📚 IB CCEA Chemistry: Typical Worked Examples | IB CCEA 化学：典型例题详解
This article presents a collection of carefully selected worked examples that bridge the core topics of IB Chemistry and CCEA GCE Chemistry. Each section targets a fundamental skill – from stoichiometry to organic mechanisms – with fully explained solutions in English and Chinese. By working through these problems, students can reinforce their conceptual understanding and sharpen problem-solving techniques essential for both qualifications.
本文精选了 IB 化学和 CCEA GCE 化学核心主题中的典型例题，逐一提供中英双语详细解析。每个小节聚焦一项基本技能——从化学计量到有机反应机理——通过全步骤解答，帮助学生巩固概念理解，并提升两类考试必备的解题能力。
1. Mole Calculations and Stoichiometry | 摩尔计算与化学计量
A sample of calcium carbonate, CaCO₃, has a mass of 5.00 g. Calculate the amount of calcium carbonate in moles and the number of oxygen atoms present.
有一份 5.00 g 的碳酸钙 (CaCO₃) 样品。计算碳酸钙的物质的量（摩尔）以及所含的氧原子数。
Molar mass of CaCO₃ = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹. Amount n = mass / M = 5.00 g / 100.1 g mol⁻¹ ≈ 0.04995 mol. Each formula unit contains 3 oxygen atoms, so moles of O atoms = 3 × 0.04995 mol = 0.14985 mol. Number of O atoms = 0.14985 mol × 6.022 × 10²³ mol⁻¹ ≈ 9.02 × 10²² atoms.
CaCO₃ 的摩尔质量 = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹。物质的量 n = 质量 / 摩尔质量 = 5.00 g / 100.1 g mol⁻¹ ≈ 0.04995 mol。每个单元含 3 个氧原子，所以氧原子的物质的量 = 3 × 0.04995 mol = 0.14985 mol。氧原子数 = 0.14985 mol × 6.022 × 10²³ mol⁻¹ ≈ 9.02 × 10²² 个。
2. Empirical and Molecular Formulae | 实验式与分子式
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is about 180 g mol⁻¹. Determine its empirical and molecular formulae.
某化合物含碳 40.0%、氢 6.7%、氧 53.3%（质量分数），其摩尔质量约为 180 g mol⁻¹。求其实验式和分子式。
Assume 100 g sample: C: 40.0 g → 40.0/12.0 = 3.33 mol; H: 6.7 g → 6.7/1.0 = 6.7 mol; O: 53.3 g → 53.3/16.0 = 3.33 mol. Divide by smallest (3.33): C: 1, H: 2, O: 1. Empirical formula = CH₂O. Empirical mass = 12.0 + 2×1.0 + 16.0 = 30.0 g mol⁻¹. Ratio of molar mass to empirical mass = 180 / 30 = 6. Molecular formula = 6 × (CH₂O) = C₆H₁₂O₆.
假设样品 100 g：C：40.0 g → 40.0/12.0 = 3.33 mol；H：6.7 g → 6.7/1.0 = 6.7 mol；O：53.3 g → 53.3/16.0 = 3.33 mol。除以最小值 (3.33)：C : 1，H : 2，O : 1。实验式 = CH₂O，实验式质量 = 30.0 g mol⁻¹。摩尔质量与实验式质量之比 = 180 / 30 = 6。分子式 = 6 × (CH₂O) = C₆H₁₂O₆。
3. Enthalpy Changes and Calorimetry | 焓变与量热法
In a calorimetry experiment, 0.0500 mol of acid is neutralised by excess alkali. The temperature of the solution rises by 4.20 °C. The total mass of the solution is 100 g and its specific heat capacity is 4.18 J g⁻¹ °C⁻¹. Calculate the enthalpy change of neutralisation in kJ mol⁻¹.
量热实验中，0.0500 mol 酸被过量的碱中和，溶液温度升高 4.20 °C。溶液总质量 100 g，比热容为 4.18 J g⁻¹ °C⁻¹。计算中和焓变 (kJ mol⁻¹)。
Heat absorbed by solution q = m × c × ΔT = 100 g × 4.18 J g⁻¹ °C⁻¹ × 4.20 °C = 1755.6 J = 1.756 kJ. This heat was released by the reaction, so q_reaction = -1.756 kJ. Moles of acid = 0.0500 mol. ΔH = q_reaction / n = -1.756 kJ / 0.0500 mol = -35.1 kJ mol⁻¹ (exothermic).
溶液吸收的热量 q = m × c × ΔT = 100 g × 4.18 J g⁻¹ °C⁻¹ × 4.20 °C = 1755.6 J = 1.756 kJ。该热量由反应放出，因此 q_reaction = -1.756 kJ。酸的物质的量 = 0.0500 mol。ΔH = q_reaction / n = -1.756 kJ / 0.0500 mol = -35.1 kJ mol⁻¹（放热）。
4. Hess’s Law | 赫斯定律
Use the following thermochemical equations to determine the enthalpy change for the reaction: C(s) + 2H₂(g) → CH₄(g).
① C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5 kJ mol⁻¹
② H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = -285.8 kJ mol⁻¹
③ CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH₃ = -890.3 kJ mol⁻¹
利用以下热化学方程式求反应 C(s) + 2H₂(g) → CH₄(g) 的焓变：
① C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5 kJ mol⁻¹
② H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = -285.8 kJ mol⁻¹
③ CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH₃ = -890.3 kJ mol⁻¹
Target: C(s) + 2H₂(g) → CH₄(g). Keep reaction ① as is: C(s) + O₂(g) → CO₂(g). Multiply reaction ② by 2: 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = 2 × (-285.8) = -571.6 kJ mol⁻¹. Reverse reaction ③: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.3 kJ mol⁻¹. Add them: C(s) + O₂(g) + 2H₂(g) + O₂(g) + CO₂(g) + 2H₂O(l) → CO₂(g) + 2H₂O(l) + CH₄(g) + 2O₂(g). Cancel common species: C(s) + 2H₂(g) → CH₄(g). ΔH = -393.5 + (-571.6) + 890.3 = -74.8 kJ mol⁻¹.
目标方程：C(s) + 2H₂(g) → CH₄(g)。保留①不变；②乘以 2：2H₂(g) + O₂(g) → 2H₂O(l) ΔH = -571.6 kJ mol⁻¹；③反转：CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.3 kJ mol⁻¹。三式相加并约去相同物质，得到目标方程，ΔH = -393.5 + (-571.6) + 890.3 = -74.8 kJ mol⁻¹。
5. Reaction Rates and Initial Rate Method | 反应速率与初速法
The reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) was studied at a constant temperature. The following initial rate data were obtained:
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.50 × 10⁻³ |
| 2 | 0.100 | 0.200 | 5.00 × 10⁻³ |
| 3 | 0.200 | 0.100 | 1.00 × 10⁻² |
Determine the rate law and calculate the rate constant.
反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) 在恒温下研究，获得以下初速数据。求速率方程并计算速率常数。
Compare expt 1 and 2: [NO] constant, [H₂] doubles → rate doubles. Hence order with respect to H₂ is 1. Compare expt 1 and 3: [H₂] constant, [NO] doubles → rate increases by factor (1.00×10⁻²)/(2.50×10⁻³)=4. Thus order with respect to NO is 2. Rate law: rate = k [NO]²[H₂]. Using expt 1: k = rate / ([NO]²[H₂]) = (2.50×10⁻³) / ((0.100)² × 0.100) = 2.50×10⁻³ / 1.00×10⁻³ = 2.5 dm⁶ mol⁻² s⁻¹.
比较实验 1 和 2：NO 浓度不变，H₂ 浓度加倍 → 速率加倍，H₂ 的级数为 1。比较实验 1 和 3：H₂ 浓度不变，NO 浓度加倍 → 速率增大为原来的 4 倍，NO 的级数为 2。速率方程：rate = k [NO]²[H₂]。代入实验 1 数据：k = (2.50×10⁻³) / ((0.100)² × 0.100) = 2.5 dm⁶ mol⁻² s⁻¹。
6. Equilibrium Constant and Le Chatelier’s Principle | 平衡常数与勒夏特列原理
For the equilibrium N₂O₄(g) ⇌ 2NO₂(g) at 298 K, the partial pressures at equilibrium are p(N₂O₄) = 0.40 atm and p(NO₂) = 0.60 atm. Calculate the equilibrium constant Kp and predict the effect of increasing total pressure on the equilibrium yield of NO₂.
对于 298 K 下的平衡 N₂O₄(g) ⇌ 2NO₂(g)，平衡时分压为 p(N₂O₄) = 0.40 atm，p(NO₂) = 0.60 atm。计算平衡常数 Kp，并预测增大总压对 NO₂ 平衡产率的影响。
Kp = [p(NO₂)]² / p(N₂O₄) = (0.60)² / 0.40 = 0.36 / 0.40 = 0.90 atm
According to Le Chatelier’s principle, increasing total pressure shifts the equilibrium towards the side with fewer gas molecules. The forward reaction (N₂O₄ → 2NO₂) increases the number of molecules (1 → 2), so high pressure favours the reverse reaction. The yield of NO₂ will decrease.
根据勒夏特列原理，增大总压使平衡向气体分子数减少的方向移动。正反应 (N₂O₄ → 2NO₂) 增加分子数 (1 → 2)，因此高压有利于逆反应。NO₂ 的产率将会降低。
7. Acid-Base Calculations: pH and pOH | 酸碱计算：pH 与 pOH
A 0.100 mol dm⁻³ solution of ethanoic acid (CH₃COOH) has a degree of dissociation of 1.34% at 25 °C. Calculate the pH of the solution and the acid dissociation constant Ka.
0.100 mol dm⁻³ 的乙酸 (CH₃COOH) 溶液在 25 °C 的电离度为 1.34%。计算溶液的 pH 和酸解离常数 Ka。
Degree of dissociation α = 1.34% = 0.0134. [H⁺] = c × α = 0.100 × 0.0134 = 1.34 × 10⁻³ mol dm⁻³. pH = -log₁₀[H⁺] = -log₁₀(1.34×10⁻³) ≈ 2.87. For weak acid HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] / [HA]. At equilibrium [H⁺] = [A⁻] = 1.34×10⁻³, [HA] ≈ 0.100 – 1.34×10⁻³ ≈ 0.0987 mol dm⁻³. Ka = (1.34×10⁻³)² / 0.0987 ≈ 1.82 × 10⁻⁵ mol dm⁻³.
电离度 α = 0.0134。 [H⁺] = c × α = 1.34 × 10⁻³ mol dm⁻³。pH = -log₁₀(1.34×10⁻³) ≈ 2.87。对于弱酸 HA ⇌ H⁺ + A⁻，Ka = [H⁺][A⁻]/[HA]。平衡时 [H⁺] = [A⁻] = 1.34×10⁻³，[HA] ≈ 0.0987 mol dm⁻³。Ka = (1.34×10⁻³)² / 0.0987 ≈ 1.82 × 10⁻⁵ mol dm⁻³。
8. Redox Titrations | 氧化还原滴定
A 25.0 cm³ sample of iron(II) sulfate solution was acidified and titrated with 0.0200 mol dm⁻³ potassium manganate(VII) solution. 22.50 cm³ of the KMnO₄ solution was required to reach the endpoint. Calculate the concentration of Fe²⁺ ions in the original solution.
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
取 25.0 cm³ 硫酸亚铁铵溶液经酸化后，用 0.0200 mol dm⁻³ 高锰酸钾溶液滴定，到达终点时消耗 22.50 cm³。计算原溶液中 Fe²⁺ 的浓度。反应式：MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
Moles of MnO₄⁻ used = concentration × volume = 0.0200 mol dm⁻³ × (22.50/1000) dm³ = 4.50 × 10⁻⁴ mol. From the equation, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺. So moles of Fe²⁺ in 25.0 cm³ = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol. [Fe²⁺] = 2.25 × 10⁻³ mol / 0.0250 dm³ = 0.0900 mol dm⁻³.
所用 MnO₄⁻ 的物质的量 = 0.0200 × 0.02250 = 4.50 × 10⁻⁴ mol。由方程式知 1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应。25.0 cm³ 溶液中 Fe²⁺ 的物质的量 = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol。[Fe²⁺] = 2.25 × 10⁻³ mol / 0.0250 dm³ = 0.0900 mol dm⁻³。
9. Organic Nomenclature and Isomerism | 有机命名与同分异构
Draw and name two branched-chain isomers of C₆H₁₄ that have exactly three methyl groups. Identify the type of isomerism between them.
画出并命名两种 C₆H₁₄ 的支链异构体，要求均恰好含有三个甲基。指出它们之间的异构类型。
One possible isomer: 2,3-dimethylbutane – structure: CH₃-CH(CH₃)-CH(CH₃)-CH₃ (two methyl branches on the main chain). This molecule has three methyl groups (two branches and one terminal). Another isomer: 3-methylpentane has only two methyl groups, so not suitable. 2,2-dimethylbutane has two methyls on carbon-2 plus one terminal methyl, total three methyls. Its structure: CH₃-C(CH₃)₂-CH₂-CH₃. The two isomers are 2,3-dimethylbutane and 2,2-dimethylbutane. They are positional isomers (or chain isomers) because they differ in the position of branching, although both have the same carbon skeleton arrangement; more precisely they are constitutional isomers with different branching patterns.
一种可能异构体：2,3-二甲基丁烷，结构为 CH₃-CH(CH₃)-CH(CH₃)-CH₃，含有三个甲基（两个支链甲基和一个端基甲基）。另一种：2,2-二甲基丁烷，结构为 CH₃-C(CH₃)₂-CH₂-CH₃，也含有三个甲基。这两种异构体分别为 2,3-二甲基丁烷和 2,2-二甲基丁烷，属于构造异构体中的位置异构（支链位置不同）。
10. Organic Reaction Mechanisms: Nucleophilic Substitution | 有机反应机理：亲核取代
Explain the mechanism of the reaction between bromoethane and aqueous sodium hydroxide, using curly arrows to show electron movement. State the type of reaction and name the organic product.
用弯箭头表示电子转移，解释溴乙烷与氢氧化钠水溶液反应的机理，指出反应类型并命名有机产物。
The reaction proceeds via an Sₙ2 mechanism. The hydroxide ion acts as a nucleophile, attacking the electrophilic carbon attached to bromine from the opposite side of the C–Br bond. A transition state forms with partial bonds to both OH and Br. As the C–O bond forms, the C–Br bond breaks, releasing bromide ion. The product is ethanol. Type: nucleophilic substitution, bimolecular.
反应按 Sₙ2 机理进行。氢氧根离子作为亲核试剂，从 C-Br 键的背面进攻与溴相连的亲电碳原子。形成过渡态，碳与 OH 和 Br 同时部分成键。随着 C-O 键的形成，C-Br 键断裂，释放溴离子。产物为乙醇。反应类型：双分子亲核取代。
11. Electrophilic Addition in Alkenes | 烯烃的亲电加成
Describe the mechanism for the reaction of ethene with hydrogen bromide (HBr). Show the electron movement and explain why Markovnikov’s rule applies when propene is used instead of ethene.
描述乙烯与溴化氢 (HBr) 反应的机理，标明电子转移，并解释若使用丙烯时为何适用马氏规则。
Ethene with HBr: The π-electrons of the C=C bond attack the slightly positive hydrogen of HBr, causing heterolytic fission of H–Br. A carbocation (ethyl carbocation, C₂H₅⁺) forms along with Br⁻. The bromide ion then attacks the carbocation to form bromoethane. With propene, the initial electrophilic attack on the double bond leads to two possible carbocations: a secondary carbocation (more stable) and a primary carbocation. The more stable secondary carbocation is preferentially formed, so Br⁻ adds to the more substituted carbon, giving 2-bromopropane as the major product – consistent with Markovnikov’s rule.
乙烯与 HBr：双键的 π 电子进攻 HBr 中稍带正电的氢，引发 H-Br 异裂，生成乙基碳正离子 (C₂H₅⁺) 和 Br⁻。溴离子随后进攻碳正离子生成溴乙烷。丙烯情况下，双键受亲电进攻后可生成两种碳正离子：稳定性更高的仲碳正离子和伯碳正离子。优先形成更稳定的仲碳正离子，因此 Br⁻ 加到取代较多的碳上，主要产物为 2-溴丙烷，符合马氏规则。
12. Mass Spectrometry and Infrared Spectroscopy | 质谱与红外光谱
An organic compound gives a molecular ion peak at m/z = 72 in its mass spectrum, and its infrared spectrum shows a strong absorption at about 1720 cm⁻¹. Suggest two possible structures for the compound and explain how you would use chemical tests to distinguish between them.
某有机化合物的质谱显示分子离子峰 m/z = 72，红外光谱在约 1720 cm⁻¹ 处有强吸收。推测两种可能结构，并说明如何用化学方法区分它们。
m/z = 72 suggests molar mass 72 g mol⁻¹. The IR absorption at 1720 cm⁻¹ indicates a carbonyl group (C=O). Possible functional groups: ketone or aldehyde. Possible structures: butanone (CH₃COCH₂CH₃) and butanal (CH₃CH₂CH₂CHO), both with formula C₄H₈O (mass 72). To distinguish: butanal is an aldehyde and will give a positive result with Tollens’ reagent (silver mirror) or Fehling’s solution, whereas butanone (a ketone) will not react. Alternatively, 2,4-DNPH test confirms carbonyl in both, followed by Tollens’ to differentiate.
m/z = 72 暗示摩尔质量为 72 g mol⁻¹。1720 cm⁻¹ 处的 IR 吸收说明含羰基 (C=O)。可能为酮或醛。可能结构：丁酮 (CH₃COCH₂CH₃) 和丁醛 (CH₃CH₂CH₂CHO)，分子式均为 C₄H₈O (质量 72)。区分方法：丁醛为醛，能与托伦斯试剂（银镜）或斐林试剂反应呈阳性，而丁酮（酮）不反应。也可先通过 2,4-二硝基苯肼确证羰基，再用托伦斯试剂区分。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导