Chemical Equilibrium: Kc, Kp and Le Chatelier’s Principle — 化学平衡：Kc、Kp与勒夏特列原理

📚 Chemical Equilibrium | 化学平衡

Chemical equilibrium is one of the most conceptually rich topics in A-Level Chemistry. It bridges thermodynamics and kinetics, explaining why some reactions never go to completion and how industrial chemists maximise yield. In this comprehensive guide, we will explore reversible reactions, the equilibrium constant (Kc and Kp), Le Chatelier’s Principle, and the factors that shift equilibrium position — all with worked examples and exam-style commentary.

化学平衡是A-Level化学中最具概念深度的主题之一。它连接了热力学和动力学，解释了为什么某些反应永远无法进行到底，以及工业化学家如何最大化产率。在本指南中，我们将探讨可逆反应、平衡常数（Kc和Kp）、勒夏特列原理以及影响平衡位置的各种因素——全部配有例题和考试风格的分析。

1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡

Many chemical reactions are reversible — the products can react together to re-form the original reactants. A reversible reaction is denoted by the ⇌ symbol. When a reversible reaction is carried out in a closed system, the forward and reverse reactions eventually proceed at the same rate. At this point, the concentrations of all reactants and products remain constant, and the system is said to have reached dynamic equilibrium. The word “dynamic” is crucial: the forward and reverse reactions have not stopped — they continue at equal rates, so there is no net change in macroscopic properties.

许多化学反应是可逆的——产物可以相互反应重新生成原始反应物。可逆反应用符号⇌表示。当可逆反应在封闭系统中进行时，正反应和逆反应最终会以相同的速率进行。此时，所有反应物和产物的浓度保持恒定，系统达到了动态平衡。”动态”这个词至关重要：正反应和逆反应并没有停止——它们以相等的速率持续进行，因此宏观性质没有净变化。

Consider the classic example of the dimerisation of nitrogen dioxide:

考虑二氧化氮二聚化的经典例子：

2NO₂(g) ⇌ N₂O₄(g)

brown gas  |  棕色气体  →  colourless gas  |  无色气体

At room temperature, the mixture appears pale brown because both NO₂ and N₂O₄ are present. If the temperature is changed, the colour intensity changes, demonstrating a shift in the equilibrium position. This is a favourite demonstration in A-Level practical assessments.

在室温下，混合物呈浅棕色，因为NO₂和N₂O₄同时存在。如果改变温度，颜色强度会发生变化，这表明平衡位置发生了移动。这是A-Level实验评估中最受欢迎的演示实验之一。

2. The Equilibrium Constant Kc | 平衡常数Kc

For a general reversible reaction at equilibrium:

对于一般可逆反应在平衡状态下：

aA + bB ⇌ cC + dD

The equilibrium constant Kc is defined as:

平衡常数Kc定义为：

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

where [X] represents the equilibrium concentration of species X in mol dm⁻³. The exponents correspond to the stoichiometric coefficients in the balanced equation.

其中[X]表示物质X在平衡时的浓度，单位为mol dm⁻³。指数对应平衡方程式中的化学计量系数。

Key Points about Kc | 关于Kc的关键点

• Kc is temperature-dependent. Changing the temperature changes Kc. For an exothermic forward reaction, increasing temperature decreases Kc. For an endothermic forward reaction, increasing temperature increases Kc.
• Kc随温度变化。改变温度会改变Kc。对于放热正反应，升高温度会降低Kc。对于吸热正反应，升高温度会增加Kc。
• Kc is independent of concentration and pressure. Adding more reactant or changing the pressure does not alter Kc. The equilibrium position shifts to restore Kc to its original value.
• Kc与浓度和压力无关。添加更多反应物或改变压力不会改变Kc。平衡位置会发生移动，使Kc恢复到原来的值。
• Catalysts do not affect Kc. A catalyst speeds up both the forward and reverse reactions equally, so it does not change the equilibrium position or the value of Kc.
• 催化剂不影响Kc。催化剂同等地加速正反应和逆反应，因此不会改变平衡位置或Kc的值。
• Solids and pure liquids are omitted from the Kc expression because their concentrations are constant. Only aqueous and gaseous species appear.
• 固体和纯液体不包含在Kc表达式中，因为它们的浓度是恒定的。只有水溶液和气态物质出现在表达式中。

Worked Example | 例题

Question: 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed at 298 K. At equilibrium, 0.30 mol of ethyl ethanoate is formed. The total volume is 1.0 dm³. Calculate Kc for the esterification reaction:

题目：在298K下，将0.50 mol的乙酸和0.50 mol的乙醇混合。平衡时，生成0.30 mol的乙酸乙酯。总体积为1.0 dm³。计算酯化反应的Kc：

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Solution | 解答：

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
Equilibrium conc. / mol dm⁻³	0.20	0.20	0.30	0.30

Kc = (0.30 × 0.30) / (0.20 × 0.20) = 0.090 / 0.040 = 2.25

Note: For this esterification reaction, water is not a solvent — it is a product — so it must be included in the Kc expression. The units of Kc in this case are (mol dm⁻³)(mol dm⁻³) / (mol dm⁻³)(mol dm⁻³), which cancel to give no units.

注意：对于这个酯化反应，水不是溶剂——它是产物——因此必须包含在Kc表达式中。在这种情况下，Kc的单位是(mol dm⁻³)(mol dm⁻³) / (mol dm⁻³)(mol dm⁻³)，相互抵消，没有单位。

3. 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 concentration, pressure, or temperature, the position of equilibrium will shift to oppose that change. This principle allows us to predict qualitatively how a system will respond to external perturbations.

勒夏特列原理指出，如果处于动态平衡的系统受到浓度、压力或温度的变化，平衡位置将发生移动以对抗这种变化。该原理使我们能够定性地预测系统将如何响应外部干扰。

It is essential to understand that Le Chatelier’s Principle describes the direction of shift, while Kc tells us about the extent of reaction. Both are needed for a complete picture.

必须理解的是，勒夏特列原理描述的是移动的方向，而Kc告诉我们反应的程度。两者结合才能获得完整的图景。

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

If the concentration of a reactant is increased, the equilibrium shifts to the right (product side) to consume the added reactant and reduce its concentration. Conversely, if a product is removed, the equilibrium also shifts to the right to produce more product. This is the basis of many industrial processes where one product is continuously removed to drive the reaction forward.

如果增加反应物的浓度，平衡将向右移动（产物侧），以消耗添加的反应物并降低其浓度。相反，如果移除产物，平衡也会向右移动以产生更多产物。这是许多工业过程的基础，在这些过程中，一种产物被持续移除以推动反应正向进行。

For example, in the Haber Process for ammonia synthesis:

例如，在哈伯法合成氨的过程中：

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

Removing ammonia as it forms shifts the equilibrium to the right, maximising the yield. This is achieved industrially by cooling the reaction mixture to liquefy and remove NH₃ while recycling unreacted N₂ and H₂.

在氨形成时将其移除，使平衡向右移动，最大化产率。在工业上，这是通过冷却反应混合物使NH₃液化并移除，同时回收未反应的N₂和H₂来实现的。

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

Pressure changes only affect equilibria involving gases where there is a difference in the total number of moles of gas on each side of the equation. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas, because this reduces the total pressure, opposing the applied change.

压力变化只影响涉及气体的平衡，且方程式两边的气体总摩尔数存在差异。增加压力会使平衡向气体摩尔数较少的一侧移动，因为这会降低总压力，对抗施加的变化。

Using the Haber Process again: there are 4 moles of gas on the left (1 N₂ + 3 H₂) and 2 moles on the right (2 NH₃). Increasing pressure shifts equilibrium to the right, increasing the yield of ammonia. This is why the Haber Process is typically run at 200 atm.

再次以哈伯法为例：左边有4摩尔气体（1 N₂ + 3 H₂），右边有2摩尔（2 NH₃）。增加压力使平衡向右移动，增加氨的产率。这就是为什么哈伯法通常在200个大气压下运行。

Important: If the number of moles of gas is the same on both sides (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)), changing pressure has no effect on the equilibrium position. The system cannot oppose the pressure change by shifting either way.

重要：如果两边气体的摩尔数相同（例如H₂(g) + I₂(g) ⇌ 2HI(g)），改变压力对平衡位置没有影响。系统无法通过向任何一侧移动来对抗压力变化。

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

Temperature is the only factor that changes the value of Kc. For an exothermic forward reaction (ΔH < 0), increasing temperature shifts the equilibrium to the left (endothermic direction) to absorb the added heat. This means Kc decreases. For an endothermic forward reaction (ΔH > 0), increasing temperature shifts equilibrium to the right and Kc increases.

温度是唯一能改变Kc值的因素。对于放热正反应（ΔH < 0），升高温度使平衡向左移动（吸热方向），以吸收增加的热量。这意味着Kc减小。对于吸热正反应（ΔH > 0），升高温度使平衡向右移动，Kc增大。

Returning to our NO₂/N₂O₄ example:

回到NO₂/N₂O₄的例子：

2NO₂(g) ⇌ N₂O₄(g)     ΔH = -57 kJ mol⁻¹

brown | 棕色            colourless | 无色

Placing a sealed tube of the equilibrium mixture in hot water makes it darker brown — equilibrium shifts left (endothermic direction), producing more NO₂. Placing it in ice water makes it paler — equilibrium shifts right (exothermic direction), producing more N₂O₄. This is a classic demonstration of Le Chatelier’s Principle.

将装有平衡混合物的密封管放入热水中，颜色变深——平衡向左移动（吸热方向），生成更多NO₂。将其放入冰水中，颜色变浅——平衡向右移动（放热方向），生成更多N₂O₄。这是勒夏特列原理的经典演示。

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. This means a catalyst:

催化剂提供了具有较低活化能的替代反应路径。关键的是，它以相同的幅度降低了正反应和逆反应的活化能。这意味着催化剂：

• Does not change the equilibrium position
• Does not change the value of Kc
• Does increase the rate at which equilibrium is reached

• 不会改变平衡位置
• 不会改变Kc的值
• 会加快达到平衡的速率

In the Haber Process, an iron catalyst is used to allow equilibrium to be reached faster at the moderate temperature of 450°C, rather than having to wait for an impractically long time at lower temperatures.

在哈伯法中，使用铁催化剂使平衡在450°C的适中温度下更快达到，而不必在较低温度下等待不切实际的长时间。

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

For reactions involving gases, it is often more convenient to use partial pressures instead of concentrations. The equilibrium constant in terms of partial pressure is denoted Kp. For the general reaction:

对于涉及气体的反应，使用分压代替浓度通常更方便。用分压表示的平衡常数记为Kp。对于一般反应：

aA(g) + bB(g) ⇌ cC(g) + dD(g)

Kp = (Pc)ᶜ(Pᴅ)ᵈ / (PA)ᵃ(PB)ᵇ

The partial pressure of a gas in a mixture is the pressure that gas would exert if it occupied the entire volume alone. It is calculated as:

混合物中气体的分压是该气体单独占据整个体积时所施加的压力。计算公式为：

Partial pressure = mole fraction × total pressure

分压 = 摩尔分数 × 总压力

Worked Example: Kp Calculation | 例题：Kp计算

Question: In the Haber Process at 450°C and 200 atm, the equilibrium mixture contains 36% NH₃ by volume. Calculate Kp. The total pressure is 200 atm.

题目：在哈伯法中，450°C和200 atm条件下，平衡混合物中含36%的NH₃（按体积计）。计算Kp。总压力为200 atm。

Solution | 解答：

For gases, volume % = mole %. NH₃ = 36%, so N₂ + H₂ = 64%.

对于气体，体积% = 摩尔%。NH₃ = 36%，因此N₂ + H₂ = 64%。

N₂ : H₂ ratio is 1:3 from the equation, so N₂ = 16%, H₂ = 48%.

从方程式可知N₂ : H₂比例为1:3，因此N₂ = 16%，H₂ = 48%。

Gas	Mole %	Mole Fraction	Partial Pressure / atm
N₂	16%	0.16	0.16 × 200 = 32
H₂	48%	0.48	0.48 × 200 = 96
NH₃	36%	0.36	0.36 × 200 = 72

Kp = (PNH₃)² / (PN₂)(PH₂)³ = (72)² / (32)(96)³ = 5184 / (32 × 884,736)

= 5184 / 28,311,552 ≈ 1.83 × 10⁻⁴ atm⁻²

Note the units: Kp has units of atm⁻² because the numerator has (atm)² and the denominator has (atm)(atm)³ = atm⁴, giving atm²⁻⁴ = atm⁻².

注意单位：Kp的单位是atm⁻²，因为分子为(atm)²，分母为(atm)(atm)³ = atm⁴，得到atm²⁻⁴ = atm⁻²。

9. Industrial Applications of Equilibrium | 平衡的工业应用

The Haber Process | 哈伯法

The Haber Process synthesises ammonia from nitrogen and hydrogen:

哈伯法从氮气和氢气合成氨：

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

Compromise conditions: Although low temperature favours the exothermic forward reaction (higher yield), the rate is too slow at low temperatures. The iron catalyst only works effectively above ~400°C. The industrial compromise is 450°C — high enough for a reasonable rate, but not so high that yield is severely compromised. High pressure (200 atm) favours the side with fewer gas moles (the product side), improving yield. The iron catalyst ensures equilibrium is reached quickly.

折中条件：虽然低温有利于放热正反应（更高产率），但低温下速率太慢。铁催化剂仅在约400°C以上才能有效工作。工业折中方案是450°C——足够高以获得合理的速率，但又不会高到严重损害产率。高压（200 atm）有利于气体摩尔数较少的一侧（产物侧），提高产率。铁催化剂确保快速达到平衡。

The Contact Process | 接触法

The Contact Process produces sulfuric acid via the oxidation of sulfur dioxide:

接触法通过二氧化硫的氧化生产硫酸：

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

Conditions: 450°C, 1-2 atm, vanadium(V) oxide (V₂O₅) catalyst. The forward reaction is exothermic, so lower temperatures favour higher yield — but again, the rate is too slow. The vanadium(V) oxide catalyst allows a compromise temperature of 450°C. Pressure of only 1-2 atm is used because the equilibrium already lies well to the right (high Kc), and higher pressure would increase costs without significant yield benefit.

条件：450°C，1-2 atm，五氧化二钒（V₂O₅）催化剂。正反应是放热的，因此较低温度有利于更高产率——但同样，速率太慢。五氧化二钒催化剂允许折中温度为450°C。仅使用1-2 atm的压力，因为平衡已经很好地偏向右侧（高Kc），更高的压力会增加成本而没有显著的产率收益。

10. Common Exam Mistakes and Tips | 常见考试错误与技巧

Mistake | 错误	Correction | 纠正
Saying “equilibrium shifts to the left/right” without explaining why in terms of opposing the change.	Always state Le Chatelier’s Principle explicitly: “The equilibrium shifts to oppose the increase in…”
Stating that a catalyst “increases yield” or “shifts equilibrium”.	A catalyst does NOT affect yield or equilibrium position. It only increases the rate at which equilibrium is reached.
Including solids or pure liquids in Kc/Kp expressions.	Only include gases (g) and aqueous (aq) species. Solids (s) and pure liquids (l) have constant concentration and are omitted.
Forgetting to calculate and state the units of Kc or Kp.	Units are derived from the balanced equation and are required for full marks in many exam boards (especially CAIE and Edexcel). Always calculate units explicitly: (mol dm⁻³)^(Δn) for Kc, atm^(Δn) for Kp.
Confusing “position of equilibrium” with “Kc”.	Concentration and pressure changes shift the position of equilibrium (the ratio of products to reactants changes temporarily) but Kc stays the same. Only temperature changes Kc.
When calculating mole fractions for Kp, forgetting that volume % equals mole % for gases.	Avogadro’s Law: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Volume % = mole % is always true for ideal gases.

11. Summary | 总结

Factor | 因素	Effect on Equilibrium Position | 对平衡位置的影响	Effect on Kc/Kp | 对Kc/Kp的影响
Increase concentration of reactant	Shifts to product side (right)	No change
Increase pressure (fewer gas moles on right)	Shifts right	No change
Increase temperature (exothermic forward)	Shifts left (endothermic direction)	Kc decreases
Increase temperature (endothermic forward)	Shifts right	Kc increases
Add a catalyst	No change	No change

Chemical equilibrium is a topic that rewards a clear, systematic approach. Remember the three golden rules: (1) Le Chatelier’s Principle predicts the direction of shift; (2) only temperature changes Kc; (3) catalysts affect rate, not position. Master these, and you will handle any equilibrium question with confidence.

化学平衡是一个需要清晰、系统方法的主题。记住三条黄金法则：（1）勒夏特列原理预测移动方向；（2）只有温度能改变Kc；（3）催化剂影响速率，不影响位置。掌握这些，你将自信地应对任何平衡问题。