📚 Combined Equilibria: Kc, Kp and Le Chatelier | 组合平衡：Kc、Kp 与勒夏特列原理
In Edexcel A-Level Chemistry, equilibrium topics appear across both AS and A2 units, often combining calculations of Kc, Kp, and applications of Le Chatelier’s principle. Many exam questions ask you to work with homogeneous and heterogeneous systems, interpret shifts, and link equilibrium constants with energetics. This article provides a structured revision guide to help you master combined equilibria, covering the core concepts, mathematical expressions, and typical pitfalls.
在Edexcel A-Level化学考试中，平衡相关题目贯穿AS和A2单元，常常需要综合运用Kc、Kp的计算以及勒夏特列原理。许多考题要求你处理均相和非均相体系、解释平衡移动，并将平衡常数与能量变化联系起来。本文提供了一份结构化的复习指南，帮助你掌握组合平衡，涵盖核心概念、数学表达式和常见易错点。
1. Dynamic Equilibrium – The Reversible Process | 动态平衡 – 可逆过程
A dynamic equilibrium is established when a reversible reaction proceeds in a closed system and the rates of the forward and reverse reactions become equal. At equilibrium, the concentrations of reactants and products remain constant, but both reactions continue to occur at the molecular level.
当可逆反应在封闭体系中进行，且正反应和逆反应的速率相等时，就建立了动态平衡。平衡时，反应物和产物的浓度保持不变，但在分子层面上两个方向的反应仍在持续进行。
For Edexcel, you must recognise that equilibrium can only be reached in a closed system. If products or reactants escape (an open system), the reaction may go to completion or never establish equilibrium. The double half-arrow symbol ⇌ is used to denote a reversible reaction.
在Edexcel考试中，你必须认识到平衡只能在封闭体系中达成。如果产物或反应物逸出（开放体系），反应可能进行到底或永远无法建立平衡。可逆反应用双半箭头符号 ⇌ 表示。
2. The Equilibrium Constant Kc – Definition and Meaning | 平衡常数 Kc – 定义与意义
For a general homogeneous equilibrium: aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc = [C]ᶜ uilibrium concentrations in mol dm⁻³. The value of Kc is constant for a given temperature.
对于一般的均相平衡：aA + bB ⇌ cC + dD，以浓度表示的平衡常数为 Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ，方括号代表平衡浓度，单位为 mol dm⁻³。在给定温度下，Kc 是一个常数。
A large Kc (much greater than 1) means the equilibrium mixture contains mostly products, while a very small Kc (much less than 1) indicates that reactants dominate. Kc has no units only when the total number of particles on each side of the equation is the same.
Kc 值很大（远大于1）意味着平衡混合物中主要是产物，而 Kc 很小（远小于1）则表明反应物占主导。只有当方程式两侧总粒子数相等时，Kc 才没有单位。
3. Writing Kc Expressions and Determining Units | 书写 Kc 表达式并确定单位
wer of their stoichiometric coefficients. For example, in the Haber process N₂(g) + 3H₂(g) ⇌ = [NH₃]² / ([N₂][H₂]³). The units are (mol dm⁻³)² / (mol dm⁻³ × (mol dm⁻³)³) = mol⁻² dm⁶.
要正确书写 Kc 表达式，总是把产物放在分子、反应物放在分母，并以化学计量系数作为指数。例如，在哈伯法 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 中，Kc = [NH₃]² / ([N₂][H₂]³)。其单位为 (mol dm⁻³)² / (mol dm⁻³ × (mol dm⁻³)³) = mol⁻² dm⁶。
Cancel units carefully: if the final expression simplifies to no net dimensions, we say Kc is dimensionless. Edexcel mark schemes often award marks for correct units, so always show your working.
仔细约去单位：如果最终表达式简化后没有量纲，则称 Kc 无量纲。Edexcel 评分标准经常对正确单位给分，因此务必展示推导过程。
4. Calculating Kc from Experimental Data – ICE Tables | 从实验数据计算 Kc – ICE 表格
When initial amounts and one equilibrium concentration are known, use an ICE table (Initial, Change, Equilibrium) to find all equilibrium concentrations. Write the balanced equation, fill in initial moles, express changes using x or stoichiometric ratios, then convert to concentrations (divide by volume in dm³) before substituting into Kc.
当已知初始量和某一个平衡浓度时，使用 ICE 表格（初始、变化、平衡）来求出所有的平衡浓度。写出配平的方程式，填入初始摩尔数，用 x 或化学计量比表示变化，然后转化为浓度（除以体积，dm³），再代入 Kc 表达式。
Consider: H₂(g) + I₂(g) ⇌ mol in a 1.0 dm³ vessel and at equilibrium 0.20 mol H₂ remains, then the change for H₂ is – mol dm⁻³, and [HI]eq = 1.60 mol dm⁻³, giving Kc = (1.60)² / (0.20 × 0.20) = 64 (no units).
例如：H₂(g) + I₂(g) ⇌ 2HI(g)。若 H₂ 和 I₂ 初始均为 1.0 mol，容器体积 1.0 dm³，平衡时剩余 0.20 mol H₂，则 H₂ 的变化为 –0.80 mol，因此 [H₂]eq = 0.20 mol dm⁻³，[I₂]eq = 0.20 mol dm⁻³，[HI]eq = 1.60 mol dm⁻³，计算 Kc = (1.60)² / (0.20 × 0.20) = 64（无单位）。
5. Homogeneous and Heterogeneous Equilibria | 均相与非均相平衡
A homogeneous equilibrium has all reactants and products in the same phase (all gases or all aqueous). All species appear in the Kc expression. A heterogeneous equilibrium involves substances in different phases, such as solids with gases. In Kc expressions, pure solids and pure liquids are omitted because their concentrations remain constant.
均相平衡中所有反应物和产物都处于同一相态（全是气体或全是溶液），所有物种都出现在 Kc 表达式中。非均相平衡涉及不同相态的物质，例如固体与气体。在 Kc 表达式中，纯固体和纯液体被省略，因为它们的浓度是常数。
Example: CaCO₃(s) ⇌ use CaCO₃ and CaO are solids. This simplification is crucial for Edexcel heterogeneous equilibrium questions, especially when combining with Kp.
示例：CaCO₃(s) ⇌ CaO(s) + CO₂(g)。Kc = [CO₂]，因为 CaCO₃ 和 CaO 是固体。这一简化对Edexcel非均相平衡题至关重要，尤其是在与 Kp 结合时。
6. The Equilibrium Constant in Terms of Pressure, Kp | 压力平衡常数 Kp
For gaseous equilibria, we often use Kp, expressed in terms of partial pressures. Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ, where pA, pB, etc., are the equilibrium partial pressures. Kp is also constant at a fixed temperature and has units of pressure raised to the power of Δn (change in moles of gas).
对于气体平衡，常使用 Kp，用分压表示。Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ，其中 pA、pB 等是平衡分压。Kp 在固定温度下也是常数，其单位是压力单位的 Δn 次方（气体摩尔数的变化量）。
moles of that gas / total moles of gas in the mixture. Edexcel often gives total pressure and initial moles, then expects you to calculate mole fractions and Kp.
气体的分压 = 摩尔分数 × 总压。摩尔分数 = 该气体的摩尔数 / 混合气体总摩尔数。Edexcel 经常给出总压和初始摩尔数，然后要求你计算摩尔分数和 Kp。
7. Kp Calculations and Worked Example | Kp 计算与典型例题
Consider the equilibrium: 2SO₂(g) + O₂(g) ⇌ 0.30+1.20) = 0.25. Partial pressure p(SO₂) = 0.25 × 200 = 50 kPa. Similarly, p(O₂) = (0.30/2.00)×200 = 30 kPa, p(SO₃) = (1.20/2.00)×200 = 120 kPa.
考虑平衡：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)。平衡时容器含 0.50 mol SO₂，0.30 mol O₂ 和 1.20 mol SO₃。总压 200 kPa。SO₂ 的摩尔分数 = 0.50 / (0.50+0.30+1.20) = 0.25。分压 p(SO₂) = 0.25 × 200 = 50 kPa。同理，p(O₂) = (0.30/2.00)×200 = 30 kPa，p(SO₃) = (1.20/2.00)×200 = 120 kPa。
Then Kp = (pSO₃)² / (pSO₂)² × pO × 30) = 14400 / 75000 = 0.192 kPa⁻¹. Edexcel expects clear substitution and unit analysis.
则 Kp = (pSO₃)² / ((pSO₂)² × pO₂) = (120)² / (50)² × 30 = 14400 / (2500 × 30) = 14400 / 75000 = 0.192 kPa⁻¹。Edexcel 要求清晰的代入和单位分析。
8. Relationship Between Kc and Kp | Kc 与 Kp 的关系
1 J mol⁻¹ K⁻¹), T is the temperature in Kelvin, and Δn = (total moles of gaseous products) – ts). This equation is given in the Edexcel data booklet, but you must be able to apply it.
对于气体反应，Kp 与 Kc 通过方程 Kp = Kc (RT)^Δn 关联，其中 R 是气体常数 (8.31 J mol⁻¹ K⁻¹)，T 是开尔文温度，Δn = (气态产物总摩尔数) – (气态反应物总摩尔数)。该公式在Edexcel数据手册中提供，但你必须会应用。
If Δn = 0, then Kp = Kc, and both are dimensionless. For example, in H₂(g) + I₂(g) ⇌ 2HI(g), Δn = 2– rt temperature to Kelvin and use consistent pressure units; typically Kp is in kPa or atm.
若 Δn = 0，则 Kp = Kc，且两者均无量纲。例如 H₂(g) + I₂(g) ⇌ 2HI(g) 中 Δn = 2–2 = 0，因此 Kp = Kc。记得将温度转换成开尔文并使用一致的压力单位；Kp 常用 kPa 或 atm。
9. Le Chatelier’s Principle – Predicting Shifts | 勒夏特列原理 – 预测平衡移动
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 shifts to counteract that change. This tool helps explain how Kc/Kp values respond or new equilibrium concentrations are established.
勒夏特列原理指出，若处于动态平衡的体系受到浓度、压力或温度的改变，平衡位置将发生移动以抵消这种改变。这一工具帮助解释 Kc/Kp 如何响应，或新的平衡浓度如何建立。
For Edexcel, you need to link the direction of shift with yield and Kc changes. A shift to the right increases product yield but does not change Kc unless temperature changes. A shift to the left reduces product yield. Only temperature alters the equilibrium constant; concentration and pressure shifts keep Kc constant.
在Edexcel考试中，你需要将移动方向与产率和 Kc 变化联系起来。向右移动提高产物产率，但不改变 Kc（除非温度改变）。向左移动降低产率。只有温度会改变平衡常数；浓度和压力的移动保持 Kc 不变。
10. Effect of Temperature on Kc and Kp – Van’t Hoff Insight | 温度对 Kc 和 Kp 的影响 – 范特霍夫视角
Increasing temperature for an endothermic reaction (ΔH > 0) causes Kc to increase because the equilibrium shifts to absorb heat (right). For an exothermic reaction (ΔH < 0), raising temperature decreases Kc as the equilibrium shifts left. This is entirely consistent with the Van’t Hoff equation, but Edexcel expects qualitative reasoning and calculations.
对于吸热反应（ΔH > 0），升高温度会使 Kc 增大，因为平衡吸热移动（向右）。对于放热反应（ΔH < 0），升高温度会使 Kc 减小，因为平衡向左移动。这与范特霍夫方程完全一致，但Edexcel要求的是定性推理和计算。
In a Kp problem, temperature change alters Kp similarly. You may be given Kp at two temperatures and asked to deduce whether the forward reaction is exothermic or endothermic. A decrease in Kp with temperature rise indicates an exothermic forward reaction.
在 Kp 问题中，温度变化同样改变 Kp。题目可能给出两个温度下的 Kp，要求你推断正反应是放热还是吸热。若 Kp 随温度升高而减小，则表明正反应放热。
11. Effect of Concentration, Pressure, and Catalysts | 浓度、压力和催化剂的影响
Adding a reactant or removing a product shifts equilibrium to the right, producing more products, but Kc remains unchanged. Increasing total pressure (by reducing volume) shifts equilibrium towards the side with fewer gas molecules, again without changing Kc or Kp. Adding an inert gas at constant volume does not alter partial pressures of reacting species, so no shift occurs.
增加反应物或移除产物会使平衡向右移动，生成更多产物，但 Kc 保持不变。增大总压（通过减小体积）使平衡向气体分子数少的一侧移动，同样不改变 Kc 或 Kp。恒容下加入惰性气体不改变反应物种的分压，因此不发生移动。
A catalyst speeds up both forward and reverse reactions equally, reducing the time to reach equilibrium but having zero effect on Kc or Kp. Equilibrium composition remains the same. Edexcel frequently probes this common misconception.
催化剂同等程度地加快正逆反应，缩短到达平衡的时间，但对 Kc 或 Kp 毫无影响，平衡组成不变。Edexcel 经常考察这一常见错误认知。
12. Combined Equilibrium Problems – Mastering the Link | 组合平衡问题 – 掌握联系
Edexcel A-Level papers love combining Kc, Kp, ICE tables, and Le Chatelier in one question. You might be given Kc and asked to calculate equilibrium moles after a pressure change, or use Kp with initial pressures to find an unknown partial pressure. The key is systematic working: write the balanced equation, construct an ICE or mole table, use the given constant, and solve for the unknown.
Edexcel A-Level 试卷喜欢将 Kc、Kp、ICE 表格和勒夏特列原理组合在一道题中。你可能拿到 Kc 值并被要求计算压力改变后的平衡摩尔数，或者利用 Kp 和初始分压求出未知分压。关键是要系统求解：写出配平方程式，构建 ICE 或摩尔表，利用给定常数，解出未知数。
Always check units, especially when converting between Kc and Kp. Convert volume to dm³, pressure to kPa or atm as specified, and temperature to K. When shifting, only temperature causes a new K value; otherwise, use the same K to find new equilibrium amounts after a concentration or pressure disturbance.
务必检查单位，尤其是在 Kc 和 Kp 换算时。体积转为 dm³，压力按题目要求使用 kPa 或 atm，温度用 K。发生移动时，仅温度改变会带来新的 K 值；否则，在浓度或压力扰动后，用同一个 K 求新的平衡量。
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