A-Level Chemistry: Chemical Equilibrium — Le Chatelier’s Principle & Equilibrium Constants | A-Level化学：化学平衡 — 勒夏特列原理与平衡常数

Chemical equilibrium is one of the most conceptually rich and mathematically demanding topics in A-Level Chemistry. Whether you’re sitting AQA, Edexcel, OCR, or CIE, a deep understanding of dynamic equilibrium, Le Chatelier’s Principle, and equilibrium constants (K_c and K_p) is essential for top marks. This article provides a comprehensive bilingual guide to the topic, covering theory, calculations, industrial applications, and exam technique.

化学平衡是A-Level化学中概念最丰富、数学要求最高的主题之一。无论你参加AQA、Edexcel、OCR还是CIE考试，深入理解动态平衡、勒夏特列原理以及平衡常数（K_c和K_p）对于取得高分至关重要。本文提供该主题的全面双语指南，涵盖理论、计算、工业应用和考试技巧。

1. Reversible Reactions & Dynamic Equilibrium | 可逆反应与动态平衡

1.1 What Is a Reversible Reaction? | 什么是可逆反应？

A reversible reaction is one in which the products can react together to re-form the original reactants. In chemical notation, we use the double-harpoon arrow (⇌) to indicate reversibility:

可逆反应是指产物可以相互反应重新生成原始反应物的反应。在化学符号中，我们使用双鱼叉箭头（⇌）来表示可逆性：

aA + bB ⇌ cC + dD

The forward reaction converts A and B into C and D, while the backward (reverse) reaction converts C and D back into A and B. Both reactions occur simultaneously when the system reaches equilibrium.

正反应将A和B转化为C和D，而逆反应将C和D转化回A和B。当系统达到平衡时，两个反应同时发生。

Key examples | 关键例子：

• N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — The Haber process for ammonia synthesis | 哈伯法合成氨
• 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) — The Contact process for sulfuric acid production | 接触法制硫酸
• H₂(g) + I₂(g) ⇌ 2HI(g) — Hydrogen iodide equilibrium | 碘化氢平衡
• CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l) — Esterification | 酯化反应

1.2 Dynamic Equilibrium — The Core Concept | 动态平衡——核心概念

Dynamic equilibrium occurs in a closed system when:

1. The rate of the forward reaction equals the rate of the backward reaction.
2. The concentrations (or partial pressures) of all reactants and products remain constant.
3. The system must be closed — no matter can enter or leave.

动态平衡发生在封闭系统中，满足以下条件时：

1. 正反应速率等于逆反应速率。
2. 所有反应物和产物的浓度（或分压）保持恒定。
3. 系统必须封闭——没有物质可以进出。

Note the word dynamic: reactions do not stop at equilibrium. Both forward and backward reactions continue at equal rates — it is a state of dynamic balance, not a static standstill. This is a common misconception that examiners love to test!

注意”动态”这个词：反应在平衡时并不停止。正反应和逆反应以相等的速率继续进行 —— 这是一个动态平衡的状态，而不是静态停滞。这是一个考官喜欢考察的常见误解！

Graphical representation | 图形表示： On a concentration–time graph, the concentrations of reactants decrease and products increase until both plateau. On a rate–time graph, the forward rate decreases while the reverse rate increases until they converge at a single value.

在浓度–时间图上，反应物浓度下降、产物浓度上升，直到两者均趋于平稳。在速率–时间图上，正反应速率下降而逆反应速率上升，直到汇聚于同一个值。

2. Le Chatelier’s Principle | 勒夏特列原理

2.1 The Principle | 原理

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

勒夏特列原理指出：如果处于动态平衡的系统受到浓度、温度或压力的变化，平衡位置将移动以抵消该变化。

Think of it as a “chemical seesaw” — push one side and the system pushes back. This is an application of the principle of minimum energy/maximum entropy that governs all natural processes.

可以把它想象成一个”化学跷跷板”——推一边，系统就推回来。这是支配所有自然过程的最小能量/最大熵原理的应用。

2.2 Effect of Concentration | 浓度的影响

Increasing concentration of a reactant: The equilibrium shifts to the right (product side) to consume the added reactant.
Increasing concentration of a product: The equilibrium shifts to the left (reactant side) to consume the added product.

增加反应物浓度：平衡右移（产物侧）以消耗添加的反应物。
增加产物浓度：平衡左移（反应物侧）以消耗添加的产物。

Example — Iron(III) thiocyanate equilibrium | 例子——硫氰酸铁(III)平衡：
Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq)
Pale yellow / 淡黄色 + Colourless / 无色 ⇌ Blood-red / 血红色

Adding Fe³⁺ ions intensifies the red colour (equilibrium shifts right). Adding thiocyanate ions also intensifies the red colour. This is a classic observable demonstration of Le Chatelier’s principle.

添加Fe³⁺离子会加深红色（平衡右移）。添加硫氰酸根离子也会加深红色。这是勒夏特列原理的一个经典可观察演示。

2.3 Effect of Pressure (Gaseous Systems Only) | 压力的影响（仅气体系统）

Pressure changes only affect equilibria involving gases where there is a difference in the number of moles of gaseous reactants and products.

压力变化只影响涉及气体的平衡，且前提是气体反应物和产物的摩尔数存在差异。

Increasing pressure: Equilibrium shifts to the side with fewer gas molecules (moles of gas).
Decreasing pressure: Equilibrium shifts to the side with more gas molecules.

增加压力：平衡向气体分子较少的一侧移动。
减少压力：平衡向气体分子较多的一侧移动。

Example — N₂(g) + 3H₂(g) ⇌ 2NH₃(g):
Left side: 1 + 3 = 4 moles of gas | 左侧：4摩尔气体
Right side: 2 moles of gas | 右侧：2摩尔气体
Increasing pressure shifts equilibrium to the right (fewer moles), producing more NH₃.

增加压力使平衡右移（更少的摩尔数），产生更多的NH₃。

⚠️ Important: If there is no change in the number of gas moles (e.g., H₂ + I₂ ⇌ 2HI — 2 moles on each side), changing pressure has no effect on the position of equilibrium. Do not fall for this exam trap!

⚠️ 重要：如果气体摩尔数没有变化（例如 H₂ + I₂ ⇌ 2HI ——每侧2摩尔），改变压力不影响平衡位置。不要落入这个考试陷阱！

2.4 Effect of Temperature | 温度的影响

Temperature is the only factor that changes the value of the equilibrium constant K_c/K_p. You must identify whether the forward reaction is exothermic or endothermic.

温度是唯一改变平衡常数K_c/K_p值的因素。你必须判断正反应是放热还是吸热。

For an exothermic forward reaction (ΔH < 0):
Increasing temperature shifts equilibrium LEFT (endothermic direction, absorbing heat).
Decreasing temperature shifts equilibrium RIGHT (exothermic direction, releasing heat).

对于放热正反应（ΔH < 0）：
升高温度使平衡左移（吸热方向，吸收热量）。
降低温度使平衡右移（放热方向，释放热量）。

For an endothermic forward reaction (ΔH > 0):
Increasing temperature shifts equilibrium RIGHT (endothermic direction).
Decreasing temperature shifts equilibrium LEFT.

对于吸热正反应（ΔH > 0）：
升高温度使平衡右移（吸热方向）。
降低温度使平衡左移。

Example — 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ mol⁻¹:
The forward reaction is exothermic. Heating favours the backward (endothermic) reaction, so the mixture turns browner (more NO₂). Cooling favours the forward (exothermic) reaction, so the mixture turns paler (more N₂O₄ which is colourless).

例子——2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ mol⁻¹：
正反应是放热的。加热有利于逆反应（吸热），所以混合物变得更棕色（更多NO₂）。冷却有利于正反应（放热），所以混合物变得更浅（更多无色N₂O₄）。

2.5 Effect of a Catalyst | 催化剂的影响

A catalyst does NOT affect the position of equilibrium. It increases the rate of both the forward and backward reactions equally by providing an alternative reaction pathway with a lower activation energy (E_a).

催化剂不影响平衡位置。它通过提供活化能（E_a）更低的替代反应路径，同等程度地增加正反应和逆反应的速率。

What a catalyst does: Equilibrium is reached faster, but the equilibrium composition is unchanged. The value of K_c/K_p remains the same. This is another favourite exam question — students often mistakenly claim that catalysts shift the equilibrium position.

催化剂的作用：平衡更快达到，但平衡组成不变。K_c/K_p的值保持不变。这是另一个常见的考试题目——学生经常错误地声称催化剂改变了平衡位置。

3. Equilibrium Constants — K_c and K_p | 平衡常数——K_c和K_p

3.1 The Equilibrium Constant K_c | 平衡常数K_c

For the general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is:

对于一般反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数为：

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

Where [X] represents the equilibrium concentration of species X in mol dm⁻³. The units of K_c depend on the stoichiometry and must be worked out for each reaction.

其中[X]表示物种X的平衡浓度，单位为mol dm⁻³。K_c的单位取决于化学计量比，每个反应都必须单独计算。

Rules for K_c expressions | K_c表达式的规则：

• Products in the numerator, reactants in the denominator. | 产物在分子，反应物在分母。
• Solids and pure liquids are omitted — their concentrations are effectively constant and are absorbed into the value of K_c. | 固体和纯液体被省略——它们的浓度实际上不变，被吸收到K_c值中。
• Water is omitted when it is the solvent (its concentration is ≈ constant), but included when it is a product in a gaseous reaction. | 水在作为溶剂时被省略（其浓度≈常数），但在气体反应中作为产物时要包括进去。

3.2 The Equilibrium Constant K_p | 平衡常数K_p

For gaseous equilibria, K_p is the equilibrium constant expressed in terms of partial pressures:

对于气体平衡，K_p是以分压表示的平衡常数：

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

Partial pressure of gas X = mole fraction of X × total pressure
气体X的分压 = X的摩尔分数 × 总压力

Mole fraction of X = (moles of X) / (total moles of all gases at equilibrium)
X的摩尔分数 = X的摩尔数 / 平衡时所有气体的总摩尔数

Units of K_p: Typically atm, kPa, or Pa — raised to the appropriate power based on the difference in moles of gas (Δn). K_p behaves analogously to K_c: only temperature changes its value.

K_p的单位：通常为atm、kPa或Pa——根据气体摩尔数的差异（Δn）取相应的幂次。K_p的行为与K_c类似：只有温度改变其值。

3.3 The Magnitude of K_c/K_p | K_c/K_p的大小

The magnitude of the equilibrium constant tells you about the position of equilibrium:

平衡常数的大小告诉你关于平衡位置的信息：

• K ≫ 1 (much greater than 1): Equilibrium lies well to the right. Products predominate at equilibrium. | 平衡明显偏右。产物在平衡时占主导。
• K ≈ 1: Significant amounts of both reactants and products present. | 反应物和产物都有显著的存在。
• K ≪ 1 (much less than 1): Equilibrium lies well to the left. Reactants predominate at equilibrium. | 平衡明显偏左。反应物在平衡时占主导。

For the Haber process at 298 K, K_c ≈ 6.0 × 10⁵ dm⁶ mol⁻² — the equilibrium lies far to the right, meaning the production of NH₃ is thermodynamically favoured at room temperature. However, the reaction is kinetically slow at low temperatures, which is why the industrial process uses elevated temperatures (≈ 450°C) with an iron catalyst.

对于哈伯法在298 K，K_c ≈ 6.0 × 10⁵ dm⁶ mol⁻²——平衡明显偏右，意味着在室温下NH₃的生成在热力学上是有利的。然而，该反应在低温下动力学上很慢，这就是为什么工业过程使用较高温度（≈ 450°C）和铁催化剂。

4. Industrial Applications | 工业应用

4.1 The Haber Process — NH₃ Production | 哈伯法——NH₃生产

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

The Haber process is the classic A-Level case study that brings together kinetics, thermodynamics, and equilibrium principles:

哈伯法是经典A-Level案例研究，汇集了动力学、热力学和平衡原理：

Condition / 条件	Choice / 选择	Reason / 原因
Temperature / 温度	400–450°C	Compromise: lower T gives higher yield but too slow; higher T reduces yield. 450°C balances rate and yield. | 折中：较低温度给出更高产率但太慢；较高温度降低产率。450°C平衡了速率和产率。
Pressure / 压力	200 atm	High pressure favours fewer moles (4→2) and increases rate. Limited by cost of reinforced vessels. | 高压有利于较少的摩尔数（4→2）并提高速率。受限于强化容器的成本。
Catalyst / 催化剂	Iron (Fe)	Lowers activation energy; does NOT affect equilibrium position — only makes it reachable faster. | 降低活化能；不影响平衡位置——只使平衡更快达到。

4.2 The Contact Process — H₂SO₄ Production | 接触法——H₂SO₄生产

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

Conditions: 450°C, 1–2 atm, vanadium(V) oxide (V₂O₅) catalyst.

条件：450°C，1–2 atm，五氧化二钒（V₂O₅）催化剂。

The forward reaction is exothermic (ΔH < 0) and reduces gas moles (3 → 2), so low temperature and high pressure would maximise yield. However, at low temperatures the rate is too slow, so a compromise temperature of 450°C and a catalyst are used. The pressure is kept near atmospheric because the equilibrium yield is already very high (≈ 98%) at 1–2 atm — spending money on high-pressure equipment is not cost-effective.

正反应是放热的（ΔH < 0）且减少气体摩尔数（3 → 2），因此低温和高压会最大化产率。然而，在低温下速率太慢，所以使用450°C的折中温度和催化剂。压力保持接近大气压，因为在1-2 atm下平衡产率已经非常高（≈ 98%）——在高压缩设备上花钱不划算。

5. Common Exam Mistakes & Tips | 常见考试错误与技巧

Mistake 1: Confusing “Rate” with “Position of Equilibrium”

错误1：混淆”速率”和”平衡位置”

A catalyst increases the rate only — it does not change the position of equilibrium or the yield. Similarly, increasing temperature increases the rate of both forward and backward reactions but shifts the equilibrium position depending on ΔH.

催化剂只增加速率——它不改变平衡位置或产率。同样，升高温度增加正逆两个反应的速率，但根据ΔH改变平衡位置。

Mistake 2: Forgetting to Omit Solids and Liquids from K_c

错误2：忘记从K_c中省略固体和液体

For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the correct K_c expression is simply K_c = [CO₂] — or K_p = P(CO₂). The solids do not appear! This is consistently tested across all exam boards.

对于 CaCO₃(s) ⇌ CaO(s) + CO₂(g)，正确的K_c表达式就是 K_c = [CO₂]——或K_p = P(CO₂)。固体不出现！这在所有考试局中都是经常考察的内容。

Mistake 3: Ignoring Units of K_c/K_p

错误3：忽略K_c/K_p的单位

Unlike pH, K_c and K_p are not dimensionless. Marks are routinely awarded for correct units. The general formula: units of K_c = (mol dm⁻³)^(Δn) where Δn = (moles of gaseous/aqueous products) − (moles of gaseous/aqueous reactants).

与pH不同，K_c和K_p不是无量纲的。正确的单位经常能得到分数。通用公式：K_c的单位 = (mol dm⁻³)^(Δn)，其中Δn =（气体/溶液中产物的摩尔数）−（气体/溶液中反应物的摩尔数）。

Mistake 4: Pressure Changes Affect Δn = 0 Equilibria

错误4：压力变化影响Δn = 0的平衡

H₂(g) + I₂(g) ⇌ 2HI(g) — both sides have 2 moles of gas. Changing pressure has no effect on equilibrium position. Always count gas moles first!

H₂(g) + I₂(g) ⇌ 2HI(g)——两侧都是2摩尔气体。改变压力不影响平衡位置。总是先数气体摩尔数！

Mistake 5: Treating Temperature Like the Other Factors

错误5：把温度当作和其他因素一样处理

Temperature is unique: it changes the value of K_c/K_p. Concentration, pressure, and catalysts do not. For exothermic reactions, K_c decreases as temperature increases; for endothermic reactions, K_c increases. This is quantifiable via the Van ‘t Hoff equation, which you may encounter in more advanced A-Level specifications.

温度是独特的：它改变K_c/K_p的值。浓度、压力和催化剂不会。对于放热反应，K_c随温度升高而降低；对于吸热反应，K_c升高。这通过范特霍夫方程可以量化，你可能在更高级的A-Level大纲中遇到。

6. Calculation Walkthrough | 计算示范

Let’s work through a full K_c calculation problem — exactly the type that appears in A-Level examinations.

让我们来完成一个完整的K_c计算问题——这正是A-Level考试中出现的题型。

Question | 问题： 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed in a sealed vessel. At equilibrium, 0.30 mol of ethyl ethanoate is formed. The total volume is 1.0 dm³. Calculate K_c for the esterification reaction:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Solution | 解答：

Step 1: Set up an ICE table (Initial, Change, Equilibrium) | 第1步：建立ICE表格（初始、变化、平衡）

Species / 物种	Initial / 初始 (mol)	Change / 变化 (mol)	Equilibrium / 平衡 (mol)	[Equilibrium] / [平衡] (mol dm⁻³)
CH₃COOH	0.50	−0.30	0.20	0.20
C₂H₅OH	0.50	−0.30	0.20	0.20
CH₃COOC₂H₅	0	+0.30	0.30	0.30
H₂O	0	+0.30	0.30	0.30

Step 2: Write the K_c expression | 第2步：写出K_c表达式

K_c = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]

Step 3: Substitute values | 第3步：代入数值

K_c = (0.30 × 0.30) / (0.20 × 0.20) = 0.090 / 0.040 = 2.25

Step 4: Determine units | 第4步：确定单位

Δn = 2 − 2 = 0, so K_c is dimensionless (no units).

Δn = 2 − 2 = 0，因此K_c无量纲（没有单位）。

Answer | 答案：K_c = 2.25 (no units | 无单位)

Interpretation: K_c > 1 means the equilibrium lies slightly to the right — as expected for an esterification reaction under these conditions.

解释：K_c > 1意味着平衡略微偏右——在这些条件下对酯化反应来说是符合预期的。

7. Summary Table | 总结表

Change / 变化	Effect on Equilibrium Position / 对平衡位置的影响	Effect on Kc/Kp / 对Kc/Kp的影响
Increase [reactant] / 增加[反应物]	Shifts right / 右移	No change / 不变
Increase [product] / 增加[产物]	Shifts left / 左移	No change / 不变
Increase pressure (more gas moles on right) / 增加压力（右侧气体摩尔数较多）	Shifts left / 左移	No change / 不变
Increase temperature (exothermic forward) / 升高温度（正反应放热）	Shifts left / 左移	Decreases / 降低
Increase temperature (endothermic forward) / 升高温度（正反应吸热）	Shifts right / 右移	Increases / 升高
Add catalyst / 添加催化剂	No effect / 无影响	No change / 不变

8. Key Vocabulary | 关键词汇

English / 英文	中文 / 中文
Dynamic equilibrium	动态平衡
Reversible reaction	可逆反应
Le Chatelier’s Principle	勒夏特列原理
Equilibrium constant	平衡常数
Position of equilibrium	平衡位置
Partial pressure	分压
Mole fraction	摩尔分数
Exothermic	放热的
Endothermic	吸热的
Activation energy	活化能
ICE table	ICE表格
Yield	产率
Compromise conditions	折中条件

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This article provides a comprehensive overview of chemical equilibrium at A-Level standard. For further practice, attempt past paper questions on K_c/K_p calculations and Le Chatelier predictions — these are among the most reliably tested topics across all exam boards. Good luck with your studies! | 本文全面概述了A-Level标准的化学平衡。如需进一步练习，请尝试关于K_c/K_p计算和勒夏特列预测的历年真题——这是所有考试局中最常考的主题之一。祝你学业顺利！