📚 Year 12 AQA Chemistry: Case Study Practical Workout | AQA 化学高一年级案例分析实战演练
Case study questions in AQA Year 12 Chemistry are designed to assess your ability to apply knowledge to unfamiliar scenarios. This guide provides a series of practical walkthroughs, each focused on a core AS topic, from redox titrations to organic synthesis and equilibrium calculations. Work through these examples to build confidence in extracting data, selecting relevant principles, and presenting clear, logical answers.
AQA 高一年级化学的案例研究题旨在考查你将知识应用于陌生情境的能力。本指南提供一系列实战演练，涵盖氧化还原滴定、有机合成、平衡计算等 AS 核心主题。通过这些例子，你将学会提炼数据、选择相关原理并给出清晰、逻辑严密的答案。
1. Understanding Case Study Questions in AQA Chemistry | 理解 AQA 化学中的案例研究问题
Case studies typically begin with a short scenario – an industrial process, an environmental problem, or a laboratory investigation. They then ask a series of linked questions. The key is to break down the information: underline numerical data, identify the relevant topic (e.g. kinetics, equilibria), and note any chemical species involved. Always answer in context, using the data provided, not just generic definitions.
案例研究通常以一个简短情境开头——工业流程、环境问题或实验室探究。随后会提出一系列关联问题。关键是把信息拆解：划出数据，确定相关主题（如动力学、平衡），并记下涉及的化学物种。回答时务必结合情境，使用所给数据，而非仅抛定义。
Typical command words include 'calculate', 'suggest', 'explain' and 'evaluate'. For 'calculate', show all steps with units. For 'suggest', think about bonding, intermolecular forces or reaction conditions. 'Evaluate' means judging reliability, identifying limitations and proposing improvements – a crucial skill for the practical endorsement.
常见指令词有“计算”“建议”“解释”和“评价”。计算题要写出完整步骤和单位；“建议”题从键合、分子间作用力或反应条件入手；“评价”意味着判断可靠性、指出局限并提出改进——这是实验技能评估的关键。
2. Case Study 1: Redox Titration of Iron(II) with Manganate(VII) | 案例研究一：用高锰酸钾滴定铁(II)的氧化还原滴定
titrated with 0.0200 mol dm⁻³ KMnO₄. The average titre was 23.50 cm³. The unbalanced half‑equations are: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; Fe²⁺ → Fe³⁺ + e⁻.
一名学生将铁片溶于稀硫酸并定容至 250.0 cm³。移取 25.0 cm³ 用 0.0200 mol dm⁻³ KMnO₄ 滴定，平均滴定体积为 23.50 cm³。未配平的半反应为：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O；Fe²⁺ → Fe³⁺ + e⁻。
First, write the full balanced equation: multiply the iron half‑ 4H₂O. Calculate moles of MnO₄⁻ used: n = c × V = 0.0200 × (23.50/1000) = 4.70 × 10⁻⁴ mol.
首先写出配平的方程式：将铁的半反应乘以 5 配平电子，得 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O。计算所用 MnO₄⁻ 的物质的量：n = c × V = 0.0200 × (23.50/1000) = 4.70 × 10⁻⁴ mol。
.35 × 10⁻² × 55.8 = 1.31 g. Common mistake: forgetting to scale up from the aliquot to the whole solution.
根据 5:1 计量比，25.0 cm³ 中 Fe²⁺ 的物质的量 = 5 × 4.70 × 10⁻⁴ = 2.35 × 10⁻³ mol。原始 250.0 cm³ 溶液中 n = 2.35 × 10⁻² mol。铁的质量 = n × Ar(Fe) = 2.35 × 10⁻² × 55.8 = 1.31 g。常见错误：忘记从等分试样放大到整个溶液。
3. Case Study 2: Determining the Rate Equation from Initial Rates Data | 案例研究二：从初始速率数据确定速率方程
the reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g), the following initial rates were obtained:
在反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) 的探究中，测得如下初始速率：
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.20 | 1.6 × 10⁻³ |
. Comparing 2 and 3: [H₂] doubles, [NO] constant, rate doubles (8.0×10⁻⁴ → 1.6×10⁻³), so order with respect to H₂ is 1. Rate equation: rate = k [NO]² [H₂].
比较实验 1 和 2：[NO] 加倍，[H₂] 不变，速率变为四倍（2.0×10⁻⁴ → 8.0×10⁻⁴），故对 NO 为二级。比较实验 2 和 3：[H₂] 加倍，[NO] 不变，速率加倍，对 H₂ 为一级。速率方程：rate = k [NO]² [H₂]。
2.0×10⁻⁴ / 1.0×10⁻³ = 0.20 dm⁶ mol⁻² s⁻¹. Always give units for k. The overall order is 3.
求 k，代入任一实验数据，如实验 1：2.0×10⁻⁴ = k (0.10)² (0.10) → k = 2.0×10⁻⁴ / 1.0×10⁻³ = 0.20 dm⁶ mol⁻² s⁻¹。务必写明 k 的单位。总反应级数为 3。
4. Case Study 3: Calculating Equilibrium Constant Kc for Esterification | 案例研究三：计算酯化反应的平衡常数 Kc
Ethanoic acid (0.50 mol) and ethanol (0.50 mol) were mixed with a little sulfuric acid catalyst and allowed to reach equilibrium at 298 K. The total volume was 100 cm³. At equilibrium, 0.30 mol of ethyl ethanoate was present. The equation is CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O.
将乙酸（0.50 mol）与乙醇（0.50 mol）在少量硫酸催化下混合，298 K 下达到平衡，总体积为 100 cm³。平衡时乙酸乙酯为 0.30 mol。反应式：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。
Change: since 0.30 mol ester formed, same moles of acid and alcohol reacted, and 0.30 mol water formed. Equilibrium: acid = 0.50 – 0.30 = 0.20 mol; alcohol = 0.20 mol; ester = 0.30 mol; water = 0.30 mol.
建立物质的量的 ICE 表（初始、变化、平衡），因体积固定。初始：酸和醇各 0.50 mol，酯和水为 0。变化：生成 0.30 mol 酯，则酸和醇消耗 0.30 mol，水生成 0.30 mol。平衡时：酸 = 0.20 mol，醇 = 0.20 mol，酯 = 0.30 mol，水 = 0.30 mol。
)/([acid][alcohol]) = (3.0×3.0)/(2.0×2.0) = 9.0/4.0 = 2.25. No units as there are equal numbers of moles on each side.
换算浓度：除以 0.100 dm³，得 [酸]=2.0，[醇]=2.0，[酯]=3.0，[水]=3.0。Kc = ([酯][水])/([酸][醇]) = (3.0×3.0)/(2.0×2.0) = 2.25，无单位（反应前后分子数相等）。
If the question asks for the effect of a catalyst on Kc, remember: a catalyst speeds up both forward and backward reactions equally, so Kc remains unchanged. Only temperature changes Kc.
若问题问及催化剂对 Kc 的影响，记住：催化剂同等加速正逆反应，Kc 不变。只有温度能改变 Kc。
5. Case Study 4: Organic Synthesis of an Alcohol from an Alkene | 案例研究四：从烯烃合成醇的有机合成路线
A chemist wants to prepare propan-2-ol from propene. She considers two routes: (A) direct hydration with steam and a phosphoric acid catalyst; (B) reaction with concentrated sulfuric acid followed by hydrolysis. Evaluate the two routes in terms of atom economy, conditions and yield.
某化学家想从丙烯制备丙-2-醇。她考虑两条路线：(A) 用蒸汽和磷酸催化剂直接水合；(B) 先与浓硫酸反应再水解。从原子经济性、条件和产率角度评价两条路线。
nomy = (mass of desired product / total mass of reactants) × 100 = 60/ (42+18) × 100 = 100%. Conditions: high temperature (~ high pressure, phosphoric acid catalyst. This is a continuous process, used industrially.
路线 A：C₃H₆ + H₂O → C₃H₇OH。原子经济性 = 60/(42+18)×100 = 100%。条件：高温（~300 °C）、高压、磷酸催化。此为连续流程，用于工业。
te incorporates sulfuric acid fragments. Overall atom economy = 60/(42+98+18) × 100 ≈ 37.5% ically gives a higher yield under mild conditions but involves handling concentrated acid.
路线 B：C₃H₆ + H₂SO₄ → C₃H₇OSO₃H，然后 C₃H₇OSO₃H + H₂O → C₃H₇OH + H₂SO₄。原子经济性较低，因中间体包含了硫酸片段。按化学计量计算，总原子经济性 ≈ 37.5%。但若硫酸可重复利用，废物会减少。路线 B 通常在温和条件下产率较高，但需要处理浓酸。
AQA examiners often ask you to suggest why hydration is preferred for large‑scale production. Answer: 100% atom economy means no by‑products, making the process more sustainable and economical in the long run, even though energy costs are high.
AQA 考官常问为何大规模生产优先选择水合。答案：100% 原子经济性意味着无副产物，尽管能耗高，但从长远看更可持续、更经济。
6. Case Study 5: Identifying an Unknown Carbonyl Compound Using IR and Chemical Tests | 案例研究五：利用红外光谱和化学测试鉴别未知羰基化合物
An unknown liquid (X) gives an orange precipitate with 2,4‑DNP, indicating a carbonyl group. It gives a silver mirror with Tollens' reagent, so it must be an aldehyde. The IR spectrum shows a sharp peak at 1720 cm⁻¹ (C=O stretch) and a broad peak around 3000–3100 cm⁻¹ (C–H in aldehyde). No broad O–H peak is seen, ruling out carboxylic acids and alcohols.
某未知液体 X 与 2,4‑DNP 反应产生橙色沉淀，表明含羰基。与 Tollens 试剂产生银镜，故为醛。红外光谱显示 1720 cm⁻¹ 处有尖峰（C=O 伸缩振动），3000–3100 cm⁻¹ 处有宽峰（醛基 C–H）。未见宽 O–H 峰，排除羧酸和醇。
ional group, the formula is likely C₄H₈O (Mr = 72). Possible structures: butanal CH₃CH₂CH₂CHO or 2‑ told that the ¹H NMR spectrum has a doublet for the aldehyde proton, indicating a neighbouring CH group – confirming 2‑methylpropanal.
质谱显示分子离子峰 m/z = 72，基峰 43。43 碎片表明可能是 CH₃CO⁺ 或 C₃H₇⁺。鉴于为醛，分子式很可能为 C₄H₈O（Mr=72）。可能结构：丁醛 CH₃CH₂CH₂CHO 或 2‑甲基丙醛 (CH₃)₂CHCHO。要区分两者，可用氧化反应（均生成酸），但 NMR 不同。在 AS 水平，题目可能给出 ¹H NMR 中醛基质子为双峰，表明相邻有 CH 基团——证实为 2‑甲基丙醛。
Always link spectroscopic evidence to specific functional groups: 1720 cm⁻¹ → C=O; Tollens' positive → aldehyde; molecular ion → molecular mass. Systematic deduction wins marks.
永远将波谱证据与特定官能团关联：1720 cm⁻¹ → C=O；Tollens 阳性 → 醛；分子离子峰 → 分子量。系统推理能得分。
7. Case Study 6: Applying Hess's Law to Find Enthalpy of Formation | 案例研究六：应用赫斯定律求生成焓
You are given the following combustion data: ΔHc° of C(s) = –394 kJ mol⁻¹, H₂(g) = –286 kJ mol⁻¹, and C₃H₈(g) = –2220 kJ mol⁻¹. Find the standard enthalpy of formation of propane.
已知以下燃烧数据：C(s) ΔHc° = –394 kJ mol⁻¹，H₂(g) = –286 kJ mol⁻¹，C₃H₈(g) = –2220 kJ mol⁻¹。求丙烷的标准生成焓。
The target equation is: 3C(s) + 4H₂(g) → C₃H₈(g). Using Hess' because forming the compound from its elements is the reverse of combusting the compound to those elements, adjusted). A cycle: ΔHf = [3ΔHc(C) + 4ΔHc(H₂)] – ΔHc(C₃H₈). Calculate: (3 × –394) + (4 × –286) = –1182 –1144 = –2326 kJ mol⁻¹. Then minus –2220 gives –2326 – (–2220) = –106 kJ mol⁻¹. So ΔHf° = –106 kJ mol⁻¹, matching the data‑book value.
目标方程：3C(s) + 4H₂(g) → C₃H₈(g)。根据赫斯定律，生成焓变等于各元素燃烧焓之和减去化合物的燃烧焓。循环：ΔHf = [3ΔHc(C) + 4ΔHc(H₂)] – ΔHc(C₃H₈)。计算：(3 × –394)+(4 × –286) = –1182 –1144 = –2326 kJ mol⁻¹。减去 –2220 得 –2326 – (–2220) = –106 kJ mol⁻¹。故 ΔHf° = –106 kJ mol⁻¹，与标准值吻合。
In case studies, you may need to interpret experimental data where combustion is incomplete or heat losses occur. If the experimental ΔHc is less negative than the true value (e.g. –2000 instead of –2220), the calculated ΔHf will be less negative, indicating a systematic error.
在案例中，你可能要解读因燃烧不完全或热量散失导致的实验误差。若实验 ΔHc 负值偏小（如 –2000 而非 –2220），算出的 ΔHf 负值也会偏小，提示系统误差。
8. Case Study 7: Environmental Chemistry – Acid Rain Formation and Limestone Reaction | 案例研究七：环境化学 – 酸雨形成与石灰石反应
Acid rain is caused by SO₂ and NOx dissolving in water. SO₂ is oxidised in cloud droplets to H₂SO₄. A sample of limestone (CaCO₃) was placed in simulated acid rain containing H₂SO₄. The mass loss was recorded. In one experiment, 0.200 g of CaCO₃ reacted with excess acid, producing 44.8 cm³ of CO₂ at RTP.
酸雨由 SO₂ 和 NOx 溶于水引起。SO₂ 在水滴中氧化生成 H₂SO₄。将一块石灰石（CaCO₃）置于含 H₂SO₄ 的模拟酸雨中，记录质量损失。某次实验中 0.200 g CaCO₃ 与过量酸反应，在室温常压下产生 44.8 cm³ CO₂。
₂ = 1.87 × 10⁻³ mol. Mass of pure CaCO₃ = 1.87 × 10⁻³ × 100.1 = 0.187 g. Purity = (0.187/0.200) × 100 = 93.5%.
CO₂ 物质的量 = 体积/24.0 dm³ mol⁻¹（RTP）= 0.0448/24.0 = 1.87 × 10⁻³ mol。由 CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂，CaCO₃ 物质的量同 CO₂ = 1.87 × 10⁻³ mol。纯 CaCO₃ 质量 = 1.87 × 10⁻³ × 100.1 = 0.187 g。纯度 = (0.187/0.200)×100 = 93.5%。
The exam might ask why a product CaSO₄ coating slows further reaction. Because CaSO₄ is sparingly soluble, it forms a protective layer on the limestone surface, reducing contact between acid and carbonate – a factor often considered when evaluating reaction rates in building stone erosion.
考试可能问为什么 CaSO₄ 覆盖层会减缓后续反应。CaSO₄ 微溶，在石灰石表面形成保护层，减少酸与碳酸盐接触——评估建筑物石材腐蚀速率时常考虑此点。
9. Case Study 8: Analysis of a Group 2 Compound – Thermal Decomposition | 案例研究八：第 2 族化合物的热分解分析
M. Equation: MCO₃(s) → MO(s) + CO₂(g). Molar mass of CO₂ = 44.0 g mol⁻¹. Moles of CO₂ = 0.44/44.0 = 0.0100 mol. Therefore moles of MCO₃ = 0.0100 mol. Original mass 1.00 g → molar mass of MCO₃ = 1.00/0.0100 = 100 g mol⁻¹. Subtract mass of CO₃²⁻ (60), gives M = 40 g mol⁻¹. Hence M is Ca (calcium).
某学生加热 1.00 g 第 2 族碳酸盐（MCO₃）至恒重，收集逸出的 CO₂。质量损失为 0.44 g。推断 M 的身份。反应：MCO₃(s) → MO(s) + CO₂(g)。CO₂ 摩尔质量 44.0 g mol⁻¹，其物质的量 = 0.44/44.0 = 0.0100 mol。故 MCO₃ 物质的量 = 0.0100 mol。原质量 1.00 g → MCO₃ 摩尔质量 = 1.00/0.0100 = 100 g mol⁻¹。减去 CO₃²⁻ 质量（60），得 M = 40 g mol⁻¹，故 M 为钙。
Explain why barium carbonate requires stronger heating than calcium carbonate. Down Group 2, the cation charge density decreases because ionic radius increases. The smaller, more highly charged Ca²⁺ polarises the carbonate ion more strongly, weakening the C–O bonds within CO₃²⁻, so decomposition happens at a lower temperature. In case studies, this trend is often linked to thermal stability and lime kiln operation.
解释为何碳酸钡比碳酸钙需要更高温度分解。沿第 2 族向下，阳离子电荷密度因半径增大而降低。较小的 Ca²⁺ 电荷密度更高，能更强地极化碳酸根离子，削弱 CO₃²⁻ 中的 C–O 键，故分解温度更低。案例研究中常将该趋势与热稳定性和石灰窑操作相联系。
10. Case Study 9: Practical Error Analysis and Evaluation | 案例研究九：实验误差分析与评估
In a calorimetry experiment to determine the enthalpy change of neutralisation, a student recorded a temperature rise of 5.2 °C using a spirit thermometer. She notes potential sources of error: heat loss to the surroundings, incomplete reaction, and parallax when reading the thermometer. Evaluate the impact and suggest improvements.
在中和焓变测定实验中，一名学生用酒精温度计测得温升 5.2 °C。她列出潜在误差源：环境散热、反应不完全、读取温度计时的视差。评价其影响并提出改进。
Heat loss means the recorded ΔT is lower than the maximum theoretical value, so the calculated ΔH becomes less negative than the literature value (systematic error). Improvement: use an expanded polystyrene cup with a lid, and record temperature over time to extrapolate the cooling curve back to
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