📚 Equilibrium | 化学平衡
Chemical equilibrium is a central idea in understanding how reversible reactions behave in closed systems. Mastering equilibrium allows chemists to predict the composition of reaction mixtures and to design industrial processes that maximise product yield. This article covers dynamic equilibrium, equilibrium constants Kc and Kp, Le Chatelier's principle, and the practical application of these concepts in industrial chemistry.
化学平衡是理解可逆反应在封闭体系中如何行为的核心概念。掌握平衡知识可以帮助化学家预测反应混合物的组成，并设计出最大化产物产率的工业流程。本文涵盖动态平衡、平衡常数 Kc 和 Kp、勒夏特列原理以及这些概念在工业化学中的实际应用。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
Many chemical reactions are reversible, proceeding in both forward and backward directions. The use of the ⇌ symbol indicates reversibility, such as in the formation of ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
许多化学反应是可逆的，可以同时向正方向和逆方向进行。使用符号 ⇌ 表示可逆性，例如氨的合成：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。
Dynamic equilibrium is reached in a closed system when the rates of the forward and reverse reactions become equal. At this point, the concentrations of reactants and products remain constant, but the reactions continue to occur at the molecular level.
当正反应和逆反应的速率相等时，封闭体系就达到了动态平衡。此时，反应物和产物的浓度保持恒定，但在分子水平上反应仍在继续进行。
It is essential to understand that equilibrium is dynamic, not static. The macroscopic properties (colour, pressure, concentration) are constant, yet the forward and reverse processes are ongoing.
必须理解平衡是动态的，而非静止的。尽管宏观性质（颜色、压力、浓度）不变，正逆过程却从未停止。
2. The Equilibrium Constant, Kc | 平衡常数 Kc
For a general homogeneous reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration, Kc, is given by the expression:
对于一般均相可逆反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数 Kc 表达式为：
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Here, square brackets denote equilibrium concentrations in mol dm⁻³. The powers a, b, c, d are the stoichiometric coefficients in the balanced equation. Kc is dimensionless only if the sum of powers on top equals the sum on the bottom. In most cases, Kc has units that depend on the stoichiometry.
式中方括号表示以 mol dm⁻³ 为单位的平衡浓度。指数 a、b、c、d 是配平方程中的化学计量数。只有当分子各项指数之和等于分母各项指数之和时，Kc 才是无单位的。在多数情况下，Kc 的单位由化学计量关系决定。
The magnitude of Kc provides information about the position of equilibrium. A very large Kc (>>1) means the equilibrium lies far to the right, favouring products. A very small Kc (<<1) means the equilibrium lies to the left, with reactants predominating.
Kc 的大小给出了关于平衡位置的信息。非常大的 Kc（远大于 1）表示平衡强烈偏向右侧，产物占优势。非常小的 Kc（远小于 1）则表示平衡偏向左侧，反应物为主。
3. Calculating Kc: Units and Examples | 计算 Kc：单位与实例
When calculating Kc, you must determine the equilibrium concentrations of all species. This often involves using initial amounts, changes, and equilibrium amounts (ICE tables). The units of Kc are derived from the concentration terms raised to their stoichiometric powers.
计算 Kc 时，必须求出所有物质的平衡浓度。这通常涉及使用初始量、变化量和平衡量（ICE 表格）。Kc 的单位由浓度项以其化学计量数为指数导出。
For example, in the esterification reaction CH₃COOH(l) + C₂H₅OH(l) ⇌ ol dm⁻³, [C₂H₅OH] = 0.20 mol dm⁻³, [ester] = 0.80 mol dm⁻³, [water] = 0.80 mol dm⁻³, then:
例如，在酯化反应 CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) 中，若平衡时浓度为：[CH₃COOH] = 0.20 mol dm⁻³，[C₂H₅OH] = 0.20 mol dm⁻³，[酯] = 0.80 mol dm⁻³，[水] = 0.80 mol dm⁻³，则：
Kc = (0.80 × 0.80) / (0.20 × 0.20) = 16
Since the number of moles on both sides of the equation is the same, Kc has no units in this case. When the sums of stoichiometric coefficients differ, units such as mol dm⁻³ or mol⁻¹ dm³ may appear.
因为方程两边物质的摩尔数相同，此例中 Kc 无单位。当化学计量数之和不同时，就可能出现如 mol dm⁻³ 或 mol⁻¹ dm³ 的单位。
Always check that you are using equilibrium concentrations, not initial ones. Solids and pure liquids are omitted from the Kc expression because their concentrations are constant and incorporated into the constant.
务必确保使用的是平衡浓度，而非初始浓度。固体和纯液体不出现在 Kc 表达式中，因为它们的浓度恒定，已并入常数。
4. The Equilibrium Constant, Kp | 平衡常数 Kp
For gaseous equilibria, it is often convenient to use partial pressures instead of concentrations. The equilibrium constant in terms of pressure, Kp, is defined for the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g) as:
对于气体平衡，通常使用分压代替浓度更为方便。对于反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)，以压力表示的平衡常数 Kp 定义为：
Kp = (PC)ᶜ (PD)ᵈ / (PA)ᵃ (PB)ᵇ
Where PA, PB, PC, PD are the partial pressures of the gases at equilibrium, usually expressed in atm, Pa, or kPa. The total pressure of the system is the sum of all partial pressures.
其中 PA、PB、PC、PD 为各气体在平衡时的分压，通常以 atm、Pa 或 kPa 表示。体系的总压等于所有分压之和。
The partial pressure of a gas is calculated using Dalton's law: PA oles of gas). Kp has units derived from the pressure terms, and like Kc, its numerical value depends on the stoichiometry and the temperature.
气体分压用道尔顿分压定律计算：PA = A 的摩尔分数 × 总压。摩尔分数 = (A 的摩尔数) / (气体总摩尔数)。Kp 的单位由压力项导出，与 Kc 类似，其数值取决于化学计量关系和温度。
ationship Kp = Kc (RT)^Δn links the two constants, where Δn is the change in moles of gas.
Kp 仅适用于均相气体平衡。任何存在的固体或液体均被排除在表达式之外，与 Kc 相同。关系式 Kp = Kc (RT)^Δn 将这两个常数联系在一起，其中 Δn 是气体摩尔数的变化量。
5. Homogeneous vs. Heterogeneous Equilibria | 均相平衡与非均相平衡
A homogeneous equilibrium is one in which all reactants and products are in the same phase, either all gases or all in solution. In such systems, all species appear in the equilibrium expression.
均相平衡是指所有反应物和产物都处于同一相，要么全是气体，要么全在溶液中。在此类体系中，所有物质都出现在平衡表达式中。
In a heterogeneous equilibrium, the substances are in different phases. For example, the thermal decomposition of calcium carbonate: CaCO₃(s) ⇌ CaO(s) + CO₂(g). In the Kc expression, only gaseous or aqueous species are included. The two solids are omitted because their concentrations are constant.
在非均相平衡中，物质处于不同的相。例如碳酸钙的热分解：CaCO₃(s) ⇌ CaO(s) + CO₂(g)。在 Kc 表达式中，只包含气态或溶液中的物种。两种固体被省略，因为它们的浓度恒定。
Thus, for CaCO₃ decomposition, Kc = [CO₂] and Kp = PCO₂. The constant value simply equals the concentration or partial pressure of the carbon dioxide produced at equilibrium.
因此，对于碳酸钙分解，Kc = [CO₂]，Kp = PCO₂。该常数值即等于平衡时产生的二氧化碳的浓度或分压。
This simplification is crucial: a solid in excess does not shift the equilibrium, and its amount does not affect the value of Kc or Kp. Only changes in the gaseous or dissolved species alter the position of equilibrium.
这种简化至关重要：过量的固体不会使平衡移动，其量也不影响 Kc 或 Kp 的值。只有气体或溶解态物种的改变才会影响平衡位置。
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, the position of equilibrium shifts to oppose the change. This is a qualitative tool to predict how equilibrium re-establishes itself.
勒夏特列原理指出，如果处于动态平衡的体系受到外界条件变化，平衡位置将移动以抵消这种改变。这是预测平衡如何重建的定性工具。
If the concentration of a reactant is increased, the system responds by shifting equilibrium to the right to consume the added substance, producing more products. Alternatively, removing a product also shifts equilibrium to the right to replenish it.
如果增加一种反应物的浓度，体系通过向右移动平衡来消耗加入的物质，生成更多产物。反之，移除一种产物也会使平衡向右移动以补充它。
Conversely, increasing the concentration of a product shifts equilibrium to the left, favouring the reverse reaction. The value of Kc remains unchanged, as concentration changes only alter the position, not the constant, provided temperature stays constant.
相反地，增加产物浓度会使平衡向左移动，促进逆反应。Kc 的值保持不变，因为浓度变化只改变平衡位置，而不改变常数本身，前提是温度不变。
In industrial processes, one reactant is often used in excess to drive the equilibrium towards products and to increase the conversion of the more expensive reactant. This is a direct application of controlling concentration.
在工业过程中，常使一种反应物过量，以驱使平衡向产物方向移动，提高较昂贵反应物的转化率。这是控制浓度的直接应用。
7. Le Chatelier's Principle: Pressure and Temperature | 勒夏特列原理：压力与温度
For gaseous equilibria, changing the total pressure (by altering volume) shifts the equilibrium to the side with fewer moles of gas, as this reduces the pressure. If the number of moles is equal on both sides, a change in pressure has no effect on the position of equilibrium.
对于气体平衡，改变总压（通过改变体积）会使平衡向气体摩尔数较少的一侧移动，因为这样可以减小压力。如果两边气体摩尔数相等，压力变化对平衡位置无影响。
For example, in N₂(g) + 3H₂(g) ⇌ 2NH₃(g), there are 4 moles on the left and 2 on the right. Increasing pressure favours the forward reaction, producing more ammonia. This is a key factor in the Haber process.
例如，在 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 中，左边有 4 摩尔气体，右边有 2 摩尔。增加压力有利于正反应，生成更多的氨。这是哈伯法中的一个关键因素。
Temperature changes affect both the position of equilibrium and the value of the equilibrium constant. For an exothermic reaction (ΔH negative), increasing temperature shifts equilibrium to the left, decreasing product yield and reducing Kc. For an endothermic reaction (ΔH positive), a temperature rise shifts equilibrium to the right and increases Kc.
温度改变既影响平衡位置，也影响平衡常数的数值。对于放热反应（ΔH 为负），升高温度使平衡向左移动，降低产物产率并减小 Kc。对于吸热反应（ΔH 为正），升温使平衡向右移动，增大 Kc。
Thus, in an exothermic process like the Contact process (2SO₂ + O₂ ⇌ ature gives a higher equilibrium yield, but in practice a compromise temperature is used to achieve a reasonable rate.
因此，在放热过程如接触法（2SO₂ + O₂ ⇌ 2SO₃，ΔH = -197 kJ mol⁻¹）中，低温可获得更高的平衡产率，但实际操作中要采用折中温度以获得适当的反应速率。
8. Catalysts and Equilibrium | 催化剂与平衡
A catalyst provides an alternative reaction pathway with a lower activation energy, equally accelerating both forward and reverse reactions. It does not, however, alter the position of equilibrium or the value of Kc or Kp.
催化剂提供了活化能较低的另一条反应路径，同等程度地加快正逆反应速率。然而，它不改变平衡位置，也不改变 Kc 或 Kp 的值。
Catalysts help a system reach equilibrium more quickly but do not change the equilibrium composition. This is why they are essential in industry: they allow a lower temperature to be used to achieve the same equilibrium yield without unduly slowing down the process.
催化剂帮助体系更快达到平衡，但不改变平衡组成。这就是它们在工业中不可或缺的原因：它们允许使用较低温度来获得相同的平衡产率，而不致使过程过于缓慢。
Remember that in a catalysed reaction, the time required to attain equilibrium is shorter, but the final concentrations of reactants and products at equilibrium remain exactly the same as in the uncatalysed reaction at the same temperature.
请记住，在催化反应中，达到平衡所需时间更短，但在相同温度下，达到平衡时的反应物和产物最终浓度与无催化反应完全相同。
9. The Reaction Quotient, Q | 反应商 Q
The reaction quotient, Q, is calculated using the same formula as Kc (or Kp), but with the concentrations (or partial pressures) measured at any point in time, not necessarily at equilibrium. It is a powerful tool to determine which direction a reaction will proceed.
反应商 Q 使用与 Kc（或 Kp）相同的公式计算，但浓度（或分压）是在任意时刻测得的，不一定处于平衡状态。它是判断反应朝哪个方向进行的强大工具。
If Q < K, the ratio of products to reactants is lower than at equilibrium. The forward reaction will be favoured to produce more products until Q equals K. If Q > K, the reverse reaction is favoured to reduce product concentration.
若 Q < K，产物与反应物的比值低于平衡时的比值，正反应将占优势以生成更多产物，直到 Q 等于 K。若 Q > K，逆反应占优势以降低产物浓度。
ilibrium and no net change occurs. The concept of Q is useful for non-equilibrium mixtures where the initial composition is known, allowing us to predict shifts.
当 Q 恰好等于 K 时，体系处于平衡，无净变化发生。Q 的概念对于已知初始组成的非平衡混合物非常有用，可以让我们预测平衡移动方向。
10. Industrial Applications: Haber Process | 工业应用：哈伯法
ammonia synthesis is a classic example of applying equilibrium principles: N₂(g) + 3H₂(g) ⇌ = -92 kJ mol⁻¹. The forward reaction is exothermic and results in a decrease in gas moles.
合成氨的哈伯法是应用平衡原理的经典实例：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92 kJ mol⁻¹。正反应放热且导致气体摩尔数减少。
High pressure (typically 200 atm) is used to favour the forward reaction, shifting equilibrium to the right according to Le Chatelier's principle. However, very high pressures are expensive and require stronger equipment, so a compromise is reached.
使用高压（通常 200 atm）有利于正反应，根据勒夏特列原理使平衡向右移动。但极高压力成本高昂且需要更坚固的设备，因此采用了折中方案。
The reaction is exothermic, so low temperatures would give a higher equilibrium yield of ammonia. However, a temperature of about 450 °C is used because a lower temperature would make the rate too slow, even with a catalyst.
该反应放热，因此低温会获得更高的氨平衡产率。然而，实际采用约 450 °C，因为即使有催化剂，更低温度会使反应速率过慢。
An iron catalyst is employed to increase the rate of attainment of equilibrium without altering the equilibrium yield. The unreacted gases are recycled to improve overall efficiency.
使用铁催化剂以加快达到平衡的速率，而不改变平衡产率。未反应的气体被循环利用，以提升整体效率。
11. Industrial Applications: Contact Process | 工业应用：接触法
The Contact process for sulfuric acid production involves the equilibrium: 2SO₂(g) + O₂(g) ⇌ Again, the forward reaction is exothermic and reduces the number of gas molecules (3 → 2).
生产硫酸的接触法涉及如下平衡：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，ΔH = -197 kJ mol⁻¹。同样，正反应放热且使气体分子数减少（3 → 2）。
A moderate pressure of around 1–2 atm is sufficient, as the equilibrium already lies far to the right under these conditions, and higher pressures are not economical. The temperature is maintained at about 450 °C with a vanadium(V) oxide catalyst, V₂O₅.
约 1–2 atm 的中等压力即已足够，因为在此条件下平衡已大幅偏向右侧，更高压力不经济。温度维持在约 450 °C，并采用五氧化二钒（V₂O₅）催化剂。
The use of excess oxygen further shifts the equilibrium to the right, converting SO₂ more completely into SO₃. This optimisation illustrates the interplay between thermodynamic equilibrium and kinetic practicalities.
使用过量的氧气进一步使平衡右移，将 SO₂ 更彻底地转化为 SO₃。这种优化展现了热力学平衡与动力学实际之间的相互作用。
12. Effect of Inert Gas Addition | 添加惰性气体的影响
Adding an inert gas (such as argon) to a gaseous equilibrium at constant volume does not change the partial pressures of the reactants or products, because it simply increases the total pressure without participating in the reaction. Thus, the position of equilibrium remains unchanged.
在恒容条件下向气体平衡中添加惰性气体（如氩气），不会改变反应物或产物的分压，因为它只是增加总压而不参与反应。因此，平衡位置保持不变。
However, if the inert gas is added at constant total pressure, the volume of the container must increase to maintain that pressure. This dilutes all gaseous species, reducing partial pressures. The equilibrium will then shift to favour the side with more moles of gas to counteract the decrease in pressure.
然而，如果在恒总压下添加惰性气体，容器的体积必须增大以维持该压力。这将稀释所有气体物种，降低分压。此时平衡将向气体摩尔数更多的一侧移动，以抵消压力的下降。
they affect equilibrium by changing the volume and thus partial pressures, unless Δn = 0.
这个区别在考试中很重要：恒容时，惰性气体无影响；恒压时，它们通过改变体积进而改变分压来影响平衡，除非 Δn = 0。
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