📚 Dynamic Equilibrium: Reversible Reactions and Le Chatelier's Principle | 动态平衡：可逆反应与勒夏特利耶原理
Dynamic equilibrium is a fundamental concept in IB Chemistry, bridging kinetics and thermodynamics. It describes the state of a reversible reaction where the rates of the forward and reverse processes are equal, resulting in constant macroscopic properties. Understanding this subtle balance is essential for predicting how chemical systems respond to changes in conditions.
动态平衡是 IB 化学中的一个基础概念，连接了动力学与热力学。它描述的是可逆反应中正逆反应速率相等的状态，此时体系的宏观性质保持恒定。理解这种微妙的平衡对于预测化学体系如何响应条件的变化至关重要。
double arrow, through the equilibrium constant and the reaction quotient, to Le Chatelier' ynamic condition ΔG = 0, and the examination traps that cost students marks every session.
本文将这一主题扩展为十二个相互衔接的小节：从双向箭头的含义，到平衡常数与反应商，再到勒夏特利耶原理、哈伯法、热力学条件 ΔG = 0，以及每场考试中让考生失分的那些"陷阱"。
1. Reversible Reactions and the Equilibrium State | 可逆反应与平衡状态
Many chemical reactions do not go to completion; instead, the products can react to reform the reactants. Such reactions are termed reversible and are denoted by a double arrow (⇌). For instance, the formation of ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The single arrow is reserved for reactions that are treated as irreversible under the stated conditions, such as the combustion of methane or the precipitation of silver chloride.
许多化学反应不会进行到底，产物会重新反应生成反应物。这类反应被称为可逆反应，用双向箭头（⇌）表示。例如氨的合成：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。单向箭头只用于在给定条件下可视为不可逆的反应，例如甲烷的燃烧或氯化银的沉淀。
A useful physical example is bromine in a sealed container: Br₂(l) ⇌ Br₂(g). At a fixed temperature, liquid bromine evaporates and gaseous bromine condenses at equal rates, so the intensity of the brown vapour stops changing even though both processes continue. The same logic applies to a saturated solution of sodium chloride in contact with undissolved solid: dissolution and crystallisation proceed at equal rates, and the concentration of the dissolved ions remains constant.
一个有用的物理实例是密闭容器中的溴：Br₂(l) ⇌ Br₂(g)。在固定温度下，液溴蒸发与溴蒸气冷凝的速率相等，因此棕色蒸气的深浅不再变化，尽管两个过程仍在继续。同样的逻辑适用于与未溶解固体共存的氯化钠饱和溶液：溶解与结晶以相同速率进行，已溶解离子的浓度保持恒定。
2. Dynamic, Not Static | 动态，而非静止
The word "dynamic" carries the whole meaning of the concept. At equilibrium both the forward and the reverse reactions are still occurring; they simply occur at identical rates. At the molecular level bonds are still breaking and forming, and individual molecules are continuously interconverting. Nothing has stopped.
"动态"二字承载了整个概念的含义。在平衡时，正反应与逆反应仍在进行，只是以完全相同的速率进行。在分子层面，化学键仍在断裂与形成，单个分子持续地相互转化。没有任何过程停止。
This is why a catalyst never changes the position of equilibrium. A catalyst lowers the activation energy of the forward and the reverse reaction by the same amount, so it accelerates both directions equally. The system reaches the same equilibrium composition faster, but the final mixture is identical to the uncatalysed one. If equilibrium were static, the presence of a catalyst would make no sense at all.
这正是催化剂从不改变平衡位置的原因。催化剂以相同的幅度降低正反应和逆反应的活化能，因此等量地加快两个方向。体系更快达到相同的平衡组成，但最终混合物与未加催化剂时完全一致。如果平衡是静止的，催化剂的存在就毫无意义了。
3. The Closed System Requirement | 封闭体系的要求
Equilibrium can only be established in a closed system, in which no matter is exchanged with the surroundings. Energy may still be exchanged as heat, but the amount of each substance present must remain available for the forward and reverse processes to balance. In an open system, a gaseous product simply escapes and can never return, so one direction is permanently suppressed.
平衡只能在封闭体系中建立，即体系与环境之间不交换物质。能量仍可以热的形式交换，但各物质的量必须保持可用于正逆两个过程相互抵消。在开放体系中，气态产物会逸散而无法返回，因此其中一个方向被永久抑制。
) + CO₂(g). Because carbon dioxide leaves the vessel, the decomposition runs to completion. Repeat the experiment in a sealed tube and the same reaction becomes reversible: CaCO₃(s) ⇌ + CO₂(g), with the partial pressure of CO₂ settling at a fixed value at each temperature.
想想在敞口坩埚中加热碳酸钙：CaCO₃(s) → CaO(s) + CO₂(g)。由于二氧化碳离开了容器，分解会进行到底。若在密封管中重复该实验，同一反应就变成可逆的：CaCO₃(s) ⇌ CaO(s) + CO₂(g)，在每个温度下二氧化碳的分压都会稳定在某个固定值。
4. Constant Macroscopic Properties and Rate Graphs | 恒定的宏观性质与速率图像
When the forward and reverse reactions proceed at exactly the same rate, the concentrations of all species stop changing. This condition is known as the equilibrium state. Because the concentrations are constant, every macroscopic, observable property is constant too: colour intensity, pressure in a constant-volume gas reaction, pH, conductivity and refractive index.
当正反应和逆反应以完全相同的速率进行时，所有物种的浓度不再变化。这一状态称为平衡状态。由于浓度恒定，所有宏观的、可观察的性质也随之恒定：颜色深浅、恒容气体反应中的压强、pH、电导率和折射率。
Two graph patterns appear repeatedly in examination questions. On a rate–time graph, the forward rate falls while the reverse rate rises from zero; the two curves meet and then run parallel, and the common value is the equilibrium rate. On a concentration–time graph, the reactant concentration falls and the product concentration rises until both level off into horizontal plateaus. Note carefully that the plateaus are not necessarily at equal heights, and the two reactants do not have to end at the same concentration.
考试题中反复出现两种图像模式。在"速率—时间"图上，正反应速率下降而逆反应速率从零上升；两条曲线相交后平行延伸，其公共值即为平衡速率。在"浓度—时间"图上，反应物浓度下降、产物浓度上升，直至两者都变成水平平台。请特别注意，平台的高度不必相等，两种反应物的最终浓度也不一定相同。
rate(forward) = rate(reverse) ⇌ constant macroscopic properties
5. The Equilibrium Constant Kc | 平衡常数 Kc
For a general reversible reaction written as aA + bB ⇌ cC + dD, the equilibrium constant is defined by the expression below, using equilibrium concentrations in mol dm⁻³. Products appear in the numerator, reactants in the denominator, and each concentration is raised to the power of its stoichiometric coefficient.
对于写作 aA + bB ⇌ cC + dD 的一般可逆反应，平衡常数由下式定义，使用以 mol dm⁻³ 表示的平衡浓度。产物在分子，反应物在分母，每个浓度的指数等于其化学计量数。
Kc = ([C]ᶜ × [D]ᵈ) ÷ ([A]ᵃ × [B]ᵇ)
Two features of Kc deserve emphasis. First, Kc depends only on temperature; changing concentration, pressure or the presence of a catalyst leaves Kc untouched. Second, the magnitude of Kc reports the position of equilibrium: a very large Kc (say 10⁵ or more) means the equilibrium mixture is product-dominated, whereas a very small Kc means the reaction barely proceeds.
Kc 有两点值得强调。第一，Kc 只取决于温度；改变浓度、压强或加入催化剂都不会改变 Kc。第二，Kc 的大小反映平衡的位置：Kc 很大（例如 10⁵ 以上）意味着平衡混合物以产物为主，而 Kc 很小则意味着反应几乎不进行。
ed out from the expression rather than memorised. For the ammonia synthesis N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = [NH₃]² ÷ ([N₂] × [H₂]³), and the units are mol⁻² dm⁶.
单位必须由表达式推演而非死记。对于氨的合成 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Kc = [NH₃]² ÷ ([N₂] × [H₂]³)，单位为 mol⁻² dm⁶。
6. Kp, the Reaction Quotient Q, and Predicting Direction | Kp、反应商 Q 与方向判断
For gaseous equilibria it is often more convenient to use partial pressures instead of concentrations. The equilibrium constant Kp is defined in exactly the same structural way, with partial pressures in place of concentrations. Partial pressure equals mole fraction multiplied by total pressure, so a change in total pressure alters the individual partial pressures even when the temperature is fixed.
对于气体平衡，用分压代替浓度往往更为方便。平衡常数 Kp 的定义结构完全相同，只是把浓度换成分压。分压等于摩尔分数乘以总压，因此在温度不变时改变总压也会改变各个分压。
p(A) = x(A) × P(total) Kp = (p(NH₃))² ÷ (p(N₂) × (p(H₂))³)
equilibrium. Comparing Q with K tells you which way the system must move: if Q < K the reaction proceeds forward to make more products; if Q > K the reaction reverses; if Q = K the system is already at equilibrium.
反应商 Q 与 Kc 或 Kp 的代数形式相同，但使用任意时刻的浓度或压强计算，不必处于平衡。比较 Q 与 K 即可判断体系必须移动的方向：若 Q < K，反应正向进行以生成更多产物；若 Q > K，反应逆向进行；若 Q = K，体系已处于平衡。
7. Le Chatelier's Principle: Concentration and Pressure | 勒夏特利耶原理：浓度与压强
Le Chatelier's principle states that when a system at equilibrium is subjected to a change, the position of equilibrium shifts in the direction that tends to oppose the change. Adding more of a reactant is opposed by consuming that reactant, so the equilibrium shifts to the right. Removing a product is opposed by making more of it, so the equilibrium also shifts to the right.
勒夏特利耶原理指出：当处于平衡的体系受到改变时，平衡位置会朝着倾向于抵消该改变的方向移动。加入更多反应物会因消耗该反应物而被抵消，因此平衡右移。移去产物会因生成更多产物而被抵消，因此平衡同样右移。
Changing pressure matters only when gases are involved and only when the number of gas moles differs between the two sides. Raising the pressure is opposed by reducing the number of gas molecules, so the equilibrium shifts towards the side with fewer moles of gas. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the left side has four moles of gas and the right side has two, so high pressure favours ammonia. If the mole counts are equal, as in H₂(g) + I₂(g) ⇌ 2HI(g), pressure has no effect on the position of equilibrium.
压强变化只有在涉及气体、且两侧气体摩尔数不同时才有影响。增大压强会因减少气体分子数而被抵消，因此平衡向气体摩尔数较少的一侧移动。对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，左侧有 4 mol 气体而右侧有 2 mol，因此高压有利于氨的生成。若两侧摩尔数相等，如 H₂(g) + I₂(g) ⇌ 2HI(g)，则压强对平衡位置没有影响。
- Adding a reactant → shifts right (forward reaction favoured) | 加入反应物 → 右移（正反应更有利）
- Removing a product → shifts right | 移去产物 → 右移
- Increasing pressure → shifts towards fewer moles of gas | 增大压强 → 向气体摩尔数少的一侧移动
- Equal moles of gas on both sides → no shift | 两侧气体摩尔数相等 → 不移动
- Adding an inert gas at constant volume → no shift | 恒容下加入惰性气体 → 不移动
8. Temperature Changes and the Sign of ΔH | 温度变化与 ΔH 的符号
Temperature is the only factor that changes the value of Kc or Kp. Raising the temperature is opposed by absorbing heat, so the equilibrium shifts in the endothermic direction. For an exothermic forward reaction, heating therefore shifts the equilibrium to the left and decreases Kc. For an endothermic forward reaction, heating shifts the equilibrium to the right and increases Kc.
温度是唯一能改变 Kc 或 Kp 数值的因素。升温会因吸热而被抵消，因此平衡向吸热方向移动。若正反应放热，升温会使平衡左移并减小 Kc。若正反应吸热，升温会使平衡右移并增大 Kc。
e is bad for the yield of ammonia, yet industry runs the reactor at around 450 °C. The reason is kinetic: at low temperature the rate is far too slow to be commercially useful, so a compromise between yield and rate is chosen.
以氨的合成为例，其正反应放热，ΔH = −92 kJ mol⁻¹。高温不利于氨的产率，然而工业上反应器仍运行在约 450 °C。原因是动力学因素：温度过低时反应速率太慢，没有商业价值，因此在产率与速率之间取折中。
| Change | 改变 | Shift | 移动方向 | Effect on K | 对 K 的影响 |
| Add reactant | 加入反应物 | Right | 右移 | Unchanged | 不变 |
| Increase pressure | 增大压强 | Fewer gas moles | 向气体少的一侧 | Unchanged | 不变 |
| Increase T (exothermic forward) | 升温（正反应放热） | Left | 左移 | K decreases | K 减小 |
| Add catalyst | 加入催化剂 | No shift | 不移动 | Unchanged | 不变 |
9. Catalysts and the Rate of Reaching Equilibrium | 催化剂与达到平衡的速率
A catalyst provides an alternative pathway with a lower activation energy. Because the same lowering applies to the reverse reaction, both rates increase by an identical factor, and the equilibrium position is unchanged. What changes is the time needed to reach that position.
催化剂提供了一条活化能更低的替代路径。由于逆向反应也获得同等幅度的降低，两个速率以相同倍数增大，平衡位置不变。真正改变的是达到该位置所需的时间。
This distinction is a common source of lost marks. A statement such as "the catalyst increases the yield of ammonia" is wrong; the correct statement is "the catalyst increases the rate at which equilibrium is reached, without changing the yield". In the Haber process, finely divided iron with potassium and aluminium oxide promoters is used precisely for this reason.
这一区别是常见的失分点。诸如"催化剂提高氨的产率"的说法是错误的；正确的表述是"催化剂加快达到平衡的速率，但不改变产率"。在哈伯法中，使用含钾和氧化铝助催化剂的细碎铁粉，正是出于这一原因。
10. The Haber Process: Applying Equilibrium Ideas | 哈伯法：平衡思想的应用
The Haber process illustrates every principle above in a single industrial context. Nitrogen from the air and hydrogen from steam reforming of methane are combined over an iron catalyst.
哈伯法在同一个工业情境中体现了上述所有原理。来自空气的氮气与来自甲烷蒸汽重整的氢气，在铁催化剂上结合。
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
- Pressure of about 200 atm: four moles of gas become two, so high pressure raises the equilibrium yield. | 约 200 atm 的压强：4 mol 气体变为 2 mol，高压提高平衡产率。
- Temperature of about 450 °C: a compromise, since the forward reaction is exothermic but the rate would otherwise be too slow. | 约 450 °C：折中温度，因为正反应放热，但温度再低速率又太慢。
- Iron catalyst: accelerates the approach to equilibrium without altering the yield. | 铁催化剂：加快趋于平衡，不改变产率。
- Removal of ammonia by liquefaction: continuous removal of product shifts the equilibrium to the right. | 通过液化移走氨：持续移去产物使平衡右移。
- Recycling of unreacted N₂ and H₂: avoids waste and improves overall efficiency. | 未反应的 N₂ 与 H₂ 循环利用：避免浪费并提高整体效率。
11. Equilibrium and Thermodynamics: ΔG = 0 | 平衡与热力学：ΔG = 0
Equilibrium has an exact thermodynamic signature: the Gibbs free energy change of the system is zero. The relationship between the free energy change at any composition and the standard value is given below, where Q is the reaction quotient.
平衡有一个精确的热力学标志：体系的吉布斯自由能变化为零。任意组成下的自由能变化与标准值之间的关系如下式所示，其中 Q 为反应商。
ΔG = ΔG° + R × T × ln Q
ollow immediately. If ΔG° is negative, then ln K is positive and K > 1, so products dominate at equilibrium. If ΔG° is positive, then K < 1 and reactants dominate. A reaction with a large negative ΔG° is described as product-favoured, but that says nothing about how fast equilibrium is reached, which is a purely kinetic question.
在平衡时 ΔG = 0 且 Q = K，由此得到 ΔG° = −R × T × ln K。立刻可以得到两个结论。若 ΔG° 为负，则 ln K 为正、K > 1，平衡时以产物为主。若 ΔG° 为正，则 K < 1，以反应物为主。ΔG° 很负的反应被称为"产物有利"，但这与达到平衡的速度快慢毫无关系，后者是纯粹的动力学问题。
12. Common Misconceptions and Exam Technique | 常见误区与考试技巧
Most lost marks in equilibrium questions come from a small set of recurring errors. Read the following list slowly and check your own answers against it.
平衡题目的失分大多来自一小撮反复出现的错误。请慢慢读下面的清单，并对照检查自己的答案。
- Equilibrium does not require equal concentrations of reactants and products; it requires equal rates. | 平衡不要求反应物与产物浓度相等，而要求速率相等。
- The reaction has not stopped. Both directions continue at the same rate. | 反应并未停止，两个方向以相同速率持续进行。
- A catalyst changes the rate, never the position of equilibrium and never Kc. | 催化剂改变速率，绝不改变平衡位置，也不改变 Kc。
- Only temperature changes the numerical value of Kc or Kp. | 只有温度会改变 Kc 或 Kp 的数值。
- Adding an inert gas at constant volume causes no shift; at constant pressure it dilutes the mixture and shifts equilibrium towards the side with more gas moles. | 恒容下加入惰性气体不引起移动；恒压下则会稀释混合物，使平衡向气体摩尔数多的一侧移动。
- Solids and pure liquids are omitted from Kc expressions because their activities are taken as 1. | 固体和纯液体不写入 Kc 表达式，因为其活度取为 1。
For structured questions, follow a fixed routine. State the direction of the shift, name the principle you are applying, and explain what is being opposed. A complete answer to "what happens when the pressure is increased?" reads: the equilibrium shifts to the right, because the forward reaction reduces the number of gas moles from four to two, which opposes the increase in pressure. Precise language, correct symbols and an explicit justification earn full marks every time.
对于结构化问答，请遵循固定套路：先说明移动方向，再点出所应用的原理，最后解释被抵消的是什么。对"增大压强会怎样"的完整答案是：平衡右移，因为正反应使气体摩尔数由 4 减少到 2，从而抵消压强的增大。精准的措辞、正确的符号和明确的理由，每次都能拿到满分。
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