Year 13 Cambridge Chemistry: Unit Test Mock Paper Walkthrough | A2化学单元测试模拟卷深度解析

📚 Year 13 Cambridge Chemistry: Unit Test Mock Paper Walkthrough | A2化学单元测试模拟卷深度解析

Welcome to this in-depth walkthrough of a unit test mock paper tailored for Year 13 Cambridge A Level Chemistry. The questions cover core A2 topics including chemical kinetics, equilibria, thermodynamics, electrochemistry, transition metal chemistry, organic mechanisms, and spectroscopy. Each worked example is carefully explained to reinforce key concepts and sharpen exam technique.

欢迎阅读这篇专为A2化学设计的单元测试模拟卷深度解析。题目涵盖了A2阶段的核心主题，包括化学动力学、平衡、热力学、电化学、过渡金属化学、有机机理和光谱学。每道例题都经过精心解析，以巩固关键概念并提升应试技巧。

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1. Rate Equation and Mechanism | 速率方程与机理

The following initial rate data were obtained for the reaction 2NO(g) + O₂(g) → 2NO₂(g) at a fixed temperature.

在固定温度下，反应 2NO(g) + O₂(g) → 2NO₂(g) 获得了下列初始速率数据。

Experiment	[NO] / mol dm⁻³	[O₂] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻⁴
2	0.20	0.10	8.0 × 10⁻⁴
3	0.20	0.05	4.0 × 10⁻⁴

Compare experiments 1 and 2: doubling [NO] while keeping [O₂] constant quadruples the rate, so the order with respect to NO is 2.

比较实验1和2：保持[O₂]不变，[NO]加倍时速率变为原来的4倍，因此对NO的级数为2。

Compare experiments 2 and 3: keeping [NO] constant, halving [O₂] halves the rate, so the order with respect to O₂ is 1.

比较实验2和3：保持[NO]不变，[O₂]减半则速率减半，因此对O₂的级数为1。

Rate = k[NO]²[O₂]

Calculate k using experiment 1: k = (2.0×10⁻⁴) / (0.10² × 0.10) = 0.20 dm⁶ mol⁻² s⁻¹.

用实验1计算k：k = (2.0×10⁻⁴) / (0.10² × 0.10) = 0.20 dm⁶ mol⁻² s⁻¹。

A proposed mechanism consistent with this rate equation involves a fast equilibrium 2NO ⇌ N₂O₂ followed by the slow step N₂O₂ + O₂ → 2NO₂. The slow step determines the rate: rate = k'[N₂O₂][O₂]. Since [N₂O₂] = K[NO]², the rate law becomes rate = k’K[NO]²[O₂].

与速率方程一致的机理会涉及快速平衡 2NO ⇌ N₂O₂，随后是慢步骤 N₂O₂ + O₂ → 2NO₂。慢步骤决定速率：rate = k'[N₂O₂][O₂]。由于[N₂O₂] = K[NO]²，速率定律变为 rate = k’K[NO]²[O₂]。

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2. Equilibrium Constant Kc and Temperature Effects | 平衡常数Kc与温度效应

For the equilibrium N₂O₄(g) ⇌ 2NO₂(g) ΔH = +58 kJ mol⁻¹, 0.40 mol of N₂O₄ was placed in a 2.0 dm³ vessel. At equilibrium, the mixture contained 0.20 mol of NO₂.

对于平衡 N₂O₄(g) ⇌ 2NO₂(g) ΔH = +58 kJ mol⁻¹，将0.40 mol N₂O₄置于2.0 dm³容器中。平衡时，混合物中含有0.20 mol NO₂。

Moles of N₂O₄ reacted = 0.10 mol; therefore equilibrium moles N₂O₄ = 0.30 mol. Concentrations: [NO₂] = 0.20/2.0 = 0.10 mol dm⁻³, [N₂O₄] = 0.30/2.0 = 0.15 mol dm⁻³.

反应的N₂O₄为0.10 mol；因此平衡时N₂O₄ = 0.30 mol。浓度：[NO₂] = 0.20/2.0 = 0.10 mol dm⁻³，[N₂O₄] = 0.30/2.0 = 0.15 mol dm⁻³。

Kc = [NO₂]² / [N₂O₄] = (0.10)² / 0.15 = 6.67 × 10⁻² mol dm⁻³

Since the forward reaction is endothermic, increasing the temperature shifts the equilibrium to the right, producing more NO₂ and raising Kc.

由于正反应吸热，升高温度会使平衡向右移动，生成更多NO₂，从而提高Kc。

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3. pH of a Weak Acid and Buffer Solution | 弱酸pH与缓冲溶液

Calculate the pH of 0.10 mol dm⁻³ ethanoic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵ mol dm⁻³).

计算0.10 mol dm⁻³ 乙酸（CH₃COOH，Ka = 1.8 × 10⁻⁵ mol dm⁻³）的pH。

For a weak acid, [H⁺] ≈ √(Ka × c) = √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³.

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) = 2.87

If 25 cm³ of 0.10 mol dm⁻³ NaOH is added to 50 cm³ of the acid, the resulting mixture contains 0.0025 mol of CH₃COONa and 0.0025 mol of unreacted CH₃COOH in 75 cm³. Using the Henderson–Hasselbalch equation: pH = pKa + log([salt]/[acid]) = −log(1.8×10⁻⁵) + log(0.0333/0.0333) = 4.74 + 0 = 4.74.

若将25 cm³ 0.10 mol dm⁻³ NaOH加到50 cm³该酸中，混合物含有0.0025 mol CH₃COONa和0.0025 mol未反应的CH₃COOH，体积为75 cm³。利用亨德森-哈塞尔巴尔赫方程：pH = pKa + log([盐]/[酸]) = −log(1.8×10⁻⁵) + log(0.0333/0.0333) = 4.74 + 0 = 4.74。

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4. Entropy and Gibbs Free Energy | 熵与吉布斯自由能

Consider the decomposition: CaCO₃(s) → CaO(s) + CO₂(g), with ΔH° = +178 kJ mol⁻¹. Standard entropies: CaCO₃(s) = 93 J K⁻¹ mol⁻¹, CaO(s) = 40 J K⁻¹ mol⁻¹, CO₂(g) = 214 J K⁻¹ mol⁻¹.

考虑分解反应：CaCO₃(s) → CaO(s) + CO₂(g)，ΔH° = +178 kJ mol⁻¹。标准熵：CaCO₃(s) = 93 J K⁻¹ mol⁻¹，CaO(s) = 40 J K⁻¹ mol⁻¹，CO₂(g) = 214 J K⁻¹ mol⁻¹。

ΔS° = [40 + 214] − 93 = 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹.

ΔS° = [40 + 214] − 93 = 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹。

ΔG° = ΔH° − TΔS° = 178 − (298 × 0.161) = 178 − 48.0 = +130 kJ mol⁻¹

At 298 K, ΔG° is positive, so decomposition is not spontaneous. The temperature at which the reaction becomes feasible is when ΔG° = 0: T = ΔH° / ΔS° = 178 / 0.161 ≈ 1105 K.

在298 K时，ΔG°为正，分解反应不能自发进行。要使反应可行，需满足ΔG° = 0：T = ΔH° / ΔS° = 178 / 0.161 ≈ 1105 K。

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5. Electrochemical Cell and Nernst Equation | 电化学电池与能斯特方程

A cell is constructed: Zn(s) | Zn²⁺ (0.10 mol dm⁻³) || Cu²⁺ (0.010 mol dm⁻³) | Cu(s). Standard electrode potentials: E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.

构建电池：Zn(s) | Zn²⁺ (0.10 mol dm⁻³) || Cu²⁺ (0.010 mol dm⁻³) | Cu(s)。标准电极电势：E°(Zn²⁺/Zn) = −0.76 V，E°(Cu²⁺/Cu) = +0.34 V。

The cell reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E°cell = +0.34 − (−0.76) = 1.10 V and n = 2.

电池反应为 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)，E°cell = +0.34 − (−0.76) = 1.10 V，n = 2。

Using the Nernst equation at 298 K: Ecell = E°cell − (0.0592/n) log Q, where Q = [Zn²⁺]/[Cu²⁺] = 0.10/0.010 = 10.

298 K下使用能斯特方程：Ecell = E°cell − (0.0592/n) log Q，其中 Q = [Zn²⁺]/[Cu²⁺] = 0.10/0.010 = 10。

Ecell = 1.10 − (0.0592/2) log 10 = 1.10 − 0.0296 =

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