Introduction to Chemical Equilibrium
Chemical equilibrium is one of the most fundamental concepts in A-Level Chemistry, bridging the gap between kinetics and thermodynamics. Understanding equilibrium allows us to predict the direction of reversible reactions, calculate reaction yields, and optimize industrial processes such as the Haber process for ammonia production. This article provides a comprehensive guide to chemical equilibrium — from the basic principles to advanced calculations — presented in both English and Chinese to support bilingual learners.
化学平衡是A-Level化学中最基础的概念之一，连接着动力学和热力学。理解平衡使我们能够预测可逆反应的方向、计算反应产率，并优化工业生产过程，如用于制氨的哈伯法。本文提供了化学平衡的全面指南——从基本原理到高级计算——以中英双语呈现，支持双语学习者。
1. What is Chemical Equilibrium? / 什么是化学平衡？
1.1 Reversible Reactions / 可逆反应
Not all chemical reactions go to completion. Many reactions are reversible, meaning the products can react to reform the original reactants. In a closed system, a reversible reaction will eventually reach a state of dynamic equilibrium.
并非所有化学反应都能进行到底。许多反应是可逆的，意味着产物可以反应重新生成原始反应物。在封闭系统中，可逆反应最终会达到动态平衡状态。
Example / 示例 — The Deacon process for chlorine production / 制氯的迪肯法：
4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)
1.2 Dynamic Equilibrium / 动态平衡
At equilibrium, the forward and reverse reactions continue to occur, but at equal rates. This is called dynamic equilibrium because the reaction has not stopped — the concentrations of reactants and products simply remain constant. Crucially, the concentrations at equilibrium are not necessarily equal; one side may be favored.
在平衡状态下，正向反应和逆向反应继续发生，但速率相等。这被称为动态平衡，因为反应并未停止——反应物和产物的浓度只是保持不变。关键的是，平衡时的浓度不一定相等；可能偏向某一侧。
The following conditions are required for dynamic equilibrium / 动态平衡需要以下条件：
- The system must be closed — no matter enters or leaves / 系统必须封闭——没有物质进出
- The temperature must be constant / 温度必须恒定
- The reaction must be reversible / 反应必须是可逆的
2. The Equilibrium Constant (Kc) / 平衡常数 (Kc)
2.1 Deriving the Expression / 推导表达式
For a general reversible reaction at a given temperature / 对于给定温度下的一般可逆反应：
aA + bB ⇌ cC + dD
The equilibrium constant in terms of concentration is / 浓度形式的平衡常数为：
Kc = [C]c [D]d / [A]a [B]b
Where [X] represents the equilibrium concentration of species X in mol dm⁻³. The key rules for writing Kc expressions are:
其中 [X] 表示物种 X 的平衡浓度，单位为 mol dm⁻³。书写 Kc 表达式的主要规则是：
- Products in the numerator, reactants in the denominator / 产物在分子，反应物在分母
- Stoichiometric coefficients become the powers / 化学计量系数变为指数
- Pure solids and pure liquids are omitted (their “concentration” is effectively constant) / 纯固体和纯液体被省略（它们的”浓度”实际上是常数）
- Kc depends only on temperature — changing concentration or pressure does not alter Kc / Kc 仅取决于温度——改变浓度或压力不会改变 Kc
2.2 Worked Example / 计算示例
Question / 问题： Consider the reaction H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium at 723 K, the concentrations are: [H₂] = 0.0240 mol dm⁻³, [I₂] = 0.0240 mol dm⁻³, [HI] = 0.168 mol dm⁻³. Calculate Kc.
Solution / 解答：
Kc = [HI]² / ([H₂] × [I₂]) = (0.168)² / (0.0240 × 0.0240) = 0.0282 / 0.000576 = 49.0
(单位 / units: dimensionless in this case because numerator and denominator powers cancel / 无量纲，因为分子分母的指数抵消)
2.3 Interpreting Kc Values / 解读 Kc 值
| Kc value / Kc 值 | Meaning / 含义 |
|---|---|
| Kc >> 1 (e.g., 10¹⁰) | Equilibrium lies far to the right — products dominate / 平衡远右移——产物占主导 |
| Kc ≈ 1 | Significant amounts of both reactants and products / 反应物和产物都有显著量 |
| Kc << 1 (e.g., 10⁻¹⁰) | Equilibrium lies far to the left — reactants dominate / 平衡远左移——反应物占主导 |
3. Le Chatelier’s Principle / 勒夏特列原理
3.1 The Principle / 原理内容
Le Chatelier’s Principle states that if a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to counteract that change. This principle is a powerful qualitative tool for predicting how equilibria respond to disturbances.
勒夏特列原理指出：如果一个处于动态平衡的系统受到条件变化的影响，平衡位置会移动以抵消该变化。该原理是预测平衡如何响应扰动的强大定性工具。
3.2 Effect of Concentration / 浓度的影响
- Adding a reactant / 加入反应物： Equilibrium shifts to the right (to consume the added reactant) / 平衡右移（消耗加入的反应物）
- Removing a product / 移除产物： Equilibrium shifts to the right (to produce more product) / 平衡右移（生成更多产物）
- Adding a product / 加入产物： Equilibrium shifts to the left (to consume the added product) / 平衡左移（消耗加入的产物）
Exam tip / 考试提示： Kc does not change when concentration is altered — only the position of equilibrium shifts. The system adjusts concentrations until the ratio once again equals Kc.
Kc 在浓度改变时不变——只有平衡位置移动。系统调整浓度直到该比值再次等于 Kc。
3.3 Effect of Pressure (Gaseous Systems) / 压力的影响（气体系统）
Pressure changes only affect equilibria where there is a change in the total number of gas molecules between reactants and products.
压力变化只影响反应物和产物之间气体分子总数发生变化的平衡。
- Increasing pressure / 增加压力： Equilibrium shifts to the side with fewer gas molecules / 平衡移向气体分子较少的一侧
- Decreasing pressure / 降低压力： Equilibrium shifts to the side with more gas molecules / 平衡移向气体分子较多的一侧
Example / 示例 — Haber Process / 哈伯法： N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Reactant side: 4 moles of gas / 反应物侧：4 摩尔气体 → Product side: 2 moles of gas / 产物侧：2 摩尔气体
Therefore, high pressure favors ammonia production (fewer gas molecules on the product side). This is why the Haber process is run at 200 atm.
因此，高压有利于氨的生成（产物侧气体分子较少）。这就是为什么哈伯法在 200 atm 下运行。
3.4 Effect of Temperature / 温度的影响
Temperature is the only factor that changes Kc itself. The direction of shift depends on whether the reaction is exothermic or endothermic.
温度是唯一改变 Kc 本身的因—素。移动方向取决于反应是放热还是吸热。
- Exothermic reaction (ΔH < 0) / 放热反应： Increasing temperature shifts equilibrium left, decreasing Kc / 升温使平衡左移，Kc 降低
- Endothermic reaction (ΔH > 0) / 吸热反应： Increasing temperature shifts equilibrium right, increasing Kc / 升温使平衡右移，Kc 升高
Why? / 为什么？ Treat heat as a “product” in exothermic reactions or a “reactant” in endothermic reactions, then apply the concentration principle. Heating an exothermic reaction is like adding a product, so equilibrium shifts left to consume the “excess heat.”
将热量视为放热反应中的”产物”或吸热反应中的”反应物”，然后应用浓度原理。加热放热反应就像加入产物，因此平衡左移以消耗”多余的热量”。
3.5 Effect of a Catalyst / 催化剂的影响
A catalyst does not affect the position of equilibrium. It speeds up both the forward and reverse reactions equally, so equilibrium is reached faster but with the same yield. Kc is unchanged. This is a common exam pitfall — students often incorrectly state that a catalyst increases yield.
催化剂不影响平衡位置。它同等加速正反应和逆反应，因此平衡更快达到，但产率不变。Kc 不变。这是常见的考试陷阱——学生常错误地声称催化剂提高产率。
4. Equilibrium Constant for Partial Pressures (Kp) / 分压平衡常数 (Kp)
4.1 Mole Fraction and Partial Pressure / 摩尔分数和分压
For gaseous equilibria, we can express the equilibrium constant in terms of partial pressures rather than concentrations. The partial pressure of a gas is the pressure it would exert if it alone occupied the container:
对于气体平衡，我们可以用分压而不是浓度来表示平衡常数。气体的分压是它单独占据容器时所施加的压力：
Partial pressure = mole fraction × total pressure
分压 = 摩尔分数 × 总压
Where / 其中：
Mole fraction = moles of gas / total moles of all gases
摩尔分数 = 该气体的摩尔数 / 所有气体的总摩尔数
4.2 Kp Expression / Kp 表达式
For the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g)：
Kp = (pC)c (pD)d / (pA)a (pB)b
Units of Kp / Kp 的单位： atmΔn, where Δn = (c + d) − (a + b), the change in moles of gas.
atmΔn，其中 Δn = (c + d) − (a + b)，即气体摩尔数的变化。
4.3 Worked Example / 计算示例
Question / 问题： For the reaction N₂O₄(g) ⇌ 2NO₂(g) at 298 K and a total pressure of 1.0 atm, at equilibrium the mixture contains 20% N₂O₄ and 80% NO₂ by moles. Calculate Kp.
Solution / 解答：
Mole fraction of N₂O₄ = 0.20, mole fraction of NO₂ = 0.80
p(N₂O₄) = 0.20 × 1.0 = 0.20 atm
p(NO₂) = 0.80 × 1.0 = 0.80 atm
Kp = (pNO₂)² / pN₂O₄ = (0.80)² / 0.20 = 0.64 / 0.20 = 3.2 atm
5. Industrial Applications / 工业应用
5.1 The Haber Process / 哈伯法
The Haber process synthesises ammonia from nitrogen and hydrogen: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Understanding equilibrium principles allows us to optimise this process:
哈伯法从氮气和氢气合成氨。理解平衡原理使我们能够优化该过程：
- Pressure / 压力： High pressure (200 atm) shifts equilibrium right (4 mol → 2 mol) — but practical limits exist due to equipment costs and safety / 高压（200 atm）使平衡右移（4 mol → 2 mol）——但设备成本和安全因素限制了实际压力
- Temperature / 温度： Compromise at 400–450°C. The reaction is exothermic, so low temperature favors yield, but too low makes the reaction too slow. An iron catalyst is used to achieve a reasonable rate / 折中温度为 400–450°C。反应放热，因此低温有利于产率，但温度太低反应过慢。使用铁催化剂以获得合理的速率
- Removing NH₃ / 移除 NH₃： Ammonia is continuously liquefied and removed, pulling the equilibrium to the right / 氨被持续液化和移除，拉动平衡向右
5.2 The Contact Process / 接触法
The Contact process produces sulfuric acid: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹. Key conditions:
接触法生产硫酸。关键条件：
- Temperature: 450°C (compromise between rate and yield) / 温度：450°C（速率和产率之间的折中）
- Pressure: 1–2 atm (equilibrium already favors products — high pressure unnecessary) / 压力：1–2 atm（平衡已有利于产物——无需高压）
- Catalyst: V₂O₅ (vanadium(V) oxide) / 催化剂：V₂O₅（五氧化二钒）
6. Common Exam Questions and Pitfalls / 常见考题和易错点
6.1 Kc Calculations from Initial Amounts / 根据初始量计算 Kc
This is the most common A-Level calculation task. The key is setting up an ICE table (Initial, Change, Equilibrium):
这是最常见的A-Level计算任务。关键是建立ICE 表格（初始、变化、平衡）：
| A | B | C | D | |
|---|---|---|---|---|
| Initial / 初始 | a₀ | b₀ | 0 | 0 |
| Change / 变化 | −x | −x | +x | +x |
| Equilibrium / 平衡 | a₀ − x | b₀ − x | x | x |
Worked example / 例题： 1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed. At equilibrium, 0.67 mol of ethyl ethanoate is formed. The volume is 250 cm³. Calculate Kc for: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
| CH₃COOH | C₂H₅OH | CH₃COOC₂H₅ | H₂O | |
|---|---|---|---|---|
| Initial / mol | 1.00 | 1.00 | 0 | 0 |
| Change | −0.67 | −0.67 | +0.67 | +0.67 |
| Eqm / mol | 0.33 | 0.33 | 0.67 | 0.67 |
| Eqm conc / mol dm⁻³ | 0.33/0.25 = 1.32 | 1.32 | 0.67/0.25 = 2.68 | 2.68 |
Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH] = (2.68 × 2.68) / (1.32 × 1.32) = 4.12 (dimensionless / 无量纲)
6.2 Top Exam Pitfalls / 主要考试易错点
- Confusing Kc change with position change / 混淆 Kc 变化与位置变化： Only temperature changes Kc. Concentration and pressure shift the position but not the constant / 只有温度改变 Kc。浓度和压力移动位置但不改变常数。
- Forgetting to convert moles to concentration / 忘记将摩尔转换为浓度： Kc uses concentrations (mol dm⁻³), not amounts (mol) — divide by volume / Kc 使用浓度而非物质的量——需除以体积。
- Including solids/liquids in Kc / 在 Kc 中包含固体/液体： Only gases and aqueous species appear in Kc and Kp expressions / 只有气体和水溶液物种出现在 Kc 和 Kp 表达式中。
- Pressure and Kp with inert gases / 惰性气体对压力和 Kp 的影响： Adding an inert gas at constant volume increases total pressure but does not change partial pressures — no shift. Adding at constant pressure (volume increases) reduces partial pressures — shifts to side with more moles / 定容加惰气增加总压但不改变分压——平衡不移。定压加惰气（体积增大）降低分压——平衡移向更多摩尔侧。
- Misreading units of Kp / 误读 Kp 的单位： Calculate Δn to determine units. A dimensional analysis mistake is a common mark-loser / 计算 Δn 确定单位。量纲分析错误是常见的失分点。
7. Summary / 总结
| Factor | Effect on Position | Effect on Kc/Kp |
|---|---|---|
| ↑ Concentration of reactant | Shifts right / 右移 | No change / 不变 |
| ↑ Concentration of product | Shifts left / 左移 | No change / 不变 |
| ↑ Pressure (fewer moles on right) | Shifts right / 右移 | No change / 不变 |
| ↑ Pressure (fewer moles on left) | Shifts left / 左移 | No change / 不变 |
| ↑ Temperature (exothermic, ΔH < 0) | Shifts left / 左移 | Decreases / 降低 |
| ↑ Temperature (endothermic, ΔH > 0) | Shifts right / 右移 | Increases / 升高 |
| Add catalyst / 加催化剂 | No shift / 不移 | No change / 不变 |
Mastering chemical equilibrium requires both qualitative understanding (Le Chatelier’s principle) and quantitative skill (Kc, Kp, and ICE table calculations). Practice with diverse examples — from simple homogeneous systems to complex industrial processes — to build confidence for the A-Level examination.
掌握化学平衡既需要定性理解（勒夏特列原理），也需要定量技能（Kc、Kp 和 ICE 表格计算）。通过多样化的例子——从简单的均相系统到复杂的工业过程——进行练习，为 A-Level 考试建立信心。
Article for aleveler.com — A-Level Chemistry revision resources / 本文为 aleveler.com 提供 — A-Level 化学复习资源
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