Pre-U CCEA Chemistry: Case Study Practical Exercises | 案例分析实战演练

📚 Pre-U CCEA Chemistry: Case Study Practical Exercises | 案例分析实战演练

Case study exercises are an integral part of the Pre-U CCEA Chemistry course, challenging students to apply theoretical knowledge to real-world scenarios. This article presents a series of carefully constructed case studies covering equilibrium, reaction kinetics, environmental chemistry, green chemistry, analytical techniques, and electrochemistry. Each case includes a scenario description, questions, detailed analysis, and calculations, with step-by-step reasoning to strengthen problem-solving skills and deepen conceptual understanding.

案例分析实战演练是Pre-U CCEA化学课程的重要组成部分，要求学生将理论知识应用于真实情境。本文提供一系列精心设计的案例，涵盖化学平衡、反应动力学、环境化学、绿色化学、分析技术和电化学等领域。每个案例包含情境描述、问题提出、详细分析和计算步骤，通过逐步推理来强化解题能力并加深对概念的理解。

1. The Haber Process – Yield and Rate Optimisation | 哈伯法——产率与速率的优化

The Haber process synthesises ammonia from nitrogen and hydrogen: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹. According to Le Chatelier’s principle, high pressure shifts the equilibrium towards the product side because there are fewer gas molecules, while low temperature favours the exothermic forward reaction. However, a low temperature drastically reduces the rate of reaction. Industrially, a compromise temperature of 400–450 °C, a pressure of about 200 atm, and an iron catalyst are used. Unreacted gases are recycled to improve overall atom economy. Consider a scenario where a reactor initially contains 100 moles of N₂ and 300 moles of H₂. At equilibrium, 160 moles of NH₃ are present. Calculate the percentage yield and explain why a catalyst is essential despite having no effect on the equilibrium position.

哈伯法由氮气和氢气合成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹。根据勒夏特列原理，高压使平衡向气体分子数较少的产物方向移动，低温则有利于放热正反应，但低温会大幅降低反应速率。工业上采用折中的400–450°C、约200 atm的压力和铁催化剂。未反应气体被循环以提高整体原子经济性。设想某反应器初始加入100摩尔N₂和300摩尔H₂，平衡时存在160摩尔NH₃。计算产率，并解释催化剂为何必不可少，尽管它不影响平衡位置。

Analysis: Based on stoichiometry, 1 mole of N₂ produces 2 moles of NH₃, therefore 100 moles of N₂ could theoretically yield 200 moles of NH₃. Percentage yield = (actual moles / theoretical moles) × 100% = (160 / 200) × 100 = 80%. The iron catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of both forward and backward reactions equally. This allows the system to reach equilibrium in a shorter time without altering the equilibrium yield. Recycling unreacted N₂ and H₂ enables nearly complete conversion over multiple cycles, significantly enhancing atom economy and reducing waste.

分析：根据化学计量关系，1摩尔N₂生成2摩尔NH₃，因此100摩尔N₂理论上可生成200摩尔NH₃。产率 = (实际摩尔 / 理论摩尔) × 100% = (160 / 200) × 100 = 80%。铁催化剂提供了活化能较低的反应路径，从而同等程度加快正逆反应速率，使体系更快达到平衡而不改变平衡产率。通过循环未反应的N₂和H₂，可在多次循环中实现接近完全的转化，大幅提高原子经济性并减少浪费。

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2. Contact Process – Optimising Sulfuric Acid Production | 接触法——硫酸生产的优化

Sulfuric acid is produced via the Contact process, with a key equilibrium step: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -196 kJ mol⁻¹. The conversion of SO₂ to SO₃ is exothermic and involves a decrease in the number of gas molecules. A plant operates under conditions of 450 °C, 2 atm pressure, and a vanadium(V) oxide catalyst. A student proposes that increasing the pressure to 100 atm would greatly increase the equilibrium yield of SO₃. Critically evaluate this proposal by considering equilibrium principles, economic costs, and safety. Also calculate the equilibrium constant expression Kc and discuss why a temperature of 450 °C is chosen despite the exothermic nature of the reaction.

硫酸通过接触法生产，其关键平衡步骤为：2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -196 kJ mol⁻¹。SO₂转化为SO₃的反应放热且气体分子数减少。某工厂在450 °C、2 atm压力及五氧化二钒催化剂下运行。一名学生提议将压力升至100 atm以大幅提高SO₃的平衡产率。请结合平衡原理、经济成本和安全因素，对这一提议进行批判性评价。同时写出平衡常数表达式Kc，并讨论为何放热反应仍选用450 °C。

Evaluation: Higher pressure indeed favours the forward reaction due to fewer moles of gas on the product side. However, at 2 atm the equilibrium already yields about 98% conversion. Increasing pressure to 100 atm would only gain a marginal extra yield, yet significantly raise energy costs, require thicker-walled reaction vessels, and introduce safety hazards. Thus the small increase in yield does not justify the huge additional expense. The catalyst allows a lower operating temperature but still requires 450 °C to achieve a commercially viable rate. Kc = [SO₃]² / ([SO₂]²[O₂]). Although low temperature favours SO₃ production, the rate would be uneconomically slow; 450 °C is an optimum compromise that balances rate and yield.

评价：高压确实因其使气体分子数减少而有利于正反应。然而，在2 atm下平衡转化率已达约98%，升至100 atm仅能微量提高产率，却会极大增加能源耗费、需要更厚的反应器壁并带来安全隐患。因此，产率的微小提升无法抵消巨大的附加成本。催化剂允许使用较低的运行温度，但仍需450 °C以保证商业可行的速率。Kc = [SO₃]² / ([SO₂]²[O₂])。尽管低温有利于SO₃生成，但反应速率将慢得不经济；450 °C是平衡速率与产率的最佳折中。

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3. Atmospheric Pollution – Acid Rain and Catalytic Converters | 大气污染——酸雨与催化转换器

Acid rain is formed when SO₂ and NOₓ from industrial emissions dissolve in cloud water and undergo oxidation to H₂SO₄ and HNO₃. A simplified pathway for SO₂ is: SO₂ + H₂O → H₂SO₃, then 2H₂SO₃ + O₂ → 2H₂SO₄. In a catalytic converter, harmful exhaust gases are converted: 2CO + 2NO → 2CO₂ + N₂. A technician analyses a car’s exhaust before and after fitting a new catalytic converter. Before: CO is 3.2% and NO is 0.18% by volume. After: CO drops to 0.3% and NO to 0.02%. Assuming the converter operates at 400 °C with a platinum-rhodium catalyst, explain the chemical principles behind these changes. Include the effect of the catalyst on activation energy and why the converter must reach its operating temperature quickly after engine start.

酸雨由工业排放的SO₂和NOₓ溶于云水并氧化为H₂SO₄和HNO₃而形成。以SO₂为例的简化途径为：SO₂ + H₂O → H₂SO₃，然后2H₂SO₃ + O₂ → 2H₂SO₄。在催化转换器中，有害废气发生转化：2CO + 2NO → 2CO₂ + N₂。某技师分析了安装新催化转换器前后的汽车尾气。之前：CO体积分数3.2%，NO 0.18%；之后：CO降至0.3%，NO降至0.02%。假设转换器在400 °C下使用铂铑催化剂，解释这些变化背后的化学原理，包括催化剂对活化能的影响，以及为何发动机启动后转换器必须迅速达到工作温度。

Explanation: The catalytic converter facilitates the redox reaction between CO and NO. The platinum-rhodium surface adsorbs CO and NO molecules, weakening their bonds and providing an alternative pathway with significantly lower activation energy. This allows the reaction to proceed rapidly at 400 °C, converting over 90% of harmful gases. When the engine starts cold, the catalyst is inactive, so efforts such as close-coupling the converter to the engine minimise the time to reach light-off temperature. The overall process drastically reduces pollutants that would otherwise contribute to acid rain and photochemical smog.

解释：催化转换器促进CO与NO之间的氧化还原反应。铂铑表面吸附CO和NO分子，削弱其化学键，并提供活化能大幅降低的替代路径，使反应在400 °C下快速进行，转化率达90%以上。发动机冷启动时催化剂尚未活化，因此常将转换器紧靠发动机安装以尽量缩短达到起燃温度的时间。整个过程大幅减少了原本会促成酸雨和光化学烟雾的污染物。

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4. Green Chemistry – Atom Economy in Ibuprofen Synthesis | 绿色化学——布洛芬合成的原子经济性

The traditional Boots synthesis of ibuprofen involved six steps with a low atom economy and large volumes of waste. The modern BHC (Boots-Hoechst-Celanese) process uses three catalytic steps, dramatically improving efficiency. For the final acylation step: C₁₀H₁₄ + C₄H₆O₃ → C₁₃H₁₈O₂ + C₂H₄O₂ (ibuprofen + acetic acid by-product). Given that the desired product is ibuprofen (C₁₃H₁₈O₂, Mr 206), calculate the atom economy for this step. Compare this with an older laboratory route using a stoichiometric reagent that generated 65 g of waste per 100 g of product. Explain how the principles of green chemistry, such as atom economy and catalysis, have transformed pharmaceutical manufacturing.

传统的Boots布洛芬合成法需要六步，原子经济性低且产生大量废弃物。现代BHC（Boots-Hoechst-Celanese）工艺采用三步催化反应，效率大幅提升。以最后酰化步骤为例：C₁₀H₁₄ + C₄H₆O₃ → C₁₃H₁₈O₂ + C₂H₄O₂（布洛芬与乙酸副产品）。已知目标产物为布洛芬（C₁₃H₁₈O₂，相对分子质量206），计算该步骤的原子经济性。并与旧实验室方法比较，旧法每生成100 g产物会产生65 g废弃物。解释原子经济性和催化等绿色化学原则如何改变了制药产业。

Calculation: Reactants are C₁₀H₁₄ (Mr 134) and C₄H₆O₃ (Mr 102). Total Mr of reactants = 134 + 102 = 236. Atom economy = (Mr of desired product / total Mr of reactants) × 100 = (206 / 236) × 100 ≈ 87.3%. This is a high atom economy, meaning most reactant mass ends up in the final product. In contrast, older multi-step routes often had atom economies below 40%, generating large volumes of salts and solvents. The BHC process also incorporates recyclable catalysts (HF or solid acid) and recovers acetic acid as a useful by-product, aligning with the principles of prevention, atom economy, and less hazardous chemical synthesis.

计算：反应物为C₁₀H₁₄（Mr 134）和C₄H₆O₃（Mr 102），反应物总Mr = 134 + 102 = 236。原子经济性 = (目标产物Mr / 反应物总Mr) × 100 = (206 / 236) × 100 ≈ 87.3%。这种高原子经济性意味着反应物的大部分质量进入最终产品。相比之下，老式多步路线的原子经济性常低于40%，产生大量盐和溶剂废弃物。BHC工艺还采用了可回收的催化剂（如HF或固体酸）并将乙酸作为有用的副产品回收，充分体现了预防、原子经济性和更安全的化学合成等原则。

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5. Redox Titration – Iron Content in Dietary Tablets | 氧化还原滴定——膳食铁片中的铁含量

Iron deficiency is treated with iron(II) sulfate tablets. The actual iron content can be determined by redox titration with standard potassium manganate(VII) in acidic medium: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. The end-point is the first permanent pink colour. In an experiment, a tablet was dissolved in dilute sulfuric acid and made up to 100 cm³. A 25.0 cm³ aliquot required 18.50 cm³ of 0.0100 mol dm⁻³ KMnO₄ solution. Calculate the mass of iron in the tablet. Discuss why an indicator is not needed and why the titration must be carried out in the absence of atmospheric oxygen.

铁缺乏症可通过硫酸亚铁片治疗。铁的实际含量可用酸性介质中高锰酸钾标准溶液的氧化还原滴定来确定：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O，终点为首次持久的粉红色。某实验将一片药片溶于稀硫酸并定容至100 cm³，移取25.0 cm³等分试样，用去18.50 cm³的0.0100 mol dm⁻³ KMnO₄溶液。计算药片中铁的质量，并讨论为何无需外加指示剂，以及为何滴定必须在隔绝空气氧气的条件下进行。

Calculation: Moles of MnO₄⁻ used = 0.01850 dm³ × 0.0100 mol dm⁻³ = 1.85 × 10⁻⁴ mol. From the equation, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺, so moles of Fe²⁺ in the 25.0 cm³ aliquot = 5 × 1.85 × 10⁻⁴ = 9.25 × 10⁻⁴ mol. In the original 100 cm³ solution, total moles of Fe²⁺ = 9.25 × 10⁻⁴ × (100/25) = 3.70 × 10⁻³ mol. Mass of iron = moles × Ar(Fe) = 3.70 × 10⁻³ × 55.8 ≈ 0.206 g. The permanganate ion acts as its own indicator; its intense purple colour disappears until all Fe²⁺ is oxidised, and one extra drop gives a lasting pink colour. Air must be excluded because oxygen can oxidise Fe²⁺ to Fe³⁺, leading to an underestimation.

计算：所用MnO₄⁻的摩尔数 = 0.01850 dm³ × 0.0100 mol dm⁻³ = 1.85 × 10⁻⁴ mol。由方程式，1 mol MnO₄⁻与5 mol Fe²⁺反应，故25.0 cm³等分试样中Fe²⁺的摩尔数 = 5 × 1.85 × 10⁻⁴ = 9.25 × 10⁻⁴ mol。原100 cm³溶液中Fe²⁺总摩尔数 = 9.25 × 10⁻⁴ × (100/25) = 3.70 × 10⁻³ mol。铁的质量 = 摩尔数 × Ar(Fe) = 3.70 × 10⁻³ × 55.8 ≈ 0.206 g。高锰酸根离子自身可作指示剂，其浓紫色在Fe²⁺被完全氧化前会褪去，再过量一滴即呈现持久的粉红色。必须隔绝空气，因氧气可将Fe²⁺氧化为Fe³⁺，导致结果偏低。

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6. Buffer Solutions – The Bicarbonate Buffer in Blood | 缓冲溶液——血液中的碳酸氢盐缓冲体系

Human blood maintains a remarkably constant pH of about 7.40, largely due to the carbonic acid–hydrogencarbonate buffer system: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq). The Henderson-Hasselbalch equation for this buffer is pH = pKa + log([HCO₃⁻]/[H₂CO₃]), where pKa for H₂CO₃ is 6.10 at body temperature. In a patient suffering from metabolic acidosis, the blood [HCO₃⁻] drops to 0.015 mol dm⁻³ while dissolved CO₂ remains equivalent to a [H₂CO₃] of 0.0012 mol dm⁻³. Calculate the resulting blood pH and suggest how the body compensates. Explain why a buffer is most effective when the ratio of conjugate base to acid lies close to 1.

人体血液将pH维持在约7.40的恒定值，主要归功于碳酸-碳酸氢根缓冲体系：H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)。该缓冲体系的Henderson-Hasselbalch方程为pH = pKa + log([HCO₃⁻]/[H₂CO₃])，其中H₂CO₃的pKa在体温下为6.10。某代谢性酸中毒患者血液[HCO₃⁻]降至0.015 mol dm⁻³，而溶解CO₂相当于[H₂CO₃] = 0.0012 mol dm⁻³。计算此时的血液pH，并说明机体如何代偿。解释为何缓冲液在共轭碱酸比例接近1时最为有效。

Calculation: pH = 6.10 + log(0.015 / 0.0012) = 6.10 + log(12.5). log(12.5) ≈ 1.10, so pH = 7.20. This is below the normal range, indicating acidosis. The body compensates by increasing breathing rate to expel CO₂, thereby lowering [H₂CO₃] and shifting the equilibrium to raise pH. The buffer’s effectiveness is maximal near its pKa because small additions of acid or base cause minimal changes in the log term. When [HCO₃⁻]/[H₂CO₃] = 1, pH = pKa, and the buffer capacity is highest.

计算：pH = 6.10 + log(0.015 / 0.0012) = 6.10 + log(12.5)，log(12.5) ≈ 1.10，则pH = 7.20。低于正常范围，表明存在酸中毒。机体通过加快呼吸速率排出CO₂来代偿，从而降低[H₂CO₃]并使平衡右移提升pH。缓冲液在共轭碱酸比例接近1时缓冲容量最大，因为此时添加少量酸或碱引起对数项的变化最小。当[HCO₃⁻]/[H₂CO₃] = 1时，pH = pKa，缓冲能力最强。

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7. Fuel Cells – Thermodynamics and Efficiency | 燃料电池——热力学与效率

A hydrogen-oxygen fuel cell operates with the overall reaction: 2H₂(g) + O₂(g) → 2H₂O(l). The standard cell potential is 1.23 V under standard

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