📚 AQA AS Chemistry Unit 4 Exam Prep: Kinetics, Equilibria & Organic Chemistry | AQA AS化学第四单元备考：动力学、平衡与有机化学
This revision guide focuses on the core topics assessed in the AQA AS Chemistry Unit 4 June 2019 insert paper, covering reaction kinetics, chemical equilibria, organic chemistry and analytical techniques. Each section pairs concise explanations with exam-focused key points to help you maximise marks.
本复习指南聚焦 AQA AS 化学第四单元（2019 年 6 月试卷插页）所考查的核心内容，涵盖反应动力学、化学平衡、有机化学与分析技术。每个小节均配有简明英文讲解与考试要点，帮助同学们高效拿分。
1. Reaction Kinetics & Rate Equations | 反应动力学与速率方程
The rate of a reaction is the change in concentration of a reactant or product per unit time, normally expressed in mol dm⁻³ s⁻¹. You can measure it from the gradient of a concentration-time graph; for a curved line, draw a tangent at the chosen time and calculate its gradient.
反应速率指单位时间内反应物或产物浓度的变化量，通常以 mol dm⁻³ s⁻¹ 表示。可通过浓度-时间曲线的斜率求得；对曲线则需在指定时间点作切线并计算其斜率。
initial rate uses the tangent at t = 0, which avoids complications from product build-up.
瞬时速率通过曲线在时间 t 处的切线求得；初始速率利用 t = 0 处的切线求出，可避开产物累积的干扰。
etermined experimentally, not from the stoichiometric equation. The overall order is m + n.
速率方程形如 rate = k[A]ᵐ[B]ⁿ，其中 m、n 为反应级数，必须由实验测定，不能由化学计量式推出。总反应级数为 m + n。
| Order | Effect on rate | 级数 | 对速率的影响 |
| Zero | Rate is independent of [A] | 零级 | 速率与 [A] 无关 |
| First | Rate ∝ [A] | 一级 | 速率 ∝ [A] |
| Second | Rate ∝ [A]² | 二级 | 速率 ∝ [A]² |
For a first-order reaction, the half-life is constant, so a constant half-life is strong evidence for first-order kinetics. The relationship is:
对一级反应而言，半衰期恒定，因此半衰期恒定是判断一级动力学的有力证据。其关系式为：
t½ = ln2 / k
The units of the rate constant k depend on the overall order: zero order gives mol dm⁻³ s⁻¹, first order gives s⁻¹, and second order gives mol⁻¹ dm³ s⁻¹.
速率常数 k 的单位取决于总反应级数：零级为 mol dm⁻³ s⁻¹，一级为 s⁻¹，二级为 mol⁻¹ dm³ s⁻¹。
2. The Arrhenius Equation & Activation Energy | 阿伦尼乌斯方程与活化能
The Arrhenius equation shows how the rate constant k varies with temperature and activation energy:
阿伦尼乌斯方程说明速率常数 k 如何随温度与活化能变化：
k = A e−Ea/RT
/(RT). Plotting ln k against 1/T gives a straight line with gradient −Ea/R and intercept ln A; you can then calculate the activation energy from the gradient.
取自然对数可得 ln k = ln A − Ea/(RT)。以 ln k 对 1/T 作图得直线，斜率为 −Ea/R、截距为 ln A，即可由斜率求算活化能。
A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the proportion of successful collisions and therefore increasing the rate without being consumed in the overall reaction.
催化剂提供了活化能较低的新反应路径，提高有效碰撞比例，从而增大反应速率，而催化剂本身在总反应中不被消耗。
Typical calculations in the exam ask for the activation energy from the gradient, or comparison of two rates at
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