📚 CIE Chemistry May 2018 Paper 5 Exam Analysis | CIE 化学：2018年5月实验卷真题解析（试卷5）
Paper 5 of CIE A Level Chemistry is a practical examination that tests planning, analysis and evaluation skills. The May 2018 session presented typical challenges: designing an experiment to investigate reaction kinetics, analysing given data to determine activation energy, and critically assessing experimental procedures. This article breaks down the key questions, highlights common student errors, and demonstrates how to structure high‑scoring answers.
CIE A Level 化学试卷 5 是考查实验设计、数据分析和评估能力的实践考试。2018 年 5 月的试卷包含了典型的挑战：设计实验探究反应动力学、分析给定数据求算活化能、以及批判性地评价实验步骤。本文拆解了核心考题，指出常见失分点，并展示如何构建高分答案。
1. Understanding the Structure of Paper 5 | 试卷 5 的结构解析
Paper 5 lasts 1 hour 15 minutes and carries 30 marks, divided equally between two questions. Question 1 focuses on experiment planning, requiring you to outline a logical method, specify apparatus, control variables, and propose a table for results. Question 2 provides raw data and asks you to process it: perform calculations, draw graphs, identify trends, and evaluate sources of error and procedural limitations.
试卷 5 时长 1 小时 15 分钟，满分 30 分，平均分配于两道题。第一题侧重实验设计，要求你概述合理的操作步骤、指定仪器、控制变量并提出结果记录表。第二题提供原始数据，要求你进行处理：计算、作图、识别趋势，并评估误差来源与操作局限性。
Many candidates lose marks by ignoring precise command words such as ‘describe how you would measure …’ or ‘evaluate the reliability of …’. The May 2018 session particularly rewarded clarity in sequential instructions and thoroughness in evaluation. Time management is critical – spend about 35 minutes on Question 1 and 40 minutes on Question 2, leaving a few minutes to check your answers.
许多考生因忽略精确的指令词（如“描述如何测量……”或“评估……的可靠性”）而丢分。2018 年 5 月考季特别看重操作步骤的清晰性和评估的条理性。时间管理至关重要——第一题用约 35 分钟，第二题用约 40 分钟，留出几分钟检查。
2. Question 1 – Planning: Determining Activation Energy | 第一题——实验设计：测定活化能
In the May 2018 Paper 5, Question 1 asked candidates to plan an experiment to determine the activation energy of the reaction between hydrochloric acid and magnesium ribbon. This is a classic kinetics investigation where the dependent variable is time (or rate) and the independent variable is temperature. You must measure the time taken for a fixed length of magnesium ribbon to react completely at five different temperatures.
在 2018 年 5 月试卷 5 中，第一题要求设计实验测定盐酸与镁条反应的活化能。这是一个经典的动力学探究，因变量是时间（或速率），自变量是温度。你必须测量固定长度的镁条在五个不同温度下完全反应所需的时间。
A clear method would include: (1) Set up a water bath at the desired temperature using a thermostatically controlled heater or a large beaker of water heated on a hot plate, monitored with a thermometer (readable to 0.5 °C). (2) Measure 25.0 cm³ of 1.0 mol dm⁻³ HCl using a volumetric pipette and transfer to a boiling tube. Immerse the boiling tube in the water bath for at least 5 minutes to equilibrate. (3) Meanwhile, clean a 5.0 cm magnesium ribbon with emery paper to remove oxide, and weigh it on a balance reading to 0.01 g if mass must be known; however, for fixed length, accurately cut a 5.0 cm piece with a ruler. (4) Add the magnesium ribbon to the acid, start a stopwatch immediately, and swirl gently. Stop timing when the ribbon fully dissolves. (5) Repeat at four other temperatures (e.g., 20 °C, 30 °C, 40 °C, 50 °C, 60 °C), ensuring the water bath is stabilised at each new temperature before adding Mg.
清晰的方法应包括：(1) 使用恒温加热器或在热板上加热的大烧杯水浴，用温度计（可读至 0.5 °C）监控至所需温度。(2) 用量液管量取 25.0 cm³ 的 1.0 mol dm⁻³ HCl 转移至沸腾管中。将沸腾管浸入水浴至少 5 分钟以达到平衡。(3) 同时，用砂纸打磨 5.0 cm 镁条去除氧化层，若需知质量，可在精度 0.01 g 的天平上称重；但如用固定长度，则用尺子精确剪裁 5.0 cm。(4) 将镁条加入酸中，立即启动秒表，轻轻旋摇。镁条完全溶解时停表。(5) 在其他四个温度下重复（如 20 °C、30 °C、40 °C、50 °C、60 °C），确保每次加镁前水浴已重新稳定。
| Temperature / °C | Time for ribbon to dissolve / s | 1/time / s⁻¹ | Temperature / K | 1/T / K⁻¹ | ln(1/time) (or ln rate) |
|---|---|---|---|---|---|
| 20 | 293 | ||||
| 30 | 303 | ||||
| 40 | 313 | ||||
| 50 | 323 | ||||
| 60 | 333 |
The candidate should state that the rate can be taken as directly proportional to 1/t because the initial amount of Mg is constant. Plot a graph of ln(1/t) (y‑axis) against 1/T (x‑axis). The gradient equals −Eₐ / R, so activation energy Eₐ = −gradient × 8.31 J mol⁻¹ K⁻¹. The candidate must specify using safety goggles and handling hot apparatus with tongs, and controlling variables: concentration and volume of acid, length (and surface cleanliness) of Mg ribbon, and ensuring the same degree of swirling each time.
考生应说明由于初始镁量恒定，速率可视为正比于 1/t。以 ln(1/t) 为 y 轴，1/T 为 x 轴作图。斜率等于 −Eₐ / R，因此活化能 Eₐ = −斜率 × 8.31 J mol⁻¹ K⁻¹。须注明佩戴护目镜、用钳子处理热仪器，并控制变量：酸的浓度体积、镁条长度（及表面清洁度），并确保每次旋摇程度相同。
3. Common Errors in Question 1 | 第一题常见错误
A typical mistake is failing to specify how temperature is controlled and measured. Saying ‘use a water bath at 30 °C’ without describing how to achieve and verify that temperature loses marks. Always mention a thermometer or temperature probe. Another error is measuring the mass of magnesium for each trial instead of using a fixed length, which introduces mass variation; if you weigh, state the balance precision and clean the surface.
一个典型错误是未能说明如何控制与测量温度。说“用 30 °C 水浴”却不描述如何达到并验证该温度会失分。务必提及温度计或温度探头。另一错误是每次试验称量镁的质量而非使用固定长度，这引入了质量差异；如果称重，须说明天平精度并清洁表面。
Many candidates forget to calculate 1/T in kelvin – using Celsius directly yields a non‑linear plot. Also, omitting the derived columns in the results table (1/t, 1/T, ln(1/t)) restricts marks. Ensure columns include units and sensible decimal places. Finally, not linking the gradient to Eₐ mathematically is a severe omission.
许多考生忘记以开尔文计算 1/T——直接使用摄氏温度会得到非线性图。此外，结果表中遗漏衍生栏（1/t、1/T、ln(1/t)）会限制得分。确保各栏包含单位与合理的小数位。最后，没有从数学上将斜率与 Eₐ 联系起来是严重的遗漏。
4. Question 2 – Data Analysis: Reaction of Iodine with Propanone | 第二题——数据分析：碘与丙酮反应
Question 2 in May 2018 presented data from an iodine‑propanone reaction catalysed by acid: CH₃COCH₃ + I₂ → CH₃COCH₂I + HI. Candidates received a table of initial concentrations and initial rates. The task was to determine the orders of reaction with respect to propanone, iodine, and H⁺, write the rate equation, calculate the rate constant, and predict the rate for a new set of concentrations.
2018 年 5 月第二题给出了酸催化碘与丙酮反应的数据：CH₃COCH₃ + I₂ → CH₃COCH₂I + HI。考生得到一份初始浓度与初始速率的数据表。任务包括确定对丙酮、碘和 H⁺ 的反应级数，写出速率方程，计算速率常数，并预测新浓度下的反应速率。
Analysis involves comparing experiments where only one concentration changes. If doubling [propanone] doubles the rate while other concentrations stay constant, the order with respect to propanone is 1. Similarly for [H⁺], the order is 1. Doubling [I₂] whilst keeping propanone and H⁺ constant shows no effect on rate, so the order with respect to iodine is 0. Therefore rate = k [CH₃COCH₃]¹ [I₂]⁰ [H⁺]¹ = k [CH₃COCH₃][H⁺].
分析方法：比较仅一个浓度改变的实验。如果 [丙酮] 加倍时速率加倍而其他浓度不变，则对丙酮为一级。同样，[H⁺] 变化也显示一级。保持丙酮和 H⁺ 不变而加倍 [I₂] 时速率不变，因此对碘为零级。所以速率方程 rate = k [CH₃COCH₃]¹ [I₂]⁰ [H⁺]¹ = k [CH₃COCH₃][H⁺]。
Candidate must then select any experimental run, substitute concentrations and rate into the rate equation to calculate k. Show the unit of k: from rate (mol dm⁻³ s⁻¹) / (mol dm⁻³)(mol dm⁻³) = dm³ mol⁻¹ s⁻¹. For the given data, a typical value was around 4.0 × 10⁻³ dm³ mol⁻¹ s⁻¹. After establishing k, use the rate equation to calculate the missing rate for the final set of concentrations, ensuring substitution is correct and the final answer has appropriate units.
考生随后须任选一组实验，代入浓度与速率求出 k。显示 k 的单位：由 rate (mol dm⁻³ s⁻¹) / (mol dm⁻³)(mol dm⁻³) 得出 dm³ mol⁻¹ s⁻¹。对于所给数据，典型值约为 4.0 × 10⁻³ dm³ mol⁻¹ s⁻¹。确定 k 后，用速率方程计算最后一组浓度的预测速率，确保代入无误且答案单位正确。
5. Graphical Work in Question 2 | 第二题的作图要求
The analysis question often requires a graph to confirm the relationship or to determine a value. In the May 2018 paper, candidates were asked to plot a graph of initial rate against [H⁺] for constant propanone and iodine concentrations. Since the reaction is first order with respect to H⁺, the plot yields a straight line through the origin, with gradient = k [CH₃COCH₃].
分析题常要求作图以验证关系或确定某值。2018 年 5 月试卷要求作初始速率对 [H⁺] 的图，保持丙酮与碘浓度不变。由于对 H⁺ 为一级，该图将是一条过原点的直线，斜率 = k [CH₃COCH₃]。
Drawing the graph demands a sharp pencil, labelled axes with units, sensible scales (avoid multiples of 3 or 7), plotting small dots in circles, and drawing a best‑fit straight line. Marks are allocated for accuracy of points and the quality of the line. Some candidates misplot by confusing the y‑axis (rate) and x‑axis ([H⁺]), or by using incorrect scale intervals.
作图需使用削尖的铅笔，标明轴名与单位，选择合理刻度（避免 3 或 7 的倍数），用小圆点标绘，并画出最佳拟合直线。点的准确性和线的质量均有给分。部分学生因混淆 y 轴（速率）和 x 轴（[H⁺]）或使用错误刻度间距而失分。
When asked to determine the gradient, construct a large triangle that covers at least half the line. Show coordinates used and compute Δy / Δx. The gradient’s unit is (mol dm⁻³ s⁻¹) / (mol dm⁻³) = s⁻¹, which matches k multiplied by concentration units – so careful unit derivation is essential.
当要求计算斜率时，应作一个大三角形覆盖至少一半的线条。显示所用坐标并计算 Δy / Δx。斜率的单位是 (mol dm⁻³ s⁻¹) / (mol dm⁻³) = s⁻¹，这与 k 乘以浓度的量纲匹配——所以仔细推导单位至关重要。
6. Calculating Activation Energy from Given Data | 由给定数据计算活化能
In some variants of the May 2018 Paper 5, a table of rate constant k at different temperatures was given, and candidates had to determine activation energy using the Arrhenius equation: ln k = ln A − Eₐ / (RT). A graph of ln k against 1/T gives a straight line of gradient −Eₐ / R.
在 2018 年 5 月试卷 5 的某些变体中，提供了不同温度下的速率常数 k 表格，要求考生用阿伦尼乌斯方程求活化能：ln k = ln A − Eₐ / (RT)。作 ln k 对 1/T 的图得一直线，斜率为 −Eₐ / R。
It is vital to convert temperature from °C to kelvin by adding 273. Then calculate 1/T and ln k for each temperature, typically to 3 significant figures. The table must be extended with these processed columns. Use the plot to find the gradient, ensuring the triangle’s coordinates are clearly marked. Then Eₐ = (−gradient) × 8.31 J mol⁻¹ K⁻¹. The final answer is usually in the range 50–100 kJ mol⁻¹, so express it in kJ mol⁻¹ by dividing by 1000.
务必把摄氏温度加 273 转换为开尔文。然后计算各温度的 1/T 和 ln k，通常保留三位有效数字。表内须增加这些处理后的数据栏。利用图线求斜率，确保三角形坐标标识清楚。然后 Eₐ = (−斜率) × 8.31 J mol⁻¹ K⁻¹。最终答案一般在 50–100 kJ mol⁻¹ 范围内，故除以 1000 以 kJ mol⁻¹ 表示。
A common slip is using the T in °C to calculate 1/T, which drastically alters the gradient. Another is misreading ln k from the calculator – confirm values are negative for k less than 1. Double‑check the units: gradient has units of K, so Eₐ ends up in J mol⁻¹.
常见失误是用摄氏温度计算 1/T，这会大幅改变斜率。另一点是从计算器中错读 ln k——当 k < 1 时 ln k 应为负值。复查单位：斜率单位是 K，故 Eₐ 最终为 J mol⁻¹。
7. Evaluation of the Experiment | 实验的评估
Evaluation is a significant part of Question 2. You must identify at least two significant sources of error and suggest realistic improvements. For the Mg/HCl experiment, thermal lag – the time for the acid to reach the water bath temperature – could cause the actual reaction temperature to be lower than intended, especially at high target temperatures. Improvement: place a thermometer directly in the acid and start timing only when the acid reaches the set temperature.
评估是第二题的重要部分。你必须至少识别两个主要误差来源并提出切实的改进措施。对于镁/盐酸实验，热滞后——酸液达到水浴温度所需的时间——可能导致实际反应温度低于预设，尤其在目标高温时。改进：将温度计直接插入酸液中，待酸液达到设定温度才开始计时。
Another error is inconsistent loss of hydrogen gas before timing starts due to delay in fitting the bung – relevant if gas collection is used, but in the dissolution method, the visual endpoint of magnesium disappearance is subjective. A student might stop the watch too early or too late. Improvement: use a colorimeter or light sensor to detect the moment the ribbon vanishes, connected to a data logger. Additionally, variation in the thickness or surface oxidation of the magnesium ribbon alters the rate; standardise by using ribbon from the same batch and sanding uniformly.
另一误差是由于塞子盖合延迟导致氢气逸出不均——这适用于气体收集法，但在溶解法中，镁条消失的视觉终点带有主观性。学生可能过早或过晚停表。改进：使用比色计或光传感器检测镁条消失的瞬间，连至数据记录仪。此外，镁条厚度或表面氧化程度的变化会改变速率；可通过使用同一批次镁带并均匀打磨来标准化。
When evaluating graphs, comment on the scatter of points about the best‑fit line. If a point is well off the line, it might be an anomalous result, possibly due to a temperature fluctuation. Suggest repeating that measurement. The evaluation should be specific – avoid vague statements like ‘human error’ or ‘use better equipment’. Instead link cause and effect.
评估图线时，要评论点分布在最佳拟合线周围的离散程度。若某点明显远离直线，它可能是异常结果，或由温度波动引起。建议重测该点。评估需具体——避免如“人为错误”或“使用更好的设备”等模糊表述，应连接原因与结果。
8. Key Skills: Units, Significant Figures, and Calculations | 关键技能：单位、有效数字与计算
Throughout Paper 5, examiners rigorously assess units and significant figures. In the planning question, list units next to each quantity in the results table. In the analysis question, express final answers to the same number of significant figures as the least precise piece of data provided – usually three significant figures. Carry intermediate calculations to at least four significant figures to avoid rounding errors.
试卷 5 通篇严格考查单位与有效数字。在设计题中，务必在结果表中每个量旁边列出单位。分析题中，最终答案的有效数字位数应与所给数据中精度最低者一致——通常为三位有效数字。中间计算至少保留四位有效数字以避免舍入误差。
For rate equations, the rate constant k must have correct units derived from the rate equation. For rate = k[A][B], the units are dm³ mol⁻¹ s⁻¹. If the order is zero for a component, its concentration term is omitted and doesn’t contribute to units. Candidates often write ‘mol⁻¹ dm³ s⁻¹’ which is acceptable, though dm³ mol⁻¹ s⁻¹ is standard. Consistency matters.
对于速率方程，速率常数 k 必须根据速率方程推导出正确单位。对于 rate = k[A][B]，单位为 dm³ mol⁻¹ s⁻¹。若某组分级数为零，其浓度项被省略且不影响单位。考生常写 ‘mol⁻¹ dm³ s⁻¹’，也是可接受的，尽管 dm³ mol⁻¹ s⁻¹ 是标准写法。一致性很重要。
When reading a graph to determine a value, state the coordinates clearly and show the substitution in any formula used. For example: ‘Gradient = (1.20 − 0.40) / (0.00330 − 0.00310) = 0.80 / 0.00020 = 4000 K. Then Eₐ = 4000 × 8.31 = 33240 J mol⁻¹ = 33.2 kJ mol⁻¹.’ This transparency lets examiners follow your logic and award partial marks even if a small slip occurs.
当读图求值时，清楚写出坐标并展示所使用公式的代入过程。例如：“斜率 = (1.20 − 0.40) / (0.00330 − 0.00310) = 0.80 / 0.00020 = 4000 K。然后 Eₐ = 4000 × 8.31 = 33240 J mol⁻¹ = 33.2 kJ mol⁻¹。” 这种透明度让考官可跟随你的逻辑，即使出现小错误也能获得部分分数。
9. Safety and Ethical Considerations | 安全与伦理考量
Marks are available in the planning question for explicitly stating safety precautions. For acid‑magnesium reactions, mention: wear safety goggles and lab coat; avoid skin contact with HCl (irritant); handle hot water and glassware with tongs or heat‑resistant gloves; tie back long hair; ensure the work area is well‑ventilated because hydrogen gas is produced.
在设计题中，明确陈述安全措施可获得分数。对于酸与镁的反应，须提及：佩戴护目镜和实验服；避免 HCl（刺激物）接触皮肤；用钳子或隔热手套处理热水与玻璃器皿；束起长发；确保工作区通风良好，因为会产生氢气。
If the experiment involves toxic substances like iodine, mention that iodine is harmful by inhalation and skin contact, so use a fume cupboard when weighing iodine solid, and wear gloves. Also refer to disposal instructions: neutralise excess acid before washing down the sink. These details demonstrate good laboratory practice and are rewarded.
如果实验涉及有毒物质如碘，须指出碘吸入及接触皮肤有害，故称量碘固体时应使用通风橱并戴手套。还要说明处置指引：在用水冲入水槽前中和过量酸。这些细节展现良好的实验习惯，并获相应分数。
10. Drawing the Required Tables in Question 1 | 第一题中绘制所需表格
The planning question always expects a well‑structured results table. Headings must include the quantity and its unit, separated by a forward slash or brackets, e.g., ‘Temperature / °C’ or ‘Time (s)’. The table should have space for all independent, dependent, and derived variables. For the activation energy experiment, include columns for Temperature / °C, Time / s, 1/time / s⁻¹, Temperature / K, 1/T / K⁻¹, and ln(1/time).
设计题总是要求一张结构严谨的结果记录表。表头必须包含物理量和单位，用斜线或括号分隔，如“温度 / °C”或“时间 (s)”。这张表应为所有自变量、因变量和衍生变量留出空间。对活化能实验，需包含温度 / °C、时间 / s、1/时间 / s⁻¹、温度 / K、1/T / K⁻¹ 以及 ln(1/时间) 等栏。
A tabular format with ruled lines is preferred, though a pencil‑drawn table without lines is accepted if neat. Ensure there are at least five rows for the five temperatures. Additional rows for repeated measurements (if time permits) add rigour, but should be indicated if planned. Explicitly note that at least three repeats are desirable to calculate a mean time and assess precision.
推荐画有线条的表格，尽管干净的无线铅笔表格也可接受。确保至少包含对应五个温度的五行。如果有重复测量（若时间允许），增加额外行可增添严谨性，但应在计划中说明。明确标注至少三次重复以计算平均时间并评估精密度。
11. Understanding the Examiner’s Mark Scheme | 理解考官评分方案
The mark scheme for Paper 5 awards marks for distinct skills: display of logical procedure (P), identification of apparatus (A), data presentation (D), manipulation and calculation (M), graphical work (G), and evaluation (E). In Question 2, marks are given for correct plot (1), suitable line (1), gradient calculation (1), activation energy (1), and evaluation points (2 + improvement). Familiarity with this scheme helps you structure answers to tick all boxes.
试卷 5 的评分方案针对不同技能给分：逻辑步骤展示（P）、仪器识别（A）、数据呈现（D）、处理与计算（M）、作图操作（G）和评估（E）。在第二题中，正确的坐标点（1）、合适的线条（1）、斜率计算（1）、活化能（1）以及评估要点（2 + 改进）均有对应分值。熟悉此方案有助于你组织答案，囊括所有采分点。
For planning, ‘P’ marks require a logical sequence that another student could follow. Avoid jumping from ‘measure 25 cm³ acid’ to ‘time the reaction’ without describing equilibration and addition steps. ‘A’ marks demand naming specific volumetric glassware – a pipette for fixed volume, a burette for variable, a measuring cylinder only where accuracy is not critical. Always choose the most precise instrument.
设计题中，“P” 类分数要求步骤逻辑连贯，另一学生也能依序操作。避免从“量取 25 cm³ 酸”直接跳到“计时反应”，而不描述平衡与加入步骤。“A” 类分数要求说出具体容量玻璃器皿——定体积用量液管、变体积用滴定管、只在精度要求不高时用量筒。务必选择最精密的仪器。
12. Summary: Strategies for Success | 总结：成功策略
To excel in CIE Chemistry Paper 5 (May 2018 style), practise writing concise, sequentially numbered steps for planning. Master rate law determination and Arrhenius plots through repeated graph drawing. Always check unit consistency and significant figures. Allocate time according to mark weight and read the question carefully to identify precisely what is being asked – plan, analyse or evaluate.
要在 CIE 化学试卷 5（2018 年 5 月类）中脱颖而出，需练习撰写简洁、按序号排列的设计步骤。通过反复作图掌握速率定律测定与阿伦尼乌斯图线。始终检查单位一致与有效数字。按分值分配时间，并仔细审题以准确辨别问题要求——是设计、分析还是评估。
Use past papers for targeted practice, initially without time constraints, then strictly timed. Mark your own attempts against the official mark scheme to internalise what gains marks. Remember that clarity, precision and thoroughness are the hallmarks of a top‑scoring script.
使用历年真题进行有针对性的练习，起初不限时，而后严格计时。对照官方评分方案批改自己的答案，内化得分要点。记住：清晰、精确与周全是一份高分答卷的标志。
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