Kinetics II: Rate Equations, Activation Energy and Reaction Mechanisms | 动力学 II：速率方程、活化能与反应机理

📚 Kinetics II: Rate Equations, Activation Energy and Reaction Mechanisms | 动力学 II：速率方程、活化能与反应机理

In Edexcel A Level Chemistry, Topic 16 (Kinetics II) moves beyond simple collision theory by introducing quantitative rate equations, the Arrhenius equation, and the link between experimental rate data and the mechanism of a reaction. It is one of the most mathematically demanding parts of the A Level specification, so clear definitions and precise graph work are essential.

在 Edexcel A Level 化学中，第 16 单元（动力学 II）从简单的碰撞理论进一步引入定量速率方程、阿伦尼乌斯方程，以及实验速率数据与反应机理之间的联系。这是 A Level 考纲中数学要求最高的内容之一，因此清晰的定义和准确的图像分析非常重要。

1. Rate Equations and Orders of Reaction | 速率方程与反应级数

The rate equation expresses how the initial rate of a reaction depends on the concentrations of the reactants. For a general reaction aA + bB → products, the rate equation takes the form:

速率方程表达了反应的初始速率如何取决于反应物的浓度。对于一般反应 aA + bB → 产物，速率方程的形式为：

rate = k[A]ᵐ[B]ⁿ

Here m is the order with respect to A, n is the order with respect to B, and the overall order is m + n. The symbol k is the rate constant.

其中 m 是相对于 A 的反应级数，n 是相对于 B 的反应级数，总反应级数为 m + n。符号 k 为速率常数。

A common misunderstanding is that m and n must equal the stoichiometric coefficients a and b. In fact, the orders are experimental quantities and can be 0, 1, 2, or even fractional. They have no necessary relationship to the balancing numbers in the chemical equation.

一个常见的误解是 m 和 n 必须等于化学计量系数 a 和 b。事实上，反应级数是实验量，可以是 0、1、2，甚至分数。它们与化学方程式中的配平系数没有必然关系。

Zero order: doubling [A] has no effect on rate.
First order: doubling [A] doubles the rate.
Second order: doubling [A] quadruples the rate.
零级：[A] 加倍对速率没有影响。
一级：[A] 加倍，速率加倍。
二级：[A] 加倍，速率变为原来的 4 倍。

2. Experimental Determination of Rate Equations | 实验测定速率方程

Rate equations can only be found by experiment. The most common approach is the initial rates method, where the initial rate is measured for several different starting concentrations. To isolate the order with respect to one reactant, you compare two experiments in which only that reactant’s concentration changes while all others are kept constant.

速率方程只能通过实验确定。最常用的方法是初始速率法，即在几种不同的起始浓度下测量初始速率。要单独确定某一反应物的级数，需要比较两个实验：只有该反应物的浓度改变，其他反应物浓度保持不变。

Clock reactions are particularly useful because the time taken to reach a fixed observable endpoint is measured. Since rate ∝ 1/time, the initial rate can be treated as proportional to 1/t for a fixed amount of reaction.

时钟反应特别有用，因为可以测量达到某一固定可见终点所需的时间。由于速率与时间成反比，对于固定的反应进度，初始速率可视为与 1/t 成正比。

For example, in the iodine clock reaction, the appearance of the blue iodine-starch complex is timed. By varying the concentrations of iodide and peroxodisulfate ions, the orders can be found, giving an experimentally determined rate equation such as rate = k[S₂O₈²⁻][I⁻].

例如，在碘时钟反应中，记录碘-淀粉蓝色配合物出现的时间。通过改变碘离子和过氧二硫酸根离子的浓度，可以确定级数，从而得到实验速率方程，例如 rate = k[S₂O₈²⁻][I⁻]。

The table below shows a simplified analysis where only [A] and [B] change:

下表展示了一个简化分析，其中仅改变 [A] 和 [B]：

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻³
2	0.20	0.10	8.0 × 10⁻³
3	0.10	0.20	4.0 × 10⁻³

Doubling [A] from 0.10 to 0.20 while [B] stays constant multiplies the rate by 4, so the order with respect to A is 2. Doubling [B] doubles the rate, so the order with respect to B is 1. Therefore rate = k[A]²[B].

在 [B] 不变时，[A] 从 0.10 增加到 0.20，速率乘以 4，因此相对于 A 的级数为 2。[B] 加倍使速率加倍，因此相对于 B 的级数为 1。所以速率方程为 rate = k[A]²[B]。

3. The Rate Constant k and Its Units | 速率常数 k 及其单位

The rate constant k is the proportionality constant in the rate equation. Its value depends on temperature, as described later by the Arrhenius equation. Importantly, the units of k vary with the overall order of reaction.

速率常数 k 是速率方程中的比例常数。其数值取决于温度，这一点稍后将由阿伦尼乌斯方程描述。重要的是，k 的单位随总反应级数而变化。

Since rate has units mol dm⁻³ s⁻¹ and concentration has units mol dm⁻³, the units of k can be worked out by rearranging the rate equation:

由于速率的单位是 mol dm⁻³ s⁻¹，浓度的单位是 mol dm⁻³，k 的单位可以通过重排速率方程求出：

Zero order overall: rate = k, so k has units mol dm⁻³ s⁻¹.
First order overall: rate = k[A], so k has units s⁻¹.
Second order overall: rate = k[A]² or k[A][B], so k has units dm³ mol⁻¹ s⁻¹.
Third order overall: k has units dm⁶ mol⁻² s⁻¹.
总级数为零级：rate = k，因此 k 的单位为 mol dm⁻³ s⁻¹。
总级数为一级：rate = k[A]，因此 k 的单位为 s⁻¹。
总级数为二级：rate = k[A]² 或 k[A][B]，因此 k 的单位为 dm³ mol⁻¹ s⁻¹。
总级数为三级：k 的单位为 dm⁶ mol⁻² s⁻¹。

In exam questions, always check that your calculated k has the correct units for the overall order. A common error is to give k without units or with units that do not match the rate equation.

在考试题中，一定要检查计算出的 k 是否具有与总级数相匹配的正确单位。一个常见错误是给出的 k 没有单位，或者单位与速率方程不匹配。

4. Concentration-Time Graphs and Half-Life | 浓度-时间图像与半衰期

Concentration-time graphs provide another way to determine reaction order. For a single reactant A, different orders give distinctive graph shapes.

浓度-时间图像提供了另一种确定反应级数的方法。对于单一反应物 A，不同级数会给出不同形状的图像。

Zero order: a plot of [A] against time is a straight line with gradient = –k.
First order: a plot of [A] against time is an exponential decay. The half-life is constant and is given by t½ = 0.693 / k.
Second order: a plot of 1/[A] against time is a straight line with gradient = k.
零级：[A] 对时间作图是一条直线，斜率为 –k。
一级：[A] 对时间作图呈指数衰减。半衰期恒定，公式为 t½ = 0.693 / k。
二级：1/[A] 对时间作图是一条直线，斜率为 k。

For a first-order reaction, the constant half-life is especially important. It means that the time taken for [A] to fall from 0.80 to 0.40 mol dm⁻³ is the same as the time taken to fall from 0.20 to 0.10 mol dm⁻³.

对于一级反应，恒定的半衰期尤其重要。这意味着 [A] 从 0.80 降到 0.40 mol dm⁻³ 所需的时间，与从 0.20 降到 0.10 mol dm⁻³ 所需的时间相同。

The half-life formula for first-order reactions is given below:

一级反应的半衰期公式如下：

t½ = ln 2 / k = 0.693 / k

This equation can be used to calculate the rate constant from an experimentally measured half-life.

该方程可用于根据实验测得的半衰期计算速率常数。

5. The Arrhenius Equation | 阿伦尼乌斯方程

The rate constant k is not independent of temperature. The Arrhenius equation shows how k depends on temperature and activation energy:

速率常数 k 并非与温度无关。阿伦尼乌斯方程展示了 k 如何随温度和活化能变化：

k = A e^(–Eₐ/RT)

In this equation, A is the pre-exponential factor, Eₐ is the activation energy, R is the gas constant (8.31 J mol⁻¹ K⁻¹), and T is the absolute temperature in kelvin. The term e^(–Eₐ/RT) represents the fraction of collisions that have sufficient energy to overcome the activation barrier.

在该方程中，A 是指前因子，Eₐ 是活化能，R 是气体常数（8.31 J mol⁻¹ K⁻¹），T 是开尔文温度。项 e^(–Eₐ/RT) 表示具有足够能量克服活化能垒的碰撞比例。

As temperature increases, the exponential term becomes larger, so k increases. As activation energy decreases, the exponential term also becomes larger, so k increases. This explains why catalysts, which lower Eₐ, cause a dramatic increase in reaction rate.

温度升高时，指数项变大，因此 k 增大。活化能降低时，指数项也会变大，因此 k 增大。这解释了为什么降低 Eₐ 的催化剂会使反应速率显著增加。

For calculations, the logarithmic form of the Arrhenius equation is more useful:

在计算中，阿伦尼乌斯方程的对数形式更为有用：

ln k = ln A – Eₐ/(RT)

6. Graphical Analysis of Arrhenius Data | 阿伦尼乌斯数据的图像分析

If ln k is plotted on the y-axis against 1/T on the x-axis, the Arrhenius equation gives a straight line:

如果以 ln k 为纵轴，以 1/T 为横轴作图，阿伦尼乌斯方程给出直线关系：

ln k = (–Eₐ/R) (1/T) + ln A

The gradient of this line is –Eₐ/R and the y-intercept is ln A. This allows the activation energy to be calculated from experimental measurements of k at different temperatures.

该直线的斜率为 –Eₐ/R，y 轴截距为 ln A。由此可以根据不同温度下 k 的实验测量值计算活化能。

To calculate Eₐ from the gradient, use:

要根据斜率计算 Eₐ，可使用：

Eₐ = –gradient × R

Remember that 1/T has units K⁻¹, so the gradient has units K. The gas constant R used is 8.31 J mol⁻¹ K⁻¹, so Eₐ is obtained in J mol⁻¹. Most exam answers require conversion to kJ mol⁻¹ by dividing by 1000.

请记住，1/T 的单位是 K⁻¹，因此斜率的单位是 K。使用的气体常数 R 为 8.31 J mol⁻¹ K⁻¹，因此得到的 Eₐ 单位为 J mol⁻¹。大多数考试答案要求除以 1000 转换为 kJ mol⁻¹。

A useful two-point form of the Arrhenius equation can also be used when measurements are made at only two temperatures:

当只在两个温度下进行测量时，阿伦尼乌斯方程还有一个实用的两点形式：

ln(k₂/k₁) = –Eₐ/R (1/T₂ – 1/T₁)

7. Reaction Mechanisms and the Rate-Determining Step | 反应机理与决速步骤

A reaction mechanism is a sequence of elementary steps that describes how reactants are converted into products at the molecular level. The slowest elementary step is called the rate-determining step because it limits the overall rate, much like the slowest section of a multi-stage journey controls the total travel time.

反应机理是描述反应物在分子水平上如何转变为产物的一系列基元步骤。最慢的基元步骤称为决速步骤，因为它限制了总反应速率，就像一段多阶段

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Kinetics II: Rate Equations, Activation Energy and Reaction Mechanisms | 动力学 II：速率方程、活化能与反应机理