📚 Chemical Equilibrium: Key Points for IB and Edexcel | 化学平衡：IB 与 Edexcel 考点精讲
Chemical equilibrium is a fundamental concept that bridges the gap between reaction kinetics and thermodynamics. For both IB and Edexcel chemistry students, mastering the ideas of reversible reactions, dynamic equilibrium, the equilibrium constant, and Le Chatelier's principle is essential for success in examinations. This guide distils the core knowledge, common pitfalls, and calculation techniques required at this level, connecting theory to industrial applications such as the Haber and Contact processes.
化学平衡是连接反应动力学与热力学的核心概念。对于 IB 与 Edexcel 化学学生来说，掌握可逆反应、动态平衡、平衡常数以及勒夏特列原理是考试成功的关键。本指南提炼了必学的核心知识、常见易错点以及计算技巧，并将理论与哈伯法、接触法等工业应用联系起来。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
A reversible reaction is one in which products can react to re-form the original reactants. When such a reaction is carried out in a closed system, it eventually reaches a state of dynamic equilibrium. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, so macroscopic properties (colour, concentration, pressure) remain constant, even though the reactions continue at the molecular level.
可逆反应是指产物能重新反应生成原反应物的反应。当这类反应在封闭系统中进行时，最终会达到动态平衡状态。在平衡时，正反应速率与逆反应速率相等，因此宏观性质（颜色、浓度、压力）保持恒定，尽管在分子层面上反应仍在不断进行。
2. The Equilibrium Constant Kc | 平衡常数 Kc
For a general homogeneous reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration, Kc, is expressed as: Kc = ([C]c[D]d) / ([A]a[B]b). The concentrations are measured in mol dm-3 and are taken at equilibrium. Kc is constant for a given reaction at a fixed temperature; changing the temperature alters Kc, but concentration changes do not.
对于一般的均相反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数 Kc 表达式为：Kc = ([C]c[D]d) / ([A]a[B]b)。浓度以 mol dm-3 为单位，并取平衡时的数值。在给定温度下，某一反应的 Kc 为常数；改变温度会改变 Kc，但浓度变化不会。
3. The Equilibrium Constant Kp | 平衡常数 Kp
When dealing with gas-phase equilibria, it is often convenient to use Kp, expressed in terms of partial pressures. For the same reaction, Kp = (pCc pDd) / (pAa pBb), where pX essure of component X. Partial pressure is calculated as mole fraction × total pressure. Kp, like Kc, is temperature dependent and has no units only if the sum of the stoichiometric coefficients of gaseous products equals that of the gaseous reactants.
处理气相平衡时，常采用以分压表示的 Kp。对于同一反应，Kp = (pCc pDd) / (pAa pBb)，其中 pX 为组分 X 的分压。分压 = 摩尔分数 × 总压。与 Kc 一样，Kp 也取决于温度，且只有当气态反应物与产物的化学计量系数之和相等时，Kp 才没有单位。
4. Writing Equilibrium Expressions | 书写平衡常数表达式
When writing Kc or Kp expressions, only gases and aqueous species appear; pure solids and pure liquids are omitted because their concentrations are essentially constant. For example, in CaCO3(s) ⇌ CaO(s) + CO2(g), the expression is simply Kp = pCO2 or Kc = [CO2]. Also remember to raise each concentration or pressure to the power of its stoichiometric coefficient.
书写 Kc 或 Kp 表达式时，只保留气体和溶液中的物种；纯固体和纯液体因其浓度基本不变而被省略。例如，对于 CaCO3(s) ⇌ CaO(s) + CO2(g)，表达式为 Kp = pCO2 或 Kc = [CO2]。同时注意每种物质的浓度或分压需以其计量系数为指数。
5. Units of Kc and Kp | Kc 和 Kp 的单位
The units of Kc depend on the sum of the stoichiometric coefficients. For a reaction with a change in the number of moles of gas (Δn ≠ 0), Kc has units of (mol dm-3)Δn. For Kp, the units are typically (atm)Δn or (Pa)Δn, depending on the pressure unit used. Examiners often ask you to derive the units from the expression; a common mistake is to forget to include them when quoting a numerical value for K.
Kc 的单位取决于化学计量系数之和。如果气体的总摩尔数发生变化（Δn ≠ 0），则 Kc 的单位为 (mol dm-3)Δn。Kp 的单位通常为 (atm)Δn 或 (Pa)Δn，视所用压力单位而定。考官常要求学生从表达式推导出单位；常见的错误是在给出 K 的数值时忘记附上单位。
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 oppose the change. This principle allows us to predict qualitatively how the equilibrium mixture responds, but it does not explain the effect on the rate of reaction or the value of K.
勒夏特列原理指出，如果处于动态平衡的系统受到浓度、压力或温度的变化，平衡位置将发生移动以削弱这种变化。运用该原理可以定性地预测平衡混合物的响应，但它不解释对反应速率或 K 值的影响。
7. Effect of Concentration Changes | 浓度变化的影响
Adding more reactant shifts the equilibrium to the right, producing more products, thereby decreasing the added reactant's concentration. Removing product also shifts the equilibrium to the right. The value of Kc remains constant because the concentration changes ultimately re-establish the same ratio. This is a common point of confusion: concentration changes do not alter the equilibrium constant, only the equilibrium position.
增加反应物会使平衡向右移动，生成更多产物，从而降低所加反应物的浓度。移除产物同样使平衡向右移动。Kc 值保持不变，因为浓度变化最终重新建立相同的比值。这是一个常见混淆点：浓度变化只改变平衡位置，不改变平衡常数。
8. Effect of Pressure Changes | 压力变化的影响
Pressure changes only affect equilibria involving gases where there is a difference in the total number of gaseous moles. Increasing the total pressure shifts the equilibrium to the side with fewer gaseous moles, reducing the pressure rise. For reactions with equal numbers of moles on both sides (e.g. H2 + I2 ⇌ 2HI), pressure changes have no effect on the equilibrium position. Again, Kp remains unchanged as long as temperature is constant.
压力变化只影响气体总摩尔数存在差异的气相平衡。增加总压会使平衡向气体摩尔数较少的方向移动，从而减弱压力的升高。对于反应前后气体摩尔数相等的情况（如 H2 + I2 ⇌ 2HI），压力变化对平衡位置无影响。同样，只要温度不变，Kp 保持不变。
9. Effect of Temperature Changes | 温度变化的影响
Unlike concentration and pressure, temperature changes do alter the equilibrium constant. For an exothermic forward reaction (ΔH < 0), increasing temperature shifts the equilibrium to the left (endothermic direction), decreasing K. For an endothermic forward reaction (ΔH > 0), raising temperature shifts the equilibrium to the right and increases K. This is the only external condition that changes the value of K.
与浓度和压力不同，温度变化会改变平衡常数。对于正向放热反应（ΔH < 0），升高温度使平衡向左（吸热方向）移动，K 值减小。对于正向吸热反应（ΔH > 0），升高温度使平衡向右移动，K 值增大。这是唯一能改变 K 值的外部条件。
10. Effect of Catalysts | 催化剂的影响
Catalysts provide an alternative reaction pathway with a lower activation energy, increasing the rates of both the forward and backward reactions equally. Consequently, a catalyst does not affect the position of equilibrium or the value of the equilibrium constant; it only helps the system reach equilibrium more quickly. This is a favourite exam trick: candidates often claim catalysts increase yield, which is incorrect.
催化剂通过提供具有较低活化能的替代反应路径，同等程度地提高正、逆反应速率。因此，催化剂不影响平衡位置，也不改变平衡常数的值；它只是帮助系统更快达到平衡。这是考试中常见的一个陷阱：考生常声称催化剂能提高产率，这是错误的。
11. Calculations Involving Kc | 涉及 Kc 的计算
Typical equilibrium calculations require an ICE (Initial, Change, Equilibrium) table. You are often given initial moles, the equilibrium moles of one species, and the volume. Deduce the changes in moles using stoichiometry, then convert to equilibrium concentrations. Substitute into the Kc expression and solve. Remember that if the volume factor cancels (equal number of moles on each side), you may work directly with moles, but careful with units. For gaseous systems, partial pressures can be found from the mole fraction and total pressure before using the Kp expression.
典型的平衡计算需要用到 ICE 表（初始、变化、平衡）。题目通常会给出初始物质的量、某一物种的平衡物质的量以及体积。利用化学计量关系推算出各物质的变化量，再转化为平衡浓度。代入 Kc 表达式求解。需注意，如果体积因子被约去（反应前后总摩尔数相等），可直接用物质的量计算，但要留心单位。对于气体系统，可先由摩尔分数和总压求出分压，再代入 Kp 表达式。
12. Industrial Applications: Haber and Contact Processes | 工业应用：哈伯法与接触法
The Haber process (N2 + 3H2 ⇌ 2NH3, ΔH = -92 kJ mol-1) uses a compromise temperature (around 450 °C), high pressure (200 atm), and an iron catalyst to achieve an economic yield. Le Chatelier's principle predicts that high pressure and low temperature favour ammonia production, but the low temperature would make the reaction too slow, hence the compromise. The Contact process (2SO2 + O2 ⇌ 2SO3, ΔH = -196 kJ mol-1) operates at moderate temperatures (~450 °C), atmospheric pressure, and uses V2O5 catalyst. These examples illustrate the interplay between equilibrium theory, kinetics, and economic factors.
哈伯法（N2 + 3H2 ⇌ 2NH3, ΔH = -92 kJ mol-1）采用折中的温度（约 450 °C）、高压（200 atm）和铁催化剂来获得经济可行的产率。勒夏特列原理指出，高压低温有利于氨的生成，但低温会使反应过慢，因此需要折中。接触法（2SO2 + O2 ⇌ 2SO3, ΔH = -196 kJ mol-1）在中温（~450 °C）、常压下运行，并采用 V2O5 催化剂。这些实例充分体现了平衡理论、动力学和经济因素之间的相互制约。
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