AQA Year 13 Chemistry: Case Study Practical Workout | AQA 高三年级化学：案例分析实战演练

📚 AQA Year 13 Chemistry: Case Study Practical Workout | AQA 高三年级化学：案例分析实战演练

In AQA Year 13 Chemistry, case study questions test your ability to apply knowledge in unfamiliar contexts. This article walks you through nine structured examples covering organic synthesis, kinetics, equilibrium, electrochemistry, transition metal chemistry, thermodynamics, spectroscopy, redox titrations, and organic mechanisms. Each case is broken down into a problem, strategy and stepwise solution, with paired English–Chinese explanations to help you build confidence and exam technique.

在 AQA 高三年级化学中，案例分析题考查你在陌生情境中应用知识的能力。本文通过九个结构化的示例，涵盖有机合成、动力学、化学平衡、电化学、过渡金属化学、热力学、波谱分析、氧化还原滴定和有机反应机理，以中英双语配对讲解，助你建立信心、掌握应试技巧。

1. Organic Synthesis Route Design | 有机合成路线设计

Problem: Propose a synthetic route from benzene to 1-phenylethanol, giving all reagents and conditions.

问题：设计一条从苯合成1-苯乙醇的路线，给出所有试剂和条件。

Retrosynthetic analysis: The target molecule is a secondary alcohol next to a benzene ring. The alcohol can be made by reduction of a ketone. The ketone, acetophenone (phenyl methyl ketone), can be prepared from benzene via Friedel–Crafts acylation.

逆合成分析：目标分子是邻接苯环的二级醇。该醇可通过酮的还原制得。酮（苯乙酮）可由苯通过傅-克酰基化制备。

Step 1: Benzene reacts with ethanoyl chloride (CH₃COCl) in the presence of anhydrous AlCl₃ catalyst under reflux. This introduces the –COCH₃ group to form acetophenone. Electrophilic substitution occurs.

步骤 1：苯与乙酰氯（CH₃COCl）在无水 AlCl₃ 催化下加热回流，发生亲电取代，引入 —COCH₃ 基团，生成苯乙酮。

Step 2: Reduce acetophenone using NaBH₄ in methanol (or LiAlH₄ in dry ether) at room temperature. The ketone is reduced to a secondary alcohol, giving 1-phenylethanol.

步骤 2：用 NaBH₄ 的甲醇溶液（或用 LiAlH₄ 的无水乙醚）在室温下还原苯乙酮，酮被还原为二级醇，得到 1-苯乙醇。

Overall: Benzene → acetophenone → 1-phenylethanol. Reagents: CH₃COCl/AlCl₃ then NaBH₄/MeOH.

总路线：苯 → 苯乙酮 → 1-苯乙醇。试剂：CH₃COCl/AlCl₃，然后 NaBH₄/MeOH。

2. Kinetics – Determining the Rate Equation from Initial Rates | 动力学——由初始速率确定速率方程

Problem: The reaction 2NO(g) + O₂(g) → 2NO₂(g) was studied at 25 °C. Initial rate data are given below. Determine the rate equation and the rate constant, k.

问题：在 25 °C 下研究反应 2NO(g) + O₂(g) → 2NO₂(g)，初始速率数据如下。求速率方程和速率常数 k。

Experiment	[NO] / mol dm⁻³	[O₂] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻³
2	0.20	0.10	8.0 × 10⁻³
3	0.20	0.20	1.6 × 10⁻²

Compare experiments 1 and 2: [O₂] is constant, [NO] doubles, and the rate increases by a factor of 4 (8.0×10⁻³ / 2.0×10⁻³ = 4). Thus the reaction is second order with respect to NO.

比较实验 1 和 2：[O₂] 不变，[NO] 加倍，速率增大到 4 倍（8.0×10⁻³ / 2.0×10⁻³ = 4），因此对 NO 为二级反应。

Compare experiments 2 and 3: [NO] is constant, [O₂] doubles, and the rate doubles (1.6×10⁻² / 8.0×10⁻³ = 2). The reaction is first order with respect to O₂.

比较实验 2 和 3：[NO] 不变，[O₂] 加倍，速率也加倍（1.6×10⁻² / 8.0×10⁻³ = 2），对 O₂ 为一级反应。

Rate = k[NO]²[O₂]

From experiment 1: 2.0×10⁻³ = k × (0.10)² × (0.10) = k × 1.0×10⁻³. Hence k = 2.0 dm⁶ mol⁻² s⁻¹.

由实验 1 计算：2.0×10⁻³ = k × (0.10)² × (0.10) = k × 1.0×10⁻³，因此 k = 2.0 dm⁶ mol⁻² s⁻¹。

3. Buffer Solutions – pH Calculation | 缓冲溶液——pH 计算

Problem: A buffer solution contains 0.10 mol dm⁻³ ethanoic acid (CH₃COOH) and 0.10 mol dm⁻³ sodium ethanoate (CH₃COONa). Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH of the buffer.

问题：某缓冲溶液含 0.10 mol dm⁻³ 乙酸和 0.10 mol dm⁻³ 乙酸钠。乙酸 Ka = 1.74 × 10⁻⁵ mol dm⁻³。计算该缓冲溶液的 pH。

For an acidic buffer, the Henderson–Hasselbalch equation is used:

对于酸性缓冲溶液，使用 Henderson–Hasselbalch 方程：

pH = pKa + log₁₀([A⁻]/[HA])

pKa = –log₁₀(1.74 × 10⁻⁵) ≈ 4.76. Since [A⁻] = [salt] = 0.10 and [HA] = [acid] = 0.10, the log term is log₁₀(1) = 0.

pKa = –log₁₀(1.74 × 10⁻⁵) ≈ 4.76。由于 [A⁻] = [盐] = 0.10，[HA] = [酸] = 0.10，对数项 log₁₀(1) = 0。

pH = 4.76 + 0 = 4.76

If the concentrations were different, the ratio would shift the pH. The buffer resists pH change when small amounts of acid or base are added.

如果浓度不同，该比值将使 pH 偏移。缓冲溶液能抵抗外加少量酸或碱引起的 pH 变化。

4. Electrochemical Cells and the Nernst Equation | 电化学电池与能斯特方程

Problem: A cell is constructed with a Zn²⁺/Zn half-cell (0.010 mol dm⁻³) and a Cu²⁺/Cu half-cell (0.10 mol dm⁻³) at 298 K. Standard electrode potentials: E°(Zn²⁺/Zn) = –0.76 V, E°(Cu²⁺/Cu) = +0.34 V. Calculate the cell EMF under these non-standard conditions.

问题：用 Zn²⁺/Zn (0.010 mol dm⁻³) 半电池和 Cu²⁺/Cu (0.10 mol dm⁻³) 半电池在 298 K 组成电池。标准电极电势：E°(Zn²⁺/Zn) = –0.76 V，E°(Cu²⁺/Cu) = +0.34 V。计算非标准条件下的电池电动势。

The cell reaction is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Standard cell EMF: E°ₓₑₗₗ = +0.34 – (–0.76) = 1.10 V. Using the Nernst equation at 298 K:

电池反应为：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。标准电池电动势 E°ₓₑₗₗ = +0.34 – (–0.76) = 1.10 V。应用 298 K 时的能斯特方程：

E = E° – (0.0592/n) log₁₀ Q

Here n = 2, and Q = [Zn²⁺]/[Cu²⁺] = 0.010 / 0.10 = 0.10.

此处 n = 2，Q = [Zn²⁺]/[Cu²⁺] = 0.010 / 0.10 = 0.10。

E = 1.10 – (0.0592/2) × log₁₀(0.10)
E = 1.10 – (0.0296) × (–1)
E = 1.10 + 0.0296 = 1.13 V

The cell EMF has increased slightly because the lower [Zn²⁺] and higher [Cu²⁺] make the reaction more spontaneous.

电池电动势略有增大，因为较低的 [Zn²⁺] 和较高的 [Cu²⁺] 使反应更自发性。

5. Transition Metal Isomerism and Colour | 过渡金属异构现象与颜色

Problem: The complex ion [CrCl₂(NH₃)₄]⁺ exists as two isomers, one green and one violet. Name the type of isomerism, draw the two isomers, and explain the origin of their colours.

问题：配离子 [CrCl₂(NH₃)₄]⁺ 存在两种异构体，一种绿色，一种紫色。指出异构体类型，画出这两种异构体，并解释其颜色的来源。

The complex has octahedral geometry with four ammonia molecules and two chloride ligands. The two isomers are cis and trans:

该配离子为八面体构型，含四个氨分子和两个氯配体。两种异构体为顺式和反式：

cis-[CrCl₂(NH₃)₄]⁺: the two Cl⁻ ligands are adjacent (90° apart).
trans-[CrCl₂(NH₃)₄]⁺: the two Cl⁻ ligands are opposite (180° apart).

顺式-[CrCl₂(NH₃)₄]⁺：两个 Cl⁻ 配体处于邻位（夹角 90°）。
反式-[CrCl₂(NH₃)₄]⁺：两个 Cl⁻ 配体处于对位（夹角 180°）。

Chromium(III) has a d³ configuration. In an octahedral field, the d orbitals split into t₂₉ and e₉ sets. Visible light absorption corresponds to electron promotion from t₂₉ to e₉. The energy gap differs slightly between cis and trans isomers because the ligand field strength of Cl⁻ is different from NH₃ and the arrangement alters the splitting. Hence, one isomer absorbs in the region complementary to green (appears violet) while the other absorbs light complementary to violet (appears green).

铬(III) 具有 d³ 电子构型。在八面体场中，d 轨道分裂为 t₂₉ 和 e₉ 两组。可见光吸收对应于电子从 t₂₉ 激发到 e₉。由于 Cl⁻ 和 NH₃ 的配位场强度不同，且排列方式影响分裂能，顺反异构体的能级差略有差异，因此一种异构体吸收互补于绿色的光而呈紫色，另一种吸收互补于紫色的光而呈绿色。

6. Thermodynamics – Entropy and Gibbs Free Energy | 热力学——熵与吉布斯自由能

Problem: For the decomposition of calcium carbonate, ΔH° = +178 kJ mol⁻¹ and ΔS° = +161 J K⁻¹ mol⁻¹ at 298 K. Calculate the temperature at which the reaction becomes feasible (ΔG ≤ 0).

问题：碳酸钙分解的 ΔH° = +178 kJ mol⁻¹，298 K 时 ΔS° = +161 J K⁻¹ mol⁻¹。计算该反应变得可行（ΔG ≤ 0）的温度。

Using ΔG° = ΔH° – TΔS°. Set ΔG° = 0 for the threshold temperature:

运用公式 ΔG° = ΔH° – TΔS°。设 ΔG° = 0 求临界温度：

0 = 178 kJ mol⁻¹ – T × 0.161 kJ K⁻¹ mol⁻¹ (converting ΔS° to kJ)

T = 178 / 0.161 ≈ 1106 K (≈ 833 °C)

At temperatures above 1106 K, TΔS° outweighs ΔH°, making ΔG° negative and the reaction thermodynamically feasible.

当温度高于 1106 K 时，TΔS° 超过 ΔH°，使 ΔG° 为负，反应在热力学上变得可行。

In reality, the decomposition of CaCO₃ in a lime kiln is carried out around 900–1000 °C to achieve a practical rate, even though the thermodynamic threshold is slightly higher.

实际石灰窑中碳酸钙分解操作温度约为 900–1000 °C，以保证实际速率，尽管热力学阈值略高。

7. NMR Spectroscopy Structure Determination | 核磁共振波谱结构解析

Problem: A compound with molecular formula C₄H₈O₂ gives the following ¹H NMR data: δ 1.2 (triplet, 3H), δ 2.3 (quartet, 2H), δ 3.7 (singlet, 3H). Deduce its structure and explain the splitting patterns.

问题：分子式为 C₄H₈O₂ 的化合物 ¹H NMR 数据如下：δ 1.2 (三重峰, 3H), δ 2.3 (四重峰, 2H), δ 3.7 (单峰, 3H)。推断其结构并解释裂分模式。

The singlet at δ 3.7 integrating for 3H suggests a methyl group next to an oxygen, typical of an ester –OCH₃ or a methoxy group. The triplet and quartet pattern is characteristic of an ethyl group (–CH₂CH₃) coupled to each other. The quartet at δ 2.3 indicates a CH₂ group adjacent to a CH₃ and also deshielded, possibly by a carbonyl. The triplet at δ 1.2 is the terminal CH₃ of the ethyl group.

δ 3.7 处的单峰积分 3H，提示是一个连氧的甲基，典型于酯的 —OCH₃。三重峰与四重峰的组合是典型的乙基（—CH₂CH₃）相互耦合模式。δ 2.3 处的四重峰表明一个与 CH₃ 相邻的 CH₂，而且受去屏蔽影响，可能连着羰基。δ 1.2 处的三重峰是乙基末端的 CH₃。

The fragments fit together as ethyl propanoate (CH₃CH₂COOCH₃) but that would give a different [OCH₃] shift; however, with C₄H₈O₂ the ester with an ethyl group attached to the carbonyl and a methoxy group is methyl propanoate: CH₃CH₂COOCH₃. The ethyl part gives quartet (CH₂) at ~2.3 ppm and triplet (CH₃) at ~1.2 ppm. The methoxy singlet appears at ~3.7 ppm.

这些碎片拼接成丙酸甲酯 (CH₃CH₂COOCH₃)。乙基部分产生约 δ 2.3 的四重峰 (CH₂) 和约 δ 1.2 的三重峰 (CH₃)。甲氧基单峰出现在约 δ 3.7 ppm。

Thus the structure is methyl propanoate. An alternative ester, ethyl ethanoate (CH₃COOCH₂CH₃), would show a singlet for CH₃CO at ~2.1 ppm and the ethyl group attached to oxygen showing a quartet near 4.1 ppm, which does not match the data.

因此结构为丙酸甲酯。另一同分异构的酯，乙酸乙酯 (CH₃COOCH₂CH₃)，会出现 ~2.1 ppm 的乙酰基单峰和连氧乙基在 ~4.1 ppm 的四重峰，与数据不符。

8. Redox Titration – Iron Tablet Analysis | 氧化还原滴定——铁片分析

Problem: A 0.250 g iron tablet was dissolved in dilute sulfuric acid, reducing all iron to Fe²⁺ ions. The solution was titrated with 0.0200 mol dm⁻³ KMnO₄, requiring 20.50 cm³ to reach a permanent pale pink endpoint. Calculate the percentage by mass of iron in the tablet.

问题：将一片 0.250 g 的铁片溶于稀硫酸，将所有铁还原为 Fe²⁺。用 0.0200 mol dm⁻³ KMnO₄ 滴定该溶液，消耗 20.50 cm³ 到达永久浅

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AQA Year 13 Chemistry: Case Study Practical Workout | AQA 高三年级化学：案例分析实战演练