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

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

In the Pre-U WJEC Chemistry syllabus, the ability to apply theoretical knowledge to unfamiliar, real‑world scenarios is essential. This article presents a series of detailed case study exercises that mirror the style of examination questions, integrating multiple areas of the specification. Each case study is followed by a thorough discussion, helping you develop the analytical skills needed for top‑grade performance.

在 Pre-U WJEC 化学大纲中，将理论知识应用于陌生的真实场景的能力至关重要。本文提供一系列模拟考试风格的详细案例分析练习，整合了考纲的多个领域。每个案例后附有深入讨论，助你培养斩获高分所需的分析能力。

1. Thermodynamics in Action: Designing a Born–Haber Cycle | 热力学实践：设计玻恩–哈伯循环

A student is asked to construct a Born– ies of Ca = +590 and +1145 kJ mol⁻¹, bond dissociation enthalpy of Cl₂ = +244 kJ mol⁻¹, electron affinity of Cl = −349 kJ mol⁻¹, and lattice enthalpy of CaCl₂ = −2258 kJ mol⁻¹. Calculate the standard enthalpy of formation of CaCl₂ and comment on the stability of the ionic compound.

学生需利用以下数据为氯化钙 (CaCl₂) 构建玻恩–哈伯循环：Ca 的原子化焓 = +178 kJ mol⁻¹，Ca 的第一和第二电离能 = +590 和 +1145 kJ mol⁻¹，Cl₂ 的键离解焓 = +244 kJ mol⁻¹，Cl 的电子亲和能 = −349 kJ mol⁻¹，CaCl₂ 的晶格焓 = −2258 kJ mol⁻¹。计算 CaCl₂ 的标准生成焓，并评论该离子化合物的稳定性。

The Born–Haber cycle links the formation of an ionic solid from its elements in their standard states to the sequential steps: atomisation of metal and non‑metal, ionisation, electron affinity and lattice formation. Applying Hess’s Law, the enthalpy of formation ΔH°f can be found from the sum of these steps. The calculation is:

玻恩–哈伯循环将离子固体从其标准状态元素生成的过程与一系列步骤联系起来：金属和非金属的原子化、电离、电子亲和和晶格形成。应用赫斯定律，生成焓 ΔH°f 可由这些步骤之和求得。计算如下：

ΔH°f = ΔH°at(Ca) + IE₁(Ca) + IE₂(Ca) + ΔH°at(Cl₂) + 2×EA(Cl) + ΔH°latt

Substituting the values: ΔH°f = (+178) + (+590) + (+1145) + (+244) + 2×(−349) + (−2258) = 178 + 590 + 1145 + 244 − 698 − 2258 = −799 kJ mol⁻¹. The negative value indicates that the formation of CaCl₂ is highly exothermic, and the large lattice enthalpy dominates, making the compound thermodynamically stable with respect to the elements. The Born– e can be represented diagrammatically, allowing visual assessment of energy contributions.

注意需要两个氯原子，因此 Cl₂ 的键离解焓提供两个 Cl 原子，电子亲和能计算两次。代入数值： +1145) + (+244) + 2×(−349) + (−2258) = 178 + 590 + 1145 + 244 − 698 − 2258 = −799 kJ mol⁻¹。负值表明 CaCl₂ 的生成强烈放热，巨大的晶格焓占主导地位，使该化合物相对于其元素在热力学上稳定。玻恩–哈伯循环可用图示表达，便于直观评估能量贡献。

2. Equilibrium and Le Chatelier’s Principle: The Haber Process | 平衡与勒夏特列原理：哈伯法

An industrial chemist examines the Haber process: N₂(g) + 3H₂(g) ⇌ perates at 450 °C and 200 atm, using an iron catalyst. Explain, using thermodynamic and kinetic principles, why these conditions are chosen rather than room temperature and atmospheric pressure, even though the equilibrium yield of ammonia would be higher at low temperature and high pressure.

工业化学家考察哈伯法：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = −92 kJ mol⁻¹。该工厂在 450 °C 和 200 atm 下运行，使用铁催化剂。请用热力学和动力学原理解释为何选择这些条件，而不是室温和常压，尽管低温高压下氨的平衡产率会更高。

igh pressure. However, at room temperature the rate of reaction is negligible because the activation energy for the N≡ viable rate; 450 °C provides a compromise between rate and equilibrium yield. Operating at 200 atm shifts the equilibrium towards the product side, increasing yield, but very high pressures would raise energy costs and require thicker, more expensive reaction vessels. The catalyst does not affect the position of equilibrium, only the speed at which it is attained. Furthermore, ammonia is continuously removed, driving the equilibrium forward. Le Chatelier’s principle predicts that the system will adjust to partially counteract a change: increasing temperature favours the endothermic reverse reaction, but the chosen temperature is a balance, and the high pressure partially compensates. The unreacted N₂ and H₂ are recycled, increasing overall efficiency.

乍看之下，放热正反应受低温促进，气体总摩尔数减少（4 mol → 2 mol）受高压有利。但室温下，由于 N≡N 三键断裂的活化能极高，反应速率微乎其微。铁催化剂降低活化能，但仍需较高的温度以实现经济可行的速率；450 °C 是速率与平衡产率间的折衷。200 atm 高压使平衡向产物方向移动，提高产率，但极高的压力会增加能源成本，并需要更厚、更贵的反应容器。催化剂不影响平衡位置，只影响达到平衡的速度。此外，氨被连续分离移走，驱动平衡正向移动。勒夏特列原理预测系统将部分抵消变化：升温有利于吸热的逆反应，但所选温度是权衡的结果，高压在一定程度上予以补偿。未反应的 N₂ 和 H₂ 循环使用，提高了整体效率。

3. Kinetics: Unravelling a Rate Equation from Initial‑Rate Data | 动力学：由初始速率数据推导速率方程

ons: S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. The following data are collected at constant temperature:

研究小组研究过氧二硫酸根离子与碘离子的反应：S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂。在恒定温度下收集到如下数据：

Experiment	[S₂O₈²⁻] / mol dm⁻³	[I⁻] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	1.6 × 10⁻⁴
2	0.20	0.10	3.2 × 10⁻⁴
3	0.10	0.20	3.2 × 10⁻⁴

Determine the rate equation and calculate the rate constant, giving its units. Suggest a possible mechanism consistent with the orders.

确定速率方程并计算速率常数，注明其单位。提出一个与级数一致的可能的反应机理。

riments 1 and 3, doubling [I⁻] while keeping [S₂O₈²⁻] constant also doubles the rate → first order with respect to I⁻. Overall order is 2. The rate equation is: Rate = k [S₂O₈²⁻][I⁻]. Using data from experiment 1: 1.6 × 10⁻⁴ = k × 0.10 × 0.10, so k = 1.6 × 10⁻² mol⁻¹ dm³ s⁻¹. The units confirm second order overall. A plausible mechanism involves a slow bimolecular rate‑ stoichiometry and the determined orders, as only one ion of each kind appears in the rate‑determining step.

对比实验1和2，保持 [I⁻] 不变而将 [S₂O₈²⁻] 加倍，速率加倍 → 对 S₂O₈²⁻ 为一级。对比实验1和3，保持 [S₂O₈²⁻] 不变而将 [I⁻] 加倍，速率也加倍 → 对 I⁻ 为一级。总级数为2。速率方程为：Rate = k [S₂O₈²⁻][I⁻]。代入实验1数据：1.6 × 10⁻⁴ = k × 0.10 × 0.10，得 k = 1.6 × 10⁻² mol⁻¹ dm³ s⁻¹。单位证实总级数为二级。可能的机理包含一个缓慢的双分子决速步：S₂O₈²⁻ + I⁻ → SO₄²⁻ + SO₄I⁻，随后中间体与另一个 I⁻ 发生快速反应。这符合化学计量比和确定的反应级数，因为决速步中每种离子都只涉及一个。

4. Organic Synthesis: Devising a Multi‑step Route | 有机合成：设计多步合成路线

Propose a synthesis of ethyl 3‑oxobutanoate starting from ethanol and any inorganic reagents. Show each step with reagents, conditions and the type of reaction. Highlight the key Claisen condensation step and explain why the desired product forms rather than self‑condensation of ethyl acetate.

从乙醇和任何无机试剂出发，提出一条合成 3-氧代丁酸乙酯的路线。写出每步的试剂、条件和反应类型。突出关键的克莱森缩合步骤，并解释为何目标产物形成而非乙酸乙酯自身缩合。

The retrosynthetic analysis shows that ethyl 3‑te can be made by a Claisen condensation of ethyl acetate with itself, but the direct self‑ thanoic acid, then converts it to acetyl chloride, and uses a mixed Claisen condensation between acetone and ethyl acetate or between two different esters. One efficient pathway: (i) ethanol is oxidised with acidified K₂Cr₂O₇ to ethanoic acid, (ii) ethanoic acid is converted to ethyl ethanoate by esterification with ethanol and H⁺ catalyst, (iii) a portion of ethyl ethanoate is reduced to ethanol and then oxidised to ethanal; alternatively, prepare acetone from propanone? Wait, 3‑ ained from ethanol via oxidation to ethanoic acid, then calcium salt pyrolysis to propanone). Simplified school route: (a) ethanol + PCl₅ → chloroethane; then (b) chloroethane + KCN → propanenitrile; (c) hydrolysis to propanoic acid; (d) oxidation of ethanol to ethanoic acid; (e) esterification to ethyl ethanoate; (f) Claisen condensation between ethyl propanoate and ethyl ethanoate? Actually, ethyl 3‑ sen condensation of two molecules of ethyl acetate under sodium ethoxide. The product is β‑keto ester. The reason it works is that the α‑ thyl acetate are acidic enough; the enolate attacks another ester molecule, and after work‑up, the β‑keto ester is favoured because its α‑ re more acidic, leading to formation of a stable enolate that drives the equilibrium. Self‑tion is exactly what is desired, but careful control of base and temperature prevents over‑ cetate is treated with sodium ethoxide in ethanol, heated, then acidified to yield ethyl 3‑ ilised by intramolecular hydrogen bonding in the enol form, contributing to its formation.

逆合成分析显示，3-氧代丁酸乙酯可由乙酸乙酯自身克莱森缩合制得，但直接缩合需强碱且常得到混合物。另一条更可控的工业路线先将乙醇氧化为乙酸，然后转化为乙酰氯，再进行混合克莱森缩合。一条可行路径：(i) 乙醇经酸化 K₂Cr₂O₇ 氧化为乙酸，(ii) 乙酸与乙醇在 H⁺催化下酯化生成乙酸乙酯，(iii) 一部分乙酸乙酯还原再氧化得到乙醛；或由乙醇制丙酮。简化校园路线：典型方法是乙酸乙酯在乙醇钠/乙醇溶液中加热，然后酸化得到3-氧代丁酸乙酯。该反应成功的关键在于两个羰基间的 α-氢酸性更强，形成的烯醇负离子稳定，拉动平衡。这就是目标产物而非其他副产物的原因——产品烯醇式存在分子内氢键，增强了稳定性。

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5. Spectroscopic Structure Elucidation: Combining NMR and IR | 波谱解析：联合 NMR 和 IR 推断结构

An unknown compound C₄H₈O₂ shows the following spectra: IR absorption at 1740 cm⁻¹ and a broad band near 3000 cm⁻¹; the ¹³C NMR spectrum has four peaks at δ 14, 22, 60 and 171 ppm; the ¹H NMR spectrum (δ ppm): 1.2 (t, 3H), 2.3 (q, 2H), 4.1 (q, 2H), and 11.0 (s, 1H, exchanges with D₂O). Deduce the structure and assign the spectral data.

未知物 C₄H₈O₂ 显示如下波谱：IR 吸收在 1740 cm⁻¹ 和 3000 cm⁻¹ 附近宽峰；¹³C NMR 有四个峰：δ 14, 22, 60, 171 ppm；¹H NMR (δ ppm)：1.2 (t, 3H), 2.3 (q, 2H), 4.1 (q, 2H) 和 11.0 (s, 1H, 与 D₂O 交换消失)。推导结构并归属波谱数据。

. The broad band near 3000 cm⁻¹ could be O−H of a carboxylic acid, but ester C−O stretching would also appear. The ¹³C peak at 171 ppm is typical of a carboxylic acid or ester carbonyl. The presence of an exchangeable ¹H signal at 11.0 ppm confirms a carboxylic acid proton. The ¹H NMR shows a triplet at 1.2 (3H) coupled to a quartet at 4.1 (2H), implying an ethyl group CH₃CH₂− attached to an electronegative atom, oxygen. The quartet at 2.3 (2H) coupled to a triplet which is the methyl? a CH₃. The triplet at 1.2 is the CH₃, and the quartet at 2.3 is the CH₂. Therefore we have an ethyl group on one side and another CH₃CH₂−? t that would give a different coupling pattern. Let’s examine: CH₃CH₂CH₂COOH would have CH₃ next to CH₂ next to CH₂ next to COOH. ¹H: CH₃ triplet at ~0.9, CH₂ (next to CH₃) multiplet, CH₂ (next to COOH) triplet at ~ , that would be an ethyl ester, but the exchangeable proton suggests acid. Ethyl ethanoate? , a quartet (CH₂), and the acid proton, consistent with the ¹H NMR: 1.2 (t, CH₃), 2.3 (q, CH₂), 11.0 (s, COOH). However, the formula has an extra oxygen? tet and another triplet, plus the acid proton. That does not match. Alternatively, it could be methyl propanoate, but no O−H exchange. So the spectra must be reinterpreted. Perhaps the broad IR band is not O−H but C−H? The ¹H signal at 11.0 is undeniable for carboxylic acid. What about HO−CH₂CH₂CH₂CHO? ehyde. The quartet at 4.1 is downfield, typical of CH₂ adjacent to oxygen, as in an ester: –CH₂–O–CO–. If it is an ester, the exchangeable proton could be from an alcohol impurity? ing C₄H₈O₂ with a COOH group: the remaining fragment is C₃H₇. An ethyl group attached to a –COOH would give propanoic acid, but that’s C₃. To have C₄, the acid could be 2‑methylpropanoic acid: (CH₃)₂CHCOOH. That would give: ¹H NMR: CH₃ doublet at ~1.2 (6H), CH septet at ~ ton. But we have two quartets, not a septet. Therefore the quartet at 4.1 must belong to a –CH₂– linked to oxygen and also to a methyl group, as in –CH₂–CH₃. The quartet at 2.3 is a CH₂ next to carbonyl and also next to a CH₃? ouring protons, so there must be a methyl attached to the CH₂ at 2.3. That would be CH₃CH₂–C(=O)–. Then we have a fragment CH₃CH₂– attached to carbonyl, leaving a CH₂? No, the rest is just –O– noate, but no acid proton. Unless the exchangeable proton is from a carboxylic acid with a –CH₂–O–? Look: 3‑hydroxybutanoic acid? CH₃CH(OH)CH₂COOH? er, the presence of an exchangeable proton at 11.0 is definitive for a carboxylic acid. Re‑examine the quartet at 4.1: that is typical of a –COO–CH₂–CH₃ ester group. In a compound that also has a COOH, we could have a half‑ester of a dicarboxylic acid, e.g. monoethyl propanedioate? But that is C₅. For C₄H₈O₂, the only acid with an ethyl ester group would be HOOC–CH₂– ill C₃. Wait: C₄H₈O₂ could be butyric acid, but its NMR is different. Actually, consider 2‑noic acid: CH₃CH₂OCH₂COOH. This has an ether and a carboxylic acid. ¹H NMR: CH₃ triplet at ~1.2, CH₂ quartet at ~3.5, CH₂ singlet at ~4.1? kably an ethyl group attached to an oxygen. The quartet at 2.3 coupled to a triplet at 1.2? triplet at 1.2, and there are two quartets: 2.3 (2H) and 4.1 (2H). This suggests two different CH₂ groups, each coupled to a CH₃? s overlapping, but the table shows only one triplet. Possibly the compound is CH₃CH₂COOCH₃? hat. Perhaps the 11.0 is due to a trace impurity, or maybe we have a different functional group: an acid with an ethyl group on the α‑carbon? 2‑ethylpropanoic acid? rbonyl, is ethoxyacetic acid: CH₃CH₂OCH₂COOH. In that case, the CH₂ attached to both oxygen and COOH would be a singlet, not a quartet. The given spectrum shows a quartet at 4.1, which means that CH₂ is coupled to three protons – so it must be an ethyl group –O–CH₂– o something else; that CH₂ is attached to the carbonyl and to a methyl, giving CH₃CH₂C(=O)– changeable proton. The ¹H NMR would show a triplet (1.2), a quartet (2.3) for the CH₃CH₂CO–, and another triplet (~1.3) and quartet (4.1) for the – gration could be 6H if the triplet accounts for both methyl groups, but the integration is given as 3H. So that’s not consistent. Given the confusion, it is more likely that the question expects 2‑methylpropanoic acid? Let’s assign the data as presented in typical pre‑U exercises: often they give a spectrum of an acid anhydride? erpretation: the quartet at 2.3 and triplet at 1.2 belong to a propionyl group CH₃CH₂C(=O)–; the quartet at 4.1 and another triplet (maybe hidden?) Actually, let’s re‑ But formula is C₄H₈O₂, not C₃. So there is an extra CH₂. If we have butanoic acid, the spectrum would be CH₃ (t, 3H) at ~0.9, CH₂ (sextet, 2H) at ~1.6, CH₂ (t, 2H) at ~ er possibility: the compound is CH₃CH₂OCH₂COOH, but the CH₂ next to ether and acid is a singlet, not a quartet. So the spectroscopic data must be re‑ haps the quartet at 4.1 is actually a CH₂ attached to an oxygen and also to a methyl group? oup. The presence of a carboxylic acid proton could be explained if the compound is a mono‑ester of a dicarboxylic acid such as ethyl hydrogen succinate? That’s C₆. For C₄, ethyl hydrogen malonate? roton is erroneous. To stay academically sound, I will adjust the unknown to be methyl propanoate: C₄H₈O₂, no exchangeable proton, IR shows C=O and C−O. Then assign: CH₃CH₂COOCH₃. ¹H NMR: triplet at 1.2 (CH₃CH₂), quartet at 2.3 (CH₂CO), singlet at 3.6 (OCH₃). The given has a second quartet at 4.1, which would be the OCH₃ if it were a quartet, but OCH₃ is singlet. Therefore, the data exactly matches ethyl ethanoate: CH₃COOCH₂CH₃, giving triplet 1.2 (CH₃ of ethyl), quartet 4.1 (CH₂ of ethyl), singlet 2.0 (CH₃CO). But the given has a quartet at 2.3, not a singlet. So it’s ethyl propanoate! t 1.2 might be 6H if the two CH₃ are isochronous. That is a common twist: the ¹H NMR shows a triplet integrating for 6H, but the student must realise it represents two methyl groups. Then the two quartets at 2.3 and 4.1 each integrate for 2H. Thus the structure is CH₃CH₂C(=O)OCH₂CH₃. And the exchangeable proton? haps it’s a red herring or due to a solvent impurity, or the compound exists as a tautomer? ; the acidic proton could be from a trace of water. However, for clarity, I’ll change the unknown to C₅H₁₀O₂ or I’ll craft a new one. To avoid confusion, I will replace the case study with a simpler one: an ester C₄H₈O₂, where NMR shows a triplet (3H) at 1.2, a singlet (3H) at 2.0, and a quartet (2H) at 4.1, IR at 1740, no O−H. Then deduce ethyl ethanoate. That is classic and fits. So I’ll rewrite that part. Since HTML has already been output, but I’m generating the whole article from scratch, I can change it. I’ll present the case as ethyl ethanoate. So the unknown is an ester, IR 1740 cm⁻¹, C−O stretches; ¹H NMR: 1.2 (t, 3H), 2.0 (s, 3H), 4.1 (q, 2H). Deduce CH₃COOCH₂CH₃. Then ¹³C: 14, 21, 60, 171. That works perfectly. So I’ll replace that. No exchangeable proton. Thus the exercise is consistent. I will rewrite accordingly. I’ll edit the article to reflect this correct data. To ensure the article is high-quality, I will now produce the final version with the correct spectroscopy case, and remove the problematic one. My output will be the complete HTML. I' w compose the entire article from the beginning to end, ensuring all cases are accurate. I' e 8 sections as above, but adjust the spectroscopy section to a straightforward example. I' ion, etc. Use proper Unicode.TITLE: Case Study Practical Exercises in Pre-U WJEC Chemistry | Pre-U WJEC 化学：案例分析实战演练