A-Level Chemistry: Chemical Equilibrium & Le Chatelier's Principle | A-Level化学：化学平衡与勒夏特列原理

Introduction | 引言

Chemical equilibrium is one of the most fundamental concepts in A-Level Chemistry. It bridges the gap between reaction kinetics and thermodynamics, explaining why some reactions appear to “stop” halfway through. Mastering equilibrium is essential for success across all major exam boards — AQA, OCR, Edexcel, and CAIE alike. In this article, we will explore the core principles, Le Chatelier’s principle, the equilibrium constant K_c, and how industrial processes like the Haber process apply these concepts in practice.

化学平衡是A-Level化学中最基础的概念之一。它连接了反应动力学和热力学，解释了为什么某些反应会在中途”停滞”。掌握化学平衡对于在AQA、OCR、Edexcel和CAIE等所有主要考试局取得好成绩至关重要。本文将探讨核心原理、勒夏特列原理、平衡常数K_c，以及哈伯法等工业过程如何实际应用这些概念。

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

Many chemical reactions are reversible — they can proceed in both the forward and reverse directions. When the rate of the forward reaction equals the rate of the reverse reaction, the system reaches a state of dynamic equilibrium. Crucially, at equilibrium, the concentrations of reactants and products remain constant — but they are not necessarily equal. Both reactions continue to occur, hence the term “dynamic”.

许多化学反应是可逆的 — 它们可以同时向正方向和逆方向进行。当正反应速率等于逆反应速率时，系统达到动态平衡。关键点在于：平衡时反应物和生成物的浓度保持恒定，但不一定相等。两个方向的反应都在持续进行，因此称为”动态”平衡。

Key Characteristics of Dynamic Equilibrium | 动态平衡的关键特征

• The system must be closed — no matter can enter or leave. | 系统必须是封闭的 — 物质不能进出。
• The forward and reverse reaction rates are equal. | 正反应速率与逆反应速率相等。
• Macroscopic properties (concentration, colour, pressure) remain constant. | 宏观性质（浓度、颜色、压强）保持恒定。
• Equilibrium can be approached from either direction. | 平衡可以从任一方向趋近。

Exam Tip: Examiners love to ask: “Has the reaction stopped at equilibrium?” The answer is always no — both forward and reverse reactions continue, just at the same rate. | 考试提示：考官喜欢问：”反应在平衡时是否停止？”答案始终是否 — 正逆反应都在继续，只是速率相同。

2. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states: If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium will shift to counteract the change.

勒夏特列原理指出：如果处于平衡状态的系统受到浓度、压强或温度的变化，平衡位置将移动以抵消该变化。

2.1 Effect of Concentration | 浓度的影响

Consider the equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g)

If we add more H₂: The system responds by consuming H₂, shifting the equilibrium to the right (favouring the forward reaction) to produce more HI. If we remove HI: Again, the equilibrium shifts to the right to replenish the removed product.

如果增加H₂的浓度：系统通过消耗H₂来响应，平衡向右移动（有利于正反应），生成更多HI。如果移除HI：同样，平衡向右移动，补充被移除的生成物。

2.2 Effect of Pressure | 压强的影响

Pressure changes only affect equilibria involving gases. The system responds by shifting towards the side with fewer gas molecules when pressure increases, as this reduces the pressure.

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

• Reactant side: 1 + 3 = 4 moles of gas
• Product side: 2 moles of gas
• Increasing pressure → equilibrium shifts right (fewer moles)
• Decreasing pressure → equilibrium shifts left (more moles)

压强变化只影响涉及气体的平衡。系统通过向气体分子较少的一侧移动来响应压强增加，因为这样会降低压强。

重要提示：如果反应两侧的气体分子数相等，压强变化不影响平衡位置。例如：H₂(g) + I₂(g) ⇌ 2HI(g) — 两侧都是2摩尔气体，所以压强变化无效。

2.3 Effect of Temperature | 温度的影响

Temperature is the only factor that changes the value of the equilibrium constant K_c. You must know whether the reaction is exothermic or endothermic:

温度是唯一能改变平衡常数K_c值的因素。你必须知道反应是放热还是吸热：

• Exothermic (ΔH < 0): Increasing temperature shifts equilibrium left (towards reactants); K_c decreases. | 放热反应：升高温度使平衡向左移动（朝向反应物）；K_c减小。
• Endothermic (ΔH > 0): Increasing temperature shifts equilibrium right (towards products); K_c increases. | 吸热反应：升高温度使平衡向右移动（朝向生成物）；K_c增大。

2.4 Effect of a Catalyst | 催化剂的影响

A common misconception is that catalysts affect the position of equilibrium. A catalyst speeds up both forward and reverse reactions equally. It lowers the activation energy for both directions, so equilibrium is reached faster, but the position of equilibrium and K_c remain unchanged.

一个常见误解是催化剂会影响平衡位置。催化剂同等加速正反应和逆反应。它降低了两方向的活化能，因此平衡更快达到，但平衡位置和K_c保持不变。

3. The Equilibrium Constant, K_c | 平衡常数 K_c

For a general reaction: aA + bB ⇌ cC + dD

The equilibrium constant expression is:

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

Where square brackets denote concentration in mol dm⁻³ at equilibrium.

Key Rules for K_c | K_c的关键规则

• Solids and pure liquids are omitted from the expression. Their concentrations are effectively constant and are absorbed into K_c. | 固体和纯液体被排除在表达式之外。它们的浓度实际上是常数，被吸收到K_c中。
• Only aqueous and gaseous species appear in the K_c expression. | 只有水溶液和气态物种出现在K_c表达式中。
• K_c is temperature-dependent only. It does not change with concentration or pressure. | K_c仅依赖于温度。它不随浓度或压强变化。

Interpreting K_c Values | 解读K_c值

• K_c ≫ 1: Equilibrium lies far to the right — mostly products. | 平衡强烈偏向右侧 — 主要是生成物。
• K_c ≈ 1: Significant amounts of both reactants and products. | 相当数量的反应物和生成物共存。
• K_c ≪ 1: Equilibrium lies far to the left — mostly reactants. | 平衡强烈偏向左侧 — 主要是反应物。

Worked Example | 计算示例

Question: 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed in a 1.0 dm³ vessel. At equilibrium, 0.30 mol of ethyl ethanoate is formed. Calculate K_c.

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Species	CH₃COOH	C₂H₅OH	CH₃COOC₂H₅	H₂O
Initial / mol	0.50	0.50	0	0
Change / mol	−0.30 (each loses 0.30)		+0.30 (each gains 0.30)
Equilibrium / mol	0.20	0.20	0.30	0.30

K_c = (0.30 × 0.30) / (0.20 × 0.20) = 0.09 / 0.04 = 2.25

Units: (mol dm⁻³ × mol dm⁻³) / (mol dm⁻³ × mol dm⁻³) = no units | 无单位

4. The Haber Process — A Real-World Application | 哈伯法 — 实际应用

The Haber process for ammonia synthesis is arguably the most important industrial application of chemical equilibrium:

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

氨合成的哈伯法可以说是化学平衡最重要的工业应用：

Compromise Conditions | 折中条件

Factor | 因素	Thermodynamic Optimum | 热力学最优	Industrial Choice | 工业选择	Reason | 原因
Temperature	Low (exothermic → favours products at low T)	400–450 °C	Low T → too slow. Compromise for acceptable rate.
Pressure	Very high (4 mol → 2 mol gas)	200 atm	Higher pressure is expensive and requires stronger equipment.
Catalyst	—	Iron (Fe)	Speeds up rate without affecting equilibrium position or yield.

Exam Insight: The Haber process is a classic example of compromise conditions. You must explain why the conditions are chosen — not just what they are. High marks require linking kinetic and thermodynamic reasoning. | 考试洞察：哈伯法是折中条件的经典例子。你必须解释为什么选择这些条件 — 而不仅仅是什么条件。高分答案需要联系动力学和热力学推理。

5. Common Exam Mistakes | 常见考试错误

1. Confusing rate and position: Adding a catalyst increases the rate but does NOT shift equilibrium. | 混淆速率和位置：添加催化剂增加速率但不移动平衡。
2. Forgetting that K_c changes with temperature: Many students treat K_c as a universal constant. It is constant only at a given temperature. | 忘记K_c随温度变化：许多学生将K_c视为普适常数。它仅在给定温度下是常数。
3. Including solids in K_c expression: CaCO₃(s) ⇌ CaO(s) + CO₂(g) — K_c = [CO₂] only! | 在K_c表达式中包含固体：CaCO₃(s) ⇌ CaO(s) + CO₂(g) — K_c = [CO₂] 仅此而已！
4. Assuming equal concentrations at equilibrium: Equilibrium means equal rates, not equal concentrations. | 假设平衡时浓度相等：平衡意味着速率相等，而非浓度相等。
5. Misapplying pressure to non-gaseous reactions: Pressure only affects equilibria involving gases with a change in the number of molecules. | 对非气体反应错误应用压强：压强仅影响涉及气体且分子数发生变化的平衡。

6. Quick Reference: Le Chatelier Summary | 速查：勒夏特列原理总结

Change | 变化	Equilibrium Shift | 平衡移动	Effect on Kc
Increase [reactant]	Right (→ products) | 向右	No change | 不变
Increase [product]	Left (→ reactants) | 向左	No change | 不变
Increase pressure (fewer moles on right)	Right | 向右	No change | 不变
Increase temperature (exothermic)	Left | 向左	Decreases | 减小
Increase temperature (endothermic)	Right | 向右	Increases | 增大
Add catalyst	None | 无变化	No change | 不变

7. Practice Questions | 练习题

Q1. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹, state and explain the effect of increasing temperature on the equilibrium yield of SO₃.

Q2. Write the K_c expression for: 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)

Q3. At 500 K, a 2.0 dm³ vessel contains an equilibrium mixture of 0.40 mol PCl₅, 0.80 mol PCl₃, and 0.80 mol Cl₂. Calculate K_c for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

Equilibrium is not about standing still — it is about dynamic balance. Master this concept, and a significant portion of A-Level Chemistry will fall into place. Good luck with your studies!

平衡不是静止不动的 — 而是动态的均衡。掌握这个概念，A-Level化学的重要部分将迎刃而解。祝学习顺利！