An original Thinka practice paper modelled on the structure and difficulty of the 2022 HKDSE Chemistry paper. Not affiliated with or reproduced from HKDSE.
卷一 甲部
回答所有選擇題。全部試題分數相同。
36 題目 · 36 分
題目 1 · 選擇題
1 分
Given the following standard enthalpy changes of combustion:
1 mark for option A. - Option B gives an incorrect sign (+124 kJ mol⁻¹). - Options C and D arise from incorrect combinations of the values.
題目 2 · 選擇題
1 分
Consider the following statements regarding butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\):
(1) It turns acidified potassium dichromate solution from orange to green when heated. (2) It can be dehydrated by heating with concentrated sulphuric acid to form a mixture of positional and geometrical isomers. (3) It reacts with ethanoic acid in the presence of concentrated sulphuric acid to form an ester with the molecular formula \(\text{C}_6\text{H}_{12}\text{O}_2\).
Which of the statements above are correct?
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: Butan-2-ol is a secondary alcohol and is oxidised to butan-2-one by acidified potassium dichromate, reducing \(\text{Cr}_2\text{O}_7^{2-}\) (orange) to \(\text{Cr}^{3+}\) (green). (2) is correct: Dehydration of butan-2-ol yields but-1-ene and but-2-ene (positional isomers), where but-2-ene exists as cis-but-2-ene and trans-but-2-ene (geometrical isomers). (3) is correct: The esterification of butan-2-ol (\(\text{C}_4\text{H}_{10}\text{O}\)) with ethanoic acid (\(\text{C}_2\text{H}_4\text{O}_2\)) produces sec-butyl ethanoate (\(\text{CH}_3\text{COOCH(CH}_3)\text{CH}_2\text{CH}_3\)), which has the molecular formula \(\text{C}_6\text{H}_{12}\text{O}_2\).
Therefore, all three statements are correct.
評分準則
1 mark for option D.
題目 3 · 選擇題
1 分
Which of the following pairs of molecules both possess a non-zero net dipole moment (i.e. both are polar molecules)?
A.\(\text{BF}_3\) and \(\text{NH}_3\)
B.\(\text{CO}_2\) and \(\text{SO}_2\)
C.\(\text{NF}_3\) and \(\text{H}_2\text{S}\)
D.\(\text{CCl}_4\) and \(\text{PCl}_3\)
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解題
A molecule has a non-zero dipole moment if its polar bonds do not cancel out due to symmetry. - \(\text{NF}_3\) has a trigonal pyramidal shape with a lone pair on the N atom; the polar N–F bond dipoles do not cancel, so it is polar. - \(\text{H}_2\text{S}\) has a V-shaped (bent) geometry with two lone pairs on the S atom; the dipoles do not cancel, so it is polar.
In the other options: - In A, \(\text{BF}_3\) is trigonal planar and non-polar. - In B, \(\text{CO}_2\) is linear and non-polar. - In D, \(\text{CCl}_4\) is tetrahedral and non-polar.
Thus, only pair C consists of two polar molecules.
評分準則
1 mark for option C.
題目 4 · 選擇題
1 分
A concentrated sodium chloride solution is electrolysed using graphite electrodes. Which of the following statements is / are correct?
(1) A gas that turns moist blue litmus paper red and then bleaches it is evolved at the anode. (2) The solution around the cathode becomes alkaline. (3) The theoretical molar ratio of the gas evolved at the anode to that at the cathode is \(1 : 1\).
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: At the anode, chloride ions are preferentially discharged to form chlorine gas: \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\). Chlorine gas is acidic and acts as a bleaching agent, so it turns moist blue litmus paper red and then bleaches it. (2) is correct: At the cathode, water/hydrogen ions are reduced: \(2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq})\). The accumulation of \(\text{OH}^-(\text{aq})\) makes the solution around the cathode alkaline. (3) is correct: For every 2 moles of electrons transferred, 1 mole of \(\text{Cl}_2(\text{g})\) is formed at the anode and 1 mole of \(\text{H}_2(\text{g})\) is formed at the cathode, giving a theoretical molar ratio of \(1 : 1\).
Hence, (1), (2), and (3) are all correct.
評分準則
1 mark for option D.
題目 5 · 選擇題
1 分
A sample of \(1.325\text{ g}\) of anhydrous sodium carbonate (\(\text{Na}_2\text{CO}_3\), molar mass \(= 106.0\text{ g mol}^{-1}\)) was dissolved in distilled water and made up to \(250.0\text{ cm}^3\) in a volumetric flask. A \(25.00\text{ cm}^3\) portion of this solution required \(20.00\text{ cm}^3\) of a hydrochloric acid solution for complete reaction.
What is the molarity of the hydrochloric acid solution?
A.\(0.0625\text{ M}\)
B.\(0.125\text{ M}\)
C.\(0.250\text{ M}\)
D.\(0.500\text{ M}\)
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解題
1. Calculate the concentration of the \(\text{Na}_2\text{CO}_3\) solution: \[\text{Number of moles of } \text{Na}_2\text{CO}_3 = \frac{1.325\text{ g}}{106.0\text{ g mol}^{-1}} = 0.01250\text{ mol}\] \[[\text{Na}_2\text{CO}_3] = \frac{0.01250\text{ mol}}{0.2500\text{ dm}^3} = 0.05000\text{ M}\]
2. Calculate the moles of \(\text{Na}_2\text{CO}_3\) in \(25.00\text{ cm}^3\): \[\text{Moles of } \text{Na}_2\text{CO}_3 = 0.05000\text{ M} \times 0.02500\text{ dm}^3 = 1.250 \times 10^{-3}\text{ mol}\]
3. Determine the moles of \(\text{HCl}\) reacting: According to the stoichiometric equation, \(1\text{ mol of } \text{Na}_2\text{CO}_3\) reacts with \(2\text{ mol of } \text{HCl}\): \[\text{Moles of } \text{HCl} = 2 \times 1.250 \times 10^{-3}\text{ mol} = 2.500 \times 10^{-3}\text{ mol}\]
4. Calculate the molarity of \(\text{HCl}\): \[[\text{HCl}] = \frac{2.500 \times 10^{-3}\text{ mol}}{0.02000\text{ dm}^3} = 0.125\text{ M}\]
Therefore, option B is correct.
評分準則
1 mark for option B. - Option A (0.0625 M) arises from neglecting the 1:2 stoichiometric mole ratio. - Option C (0.250 M) arises from an inverted mole ratio.
題目 6 · 選擇題
1 分
Consider the standard enthalpy changes of combustion (\(\Delta H_c^\circ\)) of the following substances:
What is the standard enthalpy change of formation (\(\Delta H_f^\circ\)) of methoxymethane, \(\text{CH}_3\text{OCH}_3\text{(g)}\)?
A.\(-184.0\text{ kJ mol}^{-1}\)
B.\(+184.0\text{ kJ mol}^{-1}\)
C.\(-781.1\text{ kJ mol}^{-1}\)
D.\(-2353.4\text{ kJ mol}^{-1}\)
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解題
The formation reaction of methoxymethane is: \(2\text{C(graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OCH}_3\text{(g)}\)
Using Hess's Law with standard enthalpy changes of combustion: \(\Delta H_f^\circ = \sum \Delta H_c^\circ(\text{reactants}) - \sum \Delta H_c^\circ(\text{products})\) \(\Delta H_f^\circ = [2 \times (-393.5) + 3 \times (-285.8)] - [-1460.4]\) \(\Delta H_f^\circ = [-787.0 - 857.4] + 1460.4\) \(\Delta H_f^\circ = -1644.4 + 1460.4 = -184.0\text{ kJ mol}^{-1}\)
評分準則
A (1 mark): Correct calculation of \(\Delta H_f^\circ = -184.0\text{ kJ mol}^{-1}\) using enthalpy of combustion data.
題目 7 · 選擇題
1 分
Which of the following species have a bent (V-shaped) molecular or ionic geometry?
(1) \(\text{NO}_2^-\): The central N atom has 3 electron pairs/domains (2 bonding regions and 1 lone pair), giving a trigonal planar electron geometry and a bent shape. (2) \(\text{OF}_2\): The central O atom has 4 electron domains (2 single bonds to F and 2 lone pairs), giving a tetrahedral electron geometry and a bent shape. (3) \(\text{I}_3^-\): The central I atom has 5 electron pairs (2 bonding pairs and 3 lone pairs in equatorial positions), giving a linear shape. Therefore, only (1) and (2) are bent.
評分準則
A (1 mark): Correctly identifies that \(\text{NO}_2^-\) and \(\text{OF}_2\) are bent, while \(\text{I}_3^-\) is linear.
題目 8 · 選擇題
1 分
A carbonyl compound \(W\) with molecular formula \(\text{C}_4\text{H}_8\text{O}\) forms an orange precipitate when mixed with 2,4-dinitrophenylhydrazine, but gives no observable change when warmed with Tollens' reagent. When \(W\) is treated with \(\text{NaBH}_4\), compound \(Y\) is formed. What is the systematic IUPAC name of compound \(Y\)?
A.Butan-1-ol
B.Butan-2-ol
C.2-Methylpropan-1-ol
D.2-Methylpropan-2-ol
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解題
Compound \(W\) gives a positive 2,4-DNP test, indicating it is an aldehyde or ketone. Because it gives a negative Tollens' test, it is not an aldehyde; hence \(W\) must be a ketone. With 4 carbon atoms, \(W\) is butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)). Reduction of butan-2-one with \(\text{NaBH}_4\) reduces the carbonyl group to a secondary alcohol, yielding butan-2-ol (compound \(Y\)).
評分準則
B (1 mark): Correct identification of butan-2-ol from the reduction of butan-2-one.
題目 9 · 選擇題
1 分
A concentrated aqueous solution of sodium chloride is electrolysed using graphite electrodes.
Which of the following statements concerning this process is / are correct?
(1) Chlorine gas is liberated at the anode. (2) The pH of the solution around the cathode increases. (3) If copper electrodes are used to replace the graphite electrodes, chlorine gas is still liberated at the anode.
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) Correct: In concentrated \(\text{NaCl(aq)}\), the concentration of \(\text{Cl}^-\text{(aq)}\) is high enough that \(\text{Cl}^-\text{(aq)}\) is preferentially discharged over \(\text{OH}^-\text{(aq)}\) at the graphite anode to produce \(\text{Cl}_2\text{(g)}\). (2) Correct: At the cathode, \(\text{H}^+\text{(aq)}\) ions are discharged to form \(\text{H}_2\text{(g)}\) (\(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) or \(2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-\)), resulting in an accumulation of \(\text{OH}^-\text{(aq)}\) ions, which increases the pH. (3) Incorrect: Copper is an active (reactive) anode. If copper is used as the anode, it undergoes oxidation (\(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\)) rather than discharging chloride ions, so no \(\text{Cl}_2\text{(g)}\) will be liberated.
評分準則
B (1 mark): Correctly identifies statements (1) and (2) as correct, and (3) as incorrect due to copper anode dissolution.
題目 10 · 選擇題
1 分
A section of a synthetic polymer is represented below:
\(\text{--[--NH--(CH}_2)_6\text{--NH--CO--(CH}_2)_4\text{--CO--]}_n\text{--}\ Which of the following statements concerning this polymer is / are correct? (1) It is formed by condensation polymerisation. (2) Intermolecular hydrogen bonds can form between adjacent polymer chains. (3) The repeating unit has an empirical formula of \)\text{C}_6\text{H}_{11}\text{NO}\).
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) Correct: The polymer shown is Nylon-6,6, which is formed via condensation polymerisation between hexane-1,6-diamine and hexanedioic acid with the elimination of water molecules. (2) Correct: The polymer contains amide (\(\text{--CONH--}\)) linkages with \(\text{N--H}\) donors and \(\text{C=O}\) acceptors, allowing strong intermolecular hydrogen bonds between adjacent chains. (3) Correct: The repeating unit contains: \(6 + 1 + 4 + 1 = 12\) carbon atoms, \(1 + 12 + 1 + 8 = 22\) hydrogen atoms, \(2\) nitrogen atoms, and \(2\) oxygen atoms, giving the molecular formula \(\text{C}_{12}\text{H}_{22}\text{N}_2\text{O}_2\). The simplest whole-number ratio (empirical formula) is \(\text{C}_6\text{H}_{11}\text{NO}\).
評分準則
D (1 mark): Correctly evaluates statements (1), (2), and (3) as all correct.
Standard enthalpy change of combustion of \(\text{C}_2\text{H}_6(\text{g}) = -1560\text{ kJ mol}^{-1}\) Standard enthalpy change of combustion of \(\text{H}_2(\text{g}) = -286\text{ kJ mol}^{-1}\)
What is the standard enthalpy change of combustion of ethene, \(\text{C}_2\text{H}_4(\text{g})\)?
A.\(-1983\text{ kJ mol}^{-1}\)
B.\(-1411\text{ kJ mol}^{-1}\)
C.\(-1137\text{ kJ mol}^{-1}\)
D.\(+1411\text{ kJ mol}^{-1}\)
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解題
According to Hess's Law, the enthalpy change of the hydrogenation reaction can be expressed in terms of the standard enthalpy changes of combustion (\(\Delta H_\text{c}^\theta\)) of reactants and products:
B (1 mark) - Award 1 mark for correct application of Hess's Law relating combustion enthalpies to reaction enthalpy. - Do not accept positive enthalpy values.
題目 12 · 選擇題
1 分
Consider compound \(X\) with the structural formula \(\text{CH}_3\text{CH(OH)CH}_2\text{COOCH}_3\).
Which of the following statements about compound \(X\) is/are correct?
(1) It turns acidified potassium dichromate solution from orange to green upon heating. (2) It gives an effervescence with sodium hydrogencarbonate solution at room temperature. (3) It yields two different organic products when heated under reflux with excess dilute \(\text{NaOH(aq)}\).
A.(1) only
B.(2) only
C.(1) and (3) only
D.(2) and (3) only
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解題
(1) is correct: Compound \(X\) contains a secondary alcohol group (\(-\text{CH(OH)}-\)), which can be oxidized to a ketone by acidified \(\text{K}_2\text{Cr}_2\text{O}_7\), turning the solution from orange to green. (2) is incorrect: Compound \(X\) contains an ester group and an alcohol group, but no carboxylic acid group (\(-\text{COOH}\)), so it does not react with \(\text{NaHCO}_3(\text{aq})\) to produce \(\text{CO}_2\) gas. (3) is correct: Alkaline hydrolysis of the ester group produces sodium 3-hydroxybutanoate (\(\text{CH}_3\text{CH(OH)CH}_2\text{COO}^-\text{Na}^+\)) and methanol (\(\text{CH}_3\text{OH}\)), which are two distinct organic compounds.
Therefore, (1) and (3) only are correct.
評分準則
C (1 mark) - Award 1 mark for identifying that (1) and (3) are correct while (2) is incorrect.
題目 13 · 選擇題
1 分
Which of the following pairs of species have the SAME spatial arrangement of atoms (molecular/ionic shape)?
A.\(\text{NH}_3\) and \(\text{BF}_3\)
B.\(\text{H}_3\text{O}^+\) and \(\text{NH}_3\)
C.\(\text{NH}_4^+\) and \(\text{SF}_4\)
D.\(\text{CO}_2\) and \(\text{SO}_2\)
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解題
Let us deduce the shape of each species using the VSEPR theory: - In \(\text{H}_3\text{O}^+\), oxygen has 3 bond pairs and 1 lone pair, giving a trigonal pyramidal shape. In \(\text{NH}_3\), nitrogen has 3 bond pairs and 1 lone pair, also giving a trigonal pyramidal shape. Thus, they have the same shape. - \(\text{NH}_3\) is trigonal pyramidal, whereas \(\text{BF}_3\) has 3 bond pairs and 0 lone pairs (trigonal planar). - \(\text{NH}_4^+\) is tetrahedral (4 bond pairs, 0 lone pairs), whereas \(\text{SF}_4\) has 4 bond pairs and 1 lone pair (see-saw shape). - \(\text{CO}_2\) is linear (2 double bond regions, 0 lone pairs on C), whereas \(\text{SO}_2\) is bent/V-shaped (2 bonding regions, 1 lone pair on S).
Hence, B is the correct answer.
評分準則
B (1 mark) - Award 1 mark for recognizing that both \(\text{H}_3\text{O}^+\) and \(\text{NH}_3\) have 3 bonding pairs and 1 lone pair around the central atom, resulting in a trigonal pyramidal geometry.
題目 14 · 選擇題
1 分
A chemical cell is set up by connecting a zinc electrode in \(1.0\text{ M Zn(NO}_3)_2(\text{aq})\) and a silver electrode in \(1.0\text{ M AgNO}_3(\text{aq})\) using a salt bridge containing saturated \(\text{KNO}_3(\text{aq})\).
Which of the following statements about this operating cell is/are correct?
(1) Electrons flow from the silver electrode to the zinc electrode in the external circuit. (2) Nitrate ions from the salt bridge migrate towards the zinc half-cell. (3) The mass of the silver electrode increases as the cell operates.
A.(1) only
B.(2) only
C.(1) and (3) only
D.(2) and (3) only
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解題
Zinc is higher than silver in the electrochemical series, so zinc acts as the negative electrode (anode) where oxidation occurs: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-\). Silver acts as the positive electrode (cathode) where reduction occurs: \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag(s)}\).
(1) is incorrect: Electrons flow from the zinc electrode (anode) to the silver electrode (cathode) through the external circuit. (2) is correct: As \(\text{Zn}^{2+}\) cations are generated in the zinc half-cell, anions (\(\text{NO}_3^-\)) from the salt bridge migrate into this half-cell to maintain electrical neutrality. (3) is correct: \(\text{Ag}^+\) ions in solution are reduced and deposited as metallic silver on the silver electrode, increasing its mass.
Therefore, (2) and (3) only are correct.
評分準則
D (1 mark) - Award 1 mark for correctly determining the migration of anions to the anode and the cathode deposition, while rejecting the reversed direction of electron flow.
題目 15 · 選擇題
1 分
The table below shows some physical properties of three substances, \(P\), \(Q\), and \(R\):
- Substance \(P\) has a high melting point (801 \(^\circ\text{C}\)) and conducts electricity in the molten state but not in the solid state due to the presence of mobile ions when molten. This is characteristic of a giant ionic structure. - Substance \(Q\) has a very high melting point (1610 \(^\circ\text{C}\)) and does not conduct electricity in either solid or molten states because all valence electrons are localized in strong covalent bonds. This is characteristic of a giant covalent structure (e.g. quartz, \(\text{SiO}_2\)). - Substance \(R\) has a high melting point and conducts electricity in both solid and molten states due to the presence of delocalized electrons in a giant metallic lattice. This is characteristic of a giant metallic structure (e.g. copper).
Therefore, A is the correct answer.
評分準則
A (1 mark) - Award 1 mark for matching all three substances correctly with their corresponding giant structures based on melting points and electrical conductivity.
Keep equation (2) as given: \(3\text{MnO}_2(s) \rightarrow 3\text{Mn}(s) + 3\text{O}_2(g) \quad \Delta H = +1563\text{ kJ mol}^{-1}\)
Adding these two equations gives: \(\Delta H = -3352 + (+1563) = -1789\text{ kJ mol}^{-1}\).
評分準則
A (1 mark): Correctly applies Hess's Law by doubling the enthalpy of formation of \(\text{Al}_2\text{O}_3\) and adding the decomposition enthalpy of \(3\text{MnO}_2\) to obtain \(-1789\text{ kJ mol}^{-1}\).
題目 17 · 選擇題
1 分
An organic compound has the structural formula \(\text{CH}_3\text{CH(OH)CH=CH}_2\). Which of the following statements concerning this compound are correct?
(1) It can turn acidified potassium dichromate solution from orange to green. (2) It can decolourise bromine in 1,1,1-trichloroethane in the dark. (3) It exhibits enantiomerism (optical isomerism).
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: The compound contains a secondary alcohol group (\(-\text{CH(OH)}-\)), which can be oxidised by acidified potassium dichromate solution, reducing \(\text{Cr}_2\text{O}_7^{2-}\) (orange) to \(\text{Cr}^{3+}\) (green). (2) is correct: The compound contains a carbon-carbon double bond (\(\text{C=C}\)), which undergoes electrophilic addition with bromine in the dark without requiring UV light. (3) is correct: The C-2 carbon atom (\(\text{C}^*\text{H(OH)}\)) is bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH=CH}_2\)), making it a chiral centre; thus, it exhibits enantiomerism.
評分準則
D (1 mark): All three statements (1), (2), and (3) are chemically correct.
題目 18 · 選擇題
1 分
Which of the following pairs of chemical species have the same shape?
A.\(\text{NH}_4^+\) and \(\text{CH}_4\)
B.\(\text{H}_2\text{O}\) and \(\text{CO}_2\)
C.\(\text{NH}_3\) and \(\text{BF}_3\)
D.\(\text{PCl}_3\) and \(\text{SO}_3\)
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解題
A is correct: Both \(\text{NH}_4^+\) and \(\text{CH}_4\) have 4 bond pairs and 0 lone pairs around the central atom, giving both a tetrahedral shape. B is incorrect: \(\text{H}_2\text{O}\) has 2 bond pairs and 2 lone pairs (V-shaped/bent), while \(\text{CO}_2\) has 2 double bond domains and 0 lone pairs (linear). C is incorrect: \(\text{NH}_3\) has 3 bond pairs and 1 lone pair (trigonal pyramidal), whereas \(\text{BF}_3\) has 3 bond pairs and 0 lone pairs (trigonal planar). D is incorrect: \(\text{PCl}_3\) has 3 bond pairs and 1 lone pair (trigonal pyramidal), while \(\text{SO}_3\) has 3 electron domains and 0 lone pairs (trigonal planar).
評分準則
A (1 mark): Identifies both \(\text{NH}_4^+\) and \(\text{CH}_4\) as tetrahedral.
題目 19 · 選擇題
1 分
A concentrated aqueous solution of sodium chloride is electrolysed using inert graphite electrodes. Which of the following statements concerning this process are correct?
(1) A gas that turns moist red litmus paper blue is liberated at the cathode. (2) A gas that bleaches moist blue litmus paper is liberated at the anode. (3) The pH of the solution near the cathode increases as the electrolysis proceeds.
A.(1) and (2) only
B.(1) and (3) only
C.(1) only
D.(2) and (3) only
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解題
(1) is incorrect: At the cathode, \(\text{H}^+(aq)\) ions are preferentially discharged: \(2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\). Hydrogen gas is neutral and does not change the colour of moist litmus paper. (2) is correct: At the anode, \(\text{Cl}^-(aq)\) is preferentially discharged due to its high concentration: \(2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-\). Chlorine gas turns moist blue litmus paper red and then rapidly bleaches it white. (3) is correct: As \(\text{H}^+(aq)\) is reduced at the cathode, \(\text{OH}^-(aq)\) accumulates in the cathode compartment, resulting in a higher \([\text{OH}^-]\) and an increase in pH.
評分準則
D (1 mark): Recognises that (2) and (3) are correct, while (1) is incorrect because hydrogen gas is neutral.
題目 20 · 選擇題
1 分
Consider the following dynamic equilibrium established in a closed container:
Which of the following changes would increase the equilibrium yield of \(\text{Cl}_2(g)\)?
(1) Increasing the temperature of the reaction vessel. (2) Increasing the volume of the reaction vessel at constant temperature. (3) Adding a suitable solid catalyst to the reaction mixture.
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: The forward reaction is endothermic (\(\Delta H > 0\)). According to Le Chatelier's principle, increasing the temperature shifts the equilibrium position to the right (endothermic direction), increasing the yield of \(\text{Cl}_2(g)\). (2) is correct: Increasing the volume decreases the total pressure. The system shifts to the side with a greater number of moles of gas (forward direction: 1 mole of gas \(\rightarrow\) 2 moles of gas) to counteract the decrease in pressure, thus increasing the equilibrium yield of \(\text{Cl}_2(g)\). (3) is incorrect: A catalyst increases the rates of the forward and reverse reactions equally, reducing the time to reach equilibrium without altering the equilibrium position or yield.
評分準則
A (1 mark): Identifies that both an increase in temperature and an increase in volume shift the equilibrium to the product side, while a catalyst has no effect on equilibrium yield.
(1) Both W and Y can turn acidified potassium dichromate solution from orange to green. (2) Only Y can decolourize bromine in 1,1,1-trichloroethane in the dark. (3) Tollens' reagent can be used to distinguish W from X.
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: W (propanal) is an aldehyde and is oxidized to propanoic acid; Y (prop-2-en-1-ol) has a primary alcohol group and is oxidized to an aldehyde/carboxylic acid. Both reduce orange \(\text{Cr}_2\text{O}_7^{2-}\text{(aq)}\) to green \(\text{Cr}^{3+}\text{(aq)}\). (2) is correct: Y contains a carbon-carbon double bond (\(\text{C}=\text{C}\)) and undergoes electrophilic addition with \(\text{Br}_2\) in organic solvent in the dark. W and X lack carbon-carbon multiple bonds and do not react with bromine in the dark without a catalyst. (3) is correct: W (an aldehyde) is oxidized by Tollens' reagent to give a silver mirror / grey precipitate of silver, whereas X (a ketone) does not react.
評分準則
Award 1 mark for the correct option (D). - Statement (1) evaluates oxidation of aldehydes and primary alcohols. - Statement (2) evaluates addition reaction of alkenes vs carbonyls. - Statement (3) evaluates test to distinguish aldehydes from ketones.
題目 22 · 選擇題
1 分
Given the following standard enthalpy changes of combustion:
What is the standard enthalpy change of formation of methanol, \(\Delta H_{\text{f}}^\ominus [\text{CH}_3\text{OH(l)}]\)?
A.\(-239.1\text{ kJ mol}^{-1}\)
B.\(+239.1\text{ kJ mol}^{-1}\)
C.\(-46.7\text{ kJ mol}^{-1}\)
D.\(-1405.3\text{ kJ mol}^{-1}\)
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解題
The target reaction for the standard enthalpy of formation of methanol is: \(\text{C(graphite)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\)
Award 1 mark for the correct option (A). - Distractor B (+239.1) involves a sign error. - Distractor C (-46.7) omits the stoichiometric factor of 2 for \(\text{H}_2\). - Distractor D (-1405.3) incorrectly adds all enthalpy values without subtracting the combustion of the product.
題目 23 · 選擇題
1 分
Which of the following molecules is planar in shape and possesses a permanent dipole moment?
A.\(\text{BF}_3\)
B.\(\text{CF}_4\)
C.\(\text{HCHO}\)
D.\(\text{NH}_3\)
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解題
- \(\text{BF}_3\): Central boron has 3 bond pairs and 0 lone pairs. It is trigonal planar; bond dipoles cancel out, so net dipole = 0 (non-polar). - \(\text{CF}_4\): Central carbon has 4 bond pairs and 0 lone pairs. It is tetrahedral (non-planar) and non-polar. - \(\text{HCHO}\): Central carbon has 3 electron domains (2 C-H single bonds, 1 C=O double bond) with no lone pairs. The geometry is trigonal planar. Because the C=O bond dipole is significantly larger than the C-H bond dipoles, the bond dipoles do not cancel out, giving a permanent net dipole moment. - \(\text{NH}_3\): Central nitrogen has 3 bond pairs and 1 lone pair. The geometry is trigonal pyramidal (non-planar).
評分準則
Award 1 mark for the correct option (C). Requires candidates to evaluate both the molecular geometry (VSEPR theory) and molecular polarity (vector sum of bond dipoles).
題目 24 · 選擇題
1 分
Electrolysis of concentrated sodium chloride solution, \(\text{NaCl(aq)}\), is carried out using graphite electrodes.
Which of the following statements concerning this process is/are correct?
(1) A gas that turns moist blue litmus paper red and then white is evolved at the anode. (2) The pH of the solution in the cathode compartment increases as electrolysis proceeds. (3) If the graphite cathode is replaced by a copper cathode, the cathode dissolves to form \(\text{Cu}^{2+}\text{(aq)}\).
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: At the anode, \(\text{Cl}^-\text{(aq)}\) is preferentially discharged due to its high concentration: \(2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\). \(\text{Cl}_2\) gas is acidic and bleaching; it turns moist blue litmus paper red and then bleaches it white. (2) is correct: At the cathode, \(\text{H}^+\text{(aq)}\) / \(\text{H}_2\text{O(l)}\) is discharged: \(2\text{H}_2\text{O(l)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)}\). The accumulation of \(\text{OH}^-\text{(aq)}\) increases the solution's pH. (3) is incorrect: Reduction occurs at the cathode. A copper cathode does not dissolve (metal oxidation only occurs at an active anode).
評分準則
Award 1 mark for the correct option (B). - Statement (1) tests product at the anode and chlorine gas testing. - Statement (2) tests product at the cathode and changes in [OH-]. - Statement (3) tests the concept that cathode undergoes reduction, not oxidation/dissolution.
題目 25 · 選擇題
1 分
Directions: This question consists of two separate statements. Decide whether each of the two statements is true or false; if both are true, decide whether the 2nd statement is a correct explanation of the 1st statement.
1st statement: Nylon-6,6 has a higher melting point than poly(propene) of similar relative molecular mass.
2nd statement: Extensive intermolecular hydrogen bonds exist between polymer chains in nylon-6,6, whereas only weak Van der Waals' forces exist between polymer chains in poly(propene).
A.Both statements are true and the 2nd statement is a correct explanation of the 1st statement.
B.Both statements are true but the 2nd statement is NOT a correct explanation of the 1st statement.
C.The 1st statement is true, but the 2nd statement is false.
D.The 1st statement is false, but the 2nd statement is true.
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解題
- The 1st statement is true: Nylon-6,6 melts at approximately 260 °C, which is significantly higher than that of poly(propene) (~160 °C). - The 2nd statement is true: Nylon-6,6 contains polar amide linkages (\(-\text{CONH}-\)) that allow the formation of strong intermolecular hydrogen bonds between the \(>\text{C}=\text{O}\) and \(-\text{NH}-\) groups of adjacent chains. Poly(propene) is a non-polar hydrocarbon with only weak dispersion forces between chains. - The 2nd statement provides the correct explanation for the 1st statement because stronger intermolecular hydrogen bonds require significantly more thermal energy to break during melting.
評分準則
Award 1 mark for the correct option (A). - Identifies the presence of intermolecular hydrogen bonds in polyamides vs Van der Waals' forces in polyalkenes. - Relates intermolecular force strength directly to melting point differences.
題目 26 · multiple_choice
1 分
Compound \(X\) has the structural formula \(\text{CH}_3\text{CH(OH)CH}_2\text{CHO}\).
Which of the following statements concerning Compound \(X\) is / are correct?
(1) It turns acidified potassium dichromate solution from orange to green. (2) It reacts with sodium borohydride (\(\text{NaBH}_4\)) to form butane-1,3-diol. (3) It undergoes condensation polymerisation with itself.
A. (1) and (2) only B. (1) and (3) only C. (2) and (3) only D. (1), (2) and (3)
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
Statement (1) is correct: Compound \(X\) contains a secondary alcohol group (\(-\text{CH(OH)}-\)) and an aldehyde group (\(-\text{CHO}\)), both of which are readily oxidised by acidified \(\text{K}_2\text{Cr}_2\text{O}_7\), reducing orange \(\text{Cr}_2\text{O}_7^{2-}\) to green \(\text{Cr}^{3+}\).
Statement (2) is correct: \(\text{NaBH}_4\) acts as a reducing agent that reduces the aldehyde group (\(-\text{CHO}\)) to a primary alcohol group (\(-\text{CH}_2\text{OH}\)), giving \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{OH}\), which is butane-1,3-diol.
Statement (3) is incorrect: Condensation polymerisation requires functional groups such as \(-\text{COOH}\) and \(-\text{OH}\), or \(-\text{COOH}\) and \(-\text{NH}_2\). An alcohol and an aldehyde do not undergo condensation polymerisation with each other to yield a continuous polymer backbone.
評分準則
A (1 mark): Correct identification that only statements (1) and (2) are correct.
題目 27 · multiple_choice
1 分
Consider the following thermochemical equations at \(298\text{ K}\):
What is the standard enthalpy change of combustion of ethanol, \(\text{C}_2\text{H}_5\text{OH(l)}\), in \(\text{kJ mol}^{-1}\)?
A.\(-1366.8\)
B.\(-965.7\)
C.\(+1366.8\)
D.\(-1922.0\)
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解題
The target combustion reaction of ethanol is: \(\text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}\)
A. (1) only B. (2) only C. (1) and (3) only D. (2) and (3) only
A.(1) only
B.(2) only
C.(1) and (3) only
D.(2) and (3) only
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解題
(1) \(\text{O}_3\): The central oxygen atom has two bonded atoms and one lone pair of electrons (3 electron domains), giving a bent/non-linear shape (bond angle \(\approx 117^\circ\)).
(2) \(\text{NO}_2^+\): The central nitrogen atom has two double bonds and no lone pair (2 electron domains), giving a linear shape (bond angle \(180^\circ\)).
(3) \(\text{SO}_2\): The central sulfur atom has two bonded oxygen atoms and one lone pair (3 electron domains), giving a bent/non-linear shape (bond angle \(\approx 119^\circ\)).
Thus, (1) and (3) are non-linear.
評分準則
C (1 mark): Correct deduction of molecular shapes using VSEPR theory.
題目 29 · multiple_choice
1 分
During the electrolysis of concentrated sodium chloride solution using graphite electrodes, which of the following statements is correct?
A.Sodium metal is deposited at the cathode because sodium ions are present in a higher concentration than hydrogen ions.
B.Oxygen gas is evolved at the anode because hydroxide ions are preferentially discharged over chloride ions.
C.The solution around the cathode becomes alkaline because \(\text{H}^+(aq)\) ions are discharged, leaving an excess of \(\text{OH}^-(aq)\) ions.
D.The colour of the solution gradually turns pale green due to the accumulation of dissolved \(\text{Cl}_2(g)\) in the electrolyte.
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解題
At the cathode, \(\text{H}^+(aq)\) ions from water are preferentially discharged over \(\text{Na}^+(aq)\) ions because \(\text{H}^+(aq)\) is a stronger oxidising agent: \(2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\) As \(\text{H}^+(aq)\) ions are consumed, the self-ionisation equilibrium of water produces an excess of \(\text{OH}^-(aq)\) ions around the cathode, making the region alkaline.
Option A is incorrect because \(\text{H}_2(g)\) is formed, not sodium metal. Option B is incorrect because in concentrated \(\text{NaCl}(aq)\), \(\text{Cl}^-(aq)\) is preferentially discharged due to its high concentration, forming \(\text{Cl}_2(g)\). Option D is incorrect as \(\text{Cl}_2(g)\) is released at the anode rather than remaining solely to turn the bulk electrolyte pale green.
評分準則
C (1 mark): Correct understanding of preferential discharge and ionic balance at the cathode during brine electrolysis.
題目 30 · multiple_choice
1 分
Consider the following reversible reaction at equilibrium in a closed vessel of fixed volume:
\(2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)} \quad \Delta H < 0\)
Which of the following changes will result in an increase in the number of moles of \(\text{N}_2\text{O}_4\text{(g)}\)?
(1) Decreasing the temperature of the reaction mixture (2) Adding helium gas into the vessel at constant volume (3) Decreasing the volume of the reaction vessel at constant temperature
A. (1) and (2) only B. (1) and (3) only C. (2) and (3) only D. (1), (2) and (3)
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) Decreasing the temperature shifts the equilibrium position in the exothermic direction (to the right, \(\Delta H < 0\)) to release heat, thereby increasing the number of moles of \(\text{N}_2\text{O}_4(g)\).
(2) Adding an inert gas (He) at constant volume increases the total pressure but does not change the partial pressures or molar concentrations of \(\text{NO}_2(g)\) or \(\text{N}_2\text{O}_4(g)\). Thus, the equilibrium position does not shift and the moles of \(\text{N}_2\text{O}_4(g)\) remain unchanged.
(3) Decreasing the vessel volume increases the total pressure. According to Le Chatelier's Principle, the equilibrium shifts towards the side with fewer moles of gas (the right-hand side, where 2 moles of gas become 1 mole of gas), increasing the number of moles of \(\text{N}_2\text{O}_4(g)\).
Therefore, (1) and (3) only are correct.
評分準則
B (1 mark): Correct application of Le Chatelier's Principle for temperature, volume/pressure, and inert gas additions.
By Hess's Law, the target equation is: \[ \text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)} \]
This can be constructed by combining the reactions as follows: \[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_3 - \Delta H_2 \] \[ \Delta H_{\text{rxn}} = (-1411) + (-286) - (-1560) = -1697 + 1560 = -137\text{ kJ mol}^{-1} \]
評分準則
Correct Answer: A (1 mark) - Award 1 mark for calculating \(\Delta H = \Delta H_1 + \Delta H_3 - \Delta H_2 = -137\text{ kJ mol}^{-1}\).
題目 32 · 選擇題
1 分
A haloalkane \(\text{X}\) (\(\text{C}_4\text{H}_9\text{Br}\)) undergoes elimination upon heating with concentrated ethanolic potassium hydroxide solution to produce an alkene \(\text{Y}\) (\(\text{C}_4\text{H}_8\)) as the major product. The alkene \(\text{Y}\) does NOT exhibit cis-trans isomerism.
Which of the following compounds could be \(\text{X}\)?
Let us examine each compound: (1) 1-bromobutane undergoes elimination to form but-1-ene (\(\text{CH}_2=\text{CHCH}_2\text{CH}_3\)). But-1-ene has two hydrogen atoms attached to one of the double-bonded carbon atoms, so it does not exhibit cis-trans isomerism. (2) 2-bromobutane undergoes elimination to give but-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_3\)) as the major product (Zaitsev's rule). But-2-ene exhibits cis-trans isomerism. (3) 2-bromo-2-methylpropane undergoes elimination to form 2-methylpropene (\(\text{CH}_2=\text{C(CH}_3)_2\)). It has two identical groups on each doubly-bonded carbon atom, so it does not exhibit cis-trans isomerism.
Hence, (1) and (3) only are possible structures.
評分準則
Correct Answer: C (1 mark) - (1) is correct: produces but-1-ene (no stereoisomers). - (2) is incorrect: major product is but-2-ene, which has cis-trans isomers. - (3) is correct: produces 2-methylpropene (no stereoisomers).
題目 33 · 選擇題
1 分
Which of the following statements concerning poly(ethylene terephthalate) (PET) is/are correct?
(1) It is formed by an addition polymerisation between ethane-1,2-diol and benzene-1,4-dicarboxylic acid. (2) It contains ester linkages in its repeating unit. (3) It can be hydrolysed by heating under reflux with dilute sodium hydroxide solution.
A.(1) only
B.(2) only
C.(1) and (3) only
D.(2) and (3) only
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解題
(1) is incorrect because PET is a polyester formed by condensation polymerisation with the elimination of small molecules (water), not addition polymerisation. (2) is correct because PET contains ester functional groups (\(-\text{COO}-\)) linking the monomer units. (3) is correct because polyesters can undergo alkaline hydrolysis (saponification) when heated under reflux with aqueous base.
評分準則
Correct Answer: D (1 mark) - (1) is incorrect (condensation polymerisation, not addition). - (2) is correct (contains ester linkages). - (3) is correct (ester links are susceptible to alkaline hydrolysis).
題目 34 · 選擇題
1 分
Which of the following statements concerning nitrogen trichloride (\(\text{NCl}_3\)) and boron trichloride (\(\text{BCl}_3\)) is/are correct?
(1) The central atom in \(\text{NCl}_3\) has one lone pair of electrons, whereas the central atom in \(\text{BCl}_3\) has no lone pair of electrons. (2) \(\text{NCl}_3\) has a trigonal pyramidal shape, whereas \(\text{BCl}_3\) has a trigonal planar shape. (3) Both \(\text{NCl}_3\) and \(\text{BCl}_3\) are polar molecules.
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: Nitrogen (Group 15) has 5 valence electrons, forming 3 bonding pairs and retaining 1 lone pair in \(\text{NCl}_3\). Boron (Group 13) has 3 valence electrons, all used in bonding with chlorine, leaving 0 lone pairs in \(\text{BCl}_3\). (2) is correct: According to VSEPR theory, 3 bonding pairs and 1 lone pair give \(\text{NCl}_3\) a trigonal pyramidal shape. 3 bonding pairs with 0 lone pairs give \(\text{BCl}_3\) a trigonal planar shape. (3) is incorrect: \(\text{NCl}_3\) is polar because its asymmetric trigonal pyramidal geometry leads to a non-zero net dipole moment. \(\text{BCl}_3\) is non-polar because the three polar \(\text{B}-\text{Cl}\) bonds are arranged symmetrically at \(120^\circ\), causing the individual bond dipoles to cancel each other out.
評分準則
Correct Answer: A (1 mark) - (1) is correct (N has 1 lone pair, B has 0). - (2) is correct (VSEPR geometries: trigonal pyramidal vs trigonal planar). - (3) is incorrect (\(\text{BCl}_3\) is non-polar due to symmetrical shape).
題目 35 · 選擇題
1 分
An electrolytic cell is set up to electrolyse concentrated sodium chloride solution (brine) using inert graphite electrodes.
Which of the following statements is/are correct?
(1) A yellowish-green gas that bleaches moist litmus paper is evolved at the anode. (2) Hydrogen gas is evolved at the cathode. (3) The pH of the solution around the cathode increases during the electrolysis.
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解題
(1) is correct: In concentrated \(\text{NaCl(aq)}\), \(\text{Cl}^-\text{(aq)}\) ions are at high concentration and are preferentially discharged over \(\text{OH}^-\text{(aq)}\) at the anode, releasing chlorine gas (\(\text{Cl}_2\text{(g)}\)), which turns moist blue litmus paper red and then bleaches it white. (2) is correct: At the cathode, \(\text{H}^+\text{(aq)}\) ions are discharged in preference to \(\text{Na}^+\text{(aq)}\) ions because hydrogen has a more positive reduction potential: \(2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}\) (or \(2\text{H}_2\text{O(l)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)}\)). (3) is correct: As \(\text{H}^+\text{(aq)}\) ions are removed from the solution at the cathode, an excess of hydroxide ions (\(\text{OH}^-\text{(aq)}\)) remains, forming \(\text{NaOH(aq)}\) and causing the pH to rise.
評分準則
Correct Answer: D (1 mark) - (1) is correct: \(\text{Cl}_2\text{(g)}\) is liberated at the anode due to concentration effect. - (2) is correct: \(\text{H}_2\text{(g)}\) is liberated at the cathode. - (3) is correct: \(\text{OH}^-\text{(aq)}\) accumulates near cathode, raising pH.
題目 36 · 選擇題
1 分
Which of the following reaction pathways would yield propanoic acid as the major organic product?
A.Heating 1-chloropropane with aqueous sodium hydroxide, followed by heating the resulting product with acidified potassium dichromate solution under reflux
B.Heating propan-2-ol with acidified potassium dichromate solution under reflux
C.Heating ethyl methanoate with dilute sulfuric acid under reflux
D.Reacting propene with steam in the presence of concentrated phosphoric acid at high temperature and pressure, followed by heating the major product with acidified potassium permanganate solution
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解題
In pathway A, heating 1-chloropropane with aqueous sodium hydroxide results in nucleophilic substitution to form propan-1-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)). Subsequent heating of propan-1-ol with acidified potassium dichromate solution oxidises the primary alcohol completely to propanoic acid (\( \text{CH}_3\text{CH}_2\text{COOH} \)).
In pathway B, propan-2-ol is a secondary alcohol, which is oxidised by acidified potassium dichromate solution to propanone (a ketone), not propanoic acid.
In pathway C, acid-catalysed hydrolysis of ethyl methanoate yields methanoic acid and ethanol.
In pathway D, the electrophilic addition of steam to propene in the presence of concentrated phosphoric acid follows Markovnikov's rule, yielding propan-2-ol as the major product. Subsequent oxidation of propan-2-ol produces propanone.
評分準則
A (1 mark)
Assessment point: Understanding substitution, addition, hydrolysis, and oxidation reactions of functional groups to synthesise carboxylic acids.
A student conducted a simple calorimetry experiment to determine the standard enthalpy change of combustion of propan-1-ol (\(\text{C}_3\text{H}_7\text{OH}\)).
(a) Write a balanced chemical equation for the complete combustion of propan-1-ol. (1 mark)
(b) In the experiment, \(0.552\text{ g}\) of propan-1-ol was burned completely to heat \(150.0\text{ g}\) of water in a copper beaker. The temperature of the water increased from \(21.4\,^\circ\text{C}\) to \(48.2\,^\circ\text{C}\). (Specific heat capacity of water \(= 4.18\text{ J}\,\text{g}^{-1}\,\text{K}^{-1}\); Molar mass of propan-1-ol \(= 60.0\text{ g}\,\text{mol}^{-1}\)) (i) Calculate the heat energy released in this reaction. (1 mark) (ii) Calculate the experimental standard enthalpy change of combustion of propan-1-ol in \(\text{kJ}\,\text{mol}^{-1}\). (2 marks)
(c) The theoretical value for the standard enthalpy change of combustion of propan-1-ol is \(-2021\text{ kJ}\,\text{mol}^{-1}\). State TWO major reasons why the experimental value obtained is significantly less negative than the theoretical value, assuming incomplete combustion did not occur. (2 marks)
(b)(ii) Number of moles of propan-1-ol \(n = \frac{0.552\text{ g}}{60.0\text{ g}\,\text{mol}^{-1}} = 0.00920\text{ mol}\). \(\Delta H_c = -\frac{Q}{n} = -\frac{16.8036\text{ kJ}}{0.00920\text{ mol}} = -1826.48\text{ kJ}\,\text{mol}^{-1} \approx -1830\text{ kJ}\,\text{mol}^{-1}\) (3 sig. fig.)
(c) 1. Significant heat loss to the surrounding air/environment. 2. Heat capacity of the copper container/calorimeter and thermometer was not taken into account (heat absorbed by the apparatus).
(b)(ii) Calculation of moles: \(n = 0.00920\text{ mol}\) (1* mark); Final answer: \(-1830\text{ kJ}\,\text{mol}^{-1}\) (accept \(-1826\) to \(-1830\)) with correct negative sign and units (1 mark).
(c) Any TWO reasonable sources of error (1 mark each, max 2 marks): - Heat lost to the surroundings / no draft shield or lid used. - Heat absorbed by the copper calorimeter / thermometer / tripod. - Evaporation of propan-1-ol from the wick before/after burning. (Do NOT accept 'incomplete combustion' as the stem excludes it).
題目 2 · Structured
6 分
Compound \(\mathbf{X}\) is an acyclic ester with the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\). When \(\mathbf{X}\) is refluxed with dilute sodium hydroxide solution, two organic products, \(\mathbf{Y}\) (a sodium carboxylate) and \(\mathbf{Z}\) (an alcohol), are formed.
When \(\mathbf{Z}\) is gently heated with acidified potassium dichromate solution, it turns the solution from orange to green and produces an aldehyde \(\mathbf{W}\). When \(\mathbf{Y}\) is acidified, it gives ethanoic acid.
(a) Deduce the structural formula and systematic name of alcohol \(\mathbf{Z}\). (2 marks)
(b) Draw the structure of ester \(\mathbf{X}\). (1 mark)
(c) State the type of reaction that occurs when \(\mathbf{X}\) is converted into \(\mathbf{Y}\) and \(\mathbf{Z}\). (1 mark)
(d) Describe a chemical test to distinguish between \(\mathbf{W}\) and propanone, stating the reagents, conditions, and expected observations for both compounds. (2 marks)
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解題
(a) Since \(\mathbf{Y}\) gives ethanoic acid (2 carbons), alcohol \(\mathbf{Z}\) must have \(5 - 2 = 3\) carbon atoms. \(\mathbf{Z}\) oxidises to an aldehyde \(\mathbf{W}\), so \(\mathbf{Z}\) must be a primary alcohol. Therefore, \(\mathbf{Z}\) is propan-1-ol, with structural formula \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\).
(b) \(\mathbf{X}\) is propyl ethanoate: \(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\).
(d) Reagent and conditions: Warm with Tollens' reagent / Fehling's solution / Benedict's solution (1 mark); Observations: \(\mathbf{W}\) gives silver mirror (or brick-red precipitate) AND propanone gives no observable change / remains blue (1 mark).
題目 3 · Structured
6 分
Electrolysis of concentrated sodium chloride solution (brine) using inert titanium/carbon electrodes is an important industrial process.
(a) Write the ionic half-equation for the reaction occurring at the anode. (1 mark)
(b) State the observation at the cathode during electrolysis and explain why hydrogen gas is liberated instead of sodium metal. (2 marks)
(c) As the electrolysis proceeds, the remaining electrolyte solution gradually turns strongly alkaline. (i) Explain why the solution becomes alkaline in terms of ions present and consumed. (1 mark) (ii) Suggest ONE large-scale industrial use for the alkaline substance formed in this process. (1 mark)
(d) Describe a safe chemical test to confirm the presence of the gas produced at the anode. (1 mark)
(b) Observation: Colourless gas bubbles are evolved at the cathode. Explanation: \(\text{H}^+\text{(aq)}\) ions have a higher reduction potential (are more readily reduced / lower in the electrochemical series) than \(\text{Na}^+\text{(aq)}\) ions.
(c)(i) Water auto-ionises into \(\text{H}^+\text{(aq)}\) and \(\text{OH}^-\text{(aq)}\). As \(\text{H}^+\text{(aq)}\) ions are discharged at the cathode, \(\text{OH}^-\text{(aq)}\) ions remain in the solution in excess along with \(\text{Na}^+\text{(aq)}\), producing \(\text{NaOH(aq)}\). (ii) Manufacture of soap / paper making / drain cleaner / synthesis of bleach.
(d) Hold moist blue litmus paper near the gas; it turns red and then quickly bleaches (turns white) (or moist starch-iodide paper turns blue-black).
(b) Observation: Effervescence / colourless gas bubbles (1 mark); Explanation: \(\text{H}^+\) is a stronger oxidising agent / more easily reduced / lower in electrochemical series than \(\text{Na}^+\) (1 mark).
(c)(i) \(\text{H}^+\) ions are removed, leaving an excess of \(\text{OH}^-\) ions in solution (1 mark). (c)(ii) Making soap / paper manufacture / drain cleaners / detergent / making bleach (1 mark).
(d) Moist blue litmus paper turns red then bleaches white / moist starch-iodide paper turns dark blue (1 mark).
題目 4 · Structured
6 分
Kevlar is a high-strength polyamide synthesized from benzene-1,4-dicarboxylic acid and 1,4-diaminobenzene.
(a) Draw the structural formula of the repeating unit of Kevlar. (1 mark)
(b) State the type of polymerisation involved in the formation of Kevlar and name the small molecule eliminated during this reaction. (2 marks)
(c) Explain, in terms of intermolecular forces, why Kevlar exhibits exceptionally high tensile strength and thermal stability. (2 marks)
(d) State ONE everyday application of Kevlar that takes advantage of its high tensile strength. (1 mark)
(b) Type of polymerisation: Condensation polymerisation. Small molecule eliminated: Water (\(\text{H}_2\text{O}\)).
(c) Kevlar polymer chains are rigid and planar due to the benzene rings, allowing close parallel alignment. Extensive, strong hydrogen bonds form between the carbonyl groups (\(\text{C}=\text{O}\)) and amine groups (\(\text{N}-\text{H}\)) of adjacent polymer chains, requiring a very large amount of mechanical force and thermal energy to separate them.
(a) Correct structure of repeating unit showing open bond ends and correct amide linkages (1 mark).
(b) Condensation polymerisation (1 mark); Water / \(\text{H}_2\text{O}\) (1 mark).
(c) Extensive / strong hydrogen bonding between \(\text{C}=\text{O}\) and \(\text{N}-\text{H}\) groups on adjacent chains (1 mark); Chains are linear/planar and pack closely/rigidly together (1 mark).
Consider the two molecules phosphorus trifluoride (\(\text{PF}_3\)) and boron trifluoride (\(\text{BF}_3\)).
(a) Draw the electron diagram (showing ELECTRONS IN THE OUTERMOST SHELLS only) for a molecule of \(\text{PF}_3\). (1 mark)
(b) State the shape of each molecule and state their respective bond angles. (i) \(\text{PF}_3\) (2 marks) (ii) \(\text{BF}_3\) (2 marks)
(c) Explain whether \(\text{PF}_3\) is a polar molecule. (1 mark)
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解題
(a) In \(\text{PF}_3\), phosphorus (group 15) has 5 valence electrons. It forms 3 single covalent bonds with three fluorine atoms and retains 1 lone pair. Each fluorine atom has 3 lone pairs.
(b)(i) \(\text{PF}_3\): There are 3 bonding pairs and 1 lone pair around P. Electron pairs repel to minimize repulsion. Shape: Trigonal pyramidal. Bond angle: \(107^\circ\) (or between \(100^\circ\) and \(108^\circ\)).
(b)(ii) \(\text{BF}_3\): Boron has 3 bonding pairs and 0 lone pairs. Shape: Trigonal planar. Bond angle: \(120^\circ\).
(c) Yes, \(\text{PF}_3\) is polar. The \(\text{P}-\text{F}\) bonds are polar due to the electronegativity difference between P and F. Because the molecule has a non-symmetrical trigonal pyramidal geometry, the bond dipoles do not cancel each other out, resulting in a net molecular dipole moment.
評分準則
(a) Correct electron diagram with outer shell electrons: P with 1 lone pair and 3 bonding pairs, each F with 3 lone pairs (1 mark).
(c) Polar, because it is non-symmetrical (or dipoles do not cancel out) (1 mark).
題目 6 · Short / Long 結構題
6 分
A student carried out a spirit burner experiment to determine the standard enthalpy change of combustion of propan-1-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)).
(a) Write a balanced chemical equation for the complete combustion of propan-1-ol. (1 mark)
(b) In the experiment, \( 0.450\text{ g} \) of propan-1-ol was burned to raise the temperature of \( 150.0\text{ g} \) of water in a copper calorimeter from \( 21.5\text{ }^\circ\text{C} \) to \( 45.2\text{ }^\circ\text{C} \). (i) Calculate the experimental enthalpy change of combustion of propan-1-ol in \( \text{kJ mol}^{-1} \). (Given: Specific heat capacity of water = \( 4.18\text{ J g}^{-1}\text{ K}^{-1} \); Molar mass of propan-1-ol = \( 60.1\text{ g mol}^{-1} \)) (2 marks) (ii) The theoretical literature value of the enthalpy change of combustion of propan-1-ol is \( -2021\text{ kJ mol}^{-1} \). State TWO experimental reasons why the experimental value is significantly less exothermic than the theoretical value. (2 marks)
(c) Suggest ONE modification to the experimental setup to improve the accuracy of the result. (1 mark)
(b)(i) Heat released, \( Q = mc\Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (45.2 - 21.5)\text{ K} = 14860.1\text{ J} = 14.86\text{ kJ} \) Number of moles of propan-1-ol burned, \( n = \frac{0.450\text{ g}}{60.1\text{ g mol}^{-1}} = 7.488 \times 10^{-3}\text{ mol} \) \( \Delta H_{\text{c}} = -\frac{Q}{n} = -\frac{14.8601\text{ kJ}}{7.488 \times 10^{-3}\text{ mol}} = -1984.6\text{ kJ mol}^{-1} \approx -1985\text{ kJ mol}^{-1} \) (or \( -1.98 \times 10^3\text{ kJ mol}^{-1} \) to 3 sig. fig.)
(b)(ii) 1. Significant heat loss to the surrounding air and apparatus. 2. Incomplete combustion of propan-1-ol (forming soot / carbon monoxide). (Also accept: Heat absorbed by the copper calorimeter was not included in the calculation; evaporation of propan-1-ol from the wick).
(c) Use a draught shield around the flame/calorimeter to minimise heat loss to the surroundings / cover the beaker with an insulating lid / use a bomb calorimeter with excess oxygen.
(b)(i) Calculation of \( Q \) and moles \( n \): \( Q = 14.86\text{ kJ} \) and \( n = 0.007488\text{ mol} \) (1*) Correct value with negative sign and units: \( -1985\text{ kJ mol}^{-1} \) (accept \( -1980 \text{ to } -1990\text{ kJ mol}^{-1} \)) (1)
(b)(ii) Any TWO of the following (1 mark each, max 2): - Heat loss to the surroundings - Incomplete combustion of the fuel / propan-1-ol - Heat capacity of the calorimeter was not accounted for - Evaporation of alcohol from the wick between weighing [Do not accept: inaccurate thermometer / experimental error without explanation]
(c) Any ONE of the following (1 mark): - Place a draught shield / screen around the burner - Use a lid on the calorimeter - Use a bomb calorimeter - Minimise the distance between the flame and the calorimeter
題目 7 · Short / Long 結構題
7 分
Consider the following synthetic route starting from propene (\( \text{CH}_3\text{CH}=\text{CH}_2 \)):
(a) State the name or formula of Reagent X and the reaction condition required. (1 mark)
(b) (i) Identify Reagent Y and the condition required for the conversion of 2-bromopropane to propan-2-ol. (1 mark) (ii) State the type of reaction taking place in this conversion. (1 mark)
(c) Compound W is formed when propan-2-ol is heated under reflux with acidified potassium dichromate solution (Reagent Z). (i) Give the IUPAC name and draw the structural formula of Compound W. (2 marks) (ii) State the colour change observed in the reaction mixture during this oxidation. (1 mark)
(d) Propan-2-ol reacts with ethanoic acid in the presence of concentrated sulfuric acid as a catalyst. Write the IUPAC name of the organic product formed. (1 mark)
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解題
(a) Reagent X is hydrogen bromide (\( \text{HBr} \)) or \( \text{HBr}(\text{g}) \) at room temperature. (b)(i) Reagent Y is dilute aqueous sodium hydroxide (\( \text{NaOH}(\text{aq}) \)) or aqueous potassium hydroxide (\( \text{KOH}(\text{aq}) \)), heated under reflux. (b)(ii) The reaction is a nucleophilic substitution (or substitution / hydrolysis). (c)(i) IUPAC name: Propanone (or propan-2-one). Structural formula: \( \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 \). (c)(ii) The colour changes from orange (due to \( \text{Cr}_2\text{O}_7^{2-} \)) to green (due to \( \text{Cr}^{3+} \)). (d) The esterification reaction produces 1-methylethyl ethanoate (or propan-2-yl ethanoate / isopropyl ethanoate).
(b)(i) \( \text{NaOH}(\text{aq}) \) / \( \text{KOH}(\text{aq}) \) AND heat / reflux (1) [Reject if 'ethanolic' or 'alcohol solvent' is stated] (b)(ii) Nucleophilic substitution / Substitution / Alkaline hydrolysis (1)
(c)(i) Propanone (1) Correct structural formula showing \( \text{C}=\text{O} \) carbonyl group (\( \text{CH}_3\text{COCH}_3 \)) (1) (c)(ii) Orange to green (1) [Reject: yellow to green]
Boron trifluoride (\( \text{BF}_3 \)) and nitrogen trifluoride (\( \text{NF}_3 \)) are both covalent fluoride molecules.
(a) Draw the electron diagram of \( \text{NF}_3 \), showing electrons in the outermost shells only. (1 mark)
(b) (i) Predict the shape of the \( \text{BF}_3 \) molecule and state its bond angle. (2 marks) (ii) In terms of the Valence Shell Electron Pair Repulsion (VSEPR) theory, explain why the shape of \( \text{NF}_3 \) is trigonal pyramidal rather than trigonal planar. (2 marks)
(c) Explain why \( \text{BF}_3 \) is a non-polar molecule, whereas \( \text{NF}_3 \) is a polar molecule. (1 mark)
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解題
(a) In \( \text{NF}_3 \), nitrogen has 5 valence electrons and forms 3 single covalent bonds with fluorine atoms, leaving 1 lone pair. Each fluorine has 7 valence electrons (1 bonding electron, 3 lone pairs).
(b)(i) Shape: Trigonal planar. Bond angle: \( 120^\circ \).
(b)(ii) The central N atom in \( \text{NF}_3 \) has 4 electron pairs (3 bonding pairs and 1 lone pair) in its valence shell, which adopt a tetrahedral electron-pair geometry. Since lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the bonding pairs are pushed closer together, giving a trigonal pyramidal molecular geometry.
(c) \( \text{BF}_3 \) has a symmetrical trigonal planar shape, so the three polar \( \text{B}-\text{F} \) bond dipoles cancel each other out, resulting in zero net dipole moment. \( \text{NF}_3 \) is non-symmetrical (trigonal pyramidal), so the polar \( \text{N}-\text{F} \) bond dipoles do not cancel out, resulting in a net molecular dipole moment.
評分準則
(a) Correct electron diagram with 3 \( \text{N}-\text{F} \) single shared pairs, 1 lone pair on N, and 3 lone pairs (6 electrons) on each F atom (1)
(b)(i) Trigonal planar (1) AND \( 120^\circ \) (1)
(b)(ii) Central N atom has 3 bond pairs and 1 lone pair (4 electron pairs total) (1) Repulsion between lone pair and bond pair is greater than between bond pairs, resulting in a trigonal pyramidal shape (1)
(c) \( \text{BF}_3 \) is symmetrical so its bond dipoles cancel out, while \( \text{NF}_3 \) is unsymmetrical / has a net dipole moment because the dipoles do not cancel (1)
題目 9 · Short / Long 結構題
6 分
The reaction between marble chips (\( \text{CaCO}_3 \)) and dilute hydrochloric acid (\( \text{HCl} \)) is represented by the equation:
In Experiment 1, \( 2.00\text{ g} \) of excess marble chips was reacted with \( 50.0\text{ cm}^3 \) of \( 1.00\text{ mol dm}^{-3}\text{ HCl}(\text{aq}) \) at \( 25\text{ }^\circ\text{C} \).
(a) State how the rate of this reaction can be monitored continuously during the experiment. (1 mark)
(b) In Experiment 2, the experiment was repeated using \( 50.0\text{ cm}^3 \) of \( 2.00\text{ mol dm}^{-3}\text{ HCl}(\text{aq}) \) at \( 25\text{ }^\circ\text{C} \) with the same mass of excess marble chips. (i) In terms of collision theory, explain why the initial rate of reaction in Experiment 2 is higher than that in Experiment 1. (2 marks) (ii) Calculate the volume of \( \text{CO}_2(\text{g}) \) collected at room temperature and pressure (r.t.p.) in Experiment 1 upon complete reaction. (Molar volume of gas at r.t.p. = \( 24.0\text{ dm}^3\text{ mol}^{-1} \)) (2 marks)
(c) State the effect on the activation energy (\( E_a \)) of the reaction if the temperature is increased to \( 35\text{ }^\circ\text{C} \). (1 mark)
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解題
(a) Measure the volume of \( \text{CO}_2 \) produced over time using a gas syringe / inverted burette filled with water, OR record the loss in mass of the flask and reaction contents over time using an electronic balance.
(b)(i) In Experiment 2, the concentration of \( \text{HCl} \) (or \( \text{H}^+ \) ions) is higher, so there are more reactant particles per unit volume. This increases the collision frequency between \( \text{H}^+ \) ions and \( \text{CaCO}_3 \) surface, resulting in a higher frequency of effective collisions and thus a higher initial rate.
(b)(ii) Moles of \( \text{HCl} = 1.00\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0500\text{ mol} \). Since \( \text{CaCO}_3 \) is in excess, \( \text{HCl} \) is the limiting reactant. Moles of \( \text{CO}_2 \text{ formed} = \frac{0.0500\text{ mol}}{2} = 0.0250\text{ mol} \). Volume of \( \text{CO}_2 = 0.0250\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.600\text{ dm}^3 \) (or \( 600\text{ cm}^3 \)).
(c) The activation energy (\( E_a \)) remains unchanged / no effect (temperature only affects the average kinetic energy of particles, not the activation energy barrier).
評分準則
(a) Measure the volume of \( \text{CO}_2 \) gas produced at regular time intervals using a gas syringe / Measure the decrease in mass of the reaction system over time using an electronic balance (1)
(b)(i) Higher concentration means more \( \text{H}^+ \) particles / ions per unit volume (1) Collision frequency / frequency of effective collisions between particles increases (1)
Polylactic acid (PLA) is an environmentally friendly, biodegradable polymer made from lactic acid (2-hydroxypropanoic acid, \( \text{CH}_3\text{CH(OH)COOH} \)).
(a) Write the structural formula of the repeating unit of PLA. (1 mark)
(b) (i) State the type of polymerisation involved in the formation of PLA. (1 mark) (ii) Name the small molecule eliminated during this polymerisation. (1 mark)
(c) In terms of chemical structure and bonding, explain why PLA is biodegradable under industrial composting conditions, whereas polypropene is non-biodegradable. (2 marks)
(d) Lactic acid exhibits optical isomerism (enantiomerism). (i) State what is meant by a 'chiral carbon atom'. (1 mark) (ii) Draw the three-dimensional structures of the two enantiomers of lactic acid. (1 mark)
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解題
(a) The repeating unit of PLA is \( -[\text{O}-\text{CH}(\text{CH}_3)-\text{C}(=\text{O})]- \).
(b)(i) Condensation polymerisation. (b)(ii) Water (\( \text{H}_2\text{O} \)).
(c) PLA contains ester linkages (\( -\text{COO}- \)) in its polymer backbone, which are polar and susceptible to hydrolysis by water and enzymes secreted by microorganisms. In contrast, polypropene contains only strong, non-polar \( \text{C}-\text{C} \) and \( \text{C}-\text{H} \) bonds in its backbone, which cannot be easily attacked or hydrolysed by microorganisms.
(d)(i) A chiral carbon is a carbon atom attached to four different groups/atoms.
(d)(ii) In 3D representation (wedge-and-dash), C-2 is bonded to \( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{COOH} \), arranged tetrahedrally so that the two structures are non-superimposable mirror images of each other.
評分準則
(a) Correct structure of repeating unit: \( -[\text{O}-\text{CH}(\text{CH}_3)-\text{CO}]- \) with open bonds at both ends (1)
(c) PLA contains ester linkages / \( -\text{COO}- \) groups which can undergo hydrolysis / be broken down by microorganisms or enzymes (1) Polypropene has only \( \text{C}-\text{C} \) and \( \text{C}-\text{H} \) single bonds (hydrocarbon backbone) which are non-polar / resistant to chemical or microbial attack (1)
(d)(i) A carbon atom attached to four different atoms or groups (1) (d)(ii) Correct 3D representation showing wedge, dash, and normal bonds in tetrahedral arrangement forming a pair of mirror images (1)
題目 11 · Structured
6 分
Propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) is a carboxylic acid commonly used as a food preservative.
(a) State the definition of standard enthalpy change of combustion, \(\Delta H_c^\ominus\). (1 mark)
(b) Write a thermochemical equation, including state symbols, for the reaction representing the standard enthalpy change of formation of propanoic acid, \(\Delta H_f^\ominus[\text{CH}_3\text{CH}_2\text{COOH(l)}]\). (1 mark)
(c) The standard enthalpy changes of combustion for carbon, hydrogen, and propanoic acid are given below:
Construct an enthalpy cycle and calculate the standard enthalpy change of formation of propanoic acid, \(\Delta H_f^\ominus[\text{CH}_3\text{CH}_2\text{COOH(l)}]\). (3 marks)
(d) Explain why \(\Delta H_f^\ominus[\text{CH}_3\text{CH}_2\text{COOH(l)}]\) cannot be determined directly by a simple calorimetric experiment. (1 mark)
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解題
(a) The standard enthalpy change of combustion is the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions (1 atm, 298 K).
(d) Carbon, hydrogen, and oxygen do not react directly under laboratory conditions to form propanoic acid cleanly; side products (such as CO, \(\text{H}_2\text{O}\), and other organic compounds) would form.
評分準則
(a) Enthalpy change when ONE mole of a substance is completely burned in oxygen under standard conditions (1 mark). (b) Correct balanced equation with correct state symbols: \(3\text{C(graphite)} + 3\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(l)}\) (1 mark). (c) Correctly constructed enthalpy cycle / Hess's Law expression (1 mark*); correct substitution of enthalpy values: \(3(-393.5) + 3(-285.8) - (-1527.0)\) (1 mark*); correct final answer with sign and units: \(-510.9\text{ kJ mol}^{-1}\) (1 mark). (d) Carbon, hydrogen, and oxygen do not react directly to synthesize propanoic acid / multiple competing side reactions occur (1 mark).
題目 12 · Structured
6 分
Compound \(\mathbf{X}\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\). It reacts with 2,4-dinitrophenylhydrazine to give an orange precipitate, but gives a negative result when warmed with Tollens' reagent.
(a) Deduce the structural formula and state the systematic IUPAC name of \(\mathbf{X}\). (2 marks)
(b) When \(\mathbf{X}\) is treated with \(\text{NaBH}_4\), compound \(\mathbf{Y}\) is formed. (i) Write the structural formula of \(\mathbf{Y}\). (1 mark) (ii) Compound \(\mathbf{Y}\) shows enantiomerism. Draw the three-dimensional structures of the pair of enantiomers of \(\mathbf{Y}\). (2 marks)
(c) Compound \(\mathbf{Y}\) is heated with concentrated sulfuric acid at \(170^\circ\text{C}\) to undergo dehydration. Draw the skeletal formula of the major organic product formed. (1 mark)
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解題
(a) Since \(\mathbf{X}\) reacts with 2,4-DNPH, it is a carbonyl compound (aldehyde or ketone). Since it does not react with Tollens' reagent, it is not an aldehyde, so it must be a ketone. With 4 carbons, it is butan-2-one, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).
(b) (i) Reduction of a ketone by \(\text{NaBH}_4\) yields a secondary alcohol: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) (butan-2-ol). (ii) Butan-2-ol has a chiral carbon at C2. The two non-superimposable mirror images should be drawn with wedges and dashes centered at C2.
(c) Acid-catalysed dehydration of butan-2-ol gives but-2-ene as the major product (Zaitsev's rule) rather than but-1-ene.
評分準則
(a) \(\text{CH}_3\text{COCH}_2\text{CH}_3\) / \(\text{CH}_3\text{C(=O)CH}_2\text{CH}_3\) (1 mark); butan-2-one (accept 2-butanone) (1 mark). (b)(i) \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\) (1 mark). (b)(ii) Correct 3-D representation showing tetrahedral geometry around C2 with wedge and dashed lines (1 mark); correct pair of enantiomers shown as non-superimposable mirror images (1 mark). (c) Skeletal formula of but-2-ene (\(\text{CH}_3\text{CH=CHCH}_3\), either cis or trans) (1 mark).
題目 13 · Structured
6 分
Consider nitrogen trichloride (\(\text{NCl}_3\)) and boron trichloride (\(\text{BCl}_3\)).
(a) Draw the electron diagram (showing ELECTRONS IN THE OUTERMOST SHELLS only) for a molecule of \(\text{NCl}_3\). (1 mark)
(b) Using the Valence Shell Electron Pair Repulsion (VSEPR) theory, state and explain the shape of an \(\text{NCl}_3\) molecule. (2 marks)
(c) State the molecular shape and bond angle of a \(\text{BCl}_3\) molecule. (1 mark)
(d) Explain whether \(\text{BCl}_3\) is a polar or non-polar molecule in terms of its molecular geometry and bond dipoles. (2 marks)
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解題
(a) In \(\text{NCl}_3\), the central N atom shares 3 single covalent bonds with 3 Cl atoms and possesses 1 lone pair. Each Cl atom has 3 lone pairs.
(b) Around the central nitrogen atom, there are 4 electron pairs (3 bonding pairs and 1 lone pair). To minimize repulsion, the electron pairs adopt a tetrahedral arrangement. Due to greater lone pair-bonding pair repulsion, the molecular shape is trigonal pyramidal.
(c) \(\text{BCl}_3\) has a trigonal planar shape with a bond angle of \(120^\circ\).
(d) The B-Cl bond is polar because chlorine is more electronegative than boron, creating bond dipoles. However, because \(\text{BCl}_3\) has a symmetrical trigonal planar geometry, the three B-Cl bond dipoles cancel each other out, resulting in zero net dipole moment. Hence, \(\text{BCl}_3\) is a non-polar molecule.
評分準則
(a) Correct electron diagram with 3 N-Cl shared pairs, 1 lone pair on N, and 3 lone pairs on each Cl (1 mark). (b) Central N atom has 3 bond pairs and 1 lone pair (1 mark); trigonal pyramidal (1 mark). (c) Trigonal planar AND \(120^\circ\) (1 mark). (d) B-Cl bond is polar / has a dipole due to the electronegativity difference between B and Cl (1 mark); \(\text{BCl}_3\) is symmetrical / trigonal planar, so bond dipoles cancel out completely resulting in zero overall dipole moment (1 mark).
卷二 (Electives)
在三個選修課題（工業化學、材料化學、分析化學）中選答任何兩個部分。每部分滿分為20分。
2 題目 · 40 分
題目 1 · Structured Analytical Elective
20 分
Answer ALL parts of this question.
(a) The gas-phase reaction between nitrogen monoxide (\(\text{NO}\)) and hydrogen (\(\text{H}_2\)) was investigated at \(1050\text{ K}\): $$2\text{NO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(g)}$$
The following initial rate data were obtained in a series of experiments:
(i) Deduce the order of reaction with respect to \(\text{NO}\) and \(\text{H}_2\). Hence, write the rate equation for the reaction. (3 marks)
(ii) Calculate the rate constant, \(k\), for this reaction at \(1050\text{ K}\), including appropriate units. (2 marks)
(iii) The rate constant \(k\) for this reaction was determined to be \(1.60 \times 10^{3}\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) at \(1000\text{ K}\). Using the Arrhenius equation, calculate the activation energy (\(E_a\)), in \(\text{kJ mol}^{-1}\), for this reaction. (Gas constant \(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)) (3 marks)
(iv) In the presence of a heterogeneous catalyst, the activation energy is significantly reduced. State the meaning of heterogeneous catalyst and briefly describe one step in the catalytic cycle on the solid surface. (2 marks)
(b) Methanol (\(\text{CH}_3\text{OH}\)) is produced industrially by the catalytic hydrogenation of carbon monoxide: $$\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} \quad \Delta H = -90.7\text{ kJ mol}^{-1}$$
(i) Typical industrial operating conditions are \(250\text{ }^\circ\text{C}\) and \(50\text{ atm}\) to \(100\text{ atm}\) in the presence of a \(\text{CuO/ZnO/Al}_2\text{O}_3\) catalyst. (1) Explain why a moderately high pressure is used instead of atmospheric pressure in terms of yield and reaction rate. (2 marks) (2) Explain why an operating temperature of \(250\text{ }^\circ\text{C}\) is considered a compromise between equilibrium yield and reaction rate. (2 marks)
(ii) Methanol can be converted to methyl methacrylate (MMA), a monomer for poly(methyl methacrylate) (PMMA), via two alternative routes:
(1) Calculate the atom economy of Route 1 and Route 2 with respect to methyl methacrylate (\(\text{C}_5\text{H}_8\text{O}_2\), molar mass \(= 100.1\text{ g mol}^{-1}\)). (Relative atomic masses: \(\text{H} = 1.0\), \(\text{C} = 12.0\), \(\text{N} = 14.0\), \(\text{O} = 16.0\), \(\text{S} = 32.1\)) (3 marks) (2) Based on the principles of green chemistry, give TWO reasons why Route 2 is environmentally more benign than Route 1, apart from a higher atom economy. (2 marks) (3) Suggest ONE method to manage the unreacted synthesis gas in an industrial methanol synthesis loop to enhance economic efficiency and reduce waste. (1 mark)
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解題
(a) (i) Comparing Exp 1 and Exp 2: [H2] is constant. When [NO] is doubled (from 1.50 x 10^-3 to 3.00 x 10^-3 mol dm^-3), initial rate quadruples (from 1.80 x 10^-4 to 7.20 x 10^-4 mol dm^-3 s^-1, factor of 4 = 2^2). Thus, order with respect to NO = 2. Comparing Exp 2 and Exp 3: [NO] is constant. When [H2] is tripled (from 2.00 x 10^-3 to 6.00 x 10^-3 mol dm^-3), initial rate triples (from 7.20 x 10^-4 to 2.16 x 10^-3 mol dm^-3 s^-1, factor of 3 = 3^1). Thus, order with respect to H2 = 1. Rate equation: Rate = k[NO]^2[H2]
(ii) Using Exp 1: 1.80 x 10^-4 = k (1.50 x 10^-3)^2 (2.00 x 10^-3) 1.80 x 10^-4 = k (4.50 x 10^-9) k = 4.00 x 10^4 dm^6 mol^-2 s^-1
(iii) Using the two-point Arrhenius equation: ln(k2 / k1) = -(Ea / R) * (1/T2 - 1/T1) k1 = 1.60 x 10^3 at T1 = 1000 K k2 = 4.00 x 10^4 at T2 = 1050 K ln(4.00 x 10^4 / 1.60 x 10^3) = ln(25) = 3.2189 1/T2 - 1/T1 = 1/1050 - 1/1000 = (1000 - 1050) / 1050000 = -50 / 1050000 = -4.7619 x 10^-5 K^-1 3.2189 = -(Ea / 8.314) * (-4.7619 x 10^-5) Ea = (3.2189 * 8.314) / 4.7619 x 10^-5 = 561990 J mol^-1 = 562 kJ mol^-1 (or 561.9 kJ mol^-1)
(iv) A heterogeneous catalyst is a catalyst that exists in a different phase / physical state from the reactants. Steps in catalytic cycle: Adsorption of reactant molecules onto active sites on the catalyst surface / Weakening/breaking of bonds in adsorbed species / Chemical reaction between adsorbed species / Desorption of products from the catalyst surface.
(b) (i) (1) Higher pressure shifts equilibrium position to the right (forward direction) because there are fewer moles of gas on the product side (1 mole) than reactant side (3 moles), increasing the equilibrium yield of methanol. Higher pressure also increases the concentration/frequency of collision of gaseous reactant particles, increasing the reaction rate. (2) The forward reaction is exothermic. A lower temperature favors the forward reaction giving a higher equilibrium yield. However, at lower temperatures, the reaction rate is too low. Thus, 250 °C is a compromise to achieve an acceptable reaction rate while maintaining an economically feasible yield.
(ii) (1) Route 1: Total molar mass of all reactants = (58.0 + 27.0 + 32.0 + 98.1) = 215.1 g mol^-1. Atom economy = (100.1 / 215.1) * 100% = 46.5% Route 2: Total molar mass of all reactants = (70.0 + 32.0 + 16.0) = 118.0 g mol^-1. Atom economy = (100.1 / 118.0) * 100% = 84.8%
(2) Route 2 does not use highly toxic hydrogen cyanide (HCN) / concentrated sulfuric acid (H2SO4) (safer chemicals/feedstock). Route 2 produces non-hazardous water as a byproduct instead of large amounts of corrosive/waste ammonium hydrogen sulfate (NH4HSO4).
(3) Recycle unreacted synthesis gas back into the reactor loop (to prevent wastage and increase overall conversion efficiency).
評分準則
(a) (i) - Deducing order w.r.t. NO = 2 with working (1 mark) - Deducing order w.r.t. H2 = 1 with working (1 mark) - Correct rate equation: \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\) (1 mark)
(iii) - Correct substitution into Arrhenius equation: \(\ln\left(\frac{4.00 \times 10^4}{1.60 \times 10^3}\right) = -\frac{E_a}{8.314}\left(\frac{1}{1050} - \frac{1}{1000}\right)\) (1* mark) - Correct mathematical evaluation of \(\Delta(1/T)\) and \(\ln(k_2/k_1)\) (1* mark) - Final value with units: \(+562\text{ kJ mol}^{-1}\) (accept \(561.9\text{ kJ mol}^{-1}\) to \(563\text{ kJ mol}^{-1}\)) (1 mark)
(iv) - Definition: Catalyst in a different phase from the reactants (e.g., solid catalyst with gas reactants) (1 mark) - One correct step described: Adsorption of reactants onto catalyst surface / Bond weakening / Reaction of intermediates / Desorption of products (1 mark)
(b) (i) (1) Explaining higher pressure increases yield due to fewer moles of gas on RHS (1 mark); and increases collision frequency / rate of reaction (1 mark) (2) Forward reaction is exothermic, so low temp favors yield (1 mark); higher temp needed for acceptable rate/kinetic feasibility, hence 250 °C is a compromise (1 mark)
(ii) (1) - Route 1: \(\frac{100.1}{215.1} \times 100\% = 46.5\%\) (1 mark) - Route 2: \(\frac{100.1}{118.0} \times 100\% = 84.8\%\) (1 mark) - Correct formula / setup shown (1 mark) (2) - Route 2 avoids extremely toxic HCN / avoids corrosive H2SO4 (1 mark) - Route 2 produces harmless H2O instead of hazardous/waste NH4HSO4 (1 mark) (3) - Recycling unreacted synthesis gas back into the catalytic converter (1 mark)
題目 2 · Structured Analytical Elective
20 分
Answer ALL parts of this question.
(a) Compound W is an organic compound containing carbon, hydrogen, and oxygen only. It is suspected to be an ester responsible for a fruity fragrance.
(i) Elemental analysis shows that W contains \(58.80\%\) carbon and \(9.87\%\) hydrogen by mass. Determine the empirical formula of W. (Relative atomic masses: \(\text{H} = 1.0\), \(\text{C} = 12.0\), \(\text{O} = 16.0\)) (2 marks)
(ii) The mass spectrum of W shows a molecular ion peak at \(m/z = 102\). (1) Deduce the molecular formula of W. (1 mark) (2) The mass spectrum also displays significant fragment peaks at \(m/z = 57\) and \(m/z = 43\). Write the chemical formula of the ion responsible for the peak at \(m/z = 43\). (1 mark)
(iii) The infrared (IR) spectrum of W shows a strong, sharp absorption band at \(1740\text{ cm}^{-1}\) and a strong band at \(1180\text{ cm}^{-1}\), but lacks any broad absorption in the region \(2500\text{--}3600\text{ cm}^{-1}\). (1) State the functional group indicated by these IR features. (1 mark) (2) Compound W undergoes alkaline hydrolysis with \(\text{NaOH(aq)}\) upon heating, followed by acidification, to yield ethanoic acid and an alcohol Y. Deduce the structural formula of W and give its systematic name. (2 marks) (3) Suggest a chemical test to distinguish between ethanoic acid and alcohol Y. State the reagent(s) and expected observation(s). (2 marks)
(iv) Draw the structural formula of a positional isomer of W that also hydrolyses to give a carboxylic acid and a branched primary alcohol. (1 mark)
(b) The concentration of iron(II) ions in a commercial liquid dietary supplement was determined by colorimetry using 1,10-phenanthroline. Iron(II) forms an intensely red-orange coordination complex with 1,10-phenanthroline, which has maximum light absorption at \(510\text{ nm}\).
(i) State why a calibration curve must be constructed before determining the concentration of iron in the unknown sample. (1 mark)
(ii) A standard stock solution containing \(0.100\text{ g dm}^{-3}\) of \(\text{Fe}^{2+}\) was prepared. A series of standard solutions was prepared by pipetting various volumes of the stock solution into \(100.0\text{ cm}^3\) volumetric flasks, adding excess 1,10-phenanthroline and buffer, and making up to the mark with deionised water. The absorbance of each solution was measured at \(510\text{ nm}\) using a spectrophotometer with a \(1.0\text{ cm}\) cuvette:
(1) State the mathematical relationship between absorbance and concentration demonstrated by these data. (1 mark) (2) Explain why the spectrophotometer must be zeroed using a blank solution before measuring absorbance, and state what the blank solution should contain. (2 marks)
(iii) A \(5.00\text{ cm}^3\) sample of the dietary supplement was diluted to \(250.0\text{ cm}^3\) in volumetric flask A. Then, a \(10.00\text{ cm}^3\) portion of this diluted solution was transferred into a \(100.0\text{ cm}^3\) volumetric flask B, treated with excess 1,10-phenanthroline and buffer, and made up to the mark. The absorbance of this solution was found to be \(0.646\) at \(510\text{ nm}\).
(1) Using the calibration data, determine the concentration of \(\text{Fe}^{2+}\), in \(\text{mg dm}^{-3}\), in the measured solution in flask B. (1 mark) (2) Calculate the concentration of \(\text{Fe}^{2+}\), in \(\text{g dm}^{-3}\), in the original dietary supplement. (3 marks) (3) Suggest ONE reason why colorimetry is preferred over standard redox titration (e.g. using \(\text{KMnO}_4\)) for analysing this dietary supplement sample. (1 mark)
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解題
(a) (i) Percentage of O = 100% - 58.80% - 9.87% = 31.33% Relative moles: C: 58.80 / 12.0 = 4.900 H: 9.87 / 1.0 = 9.870 O: 31.33 / 16.0 = 1.958 Dividing by 1.958: C: 4.900 / 1.958 = 2.50 H: 9.870 / 1.958 = 5.04 ≈ 5 O: 1.958 / 1.958 = 1.00 Multiplying by 2 to get integers: C5H10O2. Empirical formula = C5H10O2.
(ii) (1) Empirical formula mass of C5H10O2 = 5(12.0) + 10(1.0) + 2(16.0) = 60 + 10 + 32 = 102. Molecular ion peak m/z = 102, so relative molecular mass = 102. Molecular formula = C5H10O2. (2) CH3CO+ or C3H7+ (either [CH3CO]+ or [C3H7]+ is accepted).
(iii) (1) Ester functional group (C=O carbonyl at 1740 cm^-1 and C-O single bond at 1180 cm^-1; absence of O-H carboxylic/alcohol absorption). (2) Alkaline hydrolysis gives ethanoic acid (CH3COOH) and alcohol Y. Ethanoic acid has 2 carbons, so alcohol Y has 5 - 2 = 3 carbons, which is propan-1-ol or propan-2-ol. The ester W is propyl ethanoate (CH3COOCH2CH2CH3) or isopropyl ethanoate / 1-methylethyl ethanoate (CH3COOCH(CH3)2). Structural formula: CH3COOCH2CH2CH3 (propyl ethanoate) or CH3COOCH(CH3)2 (isopropyl ethanoate / 1-methylethyl ethanoate). (3) Add sodium hydrogencarbonate solution (NaHCO3(aq)) / sodium carbonate solution (Na2CO3(aq)) to separate samples. Ethanoic acid produces effervescence / colourless gas bubbles that turn limewater milky (CO2), whereas alcohol Y gives no observable change / no effervescence.
(iv) The required isomer hydrolyses to give a carboxylic acid and a branched primary alcohol. A 5-carbon ester with a branched primary alcohol part must be methylpropanoic acid or methylpropyl ester: specifically, since the alcohol must be branched and primary, the alcohol part is 2-methylpropan-1-ol (isobutyl alcohol, 4 carbons), so the acid part is methanoic acid (1 carbon). Formula: HCOOCH2CH(CH3)2 (2-methylpropyl methanoate / isobutyl methanoate).
(b) (i) To establish the direct relationship/proportionality between absorbance and concentration of iron(II) complex under identical experimental conditions.
(ii) (1) Beer-Lambert Law: Absorbance is directly proportional to concentration (A ∝ c). (2) The blank solution compensates for any light absorption, reflection, or scattering by the solvent, cuvette walls, and reagents other than the analyte complex. It should contain deionised water, buffer, and 1,10-phenanthroline (all reagents except the iron solution).
(iii) (1) From the table, slope = Absorbance / Concentration = 0.190 / 1.00 = 0.190 dm^3 mg^-1. Concentration in flask B = 0.646 / 0.190 = 3.40 mg dm^-3. (2) Flask B (100.0 cm^3) was prepared by diluting 10.00 cm^3 of flask A. Dilution factor from A to B = 100.0 / 10.00 = 10. Concentration of Fe2+ in flask A = 3.40 mg dm^-3 * 10 = 34.0 mg dm^-3. Flask A (250.0 cm^3) was prepared by diluting 5.00 cm^3 of the original sample. Dilution factor from original to A = 250.0 / 5.00 = 50. Concentration of Fe2+ in original sample = 34.0 mg dm^-3 * 50 = 1700 mg dm^-3 = 1.70 g dm^-3. (3) The concentration of Fe2+ in the dietary supplement is very low / trace amounts, where colorimetry provides much higher sensitivity than titration; OR the supplement may contain other reducing agents (like ascorbic acid) or coloured species that interfere with permanganate titration.
評分準則
(a) (i) - Correct calculation of mass % of oxygen (31.33%) and mole ratios: C:H:O = 4.90 : 9.87 : 1.958 = 2.5 : 5 : 1 (1* mark) - Empirical formula: \(\text{C}_5\text{H}_{10}\text{O}_2\) (1 mark)
(ii) (1) Molecular formula: \(\text{C}_5\text{H}_{10}\text{O}_2\) (1 mark) (2) \(\text{CH}_3\text{CO}^+\) or \(\text{C}_3\text{H}_7^+\) (1 mark) (must include positive charge)
(iii) (1) Ester group / \(-\text{COO}-\) (1 mark) (2) Structural formula of \(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\) or \(\text{CH}_3\text{COOCH(CH}_3\text{)}_2\) (1 mark); Systematic name: propyl ethanoate or isopropyl ethanoate / 1-methylethyl ethanoate (1 mark) (3) Reagent: \(\text{NaHCO}_3\text{(aq)}\) / \(\text{Na}_2\text{CO}_3\text{(aq)}\) (1 mark); Observation: Ethanoic acid gives effervescence / gas turns limewater milky, alcohol Y gives no observable change (1 mark)
(iv) Structural formula of 2-methylpropyl methanoate: \(\text{HCOOCH}_2\text{CH(CH}_3\text{)}_2\) (1 mark)
(b) (i) To correlate absorbance values with known concentrations / establish a standard curve for quantifying unknown (1 mark)
(ii) (1) Absorbance is directly proportional to concentration (\(A \propto c\) / Beer's Law) (1 mark) (2) Purpose: To eliminate background absorbance due to solvent / reagents / cuvette (1 mark); Composition: Deionised water, buffer, and 1,10-phenanthroline (all components except \(\text{Fe}^{2+}\)) (1 mark)
(iii) (1) Concentration in flask B = \(\frac{0.646}{0.190} = 3.40\text{ mg dm}^{-3}\) (1 mark) (2) - Dilution factor 1 (B to A): \(\times 10\) \(\rightarrow 34.0\text{ mg dm}^{-3}\) in flask A (1* mark) - Dilution factor 2 (A to original): \(\times 50\) \(\rightarrow 1700\text{ mg dm}^{-3}\) (1* mark) - Conversion to \(\text{g dm}^{-3}\): \(1.70\text{ g dm}^{-3}\) (1 mark) (3) Trace/low concentration of analyte (colorimetry is more sensitive than titration) / Colored additives or other reducing species in supplement interfere with redox titration (1 mark)
