Cambridge IAL · Thinka 原創模擬試題

2024 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

Thinka Nov 2024 (V2) Cambridge International A Level-Style Mock — Chemistry (9701)

230 345 分鐘2024
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V2) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

Paper 12

Answer all 40 multiple-choice questions. Select one alternative (A, B, C, or D) for each question.
40 題目 · 40
題目 1 · 選擇題
1
A 1.00 g sample of a mixture of calcium carbonate (\(\text{CaCO}_3\), \(M_{\text{r}} = 100.1\)) and magnesium carbonate (\(\text{MgCO}_3\), \(M_{\text{r}} = 84.3\)) is reacted with an excess of dilute hydrochloric acid. The volume of carbon dioxide gas released is \(267\text{ cm}^3\) (measured at room temperature and pressure, r.t.p.). What is the percentage by mass of \(\text{CaCO}_3\) in the mixture? [Assume 1 mol of gas occupies \(24.0\text{ dm}^3\) at r.t.p.]
  1. A.20.0%
  2. B.40.0%
  3. C.60.0%
  4. D.80.0%
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解題

1. Calculate the total number of moles of \(\text{CO}_2\) gas evolved: \(n(\text{CO}_2) = \frac{267\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.011125\text{ mol}\). 2. Both carbonates react with \(\text{HCl}\) in a 1:1 molar ratio to produce \(\text{CO}_2\): \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) and \(\text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). 3. Let \(x\) be the mass of \(\text{CaCO}_3\) in grams. The mass of \(\text{MgCO}_3\) is \(1.00 - x\) grams. 4. Set up the mole equation: \(\frac{x}{100.1} + \frac{1.00 - x}{84.3} = 0.011125\). 5. Solve for \(x\): \(0.00999x + 0.01186 - 0.01186x = 0.011125\) which gives \(-0.00187x = -0.000735\). Thus, \(x \approx 0.393\text{ g}\), which corresponds to \(40.0\%\) by mass of the \(1.00\text{ g}\) mixture.

評分準則

[1 mark] B - Correctly calculates total moles of carbon dioxide, sets up the linear equation relating mass and moles of both carbonates, and solves to find that calcium carbonate represents 40.0% of the mixture by mass.
題目 2 · 選擇題
1
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_{\text{a}} = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(25.0\text{ cm}^3\) of \(0.0800\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting buffer solution at \(298\text{ K}\)?
  1. A.4.69
  2. B.4.87
  3. C.5.05
  4. D.5.22
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解題

1. Calculate initial moles: Propanoic acid: \(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\); Sodium hydroxide: \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.0800\text{ mol dm}^{-3} = 2.00 \times 10^{-3}\text{ mol}\). 2. Calculate the moles of species present after neutralisation: \(\text{NaOH}\) reacts completely with propanoic acid to form sodium propanoate (\(\text{A}^-\)): \(n(\text{A}^-) = 2.00 \times 10^{-3}\text{ mol}\). Remaining propanoic acid (\(\text{HA}\)): \(n(\text{HA}) = 5.00 \times 10^{-3} - 2.00 \times 10^{-3} = 3.00 \times 10^{-3}\text{ mol}\). 3. Calculate the pH of the buffer: \(\text{pH} = \text{p}K_{\text{a}} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\), where \(\text{p}K_{\text{a}} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\). Therefore, \(\text{pH} = 4.87 + \log_{10}\left(\frac{2.00 \times 10^{-3}}{3.00 \times 10^{-3}}\right) = 4.87 - 0.18 = 4.69\).

評分準則

[1 mark] A - Correctly determines the moles of remaining acid and conjugate base formed, and applies the Henderson-Hasselbalch equation to find the correct pH of 4.69.
題目 3 · 選擇題
1
Which statement best explains why aqueous copper(II) ions, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), appear blue?
  1. A.The d-orbitals of the \(\text{Cu}^{2+}\) ion split into two energy levels, and the complex absorbs blue light to promote an electron to a higher level.
  2. B.The d-orbitals of the \(\text{Cu}^{2+}\) ion split into two energy levels, and the complex absorbs red-orange light to promote an electron to a higher level.
  3. C.Electrons in the d-orbitals of the \(\text{Cu}^{2+}\) ion transition to a lower energy level, emitting blue light.
  4. D.The \(\text{Cu}^{2+}\) ion has a fully filled d-subshell which reflects blue light.
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解題

When ligands bond to transition metal ions like \(\text{Cu}^{2+}\), the d-orbitals split into two groups of different energy levels. Electrons absorb a specific frequency of light corresponding to this energy difference (\(\Delta E = h\nu\)) and are promoted to the higher level (d-d transition). For the solution to appear blue, the complex must absorb light of the complementary colour (red-orange) and transmit the unabsorbed blue wavelengths.

評分準則

[1 mark] B - Correctly explains d-orbital splitting and that absorption of red-orange light (the complementary colour of blue) is responsible for the blue appearance of the complex.
題目 4 · 選擇題
1
The reaction between propanone and iodine in acidic solution is investigated: \(\text{CH}_3\text{COCH}_3 + \text{I}_2 \xrightarrow{\text{H}^+} \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^-\) The rate equation is found to be: \(\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\) Under certain conditions, the initial rate of reaction is \(1.2 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). The concentration of propanone is doubled, the concentration of \(\text{H}^+\) is halved, and the concentration of \(\text{I}_2\) is tripled. What is the new initial rate of reaction, in \(\text{mol dm}^{-3}\text{ s}^{-1}\)?
  1. A.1.2 x 10^-5
  2. B.2.4 x 10^-5
  3. C.3.6 x 10^-5
  4. D.7.2 x 10^-5
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解題

From the rate equation, \(\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\), we see that: 1. The reaction is first-order with respect to propanone, so doubling its concentration multiplies the rate by 2. 2. The reaction is first-order with respect to \(\text{H}^+\), so halving its concentration multiplies the rate by 0.5. 3. Iodine does not appear in the rate equation, meaning the reaction is zero-order with respect to \(\text{I}_2\), so changing its concentration has no effect on the rate. Overall effect on rate = \(2 \times 0.5 \times 1 = 1\). Thus, the rate remains unchanged: \(1.2 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\).

評分準則

[1 mark] A - Correctly identifies the order with respect to each reactant from the rate equation, deduces that changing iodine concentration has no effect, and calculates that the rate remains unchanged.
題目 5 · 選擇題
1
Which statement about the reaction of methylbenzene with a mixture of concentrated nitric acid and concentrated sulfuric acid is correct?
  1. A.The reaction is an addition reaction.
  2. B.The methyl group is a 3-directing group.
  3. C.The active electrophile is \(\text{NO}_2^+\).
  4. D.The sulfuric acid acts as a reducing agent.
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解題

In the nitration of methylbenzene, the concentrated sulfuric acid acts as a catalyst to generate the active electrophile, which is the nitronium ion (\(\text{NO}_2^+\)). The methyl group is an electron-donating group and therefore acts as a 2,4-directing activator. The reaction is an electrophilic substitution reaction.

評分準則

[1 mark] C - Correctly identifies that the nitronium ion (\(\text{NO}_2^+\)) is the active electrophile in this electrophilic aromatic substitution reaction.
題目 6 · 選擇題
1
The structure of 2-aminopropanoic acid (alanine) is shown: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}\) Which species is the major component present in an aqueous solution of alanine at \(\text{pH} = 12\)?
  1. A.\(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\)
  2. B.\(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\)
  3. C.\(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\)
  4. D.\(\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}\)
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解題

At a highly alkaline pH of 12, the concentration of \(\text{OH}^-\) ions is high, which deprotonates the carboxylic acid group (\(-\text{COOH}\)) to form a carboxylate anion (\(-\text{COO}^-\)). The amino group (\(-\text{NH}_2\)) remains unprotonated as \(-\text{NH}_2\). Therefore, the major species present is \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).

評分準則

[1 mark] A - Identifies that at pH = 12, the basic amino group is unprotonated and the carboxylic acid group is fully deprotonated.
題目 7 · 選擇題
1
Two Period 3 elements, \(X\) and \(Y\), form oxides \(X\text{O}_3\) and \(Y_2\text{O}_3\) respectively. Oxide \(X\text{O}_3\) reacts vigorously with water to form a solution of \(\text{pH} \approx 2\). Oxide \(Y_2\text{O}_3\) is insoluble in water but reacts with both dilute hydrochloric acid and aqueous sodium hydroxide. What are elements \(X\) and \(Y\)?
  1. A.X = phosphorus, Y = silicon
  2. B.X = sulfur, Y = aluminium
  3. C.X = silicon, Y = aluminium
  4. D.X = sulfur, Y = magnesium
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解題

- Oxide \(X\text{O}_3\) must be sulfur trioxide, \(\text{SO}_3\), which reacts vigorously with water to form sulfuric acid, a strong acid (\(\text{pH} \approx 2\)). Thus, \(X\) is sulfur. - Oxide \(Y_2\text{O}_3\) must be aluminium oxide, \(\text{Al}_2\text{O}_3\), which is insoluble in water but amphoteric, reacting with both acids (hydrochloric acid) and bases (sodium hydroxide). Thus, \(Y\) is aluminium.

評分準則

[1 mark] B - Correctly identifies sulfur from the properties of \(\text{SO}_3\) and aluminium from the amphoteric nature of \(\text{Al}_2\text{O}_3\).
題目 8 · 選擇題
1
Sulfur dioxide is oxidised to sulfur trioxide in the atmosphere, catalysed by nitrogen monoxide (\(\text{NO}\)). Which equation represents a step in this catalytic cycle?
  1. A.\(\text{SO}_2 + \text{NO} \rightarrow \text{SO} + \text{NO}_2\)
  2. B.\(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\)
  3. C.\(\text{SO}_3 + \text{NO} \rightarrow \text{SO}_2 + \text{NO}_2\)
  4. D.\(\text{NO} + \text{O}_3 \rightarrow \text{NO}_3 + \text{O}\)
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解題

In the catalytic oxidation of sulfur dioxide by nitrogen monoxide (\(\text{NO}\)), \(\text{NO}\) is first oxidised by atmospheric oxygen to form nitrogen dioxide (\(\text{NO}_2\)): \(\text{NO} + \frac{1}{2}\text{O}_2 \rightarrow \text{NO}_2\). In the second step, sulfur dioxide is oxidised to sulfur trioxide by \(\text{NO}_2\), which regenerates the \(\text{NO}\) catalyst: \(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\). Therefore, option B is correct.

評分準則

[1 mark] B - Correctly identifies the reaction step where sulfur dioxide is oxidised by nitrogen dioxide to regenerate the nitrogen monoxide catalyst.
題目 9 · 選擇題
1
A sample of \(10\text{ cm}^3\) of a gaseous hydrocarbon \(C_xH_y\) was exploded with \(80\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature, the total volume of gas remaining was \(70\text{ cm}^3\). After shaking the mixture with excess aqueous sodium hydroxide, the remaining volume of gas was \(30\text{ cm}^3\). What is the formula of the hydrocarbon?
  1. A.\(CH_4\)
  2. B.\(C_3H_8\)
  3. C.\(C_4H_4\)
  4. D.\(C_4H_{10}\)
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解題

The volume ratio of reactants and products under the same conditions of temperature and pressure is equal to their mole ratio (Avogadro's law). The volume of excess oxygen was \(80\text{ cm}^3\). Shaking with \(NaOH\) absorbs \(CO_2\), so the remaining volume of gas (\(30\text{ cm}^3\)) is the unreacted oxygen. The volume of carbon dioxide produced is \(70 - 30 = 40\text{ cm}^3\). The volume of oxygen reacted is \(80 - 30 = 50\text{ cm}^3\). Since the initial volume of hydrocarbon is \(10\text{ cm}^3\), the ratio of \(C_xH_y : CO_2 : O_2\text{ reacted} = 10 : 40 : 50 = 1 : 4 : 5\). Therefore, \(x = 4\) and \(x + y/4 = 5\), giving \(y = 4\). The molecular formula is \(C_4H_4\).

評分準則

1 mark for the correct option C. [Method: Deduce volume of CO2 (40 cm3) and reacted O2 (50 cm3); find simplest ratio of hydrocarbon to carbon dioxide (1:4) to determine x = 4; use oxygen balance to find y = 4; match with correct formula C4H4]
題目 10 · 選擇題
1
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid, \(CH_3CH_2COOH\) (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)), with \(50.0\text{ cm}^3\) of \(0.050\text{ mol dm}^{-3}\) sodium hydroxide, \(NaOH\). What is the pH of the resulting buffer solution at \(298\text{ K}\)?
  1. A.2.94
  2. B.4.57
  3. C.4.87
  4. D.5.17
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解題

First, calculate the moles of propanoic acid and sodium hydroxide. \(\text{Moles of } CH_3CH_2COOH = 0.0500 \times 0.100 = 5.00 \times 10^{-3}\text{ mol}\). \(\text{Moles of } NaOH = 0.0500 \times 0.050 = 2.50 \times 10^{-3}\text{ mol}\). The added sodium hydroxide reacts completely with the propanoic acid: \(CH_3CH_2COOH + NaOH \rightarrow CH_3CH_2COONa + H_2O\). The moles of propanoic acid remaining = \(5.00 \times 10^{-3} - 2.50 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol}\). The moles of propanoate ions formed = \(2.50 \times 10^{-3}\text{ mol}\). Since the moles of acid and conjugate base are equal, the concentration ratio is 1, and the pH is equal to the \(pK_a\) of propanoic acid: \(pH = pK_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

評分準則

1 mark for the correct option C. [Method: Calculate moles of acid and base; determine the composition of the buffer solution; recognize that [acid] = [salt] and thus pH = pKa; calculate pKa as -log10(1.35 x 10^-5) = 4.87]
題目 11 · 選擇題
1
Two ligand-exchange reactions of aqueous cobalt(II) ions are shown below. Reaction 1: \([Co(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CoCl_4]^{2-}(aq) + 6H_2O(l)\) (stability constant \(K_1\)). Reaction 2: \([Co(H_2O)_6]^{2+}(aq) + 6NH_3(aq) \rightleftharpoons [Co(NH_3)_6]^{2+}(aq) + 6H_2O(l)\) (stability constant \(K_2\)). Given that \(K_2 > K_1 > 1\\, which statement is correct?
  1. A.The addition of excess \(Cl^-(aq)\) to a solution of \([Co(NH_3)_6]^{2+}(aq)\) will easily displace the ammonia ligands to form \([CoCl_4]^{2-}(aq)\).
  2. B.\([Co(NH_3)_6]^{2+}(aq)\) is thermodynamically more stable than both \([Co(H_2O)_6]^{2+}(aq)\) and \([CoCl_4]^{2-}(aq)\).
  3. C.The value of \(K_1\) increases when the concentration of water is increased.
  4. D.\([CoCl_4]^{2-}\) has an octahedral geometry while \([Co(NH_3)_6]^{2+}\) is tetrahedral.
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解題

A larger stability constant (\(K_{\text{stab}}\) indicates a more stable complex. Since \(K_2 > K_1 > 1\), \([Co(NH_3)_6]^{2+}(aq)\) is thermodynamically more stable than \([Co(Cl)_4]^{2-}(aq)\) and \([Co(H_2O)_6]^{2+}(aq)\\. Therefore, adding chloride ions cannot easily displace ammonia from \)[Co(NH_3)_6]^{2+}\\. Water is a reactant/solvent and its concentration change doesn't alter the value of the equilibrium constant \(K_{\text{stab}}\\. \)[CoCl_4]^{2-}\) is tetrahedral, while \([Co(NH_3)_6]^{2+}\) is octahedral.

評分準則

1 mark for the correct option B. [Method: Relate stability constant magnitude to thermodynamic stability; evaluate displacement potential and complex geometries; reject A, C, and D based on chemical principles]
題目 12 · 選擇題
1
The reaction between reactants \(P\) and \(Q\) is investigated, and the initial rate of reaction is measured at \(298\text{ K}\) for various initial concentrations. Experiment 1: \([P] = 0.10\text{ mol dm}^{-3}\), \([Q] = 0.10\text{ mol dm}^{-3}\), Initial Rate = \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Experiment 2: \([P] = 0.20\text{ mol dm}^{-3}\), \([Q] = 0.10\text{ mol dm}^{-3}\), Initial Rate = \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Experiment 3: \([P] = 0.20\text{ mol dm}^{-3}\), \([Q] = 0.20\text{ mol dm}^{-3}\), Initial Rate = \(1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What are the units of the rate constant, \(k\), for this reaction?
  1. A.\(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
  2. B.\(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
  3. C.\(\text{dm}^9\text{ mol}^{-3}\text{ s}^{-1}\)
  4. D.\(\text{mol dm}^{-3}\text{ s}^{-1}\)
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解題

Comparing Experiment 1 and Experiment 2, doubling \([P]\) while keeping \([Q]\) constant quadruples the rate (\(2.0 \times 10^{-4} \rightarrow 8.0 \times 10^{-4}\)), so the reaction is second order with respect to \(P\). Comparing Experiment 2 and Experiment 3, doubling \([Q]\) while keeping \([P]\) constant doubles the rate (\(8.0 \times 10^{-4} \rightarrow 1.6 \times 10^{-3}\)), so the reaction is first order with respect to \(Q\). The overall rate equation is: \(\text{Rate} = k[P]^2[Q]\). The units of the rate constant \(k\) are: \(k = \frac{\text{Rate}}{[P]^2[Q]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

評分準則

1 mark for the correct option B. [Method: Deduce order with respect to P as 2; deduce order with respect to Q as 1; derive the rate equation; solve for the units of k]
題目 13 · 選擇題
1
Which statement about the reaction of methylbenzene with a mixture of concentrated nitric acid and concentrated sulfuric acid at \(30^\circ\text{C}\) is correct?
  1. A.The reaction requires a higher temperature than the nitration of benzene.
  2. B.The electrophile is the \(NO_2^+\) ion, which mainly attacks the 3-position (meta-position) of the methylbenzene ring.
  3. C.The methyl group is electron-donating by inductive effect, making the ring more reactive towards electrophiles than benzene.
  4. D.Sulfuric acid acts as a Bronsted-Lowry base in the generation of the electrophile.
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解題

The methyl group is an electron-donating group due to its inductive effect. It increases the electron density on the benzene ring, making methylbenzene more susceptible to electrophilic attack than benzene itself. Thus, it reacts faster and under milder conditions (lower temperatures) than benzene. It directs electrophiles to the 2- and 4-positions, and sulfuric acid acts as a Bronsted-Lowry acid by protonating nitric acid during the formation of the \(NO_2^+\) electrophile.

評分準則

1 mark for the correct option C. [Method: Identify methyl as an activating, 2,4-directing group; recognize its electron-donating inductive effect; evaluate temperature and reagent roles]
題目 14 · 選擇題
1
Alanine is 2-aminopropanoic acid. What is the major organic species present when alanine is dissolved in a buffer solution of pH 12?
  1. A.\(CH_3CH(NH_3^+)COOH\)
  2. B.\(CH_3CH(NH_3^+)COO^-\)
  3. C.\(CH_3CH(NH_2)COO^-\)
  4. D.\(CH_3CH(NH_2)COOH\)
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解題

In strongly alkaline solutions (pH 12), both the amine group and the carboxylic acid group are deprotonated. The amino group exists as the neutral \(-NH_2\) group, while the carboxylic acid group is deprotonated to form the carboxylate ion, \(-COO^-\). Therefore, the major species is \(CH_3CH(NH_2)COO^-\).

評分準則

1 mark for the correct option C. [Method: Identify functional groups of alanine; evaluate ionization states of carboxylic acid and amine groups at high pH; select the completely deprotonated anionic form]
題目 15 · 選擇題
1
Three oxides of Period 3 elements, \(X\), \(Y\), and \(Z\), have the following properties. Oxide \(X\) is insoluble in water but dissolves in both dilute hydrochloric acid and dilute sodium hydroxide. Oxide \(Y\) reacts with water to form a strongly acidic solution of pH 1-2. Oxide \(Z\) is insoluble in water and only dissolves in hot, concentrated sodium hydroxide. What are the identities of elements \(X\), \(Y\), and \(Z\)?
  1. A.\(X\) is aluminium, \(Y\) is phosphorus, \(Z\) is silicon.
  2. B.\(X\) is silicon, \(Y\) is sulfur, \(Z\) is aluminium.
  3. C.\(X\) is magnesium, \(Y\) is silicon, \(Z\) is phosphorus.
  4. D.\(X\) is aluminium, \(Y\) is sodium, \(Z\) is magnesium.
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解題

Aluminium oxide (\(Al_2O_3\)) is amphoteric, insoluble in water but reacts with both acids and alkalis; hence, \(X\) is aluminium. Phosphorus(V) oxide (\(P_4O_{10}\) or \(P_2O_5\)) reacts vigorously with water to form phosphoric(V) acid, a strong acid with pH 1-2; hence, \(Y\) is phosphorus. Silicon dioxide (\(SiO_2\)) has a giant covalent structure, is insoluble in water, and only dissolves in hot, concentrated alkalis; hence, \(Z\) is silicon.

評分準則

1 mark for the correct option A. [Method: Identify the amphoteric oxide X as Al2O3; identify the strongly acidic oxide Y as P4O10; identify the weakly acidic, giant covalent oxide Z as SiO2; deduce elements X, Y, and Z]
題目 16 · 選擇題
1
Which equation represents a step in the catalytic cycle where nitrogen monoxide, \(NO\), acts as a catalyst in the oxidation of atmospheric sulfur dioxide to sulfur trioxide?
  1. A.\(SO_2 + NO_2 \rightarrow SO_3 + NO\)
  2. B.\(SO_2 + NO \rightarrow SO_3 + \frac{1}{2} N_2\)
  3. C.\(2NO_2 + H_2O \rightarrow HNO_2 + HNO_3\)
  4. D.\(2NO + SO_2 \rightarrow N_2O + SO_3\)
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解題

In the catalysis of atmospheric sulfur dioxide oxidation to sulfur trioxide, nitrogen monoxide (\(NO\)) is first oxidized by oxygen to nitrogen dioxide (\(NO_2\)): \(2NO + O_2 \rightarrow 2NO_2\). In the subsequent step, \(NO_2\) oxidizes \(SO_2\) to \(SO_3\) and is reduced back to \(NO\): \(SO_2 + NO_2 \rightarrow SO_3 + NO\). The regenerated \(NO\) can then repeat the cycle, demonstrating its role as a catalyst.

評分準則

1 mark for the correct option A. [Method: Recall the mechanism of SO2 oxidation catalyzed by NOx; identify the step where NO2 oxidizes SO2 and regenerates NO; reject incorrect stoichiometry and products]
題目 17 · 選擇題
1
A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), was completely combusted in \(100\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature, the total volume of gas remaining was \(80\text{ cm}^3\). After passing this remaining gas through excess aqueous potassium hydroxide, the volume of gas decreased to \(40\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
  1. A.\(\text{C}_3\text{H}_6\)
  2. B.\(\text{C}_4\text{H}_6\)
  3. C.\(\text{C}_4\text{H}_8\)
  4. D.\(\text{C}_4\text{H}_{10}\)
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解題

1. The volume of \(\text{CO}_2\) produced is equal to the decrease in volume when passed through aqueous KOH: \(80 - 40 = 40\text{ cm}^3\). 2. Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produced \(40\text{ cm}^3\) of \(\text{CO}_2\), \(x = 4\) (from \(\text{C}_x\text{H}_y \to x\text{CO}_2\)). 3. The gas remaining after absorption by KOH is unreacted \(\text{O}_2\), which is \(40\text{ cm}^3\). 4. The volume of \(\text{O}_2\) reacted is \(100 - 40 = 60\text{ cm}^3\). 5. From the combustion equation: \(\text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \to x\text{CO}_2 + y/2\text{H}_2\text{O}\), the mole ratio of \(\text{O}_2\) to hydrocarbon is \(60/10 = 6\). 6. Therefore, \(x + y/4 = 6\). Substituting \(x = 4\) gives \(4 + y/4 = 6 \implies y/4 = 2 \implies y = 8\). 7. The hydrocarbon is \(\text{C}_4\text{H}_8\).

評分準則

1 mark for the correct option C.
題目 18 · 選擇題
1
A buffer solution is prepared at \(25\ ^\circ\text{C}\} by mixing \)50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) and \(50.0\text{ cm}^3\) of \(0.050\text{ mol dm}^{-3}\) sodium propanoate. What is the pH of the resulting buffer solution?
  1. A.4.57
  2. B.5.17
  3. C.4.87
  4. D.4.27
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解題

Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{salt}]}{[\text{acid}]}\). First, calculate \(\text{p}K_a\): \(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\). Since both components are mixed in equal volumes, we can use their relative moles: Moles of propanoic acid = \(0.050\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\); Moles of propanoate ions = \(0.050\text{ dm}^3 \times 0.050\text{ mol dm}^{-3} = 0.00250\text{ mol}\). Now substitute the values into the equation: \(\text{pH} = 4.87 + \log_{10} \left( \frac{0.00250}{0.00500} \right) = 4.87 + \log_{10}(0.5) = 4.87 - 0.30 = 4.57\).

評分準則

1 mark for the correct calculation and selecting option A.
題目 19 · 選擇題
1
Which statement best explains why transition metal complexes, such as aqueous cobalt(II) ions, are coloured?
  1. A.Ligands split the d-orbitals into two non-degenerate energy levels. d-Electrons absorb specific wavelengths of visible light to be promoted to a higher d-orbital.
  2. B.Ligands split the d-orbitals into two non-degenerate energy levels. d-Electrons emit visible light when they fall from a higher d-orbital to a lower d-orbital.
  3. C.Coordination of water ligands causes electrons to be promoted to 4s and 4p orbitals, absorbing visible light during the process.
  4. D.The d-orbitals of cobalt(II) are completely filled, and light absorption causes these d-electrons to be promoted to the ligand orbitals.
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解題

In transition metal complexes, the coordinate bonds formed with ligands cause the 3d orbitals to split into two non-degenerate sets with an energy difference \(\Delta E\). When white light passes through the complex, d-electrons absorb light of energy equal to \(\Delta E\) (\(\Delta E = h\nu\)) to transition from the lower energy level to the higher level. The complementary colour of the absorbed wavelength is transmitted or reflected, making the complex appear coloured.

評分準則

1 mark for identifying the correct explanation for the origin of colour in transition complexes (Option A).
題目 20 · 選擇題
1
The table shows experimental rate data for the reaction: \(2\text{A} + \text{B} + 2\text{C} \to \text{D} + 2\text{E}\). Exp 1: [A]=0.10, [B]=0.10, [C]=0.10, Rate=\(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1}\). Exp 2: [A]=0.20, [B]=0.10, [C]=0.10, Rate=\(4.0 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1}\). Exp 3: [A]=0.10, [B]=0.20, [C]=0.10, Rate=\(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1}\). Exp 4: [A]=0.10, [B]=0.10, [C]=0.30, Rate=\(1.8 \times 10^{-3}\text{ mol dm}^{-3}\text{s}^{-1}\). What is the value and units of the rate constant, \(k\), for this reaction?
  1. A.0.20\ \text{dm}^3\ \text{mol}^{-1}\ \text{s}^{-1}
  2. B.0.20\ \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}
  3. C.2.0 \times 10^{-3}\ \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}
  4. D.0.020\ \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}
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解題

1. Determine orders of reaction: From Exp 1 and Exp 2: doubling [A] doubles the rate (first order in A). From Exp 1 and Exp 3: doubling [B] has no effect on rate (zero order in B). From Exp 1 and Exp 4: tripling [C] increases rate by a factor of 9, so the reaction is second order in C. 2. The rate equation is: \(\text{Rate} = k[\text{A}][\text{C}]^2\). 3. Calculate \(k\) using Exp 1: \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{s}^{-1} = k \times (0.10\text{ mol dm}^{-3}) \times (0.10\text{ mol dm}^{-3})^2 \implies k = 0.20\). 4. Determine units: \(\text{mol dm}^{-3}\text{s}^{-1} / ((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2) = \text{dm}^6\text{mol}^{-2}\text{s}^{-1}\).

評分準則

1 mark for correct identification of the rate constant value and its units (Option B).
題目 21 · 選擇題
1
Methylbenzene undergoes nitration when reacted with a mixture of concentrated nitric acid and concentrated sulfuric acid. Which statement about this reaction is correct?
  1. A.The attacking electrophile is \(\text{NO}_2^+\) and methylbenzene reacts faster than benzene.
  2. B.The attacking electrophile is \(\text{NO}_3^-\) and methylbenzene reacts slower than benzene.
  3. C.Concentrated sulfuric acid acts as a reducing agent in the generation of the active electrophile.
  4. D.The primary organic product of mono-nitration is 3-nitromethylbenzene.
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解題

1. Nitration of benzene rings involves the electrophile nitronium ion, \(\text{NO}_2^+\). 2. The methyl group on methylbenzene is electron-donating due to positive inductive effects (+I), which increases the electron density of the benzene ring. This makes the ring more susceptible to electrophilic attack, so methylbenzene reacts faster than benzene. 3. Sulfuric acid acts as an acid catalyst to protonate nitric acid, forming \(\text{NO}_2^+\) and \(\text{HSO}_4^-\). 4. The methyl group is ortho/para-directing, so the major products are 2-nitromethylbenzene and 4-nitromethylbenzene.

評分準則

1 mark for selecting option A.
題目 22 · 選擇題
1
An aqueous solution of the amino acid alanine (2-aminopropanoic acid) is adjusted to pH 12. Which chemical species is the major form of alanine present at this pH?
  1. A.\(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\)
  2. B.\(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\)
  3. C.\(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\)
  4. D.\(\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}\)
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解題

At highly alkaline pH (pH 12), the concentration of hydroxide ions is high. Both functional groups of alanine lose protons: the carboxylic acid group (\(-\text{COOH}\)) loses a proton to become a carboxylate anion (\(-\text{COO}^-\)), and the amine group (\(-\text{NH}_3^+\) in the zwitterion form) loses a proton to become a neutral amine (\(-\text{NH}_2\)). Therefore, the predominant species is the anionic form: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).

評分準則

1 mark for the correct anionic structure of alanine in strongly basic solution (Option C).
題目 23 · 選擇題
1
Which statement correctly describes the behavior of phosphorus(V) chloride and silicon(IV) chloride when added separately to excess water at room temperature?
  1. A.Both compounds undergo rapid hydrolysis to produce strongly acidic solutions with a pH of approximately 1–2.
  2. B.Phosphorus(V) chloride hydrolyses to form an acidic solution (pH 1–2), while silicon(IV) chloride dissolves to form a neutral solution (pH 7).
  3. C.Phosphorus(V) chloride hydrolyses to form an acidic solution (pH 1–2), while silicon(IV) chloride is insoluble and does not react.
  4. D.Both compounds dissolve and hydrolyse to produce alkaline solutions with a pH of approximately 11–12.
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解題

Both silicon(IV) chloride (\(\text{SiCl}_4\)) and phosphorus(V) chloride (\(\text{PCl}_5\)) are covalent molecular chlorides that undergo vigorous hydrolysis when added to water: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \to \text{SiO}_2 + 4\text{HCl}\) and \(\text{PCl}_5 + 4\text{H}_2\text{O} \to \text{H}_3\text{PO}_4 + 5\text{HCl}\). Both reactions produce hydrochloric acid (\(\text{HCl}\)), which fully dissociates in water, resulting in highly acidic solutions with a pH of around 1 to 2.

評分準則

1 mark for correct description of the hydrolysis behavior of both Period 3 covalent chlorides (Option A).
題目 24 · 選擇題
1
Catalytic converters in car exhaust systems contain transition metal catalysts (such as Pt, Pd, and Rh) to reduce harmful emissions. Which chemical reaction does NOT take place on the surface of the catalyst in a catalytic converter?
  1. A.\(2\text{CO} + \text{O}_2 \to 2\text{CO}_2\)
  2. B.\(2\text{CO} + 2\text{NO} \to 2\text{CO}_2 + \text{N}_2\)
  3. C.\(\text{N}_2 + \text{O}_2 \to 2\text{NO}\)
  4. D.\(\text{C}_8\text{H}_{18} + 12.5\text{O}_2 \to 8\text{CO}_2 + 9\text{H}_2\text{O}\)
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解題

The reaction \(\text{N}_2 + \text{O}_2 \to 2\text{NO}\) takes place inside the high-temperature and high-pressure environment of the internal combustion engine cylinder, not inside the catalytic converter. The purpose of the catalytic converter is to remove \(\text{NO}\) by reducing it to harmless \(\text{N}_2\) gas (as shown in equation B), and to oxidise \(\text{CO}\) and unburnt hydrocarbons like \(\text{C}_8\text{H}_{18}\) (equations A and D).

評分準則

1 mark for identifying that nitrogen oxide formation happens in the engine cylinder, not the catalytic converter (Option C).
題目 25 · 選擇題
1
A sample of 1.50 g of an impure metal carbonate, \(M\text{CO}_3\) (where the \(M_r\) of \(M\text{CO}_3\) is 84.3), is reacted with excess dilute hydrochloric acid. This reaction produces 360 cm\(^3\) of carbon dioxide gas, measured at room temperature and pressure (r.t.p.). Assuming the impurity is completely unreactive, what is the percentage purity of the metal carbonate sample? [Take the molar gas volume at r.t.p. as 24.0 dm\(^3\) mol\(^{-1\)}]
  1. A.56.2%
  2. B.75.0%
  3. C.84.3%
  4. D.90.0%
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解題

1. Calculate the moles of \(\text{CO}_2\) produced: \(n = \frac{360}{24000} = 0.0150\text{ mol}\). 2. Write the balanced equation: \(M\text{CO}_3 + 2\text{HCl} \rightarrow M\text{Cl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). From the stoichiometry, 1 mole of \(M\text{CO}_3\) produces 1 mole of \(\text{CO}_2\). Therefore, the moles of pure \(M\text{CO}_3\) in the sample is also 0.0150 mol. 3. Calculate the mass of pure \(M\text{CO}_3\): \(m = 0.0150\text{ mol} \times 84.3\text{ g mol}^{-1} = 1.2645\text{ g}\). 4. Calculate the percentage purity: \(\text{Purity} = \frac{1.2645\text{ g}}{1.50\text{ g}} \times 100\% = 84.3\%\).

評分準則

Award 1 mark for the correct calculation of moles of gas, mass of pure carbonate, and percentage purity resulting in 84.3%.
題目 26 · 選擇題
1
A buffer solution is prepared by mixing 40.0 cm\(^3\) of 0.150 mol dm\(^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\) mol dm\(^{-3}\)) with 60.0 cm\(^3\) of 0.100 mol dm\(^{-3}\) sodium propanoate. What is the pH of this buffer solution?
  1. A.3.87
  2. B.4.87
  3. C.5.05
  4. D.5.17
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解題

1. Calculate the initial moles of propanoic acid: \(n = 0.0400\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 6.00 \times 10^{-3}\text{ mol}\). 2. Calculate the initial moles of propanoate ions: \(n = 0.0600\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 6.00 \times 10^{-3}\text{ mol}\). 3. Since the moles of acid equal the moles of conjugate base, the ratio \([\text{salt}]/[\text{acid}] = 1\). 4. Using the buffer equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)\). Since \(\log_{10}(1) = 0\), \(\text{pH} = \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

評分準則

Award 1 mark for calculating equal mole values and identifying that pH equals pKa, leading to 4.87.
題目 27 · 選擇題
1
Consider the two ligand exchange equilibria shown below: 1) \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CuCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)\) with stability constant \(K_1\) 2) \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{NH}_3(aq) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(aq) + 4\text{H}_2\text{O}(l)\) with stability constant \(K_2\) Given that \(K_2 \gg K_1\), which statement correctly describes what happens when a concentrated aqueous solution of ammonia is added to an equilibrium mixture containing both complexes?
  1. A.The concentration of \([\text{CuCl}_4]^{2-}\) increases because ammonia acts as a reducing agent.
  2. B.The deep blue complex \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) is preferentially formed because it is more stable than \([\text{CuCl}_4]^{2-}\).
  3. C.The value of \(K_1\) increases significantly to resist the change in ammonia concentration.
  4. D.There is no reaction because chloride ligands form a stronger coordinate bond than ammonia ligands.
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解題

Stability constants (Kstab) measure the relative stability of complex ions. Because \(K_2\) is much larger than \(K_1\), the tetraaminediaquacopper(II) complex is far more stable than the tetrachlorocuprate(II) complex. Adding ammonia will shift the system to form the highly stable deep blue amine complex, substituting other ligands.

評分準則

Award 1 mark for identifying that the amine complex is preferentially formed due to its significantly larger stability constant.
題目 28 · 選擇題
1
The reaction \(2A + B + C \rightarrow D + E\) was studied and the following initial rate data were obtained: Experiment 1: \([A] = 0.10, [B] = 0.10, [C] = 0.10\), Rate = \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\) Experiment 2: \([A] = 0.20, [B] = 0.10, [C] = 0.10\), Rate = \(4.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\) Experiment 3: \([A] = 0.10, [B] = 0.20, [C] = 0.10\), Rate = \(8.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\) Experiment 4: \([A] = 0.10, [B] = 0.10, [C] = 0.20\), Rate = \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\) What is the overall order of the reaction and the correct units of the rate constant, \(k\)?
  1. A.Overall order: 3; units of \(k\): dm\(^6\) mol\(^{-2\)} s\(^{-1\)}
  2. B.Overall order: 3; units of \(k\): dm\(^3\) mol\(^{-1\)} s\(^{-1\)}
  3. C.Overall order: 4; units of \(k\): dm\(^9\) mol\(^{-3\)} s\(^{-1\)}
  4. D.Overall order: 2; units of \(k\): dm\(^3\) mol\(^{-1\)} s\(^{-1\)}
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解題

1. Determine the order with respect to each reactant: Comparing Exp 1 and 2: doubling [A] doubles the rate, so the reaction is first-order with respect to A (order = 1). Comparing Exp 1 and 3: doubling [B] quadruples the rate, so the reaction is second-order with respect to B (order = 2). Comparing Exp 1 and 4: doubling [C] does not affect the rate, so the reaction is zero-order with respect to C (order = 0). 2. Find the overall order: \(1 + 2 + 0 = 3\). 3. Derive the units for the rate constant \(k\): \(\text{Rate} = k[A][B]^2 \Rightarrow k = \frac{\text{Rate}}{[A][B]^2}\). Units of \(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

評分準則

Award 1 mark for identifying the overall order as 3 and correctly deriving the units of k as dm6 mol-2 s-1.
題目 29 · 選擇題
1
Methylbenzene is reacted with a cold mixture of concentrated nitric acid and concentrated sulfuric acid. Which statement about this reaction is correct?
  1. A.The reaction is a nucleophilic substitution because the methyl group activates the ring.
  2. B.The active electrophile in the reaction is the \(\text{NO}_2^-\) ion.
  3. C.The main mono-nitrated products formed are 2-nitromethylbenzene and 4-nitromethylbenzene.
  4. D.Methylbenzene reacts more slowly than benzene because the methyl group is electron-withdrawing.
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解題

The methyl group in methylbenzene is electron-donating due to inductive effects. This activates the benzene ring, making it react faster than benzene in electrophilic substitution. The methyl group is a 2,4-directing group, so the primary products of mono-nitration are 2-nitromethylbenzene and 4-nitromethylbenzene. The active electrophile is the nitronium ion (\(\text{NO}_2^+\)).

評分準則

Award 1 mark for identifying that the methyl group is 2,4-directing, yielding 2-nitromethylbenzene and 4-nitromethylbenzene.
題目 30 · 選擇題
1
Alanine, \(\text{CH}_3\text{CH(NH}_2)\text{COOH}\), is an amino acid. Which structures represent the major ionic species of alanine in aqueous solutions at pH 1 and pH 13?
  1. A.pH 1: \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\); pH 13: \(\text{CH}_3\text{CH(NH}_2)\text{COO}^-\)
  2. B.pH 1: \(\text{CH}_3\text{CH(NH}_2)\text{COO}^-\); pH 13: \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\)
  3. C.pH 1: \(\text{CH}_3\text{CH(NH}_3^+)\text{COO}^-\); pH 13: \(\text{CH}_3\text{CH(NH}_3^+)\text{COO}^-\)
  4. D.pH 1: \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\); pH 13: \(\text{CH}_3\text{CH(NH}_3^+)\text{COO}^-\)
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解題

At a highly acidic pH of 1, both the amine and carboxylic acid groups are fully protonated, giving the cationic species \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\). At a highly alkaline pH of 13, both groups are deprotonated, giving the anionic species \(\text{CH}_3\text{CH(NH}_2)\text{COO}^-\).

評分準則

Award 1 mark for matching the correct fully protonated amine structure at pH 1 and deprotonated carboxyl group at pH 13.
題目 31 · 選擇題
1
An oxide of a Period 3 element, \(X\), is a white solid that reacts with both dilute hydrochloric acid and aqueous sodium hydroxide. Another Period 3 element, \(Y\), forms an oxide that reacts with water to give an acidic solution with a pH of approximately 2. What are the identities of elements \(X\) and \(Y\)?
  1. A.\(X = \text{Al}\); \(Y = \text{S}\)
  2. B.\(X = \text{Si}\); \(Y = \text{P}\)
  3. C.\(X = \text{Al}\); \(Y = \text{Na}\)
  4. D.\(X = \text{Mg}\); \(Y = \text{S}\)
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解題

An oxide that reacts with both acids and bases is amphoteric. The only amphoteric oxide in Period 3 is aluminium oxide (\(\text{Al}_2\text{O}_3\)), so \(X\) must be aluminium (\(\text{Al}\)). Sulfur forms acidic oxides, such as sulfur dioxide (\(\text{SO}_2\)), which reacts with water to form weak sulfurous acid (\(\text{H}_2\text{SO}_3\)) with a pH of approximately 2, meaning \(Y\) is sulfur (\(\text{S}\)).

評分準則

Award 1 mark for identifying the amphoteric oxide belongs to Al and the oxide that yields pH 2 belongs to S.
題目 32 · 選擇題
1
In the atmosphere, sulfur dioxide (\(\text{SO}_2\)) is oxidized to sulfur trioxide (\(\text{SO}_3\)) in a process catalyzed by nitrogen dioxide (\(\text{NO}_2\)). Which chemical equation represents the step in this catalytic cycle in which the \(\text{NO}_2\) catalyst is regenerated?
  1. A.\(\text{SO}_2(g) + \text{NO}_2(g) \rightarrow \text{SO}_3(g) + \text{NO}(g)\)
  2. B.\(2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)\)
  3. C.\(\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\text{NO}(g)\)
  4. D.\(\text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq)\)
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解題

In the catalytic oxidation of \(\text{SO}_2\) by \(\text{NO}_2\), the cycle involves: 1) The oxidation of \(\text{SO}_2\) where \(\text{NO}_2\) is reduced: \(\text{SO}_2(g) + \text{NO}_2(g) \rightarrow \text{SO}_3(g) + \text{NO}(g)\). 2) The regeneration of the catalyst where \(\text{NO}\) is oxidized back to \(\text{NO}_2\) by atmospheric oxygen: \(2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)\). Equation B correctly represents this regeneration step.

評分準則

Award 1 mark for identifying the oxidation of NO to NO2 as the regeneration step of the catalyst.
題目 33 · 選擇題
1
An anhydrous metal carbonate, \( \text{M}_2\text{CO}_3 \), where \( \text{M} \) is a Group 1 metal, has a mass of \( 1.325\text{ g} \). It is completely dissolved in water and the solution is made up to \( 250.0\text{ cm}^3 \) in a volumetric flask. A \( 25.0\text{ cm}^3 \) portion of this solution is titrated against \( 0.125\text{ mol dm}^{-3} \) hydrochloric acid, \( \text{HCl}(\text{aq}) \). It requires \( 20.0\text{ cm}^3 \) of the acid for complete neutralisation. What is the identity of metal \( \text{M} \)?
  1. A.Lithium
  2. B.Sodium
  3. C.Potassium
  4. D.Rubidium
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解題

1. Moles of \( \text{HCl} \) used in the titration:
\( n(\text{HCl}) = 0.125\text{ mol dm}^{-3} \times 0.0200\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol} \).

2. The balanced equation is:
\( \text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{CO}_2 + \text{H}_2\text{O} \).

3. Determine moles of \( \text{M}_2\text{CO}_3 \) in the titrated \( 25.0\text{ cm}^3 \) portion:
\( n(\text{M}_2\text{CO}_3) = \frac{1}{2} \times n(\text{HCl}) = 1.25 \times 10^{-3}\text{ mol} \).

4. Scale up to find the total moles in the original \( 250.0\text{ cm}^3 \) solution:
\( n_{\text{total}} = 1.25 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 0.0125\text{ mol} \).

5. Calculate the molar mass of \( \text{M}_2\text{CO}_3 \):
\( M_r = \frac{1.325\text{ g}}{0.0125\text{ mol}} = 106.0\text{ g mol}^{-1} \).

6. Find the atomic mass of \( \text{M} \):
\( 2 \times A_r(\text{M}) + 12.0 + (3 \times 16.0) = 106.0 \Rightarrow 2 \times A_r(\text{M}) + 60.0 = 106.0 \Rightarrow A_r(\text{M}) = 23.0 \).
This relative atomic mass corresponds to sodium, \( \text{Na} \).

評分準則

Award 1 mark for the correct calculations leading to the identification of sodium as metal M (Option B).
題目 34 · 選擇題
1
A monoprotic weak acid, \( \text{HA} \), has a relative molecular mass of \( 60.0 \). A solution is prepared by dissolving \( 1.80\text{ g} \) of \( \text{HA} \) in distilled water and diluting to a final volume of \( 250\text{ cm}^3 \). The pH of this solution is \( 2.88 \) at \( 298\text{ K} \). What is the acid dissociation constant, \( K_a \), of \( \text{HA} \) at \( 298\text{ K} \)?
  1. A.\( 1.45 \times 10^{-5}\text{ mol dm}^{-3} \)
  2. B.\( 5.79 \times 10^{-5}\text{ mol dm}^{-3} \)
  3. C.\( 1.10 \times 10^{-2}\text{ mol dm}^{-3} \)
  4. D.\( 1.74 \times 10^{-6}\text{ mol dm}^{-3} \)
查看答案詳解

解題

1. Determine the concentration of weak acid \( \text{HA} \):
\( n(\text{HA}) = \frac{1.80\text{ g}}{60.0\text{ g mol}^{-1}} = 0.0300\text{ mol} \)
\( [\text{HA}] = \frac{0.0300\text{ mol}}{0.250\text{ dm}^3} = 0.120\text{ mol dm}^{-3} \)

2. Find \( [\text{H}^+] \) from the pH value:
\( [\text{H}^+] = 10^{-\text{pH}} = 10^{-2.88} = 1.318 \times 10^{-3}\text{ mol dm}^{-3} \)

3. Apply the weak acid approximation:
\( K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]} = \frac{(1.318 \times 10^{-3})^2}{0.120} = \frac{1.7378 \times 10^{-6}}{0.120} = 1.45 \times 10^{-5}\text{ mol dm}^{-3} \).

評分準則

Award 1 mark for the correct calculation of Ka as 1.45 x 10^-5 mol dm^-3 (Option A).
題目 35 · 選擇題
1
Which statement correctly explains why transition metal complexes are coloured?
  1. A.Ligands cause the d-orbitals to split into two different energy levels. Light is emitted when an electron falls from a higher to a lower energy d-orbital.
  2. B.Ligands cause the d-orbitals to split into two different energy levels. Light of a specific frequency is absorbed when an electron is promoted from a lower to a higher energy d-orbital.
  3. C.Coordination of ligands causes excitation of s-electrons to d-orbitals, which absorbs visible light.
  4. D.The transition metal ion absorbs all frequencies of visible light except for the frequency corresponding to the splitting energy of its d-orbitals.
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解題

When ligands coordinate to a transition metal ion, their lone pairs repel the metal d-electrons, splitting the five d-orbitals into two groups of different energy levels. When visible light falls on the complex, an electron is promoted from a lower d-orbital to a higher d-orbital, absorbing a specific frequency of light. The complementary colour of the absorbed frequency is observed.

評分準則

Award 1 mark for selecting the option that correctly attributes the colour to d-orbital splitting and promotion of electrons absorbing light (Option B).
題目 36 · 選擇題
1
For the gaseous reaction: \( 2\text{X}(\text{g}) + \text{Y}(\text{g}) + \text{Z}(\text{g}) \rightarrow \text{products} \), the following initial rate data were obtained at constant temperature:

| Experiment | Initial \( [\text{X}] \) / \( \text{mol dm}^{-3} \) | Initial \( [\text{Y}] \) / \( \text{mol dm}^{-3} \) | Initial \( [\text{Z}] \) / \( \text{mol dm}^{-3} \) | Initial Rate / \( \text{mol dm}^{-3}\text{ s}^{-1} \) |
|---|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.10 | \( 4.0 \times 10^{-4} \) |
| 2 | 0.20 | 0.10 | 0.10 | \( 8.0 \times 10^{-4} \) |
| 3 | 0.10 | 0.20 | 0.10 | \( 1.6 \times 10^{-3} \) |
| 4 | 0.10 | 0.10 | 0.20 | \( 4.0 \times 10^{-4} \) |

What is the value and units of the rate constant, \( k \), for this reaction?
  1. A.\( 0.40\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1} \)
  2. B.\( 0.40\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \)
  3. C.\( 4.0 \times 10^{-2}\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1} \)
  4. D.\( 4.0 \\times 10^{-2}\\text{ dm}^3\\text{ mol}^{-1}\\text{ s}^{-1} \\
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解題

1. Determine reaction orders:
- Compare Exp 1 & 2: Doubling \( [\text{X}] \) doubles the rate, so the reaction is first-order with respect to \( \text{X} \).
- Compare Exp 1 & 3: Doubling \( [\text{Y}] \) quadruples the rate, so the reaction is second-order with respect to \( \text{Y} \).
- Compare Exp 1 & 4: Doubling \( [\text{Z}] \) has no effect on the rate, so the reaction is zeroth-order with respect to \( \text{Z} \).

2. Write the rate equation: \( \text{Rate} = k[\text{X}][\text{Y}]^2 \).

3. Calculate the value of \( k \) using Experiment 1:
\( 4.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k(0.10\text{ mol dm}^{-3})(0.10\text{ mol dm}^{-3})^2 \)
\( 4.0 \times 10^{-4} = k(1.0 \times 10^{-3}) \Rightarrow k = 0.40 \).

4. Determine the units of \( k \):
\( \text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

評分準則

Award 1 mark for calculating k = 0.40 and correctly identifying the unit as dm^6 mol^-2 s^-1 (Option A).
題目 37 · 選擇題
1
Which statement correctly explains both the relative rate of nitration of methylbenzene compared to benzene, and the directing effect of the methyl group?
  1. A.Methylbenzene reacts more slowly than benzene because the methyl group is electron-withdrawing, directing the incoming electrophile to the 3-position.
  2. B.Methylbenzene reacts more slowly than benzene because the methyl group is electron-donating, directing the incoming electrophile to the 2- and 4-positions.
  3. C.Methylbenzene reacts faster than benzene because the methyl group is electron-donating, directing the incoming electrophile to the 2- and 4-positions.
  4. D.Methylbenzene reacts faster than benzene because the methyl group is electron-withdrawing, directing the incoming electrophile to the 3-position.
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解題

The methyl group in methylbenzene is electron-donating due to the positive inductive effect. This increases electron density on the benzene ring, making it more susceptible to electrophilic attack, which increases the rate of nitration compared to benzene. Electron-donating groups like the methyl group direct incoming electrophiles to the 2- (ortho) and 4- (para) positions.

評分準則

Award 1 mark for identifying that methylbenzene is more reactive because the methyl group is electron-donating, and that the methyl group is 2,4-directing (Option C).
題目 38 · 選擇題
1
What is the structure of the predominant ionic species present when 2-aminobutanedioic acid (aspartic acid), \( \text{HOOC-CH}_2\text{-CH(NH}_2\text{)-COOH} \), is dissolved in an aqueous solution of pH 12?
  1. A.\( ^-\text{OOC-CH}_2\text{-CH(NH}_3^+\text{)-COO}^- \)
  2. B.\( ^-\text{OOC-CH}_2\text{-CH(NH}_2\text{)-COO}^- \)
  3. C.\( \text{HOOC-CH}_2\text{-CH(NH}_3^+\text{)-COOH} \)
  4. D.\( \text{HOOC-CH}_2\text{-CH(NH}_2\text{)-COO}^- \)
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解題

At a highly alkaline pH of 12, the concentration of hydroxide ions is high, causing proton-donating functional groups to become fully deprotonated. Both carboxylic acid groups (\( \text{-COOH} \)) lose their protons to form carboxylate anions (\( \text{-COO}^- \)), and the protonated amine group (\( \text{-NH}_3^+ \)) loses a proton to form a neutral amine (\( \text{-NH}_2 \)). This results in the species \( ^-\text{OOC-CH}_2\text{-CH(NH}_2\text{)-COO}^- \) with an overall net charge of -2.

評分準則

Award 1 mark for identifying the fully deprotonated structure of aspartic acid under basic conditions (Option B).
題目 39 · 選擇題
1
A white solid Period 3 oxide, \( \text{X} \), reacts vigorously with water to form a solution that turns blue litmus paper red. Another Period 3 oxide, \( \text{Y} \), is a white solid that is insoluble in water, but dissolves in both dilute hydrochloric acid and hot aqueous sodium hydroxide. What are the identities of \( \text{X} \\ and \) \\text{Y} \)?
  1. A.\( \text{X} = \text{SiO}_2 \); \( \text{Y} = \text{Al}_2\text{O}_3 \)
  2. B.\( \text{X} = \text{P}_4\text{O}_{10} \); \( \text{Y} = \text{Al}_2\text{O}_3 \)
  3. C.\( \text{X} = \text{SO}_3 \); \( \text{Y} = \text{SiO}_2 \)
  4. D.\( \text{X} = \text{P}_4\text{O}_{10} \); \( \text{Y} = \text{MgO} \)
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解題

- Oxide \( \text{X} \) is \( \text{P}_4\text{O}_{10} \), which is a white solid and reacts with water to form the strong phosphoric(V) acid, turning blue litmus red (acidic behavior).
- Oxide \( \text{Y} \) is \( \text{Al}_2\text{O}_3 \), which is insoluble in water but amphoteric, reacting with both acids (hydrochloric acid) and bases (sodium hydroxide) to form soluble salts.

評分準則

Award 1 mark for correctly identifying phosphorus(V) oxide (X) and aluminium oxide (Y) based on their chemical properties (Option B).
題目 40 · 選擇題
1
In the atmosphere, sulfur dioxide can be oxidised to sulfur trioxide in the presence of nitrogen dioxide.

The following reactions occur:
Reaction 1: \( \text{SO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{SO}_3(\text{g}) + \text{NO}(\text{g}) \)
Reaction 2: \( \text{NO}(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}_2(\text{g}) \)

Which statement about this process is correct?
  1. A.\( \text{NO}_2 \) acts as a heterogeneous catalyst.
  2. B.The oxidation state of nitrogen changes from +4 to +2 in Reaction 1.
  3. C.\( \text{SO}_2 \) acts as an oxidising agent in Reaction 1.
  4. D.\( \text{NO} \) is a stable catalytic intermediate that does not react further.
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解題

- In Reaction 1, the nitrogen atom in \( \text{NO}_2 \) has an oxidation state of +4. In the product \( \text{NO} \), its oxidation state is +2. This corresponds to a change from +4 to +2, making statement B correct.
- Statement A is incorrect because \( \text{NO}_2 \) is in the gas phase, the same as the reactants, so it acts as a homogeneous catalyst.
- Statement C is incorrect because \( \text{SO}_2 \) is oxidized to \( \text{SO}_3 \), so it acts as a reducing agent.
- Statement D is incorrect because \( \text{NO} \) is a reactive intermediate that is oxidized back to \( \text{NO}_2 \) by oxygen.

評分準則

Award 1 mark for identifying the correct oxidation state change of nitrogen from +4 to +2 (Option B).

Paper 22

Answer all questions. Write your answers in the spaces provided on the question paper.
4 題目 · 60
題目 1 · Structured AS
15
An organic compound, **X**, is a gaseous hydrocarbon with the empirical formula \(\text{CH}_2\). At a temperature of \(150\ ^\circ\text{C}\) and a pressure of \(1.01 \times 10^5\ \text{Pa}\), a \(0.295\ \text{g}\) sample of **X** occupies a volume of \(244\ \text{cm}^3\).

(a) Define the term *relative molecular mass*, \(M_r\). [2]

(b) (i) Use the general gas equation, \(pV = nRT\), to calculate the relative molecular mass, \(M_r\), of **X**. Show your working. \((R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1})\) [3]
(ii) Deduce the molecular formula of **X**. [1]

(c) Compound **Y** contains only carbon, hydrogen, and oxygen. Complete combustion of \(1.18\ \text{g}\) of **Y** produces \(1.76\ \text{g}\) of carbon dioxide and \(0.540\ \text{g}\) of water.
(i) Calculate the empirical formula of **Y**. Show your working. [4]
(ii) In the mass spectrum of **Y**, the molecular ion peak is at \(m/z = 118\). Deduce the molecular formula of **Y**. [1]

(d) **Y** is an aliphatic dicarboxylic acid. It reacts with excess aqueous sodium hydrogencarbonate.
(i) State the observation for this reaction. [1]
(ii) Write a balanced chemical equation for the reaction of **Y** with excess sodium hydrogencarbonate, using the molecular formula of **Y**. [2]
(iii) Identify a chemical test, including reagent and positive observation, to show that **Y** does not contain a carbon-carbon double bond. [1]
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解題

(a) Relative molecular mass, \(M_r\), is defined as the weighted average mass of a molecule of a compound relative to one-twelfth of the mass of an atom of carbon-12. [2 marks]

(b) (i) Convert units:
\(T = 150 + 273 = 423\ \text{K}\)
\(V = 244\ \text{cm}^3 = 2.44 \times 10^{-4}\ \text{m}^3\)
\(p = 1.01 \times 10^5\ \text{Pa}\)
Using \(pV = nRT\):
\(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 2.44 \times 10^{-4}}{8.31 \times 423} = 0.007011\ \text{mol}\)
\(M_r = \frac{m}{n} = \frac{0.295}{0.007011} = 42.1\ \text{g mol}^{-1}\) (accept \(42.0\) to \(42.3\)) [3 marks]

(ii) The empirical formula of **X** is \(\text{CH}_2\) (empirical formula mass \(= 14.0\)).
Since \(\frac{42.1}{14.0} \approx 3\), the molecular formula of **X** is \(\text{C}_3\text{H}_6\). [1 mark]

(c) (i) Calculate moles of each element in \(1.18\ \text{g}\) of **Y**:
- Moles of \(\text{C}\) \(= \frac{1.76}{44.0} = 0.0400\ \text{mol}\).
- Mass of \(\text{C}\) \(= 0.0400 \times 12.0 = 0.480\ \text{g}\).
- Moles of \(\text{H}\) \(= 2 \times \frac{0.540}{18.0} = 0.0600\ \text{mol}\).
- Mass of \(\text{H}\) \(= 0.0600 \times 1.0 = 0.0600\ \text{g}\).
- Mass of \(\text{O}\) \(= 1.18 - 0.480 - 0.0600 = 0.640\ \text{g}\).
- Moles of \(\text{O}\) \(= \frac{0.640}{16.0} = 0.0400\ \text{mol}\).
Find the simplest molar ratio:
\(\text{C} : \text{H} : \text{O} = 0.0400 : 0.0600 : 0.0400 = 1 : 1.5 : 1 = 2 : 3 : 2\).
Therefore, the empirical formula of **Y** is \(\text{C}_2\text{H}_3\text{O}_2\). [4 marks]

(ii) Empirical formula mass of \(\text{C}_2\text{H}_3\text{O}_2 = 59.0\).
Since \(m/z = 118\), the molecular mass is \(118\).
Ratio \(= \frac{118}{59.0} = 2\).
Therefore, the molecular formula of **Y** is \(\text{C}_4\text{H}_6\text{O}_4\). [1 mark]

(d) (i) Effervescence / bubbles of gas / fizzing. [1 mark]

(ii) Since **Y** is a dicarboxylic acid with molecular formula \(\text{C}_4\text{H}_6\text{O}_4\), it has two carboxyl groups which react with excess sodium hydrogencarbonate:
\(\text{C}_4\text{H}_6\text{O}_4 + 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{C}_4\text{H}_4\text{O}_4 + 2\text{CO}_2 + 2\text{H}_2\text{O}\) [2 marks]

(iii) Add bromine water / aqueous bromine.
Observation: The orange / brown / yellow solution remains orange / brown / yellow (does not decolourise). [1 mark]

評分準則

(a)
- Average mass of a molecule: relative to 1/12th of the mass of an atom of carbon-12 [1]
- Weighted average of isotopes [1]

(b) (i)
- Correct conversion of T to K (423 K) and V to m3 (2.44 x 10^-4 m3) [1]
- Correct rearrangement of ideal gas equation to find n = 0.00701 mol [1]
- Correct calculation of Mr = 42.1 (allow 42.0 to 42.3) [1]
(ii)
- Correct deduction of molecular formula: C3H6 [1]

(c) (i)
- Moles of C = 0.0400 mol AND moles of H = 0.0600 mol [1]
- Calculation of mass of oxygen = 0.640 g AND moles of oxygen = 0.0400 mol [1]
- Dividing by smallest to get ratio C : H : O = 1 : 1.5 : 1 [1]
- Correct empirical formula: C2H3O2 [1]
(ii)
- Correct molecular formula: C4H6O4 [1]

(d) (i)
- Effervescence / bubbling / fizzing [1]
(ii)
- Correct formulas of reactants and products [1]
- Correct balancing of equation (2NaHCO3, 2CO2, 2H2O) [1]
(iii)
- Bromine water remains orange/yellow/brown (or does not decolourise) [1]
題目 2 · Structured AS
15
The elements in Period 3 show distinct trends in their physical and chemical properties.

(a) Period 3 elements react with oxygen to form oxides.
(i) Write a balanced equation for the reaction of phosphorus with excess oxygen to form phosphorus(V) oxide. [1]
(ii) Describe the reaction of sodium oxide, \(\text{Na}_2\text{O}\), with water, including the pH of the resulting solution. [2]
(iii) Describe the reaction of phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), with water, write an equation for this reaction, and state the pH of the resulting solution. [3]

(b) (i) Aluminium oxide, \(\text{Al}_2\text{O}_3\), is described as amphoteric. Define this term by writing two equations: one for its reaction with hot aqueous hydrochloric acid, and one for its reaction with hot concentrated aqueous sodium hydroxide. [3]
(ii) Silicon dioxide, \(\text{SiO}_2\), does not react with water, but does react with hot concentrated sodium hydroxide. Explain this difference in reactivity. [2]

(c) Period 3 chlorides also show trends in their reactions with water.
(i) Contrast the behaviour of sodium chloride, \(\text{NaCl}\), and silicon tetrachloride, \(\text{SiCl}_4\), when they are separately added to water. Write a balanced chemical equation for the reaction of silicon tetrachloride with water. [3]
(ii) State and explain the difference in the pH of the resulting mixtures when \(\text{NaCl}\) and \(\text{SiCl}_4\) are separately added to water. [1]
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解題

(a) (i) Phosphorus reacts with excess oxygen to form phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\):
\(\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}\) [1 mark]

(ii) Sodium oxide reacts vigorously with water to form sodium hydroxide, which dissolves completely to form a strongly alkaline solution:
\(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\)
pH of the resulting solution is \(13\) or \(14\). [2 marks]

(iii) Phosphorus(V) oxide reacts violently with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\):
\(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\)
pH of the resulting solution is \(1\) or \(2\). [3 marks]

(b) (i) Amphoteric means the substance can react as both an acid and a base.
Reaction with \(\text{HCl}\) (acting as a base):
\(\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}\)
Reaction with \(\text{NaOH}\) (acting as an acid):
\(\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4\) [3 marks]

(ii) Silicon dioxide has a giant covalent macromolecular structure with strong silicon-oxygen covalent bonds throughout the lattice. Water molecules do not have enough energy to break these bonds, so \(\text{SiO}_2\) is insoluble and does not react. However, the strong base sodium hydroxide provides enough hydroxide ions at high temperatures to react, breaking the covalent bonds to form soluble sodium silicate and water. [2 marks]

(c) (i) Sodium chloride, \(\text{NaCl}\), is an ionic compound that simply dissolves in water to form hydrated sodium and chloride ions (neutral solution), without undergoing hydrolysis.
Silicon tetrachloride, \(\text{SiCl}_4\), is a simple covalent molecular liquid that undergoes rapid/vigorous hydrolysis in water to produce silicon dioxide (white precipitate) and fumes of hydrogen chloride gas.
Equation:
\(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) [3 marks]

(ii) \(\text{NaCl}\) solution has a pH of \(7\) because \(\text{Cl}^-\) is a very weak conjugate base and does not hydrolyse. \(\text{SiCl}_4\) hydrolysis produces the strong acid \(\text{HCl}\), which dissociates fully in water to yield a highly acidic solution with pH \(1\)–\(2\). [1 mark]

評分準則

(a) (i)
- Correct balanced equation: P4 + 5O2 -> P4O10 [1]
(ii)
- Description: reacts to form a colorless solution / dissolves [1]
- pH: 13 or 14 [1]
(iii)
- Correct equation: P4O10 + 6H2O -> 4H3PO4 [1]
- Description: reacts violently / exothermically [1]
- pH: 1 or 2 [1]

(b) (i)
- Correct equation with HCl [1]
- Correct equation with NaOH [1]
- Both equations balanced [1]
(ii)
- Mentions giant covalent structure of SiO2 and very strong covalent bonds [1]
- Explains that water cannot break these bonds, but hot NaOH is strong enough to react to form silicates [1]

(c) (i)
- NaCl dissolves but does not react / hydrolyse [1]
- SiCl4 reacts vigorously / hydrolyses to form white precipitate (SiO2) and misty fumes (HCl) [1]
- Correct balanced equation: SiCl4 + 2H2O -> SiO2 + 4HCl [1]
(ii)
- NaCl gives pH 7 because no hydrolysis occurs; SiCl4 gives pH 1 or 2 because HCl is produced [1]
題目 3 · Structured AS
15
Nitrogen and sulfur play vital roles in biological systems, but their oxides are also major environmental pollutants.

(a) Nitrogen gas, \(\text{N}_2\), is chemically unreactive under normal conditions, whereas nitrogen monoxide, \(\text{NO}\), is highly reactive.
(i) Explain the unreactivity of nitrogen gas in terms of its bonding. [2]
(ii) Explain how nitrogen oxides, \(\text{NO}_x\), are formed in internal combustion engines, and write a balanced chemical equation for the formation of nitrogen monoxide. [3]

(b) Nitrogen monoxide acts as a homogeneous catalyst in the atmospheric oxidation of sulfur dioxide, \(\text{SO}_2\), to sulfur trioxide, \(\text{SO}_3\).
(i) Write two equations to show how \(\text{NO}\) catalyses this oxidation. [2]
(ii) Explain why \(\text{NO}\) acts as a catalyst in this reaction. [1]
(iii) State one environmental consequence of acid rain on the aquatic ecosystem and one on buildings or structures. [2]

(c) Ammonium sulfate, \((\text{NH}_4)_2\text{SO}_4\), is a synthetic nitrogenous fertiliser.
(i) Write a balanced chemical equation for the preparation of ammonium sulfate from ammonia and sulfuric acid. [1]
(ii) Explain why adding calcium hydroxide (slaked lime) to soil that has recently been treated with ammonium sulfate fertiliser reduces the effectiveness of the fertiliser, and write a chemical equation for the reaction that occurs. [3]
(iii) Explain how the discharge of agricultural fertilisers into rivers can lead to eutrophication. [1]
查看答案詳解

解題

(a) (i) Nitrogen gas exists as diatomic molecules, \(\text{N}\equiv\text{N}\), containing a triple covalent bond. The bond energy of the \(\text{N}\equiv\text{N}\) bond is extremely high (\(944\ \text{kJ mol}^{-1}\)), making it highly stable and requiring a huge amount of energy to break. [2 marks]

(ii) In internal combustion engines, the extremely high temperatures (around \(2000\ ^\circ\text{C}\)) and pressures provide the required activation energy to break the strong triple bond in \(\text{N}_2\), allowing nitrogen to react with oxygen from the air:
\(\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}\) [3 marks]

(b) (i) The catalytic cycle of \(\text{NO}\) in the atmosphere:
Step 1: \(\text{NO} + \frac{1}{2}\text{O}_2 \rightarrow \text{NO}_2\)
Step 2: \(\text{NO}_2 + \text{SO}_2 \rightarrow \text{NO} + \text{SO}_3\) [2 marks]

(ii) \(\text{NO}\) acts as a catalyst because it participates in the reaction, providing an alternative pathway with a lower activation energy, and is regenerated chemically unchanged at the end of the process. [1 mark]

(iii)
- Aquatic ecosystem: Acidifies lakes and rivers, releasing toxic aluminium ions which kill fish and other aquatic organisms.
- Buildings/structures: Acid rain reacts with limestone or marble (calcium carbonate), causing chemical weathering and structural erosion. [2 marks]

(c) (i) \(2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4\) [1 mark]

(ii) Calcium hydroxide, \(\text{Ca(OH)}_2\), is basic and reacts with the ammonium ions, \(\text{NH}_4^+\), in the fertiliser in an acid-base reaction, releasing ammonia gas, \(\text{NH}_3\). The ammonia gas escapes into the atmosphere, so nitrogen is lost from the soil, reducing the effectiveness of the fertiliser.
Equation:
\((\text{NH}_4)_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2\text{NH}_3 + 2\text{H}_2\text{O}\) [3 marks]

(iii) Runoff of fertilisers causes a rapid growth of algae on the surface of water (algal bloom) which blocks light. Subsequent decomposition of dead algae by bacteria uses up dissolved oxygen, causing aquatic life to suffocate. [1 mark]

評分準則

(a) (i)
- Strong triple bond (N#N) [1]
- Extremely high bond enthalpy / activation energy to break the bond [1]
(ii)
- High temperature / pressure in the engine [1]
- Causes atmospheric N2 and O2 to react [1]
- Balanced equation: N2 + O2 -> 2NO [1]

(b) (i)
- Equation 1: NO + 1/2 O2 -> NO2 [1]
- Equation 2: NO2 + SO2 -> NO + SO3 [1]
(ii)
- It provides an alternative pathway with lower activation energy AND is regenerated [1]
(iii)
- Aquatic effect: kills fish / acidifies water [1]
- Building effect: erodes limestone / marble structures [1]

(c) (i)
- Balanced equation: 2NH3 + H2SO4 -> (NH4)2SO4 [1]
(ii)
- Ca(OH)2 reacts with NH4+ to produce ammonia gas [1]
- Ammonia gas escapes, reducing the nitrogen content of the soil [1]
- Balanced equation: (NH4)2SO4 + Ca(OH)2 -> CaSO4 + 2NH3 + 2H2O [1]
(iii)
- Mentions algal bloom, bacterial decomposition consuming oxygen, leading to death of aquatic life [1]
題目 4 · Structured AS
15
Compound **A** has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\). It exists as several structural and stereoisomers.

(a) Isomer **A1** is a straight-chain compound that reacts with:
- sodium metal to produce hydrogen gas.
- acidified potassium dichromate(VI) to form a compound that does not react with Tollens' reagent.
- alkaline aqueous iodine to give a yellow precipitate.

(i) Identify the functional group in **A1** responsible for the reaction with sodium. [1]
(ii) Identify the functional group produced when **A1** reacts with acidified potassium dichromate(VI), and explain why this product does not react with Tollens' reagent. [2]
(iii) State the identity of the yellow precipitate formed in the reaction of **A1** with alkaline aqueous iodine. [1]
(iv) Deduce the structure of **A1**. Explain your reasoning based on the reactions described. [3]

(b) Isomer **A2** is a structural isomer of **A1** that contains a chiral carbon atom. Like **A1**, it reacts with sodium metal and gives a positive reaction with alkaline aqueous iodine.
(i) Explain what is meant by a chiral carbon atom. [1]
(ii) Draw the structural formula of **A2**. Mark the chiral carbon with an asterisk (*). [1]
(iii) Draw three-dimensional structures of the two optical isomers of **A2**, showing clearly the spatial relationship between them. [2]

(c) Isomer **A3** is an ester.
(i) Draw the skeletal structure of an ester with molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) that can be hydrolysed to produce methanol as one of the products. [1]
(ii) Write an equation for the acid-catalysed hydrolysis of this ester. [2]
(iii) Name the other organic product of this hydrolysis. [1]
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解題

(a) (i) The alcohol (hydroxyl, \(\text{-OH}\)) group is responsible for the reaction with sodium metal. [1 mark]

(ii) The product contains a carboxylic acid group (\(\text{-COOH}\)) and a ketone group (which is resistant to oxidation), but does not contain an aldehyde group. Tollens' reagent only oxidises aldehydes, hence it does not react with this product. [2 marks]

(iii) Triiodomethane (or iodoform), \(\text{CHI}_3\). [1 mark]

(iv) **A1** must contain:
1. An \(\text{-OH}\) group (reaction with \(\text{Na}\)).
2. A methyl ketone group, \(\text{CH}_3\text{C}=\text{O}\) (positive iodoform test).
3. A straight chain of 4 carbon atoms.
4. The alcohol group must be a primary alcohol (since it oxidises to a carboxylic acid without an aldehyde intermediate remaining).
This gives the structure of 4-hydroxybutan-2-one: \(\text{HO-CH}_2\text{-CH}_2\text{-CO-CH}_3\). [3 marks]

(b) (i) A chiral carbon atom is a carbon atom that is bonded to four different atoms or groups of atoms. [1 mark]

(ii) Isomer **A2** is 3-hydroxybutan-2-one. Its structural formula is:
\(\text{CH}_3\text{-CO-C}^*\text{H(OH)-CH}_3\)
The chiral carbon is C3, marked with an asterisk (*). [1 mark]

(iii) The optical isomers are tetrahedral around the chiral carbon (C3) and are non-superimposable mirror images of each other. The drawings should show \(\text{-H}\), \(\text{-OH}\), \(\text{-CH}_3\), and \(\text{-COCH}_3\) arranged tetrahedrally around the central carbon atom, with a mirror plane separating them. [2 marks]

(c) (i) The ester is methyl propanoate. Skeletal structure of methyl propanoate is a 3-carbon carbonyl chain bonded to oxygen, which is bonded to a methyl group (drawn as \(\text{CH}_3\text{CH}_2\text{COOCH}_3\)). [1 mark]

(ii) Acid-catalysed hydrolysis:
\(\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{OH}\) [2 marks]

(iii) Propanoic acid. [1 mark]

評分準則

(a) (i)
- Alcohol / hydroxyl / -OH group [1]
(ii)
- Carboxylic acid / -COOH [1]
- Product has no aldehyde group / ketones and carboxylic acids cannot be further oxidised by Tollens' [1]
(iii)
- Triiodomethane / CHI3 [1]
(iv)
- Deduces straight chain of 4 carbons with -CH2OH and CH3CO- groups [1]
- Correctly identifies primary alcohol position [1]
- Structure of A1: HO-CH2-CH2-CO-CH3 [1]

(b) (i)
- Carbon atom bonded to four different groups [1]
(ii)
- Structure of A2: CH3-CO-CH(OH)-CH3 with C3 correctly marked with asterisk (*) [1]
(iii)
- Two tetrahedral representations [1]
- Drawn clearly as mirror images of each other with correct groups (-H, -OH, -CH3, -COCH3) [1]

(c) (i)
- Correct skeletal structure of methyl propanoate [1]
(ii)
- Reactants: methyl propanoate and water AND reversible arrow [1]
- Products: propanoic acid and methanol [1]
(iii)
- Propanoic acid [1]

Paper 42

Answer all structured questions. Show full working where calculations are required and write appropriate units.
8 題目 · 100
題目 1 · structured
12.5
Propanoic acid is a weak Brønsted-Lowry acid.

(a) Define a Brønsted-Lowry acid and write the expression for the acid dissociation constant, \(K_a\), of propanoic acid, \(CH_3CH_2COOH\). [2.5]

(b) Calculate the pH of a 0.150 mol dm\(^{-3}\) aqueous solution of propanoic acid. The \(K_a\) of propanoic acid is \(1.35 \times 10^{-5}\) mol dm\(^{-3}\) at 298 K. [3]

(c) A buffer solution is prepared by mixing 50.0 cm\(^3\) of 0.150 mol dm\(^{-3}\) propanoic acid with 25.0 cm\(^3\) of 0.100 mol dm\(^{-3}\) sodium hydroxide.
(i) Calculate the concentration of both propanoic acid and propanoate ions in this buffer mixture. [4]
(ii) Show by calculation that the pH of this buffer mixture is 4.57. [3]
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解題

(a) A Brønsted-Lowry acid is a proton (\(H^+\)) donor.
\[K_a = \frac{[CH_3CH_2COO^-][H^+]}{[CH_3CH_2COOH]}\]

(b) For a weak acid, \([H^+] = \sqrt{K_a \times [HA]}\)
\([H^+] = \sqrt{1.35 \times 10^{-5} \times 0.150} = \sqrt{2.025 \times 10^{-6}} = 1.423 \times 10^{-3}\) mol dm\(^{-3}\)
\(pH = -\log[H^+] = -\log(1.423 \times 10^{-3}) = 2.85\)

(c) (i) Initial moles of propanoic acid: \(n(\text{acid}) = \frac{50.0}{1000} \times 0.150 = 0.0075\) mol.
Initial moles of NaOH added: \(n(\text{NaOH}) = \frac{25.0}{1000} \times 0.100 = 0.0025\) mol.
Upon mixing, the neutralisation reaction occurs:
\(CH_3CH_2COOH + NaOH \rightarrow CH_3CH_2COONa + H_2O\)
Remaining moles of propanoic acid: \(0.0075 - 0.0025 = 0.0050\) mol.
Moles of propanoate ions formed: \(0.0025\) mol.
Total volume of the buffer: \(50.0 + 25.0 = 75.0\) cm\(^3\) = \(0.075\) dm\(^3\).
\([CH_3CH_2COOH] = \frac{0.0050}{0.075} = 0.0667\) mol dm\(^{-3}\)
\([CH_3CH_2COO^-] = \frac{0.0025}{0.075} = 0.0333\) mol dm\(^{-3}\)

(ii) Using the buffer equation:
\(pH = pK_a + \log\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}\)
\(pK_a = -\log(1.35 \times 10^{-5}) = 4.87\)
\(pH = 4.87 + \log\left(\frac{0.0333}{0.0667}\right) = 4.87 + \log(0.5) = 4.87 - 0.30 = 4.57\)

評分準則

(a) Definition of proton donor [1]
Correct expression for \(K_a\) with state symbols or standard charges [1.5]
(b) Correct calculation of \([H^+]\) [1]
Correct calculation of pH to 2 decimal places [1]
(c)(i) Moles of acid and NaOH calculated [1]
Moles of acid remaining and salt formed [1]
Total volume used correctly [1]
Both concentrations calculated correctly [1]
(c)(ii) Correct calculation of \(pK_a\) [1]
Correct buffer equation application [1]
Correct pH showing 4.57 [1]
題目 2 · structured
12.5
Transition elements form a wide variety of coloured complex ions with different geometries.

(a) Explain why transition metal complexes are coloured, making reference to d-orbital splitting, light absorption, and d-d electron transitions. [4.5]

(b) Cobalt(II) forms an octahedral complex with water, \([Co(H_2O)_6]^{2+}\), and a tetrahedral complex with chloride ions, \([CoCl_4]^{2-}\).
(i) Draw 3D diagrams of both \([Co(H_2O)_6]^{2+}\) and \([CoCl_4]^{2-}\), showing their 3D shapes clearly. [3]
(ii) State the coordination number and geometry for each complex ion. [2]

(c) Explain why cobalt(II) forms a tetrahedral complex with chloride ligands, but an octahedral complex with water ligands, in terms of ligand size and steric hindrance. [3]
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解題

(a) In a transition metal complex, the presence of ligands causes the degenerate 3d orbitals to split into two sets of non-degenerate energy levels.
When visible light is shone on the complex, an electron absorbs a specific frequency of light energy and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition).
The color observed is the complementary color of the absorbed light, which is transmitted or reflected.

(b) (i)
- Octahedral \([Co(H_2O)_6]^{2+}\): Cobalt in the center with 6 H2O molecules coordinated via dative covalent bonds. Two bonds in the plane, two wedges pointing forward, and two dashes pointing backward.
- Tetrahedral \([CoCl_4]^{2-}\): Cobalt in the center with 4 Cl\(^-\)\ ligands. One vertical bond, one bond in plane, one wedge pointing forward, and one dash pointing backward.
(ii)
- \([Co(H_2O)_6]^{2+}\): Coordination number = 6, Geometry = Octahedral
- \([CoCl_4]^{2-}\): Coordination number = 4, Geometry = Tetrahedral

(c) Chloride ions (\(Cl^-\)) are significantly larger than neutral water molecules (\(H_2O\)). Due to the larger size of the chloride ligand, there is greater steric hindrance and electrostatic repulsion between ligands around the central cobalt(II) ion. Only four chloride ligands can fit around the metal ion, resulting in a lower coordination number of 4 (tetrahedral shape).

評分準則

(a) Splitting of d-orbitals into two sets [1.5]
Absorption of light promoting an electron from lower to higher energy level / d-d transition [1.5]
Complementary color transmitted/observed [1.5]
(b)(i) Correct 3D diagram with appropriate wedges/dashes and overall charge for [Co(H2O)6]2+ [1.5]
Correct 3D diagram with wedges/dashes and overall charge for [CoCl4]2- [1.5]
(b)(ii) Coordination numbers 6 and 4 stated correctly [1]
Geometries octahedral and tetrahedral stated correctly [1]
(c) Mentioning chloride ion is larger than water molecule [1]
Steric hindrance / mutual ligand repulsion prevents 6 chlorides from fitting around Co [1]
Leads to reduced coordination number 4 [1]
題目 3 · structured
12.5
Electrochemical cells can be used to measure standard electrode potentials and predict the feasibility of reactions.

(a) Describe and draw a labeled diagram of an electrochemical cell designed to measure the standard electrode potential, \(E^\theta\), of the \(Fe^{3+}/Fe^{2+}\) half-cell relative to the standard hydrogen electrode (SHE). Label all electrodes, solution compositions, concentrations, and physical conditions. [5]

(b) The standard electrode potentials for two half-cells are given below:
- \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\) \(E^\theta = +0.80\) V
- \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq)\) \(E^\theta = +0.77\) V

(i) Write the overall ionic equation for the reaction that occurs when these two half-cells are connected under standard conditions, and calculate the standard cell potential, \(E^\theta_{\text{cell}}\). [2.5]
(ii) Use the Nernst equation, \(E = E^\theta + \frac{0.059}{z}\log\frac{[\text{oxidised}]}{[\text{reduced}]}\), to calculate the electrode potential, \(E\), of the \(Fe^{3+}/Fe^{2+}\) half-cell at 298 K when \([Fe^{3+}] = 0.010\) mol dm\(^{-3}\) and \([Fe^{2+}] = 0.500\) mol dm\(^{-3}\). [3]
(iii) Predict and explain how this change in concentration affects the cell potential, \(E_{\text{cell}}\), and the thermodynamic feasibility of the reaction in (b)(i). [2]
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解題

(a) The diagram must include:
- Standard Hydrogen Electrode (SHE): Pt electrode, 1.0 mol dm\(^{-3}\) \(H^+(aq)\), and \(H_2(g)\) bubbled at 1 bar/1 atm.
- \(Fe^{3+}/Fe^{2+}\) half-cell: Pt electrode in a solution containing 1.0 mol dm\(^{-3}\) \(Fe^{3+}(aq)\) and 1.0 mol dm\(^{-3}\) \(Fe^{2+}(aq)\).
- Connection: Voltmeter in the external circuit, and a salt bridge connecting both solutions.
- Conditions: Temperature of 298 K (25 °C).

(b) (i) Since \(E^\theta\) for \(Ag^+/Ag\) is more positive (+0.80 V) than for \(Fe^{3+}/Fe^{2+}\) (+0.77 V), \(Ag^+\) is reduced and \(Fe^{2+}\) is oxidised:
\(Ag^+(aq) + Fe^{2+}(aq) \rightarrow Ag(s) + Fe^{3+}(aq)\)
\(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = +0.80 - (+0.77) = +0.03\) V

(ii) Using Nernst: \(E = E^\theta + \frac{0.059}{z}\log\frac{[Fe^{3+}]}{[Fe^{2+}]}\) with \(z = 1\)
\(E = +0.77 + \frac{0.059}{1}\log\left(\frac{0.010}{0.500}\right) = +0.77 + 0.059\log(0.02) = +0.77 + 0.059(-1.699)\)
\(E = +0.77 - 0.100 = +0.67\) V

(iii) Since the potential of the iron half-cell decreased to +0.67 V, the cell potential becomes:
\(E_{\text{cell}} = +0.80 - (+0.67) = +0.13\) V.
Because \(E_{\text{cell}}\) becomes more positive, the overall reaction becomes more thermodynamically feasible.

評分準則

(a) Diagram showing both half-cells correctly set up [1]
Standard hydrogen electrode labeled with Pt electrode, 1.0 mol dm-3 H+, H2 gas at 1 bar [1]
Iron half-cell labeled with Pt electrode, 1.0 mol dm-3 Fe2+ and Fe3+ [1]
Salt bridge and voltmeter shown [1]
Standard temperature 298 K stated [1]
(b)(i) Correct overall equation [1.5]
Calculation of cell potential (+0.03 V) [1]
(b)(ii) Correct substitution of concentration values into Nernst equation [1]
Correct log calculation [1]
Correct final value (+0.67 V) [1]
(b)(iii) Calculation of new cell potential (+0.13 V) [1]
Explanation that positive potential means greater thermodynamic feasibility [1]
題目 4 · structured
12.5
Benzene undergoes electrophilic substitution reactions, such as nitration, due to its stable delocalised \(\pi\) system.

(a) Benzene is reacted with a mixture of concentrated nitric acid and concentrated sulfuric acid at 50 °C.
(i) Write an equation for the generation of the electrophile, \(NO_2^+\), showing the role of H\(_2\)SO\(_4\) as a catalyst. [2]
(ii) Outline the mechanism for the nitration of benzene, using curly arrows to show the movement of electron pairs. Draw the structure of the intermediate clearly. [5]

(b) Methylbenzene undergoes nitration much faster than benzene.
(i) Explain why methylbenzene is more reactive towards electrophiles than benzene. [2.5]
(ii) Identify the major organic mono-nitrated products formed when methylbenzene reacts with the nitrating mixture, and explain the directing effect of the methyl group. [3]
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解題

(a) (i) The equation for the generation of the electrophile is:
\(HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-\)
(Sulfuric acid acts as a Brønsted-Lowry acid to protonate nitric acid).

(ii) Mechanism:
- Curly arrow from the delocalised \(\pi\) ring of benzene pointing to the \(NO_2^+\) electrophile.
- Drawing the non-planar cationic intermediate: a horseshoe shape containing a positive charge, with the open end facing the tetrahedral carbon bonded to both \(-H\) and \(-NO_2\).
- Curly arrow from the C-H bond breaking, pointing back into the horseshoe of the intermediate to restore the delocalised system.
- Products: Nitrobenzene and \(H^+\) ion (which regenerates the sulfuric acid catalyst with \(HSO_4^-\)).

(b) (i) The methyl group (\(-CH_3\)) is an electron-donating group due to its positive inductive effect. This increases the electron density in the delocalised \(\pi\) system of the benzene ring. As a result, the ring in methylbenzene is more attractive to electrophiles than benzene itself.

(ii) The methyl group is a 2,4-directing group. The major organic products are 1-methyl-2-nitrobenzene (2-nitrotoluene) and 1-methyl-4-nitrobenzene (4-nitrotoluene).

評分準則

(a)(i) Equation showing generation of NO2+ [1]
Correctly showing H2SO4 acting as an acid / catalyst [1]
(a)(ii) Curly arrow from benzene ring to NO2+ [1.5]
Correct intermediate structure with positive charge in ring [2]
Curly arrow from C-H bond to reform ring [1.5]
(b)(i) Mention of the positive inductive effect of methyl group [1]
Explanation of increased electron density on the benzene ring [1]
Increased attraction to electrophiles [0.5]
(b)(ii) Correct structures/names of 2-nitrotoluene and 4-nitrotoluene [2]
Stating that methyl group is 2,4-directing [1]
題目 5 · structured
12.5
The reaction between peroxodisulfate(VI) ions, \(S_2O_8^{2-}\), and iodide ions, \(I^-\), is catalysed by \(Fe^{2+}\)(aq):
\(S_2O_8^{2-}(aq) + 2I^-(aq) \rightarrow 2SO_4^{2-}(aq) + I_2(aq)\)

(a) Initial rate experiments were carried out at 298 K, giving the following results:
- Expt 1: \([S_2O_8^{2-}] = 0.010\) mol dm\(^{-3}\), \([I^-] = 0.010\) mol dm\(^{-3}\), Initial Rate = \(1.2 \times 10^{-5}\) mol dm\(^{-3}\) s\(^{-1}\)
- Expt 2: \([S_2O_8^{2-}] = 0.020\) mol dm\(^{-3}\), \([I^-] = 0.010\) mol dm\(^{-3}\), Initial Rate = \(2.4 \times 10^{-5}\) mol dm\(^{-3}\) s\(^{-1}\)
- Expt 3: \([S_2O_8^{2-}] = 0.010\) mol dm\(^{-3}\), \([I^-] = 0.020\) mol dm\(^{-3}\), Initial Rate = \(2.4 \times 10^{-5}\) mol dm\(^{-3}\) s\(^{-1}\)

(i) Deduce the order of reaction with respect to both reactants, explaining your reasoning. [3]
(ii) Write the rate equation and calculate the rate constant, \(k\), at 298 K, including appropriate units. [3.5]

(b) (i) Define the term activation energy, \(E_a\). [1]
(ii) The rate constant \(k\) was determined at different temperatures. A plot of \(\ln k\) against \(1/T\) yielded a straight line with a gradient of \(-6130\) K. Calculate the activation energy, \(E_a\), for this reaction in kJ mol\(^{-1}\). (\(R = 8.31\) J K\(^{-1}\) mol\(^{-1}\)). [3]
(iii) Explain how the addition of \(Fe^{2+}\)(aq) catalyses the reaction. [2]
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解題

(a) (i)
- From Expt 1 to Expt 2, \([I^-]\) is constant, and \([S_2O_8^{2-}]\) is doubled. The initial rate is doubled (from \(1.2 \times 10^{-5}\) to \(2.4 \times 10^{-5}\)). Thus, the reaction is first-order with respect to \(S_2O_8^{2-}\).
- From Expt 1 to Expt 3, \([S_2O_8^{2-}]\) is constant, and \([I^-]\) is doubled. The initial rate is doubled. Thus, the reaction is first-order with respect to \(I^-\).

(ii) Rate equation:
\(\text{Rate} = k [S_2O_8^{2-}][I^-]\)
Using data from Expt 1:
\(1.2 \times 10^{-5} = k (0.010)(0.010)\)
\(k = \frac{1.2 \times 10^{-5}}{1.0 \times 10^{-4}} = 0.12\) mol\(^{-1}\) dm\(^3\) s\(^{-1}\)

(b) (i) Activation energy is the minimum energy required for a collision to result in a chemical reaction.

(ii) According to the Arrhenius equation: \(\text{gradient} = -\frac{E_a}{R}\)
\(-6130 = -\frac{E_a}{8.31}\)
\(E_a = 6130 \times 8.31 = 50940.3\) J mol\(^{-1}\)
In kJ mol\(^{-1}\):
\(E_a = 50.9\) kJ mol\(^{-1}\)

(iii) The reaction between two negative ions (\(S_2O_8^{2-}\) and \(I^-\)) has high activation energy due to strong electrostatic repulsion. \(Fe^{2+}\) acts as a homogeneous catalyst by reacting with peroxodisulfate first:
\(S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}\)
Then \(Fe^{3+}\) reacts with iodide to regenerate the catalyst:
\(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\)
These steps involve oppositely charged ions, which reduces repulsion and provides an alternative pathway with a lower activation energy.

評分準則

(a)(i) Order wrt S2O8^2- is 1 with reasoning [1.5]
Order wrt I- is 1 with reasoning [1.5]
(a)(ii) Correct rate equation [1]
Correct calculation of k value (0.12) [1.5]
Correct units (mol-1 dm3 s-1) [1]
(b)(i) Precise definition of activation energy [1]
(b)(ii) Equating gradient to -Ea/R [1]
Conversion to kJ mol-1 [1]
Correct numerical answer (50.9) [1]
(b)(iii) Explaining repulsion of like charges [1]
Showing how Fe2+ reduces activation energy by redox steps [1]
題目 6 · structured
12.5
Amino acids are bifunctional organic compounds that exhibit unique properties depending on pH.

(a) Glycine (H\(_2\)NCH\(_2\)COOH) is the simplest amino acid.
(i) Draw the zwitterionic structure of glycine. [1.5]
(ii) Explain how glycine acts as a buffer solution when small amounts of acid or alkali are added. [3]

(b) When a mixture of glycine and alanine (H\(_2\)NCH(CH\(_3\))COOH) is heated in the presence of a suitable catalyst, condensation polymerisation can occur.
(i) Draw the structures of the two possible dipeptides formed by the reaction between one molecule of glycine and one molecule of alanine. [3]
(ii) State the name of the reaction type and the name of the functional group linking the two amino acid residues. [2]

(c) Electrophoresis can be used to separate a mixture of amino acids.
Explain how the direction and rate of movement of amino acids during electrophoresis depend on the pH of the buffer solution and the structure of the amino acids. [3]
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解題

(a) (i) The zwitterion has a protonated amine group and a deprotonated carboxylic acid group:
\(^+H_3N-CH_2-COO^-\)

(ii) When an acid is added (\(H^+\)), the carboxylate group accepts the proton to minimize pH changes:
\(^+H_3N-CH_2-COO^- + H^+ \rightarrow ^+H_3N-CH_2-COOH\)
When a base is added (\(OH^-\)), the ammonium group donates a proton to neutralize the base:
\(^+H_3N-CH_2-COO^- + OH^- \rightarrow H_2N-CH_2-COO^- + H_2O\)

(b) (i)
- Dipeptide 1 (Gly-Ala): \(H_2N-CH_2-CONH-CH(CH_3)-COOH\)
- Dipeptide 2 (Ala-Gly): \(H_2N-CH(CH_3)-CONH-CH_2-COOH\)
(ii)
Reaction type: Condensation
Functional group: Amide (or peptide link)

(c)
- If the pH of the buffer is below the isoelectric point of the amino acid, the amino acid is protonated (positively charged) and moves towards the negative electrode (cathode).
- If the pH is above the isoelectric point, the amino acid is deprotonated (negatively charged) and moves towards the positive electrode (anode).
- The rate of migration depends on the net charge (larger charge = faster movement) and the relative molecular mass (smaller/lighter ions migrate faster).

評分準則

(a)(i) Correct zwitterionic structure [1.5]
(a)(ii) Equation/explanation for addition of acid [1.5]
Equation/explanation for addition of alkali [1.5]
(b)(i) Structure of Gly-Ala dipeptide [1.5]
Structure of Ala-Gly dipeptide [1.5]
(b)(ii) Reaction type stated as condensation [1]
Functional group identified as amide / peptide [1]
(c) Movement direction explained in terms of charge relative to isoelectric point [1]
Effect of net charge on rate of movement [1]
Effect of molecular size/mass on rate of movement [1]
題目 7 · structured
12.5
The properties of elements in Period 3 show distinct periodic trends across the period.

(a) Consider the three Period 3 oxides: \(Na_2O\), \(Al_2O_3\), and \(SO_3\).
(i) Describe the reaction, if any, of each oxide with water. Write chemical equations for any reactions that occur and state the approximate pH of the resulting mixture. [5]
(ii) Explain the variation in pH of these resulting solutions in terms of the structure and bonding of the three oxides. [3]

(b) (i) Describe what is observed when water is added separately to solid \(NaCl\) and liquid \(SiCl_4\). Write chemical equations for any reactions that occur. [3.5]
(ii) Explain why \(SiCl_4\) is readily hydrolysed by water, whereas \(NaCl\) simply dissolves, in terms of the electronic structure of silicon. [1.5]
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解題

(a) (i)
- \(Na_2O\) reacts vigorously with water to form sodium hydroxide:
\(Na_2O(s) + H_2O(l) \rightarrow 2NaOH(aq)\)
pH is approximately 13–14.

- \(Al_2O_3\) does not react with water as it is insoluble:
pH is approximately 7.

- \(SO_3\) reacts violently with water to form sulfuric acid:
\(SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\)
pH is approximately 1–2.

(ii)
- \(Na_2O\) has a giant ionic structure. It contains basic oxide (\(O^{2-}\)) ions which readily accept protons from water to produce hydroxide ions.
- \(Al_2O_3\) is also ionic but with very high lattice energy due to the highly charged ions (\(Al^{3+}\) and \(O^{2-}\)), making it insoluble in water.
- \(SO_3\) is a simple molecular covalent compound. Non-metal oxides are acidic; they undergo hydrolysis to generate hydronium (\(H^+\)) ions.

(b) (i)
- With \(NaCl\): The solid dissolves to form a clear, neutral solution. No gas or fumes evolved.
\(NaCl(s) \xrightarrow{H_2O} Na^+(aq) + Cl^-(aq)\)
- With \(SiCl_4\): Rapid, exothermic hydrolysis occurs. Dense white fumes of hydrogen chloride (\(HCl\)) are observed, alongside a white precipitate of silicon dioxide (\(SiO_2\)).
\(SiCl_4(l) + 2H_2O(l) \rightarrow SiO_2(s) + 4HCl(g)\) (or silicic acid formation)

(ii) Silicon has empty, relatively low-lying 3d orbitals in its valence shell. These 3d orbitals can accept a lone pair of electrons from oxygen in a water molecule, initiating the hydrolysis pathway. Sodium has no such d-orbitals available for coordination.

評分準則

(a)(i) Na2O reaction, equation, and pH 13-14 [2]
Al2O3 insolubility and pH 7 [1.5]
SO3 reaction, equation, and pH 1-2 [1.5]
(a)(ii) Ionic bonding/giant structure of Na2O and basic oxide ions [1]
High lattice energy of Al2O3 [1]
Covalent molecular nature of SO3 forming acidic species [1]
(b)(i) Dissolving of NaCl (neutral solution) [1]
Hydrolysis of SiCl4, white fumes, and precipitate [1]
Chemical equation for SiCl4 hydrolysis [1.5]
(b)(ii) Availability of d-orbitals in silicon to accept lone pair from water [1.5]
題目 8 · structured
12.5
The ideal gas equation can be used to find the relative molecular mass of volatile organic compounds.

(a) A gaseous hydrocarbon, X, has a density of 2.11 g dm\(^{-3}\) at a temperature of 323 K and a pressure of \(1.01 \times 10^5\) Pa.
(i) Use the ideal gas equation, \(pV = nRT\), to calculate the relative molecular mass, \(M_r\), of X. (\(R = 8.31\) J K\(^{-1}\) mol\(^{-1}\)). [3.5]
(ii) Combustion analysis of 1.20 g of X shows that it contains 1.03 g of carbon and the remainder is hydrogen. Calculate the empirical formula of X. [3]
(iii) Deduce the molecular formula of X. [2]

(b) (i) Write a balanced chemical equation for the complete combustion of X. [2]
(ii) Calculate the volume of carbon dioxide gas, in dm\(^3\), produced at 298 K and \(1.01 \times 10^5\) Pa when 2.50 g of X is completely combusted. (Assume that the molar volume of gas under these conditions is 24.0 dm\(^3\) mol\(^{-1}\)). [2]
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解題

(a) (i) \(pV = nRT\)
Let's assume a volume \(V = 1.0\) dm\(^{3}\) = \(1.0 \times 10^{-3}\) m\(^{3}\).
The mass of X in this volume is 2.11 g.
\(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 1.0 \times 10^{-3}}{8.31 \times 323} = \frac{101}{2684.13} = 0.03763\) mol.
Since \(n = \frac{m}{M_r}\):
\(M_r = \frac{2.11}{0.03763} = 56.1\) g mol\(^{-1}\).

(ii) Mass of carbon = 1.03 g.
Mass of hydrogen = \(1.20 - 1.03 = 0.17\) g.
Moles of Carbon: \(\frac{1.03}{12.0} = 0.0858\) mol.
Moles of Hydrogen: \(\frac{0.17}{1.0} = 0.17\) mol.
Ratio C : H = \(\frac{0.0858}{0.0858} : \frac{0.17}{0.0858} = 1 : 1.98 \approx 1 : 2\).
Therefore, the empirical formula of X is \(CH_2\).

(iii) Empirical formula mass of \(CH_2 = 12.0 + 2(1.0) = 14.0\).
Ratio of \(M_r\) to empirical formula mass = \(\frac{56.1}{14.0} \approx 4\).
Thus, the molecular formula of X is \(C_4H_8\).

(b) (i) \(C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O\)

(ii) Moles of \(C_4H_8\) in 2.50 g:
\(n(\text{hydrocarbon}) = \frac{2.50}{56.1} = 0.04456\) mol.
Moles of \(CO_2\) produced = \(4 \times 0.04456 = 0.1782\) mol.
Volume of \(CO_2\) = \(0.1782 \times 24.0 = 4.28\) dm\(^3\).

評分準則

(a)(i) Volume conversion to m3 [1]
Moles calculated as 0.0376 mol [1]
Correct calculation of Mr to 1 decimal place (56.1) [1.5]
(a)(ii) Finding mass of hydrogen (0.17 g) [1]
Moles of C and H calculated [1]
Correct empirical formula CH2 [1]
(a)(iii) Correct empirical mass calculation [1]
Correct molecular formula C4H8 [1]
(b)(i) Balanced combustion equation [2]
(b)(ii) Moles of C4H8 and CO2 calculated [1]
Correct volume of CO2 in dm3 [1]

Paper 52

Answer all planning, analysis, and evaluation questions. Pay attention to experimental detail and numerical errors.
2 題目 · 30
題目 1 · Planning, Analysis & Evaluation
15
A student conducts an experiment to determine the Faraday constant, \(F\), by electrolyzing aqueous copper(II) sulfate using copper electrodes.

(a) Draw a labelled diagram of the apparatus that the student should use to carry out this electrolysis. Your diagram should include the electrical circuit showing how current is measured and controlled. Describe how the cathode should be prepared, washed, and dried before and after the electrolysis to ensure accurate mass measurements. (5 marks)

(b) The student runs the electrolysis at a constant current of \(0.500\text{ A}\) for various time periods, \(t\). After each run, the cathode is washed, dried, and weighed to determine the mass of copper deposited, \(m\).
The table below shows the results obtained.

| Run | Time, \(t\) / s | Mass of copper deposited, \(m\) / g |
|:---:|:---:|:---:|
| 1 | 600 | 0.099 |
| 2 | 1200 | 0.198 |
| 3 | 1800 | 0.297 |
| 4 | 2400 | 0.352 |
| 5 | 3000 | 0.495 |
| 6 | 3600 | 0.594 |

(i) Plot a graph of mass of copper deposited, \(m\) (y-axis), against time, \(t\) (x-axis). Describe the shape of the graph, identify the anomalous point, and suggest a physical reason for this anomaly. (4 marks)

(ii) Use your graph to determine the gradient of the line of best fit. Show your working and state the units of the gradient. (2 marks)

(iii) The relationship between the mass of copper deposited, \(m\), and time, \(t\), is given by:
\( m = \frac{I \cdot M_r \cdot t}{z \cdot F} \)
where:
- \(I\) is the current (\(0.500\text{ A}\))
- \(M_r\) is the relative atomic mass of copper (\(63.5\))
- \(z\) is the number of electrons transferred per copper ion (\(z = 2\))
- \(F\) is the Faraday constant

Use your gradient from (b)(ii) to calculate the value of the Faraday constant, \(F\). (2 marks)

(c) Explain why using a higher current (e.g., \(5.0\text{ A}\)) in the same experimental setup might lead to a less accurate value for the Faraday constant. (2 marks)
查看答案詳解

解題

(a)
- **Diagram**: The diagram must show a DC power supply/battery, an ammeter in series, a variable resistor (rheostat) in series, and a beaker containing copper(II) sulfate solution with two copper electrodes immersed. The electrode connected to the negative terminal of the power supply must be labelled as the cathode, and the positive electrode as the anode.
- **Pre-electrolysis preparation**: Clean the copper cathode with abrasive paper (sandpaper) to remove any oxide layer or impurities. Rinse it with distilled water, then with propanone/ethanol to allow quick evaporation, and dry it thoroughly before recording its initial mass on a balance.
- **Post-electrolysis preparation**: After electrolysis, remove the cathode carefully, rinse it gently with distilled water to wash away any remaining copper(II) sulfate solution, rinse with propanone/ethanol, dry it carefully (without wiping or scraping the surface so that the deposited copper does not flake off), and weigh it.

(b)(i)
- **Graph description**: The points (except Run 4) lie on a straight line passing through the origin, showing that the mass of copper deposited is directly proportional to time.
- **Anomalous point**: Run 4 (at \(t = 2400\text{ s}\), mass is \(0.352\text{ g}\)) is anomalous because the mass of copper is significantly lower than the expected value on the line of best fit (which is \(0.396\text{ g}\)).
- **Physical reason**: Some of the copper deposited on the cathode flaked off and fell into the solution during electrolysis or washing; or there was a temporary drop in the current during this run.

(b)(ii)
- **Gradient Calculation**: Using two points on the line of best fit, such as \((0, 0)\) and \((3000, 0.495)\):
\( \text{Gradient} = \frac{0.495 - 0}{3000 - 0} = 1.65 \times 10^{-4} \)
- **Units**: \(\text{g s}^{-1}\)

(b)(iii)
- **Calculation of \(F\)**:
\( \text{Gradient} = \frac{I \cdot M_r}{z \cdot F} \implies F = \frac{I \cdot M_r}{z \cdot \text{Gradient}} \)
\( F = \frac{0.500 \times 63.5}{2 \times 1.65 \times 10^{-4}} = \frac{31.75}{3.30 \times 10^{-4}} = 96212 \approx 9.62 \times 10^4\text{ C mol}^{-1} \)
(Allow range based on gradient: \(9.40 \times 10^4\text{ C mol}^{-1}\) to \(9.80 \times 10^4\text{ C mol}^{-1}\))

(c)
- **First reason**: A higher current density leads to rapid deposition of copper, which forms a porous, spongy deposit that is poorly adherent and easily flakes off the electrode during washing, leading to an underestimate of the mass of copper.
- **Second reason**: At higher currents, secondary reactions can occur at the cathode (such as the electrolysis of water producing hydrogen gas), which lowers the current efficiency for copper deposition below 100%.

評分準則

**(a) [5 marks]**
- **M1**: Correct diagram showing DC power source, ammeter in series, variable resistor in series, and two copper electrodes in a beaker of aqueous CuSO4. [1]
- **M2**: Correctly identifies cathode (-) and anode (+) connected to the negative and positive terminals respectively. [1]
- **M3**: Clean cathode with sandpaper/abrasive, rinse with distilled water and propanone/ethanol, and dry before weighing. [1]
- **M4**: Post-electrolysis rinse with distilled water, then propanone/ethanol, and dry carefully without wiping/rubbing. [1]
- **M5**: Clearly states that rinsing with propanone/ethanol is to facilitate rapid drying/evaporation. [1]

**(b)(i) [4 marks]**
- **M6**: Correctly describes the graph as a straight line passing through the origin (excluding Run 4). [1]
- **M7**: Identifies Run 4 as the anomalous run. [1]
- **M8**: Suggests that some copper deposit flaked off/lost during washing, or that the current dropped during this run. [1]
- **M9**: Points plotted accurately with a straight line of best fit drawn that excludes the anomalous point and passes through the origin. [1]

**(b)(ii) [2 marks]**
- **M10**: Correct calculation of gradient from the graph, showing working with coordinates on the line of best fit (value should be approx. \(1.65 \times 10^{-4}\), accept range \(1.62 \times 10^{-4}\) to \(1.68 \times 10^{-4}\)). [1]
- **M11**: Correct unit for the gradient: \(\text{g s}^{-1}\). [1]

**(b)(iii) [2 marks]**
- **M12**: Correct rearrangement of the equation: \(F = \frac{I \cdot M_r}{z \cdot \text{gradient}}\). [1]
- **M13**: Correct calculation of \(F\) to 3 significant figures with units (\(\text{C mol}^{-1}\) or \(\text{A s mol}^{-1}\)), matching the gradient calculated. [1]

**(c) [2 marks]**
- **M14**: Explains that rapid deposition at high current produces a spongy, poorly adherent layer of copper that easily falls off. [1]
- **M15**: Explains that other electrode reactions (e.g. hydrogen evolution) can occur at higher currents, reducing current efficiency. [1]
題目 2 · Planning, Analysis & Evaluation
15
A student investigates the kinetics of the reaction between peroxodisulfate(VIII) ions, \(\text{S}_2\text{O}_8^{2-}\), and iodide ions, \(\text{I}^-\):
\[ \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \to 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \]
This reaction is monitored using an 'iodine clock' technique. A small, fixed amount of sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\), and starch indicator are added to the reaction mixture.
The thiosulfate ions immediately react with any iodine produced:
\[ \text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \to 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq}) \]
Once all the thiosulfate ions are consumed, the iodine remaining in solution reacts with the starch to produce a deep-blue color. The time taken, \(t\), for the blue color to appear is recorded. The rate of reaction is proportional to \(\frac{1}{t}\).

The student plans an experiment to determine the order of reaction with respect to iodide ions, \(\text{I}^-\), by varying the concentration of \(\text{I}^-\), while keeping the concentration of \(\text{S}_2\text{O}_8^{2-}\) and the total volume constant.

The student is provided with:
- \(0.200\text{ mol dm}^{-3}\) aqueous potassium iodide, \(\text{KI}(\text{aq})\)
- \(0.200\text{ mol dm}^{-3}\) aqueous potassium chloride, \(\text{KCl}(\text{aq})\)
- \(0.100\text{ mol dm}^{-3}\) aqueous ammonium peroxodisulfate, \((\text{NH}_4)_2\text{S}_2\text{O}_8(\text{aq})\)
- \(0.0050\text{ mol dm}^{-3}\) aqueous sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3(\text{aq})\)
- \(1\%\) starch indicator solution
- Distilled water

(a) Explain why potassium chloride, \(\text{KCl}(\text{aq})\), is added to the mixtures where the volume of \(\text{KI}(\text{aq})\) is reduced. (1 mark)

(b) Design a table to show the volumes of each solution that the student should mix to carry out five different trials to determine the order of reaction with respect to \(\text{I}^-\).
The total volume of each reaction mixture must be exactly \(50.0\text{ cm}^3\).
In each trial, the volume of \(0.100\text{ mol dm}^{-3}\) \((\text{NH}_4)_2\text{S}_2\text{O}_8\) must be \(10.0\text{ cm}^3\), the volume of \(0.0050\text{ mol dm}^{-3}\) \(\text{Na}_2\text{S}_2\text{O}_3\) must be \(5.0\text{ cm}^3\), and the volume of starch indicator must be \(2.0\text{ cm}^3\).
Your table must show the volumes of \(\text{KI}(\text{aq})\), \(\text{KCl}(\text{aq})\), and distilled water (if any) used for each trial. (4 marks)

(c) The results of the experiment are shown below. The initial concentration of iodide ions, \([\text{I}^-]\), is calculated for each of the five trials.

| Trial | \([\text{I}^-]\) / \(\text{mol dm}^{-3}\) | Time, \(t\) / s | \(\frac{1}{t}\) / \(\text{s}^{-1}\) |
|:---:|:---:|:---:|:---:|
| 1 | 0.080 | 25 | 0.040 |
| 2 | 0.060 | 33 | 0.030 |
| 3 | 0.040 | 50 | 0.020 |
| 4 | 0.030 | 91 | 0.011 |
| 5 | 0.020 | 100 | 0.010 |

(i) Plot a graph of \(\frac{1}{t}\) on the y-axis against \([\text{I}^-]\) on the x-axis.
Describe the relationship between \(\frac{1}{t}\) and \([\text{I}^-]\), and identify the anomalous trial. (3 marks)

(ii) Suggest a practical reason why Trial 4 might be anomalous. (1 mark)

(iii) Deduce the order of reaction with respect to iodide ions, \(\text{I}^-\). Explain your reasoning. (2 marks)

(d) In a separate experiment, the student wants to determine the activation energy, \(E_a\), of this reaction.
State the additional variable that must be measured and varied, and describe how the data obtained can be used to determine \(E_a\) graphically. (4 marks)
查看答案詳解

解題

(a) Potassium chloride is added to maintain a constant total concentration of ions (constant ionic strength) in all trials, ensuring that variations in ionic strength do not affect the reaction rate.

(b) The sum of the fixed volumes is \(10.0\text{ cm}^3\) (peroxodisulfate) + \(5.0\text{ cm}^3\) (thiosulfate) + \(2.0\text{ cm}^3\) (starch) = \(17.0\text{ cm}^3\).
To keep ionic strength constant, the total volume of halide solution (\(V_{\text{KI}} + V_{\text{KCl}}\)) is kept at a constant value, here chosen as \(20.0\text{ cm}^3\).
Therefore, the volume of distilled water is constant at \(50.0 - 17.0 - 20.0 = 13.0\text{ cm}^3\) for all trials.
Using \([\text{I}^-] = \frac{V_{\text{KI}} \times 0.200}{50.0}\), the volumes of \(0.200\text{ mol dm}^{-3}\) \(\text{KI}\) required are:
- Trial 1: \(V_{\text{KI}} = 20.0\text{ cm}^3\)
- Trial 2: \(V_{\text{KI}} = 15.0\text{ cm}^3\)
- Trial 3: \(V_{\text{KI}} = 10.0\text{ cm}^3\)
- Trial 4: \(V_{\text{KI}} = 7.5\text{ cm}^3\)
- Trial 5: \(V_{\text{KI}} = 5.0\text{ cm}^3\)

The table of volumes is:

| Trial | Vol of \(\text{KI}\) / \(\text{cm}^3\) | Vol of \(\text{KCl}\) / \(\text{cm}^3\) | Vol of water / \(\text{cm}^3\) | Vol of peroxodisulfate / \(\text{cm}^3\) | Vol of thiosulfate / \(\text{cm}^3\) | Vol of starch / \(\text{cm}^3\) | Total Vol / \(\text{cm}^3\) |
|---|---|---|---|---|---|---|---|
| 1 | 20.0 | 0.0 | 13.0 | 10.0 | 5.0 | 2.0 | 50.0 |
| 2 | 15.0 | 5.0 | 13.0 | 10.0 | 5.0 | 2.0 | 50.0 |
| 3 | 10.0 | 10.0 | 13.0 | 10.0 | 5.0 | 2.0 | 50.0 |
| 4 | 7.5 | 12.5 | 13.0 | 10.0 | 5.0 | 2.0 | 50.0 |
| 5 | 5.0 | 15.0 | 13.0 | 10.0 | 5.0 | 2.0 | 50.0 |

(c)(i)
- **Relationship**: \(\frac{1}{t}\) is directly proportional to \([\text{I}^-]\) because the plot of \(\frac{1}{t}\) against \([\text{I}^-]\) (excluding Trial 4) is a straight line passing through the origin.
- **Anomalous Trial**: Trial 4. The value of \(\frac{1}{t}\) (\(0.011\text{ s}^{-1}\)) is significantly below the line of best fit (expected is \(0.015\text{ s}^{-1}\)).

(c)(ii)
- **Reason**: The student could have added too much sodium thiosulfate solution, which took longer to be completely consumed by the iodine produced. (Alternatively, the volume of \(\text{KI}\) added was less than \(7.5\text{ cm}^3\) by mistake, or the mixture was not stirred, or the temperature dropped during Trial 4).

(c)(iii)
- **Order**: First order (1st order) with respect to \(\text{I}^-\).
- **Reasoning**: The rate of reaction (proportional to \(\frac{1}{t}\)) is directly proportional to the concentration of \(\text{I}^-\). When the concentration of \(\text{I}^-\)[\(\text{I}^-\)] is doubled (e.g., from \(0.040\) to \(0.080\text{ mol dm}^{-3}\)), the value of \(\frac{1}{t}\) doubles from \(0.020\) to \(0.040\text{ s}^{-1}\).

(d)
- **Variables to measure**: The temperature, \(T\), of the reaction mixture must be measured (in Kelvin, K) and varied.
- **Method**:
1. Perform the clock reaction at five different temperatures (e.g., using water baths from \(20^\circ\text{C}\) to \(60^\circ\text{C}\)), maintaining all concentrations constant.
2. For each trial, measure \(t\) and calculate \(\ln(\frac{1}{t})\) (or \(\ln(\text{rate})\)) and the reciprocal temperature, \(\frac{1}{T}\).
3. Plot a graph of \(\ln(\frac{1}{t})\) on the y-axis against \(\frac{1}{T}\) on the x-axis.
4. The activation energy \(E_a\) is determined from the gradient of the straight line obtained: \(\text{gradient} = -\frac{E_a}{R}\), so \(E_a = -\text{gradient} \times R\) (where \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)).

評分準則

**(a) [1 mark]**
- **M1**: To maintain a constant concentration of ions (ionic strength) in all reaction mixtures. [1]

**(b) [4 marks]**
- **M2**: Table includes appropriate headings with units (e.g. volume / \(\text{cm}^3\)) for all added components. [1]
- **M3**: Table lists correct volumes of \(0.200\text{ mol dm}^{-3}\) \(\text{KI}\) for the five trials: 20.0, 15.0, 10.0, 7.5, and 5.0 \(\text{cm}^3\). [1]
- **M4**: Table lists correct volumes of \(0.200\text{ mol dm}^{-3}\) \(\text{KCl}\) to maintain a constant total volume of halide solutions (e.g., 0.0, 5.0, 10.0, 12.5, 15.0 \(\text{cm}^3\) to keep total halide volume at \(20.0\text{ cm}^3\)). [1]
- **M5**: Table includes constant volume of distilled water (e.g. \(13.0\text{ cm}^3\)) and other fixed volumes such that the total volume of each trial is exactly \(50.0\text{ cm}^3\). [1]

**(c)(i) [3 marks]**
- **M6**: Graph of \(\frac{1}{t}\) against \([\text{I}^-]\) is plotted with correct scales and axes labelled. [1]
- **M7**: Identifies that \(\frac{1}{t}\) (rate) is directly proportional to \([\text{I}^-]\) as the points (excluding Trial 4) lie on a straight line passing through the origin. [1]
- **M8**: Identifies Trial 4 as the anomalous run. [1]

**(c)(ii) [1 mark]**
- **M9**: Suggests a valid practical reason for the anomaly (e.g., added too much sodium thiosulfate; used too little KI; temperature of Trial 4 was lower than others; insufficient stirring). [1]

**(c)(iii) [2 marks]**
- **M10**: Concludes the reaction is first-order with respect to \(\text{I}^-\). [1]
- **M11**: Explains that doubling \([\text{I}^-]\) (from 0.040 to 0.080) doubles the value of \(\frac{1}{t}\) (from 0.020 to 0.040). [1]

**(d) [4 marks]**
- **M12**: States that temperature (\(T\)) is the variable to be measured (in Kelvin) and varied. [1]
- **M13**: Explains that the reaction is run at several different temperatures while keeping concentrations constant. [1]
- **M14**: Explains that \(\ln(\frac{1}{t})\) (or \(\ln(\text{rate})\)) and \(\frac{1}{T}\) are calculated, and a graph of \(\ln(\frac{1}{t})\) against \(\frac{1}{T}\) is plotted. [1]
- **M15**: States that \(E_a\) is calculated using \(\text{gradient} = -\frac{E_a}{R}\). [1]

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