IB & WJEC Chemistry: Chemical Equilibrium – Key Points Explained | IB与WJEC化学：化学平衡考点精讲

📚 IB & WJEC Chemistry: Chemical Equilibrium – Key Points Explained | IB与WJEC化学：化学平衡考点精讲

Chemical equilibrium is a fundamental concept that bridges reaction kinetics, thermodynamics, and industrial processes. For both IB and WJEC Chemistry, a thorough understanding of dynamic equilibrium, the equilibrium law, Le Chatelier’s principle, and associated calculations is essential. This guide distils the key points, clarifies common misconceptions, and provides the rigour required for exam success.

化学平衡是连接反应动力学、热力学和工业过程的核心概念。无论对于IB还是WJEC化学课程，深刻理解动态平衡、平衡定律、勒夏特列原理以及相关计算都至关重要。本指南提炼了考试要点，澄清了常见误区，并提供了考取高分所需的严谨讲解。

1. What is Chemical Equilibrium? | 什么是化学平衡？

Chemical equilibrium is the state reached by a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. The concentrations of reactants and products remain constant, but they are not necessarily equal. This constancy is a macroscopic observation; at the molecular level, both reactions continue to occur.

化学平衡是可逆反应达到的一种状态，此时正反应速率与逆反应速率相等。反应物和产物的浓度保持恒定，但二者不一定相等。这种恒定是宏观观察结果；在分子水平上，正逆反应仍在持续进行。

For a general reaction: aA + bB ⇌ cC + dD, equilibrium is achieved in a closed system. No matter enters or leaves, so the total mass remains unchanged. Equilibrium can be approached from either direction: starting with only reactants or only products will eventually reach the same equilibrium composition under identical conditions.

对于一般反应：aA + bB ⇌ cC + dD，平衡在封闭体系中达成。没有物质进出，因此总质量不变。平衡可以从任一方向趋近：无论从纯反应物还是纯产物出发，在相同条件下最终都会达到相同的平衡组成。

2. Dynamic Nature of Equilibrium | 平衡的动态特性

Equilibrium is dynamic, not static. At equilibrium, the forward and reverse reactions do not stop; they proceed at identical rates. This dynamic character can be demonstrated using isotopic labelling. For instance, in the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), introducing radioactive iodine into a system already at equilibrium results in the isotope being incorporated into both I₂ and HI, proving that both forward and reverse reactions are active.

平衡是动态的，而非静止的。达到平衡时，正逆反应并未停止，而是以相等的速率进行。这种动态特性可用同位素标记法证明。例如，在平衡体系 H₂(g) + I₂(g) ⇌ 2HI(g) 中，将放射性碘引入已平衡的体系，会发现同位素同时出现在 I₂ 和 HI 中，这就证明了正逆反应都在进行。

A common misconception is that at equilibrium the amounts of reactants and products are equal. In fact, their concentrations become constant but can be vastly different, dictated by the equilibrium constant.

一个常见的误区是认为平衡时反应物和产物的量相等。实际上，它们的浓度保持恒定，但数值可能差异巨大，由平衡常数决定。

3. The Equilibrium Constant, Kc | 平衡常数 Kc

The equilibrium constant Kc provides a quantitative measure of the position of equilibrium for reactions in solution. For the general reaction aA + bB ⇌ cC + dD, the expression is:

平衡常数 Kc 定量描述了溶液中反应平衡的位置。对于一般反应 aA + bB ⇌ cC + dD，其表达式为：

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

Here, [X] denotes the equilibrium concentration of X in mol dm⁻³. The stoichiometric coefficients a, b, c, d appear as exponents. It is crucial to remember that pure solids and pure liquids are omitted from the Kc expression; only species in the gaseous or aqueous phase are included.

式中 [X] 表示 X 的平衡浓度，单位为 mol dm⁻³。化学计量数 a、b、c、d 作为指数出现。务必牢记，纯固体和纯液体不出现在 Kc 表达式中；只有气相或水溶液中的物种才被纳入。

Both IB and WJEC syllabuses require students to write Kc expressions, calculate Kc from given data, and deduce the units of Kc. The units depend on the sum of powers of concentration terms, and students must be able to derive them, e.g., (mol dm⁻³)ⁿ.

IB 和 WJEC 大纲均要求学生写出 Kc 表达式，根据给定数据计算 Kc，并推导 Kc 的单位。单位取决于浓度项幂次的总和，学生必须能导出如 (mol dm⁻³)ⁿ 的形式。

4. The Equilibrium Constant, Kp | 平衡常数 Kp

For gaseous equilibria, Kp is used, expressed in terms of partial pressures. The partial pressure of a gas is the pressure it would exert if it alone occupied the container. Using the same general equation, Kp is given by:

对于气体平衡，使用 Kp，以分压表示。气体的分压是指它单独占据容器时所施加的压力。沿用相同的通用方程式，Kp 表示为：

Kp = (P_C)ᶜ (P_D)ᵈ / (P_A)ᵃ (P_B)ᵇ

Partial pressure of a gas = mole fraction × total pressure. Mole fraction = (moles of that gas) / (total moles of gas). Kp is dimensionless only when the sum of the powers in the numerator equals that in the denominator; otherwise it has pressure units such as atm or Pa.

气体的分压 = 摩尔分数 × 总压。摩尔分数 = (该气体的物质的量) / (气体总物质的量)。只有当分子和分母的幂次总和相等时 Kp 才无量纲；否则带压力单位，如 atm 或 Pa。

WJEC often includes calculations of Kp from partial pressures or total pressure and composition data. IB may also link Kp to Gibbs free energy changes.

WJEC 常考查根据分压或总压与组成数据计算 Kp。IB 还可能将 Kp 与吉布斯自由能变联系起来。

5. Magnitude of K and Reaction Quotient, Q | K 值大小与反应商 Q

The magnitude of K indicates the position of equilibrium. A very large K (>>1) means the equilibrium lies far to the right, with products favoured. A very small K (<<1) means reactants are favoured. When K is close to 1, both reactants and products are present in comparable amounts.

K 值大小指示平衡的位置。K 非常大（>>1）意味着平衡强烈倾向于右侧，产物占优。K 非常小（<<1）意味着反应物占优。当 K 接近 1 时，反应物和产物的量大致相当。

The reaction quotient, Q, has the same form as the equilibrium constant expression, but uses the concentrations or pressures at any point in time, not necessarily at equilibrium. Comparing Q with K allows prediction of the direction of reaction:

反应商 Q 的表达式与平衡常数相同，但使用的是任意时刻的浓度或压力，而不一定是平衡状态。将 Q 与 K 进行比较，可以预测反应方向：

If Q < K, the forward reaction will be favoured to reach equilibrium. 如果 Q < K，正反应将更有利，直至达到平衡。
If Q > K, the reverse reaction will be favoured. 如果 Q > K，逆反应有利。
If Q = K, the system is at equilibrium. 如果 Q = K，体系处于平衡状态。

This concept is heavily tested in both IB and WJEC, particularly in predicting shifts in a disturbed equilibrium.

这一概念在 IB 和 WJEC 中都是重点考查内容，尤其用于预测平衡被扰乱时的移动方向。

6. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states that if a system at dynamic equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium will shift to counteract the change. It is a qualitative tool that predicts the direction of shift, but not the rate or the new equilibrium concentrations.

勒夏特列原理指出，如果处于动态平衡的体系受到浓度、压强或温度的改变，平衡位置会向着抵消这一改变的方向移动。这是一个定性工具，可预测移动方向，但不能预测速率或新的平衡浓度。

This principle is fundamental to understanding how chemical systems respond to external stresses. Both syllabuses require students to apply the principle to concrete examples, such as the Haber process and the Contact process, as well as to explain observed colour changes.

这一原理对于理解化学体系如何响应外界压力至关重要。两个大纲都要求学生将该原理应用于具体实例，如哈伯法和接触法，并解释观察到的颜色变化。

7. Effect of Concentration Changes | 浓度变化的影响

Increasing the concentration of a reactant shifts the equilibrium to the right, producing more products until a new equilibrium is established. Conversely, increasing the concentration of a product shifts the equilibrium to the left. Removing a substance shifts the position towards the side that replenishes it.

增大反应物浓度会使平衡右移，生成更多产物，直至建立新平衡。相反，增大产物浓度会使平衡左移。移除某种物质会使平衡向补充该物质的一侧移动。

It is important to note that while the position of equilibrium changes, the value of Kc or Kp remains constant as long as the temperature is unchanged. This is a key distinction: K is only temperature-dependent.

值得注意的是，虽然平衡位置改变了，但只要温度不变，Kc 或 Kp 的值保持不变。这是一个关键区别：K 仅依赖于温度。

8. Effect of Pressure Changes | 压强变化的影响

Pressure changes affect only gaseous equilibria where there is a change in the total number of gas molecules. Increasing pressure (by decreasing volume) favours the side with fewer moles of gas, decreasing the total number of molecules and thus reducing the pressure. Decreasing pressure favours the side with more moles of gas.

压强变化只影响气体总分子数发生改变的气体平衡。增大压强（通过缩小体积）有利于气体分子数较少的一侧，从而减少分子总数并降低压强。减小压强有利于气体分子数较多的一侧。

If the number of moles of gas is equal on both sides, a change in pressure has no effect on the position of equilibrium. The value of Kp remains unchanged with pressure alterations.

如果两边气体物质的量相等，压强变化对平衡位置无影响。压强改变不会改变 Kp 的值。

Reaction	Effect of Increasing Pressure
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (4 mol → 2 mol)	Shifts right, favouring NH₃
H₂(g) + I₂(g) ⇌ 2HI(g) (2 mol → 2 mol)	No shift in equilibrium position
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (3 mol → 2 mol)	Shifts right, favouring SO₃

Adding an inert gas at constant volume does not change partial pressures of reacting species, so equilibrium position is unaffected.

在恒容条件下加入惰性气体不会改变反应物种的分压，因此平衡位置不受影响。

9. Effect of Temperature Changes | 温度变化的影响

Temperature is the only factor that changes the value of the equilibrium constant K. For an exothermic reaction (ΔH negative), increasing temperature shifts the equilibrium to the left (favouring the endothermic direction) and K decreases. For an endothermic reaction (ΔH positive), increasing temperature shifts the equilibrium to the right, and K increases.

温度是唯一能改变平衡常数 K 值的因素。对于放热反应（ΔH 负值），升高温度使平衡左移（有利于吸热方向），K 值减小。对于吸热反应（ΔH 正值），升高温度使平衡右移，K 值增大。

This can be rationalised by treating heat as a reactant or product. In exothermic reactions, heat is a product; adding heat drives the reaction backwards. In endothermic reactions, heat is a reactant; adding heat drives the reaction forwards.

这可以通过将热量视为反应物或产物来理解。在放热反应中，热量是产物；加入热量会使反应逆向进行。在吸热反应中，热量是反应物；加入热量会使反应正向进行。

IB may also relate temperature dependence to the van’t Hoff equation and the Gibbs free energy relationship; WJEC expects qualitative application.

IB 还可能将温度依赖性联系到范特霍夫方程和吉布斯自由能关系；WJEC 则要求定性的应用。

10. Effect of Catalysts | 催化剂的影响

A catalyst provides an alternative reaction pathway with a lower activation energy. It increases the rate of both the forward and reverse reactions equally. Consequently, a catalyst does not affect the position of equilibrium or the value of K. It merely allows the system to reach equilibrium more quickly.

催化剂提供了活化能较低的另一条反应路径。它同等程度地加快正、逆反应速率。因此，催化剂不影响平衡位置，也不改变 K 值。它只是让体系更快地达到平衡。

In industrial processes, a catalyst is often used to achieve a high rate of production without sacrificing yield—temperature and pressure adjustments then manage the equilibrium position.

在工业生产中，常使用催化剂来获得高产率而不牺牲转化率——温度与压强的调整则控制平衡位置。

11. Calculating Equilibrium Concentrations – ICE Tables | 平衡浓度计算 – ICE 表格

Both IB and WJEC require systematic calculations of equilibrium concentrations using initial concentrations, changes, and equilibrium concentrations (ICE). The steps are: write the balanced equation, set up an ICE table, express the change using ‘x’, substitute into the Kc expression, solve for ‘x’, and then calculate the equilibrium concentrations.

IB 和 WJEC 都要求使用初始浓度、变化量和平衡浓度（ICE）表格体系化地计算平衡浓度。步骤为：写出配平的方程式，建立 ICE 表格，用 x 表示变化量，代入 Kc 表达式，解出 x，然后计算平衡浓度。

Example: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), if initial moles of H₂ and I₂ are 1.00 mol in a 1 dm³ vessel and Kc = 50 at a certain temperature, set up: Initial: [H₂]=1, [I₂]=1, [HI]=0; Change: -x, -x, +2x; Equilibrium: 1-x, 1-x, 2x. Then 50 = (2x)² / (1-x)², solve to find x and equilibrium composition.

示例：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，若 H₂ 和 I₂ 的初始物质的量为 1.00 mol，容器体积 1 dm³，在某温度下 Kc = 50，则可建立表格：初始 [H₂]=1, [I₂]=1, [HI]=0；变化 -x, -x, +2x；平衡 1-x, 1-x, 2x。代入得 50 = (2x)² / (1-x)²，解出 x，进而得到平衡组成。

WJEC often includes problems where units of Kc must be calculated and the approximation that x is small may be tested, while IB may involve more algebraic manipulation.

WJEC 常涉及必须计算 Kc 单位的问题，并可能考查 x 很小时的近似处理，而 IB 则可能要求更多代数操作。

12. Industrial Application: The Haber Process | 工业应用：哈伯法

The Haber process for ammonia synthesis is a classic example for applying equilibrium principles: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ mol⁻¹. The exothermic forward reaction is favoured by low temperature and high pressure (as 4 moles → 2 moles). However, a low temperature slows the rate, so a compromise temperature of about 450°C is used with an iron catalyst to give a fast reaction and an acceptable yield. Pressure is set around 200 atm to shift equilibrium right while balancing plant costs.

合成氨的哈伯法是应用平衡原理的经典实例：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92 kJ mol⁻¹。放热正向反应受低温和高压（4 mol → 2 mol）的促进。但低温会减慢速率，因此采用约450°C的折中温度并使用铁催化剂，以获得较快的反应和可接受的产率。压力设置在200 atm左右，在右移平衡的同时兼顾设备成本。

Removal of NH₃ as it forms also shifts equilibrium to the right, maximising yield. This industrial strategy is thoroughly examined in both syllabuses, linking equilibrium, kinetics, and economics.

不断将生成的 NH₃ 移除也使平衡右移，从而最大化产率。这一工业策略被两个大纲深入考查，连接了平衡、动力学和经济学。

13. Linking Equilibrium to Thermodynamics (IB focus) | 平衡与热力学的联系（IB 重点）

In IB Chemistry, the relationship between standard Gibbs free energy change and the equilibrium constant is given by: ΔG° = -RT ln K. This equation links thermodynamics and equilibrium. A negative ΔG° corresponds to K > 1, indicating a product-favoured equilibrium; a positive ΔG° gives K < 1.

IB 化学中，标准吉布斯自由能变与平衡常数的关系为：ΔG° = -RT ln K。这个公式将热力学与平衡联系起来。ΔG° 为负对应 K > 1，表示平衡有利于产物；ΔG° 为正则 K < 1。

Furthermore, substituting ΔG° = ΔH° – TΔS° yields ln K = -ΔH°/RT + ΔS°/R, which explains temperature dependence. A plot of ln K versus 1/T gives a straight line of slope -ΔH°/R, allowing experimental determination of enthalpy change. While this is primarily IB territory, it can provide deeper insight for WJEC students as well.

进一步代入 ΔG° = ΔH° – TΔS° 得到 ln K = -ΔH°/RT + ΔS°/R，这解释了温度依赖性。作 ln K 对 1/T 的图可得斜率为 -ΔH°/R 的直线，从而通过实验测定焓变。这主要是 IB 的内容，但也能为 WJEC 学生提供更深入的理解。

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IB & WJEC Chemistry: Chemical Equilibrium – Key Points Explained | IB与WJEC化学：化学平衡考点精讲