📚 Kinetics II: Rate Equations and the Arrhenius Equation | 动力学 II：速率方程与阿伦尼乌斯方程
In Kinetics II, Edexcel A-Level Chemistry moves from the qualitative ideas of collision theory to the quantitative description of reaction rates. This article covers rate equations, orders of reaction, the units of the rate constant, experimental determination of orders, rate-concentration and concentration-time graphs, the Arrhenius equation, activation energy, reaction mechanisms and the rate-determining step, and practical methods such as clock reactions and continuous monitoring.
在动力学 II 中，Edexcel A-Level 化学从碰撞理论的定性认识转向反应速率的定量描述。本文将涵盖速率方程、反应级数、速率常数的单位、级数的实验测定、速率-浓度图与浓度-时间图、阿伦尼乌斯方程、活化能、反应机理与决速步，以及时钟反应与连续监测等实验方法。
1. Rate Equations | 速率方程
species that affect it. For the general reaction aA + bB → products, the rate equation takes the form shown below, where k is the rate constant, [A] and [B] are concentrations in mol dm⁻³, and m and n are the orders of reaction with respect to A and B.
速率方程是一个由实验确定的数学关系式，把化学反应速率与影响速率的物种浓度联系起来。对于一般反应 aA + bB → 产物，速率方程的形式如下所示，其中 k 是速率常数，[A] 与 [B] 是浓度（单位 mol dm⁻³），m 与 n 分别是反应对 A 和 B 的级数。
rate = k[A]ᵐ[B]ⁿ
The orders m and n are determined experimentally, not from the balanced equation. The stoichiometric coefficients a and b normally have no direct relationship with m and n, and the orders may be zero, whole numbers, or even fractions or negative values in some complex or catalysed reactions.
级数 m 与 n 由实验确定，而不是从配平方程式中直接读出。化学计量数 a 与 b 通常与 m 和 n 没有直接关系；在某些复杂反应或催化反应中，级数可能为零、整数，甚至为分数或负值。
The rate equation only includes species whose concentration affects the rate. Solids and pure liquids are excluded because their concentration is effectively constant, and species that appear only after the rate-determining step also do not appear in the rate equation.
速率方程只包含那些浓度会影响速率的物种。固体与纯液体被排除在外，因为它们的浓度实际上是恒定的；只出现在决速步之后的物种也不会出现在速率方程中。
2. Orders of Reaction and the Rate Constant | 反应级数与速率常数
reaction with respect to a given reactant tells you how the rate depends on that reactant' espect to A: the rate is directly proportional to [A], so doubling [A] doubles the rate. If m = 2, the reaction is second order with respect to A: doubling [A] increases the rate by a factor of 2² = 4.
反应对某一反应物的级数表明了速率如何依赖于该反应物的浓度。若 m = 0，反应对 A 为零级：改变 [A] 对速率没有影响。若 m = 1，反应对 A 为一级：速率与 [A] 成正比，[A] 加倍则速率加倍。若 m = 2，反应对 A 为二级：[A] 加倍会使速率增大到 2² = 4 倍。
all order is 3: first order in A and second order in B. The overall order determines the shape of the rate-concentration graph and the units of k.
总反应级数等于各分级数之和 m + n。例如，若 rate = k[A][B]²，则总级数为 3：对 A 为一级，对 B 为二级。总级数决定了速率-浓度图的形状以及 k 的单位。
The rate constant k is independent of concentration but depends strongly on temperature. At the same concentrations, a larger k means a faster reaction. Its value also depends on the presence of a catalyst and on the solvent used, so k is only constant for a given reaction under fixed conditions.
速率常数 k 与浓度无关，但与温度密切相关。在相同浓度下，k 越大表示反应越快。k 的数值还取决于是否有催化剂以及所用溶剂，因此只有在固定条件下，k 对某一给定反应才是常数。
3. Units of the Rate Constant | 速率常数的单位
Rate is normally measured in mol dm⁻³ s⁻¹, while concentrations are in mol dm⁻³. Because the overall order changes how many concentration terms multiply together in the rate equation, the units of k also change with the overall order.
速率通常以 mol dm⁻³ s⁻¹ 为单位，而浓度以 mol dm⁻³ 为单位。由于总级数决定了速率方程中浓度项相乘的个数，所以 k 的单位随总级数而改变。
he expression below. Substituting n = 0, 1, 2 or 3 produces the units listed in the table.
对于总级数为 n 的反应，把速率方程重排即可得到下面的通式。代入 n = 0、1、2 或 3，就得到表中列出的单位。
units of k = mol¹⁻ⁿ dm³ⁿ⁻³ s⁻¹
| Overall order | 总级数 | Rate equation example | 速率方程示例 | Units of k | k 的单位 |
|---|---|---|
| 0 | rate = k | mol dm⁻³ s⁻¹ |
| 1 | rate = k[A] | s⁻¹ |
| 2 | rate = k[A]² or rate = k[A][B] | mol⁻¹ dm³ s⁻¹ |
| 3 | rate = k[A][B]² | mol⁻² dm⁶ s⁻³ |
Exam questions frequently ask you to deduce the units of k from the rate equation. Always write the rate equation first, state the overall order, then cancel the concentration units algebraically rather than memorising the table alone.
考试中经常要求由速率方程推导 k 的单位。务必先写出速率方程，说明总级数，然后用代数方法约去浓度单位，而不是单纯死记表格。
4. Determining Orders from Experimental Data | 由实验数据确定级数
Orders are found by the initial rates method. The initial rate is measured for several experiments in which the concentration of one reactant is changed while the others are kept constant. Comparing pairs of experiments shows how the rate responds to each concentration change.
级数通过初始速率法测定。在多组实验中，只改变一种反应物的浓度而保持其他反应物浓度不变，测量初始速率。通过比较成对的实验数据，可以看出速率如何随每一种浓度的变化而变化。
Consider the data below for a reaction between A and B. Between experiments 1 and 2, [A] doubles while [B] is constant, and the rate doubles, so the reaction is first order in A. Between experiments 1 and 3, [B] doubles while [A] is constant, and the rate increases by a factor of 4 (2²), so the reaction is second order in B.
考察下表中 A 与 B 反应的数据。实验 1 与 2 之间，[B] 不变而 [A] 加倍，速率也随之加倍，故反应对 A 为一级。实验 1 与 3 之间，[A] 不变而 [B] 加倍，速率增大到 4 倍（2²），故反应对 B 为二级。
| Experiment | 实验 | [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 | 4.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 8.0 × 10⁻³ |
0⁻³ ÷ (0.10 × 0.10²) = 2.0 mol⁻² dm⁶ s⁻¹. Always quote the units with the numerical value.
因此速率方程为 rate = k[A][B]²。代入实验 1 的数据可得 k = 2.0 × 10⁻³ ÷ (0.10 × 0.10²) = 2.0 mol⁻² dm⁶ s⁻¹。给出数值时必须同时写出单位。
5. Concentration-Time and Rate-Concentration Graphs | 浓度-时间图与速率-浓度图
Concentration-time graphs show how the concentration of a reactant falls as the reaction proceeds. The gradient of the tangent at any point equals the rate at that instant. A curve whose gradient decreases steadily indicates that the rate is falling as the reactant is used up.
浓度-时间图展示反应物浓度随反应进行而下降的过程。曲线上任意一点切线的斜率等于该时刻的反应速率。若曲线斜率不断减小，说明随着反应物被消耗，反应速率在降低。
For a first order reaction, the half-life is constant: the time taken for the concentration to halve is the same at every stage. This is a defining feature of first order behaviour and is related to the rate constant by the expression below.
对于一级反应，半衰期是恒定的：浓度减半所需的时间在反应的任何阶段都相同。这是一级反应的特征，并与速率常数存在下面的关系。
t½ = ln 2 ÷ k = 0.693 ÷ k
Rate-concentration graphs are plotted using initial rates. A horizontal line means zero order, a straight line through the origin means first order, and a curve that rises increasingly steeply indicates second order. For second order, a plot of rate against [A]² is a straight line through the origin.
速率-浓度图使用初始速率作数据。水平直线表示零级，过原点的直线表示一级，而上升越来越陡的曲线表示二级。对于二级反应，以 rate 对 [A]² 作图得到一条过原点的直线。
6. The Arrhenius Equation | 阿伦尼乌斯方程
The Arrhenius equation links the rate constant to temperature and activation energy. It explains why a small rise in temperature produces a large increase in rate, and it provides the quantitative basis for determining activation energy from experimental data.
阿伦尼乌斯方程把速率常数与温度和活化能联系起来。它解释了为什么温度略微升高就会使反应速率大幅增加，并为由实验数据测定活化能提供了定量基础。
k = A e−Eₐ ÷ RT
R is the gas constant 8.31 J mol⁻¹ K⁻¹, and T is the temperature in kelvin. Note that T = θ + 273 when converting from degrees Celsius.
式中，k 是速率常数，A 是前指数因子（也称频率因子），Eₐ 是活化能（单位 J mol⁻¹），R 是气体常数 8.31 J mol⁻¹ K⁻¹，T 是热力学温度（单位 K）。由摄氏度换算时，T = θ + 273。
The term e−Eₐ ÷ RT is the fraction of collisions that possess at least the activation energy. As T rises, this exponential factor increases, so k increases and the reaction is faster.
其中 e−Eₐ ÷ RT 这一项表示具有至少等于活化能的碰撞所占的比例。当 T 升高时，该指数因子增大，因此 k 增大，反应变快。
7. Activation Energy and the Exponential Factor | 活化能与指数因子
Activation energy is the minimum energy that colliding particles must possess for a successful reaction. The exponential term in the Arrhenius equation is very sensitive to Eₐ: a reaction with a high activation energy has a small exponential factor and a small rate constant at a given temperature.
活化能是发生成功反应的碰撞粒子必须具备的最低能量。阿伦尼乌斯方程中的指数项对 Eₐ 非常敏感：活化能高的反应，指数因子很小，在给定温度下的速率常数也很小。
A catalyst increases the rate by providing an alternative pathway with a lower activation energy. Because Eₐ is smaller, the exponential factor becomes larger at the same temperature, so k increases. The catalyst does not change the enthalpy change of the reaction and is not consumed overall.
催化剂通过提供一条活化能更低的替代路径来加快反应。由于 Eₐ 减小，在相同温度下指数因子变大，因此 k 增大。催化剂不改变反应的焓变，且在总反应中不被消耗。
Increasing temperature has two effects: particles collide more frequently, and a much greater proportion of collisions exceed the activation energy. The second effect dominates, which is why a rise of 10 °C can roughly double the rate of many reactions near room temperature.
升高温度有双重作用：粒子碰撞频率增大，而且超过活化能的碰撞所占比例大幅增加。第二种作用占主导，这就是为什么在室温附近温度升高 10 °C 常可使许多反应的速率大约加倍。
8. Applying the Arrhenius Equation | 阿伦尼乌斯方程的应用
Taking natural logarithms of both sides of the Arrhenius equation gives a linear form. Plotting ln k against 1 ÷ T gives a straight line with gradient −Eₐ ÷ R and intercept ln A, which is the standard graphical method for finding activation energy.
对阿伦尼乌斯方程两边取自然对数，可得到线性形式。以 ln k 对 1 ÷ T 作图得到一条直线，斜率为 −Eₐ ÷ R，截距为 ln A，这是求活化能的标准作图法。
ln k = ln A − Eₐ ÷ (RT)
If two rate constants, k₁ and k₂, are known at two temperatures, T₁ and T₂, the activation energy can be calculated directly using the two-point form below without drawing a graph.
若已知两个温度 T₁ 与 T₂ 下的两个速率常数 k₁ 与 k₂，可直接用下面的两点式求算活化能，无需作图。
ln(k₂ ÷ k₁) = (−Eₐ ÷ R) × (1 ÷ T₂ − 1 ÷ T₁)
ol⁻¹ K⁻¹, and remember that the calculated Eₐ will be in J mol⁻¹. Divide by 1000 to convert to kJ mol⁻¹, which is the unit normally expected in exam answers.
使用该式时，温度必须用 K，R 取 8.31 J mol⁻¹ K⁻¹，并注意算出的 Eₐ 单位为 J mol⁻¹。除以 1000 可换算为 kJ mol⁻¹，这也是考试答案中通常要求的单位。
9. Reaction Mechanisms and the Rate-Determining Step | 反应机理与决速步
A reaction mechanism is a proposed sequence of elementary steps that together give the overall equation. The rate-determining step is the slowest step, and it controls the overall rate because the reaction cannot proceed faster than its slowest stage.
反应机理是为得到总反应方程式而提出的若干基元步骤的组合。决速步是最慢的一步，它控制着总反应速率，因为反应不可能比它最慢的阶段更快。
The rate equation reflects the species involved in the rate-determining step and in any faster steps that precede it. Intermediates, which are formed in one step and consumed in a later one, do not appear in the rate equation. A catalyst may appear in the rate equation if it is involved in or before the rate-determining step.
速率方程反映的是参与决速步以及在此之前的任何较快步骤的物种。中间体在某一步生成、又在后续步骤中被消耗，因此不出现在速率方程中。若催化剂参与决速步或出现在其之前，则可能出现在速率方程里。
ng step must involve a single molecule of A. Any reactant whose concentration does not appear in the rate equation, or which is zero order, must react after the rate-determining step.
例如，若某反应实验测得的速率方程为 rate = k[A]，则决速步中必须只涉及一个 A 分子。凡是浓度不出现在速率方程中的反应物，或者其级数为零的反应物，必定是在决速步之后才参与反应。
Exam questions often ask you to suggest a mechanism consistent with a given rate equation, or to explain why the rate equation does not match the stoichiometry of the overall equation. In both cases, identify the rate-determining step first.
考试常要求写出与给定速率方程相符的机理，或解释为何速率方程与总反应方程式的化学计量关系不一致。这两种情况下，都应先确定决速步。
10. Practical Methods: Clock Reactions and Continuous Monitoring | 实验方法：时钟反应与连续监测
A clock reaction is used to measure an initial rate. A fixed, small amount of a limiting reagent is present, and the time taken for a visible change is recorded. The classic iodine clock uses iodide ions oxidised to iodine by peroxodisulfate(VI) ions, with a small amount of sodium thiosulfate and starch solution.
时钟反应用于测量初始速率。体系中只加入少量且固定量的限制试剂，记录出现明显颜色变化所需的时间。经典的碘钟反应中，碘离子被过二硫酸根离子氧化生成碘，同时加入少量硫代硫酸钠溶液和淀粉溶液。
As soon as all the thiosulfate has been used up, the iodine produced forms a blue-black complex with starch. The measured time is inversely proportional to the initial rate, so rate ∝ 1 ÷ t. Because only a very small fraction of the reactants has been consumed, the concentrations can be treated as the initial values.
当硫代硫酸钠恰好被消耗完，生成的碘立即与淀粉形成蓝黑色配合物。测得的时间与初始速率成反比，即 rate ∝ 1 ÷ t。由于被消耗的反应物比例极小，浓度可以视为初始浓度。
Other continuous methods include measuring the volume of gas evolved with a gas syringe, measuring mass loss as a gas escapes, following a colour change with a colorimeter, and monitoring conductivity when ions are produced or removed. Each method must be linked to the quantity that is directly proportional to concentration.
其他连续监测方法包括：用注射器测量生成气体的体积、测量气体逸出造成的质量损失、用比色计跟踪颜色变化，以及监测离子生成或消耗引起的电导率变化。每种方法都要说明所测量与浓度成正比的关系。
For a fair comparison of rates, the same temperature, the same concentrations, the same volume of solution and the same apparatus should be used. Any variation in these variables introduces uncertainty into the calculated orders or activation energy.
为了公平比较速率，必须使用相同的温度、相同的浓度、相同的溶液体积和相同的仪器。这些变量中任何一项改变，都会给算得的级数或活化能带来不确定性。
11. Exam Technique and Common Mistakes | 考试技巧与常见错误
A frequent error is reading orders directly from the balanced equation. Always state that the orders are determined experimentally, and use rate data to justify any order you quote.
常见错误之一是直接从配平方程式中读出级数。必须说明级数由实验确定，并用速率数据来论证你所给出的级数。
Another common mistake is forgetting the units of k, or using the wrong units for the rate constant. Write the rate equation first and then derive the units algebraically. In calculations involving the Arrhenius equation, convert temperatures to kelvin and keep the gas constant consistent with the energy units.
另一个常见错误是漏写 k 的单位，或用错速率常数的单位。应先写出速率方程，再用代数方法推导单位。在涉及阿伦尼乌斯方程的计算中，要把温度换算为 K，并保持气体常数与能量单位一致。
When a graph is involved, check whether the axes are ln k against 1 ÷ T or k against T. Only the linearised plot gives an accurate activation energy, because a plot of k against T is curved and does not yield a straightforward gradient.
涉及作图题时，要看清坐标轴是 ln k 对 1 ÷ T，还是 k 对 T。只有线性化的图才能给出准确的活化能，因为 k 对 T 作图是曲线，其斜率并不能直接用于求算 Eₐ。
Finally, remember that a catalyst lowers the activation energy of both the forward and the reverse reaction by the same amount, so it increases the rate
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