Mastering Equilibrium Constants: Kc and Kp | 掌握平衡常数：Kc 与 Kp

📚 Mastering Equilibrium Constants: Kc and Kp | 掌握平衡常数：Kc 与 Kp

Dynamic equilibrium is a cornerstone of physical chemistry, and the equilibrium constants Kc and Kp allow us to quantify the position of equilibrium. In Edexcel A-Level Chemistry, a thorough understanding of how to write, calculate and apply these constants is essential for success in topics covering rates, equilibria and industrial processes. This article unpacks every aspect you need, from deriving expressions to tackling tricky exam questions.

动态平衡是物理化学的基石，而平衡常数 Kc 和 Kp 使我们能够定量描述平衡的位置。在 Edexcel A-Level 化学中，全面理解如何书写、计算和应用这些常数对于掌握速率、平衡和工业过程等主题至关重要。本文将从导出表达式到解决棘手的考题，解析你需要掌握的每一个方面。

1. Revisiting Dynamic Equilibrium | 重温动态平衡

A reversible reaction reaches dynamic equilibrium when the forward and reverse reactions proceed at exactly the same rate. At this point, the concentrations of all reactants and products remain constant, but the reactions are still ongoing at a microscopic level. The system must be closed and the temperature constant for a true equilibrium to be established.

当正反应和逆反应以完全相同的速率进行时，可逆反应达到动态平衡。此时，所有反应物和产物的浓度保持恒定，但在微观层面上反应仍在进行。要建立真正的平衡，体系必须是封闭的并且温度保持不变。

The equilibrium state is characterised by a specific ratio of product to reactant concentrations, which is independent of the starting amounts but depends on temperature. This ratio is captured mathematically by the equilibrium constant.

平衡状态的特征是产物与反应物浓度的特定比值，该比值与初始量无关，但取决于温度。这个比值通过平衡常数在数学上加以描述。

2. Defining the Equilibrium Constant Kc | 定义平衡常数 Kc

For a general homogeneous reaction aA + bB ⇌ quare brackets denote equilibrium concentrations in mol dm⁻³. Note that the powers correspond to the stoichiometric coefficients in the balanced equation.

对于一般的均相反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数表达式为：Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ，其中方括号表示以 mol dm⁻³ 为单位的平衡浓度。注意，幂次与配平方程中的化学计量系数相对应。

Kc is dimensionless only if the total number of moles on both sides is equal, but for Edexcel you will often calculate a numerical value and state its units, derived by substituting concentration units into the expression, e.g. mol⁻¹ dm³ or mol dm⁻³.

当反应两边总摩尔数相等时 Kc 无量纲，但在 Edexcel 考试中，你通常需要计算数值并注明单位，单位可通过将浓度单位代入表达式导出，例如 mol⁻¹ dm³ 或 mol dm⁻³。

The magnitude of Kc gives valuable insight: a large Kc (≫ 1) means the equilibrium lies far to the right, favouring products; a very small Kc (≪ 1) indicates a reactant-favoured equilibrium.

Kc 的大小提供了有价值的信息：较大的 Kc（≫ 1）意味着平衡位置强烈偏向右边，有利于产物；非常小的 Kc（≪ 1）则表明平衡偏向反应物一侧。

3. Homogeneous vs Heterogeneous Equilibria | 均相平衡与非均相平衡

A homogeneous equilibrium is one in which all reactants and products are in the same phase, such as all gases or all aqueous species. In writing Kc expressions for heterogeneous systems, concentrations of pure solids and pure liquids are omitted because their concentrations remain essentially constant and are already incorporated into the value of Kc.

均相平衡是指所有反应物和产物都处于同一相，例如全部为气体或全部为水溶液物种。在书写非均相体系的 Kc 表达式时，纯固体和纯液体的浓度被省略，因为它们的浓度基本保持不变，并已纳入 Kc 的值中。

For example, in the equilibrium CaCO₃(s) ⇌ CO₂(g), Kc = [CO₂]; the solids do not appear. This simplification is a common exam pitfall—always check for state symbols before writing the expression.

例如，在平衡 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 中，Kc = [CO₂]；固体不出现。这种简化是考试中常见的易错点——在书写表达式前务必检查状态符号。

4. Calculating Kc from Experimental Data | 由实验数据计算 Kc

Given initial moles, the volume of the container and the equilibrium moles of one substance, you can determine all equilibrium concentrations. Set up an ICE table (Initial, Change, Equilibrium) using molar quantities, then divide by volume to get concentrations before substituting into the Kc expression.

已知初始摩尔数、容器体积和某物质的平衡摩尔数，就可以确定所有组分的平衡浓度。使用摩尔数建立 ICE 表（初始、变化、平衡），然后除以体积得到浓度，再代入 Kc 表达式。

For the reaction H₂(g) + I₂(g) ⇌ a typical calculation yields Kc = (1.6)² / (0.2 × 0.2) = 64 with no units (since Δn = 0).

对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，若在 1 dm³ 容器中混合 1.0 mol H₂ 和 1.0 mol I₂，平衡混合物含 1.6 mol HI，那么通过典型计算可得 Kc = (1.6)² / (0.2 × 0.2) = 64，无单位（因为 Δn = 0）。

Always double-check the balanced equation and the direction of change: if product is formed, reactant amounts decrease by the stoichiometric proportion. Using positive and negative signs consistently avoids errors.

务必反复检查配平方程和变化方向：如果生成了产物，反应物的量就会按照化学计量比例减少。一致地使用正负号可以避免错误。

5. Introducing Gaseous Equilibria and Kp | 气体平衡与 Kp 简介

an individual gas in a mixture and is proportional to its mole fraction: pA = (mole fraction of A) × total pressure.

对于涉及气体的反应，我们常用分压表示的平衡常数 Kp。分压是混合气体中单一气体所施加的压力，与其摩尔分数成正比：pA = （A 的摩尔分数）× 总压。

For aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp = (pC)ᶜ in units of pressure raised to the power of the change in moles of gas, e.g. atm² or Pa⁻¹.

对于 aA(g) + bB(g) ⇌ cC(g) + dD(g)，Kp = (pC)ᶜ (pD)ᵈ / (pA)ᵃ (pB)ᵇ，其中每个 p 代表平衡时的分压。Kp 的数值通常带有压力单位的幂次，该幂次对应于气体摩尔数的变化，例如 atm² 或 Pa⁻¹。

Edexcel questions often ask you to calculate Kp when total pressure and initial moles are given. The key is to find equilibrium moles, then mole fractions, then partial pressures.

Edexcel 试题经常要求你在给定总压和初始摩尔数的条件下计算 Kp。关键在于求出平衡摩尔数，然后是摩尔分数，最后是分压。

6. Expressing and Calculating Kp | Kp 的表达与计算

Worked example: For the equilibrium N₂(g) + 3H₂(g) ⇌ a total pressure of 200 atm and a certain temperature, the equilibrium mixture contains 40% ence, partial pressures: pN₂ = 30 atm, pH₂ = 90 atm, pNH₃ = 80 atm. Kp = (80)² / (30 × 90³) = 6400 / (30 × 729000) = 2.93 × 10⁻⁴ atm⁻².

计算实例：对于平衡 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，在总压 200 atm 和某温度下，平衡混合物中氨的摩尔分数为 40%。假设按化学计量比起始，1 mol N₂ 和 3 mol H₂，计算得 N₂ 的摩尔分数为 0.15，H₂ 为 0.45，NH₃ 为 0.40。因此分压：pN₂ = 30 atm，pH₂ = 90 atm，pNH₃ = 80 atm。Kp = (80)² / (30 × 90³) = 6400 / (30 × 729000) = 2.93 × 10⁻⁴ atm⁻²。

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When units are required, they are determined by (pressure)Δn, where Δn = (c+d) – (a+b). Above, Δn = 2 – 4 = –2, so unit is atm⁻². Presenting clear step-by-step working is vital for full exam marks.

当需要注明单位时，单位由 (压力)Δn 确定，其中 Δn = (c+d) – (a+b)。在上述例子中，Δn = 2 – 4 = –2，因此单位为 atm⁻²。清晰的分步计算过程对于获得满分至关重要。

7. The Relationship Between Kc and Kp | Kc 与 Kp 的关系

ing on units), T is the absolute temperature in kelvin, and Δn is the change in moles of gas. This equation arises from the ideal gas law, P = (n/V)RT = concentration × RT.

对于气体平衡，Kp 与 Kc 通过方程 Kp = Kc (RT)Δn 相关联，其中 R 是气体常数（根据单位可选用 0.0821 L atm mol⁻¹ K⁻¹ 或 8.31 J mol⁻¹ K⁻¹），T 是以开尔文为单位的绝对温度，Δn 是气体摩尔数的变化。该方程源自理想气体状态方程 P = (n/V)RT = 浓度 × RT。

umerically identical. This relationship is often tested in multiple-choice or structured questions where you convert one constant to another.

如果 Δn = 0，则 Kp = Kc，两个常数的数值相同。这种关系常在选择题或结构化问题中考查，需要你进行常数之间的换算。

8. Factors Affecting the Value of the Equilibrium Constant | 影响平衡常数值的因素

The only factor that changes the numerical value of Kc or Kp for a given reaction is temperature. Changing concentration or pressure does not alter the equilibrium constant; instead, the system shifts in composition to restore the same constant value.

对于给定反应，唯一能改变 Kc 或 Kp 数值的因素是温度。改变浓度或压强不会改变平衡常数；相反，体系会通过组成移动来恢复同一个常数值。

For exothermic forward reactions (ΔH negative), raising temperature decreases Kc/Kp because the equilibrium shifts left to absorb heat. For endothermic forward reactions (ΔH positive), increasing temperature increases the constant. This is in full agreement with Le Chatelier's principle.

对于正向放热反应（ΔH 为负），升高温度会减小 Kc/Kp，因为平衡向左移动以吸收热量。对于正向吸热反应（ΔH 为正），升高温度会使常数增大。这与勒夏特列原理完全一致。

The presence of a catalyst speeds up both forward and backward reactions equally, so it has no effect on the value of the equilibrium constant; it only helps reaching equilibrium faster.

催化剂同等程度地加快正逆反应速率，因此对平衡常数的值没有影响；它仅有助于更快地达到平衡。

9. Using Kc and Kp to Predict the Direction of Reaction | 使用 Kc 和 Kp 预测反应方向

The reaction quotient, Qc (or Qp), is calculated using the same expression as Kc but with initial or current concentrations instead of equilibrium ones. Comparing Qc with Kc tells us the direction the reaction must proceed to reach equilibrium.

反应商 Qc（或 Qp）使用与 Kc 相同的表达式计算，但代入的是初始或当前浓度而非平衡浓度。将 Qc 与 Kc 比较，就可以知道反应为达到平衡必须进行的方向。

equilibrium. This concept is powerful for predicting shifts when mixing arbitrary amounts.

若 Qc < Kc，正向反应有利，会生成更多产物；若 Qc > Kc，逆反应有利。若 Qc = Kc，体系已处于平衡。这一概念在混合任意量时预测移动方向非常有用。

10. Industrial Applications: Haber and Contact Processes | 工业应用：哈柏法和接触法

The Haber process for ammonia synthesis, N₂ + 3H₂ ⇌ 2NH₃ ΔH = – ture, while increasing rate, actually lowers Kp because the reaction is exothermic. The industrial choice of about 450 °C and 200 atm, along with an iron catalyst, balances yield, rate and economic factors.

氨合成的哈柏法 N₂ + 3H₂ ⇌ 2NH₃ ΔH = –92 kJ mol⁻¹ 体现了条件妥协。高压使平衡右移（有利于总摩尔数减少的方向），但很高的温度虽然提高速率，却会降低 Kp，因为反应是放热的。工业上选择约 450 °C 和 200 atm，并使用铁催化剂，平衡了产率、速率和经济因素。

In the Contact process, 2SO₂ + O₂ ⇌ 2SO₃ ΔH = – nough for high conversion. Understanding how Kp varies with temperature helps justify these operating conditions.

在接触法 2SO₂ + O₂ ⇌ 2SO₃ ΔH = –197 kJ mol⁻¹ 中，五氧化二钒催化剂允许使用中等温度（约 450 °C）；此时平衡常数仍然足够有利以实现高转化率。理解 Kp 随温度的变化有助于合理解释这些操作条件。

11. Common Exam Pitfalls and How to Avoid Them | 常见考试失分点与避坑指南

One common mistake is including solids or liquids in the Kc expression. Always eliminate any substance with state symbol (s) or (l). Another is forgetting to convert moles into concentrations when calculating Kc—unless the volume is 1.0 dm³ exactly.

一个常见错误是将固体或液体写入 Kc 表达式。务必省略状态符号为 (s) 或 (l) 的任何物质。另一个错误是在计算 Kc 时忘记将摩尔数转换为浓度——除非容器体积恰好为 1.0 dm³。

e = mole fraction × total pressure. Also, always state units for Kc and Kp when Δn ≠ 0; marks are routinely allocated for correct units.

对于 Kp，学生有时会误用摩尔分数或分压。记住：摩尔分数 = 该气体的摩尔数 / 气体总摩尔数。然后分压 = 摩尔分数 × 总压。此外，当 Δn ≠ 0 时，务必注明 Kc 和 Kp 的单位；正确注明单位往往会得到专门的分数。

Finally, check the temperature when comparing constants. If a problem says 'at a different temperature', the Kc values will differ; never use a Kc from one temperature for calculations at another.

最后，在比较常数时要核对温度。如果题目指出“在另一温度下”，那么 Kc 值会不同；切勿将某一温度下的 Kc 用于另一温度下的计算。

12. Summarising Key Ideas and Revision Strategy | 核心要点总结与复习策略

Mastering equilibrium constants in Edexcel A-Level Chemistry demands practice with ICE tables, unit analysis and Kp calculations. Build a routine: write the expression, plug in equilibrium values, solve and then derive units. Connect temperature changes to shifts in Kc/Kp using enthalpy sign.

在 Edexcel A-Level 化学中掌握平衡常数需要练习 ICE 表格、单位分析和 Kp 计算。建立一个常规步骤：写出表达式，代入平衡值，求解，然后导出单位。利用焓变符号将温度变化与 Kc/Kp 的移动联系起来。

A deep conceptual grasp of why only temperature changes the constant, while concentration and pressure merely shift the equilibrium position, will help you excel in explanation and prediction questions. Regular retrieval of these principles will cement your understanding for the exam.

深刻理解为什么只有温度能改变常数，而浓度和压强仅仅移动平衡位置，将有助于你在解释和预测类题目中取得优异成绩。定期回顾这些原理，能使你的理解在考试中更加牢固。

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