Equilibrium Law Expression | 平衡常数表达式

📚 Equilibrium Law Expression | 平衡常数表达式

Chemical equilibrium is a dynamic state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products. The equilibrium law expression, also known as the law of mass action, provides a quantitative relationship between the concentrations of substances at equilibrium. For IB Chemistry, mastering how to write and apply this expression is essential for solving equilibrium problems, predicting reaction behaviour, and linking to concepts such as Le Châtelier's principle and Gibbs free energy.

化学平衡是一种动态状态，此时正向和逆向反应的速率相等，反应物和产物的浓度不再发生净变化。平衡常数表达式，也称质量作用定律表达式，给出了平衡时各物质浓度之间的定量关系。在IB化学中，掌握如何书写和应用这一表达式是解答平衡问题、预测反应行为以及关联勒夏特列原理与吉布斯自由能等概念的关键。

1. Dynamic Equilibrium and the Law of Mass Action | 动态平衡与质量作用定律

A reversible reaction reaches equilibrium in a closed system when the forward and reverse processes proceed at the same rate. At this point, the macroscopic properties remain constant, but molecular collisions continue. The law of mass action, proposed by Cato Guldberg and Peter Waage in the 19th century, states that for a general reaction aA + bB ⇌ cC + dD, the ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their coefficients is constant at a given temperature.

一个可逆反应在封闭体系中达到平衡时，正向和逆向过程以相等的速率进行。此时宏观性质保持不变，但分子碰撞仍在发生。由古德贝格和瓦格在19世纪提出的质量作用定律指出，对于一般反应 aA + bB ⇌ cC + dD，在给定温度下，产物浓度以其化学计量系数为乘幂的乘积与反应物浓度以其化学计量系数为乘幂的乘积之比为一常数。

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

This expression, where square brackets denote equilibrium concentrations in mol dm⁻³, is called the equilibrium constant expression for concentration, Kc. It is valid only for systems at equilibrium and is temperature-dependent.

式中方括号表示平衡时的浓度，单位为mol dm⁻³，该表达式称为浓度平衡常数表达式Kc。它仅适用于已达平衡的体系，且与温度有关。

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2. The Equilibrium Constant Kc – Definition and Units | 平衡常数Kc的定义与单位

e powers of concentration in the numerator and denominator. For a reaction with Δn = (c+d) – (a+b), the units of Kc are (mol dm⁻³)^Δn. If Δn = 0, Kc has no units.

严格热力学意义上的平衡常数在使用活度时是无量纲的，但在IB课程中，我们使用mol dm⁻³的平衡浓度计算Kc，并经常根据浓度项的总幂次差赋予Kc单位。Kc的单位取决于分子和分母中浓度指数的代数和Δn = (c+d) – (a+b)，其单位为(mol dm⁻³)^Δn。若Δn = 0，则Kc无单位。

For example, the synthesis of ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has Δn = 2 – (1+3) = –2, so Kc units are (mol dm⁻³)⁻² or dm⁶ mol⁻².

例如合成氨反应：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Δn = 2 – (1+3) = –2，故Kc的单位为(mol dm⁻³)⁻² 或 dm⁶ mol⁻²。

Kc = [NH₃]² / ([N₂][H₂]³)

Always use equilibrium concentrations. If initial amounts are given, an ICE (Initial, Change, Equilibrium) table is necessary to find the equilibrium values.

必须始终使用平衡浓度。若给出的是初始量，则需通过ICE表（初始、变化、平衡）求出平衡值。

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3. Writing Kc for Homogeneous Reactions | 均相反应的Kc书写规则

In a homogeneous equilibrium, all reactants and products are in the same phase – typically all gases or all aqueous ions. The Kc expression includes all species present. The stoichiometric coefficients become the exponents in the expression. The balanced chemical equation is crucial; an incorrect equation leads to a wrong Kc.

在均相平衡中，所有反应物和产物处于同一相态——通常全部为气体或全部为水合离子。Kc表达式包含所有存在的物种。化学计量系数成为表达式中的指数。配平的化学方程式至关重要，错误的方程式会导致错误的Kc。

For the esterification reaction: CH₃COOH(aq) + C₂H₅OH(aq) ⇌ ][H₂O] / ([CH₃COOH][C₂H₅OH]). In IB, water is normally omitted when it is a pure liquid, but if water is a product in an aqueous equilibrium and its concentration is essentially constant, it is not included. This will be clarified in the heterogeneous section.

对于酯化反应：CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l)，但此处水不是溶质且为液态；然而，如果反应在有机溶剂中进行或认为所有物种同相（均为水溶液，忽略水是纯液体），表达式则为Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH])。在IB中，当水是纯液体时通常被省略，但如果水是水相平衡的产物且其浓度基本恒定，则不纳入表达式中。这将在非均相部分阐明。

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4. Heterogeneous Equilibria – Solids and Pure Liquids | 非均相平衡——固体与纯液体

When a reaction involves solids or pure liquids in equilibrium with gases or aqueous solutions, the concentrations of the pure solid and liquid phases are considered constant because their densities do not change. These constant values are incorporated into the equilibrium constant, so they do not appear in the Kc or Kp expression. This simplifies the expression and avoids having concentration terms for substances whose 'concentration' remains unchanged.

当反应涉及固体或纯液体与气体或水溶液成平衡时，纯固相和纯液相的浓度被认为是常数，因为其密度不变。这些常数值被并入平衡常数中，因此它们不会出现在Kc或Kp表达式中。这既简化了表达式，又避免了对“浓度”不变的物种写出浓度项。

Classic example: CaCO₃(s) ⇌ es not shift the equilibrium; it merely provides more surface, but does not affect the CO₂ pressure at a given temperature.

经典例子：CaCO₃(s) ⇌ CaO(s) + CO₂(g)。Kc表达式仅为Kc = [CO₂]，基于分压的Kp = p(CO₂)。添加更多CaCO₃不会移动平衡；它只提供更多表面，但不影响给定温度下的CO₂压力。

Therefore, H₂O(l) is omitted from Kc. For instance, the hydrolysis of an ester: CH₃COOC₂H₅ + H₂O ⇌ he Kc expression used is Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅], and water is not included.

另一个重要情况：在稀水溶液中，水是溶剂，其浓度基本恒定（约55.5 mol dm⁻³）。因此，H₂O(l) 被从Kc中省略。例如酯的水解：CH₃COOC₂H₅ + H₂O ⇌ CH₃COOH + C₂H₅OH；若水大量过量，使用的Kc表达式为Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅]，水不写入。

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5. Equilibrium Constants for Gaseous Systems – Kp | 气体体系的平衡常数Kp

sure of a gas is related to its concentration by the ideal gas law: p = (n/V)RT = [gas]RT.

对于气相反应，使用分压往往比使用浓度更方便。平衡常数Kp定义为反应物和产物的平衡分压的比值。若气体行为理想，气体的分压与其浓度由理想气体状态方程关联：p = (n/V)RT = [气体]RT。

For the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), the expression is:

对于反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)，表达式为：

Kp = (p(C)ᶜ p(D)ᵈ) / (p(A)ᵃ p(B)ᵇ)

tm, Pa, or bar. Kp, like Kc, depends solely on temperature and has units of (pressure)^Δn.

其中 p(X) 表示组分X的平衡分压，通常以atm、Pa或bar为单位。与Kc一样，Kp仅取决于温度，其单位为(压力)^Δn。

(N₂)=1.0 atm, p(H₂)=3.0 atm, then Kp = (2.0)² / (1.0 × (3.0)³) = 4.0 / 27.0 ≈ 0.148 atm⁻².

以500 K时的合成氨为例，若平衡分压为 p(NH₃)=2.0 atm，p(N₂)=1.0 atm，p(H₂)=3.0 atm，那么 Kp = (2.0)² / (1.0 × (3.0)³) = 4.0 / 27.0 ≈ 0.148 atm⁻²。

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6. Relationship Between Kp and Kc | Kp与Kc的关系

een the two constants. For a generic gas reaction with a change in moles of gas Δn = (c+d) – (a+b), the relationship is:

将[气体] = p/RT代入Kc表达式可推导出两个常数之间的联系。对于一般气体反应，气体摩尔数变化Δn = (c+d) – (a+b)，关系式为：

Kp = Kc (RT)^Δn

nd the equilibrium constant has the same numerical value regardless of the representation.

其中R是理想气体常数（如0.08206 dm³ atm mol⁻¹ K⁻¹或8.314 J mol⁻¹ K⁻¹，取决于压力单位），T是以开尔文为单位的温度。若Δn = 0，则Kp = Kc，此时两种表示法的平衡常数数值相同。

For the decomposition of phosphorus pentachloride: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Δn = (1+1) – ³ at 500 K, and R = 0.08206 dm³ atm mol⁻¹ K⁻¹, then Kp = 0.041 × 0.08206 × 500 ≈ 1.68 atm.

以五氯化磷分解为例：PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)，Δn = (1+1) – 1 = 1。因此 Kp = Kc (RT)。若在500 K时 Kc = 0.041 mol dm⁻³，R取0.08206 dm³ atm mol⁻¹ K⁻¹，则 Kp ≈ 0.041 × 0.08206 × 500 ≈ 1.68 atm。

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7. Magnitude of K and Extent of Reaction | K值的大小与反应程度

The numerical value of the equilibrium constant provides insight into the position of equilibrium. A very large K (K >> 1, e.g., >10¹⁰) indicates that at equilibrium, products are heavily favoured and the forward reaction has essentially gone to completion. A very small K (K << 1, e.g., <10⁻¹⁰) means the equilibrium mixture contains mostly reactants and very little product. Intermediate values (10⁻³ < K < 10³) signify significant amounts of both reactants and products at equilibrium.

平衡常数的数值大小可以指示平衡的位置。K值非常大（K >> 1，如 >10¹⁰）表明平衡时产物占绝对优势，正向反应基本进行完全。K值非常小（K << 1，如 <10⁻¹⁰）意味着平衡混合物中主要包含反应物，产物极少。中等数值（10⁻³ < K < 10³）表示平衡时反应物和产物都有可观的数量。

For instance, the reaction 2H₂(g) + O₂(g) ⇌ 2H₂O(g) has an extremely large K at room temperature, while the dissociation of N₂O₄(g) ⇌ 2NO₂(g) has a Kp around 0.1 at low temperatures, showing a mixture. It is important to remember that K does not give information about the rate of reaction; a thermodynamically favourable process may be extremely slow.

例如，反应 2H₂(g) + O₂(g) ⇌ 2H₂O(g) 在室温下K值极大，而 N₂O₄(g) ⇌ 2NO₂(g) 的解离在低温下Kp约为0.1，呈现混合物。要注意，K不提供反应速率信息；热力学上有利的过程可能极其缓慢。

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8. The Reaction Quotient Q – Predicting the Direction of Change | 反应商Q——预测变化方向

To determine whether a system not at equilibrium will shift toward products or reactants, the reaction quotient Q is used. Q is calculated using the same expression as Kc (or Kp), but with the current (non-equilibrium) concentrations or partial pressures. Comparing Q and K allows prediction of the net reaction direction.

为判断未处于平衡的体系将向产物还是反应物方向移动，可使用反应商Q。Q的计算表达式与Kc（或Kp）相同，但采用的是当前（非平衡）浓度或分压。通过比较Q与K即可预测净反应方向。

• If Q < K, the system will favour the forward reaction to produce more products.
• If Q > K, the reverse reaction will be favoured to produce more reactants.
• If Q = K, the system is at equilibrium and no net change occurs.

• 若 Q < K，体系将有利于正向反应，生成更多产物。
• 若 Q > K，逆向反应占优，生成更多反应物。
• 若 Q = K，体系处于平衡状态，无净变化。

This concept is extremely useful in industrial processes, such as the Haber process, where we monitor Q to adjust conditions and maintain near-equilibrium operation.

这一概念在工业过程中极为有用，例如哈伯法，通过监测Q来调整条件，维持接近平衡的操作状态。

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9. Manipulating Equilibrium Expressions | 平衡常数的运算规则

When a chemical equation is modified, the equilibrium constant changes in a predictable way. These manipulations are essential when combining reactions or analysing multi-step equilibria.

当化学方程式被修改时，平衡常数按可预见的方式变化。在合并反应或分析多步平衡时，这些运算是必不可少的。

If an equation is reversed, the new equilibrium constant K' is the reciprocal of the original: K' = 1/K. If the coefficients are multiplied by a factor n, the new constant is K' rall equilibrium constant is the product of the individual constants: K_overall = K₁ × K₂.

若方程式被反转，新的平衡常数K'为原常数的倒数：K' = 1/K。若计量系数乘以因子n，新常数为 K' = Kⁿ。当两个反应相加得到总反应时，总平衡常数为各分步常数的乘积：K_overall = K₁ × K₂。

Example: Given
(1) S(s) + O₂(g) ⇌ SO₂(g) K₁ = 4.2 × 10⁵²
(2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) K₂ = 2.8 × 10²
The overall reaction S(s) + 3/2 O₂(g) ⇌ ot directly the sum; to obtain it, we take reaction (1) plus half of reaction (2) reversed? Better: the overall can be obtained as (1) + ½×(2) in forward direction? ₂ → SO₃; net S + 1.5 O₂ → SO₃. So new K = K₁ × (K₂)^½. Calculations yield K ≈ 4.2 × 10⁵² × √2.8 × 10² ≈ 7.0 × 10⁵³.

示例：已知 (1) S(s) + O₂(g) ⇌ SO₂(g) K₁ = 4.2 × 10⁵²，(2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) K₂ = 2.8 × 10²。总反应 S(s) + 3/2 O₂(g) ⇌ SO₃(g) 可通过 (1) + ½(2) 得到。则总K = K₁ × (K₂)^1/2，计算结果约为 4.2 × 10⁵² × √2.8 × 10² ≈ 7.0 × 10⁵³。

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10. Calculations Involving Kc and Kp – ICE Table Method | 涉及Kc和Kp的计算——ICE表法

One of the most common IB problem types involves finding the equilibrium constant from initial amounts and one equilibrium concentration, or calculating equilibrium concentrations given Kc. The systematic approach uses an ICE table (Initial, Change, Equilibrium).

IB中最常见的问题类型之一是，由初始量和某一平衡浓度求平衡常数，或已知Kc计算平衡浓度。系统方法使用ICE表（初始、变化、平衡）。

Consider the reaction H₂(g) + I₂(g) ⇌ 2HI(g) at 700 K. Initially, 1.00 mol H₂ and 1.00 mol I₂ are placed in a 1.0 dm³ container. At equilibrium, the concentration of HI is found to be 1.56 mol dm⁻³. Determine Kc.

考虑反应 H₂(g) + I₂(g) ⇌ 2HI(g) 在700 K时进行。起始将1.00 mol H₂和1.00 mol I₂置于1.0 dm³容器内。平衡时测得HI浓度为1.56 mol dm⁻³。求Kc。

Set up the ICE table in terms of concentrations (mol dm⁻³):
Initial: [H₂]=1.00, [I₂]=1.00, [HI]=0
Change: Let x be the decrease in [H₂]; then [H₂] changes by –x, [I₂] by –x, [HI] by +2x.
Equilibrium: [H₂] = 1.00 – x, [I₂] = 1.00 – x, [HI] = 2x = 1.56 → x = 0.780
Thus [H₂] = [I₂] = 0.22 mol dm⁻³, [HI] = 1.56 mol dm⁻³.

建立浓度（mol dm⁻³）的ICE表：
初始：[H₂]=1.00，[I₂]=1.00，[HI]=0
变化：设[H₂]减少量为x；则[H₂]变化–x，[I₂]变化–x，[HI]变化+2x。
平衡：[H₂] = 1.00 – x，[I₂] = 1.00 – x，[HI] = 2x = 1.56 → x = 0.780
因此 [H₂] = [I₂] = 0.22 mol dm⁻³，[HI] = 1.56 mol dm⁻³。

Kc = [HI]² / ([H₂][I₂]) = (1.56)² / (0.22 × 0.22) ≈ 50.3

Thus, Kc ≈ 50 (dimensionless since Δn = 0). This method is extend

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