📚 Edexcel A-Level Chemistry Topic 16: Kinetics II | 爱德思 A-Level 化学：专题 16 动力学 II
Kinetics II builds directly on the AS-level ideas of collision theory and rates of reaction, but it moves from describing how fast a reaction goes to quantifying it mathematically. In this topic you must be able to deduce orders of reaction from experimental data, write rate equations, calculate the rate constant k together with its units, use the half-life relationship for first-order reactions, link rate equations to reaction mechanisms through the rate-determining step, and apply the Arrhenius equation to activation energy. These skills are examined through structured calculations, graph work and mechanistic reasoning, so accuracy with units and algebra matters as much as chemical understanding.
动力学 II 在 AS 阶段的碰撞理论与反应速率基础上直接延伸，但从"描述反应有多快"走向"用数学定量表达"。在本专题中，你必须能够从实验数据推导反应级数、写出速率方程、计算速率常数 k 及其单位、使用一级反应的半衰期关系、通过决速步把速率方程与反应机理联系起来，并运用阿伦尼乌斯方程处理活化能。这些技能通过结构化计算、图像分析和机理推理来考查，因此单位与代数的准确性同化学理解同样重要。
1. Rate Equations and Orders of Reaction | 速率方程与反应级数
A rate equation expresses the rate of reaction as a product of a rate constant and the concentrations of species raised to powers called orders. For a general reaction between A and B, the rate equation has the form shown below, where m is the order with respect to A, n is the order with respect to B, and k is the rate constant at that temperature.
速率方程把反应速率表示为速率常数与各物种浓度的幂（即反应级数）之积。对于 A 与 B 之间的一般反应，速率方程具有如下形式，其中 m 是关于 A 的级数，n 是关于 B 的级数，k 是该温度下的速率常数。
rate = k [A]ᵐ [B]ⁿ
The overall order of reaction is the sum m + n. A zero-order reactant has no effect on the rate, a first-order reactant changes the rate proportionally, and a second-order reactant changes the rate by the square of the concentration change. Orders may be zero, positive integers, or occasionally fractional or negative, but they are always determined experimentally.
反应的总级数是 m + n 之和。零级反应物对速率没有影响；一级反应物使速率按比例变化；二级反应物使速率按浓度变化的平方变化。级数可以是零、正整数，偶尔也可以是分数或负数，但都必须由实验测定。
e and carbon monoxide, NO₂ + CO → NO + CO₂, is second order in NO₂ and zero order in CO, giving rate = k[NO₂]². This happens because the reaction proceeds in more than one step and the rate depends only on the slowest step.
反应级数不一定等于配平方程式中的化学计量系数。例如二氧化氮与一氧化碳的反应 NO₂ + CO → NO + CO₂ 对 NO₂ 为二级、对 CO 为零级，故 rate = k[NO₂]²。这是因为反应分多步进行，速率只取决于最慢的一步。
2. Initial Rates Method | 初始速率法
ne reactant is changed while all other conditions, especially temperature, are held constant. The initial rate is used because at t = 0 the concentrations are exactly known and no products have accumulated to interfere.
初始速率法通过比较若干次实验起始时刻的速率来确定级数，每次实验只改变一种反应物的浓度，其他条件（尤其是温度）保持恒定。使用初始速率是因为在 t = 0 时浓度精确已知，且尚无产物积累造成干扰。
To find the order with respect to a reactant, look at how the initial rate responds when its concentration is multiplied by a simple factor.
要确定某反应物的级数，观察当其浓度乘以一个简单倍数时初始速率如何响应。
- If doubling the concentration has no effect on the initial rate, the order is 0.
若浓度加倍对初始速率没有影响，则级数为 0。 - If doubling the concentration doubles the initial rate, the order is 1.
若浓度加倍使初始速率也加倍，则级数为 1。 - If doubling the concentration quadruples the initial rate, the order is 2.
若浓度加倍使初始速率变为四倍，则级数为 2。 - If doubling the concentration multiplies the rate by eight, the order is 3.
若浓度加倍使速率变为八倍，则级数为 3。 - If doubling the concentration multiplies the rate by 1.4, the order is 0.5 (a fractional order).
若浓度加倍使速率变为 1.4 倍，则级数为 0.5（分数级）。
Once the orders are known, the rate constant is found by substituting the data from any one experiment into the rate equation and solving for k. Always check that k comes out with the same value (within experimental error) for every experiment; if it does not, the proposed orders are wrong.
求出级数后，把任意一次实验的数据代入速率方程解出 k，即可得到速率常数。务必检验每次实验算出的 k 在实验误差范围内一致；若不一致，说明所设级数有误。
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 only on temperature (and on the presence of a catalyst), not on concentration. Since the rate always has units of mol dm⁻³ s⁻¹ and the concentration term has units of (mol dm⁻³) raised to the overall order, the units of k must compensate, so they change with overall order.
速率常数 k 是速率方程中的比例常数。它的数值只取决于温度（以及是否有催化剂），与浓度无关。由于速率单位恒为 mol dm⁻³ s⁻¹，而浓度项单位为 (mol dm⁻³) 的总级数次幂，k 的单位必须与之相抵，因此会随总级数变化。
k = rate ÷ (concentration units)overall order
| Overall order | Units of k |
|---|---|
| 0 | mol dm⁻³ s⁻¹ |
| 1 | s⁻¹ |
| 2 | mol⁻¹ dm³ s⁻¹ |
| 3 | mol⁻² dm⁶ s⁻¹ |
red in mol dm⁻³. For example, when the overall order is 2, k = mol dm⁻³ s⁻¹ ÷ (mol dm⁻³)² = mol⁻¹ dm³ s⁻¹, which matches the table.
应记住通用规律而不是逐行背诵：总级数为 n 时，k 的单位是 mol¹⁻ⁿ dm³ⁿ⁻³ s⁻¹，其中浓度单位为 mol dm⁻³。例如总级数为 2 时，k = mol dm⁻³ s⁻¹ ÷ (mol dm⁻³)² = mol⁻¹ dm³ s⁻¹，与表格一致。
4. Continuous Monitoring and Rate–Concentration Graphs | 连续监测法与速率–浓度图
Instead of initial rates, a reaction can be followed continuously and the concentration of a reactant or product plotted against time. The rate at any instant is the gradient of the tangent to that curve at that time. Taking tangents at several points gives a set of (rate, concentration) pairs from a single experiment.
除了初始速率法，还可以连续监测反应，把某反应物或产物的浓度对时间作图。任意时刻的速率就是该点曲线切线的斜率。在多个点作切线，就能由一次实验得到一组（速率，浓度）数据。
Useful monitoring techniques include measuring the volume of gas evolved with a gas syringe, measuring the loss in mass on a balance for reactions producing a gas, following a colour change with a colorimeter, withdrawing samples and titrating them (quenching), and clock reactions such as the iodine clock or the sodium thiosulfate and hydrochloric acid cross method.
常用的监测手段包括：用气体注射器测量放出的气体体积；对产生气体的反应用天平测量质量损失；用比色计跟踪颜色变化；取样后立即淬灭并滴定；以及碘钟反应或硫代硫酸钠与盐酸的"十字法"等计时反应。
Graphs of rate against concentration then reveal the order directly.
随后把速率对浓度作图，便可直接看出级数。
- Rate against [A] horizontal line: zero order in A.
速率对 [A] 图为水平直线：对 A 为零级。 - Rate against [A] straight line through the origin: first order in A.
速率对 [A] 为过原点的直线：对 A 为一级。 - Rate against [A] upward curve that becomes a straight line when rate is plotted against [A]²: second order in A.
速率对 [A] 为向上弯曲的曲线，而速率对 [A]² 作图为直线：对 A 为二级。
5. Half-Life and First-Order Reactions | 半衰期与一级反应
The half-life, t½, is the time taken for the concentration of a reactant to fall to half of its value at that moment. For a first-order reaction the half-life is independent of concentration, so successive half-lives are equal. This constant half-life is itself evidence that a reaction is first order.
半衰期 t½ 是指某反应物浓度降到该时刻初始值一半所需的时间。对于一级反应，半衰期与浓度无关，因此逐次半衰期相等。半衰期恒定本身就是反应为一级的证据。
t½ = ln 2 ÷ k = 0.693 ÷ k
onstant to be found from a single half-life measurement. After n half-lives the remaining concentration is given by the expression below, which is useful for quick multiple-choice estimates.
该式可重排为 k = 0.693 ÷ t½，从而只需一次半衰期测量就能求得速率常数。经过 n 个半衰期后，剩余浓度可由下式给出，适合选择题中的快速估算。
[A] = [A]₀ × (½)ⁿ
rder reactions the half-life is inversely proportional to the initial concentration (it gets longer). Only the first-order case gives a constant half-life, so be careful not to apply t½ = 0.693 ÷ k to reactions of other orders.
对于零级反应，半衰期与初始浓度成正比（反应进行中半衰期越来越短）；对于二级反应，半衰期与初始浓度成反比（半衰期越来越长）。只有一级反应才有恒定半衰期，因此切勿把 t½ = 0.693 ÷ k 用于其他级数的反应。
6. Rate-Determining Step and Reaction Mechanisms | 决速步与反应机理
Most reactions occur in a series of elementary steps. The slowest step is called the rate-determining step (RDS), and it controls the overall rate because the reaction cannot proceed faster than its bottleneck. The rate equation therefore tells you which species are involved in, or formed before, the rate-determining step.
大多数反应由一系列基元步骤组成。最慢的一步称为决速步（RDS），它控制着总速率，因为反应不可能快于它的瓶颈。因此速率方程能告诉你哪些物种参与了决速步，或在决速步之前已经生成。
₂, which is first order in NO₂ and zero order in CO. A mechanism consistent with this is a slow first step in which two NO₂ molecules collide, followed by a fast second step.
以反应 NO₂ + CO → NO + CO₂ 为例，它对 NO₂ 为一级、对 CO 为零级。与之相符的机理是：第一步很慢，两个 NO₂ 分子碰撞；第二步很快。
- Step 1 (slow, rate-determining): 2NO₂ → NO₃ + NO
第 1 步（慢，决速步）：2NO₂ → NO₃ + NO - Step 2 (fast): NO₃ + CO → NO₂ + CO₂
第 2 步（快）：NO₃ + CO → NO₂ + CO₂ - Overall: NO₂ + CO → NO + CO₂, with rate = k[NO₂]²
总反应：NO₂ + CO → NO + CO₂，速率 = k[NO₂]²
The species NO₃ is an intermediate: it is produced in one step and consumed in another, so it does not appear in the overall equation or in the rate equation. A catalyst, by contrast, appears in the rate equation of the step it participates in but is regenerated at the end of the mechanism.
物种 NO₃ 是中间体：它在一步中生成、在另一步中被消耗，因此不出现在总方程或速率方程中。相反，催化剂出现在它所参与步骤的速率方程里，但在机理结束时被再生。
When the rate equation contains a species whose stoichiometric coefficient is 2, that usually means two molecules of it are involved in the slow step. When the rate equation contains a species that does not appear in the overall equation, that species must be formed in an earlier fast equilibrium step.
当速率方程中出现化学计量系数为 2 的物种时，通常意味着慢步骤中有两个该分子参与。当速率方程中出现总方程中不存在的物种时，该物种必定是在更早的快速平衡步骤中生成的。
7. 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 why catalysts, which lower the activation energy, speed up reactions.
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小幅升高会导致速率大幅增加，也解释了降低活化能的催化剂为何能加快反应。
k = A e−Eₐ ÷ RT
In this equation, A is the pre-exponential factor (related to the frequency and orientation of collisions), Eₐ is the activation energy in J mol⁻¹, R is the gas constant 8.314 J K⁻¹ mol⁻¹, and T is the temperature in kelvin. Note that Eₐ and R must be in the same energy units and that T must be converted from °C by adding 273.
式中 A 为指前因子（与碰撞频率和取向有关），Eₐ 为活化能，单位 J mol⁻¹，R 为气体常数 8.314 J K⁻¹ mol⁻¹，T 为热力学温度（K）。注意 Eₐ 与 R 必须使用相同能量单位，且 T 需由 °C 加 273 换算。
Taking natural logarithms converts the exponential equation into a straight-line form, which is the version used for graphical analysis.
取自然对数可把指数式化为直线形式，这也是作图分析所用的形式。
ln k = ln A − (Eₐ ÷ R) × (1 ÷ T)
Plotting ln k on the y-axis against 1 ÷ T on the x-axis therefore gives a straight line with gradient −Eₐ ÷ R and intercept ln A. As a rough rule of thumb, many reactions approximately double their rate for every 10 °C rise in temperature, which is a direct consequence of the exponential term.
因此，以 ln k 对 1 ÷ T 作图得到直线，斜率为 −Eₐ ÷ R，截距为 ln A。作为粗略经验规律，许多反应温度每升高 10 °C 速率约翻一倍，这正是指数项的直接结果。
8. Finding Activation Energy from Experimental Data | 由实验数据求活化能
To determine activation energy experimentally, measure the rate or the rate constant at several different temperatures, keeping all concentrations the same. Then calculate ln k and 1 ÷ T for each temperature, plot ln k against 1 ÷ T, draw a line of best fit, and find its gradient using two well-separated points on the line.
实验测定活化能的方法是：在若干不同温度下测量速率或速率常数，同时保持所有浓度不变。然后计算每个温度下的 ln k 与 1 ÷ T，作 ln k 对 1 ÷ T 的图，画出最佳拟合直线，并用线上两个相距较远的点求斜率。
gradient = −Eₐ ÷ R, so Eₐ = −gradient × R
Remember that the gradient is negative, so the two minus signs cancel and Eₐ comes out positive. A common exam error is to forget this and quote a negative activation energy. The result will be in J mol⁻¹, and it is usually sensible to convert to kJ mol⁻¹ for the final answer.
记住斜率是负值，因此两个负号相消，Eₐ 为正值。常见的考试错误是忘记这一点而写出负的活化能。结果单位为 J mol⁻¹，通常最后换算为 kJ mol⁻¹ 更为合适。
ximately 104 kJ mol⁻¹. If instead two data points are given, use the two-point form of the equation and substitute both (ln k, 1 ÷ T) pairs.
例题：ln k 对 1 ÷ T 的图斜率为 −12 500 K（1 ÷ T 的单位为 K⁻¹）。则 Eₐ = 12 500 × 8.314 = 103 925 J mol⁻¹，约为 104 kJ mol⁻¹。若题目给出两个数据点，则使用方程的两点形式，把两组（ln k, 1 ÷ T）代入即可。
9. Explaining the Effects of Temperature and Catalysts | 解释温度与催化剂的影响
Collision theory alone predicts that raising the temperature increases the frequency of collisions, but that effect is far too small to account for the observed increase in rate. The dominant factor is that a higher temperature shifts the Boltzmann distribution so that a much larger fraction of molecules possess energy greater than the activation energy.
仅用碰撞理论预测，升温会增加碰撞频率，但这一效应远不足以解释观察到的速率提升。主要因素是温度升高使玻尔兹曼分布右移，使能量大于活化能的分子比例大大增加。
A catalyst provides an alternative reaction pathway with a lower activation energy. Because Eₐ appears in the exponent of the Arrhenius equation, even a modest reduction in Eₐ produces a large increase in k. The catalyst is not used up, and it does not change the position of equilibrium or the enthalpy change of the reaction.
催化剂提供了一条活化能更低的替代反应路径。由于 Eₐ 出现在阿伦尼乌斯方程的指数中，即使适度降低 Eₐ 也会使 k 大幅增大。催化剂本身不被消耗，也不会改变平衡位置或反应的焓变。
Concentration affects the rate through the rate equation: increasing the concentration of a reactant that appears in the rate equation raises the rate because collisions involving that species become more frequent. Changing the concentration of a zero-order reactant has no effect at all, which is a strong hint that it is not involved in the rate-determining step.
浓度通过速率方程影响速率：增大出现在速率方程中的反应物浓度会提高速率，因为涉及该物种的碰撞更加频繁。改变零级反应物的浓度则毫无影响，这正是它不参与决速步的有力提示。
10. Worked Example: Orders, k and Units | 例题：级数、k 与单位
The table below gives initial rate data for the reaction between A and B at a fixed temperature. Use it to deduce the orders, write the rate equation, and calculate the rate constant with units.
下表给出固定温度下 A 与 B 反应的初始速率数据。据此推导级数、写出速率方程，并计算带单位的速率常数。
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 |
