Typical Worked Examples in IB and AQA Chemistry | IB与AQA化学典型例题详解

📚 Typical Worked Examples in IB and AQA Chemistry | IB与AQA化学典型例题详解

Mastering chemistry requires not only an understanding of concepts but also the ability to apply them to complex problems. This article presents a selection of typical worked examples that are commonly encountered in both the IB Diploma Chemistry (SL/HL) and the AQA A-level Chemistry (including International A-level) specifications. Each example is carefully solved step by step, highlighting key techniques and common pitfalls.

掌握化学不仅需要理解概念，还需要将它们应用于复杂问题的能力。本文精选了IB文凭化学（SL/HL）和AQA A-level化学（包括国际A-level）考试中常见的典型例题。每个例题都经过逐步解答，突出关键技巧和常见错误。

1. Mole Concept and Stoichiometry | 摩尔概念与化学计量学

Problem: Calculate the mass of carbon dioxide produced when 10.0 g of propane (C₃H₈) is completely burned in excess oxygen.

问题：10.0 g 丙烷（C₃H₈）在过量的氧气中完全燃烧，计算生成的二氧化碳的质量。

Step 1: Write the balanced chemical equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

步骤1：写出配平的化学方程式：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。

Step 2: Calculate moles of propane: M(C₃H₈) = 44.0 g mol⁻¹; n = m/M = 10.0 g / 44.0 g mol⁻¹ = 0.2273 mol.

步骤2：计算丙烷的物质的量：M(C₃H₈) = 44.0 g mol⁻¹；n = m/M = 10.0 g / 44.0 g mol⁻¹ = 0.2273 mol。

Step 3: Use mole ratio from equation: 1 mol C₃H₈ produces 3 mol CO₂, so moles of CO₂ = 0.2273 × 3 = 0.6818 mol.

步骤3：利用方程式中的摩尔比：1 mol C₃H₈ 生成 3 mol CO₂，因此 CO₂ 的物质的量 = 0.2273 × 3 = 0.6818 mol。

Step 4: Convert moles of CO₂ to mass: M(CO₂) = 44.0 g mol⁻¹; mass = 0.6818 mol × 44.0 g mol⁻¹ = 30.0 g (to 3 s.f.).

步骤4：将 CO₂ 的物质的量换算为质量：M(CO₂) = 44.0 g mol⁻¹；质量 = 0.6818 mol × 44.0 g mol⁻¹ = 30.0 g（保留三位有效数字）。

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2. Empirical and Molecular Formula | 经验式与分子式

Problem: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar mass is approximately 60 g mol⁻¹. Determine its empirical and molecular formulas.

问题：某化合物含碳40.0%、氢6.7%、氧53.3%（质量分数）。其摩尔质量约为60 g mol⁻¹。求其经验式和分子式。

Step 1: Assume 100 g of compound, so masses: C = 40.0 g, H = 6.7 g, O = 53.3 g.

步骤1：假设100 g 化合物，则质量：C = 40.0 g，H = 6.7 g，O = 53.3 g。

Step 2: Convert to moles: C: 40.0/12.0 = 3.33 mol; H: 6.7/1.0 = 6.7 mol; O: 53.3/16.0 = 3.33 mol.

步骤2：换算为物质的量：C: 40.0/12.0 = 3.33 mol；H: 6.7/1.0 = 6.7 mol；O: 53.3/16.0 = 3.33 mol。

Step 3: Divide by smallest mole value (3.33): C: 1, H: 2.01 ≈ 2, O: 1. Empirical formula is CH₂O.

步骤3：除以最小物质的量（3.33）：C: 1，H: 2.01 ≈ 2，O: 1。经验式为 CH₂O。

Step 4: Empirical formula mass = 12 + 2 + 16 = 30 g mol⁻¹. Since molar mass ≈ 60, n = 60/30 = 2. Molecular formula = (CH₂O)₂ = C₂H₄O₂.

步骤4：经验式质量 = 12 + 2 + 16 = 30 g mol⁻¹。摩尔质量 ≈ 60，因此 n = 60/30 = 2。分子式 = (CH₂O)₂ = C₂H₄O₂。

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3. Ideal Gas Calculations | 理想气体计算

Problem: A sample of 0.800 g of a volatile liquid is vaporised and occupies 250 cm³ at 100 °C and 101 kPa. Determine the molar mass of the liquid. (R = 8.31 J K⁻¹ mol⁻¹)

问题：0.800 g 挥发性液体气化后在100 °C、101 kPa下体积为250 cm³。求该液体的摩尔质量。(R = 8.31 J K⁻¹ mol⁻¹)

Step 1: Convert all units to SI: V = 250 cm³ = 2.50 × 10⁻⁴ m³; T = 373 K; p = 101 kPa = 1.01 × 10⁵ Pa.

步骤1：所有单位转换为国际单位：V = 250 cm³ = 2.50 × 10⁻⁴ m³；T = 373 K；p = 101 kPa = 1.01 × 10⁵ Pa。

Step 2: Use ideal gas law pV = nRT to find n: n = pV / (RT) = (1.01 × 10⁵ Pa × 2.50 × 10⁻⁴ m³) / (8.31 J K⁻¹ mol⁻¹ × 373 K) = 8.18 × 10⁻³ mol.

步骤2：利用理想气体状态方程 pV = nRT 求 n：n = pV / (RT) = (1.01 × 10⁵ Pa × 2.50 × 10⁻⁴ m³) / (8.31 J K⁻¹ mol⁻¹ × 373 K) = 8.18 × 10⁻³ mol。

Step 3: Molar mass M = mass / n = 0.800 g / 8.18 × 10⁻³ mol = 97.8 g mol⁻¹ (approx 98 g mol⁻¹).

步骤3：摩尔质量 M = 质量 / n = 0.800 g / 8.18 × 10⁻³ mol = 97.8 g mol⁻¹（约 98 g mol⁻¹）。

pV = nRT

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4. Hess’s Law and Enthalpy Changes | 赫斯定律与焓变

Problem: Given the following data: C(s) + O₂(g) → CO₂(g) ΔHᶿ = -394 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔHᶿ = -286 kJ mol⁻¹; CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔHᶿ = -890 kJ mol⁻¹. Calculate the enthalpy of formation of methane, CH₄.

问题：已知以下数据：C(s) + O₂(g) → CO₂(g) ΔHᶿ = -394 kJ mol⁻¹；H₂(g) + ½O₂(g) → H₂O(l) ΔHᶿ = -286 kJ mol⁻¹；CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔHᶿ = -890 kJ mol⁻¹。计算甲烷 CH₄ 的生成焓。

Step 1: Target equation: C(s) + 2H₂(g) → CH₄(g).

步骤1：目标方程式：C(s) + 2H₂(g) → CH₄(g)。

Step 2: Manipulate given equations: Keep reaction 1 as is; multiply reaction 2 by 2; reverse reaction 3. Thus:
(1) C(s) + O₂(g) → CO₂(g) -394
(2) 2H₂(g) + O₂(g) → 2H₂O(l) -572 (2 × -286)
(3) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) +890 (reversed, sign changed)

步骤2：调整已知方程式：保持反应1不变；反应2乘以2；反应3反转。因此：
(1) C(s) + O₂(g) → CO₂(g) -394
(2) 2H₂(g) + O₂(g) → 2H₂O(l) -572 (2 × -286)
(3) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) +890（反转，符号改变）

Step 3: Sum the enthalpies: ΔH = -394 + (-572) + 890 = -76 kJ mol⁻¹. Cancel common species: O₂, CO₂, H₂O. The sum gives C(s) + 2H₂(g) → CH₄(g). So ΔHfᶿ(CH₄) = -76 kJ mol⁻¹.

步骤3：焓变加和：ΔH = -394 + (-572) + 890 = -76 kJ mol⁻¹。消去相同物种：O₂、CO₂、H₂O。加和得到 C(s) + 2H₂(g) → CH₄(g)。因此 ΔHfᶿ(CH₄) = -76 kJ mol⁻¹。

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5. Kinetics: Rate Equations | 动力学：速率方程

Problem: For the reaction A + B → C, the following initial rate data were collected. Determine the rate equation and calculate the rate constant k.

问题：对于反应 A + B → C，收集了如下初始速率数据。确定速率方程并计算速率常数 k。

[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
0.10	0.10	2.0 × 10⁻⁴
0.20	0.10	4.0 × 10⁻⁴
0.10	0.20	8.0 × 10⁻⁴

Step 1: Compare experiments 1 and 2: [A] doubles, [B] constant, rate doubles → first order in A.

步骤1：比较实验1和2：[A] 翻倍，[B] 恒定，速率翻倍 → 对 A 为一级。

Step 2: Compare experiments 1 and 3: [B] doubles, [A] constant, rate quadruples (×4) → second order in B.

步骤2：比较实验1和3：[B] 翻倍，[A] 恒定，速率变为四倍 → 对 B 为二级。

Step 3: Rate equation: rate = k[A][B]².

步骤3：速率方程：rate = k[A][B]²。

Step 4: Using experiment 1: 2.0 × 10⁻⁴ = k (0.10)(0.10)² = k × 1.0 × 10⁻³. So k = 0.20 dm⁶ mol⁻² s⁻¹.

步骤4：使用实验1：2.0 × 10⁻⁴ = k (0.10)(0.10)² = k × 1.0 × 10⁻³。因此 k = 0.20 dm⁶ mol⁻² s⁻¹。

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6. Chemical Equilibrium and Kc | 化学平衡与Kc

Problem: 0.50 mol of ethanol, 0.50 mol of ethanoic acid, and 0.10 mol of water are mixed and allowed to reach equilibrium at 298 K. The equilibrium mixture contains 0.35 mol of ethyl ethanoate. Calculate Kc for the esterification reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O.

问题：将0.50 mol 乙醇、0.50 mol 乙酸和0.10 mol 水混合，在298 K下达到平衡。平衡混合物含有0.35 mol 乙酸乙酯。计算酯化反应 CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O 的 Kc。

Step 1: Let the change in moles be x. Since 0.35 mol of ester is formed, x = 0.35. At equilibrium: ester = 0.35, water = 0.10 + 0.35 = 0.45, acid = 0.50 – 0.35 = 0.15, ethanol = 0.50 – 0.35 = 0.15.

步骤1：设物质的量变化为 x。由于生成了0.35 mol 酯，x = 0.35。平衡时：酯 = 0.35，水 = 0.10 + 0.35 = 0.45，酸 = 0.50 – 0.35 = 0.15，乙醇 = 0.50 – 0.35 = 0.15。

Step 2: If the total volume is V dm³, concentrations are: [CH₃COOC₂H₅] = 0.35/V, [H₂O] = 0.45/V, [CH₃COOH] = 0.15/V, [C₂H₅OH] = 0.15/V.

步骤2：若总体积为 V dm³，浓度为：[CH₃COOC₂H₅] = 0.35/V，[H₂O] = 0.45/V，[CH₃COOH] = 0.15

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