An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 1
Answer all forty multiple choice questions. For each question there are four possible answers. Choose the single best option.
33 PastPaper.question · 33 PastPaper.marks
PastPaper.question 1 · multiple-choice
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How many stereoisomers exist for the organic compound 4-methylhex-2-ene?
A.2
B.3
C.4
D.8
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PastPaper.workedSolution
4-methylhex-2-ene has the structure \(CH_3-CH=CH-CH(CH_3)-CH_2-CH_3\). It contains: 1) A double bond at carbon-2 that exhibits cis/trans (E/Z) isomerism because both carbons of the double bond are attached to two different groups (\(-H\) and \(-CH_3\) on C2; \(-H\) and \(-CH(CH_3)CH_2CH_3\) on C3). 2) A chiral carbon at C4, as it is bonded to four different groups: \(-H\), \(-CH_3\), \(-CH_2CH_3\), and \(-CH=CHCH_3\). Therefore, the total number of stereoisomers is \(2^2 = 4\) (the (E,R), (E,S), (Z,R), and (Z,S) configurations).
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) and \(25.0\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\) sodium propanoate. What is the pH of this buffer solution?
1 mark: Correctly calculating the pH as 4.65 (Option A).
PastPaper.question 3 · multiple-choice
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For the reaction \(2A + B + C \rightarrow D\), the following initial rates were measured at different starting concentrations: Experiment 1: \([A] = 0.10\text{ mol dm}^{-3}\), \([B] = 0.10\text{ mol dm}^{-3}\), \([C] = 0.10\text{ mol dm}^{-3}\), Initial Rate = \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\); Experiment 2: \([A] = 0.20\text{ mol dm}^{-3}\), \([B] = 0.10\text{ mol dm}^{-3}\), \([C] = 0.10\text{ mol dm}^{-3}\), Initial Rate = \(2.4 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\); Experiment 3: \([A] = 0.10\text{ mol dm}^{-3}\), \([B] = 0.20\text{ mol dm}^{-3}\), \([C] = 0.10\text{ mol dm}^{-3}\), Initial Rate = \(4.8 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\); Experiment 4: \([A] = 0.10\text{ mol dm}^{-3}\), \([B] = 0.10\text{ mol dm}^{-3}\), \([C] = 0.20\text{ mol dm}^{-3}\), Initial Rate = \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the unit of the rate constant, \(k\)?
A.\(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
B.\(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
C.\(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)
D.\(\text{s}^{-1}\)
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PastPaper.workedSolution
Comparing Expt 1 and Expt 2: doubling \([A]\) doubles the rate, so the reaction is 1st order with respect to \(A\). Comparing Expt 1 and Expt 3: doubling \([B]\) quadruples the rate, so the reaction is 2nd order with respect to \(B\). Comparing Expt 1 and Expt 4: doubling \([C]\) has no effect on the rate, so the reaction is 0th order with respect to \(C\). The rate equation is \(\text{Rate} = k[A][B]^2\). Hence, \(k = \frac{\text{Rate}}{[A][B]^2}\). Unit of \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
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1 mark: Correctly determining the overall order is 3 and rate constant unit is dm^6 mol^-2 s^-1 (Option B).
PastPaper.question 4 · multiple-choice
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Calculate the standard enthalpy of formation of liquid hydrazine, \(N_2H_4(l)\), given the following standard enthalpy of combustion data: \(\Delta H_c^\theta[H_2(g)] = -286\text{ kJ mol}^{-1}\), \(\Delta H_c^\theta[N_2H_4(l)] = -622\text{ kJ mol}^{-1}\).
A.\(-908\text{ kJ mol}^{-1}\)
B.\(-50\text{ kJ mol}^{-1}\)
C.\(+50\text{ kJ mol}^{-1}\)
D.\(+336\text{ kJ mol}^{-1}\)
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PastPaper.workedSolution
The formation reaction of hydrazine is: \(N_2(g) + 2H_2(g) \rightarrow N_2H_4(l)\). Using Hess's law and standard enthalpies of combustion: \(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{Reactants}) - \sum \Delta H_c^\theta(\text{Products})\). The combustion of reactants is \(2 \times \Delta H_c^\theta[H_2(g)]\) (since nitrogen does not combust). Thus, \(\Delta H_f^\theta = 2(-286) - (-622) = -572 + 622 = +50\text{ kJ mol}^{-1}\).
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1 mark: Correctly calculating the enthalpy of formation as +50 kJ mol^-1 (Option C).
PastPaper.question 5 · multiple-choice
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Which set of reagents and reaction conditions is required to synthesize 4-hydroxyazobenzene starting from phenylamine?
A.1. Heat with concentrated \(HNO_3\) and concentrated \(H_2SO_4\) at \(55\ ^\circ\text{C}\); 2. Add phenol.
B.1. Add \(NaNO_2\) and dilute \(HCl\) at \(5\ ^\circ\text{C}\); 2. Add phenol dissolved in aqueous \(NaOH\).
C.1. Add \(NaNO_2\) and dilute \(HCl\) at \(5\ ^\circ\text{C}\); 2. Add phenol in concentrated \(H_2SO_4\).
D.1. Heat with \(HNO_2\) at \(80\ ^\circ\text{C}\); 2. Add chlorobenzene in alkaline solution.
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PastPaper.workedSolution
To convert phenylamine to 4-hydroxyazobenzene, the phenylamine is first diazotised using nitrous acid (generated in situ from sodium nitrite and dilute hydrochloric acid) below \(10\ ^\circ\text{C}\) (typically \(5\ ^\circ\text{C}\)) to form the benzenediazonium ion. This diazonium ion is then coupled with phenol. The coupling reaction requires alkaline conditions (phenol dissolved in aqueous \(NaOH\)) because the phenoxide ion formed is a much stronger nucleophile than neutral phenol and readily undergoes electrophilic substitution.
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1 mark: Selecting the correct reagents and conditions for diazotisation and coupling (Option B).
PastPaper.question 6 · multiple-choice
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Which statement correctly describes and explains a trend in the melting points of the elements across Period 3 (from sodium to argon)?
A.Silicon has the highest melting point because it forms a giant covalent lattice with strong covalent bonds throughout the structure.
B.Sulfur has a lower melting point than phosphorus because sulfur exists as smaller simple molecular structures.
C.Sodium has a higher melting point than magnesium because sodium has a larger cationic radius.
D.Argon has a higher melting point than chlorine because argon exists as monatomic elements with stronger instantaneous dipole-induced dipole forces.
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PastPaper.workedSolution
Across Period 3, silicon has the highest melting point due to its giant covalent (macromolecular) structure, which requires a large amount of energy to break the strong, localized covalent bonds. Sulfur (\(S_8\)) has a higher melting point than phosphorus (\(P_4\)) because it exists as larger molecules with more electrons, resulting in stronger van der Waals forces. Magnesium has a higher melting point than sodium because it has a higher charge density (\(2+\) vs \(1+\)) and more delocalised electrons. Chlorine (\(Cl_2\)) has a higher melting point than argon (\(Ar\)) because \(Cl_2\) is larger and has stronger van der Waals forces.
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1 mark: Correctly identifying the reason for silicon's high melting point (Option A).
PastPaper.question 7 · multiple-choice
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A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, X, is completely combusted in \(80\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature, the remaining gas volume is \(65\text{ cm}^3\). When this gas mixture is passed through aqueous sodium hydroxide, the volume decreases to \(25\text{ cm}^3\). What is the molecular formula of hydrocarbon X?
A.\(C_4H_{10}\)
B.\(C_4H_8\)
C.\(C_4H_6\)
D.\(C_3H_8\)
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PastPaper.workedSolution
Passing the final mixture through aqueous \(NaOH\) absorbs \(CO_2\). The decrease in volume is \(65 - 25 = 40\text{ cm}^3\), which is the volume of \(CO_2\) produced. Thus, \(10\text{ cm}^3\) of \(C_xH_y\) produces \(40\text{ cm}^3\) of \(CO_2\), meaning \(x = 4\). The remaining \(25\text{ cm}^3\) of gas is unreacted \(O_2\). The volume of \(O_2\) reacted is \(80 - 25 = 55\text{ cm}^3\). The combustion equation is \(C_4H_y + (4 + y/4)O_2 \rightarrow 4CO_2 + (y/2)H_2O\). Since \(10\text{ cm}^3\) of X reacts with \(55\text{ cm}^3\) of \(O_2\), \(4 + y/4 = 5.5\), which gives \(y/4 = 1.5\) and thus \(y = 6\). The formula of X is \(C_4H_6\).
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1 mark: Correctly determining the formula is C4H6 (Option C).
PastPaper.question 8 · multiple-choice
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Using the standard electrode potentials below, which species can be oxidised by aqueous bromine, \(Br_2(aq)\), under standard conditions, but cannot be oxidised by aqueous iodine, \(I_2(aq)\)? \(I_2(aq) + 2e^- \rightleftharpoons 2I^-(aq) \quad E^\theta = +0.54\text{ V}\); \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq) \quad E^\theta = +0.77\text{ V}\); \(Br_2(aq) + 2e^- \rightleftharpoons 2Br^-(aq) \quad E^\theta = +1.07\text{ V}\).
A.\(Fe^{3+}(aq)\)
B.\(Fe^{2+}(aq)\)
C.\(Br^-(aq)\)
D.\(I^-(aq)\)
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PastPaper.workedSolution
For an oxidation reaction to be feasible under standard conditions, the standard cell potential (\(E^\theta_{\text{cell}}\)) must be positive (\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} > 0\)). This means the oxidising agent must have a more positive \(E^\theta\) than the species being oxidised. For bromine to oxidise a species, the species' reduction potential must be less than \(+1.07\text{ V}\). For iodine to be unable to oxidise the species, the species' reduction potential must be greater than \(+0.54\text{ V}\). The \(Fe^{3+}/Fe^{2+}\) half-cell has \(E^\theta = +0.77\text{ V}\), which lies between \(+0.54\text{ V}\) and \(+1.07\text{ V}\). Thus, \(Fe^{2+}(aq)\) can be oxidised by \(Br_2(aq)\) (\(E^\theta_{\text{cell}} = 1.07 - 0.77 = +0.30\text{ V}\)) but not by \(I_2(aq)\) (\(E^\theta_{\text{cell}} = 0.54 - 0.77 = -0.23\text{ V}\)).
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1 mark: Correctly identifying Fe2+(aq) as the species that fits the potential limits (Option B).
PastPaper.question 9 · multiple-choice
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A mixture containing 1.50 g of magnesium carbonate and 1.50 g of calcium carbonate is heated strongly until no further change occurs and both carbonates are completely decomposed. What is the total volume of carbon dioxide gas, in \(cm^3\), released when measured at room temperature and pressure (r.t.p.)? [Take the molar volume of a gas at r.t.p. as 24.0 \(dm^3\) \(mol^{-1}\); Ar: Mg = 24.3, Ca = 40.1, C = 12.0, O = 16.0]
A.360 \(cm^3\)
B.427 \(cm^3\)
C.787 \(cm^3\)
D.1574 \(cm^3\)
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PastPaper.workedSolution
Molar mass of \(\text{MgCO}_3\) is 84.3 g \(mol^{-1}\) and \(\text{CaCO}_3\) is 100.1 g \(mol^{-1}\). Moles of \(\text{MgCO}_3 = 1.50 / 84.3 = 0.0178\text{ mol}\). Moles of \(\text{CaCO}_3 = 1.50 / 100.1 = 0.0150\text{ mol}\). Since 1 mole of each carbonate yields 1 mole of \(\text{CO}_2\), the total moles of \(\text{CO}_2\) produced is 0.0178 + 0.0150 = 0.0328 mol. Volume of \(\text{CO}_2\) at r.t.p. = \(0.0328 \times 24000\text{ cm}^3 = 787\text{ cm}^3\).
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1 mark for the correct calculation of total volume of carbon dioxide gas.
PastPaper.question 10 · multiple-choice
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How many stereoisomers exist for the organic compound 4-chlorohex-2-ene?
A.2
B.3
C.4
D.8
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PastPaper.workedSolution
In 4-chlorohex-2-ene, the C2=C3 double bond displays cis-trans (geometric) isomerism because each carbon in the double bond is attached to two different groups. Carbon-4 is a chiral centre because it is bonded to four different groups: -H, -Cl, -CH=CHCH3, and -CH2CH3. This gives rise to optical isomerism. The total number of stereoisomers is \(2^2 = 4\) (cis-R, cis-S, trans-R, and trans-S).
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1 mark for identifying both geometric and optical isomerism and calculating 4 stereoisomers.
PastPaper.question 11 · multiple-choice
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The gas-phase reaction \(\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}\) is allowed to reach equilibrium in a closed vessel at a constant temperature T. At equilibrium, the total pressure is \(2.00 \times 10^5\text{ Pa}\) and the mole fraction of \(\text{Cl}_2\) is 0.30. What is the value of the equilibrium constant \(K_p\) at this temperature?
A.\(4.50 \times 10^4\text{ Pa}\)
B.\(9.00 \times 10^4\text{ Pa}\)
C.\(1.50 \times 10^5\text{ Pa}\)
D.\(3.00 \times 10^5\text{ Pa}\)
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PastPaper.workedSolution
Let the initial amount of \(\text{PCl}_5\) be 1.00 mol. At equilibrium, let \(x\) mol of \(\text{PCl}_5\) decompose. Moles of \(\text{PCl}_5 = 1 - x\), \(\text{PCl}_3 = x\), and \(\text{Cl}_2 = x\). Total moles = \(1 + x\). The mole fraction of \(\text{Cl}_2 = x / (1 + x) = 0.30\), which gives \(x = 0.4286\). The mole fractions at equilibrium are: \(\text{PCl}_5 = 0.40\), \(\text{PCl}_3 = 0.30\), and \(\text{Cl}_2 = 0.30\). The partial pressures are: \(p(\text{PCl}_5) = 0.40 \times 2.00 \times 10^5 = 0.80 \times 10^5\text{ Pa}\), \(p(\text{PCl}_3) = 0.60 \times 10^5\text{ Pa}\), and \(p(\text{Cl}_2) = 0.60 \times 10^5\text{ Pa}\). Thus, \(K_p = (0.60 \times 10^5 \times 0.60 \times 10^5) / (0.80 \times 10^5) = 4.50 \times 10^4\text{ Pa}\).
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1 mark for correctly determining the partial pressures and calculating Kp.
PastPaper.question 12 · multiple-choice
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The kinetics of the reaction \(2\text{NO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(g)}\) was investigated at a constant temperature, and the following initial rate data was obtained. Expt 1: \([\text{NO}] = 0.10\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.10\text{ mol dm}^{-3}\), Initial rate = \(1.20 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). Expt 2: \([\text{NO}] = 0.20\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.10\text{ mol dm}^{-3}\), Initial rate = \(4.80 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). Expt 3: \([\text{NO}] = 0.10\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.20\text{ mol dm}^{-3}\), Initial rate = \(2.40 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the value and units of the rate constant, k, for this reaction?
Comparing Expt 1 and Expt 2: [NO] doubles while [H2] remains constant, and the rate quadruples (from 1.20 x 10^-3 to 4.80 x 10^-3). Thus, the reaction is second order with respect to NO. Comparing Expt 1 and Expt 3: [H2] doubles while [NO] remains constant, and the rate doubles (from 1.20 x 10^-3 to 2.40 x 10^-3). Thus, the reaction is first order with respect to H2. The rate equation is: \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\). Substituting values from Expt 1: \(1.20 \times 10^{-3} = k (0.10)^2 (0.10)\), which gives \(k = 1.20\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
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1 mark for determining the correct rate equation, calculating the value of k, and deriving the correct units.
PastPaper.question 13 · multiple-choice
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The standard enthalpy change of combustion of liquid hydrazine, \(\text{N}_2\text{H}_4\text{(l)}\), is \(-622\text{ kJ mol}^{-1}\), forming nitrogen gas and liquid water. The standard enthalpy change of formation of liquid water is \(-286\text{ kJ mol}^{-1}\). What is the standard enthalpy change of formation of liquid hydrazine?
A.\(-908\text{ kJ mol}^{-1}\)
B.\(-50\text{ kJ mol}^{-1}\)
C.\(+50\text{ kJ mol}^{-1}\)
D.\(+336\text{ kJ mol}^{-1}\)
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PastPaper.workedSolution
Using Hess's Law: \(\text{N}_2\text{H}_4\text{(l)} + \text{O}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(l)} \quad \Delta H_c^\theta = -622\text{ kJ mol}^{-1}\). The enthalpy change of combustion is given by: \(\Delta H_c^\theta = 2\Delta H_f^\theta(\text{H}_2\text{O(l)}) - \Delta H_f^\theta(\text{N}_2\text{H}_4\text{(l)})\). Therefore, \(-622 = 2(-286) - \Delta H_f^\theta(\text{N}_2\text{H}_4\text{(l)})\). This simplifies to \(-622 = -572 - \Delta H_f^\theta(\text{N}_2\text{H}_4\text{(l)})\), so \(\Delta H_f^\theta(\text{N}_2\text{H}_4\text{(l)}) = -572 - (-622) = +50\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark for the correct application of Hess's Law and calculation of the enthalpy of formation.
PastPaper.question 14 · multiple-choice
1 PastPaper.marks
In the synthesis of an azo dye, phenylamine is first converted to the benzenediazonium ion in Step 1, which is then coupled with phenol in Step 2. Which set of reagents and conditions is correct for these two steps?
A.Step 1: NaNO2(aq) and HCl(aq) at 5 °C; Step 2: phenol in NaOH(aq) at 5 °C
B.Step 1: NaNO2(aq) and HCl(aq) at 5 °C; Step 2: phenol in HCl(aq) at 5 °C
C.Step 1: HNO3(aq) and H2SO4(aq) at 55 °C; Step 2: phenol in NaOH(aq) at 5 °C
D.Step 1: NaNO2(aq) and HCl(aq) at 50 °C; Step 2: phenol in NaOH(aq) at 50 °C
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PastPaper.workedSolution
Step 1 involves diazotisation of phenylamine using nitrous acid (generated in situ from sodium nitrite and hydrochloric acid) at a low temperature (below 10 °C, typically 0-5 °C) to prevent the diazonium salt from decomposing. Step 2 is the coupling reaction of the diazonium salt with phenol dissolved in sodium hydroxide (forming the phenoxide ion which is more reactive towards electrophilic substitution) at a low temperature.
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1 mark for selecting the correct reagents and temperatures for both steps.
PastPaper.question 15 · multiple-choice
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Which of the following lists the species \(\text{Na}^+\), \(\text{Mg}^{2+}\), \(\text{P}^{3-}\), and \(\text{S}^{2-}\) in order of decreasing ionic radius?
All four ions are isoelectronic with either neon (10 electrons) or argon (18 electrons). The anions \(\text{P}^{3-}\) and \(\text{S}^{2-}\) have 18 electrons (outermost electrons in shell n=3) and are significantly larger than the cations \(\text{Na}^+\) and \(\text{Mg}^{2+}\) which have 10 electrons (outermost electrons in shell n=2). Within the anions, both have 18 electrons, but phosphorus has fewer protons (15) than sulfur (16), resulting in a weaker nuclear pull and a larger ionic radius: \(\text{P}^{3-} > \text{S}^{2-}\). Within the cations, both have 10 electrons, but sodium has fewer protons (11) than magnesium (12), so \(\text{Na}^+ > \text{Mg}^{2+}\). The complete order of decreasing radius is \(\text{P}^{3-} > \text{S}^{2-} > \text{Na}^+ > \text{Mg}^{2+}\).
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1 mark for correctly ordering the ions by decreasing radius based on shell number and nuclear charge.
PastPaper.question 16 · multiple-choice
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Using the standard electrode potential values provided, which reaction is thermodynamically feasible under standard conditions? \(\text{Fe}^{3+}\text{(aq)} + \text{e}^- \rightleftharpoons \text{Fe}^{2+}\text{(aq)} \quad (E^\theta = +0.77\text{ V})\); \(\text{I}_2\text{(aq)} + 2\text{e}^- \rightleftharpoons 2\text{I}^-\text{(aq)} \quad (E^\theta = +0.54\text{ V})\); \(\text{Br}_2\text{(aq)} + 2\text{e}^- \rightleftharpoons 2\text{Br}^-\text{(aq)} \quad (E^\theta = +1.07\text{ V})\); \(\text{S}_4\text{O}_6^{2-}\text{(aq)} + 2\text{e}^- \rightleftharpoons 2\text{S}_2\text{O}_3^{2-}\text{(aq)} \quad (E^\theta = +0.09\text{ V})\)
A.Iron(III) ions oxidising bromide ions to bromine.
B.Iodine oxidising thiosulfate ions to tetrathionate ions.
C.Iron(II) ions reducing iodine to iodide.
D.Bromide ions reducing tetrathionate ions to thiosulfate ions.
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PastPaper.workedSolution
For a redox reaction to be thermodynamically feasible under standard conditions, the cell potential, \(E_{\text{cell}}^\theta = E_{\text{reduction}}^\theta - E_{\text{oxidation}}^\theta\), must be positive. For Option B: Iodine is reduced to iodide (\(E^\theta = +0.54\text{ V}\)) and thiosulfate is oxidised to tetrathionate (\(E^\theta = +0.09\text{ V}\)). \(E_{\text{cell}}^\theta = +0.54 - (+0.09) = +0.45\text{ V}\), which is positive and feasible. For the other options, the cell potentials are negative: Option A (\(E_{\text{cell}}^\theta = +0.77 - 1.07 = -0.30\text{ V}\)), Option C (\(E_{\text{cell}}^\theta = +0.54 - 0.77 = -0.23\text{ V}\)), and Option D (\(E_{\text{cell}}^\theta = +0.09 - 1.07 = -0.98\text{ V}\)).
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1 mark for identifying the reaction with a positive cell potential as thermodynamically feasible.
PastPaper.question 17 · multiple-choice
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A sample of a bioactive peptide is hydrolysed to yield an amino acid with the molecular formula \(\text{C}_4\text{H}_9\text{NO}_3\). This amino acid has the IUPAC name 2-amino-3-hydroxybutanoic acid. How many chiral centres are present in one molecule of this amino acid?
A.1
B.2
C.3
D.4
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PastPaper.workedSolution
The structure of 2-amino-3-hydroxybutanoic acid is \(\text{CH}_3\text{-CH(OH)-CH(NH}_2\text{)-COOH}\). Let us identify the carbon atoms: Carbon 1 is the carboxylic carbon (\(\text{-COOH}\)), which is not chiral. Carbon 2 is attached to four different groups: \(\text{-H}\), \(\text{-NH}_2\), \(\text{-COOH}\), and \(\text{-CH(OH)CH}_3\). Thus, Carbon 2 is a chiral centre. Carbon 3 is attached to four different groups: \(\text{-H}\), \(\text{-OH}\), \(\text{-CH}_3\), and \(\text{-CH(NH}_2\text{)COOH}\). Thus, Carbon 3 is also a chiral centre. Carbon 4 is a methyl group (\(\text{-CH}_3\)), which is not chiral. Therefore, there are exactly 2 chiral centres.
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1 mark for the correct option B.
PastPaper.question 18 · multiple-choice
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How many stereoisomers exist for the compound 4-chlorohex-2-ene?
A.2
B.4
C.6
D.8
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PastPaper.workedSolution
4-chlorohex-2-ene has the structure \(\text{CH}_3\text{-CH=CH-CH(Cl)-CH}_2\text{-CH}_3\). It exhibits both geometric (cis-trans) isomerism and optical isomerism: 1. The carbon-carbon double bond at C2-C3 exhibits geometric isomerism because each carbon of the double bond is attached to two different groups. 2. Carbon-4 is a chiral centre because it is bonded to four different groups: \(\text{-H}\), \(\text{-Cl}\), \(\text{-CH}_2\text{CH}_3\), and \(\text{-CH=CHCH}_3\). Therefore, there are \(2 \times 2 = 4\) stereoisomers (cis-R, cis-S, trans-R, and trans-S).
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[1 mark] B is correct because the molecule contains one double bond showing cis-trans isomerism and one chiral centre, resulting in 4 stereoisomers.
PastPaper.question 19 · multiple-choice
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At 298 K, the solubility product, \(K_{sp}\), of calcium sulfate, \(\text{CaSO}_4\), is \(2.4 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\). A student mixes equal volumes of \(0.010\text{ mol dm}^{-3}\) \(\text{CaCl}_2\text{(aq)}\) and \(0.010\text{ mol dm}^{-3}\) \(\text{Na}_2\text{SO}_4\text{(aq)}\). What is the ionic product of \(\text{CaSO}_4\) in this mixture, and does a precipitate form?
A.\(1.0 \times 10^{-4}\text{ mol}^2\text{dm}^{-6}\); precipitate forms
B.\(2.5 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\); precipitate forms
C.\(2.5 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\); no precipitate forms
D.\(5.0 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\); no precipitate forms
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When equal volumes of two solutions are mixed, the concentration of each ion is halved due to dilution. Therefore: \([\text{Ca}^{2+}] = 0.0050\text{ mol dm}^{-3}\) and \([\text{SO}_4^{2-}] = 0.0050\text{ mol dm}^{-3}\). The ionic product is: \([\text{Ca}^{2+}][\text{SO}_4^{2-}] = (5.0 \times 10^{-3}) \times (5.0 \times 10^{-3}) = 2.5 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\). Since the ionic product (\(2.5 \times 10^{-5}\)) is greater than \(K_{sp}\) (\(2.4 \times 10^{-5}\)), a precipitate of \(\text{CaSO}_4\) forms.
PastPaper.markingScheme
[1 mark] B is correct. Calculating halved concentrations for the ionic product gives \(2.5 \times 10^{-5}\text{ mol}^2\text{dm}^{-6}\), which exceeds \(K_{sp}\), so a precipitate forms.
PastPaper.question 20 · multiple-choice
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At 298 K, the standard electrode potential for the copper half-cell is given by: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \quad E^\theta = +0.34\text{ V}\). Using the Nernst equation, \(E = E^\theta + \frac{0.059}{z} \log [\text{Cu}^{2+}(\text{aq})]\), what is the electrode potential, \(E\), of a copper electrode immersed in a \(1.0 \times 10^{-4}\text{ mol dm}^{-3}\) solution of \(\text{Cu}^{2+}(\text{aq})\)?
A.\(+0.46\text{ V}\)
B.\(+0.34\text{ V}\)
C.\(+0.22\text{ V}\)
D.\(-0.12\text{ V}\)
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Substitute the given values into the Nernst equation: \(z = 2\) because two electrons are transferred. \(E = +0.34 + \frac{0.059}{2} \log(1.0 \times 10^{-4})\). Since \(\log(1.0 \times 10^{-4}) = -4\), we have: \(E = +0.34 + 0.0295 \times (-4) = +0.34 - 0.118 = +0.222\text{ V}\), which rounds to \(+0.22\text{ V}\).
PastPaper.markingScheme
[1 mark] C is correct. Applying the Nernst equation with \(z = 2\) and \([\text{Cu}^{2+}] = 1.0 \times 10^{-4}\text{ mol dm}^{-3}\) yields \(+0.22\text{ V}\).
PastPaper.question 21 · multiple-choice
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Using the average bond energies given below, what is the enthalpy change, \(\Delta H\), for the gas-phase reaction: \(\text{CH}_4(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{CH}_3\text{Cl}(\text{g}) + \text{HCl}(\text{g})\)? Bond energies: \(\text{C-H} = 410\text{ kJ mol}^{-1}\); \(\text{Cl-Cl} = 242\text{ kJ mol}^{-1}\); \(\text{C-Cl} = 340\text{ kJ mol}^{-1}\); \(\text{H-Cl} = 431\text{ kJ mol}^{-1}\).
A.\(-119\text{ kJ mol}^{-1}\)
B.\(-59\text{ kJ mol}^{-1}\)
C.\(+119\text{ kJ mol}^{-1}\)
D.\(+771\text{ kJ mol}^{-1}\)
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\(\Delta H = \sum\text{Bond energies of reactants (bonds broken)} - \sum\text{Bond energies of products (bonds formed)}\). Bonds broken: 1 \(\text{C-H}\) bond and 1 \(\text{Cl-Cl}\) bond = \(410 + 242 = 652\text{ kJ mol}^{-1}\). Bonds formed: 1 \(\text{C-Cl}\) bond and 1 \(\text{H-Cl}\) bond = \(340 + 431 = 771\text{ kJ mol}^{-1}\). \(\Delta H = 652 - 771 = -119\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
[1 mark] A is correct. Energy required to break bonds is \(652\text{ kJ mol}^{-1}\) and energy released when bonds form is \(771\text{ kJ mol}^{-1}\). Subtracting gives \(-119\text{ kJ mol}^{-1}\).
PastPaper.question 22 · multiple-choice
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Which of the following correctly ranks the four nitrogen-containing compounds in order of decreasing basic strength (most basic first)?
Basicity depends on the availability of the lone pair on the nitrogen atom to accept a proton: 1. Alkyl groups are electron-donating (+I effect), which increases the electron density on the nitrogen atom. Diethylamine (a secondary amine) has two ethyl groups, making it more basic than ethylamine (a primary amine), which has only one. 2. Ammonia is less basic than aliphatic amines because it has no electron-donating alkyl groups. 3. Phenylamine is the least basic because the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring's \(\pi\) system, making it much less available.
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[1 mark] A is correct. The presence of electron-donating alkyl groups increases basic strength, while delocalisation in phenylamine significantly decreases it.
PastPaper.question 23 · multiple-choice
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Which statement correctly explains the trend in electrical conductivity across the Period 3 metals (sodium, magnesium, and aluminium)?
A.Electrical conductivity increases because ionic radius decreases, allowing closer packaging.
B.Electrical conductivity increases because the number of delocalised valence electrons per atom increases.
C.Electrical conductivity decreases because the increasing nuclear charge attracts outer electrons more strongly, reducing mobility.
D.Electrical conductivity remains constant because all three elements are metallic conductors.
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Sodium, magnesium, and aluminium all have metallic bonding consisting of a lattice of metal cations surrounded by a sea of delocalised electrons. From sodium to aluminium, the number of valence electrons per atom increases (Na has 1, Mg has 2, Al has 3). This increases the density of charge carriers in the lattice, leading to an increase in electrical conductivity.
PastPaper.markingScheme
[1 mark] B is correct. Electrical conductivity is determined by the density of mobile charge carriers, which increases as each successive metal atom contributes more delocalised valence electrons.
PastPaper.question 24 · multiple-choice
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An organic compound contains only carbon, hydrogen, and oxygen. Complete combustion of \(0.150\text{ g}\) of this compound produced \(0.220\text{ g}\) of \(\text{CO}_2\) and \(0.090\text{ g}\) of \(\text{H}_2\text{O}\). What is the empirical formula of this compound?
A.\(\text{CHO}\)
B.\(\text{CH}_2\text{O}\)
C.\(\text{C}_2\text{H}_4\text{O}\)
D.\(\text{CH}_3\text{O}\)
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1. Moles of carbon: \(n(\text{C}) = n(\text{CO}_2) = \frac{0.220\text{ g}}{44.0\text{ g mol}^{-1}} = 0.0050\text{ mol}\). Mass of carbon: \(0.0050\text{ mol} \times 12.0\text{ g mol}^{-1} = 0.060\text{ g}\). 2. Moles of hydrogen: \(n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{0.090\text{ g}}{18.0\text{ g mol}^{-1}} = 0.010\text{ mol}\). Mass of hydrogen: \(0.010\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.010\text{ g}\). 3. Mass of oxygen: \(0.150\text{ g} - 0.060\text{ g} - 0.010\text{ g} = 0.080\text{ g}\). Moles of oxygen: \(n(\text{O}) = \frac{0.080\text{ g}}{16.0\text{ g mol}^{-1}} = 0.0050\text{ mol}\). 4. Ratio of C : H : O = 0.0050 : 0.010 : 0.0050 = 1 : 2 : 1. The empirical formula is \(\text{CH}_2\text{O}\).
PastPaper.markingScheme
[1 mark] B is correct. Calculating the moles of carbon (0.0050 mol), hydrogen (0.010 mol), and oxygen (0.0050 mol) leads to the ratio 1 : 2 : 1.
PastPaper.question 25 · multiple-choice
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Four different halogenoalkanes are heated separately with aqueous silver nitrate in ethanol at 50 °C. Which halogenoalkane will produce a precipitate the fastest?
A.1-chlorobutane
B.2-chloro-2-methylpropane
C.1-iodobutane
D.2-iodo-2-methylpropane
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The rate of hydrolysis of halogenoalkanes depends on two factors: the strength of the C-X bond and the mechanism. 1. The C-I bond is weaker and more easily broken than the C-Cl bond, making iodoalkanes faster than chloroalkanes. 2. 2-iodo-2-methylpropane is a tertiary halogenoalkane. It undergoes hydrolysis via the \(S_N1\) mechanism, which is much faster than the \(S_N2\) mechanism of the primary halogenoalkane (1-iodobutane), due to the stability of the tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane reacts the fastest.
PastPaper.markingScheme
[1 mark] D is correct. Iodoalkanes have weaker carbon-halogen bonds than chloroalkanes, and tertiary halogenoalkanes react faster than primary ones under these conditions.
PastPaper.question 26 · multiple-choice
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An organic compound has the following structural formula:
To determine the number of stereoisomers, we identify the stereogenic centers (chiral centers and double bonds that can exhibit cis/trans isomerism): 1. The carbon atom at position 2 (\(-\text{CH(OH)}-\)) is a chiral center because it is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and the rest of the carbon chain. 2. The double bond between carbon 3 and carbon 4 (\(-\text{CH}=\text{CH}-\)) can show cis/trans (geometric) isomerism because both alkene carbons are bonded to two different groups. 3. The carbon atom at position 5 (\(-\text{CH(CH}_3)-\)) is a chiral center because it is bonded to four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and the rest of the chain.
Since the molecule is completely unsymmetrical, the total number of stereoisomers is given by \(2^n\), where \(n\) is the number of stereogenic centers: \(2^3 = 8\) stereoisomers.
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Award 1 mark for selecting correct option C. - Reject other options because they fail to account for all stereogenic centers (2 chiral centers and 1 alkene double bond).
PastPaper.question 27 · multiple-choice
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A sample of pure gas A is placed in a sealed container at a constant temperature. It establishes the following equilibrium:
Award 1 mark for the correct option C. - Method: Correctly set up the ICE table, find x from total pressure, and calculate partial pressures. - Accuracy: Correctly evaluate the Kc/Kp fraction to get 0.675.
PastPaper.question 28 · multiple-choice
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The table below shows initial rate data for the reaction between reactants X, Y, and Z:
First, determine the order of reaction with respect to each reactant: 1. Comparing Run 1 and Run 2: \([\text{Y}]\) and \([\text{Z}]\) are constant. \([\text{X}]\) doubles, and the rate doubles (\(2.4/1.2 = 2\)). Thus, the reaction is first order with respect to \([\text{X}]\). 2. Comparing Run 1 and Run 3: \([\text{X}]\) and \([\text{Z}]\) are constant. \([\text{Y}]\) doubles, and the rate increases by a factor of 4 (\(4.8/1.2 = 4\)). Thus, the reaction is second order with respect to \([\text{Y}]\). 3. Comparing Run 1 and Run 4: \([\text{X}]\) and \([\text{Y}]\) are constant. \([\text{Z}]\) doubles, and the rate remains unchanged (\(1.2/1.2 = 1\)). Thus, the reaction is zero order with respect to \([\text{Z}]\).
The rate equation is: \[\text{Rate} = k[\text{X}][\text{Y}]^2\]
Calculate the rate constant, \(k\), using the data from Run 1: \[1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3}) (0.10\text{ mol dm}^{-3})^2\] \[1.2 \times 10^{-3} = k (1.0 \times 10^{-3})\] \[k = 1.2\]
Units of \(k\): \[k = \frac{\text{Rate}}{[\text{X}][\text{Y}]^2} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\]
PastPaper.markingScheme
Award 1 mark for selecting B. - 1st order in X, 2nd order in Y, 0th order in Z leads to Rate = k[X][Y]^2. - Value is 1.2. - Units are mol^-2 dm^6 s^-1.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
The dry reforming of methane is represented by the following chemical equation:
Using the standard enthalpies of combustion, \(\Delta H^\ominus_c\), given below, what is the standard enthalpy change of this reaction, \(\Delta H^\ominus_r\)?
According to Hess's law, we can relate the enthalpy of a reaction to the standard enthalpies of combustion: \[\Delta H^\ominus_r = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})\]
Note that carbon dioxide, \(\text{CO}_2\text{(g)}\), is already a product of complete combustion, so its standard enthalpy of combustion is zero (\(\Delta H^\ominus_c [\text{CO}_2\text{(g)}] = 0\text{ kJ mol}^{-1}\)).
Award 1 mark for the correct option D. - Ensure the student correctly identifies the combustion of CO2 as 0. - Correctly apply the Hess's Law formula for enthalpies of combustion (Reactants - Products).
PastPaper.question 30 · multiple-choice
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Which option lists the following nitrogen-containing compounds in order of increasing pH of their \(0.10\text{ mol dm}^{-3}\) aqueous solutions?
The pH of an aqueous solution of a base is determined by its basic strength. A stronger base produces more hydroxide ions in solution, leading to a higher pH. Thus, we must arrange the compounds in order of increasing basicity: 1. **Phenylamine** is the weakest base because the lone pair of electrons on the nitrogen atom is delocalised into the \(\pi\)-system of the benzene ring, making it much less available to accept a proton. 2. **Ammonia** is stronger than phenylamine since there is no resonance delocalisation of its lone pair. 3. **Ethylamine** (a primary aliphatic amine) is a stronger base than ammonia because the electron-donating ethyl group (via the positive inductive effect) increases the electron density on the nitrogen atom. 4. **Diethylamine** (a secondary aliphatic amine) is stronger than ethylamine because it contains two electron-donating ethyl groups, further increasing the electron density and availability of the nitrogen lone pair.
Therefore, the order of increasing pH is: Phenylamine < Ammonia < Ethylamine < Diethylamine
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Award 1 mark for option B. - Understand that basicity determines pH for solutions of equal concentration. - Correctly identify phenylamine as the weakest and secondary aliphatic amine as the strongest.
PastPaper.question 31 · multiple-choice
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Equal amounts (in moles) of four Period 3 chlorides are separately added to equal volumes of water.
Which chloride produces the solution with the lowest pH?
A.\(\text{MgCl}_2\)
B.\(\text{AlCl}_3\)
C.\(\text{SiCl}_4\)
D.\(\text{PCl}_5\)
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Let's look at the behavior of the given Period 3 chlorides in water: - \(\text{MgCl}_2\) dissolves with slight hydrolysis, forming a weakly acidic solution (pH \(\approx 6.5\)). - \(\text{AlCl}_3\) undergoes significant hydrolysis because of the high charge-density of the \(\text{Al}^{3+}\) ion, forming an acidic solution (pH \(\approx 3\)). - \(\text{SiCl}_4\) reacts vigorously with water to form silicon dioxide and hydrogen chloride gas: \[\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}\] One mole of \(\text{SiCl}_4\) produces 4 moles of strong monobasic acid (\(\text{HCl}\)). - \(\text{PCl}_5\) reacts very violently with water to form phosphoric acid and hydrogen chloride gas: \[\text{PCl}_5\text{(s)} + 4\text{H}_2\text{O(l)} \rightarrow \text{H}_3\text{PO}_4\text{(aq)} + 5\text{HCl(aq)}\] One mole of \(\text{PCl}_5\) produces 5 moles of strong monobasic acid (\(\text{HCl}\)) in addition to the partially dissociated \(\text{H}_3\text{PO}_4\).
Thus, \(\text{PCl}_5\) produces the highest concentration of hydrogen ions and the lowest pH.
PastPaper.markingScheme
Award 1 mark for the correct option D. - Reject option C because SiCl4 only yields 4 moles of HCl per mole of reactant compared to 5 moles of HCl from PCl5.
PastPaper.question 32 · multiple-choice
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A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), was reacted with \(80\text{ cm}^3\) of oxygen (an excess).
After complete combustion and cooling to room temperature and pressure, the remaining gas volume was \(60\text{ cm}^3\). When this remaining gas was passed through aqueous sodium hydroxide, the volume decreased to \(30\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_6\)
B.\(\text{C}_3\text{H}_8\)
C.\(\text{C}_4\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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Let's analyze the volume changes step by step: 1. The initial volume of the hydrocarbon = \(10\text{ cm}^3\). 2. Initial volume of \(\text{O}_2\) = \(80\text{ cm}^3\). 3. After combustion and cooling, water condenses to liquid and has negligible volume. The remaining gas is unreacted \(\text{O}_2\) plus the \(\text{CO}_2\) produced, which totals \(60\text{ cm}^3\). 4. Aqueous sodium hydroxide absorbs \(\text{CO}_2\). The decrease in volume is therefore the volume of \(\text{CO}_2\) produced: \[V_{\text{CO}_2} = 60 - 30 = 30\text{ cm}^3\] 5. The remaining gas after \(\text{CO}_2\) absorption is the unreacted excess oxygen: \[V_{\text{excess O}_2} = 30\text{ cm}^3\] 6. The volume of oxygen that reacted is: \[V_{\text{reacted O}_2} = 80 - 30 = 50\text{ cm}^3\]
Now, we determine the mole ratio of Hydrocarbon : \(\text{CO}_2\) : \(\text{O}_2\) reacted: \[10 : 30 : 50 = 1 : 3 : 5\]
- Since 1 mole of \(\text{C}_x\text{H}_y\) gives 3 moles of \(\text{CO}_2\), we have \(x = 3\). - The combustion equation is: \[\text{C}_3\text{H}_y + 5\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\] - Balancing the oxygen atoms: \[10\text{ (on LHS)} = 6 + \frac{y}{2}\text{ (on RHS)}\] \[\frac{y}{2} = 4 \implies y = 8\]
Thus, the molecular formula is \(\text{C}_3\text{H}_8\).
PastPaper.markingScheme
Award 1 mark for the correct option B. - Check that the student correctly identifies CO2 volume as 30 cm^3 and reacted O2 as 50 cm^3. - Check that stoichiometry calculations correctly identify C3H8.
PastPaper.question 33 · multiple-choice
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When excess aqueous ammonia is added to aqueous copper(II) sulfate, the light blue precipitate initially formed dissolves to give a deep blue solution containing the complex ion \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).
Which statement correctly explains the shift in light absorption that occurs when the coordinated water ligands are replaced by ammonia ligands?
A.\(\text{NH}_3\) is a stronger field ligand than \(\text{H}_2\text{O}\), which increases the d-orbital splitting, causing light of a shorter wavelength to be absorbed.
B.\(\text{NH}_3\) is a weaker field ligand than \(\text{H}_2\text{O}\), which decreases the d-orbital splitting, causing light of a longer wavelength to be absorbed.
C.\(\text{NH}_3\) is a stronger field ligand than \(\text{H}_2\text{O}\), which decreases the d-orbital splitting, causing light of a shorter wavelength to be absorbed.
D.\(\text{NH}_3\) is a weaker field ligand than \(\text{H}_2\text{O}\), which increases the d-orbital splitting, causing light of a longer wavelength to be absorbed.
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The colour of transition metal complexes is caused by d-d transitions, where an electron is promoted from a lower energy d-orbital to a higher energy d-orbital.
The energy difference between these split d-orbitals is \(\Delta E = \frac{hc}{\lambda}\).
Ammonia (\(\text{NH}_3\)) is a stronger field ligand than water (\(\text{H}_2\O\)). This causes a greater crystal field splitting of the d-orbitals, which means \(\Delta E\) increases.
Since \(\Delta E\) is inversely proportional to the wavelength of light absorbed (\(\lambda\)), a larger \(\Delta E\) causes the complex to absorb light of a shorter wavelength (higher energy, shifting towards the yellow/orange region), resulting in a change in the complementary color seen (shifting from light blue to deep blue/violet).
PastPaper.markingScheme
Award 1 mark for selecting A. - Note that ammonia is a stronger field ligand (increases splitting) and therefore absorbs a shorter wavelength of light.
Paper 2
Answer all five structured questions on AS Level Chemistry concepts in the spaces provided.
5 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · structured-short-answer
12 PastPaper.marks
A student carries out an experiment to determine the formula of a hydrated magnesium carbonate, \(\text{MgCO}_3 \cdot n\text{H}_2\text{O}\).
A \(4.149\text{ g}\) sample of \(\text{MgCO}_3 \cdot n\text{H}_2\text{O}\) is heated gently in a crucible to drive off all the water of crystallisation. The mass of the anhydrous residue, \(\text{MgCO}_3\), is \(2.529\text{ g}\).
(a) Define the term *relative formula mass*. [1]
(b) (i) Calculate the value of \(n\) in \(\text{MgCO}_3 \cdot n\text{H}_2\text{O}\). Show your working. [3]
(ii) Write an equation, including state symbols, for the thermal decomposition of anhydrous magnesium carbonate. [2]
(iii) Calculate the volume, in \(\text{cm}^3\), of carbon dioxide gas produced when the \(2.529\text{ g}\) of anhydrous magnesium carbonate is completely decomposed. Assume \(1\text{ mol}\) of gas occupies \(24.0\text{ dm}^3\) under these conditions. [2]
(c) (i) State and explain the trend in the thermal stability of the Group 2 carbonates down the group. [3]
(ii) State how the thermal stability of calcium nitrate compares to that of magnesium nitrate. [1]
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PastPaper.workedSolution
(a) The weighted average mass of a formula unit of a compound compared to \(\frac{1}{12}\) of the mass of an atom of carbon-12.
(b) (i) Mass of water lost = \(4.149 - 2.529 = 1.620\text{ g}\) Moles of \(\text{H}_2\text{O} = \frac{1.620}{18.0} = 0.090\text{ mol}\) Moles of \(\text{MgCO}_3 = \frac{2.529}{84.3} = 0.030\text{ mol}\) Ratio \(\text{H}_2\text{O} : \text{MgCO}_3 = 0.090 : 0.030 = 3 : 1\) Therefore, \(n = 3\).
(ii) \(\text{MgCO}_3(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g})\)
(iii) Moles of \(\text{CO}_2\) produced = moles of \(\text{MgCO}_3 = 0.030\text{ mol}\). Volume of \(\text{CO}_2 = 0.030 \times 24.0\text{ dm}^3 = 0.720\text{ dm}^3 = 720\text{ cm}^3\).
(c) (i) Thermal stability increases down the group. As the group is descended, the cationic radius increases / cationic charge density decreases. Thus, the cation has less polarising power on the carbonate ion, leading to less distortion of the C-O bond and requiring more thermal energy to decompose.
(ii) Calcium nitrate is more thermally stable than magnesium nitrate.
PastPaper.markingScheme
Detailed marks: - (a) [1 mark]: Correct definition referencing average/weighted mass of a formula unit compared to 1/12th the mass of carbon-12. - (b)(i) [3 marks]: - M1: Calculation of mass (1.620 g) and moles (0.090 mol) of H2O. - M2: Calculation of moles of MgCO3 (0.030 mol). - M3: Finding the mole ratio of 3:1 to get n = 3. - (b)(ii) [2 marks]: - M1: Correct species: MgCO3 -> MgO + CO2. - M2: Correct state symbols: (s) for MgCO3 and MgO, (g) for CO2. - (b)(iii) [2 marks]: - M1: Use of 0.030 moles of MgCO3 to determine 0.030 moles of CO2. - M2: Calculation of volume = 720 cm3. - (c)(i) [3 marks]: - M1: Trend: stability increases down the group. - M2: Cationic size increases / charge density decreases. - M3: Cation has lower polarising power / distorts carbonate electron cloud less. - (c)(ii) [1 mark]: Calcium nitrate is more stable than magnesium nitrate.
PastPaper.question 2 · structured-short-answer
12 PastPaper.marks
The molecular formula \(\text{C}_3\text{H}_5\text{Cl}\) represents several structural and stereoisomers.
(a) Draw the skeletal structures of three acyclic (no rings) structural isomers with the formula \(\text{C}_3\text{H}_5\text{Cl}\). [3]
(b) One of these structural isomers exhibits stereoisomerism.
(i) Identify this isomer by its IUPAC name. [1]
(ii) Name and explain the type of stereoisomerism shown by this isomer. [3]
(iii) Draw the stereochemical structures of the two stereoisomers of this compound and label them clearly. [2]
(c) Another structural isomer of \(\text{C}_3\text{H}_5\text{Cl}\), 3-chloroprop-1-ene, reacts with cold dilute acidified potassium manganate(VII).
(i) Describe the observation for this reaction. [1]
(ii) Draw the structure of the organic product of this reaction. [2]
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PastPaper.workedSolution
(a) The three acyclic structural isomers of \(\text{C}_3\text{H}_5\text{Cl}\) are: 1. 1-chloropropene, \(\text{CH}_3\text{CH=CHCl}\) 2. 2-chloropropene, \(\text{CH}_3\text{C(Cl)=CH}_2\) 3. 3-chloropropene, \(\text{ClCH}_2\text{CH=CH}_2\)
(b) (i) 1-chloropropene (or 1-chloroprop-1-ene)
(ii) Geometrical / cis-trans (or E-Z) isomerism. Explanation: Restricted rotation about the \(\text{C=C}\) double bond due to the presence of a \(\pi\) bond, and each carbon atom in the double bond is bonded to two different atoms/groups (one is bonded to \(-\text{H}\) and \(-\text{Cl}\); the other is bonded to \(-\text{H}\) and \(-\text{CH}_3\)).
(iii) Two isomers: - Cis (or Z) isomer: Cl and CH3 are on the same side of the double bond (or both H's are on the same side). - Trans (or E) isomer: Cl and CH3 are on opposite sides of the double bond.
(c) (i) The purple potassium manganate(VII) solution is decolourised / turns colourless (or a brown precipitate is formed).
(ii) The product is 3-chloropropane-1,2-diol, with structure \(\text{CH}_2(\text{OH})\text{CH}(\text{OH})\text{CH}_2\text{Cl}\).
PastPaper.markingScheme
Detailed marks: - (a) [3 marks]: 1 mark for each correct skeletal structure of: 1-chloroprop-1-ene, 2-chloroprop-1-ene, and 3-chloroprop-1-ene. - (b)(i) [1 mark]: 1-chloropropene or 1-chloroprop-1-ene. - (b)(ii) [3 marks]: - M1: Cis-trans / geometrical / E-Z isomerism. - M2: Restricted rotation about C=C bond. - M3: Both carbons of C=C have two different groups attached. - (b)(iii) [2 marks]: 1 mark for drawing the cis/Z isomer and 1 mark for the trans/E isomer with planar, 120-degree angles around the C=C bond. - (c)(i) [1 mark]: Purple to colourless / decolourises (or brown precipitate formed). - (c)(ii) [2 marks]: Correct structure of 3-chloropropane-1,2-diol (1 mark for diol, 1 mark for keeping chlorine intact).
PastPaper.question 3 · structured-short-answer
12 PastPaper.marks
The standard enthalpy change of formation of liquid propan-1-ol, \(\text{C}_3\text{H}_7\text{OH(l)}\), can be calculated using standard enthalpy changes of combustion.
(a) Define *standard enthalpy change of combustion*. [2]
(b) (i) Write the chemical equation, including state symbols, for the reaction representing the standard enthalpy change of formation of liquid propan-1-ol, \(\text{C}_3\text{H}_7\text{OH}\). [2]
(ii) Draw a Hess's law cycle showing the relationship between the standard enthalpy change of formation of propan-1-ol and the standard enthalpy changes of combustion of carbon, hydrogen, and propan-1-ol. [2]
(iii) Calculate the standard enthalpy change of formation of liquid propan-1-ol, \(\Delta H_{\text{f}}^\ominus\). [2]
(c) The enthalpy change of combustion of propan-1-ol can be determined experimentally using a simple calorimeter. In an experiment, \(0.820\text{ g}\) of propan-1-ol is burned to heat \(150.0\text{ g}\) of water. The temperature of the water increases by \(28.4\ ^\circ\text{C}\). (Specific heat capacity of water = \(4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1}\); \(M_{\text{r}}\) of propan-1-ol = \(60.0\))
(i) Calculate the heat energy absorbed by the water, in \(\text{kJ}\). [1]
(ii) Calculate the experimental enthalpy change of combustion of propan-1-ol, in \(\text{kJ mol}^{-1}\). [2]
(iii) Suggest why this experimental value is significantly less exothermic than the standard value. [1]
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(a) The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions of 298 K and 100 kPa.
(ii) Hess's cycle: Elements in their standard states: \(3\text{C(s)} + 4\text{H}_2\text{(g)}\) reacting to form \(\text{C}_3\text{H}_7\text{OH(l)}\). Both point downwards to the combustion products: \(3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\). Left arrow: \(3\Delta H_{\text{c}}^\ominus[\text{C}] + 4\Delta H_{\text{c}}^\ominus[\text{H}_2]\) Right arrow: \(\Delta H_{\text{c}}^\ominus[\text{C}_3\text{H}_7\text{OH}]\)
(iii) Heat was lost to the surroundings (or incomplete combustion occurred, or evaporation of propan-1-ol from the wick).
PastPaper.markingScheme
Detailed marks: - (a) [2 marks]: 1 mark for 'one mole of substance completely burned in excess oxygen', 1 mark for 'under standard conditions (298 K and 1 bar/100 kPa)'. - (b)(i) [2 marks]: 1 mark for correct species and balancing, 1 mark for correct state symbols. - (b)(ii) [2 marks]: Correctly drawn Hess cycle with elements, compound, and combustion products as the intermediate, with correct arrows and stoichiometry. - (b)(iii) [2 marks]: - M1: Calculation of sum of reactants combustion (-2323.7 kJ mol-1). - M2: Subtraction of propan-1-ol combustion value to yield -302.7 kJ mol-1. - (c)(i) [1 mark]: q = 17.8 kJ (or 17.81 kJ). - (c)(ii) [2 marks]: - M1: n = 0.0137 mol. - M2: dH_c = -1303 kJ mol-1 (accept range -1300 to -1303). Minus sign must be present. - (c)(iii) [1 mark]: Heat lost to the surroundings / incomplete combustion.
PastPaper.question 4 · structured-short-answer
12 PastPaper.marks
The physical properties of the elements in Period 3 show periodic trends.
(a) The first ionisation energies of the elements in Period 3 show a general increase from sodium to argon.
(i) Explain this general increase in first ionisation energy across Period 3. [3]
(ii) Explain why the first ionisation energy of aluminium is lower than that of magnesium. [2]
(iii) Explain why the first ionisation energy of sulfur is lower than that of phosphorus. [2]
(b) The melting points of the elements in Period 3 vary significantly.
(i) Describe the differences in the structures and bonding of silicon and phosphorus. [3]
(ii) State and explain how the electrical conductivity of silicon compares to that of sodium. [2]
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(a) (i) Across Period 3, the number of protons / nuclear charge increases, while the shielding by inner shells remains approximately constant (electrons are added to the same principal shell). This results in a stronger electrostatic attraction between the nucleus and the outer electrons, pulling them closer and requiring more energy to remove an outer electron.
(ii) The outer electron of magnesium is in a 3s orbital, whereas the outer electron of aluminium is in a 3p orbital. The 3p orbital is at a higher energy level (and is more shielded by the 3s electrons), so less energy is required to remove it.
(iii) Phosphorus has three singly occupied 3p orbitals (\(3\text{p}^3\)), whereas sulfur has one pair of electrons in a 3p orbital (\(3\text{p}^4\)). The spin-pairing of electrons in the 3p orbital of sulfur causes mutual repulsion, making it easier to remove one of these electrons.
(b) (i) Silicon has a giant covalent (macromolecular) structure with strong, tetrahedral covalent bonds holding atoms together. Phosphorus has a simple molecular structure consisting of \(\text{P}_4\) molecules held together only by weak intermolecular van der Waals' forces (London dispersion forces).
(ii) Silicon has a much lower electrical conductivity than sodium. Sodium has a giant metallic structure with delocalised outer electrons that are free to move and carry a charge. In contrast, silicon has localized electrons held within strong covalent bonds, so few electrons are free to conduct electricity.
PastPaper.markingScheme
Detailed marks: - (a)(i) [3 marks]: - M1: Nuclear charge increases / proton number increases. - M2: Shielding remains constant / electrons added to the same shell. - M3: Stronger electrostatic attraction between nucleus and outer electron / outer electron closer to the nucleus. - (a)(ii) [2 marks]: - M1: Mg outer electron in 3s AND Al outer electron in 3p. - M2: 3p orbital is higher in energy / further from nucleus / shielded by 3s subshell. - (a)(iii) [2 marks]: - M1: P is 3p^3 and S is 3p^4. - M2: Repulsion between the paired electrons in the 3p orbital of S. - (b)(i) [3 marks]: - M1: Silicon has a giant covalent structure with strong covalent bonds. - M2: Phosphorus has a simple molecular structure with weak van der Waals' / London dispersion forces. - M3: Mentions P4 molecules or describes the difference in bond-breaking required (covalent bonds vs intermolecular forces). - (b)(ii) [2 marks]: - M1: Sodium has much higher conductivity / Silicon has much lower conductivity. - M2: Sodium has mobile/delocalised electrons, whereas silicon has localized valence electrons involved in covalent bonding.
(a) The rate of hydrolysis of halogenoalkanes can be determined by reacting them with aqueous silver nitrate in ethanol.
(i) Write an ionic equation, including state symbols, for the reaction that forms a precipitate when 1-bromobutane is hydrolysed in the presence of silver nitrate. [1]
(ii) State the color of the precipitate formed in (a)(i). [1]
(iii) State and explain the relative rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. [3]
(b) 2-bromo-2-methylpropane undergoes hydrolysis via an \(S_{\text{N}}1\) mechanism.
(i) Write the mechanism for this reaction with aqueous hydroxide ions. Draw the structures of the organic reactant, intermediate, and product, and use curly arrows to show the movement of electron pairs. [4]
(ii) Draw the reaction pathway diagram for this \(S_{\text{N}}1\) reaction. Label the axes, the reactants, products, activation energy (\(E_{\text{a}}\)), and the enthalpy change of the reaction (\(\Delta H\)), assuming the reaction is exothermic. [3]
(iii) Rate order: 1-iodobutane > 1-bromobutane > 1-chlorobutane (1-iodobutane hydrolyses fastest, 1-chlorobutane slowest). Explanation: The C-I bond is the weakest / has the lowest bond enthalpy, and the C-Cl bond is the strongest / has the highest bond enthalpy. The rate of reaction depends on the ease of breaking the carbon-halogen bond.
(b) (i) Mechanism: - Step 1: Slow heterolytic fission of the C-Br bond in 2-bromo-2-methylpropane to form a tertiary carbocation intermediate, \(\text{(CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\). Use a curly arrow starting from the C-Br bond pointing to the Br atom. - Step 2: Fast attack by the hydroxide ion, \(\text{OH}^-\). Use a curly arrow starting from a lone pair on the oxygen of \(\text{OH}^-\), pointing to the positively charged carbon atom of the carbocation, forming 2-methylpropan-2-ol, \(\text{(CH}_3)_3\text{COH}\).
(ii) The reaction pathway diagram should have: - Enthalpy/Energy on the y-axis, Reaction pathway/Progress of reaction on the x-axis. - A two-peak curve (representing the two-step mechanism) with a trough in the middle representing the carbocation intermediate. The first peak (representing the rate-determining step) must be higher than the second peak. - Products must be lower in energy than reactants (exothermic). - \(E_{\text{a}}\) labelled from the reactants' energy level to the top of the first peak. - \(\Delta H\) labelled as the energy difference between reactants and products (with a downward arrow).
PastPaper.markingScheme
Detailed marks: - (a)(i) [1 mark]: Ag+(aq) + Br-(aq) -> AgBr(s). Both formulas and state symbols must be correct. - (a)(ii) [1 mark]: Cream precipitate. - (a)(iii) [3 marks]: - M1: Rate order: 1-iodobutane > 1-bromobutane > 1-chlorobutane. - M2: Bond strength/enthalpy: C-I is weakest / C-Cl is strongest. - M3: Hydrolysis rate depends on ease of breaking C-X bond (bond enthalpy, not polarity). - (b)(i) [4 marks]: - M1: Correct reactant structure and polar bond dipole/arrow from C-Br bond to Br. - M2: Correct carbocation intermediate structure and Br- leaving group. - M3: Curly arrow from lone pair of OH- to positive carbon of carbocation. - M4: Correct product structure. - (b)(ii) [3 marks]: - M1: Y-axis labelled Energy/Enthalpy, X-axis labelled Reaction Progress. - M2: Two peaks with a trough, first peak higher than second, product energy level below reactants. - M3: Correctly labelled Ea (reactants to highest peak) and dH (reactants to products).
Paper 3
Carry out the practical experiments in the laboratory as detailed, recording all observations, data tables, and subsequent quantitative analysis.
3 PastPaper.question · 39.99 PastPaper.marks
PastPaper.question 1 · practical-task
13.33 PastPaper.marks
Section instructions: Carry out the practical experiments in the laboratory as detailed, recording all observations, data tables, and subsequent quantitative analysis.
In this experiment, you will determine the relative molecular mass of a dibasic organic acid, \(H_2A\), by titrating its solution with standard sodium hydroxide solution.
FB 1 is a solution containing 4.50 g of the dibasic acid \(H_2A\) per \(dm^3\). FB 2 is 0.100 \(mol\\,dm^{-3}\) sodium hydroxide, \(NaOH\). Phenolphthalein is provided as an indicator.
(a) Pipette 25.0 \(cm^3\) of FB 1 into a conical flask. Titrate with FB 2 from the burette using phenolphthalein indicator until a permanent pale pink colour is obtained. Record your titration results in a suitable tabular format to find the average titre.
(b) Show your working to calculate the concentration of \(H_2A\) in \(mol\\,dm^{-3}\) and determine the relative molecular mass, \(M_r\), of \(H_2A\). (The reaction ratio is 1 mole of \(H_2A\) reacts with 2 moles of \(NaOH\).)
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1. Moles of NaOH in average titre (assuming 25.00 cm3): \(n(NaOH) = 0.100 \times 0.02500 = 2.50 \times 10^{-3}\) mol. 2. Moles of \(H_2A\) in 25.0 cm3: \(n(H_2A) = 0.5 \times 2.50 \times 10^{-3} = 1.25 \times 10^{-3}\) mol. 3. Molar concentration of \(H_2A\): \([H_2A] = 1.25 \times 10^{-3} / 0.0250 = 0.0500\) mol \(dm^{-3}\). 4. Relative molecular mass \(M_r = 4.50 / 0.0500 = 90.0\) g \(mol^{-1}\).
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Total Marks: 13.33 - 4 marks for complete and accurate titration table with concordant titres (within 0.10 cm3). - 2 marks for calculating the correct average titre and showing the working. - 2 marks for correct calculation of moles of NaOH and moles of H2A. - 2.33 marks for calculating the molar concentration of H2A to 3 significant figures. - 3 marks for determining the relative molecular mass of H2A to an appropriate number of significant figures.
PastPaper.question 2 · practical-task
13.33 PastPaper.marks
Section instructions: Carry out the practical experiments in the laboratory as detailed, recording all observations, data tables, and subsequent quantitative analysis.
In this experiment, you will determine the enthalpy change of solution of anhydrous sodium carbonate, \(Na_2CO_3\).
FB 3 is anhydrous sodium carbonate. FB 4 is 2.00 \(mol\\,dm^{-3}\) hydrochloric acid, \(HCl\).
(a) Weigh a dry plastic cup. Record its mass. Add approximately 3.00 g of FB 3 into the cup and record the exact mass of the cup + FB 3. (b) Using a measuring cylinder, transfer 50.0 \(cm^3\) of FB 4 into a second plastic cup. Measure and record its initial temperature every minute for 3 minutes. At the 4th minute, add the weighed FB 3, stir continuously, and record the temperature at minutes 5, 6, 7, 8, 9, and 10. (c) Plot a graph of temperature against time, extrapolate to find the theoretical maximum temperature change (\(\Delta T\)). (d) Calculate the heat energy change (\(q\)) and the molar enthalpy change of the reaction (\(\Delta H\)) in \(kJ\\,mol^{-1}\). (Assume specific heat capacity of the solution is 4.18 \(J\\,g^{-1}\\,K^{-1}\) and density is 1.00 \(g\\,cm^{-3}\).)
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1. Mass of FB 3 used = 3.00 g. 2. Moles of \(Na_2CO_3 = 3.00 / 105.99 = 0.0283\) mol. 3. Heat change \(q = m c \Delta T = 50.0 \times 4.18 \times 4.5 = 940.5\) J. 4. Enthalpy change \(\Delta H = -0.9405 / 0.0283 = -33.2\) kJ \(mol^{-1}\).
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Total Marks: 13.33 - 3 marks for presenting all raw mass and temperature measurements with correct units and decimal places. - 3 marks for plotting a correct graph with temperature on y-axis, time on x-axis, and clear extrapolation lines to the 4th minute. - 2.33 marks for calculating q using the correct mass of solution (50.0 g). - 3 marks for calculating moles of Na2CO3 and standardizing the enthalpy change to kJ/mol with the correct negative sign. - 2 marks for giving the final answer to 3 significant figures.
PastPaper.question 3 · practical-task
13.33 PastPaper.marks
Section instructions: Carry out the practical experiments in the laboratory as detailed, recording all observations, data tables, and subsequent quantitative analysis.
In this experiment, you will identify the cations and anions present in two unknown solutions, FB 5 and FB 6, by performing qualitative analysis tests.
Perform the following tests on FB 5 and FB 6 and record your observations:
Test 1: Add aqueous sodium hydroxide dropwise, then in excess, to separate 2 \(cm^3\) portions of FB 5 and FB 6. Test 2: Add aqueous ammonia dropwise, then in excess, to separate 2 \(cm^3\) portions of FB 5 and FB 6. Test 3: Add acidified silver nitrate solution to separate 2 \(cm^3\) portions of FB 5 and FB 6. Test 4: Add acidified barium chloride solution to separate 2 \(cm^3\) portions of FB 5 and FB 6.
Use your observations to identify the ions in FB 5 and FB 6.
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1. FB 5 + NaOH: Green precipitate, insoluble in excess, turning brown on standing near the surface (confirms \(Fe^{2+}\)). 2. FB 5 + ammonia: Green precipitate, insoluble in excess (confirms \(Fe^{2+}\)). 3. FB 5 + silver nitrate: White precipitate soluble in dilute ammonia (confirms \(Cl^-\)). 4. FB 6 + NaOH: Pale blue precipitate, insoluble in excess (confirms \(Cu^{2+}\)). 5. FB 6 + ammonia: Pale blue precipitate, dissolving in excess to give a deep blue solution (confirms \(Cu^{2+}\)). 6. FB 6 + barium chloride: Fine white precipitate insoluble in dilute acid (confirms \(SO_4^{2-}\)).
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Total Marks: 13.33 - 4 marks for recording correct, detailed, and accurate observations for all tests with NaOH and NH3 (including precipitate colours and solubilities in excess). - 4 marks for recording correct observations for the anion tests (silver nitrate and barium chloride reactions, including testing precipitate solubility with dilute acids/ammonia where appropriate). - 2 marks for identifying the cation and anion in FB 5 with supporting evidence. - 2 marks for identifying the cation and anion in FB 6 with supporting evidence. - 1.33 marks for using correct chemical terminology throughout the observations (e.g., precipitate, soluble, excess).
Paper 4
Answer all nine structured questions covering both AS and A Level syllabus content in the spaces provided.
9 PastPaper.question · 99.99 PastPaper.marks
PastPaper.question 1 · structured-long-answer
11.11 PastPaper.marks
The reaction between reactants X, Y, and Z was investigated at 298 K: 2X + Y + Z -> Products. The initial rates of reaction were measured under different concentration conditions: (1) [X] = 0.10, [Y] = 0.10, [Z] = 0.10 mol dm-3; Initial rate = 1.2 x 10^-3 mol dm-3 s-1. (2) [X] = 0.20, [Y] = 0.10, [Z] = 0.10 mol dm-3; Initial rate = 2.4 x 10^-3 mol dm-3 s-1. (3) [X] = 0.10, [Y] = 0.20, [Z] = 0.10 mol dm-3; Initial rate = 4.8 x 10^-3 mol dm-3 s-1. (4) [X] = 0.10, [Y] = 0.10, [Z] = 0.20 mol dm-3; Initial rate = 1.2 x 10^-3 mol dm-3 s-1. (a) Deduce the order of reaction with respect to X, Y, and Z, explaining your reasoning. (b) Write the rate equation for this reaction. (c) Calculate the rate constant, k, using the data from experiment 1, including its units. (d) State and explain how the rate constant, k, is affected by an increase in temperature, referencing activation energy and the Maxwell-Boltzmann distribution. (e) Propose a two-step mechanism for this reaction where the first step is the rate-determining step.
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(a) Comparing Expt 1 and 2: [Y] and [Z] constant, [X] doubles, rate doubles, so 1st order wrt X. Comparing Expt 1 and 3: [X] and [Z] constant, [Y] doubles, rate quadruples (1.2 to 4.8 x 10^-3), so 2nd order wrt Y. Comparing Expt 1 and 4: [X] and [Y] constant, [Z] doubles, rate remains constant, so 0th order wrt Z. (b) rate = k[X][Y]^2. (c) k = rate / ([X][Y]^2) = 1.2 x 10^-3 / (0.10 * 0.10^2) = 1.2. Units: (mol dm-3 s-1) / (mol3 dm-9) = mol-2 dm6 s-1. (d) k increases with temperature. Higher temperature increases the fraction of molecules with energy greater than or equal to the activation energy (Ea), as shown by the shifted Maxwell-Boltzmann distribution, leading to more successful collisions per unit time. (e) Step 1 (slow/RDS): X + 2Y -> intermediates. Step 2 (fast): intermediates + Z -> products.
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(a) 3 marks: 1 mark for each order with correct reasoning. (b) 1 mark: correct rate equation. (c) 2 marks: 1 mark for value of 1.2, 1 mark for units mol-2 dm6 s-1. (d) 3 marks: 1 mark for stating k increases, 1 mark for mentioning more molecules have energy >= Ea, 1 mark for Maxwell-Boltzmann explanation. (e) 2 marks: 1 mark for slow step containing X and 2Y, 1 mark for overall equation adding up to the stoichiometry.
PastPaper.question 2 · structured-long-answer
11.11 PastPaper.marks
(a) Define the term optical isomers and explain the structural feature required for an organic compound to show optical isomerism. (b) 2-bromopropanoic acid, CH3CH(Br)COOH, is optically active. Draw three-dimensional structures of the two optical isomers. (c) The complex ion [Co(en)2Cl2]+ (where en is ethane-1,2-diamine) exhibits both geometrical and optical isomerism. (i) Draw the structures of the cis and trans geometrical isomers of [Co(en)2Cl2]+, using a simplified curve to represent the en ligand. (ii) State which geometrical isomer exhibits optical isomerism and draw 3D diagrams of its optical isomers.
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(a) Optical isomers are non-superimposable mirror images of each other. The compound must contain a chiral center (an asymmetric carbon atom bonded to four different groups). (b) Draw 3D structures: a central carbon bonded to -H, -CH3, -Br, -COOH, with wedge/dash bonds, and its mirror image. (c)(i) Trans isomer: octahedral, with Cl-Co-Cl angle of 180 degrees (chlorines opposite each other). Cis isomer: octahedral, with Cl-Co-Cl angle of 90 degrees (chlorines adjacent to each other). (c)(ii) The cis isomer is chiral and has no plane of symmetry, so it exhibits optical isomerism. Draw the 3D structures of the cis-enantiomers as mirror images of each other.
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(a) 2 marks: 1 mark for non-superimposable mirror images, 1 mark for chiral carbon with 4 different groups. (b) 2 marks: 1 mark for each correct 3D enantiomer structure. (c)(i) 3 marks: 1 mark for trans-isomer structure, 1 mark for cis-isomer structure, 1 mark for octahedral geometry. (c)(ii) 4 marks: 1 mark for identifying cis isomer, 3 marks for drawing two correct non-superimposable cis mirror images.
PastPaper.question 3 · structured-long-answer
11.11 PastPaper.marks
An electrochemical cell consists of a Fe3+/Fe2+ half-cell and a Ag+/Ag half-cell. Standard electrode potentials: Fe3+(aq) + e- <=> Fe2+(aq) (E^o = +0.77 V) and Ag+(aq) + e- <=> Ag(s) (E^o = +0.80 V). (a) State which species is the anode, which is the cathode, and write the overall cell equation. (b) Calculate the standard cell potential. (c) The concentration of Ag+(aq) is reduced to 0.050 mol dm-3 while other species remain at 1.00 mol dm-3. (i) Use the Nernst equation E = E^o + (0.059/z)log[oxidised] to calculate the electrode potential of the silver half-cell. (ii) Calculate the new cell potential and state if the overall reaction is more or less thermodynamically feasible. (d) Describe the role and working of a salt bridge.
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(a) Fe2+ is oxidized at the anode: Fe2+ -> Fe3+ + e-. Ag+ is reduced at the cathode: Ag+ + e- -> Ag. Overall: Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s). (b) E^o_cell = E^o_cathode - E^o_anode = +0.80 - 0.77 = +0.03 V. (c)(i) E_Ag = +0.80 + (0.059/1)log(0.050) = +0.80 - 0.077 = +0.723 V. (ii) E_cell = E_cathode - E_anode = +0.723 - 0.77 = -0.047 V. Since E_cell is negative, the reaction is not thermodynamically feasible under these conditions. (d) The salt bridge completes the electrical circuit, allowing ionic flow to maintain electrical neutrality in both half-cells without mixing the electrolyte solutions.
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(a) 3 marks: 1 mark for anode/cathode identification, 1 mark for anode half-equation, 1 mark for correct overall equation. (b) 1 mark: +0.03 V. (c)(i) 2 marks: 1 mark for substituting values, 1 mark for +0.723 V. (c)(ii) 2 marks: 1 mark for -0.047 V, 1 mark for explaining less feasible (not feasible) because E_cell is negative. (d) 3 marks: 1 mark for completing circuit, 1 mark for maintaining electrical neutrality, 1 mark for allowing ion flow / preventing mixing.
PastPaper.question 4 · structured-long-answer
11.11 PastPaper.marks
(a) Explain what is meant by a buffer solution. (b) A buffer is prepared by mixing 50.0 cm3 of 0.200 mol dm-3 propanoic acid (CH3CH2COOH) with 50.0 cm3 of 0.100 mol dm-3 NaOH. Ka of propanoic acid is 1.35 x 10^-5 mol dm-3. (i) Write the equation for the neutralization reaction. (ii) Calculate the concentrations of CH3CH2COOH and CH3CH2COO- in the buffer. (iii) Calculate the pH of the buffer. (c) Write chemical equations to show how this buffer controls pH when a small amount of strong acid is added.
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(a) A buffer solution resists changes in pH when small amounts of acid or alkali are added. (b)(i) CH3CH2COOH + NaOH -> CH3CH2COONa + H2O. (b)(ii) Initial moles: acid = 0.0500 * 0.200 = 0.0100 mol; NaOH = 0.0500 * 0.100 = 0.00500 mol. NaOH is the limiting reagent, so 0.00500 mol of salt is formed and 0.00500 mol of acid remains. Total volume = 100 cm3 = 0.100 dm3. Concentrations: [CH3CH2COOH] = 0.00500 / 0.100 = 0.0500 mol dm-3; [CH3CH2COO-] = 0.00500 / 0.100 = 0.0500 mol dm-3. (b)(iii) pH = pKa + log([salt]/[acid]). pKa = -log(1.35 x 10^-5) = 4.87. pH = 4.87 + log(0.0500/0.0500) = 4.87. (c) When H+ ions are added, they react with propanoate ions: CH3CH2COO-(aq) + H+(aq) -> CH3CH2COOH(aq).
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(a) 2 marks: 1 mark for resisting pH change, 1 mark for when small amounts of acid/alkali are added. (b)(i) 1 mark for correct equation. (b)(ii) 2 marks: 1 mark for calculating moles, 1 mark for correct concentrations. (b)(iii) 3 marks: 1 mark for pKa calculation, 1 mark for pH expression, 1 mark for final pH of 4.87. (c) 3 marks: 1 mark for identifying that H+ reacts with CH3CH2COO-, 2 marks for the correct ionic equation.
PastPaper.question 5 · structured-long-answer
11.11 PastPaper.marks
(a) Describe the reaction pathway to synthesise phenylamine starting from benzene, including reagents, conditions, and the structure of the intermediate compound. (b) Phenylamine is converted into a diazonium salt, which is then coupled with phenol to form an azo dye. (i) State the reagents and conditions to convert phenylamine to benzenediazonium chloride. (ii) State the reagents, conditions, and draw the structure of the azo dye formed by coupling benzenediazonium chloride with phenol. (c) Compare the basicity of phenylamine and ethylamine, explaining your answer.
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(a) Step 1: Benzene to nitrobenzene. Reagents: concentrated HNO3 and concentrated H2SO4. Conditions: 50-55 °C, reflux. Step 2: Nitrobenzene to phenylamine. Reagents: Sn and concentrated HCl. Conditions: heating under reflux, followed by adding NaOH(aq). (b)(i) Reagents: HNO3/HCl (or NaNO2 and excess HCl). Conditions: temperature between 0 and 5 °C. (b)(ii) Reagents: alkaline phenol (phenol in NaOH). Conditions: cold (below 10 °C). Structure of azo dye: C6H5-N=N-C6H4-OH (4-hydroxyphenylazobenzene). (c) Ethylamine is more basic than phenylamine. In ethylamine, the electron-donating ethyl group increases electron density on the nitrogen atom, making the lone pair more available to accept a proton. In phenylamine, the lone pair on nitrogen is delocalised into the benzene pi-system, making it less available to accept a proton.
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(a) 4 marks: 1 mark for nitration reagents, 1 mark for nitration conditions, 1 mark for reduction reagents/conditions, 1 mark for identifying nitrobenzene intermediate. (b)(i) 2 marks: 1 mark for NaNO2 + HCl, 1 mark for temperature 0-5 °C. (b)(ii) 3 marks: 1 mark for phenol in NaOH, 1 mark for cold conditions, 1 mark for drawing the correct azo dye structure. (c) 2 marks: 1 mark for stating ethylamine is stronger, 1 mark for explanation involving delocalisation in phenylamine vs electron donation in ethylamine.
PastPaper.question 6 · structured-long-answer
11.11 PastPaper.marks
(a) Explain what is meant by a coordinate bond and why transition metals form them. (b)(i) Draw a dot-and-cross diagram of a water molecule showing all outer shell electrons. (ii) Describe the shape and bond angles of the [Cu(H2O)6]2+ complex ion. (c) When excess concentrated HCl is added to [Cu(H2O)6]2+, a yellow-green solution of [CuCl4]2- is formed. (i) Write the equation for this reaction. (ii) State the shape and bond angles of [CuCl4]2-. (d) Explain the origin of color in transition metal complexes.
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(a) A coordinate bond is a covalent bond in which both electrons in the shared pair originate from the same atom (the ligand). Transition metals have vacant d-orbitals that can accept lone pairs from ligands. (b)(i) Dot-and-cross of H2O: Oxygen has 2 bonding pairs (O-H) and 2 lone pairs, total of 8 outer electrons around O. (b)(ii) Shape of [Cu(H2O)6]2+ is octahedral, bond angles are 90 degrees. (c)(i) [Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O. (c)(ii) Shape of [CuCl4]2- is tetrahedral, bond angle is 109.5 degrees. (d) Ligands split the degenerate d-orbitals into two groups of different energy levels. Electrons absorb light in the visible region to promote an electron from a lower to a higher d-orbital (d-d transition). The complementary color of the absorbed light is transmitted/seen.
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(a) 2 marks: 1 mark for definition of coordinate bond, 1 mark for transition metals having vacant d-orbitals. (b)(i) 2 marks: 1 mark for showing correct bonding pairs, 1 mark for showing correct lone pairs on oxygen. (b)(ii) 2 marks: 1 mark for octahedral, 1 mark for 90 degrees. (c)(i) 1 mark for correct equation. (c)(ii) 2 marks: 1 mark for tetrahedral, 1 mark for 109.5 degrees. (d) 2 marks: 1 mark for d-orbital splitting by ligands, 1 mark for promotion of electron absorbing visible light.
PastPaper.question 7 · structured-long-answer
11.11 PastPaper.marks
(a) Define the term lattice energy of an ionic solid. (b) Use the following data to calculate the lattice energy of calcium oxide, CaO(s): Enthalpy of formation of CaO(s) = -635 kJ mol-1; Enthalpy of atomisation of Ca(s) = +178 kJ mol-1; First ionisation energy of Ca(g) = +590 kJ mol-1; Second ionisation energy of Ca(g) = +1145 kJ mol-1; Enthalpy of atomisation of O2(g) (0.5 * O-O bond enthalpy) = +249 kJ mol-1; First electron affinity of O(g) = -141 kJ mol-1; Second electron affinity of O(g) = +798 kJ mol-1. (c) Compare and explain the difference between the lattice energy of CaO and NaCl (-787 kJ mol-1). (d) Explain why the second electron affinity of oxygen is endothermic while the first is exothermic.
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(a) Lattice energy is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions. (b) By Hess's Law / Born-Haber cycle: Delta H_f = Delta H_at(Ca) + IE1(Ca) + IE2(Ca) + Delta H_at(O) + EA1(O) + EA2(O) + Delta H_lattice. Therefore, -635 = 178 + 590 + 1145 + 249 - 141 + 798 + Delta H_lattice. -635 = 2819 + Delta H_lattice. Delta H_lattice = -635 - 2819 = -3454 kJ mol-1. (c) CaO has a much higher (more exothermic) lattice energy because calcium has a +2 charge and oxide has a -2 charge, whereas sodium has a +1 charge and chloride has a -1 charge. The electrostatic attraction is proportional to the product of charges. Also, Ca2+ and O2- have smaller ionic radii than Na+ and Cl- respectively, allowing ions to pack closer together, strengthening ionic bonds. (d) The first electron affinity is exothermic because of the attraction between the positive nucleus and the incoming electron. The second electron affinity is endothermic because energy must be supplied to overcome the electrostatic repulsion between the negative O- ion and the incoming negative electron.
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(a) 2 marks: 1 mark for formation of solid from gaseous ions, 1 mark for under standard conditions/one mole. (b) 4 marks: 2 marks for setting up correct cycle equation, 2 marks for correct calculation value of -3454 kJ mol-1 (deduct 1 mark for missing minus sign or math error). (c) 3 marks: 1 mark for mentioning higher ionic charges (+2/-2 vs +1/-1), 1 mark for mentioning smaller ionic radii, 1 mark for link to stronger electrostatic attraction. (d) 2 marks: 1 mark for attraction of incoming electron to nucleus in O, 1 mark for repulsion between O- and incoming electron.
PastPaper.question 8 · structured-long-answer
11.11 PastPaper.marks
(a) Describe and explain the trend in electrical conductivity of the elements across Period 3 from sodium to silicon. (b) The first ionisation energies of Period 3 elements generally increase from left to right, but there are two notable drops. (i) Identify the two elements where these drops occur (compared to the preceding element). (ii) Explain the drop in first ionisation energy between magnesium and aluminium. (iii) Explain the drop in first ionisation energy between phosphorus and sulfur.
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(a) Electrical conductivity increases from Na to Mg to Al because the number of delocalised outer-shell electrons per atom increases (Na has 1, Mg has 2, Al has 3), which increases the charge carrier density. Silicon has a much lower conductivity (it is a metalloid/semiconductor) because it has a giant covalent structure with localized electrons in covalent bonds. (b)(i) The drops occur at Aluminium (Al) and Sulfur (S). (b)(ii) Magnesium has the outer electronic configuration 3s2, while aluminium has 3s2 3p1. The electron removed from aluminium is in a 3p orbital, which is higher in energy and further from the nucleus than the 3s orbital of magnesium, and is shielded by the 3s electrons. Thus, less energy is required to remove it. (b)(iii) Phosphorus has the configuration 3s2 3p3 (with three singly-occupied 3p orbitals), whereas sulfur has 3s2 3p4 (one doubly-occupied 3p orbital). In sulfur, the repulsion between the paired electrons in the same 3p orbital makes the first electron easier to remove.
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(a) 3 marks: 1 mark for conductivity increasing Na to Al, 1 mark for explanation (more delocalised electrons per atom), 1 mark for Silicon being a giant covalent semiconductor. (b)(i) 2 marks: 1 mark for Al, 1 mark for S. (b)(ii) 3 marks: 1 mark for electron configurations, 1 mark for identifying 3p is higher in energy/shielded, 1 mark for explanation of why it is easier to remove. (b)(iii) 3 marks: 1 mark for Hund's rule/configurations, 1 mark for stating spin-pair repulsion in S, 1 mark for explaining why this makes it easier to remove.
PastPaper.question 9 · structured-long-answer
11.11 PastPaper.marks
An electrochemical cell is set up to study the effect of concentration on electrode potential. (a) Define standard electrode potential, \(E^\theta\). [2 marks] (b) Describe the components and conditions required to set up the standard \(\text{Fe}^{3+}(\text{aq}) / \text{Fe}^{2+}(\text{aq})\) half-cell, and state the reference electrode it must be connected to in order to measure its standard electrode potential. [3 marks] (c) A student sets up an electrochemical cell consisting of a standard \(\text{Ag}^{+}(\text{aq}) / \text{Ag}(\text{s})\) half-cell (\(E^\theta = +0.80\text{ V}\)) and a non-standard \(\text{Fe}^{3+}(\text{aq}) / \text{Fe}^{2+}(\text{aq})\) half-cell. In the non-standard half-cell, \([\text{Fe}^{3+}] = 0.050\text{ mol dm}^{-3}\) and \([\text{Fe}^{2+}] = 0.85\text{ mol dm}^{-3}\) at \(298\text{ K}\). The standard electrode potential for the iron system is: \(\text{Fe}^{3+}(\text{aq}) + e^{-} \rightleftharpoons \text{Fe}^{2+}(\text{aq})\quad E^\theta = +0.77\text{ V}\). (i) Write the Nernst equation for the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell. [1 mark] (ii) Calculate the electrode potential, \(E\), of this non-standard \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell at \(298\text{ K}\). Show your working. [2 marks] (iii) Calculate the cell potential, \(E_{\text{cell}}\), of the combined cell under these conditions and state the direction of electron flow in the external circuit. [1 mark] (d) Predict and explain the effect on the electrode potential of the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell if sodium hydroxide solution is added to it, forming a precipitate of \(\text{Fe(OH)}_3\). [2 marks]
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(a) Standard electrode potential, \(E^\theta\), is the electromotive force (emf) or potential difference of a half-cell coupled to a standard hydrogen electrode (SHE) under standard conditions of \(298\text{ K}\), \(1\text{ atm}\) (or \(101\text{ kPa}\)) gas pressure, and all solution concentrations at \(1.00\text{ mol dm}^{-3}\). (b) The standard \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell consists of an aqueous solution containing both \(1.00\text{ mol dm}^{-3}\ \text{Fe}^{3+}\) and \(1.00\text{ mol dm}^{-3}\ \text{Fe}^{2+}\). It requires an inert Platinum (Pt) electrode. It is connected to a Standard Hydrogen Electrode (SHE). (c)(i) The Nernst equation is: \(E = E^\theta + \frac{0.059}{1}\log\left(\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}\right)\). (c)(ii) Substituting the given values into the equation: \(E = +0.77 + 0.059\log\left(\frac{0.050}{0.85}\right) = +0.77 + 0.059\log(0.0588) = +0.77 + 0.059(-1.23) = 0.77 - 0.0726 = +0.697\text{ V}\) (or \(+0.70\text{ V}\)). (c)(iii) \(E_{\text{cell}} = E(\text{cathode}) - E(\text{anode}) = E(\text{Ag}^{+}/\text{Ag}) - E(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.80\text{ V} - (+0.697\text{ V}) = +0.103\text{ V}\) (or \(+0.10\text{ V}\)). Since the Silver electrode has the more positive potential, reduction occurs here, and oxidation occurs at the Iron half-cell. Electrons flow through the external circuit from the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell to the \(\text{Ag}^{+}/\text{Ag}\) half-cell. (d) Adding NaOH precipitates \(\text{Fe(OH)}_3\), which dramatically decreases the concentration of free \(\text{Fe}^{3+}\) ions in solution. According to the Nernst equation or Le Chatelier's principle, a decrease in \([\text{Fe}^{3+}]\) shifts the equilibrium \(\text{Fe}^{3+}(\text{aq}) + e^{-} \rightleftharpoons \text{Fe}^{2+}(\text{aq})\) to the left, which decreases the electrode potential \(E\) (making it less positive/more negative).
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(a) [1 mark]: Potential difference/emf measured relative to a standard hydrogen electrode. [1 mark]: Under standard conditions of 298 K, 1 atm pressure, and 1.00 mol dm^-3 solution concentration. (b) [1 mark]: Solution containing equal concentrations (1.00 mol dm^-3) of Fe3+(aq) and Fe2+(aq). [1 mark]: Platinum electrode. [1 mark]: Standard Hydrogen Electrode. (c)(i) [1 mark]: Correctly stated Nernst equation: E = E^o + 0.059 log([Fe3+]/[Fe2+]) (accept RT/F ln equivalents). (c)(ii) [1 mark]: Correct substitution of concentrations into the Nernst equation. [1 mark]: Correct final answer of +0.70 V (or +0.697 V). (c)(iii) [1 mark]: Correct cell potential of +0.10 V (or +0.103 V) AND direction of electrons stated from Fe3+/Fe2+ to Ag+/Ag half-cell. (d) [1 mark]: Stating that [Fe3+] decreases due to precipitation of Fe(OH)3. [1 mark]: Stating that the position of equilibrium shifts to the left, which decreases the electrode potential (makes E less positive).
Paper 5
Answer both planning, analysis and evaluation questions using the data provided.
2 PastPaper.question · 30 PastPaper.marks
PastPaper.question 1 · analytical-evaluation-task
15 PastPaper.marks
An experiment is carried out to find the formula of the complex formed between copper(II) ions, \(Cu^{2+}\), and a bidentate ligand, L, using the method of continuous variation. Solutions of \(Cu^{2+}(aq)\) and L(aq) are prepared, both having a concentration of \(0.100\text{ mol dm}^{-3}\). Varying volumes of these two solutions are mixed such that the total volume is always \(10.0\text{ cm}^3\). The absorbance of each mixture is measured using a colorimeter. The results are shown below:
(a) Identify the independent variable and the dependent variable in this experiment. [2]
(b) Describe how a student would prepare 100 cm³ of the \(0.100\text{ mol dm}^{-3}\) solution of \(Cu^{2+}\) starting from solid copper(II) sulfate pentahydrate, \(CuSO_4 \cdot 5H_2O\). Show your working. [\(M_r = 249.6\)] [3]
(c) Analyze the experimental data: (i) Identify which mixture has an anomalous absorbance value. Suggest a practical error that could account for this anomaly. [2] (ii) Describe how the student should use a plotted graph of Absorbance (y-axis) against Volume of L (x-axis) to find the formula of the complex. [2] (iii) The ideal lines of best fit are described by the equations: \(A = 0.12 V_L\) \(A = 2.40 - 0.24 V_L\) Calculate the coordinates of the intersection point of these two lines, and determine the formula of the complex, \([CuL_n]^{2+}\). [3]
(d) Copper(II) ions in aqueous solution are pale blue, and the complex formed is deep purple-blue. Explain why a colorimeter is suitable for this experiment and state which color filter (or approximate wavelength of light) should be used. [2]
(e) Suggest one precaution the student must take when using a colorimeter to ensure that the measured absorbance values are reliable. [1]
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(a) The independent variable is the volume of L (or the composition of the mixture), as it is controlled and varied by the experimenter. The dependent variable is the absorbance, which is measured in response.
(b) To prepare the solution: 1. Calculate the required mass of \(CuSO_4 \cdot 5H_2O\): \(n = c \times V = 0.100\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.0100\text{ mol}\) \(m = 0.0100\text{ mol} \times 249.6\text{ g mol}^{-1} = 2.496\text{ g}\) (accept 2.50 g). 2. Weigh 2.496 g of the solid using a balance into a clean beaker and dissolve it completely in approximately 50 cm³ of distilled water. 3. Transfer the solution quantitatively into a 100 cm³ volumetric flask, rinse the beaker and glass rod with distilled water, transferring all washings to the flask. Make up to the mark with distilled water so the bottom of the meniscus is on the graduation line, then stopper and invert several times to mix.
(c)(i) Mixture 5 is anomalous because its absorbance (0.41) lies significantly below the expected linear trend established by Mixtures 1–4. This could be due to under-pipetting the ligand L, over-dilution, incomplete mixing, or a scratch/fingerprint on the cuvette. (ii) The student should draw two straight lines of best fit: one ascending line through the points where L is limiting (excluding the anomaly) and one descending line through the points where \(Cu^{2+}\) is limiting. The intersection point of these two lines represents the maximum concentration of the complex at stoichiometric equivalence. (iii) Set the two line equations equal to find the intersection: \(0.12 V_L = 2.40 - 0.24 V_L\) \(0.36 V_L = 2.40 \Rightarrow V_L = 6.67\text{ cm}^3\) \(V_{Cu} = 10.0 - 6.67 = 3.33\text{ cm}^3\) The mole ratio of \(L : Cu^{2+}\) is \(6.67 : 3.33 = 2 : 1\). Thus, \(n = 2\) and the formula of the complex is \([CuL_2]^{2+}\).
(d) A colorimeter is suitable because the complex absorbs light in the visible region, and its absorbance is directly proportional to its concentration (Beer-Lambert Law). Since the complex is deep purple-blue, a yellow or orange filter (approx. 580–620 nm) should be used to ensure maximum absorbance by the complex and minimal absorbance by the pale blue reactant.
(e) The student must wipe the outside of the cuvette with a lint-free tissue before placing it in the colorimeter to remove fingerprints, dust, or liquid droplets that could scatter light.
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- (a) Independent and dependent variables correctly identified: 2 marks (1 mark each). - (b) Mass calculation correct: 1 mark. Practical steps (dissolving, quantitative transfer, making to mark and mixing): 2 marks. - (c)(i) Mixture 5 identified: 1 mark. Logical reason linked to lower absorbance: 1 mark. - (c)(ii) Description of two straight lines of best fit and finding the intersection point: 2 marks. - (c)(iii) Correct calculation of intersection (V_L = 6.67 cm3): 1 mark. Correct volume of Cu2+ (3.33 cm3) and mole ratio: 1 mark. Correct formula [CuL2]2+: 1 mark. - (d) Suitable explanation (absorbance proportional to concentration): 1 mark. Correct filter color (yellow/orange): 1 mark. - (e) Correct precaution (e.g., wipe cuvette / calibrate with blank): 1 mark.
PastPaper.question 2 · analytical-evaluation-task
15 PastPaper.marks
A student wants to determine the partition coefficient, \(K_{pc}\), of ethanoic acid between water and butan-1-ol. The equation for the partition coefficient is: \(K_{pc} = \frac{[CH_3COOH]_{organic}}{[CH_3COOH]_{aqueous}}\)
Procedure: 1. Mix \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) aqueous ethanoic acid with \(50.0\text{ cm}^3\) of butan-1-ol in a separating funnel. 2. Shake the mixture thoroughly at intervals over 15 minutes, releasing pressure periodically. 3. Allow the mixture to stand until the two layers separate completely. 4. Run off the lower aqueous layer into a flask. 5. Pipette a \(10.00\text{ cm}^3\) sample of the aqueous layer into a conical flask and titrate against \(0.500\text{ mol dm}^{-3}\) aqueous sodium hydroxide, \(NaOH(aq)\), using phenolphthalein indicator. 6. Pipette a \(10.00\text{ cm}^3\) sample of the organic layer into a conical flask and titrate against \(0.500\text{ mol dm}^{-3}\) \(NaOH(aq)\).
Results: - Titration of \(10.00\text{ cm}^3\) of the aqueous layer: Initial burette reading = \(0.15\text{ cm}^3\); Final burette reading = \(14.45\text{ cm}^3\) - Titration of \(10.00\text{ cm}^3\) of the organic layer: Initial burette reading = \(1.20\text{ cm}^3\); Final burette reading = \(6.90\text{ cm}^3\)
(a) State why the separating funnel must be shaken thoroughly and then allowed to stand before separating the layers. [2]
(b) The density of water is \(1.00\text{ g cm}^{-3}\) and the density of butan-1-ol is \(0.81\text{ g cm}^{-3}\). Identify which layer is the upper layer. Explain your answer. [2]
(c) Phenolphthalein is used as the indicator. State the color change at the end-point of each titration. [2]
(d) (i) Calculate the concentration of ethanoic acid in both the aqueous and organic layers. Show your working. [3] (ii) Calculate the partition coefficient, \(K_{pc}\), of ethanoic acid between butan-1-ol and water. [1]
(e) The student repeated the experiment but used \(100.0\text{ cm}^3\) of butan-1-ol instead of \(50.0\text{ cm}^3\). State and explain the effect, if any, of this change on: (i) the value of the partition coefficient, \(K_{pc}\). [2] (ii) the concentration of ethanoic acid remaining in the aqueous layer. [3]
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(a) Shaking thoroughly ensures maximum contact between the two immiscible solvents so that ethanoic acid can partition between them until dynamic equilibrium is established. Allowing it to stand ensures that the two layers separate completely so that samples of each phase can be collected without cross-contamination.
(b) The upper layer is the organic layer (butan-1-ol). This is because butan-1-ol has a density of \(0.81\text{ g cm}^{-3}\), which is less than the density of water (\(1.00\text{ g cm}^{-3}\)).
(c) Since ethanoic acid (a weak acid) is being titrated with sodium hydroxide (a strong base), the flask initially contains acid and phenolphthalein is colorless. At the end-point, the solution changes color from colorless to permanent pale pink.
(d)(i) Calculate concentration in the aqueous layer: - Titre volume = \(14.45 - 0.15 = 14.30\text{ cm}^3\) - Moles of \(NaOH\) = \(0.500\text{ mol dm}^{-3} \times 0.01430\text{ dm}^3 = 7.15 \times 10^{-3}\text{ mol}\) - Moles of \(CH_3COOH\) in 10.00 cm³ = \(7.15 \times 10^{-3}\text{ mol}\) - \([CH_3COOH]_{aqueous} = \frac{7.15 \times 10^{-3}\text{ mol}}{0.01000\text{ dm}^3} = 0.715\text{ mol dm}^{-3}\)
Calculate concentration in the organic layer: - Titre volume = \(6.90 - 1.20 = 5.70\text{ cm}^3\) - Moles of \(NaOH\) = \(0.500\text{ mol dm}^{-3} \times 0.00570\text{ dm}^3 = 2.85 \times 10^{-3}\text{ mol}\) - Moles of \(CH_3COOH\) in 10.00 cm³ = \(2.85 \times 10^{-3}\text{ mol}\) - \([CH_3COOH]_{organic} = \frac{2.85 \times 10^{-3}\text{ mol}}{0.01000\text{ dm}^3} = 0.285\text{ mol dm}^{-3}\)
(e)(i) There is no effect on the value of \(K_{pc}\) because the partition coefficient is a temperature-dependent constant and does not change with varying volumes of the solvents.
(ii) The concentration of ethanoic acid remaining in the aqueous layer will decrease. Because a larger volume of organic solvent is used, a greater total mass of ethanoic acid must partition into the organic phase to maintain the constant value of \(K_{pc}\), leaving less in the aqueous layer.
PastPaper.markingScheme
- (a) Explanation of shaking (to establish equilibrium): 1 mark. Explanation of standing (to separate layers completely): 1 mark. - (b) Identified organic layer as upper layer: 1 mark. Linked correctly to density comparison (0.81 < 1.00 g cm-3): 1 mark. - (c) Colorless to permanent pale pink (must have both): 2 marks. - (d)(i) Correct concentration of aqueous layer (0.715 mol dm-3) with working: 1 mark. Correct concentration of organic layer (0.285 mol dm-3) with working: 1 mark. Consistent use of 3 significant figures and units: 1 mark. - (d)(ii) Correct calculation of K_pc (0.399): 1 mark. - (e)(i) Value of K_pc is unchanged: 1 mark. Reason (K_pc only depends on temperature): 1 mark. - (e)(ii) Concentration of acid in water decreases: 1 mark. Reason (larger organic volume shifts equilibrium to extract more acid / mathematically shown): 2 marks.