📚 Year 12 CCEA Chemistry: Case Study Practical Exercises | Year 12 CCEA 化学：案例分析实战演练
This article presents a series of guided case study exercises that mirror the practical and data-analysis components of the Year 12 CCEA Chemistry specification. Each case focuses on a core topic – from volumetric analysis to energetics, kinetics and organic characterisation – and encourages you to apply conceptual understanding to realistic scenarios. Work through the questions methodically, paying close attention to numerical reasoning, sign conventions and the interpretation of experimental data.
本文提供一系列与 Year 12 CCEA 化学课程中实验及数据分析模块相呼应的案例实战练习。每个案例聚焦一个核心主题——从容量分析到能量学、动力学及有机表征——并鼓励你将概念性理解应用于真实情境。请有序地思考问题，特别留意数值推理、符号规范以及对实验数据的解读。
1. Back Titration of an Antacid Tablet | 抗酸片剂的返滴定
An antacid tablet containing calcium carbonate, CaCO₃, is crushed and added to 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid, HCl. After the reaction subsides, the excess acid is titrated with 0.500 mol dm⁻³ sodium hydroxide, NaOH, requiring 22.40 cm³ to reach the phenolphthalein endpoint. The relevant equations are:
将一片含碳酸钙 (CaCO₃) 的抗酸片剂碾碎，加入 50.0 cm³ 浓度为 1.00 mol dm⁻³ 的盐酸 (HCl) 中。待反应停止后，用 0.500 mol dm⁻³ 的氢氧化钠 (NaOH) 滴定过量的酸，达到酚酞终点时消耗 22.40 cm³。相关方程式为：
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
HCl + NaOH → NaCl + H₂O
es of NaOH used in back titration: n(NaOH) = 0.02240 dm³ × 0.500 mol dm⁻³ = 0.0112 mol. This equals the moles of excess HCl: n(HCl)ₑₓ = 0.0112 mol. Moles of HCl that reacted with CaCO₃: n(HCl)ᵣₑₐ = 0.0500 – 0388 = 0.0194 mol. Molar mass of CaCO₃ = 100.1 g mol⁻¹, so mass = 0.0194 × 100.1 = 1.94 g.
请计算片剂中碳酸钙的质量。首先计算最初加入的盐酸物质的量：n(HCl)ᵢ = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol。返滴定所用的 NaOH 物质的量：n(NaOH) = 0.02240 dm³ × 0.500 mol dm⁻³ = 0.0112 mol，此即过量 HCl 的物质的量：n(HCl)ₑₓ = 0.0112 mol。与 CaCO₃ 反应的 HCl 物质的量：n(HCl)ᵣₑₐ = 0.0500 – 0.0112 = 0.0388 mol。根据 1:2 的化学计量比，n(CaCO₃) = ½ × 0.0388 = 0.0194 mol。CaCO₃ 的摩尔质量为 100.1 g mol⁻¹，故质量 = 0.0194 × 100.1 = 1.94 g。
2. Enthalpy Change of Neutralisation by Calorimetry | 量热法测定中和焓变
. The maximum temperature recorded after mixing is 27.8 °C. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹ and the density is 1.00 g cm⁻³. Determine the enthalpy change of neutralisation, ΔH, giving your answer in kJ mol⁻¹. The total volume is 100.0 cm³, so mass = 100.0 g. Temperature change ΔT = 27.8 – oncentrations and volumes, n(H₂O) = n(HCl) = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol. ΔH = –q / n = –2.63 kJ / 0.0500 mol = –52.6 kJ mol⁻¹ (the negative sign indicates an exothermic reaction).
在一个简易的咖啡杯量热计中，将 50.0 cm³ 的 1.00 mol dm⁻³ HCl 与 50.0 cm³ 的 1.00 mol dm⁻³ NaOH 混合。两种溶液初始温度均为 21.5 °C，混合后记录的最高温度为 27.8 °C。假设溶液的比热容为 4.18 J g⁻¹ K⁻¹，密度为 1.00 g cm⁻³。求中和反应的焓变 ΔH，以 kJ mol⁻¹ 为单位。总体积为 100.0 cm³，故质量 m = 100.0 g。温度变化 ΔT = 27.8 – 21.5 = 6.3 °C（即 6.3 K）。释放的热量 q = mcΔT = 100.0 × 4.18 × 6.3 = 2633.4 J ≈ 2.63 kJ。生成水的物质的量：由等浓度和体积知，n(H₂O) = n(HCl) = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol。ΔH = –q / n = –2.63 kJ / 0.0500 mol = –52.6 kJ mol⁻¹（负号表示放热反应）。
3. Rate of Reaction and the Iodine Clock | 反应速率与碘时钟
The iodine clock reaction between potassium iodate(V), KIO₃, and sodium hydrogen sulfite, NaHSO₃, in acidic solution can be monitored by the sudden appearance of a blue-black colour when starch indicator is present. In one experiment, a student measured the time, t, for the colour change at different initial concentrations of iodate(V) ions while keeping all other concentrations constant. The data are shown in the table:
碘酸钾 (KIO₃) 与亚硫酸氢钠 (NaHSO₃) 在酸性溶液中的碘时钟反应，可通过在有淀粉指示剂时蓝黑色突然出现来监测。在一次实验中，一名学生测量了不同碘酸根(V)离子初始浓度下颜色变化所需的时间 t，同时保持其他所有浓度恒定。数据如下表：
| [IO₃⁻] (mol dm⁻³) | time, t (s) | 1/t (s⁻¹) |
|---|---|---|
| 0.010 | 45 | 0.0222 |
| 0.020 | 23 | 0.0435 |
| 0.030 | 15 | 0.0667 |
| 0.040 | 11 | 0.0909 |
ight line passing through the origin, indicating that the rate is directly proportional to the concentration. Therefore the reaction is first order with respect to iodate(V) ions. Write the rate equation as: Rate = k [IO₃⁻]¹ [HSO₃⁻]ᵐ, where m would be determined by an analogous experiment varying [HSO₃⁻] at constant [IO₃⁻].
推求该反应相对于 IO₃⁻ 的反应级数。在此条件下反应速率正比于 1/t。若以 1/t 对 [IO₃⁻] 作图，得到一条通过原点的直线，表明速率与浓度成正比，因此反应相对于碘酸根(V)离子为一级反应。速率方程可写为：速率 = k [IO₃⁻]¹ [HSO₃⁻]ᵐ，其中 m 可通过在恒定 [IO₃⁻] 下改变 [HSO₃⁻] 的类似实验确定。
4. Determination of an Equilibrium Constant | 平衡常数的测定
The esterification of ethanol and ethanoic acid reaches equilibrium: CH₃COOH + C₂H₅OH ⇌ lowing equilibrium to establish at a fixed temperature, titration shows that 0.33 mol of ethanoic acid remain. Calculate Kc. Initial moles: acid = 1.00, alcohol = 1.00, ester = 0, water = 0. Change: –x on reactants, +x on products, where x = 1.00 – ester][water] / [acid][alcohol] = (0.67)(0.67) / (0.33)(0.33) = (0.67/0.33)² = (2.03)² = 4.12 (no units as Δn = 0).
乙醇与乙酸的酯化反应达到平衡：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。在一个密封烧瓶中，将 1.00 mol 乙酸与 1.00 mol 乙醇混合，并加入少量硫酸催化剂。在固定温度下达到平衡后，滴定显示剩余 0.33 mol 乙酸。计算 Kc。初始物质的量：酸 = 1.00，醇 = 1.00，酯 = 0，水 = 0。变化：反应物减少 x，产物增加 x，其中 x = 1.00 – 0.33 = 0.67 mol。平衡时物质的量：酸 = 0.33，醇 = 0.33，酯 = 0.67，水 = 0.67。因总体积 V 在 Kc 表达式中被约去，Kc = [酯][水] / [酸][醇] = (0.67)(0.67) / (0.33)(0.33) = (0.67/0.33)² = (2.03)² = 4.12（无单位，因 Δn = 0）。
5. Construction and EMF of an Electrochemical Cell | 电化学电池的构建与电动势
e half-cell contains a copper rod dipping into 1.0 mol dm⁻³ CuSO₄ at 298 K. The voltmeter reads +0.34 V. Write the conventional cell representation and calculate the standard electrode potential of the Cu²⁺/Cu couple. Conventional representation: Pt | H₂ (g) | H⁺ (aq) || Cu²⁺ (aq) | Cu (s). The cell EMF, E°cell = E°(right) – duction potential than hydrogen. The spontaneous overall reaction is Cu²⁺ + H₂ → Cu + 2H⁺.
通过盐桥将标准氢电极（左侧）与 Cu²⁺/Cu 半电池（右侧）连接构成一个电池。半电池中含有一根铜棒浸在 1.0 mol dm⁻³ CuSO₄ 中，温度为 298 K。电压表读数为 +0.34 V。写出该电池的常规符号表示，并计算 Cu²⁺/Cu 电对的标准电极电势。常规符号表示：Pt | H₂ (g) | H⁺ (aq) || Cu²⁺ (aq) | Cu (s)。电池电动势 E°cell = E°(右) – E°(左)。已知 E°(左) = 0.00 V (SHE)，故 E°(Cu²⁺/Cu) = +0.34 V。正电动势证实 Cu²⁺/Cu 半电池具有比氢更正的还原电势。总的自发反应为 Cu²⁺ + H₂ → Cu + 2H⁺。
6. Percentage Yield and Atom Economy in Organic Synthesis | 有机合成中的产率与原子经济性
2-bromopropane is prepared from propan-2-ol via reaction with sodium bromide and 50% bromopropane (Mr = 122.9). Calculate the percentage yield. Moles of propan-2-ol = 5.00 / 60.0 = 0.0833 mol. Theoretical moles of 2-bromopropane = 0.0833 mol (1:1 stoichiometry). Theoretical mass = 0.0833 × 122.9 = 10.24 g. Percentage yield = (6.80 / 10.24) × 100 = 66.4% n-2-ol) + 102.9 (NaBr) + 98.1 (H₂SO₄) = 261.0. Atom economy = (122.9 / 261.0) × 100 = 47.1% w atom economy is due to the formation of sodium hydrogensulfate and water as co-products.
2-溴丙烷由丙-2-醇与溴化钠及 50% 硫酸反应制备：(CH₃)₂CHOH + NaBr + H₂SO₄ → (CH₃)₂CHBr + NaHSO₄ + H₂O。在一次实验室制备中，5.00 g 丙-2-醇（Mr = 60.0）得到 6.80 g 2-溴丙烷（Mr = 122.9）。计算产率。丙-2-醇的物质的量 = 5.00 / 60.0 = 0.0833 mol。理论上 2-溴丙烷的物质的量 = 0.0833 mol（1:1 计量比），理论质量 = 0.0833 × 122.9 = 10.24 g。产率 = (6.80 / 10.24) × 100 = 66.4%。原子经济性：目标产物的 Mr = 122.9；所有反应物的 Mr 总和 = 60.0（丙-2-醇）+ 102.9（NaBr）+ 98.1（H₂SO₄）= 261.0。原子经济性 = (122.9 / 261.0) × 100 = 47.1%。这一较低的原子经济性源于生成了共产物硫酸氢钠和水。
7. Interpreting Infrared and Mass Spectra of an Unknown Carbonyl | 未知羰基化合物的红外与质谱解析
gives the following spectrometric data: IR absorption bands at 1715 cm⁻¹ (strong) and 2950– he IR absorption at 1715 cm⁻¹ is characteristic of a carbonyl group (C=O) in a ketone or aldehyde. The absence of a broad O– fragment at 43 suggests the loss of an ethyl radical (C₂H₅, mass 29) from the molecular ion, giving the acylium ion CH₃CO⁺ (m/z = 43). This pattern is typical of butan-2-one, CH₃COCH₂CH₃, where α-cleavage produces CH₃CO⁺. The compound is therefore butan-2-one.
一种未知有机化合物给出以下波谱数据：IR 吸收带在 1715 cm⁻¹（强）和 2950–2850 cm⁻¹（中）；质谱显示分子离子峰 m/z = 72 以及一个 m/z = 43 的突出碎片峰。请利用这些线索推断其结构。1715 cm⁻¹ 处的红外吸收是酮或醛中羰基 (C=O) 的特征峰。没有宽 O–H 带可排除羧酸和醇。分子离子为 72 单位，对应于分子式 C₄H₈O（因 4×12 + 8×1 + 16 = 72）。43 的碎片意味着从分子离子失去乙基自由基 (C₂H₅，质量 29)，生成乙酰正离子 CH₃CO⁺ (m/z = 43)。此碎裂模式是丁-2-酮 CH₃COCH₂CH₃ 的典型特征，其 α-裂解产生 CH₃CO⁺。因此该化合物是丁-2-酮。
8. Redox Titration of Iron(II) with Potassium Manganate(VII) | 用高锰酸钾滴定铁(II)的氧化还原反应
ations are MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻. Determine the concentration of Fe²⁺ in the original solution. Moles of MnO₄⁻ used = 0.02010 dm³ × 0.0200 mol dm⁻³ = 4.02 × 10⁻⁴ mol. From the overall balanced equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺. So n(Fe²⁺) = 5 × 4.02 × 10⁻⁴ = 2.01 × 10⁻³ mol. This amount was in 25.0 cm³, so [Fe²⁺] = 2.01 × 10⁻³ mol / 0.0250 dm³ = 0.0804 mol dm⁻³. This direct titration is stoichiometric and gives precise results if the iron(II) solution is freshly prepared and prevented from aerial oxidation.
取 25.0 cm³ 酸化硫酸亚铁(II)溶液，用 0.0200 mol dm⁻³ KMnO₄ 滴定，终点时消耗 20.10 cm³ 出现持久的淡粉色。半反应式为 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 以及 Fe²⁺ → Fe³⁺ + e⁻。求原始溶液中 Fe²⁺ 的浓度。所用 MnO₄⁻ 的物质的量 = 0.02010 dm³ × 0.0200 mol dm⁻³ = 4.02 × 10⁻⁴ mol。由完整配平方程式：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O 可知，1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应，故 n(Fe²⁺) = 5 × 4.02 × 10⁻⁴ = 2.01 × 10⁻³ mol。此量存于 25.0 cm³ 中，因此 [Fe²⁺] = 2.01 × 10⁻³ mol / 0.0250 dm³ = 0.0804 mol dm⁻³。该直接滴定按化学计量比进行，若硫酸亚铁(II)溶液新鲜配制且防止空气氧化，可获得精确结果。
9. Thermochemical Cycle Using Born-Haber Principles | 运用玻恩-哈伯原理的热化学循环
°ₐₜ = +108; ½Cl₂(g) → Cl(g) ΔH°ₐₜ = +121; Na(g) → Na⁺(g) + e⁻ ΔH°ᵢₑ₁ = +496; Cl(g) + e⁻ → Cl⁻(g) ΔH°ₑₐ = –349; Na(s) + ½Cl₂(g) → NaCl(s) ΔH°f = – ed as: ΔH°f = ΔH°ₐₜ(Na) + ΔH°ₐₜ(Cl) + IE₁(Na) + EA(Cl) + U (lattice energy). Substituting: –411 = +108 + 121 + 496 + (–349) + U. So –411 = 376 + U, thus U = –411 – 376 = – w Hess’s law enables the determination of an energy term that cannot be measured directly.
构建氯化钠 NaCl 的玻恩-哈伯循环，并利用以下数据（单位 kJ mol⁻¹）计算晶格焓：Na(s) → Na(g) ΔH°ₐₜ = +108；½Cl₂(g) → Cl(g) ΔH°ₐₜ = +121；Na(g) → Na⁺(g) + e⁻ ΔH°ᵢₑ₁ = +496；Cl(g) + e⁻ → Cl⁻(g) ΔH°ₑₐ = –349；Na(s) + ½Cl₂(g) → NaCl(s) ΔH°f = –411。该循环可概括为：ΔH°f = ΔH°ₐₜ(Na) + ΔH°ₐₜ(Cl) + IE₁(Na) + EA(Cl) + U（晶格能）。代入得：–411 = +108 + 121 + 496 + (–349) + U。因此 –411 = 376 + U，故 U = –411 – 376 = –787 kJ mol⁻¹。较大的负值反映了离子晶格中强大的静电吸引。这一练习展示了如何通过赫斯定律确定一个无法直接测量的能量项。
10. Thin-Layer Chromatography and Rf Values of Amino Acids | 氨基酸的薄层色谱与 Rf 值
A mixture of three amino acids – glycine, alanine and leucine – taining, three spots appear with the following distances from the origin: spot A 2.1 cm, spot B 3.5 cm, spot C 5.4 cm. The solvent front moved 7.0 cm. Calculate the Rf values and identify the amino acids if the known Rf values under identical conditions are: glycine 0.30, alanine 0.50, leucine 0.77. Rf = distance moved by spot / distance moved by solvent front. For spot A: Rf = 2.1/7.0 = 0.30 (glycine); spot B: 3.5/7.0 = 0.50 (alanine); spot C: 5.4/7.0 = 0.77 (leucine). The order of increasing Rf reflects the increasing non-polar character of the side chains, as the mobile phase is relatively non-polar. This simple chromatographic technique is essential for monitoring the purity and composition of biochemical mixtures.
用硅胶板以丁-1-醇、乙酸和水的混合溶剂对甘氨酸、丙氨酸和亮氨酸的混合物进行薄层色谱 (TLC) 分析。茚三酮显色后出现三个斑点，其距原点的距离分别为：A 点 2.1 cm，B 点 3.5 cm，C 点 5.4 cm。溶剂前沿移动了 7.0 cm。计算 Rf 值，并利用相同条件下已知的 Rf 值（甘氨酸 0.30、丙氨酸 0.50、亮氨酸 0.77）进行鉴别。Rf = 斑点移动距离 / 溶剂前沿移动距离。A 点：Rf = 2.1/7.0 = 0.30（甘氨酸）；B 点：3.5/7.0 = 0.50（丙氨酸）；C 点：5.4/7.0 = 0.77（亮氨酸）。Rf 递增的顺序反映出侧链非极性特征的增强，因为流动相相对非极性。这一简单的色谱技术对监测生化混合物的纯度和组成至关重要。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图