📚 The Equilibrium Law (HL) | 平衡定律（HL）
In IB Chemistry at Higher Level, mastering the equilibrium law means far more than memorising Le Châtelier’s principle. It involves writing precise equilibrium expressions, calculating the equilibrium constant Kc from experimental data, using the reaction quotient Q to predict the direction of change, and understanding why only temperature can alter the value of Kc. This article unpacks every HL requirement clearly, with worked examples, ICE table logic and real‑world connections that will serve you in Paper 1 and Paper 2 alike.
在 IB 化学高等级课程中，掌握平衡定律远不止记住勒夏特列原理。它要求你写出准确的平衡表达式，从实验数据计算平衡常数 Kc，运用反应商 Q 预测反应方向，并理解为什么只有温度才能改变 Kc 的数值。本文清晰地分解每一项 HL 要求，提供详细例题、ICE 表格逻辑以及真实世界的联系，对试卷一和试卷二同样有帮助。
1. Dynamic Equilibrium – A Reversible Process | 动态平衡——可逆过程
Dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the macroscopic properties of the system (concentration, colour, pressure) remain constant, but on the molecular level both reactions continue simultaneously. The system must be closed so that no matter is exchanged with the surroundings, and equilibrium can only be reached if the activation energies of forward and reverse reactions are accessible under the given conditions.
动态平衡发生在一个封闭系统中，此时正反应的速率与逆反应的速率相等。这时，系统的宏观性质（浓度、颜色、压强）保持恒定，但在分子水平上，两个反应仍在同时进行。系统必须是封闭的，与环境没有物质交换，并且只有在给定条件下正反应和逆反应的活化能都可被跨越时，平衡才能达成。
Common examples include the evaporation–condensation equilibrium of water in a sealed flask, the dimerisation of nitrogen dioxide (2 NO2 ⇌ N2O4), and industrial syntheses such as the Haber process. A crucial HL point is that equilibrium is a dynamic state, not a static one.
常见例子包括密封烧瓶中水的蒸发–冷凝平衡、二氧化氮的二聚 (2 NO2 ⇌ N2O4) 以及哈伯法这样的工业合成。一个关键的 HL 要点是：平衡是一种动态状态，而不是静止状态。
2. The Equilibrium Law and Kc | 平衡定律与 Kc
The equilibrium law states that for any reversible reaction at a given temperature, the ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient, is a constant. This constant is called the equilibrium constant, denoted Kc. For the general reaction aA + bB ⇌ cC + dD, the expression is:
平衡定律指出，对于给定温度下的任何可逆反应，产物浓度与反应物浓度的比值（每种浓度以其化学计量系数为指数）是一个常数。这个常数称为平衡常数，记作 Kc。对于一般反应 aA + bB ⇌ cC + dD，表达式为：
Kc = [C]c[D]d / [A]a[B]b
Square brackets denote equilibrium concentrations in mol dm−3. The unit of Kc varies depending on the sum of stoichiometric coefficients (see Section 6). The magnitude of Kc indicates the extent of the reaction: Kc ≫ 1 implies products are favoured at equilibrium, while Kc ≪ 1 implies reactants are favoured.
方括号表示以 mol dm−3 为单位的平衡浓度。Kc 的单位随化学计量系数之和而变化（见第 6 节）。Kc 的大小反映了反应进行的程度：Kc ≫ 1 说明平衡时产物占优，Kc ≪ 1 则反应物占优。
3. Writing Equilibrium Expressions for Homogeneous and Heterogeneous Systems | 均相与多相体系的平衡表达式书写
In a homogeneous equilibrium, all reactants and products are in the same phase (e.g. gaseous or aqueous). The full concentration terms are included. For example, the equilibrium for the Haber process, N2(g) + 3 H2(g) ⇌ 2 NH3(g), gives:
在均相平衡中，所有反应物和产物处于同一相（如气相或溶液）。所有的浓度项都要写入表达式。例如哈伯法的平衡 N2(g) + 3 H2(g) ⇌ 2 NH3(g)，表达式为：
Kc = [NH3]2 / [N2][H2]3
In heterogeneous equilibria, pure solids and pure liquids are omitted from the expression because their concentrations are effectively constant. For the decomposition of calcium carbonate, CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium expression is simply Kc = [CO2]. The solids CaCO3 and CaO do not appear. This is a common source of error, so IB examiners expect you to recognise that the activity of pure solids and liquids is taken as 1.
在多相平衡中，纯固体和纯液体被省略在表达式之外，因为它们的浓度实际上是恒定的。对于碳酸钙的分解 CaCO3(s) ⇌ CaO(s) + CO2(g)，平衡表达式简化为 Kc = [CO2]。固体 CaCO3 和 CaO 不出现。这是一个常见的错误来源，因此 IB 考官希望你认识到纯固体和纯液体的活度视为 1。
4. The Reaction Quotient, Q: Predicting Direction | 反应商 Q：预测反应方向
The reaction quotient Q has exactly the same algebraic form as Kc but uses the concentrations at any point, not necessarily at equilibrium. By comparing Q with Kc at the same temperature, we can predict the direction in which a reaction will proceed to reach equilibrium. If Q < Kc, the forward reaction is favoured; the system will shift to the right to produce more products. If Q > Kc, the reverse reaction is favoured; the system will shift to the left. If Q = Kc, the system is already at equilibrium.
反应商 Q 的代数形式与 Kc 完全相同，但使用的是任意时刻的浓度，未必是平衡浓度。通过将 Q 与同温下的 Kc 比较，我们能预测反应进行的方向。若 Q < Kc，正反应被促进；系统将向右移动生成更多产物。若 Q > Kc，逆反应被促进；系统将向左移动。若 Q = Kc，系统已处于平衡。
For example, if an equilibrium mixture of PCl5 ⇌ PCl3 + Cl2 has Kc = 0.042 at a certain temperature, and we suddenly inject extra Cl2, Q becomes greater than Kc and the reverse reaction is temporarily dominant until Q drops back to Kc.
例如，在 PCl5 ⇌ PCl3 + Cl2 的平衡混合物中，某温度下 Kc = 0.042，若我们突然注入额外 Cl2，Q 变得大于 Kc，逆反应暂时占主导，直到 Q 降回 Kc。
5. Calculating Kc: ICE Tables and Worked Examples | Kc 的计算：ICE 表格与例题
IB HL students are frequently asked to determine Kc from initial and equilibrium concentrations. The most reliable method is an ICE table (Initial, Change, Equilibrium). Consider the reaction H2(g) + I2(g) ⇌ 2 HI(g). Starting with 1.00 mol of H2 and 1.00 mol of I2 in a 1.0 dm3 vessel, and finding 1.56 mol of HI at equilibrium, we can set up:
IB HL 学生常被要求从初始浓度和平衡浓度计算 Kc。最可靠的方法是使用 ICE 表格（初始、变化、平衡）。考虑反应 H2(g) + I2(g) ⇌ 2 HI(g)。在 1.0 dm3 容器中起始加入 1.00 mol H2 和 1.00 mol I2，平衡时发现 1.56 mol HI。
| Species | H2 | I2 | HI |
| Initial (mol dm−3) | 1.00 | 1.00 | 0 |
| Change | −x | −x | +2x |
| Equilibrium | 1.00 − x | 1.00 − x | 2x |
Since 2x = 1.56, x = 0.78 mol dm−3. The equilibrium concentrations are [H2] = 0.22, [I2] = 0.22, [HI] = 1.56. Then Kc = (1.56)2 cancels here because the number of moles on each side is equal. More complex problems might require solving a quadratic equation, but IB data are often designed so that the approximation that x is small can be justified, or simple algebra suffices.
由于 2x = 1.56，x = 0.78 mol dm−3。平衡浓度 [H2] = 0.22，[I2] = 0.22，[HI] = 1.56。则 Kc = (1.56)2 / (0.22 × 0.22) ≈ 50.3。注意此处单位恰好约去，因为两侧摩尔数相等。更复杂的问题可能需要解二次方程，但 IB 的数据往往设计得使 x 很小可以近似，或使用简单代数即可。
6. Units of Kc: Dependence on Stoichiometry | Kc 的单位：取决于化学计量
The unit of Kc is derived from the concentration terms: (mol dm−3)(sum of product coefficients − sum of reactant coefficients). Let us consider three classic examples:
Kc 的单位来自浓度项的幂次：(mol dm−3)(产物系数之和 − 反应物系数之和)。考虑三个经典例子：
- For N2 + 3 H2 ⇌ 2 NH3: sum products = 2, sum reactants = 4, exponent = 2 − 4 = −2, so units = mol−2 dm6.
- For H2 + I2 ⇌ 2 HI: exponent = 2 − 2 = 0, Kc has no units.
- For PCl5 ⇌ PCl3 + Cl2: exponent = 2 − 1 = 1, units = mol dm−3.
IB mark schemes expect the correct derivation and sometimes the simplification that when the number of gaseous moles is unchanged, Kc is dimensionless. Always show your working to secure full credit.
IB 的评分方案要求正确的推导，有时还要指出当气体摩尔数不变时 Kc 无量纲。记住展示推导过程以获得全部分数。
7. Le Châtelier's Principle: Effects of Concentration and Pressure | 勒夏特列原理：浓度和压强的影响
Le Châtelier's principle states that if a system at dynamic equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium will shift to oppose that change. For concentration, adding a reactant shifts equilibrium to the right (forward reaction dominates) to reduce its concentration. Removing a product also shifts equilibrium to the right. Crucially, these concentration changes alter the position of equilibrium but do not change the value of Kc; the system simply adjusts the ratio of concentrations until it matches the same Kc at the given temperature.
勒夏特列原理指出，当处于动态平衡的系统受到浓度、压强或温度的变化时，平衡的位置将发生移动以对抗这种变化。对浓度而言，增加反应物会使平衡向右移动（正反应占优）以降低其浓度；移除产物同样使平衡向右移动。关键是，这些浓度变化改变了平衡位置，但并不改变 Kc 的数值；系统只是调整浓度比值，直到与给定温度下相同的 Kc 匹配。
For pressure changes in gaseous systems, if the total pressure is increased at constant temperature, the system shifts towards the side with fewer gaseous moles. In 2 NO2 ⇌ N2O4 (2 moles → 1 mole), increasing pressure favours the forward reaction, forming more N2O4. If the number of gaseous moles is equal on both sides, pressure changes have no effect on the position of equilibrium. Adding an inert gas at constant volume does not change partial pressures of reactants, so it does not shift equilibrium.
对于气体系统的压强变化，若在恒温下增加总压强，平衡会向气态摩尔数较少的一侧移动。在 2 NO2 ⇌ N2O4 (2mol → 1mol) 中，增加压强有利于正反应，生成更多 N2O4。如果两边气态摩尔数相等，压强变化对平衡位置无影响。在恒容下加入惰性气体不会改变反应物的分压，因此不会使平衡移动。
8. Temperature Effects: The Only Factor That Changes Kc | 温度的影响：唯一能改变 Kc 的因素
Temperature is the only variable that alters the magnitude of the equilibrium constant. When the temperature is increased, the system absorbs extra energy by favouring the endothermic direction. For an exothermic forward reaction (ΔH < 0), increasing temperature shifts equilibrium to the left, reducing product concentration and thus lowering Kc. For an endothermic forward reaction (ΔH > 0), increasing temperature shifts equilibrium to the right and raises Kc. Conversely, cooling favours the exothermic direction.
温度是唯一能够改变平衡常数大小的变量。升高温度时，系统通过促进吸热方向来吸收额外能量。对于正反应放热 (ΔH < 0) 的反应，升温使平衡向左移动，产物浓度下降，因此 Kc 减小。对于正反应吸热 (ΔH > 0) 的反应，升温使平衡向右移动，Kc 增大。反之，降温有利于放热方向。
This relationship is often tested qualitatively in IB exams: you must be able to state that Kc decreases with rising temperature for the Haber process (N2 + 3 H2 ⇌ 2 NH3, ΔH = −92 kJ mol−1) and therefore lower temperatures favour a higher equilibrium yield of ammonia, even though kinetics require a compromise temperature.
这种关系在 IB 考试中经常被定性考查：你必须能陈述对于哈伯法 (N2 + 3 H2 ⇌ 2 NH3，ΔH = −92 kJ mol−1)，Kc 随温度升高而减小，因此低温有利于更高的氨平衡产率，尽管动力学要求使用折中温度。
9. Catalysts: No Effect on Position or Kc | 催化剂：对平衡位置和 Kc 无影响
A catalyst provides an alternative pathway with a lower activation energy for both forward and reverse reactions equally. Therefore, it increases the rate at which equilibrium is established but has no effect on the position of equilibrium or on the value of Kc. In IB mark schemes, a clear statement is often required: “Catalyst speeds up the rate of forward and reverse reactions equally, so equilibrium is reached faster, but the yield at equilibrium remains unchanged.”
催化剂为正向和逆向反应提供一条活化能更低的新路径，且对两者降低的程度相同。因此，它能加快达到平衡的速度，但对平衡位置或 Kc 的值没有影响。IB 评分方案中经常要求明确陈述：“催化剂同等程度地加快正逆反应速率，因此平衡能更快达到，但平衡时的产率保持不变。”
This principle explains why industrial processes use catalysts to increase productivity without sacrificing equilibrium yield. In the Contact process for sulfuric acid, V2O5 enables the SO2 oxidation to reach equilibrium rapidly, while the operating temperature is chosen as a compromise between yield and rate.
这一原理解释了为何工业过程使用催化剂提高产能而不牺牲平衡产率。在硫酸的接触法工艺中，V2O5 使 SO2 的氧化快速达到平衡，同时操作温度在产率和速率之间进行折中选择。
10. Applying the Equilibrium Law in Industry: The Haber Process | 平衡定律的工业应用：哈伯法
The synthesis of ammonia from nitrogen and hydrogen is an HL favourite for testing integrated understanding. The reaction is exothermic: N2(g) + 3 H2(g) ⇌ 2 NH3(g), ΔH = −92 kJ mol−1. According to the equilibrium law, high pressure (around 200 atm) shifts the position to the right because 4 gaseous moles become 2. However, very high pressure poses safety risks and plant capital costs.
由氮气和氢气合成氨是 HL 中考查综合理解的热门反应。该反应放热：N2(g) + 3 H2(g) ⇌ 2 NH3(g)，ΔH = −92 kJ mol−1。根据平衡定律，高压（约 200 atm）使平衡右移，因为 4 个气态分子变为 2 个。但过高的压强带来安全风险和设备投资成本。
From Section 8, we know that low temperature would give a higher Kc and thus a greater equilibrium yield of ammonia. However, a low temperature makes the reaction unacceptably slow. The compromise temperature of about 450°C, together with an iron catalyst, allows equilibrium to be reached rapidly enough while retaining an acceptable Kc value. The unreacted gases are recycled, which increases overall efficiency. This interplay of thermodynamic equilibrium and kinetic practicalities is a beautiful illustration of the equilibrium law at work.
从第 8 节可知，低温会带来更高的 Kc，从而增大氨的平衡产率。然而，低温使反应速率慢得无法接受。约 450°C 的折中温度配上铁催化剂，使平衡能足够快地达到，同时保留可接受的 Kc 值。未反应的气体循环使用，提高了总效率。这种热力学平衡与动力学现实之间的协调，正是平衡定律发挥作用的一个完美例证。
11. Common Misconceptions and Exam Tips | 常见误解与备考建议
Many HL students mistakenly believe that adding a catalyst increases equilibrium yield or that equilibrium can be reached from only one direction. Remember: equilibrium can be approached from either direction, and Kc is unaffected by concentration, pressure changes or the presence of a catalyst. Also, when writing equilibrium expressions, ensure you only include gaseous and aqueous species. Examiners frequently test the omission of solids in heterogeneous equilibria.
许多 HL 学生误以为添加催化剂能增加平衡产率，或者平衡只能从一个方向达成。请牢记：平衡可从任一方向趋近，浓度、压强变化或催化剂的存在都不会改变 Kc。此外，书写平衡表达式时，确保只包含气态和水溶物种。考官经常考察多相平衡中省略固体的情形。
For calculations, always set up an ICE table explicitly and check that your equilibrium concentrations give the correct Kc value. If you need to use a quadratic, simplify only if the approximation that x is much smaller than initial concentrations can be justified (usually when Kc is very small and initial reactant concentration is relatively large). In IB exams, be precise with units and state when Kc has no units.
计算时，务必明确列出 ICE 表格并验证你的平衡浓度能否得出正确的 Kc 值。若需使用二次方程，只有在 x 远小于初始浓度的近似合理时才可简化（通常当 Kc 非常小且初始反应物浓度相对较大时）。IB 考试中，单位要精确，并说明 Kc 何时无量纲。
12. Summary: The Power of the Equilibrium Law | 总结：平衡定律的力量
The equilibrium law is a quantitative tool that goes beyond simple qualitative predictions. By combining the expression for Kc, the ICE table methodology, the reaction quotient Q, and an understanding of the temperature dependence of Kc, you can predict and control the outcome of countless reversible reactions. Making these connections is the heart of HL Chemistry, and once mastered, it transforms how you think about chemical change—from the laboratory bench to large‑scale synthesis.
平衡定律是一个超越简单定性预测的定量工具。通过综合运用 Kc 表达式、ICE 表格方法、反应商 Q 以及对 Kc 温度依赖性的理解，你可以预测并控制无数可逆反应的结果。建立这些联系是 HL 化学的核心，一旦掌握，它将转变你对化学变化的思考方式——从实验台到大尺度合成无不如此。
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