A-Level WJEC Chemistry: Worked Examples Explained in Detail | A-Level WJEC 化学：典型例题详解

📚 A-Level WJEC Chemistry: Worked Examples Explained in Detail | A-Level WJEC 化学：典型例题详解

This article presents a collection of carefully selected worked examples covering key quantitative topics from the WJEC A-Level Chemistry specification. Each example is broken down into clear, step-by-step solutions with paired English and Chinese explanations, helping you master the essential calculation skills required for the exam.

本文精选了 WJEC A-Level 化学大纲中核心计算专题的典型例题，每个例题均以清晰的步骤逐步讲解，并配以英中双语说明，助你熟练掌握考试必备的计算技能。

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1. Moles and Empirical Formula | 摩尔与经验式计算

Example: A hydrocarbon is burned completely in excess oxygen. 1.20 g of the compound gives 3.52 g of CO₂ and 1.44 g of H₂O. Determine its empirical formula. If the molar mass of the compound is 84 g mol⁻¹, deduce its molecular formula.

例题：某碳氢化合物在过量氧气中完全燃烧，1.20 g 的该化合物生成 3.52 g CO₂ 和 1.44 g H₂O。确定其经验式；若该化合物的摩尔质量为 84 g mol⁻¹，推导其分子式。

Step 1: Calculate the amount (mol) of CO₂ and H₂O produced. Mr(CO₂) = 44.0, Mr(H₂O) = 18.0. Moles CO₂ = 3.52 g / 44.0 g mol⁻¹ = 0.0800 mol; moles H₂O = 1.44 g / 18.0 g mol⁻¹ = 0.0800 mol.

步骤1：计算生成的 CO₂ 和 H₂O 的物质的量。Mr(CO₂) = 44.0，Mr(H₂O) = 18.0。n(CO₂) = 3.52 g / 44.0 g mol⁻¹ = 0.0800 mol；n(H₂O) = 1.44 g / 18.0 g mol⁻¹ = 0.0800 mol。

Step 2: Convert moles of combustion products to moles of C and H. 1 mol CO₂ contains 1 mol C, so moles of C = 0.0800 mol. 1 mol H₂O contains 2 mol H, so moles of H = 2 × 0.0800 = 0.160 mol.

步骤2：将燃烧产物的物质的量换算为 C 和 H 的物质的量。1 mol CO₂ 含 1 mol C，故 n(C) = 0.0800 mol。1 mol H₂O 含 2 mol H，故 n(H) = 2 × 0.0800 = 0.160 mol。

Step 3: Find the simplest mole ratio. Divide by the smaller value (0.0800): C : H = 0.0800/0.0800 : 0.160/0.0800 = 1 : 2. The empirical formula is CH₂.

步骤3：求最简物质的量整数比。除以较小值 (0.0800)：C : H = 0.0800/0.0800 : 0.160/0.0800 = 1 : 2。经验式为 CH₂。

Step 4: Use the molar mass to find the molecular formula. Empirical formula mass of CH₂ = 14.0 g mol⁻¹. Given molar mass = 84 g mol⁻¹. Factor = 84 / 14.0 = 6. Therefore the molecular formula is C₆H₁₂.

步骤4：利用摩尔质量求分子式。CH₂ 式的质量为 14.0 g mol⁻¹。已知摩尔质量 = 84 g mol⁻¹。倍数因子 = 84 / 14.0 = 6。因此分子式为 C₆H₁₂。

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2. Titration Calculations | 滴定计算

Example: In a titration, 25.0 cm³ of sulphuric acid (H₂SO₄) is neutralised by 23.50 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution. Calculate the concentration of the acid.

例题：在滴定中，25.0 cm³ 的硫酸 (H₂SO₄) 被 23.50 cm³ 的 0.100 mol dm⁻³ 氢氧化钠溶液中和。计算该酸的浓度。

Step 1: Write the balanced equation. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The stoichiometry shows 1 mol H₂SO₄ reacts with 2 mol NaOH.

步骤1：写出配平方程式。H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。化学计量比表明 1 mol H₂SO₄ 与 2 mol NaOH 反应。

Step 2: Calculate moles of NaOH used. Moles = concentration × volume in dm³. Volume = 23.50 cm³ = 0.02350 dm³. n(NaOH) = 0.100 mol dm⁻³ × 0.02350 dm³ = 0.00235 mol.

步骤2：计算所用 NaOH 的物质的量。n = c × V(dm³)。体积 = 23.50 cm³ = 0.02350 dm³。n(NaOH) = 0.100 mol dm⁻³ × 0.02350 dm³ = 0.00235 mol。

Step 3: Find moles of H₂SO₄. From the equation, moles H₂SO₄ = ½ × moles NaOH = 0.00235 / 2 = 0.001175 mol.

步骤3：求 H₂SO₄ 的物质的量。根据方程式，n(H₂SO₄) = ½ × n(NaOH) = 0.00235 / 2 = 0.001175 mol。

Step 4: Calculate the concentration of the acid. Volume of acid = 25.0 cm³ = 0.0250 dm³. Concentration = moles / volume = 0.001175 mol / 0.0250 dm³ = 0.0470 mol dm⁻³.

步骤4：计算酸的浓度。酸的体积 = 25.0 cm³ = 0.0250 dm³。浓度 = n / V = 0.001175 mol / 0.0250 dm³ = 0.0470 mol dm⁻³。

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3. Enthalpy Changes using Hess’s Law | 利用赫斯定律计算焓变

Example: Calculate the standard enthalpy change for the reaction C₂H₄(g) + H₂O(g) → C₂H₅OH(l) using the following standard enthalpies of formation (ΔHf°): ΔHf°[C₂H₄(g)] = +52.3 kJ mol⁻¹, ΔHf°[H₂O(g)] = -241.8 kJ mol⁻¹, ΔHf°[C₂H₅OH(l)] = -277.7 kJ mol⁻¹.

例题：利用标准生成焓 (ΔHf°) 计算反应 C₂H₄(g) + H₂O(g) → C₂H₅OH(l) 的标准焓变。数据：ΔHf°[C₂H₄(g)] = +52.3 kJ mol⁻¹，ΔHf°[H₂O(g)] = -241.8 kJ mol⁻¹，ΔHf°[C₂H₅OH(l)] = -277.7 kJ mol⁻¹。

Step 1: Recall Hess’s law in terms of ΔHf°. ΔH° = Σ ΔHf°(products) – Σ ΔHf°(reactants).

步骤1：回顾用 ΔHf° 表示的赫斯定律表达式。ΔH° = Σ ΔHf°(生成物) – Σ ΔHf°(反应物)。

Step 2: Insert the values. Products: only C₂H₅OH contributes -277.7 kJ. Reactants: C₂H₄ (+52.3) + H₂O (-241.8) = -189.5 kJ. ΔH° = (-277.7) – (-189.5) = -88.2 kJ mol⁻¹.

步骤2：代入数值。生成物：仅 C₂H₅OH 贡献 -277.7 kJ。反应物：C₂H₄ (+52.3) + H₂O (-241.8) = -189.5 kJ。ΔH° = (-277.7) – (-189.5) = -88.2 kJ mol⁻¹。

Step 3: Interpret the sign. The negative value means the reaction is exothermic. 88.2 kJ of energy is released per mole of C₂H₄ reacted.

步骤3：解释符号。负值表明反应放热。每反应 1 摩尔 C₂H₄ 释放 88.2 kJ 能量。

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4. Bond Enthalpy Calculations | 键焓计算

Example: Estimate the enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g). Average bond enthalpies: H-H = 436 kJ mol⁻¹, Cl-Cl = 243 kJ mol⁻¹, H-Cl = 432 kJ mol⁻¹.

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

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