Mastering Calculation Questions in A-Level Chemistry Unit 4 (Jan 2021) | A-Level化学单元4计算题型突破（2021年1月卷）

📚 Mastering Calculation Questions in A-Level Chemistry Unit 4 (Jan 2021) | A-Level化学单元4计算题型突破（2021年1月卷）

The January 2021 Unit 4 paper for A-Level Chemistry features a wide range of calculation questions that test students’ quantitative skills across physical and inorganic chemistry. From thermodynamics and equilibrium to pH and electrochemistry, mastering these calculations is essential for achieving a top grade. This article will guide you through the key calculation types, with step-by-step examples and tips.

2021年1月A-Level化学第四单元试卷涵盖了物理化学和无机化学中大量的计算题型，考验学生的定量分析能力。从热力学、平衡到pH和电化学，掌握这些计算是取得高分的关键。本文将带你逐一剖析核心计算类型，结合分步例题与解题技巧。

1. Hess’s Law and Enthalpy Cycles | 赫斯定律与焓变循环

Hess’s Law states that the overall enthalpy change for a reaction is independent of the route taken. In the Jan 2021 paper, students were required to construct enthalpy cycles using enthalpies of formation or combustion. For example, given standard enthalpies of formation for CO₂ (-394 kJ mol^-1) and H₂O (-286 kJ mol^-1), and the enthalpy of combustion of methane (-890 kJ mol^-1), calculate the enthalpy of formation of methane.

赫斯定律指出反应的总焓变与途径无关。在2021年1月试卷中，学生需要利用生成焓或燃烧焓构建焓变循环。例如，已知CO₂的标准生成焓为-394 kJ mol^-1，H₂O为-286 kJ mol^-1，甲烷的燃烧焓为-890 kJ mol^-1，求甲烷的生成焓。

Using the cycle ΔH_f(CH₄) + ΔH_c(CH₄) = ΣΔH_f(products) → ΔH_f(CH₄) = [(-394) + 2(-286)] – (-890) = -76 kJ mol^-1. Always check the sign: a negative value indicates an exothermic formation.

使用循环 ΔH_f(CH₄) + ΔH_c(CH₄) = ΣΔH_f(产物)，得出 ΔH_f(CH₄) = [(-394) + 2×(-286)] – (-890) = -76 kJ mol^-1。注意检查符号：负值表示生成放热。

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2. Gibbs Free Energy and Entropy | 吉布斯自由能与熵变

The Gibbs equation ΔG = ΔH – TΔS is central to predicting feasibility. In one Jan 2021 question, students calculated the temperature at which a reaction becomes feasible by setting ΔG = 0. Given ΔH = +178 kJ mol^-1 and ΔS = +160 J K^-1 mol^-1, the temperature is T = ΔH/ΔS = 178,000 J / 160 J K^-1 = 1112.5 K.

吉布斯方程 ΔG = ΔH – TΔS 是判断反应自发性的核心。在2021年1月的一道题中，学生需通过令 ΔG = 0 来计算反应恰好可行时的温度。已知 ΔH = +178 kJ mol^-1，ΔS = +160 J K^-1 mol^-1，则 T = ΔH/ΔS = 178000 J / 160 J K^-1 = 1112.5 K。

Remember to convert kJ to J when entropy is in J. Also, if ΔG is negative at a given T, the reaction is thermodynamically feasible.

记住当熵的单位是焦耳时必须将千焦转换为焦耳。若在给定温度下 ΔG 为负，则反应在热力学上可行。

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3. Equilibrium Constant K_c | 平衡常数K_c

K_c calculations involve determining equilibrium concentrations from initial moles and a change table (ICE table). In the Jan 2021 exam, a typical problem provided initial amounts of H₂ (1.0 mol) and I₂ (1.0 mol) in a 2.0 dm³ vessel, with 0.6 mol of HI at equilibrium. Using the ICE table, we find [H₂] = [I₂] = 0.35 mol dm^-3, [HI] = 0.3 mol dm^-3. Then K_c = [HI]²/([H₂][I₂]) = (0.3)²/(0.35 × 0.35) = 0.73 (no units).

K_c计算通常需要利用初始物质的量和变化量表格（ICE表）求解平衡浓度。2021年1月试卷中一道典型题：在2.0 dm³容器中加入1.0 mol H₂和1.0 mol I₂，平衡时生成0.6 mol HI。通过ICE表可得[H₂] = [I₂] = 0.35 mol dm^-3，[HI] = 0.3 mol dm^-3。K_c = [HI]²/([H₂][I₂]) = 0.09/(0.1225) = 0.73（无单位）。

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4. Equilibrium Constant K_p | 平衡常数K_p

For gaseous equilibria, K_p uses partial pressures. In a Jan 2021 question, the reaction N₂+3H₂ ⇌ 2NH₃ had total pressure 200 atm and mole fraction of NH₃ = 0.40. Calculate mole fractions of N₂ (0.15) and H₂ (0.45) from stoichiometry. Partial pressures: p(NH₃) = 0.40 × 200 = 80 atm, p(N₂) = 0.15 × 200 = 30 atm, p(H₂) = 0.45 × 200 = 90 atm. Then K_p = (80)² / (30 × 90³) = 6400 / (30 × 729,000) = 2.93 × 10^-4 atm^-2.

对于气体平衡，K_p使用分压计算。2021年1月题中反应N₂+3H₂ ⇌ 2NH₃总压200 atm，NH₃的摩尔分数为0.40。由化学计量比求得N₂分数为0.15，H₂为0.45。分压：p(NH₃)=80 atm，p(N₂)=30 atm，p(H₂)=90 atm。K_p = (80)²/(30×90³) = 2.93×10^-4 atm^-2。

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5. pH of Strong and Weak Acids | 强酸和弱酸的pH计算

In the Jan 2021 paper, a question asked for the pH of 0.050 mol dm^-3 HCl (strong acid): pH = -log(0.050) = 1.30. For a weak acid like CH₃COOH (K_a = 1.8 × 10^-5), [H⁺] = √(K_a × C) = √(1.8×10^-5 × 0.050) = 9.49 × 10^-4 mol dm^-3, giving pH = -log(9.49×10^-4) = 3.02.

2021年1月题中要求计算0.050 mol dm^-3盐酸的pH：pH = -log(0.050) = 1.30。对于弱酸如醋酸（K_a = 1.8×10^-5），[H⁺] = √(K_a×C) = √(1.8×10^-5×0.050) = 9.49×10^-4 mol dm^-3，pH = 3.02。

Always check whether the approximation [HA] ≈ C is valid (C/K_a > 100). If not, solve the quadratic equation.

始终检查近似条件[HA]≈C是否成立（C/K_a > 100）。若不成立，则需解二次方程。

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6. Buffer Solution pH | 缓冲溶液pH值

A buffer question from the 2021 paper involved mixing CH₃COOH and CH₃COONa. Using the Henderson-Hasselbalch equation: pH = pK_a + log([salt]/[acid]). For 0.20 mol acetic acid and 0.15 mol sodium acetate in 500 cm³, [acid] = 0.40 mol dm^-3, [salt] = 0.30 mol dm^-3. pK_a = -log(1.8×10^-5) = 4.74. pH = 4.74 + log(0.30/0.40) = 4.62.

2021年试卷中的缓冲溶液题涉及醋酸与醋酸钠混合。使用汉德森公式：pH = pK_a + log([盐]/[酸])。将0.20 mol乙酸和0.15 mol乙酸钠溶于500 cm³，[酸]=0.40 mol dm^-3，[盐]=0.30 mol dm^-3。pK_a=4.74，pH = 4.74 + log(0.30/0.40) = 4.62。

When adding small amounts of H⁺ or OH^–, the ratio changes slightly; your exam may ask to calculate the new pH after such addition.

当加入少量H⁺或OH^–时，盐/酸比例会发生微小变化；试卷可能要求计算加入后的新pH值。

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7. Electrode Potentials and Cell EMF | 电极电势与电池电动势

The standard cell potential E°_cell = E°_right – E°_left. In a Jan 2021 question, a cell was set up with Zn/Zn²⁺ (E° = -0.76 V) and Cu/Cu²⁺ (E° = +0.34 V). The spontaneous direction has Cu as reduction (right) and Zn as oxidation (left). E°_cell = 0.34 – (-0.76) = 1.10 V. A negative E°_cell means the reaction is not feasible in that direction.

标准电池电动势 E°_cell = E°_右 – E°_左。2021年1月题中，用Zn/Zn²⁺（-0.76 V）和Cu/Cu²⁺（+0.34 V）组成原电池。自发方向以铜为还原极（右），锌为氧化极（左）。E°_cell = 0.34 – (-0.76) = 1.10 V。负的E°_cell表明该方向反应不可行。

If the cell diagram is given, always identify the half-cell with the higher reduction potential as the positive electrode.

若给出电池图示，始终将更高还原电位的半电池定为正极。

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

The Arrhenius equation ln k = ln A – E_a/RT often appears in the Unit 4 paper. In a Jan 2021 example, students calculated E_a from a graph of ln k vs 1/T. Given slope = -E_a/R = -1.2 × 10⁴ K, so E_a = 1.2 × 10⁴ K × 8.31 J K^-1 mol^-1 = 99.7 kJ mol^-1. Always draw the tangent or use data points to find slope.

阿伦尼乌斯方程 ln k = ln A – E_a/RT 常出现在第四单元试卷中。2021年1月例题要求学生从ln k对1/T的图中求活化能。已知斜率 = -E_a/R = -1.2×10⁴ K，则 E_a = 1.2×10⁴ K × 8.31 J K^-1 mol^-1 = 99.7 kJ mol^-1。务必通过切线或数据点求斜率。

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9. Redox Titration Calculations | 氧化还原滴定计算

Redox titrations, such as manganate(VII) with iron(II), are common. In one Jan 2021 question, 25.0 cm³ of Fe²⁺ solution required 18.5 cm³ of 0.020 mol dm^-3 KMnO₄ to reach the end point. Reaction: MnO₄^– + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Moles MnO₄^– = 0.0185 × 0.020 = 3.70 × 10^-4 mol. Moles Fe²⁺ = 5 × 3.70×10^-4 = 1.85×10^-3 mol. Concentration of Fe²⁺ = 1.85×10^-3 / 0.025 = 0.074 mol dm^-3.

氧化还原滴定如高锰酸根与亚铁离子反应非常常见。2021年1月题中，25.0 cm³ Fe²⁺溶液消耗18.5 cm³ 0.020 mol dm^-3 KMnO₄达到终点。反应：MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。MnO₄^–物质的量 = 0.0185×0.020 = 3.70×10^-4 mol，Fe²⁺物质的量 = 5×3.70×10^-4 = 1.85×10^-3 mol，Fe²⁺浓度 = 1.85×10^-3/0.025 = 0.074 mol dm^-3。

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10. Ideal Gas Equation and Molar Volume | 理想气体方程与摩尔体积

The ideal gas equation pV = nRT is frequently tested. In the Jan 2021 paper, students calculated the volume of CO₂ produced at 298 K and 100 kPa when 2.0 g CaCO₃ decomposes. Moles CaCO₃ = 2.0 / 100.1 = 0.0200 mol. Moles CO₂ = 0.0200 mol. V = nRT/p = (0.0200 × 8.31 × 298) / 100,000 = 4.95 × 10^-4 m³ = 0.495 dm³ (or 495 cm³). Remember to use SI units: m³, Pa, K.

理想气体方程 pV = nRT 是高频考点。2021年1月卷中，学生需计算2.0 g CaCO₃分解后在298 K、100 kPa下产生的CO₂体积。CaCO₃物质的量 = 2.0/100.1 = 0.0200 mol，CO₂物质的量相同。V = nRT/p = (0.0200×8.31×298)/100000 = 4.95×10^-4 m³ = 0.495 dm³（或495 cm³）。注意统一国际单位：m³，Pa，K。

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11. Faraday’s Laws of Electrolysis | 法拉第电解定律

Electrolysis calculations appear in Unit 4. In a Jan 2021 question, a current of 0.50 A was passed through AgNO₃(aq) for 30 minutes. Calculate the mass of silver deposited. Total charge Q = I × t = 0.50 × (30 × 60) = 900 C. Moles of electrons = Q/F = 900 / 96,500 = 9.33 × 10^-3 mol. Ag⁺ + e^– → Ag, so moles Ag = 9.33×10^-3 mol. Mass = 9.33×10^-3 × 107.9 = 1.01 g.

电解计算在单元四中常有涉及。2021年1月题中，0.50 A电流通过AgNO₃溶液30分钟，求析出银的质量。总电量Q = I×t = 0.50×(30×60) = 900 C。电子物质的量 = Q/F = 900/96500 = 9.33×10^-3 mol。Ag⁺ + e^– → Ag，故银的物质的量为9.33×10^-3 mol，质量 = 9.33×10^-3×107.9 = 1.01 g。

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12. Determining Order from Initial Rates | 从初始速率确定反应级数

A classic calculation in kinetics: given initial rate data, determine the rate equation. For example, in a Jan 2021 problem, doubling [A] quadrupled the initial rate (order 2 with respect to A), while doubling [B] doubled the rate (order 1). Thus rate = k[A]²[B]. Calculate k using any experiment: k = rate / ([A]²[B]). If rate = 4.0 × 10^-3 mol dm^-3 s^-1, [A] = 0.10, [B] = 0.20, then k = 4.0×10^-3 / (0.10² × 0.20) = 2.0 dm⁶ mol^-2 s^-1.

动力学经典计算：根据初始速率数据确定速率方程。2021年1月题中，[A]加倍使初速率变为4倍（对A为2级），[B]加倍使速率加倍（1级）。因此速率方程 rate = k[A]²[B]。利用任一实验求k：k = 速率/([A]²[B])。若速率=4.0×10^-3 mol dm^-3 s^-1，[A]=0.10，[B]=0.20，则k = 4.0×10^-3/(0.10²×0.20) = 2.0 dm⁶ mol^-2 s^-1。

Experiment	[A] (mol dm-3)	[B] (mol dm-3)	Initial rate (mol dm-3 s-1)
1	0.10	0.20	4.0×10-3
2	0.20	0.20	1.6×10-2
3	0.10	0.40	8.0×10-3

Published by TutorHao | Chemistry Revision Series | aleveler.com

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