📚 IB Chemistry: Chemical Equilibrium Key Points | IB 化学：化学平衡 考点精讲
Chemical equilibrium is a central concept in IB Chemistry that describes the state of a reversible reaction when the forward and reverse processes occur at identical rates. Mastering this topic requires not only a qualitative grasp of Le Chatelier's principle but also quantitative skills in writing equilibrium constant expressions, calculating Kc and Kp, and linking equilibrium to thermodynamics via Gibbs free energy. This article breaks down every key idea you need for the IB exam, from the dynamic nature of equilibrium to common pitfalls.
在 IB 化学中，化学平衡描述可逆反应在正、逆反应速率相等时的状态。掌握这一主题不仅要定性理解勒夏特列原理，还要定量掌握平衡常数表达式的书写、Kc 和 Kp 的计算，以及通过吉布斯自由能把平衡与热力学联系起来。本文将逐一剖析 IB 考试所要求的每一个核心概念——从平衡的动态本质到常见失分点，助你高效备考。
1. Introduction to Dynamic Equilibrium | 动态平衡简介
A chemical equilibrium is established in a closed system when a reversible reaction reaches the point where the rate of the forward reaction equals the rate of the reverse reaction. At this stage, the macroscopic properties—such as colour, concentration, pressure and density—remain constant, but on the molecular level both reactions are still proceeding; hence the term dynamic equilibrium. A classic IB example is the dimerisation of nitrogen dioxide: 2NO₂(g) ⇌ N₂O₄(g). In a sealed container, the brown colour eventually stops fading because the rates of formation and decomposition of N₂O₄ become equal. It is essential to remember that equilibrium can only be attained in a closed system; in an open system, products or reactants may escape, preventing a steady state.
化学平衡是可逆反应在封闭系统中达到正反应速率与逆反应速率相等时的状态。此时，颜色、浓度、压力和密度等宏观性质保持恒定，但在分子层面上两个反应仍在进行，因此被称为动态平衡。IB 考试中一个经典例子是二氧化氮的二聚：2NO₂(g) ⇌ N₂O₄(g)。在密封容器中，棕色的气体最终不再变浅，因为 N₂O₄ 的生成速率与分解速率相等。必须牢记，只有封闭系统才能达成平衡；开放系统中产物或反应物可能逸出，无法建立稳定状态。
2. The Equilibrium Constant Kc | 平衡常数 Kc
The equilibrium constant in terms of concentration, Kc, provides a quantitative measure of the position of equilibrium. For a general homogeneous reaction aA + bB ⇌ cC + dD, the equilibrium law states:
Kc = [C]c[D]d / [A]a[B]b
Here the square brackets denote equilibrium concentrations in mol dm⁻³, and the powers are the stoichiometric coefficients. Pure solids and pure liquids are omitted from the expression because their concentrations are essentially constant. The magnitude of Kc indicates which side is favoured: Kc >> 1 means products dominate at equilibrium; Kc << 1 means reactants dominate. Kc is temperature-dependent; altering concentration, pressure or adding a catalyst changes the equilibrium position but does not alter the value of Kc.
以浓度表示的平衡常数 Kc 定量地描述了平衡的位置。对于一般均相反 aA + bB ⇌ cC + dD，平衡定律表述如下：[C] 代表平衡浓度（mol dm⁻³），幂为化学计量系数。固体和纯液体不写入表达式，因为其浓度基本恒定。Kc 的大小反映平衡倾向：Kc >> 1 表示平衡时产物占优；Kc << 1 表示反应物占优。Kc 只受温度影响；改变浓度、压强或加入催化剂仅移动平衡位置，不改变 Kc 的数值。
3. The Equilibrium Constant Kp | 平衡常数 Kp
For gaseous equilibria, it is often more convenient to use partial pressures instead of concentrations. The pressure-based equilibrium constant, Kp, is constructed analogously:
Kp = (PC)c(PD)d / (PA)a(PB)b
Each Pi ressure. Like Kc, Kp is dimensionless when standard pressure (1 bar or 1 atm) is used as the reference, though IB often treats it as having units if the change in the number of gas molecules (Δn) is non-zero. Kp, too, depends solely on temperature and is independent of the total pressure or the presence of inert gases.
对于气相平衡，使用分压经常比浓度更方便。基于压强的平衡常数 Kp 构造方式类似：其中 Pi 是气体 i 在平衡时的分压，常用单位有 atm、Pa 或 kPa。分压 = 摩尔分数 × 总压。与 Kc 类似，当以标准压强（1 bar 或 1 atm）为参考时 Kp 无量纲，但若气体分子数的变化量 Δn ≠ 0，IB 常认为 Kp 带有单位。Kp 同样只依赖于温度，与总压或惰性气体的存在无关。
4. Reaction Quotient Q | 反应商 Q
The reaction quotient Q uses the same algebraic form as the equilibrium constant expression, but the concentrations (or partial pressures) are measured at any moment, not necessarily at equilibrium. Comparing Q with Kc or Kp allows prediction of the direction in which a reaction will proceed to reach equilibrium:
- Q < K: The forward reaction is favoured; the system shifts to the right to produce more products.
- Q = K: The system is at equilibrium; no net change occurs.
- Q > K: The reverse reaction is favoured; the system shifts to the left to consume excess products and form more reactants.
反应商 Q 的代数形式与平衡常数表达式相同，但浓度（或分压）取自任意时刻，不一定是平衡状态。比较 Q 与 Kc 或 Kp 可预测反应进行的方向：Q < K 时正反应占优，平衡右移；Q = K 时体系处于平衡；Q > K 时逆反应占优，平衡左移。这是 IB 考试中常用来解释外界变化引起平衡移动的定量工具。
5. Le Chatelier's Principle | 勒夏特列原理
Le Chatelier's principle states that if a system at dynamic equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium position shifts to partially counteract the imposed change. While qualitative, this principle provides a powerful tool for predicting the response. However, the IB syllabus expects you to link it with the changes in Q and K: a concentration change alters Q but not K; a pressure change may alter Q but not K; a temperature change alters K itself. Always anchor your explanation in the Kc/Kp framework for higher marks.
勒夏特列原理指出：当处于动态平衡的系统受到浓度、压强或温度的变化时，平衡位置会向削弱该变化的方向移动。该原理虽为定性描述，却是预测系统响应的有力工具。然而 IB 考纲要求你将它与 Q 和 K 的变化联系起来：浓度变化仅改变 Q 而不改变 K；压强变化可能改变 Q 而不改变 K；温度变化则直接改变 K 值。回答时请始终依托 Kc/Kp 框架，才能获得高分。
6. Effect of Concentration Changes | 浓度变化的影响
Increasing the concentration of a reactant causes Q to drop below Kc, so the forward reaction is favoured to consume the added reactant. Similarly, removing a product makes Q smaller; the system responds by producing more product. Decreasing a reactant concentration or adding a product shifts the equilibrium in the reverse direction. In aqueous solutions, diluting with water lowers the concentrations of all dissolved species, which may shift the equilibrium depending on the sum of the stoichiometric coefficients on each side. IB questions often ask you to predict the effect when water is added to an equilibrium such as [Co(H2O)6]²⁺ + 4Cl⁻ ⇌ [CoCl4]²⁻ + 6H2O.
增大反应物浓度会使 Q 降至 Kc 以下，因而加速正反应以消耗多余的反应物。类似地，移除产物使 Q 变小，系统通过生成更多产物来响应。降低反应物浓度或加入产物则使平衡逆向移动。在水溶液中，加水稀释会降低所有溶解物种的浓度，平衡移动方向取决于两侧化学计量系数之和。IB 常会要求预测向诸如 [Co(H2O)6]²⁺ + 4Cl⁻ ⇌ [CoCl4]²⁻ + 6H2O 这样的平衡中加水所产生的影响。
7. Effect of Pressure and Volume Changes | 压力与体积变化的影响
decreasing the volume) shifts the equilibrium toward the side with fewer gas molecules, reducing the pressure rise. For example, in 2NO₂(g) ⇌ the formation of colourless N₂O₄, making the mixture paler. If Δn = 0, e.g., H₂(g) + I₂(g) ⇌ reactants or products, so Q and Kp remain equal; however, adding an inert gas at constant pressure increases volume and dilutes the reactants, equivalent to lowering the partial pressures, which may shift the equilibrium if Δn ≠ 0.
压强变化只影响那些气体分子总数 Δn 发生改变的气相平衡。增大总压（即减小体积）使平衡向气体分子数较少的一侧移动，以部分抵消压强的升高。例如 2NO₂(g) ⇌ N₂O₄(g) 中 Δn = −1，压缩促使生成无色的 N₂O₄，混合气体颜色变浅。若 Δn = 0，如 H₂(g) + I₂(g) ⇌ 2HI(g)，压强变化不影响平衡位置。在恒容条件下加入惰性气体不会改变反应物和产物的分压，因此 Q 和 Kp 保持不变；但在恒压下加入惰性气体会使体积膨胀、反应物稀释，相当于降低分压，如果 Δn ≠ 0 就可能移动平衡。
8. Effect of Temperature Changes | 温度变化的影响
Temperature is the only external factor that changes the value of the equilibrium constant. If the forward reaction is exothermic (ΔH < 0), raising the temperature adds heat; the system shifts in the endothermic (reverse) direction to absorb heat, causing Kc to decrease. For an endothermic forward reaction (ΔH > 0), heating favours the forward pathway and K increases. This can be rationalised by the van ' 1/T₂ − 1/T₁). Even without calculations, IB students must be able to predict the temperature dependence of K and explain it using Le Chatelier's principle, considering heat as a 'reactant' or 'product'.
温度是唯一能改变平衡常数值的外界因素。若正反应为放热（ΔH < 0），升温相当于加入热量，系统会向吸热的逆反应方向移动以吸收热量，导致 Kc 减小。若正反应为吸热（ΔH > 0），升温则促进正反应，K 值增大。这可以用范特霍夫方程解释：ln(K₂/K₁) = −ΔH°/R (1/T₂ − 1/T₁)。即使不进行计算，IB 考生也必须能预测 K 随温度的变化，并用勒夏特列原理将热量视为“反应物”或“产物”加以阐释。
9. Effect of a Catalyst | 催化剂的影响
A catalyst provides an alternative reaction pathway with a lower activation energy, equally accelerating both the forward and reverse reactions. Consequently, a catalyst does not alter the equilibrium position, the equilibrium concentrations, or the value of Kc/Kp. Its only effect is to shorten the time needed to reach equilibrium. In the IB exam, you should never claim that a catalyst increases the yield of a product; instead, it allows the system to reach the equilibrium yield more quickly. This distinction is particularly important when discussing industrial processes like the Haber process.
催化剂通过提供一条活化能较低的替代反应路径，同等程度地加快正、逆反应速率。因此，催化剂不会改变平衡位置、平衡浓度或 Kc/Kp 的数值，唯一的效应是缩短到达平衡所需的时间。在 IB 考试中，切勿宣称催化剂提高了产物产率；正确表述是催化剂使体系更快地达到平衡产率。在讨论哈伯法等工业流程时，这一区别尤为关键。
10. Relationship between ΔG and Equilibrium Constant | 吉布斯自由能与平衡常数的关系
At constant temperature, the standard Gibbs free energy change ΔG° is linked to the equilibrium constant through the equation:
ΔG° = −RT ln K
explains why K is temperature-dependent. For a given reaction, measuring K at different temperatures allows determination of ΔH° and ΔS° from a graph of ln K versus 1/T. This is a higher-level (HL) IB topic but frequently tested.
在恒定温度下，标准吉布斯自由能变 ΔG° 与平衡常数的关系为：ΔG° = −RT ln K，其中 R 为气体常数（8.31 J K⁻¹ mol⁻¹），T 为开氏温度。若 ΔG° < 0，则 ln K > 0，K > 1，表明标准条件下反应倾向于生成产物；若 ΔG° > 0，则 K < 1，反应物占优。由于 ΔG° = ΔH° − TΔS°，该关系也解释了 K 对温度的依赖性。对于特定反应，通过在不同温度下测定 K，可由 ln K 对 1/T 作图求得 ΔH° 和 ΔS°。这属于 IB 高级（HL）内容，但考查频率相当高。
11. Industrial Application: The Haber Process | 工业应用：哈伯法
The synthesis of ammonia, N₂(g) + 3H₂(g) ⇌ e equilibrium yield, low temperature and high pressure are favoured. However, at low temperature the rate becomes too slow, so industry compromises at about 450 °C with an iron catalyst. A pressure of around 200 atm is used to shift the equilibrium right while balancing cost and safety. Continuous removal of NH₃ further drives the equilibrium forward. Students must be able to evaluate why the chosen conditions are not those that give the highest equilibrium yield, but rather an optimised combination of rate, equilibrium and economic factors.
氨的合成 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹ 是 IB 运用平衡与动力学知识的绝佳案例。正反应放热且气体分子数减少（Δn = −2）。为使平衡产率最大化，理论上是低温和高压。但低温下反应速率太慢，工业上采用约 450 °C 搭配铁催化剂作为折衷。约 200 atm 的压强在推动平衡右移的同时兼顾成本与安全。持续分离氨进一步拉动平衡正向移动。考生须能说明为何所选条件并非平衡产率最高的条件，而是速率、平衡和经济因素的最优组合。
12. Common Pitfalls and Exam Tips |
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