📚 IB CIE Chemistry: Common Mistakes and Detailed Solutions | IB CIE 化学：易错题精讲
In both IB and CIE A‑level Chemistry, students often lose marks not because they lack understanding, but because they fall into predictable traps. This article highlights ten of the most common errors seen across topics such as stoichiometry, bonding, energetics, equilibrium, and organic chemistry, providing clear corrections and exam‑focused advice. Mastering these nuances can significantly boost your grade.
在 IB 和 CIE A‑level 化学中，学生常常不是因为不懂，而是因为掉入可预见的陷阱而失分。本文梳理了十个最常见的高频易错点，涵盖化学计量、化学键、能量学、平衡和有机化学等主题，给出清晰的纠正与应试建议。吃透这些细节，能显著提升你的成绩。
1. Significant Figures and Rounding Errors | 有效数字与修约误差
Many candidates give final answers with too many, or too few, significant figures (sf). In CIE and IB, the rule is: match the sf of the given data, and never round intermediate steps before the final answer. For addition/subtraction, use decimal places, not sf.
很多考生在最终答案中给出的有效数字过多或过少。CIE 和 IB 的规则是：与题干数据的有效数字保持一致，切勿在中间步骤提前修约。加减运算要看小数点后位数，而不是有效数字。
- English: If a calculation uses 0.0250 mol (3 sf) and 50.0 cm³ (3 sf), the concentration should be reported to 3 sf, e.g. 0.500 mol·dm⁻³.
- 中文： 如果计算使用 0.0250 mol（3 位有效数字）和 50.0 cm³（3 位有效数字），浓度应保留 3 位有效数字，例如 0.500 mol·dm⁻³。
2. Balancing Redox Equations Under Acidic and Basic Conditions | 酸性/碱性条件下配平氧化还原方程式
Students often forget to add H⁺ or H₂O when balancing half‑equations in acidic solution, or they wrongly use OH⁻ instead of H⁺. In basic conditions, after balancing with H⁺ and H₂O, add OH⁻ to both sides to neutralize H⁺ and convert to water.
学生在配平酸性条件下的半反应时，经常忘记补 H⁺ 或 H₂O，或在碱性条件下错误地使用 OH⁻ 代替 H⁺。碱性条件应先按酸性配平，然后两边加上 OH⁻ 中和 H⁺ 生成水。
- English: For MnO₄⁻ → Mn²⁺ in acid, the correct half‑equation is: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Missing H⁺ or water loses the mark.
- 中文： 酸性条件下 MnO₄⁻ → Mn²⁺ 的正确半反应为：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。缺少 H⁺ 或水都会失分。
3. Interpreting ‘Limiting Reagent’ and Excess Reactant Calculations | 限量反应物与过量计算诠释错误
A frequent mistake is assuming the reactant with the smaller mass or moles is automatically the limiting reagent. The mole ratio from the balanced equation must be used. Also, after identifying the limiting reagent, all further calculations must be based on it, not the excess reactant.
常见错误是认为质量小或物质的量少的反应物就一定是限量试剂，必须用配平方程的摩尔比来判断。另外，确定限量试剂后，所有后续计算都必须基于它，而非过量的反应物。
- English: nd 2 mol O₂, H₂ is limiting because 3 mol H₂ requires only 1.5 mol O₂; O₂ is in excess. Many incorrectly pick O₂ as limiting because 2<3.
- 中文： 对于 2H₂ + O₂ → 2H₂O，如果有 3 mol H₂ 和 2 mol O₂，H₂ 是限量试剂，因为 3 mol H₂ 只需要 1.5 mol O₂；O₂ 过量。许多人因 2<3 而错选 O₂ 为限量试剂。
4. Confusing Bond Polarity with Molecular Polarity | 混淆键的极性与分子的极性
Polar bonds do not guarantee a polar molecule. In symmetrical molecules like CO₂ or CCl₄, the individual bond dipoles cancel, making the overall molecule non‑polar. Always consider molecular geometry (VSEPR) and vector addition of dipoles.
极性键并不保证分子具有极性。在对称分子如 CO₂ 或 CCl₄ 中，各键的偶极矩相互抵消，使整体分子为非极性。必须结合分子构型（VSEPR 理论）和偶极矩的矢量叠加来判定。
- English: CO₂ has two polar C=O bonds, but its linear shape cancels the dipoles, so CO₂ is non‑polar. Students frequently label it as polar.
- 中文： CO₂ 有两个极性的 C=O 键，但其直线形结构使偶极抵消，因此 CO₂ 是非极性分子。学生常常误判为极性。
5. Misapplying Le Chatelier’s Principle to Equilibrium Shifts | 勒夏特列原理应用误区
The principle predicts the direction of shift, but many students incorrectly state that ‘the position of equilibrium shifts so that the concentration of the added substance decreases back to its original value’. This is wrong: the shift only partially opposes the change; concentrations do not return to original values.
该原理预测平衡移动的方向，但许多学生错误地表述“平衡会移动，使加入物质的浓度回到初始值”。这是错误的：移动只是部分抵消变化，浓度不会回到原来的数值。
- English: Adding N₂ to the Haber process (N₂ + 3H₂ ⇌ 2NH₃) shifts equilibrium right, but the final N₂ concentration is still higher than before the addition, just lower than immediately after addition.
- 中文： 向哈伯法（N₂ + 3H₂ ⇌ 2NH₃）中加入 N₂，平衡向右移动，但最终 N₂ 的浓度仍然高于加之前的浓度，只是比刚加入后略有下降。
6. Enthalpy Change Confusion: ΔH vs. Bond Enthalpy Calculations | 焓变辨析：ΔH 与键焓计算
When using bond enthalpies, ΔH = Σ(bonds broken) – ond enthalpy values are average energies for gaseous species, so they give approximate ΔH.
使用键焓时，ΔH = Σ（断裂键的键焓）– Σ（形成键的键焓）。经典错误是符号搞反，或只关注反应物中的键。还要记住，键焓是气态物质的平均能量，因此算出的 ΔH 是近似值。
- English: For CH₄ + 2O₂ → CO₂ + 2H₂O, break 4 C–H and 2 O=O; form 2 C=O and 4 O–H. Many forget to multiply by coefficients and thus get a wrong net ΔH.
- 中文： 对于 CH₄ + 2O₂ → CO₂ + 2H₂O，断裂 4 个 C–H 和 2 个 O=O；形成 2 个 C=O 和 4 个 O–H。很多人忘记乘以系数，导致净 ΔH 算错。
7. Rate Equations and Order from Mechanisms – The Rate‑determining Step Trap | 速率方程与机理推级数——决速步陷阱
Students often deduce the rate equation directly from the stoichiometric equation, which is only valid for elementary reactions. For multi‑step mechanisms, the rate equation depends on the slowest step (rate‑determining step) and its molecularity. Species in the slow step must be either reactants from the overall equation or intermediates whose concentrations can be expressed in terms of reactants.
学生常从计量方程直接推测速率方程，但这只对基元反应成立。对于多步机理，速率方程取决于最慢的一步（决速步）及其分子数。决速步中的物种必须是总反应的反应物，或是浓度能用反应物表示的中间体。
- English: O₂], not k[NO]²[O₂] as derived from stoichiometry. The student must recognise that the slow step contains one NO and one O₂.
- 中文： 对于 2NO + O₂ → 2NO₂，若机理为慢反应：NO + O₂ → NO₃（慢），接着 NO₃ + NO → 2NO₂（快），则速率 = k[NO][O₂]，而不是按计量数写出的 k[NO]²[O₂]。学生必须识别出慢反应含 1 个 NO 和 1 个 O₂。
8. Organic Reaction Mechanisms: Curly Arrows and Regioselectivity | 有机反应机理：弯箭头与区域选择性
Curly arrows must start from a lone pair or a bond pair, and point to an atom, not a charge. A common mistake is drawing arrows from H⁺ or from a positive charge. Also, in electrophilic addition to unsymmetrical alkenes, Markovnikov’s rule is often misapplied when carbocation stability is overlooked.
弯箭头必须从孤对电子或成键电子对出发，指向原子，而不是指向电荷。常见错误是从 H⁺ 或正电荷处画箭头。此外，对于不对称烯烃的亲电加成，忽略碳正离子稳定性常常导致马尔科夫尼科夫规则用错。
- English: ene, the arrow goes from the C=C bond to the H of HBr, not from H⁺. The major product is 2‑bromopropane because the secondary carbocation is more stable than the primary.
- 中文： 在 HBr 与丙烯的加成中，箭头应从 C=C 双键指向 HBr 中的 H，而不是从 H⁺ 出发。主要产物是 2‑溴丙烷，因为二级碳正离子比一级稳定。
9. pH and pKa: Buffer Calculations and the Henderson–Hasselbalch Trap | pH 与 pKa：缓冲溶液计算与亨德森‑哈塞尔巴尔赫陷阱
]/[HA]). Students frequently use the number of moles directly instead of concentrations, overlooking that the ratio is valid only if volumes are the same. When the solution is diluted, the ratio of concentrations remains unchanged, so pH of a buffer does not change with dilution. Many forget this.
对于酸性缓冲液，pH = pKa + log([A⁻]/[HA])。学生经常直接用物质的量代替浓度，忽略了只有在体积相同时，物质的量之比才等于浓度之比。稀释缓冲溶液时，浓度比不变，因此缓冲液的 pH 不随稀释而变，这一点常被遗忘。
- English: s directly also gives the same ratio, but the concept of concentration must be clear for unequal volumes.
- 中文： 将 50 cm³ 0.1 mol·dm⁻³ HA 与 50 cm³ 0.1 mol·dm⁻³ NaA 混合，总体积为 100 cm³，因此 [HA] = [A⁻] = 0.05 mol·dm⁻³，pH = pKa。直接使用物质的量也能得到相同比值，但体积不等时必须用浓度来理解。
10. Electrochemistry: Cell Potential and the Direction of Spontaneous Reaction | 电化学：电池电势与自发反应方向
A common error is writing the cell diagram or calculating Eꝋcell by reversing the wrong half‑cell. The cell reaction must have Eꝋ aller from the larger without considering which species is oxidised. The correct method: Eꝋcell = Eꝋcathode (reduction) – Eꝋanode (reduction).
常见错误是在书写原电池符号或计算 Eꝋcell 时把半电池弄反了。自发反应必须满足 Eꝋcell > 0。学生常查表后简单用大减小，却不考虑哪种物质被氧化。正确方法：Eꝋcell = Eꝋ（阴极，还原）– Eꝋ（阳极，还原）。
- English: For Zn|Zn²⁺||Cu²⁺|Cu, Zn is the anode (oxidation) and Cu the cathode. Using Eꝋ(Zn²⁺/Zn) = –0.76 V and Eꝋ(Cu²⁺/Cu) = +0.34 V, Eꝋcell = +0.34 – (–0.76) = +1.10 V. Some candidates erroneously calculate –0.76 – 0.34 = –1.10 V and think the reaction is non‑spontaneous.
- 中文： 对于 Zn|Zn²⁺||Cu²⁺|Cu，Zn 为阳极（氧化），Cu 为阴极。Eꝋ(Zn²⁺/Zn) = –0.76 V，Eꝋ(Cu²⁺/Cu) = +0.34 V，Eꝋcell = +0.34 – (–0.76) = +1.10 V。有学生错误地算成 –0.76 – 0.34 = –1.10 V，从而认为反应非自发。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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