Year 12 Cambridge Chemistry: Interdisciplinary Integrated Question Practice | 跨学科综合题型训练

📚 Year 12 Cambridge Chemistry: Interdisciplinary Integrated Question Practice | 跨学科综合题型训练

Cross-topic questions in Cambridge AS Chemistry require you to connect concepts from different areas, such as energetics, organic chemistry, kinetics, and analytical techniques. This article provides practice with integrated problems that mirror real examination style, helping you to develop flexible thinking and problem-solving skills.

剑桥AS化学的跨学科综合题要求你将不同领域的概念联系起来，例如热能学、有机化学、动力学和分析技术。本文提供模拟真实考试风格的综合题型训练，帮助你培养灵活的思维和解决问题的能力。

1. Integrating Energetics and Organic Chemistry | 热能学与有机化学的综合

A student burns 0.92 g of a liquid alcohol X in a calorimeter. The temperature of 200 g of water rises by 13.2 °C. The specific heat capacity of water is 4.18 J g⁻¹ °C⁻¹. The alcohol has the molecular formula CₙH₂ₙ₊₁OH and contains 60.0% carbon by mass. Determine the enthalpy change of combustion per mole of X, identify X, and write a balanced equation. Also estimate the enthalpy change using the bond energy data (C–C 347, C–H 413, C–O 358, O–H 464, O=O 498, C=O 805 kJ mol⁻¹) and comment on the difference.

一名学生将0.92 g液态醇X在量热计中燃烧。200 g水的温度升高13.2 °C。水的比热容为4.18 J g⁻¹ °C⁻¹。该醇的分子式为CₙH₂ₙ₊₁OH，含碳60.0%（质量分数）。求X的摩尔燃烧焓变，鉴定X，并写出配平的方程式。同时利用键能数据（C–C 347, C–H 413, C–O 358, O–H 464, O=O 498, C=O 805 kJ mol⁻¹）估算焓变，并讨论差异。

Heat released Q = mcΔT = 200 × 4.18 × 13.2 = 11035.2 J ≈ 11.0 kJ.

释放的热量 Q = mcΔT = 200 × 4.18 × 13.2 = 11035.2 J ≈ 11.0 kJ。

Let the molar mass be M. Carbon mass % = 12n / (14n+18) = 0.60 → 12n = 0.6(14n+18) → n=3. So X is C₃H₇OH, M = 60.0 g mol⁻¹.

设摩尔质量为M。碳质量分数 = 12n/(14n+18)=0.60 → n=3。故X为C₃H₇OH，摩尔质量60.0 g mol⁻¹。

Moles of X = 0.92 / 60 = 0.01533 mol. Thus ΔHₐₒₘₒ ≈ –11.0 / 0.01533 = –718 kJ mol⁻¹ (to 3 s.f.).

X的物质的量 = 0.92 / 60 = 0.01533 mol。因此ΔHₐₒₘₒ ≈ –11.0 / 0.01533 = –718 kJ mol⁻¹（三位有效数字）。

Balanced equation: C₃H₇OH(l) + 4½ O₂(g) → 3 CO₂(g) + 4 H₂O(l). Using bond energies: bonds broken (2×C–C, 7×C–H, 1×C–O, 1×O–H, 4.5×O=O), bonds formed (6×C=O, 8×O–H). Estimated ΔH = +6481 – 8636 = –2155 kJ mol⁻¹. This is far more exothermic because bond energies are average values for gaseous species, whereas the experimental value corresponds to liquid alcohol and water, and includes intermolecular forces.

配平方程式：C₃H₇OH(l) + 4½ O₂(g) → 3 CO₂(g) + 4 H₂O(l)。利用键能：断裂的键（2×C–C, 7×C–H, 1×C–O, 1×O–H, 4.5×O=O），形成的键（6×C=O, 8×O–H）。估算ΔH= +6481 – 8636 = –2155 kJ mol⁻¹。该值远大于测量值，因为键能是气态物种的平均值，而实验值对应液态醇和水，且包含了分子间作用力的影响。

2. Electrochemistry and Equilibrium Constants | 电化学与平衡常数

A cell is constructed with Fe³⁺/Fe²⁺ and I₂/I⁻ half-cells. Standard electrode potentials: E°(Fe³⁺/Fe²⁺) = +0.77 V, E°(I₂/I⁻) = +0.54 V. Write the overall cell reaction, calculate the standard cell potential, and use the relation E°cell = (RT / nF) ln K to find the equilibrium constant at 298 K. (F = 96485 C mol⁻¹, R = 8.314 J K⁻¹ mol⁻¹).

一个由Fe³⁺/Fe²⁺和I₂/I⁻半电池构成的电池。标准电极电势：E°(Fe³⁺/Fe²⁺)=+0.77 V，E°(I₂/I⁻)=+0.54 V。写出电池总反应，计算标准电池电势，并利用关系式E°cell = (RT/nF) ln K求298 K时的平衡常数（F=96485 C mol⁻¹，R=8.314 J K⁻¹ mol⁻¹）。

Reduction occurs at the cathode: Fe³⁺ + e⁻ → Fe²⁺. Oxidation: 2I⁻ → I₂ + 2e⁻. Overall: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, n = 2.

阴极还原：Fe³⁺ + e⁻ → Fe²⁺；阳极氧化：2I⁻ → I₂ + 2e⁻。总反应：2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂，n=2。

E°cell = E°cathode – E°anode = 0.77 – 0.54 = +0.23 V.

E°cell = E°阴极 – E°阳极 = 0.77 – 0.54 = +0.23 V。

At 298 K, RT/F ≈ 0.0257 V, so 0.23 = (0.0257/2) ln K → ln K = 17.9 → K ≈ 6.0 × 10⁷. The large K indicates the reaction goes essentially to completion, confirming the positive E°cell makes it thermodynamically feasible.

在298 K，RT/F ≈ 0.0257 V，故0.23=(0.0257/2) ln K → ln K=17.9 → K≈6.0×10⁷。K值很大，表明反应几乎进行完全，印证了正的E°cell使反应在热力学上可行。

3. Organic Synthesis and Percentage Yield | 有机合成与产率计算

2.30 g of ethanol (C₂H₅OH) is reacted with excess phosphorus pentachloride (PCl₅) to produce chloroethane. The equation is: C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl. After purification, 1.95 g of chloroethane is obtained. Calculate the percentage yield and suggest two reasons for any loss.

将2.30 g乙醇（C₂H₅OH）与过量的五氯化磷（PCl₅）反应制备氯乙烷，反应式为：C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl。纯化后得到1.95 g氯乙烷。计算产率并提出两种造成损失的原因。

Molar mass of ethanol = 46.0 g mol⁻¹, so mol = 2.30/46.0 = 0.0500 mol. Theoretical yield of C₂H₅Cl (M = 64.5 g mol⁻¹) = 0.0500 × 64.5 = 3.225 g.

乙醇摩尔质量46.0 g mol⁻¹，物质的量=2.30/46.0=0.0500 mol。氯乙烷（M=64.5 g mol⁻¹）理论产量=0.0500×64.5=3.225 g。

Percentage yield = (1.95 / 3.225) × 100% = 60.5%.

产率 = (1.95/3.225) × 100% = 60.5%。

Possible reasons: incomplete reaction, loss during transfer/purification, or side reactions producing ethene via elimination.

可能原因：反应不完全、转移或纯化过程中的损失、或发生消去反应生成乙烯的副反应。

4. Kinetics and Reaction Mechanisms | 动力学与反应机理

The hydrolysis of the tertiary halogenoalkane 2-bromo-2-methylpropane, (CH₃)₃CBr, is found to be first order with respect to the halogenoalkane and zero order with respect to hydroxide ions. When the initial concentration of (CH₃)₃CBr is 0.10 mol dm⁻³, the initial rate is 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Deduce the rate equation, calculate the rate constant, and propose a mechanism consistent with the kinetics.

三级卤代烷2-溴-2-甲基丙烷（(CH₃)₃CBr）的水解反应对卤代烷为一级，对氢氧根离子为零级。当(CH₃)₃CBr初始浓度为0.10 mol dm⁻³时，初始速率为2.0×10

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