A-Level Chemistry: Unit 5 Calculation Questions from January 2020 Mark Scheme | A-Level 化学：2020年1月单元5评分方案中的计算题型

📚 A-Level Chemistry: Unit 5 Calculation Questions from January 2020 Mark Scheme | A-Level 化学：2020年1月单元5评分方案中的计算题型

Success in Unit 5 of the International A-Level Chemistry exam often hinges on mastering the calculation problems. The January 2020 mark scheme reveals the precise steps and reasoning that examiners expect, from handling thermodynamic data to manipulating equilibrium constants, pH expressions, and electrode potentials. This article unpacks the key calculation types featured in that paper, providing worked-through strategies and highlighting the mark-worthy details. Each section pairs clear English explanations with equivalent guidance in Chinese, helping you build both conceptual understanding and exam technique.

国际 A-Level 化学单元 5 考试的成功往往取决于对计算题型的掌握。2020 年 1 月的评分方案揭示了考官期望的精确步骤和推理过程，从处理热力学数据到运用平衡常数、pH 表达以及电极电势等。本文剖析了该试卷涉及的主要计算类型，提供了完整的解题策略，并突出得分要点。每一节都将清晰的英文解释与对应的中文指导相结合，帮助你建立概念理解与应考技巧。

1. Using Standard Entropy and Enthalpy Data to Calculate Gibbs Free Energy | 使用标准熵和焓数据计算吉布斯自由能

One of the first hurdles in Unit 5 is combining standard entropy (S°) and enthalpy (ΔH°) values to find the standard Gibbs free energy change (ΔG°). The mark scheme insists on using the correct equation: ΔG° = ΔH° – TΔS°. A common pitfall is failing to convert the entropy term from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹, which leads to a unit mismatch with ΔH° in kJ mol⁻¹. Always divide the total entropy change by 1000 before subtracting from ΔH°.

单元 5 的第一个难点是将标准熵 (S°) 和焓 (ΔH°) 相结合计算标准吉布斯自由能变 (ΔG°)。评分方案强调必须使用正确的方程：ΔG° = ΔH° – TΔS°。常见的错误是未将熵项从 J K⁻¹ mol⁻¹ 转换为 kJ K⁻¹ mol⁻¹，导致与以 kJ mol⁻¹ 为单位的 ΔH° 不匹配。务必将总熵变除以 1000 之后再与 ΔH° 相减。

For a reaction like N₂(g) + 3H₂(g) → 2NH₃(g), you need to sum the standard entropies of products and reactants, then compute ΔS°ₛᵧₛ = ΣS°(products) – ΣS°(reactants). The mark scheme often awards a separate mark for the correct sign of ΔS°. In the Haber process, because four moles of gas become two, ΔS° is negative. Substituting T = 298 K, you would write:

对于像 N₂(g) + 3H₂(g) → 2NH₃(g) 这样的反应，需要分别加和产物与反应物的标准熵，然后计算 ΔS°ₛᵧₛ = ΣS°(产物) – ΣS°(反应物)。评分方案通常会对 ΔS° 的正确符号单独给分。在哈伯法中，由于四摩尔气体变成两摩尔，ΔS° 为负值。代入 T = 298 K 时，应写为：

ΔG° = ΔH° – (298 × ΔS°ₛᵧₛ / 1000)

Make sure your use of significant figures matches the given data; a final answer to three significant figures is typical.

确保有效数字的位数与所给数据相符；最终结果通常保留三位有效数字。

2. Determining the Spontaneity of a Reaction | 判断反应的自发性

A calculation of ΔG° is rarely complete without interpreting its sign. A negative ΔG° indicates a thermodynamically feasible reaction under standard conditions. However, the January 2020 paper often probes whether a reaction becomes feasible only above or below a certain temperature. The mark scheme rewards setting ΔG° = 0 to find the crossover temperature: T = ΔH° / ΔS° (with ΔS° in kJ K⁻¹ mol⁻¹).

ΔG° 的计算很少不需要解释其符号。ΔG° 为负表示在标准条件下反应在热力学上可行。但 2020 年 1 月的试卷常常探究反应是否只有在高于或低于某一温度时才变得可行。评分方案奖励将 ΔG° = 0 来求解转折温度的做法：T = ΔH° / ΔS°（ΔS° 以 kJ K⁻¹ mol⁻¹ 计）。

If ΔH° is negative and ΔS° is negative, the reaction is feasible at low temperatures. The mark scheme expects a clear statement linking the calculated T to the direction of feasibility. Additionally, remember that kinetic factors may prevent a thermodynamically spontaneous reaction from occurring; the examiners like to see this qualification written explicitly.

如果 ΔH° 为负且 ΔS° 为负，则反应在低温下可行。评分方案期待一个清晰的陈述，将计算得到的 T 与可行的温度范围联系起来。此外，要记住动力学因素可能阻止热力学上自发反应的发生；考官喜欢看到这一限定句被明确写出。

3. Calculating an Equilibrium Constant Kc from Initial Amounts | 从初始量计算平衡常数 Kc

A typical Question 1 or 2 in the paper provides initial moles of all species and the equilibrium moles of one component. You are asked to find Kc. The mark scheme follows a stepwise pattern: (1) calculate the change in moles for the known species, (2) use the stoichiometric ratio to deduce changes for all others, (3) determine equilibrium moles, (4) convert to equilibrium concentrations by dividing by the total volume (in dm³), and (5) substitute into the Kc expression.

试卷中典型的第 1 或第 2 题会给出所有物质的初始摩尔数以及一种组分的平衡摩尔数，要求计算 Kc。评分方案遵循逐步模式：(1) 计算已知物质的摩尔变化量，(2) 利用化学计量比推算其他物质的变化量，(3) 确定平衡摩尔数，(4) 除以总体积（dm³）得到平衡浓度，(5) 代入 Kc 表达式。

For example, in an esterification reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. If initially 0.50 mol of acid and 0.50 mol of alcohol are mixed, and at equilibrium 0.30 mol of ester is present, the change for ester and water is +0.30, and the change for acid and alcohol is –0.30. The equilibrium amounts are acid 0.20 mol, alcohol 0.20 mol, ester 0.30 mol, water 0.30 mol. With a volume of 0.50 dm³, concentrations are 0.40, 0.40, 0.60, 0.60 mol dm⁻³, giving Kc = (0.60×0.60) / (0.40×0.40) = 2.25. The mark scheme often insists on units for Kc unless Kc is dimensionless; here, units cancel, so Kc has no units – a point frequently examined.

例如，对于酯化反应：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。如果最初混合 0.50 mol 酸和 0.50 mol 醇，平衡时存在 0.30 mol 酯，则酯和水的改变量为 +0.30，酸和醇的改变量为 –0.30。平衡量为酸 0.20 mol、醇 0.20 mol、酯 0.30 mol、水 0.30 mol。若体积为 0.50 dm³，浓度分别为 0.40、0.40、0.60、0.60 mol dm⁻³，得到 Kc = (0.60×0.60)/(0.40×0.40) = 2.25。评分方案常要求写出 Kc 的单位，除非 Kc 无量纲；在此例中单位相消，因此 Kc 无单位——此为常考点。

4. Partial Pressure and Kp Calculations | 分压和 Kp 计算

When gaseous equilibria are involved, the mark scheme guides you through mole fraction and partial pressure. Partial pressure = mole fraction × total pressure. The mole fraction is the moles of a component divided by the total moles at equilibrium. After finding all partial pressures, substitute them into the Kp expression, which is analogous to Kc but with partial pressures.

当涉及气体平衡时，评分方案引导你通过摩尔分数和分压解题。分压 = 摩尔分数 × 总压。摩尔分数等于某组分的摩尔数除以平衡时总摩尔数。求出所有分压后，代入 Kp 表达式，它与 Kc 相似，但使用的是分压。

A typical scenario: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Starting with 1.0 mol PCl₅, at equilibrium the total moles are 1.2 mol, total pressure 200 kPa. If 0.8 mol PCl₅ remains (so 0.2 mol has dissociated), then PCl₃ = 0.2 mol, Cl₂ = 0.2 mol. Mole fractions: PCl₅ = 0.8/1.2 = 0.667, PCl₃ = Cl₂ = 0.2/1.2 = 0.167. Partial pressures: PCl₅ = 0.667×200 = 133.4 kPa, PCl₃ = 33.4 kPa, Cl₂ = 33.4 kPa. Kp = (P_(PCl₃) × P_(Cl₂)) / P_(PCl₅) = (33.4×33.4)/133.4 ≈ 8.36 kPa. Always quote the unit (kPa or atm) as required; the mark scheme explicitly penalises missing units.

典型情景：PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)。初始加入 1.0 mol PCl₅，平衡时总摩尔数为 1.2 mol，总压 200 kPa。若剩余 0.8 mol PCl₅（即解离了 0.2 mol），则 PCl₃ = 0.2 mol, Cl₂ = 0.2 mol。摩尔分数：PCl₅ = 0.8/1.2 ≈ 0.667, PCl₃ = Cl₂ = 0.2/1.2 ≈ 0.167。分压：PCl₅ = 0.667×200 = 133.4 kPa, PCl₃ = 33.4 kPa, Cl₂ = 33.4 kPa。Kp = (P_PCl₃ × P_Cl₂) / P_PCl₅ = (33.4×33.4)/133.4 ≈ 8.36 kPa。务必按题目要求注明单位（kPa 或 atm）；评分方案明确扣罚遗漏单位的情况。

5. Calculating pH of a Buffer Solution | 计算缓冲溶液的 pH

Buffer calculations are a staple of Unit 5. Using the Henderson–Hasselbalch equation: pH = pKₐ + log₁₀([A⁻]/[HA]). The mark scheme rewards converting the given Kₐ to pKₐ correctly, then plugging in the concentrations of conjugate base and weak acid. Be careful to use equilibrium concentrations, but for buffers made by mixing a weak acid and its salt, we can approximate [A⁻] as the concentration of the salt and [HA] as the concentration of the acid, provided the volumes are the same or appropriately calculated after mixing.

缓冲溶液计算是单元 5 的重点题型。使用 Henderson–Hasselbalch 方程：pH = pKₐ + log₁₀([A⁻]/[HA])。评分方案奖励正确地将 Kₐ 转换为 pKₐ，然后代入共轭碱和弱酸的浓度。注意要使用平衡浓度，但对于通过混合弱酸及其盐制得的缓冲溶液，可以近似 [A⁻] 为盐的浓度，[HA] 为酸的浓度，前提是体积相同或混合后经过恰当计算。

For instance, a buffer contains 0.20 mol dm⁻³ CH₃COOH and 0.10 mol dm⁻³ CH₃COONa. Kₐ = 1.8 × 10⁻⁵. Then pKₐ = –log₁₀(1.8 × 10⁻⁵) = 4.74. pH = 4.74 + log₁₀(0.10/0.20) = 4.74 + log₁₀(0.50) = 4.74 – 0.30 = 4.44. The mark scheme often asks: ‘What is the effect of adding a small amount of acid or base?’ – you need to mention that the ratio [A⁻]/[HA] stays nearly constant, hence pH changes only slightly.

例如，某缓冲溶液含 0.20 mol dm⁻³ CH₃COOH 和 0.10 mol dm⁻³ CH₃COONa。Kₐ = 1.8 × 10⁻⁵。则 pKₐ = –log₁₀(1.8 × 10⁻⁵) = 4.74。pH = 4.74 + log₁₀(0.10/0.20) = 4.74 + log₁₀(0.50) = 4.74 – 0.30 = 4.44。评分方案常问：“加入少量酸或碱有何影响？”——需指出 [A⁻]/[HA] 比值几乎保持恒定，因此 pH 仅发生微小变化。

6. pH Changes During Acid–Base Titrations | 酸碱滴定过程中的 pH 变化

Part (c) of a pH question might involve calculating the pH at a particular point during a titration of a weak acid with a strong base. Before the equivalence point, you have a buffer; the mark scheme expects you to recognise this and use the buffer equation. At the half-equivalence point, [HA] = [A⁻] and pH = pKₐ – a classic mark-worthy observation.

pH 问题中的 (c) 部分可能涉及计算弱酸与强碱滴定过程中特定点的 pH。到达等当点之前，溶液是一个缓冲体系；评分方案期望你识别出这一点并使用缓冲方程。在半等当点处，[HA] = [A⁻]，pH = pKₐ——这是一个经典的值得得分的观察点。

At the equivalence point, only the conjugate base A⁻ is present; you must calculate its concentration taking into account the dilution, then use K₍ = Kₙ/Kₐ to find [OH⁻] and hence pH. The mark scheme explicitly checks that you use the correct K₍ expression and that the final pH is above 7 for a weak acid–strong base titration.

在等当点，只存在共轭碱 A⁻；你必须考虑稀释来计算其浓度，然后用 K₍ = Kₙ/Kₐ 求得 [OH⁻]，进而计算 pH。评分方案明确检查你是否使用了正确的 K₍ 表达式，并且对于弱酸-强碱滴定，最终 pH 应大于 7。

After the equivalence point, the pH is dominated by excess strong base [OH⁻]. Here, a simple pOH calculation suffices. The mark scheme typically awards a mark for subtracting from 14 to convert pOH to pH.

等当点之后，pH 由过量的强碱 [OH⁻] 主导。此时只需简单的 pOH 计算。评分方案通常奖励从 14 中相减以将 pOH 转化为 pH 这一步骤。

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

Cell EMF is calculated using E°_(cell) = E°_(right) – E°_(left), where both reduction potentials are written as reductions. The January 2020 paper often provides standard electrode potentials and asks you to predict spontaneity: a positive E°_(cell) implies the reaction is feasible. You must also write the overall cell reaction by reversing the left-hand half-equation and combining to cancel electrons.

电池电动势通过 E°_(电池) = E°_(右) – E°_(左) 计算，其中两边的电势均为还原电势。2020 年 1 月的试卷常提供标准电极电势并要求预测自发性：E°_(电池) 为正意味着反应可行。你还必须通过反转左侧半反应并合并以消去电子，写出总电池反应。

When concentrations are non-standard, the Nernst equation may be required: E = E° – (RT/nF) lnQ. At 298 K, this simplifies to E = E° – (0.0592/n) log₁₀Q. In the mark scheme, always show substitution of values into the Nernst equation, and ensure you note that n is the number of electrons transferred. A student who correctly reduces the log term to a numerical value gains full marks.

当浓度偏离标准状态时，可能需要能斯特方程：E = E° – (RT/nF) lnQ。在 298 K 下，可简化为 E = E° – (0.0592/n) log₁₀Q。在评分方案中，务必展示将数值代入能斯特方程的过程，并确保注明 n 为转移的电子数。正确将对数项简化为数值的学生可获得满分。

8. Rate Equations and the Arrhenius Equation | 速率方程和阿仑尼乌斯方程

The rate equation rate = k[A]ᵐ[B]ⁿ often appears in the context of determining orders from initial rate data. The mark scheme expects you to compare experiments where one reactant’s concentration changes while others are constant, then deduce the order by how the initial rate changes. Show the ratios clearly: if doubling [A] doubles the rate, m = 1; if doubling [A] quadruples the rate, m = 2.

速率方程 速率 = k[A]ᵐ[B]ⁿ 经常出现在通过初始速率数据确定反应级数的情景中。评分方案期望你比较那些只有一种反应物浓度改变而其他保持不变的实验，然后根据初始速率的变化推断级数。清晰地展示比值：若 [A] 加倍使速率加倍，则 m = 1；若 [A] 加倍使速率变为四倍，则 m = 2。

The Arrhenius equation in its logarithmic form, ln k = ln A – Eₐ/(RT), is tested either by calculating Eₐ from two rate constants at different temperatures or by interpreting a graph of ln k against 1/T. The mark scheme values correct algebraic rearrangement: Eₐ = (R ln(k₁/k₂)) / (1/T₂ – 1/T₁). Be consistent with units for R (8.31 J K⁻¹ mol⁻¹) and the activation energy (usually J mol⁻¹, then converted to kJ mol⁻¹).

阿仑尼乌斯方程的对数形式 ln k = ln A – Eₐ/(RT)，可能要求从两个不同温度下的速率常数计算 Eₐ，或解释 ln k 对 1/T 的图线。评分方案看重正确的代数变形：Eₐ = (R ln(k₁/k₂)) / (1/T₂ – 1/T₁)。注意 R 的单位要一致（8.31 J K⁻¹ mol⁻¹），活化能通常以 J mol⁻¹ 表示，然后再转换为 kJ mol⁻¹。

9. Calculating the Activation Energy from Rate Constants at Two Temperatures | 由两个温度下的速率常数计算活化能

A dedicated section is warranted for this commonly missed calculation. Suppose k₁ at T₁ = 298 K is 2.0 × 10⁻³ s⁻¹ and k₂ at T₂ = 308 K is 5.0 × 10⁻³ s⁻¹. Using the two-point form of the Arrhenius equation:

有必要为此常见易错计算专设一节。假设 T₁ = 298 K 时 k₁ = 2.0 × 10⁻³ s⁻¹，T₂ = 308 K 时 k₂ = 5.0 × 10⁻³ s⁻¹。使用阿仑尼乌斯方程的两点式：

ln(k₂/k₁) = –(Eₐ/R) × (1/T₂ – 1/T₁)

Plug in the values: ln(5.0/2.0) = ln 2.5 ≈ 0.916. The temperature term: 1/308 – 1/298 = 0.003247 – 0.003356 = –0.000109. Then Eₐ = –(8.31 × 0.916) / (–0.000109) ≈ 69,700 J mol⁻¹ or 69.7 kJ mol⁻¹. The mark scheme insists on a positive value for Eₐ and an answer to 3 significant figures unless data dictate otherwise.

代入数值：ln(5.0/2.0) = ln 2.5 ≈ 0.916。温度项：1/308 – 1/298 = 0.003247 – 0.003356 = –0.000109。则 Eₐ = –(8.31 × 0.916) / (–0.000109) ≈ 69700 J mol⁻¹ 即 69.7 kJ mol⁻¹。评分方案坚称 Eₐ 必须为正值，且结果常保留三位有效数字，除非数据另有要求。

10. Transition Metal Complex Stability and Ligand Exchange Calculations | 过渡金属配合物稳定性与配体交换计算

In Unit 5, questions on transition metals sometimes involve equilibrium constants for ligand substitution, known as stability constants (K_stab) or formation constants. The larger the log K_stab, the more stable the complex. When one ligand is replaced by another, the equilibrium constant can be derived from a combination of formation constants. The mark scheme may ask you to calculate a new equilibrium constant from given stability constants or to compare them to decide in which direction a ligand exchange lies.

在单元 5 中，过渡金属的题目有时涉及配体取代反应的平衡常数，即稳定常数 (K_stab) 或形成常数。log K_stab 越大，配合物越稳定。当一种配体被另一种取代时，平衡常数可从形成常数的组合推导得出。评分方案可能要求根据给定的稳定常数计算新的平衡常数，或比较它们以判断配体交换的方向。

For example, given [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O, the stepwise formation constants are provided, and the overall K_stab is the product of the stepwise constants. A calculation of the equilibrium concentration of a particular species often requires the use of an ICE table, exactly as with Kc, but the large magnitude of K_stab means simplifications can be made—the forward reaction may be assumed to go essentially to completion. However, the mark scheme expects you to state this assumption explicitly.

例如，已知 [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O，各级稳定常数已给出，总稳定常数 K_stab 是各级常数的乘积。计算某一物种的平衡浓度常常需要使用 ICE 表格，与处理 Kc 完全一样，但 K_stab 数值很大意味着可进行简化——可以假定正向反应基本进行完全。然而，评分方案期望你明确陈述这一假设。

Understanding how to manipulate complex equilibria is essential. Always check whether the reaction involves a stepwise displacement; the mark scheme often awards marks for correctly identifying the number of ligands replaced and writing the overall equilibrium expression.

理解如何处理配合物平衡至关重要。务必检查反应是否涉及逐级取代；评分方案通常奖励正确识别被取代配体数目及写出总平衡表达式的做法。

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