📚 AS Chemistry: Detailed Walkthrough of Typical Exam Questions | AS 化学：典型例题详解
Mastering the calculation-based and conceptual questions in AS Chemistry requires a solid grasp of core principles and plenty of worked examples. This article presents eight typical exam-style questions with step-by-step bilingual solutions, covering topics such as the mole concept, titration, energetics, equilibrium, kinetics, bonding, organic nomenclature, and redox reactions. Each worked example is designed to reinforce key skills and build confidence for high-stakes assessments.
掌握AS化学中的计算题和概念题，需要扎实理解核心原理，并通过大量典型例题加以巩固。本文精选八个常见的考试风格题目，提供逐步的中英双语解答，内容涵盖摩尔概念、滴定、能量学、平衡、动力学、化学键、有机命名以及氧化还原反应。每个例题都旨在强化关键技能，并为应对重要考试建立信心。
1. Mole Concept and Empirical Formula Calculations | 摩尔概念与经验式计算
In AS Chemistry, empirical formula questions ask you to find the simplest whole‑number ratio of atoms in a compound from mass or percentage composition data. You must be comfortable converting mass to moles and reducing ratios.
在AS化学中，经验式题目要求根据质量或百分比组成数据求出化合物中原子个数的最简整数比。你必须熟练地将质量转化为摩尔数，并会化简比例。
Example: A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. Its relative molecular mass is about 46. Determine its empirical and molecular formulae.
例题：某化合物含碳52.2%，氢13.0%，氧34.8%（按质量计），相对分子质量约为46。确定其实验式和分子式。
Step 1: Assume 100 g of the compound, so the masses are 52.2 g C, 13.0 g H, and 34.8 g O.
步骤1：假设样品质量为100 g，则各元素质量为 C 52.2 g、H 13.0 g、O 34.8 g。
Moles C = 52.2 ÷ 12.0 = 4.35 mol; H = 13.0 ÷ 1.0 = 13.0 mol; O = 34.8 ÷ 16.0 = 2.175 mol.
步骤2：使用相对原子质量 Aᵣ (C=12.0, H=1.0, O=16.0) 将质量转化为摩尔数。C: 52.2 ÷ 12.0 = 4.35 mol；H: 13.0 ÷ 1.0 = 13.0 mol；O: 34.8 ÷ 16.0 = 2.175 mol。
5 : 2.175/2.175 ≈ 2 : 5.98 : 1, which rounds to 2 : 6 : 1. The empirical formula is C₂H₆O.
步骤3：用最小摩尔数 (2.175 mol) 去除各值。比例 C:H:O = 4.35/2.175 : 13.0/2.175 : 2.175/2.175 ≈ 2 : 5.98 : 1，取整得 2 : 6 : 1。实验式为 C₂H₆O。
. Since the given relative molecular mass is 46, the molecular formula is the same as the empirical formula: C₂H₆O.
步骤4：求分子式。实验式单元 (C₂H₆O) 的相对质量 = (2×12)+(6×1)+16=46。已知相对分子质量约为46，因此分子式与实验式相同，即 C₂H₆O。
2. Titration Calculations | 滴定计算
Titration questions test your ability to use the balanced equation linking the acid and base, and to handle concentration, volume, and mole ratios. A common task is to find the unknown concentration of a solution.
滴定题目考查你运用酸碱平衡方程式以及处理浓度、体积和摩尔比的能力。常见任务是求出溶液的未知浓度。
Example: In a titration, 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide (NaOH) solution is exactly neutralised by 20.0 cm³ of sulfuric acid (H₂SO₄). Calculate the concentration of the sulfuric acid.
例题：某滴定中，25.0 cm³ 0.100 mol dm⁻³ 氢氧化钠 (NaOH) 溶液恰好被 20.0 cm³ 硫酸 (H₂SO₄) 中和。计算硫酸的浓度。
e balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The mole ratio of acid to base is 1:2.
步骤1：写出配平的化学方程式：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。酸和碱的摩尔比为 1:2。
e of NaOH = 25.0 cm³ = 0.0250 dm³. Moles NaOH = 0.100 mol dm⁻³ × 0.0250 dm³ = 0.00250 mol.
步骤2：计算所用 NaOH 的摩尔数。摩尔数 = 浓度 × 体积（以 dm³ 计）。NaOH 体积 = 25.0 cm³ = 0.0250 dm³。NaOH 摩尔数 = 0.100 mol dm⁻³ × 0.0250 dm³ = 0.00250 mol。
ion, 2 mol NaOH react with 1 mol H₂SO₄, so moles of H₂SO₄ = 0.00250 mol ÷ 2 = 0.00125 mol.
步骤3：利用摩尔比求出 H₂SO₄ 的摩尔数。由方程式知 2 mol NaOH 与 1 mol H₂SO₄ 反应，故 H₂SO₄ 的摩尔数 = 0.00250 mol ÷ 2 = 0.00125 mol。
= 0.0200 dm³. Concentration = moles ÷ volume = 0.00125 mol ÷ 0.0200 dm³ = 0.0625 mol dm⁻³.
步骤4：计算 H₂SO₄ 浓度。酸的体积 = 20.0 cm³ = 0.0200 dm³。浓度 = 摩尔数 ÷ 体积 = 0.00125 mol ÷ 0.0200 dm³ = 0.0625 mol dm⁻³。
3. Enthalpy Change Determination | 焓变测定
rred, then convert this to enthalpy change per mole. Pay attention to sign conventions (exothermic negative, endothermic positive).
量热法实验让你通过测量温度变化并使用 q = m c ΔT 计算传递的能量，进而换算成每摩尔的焓变。注意符号约定（放热为负，吸热为正）。
Example: When 2.00 g of ammonium nitrate (NH₄NO₃) is dissolved in 50.0 cm³ of water in a polystyrene cup, the temperature falls from 20.5 °C to 17.0 °C. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ °C⁻¹ and the density is 1.00 g cm⁻³. Calculate the enthalpy change of solution of NH₄NO₃ in kJ mol⁻¹.
例题：将 2.00 g 硝酸铵 (NH₄NO₃) 溶于 50.0 cm³ 水中（在聚苯乙烯杯中），温度从 20.5 °C 降至 17.0 °C。假设溶液的比热容为 4.18 J g⁻¹ °C⁻¹，密度为 1.00 g cm⁻³。计算 NH₄NO₃ 的溶解焓变，以 kJ mol⁻¹ 表示。
ulate the mass of the solution. Mass = volume × density = 50.0 cm³ × 1.00 g cm⁻³ = 50.0 g.
步骤1：计算溶液质量。质量 = 体积 × 密度 = 50.0 cm³ × 1.00 g cm⁻³ = 50.0 g。
Step 2: Determine the temperature change ΔT. Final temperature 17.0 °C – initial 20.5 °C = –3.5 °C (negative indicates a temperature drop).
步骤2：确定温度变化 ΔT。最终温度 17.0 °C – 初始 20.5 °C = –3.5 °C（负号表示温度下降）。
Calculate the heat absorbed by the solution (q). q = m c ΔT = 50.0 g × 4.18 J g⁻¹ °C⁻¹ × (–3.5 °C) = – the dissolving salt is equal in magnitude to the heat lost by the water. So q = +731.5 J.
步骤3：计算溶液吸收的热量 (q)。q = m c ΔT = 50.0 g × 4.18 J g⁻¹ °C⁻¹ × (–3.5 °C) = –731.5 J。负号表示体系从环境吸热，即过程为吸热。体系吸收的能量为 +731.5 J，因为溶解的盐吸收的热量大小等于水失去的热量。所以取 q = +731.5 J。
ar mass = 14 + (4×1) + 14 + (3×16) = 80 g mol⁻¹. Moles = 2.00 g ÷ 80 g mol⁻¹ = 0.0250 mol.
步骤4：求 NH₄NO₃ 的摩尔数。摩尔质量 = 14 + (4×1) + 14 + (3×16) = 80 g mol⁻¹。摩尔数 = 2.00 g ÷ 80 g mol⁻¹ = 0.0250 mol。
J mol⁻¹ ≈ +29.3 kJ mol⁻¹. The enthalpy change of solution is +29.3 kJ mol⁻¹ (endothermic).
步骤5：计算每摩尔焓变。ΔH = +q ÷ 摩尔数 = +731.5 J ÷ 0.0250 mol = +29260 J mol⁻¹ ≈ +29.3 kJ mol⁻¹。溶解焓变为 +29.3 kJ mol⁻¹（吸热）。
ΔHₛₒₗ = +29.3 kJ mol⁻¹
4. Equilibrium Constant Kc Calculations | 平衡常数 Kc 计算
For a reversible reaction at a given temperature, Kc expresses the ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient. You must construct the expression carefully and substitute equilibrium concentrations.
对于一定温度下的可逆反应，Kc 表示生成物浓度与反应物浓度的比值，各浓度以计量系数为指数。你需要认真构建表达式并代入平衡浓度。
Example: 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed in a flask with a small amount of acid catalyst. The total volume is 1.0 dm³. At equilibrium, 0.30 mol of ethyl ethanoate is present. Calculate Kc for: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O.
例题：将 0.50 mol 乙酸和 0.50 mol 乙醇在含少量酸催化剂的烧瓶中混合，总体积为 1.0 dm³。达到平衡时，存在 0.30 mol 乙酸乙酯。计算以下反应的 Kc：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。
CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]). The volume is constant so we can use molar amounts proportional to concentrations.
步骤1：写出 Kc 表达式。Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH])。由于体积恒定，摩尔数与浓度成正比，可直接用摩尔数代替浓度计算比例。
in ester = +0.30. Because of the 1:1:1:1 stoichiometry, the change for acid and ethanol is –0.30, and for water +0.30.
步骤2：建立 ICE 表（初始、变化、平衡）。初始摩尔数：酸 = 0.50，醇 = 0.50，酯 = 0，水 = 0。平衡时酯 = 0.30 mol，因此酯的变化 = +0.30。由于化学计量比为 1:1:1:1，酸和醇的变化为 –0.30，水为 +0.30。
Equilibrium moles: CH₃COOH = 0.50 – 0.30 = 0.20 mol; C₂H₅OH = 0.20 mol; CH₃COOC₂H₅ = 0.30 mol; H₂O = 0.30 mol.
平衡摩尔数：CH₃COOH = 0.50 – 0.30 = 0.20 mol；C₂H₅OH = 0.20 mol；CH₃COOC₂H₅ = 0.30 mol；H₂O = 0.30 mol。
/ 0.04 = 2.25. The units cancel (dm³/mol each side), so Kc is dimensionless in this case.
步骤3：因总体积为 1.0 dm³，平衡浓度数值与摩尔数相同（mol dm⁻³）。代入 Kc 表达式：Kc = (0.30 × 0.30) / (0.20 × 0.20) = 0.09 / 0.04 = 2.25。单位约去（两侧均为 dm³/mol），故此处 Kc 无单位。
Kc = 2.25 (no units)
5. Rate of Reaction and Activation Energy | 反应速率和活化能
Rate questions often require you to interpret initial‑rates data or the effect of temperature on rate. The Arrhenius relationship is often expressed through the qualitative rule that a small temperature rise causes a significant increase in rate due to the greater proportion of molecules exceeding the activation energy.
速率题常要求分析初始速率数据或温度对速率的影响。Arrhenius 关系常通过定性规律来体现：由于超过活化能的分子比例大幅增加，温度的小幅升高会导致速率显著增大。
lculate the approximate percentage increase in rate and explain why it is so large in terms of activation energy.
例题：过氧化氢的分解由碘离子催化。实验测得 20 °C 时初始速率为 2.4 × 10⁻⁵ mol dm⁻³ s⁻¹。当温度升至 30 °C 时，初始速率为 6.0 × 10⁻⁵ mol dm⁻³ s⁻¹。计算速率的大致增长率，并从活化能角度解释增长幅度为何如此之大。
Step 1: Calculate the percentage increase in rate. Increase = 6.0 × 10⁻⁵ – 2.4 × 10⁻⁵ = 3.6 × 10⁻⁵. Percentage increase = (3.6 × 10⁻⁵ / 2.4 × 10⁻⁵) × 100% = 150%.
步骤1：计算速率增长率。增加量 = 6.0 × 10⁻⁵ – 2.4 × 10⁻⁵ = 3.6 × 10⁻⁵。增长率 = (3.6 × 10⁻⁵ / 2.4 × 10⁻⁵) × 100% = 150%。
number of molecules with energy ≥ activation energy (Eₐ) increases dramatically. Even a 10‑ iple the fraction of successful collisions, as reflected in the Arrhenius equation k = A e–Eₐ/RT.
步骤2：用分子能量分布解释。温度升高时，能量分布曲线变得平坦并向右偏移。能量 ≥ 活化能 (Eₐ) 的分子数目急剧增加。即使升高 10 度，具有足够能量的碰撞分数也可能增加一倍或两倍，这反映在 Arrhenius 公式 k = A e–Eₐ/RT 中。
Step 3: Apply to this case. The 150% increase indicates that a relatively high activation energy is involved. Many more molecules can overcome the barrier, so the rate is very sensitive to temperature change. This is a strong signal for a catalysed reaction where Eₐ has been lowered but is still significant.
步骤3：联系本题。150% 的增长率表明该反应涉及较高的活化能。能够越过能垒的分子大大增多，因而速率对温度变化非常敏感。这证实了催化反应虽然活化能被降低，但仍相当可观。
6. Shapes of Molecules and Bond Angles | 分子形状和键角
VSEPR theory predicts molecular geometry based on the repulsion between electron pairs around a central atom. You must count bonding and lone pairs, then deduce the shape and typical bond angles.
VSEPR 理论基于中心原子周围电子对间的排斥作用预测分子几何构型。你需要统计成键电子对和孤对电子数，然后推导出形状和典型键角。
Example: Predict the shape and bond angle of (a) BF₃, (b) NH₃, (c) H₂O.
例题：预测 (a) BF₃、(b) NH₃、(c) H₂O 的形状和键角。
n density, the arrangement that minimises repulsion is trigonal planar. Bond angle = 120°.
(a) BF₃：硼原子有 3 个成键电子对，无孤对电子。三个电子密度区域的最佳排布为平面三角形，键角 = 120°。
(b) NH₃: Nitrogen has 3 bonding pairs and 1 lone pair. Four electron pairs give a tetrahedral electron-pair geometry, but the lone pair exerts stronger repulsion, reducing the H–N–H bond angle to approximately 107°. The molecular shape is trigonal pyramidal.
(b) NH₃：氮原子有 3 个成键电子对和 1 个孤对电子。四对电子形成四面体电子对排布，但孤对电子的排斥力更强，将 H–N–H 键角压缩至约 107°。分子形状为三角锥形。
(c) H₂O: Oxygen has 2 bonding pairs and 2 lone pairs. The electron pairs are arranged tetrahedrally, but the two lone pairs compress the bond angle to about 104.5°. The molecular shape is bent or V‑shaped.
(c) H₂O：氧原子有 2 个成键电子对和 2 个孤对电子。电子对呈四面体排布，但两对孤对电子使键角压缩至约 104.5°。分子形状为弯曲形或 V 形。
A handy summary table:
一个方便的小结表格：
| Molecule | Bonding pairs | Lone pairs | Shape | Bond angle |
|---|---|---|---|---|
| BF₃ | 3 | 0 | Trigonal planar | 120° |
| NH₃ | 3 | 1 | Trigonal pyramidal | 107° |
| H₂O | 2 | 2 | Bent / V‑shaped | 104.5° |
7.
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