Chemical Equilibrium | 化学平衡考点精讲

📚 Chemical Equilibrium | 化学平衡考点精讲

In A-Level OCR Chemistry, chemical equilibrium is a core topic that bridges kinetics, thermodynamics, and industrial processes. Understanding the dynamic nature of reversible reactions, the principles that govern the position of equilibrium, and the quantitative treatment through equilibrium constants (Kc and Kp) is essential for success. This guide breaks down every key concept, providing clear explanations, worked examples, and exam-focused tips to help you master equilibrium and tackle both structured and calculation questions with confidence.

在A-Level OCR化学中，化学平衡是一个连接动力学、热力学和工业过程的核心主题。理解可逆反应的动态本质、支配平衡位置的原则，以及通过平衡常数（Kc 和 Kp）进行的定量处理，对于取得成功至关重要。本指南分解了每一个关键概念，提供清晰的解释、实例和应试技巧，帮助你掌握平衡，自信应对结构化问题和计算题。

1. Dynamic Equilibrium | 动态平衡

In a closed system, a reversible reaction proceeds in both the forward and reverse directions. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant—but not necessarily equal. This is a dynamic equilibrium because molecular processes continue in both directions at the same speed.

在封闭系统中，可逆反应同时向正反应方向和逆反应方向进行。达到平衡时，正反应速率等于逆反应速率，反应物和产物的浓度保持不变——但两者浓度不一定相等。这是一种动态平衡，因为分子层面的正逆过程仍在以相同的速率持续进行。

It is crucial to recognise that dynamic equilibrium can only be established in a closed system, where no matter enters or leaves. The macroscopic properties (colour, pressure, concentration) do not change, yet at the molecular level, there is constant interconversion of reactants and products.

关键在于，动态平衡只能在封闭系统中建立，即没有物质进入或离开。宏观性质（颜色、压力、浓度）不变，但在分子层面，反应物和产物之间不断地相互转化。

Closed system: A system that exchanges energy but not matter with its surroundings. | 封闭系统：与环境交换能量但不交换物质的系统。
Dynamic: The forward and reverse reactions are ongoing. | 动态：正反应和逆反应持续进行。
Equilibrium condition: Rate_forward = Rate_reverse. | 平衡条件：正反应速率 = 逆反应速率。

Common examples include the Haber process (N₂ + 3H₂ ⇌ 2NH₃) and the Contact process (2SO₂ + O₂ ⇌ 2SO₃). For these reactions, equilibrium lies toward the products under certain conditions, but the reverse reaction always competes.

常见例子包括哈伯法（N₂ + 3H₂ ⇌ 2NH₃）和接触法（2SO₂ + O₂ ⇌ 2SO₃）。在这些反应中，平衡在一定条件下倾向于产物一侧，但逆反应始终与之竞争。

2. 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 shifts to oppose that change. This qualitative rule helps predict the direction of shift and the resulting effect on the yield of products.

勒夏特列原理指出，如果处于动态平衡的系统受到浓度、压力或温度变化的影响，平衡位置会发生移动以抵消该变化。这一定性规则有助于预测移动方向以及对产物产率的影响。

The principle does not explain the rate of reaction or the magnitude of the shift; it only indicates the direction in which the equilibrium will adjust. It is particularly useful for optimising industrial processes, where the aim is to maximise yield while minimising energy and material costs.

该原理不能解释反应速率或移动幅度；它只表明平衡调整的方向。它在优化工业流程中特别有用，因为目标是最大化产率，同时最小化能量和材料成本。

Change	Effect on Equilibrium Position
Increase concentration of reactant	Shifts to the right (favours forward reaction) to consume added reactant \| 增加反应物浓度 → 平衡向右移动，消耗额外的反应物
Decrease concentration of product	Shifts to the right to replace removed product \| 减少产物浓度 → 平衡向右移动，补充被移除的产物
Increase pressure (gaseous systems)	Shifts toward side with fewer moles of gas \| 增加压力 → 向气体摩尔数较少的一侧移动
Increase temperature (exothermic ΔH < 0)	Shifts to the left (favours endothermic reverse reaction) \| 放热反应升温 → 平衡向左移动（吸热方向）
Increase temperature (endothermic ΔH > 0)	Shifts to the right (favours endothermic forward reaction) \| 吸热反应升温 → 平衡向右移动
Add a catalyst	No shift in position; rates of both forward and reverse reactions increase equally, reaching equilibrium faster \| 催化剂不改变平衡位置；同等加快正逆反应速率，更快达到平衡

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

When the concentration of a reactant is increased, the system responds by favouring the forward reaction to reduce that concentration. Conversely, removing a product as it forms draws the equilibrium to the right, increasing overall yield. This is a common strategy in continuous processes where products are continuously removed to drive the reaction forward.

当反应物浓度增加时，系统通过促进正反应来降低该浓度。相反，当产物生成时将其移除，会将平衡拉向右侧，提高总产率。这是连续工艺中的常见策略，即持续移除产物以推动反应进行。

For instance, in esterification (carboxylic acid + alcohol ⇌ ester + water), adding excess alcohol or removing water (using a dehydrating agent) shifts equilibrium to the right, increasing ester yield. The effect of concentration changes does not alter the equilibrium constant (Kc); it merely adjusts the relative amounts of species until Kc is re-established.

例如，在酯化反应（羧酸 + 醇 ⇌ 酯 + 水）中，加入过量醇或移除水（使用脱水剂）会使平衡向右移动，提高酯的产率。浓度变化不会改变平衡常数（Kc）；它只是调整各组分的相对量，直到重新建立平衡常数。

Key takeaway: Only changes in concentration that affect the reacting species can shift equilibrium. Adding an inert solid or a non-reactive substance does not alter the position. In aqueous equilibrium, dilution may shift position if water is a reactant or product.

关键点：只有影响反应物种的浓度变化才能移动平衡。加入惰性固体或不反应物质不会改变平衡位置。在水溶液平衡中，若水是反应物或产物，稀释可能影响平衡。

4. Effect of Pressure Changes | 压力变化的影响

Pressure changes only affect equilibria involving gases, and only when there is a difference in the total number of gaseous moles between reactants and products. Increasing the total pressure forces the equilibrium to shift towards the side with fewer gas molecules, minimising the imposed pressure increase.

压力变化只影响涉及气体的平衡，且仅当反应物和产物之间的气体总摩尔数存在差异时。增大总压力会迫使平衡向气体分子数较少的一侧移动，以减小施加的压力增加。

Consider the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The left side has 4 moles of gas, the right side has 2 moles. High pressure favours ammonia formation. This is why the Haber process operates at around 200 atm. If the number of moles of gas is equal on both sides (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)), changing pressure has no effect on the position of equilibrium.

考虑反应：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。左边有4摩尔气体，右边有2摩尔。高压有利于氨的生成。这就是哈伯法在约200个大气压下操作的原因。如果两侧气体摩尔数相等（如H₂(g) + I₂(g) ⇌ 2HI(g)），压力变化对平衡位置没有影响。

It is often tested that adding an inert gas at constant volume does not affect the equilibrium because the partial pressures of the reacting gases remain unchanged. However, adding an inert gas at constant total pressure (increasing volume) can shift equilibrium towards the side with more moles, as partial pressures of reactants are reduced.

常考的一点是，在恒定体积下加入惰性气体不会影响平衡，因为反应气体的分压保持不变。但若在恒总压下加入惰性气体（体积增大），可能使平衡向摩尔数多的一侧移动，因为反应物分压降低。

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

Temperature is the only factor that changes the value of the equilibrium constant (Kc or Kp). According to Le Chatelier’s principle, increasing temperature shifts equilibrium in the endothermic direction to absorb the added heat. For an exothermic forward reaction (ΔH < 0), raising temperature reduces the yield of products, and Kc decreases. For an endothermic forward reaction (ΔH > 0), Kc increases with temperature.

温度是唯一改变平衡常数（Kc 或 Kp）数值的因素。根据勒夏特列原理，升高温度会使平衡向吸热方向移动，以吸收额外的热量。如果正反应放热（ΔH < 0），升温会降低产物产率，Kc 减小。如果正反应吸热（ΔH > 0），Kc 随温度升高而增大。

Consider 2NO₂(g) ⇌ N₂O₄(g) (ΔH = -58 kJ mol⁻¹). The forward reaction is exothermic. At low temperatures, the equilibrium mixture contains more N₂O₄ (colourless) and appears pale; at high temperatures, it shifts towards more NO₂ (brown), and the colour deepens. This colour change is a vivid classroom demonstration of temperature effects.

考虑 2NO₂(g) ⇌ N₂O₄(g)（ΔH = -58 kJ mol⁻¹）。正反应放热。低温下平衡混合物含更多无色 N₂O₄，颜色较浅；高温下移向更多棕色 NO₂，颜色加深。这一颜色变化是温度效应的生动课堂演示。

In exothermic reactions, a compromise temperature is often chosen industrially: low enough for a good yield, but high enough for a reasonable rate. That is why the Haber process uses ~450°C with an iron catalyst, balancing yield and kinetics.

在放热反应中，工业上通常会选择折中温度：足够低以获得良好产率，但足够高以保证合理的反应速率。这就是哈伯法采用约450°C并配合铁催化剂的原因，以平衡产率和动力学。

6. Effect of Catalysts | 催化剂的影响

A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the rate of both the forward and reverse reactions equally. It does not change the position of equilibrium or the value of the equilibrium constant. Its sole role is to enable the system to reach equilibrium more quickly.

催化剂提供了活化能较低的反应路径，同等地提高了正反应和逆反应的速率。它不会改变平衡位置或平衡常数的数值。其唯一作用是使系统更快地达到平衡。

In an industrial context, catalysts are invaluable because they allow equilibrium to be attained at lower temperatures, saving energy and improving process efficiency without sacrificing theoretical yield. For example, vanadium(V) oxide (V₂O₅) in the Contact process and iron in the Haber process both accelerate equilibration.

在工业环境中，催化剂非常宝贵，因为它们可以在较低温度下达到平衡，节省能源并提高工艺效率，而不牺牲理论产率。例如，接触法中的五氧化二钒（V₂O₅）和哈伯法中的铁都能加速平衡的建立。

Note that a catalyst does not alter ΔH or ΔS for the reaction, and therefore does not change the thermodynamics, only the kinetics. Exam questions frequently ask students to distinguish between rate effects and equilibrium position effects when a catalyst is added.

请注意，催化剂不改变反应的 ΔH 或 ΔS，因此不改变热力学，只改变动力学。考试题经常要求学生区分加入催化剂时的速率效应和平衡位置效应。

7. The Equilibrium Constant Kc | 平衡常数 Kc

For a general homogeneous equilibrium in solution: aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration, Kc, is defined as:

对于溶液中的一般均相平衡：aA + bB ⇌ cC + dD，用浓度表示的平衡常数 Kc 定义为：

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

Square brackets denote equilibrium concentrations in mol dm⁻³. Kc is dimensionless when the sum of powers of concentration in the numerator equals that in the denominator, but in OCR calculations you may be required to state units where appropriate, typically (mol dm⁻³)^(∑ν).

方括号表示平衡浓度，单位为 mol dm⁻³。当分子中浓度幂的和等于分母中浓度幂的和时，Kc 无量纲，但在 OCR 计算中有时要求注明单位，通常为 (mol dm⁻³)^(∑ν)。

Kc is constant for a given reaction at a specified temperature. Its magnitude indicates the extent of the reaction: a large Kc (>10¹⁰) means equilibrium lies far to the right (products favoured); a very small Kc (<10⁻¹⁰) means reactants dominate. Intermediate values suggest significant amounts of both reactants and products.

对于给定的反应，在指定温度下 Kc 是一个常数。其数值大小指示反应进行的程度：Kc 很大（>10¹⁰）意味着平衡远偏右（产物为主）；Kc 非常小（<10⁻¹⁰）意味着反应物占主导。中等数值则表明反应物和产物都有可观的数量。

Only species in the same phase appear in the Kc expression (homogeneous equilibria). Solids and pure liquids are omitted because their ‘concentration’ is essentially constant. For heterogeneous equilibria, only gases and aqueous species are included.

只有同一相中的物质出现在 Kc 表达式中（均相平衡）。固体和纯液体被省略，因为它们的“浓度”基本上是常数。对于多相平衡，只包括气体和水溶液中的物种。

8. The Equilibrium Constant Kp | 平衡常数 Kp

For gaseous equilibria, it is often more convenient to use partial pressures instead of concentrations. For a reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp is given by:

对于气相平衡，使用分压通常比浓度更方便。对于反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)，Kp 表达式为：

Kp = (p_C)ᶜ (p_D)ᵈ / (p_A)ᵃ (p_B)ᵇ

where p_X is the partial pressure of gas X, typically measured in atmospheres (atm), pascals (Pa), or kPa. Partial pressure of a gas = (mole fraction of that gas) × (total pressure). The mole fraction of a component is its number of moles divided by the total number of moles in the gaseous mixture.

其中 p_X 是气体 X 的分压，通常以大气压（atm）、帕斯卡（Pa）或千帕（kPa）为单位。气体的分压 = （该气体的摩尔分数）×（总压）。某组分的摩尔分数等于其摩尔数除以气体混合物中的总摩尔数。

Like Kc, Kp is constant at a given temperature and its magnitude reflects the position of equilibrium. Kp also has units unless the sum of powers of pressure in the numerator equals that in the denominator; otherwise the unit is (pressure)^(Δn), where Δn = (c+d) – (a+b).

与 Kc 类似，Kp 在给定温度下是一个常数，其数值大小反映平衡位置。Kp 也有单位，除非分子和分母中压力幂的和相等；否则单位为 (压力)^(Δn)，其中 Δn = (c+d) – (a+b)。

OCR exam questions commonly ask you to calculate partial pressures from mole fractions, then compute Kp, or to determine the effect of pressure on Kp (no effect) and on the equilibrium position via Le Chatelier.

OCR 考试题通常要求从摩尔分数计算分压，然后计算 Kp，或者确定压力对 Kp 的影响（无影响）以及通过勒夏特列原理对平衡位置的影响。

9. Calculations Involving Kc | Kc 相关计算

Typical OCR questions provide initial amounts, the volume of the container, and the equilibrium amount or concentration of one species. From this, you construct an ICE table (Initial, Change, Equilibrium) to find equilibrium concentrations of all species and then calculate Kc.

典型的 OCR 题目会给出初始量、容器体积以及某一物种的平衡量或浓度。据此，你需要构建一个 ICE 表（初始、变化、平衡），求出所有物种的平衡浓度，然后计算 Kc。

Example: 1.0 mol of PCl₅ is placed in a 2.0 dm³ vessel and heated to 250°C. At equilibrium, 0.40 mol of Cl₂ is present. Reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). First, deduce change: x = 0.40 mol. Then equilibrium moles: PCl₅ = 1.0 – 0.40 = 0.60 mol; PCl₃ = 0.40 mol; Cl₂ = 0.40 mol. Concentrations: [PCl₅]=0.60/2=0.30 M; [PCl₃]=[Cl₂]=0.20 M. Kc = (0.20×0.20)/0.30 = 0.133 mol dm⁻³.

示例：将1.0 mol PCl₅ 放入一个2.0 dm³容器中，加热至250°C。平衡时，存在0.40 mol Cl₂。反应为：PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)。首先，推断变化量：x = 0.40 mol。平衡摩尔数：PCl₅ = 1.0 – 0.40 = 0.60 mol；PCl₃ = 0.40 mol；Cl₂ = 0.40 mol。浓度：[PCl₅]=0.60/2=0.30 M；[PCl₃]=[Cl₂]=0.20 M。Kc = (0.20×0.20)/0.30 = 0.133 mol dm⁻³。

Always check that equilibrium concentrations are positive. If you obtain a negative concentration, re-evaluate the direction of change. Units for Kc in this case: mol dm⁻³.

始终检查平衡浓度是否为正值。若得到负浓度，则需重新评估变化的方向。本例中 Kc 的单位为 mol dm⁻³。

For reactions with equal numbers of moles on both sides, Kc will be dimensionless; for others, include units. OCR often expects units expressed as, for example, mol⁻¹ dm³.

对于两侧摩尔数相等的反应，Kc 无量纲；否则需写明单位。OCR常要求将单位表示为如 mol⁻¹ dm³ 等形式。

10. Calculations Involving Kp | Kp 相关计算

To find Kp, start by determining equilibrium moles, then total moles of gas. Calculate mole fractions and partial pressures using the given total pressure. Finally, substitute into the Kp expression.

要计算 Kp，先确定平衡摩尔数，再求出气体总摩尔数。计算摩尔分数，并根据给定的总压计算分压。最后代入 Kp 表达式。

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At equilibrium, moles: N₂=0.20, H₂=0.60, NH₃=0.80, total = 1.60 mol. Total pressure = 100 atm. Mole fractions: x(N₂)=0.20/1.60=0.125; x(H₂)=0.60/1.60=0.375; x(NH₃)=0.80/1.60=0.50. Partial pressures: p(N₂)=12.5 atm; p(H₂)=37.5 atm; p(NH₃)=50.0 atm. Kp = (50.0)² / (12.5 × (37.5)³) = 2500 / (12.5 × 52734.4) ≈ 2500 / 659180 ≈ 0.00380 atm⁻².

示例：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。平衡摩尔数：N₂=0.20，H₂=0.60，NH₃=0.80，总摩尔数=1.60 mol。总压=100 atm。摩尔分数：x(N₂)=0.20/1.60=0.125；x(H₂)=0.60/1.60=0.375；x(NH₃)=0.80/1.60=0.50。分压：p(N₂)=12.5 atm；p(H₂)=37.5 atm；p(NH₃)=50.0 atm。Kp = (50.0)² / (12.5 × (37.5)³) ≈ 0.00380 atm⁻²。

Notice that Kp has units atm⁻² in this case because Δn = 2 – (1+3) = -2. This matches the units derived from the expression. When total pressure changes, Kp remains constant (temperature unchanged), but the equilibrium position may shift to maintain Kp.

注意，本例中 Kp 的单位为 atm⁻²，因为 Δn = 2 – (1+3) = -2。这与表达式推导出的单位一致。当总压变化时，Kp 保持不变（温度不变），但平衡位置可能偏移以维持 Kp。

11. Factors Affecting Equilibrium Constants | 影响平衡常数的因素

The only factor that changes the numerical value of Kc or Kp is temperature. Concentration, pressure, and catalysts have no effect on the equilibrium constant itself—they only alter the position of equilibrium or the rate at which equilibrium is reached.

唯一能改变 Kc 或 Kp 数值的因素是温度。浓度、压力和催化剂对平衡常数本身没有影响——它们只改变平衡位置或达到平衡的速率。

For an exothermic reaction, Kc decreases as temperature increases because the endothermic reverse reaction is favoured. Plotting ln K against 1/T (van’t Hoff equation) gives a straight line with a positive slope for endothermic reactions and a negative slope for exothermic reactions. Although detailed van’t Hoff calculations are not always required in OCR, understanding this trend is important.

对于放热反应，随着温度升高，Kc 减小，因为吸热的逆反应受到促进。根据范特霍夫方程，绘制 ln K 对 1/T 的图，对于吸热反应斜率为正，放热反应斜率为负。尽管 OCR 不总是要求详细的范特霍夫计算，但理解这一趋势很重要。

When a question asks ‘State and explain the effect of increasing pressure on Kp’, the answer is simply ‘no effect, because Kp is only dependent on temperature.’ Similarly, a catalyst does not change Kc or Kp. This distinction is a common source of error in exams.

当题目问“陈述并解释增加压力对 Kp 的影响”，答案很简单：“无影响，因为 Kp 仅取决于温度。”同样，催化剂不改变 Kc 或 Kp。这一区别是考试中常见的错误来源。

12. Industrial Applications and Yield Optimisation | 工业应用与产率优化

The Haber process and the Contact process are classic examples of applying equilibrium principles to maximise product yield while considering economic constraints. In the Haber process, N₂ + 3H₂ ⇌ 2NH₃ (ΔH = -92 kJ mol⁻¹), high pressure favours ammonia production due to fewer moles of gas on the right. However, very high pressure is expensive and requires thick-walled vessels. A compromise of 200 atm is used.

哈伯法和接触法是应用平衡原理以最大化产率同时考虑经济约束的经典例子。在哈伯法中，N₂ + 3H₂ ⇌ 2NH₃（ΔH = -92 kJ mol⁻¹），高压有利于氨的生产，因为右侧气体摩尔数较少。然而，非常高的压力代价昂贵且需要厚壁容器。因此采用200个大气压的折中方案。

Low temperature would give a higher equilibrium yield (exothermic), but the rate would be too slow. Therefore, a moderate temperature of 400-450°C is used together with an iron catalyst. Unreacted gases are recycled to improve overall efficiency.

低温会带来更高的平衡产率（放热），但速率太慢。因此，采用400-450°C的适中温度并配合铁催化剂。未反应的气体被循环利用，以提高总体效率。

In the Contact process: 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = -197 kJ mol⁻¹). A temperature of about 450°C, pressure slightly above atmospheric (1-2 atm), and a vanadium(V) oxide catalyst are employed. High pressure would favour SO₃ but is not necessary because the equilibrium already lies well to the right at the chosen conditions.

在接触法中：2SO₂ + O₂ ⇌ 2SO₃（ΔH = -197 kJ mol⁻¹）。采用约450°C的温度、略高于大气压的压力（1-2 atm）以及五氧化二钒催化剂。高压有利于 SO₃，但在所选条件下平衡已大大偏右，故无必要。

These industrial case studies illustrate the trade-offs between thermodynamics, kinetics, and economic factors—a must-know for OCR application questions.

这些工业案例研究说明了热力学、动力学和经济因素之间的权衡——这是 OCR 应用题必考内容。

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Chemical Equilibrium | 化学平衡考点精讲