Chemical Equilibrium & Le Chatelier’s Principle | A-Level Chemistry 化学平衡与勒夏特列原理

🔬 What is Chemical Equilibrium? | 什么是化学平衡？

Chemical equilibrium is a fundamental concept in A-Level Chemistry that describes a state where the forward and reverse reactions occur at exactly the same rate in a closed system. At equilibrium, the concentrations of reactants and products remain constant — but importantly, the reactions have not stopped. This is what we call a dynamic equilibrium.

化学平衡是A-Level化学中的一个基础概念，它描述的是在一个封闭系统中，正向反应和逆向反应以完全相同速率进行的状态。在平衡状态下，反应物和产物的浓度保持不变——但关键在于，反应并没有停止。这就是我们所说的动态平衡。

Consider the classic example of the Haber Process:

以经典的哈伯法为例：

At equilibrium, nitrogen and hydrogen combine to form ammonia at the same rate that ammonia decomposes back into its constituent elements. The double arrow (⇌) is the universal symbol for a reversible reaction at equilibrium.

在平衡状态下，氮气和氢气结合生成氨的速率与氨分解回其组成元素的速率相同。双箭头（⇌）是可逆反应处于平衡状态的通用符号。

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📐 The Equilibrium Constant (Kc) | 平衡常数 (Kc)

For any reversible reaction at a given temperature, we can express the equilibrium position mathematically using the equilibrium constant, Kc.

对于任何在给定温度下的可逆反应，我们可以使用平衡常数Kc来数学化地表达平衡位置。

For a general reaction:

对于一般反应：

The equilibrium constant expression is:

平衡常数表达式为：

Where square brackets [ ] denote concentration in mol dm⁻³. The key rules to remember are:

其中方括号 [ ] 表示浓度，单位为 mol dm⁻³。需要记住的关键规则是：

• Products over reactants: Kc = [products] / [reactants] — always products on top, raised to their stoichiometric coefficients.
  产物在分子，反应物在分母：Kc = [产物] / [反应物] —— 始终是产物在分子，并以其计量系数为指数。
• Solids and pure liquids are excluded: They have constant concentration and are incorporated into the value of Kc. Only gases (g) and aqueous species (aq) appear in the expression.
  固体和纯液体被排除：它们具有恒定浓度，已包含在Kc值中。只有气体(g)和水溶液(aq)物种出现在表达式中。
• Kc is temperature-dependent only: Changing concentration or pressure does not change Kc — the system simply shifts to restore the same Kc value. Only temperature changes alter Kc.
  Kc仅依赖于温度：改变浓度或压力不会改变Kc的值——系统只是移动以恢复相同的Kc值。只有温度变化才会改变Kc。

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 — reaction goes nearly to completion 平衡远偏右——反应几乎完全进行
Kc ≈ 1	Significant amounts of both reactants and products present 反应物和产物都有显著量存在
Kc << 1 (e.g., 10⁻¹⁰)	Equilibrium lies far to the left — very little product formed 平衡远偏左——生成的产物极少

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🧭 Le Chatelier’s Principle | 勒夏特列原理

Henri Le Chatelier (1884) gave us one of the most powerful predictive tools in chemistry. The principle states:

亨利·勒夏特列（1884年）给了我们化学中最强大的预测工具之一。该原理指出：

Think of the equilibrium as a stubborn mule — push it one way, and it pushes back. Let’s examine how each type of stress affects the equilibrium position.

可以把平衡想象成一头倔驴——你往一个方向推它，它就反方向推回来。让我们逐一分析每种应力如何影响平衡位置。

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🌡️ Effect of Temperature | 温度的影响

Temperature is unique — it’s the only change that actually alters the value of Kc. All other changes simply shift the equilibrium position to restore the same Kc.

温度是独特的——它是唯一一个真正改变Kc值的变化。所有其他变化只是移动平衡位置以恢复相同的Kc值。

Exothermic Reactions (ΔH < 0) | 放热反应

In an exothermic reaction, heat is a product:

在放热反应中，热是产物：

• Increasing temperature: Equilibrium shifts LEFT (endothermic direction) to absorb the added heat. Kc decreases.
  升高温度：平衡向左移动（吸热方向）以吸收增加的热量。Kc 减小。
• Decreasing temperature: Equilibrium shifts RIGHT (exothermic direction) to release more heat. Kc increases.
  降低温度：平衡向右移动（放热方向）以释放更多热量。Kc 增大。

Endothermic Reactions (ΔH > 0) | 吸热反应

In an endothermic reaction, heat is a reactant:

在吸热反应中，热是反应物：

• Increasing temperature: Equilibrium shifts RIGHT to absorb the added heat. Kc increases.
  升高温度：平衡向右移动以吸收增加的热量。Kc 增大。
• Decreasing temperature: Equilibrium shifts LEFT. Kc decreases.
  降低温度：平衡向左移动。Kc 减小。

Exam Tip: The A-Level exam will often ask you to predict the effect of temperature on both the equilibrium position AND Kc. Always state both! First identify if the forward reaction is exothermic or endothermic, then apply Le Chatelier’s Principle.

考试提示：A-Level考试经常会要求你预测温度对平衡位置和Kc的影响。一定要两个都说明！首先判断正反应是放热还是吸热，然后应用勒夏特列原理。

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⚡ Effect of Pressure | 压力的影响

Pressure changes only affect equilibria involving gases, and only when there is a different number of moles of gas on each side of the equation.

压力变化只影响涉及气体的平衡，且仅当方程式两边气体摩尔数不同时才有影响。

Increasing pressure: Equilibrium shifts toward the side with fewer moles of gas (to reduce pressure).
升高压力：平衡向气体摩尔数较少的一侧移动（以降低压力）。

Decreasing pressure: Equilibrium shifts toward the side with more moles of gas (to increase pressure).
降低压力：平衡向气体摩尔数较多的一侧移动（以增加压力）。

Worked Example: Haber Process | 例题：哈伯法

Left side: 1 + 3 = 4 moles of gas
Right side: 2 moles of gas

左边：1 + 3 = 4摩尔气体
右边：2摩尔气体

• High pressure → shifts RIGHT (towards 2 moles → fewer moles). This is why the Haber Process operates at 200 atm!
  高压 → 向右移动（向2摩尔方向 → 较少的摩尔数）。这就是哈伯法在200个大气压下操作的原因！
• Kc does not change — pressure only shifts the position, not the constant.
  Kc 不变 —— 压力只改变位置，不改变常数。

Important: If there are equal moles of gas on both sides (e.g., H₂(g) + I₂(g) ⇌ 2HI(g): 2 moles ⇌ 2 moles), changing pressure has no effect on the equilibrium position.

重要提示：如果两边气体摩尔数相同（例如：H₂(g) + I₂(g) ⇌ 2HI(g)：2摩尔 ⇌ 2摩尔），改变压力对平衡位置没有影响。

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🧪 Effect of Concentration | 浓度的影响

Changing the concentration of a reactant or product shifts the equilibrium position — but Kc remains constant.

改变反应物或产物的浓度会移动平衡位置——但Kc保持不变。

• Adding a reactant: Equilibrium shifts RIGHT to consume the added reactant and produce more product.
  增加反应物：平衡向右移动以消耗增加的反应物并生成更多产物。
• Adding a product: Equilibrium shifts LEFT to consume the added product and regenerate reactants.
  增加产物：平衡向左移动以消耗增加的产物并重新生成反应物。
• Removing a product: Equilibrium shifts RIGHT to replace what was removed. This is an important strategy in industrial processes — continuously removing the product drives the reaction forward.
  移除产物：平衡向右移动以补充被移除的部分。这是工业过程中的一个重要策略——持续移除产物可推动反应正向进行。

Worked Example: Fe³⁺ / SCN⁻ Equilibrium | 例题：铁离子/硫氰酸根平衡

This is a classic A-Level practical experiment. The intense blood-red colour of the complex ion makes it easy to observe shifts visually:

这是经典的A-Level实验。配合离子的深血红色使得肉眼观察平衡移动变得容易：

• Add Fe³⁺ → solution turns darker red (equilibrium shifts RIGHT)
  加入Fe³⁺ → 溶液变为更深的红色（平衡向右移动）
• Add SCN⁻ → solution turns darker red (equilibrium shifts RIGHT)
  加入SCN⁻ → 溶液变为更深的红色（平衡向右移动）
• Add OH⁻ (removes Fe³⁺ as precipitate) → solution fades (equilibrium shifts LEFT)
  加入OH⁻（以沉淀形式移除Fe³⁺） → 溶液褪色（平衡向左移动）

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🧬 Effect of a Catalyst | 催化剂的影响

This is a common exam trap! A catalyst:

这是一个常见的考试陷阱！催化剂：

• ✅ Speeds up BOTH forward and reverse reactions equally
  ✅ 同等加速正向和逆向反应
• ✅ Lowers the activation energy of both reactions by providing an alternative pathway
  ✅ 通过提供替代路径降低两个反应的活化能
• ❌ Does NOT change the equilibrium position
  ❌ 不改变平衡位置
• ❌ Does NOT change the value of Kc
  ❌ 不改变Kc的值
• ✅ The only effect is that equilibrium is reached faster
  ✅ 唯一的效果是更快达到平衡

Exam Tip: If asked “what does a catalyst do to the equilibrium?”, the correct answer is “it has no effect on the position of equilibrium — it only increases the rate at which equilibrium is attained.”

考试提示：如果被问”催化剂对平衡有什么影响？”，正确答案是“它对平衡位置没有影响——它只增加达到平衡的速率。”

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🏭 Industrial Applications | 工业应用

The Haber Process (Ammonia Production) | 哈伯法（氨的生产）

Condition | 条件	Choice | 选择	Reason | 原因
Temperature | 温度	400–450°C	Compromise: low T favours yield (exothermic) but high T gives faster rate. An iron catalyst enables a moderate temperature. 折中方案：低温有利于产率（放热），但高温速率更快。铁催化剂使得中等温度成为可能。
Pressure | 压力	200 atm	High pressure favours fewer moles (4→2), shifting equilibrium right. Above 200 atm, equipment costs outweigh yield gains. 高压有利于减少摩尔数（4→2），使平衡向右移动。超过200个大气压后，设备成本超过产率收益。
Catalyst | 催化剂	Iron (Fe)	Speeds up the reaction without affecting equilibrium position. Does not change Kc. 加速反应而不影响平衡位置。不改变Kc。

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

Condition | 条件	Choice | 选择	Reason | 原因
Temperature | 温度	450°C	Moderate temperature: forward reaction is exothermic, so lower T favours yield, but the vanadium(V) oxide catalyst needs sufficient heat to work effectively. 中等温度：正反应是放热的，所以低温有利于产率，但钒氧化物催化剂需要足够的热量才能有效工作。
Pressure | 压力	1–2 atm	Equilibrium already lies far right at low pressure — 3 moles → 2 moles. High pressure is unnecessary and uneconomical. 低压下平衡已远偏右——3摩尔→2摩尔。高压不必要且不经济。
Catalyst | 催化剂	V₂O₅	Vanadium(V) oxide provides an alternative pathway with lower activation energy. 五氧化二钒提供具有更低活化能的替代路径。

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📝 Kp: Equilibrium Constant in Terms of Partial Pressure | 分压平衡常数Kp

For gas-phase reactions, it’s often more convenient to use Kp instead of Kc. Kp is expressed in terms of partial pressures rather than concentrations.

对于气相反应，通常使用Kp比Kc更方便。Kp用分压而非浓度来表示。

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

混合物中气体A的分压为：

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

其中A的摩尔分数 = A的摩尔数 / 所有气体的总摩尔数。

For: aA(g) + bB(g) ⇌ cC(g) + dD(g)

Units of Kp: Unlike Kc where we usually omit units at A-Level, Kp units must be stated. For the Haber Process: Kp has units of atm⁻² (since 2 moles of product / 4 moles of reactant = pressure⁻²).

Kp的单位：与A-Level中通常省略单位的Kc不同，Kp必须注明单位。对于哈伯法：Kp的单位是atm⁻²（因为2摩尔产物/4摩尔反应物 = pressure⁻²）。

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🎯 Common Exam Questions | 常见考题

Q1: Explain why the yield of ammonia decreases as temperature increases in the Haber Process. [3 marks]

问：解释为什么哈伯法中氨的产率随温度升高而降低。[3分]

Model Answer | 标准答案：

1. The forward reaction is exothermic / ΔH is negative (1 mark)
   正反应是放热反应 / ΔH为负值（1分）
2. Increasing temperature shifts equilibrium in the endothermic direction (left) (1 mark)
   升高温度使平衡向吸热方向（左）移动（1分）
3. To absorb the added heat / oppose the change (Le Chatelier’s Principle) (1 mark)
   以吸收增加的热量 / 抵消变化（勒夏特列原理）（1分）

Q2: State and explain the effect of a catalyst on the equilibrium yield. [2 marks]

问：陈述并解释催化剂对平衡产率的影响。[2分]

Model Answer | 标准答案：

1. A catalyst has NO effect on the equilibrium yield / position (1 mark)
   催化剂对平衡产率/位置没有影响（1分）
2. It increases the rate of BOTH forward and reverse reactions equally (1 mark)
   它同等增加正向和逆向反应的速率（1分）

Q3: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), explain why increasing pressure has no effect on the equilibrium position. [2 marks]

问：对于反应H₂(g) + I₂(g) ⇌ 2HI(g)，解释为什么增加压力对平衡位置没有影响。[2分]

Model Answer | 标准答案：

1. There are equal numbers of moles of gas on both sides of the equation (2 moles ⇌ 2 moles) (1 mark)
   方程式两边气体摩尔数相同（2摩尔⇌2摩尔）（1分）
2. Therefore, changing pressure affects both forward and reverse reactions equally (1 mark)
   因此，改变压力同等影响正向和逆向反应（1分）

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✅ Summary: Le Chatelier’s Principle at a Glance | 总结：勒夏特列原理一览

Change | 变化	Shift Direction | 移动方向	Kc?	Key Point | 关键点
↑ Temperature (exo) | 升高温度（放热）	← Left | 左	Decreases | 减小	Heat is a product | 热是产物
↑ Temperature (endo) | 升高温度（吸热）	→ Right | 右	Increases | 增大	Heat is a reactant | 热是反应物
↑ Pressure (fewer moles →) | 升高压力（摩尔数减少→）	→ Right | 右	No change | 不变	Shifts to fewer moles | 向更少摩尔方向移动
↑ [Reactant] | 增加[反应物]	→ Right | 右	No change | 不变	Consumes added reactant | 消耗增加的反应物
↑ [Product] | 增加[产物]	← Left | 左	No change | 不变	Consumes added product | 消耗增加的产物
Catalyst | 催化剂	No effect | 无影响	No change | 不变	Only speeds up rate | 仅加速反应速率

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📚 Key Definitions to Memorise | 必须记住的关键定义

Dynamic Equilibrium | 动态平衡： The state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant. The reaction has not stopped — both forward and reverse reactions continue at equal rates.

可逆反应中的一种状态，在该状态下正反应速率等于逆反应速率，反应物和产物的浓度保持不变。反应并未停止——正向和逆向反应以相等的速率继续进行。

Le Chatelier’s Principle | 勒夏特列原理： If a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the position of equilibrium will shift to oppose (counteract) that change.

如果一个处于平衡状态的系统受到温度、压力或浓度的变化影响，平衡位置将移动以抵消（对抗）该变化。

Homogeneous Equilibrium | 均相平衡： An equilibrium where all reactants and products are in the same phase (e.g., all gases or all in aqueous solution).

所有反应物和产物处于同一相的平衡（例如全部为气体，或全部在水溶液中）。

Heterogeneous Equilibrium | 非均相平衡： An equilibrium where reactants and products are in different phases (e.g., CaCO₃(s) ⇌ CaO(s) + CO₂(g)).

反应物和产物处于不同相的平衡（例如CaCO₃(s) ⇌ CaO(s) + CO₂(g)）。

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Practice makes perfect! Try predicting the shift direction for various changes on different equilibria, and always remember: the system opposes the change, but Kc only changes with temperature.

熟能生巧！尝试预测不同平衡中各种变化引起的移动方向，并始终记住：系统对抗变化，但Kc只随温度变化而改变。