📚 A-Level Chemistry: Chemical Equilibrium, Le Chatelier's Principle, Kc and Kp | A-Level化学：化学平衡、勒夏特列原理与Kc、Kp
Chemical equilibrium is one of the highest-yield topics in A-Level Chemistry. It links reaction rates, energetics and industrial chemistry, and it appears in almost every paper as a mixture of definitions, calculations and explanation questions. This guide covers the meaning of dynamic equilibrium, how to write and calculate Kc and Kp, how to apply Le Chatelier's principle to concentration, pressure and temperature changes, and the exam traps that cost students easy marks.
化学平衡是 A-Level 化学中"性价比"最高的考点之一。它把反应速率、能量变化与工业化学串联起来，几乎每套试卷都会以"定义题 + 计算题 + 解释题"的组合形式出现。本文系统梳理：动态平衡的真正含义、Kc 与 Kp 的书写与计算、如何用勒夏特列原理分析浓度/压强/温度变化，以及最容易丢分的考试陷阱。
1. What Chemical Equilibrium Really Means | 化学平衡的真正含义
A reversible reaction taking place in a closed system will eventually reach a state called dynamic equilibrium. At this point the rate of the forward reaction is exactly equal to the rate of the reverse reaction, so the concentrations of all reactants and products remain constant. The word "dynamic" is essential: both reactions are still happening at the molecular level, they simply cancel each other out in terms of observable change.
在封闭体系中进行的可逆反应，最终会达到一种状态，称为动态平衡。此时正反应速率与逆反应速率完全相等，因此所有反应物和生成物的浓度保持不变。"动态"这个词至关重要：在分子层面上两个方向的反应仍在持续进行，只是在宏观可观察的变化上互相抵消了。
Two common misconceptions appear in exams. First, equilibrium does not mean the amounts of reactants and products are equal - the equilibrium mixture may be almost entirely reactants or almost entirely products. Second, the reaction has not stopped; adding a catalyst or changing conditions can disturb the balance, and the system will respond by shifting the position of equilibrium.
考试中有两个常见误解。第一，平衡并不意味着反应物与生成物的量相等——平衡混合物可能几乎全是反应物，也可能几乎全是生成物。第二，反应并没有停止；加入催化剂或改变条件会打破平衡，体系会通过移动平衡位置来作出回应。
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹
The double arrow ⇌ eaction goes to completion, which is a different (and usually wrong) statement in this context. Equilibrium can only be established in a closed system - if a gas escapes, the reverse reaction is progressively starved of reactant and the position keeps moving.
平衡方程式中必须使用双向箭头 ⇌。单箭头 → 表示反应进行到底，在平衡语境下通常是错误的表述。平衡只能在封闭体系中建立——如果气体逸出，逆反应会逐渐缺乏反应物，平衡位置就会不断移动。
2. The Equilibrium Constant Kc | 浓度平衡常数 Kc
For a general homogeneous reaction at equilibrium, the equilibrium constant Kc is defined as the product of the equilibrium concentrations of the products, each raised to the power of its stoichiometric coefficient, divided by the corresponding product for the reactants.
对于一般的均相反应，浓度平衡常数 Kc 定义为：各生成物平衡浓度的幂之积（幂次为其化学计量数），除以各反应物平衡浓度的相应幂之积。
aA + bB ⇌ cC + dD
Kc = [C]ᶜ [D]ᵈ ÷ ( [A]ᵃ [B]ᵇ )
Square brackets mean "concentration in mol dm⁻³ at equilibrium". Notice the three key conditions: equilibrium values only (never initial values), concentrations in mol dm⁻³, and powers taken from the balanced equation. If the equation is reversed, Kc becomes 1/Kc; if it is multiplied by a factor n, Kc is raised to the power n. These two rules are frequently tested.
方括号表示"平衡时的浓度，单位为 mol dm⁻³"。请注意三个关键条件：只能代入平衡值（绝不能用初始值）、浓度单位为 mol dm⁻³、幂次取自配平后的方程式。若方程式反向书写，Kc 变为 1/Kc；若方程式整体乘以 n，则 Kc 变为原来的 n 次方。这两条规则经常被考查。
In a heterogeneous equilibrium, the concentrations of pure solids and pure liquids are treated as constant and are omitted from the expression entirely. This is why, for the decomposition of calcium carbonate, the expression contains only the gas:
在多相平衡中，纯固体和纯液体的浓度被视为常数，因此完全不出现在表达式中。这就是碳酸钙分解的 Kc 表达式中只剩下气体的原因：
CaCO₃(s) ⇌ CaO(s) + CO₂(g) Kc = [CO₂]
Kc depends only on temperature. Changing concentration, pressure, volume or adding a catalyst will shift the position of equilibrium but will not change the numerical value of Kc at a given temperature. If Kc is large, the equilibrium mixture is product-rich; if Kc is very small, the mixture is reactant-rich.
Kc 只与温度有关。改变浓度、压强、体积或加入催化剂会移动平衡位置，但在给定温度下不会改变 Kc 的数值。Kc 很大说明平衡混合物以生成物为主；Kc 极小则说明以反应物为主。
3. Calculating Kc: Worked Examples | Kc 计算实例
The standard exam method is a three-row ICE (Initial - Change - Equilibrium) table. Write the balanced equation, fill in initial moles or concentrations, work out the change using the stoichiometric ratio, then calculate the equilibrium values and substitute them into the expression.
标准解题方法是"三段式"（初始 - 变化 - 平衡）表格。先写出配平方程式，填入初始物质的量或浓度，再依据化学计量比推出变化量，最后算出平衡值并代入表达式。
Example 1 - equal volumes, no volume change. 1.00 mol of H₂ and 1.00 mol of I₂ are sealed in a 1.00 dm³ vessel and allowed to reach equilibrium at temperature T. At equilibrium, 1.56 mol of HI is present. Find Kc.
例 1——体积不变。将 1.00 mol H₂ 与 1.00 mol I₂ 密封于 1.00 dm³ 容器中，在温度 T 下达到平衡，平衡时 HI 为 1.56 mol。求 Kc。
I₂ = 0.22 mol, HI = 1.56 mol. Because the volume is 1.00 dm³, moles equal concentrations.
由于生成了 1.56 mol HI，说明已反应的 H₂ 和 I₂ 各为 0.78 mol（比例 2:1）。平衡量：H₂ = 1.00 − 0.78 = 0.22 mol，I₂ = 0.22 mol，HI = 1.56 mol。因为体积为 1.00 dm³，物质的量与浓度数值相同。
H₂(g) + I₂(g) ⇌ 2HI(g)
Kc = [HI]² ÷ ( [H₂] [I₂] ) = (1.56)² ÷ (0.22 × 0.22) = 2.4336 ÷ 0.0484 ≈ 50.3
Example 2 - esterification, equimolar. 1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed with a small amount of acid catalyst. At equilibrium, 0.67 mol of ethyl ethanoate has formed in a total volume of V dm³.
例 2——酯化反应。将 1.00 mol 乙酸与 1.00 mol 乙醇在少量酸催化下混合。平衡时生成 0.67 mol 乙酸乙酯，总体积为 V dm³。
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
67, [acid] = 0.33, [alcohol] = 0.33, giving Kc = (0.67 × 0.67) ÷ (0.33 × 0.33) ≈ 4.1. This "volume cancels" trick is extremely useful whenever Δn(gas) = 0 or when all species share the same solvent.
四种物质处于同一体积 V 中，体积可约去，因此可直接用物质的量代入：[酯] = 0.67，[水] = 0.67，[酸] = 0.33，[醇] = 0.33，得 Kc = (0.67 × 0.67) ÷ (0.33 × 0.33) ≈ 4.1。当气体 Δn = 0 或所有物质共处同一溶剂时，这个"体积约去"技巧极为实用。
Example 3 - working backwards. If Kc for a reaction is known and the initial amounts are given, define x as the change in concentration, write equilibrium terms algebraically (for example 1.00 − x and 2x), substitute into Kc, and solve. If the resulting quadratic is awkward, state clearly that you are neglecting x where it is small, and justify the approximation.
例 3——反推。若已知 Kc 和初始量，可设 x 为浓度变化量，用代数式表示平衡浓度（如 1.00 − x 和 2x），代入 Kc 求解。若得到的二次方程过于繁琐，可明确说明因 x 很小而忽略之，并给出合理性论证。
4. Kp and Partial Pressures | Kp 与分压
For gaseous equilibria it is often more convenient to use partial pressures instead of concentrations. The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the whole volume. By Dalton's law, the total pressure is the sum of the partial pressures, and each partial pressure is the mole fraction multiplied by the total pressure.
对于气体平衡，用分压代替浓度往往更方便。混合气体中某气体的分压，是它单独占据整个容器时所施加的压强。根据道尔顿分压定律，总压等于各分压之和，而每种气体的分压等于其摩尔分数乘以总压。
pₐ = xₐ × P(total) xₐ = nₐ ÷ n(total)
The Kp expression has exactly the same shape as Kc, but with partial pressures in place of concentrations:
Kp 表达式与 Kc 形式完全一致，只是把浓度换成分压：
Kp = p(C)ᶜ p(D)ᵈ ÷ ( p(A)ᵃ p(B)ᵇ )
Worked example. An equilibrium mixture for the ammonia synthesis has mole fractions of 0.20 for N₂, 0.60 for H₂ and 0.20 for NH₃, at a total pressure of 250 atm.
计算示例。合成氨的平衡混合物中，N₂、H₂、NH₃ 的摩尔分数分别为 0.20、0.60、0.20，总压为 250 atm。
p(N₂) = 0.20 × 250 = 50 atm p(H₂) = 0.60 × 250 = 150 atm p(NH₃) = 0.20 × 250 = 50 atm
Kp = (50)² ÷ (50 × 150³) = 2500 ÷ 168 750 000 ≈ 1.48 × 10⁻⁵ atm⁻²
its of Kp are atm⁻². Many exam boards accept Kp without units, but a question that supplies pressures in kPa expects kPa to be used consistently; never mix atm and kPa in the same calculation.
务必检查单位。对本反应，Δn(气) = 2 − (1 + 3) = −2，故 Kp 的单位为 atm⁻²。许多考试局允许 Kp 不写单位，但如果题目给出的压强单位是 kPa，就要全程统一使用 kPa；绝不可在同一计算中混用 atm 与 kPa。
5. Converting Between Kc and Kp | Kc 与 Kp 的换算
(n ÷ V)RT = cRT. Substituting this into the Kp expression gives the standard relationship:
Kc 与 Kp 描述的是同一个平衡，因此二者必然存在关联。由理想气体状态方程 pV = nRT 可得 p = (n ÷ V)RT = cRT。把它代入 Kp 表达式，即得到标准关系式：
Kp = Kc (RT)Δn
8.314 J K⁻¹ mol⁻¹ when pressures are in Pa, or R = 0.0821 dm³ atm K⁻¹ mol⁻¹ when pressures are in atm and concentrations in mol dm⁻³.
其中 Δn 为配平方程式中气体物质的量变化（生成物减反应物），T 为热力学温度（K），R 为气体常数。压强用 Pa 时取 R = 8.314 J K⁻¹ mol⁻¹；压强用 atm、浓度用 mol dm⁻³ 时取 R = 0.0821 dm³ atm K⁻¹ mol⁻¹。
Two special cases are worth memorising. When Δn = 0 (for example H₂ + I₂ ⇌ mperature for the same numerical convention, and the units of Kp carry a negative power of pressure.
两个特例值得牢记。当 Δn = 0（如 H₂ + I₂ ⇌ 2HI）时，Kp = Kc，数值相同，这是常见的选择题答案。当 Δn 为负（如合成氨）时，在相同温度下 Kp 与 Kc 的数值关系随单位约定而变，且 Kp 的单位带有压强的负幂次。
6. Le Chatelier's Principle: The Core Idea | 勒夏特列原理的核心思想
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that tends to oppose the change. It is a qualitative rule, but it lets you predict the direction of shift without any calculation.
勒夏特列原理指出：如果对处于平衡状态的体系施加条件改变，平衡位置将向能够抵消该改变的方向移动。这是一条定性规律，但可以让你在不做任何计算的情况下预测移动方向。
A reliable four-step routine for explanation questions: state the change that has been imposed; state the direction in which the equilibrium shifts; explain why that direction opposes the change; then state the consequence for the yield or concentration of the species named in the question. Marks are awarded for each of these steps, so never jump straight to the conclusion.
解释题可遵循固定的四步法：指出施加了什么改变；说明平衡向哪个方向移动；解释为什么该方向能抵消这一改变；最后说明对题目所指物质产率或浓度的影响。每一步都有分值，因此切勿直接跳到结论。
A useful mental model is that the system "reacts against" whatever you do to it. Squeeze it by increasing pressure and it moves to the side with fewer gas molecules, relieving the squeeze. Feed it extra reactant and it consumes that reactant by shifting forward, restoring the ratio demanded by Kc.
一个有效的思维模型是：体系会"反抗"你对它做的任何事。加压相当于挤压它，它就向气体分子数更少的一侧移动以缓解挤压；额外加入反应物，它就通过正向移动消耗掉这些反应物，以恢复 Kc 所要求的比例。
7. Effect of Concentration and Pressure | 浓度与压强的影响
Increasing the concentration (or partial pressure) of a reactant shifts the equilibrium to the right, producing more products. Increasing the concentration of a product shifts it to the left. Removing a product continuously - for example by distilling it off or precipitating it - also drives the equilibrium forwards, and this is exactly why esterification is carried out with continuous removal of the ester.
增大反应物的浓度（或分压）会使平衡向右移动，生成更多产物；增大生成物浓度则使平衡向左移动。若能不断移走生成物——例如蒸馏分离或使其沉淀——同样会推动平衡正向移动，这正是酯化反应中不断蒸出酯的原因。
es have the same number of gas moles (Δn = 0), a pressure change has no effect on the position of equilibrium at all.
关于压强变化，只有气体才起作用。增大总压使平衡向气体物质的量较少的一侧移动。若两侧气体物质的量相同（Δn = 0），则压强变化对平衡位置完全没有影响。
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Δn = 2 − 4 = −2
Therefore high pressure favours ammonia. This reaction is the classic example of a pressure-sensitive equilibrium: 4 moles of gas on the left, 2 on the right.
因此高压有利于生成氨。这个反应是压强敏感型平衡的经典例子：左侧 4 mol 气体，右侧 2 mol 气体。
A subtle but very examinable point: adding an inert gas at constant volume has no effect, because the partial pressures of the reacting gases are unchanged. Adding an inert gas at constant pressure increases the total volume, which lowers all partial pressures, so the equilibrium shifts towards the side with more gas moles - the opposite of what a pressure increase would do.
一个细微但高频考点：恒容下加入惰性气体没有影响，因为各反应气体的分压不变。恒压下加入惰性气体则使总体积增大、各分压降低，平衡会向气体物质的量较多的一侧移动——与加压的效果正好相反。
8. Effect of Temperature and the van't Hoff Link | 温度的影响与范特霍夫关系
Temperature is the only factor that changes the value of Kc or Kp. If the forward reaction is exothermic, raising the temperature shifts the equilibrium to the left (the endothermic direction) and decreases K. If the forward reaction is endothermic, raising the temperature shifts the equilibrium to the right and increases K.
温度是唯一会改变 Kc 或 Kp 数值的因素。若正反应为放热反应，升高温度使平衡向左（吸热方向）移动，K 减小；若正反应为吸热反应，升高温度使平衡向右移动，K 增大。
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
Because the forward reaction is exothermic, higher temperatures reduce the equilibrium yield of ammonia. Industry therefore compromises: a moderately high temperature (about 400-450 °C) sacrifices some yield in exchange for a much faster rate, which is essential for commercial viability.
由于正反应放热，温度越高氨的平衡产率越低。因此工业上需要折中：采用中等偏高温度（约 400–450 °C），牺牲一部分产率以换取显著更快的反应速率，这对商业可行性至关重要。
The quantitative link between temperature and K is the van't Hoff relationship, which is an extension beyond most specifications but is included in some pre-university papers:
温度与 K 之间的定量关系是范特霍夫关系式，这超出多数考试大纲，但部分大学先修类试卷会涉及：
ln(K₂ ÷ K₁) = −(ΔH ÷ R) × (1 ÷ T₂ − 1 ÷ T₁)
For an exothermic reaction ΔH is negative, so as T₂ exceeds T₁ the term on the right becomes negative and K₂ is smaller than K₁ - exactly the qualitative statement above, now with numbers.
对放热反应 ΔH 为负，当 T₂ 大于 T₁ 时右侧为负，故 K₂ 小于 K₁——这与上面的定性结论完全一致，只是变成了定量表达。
9. Catalysts and Equilibrium | 催化剂与平衡
A catalyst increases the rate of both the forward and the reverse reaction by the same factor, lowering the activation energy of each pathway. It therefore reduces the time taken to reach equilibrium, but it does not change the position of equilibrium and does not change the value of Kc or Kp.
催化剂以相同倍数提高正、逆反应速率，同时降低两条路径的活化能。因此它能缩短达到平衡所需时间，但不会改变平衡位置，也不会改变 Kc 或 Kp 的数值。
This is a favourite exam trap. A question such as "state and explain the effect of adding an iron catalyst on the equilibrium yield of ammonia" expects the answer "no effect on the yield, because the catalyst speeds up the forward and reverse reactions equally". If the question asks about the industrial advantage instead, the answer is that equilibrium is reached faster, saving time and energy costs.
这是最经典的考试陷阱之一。诸如"说明加入铁催化剂对氨平衡产率的影响"这类题目，正确回答是"对产率没有影响，因为催化剂同等加快了正逆反应速率"。如果问的是工业上的优势，答案则是更快达到平衡，节省时间与能源成本。
Catalysts do, however, allow industrial chemists to use milder conditions. Because the catalyst speeds up the reaction, a lower temperature can be used for a given production rate, and a lower temperature gives a higher equilibrium yield for an exothermic reaction. The catalyst thus improves the achievable compromise even though it does not move the equilibrium itself.
不过，催化剂确实让工业化学家可以采用更温和的条件。因为催化剂加快了反应，为达到同样的生产效率可以采用更低温度，而较低温度对放热反应意味着更高的平衡产率。因此催化剂改善的是"可行折中方案"，尽管它本身并不移动平衡。
10. Industrial Applications: Haber and Contact Processes | 工业应用：哈伯法与接触法
The Haber process and the Contact process are the two equilibrium systems you must be able to discuss with justification of every condition chosen.
哈伯法与接触法是必须掌握、且要能论证每个条件选取理由的两个平衡体系。
| System | Equation | Conditions | Reasoning |
| Haber | N₂ + 3H₂
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