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

Introduction to Chemical Equilibrium | 化学平衡导论

Chemical equilibrium is one of the most conceptually rich topics in A-Level Chemistry. It bridges thermodynamics and kinetics, explaining why reactions don’t always go to completion and how we can control the outcome of reversible processes. For A-Level students (AQA, Edexcel, OCR, CIE), mastering equilibrium concepts — especially Le Chatelier’s Principle — is essential for high marks on both structured questions and practical assessments.

化学平衡是A-Level化学中概念最丰富的主题之一。它连接了热力学和动力学，解释了为什么反应不总是进行到底，以及我们如何控制可逆过程的结果。对于A-Level学生（AQA、Edexcel、OCR、CIE各考试局），掌握平衡概念——特别是勒夏特列原理——是在结构化问题和实验评估中获得高分的关键。

What is Dynamic Equilibrium? | 什么是动态平衡？

Many chemical reactions are reversible — the products can react to re-form the reactants. When the rate of the forward reaction equals the rate of the backward reaction in a closed system, the system has reached dynamic equilibrium. At this point, the concentrations of reactants and products remain constant, but the reactions continue at the molecular level.

许多化学反应是可逆的——产物可以反应重新生成反应物。在封闭系统中，当正反应速率等于逆反应速率时，系统达到了动态平衡。此时，反应物和产物的浓度保持不变，但在分子水平上反应仍在继续。

Key characteristics of dynamic equilibrium:

动态平衡的关键特征：

• The system must be closed — no matter enters or leaves. 系统必须是封闭的——没有物质进入或离开。
• Forward and backward reaction rates are equal. 正逆反应速率相等。
• Macroscopic properties (concentration, colour, pressure) remain constant. 宏观性质（浓度、颜色、压力）保持不变。
• Equilibrium can be approached from either direction. 平衡可以从任一方向达到。

Example — The Haber Process:

示例——哈伯法：

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

This is the classic A-Level equilibrium system. The double arrow (⇌) indicates reversibility. At equilibrium, N₂, H₂, and NH₃ are all present in constant amounts.

这是经典的A-Level平衡系统。双箭头（⇌）表示可逆性。在平衡状态下，N₂、H₂和NH₃都以恒定的量存在。

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

The equilibrium constant Kc quantifies the position of equilibrium. For a general reaction:

平衡常数Kc量化了平衡的位置。对于一般反应：

aA + bB ⇌ cC + dD

The expression is:

表达式为：

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

Where square brackets denote equilibrium concentrations in mol dm⁻³.

其中方括号表示平衡浓度，单位为mol dm⁻³。

What Kc Tells Us | Kc告诉我们什么

Kc Value	Position of Equilibrium	Interpretation
Kc >> 1 (e.g., 10¹⁰)	产品一侧 / Favours products	反应几乎进行到底 / Reaction nearly goes to completion
Kc ≈ 1	中间 / Intermediate	显著量的反应物和产物都存在 / Significant amounts of both present
Kc << 1 (e.g., 10⁻¹⁰)	反应物一侧 / Favours reactants	几乎不发生反应 / Very little reaction occurs

Factors Affecting Kc | 影响Kc的因素

Temperature ONLY changes the value of Kc. Concentration, pressure, and catalysts do NOT change Kc — they only affect how quickly equilibrium is reached.

只有温度会改变Kc的值。浓度、压力和催化剂不会改变Kc——它们只影响达到平衡的快慢。

For exothermic forward reactions (ΔH < 0): Increasing temperature decreases Kc (equilibrium shifts left).

对于放热正反应（ΔH < 0）：升高温度降低Kc（平衡向左移动）。

For endothermic forward reactions (ΔH > 0): Increasing temperature increases Kc (equilibrium shifts right).

对于吸热正反应（ΔH > 0）：升高温度增大Kc（平衡向右移动）。

Le Chatelier’s Principle | 勒夏特列原理

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 (oppose) the change.

勒夏特列原理指出：如果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（对抗）这种变化。

This principle is a qualitative predictive tool — it tells you the direction of shift, not the magnitude.

该原理是一个定性的预测工具——它告诉你移动的方向，而不是幅度。

1. Effect of Concentration | 浓度的影响

Rule: Adding a reactant or removing a product shifts equilibrium to the right (favours forward reaction). Adding a product or removing a reactant shifts equilibrium to the left.

规则：增加反应物或移除产物使平衡向右移动（有利于正反应）。增加产物或移除反应物使平衡向左移动。

Worked Example — Fe³⁺/SCN⁻ equilibrium:

例题——Fe³⁺/SCN⁻平衡：

Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq)
(yellow)   (colourless)    (blood-red)

Adding more Fe³⁺ ions shifts equilibrium right → solution turns darker red. Adding more SCN⁻ has the same effect. This is a classic colour-change demonstration used in A-Level practicals.

加入更多Fe³⁺离子使平衡右移→溶液变为更深的红色。加入更多SCN⁻有相同的效果。这是A-Level实验中经典的变色演示。

2. Effect of Pressure (Gaseous Systems) | 压力的影响（气体系统）

Rule: Increasing pressure shifts equilibrium to the side with fewer gas molecules (smaller total moles of gas). Decreasing pressure shifts to the side with more gas molecules.

规则：增加压力使平衡移向气体分子总数较少的一侧。降低压力则移向气体分子较多的一侧。

Worked Example — N₂O₄/NO₂ equilibrium:

例题——N₂O₄/NO₂平衡：

N₂O₄(g) ⇌ 2NO₂(g)
(colourless)  (brown)
1 mole gas    2 moles gas

Increasing pressure → equilibrium shifts left (fewer moles, 1 vs 2) → mixture becomes paler (more colourless N₂O₄). This is visually dramatic and frequently examined.

增加压力→平衡向左移动（分子数较少，1对2）→混合物颜色变浅（更多无色的N₂O₄）。这个现象视觉冲击力强，考试中经常出现。

IMPORTANT: Pressure changes only affect equilibria where there is a difference in the number of gas molecules on each side. For H₂(g) + I₂(g) ⇌ 2HI(g) — 2 moles on each side — pressure has no effect on the position of equilibrium.

重要：压力变化只影响两侧气体分子数不同的平衡。对于H₂(g) + I₂(g) ⇌ 2HI(g)——两侧各2摩尔——压力对平衡位置没有影响。

3. Effect of Temperature | 温度的影响

Rule: Increasing temperature shifts equilibrium in the endothermic direction (the direction that absorbs heat). Decreasing temperature shifts equilibrium in the exothermic direction.

规则：升高温度使平衡向吸热方向移动（吸收热量的方向）。降低温度使平衡向放热方向移动。

Worked Example — Contact Process (SO₂/SO₃):

例题——接触法（SO₂/SO₃）：

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

The forward reaction is exothermic. Increasing temperature shifts equilibrium left → less SO₃ produced. However, industrially, a compromise temperature of ~450°C is used with a V₂O₅ catalyst — low temperature gives good yield but is too slow; high temperature is fast but poor yield. The catalyst allows a moderate temperature with acceptable rate.

正反应是放热的。升高温度使平衡向左移动→产生的SO₃减少。然而，工业上使用约450°C的折中温度和V₂O₅催化剂——低温产率高但太慢；高温快但产率差。催化剂使得在适中温度下获得可接受的速率。

This is a perfect example of the rate vs. yield compromise that examiners love to test.

这是考官喜欢考查的速率与产率权衡的完美例子。

4. Effect of a Catalyst | 催化剂的影响

A catalyst provides an alternative reaction pathway with lower activation energy. It increases the rate of both forward and backward reactions equally. Therefore, a catalyst:

催化剂提供了活化能较低的替代反应路径。它同等地增加正反应和逆反应的速率。因此，催化剂：

• Does NOT change the position of equilibrium. 不改变平衡位置。
• Does NOT change the value of Kc. 不改变Kc的值。
• Only allows equilibrium to be reached faster. 只让平衡更快达到。

Exam tip: This is one of the most common trick questions. Students often incorrectly state that a catalyst shifts equilibrium toward products because it “speeds up the forward reaction.” Remember: it speeds up both directions equally.

考试提示：这是最常见的陷阱题之一。学生经常错误地认为催化剂使平衡移向产物，因为它”加速了正反应”。记住：它同等地加速两个方向。

Equilibrium in Industry | 工业中的平衡

Understanding equilibrium is crucial for industrial chemistry. The economic viability of many processes depends on optimising equilibrium conditions.

理解平衡对工业化学至关重要。许多工艺的经济可行性取决于优化平衡条件。

The Haber Process (NH₃ production) | 哈伯法（氨的生产）

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

Condition / 条件	Industrial Choice / 工业选择	Reasoning / 理由
Temperature	400–450°C	Compromise: exothermic forward reaction favours low T for yield, but rate too slow below 400°C. 折中：放热正反应倾向低温有利产率，但400°C以下速率太慢。
Pressure	200 atm	Forward reaction reduces moles (4→2); high P favours products. Above 200 atm, equipment cost outweighs benefit. 正反应减少分子数(4→2)；高压有利产物。超过200 atm，设备成本超过收益。
Catalyst	Iron (Fe)	Speeds up both directions equally; no effect on equilibrium position. 同等加速两个方向；不影响平衡位置。

The Contact Process (H₂SO₄ production) | 接触法（硫酸生产）

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

Condition / 条件	Industrial Choice / 工业选择	Reasoning / 理由
Temperature	450°C	Compromise: low T favours products (exothermic) but V₂O₅ catalyst needs ~450°C to be active. 折中：低温有利产物（放热），但V₂O₅催化剂需要约450°C才能活化。
Pressure	1–2 atm	Forward reaction reduces moles (3→2). Equilibrium already lies far right at 1 atm (Kc is large); higher P uneconomical. 正反应减少分子数(3→2)。1 atm下平衡已大幅偏右(Kc很大)；更高压力不经济。
Catalyst	V₂O₅ (vanadium(V) oxide)	Heterogeneous catalyst; provides surface for reaction. 多相催化剂；提供反应表面。

Common Exam Questions and Pitfalls | 常见考题与陷阱

Q1: “State and explain the effect of increasing temperature on the equilibrium yield of NH₃.”

问：”陈述并解释升高温度对NH₃平衡产率的影响。”

Model Answer / 标准答案： The yield of NH₃ decreases. The forward reaction is exothermic (ΔH = −92 kJ mol⁻¹). According to Le Chatelier’s Principle, increasing temperature shifts the equilibrium in the endothermic direction to absorb the added heat — this is the backward (left) direction, so less NH₃ is produced at equilibrium.

NH₃的产率降低。正反应是放热的（ΔH = −92 kJ mol⁻¹）。根据勒夏特列原理，升高温度使平衡向吸热方向移动以吸收增加的热量——这是逆（左）方向，因此在平衡时产生的NH₃减少。

Q2: “Why is a catalyst used in the Haber Process even though it does not affect the equilibrium yield?”

问：”为什么在哈伯法中使用催化剂，即使它不影响平衡产率？”

Model Answer / 标准答案： A catalyst increases the rate of both forward and backward reactions equally by providing an alternative pathway with lower activation energy. This allows equilibrium to be reached more quickly without affecting the position or yield. In industry, faster production = more economic.

催化剂通过提供活化能较低的替代路径，同等地增加正反应和逆反应的速率。这使得平衡更快达到，而不影响位置或产率。在工业中，更快的生产=更经济。

Q3: “Explain why changing pressure has no effect on H₂(g) + I₂(g) ⇌ 2HI(g).”

问：”解释为什么改变压力对H₂(g) + I₂(g) ⇌ 2HI(g)没有影响。”

Model Answer / 标准答案： There are 2 moles of gas on each side of the equation (1 H₂ + 1 I₂ = 2 moles reactants; 2 moles HI = 2 moles products). A pressure change affects both sides equally, so the position of equilibrium does not shift.

方程式两侧各有2摩尔气体（1 H₂ + 1 I₂ = 2摩尔反应物；2摩尔HI = 2摩尔产物）。压力变化对两侧影响相同，因此平衡位置不移动。

Common Student Mistakes | 学生常见错误

1. Confusing rate and equilibrium: “Catalyst shifts equilibrium to the right” — WRONG. Catalysts affect rate, not position. 混淆速率和平衡：“催化剂使平衡右移”——错误。催化剂影响速率，不影响位置。
2. Forgetting the closed system requirement: Equilibrium only applies in closed systems. An open bottle of fizzy drink loses CO₂ — not at equilibrium. 忘记封闭系统要求：平衡仅适用于封闭系统。打开的汽水瓶会失去CO₂——不处于平衡状态。
3. Kc changes with concentration: “Adding more reactant increases Kc” — WRONG. Only temperature changes Kc. Adding reactant shifts equilibrium but Kc stays the same. Kc随浓度变化：“增加反应物会增大Kc”——错误。只有温度改变Kc。增加反应物使平衡移动，但Kc保持不变。
4. Misapplying Le Chatelier: Stating the equilibrium “opposes the change by shifting to the opposite side” without specifying the direction or linking to ΔH. Be precise. 误用勒夏特列原理：说平衡”通过移向相反一侧来对抗变化”，但没有指明方向或联系ΔH。要精确。

Kp — Equilibrium Constant for Gaseous Systems | Kp——气体系统的平衡常数

For gas-phase reactions, we can use partial pressures instead of concentrations. The equilibrium constant Kp is defined in terms of partial pressures:

对于气相反应，我们可以使用分压代替浓度。平衡常数Kp用分压定义：

For aA(g) + bB(g) ⇌ cC(g) + dD(g):

Kp = (pC)^c × (pD)^d / (pA)^a × (pB)^b

Where pA, pB, pC, pD are the equilibrium partial pressures (in atm, Pa, or kPa).

其中pA、pB、pC、pD是平衡分压（单位为atm、Pa或kPa）。

Relationship between Kp and Kc:

Kp与Kc的关系：

Kp = Kc × (RT)^Δn

Where Δn = (moles of gaseous products) − (moles of gaseous reactants), R = gas constant (8.31 J K⁻¹ mol⁻¹), T = temperature in Kelvin.

其中Δn = （气体产物的摩尔数）− （气体反应物的摩尔数），R = 气体常数（8.31 J K⁻¹ mol⁻¹），T = 开尔文温度。

Summary Table | 总结表

Change / 变化	Equilibrium Shift / 平衡移动	Kc / Kp Change? / Kc/Kp变化？
Increase [reactant] / 增加[反应物]	Right / 右移	No / 不变
Increase [product] / 增加[产物]	Left / 左移	No / 不变
Increase P (fewer gas moles on right) / 增压（右侧气体分子少）	Right / 右移	No / 不变
Increase P (fewer gas moles on left) / 增压（左侧气体分子少）	Left / 左移	No / 不变
Increase T (exothermic forward) / 升温（正反应放热）	Left / 左移	Decreases Kc / Kc减小
Increase T (endothermic forward) / 升温（正反应吸热）	Right / 右移	Increases Kc / Kc增大
Add catalyst / 加催化剂	No shift / 不移动	No / 不变

Practice Question | 练习题

Question: Consider the equilibrium: 2NO₂(g) ⇌ N₂O₄(g), ΔH = −58 kJ mol⁻¹. NO₂ is brown; N₂O₄ is colourless. A sealed syringe containing an equilibrium mixture of these gases at room temperature appears pale brown. Predict and explain what you would observe when:

问题：考虑平衡：2NO₂(g) ⇌ N₂O₄(g)，ΔH = −58 kJ mol⁻¹。NO₂是棕色的；N₂O₄是无色的。一个装有室温下这些气体平衡混合物的密封注射器呈浅棕色。预测并解释在以下情况下你会观察到什么：

(a) The plunger is pushed in rapidly (increasing pressure).
(a) 快速推入活塞（增加压力）。

(b) The syringe is placed in hot water.
(b) 将注射器放入热水中。

Answers: (a) The mixture initially darkens (compression increases [NO₂] temporarily), then pales as equilibrium shifts right (2 moles → 1 mole, fewer gas molecules) to reduce pressure. (b) The mixture darkens (turns browner) as equilibrium shifts left (endothermic direction) to absorb the added heat, producing more brown NO₂.

答案：(a) 混合物最初变深（压缩暂时增加[NO₂]），然后变浅，因为平衡向右移动（2摩尔→1摩尔，气体分子较少）以降低压力。(b) 混合物变深（变得更棕），因为平衡向左移动（吸热方向）以吸收增加的热量，产生更多棕色NO₂。

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Keywords: A-Level Chemistry, Chemical Equilibrium, Le Chatelier’s Principle, Kc, Kp, Haber Process, Contact Process, Dynamic Equilibrium, Reversible Reactions, Exam Tips

关键词：A-Level化学、化学平衡、勒夏特列原理、Kc、Kp、哈伯法、接触法、动态平衡、可逆反应、考试技巧