Year 13 CCEA Chemistry: Unit Test Mock Paper Walkthrough | Year 13 CCEA 化学：单元测试模拟卷解析

📚 Year 13 CCEA Chemistry: Unit Test Mock Paper Walkthrough | Year 13 CCEA 化学：单元测试模拟卷解析

Welcome to our in-depth walkthrough of a Year 13 CCEA Chemistry unit test mock paper. This resource is designed to help you consolidate key A2 concepts, from thermodynamics and kinetics to organic synthesis and transition metal chemistry. Each question is fully explained with step-by-step reasoning, common pitfalls, and revision tips.

欢迎阅读这份 Year 13 CCEA 化学单元测试模拟卷的详细解析。本资源旨在帮助你巩固 A2 核心概念，涵盖热力学、动力学、有机合成及过渡金属化学等。每道题都配有逐步推理、常见错误分析和复习提示，帮助你在正式考试中自信作答。

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1. Multiple Choice 1: Born–Haber Enthalpy Cycles | 选择题1：玻恩–哈伯焓循环

Question: Which of the following steps in the Born–Haber cycle for sodium chloride is exothermic?
A) Na(s) → Na(g)
B) Na(g) → Na⁺(g) + e⁻
C) Cl₂(g) → 2Cl(g)
D) Cl(g) + e⁻ → Cl⁻(g)

问题：对于氯化钠的玻恩–哈伯循环，下列哪一个步骤是放热的？
A) Na(s) → Na(g)
B) Na(g) → Na⁺(g) + e⁻
C) Cl₂(g) → 2Cl(g)
D) Cl(g) + e⁻ → Cl⁻(g)

The correct option is D. The addition of an electron to a gaseous chlorine atom corresponds to the first electron affinity, which releases energy. The other three steps – atomisation of sodium, first ionisation energy of sodium, and bond dissociation of chlorine – are all endothermic processes that require an input of energy.

正确答案为 D。一个电子加到气态氯原子上对应第一电子亲和能，这一过程会释放能量。其他三个步骤——钠的原子化、钠的第一电离能以及氯气的键断裂都是需要吸收能量的吸热过程。

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2. Multiple Choice 2: Kinetics – Determining Reaction Order | 选择题2：化学动力学 – 确定反应级数

Question: 2B → C was studied at a fixed temperature. Initial rates for three experiments are shown:
riment 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, initial rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹
Experiment 2: [A] = 0.20 mol dm⁻³, [B] = 0.10 mol dm⁻³, initial rate = 8.0 × 10⁻³
Experiment 3: [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³, initial rate = 4.0 × 10⁻³
What is the order with respect to A?

问题：在固定温度下研究了反应 A + 2B → C。三个实验的初始速率如下表所示。A 的反应级数是多少？

rder with respect to A is 2. The data also show that doubling [B] (Expt 1→3) doubles the rate, so the order with respect to B is 1. The rate equation can be written as:

Rate = k[A]²[B]

比较实验1和实验2，[A]加倍而[B]保持不变。速率从 2.0×10⁻³ 增大到 8.0×10⁻³，增大了4倍。因为 (2)² = 4，所以对 A 的反应级数为2。数据还表明[B]加倍（实验1→3）会使速率加倍，因此对 B 的级数为1。速率方程可写为速率 = k[A]²[B]。

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3. Multiple Choice 3: Writing Equilibrium Constants | 选择题3：平衡常数的书写

Question: Which expression correctly gives Kc for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g)?
A) [CaO][CO₂] / [CaCO₃]
B) [CO₂]
C) [CaO] / [CaCO₃]
D) [CO₂] × [CaO]

问题：对于反应 CaCO₃(s) ⇌ CaO(s) + CO₂(g)，哪个表达式正确给出了 Kc？

ession. Therefore Kc depends only on the concentration of the gaseous product: Kc = [CO₂].

答案：B。在多相平衡中，纯固体的活度为1，在 Kc 表达式中省略。因此 Kc 只取决于气体产物的浓度：Kc = [CO₂]。

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4. Multiple Choice 4: pH of a Weak Acid | 选择题4：弱酸pH的计算

Question: Calculate the pH of a 0.10 mol dm⁻³ solution of ethanoic acid. (Ka = 1.8 × 10⁻⁵ mol dm⁻³)
A) 1.0
B) 5.0
C) 2.9
D) 4.0

问题：计算 0.10 mol dm⁻³ 乙酸溶液的 pH。（Ka = 1.8 × 10⁻⁵ mol dm⁻³）

For a weak acid, [H⁺] = √Ka × C = √1.8 × 10⁻⁵ × 0.10 = √1.8 × 10⁻⁶ ≈ 1.34 × 10⁻³ mol dm⁻³. pH = –log(1.34 × 10⁻³) ≈ 2.87, which rounds to 2.9. Choice C is correct.

对于弱酸，[H⁺] = √Ka × C = √1.8×10⁻⁵ × 0.10 = √1.8×10⁻⁶ ≈ 1.34×10⁻³ mol dm⁻³。pH = –log(1.34×10⁻³) ≈ 2.9，因此选 C。

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5. Multiple Choice 5: Electrochemical Cells & Feasibility | 选择题5：电化学电池与反应可行性

Question: Use the standard electrode potentials: E°(Cu²⁺/Cu) = +0.34 V; E°(Zn²⁺/Zn) = –0.76 V. Which statement is correct for a cell constructed from these two half-cells?
A) Zinc is oxidised, and the cell emf is +1.10 V.
B) Copper is oxidised, and the cell emf is +1.10 V.
C) Zinc is reduced, and the cell emf is –0.42 V.
D) No reaction occurs.

问题：使用标准电极电势：E°(Cu²⁺/Cu) = +0.34 V；E°(Zn²⁺/Zn) = –0.76 V。对于用这两个半电池构建的原电池，哪项描述正确？

otential, so it will be oxidised (the anode). Copper will be reduced. E°cell = E°(cathode) – E°(anode) = (+0.34) – (–0.76) = +1.10 V. A positive cell emf confirms the reaction is spontaneous.

正确答案为 A。锌的电势更负，因此被氧化（阳极）。铜被还原。E°电池 = 正极 – 负极 = (+0.34) – (–0.76) = +1.10 V。正值电动势表明反应可行。

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6. Structured Question 1: Organic Synthesis & Yield | 结构化题1：有机合成与产率

Problem: Ethanol is heated under reflux with an excess of sodium bromide and concentrated sulfuric acid. The organic product is separated, washed and dried. 4.6 g of ethanol produced 7.2 g of pure product.

试题：将乙醇与过量的溴化钠和浓硫酸一起加热回流。分离、洗涤并干燥有机产物。4.6 g 乙醇得到了 7.2 g 纯产物。

structural formulae. The in-situ reaction generates HBr: CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O.

(a) 用结构式写出反应方程式。原位生成 HBr：CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O。

(b) Name the product and state the mechanism. The product is bromoethane. The mechanism is nucleophilic substitution, SN2, because ethanol is a primary alcohol. The hydroxide group is protonated, water leaves as a good leaving group, and bromide ion attacks the carbocation-like transition state.

(b) 命名产物并说明机理。产物是溴乙烷。机理为亲核取代（SN2），因为乙醇是伯醇。羟基被质子化，水作为良好离去基团离去，溴离子进攻碳正离子般的过渡态。

0 mol. Theoretical mass of bromoethane = 0.10 × 109 = 10.9 g. Percentage yield = (7.2 / 10.9) × 100% = 66.1%.

(c) 计算产率。乙醇摩尔质量 = 46.0 g mol⁻¹；溴乙烷摩尔质量 = 108.9 g mol⁻¹（取109）。乙醇的物质的量 = 4.6 / 46 = 0.10 mol。理论的溴乙烷质量 = 0.10 × 109 = 10.9 g。产率 = (7.2 / 10.9) × 100% = 66.1%。

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7. Structured Question 2: Buffer Solutions | 结构化题2：缓冲溶液

Problem: e (CH₃COONa). Ka for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of this buffer.

试题：将等体积的 0.20 mol dm⁻³ 乙酸 (CH₃COOH) 与 0.10 mol dm⁻³ 乙酸钠 (CH₃COONa) 混合配成缓冲溶液。乙酸的 Ka = 1.8×10⁻⁵ mol dm⁻³。计算该缓冲液的 pH。

r, because the pH of a buffer depends on the ratio of the conjugate base to the acid, the absolute concentrations are not required as long as they are in the same total volume; the ratio remains [A⁻]/[HA] = 0.050/0.10 = 0.50. Using the Henderson–Hasselbalch equation:

pH = pKa + log([A⁻] / [HA])

pKa = –log(1.8 × 10⁻⁵) ≈ 4.74. Therefore, pH = 4.74 + log(0.50) = 4.74 – cause the CH₃COO⁻ ions neutralise added H⁺ and the CH₃COOH molecules neutralise added OH⁻.

混合后，酸的浓度变为 0.10 mol dm⁻³，盐的浓度变为 0.050 mol dm⁻³（考虑体积加倍）。缓冲液的 pH 取决于共轭碱与酸的比值，此处比值为 0.050/0.10 = 0.50。利用 Henderson–Hasselbalch 方程：pH = pKa + log([A⁻] / [HA])。pKa = –log(1.8×10⁻⁵) ≈ 4.74。因此 pH = 4.74 + log(0.50) = 4.74 – 0.30 = 4.44。该缓冲液能抵抗加入少量酸或碱引起的 pH 变化，因为 CH₃COO⁻ 离子可中和外加的 H⁺，而 CH₃COOH 分子可中和外加的 OH⁻。

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8. Structured Question 3: Rate-Determining Step & Mechanism | 结构化题3：决速步与反应机理

Problem: ²[H₂]. Propose a two-step mechanism that is consistent with this rate law, clearly identifying the rate-determining step.

试题：对于反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)，实验测定的速率方程为 rate = k[NO]²[H₂]。提出一个与该速率定律一致的两步机理，并明确标出决速步骤。

A common mechanism that satisfies the rate law involves a fast pre-equilibrium followed by a slow step:

Step 1 (fast equilibrium): 2NO ⇌ N₂O₂
Step 2 (slow): N₂O₂ + H₂ → N₂ + H₂O + O (or further fast steps to complete stoichiometry).

ing gives rate = k₂ × Kc × [NO]²[H₂], i.e. rate = k[NO]²[H₂] with k = k₂Kc. Thus, the proposed mechanism is consistent with the observed rate law. The rate-determining step is Step 2.

一个符合速率定律的常见机理包含一个快速预平衡和一个慢步骤：

第一步（快速平衡）：2NO ⇌ N₂O₂
第二步（慢）：N₂O₂ + H₂ → N₂ + H₂O + O（后续的快速步骤补全化学计量数）。

慢步骤决定速率：rate = k₂[N₂O₂][H₂]。利用第一步的平衡常数 Kc = [N₂O₂] / [NO]²，可得 [N₂O₂] = Kc[NO]²。代入后得到 rate = k₂ × Kc × [NO]²[H₂]，即 rate = k[NO]²[H₂]，其中 k = k₂Kc。因此该机理与实验速率定律一致，决速步为第二步。

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9. Structured Question 4: Gibbs Free Energy & Feasibility | 结构化题4：吉布斯自由能与反应可行性

Problem: For the reaction N₂(g)

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