A-Level Chemistry: Chemical Equilibrium — Kc, Kp & Le Chatelier’s Principle | A-Level 化学：化学平衡全面指南

A-Level 化学：掌握化学平衡 — 从 Kc 计算到勒夏特列原理

This comprehensive guide covers everything you need to know about chemical equilibrium for A-Level Chemistry (AQA, Edexcel, OCR, CAIE). From dynamic equilibrium fundamentals to complex Kc and Kp calculations, Le Chatelier’s Principle applications, and industrial case studies — all explained in both English and Chinese.

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

English

Chemical equilibrium is one of the most conceptually rich topics in A-Level Chemistry. At its core, dynamic equilibrium describes a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant — but not necessarily equal.

The key word is dynamic. Unlike a static balance where nothing moves, a dynamic equilibrium features continuous forward and backward reactions occurring simultaneously at identical rates. Molecules are constantly being converted in both directions, yet the macroscopic composition of the system appears unchanged.

For equilibrium to be established, the reaction must occur in a closed system — one where no matter can enter or leave. If products escape (as gases in an open container) or reactants are added mid-reaction, the equilibrium cannot stabilise.

Consider the classic example: the Haber process for ammonia synthesis.

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

When nitrogen and hydrogen are mixed in a sealed reactor at high temperature with an iron catalyst, ammonia forms. But ammonia also decomposes back into N₂ and H₂. Eventually, the rates of formation and decomposition equalise — equilibrium is reached. The concentrations of N₂, H₂, and NH₃ remain constant from that point onward, barring any external change.

中文

化学平衡是 A-Level 化学中概念最丰富的主题之一。其核心是动态平衡，描述的是可逆反应中正反应速率等于逆反应速率且反应物与生成物浓度保持恒定（但不一定相等）的状态。

关键词是“动态”。与一切都不动的静态平衡不同，动态平衡中是正逆反应以相同速率持续进行。分子在双向不断转化，但系统的宏观组成看起来不变。

要建立平衡，反应必须发生在封闭系统中——物质不能进出。如果产物逸出（如敞口容器中的气体）或在反应中途加入反应物，平衡就无法稳定。

来看一个经典例子：哈伯法合成氨。

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

当氮气和氢气在高温密封反应器中与铁催化剂混合时，氨生成。但是氨也会分解回 N₂ 和 H₂。最终，生成与分解的速率相等——达到平衡。从那时起，N₂、H₂ 和 NH₃ 的浓度保持恒定，除非有外部变化。

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2. The Equilibrium Constant Kc | 平衡常数 Kc

English

For a general homogeneous reaction at equilibrium:

aA + bB ⇌ cC + dD

The equilibrium constant Kc (concentration-based) is defined as:

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

Where square brackets denote equilibrium concentrations in mol dm⁻³, and the lowercase letters a, b, c, d are the stoichiometric coefficients from the balanced equation.

Critical rules for Kc:

1. Temperature dependent only: Kc changes ONLY with temperature. It is NOT affected by concentration changes, pressure changes, or catalysts.
2. Pure solids and liquids are omitted: Their concentrations are effectively constant and are absorbed into the Kc value. Only gases and aqueous species appear in the expression.
3. Units vary: The units of Kc depend on the stoichiometry and are NOT fixed — always calculate them from the expression. For A-Level exams, you must derive and state the correct units.
4. Kc magnitude indicates position of equilibrium: Kc >> 1 → products favoured. Kc << 1 → reactants favoured. Kc ≈ 1 → significant amounts of both.

中文

对于一般的均相平衡反应：

aA + bB ⇌ cC + dD

平衡常数 Kc（基于浓度）定义为：

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

方括号表示以 mol dm⁻³ 为单位的平衡浓度，小写字母 a、b、c、d 是平衡方程式中的化学计量系数。

Kc 的重要规则：

1. 仅受温度影响：Kc 只随温度变化。浓度变化、压力变化或催化剂不影响 Kc。
2. 纯固体和纯液体省略：它们的浓度实际上恒定，被并入 Kc 值中。表达式中只出现气体和水溶液物种。
3. 单位不固定：Kc 的单位取决于化学计量比，不是固定的——必须从表达式中推导。A-Level 考试要求推导并写出正确单位。
4. Kc 的大小表示平衡位置：Kc >> 1 → 倾向于产物。Kc << 1 → 倾向于反应物。Kc ≈ 1 → 两者都有可观数量。

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3. Kc Calculation Walkthrough | Kc 计算示例

English

Worked Example: The esterification of ethanoic acid with ethanol:

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

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

Step 1: Set up an ICE table (Initial / Change / Equilibrium)

Species	CH₃COOH	C₂H₅OH	CH₃COOC₂H₅	H₂O
Initial (mol)	0.50	0.50	0	0
Change (mol)	−x	−x	+x	+x
Equilibrium (mol)	0.50 − x	0.50 − x	x	x

Step 2: Determine x from given data. At equilibrium, ester = 0.30 mol → x = 0.30

Eqm: CH₃COOH = 0.50 − 0.30 = 0.20 mol
Eqm: C₂H₅OH = 0.50 − 0.30 = 0.20 mol
Eqm: CH₃COOC₂H₅ = 0.30 mol
Eqm: H₂O = 0.30 mol

Step 3: Convert to concentrations (volume = 1.0 dm³, so [X] = mol)

[CH₃COOH] = 0.20   [C₂H₅OH] = 0.20   [CH₃COOC₂H₅] = 0.30   [H₂O] = 0.30

Step 4: Substitute into Kc expression

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH] = (0.30)(0.30) / (0.20)(0.20) = 0.090 / 0.040 = 2.25

Step 5: Determine units. Each concentration term is mol dm⁻³. The expression has (mol dm⁻³)² in both numerator and denominator → units cancel. Kc = 2.25 (no units)

中文

计算例题：乙酸与乙醇的酯化反应：

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

将 0.50 mol 乙酸和 0.50 mol 乙醇混合并达到平衡。平衡时有 0.30 mol 乙酸乙酯，总体积为 1.0 dm³。计算 Kc。

第 1 步：构建 ICE 表格（初始 Initial / 变化 Change / 平衡 Equilibrium）

物种	CH₃COOH	C₂H₅OH	CH₃COOC₂H₅	H₂O
初始 (mol)	0.50	0.50	0	0
变化 (mol)	−x	−x	+x	+x
平衡 (mol)	0.50 − x	0.50 − x	x	x

第 2 步：根据已知数据求 x。平衡时酯 = 0.30 mol → x = 0.30

平衡: CH₃COOH = 0.50 − 0.30 = 0.20 mol
平衡: C₂H₅OH = 0.50 − 0.30 = 0.20 mol
平衡: CH₃COOC₂H₅ = 0.30 mol
平衡: H₂O = 0.30 mol

第 3 步：转换为浓度（体积 = 1.0 dm³，所以 [X] = mol）

第 4 步：代入 Kc 表达式

第 5 步：确定单位。分子和分母各有 (mol dm⁻³)² → 单位相消。Kc = 2.25（无单位）

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4. Kp — Equilibrium Constant for Gaseous Reactions | 气体反应的平衡常数 Kp

English

For gas-phase equilibria, we use Kp, which is defined in terms of partial pressures rather than concentrations.

Partial pressure (p) of a gas in a mixture is the pressure it would exert if it alone occupied the entire volume at the same temperature. Dalton’s Law states:

p(A) = mole fraction of A × total pressure

Where:

mole fraction of A = moles of A / total moles of all gases

For the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g):

Kp = (pC)ᶜ (pD)ᵈ / (pA)ᵃ (pB)ᵇ

The units of Kp are typically atmⁿ or Paⁿ, where n = (c + d) − (a + b), i.e., the change in moles of gas.

Relationship between Kc and Kp:

Kp = Kc (RT)^(Δn)

Where R = 8.314 J K⁻¹ mol⁻¹ (or 0.0821 L atm K⁻¹ mol⁻¹), T is temperature in Kelvin, and Δn = (moles of gaseous products) − (moles of gaseous reactants).

When are Kc and Kp equal? When Δn = 0 — i.e., the number of moles of gas is the same on both sides of the equation. For example, in H₂(g) + I₂(g) ⇌ 2HI(g), Δn = 2 − 2 = 0, so Kp = Kc.

中文

对于气相平衡，我们使用 Kp，它以分压而非浓度来定义。

混合物中气体的分压 (p) 是该气体在相同温度下单独占据整个体积时所产生的压力。道尔顿定律指出：

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

对于反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)：

Kp = (pC)ᶜ (pD)ᵈ / (pA)ᵃ (pB)ᵇ

Kp 的单位通常为 atmⁿ 或 Paⁿ，其中 n = (c + d) − (a + b)，即气体摩尔数的变化。

Kc 与 Kp 的关系：

Kp = Kc (RT)^(Δn)

其中 R = 8.314 J K⁻¹ mol⁻¹，T 是开尔文温度，Δn = (气态产物摩尔数) − (气态反应物摩尔数)。

Kc 和 Kp 何时相等？当 Δn = 0 时——即方程式两边气体摩尔数相同。例如 H₂(g) + I₂(g) ⇌ 2HI(g)，Δn = 2 − 2 = 0，所以 Kp = Kc。

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5. Le Chatelier’s Principle | 勒夏特列原理

English

Le Chatelier’s Principle states:

“If a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose (partially counteract) the change.”

This principle allows us to predict the qualitative effect of changes in concentration, pressure, and temperature on the position of equilibrium.

5.1 Effect of Concentration | 浓度的影响

English: If the concentration of a reactant is increased, the equilibrium shifts to the right (towards products) to consume the added reactant. Conversely, removing a product causes the equilibrium to shift right to replenish it.

Key exam point: Kc does NOT change when concentrations are altered. Only the position of equilibrium shifts.

中文：增加反应物浓度，平衡向右移动（向产物方向）以消耗增加的反应物。反之，移除产物使平衡右移以补充它。

考试要点：改变浓度时 Kc 不变。只有平衡位置移动。

5.2 Effect of Pressure | 压力的影响

English: This only applies to gaseous equilibria where Δn ≠ 0. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas (to reduce total pressure). Decreasing pressure shifts towards the side with more moles.

Change	Shift for N₂ + 3H₂ ⇌ 2NH₃ (Δn = −2)	Reason
Increase pressure	Shifts RIGHT (fewer moles: 4 → 2)	Reduces total number of gas molecules
Decrease pressure	Shifts LEFT (more moles: 2 → 4)	Increases total number of gas molecules

Note: If Δn = 0 (equal moles on both sides), pressure changes have no effect on the position of equilibrium. Kp also does not change with pressure.

中文：这仅适用于 Δn ≠ 0 的气相平衡。增加压力使平衡向气体摩尔数较少的一侧移动（以降低总压力）。降低压力向摩尔数较多的一侧移动。

变化	N₂ + 3H₂ ⇌ 2NH₃ 的移动方向 (Δn = −2)	原因
增加压力	向右移动（摩尔数更少：4 → 2）	减少气体分子总数
降低压力	向左移动（摩尔数更多：2 → 4）	增加气体分子总数

注意：若 Δn = 0（两边摩尔数相等），压力变化不影响平衡位置。Kp 也不随压力变化。

5.3 Effect of Temperature | 温度的影响

English: This is the MOST important factor because temperature is the ONLY variable that changes the value of Kc/Kp.

• Exothermic reaction (ΔH < 0): Increasing temperature shifts equilibrium LEFT (towards reactants). Kc/Kp decreases.
• Endothermic reaction (ΔH > 0): Increasing temperature shifts equilibrium RIGHT (towards products). Kc/Kp increases.

Think of heat as a “chemical”: exothermic reactions produce heat, so adding heat (raising T) pushes it back. Endothermic reactions absorb heat, so adding heat drives it forward.

中文：这是最重要的因素，因为温度是唯一能改变 Kc/Kp 值的变量。

• 放热反应 (ΔH < 0)：升高温度，平衡向左移动（向反应物方向）。Kc/Kp 减小。
• 吸热反应 (ΔH > 0)：升高温度，平衡向右移动（向产物方向）。Kc/Kp 增大。

把热看作一种”化学物质”：放热反应产生热量，所以加入热量（升温）会将其推回。吸热反应吸收热量，所以加入热量推动其正向进行。

5.4 Effect of a Catalyst | 催化剂的影响

English: A catalyst has NO effect on the position of equilibrium or the value of Kc/Kp. It speeds up BOTH the forward and backward reactions equally, so equilibrium is reached faster, but the position does not change.

中文：催化剂不影响平衡位置或 Kc/Kp 值。它同等加速正逆反应，所以平衡更快达到，但位置不变。

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6. Industrial Applications | 工业应用

6.1 The Haber Process | 哈伯法

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

This is the classic A-Level case study. The forward reaction is exothermic and produces fewer moles of gas (4 → 2).

Condition	Compromise Value	Reasoning
Temperature	400–450 °C	Low T favours yield (exothermic) but is too slow. High T gives faster rate but lower yield. 450 °C is the compromise.
Pressure	200 atm	High P favours yield (fewer moles) but is expensive (thick pipes, energy). 200 atm is economically viable.
Catalyst	Iron (Fe)	Speeds up both directions equally, allowing lower T to be used while maintaining reasonable rate.

中文：

条件	折中值	理由
温度	400–450 °C	低温有利于产率（放热）但太慢。高温速率快但产率低。450 °C 是折中。
压力	200 atm	高压有利于产率（摩尔数更少）但成本高（厚管道、能耗）。200 atm 经济可行。
催化剂	铁 (Fe)	同等加速正逆反应，允许在较低温度下保持合理速率。

6.2 Contact Process (Sulfuric Acid) | 接触法（硫酸）

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

Conditions: 450 °C, 1–2 atm, V₂O₅ catalyst. The relatively low pressure is sufficient because the equilibrium already lies far to the right (Kc is large at 450 °C). Increasing pressure would raise costs without significant yield improvement.

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

条件：450 °C，1–2 atm，V₂O₅ 催化剂。使用相对低压是因为平衡已经大幅偏向右侧（450 °C 时 Kc 很大）。增加压力只会提高成本，而不会显著提高产率。

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7. Common Exam Mistakes | 常见考试错误

English

1. “Equilibrium means equal concentrations” — WRONG. It means equal rates. Concentrations are constant but rarely equal.
2. “Kc changes with concentration” — WRONG. Only temperature changes Kc. Concentrations change the position of equilibrium, not the constant.
3. “Increasing pressure always shifts equilibrium” — WRONG. It only shifts if there is a difference in moles of gas (Δn ≠ 0).
4. “Catalysts increase yield” — WRONG. Catalysts do not affect yield or Kc. They only increase the rate at which equilibrium is reached.
5. Forgetting to calculate Kc units — Always derive and state the units from the Kc expression.
6. Using mass instead of concentration in Kc — Kc requires concentrations (mol dm⁻³). If volume is not 1 dm³, you must divide by it.

中文

1. “平衡意味着浓度相等” — 错误。它意味着速率相等。浓度恒定但很少相等。
2. “Kc 随浓度变化” — 错误。只有温度改变 Kc。浓度改变的是平衡位置，而非常数。
3. “增加压力总是使平衡移动” — 错误。只有当气体摩尔数存在差异 (Δn ≠ 0) 时才移动。
4. “催化剂提高产率” — 错误。催化剂不影响产率或 Kc。只提高达到平衡的速率。
5. 忘记计算 Kc 单位 — 务必从 Kc 表达式中推导并写出单位。
6. Kc 中用质量而非浓度 — Kc 需要浓度 (mol dm⁻³)。如果体积不是 1 dm³，必须除以体积。

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8. Quick Revision Checklist | 快速复习清单

Topic 主题	Key Point 要点	✓
Dynamic equilibrium	Rate forward = rate backward, concentrations constant, closed system	☐
Kc expression	[products]/[reactants], omit solids and pure liquids	☐
Kc units	Derive from expression; not fixed	☐
Kc depends on	Temperature ONLY	☐
Kp & partial pressure	p(A) = mole fraction × P(total); Dalton’s Law	☐
Kp units	atm^Δn or Pa^Δn	☐
Kc ↔ Kp	Kp = Kc(RT)^Δn	☐
Le Chatelier: concentration	Shift to consume added / replenish removed	☐
Le Chatelier: pressure	Shift to side with fewer gas moles (Δn ≠ 0 only)	☐
Le Chatelier: temperature (exo)	Increase T → shift left; Kc decreases	☐
Le Chatelier: temperature (endo)	Increase T → shift right; Kc increases	☐
Catalyst	No effect on equilibrium position or Kc/Kp	☐
Haber process conditions	450 °C, 200 atm, Fe catalyst; compromise	☐
Contact process conditions	450 °C, 1–2 atm, V₂O₅; low P because high Kc	☐

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This article was created for A-Level Chemistry students studying chemical equilibrium. Master these concepts thoroughly — equilibrium questions appear in virtually every A-Level Chemistry exam paper, often worth 8–15 marks. Practice ICE table calculations, Kc/Kp derivations, and Le Chatelier predictions until they become second nature. 本文为学习化学平衡的 A-Level 化学学生编写。彻底掌握这些概念——平衡题几乎出现在每份 A-Level 化学试卷中，通常分值 8–15 分。反复练习 ICE 表格计算、Kc/Kp 推导和勒夏特列预测，直到熟练自如。