts the position of equilibrium to the right, increasing the concentration of products relative to reactants, thereby increasing the value of \(K_c\). Pressure/volume changes, catalyst additions, or concentration changes do not affect \(K_c\).
PastPaper.markingScheme
1 mark for identifying that only a temperature increase for an endothermic reaction increases \(K_c\).
PastPaper.question 23 · multiple-choice
1 PastPaper.marks
Consider the following standard enthalpy changes of formation, \(\Delta H_{\text{f}}^{\ominus}\): - \(\Delta H_{\text{f}}^{\ominus}[\text{CO}_2\text{(g)}] = -393.5\text{ kJ mol}^{-1}\) - \(\Delta H_{\text{f}}^{\ominus}[\text{H}_2\text{O(l)}] = -285.8\text{ kJ mol}^{-1}\) - \(\Delta H_{\text{f}}^{\ominus}[\text{C}_2\text{H}_5\text{OH(l)}] = -277.6\text{ kJ mol}^{-1}\)
What is the standard enthalpy change of combustion, \(\Delta H_{\text{c}}^{\ominus}\), of liquid ethanol, \(\text{C}_2\text{H}_5\text{OH(l)}\)?
A.\(-401.7\text{ kJ mol}^{-1}\)
B.\(-956.9\text{ kJ mol}^{-1}\)
C.\(-1366.8\text{ kJ mol}^{-1}\)
D.\(-1644.4\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The combustion of ethanol is represented by the equation: \(\text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}\) Using Hess's law: \(\Delta H_{\text{c}}^{\ominus} = \sum \Delta H_{\text{f}}^{\ominus}(\text{products}) - \sum \Delta H_{\text{f}}^{\ominus}(\text{reactants})\) \(\Delta H_{\text{c}}^{\ominus} = [2 \times (-393.5) + 3 \times (-285.8)] - [-277.6]\) \(\Delta H_{\text{c}}^{\ominus} = [-787.0 - 857.4] + 277.6\) \(\Delta H_{\text{c}}^{\ominus} = -1644.4 + 277.6 = -1366.8\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark for correctly applying Hess's Law and performing the arithmetic to obtain \(-1366.8\text{ kJ mol}^{-1}\).
PastPaper.question 24 · multiple-choice
1 PastPaper.marks
Three different halogenoalkanes, 1-chlorobutane, 1-bromobutane, and 1-iodobutane, are reacted separately with aqueous silver nitrate in ethanol at \(50\text{ }^{\circ}\text{C}\). Which statement correctly explains the relative rates of these reactions?
A.1-chlorobutane reacts the fastest because the \(\text{C}-\text{Cl}\) bond is the most polar.
B.1-iodobutane reacts the fastest because the \(\text{C}-\text{I}\) bond has the lowest bond enthalpy.
C.1-chlorobutane reacts the fastest because the chlorine atom has the highest electronegativity.
D.All three halogenoalkanes react at the same rate because they are all primary halogenoalkanes.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes depends on the strength (bond enthalpy) of the \(\text{C}-\text{X}\) bond. The \(\text{C}-\text{I}\) bond is the longest and weakest (lowest bond enthalpy) among the carbon-halogen bonds, so it breaks most readily. Consequently, 1-iodobutane reacts the fastest, forming a precipitate of silver iodide almost immediately. Electronegativity and bond polarity would predict the opposite trend but play a minor role compared to bond enthalpy.
PastPaper.markingScheme
1 mark for identifying that 1-iodobutane reacts fastest due to having the lowest carbon-halogen bond enthalpy.
PastPaper.question 25 · multiple-choice
1 PastPaper.marks
Dilute hydrochloric acid reacts with excess marble chips (calcium carbonate) to produce carbon dioxide gas.
Which of the following changes will increase the initial rate of the reaction while keeping the total volume of carbon dioxide gas collected constant?
A.Decreasing the temperature of the hydrochloric acid
B.Using the same volume of hydrochloric acid but at half the concentration
C.Doubling the concentration of the hydrochloric acid while halving its volume
D.Using larger pieces of marble chips of the same mass
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The total volume of carbon dioxide gas produced depends on the number of moles of the limiting reactant, which is hydrochloric acid (since marble chips are in excess).
If we double the concentration of \(\text{HCl}\) and halve its volume, the number of moles of \(\text{HCl}\) remains the same: \(n = (2c) \times \left(\frac{V}{2}\right) = cV\). Therefore, the total volume of \(\text{CO}_2\) gas collected will remain constant.
Since the concentration of \(\text{HCl}\) is doubled, the frequency of successful collisions between \(\text{H}^+\) ions and the surface of the marble chips increases, which increases the initial rate of the reaction.
PastPaper.markingScheme
1 mark: Correctly identifies that doubling concentration while halving volume keeps the moles of reactant (and thus volume of gas) constant while increasing the rate.
PastPaper.question 26 · multiple-choice
1 PastPaper.marks
An oxide of a Period 3 element, \(X\), dissolves in water to form a solution with a pH less than 3.
A chloride of a different Period 3 element, \(Y\), reacts vigorously with water to form a strongly acidic solution and a white precipitate.
What are the identities of elements \(X\) and \(Y\)?
A.\(X = \text{Al}\); \(Y = \text{P}\)
B.\(X = \text{Si}\); \(Y = \text{Al}\)
C.\(X = \text{P}\); \(Y = \text{Si}\)
D.\(X = \text{Mg}\); \(Y = \text{S}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
- Element \(X\) must be phosphorus (\(P\)) or sulfur (\(S\)) because their oxides (such as \(\text{P}_4\text{O}_{10}\) or \(\text{SO}_3\)) dissolve in water to form strongly acidic solutions (\(\text{H}_3\text{PO}_4\) or \(\text{H}_2\text{SO}_4\)) with pH values well below 3. - Element \(Y\) must be silicon (\(Si\)) because silicon tetrachloride, \(\text{SiCl}_4\), undergoes rapid hydrolysis when added to water to produce a white precipitate of silicon dioxide, \(\text{SiO}_2\), and acidic fumes of hydrogen chloride, \(\text{HCl}\): $$\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})$$ Therefore, the correct pair is \(X = \text{P}\) and \(Y = \text{Si}\).
PastPaper.markingScheme
1 mark: Correctly identifies \(X\) as phosphorus and \(Y\) as silicon based on the reaction of their oxide and chloride with water.
PastPaper.question 27 · multiple-choice
1 PastPaper.marks
A student carries out a calorimetry experiment to determine the enthalpy change of combustion of propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (\(M_r = 60.0\)).
A burner containing propan-1-ol is weighed before and after combustion.
- Mass of propan-1-ol burned = \(0.60\text{ g}\) - Mass of water in the copper calorimeter = \(100.0\text{ g}\) - Temperature rise of the water = \(12.0\text{ }^{\circ}\text{C}\) - Specific heat capacity of water = \(4.18\text{ J g}^{-1}\text{ K}^{-1}\)
What is the experimental value of the standard enthalpy change of combustion, \(\Delta H_c\), of propan-1-ol from this experiment?
A.\(-5.02\text{ kJ mol}^{-1}\)
B.\(-50.2\text{ kJ mol}^{-1}\)
C.\(-502\text{ kJ mol}^{-1}\)
D.\(-3010\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
1. Calculate the heat energy absorbed by the water: \(q = m c \Delta T\) \(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 12.0\text{ K} = 5016\text{ J} = 5.016\text{ kJ}\)
2. Calculate the number of moles of propan-1-ol burned: \(n = \frac{\text{mass}}{M_r} = \frac{0.60\text{ g}}{60.0\text{ g mol}^{-1}} = 0.010\text{ mol}\)
3. Calculate the enthalpy change of combustion (with a negative sign since the reaction is exothermic): \(\Delta H_c = -\frac{q}{n} = -\frac{5.016\text{ kJ}}{0.010\text{ mol}} = -501.6\text{ kJ mol}^{-1} \approx -502\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark: Correctly calculates heat transferred, moles of fuel, and the resulting enthalpy change of combustion with negative sign.
PastPaper.question 28 · multiple-choice
1 PastPaper.marks
How many stereoisomers exist for 3-methylhex-4-en-3-ol?
A.2
B.4
C.6
D.8
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, draw the structure of 3-methylhex-4-en-3-ol: $$\text{CH}_3-\text{CH}_2-\text{C}^*(\text{CH}_3)(\text{OH})-\text{CH}=\text{CH}-\text{CH}_3$$
Identify the sources of stereoisomerism: 1. Carbon 3 (labeled with an asterisk \(^*\)) is a chiral carbon (asymmetric carbon center) because it is bonded to four different groups: - an ethyl group (\(-\text{CH}_2\text{CH}_3\)) - a methyl group (\(-\text{CH}_3\)) - a hydroxyl group (\(-\text{OH}\)) - a prop-1-enyl group (\(-\text{CH}=\text{CHCH}_3\)) This chiral center gives rise to 2 optical configurations (R and S).
2. The carbon-carbon double bond between Carbon 4 and Carbon 5 (\(-\text{CH}=\text{CH}-\)) can exhibit geometrical (cis-trans / E-Z) isomerism because: - Carbon 4 is bonded to hydrogen and the rest of the chain. - Carbon 5 is bonded to hydrogen and a methyl group. This double bond gives rise to 2 geometric configurations (cis and trans).
Since the molecule is unsymmetrical, the total number of stereoisomers is \(2^1 \times 2^1 = 4\) (namely: R-cis, R-trans, S-cis, and S-trans).
PastPaper.markingScheme
1 mark: Correctly identifies 1 chiral center and 1 double bond capable of cis-trans isomerism, giving \(2 \times 2 = 4\) stereoisomers.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
In which of the following reactions does the oxidation number of the underlined carbon element increase by exactly 4?
Let's determine the change in oxidation number of the carbon atom in each reaction:
- In option A: In \(\text{CH}_4\), the oxidation number of C is \(-4\). In \(\text{CO}_2\), it is \(+4\). The increase is \(+8\). - In option B: In \(\text{C}_2\text{H}_5\text{OH}\), the average oxidation number of C is \(-2\). In \(\text{CO}_2\), it is \(+4\). The increase is \(+6\). - In option C: In \(\text{CH}_3\text{OH}\), the oxidation number of C is calculated as: \(x + 4(+1) + 1(-2) = 0 \Rightarrow x = -2\). In \(\text{HCOOH}\), the oxidation number of C is calculated as: \(x + 2(+1) + 2(-2) = 0 \Rightarrow x = +2\). The increase in oxidation number is \((+2) - (-2) = +4\). - In option D: In \(\text{HCHO}\), the oxidation number of C is 0. In \(\text{HCOOH}\), it is \(+2\). The increase is \(+2\).
PastPaper.markingScheme
1 mark: Correctly calculates the oxidation numbers of carbon in both methanol (\(-2\)) and methanoic acid (\(+2\)) and identifies the change as \(+4\).
PastPaper.question 30 · multiple-choice
1 PastPaper.marks
The following reaction is in dynamic equilibrium in a closed vessel at temperature \(T\):
Which of the following changes will increase the value of the equilibrium constant, \(K_c\)?
A.Adding a catalyst at constant temperature
B.Increasing the total pressure of the vessel by reducing its volume at constant temperature
C.Increasing the temperature of the reaction vessel
D.Adding more \(\text{PCl}_5(\text{g})\) at constant temperature
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The value of the equilibrium constant, \(K_c\), is solely dependent on temperature. Changes in pressure, concentration, or the addition of a catalyst will not alter the value of \(K_c\) at a constant temperature.
Since the forward reaction is endothermic (\(\Delta H > 0\)), according to Le Chatelier's principle, increasing the temperature will shift the position of the equilibrium to the right to absorb the added heat energy. This increases the concentrations of the products (\(\text{PCl}_3\) and \(\text{Cl}_2\)) and decreases the concentration of the reactant (\(\text{PCl}_5\)) at the new equilibrium state, thereby increasing the value of \(K_c\).
PastPaper.markingScheme
1 mark: Correctly identifies that only temperature changes \(K_c\) and that an increase in temperature increases \(K_c\) for an endothermic reaction.
What is the standard enthalpy change of formation, \(\Delta H_f^{\ominus}\), of dimethyl ether, \(\text{CH}_3\text{OCH}_3(\text{g})\)?
A.\(-3104.4\text{ kJ mol}^{-1}\)
B.\(-184.4\text{ kJ mol}^{-1}\)
C.\(+184.4\text{ kJ mol}^{-1}\)
D.\(+3104.4\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, write the balanced chemical equation for the complete combustion of dimethyl ether, \(\text{CH}_3\text{OCH}_3(\text{g})\): $$\text{CH}_3\text{OCH}_3(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})$$
Note that \(\Delta H_f^{\ominus}[\text{O}_2(\text{g})] = 0\) since it is an element in its standard state.
Substitute the given values into the equation: $$-1460.0 = 2(-393.5) + 3(-285.8) - \Delta H_f^{\ominus}[\text{CH}_3\text{OCH}_3(\text{g})]$$ $$-1460.0 = -787.0 - 857.4 - \Delta H_f^{\ominus}[\text{CH}_3\text{OCH}_3(\text{g})]$$ $$-1460.0 = -1644.4 - \Delta H_f^{\ominus}[\text{CH}_3\text{OCH}_3(\text{g})]$$ $$\Delta H_f^{\ominus}[\text{CH}_3\text{OCH}_3(\text{g})] = -1644.4 - (-1460.0) = -184.4\text{ kJ mol}^{-1}$$
PastPaper.markingScheme
1 mark: Correctly sets up the Hess's law equation, applies stoichiometry (2 moles of \(\text{CO}_2\) and 3 moles of \(\text{H}_2\text{O}\)), and correctly solves for the unknown enthalpy of formation.
PastPaper.question 32 · multiple-choice
1 PastPaper.marks
Four different halogenoalkanes are heated separately with aqueous silver nitrate in ethanol at \(50\text{ }^{\circ}\text{C}\).
Which halogenoalkane will produce a precipitate the fastest?
A.1-chlorobutane
B.2-chloro-2-methylpropane
C.1-iodobutane
D.2-iodo-2-methylpropane
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes depends on two key factors: 1. **The strength of the C-Halogen bond (Bond Enthalpy)**: The C-I bond is much weaker than the C-Cl bond (lower bond enthalpy), meaning that iodoalkanes hydrolyse significantly faster than chloroalkanes because the carbon-halogen bond is cleaved much more easily. 2. **The structure of the halogenoalkane**: Tertiary halogenoalkanes hydrolyse much faster than primary halogenoalkanes. This is because tertiary halogenoalkanes undergo hydrolysis via an \(\text{S}_\text{N}1\) mechanism, which involves the formation of a highly stable tertiary carbocation intermediate. Primary halogenoalkanes react via the slower \(\text{S}_\text{N}2\) pathway.
Combining both factors, **2-iodo-2-methylpropane** (a tertiary iodoalkane) will undergo hydrolysis at the fastest rate, rapidly releasing iodide ions (\(\text{I}^-\)) which react with silver ions (\(\text{Ag}^+\)) to form a yellow precipitate of silver iodide (\(\text{AgI}\)) almost immediately.
PastPaper.markingScheme
1 mark: Correctly identifies that tertiary iodoalkane has both the weakest carbon-halogen bond and forms the most stable carbocation intermediate, leading to the fastest rate of hydrolysis.
PastPaper.question 33 · multiple-choice
1 PastPaper.marks
Which statement correctly explains why an increase in temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is lowered, which increases the proportion of successful collisions.
B.The rate constant, \(k\), increases because a larger fraction of colliding molecules have energy greater than or equal to the activation energy.
C.The frequency of collisions increases, which is the sole factor responsible for the significant increase in reaction rate.
D.The average kinetic energy of the reacting gas molecules doubles when the temperature increases from \(25\ ^\circ\text{C}\) to \(50\ ^\circ\text{C}\).
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Increasing the temperature of a reaction increases the average kinetic energy of the molecules. This shifts the Maxwell-Boltzmann distribution of molecular energies to the right, meaning a much larger fraction of colliding molecules have kinetic energy greater than or equal to the activation energy, \(E \ge E_{\text{a}}\). This results in a higher frequency of successful collisions, which increases the rate constant, \(k\), and therefore the overall rate of reaction. Note that the activation energy itself remains unchanged by temperature, and the increase in collision frequency alone is not enough to account for the dramatic rate increase. The average kinetic energy is proportional to the absolute temperature in Kelvin, so raising the temperature from \(25\ ^\circ\text{C}\) (\(298\text{ K}\)) to \(50\ ^\circ\text{C}\) (\(323\text{ K}\)) does not double the kinetic energy.
PastPaper.markingScheme
1 mark: Identify that the rate constant increases due to a larger fraction of molecules possessing energy \(\ge E_{\text{a}}\).
PastPaper.question 34 · multiple-choice
1 PastPaper.marks
Two Period 3 elements, \(X\) and \(Y\), react separately with oxygen to form the oxides \(XO_2\) and \(Y_2O_3\) respectively. \(XO_2\) is a gas at room temperature that dissolves in water to form an acidic solution. \(Y_2O_3\) is a solid at room temperature that is insoluble in water.
Which statement is correct?
A.Both \(XO_2\) and \(Y_2O_3\) react with excess aqueous sodium hydroxide.
B.Both \(XO_2\) and \(Y_2O_3\) react with excess dilute hydrochloric acid.
C.The oxidation state of the element in \(XO_2\) is \(+2\), whereas in \(Y_2O_3\) it is \(+3\).
D.The bonding in \(XO_2\) is purely ionic, whereas the bonding in \(Y_2O_3\) is purely covalent.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Element \(X\) must be sulfur, forming the acidic gaseous oxide \(\text{SO}_2\). Element \(Y\) must be aluminium, forming the amphoteric oxide \(\text{Al}_2\text{O}_3\). Since \(\text{SO}_2\) is an acidic oxide, it reacts with the base \(\text{NaOH(aq)}\) to form sodium sulfite and water. Since \(\text{Al}_2\text{O}_3\) is amphoteric, it reacts with both acids (such as \(\text{HCl}\)) and bases (such as \(\text{NaOH}\)) to form soluble aluminate salts. Therefore, both oxides react with excess aqueous sodium hydroxide. \(\text{SO}_2\) does not react with hydrochloric acid (ruling out option B). The oxidation state of sulfur in \(\text{SO}_2\) is \(+4\) (ruling out option C). \(\text{SO}_2\) has a simple molecular covalent structure, whereas \(\text{Al}_2\text{O}_3\) has a giant ionic structure with significant covalent character (ruling out option D).
PastPaper.markingScheme
1 mark: Identify that both sulfur dioxide and aluminium oxide react with sodium hydroxide due to their acidic and amphoteric natures, respectively.
PastPaper.question 35 · multiple-choice
1 PastPaper.marks
In an experiment to determine the enthalpy change of combustion of methanol, \(\text{CH}_3\text{OH}\), a student burned \(0.80\text{ g}\) of methanol. The heat released was used to raise the temperature of \(100.0\text{ g}\) of water from \(20.0\ ^\circ\text{C}\) to \(45.0\ ^\circ\text{C}\).
Only \(70\%\) of the heat released by the combustion was absorbed by the water.
[Specific heat capacity of water, \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\); \(M_{\text{r}}\) of methanol = \(32.0\)]
What is the calculated experimental enthalpy change of combustion of methanol?
A.-293 kJ mol\(^{-1}\)
B.-418 kJ mol\(^{-1}\)
C.-597 kJ mol\(^{-1}\)
D.-853 kJ mol\(^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, calculate the heat energy absorbed by the water: \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (45.0 - 20.0)\text{ K} = 10450\text{ J} = 10.45\text{ kJ}\).
Since this represents only \(70\%\) of the total heat released by the combustion of methanol, the total heat released (\(Q_{\text{total}}\)) is: \(Q_{\text{total}} = \frac{10.45\text{ kJ}}{0.70} = 14.93\text{ kJ}\).
Next, find the moles of methanol burned: \(n = \frac{0.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.025\text{ mol}\).
1 mark: Calculate the correct heat absorbed (10.45 kJ), adjust for heat loss (70% efficiency to get 14.93 kJ), and divide by moles (0.025 mol) to yield -597 kJ/mol.
PastPaper.question 36 · multiple-choice
1 PastPaper.marks
An acyclic organic compound \(W\) has the molecular formula \(\text{C}_4\text{H}_7\text{Cl}\) and contains a chiral carbon atom. When \(W\) reacts with hydrogen in the presence of a nickel catalyst, compound \(Z\) is formed.
Which statement about \(W\) and \(Z\) is correct?
A.\(W\) exists as a pair of cis-trans isomers.
B.\(Z\) does not contain a chiral carbon atom.
C.\(W\) reacts with hot concentrated ethanolic sodium hydroxide to form a diene.
D.\(Z\) is achiral because the double bond has been saturated.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The acyclic isomer of \(\text{C}_4\text{H}_7\text{Cl}\) containing a chiral carbon is \(3\)-chlorobut-\(1\)-ene, \(\text{CH}_2=\text{CH}-\text{C}^*\text{H(Cl)}-\text{CH}_3\). - Option A is incorrect because the double bond is at the end of the chain (\(\text{CH}_2=\)), so it cannot show cis-trans isomerism. - When \(W\) is hydrogenated, the product \(Z\) is \(2\)-chlorobutane, \(\text{CH}_3-\text{CH}_2-\text{C}^*\text{H(Cl)}-\text{CH}_3\). This still contains a chiral carbon, so it is chiral and exists as a pair of optical isomers (ruling out options B and D). - Option C is correct: \(3\)-chlorobut-\(1\)-ene has a chlorine atom on the carbon adjacent to the double bond. Treatment with hot concentrated ethanolic sodium hydroxide causes elimination of \(\text{HCl}\) to yield a conjugated diene, buta-\(1,3\)-diene (\(\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2\)).
PastPaper.markingScheme
1 mark: Correctly identify W as 3-chlorobut-1-ene and recognize that its elimination forms buta-1,3-diene.
PastPaper.question 37 · multiple-choice
1 PastPaper.marks
An acidified solution containing \(0.020\text{ mol}\) of \(\text{XO}_4^-\) ions is completely reduced by \(0.050\text{ mol}\) of \(\text{SO}_3^{2-}\) ions. During this reaction, the sulfite ions are oxidized to sulfate ions, \(\text{SO}_4^{2-}\).
What is the final oxidation state of the element \(X\) after reduction?
A.+2
B.+3
C.+4
D.+5
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The oxidation half-equation for the conversion of sulfite to sulfate is: \(\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 2\text{H}^+ + 2\text{e}^-\) Each mole of \(\text{SO}_3^{2-}\) loses \(2\text{ mol}\) of electrons. Therefore, \(0.050\text{ mol}\) of \(\text{SO}_3^{2-}\) releases: \(0.050\text{ mol} \times 2 = 0.10\text{ mol of electrons}\).
These \(0.10\text{ mol}\) of electrons are absorbed by \(0.020\text{ mol}\) of \(\text{XO}_4^-\) ions. The number of moles of electrons absorbed per mole of \(\text{XO}_4^-\) is: \(\frac{0.10\text{ mol}}{0.020\text{ mol}} = 5\).
The initial oxidation state of \(X\) in \(\text{XO}_4^-\) is \(+7\) (since \(X + 4(-2) = -1\)). After absorbing \(5\) electrons, the oxidation state is reduced by \(5\): \(+7 - 5 = +2\).
PastPaper.markingScheme
1 mark: Calculate total moles of electrons transferred (0.10 mol), then the electron change per mole of X (5 mol), and deduce the final oxidation state (+2) from the initial state (+7).
PastPaper.question 38 · multiple-choice
1 PastPaper.marks
The reversible reaction below is in dynamic equilibrium in a sealed container of fixed volume at \(250\ ^\circ\text{C}\):
Which change would result in an increase in the partial pressure of \(\text{Cl}_2\text{(g)}\) at the new equilibrium?
A.Adding a dynamic catalyst to the system.
B.Adding helium gas to the container at constant volume.
C.Decreasing the temperature of the container.
D.Injecting more \(\text{PCl}_5\text{(g)}\) into the container.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
- Option A: A catalyst increases the rates of both forward and reverse reactions equally, leaving the equilibrium position and partial pressures unchanged. - Option B: Adding an inert gas like helium at constant volume increases the total pressure but does not affect the partial pressures of the reacting gases, so the position of equilibrium is unchanged. - Option C: Since the reaction is endothermic (\(\Delta H > 0\)), decreasing the temperature will shift the equilibrium in the exothermic (reverse) direction, decreasing the partial pressure of \(\text{Cl}_2\). - Option D: Injecting more reactant \(\text{PCl}_5\text{(g)}\) increases its concentration and partial pressure, shifting the equilibrium to the right and thus increasing the partial pressure of \(\text{Cl}_2\text{(g)}\).
PastPaper.markingScheme
1 mark: Determine that injecting a reactant shifts the equilibrium in the forward direction, increasing the amount and partial pressure of the products.
PastPaper.question 39 · multiple-choice
1 PastPaper.marks
The standard enthalpy changes of combustion, \(\Delta H_{\text{c}}^\theta\), of three substances are listed below:
What is the standard enthalpy change of formation of propanoic acid, \(\Delta H_{\text{f}}^\theta [\text{CH}_3\text{CH}_2\text{COOH(l)}]\)?
A.-3567 kJ mol\(^{-1}\)
B.-513 kJ mol\(^{-1}\)
C.+513 kJ mol\(^{-1}\)
D.+847 kJ mol\(^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The equation for the formation of propanoic acid from its elements in their standard states is: \(3\text{C(s)} + 3\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(l)}\)
1 mark: Correctly set up the Hess's law cycle incorporating stoichiometry, and calculate the standard enthalpy change of formation to be -513 kJ/mol.
PastPaper.question 40 · multiple-choice
1 PastPaper.marks
Equal amounts of \(1\)-chlorobutane, \(1\)-bromobutane, and \(1\)-iodobutane are heated separately with aqueous silver nitrate in ethanol under identical conditions.
Which row correctly identifies the fastest halogenoalkane to hydrolyze and the correct explanation?
A.Fastest: \(1\)-chlorobutane; Explanation: the \(\text{C}-\text{Cl}\) bond is the most polar bond.
B.Fastest: \(1\)-iodobutane; Explanation: the \(\text{C}-\text{I}\) bond is the weakest bond.
C.Fastest: \(1\)-iodobutane; Explanation: the iodide ion is the best nucleophile.
D.Fastest: \(1\)-chlorobutane; Explanation: the \(\text{C}-\text{Cl}\) bond has the lowest activation energy.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
During the hydrolysis of a halogenoalkane, the carbon-halogen bond is broken. The rate of this nucleophilic substitution reaction is determined by the strength of the bond (bond enthalpy), not by its polarity. Down Group 17, atomic radius increases, making the carbon-halogen bond longer and weaker (lower bond enthalpy). Thus, the \(\text{C}-\text{I}\) bond is the weakest and easiest to break, meaning \(1\)-iodobutane is hydrolyzed the fastest. The \(\text{C}-\text{Cl}\) bond is the strongest and hardest to break, meaning \(1\)-chlorobutane is hydrolyzed the slowest.
PastPaper.markingScheme
1 mark: Identify that 1-iodobutane hydrolyzes fastest because the C-I bond is the weakest bond.
Paper 23 AS Level Structured Questions
Answer all questions. Write your answers in the spaces provided. Show all your working and use appropriate units.
4 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · structured-written
15 PastPaper.marks
This question is about reaction kinetics.
(a) Define the term *activation energy*, \(E_a\). [1]
(b) Describe and explain how an increase in temperature affects the rate of a chemical reaction using the Maxwell-Boltzmann distribution of molecular energies. Your response should describe how the shape of the curve changes at a higher temperature and how this relates to successful collisions. [5]
(c) Draw a mental picture or describe the positions on a Maxwell-Boltzmann energy distribution curve of the activation energy of an uncatalysed reaction, \(E_a\), compared to a catalysed reaction, \(E_c\). Explain, in terms of activation energy and collision frequency, how a catalyst increases the rate of a chemical reaction. [4]
(d) The reaction between peroxodisulfate ions, \(\text{S}_2\text{O}_8^{2-}\), and iodide ions, \(\text{I}^-\), is very slow in the absence of a catalyst. (i) Write an ionic equation for this reaction. [1] (ii) Explain why this reaction has a high activation energy in the absence of a catalyst. [2] (iii) Aqueous iron(II) ions, \(\text{Fe}^{2+}(\text{aq})\), act as a homogeneous catalyst for this reaction. Write two equations to show how \(\text{Fe}^{2+}(\text{aq})\) acts as a catalyst in this reaction. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Activation energy is the minimum energy that colliding particles must possess in order to react.
(b) At a higher temperature (\(T_2 > T_1\)), the Maxwell-Boltzmann distribution curve changes as follows: - The peak of the curve is lower and shifts to the right (higher energy). - The curve tails off at higher energy values, meaning more molecules have high kinetic energy. - The area under the curve to the right of the activation energy (\(E_a\)) represents the number of molecules with energy greater than or equal to \(E_a\). - At a higher temperature, a significantly larger fraction of molecules have energy \(\ge E_a\). - Therefore, the frequency of successful/effective collisions increases, which results in an increased rate of reaction.
(c) On a Maxwell-Boltzmann curve: - The catalysed activation energy (\(E_c\)) is positioned to the left of the uncatalysed activation energy (\(E_a\)) on the energy axis (i.e., \(E_c < E_a\)). - A catalyst provides an alternative reaction pathway with a lower activation energy. - Because \(E_c\) is lower, a much larger proportion/fraction of colliding molecules have sufficient energy to react (\(\text{Energy} \ge E_c\)). - This leads to a higher frequency of successful collisions per unit time, thereby increasing the rate of reaction.
(d)(ii) Both the peroxodisulfate ion (\(\text{S}_2\text{O}_8^{2-}\)) and the iodide ion (\(\text{I}^-\)) are negatively charged (anions). Like charges repel each other, creating a high electrostatic repulsion barrier to bring them close enough to react, leading to a high activation energy.
(a) [1 mark]: Minimum energy required for colliding particles to react.
(b) [5 marks total]: - [1 mark]: Axis labels: y-axis as 'Number/fraction of molecules' and x-axis as 'Energy/kinetic energy'. - [1 mark]: Curve at higher temperature has a lower peak shifted to the right. - [1 mark]: Curve starts at origin and asymptotic to x-axis. - [1 mark]: Explains that area under the curve to the right of \(E_a\) represents the proportion of molecules with energy \(\ge E_a\). - [1 mark]: Explains that at higher temperature, more molecules have energy \(\ge E_a\), leading to more successful collisions per unit time.
(c) [4 marks total]: - [1 mark]: Identifies that \(E_c\) is located to the left of \(E_a\) (lower energy value). - [1 mark]: States that the catalyst provides an alternative pathway with a lower activation energy. - [1 mark]: States that a larger fraction/proportion of molecules have energy \(\ge E_c\). - [1 mark]: Concludes there is a higher frequency of successful/effective collisions.
(d) [5 marks total]: - (i) [1 mark]: Correctly balanced equation with correct chemical species: \(\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2\). - (ii) [2 marks]: [1 mark] for stating both ions are negatively charged / are anions; [1 mark] for stating they repel each other (electrostatic repulsion). - (iii) [2 marks]: [1 mark] for each correct chemical equation showing the catalytic cycle.
PastPaper.question 2 · structured-written
15 PastPaper.marks
This question is about Period 3 elements and their oxides.
(a) Describe what is observed and write a balanced chemical equation when the following elements are heated and reacted with oxygen: (i) Sodium [2] (ii) Phosphorus [2]
(b) Sodium oxide, \(\text{Na}_2\text{O}\), and sulfur trioxide, \(\text{SO}_3\), are oxides of Period 3 elements. (i) State the type of bonding present in each of these oxides. [2] (ii) Write chemical equations for the separate reactions of sodium oxide and sulfur trioxide with water. State the approximate pH of the resulting solution in each case. [4]
(c) Aluminium oxide, \(\text{Al}_2\text{O}_3\), is amphoteric. (i) State what is meant by the term *amphoteric*. [1] (ii) Write ionic equations to demonstrate the amphoteric nature of aluminium oxide. Your equations must show its reaction with an acid and its reaction with a strong alkali. [4]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a)(i) Observation: Sodium burns with a bright yellow-orange flame to produce a white solid. Equation: \(4\text{Na}(\text{s}) + \text{O}_2(\text{g}) \rightarrow 2\text{Na}_2\text{O}(\text{s})\) (or \(2\text{Na}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{Na}_2\text{O}_2(\text{s})\))
(a)(ii) Observation: Phosphorus burns with a very bright white flame to produce thick white clouds of solid phosphorus(V) oxide. Equation: \(4\text{P}(\text{s}) + 5\text{O}_2(\text{g}) \rightarrow \text{P}_4\text{O}_{10}(\text{s})\) (or \(2\text{P}_2\text{O}_5\))
(a) [4 marks total]: - (i) [1 mark]: Yellow/orange flame or white solid product. - (i) [1 mark]: Correctly balanced equation: \(4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}\). - (ii) [1 mark]: White flame or white smoke/clouds. - (ii) [1 mark]: Correctly balanced equation: \(4\text{P} + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}\).
(b) [6 marks total]: - (i) [1 mark]: Ionic for \(\text{Na}_2\text{O}\). - (i) [1 mark]: Covalent/molecular for \(\text{SO}_3\). - (ii) [1 mark]: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\). - (ii) [1 mark]: pH 13 or 14. - (ii) [1 mark]: \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\). - (ii) [1 mark]: pH 1 or 2.
(c) [5 marks total]: - (i) [1 mark]: Reacts with both acids and bases. - (ii) [2 marks]: Acid equation (1 mark for correct species, 1 mark for balancing: \(\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}\)). - (ii) [2 marks]: Base equation (1 mark for correct species, 1 mark for balancing: \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al}(\text{OH})_4]^-\)).
PastPaper.question 3 · structured-written
15 PastPaper.marks
This question is about chemical energetics and Hess's law.
(a) Define the term *standard enthalpy change of formation*, \(\Delta H_f^\ominus\). [2]
(b) The standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), at 298 K for carbon, hydrogen, and propanoic acid, \(\text{C}_2\text{H}_5\text{COOH}(\text{l})\), are given below: - Carbon, \(\text{C}(\text{s})\): \(-393.5\text{ kJ mol}^{-1}\) - Hydrogen, \(\text{H}_2(\text{g})\): \(-285.8\text{ kJ mol}^{-1}\) - Propanoic acid, \(\text{C}_2\text{H}_5\text{COOH}(\text{l})\): \(-1527.3\text{ kJ mol}^{-1}\)
(i) Write the balanced chemical equation, including state symbols, for the reaction representing the standard enthalpy change of formation of propanoic acid. [2] (ii) Draw a Hess's law cycle showing the relationship between these standard enthalpy changes of combustion and the standard enthalpy change of formation of propanoic acid. Label all enthalpy changes clearly. [2] (iii) Calculate the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of propanoic acid. [3]
(c) A student carried out a calorimetry experiment to determine the enthalpy change of neutralization. \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(\text{HCl}(\text{aq})\) was mixed with \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(\text{NaOH}(\text{aq})\) in a polystyrene cup. The initial temperature of both solutions was \(21.4\ ^\circ\text{C}\) and the maximum temperature reached was \(28.1\ ^\circ\text{C}\).
(i) Calculate the heat energy, \(q\), in kJ, released during this neutralization reaction. Show your working and state one major assumption you made in your calculation. [Assume the density of the final mixture is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\ \text{K}^{-1}\)]. [3] (ii) Calculate the standard enthalpy change of neutralization, \(\Delta H_{neu}\), in \(\text{kJ mol}^{-1}\), for this reaction. Include the appropriate sign. [3]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) The standard enthalpy change of formation, \(\Delta H_f^\ominus\), is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions of \(298\text{ K}\) and \(100\text{ kPa}\).
(b)(i) Propanoic acid, \(\text{C}_2\text{H}_5\text{COOH}\), contains 3 carbon atoms, 6 hydrogen atoms, and 2 oxygen atoms. Therefore, the formation reaction is: \(3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{COOH}(\text{l})\)
(b)(ii) The Hess's Law cycle can be represented as: Reactants: \(3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g})\) on top-left. Product: \(\text{C}_2\text{H}_5\text{COOH}(\text{l})\) on top-right. Combustion products: \(3\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})\) at the bottom. - Arrow from Reactants to Product is labelled \(\Delta H_f^\ominus\). - Arrow from Reactants to Combustion products is labelled \(3\Delta H_c^\ominus[\text{C}] + 3\Delta H_c^\ominus[\text{H}_2]\). - Arrow from Product to Combustion products is labelled \(\Delta H_c^\ominus[\text{C}_2\text{H}_5\text{COOH}]\).
(c)(i) Mass of solution, \(m = 50.0 + 50.0 = 100.0\text{ g}\). Temperature change, \(\Delta T = 28.1 - 21.4 = 6.7\ ^\circ\text{C}\) (or \(6.7\text{ K}\)). \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\ \text{K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.80\text{ kJ}\) (to 3 sig figs). Assumption: No heat energy was lost to the surroundings / the heat capacity of the polystyrene cup is negligible / density and specific heat capacity of the solution are identical to those of pure water.
(c)(ii) Moles of acid reacting: \(n(\text{HCl}) = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). Moles of alkali reacting: \(n(\text{NaOH}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). Moles of water formed in reaction: \(0.0500\text{ mol}\). \(\Delta H_{neu} = -\frac{q}{n(\text{H}_2\text{O})} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.0\text{ kJ mol}^{-1}\) (to 3 sig figs).
PastPaper.markingScheme
(a) [2 marks total]: - [1 mark]: Enthalpy change when 1 mole of a compound is formed from its constituent elements. - [1 mark]: Under standard conditions and with elements in their standard states.
(b) [7 marks total]: - (i) [2 marks]: [1 mark] for correct chemical formulas and balancing: \(3\text{C} + 3\text{H}_2 + \text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{COOH}\); [1 mark] for correct state symbols: \(\text{C}(\text{s})\), \(\text{H}_2(\text{g})\), \(\text{O}_2(\text{g})\), \(\text{C}_2\text{H}_5\text{COOH}(\text{l})\). - (ii) [2 marks]: [1 mark] for correct arrangement of reactants, products, and combustion products; [1 mark] for correct arrow directions and matching enthalpy labels. - (iii) [3 marks]: [1 mark] for correct mathematical expression; [1 mark] for correct substitution of numbers; [1 mark] for final answer of \(-510.6\text{ kJ mol}^{-1}\) (must have negative sign and unit).
(c) [6 marks total]: - (i) [3 marks]: [1 mark] for correct temperature change (\(6.7\text{ K}\)) and mass (\(100\text{ g}\)); [1 mark] for calculated heat of \(2.80\text{ kJ}\) (accept \(2.8\text{ kJ}\)); [1 mark] for stating a valid assumption (e.g., no heat lost to surroundings). - (ii) [3 marks]: [1 mark] for calculating the moles of water formed as \(0.0500\text{ mol}\); [1 mark] for dividing the heat by moles of water; [1 mark] for final answer of \(-56.0\text{ kJ mol}^{-1}\) (must have negative sign, 3 significant figures, and correct unit).
PastPaper.question 4 · structured-written
15 PastPaper.marks
This question is about halogenoalkanes and their isomerism.
(a) Compound A is a halogenoalkane with the molecular formula \(\text{C}_4\text{H}_9\text{Br}\). (i) Draw the skeletal structures of all four structural isomers of \(\text{C}_4\text{H}_9\text{Br}\). Underneath each structure, state its IUPAC name. [4] (ii) Identify the structural isomer of \(\text{C}_4\text{H}_9\text{Br}\) that exhibits optical isomerism. Explain why this isomer is optically active, and describe how the structures of its two enantiomers are related. [4]
(b) When 2-bromo-2-methylpropane is heated with aqueous sodium hydroxide, 2-methylpropan-2-ol is formed. (i) State the name of the reaction mechanism. [1] (ii) Describe this reaction mechanism using curly arrows to show the movement of electron pairs. Show the structures of the organic reactant, intermediate, and product, as well as any inorganic ions involved. [4] (iii) Explain why 2-bromo-2-methylpropane reacts via this mechanism rather than the alternative substitution mechanism favoured by primary halogenoalkanes. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a)(i) The four structural isomers of \(\text{C}_4\text{H}_9\text{Br}\) are: 1. 1-bromobutane: skeletal structure is a 4-carbon chain with Br on C1. 2. 2-bromobutane: skeletal structure is a 4-carbon chain with Br on C2. 3. 1-bromo-2-methylpropane: skeletal structure is a 3-carbon chain with a methyl branch on C2 and Br on C1. 4. 2-bromo-2-methylpropane: skeletal structure is a 3-carbon chain with a methyl branch on C2 and Br on C2.
(a)(ii) Isomer exhibiting optical isomerism: 2-bromobutane. Explanation: It has a chiral carbon (or asymmetric carbon) atom (Carbon-2), which is bonded to four different atoms/groups of atoms: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{Br}\). Relationship: The two enantiomers are non-superimposable mirror images of each other.
(b)(ii) Mechanism steps: - Step 1: Heterolytic fission of the C–Br bond. Curly arrow starts from the C–Br bond and points to the Br atom. This forms a tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\). - Step 2: Nucleophilic attack. Curly arrow starts from a lone pair on the hydroxide ion (\(\text{OH}^-\)) and points to the positively charged carbon atom of the carbocation. - Product: \((\text{CH}_3)_3\text{COH}\) (2-methylpropan-2-ol) is formed.
(b)(iii) 2-bromo-2-methylpropane is a tertiary halogenoalkane. The tertiary carbocation intermediate formed in the first step is highly stabilized by the positive inductive (+I) effect of the three electron-donating alkyl (methyl) groups. In addition, steric hindrance by the bulky methyl groups prevents direct backside attack by the nucleophile (which would occur in an \(\text{S}_\text{N}2\) mechanism).
PastPaper.markingScheme
(a) [8 marks total]: - (i) [4 marks]: [0.5 marks] for each correct skeletal structure; [0.5 marks] for each correct IUPAC name matching the structure. - 1-bromobutane - 2-bromobutane - 1-bromo-2-methylpropane - 2-bromo-2-methylpropane - (ii) [4 marks total]: - [1 mark]: Correctly identifies 2-bromobutane. - [1 mark]: Explains that it contains a chiral carbon / asymmetric carbon. - [1 mark]: Identifies the 4 different groups attached: H, methyl (\(-\text{CH}_3\)), ethyl (\(-\text{CH}_2\text{CH}_3\)), and bromine (\(-\text{Br}\)). - [1 mark]: States that the enantiomers are non-superimposable mirror images.
(b) [7 marks total]: - (i) [1 mark]: \(\text{S}_\text{N}1\). - (ii) [4 marks]: - [1 mark]: Curly arrow from C–Br bond to Br. - [1 mark]: Correct planar tertiary carbocation structure, \((\text{CH}_3)_3\text{C}^+\), and \(\text{Br}^-\). - [1 mark]: Curly arrow from lone pair on \(\text{OH}^-\) to the positive carbon atom. - [1 mark]: Correct product structure of 2-methylpropan-2-ol. - (iii) [2 marks]: - [1 mark]: The tertiary carbocation is stabilized by the positive inductive effect (+I) of 3 alkyl/methyl groups. - [1 mark]: Bulkiness of 3 methyl groups causes steric hindrance, preventing \(\text{S}_\text{N}2\) backside attack.
Paper 33 Advanced Practical Skills 1
Answer all questions. Where appropriate, prepare tables for your results. Show appropriate significant figures and working.
3 PastPaper.question · 39.99 PastPaper.marks
PastPaper.question 1 · practical-experiment
13.33 PastPaper.marks
In this experiment, you will determine the value of \( x \) in the formula of hydrated sodium thiosulfate, \( \text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O} \), by titrating a solution of this salt against iodine. Iodine is generated in situ by reacting a standard solution of potassium iodate(V), \( \text{KIO}_3 \), with excess potassium iodide, \( \text{KI} \), in acidic conditions. Equation 1: \( \text{IO}_3^-(\text{aq}) + 5\text{I}^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 3\text{I}_2(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \) and Equation 2: \( \text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq}) \). Chemicals provided: FB 1 is a solution containing \( 24.80\text{ g dm}^{-3} \) of hydrated sodium thiosulfate, \( \text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O} \). FB 2 is \( 0.0150\text{ mol dm}^{-3} \) potassium iodate(V), \( \text{KIO}_3 \). FB 3 is \( 1.0\text{ mol dm}^{-3} \) sulfuric acid, \( \text{H}_2\text{SO}_4 \). FB 4 is \( 0.5\text{ mol dm}^{-3} \) potassium iodide, \( \text{KI} \). Starch indicator. Method: 1. Fill a burette with FB 1. 2. Pipette \( 25.0\text{ cm}^3 \) of FB 2 into a conical flask. 3. Use a measuring cylinder to add about \( 10\text{ cm}^3 \) of FB 3 and \( 10\text{ cm}^3 \) of FB 4 to the flask. The solution turns dark brown. 4. Titrate the mixture with FB 1 until the brown color fades to light yellow. 5. Add a few drops of starch indicator to turn the solution blue-black. 6. Continue adding FB 1 dropwise until the blue-black color disappears, leaving a colorless solution. (a) Prepare a table for your titration results, showing clearly the initial and final burette readings, and the volume of FB 1 added for each titration. State the volume of FB 1 to be used in your calculations. (b) Calculate the number of moles of iodate(V) ions, \( \text{IO}_3^- \), in \( 25.0\text{ cm}^3 \) of FB 2. (c) Show that \( 1.00\text{ mol} \) of \( \text{IO}_3^- \) reacts to produce iodine that requires \( 6.00\text{ mol} \) of \( \text{S}_2\text{O}_3^{2-} \) for complete reaction. (d) Calculate the concentration of sodium thiosulfate in FB 1 in \( \text{mol dm}^{-3} \), assuming a mean titre of \( 22.50\text{ cm}^3 \). (e) Using the concentration calculated in (d) and the mass concentration of FB 1 (\( 24.80\text{ g dm}^{-3} \)), calculate the experimental relative formula mass of \( \text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O} \) and determine the value of \( x \). [\( A_r \): \( \text{Na} = 23.0 \), \( \text{S} = 32.1 \), \( \text{O} = 16.0 \), \( \text{H} = 1.0 \)] (f) Identify one potential source of systematic error in this practical procedure and suggest an improvement to minimize it.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) A standard titration table should show: final reading, initial reading, and titre volume, all recorded to 2 decimal places (ending in .00 or .05). The concordant titres (within \( 0.10\text{ cm}^3 \)) must be clearly indicated, and the mean titre calculated. For subsequent calculations, we assume the mean titre is \( 22.50\text{ cm}^3 \). (b) Moles of \( \text{IO}_3^- \) in \( 25.0\text{ cm}^3 \) of FB 2: \( n(\text{IO}_3^-) = \frac{25.0}{1000} \times 0.0150 = 3.75 \times 10^{-4}\text{ mol} \). (c) From Equation 1, \( 1\text{ mol of }\text{IO}_3^- \) produces \( 3\text{ mol of }\text{I}_2 \). From Equation 2, \( 1\text{ mol of }\text{I}_2 \) reacts with \( 2\text{ mol of }\text{S}_2\text{O}_3^{2-} \). Therefore, \( 3\text{ mol of }\text{I}_2 \) react with \( 6\text{ mol of }\text{S}_2\text{O}_3^{2-} \). Thus, \( 1\text{ mol of }\text{IO}_3^- \equiv 6\text{ mol of }\text{S}_2\text{O}_3^{2-} \). (d) Moles of \( \text{S}_2\text{O}_3^{2-} \) in the mean titre: \( n(\text{S}_2\text{O}_3^{2-}) = 6 \times 3.75 \times 10^{-4} = 2.25 \times 10^{-3}\text{ mol} \). Concentration of FB 1: \( [\text{S}_2\text{O}_3^{2-}] = \frac{2.25 \times 10^{-3}\text{ mol}}{22.50/1000\text{ dm}^3} = 0.100\text{ mol dm}^{-3} \). (e) Experimental relative formula mass (\( M_r \)) of \( \text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O} \): \( M_r = \frac{\text{mass concentration}}{\text{molar concentration}} = \frac{24.80\text{ g dm}^{-3}}{0.100\text{ mol dm}^{-3}} = 248.0 \). Formula mass of anhydrous \( \text{Na}_2\text{S}_2\text{O}_3 = 2(23.0) + 2(32.1) + 3(16.0) = 158.2 \). Mass of water of crystallisation: \( 248.0 - 158.2 = 89.8 \). Hence, \( x = \frac{89.8}{18.0} = 4.99 \approx 5 \). (f) Potential systematic error: Loss of volatile iodine vapor from the open flask during or before titration. Improvement: Stopper the flask immediately after adding KI and acid, or keep the flask in an ice bath to reduce iodine volatility before titrating.
PastPaper.markingScheme
[Total: 13.33 marks] (a) [2 marks] 1 mark for structured table with appropriate column headings and all readings to 2 decimal places. 1 mark for selecting concordant titres and obtaining a sensible mean value. (b) [2 marks] 1 mark for correct calculation formula, 1 mark for final answer of \( 3.75 \times 10^{-4}\text{ mol} \) to 3 significant figures. (c) [2 marks] 1 mark for relating Equation 1 stoichiometry, 1 mark for combining with Equation 2 to show the 1 to 6 mole ratio. (d) [2.33 marks] 1 mark for multiplying moles of iodate by 6, 1.33 marks for dividing by the mean titre volume to get \( 0.100\text{ mol dm}^{-3} \). (e) [3 marks] 1 mark for calculating experimental \( M_r = 248.0 \), 1 mark for working out anhydrous mass of 158.2, 1 mark for solving \( x = 5 \) (or near integer). (f) [2 marks] 1 mark for identifying iodine volatility as the error, 1 mark for suggesting the use of a stoppered flask or keeping it cold.
PastPaper.question 2 · practical-experiment
13.33 PastPaper.marks
In this experiment, you will determine the enthalpy change of hydration of anhydrous copper(II) sulfate, \( \text{CuSO}_4 \), by applying Hess's Law. You will measure the temperature changes that occur when anhydrous copper(II) sulfate and hydrated copper(II) sulfate are separately dissolved in water. Chemicals and apparatus provided: FB 1 is anhydrous copper(II) sulfate powder, \( \text{CuSO}_4 \). FB 2 is hydrated copper(II) sulfate crystals, \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \). Plastic cup (calorimeter) and thermometer. Method: Experiment 1: Dissolving FB 1. 1. Support the plastic cup in a beaker. Transfer \( 50.0\text{ cm}^3 \) of distilled water into the cup. 2. Record the temperature of the water every minute for 3 minutes. 3. At \( t = 4\text{ minutes} \), add \( 3.99\text{ g} \) of FB 1. Stir continuously. 4. Record the temperature of the solution every minute from \( t = 5\text{ minutes} \) to \( t = 10\text{ minutes} \). Experiment 2: Dissolving FB 2. Repeat the procedure but use \( 45.0\text{ cm}^3 \) of water and add \( 6.24\text{ g} \) of FB 2 at \( t = 4\text{ minutes} \). (a) Use your extrapolated temperature-time values to determine the corrected maximum temperature change, \( \Delta T_1 \), for Experiment 1 and the corrected minimum temperature change, \( \Delta T_2 \), for Experiment 2. Assume experimental values of \( \Delta T_1 = +11.5\text{ }^\circ\text{C} \) and \( \Delta T_2 = -1.5\text{ }^\circ\text{C} \). (b) Calculate the heat energy change, \( q_1 \), in joules, when FB 1 dissolves in Experiment 1. [Assume the specific heat capacity of the solution is \( 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \) and the mass of the solution is \( 50.0\text{ g} \)]. (c) Calculate the enthalpy change of solution for anhydrous copper(II) sulfate, \( \Delta H_1 \), in \( \text{kJ mol}^{-1} \). [\( M_r \) of \( \text{CuSO}_4 = 159.6 \)]. (d) Calculate the enthalpy change of solution for hydrated copper(II) sulfate, \( \Delta H_2 \), in \( \text{kJ mol}^{-1} \). [\( M_r \) of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = 249.6 \); assume mass of solution is \( 50.0\text{ g} \)]. (e) Use Hess's Law to construct an energy cycle and calculate the enthalpy change of hydration, \( \Delta H_{\text{hyd}} \), of anhydrous copper(II) sulfate: \( \text{CuSO}_4(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(\text{s}) \). (f) State one source of heat loss in this experiment and suggest a practical improvement to minimize it.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Extrapolating the temperature curves to the 4th minute (time of addition) gives the corrected temperature changes: \( \Delta T_1 = +11.5\text{ }^\circ\text{C} \) and \( \Delta T_2 = -1.5\text{ }^\circ\text{C} \). (b) Heat energy change for Experiment 1: \( q_1 = m c \Delta T_1 = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 11.5\text{ }^\circ\text{C} = 2403.5\text{ J} = 2.404\text{ kJ} \). (c) Moles of \( \text{CuSO}_4 \) in Experiment 1: \( n_1 = \frac{3.99\text{ g}}{159.6\text{ g mol}^{-1}} = 0.0250\text{ mol} \). Enthalpy change of solution, \( \Delta H_1 = -\frac{q_1}{n_1} = -\frac{2.4035\text{ kJ}}{0.0250\text{ mol}} = -96.14\text{ kJ mol}^{-1} \approx -96.1\text{ kJ mol}^{-1} \). (d) Heat energy change for Experiment 2: \( q_2 = m c \Delta T_2 = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times (-1.5\text{ }^\circ\text{C}) = -313.5\text{ J} = -0.3135\text{ kJ} \). Moles of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) in Experiment 2: \( n_2 = \frac{6.24\text{ g}}{249.6\text{ g mol}^{-1}} = 0.0250\text{ mol} \). Enthalpy change of solution, \( \Delta H_2 = -\frac{q_2}{n_2} = -\frac{-0.3135\text{ kJ}}{0.0250\text{ mol}} = +12.54\text{ kJ mol}^{-1} \approx +12.5\text{ kJ mol}^{-1} \). (e) Using Hess's Law: \( \Delta H_1 = \Delta H_{\text{hyd}} + \Delta H_2 \). Therefore, \( \Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2 = -96.14 - (+12.54) = -108.68\text{ kJ mol}^{-1} \approx -108.7\text{ kJ mol}^{-1} \). (f) Heat loss occurs to the surrounding air and through the top of the open cup. Improvement: Place a lid on the plastic cup and/or wrap the plastic cup in cotton wool/bubble wrap for insulation.
PastPaper.markingScheme
[Total: 13.33 marks] (a) [2 marks] 1 mark for correct extrapolation on the temperature-time graph to the 4th minute. 1 mark for correct determination of \( \Delta T_1 \) and \( \Delta T_2 \). (b) [2 marks] 1 mark for correct formula \( q = m c \Delta T \) and 1 mark for calculating \( 2403.5\text{ J} \) (or 2.40 kJ). (c) [2 marks] 1 mark for calculating \( n = 0.0250\text{ mol} \), 1 mark for \( \Delta H_1 = -96.1\text{ kJ mol}^{-1} \) (negative sign required). (d) [2.33 marks] 1 mark for calculating \( q_2 = -313.5\text{ J} \) (or 313.5 J absorption), 1.33 marks for \( \Delta H_2 = +12.5\text{ kJ mol}^{-1} \) (positive sign required). (e) [3 marks] 1 mark for showing a fully labelled Hess's Law cycle, 2 marks for calculating \( \Delta H_{\text{hyd}} = -108.7\text{ kJ mol}^{-1} \). (f) [2 marks] 1 mark for identifying heat loss from the top or sides of the cup, 1 mark for suggesting a lid or extra insulation.
PastPaper.question 3 · practical-experiment
13.33 PastPaper.marks
In this experiment, you will investigate how the concentration of hydrochloric acid affects the rate of its reaction with magnesium. Equation: \( \text{Mg}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2(\text{g}) \). The rate of reaction is measured by timing how long it takes for a fixed length of magnesium ribbon to completely dissolve. Chemicals provided: FB 1 is \( 2.00\text{ mol dm}^{-3} \) hydrochloric acid, \( \text{HCl} \). FB 2 is distilled water. Five \( 2.0\text{ cm} \) pieces of magnesium ribbon. Method: 1. Prepare 5 reaction mixtures with total volume \( 10.0\text{ cm}^3 \) in five test-tubes: Tube 1 (\( 10.0\text{ cm}^3 \) FB 1, \( 0.0\text{ cm}^3 \) FB 2); Tube 2 (\( 8.0\text{ cm}^3 \) FB 1, \( 2.0\text{ cm}^3 \) FB 2); Tube 3 (\( 6.0\text{ cm}^3 \) FB 1, \( 4.0\text{ cm}^3 \) FB 2); Tube 4 (\( 4.0\text{ cm}^3 \) FB 1, \( 6.0\text{ cm}^3 \) FB 2); Tube 5 (\( 2.0\text{ cm}^3 \) FB 1, \( 8.0\text{ cm}^3 \) FB 2). 2. Add one piece of Mg ribbon to Tube 1 and record the time, \( t \), in seconds for the magnesium to completely disappear. 3. Repeat for tubes 2 to 5. (a) Prepare a table to record your results, showing the volumes of FB 1 and FB 2, the concentration of HCl, the reaction time \( t \), and the rate of reaction represented by \( 1/t \). (b) Assume that for Tube 1, \( t = 24\text{ s} \), and for Tube 3, \( t = 67\text{ s} \). Calculate the concentration of \( \text{HCl} \) in Tube 1 and Tube 3. (c) Briefly describe the shape of the graph obtained when plotting \( 1/t \) against the concentration of \( \text{HCl} \). (d) Deduce the order of reaction with respect to hydrochloric acid based on the relationship between rate (\( 1/t \)) and concentration. (e) A student suggests that using a larger volume of the reaction mixture (e.g., doubling the volume of both FB 1 and FB 2, keeping the concentration the same) would decrease the reaction time. State whether the student is correct and explain your answer.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) A suitable results table should contain headings with units: Volume of FB 1 / \( \text{cm}^3 \), Volume of FB 2 / \( \text{cm}^3 \), [HCl] / \( \text{mol dm}^{-3} \), Time \( t \) / \( \text{s} \), and Rate \( 1/t \) / \( \text{s}^{-1} \). (b) Concentration of HCl in Tube 1: \( [\text{HCl}] = \frac{10.0\text{ cm}^3}{10.0\text{ cm}^3} \times 2.00\text{ mol dm}^{-3} = 2.00\text{ mol dm}^{-3} \). Concentration of HCl in Tube 3: \( [\text{HCl}] = \frac{6.0\text{ cm}^3}{10.0\text{ cm}^3} \times 2.00\text{ mol dm}^{-3} = 1.20\text{ mol dm}^{-3} \). (c) Plotting \( 1/t \) on the y-axis against \( [\text{HCl}] \) on the x-axis gives a straight line passing through the origin. (d) Since the graph of rate (represented by \( 1/t \)) against concentration of \( \text{HCl} \) is a straight line through the origin, rate is directly proportional to concentration. Therefore, the reaction is first-order with respect to \( \text{HCl} \). (e) The student is incorrect. Doubling the volume of the mixture does not change the concentration of the reactants. Since the concentration of \( \text{HCl} \) remains constant, the rate of reaction remains unchanged, meaning the time taken for the magnesium ribbon to dissolve will remain the same.
PastPaper.markingScheme
[Total: 13.33 marks] (a) [3 marks] 1 mark for clear headings and units, 1 mark for showing all concentrations calculated correctly, 1 mark for listing \( 1/t \) values to 3 significant figures. (b) [2 marks] 1 mark for Tube 1 concentration (\( 2.00\text{ mol dm}^{-3} \)), 1 mark for Tube 3 concentration (\( 1.20\text{ mol dm}^{-3} \)). (c) [2.33 marks] 1.33 marks for identifying the linear relationship, 1 mark for stating that it passes through the origin. (d) [3 marks] 1 mark for stating 'first-order', 2 marks for justifying with direct proportionality (i.e. doubling concentration doubles the rate). (e) [3 marks] 1 mark for stating the student is incorrect, 1 mark for stating that concentration is unchanged, 1 mark for explaining that rate only depends on concentration, not total volume.