📚 Year 12 Cambridge Chemistry: Case Study Practice Drill | Year 12 Cambridge 化学：案例分析实战演练
Welcome to a focused case study practice session designed for Year 12 Cambridge AS Chemistry students. This article presents real-world chemical scenarios that integrate multiple syllabus topics. Each case demonstrates how to apply core principles—from spectra interpretation and equilibrium calculations to redox and energy changes—exactly as required in exams. Use these examples to build confidence and sharpen your analytical skills.
欢迎来到专为剑桥 AS 化学（Year 12）学生设计的案例分析实战演练。本文精选了覆盖多个核心考点的真实化学情境，每个案例都展示了如何将质谱、平衡、能量变化、电化学等原理运用于数据分析和解题。通过逐题演练，你将掌握从信息提取到步步推演的完整解题逻辑，夯实考试所需的推理能力。
1. Mass Spectrometry: Determining Molecular Formula from Fragmentation Patterns | 质谱分析：从碎片峰推测分子式
A volatile organic liquid gives a mass spectrum with a molecular ion peak at m/z 88. The M⁺ and (M+2)⁺ peaks appear in a 3:1 abundance ratio. Fragments at m/z 43 and m/z 29 are also prominent. Deduce the molecular formula and suggest a possible structure.
一种挥发性有机液体的质谱显示分子离子峰位于 m/z 88，且 M⁺ 与 (M+2)⁺ 的丰度比为 3:1；同时还观察到 m/z 43 和 m/z 29 的强碎片峰。请推断分子式并提出一种可能的结构。
The 3:1 ratio of M⁺ to (M+2)⁺ is characteristic of a compound containing one chlorine atom, since ³⁵Cl and ³⁷Cl have natural abundances of approximately 75% and 25%. The molecular mass is therefore 88 g mol⁻¹ using the ³⁵Cl isotope. Subtracting 35 for chlorine leaves 53 mass units for C, H and any other atoms. Using the nitrogen rule (even molecular mass implies zero or an even number of nitrogen atoms), we try combinations of carbon and hydrogen. C₄H₅Cl gives (4×12)+5+35 = 88 exactly, matching the data.
M⁺ 与 (M+2)⁺ 的 3:1 丰度比是含一个氯原子的化合物典型特征，因为 ³⁵Cl 和 ³⁷Cl 的天然丰度分别约为 75% 和 25%。因此以 ³⁵Cl 计分子质量为 88 g mol⁻¹。减去氯原子质量 35 后，剩余碎片质量为 53。根据氮规则（偶数质量表示分子中不含氮或含偶数个氮），可尝试碳氢组合。计算得出 C₄H₅Cl：(4×12)+5+35 = 88，与数据完全吻合。
The fragment at m/z 43 can arise from loss of 45 mass units (88 – 43 = 45), consistent with the loss of CH₃CH₂• (ethyl radical) or COOH group. The fragment at m/z 29 could correspond to C₂H₅⁺. Taken together, the data suggest 2‑chlorobutane or a branched isomer, where cleavage generates stable carbocations. Thus, a plausible structure is CH₃CH₂CHClCH₃.
m/z 43 碎片可能源自分子丢失 45 质量单位（88 – 43 = 45），对应于失去 CH₃CH₂• 或 COOH 等基团。m/z 29 的碎片可归属为 C₂H₅⁺。综合来看，谱图支持 2‑氯丁烷或其同分异构体，断裂产生稳定碳正离子，因此一种合理结构为 CH₃CH₂CHClCH₃。
2. Infrared Spectroscopy: Distinguishing Structural Isomers | 红外光谱：区分同分异构体
A compound with molecular formula C₃H₆O₂ shows a strong, broad absorption around 3000 cm⁻¹ and an intense peak at 1715 cm⁻¹. The fingerprint region is complex. Use the IR data to identify the functional group and deduce whether the compound is propanoic acid, methyl ethanoate or hydroxypropanone.
分子式为 C₃H₆O₂ 的化合物在约 3000 cm⁻¹ 处显示强而宽的吸收，在 1715 cm⁻¹ 处有一个强尖峰，指纹区复杂。请利用红外数据确定官能团，并推断该化合物是丙酸、乙酸甲酯还是羟基丙酮。
The broad absorption between 2500 and 3300 cm⁻¹ is the unmistakable signature of the O–H stretch in carboxylic acids, where extensive hydrogen bonding causes band broadening. The peak at 1715 cm⁻¹ corresponds to the carbonyl C=O stretching vibration. Together, these two absorptions confirm the presence of a –COOH group. An ester would show C=O near 1735 cm⁻¹ but no broad O–H band; a hydroxyketone would show a broad O–H but a ketone C=O usually near 1705 cm⁻¹ and lacks the simultaneous acid O–H wide envelope. The data therefore point to propanoic acid, CH₃CH₂COOH.
2500–3300 cm⁻¹ 范围内的宽吸收是羧酸中 O–H 伸缩振动的特征标志，强氢键作用使得谱带大大展宽。1715 cm⁻¹ 处的峰对应羰基 C=O 伸缩振动。这两个吸收同时出现即确认了 –COOH 的存在。若是酯，C=O 伸缩通常出现在约 1735 cm⁻¹，且无宽的 O–H 带；若是羟基酮，虽然有宽 O–H，但酮羰基 C=O 通常位于 1705 cm⁻¹ 附近，且不会出现酸中典型的宽大 O–H 包络线。因此数据指向丙酸 CH₃CH₂COOH。
In exam questions, always correlate the specific wavenumber range with the bond and the class of compound. A table of IR absorptions for O–H (alcohols/hydrogen bonded 3200–3550 cm⁻¹, acids 2500–3300 cm⁻¹), C=O (acids 1700–1725 cm⁻¹, esters 1735–1750 cm⁻¹, aldehydes/ketones 1680–1750 cm⁻¹) should be memorised.
在考试中，务必将特定波数范围与键型和化合物类别相关联。需熟记：O–H（醇/氢键 3200–3550 cm⁻¹，酸 2500–3300 cm⁻¹），C=O（酸 1700–1725 cm⁻¹，酯 1735–1750 cm⁻¹，醛/酮 1680–1750 cm⁻¹）等特征区间。
3. Equilibrium Constant Calculation: The Contact Process | 平衡常数计算：接触法制硫酸
In the Contact Process, sulfur dioxide is oxidised: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = –197 kJ mol⁻¹. At a certain temperature, an equilibrium mixture in a closed vessel contains 0.60 mol SO₂, 0.40 mol O₂ and 0.80 mol SO₃. The total pressure is 2.00 atm. Calculate the equilibrium constant Kp, stating its units.
接触法中二氧化硫被氧化：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，ΔH = –197 kJ mol⁻¹。某温度下，密闭容器内的平衡混合物含有 0.60 mol SO₂、0.40 mol O₂ 和 0.80 mol SO₃，总压为 2.00 atm。计算平衡常数 Kp 并注明单位。
First, find the total number of moles at equilibrium: n_total = 0.60 + 0.40 + 0.80 = 1.80 mol. The mole fraction of SO₂ is 0.60/1.80 = 0.333, of O₂ is 0.40/1.80 = 0.222, and of SO₃ is 0.80/1.80 = 0.444. The partial pressure of each gas = mole fraction × total pressure. Thus pSO₂ = 0.333 × 2.00 = 0.666 atm, pO₂ = 0.444 atm, pSO₃ = 0.888 atm.
先求总摩尔数：n_total = 0.60 + 0.40 + 0.80 = 1.80 mol。各组分摩尔分数为 SO₂ 0.333、O₂ 0.222、SO₃ 0.444。分压 = 摩尔分数 × 总压：pSO₂ = 0.666 atm，pO₂ = 0.444 atm，pSO₃ = 0.888 atm。
Kp expression: Kp = (pSO₃)² / [(pSO₂)² (pO₂)]. Substituting values: Kp = (0.888)² / (0.666² × 0.444) = 0.7885 / (0.4436 × 0.444) ≈ 0.7885 / 0.1970 ≈ 4.00 atm⁻¹. The unit is atm⁻¹ because the pressure terms give (atm)² / (atm² × atm) = atm⁻¹.
Kp 的表达式为 Kp = (pSO₃)² / [(pSO₂)² (pO₂)]。代入数值：Kp = (0.888)² / (0.666² × 0.444) = 0.7885 / 0.1970 ≈ 4.00 atm⁻¹。单位推导：分子 (atm)²，分母 (atm)² × atm = atm³，整体 = atm⁻¹。
4. Rate Equation and Mechanism from Initial Rates | 由初速数据推断速率方程与机理
The reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) was studied at a fixed temperature. Three experiments gave the initial rates shown below. Deduce the rate equation and the order with respect to each reactant, and suggest a rate-determining step.
在固定温度下研究反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)，三组实验的初始速率如下。请推导速率方程和各组分反应级数，并提出速控步骤。
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.50 × 10⁻³ |
| 2 | 0.100 | 0.200 | 5.00 × 10⁻³ |
| 3 | 0.200 | 0.100 | 1.00 × 10⁻² |
Comparing experiments 1 and 2, [NO] is constant while [H₂] doubles, and the rate doubles (2.50×10⁻³ to 5.00×10⁻³). Hence the order with respect to H₂ is 1. Comparing experiments 1 and 3, [H₂] is constant while [NO] doubles, and the rate increases by a factor of 4 (2.50×10⁻³ to 1.00×10⁻²). Therefore the order with respect to NO is 2. The rate equation is rate = k[NO]²[H₂].
比较实验 1 和 2，当 [NO] 恒定而 [H₂] 加倍时，速率翻倍（2.50×10⁻³ → 5.00×10⁻³），因此对 H₂ 为一级。比较实验 1 和 3，[H₂] 恒定而 [NO] 加倍，速率增至 4 倍（2.50×10⁻³ → 1.00×10⁻²），故对 NO 为二级。速率方程为 rate = k[NO]²[H₂]。
The overall order is 3. Since the stoichiometric coefficients are 2 for both reactants, the rate equation indicates that the rate-determining step must involve two NO molecules and one H₂ molecule. A plausible two‑step mechanism would be: (1) 2NO ⇌ N₂O₂ (fast equilibrium), (2) N₂O₂ + H₂ → N₂ + H₂O₂ (slow), followed by a fast step H₂O₂ + H₂ → 2H₂O. This matches the derived orders.
总反应级数为 3。虽然计量系数均为 2，但速率方程显示速控步骤必须包含 2 个 NO 分子和 1 个 H₂ 分子。合理的两步机理为：(1) 2NO ⇌ N₂O₂（快平衡），(2) N₂O₂ + H₂ → N₂ + H₂O₂（慢），后续 H₂O₂ + H₂ → 2H₂O 为快步骤，与实验级数一致。
5. Le Chatelier’s Principle: Optimising Ammonia Synthesis | 勒夏特列原理：合成氨的优化
The Haber process is central to fertiliser production: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = –92 kJ mol⁻¹. Industrial conditions are 450 °C, 200–300 atm and an iron catalyst. Explain how each condition is selected as a compromise between yield and rate, and relate your reasoning to Le Chatelier’s principle.
哈伯法生产氨是肥料工业的核心：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = –92 kJ mol⁻¹。工业条件为 450°C、200–300 atm 并使用铁催化剂。请解释各项条件如何在产率和速率间取得妥协，并用勒夏特列原理加以论述。
The forward reaction is exothermic. According to Le Chatelier’s principle, lowering the temperature would shift the equilibrium to the right, increasing the equilibrium yield of ammonia. However, a low temperature results in a very slow reaction because few molecules possess the activation energy. 450 °C is a compromise: it gives a reasonably high rate while still allowing a substantial equilibrium yield.
正向反应放热。根据勒夏特列原理，降低温度会使平衡右移，提高氨的平衡产率。但低温下反应速率极慢，因为具有足够活化能的分子较少。450 °C 是一个妥协：既能保证较快的反应速率，又能维持可观的平衡产率。
The equation shows a reduction in gas moles from 4 to 2. High pressure shifts the equilibrium towards the product side (fewer moles) and also increases the collision frequency, enhancing the rate. Pressures of 200–300 atm are chosen to maximise yield; going higher would require extremely expensive reinforced equipment and increase energy costs, offering diminishing returns.
反应使气体分子总数从 4 减为 2。高压利于平衡向产物方向移动，并提高碰撞频率，加快反应速率。采用 200–300 atm 可在产率和设备成本间获得最佳性价比；更高压力使设备造价飙升且能耗剧增，收益递减。
The iron catalyst does not alter the equilibrium position but lowers the activation energy, allowing a high rate to be achieved at the moderate temperature of 450 °C. It is typically used as a finely divided solid to maximise surface area, often promoted with potassium and aluminium oxides.
铁催化剂不改变平衡位置，但通过降低活化能，使反应在 450 °C 的中等温度下即可获得高速率。通常以细粉状使用以增大表面积，并添加钾和铝的氧化物作为助催化剂。
6. Calorimetry: Determining Enthalpy Change of Neutralisation | 量热法：测定中和焓变
In a simple calorimetry experiment, a student mixes 50.0 cm³ of 1.00 mol dm⁻³ HCl with 50.0 cm³ of 1.00 mol dm⁻³ NaOH in a polystyrene cup. The temperature rises by 6.8 °C. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ °C⁻¹ and the density is 1.00 g cm⁻³. Calculate the enthalpy change of neutralisation per mole of water formed.
在一项简单的量热实验中，学生将 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，温度上升 6.8 °C。假设溶液比热容为 4.18 J g⁻¹ °C⁻¹，密度为 1.00 g cm⁻³。计算每生成 1 mol 水的中和焓变。
Total volume = 100 cm³, so mass of solution m = 100 g. Heat released Q = mcΔT = 100 × 4.18 × 6.8 = 2842.4 J ≈ 2.84 kJ. Moles of HCl = (50.0/1000) × 1.00 = 0.0500 mol, moles of NaOH are also 0.0500 mol. The reaction is 1:1, so 0.0500 mol of water is produced. Enthalpy change
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