📚 Year 11 Eduqas Chemistry: Case Study Practice | Year 11 Eduqas 化学：案例分析实战演练
In Year 11 Eduqas Chemistry, case study questions assess your ability to apply chemical knowledge to real‑world scenarios. This article provides practical drills to help you master these challenges, from ion identification to industrial processes.
在 Year 11 Eduqas 化学中，案例研究题目考察你将化学知识应用于现实情境的能力。本文提供实战演练，帮助你掌握这些挑战，从离子鉴定到工业过程。
1. Understanding the Case Study | 理解案例研究题目
Case study questions often present a short description of a chemical problem, followed by sub‑questions. Read the information carefully and underline key chemical terms such as ‘precipitate’, ‘neutralisation’, or ‘catalyst’.
案例研究题目通常先提供一段化学问题的简短描述，再给出若干小问。仔细阅读信息，并在’沉淀’、’中和’或’催化剂’等关键化学术语下划线。
Identify what the question is testing: qualitative analysis, calculations, equation writing, or evaluation of a process. Linking the scenario to your syllabus helps narrow down the required knowledge.
确定题目在考察什么：定性分析、计算、方程式书写，还是对某个过程的评价。将情境与教学大纲联系起来，有助于缩小所需知识的范围。
2. Qualitative Analysis: Identifying Ions | 定性分析：离子鉴定
Many case studies involve identifying unknown substances. For cations, flame tests and sodium hydroxide precipitation are common. For anions, specific reagents give characteristic results.
许多案例研究涉及鉴定未知物质。对于阳离子，常使用焰色反应和氢氧化钠沉淀法；对于阴离子，特定试剂会产生特征性结果。
| Ion | 离子 | Test | 检验方法 | Observation | 现象 |
|---|---|---|
| Cu²⁺ | Flame test | 焰色反应 | Blue‑green flame | 蓝绿色火焰 |
| Fe²⁺ | NaOH(aq) | Green precipitate, slowly turns brown | 绿色沉淀，缓慢变为棕色 |
| Fe³⁺ | NaOH(aq) | Red‑brown precipitate | 红棕色沉淀 |
| Cl⁻ | Add AgNO₃(aq) + dilute HNO₃ | 加入 AgNO₃(aq) 和稀 HNO₃ | White precipitate | 白色沉淀 |
| SO₄²⁻ | Add BaCl₂(aq) + dilute HCl | 加入 BaCl₂(aq) 和稀 HCl | White precipitate | 白色沉淀 |
Always provide a balanced ionic equation when asked. For example, the precipitation of silver chloride: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
要求书写时，务必给出配平的离子方程式。例如氯化银沉淀：Ag⁺(aq) + Cl⁻(aq) → AgCl(s)。
3. Quantitative Analysis: Titration Calculations | 定量分析：滴定计算
Titration case studies often ask you to determine the concentration of an unknown solution or the purity of a sample. The core relationship is moles = concentration (mol/dm³) × volume (dm³).
滴定案例研究常要求你测定未知溶液的浓度或样品的纯度。核心关系是 物质的量 = 浓度 (mol/dm³) × 体积 (dm³)。
n = c × V
You must convert volumes from cm³ to dm³ by dividing by 1000. In a typical drill, a 25.0 cm³ sample of NaOH is neutralised by 20.0 cm³ of 0.100 mol/dm³ HCl. Calculate the moles of HCl, use the mole ratio from the equation NaOH + HCl → NaCl + H₂O, and find the concentration of NaOH.
你必须将体积从 cm³ 换算为 dm³（除以 1000）。在典型演练中，25.0 cm³ 的 NaOH 样品被 20.0 cm³ 的 0.100 mol/dm³ HCl 中和。计算 HCl 的物质的量，利用方程式 NaOH + HCl → NaCl + H₂O 的物质的量之比，然后求出 NaOH 的浓度。
Step 1: n(HCl) = 0.100 × (20.0/1000) = 0.00200 mol; Step 2: mole ratio 1:1, so n(NaOH) = 0.00200 mol; Step 3: c(NaOH) = n / V = 0.00200 / (25.0/1000) = 0.0800 mol/dm³.
步骤 1：n(HCl) = 0.100 × (20.0/1000) = 0.00200 mol；步骤 2：物质的量之比 1:1，因此 n(NaOH) = 0.00200 mol；步骤 3：c(NaOH) = n / V = 0.00200 / (25.0/1000) = 0.0800 mol/dm³。
4. Environmental Chemistry: Acid Rain Formation | 环境化学：酸雨形成
Case studies on acid rain require you to explain the origin of sulfur dioxide and nitrogen oxides, their oxidation in the atmosphere, and the environmental impact.
关于酸雨的案例研究要求你解释二氧化硫和氮氧化物的来源、它们在大气中的氧化过程以及环境影响。
Fossil fuel combustion in power stations releases SO₂. In the air, SO₂ oxidises to SO₃, which dissolves in water droplets: SO₃(g) + H₂O(l) → H₂SO₄(aq). Nitrogen oxides from vehicle engines undergo similar reactions to form nitric acid.
发电厂燃烧化石燃料释放 SO₂。在空气中，SO₂ 氧化为 SO₃，SO₃ 溶解在水滴中：SO₃(g) + H₂O(l) → H₂SO₄(aq)。来自汽车发动机的氮氧化物经历类似反应形成硝酸。
You may be asked to evaluate methods for reducing acid rain, such as flue‑gas desulfurisation (using CaCO₃ or CaO) or catalytic converters. A balanced equation for desulfurisation is: CaCO₃(s) + SO₂(g) → CaSO₃(s) + CO₂(g).
你可能会被要求评价减少酸雨的方法，例如烟气脱硫（使用 CaCO₃ 或 CaO）或催化转化器。脱硫的配平方程式为：CaCO₃(s) + SO₂(g) → CaSO₃(s) + CO₂(g)。
5. Industrial Chemistry: The Haber Process | 工业化学：哈柏法
The Haber process for ammonia production is a classic case study linking equilibrium, rate, and economic considerations. The reaction is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with ΔH = –92 kJ/mol.
用于生产氨气的哈柏法是连接平衡、速率和经济考虑的经典案例研究。反应为：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = –92 kJ/mol。
Typical conditions are 450 °C, 200 atm, and an iron catalyst. Explain why a high pressure favours the forward reaction (4 moles of gas → 2 moles) but a high temperature, although it reduces yield, is used for a faster rate. The catalyst lowers activation energy without affecting equilibrium position.
典型条件是 450 °C、200 atm 和铁催化剂。解释为什么高压有利于正向反应（4 摩尔气体 → 2 摩尔），而高温虽会降低产率，但用于提高反应速率。催化剂降低活化能，但不影响平衡位置。
Case study questions often present a graph of yield vs temperature and pressure. You must read the graph, quote values, and suggest a compromise condition that balances yield, rate, and cost.
案例研究题目常给出产率随温度、压力变化的图像。你必须读取图像、引用数值，并建议一个兼顾产率、速率和成本的折中条件。
6. Electrochemistry: Cells and Corrosion | 电化学：电池与腐蚀
Eduqas case studies may include simple electrochemical cells or rusting. A cell consists of two different metals in contact with an electrolyte. The more reactive metal is the negative electrode and oxidises.
Eduqas 案例研究可能包括简单的电化学电池或铁锈。一个电池由两种不同金属与电解液接触构成。较活泼的金属作为负极，发生氧化。
For example, in a zinc‑copper cell: Zn(s) → Zn²⁺(aq) + 2e⁻ at the negative electrode; Cu²⁺(aq) + 2e⁻ → Cu(s) at the positive electrode. The overall cell reaction is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
例如，在锌‑铜电池中：负极 Zn(s) → Zn²⁺(aq) + 2e⁻；正极 Cu²⁺(aq) + 2e⁻ → Cu(s)。总电池反应为：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。
Rusting of iron involves both water and oxygen. Prevention methods include painting, galvanising (zinc coating), and sacrificial protection, where a more reactive metal corrodes instead.
铁的生锈需要水和氧气同时存在。防护方法包括涂漆、镀锌（锌层）以及牺牲保护法，即让更活泼的金属代替铁被腐蚀。
7. Organic Chemistry: Polymer Identification | 有机化学：聚合物识别
A case study might provide data about two polymers, such as their melting points, density, and biodegradability, and ask you to justify the choice of material for a specific use.
案例研究可能提供两种聚合物的数据，如熔点、密度和生物降解性，并要求你为特定用途证明材料选择的合理性。
You need to recall that addition polymers like poly(ethene) have a carbon‑chain backbone and are non‑biodegradable, while condensation polymers like polyesters contain ester linkages and can be hydrolysed.
你需要记住，加成聚合物（如聚乙烯）具有碳链骨架且不可生物降解，而缩合聚合物（如聚酯）含有酯键，可发生水解。
In a drill, you might compare poly(ethene) and PLA (a biodegradable polyester). Poly(ethene) is cheap, waterproof, and strong, but persists in the environment. PLA degrades under composting conditions but has lower mechanical strength. Justify which is better for disposable cups.
在演练中，你可能需要比较聚乙烯和 PLA（一种可生物降解聚酯）。聚乙烯便宜、防水且强度高，但会长期存留在环境中。PLA 在堆肥条件下可降解，但机械强度较低。论证哪种更适合用于一次性杯子。
8. Energy Changes: Exothermic and Endothermic Reactions | 能量变化：放热与吸热反应
Case studies often embed energy profile diagrams or temperature‑change data. You must calculate ΔH using q = m × c × ΔT and then relate the value to the number of moles reacted.
案例研究常包含能量变化图或温度变化数据。你必须使用 q = m × c × ΔT 计算热量，然后将该值与反应物物质的量联系起来。
q = m × c × ΔT
For example, when 2.0 g of NaOH is dissolved in 100 g of water, the temperature rises from 20.0 °C to 25.2 °C. Assuming c = 4.18 J/g°C, calculate q and then ΔH per mole of NaOH (Mᵣ = 40).
例如，将 2.0 g NaOH 溶解于 100 g 水中，温度从 20.0 °C 升至 25.2 °C。假定 c = 4.18 J/g°C，计算 q，再计算每摩尔 NaOH 的 ΔH（Mᵣ = 40）。
Solution: q = (100 + 2.0) × 4.18 × (25.2 – 20.0) ≈ 2220 J. Moles NaOH = 2.0 / 40 = 0.050 mol. ΔH = –2220 J / 0.050 mol = –44 400 J/mol ≈ –44.4 kJ/mol (exothermic).
解答：q = (100 + 2.0) × 4.18 × (25.2 – 20.0) ≈ 2220 J。NaOH 物质的量 = 2.0 / 40 = 0.050 mol。ΔH = –2220 J / 0.050 mol = –44 400 J/mol ≈ –44.4 kJ/mol（放热）。
9. Practical Skills: Planning an Investigation | 实验技能：设计研究方案
You may be asked to plan an investigation, for example, to determine the concentration of a salt solution by evaporation or to find the order of reactivity of metals.
你可能会被要求设计一个研究方案，例如通过蒸发测定盐溶液的浓度，或找出金属的反应活性顺序。
A good plan includes: independent, dependent, and control variables; a list of apparatus; a step‑by‑step method; a results table; and evaluation of risks. Always state how to ensure reliability, e.g., by repeating measurements and calculating a mean.
一份好的方案包括：自变量、因变量和控制变量；仪器清单；分步方法；结果表格；以及风险评估。务必说明如何保证信度，例如通过重复测量并计算平均值。
For a reactivity investigation, you might add equal lengths of magnesium, zinc, iron, and copper to test tubes containing dilute HCl of the same concentration and volume. Measure the volume of hydrogen gas produced every 10 seconds using a gas syringe.
对于反应活性研究，你可以将等长的镁、锌、铁和铜分别加入装有相同浓度和体积稀盐酸的试管中。使用气体注射器每 10 秒测量一次产生氢气的体积。
10. Common Pitfalls and Tips | 常见错误与应试技巧
Many marks are lost due to missing units, incorrect mole ratios, or failing to explain the ‘why’ behind a chosen condition. Always check your unit conversions; cm³ must be divided by 1000 for use in n = c × V.
许多失分源于遗漏单位、错误的物质的量之比，或未能解释选择某个条件背后的’原因’。务必检查单位换算；在 n = c × V 中使用时，cm³ 必须除以 1000。
When describing trends, quote specific data from the case study. For instance, ‘the yield of ammonia increases from 20% at 400 °C to 35% at 300 °C (at 200 atm)’ is far better than ‘yield increases as temperature decreases’.
在描述趋势时，引用案例研究中的具体数据。例如，’在 200 atm 下，氨的产率从 400 °C 时的 20% 增加到 300 °C 时的 35%’ 远优于’温度降低、产率增加’。
Finally, link your answers back to the real‑world context. If the scenario is about waste‑water treatment, mention how precipitation reactions remove toxic metal ions. Showing application earns top marks.
最后，将你的答案联系回现实情境。如果背景涉及废水处理，提到沉淀反应如何去除有毒金属离子。展现应用能力可获得最高分。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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