📚 Mastering Rate Equations and Activation Energy | 掌握速率方程与活化能
Kinetics is the branch of chemistry that explains how fast reactions occur. In Edexcel A-Level Chemistry, you must connect experimental rate measurements with rate equations, rate constants, activation energy and multi-step reaction mechanisms.
动力学是化学中解释反应快慢的分支。在 Edexcel A-Level Chemistry 中，你必须将实验速率测量与速率方程、速率常数、活化能以及多步反应机理联系起来。
This guide walks through every skill the examiner tests: writing rate equations, deducing orders from initial-rate data, working out the units of k, reading concentration-time graphs, applying the Arrhenius equation, and linking all of this back to collision theory and reaction mechanisms.
本文将逐一梳理考官考查的全部技能：书写速率方程、由初始速率数据推导级数、求算 k 的单位、解读浓度-时间图、应用阿伦尼乌斯方程，并把这一切与碰撞理论和反应机理联系起来。
1. Key Definitions in Kinetics | 动力学关键定义
Before using rate equations, you need secure definitions. The rate of reaction is the change in concentration of a reactant or product per unit time, with units mol dm⁻³ s⁻¹. Rate always refers to a measured quantity, not to a theoretical prediction.
在使用速率方程之前，你需要掌握基本定义。反应速率是反应物或产物浓度在单位时间内的变化，单位为 mol dm⁻³ s⁻¹。速率始终指实测量，而不是理论预测值。
The order with respect to a species is the power to which its concentration is raised in the rate equation. Overall order is the sum of all individual orders. A order of zero means the species does not appear in the rate equation at all.
对某物种的级数是指其浓度在速率方程中的幂。总级数是所有分级数之和。级数为零意味着该物种根本不出现于速率方程中。
The rate constant k is the proportionality constant that links rate to concentration. It is fixed at a given temperature and changes only when temperature or a catalyst changes. Each reaction has its own value of k.
速率常数 k 是将速率与浓度联系起来的比例常数。它在给定温度下是固定的，只有温度改变或加入催化剂时才会变化。每个反应都有自己特有的 k 值。
Rates are usually followed by measuring a physical property over time: gas volume evolved, mass lost, colour intensity (absorbance), electrical conductivity, or by titrating withdrawn samples against acid or thiosulfate.
速率通常通过随时间测量某种物理性质来跟踪：放出气体的体积、减少的质量、颜色深浅（吸光度）、电导率，或取样品用酸或硫代硫酸盐滴定。
2. Rate Equations and Orders | 速率方程与反应级数
, the rate equation is shown below. The powers m and n must be found experimentally; they are not generally the coefficients from the balanced equation.
对于反应 A + B → 产物，速率方程如下所示。幂 m 和 n 必须通过实验确定；它们通常不是配平方程中的系数。
Rate = k[A]ᵐ[B]ⁿ
The constant k is the rate constant, which changes with temperature but not with concentration. The powers m and n are the orders with respect to A and B, and m + n is the overall order of reaction.
常数 k 是速率常数，随温度变化，但不随浓度变化。幂 m 和 n 分别是 A 和 B 的级数，m + n 是反应的总级数。
The effect of doubling a concentration depends entirely on the order, which is why a single measurement is never enough to deduce a rate law.
某一浓度翻倍所带来的影响完全取决于级数，这就是为什么仅凭一次测量永远无法推导出速率方程。
| Order | 级数 | Rate equation form | 速率方程形式 | Effect of doubling [A] | [A] 翻倍的影响 |
|---|---|---|
| Zero | 零级 | Rate = k | No change | 不变 |
| First | 一级 | Rate = k[A] | Rate doubles | 速率翻倍 |
| Second | 二级 | Rate = k[A]² | Rate × 4 | 速率 × 4 |
| Third | 三级 | Rate = k[A]³ | Rate × 8 | 速率 × 8 |
Orders are usually small whole numbers, but fractional and negative orders do exist. A negative order means that increasing that concentration actually slows the reaction down, which is often seen in acid-catalysed reactions where the acid is consumed.
级数通常是小整数，但也确实存在分数级数和负级数。负级数意味着增大该浓度反而使反应变慢，这在酸催化反应中常见，因为酸会被消耗。
3. Determining Orders from Experimental Data | 通过实验数据确定级数
Use the initial rates method: keep all other concentrations constant, change the concentration of one species, measure the initial rate, and compare the changes. If doubling [A] doubles the rate, the order with respect to A is first.
使用初始速率法：保持其他所有浓度不变，只改变一种物质的浓度，测量初始速率并比较变化。如果 [A] 翻倍使速率翻倍，则对 A 的级数为一级。
If doubling [A] quadruples the rate, the order is second. If changing [A] has no effect, the order is zero. Always use initial rates because product build-up or reverse reactions can interfere later in the reaction.
如果 [A] 翻倍使速率变为四倍，则级数为二级。如果 [A] 变化对速率无影响，则级数为零级。始终使用初始速率，因为产物积累或逆反应会在反应后期产生干扰。
Consider the following data for the reaction between A and B. Comparing experiments two at a time isolates one variable and makes the arithmetic quick and reliable.
考虑下面关于 A 与 B 反应的数据。每次比较两组实验，就能隔离一个变量，使计算快捷可靠。
| Experiment | 实验 | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 1.2 × 10⁻³ |
| 2 | 0.20 | 0.10 | 2.4 × 10⁻³ |
| 3 | 0.10 | 0.20 | 4.8 × 10⁻³ |
Experiments 1 and 2: [A] doubles while [B] is constant, and the rate doubles, so the order with respect to A is 1. Experiments 1 and 3: [B] doubles while [A] is constant, and the rate becomes four times larger, so the order with respect to B is 2.
实验 1 与 2：[B] 不变而 [A] 翻倍，速率翻倍，所以对 A 的级数为 1。实验 1 与 3：[A] 不变而 [B] 翻倍，速率变为四倍，所以对 B 的级数为 2。
If the ratio is not a simple power, check whether the concentrations were actually varied by exactly a factor of two.
因此总级数为 3，速率方程为 Rate = k[A][B]²。如果比值不是简单的幂次，请检查浓度变化是否确实正好是两倍。
4. The Rate Constant k and Its Units | 速率常数 k 及其单位
e units is to rearrange the rate equation so that k is the subject: k = Rate / ([A]ᵐ[B]ⁿ).
k 的值通过将某一组完整实验数据代入速率方程求得。求其单位的快捷方法是把速率方程重排，使 k 成为主体：k = Rate / ([A]ᵐ[B]ⁿ)。
2 mol⁻² dm⁶ s⁻¹. Note that the units of k depend on the overall order, not on the individual orders.
用上面的数据：k = 1.2 × 10⁻³ ÷ (0.10 × 0.10²) = 1.2 × 10⁻³ ÷ 1.0 × 10⁻³ = 1.2 mol⁻² dm⁶ s⁻¹。注意 k 的单位取决于总级数，而非各个分级数。
The units of k are found by dividing the units of rate by the units of concentration raised to the overall order. Because rate has units mol dm⁻³ s⁻¹ and concentration has units mol dm⁻³, the powers of mol and dm cancel in a predictable pattern.
k 的单位等于速率单位除以浓度单位的总级数次方。由于速率单位为 mol dm⁻³ s⁻¹、浓度单位为 mol dm⁻³，mol 与 dm 的幂次会按可预测的规律相消。
| Overall order | 总级数 | Units of k | k 的单位 |
|---|---|
| 0 | mol dm⁻³ s⁻¹ |
| 1 | s⁻¹ |
| 2 | mol⁻¹ dm³ s⁻¹ |
| 3 | mol⁻² dm⁶ s⁻¹ |
Remember that k is a constant only at a fixed temperature. If the temperature rises, k increases, and the reaction becomes faster even though the concentrations have not changed.
请记住，k 只在固定温度下才是常数。如果温度升高，k 增大，即使浓度未变，反应也会变快。
5. Concentration-Time Graphs and Half-Life | 浓度-时间图与半衰期
A concentration-time graph shows how the concentration of a reactant falls as the reaction proceeds. The shape of the curve is a direct fingerprint of the order of reaction with respect to that reactant.
浓度-时间图显示反应物浓度随反应进行而下降的过程。曲线的形状是该反应物级数的直接“指纹”。
- Zero order: a straight line with a constant negative gradient | 零级：一条斜率恒为负的直线
- First order: a curve with a constant half-life and a constantly decreasing gradient | 一级：半衰期恒定、斜率不断减小的曲线
- Second order: a steeper initial fall with a half-life that increases as the reaction proceeds | 二级：起始下降更陡，半衰期随反应进行而变长
For a first-order reaction the half-life t½ is independent of the starting concentration, which gives the useful relationship shown below. This allows k to be calculated directly from a half-life measurement.
对于一级反应，半衰期 t½ 与起始浓度无关，由此得到如下实用关系式。这样就可以由半衰期的测量直接算出 k。
t½ = ln 2 / k ≈ 0.693 / k
A rate-concentration graph is the other powerful diagnostic tool. Plotting rate against concentration gives a horizontal line for zero order, a straight line through the origin for first order, and an upward curve for second order.
速率-浓度图是另一种有力的诊断工具。以速率对浓度作图：零级为水平线，一级为过原点的直线，二级则向上弯曲。
6. Activation Energy and the Arrhenius Equation | 活化能与阿伦尼乌斯方程
Activation energy Ea is the minimum kinetic energy that colliding particles must possess for a collision to result in a reaction. It is always positive and is measured in J mol⁻¹ or kJ mol⁻¹.
活化能 Ea 是碰撞粒子为发生反应所必须具备的最小动能。它始终为正值，单位为 J mol⁻¹ 或 kJ mol⁻¹。
Not every collision leads to reaction. Only collisions with energy greater than or equal to Ea, and with the correct orientation, are successful. This is the basis of collision theory.
并非每次碰撞都能引发反应。只有能量大于或等于 Ea、且取向正确的碰撞才有效。这就是碰撞理论的基础。
The way k varies with temperature is described by the Arrhenius equation, shown here in its exponential form. The pre-exponential factor A is related to the frequency of collisions and their orientation, and is treated as roughly constant over a small temperature range.
k 随温度变化的方式由阿伦尼乌斯方程描述，其指数形式如下。指前因子 A 与碰撞频率及取向有关，在小温度范围内可视为近似恒定。
k = A × e−Ea / RT
Taking natural logarithms of both sides converts the exponential relationship into a straight-line equation, which is far easier to handle graphically and is the form you should quote in an exam.
对两边取自然对数，可将指数关系转化为直线方程，这更适合用图解法处理，也是考试中应当引用的形式。
ln k = ln A − Ea / RT
In both equations, R is the gas constant, 8.314 J K⁻¹ mol⁻¹, and T is the temperature in kelvin. Always convert °C to K by adding 273 before using these equations.
两个方程中，R 是气体常数，8.314 J K⁻¹ mol⁻¹；T 是以开尔文表示的温度。使用这些方程前，必须先把 °C 加 273 换算成 K。
7. Using the Arrhenius Plot to Find Ea | 使用阿伦尼乌斯图求 Ea
hows that plotting ln k on the y-axis against 1/T on the x-axis gives a straight line with gradient −Ea/R and intercept ln A.
将对数形式与直线方程 y = mx + c 比较可知：以 ln k 为纵轴、1/T 为横轴作图，得到一条直线，其斜率为 −Ea/R，截距为 ln A。
Ea = − (gradient) × R
The gradient is negative because 1/T decreases as temperature rises, so ln k increases with temperature. Multiply the gradient by R and divide by 1000 to convert the result from J mol⁻¹ to kJ mol⁻¹.
斜率为负，因为温度升高时 1/T 减小，而 ln k 随温度升高而增大。将斜率乘以 R，再除以 1000，即可把结果由 J mol⁻¹ 换算为 kJ mol⁻¹。
If only two temperatures are available, combine the two Arrhenius expressions to eliminate A and solve for Ea directly. This two-point form is often quicker than plotting a full graph.
如果只有两个温度的数据，可以联立两个阿伦尼乌斯表达式消去 A，直接求 Ea。这种两点式通常比绘制完整图像更快。
ln(k₂ / k₁) = (Ea / R) × (1/T₁ − 1/T₂)
A large value of Ea gives a steep plot, meaning the rate is very sensitive to temperature. A small value of Ea gives a shallow plot, meaning the reaction is fast and only weakly temperature-dependent.
Ea 越大，图像越陡，说明速率对温度非常敏感；Ea 越小，图像越平缓，说明反应较快且对温度依赖较弱。
8. Energy Distribution and the Effect of Temperature | 能量分布与温度效应
The Maxwell-Boltzmann distribution shows the range of kinetic energies possessed by molecules in a gas or liquid at a given temperature. The area under the curve represents the total number of molecules.
麦克斯韦-玻尔兹曼分布显示了给定温度下气体或液体分子所具有的动能范围。曲线下方的面积代表分子总数。
The curve starts at the origin, rises to a peak and then tails off to the right. Only the small fraction of molecules in the tail beyond the activation energy threshold can react on collision.
曲线从原点出发，升至最高点后向右拖尾。只有超过活化能阈值的尾部那一小部分分子碰撞时才能发生反应。
Raising the temperature moves the whole distribution to higher energies, flattens the peak and broadens the curve. Crucially, the area beyond Ea increases dramatically, even though the total area stays the same.
升高温度会使整个分布向高能方向移动，峰变低、曲线变宽。关键在于，尽管总面积不变，超过 Ea 的那部分面积却大幅增加。
This explains why a modest rise in temperature, such as 10 K near room temperature, can roughly double the rate of many reactions. The increase
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