📚 Case Study Practical Exercises | 案例分析实战演练
Year 11 WJEC Chemistry often tests your ability to apply knowledge to real-world scenarios. This article presents a series of case study exercises covering key topics, each followed by a step-by-step analysis. You will learn how to interpret data, select relevant chemical principles, and construct clear, logical answers.
Year 11 WJEC 化学考试常常考查你将知识应用于真实场景的能力。本文提供一系列覆盖核心主题的案例分析练习，每个案例后都有逐步分析。你将学会如何解读数据、选择相关的化学原理，并构建清晰、有逻辑的答案。
1. Industrial Ammonia Production | 工业合成氨
A chemical plant produces ammonia via the Haber process: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g). The reaction is exothermic. The plant manager wants to maximise daily yield while minimising energy costs. Currently, the plant operates at 450 °C and 200 atm with an iron catalyst. The manager consults you about lowering the temperature to 350 °C to shift the equilibrium yield higher.
某化工厂通过哈柏法生产氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。该反应为放热反应。工厂经理想在最小化能源成本的同时实现日产最大化。目前工厂在 450 °C、200 atm 及铁催化剂下运行。经理就降温至 350 °C 以获得更高平衡产率一事咨询你的意见。
Step 1: Apply Le Chatelier's principle. Lowering temperature favours the exothermic forward reaction, so equilibrium shifts right, increasing ammonia yield. However, lower temperature also reduces the rate of reaction. This means the same amount of product might take much longer to form, potentially reducing daily output.
步骤 1：运用勒夏特列原理。降低温度有利于放热正反应，平衡右移，提高氨的产率。但较低的温度也会降低反应速率。这意味着形成同样数量的产品可能需要更长时间，从而可能降低日产量。
Step 2: Consider the catalyst. The iron catalyst works optimally within a certain temperature range. At 350 °C its activity is lower, further slowing the rate. The plant would need a larger reactor or longer residence time, raising capital and operating costs.
步骤 2：考虑催化剂。铁催化剂在特定温度范围内活性最佳。在 350 °C 下其活性较低，进一步减慢了速率。工厂需要更大的反应器或更长的停留时间，从而增加资本和运营成本。
Step 3: Balance yield and rate. The typical compromise temperature of 400–450 °C offers an acceptable equilibrium yield and a sufficiently fast rate, allowing a continuous flow process that maximises daily production. Reducing temperature without increasing catalyst volume would actually lower daily ammonia output.
步骤 3：平衡产率与速率。典型的折中温度 400–450 °C 提供了可接受的平衡产率和足够快的速率，允许连续流动工艺以实现最大日产量。在不增加催化剂装填量的情况下降低温度实际上会降低每日氨产量。
Conclusion: The manager should maintain the current temperature unless the catalyst system is redesigned. The existing conditions balance thermodynamic yield with kinetic practicality.
结论：除非重新设计催化剂系统，否则经理应维持现有温度。现有条件平衡了热力学产率与动力学实效。
2. Water Hardness Investigation | 水硬度研究
e of tap water with 0.0100 mol dm⁻³ EDTA solution, using Eriochrome Black T indicator. The titre volume is 12.40 cm³. Calculate the total hardness in mg of CaCO₃ per dm³ (CaCO₃ molar mass = 100.1).
当地水务公司报告客户抱怨水壶积垢且肥皂不起泡。分析员用 0.0100 mol dm⁻³ EDTA 溶液滴定 50.0 cm³ 自来水样本，使用铬黑 T 指示剂。滴定体积为 12.40 cm³。计算以每 dm³ 所含 CaCO₃ 毫克数表示的总硬度（CaCO₃ 摩尔质量 = 100.1）。
Step 1: Write the key relationship. . Moles of EDTA used = concentration × volume = 0.0100 × (12.40 / 1000) = 1.24 × 10⁻⁴ mol.
步骤 1：写出关键关系。EDTA 与 Ca²⁺ 和 Mg²⁺ 离子按 1:1 反应。所用 EDTA 的物质的量 = 浓度 × 体积 = 0.0100 × (12.40 / 1000) = 1.24 × 10⁻⁴ mol。
Step 2: Find moles in 1 dm³. 24 × 10⁻⁴ mol. In 1000 cm³ (1 dm³), moles = 1.24 × 10⁻⁴ × (1000 / 50.0) = 2.48 × 10⁻³ mol.
步骤 2：求 1 dm³ 中的物质的量。50.0 cm³ 中硬度离子的物质的量 = 1.24 × 10⁻⁴ mol。在 1000 cm³（1 dm³）中，物质的量 = 1.24 × 10⁻⁴ × (1000 / 50.0) = 2.48 × 10⁻³ mol。
Step 3: Convert to CaCO₃ mass. Mass = moles × molar mass = 2.48 × 10⁻³ × 100.1 ≈ 0.248 g = 248 mg.
步骤 3：换算为 CaCO₃ 质量。质量 = 物质的量 × 摩尔质量 = 2.48 × 10⁻³ × 100.1 ≈ 0.248 g = 248 mg。
Interpretation: Total hardness is 248 mg dm⁻³, which is classified as hard water. Recommendations include installing water softeners for domestic supply.
解读：总硬度为 248 mg dm⁻³，归为硬水。建议为家庭供水安装软水器。
3. Electrolysis of Brine: Choosing Electrodes | 盐水电解：选择电极
A factory manufactures chlorine and sodium hydroxide by electrolysis of concentrated brine. The engineer considers replacing graphite anodes with titanium anodes coated with ruthenium oxide. The cathode is steel. Analyse the advantages and potential drawbacks of this change, focusing on product purity and electrode lifetime.
某工厂通过电解浓盐水制造氯气和氢氧化钠。工程师考虑将石墨阳极替换为涂有氧化钌的钛阳极。阴极是钢。分析这一改变在产物纯度和电极寿命方面的优点与潜在缺点。
Advantages: Titanium anodes are dimensionally stable and resist corrosion much better than graphite. Graphite anodes gradually erode, releasing fine carbon particles that contaminate the chlorine gas and wear away, requiring frequent replacement. Coated titanium anodes last years, reduce maintenance, and keep chlorine purer.
优点：钛阳极尺寸稳定，耐腐蚀性远优于石墨。石墨阳极会逐渐腐蚀，释放出细碳颗粒污染氯气并自身损耗，需频繁更换。涂层钛阳极可用多年，减少维护，保持氯纯度。
Drawbacks: The initial cost of coated titanium electrodes is significantly higher. Also, the coating can be poisoned by impurities such as iron, so brine must be purified more rigorously, adding pre-treatment costs. However, the improved efficiency often offsets the higher capital expenditure over time.
缺点：涂层钛电极的初始成本高得多。另外，涂层易被铁等杂质毒化，因此盐水必须更严格地净化，增加预处理成本。然而，随着时间推移，效率的提高通常能抵消较高的资本支出。
Conclusion: The switch is beneficial for high-purity chlorine production despite higher upfront investment, as brine purification is already required to protect the membrane.
结论：尽管前期投资较高，这一更换对高纯度氯生产是有利的，因为保护离子膜本来就需要盐水净化。
4. Acid-Base Titration: Vinegar Analysis | 酸碱滴定：食醋分析
A food standards officer suspects that a vinegar sample labelled as containing 5.0% .05 cm³. Determine the actual percentage of ethanoic acid (density of vinegar = 1.00 g cm⁻³, Mr of CH₃COOH = 60.0).
一位食品标准官员怀疑标注含 5.0% 乙酸 (CH₃COOH) 的食醋样品被稀释了。她取 10.0 cm³ 食醋，稀释至 100 cm³ 后，用 0.100 mol dm⁻³ NaOH 滴定。平均滴定体积为 16.05 cm³。计算乙酸的实际百分含量（食醋密度 = 1.00 g cm⁻³，CH₃COOH 的 Mr = 60.0）。
Step 1: Moles of NaOH. n(NaOH) = 0.100 × (16.05 / 1000) = 1.605 × 10⁻³ mol.
步骤 1：NaOH 的物质的量。n(NaOH) = 0.100 × (16.05 / 1000) = 1.605 × 10⁻³ mol。
Step 2: Moles of ethanoic acid in the diluted sample. 100 cm³, so the original 10.0 cm³ vinegar contains 10 times this amount: 1.605 × 10⁻² mol.
步骤 2：稀释试样中乙酸的物质的量。反应按 1:1 进行，故 10.0 cm³ 稀释溶液中 n(CH₃COOH) = 1.605 × 10⁻³ mol。原 10.0 cm³ 食醋被稀释至 100 cm³，因此原 10.0 cm³ 食醋含有该量的 10 倍：1.605 × 10⁻² mol。
Step 3: Mass of ethanoic acid in 10.0 cm³. Mass = 1.605 × 10⁻² × 60.0 = 0.963 g.
步骤 3：10.0 cm³ 中乙酸的质量。质量 = 1.605 × 10⁻² × 60.0 = 0.963 g。
Step 4: Percentage mass/volume. m³ vinegar has mass 10.0 g (density 1.00 g cm⁻³). Percentage = (0.963 / 10.0) × 100 = 9.63%. This is much higher than 5.0%. The mislabelling error suggests the vinegar is actually more concentrated – the officer’s suspicion was wrong but the sample still fails the label claim.
步骤 4：质量/体积百分数。10.0 cm³ 食醋质量为 10.0 g（密度 1.00 g cm⁻³）。百分含量 = (0.963 / 10.0) × 100 = 9.63%。这远高于 5.0%。标签错误表明食醋实际浓度更高——官员的怀疑有误，但样品依然不符合标签声称。
5. Exothermic Reaction: Hand Warmer Design | 放热反应：暖手宝设计
A start-up wants to develop a reusable hand warmer based on the crystallisation of supersaturated sodium acetate trihydrate (CH₃COONa·3H₂O). The reaction is exothermic. They must select a trigger and ensure safety. Explain the chemistry and propose a suitable trigger mechanism.
一家初创公司想基于过饱和三水合醋酸钠 (CH₃COONa·3H₂O) 的结晶开发可重复使用的暖手宝。该反应放热。他们必须选择触发器并确保安全。解释其化学原理并提出合适的触发机制。
Chemical principle: The solution is supersaturated, meaning it contains more dissolved sodium acetate than stable at room temperature. Crystallisation releases the excess solute as solid, and the process is exothermic: the hydrated crystals re-form, releasing heat. The heat is the enthalpy change of solution reversed.
化学原理：该溶液为过饱和溶液，即在室温下含有比稳定状态更多的溶解醋酸钠。结晶使过量溶质以固体形式析出，同时放热：水合晶体重新形成并释放热量。该热量是溶解焓的逆过程。
Trigger: A small metal disc that can be bent or 'clicked'. The flexing creates localised shock waves and provides nucleation sites for crystal growth. Alternatively, adding a single seed crystal starts the crystallisation cascade. The metal disc is sealed inside the pouch, allowing reuse after boiling the hand warmer to re-dissolve the crystals and return to supersaturated state.
触发器：一个可弯曲或“咔哒”作响的小金属圆片。弯曲产生局部冲击波并为晶体生长提供成核位点。或者加入一颗晶种即可启动结晶级联。金属圆片密封在内袋中，用后可沸煮暖手宝使晶体重新溶解并回到过饱和状态，实现重复使用。
Safety: The temperature peaks around 50–55 °C, which is safe for skin contact. The outer material must be flexible, leak-proof, and food-grade. Overheating during recharging (boiling) must be supervised to prevent rupture.
安全性：温度峰值约 50–55 °C，对皮肤接触安全。外层材料需柔韧、防漏且食品级。充电（煮沸）时的过热须有人看管以防破裂。
6. Polymer Identification and Recycling | 聚合物鉴别与回收
A plastic recycling centre receives a batch of mixed clear bottles. Two main polymers are PET (polyethylene terephthalate) and HDPE (high-density polyethylene). PET sinks in water, HDPE floats. PET burns with a yellow, sooty flame; HDPE burns with a blue, non-sooty flame and drips. PET softens at around 260 °C, HDPE at 130 °C. Design a quick separation method and explain how the properties relate to structure.
某塑料回收中心收到一批混合透明瓶。两种主要聚合物为 PET（聚对苯二甲酸乙二酯）和 HDPE（高密度聚乙烯）。PET 在水中下沉，HDPE 上浮。PET 燃烧时产生黄色烟火焰；HDPE 燃烧时产生蓝色无烟火焰并滴落。PET 软化点约 260 °C，HDPE 约 130 °C。设计一种快速分离方法并解释性质如何与结构相关。
Separation method: Shred the bottles and place in a water-filled tank. HDPE flakes float and can be skimmed off the surface, while PET flakes sink and are collected from the bottom. This density-based sink-float separation is simple and effective.
分离方法：将瓶子粉碎后放入充满水的池中。HDPE 碎片浮于水面可被撇出，而 PET 碎片下沉并从底部收集。这种基于密度的沉浮分离简单有效。
Structure-property link: HDPE is a linear hydrocarbon with little branching, allowing chains to pack closely, giving a density just below 1 g cm⁻³. PET contains polar ester groups (–COO–) and aromatic rings, increasing intermolecular forces and density above 1.0 g cm⁻³. The higher softening temperature of PET reflects stronger dipole-dipole interactions between chains compared to the weaker van der Waals forces in HDPE.
结构与性质关联：HDPE 是支链极少的线型烃，分子链可紧密堆积，密度略低于 1 g cm⁻³。PET 含有极性酯基 (–COO–) 和芳香环，分子间力增强，密度高于 1.0 g cm⁻³。PET 较高的软化温度反映出链间偶极-偶极作用力比 HDPE 中的范德华力更强。
7. Identifying an Unknown Salt | 鉴定未知盐
A student is given a white crystalline solid, X. A flame test produces a brick-red flame. Adding dilute nitric acid and silver nitrate gives a white precipitate. Adding sodium hydroxide and warming gives no pungent gas. Identify the cation and anion, and write the formula of X.
一名学生拿到一种白色结晶固体 X。焰色反应呈砖红色。加入稀硝酸和硝酸银产生白色沉淀。加入氢氧化钠并温热未产生刺激性气体。鉴定阳离子和阴离子，写出 X 的化学式。
Flame test brick-red: This indicates Ca²⁺ ions (calcium).
焰色砖红：表明存在 Ca²⁺ 离子（钙）。
Acidified silver nitrate gives white precipitate: White precipitate that is insoluble in nitric acid confirms chloride ions, Cl⁻. (Silver sulfate is also possible but would require sulfate test; here the simplest conclusion is chloride.)
酸化硝酸银产生白色沉淀：不溶于硝酸的白色沉淀确认氯离子 Cl⁻。（硫酸银也有可能，但需要硫酸盐验证；此处最简结论为氯离子。）
No pungent gas with NaOH: No ammonium ion present. Therefore, X is calcium chloride, CaCl₂. It is often found as the hydrate CaCl₂·2H₂O.
与 NaOH 温热无刺激性气体：不存在铵离子。因此 X 为氯化钙 CaCl₂。常以二水合物 CaCl₂·2H₂O 形式存在。
8. Calculating Empirical Formula from Combustion Data | 由燃烧数据计算经验式
mine the empirical formula of Y (Mr: CO₂ = 44.0, H₂O = 18.0, C = 12.0, H = 1.0, O = 16.0).
化合物 Y 仅含碳、氢和氧。完全燃烧 1.50 g Y 生成 2.20 g CO₂ 和 0.90 g H₂O。求 Y 的经验式（相对分子质量：CO₂ = 44.0，H₂O = 18.0，C = 12.0，H = 1.0，O = 16.0）。
Step 1: Mass of carbon. CO₂ = 2.20 / 44.0 = 0.0500 mol → moles of C = 0.0500. Mass of C = 0.0500 × 12.0 = 0.600 g.
步骤 1：碳的质量。CO₂ 的物质的量 = 2.20 / 44.0 = 0.0500 mol → 碳的物质的量 = 0.0500。碳的质量 = 0.0500 × 12.0 = 0.600 g。
Step 2: Mass of hydrogen. each mole H₂O contains 2 mol H, so moles of H = 0.100. Mass of H = 0.100 × 1.0 = 0.100 g.
步骤 2：氢的质量。H₂O 的物质的量 = 0.90 / 18.0 = 0.0500 mol → 每摩尔 H₂O 含 2 mol H，故氢的物质的量 = 0.100。氢的质量 = 0.100 × 1.0 = 0.100 g。
Step 3: Mass of oxygen in Y. Mass of O = 1.50 – (0.600 + 0.100) = 0.800 g. Moles of O = 0.800 / 16.0 = 0.0500 mol.
步骤 3：Y 中氧的质量。氧的质量 = 1.50 – (0.600 + 0.100) = 0.800 g。氧的物质的量 = 0.800 / 16.0 = 0.0500 mol。
Step 4: Ratio. C : H : O = 0.0500 : 0.100 : 0.0500 = 1 : 2 : 1. Empirical formula: CH₂O.
步骤 4：比值。C : H : O = 0.0500 : 0.100 : 0.0500 = 1 : 2 : 1。经验式：CH₂O。
9. Rate of Reaction: Marble Chips and Acid | 反应速率：大理石碎片与酸
A student investigates the effect of surface area on the rate of reaction between calcium carbonate and hydrochloric acid. 5.0 g of marble chips and 5.0 g of powdered marble are each added to 50 cm³ of 2.0 mol dm⁻³ HCl. The loss of mass is recorded every 30 seconds. The powdered marble finishes in 2 minutes, the chips in 6 minutes. Explain the difference using collision theory.
一名学生研究表面积对碳酸钙与盐酸反应速率的影响。分别将 5.0 g 大理石碎片和 5.0 g 大理石粉末加入 50 cm³ 的 2.0 mol dm⁻³ HCl 中。每 30 秒记录质量损失。粉末大理石在 2 分钟内反应完毕，碎片则需 6 分钟。用碰撞理论解释差异。
Collision theory: For a reaction to occur, reactant particles must collide with sufficient energy and correct orientation. The acid particles can only react with the surface of the solid. Powdered marble has a much larger total surface area exposed to acid compared to the same mass of chips. This allows more frequent successful collisions per unit time between H⁺ ions and CaCO₃.
碰撞理论：要发生反应，反应物粒子必须以足够能量和正确取向碰撞。酸粒子只能与固体表面反应。同样质量下，粉末大理石暴露于酸的总表面积远大于碎片。这使得单位时间内 H⁺ 离子与 CaCO₃ 之间发生有效碰撞的频率更高。
Additional observation: The initial rate (slope of mass loss curve) is steeper for powder, confirming a higher rate. The final mass loss is identical because the same amounts of reactants are used, showing that surface area affects rate but not yield.
附加观察：粉末的初始速率（质量损失曲线的斜率）更陡，证实速率更高。最终质量损失相同，因为反应物用量一致，表明表面积影响速率但不影响产量。
10. Environmental Impact of Fuels: A Cost-Benefit Analysis | 燃料的环境影响：成本效益分析
A power station plans to switch from coal to natural gas (mainly methane). Coal contains 2.0% sulfur by mass. Compare the environmental advantages and challenges, including the production of SO₂, CO₂, and ash, and suggest one disadvantage of natural gas extraction.
一座发电站计划从煤改用天然气（主要成分为甲烷）。煤含硫 2.0%（质量分数）。比较环境方面的优势与挑战，包括 SO₂、CO₂ 和灰分的产生，并提出天然气开采的一个缺点。
SO₂ emissions: Burning coal releases SO₂, which causes acid rain. The sulfur content in coal (2.0%) is significant. Natural gas contains negligible sulfur, so SO₂ emissions are virtually eliminated, reducing acid deposition and respiratory problems.
SO₂ 排放：燃烧煤释放 SO₂，导致酸雨。煤中硫含量（2.0%）很高。天然气几乎不含硫，因此 SO₂ 排放几乎消除，减少酸沉降和呼吸系统问题。
CO₂ emissions: Natural gas produces less CO₂ per unit of energy released compared to coal. Methane has a higher hydrogen-to-carbon ratio, so combustion produces more H₂O and less CO₂ per joule. For equal energy output, switching reduces the carbon footprint, but CO₂ is still emitted, contributing to the greenhouse effect.
CO₂ 排放：与煤相比，天然气每单位能量释放产生的 CO₂ 更少。甲烷的氢碳比更高，因此燃烧时每焦耳产生更多 H₂O、更少 CO₂。就能量输出相等而言，这一转变降低了碳足迹，但 CO₂ 仍会排放，加剧温室效应。
Ash and particulates: Coal combustion generates fly ash and bottom ash requiring disposal and causing particulate pollution. Natural gas burns cleanly with no ash. However, extraction of natural gas via fracking can contaminate groundwater and cause tremors.
灰分与颗粒物：煤炭烧产生飞灰和底灰，需要处理并造成颗粒物污染。天然气燃烧清洁，不产生灰分。然而，通过压裂法开采天然气可能会污染地下水并引发微震。
Conclusion: Switching to natural gas offers immediate air quality benefits but is not a zero-carbon solution. Long-term strategy should include renewables.
结论：转向天然气可立即带来空气质量上的好处，但并非零碳解决方案。长期策略应包括可再生能源。
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