Year 13 OCR Chemistry: Unit Test Mock Exam Analysis | Year 13 OCR 化学：单元测试模拟卷解析

📚 Year 13 OCR Chemistry: Unit Test Mock Exam Analysis | Year 13 OCR 化学：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for Year 13 OCR A Level Chemistry, covering content from Modules 5 and 6. Each selected question is carefully explained to reinforce key concepts, highlight common pitfalls and sharpen exam technique.

本文详细讲解了一套面向 Year 13 学生的 OCR A Level 化学单元测试模拟卷，内容覆盖模块5和模块6。通过对每道典型题目的解析，帮助巩固核心概念、避开常见失分点并提升应试技巧。

1. Rate Equation Determination from Initial Rates | 从初始速率确定速率方程

a fixed temperature. The initial rates method was used, and the following data were obtained:

在一定温度下研究反应 A + 2B → C + D，使用初始速率法得到以下数据：

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	1.0 × 10⁻⁴
2	0.20	0.10	2.0 × 10⁻⁴
3	0.10	0.20	4.0 × 10⁻⁴

By comparing Experiments 1 and 2, [B] is constant while [A] doubles; the rate also doubles. This indicates that the reaction is first order with respect to A.

对比实验1和2，[B] 保持不变而 [A] 加倍，速率也加倍，表明反应对 A 为一级。

Comparing Experiments 1 and 3, [A] is constant, [B] doubles and the rate increases by a factor of 4. Hence the order with respect to B is 2.

对比实验1和3，[A] 不变，[B] 加倍，速率变为原来的4倍，因此对 B 为二级。

= (1.0 × 10⁻⁴) / (0.10 × (0.10)²) = 0.10 mol⁻² dm⁶ s⁻¹. The units of k are mol⁻² dm⁶ s⁻¹.

速率方程为 rate = k[A][B]²，总级数为3。代入实验1数据，k = (1.0 × 10⁻⁴) / (0.10 × (0.10)²) = 0.10 mol⁻² dm⁶ s⁻¹，速率常数的单位是 mol⁻² dm⁶ s⁻¹。

er. Always use the relation: units of k = mol¹⁻ⁿ dm³ⁿ⁻³ s⁻¹, where n is the overall order.

常见错误是忘记根据总级数推导 k 的单位。务必使用关系式：k 的单位 = mol¹⁻ⁿ dm³ⁿ⁻³ s⁻¹，n 为总级数。

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2. Kp Calculation for a Gaseous Equilibrium | 气态平衡的 Kp 计算

Ammonia is synthesised via the Haber process: N₂(g) + 3H₂(g) ⇌ fractions are: x(N₂) = 0.20, x(H₂) = 0.60, x(NH₃) = 0.20. Calculate Kp, stating its units.

氨的合成通过哈伯法：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。在某一温度下，平衡总压为 200 atm。各组分的摩尔分数为：x(N₂) = 0.20，x(H₂) = 0.60，x(NH₃) = 0.20。计算 Kp 并给出单位。

First, calculate partial pressures: p = x × P_total. p(N₂) = 0.20 × 200 = 40 atm, p(H₂) = 0.60 × 200 = 120 atm, p(NH₃) = 0.20 × 200 = 40 atm.

首先计算分压：p = x × P_总。p(N₂) = 0.20 × 200 = 40 atm，p(H₂) = 0.60 × 200 = 120 atm，p(NH₃) = 0.20 × 200 = 40 atm。

³) = 1600 / (40 × 1,728,000) = 1600 / 69,120,000 ≈ 2.31 × 10⁻⁵ atm⁻². The units are atm⁻².

Kp 的表达式为：Kp = (p(NH₃))² / (p(N₂) × (p(H₂))³)。代入数值： = (40)² / (40 × (120)³) = 1600 / (40 × 1,728,000) = 1600 / 69,120,000 ≈ 2.31 × 10⁻⁵ atm⁻²，单位为 atm⁻²。

Many candidates forget to raise partial pressures to the stoichiometric powers or miscalculate the units. Always check that the indices in the Kp expression match the balanced equation.

许多同学忘记将分压进行对应化学计量数的幂运算，或单位计算出错。务必确保 Kp 表达式中的指数与配平方程式一致。

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3. Buffer Solution pH Calculation | 缓冲溶液 pH 计算

olution. Given Ka for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³, calculate the pH of the buffer.

将 50 cm³ 0.20 mol dm⁻³ 的乙酸 (CH₃COOH) 与 50 cm³ 0.10 mol dm⁻³ 的氢氧化钠溶液混合制成缓冲溶液。已知乙酸的 Ka = 1.8 × 10⁻⁵ mol dm⁻³，计算该缓冲溶液的 pH。

First, calculate initial moles: n(CH₃COOH)_initial 0) = 0.005 mol. The added OH⁻ neutralises some of the acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.

首先计算初始物质的量：n(CH₃COOH)_初始 = 0.20 × (50/1000) = 0.010 mol；n(NaOH) = 0.10 × (50/1000) = 0.005 mol。加入的 OH⁻ 部分中和酸：CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O。

After neutralisation, moles of CH₃COOH remaining = 0.010 – l volume = 100 cm³ = 0.100 dm³, so [CH₃COOH] = [CH₃COO⁻] = 0.005 / 0.100 = 0.050 mol dm⁻³.

中和后剩余 CH₃COOH 的物质的量为 0.010 – 0.005 = 0.005 mol；生成的 CH₃COO⁻ 为 0.005 mol。总体积 = 100 cm³ = 0.100 dm³，因此 [CH₃COOH] = [CH₃COO⁻] = 0.005 / 0.100 = 0.050 mol dm⁻³。

Using the Henderson–Hasselbalch equation: pH = pKa + log([salt]/[acid]). pKa = –log(1.8 × 10⁻⁵) = 4.74. Since [salt] = [acid], log(1) = 0, so pH = 4.74.

使用 Henderson–Hasselbalch 方程：pH = pKa + log([盐]/[酸])。pKa = –log(1.8 × 10⁻⁵) = 4.74。因为 [盐] = [酸]，log(1) = 0，故 pH = 4.74。

In OCR examinations, you are expected to be comfortable with the assumption that the salt concentration equals the concentration of the conjugate base formed, and to use the formula directly without deriving it each time.

在 OCR 考试中，考生应能熟练设定盐的浓度即等于生成的共轭碱浓度，并直接应用公式，无需每次重新推导。

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4. Redox Titration and Concentration Calculation | 氧化还原滴定与浓度计算

II) solution. 18.5 cm³ of KMnO₄ were required. Calculate the concentration of Fe²⁺ in g dm⁻³. (Mᵣ of Fe = 55.8 g mol⁻¹)

取 25.0 cm³ 硫酸亚铁溶液，用稀硫酸酸化后，以 0.0200 mol dm⁻³ 的高锰酸钾溶液滴定，终点时消耗 KMnO₄ 溶液 18.5 cm³。计算 Fe²⁺ 的质量浓度（g dm⁻³）。(Fe 的 Mᵣ = 55.8 g mol⁻¹)

balanced redox half-equations combine to give: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O.

配平的氧化还原离子方程式为：5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O。

ol. From the stoichiometry, moles of Fe²⁺ = 5 × 3.70 × 10⁻⁴ = 1.85 × 10⁻³ mol in 25.0 cm³.

所用 MnO₄⁻ 的物质的量 = 0.0200 × (18.5/1000) = 3.70 × 10⁻⁴ mol。根据化学计量比，Fe²⁺ 的物质的量 = 5 × 3.70 × 10⁻⁴ = 1.85 × 10⁻³ mol（存在于 25.0 cm³ 中）。

1.85 × 10⁻³ / 0.0250 = 0.0740 mol dm⁻³. Mass concentration = 0.0740 × 55.8 = 4.13 g dm⁻³.

因此 [Fe²⁺] = 1.85 × 10⁻³ / 0.0250 = 0.0740 mol dm⁻³。质量浓度 = 0.0740 × 55.8 = 4.13 g dm⁻³。

An important tip is to confirm the reaction ratio: MnO₄⁻ : Fe

Published by TutorHao | Year 13 Chemistry Revision Series | aleveler.com

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