📚 Edexcel A-Level Combined Science 043: Kinetics, Equilibria and Energetics | Edexcel A-Level 综合科学 043：动力学、平衡与能量学
This revision guide covers the core physical chemistry ideas in Edexcel A-Level Combined Science topic 043: measuring rates, the rate equation, activation energy, dynamic equilibrium, equilibrium constants, Le Chatelier's principle, pH, buffers and enthalpy cycles. Understanding these ideas together is essential for quantitative and qualitative questions on Paper 1 and Paper 2, because examiners frequently combine two or three of them inside a single structured question.
本复习指南涵盖 Edexcel A-Level 综合科学主题 043 的核心物理化学概念：反应速率测量、速率方程、活化能、动态平衡、平衡常数、勒夏特列原理、pH、缓冲溶液以及焓变循环。把这些概念联系起来理解，对 Paper 1 和 Paper 2 中的定量题与定性题都至关重要，因为考官经常在同一道结构化大题里同时考查两三个知识点。
1. Measuring Reaction Rate | 反应速率的测量
The rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time. In the laboratory you can follow the rate by measuring gas volume, mass loss, colour intensity or pH change over time.
化学反应速率定义为单位时间内反应物或生成物浓度的变化。在实验室中，你可以通过测量气体体积、质量损失、颜色强度或 pH 随时间的变化来跟踪反应速率。
For a reaction A + B → C, the average rate can be written as:
对于反应 A + B → C，平均速率可以表示为：
rate = −Δ[A]/Δt = −Δ[B]/Δt = +Δ[C]/Δt
The unit of rate is mol dm⁻³ s⁻¹ when concentration is measured in mol dm⁻³ and time in seconds. Reactant terms carry a minus sign so that the calculated rate is always positive, even though the concentration of A and B is falling.
当浓度单位为 mol dm⁻³、时间单位为秒时，速率的单位是 mol dm⁻³ s⁻¹。反应物项前加负号，这样即使 A 和 B 的浓度在下降，算出的速率仍然为正值。
- Recording the tangent gradient on a concentration–time graph gives the instantaneous rate at a chosen time.
在浓度–时间图上作切线，其斜率给出某一时刻的瞬时速率。 - Initial rate is often used because concentrations are known precisely at t = 0.
常用初始速率，因为 t = 0 时浓度已知且更易控制。 - verted burette: suitable for reactions that release a gas, such as Mg + 2HCl → MgCl₂ + H₂.
气体注射器或倒置量气管：适用于放出气体的反应，例如 Mg + 2HCl → MgCl₂ + H₂。 - suitable when a gas escapes from an open flask, such as CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.
电子天平测质量损失：适用于气体从敞口瓶逸出的反应，例如 CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。 - imetry: follows the fading of a coloured species, for example I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻.
比色法：跟踪有色物质的褪色，例如 I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻。 - easurement: useful for hydrolysis reactions such as CH₃COOCH₂CH₃ + H₂O → CH₃COOH + C₂H₅OH.
pH 或电导率测量：适用于水解类反应，例如 CH₃COOCH₂CH₃ + H₂O → CH₃COOH + C₂H₅OH。 - Clock reaction with a visible end point, for example the iodine–thiosulfate 'disappearing cross' method.
带明显终点的计时反应，例如碘–硫代硫酸盐的"消失十字"法。
When you plot the results, the shape of the curve matters as much as the numbers. A steep initial section means a fast initial rate; a curve that flattens shows the rate decreasing because the reactants are being used up. Always quote the gradient with units, and remember that a tangent must touch the curve at exactly one point.
作图时，曲线形状和数值同样重要。起始段很陡说明初始速率快；曲线逐渐变平说明反应物被消耗后速率下降。写斜率时一定要带单位，并记住切线必须与曲线只接触一个点。
2. Rate Equation and Order of Reaction | 速率方程与反应级数
n A + B → products, the rate equation takes 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.
速率方程把反应速率与决速步中出现的物种浓度联系起来。对于一般反应 A + B → 生成物，速率方程形式如下，其中 m 是对 A 的级数，n 是对 B 的级数，k 是速率常数。
rate = k [A]ᵐ [B]ⁿ
The order with respect to a reactant is the power to which its concentration is raised in the experimentally determined rate equation. The overall order is the sum m + n. Orders are not the same as stoichiometric coefficients: they can be zero, fractional or negative, and they must be found by experiment.
某反应物的级数，就是实验测得的速率方程中该反应物浓度的指数。总级数是 m + n 之和。级数与化学计量数并不相同：级数可以是零、分数甚至负数，而且必须通过实验确定。
| Overall order | Rate–concentration graph | Units of k |
| 0 | Horizontal straight line | mol dm⁻³ s⁻¹ |
| 1 | Straight line through the origin | s⁻¹ |
| 2 | Upward curve through the origin | dm³ mol⁻¹ s⁻¹ |
| 3 | Steep upward curve | dm⁶ mol⁻² s⁻¹ |
n large excess such as water in an aqueous hydrolysis. A first-order reactant doubles the rate when its concentration doubles; a second-order reactant quadruples the rate, because 2² = 4.
零级反应物根本不出现在速率方程中，通常说明它不参与决速步，或者它处于大量过量状态（例如水溶液水解中的水）。一级反应物浓度加倍时速率加倍；二级反应物浓度加倍时速率变为原来的四倍，因为 2² = 4。
3. Determining Orders: Initial Rates and Graph Methods | 确定级数：初始速率法与图像法
There are three standard experimental approaches for finding orders: the initial rates method, the concentration–time graph method, and the rate–concentration graph method. All three rely on changing one concentration at a time while keeping everything else constant.
确定反应级数有三种标准实验方法：初始速率法、浓度–时间图法和速率–浓度图法。三者都要求在改变某一种浓度的同时保持其他条件不变。
In the initial rates method you run several experiments with different starting concentrations and measure the initial rate each time. Comparing two experiments in which only [A] changes lets you deduce the order with respect to A. For example, if doubling [A] doubles the initial rate, the order with respect to A is 1; if the rate is unchanged, the order is 0.
初始速率法要做多组实验，使用不同的起始浓度，并分别测出初始速率。比较只有 [A] 发生变化的两组实验，就能推断出对 A 的级数。例如 [A] 加倍而初始速率也加倍，则对 A 的级数为 1；若速率不变，则为 0。
In the graph method you plot concentration against time and inspect the shape of the curve. The half-life is the time taken for the concentration of a reactant to fall to half of its value.
图像法是作浓度对时间的图，然后观察曲线形状。半衰期指反应物浓度降到其初始值一半所需的时间。
t½ = ln 2 / k ≈ 0.693 / k
- Zero order: the concentration–time graph is a straight line with a constant negative gradient; half-life keeps decreasing.
零级：浓度–时间图为斜率恒定的直线；半衰期不断缩短。 - First order: the concentration–time graph is an exponential decay curve; the half-life is constant no matter what starting concentration you use.
一级：浓度–时间图为指数衰减曲线；无论起始浓度如何，半衰期都恒定。 - Second order: the concentration–time graph curves more sharply than for first order; half-life doubles each time the concentration halves.
二级：浓度–时间图弯曲程度比一级更大；浓度每减半一次，半衰期加倍。
A common exam trap is to assume the orders from the balanced equation. Always state that the orders are found experimentally, and quote at least one pair of rate and concentration values to justify your conclusion.
常见失分点是根据配平方程式直接推断级数。答题时务必说明级数由实验测得，并引用至少一组速率与浓度的数据来支持你的结论。
4. The Rate Constant, Arrhenius Equation and Activation Energy | 速率常数、阿伦尼乌斯方程与活化能
The rate constant k is independent of concentration but depends strongly on temperature. Its magnitude at a given temperature reflects how fast the reaction proceeds when all reactant concentrations are 1 mol dm⁻³.
速率常数 k 与浓度无关，但对温度非常敏感。在某一温度下 k 的大小，反映了当所有反应物浓度均为 1 mol dm⁻³ 时反应进行的快慢。
The temperature dependence of k is described by the Arrhenius equation, in which A is the pre-exponential factor (related to collision frequency and orientation) and Eₐ is the activation energy.
k 随温度的变化由阿伦尼乌斯方程描述，其中 A 是前指数因子（与碰撞频率和取向有关），Eₐ 是活化能。
k = A × e−Eₐ / RT
ln k = ln A − Eₐ / (R T)
Plotting ln k on the y-axis against 1/T on the x-axis gives a straight line whose gradient is −Eₐ/R and whose intercept on the y-axis is ln A. Rearranging gives the gradient relationship used in calculations:
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，纵截距为 ln A。整理后得到计算中常用的斜率关系式：
gradient = −Eₐ / R
The gas constant R is 8.31 J K⁻¹ mol⁻¹. Activation energy is normally quoted in kJ mol⁻¹, so remember to divide by 1000 after using a gradient in J mol⁻¹. A useful check is that the answer should be positive and typically between 20 and 300 kJ mol⁻¹ for a normal reaction.
气体常数 R = 8.31 J K⁻¹ mol⁻¹。活化能通常以 kJ mol⁻¹ 表示，因此用 J mol⁻¹ 的斜率算出结果后要除以 1000。一个实用的检查方法是：答案应为正值，一般反应的活化能通常落在 20 至 300 kJ mol⁻¹ 之间。
5. Collision Theory and Catalysis | 碰撞理论与催化
Collision theory states that particles must collide with energy greater than or equal to the activation energy, and with the correct orientation, before a reaction can occur. Only a small fraction of collisions satisfy both conditions, which is why most reactions are slower than a simple collision-frequency estimate would suggest.
碰撞理论认为，粒子必须以致活化能或更高能量的碰撞、并且取向正确，反应才能发生。只有一小部分碰撞同时满足这两个条件，这就是为什么大多数反应比单纯按碰撞频率估算的结果要慢。
Raising the temperature increases the rate for two reasons: particles move faster so collisions are more frequent, and a much larger fraction of collisions now exceeds the activation energy. This second effect dominates, which is why a 10 °C rise often roughly doubles the rate.
升温提高速率有两个原因：粒子运动更快，碰撞更频繁；同时超过活化能的碰撞比例大幅增加。第二个效应占主导，因此温度升高 10 °C 往往能使速率大约翻倍。
- A homogeneous catalyst is in the same phase as the reactants, for example Fe²⁺/Fe³⁺ ions in the peroxodisulfate–iodide reaction.
均相催化剂与反应物处于同一相，例如过二硫酸盐–碘离子反应中的 Fe²⁺/Fe³⁺ 离子。 - A heterogeneous catalyst is in a different phase, usually a solid with gaseous or aqueous reactants, for example iron in the Haber process.
非均相催化剂与反应物不在同一相，通常是固体催化剂配气体或溶液反应物，例如合成氨中的铁催化剂。 - A catalyst provides an alternative reaction pathway with a lower activation energy; it does not change the enthalpy change of the reaction.
催化剂提供一条活化能更低的替代反应路径；它不改变反应的焓变。 - A catalyst speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium or the value of Kc.
催化剂同等加快正、逆反应，因此不改变平衡位置，也不改变 Kc 的数值。
A Boltzmann distribution curve is often required in answers. When you sketch it, remember that raising the temperature flattens and broadens the curve, shifting the peak to higher energy, while the total area under the curve stays the same. Drawing the activation energy line at the same position shows the shaded area beyond Eₐ growing much bigger.
答题时常常需要画玻尔兹曼分布曲线。作图时记住：升温使曲线变得更平缓、更宽，峰值向高能方向移动，而曲线下总面积保持不变。在相同位置画出活化能线，可以看到 Eₐ 以外的阴影面积显著增大。
6. Dynamic Equilibrium and Le Chatelier's Principle | 动态平衡与勒夏特列原理
A reversible reaction reaches dynamic equilibrium in a closed system when the rate of the forward reaction equals the rate of the reverse reaction. At this point the concentrations of reactants and products remain constant, although both reactions continue to occur.
在封闭体系中，当正反应速率等于逆反应速率时，可逆反应达到动态平衡。此时反应物和生成物的浓度保持不变，但正、逆反应仍在持续进行。
Le Chatelier's principle states that if a system at equilibrium is subjected to a change, the position of equilibrium shifts in the direction that tends to oppose the change. This principle allows you to predict qualitatively how the yield of a product responds to changes in concentration, pressure and temperature.
勒夏特列原理指出：如果改变处于平衡状态的体系的条件，平衡将向削弱这种改变的方向移动。利用这一原理，可以定性预测产物产率如何随浓度、压强和温度的变化而改变。
| Change | Effect on equilibrium | Effect on Kc |
| Increase [reactant] | Shifts to the right | Unchanged |
| Increase pressure | Shifts to the side with fewer gas moles | Unchanged |
| Increase temperature | Shifts in the endothermic direction | Changes |
| Add a catalyst | No shift | Unchanged |
Temperature is the only factor that changes the value of the equilibrium constant. Concentration, pressure and catalysts change the position of equilibrium but leave Kc untouched, because Kc is a constant at a given temperature.
温度是唯一能改变平衡常数数值的因素。浓度、压强和催化剂只改变平衡位置，不改变 Kc，因为 Kc 在给定温度下是常数。
In industrial questions you must balance equilibrium yield against rate. In the Haber process, N₂ + 3H₂ ⇌ 2NH₃ is exothermic, so a low temperature favours a high yield, but the rate would be too slow; a compromise of about 450 °C with 200 atm and an iron catalyst is used.
在工业流程题中，必须在平衡产率与反应速率之间权衡。合成氨反应 N₂ + 3H₂ ⇌ 2NH₃ 是放热的，低温有利于提高产率，但速率太慢；因此采用约 450 °C、200 atm 并加铁催化剂的折中条件。
7. The Equilibrium Constant Kc and Kp | 平衡常数 Kc 与 Kp
For the general equilibrium aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is written as the product of the equilibrium concentrations of the products, each raised to its stoichiometric coefficient, divided by the corresponding term for the reactants.
对于一般平衡 aA + bB ⇌ cC + dD，以浓度表示的平衡常数写成：生成物平衡浓度的幂（指数为其化学计量数）之积，除以反应物的相应项。
Kc = [C]ᶜ [D]ᵈ / ( [A]ᵃ [B]ᵇ )
expression because their concentrations are effectively constant. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant is simply Kc = [CO₂].
纯固体和纯液体不写入表达式，因为它们的浓度实际上是常数。例如在 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 中，平衡常数简化为 Kc = [CO₂]。
For gaseous equilibria you can work with partial pressures instead of concentrations. The partial pressure of a gas is its mole fraction multiplied by the total pressure, and the equilibrium constant written in these terms is Kp.
对于气体平衡，可以用分压代替浓度。某气体的分压等于其摩尔分数乘以总压，以此表示的平衡常数就是 Kp。
p(A) = mole fraction of A × total pressure
Kp = p(C)ᶜ p(D)ᵈ / ( p(A)ᵃ p(B)ᵇ )
When you quote Kp you must include units, calculated from the units of pressure (kPa, atm or Pa) raised to the appropriate power. If both sides of the equation have the same number of gas moles, Kp is dimensionless, as in H₂(g) + I₂(g) ⇌ 2HI(g).
写 Kp 时必须带单位，单位由压强的单位（kPa、atm 或 Pa）按相应的指数组合得到。若方程式两边气体物质的量相同，则 Kp 无量纲，例如 H₂(g) + I₂(g) ⇌ 2HI(g)。
- A large value of Kc (much greater than 1)
广告 / SponsoredCambridge IGCSE Physics 4th edition 剑桥IGCSE物理教材 第4版 英文原版进口图…物理 备考教辅 · 正版现货 · 学习备考通用查看详情 / View on Taobao参考价 ¥340.0广告 / Sponsored新版剑桥IBDP国际文凭高中教材(2025考纲版)Cambridge Biology for the IB Diplo…生物 备考教辅 · 正版现货 · 学习备考通用查看详情 / View on Taobao参考价 ¥265.0
更多咨询请联系16621398022（同微信）
备考资讯每日更新，关注不错过考纲变化 · 科目分析 · 学习方法
在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图公众号：tutorhao