📚 AQA A Level Chemistry June 2018 Paper 4 Revision | AQA A-level化学2018年6月第四卷复习要点
This revision guide reconstructs the high-yield topics and skills assessed in the AQA A-level Chemistry June 2018 Paper 4 (Unit 4). It focuses on physical and organic chemistry, with paired English-Chinese explanations to support active recall.
本复习指南梳理 AQA A-level 化学 2018 年 6 月第四卷（Unit 4）考查的高频主题与技能，涵盖物理化学与有机化学，并以中英对照讲解帮助记忆。
1. Rate Equations, Orders and the Rate Constant | 速率方程、反应级数与速率常数
n are the orders with respect to A and B, and they must be found experimentally, not from the overall equation.
速率方程将反应速率与影响速率的物质浓度联系起来。对一般反应 A + B → C，速率方程可写为 rate = k[A]ᵐ[B]ⁿ。m 和 n 分别是对 A 和 B 的反应级数，必须由实验确定，不能从总化学方程式直接推出。
rate = k[A]ᵐ[B]ⁿ
The overall order is m + n. If doubling [A] doubles the rate, the order with respect to A is 1; if doubling [A] quadruples the rate, the order is 2; if changing [A] has no effect, the order is 0.
总反应级数为 m + n。若 [A] 加倍而速率加倍，则对 A 为一级；若 [A] 加倍而速率变为四倍，则为二级；若改变 [A] 对速率无影响，则为零级。
The units of the rate constant k depend on the overall order, as shown below.
速率常数 k 的单位取决于总反应级数，如下表所示。
| Overall order | Rate equation | Units of k |
|---|---|---|
| 0 | rate = k | mol dm⁻³ s⁻¹ |
| 1 | rate = k[A] | s⁻¹ |
| 2 | rate = k[A][B] | mol⁻¹ dm³ s⁻¹ |
| 3 | rate = k[A]²[B] | mol⁻² dm⁶ s⁻¹ |
2. Initial Rates and Concentration-Time Graphs | 初始速率法与浓度-时间图像
The initial rates method uses separate experiments in which one concentration is changed while the others are kept constant. Comparing the measured initial rates gives the order for each reagent.
初始速率法使用多组实验，每次只改变一种反应物浓度，保持其他浓度不变。比较测得的初始速率即可得出各反应物的级数。
For first order reactions, a plot of ln[A] against time is linear with slope −k. For zero order reactions, a plot of [A] against time is linear with slope −k.
一级反应的 ln[A] 对时间作图呈直线，斜率为 −k。零级反应的 [A] 对时间作图呈直线，斜率为 −k。
First order reactions have a constant half-life, while zero order reactions have a half-life that decreases as the reaction proceeds. These graphical features are frequently tested in Paper 4.
一级反应的半衰期恒定，而零级反应的半衰期随反应进行而缩短。这些图像特征是第四卷常见考点。
3. Equilibrium Constants Kc and Kp | 平衡常数 Kc 与 Kp
For a reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is written as shown below. The square brackets represent equilibrium concentrations in mol dm⁻³.
对于可逆反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数如下式所示。方括号表示以 mol dm⁻³ 为单位的平衡浓度。
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Kc is only affected by temperature. Changing concentration or pressure may shift the equilibrium position, but it does not change the value of Kc.
Kc 只受温度影响。改变浓度或压强会移动平衡位置，但不会改变 Kc 的数值。
For gas equilibria, Kp uses partial pressures. The partial pressure of a gas equals its mole fraction multiplied by the total pressure.
对气体平衡，Kp 使用分压。某气体的分压等于其摩尔分数乘以总压。
p(A) = mole fraction
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