Chemical Equilibrium & Le Chatelier’s Principle — 化学平衡与勒夏特列原理

📚 Chemical Equilibrium & Le Chatelier’s Principle | 化学平衡与勒夏特列原理

1. Introduction to Chemical Equilibrium | 化学平衡简介

A chemical reaction does not always go to completion. In many reactions, the forward and reverse processes occur simultaneously, leading to a dynamic state called chemical equilibrium. At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of all species remain constant — not because nothing is happening, but because the two opposing processes are balanced.

化学反应并不总是一进行到底。在许多反应中，正反应和逆反应同时发生，导致一种称为化学平衡的动态状态。在平衡状态下，正逆反应速率相等，所有物质的浓度保持恒定——并不是因为什么都没发生，而是因为两个相反的过程达成了一种平衡。

Consider the reversible reaction between nitrogen dioxide (NO₂) and dinitrogen tetroxide (N₂O₄):

考虑二氧化氮（NO₂）与四氧化二氮（N₂O₄）之间的可逆反应：

N₂O₄(g) ⇌ 2NO₂(g)

At room temperature, a sealed container of this mixture appears brown due to the presence of NO₂. If you cool the container, it becomes paler as the equilibrium shifts towards the colourless N₂O₄. Warm it again, and the brown colour intensifies as more NO₂ forms. This simple colour change reveals something profound: the position of equilibrium depends on temperature.

在室温下，密封容器中这种混合物由于NO₂的存在而呈现棕色。如果冷却容器，颜色变浅，因为平衡向无色的N₂O₄方向移动。再次加热，棕色加深，因为生成了更多的NO₂。这种简单的颜色变化揭示了一个深刻的道理：平衡位置取决于温度。

2. The Equilibrium Constant (Kc) | 平衡常数（Kc）

For a general reversible reaction:

对于一个一般的可逆反应：

aA + bB ⇌ cC + dD

The equilibrium constant Kc is given by:

平衡常数Kc的表达式为：

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Where the square brackets denote equilibrium concentrations in mol dm⁻³. The value of Kc is constant at a given temperature. Changing concentrations or pressures does not alter Kc — only temperature can change its value. A large Kc (much greater than 1) indicates that the equilibrium lies to the right, favouring products. A small Kc (much less than 1) indicates that the equilibrium lies to the left, favouring reactants.

方括号表示平衡浓度，单位为 mol dm⁻³。Kc的值在给定温度下是常数。改变浓度或压力不会改变Kc——只有温度才能改变它的值。大的Kc（远大于1）表示平衡偏向右边，有利于产物。小的Kc（远小于1）表示平衡偏向左边，有利于反应物。

Kc Value | Kc值	Meaning | 含义	Example | 示例
Kc >> 1 (e.g. 10⁴)	Products favoured | 产物占优	HCl formation: H₂ + Cl₂ → 2HCl
Kc ≈ 1	Significant amounts of both | 两者都显著	Esterification: acid + alcohol ⇌ ester + water
Kc << 1 (e.g. 10⁻⁴)	Reactants favoured | 反应物占优	N₂ + O₂ ⇌ 2NO at room temperature

Key exam point: Solids and pure liquids do not appear in the Kc expression because their concentrations are effectively constant. For heterogeneous equilibria, only gases and aqueous species are included. Water appears only when it is a solvent and the reaction mixture is non-aqueous.

考试重点：固体和纯液体不出现在Kc表达式中，因为它们的浓度实际上是恒定的。对于多相平衡，只包含气体和水溶液中的物种。水只有作为溶剂且反应混合物为非水溶液时才出现。

3. Le Chatelier’s Principle | 勒夏特列原理

In 1884, the French chemist Henri Louis Le Chatelier proposed a principle that has become one of the most powerful tools in chemistry: “If a system at dynamic equilibrium is subjected to a change, the position of equilibrium will shift to partially counteract that change.”

1884年，法国化学家亨利·路易·勒夏特列提出了一个原理，成为化学中最强大的工具之一：“如果一个处于动态平衡的系统受到变化的影响，平衡位置将发生移动以部分抵消这种变化。”

This principle is not a law of physics — it is a heuristic derived from the mathematics of equilibrium. Yet its predictive power is remarkable. It allows chemists to predict how a system will respond to changes in concentration, pressure, and temperature. The “partial counteract” phrasing is crucial: the shift does not fully reverse the imposed change, but it softens its effect.

这一原理不是物理定律——它是从平衡数学中推导出来的启发式规则。然而它的预测能力是非凡的。它允许化学家预测系统将如何响应浓度、压力和温度的变化。”部分抵消”这个措辞至关重要：这种移动不会完全逆转施加的变化，而是减轻其影响。

4. Effect of Concentration Changes | 浓度变化的影响

Adding more reactant pushes the equilibrium to the right, producing more product. Removing product also pushes equilibrium right. Conversely, adding product or removing reactant shifts equilibrium left.

加入更多反应物将平衡推向右侧，产生更多产物。移除产物同样将平衡推向右侧。相反，加入产物或移除反应物将使平衡向左移动。

Worked Example: Consider the equilibrium:

例题：考虑以下平衡：

Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)

(pale yellow) + (colourless) ⇌ (deep blood-red)

（浅黄色）+ （无色）⇌（深血红色）

If you add more Fe³⁺ ions (e.g., by adding FeCl₃ solution), the solution turns a deeper red — the equilibrium shifts right to consume the added Fe³⁺. If you add Na₂HPO₄, which removes Fe³⁺ by forming a colourless complex, the red colour fades. These colour changes provide direct visual evidence for Le Chatelier’s principle in action.

如果你加入更多的Fe³⁺离子（例如加入FeCl₃溶液），溶液变成更深的红色——平衡向右移动以消耗加入的Fe³⁺。如果你加入Na₂HPO₄（它通过与Fe³⁺形成无色络合物而移除Fe³⁺），红色褪去。这些颜色变化为勒夏特列原理的实际应用提供了直观的视觉证据。

Industrial relevance: In the Haber process (N₂ + 3H₂ ⇌ 2NH₃), ammonia product is continually removed by condensation. This continuous removal of NH₃ prevents equilibrium from being established and drives the reaction forward, maximising yield.

工业相关性：在哈伯法（N₂ + 3H₂ ⇌ 2NH₃）中，氨产物通过冷凝被不断移除。这种对NH₃的连续移除阻止了平衡的建立，并将反应推向前进，最大化产率。

5. Effect of Pressure Changes | 压力变化的影响

Pressure changes only affect equilibria involving gases and only when there is a difference in the number of gas molecules on each side. The system shifts to reduce the pressure change:

压力变化只影响涉及气体的平衡，并且只有当各侧气体分子数不同时才有效。系统移动以缓解压力变化：

• Increase pressure → equilibrium shifts towards the side with fewer gas molecules
• Decrease pressure → equilibrium shifts towards the side with more gas molecules
• 增加压力 → 平衡向气体分子数较少的一侧移动
• 降低压力 → 平衡向气体分子数较多的一侧移动

Example — N₂O₄ ⇌ 2NO₂ equilibrium:

示例 — N₂O₄ ⇌ 2NO₂平衡：

• Left side: 1 gas molecule (N₂O₄)
• Right side: 2 gas molecules (2NO₂)
• Increasing pressure → equilibrium shifts left (fewer molecules → lower pressure)
• Decreasing pressure → equilibrium shifts right (more molecules → higher pressure)
• 左侧：1个气体分子（N₂O₄）
• 右侧：2个气体分子（2NO₂）
• 增加压力 → 平衡向左移动（更少分子 → 更低压力）
• 降低压力 → 平衡向右移动（更多分子 → 更高压力）

Important note: Adding an inert gas (like argon) at constant volume does NOT affect the equilibrium position. Although total pressure increases, the partial pressures of the reacting gases remain unchanged. This is a common exam trap.

重要提示：在恒定体积下加入惰性气体（如氩气）不会影响平衡位置。虽然总压力增加，但反应气体的分压保持不变。这是常见的考试陷阱。

6. Effect of Temperature Changes | 温度变化的影响

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

温度是唯一改变平衡常数Kc值的因素。移动方向取决于正反应是放热还是吸热：

Forward Reaction | 正反应	Increase Temperature | 升温	Decrease Temperature | 降温	Kc Change | Kc变化	Example | 示例
Exothermic (ΔH < 0) | 放热	Shift LEFT | 向左	Shift RIGHT | 向右	Kc decreases | 减小	2SO₂ + O₂ ⇌ 2SO₃
Endothermic (ΔH > 0) | 吸热	Shift RIGHT | 向右	Shift LEFT | 向左	Kc increases | 增大	N₂ + O₂ ⇌ 2NO

Treat the heat as if it were a chemical species: for an exothermic reaction, write “heat” on the product side; for an endothermic reaction, write “heat” on the reactant side. Then apply Le Chatelier’s principle as if heat were a reactant or product:

把热量当作化学物种来处理：对于放热反应，把”热”写在产物一侧；对于吸热反应，把”热”写在反应物一侧。然后像对待反应物或产物一样应用勒夏特列原理：

Exothermic: A + B ⇌ C + D + heat

Endothermic: A + B + heat ⇌ C + D

This “heat as a species” analogy makes the direction of shift intuitive: adding heat to an exothermic equilibrium is like adding a product — the system shifts left to consume it.

这种”热量作为物种”的类比使得移动方向变得直观：向放热平衡中加入热量就像加入一个产物——系统向左移动以消耗它。

7. Effect of Catalysts | 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 the forward and reverse reactions by the same amount. Therefore:

催化剂提供了一条活化能更低的替代反应路径。关键的是，它以相同的幅度降低了正逆两个反应的活化能。因此：

• A catalyst does NOT change the position of equilibrium
• A catalyst does NOT change the value of Kc
• A catalyst DOES increase the rate at which equilibrium is reached
• 催化剂不改变平衡位置
• 催化剂不改变Kc的值
• 催化剂确实提高了达到平衡的速率

Exam trap: Students often claim a catalyst “increases yield” or “shifts equilibrium right.” This is incorrect. A catalyst only helps the system reach equilibrium faster — it cannot change where that equilibrium lies. In industry, catalysts are used to make processes economically viable at lower temperatures, but the yield is determined by the equilibrium position at that temperature.

考试陷阱：学生经常声称催化剂”提高产率”或”使平衡向右移动”。这是不正确的。催化剂只能帮助系统更快地达到平衡——它不能改变平衡的位置。在工业中，催化剂被用于使过程在较低温度下具有经济可行性，但产率由该温度下的平衡位置决定。

8. Kc Calculations — ICE Tables | Kc计算 — ICE表格法

One of the most common A-Level calculation skills involves determining Kc from initial and equilibrium data using an ICE table (Initial, Change, Equilibrium). This systematic approach prevents errors that arise from ad-hoc reasoning.

最常见的A-Level计算技能之一涉及使用ICE表格（初始Initial、变化Change、平衡Equilibrium）从初始和平衡数据确定Kc。这种系统化方法防止了由临时推理产生的错误。

Worked Example: 0.50 mol of ethanoic acid and 1.00 mol of ethanol are mixed and allowed to reach equilibrium at 298 K. At equilibrium, 0.42 mol of ethyl ethanoate is present. Calculate Kc for: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

例题：将0.50 mol乙酸和1.00 mol乙醇混合并在298 K下达到平衡。平衡时，存在0.42 mol乙酸乙酯。计算以下反应的Kc：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Species | 物种	CH₃COOH	C₂H₅OH	CH₃COOC₂H₅	H₂O	Comment | 说明
Initial	0.50	1.00	0.00	0.00	Initial moles | 初始摩尔数
Change	-0.42	-0.42	+0.42	+0.42	Reacted/Formed | 反应/生成
Equilibrium	0.08	0.58	0.42	0.42	Equilibrium moles | 平衡摩尔数

Assuming total volume = V dm³, the concentrations are n/V. Since the volume cancels (equal number of moles on both sides: 1+1 ⇌ 1+1), we can use moles directly:

假设总体积 = V dm³，浓度为 n/V。由于体积抵消（两侧摩尔数相等：1+1 ⇌ 1+1），我们可以直接使用摩尔数：

Kc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH] = (0.42)(0.42) / (0.08)(0.58) = 0.1764 / 0.0464 = 3.8

Always check: The units of Kc depend on the stoichiometry. When the total moles of gaseous/aqueous species on each side of the equation differ, Kc has units. Here, Δn = 0, so Kc is dimensionless.

始终检查：Kc的单位取决于化学计量关系。当方程式两侧气态/水溶物种的总摩尔数不同时，Kc具有单位。这里Δn = 0，所以Kc无量纲。

9. Industrial Application: The Haber Process | 工业应用：哈伯法

The Haber process for ammonia synthesis is the textbook example of how Le Chatelier’s principle is applied in industrial chemistry — and where economic reality overrides theoretical prediction:

用于氨合成的哈伯法是勒夏特列原理在工业化学中应用的教科书示例——也是经济现实超越理论预测的范例：

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

Factor | 因素	Theoretical Prediction (Le Chatelier) | 理论预测	Industrial Choice | 工业选择	Reasoning | 理由
Pressure | 压力	HIGH (4 → 2 gas molecules) | 高压	200 atm	Higher pressure gives better yield, but stronger equipment is costly and dangerous. 200 atm balances yield against cost and safety.
Temperature | 温度	LOW (exothermic, shift right) | 低温	400-450°C	Low temperature favours yield but makes the reaction prohibitively slow. 400°C is the compromise temperature where the iron catalyst is active enough for an acceptable rate.
Catalyst | 催化剂	No effect on position | 无影响	Iron (Fe)	The iron catalyst allows the reaction to proceed at a reasonable rate at the compromise temperature. Without it, even 400°C would be too slow.

The key lesson: Le Chatelier’s principle tells you the direction of the effect — it does not tell you the rate. Industrial processes must balance thermodynamic yield against kinetic feasibility and economic cost. A reaction with 95% theoretical yield at 25°C is useless if it takes 1,000 years to reach equilibrium.

关键教训：勒夏特列原理告诉你影响的方向——它不告诉速率。工业过程必须在热力学产率与动力学可行性和经济成本之间取得平衡。一个在25°C下理论产率为95%的反应，如果需要1000年才能达到平衡，那就是无用的。

10. Common Exam Pitfalls | 常见考试陷阱

10.1 Confusing Rate and Position | 混淆速率和位置

Students often think “higher temperature always increases yield” — this is true for endothermic reactions but false for exothermic ones. Higher temperature always increases rate, but for exothermic reactions it decreases equilibrium yield.

学生通常认为”更高温度总是提高产率”——这对吸热反应是正确的，但对放热反应是错误的。更高温度总是提高速率，但对放热反应而言，它降低了平衡产率。

10.2 Catalyst Misunderstandings | 催化剂的误解

Repeating for emphasis: a catalyst changes the rate of reaching equilibrium, not the position of equilibrium. In an exam, if you write “catalyst increases the yield of NH₃” you will lose the mark.

再次强调：催化剂改变达到平衡的速率，而不是平衡的位置。在考试中，如果你写”催化剂增加了NH₃的产率”，你将失分。

10.3 Forgetting Units of Kc | 忘记Kc的单位

Kc only has units when Δn ≠ 0. Calculate Δn = (moles of gaseous/aqueous products) minus (moles of gaseous/aqueous reactants). Unit: (mol dm⁻³)^Δn. Many candidates lose marks by omitting units or writing incorrect ones.

Kc只有在Δn ≠ 0时才具有单位。计算Δn =（气态/水溶产物的摩尔数）减去（气态/水溶反应物的摩尔数）。单位：(mol dm⁻³)^Δn。许多考生因遗漏单位或写错单位而失分。

10.4 Ignoring the Volume in Calculations | 在计算中忽略体积

If the volume of the container is not given but the moles of gases on each side are equal (Δn = 0), you can use moles directly since volume cancels. But when Δn ≠ 0, you must divide by the volume to get concentrations before substituting into the Kc expression.

如果未给出容器体积但两侧气体摩尔数相等（Δn = 0），可以直接使用摩尔数，因为体积抵消。但当Δn ≠ 0时，必须除以体积得到浓度后再代入Kc表达式。

10.5 Including Solids in Kc | 在Kc中包含固体

For reactions like CaCO₃(s) ⇌ CaO(s) + CO₂(g), the Kc expression is simply Kc = [CO₂]. The solids CaCO₃ and CaO have constant concentrations and are omitted.

对于CaCO₃(s) ⇌ CaO(s) + CO₂(g)这样的反应，Kc表达式就是Kc = [CO₂]。固体CaCO₃和CaO具有恒定的浓度，因此被省略。

11. Summary — The Complete Picture | 总结 — 全景图

Change | 变化	Effect on Position | 对位置的影响	Effect on Kc | 对Kc的影响	Mechanism | 机制
Add reactant | 加入反应物	Shift RIGHT | 右移	No change | 不变	System consumes added reactant. Forward rate temporarily exceeds reverse rate.
Remove product | 移除产物	Shift RIGHT | 右移	No change | 不变	System replenishes removed product. Forward rate temporarily exceeds reverse rate.
Increase pressure | 增压	Shift to fewer gas moles | 移向更少气态摩尔数	No change | 不变	System reduces total gas molecules. No effect if Δn = 0.
Increase T (exo) | 升温（放热）	Shift LEFT | 左移	Kc DECREASES | 减小	System absorbs added heat by favouring the endothermic (reverse) direction.
Increase T (endo) | 升温（吸热）	Shift RIGHT | 右移	Kc INCREASES | 增大	System absorbs added heat by favouring the endothermic (forward) direction.
Catalyst | 催化剂	NO CHANGE | 不变	No change | 不变	Lowers Ea equally for both directions. Equilibrium reached faster, not shifted.

Chemical equilibrium is not a static endpoint but a dynamic balance — a molecular tug-of-war where both sides pull with equal strength. Le Chatelier’s principle gives us the language to predict how that balance shifts when the conditions change. For the A-Level student, mastering equilibrium means understanding three things: the mathematics of Kc, the qualitative predictions of Le Chatelier, and the crucial distinction between thermodynamics (where the reaction goes) and kinetics (how fast it gets there).

化学平衡不是一个静态的终点，而是一种动态的平衡——一场分子层面的拔河比赛，双方以相同的力度拉扯。勒夏特列原理为我们提供了一种语言来预测当条件变化时这种平衡如何移动。对于A-Level学生来说，掌握平衡意味着理解三件事：Kc的数学、勒夏特列的定性预测，以及热力学（反应去向何处）与动力学（反应到达那里的速度有多快）之间的关键区别。