An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 Cambridge International A Level Chemistry (XCH11) paper. Not affiliated with or reproduced from Cambridge.
Unit 1: 甲部
Answer all multiple choice questions by crossing one box. Spend no more than 20 minutes.
20 題目 · 20 分
題目 1 · multiple_choice
1 分
An element \(X\) in Period 3 of the Periodic Table has the following successive ionization energies in \(\text{kJ mol}^{-1}\): \(IE_1 = 578\), \(IE_2 = 1817\), \(IE_3 = 2745\), \(IE_4 = 11577\), \(IE_5 = 14842\). What is the formula of the chloride of element \(X\)?
A.\(X\text{Cl}\)
B.\(X\text{Cl}_2\)
C.\(X\text{Cl}_3\)
D.\(X\text{Cl}_4\)
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解題
The successive ionization energies show a very large increase between the third and fourth ionization energies. This indicates that the fourth electron is removed from an inner shell which is closer to the nucleus and experiences less shielding. Therefore, there are 3 electrons in the outer shell of element \(X\), meaning it is in Group 13. A Group 13 element forms a \(3+\) ion, \(X^{3+}\). When reacting with chlorine which forms chloride ions, \(\text{Cl}^-\), the formula of the compound is \(X\text{Cl}_3\).
評分準則
1 mark: Correct choice C.
題目 2 · multiple_choice
1 分
What volume of carbon dioxide, measured at room temperature and pressure (r.t.p.), is produced when \(4.20\text{ g}\) of sodium hydrogencarbonate, \(\text{NaHCO}_3\), is completely thermally decomposed? [Molar volume of gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\); Molar mass of \(\text{NaHCO}_3 = 84.0\text{ g mol}^{-1}\)] Reaction equation: \(2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)
A.\(0.60\text{ dm}^3\)
B.\(1.20\text{ dm}^3\)
C.\(2.40\text{ dm}^3\)
D.\(4.80\text{ dm}^3\)
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解題
First, calculate the amount of \(\text{NaHCO}_3\) in moles: \(n(\text{NaHCO}_3) = \frac{\text{mass}}{\text{molar mass}} = \frac{4.20\text{ g}}{84.0\text{ g mol}^{-1}} = 0.050\text{ mol}\). Second, use the stoichiometric ratio from the balanced chemical equation: \(2\text{ mol}\) of \(\text{NaHCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\). Therefore, \(n(\text{CO}_2) = \frac{0.050}{2} = 0.025\text{ mol}\). Third, calculate the volume of \(\text{CO}_2\) at r.t.p.: \(V(\text{CO}_2) = n \times \text{Molar Volume} = 0.025\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.60\text{ dm}^3\).
評分準則
1 mark: Correct choice A.
題目 3 · multiple_choice
1 分
Which of the following species has a shape that is NOT based on a tetrahedral arrangement of electron pairs around the central atom?
A.\(\text{NH}_4^+\)
B.\(\text{SF}_4\)
C.\(\text{H}_3\text{O}^+\)
D.\(\text{PCl}_3\)
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解題
The arrangement of electron pairs depends on the total number of bonding pairs (BP) and lone pairs (LP) on the central atom. For \(\text{NH}_4^+\), there are 4 BP and 0 LP (tetrahedral arrangement). For \(\text{H}_3\text{O}^+\), there are 3 BP and 1 LP (tetrahedral arrangement). For \(\text{PCl}_3\), there are 3 BP and 1 LP (tetrahedral arrangement). For \(\text{SF}_4\), the sulfur atom has 6 valence electrons, forms 4 single covalent bonds with fluorine atoms, and retains 1 lone pair. This gives a total of 5 electron pairs (trigonal bipyramidal arrangement), which results in a see-saw molecular shape.
評分準則
1 mark: Correct choice B.
題目 4 · multiple_choice
1 分
In the free radical chlorination of methane, which of the following equations represents a propagation step?
A propagation step must involve a free radical as a reactant and a different free radical as a product. The reaction of a methane molecule with a chlorine radical to form a methyl radical and hydrogen chloride molecule meets this requirement: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\). Option A is an initiation step. Options B and D are termination steps because they consume two free radicals to form a stable molecule.
評分準則
1 mark: Correct choice C.
題目 5 · multiple_choice
1 分
Which of the following alkenes can exist as a pair of geometric (E/Z) isomers?
A.2-methylbut-2-ene
B.2-methylbut-1-ene
C.pent-1-ene
D.pent-2-ene
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解題
For geometric (E/Z) isomerism to occur, the molecule must contain a carbon-carbon double bond where each carbon of the double bond is attached to two different groups. In pent-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3\)), carbon-2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\), and carbon-3 is bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Both carbons satisfy the condition, so pent-2-ene exists as (E)-pent-2-ene and (Z)-pent-2-ene. In the other options, at least one of the double-bonded carbon atoms is attached to two identical groups, which prevents E/Z isomerism.
評分準則
1 mark: Correct choice D.
題目 6 · multiple_choice
1 分
Which of the following species has the electronic configuration \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\)?
A.\(\text{S}^-\)
B.\(\text{Ca}^+\)
C.\(\text{P}^{3-}\)
D.\(\text{Ar}^+\)
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解題
The electronic configuration \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6\) has a total of 18 electrons. Phosphorus (atomic number 15) has 15 electrons in its ground state. The phosphide ion, \(\text{P}^{3-}\), is formed by gaining 3 electrons, giving a total of 18 electrons, which has the electronic configuration of argon. S- has 17 electrons. Ca+ has 19 electrons. Ar+ has 17 electrons.
評分準則
1 mark: Correct choice C.
題目 7 · multiple_choice
1 分
In a titration, \(25.0\text{ cm}^3\) of a solution of a dicarboxylic acid, \(\text{H}_2\text{A}\), required \(20.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium hydroxide solution, \(\text{NaOH}\), for complete neutralization. What is the concentration of the dicarboxylic acid in \(\text{mol dm}^{-3}\)? Reaction equation: \(\text{H}_2\text{A}(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{A}(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\)
A.\(0.040\text{ mol dm}^{-3}\)
B.\(0.080\text{ mol dm}^{-3}\)
C.\(0.125\text{ mol dm}^{-3}\)
D.\(0.250\text{ mol dm}^{-3}\)
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解題
First, calculate the amount of \(\text{NaOH}\) used: \(n(\text{NaOH}) = \text{volume} \times \text{concentration} = 0.0200\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00200\text{ mol}\). Second, use the stoichiometric ratio from the equation (1 mole of \(\text{H}_2\text{A}\) reacts with 2 moles of \(\text{NaOH}\)): \(n(\text{H}_2\text{A}) = \frac{0.00200\text{ mol}}{2} = 0.00100\text{ mol}\). Third, calculate the concentration of the acid: \(c(\text{H}_2\text{A}) = \frac{n}{V} = \frac{0.00100\text{ mol}}{0.0250\text{ dm}^3} = 0.040\text{ mol dm}^{-3}\).
評分準則
1 mark: Correct choice A.
題目 8 · multiple_choice
1 分
Which of the following bonds is the most polar? (Electronegativity values: \(\text{H} = 2.1\), \(\text{C} = 2.5\), \(\text{N} = 3.0\), \(\text{O} = 3.5\), \(\text{F} = 4.0\), \(\text{S} = 2.5\))
A.\(\text{C}-\text{F}\)
B.\(\text{O}-\text{H}\)
C.\(\text{C}-\text{H}\)
D.\(\text{S}-\text{O}\)
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解題
Bond polarity is determined by the difference in electronegativity between the two bonded atoms. The greater the electronegativity difference, the more polar the bond. Let us calculate the absolute electronegativity differences: For \(\text{C}-\text{F}\), the difference is \(4.0 - 2.5 = 1.5\). For \(\text{O}-\text{H}\), the difference is \(3.5 - 2.1 = 1.4\). For \(\text{C}-\text{H}\), the difference is \(2.5 - 2.1 = 0.4\). For \(\text{S}-\text{O}\), the difference is \(3.5 - 2.5 = 1.0\). Since the \(\text{C}-\text{F}\) bond has the largest electronegativity difference, it is the most polar bond.
評分準則
1 mark: Correct choice A.
題目 9 · 選擇題
1 分
A \( 10.0\text{ cm}^3 \) sample of a gaseous hydrocarbon was completely burned in \( 70.0\text{ cm}^3 \) of oxygen (an excess). After cooling to room temperature and pressure, the remaining gas mixture had a volume of \( 55.0\text{ cm}^3 \). When this mixture was passed through aqueous sodium hydroxide, the volume decreased to \( 25.0\text{ cm}^3 \). What is the molecular formula of the hydrocarbon?
A.\( \text{C}_3\text{H}_8 \)
B.\( \text{C}_3\text{H}_6 \)
C.\( \text{C}_4\text{H}_{10} \)
D.\( \text{C}_4\text{H}_8 \)
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解題
The chemical equation for the combustion of a hydrocarbon \( \text{C}_x\text{H}_y \) is: \( \text{C}_x\text{H}_y\text{(g)} + \left(x + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow x\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O(l)} \)
Let \( V \) represent the volume in \( \text{cm}^3 \). - Volume of hydrocarbon reacted = \( 10.0\text{ cm}^3 \) - Total volume after reaction and cooling to RTP = \( 55.0\text{ cm}^3 \). This contains excess \( \text{O}_2 \) and produced \( \text{CO}_2 \). - After passing through \( \text{NaOH(aq)} \), the \( \text{CO}_2 \) is absorbed. The remaining volume is \( 25.0\text{ cm}^3 \). - Therefore, volume of excess \( \text{O}_2 = 25.0\text{ cm}^3 \). - Volume of \( \text{CO}_2 \) produced = \( 55.0 - 25.0 = 30.0\text{ cm}^3 \). - Volume of \( \text{O}_2 \) reacted = \( 70.0 - 25.0 = 45.0\text{ cm}^3 \).
Using Avogadro's Law (volume is proportional to moles): - Ratio of hydrocarbon to \( \text{CO}_2 \): \( x = \frac{30.0}{10.0} = 3 \)
- Ratio of hydrocarbon to reacted \( \text{O}_2 \): \( x + \frac{y}{4} = \frac{45.0}{10.0} = 4.5 \) \( 3 + \frac{y}{4} = 4.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6 \)
Therefore, the molecular formula is \( \text{C}_3\text{H}_6 \).
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 10 · 選擇題
1 分
The successive ionisation energies (in \( \text{kJ mol}^{-1} \)) of an element \( \text{X} \) in Period 3 of the Periodic Table are shown below:
Which of the following statements about element \( \text{X} \) is correct?
A.The formula of the oxide of \( \text{X} \) is \( \text{X}_2\text{O} \).
B.Element \( \text{X} \) belongs to Group 14 (IVA) of the Periodic Table.
C.The formula of the chloride of \( \text{X} \) is \( \text{XCl}_3 \).
D.Element \( \text{X} \) is a non-metal.
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解題
1. Identify the group of the element from the successive ionisation energies: The ionisation energies are: \( \text{IE}_1 = 578\text{ kJ mol}^{-1} \) \( \text{IE}_2 = 1817\text{ kJ mol}^{-1} \) \( \text{IE}_3 = 2745\text{ kJ mol}^{-1} \) \( \text{IE}_4 = 11577\text{ kJ mol}^{-1} \) The largest relative jump occurs between \( \text{IE}_3 \) and \( \text{IE}_4 \) (an increase of more than 4 times). This indicates that the fourth electron is being removed from a core shell closer to the nucleus, meaning the element has three valence electrons and belongs to Group 13 (III).
2. Identify the element: Since it is in Period 3 and Group 13, the element is aluminium (\( \text{Al} \)).
3. Determine the formula of its chloride: Aluminium forms a stable \( +3 \) ion (\( \text{Al}^{3+} \)) and reacts with chlorine to form aluminium chloride, which has the formula \( \text{AlCl}_3 \). Therefore, the correct statement is that the formula of the chloride of \( \text{X} \) is \( \text{XCl}_3 \).
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 11 · 選擇題
1 分
Which of the following species has a see-saw molecular geometry?
A.\( \text{SF}_4 \)
B.\( \text{CF}_4 \)
C.\( \text{XeF}_4 \)
D.\( \text{PF}_4^+ \)
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解題
- \( \text{SF}_4 \): Sulfur has 6 valence electrons and forms 4 single covalent bonds with fluorine atoms, leaving 1 lone pair. This gives a steric number of 5, leading to a trigonal bipyramidal electron geometry. To minimize repulsion, the lone pair occupies an equatorial position, resulting in a see-saw molecular shape. - \( \text{CF}_4 \): Carbon has 4 valence electrons, forms 4 bonds, and has 0 lone pairs. Shape is tetrahedral. - \( \text{XeF}_4 \): Xenon has 8 valence electrons, forms 4 bonds, and has 2 lone pairs. Shape is square planar. - \( \text{PF}_4^+ \): Phosphorus normally has 5 valence electrons, but \( \text{P}^+ \) has 4. It forms 4 bonds with 0 lone pairs. Shape is tetrahedral.
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 12 · 選擇題
1 分
What is the systematic IUPAC name of the following alkane?
To name the compound, we must first find the longest continuous carbon chain containing the maximum number of carbons: \( \text{CH}_3-\text{CH}(\text{C}_2\text{H}_5)-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \) Expanding the ethyl group (\( -\text{C}_2\text{H}_5 \)) as \( -\text{CH}_2-\text{CH}_3 \): \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \) The longest continuous carbon chain has 6 carbon atoms, so the parent alkane is hexane.
Next, number the carbon chain from the end that gives the lowest possible locants to the substituents (methyl groups): - Numbering from the right-hand side: - C1: \( \text{CH}_3 \) - C2: \( \text{CH}(\text{CH}_3) \) (methyl at position 2) - C3: \( \text{CH}_2 \) - C4: \( \text{CH}(\text{CH}_3) \) (methyl at position 4) - C5: \( \text{CH}_2 \) - C6: \( \text{CH}_3 \) This gives substituents at positions 2 and 4. - Numbering from the left-hand side would give substituents at positions 3 and 5. Since 2,4 is lower than 3,5, the correct IUPAC name is 2,4-dimethylhexane.
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 13 · 選擇題
1 分
Which of the following alkenes can exhibit stereoisomerism (E/Z isomerism)?
A.2-methylbut-2-ene
B.3-methylpent-2-ene
C.2,3-dimethylbut-2-ene
D.2-methylpent-1-ene
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解題
For an alkene to exhibit stereoisomerism (E/Z isomerism), each carbon atom of the double bond must be attached to two different groups: - 2-methylbut-2-ene: \( (\text{CH}_3)_2\text{C}=\text{CHCH}_3 \). The left-hand carbon of the double bond is attached to two identical methyl groups, so it cannot show E/Z isomerism. - 3-methylpent-2-ene: \( \text{CH}_3\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3 \). - Left-hand carbon is bonded to \( -\text{H} \) and \( -\text{CH}_3 \) (two different groups). - Right-hand carbon is bonded to \( -\text{CH}_3 \) and \( -\text{CH}_2\text{CH}_3 \) (two different groups). Therefore, it exhibits E/Z isomerism. - 2,3-dimethylbut-2-ene: \( (\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2 \). Both carbons of the double bond are attached to identical methyl groups, so it cannot show E/Z isomerism. - 2-methylpent-1-ene: \( \text{H}_2\text{C}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3 \). The terminal carbon has two identical hydrogen atoms, so it cannot show E/Z isomerism.
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 14 · 選擇題
1 分
What is the percentage atom economy by mass for the production of ethanol in the fermentation of glucose?
[Molar masses: \( \text{C}_6\text{H}_{12}\text{O}_6 = 180.0\text{ g mol}^{-1} \); \( \text{C}_2\text{H}_5\text{OH} = 46.0\text{ g mol}^{-1} \); \( \text{CO}_2 = 44.0\text{ g mol}^{-1} \)]
A.25.6%
B.51.1%
C.64.4%
D.100%
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解題
The formula for percentage atom economy is: \( \text{Percentage Atom Economy} = \frac{\text{Molar mass of desired product} \times \text{stoichiometric coefficient}}{\text{Total molar mass of all reactants}} \times 100\% \)
In the fermentation of glucose: \( \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2 \)
- Desired product: Ethanol (\( \text{C}_2\text{H}_5\text{OH} \)) - Total molar mass of desired product = \( 2 \times 46.0 = 92.0\text{ g mol}^{-1} \) - Total mass of reactant (Glucose, \( \text{C}_6\text{H}_{12}\text{O}_6 \)) = \( 180.0\text{ g mol}^{-1} \)
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 15 · 選擇題
1 分
A sample of bromine contains two isotopes, \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \), in equal abundance. What is the ratio of the heights of the peaks at \( m/z \) values of 158, 160, and 162 in the mass spectrum of molecular bromine, \( \text{Br}_2 \)?
A.\( 1 : 1 : 1 \)
B.\( 1 : 2 : 1 \)
C.\( 3 : 1 \)
D.\( 9 : 6 : 1 \)
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解題
In a sample of molecular bromine, \( \text{Br}_2 \), the molecules are formed from combinations of two bromine isotopes: \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \), which are in a 1:1 ratio (fraction of \( 0.5 \) each).
The possible combinations and their corresponding mass-to-charge (\( m/z \)) values are: 1. \( ^{79}\text{Br}^{79}\text{Br} \) with \( m/z = 79 + 79 = 158 \) - Probability = \( 0.5 \times 0.5 = 0.25 \) 2. \( ^{79}\text{Br}^{81}\text{Br} \) and \( ^{81}\text{Br}^{79}\text{Br} \) with \( m/z = 79 + 81 = 160 \) - Probability = \( (0.5 \times 0.5) + (0.5 \times 0.5) = 0.50 \) 3. \( ^{81}\text{Br}^{81}\text{Br} \) with \( m/z = 81 + 81 = 162 \) - Probability = \( 0.5 \times 0.5 = 0.25 \)
Therefore, the ratio of the heights of the peaks at \( m/z \) of 158 : 160 : 162 is \( 0.25 : 0.50 : 0.25 \), which simplifies to \( 1 : 2 : 1 \).
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 16 · 選擇題
1 分
Which of the following molecules has the largest permanent dipole moment?
A.\( \text{CF}_4 \)
B.\( \text{BF}_3 \)
C.\( \text{NF}_3 \)
D.\( \text{BeF}_2 \)
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解題
- \( \text{CF}_4 \) has a symmetrical tetrahedral geometry. Although each \( \text{C}-\text{F} \) bond is highly polar due to the large electronegativity difference, the dipoles pull in opposite directions and cancel out completely, giving a net dipole moment of 0. - \( \text{BF}_3 \) has a symmetrical trigonal planar geometry, so the individual \( \text{B}-\text{F} \) bond dipoles cancel out completely, resulting in a net dipole moment of 0. - \( \text{BeF}_2 \) has a symmetrical linear geometry, so the two polar \( \text{Be}-\text{F} \) bonds point in opposite directions and cancel out, resulting in a net dipole moment of 0. - \( \text{NF}_3 \) has a trigonal pyramidal molecular geometry due to the presence of a lone pair on the nitrogen atom. Because of this non-symmetrical shape, the individual \( \text{N}-\text{F} \) bond dipoles do not cancel out, resulting in a permanent net dipole moment.
評分準則
1 mark: Correct option chosen. 0 marks: Incorrect option chosen or left blank.
題目 17 · 選擇題
1 分
When \(3.40\text{ g}\) of sodium nitrate (\(M_r = 85.0\)) is heated, it decomposes completely according to the equation:
What is the volume of oxygen gas produced, in \(dm^3\), measured at room temperature and pressure (rtp)? (Molar volume of a gas at rtp = \(24.0\text{ dm}^3\text{ mol}^{-1}\))
A.0.240
B.0.480
C.0.960
D.9.60
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解題
First, calculate the amount of substance in moles of \(\text{NaNO}_3\): \[\text{moles of NaNO}_3 = \frac{3.40\text{ g}}{85.0\text{ g mol}^{-1}} = 0.0400\text{ mol}\]
Next, use the stoichiometric ratio from the balanced equation. Two moles of \(\text{NaNO}_3\) produce one mole of \(\text{O}_2\): \[\text{moles of O}_2 = \frac{0.0400\text{ mol}}{2} = 0.0200\text{ mol}\]
Finally, calculate the volume of oxygen gas at rtp: \[\text{Volume} = 0.0200\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.480\text{ dm}^3\]
評分準則
1 mark for the correct option (B).
Incorrect pathways: - Option A (0.240) incorrectly misses the 2:1 stoichiometry or applies a 4:1 ratio. - Option C (0.960) fails to divide the moles of nitrate by 2. - Option D (9.60) is a decimal error.
題目 18 · 選擇題
1 分
The first five successive ionization energies of an element, \(\text{X}\), in \(\text{kJ mol}^{-1}\), are shown below:
In which group of the Periodic Table is element \(\text{X}\)?
A.Group 1
B.Group 2
C.Group 3
D.Group 4
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解題
We must look for the largest ratio increase between successive ionization energies: - From 1st to 2nd: \(\frac{1817}{578} \approx 3.14\) - From 2nd to 3rd: \(\frac{2745}{1817} \approx 1.51\) - From 3rd to 4th: \(\frac{11578}{2745} \approx 4.22\) - From 4th to 5th: \(\frac{14831}{11578} \approx 1.28\)
The largest relative jump occurs between the 3rd and 4th ionization energies. This indicates that the fourth electron is being removed from an inner electron shell, which is closer to the nucleus and experiences significantly less shielding. Therefore, there are three valence electrons in the outer shell of element \(\text{X}\), placing it in Group 3.
評分準則
1 mark for the correct option (C).
題目 19 · 選擇題
1 分
Which of the following species has the smallest bond angle?
A.\(\text{NH}_4^+\)
B.\(\text{NH}_3\)
C.\(\text{NH}_2^-\)
D.\(\text{CH}_4\)
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解題
All four species are based on a tetrahedral arrangement of four electron pairs around the central atom: - \(\text{CH}_4\) and \(\text{NH}_4^+\) both have 4 bonding pairs and 0 lone pairs around the central atom, resulting in a regular tetrahedral shape with a bond angle of \(109.5^\circ\). - \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair. The stronger repulsion from the lone pair contracts the bond angle to approximately \(107^\circ\). - \(\text{NH}_2^-\) has 2 bonding pairs and 2 lone pairs on the nitrogen atom. The intense repulsion between the two lone pairs reduces the bond angle even further to approximately \(104.5^\circ\).
Therefore, \(\text{NH}_2^-\) has the smallest bond angle.
評分準則
1 mark for the correct option (C).
題目 20 · 選擇題
1 分
Which of the following alkenes can exhibit geometric (\(E\)/\(Z\)) isomerism?
A.2-methylbut-2-ene
B.but-1-ene
C.3-methylpent-2-ene
D.2,3-dimethylbut-2-ene
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解題
For an alkene to exhibit geometric (\(E\)/\(Z\)) isomerism, each carbon atom of the \(\text{C}=\text{C}\) double bond must be bonded to two different groups.
- In 2-methylbut-2-ene, \(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\), carbon-2 is bonded to two identical methyl groups, so it cannot show isomerism. - In but-1-ene, \(\text{CH}_2=\text{CHCH}_2\text{CH}_3\), carbon-1 is bonded to two identical hydrogen atoms, so it cannot show isomerism. - In 3-methylpent-2-ene, \(\text{CH}_3\text{CH}=\text{C(CH}_3)\text{CH}_2\text{CH}_3\), carbon-2 is bonded to \(\text{-H}\) and \(\text{-CH}_3\) (different), while carbon-3 is bonded to \(\text{-CH}_3\) and \(\text{-CH}_2\text{CH}_3\) (different). Thus, it exists as \(E\) and \(Z\) stereoisomers. - In 2,3-dimethylbut-2-ene, \(\text{(CH}_3)_2\text{C}=\text{C(CH}_3)_2\), both carbons are bonded to identical pairs of methyl groups, so it cannot show isomerism.
評分準則
1 mark for the correct option (C).
Unit 1: 乙部
Answer all questions in the space provided. Show clear mathematical workings.
4 題目 · 60 分
題目 1 · Structured
15 分
A student carries out investigations on a hydrated basic copper(II) carbonate, \(\text{CuCO}_3 \cdot \text{Cu(OH)}_2 \cdot x\text{H}_2\text{O}\).
(a) Write the balanced chemical equation, including state symbols, for the complete thermal decomposition of anhydrous basic copper(II) carbonate, \(\text{Cu}_2\text{CO}_3(\text{OH})_2\), to form copper(II) oxide, carbon dioxide, and steam. (2)
(b) A \(5.50\text{ g}\) sample of the hydrated salt \(\text{CuCO}_3 \cdot \text{Cu(OH)}_2 \cdot x\text{H}_2\text{O}\) was heated strongly to constant mass in a crucible. The only solid residue remaining was copper(II) oxide, \(\text{CuO}\), which had a mass of \(3.18\text{ g}\).
(i) Calculate the number of moles of \(\text{CuO}\) formed. (2)
(ii) Deduce the number of moles of the anhydrous salt, \(\text{CuCO}_3 \cdot \text{Cu(OH)}_2\), that decomposed to form this mass of \(\text{CuO}\). (2)
(iii) Calculate the relative formula mass (\(M_r\)) of the hydrated salt. (2)
(iv) Determine the value of \(x\) in the formula \(\text{CuCO}_3 \cdot \text{Cu(OH)}_2 \cdot x\text{H}_2\text{O}\), showing your working. (3)
(c) In a separate experiment, a \(0.247\text{ g}\) sample of pure anhydrous copper(II) carbonate, \(\text{CuCO}_3\), was reacted with excess dilute hydrochloric acid.
(i) Write the ionic equation, with state symbols, for this reaction. (2)
(ii) Calculate the volume, in \(\text{cm}^3\), of carbon dioxide gas produced at room temperature and pressure (r.t.p.). [Molar volume of gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\)] (2)
(b)(i) \(M_r(\text{CuO}) = 63.5 + 16.0 = 79.5\text{ g mol}^{-1}\). \(\text{Moles of CuO} = \frac{3.18}{79.5} = 0.0400\text{ mol}\).
(b)(ii) From the decomposition stoichiometry, \(1\text{ mol}\) of \(\text{CuCO}_3 \cdot \text{Cu(OH)}_2\) contains \(2\text{ mol}\) of copper atoms and thus produces \(2\text{ mol}\) of \(\text{CuO}\). Therefore, \(\text{moles of anhydrous salt} = \frac{0.0400}{2} = 0.0200\text{ mol}\).
(c)(ii) \(M_r(\text{CuCO}_3) = 63.5 + 12.0 + 48.0 = 123.5\text{ g mol}^{-1}\). \(\text{Moles of CuCO}_3 = \frac{0.247}{123.5} = 0.00200\text{ mol}\). Volume of \(\text{CO}_2 = 0.00200 \times 24000 = 48.0\text{ cm}^3\).
評分準則
(a) [2 marks] - 1 mark for correct formulas and balancing of all species. - 1 mark for correct state symbols. Reject if water is written as H2O(l) or H2O(aq).
(b)(i) [2 marks] - 1 mark for calculating correct Mr of CuO (79.5). - 1 mark for 0.0400 mol.
(b)(ii) [2 marks] - 1 mark for recognizing the 1:2 molar ratio of basic copper carbonate to CuO. - 1 mark for 0.0200 mol.
(b)(iii) [2 marks] - 1 mark for the expression: mass / moles. - 1 mark for 275 (g mol^-1).
(b)(iv) [3 marks] - 1 mark for calculating Mr of the anhydrous salt as 221. - 1 mark for finding mass of water per mole = 54 g. - 1 mark for x = 3.
(c)(i) [2 marks] - 1 mark for correct species and balancing. - 1 mark for correct state symbols (CuCO3 must be solid).
(c)(ii) [2 marks] - 1 mark for moles of CuCO3 = 0.00200 mol. - 1 mark for volume of CO2 = 48.0 cm3.
題目 2 · Structured
15 分
This question is about ionization energies and isotopes of Period 3 and Group 2 elements.
(a) Define the term first ionization energy. (3)
(b) Write an equation, with state symbols, representing the process that occurs when the second ionization energy of magnesium is measured. (2)
(c) The first ionization energies of the elements in Period 3 show a general increase from sodium to argon.
(i) Explain why there is a general increase in first ionization energy across Period 3. (3)
(ii) Explain why the first ionization energy of aluminium is lower than that of magnesium. (3)
(iii) Explain why the first ionization energy of sulfur is lower than that of phosphorus. (2)
(d) A sample of magnesium contains three isotopes: \(^{24}\text{Mg}\), \(^{25}\text{Mg}\), and \(^{26}\text{Mg}\). The relative abundance of \(^{24}\text{Mg}\) is \(78.99\%\). The relative atomic mass of this magnesium sample is \(24.31\). Calculate the percentage abundances of \(^{25}\text{Mg}\) and \(^{26}\text{Mg}\), giving your answers to two decimal places. (2)
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解題
(a) First ionization energy is the energy required to remove one mole of electrons (1) from one mole of gaseous atoms (1) to form one mole of gaseous 1+ ions (1).
(b) \(\text{Mg}^+(\text{g}) \rightarrow \text{Mg}^{2+}(\text{g}) + \text{e}^-\) (1 mark for species and balancing, 1 mark for gaseous state symbols for both Mg+ and Mg2+).
(c)(i) Across Period 3, the nuclear charge increases due to an increasing number of protons (1). Shielding remains approximately constant because the outer electrons are added to the same shell (1). Therefore, the electrostatic attraction between the nucleus and the outer electrons increases, making them harder to remove (1).
(c)(ii) The outer electron in magnesium is in a 3s orbital, whereas the outer electron in aluminium is in a 3p orbital (1). The 3p orbital is at a higher energy level than the 3s orbital (1). The 3p electron is also shielded by the 3s electrons, so less energy is needed to remove it (1).
(c)(iii) In phosphorus, the 3p sub-shell has three singly-occupied orbitals (\(3p_x^1 3p_y^1 3p_z^1\)), whereas in sulfur, one of the 3p orbitals contains a pair of electrons (\(3p_x^2 3p_y^1 3p_z^1\)) (1). The repulsion between the two paired electrons in the same 3p orbital of sulfur makes it easier to remove one of them (1).
(d) Let abundance of \(^{25}\text{Mg}\) be \(y\%\). Then abundance of \(^{26}\text{Mg}\) is \((21.01 - y)\%\). \(24.31 = \frac{24 \times 78.99 + 25y + 26(21.01 - y)}{100}\) \(2431 = 1895.76 + 25y + 546.26 - 26y\) \(2431 = 2442.02 - y\) \(y = 11.02\%\) \(^{25}\text{Mg} = 11.02\%\), \(^{26}\text{Mg} = 9.99\%\).
評分準則
(a) [3 marks] - 1 mark for energy required to remove 1 mole of electrons. - 1 mark for from 1 mole of gaseous atoms. - 1 mark for to form 1 mole of gaseous 1+ ions.
(b) [2 marks] - 1 mark for correct species and balancing. - 1 mark for gaseous state symbols on Mg+(g) and Mg2+(g).
(c)(i) [3 marks] - 1 mark for increased nuclear charge / number of protons. - 1 mark for shielding remaining constant / same shell. - 1 mark for stronger electrostatic attraction between nucleus and outer electrons.
(c)(ii) [3 marks] - 1 mark for identifying that the Mg electron is in 3s and the Al electron is in 3p. - 1 mark for stating that 3p is at a higher energy level. - 1 mark for mentioning shielding of the 3p orbital by the 3s electrons.
(c)(iii) [2 marks] - 1 mark for identifying that sulfur has a paired electron in a 3p orbital. - 1 mark for explaining that spin-pair repulsion makes it easier to remove this electron.
(d) [2 marks] - 1 mark for setting up a correct algebraic equation for the abundances. - 1 mark for calculating both percentages correctly to two decimal places (11.02% and 9.99%).
題目 3 · Structured
15 分
This question is about molecular shapes, bonding, and structures of Period 3 chlorides and oxides.
(a) Silicon tetrachloride (\(\text{SiCl}_4\)) and phosphorus trichloride (\(\text{PCl}_3\)) are covalent molecules.
(i) Draw dot-and-cross diagrams to show the outer-shell electrons in a molecule of \(\text{SiCl}_4\) and a molecule of \(\text{PCl}_3\). (2)
(ii) Use electron-pair repulsion theory (VSEPR) to predict the shape and bond angle of both \(\text{SiCl}_4\) and \(\text{PCl}_3\). Explain your predictions in terms of electron-pair repulsion. (5)
(b) Explain why the \(\text{Si}-\text{Cl}\) bond is polar, but the \(\text{SiCl}_4\) molecule as a whole is non-polar. (2)
(c) Silicon dioxide (\(\text{SiO}_2\)) has a very high melting temperature (1883 K), whereas silicon tetrachloride (\(\text{SiCl}_4\)) is a liquid at room temperature with a low boiling temperature (330 K).
(i) Describe the structure and bonding in silicon dioxide and explain why its melting temperature is so high. (3)
(ii) Describe the bonding and the nature of the forces that must be overcome when silicon tetrachloride boils. (3)
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解題
(a)(i) \(\text{SiCl}_4\): Central Si atom with 4 shared pairs of electrons with 4 Cl atoms, no lone pairs on Si. Each Cl has 3 lone pairs. \(\text{PCl}_3\): Central P atom with 3 shared pairs of electrons with 3 Cl atoms, and 1 lone pair on P. Each Cl has 3 lone pairs. (1 mark for each correct diagram showing all outer-shell electrons).
(a)(ii) \(\text{SiCl}_4\): Shape is tetrahedral, bond angle is \(109.5^\circ\) (1). Silicon has 4 bonding electron pairs and 0 lone pairs. To minimize repulsion, they arrange themselves as far apart as possible in space (1). \(\text{PCl}_3\): Shape is trigonal pyramidal, bond angle is \(107^\circ\) (1). Phosphorus has 4 electron pairs: 3 bonding pairs and 1 lone pair (1). Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, which decreases the bond angle from the tetrahedral angle (1).
(b) Chlorine is more electronegative than silicon, causing unequal sharing of the covalent electron pair and making the bond polar with a dipole (\(\text{Si}^{\delta+}-\text{Cl}^{\delta-}\)) (1). However, the tetrahedral geometry of \(\text{SiCl}_4\) is highly symmetrical, so the individual dipoles cancel each other out, leaving no net molecular dipole (1).
(c)(i) Silicon dioxide has a giant covalent / macromolecular structure (1). There are many strong covalent bonds between silicon and oxygen atoms throughout the giant 3D lattice (1). A very large amount of thermal energy is required to break these strong bonds (1).
(c)(ii) Silicon tetrachloride has a simple molecular structure with strong covalent bonds within the molecules (1) but only weak intermolecular London forces (instantaneous dipole-induced dipole forces) between the molecules (1). Very little thermal energy is needed to overcome these weak forces when it boils (1).
評分準則
(a)(i) [2 marks] - 1 mark for correct SiCl4 dot-and-cross diagram. - 1 mark for correct PCl3 dot-and-cross diagram.
(a)(ii) [5 marks] - 1 mark for predicting tetrahedral shape and 109.5 degrees for SiCl4. - 1 mark for predicting trigonal pyramidal shape and 107 degrees (accept 100-108 degrees) for PCl3. - 1 mark for stating that electron pairs repel each other to be as far apart as possible to minimize repulsion. - 1 mark for stating the number of bonding and lone pairs for both molecules. - 1 mark for stating that lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion.
(b) [2 marks] - 1 mark for identifying difference in electronegativity between Si and Cl. - 1 mark for explaining that the symmetrical tetrahedral shape causes the dipoles to cancel.
(c)(i) [3 marks] - 1 mark for identifying the giant covalent / macromolecular structure. - 1 mark for stating that many strong covalent bonds exist between atoms. - 1 mark for stating that high thermal energy is needed to break these strong covalent bonds.
(c)(ii) [3 marks] - 1 mark for simple molecular structure / covalent bonds within molecules. - 1 mark for identifying intermolecular London forces (accept van der Waals forces). - 1 mark for stating that little thermal energy is required to overcome these weak intermolecular forces (must not imply covalent bonds break).
題目 4 · Structured
15 分
This question is about alkenes, their structures, and addition reactions.
(a) But-1-ene is an unsaturated hydrocarbon.
(i) Explain what is meant by the term unsaturated hydrocarbon. (2)
(ii) Draw the skeletal formula of but-1-ene and explain why it does not exhibit geometric (E/Z) isomerism. (2)
(b) But-1-ene reacts with hydrogen bromide, \(\text{HBr}\), to form a mixture of two halogenoalkane isomers, with one being the major product and the other the minor product.
(i) Draw the mechanism for the reaction of but-1-ene with \(\text{HBr}\) to form the major product. Include curly arrows, relevant dipoles, and any lone pairs or charges. (4)
(ii) Identify the major and minor products by name. Explain, in terms of the stability of the carbocation intermediates, why the major product is formed. (3)
(c) In an industrial process, but-2-ene is reacted with steam in the presence of an acid catalyst to produce butan-2-ol:
(i) Calculate the percentage atom economy for this reaction. (1)
(ii) Explain why this reaction has the calculated atom economy, and suggest why this is highly advantageous for industrial chemical processes. (3)
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解題
(a)(i) Unsaturated: contains at least one carbon-carbon double (or triple) bond (1). Hydrocarbon: contains only carbon and hydrogen atoms (1).
(a)(ii) Skeletal formula of but-1-ene: a zigzag line of four carbon atoms with a double bond between C1 and C2 (\(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\)) (1). It does not show geometric isomerism because one of the double-bonded carbons (C1) is attached to two identical groups (hydrogen atoms) (1).
(b)(i) Mechanism steps: 1. Curly arrow from the C=C double bond of but-1-ene to the \(\text{H}^{\delta+}\) of \(\text{H}-\text{Br}\) (1). 2. Curly arrow from the H-Br bond to the \(\text{Br}^{\delta-}\) atom (1). 3. Correct structure of the secondary carbocation intermediate, \(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\), with a positive charge on C2 (1). 4. Curly arrow from a lone pair on the bromide ion (\(\text{Br}^-\)) to the carbocation carbon, forming 2-bromobutane (1).
(b)(ii) Major product is 2-bromobutane, minor product is 1-bromobutane (1). The reaction goes via a secondary carbocation intermediate (to form 2-bromobutane) and a primary carbocation (to form 1-bromobutane) (1). The secondary carbocation is more stable than the primary carbocation because of the electron-donating inductive effect of two alkyl groups compared to one (1).
(c)(i) Atom economy = 100% (1).
(c)(ii) The atom economy is 100% because there is only one product formed / all reactant atoms are incorporated into the desired product (1). High atom economy is advantageous because it minimizes the formation of waste products (1), which reduces disposal/treatment costs and conserves raw materials (1).
評分準則
(a)(i) [2 marks] - 1 mark for double C=C bond. - 1 mark for carbon and hydrogen only.
(a)(ii) [2 marks] - 1 mark for correct skeletal formula of but-1-ene. - 1 mark for explaining that C1 has two identical groups (H atoms) attached to it.
(b)(i) [4 marks] - 1 mark for curly arrow from double bond to H of H-Br with correct dipoles on H-Br. - 1 mark for arrow from H-Br bond to Br. - 1 mark for correct structure of secondary carbocation. - 1 mark for arrow from lone pair of Br- to the carbocation.
(b)(ii) [3 marks] - 1 mark for identifying 2-bromobutane as major and 1-bromobutane as minor. - 1 mark for comparing secondary and primary carbocation stability. - 1 mark for explaining stability due to the inductive effect of alkyl groups.
(c)(i) [1 mark] - 1 mark for 100%.
(c)(ii) [3 marks] - 1 mark for stating only one product is formed / no waste products. - 1 mark for saving raw materials / reducing production costs. - 1 mark for environmental benefits / less waste to treat.
Unit 2: 甲部
Answer all multiple choice questions.
20 題目 · 20 分
題目 1 · multiple_choice
1 分
Which of the following compounds has the highest boiling temperature?
Propan-1-ol (A) is capable of forming intermolecular hydrogen bonds, which are much stronger than the permanent dipole-dipole forces in methoxyethane (B) and 1-fluoropropane (C), or the London forces in butane (D). Consequently, more energy is required to overcome these forces, giving propan-1-ol the highest boiling temperature.
評分準則
1 mark: Correct answer A chosen.
題目 2 · multiple_choice
1 分
Which of the following Group 2 nitrates requires the highest temperature to undergo thermal decomposition?
A.\(\text{Mg(NO}_3)_2\)
B.\(\text{Ca(NO}_3)_2\)
C.\(\text{Sr(NO}_3)_2\)
D.\(\text{Ba(NO}_3)_2\)
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解題
Thermal stability of Group 2 nitrates increases down the group. As the cation size increases from magnesium to barium, its charge density decreases. Therefore, the larger barium cation polarises the nitrate anion to a lesser extent, making the nitrate compound more stable and requiring a higher temperature to decompose.
評分準則
1 mark: Correct answer D chosen.
題目 3 · multiple_choice
1 分
Use the following standard enthalpy changes of combustion (\(\Delta_c H^\ominus\)) to calculate the standard enthalpy change of formation of propanone, \(\text{CH}_3\text{COCH}_3\text{(l)}\). \(\Delta_c H^\ominus [\text{C(s)}] = -394\text{ kJ mol}^{-1}\), \(\Delta_c H^\ominus [\text{H}_2\text{(g)}] = -286\text{ kJ mol}^{-1}\), \(\Delta_c H^\ominus [\text{CH}_3\text{COCH}_3\text{(l)}] = -1786\text{ kJ mol}^{-1}\).
A.\(-254\text{ kJ mol}^{-1}\)
B.\(+254\text{ kJ mol}^{-1}\)
C.\(-1106\text{ kJ mol}^{-1}\)
D.\(+1106\text{ kJ mol}^{-1}\)
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解題
The equation for the formation of propanone is: \(3\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \to \text{CH}_3\text{COCH}_3\text{(l)}\). Using Hess's Law: \(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products}) = [3 \times (-394) + 3 \times (-286)] - [-1786] = [-1182 - 858] + 1786 = -2040 + 1786 = -254\text{ kJ mol}^{-1}\).
評分準則
1 mark: Correct answer A chosen.
題目 4 · multiple_choice
1 分
Equal amounts of 1-chlorobutane, 1-bromobutane and 1-iodobutane are separately warmed with aqueous silver nitrate in ethanol at 50 degrees Celsius. Which of the following statements is correct?
A.1-chlorobutane reacts the fastest because the C-Cl bond is the most polar.
B.1-iodobutane reacts the fastest because the C-I bond is the weakest.
C.1-chlorobutane reacts the fastest because the C-Cl bond has the highest activation energy.
D.1-iodobutane reacts the slowest because the iodine atom is a poor leaving group.
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解題
The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. Since the \(\text{C-I}\) bond is the weakest (lowest bond enthalpy), it is broken most easily, resulting in 1-iodobutane reacting the fastest.
評分準則
1 mark: Correct answer B chosen.
題目 5 · multiple_choice
1 分
What is the effect of adding a catalyst on the Maxwell-Boltzmann distribution curve of molecular energies and the activation energy (\(E_a\)) for a chemical reaction?
A.The peak of the distribution curve shifts to the right, and the value of \(E_a\) is unchanged.
B.The peak of the distribution curve is unchanged, and the value of \(E_a\) decreases.
C.The Maxwell-Boltzmann distribution curve is unchanged, and an alternative reaction pathway with a lower activation energy is provided.
D.The area under the distribution curve increases, and the value of \(E_a\) decreases.
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解題
A catalyst does not affect the temperature or the total number of molecules, so the Maxwell-Boltzmann distribution curve remains completely unchanged. Instead, the catalyst provides an alternative reaction pathway with a lower activation energy.
評分準則
1 mark: Correct answer C chosen.
題目 6 · multiple_choice
1 分
Which of the following statements about Group 2 compounds is correct?
A.Magnesium hydroxide and magnesium sulfate are both more soluble in water than their barium counterparts.
B.Barium hydroxide is more soluble in water than magnesium hydroxide, and magnesium sulfate is less soluble than barium sulfate.
C.Magnesium hydroxide is less soluble in water than barium hydroxide, and magnesium sulfate is more soluble than barium sulfate.
D.Barium hydroxide and barium sulfate are both more soluble in water than their magnesium counterparts.
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解題
The solubility of Group 2 hydroxides increases down the group, so magnesium hydroxide is less soluble than barium hydroxide. The solubility of Group 2 sulfates decreases down the group, so magnesium sulfate is more soluble than barium sulfate.
評分準則
1 mark: Correct answer C chosen.
題目 7 · multiple_choice
1 分
An organic compound with the molecular formula \(\text{C}_3\text{H}_6\text{O}\) is analysed. Its infrared spectrum shows a strong, sharp absorption peak at \(1715\text{ cm}^{-1}\), but no broad absorption above \(3000\text{ cm}^{-1}\). Which of the following compounds is consistent with this spectrum?
A.Propan-2-ol
B.Propanone
C.Prop-2-en-1-ol
D.Propanoic acid
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解題
The molecular formula \(\text{C}_3\text{H}_6\text{O}\) indicates one double bond equivalent. The strong, sharp peak at \(1715\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C=O}\)). The absence of a broad absorption above \(3000\text{ cm}^{-1}\) rules out any alcohol (\(\text{O-H}\) group). Propanone (B) fits this description perfectly, as it is a ketone with no \(\text{O-H}\) bond.
評分準則
1 mark: Correct answer B chosen.
題目 8 · multiple_choice
1 分
Chlorine reacts with cold, dilute aqueous sodium hydroxide to form a mixture containing sodium chloride and another chlorine-containing product, X. What is the oxidation state of chlorine in X?
A.-1
B.+1
C.+3
D.+5
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解題
The reaction of chlorine with cold, dilute sodium hydroxide is a disproportionation reaction: \(\text{Cl}_2 + 2\text{NaOH} \to \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). Product X is sodium chlorate(I), \(\text{NaClO}\). In the chlorate(I) ion (\(\text{ClO}^-\)), oxygen has an oxidation state of -2, so chlorine must have an oxidation state of +1 to give a net charge of -1.
評分準則
1 mark: Correct answer B chosen.
題目 9 · 選擇題
1 分
Which of the following lists the hydrogen halides in order of increasing boiling temperature?
Hydrogen fluoride (HF) has the highest boiling temperature because of strong hydrogen bonding between its molecules. For the other hydrogen halides (HCl, HBr, HI), the boiling temperature increases as the number of electrons increases because this leads to stronger London forces. Therefore, the correct order is HCl < HBr < HI < HF.
評分準則
1 mark: Correct option chosen.
題目 10 · 選擇題
1 分
Which of the following statements explains why barium carbonate decomposes at a higher temperature than magnesium carbonate?
A.The barium ion has a larger ionic radius than the magnesium ion and polarises the carbonate ion less.
B.The barium ion has a smaller ionic radius than the magnesium ion and polarises the carbonate ion more.
C.Barium has more shielding, making the carbon to oxygen bond in the carbonate ion stronger.
D.Barium carbonate contains stronger covalent bonds than magnesium carbonate.
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解題
As you descend Group 2, the ionic radius of the metal cation increases while the 2+ charge remains constant. This means the charge density decreases, so the larger barium ion polarises (distorts) the electron cloud of the carbonate ion less than the smaller magnesium ion. This makes the carbonate ion more stable to thermal decomposition, requiring a higher temperature.
評分準則
1 mark: Correct option chosen.
題目 11 · 選擇題
1 分
Equal amounts of 1-chlorobutane, 1-bromobutane, and 1-iodobutane are separately warmed with aqueous silver nitrate in ethanol. In which order do the precipitates of silver halide appear, from fastest to slowest?
A.1-iodobutane > 1-bromobutane > 1-chlorobutane
B.1-chlorobutane > 1-bromobutane > 1-iodobutane
C.1-bromobutane > 1-iodobutane > 1-chlorobutane
D.1-iodobutane > 1-chlorobutane > 1-bromobutane
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解題
The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond. The C-I bond has the lowest bond enthalpy and is the weakest, so it breaks most easily, reacting fastest. The C-Cl bond is the strongest and reacts slowest.
評分準則
1 mark: Correct option chosen.
題目 12 · 選擇題
1 分
How does the addition of a heterogeneous catalyst affect the Maxwell-Boltzmann distribution curve of molecular energies and the activation energy of a gas-phase reaction?
A.The distribution curve is unchanged, and the activation energy decreases.
B.The peak of the distribution curve shifts to the right, and the activation energy decreases.
C.The peak of the distribution curve shifts to the left, and the activation energy decreases.
D.The distribution curve is unchanged, and the activation energy is unchanged.
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解題
A catalyst provides an alternative reaction pathway with a lower activation energy, effectively decreasing the activation energy. Since the temperature remains constant, the Maxwell-Boltzmann distribution of molecular kinetic energies does not change.
評分準則
1 mark: Correct option chosen.
題目 13 · 選擇題
1 分
Which chlorine-containing species are formed when chlorine gas reacts with cold, dilute aqueous sodium hydroxide?
A.\( \text{NaCl} \text{ and } \text{NaClO} \)
B.\( \text{NaCl} \text{ and } \text{NaClO}_3 \)
C.\( \text{NaClO} \text{ and } \text{NaClO}_3 \)
D.\( \text{NaCl} \text{, } \text{NaClO} \text{, and } \text{NaClO}_3 \)
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解題
Chlorine undergoes disproportionation in cold, dilute sodium hydroxide according to the equation: \( \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \). Therefore, the chlorine-containing products are sodium chloride and sodium chlorate(I).
評分準則
1 mark: Correct option chosen.
題目 14 · 選擇題
1 分
An organic compound is formed by the oxidation of propan-2-ol with acidified potassium dichromate(VI). What are the main features of the infrared spectrum of the purified organic product?
A.A strong peak at \( 1700\text{--}1725\text{ cm}^{-1} \) and no broad peak at \( 3200\text{--}3750\text{ cm}^{-1} \)
B.No peak at \( 1700\text{--}1725\text{ cm}^{-1} \) and a broad peak at \( 3200\text{--}3750\text{ cm}^{-1} \)
C.A strong peak at \( 1700\text{--}1725\text{ cm}^{-1} \) and a broad peak at \( 2500\text{--}3300\text{ cm}^{-1} \)
D.A strong peak at \( 1700\text{--}1725\text{ cm}^{-1} \) and a broad peak at \( 3200\text{--}3750\text{ cm}^{-1} \)
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解題
Propan-2-ol is a secondary alcohol. Oxidation of a secondary alcohol yields a ketone (propanone). Propanone contains a carbonyl group (\( \text{C}=\text{O} \)), which gives a strong peak at \( 1700\text{--}1725\text{ cm}^{-1} \). Because it contains no hydroxyl group (\( \text{-OH} \)), there is no peak in the alcohol range \( 3200\text{--}3750\text{ cm}^{-1} \).
評分準則
1 mark: Correct option chosen.
題目 15 · 選擇題
1 分
Consider the following mean bond enthalpies: C-H is \( +413\text{ kJ mol}^{-1} \), Cl-Cl is \( +242\text{ kJ mol}^{-1} \), C-Cl is \( +346\text{ kJ mol}^{-1} \), and H-Cl is \( +432\text{ kJ mol}^{-1} \). What is the enthalpy change for the reaction: \( \text{CH}_4(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g) \)?
A.\( -123\text{ kJ mol}^{-1} \)
B.\( +123\text{ kJ mol}^{-1} \)
C.\( -321\text{ kJ mol}^{-1} \)
D.\( +321\text{ kJ mol}^{-1} \)
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解題
Energy required to break bonds = \( 1 \times (\text{C-H}) + 1 \times (\text{Cl-Cl}) = 413 + 242 = +655\text{ kJ mol}^{-1} \). Energy released in forming bonds = \( 1 \times (\text{C-Cl}) + 1 \times (\text{H-Cl}) = 346 + 432 = +778\text{ kJ mol}^{-1} \). Enthalpy change = energy broken - energy formed = \( 655 - 778 = -123\text{ kJ mol}^{-1} \).
評分準則
1 mark: Correct option chosen.
題目 16 · 選擇題
1 分
For the reversible reaction: \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \) where the forward reaction is exothermic, which of the following changes will increase both the value of the equilibrium constant, \( K_c \), and the equilibrium yield of \( \text{SO}_3 \)?
A.Decreasing the temperature
B.Increasing the temperature
C.Increasing the pressure at constant temperature
D.Adding a catalyst
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解題
The value of the equilibrium constant \( K_c \) is only changed by temperature. For an exothermic forward reaction, decreasing the temperature shifts the equilibrium to the right, which increases the concentration of the products relative to reactants, thus increasing both \( K_c \) and the equilibrium yield of \( \text{SO}_3 \).
評分準則
1 mark: Correct option chosen.
題目 17 · 選擇題
1 分
Which of the following organic compounds has the highest boiling temperature?
A.Butan-1-ol
B.Butanal
C.Butane
D.2-Methylpropan-2-ol
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解題
Butan-1-ol and 2-methylpropan-2-ol can both form hydrogen bonds because they contain a hydroxyl (\(-\text{OH}\)) group. Hydrogen bonding is the strongest type of intermolecular force. However, butan-1-ol has a straight-chain structure whereas 2-methylpropan-2-ol is branched. The straight-chain isomer allows for greater surface area contact between adjacent molecules, resulting in stronger London forces. Butanal only experiences permanent dipole-dipole forces and London forces (no hydrogen bonding), and butane only experiences London forces. Therefore, butan-1-ol requires the most energy to overcome its intermolecular forces and has the highest boiling temperature.
評分準則
[1 mark] for selecting A.
Incorrect options: - B: Butanal has a lower boiling temperature because it lacks hydrogen bonding. - C: Butane has a lower boiling temperature because it only has weak London forces. - D: 2-Methylpropan-2-ol has a lower boiling temperature than butan-1-ol due to branching reducing surface contact area and hence weaker London forces.
題目 18 · 選擇題
1 分
When solid potassium iodide reacts with concentrated sulfuric acid, several products are formed. Which of the following is a reduction product of sulfuric acid that is a yellow solid?
A.Iodine
B.Sulfur
C.Sulfur dioxide
D.Hydrogen sulfide
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解題
Iodide ions are strong reducing agents capable of reducing the sulfur in concentrated sulfuric acid (oxidation state +6) to several lower oxidation states. These reduction products include: - Sulfur dioxide, \(\text{SO}_2\) (oxidation state +4, colourless gas) - Sulfur, \(\text{S}\) (oxidation state 0, yellow solid) - Hydrogen sulfide, \(\text{H}_2\text{S}\) (oxidation state -2, colourless gas with a rotten-egg smell)
Iodine, \(\text{I}_2\), is an oxidation product formed when iodide ions (oxidation state -1) are oxidised to 0, rather than a reduction product of sulfuric acid. Therefore, the yellow solid reduction product of sulfuric acid is sulfur.
評分準則
[1 mark] for selecting B.
Incorrect options: - A: Iodine is an oxidation product of the iodide ions, not a reduction product of sulfuric acid. - C: Sulfur dioxide is a colourless gas, not a yellow solid. - D: Hydrogen sulfide is a colourless gas, not a yellow solid.
題目 19 · 選擇題
1 分
Which of the following halogenoalkanes reacts fastest when heated with aqueous silver nitrate in ethanol?
A.1-chlorobutane
B.1-bromobutane
C.2-chloro-2-methylpropane
D.2-bromo-2-methylpropane
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解題
The rate of hydrolysis of halogenoalkanes is determined by both the carbon-halogen bond strength and the mechanism of substitution.
Tertiary halogenoalkanes (such as 2-chloro-2-methylpropane and 2-bromo-2-methylpropane) react much faster than primary halogenoalkanes (such as 1-chlorobutane and 1-bromobutane) because they hydrolyse via a rapid \(S_N1\) mechanism involving a stable tertiary carbocation intermediate.
Between the two tertiary halogenoalkanes, the \(\text{C}-\text{Br}\) bond is weaker (has a lower bond enthalpy) than the \(\text{C}-\text{Cl}\) bond. This allows the \(\text{C}-\text{Br}\) bond to break more easily and quickly. Therefore, 2-bromo-2-methylpropane reacts at the fastest rate.
評分準則
[1 mark] for selecting D.
Incorrect options: - A: 1-chlorobutane is primary and has a stronger \(\text{C}-\text{Cl}\) bond, making it the slowest. - B: 1-bromobutane is primary, reacting much slower than tertiary isomers. - C: 2-chloro-2-methylpropane is tertiary, but has a stronger \(\text{C}-\text{Cl}\) bond than the \(\text{C}-\text{Br}\) bond in 2-bromo-2-methylpropane.
題目 20 · 選擇題
1 分
A reaction is carried out with and without a catalyst at a constant temperature. How do the features of the Maxwell-Boltzmann distribution of molecular energies change when a catalyst is added?
A.The peak of the curve shifts to the right and the activation energy marker shifts to the left.
B.The peak of the curve remains at the same energy, and the activation energy marker shifts to the left.
C.The peak of the curve shifts to the left and the activation energy marker remains unchanged.
D.The peak of the curve remains at the same energy, and the activation energy marker shifts to the right.
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解題
The Maxwell-Boltzmann distribution curve represents the distribution of kinetic energies of the molecules in a sample. This distribution is determined solely by the temperature. Because the temperature remains constant, the shape of the curve does not change, meaning the peak (which represents the most probable energy) remains at exactly the same energy value.
A catalyst works by providing an alternative reaction pathway with a lower activation energy. Consequently, the activation energy marker (\(E_a\)) shifts to a lower value, which is to the left on the energy axis.
評分準則
[1 mark] for selecting B.
Incorrect options: - A: The peak of the curve does not shift because temperature is constant. - C: The peak of the curve does not shift and the activation energy marker must change (shift left). - D: The activation energy decreases, so the marker must shift to the left (lower energy), not the right.
Unit 2: 乙部
Answer all structured and mathematical calculation items.
3 題目 · 39 分
題目 1 · Structured and Mathematical Analysis
13 分
A student carried out an experiment using a simple spirit burner calorimeter to determine the experimental enthalpy change of combustion of propan-1-ol, \(C_3H_7OH\).
The following experimental data were collected: - Mass of water in beaker = \(150.0 \text{ g}\) - Initial temperature of water = \(20.2 ^\circ\text{C}\) - Final temperature of water = \(48.7 ^\circ\text{C}\) - Initial mass of spirit burner + propan-1-ol = \(122.54 \text{ g}\) - Final mass of spirit burner + propan-1-ol = \(121.82 \text{ g}\)
(a) (i) Calculate the heat energy, \(q\), transferred to the water in kJ. The specific heat capacity of water is \(4.18 \text{ J g}^{-1} \text{ K}^{-1}\). (2 marks) (ii) Calculate the amount, in moles, of propan-1-ol burned. (2 marks) (iii) Calculate the experimental enthalpy change of combustion of propan-1-ol, \(\Delta H_c\), in \(\text{kJ mol}^{-1}\), including a sign. Give your answer to 3 significant figures. (2 marks)
(b) State two reasons, other than heat loss to the surroundings, why the experimental value calculated in (a)(iii) is significantly less exothermic than the standard data book value of \(-2021 \text{ kJ mol}^{-1}\). (2 marks)
(c) Use the mean bond enthalpies provided below to calculate the theoretical enthalpy change of combustion of gaseous propan-1-ol, \(C_3H_7OH(g)\). (5 marks)
(ii) Mass of propan-1-ol burned = \(122.54 - 121.82 = 0.72 \text{ g}\). Molar mass of propan-1-ol, \(M_r(C_3H_8O) = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0 \text{ g mol}^{-1}\). Moles burned = \(0.72 / 60.0 = 0.0120 \text{ mol}\).
(iii) \(\Delta H_c = -\frac{q}{\text{moles}} = -\frac{17.8695 \text{ kJ}}{0.0120 \text{ mol}} = -1489.125 \text{ kJ mol}^{-1}\). To 3 significant figures, this is \(-1490 \text{ kJ mol}^{-1}\).
(b) 1. Incomplete combustion of propan-1-ol (forming carbon soot or carbon monoxide instead of carbon dioxide). 2. Evaporation of propan-1-ol from the wick before or after combustion. 3. The heat capacity of the copper calorimeter/beaker is not accounted for (heat is absorbed by the container).
(a) (i) [2 Marks] - M1: Correct calculation of temperature change (\(\Delta T = 28.5 ^\circ\text{C}\)) and substitution: \(q = 150.0 \times 4.18 \times 28.5\) - M2: Correct calculation of energy in kJ: \(17.9 \text{ kJ}\) (or \(17.87 \text{ kJ}\))
(a) (ii) [2 Marks] - M1: Correct mass burned (\(0.72 \text{ g}\)) and molar mass of propan-1-ol (\(60.0 \text{ g mol}^{-1}\)) - M2: Correct moles calculation: \(0.0120 \text{ mol}\) (accept \(0.012\))
(a) (iii) [2 Marks] - M1: Calculation of energy divided by moles: \(17.87 / 0.012\) (or using student's values from (i) and (ii)) - M2: Correct final value of \(-1490 \text{ kJ mol}^{-1}\) (must include the negative sign and be to 3 significant figures)
(b) [2 Marks] - M1: Incomplete combustion (forming CO/C) (1) - M2: Evaporation of the alcohol from the wick / heat absorbed by the beaker/calorimeter (1) [Do not accept 'heat loss to surroundings']
(c) [5 Marks] - M1: Correct calculation of bonds broken in propan-1-ol: \(4406 \text{ kJ mol}^{-1}\) - M2: Correct calculation of bonds broken in oxygen: \(2227.5 \text{ kJ mol}^{-1}\) (Total broken = \(6633.5 \text{ kJ mol}^{-1}\)) - M3: Correct calculation of bonds formed in \(CO_2\): \(4830 \text{ kJ mol}^{-1}\) - M4: Correct calculation of bonds formed in \(H_2O\): \(3704 \text{ kJ mol}^{-1}\) (Total formed = \(8534 \text{ kJ mol}^{-1}\)) - M5: Correct subtraction and negative sign: \(-1901 \text{ kJ mol}^{-1}\) (or \(-1900.5\) / \(-1900\))
題目 2 · Structured and Mathematical Analysis
13 分
An unknown organic compound, \(X\), has the molecular formula \(C_4H_9Br\). Compound \(X\) is a secondary halogenoalkane.
(a) Draw the skeletal formula and state the systematic IUPAC name of compound \(X\). (2 marks)
(b) Compound \(X\) reacts with hot aqueous potassium hydroxide, \(KOH(aq)\), to form alcohol \(Y\) via a nucleophilic substitution reaction. (i) Write the ionic equation for this reaction. (1 mark) (ii) Outline the mechanism for this reaction, showing all necessary dipoles, lone pairs, and curly arrows. (4 marks)
(c) Alcohol \(Y\) is oxidized by heating under reflux with acidified potassium dichromate(VI) to form compound \(Z\). (i) State the color change observed in the reaction mixture during this oxidation. (1 mark) (ii) Draw the structural or displayed formula of compound \(Z\). (1 mark) (iii) Explain how infrared spectroscopy can be used to distinguish between alcohol \(Y\) and compound \(Z\). State the specific bonds and their characteristic wavenumber ranges. (4 marks)
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解題
(a) Compound X is 2-bromobutane. Its skeletal formula is a 4-carbon chain (zigzag) with a single bond to a bromine atom attached to carbon-2.
(ii) S_N2 mechanism pathway: - Show the dipole \(C^{\delta+} - Br^{\delta-}\) where the carbon bonded to Br is partially positive and the bromine is partially negative. - Draw a curly arrow starting from a lone pair on the oxygen of \(OH^-\), pointing directly to the \(C^{\delta+}\). - Draw a curly arrow starting from the C-Br bond, pointing directly to the Br atom. - Draw the structures of the final organic product (butan-2-ol) and the leaving bromide ion, \(Br^-\).
(c) (i) The reaction mixture changes color from orange to green. (ii) Compound Z is butanone (a ketone). Its structural formula is \(CH_3COCH_2CH_3\). (iii) - The infrared spectrum of alcohol Y (butan-2-ol) will show a broad absorption band for the O-H (alcohol) group in the range \(3200 - 3750 \text{ cm}^{-1}\), which will be absent in Z. - The infrared spectrum of compound Z (butanone) will show a strong, sharp absorption band for the C=O (carbonyl) group in the range \(1675 - 1750 \text{ cm}^{-1}\), which will be absent in Y.
評分準則
(a) [2 Marks] - M1: Correct skeletal formula of 2-bromobutane - M2: Correct systematic name: 2-bromobutane
(b) (ii) [4 Marks] - M1: Correct dipoles showing \(\delta+\) on C2 and \(\delta-\) on Br - M2: Curly arrow from the lone pair on the oxygen of \(OH^-\) pointing to C2 - M3: Curly arrow from the C-Br bond to the Br atom - M4: Correct products of the reaction: butan-2-ol and \(Br^-\) [Accept equivalent S_N1 mechanism steps showing carbocation intermediate for secondary halogenoalkanes]
(c) (i) [1 Mark] - M1: Orange to green (both colors needed, in this order)
(c) (ii) [1 Mark] - M1: Correct structural or displayed formula of butanone: \(CH_3COCH_2CH_3\) (or structural representation with C=O visible)
(c) (iii) [4 Marks] - M1: Alcohol Y has an O-H stretch in the range \(3200 - 3750 \text{ cm}^{-1}\) - M2: This O-H band is absent in the spectrum of Z - M3: Ketone Z has a C=O stretch in the range \(1675 - 1750 \text{ cm}^{-1}\) - M4: This C=O band is absent in the spectrum of Y
題目 3 · Structured and Mathematical Analysis
13 分
This question is about some chemical properties and reactions of Group 2 and Group 7 elements.
(a) Group 2 nitrates undergo thermal decomposition when heated. (i) Write a balanced chemical equation for the thermal decomposition of anhydrous magnesium nitrate, \(Mg(NO_3)_2\). Include state symbols. (2 marks) (ii) Describe and explain the trend in the thermal stability of Group 2 nitrates down the group from magnesium to barium. (4 marks)
(b) Chlorine gas reacts with cold, aqueous sodium hydroxide. (i) Write the balanced chemical equation for this reaction. (1 mark) (ii) Name the type of redox reaction that occurs, and explain this in terms of the oxidation numbers of chlorine. (3 marks)
(c) When concentrated sulfuric acid, \(H_2SO_4\), is added to solid potassium iodide, \(KI\), several reduction products of sulfur are formed. Identify two different reduction products of sulfur in this reaction. For each product, state one physical observation (such as color or smell) and the oxidation state of sulfur in that product. (3 marks)
(ii) Trend: Thermal stability increases down the group. Explanation: - Down the group, the ionic radius of the Group 2 cation (\(M^{2+}\)) increases, and the charge density of the cation decreases. - Consequently, the larger cation has less polarizing power and causes less polarization (distortion) of the nitrate electron cloud. - Since the N-O bonds within the nitrate ion are weakened less, more thermal energy is needed to decompose the compound.
(ii) - Reaction type: Disproportionation. - Explanation: Chlorine is simultaneously oxidized and reduced in the same reaction. - The oxidation state of chlorine changes from 0 in \(Cl_2\) to \(-1\) in \(NaCl\) (reduction) and to \(+1\) in \(NaClO\) (oxidation).
(c) Any two of the following sulfur reduction products: - Sulfur dioxide (\(SO_2\)): oxidation state of S is \(+4\); observed as a colorless gas with a pungent, choking smell. - Sulfur (\(S\)): oxidation state of S is \(0\); observed as a yellow solid. - Hydrogen sulfide (\(H_2S\)): oxidation state of S is \(-2\); observed as a colorless gas with a rotten egg smell.
評分準則
(a) (i) [2 Marks] - M1: Correct balanced chemical formulae: \(2Mg(NO_3)_2 \rightarrow 2MgO + 4NO_2 + O_2\) - M2: Correct state symbols: (s) for reactants and MgO, (g) for \(NO_2\) and \(O_2\)
(a) (ii) [4 Marks] - M1: Thermal stability increases down the group (1) - M2: Cation size / ionic radius increases OR charge density of cation decreases down the group (1) - M3: Polarizing power of the cation decreases / less polarization/distortion of the nitrate ion (1) - M4: The covalent N-O bonds in the nitrate ion are weakened less, requiring more heat energy to break (1)
(b) (ii) [3 Marks] - M1: Disproportionation (1) - M2: Oxidation state of chlorine in reactants is 0 (in \(Cl_2\)) and changes to \(-1\) (in \(NaCl\)) and \(+1\) (in \(NaClO\)) (1) - M3: Chlorine is both oxidized and reduced simultaneously (1)
(c) [3 Marks] - M1: Correct identification of first reduction product of sulfur (e.g., \(SO_2\), \(S\), or \(H_2S\)) with correct physical description (pungent gas / yellow solid / bad egg smell) (1) - M2: Correct identification of second, different reduction product of sulfur with correct physical description (1) - M3: Correct oxidation states of sulfur given for both selected products (\(SO_2 = +4\), \(S = 0\), \(H_2S = -2\)) (1)
Unit 2: 部分 C
Structured context-based question focusing on chemical logic and calculations.
1 題目 · 21 分
題目 1 · Extended Contextual Calculation
21 分
An organic compound, **X**, is a liquid halogenoalkane used in synthesis.
**(a)** Chemical analysis shows that **X** has the following percentage composition by mass: * Carbon: \(35.04\%\) * Hydrogen: \(6.57\%\) * Bromine: \(58.39\%\)
(i) Show by calculation that the empirical formula of **X** is \(\text{C}_4\text{H}_9\text{Br}\). [3 marks]
(ii) Compound **X** reacts with warm aqueous sodium hydroxide in a nucleophilic substitution reaction to form an alcohol. Write the ionic equation for this reaction, including state symbols. [2 marks]
(iii) To confirm the presence of bromide ions in the reaction mixture after the substitution is complete, the solution is acidified with dilute nitric acid and aqueous silver nitrate is added, followed by aqueous ammonia. Describe the observations that would confirm the presence of bromide ions. [2 marks]
**(b)** The rate of hydrolysis of halogenoalkanes can be studied by adding aqueous silver nitrate directly to the reaction mixture and timing the appearance of a precipitate.
(i) State and explain the relative rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Refer to carbon-halogen bond strengths in your answer. [3 marks]
(ii) 2-bromo-2-methylpropane is an isomer of **X**. Draw the skeletal formula of the organic product formed when 2-bromo-2-methylpropane is hydrolysed, and state the classification (primary, secondary, or tertiary) of this alcohol. [2 marks]
**(c)** A student decided to determine the enthalpy change of combustion of the alcohol product of the hydrolysis of **X**, which is butan-1-ol (\(\text{C}_4\text{H}_9\text{OH}\)).
The student used a spirit burner containing butan-1-ol to heat water in a copper calorimeter. The following experimental data were obtained: * Mass of water in the copper calorimeter = \(150.0\text{ g}\) * Initial mass of spirit burner + butan-1-ol = \(184.25\text{ g}\) * Final mass of spirit burner + butan-1-ol = \(183.51\text{ g}\) * Initial temperature of water = \(19.5\text{ }^\circ\text{C}\) * Final temperature of water = \(54.8\text{ }^\circ\text{C}\)
[Specific heat capacity of water, \(c = 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1}\); Molar mass of butan-1-ol = \(74.0\text{ g mol}^{-1}\)]
(i) Calculate the heat energy, \(q\), in \(\text{kJ}\), transferred to the water. [2 marks]
(ii) Calculate the experimental enthalpy change of combustion of butan-1-ol, \(\Delta_c H\), in \(\text{kJ mol}^{-1}\). Include a sign in your answer. Give your final answer to three significant figures. [3 marks]
(iii) The literature value for the standard enthalpy change of combustion of butan-1-ol is \(-2676\text{ kJ mol}^{-1}\). Suggest two reasons, other than heat loss to the surroundings, why the experimental value determined in (c)(ii) is less exothermic than the literature value. [2 marks]
(iv) Calculate the theoretical standard enthalpy change of combustion of butan-1-ol, \(\Delta_c H^\ominus\), in \(\text{kJ mol}^{-1}\), using the standard enthalpy changes of formation (\(\Delta_f H^\ominus\)) given below: * \(\Delta_f H^\ominus [\text{butan-1-ol, l}] = -327\text{ kJ mol}^{-1}\) * \(\Delta_f H^\ominus [\text{CO}_2\text{(g)}] = -394\text{ kJ mol}^{-1}\) * \(\Delta_f H^\ominus [\text{H}_2\text{O(l)}] = -286\text{ kJ mol}^{-1}\) [2 marks]
Divide each value by the smallest value (0.731): * Carbon: \(2.92 / 0.731 = 4.00\) * Hydrogen: \(6.57 / 0.731 = 8.99 \approx 9\) * Bromine: \(0.731 / 0.731 = 1.00\)
The empirical formula is \(\text{C}_4\text{H}_9\text{Br}\).
(ii) Ionic equation: \(\text{C}_4\text{H}_9\text{Br(l)} + \text{OH}^-\text{(aq)} \rightarrow \text{C}_4\text{H}_9\text{OH(aq)} + \text{Br}^-\text{(aq)}\) (Accept \(\text{C}_4\text{H}_9\text{Br(aq)}\) as ethanol is typically added as a solvent to mix the reactants.)
(iii) Adding aqueous silver nitrate yields a cream/off-white precipitate of silver bromine (\(\text{AgBr}\)). The precipitate is insoluble in dilute aqueous ammonia but dissolves when concentrated aqueous ammonia is added.
**(b)** (i) The rate of hydrolysis increases in the order: \(1\text{-chlorobutane} < 1\text{-bromobutane} < 1\text{-iodobutane}\). * The C–X bond strength (enthalpy) decreases down the group (\(\text{C–Cl} > \text{C–Br} > \text{C–I}\)). * The weaker the bond, the less energy is needed to break it, allowing the hydrolysis to proceed faster (bond enthalpy is the dominant factor, overriding bond polarity).
(ii) Hydrolysis of 2-bromo-2-methylpropane yields 2-methylpropan-2-ol. The skeletal structure is a central carbon with three methyl lines and one bond pointing to an OH group. Classification: Tertiary alcohol.
(ii) * Mass of butan-1-ol burned = \(184.25 - 183.51 = 0.74\text{ g}\) * Moles of butan-1-ol burned = \(0.74 / 74.0 = 0.010\text{ mol}\) * \(\Delta_c H = - \frac{q}{n} = - \frac{22.1382}{0.010} = -2213.82\text{ kJ mol}^{-1}\) * Rounding to 3 s.f.: \(-2210\text{ kJ mol}^{-1}\)
(iii) Any two of: * Incomplete combustion of butan-1-ol (forming soot/carbon monoxide instead of carbon dioxide). * Evaporation of the water (reducing the heat absorbed by the bulk water). * Evaporation of butan-1-ol from the wick after extinguishing but before weighing. * Non-standard conditions.
**(a) [7 marks total]** * **(i) [3 marks]** * M1: Correct division of all percentages by relevant \(A_r\) values (Carbon: \(2.92\), Hydrogen: \(6.57\), Bromine: \(0.731\)). * M2: Division by smallest value (0.731) to show a molecular ratio of \(4 : 9 : 1\). * M3: Statement/conclusion showing empirical formula is \(\text{C}_4\text{H}_9\text{Br}\). * **(ii) [2 marks]** * M1: Correct formulas: \(\text{C}_4\text{H}_9\text{Br} + \text{OH}^- \rightarrow \text{C}_4\text{H}_9\text{OH} + \text{Br}^-\) (ignore state symbols for M1). * M2: Correct state symbols on all species: \(\text{C}_4\text{H}_9\text{Br(l)}\) or \(\text{(aq)}\), \(\text{OH}^-\text{(aq)}\), \(\text{C}_4\text{H}_9\text{OH(aq)}\) or \(\text{(l)}\), \(\text{Br}^-\text{(aq)}\). * **(iii) [2 marks]** * M1: Formation of a cream / off-white precipitate. * M2: Insoluble in dilute aqueous ammonia but soluble in concentrated aqueous ammonia.
**(b) [5 marks total]** * **(i) [3 marks]** * M1: Identifies trend: rate increases in the order \(1\text{-chlorobutane} < 1\text{-bromobutane} < 1\text{-iodobutane}\) (or 1-iodobutane hydrolyses fastest and 1-chlorobutane slowest). * M2: States that C–X bond enthalpy decreases down the group: \(\text{C–Cl} > \text{C–Br} > \text{C–I}\). * M3: Explains that bond breaking (not bond polarity) is the rate-determining step, so the weaker the bond, the faster the reaction. * **(ii) [2 marks]** * M1: Skeletal structure of 2-methylpropan-2-ol: correctly drawn with a central carbon branched to three methyl lines and one OH bond. * M2: Classifies as a tertiary alcohol.
**(c) [9 marks total]** * **(i) [2 marks]** * M1: Correct temperature change calculation \(\Delta T = 35.3\text{ }^\circ\text{C}\) and use of formula \(q = m c \Delta T\). * M2: Evaluates to \(22.1\text{ kJ}\) (or \(22.138\text{ kJ}\)). * **(ii) [3 marks]** * M1: Calculates mass of alcohol burned (\(0.74\text{ g}\)) and converts to moles: \(0.010\text{ mol}\). * M2: Divides energy \(q\) by moles \(n\) to calculate experimental value. * M3: Final answer given to 3 s.f. and has a negative sign: \(-2210\text{ kJ mol}^{-1}\) (allow ECF from (c)(i)). * **(iii) [2 marks]** * M1 & M2: Any two valid reasons (1 mark each): * Incomplete combustion. * Evaporation of water from calorimeter. * Evaporation of fuel from wick after turning off. * Reaction carried out under non-standard conditions. * (Do NOT accept heat loss to the surroundings, as ruled out by the prompt). * **(iv) [2 marks]** * M1: Correct setup of Hess's Cycle or algebraic equation: \([4 \times (-394) + 5 \times (-286)] - [-327]\). * M2: Evaluates calculation to \(-2679\text{ kJ mol}^{-1}\) (must include negative sign for mark).
部分 Unit 3: Practical Chemistry
Answer all experimental laboratory skills questions.
4 題目 · 50 分
題目 1 · Structured Laboratory Skills Questions
12.5 分
A student carries out an experiment to determine the concentration of sulfamic acid, \(H_3NSO_3\) (\(M_r = 97.1\)), in a commercial household descaler.
**Step 1:** Prepare 250.0 cm³ of an aqueous solution containing exactly 2.50 g of the solid descaler. **Step 2:** Titrate 25.0 cm³ portions of this solution against 0.100 mol dm⁻³ sodium hydroxide solution, \(NaOH(aq)\).
(a) Describe how the student would prepare exactly 250.0 cm³ of the descaler solution from the weighed solid. [5 marks]
(b) The average titre of 0.100 mol dm⁻³ \(NaOH\) obtained was 21.80 cm³. Calculate the mass of sulfamic acid in the 2.50 g sample of descaler, assuming it is a monoprotic acid and the only acidic component present. [4 marks]
(c) The pipette used to measure the 25.0 cm³ portion has an uncertainty of \(\pm 0.06\text{ cm}^3\). Calculate the percentage uncertainty in this volume, and explain whether using a burette with an uncertainty of \(\pm 0.05\text{ cm}^3\) for each reading would be more precise. [1.5 marks]
(d) Explain why washing the inside of the titration flask with distilled water during the titration does not affect the value of the titre. [2 marks]
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解題
(a) 1. Weigh 2.50 g of the solid descaler in a weighing bottle or beaker using a balance and record the mass by difference. 2. Dissolve the solid in about 100 cm³ of distilled water in a beaker, stirring with a glass rod until completely dissolved. 3. Transfer the solution quantitatively to a 250 cm³ volumetric flask, rinsing the beaker, glass rod, and funnel with distilled water, adding all washings to the flask. 4. Add distilled water up to the graduation mark until the bottom of the meniscus lies exactly on the line. 5. Stopper the flask and invert several times to ensure a homogeneous solution.
(b) Moles of \(NaOH\) = \(0.100 \times \frac{21.80}{1000} = 2.18 \times 10^{-3}\text{ mol}\). Since sulfamic acid is monoprotic, moles in 25.0 cm³ = \(2.18 \times 10^{-3}\text{ mol}\). Moles in 250.0 cm³ = \(2.18 \times 10^{-3} \times 10 = 2.18 \times 10^{-2}\text{ mol}\). Mass of sulfamic acid = \(2.18 \times 10^{-2}\text{ mol} \times 97.1\text{ g mol}^{-1} = 2.11678\text{ g}\) = \(2.12\text{ g}\) (to 3 s.f.).
(c) Percentage uncertainty = \(\frac{0.06}{25.0} \times 100 = 0.24\%\). A burette requires two readings (initial and final), so total uncertainty = \(2 \times 0.05 = \pm 0.10\text{ cm}^3\). Therefore, the burette is less precise (percentage uncertainty = \(\frac{0.10}{25.0} \times 100 = 0.40\%\)).
(d) Washing with distilled water down the sides of the flask washes unreacted acid into the solution so it can react. However, it does not change the actual number of moles of sulfamic acid originally pipetted into the flask, so the volume of NaOH required (titre) remains unaffected.
評分準則
- (a) [5 marks total]: - M1: Weigh solid by difference / record initial and final mass of container (1) - M2: Dissolve in a beaker with a reasonable volume of distilled water (e.g. 50-100 cm³) and stir (1) - M3: Quantitative transfer (rinsing beaker, rod and funnel into the volumetric flask) (1) - M4: Make up to the 250 cm³ mark so bottom of meniscus is on the line (1) - M5: Stopper and invert/mix well (1) - (b) [4 marks total]: - M1: Calculates moles of NaOH = \(2.18 \times 10^{-3}\text{ mol}\) (1) - M2: Identifies 1:1 ratio so moles of acid in 25 cm³ = \(2.18 \times 10^{-3}\text{ mol}\) (1) - M3: Scales up to 250 cm³ (moles of acid = \(2.18 \times 10^{-2}\text{ mol}\)) (1) - M4: Calculates mass = \(2.12\text{ g}\) (accept 2.116 to 2.12 g) (1) - (c) [1.5 marks total]: - M1: Percentage uncertainty = \(0.24\%\) (1) - M2: Explains that burette requires two readings so total uncertainty is \(\pm 0.10\text{ cm}^3\), making it less precise (0.5) - (d) [2 marks total]: - M1: Distilled water washes reactants from the walls into the mixture so all acid reacts (1) - M2: The number of moles of acid in the flask is unchanged (1)
題目 2 · Structured Laboratory Skills Questions
12.5 分
A student prepares 1-bromobutane from butan-1-ol using sodium bromide and concentrated sulfuric acid.
(a) Write the chemical equation for the in-situ preparation of hydrogen bromide (\(HBr\)) from sodium bromide and sulfuric acid. State why \(HBr\) is generated in-situ rather than using gaseous \(HBr\) directly. [2 marks]
(b) Concentrated sulfuric acid is a hazardous reagent in this experiment. Identify one risk associated with concentrated sulfuric acid and state a suitable safety precaution. [2 marks]
(c) Draw a fully labeled diagram of the apparatus used to heat the reaction mixture under reflux. [3.5 marks]
(d) After heating, the mixture is distilled. The distillate forms two distinct liquid layers. Describe how the student would separate the 1-bromobutane layer from the aqueous layer. Identify which layer is the lower layer and justify your answer using the density values provided. (Density of 1-bromobutane = \(1.27\text{ g cm}^{-3}\), density of water = \(1.00\text{ g cm}^{-3}\)). [3 marks]
(e) Describe how the separated organic layer can be dried, and state the appearance of the product before and after drying. [2 marks]
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解題
(a) Equation: \(NaBr + H_2SO_4 \rightarrow NaHSO_4 + HBr\) (or \(2NaBr + H_2SO_4 \rightarrow Na_2SO_4 + 2HBr\)). HBr gas is highly toxic and corrosive, making it difficult and dangerous to handle as a gas directly; generating it in-situ is safer and more convenient.
(b) Risk: Concentrated sulfuric acid is highly corrosive and causes severe skin burns. Precaution: Wear safety goggles and protective gloves, or handle in a fume cupboard to avoid inhalation of fumes.
(c) Labeled diagram showing: - Round-bottomed or pear-shaped flask with heating source (e.g., heating mantle or water bath). - Condenser fitted vertically in the neck of the flask. - Water flowing into the lower condenser jacket inlet and out of the upper outlet. - Condenser open at the top (no stopper).
(d) Pour the mixture into a separating funnel, insert the stopper, and shake, periodically opening the tap to release pressure. Allow the layers to separate. The lower layer is the 1-bromobutane layer because its density (\(1.27\text{ g cm}^{-3}\)) is greater than that of water (\(1.00\text{ g cm}^{-3}\)). Run off this lower layer through the tap into a separate flask.
(e) Add an anhydrous inorganic salt, such as anhydrous calcium chloride (\(CaCl_2\)) or anhydrous sodium sulfate (\(Na_2SO_4\)), to the liquid. Swirl the mixture and let it stand. The appearance goes from cloudy/turbid (due to suspended water droplets) to clear and colorless.
評分準則
- (a) [2 marks total]: - M1: Correct chemical equation (1) - M2: Explains that HBr gas is toxic/corrosive/difficult to store, making in-situ generation safer/more practical (1) - (b) [2 marks total]: - M1: Corrosive / causes severe burns (1) - M2: Wear safety gloves / protective eyewear (1) - (c) [3.5 marks total]: - M1: Round-bottom or pear-shaped flask with a heat source (1) - M2: Vertical condenser in neck of flask (1) - M3: Correct water flow: in at bottom, out at top (1) - M4: Open top to prevent pressure build-up (0.5) - (d) [3 marks total]: - M1: Use of a separating funnel, shaking, and releasing pressure (1) - M2: Identifies 1-bromobutane is the lower layer because its density is greater than water's (1) - M3: Runs off the lower layer into a beaker (1) - (e) [2 marks total]: - M1: Adds anhydrous calcium chloride / sodium sulfate / magnesium sulfate (0.5) and swirls/decants/filters (0.5) - M2: Goes from cloudy to clear (1)
題目 3 · Structured Laboratory Skills Questions
12.5 分
A student is provided with three unlabelled bottles containing white solids, X, Y, and Z. The solids are known to be barium bromide (\(BaBr_2\)), calcium carbonate (\(CaCO_3\)), and potassium iodide (\(KI\)). The student performs a series of tests to identify them.
**Test 2: Addition of dilute hydrochloric acid to the solid** - Solid X: Bubbling observed; a colorless gas is evolved. - Solid Y: No visible reaction. - Solid Z: No visible reaction.
**Test 3: Dissolve the solid in water to make a solution, then add acidified aqueous silver nitrate** - Solution of Y: Cream precipitate formed. - Solution of Z: Yellow precipitate formed.
(a) Identify the chemical formula of solid X, and write an ionic equation (including state symbols) for its reaction with dilute hydrochloric acid. [3 marks]
(b) Identify the chemical formula of solid Y. Describe a chemical test using aqueous ammonia to confirm the identity of the anion in Y, and state the expected observations. [3.5 marks]
(c) Identify the chemical formula of solid Z. Describe two distinct observations when solid Z is heated with concentrated sulfuric acid, and name two gaseous products (other than water vapor) formed in this reaction. [4 marks]
(d) Explain the origin of the flame color observed in the flame tests in terms of electron transitions. [2 marks]
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解題
(a) Formula of X: \(CaCO_3\). Ionic equation: \(CaCO_3(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + CO_2(g) + H_2O(l)\).
(b) Formula of Y: \(BaBr_2\). To confirm the bromide ion (\(Br^-\)), add dilute aqueous ammonia to the cream precipitate (\(AgBr\)). The precipitate remains insoluble. Then add concentrated aqueous ammonia; the precipitate dissolves to form a colorless solution.
(c) Formula of Z: \(KI\). When solid potassium iodide (\(KI\)) reacts with concentrated sulfuric acid, the iodide ions are oxidized to iodine. Observations: purple vapor / dark grey solid / bad egg smell (due to \(H_2S\)) / yellow solid (sulfur) / steamy fumes (due to \(HI\)). Gaseous products: hydrogen iodide (\(HI\)), iodine (\(I_2\)), sulfur dioxide (\(SO_2\)), hydrogen sulfide (\(H_2S\)).
(d) The heat from the Bunsen burner flame promotes / excites electrons to higher energy levels. When these excited electrons drop back down to lower energy levels (ground state), they emit energy in the form of light. The wavelength/frequency of this emitted light corresponds to a specific color in the visible spectrum.
評分準則
- (a) [3 marks total]: - M1: Formula of X is \(CaCO_3\) (1) - M2: Correct species in the ionic equation: \(CaCO_3(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + CO_2(g) + H_2O(l)\) (1) - M3: Correct state symbols (1) - (b) [3.5 marks total]: - M1: Formula of Y is \(BaBr_2\) (1) - M2: Adds dilute ammonia and then concentrated ammonia (1) - M3: Observation: Insoluble in dilute ammonia (1) but soluble in concentrated ammonia (0.5) - (c) [4 marks total]: - M1: Formula of Z is \(KI\) (1) - M2: Two observations: any two from purple vapor, grey/black solid, yellow solid, smell of bad eggs, steamy fumes (1) - M3: Two gases named: any two from hydrogen iodide, iodine, sulfur dioxide, hydrogen sulfide (2, 1 mark each) - (d) [2 marks total]: - M1: Electrons are excited to higher energy levels by heat (1) - M2: Light is emitted as electrons drop back to lower energy levels (1)
題目 4 · Structured Laboratory Skills Questions
12.5 分
A student determines the enthalpy change of neutralisation by mixing 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid with 50.0 cm³ of 2.05 mol dm⁻³ sodium hydroxide solution in a polystyrene cup.
(a) Explain why the student takes temperature readings at regular intervals before and after mixing, and how they use these readings to determine an accurate maximum temperature rise (\(\Delta T\)). [3.5 marks]
(b) In this experiment, the initial temperature was 18.2 °C and the extrapolated maximum temperature was 31.8 °C. Calculate the heat energy, \(q\), in joules, released during the reaction. (Assume the density of the final mixture is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ °C}^{-1}\)). [3 marks]
(c) Calculate the enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), in kJ mol⁻¹ of water formed. Give your answer to 3 significant figures and include the appropriate sign. [4 marks]
(d) Explain the effect on the calculated value of \(\Delta H_{\text{neut}}\) if a glass beaker had been used instead of a polystyrene cup. [2 marks]
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解題
(a) Regular temperature measurements are taken to monitor the rate of heat loss to the surroundings during and after the reaction. To find \(\Delta T\), the student plots a graph of temperature against time, draws a line of best fit through the cooling points, and extrapolates this line back to the time of mixing (e.g., the 4th minute). The difference between the extrapolated temperature at the mixing time and the initial temperature represents \(\Delta T\).
(b) Total mass of solution = \(50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g}\) (since density = \(1.00\text{ g cm}^{-3}\)). \(\Delta T = 31.8 - 18.2 = 13.6\text{ °C}\). \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ °C}^{-1} \times 13.6\text{ °C} = 5684.8\text{ J}\) (or 5680 J / 5.68 kJ).
(c) Moles of \(HCl\) = \(2.00 \times 0.0500 = 0.100\text{ mol}\). Moles of \(NaOH\) = \(2.05 \times 0.0500 = 0.1025\text{ mol}\). Since \(NaOH\) is in excess, \(HCl\) is the limiting reagent. Moles of water formed = \(0.100\text{ mol}\). \(\Delta H_{\text{neut}} = -\frac{5.6848\text{ kJ}}{0.100\text{ mol}} = -56.848\text{ kJ mol}^{-1} = -56.8\text{ kJ mol}^{-1}\).
(d) Glass is a poorer thermal insulator than polystyrene, leading to greater heat loss to the surroundings during the experiment. This would result in a lower measured temperature rise (\(\Delta T\)), making the calculated value of \(\Delta H_{\text{neut}}\) less exothermic (i.e., less negative / smaller in magnitude).
評分準則
- (a) [3.5 marks total]: - M1: Explains that measurements account for heat loss to surroundings (1) - M2: Mentions plotting temperature vs time and drawing a cooling line of best fit (1) - M3: Mentions extrapolating back to the time of mixing (1) - M4: Explains \(\Delta T\) is the difference between extrapolated and initial temperature (0.5) - (b) [3 marks total]: - M1: Mass of solution = 100.0 g (1) - M2: Temperature change = 13.6 °C (1) - M3: Calculates \(q = 5684.8\text{ J}\) (allow 5680 J / 5.68 kJ) (1) - (c) [4 marks total]: - M1: Moles of acid = 0.100 mol (1) - M2: Identifies HCl as limiting and hence moles of water = 0.100 mol (1) - M3: Calculates numerical value \(\Delta H = 56.8\text{ kJ mol}^{-1}\) (1) - M4: Uses negative sign (\(-56.8\text{ kJ mol}^{-1}\)) and correct 3 s.f. (1) - (d) [2 marks total]: - M1: Glass has higher thermal conductivity / is a poor insulator so more heat is lost (1) - M2: Measured \(\Delta T\) is smaller, so calculated value is less exothermic / less negative (1)
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