📚 Case Study Practice for Year 12 CAIE Chemistry | 案例分析实战演练
Case study questions in CAIE AS Level Chemistry go beyond simple recall; they present real-world scenarios where you must apply knowledge of bonding, energetics, kinetics, equilibrium, organic chemistry and analytical techniques. This article walks you through ten worked case studies, showing how to unpack the problem, link concepts and construct full-mark answers. Each case study includes an English explanation followed by a matching Chinese version, so you can master both subject content and bilingual terminology.
CAIE AS 化学的案例分析题远不止简单的知识再现；它们给出真实情景，要求你综合运用结构、能量、动力学、平衡、有机化学和分析技术的知识。本文带你演练十个典型案例，展示如何拆解问题、串联概念并写出满分答案。每个案例均配有英文解释与对应的中文版本，助你同时掌握学科内容与双语术语。
1. Deducing a Rate Equation from Raw Data | 从原始数据推导速率方程
oubling [A] (at constant [B]) quadruples the rate, while doubling [B] (at constant [A]) doubles the rate. By comparing experiments 1 and 2, the order with respect to A is 2; by comparing experiments 2 and 3, the order with respect to B is 1. Therefore the rate equation is: rate = k[A]²[B]. Substituting values from any experiment gives the rate constant k and its unit: dm⁶ mol⁻² s⁻¹.
某学生研究反应 A + 2B → C。在三种不同起始浓度下测量初始速率。数据表显示，当 [A] 加倍（[B] 恒定）时速率为原来的 4 倍；当 [B] 加倍（[A] 恒定）时速率为原来的 2 倍。对比实验 1 和 2 可知 A 的反应级数为 2；对比实验 2 和 3 可知 B 的反应级数为 1。因此速率方程为：rate = k[A]²[B]。代入任一组实验数据即可求出速率常数 k 及其单位：dm⁶ mol⁻² s⁻¹。
2. Determining an Empirical Formula by Combustion Analysis | 通过燃烧分析确定经验式
f C = 0.468 g ÷ 44.0 g mol⁻¹ = 0.0106 mol; from the mass of H₂O, moles of H = 2 × (0.191 g ÷ 18.0 g mol⁻¹) = 0.0212 mol. Dividing by the smallest (0.0106) gives a ratio C:H = 1:2, so the empirical formula is CH₂. If the molar mass is later found to be 84.0 g mol⁻¹, the molecular formula is C₆H₁₂.
将 0.150 g 液态烃样品在过量氧气中完全燃烧，得到 0.468 g CO₂ 和 0.191 g H₂O。由 CO₂ 质量求出碳的物质的量：0.468 g ÷ 44.0 g mol⁻¹ = 0.0106 mol；由 H₂O 质量求出氢的物质的量：2 × (0.191 g ÷ 18.0 g mol⁻¹) = 0.0212 mol。除以最小值 0.0106 得 C:H = 1:2，经验式为 CH₂。若后续测得摩尔质量为 84.0 g mol⁻¹，则分子式为 C₆H₁₂。
3. Reaction Mechanism Deduction from Kinetic Evidence | 利用动力学证据推断反应机理
echanism is: Step 1 (slow): 2NO + H₂ → N₂ + H₂O₂; Step 2 (fast): H₂O₂ + H₂ → 2H₂O. The rate-determining step involves two molecules of NO and one molecule of H₂, matching the experimentally determined rate equation rate = k[NO]²[H₂]. The observed overall stoichiometry is also consistent once the fast step is added. Any intermediate such as H₂O₂ does not appear in the rate equation.
实验发现反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) 对 NO 为二级，对 H₂ 为一级。提出的两步机理为：第一步（慢）：2NO + H₂ → N₂ + H₂O₂；第二步（快）：H₂O₂ + H₂ → 2H₂O。速率控制步骤涉及两分子 NO 和一分子 H₂，这与实验速率方程 rate = k[NO]²[H₂] 吻合。加上快步骤后，总反应计量比也一致。中间体 H₂O₂ 不出现在速率方程中。
4. Bonding and Structure: Explaining the Melting Point Trend | 化学键与结构：解释熔点趋势
Sodium chloride (NaCl) melts at 801 °C, while aluminium chloride (AlCl₃) sublimes at 178 °C. In NaCl, strong ionic bonds between Na⁺ and Cl⁻ require a large amount of energy to overcome, giving a high melting point. AlCl₃, however, exists as simple covalent molecules (Al₂Cl₆) with weak van der Waals forces between molecules; only these weak intermolecular attractions need to be overcome, resulting in a much lower sublimation point. This difference illustrates how changing electronegativity and ion polarisation alters bonding character.
氯化钠 (NaCl) 熔点为 801 °C，而氯化铝 (AlCl₃) 在 178 °C 升华。NaCl 中 Na⁺ 与 Cl⁻ 之间的强离子键需要大量能量才能克服，因此熔点很高。而 AlCl₃ 以 Al₂Cl₆ 简单共价分子形式存在，分子间仅有微弱的范德华力；升华时只需克服这些分子间作用力，因此升华点低得多。这一差异说明电负性和离子极化如何改变键合特征。
5. Organic Synthesis Route Design | 有机合成路线设计
propose a two-step synthesis for propyl ethanoate (CH₃COOCH₂CH₂CH₃). Step 1: hydration of propene using steam and a phosphoric acid catalyst at 300 °C, 60 atm, to give propan-2-ol. Step 2: oxidation of propan-2-ol with acidified potassium dichromate(VI) under distillation to form propanone. However, we need propan-1-ol for esterification. A better route: Step 1: anti-Markovnikov hydration using BH₃ followed by H₂O₂/OH⁻ to give propan-1-ol. Step 2: esterification of propan-1-ol with ethanoic acid using concentrated H₂SO₄ catalyst and heating under reflux. The final product is propyl ethanoate and water.
以丙烯 (CH₃CH=CH₂) 为原料，设计两步合成乙酸丙酯 (CH₃COOCH₂CH₂CH₃)。步骤 1：丙烯与水蒸气在磷酸催化、300 °C、60 atm 条件下发生水合，得到丙-2-醇。步骤 2：丙-2-醇用酸化重铬酸钾(VI) 在蒸馏下氧化生成丙酮。但酯化需要丙-1-醇。更优路线：步骤 1：用 BH₃ 反马氏加成、再经 H₂O₂/OH⁻ 得到丙-1-醇。步骤 2：丙-1-醇与乙酸在浓硫酸催化下加热回流进行酯化，得乙酸丙酯和水。
6. Equilibrium Constant Kc Calculation | 平衡常数 Kc 的计算
For the esterification reaction CH₃COOH(l) + C₂H₅OH(l) ⇌ to contain 0.33 mol of ester. The Kc expression is [ester][water] / [acid][alcohol]. At equilibrium, moles: acid = 0.50 - 0.33 = 0.17, alcohol = 0.17, ester = 0.33, water = 0.33. Since all are liquids, we can use moles directly if volume cancels. Kc = (0.33 × 0.33) ÷ (0.17 × 0.17) ≈ 3.8 (no units because the total moles of species is unchanged).
对于酯化反应 CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)，将 0.50 mol 酸与 0.50 mol 醇混合并在 25 °C 达到平衡。测得平衡混合物中含 0.33 mol 酯。Kc 表达式为 [酯][水] / [酸][醇]。平衡时各物质物质的量：酸 = 0.50 - 0.33 = 0.17，醇 = 0.17，酯 = 0.33，水 = 0.33。由于均为液体且体积相同，可直接用物质的量计算。Kc = (0.33 × 0.33) ÷ (0.17 × 0.17) ≈ 3.8（无单位，因为总摩尔数不变）。
7. Electrochemical Cell: Predicting Feasibility | 电化学原电池：判断反应可行性
e positive emf indicates the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is thermodynamically feasible. If the concentrations are non-standard, the Nernst equation can be applied: Ecell = E°cell - (RT/nF) lnQ. At 298 K, this simplifies to Ecell = E°cell - (0.0592/n) logQ. For example, if [Zn²⁺] = 0.10 mol dm⁻³ and [Cu²⁺] = 1.0 mol dm⁻³, Q = [Zn²⁺]/[Cu²⁺] = 0.10, then Ecell = 1.10 - (0.0592/2) × log(0.10) ≈ 1.10 + 0.0296 = 1.13 V.
用 Zn²⁺/Zn 半电池 (E° = -0.76 V) 与 Cu²⁺/Cu 半电池 (E° = +0.34 V) 组成原电池。电池电动势 E°cell = +0.34 - (-0.76) = +1.10 V。正的电动势表明反应 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) 在热力学上可行。若非标准浓度，可用能斯特方程：Ecell = E°cell - (RT/nF) lnQ。在 298 K 时简化为 Ecell = E°cell - (0.0592/n) logQ。例如若 [Zn²⁺] = 0.10 mol dm⁻³，[Cu²⁺] = 1.0 mol dm⁻³，Q = 0.10，Ecell = 1.10 - (0.0592/2) × log(0.10) ≈ 1.13 V。
8. Acid-Base Titration and Indicator Choice | 酸碱滴定与指示剂选择
d with 0.100 mol dm⁻³ NaOH. The equivalence point pH is about 8.7 due to the formation of the weakly basic ethanoate ion. The titration curve shows a sharp rise in pH around the equivalence point but with a narrower range than a strong acid–strong base titration. Therefore, phenolphthalein (pH range 8.2–10.0, colourless to pink) is a suitable indicator, while methyl orange (3.1– tration curve helps visualise why the indicator must match the equivalence-point pH range.
用 0.100 mol dm⁻³ NaOH 滴定 25.0 cm³ 0.100 mol dm⁻³ 乙酸 (Ka = 1.8 × 10⁻⁵)。由于生成弱碱性的乙酸根离子，等当点 pH 约为 8.7。滴定曲线在等当点附近出现陡峭突跃，但突跃范围比强酸强碱滴定窄。因此酚酞（变色范围 pH 8.2–10.0，无色→粉红）是合适的指示剂，而甲基橙（3.1–4.4）会过早变色。绘制滴定曲线有助于理解为什么指示剂的变色区必须匹配等当点 pH。
9. Structure Elucidation Using IR and Mass Spectra | 利用红外光谱和质谱推断结构
/z = 88 in its mass spectrum. The IR spectrum shows a broad absorption around 3300 cm⁻¹ (O– the molecular formula is C₄H₈O₂ (88/44 = 2). Combining the data, the compound contains a carbonyl group and an hydroxyl group, likely a carboxylic acid or an ester with an alcohol portion. Considering the fragmentation pattern, the base peak at m/z = 45 suggests a CH₃CH₂O⁺ ion, so the structure is ethyl ethanoate (CH₃COOCH₂CH₃).
某有机物的质谱给出分子离子峰 m/z = 88。红外光谱显示约 3300 cm⁻¹ 处的宽吸收（O–H 伸缩振动）和 1720 cm⁻¹ 处的尖峰（C=O 伸缩振动）。燃烧分析给出的经验式为 C₂H₄O，经验式质量 44。因此分子式为 C₄H₈O₂ (88/44 = 2)。综合信息，分子含羰基和羟基，可能为羧酸或含醇部分的酯。根据碎片离子峰，m/z = 45 的基峰提示存在 CH₃CH₂O⁺ 离子，故结构为乙酸乙酯 (CH₃COOCH₂CH₃)。
10. Calorimetry and Enthalpy Change of Neutralisation | 量热法与中和焓变
lume is 100 cm³, assumed to have the density of water (1.00 g cm⁻³) and specific heat capacity 4.18 J g⁻¹ K⁻¹. Heat evolved, q = m c ΔT = 100 g × 4.18 J g⁻¹ K⁻¹ × 6.5 K = 2717 J. Moles of water formed = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol (from limiting reagent). ΔH = -q/n = -2717 J / 0.0500 mol = -54340 J mol⁻¹ ≈ -54.3 kJ mol⁻¹. The value is less exothermic than the standard enthalpy of neutralisation (-57.1 kJ mol⁻¹) for strong acid-strong base, likely due to heat loss and the use of a simple calorimeter.
将 50.0 cm³ 1.00 mol dm⁻³ HCl 与 50.0 cm³ 1.00 mol dm⁻³ NaOH 在聚苯乙烯杯中混合。温度由 21.0 °C 升至 27.5 °C。溶液总体积 100 cm³，假设密度与水相同 (1.00 g cm⁻³)，比热容 4.18 J g⁻¹ K⁻¹。放出的热量 q = m c ΔT = 100 g × 4.18 J g⁻¹ K⁻¹ × 6.5 K = 2717 J。生成水的物质的量（由限量试剂计算）= 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol。ΔH = -q/n ≈ -54.3 kJ mol⁻¹。该值比强酸强碱的标准中和焓 (-57.1 kJ mol⁻¹) 负得少，可能由于热量损失和简易量热计的局限。
Published by TutorHao | CAIE Chemistry Revision Series | aleveler.com
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