Chemical Equilibrium: Le Chatelier’s Principle, Kc & the Haber Process | A-Level 化学：化学平衡完全指南

Introduction | 引言

Chemical equilibrium is one of the most conceptually rich topics in A-Level Chemistry. Unlike reactions that “go to completion,” many important chemical processes are reversible — they can proceed in both the forward and backward directions simultaneously. When the rates of the forward and reverse reactions become equal, the system reaches a state of dynamic equilibrium. At this point, the concentrations of reactants and products remain constant, even though both reactions continue at the molecular level.

化学平衡是A-Level化学中概念最丰富的主题之一。与那些”进行到底”的反应不同，许多重要的化学过程是可逆的——它们可以同时向正反两个方向进行。当正反应和逆反应的速率相等时，系统达到动态平衡状态。此时，虽然两种反应仍在分子层面持续进行，但反应物和产物的浓度保持不变。

Key Concepts | 核心概念

1. Reversible Reactions and Dynamic Equilibrium | 可逆反应和动态平衡

A reversible reaction is one where the products can react to reform the original reactants. The symbol ⇌ is used to indicate reversibility. For example:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

This is the Haber process — one of the most important industrial applications of equilibrium chemistry. At equilibrium, the rate of ammonia production (forward reaction) equals the rate of ammonia decomposition (reverse reaction). Importantly, equilibrium does NOT mean the concentrations of reactants and products are equal — only that they are constant.

可逆反应是指产物可以重新反应生成原始反应物的反应。符号 ⇌ 用于表示可逆性。以上是哈伯法合成氨的反应——这是平衡化学最重要的工业应用之一。在平衡状态下，氨的生成速率（正反应）等于氨的分解速率（逆反应）。重要的是，平衡并不意味着反应物和产物的浓度相等——仅仅意味着它们是恒定的。

2. 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 will shift to counteract that change. This principle allows chemists to predict and control the outcome of reversible reactions by manipulating three key variables:

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，平衡位置将发生移动以抵消这种变化。该原理使化学家能够通过操控三个关键变量来预测和控制可逆反应的结果：

a) Effect of Concentration | 浓度的影响

If the concentration of a reactant is increased, the equilibrium shifts to the right (product side) to consume the added reactant. Conversely, if a product is removed, the equilibrium also shifts right to produce more product. This is exploited industrially — in the Haber process, ammonia is continuously removed as it liquefies, pulling the equilibrium towards more product.

如果增加反应物的浓度，平衡会向右移动（产物方向）以消耗添加的反应物。反之，如果移除产物，平衡也会向右移动以生成更多产物。这在工业中得到了应用——在哈伯法中，氨通过液化被持续移出，使平衡向生成更多产物的方向移动。

b) Effect of Pressure | 压强的影响

Changes in pressure only affect equilibria involving gases where there is a difference in the number of moles between reactants and products. Looking at the Haber process again: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), there are 4 moles of gas on the left and 2 moles on the right. Increasing pressure shifts equilibrium to the side with fewer gas molecules — in this case, the right, favouring ammonia production. This is why the Haber process operates at high pressure (typically 200 atm).

压强的变化只影响涉及气体且反应物和产物气体分子数不同的平衡。再看哈伯法：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，左侧有4摩尔气体，右侧有2摩尔气体。增加压强会使平衡向气体分子数较少的方向移动——即向右，有利于氨的生成。这就是哈伯法在高压（通常200 atm）下运行的原因。

c) Effect of Temperature | 温度的影响

Temperature is the only factor that changes the value of the equilibrium constant (Kc). An exothermic reaction (ΔH negative) releases heat, so increasing temperature shifts equilibrium to the left (endothermic direction). An endothermic reaction (ΔH positive) absorbs heat, so increasing temperature shifts equilibrium to the right. For the Haber process (ΔH = −92 kJ mol⁻¹, exothermic), higher temperatures reduce the equilibrium yield of ammonia. However, industry uses a moderate temperature (450°C) as a compromise — lower temperatures would give higher equilibrium yield but the reaction would be too slow.

温度是唯一能改变平衡常数(Kc)数值的因素。放热反应（ΔH为负）释放热量，升高温度使平衡向左移动（吸热方向）；吸热反应（ΔH为正）吸收热量，升高温度使平衡向右移动。对于哈伯法（ΔH = −92 kJ mol⁻¹，放热），较高的温度会降低氨的平衡产率。然而，工业上使用中等温度（450°C）作为折中方案——较低的温度虽然能提高平衡产率，但反应速率会过慢。

This trade-off between equilibrium yield and reaction rate is a classic A-Level exam question. Students must understand that equilibrium tells us about the maximum possible yield under given conditions, while kinetics tells us about the rate at which that yield is approached.

平衡产率与反应速率之间的这种权衡是A-Level考试的经典题目。学生必须理解，平衡告诉我们给定条件下的最大可能产率，而动力学告诉我们接近该产率的速率。

3. The Equilibrium Constant, Kc | 平衡常数 Kc

For a general reversible reaction:

aA + bB ⇌ cC + dD

The equilibrium constant expression is:

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

Where square brackets denote equilibrium concentrations in mol dm⁻³. The key rules for Kc are:

• Kc is temperature-dependent. For exothermic forward reactions, Kc decreases with increasing temperature. For endothermic forward reactions, Kc increases with increasing temperature.
• Solids and pure liquids are omitted from the Kc expression because their concentrations are essentially constant.
• Kc has no units if the total number of moles of reactants equals the total number of moles of products in the expression. Otherwise, the units depend on the stoichiometry.
• The magnitude of Kc indicates the position of equilibrium: Kc >> 1 means the equilibrium lies to the right (products favoured); Kc << 1 means equilibrium lies to the left (reactants favoured).

其关键规则为：Kc依赖于温度——对于放热正反应，Kc随温度升高而减小；对于吸热正反应，Kc随温度升高而增大。固体和纯液体从Kc表达式中省略，因为它们的浓度基本恒定。如果表达式中反应物和产物的总摩尔数相等，Kc没有单位，否则单位取决于化学计量比。Kc的大小表明平衡位置：Kc远大于1表示平衡偏右（产物为主）；Kc远小于1表示平衡偏左（反应物为主）。

4. Calculating Kc — Worked Example | Kc 计算示例

Question: 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed in a 2.0 dm³ flask at 298 K. At equilibrium, 0.30 mol of ethyl ethanoate is formed. Calculate Kc for:

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

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)	−0.30	−0.30	+0.30	+0.30
Equilibrium (mol)	0.20	0.20	0.30	0.30

Step 2 — Convert to concentrations (volume = 2.0 dm³):

[CH₃COOH] = 0.20/2.0 = 0.10 mol dm⁻³
[C₂H₅OH] = 0.20/2.0 = 0.10 mol dm⁻³
[CH₃COOC₂H₅] = 0.30/2.0 = 0.15 mol dm⁻³
[H₂O] = 0.30/2.0 = 0.15 mol dm⁻³

Step 3 — Substitute into Kc expression:

Kc = (0.15 × 0.15) / (0.10 × 0.10) = 0.0225 / 0.01 = 2.25

Since there are 2 moles on each side of the equation, Kc has no units in this case. The value of 2.25 (greater than 1) confirms that the equilibrium lies slightly to the right at this temperature.

由于方程式两边各有2摩尔，Kc在此情况下没有单位。2.25的值（大于1）确认在该温度下平衡略微偏右。

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

The Haber process synthesises ammonia from nitrogen and hydrogen:

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

The chosen industrial conditions represent a careful optimisation:

• Pressure: 200 atm — High pressure favours the forward reaction (4 moles → 2 moles), increasing yield. Beyond 200 atm, the additional cost of stronger equipment outweighs the yield benefit.
• Temperature: 450°C — A compromise. Lower temperatures favour the exothermic forward reaction, giving higher equilibrium yield, but the reaction rate becomes impractically slow. 450°C provides an acceptable rate with reasonable yield.
• Catalyst: Iron — The iron catalyst speeds up both forward and reverse reactions equally, so it does NOT affect the position of equilibrium or the value of Kc. It simply allows equilibrium to be reached faster, reducing energy costs. This is a common exam misconception — catalysts never affect equilibrium position, only the rate at which equilibrium is achieved.

所选的工业条件代表了精心的优化：压强200 atm——高压有利于正反应（4摩尔→2摩尔），提高产率。超过200 atm时，更坚固设备的额外成本超过了产率收益。温度450°C——这是一个折中方案。较低温度有利于放热正反应，产生更高的平衡产率，但反应速率会变得不切实际地慢。450°C提供了可接受的速率和合理的产率。铁催化剂——铁催化剂同等加速正反应和逆反应，因此它不会影响平衡位置或Kc的值。它只是使平衡更快达到，降低能源成本。这是常见的考试误区——催化剂从不影响平衡位置，只影响达到平衡的速率。

Common Exam Pitfalls | 常见考试陷阱

Pitfall 1: Confusing Rate and Equilibrium | 陷阱一：混淆速率和平衡

Students often write that a catalyst “increases the yield” or “shifts equilibrium to the right.” This is incorrect. A catalyst provides an alternative pathway with a lower activation energy, increasing the rate of BOTH forward and reverse reactions equally. The position of equilibrium and the value of Kc are unchanged. Only the rate at which equilibrium is established changes.

学生常写催化剂”提高产率”或”使平衡向右移动”。这是错误的。催化剂提供了一条活化能较低的替代路径，同等地增加了正反应和逆反应的速率。平衡位置和Kc的值保持不变，只有建立平衡的速率发生了变化。

Pitfall 2: Pressure Changes with Equal Moles | 陷阱二：等摩尔数时的压强变化

If a gaseous reaction has equal numbers of moles on both sides (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)), changing pressure has no effect on the position of equilibrium. This is a favourite exam trick. Always count the moles before predicting a pressure effect.

如果气体反应两边摩尔数相等（如H₂(g) + I₂(g) ⇌ 2HI(g)），改变压强对平衡位置没有影响。这是考试中常见的陷阱。在预测压强效应之前，务必先数清楚摩尔数。

Pitfall 3: Adding a Solid or Pure Liquid | 陷阱三：添加固体或纯液体

Adding more solid reactant (e.g., CaCO₃ in thermal decomposition) does NOT shift the equilibrium. Solids have constant concentration and do not appear in the Kc expression. Similarly, adding more water to an aqueous equilibrium where water is the solvent does not shift the equilibrium — water’s concentration as the solvent is effectively constant.

添加更多固体反应物（如热分解中的CaCO₃）不会移动平衡。固体浓度恒定，不出现在Kc表达式中。类似地，在水作为溶剂的溶液平衡中添加更多水不会移动平衡——作为溶剂，水的浓度实际上是恒定的。

Pitfall 4: Forgetting to Divide by Volume | 陷阱四：忘记除以体积

In Kc calculations, the values in the ICE table are in moles, but Kc requires concentrations (mol dm⁻³). Students frequently substitute mole values directly into the Kc expression without dividing by the volume. Always check: are these moles or concentrations?

在Kc计算中，ICE表中的值是摩尔，但Kc需要的是浓度（mol dm⁻³）。学生经常直接将摩尔值代入Kc表达式而没有除以体积。务必检查：这些是摩尔数还是浓度？

Exam Technique: Structuring Equilibrium Questions | 考试技巧：结构化回答平衡问题

For 6-mark questions on Le Chatelier’s Principle, use this structure:

1. State the principle: “According to Le Chatelier’s Principle, if a system at equilibrium is disturbed, the equilibrium shifts to oppose the change.”
2. Identify the change: State what variable is being altered (concentration / pressure / temperature).
3. Apply mole counts: For pressure changes, count gas moles on each side and state which side has fewer/more.
4. Predict the shift: “Therefore, the equilibrium shifts to the left/right…”
5. State the outcome: “…increasing/decreasing the yield of [product].”
6. Include ΔH context (for temperature): State whether the forward reaction is exothermic or endothermic and explain the direction of shift accordingly.

对于勒夏特列原理的6分题目，使用以下结构：陈述原理——”根据勒夏特列原理，如果平衡系统受到干扰，平衡将移动以对抗这种变化。” 识别变化——说明哪个变量正在改变。应用摩尔数——对于压强变化，计算每边的气体摩尔数。预测移动方向——”因此，平衡向左/右移动…” 陈述结果——”…增加/减少[产物]的产率。” 包含ΔH背景——说明正反应是放热还是吸热，并据此解释移动方向。

Summary | 总结

Chemical equilibrium is a cornerstone of A-Level Chemistry that bridges thermodynamics and kinetics. The key takeaways are: (1) Dynamic equilibrium means equal rates, not equal concentrations. (2) Le Chatelier’s Principle predicts the direction of shift when conditions change, but does not explain the rate at which equilibrium is re-established. (3) Kc is a temperature-dependent constant that quantifies the equilibrium position. (4) Catalysts affect the rate of reaching equilibrium, not the position itself. (5) Industrial processes like the Haber process are real-world applications where chemists optimise conditions by balancing yield, rate, and cost.

化学平衡是A-Level化学的基石，连接了热力学和动力学。关键要点是：(1) 动态平衡意味着速率相等，而不是浓度相等。(2) 勒夏特列原理预测条件变化时平衡移动的方向，但不解释重建平衡的速率。(3) Kc是一个依赖于温度的常数，量化了平衡位置。(4) 催化剂影响达到平衡的速率，而非平衡位置本身。(5) 像哈伯法这样的工业流程，是化学家在产率、速率和成本之间优化条件的实际应用。

Mastering equilibrium requires practice — work through as many Kc calculations as you can, and always draw out an ICE table before attempting any equilibrium problem. Remember: moles first, then concentrations, then Kc.

掌握平衡需要练习——尽可能多地练习Kc计算，在尝试任何平衡问题之前，务必先画出ICE表格。记住：先算摩尔数，再转换为浓度，最后计算Kc。