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 (AS MCQ)
Forty multiple-choice questions requiring candidates to select the single best option from four choices.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · MCQ
1 PastPaper.marks
Using the standard enthalpy of combustion values given below, what is the standard enthalpy of formation of liquid propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3(l)\)?
Award 1 mark for the correct calculation of \(-317.7 \text{ kJ mol}^{-1}\).
PastPaper.question 2 · MCQ
1 PastPaper.marks
An element, X, is in Period 3 of the Periodic Table. The first six successive ionisation energies (in \(\text{kJ mol}^{-1}\)) of X are shown in the table below:
Which formula represents the chloride formed when element X is reacted with excess chlorine gas?
A.\(\text{XCl}_3\)
B.\(\text{XCl}_4\)
C.\(\text{XCl}_5\)
D.\(\text{X}_2\text{Cl}_5\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
A large jump in successive ionisation energies is observed between the 5th and the 6th ionisation energies, indicating that the 6th electron is removed from a shell closer to the nucleus. Therefore, X has 5 valence electrons and is in Group 15. The Period 3 element in Group 15 is phosphorus (P). When reacted with excess chlorine gas, phosphorus forms phosphorus pentachloride, \(\text{PCl}_5\).
PastPaper.markingScheme
Award 1 mark for identifying Group 15 and selecting the formula representing the pentachloride.
PastPaper.question 3 · MCQ
1 PastPaper.marks
A sample of 0.180 g of a volatile organic compound of molecular formula \(\text{C}_n\text{H}_{2n+2}\text{O}\) is completely vaporised at a temperature of \(127^\circ\text{C}\) and a pressure of \(1.00 \times 10^5 \text{ Pa}\). The volume of vapor produced is \(99.7 \text{ cm}^3\).
What is the molecular formula of the compound? [\(R = 8.31 \text{ J K}^{-1}\text{ mol}^{-1}\)]
A.\(\text{C}_2\text{H}_6\text{O}\)
B.\(\text{C}_3\text{H}_8\text{O}\)
C.\(\text{C}_4\text{H}_{10}\text{O}\)
D.\(\text{C}_5\text{H}_{12}\text{O}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Using the ideal gas equation: \(pV = nRT = \frac{m}{M}RT \implies M = \frac{mRT}{pV}\)
For \(\text{C}_n\text{H}_{2n+2}\text{O}\): \(12n + (2n + 2) + 16 = 60 \implies 14n + 18 = 60 \implies 14n = 42 \implies n = 3\). Thus, the molecular formula is \(\text{C}_3\text{H}_8\text{O}\).
PastPaper.markingScheme
Award 1 mark for correctly using the ideal gas equation to determine the molar mass as \(60 \text{ g mol}^{-1}\) and matching it to \(\text{C}_3\text{H}_8\text{O}\).
PastPaper.question 4 · MCQ
1 PastPaper.marks
Four different halogenoalkanes are added separately to mixtures of ethanol and aqueous silver nitrate. The temperature of each mixture is maintained at \(50^\circ\text{C}\).
Which halogenoalkane produces a precipitate in the shortest time?
The rate of hydrolysis depends on two main factors: the strength of the carbon-halogen bond and the mechanism (primary vs tertiary carbocation stability). The C-I bond is much weaker than the C-Cl bond, meaning iodoalkanes hydrolyse much faster than chloroalkanes. Furthermore, tertiary halogenoalkanes undergo rapid hydrolysis via the \(\text{S}_\text{N}1\) mechanism due to the stability of the tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane, \(\text{(CH}_3\text{)}_3\text{CI}\), reacts the fastest.
PastPaper.markingScheme
Award 1 mark for choosing the option with the weakest C-halogen bond (iodide) and most stable tertiary structure.
PastPaper.question 5 · MCQ
1 PastPaper.marks
An increase in temperature from 300 K to 310 K often approximately doubles the rate of a chemical reaction.
Which statement correctly explains this observation?
A.The activation energy of the reaction decreases at higher temperatures.
B.The frequency of molecular collisions doubles.
C.The proportion of molecules with energy greater than or equal to the activation energy increases significantly.
D.The average kinetic energy of the reacting molecules doubles.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
An increase in temperature shifts the Maxwell-Boltzmann distribution of molecular energies to the right. Consequently, there is a very large and significant increase in the fraction of molecules with kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)). While collision frequency and average kinetic energy do increase, their increases are extremely small (on the order of 1-3%) and do not account for the doubling of the reaction rate. The activation energy value itself is temperature-independent.
PastPaper.markingScheme
Award 1 mark for identifying the correct explanation in terms of the Maxwell-Boltzmann distribution and the fraction of molecules exceeding the activation energy.
PastPaper.question 6 · MCQ
1 PastPaper.marks
How many structural isomers are there with the molecular formula \(\text{C}_3\text{H}_6\text{Cl}_2\)?
A.3
B.4
C.5
D.6
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
We can write down all possible arrangements of two chlorine atoms on a propane backbone: 1. 1,1-dichloropropane: \(\text{CH}_3\text{CH}_2\text{CHCl}_2\) 2. 1,2-dichloropropane: \(\text{CH}_3\text{CH(Cl)CH}_2\text{Cl}\) 3. 1,3-dichloropropane: \(\text{CH}_2\text{ClCH}_2\text{CH}_2\text{Cl}\) 4. 2,2-dichloropropane: \(\text{CH}_3\text{CCl}_2\text{CH}_3\)
There are exactly 4 structural isomers. (Note: 1,2-dichloropropane has a chiral center and exhibits stereoisomerism, but the question specifies 'structural isomers' only.)
PastPaper.markingScheme
Award 1 mark for identifying that there are exactly 4 structural isomers.
PastPaper.question 7 · MCQ
1 PastPaper.marks
A solid mixture contains 0.010 mol of anhydrous magnesium nitrate, \(\text{Mg(NO}_3\text{)}_2\), and 0.010 mol of anhydrous barium nitrate, \(\text{Ba(NO}_3\text{)}_2\). The mixture is heated strongly in an open crucible until there is no further change in mass.
What is the total mass of solid residue remaining in the crucible? [\(A_r\) values: \(\text{Mg} = 24.3\), \(\text{Ba} = 137.3\), \(\text{O} = 16.0\)]
A.1.62 g
B.1.94 g
C.3.02 g
D.4.10 g
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Both magnesium nitrate and barium nitrate undergo thermal decomposition when heated strongly to form their respective metal oxides, nitrogen dioxide, and oxygen: \(2\text{M(NO}_3\text{)}_2(s) \rightarrow 2\text{MO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)\)
Thus: 0.010 mol of \(\text{Mg(NO}_3\text{)}_2\) decomposes to yield 0.010 mol of \(\text{MgO}\). 0.010 mol of \(\text{Ba(NO}_3\text{)}_2\) decomposes to yield 0.010 mol of \(\text{BaO}\).
Award 1 mark for calculating the mass of both metal oxides correctly to get 1.94 g.
PastPaper.question 8 · MCQ
1 PastPaper.marks
In a closed container, 2.00 mol of \(\text{SO}_2(g)\) and 2.00 mol of \(\text{O}_2(g)\) are mixed at a constant temperature. The equilibrium mixture is represented by the equation:
At equilibrium, 1.60 mol of \(\text{SO}_3(g)\) is present. If the total volume of the container is \(2.00 \text{ dm}^3\), what is the value of the equilibrium constant, \(K_c\), at this temperature?
Substitute concentrations into the expression for \(K_c\): \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(0.80)^2}{(0.20)^2 \times 0.60} = \frac{0.64}{0.04 \times 0.60} = 26.67 \text{ dm}^3\text{ mol}^{-1} \approx 26.7 \text{ dm}^3\text{ mol}^{-1}\).
PastPaper.markingScheme
Award 1 mark for the correct calculation of \(K_c = 26.7 \text{ dm}^3\text{ mol}^{-1}\).
PastPaper.question 9 · MCQ
1 PastPaper.marks
A student carries out an experiment to determine the enthalpy change of combustion of pentan-1-ol (\(M_r = 88.0\)). A spirit burner containing pentan-1-ol is burned beneath a beaker containing \(150\text{ g}\) of water. The temperature of the water rises from \(20.5\text{ }^{\circ}\text{C}\) to \(44.5\text{ }^{\circ}\text{C}\). The mass of the spirit burner decreases by \(0.440\text{ g}\). The specific heat capacity of water is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). What is the enthalpy change of combustion of pentan-1-ol under these conditions?
A.\(-3010\text{ kJ mol}^{-1}\)
B.\(-1500\text{ kJ mol}^{-1}\)
C.\(-301\text{ kJ mol}^{-1}\)
D.\(+3010\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, calculate the heat energy absorbed by the water using \(q = m c \Delta T\): \(q = 150\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (44.5 - 20.5)\text{ K} = 15048\text{ J} = 15.048\text{ kJ}\).
Next, calculate the number of moles of pentan-1-ol burned: \(n = \frac{0.440\text{ g}}{88.0\text{ g mol}^{-1}} = 0.00500\text{ mol}\).
Finally, calculate the enthalpy change of combustion (which is exothermic, hence negative): \(\Delta H_c = -\frac{q}{n} = -\frac{15.048\text{ kJ}}{0.00500\text{ mol}} = -3009.6\text{ kJ mol}^{-1} \approx -3010\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark for calculating the correct enthalpy change value with the correct sign and rounding to 3 significant figures. Reject positive values or incorrect scale errors.
PastPaper.question 10 · MCQ
1 PastPaper.marks
For a particular gas-phase reaction, when the temperature is increased from \(300\text{ K}\) to \(310\text{ K}\), the rate of reaction approximately doubles. Which of the following statements correctly explains this observation?
A.The activation energy of the reaction is decreased, allowing a greater proportion of collisions to be successful.
B.The average kinetic energy of the reacting particles doubles, leading to twice as many collisions per second.
C.The frequency of successful collisions doubles because the average speed of the molecules doubles.
D.There is a large increase in the proportion of reactant molecules with energy greater than or equal to the activation energy.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
At a higher temperature, the Maxwell-Boltzmann distribution curve flattens and shifts to the right. This causes a very large (often exponential) increase in the proportion of molecules that possess energy greater than or equal to the activation energy (\(E \ge E_a\)). The average kinetic energy of the molecules and the collision frequency only increase by a very small percentage (about 3% and 1.7% respectively), which does not account for the doubling of the rate. The activation energy itself remains constant.
PastPaper.markingScheme
1 mark for identifying that the significant increase in reaction rate is due to the large increase in the proportion of molecules with \(E \ge E_a\).
PastPaper.question 11 · MCQ
1 PastPaper.marks
Which of the following species has the same number of unpaired d-electrons as the \(\text{Co}^{2+}\) ion in its ground state?
A.\(\text{Fe}^{2+}\)
B.\(\text{Ni}^{2+}\)
C.\(\text{Cr}^{3+}\)
D.\(\text{Cu}^{2+}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The electronic configuration of a cobalt atom is \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). Upon losing two electrons to form the \(\text{Co}^{2+}\) ion, the 4s electrons are lost first, leaving the configuration as \([\text{Ar}] 3\text{d}^7\). Under Hund's rule, the five d-orbitals contain two pairs and three unpaired electrons (total 3 unpaired d-electrons). Let's check the options: - \(\text{Fe}^{2+}\): \([\text{Ar}] 3\text{d}^6\) has 4 unpaired d-electrons. - \(\text{Ni}^{2+}\): \([\text{Ar}] 3\text{d}^8\) has 2 unpaired d-electrons. - \(\text{Cr}^{3+}\): \([\text{Ar}] 3\text{d}^3\) has 3 unpaired d-electrons. - \(\text{Cu}^{2+}\): \([\text{Ar}] 3\text{d}^9\) has 1 unpaired d-electron. Therefore, \(\text{Cr}^{3+}\) has the same number of unpaired d-electrons as \(\text{Co}^{2+}\).
PastPaper.markingScheme
1 mark for correctly determining the electronic configurations of both transition metal ions and matching their number of unpaired d-electrons.
PastPaper.question 12 · MCQ
1 PastPaper.marks
A student aims to synthesize propanoic acid from ethene in a three-step pathway:
Which set of reagents and conditions is most suitable for this synthesis?
A.Step 1: \(\text{HBr(g)}\) at room temperature; Step 2: \(\text{KCN}\) in ethanol, heat under reflux; Step 3: Dilute \(\text{HCl(aq)}\), heat under reflux
B.Step 1: \(\text{H}_2\text{O(g)}\), \(\text{H}_3\text{PO}_4\) catalyst, heat and pressure; Step 2: \(\text{KCN}\) in ethanol, heat under reflux; Step 3: Dilute \(\text{HCl(aq)}\), heat under reflux
C.Step 1: \(\text{HBr(g)}\) at room temperature; Step 2: \(\text{NaOH(aq)}\), heat under reflux; Step 3: Acidified \(\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}\), heat under reflux
D.Step 1: \(\text{Br}_2\text{ in CCl}_4\) at room temperature; Step 2: \(\text{KCN}\) in ethanol, heat under reflux; Step 3: Dilute \(\text{HCl(aq)}\), heat under reflux
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
To synthesize propanoic acid (3 carbons) from ethene (2 carbons), we must increase the carbon chain length by one. This is achieved in Step 2 by reacting a halogenoalkane with cyanide ions. - Step 1: React ethene with \(\text{HBr(g)}\) at room temperature to form bromoethane (Compound X). - Step 2: Heat bromoethane with \(\text{KCN}\) in ethanol under reflux to form propanenitrile (Compound Y). - Step 3: Heat propanenitrile under reflux with dilute acid, such as \(\text{HCl(aq)}\), to hydrolyze the nitrile group into a carboxylic acid group, forming propanoic acid.
PastPaper.markingScheme
1 mark for selecting the correct pathway that successfully increases the carbon chain length and yields propanoic acid with correct reagents and conditions.
PastPaper.question 13 · MCQ
1 PastPaper.marks
Consider the following heterogeneous equilibrium at a constant temperature \(T\):
The volume of the container is suddenly halved while maintaining the temperature at \(T\). Once the new equilibrium is established, how do the value of \(K_c\) and the mass of \(\text{C(s)}\) compare to their initial values?
A.\(K_c\) remains unchanged; mass of \(\text{C(s)}\) increases
B.\(K_c\) decreases; mass of \(\text{C(s)}\) increases
C.\(K_c\) remains unchanged; mass of \(\text{C(s)}\) decreases
D.\(K_c\) increases; mass of \(\text{C(s)}\) remains unchanged
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The equilibrium constant \(K_c\) is only dependent on temperature. Since the temperature is kept constant at \(T\), the value of \(K_c\) remains unchanged. Halving the volume of the container increases the pressure. According to Le Chatelier's principle, the system will shift to oppose this change by favoring the direction that produces fewer gas molecules. In this reaction, there is 1 mole of gas on the reactant side (\(\text{CO}_2\)) and 2 moles of gas on the product side (\(\text{CO}\)); the solid carbon does not contribute to the gas moles. Thus, the equilibrium shifts to the left, resulting in an increase in the mass of \(\text{C(s)}\).
PastPaper.markingScheme
1 mark for identifying that \(K_c\) is unchanged because temperature is constant, and the mass of the solid increases because the equilibrium shifts left with increased pressure.
PastPaper.question 14 · MCQ
1 PastPaper.marks
In the redox reaction between acidified potassium dichromate(VI) and tin(II) ions, the unbalanced equation is shown below:
1 mark for correctly applying Hess's Law, multiplying stoichiometric coefficients for reactants, and calculating \(+20.1\text{ kJ mol}^{-1}\).
PastPaper.question 16 · MCQ
1 PastPaper.marks
The reaction between peroxodisulfate ions, \(\text{S}_2\text{O}_8^{2-}\), and iodide ions, \(\text{I}^-\), is catalyzed by aqueous iron(II) ions, \(\text{Fe}^{2+}\)(aq):
Which statement best explains why \(\text{Fe}^{2+}\)(aq) acts as an effective homogeneous catalyst in this reaction?
A.It provides a solid surface on which both negatively charged reactant ions can adsorb, bringing them closer together.
B.It is easily oxidized to \(\text{Fe}^{3+}\) by peroxodisulfate ions, which then oxidize iodide ions to iodine, regenerating \(\text{Fe}^{2+}\).
C.It increases the kinetic energy of the reactant ions, allowing them to overcome the mutual electrostatic repulsion.
D.It reacts with both \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\) simultaneously to form a single, highly stable activated complex.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The reaction between two negatively charged ions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) has a high activation energy due to electrostatic repulsion. \(\text{Fe}^{2+}\) acts as a homogeneous catalyst because it can easily be oxidized to \(\text{Fe}^{3+}\) by peroxodisulfate ions (avoiding repulsion because they are oppositely charged), and the resulting \(\text{Fe}^{3+}\) can then easily oxidize the negative iodide ions to iodine, regenerating \(\text{Fe}^{2+}\). It works because transition metals have variable oxidation states that can participate in a multi-step pathway with lower overall activation energies.
PastPaper.markingScheme
1 mark for identifying the correct redox steps describing the homogeneous catalytic pathway of \(\text{Fe}^{2+}/\text{Fe}^{3+}\).
PastPaper.question 17 · MCQ
1 PastPaper.marks
A student burns 0.32 g of methanol (\(CH_3OH\), \(M_r = 32.0\)) to heat 150 cm³ of water in a copper beaker. The temperature of the water rises by 12.0 °C. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹. What is the experimental enthalpy change of combustion of methanol, \(\Delta H_c^\theta\), in kJ mol⁻¹?
A.-752 kJ mol⁻¹
B.-7.52 kJ mol⁻¹
C.+752 kJ mol⁻¹
D.-376 kJ mol⁻¹
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, calculate the heat energy transferred to the water using \(q = mc\Delta T\). The mass of water is 150 g (since density is 1.00 g cm⁻³). \(q = 150 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 12.0 \text{ K} = 7524 \text{ J} = 7.524 \text{ kJ}\). Next, calculate the number of moles of methanol burned: \(n = 0.32 \text{ g} / 32.0 \text{ g mol}^{-1} = 0.010 \text{ mol}\). Finally, calculate the enthalpy change of combustion per mole: \(\Delta H_c = -q / n = -7.524 \text{ kJ} / 0.010 \text{ mol} = -752.4 \text{ kJ mol}^{-1}\), which rounds to -752 kJ mol⁻¹.
PastPaper.markingScheme
1 mark for the correct calculation of standard enthalpy of combustion (-752 kJ mol⁻¹).
PastPaper.question 18 · MCQ
1 PastPaper.marks
Which statement about the Maxwell–Boltzmann energy distribution curve for a gaseous reaction is correct when the temperature is increased?
A.The peak of the curve shifts to the right and becomes higher.
B.The peak of the curve shifts to the left and becomes lower.
C.The peak of the curve shifts to the right and the total area under the curve increases.
D.The peak of the curve shifts to the right and becomes lower.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
When the temperature is increased, the average kinetic energy of the particles increases, causing the peak of the Maxwell–Boltzmann distribution curve to shift to the right (higher energy). Because the total number of molecules remains constant, the total area under the curve must also remain constant, meaning the peak must become lower to compensate for the broader spread.
PastPaper.markingScheme
1 mark for identifying that the peak shifts to the right and becomes lower.
PastPaper.question 19 · MCQ
1 PastPaper.marks
Nitrogen dioxide dissociates according to the equation: \(2\text{NO}_2(\text{g}) \rightleftharpoons 2\text{NO}(\text{g}) + \text{O}_2(\text{g})\). At a certain temperature, 2.00 mol of \(\text{NO}_2\) is placed in a closed \(2.00 \text{ dm}^3\) vessel. At equilibrium, 0.60 mol of \(\text{O}_2\) is present. What is the value of \(K_c\) at this temperature?
A.0.68 mol dm⁻³
B.1.35 mol dm⁻³
C.0.34 mol dm⁻³
D.2.70 mol dm⁻³
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Setting up the ICE table: Initial moles of \(\text{NO}_2 = 2.00\). At equilibrium, \(n(\text{O}_2) = 0.60\text{ mol}\). Change in \(n(\text{O}_2) = +0.60\text{ mol}\), change in \(n(\text{NO}) = +1.20\text{ mol}\), and change in \(n(\text{NO}_2) = -1.20\text{ mol}\). This leaves the equilibrium moles as: \(n(\text{NO}_2) = 0.80\text{ mol}\), \(n(\text{NO}) = 1.20\text{ mol}\), and \(n(\text{O}_2) = 0.60\text{ mol}\). Next, calculate equilibrium concentrations: \([\text{NO}_2] = 0.80 / 2.00 = 0.40\text{ mol dm}^{-3}\), \([\text{NO}] = 1.20 / 2.00 = 0.60\text{ mol dm}^{-3}\), and \([\text{O}_2] = 0.60 / 2.00 = 0.30\text{ mol dm}^{-3}\). Substitute these into the expression: \(K_c = \frac{[\text{NO}]^2[\text{O}_2]}{[\text{NO}_2]^2} = \frac{(0.60)^2 \times 0.30}{(0.40)^2} = 0.675\text{ mol dm}^{-3}\), which rounds to 0.68 mol dm⁻³.
PastPaper.markingScheme
1 mark for the correct calculation of equilibrium concentrations and Kc value.
PastPaper.question 20 · MCQ
1 PastPaper.marks
In the redox reaction between sodium thiosulfate and iodine, sodium tetrathionate (\(\text{Na}_2\text{S}_4\text{O}_6\)) and sodium iodide are formed: \(2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}\). What is the change in the oxidation number of sulfur in this reaction?
A.from +2 to +2.5
B.from +2 to +4
C.from +4 to +2.5
D.from +4 to +6
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
In the thiosulfate ion (\(\text{S}_2\text{O}_3^{2-}\)), the average oxidation state of sulfur is found by: \(2(x) + 3(-2) = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\). In the tetrathionate ion (\(\text{S}_4\text{O}_6^{2-}\)), the average oxidation state of sulfur is found by: \(4(y) + 6(-2) = -2 \Rightarrow 4y = +10 \Rightarrow y = +2.5\). Thus, the oxidation state of sulfur changes from +2 to +2.5.
PastPaper.markingScheme
1 mark for correctly identifying the initial and final oxidation states and the change.
PastPaper.question 21 · MCQ
1 PastPaper.marks
An organic compound X has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\). When X is warmed with acidified potassium dichromate(VI), the solution turns green. The organic product Y does not react with Fehling's solution. Which of the following is the IUPAC name of X?
A.butan-1-ol
B.butan-2-ol
C.2-methylpropan-1-ol
D.2-methylpropan-2-ol
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The transition from orange to green in the acidified potassium dichromate(VI) mixture indicates that compound X is oxidized, meaning X must be a primary or secondary alcohol. Since product Y does not react with Fehling's solution, Y is not an aldehyde but a ketone. Therefore, X must be a secondary alcohol. Among the given isomers, butan-2-ol is the only secondary alcohol with the formula \(\text{C}_4\text{H}_{10}\text{O}\).
PastPaper.markingScheme
1 mark for identifying butan-2-ol as the correct starting secondary alcohol.
PastPaper.question 22 · MCQ
1 PastPaper.marks
Bromoethane can be converted into propanoic acid in a two-step synthesis: \(\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{Step 1}} \text{Compound X} \xrightarrow{\text{Step 2}} \text{CH}_3\text{CH}_2\text{CO}_2\text{H}\). Which reagents and conditions are correct for Step 1 and Step 2?
A.Step 1: KCN in ethanol, heat under reflux; Step 2: dilute HCl(aq), heat under reflux
B.Step 1: HCN in ether, room temperature; Step 2: NaOH(aq), heat under reflux
C.Step 1: KCN in ethanol, heat under reflux; Step 2: LiAlH₄ in dry ether, room temperature
D.Step 1: NH₃ in ethanol, heated under pressure; Step 2: HNO₂, room temperature
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Step 1 is a nucleophilic substitution reaction where the bromide atom is replaced by a nitrile group. This requires heating bromoethane under reflux with \(\text{KCN}\) in ethanol to yield propanenitrile. Step 2 is an acid hydrolysis reaction where the nitrile group is hydrolyzed to a carboxylic acid group, requiring heating under reflux with a dilute aqueous acid such as \(\text{HCl}\)(aq).
PastPaper.markingScheme
1 mark for specifying the correct reagents and conditions for both steps.
PastPaper.question 23 · MCQ
1 PastPaper.marks
Two halogenoalkanes, 2-chloro-2-methylpropane and 1-chlorobutane, are hydrolyzed separately with aqueous sodium hydroxide. Which row correctly describes the predominant mechanism for each reaction?
2-chloro-2-methylpropane is a tertiary halogenoalkane and hydrolyzes predominantly via the \(S_N1\) mechanism, as it forms a highly stable tertiary carbocation intermediate. 1-chlorobutane is a primary halogenoalkane and hydrolyzes predominantly via the \(S_N2\) mechanism, since it has minimal steric hindrance and cannot easily form a stable carbocation.
PastPaper.markingScheme
1 mark for identifying the correct mechanism for both halogenoalkanes.
PastPaper.question 24 · MCQ
1 PastPaper.marks
Which statement about the trends in the properties of Group 2 elements and their compounds from magnesium to barium is correct?
A.The thermal stability of the carbonates decreases down the group.
B.The solubility of the hydroxides increases down the group.
C.The solubility of the sulfates increases down the group.
D.The first ionisation energy of the elements increases down the group.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Going down Group 2 from magnesium to barium, the solubility of the hydroxides increases (magnesium hydroxide is sparingly soluble, whereas barium hydroxide is much more soluble). The thermal stability of carbonates increases, sulfate solubility decreases, and first ionisation energy decreases.
PastPaper.markingScheme
1 mark for identifying that hydroxide solubility increases down the group.
PastPaper.question 25 · MCQ
1 PastPaper.marks
An element X in Period 3 has the following successive ionisation energies in \(\text{kJ mol}^{-1}\):
\(1^{\text{st}} = 578\)
\(2^{\text{nd}} = 1817\)
\(3^{\text{rd}} = 2745\)
\(4^{\text{th}} = 11578\)
\(5^{\text{th}} = 14831\)
What is the formula of the oxide of X?
A.\(\text{XO}\)
B.\(\text{X}_2\text{O}\)
C.\(\text{XO}_2\)
D.\(\text{X}_2\text{O}_3\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
By analyzing the successive ionisation energies, we look for the largest ratio jump.
\(\frac{4^{\text{th}}}{3^{\text{rd}}} = \frac{11578}{2745} \approx 4.22\) (This represents a very large increase, indicating the removal of an electron from a core shell closer to the nucleus).
This shows that element X has 3 valence electrons and belongs to Group 13 (III). In Period 3, this element is aluminium (\(\text{Al}\)). Elements in Group 13 form oxides with the general formula \(\text{X}_2\text{O}_3\).
PastPaper.markingScheme
Award 1 mark for selecting the correct option D. - Reject A: Corresponds to a Group 2 metal oxide. - Reject B: Corresponds to a Group 1 metal oxide. - Reject C: Corresponds to a Group 14 metal/non-metal oxide.
PastPaper.question 26 · MCQ
1 PastPaper.marks
A student uses a simple calorimeter to determine the enthalpy change of combustion of methanol (\(\text{CH}_3\text{OH}\), \(M_{\text{r}} = 32.0\)).
0.80 g of methanol is burned to heat 150 g of water from \(20.0^\circ\text{C}\) to \(41.5^\circ\text{C}\).
The specific heat capacity of water is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).
Assuming that only \(60\%\) of the heat released by the combustion of methanol is transferred to the water, what is the calculated enthalpy of combustion of methanol?
A.\(-324\text{ kJ mol}^{-1}\)
B.\(-539\text{ kJ mol}^{-1}\)
C.\(-899\text{ kJ mol}^{-1}\)
D.\(-1498\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Step 1: Calculate the heat energy absorbed by the water. \(q = m c \Delta T = 150\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (41.5 - 20.0)\text{ K}\) \(q = 150 \times 4.18 \times 21.5 = 13480.5\text{ J} = 13.48\text{ kJ}\)
Step 2: Account for the heat loss (only 60% of the total heat released is transferred). \(Q_{\text{total}} = \frac{q}{0.60} = \frac{13.4805\text{ kJ}}{0.60} = 22.4675\text{ kJ}\)
Step 3: Calculate the moles of methanol burned. \(n = \frac{0.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.025\text{ mol}\)
Award 1 mark for the correct calculation leading to option C. - Reject A: Incorrectly multiplies by 0.60 instead of dividing. - Reject B: Fails to apply the 60% efficiency adjustment. - Reject D: Uses incorrect stoichiometric/mass assumptions.
PastPaper.question 27 · MCQ
1 PastPaper.marks
For a gas-phase reaction, the temperature of the system is increased from \(T_1\) to \(T_2\) in the absence of a catalyst.
How do the most probable energy of the molecules and the activation energy, \(E_{\text{a}}\), of the reaction change?
A.The most probable energy increases, and \(E_{\text{a}}\) decreases.
B.The most probable energy increases, and \(E_{\text{a}}\) remains constant.
C.The most probable energy remains constant, and \(E_{\text{a}}\) remains constant.
D.The most probable energy decreases, and \(E_{\text{a}}\) remains constant.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
When the temperature of a gas system is increased: 1. The Maxwell-Boltzmann distribution curve flattens and shifts to the right, meaning the most probable molecular energy increases. 2. The activation energy (\(E_{\text{a}}\)) is a constant minimum energy barrier for a given reaction and does not change with temperature. It only changes in the presence of a catalyst.
PastPaper.markingScheme
Award 1 mark for selecting the correct option B. - Reject A: Temperature change does not alter the activation energy. - Reject C: The most probable energy must increase as the average kinetic energy increases. - Reject D: The most probable energy increases, not decreases.
PastPaper.question 28 · MCQ
1 PastPaper.marks
The following dynamic equilibrium is established in a closed container of fixed volume:
\[2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H < 0\]
A small amount of helium gas is injected into the container at constant volume.
How do the position of the equilibrium and the value of the equilibrium constant, \(K_{\text{c}}\), change?
A.The equilibrium shifts to the left, and \(K_{\text{c}}\) decreases.
B.The equilibrium shifts to the right, and \(K_{\text{c}}\) remains unchanged.
C.The equilibrium remains unchanged, and \(K_{\text{c}}\) remains unchanged.
D.The equilibrium shifts to the left, and \(K_{\text{c}}\) remains unchanged.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
At constant volume, adding an inert gas like helium increases the total pressure, but does not affect the partial pressures or concentrations of the reacting gases (\(\text{SO}_2\), \(\text{O}_2\), and \(\text{SO}_3\)). Therefore, the position of the equilibrium remains unchanged. Since the temperature is kept constant, the equilibrium constant \(K_{\text{c}}\) also remains unchanged.
PastPaper.markingScheme
Award 1 mark for the correct choice C. - Reject A, B, and D: Adding helium at constant volume does not change the concentrations of the species, so there is no shift in equilibrium position, and only a change in temperature can change the value of \(K_{\text{c}}\).
PastPaper.question 29 · MCQ
1 PastPaper.marks
An organic compound Y is a secondary alcohol with the molecular formula \(\text{C}_5\text{H}_{12}\text{O}\). When Y is heated with concentrated sulfuric acid, it undergoes dehydration to produce a mixture containing exactly three isomeric alkenes (including stereoisomers).
What is the IUPAC name of Y?
A.Pentan-2-ol
B.Pentan-3-ol
C.3-Methylbutan-2-ol
D.2-Methylbutan-2-ol
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Dehydration of pentan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3\)) can eliminate water in two ways: 1. Loss of H from C1 gives pent-1-ene (1 isomer). 2. Loss of H from C3 gives pent-2-ene, which exhibits cis-trans isomerism: *cis*-pent-2-ene and *trans*-pent-2-ene (2 isomers).
This yields a total of exactly three isomeric alkenes.
For comparison: - Pentan-3-ol yields only pent-2-ene (*cis* and *trans* isomers) - a total of two isomeric alkenes. - 3-Methylbutan-2-ol yields 3-methylbut-1-ene and 2-methylbut-2-ene (which does not have stereoisomers) - a total of two isomeric alkenes. - 2-Methylbutan-2-ol is a tertiary alcohol.
PastPaper.markingScheme
Award 1 mark for identifying pentan-2-ol as the correct secondary alcohol (option A). - Reject B: Only yields two isomeric alkenes. - Reject C: Only yields two isomeric alkenes. - Reject D: It is a tertiary alcohol.
PastPaper.question 30 · MCQ
1 PastPaper.marks
Equal amounts of 1-chlorobutane, 1-bromobutane, and 1-iodobutane are heated separately with aqueous silver nitrate in ethanol at \(50^\circ\text{C}\).
Which statement correctly explains the relative rate of reaction?
A.1-chlorobutane forms a precipitate fastest because the \(\text{C}-\text{Cl}\) bond is the most polar.
B.1-iodobutane forms a precipitate fastest because the \(\text{C}-\text{I}\) bond has the lowest bond enthalpy.
C.1-iodobutane forms a precipitate slowest because iodine has the largest atomic radius.
D.1-bromobutane forms a precipitate fastest because it has intermediate polarity.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. The bond enthalpies decrease in the order: \(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\). Therefore, the \(\text{C}-\text{I}\) bond is the weakest and easiest to break, meaning 1-iodobutane reacts the fastest and forms a precipitate of silver iodide first.
PastPaper.markingScheme
Award 1 mark for option B. - Reject A: Though the C-Cl bond is the most polar, bond strength (not polarity) is the dominant factor determining the rate of nucleophilic substitution here. - Reject C: 1-iodobutane is the fastest, not the slowest. - Reject D: 1-bromobutane is intermediate in rate, not the fastest.
PastPaper.question 31 · MCQ
1 PastPaper.marks
Equal masses of anhydrous calcium nitrate, \(\text{Ca(NO}_3)_2\), and anhydrous barium nitrate, \(\text{Ba(NO}_3)_2\), are heated strongly in separate test tubes until they completely decompose.
Which statement about these experiments is correct?
A.Barium nitrate decomposes at a lower temperature than calcium nitrate.
B.Calcium nitrate produces a larger total volume of gas than the same mass of barium nitrate.
C.Barium nitrate produces a larger total volume of gas than the same mass of calcium nitrate.
D.Both nitrates decompose to form the metal nitrite and oxygen gas only.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The thermal decomposition of Group 2 nitrates follows the equation: \[2\text{M(NO}_3)_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\]
1. Calcium nitrate has a smaller molar mass (\(M_{\text{r}} \approx 164.1\)) than barium nitrate (\(M_{\text{r}} \approx 261.3\)). 2. Therefore, for equal masses, there are more moles of calcium nitrate than barium nitrate. 3. Since more moles of reactant are present, calcium nitrate will produce a larger total volume of gas (\(\text{NO}_2\) and \(\text{O}_2\)) than the same mass of barium nitrate under the same conditions. 4. Additionally, calcium nitrate decomposes at a lower temperature than barium nitrate due to the higher charge density and polarising power of the smaller \(\text{Ca}^{2+}\) ion.
PastPaper.markingScheme
Award 1 mark for option B. - Reject A: Barium nitrate has a higher thermal stability and decomposes at a higher temperature than calcium nitrate. - Reject C: Calcium nitrate produces more gas because it has a lower formula mass. - Reject D: Group 2 nitrates decompose to oxides, nitrogen dioxide, and oxygen, not nitrites.
PastPaper.question 32 · MCQ
1 PastPaper.marks
Which molecule has a non-zero permanent dipole moment?
A.\(\text{BeCl}_2\)
B.\(\text{BCl}_3\)
C.\(\text{NF}_3\)
D.\(\text{CCl}_4\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
- \(\text{BeCl}_2\) is linear and symmetrical, so the individual bond dipoles cancel each other out (non-polar). - \(\text{BCl}_3\) is trigonal planar and symmetrical, so the bond dipoles cancel out (non-polar). - \(\text{NF}_3\) is trigonal pyramidal with a lone pair of electrons on the nitrogen atom. The dipoles do not cancel, resulting in a net molecular dipole (polar). - \(\text{CCl}_4\) is tetrahedral and symmetrical, so the bond dipoles cancel out (non-polar).
PastPaper.markingScheme
Award 1 mark for selecting C. - Reject A, B, and D: These molecules have highly symmetrical geometries that cause their bond dipoles to cancel, resulting in a net dipole moment of zero.
PastPaper.question 33 · MCQ
1 PastPaper.marks
In a calorimetry experiment, a sample of 2-methylpropan-1-ol (\(C_4H_{10}O\)) was burned completely in air. The heat released was used to heat water in a copper calorimeter. The experimental data collected is shown below:
* Mass of 2-methylpropan-1-ol burned = \(1.48\text{ g}\) * Mass of water in the calorimeter = \(200.0\text{ g}\) * Temperature rise of the water = \(45.0^\circ\text{C}\) * Specific heat capacity of water = \(4.18\text{ J g}^{-1}\text{ K}^{-1}\)
What is the experimental enthalpy change of combustion, \(\Delta H_c\), of 2-methylpropan-1-ol under these conditions?
A.\(-1881\text{ kJ mol}^{-1}\)
B.\(-18.8\text{ kJ mol}^{-1}\)
C.\(-37.6\text{ kJ mol}^{-1}\)
D.\(-1523\text{ kJ mol}^{-1}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, calculate the heat energy absorbed by the water using the formula \(q = m c \Delta T\):
* Option A: correct calculation of \(q\), moles, and division with correct sign. * Option B: incorrect scale factor (dividing by 2000 instead of 0.0200). * Option C: heat absorbed in kJ without converting to molar enthalpy. * Option D: incorrect calculation of moles using the molar mass of propan-1-ol (\(60.0\text{ g mol}^{-1}\)) instead of 2-methylpropan-1-ol (\(74.0\text{ g mol}^{-1}\)).
PastPaper.question 34 · MCQ
1 PastPaper.marks
When the temperature of a gaseous reaction mixture is increased from \(298\text{ K}\) to \(308\text{ K}\), the rate of reaction approximately doubles.
What is the main reason for this significant increase in reaction rate?
A.The average kinetic energy of the reactant molecules doubles.
B.The activation energy of the reaction is significantly lowered.
C.The overall frequency of collisions increases by a factor of two.
D.The fraction of molecules with energy greater than or equal to the activation energy increases significantly.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
According to the Maxwell-Boltzmann distribution, temperature is a measure of the average kinetic energy of the molecules. A modest increase in temperature (from \(298\text{ K}\) to \(308\text{ K}\)) increases the average kinetic energy of the molecules by only a small fraction (about 3%), which causes only a minor increase in collision frequency.
However, this shift in the distribution curve dramatically increases the *fraction of molecules* that possess energy greater than or equal to the activation energy (\(E \ge E_a\)). This results in a much higher frequency of successful, productive collisions, doubling the rate of reaction.
PastPaper.markingScheme
1 mark for the correct option (D).
* Option A is incorrect because average kinetic energy is proportional to absolute temperature, so it only increases by a very small percentage (\(308/298\)), not doubling. * Option B is incorrect because temperature changes do not alter the activation energy threshold itself; catalysts lower the activation energy. * Option C is incorrect because collision frequency only increases slightly with temperature (proportional to \(\sqrt{T}\)).
PastPaper.question 35 · MCQ
1 PastPaper.marks
An anhydrous Group 2 nitrate, \(X(NO_3)_2\), is heated strongly in a test-tube. A brown gas is evolved and a white solid residue remains.
When \(2.97\text{ g}\) of this anhydrous nitrate is decomposed completely, \(1.20\text{ dm}^3\) of gas (measured at room temperature and pressure, r.t.p.) is produced.
What is the identity of the metal \(X\)?
[Molar volume of a gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\)]
A.magnesium
B.calcium
C.strontium
D.barium
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Write the balanced equation for the thermal decomposition of a Group 2 nitrate:
From the equation, \(2\text{ moles}\) of \(X(NO_3)_2\) produce \(5\text{ moles}\) of gas (\(4NO_2 + 1O_2\)).
1. Calculate the total moles of gas produced: \(n(\text{gas}) = \frac{1.20\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0500\text{ mol}\)
2. Determine the moles of reactant \(X(NO_3)_2\) decomposed: \(n(X(NO_3)_2) = 0.0500\text{ mol of gas} \times \frac{2}{5} = 0.0200\text{ mol}\)
3. Calculate the molar mass of \(X(NO_3)_2\): \(M_r = \frac{2.97\text{ g}}{0.0200\text{ mol}} = 148.5\text{ g mol}^{-1}\)
4. Determine the atomic mass of \(X\): \(M_r(X) + 2 \times [14.0 + (3 \times 16.0)] = 148.5\) \(M_r(X) + 124.0 = 148.5\) \(M_r(X) = 24.5\text{ g mol}^{-1}\)
This relative atomic mass is closest to magnesium (\(24.3\)).
PastPaper.markingScheme
1 mark for the correct option (A).
* Option A: correct calculation of the molar mass of the nitrate as 148.5, yielding Magnesium. * Option B: calcium (using incorrect gas stoichiometry or molar mass calculation). * Option C: strontium (distractor). * Option D: barium (distractor).
PastPaper.question 36 · MCQ
1 PastPaper.marks
Four different halogenoalkanes are treated separately with aqueous silver nitrate in ethanol at \(50^\circ\text{C}\). The time taken for a precipitate to appear is recorded.
Which halogenoalkane produces a precipitate in the shortest time?
A.1-chlorobutane
B.1-bromobutane
C.2-chloro-2-methylpropane
D.2-bromo-2-methylpropane
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The rate of hydrolysis of halogenoalkanes is determined by two main factors:
1. **The strength of the C–Halogen bond**: The C–Br bond is weaker than the C–Cl bond (bond enthalpy of C–Br is \(285\text{ kJ mol}^{-1}\) vs \(340\text{ kJ mol}^{-1}\) for C–Cl). Thus, bromoalkanes hydrolyse faster than chloroalkanes under identical conditions. 2. **The mechanism and carbocation stability**: Tertiary halogenoalkanes undergo hydrolysis via the \(S_N1\) mechanism extremely rapidly because the intermediate tertiary carbocation is highly stable due to the inductive electron-donating effect of three alkyl groups. Primary halogenoalkanes hydrolyse much slower via the \(S_N2\) mechanism.
Comparing the options, **2-bromo-2-methylpropane** is a tertiary bromoalkane. It will hydrolyse fastest because it contains the weaker C–Br bond and undergoes rapid \(S_N1\) substitution.
PastPaper.markingScheme
1 mark for the correct option (D).
* Option A and B are primary halogenoalkanes, which hydrolyse very slowly. * Option C is a tertiary chloroalkane, which hydrolyses slower than the corresponding tertiary bromoalkane (Option D) because the C–Cl bond is stronger than the C–Br bond.
PastPaper.question 37 · MCQ
1 PastPaper.marks
An organic compound, \(Y\), reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate.
However, \(Y\) does not produce a silver mirror with Tollens' reagent, and does not form a yellow precipitate when warmed with alkaline aqueous iodine.
Which compound could be \(Y\)?
A.propanal
B.propanone
C.pentan-3-one
D.pentan-2-one
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Let's analyze the clues step-by-step:
1. **Reacts with 2,4-DNPH**: Indicates the presence of a carbonyl group (must be either an aldehyde or a ketone). 2. **Does not react with Tollens' reagent**: Rules out aldehydes. Therefore, \(Y\) must be a ketone. This eliminates propanal (Option A). 3. **Does not react with alkaline aqueous iodine (triiodomethane test)**: Rules out methyl ketones (compounds containing the \(CH_3CO-\) group).
Let's check the remaining options: * **propanone** (Option B): \(CH_3COCH_3\) is a methyl ketone (gives positive triiodomethane test). * **pentan-3-one** (Option C): \(CH_3CH_2COCH_2CH_3\) is a symmetric ketone without a \(CH_3CO-\) group (gives negative triiodomethane test). This matches all criteria. * **pentan-2-one** (Option D): \(CH_3COCH_2CH_2CH_3\) is a methyl ketone (gives positive triiodomethane test).
PastPaper.markingScheme
1 mark for the correct option (C).
* Option A is incorrect because it is an aldehyde and would react with Tollens'. * Options B and D are incorrect because they are methyl ketones and would form a yellow precipitate with alkaline iodine.
PastPaper.question 38 · MCQ
1 PastPaper.marks
The table shows the first five successive ionisation energies, in \(\text{kJ mol}^{-1}\), of an element \(Z\) in Period 3 of the Periodic Table.
To find the group of the element, look for the largest jump between successive ionisation energies:
* From 1st to 2nd: \(1817 - 578 = 1239\text{ kJ mol}^{-1}\) * From 2nd to 3rd: \(2745 - 1817 = 928\text{ kJ mol}^{-1}\) * From 3rd to 4th: \(11578 - 2745 = 8833\text{ kJ mol}^{-1}\) (a very large increase, more than 4 times the 3rd IE)
The huge jump between the 3rd and 4th ionisation energies indicates that the 4th electron is removed from a stable, inner shell closer to the nucleus. Therefore, element \(Z\) has 3 valence electrons and belongs to Group 13 (III).
In Period 3, this element is aluminium (\(Al\)), which forms an oxide with formula \(Al_2O_3\). In general, a Group 13 element \(Z\) forms an oxide with the formula \(Z_2O_3\).
PastPaper.markingScheme
1 mark for the correct option (C).
* Option A: formula for Group 1 metal oxide. * Option B: formula for Group 2 metal oxide. * Option C: correct formula for a Group 13 oxide (valency of 3+). * Option D: formula for Group 14 oxide.
PastPaper.question 39 · MCQ
1 PastPaper.marks
Phosphorus pentachloride decomposes reversibly according to the following equation:
An equilibrium mixture is established in a sealed container of volume \(V\) at temperature \(T\).
Which change will increase the value of the equilibrium constant, \(K_c\)?
A.decreasing the volume of the container at constant temperature
B.increasing the temperature of the container
C.adding a catalyst to the mixture
D.adding more \(PCl_5(g)\) at constant temperature
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The value of the equilibrium constant, \(K_c\), is constant at a constant temperature and can *only* be changed by altering the temperature.
Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature shifts the position of equilibrium to the right to absorb the extra thermal energy. This increases the equilibrium concentrations of the products (\(PCl_3\) and \(Cl_2\)) and decreases the equilibrium concentration of the reactant (\(PCl_5\)), thereby increasing the value of \(K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}\).
PastPaper.markingScheme
1 mark for the correct option (B).
* Option A, C, and D are incorrect because changes in volume, catalyst, or reactant concentration do not alter the value of the equilibrium constant \(K_c\) (they only temporarily disturb the equilibrium position, which then restores to satisfy the same \(K_c\)).
PastPaper.question 40 · MCQ
1 PastPaper.marks
Which molecule has a permanent, non-zero net dipole moment?
A.\(CF_4\)
B.\(CH_2F_2\)
C.\(BF_3\)
D.\(SF_6\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Whether a molecule has a permanent dipole moment depends on both the polarities of its individual bonds and its molecular symmetry:
* \(CF_4\): Tetrahedral and highly symmetrical. The four highly polar C–F bond dipoles point in opposite directions and cancel out perfectly. The net dipole moment is zero. * \(CH_2F_2\): Tetrahedral but asymmetric. The C–H bonds and C–F bonds have different polarities, and the dipoles do not cancel out, resulting in a permanent net dipole moment. * \(BF_3\): Trigonal planar and symmetrical. The three polar B–F bond dipoles cancel out perfectly. The net dipole moment is zero. * \(SF_6\): Octahedral and symmetrical. The six polar S–F bond dipoles cancel out perfectly. The net dipole moment is zero.
PastPaper.markingScheme
1 mark for the correct option (B).
* Options A, C, and D are incorrect because they are highly symmetrical molecules where bond dipoles cancel out completely, making them non-polar.
Paper 23 (AS Structured)
Answer all structured questions in the spaces provided on the question paper.
4 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · Structured
15 PastPaper.marks
Standard enthalpy changes are fundamental to understanding chemical thermodynamics.
(a) Define the term *standard enthalpy change of reaction*. [2]
(b) An experiment was carried out to determine the enthalpy change for the thermal decomposition of sodium hydrogencarbonate: \( 2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \) This direct enthalpy change cannot be easily measured. Instead, an indirect method was used by reacting both \( \text{NaHCO}_3(\text{s}) \) and anhydrous \( \text{Na}_2\text{CO}_3(\text{s}) \) with excess dilute hydrochloric acid.
(i) In Experiment 1, a student added 3.36 g of \( \text{NaHCO}_3(\text{s}) \) to \( 50.0 \text{ cm}^3 \) of \( 2.00 \text{ mol dm}^{-3} \) \( \text{HCl} \) (an excess) in a polystyrene cup. The temperature of the mixture decreased by \( 4.2 \text{ }^{\circ}\text{C} \). Calculate the enthalpy change, \( \Delta H_1 \), in \( \text{kJ mol}^{-1} \), for the reaction of 1 mole of \( \text{NaHCO}_3(\text{s}) \) with \( \text{HCl}(\text{aq}) \). Assume the specific heat capacity of the solution is \( 4.18 \text{ J g}^{-1}\text{ K}^{-1} \) and its density is \( 1.00 \text{ g cm}^{-3} \). Assume the mass of the solution is equal to the mass of the acid solution. Show your working. [4]
(ii) In Experiment 2, the student added 2.12 g of anhydrous \( \text{Na}_2\text{CO}_3(\text{s}) \) to \( 50.0 \text{ cm}^3 \) of \( 2.00 \text{ mol dm}^{-3} \) \( \text{HCl} \) (an excess) in a polystyrene cup. The temperature of the mixture increased by \( 3.8 \text{ }^{\circ}\text{C} \). Calculate the enthalpy change, \( \Delta H_2 \), in \( \text{kJ mol}^{-1} \), for the reaction of 1 mole of \( \text{Na}_2\text{CO}_3(\text{s}) \) with \( \text{HCl}(\text{aq}) \). Use the same assumptions as in (b)(i). Show your working. [3]
(c) Construct a Hess's Law cycle relating the decomposition of \( \text{NaHCO}_3(\text{s}) \) to its reactions with hydrochloric acid. Use your cycle and the values from (b)(i) and (b)(ii) to calculate the standard enthalpy change of reaction, \( \Delta H_{\text{r}} \), in \( \text{kJ mol}^{-1} \), for the decomposition of 2 moles of \( \text{NaHCO}_3(\text{s}) \): \( 2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \) Show your working. [4]
(d) State one reason, other than heat loss to the surroundings, why the experimental value of \( \Delta H_{\text{r}} \) determined via this indirect method might differ slightly from the literature value. [1]
(e) Predict the effect on the temperature change in Experiment 2 if a glass beaker were used instead of a polystyrene cup. Explain your answer. [1]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a)** The standard enthalpy change of reaction is the enthalpy change when the molar quantities of reactants shown in the chemical equation react together under standard conditions (298 K, 100 kPa) with all substances in their standard states.
**(b)(i)** 1. Heat energy absorbed, \( q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1}\text{ K}^{-1} \times 4.2 \text{ K} = 877.8 \text{ J} = 0.8778 \text{ kJ} \). 2. Moles of \( \text{NaHCO}_3 = \frac{3.36}{84.0} = 0.0400 \text{ mol} \) (since \( M_{\text{r}} \) of \( \text{NaHCO}_3 = 23.0 + 1.0 + 12.0 + (3 \times 16.0) = 84.0 \text{ g mol}^{-1} \)). 3. Since temperature decreased, the reaction is endothermic. \( \Delta H_1 = +\frac{0.8778}{0.0400} = +21.945 \approx +21.9 \text{ kJ mol}^{-1} \) (to 3 sig fig).
**(b)(ii)** 1. Heat energy released, \( q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1}\text{ K}^{-1} \times 3.8 \text{ K} = 794.2 \text{ J} = 0.7942 \text{ kJ} \). 2. Moles of \( \text{Na}_2\text{CO}_3 = \frac{2.12}{106.0} = 0.0200 \text{ mol} \) (since \( M_{\text{r}} \) of \( \text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1} \)). 3. Since temperature increased, the reaction is exothermic. \( \Delta H_2 = -\frac{0.7942}{0.0200} = -39.71 \approx -39.7 \text{ kJ mol}^{-1} \) (to 3 sig fig).
**(d)** Impurities in the sodium hydrogencarbonate (or anhydrous sodium carbonate) or incomplete dissolution/reaction.
**(e)** The temperature rise would be smaller because glass is a poorer thermal insulator than polystyrene, leading to faster heat loss to the surroundings during the reaction.
PastPaper.markingScheme
**(a)** [2 marks] - Enthalpy change when equation quantities of substances react [1] - Under standard conditions (298 K, 100 kPa) with all substances in standard states [1]
**(b)(i)** [4 marks] - Correct calculation of heat energy absorbed: q = 0.878 kJ [1] - Correct calculation of moles of NaHCO3 = 0.0400 mol [1] - Correct division of heat by moles [1] - Correct sign (+) and value to 3 sig fig: +21.9 kJ mol-1 [1]
**(b)(ii)** [3 marks] - Correct calculation of heat energy released: q = 0.794 kJ [1] - Correct calculation of moles of Na2CO3 = 0.0200 mol [1] - Correct sign (-) and value to 3 sig fig: -39.7 kJ mol-1 [1]
**(c)** [4 marks] - Correctly constructed Hess's Law cycle or algebraic expression representing the cycle: 2*dH1 = dHr + dH2 [2] - Correct substitution: dHr = 2*(+21.9) - (-39.7) [1] - Correct final answer with correct sign to 3 sig fig: +83.5 kJ mol-1 [1] (accept ecf from b)
**(d)** [1 mark] - Impurities in the reactants / damp NaHCO3 / incomplete reaction [1] (Do not accept 'heat loss' as it is excluded by the prompt).
**(e)** [1 mark] - The temperature change is smaller because glass has higher thermal conductivity than polystyrene / allows more heat to escape [1].
PastPaper.question 2 · Structured
15 PastPaper.marks
The rate of chemical reactions depends on factors such as temperature, concentration, and the presence of a catalyst.
(a) Define the term *activation energy*. [1]
(b) The Maxwell-Boltzmann distribution shows the distribution of molecular energies in a gas at a constant temperature.
(i) Sketch a Maxwell-Boltzmann distribution curve for a gas at a temperature \( T_1 \). On your sketch, label the axes and mark the activation energy, \( E_{\text{a}} \), of a reaction. On the same diagram, sketch the curve you would expect for the same gas at a higher temperature, \( T_2 \). [3]
(ii) Use your diagram to explain, in terms of the collision theory, why a relatively small increase in temperature leads to a very large increase in the rate of reaction. [3]
(c) Many industrial processes utilize catalysts to increase reaction rates.
(i) State what is meant by the term *homogeneous catalyst*. [1]
(ii) Explain, by referencing a reaction pathway diagram, how a catalyst increases the rate of reaction. [3]
(d) During a reaction at a constant temperature, the rate of reaction decreases as time progresses. Explain this observation in terms of collision theory. [2]
(e) Some reactions, such as the reaction between nitrogen and oxygen to form nitrogen monoxide, have extremely high activation energies.
(i) Explain why these reactions are exceptionally slow at room temperature. [1]
(ii) Suggest one source of energy in nature that provides the activation energy for this reaction. [1]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a)** Activation energy is the minimum energy colliding particles must possess for a collision to result in a chemical reaction.
**(b)(i)** - The y-axis must be labeled 'Number of molecules' or 'Fraction of molecules', and the x-axis must be labeled 'Energy' or 'Kinetic energy'. - The curve for \( T_1 \) starts at the origin, rises to a peak, and drops asymptotically to the x-axis (never touching the x-axis at high energies). - The curve for \( T_2 \) has a lower peak shifted to the right, and lies above the \( T_1 \) curve at higher energy values. - \( E_{\text{a}} \) is labeled on the x-axis, with the area under the curves to the right of \( E_{\text{a}} \) representing the molecules with sufficient energy to react.
**(b)(ii)** - At the higher temperature \( T_2 \), the average kinetic energy of the molecules increases, shifting the distribution to the right. - A significantly larger fraction (or number) of molecules now possess energy greater than or equal to the activation energy (\( E \ge E_{\text{a}} \)), shown by the larger shaded area under the \( T_2 \) curve to the right of \( E_{\text{a}} \). - This results in a much higher frequency of successful collisions per unit time, causing a large increase in the rate of reaction.
**(c)(i)** A homogeneous catalyst is a catalyst that is in the same phase (or physical state) as the reactants.
**(c)(ii)** - A catalyst provides an alternative reaction pathway. - This alternative pathway has a lower activation energy (\( E_{\text{a, catalysed}} < E_{\text{a, uncatalysed}} \)). - Therefore, a greater fraction of molecules have energy exceeding the activation energy, increasing the frequency of successful collisions and thus the reaction rate.
**(d)** As the reaction progresses, the concentration of reactant particles decreases. This means there are fewer reactant particles per unit volume, which reduces the frequency of collisions between reactant particles and thus decreases the rate.
**(e)(i)** At room temperature, the average thermal energy of the molecules is very low, meaning practically no reactant molecules have kinetic energy equal to or exceeding the extremely high activation energy.
**(e)(ii)** Lightning / electrical discharge during thunderstorms.
PastPaper.markingScheme
**(a)** [1 mark] - Minimum energy required for colliding particles to react / overcome energy barrier [1]
**(b)(i)** [3 marks] - Correctly labeled axes: y-axis as number/fraction of molecules, x-axis as energy [1] - Correct shape of T1 curve (starting at origin, asymptotic decay) and T2 curve (lower peak, shifted to the right, crossing T1 once and higher at high energy) [1] - Ea marked on x-axis with area under the curves to the right of Ea shaded/indicated [1]
**(b)(ii)** [3 marks] - Reference to a significantly larger area under the T2 curve to the right of Ea / larger fraction of molecules have energy >= Ea [1] - Increased energy of particles / shift to the right of the distribution [1] - Leads to a greater frequency of successful/effective collisions (must specify frequency or per unit time) [1]
**(c)(i)** [1 mark] - Catalyst in the same phase / state as reactants [1]
**(c)(ii)** [3 marks] - Provides an alternative reaction pathway [1] - With a lower activation energy [1] - More molecules have energy >= Ea(catalysed), leading to increased rate [1]
**(d)** [2 marks] - Reactant concentration decreases [1] - Frequency of collisions decreases [1]
**(e)(i)** [1 mark] - Almost no particles/molecules have enough energy to overcome the high activation energy at room temperature [1]
This question is about Period 3 elements and their compounds.
(a) Describe and explain the trend in electronegativity across Period 3 from sodium to chlorine. [3]
(b) Elements in Period 3 react with oxygen to form oxides.
(i) Write equations for the reactions of sodium and phosphorus with oxygen to form sodium oxide, \( \text{Na}_2\text{O} \), and phosphorus(V) oxide, \( \text{P}_4\text{O}_{10} \), respectively. [2]
(ii) State the structure and bonding of silicon dioxide, \( \text{SiO}_2 \), and sulfur trioxide, \( \text{SO}_3 \). [2]
(iii) Write an equation for the reaction of sulfur trioxide with water and state the pH of the resulting solution. [2]
(iv) Aluminium oxide, \( \text{Al}_2\text{O}_3 \), is described as amphoteric. Describe its reactions with both hydrochloric acid and aqueous sodium hydroxide, writing a balanced chemical equation for each reaction. [4]
(c) Silicon tetrachloride, \( \text{SiCl}_4 \), reacts rapidly with water, whereas carbon tetrachloride, \( \text{CCl}_4 \), does not react with water. Explain this difference in reactivity. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a)** - Electronegativity increases from sodium to chlorine. - This is due to an increase in nuclear charge (number of protons in the nucleus increases). - Shielding remains approximately constant because electrons are added to the same main energy level (third shell), resulting in a stronger electrostatic attraction between the nucleus and the bonding pair of electrons.
**(b)(ii)** - \( \text{SiO}_2 \): Giant covalent (or macromolecular) structure with strong covalent bonds between silicon and oxygen. - \( \text{SO}_3 \ shadow\): Simple molecular (or simple covalent) structure with weak intermolecular forces (van der Waals / London dispersion forces) between molecules.
**(b)(iii)** Equation: \( \text{SO}_3(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_4(\text{aq}) \) pH: Accept any value in the range 0 to 2 (e.g., pH = 1).
**(b)(iv)** - Reaction with hydrochloric acid (acts as a base): \( \text{Al}_2\text{O}_3(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \) - Reaction with sodium hydroxide (acts as an acid): \( \text{Al}_2\text{O}_3(\text{s}) + 2\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Na[Al(OH)}_4](\text{aq}) \) (Alternatively: \( \text{Al}_2\text{O}_3 + 2\text{NaOH} \rightarrow 2\text{NaAlO}_2 + \text{H}_2\text{O} \))
**(c)** - Silicon has vacant, low-lying 3d orbitals in its valence shell, which can accept a lone pair of electrons from a water molecule, initiating nucleophilic attack and hydrolysis. - Carbon does not have d-orbitals in its valence shell (it is in Period 2), so it cannot coordinate with water molecules; additionally, the carbon atom is sterically shielded by the four large chlorine atoms.
PastPaper.markingScheme
**(a)** [3 marks] - Electronegativity increases [1] - Increase in nuclear charge / proton number [1] - Shielding is similar / same shell, so stronger attraction of bonding pair of electrons [1]
**(b)(i)** [2 marks] - Correct equation for sodium oxide formation [1] - Correct equation for phosphorus(V) oxide formation [1]
**(b)(iv)** [4 marks] - Equation with HCl: Al2O3 + 6HCl -> 2AlCl3 + 3H2O [1] - Correct species in NaOH equation [1] - Correctly balanced NaOH equation (e.g., Al2O3 + 2NaOH + 3H2O -> 2Na[Al(OH)4] or Al2O3 + 2NaOH -> 2NaAlO2 + H2O) [1] - Statement that Al2O3 acts as a base with acid and as an acid with base / amphoteric [1]
**(c)** [2 marks] - Silicon has vacant d-orbitals (to accept lone pairs from water) [1] - Carbon does not have available d-orbitals [1]
PastPaper.question 4 · Structured
15 PastPaper.marks
Alcohols, carbonyl compounds, and halogenoalkanes are key intermediates in organic synthesis.
(a) Compound A is propan-1-ol, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \).
(i) State the class of alcohol to which propan-1-ol belongs. [1]
(ii) Propan-1-ol can be dehydrated to form an alkene, Compound D. State a suitable reagent and condition for this reaction, and name the alkene D formed. [2]
(b) Propan-1-ol can be oxidized to form different products depending on the reaction conditions.
(i) State the reagents and conditions required to obtain a good yield of propanal from propan-1-ol. [2]
(ii) State the reagents and conditions required to obtain propanoic acid from propan-1-ol. [2]
(iii) State the color change of the oxidizing agent observed during these oxidation reactions. [1]
(iv) Suggest a chemical test that could distinguish between propanal and propanoic acid. State the reagent and the observations for both compounds. [2]
(c) Propan-1-ol can be converted into 1-chloropropane, Compound E, by reaction with phosphorus trichloride, \( \text{PCl}_3 \).
(i) Write a balanced chemical equation for this reaction. [1]
(ii) Compound E reacts with hot aqueous sodium hydroxide to reform propan-1-ol via an \( \text{S}_{\text{N}}2 \) mechanism. Draw the mechanism for this reaction. Your diagram must show relevant dipoles, lone pairs, curly arrows, the structure of the transition state, and any charges. [4]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a)(i)** Primary alcohol.
**(a)(ii)** - Reagent and condition: Concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) and heat (approx. 170\( ^{\circ}\text{C} \)) or pass vapor over hot aluminium oxide catalyst (\( \text{Al}_2\text{O}_3 \)) at 300\( ^{\circ}\text{C} \). - Name of alkene D: Propene.
**(b)(i)** - Reagents: Acidified potassium dichromate(VI) (\( \text{K}_2\text{Cr}_2\text{O}_7\text{/H}^+ \)). - Conditions: Warm gently and distil off the product (propanal) as it forms.
**(b)(iv)** - Reagent: Tollens' reagent (or Fehling's solution, or a carbonate/hydrogencarbonate). - Observations: With Tollens': Propanal forms a silver mirror; propanoic acid shows no change. With Carbonate (e.g., \( \text{Na}_2\text{CO}_3 \)): Propanoic acid causes effervescence (bubbles of gas); propanal shows no reaction.
**(c)(ii)** The \( \text{S}_{\text{N}}2 \) mechanism is drawn as follows: 1. Draw 1-chloropropane showing the \( \text{C}-\text{Cl} \) bond with \( \text{C}^{\delta+} \) and \( \text{Cl}^{\delta-} \). 2. Draw the hydroxide ion, \( \text{OH}^- \), with a lone pair on the oxygen atom. 3. Draw a curly arrow from the lone pair on the oxygen atom of \( \text{OH}^- \) to the \( \text{C}^{\delta+} \) carbon atom. 4. Draw a curly arrow from the \( \text{C}-\text{Cl} \) single bond to the chlorine atom. 5. Draw the transition state in square brackets with a negative charge superscript outside the brackets. The central carbon should have partial (dotted) bonds to both the incoming \( \text{OH} \) group and the outgoing \( \text{Cl} \) group, with the rest of the propyl chain and hydrogen atoms attached. 6. Draw the products: propan-1-ol and a chloride ion (\( \text{Cl}^- \)).
PastPaper.markingScheme
**(a)(i)** [1 mark] - Primary alcohol [1]
**(a)(ii)** [2 marks] - Reagent: conc H2SO4 AND heat OR Al2O3 AND heat [1] - Name: propene [1]
**(b)(iv)** [2 marks] - Suitable test reagent: Tollens' reagent OR Fehling's solution OR Na2CO3/NaHCO3 [1] - Correct observations for both: Tollens' -> silver mirror with propanal, no change with propanoic acid; Carbonate -> effervescence/fizzing with propanoic acid, no reaction with propanal [1]
**(c)(ii)** [4 marks] - Dipoles on C-Cl bond (C\delta+ and Cl\delta-) AND curly arrow from lone pair on OH- to C\delta+ [1] - Curly arrow from C-Cl bond to Cl [1] - Correct transition state structure showing partial bonds to OH and Cl, with negative charge outside brackets [1] - Correct structures of products: propan-1-ol and Cl- [1]
Paper 33 (Practical Skills)
Perform quantitative analyses, enthalpy measurements, and qualitative chemical tests as directed.
3 PastPaper.question · 39.99 PastPaper.marks
PastPaper.question 1 · Practical
13.33 PastPaper.marks
FA 1 is an aqueous solution containing \(8.40\text{ g dm}^{-3}\) of an impure sample of hydrated ethanedioic acid, \(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\). FA 2 is \(0.0200\text{ mol dm}^{-3}\) potassium manganate(VII), \(\text{KMnO}_4\). Dilute sulfuric acid is also provided.
(a) You are to determine the concentration of ethanedioic acid by titration. Pipette \(25.0\text{ cm}^3\) of FA 1 into a conical flask. Use a measuring cylinder to add \(20\text{ cm}^3\) of dilute sulfuric acid. Heat the mixture to approximately \(60^\circ\text{C}\), then titrate with FA 2 until a permanent pale pink color is reached. Repeat the titration until concordant results are achieved. Record your titration results in a standard table. (For calculations, assume your average concordant titre of FA 2 is \(20.00\text{ cm}^3\).)
(b) Calculate the number of moles of \(\text{KMnO}_4\) present in the average titre of FA 2.
(c) Using the equation: \(2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\), calculate the concentration, in \(\text{mol dm}^{-3}\), of ethanedioic acid in FA 1.
(d) Calculate the percentage purity of the \(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\) sample (\(M_r = 126.0\)).
(e) Explain why dilute hydrochloric acid cannot be used to acidify this titration mixture instead of dilute sulfuric acid.
(c) From the stoichiometric ratio, 2 moles of \(\text{MnO}_4^-\text{ react with }5\text{ moles of }\text{H}_2\text{C}_2\text{O}_4\). Moles of \(\text{H}_2\text{C}_2\text{O}_4\) in \(25.0\text{ cm}^3 = 4.00 \times 10^{-4} \times \frac{5}{2} = 1.00 \times 10^{-3}\text{ mol}\). Concentration of \(\text{H}_2\text{C}_2\text{O}_4\) in FA 1 \(= \frac{1.00 \times 10^{-3}}{0.0250} = 0.0400\text{ mol dm}^{-3}\).
(d) Mass of pure \(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\) in \(1\text{ dm}^3 = 0.0400\text{ mol} \times 126.0\text{ g mol}^{-1} = 5.04\text{ g}\). Percentage purity \(= \frac{5.04}{8.40} \times 100\% = 60.0\%\).
(e) Hydrochloric acid contains chloride ions (\(\text{Cl}^-\)), which are oxidized to toxic chlorine gas (\(\text{Cl}_2\)) by the strong oxidizing agent potassium manganate(VII). This would lead to an artificially high titre volume.
PastPaper.markingScheme
Total: 13.33 marks. - Titration table & Concordant Titres (Accuracy): 4.00 marks (1 mark for table structure, 1 mark for recording all burette readings to 0.05 cm3, 2 marks for obtaining titres within 0.10 cm3 of each other). - Part (b) Moles of KMnO4 calculation: 1.00 mark. - Part (c) Moles of H2C2O4 and Concentration calculation: 2.00 marks (1 mark for correct mole ratio, 1 mark for correct final concentration value). - Part (d) Mass of pure acid and Percentage Purity calculation: 2.00 marks (1 mark for multiplying concentration by 126.0, 1 mark for correct final percentage to 3 sig figs). - Part (e) Acid explanation: 4.33 marks (2 marks for identifying that chloride ions / HCl are oxidized, 2.33 marks for identifying that this consumes extra KMnO4 / produces toxic Cl2 gas).
PastPaper.question 2 · Practical
13.33 PastPaper.marks
In this experiment, you will determine the enthalpy change of decomposition of calcium carbonate indirectly using Hess's Law. You are provided with: FA 3: \(3.00\text{ g}\) of calcium carbonate, \(\text{CaCO}_3\) (\(M_r = 100.1\)) FA 4: \(1.40\text{ g}\) of calcium oxide, \(\text{CaO}\) (\(M_r = 56.1\)) Dilute hydrochloric acid, \(\text{HCl}\), of concentration \(2.0\text{ mol dm}^{-3}\) (excess)
(a) Reaction 1: Pour \(50.0\text{ cm}^3\) of the acid into a polystyrene cup. Record the initial temperature \(T_1\). Add all the FA 3, stir continuously, and record the maximum temperature achieved \(T_2\). (Assume: \(T_1 = 22.0^\circ\text{C}\), \(T_2 = 24.3^\circ\text{C}\).) Calculate the heat energy released, \(q_1\), in kJ, and the enthalpy change, \(\Delta H_1\), in \(\text{kJ mol}^{-1}\), for the reaction of \(1\text{ mol}\) of \(\text{CaCO}_3\) with \(\text{HCl}\). (Assume density of solution is \(1.00\text{ g cm}^{-3}\) and specific heat capacity \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)).
(b) Reaction 2: Repeat the procedure using \(50.0\text{ cm}^3\) of acid and all the FA 4. Record the initial temperature \(T_3\) and maximum temperature \(T_4\). (Assume: \(T_3 = 21.5^\circ\text{C}\), \(T_4 = 43.8^\circ\text{C}\).) Calculate the heat energy released, \(q_2\), in kJ, and the enthalpy change, \(\Delta H_2\), in \(\text{kJ mol}^{-1}\), for the reaction of \(1\text{ mol}\) of \(\text{CaO}\) with \(\text{HCl}\).
(c) Use your calculated values of \(\Delta H_1\) and \(\Delta H_2\) to determine the enthalpy change, \(\Delta H_r\), for the decomposition: \(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\)
(d) State one significant source of thermal energy loss in this experiment and suggest a specific practical improvement to reduce it.
(c) The Hess's Law cycle is: \(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\) Reacting both with \(\text{HCl}\) gives: \(\Delta H_r = \Delta H_1 - \Delta H_2\) \(\Delta H_r = -16.0 - (-186.4) = +170.4\text{ kJ mol}^{-1}\).
(d) Heat is lost from the open top of the polystyrene cup to the surrounding air. This can be reduced by placing a plastic lid with a small hole for the thermometer over the cup.
PastPaper.markingScheme
Total: 13.33 marks. - Part (a): 3.00 marks (1 mark for correct calculation of q1, 1 mark for moles of CaCO3, 1 mark for correct Delta H1 with negative sign). - Part (b): 3.00 marks (1 mark for correct calculation of q2, 1 mark for moles of CaO, 1 mark for correct Delta H2 with negative sign). - Part (c): 4.00 marks (2 marks for drawing or describing a valid Hess's Law cycle, 2 marks for the correct value and sign of Delta Hr). - Part (d): 3.33 marks (1.33 marks for identifying heat loss to surroundings / air, 2.00 marks for proposing a lid, double cups, or vacuum flask).
PastPaper.question 3 · Practical
13.33 PastPaper.marks
You are provided with three solutions: FA 5, FA 6, and FA 7. Each solution contains one cation and one anion from the standard Qualitative Analysis Notes. Carry out the following tests and record your observations.
(a) Test 1: Add aqueous sodium hydroxide, \(\text{NaOH}(aq)\), dropwise then in excess to separate portions of FA 5, FA 6, and FA 7. Test 2: Add aqueous ammonia, \(\text{NH}_3(aq)\), dropwise then in excess to separate portions of FA 5, FA 6, and FA 7.
(b) Test 3: Add acidified barium nitrate, \(\text{Ba(NO}_3)_2(aq)\), to FA 5. Add acidified silver nitrate, \(\text{AgNO}_3(aq)\), to FA 6. Add sodium hydroxide and a piece of aluminium foil, then heat gently, to FA 7, testing any gas evolved with damp red litmus paper.
(c) Identify the cations and anions in FA 5, FA 6, and FA 7 based on your tests.
(d) Write the ionic equation (including state symbols) for the reaction of FA 6 with acidified silver nitrate.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
FA 5 contains \(\text{Cr}^{3+}\) and \(\text{SO}_4^{2-}\): - Test 1: Grey-green precipitate, soluble in excess to a dark green solution. - Test 2: Grey-green precipitate, insoluble in excess. - Test 3: White precipitate of \(\text{BaSO}_4\) forms which is insoluble in dilute nitric acid.
FA 6 contains \(\text{Fe}^{2+}\) and \(\text{Cl}^-\): - Test 1: Green precipitate, insoluble in excess, turning brown at the surface on standing. - Test 2: Green precipitate, insoluble in excess. - Test 3: White precipitate of \(\text{AgCl}\) forms, which is soluble in dilute ammonia.
FA 7 contains \(\text{Zn}^{2+}\) and \(\text{NO}_3^-\): - Test 1: White precipitate, soluble in excess to give a colorless solution. - Test 2: White precipitate, soluble in excess to give a colorless solution. - Test 3: Ammonia gas is evolved which turns damp red litmus paper blue, confirming nitrate.
(d) The ionic equation is: \(\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)\).
PastPaper.markingScheme
Total: 13.33 marks. - Part (a) & (b) Observations: 6.00 marks (1 mark for each fully correct set of observations for FA 5, FA 6, and FA 7 in NaOH and NH3; 1 mark for each confirmatory anion test observation in Test 3). - Part (c) Identifications: 3.00 marks (0.5 marks for each correctly identified cation and anion). - Part (d) Ionic Equation: 2.33 marks (1.33 marks for correct balanced ionic equation, 1.00 mark for correct state symbols). - Chemical reasoning: 2.00 marks (explaining how Zn2+ was distinguished from Al3+ by its solubility in excess NH3).
Paper 43 (A Level Structured)
Answer all structured questions addressing both core and optional A Level topics.
10 PastPaper.question · 111.1 PastPaper.marks
PastPaper.question 1 · Structured
11.11 PastPaper.marks
This question is about Born-Haber cycles and the lattice energy of barium sulfide, BaS. The following standard enthalpy change values are given: Enthalpy change of formation of BaS(s): \(-460\text{ kJ mol}^{-1}\), Enthalpy change of atomisation of Ba(s): \(+180\text{ kJ mol}^{-1}\), First ionisation energy of Ba(g): \(+503\text{ kJ mol}^{-1}\), Second ionisation energy of Ba(g): \(+965\text{ kJ mol}^{-1}\), Enthalpy change of atomisation of S(s): \(+279\text{ kJ mol}^{-1}\), First electron affinity of S(g): \(-200\text{ kJ mol}^{-1}\), Second electron affinity of S(g): \(+640\text{ kJ mol}^{-1}\). (a) Define the term lattice energy. [2] (b) Use the given data to calculate the lattice energy of BaS(s). Show your working. [3] (c) Suggest how the lattice energy of MgS(s) compares with that of BaS(s). Explain your answer. [3] (d) Explain why the second electron affinity of sulfur is endothermic, whereas the first electron affinity is exothermic. [3]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Lattice energy is the enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent gaseous ions under standard conditions. (b) Standard enthalpy of formation is equal to the sum of all individual steps in the Born-Haber cycle: \(\Delta H_{\text{f}}^{\ominus}[\text{BaS(s)}] = \Delta H_{\text{at}}^{\ominus}[\text{Ba(s)}] + IE_1[\text{Ba(g)}] + IE_2[\text{Ba(g)}] + \Delta H_{\text{at}}^{\ominus}[\text{S(s)}] + EA_1[\text{S(g)}] + EA_2[\text{S(g)}] + \Delta H_{\text{latt}}^{\ominus}[\text{BaS(s)}]\). Substituting the values: \(-460 = 180 + 503 + 965 + 279 - 200 + 640 + \Delta H_{\text{latt}}^{\ominus}\). \(-460 = 2367 + \Delta H_{\text{latt}}^{\ominus}\). \(\Delta H_{\text{latt}}^{\ominus} = -460 - 2367 = -2827\text{ kJ mol}^{-1}\). (c) The lattice energy of MgS(s) will be more exothermic (more negative) than that of BaS(s). This is because the magnesium ion (\(\text{Mg}^{2+}\)) is smaller than the barium ion (\(\text{Ba}^{2+}\)), resulting in a higher charge density. Consequently, there is a stronger electrostatic force of attraction between the oppositely charged ions (\(\text{Mg}^{2+}\) and \(\text{S}^{2-}\)), which releases more energy when the lattice is formed. (d) The first electron affinity of sulfur is exothermic because there is an attractive force between the neutral sulfur atom and the incoming electron. The second electron affinity is endothermic because the incoming electron is being added to a negatively charged \(\text{S}^-\) ion. Energy must be supplied to overcome the electrostatic repulsion between the negative ion and the incoming negative electron.
PastPaper.markingScheme
M1: Lattice energy definition - enthalpy change when 1 mole of ionic solid is formed [1] from its constituent gaseous ions [1]. M2: Correct equation or Born-Haber cycle setup [1]. M3: Correct substitution of values [1]. M4: Calculation giving -2827 kJ mol^-1 [1]. M5: MgS is more exothermic/negative [1]. M6: Mg2+ is smaller / has a higher charge density than Ba2+ [1]. M7: Stronger electrostatic attraction between Mg2+ and S2- [1]. M8: First EA is exothermic because of attraction of nucleus to electron [1]. M9: Second EA is endothermic because of repulsion [1] between the incoming electron and the negatively charged S- ion [1].
PastPaper.question 2 · Structured
11.11 PastPaper.marks
An electrochemical cell is set up consisting of a standard \(\text{Fe}^{3+}(\text{aq})/\text{Fe}^{2+}(\text{aq})\) half-cell connected to a non-standard \(\text{Ag}^{+}(\text{aq})/\text{Ag}(\text{s})\) half-cell. The standard electrode potentials are: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\) \(E^{\ominus} = +0.77\text{ V}\); \(\text{Ag}^{+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s})\) \(E^{\ominus} = +0.80\text{ V}\). In the non-standard silver half-cell, \([\text{Ag}^{+}] = 0.025\text{ mol dm}^{-3}\) at \(298\text{ K}\). (a) Draw a fully labelled diagram of this electrochemical cell. Include details of the electrodes, solutions, and salt bridge. [4] (b) State the function of the salt bridge and suggest a suitable electrolyte to prepare it. [2] (c) Calculate the electrode potential, \(E\), of the silver half-cell under these non-standard conditions using the Nernst equation. [2] (d) Calculate the overall cell potential, \(E_{\text{cell}}\), under these conditions and write the overall ionic equation for the cell reaction. [3]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) A labeled diagram must show a platinum electrode dipping into a solution containing 1.0 mol dm^-3 of both Fe2+(aq) and Fe3+(aq), connected via a voltmeter and wire to a silver electrode dipping into 0.025 mol dm^-3 Ag+(aq) solution. A salt bridge must connect both beakers. (b) The function of the salt bridge is to complete the electrical circuit and maintain electrical neutrality in both half-cells by allowing ion migration. Potassium nitrate (KNO3) is a suitable electrolyte. (c) Using the Nernst equation: \(E = E^{\ominus} + 0.059 \log [\text{Ag}^{+}] = 0.80 + 0.059 \log(0.025) = 0.80 - 0.0945 = +0.706\text{ V} \approx +0.71\text{ V}\). (d) The silver half-cell has a lower electrode potential (+0.71 V) than the iron half-cell (+0.77 V). Therefore, oxidation occurs at the silver electrode and reduction at the iron electrode. \(E_{\text{cell}} = 0.77 - 0.706 = +0.064\text{ V} \approx +0.06\text{ V}\). The overall ionic equation is: \(\text{Ag}(\text{s}) + \text{Fe}^{3+}(\text{aq}) \rightarrow \text{Ag}^{+}(\text{aq}) + \text{Fe}^{2+}(\text{aq})\).
PastPaper.markingScheme
M1: Draw Pt electrode in Fe3+/Fe2+ solution and Ag electrode in Ag+ solution [1]. M2: Standard concentrations for iron half-cell (1.0 M) and non-standard for silver (0.025 M) [1]. M3: Voltmeter and salt bridge labeled [1]. M4: All components correctly connected [1]. M5: Role of salt bridge described (neutrality/migration) [1]. M6: Suggest KNO3 or NH4NO3 [1]. M7: Correct substitution into Nernst equation [1]. M8: Obtain E = +0.71 V [1]. M9: Calculate cell potential = +0.06 V [1]. M10: Correct overall equation: Ag(s) + Fe3+(aq) -> Ag+(aq) + Fe2+(aq) [1]. M11: Correct state symbols [1].
PastPaper.question 3 · Structured
11.11 PastPaper.marks
Nitric oxide, \(\text{NO}\), reacts with hydrogen gas according to the equation: \(2\text{NO}(\text{g}) + 2\text{H}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g})\). The following initial rate data were obtained at \(800\text{ K}\): Experiment 1: \([\text{NO}] = 0.010\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.010\text{ mol dm}^{-3}\), Initial Rate = \(1.2 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\); Experiment 2: \([\text{NO}] = 0.020\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.010\text{ mol dm}^{-3}\), Initial Rate = \(4.8 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\); Experiment 3: \([\text{NO}] = 0.010\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.020\text{ mol dm}^{-3}\), Initial Rate = \(2.4 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). (a) Deduce the order of reaction with respect to both \(\text{NO}\) and \(\text{H}_2\). Show your reasoning. [4] (b) Write the rate equation for the reaction, and calculate the value of the rate constant, \(k\), including its units. [3] (c) Suggest a multi-step reaction mechanism consistent with this rate equation, indicating which step is the rate-determining step. [4]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Comparing Exp 1 and Exp 2: \([\text{H}_2]\) is constant. Doubling \([\text{NO}]\) (from 0.010 to 0.020) causes the rate to increase by a factor of 4 (from \(1.2 \times 10^{-5}\) to \(4.8 \times 10^{-5}\)). Since \(2^2 = 4\), the order with respect to \(\text{NO}\) is 2. Comparing Exp 1 and Exp 3: \([\text{NO}]\) is constant. Doubling \([\text{H}_2]\) (from 0.010 to 0.020) causes the rate to double (from \(1.2 \times 10^{-5}\) to \(2.4 \times 10^{-5}\)). Since \(2^1 = 2\), the order with respect to \(\text{H}_2\) is 1. (b) Rate equation: \(\text{rate} = k[\text{NO}]^2[\text{H}_2]\). Substituting Exp 1 data: \(1.2 \times 10^{-5} = k (0.010)^2 (0.010)\), which gives \(k = 12\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). (c) Proposed mechanism: Step 1: \(\text{NO} + \text{NO} \rightleftharpoons \text{N}_2\text{O}_2\) (fast equilibrium); Step 2: \(\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}\) (slow, rate-determining step); Step 3: \(\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}\) (fast). The slow step has the rate law \(\text{rate} = k'[\text{N}_2\text{O}_2][\text{H}_2]\). Substituting \([\text{N}_2\text{O}_2] = K_{\text{eq}}[\text{NO}]^2\) gives \(\text{rate} = k' K_{\text{eq}}[\text{NO}]^2[\text{H}_2]\), which matches the experimental rate law.
PastPaper.markingScheme
M1: Order with respect to NO is 2 [1]. M2: Reasoning for NO order (rate quadrupled when concentration doubled) [1]. M3: Order with respect to H2 is 1 [1]. M4: Reasoning for H2 order (rate doubled when concentration doubled) [1]. M5: Rate equation: rate = k[NO]^2[H2] [1]. M6: Value of k = 12 [1]. M7: Units of k = dm6 mol-2 s-1 [1]. M8: Step 1: equilibrium forming an intermediate like N2O2 [1]. M9: Step 2: slow step involving H2 and the intermediate [1]. M10: Step 3: fast step resulting in final products [1]. M11: Explanation connecting slow step molecularity to the rate law [1].
PastPaper.question 4 · Structured
11.11 PastPaper.marks
Aqueous copper(II) ions exist as \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\). When excess aqueous ammonia is added, ligand substitution occurs: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\) for which \(K_{\text{stab}} = 2.1 \times 10^{13}\text{ dm}^{12}\text{ mol}^{-4}\). When 1,2-diaminoethane (en) is added, it forms a tris-chelate complex: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightleftharpoons [\text{Cu}(\text{en})_3]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\) for which \(K_{\text{stab}} = 2.0 \times 10^{18}\text{ dm}^9\text{ mol}^{-3}\). (a) Write the expression for the stability constant, \(K_{\text{stab}}\), for the formation of \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\). [2] (b) Explain why the tris-chelate complex is much more stable than the ammonia complex, referencing entropy changes, \(\Delta S^{\ominus}\). [4] (c) Describe the color change when excess ammonia is added to aqueous copper(II) sulfate, and state the geometry of the resulting complex. [3] (d) Draw the structure of one of the optical isomers of \([\text{Cu}(\text{en})_3]^{2+}\), showing its three-dimensional shape. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(K_{\text{stab}} = \frac{[ [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} ]}{[ [\text{Cu}(\text{H}_2\text{O})_6]^{2+} ] [\text{NH}_3]^4}\). (Water is omitted as it is the solvent). (b) The tris-chelate complex formation is driven by a highly positive entropy change (\(\Delta S^{\ominus} > 0\)). The reaction of 1 complex ion with 3 bidentate ligands (total 4 particles) produces 1 complex ion and 6 water molecules (total 7 particles), causing a significant increase in disorder. In contrast, the ammonia reaction goes from 5 particles to 5 particles, resulting in little entropy change. Since \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\), the more positive entropy change makes \(\Delta G^{\ominus}\) more negative, indicating a much more stable complex. (c) The color changes from pale blue to deep blue. The geometry of the resulting complex is octahedral. (d) The drawing must show central Cu2+ in an octahedral environment connected to three 'en' (NH2-CH2-CH2-NH2) bidentate ligands, forming a non-superimposable mirror image.
PastPaper.markingScheme
M1: Correct Kstab expression with correct charges [1]. M2: Correctly omit water from the denominator [1]. M3: Identify increase in number of particles from 4 to 7 in the chelation reaction [1]. M4: State that this leads to a highly positive standard entropy change (disorder increases) [1]. M5: Link positive entropy change to more negative Gibbs free energy [1]. M6: Use the term 'chelate effect' [1]. M7: Pale blue [1] to deep blue solution [1]. M8: Octahedral geometry [1]. M9: Draw central Cu with 6 coordinate bonds shown in 3D (wedges/dashes) [1]. M10: Correctly link three pairs of adjacent positions with 'en' bridges showing a 3D optical isomer [1].
PastPaper.question 5 · Structured
11.11 PastPaper.marks
An organic acid, X, is soluble in both water and ethoxyethane (ether). The partition coefficient, \(K_{\text{pc}}\), of X between ether and water is defined as: \(K_{\text{pc}} = \frac{[\text{X}]_{\text{ether}}}{[\text{X}]_{\text{water}}} = 4.50\). A solution is prepared by dissolving \(10.0\text{ g}\) of X in \(100\text{ cm}^3\) of water. (a) State the definition of partition coefficient. [2] (b) Calculate the mass of X extracted from this aqueous solution if it is shaken with: (i) One single portion of \(60\text{ cm}^3\) of ether. [3] (ii) Two successive portions of \(30\text{ cm}^3\) of ether. [4] (c) Compare the efficiency of the extraction in (b)(i) and (b)(ii) and state a general rule for solvent extraction. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) The partition coefficient is the ratio of the concentrations of a solute in two immiscible solvents at equilibrium at a specified temperature. (b)(i) Let \(m\) be the mass of X extracted into 60 cm^3 of ether. Mass left in water is \(10.0 - m\). \(K_{\text{pc}} = \frac{m / 60}{(10.0 - m) / 100} = 4.50\). Rearranging: \(\frac{100m}{60(10.0 - m)} = 4.50 \implies 5m = 13.5(10.0 - m) \implies 18.5m = 135 \implies m = 7.30\text{ g}\). (b)(ii) First portion: let \(m_1\) be the mass extracted. \(\frac{m_1 / 30}{(10.0 - m_1) / 100} = 4.50 \implies \frac{10m_1}{3(10.0 - m_1)} = 4.50 \implies 23.5m_1 = 135 \implies m_1 = 5.74\text{ g}\). Mass remaining in water is \(10.0 - 5.74 = 4.26\text{ g}\). Second portion: let \(m_2\) be the mass extracted. \(\frac{m_2 / 30}{(4.26 - m_2) / 100} = 4.50 \implies \frac{10m_2}{3(4.26 - m_2)} = 4.50 \implies 23.5m_2 = 57.51 \implies m_2 = 2.45\text{ g}\). Total mass extracted = \(5.74 + 2.45 = 8.19\text{ g}\). (c) Shaking with two portions of 30 cm^3 of ether extracts more acid (8.19 g) than with one 60 cm^3 portion (7.30 g). General rule: Successive extractions with smaller volumes of solvent are more efficient than a single extraction with the same total volume of solvent.
PastPaper.markingScheme
M1: Ratio of concentrations of a solute in two immiscible solvents [1] at equilibrium at a constant temperature [1]. M2: Correct setup of the fraction for the first extraction [1]. M3: Correct algebraic step to solve for m [1]. M4: First extraction mass = 7.30 g [1]. M5: Correct setup for first portion of the second extraction [1]. M6: First portion mass = 5.74 g and remaining mass = 4.26 g [1]. M7: Correct setup for second portion [1]. M8: Total mass extracted = 8.19 g [1]. M9: Clearly states that two smaller portions extract more than one larger portion [1]. M10: General rule of multiple extraction efficiency stated [1].
PastPaper.question 6 · Structured
11.11 PastPaper.marks
A synthesis of 4-aminobenzoic acid (PABA) starting from methylbenzene is outlined below: Methylbenzene \(\xrightarrow{\text{Step 1}} \) 4-nitromethylbenzene \(\xrightarrow{\text{Step 2}} \) 4-nitrobenzoic acid \(\xrightarrow{\text{Step 3}} \) 4-aminobenzoic acid. (a) Give the reagents and conditions for Step 1. State the role of sulfuric acid in this step and write an equation for the formation of the electrophile. [4] (b) Draw the structures of 4-nitromethylbenzene and 4-nitrobenzoic acid. [2] (c) State the reagents and conditions required for: (i) Step 2 [2] (ii) Step 3 [1] (d) State how methylbenzene and 4-aminobenzoic acid can be distinguished using a simple chemical test, describing the observations. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Reagents: concentrated HNO3 and concentrated H2SO4. Conditions: warm at 50-60 °C. Role of H2SO4: acts as an acid catalyst to produce the electrophile. Equation: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^{+} + \text{H}_3\text{O}^{+} + 2\text{HSO}_4^{-}\). (b) 4-nitromethylbenzene consists of a benzene ring with a -CH3 group at C1 and a -NO2 group at C4. 4-nitrobenzoic acid has a -COOH group at C1 and a -NO2 group at C4. (c)(i) Step 2: Alkaline KMnO4(aq), heat under reflux, followed by adding dilute acid (HCl or H2SO4). (ii) Step 3: Sn and concentrated HCl, heat under reflux, followed by adding NaOH(aq). (d) Test: Add Na2CO3(aq) or NaHCO3(aq). Observation: Methylbenzene shows no reaction. 4-aminobenzoic acid reacts to produce effervescence/bubbles of CO2 gas.
PastPaper.markingScheme
M1: Step 1 reagents: concentrated HNO3 and concentrated H2SO4 [1]. M2: Step 1 conditions: 50-60 °C [1]. M3: Role of H2SO4: catalyst / protonates HNO3 [1]. M4: Formation of NO2+ equation: HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4- [1]. M5: Draw correct structural formula of 4-nitromethylbenzene [1]. M6: Draw correct structural formula of 4-nitrobenzoic acid [1]. M7: Step 2: alkaline KMnO4, heat under reflux [1], then add acid / H+ [1]. M8: Step 3: Sn and concentrated HCl, heat, followed by NaOH(aq) [1]. M9: Test: add Na2CO3 or NaHCO3 [1]. M10: Observation: methylbenzene shows no reaction; 4-aminobenzoic acid shows effervescence [1].
PastPaper.question 7 · Structured
11.11 PastPaper.marks
Calcium carbonate decomposes at high temperatures according to the following equation: \(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\). The following thermodynamic data at \(298\text{ K}\) are given: \(\Delta H_{\text{f}}^{\ominus}[\text{CaCO}_3(\text{s})] = -1207\text{ kJ mol}^{-1}\), \(\Delta H_{\text{f}}^{\ominus}[\text{CaO}(\text{s})] = -635\text{ kJ mol}^{-1}\), \(\Delta H_{\text{f}}^{\ominus}[\text{CO}_2(\text{g})] = -394\text{ kJ mol}^{-1}\). Standard entropies, \(S^{\ominus}\): \(\text{CaCO}_3(\text{s}) = 92.9\text{ J K}^{-1}\text{ mol}^{-1}\), \(\text{CaO}(\text{s}) = 39.7\text{ J K}^{-1}\text{ mol}^{-1}\), \(\text{CO}_2(\text{g}) = 213.6\text{ J K}^{-1}\text{ mol}^{-1}\). (a) Define the term standard entropy of a substance. [2] (b) Calculate the standard enthalpy change, \(\Delta H^{\ominus}\), for this reaction at \(298\text{ K}\). [2] (c) Calculate the standard entropy change, \(\Delta S^{\ominus}\), for this reaction at \(298\text{ K}\). [2] (d) Calculate the standard Gibbs free energy change, \(\Delta G^{\ominus}\), for this reaction at \(298\text{ K}\). State and explain whether the reaction is spontaneous at this temperature. [3] (e) Determine the temperature (in K) above which this decomposition reaction becomes feasible. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Standard entropy is the entropy of one mole of a substance under standard conditions (a pressure of 1 bar / 100 kPa and temperature of 298 K). (b) \(\Delta H^{\ominus} = \Sigma \Delta H_{\text{f}}^{\ominus}(\text{products}) - \Sigma \Delta H_{\text{f}}^{\ominus}(\text{reactants}) = [(-635) + (-394)] - [-1207] = -1029 + 1207 = +178\text{ kJ mol}^{-1}\). (c) \(\Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants}) = [39.7 + 213.6] - 92.9 = 253.3 - 92.9 = +160.4\text{ J K}^{-1}\text{ mol}^{-1}\). (d) \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = +178 - (298 \times \frac{160.4}{1000}) = 178 - 47.8 = +130.2\text{ kJ mol}^{-1}\). Since \(\Delta G^{\ominus}\) is positive, the reaction is not spontaneous at 298 K. (e) Feasibility requires \(\Delta G \le 0 \implies T \ge \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \frac{178 \times 1000}{160.4} = 1109.7\text{ K} \approx 1110\text{ K}\).
PastPaper.markingScheme
M1: Entropy of 1 mole of substance [1] under standard conditions (1 bar / 100 kPa and 298 K) [1]. M2: Correct equation for enthalpy of reaction and substitution [1]. M3: Correct value of \(\Delta H^{\ominus} = +178\text{ kJ mol}^{-1}\) [1]. M4: Correct equation for entropy of reaction and substitution [1]. M5: Correct value of \(\Delta S^{\ominus} = +160.4\text{ J K}^{-1}\text{ mol}^{-1}\) [1]. M6: Substitution into Gibbs equation [1]. M7: Correct value of \(\Delta G^{\ominus} = +130.2\text{ kJ mol}^{-1}\) [1]. M8: Explaining that since \(\Delta G > 0\), it is not spontaneous [1]. M9: Condition for feasibility: \(\Delta G \le 0\) or \(T = \Delta H / \Delta S\) [1]. M10: Calculation: T = 1110 K [1].
PastPaper.question 8 · Structured
11.11 PastPaper.marks
This question is about transition elements and their physical and chemical properties. (a) Explain in detail why transition metal complex ions are typically colored. Your answer should make reference to: d-orbitals and their splitting, the role of ligands, energy absorption and transition of electrons. [5] (b) When concentrated hydrochloric acid is added to a pale blue solution of aqueous copper(II) sulfate, the solution changes color to yellow-green. Write an ionic equation for this reaction and explain the color change in terms of ligand exchange and the size of the ligands. [4] (c) Suggest why solutions of \(\text{Sc}^{3+}\) and \(\text{Zn}^{2+}\) ions are completely colorless. [2]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) In an isolated transition metal ion, the five d-orbitals are degenerate. When ligands approach to form coordinate bonds, electrostatic repulsion splits the d-orbitals into two groups of different energy levels. The energy difference is \(\Delta E\). When light shines on the complex, an electron in a lower-energy d-orbital absorbs a photon of light of specific frequency/energy (\(\Delta E = h\nu\)) and is promoted to a higher-energy d-orbital (d-d transition). The light that is not absorbed is transmitted, and the observer sees the complementary color of the absorbed light. (b) Equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^{-}(\text{aq}) \rightleftharpoons [\text{Cu}\text{Cl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\). Explanation: Chloride ligands replace water ligands. Chloride ions are larger than water molecules and negatively charged, causing steric hindrance and electrostatic repulsion. Consequently, only four chloride ligands can fit, changing coordination number from 6 (octahedral) to 4 (tetrahedral). The change in ligands and geometry changes \(\Delta E\), which alters the absorbed light wavelength and the observed color. (c) Sc3+ has the electron configuration [Ar] 3d0. Since there are no electrons in the d-orbitals, d-d transitions cannot occur. Zn2+ has the electron configuration [Ar] 3d10. Since the d-orbitals are completely filled, there are no empty d-orbitals available for promotion of electrons. Hence, both solutions are colorless.
PastPaper.markingScheme
M1: Degenerate d-orbitals split into two different energy levels [1] in the presence of ligands [1]. M2: Electron is promoted from a lower to a higher d-orbital level [1]. M3: Absorption of specific wavelength of visible light [1]. M4: Color observed is the complementary color of absorbed light [1]. M5: Correct equation: [Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O [1]. M6: State change in coordination number from 6 to 4 / geometry octahedral to tetrahedral [1]. M7: Explain that Cl- is larger / negatively charged, causing steric hindrance [1]. M8: Changing ligands changes d-orbital splitting energy, changing absorbed light color [1]. M9: Sc3+ has 3d0, no d-electrons, so no transitions possible [1]. M10: Zn2+ has 3d10, full d-orbitals, so no transitions possible [1].
PastPaper.question 9 · Structured
11.11 PastPaper.marks
(a) (i) Define the term *stability constant*, \(K_{\text{stab}}\), of a complex ion.
(ii) Write an expression for the stability constant, \(K_{\text{stab}}\), of the following ligand exchange reaction, and state its units.
(b) Cobalt(II) ions exist in aqueous solution as the pink hexaaquacobalt(II) ion, \[[\text{Co}(\text{H}_2\text{O})_6]^{2+}\]. When concentrated hydrochloric acid is added, a ligand exchange reaction occurs to form a blue tetrahedral complex:
The stability constant, \(K_{\text{stab}}\), for this equilibrium is \(3.2 \times 10^1\text{ dm}^{12}\text{ mol}^{-4}\).
At equilibrium, a solution of cobalt(II) chloride in concentrated hydrochloric acid has a total concentration of cobalt species of \(0.10\text{ mol dm}^{-3}\) and a free chloride ion concentration, \[[\text{Cl}^-]\], of \(2.5\text{ mol dm}^{-3}\).
Calculate the equilibrium concentration of the remaining hexaaquacobalt(II) ions, \[[[\text{Co}(\text{H}_2\text{O})_6]^{2+}]\).
(c) The bidentate ligand 1,2-diaminoethane is represented by \(\text{en}\).
The reaction of hexaaquacobalt(II) with \(\text{en}\) has a very large stability constant:
Explain this difference in the stability of the two complexes in terms of entropy change, \(\Delta S^\theta\), and Gibbs free energy change, \(\Delta G^\theta\).
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a) (i)**
The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion from its constituent ligands and parent metal ion in solvent (usually water).
2. Define variables based on the total cobalt concentration of \(0.10\text{ mol dm}^{-3}\): Let \([[\text{Co}(\text{H}_2\text{O})_6]^{2+}] = y\text{ mol dm}^{-3}\). Therefore, \([[\text{CoCl}_4]^{2-}] = (0.10 - y)\text{ mol dm}^{-3}\).
1. In the reaction with \(\text{en}\), 4 reactant particles react to form 7 product particles. This results in an increase in the number of particles in solution, leading to a significant increase in disorder and a highly positive entropy change (\(\Delta S^\theta \gg 0\)). 2. In the reaction with \(\text{NH}_3\), 7 reactant particles react to form 7 product particles. Thus, there is little to no change in the number of particles, so the entropy change is close to zero (\(\Delta S^\theta \approx 0\)). 3. Since both reactions involve the formation of similar metal-nitrogen (Co–N) coordinate bonds, the enthalpy change (\(\Delta H^\theta\)) for both reactions is very similar. 4. According to \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), the highly positive \(\Delta S^\theta\) for the bidentate \(\text{en}\) reaction makes its \(\Delta G^\theta\) much more negative than that for the ammonia reaction. 5. A much more negative \(\Delta G^\theta\) corresponds to a significantly larger stability constant (\(K_{\text{stab}}\)).
PastPaper.markingScheme
**(a) (i)** - **1 mark**: Define \(K_{\text{stab}}\) as the equilibrium constant for the formation of a complex ion from its constituent ions / ligands in aqueous solution (accept formula/expression definition if terms are clearly defined).
**(a) (ii)** - **1 mark**: Correct \(K_{\text{stab}}\) expression: \(K_{\text{stab}} = \frac{[[\text{Fe}(\text{SCN})(\text{H}_2\text{O})_5]^{2+}]}{[[\text{Fe}(\text{H}_2\text{O})_6]^{3+}][\text{SCN}^-]}\) (ignore state symbols, do not include water). - **1 mark**: Units: \(\text{dm}^3\text{ mol}^{-1}\) (allow \(\text{mol}^{-1}\text{ dm}^3\)).
**(b)** - **1 mark**: Correct substitution of values into a valid equilibrium expression (e.g., \(32 = \frac{x}{y \times 2.5^4}\)). - **1 mark**: Use of conservation of cobalt mass: \([[\text{CoCl}_4]^{2-}] = 0.10 - [[\text{Co}(\text{H}_2\text{O})_6]^{2+}]\). - **1 mark**: Correct mathematical rearrangement showing \(1251 y = 0.10\) (or equivalent algebraic step). - **1 mark**: Correct calculation of final concentration: \(8.0 \times 10^{-5}\text{ mol dm}^{-3}\) (or \(7.99 \times 10^{-5}\)), including correct unit (\(\text{mol dm}^{-3}\)). Allow 2 or 3 significant figures.
**(c)** - **1 mark**: Identifies that the reaction with \(\text{en}\) increases the number of particles in solution (4 molecules/ions to 7), leading to a positive/increased entropy change (\(\Delta S^\theta > 0\)). - **1 mark**: Identifies that the reaction with \(\text{NH}_3\) has no change in the number of particles (7 to 7), so the entropy change is minimal/close to zero. - **1 mark**: Mentions that \(\Delta H^\theta\) is similar for both reactions (as both involve the formation of six Co–N bonds). - **1 mark**: Links the highly positive \(\Delta S^\theta\) for \(\text{en}\) to a much more negative \(\Delta G^\theta\) (via \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)), which leads to a larger stability constant.
PastPaper.question 10 · Structured
11.11 PastPaper.marks
(a) (i) Define the term *stability constant*, \(K_{\text{stab}}\), of a complex ion.
(ii) Write an expression for the stability constant, \(K_{\text{stab}}\), of the following ligand exchange reaction, and state its units.
(b) Cobalt(II) ions exist in aqueous solution as the pink hexaaquacobalt(II) ion, \[[\text{Co}(\text{H}_2\text{O})_6]^{2+}\]. When concentrated hydrochloric acid is added, a ligand exchange reaction occurs to form a blue tetrahedral complex:
The stability constant, \(K_{\text{stab}}\), for this equilibrium is \(3.2 \times 10^1\text{ dm}^{12}\text{ mol}^{-4}\).
At equilibrium, a solution of cobalt(II) chloride in concentrated hydrochloric acid has a total concentration of cobalt species of \(0.10\text{ mol dm}^{-3}\) and a free chloride ion concentration, \[[\text{Cl}^-]\], of \(2.5\text{ mol dm}^{-3}\).
Calculate the equilibrium concentration of the remaining hexaaquacobalt(II) ions, \[[[\text{Co}(\text{H}_2\text{O})_6]^{2+}]\).
(c) The bidentate ligand 1,2-diaminoethane is represented by \(\text{en}\).
The reaction of hexaaquacobalt(II) with \(\text{en}\) has a very large stability constant:
Explain this difference in the stability of the two complexes in terms of entropy change, \(\Delta S^\theta\), and Gibbs free energy change, \(\Delta G^\theta\).
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a) (i)**
The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion from its constituent ligands and parent metal ion in solvent (usually water).
2. Define variables based on the total cobalt concentration of \(0.10\text{ mol dm}^{-3}\): Let \([[\text{Co}(\text{H}_2\text{O})_6]^{2+}] = y\text{ mol dm}^{-3}\). Therefore, \([[\text{CoCl}_4]^{2-}] = (0.10 - y)\text{ mol dm}^{-3}\).
1. In the reaction with \(\text{en}\), 4 reactant particles react to form 7 product particles. This results in an increase in the number of particles in solution, leading to a significant increase in disorder and a highly positive entropy change (\(\Delta S^\theta \gg 0\)). 2. In the reaction with \(\text{NH}_3\), 7 reactant particles react to form 7 product particles. Thus, there is little to no change in the number of particles, so the entropy change is close to zero (\(\Delta S^\theta \approx 0\)). 3. Since both reactions involve the formation of similar metal-nitrogen (Co–N) coordinate bonds, the enthalpy change (\(\Delta H^\theta\)) for both reactions is very similar. 4. According to \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), the highly positive \(\Delta S^\theta\) for the bidentate \(\text{en}\) reaction makes its \(\Delta G^\theta\) much more negative than that for the ammonia reaction. 5. A much more negative \(\Delta G^\theta\) corresponds to a significantly larger stability constant (\(K_{\text{stab}}\)).
PastPaper.markingScheme
**(a) (i)** - **1 mark**: Define \(K_{\text{stab}}\) as the equilibrium constant for the formation of a complex ion from its constituent ions / ligands in aqueous solution (accept formula/expression definition if terms are clearly defined).
**(a) (ii)** - **1 mark**: Correct \(K_{\text{stab}}\) expression: \(K_{\text{stab}} = \frac{[[\text{Fe}(\text{SCN})(\text{H}_2\text{O})_5]^{2+}]}{[[\text{Fe}(\text{H}_2\text{O})_6]^{3+}][\text{SCN}^-]}\) (ignore state symbols, do not include water). - **1 mark**: Units: \(\text{dm}^3\text{ mol}^{-1}\) (allow \(\text{mol}^{-1}\text{ dm}^3\)).
**(b)** - **1 mark**: Correct substitution of values into a valid equilibrium expression (e.g., \(32 = \frac{x}{y \times 2.5^4}\)). - **1 mark**: Use of conservation of cobalt mass: \([[\text{CoCl}_4]^{2-}] = 0.10 - [[\text{Co}(\text{H}_2\text{O})_6]^{2+}]\). - **1 mark**: Correct mathematical rearrangement showing \(1251 y = 0.10\) (or equivalent algebraic step). - **1 mark**: Correct calculation of final concentration: \(8.0 \times 10^{-5}\text{ mol dm}^{-3}\) (or \(7.99 \times 10^{-5}\)), including correct unit (\(\text{mol dm}^{-3}\)). Allow 2 or 3 significant figures.
**(c)** - **1 mark**: Identifies that the reaction with \(\text{en}\) increases the number of particles in solution (4 molecules/ions to 7), leading to a positive/increased entropy change (\(\Delta S^\theta > 0\)). - **1 mark**: Identifies that the reaction with \(\text{NH}_3\) has no change in the number of particles (7 to 7), so the entropy change is minimal/close to zero. - **1 mark**: Mentions that \(\Delta H^\theta\) is similar for both reactions (as both involve the formation of six Co–N bonds). - **1 mark**: Links the highly positive \(\Delta S^\theta\) for \(\text{en}\) to a much more negative \(\Delta G^\theta\) (via \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)), which leads to a larger stability constant.
Paper 53 (Planning & Evaluation)
Answer all questions focusing on experimental planning, data analysis, and error reduction.
2 PastPaper.question · 30 PastPaper.marks
PastPaper.question 1 · Planning
15 PastPaper.marks
A student wants to determine the activation energy, \(E_a\), of the reaction between peroxodisulfate ions and iodide ions: \[ \text{S}_2\text{O}_8^{2-}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{SO}_4^{2-}(aq) + \text{I}_2(aq) \]
The rate of this reaction can be monitored using a 'clock reaction' by adding a small, fixed amount of sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\), and starch indicator.
The thiosulfate ions instantly reduce any iodine formed back to iodide: \[ \text{I}_2(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 2\text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq) \] Once all the thiosulfate is consumed, the remaining iodine reacts with starch, turning the solution blue-black. The time, \(t\), taken for the blue-black color to appear is measured. The rate of reaction can be assumed to be proportional to \(1/t\).
**(a)** Identify: (i) the independent variable. (ii) the dependent variable. (iii) two variables that must be kept constant. [3 marks]
**(b)** Design a detailed experimental procedure to investigate how the rate of this reaction varies with temperature in the range \(20\ ^\circ\text{C}\) to \(60\ ^\circ\text{C}\). Your plan should include: - details of how the temperature will be varied and controlled. - the volumes and concentrations of reagents you would use (suggested reagents: \(0.100\text{ mol dm}^{-3}\ \text{K}_2\text{S}_2\text{O}_8\), \(0.200\text{ mol dm}^{-3}\ \text{KI}\), \(0.0100\text{ mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3\), and starch solution). Total volume of the reaction mixture should be \(50.0\text{ cm}^3\). - how you will ensure that the reagents are at the correct temperature before mixing. [5 marks]
**(c)** The Arrhenius equation is: \[ \ln k = -\frac{E_a}{RT} + \text{constant} \] Since rate \(\propto 1/t\), the rate constant \(k\) can be approximated by \(1/t\), giving: \[ \ln\left(\frac{1}{t}\right) = -\frac{E_a}{RT} + \text{constant} \] Draw a template table that the student would use to record raw and processed data to plot a straight-line graph to determine \(E_a\). Include appropriate column headings with units. [3 marks]
**(d)** Describe how the student would use the gradient of the graph to determine the activation energy, \(E_a\), in \(\text{kJ mol}^{-1}\). State the value of any constant(s) used. [2 marks]
**(e)** A student carried out the experiment and obtained the following data point: At temperature \(\theta = 45.0\ ^\circ\text{C}\), the time taken for the blue-black color to appear was \(t = 34.0\text{ s}\). Calculate the value of: (i) \(1/T\) in \(\text{K}^{-1}\) (ii) \(\ln(1/t)\) Show your working and express your answers to an appropriate number of significant figures. [2 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a)** (i) Independent variable: Temperature, \(T\). (ii) Dependent variable: Time, \(t\), taken for the blue-black color to appear. (iii) Controlled variables: Volume and concentration of potassium peroxodisulfate, potassium iodide, sodium thiosulfate, and starch; total volume of the reaction mixture.
**(b)** 1. Set up a water bath (thermostatically controlled or heated beaker of water) to the target temperature (e.g., \(20, 30, 40, 50, 60\ ^\circ\text{C}\)). 2. Prepare Tube A with \(15.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\ \text{KI}\), \(10.0\text{ cm}^3\) of \(0.0100\text{ mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3\), and \(5.0\text{ cm}^3\) of starch indicator solution. 3. Prepare Tube B with \(20.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\ \text{K}_2\text{S}_2\text{O}_8\) (total volume = \(50.0\text{ cm}^3\)). 4. Place both tubes in the water bath and leave them for at least 5 minutes to reach thermal equilibrium. 5. Record the initial temperature of the solutions using a thermometer. Mix the contents of Tube A and Tube B, start the stopwatch immediately, and keep the reaction tube in the water bath. 6. Stop the stopwatch when the blue-black color appears.
**(c)** Columns needed in the table: 1. Temperature, \(\theta\ /\ ^\circ\text{C}\) 2. Temperature, \(T\ /\ \text{K}\) 3. Reciprocal of Temperature, \(1/T\ /\ \text{K}^{-1}\) 4. Time, \(t\ /\ \text{s}\) 5. \(\ln(1/t)\)
**(d)** Plot \(\ln(1/t)\) on the y-axis against \(1/T\) on the x-axis. The gradient of the resulting straight line is given by: \[ \text{Gradient} = -\frac{E_a}{R} \] Thus, \(E_a = -\text{Gradient} \times R\), where \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\). Divide by 1000 to convert \(\text{J mol}^{-1}\) to \(\text{kJ mol}^{-1}\).
- **(a)** [3 marks total]: - 1 mark for identifying the independent variable as temperature. - 1 mark for identifying the dependent variable as time (or rate / 1/time). - 1 mark for identifying any two controlled variables: concentration/volume of any reagent, total volume. - **(b)** [5 marks total]: - 1 mark for stating that a thermostatically controlled water bath or beaker water bath with burner is used to vary/control temperature. - 1 mark for proposing volumes of reagents that sum to exactly \(50.0\text{ cm}^3\) with thiosulfate as limiting reactant. - 1 mark for describing at least 5 different temperatures covering the range \(20-60\ ^\circ\text{C}\). - 1 mark for placing reactants in separate boiling tubes in the water bath before mixing. - 1 mark for checking and recording the temperature of the reactants immediately before mixing. - **(c)** [3 marks total]: - 1 mark for temperature columns (with \(^\circ\text{C}\) and \(\text{K}\) units properly stated). - 1 mark for reciprocal temperature \(1/T\) with unit \(\text{K}^{-1}\). - 1 mark for time \(t\ /\ \text{s}\) and \(\ln(1/t)\) (with no units). - **(d)** [2 marks total]: - 1 mark for stating that \(E_a = -\text{gradient} \times R\). - 1 mark for specifying the value of \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\) and stating that the calculated value must be divided by 1000 to get \(\text{kJ mol}^{-1}\). - **(e)** [2 marks total]: - 1 mark for \(1/T = 3.14 \times 10^{-3}\text{ K}^{-1}\) (allow \(3.14 \times 10^{-3}\) to \(3.15 \times 10^{-3}\)). - 1 mark for \(\ln(1/t) = -3.53\) (must be negative and to 3 sig figs/2 decimal places).
PastPaper.question 2 · Planning
15 PastPaper.marks
A student wants to determine the enthalpy change of hydration of anhydrous magnesium chloride, \(\text{MgCl}_2(s)\): \[ \text{MgCl}_2(s) + 6\text{H}_2\text{O}(l) \rightarrow \text{MgCl}_2\cdot 6\text{H}_2\text{O}(s) \quad \Delta H_{hyd} \]
To do this, the student plans to determine the enthalpy changes of solution of both anhydrous magnesium chloride and hydrated magnesium chloride.
**(a)** The student first plans to measure the enthalpy change of solution of anhydrous \(\text{MgCl}_2\). (i) Describe how the student would weigh the anhydrous \(\text{MgCl}_2\) to ensure the exact mass added to the water is known. [1 mark] (ii) Show by calculation that if \(50.0\text{ cm}^3\) of water is used, a mass of approximately \(1.50\text{ g}\) of anhydrous \(\text{MgCl}_2\) will produce a temperature rise of at least \(5.0\ ^\circ\text{C}\). [Assume density of water is \(1.00\text{ g cm}^{-3}\), specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\), \(M_r\) of \(\text{MgCl}_2 = 95.3\text{ g mol}^{-1}\), and approximate \(\Delta H_{sol} \approx -155\text{ kJ mol}^{-1}\)]. [3 marks] (iii) In the experiment, the mixture is stirred continuously, and the temperature is recorded at regular intervals. Explain why the maximum recorded temperature is not the theoretical maximum temperature change, and state how the student can find the correct maximum temperature change. [2 marks]
**(b)** Draw a Hess's Law cycle that allows the calculation of the enthalpy change of hydration, \(\Delta H_{hyd}\), of anhydrous magnesium chloride using the enthalpy changes of solution of anhydrous and hydrated magnesium chloride. Write an equation relating these three enthalpy changes. [2 marks]
**(c)** In the second experiment, the student dissolves \(10.15\text{ g}\) of hydrated magnesium chloride, \(\text{MgCl}_2\cdot 6\text{H}_2\text{O}\), in \(50.0\text{ cm}^3\) of water in a polystyrene cup. The temperature of the water decreased by \(1.80\ ^\circ\text{C}\). (i) Calculate the number of moles of \(\text{MgCl}_2\cdot 6\text{H}_2\text{O}\) used. [\(M_r\) of \(\text{MgCl}_2\cdot 6\text{H}_2\text{O} = 203.3\text{ g mol}^{-1}\)] [1 mark] (ii) Calculate the enthalpy change of solution of \(\text{MgCl}_2\cdot 6\text{H}_2\text{O}(s)\) in \(\text{kJ mol}^{-1}\). Include the correct sign. [3 marks]
**(d)** Using your answer to **(c)(ii)** and a value of \(\Delta H_{sol}(\text{MgCl}_2(s)) = -160.0\text{ kJ mol}^{-1}\), calculate the enthalpy change of hydration, \(\Delta H_{hyd}\), of anhydrous magnesium chloride. [2 marks]
**(e)** A second student used a weighing bottle containing anhydrous magnesium chloride. After transferring the solid, they noticed that the inside of the weighing bottle was wet. Predict and explain the effect of this on the calculated enthalpy change of solution of the anhydrous magnesium chloride. [1 mark]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
**(a)** (i) The student should weigh by difference: Weigh the weighing bottle containing the anhydrous solid, transfer the solid into the polystyrene cup, and then reweigh the empty weighing bottle. The difference between the two mass readings is the exact mass added. (ii) Moles of \(\text{MgCl}_2 = \frac{1.50}{95.3} = 0.01574\text{ mol}\). Heat released \(q = n \times |\Delta H_{sol}| = 0.01574\text{ mol} \times 155,000\text{ J mol}^{-1} = 2440\text{ J}\). Using \(q = m c \Delta T\) where \(m = 50.0\text{ g}\): \(\Delta T = \frac{2440}{50.0 \times 4.18} = 11.7\ ^\circ\text{C}\). Since \(11.7\ ^\circ\text{C} \ge 5.0\ ^\circ\text{C}\), a mass of \(1.50\text{ g}\) is sufficient. (iii) Heat is lost to the surroundings while the solid is dissolving, preventing the solution from reaching the theoretical maximum temperature. The correct maximum temperature can be found by plotting temperature against time, drawing a cooling curve, and extrapolating back to the time of mixing.
**(c)** (i) Moles of \(\text{MgCl}_2\cdot 6\text{H}_2\text{O} = \frac{10.15}{203.3} = 0.0499\text{ mol}\) (or \(0.0500\text{ mol}\)). (ii) Assuming the mass of solution is the mass of water (\(50.0\text{ g}\)): \(q = 50.0 \times 4.18 \times 1.80 = 376.2\text{ J} = 0.3762\text{ kJ}\). \(\Delta H_{sol}(\text{hydrated}) = +\frac{0.3762\text{ kJ}}{0.0499\text{ mol}} = +7.54\text{ kJ mol}^{-1}\). (Note: If total mass of solution, \(50.0 + 10.15 = 60.15\text{ g}\), is used: \(q = 60.15 \times 4.18 \times 1.80 = 452.8\text{ J} = 0.4528\text{ kJ}\). \(\Delta H_{sol}(\text{hydrated}) = +\frac{0.4528}{0.0499} = +9.07\text{ kJ mol}^{-1}\). Both methods are acceptable).
**(d)** Using \(\Delta H_{sol}(\text{hydrated}) = +7.54\text{ kJ mol}^{-1}\): \(\Delta H_{hyd} = -160.0 - (+7.54) = -167.5\text{ kJ mol}^{-1}\). (Or if using \(+9.07\text{ kJ mol}^{-1}\), \(\Delta H_{hyd} = -169.1\text{ kJ mol}^{-1}\)).
**(e)** If the bottle is wet, some anhydrous \(\text{MgCl}_2\) reacts with water inside the weighing bottle before being transferred. This releases heat inside the bottle rather than in the cup. Consequently, less anhydrous salt enters the cup, resulting in a lower temperature rise and a calculated enthalpy change of solution that is less exothermic (more positive / less negative).
PastPaper.markingScheme
- **(a)** [6 marks total]: - **(a)(i)** [1 mark]: For weighing by difference. - **(a)(ii)** [3 marks]: - 1 mark for calculating the correct number of moles (\(0.0157\text{ mol}\)). - 1 mark for calculating the heat change (\(2440\text{ J}\)). - 1 mark for calculating \(\Delta T = 11.7\ ^\circ\text{C}\) and concluding that it is \(\ge 5.0\ ^\circ\text{C}\). - **(a)(iii)** [2 marks]: - 1 mark for explaining that heat loss to the surroundings occurs during dissolution. - 1 mark for describing the extrapolation of the cooling curve back to the time of mixing. - **(b)** [2 marks total]: - 1 mark for a correct Hess's Law Cycle diagram. - 1 mark for the correct equation: \(\Delta H_{hyd} = \Delta H_{sol}(\text{anhydrous}) - \Delta H_{sol}(\text{hydrated})\). - **(c)** [4 marks total]: - **(c)(i)** [1 mark]: For calculating \(0.0499\text{ mol}\) (or \(0.0500\text{ mol}\)). - **(c)(ii)** [3 marks]: - 1 mark for calculating heat change \(q = 376.2\text{ J}\) (using water mass) or \(452.8\text{ J}\) (using solution mass). - 1 mark for dividing heat by moles to obtain the correct magnitude (e.g., \(7.54\) or \(9.07\)). - 1 mark for the correct units (\(\text{kJ mol}^{-1}\)) and positive sign (\(+\)). - **(d)** [2 marks total]: - 1 mark for substituting values correctly into the Hess's Law equation. - 1 mark for the correct final answer (e.g., \(-167.5\text{ kJ mol}^{-1}\) or \(-169.1\text{ kJ mol}^{-1}\)) with correct sign and units. - **(e)** [1 mark total]: - 1 mark for predicting that the calculated value would be less exothermic (less negative) because some of the salt reacted/hydrated prematurely in the bottle.