Le Chatelier’s Principle and Chemical Equilibrium — A Complete Guide for A-Level Chemistry | 勒夏特列原理与化学平衡完全指南

Introduction to Chemical Equilibrium

Chemical equilibrium is one of the most important and frequently examined topics in A-Level Chemistry. It bridges the gap between reaction kinetics and thermodynamics, explaining why reactions reach a steady state rather than proceeding to completion. Understanding equilibrium is essential not only for exams but also for grasping real-world chemical processes — from industrial ammonia synthesis to the carbon dioxide balance in our blood.

化学平衡是 A-Level 化学中最重要且最常考的主题之一。它连接了反应动力学和热力学，解释了为什么化学反应会达到稳定状态而非进行到底。理解平衡不仅是考试需要，对于掌握现实世界的化学过程也至关重要——从工业合成氨到我们血液中的二氧化碳平衡。

---

What Is Dynamic Equilibrium?

A dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the backward reaction. The key word is “dynamic” — reactions do not stop. Instead, the forward and reverse reactions continue at equal rates, meaning that the macroscopic concentrations of all species remain constant while molecular-level processes continue uninterrupted.

Consider the reversible decomposition of dinitrogen tetroxide:

N₂O₄(g) ⇌ 2NO₂(g)

At equilibrium, N₂O₄ molecules continue to dissociate into NO₂ at the same rate that NO₂ molecules recombine to form N₂O₄. The brown colour of NO₂ reaches a constant intensity — visually confirming that equilibrium has been established.

动态平衡

当正向反应速率等于逆向反应速率时，封闭系统中就会出现动态平衡。关键词是”动态”——反应并没有停止。相反，正反应和逆反应以相等的速率继续进行，这意味着所有物质的宏观浓度保持不变，而分子层面的过程持续不断。

以四氧化二氮的可逆分解为例：N₂O₄(g) ⇌ 2NO₂(g)。在平衡状态下，N₂O₄ 分子继续分解为 NO₂，而 NO₂ 分子以相同的速率重新结合形成 N₂O₄。NO₂ 的棕色达到恒定强度——直观地确认了平衡的建立。

---

Le Chatelier’s Principle

When a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose that change. This is Le Chatelier’s Principle, named after the French chemist Henry Louis Le Chatelier who formulated it in 1884. It is a qualitative tool that predicts the direction of equilibrium shifts without requiring quantitative data.

勒夏特列原理

当一个处于动态平衡的系统受到条件变化的影响时，平衡位置将移动以抵消这种变化。这就是勒夏特列原理，以1884年提出该原理的法国化学家亨利·路易·勒夏特列命名。它是一个定性工具，可以在不需要定量数据的情况下预测平衡移动的方向。

The principle does NOT explain why the shift occurs thermodynamically — it is merely a predictive rule. The thermodynamic justification lies in the fact that systems evolve towards minimum Gibbs free energy.

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 (towards products) to consume the added reactant. Conversely, removing a product shifts the equilibrium to the right to replenish it.

Example: For the reaction Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq), adding more Fe³⁺ ions deepens the blood-red colour as more [FeSCN]²⁺ is formed. Removing SCN⁻ (e.g., by precipitation) shifts equilibrium left, fading the colour.

实例：反应 Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq) 中，加入更多 Fe³⁺ 离子会使血红色加深，因为生成了更多 [FeSCN]²⁺。除去 SCN⁻（例如通过沉淀）会使平衡左移，颜色变浅。

Effect of Pressure Changes | 压力变化的影响

Pressure changes only affect equilibria involving gases where there is a change in the number of gas molecules. When pressure is increased, the equilibrium shifts towards the side with fewer gas molecules to reduce pressure.

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

实例：在哈伯法中，N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，左边有4摩尔气体，右边有2摩尔。增加压力使平衡右移，有利于氨的生成。这就是为什么哈伯法在高压（约200 atm）下运行。

For reactions with equal moles of gas on both sides, such as H₂(g) + I₂(g) ⇌ 2HI(g), pressure changes have no effect on the equilibrium position.

Effect of Temperature Changes | 温度变化的影响

Temperature is the only condition that changes the value of the equilibrium constant K_c. For an exothermic reaction (ΔH < 0), increasing temperature shifts equilibrium to the left (endothermic direction) to absorb the added heat. For an endothermic reaction (ΔH > 0), increasing temperature shifts equilibrium to the right.

温度是唯一改变平衡常数 Kc 值的条件。对于放热反应（ΔH < 0），升温会使平衡向左（吸热方向）移动以吸收增加的热量。对于吸热反应（ΔH > 0），升温会使平衡向右移动。

Exam tip: Always identify whether the forward reaction is exothermic or endothermic before predicting the temperature effect. A common mistake is confusing “exothermic” with “shifts right” — the shift depends on whether you are heating or cooling the system.

Effect of Catalysts | 催化剂的影响

A catalyst does NOT affect the position of equilibrium. It lowers the activation energy equally for both the forward and reverse reactions, so the system reaches equilibrium faster but the equilibrium composition remains unchanged. This is a critical distinction that examiners love to test.

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，因此系统更快达到平衡，但平衡组成保持不变。这是考官喜欢考察的一个关键区别。

---

The Equilibrium Constant — K_c and K_p

Quantitative treatment of equilibrium requires equilibrium constants. For reactions in solution, we use K_c (concentration-based). For gas-phase reactions, we use K_p (pressure-based).

平衡的定量处理需要平衡常数。对于溶液中的反应，我们使用 Kc（基于浓度）。对于气相反应，我们使用 Kp（基于分压）。

K_c Expression | Kc 表达式

For a general reaction aA + bB ⇌ cC + dD:

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

Square brackets denote equilibrium concentrations in mol dm⁻³. The stoichiometric coefficients become exponents. Pure solids and liquids are excluded from the expression because their concentrations are effectively constant.

K_p Expression | Kp 表达式

For gas-phase reactions, K_p uses partial pressures instead of concentrations:

K_p = (p_C)^c(p_D)^d / (p_A)^a(p_B)^b

Where partial pressure p_A = mole fraction of A × total pressure. These are typically expressed in atmospheres (atm) or pascals (Pa).

其中分压 pA = A的摩尔分数 × 总压。通常以大气压（atm）或帕斯卡（Pa）表示。

Interpreting the Magnitude of K | 解读 K 值的大小

Kc Value	Interpretation | 解释	Example
Kc >> 1 (e.g., 1010)	Equilibrium lies far to the right; reaction essentially goes to completion | 平衡大幅右移；反应基本进行到底	HCl + NaOH → NaCl + H2O
Kc ≈ 1	Significant amounts of both reactants and products present | 反应物和产物都存在显著量	Esterification reactions
Kc << 1 (e.g., 10−10)	Equilibrium lies far to the left; very little product formed | 平衡大幅左移；产物很少	N2 + O2 ⇌ 2NO at 25°C

---

Calculations with K_c and K_p

Type 1: Calculating K_c from Equilibrium Data

Question: 0.50 mol of PCl₅ is placed in a 2.0 dm³ sealed container and heated to 250°C. At equilibrium, 0.30 mol of PCl₅ remains. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Calculate K_c.

Step-by-step:

1. Determine amounts at equilibrium:
   Initial PCl₅ = 0.50 mol. Reacted = 0.50 − 0.30 = 0.20 mol.
   From stoichiometry: PCl₃ formed = Cl₂ formed = 0.20 mol.
   Equilibrium amounts: PCl₅ = 0.30, PCl₃ = 0.20, Cl₂ = 0.20 mol.
2. Calculate concentrations:
   [PCl₅] = 0.30 / 2.0 = 0.15 mol dm⁻³
   [PCl₃] = [Cl₂] = 0.20 / 2.0 = 0.10 mol dm⁻³
3. Apply K_c expression:
   K_c = [PCl₃][Cl₂] / [PCl₅] = (0.10 × 0.10) / 0.15 = 0.067 mol dm⁻³

Type 2: Calculating K_p from Partial Pressures

Question: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At equilibrium at 400°C, the total pressure is 200 atm and the partial pressures are: p_N₂ = 50 atm, p_H₂ = 150 atm, p_NH₃ = 0 atm? That’s impossible — let’s use real data: At equilibrium, the mole fractions are N₂: 0.20, H₂: 0.60, NH₃: 0.20. Total pressure = 200 atm.

p_N₂ = 0.20 × 200 = 40 atm; p_H₂ = 0.60 × 200 = 120 atm; p_NH₃ = 0.20 × 200 = 40 atm

K_p = (p_NH₃)² / [p_N₂ × (p_H₂)³] = (40)² / [40 × (120)³] = 1600 / [40 × 1,728,000] = 1600 / 69,120,000 = 2.31 × 10⁻⁵ atm⁻²

🚨 Common mistake: Forgetting to include units with K_p. The units depend on the stoichiometry: for this reaction, units are atm⁻² because the sum of exponents in numerator (2) minus denominator (1+3=4) gives −2.

考试计算要点 | Exam Calculation Tips

• Always start with a balanced equation — stoichiometry errors cascade
• Use an ICE table (Initial, Change, Equilibrium) for systematic tracking
• Round to 3 significant figures unless stated otherwise
• For K_p calculations, ensure your mole fractions sum to 1.0
• Double-check units: K_c has units of (mol dm⁻³)^Δn; K_p has units of (atm or Pa)^Δn

---

The Relationship Between K and Temperature

The van ‘t Hoff equation quantifies how K varies with temperature:

ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁)

Where ΔH° is the standard enthalpy change and R = 8.314 J K⁻¹ mol⁻¹.

Van ‘t Hoff 方程量化了 K 如何随温度变化：ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁)。其中 ΔH° 是标准焓变，R = 8.314 J K⁻¹ mol⁻¹。

This equation formalises Le Chatelier’s Principle: for exothermic reactions (ΔH° < 0), increasing T makes ln(K₂/K₁) positive (K decreases). For endothermic reactions (ΔH° > 0), increasing T increases K.

---

Industrial Applications | 工业应用

The Haber Process | 哈伯法

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

Compromise conditions: 400–450°C, 200 atm, iron catalyst. Lower temperature favours higher yield (exothermic) but slows the rate — hence the compromise. Produces ~150 million tonnes of ammonia annually for fertilisers.

妥协条件：400–450°C，200 atm，铁催化剂。较低温度有利于更高产率（放热反应），但会减慢速率——因此需要妥协。每年生产约1.5亿吨氨用于肥料。

The Contact Process | 接触法

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

Conditions: 450°C, 1–2 atm, V₂O₅ catalyst. Used to produce sulfuric acid, the world’s most manufactured chemical.

条件：450°C，1–2 atm，V₂O₅ 催化剂。用于生产硫酸——世界上制造量最大的化学品。

---

Common Exam Questions and Pitfalls

Q1: “State the effect of adding a catalyst on the equilibrium yield.”
A: No effect. A catalyst increases the rate of both forward and reverse reactions equally, so equilibrium position is unchanged. The system reaches equilibrium faster but the composition at equilibrium is identical.

Q1: “陈述加入催化剂对平衡产率的影响。”
答：无影响。催化剂同等程度地增加正反应和逆反应的速率，因此平衡位置不变。系统更快达到平衡，但平衡组成不变。

Q2: “Explain why K_c has no units for this reaction.”
A: When the sum of stoichiometric coefficients on both sides of the equation is equal (Δn = 0), the concentration terms in the numerator and denominator cancel dimensionally.

Q2: “解释为什么该反应的 Kc 没有单位。”
答：当方程两边的化学计量系数之和相等时（Δn = 0），分子和分母中的浓度项在量纲上相消。

Q3: “A student claims that at equilibrium, the concentrations of reactants and products are equal. Explain why this is incorrect.”
A: At equilibrium, the rates of forward and reverse reactions are equal — not the concentrations. The actual concentrations depend on the value of K_c. Unless K_c = 1, the concentrations are different.

Q3: “某学生声称在平衡状态下，反应物和产物的浓度相等。解释为什么这是错误的。”
答：在平衡状态下，正反应和逆反应的速率相等——而非浓度。实际浓度取决于 Kc 的值。除非 Kc = 1，否则浓度是不同的。

---

Summary: Key Concepts Checklist

Concept | 概念	Key Point | 要点
Dynamic equilibrium 动态平衡	Rate forward = Rate reverse; concentrations constant but not necessarily equal
Le Chatelier’s Principle 勒夏特列原理	System shifts to oppose imposed change; qualitative only
Kc and Kp	Kc uses concentrations; Kp uses partial pressures; include units
Temperature 温度	Only condition that changes K value; van ‘t Hoff equation quantifies it
Catalyst 催化剂	No effect on position or K; only increases rate of reaching equilibrium
Haber Process 哈伯法	Compromise T (450°C), high P (200 atm), Fe catalyst; exothermic

---

Practice Questions | 练习题

1. For the reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹, predict and explain the effect of: (a) increasing pressure, (b) increasing temperature, (c) adding a catalyst on the equilibrium yield of methanol.

2. 0.80 mol of SO₂ and 0.40 mol of O₂ are mixed in a 5.0 dm³ vessel at 700 K. At equilibrium, 0.30 mol of SO₃ is formed. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), calculate K_c and its units.

3. The equilibrium constant K_c for the esterification of ethanoic acid with ethanol is 4.0 at 25°C. If 1.0 mol of acid and 1.0 mol of alcohol are mixed, calculate the equilibrium yield of ester.

---

This guide covers all major A-Level Chemistry specifications including AQA, Edexcel, OCR, and CIE. For past paper practice and more detailed explanations, browse the Chemistry section on aleveler.com. Good luck with your studies!

本指南涵盖所有主要 A-Level 化学考试局，包括 AQA、Edexcel、OCR 和 CIE。如需历年真题练习和更详细的解释，请浏览 aleveler.com 的化学板块。祝学习顺利！