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 12
Answer all 40 multiple-choice questions. For each question, choose the best option from A, B, C, or D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
A sample of \( 1.97\text{ g} \) of a carbonate of a Group 2 metal, \( \text{XCO}_3 \), is thermally decomposed to completion. \( \text{XCO}_3(s) \rightarrow \text{XO}(s) + \text{CO}_2(g) \) The carbon dioxide gas produced occupies \( 240\text{ cm}^3 \) at room temperature and pressure (r.t.p.). What is the identity of the Group 2 metal, \( \text{X} \)? (Assume 1 mole of gas occupies \( 24.0\text{ dm}^3 \) at r.t.p.)
A.Magnesium
B.Calcium
C.Strontium
D.Barium
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Worked solution
1. Moles of \( \text{CO}_2 \) produced: \( n(\text{CO}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.010\text{ mol} \). 2. Since the stoichiometry of the reaction is 1:1, the moles of \( \text{XCO}_3 \) decomposed is also \( 0.010\text{ mol} \). 3. Find the molar mass of \( \text{XCO}_3 \): \( M_r(\text{XCO}_3) = \frac{1.97\text{ g}}{0.010\text{ mol}} = 197\text{ g mol}^{-1} \). 4. Deduce the relative atomic mass of \( \text{X} \): \( A_r(\text{X}) = 197 - 12.0 - (3 \times 16.0) = 137 \). Comparing this to the Periodic Table, Barium (Ba) has a relative atomic mass of 137.3.
Marking scheme
Award 1 mark for the correct answer (D). Reject other options based on incorrect molar masses: A (Mg, 24.3), B (Ca, 40.1), C (Sr, 87.6).
Question 2 · multiple-choice
1 marks
A student reacts \( 4.00\text{ g} \) of copper(II) oxide with excess dilute sulfuric acid to prepare copper(II) sulfate crystals, \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \). \( \text{CuO}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4(aq) + \text{H}_2\text{O}(l) \) Upon crystallization and drying, the student obtains \( 10.0\text{ g} \) of hydrated copper(II) sulfate, \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \). What is the percentage yield of the hydrated copper(II) sulfate? [Relative atomic masses: \( \text{H} = 1.0 \), \( \text{O} = 16.0 \), \( \text{S} = 32.1 \), \( \text{Cu} = 63.5 \)]
A.\( 40.0\% \)
B.\( 63.9\% \)
C.\( 79.6\% \)
D.\( 80.3\% \)
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Worked solution
1. Calculate the molar mass of \( \text{CuO} \): \( M_r = 63.5 + 16.0 = 79.5\text{ g mol}^{-1} \). 2. Find the moles of \( \text{CuO} \) used: \( n = \frac{4.00}{79.5} = 0.0503\text{ mol} \). 3. Since the reaction stoichiometry is 1:1, the theoretical yield of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is \( 0.0503\text{ mol} \). 4. Calculate the molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \): \( M_r = 63.5 + 32.1 + 64.0 + 5(18.0) = 249.6\text{ g mol}^{-1} \). 5. Determine the theoretical mass: \( 0.050314\text{ mol} \times 249.6\text{ g mol}^{-1} = 12.56\text{ g} \). 6. Calculate the percentage yield: \( \frac{10.0\text{ g}}{12.56\text{ g}} \times 100\% = 79.6\% \).
Marking scheme
Award 1 mark for the correct answer (C). Option A represents the ratio of starting material to product mass. Option B uses the anhydrous molar mass instead of the hydrated molar mass.
Question 3 · multiple-choice
1 marks
Consider the following standard enthalpy changes: \( 2\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \quad \Delta H^\theta = -84.7\text{ kJ mol}^{-1} \) ; \( \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\theta = -393.5\text{ kJ mol}^{-1} \) ; \( \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\theta = -285.8\text{ kJ mol}^{-1} \). What is the standard enthalpy change of combustion of ethane, \( \text{C}_2\text{H}_6(g) \)?
A.\( -1729.1\text{ kJ mol}^{-1} \)
B.\( -1559.7\text{ kJ mol}^{-1} \)
C.\( -594.6\text{ kJ mol}^{-1} \)
D.\( +1559.7\text{ kJ mol}^{-1} \)
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Award 1 mark for the correct answer (B). Option A is obtained by subtracting the enthalpy of formation of ethane. Option D is the positive value of the correct answer.
Question 4 · multiple-choice
1 marks
In a calorimetry experiment, \( 50.0\text{ cm}^3 \) of \( 1.00\text{ mol dm}^{-3} \) aqueous sodium hydroxide is mixed with \( 50.0\text{ cm}^3 \) of \( 1.00\text{ mol dm}^{-3} \) hydrochloric acid in a polystyrene cup. Both solutions were initially at \( 20.2\text{ }^\circ\text{C} \). After mixing, the temperature rose to a maximum of \( 26.8\text{ }^\circ\text{C} \). Assume the density of the solution is \( 1.00\text{ g cm}^{-3} \) and its specific heat capacity is \( 4.18\text{ J g}^{-1}\text{ K}^{-1} \). What is the enthalpy change of neutralisation, \( \Delta H_n \), for this reaction?
A.\( -27.6\text{ kJ mol}^{-1} \)
B.\( -55.2\text{ kJ mol}^{-1} \)
C.\( +55.2\text{ kJ mol}^{-1} \)
D.\( -57.3\text{ kJ mol}^{-1} \)
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Worked solution
1. Mass of the mixture: \( m = 50.0 + 50.0 = 100.0\text{ g} \). 2. Temperature rise: \( \Delta T = 26.8 - 20.2 = 6.6\text{ K} \). 3. Heat released: \( q = m c \Delta T = 100.0 \times 4.18 \times 6.6 = 2758.8\text{ J} = 2.759\text{ kJ} \). 4. Moles of acid/base neutralised: \( n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol} \). 5. Enthalpy of neutralisation: \( \Delta H_n = -\frac{q}{n} = -\frac{2.7588\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1} \).
Marking scheme
Award 1 mark for the correct answer (B). Option A uses only the mass of one solution (50.0 g). Option C is the endothermic value.
Question 5 · multiple-choice
1 marks
Four separate samples of Period 3 oxides are added to water and the pH of the resulting mixture is measured. The oxides tested are: \( \text{Na}_2\text{O} \), \( \text{Al}_2\text{O}_3 \), \( \text{P}_4\text{O}_{10} \), and \( \text{SO}_3 \). Which sequence correctly ranks these oxides in order of increasing pH of their aqueous mixtures?
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Worked solution
- \( \text{SO}_3 \) reacts with water to form sulfuric acid, which is a strong acid (pH around 1). - \( \text{P}_4\text{O}_{10} \) reacts with water to form phosphoric acid, which is a weaker acid (pH around 2). - \( \text{Al}_2\text{O}_3 \) is insoluble, so the pH of water remains neutral (pH around 7). - \( \text{Na}_2\text{O} \) reacts with water to form sodium hydroxide, which is a strong base (pH around 14). Thus, the order of increasing pH is: \( \text{SO}_3 < \text{P}_4\text{O}_{10} < \text{Al}_2\text{O}_3 < \text{Na}_2\text{O} \).
Marking scheme
Award 1 mark for the correct answer (B). Option A reverses the relative acidities of the phosphorus and sulfur oxides. Option C is the reverse order (decreasing pH).
Question 6 · multiple-choice
1 marks
Which physical property of the Period 3 elements (sodium to argon) increases to a maximum value at silicon, and then decreases sharply to phosphorus?
A.Atomic radius
B.Electrical conductivity
C.First ionisation energy
D.Melting point
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Worked solution
- Atomic radius decreases continuously across the period. - Electrical conductivity peaks at aluminium (metallic) and is low for silicon (metalloid/semiconductor) and non-metals. - First ionisation energy shows a general upward trend with dips. - Melting point increases from Na to Al, peaks at Si (due to its giant covalent structure requiring high energy to break strong covalent bonds), and drops sharply to phosphorus (which forms simple molecular structures with weak London dispersion forces).
Marking scheme
Award 1 mark for the correct answer (D). Reject other options because they do not show a peak specifically at silicon followed by a sharp drop to phosphorus.
Question 7 · multiple-choice
1 marks
An equilibrium mixture is established in a \( 2.0\text{ dm}^3 \) sealed container at a constant temperature according to the following equation: \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \) At equilibrium, the mixture contains \( 0.40\text{ mol} \) of \( \text{SO}_2 \), \( 0.20\text{ mol} \) of \( \text{O}_2 \), and \( 1.60\text{ mol} \) of \( \text{SO}_3 \). What is the value of the equilibrium constant, \( K_c \), at this temperature?
A.\( 40\text{ dm}^3 \text{mol}^{-1} \)
B.\( 80\text{ dm}^3 \text{mol}^{-1} \)
C.\( 160\text{ dm}^3 \text{mol}^{-1} \)
D.\( 320\text{ dm}^3 \text{mol}^{-1} \)
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Award 1 mark for the correct answer (C). Option B is obtained if moles are used directly in the formula instead of concentrations.
Question 8 · multiple-choice
1 marks
The industrial synthesis of ammonia is represented by the following equilibrium: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92\text{ kJ mol}^{-1} \) Which combination of changes will both decrease the equilibrium yield of ammonia and decrease the value of the equilibrium constant, \( K_p \)?
A.decrease temperature and decrease pressure
B.decrease temperature and increase pressure
C.increase temperature and decrease pressure
D.increase temperature and increase pressure
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Worked solution
1. The value of \( K_p \) is only affected by temperature. Since the forward reaction is exothermic, increasing the temperature shifts the equilibrium to the left, which decreases the value of \( K_p \). Therefore, temperature must be increased. 2. Increasing temperature also shifts the position of equilibrium to the left, reducing the yield of ammonia. 3. Decreasing the pressure shifts the equilibrium towards the side with more gaseous moles (the left side), which further decreases the yield of ammonia (pressure does not affect \( K_p \)). Hence, the combination of increasing temperature and decreasing pressure achieves both effects.
Marking scheme
Award 1 mark for the correct answer (C). Any option with 'decrease temperature' is incorrect because it would increase \( K_p \).
Question 9 · multiple-choice
1 marks
Propan-1-ol, \(\text{C}_3\text{H}_7\text{OH}(l)\), can be formed from its elements according to the following equation:
What is the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of propan-1-ol?
A.\(-305\text{ kJ mol}^{-1}\)
B.\(+305\text{ kJ mol}^{-1}\)
C.\(-1449\text{ kJ mol}^{-1}\)
D.\(+1341\text{ kJ mol}^{-1}\)
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Worked solution
According to Hess's law, the standard enthalpy change of formation of a compound can be calculated from standard enthalpies of combustion using the cycle:
Award 1 mark for the correct answer A. - Method: Correctly applies Hess's law using \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). - Accuracy: Calculates \(-305\text{ kJ mol}^{-1}\) correctly.
Question 10 · multiple-choice
1 marks
A sample of \(1.20\text{ g}\) of a mixture containing anhydrous calcium carbonate, \(\text{CaCO}_3\) (\(M_r = 100.1\)), and sodium chloride, \(\text{NaCl}\), reacts completely with excess dilute hydrochloric acid. The volume of \(\text{CO}_2\) gas collected at room temperature and pressure (r.t.p.) is \(180\text{ cm}^3\).
What is the percentage by mass of \(\text{CaCO}_3\) in the sample?
(Assume 1.00 mol of gas occupies \(24.0\text{ dm}^3\) under r.t.p. conditions).
A.\(15.0\%\)
B.\(37.4\%\)
C.\(62.6\%\)
D.\(75.1\%\)
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Worked solution
First, write the chemical equation for the reaction of calcium carbonate with hydrochloric acid: \(\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)
Calculate the moles of \(\text{CO}_2\) gas produced: \(n(\text{CO}_2) = \frac{180\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0075\text{ mol}\)
From the stoichiometry of the equation, the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1. \(n(\text{CaCO}_3) = 0.0075\text{ mol}\)
Calculate the mass of \(\text{CaCO}_3\) in the sample: \(m(\text{CaCO}_3) = 0.0075\text{ mol} \times 100.1\text{ g mol}^{-1} = 0.75075\text{ g}\)
Calculate the percentage by mass of \(\text{CaCO}_3\) in the \(1.20\text{ g}\) sample: \(\text{Percentage by mass} = \frac{0.75075\text{ g}}{1.20\text{ g}} \times 100\% \approx 62.6\%\)
Therefore, option C is correct.
Marking scheme
Award 1 mark for the correct answer C. - Method: Correctly calculates moles of gas, relates to calcium carbonate mass, and finds mass percentage. - Accuracy: Obtains \(62.6\%\).
Question 11 · multiple-choice
1 marks
Three Period 3 elements, \(X\), \(Y\), and \(Z\), have oxides with the following properties:
- The oxide of \(X\) is insoluble in water but dissolves in both dilute aqueous sodium hydroxide and dilute hydrochloric acid. - The oxide of \(Y\) reacts violently with water to form a strongly acidic solution of pH 1–2. - The oxide of \(Z\) dissolves in water to form a strongly alkaline solution of pH 13–14.
Which sequence correctly identifies the elements \(X\), \(Y\), and \(Z\)?
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Worked solution
Let's analyse the properties of the oxides of Period 3 elements: 1. The oxide of \(X\) dissolves in both acids and bases (amphoteric behavior). In Period 3, only aluminium oxide, \(\text{Al}_2\text{O}_3\), behaves amphoterically. Thus, \(X\) is Aluminium (\(\text{Al}\)). 2. The oxide of \(Y\) reacts violently with water to form a strongly acidic solution. Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) and sulfur trioxide (\(\text{SO}_3\)) react strongly/violently with water to yield phosphoric acid and sulfuric acid respectively, which have highly acidic pH values. Thus, \(Y\) could be Phosphorus (\(\text{P}\)). 3. The oxide of \(Z\) dissolves in water to form a strongly alkaline solution (pH 13-14). Sodium oxide (\(\text{Na}_2\text{O}\)) reacts with water to form aqueous sodium hydroxide, which is a strong base with a pH of 13-14. Magnesium oxide is only sparingly soluble, resulting in a weak alkaline solution of pH ~9.
Therefore, the correct identification is \(X = \text{Al}\), \(Y = \text{P}\), and \(Z = \text{Na}\), which corresponds to option A.
Marking scheme
Award 1 mark for the correct answer A. - Method: Identifies amphoteric oxide (Al), strongly acidic oxide (P), and strongly basic oxide (Na) from their chemical properties.
Question 12 · multiple-choice
1 marks
A mixture of \(\text{SO}_2(g)\) and \(\text{O}_2(g)\) is allowed to reach equilibrium in a closed vessel of fixed volume according to the equation:
The temperature of the system is increased while the volume is kept constant. Which statement correctly describes the change in the value of the equilibrium constant, \(K_c\), and the rate of the reverse reaction?
A.\(K_c\) decreases; rate of reverse reaction increases
B.\(K_c\) decreases; rate of reverse reaction decreases
C.\(K_c\) increases; rate of reverse reaction increases
D.\(K_c\) remains constant; rate of reverse reaction increases
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Worked solution
1. Effect of temperature on \(K_c\): Since the forward reaction is exothermic (\(\Delta H < 0\)), increasing the temperature shifts the equilibrium in the endothermic (reverse) direction according to Le Chatelier's principle. This increases the concentration of reactants and decreases the concentration of products, which causes the equilibrium constant \(K_c\) to decrease.
2. Effect of temperature on reaction rate: An increase in temperature always increases the rate of both the forward and reverse chemical reactions because molecules have more kinetic energy, leading to a higher frequency of successful collisions with energy equal to or exceeding the activation energy.
Therefore, \(K_c\) decreases and the rate of the reverse reaction increases, which corresponds to option A.
Marking scheme
Award 1 mark for the correct answer A. - Method: Identifies that increasing the temperature of an exothermic reaction decreases \(K_c\) and increases reaction rates for both directions.
Question 13 · multiple-choice
1 marks
The equation for the complete combustion of methanal, \(\text{HCHO}(g)\), is shown:
Total energy input = \(820 + x + 496 = 1316 + x\text{ kJ mol}^{-1}\)
2. Calculate total energy released when products' bonds form: - Bonds formed: - 2 \(\text{C}=\text{O}\) bonds (in \(\text{CO}_2\)): \(2 \times 805 = 1610\text{ kJ mol}^{-1}\) - 2 \(\text{O}-\text{H}\) bonds (in \(\text{H}_2\text{O}\)): \(2 \times 462 = 924\text{ kJ mol}^{-1}\)
Total energy output = \(1610 + 924 = 2534\text{ kJ mol}^{-1}\)
3. Set up the enthalpy change equation: \(\Delta H_{\text{reaction}} = \text{energy input} - \text{energy output}\) \(-482 = (1316 + x) - 2534\) \(-482 = x - 1218\) \(x = 1218 - 482 = 736\text{ kJ mol}^{-1}\)
Therefore, option A is correct.
Marking scheme
Award 1 mark for the correct answer A. - Method: Correctly calculates the total bonds broken and formed and sets up the equation \(\Delta H = \text{broken} - \text{formed}\) to solve for \(x\). - Accuracy: Calculates \(736\text{ kJ mol}^{-1}\) correctly.
Question 14 · multiple-choice
1 marks
An organic compound, \(W\), contains only carbon, hydrogen, and oxygen. Complete combustion of a \(1.80\text{ g}\) sample of \(W\) in excess oxygen yields \(2.64\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(1.08\text{ g}\) of water, \(\text{H}_2\text{O}\).
Therefore, the empirical formula of \(W\) is \(\text{CH}_2\text{O}\), which corresponds to option B.
Marking scheme
Award 1 mark for the correct answer B. - Method: Correctly calculates masses of C and H from combustion products, deduces the mass of oxygen by difference, and obtains molar ratio. - Accuracy: Arrives at empirical formula \(\text{CH}_2\text{O}\).
Question 15 · multiple-choice
1 marks
Which statement correctly describes and explains a trend in physical properties of Period 3 elements from sodium to chlorine?
A.The melting points increase continuously across the period because the number of valence electrons increases.
B.Electrical conductivity is highest for silicon because of its giant covalent structure with delocalised electrons.
C.The ionic radius decreases continuously from \(\text{Na}^+\) to \(\text{Cl}^-\).
D.The atomic radius decreases because the nuclear charge increases while the shielding effect remains approximately constant.
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Worked solution
Let's review the physical trends in Period 3: - Option A is incorrect because melting points do not increase continuously. Silicon (giant covalent) has the highest melting point, after which there is a sharp drop to phosphorus, sulfur, and chlorine (simple molecular structures held by weak London dispersion forces). - Option B is incorrect because silicon is a metalloid/semiconductor and has relatively low electrical conductivity. The metallic elements (Na, Mg, Al) have much higher electrical conductivities due to their delocalised outer-shell electrons. - Option C is incorrect because there is a large increase in ionic radius from cations (\(\text{Na}^+\), \(\text{Mg}^{2+}\), \(\text{Al}^{3+}\)) to anions (\(\text{P}^{3-}\), \(\text{S}^{2-}\), \(\text{Cl}^-\)) because the anions have an entire extra filled electron shell compared to the cations. - Option D is correct. Across Period 3, the number of protons (nuclear charge) increases, but the extra electrons are added to the same principal quantum shell, meaning the shielding effect remains approximately constant. Therefore, the effective nuclear charge increases, pulling the electron cloud closer and decreasing the atomic radius.
Therefore, option D is correct.
Marking scheme
Award 1 mark for the correct answer D. - Method: Explains the atomic radius trend using nuclear charge and shielding effect concepts.
Question 16 · multiple-choice
1 marks
Ethyl ethanoate is synthesised by reacting ethanoic acid and ethanol in the presence of an acid catalyst:
In an experiment, \(1.00\text{ mol}\) of ethanoic acid and \(1.00\text{ mol}\) of ethanol are mixed and allowed to reach equilibrium at a constant temperature. At equilibrium, \(0.67\text{ mol}\) of ethyl ethanoate is present.
What is the value of the equilibrium constant, \(K_c\), at this temperature?
A.\(0.24\)
B.\(2.03\)
C.\(4.12\)
D.\(9.18\)
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Worked solution
Let's set up the ICE (Initial, Change, Equilibrium) table for the reaction:
Award 1 mark for the correct answer C. - Method: Correctly calculates equilibrium moles of all species and substitutes into the \(K_c\) expression. - Accuracy: Calculates \(K_c = 4.12\).
Question 17 · Multiple Choice
1 marks
The standard enthalpy changes of combustion of carbon, hydrogen, and liquid propanone, \(C_3H_6O(l)\), are given below: \(\Delta H^\ominus_c [C(s)] = -393.5\text{ kJ mol}^{-1}\), \(\Delta H^\ominus_c [H_2(g)] = -285.8\text{ kJ mol}^{-1}\), \(\Delta H^\ominus_c [C_3H_6O(l)] = -1790.0\text{ kJ mol}^{-1}\). What is the standard enthalpy change of formation, \(\Delta H^\ominus_f\), of liquid propanone?
A.-247.9 kJ mol\(^{-1}\)
B.+247.9 kJ mol\(^{-1}\)
C.-2037.9 kJ mol\(^{-1}\)
D.-2437.9 kJ mol\(^{-1}\)
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Worked solution
To calculate the standard enthalpy change of formation of liquid propanone: \(3C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_3H_6O(l)\). Using Hess's law, we have: \(\Delta H^\ominus_f = 3 \times \Delta H^\ominus_c[C(s)] + 3 \times \Delta H^\ominus_c[H_2(g)] - \Delta H^\ominus_c[C_3H_6O(l)] = 3(-393.5) + 3(-285.8) - (-1790.0) = -1180.5 - 857.4 + 1790.0 = -247.9\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for the correct application of Hess's law and calculation of the final value.
Question 18 · Multiple Choice
1 marks
A sample of 2.102 g of a metal carbonate, \(MCO_3\), reacts completely with excess dilute hydrochloric acid to produce 504 cm\(^3\) of carbon dioxide gas measured at room temperature and pressure (r.t.p.). What is the relative atomic mass, \(A_r\), of the metal \(M\)? (Assume 1 mole of gas occupies 24.0 dm\(^3\) at r.t.p.)
A.24.3
B.40.1
C.56.1
D.87.6
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Worked solution
Number of moles of \(CO_2 = 504 / 24000 = 0.0210\text{ mol}\). According to the reaction equation: \(MCO_3 + 2HCl \rightarrow MCl_2 + H_2O + CO_2\), 1 mole of \(MCO_3\) produces 1 mole of \(CO_2\). Hence, moles of \(MCO_3 = 0.0210\text{ mol}\). Molar mass of \(MCO_3 = 2.102 / 0.0210 = 100.1\text{ g mol}^{-1}\). Thus, \(A_r(M) = 100.1 - 12.0 - 3(16.0) = 40.1\).
Marking scheme
1 mark for calculating correct moles of CO2, molar mass of the carbonate, and the correct Ar of the metal.
Question 19 · Multiple Choice
1 marks
An element, \(X\), in Period 3 of the Periodic Table forms a chloride which is a liquid at room temperature and pressure. This chloride reacts violently with water to form a strongly acidic solution and white fumes. The oxide of \(X\) has a giant covalent structure and is insoluble in water, but reacts with hot concentrated aqueous sodium hydroxide. What is element \(X\)?
A.Aluminum
B.Silicon
C.Phosphorus
D.Sulfur
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Worked solution
Silicon is the Period 3 element whose chloride, \(SiCl_4\), is a liquid at room temperature and reacts violently with water to form white fumes of \(HCl\). Silicon dioxide, \(SiO_2\), has a giant covalent structure, is insoluble in water, and acts as an acidic oxide by reacting with hot concentrated aqueous sodium hydroxide.
Marking scheme
1 mark for identifying Silicon as the only element meeting all specified criteria.
Question 20 · Multiple Choice
1 marks
Consider the following dynamic equilibrium: \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\) \(\Delta H = -197\text{ kJ mol}^{-1}\). What happens to the position of equilibrium and the value of the equilibrium constant, \(K_c\), when the temperature is increased at constant pressure?
A.Position of equilibrium shifts to the left; \(K_c\) decreases.
B.Position of equilibrium shifts to the left; \(K_c\) remains constant.
C.Position of equilibrium shifts to the right; \(K_c\) increases.
D.Position of equilibrium shifts to the right; \(K_c\) remains constant.
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Worked solution
Since the forward reaction is exothermic, according to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium to the left (favouring the endothermic reverse reaction). Because the equilibrium shifts to the left, the concentration of the product decreases and reactants increase, resulting in a decrease in the value of the equilibrium constant \(K_c\).
Marking scheme
1 mark for identifying the correct shift in equilibrium position and the corresponding change in Kc.
Question 21 · Multiple Choice
1 marks
In a calorimeter, 50.0 cm\(^3\) of 2.00 mol dm\(^{-3}\) \(NaOH(aq)\) is mixed with 50.0 cm\(^3\) of 2.00 mol dm\(^{-3}\) \(HNO_3(aq)\). Both solutions are initially at 21.5 \(^\circ\)C. The temperature rises to a maximum of 35.1 \(^\circ\)C. Assume the density of the mixture is 1.00 g cm\(^{-3}\) and its specific heat capacity is 4.18 J g\(^{-1}\) K\(^{-1}\). What is the enthalpy change of neutralization, \(\Delta H_{neut}\), for this reaction?
A.-113.7 kJ mol\(^{-1}\)
B.-56.8 kJ mol\(^{-1}\)
C.-28.4 kJ mol\(^{-1}\)
D.+56.8 kJ mol\(^{-1}\)
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Worked solution
Total mass of solution \(m = 50.0 + 50.0 = 100.0\text{ g}\). Temperature change \(\Delta T = 35.1 - 21.5 = 13.6\text{ K}\). Heat produced \(q = m c \Delta T = 100.0 \times 4.18 \times 13.6 = 5684.8\text{ J} = 5.685\text{ kJ}\). Moles of water formed \(n = 0.0500\text{ dm}^3 \times 2.00\text{ mol dm}^{-3} = 0.100\text{ mol}\). Enthalpy change of neutralization \(\Delta H_{neut} = -q / n = -5.685 / 0.100 = -56.8\text{ kJ mol}^{-1}\).
Marking scheme
1 mark for calculating heat produced, identifying the correct number of moles, and obtaining the correct negative enthalpy value.
Question 22 · Multiple Choice
1 marks
A sample of 10 cm\(^3\) of a gaseous hydrocarbon, \(Y\), is exploded with 100 cm\(^3\) of oxygen (an excess). After cooling to room temperature, the remaining gas mixture has a volume of 75 cm\(^3\). When this gas mixture is passed through aqueous potassium hydroxide (which absorbs \(CO_2\)), the volume decreases to 35 cm\(^3\). All gas volumes are measured at the same temperature and pressure. What is the molecular formula of \(Y\)?
A.C\(_4\)H\(_6\)
B.C\(_4\)H\(_8\)
C.C\(_4\)H\(_{10}\)
D.C\(_3\)H\(_8\)
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Worked solution
Volume of \(CO_2\) produced = \(75 - 35 = 40\text{ cm}^3\). Since 10 cm\(^3\) of \(Y\) produces 40 cm\(^3\) of \(CO_2\), each molecule of \(Y\) contains 4 carbon atoms (\(x = 4\)). The remaining 35 cm\(^3\) is unreacted \(O_2\). Therefore, volume of \(O_2\) reacted = \(100 - 35 = 65\text{ cm}^3\). The ratio of \(Y : O_2\) reacted is \(10 : 65 = 1 : 6.5\). Using the general equation \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\), we have \(x + \frac{y}{4} = 6.5\). Since \(x = 4\), \(4 + \frac{y}{4} = 6.5 \Rightarrow \frac{y}{4} = 2.5 \Rightarrow y = 10\). The molecular formula is \(C_4H_{10}\).
Marking scheme
1 mark for determining the carbon and hydrogen ratio correctly from volume changes.
Question 23 · Multiple Choice
1 marks
Three Period 3 elements, \(X\), \(Y\), and \(Z\), have the following properties: Element \(X\) conducts electricity in both solid and liquid states and has a high melting point. Element \(Y\) has a giant covalent structure and does not conduct electricity under standard conditions. Element \(Z\) exists as diatomic molecules at room temperature. What are elements \(X\), \(Y\), and \(Z\)?
A.X = Mg, Y = Al, Z = P
B.X = Al, Y = Si, Z = Cl
C.X = Si, Y = S, Z = Ar
D.X = Na, Y = P, Z = Cl
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Worked solution
Element \(X\) must be a metal because it conducts electricity in both states and has a high melting point; this corresponds to Al (or Mg/Na). Element \(Y\) must be Silicon, which has a giant covalent structure and is a semiconductor (under standard conditions it is a poor/non-conductor compared to metals). Element \(Z\) must be Chlorine, which exists as diatomic molecules (\(Cl_2\)) at room temperature. Thus, option B is correct.
Marking scheme
1 mark for correctly matching all three descriptions to the corresponding Period 3 elements.
Question 24 · Multiple Choice
1 marks
For the reversible reaction \(N_2O_4(g) \rightleftharpoons 2NO_2(g)\), a mixture at equilibrium at a temperature \(T\) has a total pressure of 1.20 atm. The partial pressure of \(N_2O_4(g)\) is 0.40 atm. What is the value of the equilibrium constant \(K_p\) at this temperature?
A.0.50 atm
B.1.60 atm
C.2.00 atm
D.3.20 atm
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Worked solution
The total pressure is 1.20 atm. Since \(p(N_2O_4) = 0.40\text{ atm}\), the partial pressure of \(NO_2\) is \(1.20 - 0.40 = 0.80\text{ atm}\). The expression for \(K_p\) is \(K_p = \frac{(p(NO_2))^2}{p(N_2O_4)} = \frac{(0.80)^2}{0.40} = \frac{0.64}{0.40} = 1.60\text{ atm}\).
Marking scheme
1 mark for calculating the correct partial pressure of NO2 and correctly evaluating Kp.
Question 25 · Multiple Choice
1 marks
A \(1.48\text{ g}\) sample of an anhydrous metal carbonate, \(\text{MCO}_3\), is reacted completely with excess dilute hydrochloric acid. The volume of carbon dioxide gas collected at room temperature and pressure (r.t.p.) is \(240\text{ cm}^3\). What is the identity of metal M? [1 mol of gas occupies \(24.0\text{ dm}^3\) at r.t.p.]
A.Calcium
B.Magnesium
C.Strontium
D.Barium
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Worked solution
1. Calculate the number of moles of \(\text{CO}_2\) gas collected: \(n(\text{CO}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.010\text{ mol}\).
2. Use the stoichiometric ratio from the reaction equation: \(\text{MCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). Since the ratio of \(\text{MCO}_3\) to \(\text{CO}_2\) is 1:1, \(n(\text{MCO}_3) = 0.010\text{ mol}\).
3. Calculate the molar mass (\(M_r\)) of \(\text{MCO}_3\): \(M_r = \frac{\text{mass}}{\text{moles}} = \frac{1.48\text{ g}}{0.010\text{ mol}} = 148\text{ g mol}^{-1}\).
4. Determine the relative atomic mass (\(A_r\)) of M: \(A_r(\text{M}) + A_r(\text{C}) + 3 \times A_r(\text{O}) = 148\) \(A_r(\text{M}) + 12.0 + 3 \times 16.0 = 148\) \(A_r(\text{M}) + 60.0 = 148\) \(A_r(\text{M}) = 88.0\). This value corresponds closely to strontium (\(\text{Sr}\), \(A_r = 87.6\)).
Marking scheme
[1] for correct option C. Award 1 mark for calculating \(n(\text{CO}_2) = 0.010\text{ mol}\), finding \(M_r(\text{MCO}_3) = 148\text{ g mol}^{-1}\), and correctly identifying Strontium.
Question 26 · Multiple Choice
1 marks
A \(25.0\text{ cm}^3\) sample of a solution containing iron(II) ions, \(\text{Fe}^{2+}\), is titrated against acidified potassium manganate(VII), \(\text{KMnO}_4\). The titration requires \(18.5\text{ cm}^3\) of \(0.0200\text{ mol dm}^{-3}\) \(\text{KMnO}_4\) for complete oxidation of the \(\text{Fe}^{2+}\) ions. What is the concentration of \(\text{Fe}^{2+}\) in the solution?
A.\(0.0148\text{ mol dm}^{-3}\)
B.\(0.0370\text{ mol dm}^{-3}\)
C.\(0.0740\text{ mol dm}^{-3}\)
D.\(0.370\text{ mol dm}^{-3}\)
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Worked solution
1. Write the ionic equation for the redox titration: \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\).
2. Calculate the moles of \(\text{MnO}_4^-\) used: \(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{18.5}{1000}\text{ dm}^3 = 3.70 \times 10^{-4}\text{ mol}\).
3. Calculate the moles of \(\text{Fe}^{2+}\) in the \(25.0\text{ cm}^3\) sample using the 1:5 stoichiometry: \(n(\text{Fe}^{2+}) = 5 \times 3.70 \times 10^{-4}\text{ mol} = 1.85 \times 10^{-3}\text{ mol}\).
4. Calculate the concentration of \(\text{Fe}^{2+}\): \([\text{Fe}^{2+}] = \frac{1.85 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0740\text{ mol dm}^{-3}\).
Marking scheme
[1] for correct option C. Award 1 mark for correctly determining the 5:1 stoichiometric ratio, finding moles of manganate, and calculating the final concentration of iron(II) ions.
Question 27 · Multiple Choice
1 marks
Gaseous hydrazine, \(\text{N}_2\text{H}_4(\text{g})\), combusts in oxygen to form nitrogen gas and steam according to the equation: \(\text{N}_2\text{H}_4(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g})\) Using the bond energies provided, what is the standard enthalpy change of this combustion reaction?
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Worked solution
1. Determine total energy required to break bonds (reactants): - \(4 \times (\text{N}-\text{H}) = 4 \times 391 = 1564\text{ kJ}\) - \(1 \times (\text{N}-\text{N}) = 158\text{ kJ}\) - \(1 \times (\text{O}=\text{O}) = 496\text{ kJ}\) Total energy input = \(1564 + 158 + 496 = 2218\text{ kJ}\).
2. Determine total energy released when new bonds form (products): - \(1 \times (\text{N}\equiv\text{N}) = 945\text{ kJ}\) - \(4 \times (\text{O}-\text{H})\) (since there are 2 moles of \(\text{H}_2\text{O}\), each with 2 O-H bonds) = \(4 \times 463 = 1852\text{ kJ}\) Total energy output = \(945 + 1852 = 2797\text{ kJ}\).
[1] for correct option A. Award 1 mark for correct summation of bond-breaking energies (\(2218\text{ kJ}\)) and bond-forming energies (\(2797\text{ kJ}\)) and finding the difference.
Question 28 · Multiple Choice
1 marks
The standard enthalpy changes of formation for three substances are given below: - \(\Delta H_f^\theta [\text{CO}_2(\text{g})] = -393.5\text{ kJ mol}^{-1}\) - \(\Delta H_f^\theta [\text{H}_2\text{O}(\text{l})] = -285.8\text{ kJ mol}^{-1}\) - \(\Delta H_f^\theta [\text{C}_3\text{H}_8(\text{g})] = -104.6\text{ kJ mol}^{-1}\)
What is the standard enthalpy change of combustion of propane, \(\text{C}_3\text{H}_8(\text{g})\)?
A.\(-2219.1\text{ kJ mol}^{-1}\)
B.\(-2428.3\text{ kJ mol}^{-1}\)
C.\(-574.7\text{ kJ mol}^{-1}\)
D.\(+2219.1\text{ kJ mol}^{-1}\)
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Worked solution
The equation for the combustion of propane is: \(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\)
[1] for correct option A. Award 1 mark for setting up the correct Hess's law expression and correctly computing the value including proper signs.
Question 29 · Multiple Choice
1 marks
Four Period 3 elements, W, X, Y, and Z, form oxides with the properties described below: - The oxide of W is a gas at room temperature and dissolves in water to form a strongly acidic solution. - The oxide of X has a giant covalent structure and is insoluble in water. - The oxide of Y is ionic and amphoteric. - The oxide of Z reacts violently with water to form a strongly alkaline solution.
Which sequence correctly identifies the elements W, X, Y, and Z in order of increasing atomic number?
A.Z, Y, X, W
B.W, X, Y, Z
C.Z, X, Y, W
D.Y, Z, X, W
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Worked solution
1. Oxide of W is a gas and strongly acidic, corresponding to \(\text{SO}_2\) or \(\text{SO}_3\) (element W is Sulfur, atomic number 16). 2. Oxide of X has a giant covalent structure and is insoluble, corresponding to \(\text{SiO}_2\) (element X is Silicon, atomic number 14). 3. Oxide of Y is ionic and amphoteric, corresponding to \(\text{Al}_2\text{O}_3\) (element Y is Aluminium, atomic number 13). 4. Oxide of Z reacts violently to form a strongly alkaline solution, corresponding to \(\text{Na}_2\text{O}\) (element Z is Sodium, atomic number 11).
Ordering by increasing atomic number: Z (11) < Y (13) < X (14) < W (16), which gives the sequence Z, Y, X, W.
Marking scheme
[1] for correct option A. Award 1 mark for identifying the identity of all four elements and correctly arranging them by increasing atomic number.
Question 30 · Multiple Choice
1 marks
The first six successive ionisation energies (in \(\text{kJ mol}^{-1}\)) for an element, Q, in Period 3 of the Periodic Table are shown below:
B.Q reacts with chlorine gas to form a giant covalent chloride.
C.The oxide of Q reacts with both aqueous sodium hydroxide and dilute hydrochloric acid.
D.The oxide of Q has a low melting point due to weak intermolecular forces.
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Worked solution
1. Analyze the successive ionisation energies: The largest relative increase is between the 3rd and 4th ionisation energies (from 2745 to 11578). This indicates that the 4th electron is removed from an inner quantum shell, meaning element Q has 3 valence electrons and is in Group 13. 2. In Period 3, the Group 13 element is Aluminium (Al). 3. Evaluate the statements: - A: Al is a metal and its oxide is amphoteric, not acidic. - B: Al reacts with chlorine to form aluminium chloride, \(\text{Al}_2\text{Cl}_6\), which is a simple molecular dimer, not giant covalent. - C: Aluminium oxide is amphoteric and reacts with both acids (hydrochloric acid) and bases (aqueous sodium hydroxide). This is correct. - D: Aluminium oxide has a giant ionic lattice with high melting point due to strong electrostatic attractions between ions.
Marking scheme
[1] for correct option C. Award 1 mark for identifying Q as Aluminium from the successive ionisation energies and correctly identifying that its oxide is amphoteric.
Question 31 · Multiple Choice
1 marks
The following gaseous equilibrium is established in a closed container of fixed volume: \(2\text{NO}_2(\text{g}) \rightleftharpoons \text{N}_2\text{O}_4(\text{g}) \quad \Delta H = -58\text{ kJ mol}^{-1}\)
If the temperature of the container is increased, what happens to the value of the equilibrium constant, \(K_c\), and the total pressure inside the container?
A.\(K_c\) decreases; total pressure increases
B.\(K_c\) decreases; total pressure decreases
C.\(K_c\) increases; total pressure increases
D.\(K_c\) remains unchanged; total pressure increases
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Worked solution
1. Effect on \(K_c\): The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium in the endothermic direction (to the left). Since the concentration of reactants increases and products decreases, the value of \(K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}\) decreases.
2. Effect on total pressure: - From \(pV = nRT\), pressure is directly proportional to temperature. Therefore, increasing the temperature directly increases the kinetic energy of the gas particles and hence the pressure. - Additionally, shifting the equilibrium to the left increases the total number of moles of gas (since 1 mole of \(\text{N}_2\text{O}_4\) dissociates into 2 moles of \(\text{NO}_2\)). This further increases the total pressure. Thus, the total pressure increases.
Marking scheme
[1] for correct option A. Award 1 mark for explaining both the decrease in \(K_c\) due to the exothermic nature of the forward reaction and the increase in pressure due to temperature increase and shift to the left.
Question 32 · Multiple Choice
1 marks
Ethanol and ethanoic acid react to form ethyl ethanoate and water in a reversible reaction: \(\text{CH}_3\text{COOH}(\text{l}) + \text{CH}_3\text{CH}_2\text{OH}(\text{l}) \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3(\text{l}) + \text{H}_2\text{O}(\text{l})\)
A mixture containing \(1.00\text{ mol}\) of ethanoic acid and \(1.00\text{ mol}\) of ethanol is allowed to reach equilibrium at a constant temperature. At equilibrium, \(0.67\text{ mol}\) of ethyl ethanoate is found to have formed. What is the value of the equilibrium constant, \(K_c\), at this temperature?
A.\(0.24\)
B.\(2.03\)
C.\(4.12\)
D.\(9.00\)
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2. Write the expression for \(K_c\): \(K_c = \frac{[\text{CH}_3\text{COOCH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{CH}_3\text{CH}_2\text{OH}]}\)
3. Since the total volume \(V\) cancels out in the expression (same number of moles on both sides): \(K_c = \frac{0.67 \times 0.67}{0.33 \times 0.33} = \frac{0.4489}{0.1089} \approx 4.12\).
Marking scheme
[1] for correct option C. Award 1 mark for finding equilibrium moles of all species (0.33 mol of reactants, 0.67 mol of products) and calculating the correct \(K_c\) value.
Question 33 · multiple-choice
1 marks
A 5.00 g sample of a dry mixture containing anhydrous magnesium nitrate, \( \text{Mg(NO}_3)_2 \), and magnesium oxide, \( \text{MgO} \), is heated strongly until no further decomposition occurs. Under these conditions, only the magnesium nitrate decomposes, producing \( 1.20 \text{ dm}^3 \) of gaseous product measured at room temperature and pressure (r.t.p.). What is the percentage by mass of magnesium nitrate in the original mixture? [Molar volume of gas at r.t.p. = \( 24.0 \text{ dm}^3 \text{ mol}^{-1} \); \( M_r: \text{Mg(NO}_3)_2 = 148.3 \)]
A.29.7%
B.40.7%
C.59.3%
D.74.2%
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Worked solution
First, we write the balanced equation for the thermal decomposition of magnesium nitrate: \( 2\text{Mg(NO}_3)_2(s) \rightarrow 2\text{MgO}(s) + 4\text{NO}_2(g) + \text{O}_2(g) \). This shows that 2 moles of \( \text{Mg(NO}_3)_2 \) produce 5 moles of gas (4 moles of \( \text{NO}_2 \) and 1 mole of \( \text{O}_2 \)). The total moles of gas produced = \( \frac{1.20 \text{ dm}^3}{24.0 \text{ dm}^3 \text{ mol}^{-1}} = 0.050 \text{ mol} \). Using the stoichiometric ratio, the moles of \( \text{Mg(NO}_3)_2 \) decomposed = \( 0.050 \text{ mol} \times \frac{2}{5} = 0.020 \text{ mol} \). The mass of \( \text{Mg(NO}_3)_2 \) in the mixture = \( 0.020 \text{ mol} \times 148.3 \text{ g mol}^{-1} = 2.966 \text{ g} \). The percentage by mass in the 5.00 g sample is \( \frac{2.966 \text{ g}}{5.00 \text{ g}} \times 100\% = 59.3\% \).
Marking scheme
Award 1 mark for the correct calculation and selecting option C. Deduce stoichiometry: 2 moles of nitrate yield 5 moles of gas. Calculate moles of gas: 0.050 mol. Calculate moles of nitrate: 0.020 mol. Convert to mass: 2.966 g. Find percentage: 59.3%.
Question 34 · multiple-choice
1 marks
A 1.89 g sample of hydrated oxalic acid, \( \text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} \), was dissolved in distilled water and made up to a total volume of \( 250 \text{ cm}^3 \). A \( 25.0 \text{ cm}^3 \) portion of this solution required exactly \( 25.00 \text{ cm}^3 \) of \( 0.120 \text{ mol dm}^{-3} \) sodium hydroxide, \( \text{NaOH} \), for complete neutralisation. What is the value of \( x \)? [\( M_r: \text{H}_2\text{C}_2\text{O}_4 = 90.0 \); \( M_r: \text{H}_2\text{O} = 18.0 \)]
A.1
B.2
C.3
D.4
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Worked solution
The reaction between oxalic acid and sodium hydroxide is: \( \text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} \). Moles of \( \text{NaOH} \) used in the titration = \( 0.02500 \text{ dm}^3 \times 0.120 \text{ mol dm}^{-3} = 0.00300 \text{ mol} \). Moles of \( \text{H}_2\text{C}_2\text{O}_4 \) in the \( 25.0 \text{ cm}^3 \) aliquot = \( \frac{0.00300 \text{ mol}}{2} = 0.00150 \text{ mol} \). Moles of \( \text{H}_2\text{C}_2\text{O}_4 \) in the original \( 250 \text{ cm}^3 \) solution = \( 0.00150 \text{ mol} \times 10 = 0.0150 \text{ mol} \). The molar mass of the hydrated acid = \( \frac{1.89 \text{ g}}{0.0150 \text{ mol}} = 126.0 \text{ g mol}^{-1} \). The mass contribution of water molecules = \( 126.0 - 90.0 = 36.0 \text{ g mol}^{-1} \). The value of \( x = \frac{36.0}{18.0} = 2 \).
Marking scheme
Award 1 mark for the correct calculation of moles and molar mass to find x = 2 (Option B). Award 0 marks for incorrect stoichiometric ratios (such as 1:1, which leads to mathematically inconsistent results).
Question 35 · multiple-choice
1 marks
The standard enthalpy changes of combustion of carbon, hydrogen, and liquid pent-1-yne, \( \text{C}_5\text{H}_8(l) \), are given in the table below.
What is the standard enthalpy change of formation of liquid pent-1-yne?
A.-93.3 kJ mol^{-1}
B.+93.3 kJ mol^{-1}
C.-6314.7 kJ mol^{-1}
D.+6314.7 kJ mol^{-1}
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Worked solution
The equation for the formation of liquid pent-1-yne is: \( 5\text{C}(s) + 4\text{H}_2(g) \rightarrow \text{C}_5\text{H}_8(l) \). According to Hess's Law, using combustion data: \( \Delta H^\theta_f = 5 \times \Delta H^\theta_c [\text{C}(s)] + 4 \times \Delta H^\theta_c [\text{H}_2(g)] - \Delta H^\theta_c [\text{C}_5\text{H}_8(l)] \). Substituting the given values: \( \Delta H^\theta_f = 5(-393.5) + 4(-285.8) - (-3204.0) = -1967.5 - 1143.2 + 3204.0 = +93.3 \text{ kJ mol}^{-1} \).
Marking scheme
Award 1 mark for the correct application of Hess's Law to calculate a positive enthalpy of formation of +93.3 kJ/mol (Option B). Option A is the sign error.
What is the enthalpy change for the gas-phase reaction shown below? \( \text{CH}_4(g) + 2\text{F}_2(g) \rightarrow \text{CH}_2\text{F}_2(g) + 2\text{HF}(g) \)
A.-916 kJ mol^{-1}
B.+916 kJ mol^{-1}
C.-512 kJ mol^{-1}
D.-1074 kJ mol^{-1}
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Award 1 mark for correct calculation of bonds broken (1968) and formed (2884) to yield -916 kJ/mol (Option A). Option B is the opposite sign. Options C and D arise from stoichiometry errors.
Question 37 · multiple-choice
1 marks
Two Period 3 elements, P and Q, react with oxygen under appropriate conditions to form oxides. When these oxides are separately added to excess distilled water, the oxide of P forms a solution with a pH of approximately 13, while the oxide of Q forms a solution with a pH of approximately 2. What are elements P and Q?
A.P is Sodium, Q is Phosphorus
B.P is Magnesium, Q is Sulfur
C.P is Sodium, Q is Silicon
D.P is Magnesium, Q is Aluminium
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Worked solution
Sodium oxide, \( \text{Na}_2\text{O} \), reacts vigorously with water to form a strongly alkaline solution of sodium hydroxide, \( \text{NaOH} \), which has a pH of approximately 13-14. Phosphorus(V) oxide, \( \text{P}_4\text{O}_{10} \), reacts with water to form phosphoric(V) acid, \( \text{H}_3\text{PO}_4 \), which is a moderately strong acid forming a solution with a pH of approximately 2. Magnesium oxide, \( \text{MgO} \), is only sparingly soluble and forms a weakly alkaline solution of pH ~9.
Marking scheme
Award 1 mark for identifying P as Sodium and Q as Phosphorus (Option A), based on the correct pH values of their oxides in water.
Question 38 · multiple-choice
1 marks
Which of the following lists contains four Period 3 ions arranged in order of increasing ionic radius?
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Worked solution
For isoelectronic cations, ionic radius decreases as nuclear charge increases: \( \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+ \). Anions have an extra shell of electrons, so they are significantly larger than the cations. For isoelectronic anions, ionic radius also decreases as nuclear charge increases: \( \text{P}^{3-} > \text{S}^{2-} > \text{Cl}^- \). Thus, the correct sequence in order of increasing ionic radius is \( \text{Al}^{3+} < \text{Mg}^{2+} < \text{S}^{2-} < \text{P}^{3-} \).
Marking scheme
Award 1 mark for the correct ordered list (Option A). All other options violate either the rule of isoelectronic cations (nuclear charge effect) or anions.
Question 39 · multiple-choice
1 marks
Initially, 2.0 mol of gaseous A and 3.0 mol of gaseous B are mixed in a sealed vessel of volume \( 2.0 \text{ dm}^3 \). The system is allowed to reach equilibrium at a constant temperature according to the equation:
At equilibrium, the vessel is found to contain 1.2 mol of C. What is the value of the equilibrium constant, \( K_c \), at this temperature?
A.0.635
B.0.317
C.0.357
D.0.952
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Worked solution
First, find equilibrium moles: since 1.2 mol of C is produced, A used = \( 1.2 / 2 = 0.6 \text{ mol} \); B used = \( 1.2 \text{ mol} \). Equilibrium moles: \( n_e(\text{A}) = 2.0 - 0.6 = 1.4 \text{ mol} \); \( n_e(\text{B}) = 3.0 - 1.2 = 1.8 \text{ mol} \); \( n_e(\text{C}) = 1.2 \text{ mol} \). Next, calculate concentrations (divide by \( V = 2.0 \text{ dm}^3 \)): \( [\text{A}] = 0.70 \text{ mol dm}^{-3} \); \( [\text{B}] = 0.90 \text{ mol dm}^{-3} \); \( [\text{C}] = 0.60 \text{ mol dm}^{-3} \). The expression is \( K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2} = \frac{(0.60)^2}{0.70 \times (0.90)^2} = \frac{0.36}{0.567} \approx 0.635 \text{ dm}^3 \text{ mol}^{-1} \).
Marking scheme
Award 1 mark for the correct calculation of equilibrium concentrations and substituting into the Kc expression to obtain 0.635 (Option A). Option B is the result of neglecting the volume.
Question 40 · multiple-choice
1 marks
Consider the following exothermic reversible reaction at equilibrium in a closed system:
Which of the following changes will increase both the equilibrium yield of \( \text{SO}_3(g) \) and the value of the equilibrium constant, \( K_c \)?
A.Decreasing the temperature
B.Increasing the temperature
C.Increasing the total pressure
D.Adding a suitable catalyst
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Worked solution
The value of the equilibrium constant \( K_c \) is only affected by temperature. Since the forward reaction is exothermic, decreasing the temperature will shift the equilibrium position to the right to oppose the change, thereby increasing the equilibrium concentration of the product, \( \text{SO}_3(g) \), and consequently increasing both the yield and the value of \( K_c \). Increasing pressure shifts the equilibrium to the right but does not alter the value of \( K_c \). Catalysts do not change the yield or \( K_c \).
Marking scheme
Award 1 mark for selecting option A. Candidates must recognize that temperature is the only variable that changes Kc, and since the reaction is exothermic, lowering the temperature increases Kc and yield.
Paper 22
Answer all structured questions in the spaces provided. Show all working and appropriate units.
5 Question · 60 marks
Question 1 · Structured
12 marks
An anhydrous mixture containing sodium carbonate, \( \text{Na}_2\text{CO}_3 \), and sodium hydrogencarbonate, \( \text{NaHCO}_3 \), has a total mass of \( 2.50\text{ g} \). This mixture is heated until no further reaction occurs.
The thermal decomposition of sodium hydrogencarbonate produces sodium carbonate, carbon dioxide, and water vapor as shown below: \[ 2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g}) \]
(a) (i) Explain why sodium carbonate does not decompose under the same conditions. [1] (ii) Suggest why the solid residue after heating has a lower mass than the original mixture. [1]
(b) The gaseous products, \( \text{CO}_2 \) and \( \text{H}_2\text{O} \), are collected. After removing the water vapor, the volume of dry carbon dioxide gas collected is found to be \( 180\text{ cm}^3 \) measured at \( 298\text{ K} \) and \( 101\text{ kPa} \).
(i) Calculate the amount, in moles, of carbon dioxide gas collected. Treat carbon dioxide as an ideal gas. [2] (ii) Hence, calculate the mass of \( \text{NaHCO}_3 \) present in the original \( 2.50\text{ g} \) mixture. [2] (iii) Calculate the percentage by mass of \( \text{Na}_2\text{CO}_3 \) in the original mixture. [2]
(c) In a separate experiment, a \( 1.45\text{ g} \) sample of hydrated sodium carbonate, \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \), is dissolved in water to make \( 250.0\text{ cm}^3 \) of solution. A \( 25.0\text{ cm}^3 \) portion of this solution requires \( 20.25\text{ cm}^3 \) of \( 0.0500\text{ mol dm}^{-3} \) hydrochloric acid for complete neutralisation. \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]
(i) Calculate the amount, in moles, of \( \text{HCl} \) used in the titration. [1] (ii) Calculate the amount, in moles, of sodium carbonate present in the entire \( 250.0\text{ cm}^3 \) flask. [2] (iii) Determine the value of \( x \) in \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \). [1]
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Worked solution
(a) (i) Sodium carbonate requires a much higher temperature to decompose because of the higher stability of the carbonate anion in the presence of the sodium cation (no decomposition at this temperature). (ii) Carbon dioxide and water vapor are lost to the atmosphere as gases.
(b) (i) Using \( pV = nRT \): \( p = 101 \times 10^3\text{ Pa} \) \( V = 180 \times 10^{-6}\text{ m}^3 \) \( T = 298\text{ K} \) \( n = \frac{101000 \times 1.80 \times 10^{-4}}{8.31 \times 298} = 7.34 \times 10^{-3}\text{ mol} \)
(ii) From the stoichiometry of the equation, \( 2\text{ mol} \) of \( \text{NaHCO}_3 \) yields \( 1\text{ mol} \) of \( \text{CO}_2 \). \( n(\text{NaHCO}_3) = 2 \times 7.341 \times 10^{-3} = 0.01468\text{ mol} \) \( \text{Mass of NaHCO}_3 = 0.01468 \times 84.0 = 1.233\text{ g} \) (or \( 1.23\text{ g} \))
(a) (i) Sodium carbonate is thermally stable / does not decompose under these conditions. [1] (ii) Carbon dioxide and water vapor are lost / escape as gases. [1]
(b) (i) Correct use of pV=nRT equation with appropriate units: [1] Correct calculation of moles = 7.34 x 10^-3 mol: [1] (ii) Correct mole ratio used (multiplied by 2): [1] Correct mass of NaHCO3 = 1.23 g (or 1.233 g): [1] (iii) Correct subtraction to find mass of Na2CO3: [1] Correct percentage calculation to 3 s.f. = 50.7%: [1]
(c) (i) Correct calculation of moles of HCl = 1.01 x 10^-3 mol: [1] (ii) Moles of Na2CO3 in 25 cm3 (divided by 2): [1] Moles of Na2CO3 in 250 cm3 (multiplied by 10) = 5.06 x 10^-3 mol: [1] (iii) Correct calculation of Mr of hydrated salt and final determination of x = 10: [1]
Question 2 · Structured
12 marks
A student carries out an experiment to determine the enthalpy change of combustion, \( \Delta H_c \), of liquid propan-1-ol, \( \text{C}_3\text{H}_7\text{OH} \).
A sample of propan-1-ol is burned in a spirit burner to heat \( 150\text{ g} \) of water in a copper calorimeter. The following results are obtained: - Mass of spirit burner before combustion = 124.85 g - Mass of spirit burner after combustion = 124.35 g - Temperature of water before heating = \( 21.5^\circ\text{C} \) - Temperature of water after heating = \( 43.5^\circ\text{C} \) - Specific heat capacity of water, \( c = 4.18\text{ J g}^{-1}\text{ K}^{-1} \)
(a) (i) Calculate the heat energy, in kJ, absorbed by the water. [1] (ii) Calculate the amount, in moles, of propan-1-ol burned. [2] (iii) Calculate the experimental enthalpy change of combustion, \( \Delta H_c \), of propan-1-ol. Include a sign and appropriate units. [2]
(b) Suggest two reasons why the experimental value of \( \Delta H_c \) calculated in (a)(iii) is much less exothermic than the standard data book value of \( -2021\text{ kJ mol}^{-1} \). [2]
(c) The theoretical standard enthalpy change of combustion of propan-1-ol can be calculated using standard enthalpy changes of formation, \( \Delta H^\theta_f \): - \( \Delta H^\theta_f [\text{C}_3\text{H}_7\text{OH(l)}] = -303\text{ kJ mol}^{-1} \) - \( \Delta H^\theta_f [\text{CO}_2\text{(g)}] = -394\text{ kJ mol}^{-1} \) - \( \Delta H^\theta_f [\text{H}_2\text{O(l)}] = -286\text{ kJ mol}^{-1} \)
(i) Write the balanced chemical equation, including state symbols, for the complete combustion of propan-1-ol. [1] (ii) Calculate the theoretical standard enthalpy change of combustion, \( \Delta H^\theta_c \), of propan-1-ol using the provided \( \Delta H^\theta_f \) values. [2]
(d) Define the term *standard enthalpy change of formation*. [2]
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(d) Enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K and 100 kPa).
Marking scheme
(a) (i) Correct calculation of energy: 13.8 kJ (or 13.794 kJ) [1] (ii) Correct determination of mass burned (0.50 g): [1] Correct calculation of moles = 8.33 x 10^-3 mol (accept 0.0083 mol): [1] (iii) Enthalpy value calculated correctly: [1] Negative sign and correct units (kJ mol^-1): [1]
(b) Accept any two of: Heat loss to surroundings, incomplete combustion, evaporation of fuel, heat capacity of calorimeter not included. [2, 1 mark each]
(c) (i) Correct balanced equation with state symbols: [1] (ii) Correct Hess's Law expression: [1] Correct final answer: -2023 kJ mol^-1: [1]
(d) Enthalpy change when 1 mole of a compound is formed: [1] From its constituent elements in their standard states under standard conditions: [1]
Question 3 · Structured
12 marks
The elements in Period 3 show characteristic trends in their physical and chemical properties.
(a) Phosphorus and sulfur form the oxides \( \text{P}_4\text{O}_{10} \) and \( \text{SO}_3 \) respectively.
(i) State the structure and bonding of phosphorus(V) oxide, \( \text{P}_4\text{O}_{10} \), and sulfur trioxide, \( \text{SO}_3 \). [2] (ii) Write equations for the separate reactions of \( \text{P}_4\text{O}_{10} \) and \( \text{SO}_3 \) with water, and state the approximate pH of each resulting solution. [4]
(b) Silicon tetrachloride, \( \text{SiCl}_4 \), and sodium chloride, \( \text{NaCl} \), behave differently when added to water.
(i) Write the equation for the reaction of \( \text{SiCl}_4 \) with water, and state the approximate pH of the solution. [2] (ii) Explain, in terms of structure and the availability of orbitals, why \( \text{SiCl}_4 \) reacts vigorously with water whereas \( \text{NaCl} \) simply dissolves to form a neutral solution. [2]
(c) Aluminium oxide, \( \text{Al}_2\text{O}_3 \), is amphoteric.
Write two ionic equations to demonstrate the amphoteric behavior of aluminium oxide. [2]
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Worked solution
(a) (i) Both \( \text{P}_4\text{O}_{10} \) and \( \text{SO}_3 \) have simple molecular / molecular structures with covalent bonding.
(ii) Reaction of \( \text{P}_4\text{O}_{10} \) with water: \( \text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 \) pH = 1 or 2
Reaction of \( \text{SO}_3 \) with water: \( \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \) pH = 0 to 2
(ii) Silicon has low-lying empty 3d orbitals which can accept lone pairs of electrons from water molecules to initiate hydrolysis/reaction. NaCl is an ionic compound containing sodium ions which have no available low-energy orbitals (and low charge density), resulting only in the hydration of ions rather than hydrolysis.
(a) (i) Simple molecular structure: [1] Covalent bonding in both: [1] (ii) Equation for P4O10: [1] pH = 1 or 2: [1] Equation for SO3: [1] pH = 0, 1 or 2: [1]
(b) (i) Correct equation (forming SiO2 or Si(OH)4 and HCl): [1] pH = 1 or 2: [1] (ii) Mention of low-lying, vacant 3d orbitals in Si: [1] NaCl is ionic and does not undergo covalent bond cleavage / hydration only occurs: [1]
(c) Correct ionic equation with H+ (aq): [1] Correct ionic equation with OH- (aq): [1]
Question 4 · Structured
12 marks
Methanol is produced industrially by the reversible reaction of carbon monoxide with hydrogen in the presence of a catalyst:
The forward reaction is exothermic with \( \Delta H = -91\text{ kJ mol}^{-1} \).
(a) Write the expression for the equilibrium constant, \( K_c \), for this reaction, and state its units. [2]
(b) A mixture of \( 1.20\text{ mol} \) of \( \text{CO} \) and \( 2.40\text{ mol} \) of \( \text{H}_2 \) is sealed in a \( 5.00\text{ dm}^3 \) flask at a constant temperature of \( 500\text{ K} \). When equilibrium is reached, \( 0.40\text{ mol} \) of \( \text{CH}_3\text{OH} \) has been formed.
(i) Calculate the equilibrium amounts, in moles, of \( \text{CO} \) and \( \text{H}_2 \). [2] (ii) Calculate the value of \( K_c \) at \( 500\text{ K} \). [3]
(c) The temperature of the flask is now increased to \( 600\text{ K} \) while keeping the volume constant.
(i) State and explain the effect of this temperature increase on the value of \( K_c \). [2] (ii) State and explain the effect of this temperature increase on the rate of the forward reaction. [2]
(d) Explain why the addition of a catalyst does not change the position of the equilibrium. [1]
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(c) (i) \( K_c \) decreases. The forward reaction is exothermic, so increasing the temperature shifts the equilibrium to the left (in the endothermic direction) to absorb the added heat. (ii) The rate increases. Higher temperature gives molecules more kinetic energy, so a larger fraction of reactant molecules have collision energy \( \ge E_a \) (and collision frequency increases).
(d) A catalyst increases the rates of both the forward and reverse reactions by the same proportion, so the position of equilibrium is unchanged.
(b) (i) n(CO) = 0.80 mol: [1] n(H2) = 1.60 mol: [1] (ii) Conversion of moles to concentrations by dividing by 5.00: [1] Substitution of concentrations into Kc expression: [1] Correct final answer (4.88 or 4.9): [1]
(c) (i) Kc decreases: [1] Forward reaction is exothermic, equilibrium shifts to the left / endothermic direction: [1] (ii) Forward rate increases: [1] Greater proportion of particles with energy equal to or exceeding activation energy: [1]
(d) A catalyst increases the rate of both forward and reverse reactions equally. [1]
Question 5 · Structured
12 marks
Halogenoalkanes are important intermediates in organic synthesis. Two structural isomers with the molecular formula \( \text{C}_4\text{H}_9\text{Br} \) are: - Isomer A: 1-bromobutane, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \) - Isomer B: 2-bromo-2-methylpropane, \( (\text{CH}_3)_3\text{CBr} \)
(a) (i) Classify Isomer A and Isomer B as primary, secondary, or tertiary halogenoalkanes. [2] (ii) Describe the mechanism for the reaction of Isomer B with aqueous sodium hydroxide, \( \text{NaOH(aq)} \). Use curly arrows to show the movement of electron pairs, and include any relevant partial charges, lone pairs, and intermediates. [4]
(b) The relative rate of hydrolysis of these halogenoalkanes can be investigated by heating each isomer separately with aqueous silver nitrate in ethanol.
(i) State the colour of the precipitate observed when the halogenoalkanes react. [1] (ii) Explain which isomer, A or B, reacts faster in this reaction. Refer to the mechanism of the reaction in your answer. [2]
(c) Under different conditions, Isomer A can undergo an elimination reaction to form an alkene.
(i) Identify a suitable reagent and solvent for this elimination reaction. [2] (ii) State the IUPAC name of the organic product formed from this elimination reaction of Isomer A. [1]
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Worked solution
(a) (i) Isomer A: primary halogenoalkane. Isomer B: tertiary halogenoalkane. (ii) Mechanism is \( S_N1 \): - Step 1: Heterolytic fission of the \( \text{C}-\text{Br} \) bond. Carbon atom in \( (\text{CH}_3)_3\text{C}-\text{Br} \) has a \( \delta^+ \) charge and bromine has a \( \delta^- \) charge. A curly arrow starts from the \( \text{C}-\text{Br} \) bond and points to the Br atom. This yields a stable tertiary carbocation intermediate, \( (\text{CH}_3)_3\text{C}^+ \), and \( \text{Br}^- \). - Step 2: Nucleophilic attack. A curly arrow starts from a lone pair on the oxygen of the hydroxide ion, \( \text{OH}^- \), and points to the carbocation carbon, forming the product \( (\text{CH}_3)_3\text{COH} \).
(b) (i) Cream precipitate (silver bromide, \( \text{AgBr} \)). (ii) Isomer B reacts faster. It undergoes hydrolysis via an \( S_N1 \) mechanism which involves the formation of a highly stable tertiary carbocation intermediate. This intermediate forms much more rapidly than the transition state of the \( S_N2 \) mechanism of Isomer A.
(a) (i) Isomer A is primary [1] and Isomer B is tertiary [1] (ii) Heterolytic cleavage: draw structure with delta+ and delta- charges on carbon and bromine, and curly arrow from C-Br bond to Br: [1] Correct drawing of carbocation intermediate: [1] Curly arrow from lone pair of OH- to positive carbon of carbocation: [1] Correct final structure of tertiary alcohol: [1]
(b) (i) Cream precipitate: [1] (ii) Isomer B reacts faster: [1] Due to stable tertiary carbocation / SN1 mechanism is faster than SN2: [1]
Complete all practical experiments. Record all results to the appropriate precision and complete the calculations.
3 Question · 39.99 marks
Question 1 · Practical
13.33 marks
### Determination of the percentage by mass of sodium carbonate in a mixture
You are to determine the percentage by mass of sodium carbonate, \(\text{Na}_2\text{CO}_3\), in a solid mixture of sodium carbonate and sodium chloride, **FA 1**. You will prepare a solution of the mixture and titrate it against a standard solution of hydrochloric acid, **FA 2**, of concentration \(0.115\text{ mol dm}^{-3}\).
The equation for the reaction is: \[\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\]
#### Procedure 1. Weigh a clean, dry weighing bottle. 2. Add between 2.20 g and 2.60 g of **FA 1** to the bottle and record the mass. 3. Transfer the solid to a beaker, reweigh the empty weighing bottle, and calculate the mass of **FA 1** transferred. 4. Dissolve the solid in about 100 cm3 of distilled water, transfer the solution quantitatively to a 250 cm3 volumetric flask, make up to the mark with distilled water, and mix thoroughly. Label this solution **FA 3**. 5. Pipette 25.0 cm3 of **FA 3** into a conical flask. Add 3–4 drops of methyl orange indicator. 6. Titrate **FA 3** with the standard hydrochloric acid, **FA 2**, contained in a burette. 7. Record your titration results in a suitable table in the space below. Repeat the titration until you obtain concordant results (within \(0.10\text{ cm}^3\)).
#### Questions **(a)** Present your titration results in a clear table, including all raw burette readings and volumes of **FA 2** added. (4.00 marks)
**(b)** Show which results you used to calculate the average titre and write down this value. (1.33 marks)
**(c)** Calculate: - (i) the number of moles of \(\text{HCl}\) in the average titre volume of **FA 2**. - (ii) the number of moles of \(\text{Na}_2\text{CO}_3\) in 25.0 cm3 of **FA 3**. - (iii) the number of moles of \(\text{Na}_2\text{CO}_3\) in the 250.0 cm3 volumetric flask. - (iv) the percentage by mass of \(\text{Na}_2\text{CO}_3\) in the original solid mixture **FA 1**. (6.00 marks)
**(d)** Calculate the percentage error in the volume of **FA 2** delivered from a Class B burette (uncertainty of \(\pm0.05\text{ cm}^3\) for each reading) for your average titre. Suggest one modification to the experimental procedure (without changing the apparatus or reactants) that would reduce the percentage error in the titration volume. (2.00 marks)
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Worked solution
Based on typical experimental measurements: 1. Mass of FA 1 = 2.45 g 2. Average titre = 24.10 cm3
- (c)(iv) Mass and percentage by mass of \(\text{Na}_2\text{CO}_3\): \(M_r(\text{Na}_2\text{CO}_3) = 105.99\text{ g mol}^{-1}\) \(m(\text{Na}_2\text{CO}_3) = 1.3858 \times 10^{-2} \times 105.99 = 1.4688\text{ g}\) \(\text{Percentage by mass} = \frac{1.4688}{2.45} \times 100\% = 59.95\% \approx 60.0\%\)
- (d) Percentage error and modification: \(\text{Percentage error} = \frac{2 \times 0.05}{24.10} \times 100\% = 0.41\%\) Modification: Use a larger starting mass of the mixture **FA 1** to prepare **FA 3** (or use a larger pipetted volume of **FA 3**, e.g., 50.0 cm3). This will increase the average volume of **FA 2** needed, which directly decreases the percentage error.
Marking scheme
**(a) Titration Table [Total 4.00 marks]** - 1 mark: Table is presented with appropriate column headings and units (e.g., initial and final burette readings / \(\text{cm}^3\), volume added / \(\text{cm}^3\)). - 1 mark: All burette readings (initial and final) are recorded to the nearest 0.05 \(\text{cm}^3\) (ending in .00 or .05). - 2 marks: Two accurate titrations are within 0.10 \(\text{cm}^3\) (concordant results).
**(b) Average Titre [Total 1.33 marks]** - 1.33 marks: Correctly calculates the mean titre showing the working (values used) and states the answer to 2 decimal places.
**(c) Calculations [Total 6.00 marks]** - (i) 1.50 marks: Moles of \(\text{HCl}\) correctly calculated with 3 significant figures: \(\text{Titre} \times 10^{-3} \times 0.115\). - (ii) 1.50 marks: Moles of \(\text{Na}_2\text{CO}_3\) in 25.0 cm3 is \(\text{moles of HCl} / 2\). - (iii) 1.50 marks: Moles of \(\text{Na}_2\text{CO}_3\) in 250.0 cm3 is \(\text{moles of Na}_2\text{CO}_3 \text{ in 25.0 cm}^3 \times 10\). - (iv) 1.50 marks: Correctly calculates mass of \(\text{Na}_2\text{CO}_3\) using \(M_r = 106.0\) (or 105.99) and calculates the percentage by mass based on the student's mass of **FA 1** to 3 s.f.
**(d) Error Analysis and Improvement [Total 2.00 marks]** - 1 mark: Correct calculation of percentage error using \((0.10 / \text{mean titre}) \times 100\%\). - 1 mark: Suggests preparing a more concentrated solution of **FA 3** / using a larger mass of **FA 1**, which increases the required titre volume and thus reduces the percentage error.
Question 2 · Practical
13.33 marks
### Enthalpy change of hydration of anhydrous copper(II) sulfate
You will determine the enthalpy change of solution of anhydrous copper(II) sulfate and hydrated copper(II) sulfate, and then use Hess's Law to calculate the enthalpy change of hydration of anhydrous copper(II) sulfate.
**FA 4** is anhydrous copper(II) sulfate, \(\text{CuSO}_4\). **FA 5** is hydrated copper(II) sulfate, \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\).
#### Procedure
##### Experiment 1: Enthalpy change of solution of anhydrous copper(II) sulfate 1. Support a plastic cup in a 250 cm3 beaker. 2. Use a measuring cylinder to transfer 50.0 cm3 of distilled water into the plastic cup. 3. Measure and record the initial temperature of the water to the nearest \(0.2\ ^\circ\text{C}\). 4. Weigh a weighing bottle containing between 3.80 g and 4.20 g of anhydrous copper(II) sulfate, **FA 4**. Record the mass. 5. Add the **FA 4** to the water in the plastic cup, stir continuously, and record the maximum temperature achieved. 6. Reweigh the weighing bottle to determine the mass of **FA 4** transferred.
##### Experiment 2: Enthalpy change of solution of hydrated copper(II) sulfate 1. Empty and dry the plastic cup. 2. Repeat the procedure using 50.0 cm3 of distilled water and between 5.80 g and 6.20 g of hydrated copper(II) sulfate, **FA 5**. 3. Record the initial temperature and the minimum temperature achieved.
#### Questions **(a)** Present your results for both Experiment 1 and Experiment 2 in clear tables. Include all mass and temperature measurements with their appropriate units. (4.00 marks)
**(b)** Calculate: - (i) the heat energy change, \(q_1\), for Experiment 1 (assume density of water = \(1.00\text{ g cm}^{-3}\) and specific heat capacity \(c = 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1}\)). - (ii) the enthalpy change of solution, \(\Delta H_1\), of anhydrous \(\text{CuSO}_4\) in \(\text{kJ mol}^{-1}\). - (iii) the heat energy change, \(q_2\), for Experiment 2. - (iv) the enthalpy change of solution, \(\Delta H_2\), of hydrated \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\) in \(\text{kJ mol}^{-1}\). (5.33 marks)
**(c)** Draw a Hess's Law cycle showing how \(\Delta H_1\), \(\Delta H_2\), and the enthalpy change of hydration, \(\Delta H_{\text{hyd}}\), of anhydrous copper(II) sulfate are related. Use your cycle to calculate \(\Delta H_{\text{hyd}}\). (2.00 marks)
**(d)** State how heat loss to the surroundings would affect the calculated value of \(\Delta H_1\) (make it more negative or less negative) and suggest one improvement to the apparatus to reduce heat loss. (2.00 marks)
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Worked solution
Let's work through the calculations with typical measurements:
##### Experiment 1 (FA 4 - Anhydrous) - Mass of weighing bottle + solid = 14.85 g - Mass of weighing bottle + residue = 10.85 g - Mass of **FA 4** transferred, \(m_1 = 4.00\text{ g}\) - Initial temperature, \(T_{\text{initial}} = 21.5\ ^\circ\text{C}\) - Maximum temperature, \(T_{\text{max}} = 29.4\ ^\circ\text{C}\) - Temperature change, \(\Delta T_1 = +7.9\ ^\circ\text{C}\)
##### Experiment 2 (FA 5 - Hydrated) - Mass of weighing bottle + solid = 17.05 g - Mass of weighing bottle + residue = 10.85 g - Mass of **FA 5** transferred, \(m_2 = 6.20\text{ g}\) - Initial temperature, \(T_{\text{initial}} = 21.5\ ^\circ\text{C}\) - Minimum temperature, \(T_{\text{min}} = 20.1\ ^\circ\text{C}\) - Temperature change, \(\Delta T_2 = -1.4\ ^\circ\text{C}\)
##### Calculations: - (b)(i) \(q_1 = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 7.9\ ^\circ\text{C} = 1651.1\text{ J} \approx 1.65\text{ kJ}\)
- (d) Discussion of error: Heat loss to the surroundings reduces the measured maximum temperature change (makes \(\Delta T_1\) smaller than the theoretical value). This makes the calculated heat energy \(q_1\) lower and thus \(\Delta H_1\) less negative (less exothermic). An experimental improvement would be to use a lid on the polystyrene cup to reduce heat loss by convection, or nest the cup inside another polystyrene cup to provide better insulation.
Marking scheme
**(a) Presentation of Results [Total 4.00 marks]** - 1 mark: Results of both experiments are clearly presented in tables with clear column headings and units: mass / g, temperature / °C. - 1 mark: All temperatures are recorded to the nearest 0.1 °C or 0.2 °C, and all masses to 2 or 3 decimal places. - 2 marks: Reasonable temperature differences are obtained (Experiment 1 is exothermic / temperature rises; Experiment 2 is endothermic / temperature falls).
**(b) Calculations [Total 5.33 marks]** - (i) 1.33 marks: Correct calculation of \(q_1 = m c \Delta T\) using 50.0 g and correct temperature rise. - (ii) 1.33 marks: Correct calculation of \(n(\text{CuSO}_4)\) using \(M_r = 159.62\), and \(\Delta H_1\) with a negative sign and correct unit (\(\text{kJ mol}^{-1}\)). - (iii) 1.33 marks: Correct calculation of \(q_2\) using 50.0 g and correct temperature drop. - (iv) 1.33 marks: Correct calculation of \(n(\text{CuSO}_4\cdot5\text{H}_2\text{O})\) using \(M_r = 249.72\), and \(\Delta H_2\) with a positive sign and correct unit (\(\text{kJ mol}^{-1}\)).
**(c) Hess's Law Cycle and Calculation [Total 2.00 marks]** - 1 mark: Draw a fully labeled Hess's Law cycle relating the reactants, products, and solutions. - 1 mark: Calculate \(\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2\) correctly using the student's values.
**(d) Sources of Error and Improvement [Total 2.00 marks]** - 1 mark: Correctly identifies that heat loss makes the measured temperature change smaller, making \(\Delta H_1\) less negative. - 1 mark: Suggests a lid / double-cupping to reduce heat loss.
Question 3 · Practical
13.33 marks
### Qualitative Analysis: Identification of unknown ions
You will carry out tests to identify the cations and anions present in three unknown aqueous solutions, **FA 6**, **FA 7**, and **FA 8**.
#### Procedure
Perform the following tests on each solution and record your observations in the spaces provided.
| Test | Instructions | | :--- | :--- | | **Test 1** | Add aqueous sodium hydroxide, \(\text{NaOH}(\text{aq})\), dropwise and then in excess to 1 cm3 of each solution in a test-tube. Warm the mixture with **FA 8** carefully, and test any gas evolved with damp red litmus paper. | | **Test 2** | Add aqueous ammonia, \(\text{NH}_3(aq)\), dropwise and then in excess to 1 cm3 of each solution in a test-tube. | | **Test 3** | Add 1 cm3 of aqueous barium chloride, \(\text{BaCl}_2(\text{aq})\) (or barium nitrate), to 1 cm3 of each solution, followed by dilute nitric acid. | | **Test 4** | Add 1 cm3 of aqueous silver nitrate, \(\text{AgNO}_3(\text{aq})\), to 1 cm3 of each solution, followed by dilute aqueous ammonia. |
#### Questions **(a)** Record your observations for **Test 1** and **Test 2** in a suitable table. (4.00 marks)
**(b)** Record your observations for **Test 3** and **Test 4** in a suitable table. (4.00 marks)
**(c)** Identify the cations and anions present in each solution. Support each identification with a brief explanation referencing your observations. - **FA 6**: Cation = \_\_\_\_\_\_\_\_, Anion = \_\_\_\_\_\_\_\_ - **FA 7**: Cation = \_\_\_\_\_\_\_\_, Anion = \_\_\_\_\_\_\_\_ - **FA 8**: Cation = \_\_\_\_\_\_\_\_, Anion = \_\_\_\_\_\_\_\_ (3.33 marks)
**(d)** Write ionic equations for: - (i) the formation of the precipitate when **FA 6** reacts with aqueous sodium hydroxide. - (ii) the reaction that occurs when excess aqueous sodium hydroxide is added to **FA 7** and the precipitate dissolves. (2.00 marks)
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Worked solution
Let's summarize the expected results and explanations:
##### Observations: - **FA 6** contains \(\text{MgCl}_2\): - Test 1 (\(\text{NaOH}\)): White precipitate formed, insoluble in excess. - Test 2 (\(\text{NH}_3\)): White precipitate formed, insoluble in excess. - Test 3 (\(\text{Ba}^{2+}\)): No precipitate / remains a colorless solution. - Test 4 (\(\text{Ag}^+\)): White precipitate formed, which dissolves in dilute ammonia to form a colorless solution.
- **FA 7** contains \(\text{Al}_2(\text{SO}_4)_3\): - Test 1 (\(\text{NaOH}\)): White precipitate formed, which dissolves in excess \(\text{NaOH}\) to form a colorless solution. - Test 2 (\(\text{NH}_3\)): White precipitate formed, insoluble in excess. - Test 3 (\(\text{Ba}^{2+}\)): Thick white precipitate formed, insoluble in dilute nitric acid. - Test 4 (\(\text{Ag}^+\)): No precipitate / remains a colorless solution.
- **FA 8** contains \(\text{NH}_4\text{I}\): - Test 1 (\(\text{NaOH}\)): No precipitate. On warming, a gas is evolved with a pungent smell that turns damp red litmus paper blue (ammonia gas). - Test 2 (\(\text{NH}_3\)): No precipitate. - Test 3 (\(\text{Ba}^{2+}\)): No precipitate. - Test 4 (\(\text{Ag}^+\)): Pale yellow precipitate formed, insoluble in dilute ammonia.
##### Identifications: - **FA 6**: Cation is \(\text{Mg}^{2+}\) (white ppt with both \(\text{NaOH}\) and \(\text{NH}_3\) insoluble in excess); Anion is \(\text{Cl}^-\) (white precipitate with \(\text{Ag}^+\) soluble in NH₃). - **FA 7**: Cation is \(\text{Al}^{3+}\) (white ppt with \(\text{NaOH}\) soluble in excess, white ppt with \(\text{NH}_3\) insoluble in excess); Anion is \(\text{SO}_4^{2-}\) (thick white precipitate with barium ions insoluble in strong acid). - **FA 8**: Cation is \(\text{NH}_4^+\) (evolution of ammonia gas upon warming with aqueous sodium hydroxide); Anion is \(\text{I}^-\) (yellow precipitate with silver ions insoluble in ammonia).
**(a) Test 1 and Test 2 Table [Total 4.00 marks]** - 2 marks: Correct observations for **FA 6** (white ppt with both NaOH and NH₃, insoluble in excess). - 1 mark: Correct observations for **FA 7** (white ppt with NaOH soluble in excess; white ppt with NH₃ insoluble in excess). - 1 mark: Correct observations for **FA 8** (no ppt; gas evolved on warming with NaOH turns damp red litmus paper blue).
**(b) Test 3 and Test 4 Table [Total 4.00 marks]** - 1.5 marks: Correct observations for **FA 6** (no ppt with Ba²⁺; white ppt with Ag⁺, which dissolves in dilute NH₃). - 1.5 marks: Correct observations for **FA 7** (white ppt with Ba²⁺, insoluble in HNO₃; no ppt with Ag⁺). - 1 mark: Correct observations for **FA 8** (no ppt with Ba²⁺; yellow ppt with Ag⁺, insoluble in NH₃).
**(c) Identifications and Evidence [Total 3.33 marks]** - 1.11 marks: **FA 6** identified as \(\text{Mg}^{2+}\) and \(\text{Cl}^-\). - 1.11 marks: **FA 7** identified as \(\text{Al}^{3+}\) and \(\text{SO}_4^{2-}\). - 1.11 marks: **FA 8** identified as \(\text{NH}_4^+\) and \(\text{I}^-\). All identifications must be supported by valid experimental evidence from parts (a) and (b).
**(d) Ionic Equations [Total 2.00 marks]** - 1 mark: \(\text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s})\) (including correct state symbols). - 1 mark: \(\text{Al(OH)}_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Al(OH)}_4]^-(\text{aq})\) (or balanced \(\text{Al}^{3+}\) equation; must include correct state symbols).
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