IB & AQA Chemistry: Chemical Equilibrium Key Concepts | IB AQA 化学：化学平衡考点精讲

📚 IB & AQA Chemistry: Chemical Equilibrium Key Concepts | IB AQA 化学：化学平衡考点精讲

Chemical equilibrium is a central topic in both IB and AQA A-Level Chemistry, bridging kinetics and thermodynamics. Mastering it is essential for solving problems involving reversible reactions, predicting shifts, and calculating equilibrium constants. This article synthesises the key concepts, formulas, and exam techniques to help you tackle equilibrium questions confidently.

化学平衡是IB和AQA A-Level化学的核心主题，连接着动力学与热力学。掌握它对于解决可逆反应问题、预测平衡移动以及计算平衡常数至关重要。本文综合关键概念、公式与应试技巧，助你自信应对平衡考题。

1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡

Many reactions are reversible, indicated by the ⇌ symbol. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The concentrations of reactants and products remain constant, but not necessarily equal. This is a dynamic equilibrium at the molecular level – reactions continue in both directions.

许多反应是可逆的，用⇌符号表示。在平衡状态下，正反应速率等于逆反应速率。反应物与产物的浓度保持恒定，但未必相等。在分子水平上，这是一个动态平衡——两个方向的反应仍在进行。

Equilibrium can only be established in a closed system at constant temperature. 平衡只能在恒温的封闭系统中建立。
At equilibrium, macroscopic properties (colour, pressure, concentration) remain unchanged. 平衡时，宏观性质（颜色、压强、浓度）保持不变。
The dynamic nature can be demonstrated by isotopic labelling. 可通过同位素标记证明其动态性。

2. Equilibrium Constant Kc | 平衡常数 Kc

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is:

对于一般反应 aA + bB ⇌ cC + dD，用浓度表示的平衡常数为：

Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ

Square brackets denote equilibrium concentrations in mol·dm⁻³. Kc is dimensionless only if the sum of stoichiometric coefficients is equal on both sides; otherwise, it has units. Pure solids and liquids are omitted from the expression because their concentrations are constant.

方括号表示平衡浓度，单位为 mol·dm⁻³。仅当两边化学计量数之和相等时Kc无量纲；否则具有单位。纯固体和液体因其浓度恒定而不出现在表达式中。

Kc > 1: product-favoured equilibrium at given temperature. Kc > 1：在该温度下有利于产物的平衡。
Kc < 1: reactant-favoured equilibrium. Kc < 1：有利于反应物的平衡。
Kc depends only on temperature. Kc 仅取决于温度。

3. Equilibrium Constant Kp | 平衡常数 Kp

For gaseous equilibria, the equilibrium constant can be expressed in terms of partial pressures:

对于气体平衡，平衡常数可用分压表示：

Kp = (p(C))ᶜ (p(D))ᵈ / (p(A))ᵃ (p(B))ᵇ

where p(X) is the partial pressure of species X at equilibrium, usually measured in Pa, atm or bar. Total pressure P_total = Σ p(i). Mole fraction of i = n_i / n_total, and partial pressure p(i) = mole fraction × P_total.

其中 p(X) 是物种 X 在平衡时的分压，常用 Pa、atm 或 bar 表示。总压 P_total = Σ p(i)。摩尔分数 = n_i / n_total，分压 p(i) = 摩尔分数 × P_total。

Kp, like Kc, is temperature-dependent and has units when Δn(gas) ≠ 0. Do not include solids or liquids in Kp expressions.

Kp 与 Kc 一样，依赖温度，当气体摩尔数变化 Δn(gas) ≠ 0 时有单位。Kp 表达式中不要包含固体或液体。

4. Reaction Quotient Q | 反应商 Q

The reaction quotient Q has the same form as K but uses initial or non-equilibrium concentrations (or pressures). By comparing Q with K, you can predict the direction of the reaction to reach equilibrium.

反应商 Q 的形式与 K 相同，但代入的是初始或非平衡浓度（或压力）。通过比较 Q 与 K，可判断反应进行的方向。

If Q < K: forward reaction is favoured, shift right. 若 Q < K：正反应有利，平衡向右移动。
If Q > K: reverse reaction is favoured, shift left. 若 Q > K：逆反应有利，平衡向左移动。
If Q = K: system is at equilibrium. 若 Q = K：系统处于平衡状态。

5. Le Chatelier’s Principle: Effect of Concentration | 勒夏特列原理：浓度的影响

If the concentration of a reactant or product changes, the equilibrium shifts to counteract that change. Adding a reactant shifts equilibrium toward products; removing a product shifts equilibrium toward products as well.

若改变反应物或产物的浓度，平衡会向抵消这一改变的方向移动。增加反应物，平衡向产物方向移动；移除产物也会使平衡向产物方向移动。

This principle is used to maximise yield in industrial processes, e.g. adding excess N₂ or H₂ in the Haber process to increase NH₃ formation.

此原理用于工业生产中提高产率，例如在哈伯法中通入过量 N₂ 或 H₂ 以增加 NH₃ 生成。

6. Le Chatelier’s Principle: Effect of Pressure | 勒夏特列原理：压力的影响

Changing total pressure mainly affects equilibria involving gases with a change in the number of gas molecules, Δn(gas). Increasing pressure shifts equilibrium toward the side with fewer gas molecules to reduce pressure.

改变总压主要影响气体分子数有变化（即 Δn(gas) ≠ 0）的气体平衡。加压会使平衡向气体分子数较少的方向移动，以降低压强。

For example, N₂(g) + 3H₂(g) ⇌ 2NH₃(g): 4 moles on left, 2 moles on right. High pressure favours NH₃ production. If Δn(gas) = 0, pressure changes have no effect on equilibrium position.

例如 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)：左侧 4 mol，右侧 2 mol，高压有利于 NH₃ 生成。若 Δn(gas) = 0，压力变化对平衡位置无影响。

Note: this does not alter Kc or Kp; only temperature can do that.

注意：这不会改变 Kc 或 Kp；只有温度可改变平衡常数。

7. Le Chatelier’s Principle: Effect of Temperature | 勒夏特列原理：温度的影响

Temperature is the only factor that changes the equilibrium constant. Treat heat as a reactant (endothermic) or product (exothermic):

温度是唯一能改变平衡常数的因素。将热视为反应物（吸热反应）或产物（放热反应）：

Endothermic (ΔH > 0): increasing temperature shifts equilibrium to the right, K increases. 吸热反应 (ΔH > 0)：升温平衡右移，K 增大。
Exothermic (ΔH < 0): increasing temperature shifts equilibrium to the left, K decreases. 放热反应 (ΔH < 0)：升温平衡左移，K 减小。

This is consistent with the van ‘t Hoff equation. For IB students, the relationship ΔG° = -RT ln K may be required; for AQA, a qualitative Le Chatelier approach is sufficient, but linking temperature to K is expected.

这与范特霍夫方程一致。对于IB学生，可能需要关系式 ΔG° = -RT ln K；对AQA，定性运用勒夏特列原理即可，但仍需将温度变化与 K 关联。

8. Role of Catalysts | 催化剂的作用

Catalysts provide an alternative pathway with lower activation energy, thereby increasing the rates of both forward and reverse reactions equally. They do not affect the equilibrium position, yield, or the value of K.

催化剂提供了一条活化能更低的替代路径，等比例地加快正逆反应速率。它们不影响平衡位置、产率或 K 值。

Catalysts only help the system reach equilibrium faster, which is crucial in industry to achieve economic production rates, e.g. iron in Haber process, V₂O₅ in Contact process.

催化剂仅帮助系统更快达到平衡，这在工业中对于实现经济的生产速率至关重要，例如哈伯法中的铁触媒、接触法中的 V₂O₅。

9. Dependence of Equilibrium Constant on Temperature | 平衡常数与温度的关系

Quantitatively, the temperature dependence can be described by the van ‘t Hoff equation (often introduced at IB HL). Qualitatively: for an endothermic forward reaction, K increases with T; for exothermic, K decreases. Plotting ln K vs 1/T gives a straight line with slope = –ΔH°/R.

从定量角度看，温度依赖关系可由范特霍夫方程描述（IB HL 常涉及）。定性而言：正向吸热，K 随 T 升高而增大；正向放热，K 随 T 升高而减小。绘制 ln K 对 1/T 图，可得斜率为 –ΔH°/R 的直线。

Exam questions may ask you to calculate K at a different temperature using the equation: ln(K₂/K₁) = –ΔH°/R (1/T₂ – 1/T₁). Ensure ΔH° is in J mol⁻¹ to match R = 8

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IB & AQA Chemistry: Chemical Equilibrium Key Concepts | IB AQA 化学：化学平衡考点精讲