📚 Case Study Practical Drills for Year 12 CAIE Chemistry | Year 12 CAIE 化学：案例分析实战演练
Mastering Year 12 CAIE Chemistry requires the ability to tackle unfamiliar data, perform multi-step calculations, and apply concepts to real-world scenarios. This article presents a series of carefully designed case studies that mirror the style of Paper 2 and Paper 4 questions. Each case study guides you through the thought process, from identifying the relevant concept to executing calculations and justifying your answer. By working through these examples, you will build confidence and accuracy for your examinations.
掌握 Year 12 CAIE 化学需要具备处理陌生数据、进行多步计算以及将概念应用于实际情景的能力。本文提供了一系列精心设计的案例分析，它们模仿了 Paper 2 和 Paper 4 问题的风格。每个案例引导你经历从识别相关概念到执行计算和合理论证的全过程。通过这些练习，你将增强应对考试的信心与准确性。
1. Case Study 1: Empirical Formula from Combustion Data | 案例1：由燃烧数据确定经验式
A hydrocarbon containing only carbon and hydrogen was analysed by combustion. When 0.250 g of the compound was burnt in excess oxygen, 0.785 g of CO₂ and 0.321 g of H₂O were collected. Determine the empirical formula.
某仅含碳和氢的烃类化合物通过燃烧进行分析。将 0.250 g 该化合物在过量的氧气中燃烧，收集到 0.785 g CO₂ 和 0.321 g H₂O。试确定其经验式。
Step 1: Calculate the moles of carbon from the mass of CO₂ produced.
步骤1：从产生的 CO₂ 质量计算碳的摩尔数。
n(CO₂) = 0.785 g / 44.0 g mol⁻¹ = 0.01784 mol. Each mole of CO₂ contains one mole of carbon, so n(C) = 0.01784 mol.
n(CO₂) = 0.785 g / 44.0 g mol⁻¹ = 0.01784 mol。每摩尔 CO₂ 含一摩尔碳，所以 n(C) = 0.01784 mol。
Mass of carbon = 0.01784 mol × 12.0 g mol⁻¹ = 0.2141 g.
碳的质量 = 0.01784 mol × 12.0 g mol⁻¹ = 0.2141 g。
Step 2: Calculate the moles of hydrogen from the water produced.
步骤2：从产生的水计算氢的摩尔数。
n(H₂O) = 0.321 g / 18.0 g mol⁻¹ = 0.01783 mol. Each mole of H₂O contains two moles of H, so n(H) = 0.01783 × 2 = 0.03567 mol.
n(H₂O) = 0.321 g / 18.0 g mol⁻¹ = 0.01783 mol。每摩尔 H₂O 含两摩尔氢，所以 n(H) = 0.03567 mol。
Mass of hydrogen = 0.03567 mol × 1.0 g mol⁻¹ = 0.03567 g.
氢的质量 = 0.03567 g。
Step 3: Confirm that the sample contains only carbon and hydrogen by summing the masses (0.2141 + 0.03567 = 0.2498 g ≈ 0.250 g).
步骤3：通过质量加和（0.2141 + 0.03567 = 0.2498 g ≈ 0.250 g）确认样品仅含碳和氢。
Step 4: Determine the simplest whole-number ratio of elements.
步骤4：确定元素的最简整数比。
Mole ratio C : H = 0.01784 : 0.03567. Divide by the smallest value (0.01784): C = 1, H = 2.00. The empirical formula is CH₂.
摩尔比 C : H = 0.01784 : 0.03567。除以最小数值（0.01784）：C = 1，H = 2.00，经验式为 CH₂。
Extension: If the relative molecular mass is found to be 56.0, the molecular formula is determined by n = Mr(compound) / Mr(empirical) = 56.0 / 14.0 = 4, giving C₄H₈.
延伸：若相对分子质量为 56.0，则分子式由 n = Mr(化合物) / Mr(经验式) = 56.0 / 14.0 = 4 得出，即 C₄H₈。
2. Case Study 2: Using Molar Volume in Gas Reactions | 案例2：在气体反应中运用摩尔体积
Calcium carbonate reacts with excess hydrochloric acid according to the equation: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). When 2.00 g of CaCO₃ (Mr = 100.1) is used, calculate the volume of CO₂ produced at room conditions (25 °C, 1 atm), assuming the molar volume of a gas is 24.0 dm³ mol⁻¹.
碳酸钙与过量盐酸反应，方程式为：CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)。当使用 2.00 g CaCO₃（Mr = 100.1）时，计算在常温常压下（25 °C, 1 atm）产生的 CO₂ 的体积，假设气体摩尔体积为 24.0 dm³ mol⁻¹。
Step 1: Calculate the moles of CaCO₃ used.
步骤1：计算所用 CaCO₃ 的摩尔数。
n(CaCO₃) = mass / Mr = 2.00 g / 100.1 g mol⁻¹ ≈ 0.01998 mol.
n(CaCO₃) = 质量 / 相对分子质量 = 2.00 g / 100.1 g mol⁻¹ ≈ 0.01998 mol。
Step 2: Deduce the moles of CO₂ produced from the stoichiometry.
步骤2：根据化学计量比推导产生的 CO₂ 摩尔数。
The equation shows a 1 : 1 mole ratio between CaCO₃ and CO₂. Therefore, n(CO₂) = 0.01998 mol.
方程式显示 CaCO₃ 与 CO₂ 的摩尔比为 1 : 1，因此 n(CO₂) = 0.01998 mol。
Step 3: Convert moles of gas to volume using the molar volume.
步骤3：用摩尔体积将气体摩尔数转换为体积。
Volume of CO₂ = n(CO₂) × Vm = 0.01998 mol × 24.0 dm³ mol⁻¹ = 0.480 dm³ (or 480 cm³).
CO₂ 体积 = n(CO₂) × Vm = 0.01998 mol × 24.0 dm³ mol⁻¹ = 0.480 dm³（或 480 cm³）。
Interpretation: The answer highlights the direct use of the ideal gas molar volume, a concept tested frequently in AS Chemistry.
解读：该答案直接运用了理想气体摩尔体积，这是 AS 化学中频繁考查的概念。
3. Case Study 3: Applying Hess’s Law to Find Enthalpy of Formation | 案例3：运用赫斯定律求生成焓
Given the following standard enthalpy changes of combustion (ΔHc°): C(s) = -393.5 kJ mol⁻¹, H₂(g) = -285.8 kJ mol⁻¹, C₂H₅OH(l) = -1367 kJ mol⁻¹, determine the standard enthalpy of formation of ethanol, ΔHf°(C₂H₅OH).
已知下列标准燃烧焓变（ΔHc°）：C(s) = -393.5 kJ mol⁻¹，H₂(g) = -285.8 kJ mol⁻¹，C₂H₅OH(l) = -1367 kJ mol⁻¹，求乙醇的标准生成焓 ΔHf°(C₂H₅OH)。
Step 1: Write the target formation equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).
步骤1：写出目标生成方程式：2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)。
Step 2: Construct a Hess cycle using combustion data or use the formula ΔHf° = Σ ΔHc°(reactants) – Σ ΔHc°(products).
步骤2：利用燃烧数据构建赫斯循环，或使用公式 ΔHf° = Σ ΔHc°(反应物) – Σ ΔHc°(生成物)。
Σ ΔHc°(reactants) = 2 × (-393.5) + 3 × (-285.8) = -787.0 – 857.4 = -1644.4 kJ mol⁻¹.
Σ ΔHc°(反应物) = 2 × (-393.5) + 3 × (-285.8) = -787.0 – 857.4 = -1644.4 kJ mol⁻¹。
Σ ΔHc°(products) = ΔHc°(C₂H₅OH) = -1367 kJ mol⁻¹.
Σ ΔHc°(生成物) = -1367 kJ mol⁻¹。
Step 3: ΔHf° = -1644.4 – (-1367) = -277.4 kJ mol⁻¹ (values may vary slightly due to rounding).
步骤3：ΔHf° = -1644.4 – (-1367) = -277.4 kJ mol⁻¹（四舍五入可能导致微小差异）。
Check: The typical literature value for ΔHf° of ethanol is around -278 kJ mol⁻¹, confirming the calculation.
验证：乙醇的文献 ΔHf° 约为 -278 kJ mol⁻¹，证实了计算结果。
4. Case Study 4: Equilibrium Constant Kc for an Esterification | 案例4：酯化反应的平衡常数 Kc
Ethanoic acid (0.10 mol) and ethanol (0.10 mol) are mixed in a sealed flask with a small amount of acid catalyst. At equilibrium, two-thirds of the acid has reacted. Determine the equilibrium constant Kc for the reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Assume the total volume V remains constant and all components are in the liquid phase.
将 0.10 mol 乙酸和 0.10 mol 乙醇与少量酸催化剂在密封烧瓶中混合。达到平衡时，三分之二的酸已反应。确定反应 CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O 的平衡常数 Kc。假设总容积 V 恒定且所有组分均为液相。
Step 1: Calculate the equilibrium amounts (mol). Initial: acid 0.10, alcohol 0.10, ester 0, water 0.
步骤1：计算平衡时各物质的量（mol）。初始：酸 0.10，醇 0.10，酯 0，水 0。
Change: x = (2/3) × 0.10 = 0.0667 mol of acid consumed. The same amount of alcohol reacts, and producing 0.0667 mol each of ester and water.
变化：x = (2/3) × 0.10 = 0.0667 mol 酸被消耗。相同量的醇反应，生成酯和水各 0.0667 mol。
Equilibrium amounts: acid = 0.10 – 0.0667 = 0.0333 mol, alcohol = 0.0333 mol, ester = 0.0667 mol, water = 0.0667 mol.
平衡时物质的量：酸 = 0.0333 mol，醇 = 0.0333 mol，酯 = 0.0667 mol，水 = 0.0667 mol。
Step 2: Express Kc in terms of concentrations (mol dm⁻³). Concentration = amount / V.
步骤2：用浓度（mol dm⁻³）表示 Kc。浓度 = 物质的量 / V。
[acid] = 0.0333/V, [alcohol] = 0.0333/V, [ester] = 0.0667/V, [water] = 0.0667/V.
[酸] = 0.0333/V，[醇] = 0.0333/V，[酯] = 0.0667/V，[水] = 0.0667/V。
Step 3: Kc = [ester][water] / ([acid][alcohol]) = (0.0667/V × 0.0667/V) / (0.0333/V × 0.0333/V).
步骤3：Kc = [酯][水] / ([酸][醇]) = (0.0667/V × 0.0667/V) / (0.0333/V × 0.0333/V)。
The V terms cancel: Kc = (0.0667)² / (0.0333)² = (0.0667/0.0333)² = (2.00)² = 4.0 (no units as number of moles is equal on both sides).
V 项消去：Kc = (0.0667)² / (0.0333)² = (2.00)² = 4.0（无单位，因为两边摩尔数相等）。
Interpretation: A Kc of 4 indicates that the equilibrium lies moderately in favour of products.
解读：Kc = 4 表示平衡略微偏向生成物方向。
5. Case Study 5: Determining Rate Equation from Initial Rates | 案例5：由初始速率确定速率方程
The reaction 2A + B → products was studied at constant temperature. The following initial rate data were collected:
反应 2A + B → 产物在恒温下进行研究。收集到下列初始速率数据：
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