📚 Equilibrium Expressions and the Equilibrium Constant, Kc | 平衡表达式与平衡常数 Kc
Equilibrium expressions are a central quantitative tool in A-Level Chemistry. They allow chemists to describe the exact composition of a reaction mixture at equilibrium using the equilibrium constant, Kc. Understanding how to write, calculate and interpret Kc is essential for Cambridge A-Level examinations.
平衡表达式是 A-Level 化学中一个核心的定量工具。它们使化学家能够利用平衡常数 Kc 描述平衡反应混合物的精确组成。掌握如何书写、计算和解释 Kc 对剑桥 A-Level 考试至关重要。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
Many chemical reactions are reversible: products can collide and react to re-form the original reactants. In a closed system, the forward and reverse reactions proceed until their rates become equal. At this point a dynamic equilibrium is established, meaning both reactions continue microscopically but there is no net change in the macroscopic concentrations of reactants and products.
许多化学反应是可逆的：产物可以碰撞并重新反应生成原来的反应物。在封闭系统中，正反应和逆反应会一直进行，直到两者速率相等。此时建立动态平衡，意味着微观上正逆反应仍在继续，但宏观上反应物和产物的浓度不再发生净变化。
2. The Equilibrium Law and Writing Kc Expressions | 平衡定律与 Kc 表达式的书写
The equilibrium constant expression is written from the balanced chemical equation for a homogeneous reaction. For a general reversible reaction:
平衡常数表达式根据均相反应的配平化学方程式书写。对于一般的可逆反应：
aA + bB ⇌ cC + dD
The equilibrium constant expression is:
平衡常数表达式为：
Kc = [C]c [D]d / ([A]a [B]b)
Square brackets represent equilibrium concentrations in mol dm⁻³. The stoichiometric coefficients in the balanced equation become powers in the Kc expression. Products appear in the numerator and reactants in the denominator.
方括号表示以 mol dm⁻³ 为单位的平衡浓度。配平方程中的化学计量数在 Kc 表达式中成为幂指数。产物出现在分子，反应物出现在分母。
3. Homogeneous vs Heterogeneous Equilibria | 均相平衡与非均相平衡
In a homogeneous equilibrium, all reacting species are in the same phase, usually all gases or all aqueous ions. In a heterogeneous equilibrium, more than one phase is present. For Kc expressions, pure solids and pure liquids are omitted because their concentrations remain effectively constant throughout the reaction.
在均相平衡中，所有反应物种处于同一相，通常全是气体或全是水溶液离子。在非均相平衡中，存在多个相。书写 Kc 表达式时，纯固体和纯液体被省略，因为它们的浓度在整个反应过程中基本保持恒定。
For example, in the decomposition of calcium carbonate:
例如，在碳酸钙的分解反应中：
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
The Kc expression only includes the gaseous product:
Kc 表达式只包含气体产物：
Kc = [CO₂]
This simplification is important when writing expressions for reactions involving solids such as metal carbonates or hydroxides.
这种简化在书写涉及金属碳酸盐或氢氧化物等固体的反应表达式时非常重要。
4. Units of Kc | Kc 的单位
The units of Kc depend on the powers of the concentration terms in the expression. Since concentrations are usually measured in mol dm⁻³, substituting units into the expression gives Kc units such as mol dm⁻³, mol⁻¹ dm³, mol² dm⁻⁶ or no units if the powers cancel.
Kc 的单位取决于表达式中浓度项的幂次。由于浓度通常以 mol dm⁻³ 为单位，将单位代入表达式后，Kc 的单位可能是 mol dm⁻³、mol⁻¹ dm³、mol² dm⁻⁶，或者如果幂次相消则无单位。
The table below summarises common cases:
下表总结了常见情况：
| Equilibrium system | Kc expression | Units |
|---|---|---|
| N₂ + 3H₂ ⇌ 2NH₃ | [NH₃]² / ([N₂][H₂]³) | mol⁻² dm⁶ |
| H₂ + I₂ ⇌ 2HI | [HI]² / ([H₂][I₂]) | no units |
| PCl₅ ⇌ PCl₃ + Cl₂ | [PCl₃][Cl₂] / [PCl₅] | mol dm⁻³ |
Always derive the units by substituting mol dm⁻³ for each concentration term and cancelling powers. Never assume that Kc is dimensionless.
务必通过将每个浓度项替换为 mol dm⁻³ 并约去幂次来推导单位。不要假设 Kc 总是无量纲。
5. Writing Kc Expressions from Balanced Equations | 根据配平方程式书写 Kc 表达式
Examiners often ask you to write the Kc expression for a given equation. The first step is to ensure the equation is balanced. For example, for the Haber process:
考官经常要求根据给定方程式书写 Kc 表达式。第一步是确保方程式已配平。例如，对于哈伯法：
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
The Kc expression is:
其 Kc 表达式为：
Kc = [NH₃]² / ([N₂][H₂]³)
The coefficient 2 for NH₃ becomes the power 2, the coefficient 1 for N₂ is not normally written, and the coefficient 3 for H₂ becomes the power 3. The units are mol⁻² dm⁶.
NH₃ 的系数 2 成为幂次 2，N₂ 的系数 1 通常不写，H₂ 的系数 3 成为幂次 3。Kc 的单位是 mol⁻² dm⁶。
6. Calculating Kc from Equilibrium Concentrations | 根据平衡浓度计算 Kc
If equilibrium concentrations are given, you can calculate Kc by direct substitution. For the reaction:
如果已知平衡浓度，你可以直接代入计算 Kc。对于反应：
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Suppose at equilibrium [SO₂] = 0.20 mol dm⁻³, [O₂] = 0.10 mol dm⁻³ and [SO₃] = 0.40 mol dm⁻³. The expression is:
假设在平衡时 [SO₂] = 0.20 mol dm⁻³，[O₂] = 0.10 mol dm⁻³，[SO₃] = 0.40 mol dm⁻³。其表达式为：
Kc = [SO₃]² / ([SO₂]² [O₂])
Substituting gives:
代入可得：
Kc = (0.40)² / ((0.20)² × 0.10) = 40 mol⁻¹ dm³
Always use equilibrium concentrations, not initial concentrations, and include units in your final answer unless the calculation shows they cancel.
务必使用平衡浓度，而不是初始浓度，并在最终答案中注明单位，除非计算表明单位已相消。
7. Using ICE Tables to Find Kc | 使用 ICE 表格计算 Kc
Many examination questions provide initial amounts and one equilibrium amount, requiring you to work out the other equilibrium concentrations. The ICE method uses Initial, Change and Equilibrium rows. For example, 1.00 mol of PCl₅ is placed in a 1.00 dm³ vessel and allowed to reach equilibrium. At equilibrium 0.40 mol of PCl₅ remains:
许多试题给出初始量和某一个平衡量，要求你算出其他平衡浓度。ICE 法使用初始、变化和平衡三行。例如，将 1.00 mol 的 PCl₅ 放入 1.00 dm³ 的容器中达到平衡，平衡时剩余 0.40 mol PCl₅：
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
- Initial: [PCl₅] = 1.00 mol dm⁻³, [PCl₃] = 0, [Cl₂] = 0
- Change: [PCl₅] decreases by 0.60, so [PCl₃] and [Cl₂] each increase by 0.60
- Equilibrium: [PCl₅] = 0.40, [PCl₃] = 0.60, [Cl₂] = 0.60 mol dm⁻³
Then:
然后：
Kc = [PCl₃][Cl₂] / [PCl₅] = (0.60 × 0.60) / 0.40 = 0.90 mol dm⁻³
The ICE table method is systematic and reduces errors in calculations involving changes in amounts.
ICE 表格法具有系统性，可在涉及量变化的计算中减少错误。
8. Reaction Quotient Qc and Predicting Direction | 反应商 Qc 与方向判断
The reaction quotient Qc uses the same expression as Kc, but the concentrations are not necessarily equilibrium values. By comparing Qc with Kc, you can predict the direction the reaction must proceed to reach equilibrium.
反应商 Qc 使用的表达式与 Kc 相同，但浓度不一定是平衡值。通过比较 Qc 与 Kc，你能预测反应必须朝哪个方向进行才能达到平衡。
- If Qc < Kc, the forward reaction is favoured and products will form.
- If Qc > Kc, the reverse reaction is favoured and reactants will form.
- If Qc = Kc, the system is already at equilibrium.
中文要点：
- 若 Qc < Kc，正反应占优势，将生成产物。
- 若 Qc > Kc，逆反应占优势，将生成反应物。
- 若 Qc = Kc，系统已经处于平衡状态。
This comparison is particularly useful when a system is disturbed by adding or removing a species.
当系统因加入或移走某种物质而受到扰动时，这种比较特别有用。
9. Temperature Dependence of Kc | Kc 对温度的依赖性
Of all the changes you can make to a system at equilibrium, only temperature changes the value of Kc. Concentration, pressure and catalysts do not change Kc; they only affect the position of equilibrium.
在对平衡系统所做的所有改变中，只有温度会改变 Kc 的值。浓度、压强和催化剂不会改变 Kc，它们只影响平衡位置。
For an endothermic forward reaction, increasing the temperature shifts the equilibrium to the right and increases Kc because the system absorbs extra heat. For an exothermic forward reaction, increasing the temperature shifts the equilibrium to the left and decreases Kc.
对于正向吸热的反应，升高温度会使平衡向右移动并增大 Kc，因为系统吸收额外的热量。对于正向放热的反应，升高温度会使平衡向左移动并减小 Kc。
Always link the temperature change to the sign of the forward reaction’s enthalpy change.
务必把温度变化与正反应焓变的符号联系起来。
10. Effects of Concentration, Pressure and Catalyst | 浓度、压强和催化剂的影响
Changing the concentration of a reactant or product shifts the position of equilibrium according to Le Chatelier’s principle, but it does not change Kc as long as temperature is constant. The system adjusts concentrations until the same Kc value is restored.
改变反应物或产物的浓度会根据勒夏特列原理移动平衡位置，但只要温度恒定，它不会改变 Kc。系统会调整浓度，直到恢复相同的 Kc 值。
Changing the pressure only affects gaseous equilibria in which the total number of gas molecules changes. It shifts the position but again does not alter Kc. A catalyst speeds up both forward and reverse reactions equally, so it shortens the time to reach equilibrium without changing position or Kc.
改变压强只会影响气体总分子数发生变化的那些气体平衡。它移动平衡位置，但同样不改变 Kc。催化剂同等程度地加快正逆反应速率，因此它缩短达到平衡的时间，但不改变平衡位置或 Kc。
11. Kc in Gaseous Systems and Its Relation to Kp | 气体系统中的 Kc 及其与 Kp 的关系
For gaseous reactions, Kc uses concentrations in mol dm⁻³, while Kp uses partial pressures in atm, Pa or kPa. At A-Level Cambridge, Kc is the main equilibrium constant tested; Kp is introduced separately for gaseous systems. The two are related by the equation:
对于气体反应，Kc 使用以 mol dm⁻³ 为单位的浓度，而 Kp 使用以 atm、Pa 或 kPa 为单位的分压。在剑桥 A-Level 中，Kc 是考查的主要平衡常数；Kp 在气体系统中单独引入。两者之间的关系为：
Kp = Kc (RT)^Δn
where R is the gas constant, T is the absolute temperature and Δn is the change in moles of gas. When Δn = 0, Kp equals Kc numerically.
其中 R 是气体常数，T 是热力学温度，Δn 是气体摩尔数的变化量。当 Δn = 0 时，Kp 与 Kc 在数值上相等。
12. Common Exam Pitfalls and Tips | 常见考试易错点与提示
Avoid these frequent mistakes when answering Kc questions:
在回答 Kc 题目时，要避免以下常见错误：
- Forgetting to omit pure solids and pure liquids from the expression.
- Using initial concentrations instead of equilibrium concentrations.
- Writing the expression before balancing the chemical equation.
- Stating incorrect or missing units for Kc.
- Confusing Qc with Kc when predicting the direction of shift.
- Saying that a catalyst or pressure change changes the value of Kc when only temperature does.
中文要点：
- 忘记从表达式中省略纯固体和纯液体。
- 使用初始浓度而不是平衡浓度。
- 在配平化学方程式之前就书写表达式。
- Kc 的单位写错或漏写。
- 在预测移动方向时混淆 Qc 与 Kc。
- 说催化剂或压强变化会改变 Kc 的值，实际上只有温度会改变 Kc。
A careful, systematic approach using balanced equations, correct expressions and ICE tables will give you the best chance of full marks.
采用配平方程式、正确表达式和 ICE 表格的严谨系统方法，将让你最有可能获得满分。
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