SAT II Chemistry: Key Difficult Sample Questions Analysis | SAT II 化学考前重难点例题解析

📚 SAT II Chemistry: Key Difficult Sample Questions Analysis | SAT II 化学考前重难点例题解析

SAT II Chemistry tests a wide range of topics with tricky twists on fundamental concepts. This article walks through ten carefully selected sample questions that cover the most challenging areas—stoichiometry, gas deviations, transition-metal electron configurations, hybridization, thermochemistry, kinetics, equilibrium, weak acid/base pH, electrochemistry, and organic functional groups. Each question is followed by step-by-step bilingual explanations to sharpen your problem-solving skills.

SAT II 化学考试知识点覆盖面广，常在最基础的概念上设置陷阱。本文精选十道高频重难点例题，涵盖化学计量、气体偏离、过渡金属电子排布、杂化、热化学、动力学、平衡、弱酸弱碱 pH、电化学和有机官能团识别。每道题配有逐步中英文解析，帮助考生精准掌握解题思路。

1. Stoichiometry Limiting Reagent | 化学计量与限量试剂

0.0 g of C₂H₆ (molar mass 30.0 g·mol⁻¹) react with 50.0 g of O₂ (molar mass 32.0 g·mol⁻¹), what mass of CO₂ is produced?

例题：已知乙烷燃烧反应 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O。若 20.0 g C₂H₆（摩尔质量 30.0 g·mol⁻¹）与 50.0 g O₂（摩尔质量 32.0 g·mol⁻¹）反应，生成多少克 CO₂？

Step 1 — es. n(C₂H₆) = 20.0 g / 30.0 g·mol⁻¹ = 0.667 mol. n(O₂) = 50.0 g / 32.0 g·mol⁻¹ = 1.56 mol.

步骤 1 — 质量转化为摩尔。n(C₂H₆) = 20.0 g / 30.0 g·mol⁻¹ = 0.667 mol。n(O₂) = 50.0 g / 32.0 g·mol⁻¹ = 1.56 mol。

Step 2 — 667 mol C₂H₆ = 0.667 × (7/2) = 2.33 mol. Available O₂ is only 1.56 mol, so O₂ is limiting.

步骤 2 — 判断限量试剂。根据反应式，2 mol C₂H₆ 需 7 mol O₂。0.667 mol C₂H₆ 需 O₂ 0.667 × (7/2) = 2.33 mol。现有 O₂ 仅 1.56 mol，故 O₂ 是限量试剂。

Step 3 — of CO₂ = 1.56 mol O₂ × (4/7) = 0.891 mol. Mass of CO₂ = 0.891 mol × 44.0 g·mol⁻¹ = 39.2 g.

步骤 3 — 根据限量试剂求 CO₂。根据反应式，7 mol O₂ 生成 4 mol CO₂。CO₂ 摩尔数 = 1.56 mol × (4/7) = 0.891 mol。质量 CO₂ = 0.891 mol × 44.0 g·mol⁻¹ = 39.2 g。

Common pitfall: Students often take C₂H₆ as the limiting reagent because its mass looks smaller, but moles must be compared using stoichiometric ratios.

常见易错点：考生常因 C₂H₆ 质量看起来更小而认为它是限量试剂，但必须用摩尔数并按化学计量比比较。

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2. Ideal Gas Law Deviations | 理想气体定律偏离

Question: Which gas would show the largest deviation from ideal behavior at high pressure and low temperature: He, N₂, CO₂, or CH₄?

例题：下列哪种气体在高压低温下偏离理想气体行为最明显：He、N₂、CO₂ 还是 CH₄？

ally London dispersion forces and dipole-dipole interactions) are strong, and particles are large. CO₂ has the strongest London forces among these because of its higher molar mass and linear shape with more electrons; it also has a small dipole moment on each C=O bond, but the molecule is nonpolar overall. Still, its larger electron cloud increases polarizability and thus attractive forces. Additionally, CO₂ has the largest molecular size, making volume exclusion significant. Consequently, CO₂ deviates most.

理想气体行为假设分子间无作用力且粒子体积可忽略。实际气体在分子间作用力（尤其是伦敦色散力和偶极‑偶极作用）较强、粒子体积较大时偏离最明显。在这些气体中，CO₂ 的摩尔质量最大、电子数最多、极化率最高，因此伦敦力最强；虽整体无极性，但 C=O 键矩仍有一定影响。同时 CO₂ 分子体积最大，排斥体积效应也较显著，故偏离最大。

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3. Electron Configuration of Transition Metals | 过渡金属电子排布

Question: What is the electron configuration of Cr³⁺? (Atomic number Cr = 24)

例题：Cr³⁺ 的电子排布是什么？（Cr 原子序数 = 24）

Neutral Cr has configuration [Ar] 4s¹ 3d⁵ (exceptional stability of half‑filled d sublevel). When forming Cr³⁺, electrons are removed first from the highest energy level, which is 4s, then from 3d. Thus, remove the 4s¹ electron and two 3d electrons: [Ar] 3d³. Many students mistakenly write [Ar] 4s² 3d¹, forgetting that transition metals lose ns electrons before (n−1)d electrons. The correct answer is [Ar] 3d³.

Cr 原子的电子排布为 [Ar] 4s¹ 3d⁵（半充满 d 亚层带来特殊稳定性）。形成 Cr³⁺ 时，电子先从最高能级 4s 失去，再从 3d 失去。因此失去 1 个 4s 电子和 2 个 3d 电子，得到 [Ar] 3d³。常见错误是写成 [Ar] 4s² 3d¹，忘记了过渡金属阳离子先失去 ns 电子再失去 (n−1)d 电子。正确答案是 [Ar] 3d³。

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4. Hybridization and Molecular Geometry | 杂化与分子几何

Question: What is the hybridization of the central atom and the molecular shape of XeF₄?

例题：XeF₄ 的中心原子杂化类型和分子构型分别是什么？

Xenon has 8 valence electrons. Four are used in Xe– iring sp³d² hybridization. The electron geometry is octahedral. With two lone pairs occupying opposite positions to minimize repulsion, the molecular shape is square planar. Thus, hybridization: sp³d²; shape: square planar.

Xe 有 8 个价电子。4 个用于形成 Xe–F 键，剩下 4 个电子形成两对孤对电子。总电子域 = 4 个成键域 + 2 个孤对域 = 6，因此需 sp³d² 杂化。电子对排布为八面体。两对孤对电子占据对位以减少排斥，分子构型为平面正方形。杂化：sp³d²；构型：平面正方形。

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5. Thermochemical Hess’s Law | 热化学与盖斯定律

Question: Given: (1) 2 CO(g) + O₂(g) → 2 CO₂(g) ΔH° = −566.0 kJ; (2) 2 CO(g) → 2 C(s) + O₂(g) ΔH° = +221.0 kJ. Calculate ΔH° for C(s) + O₂(g) → CO₂(g).

例题：已知：(1) 2 CO(g) + O₂(g) → 2 CO₂(g) ΔH° = −566.0 kJ；(2) 2 CO(g) → 2 C(s) + O₂(g) ΔH° = +221.0 kJ。求 C(s) + O₂(g) → CO₂(g) 的 ΔH°。

1): (2 C(s) + O₂(g) → 2 CO(g)) + (2 CO(g) + O₂(g) → 2 CO₂(g)) gives 2 C(s) + 2 O₂(g) → 2 CO₂(g). The sum of ΔH° = −221.0 kJ + (−566.0 kJ) = −787.0 kJ. Divide by 2 to get per mole: C(s) + O₂(g) → CO₂(g) ΔH° = −393.5 kJ.

需通过组合已知反应得到目标反应。将反应 (2) 反转：2 C(s) + O₂(g) → 2 CO(g)，ΔH° = −221.0 kJ。与反应 (1) 相加：2 C(s) + 2 O₂(g) → 2 CO₂(g)，ΔH° = −221.0 kJ + (−566.0 kJ) = −787.0 kJ。除以 2 即得 1 mol 反应：C(s) + O₂(g) → CO₂(g) ΔH° = −393.5 kJ。

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6. Rate Law Determination from Data | 由数据确定速率方程

ree experiments gave: [A]₀ (M): 0.10, 0.20, 0.10; [B]₀ (M): 0.10, 0.10, 0.20; Initial rate (M·s⁻¹): 0.015, 0.060, 0.015. Determine the rate law.

例题：反应 A + B → C 的三组初始浓度和初始速率如下：[A]₀ (M): 0.10, 0.20, 0.10；[B]₀ (M): 0.10, 0.10, 0.20；初速率 (M·s⁻¹): 0.015, 0.060, 0.015。确定速率方程。

ments 1 and 3: [A] constant, [B] doubles (0.10→0.20), rate stays same (0.015→0.015), so order with respect to B is 0. Rate law: rate = k[A]². Using first experiment: 0.015 = k (0.10)² → k = 1.5 M⁻¹·s⁻¹.

对比实验 1 和 2：[B] 不变，[A] 加倍（0.10→0.20），速率变为 4 倍（0.015→0.060），故对 A 为 2 级。对比实验 1 和 3：[A] 不变，[B] 加倍（0.10→0.20），速率不变，故对 B 为 0 级。速率方程：rate = k[A]²。代入实验 1：0.015 = k (0.10)² → k = 1.5 M⁻¹·s⁻¹。

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7. Equilibrium Constant and Reaction Quotient Q | 平衡常数与反应商

Question: For the reaction N₂(g) + 3 H₂(g) ⇌ .0 L flask contains 1.0 mol N₂, 1.0 mol H₂, and 0.50 mol NH₃, is the system at equilibrium? If not, which direction will it shift?

例题：反应 N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) 在 400°C 时 Kc = 0.50。将 1.0 mol N₂、1.0 mol H₂ 和 0.50 mol NH₃ 放入 1.0 L 容器中，体系是否处于平衡？若否，向哪个方向移动？

(0.25) < Kc (0.50), the reaction will shift to the right (toward products) to reach equilibrium.

计算反应商 Qc = [NH₃]² / ([N₂][H₂]³) = (0.50)² / (1.0 × 1.0³) = 0.25。Qc (0.25) < Kc (0.50)，因此反应正向移动（向生成 NH₃ 方向）。

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8. pH of Weak Acid/Base Mixtures | 弱酸弱碱混合物的pH

0⁻⁵) and 0.15 mol of sodium acetate in enough water to make 1.0 L solution. Calculate the pH.

例题：由 0.20 mol 醋酸（Ka = 1.8 × 10⁻⁵）和 0.15 mol 醋酸钠配制 1.0 L 缓冲溶液，求 pH。

Use Henderson– 0.15 M, [HA] = 0.20 M. pH = 4.74 + log(0.15/0.20) = 4.74 + log(0.75) = 4.74 − 0.12 = 4.62.

应用 Henderson–Hasselbalch 方程：pH = pKa + log([A⁻]/[HA])。pKa = −log(1.8 × 10⁻⁵) ≈ 4.74。[A⁻] = 0.15 M，[HA] = 0.20 M。pH = 4.74 + log(0.15/0.20) = 4.74 + log(0.75) = 4.74 − 0.12 = 4.62。

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9. Redox and Electrochemistry | 氧化还原与电化学

ion: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺. How many moles of Fe²⁺ can be oxidized by 1 mole of MnO₄⁻?

例题：在酸性溶液中配平反应：MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺。1 mol MnO₄⁻ 能氧化多少摩尔的 Fe²⁺？

Add: MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺. Thus, 1 mol MnO₄⁻ oxidizes 5 mol Fe²⁺.

半反应：还原：MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O。氧化：Fe²⁺ → Fe³⁺ + e⁻。氧化半反应用 5 乘：5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻。相加得：MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺。因此 1 mol MnO₄⁻ 可氧化 5 mol Fe²⁺。

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10. Organic Functional Group Identification | 有机官能团识别

Question: A compound with molecular formula C₃H₆O gives a positive Tollens’ test and reacts with 2,4-DNPH to form a yellow-orange precipitate. Identify the compound and its functional group.

例题：分子式为 C₃H₆O 的某化合物能使 Tollens 试剂产生银镜，并能与 2,4-二硝基苯肼反应生成黄橙色沉淀。试推断该化合物及官能团。

Positive Tollens’ test indicates an aldehyde (– an aldehyde group eliminates other possibilities like ketone (which gives negative Tollens) or alcohol. The only aldehyde with formula C₃H₆O is propanal, CH₃CH₂CHO. Thus, the compound is propanal, functional group: aldehyde.

Tollens 试验阳性表明含醛基 (–CHO)；与 2,4-二硝基苯肼反应证实有羰基 (C=O)。分子式 C₃H₆O 含醛基只能是丙醛 CH₃CH₂CHO，因为酮不会给出 Tollens 阳性结果，而醇不含羰基。因此化合物为丙醛，官能团为醛基。

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