A-Level Chemistry: Determining Reaction Orders from Experimental Data | A-Level化学：由实验数据推导反应级数

📚 A-Level Chemistry: Determining Reaction Orders from Experimental Data | A-Level化学：由实验数据推导反应级数

In chemical kinetics, the reaction order is a set of numbers that describes how the rate of a reaction depends on the concentration of each reactant. These orders cannot be found from the balanced chemical equation; they must be determined from experimental data. In this article we will explore the most reliable methods for deriving reaction orders, including the method of initial rates, concentration-time graphs, and half-life analysis.

在化学动力学中，反应级数是一组用来描述反应速率如何随各反应物浓度变化的数字。反应级数不能从配平的化学方程式中直接得到，而必须通过实验数据来确定。本文将介绍推导反应级数最可靠的方法，包括初始速率法、浓度-时间图像法以及半衰期法。

1. The Rate Equation and Reaction Order | 速率方程与反应级数

For a reaction in which the rate depends on two reactants A and B, the general rate equation is usually written as:

对于一个速率取决于两种反应物 A 和 B 的反应，其速率方程通常写成：

rate = k[A]ᵐ[B]ⁿ

Here, k is the rate constant, square brackets denote concentration in mol dm⁻³, and m and n are the orders of reaction with respect to A and B. The overall order is m + n. The orders may be 0, 1, 2, or sometimes fractional, but in CIE A-Level chemistry they are usually simple integers.

其中 k 是速率常数，方括号表示浓度，单位为 mol dm⁻³，m 和 n 分别是反应物 A 和 B 的反应级数。总反应级数为 m + n。反应级数可以是 0、1、2，有时也可以是小数，但在 CIE A-Level 化学中通常是简单整数。

2. What Do Orders of 0, 1 and 2 Mean? | 0级、1级和2级反应级数的含义

If the order with respect to A is zero, changing [A] has no effect on the rate:

如果反应物 A 的级数为 0，则改变 [A] 不会影响反应速率：

rate = k

If the order is first, rate is directly proportional to [A]:

如果级数为 1，则速率与 [A] 成正比：

rate = k[A]

If the order is second, rate is proportional to [A]²:

如果级数为 2，则速率与 [A]² 成正比：

rate = k[A]²

For example, doubling [A] has no effect for zero order, doubles the rate for first order, and quadruples the rate for second order.

例如，将 [A] 加倍：对零级反应无影响，对一级反应速率翻倍，对二级反应速率变为原来的四倍。

3. The Method of Initial Rates | 初始速率法

The initial rate is the instantaneous rate at the moment the reactants are first mixed, before any significant amount of product has formed. To find orders by this method, we run several experiments at the same temperature and measure the initial rate for different initial concentrations.

初始速率是指反应物刚混合时、尚未生成大量产物之前的瞬时速率。为了用这种方法求反应级数，需要在相同温度下进行多组实验，并在不同初始浓度下测量初始速率。

The key steps are:

关键步骤如下：

Measure the initial rate for a known set of initial concentrations.

在已知初始浓度条件下测量初始速率。
Change the concentration of one reactant while keeping all other concentrations constant.

只改变一种反应物的浓度，同时保持其他反应物浓度不变。
Observe how the initial rate changes, and repeat for each reactant.

观察初始速率如何变化，并对每种反应物重复该操作。

4. Worked Example 1: Two Reactants | 例1：两种反应物

Consider the following data for the reaction A + B → products.

考虑反应 A + B → 产物的下列实验数据。

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

Compare experiments 1 and 2. When [B] is constant and [A] is doubled, the rate increases from 2.0 × 10⁻³ to 8.0 × 10⁻³, a factor of 4. Therefore:

比较实验1和实验2。当 [B] 不变而 [A] 加倍时，速率从 2.0 × 10⁻³ 增加到 8.0 × 10⁻³，即变为原来的 4 倍。因此：

2ᵐ = 4, so m = 2

The order with respect to A is 2. Now compare experiments 1 and 3. When [A] is constant and [B] is doubled, the rate doubles from 2.0 × 10⁻³ to 4.0 × 10⁻³. Therefore:

所以 A 的反应级数为 2。现在比较实验1和实验3。当 [A] 不变而 [B] 加倍时，速率从 2.0 × 10⁻³ 加倍为 4.0 × 10⁻³。因此：

2ⁿ = 2, so n = 1

The rate equation is:

速率方程为：

rate = k[A]²[B]

The overall order is 2 + 1 = 3.

总反应级数为 2 + 1 = 3。

5. Finding the Rate Constant k | 求速率常数 k

Once the rate equation is known, the rate constant can be calculated by substituting values from any experiment. Using experiment 1:

一旦知道了速率方程，就可以将任意一组实验数据代入，计算出速率常数。使用实验1的数据：

2.0 × 10⁻³ = k × (0.10)² × (0.10)

So:

因此：

k = 2.0 × 10⁻³ / 0.001 = 2.0 mol⁻² dm⁶ s⁻¹

The units of k come from the overall order. For an overall third-order reaction, the units are mol⁻² dm⁶ s⁻¹. The value of k depends on temperature and, for heterogeneous reactions, on the presence of a catalyst, but it does not change with concentration.

k 的单位取决于总反应级数。对于总反应级数为 3 的反应，单位是 mol⁻² dm⁶ s⁻¹。k 的数值取决于温度，对于多相反应还取决于是否存在催化剂，但它不随浓度改变。

6. Using Ratios When Concentrations Are Not Exact Multiples | 当浓度不是刚好成倍变化时的方法

Sometimes the experimental concentrations do not change by a simple factor such as 2 or 3. In that case, use the ratio equation. For a change in [A] while [B] is constant:

有时实验中的浓度变化不是简单的 2 倍或 3 倍。此时可以使用比值方程。在 [B] 恒定时，若只改变 [A]：

rate₂ / rate₁ = ([A]₂ / [A]₁)ᵐ

Taking natural logarithms gives:

两边取自然对数得到：

m = ln(rate₂ / rate₁) / ln([A]₂ / [A]₁)

For example, if [A] changes from 0.010 to 0.015 mol dm⁻³ and the rate changes from 3.6 × 10⁻⁵ to 8.1 × 10⁻⁵ mol dm⁻³ s⁻¹, then:

例如，如果 [A] 从 0.010 mol dm⁻³ 变为 0.015 mol dm⁻³，速率从 3.6 × 10⁻⁵ 变为 8.1 × 10⁻⁵ mol dm⁻³ s⁻¹，则：

m = ln(8.1 / 3.6) / ln(0.015 / 0.010) = 2.00

The order with respect to A is 2. This log-ratio method is useful for non-integer or awkward data.

因此 A 的级数为 2。这种对数比值法对于处理非整数或不太规整的数据非常有用。

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

Another approach is to measure concentration at regular time intervals and then test which mathematical plot gives a straight line.

另一种方法是每隔一定时间测量浓度，然后检测哪一种数学图像能给出直线。

For zero order, a graph of [A] against time is linear, and the half-life becomes shorter as the reaction proceeds.

对零级反应，[A] 对时间作图得到直线，且随着反应进行，半衰期会越来越短。
For first order, a graph of ln[A] against time is linear, and the half-life is constant.

对一级反应，ln[A] 对时间作图得到直线，且半衰期恒定。
For second order, a graph of 1/[A] against time is linear, and the half-life becomes longer as the reaction proceeds.

对二级反应，1/[A] 对时间作图得到直线，且随着反应进行，半衰期会越来越长。

The relevant half-life equations are:

8. Worked Example 2: Using Half-Life | 例2：利用半衰期求反应级数

The decomposition of dinitrogen pentoxide can be followed by measuring [N₂O₅] over time:

五氧化二氮的分解反应可以通过随时间测量 [N₂O₅] 来跟踪：

2N₂O₅(g) → 4NO₂(g) + O₂(g)

t / s	0	100	200	300	400
[N₂O₅] / mol dm⁻³	0.500	0.354	0.250	0.177	0.125

From 0.500 to 0.250 mol dm⁻³ takes 200 s. From 0.250 to 0.125 mol dm⁻³ also takes 200 s. The half-life is constant, so the reaction is first order with respect to N₂O₅.

从 0.500 mol dm⁻³ 下降到 0.250 mol dm⁻³ 需要 200 s。从 0.250 下降到 0.125 mol dm⁻³ 同样需要 200 s。半衰期恒定，因此该反应对 N₂O₅ 为一级反应。

The rate constant is then:

于是速率常数为：

k = ln 2 / t½ = 0.693 / 200 = 3.47 × 10⁻³ s⁻¹

Notice that the units for a first-order rate constant are s⁻¹.

注意，一级反应的速率常数单位是 s⁻¹。

9. Rate-Concentration Graphs | 速率-浓度图像

If the initial rate is measured for several different initial concentrations of a single reactant, the shape of the rate-concentration graph allows us to identify the order directly.

如果对同一反应物在多个不同初始浓度下测量初始速率，则速率-浓度图像的形状可以直接告诉我们反应级数。

Order	Graph of initial rate against [A]
0	Horizontal line: rate is constant
1	Straight line passing through the origin
2	Upward curve, with rate increasing as [A] increases

For a first-order reaction, doubling [A] doubles the rate. For a second-order reaction, doubling [A] quadruples the rate. These graphical patterns are a fast way to communicate reaction order in exam answers.

对于

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

A-Level Chemistry: Determining Reaction Orders from Experimental Data | A-Level化学：由实验数据推导反应级数