📚 Chemical Equilibrium | 化学平衡
Many chemical reactions do not go to completion; instead, they reach a state where the forward and reverse reactions occur at the same rate. This dynamic equilibrium is central to understanding how chemical systems respond to changing conditions, and it forms the foundation of industrial processes such as the Haber and Contact processes. In A‑Level Chemistry, a firm grasp of equilibrium principles, equilibrium constants, and Le Chatelier's principle is essential for predicting the outcome of reversible reactions.
许多化学反应并不会进行到底，而是会到达正向反应和逆向反应速率相等的状态。这种动态平衡是理解化学系统如何响应条件变化的核心，也是哈伯法和接触法等工业过程的基础。在A-Level化学学习中，扎实掌握平衡原理、平衡常数以及勒夏特列原理对预测可逆反应的结果至关重要。
1. Introduction to Dynamic Equilibrium | 动态平衡简介
A dynamic equilibrium is established in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time.
在密闭系统中，当正向反应速率等于逆向反应速率，且反应物和产物的浓度随时间保持不变时，即建立了动态平衡。
This state is 'dynamic' because both forward and reverse reactions continue to occur at the molecular level, even though there is no net change in macroscopic properties such as colour, pressure, or concentration.
这种状态之所以称为“动态”，是因为在分子水平上正向和逆向反应仍在持续进行，尽管颜色、压强或浓度等宏观性质没有净变化。
Equilibrium can only be achieved in a closed system where no matter can escape; in an open system, products or reactants may be lost, preventing the establishment of a true equilibrium.
平衡只能在物质无法逸出的密闭系统中达到；在开放系统中，产物或反应物可能会流失，从而无法建立真正的平衡。
2. Reversible Reactions and the Equilibrium State | 可逆反应与平衡状态
Reversible reactions are indicated by the symbol '⇌'. At the start, only reactants are present and the forward reaction is fast, but as products build up, the reverse reaction speeds up until the two rates become identical.
可逆反应用符号“⇌”表示。反应开始时只有反应物存在，正向反应较快，但随着产物的累积，逆向反应逐渐加速，直至两个速率相等。
A system at equilibrium appears static, but it is balanced by opposing processes. For example, in the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), nitrogen and hydrogen molecules continue to combine into ammonia while ammonia decomposes back into its elements at the same overall rate.
处于平衡状态的系统看似静止，实则是被相反的过程所抵消。例如在反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 中，氮分子和氢分子仍在不断结合成氨，而氨也以相同的总速率分解回其单质。
It is important to remember that equilibrium does not mean the concentrations of reactants and products are equal; rather, their concentrations have stabilised to values that give equal reaction rates in both directions.
需要记住的是，平衡并不意味着反应物和产物的浓度相等；而是说它们的浓度已经稳定在使正逆反应速率相等的数值。
3. The Equilibrium Constant Kc | 平衡常数 Kc
For a general reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc is given by the expression:
对于一般的可逆反应 aA + bB ⇌ cC + dD，平衡常数 Kc 的表达式为：
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
where square brackets denote the equilibrium concentration in mol dm⁻³. The value of Kc is constant at a given temperature, regardless of initial concentrations.
其中方括号表示以 mol dm⁻³ 为单位的平衡浓度。在给定温度下，Kc 的值是常数，与初始浓度无关。
A large Kc (much greater than 1) indicates that the equilibrium position lies far to the right, favouring products. A very small Kc (much less than 1) means the reaction favours reactants.
Kc 值很大（远大于 1）表明平衡位置偏右，倾向于产物；Kc 值非常小（远小于 1）则意味着反应倾向于反应物。
When writing Kc expressions, pure solids and pure liquids are omitted because their concentrations are effectively constant. Only aqueous solutions and gases appear in the Kc expression.
书写 Kc 表达式时，纯固体和纯液体因其浓度可视为常数而被省略，只有水溶液和气体才出现在 Kc 表达式中。
4. Calculating Kc and Equilibrium Concentrations | 计算 Kc 和平衡浓度
To calculate Kc, you first need the equilibrium concentrations of all species. Often exam problems provide initial amounts and the equilibrium amount of one substance, allowing an ICE (Initial, Change, Equilibrium) table to be used.
要计算 Kc，首先需要各物质的平衡浓度。考试题目通常会给出初始量及某一物质的平衡量，从而允许使用 ICE 表（初始、变化、平衡）来求解。
The change row reflects the stoichiometry of the reaction: for every x moles of reactant consumed, a proportional amount of product is formed. After determining equilibrium concentrations, they are substituted into the Kc expression along with any necessary powers.
变化行反映反应的化学计量关系：每消耗 x 摩尔的反应物，会生成相应比例的产物。确定平衡浓度后，将它们连同必要的幂次代入 Kc 表达式。
Units of Kc depend on the stoichiometry. For example, for N₂ + 3H₂ ⇌ al concentration powers in numerator minus denominator yields (mol dm⁻³)⁽²⁻⁴⁾ = mol⁻² dm⁶.
Kc 的单位取决于化学计量数。例如，对于 N₂ + 3H₂ ⇌ 2NH₃，Kc 的单位是 dm⁶ mol⁻²，因为分子与分母的浓度幂次之差得到 (mol dm⁻³)⁽²⁻⁴⁾ = mol⁻² dm⁶。
You may also be asked to find an unknown equilibrium concentration when Kc and other concentrations are given. Rearranging the Kc expression algebraically is a common task.
也可能要求已知 Kc 和其他浓度，求某一未知平衡浓度。对 Kc 表达式进行代数变换是常见的考查方式。
5. The Equilibrium Constant Kp for Gaseous Reactions | 气相反应的平衡常数 Kp
For reactions involving gases, it is often more convenient to use partial pressures. The equilibrium constant Kp is defined in terms of the partial pressures of the gases, raised to their stoichiometric coefficients.
对于涉及气体的反应，使用分压往往更方便。平衡常数 Kp 根据气体的分压定义，并以其化学计量系数为幂次。
Kp = (pC)ᶜ (pD)ᵈ / (pA)ᵃ (pB)ᵇ
where pA, pB etc. are the partial pressures at equilibrium, usually expressed in atm, Pa or kPa. Like Kc, Kp is temperature-dependent.
其中 pA、pB 等为平衡时的分压，通常以 atm、Pa 或 kPa 表示。与 Kc 一样，Kp 也依赖于温度。
Mole fraction = number of moles of that gas / total number of moles of all gases present.
分压等于该气体的摩尔分数乘以总压。摩尔分数 = 该气体的物质的量 / 所有气体物质的量的总和。
Heterogeneous equilibria (where substances are in different phases) are expressed similarly, but solids and liquids are excluded from Kp. Only gaseous species contribute to the equilibrium expression.
多相平衡（物质处于不同相态）的写法类似，但固体和液体被排除在 Kp 表达式之外，只有气态物种参与其中。
6. Relationship Between Kp and Kc | Kp 与 Kc 的关系
nt, T is the absolute temperature in Kelvin, and Δn is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants).
Kp 与 Kc 通过关系式 Kp = Kc (RT)^Δn 联系起来，其中 R 为气体常数，T 为开尔文温度，Δn 是气体物质的量的变化（气态产物的物质的量减去气态反应物的物质的量）。
s are dimensionless in terms of pressure. This occurs when the number of moles of gas is the same on both sides of the equation.
如果 Δn = 0，则 Kp = Kc，且这两个常数在压力项上无量纲。当方程式两边气体物质的量相同时，就会出现这种情况。
This relationship is fundamental for converting between Kc and Kp in gaseous equilibria, and it helps explain why the numerical values can differ even when the equilibrium position is the same at a given temperature.
该关系式对在气态平衡中进行 Kc 与 Kp 的换算至关重要，也有助于理解为何在相同温度下即使平衡位置相同，两者的数值也可能不同。
7. Le Chatelier's Principle: Effect of Concentration Changes | 勒夏特列原理：浓度变化的影响
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift to partially counteract the imposed change.
勒夏特列原理指出，如果平衡系统受到浓度、温度或压强的改变，平衡将朝着部分抵消该改变的方向移动。
If the concentration of a reactant is increased, the system shifts to the right to consume the added reactant, producing more products. Conversely, removing a product causes the forward reaction to be favoured to replenish it.
若增加反应物的浓度，平衡向右移动以消耗掉添加的反应物，生成更多产物。相反，移除产物会使正向反应得到促进，以补充被移除的产物。
Dilution of an aqueous equilibrium (adding water) generally shifts the position towards the side with the greater number of aqueous ions or molecules. This can be rationalised by considering the effect on relative concentrations.
稀释水溶液平衡（加水）通常会使平衡向含有更多水合离子或分子的一侧移动。这可以通过考虑相对浓度的影响来理解。
Changes in concentration do not alter the value of Kc or Kp; the system re‑adjusts concentrations to restore the same ratio as expressed by the equilibrium constant.
浓度变化不会改变 Kc 或 Kp 的值；系统会重新调整浓度，以恢复由平衡常数所表达的同一比值。
8. Le Chatelier's Principle: Effect of Pressure Changes | 勒夏特列原理：压强变化的影响
Pressure changes only affect equilibria involving gases where there is a difference in the number of moles of gas on each side of the equation.
压强变化只影响方程式两边气体物质的量不相等的、涉及气体的平衡体系。
with fewer gas molecules, as this helps to reduce the pressure. For instance, in N₂ + 3H₂ ⇌ 2NH₃, high pressure favours the forward reaction (4 mol → 2 mol).
增大总压（通过减小体积）会使平衡向气体分子数较少的一侧移动，因为这有助于降低压强。例如在 N₂ + 3H₂ ⇌ 2NH₃ 中，高压有利于正向反应（4 mol → 2 mol）。
e unaffected by pressure changes, as both sides have the same number of gaseous particles.
减小压强会使平衡向气体分子数较多的一侧移动。Δn = 0 的反应不受压强变化的影响，因为两侧的气态粒子数相同。
It is crucial to remember that while pressure shifts alter equilibrium yields, Kp itself remains constant at constant temperature; only the individual partial pressures adjust to maintain the constant Kp value.
必须记住，虽然压强变化会改变平衡产率，但 Kp 本身在恒温下保持不变；只有各个分压发生调整以维持 Kp 值的恒定。
9. Le Chatelier's Principle: Effect of Temperature Changes | 勒夏特列原理：温度变化的影响
Unlike concentration and pressure, temperature changes do affect the value of the equilibrium constant. The direction of the shift depends on whether the forward reaction is exothermic or endothermic.
与浓度和压强不同，温度变化确实会影响平衡常数的数值。移动的方向取决于正向反应是放热还是吸热。
For an exothermic forward reaction (ΔH negative), increasing the temperature shifts equilibrium to the left (endothermic direction) to absorb the extra heat, so Kc or Kp decreases. Lowering the temperature favours the exothermic direction, increasing the equilibrium constant.
对于正向放热反应（ΔH 为负），升高温度使平衡向左（吸热方向）移动以吸收额外热量，因此 Kc 或 Kp 减小。降低温度有利于放热方向，平衡常数增大。
Conversely, for an endothermic forward reaction (ΔH positive), increasing temperature increases the value of Kc/Kp, favouring product formation. This is why some reactions are carried out at high temperatures despite endothermicity.
相反，对于正向吸热反应（ΔH 为正），升高温度会增大 Kc/Kp 值，有利于产物的生成。这就是为什么某些反应虽然吸热但仍需在高温下进行。
Quantitatively, the temperature dependence of K is described by the van 't Hoff equation, which links ln K to 1/T. A plot of ln K against 1/T gives a straight line with slope –ΔH/R.
定量地讲，平衡常数 K 对温度的依赖性由范特霍夫方程描述，该方程将 ln K 与 1/T 联系起来。以 ln K 对 1/T 作图将得到一条斜率为 –ΔH/R 的直线。
10. Effect of Catalysts on Equilibrium | 催化剂对平衡的影响
Catalysts provide an alternative pathway with a lower activation energy, increasing the rate of both forward and reverse reactions by the same factor.
催化剂提供一条活化能较低的替代途径，以相同的倍数加快正向和逆向反应的速率。
Because a catalyst accelerates both directions equally, it does not alter the position of equilibrium or the value of the equilibrium constant. It simply helps the system reach equilibrium faster.
由于催化剂同等程度地加速正逆反应，它不会改变平衡位置或平衡常数的数值，只是帮助系统更快地达到平衡。
In an uncatalysed reaction, equilibrium may be established very slowly. Adding a catalyst is essential in industrial processes to achieve acceptable rates without changing the equilibrium yield—temperature and pressure are then chosen to optimise position and rate.
在非催化反应中，平衡可能建立得非常缓慢。在工业过程中加入催化剂对于在不改变平衡产率的前提下获得可接受的速率至关重要——温度和压强的选择则用以优化平衡位置和速率。
11. Industrial Applications: Haber Process and Contact Process | 工业应用：哈伯法与接触法
The Haber process for ammonia synthesis, N₂ + 3H₂ ⇌ 2NH₃ (ΔH = – ut 200 atm) favours the forward reaction because the product side has fewer gas molecules.
合成氨的哈伯法，N₂ + 3H₂ ⇌ 2NH₃ (ΔH = –92 kJ mol⁻¹)，是折中条件的经典例子。高压（约 200 atm）有利于正向反应，因为产物一侧气体分子更少。
The forward reaction is exothermic, so low temperature favours a higher equilibrium yield. However, a very low temperature makes the rate impractically slow. A compromise temperature of around 400–450 °C is used, together with an iron catalyst.
正向反应是放热的，因此低温有利于更高的平衡产率。但过低的温度会使反应速率慢得不切实际。实际中使用约 400–450 °C 的折中温度，并配合铁催化剂。
The Contact process for sulfuric acid production involves the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (ΔH = – kJ mol⁻¹). A vanadium(V) oxide catalyst is used at around 450 °C and moderate pressure (1– ause the equilibrium already lies strongly to the right, making high pressure unnecessary.
硫酸生产的接触法涉及平衡 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (ΔH = –197 kJ mol⁻¹)。使用五氧化二钒催化剂，在约 450 °C 和常压略高（1–2 atm）下进行，因为平衡已强烈偏右，无需高压。
In both processes, economic factors (energy cost, plant safety, and catalyst lifetime) are balanced against chemical principles, illustrating the real‑world significance of equilibrium theory.
在这两种工艺中，经济因素（能源成本、工厂安全性和催化剂寿命）与化学原理相互权衡，体现了平衡理论在现实世界中的重要意义。
12. Summary and Exam Tips | 总结与考试技巧
Equilibrium is a dynamic state with equal forward and reverse rates. The equilibrium constant (Kc or Kp) is temperature-dependent and has a fixed form derived from the balanced equation. Le Chatelier's principle allows qualitative prediction of shifts, while calculations with ICE tables and K expressions provide quantitative insight.
平衡是一种正逆速率相等的动态状态。平衡常数（Kc 或 Kp）依赖于温度，并具有源于配平方程式的固定形式。勒夏特列原理允许对平衡移动进行定性预测，而利用 ICE 表和平衡常数表达式进行的计算则提供定量的认识。
rease in K with increased T), and incorrectly applying pressure shifts when Δn = 0. Always check whether you are dealing with Kc or Kp and express units clearly except when they cancel out.
常见错误包括：忘记在 K 表达式中省略固体/液体、混淆温度效应的方向（放热 ↔ 随 T 升高 K 减小）、以及在 Δn = 0 时错误地应用压强变化。务必检查处理的是 Kc 还是 Kp，并清晰注明单位（除非单位相互抵消）。
Use past-paper questions to practise constructing ICE tables for both concentration and partial pressure. Remember that a catalyst saves time but does not alter the final equilibrium composition.
通过历年真题练习构建浓度和分压的 ICE 表。记住，催化剂节省时间但不会改变最终的平衡组成。
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