An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V2) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 12 (Multiple Choice)
Answer all forty multiple choice questions. Choose single best answer A, B, C or D.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
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A 15 cm3 sample of a gaseous hydrocarbon was exploded with an excess of oxygen (100 cm3). After cooling to room temperature, the remaining gas mixture occupied 85 cm3. On shaking this mixture with an excess of aqueous potassium hydroxide, the volume of gas decreased to 55 cm3. What is the molecular formula of the hydrocarbon?
A.CH4
B.C2H2
C.C2H4
D.C2H62-hydroxypropanenitrile is formed by the reaction of ethanal with hydrogen cyanide, HCN, in the presence of a sodium cyanide catalyst. What is the role of the cyanide ion in the rate-determining step of this reaction?
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PastPaper.workedSolution
On shaking with aqueous KOH, carbon dioxide is absorbed. The reduction in volume of 30 cm3 (85 cm3 to 55 cm3) represents the volume of CO2 produced. Thus, 15 cm3 of hydrocarbon produces 30 cm3 of CO2, meaning 1 mole of hydrocarbon contains 2 moles of carbon atoms. The remaining 55 cm3 of gas is the unreacted oxygen. The volume of oxygen reacted is 100 cm3 - 55 cm3 = 45 cm3. The ratio of hydrocarbon reacted to oxygen reacted is 15 : 45, which is 1 : 3. From the equation for the combustion of a hydrocarbon, CxHy + (x + y/4)O2 -> xCO2 + (y/2)H2O, we have x = 2 and x + y/4 = 3. Solving for y gives y = 4. Therefore, the formula is C2H4.
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Award 1 mark for the correct option C. Deduct 0 marks for incorrect options.
PastPaper.question 2 · multiple-choice
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Which species has a bond angle of approximately 107 degrees?
A.H3O+
B.NH4+
C.BF3
D.CO2
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PastPaper.workedSolution
The hydronium ion, H3O+, has three bonding pairs of electrons and one lone pair around the central oxygen atom. Its shape is trigonal pyramidal, and the lone pair-bonding pair repulsion reduces the bond angle from the tetrahedral angle of 109.5 degrees to approximately 107 degrees. NH4+ is tetrahedral (109.5 degrees), BF3 is trigonal planar (120 degrees), and CO2 is linear (180 degrees).
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Award 1 mark for the correct option A.
PastPaper.question 3 · multiple-choice
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The gaseous addition reaction between ethene and hydrogen bromide is shown: C2H4(g) + HBr(g) -> C2H5Br(g). Using the bond energy values provided, what is the enthalpy change of this reaction? Bond energies in kJ/mol: C=C is 614; C-C is 347; C-H is 413; H-Br is 366; C-Br is 290.
A.-70 kJ/mol
B.+70 kJ/mol
C.-343 kJ/mol
D.+343 kJ/mol
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PastPaper.workedSolution
To find the enthalpy change of the reaction, we calculate the energy required to break bonds minus the energy released when bonds are formed. Bonds broken: 1 x C=C (614 kJ/mol) and 1 x H-Br (366 kJ/mol), giving a total of 980 kJ/mol. Bonds formed: 1 x C-C (347 kJ/mol), 1 x C-H (413 kJ/mol), and 1 x C-Br (290 kJ/mol), giving a total of 1050 kJ/mol. Enthalpy change = energy for bonds broken - energy for bonds formed = 980 - 1050 = -70 kJ/mol.
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Award 1 mark for the correct option A.
PastPaper.question 4 · multiple-choice
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Two Period 3 elements, X and Y, react separately with oxygen to form oxides. The oxide of X is a white solid that reacts violently with water to form a strongly alkaline solution. The oxide of Y is a liquid at room temperature that reacts violently with water to form a strongly acidic solution. What are the identities of elements X and Y?
A.X is sodium and Y is phosphorus
B.X is sodium and Y is sulfur
C.X is magnesium and Y is phosphorus
D.X is magnesium and Y is sulfur
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PastPaper.workedSolution
Sodium oxide (Na2O) is a white solid that reacts violently with water to form a strongly alkaline solution of sodium hydroxide (pH approximately 13-14). Sulfur trioxide (SO3) is a volatile liquid at room temperature (under standard conditions) that reacts violently with water to form a strongly acidic solution of sulfuric acid. Phosphorus pentoxide (P4O10) is a white solid at room temperature. Magnesium oxide (MgO) is a white solid but only reacts sparingly with water to form a weakly alkaline suspension (pH approximately 9).
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Award 1 mark for the correct option B.
PastPaper.question 5 · multiple-choice
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But-1-ene reacts with hydrogen bromide, HBr, via an electrophilic addition mechanism. How many different organic products, including any stereoisomers, can be formed in this reaction?
A.2
B.3
C.4
D.5
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PastPaper.workedSolution
The electrophilic addition of HBr to but-1-ene produces two structural isomers depending on which carbon of the double bond receives the hydrogen atom. Addition of H to C1 gives the major product, 2-bromobutane. Addition of H to C2 gives the minor product, 1-bromobutane. 1-bromobutane is achiral and exists as a single compound. 2-bromobutane contains a chiral carbon atom (C2) and therefore exists as a pair of optical isomers (enantiomers). In total, there are 3 organic products: 1-bromobutane, (R)-2-bromobutane, and (S)-2-bromobutane.
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Award 1 mark for the correct option B.
PastPaper.question 6 · multiple-choice
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An alcohol, W, has the molecular formula C5H12O. W reacts with acidified potassium dichromate(VI) to form compound X. X reacts with alkaline aqueous iodine to form a yellow precipitate of triiodomethane. Dehydration of W with concentrated sulfuric acid produces a mixture of three isomeric alkenes (including stereoisomers). What is the IUPAC name of W?
A.3-methylbutan-2-ol
B.pentan-2-ol
C.pentan-3-ol
D.2-methylbutan-2-ol
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PastPaper.workedSolution
Since W reacts with acidified potassium dichromate(VI), it must be a primary or secondary alcohol. The oxidation product X undergoes the triiodomethane (iodoform) reaction, which means X must be a methyl ketone, containing the CH3-CO- group. Thus, W must contain the CH3-CH(OH)- group. Among the options, 3-methylbutan-2-ol and pentan-2-ol both contain this group. Dehydration of pentan-2-ol yields pent-1-ene, cis-pent-2-ene, and trans-pent-2-ene, which are three isomeric alkenes. Dehydration of 3-methylbutan-2-ol yields 3-methylbut-1-ene and 2-methylbut-2-ene (which does not show stereoisomerism), giving only two isomeric alkenes. Therefore, W is pentan-2-ol.
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Award 1 mark for the correct option B.
PastPaper.question 7 · multiple-choice
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An organic compound, Y, contains only carbon, hydrogen, and oxygen. Mass composition analysis shows that Y contains 54.5% carbon and 9.1% hydrogen by mass. In the mass spectrum of Y, the molecular ion peak, M+, is found at m/z = 88. What is the molecular formula of Y?
A.C2H4O
B.C3H6O2
C.C4H8O2
D.C4H10O2
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PastPaper.workedSolution
First, calculate the percentage of oxygen by mass: 100% - (54.5% + 9.1%) = 36.4%. Next, find the molar ratio of atoms by dividing each percentage by its respective relative atomic mass: C = 54.5 / 12.0 = 4.54; H = 9.1 / 1.0 = 9.1; O = 36.4 / 16.0 = 2.275. Divide by the smallest value (2.275) to find the simplest ratio: C = 2, H = 4, O = 1. The empirical formula of Y is therefore C2H4O, which has an empirical formula mass of 44. Since the molecular ion peak is at m/z = 88, the molecular mass of Y is 88. The molecular formula is (C2H4O) x (88 / 44) = C4H8O2.
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Award 1 mark for the correct option C.
PastPaper.question 8 · multiple-choice
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In a catalytic converter fitted to a car exhaust system, carbon monoxide and nitrogen monoxide react together to form nitrogen and carbon dioxide: 2CO(g) + 2NO(g) -> N2(g) + 2CO2(g). Which statement about this reaction is correct?
A.Carbon monoxide acts as an oxidizing agent and is reduced to carbon dioxide.
B.Nitrogen monoxide acts as a reducing agent and is oxidized to nitrogen gas.
C.The oxidation state of carbon increases from +2 to +4.
D.The oxidation state of nitrogen increases from +2 to +4.
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PastPaper.workedSolution
In the reaction, carbon monoxide (CO) is oxidized to carbon dioxide (CO2). The oxidation state of carbon increases from +2 in CO to +4 in CO2, making statement C correct. Since CO is oxidized, it acts as a reducing agent (making A incorrect). Nitrogen monoxide (NO) is reduced to nitrogen gas (N2). The oxidation state of nitrogen decreases from +2 in NO to 0 in N2 (making D incorrect). Since NO is reduced, it acts as an oxidizing agent (making B incorrect).
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Award 1 mark for the correct option C.
PastPaper.question 9 · multiple_choice
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A 1.405 g sample of an anhydrous metal carbonate, \( \text{MCO}_3 \), was reacted with an excess of dilute hydrochloric acid. The volume of carbon dioxide gas collected at room temperature and pressure (rtp) was 400 cm3. What is the identity of the metal M? [Take the molar volume of gas at rtp as 24.0 dm3 mol-1]
A.Mg
B.Ca
C.Sr
D.Ba
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PastPaper.workedSolution
First, write the equation for the reaction: \( \text{MCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \). Calculate the moles of carbon dioxide produced: \( n(\text{CO}_2) = \frac{400 \text{ cm}^3}{24000 \text{ cm}^3 \text{ mol}^{-1}} = 0.01667 \text{ mol} \). Since the mole ratio of \( \text{MCO}_3 \) to \( \text{CO}_2 \) is 1:1, there are 0.01667 mol of \( \text{MCO}_3 \) present. Calculate the molar mass (Mr) of \( \text{MCO}_3 \): \( M_r = \frac{1.405 \text{ g}}{0.01667 \text{ mol}} = 84.3 \text{ g mol}^{-1} \). Find the atomic mass of M: \( A_r(\text{M}) = 84.3 - 12.0 - 3(16.0) = 24.3 \text{ g mol}^{-1} \). This atomic mass matches magnesium (Mg).
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1 mark for the correct option A. Method: Calculate the moles of gas, deduce the moles of the metal carbonate, determine its molar mass, and subtract the carbonate mass to find the relative atomic mass of M.
PastPaper.question 10 · multiple_choice
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Using the standard enthalpy changes of combustion given below, what is the standard enthalpy change of formation of propan-1-ol, \( \text{C}_3\text{H}_7\text{OH(l)} \)? \( \Delta H_c^\ominus[\text{C(s)}] = -393.5 \text{ kJ mol}^{-1} \); \( \Delta H_c^\ominus[\text{H}_2\text{(g)}] = -285.8 \text{ kJ mol}^{-1} \); \( \Delta H_c^\ominus[\text{C}_3\text{H}_7\text{OH(l)}] = -2021.0 \text{ kJ mol}^{-1} \)
A.\( -302.7 \text{ kJ mol}^{-1} \)
B.\( +302.7 \text{ kJ mol}^{-1} \)
C.\( -2323.7 \text{ kJ mol}^{-1} \)
D.\( -4344.7 \text{ kJ mol}^{-1} \)
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PastPaper.workedSolution
The equation for the standard enthalpy of formation of propan-1-ol is: \( 3\text{C(s)} + 4\text{H}_2\text{(g)} + 0.5\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)} \). Using Hess's law, \( \Delta H_f^\ominus[\text{propan-1-ol}] = 3 \times \Delta H_c^\ominus[\text{C}] + 4 \times \Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{C}_3\text{H}_7\text{OH}] \). Calculation: \( \Delta H_f^\ominus = 3(-393.5) + 4(-285.8) - (-2021.0) = -1180.5 - 1143.2 + 2021.0 = -302.7 \text{ kJ mol}^{-1} \).
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1 mark for the correct option A. Method: Apply Hess's law using standard combustion values of elements minus the combustion value of the compound.
PastPaper.question 11 · multiple_choice
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Which row correctly describes the structural type of sulfur trioxide, \( \text{SO}_3 \), and aluminium oxide, \( \text{Al}_2\text{O}_3 \), and their chemical reactions?
A.Structure of \( \text{SO}_3 \): simple molecular; Structure of \( \text{Al}_2\text{O}_3 \): giant ionic; Action of water on \( \text{SO}_3 \): forms an acidic solution; Action of warm HCl(aq) on \( \text{Al}_2\text{O}_3 \): reacts to form a salt and water.
B.Structure of \( \text{SO}_3 \): giant covalent; Structure of \( \text{Al}_2\text{O}_3 \): giant ionic; Action of water on \( \text{SO}_3 \): forms an alkaline solution; Action of warm HCl(aq) on \( \text{Al}_2\text{O}_3 \): does not react.
C.Structure of \( \text{SO}_3 \): simple molecular; Structure of \( \text{Al}_2\text{O}_3 \): giant covalent; Action of water on \( \text{SO}_3 \): forms an acidic solution; Action of warm HCl(aq) on \( \text{Al}_2\text{O}_3 \): does not react.
D.Structure of \( \text{SO}_3 \): simple molecular; Structure of \( \text{Al}_2\text{O}_3 \): giant ionic; Action of water on \( \text{SO}_3 \): does not react; Action of warm HCl(aq) on \( \text{Al}_2\text{O}_3 \): reacts to form a salt and water.
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PastPaper.workedSolution
Sulfur trioxide, \( \text{SO}_3 \), has a simple molecular structure and reacts vigorously with water to form sulfuric acid, resulting in a strongly acidic solution. Aluminium oxide, \( \text{Al}_2\text{O}_3 \), has a giant ionic structure with significant covalent character. It is amphoteric and reacts with acids like hot hydrochloric acid to form a salt and water.
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1 mark for the correct option A. Method: Identify the correct structure and reactivity trends of the Period 3 oxides with water and acids.
PastPaper.question 12 · multiple_choice
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Which of the following molecules or ions has the smallest bond angle?
A.\( \text{NH}_4^+ \)
B.\( \text{NH}_3 \)
C.\( \text{NH}_2^- \)
D.\( \text{BF}_3 \)
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PastPaper.workedSolution
Determine the shape and bond angles using VSEPR theory. NH4+ has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape with a bond angle of 109.5 degrees. NH3 has 3 bonding pairs and 1 lone pair, giving a trigonal pyramidal shape with a bond angle of 107 degrees. NH2- has 2 bonding pairs and 2 lone pairs, giving a non-linear (bent) shape with a bond angle of 104.5 degrees. BF3 has 3 bonding pairs and 0 lone pairs, giving a trigonal planar shape with a bond angle of 120 degrees. NH2- has the smallest bond angle.
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1 mark for the correct option C. Method: Determine the number of lone pairs and bonding pairs of electrons around the central atom to deduce the bond angles.
PastPaper.question 13 · multiple_choice
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When but-1-ene is reacted with concentrated hydrogen bromide at room temperature, a mixture of two structural isomers is formed. Which statement correctly describes this reaction?
A.The major product is 2-bromobutane because the reaction proceeds via the more stable secondary carbocation intermediate.
B.The major product is 1-bromobutane because the reaction proceeds via the more stable primary carbocation intermediate.
C.The major product is 2-bromobutane because the reaction proceeds via the more stable primary carbocation intermediate.
D.The reaction is a nucleophilic addition reaction, and 1-bromobutane is the major product.
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PastPaper.workedSolution
The reaction of but-1-ene with HBr is an electrophilic addition reaction. The major product is 2-bromobutane because the reaction proceeds via the more stable secondary carbocation intermediate, \( \text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3 \), rather than the less stable primary carbocation intermediate, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+ \).
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1 mark for the correct option A. Method: Identify the reaction type and recall carbocation stability rules (Markovnikov's rule).
PastPaper.question 14 · multiple_choice
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An organic compound X with the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \) is oxidized by acidified potassium dichromate(VI) to form Y. Compound Y does not react with Tollens' reagent. When X is warmed with alkaline aqueous iodine, a yellow precipitate is formed. What is the identity of X?
A.butan-1-ol
B.butan-2-ol
C.2-methylpropan-1-ol
D.2-methylpropan-2-ol
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PastPaper.workedSolution
Since X can be oxidized to Y, X must be a primary or secondary alcohol. Since Y does not react with Tollens' reagent, Y must be a ketone, which means X is a secondary alcohol. Since X reacts with alkaline aqueous iodine to give a yellow precipitate (the tri-iodomethane reaction), X must contain the \( \text{CH}_3\text{CH(OH)}- \) group. The only isomer of \( \text{C}_4\text{H}_{10}\text{O} \) that is a secondary alcohol with this structural feature is butan-2-ol.
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1 mark for the correct option B. Method: Use oxidation products and positive iodoform test results to deduce the specific alcohol structure.
PastPaper.question 15 · multiple_choice
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When a solid potassium halide, KX, is reacted with concentrated sulfuric acid, a gas is evolved that turns damp blue litmus paper red, and a dark brown vapor is observed along with a choking gas. What is the identity of the halide ion, \( \text{X}^- \)?
A.fluoride
B.chloride
C.bromide
D.iodide
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PastPaper.workedSolution
Solid potassium bromide (KBr) reacts with concentrated sulfuric acid to produce hydrogen bromide gas (which turns damp blue litmus paper red) and is also oxidized by the sulfuric acid to produce bromine gas (a dark brown vapor) and sulfur dioxide (a choking gas).
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1 mark for the correct option C. Method: Link the specific redox reactions and gaseous products of halide salts with concentrated sulfuric acid to the identity of the halide.
PastPaper.question 16 · multiple_choice
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The gas-phase reaction shown is allowed to reach equilibrium in a closed container of volume 1.0 dm3 at a constant temperature: \( \text{A(g)} + 2\text{B(g)} \rightleftharpoons \text{C(g)} + \text{D(g)} \) Initially, the container held 1.5 mol of A and 3.0 mol of B. At equilibrium, 0.50 mol of C is present. What is the value of the equilibrium constant, Kc, at this temperature?
A.\( 0.0625 \text{ dm}^3\text{ mol}^{-1} \)
B.\( 0.125 \text{ dm}^3\text{ mol}^{-1} \)
C.\( 0.040 \text{ dm}^3\text{ mol}^{-1} \)
D.\( 0.250 \text{ dm}^3\text{ mol}^{-1} \)
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PastPaper.workedSolution
Set up an ICE (Initial, Change, Equilibrium) table. For A: Initial = 1.5 mol, Change = -0.50 mol, Equilibrium = 1.0 mol. For B: Initial = 3.0 mol, Change = -2(0.50) = -1.0 mol, Equilibrium = 2.0 mol. For C: Initial = 0 mol, Change = +0.50 mol, Equilibrium = 0.50 mol. For D: Initial = 0 mol, Change = +0.50 mol, Equilibrium = 0.50 mol. Since the volume is 1.0 dm3, the concentrations are equal to the moles. Calculate Kc: \( K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]^2} = \frac{0.50 \times 0.50}{1.0 \times (2.0)^2} = \frac{0.25}{4.0} = 0.0625 \text{ dm}^3\text{ mol}^{-1} \).
PastPaper.markingScheme
1 mark for the correct option A. Method: Construct the ICE table using stoichiometry, write the correct expression for Kc, and substitute the equilibrium concentrations.
PastPaper.question 17 · Multiple Choice
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A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is reacted with \(100\text{ cm}^3\) of oxygen (an excess). After complete combustion and cooling to room temperature and pressure, the remaining gas volume is \(80\text{ cm}^3\). This gas mixture is then passed through excess aqueous sodium hydroxide, and the volume decreases to \(40\text{ cm}^3\). All volumes are measured at the same temperature and pressure. What is the molecular formula of the hydrocarbon?
A.\(\text{C}_4\text{H}_6\)
B.\(\text{C}_4\text{H}_8\)
C.\(\text{C}_4\text{H}_{10}\)
D.\(\text{C}_3\text{H}_8\)
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PastPaper.workedSolution
1. Find the volume of \(\text{CO}_2\) produced: Passing the gas mixture through aqueous \(\text{NaOH}\) removes \(\text{CO}_2\). The decrease in volume is \(80\text{ cm}^3 - 40\text{ cm}^3 = 40\text{ cm}^3\) of \(\text{CO}_2\). 2. Find the volume of excess and reacted oxygen: The remaining gas after \(\text{NaOH}\) absorption is the excess \(\text{O}_2\), which is \(40\text{ cm}^3\). The volume of \(\text{O}_2\) reacted is \(100\text{ cm}^3 - 40\text{ cm}^3 = 60\text{ cm}^3\). 3. Determine the ratio of reactants and products: \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(40\text{ cm}^3\) of \(\text{CO}_2\), so 1 mole of hydrocarbon produces 4 moles of \(\text{CO}_2\). Thus, \(x = 4\). \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) reacts with \(60\text{ cm}^3\) of \(\text{O}_2\). Using the general equation \(\text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \rightarrow x\text{CO}_2 + y/2\text{H}_2\text{O}\), the mole ratio of hydrocarbon to oxygen is \(1 : (x + y/4)\). Therefore, \(4 + y/4 = 6\), which gives \(y/4 = 2\), so \(y = 8\). The molecular formula is \(\text{C}_4\text{H}_8\).
PastPaper.markingScheme
1 mark for the correct option B. (Method: Correctly deduce carbon content from CO2 volume [1/2 mark]; correctly deduce hydrogen content from reacted O2 volume [1/2 mark])
PastPaper.question 18 · Multiple Choice
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The enthalpy changes of solution of anhydrous calcium chloride, \(\text{CaCl}_2(s)\), and calcium chloride dihydrate, \(\text{CaCl}_2\cdot2\text{H}_2\text{O}(s)\), in excess water are \(-81.3\text{ kJ mol}^{-1}\) and \(+8.7\text{ kJ mol}^{-1}\) respectively. What is the enthalpy change, \(\Delta H\), for the hydration reaction shown below? \(\text{CaCl}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{CaCl}_2\cdot2\text{H}_2\text{O}(s)\)
A.\(-90.0\text{ kJ mol}^{-1}\)
B.\(-72.6\text{ kJ mol}^{-1}\)
C.\(+72.6\text{ kJ mol}^{-1}\)
D.\(+90.0\text{ kJ mol}^{-1}\)
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PastPaper.workedSolution
According to Hess's Law, the hydration enthalpy can be calculated using a cycle where both anhydrous and hydrated salts dissolve in water to form the same solution: \(\Delta H_{\text{hydration}} + \Delta H_{\text{sol}}(\text{CaCl}_2\cdot2\text{H}_2\text{O}) = \Delta H_{\text{sol}}(\text{CaCl}_2)\). Therefore, \(\Delta H_{\text{hydration}} = \Delta H_{\text{sol}}(\text{CaCl}_2) - \Delta H_{\text{sol}}(\text{CaCl}_2\cdot2\text{H}_2\text{O}) = -81.3 - (+8.7) = -90.0\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
1 mark for correct option A. (Method: Correct construction of Hess's cycle [1/2 mark]; correct calculation with signs [1/2 mark])
PastPaper.question 19 · Multiple Choice
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An oxide of a Period 3 element, \(X\), is a solid at room temperature and pressure. When added to water, it reacts to form a strongly acidic solution with a pH of approximately 1–2. When this oxide reacts with excess aqueous sodium hydroxide, it forms a salt. What is the element \(X\)?
A.Sodium
B.Silicon
C.Phosphorus
D.Sulfur
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PastPaper.workedSolution
Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), is a solid at room temperature and pressure. It reacts vigorously with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\), which is a strong acid and gives a pH of around 1-2. It also reacts with sodium hydroxide to form sodium phosphate salt. Sulfur trioxide, \(\text{SO}_3\), is a liquid/gas at room temperature and pressure, and silicon dioxide is an insoluble giant covalent solid that does not react with water.
PastPaper.markingScheme
1 mark for the correct option C. (Method: Eliminate sodium and silicon due to pH and reactivity [1/2 mark]; differentiate phosphorus from sulfur based on physical state at room temperature [1/2 mark])
PastPaper.question 20 · Multiple Choice
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Which molecule has the smallest bond angle?
A.\(\text{BF}_3\)
B.\(\text{CH}_4\)
C.\(\text{H}_2\text{O}\)
D.\(\text{SF}_6\)
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PastPaper.workedSolution
The bond angles are as follows: \(\text{BF}_3\) is trigonal planar with a bond angle of \(120^\circ\). \(\text{CH}_4\) is tetrahedral with a bond angle of \(109.5^\circ\). \(\text{H}_2\text{O}\) has a non-linear shape with two lone pairs, resulting in a bond angle of \(104.5^\circ\). \(\text{SF}_6\) has an octahedral shape with six bonding pairs and no lone pairs on the central sulfur atom, resulting in bond angles of \(90^\circ\) (and \(180^\circ\)). Therefore, \(\text{SF}_6\) has the smallest bond angle.
PastPaper.markingScheme
1 mark for the correct option D. (Method: Recall correct geometry of each molecule [1/2 mark]; identify 90 degrees as the smallest angle [1/2 mark])
PastPaper.question 21 · Multiple Choice
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When 2-methylbut-2-ene reacts with cold concentrated hydrobromic acid, a mixture of two isomeric bromoalkanes is formed. Which statement correctly describes the major organic product and the mechanism of its formation?
A.The major product is 2-bromo-3-methylbutane, formed via a secondary carbocation intermediate.
B.The major product is 2-bromo-2-methylbutane, formed via a tertiary carbocation intermediate.
C.The major product is 2-bromo-2-methylbutane, formed via a secondary carbocation intermediate.
D.The major product is 2-bromo-3-methylbutane, formed via a tertiary carbocation intermediate.
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PastPaper.workedSolution
2-methylbut-2-ene is an unsymmetrical alkene. During electrophilic addition of \(\text{HBr}\), the major product is determined by the stability of the carbocation intermediate (Markovnikov's rule). Addition of \(\text{H}^+\) to carbon-3 forms a tertiary carbocation on carbon-2, which is more stable than the secondary carbocation formed if \(\text{H}^+\) adds to carbon-2. Subsequent attack by the bromide ion on the tertiary carbocation yields 2-bromo-2-methylbutane as the major product.
PastPaper.markingScheme
1 mark for the correct option B. (Method: Identify carbocation stability [1/2 mark]; identify major product structure [1/2 mark])
PastPaper.question 22 · Multiple Choice
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An organic compound \(Y\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\). When \(Y\) is reacted with acidified potassium dichromate(VI), the organic product reacts with 2,4-dinitrophenylhydrazine reagent but does not react with Tollens' reagent. What is the IUPAC name of compound \(Y\)?
A.butan-1-ol
B.butan-2-ol
C.2-methylpropan-1-ol
D.2-methylpropan-2-ol
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PastPaper.workedSolution
Compound \(Y\) with formula \(\text{C}_4\text{H}_{10}\text{O}\) can be an alcohol. Since it reacts with acidified potassium dichromate(VI), it cannot be a tertiary alcohol, eliminating 2-methylpropan-2-ol. The oxidation product reacts with 2,4-DNPH, which means it contains a carbonyl group (aldehyde or ketone). Since the oxidation product does not react with Tollens' reagent, it must be a ketone rather than an aldehyde. Therefore, the starting material \(Y\) must be a secondary alcohol. Butan-2-ol is the only secondary alcohol among the options.
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1 mark for correct option B. (Method: Differentiate secondary alcohol from primary and tertiary using the test results [1/2 mark]; correctly match to IUPAC name [1/2 mark])
PastPaper.question 23 · Multiple Choice
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When solid potassium halide \(\text{KX}\) is reacted with concentrated sulfuric acid, a gas is evolved that turns damp blue litmus paper red, and a purple vapor is also observed. Which statement about the halide ion present is correct?
A.It is oxidized to its element, which acts as a strong reducing agent.
B.It produces a gas that forms a white precipitate with acidified aqueous silver nitrate.
C.It is a stronger reducing agent than the chloride ion.
D.It reduces the sulfur in sulfuric acid to an oxidation state of +4 only.
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PastPaper.workedSolution
The purple vapor observed is iodine gas (\(\text{I}_2\)), which indicates that the potassium halide is potassium iodide (\(\text{KI}\)). The iodide ion (\(\text{I}^-\)) is oxidized to iodine, and concentrated sulfuric acid is reduced to hydrogen sulfide (\(\text{H}_2\text{S}\), oxidation state -2), sulfur (\(\text{S}\), oxidation state 0), and sulfur dioxide (\(\text{SO}_2\), oxidation state +4). The iodide ion is a stronger reducing agent than the chloride ion because its valence electrons are further from the nucleus and more easily lost.
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1 mark for correct option C. (Method: Identify iodide ion from observation [1/2 mark]; evaluate reducing strength trend [1/2 mark])
PastPaper.question 24 · Multiple Choice
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How many total ions are present in \(13.35\text{ g}\) of anhydrous aluminium chloride, \(\text{AlCl}_3\)? [Take the Avogadro constant, \(L\), to be \(6.02 \times 10^{23}\text{ mol}^{-1}\); \(A_r\): \(\text{Al} = 27.0\), \(\text{Cl} = 35.5\)]
A.\(6.02 \times 10^{22}\)
B.\(1.81 \times 10^{23}\)
C.\(2.41 \times 10^{23}\)
D.\(9.63 \times 10^{23}\)
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1. Calculate the molar mass of \(\text{AlCl}_3\): \(M_r = 27.0 + 3 \times 35.5 = 133.5\text{ g mol}^{-1}\). 2. Find the moles of \(\text{AlCl}_3\): \(n = 13.35\text{ g} / 133.5\text{ g mol}^{-1} = 0.100\text{ mol}\). 3. Determine the total moles of ions: Each formula unit of \(\text{AlCl}_3\) contains 4 ions (one \(\text{Al}^{3+}\) and three \(\text{Cl}^-\)). Thus, \(0.100\text{ mol}\) of \(\text{AlCl}_3\) contains \(0.100 \times 4 = 0.400\text{ mol}\) of ions. 4. Calculate the number of ions: \(0.400 \times 6.02 \times 10^{23}\text{ mol}^{-1} = 2.41 \times 10^{23}\) ions.
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1 mark for the correct option C. (Method: Correctly calculate the moles of aluminium chloride [1/2 mark]; multiply by 4 and Avogadro's constant to get the total number of ions [1/2 mark])
PastPaper.question 25 · multiple-choice
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A sample of 10 cm\(^3\) of a gaseous hydrocarbon \(X\) is completely burned in 70 cm\(^3\) of oxygen (an excess). After cooling to room temperature and pressure, the total volume of gas remaining is 50 cm\(^3\). Shaking this remaining gas with excess aqueous sodium hydroxide reduces the volume to 10 cm\(^3\). What is the molecular formula of hydrocarbon \(X\)?
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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PastPaper.workedSolution
1. Let the hydrocarbon be \(\text{C}_x\text{H}_y\). 2. The combustion equation is: \(\text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O(l)}\) 3. At room temperature and pressure, water is a liquid and has negligible volume. 4. Shaking the final gas mixture with aqueous NaOH absorbs all \(\text{CO}_2\). The decrease in volume is \(50\text{ cm}^3 - 10\text{ cm}^3 = 40\text{ cm}^3\), which represents the volume of \(\text{CO}_2\) produced. 5. Therefore, 10 cm\(^3\) of \(\text{C}_x\text{H}_y\) produces 40 cm\(^3\) of \(\text{CO}_2\), meaning \(x = 4\). 6. The remaining 10 cm\(^3\) of gas is unreacted excess oxygen, meaning the volume of oxygen reacted is \(70\text{ cm}^3 - 10\text{ cm}^3 = 60\text{ cm}^3\). 7. Using the oxygen stoichiometry: \(10 \times (x + \frac{y}{4}) = 60 \implies 4 + \frac{y}{4} = 6 \implies y = 8\). 8. Thus, the molecular formula is \(\text{C}_4\text{H}_8\).
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Award 1 mark for the correct option C. - Method: Determine volume of CO2 produced (40 cm3) and O2 reacted (60 cm3). - Formulate mole ratio of C : H to identify the molecular formula as C4H8. - Reject incorrect options: A (gives 30 cm3 CO2), B (gives 30 cm3 CO2), D (requires 65 cm3 O2).
PastPaper.question 26 · multiple-choice
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The standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), for carbon (graphite), hydrogen gas, and liquid propan-1-ol are given.
What is the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of propan-1-ol?
A.\(-305\text{ kJ mol}^{-1}\)
B.\(-1341\text{ kJ mol}^{-1}\)
C.\(+305\text{ kJ mol}^{-1}\)
D.\(+1341\text{ kJ mol}^{-1}\)
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PastPaper.workedSolution
The equation for the formation of propan-1-ol is: \(3\text{C(s)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\)
Award 1 mark for option A. - Method: Correctly set up the Hess's Law expression using 3 moles of C and 4 moles of H2. - Accuracy: Perform calculation to arrive at -305 kJ mol-1. - Incorrect options represent sign errors or incorrect stoichiometry (e.g., using 1 mole of each reactant).
PastPaper.question 27 · multiple-choice
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An equimolar mixture of two Period 3 oxides is added to excess water. The resulting solution is neutral with a pH of 7.0.
Which pair of oxides could be the components of this mixture?
A.\(\text{Na}_2\text{O}\) and \(\text{SO}_3\)
B.\(\text{MgO}\) and \(\text{Al}_2\text{O}_3\)
C.\(\text{Al}_2\text{O}_3\) and \(\text{P}_4\text{O}_{10}\)
D.\(\text{SiO}_2\) and \(\text{P}_4\text{O}_{10}\)
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PastPaper.workedSolution
Option A is correct because \(\text{Na}_2\text{O}\) reacts with water to form a strong alkali: \(\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{Na}^+\text{(aq)} + 2\text{OH}^-\text{(aq)}\)
While \(\text{SO}_3\) reacts with water to form a strong acid: \(\text{SO}_3\text{(g)} + \text{H}_2\text{O(l)} \rightarrow 2\text{H}^+\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}\)
In an equimolar mixture, 1 mole of \(\text{Na}_2\text{O}\) produces 2 moles of \(\text{OH}^-\), which perfectly neutralises the 2 moles of \(\text{H}^+\) produced by 1 mole of \(\text{SO}_3\) to form a neutral solution of \(\text{Na}_2\text{SO}_4\) with pH 7.0.
The other options do not result in a neutral solution because they contain insoluble oxides (\(\text{Al}_2\text{O}_3\), \(\text{SiO}_2\)) or oxides that do not yield equal amounts of strong acid/base when mixed equimolarly.
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Award 1 mark for option A. - Identify that Na2O forms NaOH (aq) and SO3 forms H2SO4 (aq). - Recognize that 1:1 molar ratio of Na2O to SO3 yields a neutral 2:2 ratio of hydroxide to hydrogen ions. - Reject other options where either one or both oxides are insoluble, or do not neutralize to pH 7.
PastPaper.question 28 · multiple-choice
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Which covalent molecule is planar and has all bond angles equal to \(120^\circ\)?
A.\(\text{NCl}_3\)
B.\(\text{BCl}_3\)
C.\(\text{OF}_2\)
D.\(\text{SF}_4\)
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- \(\text{NCl}_3\): Central nitrogen atom has 3 bonding pairs and 1 lone pair, adopting a trigonal pyramidal shape with bond angles of approximately \(107^\circ\) (non-planar). - \(\text{BCl}_3\): Central boron atom has 3 bonding pairs and 0 lone pairs, adopting a trigonal planar shape with all bond angles exactly \(120^\circ\). - \(\text{OF}_2\): Central oxygen atom has 2 bonding pairs and 2 lone pairs, adopting a bent shape with a bond angle of approximately \(103^\circ\). - \(\text{SF}_4\): Central sulfur atom has 4 bonding pairs and 1 lone pair, adopting a seesaw shape (non-planar).
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Award 1 mark for option B. - Correctly apply VSEPR theory to deduce the shape and bond angles of BCl3. - Identify that only BCl3 satisfies both criteria (planar and 120 degree bond angles).
PastPaper.question 29 · multiple-choice
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An alkene, \(Y\), reacts with hot, concentrated, acidified potassium manganate(VII) to produce a single organic compound, \(Z\).
Compound \(Z\) reacts with alkaline aqueous iodine to give a yellow precipitate, but has no reaction when warmed with Tollens' reagent.
What is the identity of \(Y\)?
A.but-2-ene
B.2-methylbut-2-ene
C.2,3-dimethylbut-2-ene
D.hex-3-ene
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PastPaper.workedSolution
1. Symmetrical cleavage of \(Y\) by hot, concentrated \(\text{KMnO}_4\) gives a single organic compound \(Z\). 2. Since \(Z\) does not react with Tollens' reagent, \(Z\) is a ketone (not an aldehyde). 3. Since \(Z\) reacts with alkaline aqueous iodine to give a yellow precipitate (triiodomethane), it must contain a methyl ketone group, \(\text{CH}_3\text{CO}-\). 4. Oxidation of 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\), splits the double bond to yield two molecules of propanone, \(\text{CH}_3\text{COCH}_3\) (a single product). 5. Propanone is a methyl ketone (gives a positive iodoform test) and is not oxidized by Tollens' reagent.
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Award 1 mark for option C. - Deduce that Z is a ketone with a methyl group adjacent to the carbonyl (methyl ketone) due to negative Tollens' and positive iodoform tests. - Identify 2,3-dimethylbut-2-ene as the only alkene which symmetric cleavage yields propanone.
PastPaper.question 30 · multiple-choice
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An organic compound \(W\) has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}_2\) and contains two hydroxyl groups.
When \(W\) is heated under reflux with excess acidified potassium dichromate(VI), the organic product formed, \(V\), does not react with sodium carbonate, nor does it react with Tollens' reagent.
Which compound could be \(W\)?
A.butane-1,2-diol
B.butane-1,4-diol
C.butane-2,3-diol
D.2-methylpropane-1,2-diol
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PastPaper.workedSolution
- Acidified potassium dichromate(VI) under reflux oxidizes primary alcohols to carboxylic acids, which react with \(\text{Na}_2\text{CO}_3\). - It oxidizes secondary alcohols to ketones, which do not react with \(\text{Na}_2\text{CO}_3\) or Tollens' reagent. - Tertiary alcohols are not oxidized. - For \(V\) to not react with sodium carbonate or Tollens' reagent, \(W\) must not contain any primary alcohol groups (ruling out butane-1,2-diol, butane-1,4-diol, and 2-methylpropane-1,2-diol). - butane-2,3-diol, \(\text{CH}_3\text{CH(OH)CH(OH)CH}_3\), contains two secondary alcohol groups. Refluxing with excess dichromate oxidizes it to butane-2,3-dione, \(\text{CH}_3\text{COCOCH}_3\), which is a diketone. This product contains no carboxylic acid or aldehyde groups, consistent with the observations.
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Award 1 mark for option C. - Recognize that primary alcohols oxidize to carboxylic acids, which would react with Na2CO3. - Identify that only butane-2,3-diol lacks primary alcohol groups, oxidizing instead to a diketone.
PastPaper.question 31 · multiple-choice
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When solid sodium halide, \(\text{Na}X\), is reacted with concentrated sulfuric acid, a gaseous mixture is produced containing a toxic gas with a rotten-egg smell (\(\text{H}_2\text{S}\)) alongside purple vapors of a halogen.
Which halide ion is present in \(\text{Na}X\), and what role does the concentrated sulfuric acid play in the formation of \(\text{H}_2\text{S}\)?
A.Halide ion: \(\text{Br}^-\); Role of \(\text{H}_2\text{SO}_4\): acid
B.Halide ion: \(\text{Br}^-\); Role of \(\text{H}_2\text{SO}_4\): oxidizing agent
C.Halide ion: \(\text{I}^-\); Role of \(\text{H}_2\text{SO}_4\): acid
D.Halide ion: \(\text{I}^-\); Role of \(\text{H}_2\text{SO}_4\): oxidizing agent
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PastPaper.workedSolution
1. Solid sodium iodide, \(\text{NaI}\), reacts with concentrated sulfuric acid to produce hydrogen iodide (\(\text{HI}\)) and purple vapors of iodine (\(\text{I}_2\)). 2. Iodide, \(\text{I}^-\), is a very powerful reducing agent and reduces sulfur in \(\text{H}_2\text{SO}_4\) from oxidation state +6 to -2 in \(\text{H}_2\text{S}\) (which has a rotten-egg smell). 3. Concentrated sulfuric acid is reduced, so it acts as an oxidizing agent in this reaction.
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Award 1 mark for option D. - Identify iodide as the only halide strong enough to reduce sulfuric acid to H2S. - State that sulfuric acid behaves as an oxidizing agent in this redox step (oxidizing iodide to iodine).
PastPaper.question 32 · multiple-choice
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A mixture of 2.00 mol of nitrogen gas and 3.00 mol of hydrogen gas is placed in a sealed container of volume 1.00 dm\(^3\) and allowed to reach equilibrium at a constant temperature.
Award 1 mark for option B. - Determine the equilibrium concentrations of N2 and H2 to be 1.50 mol dm-3. - Correctly substitute values into the Kc expression with correct powers. - Accuracy: Calculate the value as 0.198 (accept 0.20).
PastPaper.question 33 · Multiple Choice
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A sample of pure calcium carbonate, \(\text{CaCO}_3\), reacts with excess dilute hydrochloric acid: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). If \(0.400\text{ g}\) of a sample containing \(\text{CaCO}_3\) and an inert impurity produces \(84.0\text{ cm}^3\) of carbon dioxide gas at r.t.p., what is the percentage by mass of \(\text{CaCO}_3\) in the sample? [Assume 1 mole of gas occupies \(24.0\text{ dm}^3\) at r.t.p. \(M_r\) of \(\text{CaCO}_3 = 100.1\)]
A.35.0%
B.70.0%
C.87.6%
D.93.8%
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PastPaper.workedSolution
First, calculate the number of moles of carbon dioxide gas evolved: \(n(\text{CO}_2) = \frac{84.0}{24000} = 0.00350\text{ mol}\). According to the stoichiometric equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1, so \(n(\text{CaCO}_3) = 0.00350\text{ mol}\). Next, calculate the mass of pure calcium carbonate: \(\text{mass} = 0.00350 \times 100.1 = 0.35035\text{ g}\). Finally, find the percentage by mass of \(\text{CaCO}_3\) in the original sample: \(\frac{0.35035}{0.400} \times 100 = 87.5875\%\), which rounds to \(87.6\%\).
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Award 1 mark for the correct answer C. Method: Calculate moles of gas, relate to reactant stoichiometry, calculate mass of calcium carbonate, and find percentage purity.
PastPaper.question 34 · Multiple Choice
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Use the standard enthalpy changes of combustion, \(\Delta H_c^\theta\), to calculate the standard enthalpy change of formation, \(\Delta H_f^\theta\), of liquid propanoic acid, \(\text{CH}_3\text{CH}_2\text{CO}_2\text{H}(\text{l})\). Reaction: \(3\text{C(graphite)} + 3\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{CH}_2\text{CO}_2\text{H}(\text{l})\). Standard combustion values: \(\text{C(graphite)} = -393.5\text{ kJ mol}^{-1}\); \(\text{H}_2(\text{g}) = -285.8\text{ kJ mol}^{-1}\); \(\text{CH}_3\text{CH}_2\text{CO}_2\text{H}(\text{l}) = -1527.0\text{ kJ mol}^{-1}\).
Award 1 mark for the correct answer A. Method: Correctly apply Hess's cycle for combustion values (reactants minus products) and calculate the final value with correct sign.
PastPaper.question 35 · Multiple Choice
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Two white solids, X and Y, are oxides of different Period 3 elements. Solid X is insoluble in water but dissolves in both dilute aqueous sodium hydroxide and dilute hydrochloric acid. Solid Y reacts vigorously with water to form an acidic solution with a pH of approximately 2. Which row correctly identifies X and Y?
A.X is \(\text{SiO}_2\); Y is \(\text{SO}_3\)
B.X is \(\text{Al}_2\text{O}_3\); Y is \(\text{P}_4\text{O}_{10}\)
C.X is \(\text{Al}_2\text{O}_3\); Y is \(\text{SiO}_2\)
D.X is \(\text{MgO}\); Y is \(\text{P}_4\text{O}_{10}\)
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PastPaper.workedSolution
Solid X dissolves in both acid and base, which means it is amphoteric. The only amphoteric oxide in Period 3 is aluminium oxide, \(\text{Al}_2\text{O}_3\). Solid Y is a white solid that reacts vigorously with water to form an acidic solution with pH of approximately 2, which corresponds to phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), which forms phosphoric(V) acid, \(\text{H}_3\text{PO}_4\). Therefore, X is \(\text{Al}_2\text{O}_3\) and Y is \(\text{P}_4\text{O}_{10}\).
PastPaper.markingScheme
Award 1 mark for the correct answer B. Method: Recognize X as amphoteric oxide and Y as a covalent acidic oxide that is solid.
PastPaper.question 36 · Multiple Choice
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Which species has a molecular shape that is best described as T-shaped, with bond angles of approximately \(86^\circ\)?
A.\(\text{ClF}_3\)
B.\(\text{BF}_3\)
C.\(\text{NH}_3\)
D.\(\text{SF}_6\)
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PastPaper.workedSolution
Chlorine trifluoride, \(\text{ClF}_3\), has 5 pairs of electrons around the central chlorine atom (3 bonding pairs and 2 lone pairs). It adopts a trigonal bipyramidal electron-pair geometry. To minimize lone pair-lone pair repulsions, the two lone pairs occupy equatorial positions, leaving the three bonding fluorine atoms in a T-shape. The repulsion from the lone pairs compresses the axial-equatorial bond angles to approximately \(86^\circ\).
PastPaper.markingScheme
Award 1 mark for the correct answer A. Method: Determine number of bonding and lone pairs for each molecule and apply VSEPR theory.
PastPaper.question 37 · Multiple Choice
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An organic compound W is treated with hot, concentrated, acidified potassium manganate(VII). The reaction yields propanone and carbon dioxide as the only carbon-containing products. What is the structural formula of W?
A.\(\text{CH}_3\text{CH}=\text{CHCH}_3\)
B.\((\text{CH}_3)_2\text{C}=\text{CH}_2\)
C.\((\text{CH}_3)_2\text{C}=\text{CHCH}_3\)
D.\(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\)
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PastPaper.workedSolution
Hot, concentrated, acidified potassium manganate(VII) cleaves the double bond of alkenes. A disubstituted alkene carbon (\(=\text{CR}_1\text{R}_2\)) is oxidized to a ketone, while a terminal carbon (\(=\text{CH}_2\)) is oxidized to carbon dioxide. Since the products are propanone, \((\text{CH}_3)_2\text{C}=\text{O}\), and carbon dioxide, \(\text{CO}_2\), the original alkene must have been 2-methylpropene, \((\text{CH}_3)_2\text{C}=\text{CH}_2\).
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Award 1 mark for the correct answer B. Method: Identify the structural segments of the alkene by working backward from the oxidation cleavage products.
PastPaper.question 38 · Multiple Choice
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An alcohol Z has the molecular formula \(\text{C}_5\text{H}_{12}\text{O}\). When Z is warmed with acidified potassium dichromate(VI), the solution changes color from orange to green. When Z is warmed with alkaline aqueous iodine, a yellow precipitate of triiodomethane is formed. How many possible structural isomers of Z fit this description?
A.1
B.2
C.3
D.4
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PastPaper.workedSolution
The molecular formula \(\text{C}_5\text{H}_{12}\text{O}\) represents a saturated monohydric alcohol. The oxidation by acidified potassium dichromate(VI) (orange to green) rules out tertiary alcohols, meaning Z must be a primary or secondary alcohol. The positive triiodomethane (iodoform) test indicates the presence of the \(\text{CH}_3\text{CH}(\text{OH})-\) group, which is characteristic of secondary methyl alcohols. To have 5 carbons in total, the alkyl group attached to the \(\text{CH}_3\text{CH}(\text{OH})-\) carbon must have 3 carbons. There are only two possible 3-carbon alkyl groups: the propyl group, yielding pentan-2-ol: \(\text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_2\text{CH}_3\), and the isopropyl group, yielding 3-methylbutan-2-ol: \(\text{CH}_3\text{CH}(\text{OH})\text{CH}(\text{CH}_3)_2\). Thus, there are exactly 2 structural isomers.
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Award 1 mark for the correct answer B. Method: Deduce the functional group requirements (secondary methyl alcohol) and count the possible structural isomers with 5 carbon atoms.
PastPaper.question 39 · Multiple Choice
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A sample of hydrated magnesium sulfate, \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\), has a mass of \(4.93\text{ g}\). After heating to constant mass to remove all water of crystallisation, the remaining anhydrous magnesium sulfate has a mass of \(2.41\text{ g}\). What is the value of \(x\)? [\(A_r\): \(\text{H} = 1.0\), \(\text{O} = 16.0\), \(\text{Mg} = 24.3\), \(\text{S} = 32.1\)]
A.2
B.5
C.7
D.10
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PastPaper.workedSolution
First, calculate the molar mass of anhydrous \(\text{MgSO}_4\): \(24.3 + 32.1 + (4 \times 16.0) = 120.4\text{ g mol}^{-1}\). The moles of anhydrous \(\text{MgSO}_4\) are: \(\frac{2.41}{120.4} = 0.0200\text{ mol}\). The mass of water lost during heating is: \(4.93 - 2.41 = 2.52\text{ g}\). The moles of water are: \(\frac{2.52}{18.0} = 0.140\text{ mol}\). The mole ratio of water to anhydrous salt is: \(x = \frac{0.140}{0.0200} = 7\).
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Award 1 mark for the correct answer C. Method: Calculate moles of anhydrous salt and water, and find the simplest integer ratio.
PastPaper.question 40 · Multiple Choice
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In an experiment, \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) hydrochloric acid is mixed with \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) sodium hydroxide in a polystyrene cup. The initial temperature of both solutions was \(21.5\text{ }^\circ\text{C}\), and the maximum temperature reached was \(28.2\text{ }^\circ\text{C}\). Assume the density of the mixture is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). What is the enthalpy change of neutralisation, \(\Delta H_{neut}\), in \(\text{kJ mol}^{-1}\)?
A.\(-28.0\text{ kJ mol}^{-1}\)
B.\(-56.0\text{ kJ mol}^{-1}\)
C.\(-112\text{ kJ mol}^{-1}\)
D.\(+56.0\text{ kJ mol}^{-1}\)
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PastPaper.workedSolution
First, find the total mass of the mixture: \(m = 50.0 + 50.0 = 100.0\text{ g}\). The change in temperature is \(\Delta T = 28.2 - 21.5 = 6.7\text{ K}\). The heat energy released is: \(q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J} = 2.801\text{ kJ}\). The number of moles of water formed is: \(n = 1.00 \times \frac{50.0}{1000} = 0.0500\text{ mol}\). The enthalpy change of neutralisation is \(\Delta H_{neut} = -\frac{q}{n} = -\frac{2.801}{0.0500} = -56.0\text{ kJ mol}^{-1}\).
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Award 1 mark for the correct answer B. Method: Correctly calculate the heat released using q = mcΔT, calculate the moles of water formed, and then calculate the molar enthalpy change with the correct negative sign.
Paper 22 (AS Structured)
Answer all structured questions in spaces provided. Show clear working for calculations.
5 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · Structured
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A student analyzes a sample of hydrated sodium carbonate, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\).
(a) A \(3.58\text{ g}\) sample of hydrated sodium carbonate is dissolved in distilled water and made up to \(250.0\text{ cm}^3\) in a volumetric flask. A \(25.0\text{ cm}^3\) portion of this solution is titrated against \(0.125\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl(aq)}\), requiring \(20.00\text{ cm}^3\) of the acid for complete neutralisation.
(i) Write a balanced chemical equation for the reaction between sodium carbonate and hydrochloric acid.
(ii) Calculate the amount, in moles, of \(\text{HCl}\) used in the titration.
(iii) Calculate the amount, in moles, of \(\text{Na}_2\text{CO}_3\) in the original \(250.0\text{ cm}^3\) solution.
(iv) Calculate the molar mass of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\) and determine the value of \(x\).
(b) Another sample containing anhydrous calcium carbonate, \(\text{CaCO}_3\), and an inert impurity is analyzed. When \(1.50\text{ g}\) of this mixture is reacted with excess dilute hydrochloric acid, dry carbon dioxide gas is collected.
(i) State the ionic equation, including state symbols, for the reaction between solid calcium carbonate and aqueous hydrogen ions.
(ii) At \(298\text{ K}\) and \(101\text{ kPa}\) pressure, the volume of dry carbon dioxide gas collected is \(285\text{ cm}^3\). Calculate the percentage by mass of \(\text{CaCO}_3\) in the mixture. [Gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\), \(M_r(\text{CaCO}_3) = 100.1\)]
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(a)(i) \(\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) (ii) Moles of \(\text{HCl} = \text{concentration} \times \text{volume} = 0.125\text{ mol dm}^{-3} \times \frac{20.00}{1000}\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\). (iii) From the reaction stoichiometry, \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\). Moles of \(\text{Na}_2\text{CO}_3\) in the titrated \(25.0\text{ cm}^3\) portion = \frac{2.50 \times 10^{-3}\text{ mol}}{2} = 1.25 \times 10^{-3}\text{ mol}\). Moles of \(\text{Na}_2\text{CO}_3\) in the original \(250.0\text{ cm}^3\) solution = 1.25 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 1.25 \times 10^{-2}\text{ mol}\). (iv) Molar mass of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} = \frac{\text{mass}}{\text{moles}} = \frac{3.58\text{ g}}{1.25 \times 10^{-2}\text{ mol}} = 286.4\text{ g mol}^{-1}\). \(M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}\). Mass of water of crystallisation per mole = \(286.4 - 106.0 = 180.4\text{ g mol}^{-1}\). \(x = \frac{180.4}{18.0} = 10.02 \approx 10\).
(b)(i) \(\text{CaCO}_3\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) (ii) Using the ideal gas equation, \(pV = nRT\): \(p = 101 \times 10^3\text{ Pa}\) \(V = 285 \times 10^{-6}\text{ m}^3\) \(n = \frac{101 \times 10^3\text{ Pa} \times 285 \times 10^{-6}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 298\text{ K}} = 0.01162\text{ mol of } \text{CO}_2\). Since \(1\text{ mol}\) of \(\text{CaCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\), moles of \(\text{CaCO}_3\) in the sample = \(0.01162\text{ mol}\). Mass of \(\text{CaCO}_3 = 0.01162\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.163\text{ g}\). Percentage by mass = \(\frac{1.163\text{ g}}{1.50\text{ g}} \times 100 = 77.5\%\).
PastPaper.markingScheme
(a)(i) Correct balanced equation: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\) [1] (ii) Moles of \(\text{HCl} = 2.50 \times 10^{-3}\text{ mol}\) [1] (iii) Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = 1.25 \times 10^{-3}\text{ mol}\) (method mark) [1] Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3 = 1.25 \times 10^{-2}\text{ mol}\) (accuracy mark) [1] (iv) Correct calculation of molar mass of hydrate \(= 286.4\text{ g mol}^{-1}\) [1] Correct calculation of total mass of water \(= 180.4\text{ g mol}^{-1}\) [1] Correct value of \(x = 10\) (must be an integer) [1]
(b)(i) Correct species: \(\text{CaCO}_3\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)[1] Correct state symbols on all five species [1] (ii) Correct conversion of units (V to \(\text{m}^3\) and p to \(\text{Pa}\)) and calculation of moles of \(\text{CO}_2 = 0.01162\text{ mol}\) [1] Correct calculation of mass of \(\text{CaCO}_3 = 1.163\text{ g}\) [1] Correct calculation of percentage by mass \(= 77.5\%\) (accept range 77.5% - 78.0%) [1]
PastPaper.question 2 · Structured
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Liquid pentan-1-ol, \(\text{C}_5\text{H}_{11}\text{OH}\), is a promising renewable biofuel.
(a) Define the term *standard enthalpy change of combustion*, \(\Delta H^\ominus_c\).
(b) A student performed a calorimetry experiment to determine the enthalpy change of combustion of pentan-1-ol. The following results were obtained: - Mass of water in copper calorimeter = \(150.0\text{ g}\) - Initial temperature of water = \(21.5\text{ }^\circ\text{C}\) - Final temperature of water = \(58.2\text{ }^\circ\text{C}\) - Mass of pentan-1-ol burner before combustion = \(224.85\text{ g}\) - Mass of pentan-1-ol burner after combustion = \(224.21\text{ g}\)
(i) Calculate the heat energy, \(q\), in \(\text{kJ}\), released during this combustion. Assume the specific heat capacity of water is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).
(ii) Calculate the experimental enthalpy change of combustion of pentan-1-ol, in \(\text{kJ mol}^{-1}\). [\(M_r(\text{C}_5\text{H}_{11}\text{OH}) = 88.0\)]
(iii) State two reasons why the experimental value obtained in (b)(ii) is significantly less exothermic than the standard data book value.
(c) The standard enthalpy changes of combustion of carbon, hydrogen, and pentan-1-ol are given in the table below: - \(\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H^\ominus_c = -393.5\text{ kJ mol}^{-1}\) - \(\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(l)} \quad \Delta H^\ominus_c = -285.8\text{ kJ mol}^{-1}\) - \(\text{C}_5\text{H}_{11}\text{OH(l)} + 7.5\text{O}_2\text{(g)} \rightarrow 5\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} \quad \Delta H^\ominus_c = -3329.0\text{ kJ mol}^{-1}\)
(i) Write the balanced chemical equation, including state symbols, for the reaction representing the standard enthalpy change of formation of liquid pentan-1-ol.
(ii) Use Hess’s Law and the data provided to calculate the standard enthalpy change of formation, \(\Delta H^\ominus_f\), of liquid pentan-1-ol.
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(a) The standard enthalpy change of combustion is the enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions (298 K, 100 kPa).
(b)(i) \(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (58.2 - 21.5)\text{ K} = 23011.8\text{ J} = 23.01\text{ kJ}\) (or \(23.0\text{ kJ}\)). (ii) Mass of pentan-1-ol burned = \(224.85 - 224.21 = 0.64\text{ g}\). Moles of pentan-1-ol burned = \(\frac{0.64\text{ g}}{88.0\text{ g mol}^{-1}} = 7.273 \times 10^{-3}\text{ mol}\). Enthalpy of combustion, \(\Delta H_c = -\frac{23.0118\text{ kJ}}{7.273 \times 10^{-3}\text{ mol}} = -3164\text{ kJ mol}^{-1}\) (or \(-3160\text{ kJ mol}^{-1}\) to 3 significant figures). (iii) Any two of the following: 1. Heat energy lost to the surrounding air or draughts. 2. Incomplete combustion of pentan-1-ol (indicated by soot/carbon monoxide formation). 3. Heat capacity of the copper calorimeter itself was not included in the calculation. 4. Evaporation of alcohol from the wick after extinguishing.
(a) Enthalpy change when 1 mole of a substance [1] Is burned completely in excess oxygen under standard conditions [1]
(b)(i) Correct calculation of \(q = 23.0\text{ kJ}\) (or \(23.01\text{ kJ}\)) [1] (ii) Mass of fuel = 0.64 g and moles of pentan-1-ol = \(7.27 \times 10^{-3}\text{ mol}\) [1] Enthalpy calculation: \(\frac{q}{\text{moles}}\) [1] Correct final value with a negative sign: \(-3164\text{ kJ mol}^{-1}\) (or \(-3160\text{ kJ mol}^{-1}\)) [1] (iii) Any two valid experimental sources of heat loss/inefficiency [2] - Heat loss to surroundings / draughts - Incomplete combustion - Heat capacity of copper calorimeter neglected - Evaporation of fuel
(c)(i) Balanced chemical equation for formation of pentan-1-ol: \(5\text{C} + 6\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{C}_5\text{H}_{11}\text{OH}\) [1] All state symbols correct: \(\text{C(s)}\), \(\text{H}_2\text{(g)}\), \(\text{O}_2\text{(g)}\), \(\text{C}_5\text{H}_{11}\text{OH(l)}\) [1] (ii) Correct expression of Hess's Law [1] Correct calculation: \(-353.3\text{ kJ mol}^{-1}\) [1]
PastPaper.question 3 · Structured
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This question concerns the elements in Period 3 of the Periodic Table (sodium to argon) and their compounds.
(a) Period 3 elements show distinct trends in their reactions with oxygen and chlorine.
(i) Phosphorus reacts with excess chlorine gas when heated. Write a balanced chemical equation for this reaction and state the oxidation state of phosphorus in the product.
(ii) Write the equation for the reaction of sodium with oxygen. State the type of bonding present in the oxide formed.
(b) Two of the Period 3 oxides are sulfur dioxide, \(\text{SO}_2\), and aluminium oxide, \(\text{Al}_2\text{O}_3\).
(i) Sulfur dioxide is an acidic oxide. Write a balanced equation to show its reaction with an excess of aqueous sodium hydroxide.
(ii) Aluminium oxide is described as amphoteric. - Define the term *amphoteric*. - Write two equations to demonstrate the amphoteric nature of aluminium oxide, one with an acid and one with a base.
(c) The chlorides of Period 3 show differing behavior when added to water.
(i) Describe what you would observe when silicon(IV) chloride, \(\text{SiCl}_4\), is added to water. Explain this behavior in terms of structure and bonding, and write a balanced equation for the reaction.
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(a)(i) Equation: \(\text{P}_4\text{(s)} + 10\text{Cl}_2\text{(g)} \rightarrow 4\text{PCl}_5\text{(s)}\) (or \(\text{P} + 2.5\text{Cl}_2 \rightarrow \text{PCl}_5\)). Oxidation state of phosphorus in \(\text{PCl}_5\) = +5. (ii) Equation: \(4\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Na}_2\text{O(s)}\). Bonding: Ionic bonding.
(b)(i) \(\text{SO}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_3\text{(aq)} + \text{H}_2\text{O(l)}\) (ii) - Amphoteric: A substance that can react/behave as both an acid and a base. - Reacting with acid: \(\text{Al}_2\text{O}_3\text{(s)} + 6\text{HCl(aq)} \rightarrow 2\text{AlCl}_3\text{(aq)} + 3\text{H}_2\text{O(l)}\) (or ionic: \(\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}\)). - Reacting with base: \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{Na}[\text{Al(OH)}_4]\text{(aq)}\) (or ionic: \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-\)).
(c)(i) Observations: Vigorous reaction, evolution of misty white fumes (of \(\text{HCl}\) gas), and formation of a white precipitate/solid of silicon dioxide (or hydrated silica). Explanation: Silicon(IV) chloride is a simple molecular structure with covalent molecular bonding. Silicon has vacant, low-energy \(3\text{d}\) orbitals that allow it to coordinate with water molecules, facilitating rapid hydrolysis (unlike carbon in \(\text{CCl}_4\)). Equation: \(\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(g)}\) (or forming silicic acid \(\text{H}_4\text{SiO}_4\)).
PastPaper.markingScheme
(a)(i) Correct balanced equation for formation of \(\text{PCl}_5\) [1] Correct oxidation state of phosphorus: +5 [1] (ii) Correct equation for sodium oxide formation [1] Correct identification of bonding: Ionic [1]
(b)(i) Correct equation: \(\text{SO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_3 + \text{H}_2\text{O}\) [1] (ii) Definition: A substance that reacts with both acids and bases [1] Correct equation showing reaction of \(\text{Al}_2\text{O}_3\) with an acid [1] Correct equation showing reaction of \(\text{Al}_2\text{O}_3\) with a base [1]
(c)(i) Observation: Misty white fumes / white precipitate [1] Structure & Bonding: Simple covalent molecular [1] Explanation: Silicon has vacant low-lying \(3\text{d}\) orbitals that can accept a lone pair from water to undergo hydrolysis [1] Equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) (or balanced equation forming \(\text{Si(OH)}_4\)) [1]
PastPaper.question 4 · Structured
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Compound **A** is an alkene with the molecular formula \(\text{C}_6\text{H}_{12}\) and name 3-methylpent-2-ene.
(a) Compound **A** exhibits stereoisomerism.
(i) Draw the structures of the \((E)\)- and \((Z)\)- stereoisomers of Compound **A** and label each clearly.
(ii) Explain why alkenes can exhibit stereoisomerism of this type, whereas alkanes cannot.
(b) When Compound **A** is reacted with hydrogen bromide, \(\text{HBr}\), two bromoalkane products, **B** (major) and **C** (minor), are formed.
(i) Draw the structures of both **B** and **C** and state their IUPAC names.
(ii) Outline the mechanism for the reaction of Compound **A** with \(\text{HBr}\) to form the major product **B**. Include curly arrows, relevant dipoles, and any intermediate structures.
(iii) Explain, in terms of the stability of intermediates, why **B** is formed as the major product rather than **C**.
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(a)(i) Compound **A** is 3-methylpent-2-ene, \(\text{CH}_3\text{CH}=\text{C(CH}_3)\text{CH}_2\text{CH}_3\). - In the \((E)\)-isomer, the high priority groups (the methyl group on C2 and the ethyl group on C3) lie on opposite sides of the \(\text{C}=\text{C}\) double bond. - In the \((Z)\)-isomer, these same high priority groups lie on the same side of the \(\text{C}=\text{C}\) double bond. [Structures must show the carbon-carbon double bond with four substituents correctly arranged in space.]
(ii) Alkenes contain a \(\pi\) bond in the \(\text{C}=\text{C}\) double bond, which restricts rotation about the carbon-carbon bond axis. Additionally, each carbon atom involved in the double bond must be bonded to two different groups. Alkanes only have \(\sigma\) bonds, which allow free rotation, and thus do not show cis-trans/E-Z isomerism.
(b)(i) Major product **B** is 3-bromo-3-methylpentane. Structure: \(\text{CH}_3-\text{CH}_2-\text{C(Br)(CH}_3)-\text{CH}_2-\text{CH}_3\). Minor product **C** is 2-bromo-3-methylpentane. Structure: \(\text{CH}_3-\text{CH(Br)}-\text{CH(CH}_3)-\text{CH}_2-\text{CH}_3\).
(ii) Mechanism of electrophilic addition: - Step 1: Hydrogen bromide is polar: \(\text{H}^{\delta+}-\text{Br}^{\delta-}\). A curly arrow is drawn from the \(\pi\) bond of the double bond to the \(\text{H}\) atom of \(\text{H}-\text{Br}\). Simultaneously, a curly arrow is drawn from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom. - Intermediate: A stable tertiary carbocation is formed: \(\text{CH}_3-\text{CH}_2-\text{C}^+\text{(CH}_3)-\text{CH}_2-\text{CH}_3\) alongside a bromide ion, \(:\text{Br}^-\). - Step 2: A curly arrow is drawn from the lone pair on the bromide ion, \(:\text{Br}^-\), to the positively charged carbon atom of the carbocation, yielding 3-bromo-3-methylpentane.
(iii) Product **B** is formed via a tertiary carbocation intermediate, whereas product **C** is formed via a secondary carbocation intermediate. The tertiary carbocation is more stable than the secondary carbocation because it has three electron-releasing alkyl groups attached to the positively charged carbon, which disperse the positive charge more effectively through the positive inductive effect.
PastPaper.markingScheme
(a)(i) Correct structure of \((E)\)-3-methylpent-2-ene [1] Correct structure of \((Z)\)-3-methylpent-2-ene [1] Correct assignment of \((E)\) and \((Z)\) configurations based on group priorities [1] (ii) Identifies that rotation is restricted due to the presence of a \(\pi\) bond [1] Explains that each double-bonded carbon must be bonded to two different atoms/groups [1]
(b)(i) Correct structure and IUPAC name for major product **B** (3-bromo-3-methylpentane) [1] Correct structure and IUPAC name for minor product **C** (2-bromo-3-methylpentane) [1] (ii) Mechanism: - Curly arrow from \(\text{C}=\text{C}\) double bond to \(\text{H}\) of \(\text{H-Br}\) and correct dipole on \(\text{H}-\text{Br}\) [1] - Curly arrow from \(\text{H}-\text{Br}\) bond to \(\text{Br}\) [1] - Correct structure of the tertiary carbocation intermediate [1] - Curly arrow from lone pair on \(\text{Br}^-\) to the carbocation carbon [1] (iii) Explains that the tertiary carbocation intermediate is more stable than the secondary carbocation intermediate [1] due to the greater inductive effect of three electron-releasing alkyl groups [1]
PastPaper.question 5 · Structured
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Halogens and their halide ions display distinctive chemical properties and trends.
(a) Sodium halides react with concentrated sulfuric acid to produce various products depending on the identity of the halide.
(i) When concentrated sulfuric acid is added to solid sodium chloride, \(\text{NaCl}\), a misty gas is evolved. Identify this gas and write a balanced equation for the reaction.
(ii) When concentrated sulfuric acid is added to solid sodium iodide, \(\text{NaI}\), several products are observed, including a dark grey solid, a yellow solid, and a choking gas. - Identify the dark grey solid and the yellow solid. - Explain, in terms of the reducing power of the halide ions, why sodium iodide reacts differently from sodium chloride.
(b) Aqueous solutions of three unknown sodium halides, **X**, **Y**, and **Z**, are tested using aqueous silver nitrate, followed by the addition of ammonia solution. - Halide **X** forms a cream precipitate with aqueous silver nitrate, which is insoluble in dilute aqueous ammonia but soluble in concentrated aqueous ammonia. - Halide **Y** forms a white precipitate with aqueous silver nitrate, which is soluble in dilute aqueous ammonia. - Halide **Z** forms a yellow precipitate with aqueous silver nitrate, which is insoluble in both dilute and concentrated aqueous ammonia.
(i) Identify the halide ions present in solutions **X**, **Y**, and **Z**.
(ii) Write an ionic equation, including state symbols, for the reaction of halide **Y** with silver nitrate.
(c) Chlorine is added to cold, dilute sodium hydroxide solution to form a useful bleaching agent.
(i) Write a balanced chemical equation for this reaction.
(ii) Explain why this reaction is described as a disproportionation reaction by referring to the oxidation numbers of chlorine.
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(a)(i) Misty gas: Hydrogen chloride, \(\text{HCl}\). Equation: \(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\) (or \(2\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Na}_2\text{SO}_4\text{(s)} + 2\text{HCl(g)}\)). (ii) - Dark grey solid: Iodine, \(\text{I}_2\). - Yellow solid: Sulfur, \(\text{S}\). Explanation: Iodide ions (\(\text{I}^-\)) are much stronger reducing agents than chloride ions (\(\text{Cl}^-\)). The ionic radius of the iodide ion is larger than that of the chloride ion. Consequently, the valence electrons in \(\text{I}^-\) are further from the nucleus and experience more shielding, meaning they are lost more easily to reduce sulfuric acid.
(c)(i) \(\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}\). (ii) Disproportionation is a reaction in which the same element is simultaneously oxidized and reduced. The oxidation state of chlorine in \(\text{Cl}_2\) is 0. It is reduced to -1 in \(\text{NaCl}\) and oxidized to +1 in \(\text{NaClO}\).
PastPaper.markingScheme
(a)(i) Gas: Hydrogen chloride / \(\text{HCl}\) [1] Equation: \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) [1] (ii) Dark grey solid: Iodine / \(\text{I}_2\) [1] Yellow solid: Sulfur / \(\text{S}\) [1] Explains that iodide is a stronger reducing agent than chloride [1] Explains reducing power in terms of larger ionic radius / valence electrons further from nucleus / more shielded / more easily lost [1]
(b)(i) Identifies **X** as bromide / \(\text{Br}^-\), **Y** as chloride / \(\text{Cl}^-\), and **Z** as iodide / \(\text{I}^-\)[3] (1 mark for each correct identification) (ii) Ionic equation: \(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\) [1] (Must include state symbols and be balanced)
(c)(i) Balanced equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) [1] (ii) Explanation: Chlorine is oxidized (from 0 in \(\text{Cl}_2\) to +1 in \(\text{NaClO}\)) and reduced (from 0 to -1 in \(\text{NaCl}\)) simultaneously [1]
Paper 32 (Practical)
Perform physical laboratory experiments. Record values with correct significant figures and answer matching questions.
### Determination of the percentage purity of an impure sample of potassium iodate(V)
You are provided with: * **FB 1:** \(0.100\text{ mol dm}^{-3}\) aqueous sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\) * **FB 2:** An aqueous solution prepared by dissolving \(3.75\text{ g}\) of an impure sample of potassium iodate(V), \(\text{KIO}_3\), in water and making it up to \(1.00\text{ dm}^3\) * **FB 3:** \(1.0\text{ mol dm}^{-3}\) sulfuric acid, \(\text{H}_2\text{SO}_4\) * **FB 4:** \(0.5\text{ mol dm}^{-3}\) potassium iodide, \(\text{KI}\) * Starch indicator
**Procedure:** 1. Fill the burette with **FB 1**. 2. Pipette \(25.0\text{ cm}^3\) of **FB 2** into a conical flask. 3. Use a measuring cylinder to add \(10\text{ cm}^3\) of **FB 3** followed by \(10\text{ cm}^3\) of **FB 4** to the conical flask. The solution will turn dark brown due to liberated iodine. 4. Titrate the mixture with **FB 1** until the dark brown color fades to a light yellow. 5. Add a few drops of starch indicator. The mixture will turn blue-black. 6. Continue titrating dropwise with **FB 1** until the blue-black color completely disappears, leaving a colorless solution. 7. Repeat the titration as necessary to obtain concordant results.
**(a)** Present your titration results in a clearly labeled table, including your initial and final burette readings, and the volume of **FB 1** added for each titration. State the volume of **FB 1** to be used in your calculations.
**(b)** From your titration results, calculate: (i) the number of moles of sodium thiosulfate used in your average titre. (ii) the number of moles of potassium iodate(V) present in \(25.0\text{ cm}^3\) of **FB 2**, using the equations: \[\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}\] \[\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\] (iii) the concentration, in \(\text{mol dm}^{-3}\), of \(\text{KIO}_3\) in **FB 2**. (iv) the percentage purity by mass of the potassium iodate(V) sample in **FB 2**. [Assume \(M_r\) of \(\text{KIO}_3 = 214.0\)]
**(c)** Suggest one significant source of error in this procedure (excluding human/reading errors) and describe how the procedure could be modified to reduce this error.
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**(a) Titration Table:** Students should produce a table with appropriate headings and units (e.g., /cm³). Assuming experimental concordant titres are obtained, e.g., Titre 1 = 23.45 cm³, Titre 2 = 23.35 cm³. Average titre = \((23.45 + 23.35) / 2 = 23.40\text{ cm}^3\).
**(b) Calculations:** (i) Moles of \(\text{Na}_2\text{S}_2\text{O}_3\) used: \(0.100\text{ mol dm}^{-3} \times \frac{23.40}{1000}\text{ dm}^3 = 2.34 \times 10^{-3}\text{ mol}\). (ii) From the equations, \(1\text{ mol of }\text{IO}_3^-\) produces \(3\text{ mol of }\text{I}_2\), and each mole of \(\text{I}_2\) reacts with \(2\text{ mol of }\text{S}_2\text{O}_3^{2-}\). Thus, \(1\text{ mol of }\text{IO}_3^- \equiv 6\text{ mol of }\text{S}_2\text{O}_3^{2-}\). Moles of \(\text{KIO}_3\) in \(25.0\text{ cm}^3\) = \(\frac{2.34 \times 10^{-3}}{6} = 3.90 \times 10^{-4}\text{ mol}\). (iii) Concentration of \(\text{KIO}_3\) in **FB 2**: \(3.90 \times 10^{-4}\text{ mol} \times \frac{1000}{25.0} = 0.0156\text{ mol dm}^{-3}\). (iv) Mass of \(\text{KIO}_3\) in \(1.00\text{ dm}^3\) of **FB 2**: \(0.0156\text{ mol} \times 214.0\text{ g mol}^{-1} = 3.338\text{ g}\). Percentage purity = \(\frac{3.338}{3.75} \times 100\\% = 89.0\\%\).
**(c) Source of Error & Modification:** * **Error:** Iodine is volatile and some may escape by evaporation from the conical flask before or during the titration, leading to a smaller titre and underestimating the purity. * **Modification:** Cover the conical flask with a watch glass immediately after adding potassium iodide, or carry out the reaction in a stoppered flask and titrate immediately.
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**Part (a): [5 marks]** * **1 mark:** Table constructed with clear headings and correct units (e.g., Initial / Final burette reading / \(\text{cm}^3\)). * **1 mark:** All burette readings recorded to the nearest \(0.05\text{ cm}^3\). * **2 marks:** Two concordant titres within \(0.10\text{ cm}^3\) obtained (deduct 1 mark if within \(0.20\text{ cm}^3\)). * **1 mark:** Correct calculation of the average titre, showing the working or selecting concordant values.
**Part (b): [5 marks]** * **1 mark:** Correct calculation of thiosulfate moles in (i). * **1 mark:** Correct stoichiometric ratio \(1:6\) used to find moles of \(\text{KIO}_3\) in (ii). * **1 mark:** Correct concentration calculation in (iii). * **1 mark:** Correct percentage purity calculation in (iv) using \(M_r = 214.0\). * **1 mark:** All final numerical answers in (b) expressed to 3 significant figures.
**Part (c): [3 marks]** * **1 mark:** Identifies loss of volatile iodine by evaporation as the source of error. * **1 mark:** Proposes an appropriate modification (e.g., covering the flask with a watch glass or using a stoppered flask). * **1 mark:** Explains that this keeps the iodine in the flask, preventing underestimation of the titre.
PastPaper.question 2 · Practical
13.3 PastPaper.marks
### Determination of the enthalpy change of neutralization of chloroacetic acid
You are provided with: * **FB 5:** \(2.00\text{ mol dm}^{-3}\) aqueous chloroacetic acid, \(\text{CH}_2\text{ClCOOH}\) * **FB 6:** \(2.00\text{ mol dm}^{-3}\) aqueous sodium hydroxide, \(\text{NaOH}\) * A plastic cup, a beaker, and a thermometer
**Procedure:** 1. Support the plastic cup inside a \(250\text{ cm}^3\) beaker to provide insulation. 2. Use a measuring cylinder to transfer \(25.0\text{ cm}^3\) of **FB 5** into the plastic cup. 3. Measure the temperature of **FB 5** every minute for \(3\text{ minutes}\), stirring gently to ensure temperature uniformity. 4. At the \(4^{\text{th}}\) minute, use a second measuring cylinder to rapidly add \(25.0\text{ cm}^3\) of **FB 6** (which has been allowed to stand to reach the same initial temperature). Do not record the temperature at the \(4^{\text{th}}\) minute. 5. Stir the mixture thoroughly. 6. Measure and record the temperature of the mixture at \(1\text{ minute}\) intervals from \(5\text{ minutes}\) to \(10\text{ minutes}\).
**(a)** Present your temperature and time readings in a single, clearly structured table. Plot a graph of temperature (on the y-axis) against time (on the x-axis) on graph paper. Draw two best-fit straight lines (one for cooling and one for heating before mixing) and extrapolate them to determine the maximum theoretical temperature rise, \(\Delta T\), at the \(4^{\text{th}}\) minute. State your value of \(\Delta T\).
**(b)** Using your value of \(\Delta T\), calculate: (i) the heat energy, \(q\), released in the reaction. [Assume the specific heat capacity of the mixture is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\) and its density is \(1.00\text{ g cm}^{-3}\).] (ii) the number of moles of water formed during the neutralization. (iii) the enthalpy change of neutralization, \(\Delta H_{\text{neut}}\), in \(\text{kJ mol}^{-1}\). Include the appropriate sign and express your answer to 3 significant figures.
**(c)** (i) The standard enthalpy change of neutralization of a strong monoprotic acid by a strong base is \(-57.2\text{ kJ mol}^{-1}\). State and explain how your experimental value of \(\Delta H_{\text{neut}}\) for chloroacetic acid compares with this value. (ii) Calculate the percentage uncertainty in the temperature change, \(\Delta T\), if the uncertainty of each temperature reading of the thermometer is \(\pm 0.2\text{ }^\circ\text{C}\).
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PastPaper.workedSolution
**(a) Temperature and Time Readings & Graph Extrapolation:** Assume a set of experimental results where the initial temperature at \(t = 0, 1, 2, 3\text{ min}\) is constant at \(21.5\text{ }^\circ\text{C}\). After mixing at \(t = 4\text{ min}\), temperature readings are: * \(5\text{ min}: 33.8\text{ }^\circ\text{C}\) * \(6\text{ min}: 33.5\text{ }^\circ\text{C}\) * \(7\text{ min}: 33.2\text{ }^\circ\text{C}\) * \(8\text{ min}: 32.9\text{ }^\circ\text{C}\) * \(9\text{ min}: 32.6\text{ }^\circ\text{C}\) * \(10\text{ min}: 32.3\text{ }^\circ\text{C}\) By plotting and extrapolating back to the \(4^{\text{th}}\) minute, the maximum theoretical temperature is determined to be \(34.2\text{ }^\circ\text{C}\). \(\Delta T = 34.2 - 21.5 = 12.7\text{ }^\circ\text{C}\).
**(b) Calculations:** (i) Mass of mixture = \((25.0 + 25.0)\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\). \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 12.7\text{ K} = 2654.3\text{ J} = 2.654\text{ kJ}\). (ii) Moles of acid used = \(2.00\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.0500\text{ mol}\). Moles of base used = \(2.00\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.0500\text{ mol}\). Moles of water formed = \(0.0500\text{ mol}\). (iii) \(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{2.654\text{ kJ}}{0.0500\text{ mol}} = -53.1\text{ kJ mol}^{-1}\) (3 sig figs).
**(c) Analysis:** (i) The experimental enthalpy change of neutralization (\(-53.1\text{ kJ mol}^{-1}\)) is less exothermic than the standard strong acid value (\(-57.2\text{ kJ mol}^{-1}\)). Chloroacetic acid is a weak acid and is only partially dissociated in solution. Energy is absorbed to break the covalent \(\text{O}-\text{H}\) bonds in the undissociated acid molecules during dissociation, making the overall process less exothermic. (ii) Since there are two temperature readings (initial and maximum), the total uncertainty in \(\Delta T\) is \(2 \times (\pm 0.2) = \pm 0.4\text{ }^\circ\text{C}\). Percentage uncertainty = \(\frac{0.4}{12.7} \times 100\\% = 3.15\\%\).
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**Part (a): [5 marks]** * **1 mark:** Table formatted clearly with column headers including units (Time/min, Temperature/°C). * **1 mark:** All temperatures recorded to \(0.1\text{ }^\circ\text{C}\). * **2 marks:** Plotting: Axes correctly labeled with linear scale, points plotted accurately, and two distinct straight lines drawn with proper extrapolation back to the 4th minute. * **1 mark:** Correct value of \(\Delta T\) determined and stated.
**Part (b): [4 marks]** * **1 mark:** Correct calculation of heat energy \(q\) in (i). * **1 mark:** Correct calculation of moles of water in (ii). * **1 mark:** Correct calculation of \(\Delta H_{\text{neut}}\) in (iii). * **1 mark:** Negative sign shown and final answer expressed to 3 significant figures.
**Part (c): [4 marks]** * **1 mark:** Mentions that the experimental value is less exothermic than the strong acid value. * **1 mark:** Explains that chloroacetic acid is a weak acid (partially dissociated). * **1 mark:** Mentions that energy is absorbed to dissociate the weak acid fully. * **1 mark:** Correct calculation of percentage uncertainty in \(\Delta T\) in (ii) using total uncertainty of \(0.4\text{ }^\circ\text{C}\).
PastPaper.question 3 · Practical
13.3 PastPaper.marks
### Qualitative analysis of inorganic solutions
You are provided with three solutions: **FB 7**, **FB 8**, and **FB 9**. Each solution contains one cation and one anion from the qualitative analysis notes.
**Procedure:** Perform the following tests on separate portions of each solution and record all your observations in a suitable table.
* **Test 1:** Add aqueous sodium hydroxide dropwise, then in excess, to a \(1\text{ cm}\) depth of the solution in a test-tube. If a precipitate is formed, heat the mixture gently and test any gas evolved with damp litmus paper. * **Test 2:** Add aqueous ammonia dropwise, then in excess, to a \(1\text{ cm}\) depth of the solution in a test-tube. * **Test 3:** Add dilute nitric acid followed by aqueous silver nitrate to a \(1\text{ cm}\) depth of the solution in a test-tube. Then add aqueous ammonia (dilute or concentrated) to any precipitate formed. * **Test 4:** Add dilute hydrochloric acid followed by aqueous barium chloride to a \(1\text{ cm}\) depth of the solution in a test-tube.
**(a)** Present your observations in a single, clearly structured table.
**(b)** From your observations, identify the ions present in **FB 7**, **FB 8**, and **FB 9**. Support each identification with a specific piece of evidence from your observations. * **FB 7:** Cation and Anion. Evidence. * **FB 8:** Cation and Anion. Evidence. * **FB 9:** Cation and Anion. Evidence.
**(c)** Write the ionic equation (including state symbols) for: (i) the reaction between the cation in **FB 9** and aqueous sodium hydroxide. (ii) the reaction between the anion in **FB 8** and aqueous silver nitrate.
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**(a) Table of Observations:**
| Test | FB 7 | FB 8 | FB 9 | |---|---|---|---| | **Test 1 (NaOH)** | No precipitate. On heating, a pungent gas is evolved that turns damp red litmus paper blue. | Off-white/beige precipitate, insoluble in excess, turning brown on standing. | Red-brown precipitate, insoluble in excess. | | **Test 2 (NH₃)** | No precipitate/no reaction. | Off-white/beige precipitate, insoluble in excess, turning brown on standing. | Red-brown precipitate, insoluble in excess. | | **Test 3 (HNO₃ + AgNO₃)** | Cream precipitate formed. Precipitate is insoluble in dilute ammonia but soluble in concentrated ammonia. | White precipitate formed. Precipitate is soluble in dilute ammonia. | No precipitate/no reaction. | | **Test 4 (HCl + BaCl₂)** | No precipitate/no reaction. | No precipitate/no reaction. | White precipitate formed, insoluble in dilute hydrochloric acid. |
**(b) Identifications & Evidence:** * **FB 7:** Contains \(\text{NH}_4^+\) and \(\text{Br}^-\). * *Evidence for \(\text{NH}_4^+\):* No precipitate with NaOH, but heating produces a gas (ammonia) that turns damp red litmus blue. * *Evidence for \(\text{Br}^-\):* Gives a cream precipitate with silver nitrate that is soluble only in concentrated ammonia. * **FB 8:** Contains \(\text{Mn}^{2+}\) and \(\text{Cl}^-\). * *Evidence for \(\text{Mn}^{2+}\):* Gives an off-white/beige precipitate with NaOH and NH₃ that darkens on contact with air. * *Evidence for \(\text{Cl}^-\):* Gives a white precipitate with silver nitrate that dissolves easily in dilute ammonia. * **FB 9:** Contains \(\text{Fe}^{3+}\) and \(\text{SO}_4^{2-}\). * *Evidence for \(\text{Fe}^{3+}\):* Gives a characteristic red-brown precipitate with both NaOH and NH₃. * *Evidence for \(\text{SO}_4^{2-}\):* Gives a thick white precipitate with barium chloride in acidic conditions.
**Part (a): [6 marks]** * **1 mark:** Table is fully organized with all test descriptions and columns for FB 7, FB 8, and FB 9. * **1 mark:** FB 7 observations are complete (no ppt with NaOH/NH₃, ammonia gas tested on heating, cream ppt with AgNO₃). * **1 mark:** FB 8 observations are complete (off-white ppt turning brown in air for NaOH/NH₃, white ppt with AgNO₃ dissolving in dilute NH₃). * **1 mark:** FB 9 observations are complete (red-brown ppt with NaOH/NH₃, white ppt with BaCl₂).
**Part (b): [5 marks]** * **1 mark:** Correct identification of FB 7 as containing \(\text{NH}_4^+\) and \(\text{Br}^-\). * **1 mark:** Correct identification of FB 8 as containing \(\text{Mn}^{2+}\) and \(\text{Cl}^-\). * **1 mark:** Correct identification of FB 9 as containing \(\text{Fe}^{3+}\) and \(\text{SO}_4^{2-}\). * **2 marks:** Clear and logical supporting evidence linked directly to observations (1 mark for cations, 1 mark for anions).
**Part (c): [2 marks]** * **1 mark:** Correct balanced ionic equation with state symbols in (i). * **1 mark:** Correct balanced ionic equation with state symbols in (ii).