A-Level Chemistry: Chemical Equilibrium — Complete Guide | A-Level 化学：化学平衡完全指南

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English Version

Introduction to Chemical Equilibrium

Chemical equilibrium is one of the most fundamental concepts in A-Level Chemistry. It describes the state in a reversible reaction where the forward and reverse reactions proceed at exactly the same rate, resulting in no net change in the concentrations of reactants and products. Understanding equilibrium is essential not only for exam success but also for grasping how real-world chemical processes — from the Haber process to biological systems — operate under dynamic balance.

In A-Level specifications (AQA, Edexcel, OCR, and CIE), equilibrium appears across multiple topics: physical chemistry, industrial chemistry, and even organic reaction mechanisms. Students are expected to master both qualitative reasoning (Le Chatelier’s Principle) and quantitative calculations (Kc, Kp, and the reaction quotient Q).

1. Dynamic Equilibrium: The Core Concept

A reversible reaction is one that can proceed in both the forward and backward directions. Consider the general reaction:

aA + bB ⇌ cC + dD

At the start, only reactants A and B are present, so the forward reaction rate is high while the reverse rate is zero. As products C and D form, the forward rate gradually decreases (because reactant concentrations drop), while the reverse rate increases (because product concentrations rise). Eventually, the two rates become equal — this is the dynamic equilibrium state.

Key characteristics of dynamic equilibrium:

• Closed system required: No matter can enter or leave the system. If products escape (e.g., as a gas), equilibrium cannot be established.
• Macroscopic properties are constant: Concentrations, pressure, and colour appear unchanged at the macroscopic level.
• Microscopic changes continue: At the molecular level, both forward and reverse reactions are still occurring — hence “dynamic.”
• Can be approached from either direction: The same equilibrium mixture results whether you start with pure reactants or pure products (provided the same total amounts of atoms are present).

2. Le Chatelier’s Principle

Henri Louis Le Chatelier (1884) stated: “If a system at dynamic equilibrium is subjected to a change, the position of equilibrium will shift to oppose that change.”

This principle is not a rigorous thermodynamic law but an incredibly useful heuristic for predicting equilibrium shifts. Let’s examine each type of perturbation:

2.1 Effect of Concentration Changes

If the concentration of a reactant is increased, the equilibrium shifts to the right (towards products) to consume the added reactant. Conversely, increasing product concentration shifts equilibrium to the left.

Example — Fe³⁺ / SCN⁻ equilibrium:

Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq)

(yellow) + (colourless) ⇌ (blood-red)

Adding more Fe³⁺ or SCN⁻ intensifies the blood-red colour (shift right). Adding a reagent that removes Fe³⁺ (e.g., F⁻ to form stable [FeF₆]³⁻) causes the colour to fade (shift left).

2.2 Effect of Pressure Changes (Gaseous Systems Only)

Increasing pressure favours the side with fewer gas molecules (moles). Decreasing pressure favours the side with more gas molecules.

Example — N₂O₄ ⇌ 2NO₂ equilibrium:

N₂O₄(g) ⇌ 2NO₂(g)

(colourless, 1 mol gas) ⇌ (brown, 2 mol gas)

Increasing pressure shifts equilibrium left — the colour fades as brown NO₂ converts to colourless N₂O₄. Decreasing pressure shifts right — the mixture darkens.

Important: If both sides have equal numbers of gas molecules (e.g., H₂ + I₂ ⇌ 2HI), pressure changes have no effect on equilibrium position — only on the rate at which equilibrium is reached.

2.3 Effect of Temperature Changes

Temperature is the only factor that changes the value of the equilibrium constant (Kc or Kp). The direction of shift depends on whether the forward reaction is exothermic or endothermic:

• Exothermic forward reaction (ΔH < 0): Increasing temperature shifts equilibrium left (towards reactants). The system absorbs heat by favouring the endothermic reverse reaction.
• Endothermic forward reaction (ΔH > 0): Increasing temperature shifts equilibrium right (towards products).

Example — The Haber Process:

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

Since the forward reaction is exothermic, lower temperatures favour ammonia production. However, this must be balanced against kinetics — at very low temperatures, the reaction is impractically slow. The industrial compromise is ~450°C with an iron catalyst.

2.4 Effect of Catalysts

A catalyst does not affect the position of equilibrium. It lowers the activation energy for both the forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium mixture composition is unchanged. This is a common exam trap — students often incorrectly claim that catalysts shift equilibrium.

3. The Equilibrium Constant: Kc

For a homogeneous reaction in solution at constant temperature:

aA + bB ⇌ cC + dD

The equilibrium constant in terms of concentration is:

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Where square brackets denote equilibrium concentrations in mol dm⁻³.

3.1 Rules for Writing Kc Expressions

• Products in numerator, reactants in denominator.
• Stoichiometric coefficients become exponents.
• Pure solids and pure liquids are omitted — their “concentrations” are effectively constant and are absorbed into Kc.
• Water is omitted in aqueous systems unless it is a reactant/product in a non-aqueous solvent or the solvent itself is part of the equilibrium.

3.2 Worked Example: Esterification

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

At equilibrium, the concentrations are: [CH₃COOH] = 0.17, [C₂H₅OH] = 0.17, [CH₃COOC₂H₅] = 0.33, [H₂O] = 0.33 (all in mol dm⁻³).

Kc = (0.33 × 0.33) / (0.17 × 0.17) = 0.1089 / 0.0289 = 3.77

Units: (mol dm⁻³)² / (mol dm⁻³)² = no units. Kc = 3.77 (dimensionless).

Kc values much greater than 1 indicate the equilibrium lies far to the right (products favoured). Kc values much less than 1 indicate reactants are favoured.

3.3 Determining Kc Experimentally

The classic A-Level practical involves the esterification equilibrium above. Known amounts of ethanoic acid and ethanol are mixed with a strong acid catalyst (H₂SO₄). After reaching equilibrium (typically 48+ hours at room temperature, or 1 hour at reflux), the mixture is titrated against standard NaOH to determine the remaining acid concentration. From this, all equilibrium concentrations can be calculated using an ICE table (Initial, Change, Equilibrium).

4. The Equilibrium Constant: Kp (Gaseous Systems)

For gaseous equilibria, it is often more convenient to use partial pressures instead of concentrations:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

Kp = (p_C)ᶜ(p_D)ᵈ / (p_A)ᵃ(p_B)ᵇ

Where p_X is the partial pressure of gas X at equilibrium, typically expressed in atm, kPa, or Pa.

4.1 Partial Pressure and Mole Fraction

The partial pressure of a gas in a mixture is:

p_A = mole fraction of A × total pressure

Mole fraction of A = (moles of A at equilibrium) / (total moles of all gases at equilibrium).

4.2 Units of Kp

Like Kc, Kp may or may not have units depending on the stoichiometry. For N₂ + 3H₂ ⇌ 2NH₃:

Kp units = (pressure)² / (pressure × pressure³) = pressure⁻² (e.g., atm⁻² or kPa⁻²).

4.3 Relationship Between Kc and Kp

Kp = Kc (RT)^(Δn), where Δn = (moles of gaseous products) − (moles of gaseous reactants), R is the gas constant (8.314 J K⁻¹ mol⁻¹), and T is temperature in Kelvin. This equation is useful when converting between concentration-based and pressure-based equilibrium constants.

5. The Reaction Quotient: Q

The reaction quotient Q has the same mathematical form as Kc but uses current concentrations (not necessarily at equilibrium):

• Q < Kc: Reaction proceeds forward (towards products) to reach equilibrium.
• Q = Kc: System is at equilibrium.
• Q > Kc: Reaction proceeds backward (towards reactants) to reach equilibrium.

This is an extremely powerful diagnostic tool — by calculating Q at any moment, you can predict which direction a reaction must proceed to achieve equilibrium.

6. Industrial Applications

6.1 The Haber Process (NH₃ Synthesis)

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

Conditions chosen: 450°C, 200 atm, iron catalyst.

Compromise reasoning: Low temperature favours yield (exothermic forward reaction) but reduces rate. High pressure favours yield (4 mol → 2 mol gas) but increases plant cost and safety concerns. The iron catalyst allows a moderate temperature while maintaining an acceptable rate.

Yield: ~15-20% per pass. Unreacted N₂ and H₂ are recycled.

6.2 The Contact Process (H₂SO₄ Production)

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

Conditions: 450°C, 1-2 atm, V₂O₅ catalyst. High conversion (~99%) is achieved using excess oxygen and multi-stage reactors.

6.3 Methanol Synthesis

CO(g) + 2H₂(g) ⇌ CH₃OH(g)    ΔH = −91 kJ mol⁻¹

Conditions: 250°C, 50-100 atm, Cu/ZnO/Al₂O₃ catalyst. Methanol is a key feedstock for plastics, paints, and fuels.

7. Common Exam Pitfalls

1. Claiming catalysts shift equilibrium: They don’t — they only speed up the rate at which equilibrium is reached.
2. Forgetting to include stoichiometric coefficients as exponents in Kc/Kp expressions.
3. Confusing rate and equilibrium: A fast reaction does not necessarily have a large Kc. Rate and equilibrium position are independent.
4. Including solids/liquids in Kc: Only aqueous and gaseous species appear in equilibrium expressions.
5. Using moles instead of concentrations when calculating Kc — always divide by volume first (unless V = 1 dm³).
6. Forgetting to check Kc/Kp units: Marks are routinely awarded for correct units in A-Level exams.

8. Practice Questions

1. For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), write the expression for Kp and state its units when pressure is measured in atm.
2. At 500 K, Kc = 0.25 for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). If 2.0 mol of PCl₅ is placed in a 4.0 dm³ vessel, calculate the equilibrium concentrations of all species.
3. Explain, using Le Chatelier’s Principle, why the Haber process uses high pressure but not extremely high pressure (e.g., 1000 atm).
4. The equilibrium N₂O₄(g) ⇌ 2NO₂(g) has ΔH = +58 kJ mol⁻¹. Predict and explain the effect of (a) increasing temperature, (b) increasing pressure, on the colour of the mixture.

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中文版

化学平衡导论

化学平衡是A-Level化学中最基础的概念之一。它描述了可逆反应中正反应和逆反应速率相等、反应物和生成物浓度不再发生净变化的状态。理解平衡不仅对考试成功至关重要，对于掌握现实世界中的化学过程——从哈伯法到生物系统——如何在动态平衡下运行也同样关键。

在A-Level各考试局的考纲中（AQA、Edexcel、OCR和CIE），平衡出现在多个主题中：物理化学、工业化学，甚至有机反应机理。学生需要同时掌握定性推理（勒夏特列原理）和定量计算（Kc、Kp和反应商Q）。

1. 动态平衡：核心概念

可逆反应是指可以同时向正方向和逆方向进行的反应。考虑以下通式：

aA + bB ⇌ cC + dD

反应开始时，只有反应物A和B存在，因此正反应速率高而逆反应速率为零。随着产物C和D的生成，正反应速率逐渐降低（因为反应物浓度下降），而逆反应速率增加（因为产物浓度上升）。最终，两个速率相等——这就是动态平衡状态。

动态平衡的关键特征：

• 需要封闭系统：物质不能进入或离开系统。如果产物逸出（例如以气体形式），平衡无法建立。
• 宏观性质恒定：浓度、压力和颜色在宏观层面上看起来不变。
• 微观变化持续：在分子层面，正反应和逆反应仍在进行——因此称为”动态”。
• 可从任一方向接近：无论从纯反应物还是纯产物开始，只要原子总量相同，最终都会得到相同的平衡混合物。

2. 勒夏特列原理

亨利·路易·勒夏特列（1884年）指出：“如果一个处于动态平衡的系统受到外界条件的改变，平衡将向减弱这种改变的方向移动。”

这个原理不是一个严格的热力学定律，而是一个极其有用的启发式工具，用于预测平衡移动。让我们逐一分析每种扰动类型：

2.1 浓度变化的影响

如果增加反应物的浓度，平衡向右移动（向产物方向）以消耗新增的反应物。相反，增加产物浓度会使平衡向左移动。

示例 — Fe³⁺ / SCN⁻ 平衡：

Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq)

（黄色）+（无色）⇌（血红色）

加入更多Fe³⁺或SCN⁻会使血红色加深（向右移动）。加入能去除Fe³⁺的试剂（如F⁻形成稳定的[FeF₆]³⁻）则导致颜色褪去（向左移动）。

2.2 压强变化的影响（仅限气体系统）

增加压强有利于气体分子数较少的一侧。降低压强有利于气体分子数较多的一侧。

示例 — N₂O₄ ⇌ 2NO₂ 平衡：

N₂O₄(g) ⇌ 2NO₂(g)

（无色，1 mol气体）⇌（棕色，2 mol气体）

增加压强使平衡向左移动——随着棕色NO₂转化为无色N₂O₄，颜色褪去。降低压强使平衡向右移动——混合物颜色加深。

重要提示：如果两侧气体分子数相等（如H₂ + I₂ ⇌ 2HI），压强变化对平衡位置没有影响——只影响达到平衡的速率。

2.3 温度变化的影响

温度是唯一能改变平衡常数（Kc或Kp）数值的因素。移动方向取决于正反应是放热还是吸热：

• 正反应放热（ΔH < 0）：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热量。
• 正反应吸热（ΔH > 0）：升高温度使平衡向右移动（向产物方向）。

示例 — 哈伯法：

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

由于正反应放热，较低温度有利于氨的生成。然而，这必须与动力学平衡——在非常低的温度下，反应速度慢得不切实际。工业上的折中方案是约450°C并使用铁催化剂。

2.4 催化剂的影响

催化剂不影响平衡位置。它同等程度地降低正反应和逆反应的活化能，因此平衡更快达到，但平衡混合物的组成不变。这是常见的考试陷阱——学生经常错误地声称催化剂会移动平衡。

3. 平衡常数：Kc

对于恒温下溶液中的均相反应：

aA + bB ⇌ cC + dD

以浓度表示的平衡常数为：

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

其中方括号表示平衡浓度，单位为mol dm⁻³。

3.1 书写Kc表达式的规则

• 产物在分子，反应物在分母。
• 化学计量系数变为指数。
• 纯固体和纯液体省略——它们的”浓度”实际上是常数，被吸收到Kc中。
• 水在水溶液系统中省略，除非它是非水溶剂中的反应物/产物，或者溶剂本身参与了平衡。

3.2 计算示例：酯化反应

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

平衡时浓度：[CH₃COOH] = 0.17，[C₂H₅OH] = 0.17，[CH₃COOC₂H₅] = 0.33，[H₂O] = 0.33（单位均为mol dm⁻³）。

Kc = (0.33 × 0.33) / (0.17 × 0.17) = 0.1089 / 0.0289 = 3.77

单位：(mol dm⁻³)² / (mol dm⁻³)² = 无单位。Kc = 3.77（无量纲）。

Kc值远大于1表示平衡远在右侧（产物占优）。Kc值远小于1表示反应物占优。

3.3 实验测定Kc

经典的A-Level实验涉及上述酯化平衡。将已知量的乙酸和乙醇与强酸催化剂（H₂SO₄）混合。达到平衡后（通常室温下48小时以上，或回流1小时），用标准NaOH溶液滴定混合物以确定剩余酸浓度。由此，可以使用ICE表格（初始、变化、平衡）计算所有物种的平衡浓度。

4. 平衡常数：Kp（气体系统）

对于气体平衡，使用分压比使用浓度更方便：

aA(g) + bB(g) ⇌ cC(g) + dD(g)

Kp = (p_C)ᶜ(p_D)ᵈ / (p_A)ᵃ(p_B)ᵇ

其中p_X是气体X在平衡时的分压，通常以atm、kPa或Pa表示。

4.1 分压与摩尔分数

混合气体中某气体的分压为：

p_A = A的摩尔分数 × 总压强

A的摩尔分数 = （平衡时A的摩尔数）/（平衡时所有气体的总摩尔数）。

4.2 Kp的单位

与Kc一样，Kp可能有也可能没有单位，取决于化学计量关系。对于N₂ + 3H₂ ⇌ 2NH₃：

Kp单位 = (压强)² / (压强 × 压强³) = 压强⁻²（如atm⁻²或kPa⁻²）。

4.3 Kc与Kp的关系

Kp = Kc (RT)^(Δn)，其中Δn = （气体产物的摩尔数）−（气体反应物的摩尔数），R是气体常数（8.314 J K⁻¹ mol⁻¹），T是开尔文温度。此公式在基于浓度和基于压强的平衡常数之间进行转换时非常有用。

5. 反应商：Q

反应商Q的数学形式与Kc相同，但使用的是当前浓度（不一定是平衡时的浓度）：

• Q < Kc：反应正向进行（向产物方向）以达到平衡。
• Q = Kc：系统处于平衡状态。
• Q > Kc：反应逆向进行（向反应物方向）以达到平衡。

这是一个极其强大的诊断工具——通过在任何时刻计算Q，你可以预测反应必须朝哪个方向进行才能达到平衡。

6. 工业应用

6.1 哈伯法（合成氨）

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

选择的条件：450°C，200 atm，铁催化剂。

折中理由：低温有利于产率（放热正反应）但降低速率。高压有利于产率（4 mol → 2 mol气体）但增加设备成本和安全问题。铁催化剂在保持可接受速率的同时允许使用适中的温度。

产率：每次通过约15-20%。未反应的N₂和H₂循环使用。

6.2 接触法（硫酸生产）

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

条件：450°C，1-2 atm，V₂O₅催化剂。通过使用过量氧气和多级反应器实现高转化率（~99%）。

6.3 甲醇合成

CO(g) + 2H₂(g) ⇌ CH₃OH(g)    ΔH = −91 kJ mol⁻¹

条件：250°C，50-100 atm，Cu/ZnO/Al₂O₃催化剂。甲醇是塑料、涂料和燃料的关键原料。

7. 常见考试陷阱

1. 声称催化剂移动平衡：催化剂不会移动平衡——它们只加快达到平衡的速率。
2. 忘记将化学计量系数作为指数写入Kc/Kp表达式。
3. 混淆速率和平衡：快速反应不一定有大的Kc。速率和平衡位置是独立的。
4. 在Kc中包含固体/液体：只有水溶液和气态物种出现在平衡表达式中。
5. 计算Kc时使用摩尔数而非浓度——始终先除以体积（除非V = 1 dm³）。
6. 忘记检查Kc/Kp单位：在A-Level考试中，正确单位通常会获得专门分数。

8. 练习题

1. 对于平衡2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，写出Kp表达式，并说明当压强以atm为单位时Kp的单位。
2. 在500 K时，PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)的Kc = 0.25。如果将2.0 mol PCl₅放入4.0 dm³容器中，计算所有物种的平衡浓度。
3. 用勒夏特列原理解释为什么哈伯法使用高压但不是极高压力（如1000 atm）。
4. 平衡N₂O₄(g) ⇌ 2NO₂(g)的ΔH = +58 kJ mol⁻¹。预测并解释(a)升高温度，(b)增加压强对混合物颜色的影响。

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