📚 Year 12 SQA Chemistry: Mock Unit Test Walkthrough | 12年级 SQA 化学：单元测试模拟卷解析
This walkthrough takes a full SQA Higher Chemistry mock unit test and works through it question by question, showing the calculation steps, the reasoning and the mark-scheme habits that turn a correct answer into full marks. It covers enthalpy of combustion, rate equations from initial rates, energy distribution, equilibrium, acids and bases, redox, bonding, organic chemistry and volumetric calculations - the areas that appear most often in Unit 1 and Unit 2 assessments.
本文逐题解析一份完整的 SQA 高等化学模拟单元测试，展示计算步骤、推理过程以及把"会做"变成"满分"的评分习惯。内容涵盖燃烧焓、由初始速率推导速率方程、能量分布、平衡、酸碱、氧化还原、化学键、有机化学与滴定计算——这些板块在 Unit 1 与 Unit 2 的测评中出现频率最高。
1. How to Use This Walkthrough | 如何使用这份解析
The mock is written to the standard Higher Chemistry format: 100 marks in 2 hours 30 minutes, split into a multiple-choice section and a written section. Sit the paper first with a timer and no notes, then use this article to mark it. Each question below is followed by the method, the final answer with units, and the mistakes that most commonly cost marks.
这份模拟卷按高等化学标准格式编写：满分 100 分，时长 2 小时 30 分钟，分为选择题部分与书面作答部分。请先用计时器、在无笔记的情况下完成，再用本文批改。下面每道题都配有解题方法、带单位的最终答案，以及最常见的失分点。
A useful habit is to write down the data you are given in symbols before touching the calculator. In SQA papers, roughly half of the lost marks in calculation questions come from using the wrong quantity rather than from arithmetic errors.
一个有用的习惯是：在动笔计算前，先用符号把已知数据整理出来。在 SQA 试卷中，计算题失分大约有一半来自用错了量，而不是算错了数。
2. Question 1: Enthalpy of Combustion | 第1题：燃烧焓
Question: 47.4 °C. Calculate the enthalpy of combustion of ethanol. (Specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹; molar mass of ethanol = 46.0 g mol⁻¹)
题目：某学生用酒精灯燃烧 0.92 g 乙醇 (C₂H₅OH)，加热 100 g 水，水温从 21.0 °C 升至 47.4 °C。计算乙醇的燃烧焓。(水的比热容 = 4.18 J g⁻¹ °C⁻¹；乙醇摩尔质量 = 46.0 g mol⁻¹)
Step 1: Calculate the heat absorbed by the water using the relationship between heat, mass, specific heat capacity and temperature change.
步骤1：计算水吸收的热量，使用热量、质量、比热容与温度变化之间的关系式。
q = m c ΔT
at ΔT is the temperature rise, not the final temperature, and that the answer must be converted to kJ before it is divided by moles.
代入数据：q = 100 g × 4.18 J g⁻¹ °C⁻¹ × (47.4 - 21.0) °C = 100 × 4.18 × 26.4 = 11 035.2 J ≈ 11.04 kJ。注意 ΔT 是温度升高值而不是末温，并且在除以物质的量之前必须把单位换算成 kJ。
Step 2: Calculate the number of moles of ethanol burned.
步骤2：计算燃烧的乙醇的物质的量。
n = mass ÷ M = 0.92 g ÷ 46.0 g mol⁻¹ = 0.0200 mol
Step 3: Calculate the enthalpy of combustion. Combustion releases energy to the surroundings, so the sign of ΔH is negative.
步骤3：计算燃烧焓。燃烧向环境释放能量，因此 ΔH 为负号。
ΔH = -q ÷ n = -11.04 kJ ÷ 0.0200 mol = -552 kJ mol⁻¹
The accepted data-book value is far more negative because heat is lost to the air, to the burner and to the apparatus rather than to the water. In an exam you must credit the calculated value; never replace it with the data-book value.
数据手册中的公认值要比这个负得多，因为部分热量散失到空气、灯体和仪器上，而没有传给水。考试中必须写计算所得的值，绝不能直接替换成数据手册的值。
3. Question 2: Rate Equation from Initial Rates | 第2题：由初始速率推导速率方程
Question: 2H₂O(g) was studied at a fixed temperature. The table shows the initial concentrations and the initial rates. Determine the rate equation and calculate the rate constant, giving its units.
题目：研究反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) 在一定温度下的动力学。下表给出初始浓度与初始速率。推导速率方程并计算速率常数及其单位。
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.5 × 10⁻³ |
| 2 | 0.200 | 0.100 | 1.0 × 10⁻² |
| 3 | 0.100 | 0.200 | 5.0 × 10⁻³ |
Order with respect to NO: ate by four (2.5 × 10⁻³ → 1.0 × 10⁻²), so the reaction is second order with respect to NO.
关于 NO 的级数：比较实验 1 与 2，此时 [H₂] 恒定。[NO] 加倍使速率变为四倍 (2.5 × 10⁻³ → 1.0 × 10⁻²)，所以反应对 NO 为二级。
Order with respect to H₂: les the rate (2.5 × 10⁻³ → 5.0 × 10⁻³), so the reaction is first order with respect to H₂.
关于 H₂ 的级数：比较实验 1 与 3，此时 [NO] 恒定。[H₂] 加倍使速率加倍 (2.5 × 10⁻³ → 5.0 × 10⁻³)，所以反应对 H₂ 为一级。
rate = k [NO]² [H₂]
The overall order is three. Note carefully that these orders cannot be read from the balanced equation: the coefficients 2 and 2 do not give the orders 2 and 1. Orders are experimental quantities and must always be deduced from rate data.
总级数为三。务必注意：这些级数不能从配平方程式中读出。系数 2 和 2 并不给出 2 级和 1 级。级数是实验量，必须从速率数据中推导。
Calculating k: rearrange and substitute the data from experiment 1.
计算 k：整理公式并代入实验 1 的数据。
([NO]²[H₂]) = 2.5 × 10⁻³ ÷ (0.100² × 0.100) = 2.5 × 10⁻³ ÷ 1.00 × 10⁻³ = 2.5 mol⁻² dm⁶ s⁻¹
Units are found by rearranging the rate equation: rate has units mol dm⁻³ s⁻¹ and the concentration term has units (mol dm⁻³)³, so k has units mol⁻² dm⁶ s⁻¹. Repeating the calculation with experiments 2 and 3 gives the same value, which is a quick way to check your work.
单位的求法是把速率方程移项：速率的单位是 mol dm⁻³ s⁻¹，浓度项的单位是 (mol dm⁻³)³，因此 k 的单位是 mol⁻² dm⁶ s⁻¹。用实验 2 和 3 重复计算可得到相同数值，这是快速检验的方法。
4. Question 3: Energy Distribution and Collision Theory | 第3题：能量分布与碰撞理论
Question: Sketch the distribution of molecular energies for a gas at temperature T₁, mark the activation energy, and explain why a small increase in temperature produces a large increase in reaction rate.
题目：画出某气体在温度 T₁ 下的分子能量分布曲线，标出活化能，并解释为什么温度略微升高会使反应速率大幅增加。
The curve starts at the origin, rises steeply to a peak and then tails off to the right without touching the energy axis. The activation energy Eₐ is marked as a vertical line to the right of the peak. Only the molecules in the shaded area to the right of that line have enough energy to react when they collide.
曲线从原点出发，陡峭上升至峰值，再向右侧拖尾延伸，且不与能量轴相交。活化能 Eₐ 以峰值右侧的一条竖直线标出。只有该线右侧阴影区域内的分子，在碰撞时才具有足够的能量发生反应。
Raising the temperature moves the whole curve to the right and flattens the peak, so a much larger fraction of molecules lies beyond the Eₐ line. The rate therefore rises sharply, even though the average energy has only increased slightly.
升高温度会使整条曲线右移、峰值降低，因此位于 Eₐ 线右侧的分子比例大幅增加。所以即便平均能量只略微升高，反应速率也会显著加快。
Two further points earn marks. First, a catalyst lowers the activation energy, so the same curve then has a larger area beyond the new line. Second, rate also depends on collision frequency and on the correct orientation of colliding particles, so not every collision with sufficient energy leads to reaction.
另外两点也能得分。其一，催化剂降低活化能，同一条曲线在新直线右侧的面积因此变大。其二，速率还取决于碰撞频率与碰撞粒子的正确取向，因此并非每一次能量足够的碰撞都能引发反应。
5. Question 4: Chemical Equilibrium and Le Chatelier | 第4题：化学平衡与勒沙特列原理
Question: For the reversible reaction N₂(g) + 3H₂(g) ⇌ xplain the effect on the position of equilibrium of (a) increasing the pressure, (b) increasing the temperature, and (c) adding a catalyst.
题目：对于可逆反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92 kJ mol⁻¹，说明并解释下列变化对平衡位置的影响：(a) 增大压强；(b) 升高温度；(c) 加入催化剂。
(a) Increasing the pressure shifts the equilibrium to the right. There are four moles of gas on the left and two on the right, so the forward reaction reduces the number of gas molecules and relieves the increase in pressure.
(a) 增大压强使平衡右移。左侧有 4 mol 气体，右侧有 2 mol，正反应减少气体分子数，从而缓解压强升高。
(b) Increasing the temperature shifts the equilibrium to the left, because the forward reaction is exothermic. The system absorbs the added heat by favouring the endothermic reverse reaction. Note that a higher temperature also increases the rate of both reactions, but it does not change the equilibrium constant in favour of the products.
(b) 升高温度使平衡左移，因为正反应是放热的。体系通过有利于吸热的逆反应来吸收额外热量。注意：升温同时加快正逆反应速率，但不会使平衡常数向产物方向改变。
(c) A catalyst has no effect on the position of equilibrium. It speeds up the forward and reverse reactions equally, so equilibrium is simply reached more quickly; it does not change the yield at equilibrium, although the industrial process still uses one for economic reasons.
(c) 催化剂对平衡位置没有影响。它同等加快正逆反应，因此只是更快达到平衡；它不改变平衡时的产率，但工业流程出于经济考虑仍会使用催化剂。
Calculating K: at equilibrium [N₂] = 0.50 mol dm⁻³, [H₂] = 1.5 mol dm⁻³ and [NH₃] = 0.30 mol dm⁻³, then:
计算 K：若平衡时 [N₂] = 0.50 mol dm⁻³，[H₂] = 1.5 mol dm⁻³，[NH₃] = 0.30 mol dm⁻³，则：
K = [NH₃]² ÷ ([N₂][H₂]³) = 0.30² ÷ (0.50 × 1.5³) = 0.090 ÷ 1.6875 = 0.053 mol⁻² dm⁶
Only temperature changes the value of K. Pressure and concentration changes move the position of equilibrium along the same K value; that distinction is a very common exam question.
只有温度会改变 K 值。压强和浓度的变化只是让平衡位置在同一個 K 值下移动；这一区别是极常见的考点。
6. Question 5: Acids, Bases and pH | 第5题：酸碱与 pH
Question: Calculate the pH of (a) 0.0100 mol dm⁻³ hydrochloric acid and (b) 0.0500 mol dm⁻³ sodium hydroxide solution at 25 °C.
题目：计算 25 °C 时 (a) 0.0100 mol dm⁻³ 盐酸和 (b) 0.0500 mol dm⁻³ 氢氧化钠溶液的 pH。
loric acid is a strong monobasic acid, so it is fully dissociated: [H⁺] = 0.0100 mol dm⁻³.
盐酸是强一元酸，完全电离：[H⁺] = 0.0100 mol dm⁻³。
pH = -log₁₀[H⁺] = -log₁₀(0.0100) = 2.00
For the alkali, use the ionic product of water, Kw ] = 1 × 10⁻¹⁴ at 25 °C. Sodium hydroxide is fully dissociated, so [OH⁻] = 0.0500 mol dm⁻³.
对于碱，使用水的离子积 Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ (25 °C)。氢氧化钠完全电离，所以 [OH⁻] = 0.0500 mol dm⁻³。
[H⁺] = 1 × 10⁻¹⁴ ÷ 0.0500 = 2.0 × 10⁻¹³ mol dm⁻³, so pH = 12.70
A weak acid such as ethanoic acid is only partially dissociated, so its pH cannot be calculated from the concentration alone; the acid dissociation constant Ka is needed, using Ka cid. Two decimal places are expected in pH answers, and the final answer must be a number without units.
像乙酸这样的弱酸只部分电离，因此仅凭浓度无法算出 pH，需要酸电离常数 Ka；对一元弱酸可用 Ka = [H⁺]² ÷ [酸]。pH 答案应保留两位小数，且最终结果是不带单位的数值。
7. Question 6: Redox and Electrochemical Cells | 第6题：氧化还原与原电池
Question: (a) Write the ion-electron equation for the reduction of manganate(VII) ions in acidic solution. (b) Calculate the cell potential
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