CIE A-Level Chemistry: Chemical Equilibrium & Le Chatelier’s Principle

Chemical equilibrium is one of the most conceptually rich topics in CIE A-Level Chemistry (9701). Understanding how reversible reactions reach a dynamic balance and how that balance responds to external stresses is essential for mastering Paper 4 and the practical exam. This article provides a thorough walkthrough of equilibrium theory, Le Chatelier’s principle, the equilibrium constant Kc and Kp, and the Haber and Contact processes — with fully worked examples.

化学平衡是CIE A-Level化学(9701)中概念最丰富的主题之一。 理解可逆反应如何达到动态平衡以及这种平衡如何响应外部压力，对于掌握Paper 4和实验考试至关重要。本文全面讲解平衡理论、勒夏特列原理、平衡常数Kc和Kp，以及哈伯法和接触法工艺——配有完整解题示例。

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

A reversible reaction is one in which the products can react to reform the original reactants. When the rate of the forward reaction equals the rate of the reverse reaction, the system has reached dynamic equilibrium. At this point, the concentrations of all species remain constant — but the reactions have not stopped. Both forward and reverse reactions continue at equal rates.

可逆反应是指产物可以重新反应生成原始反应物的反应。当正反应速率等于逆反应速率时，系统达到动态平衡。此时，所有物质的浓度保持恒定——但反应并未停止。正逆反应以相等的速率继续进行。

Dynamic equilibrium can only be established in a closed system — one where no matter enters or leaves. If gases escape or reactants are added midway, the system is no longer closed and equilibrium cannot be maintained. This is a common exam trap: students often forget to specify “closed system” when defining equilibrium.

动态平衡只能在封闭系统中建立——即没有物质进入或离开的系统。如果气体逸出或在过程中添加反应物，系统就不再是封闭的，平衡无法维持。这是常见的考试陷阱：学生在定义平衡时经常忘记指定”封闭系统”。

Key characteristics of a system at equilibrium:

平衡系统的主要特征：

• Macroscopic properties are constant — no visible change in colour, pressure, or concentration. / 宏观性质恒定——颜色、压强或浓度没有可见变化。
• Rate forward = Rate reverse — the defining criterion of equilibrium. / 正反应速率 = 逆反应速率——平衡的定义标准。
• Requires a closed system — no exchange of matter with surroundings. / 需要封闭系统——与环境没有物质交换。
• Can be approached from either direction — starting with reactants or products yields the same equilibrium mixture. / 可从任一方向达到——从反应物或产物开始都会得到相同的平衡混合物。

2. Le Chatelier’s Principle / 勒夏特列原理

Le Chatelier’s principle states: If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium shifts to counteract that change. This is not just a qualitative rule — it is a powerful predictive tool for industrial chemistry and exam problem-solving.

勒夏特列原理指出：如果处于平衡状态的系统受到浓度、压强或温度的变化，平衡位置会移动以抵消该变化。 这不仅是一个定性规则——它是工业化学和考试解题的强大预测工具。

2.1 Effect of Concentration Changes / 浓度变化的影响

When the concentration of a reactant is increased, the equilibrium shifts to the right (towards products) to consume the added reactant. Conversely, increasing a product concentration shifts equilibrium to the left. Removing a species causes the equilibrium to shift towards the side that produces more of that species.

当反应物浓度增加时，平衡向右移动（朝向产物）以消耗增加的反应物。相反，增加产物浓度会使平衡向左移动。移除某种物质会导致平衡向产生更多该物质的方向移动。

Consider the esterification reaction as a classic example:

以酯化反应作为经典例子：

CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

• Adding more ethanol (C2H5OH) shifts equilibrium right → more ester produced. / 添加更多乙醇使平衡向右移动→生成更多酯。
• Removing water (using a drying agent) also shifts equilibrium right. / 移除水（使用干燥剂）也使平衡向右移动。
• Adding more ethyl ethanoate shifts equilibrium left → more reactants formed. / 添加更多乙酸乙酯使平衡向左移动→生成更多反应物。

Exam tip: Catalysts have NO EFFECT on the position of equilibrium. They only increase the rate at which equilibrium is reached by lowering the activation energy for both forward and reverse reactions equally. This is one of the most frequently tested concepts in CIE papers.

考试提示：催化剂对平衡位置没有影响。它们只是通过同等地降低正逆反应的活化能来加快达到平衡的速率。这是CIE试卷中最常考的概念之一。

2.2 Effect of Pressure Changes / 压强变化的影响

Pressure changes only affect equilibria involving gases where there is a difference in the number of moles of gas on each side of the equation. Increasing pressure shifts equilibrium towards the side with fewer gas molecules (lower volume), while decreasing pressure favours the side with more gas molecules.

压强变化只影响涉及气体且方程式两侧气体摩尔数不同的平衡。增加压强使平衡向气体分子较少（体积较小）的一侧移动，而降低压强有利于气体分子较多的一侧。

Example — the Haber process:

示例——哈伯法：

N2(g) + 3H2(g) ⇌ 2NH3(g)

• Left side: 1 + 3 = 4 moles of gas / 左侧：1 + 3 = 4摩尔气体
• Right side: 2 moles of gas / 右侧：2摩尔气体
• Increasing pressure shifts equilibrium → right (fewer gas molecules) / 增加压强使平衡→向右移动（气体分子较少）
• Decreasing pressure shifts equilibrium → left / 降低压强使平衡→向左移动

If the number of gas moles is the same on both sides (e.g., H2 + I2 ⇌ 2HI), pressure changes have no effect on the position of equilibrium, though they do increase the rate at which equilibrium is reached.

如果两侧气体摩尔数相同（例如H2 + I2 ⇌ 2HI），压强变化对平衡位置没有影响，尽管它们确实加快了达到平衡的速率。

2.3 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 or Kp). Increasing temperature favours the endothermic direction, while decreasing temperature favours the exothermic direction.

温度是唯一改变平衡常数值（Kc或Kp）的因素。升高温度有利于吸热方向，而降低温度有利于放热方向。

For an exothermic forward reaction (ΔH negative):

对于放热正反应（ΔH为负）：

• Increasing temperature: equilibrium shifts left (endothermic reverse direction), Kc/Kp decreases. / 升高温度：平衡向左移动（吸热的逆反应方向），Kc/Kp减小。
• Decreasing temperature: equilibrium shifts right, Kc/Kp increases. / 降低温度：平衡向右移动，Kc/Kp增大。

For an endothermic forward reaction (ΔH positive):

对于吸热正反应（ΔH为正）：

• Increasing temperature: equilibrium shifts right, Kc/Kp increases. / 升高温度：平衡向右移动，Kc/Kp增大。
• Decreasing temperature: equilibrium shifts left, Kc/Kp decreases. / 降低温度：平衡向左移动，Kc/Kp减小。

3. The Equilibrium Constant (Kc and Kp) / 平衡常数（Kc和Kp）

3.1 Kc — Equilibrium Constant in Terms of Concentration / Kc——基于浓度的平衡常数

For a general reaction: aA + bB ⇌ cC + dD

对于一般反应：aA + bB ⇌ cC + dD

The equilibrium constant Kc is expressed as:

平衡常数Kc表示为：

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

Where square brackets denote equilibrium concentrations in mol/dm³. The stoichiometric coefficients (a, b, c, d) become the powers in the expression.

其中方括号表示以mol/dm³为单位的平衡浓度。化学计量系数（a, b, c, d）成为表达式中的幂。

Important rules for Kc:

Kc的重要规则：

• Solids and pure liquids do NOT appear in the Kc expression — their concentrations are effectively constant. / 固体和纯液体不出现在Kc表达式中——它们的浓度实际上是常数。
• Water does NOT appear when it is the solvent (very large excess). / 水不出现当它是溶剂时（大量过量）。
• The units of Kc depend on the stoichiometry and must be calculated for each reaction. / Kc的单位取决于化学计量学，必须为每个反应计算。
• Kc is temperature-dependent only — concentration, pressure, and catalysts do not change Kc. / Kc只依赖于温度——浓度、压强和催化剂不改变Kc。

3.2 Kp — Equilibrium Constant in Terms of Partial Pressure / Kp——基于分压的平衡常数

For gas-phase reactions, Kp uses partial pressures instead of concentrations. The partial pressure of a gas (pA) is given by:

对于气相反应，Kp使用分压而不是浓度。气体的分压（pA）由下式给出：

pA = (moles of A / total moles) × total pressure

广告 / Sponsored

The Kp expression mirrors Kc but uses partial pressures:

Kp表达式与Kc类似，但使用分压：

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

The units of Kp are typically atm^Δn where Δn is the change in moles of gas (products minus reactants).

Kp的单位通常是atm^Δn，其中Δn是气体摩尔数的变化（产物减反应物）。

3.3 Worked Example: Calculating Kc / 解题示例：计算Kc

Question: 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed and allowed to reach equilibrium at 298 K. At equilibrium, 0.20 mol of ethanoic acid remains. The total volume is 1.0 dm³. Calculate Kc for the esterification reaction:

题目：将0.50 mol乙酸和0.50 mol乙醇混合，在298 K下达到平衡。平衡时，剩余0.20 mol乙酸。总体积为1.0 dm³。计算酯化反应的Kc：

CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

Step 1 — Set up the ICE table (Initial, Change, Equilibrium):

第1步——建立ICE表格（初始、变化、平衡）：

• Initial: [CH3COOH] = 0.50, [C2H5OH] = 0.50, [CH3COOC2H5] = 0, [H2O] = 0 / 初始
• Change: Reacted = 0.50 – 0.20 = 0.30 mol of CH3COOH consumed / 变化：消耗了0.30 mol CH3COOH
• Equilibrium: [CH3COOH] = 0.20/1.0 = 0.20 mol/dm³ / 平衡
• Equilibrium: [C2H5OH] = (0.50-0.30)/1.0 = 0.20 mol/dm³ / 平衡
• Equilibrium: [CH3COOC2H5] = 0.30/1.0 = 0.30 mol/dm³ / 平衡
• Equilibrium: [H2O] = 0.30/1.0 = 0.30 mol/dm³ / 平衡

Step 2 — Write the Kc expression:

第2步——写出Kc表达式：

Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]

Step 3 — Substitute and calculate:

第3步——代入并计算：

Kc = (0.30 × 0.30) / (0.20 × 0.20) = 0.09 / 0.04 = 2.25

The units cancel completely (same number of concentration terms top and bottom), so Kc is dimensionless in this case.

单位完全抵消（分子和分母中浓度项的数量相同），因此在这种情况下Kc是无量纲的。

4. Industrial Applications / 工业应用

4.1 The Haber Process / 哈伯法

The Haber process produces ammonia (NH3) from nitrogen and hydrogen:

哈伯法从氮气和氢气生产氨（NH3）：

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

The forward reaction is exothermic (ΔH negative) and produces fewer gas molecules (4 mol → 2 mol). Industry uses a compromise between rate and yield:

正反应是放热的（ΔH为负）并产生较少的气体分子（4 mol → 2 mol）。工业在速率和产率之间采用折中方案：

• Temperature: 400-450°C — Low temperature favours higher yield (exothermic forward reaction favoured), but the rate would be too slow. The compromise temperature gives a reasonable rate with acceptable yield. / 温度：400-450°C——低温有利于较高产率（放热正反应被促进），但速率太慢。折中温度给出了合理速率和可接受的产率。
• Pressure: 200 atm — High pressure favours the side with fewer gas molecules (product side, 2 mol vs 4 mol). Higher pressures give higher yields but require more expensive equipment. / 压强：200 atm——高压有利于气体分子较少的一侧（产物侧，2 mol对4 mol）。更高压强给出更高产率但需要更昂贵的设备。
• Catalyst: Finely divided iron — Speeds up the attainment of equilibrium without affecting the position or the yield. / 催化剂：细碎铁粉——加速达到平衡而不影响平衡位置或产率。

4.2 The Contact Process / 接触法

The Contact process produces sulfuric acid via the oxidation of SO2 to SO3:

接触法通过SO2氧化为SO3来生产硫酸：

2SO2(g) + O2(g) ⇌ 2SO3(g)     ΔH = -197 kJ/mol

Again, the forward reaction is exothermic and reduces the number of gas molecules (3 mol → 2 mol). Industrial conditions:

同样，正反应是放热的并减少气体分子数（3 mol → 2 mol）。工业条件：

• Temperature: 450°C — Compromise between rate and equilibrium yield. / 温度：450°C——速率和平衡产率之间的折中。
• Pressure: 1-2 atm — Surprisingly low! At this temperature, the equilibrium already lies far to the right (Kp is very large), so high pressure is unnecessary and would add cost. / 压强：1-2 atm——出人意料地低！在此温度下，平衡已经大幅偏右（Kp非常大），因此高压是不必要的且会增加成本。
• Catalyst: Vanadium(V) oxide (V2O5) — Heterogeneous catalyst that provides an alternative pathway with lower activation energy. / 催化剂：五氧化二钒(V2O5)——提供较低活化能的替代路径的多相催化剂。

5. Common Exam Mistakes and Pitfalls / 常见考试错误和陷阱

After marking hundreds of CIE Chemistry scripts, here are the most frequent errors students make on equilibrium questions:

在批改数百份CIE化学试卷后，以下是学生在平衡问题上最常犯的错误：

1. Forgetting to state “closed system” when defining dynamic equilibrium. Marks are routinely lost for this omission. / 定义动态平衡时忘记说明”封闭系统”。这个遗漏经常导致失分。
2. Claiming that a catalyst increases yield — it does not. A catalyst only increases the rate of attainment of equilibrium; it has zero effect on the equilibrium position or the value of Kc/Kp. / 声称催化剂增加产率——它不会。催化剂只增加达到平衡的速率；它对平衡位置或Kc/Kp的值没有影响。
3. Confusing rate and equilibrium — Increasing temperature always increases rate, but its effect on equilibrium position depends on whether the reaction is endothermic or exothermic. / 混淆速率和平衡——升高温度总是增加速率，但它对平衡位置的影响取决于反应是吸热还是放热。
4. Incorrect Kc units — Always work out the units from the expression. Do not assume Kc is dimensionless. / Kc单位错误——始终从表达式推导单位。不要假设Kc是无量纲的。
5. Including solids or pure liquids in Kc/Kp — These have constant concentration/activity and are omitted from the expression. / 在Kc/Kp中包含固体或纯液体——它们具有恒定的浓度/活度，应从表达式中省略。
6. Using initial concentrations instead of equilibrium concentrations — Kc/Kp is calculated using concentrations at equilibrium, not the starting amounts. / 使用初始浓度而不是平衡浓度——Kc/Kp使用平衡时的浓度计算，而不是起始量。
7. Confusing the sign of ΔH — For exothermic reactions (ΔH < 0), increasing temperature decreases Kc. The opposite is true for endothermic reactions. / 混淆ΔH的符号——对于放热反应（ΔH < 0），升高温度降低Kc。吸热反应则相反。

6. Practice Questions / 练习题

Test your understanding with these CIE-style questions:

用这些CIE风格的题目测试你的理解：

Q1. For the reaction 2NO2(g) ⇌ plain the effect on the equilibrium position of: (a) increasing pressure, (b) increasing temperature, (c) adding a catalyst.

Q1. 对于反应2NO2(g) ⇌ N2O4(g)，ΔH = -57 kJ/mol。说明并解释以下操作对平衡位置的影响：(a)增加压强，(b)升高温度，(c)添加催化剂。

Q2. At 500 K, 1.0 mol of PCl5 is placed in a 2.0 dm³ container. At equilibrium, 0.60 mol of PCl5 has decomposed according to: PCl5(g) ⇌ PCl3(g) + Cl2(g). Calculate Kc and state its units.

Q2. 在500 K下，将1.0 mol PCl5置于2.0 dm³容器中。平衡时，0.60 mol PCl5已按以下反应分解：PCl5(g) ⇌ PCl3(g) + Cl2(g)。计算Kc并说明其单位。

Q3. The Contact process uses a temperature of 450°C and near-atmospheric pressure. Explain why these conditions are chosen, referring to both equilibrium and rate considerations.

Q3. 接触法使用450°C的温度和接近大气压的压强。解释为什么选择这些条件，参考平衡和速率两方面的考虑。

7. Summary / 总结

Chemical equilibrium is the bridge between thermodynamic feasibility and kinetic reality. Mastering this topic requires not just memorising Le Chatelier’s principle, but understanding the quantitative framework of Kc and Kp, and being able to apply it in unfamiliar contexts — exactly what CIE examiners test. The key takeaways:

化学平衡是热力学可行性和动力学现实之间的桥梁。掌握这个主题不仅需要记住勒夏特列原理，还需要理解Kc和Kp的定量框架，并能够在陌生情境中应用它——这正是CIE考官所测试的。关键要点：

• Dynamic equilibrium requires a closed system, equal forward/reverse rates, and constant macroscopic properties. / 动态平衡需要封闭系统、相等的正逆反应速率和恒定的宏观性质。
• Le Chatelier’s principle predicts the direction of shift but does not explain why — link your answer to rates or Kc values for full marks. / 勒夏特列原理预测移动方向但不解释原因——将你的答案与速率或Kc值联系起来以获得满分。
• Only temperature changes the value of Kc or Kp. / 只有温度改变Kc或Kp的值。
• Catalysts affect rate, not position — this is examined relentlessly. / 催化剂影响速率，不影响位置——这一点被反复考查。
• Industrial processes use compromise conditions balancing rate, yield, safety, and cost. / 工业过程使用平衡速率、产率、安全性和成本的折中条件。
• Always derive Kc/Kp units from the balanced equation — never assume. / 始终从配平方程式推导Kc/Kp单位——永远不要假设。

更多咨询请联系16621398022（同微信）