Deducing Order of Reaction from Raw Data | 从原始数据推断反应级数

📚 Deducing Order of Reaction from Raw Data | 从原始数据推断反应级数

In A-level chemistry, raw kinetic data usually appear either as tables of initial concentrations and initial rates, or as concentration-time readings for a single reaction mixture. The central skill is to move from these numbers to a rate equation and to state the order with respect to each reactant. This article explains the main methods used in Cambridge A-level Chemistry, including the initial rates comparison, concentration-time graphs, half-life analysis and logarithmic plots.

在 A-level 化学中，原始动力学数据通常以初始浓度和初始速率的表格出现，或以某一反应混合物的浓度-时间读数出现。核心技能是根据这些数字推出速率方程，并说明对各反应物的级数。本文介绍剑桥 A-level 化学中使用的主要方法，包括初始速率比较、浓度-时间图、半衰期分析和对数图法。

---

1. The Meaning of Reaction Order | 反应级数的含义

The order of reaction with respect to a reactant is the power to which its concentration is raised in the experimentally determined rate equation. It is not automatically equal to the stoichiometric coefficient unless the reaction is known to be an elementary step.

反应对某一反应物的级数，是指实验测得的速率方程中该反应物浓度的幂指数。除非已知反应是基元步骤，否则级数并不自动等于化学计量系数。

Orders are usually small integers: 0, 1, 2, or occasionally 3. The overall order is the sum of the individual orders, and it controls the units of the rate constant.

级数通常为小整数：0、1、2，偶尔为 3。总级数是各分级数之和，它决定速率常数的单位。

---

2. Rate Equation, Rate Constant and Units | 速率方程、速率常数与单位

For a reaction A + B → products, the rate equation has the general form rate = k[A]ᵐ[B]ⁿ, where m is the order with respect to A, n is the order with respect to B, and k is the rate constant at a fixed temperature.

对于反应 A + B → 产物，速率方程的一般形式为 rate = k[A]ᵐ[B]ⁿ，其中 m 是对 A 的级数，n 是对 B 的级数，k 是固定温度下的速率常数。

The units of k depend on the overall order. For overall order 0, k has units mol dm⁻³ s⁻¹. For order 1, k has units s⁻¹. For order 2, the units are dm³ mol⁻¹ s⁻¹. For overall order 3, the units become dm⁶ mol⁻² s⁻¹.

k 的单位取决于总级数。总级数为 0 时，k 的单位为 mol dm⁻³ s⁻¹。一级时，k 的单位为 s⁻¹。二级时，单位为 dm³ mol⁻¹ s⁻¹。总级数为 3 时，单位变为 dm⁶ mol⁻² s⁻¹。

When you calculate k from raw data, always include the units. The units provide a quick way to check whether the overall order you have inferred is sensible.

当你从原始数据计算 k 时，务必写出单位。单位可以快速检查你所推断的总级数是否合理。

---

3. Why Initial Rates Are Preferred | 为什么优先使用初始速率

Raw data are most useful when they record the initial rate of reaction. At time zero, the concentrations of all reactants are known exactly, and the products have not yet built up. This avoids interference from the reverse reaction or from product inhibition.

原始数据在记录初始反应速率时最有用。在时间为零时，所有反应物的浓度都精确已知，产物尚未积累。这避免了逆反应或产物抑制带来的干扰。

If the raw data are concentration-time readings, the initial rate can be estimated by drawing a tangent to the concentration-time curve at t = 0. The slope of that tangent is the initial rate.

如果原始数据是浓度-时间读数，则可以通过在 t = 0 处作曲线的切线来估算初始速率。该切线的斜率就是初始速率。

Using initial rates is important because later on the reactant concentrations change, and the rate at any later point depends on those changed concentrations. Comparing rates at different times without correcting for concentration changes is not valid for determining orders.

使用初始速率很重要，因为随后反应物浓度会改变，而任意后续时刻的速率取决于这些已改变的浓度。如果不校正浓度变化，比较不同时刻的速率不能用于确定级数。

---

4. Isolating Variables: The Core Strategy | 隔离变量：核心策略

The simplest way to find orders from a raw data table is to compare two experiments in which only one reactant concentration changes while all other concentrations are kept constant. This isolates the effect of that one reactant.

从原始数据表求级数的最简单方法，是比较两个实验：其中只有一种反应物的浓度改变，其他浓度保持不变。这样可以隔离该反应物的影响。

If doubling [A] doubles the rate, the order with respect to A is 1. If doubling [A] has no effect on the rate, the order is 0. If doubling [A] quadruples the rate, the order is 2.

如果 [A] 加倍使速率加倍，则对 A 的级数为 1。如果 [A] 加倍对速率无影响，则级数为 0。如果 [A] 加倍使速率变为四倍，则级数为 2。

Use the ratio form when [B] is constant:

当 [B] 恒定时，使用比值形式：

rate₂ / rate₁ = ([A]₂ / [A]₁)ᵐ

This equation lets you handle non-integer multiples confidently. For example, if [A] increases by a factor of 1.5 and the rate increases by a factor of 2.25, then 2.25 = (1.5)², so m = 2.

该方程可让你自信地处理非整数倍数。例如，如果 [A] 增加了 1.5 倍，而速率增加了 2.25 倍，那么 2.25 = (1.5)²，因此 m = 2。

---

5. Worked Example: Initial Rates Table | 例题：初始速率表

The table below shows raw data for the reaction P + 2Q → R.

下表显示了反应 P + 2Q → R 的原始数据。

Experiment	[P] / mol dm⁻³	[Q] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0 × 10⁻⁴
2	0.20	0.10	4.0 × 10⁻⁴
3	0.10	0.30	1.8 × 10⁻³

Comparing experiments 1 and 2, [P] doubles while [Q] is constant. The rate also doubles from 2.0 × 10⁻⁴ to 4.0 × 10⁻⁴, so the reaction is first order with respect to P.

比较实验 1 和 2，[P] 加倍而 [Q] 不变。速率也从 2.0 × 10⁻⁴ 加倍到 4.0 × 10⁻⁴，因此反应对 P 为一级。

Comparing experiments 1 and 3, [P] is constant while [Q] triples. The rate increases from 2.0 × 10⁻⁴ to 1.8 × 10⁻³, which is a factor of 9. Since 3² = 9, the reaction is second order with respect to Q.

比较实验 1 和 3，[P] 不变而 [Q] 增至三倍。速率从 2.0 × 10⁻⁴ 增加到 1.8 × 10⁻³，倍数为 9。因为 3² = 9，反应对 Q 为

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

更多咨询请联系16621398022（同微信）