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

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

This walkthrough covers a full mock paper designed for Year 13 OCR Chemistry, spanning core topics from physical chemistry (thermodynamics, kinetics, equilibrium, redox, electrochemistry and transition elements) and organic chemistry (aromatic chemistry, carbonyls, carboxylic acid derivatives, nitrogen compounds, synthesis and analysis). Each question is followed by a detailed explanation to sharpen your problem-solving skills and reinforce key concepts before the final examination.

本解析覆盖一份为Year 13 OCR化学设计的完整模拟卷，涉及物理化学（热力学、动力学、平衡、氧化还原、电化学和过渡元素）和有机化学（芳香化学、羰基化合物、羧酸衍生物、含氮化合物、合成与分析）的核心主题。每道题后都附有详细解析，以锻炼你的解题能力并在期末考试前强化关键概念。

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

A proposed reaction for ethanol production is: C₂H₄(g) + H₂O(g) → C₂H₅OH(l). At 298 K, ΔH = -45 kJ mol^-1 and ΔS_system = -126 J K^-1 mol^-1. Calculate ΔG and determine whether the reaction is feasible at this temperature.

一个制备乙醇的提议反应为：C₂H₄(g) + H₂O(g) → C₂H₅OH(l)。在298 K时，ΔH = -45 kJ mol⁻¹，ΔSₛₙₛₜₑₘ = -126 J K⁻¹ mol⁻¹。计算ΔG并判断该温度下反应是否可行。

ΔG = ΔH – TΔS

First, convert ΔS to kJ: ΔS = -126 J K^-1 mol^-1 = -0.126 kJ K^-1 mol^-1. Then ΔG = -45 – (298 × -0.126) = -45 + 37.5 = -7.5 kJ mol^-1. Since ΔG is negative, the reaction is thermodynamically feasible at 298 K. However, the negative entropy change (gas → liquid) makes the –TΔS term positive; the exothermic enthalpy drives feasibility.

首先将ΔS换算为kJ：ΔS = -126 J K⁻¹ mol⁻¹ = -0.126 kJ K⁻¹ mol⁻¹。则ΔG = -45 – (298 × -0.126) = -45 + 37.5 = -7.5 kJ mol⁻¹。由于ΔG为负值，该反应在298 K时热力学上可行。但熵减（气体→液体）使–TΔS项为正；放热的焓变驱动了可行性。

2. Born–Haber Cycle Calculations | 玻恩–哈伯循环计算

Use the following data to calculate the lattice enthalpy of magnesium chloride, MgCl₂:

• Standard enthalpy of formation of MgCl₂(s) = -641 kJ mol^-1
• Enthalpy of atomisation of Mg(s) = +148 kJ mol^-1
• First ionisation energy of Mg = +738 kJ mol^-1
• Second ionisation energy of Mg = +1451 kJ mol^-1
• Enthalpy of atomisation of Cl₂(g) = +244 kJ mol^-1
• Electron affinity of Cl(g) = -349 kJ mol^-1

使用下列数据计算氯化镁MgCl₂的晶格焓：

• MgCl₂(s)标准生成焓 = -641 kJ mol⁻¹
• Mg(s)原子化焓 = +148 kJ mol⁻¹
• Mg第一电离能 = +738 kJ mol⁻¹
• Mg第二电离能 = +1451 kJ mol⁻¹
• Cl₂(g)原子化焓 = +244 kJ mol⁻¹
• Cl(g)电子亲和势 = -349 kJ mol⁻¹

ΔH_f = Σ(atomisation) + Σ(ionisation energies) + Σ(electron affinities) + LE

Construct the cycle: Mg(s) → Mg(g) (+148); Mg(g) → Mg⁺(g) (+738) → Mg²⁺(g) (+1451); Cl₂(g) → 2Cl(g) (+244); 2Cl(g) + 2e^– → 2Cl^–(g) (2 × -349 = -698). Sum of these steps = 148 + 738 + 1451 + 244 – 698 = +1883 kJ mol^-1. Then ΔH_f = Σ + LE → -641 = 1883 + LE → LE = -641 – 1883 = -2524 kJ mol^-1. The lattice enthalpy is -2524 kJ mol^-1, indicating a highly exothermic process and strong ionic bonding.

构建循环：Mg(s) → Mg(g) (+148)；Mg(g) → Mg⁺(g) (+738) → Mg²⁺(g) (+1451)；Cl₂(g) → 2Cl(g) (+244)；2Cl(g) + 2e⁻ → 2Cl⁻(g) (2 × -349 = -698)。这些步骤之和 = 148 + 738 + 1451 + 244 – 698 = +1883 kJ mol⁻¹。由ΔHf = Σ + LE，得 -641 = 1883 + LE → LE = -641 – 1883 = -2524 kJ mol⁻¹。晶格焓为-2524 kJ mol⁻¹，表明过程高度放热且离子键极强。

3. Buffer Solution pH Calculation | 缓冲溶液pH计算

A buffer is made by mixing ethanoic acid (0.20 mol dm^-3) and sodium ethanoate (0.10 mol dm^-3). Given K_a of ethanoic acid = 1.8 × 10^-5 mol dm^-3, calculate the pH of the buffer.

由0.20 mol dm⁻³的乙酸和0.10 mol dm⁻³的乙酸钠混合制成缓冲溶液。已知乙酸的Kₐ = 1.8 × 10⁻⁵ mol dm⁻³，计算缓冲溶液的pH。

[H⁺] = K_a × [HA] / [A^–]

[H⁺] = 1.8 × 10^-5 × (0.20 / 0.10) = 3.6 × 10^-5 mol dm^-3. pH = –log(3.6 × 10^-5) ≈ 4.44. Alternatively, using the Henderson–Hasselbalch equation: pK_a = –log(1.8 × 10^-5) = 4.74; pH = pK_a + log([A^–]/[HA]) = 4.74 + log(0.10/0.20) = 4.74 – 0.30 = 4.44. The buffer resists pH changes because the weak acid neutralises added base and the conjugate base neutralises added acid.

[H⁺] = 1.8 × 10⁻⁵ × (0.20 / 0.10) = 3.6 × 10⁻⁵ mol dm⁻³，pH = –log(3.6 × 10⁻⁵) ≈ 4.44。或用Henderson–Hasselbalch方程：pKₐ = –log(1.8 × 10⁻⁵) = 4.74；pH = pKₐ + log([A⁻]/[HA]) = 4.74 + log(0.10/0.20) = 4.74 – 0.30 = 4.44。该缓冲液因弱酸中和加入的碱、共轭碱中和加入的酸而具有抵抗pH变化的能力。

4. Redox Titration: Manganate(VII) and Iron(II) | 氧化还原滴定：高锰酸根与铁(II)

A 1.50 g sample of iron ore is dissolved in sulfuric acid and made up to 250 cm³. A 25.0 cm³ aliquot required 24.50 cm³ of 0.0200 mol dm^-3 KMnO₄ for complete reaction. The equation is: MnO₄^– + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Determine the percentage by mass of iron in the ore.

取1.50 g铁矿石溶于硫酸并配制为250 cm³溶液。移取25.0 cm³，用0.0200 mol dm⁻³ KMnO₄滴定至终点，消耗24.50 cm³。反应式：MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺。计算矿石中铁的质量分数。

Moles of MnO₄^– used = 0.0200 × 24.50 / 1000 = 4.90 × 10^-4 mol. Moles of Fe²⁺ in 25.0 cm³ = 5 × 4.90 × 10^-4 = 2.45 × 10^-3 mol. Moles of Fe²⁺ in 250 cm³ = 2.45 × 10^-3 × (250/25.0) = 2.45 × 10^-2 mol. Mass of Fe = 2.45 × 10^-2 × 55.85 = 1.368 g. Percentage by mass = (1.368 / 1.50) × 100% ≈ 91.2%. The high percentage indicates a rich iron ore, such as haematite, and the titration end point is marked by the first permanent pink colour of excess MnO₄^–.

所用MnO₄⁻的物质的量 = 0.0200 × 24.50 / 1000 = 4.90 × 10⁻⁴ mol。25.0 cm³中Fe²⁺的物质的量 = 5 × 4.90 × 10⁻⁴ = 2.45 × 10⁻³ mol。250 cm³中Fe²⁺总量 = 2.45 × 10⁻³ × (250/25.0) = 2.45 × 10⁻² mol。铁的质量 = 2.45 × 10⁻² × 55.85 = 1.368 g。质量分数 = (1.368 / 1.50) × 100% ≈ 91.2%。高含量表明是富铁矿（如赤铁矿）；滴定终点由过量MnO₄⁻的首次持久粉红色指示。

5. Standard Electrode Potentials and Cell EMF | 标准电极电势与电池电动势

A cell is set up as: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Given E°(Zn²⁺/Zn) = –0.76 V and E°(Cu²⁺/Cu) = +0.34 V. Calculate the standard cell EMF, write the overall cell reaction and comment on feasibility.

构建电池：Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)。已知E°(Zn²⁺/Zn) = –0.76 V，E°(Cu²⁺/Cu) = +0.34 V。计算标准电池电动势，写出总反应并评论其可行性。

E°_cell = E°_cathode – E°_anode

The more positive electrode is the cathode (reduction): Cu²⁺ + 2e^– → Cu. The anode (oxidation): Zn → Zn²⁺ + 2e^–. E°_cell = +

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