Chemical Equilibrium & Le Chatelier’s Principle — 化学平衡与勒夏特列原理

📚 Chemical Equilibrium & Le Chatelier’s Principle | 化学平衡与勒夏特列原理

Chemical equilibrium is one of the most conceptually rich topics in A-Level Chemistry. It bridges the gap between reaction kinetics and thermodynamics, explaining not just how fast reactions proceed but how far they go. Understanding equilibrium is essential for topics ranging from industrial process design (the Haber and Contact processes) to acid-base chemistry and biochemical pathways.

化学平衡是A-Level化学中概念最丰富的主题之一。它连接了反应动力学和热力学，不仅解释了反应进行得有多快，还解释了反应能进行到何种程度。理解平衡对于从工业过程设计（哈柏法和接触法）到酸碱化学和生化途径等主题至关重要。

1. The Concept of Dynamic Equilibrium | 动态平衡的概念

Many chemical reactions are reversible — the products can react to reform the reactants. When a reversible reaction is carried out in a closed system, it eventually reaches a state of dynamic equilibrium where the forward and reverse reactions proceed at exactly the same rate. At this point, the concentrations of all species remain constant, but the reactions have not stopped.

许多化学反应是可逆的——产物可以反应重新生成反应物。当可逆反应在封闭系统中进行时，最终会达到动态平衡状态，此时正反应和逆反应以完全相同的速率进行。在这一点上，所有物质的浓度保持不变，但反应并未停止。

Key characteristics of dynamic equilibrium:

动态平衡的主要特征：

• The system must be closed — no matter can enter or leave. If a gaseous product escapes, equilibrium cannot be established.
• 系统必须是封闭的——物质不能进入或离开。如果气体产物逸出，就无法建立平衡。
• Macroscopic properties are constant — colour, pressure, concentration, and density do not change.
• 宏观性质保持不变——颜色、压力、浓度和密度不发生变化。
• Microscopic processes continue — at the molecular level, both forward and reverse reactions are still occurring.
• 微观过程仍在进行——在分子层面上，正反应和逆反应都在发生。
• Equilibrium can be approached from either direction — starting with only reactants or only products yields the same equilibrium mixture (provided the same conditions).
• 平衡可以从任一方向达到——从纯反应物或纯产物开始，在相同条件下会得到相同的平衡混合物。

2. The Equilibrium Constant: Kc | 平衡常数：Kc

For a general reversible reaction:

对于一般的可逆反应：

aA + bB ⇌ cC + dD

The equilibrium constant in terms of concentration, Kc, is defined as:

浓度平衡常数Kc定义为：

Kc = [C]^c[D]^d / [A]^a[B]^b

Where square brackets denote equilibrium concentrations in mol dm^-3.

其中方括号表示以mol dm^-3为单位的平衡浓度。

2.1 What Kc Tells Us | Kc的含义

Kc Value | Kc值	Interpretation | 解读
Kc >> 1 (e.g., 1010)	Equilibrium lies far to the right — products dominate. The reaction essentially goes to completion.
Kc ≈ 1	Significant amounts of both reactants and products present at equilibrium.
Kc << 1 (e.g., 10-10)	Equilibrium lies far to the left — reactants dominate. Very little product forms.

Important: Kc is constant for a given reaction at a given temperature. If the temperature changes, Kc changes. However, changing concentration or pressure does not change Kc — the equilibrium position shifts but the value of the constant remains the same.

重要：Kc在给定温度下对于给定反应是常数。如果温度改变，Kc也会改变。然而，改变浓度或压力不会改变Kc——平衡位置会移动，但常数的值保持不变。

2.2 Calculating Kc | Kc的计算

Worked Example | 例题：

Ethanoic acid reacts with ethanol to form ethyl ethanoate and water:

乙酸与乙醇反应生成乙酸乙酯和水：

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed. At equilibrium, 0.30 mol of ethyl ethanoate is present in a total volume of 1.0 dm³. Calculate Kc.

将0.50 mol乙酸与0.50 mol乙醇混合。平衡时，在总体积为1.0 dm³的容器中有0.30 mol乙酸乙酯。计算Kc。

Solution | 解答：

At equilibrium, [CH₃COOC₂H₅] = [H₂O] = 0.30 mol dm^-3 (1:1 ratio from the equation). Since 0.30 mol of each product formed, 0.30 mol of each reactant was consumed. Remaining: 0.50 – 0.30 = 0.20 mol dm^-3 for each reactant.

平衡时，[CH₃COOC₂H₅] = [H₂O] = 0.30 mol dm^-3（根据方程式1:1比例）。由于每种产物生成了0.30 mol，每种反应物消耗了0.30 mol。剩余：每种反应物0.50 – 0.30 = 0.20 mol dm^-3。

Kc = (0.30)(0.30) / (0.20)(0.20) = 0.090 / 0.040 = 2.25

Units: In this case, Kc has no units because the total number of moles is the same on both sides of the equation (2 vs 2).

单位：在这种情况下，Kc没有单位，因为方程式两边的总摩尔数相同（2对2）。

3. Gaseous Equilibria: Kp | 气体平衡：Kp

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

对于涉及气体的反应，使用分压代替浓度通常更方便。分压平衡常数Kp的定义类似：

Kp = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b

The partial pressure of a gas is the pressure that gas would exert if it alone occupied the container. It is calculated using the mole fraction:

气体的分压是该气体单独占据容器时所施加的压力。它使用摩尔分数计算：

Partial Pressure = Mole Fraction × Total Pressure

分压 = 摩尔分数 × 总压力

Where mole fraction of gas A = n(A) / n(total).

其中气体A的摩尔分数 = n(A) / n(总)。

4. Le Chatelier’s Principle | 勒夏特列原理

Henri Louis Le Chatelier stated in 1884: “If a system at dynamic equilibrium is subjected to a change, the position of equilibrium will shift to oppose that change.”

亨利·路易·勒夏特列于1884年提出：“如果一个处于动态平衡的系统受到改变，平衡位置将移动以对抗这种改变。”

This principle allows us to predict how equilibrium responds to changes in concentration, pressure, and temperature.

这个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。

4.1 Effect of Concentration | 浓度的影响

Increasing the concentration of a reactant shifts the equilibrium to the right (towards products) to use up the added reactant. Increasing the concentration of a product shifts equilibrium to the left.

增加反应物的浓度会使平衡向右移动（朝向产物），以消耗添加的反应物。增加产物的浓度会使平衡向左移动。

Laboratory example | 实验示例： The Fe³⁺/SCN^– equilibrium:

Fe³⁺(aq) + SCN^–(aq) ⇌ [FeSCN]²⁺(aq)

(pale yellow | 浅黄色) + (colourless | 无色) ⇌ (blood-red | 血红色)

Adding more Fe³⁺ or SCN^– makes the solution turn a deeper red. Adding a reagent that removes Fe³⁺ (such as F^– ions which form a stable complex) causes the red colour to fade.

添加更多Fe³⁺或SCN^–会使溶液变成更深的红色。添加能去除Fe³⁺的试剂（如能形成稳定络合物的F^–离子）会导致红色消退。

4.2 Effect of Pressure | 压力的影响

Pressure changes only affect equilibria involving gases where there is a difference in the number of moles on each side of the equation.

压力变化仅影响涉及气体且方程式两边摩尔数不同的平衡。

• Increasing pressure shifts equilibrium to the side with fewer gaseous moles (to reduce pressure).
• 增加压力使平衡移向气体摩尔数较少的一侧（以降低压力）。
• Decreasing pressure shifts equilibrium to the side with more gaseous moles.
• 降低压力使平衡移向气体摩尔数较多的一侧。
• If the number of moles is the same on both sides, pressure has no effect on the equilibrium position.
• 如果两边摩尔数相同，压力对平衡位置没有影响。

Example | 示例：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

4 moles gaseous reactants ⇌ 2 moles gaseous product

High pressure favours the forward reaction (fewer moles). This is why the Haber Process operates at high pressure (around 200 atm).

高压有利于正反应（更少的摩尔数）。这就是哈柏法在高压（约200 atm）下操作的原因。

4.3 Effect of Temperature | 温度的影响

Temperature is the only factor that changes the value of Kc. The direction of the shift depends on whether the forward reaction is exothermic or endothermic:

温度是唯一能改变Kc值的因素。移动的方向取决于正反应是放热还是吸热：

• Increasing temperature shifts equilibrium in the endothermic direction (to absorb the added heat).
• 升高温度使平衡向吸热方向移动（以吸收添加的热量）。
• Decreasing temperature shifts equilibrium in the exothermic direction (to release heat).
• 降低温度使平衡向放热方向移动（以释放热量）。

For an exothermic forward reaction (ΔH < 0): Increasing temperature shifts equilibrium left, decreasing Kc. Decreasing temperature shifts equilibrium right, increasing Kc.

对于放热正反应（ΔH < 0）：升高温度使平衡向左移动，Kc减小。降低温度使平衡向右移动，Kc增大。

For an endothermic forward reaction (ΔH > 0): Increasing temperature shifts equilibrium right, increasing Kc.

对于吸热正反应（ΔH > 0）：升高温度使平衡向右移动，Kc增大。

4.4 Effect of a Catalyst | 催化剂的影响

A catalyst does not affect the position of equilibrium or the value of Kc. It speeds up both the forward and reverse reactions equally, so equilibrium is reached more quickly but the equilibrium composition is unchanged.

催化剂不影响平衡位置也不影响Kc的值。它同等地加速正反应和逆反应，因此平衡更快达到，但平衡组成不变。

5. Industrial Applications | 工业应用

5.1 The Haber Process | 哈柏法

The synthesis of ammonia is one of the most important industrial chemical reactions, producing over 150 million tonnes annually for fertilisers:

氨的合成是最重要的工业化学反应之一，每年生产超过1.5亿吨用于肥料：

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

Condition | 条件	Compromise Value | 折衷值	Reasoning | 理由
Temperature | 温度	400-450°C	Forward reaction is exothermic — low temperature favours higher yield but makes the reaction too slow. A moderate temperature balances rate and yield.
Pressure | 压力	200 atm	High pressure favours the side with fewer moles (products). Higher pressure = higher yield + faster rate. Limited by equipment cost and safety.
Catalyst | 催化剂	Iron (Fe) | 铁	Speeds up the reaction without affecting yield. Allows lower temperature to be used while maintaining economic rate.

5.2 The Contact Process | 接触法

Production of sulfuric acid, the world’s most produced chemical:

硫酸的生产，世界上产量最大的化学品：

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

The key equilibrium step is carried out at 450°C, 1-2 atm pressure, with a vanadium(V) oxide (V₂O₅) catalyst. The moderate temperature is again a compromise — the forward reaction is exothermic, and although SO₃ yield would be higher at lower temperatures, the rate would be uneconomically slow.

关键的平衡步骤在450°C、1-2 atm压力下进行，使用五氧化二钒（V₂O₅）催化剂。适中的温度同样是一个折衷——正反应是放热的，虽然在较低温度下SO₃的产率会更高，但速率会慢得不经济。

6. Acid-Base Equilibria | 酸碱平衡

Equilibrium concepts extend naturally to acid-base chemistry. For a weak acid HA:

平衡概念自然延伸到酸碱化学。对于弱酸HA：

HA(aq) ⇌ H⁺(aq) + A^–(aq)

Ka = [H⁺][A^–] / [HA]

The acid dissociation constant Ka quantifies acid strength. A larger Ka means a stronger acid (more dissociation at equilibrium). The logarithmic scale pKa = -log₁₀(Ka) is often used — a smaller pKa means a stronger acid.

酸解离常数Ka量化了酸的强度。Ka越大意味着酸越强（平衡时解离更多）。通常使用对数标度pKa = -log₁₀(Ka)——pKa越小意味着酸越强。

The ionic product of water, Kw, is another crucial equilibrium constant:

水的离子积Kw是另一个关键的平衡常数：

H₂O(l) ⇌ H⁺(aq) + OH^–(aq)

Kw = [H⁺][OH^–] = 1.0 × 10^-14 mol² dm^-6 at 298 K

Relationship | 关系	Formula | 公式
Ka × Kb = Kw (for conjugate acid-base pair)	pKa + pKb = pKw = 14.00 (at 298 K)
pH of a weak acid	[H+] = √(Ka × [HA]) (approximation when dissociation is small)

7. Exam Tips & Common Mistakes | 考试技巧与常见错误

Common Mistakes | 常见错误

1. Confusing rate and equilibrium position: A catalyst speeds up the rate but does not change the equilibrium position. Students often write that a catalyst increases yield — it does not.
2. 混淆速率和平衡位置：催化剂加快速率但不改变平衡位置。学生经常写催化剂提高产率——实际上不会。
3. Forgetting that Kc changes with temperature: Only temperature changes Kc. Concentration and pressure shift the position without changing Kc.
4. 忘记Kc随温度变化：只有温度改变Kc。浓度和压力移动位置但不改变Kc。
5. Including solids and pure liquids in Kc expressions: The concentration of a solid or pure liquid is constant and is incorporated into Kc — they do not appear in the expression.
6. 在Kc表达式中包含固体和纯液体：固体或纯液体的浓度是常数，已被纳入Kc——它们不出现在表达式中。
7. Missing units on Kc: Always calculate and state the units of Kc, which depend on the stoichiometry.
8. 遗漏Kc的单位：始终计算并说明Kc的单位，这取决于化学计量比。
9. Misapplying Le Chatelier to pressure: Pressure only matters when there are gases and when the number of moles differs. Adding an inert gas at constant volume does not shift equilibrium.
10. 错误地将勒夏特列原理应用于压力：压力只有在有气体且摩尔数不同时才重要。在恒定体积下添加惰性气体不会移动平衡。

Key Definitions for the Exam | 考试关键定义

Term | 术语	Definition | 定义
Dynamic Equilibrium | 动态平衡	The state in a closed system where the rates of the forward and reverse reactions are equal and the concentrations of reactants and products remain constant.
Le Chatelier’s Principle | 勒夏特列原理	If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium shifts to oppose that change.
Kc | 浓度平衡常数	The ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient, at equilibrium at a given temperature.
Homogeneous Equilibrium | 均相平衡	An equilibrium where all reactants and products are in the same phase (all gases or all aqueous).
Heterogeneous Equilibrium | 多相平衡	An equilibrium where reactants and products are in different phases (e.g., CaCO3(s) ⇌ CaO(s) + CO2(g)).

8. Practice Questions | 练习题

Q1: For the reaction 2NO₂(g) ⇌ N₂O₄(g), ΔH = -57 kJ mol^-1, predict and explain the effect of (a) increasing temperature, (b) increasing pressure, (c) adding a catalyst.

问题1：对于反应2NO₂(g) ⇌ N₂O₄(g)，ΔH = -57 kJ mol^-1，预测并解释(a)升高温度、(b)增加压力、(c)添加催化剂的影响。

Q2: 0.20 mol of PCl₅ is placed in a 2.0 dm³ container and heated. At equilibrium, 40% of the PCl₅ has dissociated: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Calculate Kc with units.

问题2：将0.20 mol PCl₅放入2.0 dm³容器中加热。平衡时，40%的PCl₅已解离：PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)。计算Kc及其单位。

Q3: Explain why the Haber Process uses a temperature of 400-450°C rather than room temperature, even though the forward reaction is exothermic.

问题3：解释为什么哈柏法使用400-450°C而不是室温，尽管正反应是放热的。

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Understanding chemical equilibrium is fundamental to mastering A-Level Chemistry. The interplay of Kc, Kp, and Le Chatelier’s Principle gives chemists powerful tools to predict and control the outcome of reversible reactions — from industrial ammonia synthesis to the delicate pH regulation in living cells. Practice writing Kc expressions, performing equilibrium calculations, and applying Le Chatelier’s Principle to unfamiliar reactions to build confidence for the exam.

理解化学平衡是掌握A-Level化学的基础。Kc、Kp和勒夏特列原理的相互作用为化学家提供了强大的工具，用于预测和控制可逆反应的结果——从工业氨合成到活细胞中精密的pH调节。练习书写Kc表达式、进行平衡计算以及将勒夏特列原理应用于不熟悉的反应，以建立考试信心。