An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 12
Forty multiple-choice questions. For each question, there are four possible answers. Answer all questions.
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PastPaper.question 1 · Multiple Choice
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A \( 10\text{ cm}^3 \) sample of a gaseous hydrocarbon was exploded with \( 80\text{ cm}^3 \) (an excess) of oxygen. After cooling to room temperature, the remaining gas volume was \( 65\text{ cm}^3 \). This gas mixture was then shaken with aqueous sodium hydroxide, and the volume decreased to \( 35\text{ cm}^3 \). All gas volumes are measured at room temperature and pressure. What is the molecular formula of the hydrocarbon?
A.\( \text{C}_3\text{H}_4 \)
B.\( \text{C}_3\text{H}_6 \)
C.\( \text{C}_3\text{H}_8 \)
D.\( \text{C}_4\text{H}_{10} \)
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PastPaper.workedSolution
Let the hydrocarbon be \( \text{C}_x\text{H}_y \). Shaking with \( \text{NaOH} \) absorbs the \( \text{CO}_2 \) gas produced. The decrease in volume is \( 65\text{ cm}^3 - 35\text{ cm}^3 = 30\text{ cm}^3 \). Thus, \( 10\text{ cm}^3 \) of hydrocarbon produces \( 30\text{ cm}^3 \) of \( \text{CO}_2 \), giving \( x = 3 \). The remaining \( 35\text{ cm}^3 \) gas is unreacted \( \text{O}_2 \). The volume of \( \text{O}_2 \) reacted is \( 80\text{ cm}^3 - 35\text{ cm}^3 = 45\text{ cm}^3 \). From the general combustion equation, \( 10(x + y/4) = 45 \). Substituting \( x = 3 \) gives \( 3 + y/4 = 4.5 \), which simplifies to \( y = 6 \). Therefore, the molecular formula is \( \text{C}_3\text{H}_6 \).
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PastPaper.question 2 · Multiple Choice
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A mixture of anhydrous calcium carbonate (\( \text{CaCO}_3 \), \( M_r = 100.1 \)) and anhydrous magnesium carbonate (\( \text{MgCO}_3 \), \( M_r = 84.3 \)) has a mass of \( 10.00\text{ g} \). The mixture is heated strongly until there is no further change in mass. The remaining solid residue has a mass of \( 5.20\text{ g} \). What is the percentage by mass of \( \text{MgCO}_3 \) in the original mixture?
A.24.0%
B.45.4%
C.49.1%
D.50.9%
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PastPaper.workedSolution
Let the mass of \( \text{MgCO}_3 \) be \( m\text{ g} \), so the mass of \( \text{CaCO}_3 \) is \( (10.00 - m)\text{ g} \). The mass loss is due to the release of \( \text{CO}_2 \) gas: \( 10.00\text{ g} - 5.20\text{ g} = 4.80\text{ g} \). Total moles of \( \text{CO}_2 \) evolved is \( 4.80 / 44.0 = 0.1091\text{ mol} \). Since both carbonates decompose with a 1:1 molar ratio to \( \text{CO}_2 \), we have: \( m/84.3 + (10.00 - m)/100.1 = 0.1091 \). Solving for \( m \): \( m(0.01186 - 0.00999) + 0.0999 = 0.1091 \) which gives \( 0.00187 m = 0.0092 \), so \( m = 4.91\text{ g} \). The percentage by mass of \( \text{MgCO}_3 \) is \( (4.91 / 10.00) \times 100\% = 49.1\% \).
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PastPaper.question 3 · Multiple Choice
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Propylbenzene, \( \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_3 \), is reacted with chlorine in the presence of ultraviolet light at boiling point. Which chlorinated organic compound is formed in the greatest yield as the major mono-chlorinated product?
A.(1-chloropropyl)benzene
B.(2-chloropropyl)benzene
C.(3-chloropropyl)benzene
D.1-chloro-4-propylbenzene
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PastPaper.workedSolution
Under ultraviolet light and at boiling point, free-radical substitution occurs on the alkyl side-chain of propylbenzene. The most stable free radical intermediate is the benzylic radical, \( \text{C}_6\text{H}_5\dot{\text{C}}\text{HCH}_2\text{CH}_3 \), where the unpaired electron is highly delocalised into the aromatic \( \pi \) ring system. Thus, substitution occurs preferentially at the C1 position of the side chain, forming (1-chloropropyl)benzene as the major product.
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PastPaper.question 4 · Multiple Choice
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In the electrophilic substitution of benzene to form nitrobenzene, a reaction intermediate is formed. Which statement about this intermediate is correct?
A.The intermediate has a fully delocalised ring of six \(\pi\) electrons.
B.The intermediate contains a carbon atom with \(\text{sp}^3\) hybridisation.
C.The intermediate is a free radical.
D.The overall charge of the intermediate is negative.
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PastPaper.workedSolution
In the electrophilic substitution mechanism, the electrophile \( \text{NO}_2^+ \) attacks a carbon atom in the benzene ring to form a covalent bond. This carbon atom becomes bonded to four single bonds (to hydrogen, to the nitro group, and to two adjacent ring carbon atoms), changing its hybridisation from \( \text{sp}^2 \) to \( \text{sp}^3 \). The ring loses its fully delocalised six-electron system, instead holding a positive charge over the remaining five carbon atoms.
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PastPaper.question 5 · Multiple Choice
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The standard electrode potentials for several half-reactions are shown below. \( \text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\theta = +0.77\text{ V} \) \( \text{I}_2(\text{aq}) + 2\text{e}^- \rightleftharpoons 2\text{I}^-(\text{aq}) \quad E^\theta = +0.54\text{ V} \) \( \text{Fe}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Fe}(\text{s}) \quad E^\theta = -0.44\text{ V} \) \( \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \quad E^\theta = +0.34\text{ V} \) Which of the following reactions is feasible under standard conditions?
A reaction is feasible under standard conditions if its cell potential \( E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} \) is positive. For option C: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \), we have reduction: \( \text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+} \) (\( E^\theta = +0.77\text{ V} \)) and oxidation: \( 2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^- \) (\( E^\theta = +0.54\text{ V} \)). Therefore, \( E^\theta_{\text{cell}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V} \), which is positive and feasible. The other options yield negative cell potentials.
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PastPaper.question 6 · Multiple Choice
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A standard cell is set up using the two half-cells: \( \text{Ag}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s}) \quad E^\theta = +0.80\text{ V} \) \( \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \quad E^\theta = +0.34\text{ V} \) The overall cell equation is: \( \text{Cu}(\text{s}) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Ag}(\text{s}) \quad E^\theta_{\text{cell}} = +0.46\text{ V} \) The concentration of \( \text{Ag}^+(\text{aq}) \) is decreased from \( 1.0\text{ mol dm}^{-3} \) to \( 0.001\text{ mol dm}^{-3} \) while keeping the concentration of \( \text{Cu}^{2+}(\text{aq}) \) at \( 1.0\text{ mol dm}^{-3} \) and temperature at \( 298\text{ K} \). Which statement correctly describes the effect of this change on the electrode potential of the silver half-cell, \( E_{\text{Ag}^+/\text{Ag}} \), and the cell potential, \( E_{\text{cell}} \)?
A.Both \( E_{\text{Ag}^+/\text{Ag}} \) and \( E_{\text{cell}} \) increase.
B.Both \( E_{\text{Ag}^+/\text{Ag}} \) and \( E_{\text{cell}} \) decrease.
C.\( E_{\text{Ag}^+/\text{Ag}} \) decreases, but \( E_{\text{cell}} \) increases.
D.\( E_{\text{Ag}^+/\text{Ag}} \) increases, but \( E_{\text{cell}} \) decreases.
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Applying the Nernst equation for the silver electrode: \( E_{\text{Ag}^+/\text{Ag}} = E^\theta_{\text{Ag}^+/\text{Ag}} + 0.059 \log [\text{Ag}^+] \). Decreasing \( [\text{Ag}^+] \) to \( 0.001\text{ mol dm}^{-3} \) decreases the electrode potential (becomes less positive). For the overall cell reaction, decreasing the concentration of the reactant \( \text{Ag}^+ \) shifts the equilibrium position to the left, which decreases the cell potential \( E_{\text{cell}} \). Therefore, both values decrease.
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PastPaper.question 7 · Multiple Choice
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The initial rate of the reaction \( 2\text{A} + \text{B} + 2\text{C} \rightarrow \text{products} \) was measured at constant temperature: Experiment 1: \( [\text{A}] = 0.10 \), \( [\text{B}] = 0.10 \), \( [\text{C}] = 0.10 \); Rate = \( 1.2 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1} \). Experiment 2: \( [\text{A}] = 0.20 \), \( [\text{B}] = 0.10 \), \( [\text{C}] = 0.10 \); Rate = \( 2.4 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1} \). Experiment 3: \( [\text{A}] = 0.10 \), \( [\text{B}] = 0.20 \), \( [\text{C}] = 0.10 \); Rate = \( 4.8 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1} \). Experiment 4: \( [\text{A}] = 0.10 \), \( [\text{B}] = 0.10 \), \( [\text{C}] = 0.20 \); Rate = \( 1.2 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1} \). What are the value and units of the rate constant, \( k \), for this reaction?
1. Determine order of reaction: - Comparing Exp 1 & 2: doubling \( [\text{A}] \) doubles the rate, so order with respect to \( \text{A} \) is 1. - Comparing Exp 1 & 3: doubling \( [\text{B}] \) quadruples the rate, so order with respect to \( \text{B} \) is 2. - Comparing Exp 1 & 4: doubling \( [\text{C}] \) does not change the rate, so order with respect to \( \text{C} \) is 0. 2. Rate Equation: \( \text{Rate} = k[\text{A}][\text{B}]^2 \). 3. Calculate \( k \) using Exp 1: \( 1.2 \times 10^{-3} = k (0.10)(0.10)^2 \implies k = 1.2 \). 4. Calculate units: \( \text{units of } k = \text{mol dm}^{-3}\text{ s}^{-1} / ((\text{mol dm}^{-3})(\text{mol dm}^{-3})^2) = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1} \).
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PastPaper.question 8 · Multiple Choice
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The reaction between nitrogen monoxide and hydrogen has the overall equation: \( 2\text{NO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(g)} \) and experimentally determined rate equation: \( \text{Rate} = k[\text{NO}]^2[\text{H}_2] \). A proposed mechanism is: Step 1: \( 2\text{NO(g)} \rightleftharpoons \text{N}_2\text{O}_2\text{(g)} \) (fast equilibrium); Step 2: \( \text{N}_2\text{O}_2\text{(g)} + \text{H}_2\text{(g)} \rightarrow \text{N}_2\text{O(g)} + \text{H}_2\text{O(g)} \) (slow); Step 3: \( \text{N}_2\text{O(g)} + \text{H}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + \text{H}_2\text{O(g)} \) (fast). Which statement about this mechanism is correct?
A.Step 1 is the rate-determining step.
B.The species \( \text{N}_2\text{O}_2 \) acts as a catalyst in this reaction.
C.The proposed mechanism is consistent with the rate equation because Step 2 is the rate-determining step.
D.Increasing the concentration of \( \text{H}_2 \) has no effect on the rate of Step 2.
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PastPaper.workedSolution
Step 2 is the slowest step, so it is the rate-determining step, giving: \( \text{Rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2] \). Since \( \text{N}_2\text{O}_2 \) is a fast-equilibrium intermediate, from Step 1 we have \( [\text{N}_2\text{O}_2] = K_c [\text{NO}]^2 \). Substituting this into the rate-determining step rate expression yields \( \text{Rate} = k_2 K_c [\text{NO}]^2 [\text{H}_2] \), which is equivalent to the experimental rate equation. Therefore, the proposed mechanism is consistent with the rate equation because Step 2 is the rate-determining step.
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PastPaper.question 9 · Multiple Choice
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A 10.0 g sample of a mixture containing only calcium carbonate, \(CaCO_3\), and calcium oxide, \(CaO\), is heated strongly until constant mass is achieved. The mass of the solid residue remaining after heating is 7.8 g. What is the percentage by mass of \(CaCO_3\) in the original mixture? [Ar values: \(Ca = 40.1\), \(C = 12.0\), \(O = 16.0\)]
A.22.0%
B.50.1%
C.78.0%
D.89.3%
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PastPaper.workedSolution
The thermal decomposition of calcium carbonate is: \(CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\). The mass loss of \(2.2\text{ g}\) (from \(10.0\text{ g} - 7.8\text{ g}\)) is entirely due to the loss of carbon dioxide gas. The number of moles of \(CO_2\) lost is \(2.2\text{ g} / 44.0\text{ g mol}^{-1} = 0.050\text{ mol}\). Because the mole ratio of \(CaCO_3\) to \(CO_2\) is 1:1, there was originally \(0.050\text{ mol}\) of \(CaCO_3\). The molar mass of \(CaCO_3\) is \(100.1\text{ g mol}^{-1}\), so the mass of \(CaCO_3\) in the mixture is \(0.050\text{ mol} \times 100.1\text{ g mol}^{-1} = 5.005\text{ g}\). The percentage by mass in the \(10.0\text{ g}\) mixture is \((5.005\text{ g} / 10.0\text{ g}) \times 100\% = 50.1\%\).
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Award 1 mark for calculating the mass loss, converting to moles of \(CO_2\), determining the stoichiometric amount of \(CaCO_3\), and obtaining the final percentage of 50.1%.
PastPaper.question 10 · Multiple Choice
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A piece of magnesium ribbon of mass \(0.120\text{ g}\) is added to \(50.0\text{ cm}^3\) of \(0.0500\text{ mol dm}^{-3}\) hydrochloric acid, \(HCl(aq)\). What is the maximum volume of hydrogen gas, in \(\text{cm}^3\), collected at room temperature and pressure (r.t.p.)? [Molar volume of gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\), \(A_r(Mg) = 24.3\)]
A.30.0 \(\text{cm}^3\)
B.60.0 \(\text{cm}^3\)
C.119 \(\text{cm}^3\)
D.237 \(\text{cm}^3\)
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PastPaper.workedSolution
The balanced chemical equation is: \(Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)\). First, calculate the moles of each reactant: \(n(Mg) = 0.120\text{ g} / 24.3\text{ g mol}^{-1} = 0.00494\text{ mol}\) and \(n(HCl) = 0.0500\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.00250\text{ mol}\). Since 1 mole of \(Mg\) reacts with 2 moles of \(HCl\), the stoichiometric requirement of \(HCl\) for complete reaction of the \(Mg\) is \(2 \times 0.00494 = 0.00988\text{ mol}\). Since we only have \(0.00250\text{ mol}\) of \(HCl\), \(HCl\) is the limiting reactant. The moles of \(H_2\) produced is \(0.00250 / 2 = 0.00125\text{ mol}\). The volume of \(H_2\) at r.t.p. is \(0.00125\text{ mol} \times 24000\text{ cm}^3\text{ mol}^{-1} = 30.0\text{ cm}^3\).
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Award 1 mark for identifying \(HCl\) as the limiting reactant and calculating the maximum volume of hydrogen gas as 30.0 \(\text{cm}^3\).
PastPaper.question 11 · Multiple Choice
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Methylbenzene is treated with \(Cl_2\) in the presence of anhydrous \(AlCl_3\) to produce a major monosubstituted product X. Separately, methylbenzene is heated with alkaline \(KMnO_4\) and then acidified to yield compound Y. When Y is treated with \(Cl_2\) in the presence of anhydrous \(AlCl_3\), the major monosubstituted product Z is obtained. What are the positions of the chlorine atom relative to the pre-existing substituent in X and Z?
A.X: 2- or 4-position; Z: 3-position
B.X: 3-position; Z: 2- or 4-position
C.X: 2- or 4-position; Z: 2- or 4-position
D.X: 3-position; Z: 3-position
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PastPaper.workedSolution
For product X, methylbenzene undergoes electrophilic aromatic substitution. The methyl group is electron-releasing and directs the incoming chlorine electrophile to the ortho (2-) and para (4-) positions, making X a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene. Heating methylbenzene with alkaline \(KMnO_4\) followed by acidification oxidizes the methyl side chain to a carboxylic acid group, producing benzoic acid (Y). The \(-COOH\) group in benzoic acid is electron-withdrawing and deactivating, which directs the incoming electrophile to the meta (3-) position. Thus, chlorination of benzoic acid yields 3-chlorobenzoic acid (Z).
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Award 1 mark for correctly identifying that the methyl group directs to 2- or 4-positions (for X) and the carboxylic acid group directs to the 3-position (for Z).
PastPaper.question 12 · Multiple Choice
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Under standard nitration conditions, methylbenzene reacts significantly faster than benzene. Which statement correctly explains this observation?
A.The methyl group is electron-withdrawing by inductive effect, which destabilizes the intermediate carbocation.
B.The methyl group is electron-releasing by inductive effect, which increases the electron density on the benzene ring.
C.The methyl group acts as a heterogeneous catalyst, lowering the activation energy for the nitration reaction.
D.The methyl group prevents electrophilic attack at the meta-positions, forcing the reaction to go faster at ortho- and para-positions.
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PastPaper.workedSolution
The methyl group attached to the benzene ring is electron-releasing by inductive effect. This pushes electron density into the delocalized pi-system of the benzene ring, making the ring more attractive to the electrophile (the nitronium ion, \(NO_2^+\)) and stabilizing the transition state/carbocation intermediate, thereby increasing the reaction rate compared to benzene.
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Award 1 mark for identifying the electron-releasing inductive effect of the methyl group as the correct cause of increased ring reactivity.
PastPaper.question 13 · Multiple Choice
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Use the standard electrode potentials provided below to determine which of the reactions will NOT occur spontaneously under standard conditions: \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq)\) \(E^\ominus = +0.77\text{ V}\); \(I_2(aq) + 2e^- \rightleftharpoons 2I^-(aq)\) \(E^\ominus = +0.54\text{ V}\); \(Br_2(aq) + 2e^- \rightleftharpoons 2Br^-(aq)\) \(E^\ominus = +1.07\text{ V}\); \(SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightleftharpoons SO_2(g) + 2H_2O(l)\) \(E^\ominus = +0.17\text{ V}\).
A redox reaction is spontaneous (feasible) under standard conditions if the standard cell potential, \(E^\ominus_{\text{cell}}\), is positive. \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\). For reaction C, the reduction is \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) (\(E^\ominus = +0.77\text{ V}\)) and the oxidation is \(2Br^- \rightarrow Br_2 + 2e^-\). Therefore, \(E^\ominus_{\text{cell}} = +0.77\text{ V} - (+1.07\text{ V}) = -0.30\text{ V}\). Since this cell potential is negative, the reaction is non-spontaneous under standard conditions. For comparison, reaction A cell potential is \(+0.77 - (+0.54) = +0.23\text{ V}\), reaction B is \(+0.54 - (+0.17) = +0.37\text{ V}\), and reaction D is \(+1.07 - (+0.54) = +0.53\text{ V}\), all of which are positive and therefore spontaneous.
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Award 1 mark for the correct application of the cell potential formula to deduce that reaction C is non-spontaneous due to a negative cell potential.
PastPaper.question 14 · Multiple Choice
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An electrochemical cell is set up under standard conditions using the half-cells: \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\) \(E^\ominus = +0.80\text{ V}\) and \(Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)\) \(E^\ominus = +0.34\text{ V}\). If the concentration of \(Ag^+(aq)\) in the silver half-cell is decreased to \(0.01\text{ mol dm}^{-3}\) while all other conditions are kept standard, how do the electrode potential of the silver half-cell and the overall cell potential change?
According to Le Chatelier's principle, for the reduction half-reaction \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\), reducing the concentration of the reactant \(Ag^+(aq)\) shifts the position of equilibrium to the left, which decreases the tendency for reduction to occur. This decreases the electrode potential of the silver half-cell (making it less positive). Since the silver electrode acts as the cathode (reduction electrode) and the copper electrode acts as the anode, the overall cell potential is \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = E(Ag^+/Ag) - E(Cu^{2+}/Cu)\). A decrease in \(E(Ag^+/Ag)\) directly reduces the overall cell potential.
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Award 1 mark for correctly determining that both the silver half-cell potential and the overall cell potential decrease when the silver ion concentration is lowered.
PastPaper.question 15 · Multiple Choice
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For the reaction \(P + Q + R \rightarrow\text{products}\), initial rate experiments at constant temperature yielded the following data: Run 1: \([P] = 0.10\), \([Q] = 0.10\), \([R] = 0.10\text{ mol dm}^{-3}\), Rate = \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\); Run 2: \([P] = 0.20\), \([Q] = 0.10\), \([R] = 0.10\text{ mol dm}^{-3}\), Rate = \(2.4 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\); Run 3: \([P] = 0.10\), \([Q] = 0.20\), \([R] = 0.10\text{ mol dm}^{-3}\), Rate = \(4.8 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\); Run 4: \([P] = 0.10\), \([Q] = 0.10\), \([R] = 0.20\text{ mol dm}^{-3}\), Rate = \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the overall order of the reaction and the unit of the rate constant, \(k\)?
A.Overall order = 3; unit of \(k\) = \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
B.Overall order = 3; unit of \(k\) = \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
C.Overall order = 2; unit of \(k\) = \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
D.Overall order = 4; unit of \(k\) = \(\text{dm}^9\text{ mol}^{-3}\text{ s}^{-1}\)
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PastPaper.workedSolution
Comparing Run 1 and Run 2: doubling the concentration of P (with others kept constant) doubles the rate from \(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\), showing first-order dependency on P. Comparing Run 1 and Run 3: doubling the concentration of Q (with others kept constant) quadruples the rate from \(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\), showing second-order dependency on Q. Comparing Run 1 and Run 4: doubling the concentration of R (with others kept constant) does not change the rate, showing zero-order dependency on R. The rate equation is \(Rate = k[P]^1[Q]^2\). The overall order of reaction is \(1 + 2 + 0 = 3\). The unit of \(k\) is given by: \(k = Rate / ([P][Q]^2) = (\text{mol dm}^{-3}\text{ s}^{-1}) / (\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2) = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
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Award 1 mark for finding the correct reaction order for each reactant, determining the overall order as 3, and calculating the correct unit of the rate constant.
PastPaper.question 16 · Multiple Choice
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A first-order reaction has a rate constant \(k = 4.50 \times 10^{-4}\text{ s}^{-1}\). If the initial concentration of the reactant is \(0.800\text{ mol dm}^{-3}\), how long will it take for the concentration of the reactant to fall to \(0.100\text{ mol dm}^{-3}\)?
A.1.54 \(\times 10^3\text{ s}\)
B.3.08 \(\times 10^3\text{ s}\)
C.4.62 \(\times 10^3\text{ s}\)
D.6.16 \(\times 10^3\text{ s}\)
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PastPaper.workedSolution
For a first-order reaction, the half-life is given by \(t_{1/2} = \ln(2) / k = 0.693 / (4.50 \times 10^{-4}\text{ s}^{-1}) = 1540\text{ s}\). To reduce the concentration from \(0.800\text{ mol dm}^{-3}\) to \(0.100\text{ mol dm}^{-3}\), the reaction must proceed through exactly three half-lives: \(0.800 \rightarrow 0.400 \rightarrow 0.200 \rightarrow 0.100\text{ mol dm}^{-3}\). The total time is \(3 \times t_{1/2} = 3 \times 1540\text{ s} = 4.62 \times 10^3\text{ s}\). Alternatively, using the integrated first-order rate law: \(\ln([A]_0 / [A]_t) = k t \Rightarrow \ln(0.800 / 0.100) = (4.50 \times 10^{-4}) t \Rightarrow \ln(8) = 2.079 = (4.50 \times 10^{-4}) t \Rightarrow t = 4620\text{ s} = 4.62 \times 10^3\text{ s}\).
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Award 1 mark for the correct calculation of time using either the half-life relationship or the integrated rate law.
PastPaper.question 17 · Multiple Choice
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A sample of \( 0.250\text{ g} \) of a carbonate of a Group 2 metal, \( \text{MCO}_3 \), reacted completely with excess dilute hydrochloric acid. The carbon dioxide gas evolved was collected and measured at room temperature and pressure (r.t.p.) to be \( 60.0\text{ cm}^3 \). What is the identity of the Group 2 metal, \( \text{M} \)? [Take the molar volume of a gas at r.t.p. as \( 24.0\text{ dm}^3\text{ mol}^{-1} \).]
A.Magnesium
B.Calcium
C.Strontium
D.Barium
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First, calculate the number of moles of \( \text{CO}_2 \) evolved: \( n(\text{CO}_2) = \frac{60.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.00250\text{ mol} \)
From the chemical equation: \( \text{MCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \) The mole ratio of \( \text{MCO}_3 \) to \( \text{CO}_2 \) is \( 1 : 1 \). Therefore: \( n(\text{MCO}_3) = 0.00250\text{ mol} \)
Now, calculate the molar mass of \( \text{MCO}_3 \): \( M_r(\text{MCO}_3) = \frac{0.250\text{ g}}{0.00250\text{ mol}} = 100.0\text{ g mol}^{-1} \)
Calculate the relative atomic mass of \( \text{M} \): \( A_r(\text{M}) + 12.0 + 3(16.0) = 100.0 \) \( A_r(\text{M}) + 60.0 = 100.0 \) \( A_r(\text{M}) = 40.0 \)
This relative atomic mass corresponds to calcium (\( A_r = 40.1 \)).
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1 mark for the correct option B. - Award 1 mark for correct determination of moles of gas and calculating molar mass of the Group 2 carbonate to identify Ca. - Reject options A (gives 71.2 cm3), C (gives 40.7 cm3), and D (gives 30.4 cm3).
PastPaper.question 18 · Multiple Choice
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When 10 cm\(^3\) of a gaseous hydrocarbon is completely combusted in 100 cm\(^3\) of oxygen (which is in excess), the total volume of gas remaining after cooling to room temperature and pressure is 85 cm\(^3\). On passing this gaseous mixture through concentrated aqueous sodium hydroxide, the volume decreases to 55 cm\(^3\). What is the molecular formula of the hydrocarbon? [Assume all gas volumes are measured at the same temperature and pressure.]
A.\( \text{C}_3\text{H}_4 \)
B.\( \text{C}_3\text{H}_6 \)
C.\( \text{C}_3\text{H}_8 \)
D.\( \text{C}_4\text{H}_8 \)
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The general combustion equation for a hydrocarbon is: \( \text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \rightarrow x\text{CO}_2 + y/2\text{H}_2\text{O} \) After cooling, the water is liquid and occupies negligible volume. The remaining 85 cm\(^3\) of gas consists of excess \( \text{O}_2 \) and produced \( \text{CO}_2 \). Passing the gas through aqueous NaOH removes all \( \text{CO}_2 \). The decrease in volume is the volume of \( \text{CO}_2 \): \( V(\text{CO}_2) = 85 - 55 = 30\text{ cm}^3 \) Since 10 cm\(^3\) of \( \text{C}_x\text{H}_y \) produces 30 cm\(^3\) of \( \text{CO}_2 \), the number of carbon atoms per molecule is: \( x = 30 / 10 = 3 \) The remaining 55 cm\(^3\) of gas is unreacted \( \text{O}_2 \). Thus, the volume of reacted \( \text{O}_2 \) is: \( 100 - 55 = 45\text{ cm}^3 \) The mole ratio of hydrocarbon to reacted \( \text{O}_2 \) is \( 10 : 45 = 1 : 4.5 \). Therefore: \( x + y/4 = 4.5 \) Substituting \( x = 3 \) into the equation: \( 3 + y/4 = 4.5 \Rightarrow y/4 = 1.5 \Rightarrow y = 6 \) The molecular formula is therefore \( \text{C}_3\text{H}_6 \).
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1 mark for the correct option B. - Award 1 mark for calculating the volume of CO2 formed (30 cm3) and volume of O2 reacted (45 cm3) to deduce x = 3 and y = 6.
PastPaper.question 19 · Multiple Choice
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Methylbenzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid at \( 55\text{ }^\circ\text{C} \) to form a mixture of monosubstituted nitro-isomers. Which statement correctly explains why the reaction of methylbenzene is faster than that of benzene under the same conditions?
A.The methyl group is electron-withdrawing by inductive effect, which destabilises the carbocation intermediate.
B.The methyl group is electron-donating by inductive effect, increasing the electron density of the benzene ring.
C.The methyl group increases the activation energy for the electrophilic substitution reaction.
D.The sulfuric acid acts as a stronger nucleophile when reacting with methylbenzene than with benzene.
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The methyl group is an electron-donating group due to the inductive effect. This donation increases the electron density on the benzene ring compared to benzene itself. Consequently, the aromatic ring of methylbenzene is more nucleophilic and is attacked more rapidly by the electrophilic nitronium ion (\( \text{NO}_2^+ \)). It also stabilizes the positively charged carbocation intermediate, thereby lowering the activation energy of the rate-determining step.
PastPaper.markingScheme
1 mark for the correct option B. - Reject A: The methyl group is electron-donating, not electron-withdrawing. - Reject C: The methyl group decreases the activation energy, making the reaction faster. - Reject D: Sulfuric acid acts as a catalyst/Brønsted-Lowry acid to produce the electrophile, not as a nucleophile.
PastPaper.question 20 · Multiple Choice
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A student carries out the halogenation of methylbenzene using chlorine gas. Depending on the conditions used, different products are formed. Under which set of conditions is (chloromethyl)benzene, \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \), the major organic product?
A.\( \text{Cl}_2(\text{g}) \) in the presence of an \( \text{AlCl}_3 \) catalyst at room temperature in the dark
B.\( \text{Cl}_2(\text{g}) \) in the presence of an \( \text{Fe} \) catalyst with heating
C.\( \text{Cl}_2(\text{g}) \) passed into boiling methylbenzene in the presence of ultraviolet light
D.Concentrated hydrochloric acid at room temperature
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Substitution on the alkyl side chain of methylbenzene (to form (chloromethyl)benzene) occurs via a free-radical mechanism. This pathway requires chlorine gas and ultraviolet (UV) light and/or high temperature (boiling methylbenzene). Electrophilic substitution on the aromatic ring occurs instead if a halogen carrier catalyst (such as AlCl\(_3\) or Fe) is present at room temperature, which would produce 2-chlorotoluene and 4-chlorotoluene.
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1 mark for the correct option C. - Reject A and B: These conditions promote ring substitution via electrophilic aromatic substitution. - Reject D: Methylbenzene does not undergo substitution with hydrochloric acid.
PastPaper.question 21 · Multiple Choice
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The standard electrode potentials for three redox systems are shown below. \( \text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\ominus = +0.77\text{ V} \) \( \text{I}_2(\text{aq}) + 2\text{e}^- \rightleftharpoons 2\text{I}^-(\text{aq}) \quad E^\ominus = +0.54\text{ V} \) \( \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{e}^- \rightleftharpoons 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \quad E^\ominus = +0.09\text{ V} \)
Which species is the strongest reducing agent under standard conditions?
A.\( \text{Fe}^{3+}(\text{aq}) \)
B.\( \text{Fe}^{2+}(\text{aq}) \)
C.\( \text{I}^-(\text{aq}) \)
D.\( \text{S}_2\text{O}_3^{2-}(\text{aq}) \)
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A reducing agent is a species that undergoes oxidation (loses electrons) in a chemical reaction. The standard oxidation potential is the negative of the standard reduction potential. The species that has the most negative standard reduction potential (least positive) will be the easiest to oxidize, representing the strongest reducing agent.
The thiosulfate system has the least positive standard reduction potential. Its oxidation half-reaction is: \( 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{e}^- \quad E^\ominus = -0.09\text{ V} \) This is the most thermodynamically favorable oxidation half-reaction under standard conditions. Thus, \( \text{S}_2\text{O}_3^{2-}(\text{aq}) \) is the strongest reducing agent.
PastPaper.markingScheme
1 mark for the correct option D. - Reject A: Fe3+(aq) is an oxidizing agent. - Reject B and C: Both Fe2+(aq) and I-(aq) are weaker reducing agents than thiosulfate because their corresponding systems have more positive reduction potentials (+0.77 V and +0.54 V respectively).
PastPaper.question 22 · Multiple Choice
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An electrochemical cell is set up under standard conditions using a standard hydrogen electrode (SHE) as one half-cell and a standard copper half-cell, \( \text{Cu}^{2+}(\text{aq}) | \text{Cu}(\text{s}) \), as the other. \( \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \quad E^\ominus = +0.34\text{ V} \)
Which statement about this cell is correct when it is operating to deliver current?
A.Electrons flow through the external circuit from the copper electrode to the platinum electrode.
B.The concentration of \( \text{H}^+(\text{aq}) \) ions in the standard hydrogen electrode increases.
C.The mass of the copper electrode decreases.
D.The copper half-cell acts as the anode where oxidation takes place.
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The standard reduction potential for the copper half-cell is \( E^\ominus = +0.34\text{ V} \), which is more positive than the standard hydrogen electrode (\( E^\ominus = 0.00\text{ V} \)). When the cell is operating: - Reduction occurs at the more positive electrode (copper electrode), making it the cathode: \( \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s}) \) - Oxidation occurs at the less positive electrode (standard hydrogen electrode), making it the anode: \( \text{H}_2(\text{g}) \rightarrow 2\text{H}^+(\text{aq}) + 2\text{e}^- \)
Therefore: - Electrons flow from the SHE (anode, Pt electrode) to the copper electrode (cathode) through the external circuit. Thus, option A is incorrect. - \( \text{H}_2 \) is oxidized to \( \text{H}^+ \), so the concentration of \( \text{H}^+(\text{aq}) \) ions in the SHE increases. Option B is correct. - Copper ions are deposited onto the copper electrode, so its mass increases. Option C is incorrect. - The copper half-cell acts as the cathode. Option D is incorrect.
PastPaper.markingScheme
1 mark for the correct option B. - Reject A: Electrons flow from anode (Pt) to cathode (Cu). - Reject C: The mass of the copper electrode increases due to copper deposition. - Reject D: Reduction occurs at the copper electrode, meaning it acts as the cathode.
PastPaper.question 23 · Multiple Choice
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For the reaction \( 2\text{A} + \text{B} \rightarrow \text{C} + \text{D} \), the rate equation is determined experimentally to be: \[ \text{rate} = k[\text{A}][\text{B}]^2 \]
When the initial concentration of \( \text{A} \) is doubled and the initial concentration of \( \text{B} \) is halved, how does the initial rate of reaction change?
A.It decreases by a factor of 2.
B.It increases by a factor of 2.
C.It remains unchanged.
D.It decreases by a factor of 4.
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Let the initial rate of the reaction be: \( \text{rate}_1 = k[\text{A}][\text{B}]^2 \)
When the concentration of \( \text{A} \) is doubled (\( 2[\text{A}] \)) and the concentration of \( \text{B} \) is halved (\( 0.5[\text{B}] \)), the new rate of reaction \( \text{rate}_2 \) is: \( \text{rate}_2 = k(2[\text{A}])(0.5[\text{B}])^2 \) \( \text{rate}_2 = k(2[\text{A}])(0.25[\text{B}]^2) \) \( \text{rate}_2 = 0.5 k[\text{A}][\text{B}]^2 = 0.5 \text{ rate}_1 \)
Therefore, the new initial rate decreases by a factor of 2 (it is halved).
PastPaper.markingScheme
1 mark for the correct option A. - Reject B, C, and D based on the calculated mathematical factor of 0.5.
PastPaper.question 24 · Multiple Choice
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The table below shows initial rate data for the reaction: \[ \text{P} + \text{Q} + \text{R} \rightarrow \text{products} \]
To find the order with respect to each reactant: 1. Compare Experiments 1 and 2: The concentration of \( \text{P} \) is doubled, while the concentrations of \( \text{Q} \) and \( \text{R} \) are kept constant. The rate doubles (from \( 1.2 \times 10^{-3} \) to \( 2.4 \times 10^{-3} \)). This shows the reaction is first-order with respect to \( \text{P} \) (order = 1). 2. Compare Experiments 1 and 3: The concentration of \( \text{Q} \) is doubled, while the concentrations of \( \text{P} \) and \( \text{R} \) are kept constant. The rate quadruples (from \( 1.2 \times 10^{-3} \) to \( 4.8 \times 10^{-3} \)). Since \( 2^2 = 4 \), the reaction is second-order with respect to \( \text{Q} \) (order = 2). 3. Compare Experiments 1 and 4: The concentration of \( \text{R} \) is doubled, while the concentrations of \( \text{P} \) and \( \text{Q} \) are kept constant. The rate remains unchanged (\( 1.2 \times 10^{-3} \)). This indicates the reaction is zero-order with respect to \( \text{R} \) (order = 0).
The overall order of reaction is the sum of the individual orders: \( \text{Overall order} = 1 + 2 + 0 = 3 \)
PastPaper.markingScheme
1 mark for the correct option C. - Reject A, B, and D: Calculated overall order of reaction is 3.
PastPaper.question 25 · multiple-choice
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\( 10\text{ cm}^3 \) of a gaseous hydrocarbon is reacted with \( 70\text{ cm}^3 \) of oxygen (an excess). After complete combustion and cooling to room temperature, the total volume of dry gas remaining is \( 50\text{ cm}^3 \). Passing this remaining gas through aqueous sodium hydroxide reduces its volume to \( 20\text{ cm}^3 \).
What is the molecular formula of the hydrocarbon?
A.\( \text{C}_3\text{H}_6 \)
B.\( \text{C}_3\text{H}_8 \)
C.\( \text{C}_4\text{H}_{10} \)
D.\( \text{C}_4\text{H}_8 \)
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PastPaper.workedSolution
1. Volume of \( \text{CO}_2 \) produced is the volume absorbed by \( \text{NaOH} \): \( 50\text{ cm}^3 - 20\text{ cm}^3 = 30\text{ cm}^3 \). 2. Since \( 10\text{ cm}^3 \) of \( \text{C}_x\text{H}_y \) yields \( 30\text{ cm}^3 \) of \( \text{CO}_2 \), \( x = 3 \). 3. The remaining \( 20\text{ cm}^3 \) of gas is unreacted \( \text{O}_2 \). 4. Volume of \( \text{O}_2 \) reacted = \( 70\text{ cm}^3 - 20\text{ cm}^3 = 50\text{ cm}^3 \). 5. From the combustion equation: \( \text{C}_3\text{H}_y + (3 + \frac{y}{4})\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} \). 6. Volume ratio of hydrocarbon to oxygen is \( 1 : (3 + \frac{y}{4}) \). 7. \( 10 \times (3 + \frac{y}{4}) = 50 \Rightarrow 3 + \frac{y}{4} = 5 \Rightarrow y = 8 \). Therefore, the molecular formula is \( \text{C}_3\text{H}_8 \).
PastPaper.markingScheme
Award 1 mark for the correct option (B).
- Reject all other choices.
PastPaper.question 26 · multiple-choice
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A \( 5.00\text{ g} \) sample of a mixture containing anhydrous sodium carbonate (\( \text{Na}_2\text{CO}_3 \), \( M_{\text{r}} = 106.0 \)) and sodium chloride (\( \text{NaCl} \), \( M_{\text{r}} = 58.5 \)) is reacted completely with excess dilute hydrochloric acid. The volume of carbon dioxide gas collected at room temperature and pressure (r.t.p.) is \( 840\text{ cm}^3 \).
What is the percentage by mass of sodium chloride in the original sample?
[Assume 1 mol of gas occupies \( 24.0\text{ dm}^3 \) at r.t.p.]
A.\( 25.8\% \)
B.\( 74.2\% \)
C.\( 37.1\% \)
D.\( 12.9\% \)
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PastPaper.workedSolution
1. Equation of reaction: \( \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \). 2. Moles of \( \text{CO}_2 = \frac{840\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0350\text{ mol} \). 3. Moles of \( \text{Na}_2\text{CO}_3 = 0.0350\text{ mol} \). 4. Mass of \( \text{Na}_2\text{CO}_3 = 0.0350\text{ mol} \times 106.0\text{ g mol}^{-1} = 3.71\text{ g} \). 5. Mass of \( \text{NaCl} \) in the mixture = \( 5.00\text{ g} - 3.71\text{ g} = 1.29\text{ g} \). 6. Percentage by mass of \( \text{NaCl} = \frac{1.29}{5.00} \times 100\% = 25.8\% \).
PastPaper.markingScheme
Award 1 mark for the correct option (A).
- Option B is incorrect as it is the percentage of sodium carbonate. - Reject B, C, and D.
PastPaper.question 27 · multiple-choice
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Which of the following compounds undergoes electrophilic substitution of the aromatic ring at the fastest rate when reacted with chlorine in the presence of an anhydrous iron(III) chloride catalyst?
A.chlorobenzene
B.methylbenzene
C.nitrobenzene
D.benzene
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PastPaper.workedSolution
1. The rate of electrophilic substitution depends on the electron density of the benzene ring. 2. The methyl group (\( -\text{CH}_3 \)) in methylbenzene is electron-donating due to positive inductive effects (+I), which increases the electron density on the benzene ring and activates it towards electrophilic attack. 3. Chlorine in chlorobenzene is weakly deactivating due to its electronegativity (negative inductive effect dominates). 4. The nitro group (\( -\text{NO}_2 \)) in nitrobenzene is strongly deactivating. 5. Therefore, methylbenzene reacts at the fastest rate.
PastPaper.markingScheme
Award 1 mark for selecting B.
- Reject other options because methylbenzene is activated compared to benzene, whereas chlorobenzene and nitrobenzene are deactivated.
PastPaper.question 28 · multiple-choice
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What are the major organic products formed when propylbenzene is treated under the following reaction conditions?
Reaction 1: heating with alkaline \( \text{KMnO}_4 \), followed by acidification Reaction 2: boiling with \( \text{Cl}_2 \) in the presence of UV light
1. Heating an alkylbenzene with alkaline \( \text{KMnO}_4 \) followed by acidification oxidizes the side chain down to a carboxylic acid group, producing benzoic acid, regardless of the length of the alkyl chain. 2. Boiling with \( \text{Cl}_2 \) in the presence of UV light leads to free radical substitution on the side chain. The most stable intermediate radical is the benzylic radical (stabilized by resonance with the benzene ring), so chlorine preferentially substitutes at the C-1 position of the propyl chain to form (1-chloropropyl)benzene.
PastPaper.markingScheme
Award 1 mark for selecting A.
- Reject C and D because side chain oxidation of propylbenzene yields benzoic acid, not propanoic acid. - Reject B because UV light/chlorine results in side-chain substitution, not ring substitution.
PastPaper.question 29 · multiple-choice
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The following standard electrode potentials are given:
A half-cell is set up containing a silver electrode in contact with a solution of silver ions, \( \text{Ag}^+(\text{aq}) \). The standard electrode potential is:
At \( 298\text{ K} \, the measured electrode potential of this half-cell is \) +0.68\text{ V} \). What is the concentration of \( \text{Ag}^+(\text{aq}) \) in the solution?
A.\( 1.0 \times 10^{-2}\text{ mol dm}^{-3} \)
B.\( 1.0 \times 10^{-1}\text{ mol dm}^{-3} \)
C.\( 2.0 \times 10^{-2}\text{ mol dm}^{-3} \)
D.\( 4.0 \times 10^{-3}\text{ mol dm}^{-3} \)
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PastPaper.workedSolution
1. Use the Nernst equation: \( E = E^\ominus + \frac{0.059}{z}\log[\text{Ag}^+] \). 2. For \( \text{Ag}^+/\text{Ag} \), \( z = 1 \), so \( E = E^\ominus + 0.059\log[\text{Ag}^+] \). 3. Substitute the values: \( 0.68 = 0.80 + 0.059\log[\text{Ag}^+] \). 4. \( -0.12 = 0.059\log[\text{Ag}^+] \). 5. \( \log[\text{Ag}^+] = \frac{-0.12}{0.059} \approx -2.03 \). 6. \( [\text{Ag}^+] = 10^{-2.03} \approx 1.0 \times 10^{-2}\text{ mol dm}^{-3} \).
PastPaper.markingScheme
Award 1 mark for the correct option (A).
- Reject B, C, and D.
PastPaper.question 31 · multiple-choice
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The initial rates of the reaction \( \text{A} + 2\text{B} + \text{C} \rightarrow \text{Products} \) were measured at different concentrations of reactants at a constant temperature:
1. Determine the reaction order for each reactant: - From Exps 1 and 2: doubling \( [\text{A}] \) doubles the rate, so the reaction is 1st order with respect to A. - From Exps 1 and 3: doubling \( [\text{B}] \) quadruples the rate (\( 1.2 \times 10^{-3} \rightarrow 4.8 \times 10^{-3} \)), so the reaction is 2nd order with respect to B. - From Exps 1 and 4: doubling \( [\text{C}] \) has no effect on the rate, so the reaction is 0th order with respect to C. 2. Write the rate equation: \( \text{Rate} = k[\text{A}][\text{B}]^2 \). 3. Find the units of \( k \): \( k = \frac{\text{Rate}}{[\text{A}][\text{B}]^2} \Rightarrow \text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).
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Award 1 mark for the correct option (A).
- Reject incorrect rate orders or calculation errors leading to options B, C, or D.
PastPaper.question 32 · multiple-choice
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A chemical reaction is first-order with respect to a reactant P. When the initial concentration of P is \( 1.60\text{ mol dm}^{-3} \), it takes exactly \( 40\text{ minutes} \) for the concentration of P to fall to \( 0.20\text{ mol dm}^{-3} \).
Under the same conditions, how long will it take for the concentration of P to fall from \( 0.80\text{ mol dm}^{-3} \) to \( 0.10\text{ mol dm}^{-3} \)?
A.\( 20\text{ minutes} \)
B.\( 30\text{ minutes} \)
C.\( 40\text{ minutes} \)
D.\( 80\text{ minutes} \)
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PastPaper.workedSolution
1. For a first-order reaction, the half-life \( t_{1/2} \) is constant and independent of the initial concentration. 2. The concentration change from \( 1.60\text{ mol dm}^{-3} \) to \( 0.20\text{ mol dm}^{-3} \) represents a reduction to \( \frac{1}{8} \) of the initial value (which corresponds to exactly 3 half-lives: \( 1.60 \rightarrow 0.80 \rightarrow 0.40 \rightarrow 0.20 \)). 3. Therefore, \( 3 \times t_{1/2} = 40\text{ minutes} \). 4. The concentration change from \( 0.80\text{ mol dm}^{-3} \) to \( 0.10\text{ mol dm}^{-3} \) also represents a reduction to \( \frac{1}{8} \) of its initial value (which also corresponds to exactly 3 half-lives: \( 0.80 \rightarrow 0.40 \rightarrow 0.20 \rightarrow 0.10 \)). 5. Since the number of half-lives is identical, the time required remains exactly the same, \( 40\text{ minutes} \).
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Award 1 mark for selecting C.
- Reject all other choices.
PastPaper.question 33 · multiple_choice
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A 1.20 g sample of a Group 2 metal carbonate, \(MCO_3\), is heated strongly until it completely decomposes to form the metal oxide and carbon dioxide. The volume of carbon dioxide gas collected at room temperature and pressure (r.t.p.) is 288 cm\(^3\). What is the identity of the metal \(M\)? [Molar volume of gas at r.t.p. = 24.0 dm\(^3\) mol\(^{-1\)}]
A.magnesium
B.calcium
C.strontium
D.barium
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PastPaper.workedSolution
Step 1: Calculate the moles of \(CO_2\) gas collected: moles of \(CO_2\) = 288 cm\(^3\) / 24000 cm\(^3\) mol\(^{-1\)} = 0.0120 mol. Step 2: From the thermal decomposition equation: \(MCO_3(s) \rightarrow MO(s) + CO_2(g)\), one mole of \(MCO_3\) produces one mole of \(CO_2\). Therefore, the moles of \(MCO_3\) is also 0.0120 mol. Step 3: Calculate the molar mass of \(MCO_3\): \(M_{\text{r}}(MCO_3) = \text{mass} / \text{moles} = 1.20\text{ g} / 0.0120\text{ mol} = 100\text{ g mol}^{-1}\). Step 4: Find the relative atomic mass of the metal \(M\): \(A_{\text{r}}(M) + 12.0 + 3 \times 16.0 = 100 \implies A_{\text{r}}(M) + 60.0 = 100 \implies A_{\text{r}}(M) = 40.0\). This value corresponds closely to the relative atomic mass of calcium (40.1).
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1 mark for the correct option B. Reject all other options.
PastPaper.question 34 · multiple_choice
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A 25.0 cm\(^3\) sample of 0.0200 mol dm\(^{-3}\) of a metal ion \(M^{n+}\) reacts completely with exactly 10.0 cm\(^3\) of 0.0200 mol dm\(^{-3}\) of acidified potassium manganate(VII), \(KMnO_4\). In this reaction, the manganate(VII) ions are reduced to \(Mn^{2+}\) ions, and the metal ions are oxidized to \(M^{5+}\) ions. What is the initial oxidation state, \(n\), of the metal ion?
A.+1
B.+2
C.+3
D.+4
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Step 1: Calculate the moles of reactants: moles of \(KMnO_4 (MnO_4^-) = 10.0 \times 10^{-3} \text{ dm}^3 \times 0.0200 \text{ mol dm}^{-3} = 2.00 \times 10^{-4}\text{ mol}\); moles of \(M^{n+} = 25.0 \times 10^{-3} \text{ dm}^3 \times 0.0200 \text{ mol dm}^{-3} = 5.00 \times 10^{-4}\text{ mol}\). Step 2: Determine the reacting ratio: \(\text{moles of } M^{n+} / \text{moles of } MnO_4^- = 5.00 \times 10^{-4} / 2.00 \times 10^{-4} = 5/2\). This means 5 moles of \(M^{n+}\) react with 2 moles of \(MnO_4^-\). Step 3: Set up the electron balance: Each \(MnO_4^-\) ion (where Mn is in +7 oxidation state) is reduced to \(Mn^{2+}\) (+2 oxidation state), gaining 5 electrons. For 2 moles of \(MnO_4^-\), total electrons gained = \(2 \times 5 = 10\) moles of electrons. Therefore, 5 moles of \(M^{n+}\) must lose a total of 10 moles of electrons during oxidation to \(M^{5+}\). Each \(M^{n+}\) ion loses \(10 / 5 = 2\) electrons. Step 4: Solve for \(n\): \(5 - n = 2 \implies n = 3\). Thus, the initial oxidation state of the metal ion is +3.
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PastPaper.question 35 · multiple_choice
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When a dimethylbenzene isomer is reacted with bromine in the presence of an anhydrous \(AlBr_3\) catalyst, only one monobrominated organic product is formed as the major product. Which compound is this starting isomer?
A.1,2-dimethylbenzene
B.1,3-dimethylbenzene
C.1,4-dimethylbenzene
D.methylbenzene
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Analyze the symmetry and possible monobromination positions of the dimethylbenzene isomers: In 1,4-dimethylbenzene, all four remaining positions on the benzene ring are chemically equivalent because of the highly symmetrical para-substituted structure. Electrophilic substitution at any of these positions produces 2-bromo-1,4-dimethylbenzene. Thus, only one monobrominated isomer is formed. In contrast, 1,2-dimethylbenzene can give two different monobrominated products (3-bromo and 4-bromo), and 1,3-dimethylbenzene can give three different monobrominated products. Methylbenzene is not a dimethylbenzene and typically gives a mixture of ortho and para monobrominated isomers.
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PastPaper.question 36 · multiple_choice
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An unknown hydrocarbon with the molecular formula \(C_9H_{12}\) is heated under reflux with excess alkaline potassium manganate(VII). After acidification of the reaction mixture, a single organic product, benzoic acid (\(C_6H_5COOH\)), is obtained as a white crystalline solid. Which structure could represent the starting hydrocarbon?
A.1,2,3-trimethylbenzene
B.1-ethyl-4-methylbenzene
C.propylbenzene
D.1,3,5-trimethylbenzene
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When alkylbenzenes are oxidized by heating under reflux with alkaline \(KMnO_4\) followed by acidification, each alkyl side chain with at least one benzylic hydrogen atom is oxidized to a carboxylic acid (\(-COOH\)) group. If the starting material has more than one alkyl side chain, a polycarboxylic acid will be formed. 1,2,3-Trimethylbenzene and 1,3,5-trimethylbenzene have three alkyl groups, which oxidize to benzene-1,2,3-tricarboxylic acid and benzene-1,3,5-tricarboxylic acid, respectively. 1-ethyl-4-methylbenzene has two alkyl groups and will oxidize to benzene-1,4-dicarboxylic acid. Propylbenzene (\(C_6H_5CH_2CH_2CH_3\)), which has the molecular formula \(C_9H_{12}\), has only one alkyl side chain, so it oxidizes to benzoic acid (\(C_6H_5COOH\)).
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PastPaper.question 37 · multiple_choice
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The standard electrode potentials for three redox systems are shown below: 1. \(Ag^+(aq) + e^- \rightleftharpoons Ag(s) \quad E^\theta = +0.80\text{ V}\) 2. \(Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s) \quad E^\theta = +0.34\text{ V}\) 3. \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq) \quad E^\theta = +0.77\text{ V}\) Which reaction is thermodynamically feasible under standard conditions?
To determine if a reaction is thermodynamically feasible under standard conditions, we calculate the standard cell potential, \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\). For reaction A: \(E^\theta_{\text{cell}} = E^\theta(Fe^{3+}/Fe^{2+}) - E^\theta(Ag^+/Ag) = +0.77\text{ V} - (+0.80\text{ V}) = -0.03\text{ V}\) (Not feasible). For reaction B: \(E^\theta_{\text{cell}} = E^\theta(Cu^{2+}/Cu) - E^\theta(Fe^{3+}/Fe^{2+}) = +0.34\text{ V} - (+0.77\text{ V}) = -0.43\text{ V}\) (Not feasible). For reaction C: \(E^\theta_{\text{cell}} = E^\theta(Ag^+/Ag) - E^\theta(Cu^{2+}/Cu) = +0.80\text{ V} - (+0.34\text{ V}) = +0.46\text{ V}\) (Feasible since \(E^\theta_{\text{cell}} > 0\)). For reaction D: \(E^\theta_{\text{cell}} = E^\theta(Cu^{2+}/Cu) - E^\theta(Ag^+/Ag) = +0.34\text{ V} - (+0.80\text{ V}) = -0.46\text{ V}\) (Not feasible).
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PastPaper.question 38 · multiple_choice
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The standard electrode potential, \(E^\theta\), for the half-cell \(Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)\) is +0.34 V at 298 K. The Nernst equation for this half-cell is: \(E = E^\theta + \frac{0.059}{z}\log[Cu^{2+}(aq)]\). What is the electrode potential of this half-cell when the concentration of \(Cu^{2+}(aq)\) is \(1.0 \times 10^{-3}\text{ mol dm}^{-3}\) at 298 K?
A.+0.16 V
B.+0.25 V
C.+0.34 V
D.+0.43 V
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Step 1: Identify the number of electrons transferred, \(z\), in the half-reaction: \(Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)\) implies \(z = 2\). Step 2: Substitute the given values into the Nernst equation: \(E = +0.34 + \frac{0.059}{2}\log(1.0 \times 10^{-3})\). Step 3: Calculate the logarithm term: \(\log(1.0 \times 10^{-3}) = -3\). Step 4: Solve for \(E\): \(E = +0.34 + 0.0295 \times (-3) = +0.34 - 0.0885 = +0.2515\text{ V}\), which rounds to +0.25 V.
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PastPaper.question 39 · multiple_choice
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The reaction between peroxodisulfate ions, \(S_2O_8^{2-}\), and iodide ions, \(I^-\), is represented by the equation: \(S_2O_8^{2-}(aq) + 2I^-(aq) \rightarrow 2SO_4^{2-}(aq) + I_2(aq)\). Initial rate data for this reaction at a constant temperature are shown below. Exp 1: \([S_2O_8^{2-}] = 0.10\text{ mol dm}^{-3}\), \([I^-] = 0.10\text{ mol dm}^{-3}\), rate = \(2.2 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Exp 2: \([S_2O_8^{2-}] = 0.20\text{ mol dm}^{-3}\), \([I^-] = 0.10\text{ mol dm}^{-3}\), rate = \(4.4 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Exp 3: \([S_2O_8^{2-}] = 0.20\text{ mol dm}^{-3}\), \([I^-] = 0.20\text{ mol dm}^{-3}\), rate = \(8.8 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). What are the orders of reaction with respect to \(S_2O_8^{2-}\) and \(I^-\)?
A.1 with respect to S2O8(2-), 2 with respect to I-
B.2 with respect to S2O8(2-), 1 with respect to I-
C.1 with respect to S2O8(2-), 1 with respect to I-
D.2 with respect to S2O8(2-), 2 with respect to I-
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Step 1: To find the order with respect to \(S_2O_8^{2-}\), compare Experiments 1 and 2 where \([I^-]\) is constant: \([S_2O_8^{2-}]\) doubles (from 0.10 to 0.20), and the initial rate doubles (from \(2.2 \times 10^{-4}\) to \(4.4 \times 10^{-4}\)). Since \(2^x = 2\), the reaction is first order (\(x = 1\)) with respect to \(S_2O_8^{2-}\). Step 2: To find the order with respect to \(I^-\), compare Experiments 2 and 3 where \([S_2O_8^{2-}]\) is constant: \([I^-]\) doubles (from 0.10 to 0.20), and the initial rate doubles (from \(4.4 \times 10^{-4}\) to \(8.8 \times 10^{-4}\)). Since \(2^y = 2\), the reaction is first order (\(y = 1\)) with respect to \(I^-\).
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PastPaper.question 40 · multiple_choice
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A chemical reaction follows the rate equation: \(\text{rate} = k[X][Y]^2\). In an experiment, when the concentration of \(X\) is \(0.40\text{ mol dm}^{-3}\) and the concentration of \(Y\) is \(0.50\text{ mol dm}^{-3}\), the rate of reaction is \(1.5 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the value and units of the rate constant \(k\)?
Step 1: Rearrange the rate equation to solve for \(k\): \(k = \text{rate} / ([X][Y]^2)\). Step 2: Substitute the given values into the equation: \(k = (1.5 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}) / ((0.40\text{ mol dm}^{-3}) \times (0.50\text{ mol dm}^{-3})^2) = 1.5 \times 10^{-3} / (0.40 \times 0.25) = 1.5 \times 10^{-3} / 0.10 = 1.5 \times 10^{-2}\) (or 0.015). Step 3: Determine the units of \(k\): Units of \(k\) = \((\text{mol dm}^{-3}\text{ s}^{-1}) / ((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2) = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). Thus, the rate constant is \(1.50 \times 10^{-2}\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
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Paper 22
Answer all structured questions in the spaces provided. Show all working and use appropriate units.
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PastPaper.question 1 · structured
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A sample of 5.62 g of a hydrated transition metal salt, MSO4.7H2O, is heated to constant mass, driving off all water of crystallisation. 3.10 g of anhydrous MSO4 is obtained. (a)(i) Calculate the number of moles of water lost during heating. (ii) Calculate the molar mass of the anhydrous salt, MSO4, and hence identify the transition metal M. (iii) Write the full chemical formula of the hydrated transition metal salt. (b) A solution of this salt, MSO4(aq), reacts with aqueous sodium carbonate, Na2CO3(aq), to form a precipitate of MCO3. (i) Write an ionic equation, with state symbols, for this precipitation reaction. (ii) Calculate the mass of dry precipitate obtained when 50.0 cm3 of 0.250 mol dm-3 MSO4(aq) is reacted with excess sodium carbonate solution. (c)(i) In another reaction, MCO3(s) is heated to decompose into MO(s) and CO2(g). Calculate the volume of CO2(g), in dm3, produced at r.t.p. when 2.38 g of MCO3 is completely decomposed. [Take the volume of 1 mole of gas at r.t.p. to be 24.0 dm3] (ii) Explain how the thermal stability of Group 2 carbonates compares with that of transition metal carbonates like MCO3.
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(a)(i) Mass of water lost = 5.62 g - 3.10 g = 2.52 g. Moles of water lost = 2.52 g / 18.0 g mol-1 = 0.140 mol. (ii) Moles of anhydrous MSO4 = Moles of water lost / 7 = 0.140 / 7 = 0.0200 mol. Molar mass of MSO4 = 3.10 g / 0.0200 mol = 155 g mol-1. Molar mass of metal M = 155 - (32.1 + 4 * 16.0) = 155 - 96.1 = 58.9 g mol-1. Metal M is cobalt (Co). (iii) The formula of the hydrated salt is CoSO4.7H2O. (b)(i) Co2+(aq) + CO3 2-(aq) -> CoCO3(s). (ii) Moles of Co2+ = concentration * volume = 0.250 mol dm-3 * 0.0500 dm3 = 0.0125 mol. Molar mass of CoCO3 = 58.9 + 12.0 + 3 * 16.0 = 118.9 g mol-1. Mass of CoCO3 precipitate = 0.0125 mol * 118.9 g mol-1 = 1.49 g. (c)(i) Moles of CoCO3 = 2.38 g / 118.9 g mol-1 = 0.0200 mol. Moles of CO2 produced = 0.0200 mol. Volume of CO2 = 0.0200 mol * 24.0 dm3 mol-1 = 0.480 dm3. (ii) Group 2 carbonates are more thermally stable than transition metal carbonates like CoCO3 of similar ionic radius because transition metal ions have greater polarizing power, weakening the C-O bond in the carbonate ion more effectively.
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(a)(i) [2 marks] M1: Calculate mass of water lost = 2.52 g. M2: Moles of water = 2.52 / 18.0 = 0.140 mol. (ii) [3 marks] M1: Moles of MSO4 = 0.0200 mol (by dividing water moles by 7). M2: Molar mass of MSO4 = 3.10 / 0.0200 = 155 g mol-1. M3: Identifies metal M as cobalt / Co by subtracting 96.1 to get Ar = 58.9. (iii) [1 mark] CoSO4.7H2O. (b)(i) [2 marks] M1: Correct species: Co2+(aq) + CO3 2-(aq) -> CoCO3(s). M2: Correct state symbols. (ii) [3 marks] M1: Moles of Co2+ = 0.0125 mol. M2: Molar mass of CoCO3 = 118.9 g mol-1. M3: Mass of CoCO3 = 1.49 g (accept 1.486 g). (c)(i) [3 marks] M1: Moles of CoCO3 = 0.0200 mol. M2: Molar ratio is 1:1, so moles of CO2 = 0.0200 mol. M3: Volume of CO2 = 0.480 dm3. (ii) [1 mark] Group 2 carbonates are more thermally stable than transition metal carbonates.
PastPaper.question 2 · structured
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Benzene can be converted into methylbenzene, which can then undergo further reactions to produce a variety of organic compounds. (a) Write an equation for the reaction of benzene with a suitable reagent to produce methylbenzene, stating the catalyst used. (b)(i) Describe the mechanism for the electrophilic substitution reaction of benzene with chloromethane in the presence of an aluminium chloride catalyst. Your answer should show the generation of the electrophile, the structure of the intermediate, and the recovery of the catalyst. (ii) Explain why methylbenzene reacts faster than benzene in electrophilic substitution reactions. (c) Methylbenzene can be converted into other organic compounds. (i) State the reagents and conditions required to oxidise methylbenzene to benzoic acid. (ii) Complete the structure of the organic product formed when methylbenzene is reacted with a mixture of concentrated nitric acid and concentrated sulfuric acid at 30 degrees Celsius, and state the type of reaction. (iii) Draw the skeletal structure of the product formed when methylbenzene reacts with chlorine in the presence of ultraviolet light.
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(a) Equation: C6H6 + CH3Cl -> C6H5CH3 + HCl. Catalyst: AlCl3 (or FeCl3). (b)(i) 1. Generation of electrophile: CH3Cl + AlCl3 -> CH3+ + AlCl4-. 2. Curly arrow from benzene pi-ring to CH3+ electrophile. 3. Positive intermediate drawn showing a horseshoe open to the sp3 carbon, with the positive charge inside. 4. Curly arrow from the C-H bond back into the ring to restore aromaticity. 5. Recovery of catalyst: H+ + AlCl4- -> AlCl3 + HCl. (ii) The methyl group is an electron-donating group by a positive inductive effect. This increases electron density in the benzene ring, making it more susceptible to electrophilic attack. (c)(i) Reagents: Alkaline potassium manganate(VII), KMnO4(aq), followed by dilute acid (e.g. H2SO4). Conditions: Heat under reflux. (ii) Product: 2-nitromethylbenzene (or 4-nitromethylbenzene). Type of reaction: Electrophilic substitution. (iii) Product: (chloromethyl)benzene, represented as a benzene ring with a -CH2Cl substituent group.
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(a) [2 marks] M1: Equation: C6H6 + CH3Cl -> C6H5CH3 + HCl (or structural equivalent). M2: Catalyst: AlCl3 or FeCl3. (b)(i) [5 marks] M1: CH3Cl + AlCl3 -> CH3+ + AlCl4-. M2: Curly arrow from benzene ring to CH3+. M3: Correctly drawn intermediate showing positive charge and open horseshoe. M4: Curly arrow from C-H bond restoring ring. M5: H+ + AlCl4- -> AlCl3 + HCl. (ii) [2 marks] M1: Methyl group is electron-donating (positive inductive effect). M2: Increases ring electron density, attracting electrophiles more strongly. (c)(i) [2 marks] M1: KMnO4(aq) (alkaline) then acid. M2: Heat under reflux. (ii) [2 marks] M1: Structure of 2-nitromethylbenzene or 4-nitromethylbenzene. M2: Electrophilic substitution. (iii) [2 marks] M1: Structure showing chloromethyl group on benzene ring. M2: Skeletal formula correctly represented.
PastPaper.question 3 · structured
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Standard electrode potentials, E-standard, are useful for predicting redox processes. (a) Define the term standard electrode potential. (b) An electrochemical cell is constructed from the following half-cells: Half-cell A: A copper electrode in Cu2+(aq). [Cu2+(aq) + 2e- <=> Cu(s), E-standard = +0.34 V] Half-cell B: A platinum electrode in Fe3+(aq) and Fe2+(aq). [Fe3+(aq) + e- <=> Fe2+(aq), E-standard = +0.77 V] (i) State the material used as the inert electrode in Half-cell B, and describe the standard conditions required for this half-cell. (ii) Write the overall cell reaction that occurs when this cell operates. (iii) Calculate the standard cell potential, E-standard of the cell. (iv) Draw a fully labelled diagram of this electrochemical cell, showing the direction of electron flow in the external circuit. (c) The Nernst equation can be used to calculate electrode potentials under non-standard conditions: E = E-standard + (0.059/z) * log([oxidised species]/[reduced species]). Calculate the electrode potential, E, for Half-cell A if the concentration of Cu2+(aq) is decreased to 0.0150 mol dm-3.
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(a) Standard electrode potential is the electromotive force (emf) of a cell consisting of the half-cell connected to a standard hydrogen electrode, measured under standard conditions of 298 K, 1 atm gas pressure, and 1.00 mol dm-3 solution concentrations, with all reactants and products in their standard states. (b)(i) Inert electrode: Platinum (Pt). Standard conditions: Temperature of 298 K, and concentrations of both Fe3+(aq) and Fe2+(aq) at 1.00 mol dm-3. (ii) 2Fe3+(aq) + Cu(s) -> 2Fe2+(aq) + Cu2+(aq). (iii) E-cell = E-reduction - E-oxidation = +0.77 V - (+0.34 V) = +0.43 V. (iv) The diagram must include: 1. A copper electrode in a beaker of Cu2+(aq) and a platinum electrode in a beaker containing both Fe2+(aq) and Fe3+(aq). 2. A salt bridge connecting the two solutions. 3. A voltmeter connected via wires to both electrodes. 4. An arrow showing electron flow in the external wire from the copper electrode to the platinum electrode. (c) Using the Nernst equation for Cu2+ + 2e- <=> Cu: z = 2. [oxidised] = [Cu2+] = 0.0150. [reduced] = [Cu] = 1 (constant for solid). E = +0.34 + (0.059/2) * log(0.0150) = +0.34 + 0.0295 * (-1.8239) = +0.34 - 0.0538 = +0.286 V (or +0.29 V).
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(a) [3 marks] M1: EMF/potential of half-cell relative to a standard hydrogen electrode. M2: Standard conditions: 298 K and 1 atm pressure. M3: Concentration of 1.00 mol dm-3 for all aqueous species. (b)(i) [2 marks] M1: Platinum / Pt electrode. M2: 1.00 mol dm-3 Fe3+ and 1.00 mol dm-3 Fe2+ at 298 K. (ii) [2 marks] M1: Correct reactants and products: 2Fe3+ + Cu -> 2Fe2+ + Cu2+. M2: Correct state symbols. (iii) [1 mark] +0.43 V (or 0.43 V). (iv) [4 marks] M1: Correct electrodes and solutions drawn and labelled. M2: Salt bridge connecting the beakers. M3: Voltmeter in the external circuit. M4: Electron flow shown from Cu to Pt. (c) [3 marks] M1: Recognising z = 2. M2: Correct substitution into equation: E = 0.34 + (0.059/2) * log(0.0150). M3: Correct evaluation of E = +0.286 V (accept +0.29 V).
PastPaper.question 4 · structured
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The reaction between peroxodisulfate ions, S2O8 2-(aq), and iodide ions, I-(aq), is represented by: S2O8 2-(aq) + 2I-(aq) -> 2SO4 2-(aq) + I2(aq). Experimental data at constant temperature: Experiment 1: [S2O8 2-] = 0.0500 mol dm-3, [I-] = 0.0800 mol dm-3, Initial Rate = 1.10 x 10^-4 mol dm-3 s-1. Experiment 2: [S2O8 2-] = 0.1000 mol dm-3, [I-] = 0.0800 mol dm-3, Initial Rate = 2.20 x 10^-4 mol dm-3 s-1. Experiment 3: [S2O8 2-] = 0.0500 mol dm-3, [I-] = 0.1600 mol dm-3, Initial Rate = 2.20 x 10^-4 mol dm-3 s-1. (a)(i) Deduce the order of reaction with respect to S2O8 2-. Explain your reasoning. (ii) Deduce the order of reaction with respect to I-. Explain your reasoning. (iii) Write the rate equation for this reaction. (iv) Calculate the rate constant, k, for this reaction, using the data from Experiment 1. State its units. (b)(i) Explain why this reaction is very slow in the absence of a catalyst. (ii) Write two equations to show how Fe2+(aq) acts as a homogeneous catalyst for this reaction. (iii) Explain why the catalyst in this reaction is described as homogeneous. (c)(i) Sketch a Boltzmann distribution curve on appropriate axes to show how an increase in temperature increases the rate of reaction. Label your curves T1 and T2, where T2 > T1, and label the activation energy, Ea.
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(a)(i) Order with respect to S2O8 2- is 1. Comparing Experiment 1 and 2, when [S2O8 2-] is doubled while [I-] is kept constant, the rate doubles from 1.10 x 10^-4 to 2.20 x 10^-4. (ii) Order with respect to I- is 1. Comparing Experiment 1 and 3, when [I-] is doubled while [S2O8 2-] is kept constant, the rate doubles from 1.10 x 10^-4 to 2.20 x 10^-4. (iii) Rate = k[S2O8 2-][I-]. (iv) k = Rate / ([S2O8 2-][I-]) = 1.10 x 10^-4 / (0.0500 * 0.0800) = 0.0275. Units: dm3 mol-1 s-1. (b)(i) Both S2O8 2- and I- ions are negatively charged and repel each other, resulting in a very high activation energy. (ii) Equation 1: 2Fe2+(aq) + S2O8 2-(aq) -> 2Fe3+(aq) + 2SO4 2-(aq). Equation 2: 2Fe3+(aq) + 2I-(aq) -> 2Fe2+(aq) + I2(aq). (iii) The catalyst is in the same phase / physical state (aqueous solution) as the reactants. (c)(i) The graph has 'Number of molecules' on the y-axis and 'Kinetic energy' on the x-axis. The T1 curve starts at the origin, rises to a peak, and asymptotes to the x-axis. The T2 curve (higher temperature) has a lower peak shifted to the right, and lies above the T1 curve at high energies. A vertical line labelled Ea shows that at T2, a greater fraction of molecules has energy >= Ea.
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(a)(i) [2 marks] M1: Order is 1. M2: Explanation: comparing Exp 1 and 2, doubling [S2O8 2-] doubles the rate. (ii) [2 marks] M1: Order is 1. M2: Explanation: comparing Exp 1 and 3, doubling [I-] doubles the rate. (iii) [1 mark] Rate = k[S2O8 2-][I-]. (iv) [3 marks] M1: Calculation of k value = 0.0275. M2: Value given to 3 significant figures. M3: Units: dm3 mol-1 s-1. (b)(i) [1 mark] Repulsion between two negatively charged ions leads to high activation energy. (ii) [2 marks] M1: 2Fe2+ + S2O8 2- -> 2Fe3+ + 2SO4 2-. M2: 2Fe3+ + 2I- -> 2Fe2+ + I2. (iii) [1 mark] Catalyst and reactants are in the same phase/state. (c)(i) [3 marks] M1: Correct axes labels (y-axis: Fraction/number of molecules, x-axis: Kinetic energy) and curves starting at origin. M2: T2 curve has a lower peak shifted to the right compared to T1. M3: Ea marked with more area under T2 curve than T1 curve to the right of Ea.
Paper 42
Answer all structured questions in the spaces provided. Show all working and use appropriate units.
PastPaper.question 1 · Structured A Level Questions
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A sample of 2.50 g of a mixture of calcium carbonate, \( \text{CaCO}_3 \), and inert silica, \( \text{SiO}_2 \), is treated with 50.0 cm\(^3\) of 1.00 mol dm\(^{-3}\) hydrochloric acid (an excess). The resulting solution is made up to 250 cm\(^3\) in a volumetric flask. A 25.0 cm\(^3\) portion of this diluted solution requires 18.40 cm\(^3\) of 0.100 mol dm\(^{-3}\) sodium hydroxide for complete neutralisation. (a) Write a balanced chemical equation for the reaction of calcium carbonate with hydrochloric acid. (b) Calculate the number of moles of sodium hydroxide used in the titration. (c) Calculate the number of moles of excess hydrochloric acid in the 250 cm\(^3\) volumetric flask. (d) Determine the percentage by mass of \( \text{CaCO}_3 \) in the original mixture. Show all your working.
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(a) \( \text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \). (b) Moles of NaOH = \( \frac{18.40}{1000} \times 0.100 = 1.84 \times 10^{-3}\text{ mol} \). (c) Moles of excess HCl in 25.0 cm\(^3\) titration portion = \( 1.84 \times 10^{-3}\text{ mol} \). Thus, moles of excess HCl in the 250 cm\(^3\) volumetric flask = \( 1.84 \times 10^{-3} \times \frac{250}{25.0} = 1.84 \times 10^{-2}\text{ mol} \). (d) Initial moles of HCl added = \( \frac{50.0}{1000} \times 1.00 = 5.00 \times 10^{-2}\text{ mol} \). Moles of HCl that reacted with \( \text{CaCO}_3 \) = \( (5.00 \times 10^{-2}) - (1.84 \times 10^{-2}) = 3.16 \times 10^{-2}\text{ mol} \). Moles of \( \text{CaCO}_3 \) reacted = \( \frac{3.16 \times 10^{-2}}{2} = 1.58 \times 10^{-2}\text{ mol} \). Mass of \( \text{CaCO}_3 \) = \( 1.58 \times 10^{-2}\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.582\text{ g} \). Percentage by mass of \( \text{CaCO}_3 \) = \( \frac{1.582}{2.50} \times 100\% = 63.3\% \).
PastPaper.markingScheme
Part (a): 1 mark for correct equation with state symbols. Part (b): 1 mark for moles of NaOH = 1.84 x 10^-3 mol. Part (c): 2 marks for scaling up by 10 to get 1.84 x 10^-2 mol (1 mark for intermediate). Part (d): 1 mark for initial moles of HCl = 0.050 mol, 1 mark for moles of reacted HCl = 0.0316 mol, 1 mark for moles of CaCO3 = 0.0158 mol, 1 mark for mass of CaCO3 = 1.582 g, 1 mark for final percentage of 63.3% (accept 63.2% to 63.4%).
PastPaper.question 2 · Structured A Level Questions
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Methylbenzene is an important starting material in organic synthesis. (a) Write an equation for the generation of the nitronium ion electrophile, \( \text{NO}_2^+ \), from a mixture of concentrated nitric acid and concentrated sulfuric acid. (b) Explain why methylbenzene is nitrated much more rapidly than benzene. (c) Methylbenzene can be converted into 3-nitrobenzoic acid. State the reagents and conditions for each step of this two-step synthesis, and explain why the order of these steps is crucial.
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(a) \( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+ \). (b) The methyl group is electron-donating by positive inductive effect. This increases the electron density of the benzene ring, making it more nucleophilic and more susceptible to electrophilic attack than benzene. (c) Step 1: React methylbenzene with alkaline potassium manganate(VII), \( \text{KMnO}_4 \), heat under reflux, followed by acidification with dilute HCl or \( \text{H}_2\text{SO}_4 \) to form benzoic acid. Step 2: React benzoic acid with concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \) at 50 to 60 degrees Celsius to form 3-nitrobenzoic acid. Crucial order: The methyl group is 2,4-directing, whereas the carboxylic acid group (\( -\text{COOH} \)) is 3-directing. If nitration is done first, we obtain 2-nitromethylbenzene or 4-nitromethylbenzene. Therefore, oxidation to \( -\text{COOH} \) must be performed first to direct the subsequent nitro group to the 3-position.
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Part (a): 2 marks for the correct equation. Part (b): 2 marks (1 mark for identifying the electron-donating methyl group, 1 mark for stating this increases electron density on the ring/increases nucleophilicity). Part (c): 4 marks for reagents and conditions (2 marks for alkaline KMnO4, heat, and acidification; 2 marks for conc HNO3 + conc H2SO4, 50-60C). 2 marks for explaining directing effects (1 mark for methyl being 2,4-directing, 1 mark for -COOH being 3-directing).
PastPaper.question 3 · Structured A Level Questions
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An electrochemical cell is set up at 298 K 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. Given standard electrode potentials: \( \text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq}) \quad E^\ominus = +0.77\text{ V} \) and \( \text{Ag}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s}) \quad E^\ominus = +0.80\text{ V} \). (a) Calculate the standard cell potential, \( E^\ominus_{\text{cell}} \). (b) Write the overall ionic equation for the cell reaction that occurs when current flows under standard conditions. (c) The concentration of \( \text{Ag}^+ \) in the silver half-cell is decreased to \( 0.025\text{ mol dm}^{-3} \), while the \( \text{Fe}^{3+}/\text{Fe}^{2+} \) half-cell remains under standard conditions. Use the Nernst equation, \( E = E^\ominus + 0.059 \log [\text{Ag}^+] \), to calculate the non-standard electrode potential of the \( \text{Ag}^+/\text{Ag} \) half-cell. (d) Determine the cell potential, \( E_{\text{cell}} \), under these non-standard conditions and state whether the direction of electron flow in the external circuit is the same or has reversed.
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(a) \( E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +0.80 - (+0.77) = +0.03\text{ V} \). (b) Since silver has a more positive standard electrode potential, it undergoes reduction while iron(II) undergoes oxidation: \( \text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s}) \). (c) Non-standard potential of \( \text{Ag}^+/\text{Ag} \): \( E = +0.80 + 0.059 \log(0.025) = +0.80 + 0.059 \times (-1.602) = +0.80 - 0.095 = +0.705\text{ V} \). (d) Under these conditions, the \( \text{Fe}^{3+}/\text{Fe}^{2+} \) half-cell has a higher potential (\( +0.77\text{ V} \)) than the \( \text{Ag}^+/\text{Ag} \) half-cell (\( +0.705\text{ V} \)). Therefore, \( \text{Fe}^{3+} \) is reduced and \( \text{Ag} \) is oxidised. \( E_{\text{cell}} = 0.77 - 0.705 = +0.065\text{ V} \). Since the cathode and anode have switched roles, the direction of electron flow in the external circuit is reversed.
PastPaper.markingScheme
Part (a): 1 mark for +0.03 V. Part (b): 2 marks for overall ionic equation (1 mark for correct formulas, 1 mark for correct state symbols). Part (c): 3 marks for Nernst calculation (1 mark for substituting values, 1 mark for log calculation, 1 mark for final value of +0.705 V or +0.71 V). Part (d): 3 marks for cell potential and direction (1 mark for E_cell = +0.065 V, 1 mark for stating that electron flow is reversed, 1 mark for explaining that Fe3+/Fe2+ now has a higher reduction potential).
PastPaper.question 4 · Structured A Level Questions
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The reaction between peroxodisulfate ions, \( \text{S}_2\text{O}_8^{2-} \), and iodide ions, \( \text{I}^- \), is represented by the equation: \( \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \). Rate data at a constant temperature: Exp 1: \( [\text{S}_2\text{O}_8^{2-}] = 0.010 \), \( [\text{I}^-] = 0.020 \), Rate = \( 1.50 \times 10^{-4} \); Exp 2: \( [\text{S}_2\text{O}_8^{2-}] = 0.020 \), \( [\text{I}^-] = 0.020 \), Rate = \( 3.00 \times 10^{-4} \); Exp 3: \( [\text{S}_2\text{O}_8^{2-}] = 0.010 \), \( [\text{I}^-] = 0.040 \), Rate = \( 3.00 \times 10^{-4} \). All concentrations in mol dm\(^{-3}\), rates in mol dm\(^{-3}\) s\(^{-1}\). (a) Determine the order of reaction with respect to \( \text{S}_2\text{O}_8^{2-} \) and \( \text{I}^- \). Explain your reasoning. (b) Write the rate equation. (c) Calculate the rate constant, \( k \), and state its units. (d) Describe how \( \text{Fe}^{2+} \) ions act as a homogeneous catalyst for this reaction, including two chemical equations.
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(a) Comparing Exp 1 and 2: \( [\text{S}_2\text{O}_8^{2-}] \) doubles while \( [\text{I}^-] \) remains constant, and the rate doubles. Thus, the reaction is first order with respect to \( [\text{S}_2\text{O}_8^{2-}] \). Comparing Exp 1 and 3: \( [\text{I}^-] \) doubles while \( [\text{S}_2\text{O}_8^{2-}] \) remains constant, and the rate doubles. Thus, the reaction is first order with respect to \( [\text{I}^-] \). (b) Rate = \( k[\text{S}_2\text{O}_8^{2-}][\text{I}^-] \). (c) Using Exp 1: \( 1.50 \times 10^{-4} = k \times 0.010 \times 0.020 \implies k = 0.75 \). Units: \( \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1} \). (d) \( \text{Fe}^{2+} \) acts as a catalyst because it provides an alternative pathway with lower activation energy. The uncatalysed reaction involves the collision of two negatively charged ions (repulsion), whereas the catalysed pathway involves oppositely charged ions: Step 1: \( \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq}) \). Step 2: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \).
PastPaper.markingScheme
Part (a): 2 marks for explaining first-order with respect to S2O8^2-, 2 marks for explaining first-order with respect to I-. Part (b): 1 mark for correct rate equation. Part (c): 2 marks for calculation of k = 0.75, 1 mark for units (dm3 mol-1 s-1). Part (d): 2 marks for the two correct equations, 1 mark for explaining why catalysis occurs (opposite charge attraction instead of like-charge repulsion).
PastPaper.question 5 · Structured A Level Questions
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Hydrocarbons can be characterised by their molecular formulas. (a) A sample of 0.220 g of a liquid alkane, X, was vaporised at 100 degrees Celsius and \( 1.01 \times 10^5 \text{ Pa} \). The volume of vaporised hydrocarbon X was 78.5 cm\(^3\). (i) Calculate the relative molecular mass, \( M_{\text{r}} \), of X using the ideal gas equation. (ii) Determine the molecular formula of X. (b) A 10.0 cm\(^3\) sample of a gaseous hydrocarbon, Y, of molecular formula \( \text{C}_x\text{H}_y \), was mixed with 80.0 cm\(^3\) of \( \text{O}_2 \) (an excess) and exploded. After cooling to room temperature, the total gas volume was 65.0 cm\(^3\). Passing this mixture through aqueous NaOH reduced the volume to 35.0 cm\(^3\). Determine the molecular formula of Y. Show all working.
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(a)(i) Ideal gas equation: \( pV = nRT = \frac{m}{M_{\text{r}}}RT \implies M_{\text{r}} = \frac{mRT}{pV} \). Here, \( m = 0.220\text{ g} \), \( R = 8.31\text{ J K}^{-1}\text{ mol}^{-1} \), \( T = 373\text{ K} \), \( p = 1.01 \times 10^5\text{ Pa} \), \( V = 7.85 \times 10^{-5}\text{ m}^3 \). \( M_{\text{r}} = \frac{0.220 \times 8.31 \times 373}{1.01 \times 10^5 \times 7.85 \times 10^{-5}} = \frac{681.89}{7.9285} = 86.0\text{ g mol}^{-1} \). (ii) For an alkane, \( 14n + 2 = 86 \implies 14n = 84 \implies n = 6 \). Molecular formula of X is \( \text{C}_6\text{H}_{14} \). (b) Volume of \( \text{CO}_2 \) produced = \( 65.0 - 35.0 = 30.0\text{ cm}^3 \). The remaining 35.0 cm\(^3\) is unreacted \( \text{O}_2 \). Volume of \( \text{O}_2 \) reacted = \( 80.0 - 35.0 = 45.0\text{ cm}^3 \). Since 10.0 cm\(^3\) of Y produced 30.0 cm\(^3\) of \( \text{CO}_2 \), \( x = 3 \). Volume ratio of hydrocarbon to \( \text{O}_2 \) reacted is \( 1 : (x + y/4) \implies 10 : 45 = 1 : 4.5 \implies x + y/4 = 4.5 \implies 3 + y/4 = 4.5 \implies y/4 = 1.5 \implies y = 6 \). Molecular formula of Y is \( \text{C}_3\text{H}_6 \).
PastPaper.markingScheme
Part (a)(i): 2 marks for Mr calculation (1 mark for volume conversion, 1 mark for correct calculation to 86.0). Part (a)(ii): 1 mark for using 14n+2 formula, 1 mark for C6H14. Part (b): 1 mark for CO2 volume = 30.0 cm3, 1 mark for finding x = 3, 1 mark for O2 reacted = 45.0 cm3, 2 marks for finding y = 6, 1 mark for final molecular formula C3H6.
PastPaper.question 6 · Structured A Level Questions
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Phenol, \( \text{C}_6\text{H}_5\text{OH} \), is a weak acid. (a) Arrange the following compounds in order of increasing acidity: ethanol, phenol, water. Explain this trend by comparing the stability of their conjugate bases. (b) State what is observed when phenol reacts with: (i) bromine water at room temperature, (ii) dilute nitric acid at room temperature. (c) Suggest a two-step synthesis to convert phenylamine, \( \text{C}_6\text{H}_5\text{NH}_2 \), into phenol. Include all reagents and conditions.
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(a) Order of increasing acidity: ethanol < water < phenol. Explanation: The ethoxide ion is destabilised by the electron-donating ethyl group (+I effect), which concentrates negative charge on the oxygen. In water, there is no such inductive group. In the phenoxide ion, the lone pair of electrons on the oxygen atom overlaps with the pi-delocalised ring. This delocalisation of negative charge stabilises the phenoxide ion, making it the most stable conjugate base and thus phenol is the strongest acid. (b)(i) White precipitate is formed and the orange/brown bromine water is decolourised. (ii) Solution turns brown/yellow as a mixture of 2-nitrophenol and 4-nitrophenol is formed. (c) Step 1: React phenylamine with nitrous acid (prepared from \( \text{NaNO}_2 \) and dilute HCl) at a temperature between 0 and 10 degrees Celsius to form benzenediazonium chloride. Step 2: Warm the resulting benzenediazonium chloride solution with water to produce phenol and nitrogen gas.
PastPaper.markingScheme
Part (a): 1 mark for correct order of acidity. 2 marks for explaining ethoxide destabilisation (ethyl group +I effect). 2 marks for explaining phenoxide stabilisation (delocalisation of lone pair into pi-ring). Part (b)(i): 2 marks (1 mark for white precipitate, 1 mark for decolourisation of bromine water). Part (b)(ii): 1 mark for brown/yellow solution or nitration product. Part (c): 2 marks for Step 1 reagents and conditions (HNO2/NaNO2 + HCl, temperature < 10 degrees Celsius). 1 mark for Step 2 (warm with water).
PastPaper.question 7 · Structured A Level Questions
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The thermal decomposition of calcium carbonate is represented by: \( \text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \). Standard thermodynamic data at 298 K: \( \text{CaCO}_3(\text{s}) \): \( \Delta H^\ominus_{\text{f}} = -1207\text{ kJ mol}^{-1} \), \( S^\ominus = 93\text{ J K}^{-1}\text{ mol}^{-1} \); \( \text{CaO}(\text{s}) \): \( \Delta H^\ominus_{\text{f}} = -635\text{ kJ mol}^{-1} \), \( S^\ominus = 40\text{ J K}^{-1}\text{ mol}^{-1} \); \( \text{CO}_2(\text{g}) \): \( \Delta H^\ominus_{\text{f}} = -394\text{ kJ mol}^{-1} \), \( S^\ominus = 214\text{ J K}^{-1}\text{ mol}^{-1} \). (a) Explain why the standard entropy change of this reaction is positive. (b) Calculate the standard enthalpy change, \( \Delta H^\ominus_{\text{reaction}} \), in \( \text{kJ mol}^{-1} \). (c) Calculate the standard entropy change, \( \Delta S^\ominus_{\text{reaction}} \), in \( \text{J K}^{-1}\text{ mol}^{-1} \). (d) Determine the minimum temperature, in Kelvin, at which the decomposition of calcium carbonate becomes thermodynamically feasible. State any assumption made in your calculation.
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(a) A solid reactant decomposes to form a solid and a gas. Gases have a much higher degree of disorder and freedom of movement than solids, resulting in an increase in entropy. (b) \( \Delta H^\ominus = [-635 + (-394)] - [-1207] = -1029 + 1207 = +178\text{ kJ mol}^{-1} \). (c) \( \Delta S^\ominus = [40 + 214] - 93 = 254 - 93 = +161\text{ J K}^{-1}\text{ mol}^{-1} \). (d) For the reaction to be feasible, \( \Delta G \le 0 \). At the threshold, \( \Delta G = \Delta H - T\Delta S = 0 \implies T = \frac{\Delta H}{\Delta S} \). Converting \( \Delta H \) to \( \text{J mol}^{-1} \): \( \Delta H = 178 \times 10^3\text{ J mol}^{-1} \). \( T = \frac{178 \times 10^3}{161} = 1105.6\text{ K} \) (approx 1106 K). Assumption: Both \( \Delta H \) and \( \Delta S \) are independent of temperature.
PastPaper.markingScheme
Part (a): 1 mark for identifying the production of gas from a solid. Part (b): 2 marks for +178 kJ mol-1 (1 mark for correct method). Part (c): 2 marks for +161 J K-1 mol-1 (1 mark for correct method). Part (d): 3 marks for temperature calculation (1 mark for Delta G = 0 formula, 1 mark for converting kJ to J, 1 mark for 1106 K or 1110 K). 1 mark for assumption (Delta H and Delta S are independent of temperature).
PastPaper.question 8 · Structured A Level Questions
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Transition metal complexes display a wide range of structures and isomerism. (a) Define the term bidentate ligand, and name a neutral bidentate ligand. (b) The octahedral complex \( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \) (where en is 1,2-diaminoethane) has cis and trans stereoisomers. (i) Describe the structural difference between the cis and trans isomers. (ii) State which of these two isomers is optically active, and explain why. (c) Describe how the d-orbitals of a transition metal ion split in an octahedral complex, and explain how this splitting leads to the complex being coloured.
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(a) A bidentate ligand is a species that donates two lone pairs of electrons to form two coordinate bonds to the central metal ion. Example: 1,2-diaminoethane (en). (b)(i) In the trans-isomer, the two chloride ligands are located opposite to each other (at an angle of 180 degrees). In the cis-isomer, the two chloride ligands are located adjacent to each other (at an angle of 90 degrees). (ii) The cis-isomer is optically active because it lacks a plane of symmetry and is non-superimposable on its mirror image (is chiral). (c) When ligands approach octahedrally, the repulsion between ligand lone pairs and d-orbital electrons causes the d-orbitals to split into two energy levels: three lower-energy d-orbitals and two higher-energy d-orbitals. When visible light is shone on the complex, an electron absorbs a photon of light of a specific frequency and is promoted from a lower-energy d-orbital to a higher-energy d-orbital (d-to-d transition). The remaining non-absorbed wavelengths of visible light are transmitted or reflected, which we perceive as the complementary colour.
PastPaper.markingScheme
Part (a): 2 marks (1 mark for definition of bidentate, 1 mark for 1,2-diaminoethane). Part (b)(i): 2 marks (1 mark for trans being 180 degrees/opposite, 1 mark for cis being 90 degrees/adjacent). Part (b)(ii): 3 marks (1 mark for cis-isomer, 1 mark for explaining lack of symmetry/chiral, 1 mark for stating non-superimposable mirror images). Part (c): 4 marks (1 mark for d-orbitals splitting into two energy levels, 1 mark for three lower and two higher energy orbitals, 1 mark for excitation of electron by absorbing visible light photon / d-to-d transition, 1 mark for light not absorbed being transmitted/reflected as complementary colour).
PastPaper.question 9 · structured
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**Part (a)**
Define the term *standard electrode potential*, \(E^\ominus\). [2]
**Part (b)**
An electrochemical cell is set up consisting of a standard \(\text{Cu}^{2+}(aq)/\text{Cu}(s)\) half-cell and a standard \(\text{Fe}^{3+}(aq)/\text{Fe}^{2+}(aq)\) half-cell.
(i) Write the two reduction half-equations that represent these standard half-cells and state their standard electrode potentials, \(E^\ominus\). [2]
(ii) Deduce the equation for the overall cell reaction when the cell is discharging and calculate the standard cell potential, \(E^\ominus_{\text{cell}}\). [2]
**Part (c)**
The concentration of the species in the iron half-cell are adjusted such that \([\text{Fe}^{3+}(aq)] = 0.85\text{ mol dm}^{-3}\) and \([\text{Fe}^{2+}(aq)] = 0.015\text{ mol dm}^{-3}\).
Calculate the electrode potential, \(E\), of this non-standard iron half-cell at \(298\text{ K}\) using the Nernst equation:
The copper half-cell concentration of \(\text{Cu}^{2+}(aq)\) is now decreased to \(0.050\text{ mol dm}^{-3}\).
Explain, with reference to the \(\text{Cu}^{2+}/\text{Cu}\) half-cell electrode potential and the overall cell system, how this change affects the value of the cell potential, \(E_{\text{cell}}\). [2.1]
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**Part (a)** The standard electrode potential, \(E^\ominus\), is defined as the voltage / potential difference / electromotive force (emf) of a cell consisting of the half-cell of interest connected to a standard hydrogen electrode (SHE) under standard conditions of \(298\text{ K}\), \(1\text{ atm}\) (or \(100\text{ kPa}\)) gas pressure, and all solutions at a concentration of \(1.00\text{ mol dm}^{-3}\).
**Part (b)** (i) The two reduction half-equations and their standard electrode potentials from standard data are: 1. \(\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)\) \(E^\ominus = +0.34\text{ V}\) 2. \(\text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq)\) \(E^\ominus = +0.77\text{ V}\)
(ii) The half-cell with the more positive \(E^\ominus\) value undergoes reduction, while the other is forced to undergo oxidation. Thus, \(\text{Fe}^{3+}\) is reduced and \(\text{Cu}\) is oxidised: Reduction: \(2\text{Fe}^{3+}(aq) + 2e^- \rightarrow 2\text{Fe}^{2+}(aq)\) Oxidation: \(\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-\) Overall equation: \(2\text{Fe}^{3+}(aq) + \text{Cu}(s) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Cu}^{2+}(aq)\)
The standard cell potential is: \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.77\text{ V} - (+0.34\text{ V}) = +0.43\text{ V}\)
**Part (c)** For the iron half-cell, \(\text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq)\), the number of electrons transferred, \(z\), is 1. The oxidised species is \(\text{Fe}^{3+}\) and the reduced species is \(\text{Fe}^{2+}\). Substitute these into the Nernst equation: \(E = +0.77 + \frac{0.059}{1}\log\left(\frac{0.85}{0.015}\right)\) \(E = +0.77 + 0.059\log(56.67)\) \(E = +0.77 + 0.059(1.753) = +0.77 + 0.103 = +0.873\text{ V}\) (or \(+0.87\text{ V}\))
**Part (d)** Decreasing the concentration of \(\text{Cu}^{2+}(aq)\) below the standard value of \(1.0\text{ mol dm}^{-3}\) shifts the position of the equilibrium \(\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)\) to the left (by Le Chatelier's principle). This makes the electrode potential, \(E\), for the copper half-cell less positive (i.e., decreases from \(+0.34\text{ V}\)). Because the overall cell potential is calculated as \(E_{\text{cell}} = E(\text{Fe}^{3+}/\text{Fe}^{2+}) - E(\text{Cu}^{2+}/\text{Cu})\), a smaller value of \(E(\text{Cu}^{2+}/\text{Cu})\) being subtracted will result in an increased overall standard/non-standard cell potential, \(E_{\text{cell}}\).
PastPaper.markingScheme
**Part (a) [2 marks]** - 1 mark: emf / potential difference / voltage of a half-cell connected to a standard hydrogen electrode (SHE). - 1 mark: standard conditions specified: temperature of \(298\text{ K}\) AND pressure of \(1\text{ bar}\) / \(1\text{ atm}\) / \(100\text{ kPa}\) AND solutions at \(1.00\text{ mol dm}^{-3}\).
**Part (b)(i) [2 marks]** - 1 mark: both reduction half-equations correct: \(\text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu}\) and \(\text{Fe}^{3+} + e^- \rightleftharpoons \text{Fe}^{2+}\) (accept single arrow). - 1 mark: both \(E^\ominus\) values correct with signs: \(+0.34\text{ V}\) and \(+0.77\text{ V}\).
**Part (b)(ii) [2 marks]** - 1 mark: overall reaction equation correctly balanced: \(2\text{Fe}^{3+}(aq) + \text{Cu}(s) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Cu}^{2+}(aq)\) (state symbols not strictly required unless requested, but must be chemically correct). - 1 mark: standard cell potential calculation: \(E^\ominus_{\text{cell}} = +0.43\text{ V}\).
**Part (c) [3 marks]** - 1 mark: correctly identifying \(z = 1\). - 1 mark: correct substitution of concentrations into the Nernst equation expression: \(\log(0.85 / 0.015)\). - 1 mark: correct calculation of potential: \(+0.87\text{ V}\) or \(+0.873\text{ V}\) (must show sign and units).
**Part (d) [2.1 marks]** - 1.1 marks: explaining that decreasing \([\text{Cu}^{2+}(aq)]\) shifts the copper equilibrium to the left, which decreases the electrode potential of the copper half-cell (making it less positive). - 1 mark: explaining that since \(E_{\text{cell}} = E(\text{Fe}^{3+}/\text{Fe}^{2+}) - E(\text{Cu}^{2+}/\text{Cu})\), a lower value for the copper half-cell electrode potential results in an increased overall \(E_{\text{cell}}\).
Paper 52
Answer all planning, analysis, and evaluation questions in the spaces provided.
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PastPaper.question 1 · practical
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A student investigates the kinetics of the reaction between peroxodisulfate ions, \( \text{S}_2\text{O}_8^{2-}(\text{aq}) \), and iodide ions, \( \text{I}^-(\text{aq}) \): \( \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \). This reaction is studied using a clock reaction method. A small, constant amount of sodium thiosulfate, \( \text{Na}_2\text{S}_2\text{O}_3 \), and starch indicator are added to the reaction mixture. The time, \( t \), taken for the mixture to turn blue-black is recorded. Five experiments are carried out at 298 K. In each experiment, the volume of \( 0.100 \text{ mol dm}^{-3} \) \( \text{K}_2\text{S}_2\text{O}_8 \) is kept constant at \( 20.0 \text{ cm}^3 \), and the volume of \( 0.010 \text{ mol dm}^{-3} \) \( \text{Na}_2\text{S}_2\text{O}_3 \) is kept constant at \( 5.0 \text{ cm}^3 \). The volume of \( 0.200 \text{ mol dm}^{-3} \) \( \text{KI} \), \( x \), is varied, and distilled water is added to keep the total volume of \( \text{KI} \) and water constant at \( 25.0 \text{ cm}^3 \). Starch indicator (\( 1.0 \text{ cm}^3 \)) is added to each mixture, making the total volume of the mixture \( 51.0 \text{ cm}^3 \) in all cases. (a) (i) Explain why the total volume of the reaction mixture must be kept constant at \( 51.0 \text{ cm}^3 \). (ii) Calculate the concentration of \( \text{I}^-(\text{aq}) \) in the reaction mixture for Experiment 2, where \( x = 10.0 \text{ cm}^3 \). (b) The times, \( t \), recorded for the five experiments are shown below: Experiment 1: \( x = 5.0 \text{ cm}^3 \), \( t = 120.0 \text{ s} \); Experiment 2: \( x = 10.0 \text{ cm}^3 \), \( t = 60.0 \text{ s} \); Experiment 3: \( x = 15.0 \text{ cm}^3 \), \( t = 50.0 \text{ s} \); Experiment 4: \( x = 20.0 \text{ cm}^3 \), \( t = 30.0 \text{ s} \); Experiment 5: \( x = 25.0 \text{ cm}^3 \), \( t = 24.0 \text{ s} \). (i) Calculate the value of \( 1/t \) for each of the five experiments. Identify which experiment gave an anomalous result and suggest a reason for this anomaly. (ii) State and explain how the student should treat this anomalous result when analyzing the data graphically. (iii) Use the relationship between \( 1/t \) and \( x \) to deduce the order of reaction with respect to \( \text{I}^-(\text{aq}) \). (iv) Calculate the percentage uncertainty in the time measurement for Experiment 5, given that the uncertainty of the stopwatch is \( \pm 0.2 \text{ s} \). (c) Describe how the student could modify this experimental procedure to determine the activation energy, \( E_a \), of the reaction. State the measurements that would be recorded and how the data would be processed graphically.
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Part (a)(i): The total volume must be kept constant so that the volume of KI added (x) is directly proportional to its concentration in the final reaction mixture. Part (a)(ii): Moles of I^- = 0.200 mol dm^-3 * 0.0100 dm^3 = 2.00 * 10^-3 mol. Total volume = 51.0 cm^3 = 0.0510 dm^3. [I^-] = 2.00 * 10^-3 mol / 0.0510 dm^3 = 0.0392 mol dm^-3. Part (b)(i): Values of 1/t (s^-1) are: Exp 1: 0.00833, Exp 2: 0.0167, Exp 3: 0.0200, Exp 4: 0.0333, Exp 5: 0.0417. Experiment 3 is anomalous because its rate (0.0200 s^-1) is lower than expected (expected rate is 0.0250 s^-1 for x = 15.0 cm^3). Possible reason: the student added too much water, lowering the concentration of KI, or started the timer late, or some thiosulfate was spilled. Part (b)(ii): The student should ignore the anomalous point when drawing the line of best fit and when making calculations. Part (b)(iii): Since 1/t is directly proportional to x (and thus to [I^-]), as doubling x from 10.0 to 20.0 cm^3 doubles the rate (1/t) from 0.0167 to 0.0333 s^-1, the reaction is first-order with respect to I^-. Part (b)(iv): Percentage uncertainty = (0.2 / 24.0) * 100% = 0.83% (or if considering start/stop uncertainties as +/- 0.4 s, (0.4 / 24.0) * 100% = 1.67%). Part (c): 1. Prepare the reaction mixtures at five different temperatures (e.g., in the range 20 - 60 degrees C) using a thermostatically controlled water bath. 2. Keep the concentration of all reactants constant across all trials. 3. Record the temperature T (in Kelvin) and the time t taken for the blue-black color to appear. 4. Calculate ln(1/t) as a measure of ln(k). 5. Plot ln(1/t) against 1/T. The gradient of this graph is equal to -Ea/R, allowing Ea to be calculated.
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Part (a)(i): [1 mark] State that keeping total volume constant ensures that the concentration of I^-(aq) is directly proportional to the volume of KI(aq) added. Part (a)(ii): [2 marks] Moles of KI = 0.00200 mol [1 mark], Concentration = 0.0392 mol dm^-3 [1 mark] (accept 0.039). Part (b)(i): [2 marks] All 1/t values calculated correctly to 3 sig figs [1 mark]. State Experiment 3 is anomalous and suggest a valid source of experimental error (e.g., incorrect volume of water/KI, temperature fluctuation, delayed starting of stopwatch) [1 mark]. Part (b)(ii): [1 mark] State that the anomaly must be excluded/ignored when drawing the line of best fit. Part (b)(iii): [1 mark] Identify that rate is directly proportional to volume of KI (or concentration of I^-) and therefore first-order. Part (b)(iv): [1 mark] Calculate percentage uncertainty correctly: 0.83% (accept 1.67% if 0.4 s total uncertainty is used). Part (c): [5 marks] 1. Use of a water bath to vary and maintain constant temperature [1 mark]. 2. Carry out the reaction at a minimum of 5 different temperatures [1 mark]. 3. State that concentrations must be kept constant [1 mark]. 4. Measure temperature (in K) and time t [1 mark]. 5. Plot ln(1/t) vs 1/T and state that gradient = -Ea/R [1 mark].
PastPaper.question 2 · practical
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An electrochemical cell is set up to investigate the Nernst equation for the \( \text{Cu}^{2+}(\text{aq})/\text{Cu}(\text{s}) \) half-cell. The cell is represented as: \( \text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}, 1.00 \text{ mol dm}^{-3}) \parallel \text{Cu}^{2+}(\text{aq}, C \text{ mol dm}^{-3}) \mid \text{Cu}(\text{s}) \). The standard cell potential, \( E^{\ominus}_{\text{cell}} \), is \( 1.10 \text{ V} \) at 298 K. The cell potential, \( E_{\text{cell}} \), is measured for various concentrations, \( C \), of \( \text{Cu}^{2+}(\text{aq}) \) at 298 K. (a) (i) Draw a fully labelled diagram of the experimental setup used to measure \( E_{\text{cell}} \). (ii) State the role of the salt bridge and explain why potassium nitrate, \( \text{KNO}_3 \), is a suitable electrolyte for it. (b) Describe how a student would prepare \( 100 \text{ cm}^3 \) of \( 0.0100 \text{ mol dm}^{-3} \) \( \text{CuSO}_4(\text{aq}) \) from a stock solution of \( 1.00 \text{ mol dm}^{-3} \) \( \text{CuSO}_4(\text{aq}) \). State the names and capacities of the volumetric apparatus required. (c) The table below shows the results of the experiment: \( [\text{Cu}^{2+}] \) in \( \text{mol dm}^{-3} \): \( 1.00, 0.100, 0.0100, 0.00100, 0.000100 \). Measured \( E_{\text{cell}} \) in \( \text{V} \): \( 1.10, 1.07, 1.04, 1.01, 0.98 \). (i) Calculate the value of \( \log[\text{Cu}^{2+}] \) for each concentration and describe the trend in \( E_{\text{cell}} \) as \( \log[\text{Cu}^{2+}] \) decreases. (ii) Determine the gradient of the graph of \( E_{\text{cell}} \) against \( \log[\text{Cu}^{2+}] \). Show your working. (iii) The theoretical gradient of this graph is equal to \( \frac{2.303RT}{zF} \), where \( R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \), \( T = 298 \text{ K} \), \( F = 96500 \text{ C mol}^{-1} \), and \( z \) is the number of electrons transferred per copper ion. Use your calculated gradient to determine the experimental value of \( z \) to two significant figures. (d) Predict how the measured cell potential, \( E_{\text{cell}} \), would change, if at all, under the following conditions. Explain your predictions. (i) The concentration of \( \text{Zn}^{2+}(\text{aq}) \) in the zinc half-cell is increased to \( 2.00 \text{ mol dm}^{-3} \), while \( [\text{Cu}^{2+}(\text{aq})] \) is kept at \( 1.00 \text{ mol dm}^{-3} \). (ii) The copper electrode in the copper half-cell is replaced with a copper sheet of twice the surface area.
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Part (a)(i): The diagram must include: a beaker with a zinc electrode in 1.00 mol dm^-3 Zn^2+(aq) solution; a beaker with a copper electrode in Cu^2+(aq) solution; a salt bridge connecting the two solutions; a high-resistance voltmeter connected in series between the zinc and copper electrodes with wires. Part (a)(ii): The salt bridge completes the electrical circuit and maintains electrical neutrality in the half-cells by allowing ion flow. KNO3 is suitable because potassium and nitrate ions do not react or form precipitates with Zn^2+, Cu^2+, SO4^2-, or any other ions in the cell. Part (b): Volume of stock solution required: C1V1 = C2V2 => 1.00 * V1 = 0.0100 * 100 => V1 = 1.00 cm^3. Use a 1.0 cm^3 pipette to measure 1.00 cm^3 of 1.00 mol dm^-3 CuSO4 stock solution. Transfer it to a 100 cm^3 volumetric flask. Fill the flask with distilled water up to the graduation mark, insert the stopper, and invert several times to mix thoroughly. Part (c)(i): log[Cu^2+] values: 0.00, -1.00, -2.00, -3.00, -4.00. As log[Cu^2+] decreases, E_cell decreases linearly. Part (c)(ii): Gradient = (E2 - E1) / (log[Cu^2+]2 - log[Cu^2+]1) = (1.10 - 0.98) / (0.00 - (-4.00)) = 0.12 / 4.00 = 0.030 V. Part (c)(iii): Theoretical gradient = 2.303 * 8.31 * 298 / (z * 96500) = 0.0591 / z. Setting experimental gradient equal to theoretical: 0.030 = 0.0591 / z => z = 0.0591 / 0.030 = 1.97, which rounds to 2.0 (to 2 significant figures). Part (d)(i): The cell potential E_cell will decrease. According to Le Chatelier's principle, increasing the concentration of the product ion Zn^2+ in the cell reaction Zn(s) + Cu^2+(aq) <=> Zn^2+(aq) + Cu(s) shifts the equilibrium to the left, decreasing the driving force of the reaction. Part (d)(ii): No change in E_cell because copper is a solid reactant, and the surface area of a solid does not affect the position of chemical equilibrium or the thermodynamic cell potential.
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Part (a)(i): [3 marks] 1. Correctly drawn and labelled half-cells (Zn in Zn^2+, Cu in Cu^2+) [1 mark]. 2. Salt bridge correctly positioned and labelled [1 mark]. 3. High-resistance voltmeter connected across the electrodes [1 mark]. Part (a)(ii): [1 mark] Explains that KNO3 does not react with or precipitate ions in the half-cells. Part (b): [3 marks] 1. Calculate the volume of stock solution as 1.00 cm^3 [1 mark]. 2. Mention using a 1.0 cm^3 pipette and transferring to a 100 cm^3 volumetric flask [1 mark]. 3. Describe filling to the graduation mark with distilled water and inverting to mix [1 mark]. Part (c)(i): [1 mark] Correctly calculate log values (0.00, -1.00, -2.00, -3.00, -4.00) and state that E_cell decreases linearly as log[Cu^2+] decreases. Part (c)(ii): [2 marks] Gradient calculation showing working: change in y / change in x = 0.030 V [2 marks] (1 mark for correct working, 1 mark for correct value with units). Part (c)(iii): [2 marks] Rearrange formula to z = 0.0591 / gradient [1 mark], calculate z = 2.0 (must be 2 significant figures) [1 mark]. Part (d)(i): [2 marks] State that E_cell decreases [1 mark] and explain using Le Chatelier's principle (shift to the left) or the Nernst equation (increase in Q reduces E) [1 mark]. Part (d)(ii): [1 mark] State that E_cell remains unchanged because solids do not affect the reaction quotient / concentration is constant.