Equilibrium | 化学平衡

📚 Equilibrium | 化学平衡

Chemical equilibrium is one of the most fundamental concepts in physical chemistry. It describes the state in which a reversible reaction has reached a balance between the forward and reverse processes, allowing chemists to predict the direction of reactions and optimise industrial conditions for maximum yield.

化学平衡是物理化学中最基本的概念之一。它描述了可逆反应在正向和反向过程之间达到平衡的状态，使化学家能够预测反应方向并优化工业条件以获得最大产率。

1. Dynamic Equilibrium | 动态平衡

A chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant over time, but the system is not static — reactions continue to occur at the molecular level in both directions simultaneously.

化学平衡是指正反应速率等于逆反应速率的状态。在平衡时，反应物和生成物的浓度随时间保持不变，但系统并非静止——分子层面的正逆反应仍在同时持续进行。

Dynamic equilibrium is achieved only in closed systems where no substances can enter or leave. If the system is open, products or reactants may escape, preventing equilibrium from being established. Additionally, equilibrium can be approached from either direction — starting with pure reactants or pure products.

动态平衡仅在封闭系统中才能实现，因为封闭系统内没有物质可以进出。如果系统是开放的，反应物或生成物可能逸出，从而无法建立平衡。此外，平衡可以从任一方向趋近——无论是从纯反应物还是纯生成物开始。

Key characteristics of dynamic equilibrium:

动态平衡的主要特征：

The rate of the forward reaction equals the rate of the reverse reaction | 正反应速率等于逆反应速率
The concentrations of all species remain constant | 所有物质的浓度保持不变
The equilibrium is dynamic, not static | 平衡是动态的，而非静态的
Equilibrium can be approached from either direction | 平衡可以从任一方向趋近

2. 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 conditions, the system will respond by shifting the equilibrium position to counteract the change and minimise its effect. This principle is the cornerstone for predicting how a system responds to external disturbances.

勒夏特列原理指出：如果对处于平衡状态的系统施加条件变化，系统将通过移动平衡位置来抵消这种变化并将其影响降至最低。该原理是预测系统如何响应外部干扰的基石。

This principle allows chemists to predict how changing concentration, pressure, or temperature will affect the position of equilibrium in a reversible reaction. It is important to note that the equilibrium position shifts, not the equilibrium constant — Kc and Kp remain unchanged unless temperature changes.

该原理使化学家能够预测浓度、压力或温度的变化将如何影响可逆反应的平衡位置。需要特别注意的是，移动的是平衡位置，而非平衡常数——除非温度改变，否则 Kc 和 Kp 保持不变。

The effect of any change in conditions can be summarised as: the system always acts to oppose the change imposed upon it.

任何条件变化的影响可以总结为：系统总是趋向于抵消外界施加的变化。

3. Effect of Concentration Changes | 浓度变化的影响

Increasing the concentration of a reactant shifts the equilibrium to the right, favouring the forward reaction, in order to consume the added reactant. Conversely, decreasing the concentration of a product also shifts the equilibrium to the right, as the system attempts to produce more product to replace what was removed.

增加反应物浓度会使平衡向右移动，有利于正反应，以消耗额外加入的反应物。反之，降低生成物浓度同样会使平衡向右移动，因为系统试图生成更多产物来补充被移除的部分。

Similarly, increasing product concentration or decreasing reactant concentration shifts the equilibrium to the left. The system always moves in the direction that opposes the imposed change.

类似地，增加生成物浓度或减少反应物浓度会使平衡向左移动。系统总是朝着抵消外加变化的方向运动。

In industrial practice, the product is often removed continuously from the reaction mixture to drive the equilibrium towards the product side, a strategy that is both economical and efficient.

在工业实践中，通常连续从反应混合物中移除产物，以推动平衡向产物方向移动，这一策略既经济又高效。

4. Effect of Pressure Changes | 压力变化的影响

For reactions involving gases, changes in pressure can significantly affect the equilibrium position. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas, while decreasing pressure shifts it towards the side with more moles of gas.

对于涉及气体的反应，压力变化会显著影响平衡位置。增加压力使平衡向气体物质的量较少的一侧移动，而降低压力使平衡向气体物质的量较多的一侧移动。

Consider the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The left side has 4 moles of gas, while the right side has 2 moles. Increasing pressure therefore favours the forward reaction and increases the yield of ammonia.

以哈伯法为例：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。左侧有 4 mol 气体，右侧有 2 mol 气体。因此增加压力有利于正反应，提高氨的产率。

If the number of moles of gas is equal on both sides of the equation, pressure changes have no effect on the equilibrium position. For example, H₂(g) + I₂(g) ⇌ 2HI(g) is unaffected by pressure changes.

如果方程式两边气体的物质的量相等，则压力变化对平衡位置没有影响。例如，H₂(g) + I₂(g) ⇌ 2HI(g) 不受压力变化的影响。

5. Effect of Temperature Changes | 温度变化的影响

Increasing temperature shifts the equilibrium in the endothermic (positive ΔH) direction, while decreasing temperature shifts it in the exothermic (negative ΔH) direction. This behaviour is entirely consistent with Le Chatelier’s Principle.

升高温度使平衡向吸热方向移动，降低温度使平衡向放热方向移动。这一行为与勒夏特列原理完全一致。

For the forward reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92.2 kJ mol⁻¹, the forward direction is exothermic. Increasing the temperature therefore favours the reverse (endothermic) reaction, decreasing the equilibrium yield of ammonia.

对于正反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92.2 kJ mol⁻¹，正方向为放热方向。因此升高温度有利于逆反应（吸热方向），降低氨的平衡产率。

It is crucial to understand that temperature is the only factor that changes the value of the equilibrium constant itself. A change in temperature alters Kc and Kp; a change in concentration or pressure alters only the position of equilibrium, not the constant.

必须理解，温度是唯一能改变平衡常数本身数值的因素。温度变化会改变 Kc 和 Kp；而浓度或压力变化只会改变平衡位置，不会改变平衡常数的值。

6. Effect of Catalysts | 催化剂的影响

A catalyst speeds up both the forward and reverse reactions to the same extent by providing an alternative pathway with lower activation energy. As a result, a catalyst does not alter the position of equilibrium, nor does it change the value of the equilibrium constant.

催化剂通过提供活化能较低的替代反应途径，同等程度地加快正反应和逆反应。因此，催化剂不改变平衡位置，也不改变平衡常数的数值。

While catalysts do not change the equilibrium yield, they enable the system to reach equilibrium more quickly. This is of great economic significance in industrial chemistry, where time is money and faster equilibration means higher throughput.

虽然催化剂不改变平衡产率，但能使系统更快地达到平衡。这在工业化学中具有重要的经济意义——时间就是金钱，更快的平衡意味着更高的产量。

For instance, in the Haber process, an iron catalyst is used to accelerate the attainment of equilibrium, allowing ammonia to be produced at a commercially viable rate.

例如，在哈伯法中，使用铁催化剂加速平衡的到达，使氨能够以商业可行的速率生产。

7. The Equilibrium Constant Kc | 平衡常数 Kc

For a general reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc is expressed in terms of molar concentrations:

对于一般可逆反应 aA + bB ⇌ cC + dD，平衡常数 Kc 以摩尔浓度表示：

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

The square brackets denote molar concentration in mol dm⁻³. Importantly, pure solids and pure liquids do not appear in the expression for Kc because their concentrations are effectively constant.

方括号表示摩尔浓度，单位为 mol dm⁻³。重要的是，纯固体和纯液体不出现在 Kc 的表达式中，因为它们的浓度实际上是恒定的。

Kc is constant at a given temperature. The value of Kc indicates the extent of the reaction: a large Kc (≫ 1) means the equilibrium lies to the right, favouring products, while a small Kc (≪ 1) means the equilibrium lies to the left, favouring reactants.

在给定温度下，Kc 为常数。Kc 的值反映反应进行的程度：Kc 很大（≫ 1）表示平衡偏向右侧，有利于生成物；Kc 很小（≪ 1）表示平衡偏向左侧，有利于反应物。

The units of Kc depend on the stoichiometry of the reaction. For example, for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the units are mol⁻² dm⁶.

Kc 的单位取决于反应的化学计量数。例如，对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，单位为 mol⁻² dm⁶。

8. The Equilibrium Constant Kp | 平衡常数 Kp

Kp is the equilibrium constant expressed in terms of partial pressures for reactions involving gases. For aA(g) + bB(g) ⇌ cC(g) + dD(g):

Kp 是对气体反应以分压形式表达的平衡常数。对于 aA(g) + bB(g) ⇌ cC(g) + dD(g)：

Kp = (pCᶜ × pDᵈ) / (pAᵃ × pBᵇ)

The partial pressure of a gas in a mixture is its individual contribution to the total pressure, calculated as the product of its mole fraction and the total pressure:

气体在混合物中的分压是其对总压的单独贡献，等于其摩尔分数乘以总压：

pᵢ = xᵢ × P_total

where xᵢ = nᵢ / n_total is the mole fraction and P_total is the total pressure. Partial pressures are measured in pascals (Pa) or atmospheres (atm).

其中 xᵢ = nᵢ / n_total 为摩尔分数，P_total 为总压。分压的单位为帕斯卡（Pa）或大气压（atm）。

Kp is temperature-dependent, and like Kc, pure solids and pure liquids do not appear in its expression.

Kp 与温度有关，与 Kc 类似，纯固体和纯液体不出现在其表达式中。

9. Relationship Between Kc and Kp | Kc 与 Kp 的关系

The relationship between Kc and Kp for gas-phase equilibria is given by the following equation:

气相平衡中 Kc 与 Kp 的关系由以下公式给出：

Kp = Kc(RT)^Δn

where Δn = (c + d) − (a + b), representing the change in the number of moles of gas from reactants to products. R is the universal gas constant (8.31 J mol⁻¹ K⁻¹), and T is the temperature in Kelvin, K.

其中 Δn = (c + d) − (a + b)，表示从反应物到生成物气体物质的量的变化。R 为通用气体常数（8.31 J mol⁻¹ K⁻¹），T 为热力学温度，单位为开尔文（K）。

When Δn = 0, the number of moles of gas is unchanged and Kp = Kc. For example, in H₂(g) + I₂(g) ⇌ 2HI(g), Δn = 0 and therefore Kp equals Kc.

当 Δn = 0 时，气体物质的量不变，Kp = Kc。例如，在 H₂(g) + I₂(g) ⇌ 2HI(g) 中，Δn = 0，因此 Kp 等于 Kc。

10. The Reaction Quotient Q | 反应商 Q

The reaction quotient Q has the same mathematical form as the equilibrium constant Kc, but it uses the concentrations of species at any moment in time, not necessarily at equilibrium. Comparing Q with Kc enables chemists to predict the direction in which a reaction will proceed.

反应商 Q 的数学表达式与平衡常数 Kc 相同，但它使用的是任意时刻的浓度，而不一定是平衡浓度。比较 Q 与 Kc 可以预测反应进行的方向。

There are three possible cases:

有三种可能的情况：

If Q < Kc, the reaction will proceed in the forward direction to reach equilibrium | 如果 Q < Kc，反应将沿正方向进行以达到平衡
If Q > Kc, the reaction will proceed in the reverse direction to reach equilibrium | 如果 Q > Kc，反应将沿逆方向进行以达到平衡
If Q = Kc, the system is already at equilibrium | 如果 Q = Kc，系统已处于平衡状态

The reaction quotient is an extremely useful diagnostic tool in both laboratory analysis and industrial process control, allowing chemists to determine instantly whether a reacting mixture has reached equilibrium.

反应商在实验室分析和工业过程控制中都是极为有用的诊断工具，它使化学家能够即时判断反应混合物是否已达到平衡。

11. Industrial Application: The Haber Process | 工业应用：哈伯法

The Haber process is used to manufacture ammonia on an industrial scale: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.2 kJ mol⁻¹. Nitrogen is obtained from the air and hydrogen from natural gas or steam reforming.

哈伯法用于工业规模合成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = −92.2 kJ mol⁻¹。氮气来自空气，氢气来自天然气或蒸汽重整。

Although the forward reaction is exothermic and would theoretically favour low temperatures, a moderate temperature of 400−450 °C is used as a compromise between a high rate of reaction and a reasonable equilibrium yield. An iron catalyst accelerates the rate at which equilibrium is achieved.

尽管正反应放热，理论上有利于低温，但仍使用 400−450 °C 的中等温度，这是高反应速率与合理平衡产率之间的折衷选择。铁催化剂加速平衡到达的速率。

To maximise the yield, a high pressure of approximately 200 atm is employed, which shifts the equilibrium to the side with fewer moles of gas (the product side). Ammonia is continuously liquefied and removed, driving the equilibrium further to the right while allowing unreacted gases to be recycled.

为了最大化产率，采用约 200 atm 的高压，使平衡向气体物质的量较少的一侧（产物侧）移动。氨被连续液化并移走，进一步推动平衡右移，未反应的气体则循环使用。

12. Industrial Application: The Contact Process | 工业应用：接触法

The Contact process is the principal industrial method for producing sulphuric acid. The key equilibrium step is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ mol⁻¹. Vanadium(V) oxide, V₂O₅, serves as the catalyst and operates at a temperature of approximately 450 °C.

接触法是生产硫酸的主要工业方法。关键的平衡步骤为：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，ΔH = −196 kJ mol⁻¹。五氧化二钒（V₂O₅）作为催化剂，操作温度约为 450 °C。

Since the forward reaction is exothermic, a lower temperature would favour a higher equilibrium yield; however, the rate would become impractically slow. The temperature of 450 °C represents the optimal balance between rate and yield for this exothermic equilibrium.

由于正反应为放热反应，较低温度有利于更高的平衡产率；但反应速率会变得过低而不切实际。450 °C 的温度代表了该放热平衡中速率与产率之间的最佳平衡点。

A pressure of only 1−2 atm is used because at this pressure the conversion of SO₂ to SO₃ already exceeds 95%, rendering higher pressures economically unjustified for the marginal gain in yield.

仅使用 1−2 atm 的压力，因为在该压力下 SO₂ 转化为 SO₃ 的转化率已超过 95%，更高压力带来的产率边际收益不足以弥补其经济成本。

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Equilibrium | 化学平衡