📚 IB & WJEC Chemistry: Worked Examples Explained | IB 与 WJEC 化学：典型例题详解
This article provides detailed step-by-step solutions to typical IB and WJEC A-level chemistry problems, covering stoichiometry, titrations, thermochemistry, kinetics, equilibrium, buffer calculations, organic mechanisms and electrochemistry. Each worked example is presented with parallel explanations in English and Chinese to reinforce understanding of key concepts and exam techniques.
本文通过详尽的步骤解析 IB 与 WJEC 化学典型例题，涵盖化学计量、滴定、热化学、动力学、平衡、缓冲溶液计算、有机反应机理和电化学。每个例题均以中英对照的方式逐步讲解，帮助读者巩固核心概念和应试技巧。
1. Stoichiometry and Mole Calculations | 化学计量与摩尔计算
Question: Calculate the mass of water produced when 50.0 g of propane (C₃H₈) undergoes complete combustion in excess oxygen. Assume atomic masses: C = 12.0, H = 1.0, O = 16.0 g mol⁻¹.
问题：计算 50.0 g 丙烷 (C₃H₈) 在过量氧气中完全燃烧时生成的水的质量。假设原子量：C = 12.0, H = 1.0, O = 16.0 g mol⁻¹。
Step 1: Write the balanced chemical equation for the combustion of propane.
步骤 1：写出丙烷完全燃烧的配平化学方程式。
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Step 2: Calculate the molar mass of propane (C₃H₈). M = (3 × 12.0) + (8 × 1.0) = 36.0 + 8.0 = 44.0 g mol⁻¹.
步骤 2：计算丙烷 (C₃H₈) 的摩尔质量。M = (3 × 12.0) + (8 × 1.0) = 36.0 + 8.0 = 44.0 g mol⁻¹。
Step 3: Determine the number of moles of propane. n(C₃H₈) = mass / M = 50.0 g / 44.0 g mol⁻¹ = 1.136 mol.
步骤 3：求丙烷的物质的量。n(C₃H₈) = 质量 / M = 50.0 g / 44.0 g mol⁻¹ = 1.136 mol。
Step 4: Use the stoichiometric ratio from the balanced equation. 1 mol C₃H₈ produces 4 mol H₂O, therefore moles of H₂O = 1.136 × 4 = 4.545 mol.
步骤 4：利用方程式中的化学计量比。1 mol C₃H₈ 生成 4 mol H₂O，所以 H₂O 的物质的量 = 1.136 × 4 = 4.545 mol。
Step 5: Convert moles of water to mass. M(H₂O) = (2 × 1.0) + 16.0 = 18.0 g mol⁻¹. Mass = n × M = 4.545 mol × 18.0 g mol⁻¹ = 81.8 g.
步骤 5：将水的物质的量转化为质量。M(H₂O) = (2 × 1.0) + 16.0 = 18.0 g mol⁻¹。质量 = n × M = 4.545 mol × 18.0 g mol⁻¹ = 81.8 g。
Mass of H₂O = 81.8 g (3 significant figures).
2. Titration and Concentration | 滴定与浓度计算
Question: 25.0 cm³ of hydrochloric acid is titrated against 0.100 mol dm⁻³ sodium hydroxide solution. The endpoint is reached after adding 30.0 cm³ of NaOH. Determine the concentration of the HCl solution.
问题：用 0.100 mol dm⁻³ 氢氧化钠溶液滴定 25.0 cm³ 盐酸，滴定终点时消耗 NaOH 溶液 30.0 cm³。求盐酸的浓度。
Step 1: Write the neutralisation equation. HCl + NaOH → NaCl + H₂O. The stoichiometry is 1:1.
步骤 1：写出中和反应方程式。HCl + NaOH → NaCl + H₂O。化学计量比为 1:1。
Step 2: Calculate the moles of NaOH used. n(NaOH) = concentration × volume (in dm³) = 0.100 mol dm⁻³ × (30.0 / 1000) dm³ = 0.00300 mol.
步骤 2：计算所用 NaOH 的物质的量。n(NaOH) = 浓度 × 体积（以 dm³ 计）= 0.100 mol dm⁻³ × (30.0 / 1000) dm³ = 0.00300 mol。
Step 3: From the 1:1 ratio, n(HCl) = n(NaOH) = 0.00300 mol.
步骤 3：根据 1:1 计量比，n(HCl) = n(NaOH) = 0.00300 mol。
Step 4: Calculate the concentration of HCl. c(HCl) = n / V = 0.00300 mol / (25.0 / 1000) dm³ = 0.120 mol dm⁻³.
步骤 4：计算 HCl 的浓度。c(HCl) = n / V = 0.00300 mol / (25.0 / 1000) dm³ = 0.120 mol dm⁻³。
Concentration of HCl = 0.120 mol dm⁻³
3. Enthalpy Changes and Hess’ s Law | 焓变与赫斯定律
Question: Calculate the standard enthalpy change for the formation of ethyne (C₂H₂) from its elements: 2C(s) + H₂(g) → C₂H₂(g). Use the following data:
问题：计算由单质生成乙炔 (C₂H₂) 的标准焓变：2C(s) + H₂(g) → C₂H₂(g)。已知数据如下：
- C(s) + O₂(g) → CO₂(g) ΔH° = -394 kJ mol⁻¹
- H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -286 kJ mol⁻¹
- 2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l) ΔH° = -2600 kJ mol⁻¹
English: C(s) + O₂(g) → CO₂(g) ΔH° = -394 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -286 kJ mol⁻¹; 2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l) ΔH° = -2600 kJ mol⁻¹.
中文：C(s) + O₂(g) → CO₂(g) ΔH° = -394 kJ mol⁻¹；H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -286 kJ mol⁻¹；2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l) ΔH° = -2600 kJ mol⁻¹。
Step 1: We need to express the target reaction as a combination of the given equations. The target is 2C(s) + H₂(g) → C₂H₂(g).
步骤 1：目标反应需用已知方程式组合表示。目标为 2C(s) + H₂(g) → C₂H₂(g)。
Step 2: Multiply the first equation by 2 to give 2C(s) + 2O₂(g) → 2CO₂(g) ΔH° = 2 × (-394) = -788 kJ mol⁻¹.
步骤 2：将第一个方程乘以 2，得到 2C(s) + 2O₂(g) → 2CO₂(g) ΔH° = 2 × (-394) = -788 kJ mol⁻¹。
Step 3: Keep the second equation as is: H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -286 kJ mol⁻¹.
步骤 3：第二个方程保持不变：H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -286 kJ mol⁻¹。
Step 4: Reverse the third equation and divide by 2 to get C₂H₂ on the product side: 2CO₂(g) + H₂O(l) → C₂H₂(g) + 2.5O₂(g). ΔH° becomes +2600/2? Wait: The given third reaction is for 2 mol C₂H₂. Reverse and halve it: 4CO₂ + 2H₂O → 2C₂H₂ + 5O₂, ΔH° = +2600 kJ. Then divide by 2: 2CO₂(g) + H₂O(l) → C₂H₂(g) + 2.5O₂(g) ΔH° = +1300 kJ mol⁻¹.
步骤 4：将第三个方程反转并除以 2，使 C₂H₂ 出现在生成物一侧：2CO₂(g) + H₂O(l) → C₂H₂(g) + 2.5O₂(g)，ΔH° 变为 +1300 kJ mol⁻¹。解释：原方程 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O ΔH = -2600。反转：4CO₂ + 2H₂O → 2C₂H₂ + 5O₂ ΔH = +2600。除以2得：2CO₂ + H₂O → C₂H₂ + 2.5O₂ ΔH = +1300 kJ mol⁻¹。
Step 5: Add the modified equations. (2C + 2O₂ → 2CO₂) + (H₂ + 0.5O₂ → H₂O) + (2CO₂ + H₂O → C₂H₂ + 2.5O₂) yields 2C + H₂ + 2O₂ + 0.5O₂ + 2CO₂ + H₂O → 2CO₂ + H₂O + C₂H₂ + 2.5O₂. Cancelling gives 2C(s) + H₂(g) → C₂H₂(g). Sum the ΔH values: -788 + (-286) + 1300 = +226 kJ mol⁻¹.
步骤 5：将变形后的方程式相加。（2C + 2O₂ → 2CO₂）+（H₂ + 0.5O₂ → H₂O）+（2CO₂ + H₂O → C₂H₂ + 2.5O₂）得到 2C + H₂ + 2O₂ + 0.5O₂ + 2CO₂ + H₂O → 2CO₂ + H₂O + C₂H₂ + 2.5O₂。消去相同项后得到 2C(s) + H₂(g) → C₂H₂(g)。ΔH 加和：-788 + (-286) + 1300 = +226 kJ mol⁻¹。
ΔH°f (C₂H₂) = +226 kJ mol⁻¹
4. Reaction Kinetics and Rate Laws | 反应动力学与速率方程
Question: The following initial rate data were collected for the reaction 2NO(g) + Cl₂(g) → 2NOCl(g) at a fixed temperature. Determine the rate law and the rate constant k.
问题：在固定温度下，对反应 2NO(g) + Cl₂(g) → 2NOCl(g) 测得以下初始速率数据。确定速率方程和速率常数 k。
| Experiment | [NO] / mol dm⁻³ | [Cl₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
| 1 | 0.10 | 0.10 | 3.0 × 10⁻³ |
| 2 | 0.10 | 0.20 | 6.0 × 10⁻³ |
| 3 | 0.20 | 0.20 | 2.4 × 10⁻² |
Step 1: Assume the rate law is rate = k[NO]ᵐ[Cl₂]ⁿ. Find m and n by comparing experiments.
步骤 1：假设速率方程为 rate = k[NO]ᵐ[Cl₂]ⁿ。通过比较实验数据求出 m 和 n。
Step 2: Compare Exp 1 and 2: [NO] constant, [Cl₂] doubles from 0.10 to 0.20 mol dm⁻³. Rate doubles from 3.0 × 10⁻³ to 6.0 × 10⁻³. Therefore, n = 1 (first order with respect to Cl₂).
步骤 2：比较实验 1 和 2：[NO] 恒定，[Cl₂] 从 0.10 倍增至 0.20 mol dm⁻³，速率从 3.0 × 10⁻³ 倍增至 6.0 × 10⁻³。因此 n = 1（对 Cl₂ 为一级反应）。
Step 3: Compare Exp 2 and 3: [Cl₂] constant at 0.20, [NO] doubles from 0.10 to 0.20 mol dm⁻³. Rate goes from 6.0 × 10⁻³ to 2.4 × 10⁻², a factor of 4. So 2ᵐ = 4 → m = 2 (second order with respect to NO).
步骤 3：比较实验 2 和 3：[Cl₂] 保持在 0.20，[NO] 从 0.10 倍增至 0.20 mol dm⁻³，速率从 6.0 × 10⁻³ 增至 2.4 × 10⁻²，变为 4 倍。因此 2ᵐ = 4 → m = 2（对 NO 为二级反应）。
Step 4: Rate law: rate = k[NO]²[Cl₂]. Use data from Exp 1 to calculate k.
步骤 4：速率方程为 rate = k[NO]²[Cl₂]。使用实验 1 数据计算 k。
k = rate / ([NO]²[Cl₂]) = 3.0 × 10⁻³ / (0.10² × 0.10) = 3.0 × 10⁻³ / 0.001 = 3.0 dm⁶ mol⁻² s⁻¹.
k = 3.0 dm⁶ mol⁻² s⁻¹ (units derived from rate / concentration terms).
k = 3.0 dm⁶ mol⁻² s⁻¹（单位由速率除以浓度项导出）。
5. Equilibrium Constant Kc | 平衡常数 Kc
Question: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), at equilibrium the concentrations are [H₂] = 0.20 mol dm⁻³, [I₂] = 0.20 mol dm⁻³ and [HI] = 1.60 mol dm⁻³. Calculate the value of Kc and state its units.
问题：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，平衡时各物质浓度为 [H₂] = 0.20 mol dm⁻³，[I₂] = 0.20 mol dm⁻³，[HI] = 1.60 mol dm⁻³。计算 Kc 值并写出单位。
Step 1: Write the expression for Kc: Kc = [HI]² / ([H₂][I₂]).
步骤 1：写出 Kc 的表达式：Kc = [HI]² / ([H₂][I₂])。
Step 2: Substitute equilibrium concentrations: Kc = (1.60)² / (0.20 × 0.20) = 2.56 / 0.04 = 64.0.
步骤 2：代入平衡浓度：Kc = (1.60)² / (0.20 × 0.20) = 2.56 / 0.04 = 64.0。
Step 3: Determine units. (mol dm⁻³)² / (mol dm⁻³ × mol dm⁻³) = no units, so Kc is dimensionless. This is typical when the number of moles on each side of the equation is equal.
步骤 3：确定单位。(mol dm⁻³)² / (mol dm⁻³ × mol dm⁻³) 无单位，因此 Kc 无量纲。这是反应前后气体分子数相同时的典型情况。
Kc = 64.0 (no units)
6. Buffer Solutions and pH | 缓冲溶液与 pH 计算
Question: A buffer is prepared by mixing 50.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (CH₃COOH) with 50.0 cm³ of 0.100 mol dm⁻³ sodium ethanoate (CH₃COONa). Given Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³, calculate the pH of the buffer.
问题：将 50.0 cm³ 0.100 mol dm⁻³ 乙酸 (CH₃COOH) 与 50.0 cm³ 0.100 mol dm⁻³ 乙酸钠 (CH₃COONa) 混合制备缓冲溶液。已知乙酸的 Ka = 1.74 × 10⁻⁵ mol dm⁻³，计算该缓冲溶液的 pH。
Step 1: After mixing, total volume = 100.0 cm³. The concentrations become halved: [CH₃COOH] = 0.050 mol dm⁻³, [CH₃COO⁻] = 0.050 mol dm⁻³.
步骤 1：混合后总体积为 100.0 cm³，浓度减半：[CH₃COOH] = 0.050 mol dm⁻³，[CH₃COO⁻] = 0.050 mol dm⁻³。
Step 2: Use the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]).
步骤 2：使用 Henderson-Hasselbalch 方程：pH = pKa + log([A⁻]/[HA])。
Step 3: Calculate pKa = -log(Ka) = -log(1.74 × 10⁻⁵) = 4.76 (approx).
步骤 3：计算 pKa = -log(Ka) = -log(1.74 × 10⁻⁵) = 4.76（近似值）。
Step 4: Since [A⁻] = [HA], log(0.050/0.050) = log 1 = 0. Therefore, pH = pKa = 4.76.
步骤 4：由于 [A⁻] = [HA]，log(0.050/0.050) = log 1 = 0，因此 pH = pKa = 4.76。
pH = 4.76
7. Organic Reaction Mechanisms | 有机反应机理
Question: 2-bromo-2-methylpropane
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