📚 IB Chemistry: Core Concepts of Chemical Equilibrium | IB化学：化学平衡核心考点梳理
Chemical equilibrium is one of the most fundamental topics in IB Chemistry. It explains how reversible reactions reach a state where the rates of the forward and reverse reactions are equal, and it forms the basis for understanding acid-base, redox, and solubility equilibria. This article summarizes the core concepts and problem-solving strategies you need for your exams.
化学平衡是IB化学中最基础且最重要的主题之一。它解释了可逆反应如何达到正逆反应速率相等的状态，也是理解酸碱平衡、氧化还原平衡和溶解平衡的基础。本文为你梳理考试中必备的核心概念与解题策略。
1. Dynamic Equilibrium | 动态平衡
A reversible reaction reaches dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant, but both reactions are still occurring continuously. Macroscopic properties such as color, pressure, and concentration no longer change.
当正反应速率等于逆反应速率时，可逆反应达到动态平衡。平衡时，反应物和产物的浓度保持不变，但正逆反应仍在持续进行。宏观性质如颜色、压力和浓度不再变化。
- Equilibrium is dynamic at the molecular level, not static.
- It can only occur in a closed system.
- Equilibrium can be approached from either direction.
- 平衡在分子层面是动态的，而不是静态的。
- 平衡只能在封闭体系中建立。
- 平衡可以从正反应或逆反应方向达到。
2. Equilibrium Constant Kc | 平衡常数 Kc
For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentrations is written as:
对于一般反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数写作：
Kc = [C]ᶜ [D]ᵈ / ([A]ᵃ [B]ᵇ)
Square brackets denote molar concentrations at equilibrium. Pure solids and pure liquids are omitted from the expression. The value of Kc depends only on temperature, not on initial concentrations or the presence of a catalyst.
方括号表示平衡时的摩尔浓度。纯固体和纯液体不出现在表达式中。Kc 只取决于温度，与初始浓度或催化剂的存在无关。
- Kc > 10³: equilibrium lies far to the right (product-favoured).
- Kc < 10⁻³: equilibrium lies far to the left (reactant-favoured).
- 10⁻³ < Kc < 10³: significant amounts of both reactants and products.
- Kc > 10³：平衡强烈向右移动，有利于产物。
- Kc < 10⁻³：平衡强烈向左移动，有利于反应物。
- 10⁻³ < Kc < 10³：反应物和产物都有显著存在。
3. Kc and Kp for Gases | 气体反应的 Kc 与 Kp
For reactions involving gases, the equilibrium constant can also be expressed using partial pressures. For aA + bB ⇌ cC + dD, the pressure equilibrium constant is:
对于涉及气体的反应，平衡常数也可以用分压表示。对于 aA + bB ⇌ cC + dD，压力平衡常数为：
Kp = (PC)ᶜ (PD)ᵈ / ((PA)ᵃ (PB)ᵇ)
Partial pressures are usually measured in atm or kPa. For the same reaction at the same temperature, Kp and Kc are related by the equation:
分压通常以 atm 或 kPa 为单位。在同一温度下，同一反应的 Kp 与 Kc 的关系为：
Kp = Kc (RT)^Δngas
where Δngas the number of moles of gas. R is the ideal gas constant, and T is the absolute temperature.
其中 Δngas = (c + d) − (a + b)，即气体物质的量变化值。R 是理想气体常数，T 是绝对温度。
4. Reaction Quotient Q | 反应商 Q
The reaction quotient Q has the same mathematical form as Kc, but uses concentrations at any moment, not necessarily at equilibrium. Comparing Q with Kc tells us which direction the reaction will shift to reach equilibrium.
反应商 Q 与 Kc 的数学表达式相同，但使用的是任意时刻的浓度，不一定处于平衡。比较 Q 与 Kc 的大小可判断反应将向哪个方向移动以达到平衡。
| Condition | Direction of shift |
| Q < Kc | Forward (towards products) |
| Q = Kc | At equilibrium |
| Q > Kc | Reverse (towards reactants) |
| 条件 | 反应方向 |
| Q < Kc | 正向移动（生成产物） |
| Q = Kc | 处于平衡 |
| Q > Kc | 逆向移动（生成反应物） |
5. Le Chatelier's Principle | 勒夏特列原理
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. This principle allows us to predict the qualitative effect of changes in concentration, pressure, and temperature.
勒夏特列原理指出：当动态平衡受到条件改变的干扰时，平衡位置会向抵消该改变的方向移动。该原理帮助我们定性预测浓度、压力和温度变化的影响。
- Adding a reactant shifts equilibrium to the right.
- Removing a product also shifts equilibrium to the right.
- Changes in pressure only affect reactions with unequal numbers of gas moles.
- Temperature changes alter the value of K itself.
- 增加反应物，平衡向右移动。
- 减少产物，平衡向右移动。
- 压力变化只影响反应前后气体物质的量不等的反应。
- 温度变化会改变 K 值本身。
6. Effect of Concentration Changes | 浓度变化的影响
When the concentration of a species is changed, Q no longer equals Kc. The system responds by shifting in the direction that consumes the added species or replenishes the removed species. For example, in the reaction N₂ + 3H₂ ⇌ 2NH₃, adding N₂ increases Q, causing a shift to the right to produce more NH₃.
当某物种的浓度改变时，Q 不再等于 Kc。体系会朝着消耗新增物种或补充被移除物种的方向移动。例如，在 N₂ + 3H₂ ⇌ 2NH₃ 中，增加 N₂ 会使 Q 暂时变小（或增加分母），促使平衡向右移动以生成更多 NH₃。
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
If NH₃ is removed continuously from the system, the equilibrium shifts right, increasing the yield of ammonia. This is the principle behind industrial ammonia synthesis.
如果不断从体系中移走 NH₃，平衡将向右移动，提高氨的产率。这是工业合成氨的原理之一。
7. Effect of Pressure and Volume | 压力与体积的影响
For reactions involving gases, changing pressure or volume affects equilibrium only if Δngas ≠ 0. Increasing pressure (decreasing volume) shifts the equilibrium toward the side with fewer moles of gas. Decreasing pressure (increasing volume) shifts it toward the side with more moles of gas.
对于涉及气体的反应，只有当 Δngas ≠ 0 时，改变压力或体积才会影响平衡。增大压力（减小体积）会使平衡向气体物质的量较少的一侧移动；减小压力（增大体积）则向气体物质的量较多的一侧移动。
- N₂ + 3H₂ ⇌ 2NH₃: increasing pressure favours NH₃ (4 mol → 2 mol).
- PCl₅ ⇌ PCl₃ + Cl₂: increasing pressure favours PCl₅ (1 mol → 2 mol).
- Adding an inert gas at constant volume does not shift equilibrium.
- N₂ + 3H₂ ⇌ 2NH₃：增大压力有利于 NH₃（4 mol → 2 mol）。
- PCl₅ ⇌ PCl₃ + Cl₂：增大压力有利于 PCl₅（1 mol → 2 mol）。
- 恒容条件下加入惰性气体不会使平衡移动。
Note: Adding an inert gas at constant pressure increases the total volume, thereby decreasing the partial pressures of reactants and products. If Δngas ≠ 0, the equilibrium will shift in the direction that produces more gas moles.
注意：恒压条件下加入惰性气体会增大总体积，从而降低反应物和产物的分压。若 Δngas ≠ 0，平衡将向产生更多气体物质的量的方向移动。
8. Effect of Temperature and van't Hoff Equation | 温度的影响与范特霍夫方程
Temperature is the only factor that changes the value of K. For an exothermic forward reaction, increasing temperature shifts equilibrium to the left and decreases Kc. For an endothermic forward reaction, increasing temperature shifts equilibrium to the right and increases Kc.
温度是唯一能改变 K 值的因素。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；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。
The temperature dependence of K is given by the van't Hoff equation:
K 对温度的依赖关系由范特霍夫方程给出：
ln(K₂/K₁) = -ΔH°/R × (1/T₂ - 1/T₁)
K⁻¹, and T₁ and T₂ are absolute temperatures. This equation is used to calculate K at a different temperature if K at one temperature and ΔH° are known.
其中 ΔH° 是反应的标准焓变，R = 8.314 J mol⁻¹ K⁻¹，T₁ 和 T₂ 是绝对温度。已知某一温度下的 K 和 ΔH° 时，可直接用该方程计算另一温度下的 K。
9. Catalysts and Equilibrium | 催化剂与平衡
A catalyst speeds up both the forward and reverse reactions equally, so it does not change the position of equilibrium. It only helps the system reach equilibrium faster. Catalysts do not affect Kc or Kp values.
催化剂同时等程度地加快正逆反应速率，因此不改变平衡位置。它只是帮助体系更快地达到平衡。催化剂不影响 Kc 或 Kp 值。
- Catalysts lower the activation energy for both directions.
- Catalysts do not alter the yield at equilibrium.
- Catalysts are important for industrial processes to reduce time and energy costs.
- 催化剂降低正逆两个方向的活化能。
- 催化剂不改变平衡时的产率。
- 催化剂在工业中用于缩短时间、降低能耗。
10. Equilibrium Calculations using ICE Table | 用 ICE 表进行平衡计算
The ICE table (Initial, Change, Equilibrium) is a systematic method for equilibrium problems. For the reaction A + B ⇌ C, assume initial concentrations [A]₀, [B]₀, [C]₀, and let x be the change in concentration of C. Then the change in A is −x, and for B is −x (stoichiometric coefficients must be accounted for).
ICE 表（初始、变化、平衡）是解决平衡问题的系统方法。对于反应 A + B ⇌ C，设初始浓度为 [A]₀、[B]₀、[C]₀，令 x 表示 C 的浓度变化，则 A 的变化为 −x，B 的变化为 −x（需考虑化学计量系数）。
| A | B | C | |
| Initial | a | b | 0 |
| Change | −x | −x | +x |
| Equilibrium | a−x | b−x | x |
Substitute the equilibrium row into the Kc expression and solve for x. For quadratic equations, use the quadratic formula. Always check which root is chemically meaningful (concentrations cannot be negative).
将平衡行代入 Kc 表达式并解出 x。若出现二次方程，使用求根公式。务必检查哪个根在化学上有意义（浓度不能为负）。
Kc = x / ((a - x)(b - x))
11. Heterogeneous Equilibria and Ksp | 多相平衡与溶度积
s defined as 1, so they are omitted from equilibrium expressions. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant is simply Kp = P(CO₂).
对于涉及固体或纯液体的多相平衡，纯固体和纯液体的活度定义为 1，因此不出现在平衡表达式中。例如，CaCO₃(s) ⇌ CaO(s) + CO₂(g) 的平衡常数简化为 Kp = P(CO₂)。
The solubility product Ksp is the equilibrium constant for dissolving a sparingly soluble salt. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq):
溶度积 Ksp 是难溶盐溶解过程的平衡常数。对于 AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)：
Ksp = [Ag⁺][Cl⁻]
Ksp can be used to predict precipitation: if Q > Ksp, precipitation occurs; if Q < Ksp, the solution is unsaturated and no precipitate forms.
Ksp 可用于判断沉淀的生成：若 Q > Ksp，则产生沉淀；若 Q < Ksp，则溶液未饱和，无沉淀生成。
12. Key Exam Tips and Common Mistakes | 考试要点与常见错误
In IB exams, students often lose marks because of missing units, forgetting to omit solids/liquids, or using initial concentrations instead of equilibrium concentrations. Always write the K expression before plugging in numbers, and state whether Kc has units (the IB usually accepts units only if derived from the expression).
在IB考试中，学生常因漏写单位、忘记省略固体/液体，或将初始浓度当作平衡浓度而失分。代入数值前务必先写出 K 的表达式，并说明 Kc 是否有单位（IB通常只有在由表达式导出时才承认单位）。
- Always use the correct stoichiometric coefficients as exponents.
- Check the direction: if Q < K, forward shift; if Q > K, reverse shift.
- Only temperature can change K.
- Do not include solids, pure liquids, or inert gases in K expressions.
- 始终使用正确的化学计量系数作为指数。
- 判断方向：Q < K 正向移动；Q > K 逆向移动。
- 只有温度能改变 K。
- 固体、纯液体、惰性气体不写入 K 表达式。
Published by TutorHao | IB Chemistry Revision Series | aleveler.com
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