📚 Common Mistake Questions in IGCSE AQA Chemistry | IGCSE AQA 化学易错题精讲
Many students lose marks in IGCSE AQA Chemistry not because they don’t understand the concepts, but because they fall into classic traps set by examiners. This article picks out the most frequently misunderstood topics and common mistake questions, illustrating with worked examples and step-by-step corrections. Mastering these will sharpen your exam technique and boost your grade.
许多学生在 IGCSE AQA 化学考试中失分，并不是因为不理解概念，而是因为掉进了考官设置的经典陷阱。本文挑选了最常被误解的主题和易错题，通过范例和逐步纠错进行讲解。掌握这些内容将提高你的应试技巧，提升成绩。
1. Mole Calculations with Gas Volumes | 气体体积的摩尔计算
Mistake example: A student calculates the volume of 0.5 mol of CO2 at room temperature and pressure (rtp) as 12 dm3 using the formula ‘volume = moles × 24’. The answer is correct, but then the student states the volume of 0.5 mol of H2 under the same conditions is also 12 dm3 and writes that it is heavier. Where is the misconception?
易错案例：某学生计算室温常压下 0.5 mol CO2 的体积，用 “体积 = 摩尔数 × 24” 得到 12 dm3，答案正确。接着该学生说相同条件下 0.5 mol H2 的体积也是 12 dm3，但认为它更重。误解在哪？
The trap: many learners confuse volume with mass. At the same temperature and pressure, one mole of any gas occupies the same volume (24 dm3 at rtp). So 0.5 mol of any gas will indeed have a volume of 12 dm3. However, the mass differs because molar mass differs. The student mistakenly thought that equal volume means equal mass, which is wrong. The 0.5 mol of CO2 has a mass of 0.5 × 44 = 22 g, while 0.5 mol of H2 has a mass of only 1 g. Understanding that ‘moles’ link mass to volume is crucial.
陷阱：许多学习者混淆了体积和质量。在相同温度和压力下，1 摩尔任何气体的体积都相同（rtp 下为 24 dm3）。因此 0.5 mol 任何气体确实体积都是 12 dm3。但质量不同，因为摩尔质量不同。学生错误地认为等体积意味着等质量，这是错误的。0.5 mol CO2 的质量是 0.5 × 44 = 22 g，而 0.5 mol H2 的质量只有 1 g。理解 “摩尔” 能把质量和体积联系起来至关重要。
Another common slip: using 22.4 dm3 for molar volume at rtp. In AQA, rtp is 20 °C and 1 atm, giving 24 dm3/mol. Only use 22.4 dm3 at standard temperature and pressure (0 °C). Always read the question conditions carefully.
另一个常见错误：在 rtp 下使用 22.4 dm3 作为摩尔体积。在 AQA 体系中，rtp 是 20 °C 和 1 个大气压，摩尔体积为 24 dm3/mol。只有在标准状况（0 °C）下才用 22.4 dm3/mol。务必仔细阅读题目给出的条件。
2. Writing Ionic Half-Equations in Electrolysis | 电解中的离子半方程式
Mistake example: When asked to write the half-equation at the cathode during electrolysis of molten lead(II) bromide, a student writes: Pb2+ + 2Br– → Pb. What is wrong?
易错案例：题目要求写出电解熔融溴化铅时阴极的半反应方程式，一个学生写：Pb2+ + 2Br– → Pb。错在哪里？
The error is mixing species from both electrodes. A half-equation only shows what happens at one electrode. The cathode attracts Pb2+ ions, and they gain electrons. The correct half-equation is: Pb2+ + 2e– → Pb. The bromide ions are not involved at the cathode; they are oxidised at the anode: 2Br– → Br2 + 2e–. Always remember: reduction (gain of electrons) at the cathode, oxidation (loss of electrons) at the anode.
错误在于混入了两个电极的物质。半方程式只表示某一个电极上发生的反应。阴极吸引 Pb2+ 离子，它们得到电子。正确的半方程式是：Pb2+ + 2e– → Pb。溴离子并不在阴极参与反应，它们在阳极被氧化：2Br– → Br2 + 2e–。始终记住：阴极发生还原（得电子），阳极发生氧化（失电子）。
For aqueous solutions, another common pitfall is forgetting that water can also be discharged. In the electrolysis of dilute NaCl solution, the cathode product is hydrogen gas, not sodium, because H+ ions from water are more easily reduced than Na+ ions. Students often blindly write Na+ + e– → Na, which is incorrect for aqueous solutions. Always check the reactivity series and the ion present in water.
对于水溶液，另一个常见陷阱是忘记水也可以放电。在电解稀 NaCl 溶液时，阴极产物是氢气，而不是钠，因为水中的 H+ 离子比 Na+ 更容易被还原。学生常常盲目地写 Na+ + e– → Na，这对水溶液来说是错误的。一定要结合金属活动性顺序和水中存在的离子来判断。
3. Energy Profile Diagrams: Exothermic vs Endothermic | 能量变化图：放热与吸热
Mistake example: A student labels the enthalpy change ΔH on an exothermic reaction profile as the difference between the energy of reactants and the peak of the curve. Is this correct?
易错案例：学生在放热反应能量图上，将焓变 ΔH 标注为反应物能量与曲线顶峰之间的差值。这样对吗？
No, this is a very common mistake. The peak represents the activation energy barrier, not the enthalpy change. ΔH is the energy difference between the products and the reactants. For an exothermic reaction, the products are at a lower energy level than the reactants, so ΔH is negative and is measured as the vertical gap between reactants and products, not the peak. The student confused activation energy with enthalpy change.
不对，这是一个非常普遍的误解。顶峰代表活化能势垒，而不是焓变。ΔH 是生成物与反应物之间的能量差。对于放热反应，生成物的能级低于反应物，因此 ΔH 为负值，测量的是反应物与生成物之间的垂直差距，而不是到顶峰的距离。学生混淆了活化能和焓变。
In an endothermic profile, the products are higher than the reactants, so ΔH is positive. Examiners often ask to calculate the activation energy using the energy values given on the diagram. Remember that activation energy is measured from the reactants’ energy to the top of the curve, not from the products. A common trap is to use the difference between the curve peak and the products’ energy, which gives a smaller value and is incorrect.
在吸热反应图中，生成物的能量高于反应物，所以 ΔH 为正值。考官常要求利用图中给出的能量值计算活化能。记住活化能是从反应物的能量水平量到曲线顶端，而不是从生成物量起。一个常见陷阱是用曲线顶峰与生成物能量之差来计算，这样得到的是一个较小的数值，是错误的。
4. Dynamic Equilibrium and Le Chatelier’s Principle | 动态平衡与勒夏特列原理
Mistake example: For the reaction 3H2(g) + N2(g) ⇌ 2NH3(g) (ΔH = -92 kJ/mol), a student states that increasing the temperature will increase the yield of ammonia because it speeds up the forward reaction. Comment on this.
易错案例：对于反应 3H2(g) + N2(g) ⇌ 2NH3(g) (ΔH = -92 kJ/mol)，某学生说升高温度能提高氨的产率，因为它加快了正反应速率。请评价。
This is a misunderstanding of equilibrium. While increasing temperature increases the rate of both forward and backward reactions, the key effect is that the equilibrium position shifts to oppose the change. Since the forward reaction is exothermic, increasing the temperature favours the endothermic direction, which is the reverse reaction. Therefore, the equilibrium shifts to the left, decreasing the yield of ammonia. The student confused ‘rate’ with ‘yield’. Higher temperature does make the reaction go faster, but it reduces the maximum possible amount of ammonia at equilibrium. Industrial conditions (450 °C) are a compromise between rate and yield.
这是对平衡的误解。虽然升高温度会同时加快正逆反应速率，但关键影响是平衡位置会向削弱这一改变的方向移动。由于正反应是放热的，升高温度有利于吸热的方向，也就是逆反应。因此平衡向左移动，氨的产率降低。该学生混淆了 “速率” 和 “产率”。更高的温度确实会使反应更快，但却降低了平衡时氨的最大可能产量。工业条件（450 °C）是速率和产率之间的折衷方案。
Similarly, for pressure changes, students often say increasing pressure increases the rate, so yield increases. In equilibrium, you must count the number of gas molecules on each side. In this reaction, 4 moles on the left and 2 on the right; higher pressure shifts equilibrium to the right, increasing yield. But that is due to the position shift, not simply the rate increase. Always apply Le Chatelier’s principle to predict the shift direction.
类似地，对于压强变化，学生常说增大压强会提高速率，所以产率增加。在平衡中，必须数两边气体分子数。该反应左边 4 摩尔，右边 2 摩尔；更高的压强会使平衡向右移动，增加产率。但这归因于平衡位置移动，而不是单纯的速率增加。始终应用勒夏特列原理来预测移动方向。
5. Naming Organic Compounds: Isomers and Functional Groups | 有机化合物命名：同分异构体和官能团
Mistake example: A student names CH3CH2CH2OH as propan-3-ol. Is this correct?
易错案例：学生将 CH3CH2CH2OH 命名为 3-丙醇。这个命名正确吗？
No. The longest continuous carbon chain has three carbons, so the parent name is propan-. The -OH group is on the terminal carbon. However, numbering should give the functional group the lowest possible number. Counting from the right as carbon 1 gives propan-1-ol, not propan-3-ol. The name propan-3-ol implies numbering from the opposite end, but the rules demand the lowest locant for the principal functional group. So the correct name is propan-1-ol. Many students forget to minimise the numbers and lose easy marks.
不正确。最长的碳链有三个碳，母体名称为丙烷。–OH 基团在末端碳上。然而编号应使官能团的位置号尽可能小。从右端编号碳 1 得到 1-丙醇，而不是 3-丙醇。3-丙醇的名称意味着从另一端编号，但命名规则要求主官能团的位次最小。因此正确名称是 1-丙醇。许多学生忘记最小化数字，轻易失分。
Another typical error is with esters. When given the structural formula CH3COOCH2CH3, students may call it ethyl ethanoate but then incorrectly draw it with the acid and alcohol parts reversed. Remember that the name ester is formed from the alcohol part first (ethyl from ethanol) and the carboxylic acid part second (ethanoate from ethanoic acid). The structure must show the acid part providing the C=O bond and the alkyl from alcohol attached to the oxygen. Mixing up the order leads to a completely different compound.
另一个典型错误是酯类。当给出结构式 CH3COOCH2CH3 时，学生可能会叫它乙酸乙酯，但画图时却把酸和醇的部分弄反了。记住酯的名称是醇的部分在前（乙基来自乙醇），羧酸的部分在后（乙酸根来自乙酸）。结构上必须显示酸的部分提供 C=O 键，醇的烷基连接在氧原子上。弄错顺序会导致完全不同的化合物。
6. Diamond vs Graphite: Giant Covalent Structures | 金刚石和石墨：巨型共价结构
Mistake example: A student writes that graphite conducts electricity because it has free electrons from a sea of delocalised electrons like in metals. Is the reasoning fully correct?
易错案例：某学生写道：石墨能导电，因为它像金属一样拥有离域电子海洋中的自由电子。这个理由完全正确吗？
Partially correct but misses the crucial structural detail. In graphite, each carbon atom is bonded to three others, forming layers of hexagonal rings. The fourth outer electron of each carbon is delocalised between these layers, and it is these delocalised electrons that can move and carry charge. However, unlike metals, the delocalisation is only along the layers, not in a 3D ‘sea’. So the student should mention that the electrons are free to move parallel to the layers, which is why graphite conducts only along the planes. That nuance often appears in mark schemes.
部分正确，但漏掉了关键的结构细节。在石墨中，每个碳原子与另外三个碳原子键合，形成六角形环的层状结构。每个碳的第四个外层电子在这些层间离域，这些离域电子可以移动并携带电荷。然而，与金属不同的是，离域只发生在层间，而非三维的 “海洋”。因此学生应该提到电子可以平行于层面自由移动，这就是石墨只能沿层面导电的原因。这个细微差别常出现在评分方案中。
Another common error is comparing hardness. Students say ‘diamond is hard because it has strong covalent bonds,’ which is true, but they may also say ‘graphite is soft because it has weak intermolecular forces.’ Actually, graphite also has strong covalent bonds within the layers; it is soft because the layers can slide over each other due to the weak intermolecular forces between them. So correctly: diamond is hard due to the 3D network of strong covalent bonds; graphite is slippery because of the weak forces between layers, not because the whole structure is weak.
另一个常见错误是比较硬度。学生说 “金刚石很硬，因为它有很强的共价键”，这没错，但他们可能也会说 “石墨很软，因为它分子间作用力弱”。实际上，石墨的层内也有很强的共价键；它之所以软，是因为层与层之间可以滑动，因为层间的分子间作用力较弱。所以正确表述是：金刚石因三维网状强共价键而坚硬；石墨因层间弱作用力而润滑，并非整个结构都弱。
7. Strong and Weak Acids: Degree of Ionisation | 强酸与弱酸：电离程度
Mistake example: A student claims that dilute hydrochloric acid is a weak acid because it contains a lot of water. Analyse this statement.
易错案例：学生声称稀盐酸是弱酸，因为它含有很多水。分析这个说法。
Confusion between ‘concentration’ and ‘strength’ is a top AQA pitfall. The terms strong and weak refer to the degree of dissociation (ionisation) in water, not how much acid is dissolved. Hydrochloric acid is a strong acid because it fully dissociates into H+ and Cl– ions, regardless of dilution. Even dilute HCl still has 100% dissociation. By contrast, ethanoic acid is a weak acid because it only partially dissociates, even in concentrated solution. Dilute vs concentrated describes the amount of solute per volume; strong vs weak describes the extent of ionisation.
混淆 “浓度” 和 “强度” 是 AQA 中最大的陷阱之一。强和弱指的是在水中的电离程度，而不是溶解了多少酸。盐酸是强酸，因为它完全电离成 H+ 和 Cl– 离子，无论浓稀。即使是稀盐酸，其电离度仍是 100%。相反，乙酸是弱酸，因为它只能部分电离，即便在浓溶液中也是如此。稀与浓描述的是单位体积溶质的量；强与弱描述的是电离的程度。
A follow-up mistake: comparing pH of strong and weak acids of the same concentration. A student might say ‘0.1 mol/dm3 HCl has pH 1, so 0.1 mol/dm3 ethanoic acid also has pH 1.’ This is wrong because the weak acid only partially dissociates, producing a lower concentration of H+ ions, so its pH is higher (e.g., around 3). Always remember: for the same concentration, strong acids have lower pH values than weak acids due to complete dissociation.
一个连带错误：比较相同浓度的强酸和弱酸的 pH 值。学生可能会说 “0.1 mol/dm3 HCl 的 pH 是 1，因此 0.1 mol/dm3 乙酸的 pH 也是 1。” 这是错误的，因为弱酸只部分电离，产生的 H+ 浓度较低，因此 pH 较高（例如约为 3）。始终记住：相同浓度下，强酸由于完全电离，其 pH 值比弱酸更低。
8. Interpreting Rate of Reaction Graphs | 解读反应速率图
Mistake example: A student draws a graph of volume of gas produced against time for a reaction using powdered marble and another using marble chips, both with excess acid. The student makes the curve for chips steeper at the start and ending at a lower final volume. Why is this wrong?
易错案例：一个学生在画气体体积–时间图时，分别绘制了粉末状大理石和块状大理石（均与过量酸反应）的曲线。他让块状大理石的曲线起始更陡，且最终体积更低。为什么错了？
Powdered marble has a larger surface area, so the rate is faster, and its curve should be steeper initially. The student reversed this. Moreover, since the same mass of marble (same amount of CaCO3) is used and acid is in excess, the total volume of CO2 produced should be exactly the same for both; only the time taken differs. Therefore, both curves should eventually level off at the same final volume. The student mistakenly thought chips produce less gas, perhaps confusing ‘rate’ with ‘extent of reaction’.
粉末状大理石表面积更大，所以速率更快，其曲线起始应该更陡。学生把这个弄反了。此外，由于使用相同质量的大理石（等量 CaCO3），且酸过量，两者产生的 CO2 总体积完全相同；只是所需时间不同。因此，两条曲线最终应在同一终体积处趋于平缓。学生错误地认为块状产生的气体更少，可能混淆了 “速率” 和 “反应程度”。
A common extension: explaining the shape of the curve. It becomes less steep over time because the reactant concentration (acid) decreases as it is used up, so the frequency of successful collisions drops. Students sometimes say ‘because the marble is used up’, but if acid is in excess, the marble is the limiting reactant and is indeed used up at the end. The shape during the reaction, however, is controlled by the falling acid concentration. Be precise with explanations.
常见的延伸考点：解释曲线形状。曲线随时间推移变得平缓，是因为反应物浓度（酸）随着消耗而降低，导致有效碰撞频率下降。学生有时会说 “因为大理石用完了”，但如果酸过量，大理石是限制反应物，确实在终点用完。然而反应过程中的曲线形状是由酸浓度的下降所控制的。解释时务必精确。
9. Chemical Analysis: Flame Tests and Precipitate Reactions | 化学分析：焰色反应和沉淀反应
Mistake example: A student says that sodium ions give a yellow flame, and potassium ions give a lilac flame, but then uses a flame test to distinguish between sodium chloride and sodium sulfate, expecting different colours. Is the test suitable?
易错案例：一个学生说钠离子产生黄色火焰，钾离子产生淡紫色火焰，但接着却用焰色反应来区分氯化钠和硫酸钠，并期待看到不同颜色。这个测试合适吗？
No, flame tests identify the metal cation, not the anion or the whole compound. Both NaCl and Na2SO4 contain Na+ ions, so both would give the same yellow colour. To distinguish between them, you would test for the anion, e.g., use acidified silver nitrate for chloride ions (white precipitate) or barium chloride for sulfate ions (white precipitate). The student misunderstood that the flame colour depends solely on the metal ion present.
不合适，焰色反应鉴定的是金属阳离子，而不是阴离子或整个化合物。NaCl 和 Na2SO4 都含有 Na+，因此都会呈现相同的黄色。要区分它们，应测试阴离子，例如用酸化硝酸银检测氯离子（白色沉淀）或用氯化钡检测硫酸根离子（白色沉淀）。学生误以为焰色取决于化合物整体，而实际只取决于金属离子。
Another common error: describing the precipitate results imprecisely. For instance, adding NaOH solution to Cu2+ gives a blue precipitate, but students often just say ‘solution turns blue.’ It is a precipitate, so the correct observation is a blue precipitate formed. Similarly, for Fe2+ it is a green precipitate, and Fe3+ a brown precipitate. Ensure you use the word ‘precipitate’ and the exact colour description as per AQA specification.
另一个常见错误：描述沉淀结果不严谨。例如，向 Cu2+ 中加入 NaOH 溶液产生蓝色沉淀，但学生常常只说 “溶液变蓝”。这是沉淀，因此正确的观察是形成蓝色沉淀。类似地，Fe2+ 生成的沉淀是绿色的，Fe3+ 是棕色的。务必使用 “沉淀” 一词，并采用 AQA 考纲要求的准确颜色描述。
10. Balancing Equations and Conservation of Mass | 配平方程式与质量守恒
Mistake example: A student balances the equation for the thermal decomposition of CaCO3 as: CaCO3 → CaO + CO2. It is already balanced, so the student thinks no further work is needed. Yet when asked to calculate the mass of CaO produced from 50 g of CaCO3, the student mistakenly uses the atomic mass of Ca as 20 instead of 40, leading to a wrong answer. What is the learning point?
易错案例：学生配平碳酸钙热分解的方程式为：CaCO3 → CaO + CO2，已经配平，学生认为无需再做任何工作。然而当被要求计算 50 g CaCO3 产生 CaO 的质量时，该学生错误地将 Ca 的原子量当作 20 而不是 40，导致答案出错。学习要点是什么？
The main lesson is that accuracy in relative atomic masses from the Periodic Table is essential. Many marks are lost due to misreading Mr values. The correct Mr of CaCO3 is 40 + 12 + (3 × 16) = 100. Then moles of CaCO3 = 50/100 = 0.5 mol. Since the mole ratio is 1:1, moles of CaO = 0.5 mol. Mr of CaO = 40 + 16 = 56, so mass = 0.5 × 56 = 28 g. The student’s error gave a totally different result. Always double-check atomic numbers and the formula of the compound.
主要学习要点是从周期表中准确读取相对原子质量至关重要。很多失分都是由于看错 Mr 数值。CaCO3 的正确 Mr = 40 + 12 + (3 × 16) = 100。那么 CaCO3 的摩尔数 = 50/100 = 0.5 mol。因为摩尔比是 1:1，CaO 的摩尔数也是 0.5 mol。CaO 的 Mr = 40 + 16 = 56，所以质量 = 0.5 × 56 = 28 g。学生的错误导致了完全不同的结果。一定要反复核对原子序数和化合物分子式。
Also, some students add subscripts when balancing, altering the substances. For example, changing H2O to H2O2 to balance oxygen atoms. That is strictly forbidden. Only coefficients (big numbers in front) can be added. The chemical formulas themselves must remain unchanged. This fundamental rule is often overlooked in the rush to balance.
此外，有些学生在配平时添加下标，改变了物质本身。例如，为平衡氧原子而将 H2O 改为 H2O2。这是严格禁止的。只能添加系数（前面的大数字）。化学式本身必须保持不变。这一基本规则常在匆忙配平时被忽略。
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