📚 Case Study Practical Drills for AQA GCSE Chemistry | AQA化学案例分析实战演练
Case study questions are an essential part of AQA GCSE Chemistry exams. They present real-world scenarios and require you to apply chemical principles, analyse data and justify your answers. This article works through ten carefully designed practical case studies, each targeting key topics from the Year 11 specification. By reading the paired English and Chinese explanations, you will strengthen both your subject knowledge and exam technique.
案例分析题是 AQA GCSE 化学考试的重要组成部分。它们呈现真实世界的情境，要求你运用化学原理、分析数据并论证你的答案。本文通过十个精心设计的实战案例，每个都针对 Year 11 考纲中的核心主题。通过阅读配对的英文与中文解析，你将同时强化学科知识和应试技巧。
1. Identifying an Unknown Salt | 鉴定未知盐
A student was given two white solids, X and Y. She carried out a flame test on X, which produced a brick-red flame. When sodium hydroxide solution was added to a solution of X, a white precipitate formed that did not dissolve in excess. For sample Y, the flame test gave a lilac flame, and adding dilute nitric acid and silver nitrate solution produced a yellow precipitate.
一名学生拿到两种白色固体 X 和 Y。她对 X 进行焰色试验，产生了砖红色火焰。向 X 的溶液中加入氢氧化钠溶液后，生成不溶于过量的白色沉淀。对于样品 Y，焰色试验呈紫色，加入稀硝酸和硝酸银溶液后产生黄色沉淀。
Brick-red flame indicates Ca²⁺ ions; the white precipitate with NaOH that is insoluble in excess confirms Ca²⁺ (and rules out Zn²⁺ or Al³⁺). X is a calcium salt. The lilac flame points to K⁺, and the yellow precipitate with silver nitrate is AgI, so the anion is iodide, I⁻. Hence Y is a potassium salt containing iodide, such as KI. When writing answers, always mention the specific observations and link them to the expected ion.
砖红色火焰表明存在 Ca²⁺ 离子；加入氢氧化钠后生成不溶于过量 NaOH 的白色沉淀进一步确认为 Ca²⁺（排除了 Zn²⁺ 或 Al³⁺）。X 是一种钙盐。紫色火焰指向 K⁺，与硝酸银生成的黄色沉淀是 AgI，因此阴离子为碘离子 I⁻。所以 Y 是一种含碘的钾盐，例如 KI。作答时，务必描述具体现象并将其与对应的离子联系起来。
2. Rate of Reaction – Surface Area | 反应速率——表面积的影响
An experiment was carried out using 2.0 g of marble chips (large chunks) and 50 cm³ of 1.0 mol/dm³ hydrochloric acid. The carbon dioxide produced was collected in a gas syringe, and the volume was recorded every 30 seconds. The same experiment was repeated with the same mass of powdered marble. The results are shown below.
一次实验使用 2.0 g 大理石碎片（大块）和 50 cm³ 的 1.0 mol/dm³ 盐酸。产生的二氧化碳用气体注射器收集，每 30 秒记录体积。然后用相同质量的大理石粉末重复实验。结果如下表所示。
| Time / s | Vol. of gas (large) / cm³ | Vol. of gas (powder) / cm³ |
|---|---|---|
| 0 | 0 | 0 |
| 30 | 14 | 28 |
| 60 | 26 | 48 |
| 90 | 36 | 60 |
| 120 | 44 | 68 |
Calculate the average rate of reaction in the first 30 seconds for each form of marble.
计算每种大理石前 30 秒内的平均反应速率。
Average rate = ΔV / Δt
平均速率 = 体积变化 / 时间
For large chips: (14 – 0) / 30 = 0.47 cm³/s. For powder: (28 – 0) / 30 = 0.93 cm³/s. The powdered marble reacts faster because it has a much larger surface area, leading to more frequent collisions between reactant particles. The final volume of gas collected differs because the same mass of marble is used, but excess acid ensures all CaCO₃ reacts; the powder reaction simply reaches completion sooner.
大块大理石：(14–0)/30 = 0.47 cm³/s。粉末：(28–0)/30 = 0.93 cm³/s。大理石粉末反应更快，因为其表面积大得多，反应物粒子间的碰撞更加频繁。最终收集到的气体体积不同吗？由于使用相同质量的大理石，且酸过量，所有 CaCO₃ 都完全反应；粉末反应只是更早达到终点。
3. Electrolysis – Purifying Copper | 电解——精炼铜
In industry, impure copper is purified by electrolysis. The impure copper is made the anode, and a pure copper sheet is the cathode, both placed in copper(II) sulfate solution. When a direct current is applied, the anode dissolves and pure copper deposits on the cathode.
工业上通过电解精炼不纯的铜。不纯铜作为阳极，纯铜片作为阴极，一同放入硫酸铜溶液中。通电后，阳极溶解，纯铜在阴极上沉积。
Anode reaction: Cu(s) → Cu²⁺(aq) + 2e⁻. Cathode reaction: Cu²⁺(aq) + 2e⁻ → Cu(s). Impurities such as silver and gold do not oxidise and fall to the bottom as anode sludge. If a current is passed until 0.10 mol of electrons has been transferred, calculate the mass of copper deposited at the cathode. (Cu: 63.5)
阳极反应：Cu(s) → Cu²⁺(aq) + 2e⁻。阴极反应：Cu²⁺(aq) + 2e⁻ → Cu(s)。银、金等杂质不被氧化，沉落底部成为阳极泥。若通电至转移了 0.10 mol 电子，求阴极沉积铜的质量。（Cu: 63.5）
Moles of Cu = moles of electrons / 2 = 0.10 / 2 = 0.050 mol
Cu 的物质的量 = 电子物质的量 / 2 = 0.10 / 2 = 0.050 mol
Mass = 0.050 × 63.5 = 3.175 g ≈ 3.18 g
The calculated mass would be about 3.18 g if the current efficiency is 100%. This type of calculation often combines electrolysis half-equations with molar mass, so always check the ratio of electrons to metal.
若电流效率为 100%，计算出的质量约为 3.18 g。此类计算常将电解半反应与摩尔质量结合，因此务必核对电子与金属的物质的量之比。
4. Energy Changes – Exothermic and Endothermic | 能量变化——放热与吸热
A student dissolved 5.0 g of anhydrous calcium chloride in 50 cm³ of water and recorded a temperature rise from 20°C to 35°C. She then repeated the procedure using 5.0 g of ammonium nitrate, and the temperature fell from 20°C to 12°C. Classify each process and explain the molecular-level changes.
学生将 5.0 g 无水氯化钙溶于 50 cm³ 水中，记录到温度从 20°C 升至 35°C。然后她用 5.0 g 硝酸铵重复操作，温度从 20°C 降至 12°C。判断两个过程的类型，并解释分子层面的变化。
Dissolving calcium chloride is exothermic because breaking the ionic lattice requires less energy than the energy released when ions are hydrated. The reverse is true for ammonium nitrate: lattice energy is higher than hydration energy, so the overall process is endothermic. Exothermic reactions increase the temperature of the surroundings, while endothermic ones decrease it. Such temperature changes can be used in hand warmers (CaCl₂) or instant cold packs (NH₄NO₃).
氯化钙溶解是放热过程，因为破坏离子晶格所需的能量小于离子水合时释放的能量。硝酸铵情况恰好相反：晶格能大于水合能，因此整体表现为吸热。放热反应使环境温度升高，吸热反应使温度降低。这些温度变化可用于暖手宝（CaCl₂）或瞬时冰袋（NH₄NO₃）。
5. Reversible Reactions – The Haber Process | 可逆反应——哈伯法
The Haber process synthesises ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol. The industrial conditions are 450 °C, 200 atm and an iron catalyst. Explain why a lower temperature is not used despite the forward reaction being exothermic, and why a pressure of 200 atm is chosen rather than an even higher pressure.
哈伯法合成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol。工业条件为 450 °C、200 atm 和铁催化剂。解释为何正向放热却不使用更低温度，以及为何选择 200 atm 而不是更高压强。
A lower temperature would favour the forward exothermic reaction and give a higher equilibrium yield of ammonia, but the rate would be too slow to be economical. 450 °C is a compromise that gives a reasonable rate while still maintaining a useful yield. A higher pressure would further increase the yield, but safety concerns, plant costs and the energy required to compress gases make pressures above about 200 atm impractical. The iron catalyst increases the rate but does not affect the equilibrium position.
更低温度有利于放热正反应进行，从而提高氨的平衡产率，但反应速率会过慢而不经济。450 °C 是一个折衷温度，既可获得可接受的速率，又能保持实用的产率。更高压强可进一步提高产率，但安全考虑、设备成本以及压缩气体所需的能量使得超过约 200 atm 的压强不切实际。铁催化剂可加快反应速率，但不影响平衡位置。
At a certain temperature, an equilibrium mixture was found to contain 0.20 mol N₂, 0.60 mol H₂ and 0.40 mol NH₃ in a 1.0 dm³ reactor. Calculate Kc, including its units.
在某温度下，1.0 dm³ 反应器中平衡混合物含 0.20 mol N₂、0.60 mol H₂ 和 0.40 mol NH₃。计算 Kc 及其单位。
Kc = [NH₃]² / ([N₂][H₂]³)
Kc = (0.40)² / (0.20 × (0.60)³) = 0.16 / (0.20 × 0.216) = 0.16 / 0.0432 ≈ 3.70
Units: (mol/dm³)² / (mol/dm³)⁴ = mol⁻² dm⁶. Therefore Kc = 3.70 mol⁻² dm⁶. Always remember to work out units by substituting mol/dm⁻³ into the expression.
单位：(mol/dm³)² / (mol/dm³)⁴ = mol⁻² dm⁶。因此 Kc = 3.70 mol⁻² dm⁶。始终记住将 mol/dm⁻³ 代入表达式来推导单位。
6. Water Treatment – From Reservoir to Tap | 水处理——从水库到水龙头
Water from a reservoir contains suspended solids, bacteria and dissolved substances. The treatment process involves adding aluminium sulfate and calcium hydroxide, which react to form a gelatinous precipitate of aluminium hydroxide. The water then passes through a sand filter before chlorine gas is added. Explain the purpose of each stage.
水库水含有悬浮固体、细菌和溶解物质。处理过程包括加入硫酸铝和氢氧化钙，反应生成凝胶状氢氧化铝沉淀。然后水通过砂滤器，最后加入氯气。解释每个阶段的目的。
Aluminium hydroxide acts as a flocculant: it traps fine particles and forms larger clumps that settle out. Sand filtration removes remaining small particles. Chlorination kills bacteria and other pathogens to make water safe to drink. The relevant ionic equation for precipitation is Al³⁺ + 3OH⁻ → Al(OH)₃(s). Chlorine reacts with water to form hypochlorous acid (HOCl) and HCl, which are effective disinfectants. Distillation would produce very pure water, but it is far too energy-intensive for municipal supply.
氢氧化铝作为絮凝剂：它能捕获细小颗粒并形成较大的团块沉降下来。砂滤可去除残余的微小颗粒。氯化消毒能杀灭细菌和其他病原体，使水可安全饮用。沉淀反应的离子方程式为 Al³⁺ + 3OH⁻ → Al(OH)₃(s)。氯气与水反应生成次氯酸（HOCl）和盐酸，都是有效的消毒剂。蒸馏可制得高纯水，但对于市政供水来说能耗过高。
7. Electrochemical Cells – Making a Battery | 电化学电池——制作一个电池
A simple cell was made by placing a zinc strip and a copper strip into a beaker of dilute sulfuric acid, connected by a wire and a voltmeter. The voltmeter reads 1.1 V. The zinc electrode loses mass, while the copper electrode gains mass. Explain the reactions and why the voltage arises.
一个简单电池由锌片和铜片浸在稀硫酸中并用导线和电压表连接而成。电压表读数为 1.1 V。锌电极质量减小，铜电极质量增加。解释其中的反应以及电压产生的原因。
Zinc is more reactive than copper, so it oxidises more readily: Zn(s) → Zn²⁺(aq) + 2e⁻. Electrons flow through the external circuit to the copper electrode, where H⁺ ions from the acid are reduced: 2H⁺ + 2e⁻ → H₂(g). The voltage results from the difference in reactivity (tendency to lose electrons) between the two metals. If zinc is replaced by magnesium, the voltage would be larger because magnesium is even more reactive. In cells without a salt bridge, the voltage drops quickly because gas bubbles build up on the electrodes, increasing internal resistance.
锌比铜活泼，因此更易被氧化：Zn(s) → Zn²⁺(aq) + 2e⁻。电子沿外电路流向铜电极，溶液中的 H⁺ 离子被还原：2H⁺ + 2e⁻ → H₂(g)。电压源于两种金属活泼性（失电子倾向）的差异。若用镁代替锌，电压会更大，因为镁更活泼。在没有盐桥的电池中，气泡在电极上积聚导致内阻增大，电压会迅速下降。
8. Life Cycle Assessment – Plastic vs Paper Bags | 生命周期评估——塑料袋与纸袋
A supermarket is choosing between high-density polyethylene (HDPE) plastic bags and paper bags. HDPE bags are made from crude oil, require less energy to manufacture, and are lightweight but do not biodegrade. Paper bags come from trees, use more water and energy in production, produce more solid waste, but are biodegradable and can be recycled. Analyse the environmental impact at each life cycle stage.
一家超市要在高密度聚乙烯（HDPE）塑料袋和纸袋之间做出选择。HDPE 袋由原油制造，生产能耗较低，重量轻但不生物降解。纸袋来自树木，生产过程耗水耗能较多，产生更多固体废物，但可生物降解且能回收。分析各生命周期阶段的环境影响。
Raw material extraction: HDPE uses non-renewable petroleum; paper uses renewable wood but may contribute to deforestation. Manufacturing: paper production emits more CO₂ and requires more water. Distribution: HDPE bags are lighter, reducing transport emissions. Disposal: HDPE persists in landfills for centuries, while paper decomposes but generates methane in anaerobic landfill conditions. Reuse and recycling habits significantly alter the overall impact. A full LCA would also consider acidification, eutrophication and toxicity, making the ‘better’
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