📚 Equilibrium Constant and Equilibrium Law Calculations in IB Chemistry HL | IB化学HL：平衡常数与平衡定律计算
The equilibrium constant is one of the most powerful concepts in chemistry. It tells us the extent to which a reaction proceeds, predicts the direction of a reaction, and enables quantitative analysis of equilibrium systems. In IB Chemistry HL, mastering equilibrium constant calculations is essential for success in both Paper 1 and Paper 2.
平衡常数是化学中最有力的概念之一。它告诉我们反应进行的程度，预测反应的方向，并能够对平衡体系进行定量分析。在IB化学HL中，掌握平衡常数的计算是Paper 1和Paper 2取得好成绩的关键。
1. The Concept of Chemical Equilibrium | 化学平衡的概念
A chemical equilibrium is established when the rates of the forward and reverse reactions become equal, and the concentrations of reactants and products remain constant over time. The system is said to be in a state of dynamic equilibrium because both forward and reverse reactions continue to occur at the same rate.
当正反应和逆反应的速率相等，反应物和生成物的浓度随时间保持不变时，就建立了化学平衡。该系统处于动态平衡状态，因为正反应和逆反应仍在以相同的速率持续进行。
Equilibrium is only possible in closed systems.
平衡只在封闭体系中才能实现。
Equilibrium can be approached from either direction — starting with only reactants or only products.
平衡可以从任意方向趋近——无论是只从反应物开始还是只从生成物开始。
Equilibrium is dynamic, not static — molecules constantly interconvert.
平衡是动态的而非静态的——分子在不断相互转化。
2. The Law of Mass Action | 质量作用定律
The law of mass action, formulated by Guldberg and Waage in 1864, states that at a given temperature, the rate of a chemical reaction is proportional to the product of the active masses (concentrations) of the reactants, each raised to the power of its stoichiometric coefficient. For a general reaction at equilibrium:
质量作用定律由古德贝格和瓦格于1864年提出，该定律指出在给定温度下，化学反应速率与各反应物活度（浓度）以其化学计量系数为幂次的乘积成正比。对于达到平衡的通用反应：
aA + bB ⇌ cC + dD
The equilibrium expression is written as:
平衡表达式写为：
Kc = [C]c[D]d / [A]a[B]b
Kc is the equilibrium constant expressed in terms of molar concentrations (mol dm⁻³).
Kc 是以物质的量浓度（mol dm⁻³）表示的平衡常数。
Square brackets [ ] denote molar concentration.
方括号 [ ] 表示摩尔浓度。
Only aqueous species and gases appear in the equilibrium expression; pure solids and pure liquids are excluded.
只有水溶液中的物种和气体出现在平衡表达式中；纯固体和纯液体不包括在内。
3. The Equilibrium Constant Kc | 平衡常数 Kc
The equilibrium constant Kc is a dimensionless quantity that characterises the position of equilibrium at a specific temperature. A large Kc (Kc >> 1) indicates that the equilibrium lies far to the right, favouring products. A small Kc (Kc << 1) indicates that the equilibrium lies to the left, favouring reactants.
平衡常数Kc是一个无量纲的量，它表征在特定温度下平衡的位置。大的Kc（Kc >> 1）表示平衡强烈偏向右边，有利于生成物；小的Kc（Kc << 1）表示平衡偏向左边，有利于反应物。
The equilibrium constant depends only on temperature. It is unaffected by changes in concentration, pressure, or the presence of a catalyst. This is a crucial point frequently tested in IB exams.
平衡常数仅取决于温度。它不受浓度、压力变化或催化剂存在的影响。这是IB考试中经常考查的一个关键点。
An exception: if the reaction is reversed, the new Kc is the reciprocal of the original. If the reaction is multiplied by a factor n, the new Kc is raised to the power n. If two reactions are added, their K values are multiplied.
一个例外：如果反应方向反转，新的Kc是原来Kc的倒数。如果反应方程乘以因子n，新的Kc将取n次幂。如果两个反应相加，它们的K值相乘。
4. Kp: The Equilibrium Constant for Gases | Kp：气态平衡常数
For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures. For the general reaction:
对于涉及气体的反应，平衡常数可以用分压来表示。对于通用反应：
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Kp = (PC)c(PD)d / (PA)a(PB)b
where PA is the partial pressure of gas A. Partial pressure is calculated by multiplying the mole fraction of a gas by the total pressure of the system:
其中PA是气体A的分压。分压通过将气体的摩尔分数乘以体系的总压来计算：
Partial pressure of A = mole fraction of A × total pressure
A的分压 = A的摩尔分数 × 总压
PA = (nA/ntotal) × Ptotal
PA = (nA/ntotal) × Ptotal
The relationship between Kp and Kc is given by:
Kp与Kc之间的关系为：
Kp = Kc(RT)Δn
+ b), R is the ideal gas constant (8.31 J mol⁻¹ K⁻¹), and T is the temperature in kelvin.
其中 Δn = (c + d) − (a + b)，R是理想气体常数（8.31 J mol⁻¹ K⁻¹），T是以开尔文为单位的温度。
5. The Reaction Quotient Q | 反应商 Q
The reaction quotient Q has the same mathematical form as the equilibrium constant, but it is calculated using concentrations or pressures that are not necessarily at equilibrium. It is used to predict the direction in which a reaction will proceed to reach equilibrium.
反应商Q与平衡常数具有相同的数学形式，但它使用的是不一定处于平衡状态的浓度或压力。它用于预测反应向哪个方向进行以达到平衡。
If Q < K: the reaction will proceed in the forward direction to form more products.
如果Q < K：反应将向正方向进行，生成更多产物。
If Q = K: the system is at equilibrium.
如果Q = K：体系处于平衡状态。
If Q > K: the reaction will proceed in the reverse direction to form more reactants.
如果Q > K：反应将向逆方向进行，生成更多反应物。
6. Stoichiometric Setup for Equilibrium Calculations | 平衡计算的化学计量设列
Equilibrium calculation problems typically require an ICE table (Initial, Change, Equilibrium). This systematic approach is crucial for solving IB equilibrium problems accurately.
平衡计算问题通常需要使用ICE表格（初始、变化、平衡）。这种系统性的方法对于准确解决IB平衡问题至关重要。
| Species 物种 |
Initial (mol dm⁻³) 初始浓度 |
Change (mol dm⁻³) 变化量 |
Equilibrium (mol dm⁻³) 平衡浓度 |
| A | a | −mx | a − mx |
| B | b | −nx | b − nx |
| C | c | +px | c + px |
| D | d | +qx | d + qx |
The coefficients m, n, p, q are the stoichiometric coefficients from the balanced equation. The variable x represents the extent of reaction in terms of concentration change. Always use the mole ratio from the balanced equation to determine the change row.
系数m、n、p、q来自配平方程式的化学计量系数。变量x表示反应进行程度（以浓度变化表示）。始终使用配平方程式中的摩尔比来确定变化行。
7. Worked Example: Calculating Kc from Initial and Equilibrium Concentrations | 例题：从初始浓度和平衡浓度计算Kc
Problem: 0.50 mol of H₂ and 0.50 mol of I₂ are placed in a 1.00 dm³ flask at 700 K. At equilibrium, 0.78 mol of HI is present. Calculate Kc.
题目：将0.50 mol H₂和0.50 mol I₂放入1.00 dm³的烧瓶中，温度为700 K。平衡时，有0.78 mol HI存在。计算Kc。
Solution: The balanced equation is:
解答：配平的方程式为：
H₂(g) + I₂(g) ⇌ 2HI(g)
³, [I₂]₀ = 0.50 mol dm⁻³, [HI]₀ = 0. The equilibrium concentration of HI is 0.78 mol dm⁻³.
由于体积为1.00 dm³，初始浓度为：[H₂]₀ = 0.50 mol dm⁻³，[I₂]₀ = 0.50 mol dm⁻³，[HI]₀ = 0。HI的平衡浓度为0.78 mol dm⁻³。
From the stoichiometry, 2 moles of HI are produced for every 1 mole of H₂ consumed. Therefore:
根据化学计量关系，每消耗1摩尔H₂会产生2摩尔HI。因此：
[H₂] consumed = [I₂] consumed = 0.78/2 = 0.39 mol dm⁻³
[H₂] = 0.50 − 0.39 = 0.11 mol dm⁻³
[I₂] = 0.50 − 0.39 = 0.11 mol dm⁻³
[HI] = 0.78 mol dm⁻³
Now substitute into the equilibrium expression:
现在代入平衡表达式：
Kc = [HI]² / ([H₂][I₂]) = (0.78)² / (0.11 × 0.11) = 0.6084 / 0.0121 = 50.3
Thus Kc = 50.3 (no units). Note that the magnitude of Kc > 1 confirms the equilibrium favours products.
因此Kc = 50.3（无单位）。注意Kc > 1的量级确认平衡有利于生成物。
8. Worked Example: Calculating Extent of Reaction from Kc | 例题：从Kc计算反应进行程度
Problem: For the reaction N₂O₄(g) ⇌ 2NO₂(g), Kc N₂O₄ is placed in a 2.00 dm³ container, calculate the concentration of NO₂ at equilibrium.
题目：对于反应N₂O₄(g) ⇌ 2NO₂(g)，在300 K时Kc = 0.210。如果将1.00 mol N₂O₄放入2.00 dm³容器中，计算平衡时NO₂的浓度。
Solution: O₄ = 1.00/2.00 = 0.500 mol dm⁻³. Let x be the concentration of N₂O₄ that decomposes. Then:
解答：N₂O₄的初始浓度 = 1.00/2.00 = 0.500 mol dm⁻³。设x为分解的N₂O₄的浓度。则：
[N₂O₄] = 0.500 − x
[NO₂] = 2x
Substituting into the equilibrium expression:
代入平衡表达式：
Kc = (2x)² / (0.500 − x) = 0.210
4x² = 0.210(0.500 − x)
4x² = 0.105 − 0.210x
4x² + 0.210x − 0.105 = 0
Using the quadratic formula:
使用二次方程求根公式：
x = [−0.210 + √0.210² + 4 × 4 × 0.105] / (2 × 4)
x = [−0.210 + √0.0441 + 1.68] / 8 = [−0.210 + 1.313] / 8 = 0.138
The positive root is taken. Therefore [NO₂] = 2x = 2 × 0.138 = 0.276 mol dm⁻³.
取正根。因此[NO₂] = 2x = 2 × 0.138 = 0.276 mol dm⁻³。
labus requires you to know when such approximation is valid and when it is not. Always check whether the approximation error is acceptable.
注意：当K值很小且初始浓度较大时，可以近似认为0.500 − x ≈ 0.500，但IB大纲要求你知道何时这种近似是有效的，何时不是。始终检查近似误差是否可接受。
9. Equilibrium Calculations with Halogens: A Classic IB Problem | 卤素平衡计算：经典IB题型
Consider the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with Kc 0.100 mol of PCl₅ is placed in a 1.00 dm³ flask, find the equilibrium concentrations of all species.
考虑反应PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)，在某温度下Kc = 0.0411。如果将0.100 mol PCl₅放入1.00 dm³烧瓶中，求所有物种的平衡浓度。
Solution: ICE Table (all concentrations in mol dm⁻³):
解答：ICE表格（所有浓度单位为mol dm⁻³）：
| PCl₅ | PCl₃ | Cl₂ | |
| Initial | 0.100 | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | 0.100 − x | x | x |
Kc = [PCl₃][Cl₂] / [PCl₅] = x² / (0.100 − x) = 0.0411
Using the quadratic formula:
使用二次方程求根公式：
x² + 0.0411x − 0.00411 = 0
x = [−0.0411 + √0.00169 + 0.01644] / 2 = [−0.0411 + 0.1346] / 2 = 0.04675
Therefore: [PCl₃] = [Cl₂] = 0.0468 mol dm⁻³, [PCl₅] = 0.100 − 0.0468 = 0.0532 mol dm⁻³.
因此：[PCl₃] = [Cl₂] = 0.0468 mol dm⁻³，[PCl₅] = 0.100 − 0.0468 = 0.0532 mol dm⁻³。
You can verify by substituting back: 0.0468²/0.0532 = 0.0412 ≈ 0.0411. ✔
你可以代回验证：0.0468²/0.0532 = 0.0412 ≈ 0.0411。✔
10. Factors Affecting the Equilibrium Constant | 影响平衡常数的因素
The only factor that changes the value of Kc or Kp is temperature. The effect follows the principle that equilibrium shifts in the endothermic direction when temperature increases, and in the exothermic direction when temperature decreases.
唯一能改变Kc或Kp数值的因素是温度。这种效应遵循如下原理：温度升高时平衡向吸热方向移动，温度降低时平衡向放热方向移动。
For an endothermic forward reaction (ΔH > 0): increasing T increases K.
对于正向吸热的反应（ΔH > 0）：升高温度会增大K。
For an exothermic forward reaction (ΔH < 0): increasing T decreases K.
对于正向放热的反应（ΔH < 0）：升高温度会减小K。
The van't Hoff equation relates K to temperature quantitatively:
范特霍夫方程将K与温度定量关联：
ln(K₂/K₁) = −ΔH°/R × (1/T₂ − 1/T₁)
Although the van't Hoff equation is not explicitly required in the IB HL syllabus for calculation, understanding the qualitative relationship between K and temperature is essential for multiple-choice and short-answer questions.
虽然IB HL大纲不要求用范特霍夫方程进行计算，但理解K与温度之间的定性关系对于选择题和简答题至关重要。
11. Common Pitfalls and Examination Tips | 常见错误与考试技巧
Students often lose marks in equilibrium calculations due to avoidable mistakes. Here are the most common pitfalls and how to avoid them:
学生在平衡计算中常常因可避免的错误而丢分。以下是最常见的陷阱以及如何避免它们：
Omitting pure solids and liquids: Species in solid or pure liquid state are excluded from the equilibrium expression. Only gases and aqueous species appear in K in terms of concentration.
忽略纯固体和纯液体：固态和纯液态物种不包括在平衡表达式中。在浓度平衡常数中，只有气体和水溶液物种出现。
Incorrect stoichiometric coefficients: The powers in the equilibrium expression must match the coefficients in the balanced equation exactly.
错误的化学计量系数：平衡表达式中的幂次必须与配平方程式中的系数完全一致。
Forgetting units: In IB, K values are generally treated as dimensionless; however, you should be aware that the equilibrium expression yields units when the sum of coefficients differs between products and reactants.
忘记单位：在IB中，K值通常被视为无量纲；然而，你应该意识到当产物和反应物的系数总和不同时，平衡表达式会产生单位。
Choosing the wrong root of the quadratic: Always select the positive root that gives physically meaningful (positive) concentrations.
选择二次方程错误的根：始终选择能给出物理上有意义（正值）浓度的正根。
Confusing Q with K: Q is calculated at any point in time; K only at equilibrium. Using Q when K is required (or vice versa) is a fatal error.
混淆Q与K：Q在任何时间点都可计算；K只在平衡时存在。在需要K时使用Q（或反之）是致命错误。
12. Summary and Examination Strategy | 总结与应试策略
Equilibrium constant calculations follow a predictable pattern. To maximise your marks, follow this systematic approach:
平衡常数的计算遵循可预测的规律。为了最大化你的分数，请遵循以下系统化步骤：
Write the balanced chemical equation.
写出配平的化学方程式。
Identify the phase of each species — exclude solids and pure liquids from the K expression.
确定每种物质的相态——从K表达式中排除固体和纯液体。
Write down the equilibrium expression — check the stoichiometric coefficients.
写出平衡表达式——检查化学计量系数。
Set up an ICE table with all concentrations in the same units (usually mol dm⁻³).
建立ICE表格，确保所有浓度使用相同单位（通常为mol dm⁻³）。
Substitute equilibrium concentrations into the K expression.
将平衡浓度代入K表达式。
Solve for x — use the quadratic formula if necessary; choose the physically meaningful root.
求解x——必要时使用二次方程求根公式；选择物理上有意义的根。
Verify your answer: substitute back into the equilibrium expression to confirm K matches.
验证答案：代回平衡表达式以确认K值吻合。
State the answer with correct significant figures and appropriate units (if any).
以正确的有效数字和适当的单位（如有）表述答案。
In Paper 2, equilibrium calculation questions typically carry 4–7 marks. The marks are allocated for each step: balanced equation, K expression, ICE setup, algebraic solution, and final answer with verification. Show all your working — partial credit is awarded generously.
在Paper 2中，平衡计算题通常占4–7分。分数分配到每个步骤：方程式、K表达式、ICE建立、代数求解以及带验证的最终答案。展示你所有的解题过程——部分分是会慷慨给出的。
Mastering equilibrium calculations requires practice. Work through as many past-paper questions as possible, paying attention to the mark schemes to understand how marks are allocated.
掌握平衡计算需要大量练习。尽量多做历年真题，关注评分标准以理解分数如何分配。
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