📚 Edexcel A-Level Combined Science 024: Rates, Equilibrium and Energetics | Edexcel A-Level 综合科学 024：反应速率、化学平衡与能量学
This revision guide covers the core physical chemistry ideas in Edexcel A-Level Combined Science 024: how fast reactions occur, how far they go, and the energy changes that drive them. It links rate equations, equilibrium constants, enthalpy, entropy and Gibbs free energy into one coherent framework.
本复习指南涵盖 Edexcel A-Level 综合科学 024 的核心物理化学内容：反应进行得多快、达到什么程度，以及推动反应的能量变化。它将速率方程、平衡常数、焓变、熵和吉布斯自由能整合为一个完整的知识框架。
Mastering these topics requires more than memorising definitions. You must be able to interpret experimental data, calculate constants from graphs, and apply principles such as Le Chatelier’s principle and Hess’s law to unfamiliar situations.
掌握这些主题不仅仅是背诵定义。你必须能够解释实验数据、通过图像计算常数，并能在陌生情境中应用勒夏特列原理和赫斯定律等原理。
1. Rate Equations and Order of Reaction | 速率方程与反应级数
For a reaction A + B → products, the rate equation has the general form rate = k[A]ᵐ[B]ⁿ. Here k is the rate constant, while m and n are the orders of reaction with respect to A and B. The overall order is m + n.
对于反应 A + B → 产物，速率方程的一般形式为 rate = k[A]ᵐ[B]ⁿ。其中 k 是速率常数，m 和 n 分别是反应物 A 和 B 的反应级数，总级数为 m + n。
Orders are not the same as stoichiometric coefficients unless the reaction is an elementary step. They must be determined experimentally by comparing initial rates while changing one concentration at a time.
除非反应是基元步骤，否则级数不一定等于化学计量数。级数必须通过实验测定，通常在一次只改变一种物质浓度的条件下比较初始速率。
For zero order, doubling [A] leaves the rate unchanged; for first order, doubling [A] doubles the rate; for second order, doubling [A] quadruples the rate.
零级反应中，[A] 加倍速率不变；一级反应中，[A] 加倍速率加倍；二级反应中，[A] 加倍速率变为原来的四倍。
Rate constants have units that depend on the overall order. For a zero-order reaction k has units mol dm⁻³ s⁻¹; for first order s⁻¹; for second order dm³ mol⁻¹ s⁻¹.
速率常数的单位取决于总级数。零级反应 k 的单位为 mol dm⁻³ s⁻¹，一级反应为 s⁻¹，二级反应为 dm³ mol⁻¹ s⁻¹。
| Order | Effect of doubling concentration | Units of k |
|---|---|---|
| 0 | No change | mol dm⁻³ s⁻¹ |
| 1 | Rate doubles | s⁻¹ |
| 2 | Rate quadruples | dm³ mol⁻¹ s⁻¹ |
When you are given a table of initial rates, choose two experiments where only one reactant concentration changes. Then divide the rate equations to cancel k and solve for the order.
当给定初始速率数据表时，选择只有一种反应物浓度变化的两组实验。然后将速率方程相除以约去 k，并解出反应级数。
2. Experimental Methods for Following Reaction Rates | 跟踪反应速率的实验方法
Common A-Level methods to monitor a reaction include collecting gas volume with a gas syringe, measuring mass loss for reactions that produce a gas, using colorimetry for coloured species, monitoring conductivity for ionic reactions, measuring pH for acid-base reactions, and sampling by titration after quenching.
A-Level 中常见的反应监测方法包括：用气体注射器收集气体体积、测量产生气体的反应的质量损失、用比色法监测有色物质、监测离子反应的电导率、测量酸碱反应的 pH，以及淬灭后用滴定法取样分析。
The iodine clock reaction is a useful way to investigate rate. Hydrogen peroxide, iodide ions and starch are mixed with thiosulfate, and the sudden blue-black colour appears after a fixed amount of iodine has been produced. The reciprocal of the time taken gives a measure of the initial rate.
碘钟反应是研究速率的有效方法。过氧化氢、碘离子和淀粉与硫代硫酸盐混合，当生成固定量的碘后会出现蓝黑色突现。所用时间的倒数可以用来衡量初始速率。
When using continuous monitoring, plot concentration against time. The gradient of a tangent at t = 0 gives the initial rate, while a straight-line graph of concentration against time indicates zero order.
使用连续监测法时，绘制浓度-时间图。t = 0 时切线的斜率给出初始速率，而浓度-时间图为直线则表明该反应为零级。
For first-order reactions, a plot of ln[A] against time gives a straight line with gradient −k. For second order, a plot of 1/[A] against time gives a straight line with gradient +k.
对于一级反应，ln[A] 对时间作图得到斜率为 −k 的直线；对于二级反应，1/[A] 对时间作图得到斜率为 +k 的直线。
3. Rate Constants, Temperature and Catalysts | 速率常数、温度与催化剂
Increasing temperature increases the rate constant k, even though concentrations may remain the same. This happens because more molecules have energy greater than or equal to the activation energy, so a greater fraction of collisions are successful.
升高温度会增大速率常数 k，即使浓度保持不变也会如此。这是因为有更多分子的能量大于或等于活化能，成功碰撞的比例增大。
The Arrhenius equation links k to temperature and activation energy:
阿伦尼乌斯方程将 k 与温度和活化能联系起来：
k = Ae^(−Eₐ/(RT))
Here A is the pre-exponential factor, Eₐ is the activation energy, R is the gas constant, and T is the absolute temperature in kelvin.
其中 A 是指前因子，Eₐ 是活化能，R 是气体常数，T 是开尔文绝对温度。
Taking natural logarithms gives a linear form: ln k = −Eₐ/(RT) + ln A. A plot of ln k against 1/T has gradient −Eₐ/R and intercept ln A.
取自然对数后得到线性形式：ln k = −Eₐ/(RT) + ln A。以 ln k 对 1/T 作图，斜率为 −Eₐ/R，截距为 ln A。
Catalysts increase the rate by providing an alternative reaction pathway with a lower activation energy. This increases the value of k without being used up in the reaction.
催化剂通过提供活化能较低的替代反应路径来提高速率。它增大了 k 值，而自身在反应中不被消耗。
4. Equilibrium Constants Kc and Kp | 平衡常数 Kc 与 Kp
For a homogeneous equilibrium aA + bB ⇌ cC + dD, the concentration equilibrium constant is written as:
对于均相平衡 aA + bB ⇌ cC + dD，浓度平衡常数写作：
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Square brackets represent equilibrium concentrations in mol dm⁻³. Kc is constant only if temperature is constant, and its units vary according to the stoichiometry.
方括号表示以 mol dm⁻³ 为单位的平衡浓度。只有温度不变时 Kc 才恒定，其单位随化学计量数而变化。
For gas-phase equilibria, Kp uses partial pressures instead of concentrations. Each partial pressure is the mole fraction of the gas multiplied by the total pressure.
对于气相平衡，Kp 使用分压而非浓度。每个分压等于该气体的摩尔分数乘以总压力。
Kp = pCᶜ pDᵈ / pAᵃ pBᵇ
Remember to divide the number of moles of a gas by the total moles of gas to find its mole fraction. Then multiply by the total pressure to obtain the partial pressure.
计算时，先用某气体的物质的量除以气体总物质的量得到摩尔分数，再乘以总压力得到分压。
Only changes in temperature alter the value of Kc or Kp. Concentration and pressure changes shift the position of equilibrium but do not change the equilibrium constant.
只有温度变化才会改变 Kc 或 Kp 的值。浓度和压力变化只会移动平衡位置，不会改变平衡常数。
5. Le Chatelier’s Principle and Industrial Applications | 勒夏特列原理与工业应用
Le Chatelier’s principle states that if a system at equilibrium is disturbed, the position of equilibrium shifts to oppose the change. This helps predict the effect of concentration, pressure and temperature changes.
勒夏特列原理指出，如果平衡体系受到扰动，平衡位置会向削弱该扰动的方向移动。这有助于预测浓度、压力和温度变化的影响。
Increasing the concentration of a reactant shifts equilibrium to the right to use up the added reactant. Increasing the pressure shifts equilibrium towards the side with fewer gas moles; if both sides have equal moles, pressure has no effect.
增大反应物浓度会使平衡向右移动，以消耗加入的反应物。增大压力会使平衡向气体摩尔数较少的一侧移动；如果两侧气体摩尔数相等，则压力无影响。
Increasing temperature shifts equilibrium in the endothermic direction. For an exothermic forward reaction, raising temperature shifts equilibrium left and reduces Kc; for an endothermic forward reaction, Kc increases.
升高温度会使平衡向吸热方向移动。若正向放热，升温使平衡左移并减小 Kc；若正向吸热，则 Kc 增大。
The Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), is exothermic in the forward direction. It is run at high pressure to favour fewer gas moles and at moderate temperature as a compromise between equilibrium yield and rate.
哈伯法 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 正向放热。工业上采用高压以促进气体摩尔数减少的方向，并使用适中温度以在平衡产率和速率之间取得折中。
The Contact process for sulfuric acid uses 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) with vanadium(V) oxide as a catalyst. A catalyst does not change equilibrium position or Kc but allows equilibrium to be reached faster.
接触法生产硫酸中 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 使用五氧化二钒作催化剂。催化剂不会改变平衡位置或 Kc，但能更快达到平衡。
6. Enthalpy Changes: Definitions and Hess’s Law | 焓变：定义与赫斯定律
Standard enthalpy change of reaction, ΔH°, is the heat energy change under standard conditions, pressure 100 kPa and stated temperature, usually 298 K, with substances in their standard states.
标准反应焓变 ΔH° 是在标准条件（压力 100 kPa，通常温度 298 K，物质处于标准状态）下的热能量变化。
| Enthalpy change | Definition |
|---|---|
| ΔH°f | Enthalpy change when 1 mol of compound is formed from its elements |
| ΔH°c | Enthalpy change when 1 mol of substance is completely burned in oxygen |
| ΔH°neut | Enthalpy change when 1 mol of water is formed from acid and alkali |
中文表中：ΔH°f 表示由元素生成 1 mol 化合物时的焓变；ΔH°c 表示 1 mol 物质在氧气中完全燃烧时的焓变；ΔH°neut 表示由酸和碱生成 1 mol 水时的焓变。
Hess’s law states that the total enthalpy change for a reaction is independent of the route taken. This allows unknown ΔH values to be
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