📚 Equilibrium in A-Level Chemistry | A-Level 化学平衡
Equilibrium is a central topic in Edexcel A-Level Chemistry, especially in Topics 10 and 11. It connects reversible reactions, reaction rates, quantitative calculations with Kc and Kp, and industrial processes such as the Haber and Contact processes. This guide explains the key principles, equilibrium constant expressions, and common exam-style applications.
化学平衡是 Edexcel A-Level 化学的核心主题，尤其集中在 Topic 10 与 Topic 11。它把可逆反应、反应速率、Kc 与 Kp 的定量计算以及哈伯法、接触法等工业过程联系起来。本指南讲解关键原理、平衡常数表达式以及常见考试应用。
1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡
A reversible reaction is one that can proceed in both the forward and backward directions under the same conditions. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, so the concentrations of reactants and products remain constant. The system is dynamic because both reactions continue to occur, but there is no net change in concentrations. The same equilibrium mixture can be reached from either the reactants or the products.
可逆反应是在相同条件下既能向正方向也能向逆方向进行的反应。达到平衡时，正反应速率等于逆反应速率，因此反应物和生成物的浓度保持恒定。体系是动态的，因为两个方向的反应仍在继续，只是浓度没有净变化。无论从反应物还是从生成物开始，最终都能达到相同的平衡混合物。
2. Le Chatelier’s Principle | 勒夏特列原理
Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the position of equilibrium shifts in the direction that tends to oppose that change. It is a qualitative rule used to predict the direction of shift, not the rate of reaction or the numerical value of the equilibrium constant.
勒夏特列原理指出：若一个处于平衡的体系受到浓度、压强或温度的改变，平衡位置会向削弱这种改变的方向移动。这是一条用于预测移动方向的定性规则，不能用于预测反应速率，也不能给出平衡常数的数值。
3. Effect of Concentration Changes | 浓度变化的影响
Increasing the concentration of a reactant shifts the equilibrium position towards the products, consuming the added reactant. Decreasing the concentration of a product also shifts the equilibrium to the right, producing more product. In industry, removing a product or adding an excess of a cheap reactant can drive a reversible reaction towards a higher yield. However, the value of Kc remains unchanged because the concentration ratio re-adjusts to the same constant at a given temperature.
增大反应物浓度会使平衡位置向生成物方向移动，从而消耗掉加入的反应物。降低生成物浓度也会使平衡向右移动，生成更多产物。在工业中，移除产物或加入过量廉价反应物可以推动可逆反应获得更高产率。但 Kc 的值不变，因为浓度比会在同一温度下重新调整到同一个常数。
4. Effect of Pressure Changes | 压强变化的影响
For gaseous equilibria, increasing pressure favours the side with fewer gas molecules because that reduces the total number of molecules and therefore opposes the pressure increase. Decreasing pressure favours the side with more gas molecules. If the total number of gas molecules is the same on both sides of the equation, changing pressure has no effect on the equilibrium position. Adding an inert gas at constant volume does not change the partial pressures of the reacting gases, so the equilibrium position is unaffected.
对于气体平衡，增大压强有利于气体分子数较少的一侧，因为这会减少总分子数，从而削弱压强增大。减小压强有利于气体分子数较多的一侧。如果反应方程式两边气体分子总数相同，改变压强不会影响平衡位置。在恒容条件下加入惰性气体不会改变反应气体的分压，因此平衡位置不受影响。
5. Effect of Temperature Changes | 温度变化的影响
Temperature is the only condition that changes the numerical value of Kc and Kp. For an exothermic forward reaction, increasing temperature shifts the equilibrium to the left, towards the reactants, so the value of K decreases. For an endothermic forward reaction, increasing temperature shifts the equilibrium to the right, towards the products, so K increases. Heat can be treated as a product in an exothermic forward reaction and as a reactant in an endothermic forward reaction.
温度是唯一能改变 Kc 和 Kp 数值的条件。对于正反应放热的反应，升高温度会使平衡向左移动，即向反应物方向移动，因此 K 减小。对于正反应吸热的反应，升高温度使平衡向右移动，即向生成物方向移动，因此 K 增大。在正反应放热时，热可看作生成物；在正反应吸热时，热可看作反应物。
6. Catalysts and Equilibrium | 催化剂与平衡
A catalyst speeds up both the forward and backward reactions equally by providing an alternative pathway with lower activation energy. It therefore shortens the time needed to reach equilibrium but does not change the equilibrium position or the value of Kc or Kp. In industrial processes, catalysts allow lower temperatures to be used while maintaining a useful reaction rate, which improves economic efficiency even though the theoretical yield is unchanged.
催化剂通过提供活化能较低的替代路径，同等程度地加快正、逆反应速率。因此它只缩短达到平衡所需的时间，不改变平衡位置，也不改变 Kc 或 Kp 的数值。在工业过程中，催化剂允许在较低温度下仍保持有用反应速率，从而改善经济效益，即使理论产率不变。
7. Equilibrium Constant Kc | 平衡常数 Kc
For a homogeneous reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc is written using equilibrium concentrations. Only gases and aqueous species appear in the expression; pure solids and pure liquids are omitted because their concentrations are effectively constant.
对于均相反应 aA + bB ⇌ cC + dD，平衡常数 Kc 用平衡浓度表示。表达式中只包含气体和溶液中的物种；纯固体和纯液体因浓度几乎恒定而被省略。
Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
The units of Kc depend on the stoichiometry and must be derived by substituting concentration units, usually mol dm⁻³, into the expression. For example, for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the Kc expression is Kc = [NH₃]² / ([N₂][H₂]³). If at equilibrium [N₂] = 0.20 mol dm⁻³, [H₂] = 0.30 mol dm⁻³ and [NH₃] = 0.40 mol dm⁻³, then Kc = (0.40)² / (0.20 × 0.30³) = 29.6 mol⁻² dm⁶.
Kc 的单位取决于计量数，必须通过将浓度单位（通常为 mol dm⁻³）代入表达式来推导。例如，对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Kc 表达式为 Kc = [NH₃]² / ([N₂][H₂]³)。若平衡时 [N₂] = 0.20 mol dm⁻³、[H₂] = 0.30 mol dm⁻³、[NH₃] = 0.40 mol dm⁻³，则 Kc = (0.40)² / (0.20 × 0.30³) = 29.6 mol⁻² dm⁶。
8. Equilibrium Constant Kp | 平衡常数 Kp
For gas-phase equilibria, Kp is expressed in terms of partial pressures. The partial pressure of a gas A, pA, is its mole fraction multiplied by the total pressure: pA = xA × Ptotal. For a reaction aA + bB ⇌ cC + dD, the Kp expression is:
对于气相平衡，Kp 用分压表示。气体 A 的分压 pA 等于其摩尔分数乘以总压：pA = xA × Ptotal。对于反应 aA + bB ⇌ cC + dD，Kp 表达式为：
Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ
Kp has no units only if the total number of gas moles is the same on both sides. Otherwise its units depend on the pressure unit used, usually atm or Pa. Mole fraction has no units, and the sum of all mole fractions in a mixture is always 1.
仅当反应两边气体总摩尔数相等时，Kp 才没有单位。否则其单位取决于所用的压强单位，通常为 atm 或 Pa。摩尔分数没有单位，混合物中所有组分的摩尔分数之和始终为 1。
9. Factors Affecting Equilibrium Constants | 影响平衡常数的因素
Kc and Kp are constant at a given temperature. Changing concentration or pressure or adding a catalyst does not change the value of K. Only temperature changes K. For an exothermic forward
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