A-Level Chemistry: Chemical Equilibrium Guide

What is Chemical Equilibrium?

Chemical equilibrium is a fundamental concept in A-Level Chemistry that describes the state of a reversible reaction when the forward and backward reactions proceed at the same rate. At equilibrium, the concentrations of reactants and products remain constant — but importantly, the reaction has not stopped. Instead, both forward and reverse reactions continue at equal rates, making equilibrium a dynamic rather than static condition.

化学平衡是A-Level化学中的一个基本概念，描述的是可逆反应达到正反应和逆反应速率相等的状态。在平衡状态下，反应物和产物的浓度保持恒定——但重要的是，反应并没有停止。相反，正反应和逆反应以相同的速率继续进行，使平衡成为一种动态而非静态的状态。

A reversible reaction is denoted by the ⇌ symbol (double half-arrow). For example, the reaction between nitrogen and hydrogen to form ammonia is reversible:

可逆反应用⇌符号（双半箭头）表示。例如，氮气与氢气合成氨的反应就是可逆的：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

This means ammonia can decompose back into nitrogen and hydrogen under certain conditions. At equilibrium, both the forward reaction (N₂ + 3H₂ → 2NH₃) and the backward reaction (2NH₃ → N₂ + 3H₂) occur at identical rates.

这意味着氨在某些条件下可以分解回氮气和氢气。在平衡状态下，正反应（N₂ + 3H₂ → 2NH₃）和逆反应（2NH₃ → N₂ + 3H₂）以相同的速率发生。

Dynamic Equilibrium: The Key Idea

Dynamic equilibrium is the state where the forward and reverse reactions occur at exactly the same rate, resulting in no net change in the concentrations of reactants and products. Three essential conditions must be met for dynamic equilibrium:

动态平衡是正反应和逆反应以完全相同的速率进行的状态，导致反应物和产物的浓度没有净变化。达到动态平衡必须满足三个基本条件：

• Closed system: No matter can enter or leave the system. In an open system, reactants or products could escape, preventing equilibrium from being established. / 封闭系统：物质不能进入或离开系统。在开放系统中，反应物或产物可能逸出，阻止平衡的建立。
• Reversible reaction: The reaction must be able to proceed in both directions. Irreversible reactions go to completion and never reach equilibrium. / 可逆反应：反应必须能够向两个方向进行。不可逆反应会进行完全，永远不会达到平衡。
• Constant temperature: Temperature affects both the position of equilibrium and the value of the equilibrium constant. Equilibrium can only be discussed at a specific temperature. / 恒温：温度影响平衡位置和平衡常数的值。平衡只能在特定温度下讨论。

A helpful analogy is a full bathtub with the drain open while the tap runs at the same rate — the water level stays constant despite continuous inflow and outflow. Similarly, at equilibrium, molecules constantly convert back and forth, but the macroscopic concentrations appear unchanged.

一个有用的比喻是：一个装满水的浴缸，排水口打开的同时水龙头以相同的速率放水——尽管持续有水流进流出，水位却保持不变。类似地，在平衡状态下，分子不断地来回转化，但宏观浓度看起来不变。

The Equilibrium Constant Kc

The equilibrium constant Kc provides a quantitative measure of the position of equilibrium. For a general reversible reaction:

平衡常数Kc提供了对平衡位置的定量测量。对于一般的可逆反应：

aA + bB ⇌ cC + dD

The expression for Kc is:

Kc的表达式为：

Kc = [C]ᶜ × [D]ᵈ / [A]ᵃ × [B]ᵇ

Where [A], [B], [C], and [D] represent the equilibrium concentrations of each species in mol/dm³. The powers a, b, c, and d are the stoichiometric coefficients from the balanced equation.

其中[A]、[B]、[C]和[D]代表每种物质在平衡时的浓度，单位为mol/dm³。指数a、b、c和d是配平方程中的化学计量系数。

Key points about Kc:

• Kc is temperature-dependent: Changing the temperature changes Kc. For an exothermic reaction (ΔH negative), increasing temperature decreases Kc. For an endothermic reaction (ΔH positive), increasing temperature increases Kc. / Kc依赖于温度：改变温度会改变Kc。对于放热反应，升高温度会降低Kc。对于吸热反应，升高温度会增加Kc。
• Kc is independent of concentration and pressure: Changing concentration or pressure shifts the position of equilibrium but does NOT change Kc at constant temperature. / Kc与浓度和压力无关：改变浓度或压力会移动平衡位置，但在恒温下不会改变Kc。
• Catalysts do not affect Kc: A catalyst speeds up both forward and reverse reactions equally, so Kc remains unchanged. The catalyst only helps reach equilibrium faster. / 催化剂不影响Kc：催化剂同等程度地加速正反应和逆反应，因此Kc保持不变。催化剂只是帮助更快地达到平衡。
• Interpreting Kc values: A large Kc (> 10³) means the equilibrium lies far to the right — products are favoured. A small Kc (< 10⁻³) means the equilibrium lies far to the left — reactants are favoured. An intermediate Kc (~1) indicates significant amounts of both reactants and products. / 解读Kc值：大的Kc意味着平衡位置偏向右侧——产物占优势。小的Kc意味着平衡位置偏向左侧——反应物占优势。中等Kc表示反应物和产物都有显著含量。

Calculating Kc from Experimental Data

A typical A-Level exam question provides initial amounts and one equilibrium amount, and asks you to calculate Kc. The standard approach uses an ICE table (Initial, Change, Equilibrium):

典型的A-Level考试题目会给出初始量和某一个平衡量，要求你计算Kc。标准方法使用ICE表（初始Initial、变化Change、平衡Equilibrium）：

Example: 2.0 mol of H₂ and 1.0 mol of I₂ are placed in a 1.0 dm³ vessel at 450°C. At equilibrium, 1.56 mol of HI is found. Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g).

例子：在450°C下，将2.0 mol H₂和1.0 mol I₂放入1.0 dm³容器中。平衡时发现1.56 mol HI。计算H₂(g) + I₂(g) ⇌ 2HI(g)的Kc。

1. Set up the ICE table with mole ratios. Since 1.56 mol of HI is produced, from the stoichiometry, 0.78 mol of H₂ and 0.78 mol of I₂ must have reacted. / 建立ICE表，使用摩尔比。由于产生了1.56 mol HI，根据化学计量关系，必定有0.78 mol H₂和0.78 mol I₂反应了。
2. Equilibrium amounts: H₂ = 2.0 – 0.78 = 1.22 mol, I₂ = 1.0 – 0.78 = 0.22 mol, HI = 1.56 mol. / 平衡量：H₂ = 2.0 – 0.78 = 1.22 mol，I₂ = 1.0 – 0.78 = 0.22 mol，HI = 1.56 mol。
3. Convert to concentrations (divide by 1.0 dm³): [H₂] = 1.22, [I₂] = 0.22, [HI] = 1.56 mol/dm³. / 转换为浓度（除以1.0 dm³）：[H₂] = 1.22，[I₂] = 0.22，[HI] = 1.56 mol/dm³。
4. Kc = [HI]² / ([H₂][I₂]) = (1.56)² / (1.22 × 0.22) = 2.4336 / 0.2684 ≈ 9.07 (no units as moles cancel). / Kc = [HI]² / ([H₂][I₂]) = (1.56)² / (1.22 × 0.22) = 2.4336 / 0.2684 ≈ 9.07（无单位，因摩尔数抵消）。

Le Chatelier’s Principle

Le Chatelier’s Principle is one of the most powerful predictive tools in chemistry. It states:

勒夏特列原理是化学中最强大的预测工具之一。其内容为：

If a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose that change.

如果处于动态平衡的系统受到条件改变的影响，平衡位置会移动以抵消这种改变。

This principle allows chemists to predict how equilibrium responds to changes in concentration, pressure, and temperature. It is not a theoretical explanation of why equilibrium shifts — it is a predictive rule based on countless experimental observations. The molecular explanation comes from collision theory and the effect of conditions on forward and reverse reaction rates.

这个原理使化学家能够预测平衡如何响应浓度、压力和温度的变化。这不是对平衡为什么会移动的理论解释——它是一个基于无数实验观察的预测规则。分子层面的解释来自碰撞理论以及条件对正反应和逆反应速率的影响。

Effect of Concentration Changes

When the concentration of a reactant or product is changed, the equilibrium shifts to oppose that change:

当反应物或产物的浓度改变时，平衡会移动以抵消这种改变：

• Adding more reactant: The equilibrium shifts to the right (towards products) to consume some of the added reactant. / 添加更多反应物：平衡向右移动（向产物方向），以消耗部分添加的反应物。
• Adding more product: The equilibrium shifts to the left (towards reactants) to consume some of the added product. / 添加更多产物：平衡向左移动（向反应物方向），以消耗部分添加的产物。
• Removing reactant: The equilibrium shifts to the left to produce more of the depleted reactant. / 移除反应物：平衡向左移动，以产生更多被消耗的反应物。
• Removing product: The equilibrium shifts to the right to produce more product. / 移除产物：平衡向右移动，以产生更多产物。

Example with the Fe³⁺/SCN⁻ equilibrium system: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), which appears blood-red. Adding KSCN (increasing [SCN⁻]) deepens the colour — the equilibrium shifts right, producing more FeSCN²⁺. Adding Fe³⁺ produces a similar effect. Adding F⁻ (which removes Fe³⁺ by forming a colourless complex) lightens the colour — equilibrium shifts left.

以Fe³⁺/SCN⁻平衡体系为例：Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)，呈现血红色。加入KSCN（增加[SCN⁻]）会使颜色加深——平衡向右移动，产生更多FeSCN²⁺。加入Fe³⁺产生类似效果。加入F⁻（通过与Fe³⁺形成无色配合物将其移除）会使颜色变浅——平衡向左移动。

Effect of Temperature Changes

Temperature changes affect equilibrium differently depending on whether the forward reaction is exothermic or endothermic:

温度变化对平衡的影响取决于正反应是放热还是吸热：

• Exothermic forward reaction (ΔH negative): Increasing temperature shifts equilibrium to the LEFT (towards reactants). The system absorbs the added heat by favouring the endothermic backward reaction. Decreasing temperature shifts equilibrium to the RIGHT. / 正反应放热（ΔH为负）：升高温度使平衡向左移动（向反应物方向）。系统通过促进吸热的逆反应来吸收增加的热量。降低温度使平衡向右移动。
• Endothermic forward reaction (ΔH positive): Increasing temperature shifts equilibrium to the RIGHT (towards products). The system absorbs the added heat by favouring the endothermic forward reaction. Decreasing temperature shifts equilibrium to the LEFT. / 正反应吸热（ΔH为正）：升高温度使平衡向右移动（向产物方向）。系统通过促进吸热的正反应来吸收增加的热量。降低温度使平衡向左移动。

A classic demonstration is the N₂O₄ ⇌ 2NO₂ equilibrium. N₂O₄ is colourless while NO₂ is brown. The forward reaction is endothermic. Heating a sealed tube of the equilibrium mixture intensifies the brown colour (equilibrium shifts right). Placing it in ice water lightens the colour (equilibrium shifts left).

一个经典演示是N₂O₄ ⇌ 2NO₂平衡。N₂O₄无色而NO₂为棕色。正反应是吸热的。加热装有平衡混合物的密封管，棕色加深（平衡向右移动）。将其放入冰水中，颜色变浅（平衡向左移动）。

Effect of Pressure Changes (Gaseous Systems Only)

Pressure changes only affect equilibria involving gases where there is a difference in the number of gas molecules on each side. The principle is:

压力变化仅影响涉及气体且两侧气体分子数不同的平衡。原理是：

• Increasing pressure: Equilibrium shifts to the side with FEWER gas molecules to reduce the pressure. / 增加压力：平衡向气体分子数较少的一侧移动，以降低压力。
• Decreasing pressure: Equilibrium shifts to the side with MORE gas molecules to increase the pressure. / 降低压力：平衡向气体分子数较多的一侧移动，以增加压力。

If the number of gas molecules is equal on both sides, pressure changes have NO effect on the position of equilibrium (e.g., H₂(g) + I₂(g) ⇌ 2HI(g) — 2 molecules on each side). Kc remains unchanged because Kc is defined in terms of concentrations, not partial pressures. However, note that for Kp (the equilibrium constant in terms of partial pressure), the analysis is different for heterogeneous systems.

如果两侧气体分子数相等，压力变化对平衡位置没有影响（例如H₂(g) + I₂(g) ⇌ 2HI(g)——每侧2个分子）。Kc保持不变，因为Kc是用浓度而不是分压定义的。然而需要注意的是，对于Kp（用分压表示的平衡常数），对非均相体系的分析是不同的。

Effect of Catalysts

A catalyst provides an alternative reaction pathway with a lower activation energy. Crucially, it lowers the activation energy of BOTH the forward and reverse reactions by exactly the same amount:

催化剂提供了具有较低活化能的替代反应途径。关键的是，它将正反应和逆反应的活化能降低了完全相同的量：

• A catalyst does NOT change the position of equilibrium. / 催化剂不改变平衡位置。
• A catalyst does NOT change the value of Kc. / 催化剂不改变Kc的值。
• A catalyst ONLY increases the rate at which equilibrium is reached. / 催化剂只提高达到平衡的速率。

This is why catalysts are so valuable in industrial processes — they allow equilibrium to be reached faster at lower temperatures, improving economic efficiency without affecting the equilibrium yield. They are not a “magic bullet” for increasing product yield; they only affect kinetics, not thermodynamics.

这就是催化剂在工业过程中如此宝贵的原因——它们使平衡在较低温度下更快达到，提高经济效率而不影响平衡产率。它们不是增加产物产率的”灵丹妙药”；它们只影响动力学，不影响热力学。

Industrial Application: The Haber Process

The Haber Process for ammonia synthesis is the textbook example of applying Le Chatelier’s Principle in industry:

哈伯法合成氨是勒夏特列原理在工业中应用的教科书范例：

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

• 4 gas molecules on the left, 2 on the right — high pressure favours the forward reaction. / 左侧4个气体分子，右侧2个——高压有利于正反应。
• Exothermic forward reaction — low temperature favours the forward reaction. / 正反应放热——低温有利于正反应。

Based on Le Chatelier’s Principle alone, the ideal conditions would be extremely high pressure and very low temperature. However, in practice, the Haber Process uses:

仅基于勒夏特列原理，理想条件应该是极高压力和非常低的温度。然而，实际上哈伯法使用：

• Pressure: 200 atm — Higher pressures would increase yield but are too expensive and dangerous (equipment costs, safety risks). The marginal benefit above 200 atm does not justify the cost. / 压力：200 atm——更高的压力会增加产率，但成本太高且危险性大（设备成本、安全风险）。200 atm以上的边际效益不值得增加的成本。
• Temperature: 400-450°C — Lower temperatures would give a higher equilibrium yield, but the reaction would be far too slow to be economical. An iron catalyst is used to speed up the reaction at this moderate temperature. / 温度：400-450°C——较低温度会给出更高的平衡产率，但反应速度会太慢，不经济。使用铁催化剂在此中等温度下加速反应。
• Catalyst: Finely divided iron (with K₂O and Al₂O₃ promoters) — Speeds up reaction without affecting yield. The promoters increase the surface area and electronic properties of the iron catalyst. / 催化剂：细碎铁粉（含K₂O和Al₂O₃助催化剂）——加速反应而不影响产率。助催化剂增加铁催化剂的表面积和电子性能。
• Continuous removal of NH₃ — Shifts equilibrium right by constantly removing product (Le Chatelier’s Principle in action). Unreacted N₂ and H₂ are recycled. / 连续移除NH₃——通过不断移除产物使平衡向右移动（勒夏特列原理的实际应用）。未反应的N₂和H₂被循环利用。

This illustrates a crucial exam point: compromise conditions are used in industry, balancing thermodynamic yield against kinetic rate and economic practicality.

这说明了一个关键的考试要点：工业中使用的是妥协条件，在热力学产率与动力学速率以及经济实用性之间取得平衡。

Common Exam Problems and Tips

A-Level exam questions on equilibrium typically fall into these categories:

A-Level关于平衡的考试题通常分为以下几类：

1. Writing Kc expressions: Remember that pure solids and pure liquids are omitted from the expression because their “concentration” is effectively constant. Only include aqueous (aq) and gaseous (g) species. / 写Kc表达式：记住纯固体和纯液体从表达式中省略，因为它们的”浓度”实际上是恒定的。只包括水溶液(aq)和气体(g)物种。
2. Calculating Kc from ICE tables: Always check units — Kc may or may not have units depending on the stoichiometry. If the sum of powers in numerator equals the sum in denominator, Kc has no units. / 使用ICE表计算Kc：总是检查单位——Kc可能有单位，也可能没有，取决于化学计量关系。如果分子中指数之和等于分母中指数之和，Kc没有单位。
3. Predicting shifts using Le Chatelier’s Principle: Always state the direction of shift (left or right), the reason (which side is favoured by the change), and the observable outcome (colour change, yield change, etc.). / 使用勒夏特列原理预测移动：总是陈述移动方向（左或右）、原因（哪一侧受变化影响更有利）以及可观察的结果（颜色变化、产率变化等）。
4. Interpreting yield vs temperature graphs: For exothermic reactions, yield decreases with temperature. For endothermic, yield increases. Memorise: exothermic = product yield drops as temperature rises. / 解读产率-温度图：对于放热反应，产率随温度升高而降低。对于吸热反应，产率随温度升高而增加。记住：放热反应——产物产率随温度升高而下降。
5. Explaining compromise conditions: Always discuss both thermodynamic (equilibrium yield) AND kinetic (rate) factors. A common mark-scheme error is to discuss only one of these. / 解释妥协条件：总是讨论热力学（平衡产率）和动力学（速率）两个因素。常见的评分方案错误是只讨论其中之一。

Summary of Key Concepts

To master equilibrium for A-Level Chemistry, you should be able to:

要掌握A-Level化学中的平衡，你应该能够：

• Define dynamic equilibrium and state the three necessary conditions (closed system, reversible reaction, constant temperature). / 定义动态平衡并陈述三个必要条件（封闭系统、可逆反应、恒温）。
• Write and use Kc expressions correctly, omitting solids and liquids. / 正确写出并使用Kc表达式，省略固体和液体。
• Calculate Kc from experimental data using ICE tables. / 使用ICE表从实验数据计算Kc。
• Apply Le Chatelier’s Principle to predict the effect of concentration, temperature, and pressure changes. / 应用勒夏特列原理预测浓度、温度和压力变化的影响。
• Explain why catalysts affect the rate of reaching equilibrium but not the equilibrium position or Kc. / 解释为什么催化剂影响达到平衡的速率但不影响平衡位置或Kc。
• Analyse industrial processes (Haber, Contact) in terms of compromise conditions. / 从妥协条件的角度分析工业过程（哈伯法、接触法）。
• Distinguish between thermodynamic control (position of equilibrium, Kc) and kinetic control (rate of reaction). / 区分热力学控制（平衡位置、Kc）和动力学控制（反应速率）。

Chemical equilibrium is not just an isolated topic — it connects to energetics, kinetics, and organic chemistry throughout the A-Level syllabus. A solid understanding of equilibrium will serve you well across multiple exam papers and is a foundation for university-level chemistry.

化学平衡不仅仅是一个孤立的主题——它贯穿A-Level教学大纲，与能量学、动力学和有机化学相互联系。扎实理解平衡概念将帮助你在多份考试卷中取得好成绩，也是大学化学的基础。