A-Level Chemistry: Chemical Equilibrium & Le Chatelier’s Principle — Complete Guide | A-Level 化学：化学平衡与勒夏特列原理 — 完整指南

Introduction / 导言

Chemical equilibrium is one of the most fundamental concepts in A-Level Chemistry. It bridges the gap between the macroscopic observations we make in the laboratory and the molecular-level understanding of what is actually happening during a chemical reaction. Mastering equilibrium is essential not only for scoring well on your exams but also for building a genuine intuition about how chemical systems behave — an intuition that will serve you well in university-level chemistry and beyond.

化学平衡是A-Level化学中最基本的概念之一。它连接了我们在实验室中观察到的宏观现象与在分子层面上的真实反应过程。掌握化学平衡不仅对考试取得好成绩至关重要，还能帮助你建立对化学体系行为的真正直觉——这种直觉将在大学化学乃至更远的学习中让你受益匪浅。

In this comprehensive guide, we will explore reversible reactions, dynamic equilibrium, Le Chatelier’s Principle, the equilibrium constant K_c, and how to apply these concepts to solve exam-style problems. Whether you are studying for CIE, Edexcel, AQA, or OCR, the principles covered here are universal across all A-Level Chemistry specifications.

在这本完整指南中，我们将深入探讨可逆反应、动态平衡、勒夏特列原理、平衡常数 K_c，以及如何应用这些概念来解决考试题型。无论你正在备考CIE、Edexcel、AQA还是OCR考试局，这里涵盖的原理在所有A-Level化学大纲中都是通用的。

---

1. Reversible Reactions / 可逆反应

1.1 What Is a Reversible Reaction? / 什么是可逆反应？

A reversible reaction is one in which the products can react together to re-form the original reactants. Unlike irreversible reactions — such as the combustion of hydrocarbons, where the products cannot easily be converted back — reversible reactions proceed in both directions simultaneously under the same conditions.

可逆反应是指产物可以重新反应生成原始反应物的化学反应。与不可逆反应（例如碳氢化合物的燃烧，其产物无法轻易转化回来）不同，可逆反应在相同条件下同时在两个方向上进行。

The classic example studied at A-Level is the Haber process for ammonia synthesis:

A-Level学习的经典例子是合成氨的哈伯过程：

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

The double arrow (⇌) is the universal symbol indicating reversibility. It tells us that nitrogen and hydrogen react to form ammonia, but ammonia also decomposes back into nitrogen and hydrogen — all within the same reaction vessel.

双向箭头（⇌）是表示可逆性的通用符号。它告诉我们，氮气和氢气反应生成氨，但氨也会分解回氮气和氢气——这一切都在同一个反应容器中发生。

1.2 Other Important Reversible Reactions / 其他重要的可逆反应

Several reversible reactions feature prominently in A-Level syllabi. Familiarising yourself with these examples now will save you time during revision:

以下是A-Level大纲中经常出现的几个可逆反应，提前熟悉它们将为你的复习节省大量时间：

• Contact Process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) — used in sulfuric acid production / 用于硫酸生产
• Esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O — carboxylic acid + alcohol ⇌ ester + water / 羧酸 + 醇 ⇌ 酯 + 水
• Nitrogen dioxide dimerisation: 2NO₂(g) ⇌ N₂O₄(g) — a beautiful colour-change demonstration (brown ⇌ colourless) / 一个美丽的颜色变化实验（棕红色 ⇌ 无色）
• Iodine-hydrogen equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g) — often used to illustrate K_c calculations / 常用于说明K_c计算

---

2. Dynamic Equilibrium / 动态平衡

2.1 The Concept / 概念

When a reversible reaction is left in a closed system, the forward and reverse reactions eventually reach a state of dynamic equilibrium. This is one of the most commonly misunderstood concepts in A-Level Chemistry, so let us be precise about what it means:

当可逆反应在封闭体系中进行时，正向反应和逆向反应最终会达到动态平衡状态。这是A-Level化学中最容易被误解的概念之一，让我们精确地定义它的含义：

1. Dynamic means the reactions have not stopped. Both the forward and reverse reactions continue to occur at the molecular level. / 动态意味着反应并没有停止。正向反应和逆向反应在分子层面仍在持续进行。
2. Equilibrium means the rates of the forward and reverse reactions are equal. / 平衡意味着正向反应和逆向反应的速率相等。
3. Because the rates are equal, the concentrations of all reactants and products remain constant (but they are not necessarily equal to each other). / 由于速率相等，所有反应物和产物的浓度保持恒定（但彼此之间不一定相等）。

A useful analogy is a person walking up a “down” escalator at exactly the same speed the escalator is moving down. The person appears stationary to an external observer, but internally, both movements are still happening. Similarly, at equilibrium, molecules are continuously reacting in both directions, but there is no net change in the macroscopic composition of the system.

一个有用的类比是：一个人以与下行自动扶梯完全相同的速度向上行走。对外部观察者来说，这个人看起来是静止的，但实际上，两个运动都在持续进行。同样地，在平衡状态下，分子在两个方向上持续反应，但体系的宏观组成没有净变化。

2.2 Critical Conditions for Equilibrium / 平衡的关键条件

For a dynamic equilibrium to be established, three conditions must be met. Exam questions frequently test your understanding of these prerequisites:

要建立动态平衡，必须满足三个条件。考试题目经常会测试你对这些前提条件的理解：

Condition / 条件	Explanation / 说明
Closed System / 封闭体系	No matter (reactants or products) can escape. Energy, however, may be exchanged with the surroundings. If a gaseous product escapes, equilibrium can never be reached because the reverse reaction is prevented. / 物质（反应物或产物）不能逸出。但能量可以与周围环境交换。如果气体产物逸出，由于逆向反应被阻止，平衡永远无法达到。
Constant Temperature / 恒温	Temperature affects both the rate and the position of equilibrium. A fluctuating temperature means the equilibrium position is constantly shifting. / 温度同时影响反应速率和平衡位置。波动的温度意味着平衡位置在不断变化。
Reversible Reaction / 可逆反应	The reaction must be capable of proceeding in both directions under the given conditions. Some reactions are effectively irreversible under normal laboratory conditions (e.g., combustion). / 反应必须在给定条件下能够沿两个方向进行。某些反应在正常实验室条件下实际上是不可逆的（例如燃烧反应）。

---

3. Le Chatelier’s Principle / 勒夏特列原理

3.1 Statement of the Principle / 原理表述

Le Chatelier’s Principle is the cornerstone of equilibrium analysis. It states:

勒夏特列原理是平衡分析的基石。其表述如下：

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

如果处于动态平衡的体系受到外界条件的改变，平衡位置会向着对抗（削弱）该改变的方向移动。

This principle is remarkably powerful because it allows you to predict the qualitative effect of any perturbation — concentration changes, pressure changes, temperature changes — without needing to calculate anything. Simply ask yourself: “How can the system respond to undo what I just did?” The answer is the direction of shift.

这个原理之所以强大，是因为它使你无需任何计算就能预测任何扰动——浓度变化、压强变化、温度变化——对平衡的定性影响。只需问自己：”体系如何响应才能抵消我刚刚施加的改变？” 答案就是平衡移动的方向。

3.2 Concentration Changes / 浓度变化

If you increase the concentration of a reactant, the system will try to “use up” the extra reactant — it shifts to the right (product side). If you increase the concentration of a product, the system shifts to the left (reactant side).

如果你增加反应物的浓度，体系会试图”消耗掉”多余的反应物——平衡向右（产物方向）移动。如果你增加产物的浓度，体系会向左（反应物方向）移动。

Worked Example / 解题示例：

Consider the esterification equilibrium: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

• Adding more ethanol (a reactant): Equilibrium shifts → right, more ester formed. / 加入更多乙醇（反应物）：平衡向右移动，生成更多酯。
• Removing water (a product) as it forms: Equilibrium shifts → right, more ester formed. This is why esterification is often carried out with a dehydrating agent or by distillation. / 在生成时移除水（产物）：平衡向右移动，生成更多酯。这就是为什么酯化反应通常使用脱水剂或通过蒸馏进行。
• Adding more ester: Equilibrium shifts → left, more reactants re-formed. / 加入更多酯：平衡向左移动，重新生成更多反应物。

3.3 Pressure Changes (Gaseous Systems Only) / 压强变化（仅适用于气体体系）

Pressure changes only affect equilibria involving gases, and only when there is a difference in the total number of gaseous moles between the two sides of the equation.

压强的变化只影响涉及气体的平衡，且仅当方程式两边气体总摩尔数不同时才产生影响。

The Rule / 规则：

• Increase pressure → equilibrium shifts to the side with fewer gas molecules (to reduce pressure). / 增加压强 → 平衡向气体分子数较少的一侧移动（以降低压强）。
• Decrease pressure → equilibrium shifts to the side with more gas molecules (to increase pressure). / 降低压强 → 平衡向气体分子数较多的一侧移动（以增加压强）。

Haber Process Example / 哈伯过程示例：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Left side: 1 + 3 = 4 moles of gas / 左侧：4摩尔气体
Right side: 2 moles of gas / 右侧：2摩尔气体

Increasing pressure favours the forward reaction (producing more NH₃) because the system tries to reduce pressure by moving to the side with fewer gas molecules. This is exactly why the Haber process is carried out at high pressure (typically 200 atm).

增加压强有利于正向反应（生成更多NH₃），因为体系试图通过向气体分子数较少的一侧移动来降低压强。这正是为什么哈伯过程在高压（通常200 atm）下进行。

Important Caution / 重要警示： If the number of gas moles is the same on both sides (e.g., H₂ + I₂ ⇌ 2HI — 2 moles on each side), pressure changes have no effect on the equilibrium position. This is a classic exam trap!

如果两边气体摩尔数相同（例如 H₂ + I₂ ⇌ 2HI——两边各2摩尔），压强变化对平衡位置没有影响。这是一个经典的考试陷阱！

3.4 Temperature Changes / 温度变化

Temperature is unique among the Le Chatelier factors because it actually changes the value of the equilibrium constant K_c. Concentration and pressure changes do not affect K_c — they only shift the position of equilibrium. Temperature changes do both.

温度在勒夏特列因素中具有独特性，因为它实际上会改变平衡常数K_c的值。浓度和压强的变化不会影响K_c——它们只改变平衡位置。而温度的改变则同时影响两者。

The Rule / 规则：

• Exothermic reaction (ΔH < 0): Increasing temperature shifts equilibrium left (endothermic direction). K_c decreases. / 放热反应（ΔH < 0）：升高温度使平衡向左（吸热方向）移动。K_c减小。
• Endothermic reaction (ΔH > 0): Increasing temperature shifts equilibrium right (endothermic direction). K_c increases. / 吸热反应（ΔH > 0）：升高温度使平衡向右（吸热方向）移动。K_c增大。

Think of “heat” as a chemical — in an exothermic reaction, heat is a product. Adding heat (raising temperature) is like adding a product, so the equilibrium shifts left to consume it. In an endothermic reaction, heat is a reactant. Adding heat shifts the equilibrium right to consume it.

将”热量”视为一种化学物质——在放热反应中，热量是产物。加入热量（升高温度）就像加入产物一样，因此平衡向左移动以消耗它。在吸热反应中，热量是反应物。加入热量使平衡向右移动以消耗它。

NO₂/N₂O₄ Demonstration / NO₂/N₂O₄实验：

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

NO₂ is brown; N₂O₄ is colourless. The forward reaction is exothermic. / NO₂为棕红色；N₂O₄为无色。正向反应是放热的。

• Placing the sealed tube in hot water: equilibrium shifts left (endothermic direction). The mixture becomes darker brown (more NO₂). / 将密封管放入热水中：平衡向左（吸热方向）移动。混合物变为更深的棕红色（更多NO₂）。
• Placing the sealed tube in ice water: equilibrium shifts right (exothermic direction). The mixture becomes paler (more N₂O₄). / 将密封管放入冰水中：平衡向右（放热方向）移动。混合物颜色变浅（更多N₂O₄）。

3.5 Catalysts / 催化剂

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. Therefore:

催化剂提供了一个活化能更低的替代反应路径。关键的是，它以相同的幅度降低了正向和逆向两个反应的活化能。因此：

• A catalyst does not affect the position of equilibrium — it cannot shift the equilibrium left or right. / 催化剂不会影响平衡位置——它不能使平衡向左或向右移动。
• A catalyst does not change the value of K_c. / 催化剂不会改变K_c的值。
• A catalyst does allow equilibrium to be reached more quickly because it speeds up both forward and reverse reactions equally. / 催化剂确实能使平衡更快达到，因为它同等地加快了正向和逆向反应。

This is a very common exam question. Many students incorrectly claim that a catalyst “increases the yield” or “shifts the equilibrium to the right.” A catalyst only affects the rate at which equilibrium is established, never the equilibrium composition itself.

这是一个非常常见的考试问题。许多学生错误地声称催化剂”提高了产率”或”使平衡向右移动”。催化剂只影响达到平衡的速率，绝不会影响平衡组成本身。

---

4. The Equilibrium Constant (K_c) / 平衡常数

4.1 Definition and Expression / 定义与表达式

For a general reversible reaction at equilibrium:

对于达到平衡的一般可逆反应：

aA + bB ⇌ cC + dD

The equilibrium constant K_c is defined as:

平衡常数K_c的定义为：

K_c = [C]^c[D]^d / [A]^a[B]^b

Where [X] represents the equilibrium concentration of species X in mol dm⁻³. This expression is sometimes memorised as “products over reactants,” with each concentration raised to the power of its stoichiometric coefficient.

其中 [X] 表示物质X在平衡时的浓度，单位为 mol dm⁻³。这个表达式可以记为”产物在分子，反应物在分母“，每种物质的浓度以其化学计量系数为指数。

4.2 Rules for Writing K_c Expressions / 书写K_c表达式的规则

1. Only include gases and aqueous species. Pure solids and pure liquids have an activity of 1 and are omitted from the K_c expression. / 只包括气体和水溶液中的物质。纯固体和纯液体的活度为1，在K_c表达式中被省略。
2. Water as a solvent is omitted when it is in large excess (its concentration is essentially constant). / 当水作为溶剂大量过量时，水被省略（其浓度基本恒定）。
3. Concentrations must be equilibrium concentrations, not initial concentrations. This is the single most common mistake students make in K_c calculations. / 浓度必须是平衡时的浓度，而不是初始浓度。这是学生在K_c计算中最常见的错误。

4.3 Worked K_c Calculation / K_c计算示例

Question / 题目： 0.50 mol of ethanoic acid and 0.50 mol of ethanol are mixed in a 1.0 dm³ vessel at 298 K. At equilibrium, 0.33 mol of ethyl ethanoate is present. Calculate K_c.

将0.50 mol乙酸和0.50 mol乙醇在1.0 dm³容器中混合，温度为298 K。达到平衡时，存在0.33 mol乙酸乙酯。计算K_c。

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Step 1 — Set up an ICE table (Initial / Change / Equilibrium) / 建立ICE表格（初始/变化/平衡）：

	CH3COOH	C2H5OH	CH3COOC2H5	H2O
Initial / mol	0.50	0.50	0	0
Change / mol	−0.33	−0.33	+0.33	+0.33
Equilibrium / mol	0.17	0.17	0.33	0.33

Step 2 — Calculate K_c:

Since the volume is 1.0 dm³, the equilibrium concentrations in mol dm⁻³ are numerically equal to the equilibrium amounts in moles:

由于体积为1.0 dm³，平衡浓度（mol dm⁻³）在数值上等于平衡时物质的量（mol）：

K_c = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]

K_c = (0.33 × 0.33) / (0.17 × 0.17) = 0.1089 / 0.0289 = 3.8

Answer / 答案： K_c = 3.8 (no units, as the total number of moles on each side is equal: 2 ⇌ 2)

4.4 Interpreting K_c Values / K_c值的解读

• K_c ≫ 1: Equilibrium lies far to the right — the reaction effectively goes to completion. Products dominate. / 平衡远在右侧——反应几乎完全进行。产物占主导。
• K_c ≈ 1: Significant amounts of both reactants and products present at equilibrium. / 平衡时反应物和产物均有显著量存在。
• K_c ≪ 1: Equilibrium lies far to the left — very little product is formed. Reactants dominate. / 平衡远在左侧——几乎不形成产物。反应物占主导。

---

5. Industrial Applications / 工业应用

Understanding Le Chatelier’s Principle and equilibrium constants is not just an academic exercise — it is the intellectual foundation of the entire chemical industry. Here are the key industrial processes you are expected to know:

理解勒夏特列原理和平衡常数不仅仅是学术练习——它是整个化学工业的智力基础。以下是需要掌握的关键工业过程：

5.1 The Haber Process / 哈伯过程

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

Condition / 条件	Typical Value / 典型值	Reason / 原因
Pressure / 压强	200 atm	High pressure favours the forward reaction (4 → 2 gas moles). Higher pressures would increase yield further but cost more and require stronger equipment. / 高压有利于正向反应（4 → 2气体摩尔）。更高压强会进一步提高产率，但成本更高且需要更坚固的设备。
Temperature / 温度	400–450 °C	A compromise. Low temperature favours the exothermic forward reaction (higher yield), but the rate would be too slow. 400–450 °C balances yield against rate. / 一个折衷方案。低温有利于放热正向反应（更高产率），但速率太慢。400–450 °C在产率和速率之间取得平衡。
Catalyst / 催化剂	Iron (Fe)	Speeds up the attainment of equilibrium without affecting yield. / 加快达到平衡而不影响产率。

5.2 The Contact Process / 接触法

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

Conditions: 450 °C, 1–2 atm, vanadium(V) oxide (V₂O₅) catalyst. The pressure is low (only 1–2 atm) because the equilibrium already lies far to the right at this temperature — a K_c value that is large enough to give a ~99% conversion without needing high pressure.

条件：450 °C，1–2 atm，五氧化二钒（V₂O₅）催化剂。压强较低（仅1–2 atm），因为在该温度下平衡已经远在右侧——K_c值足够大，无需高压即可获得约99%的转化率。

---

6. Common Exam Mistakes and How to Avoid Them / 常见考试错误及避免方法

Mistake 1: Confusing Rate and Equilibrium Position / 混淆速率与平衡位置

Many students write that “increasing temperature increases the yield because the particles have more kinetic energy.” This is incorrect reasoning. Temperature affects both rate and equilibrium position for different reasons. Rate increases because more particles have E ≥ E_a (kinetic argument). Equilibrium position shifts according to Le Chatelier’s Principle (thermodynamic argument). These are separate concepts — do not conflate them in your answer.

许多学生写道”升高温度提高了产率，因为粒子具有更多的动能”。这是不正确的推理。温度影响速率和平衡位置的原因是不同的。速率增加是因为更多粒子具有E ≥ E_a（动力学论证）。平衡位置根据勒夏特列原理移动（热力学论证）。这是两个独立的概念——不要在答案中将它们混为一谈。

Mistake 2: Forgetting That K_c Depends Only on Temperature / 忘记K_c只取决于温度

Adding more reactant, changing pressure, or adding a catalyst does not change K_c. The equilibrium constant is a thermodynamic quantity — it changes only when temperature changes. If a question asks “What happens to K_c when the pressure is increased?” the correct answer is “No change.”

加入更多反应物、改变压强或加入催化剂都不会改变K_c。平衡常数是一个热力学量——它只在温度变化时改变。如果题目问”增加压强后K_c会怎样？”，正确答案是”不变”。

Mistake 3: Claiming a Catalyst Shifts Equilibrium / 声称催化剂改变平衡

A catalyst provides an alternative pathway with lower activation energy for both directions equally. It does not alter the relative stability of reactants and products, so it cannot shift the equilibrium position. It only reduces the time needed to reach equilibrium.

催化剂为两个方向同等地提供了一个活化能更低的替代路径。它不会改变反应物和产物的相对稳定性，因此不能改变平衡位置。它只减少达到平衡所需的时间。

Mistake 4: Using Initial Concentrations in K_c Expressions / 在K_c表达式中使用初始浓度

K_c is calculated using equilibrium concentrations only. If a question gives you initial amounts and an equilibrium amount, you must construct an ICE table to find all equilibrium concentrations before substituting into the K_c expression.

K_c仅使用平衡浓度计算。如果题目给出了初始量和某个物质的平衡量，你必须建立ICE表格，求出所有物质的平衡浓度，然后才能代入K_c表达式。

Mistake 5: Omitting Units from K_c / 忘记K_c的单位

K_c may or may not have units depending on the stoichiometry. Calculate the units by substituting mol dm⁻³ into the K_c expression. For example, for 2SO₂ + O₂ ⇌ 2SO₃, the units of K_c are (mol dm⁻³)² / (mol dm⁻³)³ = mol⁻¹ dm³. Always include units in your final answer unless they cancel to give a dimensionless K_c.

K_c可能带有单位，也可能没有单位，这取决于化学计量比。通过将mol dm⁻³代入K_c表达式来计算单位。例如，对于2SO₂ + O₂ ⇌ 2SO₃，K_c的单位为(mol dm⁻³)² / (mol dm⁻³)³ = mol⁻¹ dm³。除非单位互相抵消得到无量纲的K_c，否则始终在最终答案中包含单位。

---

7. Summary Table / 总结表格

Here is a comprehensive summary of how each factor affects equilibrium:

以下是每种因素如何影响平衡的全面总结：

Factor / 因素	Effect on Equilibrium Position / 对平衡位置的影响	Effect on Kc / 对Kc的影响
Increase reactant concentration / 增加反应物浓度	Shifts right / 向右移动	No change / 不变
Increase product concentration / 增加产物浓度	Shifts left / 向左移动	No change / 不变
Increase pressure (fewer gas moles on right) / 增加压强（右侧气体摩尔数更少）	Shifts right / 向右移动	No change / 不变
Increase pressure (equal gas moles) / 增加压强（两边气体摩尔数相同）	No shift / 不移动	No change / 不变
Increase temperature (exothermic ΔH < 0) / 升高温度（放热反应）	Shifts left / 向左移动	Decreases / 减小
Increase temperature (endothermic ΔH > 0) / 升高温度（吸热反应）	Shifts right / 向右移动	Increases / 增大
Add a catalyst / 加入催化剂	No shift / 不移动	No change / 不变

---

8. Practice Questions / 练习题

Test your understanding with these exam-style questions. Try to answer them before looking at the solutions below.

用这些考试风格的题目测试你的理解。在看下面的解答之前，请先尝试自己回答。

Q1. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), state and explain the effect of increasing the pressure on (a) the equilibrium position, (b) the value of K_c, and (c) the rate of attainment of equilibrium.

Q2. At 500 K, 1.00 mol of PCl₅ is placed in a 2.00 dm³ vessel. At equilibrium, 0.30 mol of PCl₅ remains. Calculate K_c for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

Q3. Explain why a low temperature is thermodynamically favourable for the Haber process but is not used in practice.

Solutions / 解答：

A1. (a) Equilibrium shifts right — the forward reaction produces fewer gas molecules (3 → 2), opposing the pressure increase. (b) K_c is unchanged — only temperature changes can alter K_c. (c) The rate of attainment increases — higher pressure means more frequent collisions between reactant molecules.

(a) 平衡向右移动——正向反应产生更少的气体分子（3 → 2），对抗压强的增加。(b) K_c不变——只有温度变化才能改变K_c。(c) 达到平衡的速率增加——更高的压强意味着反应物分子之间碰撞更频繁。

A2. PCl₅ decomposed = 1.00 − 0.30 = 0.70 mol. So [PCl₃]_eq = [Cl₂]_eq = 0.70 / 2.00 = 0.35 mol dm⁻³. [PCl₅]_eq = 0.30 / 2.00 = 0.15 mol dm⁻³. K_c = (0.35 × 0.35) / 0.15 = 0.82 mol dm⁻³.

A3. A low temperature favours the exothermic forward reaction, giving a higher equilibrium yield of ammonia. However, at low temperatures, the rate of reaction is too slow to be economically viable. The industrial temperature of 400–450 °C is a compromise between yield (thermodynamics) and rate (kinetics). The iron catalyst further increases the rate without affecting the equilibrium position.

低温有利于放热正向反应，给出更高的氨平衡产率。然而，在低温下，反应速率太慢，不具备经济可行性。400–450 °C的工业温度是产率（热力学）和速率（动力学）之间的折衷。铁催化剂进一步提高了速率，而不影响平衡位置。

---

Conclusion / 总结

Chemical equilibrium is a topic that rewards systematic thinking. If you can consistently apply Le Chatelier’s Principle — identifying the change, determining how the system will oppose it, and predicting the direction of shift — you will be able to handle the vast majority of A-Level equilibrium questions with confidence. Combine this with careful ICE-table construction for K_c calculations, and you have a complete toolkit for this topic.

化学平衡是一个奖励系统性思维的课题。如果你能够持续应用勒夏特列原理——识别变化、判断体系如何对抗该变化、预测移动方向——你就能自信地处理绝大多数A-Level平衡问题。将此与K_c计算中精心构建ICE表格相结合，你就拥有了处理这个主题的完整工具箱。

Remember the three golden rules: (1) Le Chatelier predicts the direction of shift, (2) only temperature changes K_c, and (3) catalysts affect rate, not position. Master these, and equilibrium will never trouble you again.

记住三条黄金法则：(1) 勒夏特列预测移动方向，(2) 只有温度改变K_c，(3) 催化剂影响速率而非位置。掌握这些，化学平衡将再也难不倒你。

— End / 完 —