Chemical Equilibrium Le Chateliers Principle

What Is Chemical Equilibrium?

Chemical equilibrium is a state in a reversible reaction where the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant over time. It is important to understand that equilibrium does not mean the reaction has stopped — both the forward and reverse reactions continue to occur, just at the same rate. This is why we call it a dynamic equilibrium.

化学平衡是可逆反应中的一种状态，此时正反应和逆反应的速率相等，反应物和生成物的浓度随时间保持不变。重要的是要理解，平衡并不意味着反应已经停止——正反应和逆反应都在继续进行，只是速率相同。这就是为什么我们称之为动态平衡。

Consider the general reversible reaction: aA + bB ⇌ cC + dD. At equilibrium, the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients, is constant at a given temperature. This constant is known as the equilibrium constant, Kc.

考虑一般的可逆反应：aA + bB ⇌ cC + dD。在平衡状态下，生成物浓度与反应物浓度的比值（各自以其化学计量系数为指数）在给定温度下是恒定的。这个常数被称为平衡常数 Kc。

The Equilibrium Constant: Kc and Kp

For reactions in solution, we use Kc, which is expressed in terms of concentration (mol dm⁻³). The expression for the general reaction above is: Kc = [C]^c [D]^d / [A]^a [B]^b. The square brackets denote equilibrium concentrations. A large Kc value (Kc >> 1) indicates that the equilibrium lies to the right, favouring products. A small Kc value (Kc << 1) indicates that the equilibrium lies to the left, favouring reactants.

对于溶液中的反应，我们使用 Kc，它以浓度（mol dm⁻³）表示。上述一般反应的表达式为：Kc = [C]^c [D]^d / [A]^a [B]^b。方括号表示平衡浓度。较大的 Kc 值（Kc >> 1）表明平衡位置偏右，有利于生成物。较小的 Kc 值（Kc << 1）表明平衡位置偏左，有利于反应物。

For gas-phase reactions, we use Kp, which is expressed in terms of partial pressures. The expression takes the same form: Kp = (pC)^c (pD)^d / (pA)^a (pB)^b, where pX is the partial pressure of gas X. The relationship between Kp and Kc is given by: Kp = Kc (RT)^Δn, where Δn is the change in the number of moles of gas (products minus reactants), R is the gas constant (8.31 J K⁻¹ mol⁻¹), and T is the temperature in Kelvin.

对于气相反应，我们使用 Kp，它以分压表示。表达式形式相同：Kp = (pC)^c (pD)^d / (pA)^a (pB)^b，其中 pX 是气体 X 的分压。Kp 和 Kc 之间的关系为：Kp = Kc (RT)^Δn，其中 Δn 是气体摩尔数的变化（生成物减去反应物），R 是气体常数（8.31 J K⁻¹ mol⁻¹），T 是以开尔文为单位的温度。

Key Rules for Equilibrium Calculations

When working with equilibrium calculations, there are several important rules to remember:

在进行平衡计算时，有几个重要的规则需要记住：

• Solids and pure liquids are omitted from Kc and Kp expressions because their concentrations (or activities) are essentially constant. Only aqueous solutions and gases appear in the equilibrium expression.
• 固体和纯液体在 Kc 和 Kp 表达式中被省略，因为它们的浓度（或活度）基本上是恒定的。只有水溶液和气体出现在平衡表达式中。
• Kc and Kp are temperature-dependent. Changing the temperature changes the value of the equilibrium constant. A change in concentration or pressure does NOT change Kc or Kp — it only shifts the position of equilibrium.
• Kc 和 Kp 与温度有关。改变温度会改变平衡常数的值。改变浓度或压力不会改变 Kc 或 Kp——它只会改变平衡的位置。
• The units of Kc depend on the stoichiometry of the reaction. They are derived from the expression: (mol dm⁻³)^(sum of product coefficients) / (mol dm⁻³)^(sum of reactant coefficients). Always calculate and state the units.
• Kc 的单位取决于反应的化学计量。它们从表达式中推导出来：(mol dm⁻³)^(生成物系数之和) / (mol dm⁻³)^(反应物系数之和)。始终计算并注明单位。
• For Kp, partial pressures are typically expressed in atm, Pa, or kPa. Make sure all pressures use the same unit when calculating Kp.
• 对于 Kp，分压通常以 atm、Pa 或 kPa 表示。计算 Kp 时确保所有压力使用相同的单位。

Le Chatelier’s Principle

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 us to predict qualitatively how a system will respond to changes in concentration, pressure, and temperature.

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，平衡位置将移动以抵消这种变化。这个原理使我们能够定性地预测系统将如何响应浓度、压力和温度的变化。

Effect of Concentration Changes

If the concentration of a reactant is increased, the equilibrium shifts to the right to use up the added reactant and produce more products. If the concentration of a product is increased, the equilibrium shifts to the left to consume the added product and produce more reactants. This is directly analogous to the common ion effect in solubility equilibria.

如果增加反应物的浓度，平衡向右移动，消耗掉添加的反应物并生成更多的生成物。如果增加生成物的浓度，平衡向左移动，消耗掉添加的生成物并生成更多的反应物。这与溶解度平衡中的同离子效应直接相似。

Example: For the reaction Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), adding more Fe³⁺ ions shifts the equilibrium to the right, producing more of the deep red FeSCN²⁺ complex. This is a classic demonstration experiment in A-Level chemistry.

例如：对于反应 Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)，加入更多的 Fe³⁺ 离子会使平衡向右移动，产生更多的深红色 FeSCN²⁺ 络合物。这是 A-Level 化学中的一个经典演示实验。

Effect of Pressure Changes

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. If the pressure is increased, the equilibrium shifts to the side with fewer moles of gas to reduce the pressure. If the pressure is decreased, the equilibrium shifts to the side with more moles of gas.

压力变化只影响涉及气体的平衡，且方程两边气体摩尔数存在差异时才会发生。如果增加压力，平衡向气体摩尔数较少的一侧移动以降低压力。如果降低压力，平衡向气体摩尔数较多的一侧移动。

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). On the left, there are 4 moles of gas (1 N₂ + 3 H₂). On the right, there are 2 moles of gas (2 NH₃). Increasing the pressure shifts the equilibrium to the right, favouring ammonia production. This is exactly why the Haber process is carried out at high pressure (around 200 atm).

例如：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。左边有 4 摩尔气体（1 N₂ + 3 H₂）。右边有 2 摩尔气体（2 NH₃）。增加压力使平衡向右移动，有利于氨的生成。这正是哈伯法在高压（约 200 atm）下进行的原因。

Important note: Adding an inert gas at constant volume does NOT change the partial pressures of the reacting gases, so it has no effect on the position of equilibrium. However, adding an inert gas at constant pressure increases the total volume, which decreases the partial pressures of all gases — this is equivalent to decreasing the pressure.

重要提示：在恒容条件下加入惰性气体不会改变反应气体的分压，因此对平衡位置没有影响。然而，在恒压条件下加入惰性气体会增加总体积，从而降低所有气体的分压——这相当于降低压力。

Effect of Temperature Changes

Temperature is the only factor that changes the value of the equilibrium constant. For an exothermic reaction (ΔH < 0), increasing the temperature shifts the equilibrium to the left (favouring reactants), and Kc decreases. For an endothermic reaction (ΔH > 0), increasing the temperature shifts the equilibrium to the right (favouring products), and Kc increases.

温度是唯一能改变平衡常数值的因素。对于放热反应（ΔH < 0），升高温度使平衡向左移动（有利于反应物），Kc 减小。对于吸热反应（ΔH > 0），升高温度使平衡向右移动（有利于生成物），Kc 增大。

This can be understood by treating heat as a reactant or product. In an exothermic reaction, heat is a product: A + B ⇌ C + D + heat. Adding heat (increasing temperature) shifts the equilibrium left. In an endothermic reaction, heat is a reactant: A + B + heat ⇌ C + D. Adding heat shifts the equilibrium right.

这可以通过将热量视为反应物或生成物来理解。在放热反应中，热量是生成物：A + B ⇌ C + D + heat。加入热量（升高温度）使平衡向左移动。在吸热反应中，热量是反应物：A + B + heat ⇌ C + D。加入热量使平衡向右移动。

Effect of Catalysts

A catalyst provides an alternative reaction pathway with a lower activation energy. It speeds up BOTH the forward and reverse reactions equally. Therefore, a catalyst does NOT affect the position of equilibrium and does NOT change the value of Kc or Kp. A catalyst simply allows equilibrium to be reached faster.

催化剂提供了具有较低活化能的替代反应路径。它同等地加速正反应和逆反应。因此，催化剂不会影响平衡位置，也不会改变 Kc 或 Kp 的值。催化剂只是使平衡更快地达到。

Industrial Applications: The Haber Process

The Haber process for ammonia synthesis is the perfect case study for applying equilibrium principles to industrial chemistry: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ mol⁻¹.

哈伯法合成氨是将平衡原理应用于工业化学的完美案例研究：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92 kJ mol⁻¹。

The reaction is exothermic, so lower temperatures favour a higher equilibrium yield of ammonia. However, at low temperatures, the rate of reaction is too slow to be economical. The industrial compromise is to use a temperature of around 400-450°C, which gives a reasonable rate while still producing an acceptable yield.

该反应是放热的，因此较低的温度有利于更高的氨平衡产率。然而，在低温下，反应速率太慢，不经济。工业上的折中方案是使用约 400-450°C 的温度，这提供了合理的速率，同时仍能产生可接受的产率。

High pressure favours the forward reaction (4 moles → 2 moles of gas), so the process is carried out at around 200 atm. An iron catalyst is used to speed up the reaction without affecting the equilibrium position. The ammonia is continuously removed by condensation, which shifts the equilibrium further to the right according to Le Chatelier’s Principle.

高压有利于正反应（4 摩尔 → 2 摩尔气体），因此该过程在约 200 atm 下进行。使用铁催化剂加速反应而不影响平衡位置。氨通过冷凝不断被移除，根据勒夏特列原理，这使平衡进一步向右移动。

The Contact Process: Sulfuric Acid Production

Another important industrial equilibrium is the Contact Process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = -197 kJ mol⁻¹. This is the key step in sulfuric acid production.

另一个重要的工业平衡是接触法：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，ΔH = -197 kJ mol⁻¹。这是硫酸生产的关键步骤。

The reaction is exothermic, so lower temperatures give a higher equilibrium yield (Kp decreases as temperature increases). However, the rate is too slow below 400°C. The industrial compromise uses a temperature of about 450°C with a vanadium(V) oxide (V₂O₅) catalyst. The pressure used is only 1-2 atm because the equilibrium already lies far to the right — higher pressure would add cost without significant benefit.

该反应是放热的，因此较低温度给出更高的平衡产率（Kp 随温度升高而降低）。然而，低于 400°C 时速率太慢。工业折中方案使用约 450°C 的温度和五氧化二钒（V₂O₅）催化剂。使用的压力仅为 1-2 atm，因为平衡已经大幅偏右——更高的压力会增加成本而没有显著的收益。

Equilibrium Calculations: ICE Tables

ICE tables (Initial, Change, Equilibrium) are the standard method for solving equilibrium problems in A-Level chemistry. The approach is systematic and works for almost any equilibrium calculation.

ICE 表格（初始 Initial、变化 Change、平衡 Equilibrium）是解决 A-Level 化学中平衡问题的标准方法。这种方法是系统性的，适用于几乎任何平衡计算。

Step-by-step method:

逐步方法：

1. Write the balanced equation and the Kc expression.
   写出配平的方程式和 Kc 表达式。
2. Set up the ICE table with rows for Initial concentrations, Change, and Equilibrium concentrations.
   建立 ICE 表格，包含初始浓度、变化和平衡浓度的行。
3. Fill in known values. Enter the initial concentrations of all species. If starting with only reactants, product initial concentrations are zero.
   填入已知值。输入所有物种的初始浓度。如果仅从反应物开始，生成物的初始浓度为零。
4. Express changes using x. Let x moles of reactant react per dm³. The changes are proportional to the stoichiometric coefficients.
   用 x 表示变化。设每 dm³ 有 x 摩尔反应物反应。变化与化学计量系数成比例。
5. Write equilibrium concentrations in terms of x and substitute into the Kc expression.
   用 x 表示平衡浓度并代入 Kc 表达式。
6. Solve for x and then calculate all equilibrium concentrations.
   解出 x，然后计算所有平衡浓度。

Example: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), if we start with 1.0 mol of H₂ and 1.0 mol of I₂ in a 1 dm³ vessel, and Kc = 50 at the given temperature, find the equilibrium concentrations.

例如：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，如果我们在 1 dm³ 容器中以 1.0 mol H₂ 和 1.0 mol I₂ 开始，在给定温度下 Kc = 50，求平衡浓度。

         H₂      I₂      2HI
I:       1.0     1.0     0
C:       -x      -x      +2x
E:       1.0-x   1.0-x   2x

Kc = [HI]² / ([H₂][I₂]) = (2x)² / ((1.0-x)(1.0-x)) = 4x² / (1.0-x)² = 50
√50 = 2x / (1.0-x) → 7.07 = 2x / (1.0-x) → 7.07(1.0-x) = 2x → 7.07 - 7.07x = 2x → 7.07 = 9.07x → x = 0.78

[H₂] = [I₂] = 1.0 - 0.78 = 0.22 mol dm⁻³
[HI] = 2 × 0.78 = 1.56 mol dm⁻³

Common Exam Mistakes to Avoid

Students frequently make these errors in equilibrium questions. Being aware of them can help you avoid losing marks:

学生在平衡问题中经常犯以下错误。了解这些可以帮助你避免失分：

• Forgetting to state the units of Kc. If the question asks for Kc, you must include the correct units. If the powers in the numerator and denominator cancel out, state that Kc has no units.
• 忘记说明 Kc 的单位。如果题目要求计算 Kc，你必须包含正确的单位。如果分子和分母中的指数相互抵消，说明 Kc 没有单位。
• Confusing the effect of pressure and concentration changes on Kc. Remember: only temperature changes the value of Kc. Pressure and concentration changes shift the position of equilibrium but Kc remains the same.
• 混淆压力和浓度变化对 Kc 的影响。记住：只有温度改变 Kc 的值。压力和浓度的变化改变平衡位置，但 Kc 保持不变。
• Including solids or pure liquids in the Kc expression. They should be omitted because their effective concentration is constant.
• 在 Kc 表达式中包含固体或纯液体。它们应该被省略，因为它们有效浓度是恒定的。
• Incorrectly applying stoichiometric coefficients as exponents in the Kc expression. The coefficient becomes the power, not a multiplier.
• 错误地将化学计量系数用作 Kc 表达式中的指数。系数变成幂，而不是乘数。
• Using initial concentrations instead of equilibrium concentrations in the Kc expression. Always use the equilibrium row of the ICE table.
• 使用初始浓度而不是平衡浓度代入 Kc 表达式。始终使用 ICE 表格的平衡行。
• Incorrect sign for ΔH when applying Le Chatelier’s Principle to temperature changes. For an exothermic reaction, increasing temperature favours the reverse (endothermic) reaction.
• 在应用勒夏特列原理处理温度变化时，ΔH 符号错误。对于放热反应，升高温度有利于逆（吸热）反应。

Acid-Base Equilibria: A Special Case

A particularly important application of equilibrium principles in A-Level chemistry is acid-base equilibria. Weak acids and bases only partially dissociate in water, establishing an equilibrium. For a weak acid HA: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant Ka = [H⁺][A⁻] / [HA].

A-Level 化学中平衡原理的一个特别重要的应用是酸碱平衡。弱酸和弱碱仅在水溶液中部分解离，建立平衡。对于弱酸 HA：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。酸解离常数 Ka = [H⁺][A⁻] / [HA]。

The larger the Ka value, the stronger the acid. For convenience, we often use pKa = -log₁₀(Ka). A smaller pKa indicates a stronger acid. This is directly analogous to the pH scale, where pH = -log₁₀[H⁺].

Ka 值越大，酸越强。为方便起见，我们通常使用 pKa = -log₁₀(Ka)。较小的 pKa 表示较强的酸。这直接类似于 pH 标度，其中 pH = -log₁₀[H⁺]。

Buffer solutions are a key application of acid-base equilibria. A buffer resists changes in pH when small amounts of acid or base are added. An acidic buffer consists of a weak acid and its conjugate base (e.g., ethanoic acid and sodium ethanoate). The pH of a buffer can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]).

缓冲溶液是酸碱平衡的一个关键应用。缓冲液在加入少量酸或碱时抵抗 pH 的变化。酸性缓冲液由弱酸及其共轭碱组成（例如，乙酸和乙酸钠）。缓冲液的 pH 可以使用亨德森-哈塞尔巴尔赫方程计算：pH = pKa + log₁₀([A⁻]/[HA])。

Summary: Key Points for Exam Success

Chemical equilibrium is a central topic in A-Level Chemistry that connects to kinetics, thermodynamics, and industrial chemistry. The core ideas are:

化学平衡是 A-Level 化学的核心主题，连接着动力学、热力学和工业化学。核心思想是：

• Dynamic equilibrium: forward and reverse rates are equal, concentrations remain constant.
• 动态平衡：正逆反应速率相等，浓度保持恒定。
• Kc and Kp quantify the position of equilibrium and are only affected by temperature.
• Kc 和 Kp 量化了平衡位置，仅受温度影响。
• Le Chatelier’s Principle predicts the direction of shift when conditions change.
• 勒夏特列原理预测条件改变时平衡移动的方向。
• ICE tables provide a systematic method for calculating equilibrium concentrations.
• ICE 表格提供了计算平衡浓度的系统方法。
• Industrial processes like the Haber and Contact processes represent compromises between equilibrium yield and reaction rate.
• 哈伯法和接触法等工业过程代表了平衡产率和反应速率之间的折中。
• Acid-base equilibria extend these principles to weak acids, weak bases, and buffer solutions.
• 酸碱平衡将这些原理扩展到弱酸、弱碱和缓冲溶液。

Mastering chemical equilibrium requires both qualitative understanding (Le Chatelier’s Principle) and quantitative skills (Kc/Kp calculations using ICE tables). Practice with a wide variety of problems, and always pay attention to units and significant figures in your calculations.

掌握化学平衡需要定性理解（勒夏特列原理）和定量技能（使用 ICE 表格进行 Kc/Kp 计算）。通过大量不同类型的问题进行练习，并始终注意计算中的单位和有效数字。