Equilibrium Constant Kp for Homogeneous Systems | 均相体系平衡常数 Kp

📚 Equilibrium Constant Kp for Homogeneous Systems | 均相体系平衡常数 Kp

In A-level Chemistry, the equilibrium constant Kp extends the idea of Kc to reactions involving gases. For a homogeneous system — where all reactants and products are in the same phase, typically gaseous — Kp is expressed in terms of partial pressures. This article explains the definition, calculation, units, and applications of Kp, with clear worked examples aligned to the AQA specification.

在 A-Level 化学中，平衡常数 Kp 将 Kc 的概念推广到涉及气体的反应。对于均相体系——即所有反应物和产物都处于同一相态，通常是气相——Kp 用分压来表示。本文将解释 Kp 的定义、计算、单位及应用，并结合 AQA 考纲提供清晰的例题讲解。

1. What Is a Homogeneous Equilibrium? | 什么是均相平衡？

A homogeneous equilibrium is one in which all species involved in the reaction are in the same physical state. For example, the reaction between hydrogen and iodine to form hydrogen iodide is a homogeneous gaseous equilibrium:

均相平衡是指反应中所有参与物种都处于同一物理状态。例如，氢气和碘反应生成碘化氢就是一个均相气相平衡：

H₂(g) + I₂(g) ⇌ 2HI(g)

Because all reactants and products are gases, we can describe their concentrations using partial pressures. Kp is the equilibrium constant defined for such gas-phase reactions. The word ‘homogeneous’ emphasises that there is only one phase present, so no surface effects or boundary terms need to be considered.

由于所有反应物和产物都是气体，我们可以使用分压来描述它们的浓度。Kp 就是为此类气相反应定义的平衡常数。“均相”一词强调体系中只存在一个相，因此无需考虑表面效应或界面项。

2. Partial Pressure and Mole Fractions | 分压与摩尔分数

In a mixture of gases, each gas exerts a pressure as if it alone occupied the container. This is called its partial pressure. The total pressure P_total is the sum of the partial pressures of all gases present:

在气体混合物中，每种气体都表现出一种压力，就好像它单独占据容器一样。这个压力称为分压。总压 P_total 是所有存在气体的分压之和：

P_total = P_A + P_B + P_C + …

The partial pressure of a gas A is related to its mole fraction x_A by:

气体 A 的分压与其摩尔分数 x_A 的关系为：

P_A = x_A × P_total

where x_A = (moles of A) / (total moles of gas). For example, if a mixture contains 2 mol N₂ and 3 mol H₂ at a total pressure of 5 atm, the mole fraction of N₂ is 2/5 = 0.4, so its partial pressure is 0.4 × 5 = 2 atm.

其中 x_A = (A 的物质的量) / (气体总物质的量)。例如，若某混合物含 2 mol N₂ 和 3 mol H₂，总压为 5 atm，则 N₂ 的摩尔分数为 2/5 = 0.4，其分压为 0.4 × 5 = 2 atm。

3. Writing the Kp Expression | 书写 Kp 表达式

For a general homogeneous gaseous reaction:

对于一般的均相气相反应：

aA(g) + bB(g) ⇌ cC(g) + dD(g)

the equilibrium constant Kp is written as:

平衡常数 Kp 写作：

Kp = (P_Cᶜ × P_Dᵈ) / (P_Aᵃ × P_Bᵇ)

Here, P_A, P_B, P_C, P_D are the partial pressures of the gases at equilibrium, and a, b, c, d are the stoichiometric coefficients from the balanced equation. Note that the expression is ‘products over reactants’, with each partial pressure raised to the power equal to its coefficient.

其中 P_A、P_B、P_C、P_D 是平衡时各气体的分压，a、b、c、d 是平衡方程式中的化学计量系数。注意表达式是“产物除以反应物”，每个分压的幂等于其系数。

For example, for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the Kp expression is:

例如，对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Kp 表达式为：

Kp = (P_NH₃²) / (P_N₂ × P_H₂³)

4. Units of Kp | Kp 的单位

The units of Kp depend on the stoichiometry of the reaction. They are derived by substituting pressure units (usually Pa, kPa, atm, or bar) into the Kp expression and cancelling. For the Haber process example above:

Kp 的单位取决于反应的化学计量关系。它们通过将压力单位（通常为 Pa、kPa、atm 或 bar）代入 Kp 表达式并相消得出。对于上述哈伯法例子：

Kp = (pressure)² / (pressure × pressure³) = 1 / (pressure)² = pressure⁻²

So if pressure is measured in kPa, the unit of Kp is kPa⁻². In general, the exponent of the unit is the sum of coefficients of products minus the sum of coefficients of reactants (Δn). When Δn = 0, Kp has no units.

因此，若压力以 kPa 为单位，则 Kp 的单位为 kPa⁻²。一般来说，单位的指数是产物系数之和减去反应物系数之和（Δn）。当 Δn = 0 时，Kp 没有单位。

For example, the reaction H₂(g) + I₂(g) ⇌ 2HI(g) has Δn = 2 – 2 = 0, so Kp is dimensionless. The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) has Δn = 2 – 3 = -1, so Kp has units of pressure⁻¹.

例如，反应 H₂(g) + I₂(g) ⇌ 2HI(g) 的 Δn = 2 – 2 = 0，所以 Kp 无单位。反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 的 Δn = 2 – 3 = -1，所以 Kp 的单位为压力⁻¹。

5. Calculating Kp from Equilibrium Data | 由平衡数据计算 Kp

To calculate Kp, you need the partial pressures (or enough information to find them) at equilibrium. A typical problem provides the number of moles of each gas at equilibrium and the total pressure. Follow these steps:

要计算 Kp，你需要平衡时的分压（或足以求出分压的信息）。典型题目会给出平衡时各气体的物质的量和总压。按以下步骤进行：

Calculate the total number of moles of gas at equilibrium.
计算平衡时气体总物质的量。
Determine the mole fraction of each gas.
确定各气体的摩尔分数。
Multiply each mole fraction by the total pressure to obtain the partial pressure of each gas.
将每个摩尔分数乘以总压，得到各气体的分压。
Substitute the partial pressures into the Kp expression and calculate the value.
将分压代入 Kp 表达式，计算出数值。
State the units of Kp.
写出 Kp 的单位。

Let’s work through an example. In the reaction N₂O₄(g) ⇌ 2NO₂(g), suppose 2.00 mol of N₂O₄ is placed in a flask and heated to a certain temperature. At equilibrium, 1.20 mol of NO₂ is present. The total pressure is 5.00 atm.

我们来看一个例子。在反应 N₂O₄(g) ⇌ 2NO₂(g) 中，假设将 2.00 mol N₂O₄ 放入烧瓶并加热至某温度。平衡时，有 1.20 mol NO₂ 存在。总压为 5.00 atm。

First, determine the amount of N₂O₄ remaining. Since each 2 mol of NO₂ formed consumes 1 mol of N₂O₄, 1.20 mol of NO₂ means 0.60 mol of N₂O₄ reacted. Therefore, remaining N₂O₄ = 2.00 – 0.60 = 1.40 mol. Total moles = 1.40 + 1.20 = 2.60 mol.

首先，确定剩余的 N₂O₄ 的量。由于每生成 2 mol NO₂ 消耗 1 mol N₂O₄，所以生成 1.20 mol NO₂ 意味着有 0.60 mol N₂O₄ 发生了反应。因此，剩余的 N₂O₄ = 2.00 – 0.60 = 1.40 mol。总物质的量 = 1.40 + 1.20 = 2.60 mol。

Mole fraction of N₂O₄ = 1.40 / 2.60 = 0.538; mole fraction of NO₂ = 1.20 / 2.60 = 0.462.

N₂O₄ 的摩尔分数 = 1.40 / 2.60 = 0.538；NO₂ 的摩尔分数 = 1.20 / 2.60 = 0.462。

Partial pressures: P(N₂O₄) = 0.538 × 5.00 = 2.69 atm; P(NO₂) = 0.462 × 5.00 = 2.31 atm.

分压：P(N₂O₄) = 0.538 × 5.00 = 2.69 atm；P(NO₂) = 0.462 × 5.00 = 2.31 atm。

Kp = P(NO₂)² / P(N₂O₄) = (2.31)² / 2.69 = 5.34 / 2.69 = 1.98 atm.

Kp = P(NO₂)² / P(N₂O₄) = (2.31)² / 2.69 = 5.34 / 2.69 = 1.98 atm。

6. Kp and the Extent of Reaction | Kp 与反应程度

The magnitude of Kp indicates the position of equilibrium at a given temperature. A large Kp (>> 1) means the equilibrium lies far to the right — products dominate. A small Kp (<< 1) means the equilibrium lies to the left — reactants dominate. A Kp around 1 suggests a significant mixture of both.

Kp 的大小表明在给定温度下平衡的位置。Kp 很大（远大于 1）意味着平衡强烈向右——产物占主导。Kp 很小（远小于 1）意味着平衡向左——反应物占主导。Kp 接近 1 表明反应物和产物都有显著存在。

However, Kp only depends on temperature. Changing pressure, adding an inert gas, or changing the amounts of substances does not change the value of Kp at a fixed temperature — although it may shift the position of equilibrium if the total number of gas moles changes.

然而，Kp 只取决于温度。改变压力、加入惰性气体或改变物质的量，在固定温度下不会改变 Kp 的值——尽管如果气体总物质的量发生变化，可能会移动平衡的位置。

7. Effect of Changing Pressure on Equilibrium Position | 改变压力对平衡位置的影响

Since Kp is constant at a given temperature, a change in total pressure will cause the equilibrium position to shift so that Kp remains constant. This is in line with Le Chatelier’s principle. If Δn > 0 (more gas molecules on the product side), increasing pressure shifts equilibrium to the left. If Δn < 0, increasing pressure shifts equilibrium to the right. If Δn = 0, pressure has no effect on the position.

由于在给定温度下 Kp 为常数，改变总压会使平衡位置发生移动，以保持 Kp 不变。这与勒夏特列原理一致。如果 Δn > 0（产物侧气体分子数更多），增大压力会使平衡向左移动。如果 Δn < 0，增大压力会使平衡向右移动。如果 Δn = 0，压力对平衡位置没有影响。

For the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = 2 – 4 = -2. Increasing the pressure therefore shifts the equilibrium toward ammonia, increasing the yield. This is why industrial processes often use high pressures.

对于哈伯法 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Δn = 2 – 4 = -2。因此增大压力会使平衡向生成氨的方向移动，提高产率。这就是工业过程常采用高压的原因。

8. The Reaction Quotient Qp | 反应商 Qp

The reaction quotient Qp has the same expression as Kp, but it is calculated using partial pressures that are not necessarily at equilibrium. Comparing Qp with Kp tells you the direction in which a reaction will proceed:

反应商 Qp 与 Kp 具有相同的表达式，但它是用未必处于平衡的分压来计算的。比较 Qp 与 Kp 可以判断反应进行的方向：

If Qp < Kp: the reaction will proceed in the forward direction (toward products) to reach equilibrium.
如果 Qp < Kp：反应将正向（向产物方向）进行以达到平衡。
If Qp > Kp: the reaction will proceed in the reverse direction (toward reactants).
如果 Qp > Kp：反应将逆向（向反应物方向）进行。
If Qp = Kp: the system is already at equilibrium.
如果 Qp = Kp：体系已经处于平衡。

9. Kp vs Kc | Kp 与 Kc 的比较

Both Kp and Kc describe the same equilibrium, but they use different measures of concentration. Kc uses molar concentrations (mol dm⁻³), while Kp uses partial pressures. They are related by the equation:

Kp 和 Kc 都描述同一个平衡，但使用不同的浓度度量。Kc 使用摩尔浓度（mol dm⁻³），而 Kp 使用分压。它们之间的关系式为：

Kp = Kc (RT)^Δn

where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas (moles of gaseous products − moles of gaseous reactants). For reactions with Δn = 0, Kp = Kc.

其中 R 是气体常数，T 是以开尔文为单位的温度，Δn 是气体物质的量的变化（气态产物的物质的量 − 气态反应物的物质的量）。对于 Δn = 0 的反应，Kp = Kc。

When using this equation, make sure the units of pressure and concentration are consistent. For example, if pressure is in atm and concentration in mol dm⁻³, R = 0.0821 dm³ atm mol⁻¹ K⁻¹.

使用此方程时，确保压力和浓度的单位一致。例如，若压力以 atm 为单位、浓度以 mol dm⁻³ 为单位，则 R = 0.0821 dm³ atm mol⁻¹ K⁻¹。

10. Worked Example: Calculating Kp from Initial Moles | 例题：由初始物质的量计算 Kp

A mixture of 2.00 mol of SO₂ and 1.00 mol of O₂ is allowed to reach equilibrium in a container at a constant temperature and a total pressure of 3.00 atm. At equilibrium, 1.20 mol of SO₃ is formed. Find Kp for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

将 2.00 mol SO₂ 和 1.00 mol O₂ 的混合物置于容器中，在恒定温度和 3.00 atm 总压下达到平衡。平衡时生成了 1.20 mol SO₃。求反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 的 Kp。

Step 1 — Use stoichiometry to find equilibrium moles:

步骤 1 — 利用化学计量关系求平衡时的物质的量：

Species	Initial / mol	Change / mol	Equilibrium / mol
SO₂	2.00	−1.20	0.80
O₂	1.00	−0.60	0.40
SO₃	0	+1.20	1.20

Total moles at equilibrium = 0.80 + 0.40 + 1.20 = 2.40 mol.

平衡时总物质的量 = 0.80 + 0.40 + 1.20 = 2.40 mol。

Step 2 — Calculate partial pressures:

步骤 2 — 计算分压：

P(SO₂) = (0.80/2.40) × 3.00 = 1.00 atm

P(O₂) = (0.40/2.40) × 3.00 = 0.50 atm

P(SO₃) = (1.20/2.40) × 3.00 = 1.50 atm

Step 3 — Write Kp expression and substitute:

步骤 3 — 写出 Kp 表达式并代入：

Kp = P(SO₃)² / [P(SO₂)² P(O₂)] = (1.50)² / [(1.00)² × (0.50)] = 2.25 / 0.50 = 4.5

Since Δn = 2 – 3 = -1, Kp has units of atm⁻¹. Therefore Kp = 4.5 atm⁻¹.

由于 Δn = 2 – 3 = -1，Kp 的单位为 atm⁻¹。因此 Kp = 4.5 atm⁻¹。

11. Common Exam Pitfalls | 常见考试易错点

Students frequently lose marks on Kp questions for the following reasons:

学生在 Kp 题目中经常因以下原因失分：

Using initial moles instead of equilibrium moles when calculating partial pressures. Always construct an ICE table (Initial, Change, Equilibrium).
计算分压时使用初始物质的量而非平衡物质的量。务必建立 ICE 表格（初始、变化、平衡）。
Forgetting to raise each partial pressure to the correct power equal to its stoichiometric coefficient.
忘记将每个分压提升到与其化学计量系数相等的正确幂次。
Omitting units or stating incorrect units for Kp.
漏写 Kp 的单位或写错单位。
Confusing Kp with Kc, or using concentration values in a Kp expression.
混淆 Kp 与 Kc，或在 Kp 表达式中使用浓度值。
Forgetting to convert temperatures to Kelvin when using the Kp = Kc(RT)^Δn relationship.
使用 Kp = Kc(RT)^Δn 关系式时忘记将温度转换为开尔文。

Always check whether the reaction is homogeneous and whether all species are gaseous before writing Kp. If any species is a solid or liquid, it is not included in the expression.

在书写 Kp 之前，务必检查反应是否为均相以及所有物种是否均为气体。如果任何物种是固体或液体，则不应包含在表达式中。

12. Summary and Key Points | 总结与要点

Kp is a powerful tool for quantifying gas-phase equilibria. Remember the key ideas:

Kp 是量化气相平衡的有力工具。记住以下关键要点：

Kp is used only for homogeneous gaseous equilibria. The expression is products over reactants, each raised to the power of its stoichiometric coefficient.
Kp 仅用于均相气相平衡。表达式为产物除以反应物，每个分压的幂等于其化学计量系数。
Partial pressure = mole fraction × total pressure.
分压 = 摩尔分数 × 总压。
Units of Kp depend on Δn; if Δn = 0, Kp is unitless.
Kp 的单位取决于 Δn；若 Δn = 0，则 Kp 无单位。
Kp depends only on temperature, not on pressure or amounts.
Kp 只取决于温度，与压力或物质的量无关。
Use Qp to predict the direction of reaction toward equilibrium.
使用 Qp 预测反应朝平衡方向移动的方向。
Kp relates to Kc through Kp = Kc(RT)^Δn.
Kp 与 Kc 通过 Kp = Kc(RT)^Δn 相关联。

Mastering these concepts will help you tackle equilibrium problems confidently in your AQA A-level Chemistry exam.

掌握这些概念将帮助你在 AQA A-Level 化学考试中自信地应对平衡问题。

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Equilibrium Constant Kp for Homogeneous Systems | 均相体系平衡常数 Kp