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

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

Mock unit tests are one of the most effective ways to prepare for SQA Higher Chemistry assessments. They help consolidate knowledge across the mandatory units and familiarise you with the style of multi-step problem solving required for the final exam. In this walkthrough we analyse ten carefully chosen questions that mirror the difficulty and breadth of a typical Year 13 unit test. Each problem is broken down with a full explanation and key revision pointers.

单元模拟卷是备战 SQA Higher 化学考试最有效的方式之一。它能帮助你巩固各必修单元的知识，并熟悉期末考试中常见的多步骤解题套路。本文精选了十道模拟题，覆盖典型单元测试的难度和广度，每题都配有详细解析和关键考点提示。

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1. Polarity, Hydrogen Bonding and Solubility | 极性、氢键与溶解度

Question: Ethanol (C₂H₅OH) is completely miscible with water, whereas pentan-1-ol (C₅H₁₁OH) has a solubility of only 2.7 g per 100 g of water. Explain this difference by referring to the intermolecular forces involved.

题目： 乙醇 (C₂H₅OH) 与水可以任意比例混溶，而 1-戊醇 (C₅H₁₁OH) 在 100 g 水中的溶解度仅为 2.7 g。请用分子间作用力解释这一差异。

Both molecules contain the hydroxyl (–OH) functional group, which allows them to form hydrogen bonds with water molecules. Hydrogen bonding is the strongest type of intermolecular force and dominates the interaction between alcohol and water.

两种分子都含有羟基 (-OH) 官能团，这使得它们能与水分子形成氢键。氢键是分子间作用力中最强的一种，主导着醇与水之间的相互作用。

However, pentan-1-ol has a much longer non-polar hydrocarbon chain than ethanol. The long pentyl chain cannot form hydrogen bonds with water; instead, it disrupts the existing water-water hydrogen bonds and only experiences weak London dispersion forces with water. The large non-polar region increases the energy cost of dissolving and reduces overall solubility.

然而，1-戊醇的非极性碳氢链比乙醇长得多。长戊基链无法与水形成氢键；相反，它会破坏水分子之间原有的氢键网络，自身与水之间仅存在微弱的伦敦色散力。较大的非极性区域增加了溶解所需的能量代价，从而降低了整体溶解度。

In ethanol, the ethyl group is fairly small, so the hydrophilic –OH group can still dominate, keeping the molecule fully soluble.

在乙醇中，乙基较小，亲水的 -OH 基团仍能起主导作用，使得分子完全溶解。

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2. Calculating Enthalpy Change Using Bond Enthalpies | 利用平均键焓计算焓变

The reaction of hydrogen with chlorine forms hydrogen chloride:

氢气与氯气反应生成氯化氢：

H–H + Cl–Cl → 2 H–Cl

Use the average bond enthalpies in the table below to calculate ΔH for this reaction.

利用下表中的平均键焓计算该反应的 ΔH。

Bond	Average bond enthalpy / kJ mol⁻¹
H–H	436
Cl–Cl	243
H–Cl	432

Energy is absorbed to break bonds (endothermic) and released when new bonds form (exothermic). ΔH is the sum of bond enthalpies broken minus the sum of bond enthalpies formed.

断裂化学键需要吸收能量（吸热），生成新化学键则释放能量（放热）。ΔH 等于断裂键的键焓总和减去生成键的键焓总和。

Bonds broken: one H–H (436) and one Cl–Cl (243), total absorbed = 436 + 243 = 679 kJ.

断裂的键：一个 H–H (436) 和一个 Cl–Cl (243)，总吸热量 = 436 + 243 = 679 kJ。

Bonds formed: two H–Cl bonds, each 432, total released = 2 × 432 = 864 kJ.

形成的键：两个 H–Cl 键，每个 432，总放热量 = 2 × 432 = 864 kJ。

Applying the equation: ΔH = Σ(bond enthalpies broken) – Σ(bond enthalpies formed) = 679 – 864 = –185 kJ mol⁻¹. The negative sign indicates an exothermic reaction.

代入公式：ΔH = 断裂键焓之和 – 生成键焓之和 = 679 – 864 = –185 kJ mol⁻¹。负值表示该反应为放热反应。

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3. Hess's Law Energy Cycle – Using Enthalpies of Combustion | 赫斯定律能量循环 – 利用燃烧焓

Given the following standard enthalpies of combustion:

已知以下标准燃烧焓：

C(s) + O₂(g) → CO₂(g) ΔH_c = –394 kJ mol⁻¹

H₂(g) + ½O₂(g) → H₂O(l) ΔH_c = –286 kJ mol⁻¹

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH_c = –890 kJ mol⁻¹

Calculate the standard enthalpy of formation of methane, ΔH_f of CH₄(g).

请计算甲烷的标准生成焓 ΔH_f (CH₄(g))。

gy cycle using combustion data: the total enthalpy change when reactants are burned and when product is burned to the same combustion products.

生成反应为：C(s) + 2H₂(g) → CH₄(g)。我们可以利用燃烧数据构建能量循环：反应物完全燃烧的总焓变与产物燃烧至相同燃烧产物时的总焓变。

By Hess's law: ΔH_f = ΣΔH_c(reactants) – ΣΔH_c(products).

根据赫斯定律：ΔH_f = 反应物的燃烧焓之和 – 产物的燃烧焓之和。

Burning the elements: C(s) gives –394; 2H₂(g) gives 2 × (–286) = –572. Total = –394 – 572 = –966 kJ mol⁻¹. Burning the product CH₄ gives –890 kJ mol⁻¹.

元素燃烧：C(s) 产生 –394；2H₂(g) 产生 2 × (–286) = –572。总和 = –394 – 572 = –966 kJ mol⁻¹。产物 CH₄ 燃烧产生 –890 kJ mol⁻¹。

ΔH_f = (–966) – (–890) = –76 kJ mol⁻¹. This is the accepted value for the enthalpy of formation of methane.

ΔH_f = (–966) – (–890) = –76 kJ mol⁻¹。这正是公认的甲烷生成焓。

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4. Equilibrium Constant Kc and the Effect of Temperature | 平衡常数 Kc 与温度的影响

The reaction H₂(g) + I₂(g) ⇌ ations were found to be: [H₂] = 0.086 mol l⁻¹, [I₂] = 0.012 mol l⁻¹, [HI] = 0.064 mol l⁻¹.

在密封容器中，反应 H₂(g) + I₂(g) ⇌ 2HI(g) 于 700 K 达到平衡。测得平衡浓度为：[H₂] = 0.086 mol l⁻¹，[I₂] = 0.012 mol l⁻¹，[HI] = 0.064 mol l⁻¹。

Write the expression for Kc and calculate its value at this temperature.

写出 Kc 的表达式，并计算在该温度下的数值。

Kc = [HI]² / ([H₂][I₂])

es: Kc = (0.064)² / (0.086 × 0.012) = 0.004096 / 0.001032 ≈ 3.97 (units cancel, no units).

代入数值：Kc = (0.064)² / (0.086 × 0.012) = 0.004096 / 0.001032 ≈ 3.97（单位相消，无单位）。

The forward reaction is exothermic. Predict what happens to the value of Kc if the temperature is increased to 800 K. Explain your answer.

正向反应为放热反应。请预测如果将温度升至 800 K，Kc 的值将如何变化，并解释原因。

For an exothermic reaction, increasing temperature shifts the equilibrium to the left (endothermic direction) to absorb the added heat. This decreases the concentration of HI and increases H₂ and I₂, so the value of Kc decreases.

对于放热反应，升高温度会使平衡向左（吸热方向）移动，以吸收新增的热量。这会导致 HI 浓度降低而 H₂ 和 I₂ 浓度升高，因此 Kc 值减小。

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5. Effect of Temperature on Reaction Rate – Maxwell-Boltzmann Distribution | 温度对反应速率的影响 – 麦克斯韦-玻尔兹曼分布

A sample of gas is heated from 300 K to 350 K. Sketch the Maxwell-Boltzmann distribution of molecular kinetic energies at both temperatures on the same axes, and explain why the rate of reaction increases significantly, even though the average energy increase is modest.

某气体样品从 300 K 加热至 350 K。在同一坐标图中画出两种温度下分子动能的麦克斯韦-玻尔兹曼分布曲线，并解释为何尽管平均能量增幅有限，反应速率却显著提高。

At higher temperature, the distribution flattens and shifts to the right, with the peak moving to a slightly higher energy and becoming lower in height. The crucial feature is that the area under the curve beyond the activation energy (Eₐ) threshold increases dramatically.

温度升高时，分布曲线变得扁平并右移，峰值略微移向更高能量且高度降低。关键特征是曲线超出活化能 (Eₐ) 阈值的面积大幅增加。

react, the number of successful collisions rises sharply even if the average kinetic energy rises only a little. This is because the high-energy 'tail' of the distribution is highly sensitive to temperature.

因为只有动能 E ≥ Eₐ 的分子才能发生反应，即使平均动能仅小幅上升，能够发生有效碰撞的分子数量也会急剧增加。这是由于分布曲线的高能“尾部”对温度极为敏感。

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6. Organic Nomenclature and Constitutional Isomerism | 有机命名与构造异构

Draw and name all the constitutionally isomeric alcohols and ethers with the molecular formula C₄H₁₀O.

画出分子式为 C₄H₁₀O 的所有构造异构的醇与醚，并进行命名。

There are four alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol. The three ethers are: methoxypropane (methyl propyl ether), ethoxyethane (diethyl ether) and 2-methoxypropane (methyl isopropyl ether).

共有四种醇：丁-1-醇、丁-2-醇、2-甲基丙-1-醇和 2-甲基丙-2-醇。三种醚：甲氧基丙烷（甲基丙基醚）、乙氧基乙烷（二乙醚）和 2-甲氧基丙烷（甲基异丙基醚）。

For systematic naming, identify the longest carbon chain and priority: –OH takes precedence in numbering for alcohols. In ethers, the larger alkyl group forms the root, and the other is named as an alkoxy substituent.

系统命名时，需确定最长碳链及优先级：醇的编号以 -OH 位次最低为准。在醚中，较大的烷基作为母体，较小的作为烷氧基取代基。

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7. Nucleophilic Substitution – Haloalkane Hydrolysis | 亲核取代 – 卤代烷水解

Bromoethane reacts with aqueous sodium hydroxide under reflux to form ethanol. Write the overall equation, describe the mechanism (including the role of the nucleophile) and state the condition required.

溴乙烷与氢氧化钠水溶液在回流条件下反应生成乙醇。写出总反应方程式，描述反应机理（包括亲核试剂的作用），并指出所需条件。

CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻

This is a nucleophilic substitution. The hydroxide ion, OH⁻, acts as the nucleophile because it has a lone pair of electrons and is attracted to the electron-deficient carbon atom bonded to bromine. The carbon is δ+ due to the electronegativity difference.

这是一个亲核取代反应。氢氧根离子 OH⁻ 作为亲核试剂，因为它具有孤对电子，会被与溴相连的缺电子碳原子吸引。由于电负性差异，该碳原子带 δ+。

The mechanism involves the nucleophile attacking the carbon from the side opposite the leaving group, forming a new C–O bond while the C–Br bond breaks heterolytically, releasing Br⁻.

反应机理为亲核试剂从离去基团的反侧进攻碳原子，形成新的 C–O 键，同时 C–Br 键异裂，释放出 Br⁻。

The reaction requires aqueous sodium hydroxide and heating under reflux to ensure complete conversion.

该反应需要氢氧化钠水溶液并加热回流，以确保转化完全。

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8. Redox Titration – Potassium Manganate(VII) with Iron(II) | 氧化还原滴定 – 高锰酸钾与铁(II)

quired to reach the end-point. The reaction is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

将含有 Fe²⁺ 离子的 25.0 cm³ 溶液酸化，用 0.0200 mol l⁻¹ KMnO₄ 溶液滴定，达到终点时消耗了 21.3 cm³ 高锰酸钾溶液。反应为：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。

Calculate the concentration of Fe²⁺ in the original solution.

计算原溶液中 Fe²⁺ 的浓度。

f MnO₄⁻ used = concentration × volume (in dm³) = 0.0200 × (21.3 / 1000) = 4.26 × 10⁻⁴ mol.

所用 MnO₄⁻ 的物质的量 = 浓度 × 体积 (dm³) = 0.0200 × (21.3 / 1000) = 4.26 × 10⁻⁴ mol。

on, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺. Moles of Fe²⁺ = 5 × 4.26 × 10⁻⁴ = 2.13 × 10⁻³ mol.

由配平方程式，1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应。故 Fe²⁺ 的物质的量 = 5 × 4.26 × 10⁻⁴ = 2.13 × 10⁻³ mol。

Concentration of Fe²⁺ = moles / volume = 2.13 × 10⁻³ / (25.0 / 1000) = 0.0852 mol l⁻¹.

Fe²⁺ 浓度 = 物质的量 / 体积 = 2.13 × 10⁻³ / (25.0 / 1000) = 0.0852 mol l⁻¹。

Always ensure the titration mixture is well acidified with dilute sulfuric acid; otherwise, MnO₂ may form, causing a brown precipitate and invalidating the result.

务必确保滴定混合液用稀硫酸充分酸化，否则可能生成 MnO₂ 棕色沉淀，导致结果无效。

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9. Percentage Yield in Esterification | 酯化反应的产率计算

Ethyl ethanoate was prepared by reacting 12.0 g of ethanoic acid with excess ethanol, using a sulfuric acid catalyst. After purification, 8.50 g of pure ester was obtained. Calculate the percentage yield.

用 12.0 g 乙酸与过量乙醇在硫酸催化下制备乙酸乙酯。纯化后得到纯酯 8.50 g。请计算产率。

The equation is: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O.

反应方程式：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。

ic acid = (2×12) + (4×1) + (2×16) = 60 g mol⁻¹. Moles of acid used = 12.0 / 60 = 0.20 mol.

乙酸的摩尔质量 = (2×12) + (4×1) + (2×16) = 60 g mol⁻¹。所用乙酸的物质的量 = 12.0 / 60 = 0.20 mol。

l ethanoate = (4×12) + (8×1) + (2×16) = 88 g mol⁻¹. Theoretical mass = 0.20 × 88 = 17.6 g.

根据 1:1 计量比，理论生成的酯的物质的量 = 0.20 mol。乙酸乙酯摩尔质量 = (4×12) + (8×1) + (2×16) = 88 g mol⁻¹。理论质量 = 0.20 × 88 = 17.6 g。

Percentage yield = (actual mass / theoretical mass) × 100 = (8.50 / 17.6) × 100 ≈ 48.3%.

产率 = (实际质量 / 理论质量) × 100 = (8.50 / 17.6) × 100 ≈ 48.3%。

The yield is less than 100% mainly because the esterification is reversible and reaches equilibrium. Transfer losses during purification also contribute.

产率低于 100% 主要是因为酯化反应是可逆的，会达到平衡；纯化过程中的转移损失也会降低收率。

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10. Thin-Layer Chromatography and Rf Values | 薄层色谱与 Rf 值

Three amino acid samples (A, B, C) along with two known standards were spotted on a TLC plate and developed. The solvent front moved 8.0 cm from the baseline. The spot distances travelled from the baseline are: standard valine 4.2 cm, standard glycine 2.0 cm, sample A 4.2 cm, sample B 2.0 cm, sample C 5.6 cm. Identify the amino acids present in the samples and calculate the Rf value of sample C.

将三种氨基酸样品 (A, B, C) 与两种已知标样点在 TLC 板上并展开。溶剂前沿距基线 8.0 cm。各斑点距基线的距离为：标样缬氨酸 4.2 cm，标样甘氨酸 2.0 cm，样品 A 4.2 cm，样品 B 2.0 cm，样品 C 5.6 cm。请确定样品中含有哪些氨基酸，并计算样品 C 的 Rf 值。

/ distance travelled by solvent front. By comparison, sample A has the same distance as valine, so it contains valine. Sample B matches glycine.

Rf = 组分移动距离 / 溶剂前沿距离。通过比较，样品 A 的距离与缬氨酸一致，故含有缬氨酸；样品 B 与甘氨酸匹配。

Rf of sample C =

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