📚 Chemical Equilibrium for A-Level WJEC Chemistry | A-Level WJEC 化学：化学平衡 考点精讲
Chemical equilibrium is one of the most conceptually rich and mathematically tested topics in the WJEC A-Level Chemistry specification. A deep understanding of dynamic equilibrium, equilibrium constants (Kc and Kp), Le Chatelier’s principle, and the reaction quotient is essential for success. This guide breaks down every key idea, formula, and exam technique you need to master equilibria with confidence.
化学平衡是 WJEC A-Level 化学考纲中概念最丰富、计算考查最多的主题之一。要取得高分，你必须透彻理解动态平衡、平衡常数 (Kc 和 Kp)、勒夏特列原理以及反应商。本篇指南将拆解每一个关键概念、公式和应试技巧，帮助你自信掌握平衡的全部考点。
1. Dynamic Equilibrium | 动态平衡
A reversible reaction reaches a state of dynamic equilibrium when it is carried out in a closed system and the rate of the forward reaction becomes exactly equal to the rate of the backward reaction. At equilibrium, the macroscopic properties (concentrations, pressure, colour) remain constant, but on the molecular scale both reactions continue to occur.
当可逆反应在封闭体系中进行，且正反应速率恰好等于逆反应速率时，体系便达到动态平衡。平衡时，宏观性质（浓度、压强、颜色）保持不变，但在分子层面上，正逆两个反应仍在持续进行。
The equilibrium position describes the relative amounts of reactants and products. If the equilibrium mixture contains more products, the position lies to the right; if more reactants are present, it lies to the left.
平衡位置描述的是反应物和产物的相对含量。若平衡混合物中产物更多，则平衡位置靠右；若反应物更多，则靠左。
It is vital to remember that equilibrium can only be established in a closed system where no matter can enter or leave. An open system will continuously lose products or reactants, preventing the attainment of a steady state.
务必记住，只有在没有物质进出的封闭体系中才可能建立平衡。开放体系会不断流失产物或反应物，无法达到稳态。
2. The Equilibrium Constant, Kc | 平衡常数 Kc
For the general homogeneous reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is given by:
对于一般均相反应 aA + bB ⇌ cC + dD，基于浓度的平衡常数表达式为：
Kc = [C]c [D]d / [A]a [B]b
The square brackets denote equilibrium concentrations in mol dm−3. Kc has units that depend on the stoichiometric coefficients; you must calculate units carefully by cancelling powers of mol dm−3.
方括号表示平衡浓度，单位为 mol dm−3。Kc 的单位取决于化学计量数；你必须通过约掉 mol dm−3 的幂次来仔细计算单位。
A large Kc (>> 1) implies the equilibrium position lies heavily to the right, favouring products. A small Kc (<< 1) indicates the equilibrium favours reactants.
Kc 很大（远大于 1）表明平衡位置强烈偏右，有利于产物；Kc 很小（远小于 1）则表明平衡有利于反应物。
Only temperature affects the value of Kc; changes in concentration or pressure shift the equilibrium position but do not alter the numerical value of Kc at a given temperature.
只有温度才会改变 Kc 的数值；浓度或压强的变化只会移动平衡位置，但不会改变给定温度下的 Kc 值。
3. Calculating Kc from Experimental Data | 根据实验数据计算 Kc
WJEC exam questions frequently provide initial amounts and equilibrium amounts of one component. You must construct an ICE (Initial, Change, Equilibrium) table to work out all equilibrium concentrations before substituting into the Kc expression.
WJEC 考试经常给出初始量以及某一组分的平衡量。你必须先构建 ICE（初始、变化、平衡）表格求出所有平衡浓度，再代入 Kc 表达式。
Always convert moles to concentrations by dividing by the volume of the container. For homogeneous liquid‑phase or solution equilibria, the total volume is constant for all species.
始终将物质的量除以容器体积以得到浓度。对于均相液相或溶液平衡，所有物种所处的总体积相同。
Example: 2.0 mol of A and 1.0 mol of B are placed in a 2.0 dm³ vessel. At equilibrium 0.4 mol of D is present for A + 2B ⇌ C + D. Use the stoichiometry to determine changes, calculate equilibrium moles, convert to concentrations, and then evaluate Kc.
示例：将 2.0 mol A 和 1.0 mol B 放入 2.0 dm³ 容器，对反应 A + 2B ⇌ C + D，平衡时存在 0.4 mol D。利用化学计量数确定变化量，计算平衡时各物质的量，换算为浓度后计算 Kc。
4. Gaseous Equilibria and Kp | 气体平衡与 Kp
For gas‑phase reactions, the equilibrium constant can be expressed in terms of partial pressures. For aA(g) + bB(g) ⇌ cC(g) + dD(g):
对于气相反应，平衡常数可用分压表示。对 aA(g) + bB(g) ⇌ cC(g) + dD(g)：
Kp = (pC)c (pD)d / (pA)a (pB)b
Partial pressure is calculated using mole fraction: pA = mole fraction of A × total pressure. Mole fraction = moles of A / total moles of all gases present at equilibrium.
分压通过摩尔分数计算：pA = A 的摩尔分数 × 总压。摩尔分数 = A 的物质的量 / 平衡时所有气体总物质的量。
Kp has units of pressure raised to a power (often atm or Pa) and, like Kc, only changes with temperature. When total pressure is changed, the values of individual partial pressures adjust but Kp remains constant at constant temperature.
Kp 的单位是压强的某次幂（通常为 atm 或 Pa），与 Kc 一样，它只随温度改变。当总压改变时，各分压的数值会相应调整，但温度不变时 Kp 保持不变。
5. Relationship Between Kp and Kc | Kp 与 Kc 的关系
For gaseous equilibria, Kp and Kc are linked by the equation Kp = Kc (RT)Δn, where Δn = (c+d) – (a+b), the change in moles of gas. R is the gas constant (8.31 J K−1 mol−1) and T is the absolute temperature in Kelvin.
对于气体平衡，Kp 与 Kc 通过公式 Kp = Kc (RT)Δn 相关联，其中 Δn = (c+d) – (a+b)，即气体物质的量的变化。R 为气体常数（8.31 J K−1 mol−1），T 为开尔文温度。
If Δn = 0, then Kp = Kc and the equilibrium constants are numerically equal (though units will differ if Kc has units). This relationship is often tested in WJEC questions that require conversion between Kc and Kp.
若 Δn = 0，则 Kp = Kc，两者数值相等（但如果 Kc 有单位，两者的单位会不同）。WJEC 试题常考查这一关系，要求你在 Kc 和 Kp 之间进行转换。
6. 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, temperature, or pressure), the position of equilibrium will shift to oppose that change. This principle allows you to predict qualitatively how an equilibrium responds to external perturbations.
勒夏特列原理指出，若处于动态平衡的体系受到外界条件（浓度、温度或压强）的改变，平衡位置将向削弱这种改变的方向移动。这一原理能够帮助你定性预测平衡如何响应外部干扰。
It is crucial to distinguish between a shift in equilibrium position and a change in the equilibrium constant. A concentration or pressure change shifts the position but keeps Kc or Kp unchanged; a temperature change shifts the position and alters the value of the equilibrium constant.
区分平衡位置的移动与平衡常数的变化至关重要。浓度或压强的改变会移动平衡位置，但 Kc 或 Kp 保持不变；温度的改变则既移动平衡位置，也改变平衡常数的数值。
7. Effect of Temperature Changes | 温度变化的影响
Raising the temperature of an equilibrium mixture causes the system to shift in the endothermic direction (ΔH positive) to absorb the added heat. Lowering the temperature shifts the equilibrium in the exothermic direction (ΔH negative) to release heat.
升高平衡体系的温度会使平衡向吸热方向（ΔH 为正）移动，以吸收多余的热量。降低温度则使平衡向放热方向（ΔH 为负）移动，以释放热量。
For an exothermic forward reaction, increasing temperature decreases the value of Kc and Kp because the equilibrium moves left, reducing the product‑to‑reactant ratio. For an endothermic forward reaction, increasing temperature increases the equilibrium constant.
对于正反应放热的反应，升温会降低 Kc 和 Kp 的值，因为平衡向左移动，产物与反应物的比值下降。对于正反应吸热的反应，升温则会使平衡常数增大。
In WJEC data analysis questions, you may be asked to use the change in Kc with temperature to deduce the sign of ΔH, or to match a temperature profile to an equilibrium shift.
在 WJEC 的数据分析题中，可能会让你根据 Kc 随温度的变化来推断 ΔH 的符号，或将温度变化趋势与平衡移动方向进行匹配。
8. Effect of Pressure and Volume Changes | 压力与体积变化的影响
Changing the pressure only affects equilibria involving gases when there is a change in the total number of gaseous moles (Δn ≠ 0). Increasing pressure (or decreasing volume) shifts the equilibrium toward the side with fewer gas molecules. Decreasing pressure shifts it toward the side with more gas molecules.
只有涉及气体且气体总物质的量发生变化（Δn ≠ 0）的平衡才会受压强变化的影响。增大压强（或减小体积）使平衡向气体分子数较少的一侧移动；减小压强则向气体分子数较多的一侧移动。
If Δn = 0, a pressure change has no effect on the equilibrium position because both sides contain the same number of gas molecules; the system cannot oppose the change by shifting in either direction.
若 Δn = 0，压强变化对平衡位置无影响，因为两侧气体分子数相等，体系无法通过向任一方向移动来抵消压强的改变。
When volume is decreased, the concentrations of all gaseous species increase instantaneously. The equilibrium then readjusts to re‑establish Kc (or Kp). Remember that individual partial pressures change, but Kp stays constant at constant temperature.
当体积减小时，所有气体物种的浓度瞬间增大，随后平衡将重新调整以重建 Kc（或 Kp）。请记住，各分压会变化，但在温度不变时 Kp 保持不变。
9. Effect of Concentration Changes | 浓度变化的影响
If the concentration of a reactant is increased at constant temperature, the equilibrium shifts to the right to consume the added reactant and produce more product. If a product is removed from the equilibrium mixture, the position shifts to the right to restore the product concentration.
在温度不变时，若增加某一反应物的浓度，平衡将向右移动以消耗额外的反应物并生成更多产物。若从平衡混合物中移除产物，平衡位置同样向右移动以恢复产物的浓度。
Adding a solid or a pure liquid does not affect the equilibrium position of a heterogeneous system because their concentrations remain constant and do not appear in the Kc or Kp expression.
加入固体或纯液体不会影响多相体系的平衡位置，因为它们的浓度保持恒定，且不出现在 Kc 或 Kp 的表达式中。
Always return to the Kc expression: adding reactant instantaneously increases the reactant term, so the ratio drops below Kc. The forward reaction is then favoured until the ratio equals Kc again.
任何时候都要回到 Kc 表达式：增加反应物会瞬间增大反应物项，使比值低于 Kc，随后正反应被促进，直至比值重新等于 Kc。
10. Effect of a Catalyst | 催化剂的影响
A catalyst provides an alternative reaction pathway with lower activation energy for both the forward and backward reactions. It increases the rate of both reactions equally, causing the system to reach equilibrium more quickly but having no effect on the equilibrium position or the value of Kc or Kp.
催化剂为正逆两个反应提供活化能更低的替代路径，同等程度地加快正逆反应速率，使体系更快达到平衡，但不会影响平衡位置，也不会改变 Kc 或 Kp 的数值。
In an industrial context, catalysts are invaluable because they allow lower operating temperatures and pressures to be used while still achieving a high rate of production. They never alter the maximum theoretical yield dictated by thermodynamics.
在工业背景下，催化剂极为重要，因为它允许在较低温度和压强下仍然获得高反应速率。但催化剂绝不会改变由热力学决定的最大理论产率。
11. The Reaction Quotient, Q | 反应商 Q
The reaction quotient, usually denoted as Qc or Qp, uses the same expression as Kc or Kp but with concentrations or partial pressures that are not necessarily at equilibrium. Comparing Q with K tells you the direction in which the reaction will proceed.
反应商，常记作 Qc 或 Qp，其表达式与 Kc 或 Kp 相同，但使用的浓度或分压不一定是平衡时的数值。通过比较 Q 与 K，可以判断反应进行的方向。
If Q < K, the system is not yet at equilibrium; the forward reaction is favoured to increase products until Q = K. If Q > K, the reverse reaction is favoured to reduce products. If Q = K, the system is exactly at equilibrium.
若 Q < K，体系尚未达到平衡，正反应被促进以增加产物，直至 Q = K。若 Q > K，逆反应被促进以减少产物。若 Q = K，体系恰好处于平衡状态。
This concept is extremely useful in explaining the effect of sudden concentration or pressure changes. A sharp spike in a reactant’s concentration instantly forces Q below K, triggering a net forward reaction.
这一概念在解释突然改变浓度或压强的效果时极为有用。某种反应物浓度骤然升高会使 Q 瞬间低于 K，从而引发净正反应。
12. Industrial Application: The Haber Process | 工业应用：哈伯法
The Haber process for ammonia synthesis is the classic WJEC example linking equilibrium theory to industrial reality: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with ΔH = −92 kJ mol−1. The forward reaction is exothermic and involves a decrease in moles of gas (Δn = −2).
哈伯法合成氨是 WJEC 将平衡理论与工业现实联系起来的经典案例：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = −92 kJ mol−1。正反应放热且气体物质的量减少（Δn = −2）。
Applying Le Chatelier’s principle, high pressure favours the forward reaction (fewer gas molecules) and low temperature favours the exothermic forward reaction. However, a compromise temperature (around 400‑450 °C) and high pressure (around 200 atm) are used, together with an iron catalyst, to achieve a viable rate and acceptable yield.
根据勒夏特列原理，高压有利于正反应（气体分子数减少），低温有利于放热正反应。但实际生产中采用折中温度（约 400–450 °C）和高压（约 200 atm）并搭配铁催化剂，以在产率和速率之间取得平衡。
This exemplifies the trade‑off between thermodynamic favourability and kinetic feasibility. The catalyst speeds up the approach to equilibrium without changing the equilibrium yield.
这完美体现了热力学有利性和动力学可行性之间的权衡。催化剂加速到达平衡，但不改变平衡产率。
WJEC questions often ask you to explain the choice of conditions using rate, equilibrium and economic arguments, so be prepared to integrate all three perspectives.
WJEC 试题常要求你从速率、平衡和经济效益三个角度解释条件的选择，因此要准备好综合运用这三方面的论据。
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