A-Level化学化学平衡勒夏特列原理 Kc Kp

Introduction to Chemical Equilibrium 化学平衡导论

Chemical equilibrium is a fundamental concept in A-Level Chemistry that describes the state of a reversible reaction when the rates of the forward and reverse reactions become equal. At this point, the concentrations of reactants and products remain constant over time, even though both reactions continue to occur at the molecular level. This dynamic nature is crucial to understanding why equilibrium systems respond to external changes in predictable ways. 化学平衡是A-Level化学中的一个基本概念，描述了可逆反应在正向和逆向反应速率相等时的状态。此时，反应物和产物的浓度随时间保持恒定，尽管在分子层面上两个反应仍在持续进行。这种动态特性对于理解平衡系统为何以可预测的方式应对外部变化至关重要。

Unlike complete reactions that go to 100% conversion, equilibrium reactions reach a state where all species coexist. The position of equilibrium indicates whether reactants or products are favoured and is influenced by temperature, pressure, and concentration. Mastering these principles allows chemists to optimise industrial processes like the Haber Process for ammonia production, which sustains global food supply through nitrogen-based fertilisers. 与达到100%转化的完全反应不同，平衡反应达到所有物质共存的状态。平衡位置表明反应物或产物哪一方占优势，并受温度、压力和浓度的影响。掌握这些原理使化学家能够优化工业过程，例如为氨生产设计的哈伯法，该过程通过氮基肥料维持全球粮食供应。

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 and restore a new equilibrium. This principle is qualitative and predictive： it tells you the direction of shift without requiring numerical calculations. For temperature changes, the system absorbs or releases heat accordingly; for pressure changes, it shifts toward the side with fewer gas molecules; for concentration changes, it consumes the added substance or replenishes the removed one. 勒夏特列原理指出，如果动态平衡因条件改变而受到扰动，平衡位置会移动以抵消该变化并建立新的平衡。这一原理是定性的预测性工具：它告诉你移动的方向而无需进行数值计算。对于温度变化，系统相应地吸收或释放热量；对于压力变化，它向气体分子较少的一侧移动；对于浓度变化，它会消耗添加的物质或补充移除的物质。

A common exam question asks students to predict and explain the effect of various disturbances on an equilibrium system. For example, in the exothermic reaction N₂ + 3H₂ ⇌ 2NH₃ (ΔH = -92 kJ mol⁻¹), increasing temperature shifts equilibrium left toward reactants because the system absorbs heat to oppose the temperature rise. Decreasing pressure shifts equilibrium left toward the side with more gas molecules (4 vs 2). Adding more N₂ shifts equilibrium right to consume the added reactant. 一个常见的考试问题要求学生预测并解释各种扰动对平衡系统的影响。例如，在放热反应 N₂ + 3H₂ ⇌ 2NH₃ (ΔH = -92 kJ mol⁻¹) 中，升高温度会使平衡向左移动（向反应物方向），因为系统吸收热量以抵抗温度上升。降低压力会使平衡向左移动（向气体分子较多的一侧，4对2）。添加更多的 N₂ 会使平衡向右移动以消耗添加的反应物。

Equilibrium Constant Kc 平衡常数 Kc

The equilibrium constant Kc expresses the relationship between the concentrations of products and reactants at equilibrium at a given temperature. For the general reaction aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, where square brackets denote equilibrium concentrations in mol dm⁻³. The magnitude of Kc indicates the position of equilibrium： a large Kc (>>1) means the equilibrium lies to the right with products favoured, while a small Kc (<<1) means reactants are favoured. 平衡常数 Kc 表达了在给定温度下产物和反应物平衡浓度之间的关系。对于一般反应 aA + bB ⇌ cC + dD，Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ，其中方括号表示以 mol dm⁻³ 为单位的平衡浓度。Kc 的大小表示平衡位置：大的 Kc (>>1) 意味着平衡位于右侧，产物占优势；小的 Kc (<<1) 意味着反应物占优势。

The only factor that changes the value of Kc is temperature. Adding a catalyst, changing pressure, or altering concentration does NOT change Kc ： these factors only shift the position of equilibrium without affecting the constant itself. This is a common exam trap： students often confuse the effect of a catalyst on the rate (it speeds up both forward and reverse reactions equally) with its effect on Kc (none). For an exothermic reaction, increasing temperature decreases Kc because the equilibrium shifts left toward reactants. 唯一改变 Kc 数值的因素是温度。添加催化剂、改变压力或改变浓度不会改变 Kc ：这些因素只改变平衡位置而不影响常数本身。这是一个常见的考试陷阱：学生常常混淆催化剂对速率的影响（它同等加速正逆反应）与其对 Kc 的影响（无影响）。对于放热反应，升高温度会降低 Kc，因为平衡向左移动（向反应物方向）。

Equilibrium Constant Kp and Partial Pressure 平衡常数 Kp 与分压

For gas-phase equilibria, the equilibrium constant Kp is used instead of Kc. Kp is expressed in terms of partial pressures rather than concentrations. The partial pressure of a gas is the pressure it would exert if it alone occupied the entire volume, and it is calculated as： partial pressure = mole fraction × total pressure. Mole fraction = moles of that gas / total moles of all gases in the mixture. 对于气相平衡，使用平衡常数 Kp 而非 Kc。Kp 用分压而不是浓度来表示。气体的分压是该气体单独占据整个体积时所施加的压力，其计算公式为：分压 = 摩尔分数 × 总压。摩尔分数 = 该气体的摩尔数 / 混合物中所有气体的总摩尔数。

Kp has the same form as Kc but uses partial pressures： Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ. Like Kc, Kp is only affected by temperature. The units of Kp depend on the stoichiometry of the reaction and are expressed in atm, Pa, or kPa raised to the appropriate power. A classic exam calculation involves determining Kp for the Haber Process N₂(g) + 3H₂(g) ⇌ 2NH₃(g) given total pressure and equilibrium mole fractions. Kp 的形式与 Kc 相同但使用分压：Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ。与 Kc 一样，Kp 只受温度影响。Kp 的单位取决于反应的化学计量，以 atm、Pa 或 kPa 的适当幂次表示。一个经典的考试计算题涉及根据总压和平衡摩尔分数计算哈伯法 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 的 Kp。

Calculating Kc and Kp： Worked Examples 计算 Kc 与 Kp：例题详解

A typical A-Level calculation： 2.00 moles of PCl₅ are placed in a 5.00 dm³ vessel at 500 K. At equilibrium, 1.20 moles of PCl₅ remain. Calculate Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Step 1： initial moles PCl₅ = 2.00, PCl₃ = 0, Cl₂ = 0. Step 2： change ： PCl₅ decomposed = 2.00 – 1.20 = 0.80 mol, so PCl₃ formed = 0.80 mol, Cl₂ formed = 0.80 mol. Step 3： equilibrium moles ： PCl₅ = 1.20, PCl₃ = 0.80, Cl₂ = 0.80. Step 4： concentrations ： [PCl₅] = 1.20/5.00 = 0.240 mol dm⁻³, [PCl₃] = 0.80/5.00 = 0.160 mol dm⁻³, [Cl₂] = 0.80/5.00 = 0.160 mol dm⁻³. Step 5： Kc = [PCl₃][Cl₂]/[PCl₅] = (0.160)(0.160)/0.240 = 0.107 mol dm⁻³. 一个典型的A-Level计算：将 2.00 mol PCl₅ 放入 5.00 dm³ 容器中，在 500 K 下达到平衡时剩余 1.20 mol PCl₅。计算 PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) 的 Kc。步骤1：初始摩尔数 PCl₅ = 2.00，PCl₃ = 0，Cl₂ = 0。步骤2：变化：分解的 PCl₅ = 2.00 – 1.20 = 0.80 mol，因此生成的 PCl₃ = 0.80 mol，Cl₂ = 0.80 mol。步骤3：平衡摩尔数： PCl₅ = 1.20，PCl₃ = 0.80，Cl₂ = 0.80。步骤4：浓度： [PCl₅] = 1.20/5.00 = 0.240 mol dm⁻³，[PCl₃] = 0.80/5.00 = 0.160 mol dm⁻³，[Cl₂] = 0.80/5.00 = 0.160 mol dm⁻³。步骤5：Kc = [PCl₃][Cl₂]/[PCl₅] = (0.160)(0.160)/0.240 = 0.107 mol dm⁻³。

For Kp with the Haber Process, a common problem gives： total pressure = 1.00 × 10⁷ Pa, equilibrium mixture contains 25.0% NH₃ by volume. Since volume % = mole % for gases, mole fraction of NH₃ = 0.250. The remaining 75.0% is N₂ and H₂ in 1：3 ratio, so mole fraction N₂ = 0.250 × 0.750 = 0.1875, mole fraction H₂ = 0.750 × 0.750 = 0.5625. Partial pressures： pNH₃ = 0.250 × 1.00×10⁷ = 2.50×10⁶ Pa, pN₂ = 0.1875 × 1.00×10⁷ = 1.875×10⁶ Pa, pH₂ = 0.5625 × 1.00×10⁷ = 5.625×10⁶ Pa. Kp = (pNH₃)²/(pN₂)(pH₂)³ = (2.50×10⁶)²/[(1.875×10⁶)(5.625×10⁶)³] = 6.25×10⁻¹⁴ Pa⁻². This tiny Kp reflects the unfavourable equilibrium position at high temperature ： the industrial compromise accepts lower yield for faster rate. 对于哈伯法的 Kp，一个常见问题给出：总压 = 1.00 × 10⁷ Pa，平衡混合物中 NH₃ 体积占比为 25.0%。由于气体的体积百分比 = 摩尔百分比，NH₃ 的摩尔分数 = 0.250。剩余 75.0% 是 N₂ 和 H₂，比例为 1：3，因此 N₂ 摩尔分数 = 0.250 × 0.750 = 0.1875，H₂ 摩尔分数 = 0.750 × 0.750 = 0.5625。分压：pNH₃ = 0.250 × 1.00×10⁷ = 2.50×10⁶ Pa，pN₂ = 0.1875 × 1.00×10⁷ = 1.875×10⁶ Pa，pH₂ = 0.5625 × 1.00×10⁷ = 5.625×10⁶ Pa。Kp = (pNH₃)²/(pN₂)(pH₂)³ = (2.50×10⁶)²/[(1.875×10⁶)(5.625×10⁶)³] = 6.25×10⁻¹⁴ Pa⁻²。这个极小的 Kp 反映了高温下不利的平衡位置：工业折衷方案接受较低的产率以换取较快的速率。

The Haber Process： An Industrial Application 哈伯法：工业应用

The Haber Process is the most important industrial application of equilibrium principles in A-Level Chemistry. It produces ammonia from nitrogen and hydrogen： N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ mol⁻¹. The forward reaction is exothermic and produces fewer gas molecules (4 = 2), which creates a tension between yield and rate. Low temperature favours high equilibrium yield (exothermic reaction shifts right at low T), but low temperature makes the reaction unacceptably slow. High pressure favours both yield and rate (fewer gas molecules on product side), but very high pressure is expensive and requires costly equipment. 哈伯法是A-Level化学中平衡原理最重要的工业应用。它从氮气和氢气生产氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92 kJ mol⁻¹。正向反应是放热的且产生较少气体分子（4 = 2），这在产率和速率之间造成了张力。低温有利于高平衡产率（放热反应在低温下向右移动），但低温会使反应速度慢得无法接受。高压对产率和速率都有利（产物侧气体分子较少），但非常高的压力成本高昂且需要昂贵的设备。

The industrial compromise uses： temperature of 400-450°C (moderate, with an iron catalyst to compensate for the slower rate), pressure of 200 atm (high but not extreme), and an iron catalyst (magnetite, Fe₃O₄, with potassium hydroxide and alumina promoters). The ammonia is continuously removed by condensation, which shifts the equilibrium right by removing the product ： an application of Le Chatelier’s Principle. Unreacted N₂ and H₂ are recycled, giving an overall conversion of about 97%. 工业折衷方案使用：温度 400-450°C（中等，配合铁催化剂补偿较慢的速率），压力 200 atm（高但非极端），以及铁催化剂（磁铁矿 Fe₃O₄，含有氢氧化钾和氧化铝助催化剂）。氨通过冷凝被连续移除，这通过移除产物使平衡向右移动：这是勒夏特列原理的一个应用。未反应的 N₂ 和 H₂ 被循环利用，使总体转化率达到约 97%。

Exam Techniques and Common Pitfalls 考试技巧与常见误区

When answering equilibrium questions, always state the direction of shift FIRST, then explain WHY using Le Chatelier’s Principle. Use precise language： “the equilibrium shifts to the left/right” rather than “the reaction goes backwards/forwards”. Always specify that at the new equilibrium, the concentrations are different from the original equilibrium ： they do not return to the original values. For Kc/Kp calculations, remember to divide moles by volume (for Kc) or calculate mole fractions (for Kp) before substituting into the expression. 回答平衡问题时，首先陈述移动方向，然后使用勒夏特列原理解释原因。使用精确的语言：”平衡向左/右移动”而非”反应朝后/前进”。始终说明在新的平衡状态下，浓度与原始平衡不同：它们不会回到原始值。对于 Kc/Kp 计算，记得在代入表达式之前先将摩尔数除以体积（对于 Kc）或计算摩尔分数（对于 Kp）。

A frequent mistake is stating that a catalyst “increases the yield” of a reaction. A catalyst increases the RATE at which equilibrium is reached but does NOT alter the equilibrium position or the value of Kc/Kp. Another trap involves the effect of adding an inert gas at constant volume： it increases total pressure but does NOT change partial pressures of reactants and products, so there is NO shift in equilibrium. However, adding an inert gas at constant pressure increases volume and decreases partial pressures, which shifts equilibrium toward the side with more gas molecules. 一个常见错误是声称催化剂”增加了反应产率”。催化剂增加了达到平衡的速率，但不改变平衡位置或 Kc/Kp 的值。另一个陷阱涉及在恒容条件下添加惰性气体的影响：它增加了总压但不改变反应物和产物的分压，因此平衡没有移动。然而，在恒压条件下添加惰性气体会增加体积并降低分压，这会使平衡向气体分子较多的一侧移动。

Key Bilingual Terms 关键双语术语

A-Level化学 化学平衡 勒夏特列原理 Kc Kp