A-Level Chemistry: Calculation Drills & Exercises | A-Level 化学：计算题专项训练

📚 A-Level Chemistry: Calculation Drills & Exercises | A-Level 化学：计算题专项训练

Calculation is a core skill in A-Level Chemistry. Mastering numerical problems helps students apply concepts from stoichiometry, energetics, kinetics, and equilibrium. This article provides a structured set of calculation drills, covering essential question types with worked examples and key formulas. Use these exercises to strengthen your problem-solving ability and boost exam confidence.

计算是 A-Level 化学的核心技能。掌握数值问题有助于学生将化学计量学、能量学、动力学和平衡等概念应用于实际。本文提供了一套结构化的计算训练，涵盖基本题型、示例演算和关键公式。通过这些练习，你可以提升解题能力，增强考试信心。

1. Mole Calculations | 摩尔计算

The mole is the central unit in quantitative chemistry. The number of moles n is found by dividing the mass of a substance (in grams) by its molar mass M (g mol⁻¹). Always ensure the mass used is the pure substance mass, and check that units are consistent.

摩尔是定量化学的核心单位。物质的量 n 可通过物质的质量（克）除以其摩尔质量 M（g mol⁻¹）求得。务必确保所用质量为纯物质质量，并检查单位是否一致。

n = m ÷ M

Worked example: Calculate the number of moles in 8.0 g of sodium hydroxide, NaOH (M = 40.0 g mol⁻¹).

示例：计算 8.0 g 氢氧化钠 NaOH 的物质的量（M = 40.0 g mol⁻¹）。

Solution: n = 8.0 ÷ 40.0 = 0.20 mol. For atoms or molecules, you may also relate moles to the Avogadro constant (6.02 × 10²³ mol⁻¹) when particle counts are required.

解答：n = 8.0 ÷ 40.0 = 0.20 mol。对于原子或分子，当需要计算粒子数时，还可借助阿伏伽德罗常数（6.02 × 10²³ mol⁻¹）建立联系。

2. Concentration and Dilution | 浓度与稀释

Concentration is usually expressed in mol dm⁻³. The amount of solute n in a solution of volume V (dm³) is given by n = c × V. When diluting a stock solution, the number of moles stays constant, leading to the dilution formula c₁V₁ = c₂V₂.

浓度通常以 mol dm⁻³ 表示。体积为 V（dm³）的溶液中溶质的物质的量 n 为 n = c × V。稀释储备液时，物质的量保持不变，由此得到稀释公式 c₁V₁ = c₂V₂。

c₁V₁ = c₂V₂

Example: You have 250 cm³ of 2.0 mol dm⁻³ HCl. What volume of water must be added to obtain a 0.50 mol dm⁻³ solution? First, convert 250 cm³ to 0.250 dm³. Then (2.0 × 0.250) = (0.50 × V₂). Solving gives V₂ = 1.0 dm³, so add (1.0 − 0.250) = 0.75 dm³ of water.

示例：现有 250 cm³ 浓度为 2.0 mol dm⁻³ 的 HCl，需加入多少水才能配成 0.50 mol dm⁻³ 的溶液？先将 250 cm³ 转换为 0.250 dm³。然后 (2.0 × 0.250) = (0.50 × V₂)。解得 V₂ = 1.0 dm³，因此需加入水 1.0 − 0.250 = 0.75 dm³。

3. Molar Volume of Gases | 气体摩尔体积

At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³. At standard temperature and pressure (STP, 0 °C and 1 atm), the molar volume is 22.7 dm³. Always check which conditions are specified in the exam question.

在室温和常压（RTP，20 °C、1 atm）下，1 mol 任何气体的体积为 24 dm³。在标准温度和压力（STP，0 °C、1 atm）下，摩尔体积为 22.7 dm³。解题时务必看清题目给定的是哪种条件。

Formula: volume of gas V = n × molar volume. To calculate moles from a gas volume, use n = V ÷ molar volume.

公式：气体体积 V = n × 摩尔体积。由气体体积求物质的量，则用 n = V ÷ 摩尔体积。

Drill: What is the mass of 1.20 dm³ of carbon dioxide at RTP? (Mᵣ of CO₂ = 44.0) Moles = 1.20 ÷ 24 = 0.0500 mol; mass = 0.0500 × 44.0 = 2.20 g. Practice similar calculations with O₂, H₂, and NH₃.

练习：在 RTP 下，1.20 dm³ 二氧化碳的质量是多少？（CO₂ 相对分子质量 = 44.0）物质的量 = 1.20 ÷ 24 = 0.0500 mol；质量 = 0.0500 × 44.0 = 2.20 g。可用 O₂、H₂ 和 NH₃ 进行类似计算。

4. Titration and Back Titration | 滴定与反滴定

Titration calculations link the concentration of an unknown solution to a standard solution via the reaction stoichiometry. Always write a balanced equation first. At the end point, n(unknown) is found from the titre volume and concentration of the known solution.

滴定计算通过反应计量关系将未知溶液的浓度与标准溶液联系起来。务必先写出配平的化学方程式。在终点处，可根据标准溶液的体积和浓度求得未知物的物质的量。

Titration of 25.0 cm³ NaOH with 0.100 mol dm⁻³ HCl	Trial 1	Trial 2	Trial 3
Final burette reading / cm³	24.10	47.05	23.55
Initial reading / cm³	0.00	24.10	0.00
Titre / cm³	24.10	22.95 (reject)	23.55

From concordant titres (Trial 1 and 3), mean volume = 23.83 cm³ = 0.02383 dm³. Moles HCl = 0.100 × 0.02383 = 0.002383 mol. HCl + NaOH → NaCl + H₂O, so NaOH moles identical. Concentration NaOH = 0.002383 ÷ 0.0250 = 0.0953 mol dm⁻³.

从吻合的滴定值（实验1和3）得平均体积 = 23.83 cm³ = 0.02383 dm³。HCl 的物质的量 = 0.100 × 0.02383 = 0.002383 mol。反应 HCl + NaOH → NaCl + H₂O，因此 NaOH 的物质的量相同。NaOH 浓度 = 0.002383 ÷ 0.0250 = 0.0953 mol dm⁻³。

Back titration is used when the substance is insoluble or volatile. An excess of reagent is added, and the remaining excess is titrated. Calculate total moles added, subtract moles reacted with the substance, and use the difference to find the quantity of the unknown.

反滴定适用于不溶或易挥发的物质。加入过量试剂，再滴定剩余量。计算加入的总物质的量，减去与物质反应的量，差值即用于求算未知物含量。

5. Percentage Yield and Atom Economy | 产率与原子经济性

The percentage yield compares the actual yield of a product to the theoretical yield. It reflects practical losses. Atom economy measures how efficiently atoms in the reactants are incorporated into the desired product; it is calculated from the balanced equation.

产率将实际产量与理论产量进行比较，反映了实际操作中的损失。原子经济性衡量反应物中的原子有多少并入目标产物，它由配平的化学方程式计算得出。

% Yield = (actual yield ÷ theoretical yield) × 100

% 产率 = (实际产量 ÷ 理论产量) × 100

% Atom economy = (Mᵣ desired product ÷ sum of Mᵣ of all reactants) × 100

% 原子经济性 = (目标产物 Mᵣ ÷ 所有反应物 Mᵣ 之和) × 100

Example: In the synthesis of aspirin, 5.0 g of salicylic acid (Mᵣ = 138) produced 4.8 g of aspirin (Mᵣ = 180). Theoretical moles of aspirin = 5.0 ÷ 138 = 0.0362 mol, theoretical mass = 0.0362 × 180 = 6.52 g. % yield = (4.8 ÷ 6.52) × 100 = 73.6%. Always assess whether the atom economy is high (typical for addition reactions) or low (substitution/elimination).

示例：在阿司匹林合成中，5.0 g 水杨酸（Mᵣ = 138）制得 4.8 g 阿司匹林（Mᵣ = 180）。理论阿司匹林物质的量 = 5.0 ÷ 138 = 0.0362 mol，理论质量 = 0.0362 × 180 = 6.52 g。产率 = (4.8 ÷ 6.52) × 100 = 73.6%。注意评估原子经济性高低：加成反应通常原子经济性高，取代/消除反应则较低。

6. Enthalpy Changes and Hess’s Law | 焓变与盖斯定律

Hess’s Law states that the total enthalpy change for a reaction is independent of the pathway taken. It allows enthalpy changes of formation and combustion to be combined to find unknown ΔH values. Use ΔH = ΣΔHf°(products) − ΣΔHf°(reactants), or construct energy cycles.

盖斯定律指出，反应的总焓变与路径无关。它使得可以利用生成焓或燃烧焓数据求算未知的 ΔH。可使用公式 ΔH = ΣΔHf°(产物) − ΣΔHf°(反应物)，或者绘制能量循环图。

Drill: Calculate ΔH for the reaction: 2CO(g) + O₂(g) → 2CO₂(g). Given: ΔHf°(CO₂) = −394 kJ mol⁻¹; ΔHf°(CO) = −111 kJ mol⁻¹. ΔH = [2(−394)] − [2(−111) + 0] = −788 + 222 = −566 kJ mol⁻¹. Practice with combustion data and bond enthalpies; remember bonds broken (+) and bonds formed (−).

练习：计算反应 2CO(g) + O₂(g) → 2CO₂(g) 的 ΔH。已知 ΔHf°(CO₂) = −394 kJ mol⁻¹；ΔHf°(CO) = −111 kJ mol⁻¹。ΔH = [2(−394)] − [2(−111) + 0] = −788 + 222 = −566 kJ mol⁻¹。用燃烧热数据处理类似问题，也可练习键焓计算，记住断键吸热（+）和成键放热（−）。

7. Equilibrium Constants Kc and Kp | 平衡常数 Kc 与 Kp

Kc and Kp expressions are built from the balanced equation. For a generic reaction aA + bB &rightleftharpoons; cC + dD, Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ). Only gases and aqueous species appear; pure solids/liquids are omitted. Units of Kc depend on the sum of stoichiometric coefficients.

Kc 与 Kp 表达式由配平的方程式构建。对于一般反应 aA + bB &rightleftharpoons; cC + dD，Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)。只包括气体与溶液物种，纯固体和液体不出现在表达式中。Kc 的单位取决于化学计量数之和。

To find Kc from experimental data, set up an Initial/Change/Equilibrium (ICE) table. Use concentration in mol dm⁻³ and solve for unknown changes. Kp uses partial pressures: pA = mole fraction × total pressure.

要从实验数据求 Kc，可建立初始/变化/平衡（ICE）表格。浓度使用 mol dm⁻³，求解未知变化量。Kp 使用分压：分压 = 摩尔分数 × 总压。

Example: H₂(g) + I₂(g) &rightleftharpoons; 2HI(g) starts with 1.0 mol H₂ and 1.0 mol I₂ in 1 dm³ vessel. At equilibrium, 0.20 mol H₂ remains. Then reacted = 0.80 mol, producing 1.60 mol HI. Equilib conc: [H₂]=0.20, [I₂]=0.20, [HI]=1.60. Kc = (1.60)² ÷ (0.20 × 0.20) = 64. Calculate units: (mol dm⁻³)² ÷ (mol dm⁻³)² = no units. Always state units carefully.

示例：H₂(g) + I₂(g) &rightleftharpoons; 2HI(g) 初始在 1 dm³ 容器中加入 1.0 mol H₂ 和 1.0 mol I₂。平衡时剩余 0.20 mol H₂。则反应了 0.80 mol，生成 1.60 mol HI。平衡浓度：[H₂]=0.20, [I₂]=0.20, [HI]=1.60。Kc = (1.60)² ÷ (0.20 × 0.20) = 64。单位：(mol dm⁻³)² ÷ (mol dm⁻³)² = 无单位。请务必小心注明单位。

8. Rate Equations and Arrhenius | 速率方程与阿伦尼乌斯

The rate equation shows how rate depends on concentrations: rate = k[A]ᵐ[B]ⁿ, where m and n are orders determined experimentally. The rate constant k changes with temperature according to the Arrhenius equation.

速率方程表明反应速率与浓度的关系：速率 = k[A]ᵐ[B]ⁿ，其中 m、n 是通过实验测定的反应级数。速率常数 k 随温度变化，遵循阿伦尼乌斯方程。

k = A e^( −Ea / RT ) or ln k = ln A − Ea / (RT)

k = A e^( −Ea / RT ) 或 ln k = ln A − Ea / (RT)

To find activation energy Ea, plot ln k against 1/T. The slope = −Ea/R. Use R = 8.31 J K⁻¹ mol⁻¹. For two temperatures, ln (k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁). Practice constructing rate equations from concentration-time data and initial rate tables.

要求活化能 Ea，可绘制 ln k 对 1/T 的图，斜率 = −Ea/R。使用 R = 8.31 J K⁻¹ mol⁻¹。对于两个温度，可使用 ln (k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁)。多加练习从浓度–时间数据和初始速率表格推导速率方程。

9. Electrochemistry and Cell Potentials | 电化学与电池电势

The standard cell potential E°cell = E°(right half-cell) − E°(left half-cell). A positive E°cell indicates a thermodynamically feasible reaction. Use the relationship ΔG° = −nFE°cell, where F = 96 500 C mol⁻¹.

标准电池电动势 E°cell = E°(右半电池) − E°(左半电池)。E°cell 为正值表示反应热力学可行。使用关系式 ΔG° = −nFE°cell，其中 F = 96 500 C mol⁻¹。

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A-Level Chemistry: Calculation Drills & Exercises | A-Level 化学：计算题专项训练