📚 Mastering AS Physical Chemistry Unit 2: Energetics, Kinetics and Equilibria | 掌握AS物理化学第二单元：能量学、动力学与平衡
This article covers the physical chemistry topics in OxfordAQA International AS Chemistry Unit 2: enthalpy changes, Hess cycles, bond enthalpies, collision theory, Maxwell-Boltzmann distributions, catalysts, dynamic equilibrium and Kc calculations. Use it as a topic test revision guide.
本文涵盖 OxfordAQA 国际 AS 化学第二单元中的物理化学内容：焓变、盖斯循环、键焓、碰撞理论、麦克斯韦-玻尔兹曼分布、催化剂、动态平衡和 Kc 计算。可用作专题测验复习指南。
1. Enthalpy Changes and Thermochemistry | 焓变与热化学
Enthalpy, H, is the heat content of a system at constant pressure. An enthalpy change, ΔH, is the heat transferred in a reaction at constant pressure, measured in kJ mol⁻¹.
焓 H 是体系在恒压下的热含量。焓变 ΔH 是反应在恒压下的热传递，单位为 kJ mol⁻¹。
In an exothermic reaction, ΔH is negative because heat is released to the surroundings. In an endothermic reaction, ΔH is positive because heat is absorbed from the surroundings.
放热反应的 ΔH 为负值，因为热量释放到环境中；吸热反应的 ΔH 为正值，因为热量从环境中吸收。
Standard enthalpy changes are measured under standard conditions: 100 kPa, 298 K, and solutions at 1 mol dm⁻³. Standard states must be used.
标准焓变在标准条件下测量：100 kPa、298 K、溶液浓度 1 mol dm⁻³，并且必须使用标准状态。
| Standard enthalpy change | Definition | 中文定义 |
|---|---|---|
| Combustion | Enthalpy change when 1 mole of a substance burns completely in oxygen under standard conditions. | 1 mol 物质在标准条件下完全燃烧的焓变。 |
| Formation | Enthalpy change when 1 mole of a compound is formed from its elements in their standard states. | 由标准状态下的元素生成 1 mol 化合物的焓变。 |
| Neutralisation | Enthalpy change when an acid and a base react to form 1 mole of water under standard conditions. | 酸碱反应生成 1 mol 水的焓变。 |
2. Calorimetry and Measuring Enthalpy Changes | 量热法测量焓变
Calorimetry measures heat changes using the relationship q = mcΔT, where q is heat energy, m is mass of water, c is specific heat capacity, and ΔT is temperature change.
量热法利用 q = mcΔT 测量热量变化，其中 q 为热能，m 为水的质量，c 为比热容，ΔT 为温度变化。
For a reaction carried out in solution, the specific heat capacity is usually taken as 4.18 J g⁻¹ K⁻¹ and the mass is the mass of the solution, often approximated as the mass of water.
对于在溶液中进行的反应，比热容通常取 4.18 J g⁻¹ K⁻¹，质量取溶液的质量，通常近似为水的质量。
To calculate the molar enthalpy change, divide the heat energy by the number of moles of limiting reactant and adjust the sign. If the temperature increases, the reaction is exothermic so ΔH = −q/n.
计算摩尔焓变时，用热能除以限制反应物的物质的量并调整符号。如果温度升高，反应放热，因此 ΔH = −q/n。
Common errors in calorimetry include heat loss to the surroundings, incomplete combustion, and assuming the solution has the same density and specific heat capacity as water.
量热法中常见误差包括热量散失到环境中、燃烧不完全，以及假设溶液的密度和比热容与水相同。
3. Hess’s Law and Enthalpy Cycles | 盖斯定律与焓循环
Hess’s law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
盖斯定律指出，只要反应的始态和终态相同，总焓变与反应途径无关。
Using standard enthalpies of formation, the enthalpy change of a reaction is calculated as: ΔH = ΣΔHf(products) − ΣΔHf(reactants).
使用标准生成焓，反应的焓变计算为：ΔH = ΣΔHf(生成物) − ΣΔHf(反应物)。
Using standard enthalpies of combustion, the enthalpy change is calculated as: ΔH = ΣΔHc(reactants) − ΣΔHc(products).
使用标准燃烧焓，焓变计算为：ΔH = ΣΔHc(反应物) − ΣΔHc(生成物)。
Drawing a Hess cycle with arrows helps avoid sign errors. Always label the known values and the unknown ΔH clearly before substituting numbers.
绘制盖斯循环并用箭头标明方向有助于避免符号错误。在代入数字前，始终清楚地标出已知值和未知 ΔH。
4. Bond Enthalpies and Mean Bond Energy | 键焓与平均键能
Bond breaking is endothermic and bond forming is exothermic. The enthalpy change for a reaction can be estimated using mean bond enthalpies: ΔH = Σ(bonds broken) − Σ(bonds formed).
断裂化学键需要吸热，形成化学键会放热。反应的焓变可用平均键焓估算：ΔH = Σ(断裂的键) − Σ(形成的键)。
Mean bond enthalpy is the average energy required to break one mole of a given covalent bond in the gaseous state, averaged over different compounds.
平均键焓是在气态下断裂 1 mol 某共价键所需的平均能量，取不同化合物中的平均值。
Calculated ΔH values from mean bond enthalpies are less accurate than Hess’s law calculations because they use average values rather than compound-specific bond strengths.
用平均键焓计算的 ΔH 值不如盖斯定律计算准确，因为平均键焓使用平均值而不是特定化合物中的键强度。
5. Collision Theory and Reaction Rates | 碰撞理论与反应速率
For a reaction to occur, particles must collide with energy equal to or greater than the activation energy, and they must collide in the correct orientation.
要使反应发生，粒子必须以等于或大于活化能的能量碰撞，并且必须以正确的方向碰撞。
Factors that increase reaction rate include increasing concentration, increasing pressure for gases, increasing temperature, increasing surface area, and adding a catalyst.
提高反应速率的因素包括增加浓度、增加气体压强、升高温度、增大表面积以及加入催化剂。
Increasing concentration or pressure increases the number of particles per unit volume, so successful collisions occur more frequently.
增加浓度或压强会增加单位体积内的粒子数量，因此成功碰撞发生得更频繁。
Increasing temperature increases the average kinetic energy of particles, so a greater proportion of collisions have energy above the activation energy.
升高温度会增加粒子的平均动能，因此更高比例的碰撞具有超过活化能的能量。
6. Maxwell-Boltzmann Distribution | 麦克斯韦-玻尔兹曼分布
The Maxwell-Boltzmann distribution shows the spread of molecular kinetic energies in a gas at a given temperature. The curve starts at the origin, rises to a peak, and then decreases to the right.
麦克斯韦-玻尔兹曼分布显示在给定温度下气体分子动能的分布。曲线从原点开始，上升到峰值，然后向右下降。
The area under the curve represents the total number of molecules. The shaded area to the right of the activation energy represents molecules with enough energy to react.
曲线下的面积代表分子总数。活化能右侧的阴影面积代表具有足够能量反应的分子。
When temperature increases, the curve peak moves to the right and becomes lower and broader. The total area remains the same, but the area beyond the activation energy increases significantly.
当温度升高时，曲线峰值向右移动，变低变宽。总面积保持不变，但活化能右侧的面积显著增加。
A catalyst lowers the activation energy, so the activation energy line shifts to the left. This increases the area beyond the activation energy without changing the shape of the curve.
催化剂降低活化能，因此活化能线向左移动。这增加了活化能右侧的面积，而不改变曲线形状。
7. Catalysts and Activation Energy | 催化剂与活化能
A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. It is chemically unchanged at the end of the reaction.
催化剂通过提供活化能较低的替代反应路径来提高反应速率，并在反应结束时化学性质不变。
Homogeneous catalysts are in the same phase as the reactants, while heterogeneous catalysts are in a different phase. Heterogeneous catalysts usually involve adsorption onto a solid surface.
均相催化剂与反应物处于同一相，而非均相催化剂处于不同相。非均相催化剂通常涉及在固体表面上的吸附。
Catalysts do not affect the equilibrium position or the yield of a reversible reaction. They only speed up the attainment of equilibrium by increasing the rates of both forward and backward reactions equally.
催化剂不影响可逆反应的平衡位置或产率。它们只通过同样加快正逆反应速率来缩短达到平衡所需的时间。
8. Dynamic Equilibrium and Reversible Reactions | 动态平衡与可逆反应
Many chemical reactions are reversible, represented by the symbol ⇌. In a closed system, a reversible reaction can reach dynamic equilibrium.
许多化学反应是可逆的，用符号 ⇌ 表示。在封闭体系中，可逆反应可以达到动态平衡。
At dynamic equilibrium, the rate of the forward reaction equals the rate of the backward reaction. The concentrations of reactants and products remain constant, but not necessarily equal.
在动态平衡时，正反应速率等于逆反应速率。反应物和生成物的浓度保持不变，但不一定相等。
Equilibrium can only be established in a closed system so that no substances can escape. Macroscopic properties such as colour, pressure and concentration stay constant.
只有在封闭体系中才能建立平衡，因为物质无法逸出。颜色、压强和浓度等宏观性质保持恒定。
9. Le Chatelier’s Principle | 勒夏特列原理
Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium position shifts to oppose that change.
勒夏特列原理指出，如果平衡体系受到浓度、压强或温度的变化，平衡位置会向削弱该变化的方向移动。
Increasing the concentration of a reactant shifts equilibrium to the right to produce more product. Removing a product also shifts equilibrium to the right.
增加反应物浓度会使平衡向右移动，生成更多产物。移除产物也会使平衡向右移动。
For gas-phase reactions, increasing pressure shifts equilibrium towards the side with fewer moles of gas. If the number of moles is equal on both sides, pressure has no effect.
对于气相反应，增加压强会使平衡向气体物质的量较少的一侧移动。如果两侧气体物质的量相等，则压强变化没有影响。
Increasing temperature shifts equilibrium in the endothermic direction. Decreasing temperature shifts equilibrium in the exothermic direction.
升高温度会使平衡向吸热方向移动。降低温度会使平衡向放热方向移动。
Adding a catalyst does not shift the equilibrium position because it speeds up both forward and backward reactions equally.
加入催化剂不会使平衡位置移动，因为它同等程度地加快正逆反应。
10. Equilibrium Constant Kc | 平衡常数 Kc
For a general homogeneous equilibrium aA + bB ⇌ cC + dD, the equilibrium constant is expressed as:
对于一般的均相平衡 aA + bB ⇌ cC + dD，平衡常数表达式为：
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Only species in the gas phase or aqueous solution appear in the Kc expression. Solids and liquids are omitted because their concentrations are unchanged.
只有气相或溶液中的物种出现在 Kc 表达式中。固体和液体因浓度不变而被省略。
Kc has units that depend on the stoichiometry of the reaction. You must cancel the concentration units mol dm⁻³ to find the correct overall units.
Kc 的单位取决于反应的化学计量数。必须约去浓度单位 mol dm⁻³ 才能得到正确的总体单位。
The value of Kc is constant at a given temperature. Temperature is the only factor that changes Kc; concentration, pressure and catalysts do not.
Kc 的值在给定温度下恒定。只有温度能改变 Kc；浓度、压强和催化剂不能。
11. Industrial Applications of Equilibria | 平衡的工业应用
The Haber process produces ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The forward reaction is exothermic, so a low temperature favours a high yield, but the rate becomes too slow.
哈伯法生产氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。正反应放热，因此低温有利于高产率，但反应速率会太慢。
A compromise temperature of about 400–450 °C and a high pressure of about 200 atm are used, together with an iron catalyst. High pressure favours the side with fewer gas moles.
工业上使用约 400–450 °C 的折中温度和约 200 atm 的高压，并配合铁催化剂。高压有利于气体物质的量较少的一侧。
The catalyst does not increase the yield but allows lower temperatures to be used while maintaining an acceptable rate, saving energy and costs.
催化剂不会提高产率，但可以在保持可接受速率的同时使用更低温度，从而节省能源和成本。
12. Exam Tips and Common Misconceptions | 考试技巧与常见误区
Always check the sign of ΔH. Exothermic reactions have negative ΔH and endothermic reactions have positive ΔH. In calorimetry, use the correct sign when converting q to ΔH.
务必检查 ΔH 的符号。放热反应的 ΔH 为负，吸热反应的 ΔH 为正。在量热法中，将 q 转换为 ΔH 时要使用正确的符号。
Write units for Kc carefully. Cancel mol dm⁻³ before giving the final unit. For example, N₂ + 3H₂ ⇌ 2NH₃ gives Kc units of dm⁶ mol⁻².
仔细书写 Kc 的单位。在给出最终单位前先约去 mol dm⁻³。例如，N₂ + 3H₂ ⇌ 2NH₃ 的 Kc 单位为 dm⁶ mol⁻²。
Do not include solids or pure liquids in Kc expressions. Only aqueous and gaseous species appear.
不要把固体或纯液体写入 Kc 表达式。只有水溶液和气态物种才出现。
Remember that a catalyst lowers activation energy but does not change ΔH, equilibrium position, or Kc. It only increases the rate.
记住催化剂降低活化能，但不改变 ΔH、平衡位置或 Kc。它只加快反应速率。
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