📚 A-Level Edexcel Chemistry: Typical Example Questions Explained | A-Level Edexcel 化学：典型例题详解
This article walks you through a selection of typical A-Level Edexcel Chemistry problems, covering key quantitative and qualitative topics. Each worked example is broken down into logical steps with paired English and Chinese explanations, making it ideal for revision and exam practice.
本文精选了 A-Level Edexcel 化学中典型的例题，覆盖重要的定量与定性主题。每一道例题都分解为清晰的解题步骤，并配以中英双语解释，非常适合复习和应试训练。
1. Redox Titration – Determination of Iron | 氧化还原滴定 – 铁含量的测定
A sample of iron(II) sulfate is dissolved in dilute sulfuric acid and titrated with 0.0200 mol dm⁻³ potassium manganate(VII) solution. 25.0 cm³ of the iron(II) solution required 23.40 cm³ of KMnO₄ solution to reach a permanent pink end-point. Calculate the concentration of Fe²⁺ ions in the original solution.
将一份硫酸亚铁(II)样品溶于稀硫酸中，用 0.0200 mol dm⁻³ 的高锰酸钾(VII)溶液滴定。25.0 cm³ 的铁(II)溶液需要 23.40 cm³ 的 KMnO₄ 溶液达到持久粉红色终点。计算原溶液中 Fe²⁺ 离子的浓度。
The relevant half-equations are: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻. The overall stoichiometry is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Therefore, 1 mole of MnO₄⁻ reacts with 5 moles of Fe²⁺.
相关的半反应式为：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 以及 Fe²⁺ → Fe³⁺ + e⁻。总反应计量比为：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。因此，1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应。
Moles of MnO₄⁻ used = 0.0200 × (23.40 / 1000) = 4.68 × 10⁻⁴ mol. Hence, moles of Fe²⁺ in 25.0 cm³ = 5 × 4.68 × 10⁻⁴ = 2.34 × 10⁻³ mol. The concentration of Fe²⁺ = (2.34 × 10⁻³) / (25.0 / 1000) = 0.0936 mol dm⁻³.
所用 MnO₄⁻ 的物质的量 = 0.0200 × (23.40 / 1000) = 4.68 × 10⁻⁴ mol。因此，25.0 cm³ 中 Fe²⁺ 的物质的量 = 5 × 4.68 × 10⁻⁴ = 2.34 × 10⁻³ mol。Fe²⁺ 的浓度 = (2.34 × 10⁻³) / (25.0 / 1000) = 0.0936 mol dm⁻³。
2. Nucleophilic Substitution Mechanism (SN2) | 亲核取代反应机理 (SN2)
Explain the mechanism of the reaction between bromoethane and aqueous hydroxide ions, including the rate equation and stereochemical outcome. Use curly arrows to show electron movement.
解释溴乙烷与氢氧根离子水溶液反应的机理，包括速率方程和立体化学结果。用弯箭头表示电子转移。
The reaction CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ proceeds via an SN2 mechanism, which is a single-step bimolecular process. The nucleophile OH⁻ attacks the electrophilic carbon from the opposite side of the leaving group Br⁻. A transition state forms with a partially formed C–O bond and a partially broken C–Br bond.
反应 CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ 通过一个 SN2 机理进行，即单步双分子过程。亲核试剂 OH⁻ 从离去基团 Br⁻ 的背面进攻亲电碳原子。形成一个过渡态，其中 C–O 键部分形成，C–Br 键部分断裂。
Curly arrow from the lone pair on OH⁻ goes to the carbon, and another curly arrow from the C–Br bond goes to the Br atom. The rate equation is: rate = k[CH₃CH₂Br][OH⁻], consistent with both reactants appearing in the rate-determining step. The stereochemistry undergoes inversion, like an umbrella turning inside out.
OH⁻ 上孤对电子的弯箭头指向碳原子，C–Br 键的弯箭头指向 Br 原子。速率方程为：rate = k[CH₃CH₂Br][OH⁻]，与两种反应物均出现在决速步中一致。立体化学发生翻转，像雨伞内翻一般。
3. Born-Haber Cycle for Sodium Chloride | 氯化钠的 Born-Haber 循环
Use the following data to construct a Born-Haber cycle and calculate the lattice energy of NaCl(s). Values in kJ mol⁻¹: ΔHₐ(Na) = +108, IE₁(Na) = +496, ΔHₐ(½Cl₂) = +121, EA₁(Cl) = -349, ΔHf(NaCl) = -411.
使用下列数据构建 Born-Haber 循环并计算 NaCl(s) 的晶格能。数值单位 kJ mol⁻¹：ΔHₐ(Na) = +108，IE₁(Na) = +496，ΔHₐ(½Cl₂) = +121，EA₁(Cl) = -349，ΔHf(NaCl) = -411。
The Born-Haber cycle relates the enthalpy of formation to the sum of enthalpy changes along an alternative route: atomisation of sodium, ionisation of sodium, atomisation of chlorine, electron affinity of chlorine, and lattice energy. By Hess’s law: ΔHf(NaCl) = ΔHₐ(Na) + IE₁(Na) + ΔHₐ(½Cl₂) + EA₁(Cl) + U.
Born-Haber 循环将生成焓与另一条路径的焓变总和联系起来：钠的原子化、钠的电离、氯的原子化、氯的电子亲和势以及晶格能。根据盖斯定律：ΔHf(NaCl) = ΔHₐ(Na) + IE₁(Na) + ΔHₐ(½Cl₂) + EA₁(Cl) + U。
Substituting: -411 = 108 + 496 + 121 + (-349) + U. So U = -411 – (108 + 496 + 121 – 349) = -411 – 376 = -787 kJ mol⁻¹. The lattice energy is -787 kJ mol⁻¹.
代入：-411 = 108 + 496 + 121 + (-349) + U。因此 U = -411 – (108 + 496 + 121 – 349) = -411 – 376 = -787 kJ mol⁻¹。晶格能为 -787 kJ mol⁻¹。
4. Equilibrium Constant Kc Calculation | 平衡常数 Kc 的计算
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), at a certain temperature the equilibrium concentrations are: [N₂] = 0.60 mol dm⁻³, [H₂] = 1.80 mol dm⁻³, [NH₃] = 0.80 mol dm⁻³. Calculate Kc and state its units.
对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，在某一温度下平衡浓度分别为：[N₂] = 0.60 mol dm⁻³，[H₂] = 1.80 mol dm⁻³，[NH₃] = 0.80 mol dm⁻³。计算 Kc 并标明其单位。
The equilibrium expression is: Kc = [NH₃]² / ([N₂][H₂]³). Plug in values: [NH₃]² = (0.80)² = 0.64; [N₂] = 0.60; [H₂]³ = (1.80)³ = 5.832. So Kc = 0.64 / (0.60 × 5.832) = 0.64 / 3.4992 ≈ 0.183.
平衡表达式为：Kc = [NH₃]² / ([N₂][H₂]³)。代入数值：[NH₃]² = (0.80)² = 0.64；[N₂] = 0.60；[H₂]³ = (1.80)³ = 5.832。因此 Kc = 0.64 / (0.60 × 5.832) = 0.64 / 3.4992 ≈ 0.183。
Units: (mol dm⁻³)² / (mol dm⁻³ × (mol dm⁻³)³) = (mol dm⁻³)² / (mol⁴ dm⁻¹²) = mol⁻² dm⁶. So Kc = 0.183 mol⁻² dm⁶.
单位：(mol dm⁻³)² / (mol dm⁻³ × (mol dm⁻³)³) = (mol dm⁻³)² / (mol⁴ dm⁻¹²) = mol⁻² dm⁶。因此 Kc = 0.183 mol⁻² dm⁶。
5. Electrochemical Cells and Standard EMF | 电化学电池与标准电动势
A cell is constructed with Zn²⁺/Zn and Cu²⁺/Cu half-cells under standard conditions. Standard electrode potentials: E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V. Write the cell diagram, calculate E°cell, and identify the positive electrode.
在标准条件下，用 Zn²⁺/Zn 和 Cu²⁺/Cu 半电池构建一个电池。标准电极电势：E°(Zn²⁺/Zn) = -0.76 V，E°(Cu²⁺/Cu) = +0.34 V。书写电池图式，计算 E°cell，并指出正极。
The more negative electrode (Zn) undergoes oxidation, so the cell diagram is: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). E°cell = E°(right) – E°(left) = +0.34 – (-0.76) = +1.10 V. The positive electrode is copper, where reduction occurs.
电势更负的电极 (Zn) 发生氧化，因此电池图式为：Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)。E°cell = E°(右) – E°(左) = +0.34 – (-0.76) = +1.10 V。正极是铜极，发生还原反应。
The cell reaction is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Electrons flow from zinc to copper through the external circuit, and the salt bridge completes the circuit. A positive E°cell indicates a feasible reaction under standard conditions.
电池反应为：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。电子经外电路从锌流向铜，盐桥导通电路。正值的 E°cell 表明该反应在标准条件下可行。
6. Determining Rate Equation from Initial Rates | 由初始速率法确定速率方程
For the reaction A + B → C, the following initial rate data were collected. Experiment 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, initial rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Experiment 2: [A] = 0.20, [B] = 0.10, rate = 8.0 × 10⁻⁴. Experiment 3: [A] = 0.20, [B] = 0.20, rate = 1.6 × 10⁻³. Determine the rate equation and calculate the rate constant k, giving its units.
对于反应 A + B → C，收集了如下初始速率数据。实验1：[A] = 0.10 mol dm⁻³，[B] = 0.10 mol dm⁻³，初始速率 = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹。实验2：[A] = 0.20，[B] = 0.10，速率 = 8.0 × 10⁻⁴。实验3：[A] = 0.20，[B] = 0.20，速率 = 1.6 × 10⁻³。确定速率方程并计算速率常数 k，标明单位。
Compare Expts 1 and 2: [A] doubles, [B] constant, rate increases by 4 times (8.0/2.0 = 4). Therefore, reaction is second order with respect to A. Compare Expts 2 and 3: [B] doubles, [A] constant, rate doubles (1.6/8.0 = 2). Reaction is first order with respect to B. Rate equation: rate = k[A]²[B].
比较实验1和2：[A] 加倍，[B] 不变，速率增加 4 倍 (8.0/2.0 = 4)。因此反应对 A 为二级。比较实验2和3：[B] 加倍，[A] 不变，速率加倍 (1.6/8.0 = 2)。反应对 B 为一级。速率方程：rate = k[A]²[B]。
Using Expt 1: k = rate / ([A]²[B]) = 2.0 × 10⁻⁴ / (0.10² × 0.10) = 2.0 × 10⁻⁴ / (0.001) = 0.20. Units: mol dm⁻³ s⁻¹ / (mol² dm⁻⁶ × mol dm⁻³) = mol⁻² dm⁶ s⁻¹. So k = 0.20 mol⁻² dm⁶ s⁻¹.
使用实验1：k = rate / ([A]²[B]) = 2.0 × 10⁻⁴ / (0.10² × 0.10) = 2.0 × 10⁻⁴ / (0.001) = 0.20。单位：mol dm⁻³ s⁻¹ / (mol² dm⁻⁶ × mol dm⁻³) = mol⁻² dm⁶ s⁻¹。因此 k = 0.20 mol⁻² dm⁶ s⁻¹。
7. Transition Metal Complex – Colour and Isomerism | 过渡金属配合物 – 颜色与异构
Explain why an aqueous solution of copper(II) sulfate appears blue, while the addition of excess ammonia produces a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺. Also, state the type of isomerism possible for this complex and draw the two isomers.
解释为什么硫酸铜(II)水溶液呈蓝色，而加入过量氨水后产生深蓝色的 [Cu(NH₃)₄(H₂O)₂]²⁺ 溶液。并说明该配合物可能的异构类型，画出两种异构体。
In [Cu(H₂O)₆]²⁺, the central Cu²⁺ ion has a 3d⁹ configuration. The d orbitals are split by the octahedral ligand field. Visible light promotes an electron from the lower energy d orbitals to the higher energy ones, absorbing orange-red light and transmitting blue. When excess NH₃ replaces four water ligands, a stronger ligand field is created, increasing the d–d splitting and shifting the absorption, resulting in a deeper blue colour.
在 [Cu(H₂O)₆]²⁺ 中，中心 Cu²⁺ 离子具有 3d⁹ 构型。d 轨道在八面体配位场中分裂。可见光将电子从低能 d 轨道激发到高能 d 轨道，吸收橙红光而透过蓝色。当过量 NH₃ 取代四个水配体后，形成更强的配位场，增大 d–d 分裂，吸收波长改变，产生更深的蓝色。
The complex [Cu(NH₃)₄(H₂O)₂]²⁺ can exhibit geometric (cis-trans) isomerism. The cis isomer has the two water ligands adjacent (90° apart), while the trans isomer has them opposite (180° apart). These isomers have different physical properties and sometimes different colours.
配合物 [Cu(NH₃)₄(H₂O)₂]²⁺ 可表现出几何 (顺反) 异构。顺式异构体中两个水配体相邻 (呈 90°)，反式异构体中它们相对 (呈 180°)。这些异构体具有不同的物理性质，有时颜色也不同。
8. Acid-Base Titration and Buffer pH Calculation | 酸碱滴定与缓冲液 pH 计算
25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³) is titrated with 0.100 mol dm⁻³ NaOH. Calculate the pH after adding 12.5 cm³ of NaOH, and identify a suitable indicator for the full titration.
用 0.100 mol dm⁻³ NaOH 滴定 25.0 cm³ 的 0.100 mol dm⁻³ 乙酸 (Ka = 1.74 × 10⁻⁵ mol dm⁻³)。计算加入 12.5 cm³ NaOH 后的 pH，并为全滴定选择合适的指示剂。
After adding 12.5 cm³ NaOH, exactly half of the acid has been neutralised, leaving a buffer solution containing equal concentrations of CH₃COOH and CH₃COO⁻. Using the Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]) = pKa + log(1) = pKa. pKa = -log(1.74 × 10⁻⁵) ≈ 4.76. Therefore pH = 4.76.
加入 12.5 cm³ NaOH 后，恰好一半的酸被中和，形成含有等浓度 CH₃COOH 和 CH₃COO⁻ 的缓冲溶液。使用 Henderson-Hasselbalch 方程：pH = pKa + log([盐]/[酸]) = pKa + log(1) = pKa。pKa = -log(1.74 × 10⁻⁵) ≈ 4.76。因此 pH = 4.76。
The equivalence point lies above pH 7 due to the formation of the weak base CH₃COO⁻; the pH jump occurs around 8–10. A suitable indicator is phenolphthalein (pH range 8.3–10.0) because its colour change falls within the steep part of the titration curve.
等当点在 pH 7 以上，因为生成了弱碱 CH₃COO⁻；pH 突跃范围大约在 8–10。合适的指示剂是酚酞 (pH 范围 8.3–10.0)，因为其变色范围落在滴定曲线的陡峭部分。
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