📚 AQA International AS Chemistry CH02 Unit 2 Example Responses | AQA 国际 AS 化学第二单元示例作答
This revision guide walks through example responses for AQA International AS Chemistry Unit 2 (CH02). Use the model answers to understand how marks are awarded, how to structure calculations, and how to phrase explanations concisely for the written paper.
本复习指南梳理 AQA 国际 AS 化学第二单元（CH02）的示例作答。通过这些示范答案，你可以理解评分方式、掌握计算题的书写结构以及笔试中简洁表达解释的方法。
1. How to Use Example Responses | 如何使用示例作答
Example responses are most useful when you compare them against the mark scheme and identify exactly where each mark is gained. Do not memorise whole answers; instead, learn the key phrases, equations and definitions that examiners expect to see.
示例作答最有用的方式是与评分方案对照，找出每个得分点的来源。不要整段背诵答案，而要学习考官希望看到的关键短语、方程式和定义。
In CH02, marks are often split between correct working and final answer. Always show your working, give units, and quote answers to the appropriate number of significant figures.
在 CH02 中，分数通常分为计算过程和最终答案两部分。始终写出计算过程、标明单位，并使用适当的有效数字给出答案。
2. Energetics: Calorimetry and Enthalpy | 能量学：量热法与焓变
Example question: 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.8 °C. Assume the density of the solution is 1.00 g cm⁻³ and the specific heat capacity is 4.18 J g⁻¹ K⁻¹. Calculate the enthalpy change of neutralisation.
示例问题：将 50.0 cm³ 的 1.00 mol dm⁻³ HCl 与 50.0 cm³ 的 1.00 mol dm⁻³ NaOH 混合。温度上升 6.8 °C。假设溶液密度为 1.00 g cm⁻³，比热容为 4.18 J g⁻¹ K⁻¹。计算中和反应的焓变。
Model response: Total mass = 100 cm³ × 1.00 g cm⁻³ = 100 g. Heat absorbed, q = mcΔT = 100 g × 4.18 J g⁻¹ K⁻¹ × 6.8 K = 2842.4 J = 2.84 kJ. Moles of HCl = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol. ΔH = −q ÷ n = −2.84 kJ ÷ 0.0500 mol = −56.8 kJ mol⁻¹.
示范答案：总质量 = 100 cm³ × 1.00 g cm⁻³ = 100 g。吸收的热量 q = mcΔT = 100 g × 4.18 J g⁻¹ K⁻¹ × 6.8 K = 2842.4 J = 2.84 kJ。HCl 的物质的量 = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol。ΔH = −q ÷ n = −2.84 kJ ÷ 0.0500 mol = −56.8 kJ mol⁻¹。
Always include the negative sign because the temperature rise shows the reaction is exothermic. A common error is forgetting to convert cm³ to dm³ when calculating moles.
一定要加上负号，因为温度升高表明反应放热。常见错误是在计算物质的量时忘记将 cm³ 转换为 dm³。
3. Hess’s Law Cycles | 赫斯定律循环
Example question: Given C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ mol⁻¹ and CO(g) + ½O₂(g) → CO₂(g) ΔH = −283.0 kJ mol⁻¹, calculate the standard enthalpy of formation of CO(g).
示例问题：已知 C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ mol⁻¹ 和 CO(g) + ½O₂(g) → CO₂(g) ΔH = −283.0 kJ mol⁻¹，计算 CO(g) 的标准生成焓。
Model response: The target equation is C(s) + ½O₂(g) → CO(g). Use the cycle: ΔHf(CO) = ΔHf(CO₂) − ΔHc(CO) = −393.5 kJ mol⁻¹ − (−283.0 kJ mol⁻¹) = −110.5 kJ mol⁻¹.
示范答案：目标方程式为 C(s) + ½O₂(g) → CO(g)。利用循环：ΔHf(CO) = ΔHf(CO₂) − ΔHc(CO) = −393.5 kJ mol⁻¹ − (−283.0 kJ mol⁻¹) = −110.5 kJ mol⁻¹。
Drawing a labelled Hess cycle can help you avoid sign errors. Check that the arrow directions match the data you have been given.
画出带标注的赫斯循环有助于避免符号错误。检查箭头方向是否与题目给出的数据一致。
4. Kinetics: Maxwell-Boltzmann Distribution | 动力学：麦克斯韦-玻尔兹曼分布
Example question: Explain why increasing temperature increases the rate of a reaction.
示例问题：解释为什么升高温度会加快反应速率。
Model response: At higher temperature, the average kinetic energy of particles increases. The Maxwell-Boltzmann distribution shifts to the right and flattens, so a much larger proportion of particles have energy greater than or equal to the activation energy. This increases the frequency of successful collisions.
示范答案：温度升高时，粒子的平均动能增大。麦克斯韦-玻尔兹曼分布曲线向右移动并趋于平缓，因此能量大于或等于活化能的粒子比例大幅增加。成功碰撞的频率随之增大。
A catalyst increases rate by providing an alternative reaction pathway with lower activation energy, so more particles have sufficient energy without changing the temperature.
催化剂通过提供活化能较低的替代反应路径来加快反应速率，因此无需改变温度就有更多粒子具有足够能量。
5. Chemical Equilibria and Kc | 化学平衡与 Kc
Example question: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium concentrations are [N₂] = 0.50 mol dm⁻³, [H₂] = 0.20 mol dm⁻³ and [NH₃] = 0.10 mol dm⁻³. Calculate Kc and state its units.
示例问题：对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，平衡浓度分别为 [N₂] = 0.50 mol dm⁻³、[H₂] = 0.20 mol dm⁻³、[NH₃] = 0.10 mol dm⁻³。计算 Kc 并写出其单位。
Model response: Kc = [NH₃]² ÷ ([N₂][H₂]³) = (0.10)² ÷ (0.50 × 0.20³) = 0.010 ÷ 0.0040 = 2.5. Units: mol⁻² dm⁶.
示范答案：Kc = [NH₃]² ÷ ([N₂][H₂]³) = (0.10)² ÷ (0.50 × 0.20³) = 0.010 ÷ 0.0040 = 2.5。单位：mol⁻² dm⁶。
For Le Chatelier questions, state the change, the shift, and the reason. Increasing pressure shifts this equilibrium to the right because there are fewer moles of gas on the right-hand side.
对于勒夏特列原理的问题，要说明变化、平衡移动方向和原因。增大压力会使该平衡向右移动，因为右侧气体物质的量更少。
6. Redox and Oxidation States | 氧化还原与氧化态
Example question: Determine the oxidation state of manganese in MnO₄⁻ and write the half-equation for its reduction in acidic solution.
示例问题：确定 MnO₄⁻ 中锰的氧化态，并写出其在酸性溶液中还原的半反应方程式。
Model response: Oxygen is −2, so four O atoms contribute −8. The overall charge is −1, so Mn + (−8) = −1, giving Mn = +7. Half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
示范答案：氧的氧化态为 −2，四个氧原子共为 −8。总电荷为 −1，因此 Mn + (−8) = −1，得到 Mn = +7。半反应方程式：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。
Always balance atoms first, then charge using electrons. Check that the total charge is equal on both sides of a half-equation.
先配平原子，再用电子配平电荷。检查半反应方程式两边总电荷是否相等。
7. Organic Nomenclature and Isomerism | 有机命名与同分异构
Example question: Name the compound CH₃CH₂CH(CH₃)CH₂OH and draw one positional isomer.
示例问题：命名化合物 CH₃CH₂CH(CH₃)CH₂OH，并画出一种位置异构体。
Model response: The longest chain has four carbons, with a methyl group on carbon 2 and an −OH on carbon 1, so the name is 2-methylbutan-1-ol. A positional isomer is 3-methylbutan-1-ol, where the methyl group is on carbon 3 instead of carbon 2.
示范答案：最长碳链有四个碳，甲基位于 2 号碳，羟基位于 1 号碳，因此名称为 2-甲基丁-1-醇。一种位置异构体是 3-甲基丁-1-醇，甲基位于 3 号碳而不是 2 号碳。
Chain, position and functional group isomerism are all common in CH02. Always number the longest carbon chain so that the functional group gets the lowest possible locant.
碳链异构、位置异构和官能团异构在 CH02 中都很常见。给最长碳链编号时，应使官能团获得尽可能小的位次。
8. Free Radical Substitution of Alkanes | 烷烃自由基取代
Example question: Write equations for the initiation and propagation steps in the reaction of methane with chlorine under ultraviolet light.
示例问题：写出甲烷与氯气在紫外光下反应的引发步骤和增长步骤方程式。
Model response: Initiation: Cl₂ → 2Cl•. Propagation: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•.
示范答案：引发：Cl₂ → 2Cl•。增长：Cl• + CH₄ → •CH₃ + HCl，然后 •CH₃ + Cl₂ → CH₃Cl + Cl•。
The propagation steps produce the organic product and regenerate the chlorine radical, so the chain reaction continues. Termination steps consume radicals.
增长步骤生成有机产物并再生成氯自由基，因此链反应可以持续进行。终止步骤消耗自由基。
9. Electrophilic Addition of Alkenes | 烯烃亲电加成
Example question: Describe the mechanism for the reaction of ethene with hydrogen bromide to form bromoethane.
示例问题：描述乙烯与溴化氢反应生成溴乙烷的机理。
Model response: The C=C double bond attacks the hydrogen atom in HBr, using a curly arrow from the double bond to the H atom. The H−Br bond breaks heterolytically, with the electron pair moving to Br, forming Br⁻ and a carbocation CH₃CH₂⁺. The Br⁻ then attacks the carbocation to form CH₃CH₂Br.
示范答案：C=C 双键进攻 HBr 中的氢原子，用弯箭头从双键指向氢原子。H−Br 键发生异裂，电子对移向溴，形成 Br⁻ 和碳正离子 CH₃CH₂⁺。然后 Br⁻ 进攻碳正离子生成 CH₃CH₂Br。
Markovnikov’s rule applies when the alkene is unsymmetrical: the major product comes from the more stable carbocation intermediate.
当烯烃不对称时，马氏规则适用：主要产物来自更稳定的碳正离子中间体。
10. Alcohols: Oxidation and Dehydration | 醇的氧化与脱水
Example question: Describe how ethanol can be oxidised to ethanal and then to ethanoic acid, including reagents and conditions.
示例问题：描述乙醇如何被氧化为乙醛，再氧化为乙酸，包括试剂和条件。
Model response: Heat ethanol with acidified potassium dichromate(VI). To obtain ethanal, use distillation so the aldehyde is removed as it forms. To obtain ethanoic acid, use reflux with excess oxidising agent so oxidation goes to completion. The orange dichromate ion is reduced to green Cr³⁺.
示范答案：将乙醇与酸化重铬酸钾(VI)一起加热。要得到乙醛，使用蒸馏法使醛生成后即被分离。要得到乙酸，使用回流并加入过量氧化剂使氧化反应进行完全。橙色的重铬酸根离子被还原为绿色的 Cr³⁺。
Dehydration of ethanol to ethene can be achieved using hot aluminium oxide or concentrated sulfuric acid. Always state the conditions clearly in organic preparation questions.
乙醇脱水生成乙烯可使用热的氧化铝或浓硫酸。在有机制备题中，务必清楚写出条件。
11. Organic Analysis: IR Spectroscopy | 有机分析：红外光谱
Example question: Explain how infrared spectroscopy can distinguish between ethanol and ethanoic acid.
示例问题：解释红外光谱如何区分乙醇和乙酸。
Model response: Ethanol shows a broad O−H absorption at about 3200–3550 cm⁻¹ and a C−O absorption around 1000–1300 cm⁻¹. Ethanoic acid also shows an O−H absorption, but it has a characteristic C=O absorption at about 1680–1750 cm⁻¹, which ethanol lacks.
示范答案：乙醇在约 3200–3550 cm⁻¹ 处有宽 O−H 吸收峰，在约 1000–1300 cm⁻¹ 处有 C−O 吸收峰。乙酸也有 O−H 吸收峰，但它在约 1680–1750 cm⁻¹ 处有特征的 C=O 吸收峰，而乙醇没有。
You should quote absorption ranges, not just peak names, and link them to the functional group present.
答题时应写出吸收范围，而不只是峰的称呼，并将其与所含官能团联系起来。
12. Common Pitfalls in CH02 | CH02 常见失分点
Many marks are lost through missing units, incorrect significant figures, or forgetting the minus sign in exothermic enthalpy changes. Always check these before moving on.
许多失分来自遗漏单位、有效数字错误或放热焓变缺少负号。答题后务必检查这些项目。
In organic mechanisms, examiners look for the correct direction of curly arrows and the correct charges on intermediates. In equilibrium questions, make sure Kc expressions match the balanced equation and include units where required.
在有机机理题中，考官关注弯箭头方向和中间体电荷是否正确。在平衡题中，确保 Kc 表达式与配平方程式一致，并在需要时写出单位。
Practise writing model responses under timed conditions, then mark them using the AQA mark scheme. This will help you spot exactly where your answers could gain or lose marks in CH02 Unit 2.
在限时条件下练习书写示范答案，然后对照 AQA 评分方案自行批改。这样可以准确发现在 CH02 第二单元中哪些地方可能得分或失分。
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