📚 Typical Example Questions Explained for CCEA A-Level Chemistry | CCEA A-Level 化学典型例题详解
This revision guide presents a selection of typical worked examples covering core topics from the CCEA A-Level Chemistry specification. Each problem is solved step by step with clear reasoning, highlighting common pitfalls and examination technique. The bilingual format helps reinforce both chemical concepts and precise scientific language.
本复习指南精选了 CCEA A-Level 化学考纲核心主题的典型例题，通过分步解析展示清晰的推理过程，突出常见易错点与应试技巧。中英对照的形式有助于巩固化学概念和准确的科学表达。
1. Mole Concept and Molar Mass Calculations | 摩尔概念与摩尔质量计算
Example: Calculate the amount of substance (in mol) present in 8.20 g of calcium carbonate, CaCO₃. (Ar values: Ca = 40.1, C = 12.0, O = 16.0)
例题：计算 8.20 g 碳酸钙 (CaCO₃) 的物质的量（mol）。（相对原子质量：Ca = 40.1, C = 12.0, O = 16.0）
Solution: First determine the molar mass, M, of CaCO₃. M = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹. Then use moles = mass ÷ M: n = 8.20 g / 100.1 g mol⁻¹ ≈ 0.0819 mol. Always show the unit and round to three significant figures where appropriate.
解答：首先求出 CaCO₃ 的摩尔质量 M = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹。然后使用公式物质的量 = 质量 ÷ 摩尔质量：n = 8.20 g / 100.1 g mol⁻¹ ≈ 0.0819 mol。注意单位并合理保留三位有效数字。
2. Empirical and Molecular Formulae | 实验式与分子式
Example: A hydrocarbon contains 85.7% carbon by mass and has a relative molecular mass of 56.0. Determine its empirical formula and molecular formula.
例题：某碳氢化合物含碳 85.7%（质量分数），其相对分子质量为 56.0。求其实验式与分子式。
Solution: Assume 100 g of compound: mass of C = 85.7 g, mass of H = 14.3 g. Moles of C = 85.7 / 12.0 = 7.14 mol; moles of H = 14.3 / 1.0 = 14.3 mol. Divide by smallest (7.14): C = 1, H ≈ 2. So empirical formula is CH₂. Empirical formula mass = 12.0 + 2.0 = 14.0. Since Mr = 56.0, the multiplier is 56.0 / 14.0 = 4. Molecular formula = C₄H₈.
解答：假设样品 100 g，m(C) = 85.7 g，m(H) = 14.3 g。n(C) = 85.7 / 12.0 = 7.14 mol，n(H) = 14.3 / 1.0 = 14.3 mol。除以较小值 7.14，得 C : H = 1 : 2，实验式为 CH₂。实验式量 = 14.0，由相对分子质量 56.0 得倍数 4，故分子式为 C₄H₈。
3. Gas Calculations Using the Ideal Gas Equation | 运用理想气体方程的气体计算
Example: Calculate the volume occupied by 0.500 mol of nitrogen gas at 298 K and 100 kPa. (R = 8.31 J K⁻¹ mol⁻¹)
例题：计算 0.500 mol 氮气在 298 K、100 kPa 下所占的体积。（R = 8.31 J K⁻¹ mol⁻¹）
Solution: Use pV = nRT. Convert pressure to Pa: 100 kPa = 100 × 10³ Pa = 1.00 × 10⁵ Pa. Rearrange: V = nRT / p = (0.500 mol × 8.31 J K⁻¹ mol⁻¹ × 298 K) / (1.00 × 10⁵ Pa). Calculate numerator = 0.500 × 8.31 × 298 ≈ 1238 J. Since 1 J = 1 Pa m³, V ≈ 1238 / 1.00 × 10⁵ = 0.01238 m³ = 12.4 dm³ (or 12.4 L). Common mistake: forgetting to convert kPa to Pa or cm³ to m³.
解答：运用 pV = nRT，将压强换算为 Pa：100 kPa = 1.00 × 10⁵ Pa。变形得 V = nRT / p = (0.500 × 8.31 × 298) / (1.00 × 10⁵)。分子计算约得 1238 J，1 J = 1 Pa m³，故 V = 0.01238 m³ = 12.4 dm³。常见错误：未将 kPa 转为 Pa 或未统一体积单位。
4. Enthalpy Changes and Hess’s Law | 焓变与盖斯定律
Example: Given the following data: C(s) + O₂(g) → CO₂(g) ΔH = -394 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔH = -286 kJ mol⁻¹; CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = -890 kJ mol⁻¹. Calculate the standard enthalpy of formation of methane, CH₄.
例题：已知以下数据：C(s) + O₂(g) → CO₂(g) ΔH = -394 kJ mol⁻¹；H₂(g) + ½O₂(g) → H₂O(l) ΔH = -286 kJ mol⁻¹；CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = -890 kJ mol⁻¹。计算甲烷 CH₄ 的标准生成焓。
Solution: The formation reaction is C(s) + 2H₂(g) → CH₄(g). Use Hess’s Law to combine given equations. Route: target = CO₂ + 2H₂O → CH₄ + 2O₂ (reverse of combustion) plus C + O₂ → CO₂ and 2×(H₂ + ½O₂ → H₂O). ΔH_f = ΔH_combustion(C) + 2×ΔH_combustion(H₂) – ΔH_combustion(CH₄) = (-394) + 2×(-286) – (-890) = -394 -572 +890 = -76 kJ mol⁻¹. Thus the standard enthalpy of formation of methane is -76 kJ mol⁻¹.
解答：生成反应为 C(s) + 2H₂(g) → CH₄(g)。运用盖斯定律：目标反应可通过燃烧反应逆向和单质燃烧组合。ΔH_f = ΔH_c(C) + 2×ΔH_c(H₂) – ΔH_c(CH₄) = (-394) + 2×(-286) – (-890) = -76 kJ mol⁻¹。故甲烷的标准生成焓为 -76 kJ mol⁻¹。
5. Chemical Equilibrium and Kc Calculations | 化学平衡与 Kc 计算
Example: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium concentrations at a certain temperature are: [N₂] = 0.40 mol dm⁻³, [H₂] = 0.10 mol dm⁻³, [NH₃] = 0.20 mol dm⁻³. Calculate the equilibrium constant Kc and state its units.
例题：对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，某温度下的平衡浓度分别为 [N₂] = 0.40 mol dm⁻³，[H₂] = 0.10 mol dm⁻³，[NH₃] = 0.20 mol dm⁻³。计算平衡常数 Kc 并注明单位。
Solution: The Kc expression is:
Kc = [NH₃]² / ([N₂][H₂]³)
Substitute: Kc = (0.20)² / (0.40 × (0.10)³) = 0.040 / (0.40 × 0.0010) = 0.040 / 0.00040 = 100. Unit analysis: (mol dm⁻³)² / (mol dm⁻³)(mol dm⁻³)³ = mol⁻² dm⁶. So Kc = 100 mol⁻² dm⁶.
解答：Kc 表达式如上。代入数值：Kc = (0.20)² / (0.40 × (0.10)³) = 0.040 / 0.00040 = 100。量纲分析：分子 (mol dm⁻³)²，分母 (mol dm⁻³)¹⁺³ = mol⁴ dm⁻¹²，结果为 mol⁻² dm⁶。故 Kc = 100 mol⁻² dm⁶。
6. Acid–Base Equilibria and pH | 酸碱平衡与 pH
Example (a): Calculate the pH of a 0.0500 mol dm⁻³ solution of HCl. (b) For a 0.100 mol dm⁻³ solution of ethanoic acid (CH₃COOH, Ka = 1.74 × 10⁻⁵ mol dm⁻³), calculate the pH. State any assumptions made.
例题（a）：计算 0.0500 mol dm⁻³ HCl 溶液的 pH。（b）对于 0.100 mol dm⁻³ 的乙酸溶液（CH₃COOH，Ka = 1.74 × 10⁻⁵ mol dm⁻³），计算其 pH，并说明所作的假设。
Solution (a): HCl is a strong acid, fully dissociated. [H⁺] = 0.0500 mol dm⁻³. pH = -log(0.0500) = 1.30.
Solution (b): CH₃COOH ⇌ CH₃COO⁻ + H⁺. Ka = [H⁺]² / [CH₃COOH] assuming [H⁺] << initial acid concentration. Then [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) ≈ 1.32 × 10⁻³ mol dm⁻³. pH = -log(1.32 × 10⁻³) = 2.88. The approximation is valid because [H⁺] is less than 5% of 0.100.
解答（a）：HCl 为强酸，完全电离。[H⁺] = 0.0500 mol dm⁻³，pH = 1.30。
解答（b）：乙酸为弱酸，建立平衡。假设 [H⁺] 远小于初始浓度，则 [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) ≈ 1.32 × 10⁻³ mol dm⁻³，pH = 2.88。该近似成立，因为 [H⁺] 小于 0.100 的 5%。
7. Redox Titration – Manganate(VII) with Iron(II) | 氧化还原滴定—高锰酸钾与亚铁离子
Example: In a titration, 25.0 cm³ of 0.0200 mol dm⁻³ KMnO₄ solution (acidified with dilute H₂SO₄) reacted completely with 20.0 cm³ of a solution of Fe²⁺ ions. The equation for the reaction is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Calculate the concentration of Fe²⁺ ions in the original solution.
例题：在一次滴定中，25.0 cm³ 0.0200 mol dm⁻³ 的酸性 KMnO₄ 溶液与 20.0 cm³ Fe²⁺ 溶液完全反应。反应方程式为：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。计算原溶液中 Fe²⁺ 的浓度。
Solution: Moles of MnO₄⁻ used = concentration × volume (in dm³) = 0.0200 × (25.0/1000) = 5.00 × 10⁻⁴ mol. From the equation, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺. Moles of Fe²⁺ = 5 × 5.00 × 10⁻⁴ = 2.50 × 10⁻³ mol. Concentration of Fe²⁺ = moles / volume (dm³) = (2.50 × 10⁻³) / (20.0/1000) = 0.125 mol dm⁻³. Ensure volumes are converted to dm³ consistently.
解答：KMnO₄ 的物质的量 = 0.0200 × (25.0/1000) = 5.00 × 10⁻⁴ mol。由方程式知 MnO₄⁻ 与 Fe²⁺ 的物质的量之比为 1:5，故 Fe²⁺ 的物质的量 = 5 × 5.00 × 10⁻⁴ = 2.50 × 10⁻³ mol。Fe²⁺ 浓度 = 2.50 × 10⁻³ / (20.0/1000) = 0.125 mol dm⁻³。注意统一体积单位为 dm³。
8. Organic Reaction Mechanisms – Nucleophilic Substitution | 有机反应机理—亲核取代
Example: Describe the mechanism for the reaction of bromoethane with aqueous sodium hydroxide, naming the organic product. Include curly arrows and relevant conditions.
例题：描述溴乙烷与氢氧化钠水溶液反应的机理，命名有机产物。需包括弯箭头及相关反应条件。
Solution: The reaction is a nucleophilic substitution (SN2). Conditions: warm under reflux with aqueous NaOH. The hydroxide ion, OH⁻, acts as a nucleophile. It attacks the slightly positive carbon (δ⁺) of the C–Br bond from the opposite side to the bromine atom. A curly arrow is drawn from the lone pair on OH⁻ to the δ⁺ carbon. Simultaneously, the C–Br bond breaks heterolytically, with both electrons moving to the bromine, shown by a curly arrow from the bond to Br. The transition state involves a pentacoordinate carbon. The organic product is ethanol, CH₃CH₂OH. The bromide ion is released.
解答：该反应为亲核取代（SN2 机理）。条件：氢氧化钠水溶液，加热回流。OH⁻ 作为亲核试剂，攻击 C–Br 键中带部分正电荷的碳（δ⁺），从溴原子的背面进攻。弯箭头从 OH⁻ 的孤对电子指向 δ⁺ 碳；同时 C–Br 键异裂，一对电子移向溴原子（弯箭头从键指向 Br）。过渡态为五价碳中间体。有机产物是乙醇 CH₃CH₂OH，并释放溴离子。
9. Rates of Reaction – Determining Rate Equations | 反应速率—确定速率方程
Example: The reaction 2NO(g) + O₂(g) → 2NO₂(g) was studied at a fixed temperature. The initial rate data are tabulated below. Determine the rate equation and calculate the rate constant, k.
| Experiment | [NO] / mol dm⁻³ | [O₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻³ |
例题：反应 2NO(g) + O₂(g) → 2NO₂(g) 在恒温下研究，初始速率数据见上表。确定速率方程并计算速率常数 k。
Solution: Compare experiments 1 and 2: [NO] doubles, [O₂] constant, rate increases from 2.0×10⁻³ to 8.0×10⁻³, i.e. quadruples. So rate ∝ [NO]² (order 2 with respect to NO). Compare experiments 1 and 3: [O₂] doubles, [NO] constant, rate doubles from 2.0×10⁻³ to 4.0×10⁻³. Hence rate ∝ [O₂]¹ (first order). Rate equation: rate = k [NO]² [O₂]. To find k, use data from expt 1: 2.0×10⁻³ = k × (0.10)² × (0.10) → k = 2.0×10⁻³ / 1.0×10⁻³ = 2.0. Units: mol⁻² dm⁶ s⁻¹. So k = 2.0 mol⁻² dm⁶ s⁻¹.
解答：对比实验 1 和 2：[NO] 加倍，[O₂] 不变，速率增大为四倍，故对 NO 为二级。对比实验 1 和 3：[O₂] 加倍，速率加倍，对 O₂ 为一级。速率方程：rate = k [NO]² [O₂]。代入实验 1 数据：2.0×10⁻³ = k × (0.10)² × 0.10，得 k = 2.0 mol⁻² dm⁶ s⁻¹。
10. Electrochemical Cells and Standard Electrode Potentials | 电化学电池与标准电极电势
Example: A cell is made from a Zn²⁺/Zn half-cell (E° = -0.76 V) and a Cu²⁺/Cu half-cell (E° = +0.34 V). Write the cell diagram and calculate the standard emf of the cell. Identify which electrode is the anode and write the overall cell reaction.
例题：由 Zn²⁺/Zn 半电池（E° = -0.76 V）和 Cu²⁺/Cu 半电池（E° = +0.34 V）组成电池。画出电池图式，计算标准电动势，指出哪一极为负极并写出电池总反应。
Solution: The more negative electrode is the anode (oxidation occurs). Here Zn²⁺/Zn is more negative, so Zn electrode is the anode. Cell diagram: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Standard emf E°cell = E°cathode – E°anode = +0.34 – (-0.76) = +1.10 V. Anode half-reaction (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻. Cathode half-reaction (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s). Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
解答：电极电势更负的一极作负极，发生氧化。故锌电极为负极。电池图式：Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)。电动势 E°cell = +0.34 – (-0.76) = +1.10 V。负极半反应：Zn → Zn²⁺ + 2e⁻；正极半反应：Cu²⁺ + 2e⁻ → Cu。总反应：Zn + Cu²⁺ → Zn²⁺ + Cu。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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