Worked Examples in IB and CIE Chemistry | IB CIE 化学：典型例题详解

📚 Worked Examples in IB and CIE Chemistry | IB CIE 化学：典型例题详解

This article provides a carefully selected set of worked examples covering core topics common to both IB and CIE chemistry syllabi. Each example is solved step by step, with commentary in both English and Chinese, to help you master problem-solving techniques and excel in your examinations.

本文精选了 IB 与 CIE 化学课程中共同的核心主题典型例题，逐一详细解析。每道题目均提供英中双语分步解答，帮助你掌握解题技巧，在考试中脱颖而出。

1. Stoichiometry and Molar Calculations | 化学计量与摩尔计算

A 0.500 g sample of impure calcium carbonate is reacted with excess hydrochloric acid. The carbon dioxide produced is collected and measured at room temperature and pressure (RTP). The volume of CO₂ collected is 96.0 cm³. Calculate the percentage purity of the calcium carbonate sample. (Molar volume at RTP = 24.0 dm³ mol⁻¹, Mᵣ of CaCO₃ = 100.1)

取 0.500 g 不纯的碳酸钙样品与过量盐酸反应。在室温和常压 (RTP) 下收集生成的二氧化碳，体积为 96.0 cm³。计算碳酸钙样品的纯度百分比。(RTP 下摩尔体积 = 24.0 dm³ mol⁻¹，CaCO₃ 相对分子质量 = 100.1)

Solution: Moles of CO₂ = volume / molar volume = 96.0 cm³ / 24000 cm³ mol⁻¹ = 0.00400 mol. The reaction is CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so mole ratio CaCO₃:CO₂ = 1:1. Moles of CaCO₃ = 0.00400 mol. Mass of pure CaCO₃ = 0.00400 mol × 100.1 g mol⁻¹ = 0.4004 g. Percentage purity = (0.4004 g / 0.500 g) × 100% = 80.1%.

解答：CO₂ 的物质的量 = 体积 / 摩尔体积 = 96.0 cm³ / 24000 cm³ mol⁻¹ = 0.00400 mol。反应为 CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O，CaCO₃ 与 CO₂ 的物质的量比为 1:1。故 CaCO₃ 的物质的量 = 0.00400 mol。纯 CaCO₃ 的质量 = 0.00400 mol × 100.1 g mol⁻¹ = 0.4004 g。纯度百分比 = (0.4004 g / 0.500 g) × 100% = 80.1%。

2. Limiting Reactant Problems | 限量反应物问题

2.00 g of aluminium is heated with 8.00 g of iron(III) oxide according to the thermite reaction: 2Al + Fe₂O₃ → 2Fe + Al₂O₃. Determine the limiting reactant and the maximum mass of iron that can be produced. (Aᵣ: Al = 27.0, Fe = 55.8, O = 16.0)

将 2.00 g 铝与 8.00 g 氧化铁(III) 按铝热反应加热：2Al + Fe₂O₃ → 2Fe + Al₂O₃。确定限量反应物及最多能生成铁的质量。(相对原子质量：Al = 27.0, Fe = 55.8, O = 16.0)

Solution: Moles of Al = 2.00 g / 27.0 g mol⁻¹ = 0.0741 mol. Moles of Fe₂O₃ = 8.00 g / (2×55.8 + 3×16.0) g mol⁻¹ = 8.00 / 159.6 = 0.0501 mol. From 2Al + Fe₂O₃ → 2Fe + Al₂O₃, 2 mol Al react with 1 mol Fe₂O₃. Required mol of Al for 0.0501 mol Fe₂O₃ = 0.1002 mol. Available Al = 0.0741 mol, so Al is the limiting reactant. Moles of Fe produced = mole ratio from Al (2:2) = 0.0741 mol. Mass of Fe = 0.0741 mol × 55.8 g mol⁻¹ = 4.13 g.

解答：Al 的物质的量 = 2.00 g / 27.0 g mol⁻¹ = 0.0741 mol。Fe₂O₃ 的物质的量 = 8.00 g / (2×55.8+3×16.0) g mol⁻¹ = 8.00 / 159.6 = 0.0501 mol。根据 2Al + Fe₂O₃ → 2Fe + Al₂O₃，2 mol Al 与 1 mol Fe₂O₃ 反应。与 0.0501 mol Fe₂O₃ 反应所需 Al 的物质的量 = 0.1002 mol。实际 Al 为 0.0741 mol，因此 Al 为限量反应物。生成 Fe 的物质的量 = 与 Al 的比 2:2 = 0.0741 mol。Fe 的质量 = 0.0741 mol × 55.8 g mol⁻¹ = 4.13 g。

3. Gas Laws and Molar Volume | 气体定律与摩尔体积

A gaseous hydrocarbon weighing 0.290 g occupies a volume of 112 cm³ at 100 °C and 101 kPa. Determine the empirical and molecular formula of the compound if its empirical formula is CH₂. (R = 8.31 J K⁻¹ mol⁻¹; Mᵣ CH₂ = 14.0)

0.290 g 某气态烃在 100 °C 和 101 kPa 下的体积为 112 cm³。已知其实验式为 CH₂，求该化合物的实验式和分子式。(R = 8.31 J K⁻¹ mol⁻¹; CH₂ 相对质量 = 14.0)

Solution: Use PV = nRT. P = 101 kPa = 101000 Pa, V = 112 cm³ = 1.12 × 10⁻⁴ m³, T = 100 + 273 = 373 K. n = PV/RT = (101000 × 1.12 × 10⁻⁴) / (8.31 × 373) = 11.312 / 3099.63 = 0.00365 mol. Molar mass M = mass / n = 0.290 g / 0.00365 mol = 79.5 g mol⁻¹. Empirical formula mass = 14.0. n = 79.5 / 14.0 = 5.68 ≈ 6. Thus molecular formula is (CH₂)₆ = C₆H₁₂.

解答：使用 PV = nRT。P = 101 kPa = 101000 Pa，V = 112 cm³ = 1.12 × 10⁻⁴ m³，T = 100 + 273 = 373 K。n = PV/RT = (101000 × 1.12 × 10⁻⁴) / (8.31 × 373) = 11.312 / 3099.63 = 0.00365 mol。摩尔质量 M = 质量 / n = 0.290 g / 0.00365 mol = 79.5 g mol⁻¹。实验式质量 = 14.0。n = 79.5 / 14.0 = 5.68 ≈ 6。因此分子式为 (CH₂)₆ = C₆H₁₂。

4. Equilibrium Constant Calculations | 平衡常数计算

For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), 0.500 mol of H₂ and 0.500 mol of I₂ are mixed in a 2.00 dm³ container at 440 °C. At equilibrium, 0.780 mol of HI is present. Calculate the equilibrium constant Kc at this temperature.

对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，在 440 °C 将 0.500 mol H₂ 和 0.500 mol I₂ 混合于 2.00 dm³ 容器中。达到平衡时，存在 0.780 mol HI。计算该温度下的平衡常数 Kc。

Solution: Initial amounts: H₂ = 0.500 mol, I₂ = 0.500 mol, HI = 0. Change: HI formed = 0.780 mol; from stoichiometry, H₂ reacted = I₂ reacted = 0.390 mol. Equilibrium amounts: H₂ = 0.500 – 0.390 = 0.110 mol, I₂ = 0.110 mol, HI = 0.780 mol. Concentrations (mol dm⁻³): [H₂] = 0.110/2.00 = 0.0550, [I₂] = 0.0550, [HI] = 0.780/2.00 = 0.390. Kc = [HI]²/([H₂][I₂]) = (0.390)²/(0.0550 × 0.0550) = 0.1521 / 0.003025 = 50.3. Units: (mol dm⁻³)²/(mol dm⁻³)² = no unit; so Kc = 50.3.

解答：初始物质的量：H₂ = 0.500 mol，I₂ = 0.500 mol，HI = 0。变化量：生成 HI = 0.780 mol；根据化学计量，反应的 H₂ = 反应的 I₂ = 0.390 mol。平衡时物质的量：H₂ = 0.500 – 0.390 = 0.110 mol，I₂ = 0.110 mol，HI = 0.780 mol。浓度 (mol dm⁻³)：[H₂] = 0.110/2.00 = 0.0550，[I₂] = 0.0550，[HI] = 0.780/2.00 = 0.390。Kc = [HI]²/([H₂][I₂]) = (0.390)²/(0.0550 × 0.0550) = 0.1521 / 0.003025 = 50.3。单位：(mol dm⁻³)²/(mol dm⁻³)² 无量纲，所以 Kc = 50.3。

5. Acid-Base Titration Curves | 酸碱滴定曲线

A 25.0 cm³ sample of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵) is titrated with 0.100 mol dm⁻³ NaOH. Calculate the pH after adding 12.5 cm³ of NaOH, and identify the region of the titration curve.

用 0.100 mol dm⁻³ NaOH 滴定 25.0 cm³ 0.100 mol dm⁻³ 乙酸 (Ka = 1.8 × 10⁻⁵)。计算加入 12.5 cm³ NaOH 后的 pH，判断此时处于滴定曲线的哪个区域。

Solution: Half-neutralisation point. Moles of acid initially = 0.0250 dm³ × 0.100 mol dm⁻³ = 0.00250 mol. Moles of OH⁻ added = 0.0125 dm³ × 0.100 = 0.00125 mol. Buffer system: CH₃COOH remaining = 0.00125 mol; CH₃COONa formed = 0.00125 mol. By Henderson-Hasselbalch: pH = pKa + log([salt]/[acid]) = –log(1.8×10⁻⁵) + log(1) = 4.74. This is the buffer region, exactly the half-equivalence point where pH = pKa.

解答：半中和点。酸初始物质的量 = 0.0250 dm³ × 0.100 mol dm⁻³ = 0.00250 mol。加入 OH⁻ 的物质的量 = 0.0125 dm³ × 0.100 = 0.00125 mol。缓冲体系：剩余 CH₃COOH = 0.00125 mol；生成 CH₃COONa = 0.00125 mol。按亨德森-哈塞尔巴尔赫方程：pH = pKa + log([盐]/[酸]) = –log(1.8×10⁻⁵) + log(1) = 4.74。此处为缓冲区域，正是半等当点，pH = pKa。

6. Organic Reaction Mechanisms | 有机反应机理

Outline the mechanism for the electrophilic addition of HBr to propene, CH₃–CH=CH₂, describing the formation of the major and minor products. Use curly arrows to show electron movement.

描绘 HBr 与丙烯 CH₃–CH=CH₂ 的亲电加成反应机理，说明主产物与次产物的形成过程。用弯箭头表示电子转移。

Solution: Step 1: Electrophilic attack. The π-bond electrons of propene attack the partially positive H of HBr. Heterolytic fission of H–Br occurs, forming a bromide ion and a carbocation. There are two possible carbocations: secondary (CH₃–CH⁺–CH₃) and primary (CH₃–CH₂–CH₂⁺). The secondary carbocation is more stable (due to inductive effect and hyperconjugation), so it forms predominantly. Step 2: Nucleophilic attack. The bromide ion Br⁻ attacks the secondary carbocation to give 2-bromopropane (major product). Minor product: attack on primary carbocation gives 1-bromopropane. Overall major product follows Markovnikov’s rule: H adds to the less substituted carbon.

解答：第一步：亲电进攻。丙烯的 π 键电子进攻 HBr 中带部分正电荷的 H。H–Br 发生异裂，生成溴离子和碳正离子。可能有两种碳正离子：仲碳正离子 (CH₃–CH⁺–CH₃) 和伯碳正离子 (CH₃–CH₂–CH₂⁺)。仲碳正离子更稳定（由于诱导效应与超共轭），因此主要形成它。第二步：亲核进攻。溴离子 Br⁻ 进攻仲碳正离子得到 2-溴丙烷（主产物）。次要产物：进攻伯碳正离子得到 1-溴丙烷。总主产物符合马氏规则：H 加在含氢较多的碳上。

7. Energetics and Hess’s Law | 热力学与赫斯定律

Use the following data to calculate the standard enthalpy change of formation of methane, CH₄, in kJ mol⁻¹. ΔH°c(C) = –394 kJ mol⁻¹, ΔH°c(H₂) = –286 kJ mol⁻¹, ΔH°c(CH₄) = –890 kJ mol⁻¹.

利用下列数据计算甲烷 CH₄ 的标准生成焓变（kJ mol⁻¹）。ΔH°c(C) = –394 kJ mol⁻¹，ΔH°c(H₂) = –286 kJ mol⁻¹，ΔH°c(CH₄) = –890 kJ mol⁻¹。

Solution: Enthalpy cycle: target reaction: C(s) + 2H₂(g) → CH₄(g). Route via combustion: C(s) + O₂(g) → CO₂(g) ΔH = –394; 2[H₂(g) + ½O₂(g) → H₂O(l)] ΔH = 2(–286) = –572; CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = –890. By Hess’s Law: ΔH°f(CH₄) = ΔH°c(C) + 2ΔH°c(H₂) – ΔH°c(CH₄) = –394 + (–572) – (–890) = –76 kJ mol⁻¹.

解答：焓循环：目标反应：C(s) + 2H₂(g) → CH₄(g)。经燃烧路径：C(s) + O₂(g) → CO₂(g) ΔH = –394；2[H₂(g) + ½O₂(g) → H₂O(l)] ΔH = 2(–286) = –572；CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = –890。根据赫斯定律：ΔH°f(CH₄) = ΔH°c(C) + 2ΔH°c(H₂) – ΔH°c(CH₄) = –394 + (–572) – (–890) = –76 kJ mol⁻¹。

8. Electrochemical Cells and Standard Potentials | 电化学电池与标准电势

A voltaic cell is constructed with Zn²⁺/Zn and Cu²⁺/Cu half-cells under standard conditions. E°(Zn²⁺/Zn) = –0.76 V, E°(Cu²⁺/Cu) = +0.34 V. Calculate the standard cell potential E°cell and write the overall spontaneous reaction.

用 Zn²⁺/Zn 和 Cu²⁺/Cu 半电池在标准条件下构建伏打电池。E°(Zn²⁺/Zn) = –0.76 V，E°(Cu²⁺/Cu) = +0.34 V。计算标准电池电动势 E°cell，并写出自发总反应。

Solution: More negative half-cell undergoes oxidation. Zn has more negative potential, so Zn(s) → Zn²⁺(aq) + 2e⁻ (oxidation). Cu²⁺ has more positive potential, so Cu²⁺(aq) + 2e⁻ → Cu(s) (reduction). E°cell = E°reduction – E°oxidation = +0.34 V – (–0.76 V) = 1.10 V. Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

解答：电极电势较负的半电池发生氧化。Zn 的电势更负，因此 Zn(s) → Zn²⁺(aq) + 2e⁻ (氧化)。Cu²⁺ 的电势更正，因此 Cu²⁺(aq) + 2e⁻ → Cu(s) (还原)。E°cell = E°还原 – E°氧化 = +0.34 V – (–0.76 V) = 1.10 V。总反应：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。

9. Spectroscopy and Structure Determination | 光谱与结构测定

An organic compound with molecular formula C₄H₈O₂ gives the following spectra: IR absorption at 1735 cm⁻¹ (strong) and 1240 cm⁻¹; ¹H NMR: δ 1.3 (3H, triplet), δ 2.3 (2H, quartet), δ 4.1 (2H, quartet), δ 3.7 (3H, singlet). Deduce the structure.

某有机化合物分子式为 C₄H₈O₂，谱图如下：IR 吸收在 1735 cm⁻¹（强）和 1240 cm⁻¹；¹H NMR：δ 1.3 (3H, 三重峰)，δ 2.3 (2H, 四重峰)，δ 4.1 (2H, 四重峰)，δ 3.7 (3H, 单峰)。推断其结构。

Solution: IR 1735 cm⁻¹ indicates C=O stretch (ester or carbonyl), 1240 cm⁻¹ suggests C–O stretch. The NMR shows an ethyl group: CH₃ triplet at 1.3 coupled to a CH₂ quartet at 2.3 or 4.1? The coupling pattern and integration: CH₃ triplet (1.3) and CH₂ quartet (2.3) are coupled (ethyl attached to carbonyl, typically CH₂–C=O appears ~2.3). Another CH₂ quartet at 4.1 is deshielded, attached to oxygen (O–CH₂–). Singlet at 3.7 (3H) is O–CH₃. So combining: CH₃–CH₂–CO–O–CH₃? However, integration: CH₃CH₂COOCH₃ would give CH₃ triplet, CH₂ quartet at 2.3, OCH₃ singlet at 3.7, and OCH₂? No, methyl propanoate is CH₃CH₂COOCH₃: OCH₃ (singlet, 3H), CH₂ (quartet at 2.3, 2H), CH₃ (triplet at 1.3, 3H). That matches all signals? But the 4.1 quartet is missing; wait, methyl propanoate does not have an O–CH₂– group. The formula C₄H₈O₂ could be methyl propanoate (CH₃CH₂COOCH₃) or ethyl ethanoate (CH₃COOCH₂CH₃). For ethyl ethanoate: CH₃CO– (singlet, 3H, ~2.0-2.2), –OCH₂CH₃ (CH₂ quartet 4.1, CH₃ triplet 1.3). Here we have a singlet at 3.7 (OCH₃) but also a quartet at 4.1. That suggests both –OCH₃ and –OCH₂–? The total hydrogen count: triplet (3H), quartet (2H) at 2.3, quartet (2H) at 4.1, singlet (3H) = 10 H? But formula C₄H₈O₂ only has 8 H. Recalculate: the problem says: δ 1.3 (3H, triplet), δ 2.3 (2H, quartet), δ 4.1 (2H, quartet), δ 3.7 (3H, singlet). That sums to 10 H? Wait: 3 + 2 + 2 + 3 = 10. But C₄H₈O₂ has 8 H. There is an inconsistency. Possibly I mis-wrote example. Let me adjust: for ethyl ethanoate: CH₃COOCH₂CH₃: CH₃COO (singlet 3H at ~2.0), OCH₂ (quartet 4.1, 2H), CH₃ (triplet 1.3, 3H). That’s 3+2+3=8 H, matches. Then the quartet at 2.3 is not present. So my example must be something else: methyl propanoate: CH₃CH₂COOCH₃: CH₃ triplet 1.1, CH₂ quartet 2.3, OCH₃ singlet 3.6, no peak at 4.1. That’s 3+2+3=8 H. So the given peaks don’t match either pure ester. Perhaps the compound is ethanoic acid? CH₃COOH: O–H at ~12, CH₃ singlet 2.1, no 4.1. So to be consistent, I need to design a problem that works. Let’s change the example to a ketone or something. Better: I’ll use C₄H₈O₂ for ethyl ethanoate, but only provide relevant NMR: δ 1.3 (3H, t), δ 2.0 (3H, s), δ 4.1 (2H, q). Then solution would be ethyl ethanoate. But I’ll rephrase: Provided NMR: δ 1.25 (3H, triplet), δ 2.05 (3H, singlet), δ 4.12 (2H, quartet). IR 1740, 1240. Deduce structure. That’s typical IB/CIE question. So I’ll adjust. But I already wrote the problem with four signals. Let’s just rewrite the example with correct signals. I will modify the English and Chinese. Choose: ethyl ethanoate. NMR: δ 1.3 (3H, triplet), δ 2.0 (3H, singlet), δ 4.1 (2H, quartet). IR: 1740, 1240. So in the next paragraph I’ll correct. Actually, I’ll rewrite the entire section. I’ll output accordingly.

I’ll replace:

“An organic compound with molecular formula C₄H₈O₂ gives the following spectra: IR absorption at 1735 cm⁻¹ (strong) and 1240 cm⁻¹; ¹H NMR: δ 1.3 (3H, triplet), δ 2.0 (3H, singlet), δ 4.1 (2H, quartet). Deduce the structure.”

Then solution: IR 1735 indicates C=O (ester), 1240 C–O. NMR: triplet at 1.3 and quartet at 4.1 are an ethyl group attached to oxygen (–OCH₂CH₃). Singlet at 2.0 corresponds to a methyl attached to carbonyl (CH₃CO–). Thus the structure is ethyl ethanoate, CH₃COOCH₂CH₃.

Match Chinese accordingly. I’ll now compose sections 9 and 10, ensuring consistency.

9. Spectroscopy and Structure Determination | 光谱与结构测定

An organic compound with molecular formula C₄H₈O₂ gives the following spectra: IR absorption at 1735 cm⁻¹ (strong) and 1240 cm⁻¹; ¹H NMR: δ 1.3 (3H, triplet), δ 2.0 (3H, singlet), δ 4.1 (2H, quartet). Deduce the structure.

某有机物分子式为 C₄H₈O₂，给出以下谱图：IR 吸收 1735 cm⁻¹ (强) 和 1240 cm⁻¹；¹H NMR：δ 1.3 (3H, 三重峰), δ 2.0 (3H, 单峰), δ 4.1 (2H, 四重峰)。推断其结构。

Solution: IR 1735 cm⁻¹ corresponds to C=O stretch (ester or ketone), 1240 cm⁻¹ indicates C–O stretch. In NMR, the triplet at δ 1.3 coupled with quartet at 4.1 reveals an –O–CH₂–CH₃ ethyl group. The singlet at δ 2.0 is characteristic of a methyl group adjacent to carbonyl (CH₃–C=O). Combining these fragments yields CH₃COOCH₂CH₃, ethyl ethanoate. The integration (3:3:2) matches the total 8 hydrogens.

解答：IR 中 1735 cm⁻¹ 对应 C=O 伸缩振动（酯或酮），1240 cm⁻¹ 表明 C–O 伸缩。NMR 中 δ 1.3 的三重峰与 δ 4.1 的四重峰耦合，揭示 –O–CH₂–CH₃ 乙氧基。δ 2.0 的单峰是连接在羰基旁的甲基 (CH₃–C=O)。组合得到 CH₃COOCH₂CH₃，乙酸乙酯。积分比 3:3:2 符合总氢数 8。

10. Kinetics and Rate Law Determination | 动力学与速率定律确定

The following data were collected for the reaction A + B → products. Initial rate method: Trial 1: [A]⁰ = 0.10 mol dm⁻³, [B]⁰ = 0.10 mol dm⁻³, initial rate = 0.020 mol dm⁻³ s⁻¹. Trial 2: [A]⁰ = 0.20, [B]⁰ = 0.10, rate = 0.080. Trial 3: [A]⁰ = 0.10, [B]⁰ = 0.20, rate = 0.020. Determine the rate law and the rate constant k.

对于反应 A + B → 产物收集了以下数据。初速率法：实验1：[A]⁰ = 0.10 mol dm⁻³，[B]⁰ = 0.10 mol dm⁻³，初速率 = 0.020 mol dm⁻³ s⁻¹。实验2：[A]⁰ = 0.20，[B]⁰ = 0.10，速率 = 0.080。实验3：[A]⁰ = 0.10，[B]⁰ = 0.20，速率 = 0.020。确定速率定律和速率常数 k。

Solution: Compare trials 1 and 2: [B] constant, [A] doubles, rate rises from 0.020 to 0.080 (factor 4). So order with respect to A is 2 (since rate ∝ [A]²). Compare trials 1 and 3: [A] constant, [B] doubles, rate unchanged (0.020). So order with respect to B is 0. Rate law: rate = k[A]²[B]⁰ = k[A]². Using trial 1: 0.020 = k (0.10)² → k = 0.020 / 0.010 = 2.0 dm³ mol⁻¹ s⁻¹.

解答：比较实验1和2：[B] 恒定，[A] 加倍，速率从 0.020 升至 0.080（4倍）。因此对 A 的反应级数为 2。比较实验1和3：[A] 恒定，[B] 加倍，速率不变 (0.020)。因此对 B 的反应级数为 0。速率方程：rate = k[A]²[B]⁰ = k[A]²。利用实验1：0.020 = k (0.10)² → k = 0.020 / 0.010 = 2.0 dm³ mol⁻¹ s⁻¹。

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Worked Examples in IB and CIE Chemistry | IB CIE 化学：典型例题详解