Understanding and Calculating the Equilibrium Constant K | 平衡常数K的意义与计算

📚 Understanding and Calculating the Equilibrium Constant K | 平衡常数K的意义与计算

The equilibrium constant, K, is one of the most powerful concepts in chemical thermodynamics. It tells us exactly where a reaction “settles” at equilibrium, and allows us to predict the direction of change when conditions are disturbed. Mastering K is not just about plugging numbers into a formula — it requires a deep understanding of what the value actually means, how it is derived, and why it matters in real chemical systems.

平衡常数 K 是化学热力学中最强大的概念之一。它精确地告诉我们反应在平衡时”停在”哪里，并使我们能够预测当条件被改变时反应会向哪个方向移动。掌握 K 不仅仅是套公式代数字，更需要深入理解这个数值的真正含义、它的推导过程，以及它在真实化学体系中的意义。

1. What is the Equilibrium Constant? | 什么是平衡常数？

The equilibrium constant, K, is a dimensionless number that relates the concentrations (or partial pressures) of reactants and products at equilibrium for a given reaction at a constant temperature. For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration (Kc) is written using the law of mass action.

平衡常数 K 是一个无量纲的数，它关联了在恒定温度下给定的反应达到平衡时，反应物与产物的浓度（或分压）。对于一般反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数（Kc）依据质量作用定律写出。

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

Here, square brackets denote molar concentration in mol dm⁻³, and the exponents correspond to the stoichiometric coefficients in the balanced chemical equation. For gaseous reactions, Kp uses partial pressures in atmospheres or kPa instead.

这里的方括号表示物质的量浓度，单位 mol dm⁻³，指数对应化学方程式中的化学计量系数。对于气相反应，Kp 使用以 atm 或 kPa 表示的分压来代替浓度。

2. The Meaning Behind the Magnitude | K 数值大小的意义

The value of K gives a direct measure of how far a reaction proceeds toward products at equilibrium. A large K (much greater than 1) indicates that at equilibrium, the concentration of products dominates — the reaction lies “to the right”. A small K (much less than 1) indicates that reactants dominate at equilibrium — the reaction lies “to the left”.

K 的数值直接衡量了反应在平衡时向产物方向进行的程度。若 K 很大（远大于 1），表明平衡时产物浓度占主导，反应”偏右”；若 K 很小（远小于 1），表明平衡时反应物占主导，反应”偏左”。

K >> 1: products favoured; reaction essentially goes to completion.
K << 1: reactants favoured; reaction barely occurs.
K ≈ 1: significant amounts of both reactants and products coexist.

Importantly, K does not tell us how fast equilibrium is reached — that is the role of kinetics. A reaction with a very large K may still be extremely slow without a catalyst.

必须注意，K 并不能告诉我们到达平衡的速度有多快——那是化学动力学的范畴。一个 K 极大的反应若没有催化剂，仍然可能进行得非常缓慢。

K >> 1：产物占优，反应基本进行完全。
K << 1：反应物占优，反应几乎不发生。
K ≈ 1：反应物与产物大量共存。

3. Kc vs Kp | Kc 与 Kp 的区别

When dealing with gaseous equilibria, we can express the equilibrium constant in two equivalent ways. Kc uses molar concentrations, while Kp uses partial pressures. For the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g):

在处理气相平衡时，我们可以用两种等价的方式表达平衡常数。Kc 使用摩尔浓度，而 Kp 使用分压。对于反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)：

Kp = (p_Cᶜ · p_Dᵈ) / (p_Aᵃ · p_Bᵇ)

The two constants are related by the equation Kp = Kc(RT)^Δn, where Δn is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants), R is the gas constant, and T is the temperature in Kelvin.

两者通过公式 Kp = Kc(RT)^Δn 联系，其中 Δn 是气体摩尔数的变化量（气态产物摩尔数减去气态反应物摩尔数），R 是气体常数，T 是开尔文温度。

4. Writing Equilibrium Expressions Correctly | 正确书写平衡表达式

Writing the correct expression for K requires strict attention to three rules. First, only species in the homogeneous phase appear in the expression — pure solids and pure liquids are omitted because their concentrations are effectively constant. Second, the exponents must match the exact stoichiometric coefficients from the balanced equation. Third, the expression must correctly account for whether the species is a product or reactant.

书写 K 的表达式需要严格遵守三条规则。第一，只有均相中的物种才能出现在表达式中——纯固体和纯液体被省略，因为它们的浓度实际上是常数。第二，指数必须与配平方程式中的化学计量系数完全一致。第三，表达式必须正确体现物种是产物还是反应物。

For example, consider the heterogeneous equilibrium of thermal decomposition of calcium carbonate:

例如，考虑碳酸钙热分解的非均相平衡：

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Kc = [CO₂] (solids omitted)

Similarly, in the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), even though there are three gaseous species, the expression is written correctly as:

类似地，在 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 的平衡中，即使有三种气态物质，表达式仍应正确地写作：

Kc = [SO₃]² / ([SO₂]²[O₂])

5. Calculating K from Equilibrium Concentrations | 从平衡浓度计算 K

The most direct method of calculating K involves substituting experimentally measured equilibrium concentrations directly into the expression. A common A-Level question provides initial amounts and asks you to determine equilibrium amounts using an ICE table (Initial, Change, Equilibrium).

计算 K 最直接的方法是将实验测得的平衡浓度直接代入表达式。A-Level 常见题型是给出初始量，要求你使用 ICE 表（初始、变化、平衡）来确定平衡量。

Worked example: 1.00 mol of H₂ and 1.00 mol of I₂ are placed in a 1.00 dm³ vessel at 450°C. At equilibrium, 1.56 mol of HI has formed. Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g).

实例演示：将 1.00 mol 的 H₂ 和 1.00 mol 的 I₂ 放入 1.00 dm³ 的容器中，在 450°C 下，平衡时生成了 1.56 mol 的 HI。计算 H₂(g) + I₂(g) ⇌ 2HI(g) 的 Kc。

Species	H₂	I₂	HI
Initial (mol)	1.00	1.00	0
Change (mol)	-0.78	-0.78	+1.56
Equilibrium (mol)	0.22	0.22	1.56

Since volume = 1.00 dm³, equilibrium concentrations are equal to the number of moles. Substituting into the Kc expression:

由于体积 = 1.00 dm³，平衡浓度在数值上等于摩尔数。代入 Kc 表达式：

Kc = (1.56)² / (0.22 × 0.22) = 2.4336 / 0.0484 ≈ 50.3 (no units)

Note that for this reaction, Δn = 0, so Kc is dimensionless. Whenever Δn ≠ 0, units must be included.

注意，对于这个反应，Δn = 0，因此 Kc 无量纲。当 Δn ≠ 0 时，必须带上单位。

6. Using K to Predict Reaction Direction: The Reaction Quotient Q | 用 K 预测反应方向：反应商 Q

The reaction quotient Q has exactly the same form as K, but it is evaluated using concentrations at any moment in time, not necessarily at equilibrium. Comparing Q to K tells us the direction in which the reaction will proceed to reach equilibrium.

反应商 Q 的形式与 K 完全相同，但它是在任意时刻（不一定是平衡时）用当时浓度计算的。比较 Q 与 K 可以告诉我们反应将朝哪个方向进行以达到平衡。

If Q < K: the reaction proceeds forward (toward products) to reach equilibrium.
If Q > K: the reaction proceeds in reverse (toward reactants) to reach equilibrium.
If Q = K: the system is already at equilibrium — no net change occurs.

This predictive power makes Q an indispensable tool in industrial process control, where chemists constantly monitor concentrations and adjust conditions to keep Q moving toward K.

这种预测能力使 Q 成为工业过程控制中不可或缺的工具。化学家不断监测浓度并调整条件，使 Q 始终向 K 靠拢。

若 Q < K：反应正向进行（向产物方向）以达平衡。
若 Q > K：反应逆向进行（向反应物方向）以达平衡。
若 Q = K：系统已达平衡，无净变化。

7. The Temperature Dependence of K | K 对温度的依赖

Changing the temperature is the only factor that can change the value of K itself. Adding or removing a catalyst does not change K at all — it merely helps equilibrium be reached more quickly. Changing concentrations or pressures shifts the position of equilibrium, but the value of K at that temperature remains fixed.

改变温度是唯一能改变 K 本身数值的因素。添加或移除催化剂完全不会改变 K——它只是帮助更快到达平衡。改变浓度或压力会移动平衡位置，但该温度下 K 的数值保持恒定。

The relationship between K and temperature is given by the van’t Hoff equation:

K 与温度的关系由范特霍夫方程给出：

ln(K₂/K₁) = -ΔH°/R × (1/T₂ – 1/T₁)

For exothermic reactions (ΔH° < 0), increasing temperature decreases K; for endothermic reactions (ΔH° > 0), increasing temperature increases K. This is consistent with Le Chatelier’s principle: exothermic reactions are favoured by cooling, endothermic reactions by heating.

对于放热反应（ΔH° < 0），升高温度会使 K 减小；对于吸热反应（ΔH° > 0），升高温度会使 K 增大。这与勒夏特列原理一致：放热反应偏好冷却，吸热反应偏好加热。

8. Worked Example: Calculating Kp from Total Pressure | 实例：从总压计算 Kp

A classic examination question involves determining Kp from given equilibrium data and total pressure.

考试中一类经典问题是根据给定的平衡数据和总压求 Kp。

Problem: 2.00 mol of PCl₅ are heated at 250°C in a closed 2.00 dm³ vessel. At equilibrium, 0.40 mol of Cl₂ is present. Calculate Kc and Kp for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

题目：在 250°C 时，将 2.00 mol 的 PCl₅ 置于密闭的 2.00 dm³ 容器中加热。平衡时存在 0.40 mol 的 Cl₂。计算 PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) 的 Kc 和 Kp。

From the stoichiometry, 0.40 mol of PCl₃ is also formed, and 0.40 mol of PCl₅ is consumed, leaving 1.60 mol PCl₅ at equilibrium:

根据化学计量关系，同时生成 0.40 mol PCl₃，消耗 0.40 mol PCl₅，平衡时剩余 1.60 mol PCl₅：

Kc = ([PCl₃][Cl₂]) / [PCl₅] = (0.20 × 0.20) / 0.80 = 0.050 mol dm⁻³

To find Kp, we first determine mole fractions. Total moles at equilibrium = 1.60 + 0.40 + 0.40 = 2.40 mol.

求 Kp 时，首先计算摩尔分数。平衡时总摩尔数 = 1.60 + 0.40 + 0.40 = 2.40 mol。

P_total = nRT/V = (2.40 × 0.0821 × 523) / 2.00 ≈ 51.5 atm

The partial pressure of each gas equals its mole fraction × total pressure:

每种气体的分压等于其摩尔分数 × 总压：

Kp = (p_PCl₃ · p_Cl₂) / p_PCl₅ ≈ 2.14 atm

This example illustrates the complete workflow: ICE table, Kc calculation, then conversion to Kp using the ideal gas law.

这个例子展示了完整流程：ICE 表、Kc 计算，然后通过理想气体定律换算为 Kp。

9. Qualitative Reasoning: Predicting the Effect of Changes | 定性推理：预测条件变化的影响

A common error students make is claiming that changing concentration or pressure “changes K”. It does not. What changes is the position of equilibrium — the system responds by shifting to reduce the imposed stress while K at that temperature stays constant.

学生常犯的错误是声称改变浓度或压力”会改变 K”。事实并非如此。改变的是平衡位置——系统通过移动来减小外部施加的应力，而该温度下的 K 保持恒定。

For example, in the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), adding more N₂ increases the denominator in the Q expression, making Q < K. The system shifts right to reduce the excess N₂, producing more NH₃ until Q again equals K. The numerical value of K remains unchanged.

例如，在哈伯法中 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，增加 N₂ 会增大 Q 表达式中的分母，使 Q < K。系统右移以消耗过量 N₂，生成更多 NH₃，直到 Q 重新等于 K。K 的数值始终保持不变。

10. Advanced Applications: Multiple Equilibria and Combined Reactions | 进阶应用：多重平衡与反应组合

A sophisticated but highly testable concept is that when chemical equations are manipulated, the corresponding K values transform in predictable ways. If a reaction is reversed, K becomes its reciprocal (1/K). If a reaction is multiplied by a coefficient n, K is raised to the power n. When two reactions are added, their K values are multiplied.

一个进阶且高频率被考察的概念是：当化学反应方程式被操作时，对应的 K 值会以可预期的方式变化。若反应反向，K 变为其倒数（1/K）；若反应乘以系数 n，K 取 n 次幂；若两个反应相加，其 K 值相乘。

Consider the following equilibria at the same temperature:

考虑同一温度下的以下平衡：

2NO(g) ⇌ N₂(g) + O₂(g), K₁ = 2.4 × 10³⁰
2NO₂(g) ⇌ 2NO(g) + O₂(g), K₂ = 4.0 × 10⁻¹³

Adding the two equations gives: 2NO₂(g) ⇌ N₂(g) + 2O₂(g), with K₃ = K₁ × K₂ = 9.6 × 10¹⁷.

将两式相加得：2NO₂(g) ⇌ N₂(g) + 2O₂(g)，其 K₃ = K₁ × K₂ = 9.6 × 10¹⁷。

This manipulation rule is extremely useful when directly determining an equilibrium constant is experimentally challenging. Chemists can instead measure K for a series of simpler reactions and combine them mathematically.

当直接测定某个平衡常数在实验上存在困难时，这种操作规则非常实用。化学家可以改测一系列简单反应的 K，然后通过数学方法进行组合。

11. Common Pitfalls and Exam Strategies | 常见误区与应试策略

Several recurring pitfalls cause students to lose marks on equilibrium constant questions. Missing units when Δn ≠ 0, omitting solid or liquid species from the expression, forgetting to convert initial moles to concentration using the vessel volume, and incorrectly using initial concentrations instead of equilibrium concentrations are among the most frequent errors.

有几个反复出现的误区会导致学生在平衡常数题目上失分。当 Δn ≠ 0 时漏写单位、表达式中省略固体或液体、忘记用容器体积将摩尔数转换为浓度、错误地使用初始浓度而非平衡浓度，这些都是最常见的错误。

Mistake	Correct Approach
Using initial moles in expression	Always use equilibrium concentrations only
Omitting units when Δn ≠ 0	Include mol/dm³ with appropriate powers
Forgetting conversion by volume	Divide moles by volume (dm³) first
Substituting Q for K	Distinguish: K uses equilibrium values only

A reliable exam strategy is always to set up an ICE table explicitly before writing the expression, and to explicitly state whether Δn = 0 to justify the presence or absence of units.

可靠得分的应试策略是：始终在写表达式前明确列出 ICE 表，并明确判断 Δn 是否为零，以充分论证带或不带单位。

常见错误	正确做法
表达式中使用了初始摩尔数	一律只用平衡浓度
Δn ≠ 0 时漏写单位	以 mol/dm³ 为底按幂次标注单位
忘记除以体积	先摩尔数除以体积（dm³）
将 Q 误当作 K 代入	区分：K 只能用平衡时的数值

12. Summary and Mastery Checklist | 总结与掌握清单

Mastering the equilibrium constant requires both conceptual clarity and computational fluency. K quantifies the position of equilibrium; Q predicts the direction of change; temperature alone alters the value of K; catalysts do not affect K; concentrations and pressures merely shift the equilibrium position.

掌握平衡常数既需要概念清晰，也需要计算流畅。K 量化了平衡位置；Q 预测了变化方向；只有温度能改变 K 的数值；催化剂不影响 K；浓度和压力只是移动平衡位置。

To confirm mastery, you should be able to:

为确认掌握程度，你应该能够做到以下几点：

Write correct Kc and Kp expressions for any balanced equation, handling solids and liquids correctly.
Complete ICE tables to determine equilibrium concentrations from initial data.
Calculate Kc and Kp, including unit conversions and ideal gas law applications.
Use Q vs K comparison to predict the direction of reaction.
Explain how temperature, pressure, concentration, and catalysts affect K and the position of equilibrium.
Combine K values for reversed, scaled, and summed chemical equations.

为任何配平方程式写出正确的 Kc 和 Kp 表达式，正确处理固体和液体。
完成 ICE 表，从初始数据推导平衡浓度。
计算 Kc 和 Kp，包括单位换算和理想气体定律的应用。
通过 Q 与 K 的比较预测反应方向。
解释温度、压力、浓度和催化剂如何影响 K 以及平衡位置。
组合反向、缩放和相加的化学方程式的 K 值。

With these abilities, you are fully equipped to tackle any equilibrium constant question in the A-Level examination.

掌握了这些能力，你就能从容应对 A-Level 考试中任何关于平衡常数的考题。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Understanding and Calculating the Equilibrium Constant K | 平衡常数K的意义与计算