📚 AQA A Level Chemistry 7405/2 June 2018 Mark Scheme: Key Points | AQA A Level 化学 7405/2 2018年6月评分标准：核心要点
The June 2018 AQA A Level Chemistry Paper 2 mark scheme shows exactly how examiners award marks for method, equations, units, state symbols and significant figures. This revision guide summarises the main mark scheme demands and common errors, without reproducing the paper.
2018年6月 AQA A Level 化学卷二评分标准展示了考官如何根据计算过程、方程式、单位、状态符号和有效数字给分。本复习指南总结了评分标准的主要要求和常见错误，不复制原试卷。
1. Why Mark Scheme Analysis Matters | 为什么要分析评分标准
Mark schemes are not just answer lists; they reveal the exact wording, working and observations that gain marks. In Paper 2, many marks are attached to definitions, state symbols and clear step-by-step calculations rather than the final answer alone.
评分标准不只是答案列表；它揭示了能够得分的准确表述、计算步骤和实验现象。在卷二，很多分数取决于定义、状态符号和清晰的分步计算，而不仅仅是最终结果。
Common command words such as state, give, name, calculate, deduce and explain require different levels of detail. The June 2018 mark scheme penalises vague explanations that do not refer to structure, energy or equilibrium position where required.
常见的指令词如 state、give、name、calculate、deduce 和 explain 需要不同的作答详细程度。2018年6月的评分标准会扣掉那些未根据要求提及结构、能量或平衡位置的模糊解释。
2. Paper 2 Overview and Assessment Focus | 卷二概述与考查重点
Paper 2 assesses physical chemistry topics including amount of substance, energetics, kinetics, equilibria, acids and bases, plus organic chemistry such as mechanisms, functional group tests and spectroscopy. The June 2018 paper mixed short structured questions with multi-step calculations.
卷二考查物理化学主题，包括物质的量、能量学、动力学、平衡、酸和碱，以及有机化学中的机理、官能团检验和光谱学。2018年6月的试卷将简短结构化问题与多步骤计算题结合起来。
Questions often begin with basic recall, then ask you to apply knowledge to unfamiliar compounds or data. The mark scheme rewards the use of data from the question, especially when deducing orders, calculating pH or identifying an organic structure.
题目通常从基础记忆开始，然后要求将知识应用于陌生的化合物或数据。评分标准认可使用题目所给数据，尤其是在推导反应级数、计算 pH 或确定有机结构时。
3. Amount of Substance and Gas Calculations | 物质的量与气体计算
Paper 2 consistently covers moles, empirical formulae, ideal gas equation and percentage yield. The mark scheme rewards converting mass or gas volume to moles first, then applying the balanced equation ratio, and finally giving answers to 3 significant figures unless stated otherwise.
卷二一贯考查摩尔、经验式、理想气体状态方程和百分产率。评分标准奖励先将质量或气体体积换算为物质的量，再使用配平方程式的化学计量比，最后除非另有说明，答案保留三位有效数字。
pV = nRT
For gas calculations, pressure must be in kPa or Pa consistently with the value of R, temperature must be in kelvin, and volume in m³ when using R = 8.31 J K⁻¹ mol⁻¹. The mark scheme usually awards a separate mark for the correct unit of the final quantity.
气体计算中，压强必须与 R 的取值一致（kPa 或 Pa），温度必须为开尔文，使用 R = 8.31 J K⁻¹ mol⁻¹ 时体积必须为 m³。评分标准通常为最终量的正确单位单独给分。
A very common mark scheme trap is forgetting to convert cm³ to m³ by dividing by 10⁶ or converting °C to K by adding 273. The mark scheme may give an answer mark for the correct value but withhold the unit mark if the conversion is missing.
一个非常常见的评分标准陷阱是忘记将 cm³ 除以 10⁶ 换算为 m³，或忘记将 °C 加上 273 换算为 K。即使数值正确，如果缺少换算步骤，评分标准也可能不给予单位分。
4. Hess’s Law, Born-Haber Cycles and Enthalpy of Solution | 赫斯定律、玻恩-哈伯循环与溶解焓
Born-Haber cycle questions in June 2018 required correct species with state symbols, for example Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s), and correct labelling of enthalpy changes such as lattice enthalpy and enthalpy of formation.
2018年6月的玻恩-哈伯循环题要求正确的物种及状态符号，例如 Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)，并正确标注晶格焓、生成焓等焓变。
The mark scheme credits arrows that point in the correct direction and allows enthalpy of solution to be calculated using a Hess cycle. The equation below summarises the relationship for an ionic solid dissolving in water:
评分标准认可方向正确的箭头，并允许使用赫斯循环计算溶解焓。以下方程总结了离子固体溶于水的关系：
ΔH(solution) = ΔH(lattice) + ΣΔH(hydration)
Definitions are often worth a mark: lattice enthalpy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. Missing ‘gaseous ions’ or ‘one mole’ usually loses the mark.
定义经常占一分：晶格焓是在标准条件下，由气态离子生成一摩尔离子化合物时的焓变。漏写’气态离子’或’一摩尔’通常会被扣分。
5. Gibbs Free Energy and Feasibility | 吉布斯自由能与反应可行性
Gibbs free-energy questions require consistent units. If ΔS is given in J K⁻¹ mol⁻¹, then ΔH must be converted to J mol⁻¹ before substitution. The mark scheme frequently penalises mixing kJ and J.
吉布斯自由能题目要求单位一致。如果 ΔS 的单位为 J K⁻¹ mol⁻¹，那么代入前必须将 ΔH 换算为 J mol⁻¹。评分标准经常因混用 kJ 和 J 而扣分。
ΔG = ΔH − TΔS
To find the temperature at which the reaction becomes feasible, set ΔG = 0. This gives T = ΔH/ΔS, and the temperature must be converted to °C if the question asks for it in Celsius. A common error is to leave T in kelvin after the question asks for °C.
求反应刚好可行时的温度，令 ΔG = 0，得到 T = ΔH/ΔS。如果题目要求以摄氏度作答，必须将温度换算为 °C。常见错误是题目要求 °C 时答案仍保留开尔文。
The mark scheme often awards a mark for stating that a reaction is feasible when ΔG is negative, and another for correctly interpreting the temperature range. A negative ΔH and positive ΔS does not automatically guarantee feasibility at all temperatures unless stated.
评分标准通常会对’ΔG 为负时反应可行’这一表述给分，并对正确分析温度范围再给一分。ΔH 为负且 ΔS 为正并不自动保证所有温度下都可行，除非题目说明。
6. Gas Equilibria and Kp | 气体平衡与 Kp
For Kp calculations, the mark scheme expects an expression with partial pressures raised to the powers of the balanced equation. For example, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) gives Kp = p(NH₃)² / [p(N₂) × p(H₂)³].
对于 Kp 计算，评分标准要求根据配平方程式写出分压的幂表达式。例如 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 的 Kp = p(NH₃)² / [p(N₂) × p(H₂)³]。
p(A) = x(A) × P(total)
The mark scheme rewards a clear two-step process: first calculate mole fractions, then multiply by total pressure. If total pressure is in kPa, all partial pressures must be in kPa before substitution into the Kp expression.
评分标准认可清晰的二步过程：先计算摩尔分数，再乘以总压。如果总压单位为 kPa，代入 Kp 表达式前所有分压都必须以 kPa 为单位。
| Species | Mole fraction | Partial pressure |
| N₂ | 0.20 | 0.20 × P(total) |
| H₂ | 0.60 | 0.60 × P(total) |
| NH₃ | 0.20 | 0.20 × P(total) |
Units are not always required for Kp, but when they are, they depend on the powers in the expression. The June 2018 mark scheme allowed omission of units if no unit mark was allocated, but always check the question wording.
Kp 的单位并非总是要求，但如果题目要求，单位取决于表达式中的幂次。2018年6月的评分标准在未分配单位分时允许省略单位，但务必检查题目措辞。
7. pH, Ka and Buffer Calculations | pH、Ka 与缓冲溶液计算
Weak acid pH calculations use the acid dissociation constant. For a pure weak acid, the mark scheme accepts the approximation below, provided the degree of dissociation is small:
弱酸 pH 计算使用酸解离常数。对于纯弱酸，只要解离度很小，评分标准接受以下近似：
[H⁺] ≈ √(Ka × [HA])
For buffer solutions, the mark scheme usually expects the equation below after calculating the amounts of acid and conjugate base in the mixture. A separate mark is often given for correctly deducing the ratio of [HA] to [A⁻] from the given concentrations and volumes.
对于缓冲溶液，评分标准通常要求在计算出混合物中酸和共轭碱的物质的量后使用以下方程。正确根据给定浓度和体积推导 [HA] 与 [A⁻] 的比值通常单独给分。
[H⁺] = Ka × [HA]/[A⁻]
The mark scheme in June 2018 often awarded one mark for the pH conversion pH = −log₁₀[H⁺], and another for the final pH quoted to 2 decimal places. Forgetting the negative sign or rounding too early were common errors.
2018年6月的评分标准经常对 pH = −log₁₀[H⁺] 的换算给一分
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