Year 12 OCR Chemistry: Case Study Practice and Problem-Solving | OCR 化学 Year 12：案例分析实战演练

📚 Year 12 OCR Chemistry: Case Study Practice and Problem-Solving | OCR 化学 Year 12：案例分析实战演练

OCR Year 12 Chemistry exams often present case study questions that weave together knowledge from Modules 2, 3 and 4. These scenarios require you to interpret data, perform multi-step calculations and draw links between topics such as organic chemistry, energetics, equilibrium and periodicity. This article guides you through ten realistic case studies, showing exactly how to apply AS-level concepts to unfamiliar contexts and maximise your marks.

OCR 化学 Year 12 试卷中常出现案例分析题，将模块 2、3、4 的知识融合在陌生情境中。这些问题要求你解读数据、完成多步计算，并在有机化学、能量学、平衡和周期性等主题之间建立联系。本文带你走过十个真实的案例研究，展示如何将 AS 阶段的概念应用于新情境，从而拿下高分。

1. Understanding the Case Study Approach | 理解案例分析的方法

A successful case study response begins with careful reading. Identify which modules are being tested, highlight the data provided and note the exact question being asked. Always show your working systematically and link your answer back to chemical principles rather than just quoting memorised facts.

成功解答案例分析的第一步是仔细阅读。找出题目考查的模块，圈出给出的数据，明确问题究竟在问什么。解答时要系统展示计算步骤，并将答案与化学原理联系起来，而不是单纯复述记忆的内容。

In OCR exams, marks are awarded for logical reasoning, correct use of units and significant figures, and for evaluating experimental limitations. Practise by working through the case studies below, and you will build confidence in tackling even the most complex synoptic questions.

在 OCR 考试中，逻辑推理、单位的正确使用、有效数字以及实验局限性的评价都是得分点。通过练习下面的案例研究，你将积累信心，从容应对最复杂的综合性问题。

2. Case Study 1: Identifying an Unknown Organic Compound | 案例 1：未知有机化合物的鉴定

An organic compound A (0.150 g) undergoes complete combustion, producing 0.357 g of CO₂ and 0.182 g of H₂O. The mass spectrum shows a molecular ion peak at m/z = 74. The IR spectrum displays a broad absorption at approximately 3300 cm⁻¹ and a strong peak at 1050 cm⁻¹.

某有机化合物 A（0.150 g）完全燃烧，生成 0.357 g CO₂ 和 0.182 g H₂O。其质谱显示分子离子峰在 m/z = 74。红外光谱在约 3300 cm⁻¹ 处有宽吸收，并在 1050 cm⁻¹ 处有强吸收峰。

First, calculate the mass of carbon: mass C = 0.357 g × (12.0 / 44.0) = 0.0974 g. Next, hydrogen: mass H = 0.182 g × (2.0 / 18.0) = 0.0202 g. The remainder is oxygen: mass O = 0.150 – 0.0974 – 0.0202 = 0.0324 g.

首先，计算碳的质量：碳的质量 = 0.357 g × (12.0 / 44.0) = 0.0974 g。接着算氢：氢的质量 = 0.182 g × (2.0 / 18.0) = 0.0202 g。剩余的是氧：氧的质量 = 0.150 – 0.0974 – 0.0202 = 0.0324 g。

Convert masses to moles: C: 0.0974 / 12.0 = 0.00812 mol, H: 0.0202 / 1.0 = 0.0202 mol, O: 0.0324 / 16.0 = 0.002025 mol. Divide by the smallest: C : H : O = 4.01 : 10.0 : 1.00. The empirical formula is C₄H₁₀O.

将质量换算为物质的量：C：0.0974/12.0 = 0.00812 mol，H：0.0202/1.0 = 0.0202 mol，O：0.0324/16.0 = 0.002025 mol。除以最小值得 C : H : O = 4.01 : 10.0 : 1.00。实验式为 C₄H₁₀O。

The empirical formula mass is 74 g mol⁻¹, matching the mass spectrum molecular ion. The IR broad band at 3300 cm⁻¹ indicates an O–H group, while the peak at 1050 cm⁻¹ corresponds to a C–O stretch. Therefore, compound A is a saturated alcohol with formula C₄H₁₀O, such as butan-1-ol or butan-2-ol.

实验式式量为 74 g mol⁻¹，与质谱中的分子离子峰相符。红外光谱中 3300 cm⁻¹ 处的宽吸收表明存在 O–H 键，而 1050 cm⁻¹ 处的峰对应 C–O 伸缩振动。因此，化合物 A 是饱和醇，分子式为 C₄H₁₀O，例如丁-1-醇或丁-2-醇。

3. Case Study 2: Enthalpy Change of Neutralisation | 案例 2：中和焓变的测定

In a calorimetry experiment, 50.0 cm³ of 2.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 2.00 mol dm⁻³ NaOH. The initial temperature is 21.0 °C and the maximum temperature reached is 34.5 °C. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹ and the density is 1.00 g cm⁻³.

在一次量热实验中，将 50.0 cm³ 2.00 mol dm⁻³ 的盐酸与 50.0 cm³ 2.00 mol dm⁻³ 的氢氧化钠溶液混合。初始温度为 21.0 °C，达到的最高温度为 34.5 °C。假设溶液的比热容为 4.18 J g⁻¹ K⁻¹，密度为 1.00 g cm⁻³。

Total volume = 100.0 cm³, so mass m = 100.0 g. Temperature rise ΔT = 34.5 – 21.0 = 13.5 °C (or K). Heat released q = m × c × ΔT = 100.0 × 4.18 × 13.5 = 5643 J = 5.643 kJ.

总体积为 100.0 cm³，因此质量 m = 100.0 g。温升 ΔT = 34.5 – 21.0 = 13.5 °C（或 K）。释放的热量 q = m × c × ΔT = 100.0 × 4.18 × 13.5 = 5643 J = 5.643 kJ。

Moles of HCl = 0.0500 dm³ × 2.00 mol dm⁻³ = 0.100 mol; same for NaOH. The reaction is 1:1, so 0.100 mol of water forms. ΔH neut = -q / n = -5.643 kJ / 0.100 mol = -56.4 kJ mol⁻¹. This is close to the accepted value of -57.1 kJ mol⁻¹; small differences arise from heat loss, the assumption of specific heat capacity, and incomplete mixing.

HCl 物质的量 = 0.0500 dm³ × 2.00 mol dm⁻³ = 0.100 mol；NaOH 相同。反应为 1:1，因此生成 0.100 mol 水。中和焓 ΔH = -q / n = -5.643 kJ / 0.100 mol = -56.4 kJ mol⁻¹。这接近于公认值 -57.1 kJ mol⁻¹；微小的差异源于热量散失、比热容假设以及混合不充分。

4. Case Study 3: Analysing a Reaction Mechanism – Halogenoalkane Hydrolysis | 案例 3：反应机理分析——卤代烷水解

Two halogenoalkanes, 1-bromobutane (primary) and 2-bromo-2-methylpropane (tertiary), are warmed with aqueous sodium hydroxide. Silver nitrate solution is added to monitor the rate of bromide ion formation. The tertiary halogenoalkane produces a precipitate almost instantly, while the primary one reacts much more slowly.

将 1-溴丁烷（伯卤代烷）和 2-溴-2-甲基丙烷（叔卤代烷）分别与氢氧化钠水溶液共热。加入硝酸银溶液以监测溴离子的生成速率。叔卤代

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