📚 A-Level Chemistry: Establishing and Applying Rate Equations | A-Level化学：速率方程式的建立与应用
The rate equation is one of the most powerful tools in chemical kinetics. For CIE A-Level Chemistry, it is not enough to simply memorise the formula; you must be able to deduce orders from experimental data, calculate the rate constant with the correct units, and connect the rate equation to a plausible reaction mechanism. This article breaks down every aspect of rate equations, from the basic definitions to advanced exam-style reasoning.
速率方程式是化学动力学中最有力的工具之一。对于CIE A-Level化学而言，仅仅背诵公式是不够的；你必须能够从实验数据中推断反应级数、计算带有正确单位的速率常数，并将速率方程式与合理的反应机理联系起来。本文将从基本定义到考试风格的深入推理，逐一剖析速率方程式的各个方面。
1. The Form of the Rate Equation | 速率方程式的形式
For a general reaction aA + bB → products, the rate equation is written as rate = k[A]m[B]n. It is important to realise that the powers m and n are not the stoichiometric coefficients a and b; they must be determined experimentally.
对于一个一般反应 aA + bB → 产物，速率方程式写作 rate = k[A]m[B]n。必须注意，幂次m和n并不是化学计量数a和b；它们必须通过实验来确定。
rate = k[A]m[B]n
Here, k is the rate constant, [A] and [B] are reactant concentrations in mol dm⁻³, and m and n are the orders of reaction with respect to A and B respectively. The overall order of reaction is m + n. The rate equation applies at a fixed temperature because k is temperature-dependent.
式中，k为速率常数，[A]和[B]是反应物浓度（单位mol dm⁻³），m和n分别是反应对A和B的级数。总反应级数为m + n。由于k与温度有关，速率方程式只在固定温度下成立。
2. Significance of Reaction Orders | 反应级数的含义
The order with respect to a reactant tells us how the initial rate changes as that reactant’s concentration changes, assuming all other concentrations are held constant.
对某一反应物的级数告诉我们：在保持其他物质浓度不变的条件下，初始速率如何随该反应物浓度的变化而变化。
| Order | Double [A] | Triple [A] |
| 0 | rate unchanged | rate unchanged |
| 1 | rate doubles | rate triples |
| 2 | rate quadruples | rate becomes 9 times |
For zero-order, the reactant concentration does not appear in the rate equation. For first-order, rate ∝ [A]. For second-order, rate ∝ [A]2. A solid grasp of these proportions allows you to interpret data tables quickly.
对于零级反应，反应物浓度不出现在速率方程中。对于一级反应，rate ∝ [A]；对于二级反应，rate ∝ [A]2。扎实掌握这些比例关系能帮助你快速解读数据表。
3. The Initial Rates Method | 初始速率法
The initial rate is the instantaneous rate at the moment the reactants are first mixed, when concentrations are still effectively equal to their starting values. Experimentally, we measure the gradient of the concentration-time curve at t = 0, then repeat the experiment with different initial concentrations.
初始速率是反应物刚混合时、各物质浓度仍等于起始值时的瞬时速率。实验上，我们测量浓度-时间曲线在t = 0处的斜率，然后用不同的初始浓度重复实验。
Consider the data below for the reaction A + B → products, with [B] held constant:
考虑以下反应 A + B → 产物 的数据，其中[B]保持不变：
| Experiment | [A] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
| 1 | 0.10 | 0.02 |
| 2 | 0.20 | 0.08 |
When [A] doubles from 0.10 to 0.20 mol dm⁻³, the initial rate increases from 0.02 to 0.08 mol dm⁻³ s⁻¹, which is a factor of 4. Since 22 = 4, the reaction is second order with respect to A. Using rate₂/rate₁ = ([A]₂/[A]₁)m gives m = 2.
当[A]从0.10加倍到0.20 mol dm⁻³时，初始速率从0.02增大到0.08 mol dm⁻³ s⁻¹，即变为原来的4倍。因为2² = 4，反应对A为二级。利用 rate₂/rate₁ = ([A]₂/[A]₁)m 可得 m = 2。
4. Concentration-Time Graphs and Half-Life | 浓度-时间图与半衰期
Plotting the concentration of a reactant against time gives a curve whose shape depends on the order. For a zero-order reaction, [A] decreases linearly with time, with a constant gradient equal to −k.
将反应物浓度对时间作图，得到的曲线形状取决于反应级数。对于零级反应，[A]随时间线性下降，斜率为常数且等于−k。
For a first-order reaction, the concentration decays exponentially: [A] = [A]0e−kt. The half-life t₁/₂ is independent of the initial concentration, so equal time intervals correspond to equal fractional decreases. For a second-order reaction, the curve is steeper at the start and flattens more gradually; the half-life increases as the reaction proceeds.
对于一级反应，浓度呈指数衰减：[A] = [A]0e−kt。其半衰期t₁/₂与初始浓度无关，因此相同的时间间隔对应相同的浓度下降比例。对于二级反应，曲线起始较陡，随后逐渐平缓，半衰期随反应进行而增大。
In CIE exams, you may be asked to identify the order from a concentration-time graph by measuring successive half-lives. Constant half-lives indicate first order; an increasing half-life suggests second order.
在CIE考试中，你可能会被要求通过测量连续半衰期来判断反应级数。半衰期恒定说明是一级反应；半衰期逐渐增大则提示二级反应。
5. Rate-Concentration Graphs | 速率-浓度图
If initial rates are measured at various initial concentrations, a plot of initial rate against concentration reveals the order directly:
如果在不同初始浓度下分别测定初始速率，再以初始速率对浓度作图，可直接看出反应级数：
- Zero order: a horizontal line (rate is constant).
- First order: a straight line through the origin (rate ∝ [A]).
- Second order: a curve through the origin that becomes steeper as [A] increases (rate ∝ [A]2).
对于零级反应：水平直线，速率恒定；一级反应：过原点的直线，速率∝[A]；二级反应：过原点的上升曲线，速率∝[A]²。
A rate-concentration graph can also be produced by measuring tangents at different points along a single concentration-time curve. This method is especially useful when preparing a clock reaction or when repeated experiments are impractical.
速率-浓度图也可以通过在同一条浓度-时间曲线上不同点作切线来获得。当使用钟表反应或重复实验不可行时，这种方法尤为有用。
6. Units of the Rate Constant | 速率常数的单位
The units of k depend on the overall order of the reaction. Since rate has units mol dm⁻³ s⁻¹, we can derive the units of k from rate = k[reactant]n.
速率常数k的单位取决于总反应级数。由于速率的单位是mol dm⁻³ s⁻¹，我们可以从 rate = k[反应物]n 推导k的单位。
units of k = (mol dm⁻³)1−n s⁻¹
- Zero order (n = 0): mol dm⁻³ s⁻¹
- First order (n = 1): s⁻¹
- Second order (n = 2): mol⁻¹ dm³ s⁻¹ (also written dm³ mol⁻¹ s⁻¹)
零级（n = 0）：mol dm⁻³ s⁻¹；一级（n = 1）：s⁻¹；二级（n = 2）：mol⁻¹ dm³ s⁻¹（也写作dm³ mol⁻¹ s⁻¹）。
Always check your unit derivation in exam answers; forgetting to include the correct units for k is a common source of lost marks.
在考试答案中务必检查单位推导；忘记写出k的正确单位是常见的失分原因之一。
7. First-Order Half-Life and Its Uses | 一级反应的半衰期及其应用
For a first-order reaction, the half-life is linked to k by a simple expression:
对于一级反应，半衰期与k之间有一个简单的关系式：
t₁/₂ = ln 2 / k = 0.693 / k
This equation allows chemists to find k directly from a concentration-time graph without measuring tangents. For example, if a first-order reaction shows [A] falling from 0.80 mol dm⁻³ to 0.40 mol dm⁻³ in 35 s, then t₁/₂ = 35 s and k = 0.693 / 35 = 0.0198 s⁻¹.
该公式使化学家无需作切线即可直接从浓度-时间图中求出k。例如，若一级反应的[A]从0.80 mol dm⁻³降到0.40 mol dm⁻³用时35 s，则t₁/₂ = 35 s，因此k = 0.693 / 35 = 0.0198 s⁻¹。
The constancy of half-life for a first-order process is also the basis of radiocarbon dating and drug clearance in pharmacokinetics, which makes this concept practically relevant beyond the exam.
一级过程半衰期的恒定性也是放射性碳定年和药物代谢动力学中药物清除速率的理论基础，因此这一概念在考试之外也具有实际意义。
8. Rate-Determining Step and Reaction Mechanism | 速率决定步骤与反应机理
In a multi-step reaction, the overall rate is controlled by the slowest elementary step, called the rate-determining step (RDS). The rate equation for the overall reaction can be written from the molecularity of this slow step.
在多步反应中，总反应速率受最慢的基元步骤控制，该步骤称为速率决定步骤（RDS）。总反应的速率方程可以根据这个慢步骤的分子数写出。
For example, the overall reaction 2NO + O₂ → 2NO₂ is believed to proceed as:
例如，总反应 2NO + O₂ → 2NO₂ 被认为按以下步骤进行：
- Step 1 (slow): NO + NO → N₂O₂
- Step 2 (fast): N₂O₂ + O₂ → 2NO₂
Since Step 1 is the slowest, the rate equation is rate = k[NO]2, and O₂ does not appear in the rate equation. This matches many experimental observations for this reaction.
由于第一步最慢，速率方程式为 rate = k[NO]2，而O₂不出现于速率方程中。这与许多实验观察结果一致。
You should not write the rate equation from the balanced equation. Instead, use experimental orders to infer which species are involved in or before the rate-determining step, including any reaction intermediates formed in fast prior equilibria.
你不应该根据化学平衡方程式写出速率方程。相反，应利用实验级
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