Year 12 AQA Chemistry: Mock Test Walkthrough | AQA 化学单元测试模拟卷解析

📚 Year 12 AQA Chemistry: Mock Test Walkthrough | AQA 化学单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test designed for Year 12 AQA Chemistry. Each section presents a typical exam question followed by a clear, step-by-step analysis in both English and Chinese. The topics cover atomic structure, bonding, calculations, energetics, kinetics, equilibrium, redox, and organic chemistry -- reinforcing key concepts and exam techniques to help you build confidence and avoid common pitfalls.

本文为AQA化学Year 12单元测试模拟卷的详细解析。每个小节给出一道典型试题，并以中英双语逐步分析解答。内容涵盖原子结构、化学键、计算、能量学、动力学、平衡、氧化还原以及有机化学，旨在巩固核心概念和应试技巧，帮助你树立信心，避开常见失分点。

1. Successive Ionisation Energies | 逐级电离能

dentify the group of element X in the Periodic Table and explain your reasoning. (3 marks)

题目：元素X的逐级电离能（单位 kJ mol⁻¹）为：第一电离能590，第二1145，第三4912，第四6474。判断元素X在周期表中所属的族，并解释你的推理。(3分)

Answer guidance: The enormous increase between the 2nd and 3rd ionisation energies shows that the third electron is being removed from a new, inner principle energy level. The first two electrons are removed relatively easily from the outer shell, but the third electron experiences a much greater effective nuclear attraction and less shielding because it is in a shell closer to the nucleus. This pattern is characteristic of an element with two electrons in its outermost shell. Therefore, element X belongs to Group 2.

答案指导：第二与第三电离能之间的剧增表明第三个电子是从一个全新的、内层电子主能级中移除的。前两个电子从外层移除相对容易，但第三个电子受到的有效核引力大得多，且屏蔽作用更弱，因为它处于更靠近原子核的内层。这种模式正是最外层有两个电子的元素的典型特征。因此，元素X属于第2族。

Key phrases for full marks: 'large jump after removal of two electrons', 'third electron from inner shell', 'greater nuclear attraction / less shielding'. The mark scheme rewards linking the jump in energy to a change in principal quantum number and the number of outer electrons.

获得满分的要点：'失去两个电子后出现大跳'、'第三个电子来自内层'、'核引力更大/屏蔽更弱'。评分标准要求将电离能的跃升与主量子数的变化以及外层电子数联系起来。

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2. Shapes of Molecules and Bond Angles | 分子形状与键角

Question: Predict the shape and the bond angle of a BF₃ molecule. Explain your answer with reference to electron pairs around the central atom. (2 marks)

题目：预测BF₃分子的形状和键角，并结合中心原子周围的电子对加以解释。(2分)

Answer: The shape is trigonal planar and the bond angle is 120°. The central boron atom has three bonding pairs of electrons and no lone pairs in the valence shell. The three bonding pairs repel one another equally and position themselves as far apart as possible, giving a planar triangular arrangement with 120° between each B-F bond.

答案：形状为平面三角形，键角120°。中心硼原子的价层有三对成键电子，没有孤对电子。这三对成键电子彼此间以相等斥力尽可能排布得最远，形成平面三角形排布，B-F键之间夹角均为120°。

Extension: If a fluoride ion donates a lone pair to BF₃, the adduct BF₄⁻ is formed. The shape then becomes tetrahedral with a bond angle of approximately 109.5°, because there are four bonding pairs and no lone pairs around the central boron.

延伸：若一个氟离子提供一对孤对电子给BF₃，会生成加合物BF₄⁻。此时形状变为四面体，键角约为109.5°，因为中心硼周围有四对成键电子，没有孤对电子。

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3. Mole Calculations and Empirical Formula | 摩尔计算与经验式

Question: Complete combustion of 0.50 g of a liquid hydrocarbon produced 1.65 g of CO₂ and 0.81 g of H₂O. Calculate the empirical formula of the hydrocarbon. (3 marks)

题目：0.50 g 液态烃完全燃烧，生成1.65 g CO₂ 和 0.81 g H₂O。计算该烃的经验式。(3分)

Mass of C = 0.0375 mol × 12.0 g mol⁻¹ = 0.450 g. Next, moles of hydrogen: moles of H₂O = 0.81 g / 18.0 g mol⁻¹ = 0.045 mol. Each H₂O contains 2 H atoms, so moles of H = 0.045 × 2 = 0.090 mol. Mass of H = 0.090 g. Divide by smallest number of moles (0.0375): C = 1, H = 0.090 / 0.0375 = 2.4. To obtain whole numbers, multiply by 5: C = 5, H = 12. Empirical formula is C₅H₁₂.

分步解答：首先求碳的物质的量：n(CO₂) = 1.65 g ÷ 44.0 g mol⁻¹ = 0.0375 mol。每个CO₂含1个C，n(C) = 0.0375 mol。碳的质量 = 0.0375 × 12.0 = 0.450 g。再求氢：n(H₂O) = 0.81 g ÷ 18.0 g mol⁻¹ = 0.045 mol，每个水分子含2个H，n(H) = 0.090 mol。氢的质量 = 0.090 g。除以最小物质的量(0.0375)，得比例C : H = 1 : 2.4。化为整数乘以5，得C₅H₁₂。经验式为C₅H₁₂。

mately the original sample mass (0.450 + 0.090 = 0.540 g ≈ 0.50 g here; slight combustion errors are acceptable in exam questions).

常见错误：忘记将水的物质的量乘以2以获得氢原子的物质的量。务必检查碳和氢的质量之和是否约等于原始样品质量（此处0.450 + 0.090 = 0.540 g，略大于0.50 g，在考题中轻微燃烧误差是可接受的）。

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4. Hess's Law and Enthalpy Changes | 赫斯定律与焓变

g) + 6H₂O(g). Data: ΔHf° [NH₃(g)] = -46.2 kJ mol⁻¹, ΔHf° [NO(g)] = +90.4 kJ mol⁻¹, ΔHf° [H₂O(g)] = -242 kJ mol⁻¹. (3 marks)

题目：利用以下标准生成焓，计算反应 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) 的标准焓变。数据：f° [NH₃(g)] = -46.2 kJ mol⁻¹, ΔHf° [NO(g)] = +90.4 kJ mol⁻¹, ΔHf° [H₂O(g)] = -242 kJ mol⁻¹。(3分)

Solution: Apply Hess' ) is zero. ΔH° = [4 × (+90.4) + 6 × (-242)] - [4 × (-46.2) + 5 × 0] = [361.6 - 1452] - [-184.8] = -1090.4 + 184.8 = -905.6 kJ mol⁻¹ (for the reaction as written, so units are kJ per mole of reaction). Final answer: -906 kJ mol⁻¹ (to 3 significant figures).

解答：运用赫斯定律：ΔH° = Σ ΔHf°(生成物) - Σ ΔHf°(反应物)。注意任何元素在其标准状态下的标准生成焓为零（此处O₂(g)为0）。 )] - [4 × (-46.2) + 5 × 0] = [361.6 - 1452] - [-184.8] = -1090.4 + 184.8 = -905.6 kJ mol⁻¹ (为所写方程的反应焓变，单位是每摩尔反应)。最终答案：-906 kJ mol⁻¹ (保留三位有效数字)。

Exam tip: Always show the expression clearly and watch signs. A negative value indicates an exothermic reaction, which is expected for combustion-like processes. Ensure the coefficients from the balanced equation are used when multiplying the enthalpy values.

考试技巧：务必清晰写出表达式并注意正负号。负值表示反应放热，这与类燃烧反应相符。确保乘以焓值时使用配平方程中的系数。

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5. Maxwell-Boltzmann Distribution and Catalysts | 麦克斯韦-玻尔兹曼分布与催化剂

Question: Sketch the Maxwell-Boltzmann distribution of molecular energies for a gas at a given temperature. On the same axes, draw the distribution at a higher temperature and indicate how the position of the activation energy (Eₐ) changes when a catalyst is added. Explain why a small increase in temperature leads to a large increase in reaction rate. (4 marks)

题目：绘制某一温度下气体分子能量的麦克斯韦-玻尔兹曼分布曲线。在同一坐标轴上画出更高温度下的分布，并标示出加入催化剂后活化能(Eₐ)位置的移动。解释为什么温度略微升高会导致反应速率大幅增加。(4分)

ly higher and the number of molecules with very high energy increasing significantly. The area under the curve beyond Eₐ represents the fraction of molecules with energy ≥ Eₐ. A catalyst provides an alternative pathway with a lower Eₐ (Eₐ' reatly increases the number of molecules with energy beyond Eₐ because the high-energy tail of the distribution is very sensitive to temperature, leading to a dramatic increase in successful collisions.

答案描述：温度升高时，分布曲线变平并右移，最概然能量略有升高，具有很高能量的分子数显著增加。曲线下超过Eₐ的区域代表能量 ≥ Eₐ 的分子分数。催化剂提供了另一条反应途径，其活化能更低（Eₐ' 位于原始Eₐ的左侧），因此更多比例的分子超过活化能。即使温度只有小幅升高，由于分布曲线的高能末端对温度极为敏感，能量超过Eₐ的分子数也会大幅增加，从而使有效碰撞频率急剧上升。

Mark scheme essentials: curves must start at origin, higher T curve must be broader and lower at the peak, Eₐ line drawn and labelled, and the catalyst Eₐ' shown to the left. Explanation must link increased area beyond Eₐ to more frequent successful collisions.

评分要点：曲线必须从原点出发，较高温度的曲线应更宽、峰更低，标出Eₐ线，并画出催化剂降低后的Eₐ' 位于左侧。解释必须将超过Eₐ的面积增大与更频繁的有效碰撞联系起来。

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6. Equilibrium Constant Kc | 平衡常数Kc

Question: In a closed container at 700 K, 1.00 mol of H₂ and 1.00 mol of I₂ are allowed to reach equilibrium according to H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium, 0.20 mol of H₂ remains. The total volume is 2.0 dm³. Calculate a value for the equilibrium constant Kc and determine its units. (4 marks)

题目：在700 K的密闭容器中，1.00 mol H₂ 与 1.00 mol I₂ 反应建立平衡 H₂(g) + I₂(g) ⇌ 2HI(g)。平衡时剩余0.20 mol H₂。容器总体积为2.0 dm³。计算平衡常数Kc的数值并确定其单位。(4分)

creases by 2 × 0.80 = 1.60 mol. Equilibrium moles: H₂ = 0.20, I₂ = 0.20, HI = 1.60. Equilibrium concentrations (in mol dm⁻³): [H₂] = 0.20 / 2.0 = 0.10, [I₂] = 0.10, [HI] = 1.60 / 2.0 = 0.80. Kc = [HI]² / ([H₂][I₂]) = (0.80)² / (0.10 × 0.10) = 0.64 / 0.01 = 64. The units: (mol dm⁻³)² / (mol dm⁻³)(mol dm⁻³) = no units (the concentration terms cancel). Kc = 64 with no units.

答案：初始物质的量：H₂ 1.00, I₂ 1.00, HI 0。平衡时 H₂ = 0.20 mol，H₂的变化量 = -0.80 mol。因此 I₂ 也减少0.80 mol，HI 增加 2 × 0.80 = 1.60 mol。平衡时的量：H₂ 0.20, I₂ 0.20, HI 1.60。平衡浓度(mol dm⁻³)：[H₂] = 0.20/2.0 = 0.10, [I₂] = 0.10, [HI] = 1.60/2.0 = 0.80。Kc = [HI]²/([H₂][I₂]) = (0.80)²/(0.10×0.10) = 64。单位：(mol dm⁻³)²/(mol dm⁻³)(mol dm⁻³) 相消，无单位。Kc = 64，无单位。

Common error: Using equilibrium moles instead of concentrations. Always divide by volume first. Also note that Kc has no units when the number of moles of gaseous reactants equals the number of moles of gaseous products.

常见错误：直接使用平衡时物质的量而忽略体积。务必先除以体积得到浓度。还需注意，当气体反应物与气体生成物的总物质的量相等时，Kc是无单位的。

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7. Redox Reactions and Oxidation States | 氧化还原反应与氧化数

l⁻(aq) + ClO⁻(aq) + 2H⁺(aq), deduce the oxidation state of chlorine in Cl₂, Cl⁻, and ClO⁻, and explain why this is a disproportionation reaction. Write the half-equation for the reduction process. (3 marks)

题目：对于氯气与水的反应：Cl₂(g) + H₂O(l) → Cl⁻(aq) + ClO⁻(aq) + 2H⁺(aq)，推出Cl₂、Cl⁻ 及 ClO⁻ 中氯的氧化数，并解释为何这是一个歧化反应。写出该反应中还原过程的半方程式。(3分)

portionation means the same element is simultaneously oxidised and reduced. Here, chlorine' (oxidation), so it is a disproportionation. The reduction half-equation: Cl₂ + 2e⁻ → 2Cl⁻.

答案：Cl₂中Cl的氧化数 = 0。Cl⁻中为 -1。ClO⁻中：氧为 -2，离子总电荷 -1，故Cl为 +1 [x + (-2) = -1 => x = +1]。歧化反应指同一元素同时被氧化和还原。该反应中，氯的氧化数由0变为-1（还原），同时由0变为+1（氧化），故为歧化反应。还原半方程式：Cl₂ + 2e⁻ → 2Cl⁻。

Exam tip: Always assign oxidation states using the rules, and clearly state that both oxidation and reduction of the same species occur for disproportionation. Half-equations should balance atoms and charge.

应试技巧：根据规则确定氧化数，并明确说明歧化反应中同一物种既被氧化又被还原。半方程式必须同时配平原子与电荷。

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8. Electrophilic Addition in Alkenes | 烯烃的亲电加成

Question: Propene reacts with hydrogen bromide to form two possible products. Outline the mechanism for the reaction, name the type of mechanism, and explain why 2-bromopropane is the major product. (6 marks)

题目：丙烯与溴化氢反应可生成两种产物。概述该反应的机理，指出机理类型，并解释为何2-溴丙烷是主要产物。(6分)

ive hydrogen, forming a carbocation and a bromide ion. Two carbocations are possible: a secondary carbocation (CH₃-⁺CH-CH₃) from H attaching to the terminal CH₂, and a primary carbocation (⁺CH₂-CH₂-CH₃) from H attaching to the central CH. Step 2: The bromide ion attacks the carbocation to form the final haloalkane. The secondary carbocation is more stable than the primary because alkyl groups donate electron density, stabilising the positive charge. Therefore, 2-bromopropane (from the secondary carbocation) is the major product. Overall reaction: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (major) + CH₃CH₂CH₂Br (minor).

答案：机理为亲电加成。第一步：H-Br键极化；丙烯中富电子的π键进攻略带正电的氢，形成碳正离子和溴离子。可能生成两种碳正离子：二级碳正离子(CH₃-⁺CH-CH₃)由H连接在末端CH₂形成，和一级碳正离子(⁺CH₂-CH₂-CH₃)由H连接在中间的CH形成。第二步：溴离子进攻碳正离子生成最终卤代烷。二级碳正离子比一级更稳定，因为烷基可以供电子以稳定正电荷。因此，2-溴丙烷（来自二级碳正离子）为主要产物。总反应：CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (主产物) + CH₃CH₂CH₂Br (副产物)。

Mechanism details to include: Use curly arrows showing electron pair movement from the π-bond to H, from H-Br bond to Br to make Br⁻, and finally from Br⁻ lone pair to the positive carbon. State Markovnikov's rule: the hydrogen adds to the carbon with more hydrogens to form the more stable carbocation.

需包含的机理细节：用弯箭头表示电子对从π键移向H，从H-Br键移向Br以生成Br⁻，最后从Br⁻的孤对电子移向正碳离子。陈述马氏规则：氢加到含氢较多的碳上，以形成更稳定的碳正离子。

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