📚 IB CCEA Chemistry: Chemical Equilibrium – Key Concepts | IB CCEA 化学：化学平衡考点精讲
Chemical equilibrium is a cornerstone of physical chemistry, appearing in both IB and CCEA specifications. It explains why many important reactions do not go to completion and allows chemists to predict and optimise yields. This article distils the essential concepts, worked examples, and common pitfalls to help you master equilibrium for your exams.
化学平衡是物理化学的基石，在 IB 和 CCEA 考试大纲中均有出现。它解释了为什么许多重要的反应不会进行到底，并使化学家能够预测和优化产率。本文提炼了核心概念、典型示例和常见误区，助你扎实掌握平衡考点，轻松应试。
1. Introduction to Reversible Reactions | 可逆反应简介
A reversible reaction is one that can proceed in both the forward and reverse directions under the same conditions. It is represented by the double arrow ⇌.
可逆反应是指在同一条件下既能正向进行又能逆向进行的反应，用双箭头 ⇌ 表示。
For example, the thermal decomposition of ammonium chloride: NH4Cl(s) ⇌ NH3(g) + HCl(g). In a closed system, the reaction never reaches zero reactants; instead, it establishes a state of dynamic balance.
例如，氯化铵的热分解：NH4Cl(s) ⇌ NH3(g) + HCl(g)。在密闭体系中，反应物永远不会耗尽，而是会建立一个动态平衡状态。
2. Dynamic Equilibrium | 动态平衡
Dynamic equilibrium is the state in which the rate of the forward reaction equals the rate of the reverse reaction. Macroscopic properties such as colour, concentration, and pressure remain constant, but at the molecular level, both reactions continue to occur.
动态平衡是指正反应速率与逆反应速率相等的状态。宏观性质（如颜色、浓度、压强）保持恒定，但在分子尺度上，两个反应仍在持续进行。
Equilibrium can only be reached in a closed system where no matter can escape. Open systems allow products or reactants to leave, preventing equilibrium from being established.
平衡只能在物质无法逸出的密闭体系中达到。敞开体系会让产物或反应物离开，导致无法建立平衡状态。
3. The Equilibrium Constant Kc | 平衡常数 Kc
For a general reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is expressed as:
对于一般的可逆反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数表达式为：
Kc = [C]c[D]d / [A]a[B]b
The square brackets denote equilibrium concentrations in mol dm-3. Only gases and aqueous species appear in the expression; pure solids and pure liquids are omitted because their effective concentrations remain constant.
方括号表示平衡浓度（mol dm-3）。只有气体和溶液中的物质会出现在表达式中，纯固体和纯液体因有效浓度恒定而被省略。
Kc is temperature dependent. For a given reaction at a specified temperature, Kc has a fixed value regardless of the initial concentrations used.
Kc 与温度有关。在给定温度下，指定反应的 Kc 为定值，与初始浓度无关。
4. Equilibrium Constant in Terms of Pressure Kp | 分压平衡常数 Kp
When all reactants and products are gases, the equilibrium constant may be expressed using partial pressures as Kp. For aA(g) + bB(g) ⇌ cC(g) + dD(g):
当所有反应物和产物均为气体时，平衡常数可用分压表示为 Kp。对于 aA(g) + bB(g) ⇌ cC(g) + dD(g)：
Kp = (PC)c(PD)d / (PA)a(PB)b
Partial pressure Pi = mole fraction of i × total pressure. Kp has no units, but it is common to express pressures in atm or kPa and then drop units in the equilibrium expression.
分压 Pi = i 的摩尔分数 × 总压。Kp 没有单位，但计算时通常以 atm 或 kPa 表示压力，然后在平衡表达式中省略单位。
5. Magnitude of Equilibrium Constant | 平衡常数的大小意义
The size of the equilibrium constant indicates the position of equilibrium. If Kc ≫ 1 (e.g., 103 or greater), the equilibrium lies far to the right, favouring products. If Kc ≪ 1 (e.g., 10-3 or smaller), equilibrium favours reactants.
平衡常数的大小指示了平衡的位置。若 Kc 远大于 1（如 103 或更大），平衡强烈偏向产物一侧。若 Kc 远小于 1（如 10-3 或更小），平衡偏向反应物。
When Kc is close to 1, significant amounts of both reactants and products are present at equilibrium. Remember: the magnitude of K says nothing about the rate at which equilibrium is reached.
当 Kc 接近 1 时，平衡时反应物和产物都有可观的数量。请记住：K 值的大小与达到平衡的速率无关。
6. Le Chatelier’s Principle | 勒夏特列原理
If a system at dynamic equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium shifts to counteract the imposed change.
如果处于动态平衡的体系受到浓度、压力或温度的改变，平衡的位置将朝着减弱这种改变的方向移动。
This principle provides a qualitative way to predict the direction of shift without recalculating K. It applies to all equilibrium systems and is particularly useful for analysing industrial processes.
这一原理提供了无需重新计算 K 即可定性预测移动方向的方法。它适用于所有平衡体系，在分析工业流程时格外有用。
7. Effect of Concentration Changes | 浓度变化的影响
Increasing the concentration of a reactant shifts equilibrium to the right, increasing product yield. Increasing the concentration of a product shifts equilibrium to the left, favouring reactants.
增加反应物的浓度会使平衡向右移动，提高产物产率。增加产物的浓度则使平衡向左移动，有利于反应物。
Removing a product continuously, for example by precipitation or distillation, drives the reaction towards products. This is a common strategy to improve yield in reversible processes.
持续移除产物（例如通过沉淀或蒸馏）会促使反应向产物方向进行。这是提高可逆过程产率的常用策略。
Adding a solid or liquid does not shift the equilibrium because its effective concentration remains constant.
加入固体或液体不会引起平衡移动，因为它们的有效浓度保持恒定。
8. Effect of Pressure Changes | 压力变化的影响
Pressure changes only affect gaseous equilibria. Increasing the total pressure shifts the equilibrium towards the side with fewer moles of gas molecules. Decreasing pressure favours the side with more moles of gas.
压力变化只影响气体参与的平衡。增大总压会使平衡向气体分子总物质的量较少的一侧移动。减小压力则有利于气体分子总物质的量较多的一侧。
For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), increasing pressure produces more SO3 because the forward reaction reduces gas moles from 3 to 2.
对于反应 2SO2(g) + O2(g) ⇌ 2SO3(g)，增大压力会生成更多 SO3，因为正反应将气体物质的量从 3 减少到 2。
If the number of gas moles is the same on both sides, pressure changes have no effect on the equilibrium position. Equilibrium constant Kp itself is unaffected by pressure; only the amounts adjust to maintain Kp.
若反应前后气体分子总数相同，压力改变对平衡位置无影响。注意 Kp 本身不受压强影响，改变的只是各组分的量来维持 Kp 不变。
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. For an exothermic forward reaction (∆H < 0), increasing temperature decreases K, shifting equilibrium to the left. For an endothermic forward reaction (∆H > 0), increasing temperature increases K, shifting equilibrium to the right.
温度是唯一能改变平衡常数值的因素。对于正向放热反应（∆H < 0），升高温度会减小 K，使平衡左移。对于正向吸热反应（∆H > 0），升高温度会增大 K，使平衡右移。
Cooling an exothermic equilibrium favours the forward reaction, releasing more heat. Thus, low temperatures improve yield for exothermic reactions, although they may slow the rate.
对放热平衡进行冷却有利于正向反应，从而释放更多热量。因此，低温对放热反应的产率有利，但可能会减慢速率。
10. Effect of Catalyst | 催化剂的影响
A catalyst provides an alternative reaction pathway with lower activation energy for both the forward and reverse reactions. It increases the rate of both directions equally, therefore it does not shift the position of equilibrium or alter the value of Kc or Kp.
催化剂为正向和逆向反应提供了一条活化能更低的替代路径。它同等程度地提高正逆反应速率，因此不会移动平衡位置，也不会改变 Kc 或 Kp 的数值。
The key benefit of a catalyst in equilibrium is that it allows equilibrium to be reached more quickly. In industry, this means higher productivity without sacrificing yield.
催化剂在平衡中的关键好处是能让体系更快达到平衡。在工业上，这意味着在不牺牲产率的同时提高生产效率。
11. Industrial Application: Haber Process | 工业应用：哈伯法
The Haber process for ammonia synthesis illustrates the compromise between equilibrium yield and reaction rate: N2(g) + 3H2(g) ⇌ 2NH3(g) (exothermic, ΔH = -92 kJ mol-1).
合成氨的哈伯法体现了平衡产率与反应速率之间的妥协：N2(g) + 3H2(g) ⇌ 2NH3(g)（放热，ΔH = -92 kJ mol-1）。
| Condition / 条件 | Chosen Value / 选用值 | Reason / 理由 |
|---|---|---|
| Temperature | ~450 °C | Compromise: lower temperature improves equilibrium yield but gives a slow rate; higher temperature gives faster rate but reduces yield. |
| Pressure | 200 atm | High pressure shifts equilibrium right (4 moles gas → 2 moles) and increases rate. Very high pressures are too expensive and require stronger equipment. |
| Catalyst | Iron | Increases rate without affecting equilibrium position. Allows a lower temperature to be used economically. |
Unused N2 and H2 are recycled, and ammonia is continuously removed by condensation. This pushes the equilibrium towards products and achieves a viable industrial yield.
未反应的 N2 和 H2 被循环利用，而氨通过冷凝持续移出。这样不断拉动平衡向产物一侧移动，从而获得可观的工业产率。
12. Common Exam Pitfalls | 常见考试误区
- Forgetting to omit solids and liquids from K expressions: Never include solids or pure liquids. Only aqueous and gaseous species are written.
- 混淆 Kc 和 Kp 表达式：书写时务必排除固体和纯液体，仅包含溶液和气体物种。
- Assuming catalyst changes K: A catalyst never alters K or equilibrium position; it only speeds up attainment of equilibrium.
- 误认为催化剂改变 K 值：催化剂绝不改变平衡常数或平衡位置，仅加速平衡的达到。
- Using Le Chatelier’s principle to explain the effect of a catalyst or inert gas addition at constant volume: Neither shifts equilibrium. Inert gas at constant volume changes total pressure but not partial pressures, so no shift occurs.
- 错误使用勒夏特列原理解释催化剂或恒容下加入惰性气体的影响：这两种情况都不会引起平衡移动。恒容下加入惰性气体虽改变总压，但分压不变，故平衡不移动。
- Mixing up K and rate: A large K does not mean a fast reaction. Equilibrium and kinetics are separate concepts.
- 将平衡常数与速率混为一谈：K 值大并不意味着反应快。平衡和动力学是两个独立的概念。
- Temperature vs. exothermic reactions: When temperature increases, K decreases for an exothermic reaction and increases for an endothermic reaction — remember to link the shift to heat absorbed or released.
- 温度与放热反应的关系：升温时，放热反应的 K 值减小，吸热反应的 K 值增大——要记得将移动方向与吸热或放热联系起来。
By consistently practicing Kc and Kp calculations, interpreting K magnitudes, and applying Le Chatelier’s principle logically, you will be well prepared for any equilibrium question on IB or CCEA papers.
通过持续练习 Kc 和 Kp 的计算、理解 K 值的含义并有逻辑地运用勒夏特列原理，你将能从容应对 IB 或 CCEA 试卷中的任何平衡问题。
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