Cambridge IAL · Thinka 原創模擬試題

2025 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

Thinka Jun 2025 (V1) Cambridge International A Level-Style Mock — Chemistry (9701)

270 分465 分鐘2025

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V1) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

Paper 11 (選擇題)

Answer all 40 questions. Choose the single best answer for each question.

40 題目 · 40 分

題目 1 · 選擇題

1 分

A 2.30 g sample of an organic compound, containing only carbon, hydrogen, and oxygen, is completely burned in excess oxygen. This produces 4.40 g of carbon dioxide and 2.70 g of water.

What is the empirical formula of the compound?

[Relative atomic masses, \(A_r\): \(H = 1.0\), \(C = 12.0\), \(O = 16.0\)]

A.\(CH_3O\)
B.\(C_2H_6O\)
C.\(C_3H_8O\)
D.\(C_4H_{10}O_2\)

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解題

1. **Find the mass and moles of carbon:**
The molar mass of \(CO_2\) is \(44.0\text{ g mol}^{-1}\).
\(m(C) = 4.40\text{ g} \times \frac{12.0}{44.0} = 1.20\text{ g}\)
\(n(C) = \frac{1.20\text{ g}}{12.0\text{ g mol}^{-1}} = 0.10\text{ mol}\)

2. **Find the mass and moles of hydrogen:**
The molar mass of \(H_2O\) is \(18.0\text{ g mol}^{-1}\).
\(m(H) = 2.70\text{ g} \times \frac{2.0}{18.0} = 0.30\text{ g}\)
\(n(H) = \frac{0.30\text{ g}}{1.0\text{ g mol}^{-1}} = 0.30\text{ mol}\)

3. **Find the mass and moles of oxygen:**
The total mass of the compound is 2.30 g.
\(m(O) = 2.30\text{ g} - (1.20\text{ g} + 0.30\text{ g}) = 0.80\text{ g}\)
\(n(O) = \frac{0.80\text{ g}}{16.0\text{ g mol}^{-1}} = 0.050\text{ mol}\)

4. **Determine the simplest whole number ratio:**
\(C : H : O = 0.10 : 0.30 : 0.050\)
Dividing each by 0.050 gives:
\(C : H : O = 2 : 6 : 1\)

Therefore, the empirical formula of the compound is \(C_2H_6O\).

評分準則

[1 mark] Correctly calculate the moles of C, H, and O from the given combustion data and find the empirical ratio 2:6:1, corresponding to option B.
[0 marks] Incorrect calculations or choice of any other option.

題目 2 · 選擇題

1 分

An aqueous solution of cobalt(II) chloride is treated with excess concentrated hydrochloric acid, causing a color change from pink to blue. Water is then added to the resulting mixture, which restores the original pink color.

Which row correctly describes the geometry and color of the cobalt complexes present in the initial aqueous solution and after adding excess concentrated hydrochloric acid?

A.Initial complex: octahedral and pink; complex after excess HCl: tetrahedral and blue
B.Initial complex: octahedral and blue; complex after excess HCl: square planar and pink
C.Initial complex: tetrahedral and pink; complex after excess HCl: octahedral and blue
D.Initial complex: octahedral and pink; complex after excess HCl: octahedral and blue

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解題

The initial aqueous solution of cobalt(II) chloride contains the hexaaquacobalt(II) complex, \([Co(H_2O)_6]^{2+}\), which is octahedral and pink in color.

When excess concentrated hydrochloric acid (containing high concentration of halide ligands, \(Cl^-\)) is added, a ligand exchange reaction occurs to form the tetrachorocobaltate(II) complex, \([CoCl_4]^{2-}\):

\([Co(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CoCl_4]^{2-}(aq) + 6H_2O(l)\)

This tetrachlorocobaltate(II) complex is tetrahedral and blue in color. Adding water shifts the equilibrium position back to the left, regenerating the pink octahedral hexaaquacobalt(II) complex.

評分準則

[1 mark] Correctly identify the initial complex as octahedral and pink, and the chloride complex as tetrahedral and blue, choosing option A.
[0 marks] For any other combination.

題目 3 · 選擇題

1 分

The reaction between reactant P and reactant Q was studied at constant temperature and the following initial rates of reaction were obtained:

* **Experiment 1:** \([P] = 0.10\text{ mol dm}^{-3}\), \([Q] = 0.10\text{ mol dm}^{-3}\), \(\text{Initial rate} = 2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)
* **Experiment 2:** \([P] = 0.20\text{ mol dm}^{-3}\), \([Q] = 0.10\text{ mol dm}^{-3}\), \(\text{Initial rate} = 8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)
* **Experiment 3:** \([P] = 0.20\text{ mol dm}^{-3}\), \([Q] = 0.20\text{ mol dm}^{-3}\), \(\text{Initial rate} = 1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)

What is the rate equation and the overall order of this reaction?

A.\(\text{Rate} = k[P][Q]\); overall order is 2
B.\(\text{Rate} = k[P]^2[Q]\); overall order is 3
C.\(\text{Rate} = k[P][Q]^2\); overall order is 3
D.\(\text{Rate} = k[P]^2[Q]^2\); overall order is 4

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解題

1. **Determine the order with respect to P:**
Compare Experiment 1 and Experiment 2 where \([Q]\) is kept constant at \(0.10\text{ mol dm}^{-3}\):
\([P]\) is doubled from \(0.10\) to \(0.20\text{ mol dm}^{-3}\).
The initial rate increases from \(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\), which is an increase by a factor of \(\frac{8.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 4 = 2^2\).
Therefore, the reaction is second order with respect to P.

2. **Determine the order with respect to Q:**
Compare Experiment 2 and Experiment 3 where \([P]\) is kept constant at \(0.20\text{ mol dm}^{-3}\):
\([Q]\) is doubled from \(0.10\) to \(0.20\text{ mol dm}^{-3}\).
The initial rate increases from \(8.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\), which is an increase by a factor of \(\frac{1.6 \times 10^{-3}}{8.0 \times 10^{-4}} = 2 = 2^1\).
Therefore, the reaction is first order with respect to Q.

3. **Combine to find the rate equation and overall order:**
\(\text{Rate} = k[P]^2[Q]\)
\(\text{Overall order} = 2 + 1 = 3\).

評分準則

[1 mark] Deduce that the order with respect to P is 2 and Q is 1 to select option B.
[0 marks] Incorrect deduction of rate equation or overall order.

題目 4 · 選擇題

1 分

But-2-ene can be converted into butanone in a three-step synthetic pathway:

\(\text{But-2-ene} \xrightarrow{\text{Step 1}} \text{Compound A} \xrightarrow{\text{Step 2}} \text{Compound B} \xrightarrow{\text{Step 3}} \text{Butanone}\)

Which of the following describes the correct set of reagents and conditions for Step 1, Step 2, and Step 3?

A.Step 1: \(HCl(g)\) at room temperature; Step 2: \(NaOH\) in ethanol, heat under reflux; Step 3: Acidified \(K_2Cr_2O_7\), heat under reflux
B.Step 1: \(HCl(g)\) at room temperature; Step 2: \(NaOH(aq)\), heat under reflux; Step 3: Acidified \(K_2Cr_2O_7\), heat under reflux
C.Step 1: \(Cl_2(g)\), UV light; Step 2: \(NaOH(aq)\), room temperature; Step 3: Tollens' reagent, warm
D.Step 1: \(Cl_2(aq)\), room temperature; Step 2: \(NaOH\) in ethanol, heat under reflux; Step 3: Acidified \(K_2Cr_2O_7\), distill

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解題

Let's trace the reactions step-by-step to form butanone (a ketone) starting from the symmetrical alkene but-2-ene:

* **Step 1:** The electrophilic addition of hydrogen chloride gas (\(HCl(g)\)) at room temperature to but-2-ene produces the halogenoalkane **2-chlorobutane** (Compound A).
* **Step 2:** The nucleophilic substitution of 2-chlorobutane with aqueous sodium hydroxide (\(NaOH(aq)\)) under heating with reflux produces the secondary alcohol **butan-2-ol** (Compound B). Note that using NaOH in ethanol would instead cause elimination, forming but-2-ene back.
* **Step 3:** The oxidation of the secondary alcohol butan-2-ol using acidified potassium dichromate(VI) (\(K_2Cr_2O_7 / H^+\)) under reflux produces **butanone**.

This sequence of steps corresponds perfectly to Option B.

評分準則

[1 mark] Correctly identify the reagents and conditions for electrophilic addition, nucleophilic substitution, and oxidation to choose option B.
[0 marks] Any incorrect step (e.g., elimination instead of substitution, or wrong oxidation reagents).

題目 5 · 選擇題

1 分

Under suitable conditions, methylbenzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid to form a mixture of mono-nitrated products.

Which statement about this reaction and its products is correct?

A.The active electrophile is the \(NO_2^-\rightleftharpoons\) ion.
B.The methyl group acts as a deactivating and 3-directing group.
C.Methylbenzene reacts faster than benzene in this nitration reaction because the methyl group increases the electron density of the benzene ring.
D.The reaction is classified as a nucleophilic substitution reaction.

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解題

Let's analyze the statements:

* **A is incorrect:** The active electrophile in this electrophilic aromatic substitution reaction is the nitronium ion, \(NO_2^+\), not the nitrite ion \(NO_2^-\).
* **B is incorrect:** The methyl group (\(-CH_3\)) on the benzene ring is electron-donating due to the inductive effect. It is a 2,4-directing (ortho/para-directing) group, not a 3-directing (meta-directing) group.
* **C is correct:** Because the methyl group is an electron-donating group, it increases the electron density on the benzene ring. This makes the ring more nucleophilic and highly susceptible to electrophilic attack, meaning methylbenzene undergoes nitration at a faster rate than benzene under identical conditions.
* **D is incorrect:** The reaction mechanism is electrophilic substitution (specifically, electrophilic aromatic substitution), not nucleophilic substitution.

評分準則

[1 mark] Correctly identify that the electron-donating methyl group increases the reactivity of the ring towards electrophilic attack, leading to a faster rate than benzene (Option C).
[0 marks] Incorrect identification of mechanism, electrophile, or directing effect.

題目 6 · 選擇題

1 分

The homogeneous aqueous reaction between peroxodisulfate ions, \(S_2O_8^{2-}(aq)\), and iodide ions, \(I^-(aq)\), has a high activation energy because both reactants are negatively charged. It is catalyzed by aqueous iron(III) ions, \(Fe^{3+}(aq)\).

Which equation represents a feasible step in the mechanism of this catalyzed reaction?

A.\(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\)
B.\(2Fe^{3+}(aq) + S_2O_8^{2-}(aq) \rightarrow 2Fe^{2+}(aq) + 2SO_4^{2-}(aq)\)
C.\(Fe^{2+}(aq) + 2I^-(aq) \rightarrow Fe^{3+}(aq) + I_2(aq)\)
D.\(2Fe^{2+}(aq) + 2I^-(aq) \rightarrow 2Fe^{3+}(aq) + I_2(aq)\)

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解題

The reaction catalyzed by \(Fe^{3+}\) operates through two feasible redox steps because \(Fe^{3+}/Fe^{2+}\) can act as an intermediate electron carrier:

* **Step 1:** The iron(III) ions oxidize iodide ions to iodine, while being reduced to iron(II) ions:
\(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\)
*Standard cell potential:* \(E^\ominus_{\text{cell}} = E^\ominus(Fe^{3+}/Fe^{2+}) - E^\ominus(I_2/I^-) = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\) (feasible).

* **Step 2:** The iron(II) ions are then re-oxidized back to iron(III) ions by the peroxodisulfate ions:
\(2Fe^{2+}(aq) + S_2O_8^{2-}(aq) \rightarrow 2Fe^{3+}(aq) + 2SO_4^{2-}(aq)\)
*Standard cell potential:* \(E^\ominus_{\text{cell}} = E^\ominus(S_2O_8^{2-}/SO_4^{2-}) - E^\ominus(Fe^{3+}/Fe^{2+}) = +2.01\text{ V} - (+0.77\text{ V}) = +1.24\text{ V}\) (feasible).

Option A represents Step 1 of this mechanism. Options B, C, and D are stoichiometrically or thermodynamically incorrect.

評分準則

[1 mark] Correctly identify the thermodynamically feasible step involving reduction of \(Fe^{3+}\) and oxidation of \(I^-\), selecting option A.
[0 marks] Incorrect step selection.

題目 7 · 選擇題

1 分

Benzene can be converted into 3-bromobenzoic acid in a three-step synthesis:

\(\text{Benzene} \xrightarrow{\text{Step 1}} \text{Methylbenzene} \xrightarrow{\text{Step 2}} \text{Benzoic acid} \xrightarrow{\text{Step 3}} \text{3-bromobenzoic acid}\)

What are the correct reagents for Step 1, Step 2, and Step 3?

A.Step 1: \(CH_3Cl\) and anhydrous \(AlCl_3\); Step 2: Acidified \(K_2Cr_2O_7\), heat under reflux; Step 3: \(Br_2(aq)\) at room temperature
B.Step 1: \(CH_3Cl\) and anhydrous \(AlCl_3\); Step 2: Alkaline \(KMnO_4\), heat under reflux, followed by dilute acid; Step 3: \(Br_2\) and anhydrous \(FeBr_3\)
C.Step 1: \(CH_3OH\) and concentrated \(H_2SO_4\); Step 2: Alkaline \(KMnO_4\), heat under reflux, followed by dilute acid; Step 3: \(HBr(g)\) at room temperature
D.Step 1: \(CH_3Cl\) and anhydrous \(AlCl_3\); Step 2: \(LiAlH_4\) in dry ether; Step 3: \(Br_2\) and anhydrous \(FeBr_3\)

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解題

Let's analyze each step of the synthesis:

* **Step 1:** Friedel-Crafts alkylation of benzene to form methylbenzene.
Reagents: Chloromethane (\(CH_3Cl\)) and an anhydrous aluminium chloride (\(AlCl_3\)) catalyst.
* **Step 2:** Oxidation of the methyl side-chain to a carboxylic acid group.
Reagents: Alkaline potassium manganate(VII) (\(KMnO_4\)), heated under reflux, followed by addition of dilute acid (e.g., dilute \(H_2SO_4\)) to protonate the benzoate ion.
*Note:* Acidified potassium dichromate(VI) (\(K_2Cr_2O_7\)) is not strong enough to oxidize alkyl side-chains on a benzene ring.
* **Step 3:** Electrophilic bromination of the aromatic ring.
The carboxylic acid group (\(-COOH\)) is electron-withdrawing and acts as a 3-directing (meta-directing) group. Thus, brominating benzoic acid will direct the incoming bromine to the 3-position.
Reagents: Bromine (\(Br_2\)) and an anhydrous iron(III) bromide (\(FeBr_3\)) catalyst.

Therefore, the correct set of reagents is given in option B.

評分準則

[1 mark] Correctly identify the reagents/catalysts for Friedel-Crafts alkylation, side-chain oxidation, and ring bromination of meta-directed benzoic acid, matching option B.
[0 marks] Selecting an option with incorrect reagents.

題目 8 · 選擇題

1 分

An electrochemical cell is set up under standard conditions using two half-cells:

* **Half-cell 1:** \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq)\) \(E^\ominus = +0.77\text{ V}\)
* **Half-cell 2:** \(Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightleftharpoons 2Cr^{3+}(aq) + 7H_2O(l)\) \(E^\ominus = +1.33\text{ V}\)

Which statement about this cell is correct when it is operating and generating current?

A.The standard cell potential, \(E^\ominus_{\text{cell}}\), is \(+2.10\text{ V}\).
B.Electrons flow from the dichromate half-cell to the iron half-cell through the external circuit.
C.The pH of the solution in the dichromate half-cell increases.
D.The concentration of \(Fe^{2+}(aq)\) ions increases.

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解題

Let's analyze the behavior of the electrochemical cell:

1. **Determine the direction of the reactions:**
The dichromate half-cell (Half-cell 2) has a more positive standard electrode potential (\(E^\ominus = +1.33\text{ V}\)) than the iron half-cell (Half-cell 1, \(E^\ominus = +0.77\text{ V}\)).
Therefore, Half-cell 2 undergoes reduction (acts as the cathode):
\(Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)\)
Half-cell 1 undergoes oxidation (acts as the anode):
\(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-\)

2. **Evaluate the options:**
* **A is incorrect:** \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +1.33\text{ V} - (+0.77\text{ V}) = +0.56\text{ V}\).
* **B is incorrect:** Electrons flow from the negative electrode (anode, where oxidation happens: iron half-cell) to the positive electrode (cathode, where reduction happens: dichromate half-cell) through the external circuit.
* **C is correct:** The reduction reaction consumes hydrogen ions (\(H^+(aq)\)), decreasing their concentration in the dichromate half-cell. Since \(pH = -\log_{10}[H^+]\), a decrease in \([H^+]\) causes the pH of the solution to increase.
* **D is incorrect:** \(Fe^{2+}\) is oxidized to \(Fe^{3+}\), so the concentration of \(Fe^{2+}(aq)\) decreases.

評分準則

[1 mark] Deduce the correct half-cell reactions based on standard electrode potentials, recognize that the consumption of \(H^+\) ions in the reduction half-cell increases the pH, and select option C.
[0 marks] Incorrect deduction or choice of another option.

題目 9 · 選擇題

1 分

A mixture of \( 10\text{ cm}^3 \) of a gaseous hydrocarbon and \( 70\text{ cm}^3 \) of oxygen (an excess) is exploded in a sealed tube. After cooling to room temperature, the total gas volume is \( 50\text{ cm}^3 \). On treatment with excess aqueous sodium hydroxide, the volume decreases to \( 20\text{ cm}^3 \). What is the molecular formula of the hydrocarbon?

A.\( \text{C}_3\text{H}_6 \)
B.\( \text{C}_3\text{H}_8 \)
C.\( \text{C}_4\text{H}_8 \)
D.\( \text{C}_4\text{H}_{10} \)

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解題

First, find the volume of carbon dioxide produced and oxygen consumed. The reduction in gas volume on adding NaOH is due to the absorption of \( \text{CO}_2 \). Thus, volume of \( \text{CO}_2 \text{ produced} = 50\text{ cm}^3 - 20\text{ cm}^3 = 30\text{ cm}^3 \). The remaining gas volume of \( 20\text{ cm}^3 \) is the excess, unreacted oxygen. The volume of oxygen reacted = \( 70\text{ cm}^3 - 20\text{ cm}^3 = 50\text{ cm}^3 \). Let the hydrocarbon formula be \( \text{C}_x\text{H}_y \). 1 volume of \( \text{C}_x\text{H}_y \) produces x volumes of \( \text{CO}_2 \). Given \( 10\text{ cm}^3 \) of hydrocarbon produces \( 30\text{ cm}^3 \) of \( \text{CO}_2 \), we have x = 3. The equation for complete combustion of \( \text{C}_x\text{H}_y \) is: \( \text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} \). The volume ratio of hydrocarbon to reacted oxygen is 10 : 50 = 1 : 5. Therefore, \( x + \frac{y}{4} = 5 \). Since x = 3, then \( 3 + \frac{y}{4} = 5 \) which simplifies to \( y = 8 \). Thus, the hydrocarbon is \( \text{C}_3\text{H}_8 \).

評分準則

1 mark for the correct answer B.

題目 10 · 選擇題

1 分

Which statement best explains why aqueous copper(II) ions, \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \), are blue, whereas aqueous zinc(II) ions, \( [\text{Zn}(\text{H}_2\text{O})_6]^{2+} \), are colourless?

A.Copper(II) has an incomplete d-subshell that splits into two energy levels, allowing d-d electronic transitions by absorbing light from the visible spectrum. Zinc(II) has a fully filled d-subshell, so d-d transitions cannot occur.
B.Zinc(II) ions have a higher charge density than copper(II) ions, which prevents the splitting of the 3d orbitals in the presence of water ligands.
C.In copper(II) complexes, blue light is emitted when excited 3d electrons fall back to the ground state, whereas in zinc(II) complexes, the emitted light falls outside the visible region.
D.The ligands in \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \) undergo a ligand exchange reaction that produces a colored precipitate, whereas no such reaction occurs for \( [\text{Zn}(\text{H}_2\text{O})_6]^{2+} \).

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解題

Copper(II) has a \( 3\text{d}^9 \) electronic configuration, which is an incomplete d-subshell. In the presence of water ligands, the 3d orbitals split into two energy levels with a small energy gap. Visible light is absorbed to promote an electron from the lower to the higher d-orbitals (a d-d electronic transition), and the complementary blue colour is transmitted. Zinc(II) has a \( 3\text{d}^{10} \) configuration with a completely filled d-subshell, so d-d transitions are not possible, making it colourless.

評分準則

1 mark for the correct answer A.

題目 11 · 選擇題

1 分

The reaction between three reactants, P, Q and R, is investigated at a constant temperature. The initial rates of reaction are: Experiment 1: \( [\text{P}] = 0.10\text{ mol dm}^{-3} \), \( [\text{Q}] = 0.10\text{ mol dm}^{-3} \), \( [\text{R}] = 0.10\text{ mol dm}^{-3} \), Initial rate = \( 1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} \); Experiment 2: \( [\text{P}] = 0.20\text{ mol dm}^{-3} \), \( [\text{Q}] = 0.10\text{ mol dm}^{-3} \), \( [\text{R}] = 0.10\text{ mol dm}^{-3} \), Initial rate = \( 2.4 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} \); Experiment 3: \( [\text{P}] = 0.10\text{ mol dm}^{-3} \), \( [\text{Q}] = 0.30\text{ mol dm}^{-3} \), \( [\text{R}] = 0.10\text{ mol dm}^{-3} \), Initial rate = \( 1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} \); Experiment 4: \( [\text{P}] = 0.10\text{ mol dm}^{-3} \), \( [\text{Q}] = 0.10\text{ mol dm}^{-3} \), \( [\text{R}] = 0.20\text{ mol dm}^{-3} \), Initial rate = \( 4.8 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} \). What is the correct rate equation for this reaction?

A.\( \text{Rate} = k[\text{P}][\text{R}]^2 \)
B.\( \text{Rate} = k[\text{P}]^2[\text{Q}][\text{R}] \)
C.\( \text{Rate} = k[\text{P}][\text{Q}][\text{R}]^2 \)
D.\( \text{Rate} = k[\text{P}][\text{R}] \)

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解題

Let us determine the order of reaction with respect to each reactant. Comparing Experiment 1 and 2: when \( [\text{P}] \) is doubled while \( [\text{Q}] \) and \( [\text{R}] \) remain constant, the rate doubles. Therefore, the reaction is first-order with respect to P. Comparing Experiment 1 and 3: when \( [\text{Q}] \) is tripled while \( [\text{P}] \) and \( [\text{R}] \) remain constant, the rate is unchanged. Therefore, the reaction is zero-order with respect to Q. Comparing Experiment 1 and 4: when \( [\text{R}] \) is doubled while \( [\text{P}] \) and \( [\text{Q}] \) remain constant, the rate increases by a factor of 4. Therefore, the reaction is second-order with respect to R. Combining these, the rate equation is \( \text{Rate} = k[\text{P}][\text{R}]^2 \).

評分準則

1 mark for the correct answer A.

題目 12 · 選擇題

1 分

An organic synthesis involves the following two-step pathway starting from propan-1-ol: propan-1-ol is heated with concentrated sulfuric acid to form Compound Y, which is then reacted with cold, dilute, alkaline potassium manganate(VII) to form Compound Z. What are the structural formulas of Compound Y and Compound Z?

A.\( \text{Compound Y} = \text{CH}_3\text{CH}=\text{CH}_2 \); \( \text{Compound Z} = \text{CH}_3\text{CH(OH)CH}_2\text{OH} \)
B.\( \text{Compound Y} = \text{CH}_3\text{CH}=\text{CH}_2 \); \( \text{Compound Z} = \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)
C.\( \text{Compound Y} = \text{CH}_3\text{CH}_2\text{CH}_2\text{OSO}_3\text{H} \); \( \text{Compound Z} = \text{CH}_3\text{CH(OH)CH}_3 \)
D.\( \text{Compound Y} = \text{CH}_3\text{CH}_2\text{CHO} \); \( \text{Compound Z} = \text{CH}_3\text{CH}_2\text{COOH} \)

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解題

Step 1 is an acid-catalyzed dehydration of propan-1-ol, which eliminates a water molecule to form propene, \( \text{CH}_3\text{CH}=\text{CH}_2 \) (Compound Y). Step 2 is a mild oxidation of propene using cold, dilute, alkaline potassium manganate(VII) to form a diol, propane-1,2-diol, \( \text{CH}_3\text{CH(OH)CH}_2\text{OH} \) (Compound Z).

評分準則

1 mark for the correct answer A.

題目 13 · 選擇題

1 分

Methylbenzene is treated with a mixture of concentrated nitric acid and concentrated sulfuric acid at \( 30\ ^\circ\text{C} \). What is the major organic product and what is the electrophile involved in this reaction?

A.Major organic product: 4-nitromethylbenzene; Electrophile: \( \text{NO}_2^+ \)
B.Major organic product: 3-nitromethylbenzene; Electrophile: \( \text{NO}_2^+ \)
C.Major organic product: 4-nitromethylbenzene; Electrophile: \( \text{NO}_2^- \)
D.Major organic product: 3-nitromethylbenzene; Electrophile: \( \text{NO}^+ \)

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解題

The methyl group on the benzene ring is electron-donating by induction, making it activating and 2,4-directing. Mononitration of methylbenzene at \( 30\ ^\circ\text{C} \) yields a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene. Among these, 4-nitromethylbenzene is less sterically hindered and is a major organic product. The active electrophile in this electrophilic aromatic substitution is the nitronium ion, \( \text{NO}_2^+ \), generated by the reaction between concentrated nitric and sulfuric acids.

評分準則

1 mark for the correct answer A.

題目 14 · 選擇題

1 分

Consider the ligand exchange reaction: \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \). What is the correct expression for the stability constant, \( K_{\text{stab}} \), and its corresponding units?

A.\( K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4} \) with unit \( \text{mol}^{-4}\text{ dm}^{12} \)
B.\( K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4} \) with unit \( \text{mol}^{-3}\text{ dm}^9 \)
C.\( K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}][\text{H}_2\text{O}]^4}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4} \) with unit \( \text{mol}^{-4}\text{ dm}^{12} \)
D.\( K_{\text{stab}} = \frac{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4}{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]} \) with unit \( \text{mol}^4\text{ dm}^{-12} \)

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解題

For ligand exchange reactions in aqueous solutions, the concentration of water, \( [\text{H}_2\text{O}] \), is omitted from the stability constant expression because it is the solvent and its concentration remains virtually constant. Therefore, \( K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4} \). The units are calculated as \( \frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^4} = \text{mol}^{-4}\text{ dm}^{12} \).

評分準則

1 mark for the correct answer A.

題目 15 · 選擇題

1 分

The temperature of a reaction mixture is increased from \( 298\text{ K} \) to \( 308\text{ K} \). Which statement correctly describes the changes to the Maxwell-Boltzmann distribution curve and the primary cause of the increased rate of reaction?

A.The peak of the curve shifts to the right and becomes lower; there is a significant increase in the fraction of molecules with energy greater than or equal to the activation energy.
B.The peak of the curve shifts to the left and becomes higher; there is a decrease in the activation energy of the reaction.
C.The peak of the curve shifts to the right and becomes higher; molecules collide with a much greater frequency.
D.The total area under the curve increases; there is a significant increase in the activation energy of the reaction.

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解題

As temperature increases, the average kinetic energy of the molecules increases, causing the Maxwell-Boltzmann distribution curve to flatten: the peak shifts to the right (higher energy) and downwards (fewer molecules have the average energy). The total area under the curve remains constant because the total number of particles is unchanged. The primary cause of the increased rate is that a much larger fraction of molecules now possess kinetic energy greater than or equal to the activation energy (\( E \ge E_a \)), leading to a higher frequency of successful collisions.

評分準則

1 mark for the correct answer A.

題目 16 · 選擇題

1 分

A sample of hydrated magnesium sulfate, \( \text{MgSO}_4 \cdot x\text{H}_2\text{O} \), has an initial mass of \( 4.93\text{ g} \). The sample is heated strongly to constant mass in a crucible to remove all water of crystallisation. The remaining anhydrous solid has a mass of \( 2.41\text{ g} \). What is the value of \( x \)? [Relative atomic masses, \( A_r \): \( \text{H} = 1.0 \); \( \text{O} = 16.0 \); \( \text{Mg} = 24.3 \); \( \text{S} = 32.1 \)]

A.2
B.5
C.7
D.10

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解題

Calculate the molar mass of anhydrous magnesium sulfate, \( \text{MgSO}_4 \): \( M_r(\text{MgSO}_4) = 24.3 + 32.1 + (4 \times 16.0) = 120.4\text{ g mol}^{-1} \). Calculate the moles of anhydrous \( \text{MgSO}_4 \): \( n(\text{MgSO}_4) = \frac{2.41\text{ g}}{120.4\text{ g mol}^{-1}} \approx 0.0200\text{ mol} \). Calculate the mass of water lost: \( 4.93\text{ g} - 2.41\text{ g} = 2.52\text{ g} \). Calculate the moles of water lost: \( n(\text{H}_2\text{O}) = \frac{2.52\text{ g}}{18.0\text{ g mol}^{-1}} = 0.140\text{ mol} \). Determine the mole ratio, \( x \): \( x = \frac{n(\text{H}_2\text{O})}{n(\text{MgSO}_4)} = \frac{0.140}{0.0200} = 7 \).

評分準則

1 mark for the correct answer C.

題目 17 · multiple_choice

1 分

A \(10.0\text{ cm}^3\) sample of a gaseous hydrocarbon, \(C_xH_y\), was exploded with \(70.0\text{ cm}^3\) of oxygen gas (an excess). After cooling to room temperature, the total volume of gas remaining was \(55.0\text{ cm}^3\). This gas mixture was then passed through concentrated aqueous sodium hydroxide, which reduced the gas volume to \(25.0\text{ cm}^3\). All gas volumes were measured at room temperature and pressure.

What is the molecular formula of the hydrocarbon?

A.\(C_3H_4\)
B.\(C_3H_6\)
C.\(C_3H_8\)
D.\(C_4H_8\)

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解題

The equation for the complete combustion of a hydrocarbon is:
\(C_xH_y(g) + \left(x + \frac{y}{4}\right)O_2(g) \rightarrow xCO_2(g) + \frac{y}{2}H_2O(l)\)

Since the final volumes are measured at room temperature and pressure, the water produced is in the liquid state and its volume is negligible.

The sodium hydroxide absorbs \(CO_2\) gas.
Volume of \(CO_2\) produced = \(55.0\text{ cm}^3 - 25.0\text{ cm}^3 = 30.0\text{ cm}^3\).
The remaining \(25.0\text{ cm}^3\) is the unreacted excess oxygen.
Volume of oxygen reacted = \(70.0\text{ cm}^3 - 25.0\text{ cm}^3 = 45.0\text{ cm}^3\).

Since the initial volume of the hydrocarbon is \(10.0\text{ cm}^3\):
\(x = \frac{\text{Volume of } CO_2}{\text{Volume of } C_xH_y} = \frac{30.0}{10.0} = 3\).

Using the reacting volume of oxygen:
\(x + \frac{y}{4} = \frac{\text{Volume of } O_2 \text{ reacted}}{\text{Volume of } C_xH_y} = \frac{45.0}{10.0} = 4.5\).

Substituting \(x = 3\):
\(3 + \frac{y}{4} = 4.5 \implies \frac{y}{4} = 1.5 \implies y = 6\).

Therefore, the molecular formula of the hydrocarbon is \(C_3H_6\).

評分準則

1 mark for the correct answer (B).
[1] Method: Deducing the volume of carbon dioxide (30.0 cm^3) and the volume of oxygen reacted (45.0 cm^3).
[1] Accuracy: Finding x = 3 and y = 6 to give C3H6.

題目 18 · multiple_choice

1 分

An anhydrous metal chloride with the formula \(MCl_n\) contains 44.4% by mass of chlorine. A \(1.60\text{ g}\) sample of this chloride is completely dissolved in water and treated with an excess of aqueous silver nitrate, \(AgNO_3\). A white precipitate of silver chloride, \(AgCl\), is formed.

What is the mass of the dry precipitate obtained?
(\(A_r\): \(Cl = 35.5\), \(Ag = 107.9\))

A.\(0.71\text{ g}\)
B.\(1.43\text{ g}\)
C.\(2.87\text{ g}\)
D.\(5.74\text{ g}\)

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解題

1. Calculate the mass of chlorine in the \(1.60\text{ g}\) sample of \(MCl_n\):
\(\text{Mass of } Cl = 1.60\text{ g} \times 44.4\% = 0.7104\text{ g}\).

2. Calculate the number of moles of chloride ions, \(Cl^-\):
\(n(Cl^-) = \frac{0.7104\text{ g}}{35.5\text{ g mol}^{-1}} = 0.0200\text{ mol}\).

3. According to the precipitation reaction:
\(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\)
The mole ratio of \(Cl^-\) to \(AgCl\) is 1:1.
Therefore, \(n(AgCl) = 0.0200\text{ mol}\).

4. Calculate the molar mass of \(AgCl\):
\(M_r(AgCl) = 107.9 + 35.5 = 143.4\text{ g mol}^{-1}\).

5. Calculate the mass of \(AgCl\) precipitate:
\(\text{Mass} = 0.0200\text{ mol} \times 143.4\text{ g mol}^{-1} = 2.87\text{ g}\).

評分準則

1 mark for the correct option (C).
[1] Method: Determine moles of chloride ions from percentage mass and sample mass, and use stoichiometric equivalence to calculate the mass of AgCl precipitate.

題目 19 · multiple_choice

1 分

Aqueous solutions of scandium(III) salts, such as \([Sc(H_2O)_6]^{3+}\), and zinc(II) salts, such as \([Zn(H_2O)_6]^{2+}\), are both colourless.

Which statement correctly explains this observation?

(\(Sc\) has proton number 21; \(Zn\) has proton number 30)

A.The d-orbitals in both complex ions are not split into different energy levels by the water ligands.
B.In \(Sc^{3+}\) the 3d subshell is completely empty, while in \(Zn^{2+}\) the 3d subshell is completely full.
C.Both ions absorb light in the visible region of the electromagnetic spectrum, but the light is reflected as white light.
D.The ligands in both complexes are too weak to cause any electronic transitions between d-orbitals.

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解題

The colour in transition metal complexes arises from the absorption of specific wavelengths of visible light when d-electrons are promoted from lower-energy d-orbitals to higher-energy d-orbitals (d-d transitions).

- Scandium has the electron configuration \([Ar] 3d^1 4s^2\). In the \(Sc^{3+}\) ion, the configuration is \([Ar]\) (with an empty 3d subshell). Since there are no d-electrons, d-d transitions cannot occur.
- Zinc has the electron configuration \([Ar] 3d^{10} 4s^2\). In the \(Zn^{2+}\) ion, the configuration is \([Ar] 3d^{10}\) (with a completely filled 3d subshell). Since there are no vacant d-orbitals available for an electron to be promoted into, d-d transitions cannot occur.

Therefore, both aqueous ions are colourless.

評分準則

1 mark for the correct option (B).
[1] Method: Relate the electronic configurations of Sc3+ and Zn2+ to the vacancy/presence of d-electrons required for d-d transitions.

題目 20 · multiple_choice

1 分

The stability constants, \(K_{stab}\), of three copper(II) complexes in aqueous solution at \(298\text{ K}\) are given below:

1. \([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O \quad K_{stab} = 5.6 \times 10^1\text{ dm}^{12}\text{ mol}^{-4}\)
2. \([Cu(H_2O)_6]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O \quad K_{stab} = 1.2 \times 10^{13}\text{ dm}^{12}\text{ mol}^{-4}\)
3. \([Cu(H_2O)_6]^{2+} + 2en \rightleftharpoons [Cu(en)_2(H_2O)_2]^{2+} + 4H_2O \quad K_{stab} = 1.0 \times 10^{20}\text{ dm}^6\text{ mol}^{-2}\)
(where \(en\) is the bidentate ligand ethylenediamine, \(H_2NCH_2CH_2NH_2\))

Which statement is correct?

A.The complex \([CuCl_4]^{2-}\) is thermodynamically more stable than \([Cu(NH_3)_4(H_2O)_2]^{2+}\).
B.Adding concentrated aqueous ammonia to a solution of \([Cu(en)_2(H_2O)_2]^{2+}\) will readily displace the ethylenediamine ligands.
C.The stability of the complex ions increases in the order: \([CuCl_4]^{2-} < [Cu(NH_3)_4(H_2O)_2]^{2+} < [Cu(en)_2(H_2O)_2]^{2+}\).
D.The units of \(K_{stab}\) for equilibrium 3 are \(dm^{12}\text{ mol}^{-4}\).

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解題

The stability constant, \(K_{stab}\), is an equilibrium constant for the formation of a complex ion. A higher value of \(K_{stab}\) indicates a more stable complex.

- Option A is incorrect because \(K_{stab}\) of \([CuCl_4]^{2-}\) (\(5.6 \times 10^1\)) is much smaller than that of \([Cu(NH_3)_4(H_2O)_2]^{2+}\) (\(1.2 \times 10^{13}\)).
- Option B is incorrect because \([Cu(en)_2(H_2O)_2]^{2+}\) has a much higher \(K_{stab}\) (\(1.0 \times 10^{20}\)) than the ammonia complex, so ammonia cannot easily displace the ethylenediamine ligands (which is also supported by the chelate effect).
- Option C is correct because the values of \(K_{stab}\) increase in the order: \(5.6 \times 10^1 < 1.2 \times 10^{13} < 1.0 \times 10^{20}\).
- Option D is incorrect. For equilibrium 3, \(K_{stab} = \frac{[[Cu(en)_2(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][en]^2}\), giving units of \(\text{dm}^6\text{ mol}^{-2}\).

評分準則

1 mark for the correct option (C).
[1] Method: Interpret the magnitude of the stability constant (Kstab) as a measure of complex stability and identify the correct trend.

題目 21 · multiple_choice

1 分

The Maxwell-Boltzmann distribution of molecular energies of a gaseous reactant is analyzed at two different temperatures, \(T_1\) and \(T_2\), where \(T_2 > T_1\). The activation energy of the reaction is \(E_a\).

Which statement correctly describes how the molecular energy properties change when the temperature is increased from \(T_1\) to \(T_2\)?

A.The most probable energy of the molecules decreases, and the activation energy, \(E_a\), decreases.
B.The most probable energy of the molecules increases, and the activation energy, \(E_a\), remains unchanged.
C.The total area under the Maxwell-Boltzmann curve increases, and the rate constant, \(k\), increases.
D.The activation energy, \(E_a\), increases, and the rate constant, \(k\), increases.

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解題

When the temperature is increased from \(T_1\) to \(T_2\):
1. The Maxwell-Boltzmann distribution curve flattens and shifts to the right. The peak of the curve, which corresponds to the most probable energy of the molecules, moves to a higher energy.
2. The activation energy, \(E_a\), is the minimum energy required for a collision to result in a reaction. It depends only on the reaction pathway and is not affected by temperature (it remains unchanged).
3. The total area under the curve represents the total number of particles, which remains constant.
4. The fraction of molecules with energy greater than or equal to \(E_a\) increases significantly, leading to an increase in the rate of reaction (and an increase in the rate constant, \(k\)).

Therefore, statement B is correct.

評分準則

1 mark for the correct option (B).
[1] Method: Understand that increasing temperature shifts the peak of the Maxwell-Boltzmann distribution to higher energy (increasing the most probable energy) but does not alter the activation energy Ea.

題目 22 · multiple_choice

1 分

For the reaction between reactants \(A\), \(B\), and \(C\):

\(2A + B + 2C \rightarrow \text{products}\)

The initial rates of reaction were measured at various concentrations at constant temperature:

| Experiment | \([A] / \text{mol dm}^{-3}\) | \([B] / \text{mol dm}^{-3}\) | \([C] / \text{mol dm}^{-3}\) | Initial rate / \text{mol dm}^{-3}\text{ s}^{-1} |
|:---:|:---:|:---:|:---:|:---:|
| 1 | 0.10 | 0.10 | 0.10 | \(1.2 \times 10^{-3}\) |
| 2 | 0.20 | 0.10 | 0.10 | \(2.4 \times 10^{-3}\) |
| 3 | 0.10 | 0.20 | 0.10 | \(4.8 \times 10^{-3}\) |
| 4 | 0.10 | 0.10 | 0.20 | \(1.2 \times 10^{-3}\) |

What is the overall order of the reaction and the units of the rate constant, \(k\)?

A.overall order = 3; units of \(k\) = \(dm^{6}\text{ mol}^{-2}\text{ s}^{-1}\)
B.overall order = 3; units of \(k\) = \(dm^{3}\text{ mol}^{-1}\text{ s}^{-1}\)
C.overall order = 5; units of \(k\) = \(dm^{12}\text{ mol}^{-4}\text{ s}^{-1}\)
D.overall order = 2; units of \(k\) = \(dm^{3}\text{ mol}^{-1}\text{ s}^{-1}\)

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解題

Let the rate equation be: \(\text{Rate} = k [A]^x [B]^y [C]^z\).

- To find the order with respect to \(A\) (value of \(x\)), compare Experiments 1 and 2:
\([B]\) and \([C]\) are kept constant. \([A]\) is doubled (from 0.10 to 0.20), and the initial rate doubles (from \(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\)).
So, \(x = 1\) (first order with respect to \(A\)).

- To find the order with respect to \(B\) (value of \(y\)), compare Experiments 1 and 3:
\([A]\) and \([C]\) are kept constant. \([B]\) is doubled (from 0.10 to 0.20), and the initial rate quadruples (from \(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3}\)).
So, \(2^y = 4 \implies y = 2\) (second order with respect to \(B\)).

- To find the order with respect to \(C\) (value of \(z\)), compare Experiments 1 and 4:
\([A]\) and \([B]\) are kept constant. \([C]\) is doubled (from 0.10 to 0.20), and the initial rate remains constant (\(1.2 \times 10^{-3}\)).
So, \(z = 0\) (zero order with respect to \(C\)).

Overall order of the reaction = \(x + y + z = 1 + 2 + 0 = 3\).

The rate equation is: \(\text{Rate} = k [A][B]^2\).

To find the units of \(k\):
\(k = \frac{\text{Rate}}{[A][B]^2}\)
Units of \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{\text{s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

評分準則

1 mark for the correct option (A).
[1] Method: Deduce the reaction order with respect to each reactant, sum them to find the overall order, and derive the corresponding units of the rate constant.

題目 23 · multiple_choice

1 分

When methylbenzene is reacted with a mixture of concentrated nitric acid and concentrated sulfuric acid at \(55^\circ\text{C}\), it undergoes electrophilic substitution.

Which row correctly identifies the electrophile in this reaction and the role played by concentrated sulfuric acid in the generation of this electrophile?

A.electrophile: \(NO_2^+\); role of \(H_2SO_4\): Brønsted-Lowry acid
B.electrophile: \(NO_2^+\); role of \(H_2SO_4\): Brønsted-Lowry base
C.electrophile: \(NO_2^-\); role of \(H_2SO_4\): Brønsted-Lowry acid
D.electrophile: \(NO_2\); role of \(H_2SO_4\): heterogeneous catalyst

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解題

During the nitration of arenes, the electrophile is the nitronium ion, \(NO_2^+\).

This electrophile is generated by the reaction between concentrated nitric acid (\(HNO_3\)) and concentrated sulfuric acid (\(H_2SO_4\)):
\(HNO_3 + 2H_2SO_4 \rightleftharpoons NO_2^+ + 2HSO_4^- + H_3O^+\)

In this reaction, sulfuric acid (\(H_2SO_4\)) donates protons (\(H^+\)) to nitric acid, which acts as a base by accepting the proton before losing water. Because sulfuric acid is a proton donor, it acts as a Brønsted-Lowry acid.

Therefore, row A is correct.

評分準則

1 mark for the correct option (A).
[1] Method: Identify NO2+ as the electrophile and recognize that sulfuric acid donates a proton to nitric acid, thereby acting as a Brønsted-Lowry acid.

題目 24 · multiple_choice

1 分

An organic reaction sequence is shown below:

\(\text{propene } (CH_3CH=CH_2) \xrightarrow{\text{Step 1}} Y \xrightarrow{\text{Step 2}} \text{propanone } (CH_3COCH_3)\)

Which reagents and conditions are most suitable for Step 1 and Step 2?

A.Step 1: \(H_2O\text{(g)}\) and \(H_3PO_4\) catalyst; Step 2: acidified \(K_2Cr_2O_7\), heat under reflux
B.Step 1: \(H_2O\text{(g)}\) and \(H_3PO_4\) catalyst; Step 2: \(LiAlH_4\) in dry ether
C.Step 1: acidified \(KMnO_4\), cold; Step 2: acidified \(K_2Cr_2O_7\), heat under reflux
D.Step 1: \(HBr\text{(g)}\); Step 2: \(NaOH\text{(aq)}\), heat

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解題

To convert propene into propanone in two steps:
- Step 1: Hydrate propene to form propan-2-ol (intermediate \(Y\)). The reagents and conditions for hydration of alkenes are steam (\(H_2O(g)\)) in the presence of a phosphoric acid catalyst (\(H_3PO_4\)) at high temperature and pressure.
- Step 2: Oxidize the secondary alcohol propan-2-ol to the ketone propanone. The suitable reagent is acidified potassium dichromate(VI) (\(K_2Cr_2O_7\)), heated under reflux.

Let us analyze why the other options are incorrect:
- Option B: \(LiAlH_4\) is a reducing agent, not an oxidizing agent, and would not oxidize propan-2-ol to propanone.
- Option C: Cold acidified \(KMnO_4\) reacts with propene to form a diol (propane-1,2-diol), not propan-2-ol.
- Option D: Reacting propene with \(HBr\) yields 2-bromopropane (major product). Reacting 2-bromopropane with aqueous \(NaOH\) yields propan-2-ol (not propanone).

評分準則

1 mark for the correct option (A).
[1] Method: Identify the intermediate Y as propan-2-ol, and determine the correct reagents and reaction conditions for the hydration of alkenes (Step 1) and the oxidation of secondary alcohols (Step 2).

題目 25 · 選擇題

1 分

A student reacts a sample of a Group 2 carbonate, \(M\text{CO}_3\), with excess hydrochloric acid:

\(M\text{CO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow M\text{Cl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

A \(0.375\text{ g}\) sample of the carbonate produces \(90.0\text{ cm}^3\) of carbon dioxide gas at room temperature and pressure (r.t.p.).

[Assume \(1\text{ mol}\) of gas occupies \(24.0\text{ dm}^3\) at r.t.p.]

What is the identity of the metal \(M\)?

A.Magnesium
B.Calcium
C.Strontium
D.Barium

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解題

1. Calculate the number of moles of \(\text{CO}_2\) gas formed:
\(n(\text{CO}_2) = \frac{90.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.00375\text{ mol}\)

2. Determine the number of moles of \(M\text{CO}_3\):
From the balanced equation, \(1\text{ mol}\) of \(M\text{CO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\).
\(n(M\text{CO}_3) = 0.00375\text{ mol}\)

3. Calculate the molar mass of \(M\text{CO}_3\):
\(M_r(M\text{CO}_3) = \frac{0.375\text{ g}}{0.00375\text{ mol}} = 100.0\text{ g mol}^{-1}\)

4. Find the relative atomic mass of \(M\):
\(M_r(M\text{CO}_3) = A_r(M) + 12.0 + 3(16.0) = 100.0\)
\(A_r(M) + 60.0 = 100.0 \Rightarrow A_r(M) = 40.0\)

This relative atomic mass corresponds to calcium (\(A_r = 40.1\)).

評分準則

[1 mark] B is the correct answer. Award 1 mark for calculating the correct molar mass (100 g/mol) and matching to Calcium.

題目 26 · 選擇題

1 分

Consider the complex ion \([Co(en)_2Cl_2]^+\), where \(en\) is the bidentate ligand ethane-1,2-diamine. Which statement about the isomers of \([Co(en)_2Cl_2]^+\) is correct?

A.There are two geometrical isomers, and both are optically active.
B.There are two geometrical isomers, and only the *cis* isomer is optically active.
C.There are three geometrical isomers, and none are optically active.
D.There are three geometrical isomers, and only the *trans* isomer is optically active.

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解題

The complex \([Co(en)_2Cl_2]^+\) exists as two geometrical isomers: *cis* and *trans*. The *trans* isomer is symmetrical (has a plane of symmetry) and is therefore optically inactive. The *cis* isomer is chiral (lacks a plane of symmetry) and exists as a pair of non-superimposable mirror images (enantiomers), which are optically active.

評分準則

[1 mark] B is correct. Award 1 mark for correctly identifying that only the *cis* isomer lacks symmetry and is optically active.

題目 27 · 選擇題

1 分

For a reaction \(2\text{A} + \text{B} \rightarrow \text{C}\), the rate equation is determined to be:

\(\text{rate} = k[\text{A}][\text{B}]^2\)

If the concentration of \(\text{A}\) is doubled and the concentration of \(\text{B}\) is halved, how does the initial rate of reaction change?

A.It decreases by a factor of 2.
B.It increases by a factor of 2.
C.It decreases by a factor of 4.
D.It remains unchanged.

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解題

Let the initial rate be \(R_1 = k[\text{A}][\text{B}]^2\).
When \([\text{A}]\) is doubled to \(2[\text{A}]\) and \([\text{B}]\) is halved to \(0.5[\text{B}]\), the new rate \(R_2\) is:
\(R_2 = k(2[\text{A}])(0.5[\text{B}])^2 = k \times 2[\text{A}] \times 0.25[\text{B}]^2 = 0.5 k[\text{A}][\text{B}]^2 = 0.5 R_1\).
Therefore, the rate decreases by a factor of 2.

評分準則

[1 mark] A is correct. Award 1 mark for calculating that the rate becomes 0.5 times the original rate.

題目 28 · 選擇題

1 分

A student wants to synthesize propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\), starting from ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\), in a three-step reaction pathway.

Which series of reagents and conditions would achieve this synthesis successfully?

A.1. \(\text{NaBH}_4\); 2. \(\text{KCN}\) in ethanol, heat under reflux; 3. \(\text{dilute HCl(aq)}\), heat under reflux
B.1. \(\text{H}_2\text{SO}_4\text{(conc)}\) at \(170\ ^\circ\text{C}\); 2. \(\text{HBr(g)}\); 3. \(\text{KCN}\) in ethanol, heat under reflux
C.1. \(\text{PCl}_5\); 2. \(\text{KCN}\) in ethanol, heat under reflux; 3. \(\text{dilute HCl(aq)}\), heat under reflux
D.1. \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\), heat under reflux; 2. \(\text{PCl}_5\); 3. \(\text{NH}_3\text{(aq)}\)

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解題

Step 1: Reaction of ethanol with \(\text{PCl}_5\) produces chloroethane, \(\text{CH}_3\text{CH}_2\text{Cl}\).
Step 2: Nucleophilic substitution of chloroethane with \(\text{KCN}\) in ethanol under reflux yields propanenitrile, \(\text{CH}_3\text{CH}_2\text{CN}\), extending the carbon chain by one.
Step 3: Acid hydrolysis of propanenitrile with dilute \(\text{HCl(aq)}\) under reflux yields propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).

評分準則

[1 mark] C is correct. Award 1 mark for identifying the correct 3-step pathway of chlorination, nitrile formation, and nitrile hydrolysis.

題目 29 · 選擇題

1 分

Methylbenzene is reacted with a mixture of concentrated nitric acid and concentrated sulfuric acid at \(50\ ^\circ\text{C}\) to form mononitrated products.

Which isomer is the major organic product of this reaction, and what is the role of sulfuric acid in the reaction mechanism?

A.3-nitromethylbenzene; catalyst
B.3-nitromethylbenzene; oxidant
C.4-nitromethylbenzene; catalyst
D.4-nitromethylbenzene; oxidant

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解題

The methyl group in methylbenzene is electron-donating and directs incoming electrophiles to the 2- (ortho) and 4- (para) positions. Thus, 4-nitromethylbenzene is a major product, while 3-nitromethylbenzene is a minor product. Concentrated sulfuric acid reacts with concentrated nitric acid to generate the active electrophile (\(\text{NO}_2^+\)) and is regenerated at the end, acting as a catalyst.

評分準則

Paper 21 (AS Level Structured)

Answer all questions in the spaces provided.

Part (a):
- [1 Mark] Mentions heat energy released when 1 mole of substance is completely burned in oxygen.
- [1 Mark] Mentions standard conditions (298 K, 100 kPa / 1 bar) and states.

Part (b)(i):
- [1 Mark] Correct calculation of temperature change = 24.0 degrees C / K.
- [1 Mark] Heat energy calculation: q = 15048 J or 15.05 kJ (accept 15.0 kJ).

Part (b)(ii):
- [1 Mark] Moles of methanol = 0.0250 (or 0.025) mol.

Part (b)(iii):
- [1 Mark] Divides q by moles calculated in (ii).
- [1 Mark] Correct negative sign and value: -602 kJ mol^-1 (accept -601.9 kJ mol^-1).

Part (c)(i):
- [1 Mark] Correct chemical equation with balancing and all correct state symbols: CH3OH(l) + 1.5O2(g) -> CO2(g) + 2H2O(l).

Part (c)(ii):
- [1 Mark] Correct application of Hess's Law cycle or expression: [-394 + 2(-286)] - [-239].
- [1 Mark] Final calculated value of -727 kJ mol^-1 (must include minus sign).

Paper 31 (Practical Skills)

Carry out all practical experiments and record your findings.

- Test (i): Correctly reports red-brown and blue precipitates insoluble in excess NaOH (1.33 marks)
- Test (ii): Correctly reports red-brown and blue precipitates, with the blue precipitate dissolving in excess NH3 to give a deep blue solution (1.5 marks)
- Test (iii): Correctly reports formation of a brown precipitate/solution (1.5 marks)
- Test (iv): Correctly reports no precipitate/no reaction (1.5 marks)
- Test (v): Correctly reports white precipitate insoluble in dilute acid (1.5 marks)
- Correctly identifies Cu2+ and explains based on blue ppt dissolving to a deep blue solution in excess NH3 (1.5 marks)
- Correctly identifies Fe3+ and explains based on red-brown ppt insoluble in excess NH3 and NaOH (1.5 marks)
- Correctly identifies SO4^2- and explains based on white ppt with Ba2+ insoluble in acid (1 mark)
- Balanced ionic equation with correct state symbols for either Cu2+ or Fe3+ reaction with iodide ions (2 marks)

Paper 41 (A Level Structured)

Answer all structured questions in the spaces provided.

Part (a) [4 marks]:
- M1: Step 1 reagents: Conc. HNO3 and conc. H2SO4. [1]
- M2: Step 1 conditions: Temperature between 50 and 55 degrees Celsius (Reject if > 60 degrees Celsius). [1]
- M3: Step 2 reagents: Tin (Sn) and conc. HCl. [1]
- M4: Step 2 conditions: Heat / reflux, followed by adding NaOH(aq). [1]

Part (b) [4 marks]:
- M5: Preparation of HNO2: Sodium nitrite (NaNO2) and hydrochloric acid (HCl) (or any dilute acid). [1]
- M6: Correct structure of 4-(phenylazo)phenol showing both benzene rings joined by -N=N- and the -OH group in the para position relative to the azo group. [1]
- M7: Temperature explanation: Diazonium salt is unstable and decomposes above 10 degrees Celsius. [1]
- M8: Observation: Effervescence / bubbles of nitrogen gas (and formation of phenol). [1]

Part (c) [2 marks]:
- M9: Correct order of basicity: phenylamine < ammonia < ethylamine. [1]
- M10: Explanation: Lone pair on nitrogen in phenylamine delocalises into the benzene pi-cloud, making it less available to accept a proton. [1]

Paper 51 (Planning & Evaluation)

Answer all planning, analysis and evaluation questions.

2 題目 · 30 分

題目 1 · Planning & Evaluation

15 分

A student plans to investigate the stoichiometry of the colored complex formed between iron(III) ions, \(\text{Fe}^{3+}\), and sulfosalicylate ligands, \(\text{L}^-\), in aqueous solution using Job's method of continuous variation. This method involves mixing varying volumes of \(0.00200\text{ mol dm}^{-3}\) \(\text{Fe}^{3+}(aq)\) and \(0.00200\text{ mol dm}^{-3}\) \(\text{L}^-(aq)\) solutions, keeping the total volume constant at \(10.0\text{ cm}^3\), and measuring the absorbance of each mixture using a colorimeter.

(a) Identify the independent variable and the dependent variable in this experiment. [2]

(b) The student is provided with solid iron(III) ammonium sulfate dodecahydrate, \(\text{NH}_4\text{Fe}(\text{SO}_4)_2 \cdot 12\text{H}_2\text{O}\) (\(M_r = 482.2\)).
(i) Calculate the mass of the solid required to prepare \(250\text{ cm}^3\) of a \(0.0200\text{ mol dm}^{-3}\) stock solution of \(\text{Fe}^{3+}(aq)\). [1]
(ii) Describe how the student should dilute this stock solution to prepare \(100\text{ cm}^3\) of a \(0.00200\text{ mol dm}^{-3}\) \(\text{Fe}^{3+}(aq)\) solution. State the names and capacities of the apparatus used. [3]

(c) Iron(III) solutions are highly acidic and can cause skin irritation. State one appropriate safety precaution, other than wearing safety goggles, that the student should take when handling the solid or solution. [1]

(d) The student mixes the \(0.00200\text{ mol dm}^{-3}\) solutions of \(\text{Fe}^{3+}(aq)\) and \(\text{L}^-(aq)\) in different volume ratios. The absorbance of each mixture is measured and recorded in the table below:

| Volume of \(0.00200\text{ mol dm}^{-3}\) \(\text{Fe}^{3+}(aq)\) / \(\text{cm}^3\) | Volume of \(0.00200\text{ mol dm}^{-3}\) \(\text{L}^-(aq)\) / \(\text{cm}^3\) | Absorbance / arbitrary units |
| :--- | :--- | :--- |
| 1.0 | 9.0 | 0.24 |
| 2.0 | 8.0 | 0.48 |
| 3.0 | 7.0 | 0.55 |
| 4.0 | 6.0 | 0.47 |
| 5.0 | 5.0 | 0.25 |
| 6.0 | 4.0 | 0.31 |
| 7.0 | 3.0 | 0.23 |
| 8.0 | 2.0 | 0.15 |
| 9.0 | 1.0 | 0.07 |

(i) Identify the anomalous absorbance reading from the table. [1]
(ii) Suggest a practical reason, other than a mistake in measuring volumes, that could explain this anomalous reading. [1]

(e) By ignoring the anomalous point:
(i) Calculate the mathematical equations of the two intersecting straight lines of best fit (absorbance vs volume of \(\text{Fe}^{3+}\)). [2]
(ii) Use these equations to determine the precise volume of \(\text{Fe}^{3+}\) at the point of intersection. [1]

(f) (i) Deduce the mole ratio of \(\text{Fe}^{3+} : \text{L}^-\) at the intersection and hence suggest the formula of the complex. [2]
(ii) Explain why Job's method of continuous variation is not suitable if the stability constant, \(K_{\text{stab}}\), of the complex is extremely low. [1]

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解題

(a) Independent variable: The volume ratio of \(\text{Fe}^{3+}\) to \(\text{L}^-\). Dependent variable: The absorbance of the mixture.

(b) (i) \(n = 0.0200\text{ mol dm}^{-3} \times 0.250\text{ dm}^3 = 0.00500\text{ mol}\). Mass \(= 0.00500\text{ mol} \times 482.2\text{ g mol}^{-1} = 2.41\text{ g}\) (or \(2.411\text{ g}\)).
(ii) Pipette \(10.0\text{ cm}^3\) of the stock solution using a volumetric pipette. Transfer the pipette contents into a \(100\text{ cm}^3\) volumetric flask. Fill the flask with distilled water up to the graduation mark and shake to mix.

(c) Wear chemical-resistant gloves or a lab coat to avoid skin contact with the solution.

(d) (i) The mixture with \(5.0\text{ cm}^3\) of \(\text{Fe}^{3+}\) (absorbance of 0.25).
(ii) The outside of the cuvette was dirty/smudged with fingerprints, or the cuvette was not rinsed/dried properly, leaving residual solvent, or the colorimeter was not zeroed correctly.

(e) (i) Line 1 passes through (0,0), (1.0, 0.24), and (2.0, 0.48). Gradient \(= 0.24\). Equation: \(\text{Absorbance} = 0.24 \times V(\text{Fe}^{3+})\).
Line 2 passes through remaining points except the anomalous one. Gradient \(= \frac{0.07 - 0.55}{9.0 - 3.0} = -0.08\). Using \(y - y_1 = m(x - x_1)\): \(\text{Absorbance} - 0.07 = -0.08(V(\text{Fe}^{3+}) - 9.0) \implies \text{Absorbance} = -0.08 \times V(\text{Fe}^{3+}) + 0.79\).
(ii) At the intersection: \(0.24 \times V = -0.08 \times V + 0.79 \implies 0.32 \times V = 0.79 \implies V = 2.47\text{ cm}^3\) (accept range \(2.45 - 2.50\text{ cm}^3\)).

(f) (i) Volume of \(\text{Fe}^{3+} = 2.47\text{ cm}^3\); Volume of \(\text{L}^- = 10.0 - 2.47 = 7.53\text{ cm}^3\). Ratio of \(\text{Fe}^{3+} : \text{L}^- = 2.47 : 7.53 \approx 1 : 3\). The formula of the complex is \([\text{FeL}_3]\).
(ii) If \(K_{\text{stab}}\) is low, the complex dissociates significantly. The graph would show a broad curve rather than two intersecting straight lines, making it impossible to identify the correct intersection point.

評分準則

Part (a):
- 1 mark for identifying the independent variable as volume ratio or volumes of Fe3+ / L-.
- 1 mark for identifying the dependent variable as absorbance.

Part (b):
- 1 mark for calculating mass = 2.41 g (or 2.411 g).
- 1 mark for stating the use of a 10.0 cm3 volumetric pipette.
- 1 mark for stating the use of a 100 cm3 volumetric flask.
- 1 mark for describing making up to the mark with distilled water and mixing.

Part (c):
- 1 mark for specifying chemical-resistant gloves or lab coat (reject safety goggles as it is in the question stem).

Part (d):
- 1 mark for identifying the anomalous reading at 5.0 cm3 of Fe3+.
- 1 mark for a correct practical reason (e.g., fingerprints on cuvette, poor rinsing, bubble in solution).

Part (e):
- 1 mark for calculating Equation 1: Absorbance = 0.24 * V.
- 1 mark for calculating Equation 2: Absorbance = -0.08 * V + 0.79.
- 1 mark for calculating the intersection volume V = 2.47 cm3 (allow 2.45 to 2.50 cm3).

Part (f):
- 1 mark for finding the mole ratio of 1 : 3.
- 1 mark for deducing the formula [FeL3].
- 1 mark for explaining that significant dissociation due to low stability constant produces a broad curve instead of distinct intersecting lines.

題目 2 · Planning & Evaluation

15 分

The thermal decomposition of aqueous benzenediazonium chloride, \(\text{C}_6\text{H}_5\text{N}_2\text{Cl}\), proceeds according to the equation:
\[\text{C}_6\text{H}_5\text{N}_2^+(aq) + \text{H}_2\text{O}(l) \to \text{C}_6\text{H}_5\text{OH}(aq) + \text{N}_2(g) + \text{H}^+(aq)\]
The reaction is first-order with respect to \(\text{C}_6\text{H}_5\text{N}_2^+\). A student wants to determine the activation energy, \(E_a\), of this reaction by measuring the first-order rate constant, \(k\), at several different temperatures.

(a) Draw a clearly labeled diagram of the experimental setup that can be used to monitor the progress of this reaction at an elevated temperature by measuring the volume of gas evolved over time. [2]

(b) The experiment is performed at temperatures between \(35.0^{\circ}\text{C}\) and \(55.0^{\circ}\text{C}\).
(i) Describe how the temperature of the reaction mixture should be maintained constant during a run. [1]
(ii) Explain why heating the reaction flask directly with a Bunsen burner is not appropriate for this investigation. [1]

(c) Explain how the student can use the volumes of nitrogen gas collected at various times, \(V_t\), and the final volume of gas collected at completion, \(V_{\infty}\), to show that the reaction is first-order and to determine the rate constant, \(k\), at a single temperature. Your explanation must include the specific graph that should be plotted and how \(k\) is obtained from it. [3]

(d) The student determined the rate constant, \(k\), at five different temperatures. The experimental results are shown in the table:

| Temperature / \(^{\circ}\text{C}\) | Rate constant, \(k\) / \(\text{s}^{-1}\) | Temperature, \(T\) / \(\text{K}\) | \(\frac{1}{T} \times 10^3\) / \(\text{K}^{-1}\) | \(\ln(k / \text{s}^{-1})\) |
| :--- | :--- | :--- | :--- | :--- |
| 35.0 | \(6.41 \times 10^{-5}\) | | | |
| 40.0 | \(1.31 \times 10^{-4}\) | | | |
| 45.0 | \(2.62 \times 10^{-4}\) | | | |
| 50.0 | \(5.15 \times 10^{-4}\) | | | |
| 55.0 | \(9.88 \times 10^{-4}\) | | | |

(i) Complete the table. Record your values of \(T\) to 1 decimal place, \(\frac{1}{T} \times 10^3\) to 3 decimal places, and \(\ln(k / \text{s}^{-1})\) to 3 decimal places. [2]
(ii) Calculate the gradient of the Arrhenius plot of \(\ln(k / \text{s}^{-1})\) against \(\frac{1}{T}\) using your completed data. Show your working clearly. [2]

(e) Use your gradient from (d)(ii) to calculate the activation energy, \(E_a\), of the reaction in \(\text{kJ mol}^{-1}\). (Gas constant, \(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)). [2]

(f) (i) Identify a potential source of systematic error in this experimental method of gas collection. [1]
(ii) Suggest a modification to the procedure or apparatus to minimize this error. [1]

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解題

(a) The diagram must show: a sealed reaction vessel (conical flask or round-bottomed flask) placed inside a water bath; a thermometer in the reaction vessel or water bath; a delivery tube connecting the reaction vessel to a gas collection apparatus (either a gas syringe or a graduated cylinder inverted over water).

(b) (i) Place the flask in a thermostatically controlled water bath.
(ii) Direct heating with a Bunsen burner leads to local overheating, making it impossible to control the temperature evenly. In addition, rapid decomposition or boiling of the solvent may occur.

(c) The concentration of reactant at time \(t\) is proportional to the remaining volume of gas to be evolved, \((V_{\infty} - V_t)\). For a first-order reaction, \(\ln(V_{\infty} - V_t) = -kt + \ln(V_{\infty})\). Thus, a plot of \(\ln(V_{\infty} - V_t)\) against time \(t\) yields a straight line with a constant negative gradient. The rate constant \(k\) is equal to \(-\text{gradient}\).

(d) (i) The completed table values:
- 35.0: \(T = 308.2\text{ K}\), \(\frac{1}{T} \times 10^3 = 3.245\text{ K}^{-1}\), \(\ln(k) = -9.655\)
- 40.0: \(T = 313.2\text{ K}\), \(\frac{1}{T} \times 10^3 = 3.193\text{ K}^{-1}\), \(\ln(k) = -8.940\)
- 45.0: \(T = 318.2\text{ K}\), \(\frac{1}{T} \times 10^3 = 3.143\text{ K}^{-1}\), \(\ln(k) = -8.247\)
- 50.0: \(T = 323.2\text{ K}\), \(\frac{1}{T} \times 10^3 = 3.094\text{ K}^{-1}\), \(\ln(k) = -7.571\)
- 55.0: \(T = 328.2\text{ K}\), \(\frac{1}{T} \times 10^3 = 3.047\text{ K}^{-1}\), \(\ln(k) = -6.920\)

(ii) Choosing two points, e.g., (3.245, -9.655) and (3.047, -6.920):
\(\text{Gradient} = \frac{-6.920 - (-9.655)}{(3.047 - 3.245) \times 10^{-3}} = \frac{2.735}{-0.198 \times 10^{-3}} = -13813\text{ K}\) (accept range \(-13700\text{ K}\) to \(-13900\text{ K}\)).

(e) Arrhenius equation: \(\ln k = -\frac{E_a}{RT} + \ln A\), so \(\text{Gradient} = -\frac{E_a}{R}\).
\(E_a = -\text{Gradient} \times R = 13813 \times 8.314 = 114841\text{ J mol}^{-1} \approx 115\text{ kJ mol}^{-1}\) (accept range \(113 - 116\text{ kJ mol}^{-1}\)).

(f) (i) Nitrogen is slightly soluble in water, leading to a loss of recorded gas volume.
(ii) Use a gas syringe instead of collecting over water, or pre-saturate the water in the trough with nitrogen gas.

評分準則

Part (a):
- 1 mark for drawing a closed reaction flask in a water bath with a thermometer.
- 1 mark for showing a connection via a delivery tube to a gas syringe or a graduated cylinder inverted over water.

Part (b):
- 1 mark for describing the use of a thermostatically controlled water bath.
- 1 mark for explaining that a Bunsen burner causes uneven/local overheating and poor temperature control.

Part (c):
- 1 mark for stating that concentration at time t is proportional to (V_infinity - V_t).
- 1 mark for stating that a plot of ln(V_infinity - V_t) against time t is plotted to give a straight line.
- 1 mark for stating that the gradient of this line is equal to -k.

Part (d):
- 2 marks for completing all table values correctly (1 mark if 1-2 errors in calculation or decimal places).
- 2 marks for calculating the gradient showing clear working (e.g., -13813 K, accept -13700 to -13900 K).

Part (e):
- 1 mark for equating gradient = -Ea / R.
- 1 mark for calculating Ea in kJ mol-1 with a value consistent with their gradient (e.g., 115 kJ mol-1).

Part (f):
- 1 mark for identifying nitrogen solubility in water or initial air expansion in flask.
- 1 mark for a matching modification (e.g., using a gas syringe or pre-saturating the water).

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