📚 Year 13 WJEC Chemistry: Case Study Practical Drills | 13年级 WJEC 化学：案例分析实战演练
Mastering Year 13 Chemistry goes beyond memorising facts; it requires the ability to apply concepts to real-world scenarios. WJEC examinations increasingly demand that students tackle multi-step problems drawn from industrial processes, biological systems, and analytical challenges. This article presents ten carefully constructed case studies that mirror the depth and style of assessment material, each designed to strengthen your problem-solving toolkit. Work through them systematically, checking your reasoning against the paired explanations.
掌握 13 年级化学不只是记忆知识点，更需要将概念应用到真实世界的情境中。WJEC 考试越来越要求学生解决来自工业流程、生物体系和分析挑战的多步骤问题。本文带来十个精心构建的案例研究，它们模拟了考试材料的深度与风格，旨在强化你的解题工具箱。请系统性地逐一完成，并用配对的中文讲解检验你的推理过程。
1. Arrhenius Equation and Drug Shelf Life | 案例一：阿伦尼乌斯方程与药物保质期
The degradation of aspirin in tablets follows first‑ tion energy Eₐ for the hydrolysis reaction is 75 kJ mol⁻¹. A shelf life is defined as the time for 10% decomposition. Using the Arrhenius equation in its two‑ mine the shelf life at 35 °C and explain why the drug must be stored in a cool, dry place.
阿司匹林药片的降解遵循一级动力学。25 °C 时，速率常数 k₁ = 2.0 × 10⁻³ 天⁻¹。水解反应的活化能 Eₐ 为 75 kJ mol⁻¹。保质期定义为药物分解 10% 所需的时间。利用两点式阿伦尼乌斯方程，求 35 °C 时的保质期，并解释药品为何需在阴凉干燥处保存。
The two‑point form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ – .31 J mol⁻¹ K⁻¹, T₁ = 298 K, T₂ = 308 K. Plugging in: ln(k₂/2.0×10⁻³) = (75000/8.31)(1/298 – 1/308) = 9025 × (0.003356 – 47) = 9025 × 0.000109 ≈ 0.983. Thus k₂ = 2.0×10⁻³ × e⁰·⁹⁸³ ≈ 5.3 × 10⁻³ day⁻¹. For a first‑order reaction, t = ln([A]₀/[A])/k. When 10% has reacted, [A]/[A]₀ = 0.9. tshelf₃₅ = ln(1/0.9) / 5.3×10⁻³ = 0.1054 / 0.0053 ≈ 19.9 days. At 25 °C, tshelf he shelf life halves with a 10 °C rise, demonstrating the strong temperature dependence. Storing at low temperature reduces k dramatically; humidity accelerates acid‑catalysed hydrolysis, so a dry environment is essential.
两点公式：ln(k₂/k₁) = (Eₐ/R)(1/T₁ – 1/T₂)。取 R = 8.31 J mol⁻¹ K⁻¹，T₁ = 298 K，T₂ = 308 K。代入得：ln(k₂/2.0×10⁻³) = (75000/8.31)(1/298 – 1/308) = 9025 × (0.003356 – 0.003247) = 9025 × 0.000109 ≈ 0.983。于是 k₂ = 2.0×10⁻³ × e⁰·⁹⁸³ ≈ 5.3 × 10⁻³ 天⁻¹。对一级反应，t = ln([A]₀/[A])/k。当反应掉 10% 时，[A]/[A]₀ = 0.9。t_保质期₃₅ = ln(1/0.9) / 5.3×10⁻³ = 0.1054 / 0.0053 ≈ 19.9 天。25 °C 时，t_保质期₂₅ = 52.7 天。温度升高 10 °C，保质期几乎减半，表明强烈的温度依赖性。低温储存可大幅降低 k；湿气会加速酸催化水解，因此干燥环境至关重要。
2. Haber Process: Optimising Equilibrium Yield | 案例二：哈伯法——平衡产率的优化
The synthesis of ammonia is an exothermic equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹. A plant operates at 400 °C and 200 atm. The actual yield is about 40% ions and use Kₚ expressions to calculate the equilibrium mole fraction of NH₃ given that Kₚ = 4.5 × 10⁻⁵ atm⁻² at 400 °C.
合成氨是一个放热可逆反应：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹。某工厂在 400 °C 和 200 atm 下操作，单程产率约 40%。勒夏特列原理预测低温和高压有利于正反应。请解释条件选择的妥协策略，并利用 Kₚ 表达式计算 400 °C 时 NH₃ 的平衡摩尔分数，已知 Kₚ = 4.5 × 10⁻⁵ atm⁻²。
alyst at moderate 400 °C. High pressure also shifts equilibrium towards fewer gas moles (2 versus 4), improving yield; 200 atm is a balance between yield gains and plant costs. Let the initial mole ratio be 1 N₂ : 3 H₂, total initial moles = 4. Let x moles of N₂ react. At equilibrium: N₂ = (1–x), H₂ = 3(1–x), NH₃ = 2x. Total moles = 4 – 2x. Partial pressures: pN₂ = (1–x)/(4–2x) × P, pH₂ = 3(1–x)/(4–2x) × P, pNH₃ = 2x/(4–2x) × P. Kₚ = (pNH₃²) / (pN₂ × pH₂³) = [ (2x/(4–2x))² P² ] / [ (1–x)/(4–2x) P × (3(1–x)/(4–2x))³ P³ ] = [4x² P² / (4–2x)²] × [ (4–2x)⁴ / (27(1–x)⁴ P⁴) ] = (4x² (4–2x)² P²) / (27(1–x)⁴ P⁴) = (4x² (4–2x)²) / (27(1– Kₚ = 4.5×10⁻⁵, P=200. After trial, x ≈ 0.66 gives Kₚ ≈ 4.5×10⁻⁵. Mole fraction NH₃ = 2x/(4–2x) = 1.32/2.68 ≈ 0.49, or 49%. In practice, incomplete conversion and separator efficiency lower the single‑pass yield to around 40%, showing good agreement.
根据勒夏特列原理：降低温度使平衡右移，Kₚ 增大，但低温反应太慢，经济上需要借助铁催化剂在适中 400 °C 下进行。高压也有利于生成气体摩尔数少的一方（2 对 4），提高产率；200 atm 是在增产与设备成本之间的平衡。设初始摩尔比 N₂:H₂ = 1:3，总初摩尔数为 4。令 x 摩尔 N₂ 反应。平衡时：N₂ = (1–x)，H₂ = 3(1–x)，NH₃ = 2x。总摩尔数 = 4–2x。分压：pN₂ = (1–x)/(4–2x)×P，pH₂ = 3(1–x)/(4–2x)×P，pNH₃ = 2x/(4–2x)×P。Kₚ = (pNH₃²) / (pN₂ pH₂³) = [ (2x/(4–2x))² P² ] / [ (1–x)/(4–2x) P × (3(1–x)/(4–2x))³ P³ ] = (4x² (4–2x)²) / (27(1–x)⁴ P²)。代入 Kₚ = 4.5×10⁻⁵，P=200。试算得到 x ≈ 0.66，此时 Kₚ ≈ 4.5×10⁻⁵。NH₃ 摩尔分数 = 2x/(4–2x) = 1.32/2.68 ≈ 0.49，即 49%。实际单程产率约 40%，系因未完全转化及分离效率有限，两者吻合良好。
3. Blood Buffer and Henderson–Hasselbalch | 案例三：血液缓冲与 Henderson–Hasselbalch 方程
Human blood pH is maintained at 7.40 ± 0.05 largely by the carbonic acid–hydrogencarbonate buffer. The relevant equilibrium is H₂CO₃ ⇌ abolic acidosis adds 0.5 mmol H⁺ per litre of blood, assuming initial buffering only by this system.
人体血液 pH 维持在 7.40 ± 0.05，主要依靠碳酸–碳酸氢盐缓冲对。相关平衡为 H₂CO₃ ⇌ H⁺ + HCO₃⁻，体温下 pKₐ ≈ 6.1。已知 [HCO₃⁻] = 24 mmol dm⁻³，计算维持 pH 7.40 所需的 [H₂CO₃]。再假设代谢性酸中毒向每升血液加入 0.5 mmol H⁺，且仅由此缓冲系统缓冲，评估 pH 变化。
Henderson– 0.0 → [H₂CO₃] = 1.2 mmol dm⁻³. Adding 0.5 mmol H⁺ per litre: H⁺ reacts with HCO₃⁻ forming H₂CO₃, so [HCO₃⁻] drops to 23.5 mmol dm⁻³ and [H₂CO₃] rises to 1.7 mmol dm⁻³. New pH = 6.1 + log(23.5/1.7) → log(13.82) ≈ 1.14 → pH ≈ 7.24. The drop to 7.24 remains within survivable limits, illustrating the buffer’s capacity, but further acid load would exhaust it, causing acidosis. Clinically, respiration adjusts CO₂ to modulate [H₂CO₃].
利用 Henderson–Hasselbalch 公式：pH = pKₐ + log([HCO₃⁻]/[H₂CO₃])。代入得 24/[H₂CO₃]) → log(24/[H₂CO₃]) = 1.30 → 24/[H₂CO₃] = 10¹·³ ≈ 20.0 → [H₂CO₃] = 1.2 mmol dm⁻³。加入 0.5 mmol H⁺/L 后，H⁺ 与 HCO₃⁻ 反应生成 H₂CO₃，[HCO₃⁻] 降至 23.5 mmol dm⁻³，[H₂CO₃] 升至 1.7 mmol dm⁻³。新 pH = 6.1 + log(23.5/1.7) → log(13.82) ≈ 1.14 → pH ≈ 7.24。降至 7.24 仍在可存活范围内，体现了缓冲能力，但持续的酸负荷会耗尽缓冲，导致酸中毒。临床上通过呼吸调节 CO₂ 来调控 [H₂CO₃]。
4. Gibbs Free Energy and Spontaneity of Aluminium Smelting | 案例四：吉布斯自由能与铝冶炼的自发性
erature), ΔH = +2170 kJ mol⁻¹ and ΔS = +635 J mol⁻¹ K⁻¹. Calculate ΔG at 1000 °C and comment on why the process is non‑ itions yet proceeds in electrolysis. Discuss the role of an external electrical potential.
用碳阳极从 Al₂O₃ 中提取铝可表示为：2Al₂O₃(s) + 3C(s) → 4Al(s) + 3CO₂(g)。在典型电解槽温度 1000 °C 下，ΔH = +2170 kJ mol⁻¹，ΔS = +635 J mol⁻¹ K⁻¹。计算 1000 °C 时的 ΔG，并说明为何标准状态下该反应非自发，却在电解中能够进行。讨论外加电势的作用。
ΔG = ΔH – TΔS. T = 1273 K, ΔS must be in kJ: 0.635 kJ mol⁻¹ K⁻¹. ΔG = +2170 – (1273 × 0.635) = +2170 – 808 = +1362 kJ mol⁻¹. Since ΔG > 0, the reaction is thermodynamically non‑spontaneous. In the Hall‑Hé rical power supply drives the reaction by providing electrical work. The relationship ΔG = – e minimum voltage required: 1362 kJ is per reaction as written, which involves n electrons? The half‑reaction: Al³⁺ + 3e⁻ → Al, and for 4Al, n = 12 mol electrons. Thus E° = –ΔG/(nF) = –1362 × 10³ J / (12 × 96485 C) ≈ –1.18 V. So an applied voltage exceeding +1.18 V (plus overpotentials) makes the non‑ . The positive entropy change from solid to gas helps reduce ΔG but is insufficient alone.
ΔG = ΔH – TΔS。T = 1273 K，ΔS 需用 kJ 单位：0.635 kJ mol⁻¹ K⁻¹。ΔG = +2170 – (1273 × 0.635) = +2170 – 808 = +1362 kJ mol⁻¹。因为 ΔG > 0，该反应热力学非自发。在 Hall‑Héroult 电解槽中，外部电源通过提供电功驱动反应。由 ΔG = –nFE°cell 可计算最小所需电压：对于方程中的 4 mol Al，转移电子数 n = 12 mol。E° = –ΔG/(nF) = –1362×10³ J / (12×96485 C) ≈ –1.18 V。因此外加电压超过 +1.18 V（加上过电位）即可使非自发反应发生。固相到气相的正熵变有助于降低 ΔG，但本身并不足以逆转自发性。
5. Lithium‑Ion Cell Electrochemistry | 案例五：锂离子电池的电化学
A modern lithium‑ion cell uses graphite anode and LiCoO₂ cathode. During discharge, Li⁺ ions de‑ate from graphite and intercalate into the cobalt oxide. The cell reaction is: LiC₆ + CoO₂ ⇌ C₆ + LiCoO₂. Given standard electrode potentials: Li⁺/Li = – ulate the standard cell emf. Explain why the actual voltage is higher than the simple half‑cell difference, and discuss safety risks due to overcharging.
现代锂离子电池采用石墨负极和 LiCoO₂ 正极。放电时，Li⁺ 从石墨中脱嵌并嵌入钴氧化物。电池反应为：LiC₆ + CoO₂ ⇌ C₆ + LiCoO₂。已知标准电极电势：Li⁺/Li = –3.04 V，而 LiCoO₂/CoO₂ 的 E° 约简化取 +1.0 V。计算标准电池电动势。解释为何实际电压高于简单半电池电势差，并讨论过充电引发的安全风险。
E°cell = E°cathode – Li⁺ + e⁻, which is the reverse of Li⁺ + e⁻ → Li(s) on graphite, with a potential close to –3.0 V vs Li/Li⁺. Thus E°cell = +1.0 – (–3.0) = +4.0 V. However, typical commercial cells deliver 3.6– d pure Li. During overcharging, Li⁺ can deposit as metallic lithium dendrites, which grow through the separator, causing short circuits and thermal runaway. The electrolyte is also oxidised at high voltage, releasing gases and potentially leading to fire.
电池标准电动势 E°cell = E°正极 – E°负极。正极还原：CoO₂ + Li⁺ + e⁻ → LiCoO₂，E° ≈ +1.0 V；负极氧化为 LiC₆ → C₆ + Li⁺ + e⁻，相当于 Li⁺/Li 沉积的逆过程，其在石墨中的电势约 –3.0 V（对 Li/Li⁺）。故 E°cell = +1.0 – (–3.0) = +4.0 V。但市售电池实际电压为 3.6–3.7 V，因为嵌入材料并不处于标准态；LiC₆ 比纯锂电位更正，从而降低了端电压。过充电时，Li⁺ 可能以金属锂枝晶形式析出，刺穿隔膜导致短路和热失控。高电压下电解液也会氧化，释放气体，可能引发起火。
6. Transition Metal Catalysis: Catalytic Converter | 案例六：过渡金属催化——汽车催化转化器
A three‑ O₂(g) + N₂(g). The uncatalysed activation energy is 250 kJ mol⁻¹; with Rh catalyst it drops to 120 kJ mol⁻¹. Using the Arrhenius factor change, estimate how many times faster the catalysed reaction is at 500 K. Also write the elementary steps involving NO dissociation and CO oxidation on the metal surface.
三效催化转化器利用 Pt、Pd 和 Rh 同时去除 CO、未燃碳氢化合物和 NOₓ。考虑反应：2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)。无催化剂时活化能为 250 kJ mol⁻¹，有 Rh 催化剂时降至 120 kJ mol⁻¹。利用阿伦尼乌斯因子变化，估算 500 K 时催化反应比非催化快多少倍。并写出所涉及的 NO 解离和 CO 在金属表面氧化的基元步骤。
Rate ratio = exp[(Euncat – Ecat)/RT] = exp[(250000– exp(130000/4155) = exp(31.3) ≈ 3.3 × 10¹³. This enormous acceleration makes conversion >99% en at moderate temperatures. Elementary steps on Rh: (1) NO(g) adsorbs and dissociates: NO ⇌ Nads + Oads. (2) CO adsorbs: CO(g) ⇌ CO_ads. (3) Surface reaction: CO_ads + Oads ads → N₂(g). The metal provides a pathway of much lower Eₐ via adsorption weakening bonds.
速率比 = exp[(E_无催 – E_催)/RT] = exp[(250000–120000)/(8.31×500)] = exp(130000/4155) = exp(31.3) ≈ 3.3 × 10¹³。这一巨大加速使在中等温度下数毫秒内实现 >99% 的转化成为可能。在 Rh 上的基元步骤：(1) NO 吸附并解离：NO ⇌ N_吸 + O_吸。(2) CO 吸附：CO(g) ⇌ CO_吸。(3) 表面反应：CO_吸 + O_吸 → CO₂(g) + 2 个空位。(4) N 原子结合：2N_吸 → N₂(g)。金属通过吸附削弱化学键，提供了活化能低得多的路径。
7. Organic Synthesis Route Design: Paracetamol | 案例七：有机合成路线设计——对乙酰氨基酚（扑热息痛）
Paracetamol (4‑hydroxyacetanilide) can be synthesised from phenol. Outline a three‑step route: nitration, reduction, and acetylation. For each step, give reagents, conditions, and equations. Discuss why the nitration must be carefully controlled, and how you would confirm the product identity using a combination of IR and ¹H NMR spectroscopy.
对乙酰氨基酚（4-羟基乙酰苯胺）可由苯酚出发合成。设计一条三步路线：硝化、还原、乙酰化。分别给出每步的试剂、条件和化学方程式。讨论为何硝化步骤须严格控制，以及如何联用 IR 和 ¹H NMR 确证产物结构。
Step 1: Nitration. Phenol is treated with dilute HNO₃ at low temperature (ca. 0–5 °C) because phenol is highly activated and concentrated acid/heat would cause poly‑nitration and oxidation. Product: a mixture of ortho‑ and para‑nitrophenol; para‑mer is separated by steam distillation or recrystallisation. Equation: C₆H₅OH + HNO₃ → O₂N–C₆H₄–OH (para). Step 2: Reduction. Para‑nol is reduced with tin (Sn) and concentrated HCl, then neutralised with NaOH to give para‑aminophenol. Alternatively, catalytic hydrogenation with H₂/Pd‑C. Equation: O₂N–C₆H₄–OH + 6[H] → H₂N–C₆H₄–OH + 2H₂O. Step 3: Acetylation. React para‑ th acetic anhydride (excess) under mild heating (40 °C) to form paracetamol. Equation: H₂N–C₆H₄–OH + (CH₃CO)₂O → CH₃CONH–C₆H₄–OH + CH₃COOH. IR confirmation: broad O–H stretch near 3300–3500 cm⁻¹, amide N–H around 3300 cm⁻¹, C=O stretch (amide I) at ~1650 cm⁻¹, aromatic C=C around 1600, 1500 cm⁻¹. ¹H NMR: singlet for CH₃–CO at δ 2.1 (3H), broad singlet for NH and OH around δ 9–10 (exchangeable with D₂O), two doublets for symmetrical aromatic protons (AA'BB' pattern) at δ 6.8 and 7.4 (each 2H).
第一步：硝化。苯酚用稀硝酸在低温（约 0–5 °C）处理，因为苯酚高度活化，浓酸/加热会导致多硝化及氧化。产物为邻和对硝基苯酚混合物；对位异构体可通过水蒸气蒸馏或重结晶分离。方程式：C₆H₅OH + HNO₃ → O₂N–C₆H₄–OH（对位）。第二步：还原。对硝基苯酚与锡（Sn）和浓盐酸加热还原，再用 NaOH 中和得到对氨基苯酚。也可用 H₂/Pd‑C 催化氢化。方程式：O₂N–C₆H₄–OH + 6[H] → H₂N–C₆H₄–OH + 2H₂O。第三步：乙酰化。对氨基苯酚与过量乙酸酐微热（40 °C）反应生成扑热息痛。方程式：H₂N–C₆H₄–OH + (CH₃CO)₂O → CH₃CONH–C₆H₄–OH + CH₃COOH。IR 确证：约 3300–3500 cm⁻¹ 的宽 O–H 伸缩振动，酰胺 N–H 约 3300 cm⁻¹，酰胺 I 带 C=O 伸缩振动约 1650 cm⁻¹，芳环 C=C 约 1600 和 1500 cm⁻¹。¹H NMR：δ 2.1 处单峰 (3H, CH₃–CO)，δ 9–10 处宽单峰（NH 和 OH，加 D₂O 后消失），δ 6.8 和 7.4 处两组双重峰（对位二取代苯的 AA'BB' 体系，各 2H）。
8. Integrated Spectroscopic Analysis: An Unknown Compound | 案例八：综合波谱解析——一个未知化合物
ak at 1740 cm⁻¹, no broad OH; ¹H NMR: triplet (3H) at δ 1.2, quartet (2H) at δ 4.1, singlet (3H) at δ 2.0. Determine the structure, name it, and assign all spectral features.
某无色果香液体给出如下数据：质谱分子离子峰 m/z = 88，基峰 m/z = 43；IR：1740 cm⁻¹ 强吸收，无宽 OH 峰；¹H NMR：δ 1.2 (三重峰, 3H), δ 4.1 (四重峰, 2H), δ 2.0 (单峰, 3H)。试确定结构并命名，归属所有谱学特征。
88 suggests formula C₄H₈O₂ (4×12+8+32=88). IR peak at 1740 cm⁻¹ indicates an ester C=O (no –OH). NMR: singlet 3H at 2.0 ppm is a CH₃–CO– group; triplet+quartet (3H, 2H) is an ethyl group –CH₂CH₃ attached to oxygen, with chemical shift ~4.1 for –O–CH₂– tate, CH₃COOCH₂CH₃. Mass spectrum: m/z 88 is M⁺; base peak 43 arises from acyl cation CH₃C≡ rbonyl at 1740 cm⁻¹, absence of OH rules out acid. All data perfectly match ethyl acetate.
分子离子 88 提示分子式 C₄H₈O₂（4×12+8+32=88）。IR 1740 cm⁻¹ 为酯羰基（无 –OH）。NMR：δ 2.0 单峰 3H 为 CH₃–CO–；δ 1.2 三重峰 (3H) 和 δ 4.1 四重峰 (2H) 构成一个连氧的乙基 –O–CH₂CH₃，其中亚甲基因与氧相连而低场至约 4.1。故结构为乙酸乙酯，CH₃COOCH₂CH₃。质谱：m/z 88 为分子离子峰；基峰 43 来自乙酰正离子 CH₃C≡O⁺，为乙酸酯特征。IR 酯羰基 1740 cm⁻¹，无 OH 排除酸。所有数据吻合乙酸乙酯。
9. Rate Equation from Initial‑Rate Data: Iodine Clock Variant | 案例九：由初速数据求速率方程——碘钟反应变体
The persulfate– ⁻ + I₂. Using initial concentrations and initial rates below, determine the rate law and the rate constant k, including units.
| Experiment | [S₂O₈²⁻] / mol dm⁻³ | [I⁻] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.020 | 0.020 | 1.4 × 10⁻⁵ |
| 2 | Published by TutorHao | Year 13 Chemistry Revision Series | aleveler.com
