📚 Case Study Practice in Year 12 CIE Chemistry | Year 12 CIE 化学案例分析实战演练
In CIE Year 12 Chemistry, case studies are not just about recalling facts but about applying core principles to unfamiliar situations. This article presents a series of structured practice scenarios covering reaction kinetics, equilibrium, thermochemistry, electrochemistry, and organic chemistry. Each case study is accompanied by step‑by‑step analysis, typical exam‑style questions, and revision of key concepts. Working through these will sharpen your data interpretation, calculation, and explanation skills – exactly what is needed for high marks in Paper 2 and Paper 3.
在 CIE Year 12 化学中，案例分析不只是背诵知识点，而是将核心原理应用于陌生情境。本文提供了一系列结构化实战场景，涵盖反应动力学、化学平衡、热化学、电化学和有机化学。每个案例都配有逐步分析、典型考题风格的问题以及关键概念回顾。通过这些练习，你将提升数据解读、计算和解释的能力——这正是冲击试卷二和试卷三高分的必备技能。
1. Case Study: Rate of Reaction and Activation Energy | 反应速率与活化能案例
A student investigates the reaction between magnesium ribbon and excess dilute sulfuric acid. The volume of hydrogen gas produced is recorded every 10 seconds at 25°C. The data are shown in the table below.
一名学生研究了镁条与过量稀硫酸的反应。在 25°C 下，每 10 秒记录一次产生的氢气体积。数据如下表所示。
| Time / s | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| Volume of H₂ / cm³ | 0 | 14 | 26 | 36 | 44 | 49 | 50 |
(a) Plot a graph of volume of H₂ against time. Use your graph to determine the initial rate of reaction. (b) Explain how the shape of the curve shows the change in rate as the reaction proceeds. (c) The student repeats the experiment at 35°C. Predict and explain how the plot would differ.
(a) 绘制氢气体积对时间的图像。利用图像确定反应的初始速率。(b) 解释曲线的形状如何体现反应进行中速率的变化。(c) 该学生在 35°C 下重复实验。预测并解释图像会发生怎样的变化。
Analysis: e : The initial rate is the gradient of the tangent at t = 0. Using data, the gradient from 0 to 20 s is roughly (26‑0)/(20‑ urve plateaus when magnesium is used up. At a higher temperature, the curve rises more steeply initially, giving a larger initial rate, and reaches the same final volume faster because the same amount of magnesium is consumed but with greater average kinetic energy and more frequent successful collisions.
分析：初始速率是 t = 0 处切线的斜率。利用数据，0 到 20 秒的斜率约为 (26‑0)/(20‑0) = 1.3 cm³ s⁻¹。随着反应进行，斜率减小，因为硫酸浓度下降（镁过量）。当镁耗尽时出现平台。在较高温度下，曲线初始上升更陡，初始速率更大，更快达到相同最终体积，因为相同量的镁被消耗，但平均动能更高且成功碰撞更频繁。
2. Case Study: Interpreting a Maxwell‑Boltzmann Distribution | 麦克斯韦‑玻尔兹曼分布解读
The diagram below (not shown) represents the Maxwell‑Boltzmann distribution of molecular energies for a gaseous reaction at temperature T. The activation energy, Eₐ, is marked. (a) State what the area under the curve represents. (b) Explain why raising the temperature increases the rate of reaction. Use the distribution to support your answer. (c) A catalyst is added. Explain, with reference to the distribution, how the rate increases.
下图（未显示）代表温度 T 下气体反应的分子能量麦克斯韦‑玻尔兹曼分布。活化能 Eₐ 已标出。(a) 说明曲线下面积代表什么。(b) 解释升高温度为何加快反应速率，并利用分布图来支持你的答案。(c) 加入催化剂。参照分布图，解释速率如何增加。
Analysis: tens and shifts to the right; the peak moves to higher energy. The proportion of molecules with E ≥ Eₐ increases dramatically (the shaded area beyond Eₐ enlarges), so the frequency of successful collisions increases. For a catalyst, an alternative pathway with lower activation energy Ec is provided. The new curve is the same shape but more molecules have energy ≥ Ec, increasing the number of effective collisions without raising temperature.
分析：曲线下总面积等于分子总数。只有能量 ≥ Eₐ 的分子才能反应。升高温度时，曲线变得扁平并右移；峰顶移向更高能量。能量 ≥ Eₐ 的分子比例急剧增加（Eₐ 右侧阴影面积扩大），因此成功碰撞频率增加。对于催化剂，提供了活化能 Ec 更低的替代路径。分布形状不变，但有更多分子能量 ≥ Ec，在不升高温度的情况下增加有效碰撞数。
3. Case Study: Determining Enthalpy Change of Neutralisation | 中和反应焓变的测定
l dm⁻³ sodium hydroxide is added. The mixture is stirred and temperatures are recorded until constant. The temperature rise ΔT is found by extrapolation to be 13.2°C. Calculate the enthalpy change of neutralisation per mole of water formed. (Assume density = 1.00 g cm⁻³, specific heat capacity = 4.18 J g⁻¹ K⁻¹.)
将 25.0 cm³ 的 2.00 mol dm⁻³ 盐酸放入聚苯乙烯杯中。每分钟记录一次温度。在第 4 分钟加入 25.0 cm³ 的 2.00 mol dm⁻³ 氢氧化钠溶液。搅拌混合物并持续记录温度至恒定。通过外推法得出温升 ΔT 为 13.2°C。计算每生成 1 mol 水的中和焓变。（假设密度 = 1.00 g cm⁻³，比热容 = 4.18 J g⁻¹ K⁻¹。）
Analysis: 0500 mol; moles of NaOH = 0.0500 mol, both limit. Moles of water formed = 0.0500 mol. ΔH = ‑q / n = ‑2.76 kJ / 0.0500 mol = ‑ cup is a good insulator. Incomplete mixing, heat capacity of apparatus ignored, and assumptions affect accuracy.
分析：总体积 = 50.0 cm³，质量 = 50.0 g。释放热量 q = mcΔT = 50.0 × 4.18 × 13.2 = 2758.8 J ≈ 2.76 kJ。HCl 物质的量 = 0.0250 dm³ × 2.00 mol dm⁻³ = 0.0500 mol；NaOH 物质的量 = 0.0500 mol，两者恰好完全反应。生成水的物质的量 = 0.0500 mol。ΔH = ‑q / n = ‑2.76 kJ / 0.0500 mol = ‑55.2 kJ mol⁻¹。负号表示放热。实验采用外推法补偿热量损失；聚苯乙烯杯是良好的绝热体。混合不充分、忽略装置热容以及假设会影响准确度。
4. Case Study: Le Chatelier’s Principle and Yield Optimisation | 勒夏特列原理与产率优化
The Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = ‑ ow temperature would give a higher equilibrium yield but is not used. (c) Explain the role of the catalyst on the equilibrium composition.
哈伯法：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = ‑92 kJ mol⁻¹。通常使用铁催化剂，在 450°C 和 200 atm 下操作。(a) 解释高压为何有利于氨的产率。(b) 解释较低温度可获得更高平衡产率，但实际并不采用的原因。(c) 说明催化剂对平衡组成的作用。
Analysis: (a) 4 gas moles on left, 2 on right. By Le Chatelier’s principle, increasing pressure favours the forward reaction that produces fewer moles, shifting equilibrium to the right, giving a higher yield of NH₃. (b) The forward reaction is exothermic. Low temperature shifts equilibrium to the right, increasing yield. However, a low temperature makes the rate unacceptably slow. 450°C is a compromise: a reasonable rate at acceptable yield. (c) A catalyst provides an alternative pathway with lower activation energy, speeding up both forward and reverse reactions equally. It does not affect the position of equilibrium or yield; it only helps the system reach equilibrium faster.
分析：(a) 左边 4 mol 气体，右边 2 mol。根据勒夏特列原理，增大压强有利于气体分子数减少的方向，即正向反应，平衡右移，氨产率提高。(b) 正向反应放热。低温使平衡右移，提高产率。但低温导致反应速率过慢。450°C 是一个折衷方案：在可接受的产率下维持合理的速率。(c) 催化剂提供活化能较低的替代路径，同等程度地加快正、逆反应速率。它不影响平衡位置或产率；仅使体系更快达到平衡。
5. Case Study: Electrochemical Cell and Standard Electrode Potentials | 电化学电池与标准电极电势
A student sets up a cell using Cu²⁺/Cu and Zn²⁺/Zn half‑cells connected by a salt bridge. The standard electrode potentials are: E°(Zn²⁺/Zn) = ‑0.76 V, E°(Cu²⁺/Cu) = +0.34 V. (a) Write half‑ direction of electron flow in the external circuit. (c) The student replaces the zinc half‑cell with an Fe²⁺/Fe electrode (E° = ‑0.44 V). Predict the new cell potential.
一名学生使用 Cu²⁺/Cu 和 Zn²⁺/Zn 半电池通过盐桥构成原电池。标准电极电势为：E°(Zn²⁺/Zn) = ‑0.76 V，E°(Cu²⁺/Cu) = +0.34 V。(a) 写出每个电极上的半反应方程式，标明阳极和阴极，并计算标准电池电动势。(b) 说明外电路中电子的流动方向。(c) 该学生用 Fe²⁺/Fe 电极（E° = ‑0.44 V）替换锌半电池。预测新的电池电动势。
Analysis: 2e⁻ at anode. Reduction at cathode: Cu²⁺ + 2e⁻ → Cu. Cell potential E°_cell = E°(cathode) ‑ E°(anode) = 0.34 ‑ (‑ With Fe²⁺/Fe and Cu²⁺/Cu, Fe is more negative so oxidised: Fe → Fe²⁺ + 2e⁻. E°_cell = 0.34 ‑ (‑0.44) = +0.78 V. A smaller cell potential indicates a smaller 'driving force' for the reaction.
分析：(a) 电势更负的电极为氧化极：Zn → Zn²⁺ + 2e⁻（阳极）。阴极还原：Cu²⁺ + 2e⁻ → Cu。电池电动势 E°_cell = E°(阴极) ‑ E°(阳极) = 0.34 ‑ (‑0.76) = +1.10 V。(b) 电子从锌阳极通过外电路流向铜阴极。(c) 对于 Fe²⁺/Fe 和 Cu²⁺/Cu，Fe 更负所以被氧化：Fe → Fe²⁺ + 2e⁻。E°_cell = 0.34 ‑ (‑0.44) = +0.78 V。较小的电动势表明反应的“驱动力”更小。
6. Case Study: Interpreting an Electrolysis Setup with Active Electrodes | 活性电极电解案例分析
A student electrolyses aqueous copper(II) sulfate using copper electrodes. (a) Write the half‑equations for the reactions at the anode and cathode. (b) Explain why the deep blue colour of the electrolyte remains unchanged. (c) Predict what would happen if platinum electrodes were used instead. Include any observations.
学生用铜电极电解硫酸铜溶液。(a) 写出阳极和阴极的半反应方程式。(b) 解释为何电解质溶液的深蓝色不发生变化。(c) 若换用铂电极，预测会发生什么现象，并说明观察结果。
Analysis: re replaced by Cu²⁺ from the anode dissolving. The concentration of Cu²⁺ stays constant, so the colour intensity does not change. This is the principle of electroplating/electrorefining. (c) With inert platinum electrodes, OH⁻ and SO₄²⁻ are present. At the anode, O₂ gas is produced (4OH⁻ → O₂ + 2H₂O + 4e⁻). At the cathode, Cu²⁺ is discharged forming a pink‑ lue colour fades, and the solution may eventually turn colourless if all Cu²⁺ is consumed.
分析：(a) 阳极（氧化）：Cu(s) → Cu²⁺(aq) + 2e⁻；阴极（还原）：Cu²⁺(aq) + 2e⁻ → Cu(s)。(b) 阴极放电消耗的 Cu²⁺ 被阳极溶解产生的 Cu²⁺ 补充。Cu²⁺ 浓度保持恒定，因此颜色深度不变。这就是电镀/电解精炼的原理。(c) 使用惰性铂电极时，溶液中存在 OH⁻ 和 SO₄²⁻。阳极生成 O₂ 气体（4OH⁻ → O₂ + 2H₂O + 4e⁻）。阴极析出 Cu²⁺，形成粉棕色固体。Cu²⁺ 浓度降低，蓝色变浅，最终若所有 Cu²⁺ 耗尽溶液变为无色。
7. Case Study: Organic Reaction Pathways and Functional Group Tests | 有机反应路线与官能团检验
A student is given three colourless liquids labelled A, B and C. They are known to be ethanol, ethanal and ethanoic acid, but not in that order. The student carries out the following tests: (i) A drop of each is added to a separate solution of 2,4‑dinitrophenylhydrazine (2,4‑DNPH). Only B gives an orange precipitate. (ii) To fresh samples, a piece of sodium is added. A and C produce effervescence. (iii) A small amount of each is warmed with acidified potassium dichromate(VI). A turns green; C shows no change. Identify A, B and C and explain your reasoning.
学生拿到三种无色液体，标记为 A、B 和 C。它们已知为乙醇、乙醛和乙酸，但顺序未知。学生进行了以下测试：(i) 各取一滴分别加入 2,4‑二硝基苯肼（2,4‑DNPH）溶液中。只有 B 产生橙色沉淀。(ii) 另取新样品，加入一小块金属钠。A 和 C 产生气泡。(iii) 取少量各样品与酸化重铬酸钾(VI) 加热。A 变为绿色；C 无变化。确定 A、B 和 C 并解释你的推理。
Analysis: 2,4‑ . B gives a positive result, so B is ethanal. Sodium reacts with acidic protons: ethanol (‑OH) and ethanoic acid (‑ o carboxylic acids, turning from orange to green. A turns green, so A is ethanol (oxidised to ethanoic acid). C shows no change because ethanoic acid cannot be further oxidised. Thus A = ethanol, B = ethanal, C = ethanoic acid.
分析：2,4‑DNPH 检验针对羰基化合物（醛/酮）。B 呈阳性，故 B 是乙醛。钠与酸性氢反应：乙醇（‑OH）和乙酸（‑COOH）产生氢气；醛不反应。因此 A 和 C 是乙醇和乙酸。酸化 K₂Cr₂O₇ 可将一级醇和醛氧化为羧酸，由橙色变为绿色。A 变绿，故 A 是乙醇（氧化为乙酸）。C 无变化，因为乙酸不能再被氧化。由此 A = 乙醇，B = 乙醛，C = 乙酸。
8. Case Study: Hess’s Law and Born‑Haber Cycle Calculation | 盖斯定律与玻恩‑哈伯循环计算
Using the following data, construct a Born‑ t ionisation energy Mg = +738 kJ mol⁻¹; second ionisation energy Mg = +1451 kJ mol⁻¹; first electron affinity O = ‑141 kJ mol⁻¹; second electron affinity O = +798 kJ mol⁻¹; enthalpy of formation MgO = ‑602 kJ mol⁻¹.) Show all steps and apply Hess’s Law.
利用以下数据，构建氧化镁的玻恩‑哈伯循环并计算晶格能。（数据：Mg 的原子化焓 = +148 kJ mol⁻¹；O₂ 原子化焓 = +249 kJ mol⁻¹；Mg 第一电离能 = +738 kJ mol⁻¹；Mg 第二电离能 = +1451 kJ mol⁻¹；O 第一电子亲合能 = ‑141 kJ mol⁻¹；O 第二电子亲合能 = +798 kJ mol⁻¹；MgO 生成焓 = ‑602 kJ mol⁻¹。）展示全部步骤并应用盖斯定律。
Analysis: Born‑Haber cycle: ΔHf(MgO) = ΔHatom(Mg) + ΔHatom(½O₂) + IE₁(Mg) + IE₂(Mg) + (EA₁(O) + EA₂(O)) + ΔH_lattice. Note: atomisation of O₂ uses half a mole for one O atom: ½ × 249 = +124.5 kJ mol⁻¹. So, ‑602 = 148 + 124.5 + 738 + 1451 + (‑141 + 798) + ΔH_lattice. Sum of endothermic terms: 148 + 124.5 + 738 + 1451 + 657 = 3118.5 kJ mol⁻¹. Then ‑602 = 3118.5 + ΔH_lattice → ΔH_lattice = ‑602 ‑ 3118.5 = ‑3720.5 kJ mol⁻¹ (approx ‑3721 kJ mol⁻¹). ons must be followed: ionisation is endothermic (positive), electron affinities can be exo‑ or endothermic, formation exothermic (negative).
分析：玻恩‑哈伯循环： ΔHf(MgO) = ΔHatom(Mg) + ΔHatom(½O₂) + IE₁(Mg) + IE₂(Mg) + (EA₁(O) + EA₂(O)) + ΔH_lattice。 注意：一个氧原子需要半个 O₂ 的原子化焓：½ × 249 = +124.5 kJ mol⁻¹。 因此，‑602 = 148 + 124.5 + 738 + 1451 + (‑141 + 798) + ΔH_lattice。 吸热项之和：148 + 124.5 + 738 + 1451 + 657 = 3118.5 kJ mol⁻¹。 则 ‑602 = 3118.5 + ΔH_lattice → ΔH_lattice = ‑602 ‑ 3118.5 = ‑3720.5 kJ mol⁻¹（约 ‑3721 kJ mol⁻¹）。 该大负值反映了离子晶格中强静电吸引力。必须遵守所有符号规则：电离吸热（正），电子亲和能放热也可吸热，生成焓放热（负）。
9. Case Study: Cracking and Isomers in Organic Chemistry | 裂解与有机化学中的同分异构体
Decane, C₁₀H₂₂, is cracked to produce ethene, C₂H₄, and one other hydrocarbon. (a) Write a balanced equation for the reaction. (b) The other hydrocarbon produced shows chain isomerism. Draw and name two possible structural isomers. (c) Ethene is used to make poly(ethene). State the type of polymerisation and draw two repeating units to show the structure of the polymer.
癸烷 (C₁₀H₂₂) 经裂解生成乙烯 (C₂H₄) 和另一种烃。(a) 写出该反应的配平方程式。(b) 生成的另一种烃具有碳链异构现象。画出并命名两种可能的结构异构体。(c) 乙烯用于制造聚乙烯。指出聚合类型，并画出两个重复单元以表示聚合物的结构。
Analysis: (a) C₁₀H₂₂ → C₂H₄ + C₈H₁₈. (b) C₈H₁₈ can be octane with various chain isomers, e.g., 2‑methylheptane, 3‑ethyl‑2‑ , different arrangement of carbon skeleton. (c) Addition polymerisation. Repeating unit: [‑CH₂‑CH₂‑]ₙ. Two units: ‑CH₂‑CH₂‑CH₂‑CH₂‑. No small molecule eliminated, distinguishing it from condensation polymerisation.
分析：(a) C₁₀H₂₂ → C₂H₄ + C₈H₁₈。(b) C₈H₁₈ 可以是辛烷的多种碳链异构体，例如 2‑甲基庚烷、3‑乙基‑2‑甲基戊烷（任何正确的名称和结构均可）。碳链异构：分子式相同，碳骨架排列不同。(c) 加成聚合。重复单元：[‑CH₂‑CH₂‑]ₙ。两个单元：‑CH₂‑CH₂‑CH₂‑CH₂‑。无小分子脱去，这是区别于缩聚的特征。
10. Case Study: Determining an Unknown Concentration by Titration | 通过滴定测定未知浓度
A student titrates 25.0 cm³ of a solution of sodium hydroxide (unknown concentration) against 0.100 mol dm⁻³ sulfuric acid. The average titre is 21.40 cm³. (a) Write the balanced equation. (b) Calculate the concentration of the NaOH solution. (c) The student suspects that some NaOH has reacted with CO₂ from air. What effect would this have on the calculated concentration? Explain.
学生用 0.100 mol dm⁻³ 硫酸滴定 25.0 cm³ 未知浓度的氢氧化钠溶液。平均滴定体积为 21.40 cm³。(a) 写出配平方程式。(b) 计算 NaOH 溶液的浓度。(c) 学生怀疑部分 NaOH 已与空气中的 CO₂ 反应。这会对计算出的浓度产生什么影响？请解释。
Analysis: 0.00428 mol in 25.0 cm³. Concentration = 0.00428 / 0.0250 = 0.171 mol dm⁻³. (c) NaOH + CO₂ → Na₂CO₃ + H₂O (or 2NaOH + CO₂ → Na₂CO₃ + H₂O). Some NaOH is converted to carbonate, which still reacts with acid but requires 2 moles of acid per mole of carbonate if neutralised. However, the titration using phenolphthalein (or methyl orange) would give a different mole ratio. Often this leads to a lower apparent concentration of NaOH because the amount of base titrated against acid appears reduced if the indicator is not chosen appropriately, or it causes the endpoint to be less sharp. In a typical scenario, the calculated concentration would be lower than the true value because the reacted NaOH is no longer free but still consumes acid; careful reasoning about the indicator used is required. A common exam response: CO₂ turns some NaOH into Na₂CO₃, which reacts with acid but in a 1:2 ratio (Na₂CO₃ + 2HCl → ...), causing a smaller titre and leading to underestimated concentration (if using phenolphthalein, the endpoint occurs when carbonate is only half‑neutralised to bicarbonate, producing an error).
分析：(a) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。(b) H₂SO₄ 物质的量 = 0.02140 dm³ × 0.100 = 0.00214 mol。由方程式知 NaOH : H₂SO₄ = 2:1，所以 25.0 cm³ 溶液中 NaOH 物质的量 = 2 × 0.00214 = 0.00428 mol。浓度 = 0.00428 / 0.0250 = 0.171 mol dm⁻³。(c) NaOH + CO₂ → Na₂CO₃ + H₂O。部分 NaOH 转化为碳酸钠，碳酸钠仍与酸反应，但计量比不同。使用酚酞时，碳酸盐只被中和至碳酸氢盐，滴定体积偏小，导致计算浓度偏低。这体现了实际分析中可能的系统误差。
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