Year 13 AQA Chemistry: Unit Test Mock Paper Walkthrough | 英国AQA化学13年级单元测试模拟卷解析

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

This article provides a detailed walkthrough of a mock unit test for Year 13 AQA Chemistry. The questions cover thermodynamics, rates, equilibrium, acids and bases, electrochemistry, transition metals, and organic synthesis with spectroscopy. Each question is explained step by step, with key equations and reasoning, followed by its Chinese translation to support bilingual learning.

本文详细解析了一份AQA化学13年级单元测试模拟卷。试题涵盖热力学、反应速率、化学平衡、酸碱、电化学、过渡金属以及有机合成与波谱分析等内容。每题均提供逐步解析、关键方程式与解题思路，并附中文翻译，助力双语学习。

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1. Born-Haber Cycle and Lattice Enthalpy | 第1题：玻恩-哈伯循环与晶格焓

The question provided the following data: enthalpy of atomisation of Mg(s) = +148 kJ mol⁻¹, first ionisation energy of Mg(g) = +738 kJ mol⁻¹, second ionisation energy of Mg(g) = +1451 kJ mol⁻¹, bond dissociation enthalpy of O₂(g) = +498 kJ mol⁻¹, first electron affinity of O(g) = -141 kJ mol⁻¹, second electron affinity of O(g) = +798 kJ mol⁻¹, and the standard enthalpy of formation of MgO(s) = -602 kJ mol⁻¹. Use these to calculate the lattice enthalpy of MgO(s).

题目给出如下数据：Mg(s)的原子化焓 = +148 kJ mol⁻¹，Mg(g)的第一电离能 = +738 kJ mol⁻¹，第二电离能 = +1451 kJ mol⁻¹，O₂(g)的键解离焓 = +498 kJ mol⁻¹，O(g)的第一电子亲和能 = -141 kJ mol⁻¹，第二电子亲和能 = +798 kJ mol⁻¹，以及MgO(s)的标准生成焓 = -602 kJ mol⁻¹。利用这些数据计算MgO(s)的晶格焓。

Step 1: Write the Born-Haber cycle equation applying Hess’s law. The formation enthalpy of MgO(s) equals the sum of the enthalpy changes along the alternative route: atomisation of Mg(s), ionisation energies of Mg(g) to Mg²⁺(g), atomisation of ½O₂(g) to O(g), addition of two electrons to form O²⁻(g), and the lattice enthalpy (from gaseous ions to solid). The lattice enthalpy is defined as the enthalpy change when 1 mole of an ionic solid is formed from its gaseous ions.

步骤1：写出玻恩-哈伯循环方程，应用盖斯定律。MgO(s)的生成焓等于另一条路径的焓变总和：Mg(s)的原子化、Mg(g)电离为Mg²⁺(g)、½O₂(g)解离为O(g)、加两个电子生成O²⁻(g)，以及晶格焓（由气态离子形成固体）。晶格焓定义为1摩尔离子固体由其气态离子形成时的焓变。

Step 2: Set up the equation: ΔH°f(MgO) = ΔH°atom(Mg) + IE₁(Mg) + IE₂(Mg) + ½ΔH°diss(O₂) + EA₁(O) + EA₂(O) + ΔH°lattice. Substitute the values: -602 = 148 + 738 + 1451 + ½(498) + (-141) + 798 + ΔH°lattice. Note that ½ × 498 = 249. Then sum the known positive terms: 148 + 738 + 1451 + 249 + 798 = 3384; add EA₁(-141) gives 3243. Then -602 – 3243 = ΔH°lattice, so ΔH°lattice = -3845 kJ mol⁻¹ (approximately).

步骤2：建立方程：ΔH°f(MgO) = ΔH°atom(Mg) + IE₁(Mg) + IE₂(Mg) + ½ΔH°diss(O₂) + EA₁(O) + EA₂(O) + ΔH°lattice。代入数值：-602 = 148 + 738 + 1451 + ½(498) + (-141) + 798 + ΔH°lattice。注意½×498 = 249。正项总和：148+738+1451+249+798 = 3384；加上EA₁(-141)得3243。因此ΔH°lattice = -602 – 3243 = -3845 kJ mol⁻¹（约）。

The large negative value indicates a strongly exothermic lattice formation, typical for doubly charged ions with relatively small radii.

该数值为很大的负值，表明晶格形成过程强烈放热，这对于半径较小的双电荷离子是典型现象。

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2. Entropy and Gibbs Free Energy | 第2题：熵与吉布斯自由能

The question asks: calculate the temperature at which the decomposition of calcium carbonate becomes feasible: CaCO₃(s) → CaO(s) + CO₂(g). Given standard enthalpy change ΔH° = +178 kJ mol⁻¹ and standard entropy change ΔS° = +161 J K⁻¹ mol⁻¹.

题目要求：计算碳酸钙分解反应可行的温度：CaCO₃(s) → CaO(s) + CO₂(g)。已知标准焓变 ΔH° = +178 kJ mol⁻¹，标准熵变 ΔS° = +161 J K⁻¹ mol⁻¹。

Step 1: For a reaction to be feasible, ΔG° ≤ 0. The Gibbs equation is ΔG° = ΔH° – TΔS°. At the temperature where the reaction just becomes feasible, set ΔG° = 0, giving T = ΔH° / ΔS°. Convert ΔS° to kJ K⁻¹ mol⁻¹: 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹. Then T = 178 / 0.161 ≈ 1106 K. In Celsius, this is about 833 °C.

步骤1：反应可行需要 ΔG° ≤ 0。吉布斯方程为 ΔG° = ΔH° – TΔS°。在反应刚好可行的温度，设 ΔG° = 0，得 T = ΔH° / ΔS°。将 ΔS° 换算为 kJ K⁻¹ mol⁻¹：161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹。则 T = 178 / 0.161 ≈ 1106 K，换算为摄氏度约为 833 °C。

Step 2: Above this temperature, the positive entropy term TΔS° outweighs the positive enthalpy, making ΔG° negative and the decomposition spontaneous. At lower temperatures, ΔG° > 0, and the reaction is not thermodynamically feasible.

步骤2：高于此温度时，正的熵项 TΔS° 压倒正的焓项，使 ΔG° 为负，分解反应自发进行。低于此温度时 ΔG° > 0，反应在热力学上不可行。

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3. Rate Equation and Order of Reaction | 第3题：速率方程与反应级数

The following initial rate data were obtained for the reaction A + 2B → C at constant temperature. Determine the rate equation and calculate the rate constant k, with its units.

在恒定温度下，对反应 A + 2B → C 测得如下初始速率数据。确定速率方程并计算速率常数 k 及其单位。

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻⁴
2	0.20	0.10	4.0 × 10⁻⁴
3	0.10	0.20	8.0 × 10⁻⁴

Step 1: By comparing experiments 1 and 2, keeping [B] constant, doubling [A] doubles the rate, so the reaction is first order with respect to A.

步骤1：对比实验1和2，保持 [B] 不变，[A] 加倍，速率加倍，说明对 A 为一级反应。

Step 2: Comparing experiments 1 and 3, keeping [A] constant, doubling [B] increases the rate by a factor of 4 (from 2.0 to 8.0 × 10⁻⁴), so the order with respect to B is second order.

步骤2：对比实验1和3，保持 [A] 不变，[B] 加倍，速率增大4倍（从 2.0 到 8.0 × 10⁻⁴），因此对 B 为二级反应。

Step 3: The rate equation is rate = k[A][B]². Using data from experiment 1: 2.0 × 10⁻⁴ = k × 0.10 × (0.10)² = k × 1.0 × 10⁻³. Thus k = 0.20 mol⁻² dm⁶ s⁻¹. The units of k are derived from the requirement: rate (mol dm⁻³ s⁻¹) = k × (mol dm⁻³)³, so k = mol⁻² dm⁶ s⁻¹.

步骤3：速率方程为 rate = k[A][B]²。代入实验1数据：2.0 × 10⁻⁴ = k × 0.10 × (0.10)² = k × 1.0 × 10⁻³。得 k = 0.20 mol⁻² dm⁶ s⁻¹。k 的单位由量纲导出：速率 (mol dm⁻³ s⁻¹) = k × (mol dm⁻³)³，因此 k = mol⁻² dm⁶ s⁻¹。

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4. Equilibrium Constant Kp | 第4题：气相平衡常数 Kp

The question gives the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At a certain temperature, the equilibrium partial pressures are: p(N₂) = 2.50 MPa, p(H₂) = 1.30 MPa, and p(NH₃) = 3.70 MPa. Calculate Kp and state its units.

题目给出反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。在一定温度下，平衡分压为：p(N₂) = 2.50 MPa，p(H₂) = 1.30 MPa，p(NH₃) = 3.70 MPa。计算 Kp 并给出单位。

Step 1: Write the expression Kp = p(NH₃)² / [p(N₂) × p(H₂)³]. Substitute the values: Kp = (3.70)² / [2.50 × (1.30)³]. Carrying out the calculation: 3.70² = 13.69; 1.30³ = 2.197; denominator = 2.50 × 2.197 = 5.4925. Therefore Kp = 13.69 / 5.4925 ≈ 2.49.

步骤1：写出表达式 Kp = p(NH₃)² / [p(N₂) × p(H₂)³]。代入数值：Kp = (3.70)² / [2.50 × (1.30)³]。计算：3.70² = 13.69；1.30³ = 2.197；分母 = 2.50 × 2.197 = 5.4925。因此 Kp ≈ 13.69 / 5.4925 ≈ 2.49。

Step 2: Determine the units. The numerator has units MPa², denominator MPa × MPa³ = MPa⁴, giving overall units of MPa⁻². So Kp = 2.49 MPa⁻².

步骤2：确定单位。分子的单位是 MPa²，分母为 MPa × MPa³ = MPa⁴，总单位为 MPa⁻²。因此 Kp = 2.49 MPa⁻²。

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5. Buffer Solution pH Calculation | 第5题：缓冲溶液pH计算

A buffer solution is made by mixing 50.0 cm³ of 0.100 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.100 mol dm⁻³ sodium ethanoate. Calculate the pH of the buffer. Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³.

将 50.0 cm³ 0.100 mol dm⁻³ 的乙酸与 50.0 cm³ 0.100 mol dm⁻³ 的乙酸钠混合制成缓冲溶液。计算此缓冲溶液的 pH。乙酸的 Ka = 1.74 × 10⁻⁵ mol dm⁻³。

Step 1: After mixing, the total volume is 100.0 cm³, so the concentration of the weak acid [HA] and the conjugate base [A⁻] are both halved: [HA] = [A⁻] = (0.100 × 50.0/100.0) = 0.0500 mol dm⁻³.

步骤1：混合后总体积为 100.0 cm³，因此弱酸 [HA] 和共轭碱 [A⁻] 的浓度均减半：[HA] = [A⁻] = (0.100 × 50.0/100.0) = 0.0500 mol dm⁻³。

Step 2: Use the Henderson-Hasselbalch equation or the Ka expression directly. Since [HA] = [A⁻], [H⁺] = Ka × [HA]/[A⁻] = Ka = 1.74 × 10⁻⁵ mol dm⁻³. Thus pH = -log₁₀(1.74 × 10⁻⁵) = 4.76 (to 3 s.f.).

步骤2：使用 Henderson-Hasselbalch 方程或直接利用 Ka 表达式。由于 [HA] = [A⁻]，[H⁺] = Ka × [HA]/[A⁻] =

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