Chemical Equilibrium — 化学平衡：可逆反应与动态平衡

📚 Chemical Equilibrium | 化学平衡

1. Introduction to Reversible Reactions | 可逆反应简介

Many chemical reactions proceed in only one direction — reactants are converted into products until one of the reactants is completely used up. These are called irreversible reactions. For example, when magnesium burns in oxygen, it forms magnesium oxide and cannot spontaneously revert back to magnesium and oxygen under normal conditions.

许多化学反应只朝一个方向进行——反应物转化为产物，直到某一反应物完全耗尽。这类反应称为不可逆反应。例如，镁在氧气中燃烧生成氧化镁，在正常条件下无法自发还原为镁和氧气。

However, some reactions are reversible. In a reversible reaction, the products can react together to reform the original reactants. The reaction proceeds in both the forward and backward directions simultaneously. A classic example is the Haber process:

然而，有些反应是可逆的。在可逆反应中，产物可以相互反应重新生成原始反应物。反应同时向正反应方向和逆反应方向进行。一个经典例子是哈伯法合成氨：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = −92 kJ mol⁻¹

The double arrow (⇌) is the symbol used to indicate a reversible reaction. The forward reaction is exothermic (releases heat), while the reverse reaction is endothermic (absorbs heat). Understanding this interplay is at the heart of chemical equilibrium.

双箭头符号（⇌）用于表示可逆反应。正反应是放热反应（释放热量），而逆反应是吸热反应（吸收热量）。理解这种相互作用是化学平衡的核心。

2. What Is Dynamic Equilibrium? | 什么是动态平衡？

Dynamic equilibrium is a state reached in a closed system when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant — but this does NOT mean the reaction has stopped. Both forward and reverse reactions continue at equal rates, hence the term dynamic equilibrium.

动态平衡是在封闭系统中，当正反应速率等于逆反应速率时所达到的状态。平衡时，反应物和产物的浓度保持恒定——但这并不意味着反应停止了。正反应和逆反应以相等的速率继续进行，因此称之为动态平衡。

Three key conditions must be met for dynamic equilibrium to exist:

动态平衡的存在必须满足三个关键条件：

1. Closed system: No matter can enter or leave the system. If a gaseous product escapes, equilibrium cannot be established. 封闭系统：物质不能进入或离开系统。如果气体产物逸出，则无法建立平衡。

2. Constant temperature: The equilibrium position depends on temperature. Changing the temperature shifts the equilibrium. 恒温：平衡位置取决于温度。改变温度会使平衡发生移动。

3. Macroscopic properties remain constant: Observable properties such as colour, pressure, and concentration do not change over time. 宏观性质保持不变：可观察的性质如颜色、压力和浓度不随时间变化。

It is important to distinguish between a static equilibrium (where nothing is happening) and a dynamic equilibrium (where opposing processes occur at equal rates). Chemical equilibrium is always dynamic.

区分静态平衡（没有任何事情发生）和动态平衡（对立过程以相等速率发生）是很重要的。化学平衡始终是动态的。

3. The Equilibrium Constant — Kc | 平衡常数 — Kc

For any reversible reaction at a given temperature, the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficient, is a constant. This is the equilibrium constant, Kc.

对于给定温度下的任何可逆反应，产物浓度与反应物浓度之比（各浓度以其化学计量系数为幂）是一个常数。这就是平衡常数 Kc。

For the general reaction:

对于一般反应：

aA + bB ⇌ cC + dD

The equilibrium expression is:

平衡表达式为：

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Where square brackets denote concentrations in mol dm⁻³. Key points about Kc:

其中方括号表示浓度，单位为 mol dm⁻³。关于 Kc 的关键要点：

• Kc is temperature-dependent only. Changing concentration or pressure does NOT change Kc (though it may shift the equilibrium position). Kc 仅取决于温度。改变浓度或压力不会改变 Kc（虽然可能使平衡位置发生移动）。

• A large Kc (>> 1) means the equilibrium lies to the right — products are favoured. 大的 Kc（远大于 1）意味着平衡偏向右侧——有利于产物。

• A small Kc (<< 1) means the equilibrium lies to the left — reactants are favoured. 小的 Kc（远小于 1）意味着平衡偏向左侧——有利于反应物。

• Pure solids and pure liquids do not appear in the Kc expression — their concentrations are effectively constant. 纯固体和纯液体不出现在 Kc 表达式中——它们的浓度实际上是恒定的。

4. The Equilibrium Constant — Kp (for Gaseous Reactions) | 平衡常数 — Kp（用于气体反应）

For reactions involving gases, it is often more convenient to use partial pressures instead of concentrations. The equilibrium constant in terms of pressure is Kp.

对于涉及气体的反应，通常使用分压代替浓度更为方便。以压力表示的平衡常数是 Kp。

Partial pressure is the pressure that an individual gas would exert if it alone occupied the entire volume at the same temperature. The partial pressure of gas A is given by:

分压是单个气体在相同温度下单独占据整个体积时所产生的压力。气体 A 的分压由下式计算：

pA = (moles of A / total moles) × total pressure = mole fraction of A × P

For the Haber process, the Kp expression is:

对于哈伯法，Kp 表达式为：

Kp = (pNH₃)² / (pN₂)(pH₂)³

The units of Kp depend on the stoichiometry of the reaction and are typically expressed in atm, Pa, or kPa raised to the appropriate power. Like Kc, Kp is constant at a given temperature.

Kp 的单位取决于反应的化学计量关系，通常以 atm、Pa 或 kPa 的适当次幂表示。与 Kc 一样，Kp 在给定温度下为常数。

5. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s Principle states that if a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium will shift to counteract (oppose) the imposed change.

勒夏特列原理指出，如果处于动态平衡的系统受到条件变化的影响，平衡位置将发生移动以抵消（对抗）所施加的变化。

This principle allows us to predict how equilibrium responds to changes in concentration, pressure, and temperature. Let us examine each factor in detail.

这一原理使我们能够预测平衡如何响应浓度、压力和温度的变化。让我们详细分析每个因素。

6. Effect of Changing Concentration | 改变浓度的影响

If the concentration of a reactant is increased, the system shifts to oppose this increase by converting some of the added reactant into product. The equilibrium shifts to the right. Conversely, if a product is removed, the system shifts to produce more product — again shifting to the right.

如果反应物的浓度增加，系统通过将部分新增反应物转化为产物来抵消这种增加。平衡向右移动。反之，如果产物被移除，系统会生成更多产物——同样向右移动。

Example — the Fe³⁺ / SCN⁻ equilibrium:

示例 — Fe³⁺ / SCN⁻ 平衡：

Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq)

(pale yellow 淡黄色)   (colourless 无色)   (blood-red 血红色)

Adding more Fe³⁺ ions shifts equilibrium to the right — the solution turns a deeper red. Adding a reagent that removes Fe³⁺ (such as F⁻ which forms a stable complex) shifts equilibrium to the left — the red colour fades.

加入更多 Fe³⁺ 离子使平衡向右移动——溶液变成更深的红色。加入移除 Fe³⁺ 的试剂（如形成稳定络合物的 F⁻）使平衡向左移动——红色褪去。

Importantly, changing concentration does not change the value of Kc — the ratio of concentrations at the new equilibrium position is the same as before.

重要的是，改变浓度不会改变 Kc 的值——新平衡位置下的浓度比与之前相同。

7. Effect of Changing Pressure (Gaseous Systems) | 改变压力的影响（气体系统）

Pressure changes only affect equilibria involving gases where there is a difference in the number of moles of gas on each side of the equation. If there are more moles of gas on the reactant side than the product side, increasing the pressure shifts the equilibrium to the side with fewer moles of gas to reduce the pressure.

压力的变化仅影响方程式两侧气体摩尔数存在差异的气体平衡。如果反应物侧的气体摩尔数多于产物侧，增加压力会使平衡移向气体摩尔数较少的一侧以降低压力。

Example — the Haber process:

示例 — 哈伯法：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

4 moles of gas   ⇌   2 moles of gas

4 摩尔气体   ⇌   2 摩尔气体

Increasing pressure favours the forward reaction (fewer moles, 2 vs 4), shifting equilibrium to the right and increasing the yield of ammonia. This is why the Haber process is carried out at high pressure (typically 200 atm).

增加压力有利于正反应（摩尔数较少，2 对比 4），使平衡向右移动，增加氨的产率。这就是哈伯法在高压（通常为 200 atm）下进行的原因。

When there are equal numbers of moles of gas on both sides, changing pressure has no effect on the equilibrium position. For example:

当两侧气体摩尔数相等时，改变压力对平衡位置没有影响。例如：

H₂(g) + I₂(g) ⇌ 2HI(g)

2 moles   ⇌   2 moles

Pressure has no effect here because the forward and reverse reactions are affected equally.

这里压力没有影响，因为正反应和逆反应受到同等影响。

8. Effect of Changing Temperature | 改变温度的影响

Unlike concentration and pressure changes, changing temperature does change the value of Kc and Kp. The direction of the shift depends on whether the forward reaction is exothermic or endothermic.

与浓度和压力的变化不同，改变温度确实会改变 Kc 和 Kp 的值。移动的方向取决于正反应是放热还是吸热。

• Exothermic forward reaction (ΔH < 0): Increasing temperature shifts equilibrium to the left (endothermic direction) to absorb the added heat. Kc decreases. 正反应放热（ΔH < 0）：升高温度使平衡向左移动（吸热方向）以吸收增加的热量。Kc 减小。

• Endothermic forward reaction (ΔH > 0): Increasing temperature shifts equilibrium to the right (endothermic direction) to absorb the added heat. Kc increases. 正反应吸热（ΔH > 0）：升高温度使平衡向右移动（吸热方向）以吸收增加的热量。Kc 增大。

Example — the Haber process (ΔH = −92 kJ mol⁻¹, exothermic):

示例 — 哈伯法（ΔH = −92 kJ mol⁻¹，放热）：

Although low temperature would favour a higher equilibrium yield of ammonia, the rate of reaction would be too slow. A compromise temperature of around 400–450 °C is used, along with an iron catalyst to increase the rate without affecting the equilibrium position.

虽然低温有利于提高氨的平衡产率，但反应速率会过慢。实际使用约 400–450 °C 的折中温度，并配合铁催化剂以提高速率而不影响平衡位置。

9. Catalysts and Equilibrium | 催化剂与平衡

A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount. As a result:

催化剂以相同幅度降低正反应和逆反应的活化能。因此：

• A catalyst does NOT affect the position of equilibrium. 催化剂不影响平衡位置。

• A catalyst does NOT change the value of Kc or Kp. 催化剂不改变 Kc 或 Kp 的值。

• A catalyst increases the rate at which equilibrium is reached. 催化剂提高达到平衡的速率。

This is a common exam misconception. Students often think a catalyst increases yield — it does not. It simply allows equilibrium to be attained more quickly.

这是一个常见的考试误区。学生常认为催化剂能提高产率——它并不能。它只是使平衡更快达到。

10. Industrial Application — The Haber Process | 工业应用 — 哈伯法

The Haber process is the most important industrial application of equilibrium principles. It produces ammonia from nitrogen and hydrogen, which is then used to manufacture fertilisers, explosives, and other chemicals. Over 150 million tonnes of ammonia are produced annually worldwide.

哈伯法是平衡原理最重要的工业应用。它从氮气和氢气生产氨，氨随后用于制造化肥、炸药和其他化学品。全球每年生产超过 1.5 亿吨氨。

Reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = −92 kJ mol⁻¹

Compromise conditions used in industry:

工业中使用的折中条件：

Factor 因素	Optimal for Yield 最优产率	Compromise Used 实际折中	Reason 原因
Temperature	Low (exothermic forward)	400–450 °C	Low temp = slow rate; compromise needed
Pressure	High (fewer moles of gas on right)	200 atm	High pressure is expensive and dangerous
Catalyst	N/A (does not affect yield)	Iron (Fe)	Speeds up attainment of equilibrium

中文翻译：温度 — 低温（正反应放热）→ 实际 400–450 °C（低温导致速率慢，需要折中）；压力 — 高压（右侧气体摩尔数少）→ 实际 200 atm（高压昂贵且危险）；催化剂 — 不影响产率 → 铁催化剂（加速达到平衡）。

The raw materials are sourced as follows: nitrogen is obtained from the fractional distillation of liquid air, and hydrogen is primarily obtained from the steam reforming of methane (natural gas): CH₄ + H₂O → CO + 3H₂.

原材料的来源如下：氮气通过液态空气的分馏获得，氢气主要通过甲烷（天然气）的蒸汽重整获得：CH₄ + H₂O → CO + 3H₂。

11. Industrial Application — The Contact Process | 工业应用 — 接触法

The Contact Process is used to manufacture sulfuric acid (H₂SO₄), one of the most important industrial chemicals. It involves a key equilibrium step:

接触法用于制造硫酸（H₂SO₄），这是最重要的工业化学品之一。其中包含一个关键的平衡步骤：

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)   ΔH = −197 kJ mol⁻¹

The oxidation of sulfur dioxide to sulfur trioxide is exothermic and involves a decrease in the number of moles of gas (3 moles → 2 moles). Conditions used:

二氧化硫氧化为三氧化硫是放热反应，且气体摩尔数减少（3 摩尔 → 2 摩尔）。使用的条件：

Condition 条件	Value 值	Reason 原因
Temperature	450 °C	Compromise: low temp = good yield but slow rate
Pressure	1–2 atm	Yield is already ~99% at low pressure; high pressure not needed
Catalyst	V₂O₅ (vanadium(V) oxide)	Heterogeneous catalyst; speeds up attainment of equilibrium

中文翻译：温度 450 °C（折中：低温产率好但速率慢）；压力 1–2 atm（低压下产率已达约 99%，无需高压）；催化剂 V₂O₅ 五氧化二钒（多相催化剂，加速达到平衡）。

12. Calculating Equilibrium Concentrations | 计算平衡浓度

A classic exam question involves calculating Kc from initial amounts and one equilibrium amount. The approach uses an ICE table (Initial, Change, Equilibrium).

经典考试题型涉及从初始量和某一平衡量计算 Kc。方法使用ICE 表格（初始 Initial、变化 Change、平衡 Equilibrium）。

Worked Example:

计算示例：

0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed in a 2.0 dm³ flask and allowed to reach equilibrium. At equilibrium, 0.30 mol of ethyl ethanoate is present. Calculate Kc for the esterification reaction:

将 0.50 mol 乙酸和 0.50 mol 乙醇在 2.0 dm³ 烧瓶中混合并使其达到平衡。平衡时存在 0.30 mol 乙酸乙酯。计算酯化反应的 Kc：

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

	CH₃COOH	C₂H₅OH	CH₃COOC₂H₅	H₂O
Initial (mol)	0.50	0.50	0	0
Change (mol)	−0.30	−0.30	+0.30	+0.30
Equilibrium (mol)	0.20	0.20	0.30	0.30
Equilibrium conc. (mol dm⁻³)	0.10	0.10	0.15	0.15

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH] = (0.15)(0.15) / (0.10)(0.10) = 2.25 (no units, as number of moles is equal on both sides).

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH] = (0.15)(0.15) / (0.10)(0.10) = 2.25（无单位，因为两侧摩尔数相等）。

13. Common Exam Pitfalls | 常见考试误区

• Forgetting to convert moles to concentrations: Kc requires concentrations (mol dm⁻³), not amounts (mol). Divide moles by the volume of the container. 忘记将摩尔数转换为浓度：Kc 需要浓度（mol dm⁻³）而非物质的量（mol）。将摩尔数除以容器体积。

• Including solids or liquids in Kc: Only aqueous and gaseous species appear in the equilibrium expression. Pure solids and liquids have constant concentration. 在 Kc 中包含固体或液体：只有水溶液和气体物种出现在平衡表达式中。纯固体和液体具有恒定浓度。

• Confusing Kc with equilibrium position: Kc is a constant at a given temperature; the equilibrium position can shift while Kc stays the same (concentration and pressure changes). 混淆 Kc 与平衡位置：Kc 在给定温度下是常数；平衡位置可以移动而 Kc 保持不变（浓度和压力变化）。

• Thinking catalysts increase yield: Catalysts affect only the rate, never the position of equilibrium or the value of Kc. 认为催化剂提高产率：催化剂只影响速率，从不影响平衡位置或 Kc 的值。

• Stating pressure shift without checking moles: If the number of moles of gas is equal on both sides, pressure has no effect. Always count the moles first. 未检查摩尔数就陈述压力移动：如果两侧气体摩尔数相等，压力没有影响。始终先数摩尔数。

14. Summary | 总结

Chemical equilibrium is a fundamental concept that underpins much of A-Level Chemistry. The key takeaways are:

化学平衡是支撑 A-Level 化学大部分内容的基本概念。关键要点如下：

Concept 概念	Key Point 要点
Dynamic Equilibrium	Forward rate = reverse rate; concentrations constant but not equal
Kc and Kp	Constant at a given temperature; temperature-dependent only
Le Chatelier’s Principle	System shifts to oppose imposed changes
Concentration change	Shifts equilibrium position; Kc unchanged
Pressure change	Only affects systems with unequal gas moles; Kc unchanged
Temperature change	Shifts equilibrium position AND changes Kc/Kp value
Catalyst	No effect on position or Kc; only speeds up attainment of equilibrium

中文总结：动态平衡 — 正反应速率 = 逆反应速率，浓度恒定但不相等；Kc 和 Kp — 给定温度下为常数，仅依赖温度；勒夏特列原理 — 系统移动以对抗外加变化；浓度变化 — 移动平衡位置，Kc 不变；压力变化 — 仅影响气体摩尔数不等的系统，Kc 不变；温度变化 — 移动平衡位置且改变 Kc/Kp 值；催化剂 — 不影响位置或 Kc，仅加速达到平衡。

Understanding equilibrium requires practice with calculations and an appreciation of how industrial chemists balance thermodynamic favourability against kinetic practicality. Master these concepts, and you will have a solid foundation for both your examinations and further study in chemistry.

理解平衡需要对计算进行练习，并理解工业化学家如何在热力学有利性与动力学可行性之间取得平衡。掌握这些概念，你将为考试和进一步的化学学习奠定坚实基础。