A-Level Chemistry: Chemical Equilibrium Complete Guide | A-Level 化学：化学平衡完全指南

Chemical equilibrium is one of the most conceptually rich and frequently examined topics in A-Level Chemistry. It bridges thermodynamics, kinetics, and industrial chemistry — making it essential for both the written papers and the practical endorsement. This bilingual guide covers everything from the foundational principles of dynamic equilibrium to Le Chatelier’s Principle, the equilibrium constant Kc, Kp for gaseous systems, and the industrial applications that examiners love to test. Whether you are studying under AQA, OCR, Edexcel, or CIE, this guide will help you master equilibrium with confidence.

化学平衡（Chemical Equilibrium）是 A-Level 化学中概念最丰富、考试频率最高的主题之一。它连接了热力学、动力学和工业化学——因此无论是对笔试还是实验考核都至关重要。本双语指南涵盖了从动态平衡的基础原理到勒夏特列原理（Le Chatelier’s Principle）、平衡常数 Kc、气体体系的 Kp，以及考官最喜欢考察的工业应用等所有内容。无论你学习的是 AQA、OCR、Edexcel 还是 CIE 课程体系，本指南都将帮助你充满信心地掌握化学平衡。

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

At the heart of chemical equilibrium lies a single critical concept: dynamic equilibrium. Unlike static equilibrium — where nothing appears to happen — dynamic equilibrium describes a state in which the forward and reverse reactions continue to occur simultaneously at exactly the same rate.

Consider a simple reversible reaction:

aA + bB ⇌ cC + dD

At the start, only reactants A and B are present, so the forward reaction rate is high. As products C and D accumulate, the reverse reaction begins. Over time, the forward rate decreases (as reactants are consumed) and the reverse rate increases (as products accumulate). When the two rates become equal, the system has reached dynamic equilibrium. At this point:

• The concentrations of all species remain constant (but are NOT necessarily equal)
• The forward and reverse reactions continue to occur — hence “dynamic”
• The system is closed — no matter enters or leaves
• The equilibrium can be approached from either direction

化学平衡的核心在于一个关键概念：动态平衡。与静态平衡（表面上看不到任何变化）不同，动态平衡描述的是一种状态，在这种状态下，正向反应和逆向反应以完全相同的速率同时进行。

考虑一个简单的可逆反应：aA + bB ⇌ cC + dD。开始时，只有反应物 A 和 B 存在，因此正向反应速率很高。随着产物 C 和 D 的积累，逆向反应开始进行。随着时间的推移，正向速率下降（因为反应物被消耗），逆向速率上升（因为产物积累）。当两个速率相等时，系统就达到了动态平衡。此时：

• 所有物种的浓度保持恒定（但未必相等）
• 正向反应和逆向反应持续进行——因此称为”动态”
• 系统是封闭的——没有物质进出
• 平衡可以从任意方向达到

Exam Tip: A classic A-Level question asks you to explain why a particular graph shows that equilibrium has been reached. The answer: the concentrations of all species have become constant over time, indicating that the forward and reverse rates are equal.

考试提示：经典的 A-Level 题目会要求你解释某张图如何表明已经达到平衡。答案是：所有物种的浓度随时间保持不变，表明正逆反应速率相等。

2. Le Chatelier’s Principle / 勒夏特列原理

Henri Louis Le Chatelier formulated one of the most powerful predictive tools in chemistry. His principle states:

“If a system at dynamic equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium will shift to oppose that change.”

亨利·路易·勒夏特列提出了化学中最强大的预测工具之一。他的原理指出：“如果一个处于动态平衡的系统受到浓度、压力或温度的变化，平衡位置将移动以抵消这种变化。”

2.1 Effect of Concentration / 浓度的影响

If you increase the concentration of a reactant, the equilibrium shifts to the right (towards products) to consume the added reactant. Conversely, if you remove a product, the equilibrium also shifts right to produce more of that product. If you increase the concentration of a product, the equilibrium shifts left (towards reactants).

如果你增加了反应物的浓度，平衡将向右移动（向产物方向），以消耗新增的反应物。反之，如果你移除了某种产物，平衡同样向右移动以产生更多该产物。如果你增加了产物的浓度，平衡将向左移动（向反应物方向）。

Example — The Fe(SCN)²⁺ equilibrium:

Fe³⁺(aq) + SCN⁻(aq) ⇌ Fe(SCN)²⁺(aq) (blood-red)

Adding more Fe³⁺ or SCN⁻ intensifies the red color (shifts right). Adding a reagent that removes Fe³⁺ (such as F⁻, which forms a stable complex) causes the red color to fade (shifts left).

示例——Fe(SCN)²⁺ 平衡：Fe³⁺(aq) + SCN⁻(aq) ⇌ Fe(SCN)²⁺(aq)（血红色）。加入更多 Fe³⁺ 或 SCN⁻ 会使红色加深（向右移动）。加入能去除 Fe³⁺ 的试剂（如 F⁻，它会形成稳定络合物）会使红色褪去（向左移动）。

2.2 Effect of Pressure / 压力的影响

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

• Increasing pressure (by decreasing volume) shifts equilibrium towards the side with fewer moles of gas
• Decreasing pressure (by increasing volume) shifts equilibrium towards the side with more moles of gas
• If the number of moles of gas is the same on both sides, pressure has no effect

压力的变化只影响涉及气体且反应物和产物之间气体摩尔数发生变化的平衡。

• 增加压力（通过减小体积）使平衡向气体摩尔数较少的一侧移动
• 减小压力（通过增大体积）使平衡向气体摩尔数较多的一侧移动
• 如果两侧的气体摩尔数相同，压力没有影响

Example — The Haber Process:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Left side: 1 + 3 = 4 moles of gas. Right side: 2 moles of gas. Increasing pressure shifts equilibrium to the right, favouring ammonia production. This is why the Haber process is carried out at high pressure (typically 200 atm).

示例——哈伯法合成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。左侧：1 + 3 = 4 摩尔气体。右侧：2 摩尔气体。增加压力使平衡向右移动，有利于氨的生成。这就是哈伯法在高压（通常 200 atm）下进行的原因。

Common Pitfall: Students often forget that adding an inert gas at constant volume does NOT change partial pressures of reactants, so it has no effect on equilibrium position. Adding an inert gas at constant pressure, however, increases volume and thus decreases partial pressures — this does shift equilibrium towards the side with more moles of gas.

常见误区：学生常常忘记在恒定体积下加入惰性气体不会改变反应物的分压，因此对平衡位置没有影响。然而，在恒定压力下加入惰性气体会增加体积，从而降低分压——这确实会使平衡向气体摩尔数更多的一侧移动。

2.3 Effect of Temperature / 温度的影响

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

• Exothermic forward reaction (ΔH < 0): Increasing temperature shifts equilibrium left (endothermic direction). Kc decreases.
• Endothermic forward reaction (ΔH > 0): Increasing temperature shifts equilibrium right (endothermic direction). Kc increases.

温度是唯一会改变平衡常数（Kc 或 Kp）数值的因素。移动方向取决于正向反应是放热还是吸热：

• 正向放热反应（ΔH < 0）：升高温度使平衡向左移动（吸热方向），Kc 减小。
• 正向吸热反应（ΔH > 0）：升高温度使平衡向右移动（吸热方向），Kc 增大。

Critical distinction: Temperature changes Kc/Kp. Concentration and pressure changes do NOT — they only shift the position of equilibrium while Kc/Kp remains constant (provided temperature is constant).

关键区别：温度会改变 Kc/Kp。浓度和压力的变化不会——它们只改变平衡位置，而 Kc/Kp 保持不变（前提是温度恒定）。

2.4 Effect of a Catalyst / 催化剂的影响

A catalyst provides an alternative reaction pathway with a lower activation energy. Crucially, it lowers the activation energy for both forward and reverse reactions equally. Therefore:

• A catalyst has NO effect on the position of equilibrium
• A catalyst has NO effect on the value of Kc
• A catalyst speeds up the rate at which equilibrium is reached

催化剂提供了一条活化能更低的替代反应途径。关键的是，它同等地降低了正向反应和逆向反应的活化能。因此：

• 催化剂对平衡位置没有影响
• 催化剂对 Kc 的值没有影响
• 催化剂加快了达到平衡的速率

In the Haber process, an iron catalyst is used not to increase yield (pressure and temperature do that), but to allow the reaction to reach equilibrium faster at a given temperature — making the process economically viable.

在哈伯法中，使用铁催化剂的目的不是提高产率（压力和温度已经做到了这一点），而是使反应在给定温度下更快地达到平衡——使该工艺在经济上可行。

3. The Equilibrium Constant Kc / 平衡常数 Kc

For a general homogeneous reaction in solution:

aA + bB ⇌ cC + dD

The equilibrium constant Kc is defined as:

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

Where square brackets denote equilibrium concentrations in mol dm⁻³. The units of Kc depend on the stoichiometry of the specific reaction and are derived by cancelling the units in the expression.

对于一个均相溶液反应 aA + bB ⇌ cC + dD，平衡常数 Kc 定义为：Kc = [C]^c [D]^d / [A]^a [B]^b。其中方括号表示以 mol dm⁻³ 为单位的平衡浓度。Kc 的单位取决于具体反应的化学计量比，通过对表达式中的单位进行约分得出。

3.1 What Kc Tells Us / Kc 告诉我们什么

• Kc >> 1: Equilibrium lies far to the right. Products are strongly favoured. The reaction essentially goes to completion.
• Kc ≈ 1: Significant amounts of both reactants and products are present at equilibrium.
• Kc << 1: Equilibrium lies far to the left. Reactants are strongly favoured. Very little product forms.

Kc >> 1：平衡远远偏向右侧，强烈倾向于生成产物，反应基本进行到底。Kc ≈ 1：平衡时存在显著量的反应物和产物。Kc << 1：平衡远远偏向左侧，强烈倾向于保留反应物，只生成很少的产物。

3.2 Calculating Kc — Worked Example / Kc 计算——示例

Question: For the esterification reaction:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed in a 1.00 dm³ vessel. At equilibrium, 0.66 mol of ethyl ethanoate is present. Calculate Kc.

Solution:

Initial amounts: [CH₃COOH] = 1.00, [C₂H₅OH] = 1.00, [CH₃COOC₂H₅] = 0, [H₂O] = 0

Change: Since 0.66 mol of ester is formed, 0.66 mol of acid and alcohol are consumed, and 0.66 mol of water is formed.

Equilibrium concentrations: [CH₃COOH] = 1.00 – 0.66 = 0.34, [C₂H₅OH] = 0.34, [CH₃COOC₂H₅] = 0.66, [H₂O] = 0.66

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH] = (0.66 × 0.66) / (0.34 × 0.34) = 0.4356 / 0.1156 = 3.77

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

问题：对于酯化反应 CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O，将 1.00 mol 乙酸和 1.00 mol 乙醇混合在 1.00 dm³ 容器中。平衡时存在 0.66 mol 乙酸乙酯。计算 Kc。

解答：初始：[CH₃COOH] = 1.00，[C₂H₅OH] = 1.00，[CH₃COOC₂H₅] = 0，[H₂O] = 0。变化：生成 0.66 mol 酯，消耗 0.66 mol 酸和醇，生成 0.66 mol 水。平衡浓度：[CH₃COOH] = 0.34，[C₂H₅OH] = 0.34，[CH₃COOC₂H₅] = 0.66，[H₂O] = 0.66。Kc = (0.66 × 0.66) / (0.34 × 0.34) = 3.77，无单位。

Exam technique: Always set up an ICE table (Initial, Change, Equilibrium) when solving Kc problems. This systematic approach prevents errors and earns method marks even if the final answer is slightly off.

考试技巧：在解 Kc 题目时，始终建立 ICE 表格（Initial 初始、Change 变化、Equilibrium 平衡）。这种系统方法可以防止错误，即使最终答案略有偏差也能获得方法分。

4. Kp — Equilibrium Constant for Gaseous Systems / Kp——气体体系的平衡常数

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

对于涉及气体的反应，使用分压代替浓度通常更为方便。以分压表示的平衡常数记作 Kp。

4.1 Mole Fraction and Partial Pressure / 摩尔分数和分压

The partial pressure of a gas A in a mixture is:

p(A) = Mole fraction of A × Total pressure

Where: Mole fraction of A = Moles of A / Total moles of all gases

混合物中气体 A 的分压为：p(A) = A 的摩尔分数 × 总压力。其中：A 的摩尔分数 = A 的摩尔数 / 所有气体的总摩尔数。

The Kp expression follows the same pattern as Kc:

Kp = p(C)^c × p(D)^d / p(A)^a × p(B)^b

The units of Kp are typically in atm, Pa, or kPa raised to an appropriate power, depending on the change in moles of gas.

Kp 的单位通常是 atm、Pa 或 kPa 的某次幂，取决于气体摩尔数的变化。

4.2 Kp Worked Example / Kp 计算示例

Question: For the reaction N₂O₄(g) ⇌ 2NO₂(g) at 298 K, 1.00 mol of N₂O₄ is placed in a vessel. At equilibrium, the total pressure is 150 kPa and the mixture contains 0.40 mol of N₂O₄. Calculate Kp.

Solution:

N₂O₄ decomposed = 1.00 – 0.40 = 0.60 mol. From the stoichiometry, NO₂ formed = 2 × 0.60 = 1.20 mol.

Total moles at equilibrium = 0.40 + 1.20 = 1.60 mol.

Mole fraction of N₂O₄ = 0.40/1.60 = 0.25. Mole fraction of NO₂ = 1.20/1.60 = 0.75.

p(N₂O₄) = 0.25 × 150 = 37.5 kPa. p(NO₂) = 0.75 × 150 = 112.5 kPa.

Kp = p(NO₂)² / p(N₂O₄) = (112.5)² / 37.5 = 12656.25 / 37.5 = 337.5 kPa

问题：对于反应 N₂O₄(g) ⇌ 2NO₂(g)，在 298 K 下将 1.00 mol N₂O₄ 放入容器中。平衡时总压力为 150 kPa，混合物含 0.40 mol N₂O₄。计算 Kp。

解答：分解的 N₂O₄ = 1.00 – 0.40 = 0.60 mol。根据化学计量比，生成的 NO₂ = 2 × 0.60 = 1.20 mol。平衡时总摩尔数 = 0.40 + 1.20 = 1.60 mol。N₂O₄ 的摩尔分数 = 0.25，NO₂ 的摩尔分数 = 0.75。p(N₂O₄) = 37.5 kPa，p(NO₂) = 112.5 kPa。Kp = (112.5)² / 37.5 = 337.5 kPa。

5. Industrial Applications / 工业应用

A-Level examiners frequently test your understanding of equilibrium through industrial contexts. Here are the three most important processes:

A-Level 考官经常通过工业背景来考察你对平衡的理解。以下是三个最重要的工艺：

5.1 The Haber Process — Ammonia Synthesis / 哈伯法——合成氨

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

Conditions: 400-450 °C, 200 atm, iron catalyst

This is an exothermic reaction that produces fewer moles of gas (4 mol → 2 mol). According to Le Chatelier’s Principle, a high pressure and low temperature should favour the forward reaction and maximize yield. However, the actual conditions represent a compromise:

• Pressure (200 atm): High pressure favours the forward reaction (fewer moles), but higher pressures are expensive (stronger equipment, higher energy costs). 200 atm is the economic compromise.
• Temperature (400-450 °C): Low temperature favours the exothermic forward reaction for higher yield, but the reaction rate would be too slow. Higher temperatures increase the rate (and the catalyst is only active above ~400 °C). The yield is lower at 450 °C than at room temperature, but the rate is commercially viable.
• Iron catalyst: Speeds up the approach to equilibrium without affecting position or yield.

条件：400-450 °C，200 atm，铁催化剂。

这是一个放热反应，产物气体摩尔数更少（4 mol → 2 mol）。根据勒夏特列原理，高压和低温应该有利于正向反应并最大化产率。然而，实际条件代表了一种折中：

• 压力（200 atm）：高压有利于正向反应（摩尔数更少），但更高的压力成本高昂（更强的设备、更高的能源成本）。200 atm 是经济折中点。
• 温度（400-450 °C）：低温有利于放热正向反应以获得更高的产率，但反应速率会太慢。较高的温度增加了速率（且催化剂仅在约 400 °C 以上才有活性）。在 450 °C 时的产率虽然低于室温，但速率具有商业可行性。
• 铁催化剂：加快接近平衡的速率，不影响平衡位置或产率。

5.2 The Contact Process — Sulfuric Acid / 接触法——硫酸

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

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

Here, the pressure is kept relatively low (1-2 atm) because the equilibrium already lies far to the right at atmospheric pressure — the Kp value is so large that using high pressure would add cost without significant yield improvement. The temperature is a compromise between rate and yield, similar to the Haber process.

条件：450 °C，1-2 atm，五氧化二钒（V₂O₅）催化剂。

在这里，压力保持相对较低（1-2 atm），因为在常压下平衡已经远远偏向右侧——Kp 值非常大，使用高压会增加成本而不会显著提高产率。温度是速率和产率之间的折中，与哈伯法类似。

5.3 Methanol Production / 甲醇生产

CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = -91 kJ mol⁻¹

Conditions: 250 °C, 50-100 atm, Cu/ZnO/Al₂O₃ catalyst

3 moles of gas on the left, 1 mole on the right — so high pressure favours methanol production. Again, temperature is a compromise: low for high yield, high for fast rate.

左侧 3 摩尔气体，右侧 1 摩尔——因此高压有利于甲醇的生成。同样，温度是一个折中：低温有利于高产率，高温有利于快速速率。

7. Common Exam Mistakes and How to Avoid Them / 常见考试错误及避免方法

Mistake 1: Confusing Rate and Equilibrium / 错误 1：混淆速率和平衡

Students often say “increasing temperature increases the rate, so equilibrium shifts right.” This is only true for ENDOthermic forward reactions. For exothermic reactions, increasing temperature increases the rate of both forward and reverse reactions, but the reverse (endothermic) reaction speeds up MORE — so equilibrium shifts left. Always check the sign of ΔH.

学生常说”升高温度增加速率，所以平衡向右移动。”这仅在正向吸热反应中成立。对于放热反应，升高温度会同时增加正逆反应速率，但逆向（吸热）反应加速更多——因此平衡向左移动。始终检查 ΔH 的符号。

Mistake 2: Forgetting Units of Kc / 错误 2：忘记 Kc 的单位

A common mark-loser is omitting the units of Kc or giving them incorrectly. The units are derived algebraically from the Kc expression. For a reaction where the number of moles of products minus reactants in the numerator is Δn, the units are (mol dm⁻³)^(Δn).

一个常见的丢分点是遗漏 Kc 的单位或给出错误的单位。单位是从 Kc 表达式代数推导出来的。对于一个反应，如果分子中产物摩尔数减去反应物摩尔数为 Δn，则单位为 (mol dm⁻³)^(Δn)。

Mistake 4: Misapplying Le Chatelier to Catalysts / 错误 4：将勒夏特列原理误用于催化剂

A catalyst does NOT shift the equilibrium position. It only increases the rate at which equilibrium is attained. This is a standard exam question — do not be caught out.

催化剂不会改变平衡位置。它只会加快达到平衡的速率。这是标准的考题——不要被难倒。

1. Write the balanced equation — identify the number of moles of gas on each side
2. Check ΔH — is the forward reaction exothermic or endothermic?
3. Apply Le Chatelier — what happens to the position of equilibrium?
4. Set up ICE table — for quantitative Kc/Kp problems
5. Write Kc/Kp expression — omit solids and liquids
6. Substitute values and calculate — include units
7. Interpret the result — what does Kc/Kp tell you about the extent of the reaction?

分析任何平衡问题时，遵循以下系统方法：

1. 写出配平方程式——确定每侧的气体摩尔数
2. 检查 ΔH——正向反应是放热还是吸热？
3. 应用勒夏特列原理——平衡位置会发生什么变化？
4. 建立 ICE 表格——用于定量的 Kc/Kp 问题
5. 写出 Kc/Kp 表达式——省略固体和液体
6. 代入数值并计算——包含单位
7. 解释结果——Kc/Kp 告诉你关于反应程度的什么信息？

---

Final Thoughts: Chemical equilibrium is more than just a chapter in your textbook — it is a fundamental principle that governs everything from industrial chemical manufacturing to the biochemistry inside your own cells. Master the concepts in this guide, practise ICE table calculations until they become second nature, and always remember: equilibrium is dynamic, not static. The reactions never stop — they just reach a state of balance.

最后的话：化学平衡不仅仅是你教科书中的一章——它是一项基本原理，支配着从工业化学品制造到你自身细胞内生物化学的一切。掌握本指南中的概念，练习 ICE 表格计算直到它们成为第二天性，并始终记住：平衡是动态的，而非静态的。反应永远不会停止——它们只是达到了一种平衡状态。

For more A-Level Chemistry resources, study guides, and past paper walkthroughs, explore the A-Level Chemistry section on aleveler.com.

更多 A-Level 化学资源、学习指南和真题讲解，请访问 aleveler.com 的 A-Level Chemistry 专区。