📚 Year 11 SQA Chemistry: Case Study Analysis Practice | SQA化学案例分析实战演练
Case study questions are a distinctive and challenging part of SQA Chemistry assessments at National 5 and Higher levels. They present a realistic scientific scenario, often backed by data, graphs or experimental notes, and ask you to apply your chemical knowledge to solve problems, make predictions, and evaluate conclusions. Mastering case studies is not just about memorising facts – it is about learning to think like a chemist. This article walks you through a practical, step‑by‑step approach to case study analysis, using seven fully worked examples that mirror the style of SQA examination papers.
案例分析题是 SQA 化学（National 5 和 Higher）中独特且具有挑战性的部分。题目给出一个真实的科学场景，通常配有数据、图表或实验记录，要求你运用化学知识去解决问题、做出预测并评价结论。掌握案例分析不只是记忆事实，而是学会像化学家一样思考。本文将通过七个与 SQA 考试风格高度相似的完整示例，带你一步步演练案例分析的方法，帮助你在考场上从容应对。
1. Understanding Case Study Questions in SQA Chemistry | 理解SQA化学中的案例分析题
Unlike straightforward recall questions, a case study weaves several topics together. You might encounter a passage about an industrial process, a titration experiment, a pollution incident, or a new material. The question will then ask you to interpret data, explain observations using collision theory or equilibrium principles, perform calculations, and sometimes evaluate the reliability of the method. The key is to recognise that the scenario is just a wrapper – the chemistry underneath is the same principles you have studied in class.
与直接回忆知识的题目不同，案例分析会把多个主题编织在一起。你可能会读到一段关于工业流程、滴定实验、污染事件或新材料的文字。随后的问题会要求你解读数据、用碰撞理论或平衡原理解释观察到的现象、进行计算，有时还要评价方法的可靠性。关键在于要意识到，情景只是一层包装纸——底下的化学知识依然是你课堂上学过的那些原理。
2. A Step‑by‑Step Strategy for Tackling Case Studies | 攻克案例分析的逐步策略
Before diving into examples, adopt a clear strategy. First, read the scenario and all questions carefully, highlighting key numbers, units, and substances. Second, identify the core chemical concepts being tested – is it bonding, rates, acids and bases, or organic chemistry? Third, extract relevant data and organise it in a table or simple diagram if needed. Fourth, answer in logical steps, always linking an observation to a chemical explanation. Fifth, check units, significant figures, and whether your answer makes chemical sense. This approach saves time and reduces careless errors.
在进入案例之前，先要有一套清晰的策略。第一，仔细阅读情景和所有问题，标出关键的数字、单位与物质。第二，识别题目考查的核心化学概念——是结构、速率、酸碱还是有机化学？第三，提取相关数据，如有需要可用表格或简图加以整理。第四，按逻辑分步作答，始终把观察点与化学解释联系起来。第五，检查单位、有效数字，以及答案在化学上是否合理。这套方法能节省时间并减少粗心失误。
3. Case Study 1: Reaction Rates and Collision Theory | 案例1：反应速率与碰撞理论
A student investigated the reaction between marble chips (calcium carbonate) and excess hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). The volume of carbon dioxide produced was recorded every 20 seconds. Results are shown in the table below.
一名学生研究了石灰石碎片（碳酸钙）与过量盐酸的反应：CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)。每20秒记录一次生成的二氧化碳体积。结果如下表所示。
| Time (s) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Volume of CO₂ (cm³) | 0 | 32 | 55 | 70 | 78 | 82 |
Question: Calculate the average rate of reaction in the first 60 seconds. Explain, using collision theory, why the rate decreases over time.
问题：计算前60秒内的平均反应速率。用碰撞理论解释为何反应速率随时间逐渐减小。
Answer – Calculation: Average rate = change in volume / change in time = (70 – 0) cm³ / (60 – 0) s = 70 cm³ / 60 s = 1.17 cm³ s⁻¹ (to 3 significant figures).
答案——计算：平均速率 = 体积变化 / 时间变化 = (70 – 0) cm³ / (60 – 0) s = 70 cm³ / 60 s = 1.17 cm³ s⁻¹（3位有效数字）。
Explanation: As the reaction proceeds, the concentration of hydrochloric acid decreases because it is being used up. Marble chips also become smaller, reducing total surface area. According to collision theory, fewer reactant particles per unit volume mean a lower frequency of successful collisions. Consequently, less carbon dioxide is produced per unit time, so the rate falls.
解释：随着反应进行，盐酸因被消耗而浓度下降。石灰石碎片也逐渐变小，总表面积减少。根据碰撞理论，单位体积内反应物粒子数减少，导致成功碰撞的频率降低。因此，单位时间生成的二氧化碳减少，速率下降。
4. Case Study 2: Chemical Equilibrium in Action | 案例2：化学平衡的实际应用
The Haber process is carried out at about 450 °C and 200 atmospheres: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹. An engineer notices that when the reactor temperature accidentally rises to 550 °C, the yield of ammonia drops significantly. She also tests the effect of adding more nitrogen to a system already at equilibrium.
哈柏法在大约450 °C和200个大气压下进行：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹。工程师注意到，当反应器温度意外升至550 °C时，氨的产率明显下降。她还测试了在已达平衡的体系中添加更多氮气带来的影响。
Question: (a) Explain why the yield decreases at the higher temperature. (b) Predict and explain what happens to the concentration of hydrogen when extra nitrogen is injected.
问题：(a) 解释为何温度升高时产率下降。(b) 预测并解释注入额外氮气后氢气浓度会发生什么变化。
Answer (a): The forward reaction is exothermic (ΔH negative). According to Le Chatelier’s principle, increasing the temperature shifts the position of equilibrium in the endothermic direction to absorb the added heat. The reverse reaction (decomposition of ammonia) is endothermic, so equilibrium shifts to the left, reducing the equilibrium yield of ammonia.
答案(a)：正反应是放热反应（ΔH为负值）。根据勒夏特列原理，升高温度会使平衡向吸热方向移动以吸收添加的热量。逆反应（氨的分解）是吸热的，因此平衡向左移动，氨的平衡产率下降。
Answer (b): Adding nitrogen increases the concentration of N₂. Le Chatelier’s principle states that the system responds to oppose the change, so the equilibrium shifts to the right to use up the extra nitrogen. This causes more hydrogen to react, so the concentration of H₂ decreases until a new equilibrium is established.
答案(b)：添加氮气提高了N₂的浓度。勒夏特列原理指出，体系会朝着削弱这一改变的方向移动，即平衡向右移动以消耗多余的氮气。这样一来，更多的氢气参与反应，H₂浓度下降，直至建立新的平衡。
5. Case Study 3: Acid‑Base Titration and Analysis | 案例3：酸碱滴定与分析
To determine the concentration of a diluted vinegar solution (ethanoic acid, CH₃COOH), a student titrates 25.0 cm³ portions with 0.100 mol dm⁻³ sodium hydroxide solution using phenolphthalein indicator. The average titre is 22.40 cm³. The equation: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).
为了测定一份稀释白醋（乙酸，CH₃COOH）的浓度，一名学生用0.100 mol dm⁻³ 氢氧化钠溶液滴定25.0 cm³ 样品，以酚酞为指示剂。平均滴定体积为22.40 cm³。反应方程式：CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)。
Question: Calculate the concentration of ethanoic acid in the diluted vinegar. The student repeated the experiment but forgot to add the indicator before the titration. Suggest one possible consequence.
问题：计算稀释白醋中乙酸的浓度。该学生又重复了实验，但在滴定前忘记加入指示剂。说明一个可能的后果。
Calculation: Moles of NaOH used = c × V = 0.100 mol dm⁻³ × (22.40 / 1000) dm³ = 0.00224 mol. From the 1:1 mole ratio, moles of CH₃COOH in 25.0 cm³ = 0.00224 mol. Concentration of CH₃COOH = moles / volume (dm³) = 0.00224 mol / (25.0/1000) dm³ = 0.0896 mol dm⁻³.
计算：所用NaOH的物质的量 = c × V = 0.100 mol dm⁻³ × (22.40 / 1000) dm³ = 0.00224 mol。由1:1摩尔比，25.0 cm³样品中CH₃COOH的物质的量 = 0.00224 mol。CH₃COOH浓度 = 物质的量 / 体积(dm³) = 0.00224 mol / (25.0/1000) dm³ = 0.0896 mol dm⁻³。
Consequence: Without an indicator, it would be impossible to see a sharp colour change at the end‑point. The student would likely overshoot the endpoint, adding too much NaOH and making the result inaccurate. This would lead to an apparent higher concentration of acid.
后果：没有指示剂就无法观察到终点的明显颜色变化。学生很可能会滴过头，加入过多的NaOH，导致结果不准确。这样会使计算出的酸浓度偏高。
6. Case Study 4: Organic Structure Elucidation | 案例4：有机结构推断
A compound has the molecular formula C₄H₁₀O. It does not react with sodium carbonate, but it effervesces gently with sodium metal. When warmed with acidified potassium dichromate solution, the compound turns the solution from orange to green, and the product gives a positive result with Tollens’ reagent.
一种化合物的分子式为C₄H₁₀O。它不与碳酸钠反应，但与金属钠缓慢产生气泡。当与酸化重铬酸钾溶液一起加热时，该化合物使溶液由橙色变为绿色，产物与托伦斯试剂反应呈阳性。
Question: Deduce the structural formula and name of the original compound. Explain your reasoning using all the observations.
问题：推断原化合物的结构式和名称。利用所有观察结果说明你的推理。
Reasoning: Molecular formula C₄H₁₀O suggests either an alcohol or an ether. No reaction with Na₂CO₃ rules out a carboxylic acid. Reaction with sodium metal indicates an –OH group, so it is an alcohol. Oxidation by acidified dichromate (orange → green) means a primary or secondary alcohol is present – tertiary alcohols are not oxidised. The product gives a positive Tollens’ test, which is characteristic of an aldehyde. Therefore, the alcohol must be a primary alcohol that is oxidised to an aldehyde. The only straight‑chain primary alcohol with four carbons is butan‑1‑ol; another possibility is 2‑methylpropan‑1‑ol. Both are consistent with the data. Any valid structure with an –OH group on a terminal carbon of a C₄ chain, and no branching that would create a tertiary alcohol, is acceptable. For simplicity: butan‑1‑ol, CH₃CH₂CH₂CH₂OH.
推理：分子式C₄H₁₀O可能是醇或醚。不与Na₂CO₃反应排除羧酸。与金属钠反应表明含有–OH基团，因此是醇。酸化重铬酸钾氧化（橙→绿）说明存在伯醇或仲醇——叔醇不被氧化。产物托伦斯试验呈阳性，这是醛的特征。因此，该醇必定是被氧化成醛的伯醇。四个碳的直链伯醇是丁‑1‑醇；另外可能是2‑甲基丙‑1‑醇。两者都与数据相符。任何在C₄链的末端碳上带有–OH且没有形成叔醇分支的结构均可。简答：丁‑1‑醇，CH₃CH₂CH₂CH₂OH。
7. Case Study 5: Electrochemistry and the Electrochemical Series | 案例5：电化学与电化序
A student sets up a cell using a zinc electrode dipped in ZnSO₄(aq) and a copper electrode dipped in CuSO₄(aq). The two half‑cells are connected by a salt bridge. The voltmeter reads 1.10 V. The student then replaces copper with a silver electrode in AgNO₃(aq). The voltage increases. Standard reduction potentials: Zn²⁺/Zn = –0.76 V; Cu²⁺/Cu = +0.34 V; Ag⁺/Ag = +0.80 V.
一名学生用锌电极浸在ZnSO₄(aq)中，铜电极浸在CuSO₄(aq)中，搭建了一个原电池。两个半电池通过盐桥连接，电压表读数为1.10 V。随后学生将铜电极换成银电极并浸在AgNO₃(aq)中，电压增大。已知标准还原电位：Zn²⁺/Zn = –0.76 V；Cu²⁺/Cu = +0.34 V；Ag⁺/Ag = +0.80 V。
Question: (a) Write the half‑equations and overall redox equation for the Zn–Cu cell. (b) Explain why swapping copper for silver increases the cell voltage.
问题：(a) 写出锌–铜电池的半反应式和总氧化还原方程式。(b) 解释为何将铜换成银后电池电压增大。
Answer (a): Oxidation (anode): Zn(s) → Zn²⁺(aq) + 2e⁻; Reduction (cathode): Cu²⁺(aq) + 2e⁻ → Cu(s); Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
答案(a)：氧化（阳极）：Zn(s) → Zn²⁺(aq) + 2e⁻；还原（阴极）：Cu²⁺(aq) + 2e⁻ → Cu(s)；总反应：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。
Answer (b): Cell voltage is calculated from the difference in reduction potentials: E°cell = E°(cathode) – E°(anode). For Zn–Cu: +0.34 V – (–0.76 V) = 1.10 V. For Zn–Ag: +0.80 V – (–0.76 V) = 1.56 V. Silver has a more positive reduction potential than copper, meaning Ag⁺ is a stronger oxidising agent. The greater difference between the two half‑cells results in a larger driving force for electron flow, hence a higher voltage.
答案(b)：电池电压由两电极还原电位之差决定：E°电池 = E°(阴极) – E°(阳极)。锌–铜电池：+0.34 V – (–0.76 V) = 1.10 V。锌–银电池：+0.80 V – (–0.76 V) = 1.56 V。银的还原电位比铜更正，意味着Ag⁺是更强的氧化剂。两个半电池之间的电位差更大，导致电子流动的驱动力更大，因此电压更高。
8. Case Study 6: Quantitative Chemistry and Yield Calculations | 案例6：定量化学与产率计算
In a lab, 2.00 g of calcium carbonate is strongly heated: CaCO₃(s) → CaO(s) + CO₂(g). The mass of solid residue after heating is 1.12 g. (Molar masses: CaCO₃ = 100 g mol⁻¹; CaO = 56 g mol⁻¹).
在实验室中，将2.00 g碳酸钙强热分解：CaCO₃(s) → CaO(s) + CO₂(g)。加热后残余固体的质量为1.12 g。（摩尔质量：CaCO₃ = 100 g mol⁻¹；CaO = 56 g mol⁻¹）。
Question: Calculate the percentage yield of calcium oxide. Suggest a reason why the yield is not 100 %.
问题：计算氧化钙的产率。举出一个产率未达100%的原因。
Calculation: Theoretical moles of CaCO₃ = 2.00 g / 100 g mol⁻¹ = 0.0200 mol. 1:1 mole ratio gives theoretical moles of CaO = 0.0200 mol. Theoretical mass of CaO = 0.0200 mol × 56 g mol⁻¹ = 1.12 g. But wait: the experimental residue mass is 1.12 g, which would suggest 100% yield. Possibly the student has misread the question – perhaps the initial mass was different. Let’s re‑work with a more realistic scenario: Suppose initial CaCO₃ mass was 3.00 g, residue 1.12 g. Then theoretical CaO mass = 3.00/100 × 56 = 1.68 g. Percentage yield = (actual / theoretical) × 100 = (1.12 g / 1.68 g) × 100 = 66.7%. This makes a better teaching point. I will adjust: “2.50 g of calcium carbonate gives 1.12 g of residue.” Theoretical mass = (2.50/100)×56 = 1.40 g. Yield = (1.12/1.40)×100 = 80.0%.
为了教学目的，调整数据：2.50 g碳酸钙得到1.12 g残余固体。理论CaCO₃物质的量 = 2.50 g / 100 g mol⁻¹ = 0.0250 mol，理论CaO质量 = 0.0250 mol × 56 g mol⁻¹ = 1.40 g。产率 = (1.12 g / 1.40 g) × 100 = 80.0%。
Reason for less than 100 %: Some calcium carbonate may not have fully decomposed because heating was insufficient, or some CaO powder was lost during transfer and weighing. Also, if the crucible was not properly sealed, CO₂ could escape before complete decomposition, but that does not affect residue mass directly. Realistically, incomplete reaction is the most common cause.
产率不足100%的原因：部分碳酸钙可能因加热不充分而未完全分解，或转移和称量过程中有部分氧化钙粉末损失。另外，坩埚未盖好导致产品飘散也会造成损失。最实际的原因通常是反应不完全。
9. Case Study 7: Environmental Chemistry – Acid Rain | 案例7：环境化学——酸雨
An environmental report states that rain in a certain region has a pH of 4.2 because of dissolved SO₂ from factory emissions. The acid rain reacts with marble statues (mainly CaCO₃) to cause erosion: CaCO₃(s) + H₂SO₃(aq) → CaSO₃(s) + H₂O(l) + CO₂(g). Over time, exposed limestone surfaces lose 15.0 g of mass.
一份环境报告指出，某地区因工厂排放的SO₂溶解在雨水中，雨水pH为4.2。酸雨与大理石雕像（主要成分CaCO₃）反应造成侵蚀：CaCO₃(s) + H₂SO₃(aq) → CaSO₃(s) + H₂O(l) + CO₂(g)。随时间推移，暴露的石灰岩表面损失了15.0 g的质量。
Question: (a) Calculate the mass of H₂SO₃ that reacted, assuming all mass loss is due to the reaction. (Molar masses: CaCO₃ = 100 g mol⁻¹; H₂SO₃ = 82 g mol⁻¹). (b) Name one environmental effect of acid rain other than damage to stonework.
问题：(a) 假设质量损失全部由该反应引起，计算反应了的H₂SO₃的质量。（摩尔质量：CaCO₃ = 100 g mol⁻¹；H₂SO₃ = 82 g mol⁻¹）。(b) 除损坏石质文物外，举出酸雨对环境的另一种影响。
Calculation: Moles of CaCO₃ lost = 15.0 g / 100 g mol⁻¹ = 0.150 mol. From the 1:1 equation, moles of H₂SO₃ needed = 0.150 mol. Mass of H₂SO₃ = 0.150 mol × 82 g mol⁻¹ = 12.3 g.
计算：损失的CaCO₃物质的量 = 15.0 g / 100 g mol⁻¹ = 0.150 mol。根据1:1方程式，所需H₂SO₃的物质的量 = 0.150 mol，质量 = 0.150 mol × 82 g mol⁻¹ = 12.3 g。
Environmental effect: Acid rain lowers the pH of lakes and rivers, making the water toxic to aquatic life, particularly fish eggs and shellfish. It also leaches aluminium from soil into waterways, harming organisms.
环境影响：酸雨降低湖泊与河流的pH，使水体对水生生物（尤其是鱼卵和贝类）产生毒性。它还会将土壤中的铝淋滤到水体中，危害生物。
10. Common Pitfalls and How to Avoid Them | 常见陷阱及应对方法
Many students lose marks by misreading units, forgetting to convert cm³ to dm³ in titration calculations, or misidentifying the limiting reactant. Another frequent error is ignoring the states given in an equilibrium system – a change in pressure only affects gases. In organic case studies, make sure you link each test result to a specific functional group, rather than guessing. Practice the habit of writing “moles of … = …” before jumping to a final answer, and always ask yourself: “Does this fit the chemical story?”
许多学生因读错单位、在滴定计算中忘了将cm³转换为dm³，或错误判断限制反应物而失分。另一个常见错误是忽略平衡体系中给出的状态——压力的改变只影响气体。在有机案例中，要确保将每个测试结果与特定的官能团联系起来，而不是胡乱猜测。养成在得出最终答案前先写出“XXX的物质的量 = ……”的习惯，并始终问自己：“这符合化学情景吗？”
11. Exam Tips for Maximising Your Marks | 最大化得分的考试技巧
Begin each case study by scanning the questions first so you know what to look for in the text. Always show your working step by step – SQA examiners award marks for method, not just the final answer. If you are asked to “explain”, use a scientifically precise sentence that states cause and effect, and include relevant keywords such as “collision frequency”, “activation energy”, or “position of equilibrium”. For calculations, keep all numbers on your calculator until the end to avoid rounding errors, then round your final answer to the appropriate number of significant figures. If a calculation seems too high or too low, re‑check the logic.
开始做案例分析时，先快速浏览问题，这样你就能知道在文章中需要提取什么信息。务必一步步写出解题过程——SQA考官是按步骤给分，而不只看最后答案。如果要求“解释”，要用科学严谨的句子陈述因果关系，并包含相关的关键词，如“碰撞频率”、“活化能”或“平衡位置”。计算时，把数字保留在计算器中直到最后再约分，以减小舍入误差，然后将最终答案约至合适的有效数字。如果计算结果看起来过大或过小，重新检查逻辑。
12. Conclusion: Practice Makes Perfect | 结语：熟能生巧
Case study analysis is a skill that develops with practice. Every time you work through a scenario, you train your brain to link theoretical knowledge with practical applications – exactly what SQA Chemistry aims to assess. Re‑work the examples above, then seek out past‑paper case studies and apply the same structured approach. The more you practise, the more confident you will become in unpacking unfamiliar situations and turning them into logical, chemical solutions.
案例分析是一项通过练习才能养成的技能。每当你分析一个情景，你就在训练大脑将理论知识与实际应用联系起来——这正是SQA化学所要考查的能力。重新演练上述例题，然后寻找历年真题中的案例，运用同样的结构化方法加以解答。练习得越多，你就越能自信地解析陌生的情景，并将其转化为条理清晰、符合化学原理的答案。
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