📚 Equilibrium | 化学平衡
Chemical equilibrium is a fundamental concept describing the state in which the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of reactants and products. Understanding equilibrium allows chemists to control reaction conditions to maximise yield in industrial processes and to predict how systems respond to changes.
化学平衡是一个基本概念，描述正反应和逆反应以相同速率进行，从而导致反应物和产物浓度不再发生净变化的状态。理解平衡有助于化学家在工业过程中控制反应条件以最大化产率，并预测系统如何响应外部变化。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
A reversible reaction is one that can proceed in both forward and reverse directions. When a reversible reaction is carried out in a closed system, the rates of the forward and reverse reactions eventually become equal, and the concentrations of all species remain constant over time. This condition is known as dynamic equilibrium; the reaction has not stopped but continues with the forward and reverse processes balancing each other.
可逆反应是指既能正向进行又能逆向进行的反应。当可逆反应在一个封闭体系中进行时，正反应和逆反应的速率最终会相等，所有物种的浓度将会保持不变。这种状态称为动态平衡；反应并没有停止，而是保持着正逆过程互相抵消的持续状态。
At the macroscopic level, no observable change occurs, but at the molecular level, both reactions are still taking place. For the general reaction aA + bB ⇌ cC + dD, the double arrow (⇌) is used to indicate reversibility.
在宏观层面上，观察不到任何变化，但在分子水平上，两个反应仍在同时进行。对于一般的反应 aA + bB ⇌ cC + dD，使用双向箭头（⇌）来表示可逆性。
2. The Equilibrium Constant Kc | 平衡常数 Kc
The equilibrium constant expressed in terms of concentration, Kc, provides a quantitative measure of the position of equilibrium. For the reaction aA + bB ⇌ cC + dD, the equilibrium constant is defined as:
以浓度表示的平衡常数 Kc 为平衡位置提供了定量的量度。对于反应 aA + bB ⇌ cC + dD，平衡常数定义为：
Kc = [C]c[D]d / [A]a[B]b
The square brackets denote equilibrium concentrations in mol dm-3. Kc is dimensionless only when the sum of the stoichiometric coefficients of products equals that of reactants; otherwise units must be specified.
方括号表示以 mol dm-3 为单位的平衡浓度。只有当反应物与产物化学计量系数之和相等时 Kc 才没有量纲；否则必须标明单位。
3. Writing Equilibrium Expressions | 书写平衡表达式
When writing the expression for Kc, only gases and aqueous species appear; pure solids and pure liquids have constant concentration and are omitted because their activities are taken as 1. For heterogeneous equilibria, only the concentrations of species in the gas phase or in solution are included.
书写 Kc 表达式时，只有气体和溶液中的物种会出现；纯固体和纯液体具有恒定的浓度，因其活度视为 1 而被省略。对于多相平衡，只包括气相或溶液中的物种的浓度。
For example, in the decomposition of calcium carbonate: CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium expression is simply Kc = [CO2]. Solids do not contribute.
例如，碳酸钙分解：CaCO3(s) ⇌ CaO(s) + CO2(g)，平衡表达式仅为 Kc = [CO2]。固体不参与表达式。
4. Magnitude of Kc and Position of Equilibrium | Kc 的大小与平衡位置
The numerical value of Kc indicates the position of equilibrium. If Kc is very large (Kc ≫ 1), the equilibrium lies far to the right, meaning product formation is highly favoured and the reaction goes nearly to completion. If Kc is very small (Kc ≪ 1), the equilibrium lies far to the left, with very little product formed at equilibrium.
Kc 的数值大小表明平衡的位置。如果 Kc 非常大（Kc ≫ 1），平衡远偏向右边，意味着强烈倾向于生成产物，反应几乎进行到底。如果 Kc 非常小（Kc ≪ 1），平衡远偏向左边，平衡时生成的产物很少。
It is important to note that Kc is constant at a given temperature; changing concentrations or pressure does not alter Kc, only temperature can change its value.
重要的是，在给定温度下 Kc 是常数；改变浓度或压力并不会改变 Kc，只有温度能改变其数值。
5. Equilibrium Constant in Terms of Pressure, Kp | 压强平衡常数 Kp
For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures, denoted Kp. For the general gas-phase reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp is defined as:
对于涉及气体的反应，平衡常数可以用分压表示，记作 Kp。对于一般的气相反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)，Kp 定义为：
Kp = (PCc PDd) / (PAa PBb)
Here PX represents the partial pressure of gas X. The unit of Kp depends on the change in the number of moles of gas, Δn, and is typically expressed in units of pressure raised to Δn, e.g., atmΔn or PaΔn.
这里 PX 表示气体 X 的分压。Kp 的单位取决于气体摩尔数的变化 Δn，通常表示为压力单位的 Δn 次方，例如 atmΔn 或 PaΔn。
The relationship between Kp and Kc is given by Kp = Kc (RT)Δn, where R is the gas constant and T is the absolute temperature. This equation is essential when converting between the two constants.
Kp 与 Kc 的关系为 Kp = Kc (RT)Δn，其中 R 是气体常数，T 是绝对温度。当需要在两者间转换时，该方程至关重要。
6. Le Chatelier's Principle: Concentration Changes | 勒夏特列原理：浓度变化
Le Chatelier'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 counteract the imposed change and restore a new equilibrium. If the concentration of a reactant is increased, the system shifts to consume the added species, favouring the forward reaction.
勒夏特列原理指出，如果动态平衡的系统受到浓度、压力或温度的改变，平衡位置将会移动以抵消这种外加的变化，并建立新的平衡。如果增加反应物的浓度，系统将移向消耗所加物种的方向，有利于正反应。
Conversely, removing a product will also shift equilibrium to the right to produce more product. These shifts alter the equilibrium concentrations but do not change the value of Kc or Kp, provided temperature remains constant.
相反，移除产物也会使平衡向右移动以生成更多的产物。这些移动改变了平衡浓度，但只要温度不变，不会改变 Kc 或 Kp 的值。
7. Le Chatelier's Principle: Pressure Changes | 勒夏特列原理：压强变化
Pressure changes only affect equilibria involving gases where there is a difference in the total number of gaseous moles on each side of the equation. Increasing the total pressure (by decreasing volume) shifts the equilibrium towards the side with fewer gas molecules, thus reducing the pressure.
压力变化只影响那些方程式两边气体总摩尔数存在差异的气相平衡。增加总压强（通过减小体积）将使平衡移向气体分子数较少的一边，从而降低压力。
For example, N2(g) + 3H2(g) ⇌ 2NH3(g) has 4 moles of gas on the left and 2 on the right; increasing pressure shifts the equilibrium to the right, improving ammonia yield. If the numbers of moles of gas are equal on both sides, pressure has no effect on the position of equilibrium.
例如，N2(g) + 3H2(g) ⇌ 2NH3(g) 左边有 4 摩尔气体，右边有 2 摩尔；增加压强使平衡向右移动，提高氨的产率。如果两边气体摩尔数相等，压强对平衡位置没有影响。
8. Le Chatelier's Principle: Temperature Changes | 勒夏特列原理：温度变化
Temperature is the only factor that changes the value of the equilibrium constant. For an exothermic forward reaction (ΔH < 0), increasing temperature adds heat to the system; equilibrium shifts to favour the endothermic reverse reaction, decreasing Kc. For an endothermic forward reaction (ΔH > 0), increasing temperature shifts equilibrium to the right, increasing Kc.
温度是唯一改变平衡常数值的因素。对于正向放热反应（ΔH < 0），升高温度相当于向系统增加热量；平衡移向吸热的逆反应方向，Kc 减小。对于正向吸热反应（ΔH > 0），升高温度使平衡向右移动，Kc 增大。
This can be rationalised by treating heat as a component: in exothermic reactions, heat is a product; adding heat shifts equilibrium left. Students must be able to predict the sign of ΔH from the effect of temperature on Kc or equilibrium yield.
这可以通过将热量视为一种组分来解释：在放热反应中，热量是产物；增加热量使平衡左移。学生必须能够根据温度对 Kc 或平衡产率的影响推断 ΔH 的符号。
9. Effect of a Catalyst on Equilibrium | 催化剂对平衡的影响
Adding a catalyst speeds up both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy. It does not affect the position of equilibrium or the value of the equilibrium constant; a catalyst simply allows the system to reach equilibrium more quickly.
加入催化剂通过提供一条活化能较低的替代反应路径，同等程度地加快正反应和逆反应的速率。催化剂不影响平衡位置，也不改变平衡常数值；它只是让系统更快地达到平衡。
In industrial processes, catalysts are vital for economic reasons, even though they do not increase the equilibrium yield. For example, iron in the Haber process and vanadium(V) oxide in the Contact process enable reactions to proceed at lower temperatures while maintaining acceptable rates.
在工业过程中，催化剂出于经济原因至关重要，尽管它们不会提高平衡产率。例如，哈伯法中的铁和接触法中的五氧化二钒使反应能在较低温度下进行，同时保持可接受的速率。
10. Industrial Applications: The Haber Process | 工业应用：哈伯法
The Haber process for ammonia synthesis, N2(g) + 3H2(g) ⇌ 2NH3(g) with ΔH = -92 kJ mol-1 s too slow. High pressure favours the forward reaction (4 → 2 gas moles) but is expensive.
合成氨的哈伯法 N2(g) + 3H2(g) ⇌ 2NH3(g)，ΔH = -92 kJ mol-1，体现了热力学与动力学考量之间的折衷。该反应放热，因此较低的温度有利于提高氨的平衡产率，但速率会变得太慢。高压有利于正反应（4 → 2 气体分子），但成本高昂。
The typical industrial conditions are a pressure of about 200 atm, a temperature of around 450°C, and an iron catalyst. These conditions provide an optimum balance between rate, yield, and economic cost, yielding about 15% ammonia per pass, with unreacted gases recycled.
典型的工业条件是约 200 atm 的压强、450°C 左右的温度和铁催化剂。这些条件在速率、产率和经济成本之间提供了最佳平衡，单程氨产率约为 15%，未反应的气体循环使用。
11. Calculations Involving Kc and Equilibrium Concentrations | Kc 与平衡浓度的计算
Typical problems involve finding Kc from equilibrium concentrations or using the initial amounts and the balanced equation to determine equilibrium concentrations. An ICE table (Initial, Change, Equilibrium) is a powerful tool. For the reaction H2(g) + I2(g) ⇌ 2HI(g), if 1.0 mol of H2 and 1.0 mol of I2 are placed in a 1 dm3 vessel and at equilibrium 0.40 mol of H2 remains, the change is 0.60 mol. The equilibrium concentrations are [H2] = 0.40 M, [I2] = 0.40 M, [HI] = 1.20 M. Then Kc = (1.20)2/(0.40×0.40) = 9.0.
典型的题目包括由平衡浓度求 Kc 或利用初始量和方程式确定平衡浓度。ICE 表（初始、变化、平衡）是强有力的工具。对于反应 H2(g) + I2(g) ⇌ 2HI(g)，若将 1.0 mol H2 和 1.0 mol I2 置于 1 dm3 容器中，平衡时剩余 0.40 mol H2，则变化量为 0.60 mol。平衡浓度为 [H2] = 0.40 M，[I2] = 0.40 M，[HI] = 1.20 M。然后 Kc = (1.20)2/(0.40×0.40) = 9.0。
When solving quadratic equations arise, approximations can often be made if Kc is very small, simplifying the calculation by assuming the change in reactant concentration is negligible compared with the initial concentration.
当需要求解二次方程时，如果 Kc 非常小，通常可以假定反应物浓度的变化相对于初始浓度可忽略不计，从而简化计算。
12. Summary and Common Misconceptions | 总结与常见误区
Chemical equilibrium is a dynamic balance between forward and reverse rates. The equilibrium constant Kc or Kp is dependent only on temperature. Le Chatelier's principle helps predict the direction of shift but does not explain the rate at which new equilibrium is established. A common mistake is to think catalysts increase yield; they do not. Another is confusing the effect of temperature on rate and on equilibrium position: temperature always increases rate, but its effect on equilibrium depends on ΔH.
化学平衡是正逆反应速率的动态平衡。平衡常数 Kc 或 Kp 只依赖于温度。勒夏特列原理有助于预测移动的方向，但不能解释建立新平衡的速率。常见的错误是认为催化剂能提高产率，实际上并不可以。另一个误区是混淆温度对速率和平衡位置的影响：温度总会提高速率，但对平衡的影响取决于 ΔH。
Mastering equilibrium requires careful interpretation of concentration, pressure, and temperature effects, and competence in both qualitative predictions and quantitative calculations. It forms the foundation for understanding acid-base equilibria, solubility, and redox systems, all of which are central to the IB and Cambridge chemistry curriculum.
掌握化学平衡需要仔细理解浓度、压力和温度的影响，并具备定性预测和定量计算的能力。它构成了理解酸碱平衡、溶解平衡和氧化还原体系的基础，这些都是 IB 和剑桥化学课程的核心内容。
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