A-Level化学 化学平衡 勒夏特列原理 Kc Kp

A-Level化学 化学平衡 勒夏特列原理 Kc Kp

What is Chemical Equilibrium？ / 什么是化学平衡？

Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. It is important to understand that equilibrium is dynamic ： both forward and reverse reactions continue to occur simultaneously, but there is no net change in the macroscopic properties of the system. / 化学平衡是可逆反应中正反应速率等于逆反应速率的状态，此时反应物和生成物的浓度随时间保持不变。重要的是要理解平衡是动态的 ： 正反应和逆反应同时持续进行，但系统的宏观性质没有净变化。

Dynamic Equilibrium： The Molecular Picture / 动态平衡：分子层面的图景

At the molecular level, dynamic equilibrium means that individual reactant molecules are constantly being converted into product molecules, while product molecules are simultaneously breaking down back into reactants. The forward and reverse reactions never stop. The system appears static from the outside because the rates are equal, but at the microscopic level, molecules are in constant flux. This concept is fundamental to understanding why changing conditions can shift the position of equilibrium. / 在分子层面，动态平衡意味着单个反应物分子不断转化为生成物分子，同时生成物分子也在分解回反应物。正反应和逆反应从未停止。从外部看系统似乎是静止的，因为速率相等，但在微观层面，分子处于持续变化之中。这一概念对于理解为什么改变条件可以移动平衡位置至关重要。

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 position of equilibrium will shift to oppose ： and partially counteract ： the imposed change. This principle is a powerful predictive tool that allows chemists to anticipate how an equilibrium system will respond to external perturbations. / 勒夏特列原理指出，如果处于动态平衡的系统受到浓度、压力或温度的变化，平衡位置将移动以对抗并部分抵消所施加的变化。这一原理是一个强大的预测工具，使化学家能够预判平衡系统对外部扰动的响应。

Effect of Concentration Changes / 浓度变化的影响

When the concentration of a reactant is increased, the equilibrium shifts to the right (towards products) to consume the added reactant. Conversely, removing a product shifts the equilibrium to the right to replace the removed species. If a product is added, the equilibrium shifts to the left. For example, in the esterification reaction RCOOH + R’OH ⇌ RCOOR’ + H₂O, adding more alcohol drives the equilibrium towards producing more ester. / 当反应物浓度增加时，平衡向右移动（朝向生成物）以消耗添加的反应物。相反，移除生成物会使平衡向右移动以补充被移除的物质。如果添加生成物，平衡会向左移动。例如，在酯化反应 RCOOH + R’OH ⇌ RCOOR’ + H₂O 中，添加更多醇会推动平衡产生更多酯。

Effect of Pressure Changes / 压力变化的影响

Pressure changes affect only gaseous equilibria where there is a difference in the number of moles of gas between reactants and products. Increasing pressure shifts the equilibrium towards the side with fewer gas molecules, reducing the total pressure. Decreasing pressure shifts the equilibrium towards the side with more gas molecules. If the number of gas molecules is the same on both sides, pressure changes have no effect on the position of equilibrium. / 压力变化只影响反应物和生成物之间气体摩尔数不同的气相平衡。增加压力会使平衡向气体分子较少的一侧移动，从而降低总压力。减少压力会使平衡向气体分子较多的一侧移动。如果两侧的气体分子数相同，压力变化对平衡位置没有影响。

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

The Haber process for ammonia synthesis provides an excellent illustration of Le Chatelier’s Principle in action. The reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has 4 moles of gas on the left and 2 on the right. High pressure favours the forward reaction, shifting equilibrium towards ammonia. However, the reaction is exothermic (ΔH = -92 kJ mol⁻¹), so low temperatures favour product formation. In practice, a compromise temperature of 400-450°C is used together with an iron catalyst, and pressures of 150-250 atm, to achieve an economically viable rate while maintaining a reasonable yield. / 哈伯法合成氨为勒夏特列原理的实际应用提供了极好的例证。反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 左侧有 4 摩尔气体，右侧有 2 摩尔。高压有利于正反应，使平衡向氨的方向移动。然而，该反应是放热的 (ΔH = -92 kJ mol⁻¹)，所以低温有利于产物生成。在实践中，使用 400-450°C 的折中温度配合铁催化剂，以及 150-250 atm 的压力，以达到经济可行的速率同时保持合理的产率。

Effect of Temperature Changes / 温度变化的影响

Temperature is the only factor that changes the value of the equilibrium constant. For an exothermic reaction (ΔH negative), increasing temperature shifts the equilibrium to the left, decreasing the yield of products and reducing Kc. For an endothermic reaction (ΔH positive), increasing temperature shifts the equilibrium to the right, increasing product yield and raising Kc. This is because the system absorbs or releases heat to oppose the temperature change, consistent with Le Chatelier’s Principle. / 温度是唯一改变平衡常数值的因素。对于放热反应（ΔH 为负），升高温度会使平衡向左移动，降低产物产率并减小 Kc。对于吸热反应（ΔH 为正），升高温度会使平衡向右移动，增加产物产率并增大 Kc。这是因为系统吸收或释放热量以对抗温度变化，与勒夏特列原理一致。

Effect of Catalysts / 催化剂的影响

A catalyst has no effect on the position of equilibrium. It increases the rates of both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy. A catalyst allows equilibrium to be reached more quickly but does not alter the equilibrium composition. This is a common exam pitfall ： students often incorrectly claim that catalysts shift equilibrium towards products. / 催化剂对平衡位置没有影响。它通过提供具有较低活化能的替代反应路径，同等程度地加快正反应和逆反应的速率。催化剂使平衡更快达到，但不改变平衡组成。这是一个常见的考试陷阱 ： 学生经常错误地声称催化剂使平衡向产物方向移动。

The Equilibrium Constant Kc / 平衡常数 Kc

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc = [C]^c [D]^d / [A]^a [B]^b, where square brackets denote equilibrium concentrations in mol dm⁻³. Kc has a fixed value at a given temperature. A large Kc value (>> 1) indicates the equilibrium lies to the right, favouring products. A small Kc value (<< 1) indicates the equilibrium lies to the left, favouring reactants. Solids and pure liquids are omitted from the Kc expression because their concentrations are effectively constant. / 对于一般反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数为 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 在给定温度下具有固定值。较大的 Kc 值（>> 1）表示平衡偏向右侧，有利于产物。较小的 Kc 值（<< 1）表示平衡偏向左侧，有利于反应物。固体和纯液体从 Kc 表达式中省略，因为它们的浓度实际上是恒定的。

The ICE Table Method / ICE 表格法

ICE stands for Initial, Change, and Equilibrium. This systematic approach is essential for solving equilibrium problems where you know the initial amounts and the equilibrium concentration of one species. Step 1: Write the balanced equation and the Kc expression. Step 2: Set up the ICE table with rows for Initial (mol), Change (mol), and Equilibrium (mol). Step 3: Use the stoichiometric ratios from the balanced equation to express all changes in terms of a single variable x. Step 4: Convert equilibrium moles to concentrations by dividing by volume. Step 5: Substitute into the Kc expression and solve for x. This method is particularly powerful when combined with degree of dissociation or percentage conversion data. / ICE 代表 Initial（初始）、Change（变化）和 Equilibrium（平衡）。这种系统化方法对于解决已知初始量和一种物质平衡浓度的平衡问题至关重要。第 1 步：写出配平的方程式和 Kc 表达式。第 2 步：建立 ICE 表格，包含 Initial (mol)、Change (mol) 和 Equilibrium (mol) 行。第 3 步：使用平衡方程式中的化学计量比，用单一变量 x 表示所有变化。第 4 步：通过除以体积将平衡摩尔数转换为浓度。第 5 步：代入 Kc 表达式并求解 x。当与解离度或转化率数据结合使用时，这种方法特别有力。

Calculating Kc： A Worked Example / 计算 Kc：例题

Consider the esterification reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. At equilibrium, a 1.0 dm³ flask contains 0.30 mol of ethanoic acid, 0.30 mol of ethanol, 0.40 mol of ethyl ethanoate, and 0.40 mol of water. Kc = (0.40 × 0.40) / (0.30 × 0.30) = 0.16 / 0.09 = 1.78. Units: mol⁻¹ dm³ × dm³ mol⁻¹ = no units in this case because the total number of moles is equal on both sides. / 考虑酯化反应：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。在平衡时，一个 1.0 dm³ 的烧瓶中含有 0.30 mol 乙 酸、0.30 mol 乙醇、0.40 mol 乙酸乙酯和 0.40 mol 水。Kc = (0.40 × 0.40) / (0.30 × 0.30) = 0.16 / 0.09 = 1.78。单位：mol⁻¹ dm³ × dm³ mol⁻¹ = 本例中无单位，因为两侧的总摩尔数相等。

ICE Table： A Complete Worked Example / ICE 表格：完整例题

In a 2.0 dm³ vessel, 0.80 mol of HI gas is heated. The equilibrium 2HI(g) ⇌ H₂(g) + I₂(g) is established. At equilibrium, 0.10 mol of I₂ is present. Set up the ICE table: Initial (mol): HI = 0.80, H₂ = 0, I₂ = 0. Change: HI = -2x, H₂ = +x, I₂ = +x. Equilibrium: HI = 0.80 – 2x, H₂ = x, I₂ = x. Since x = 0.10 (from the equilibrium I₂), we have HI = 0.80 – 0.20 = 0.60 mol, H₂ = 0.10 mol, I₂ = 0.10 mol. Concentrations: [HI] = 0.60/2.0 = 0.30, [H₂] = 0.10/2.0 = 0.050, [I₂] = 0.10/2.0 = 0.050 mol dm⁻³. Kc = (0.050 × 0.050) / (0.30)² = 0.0025 / 0.090 = 0.0278. Units: (mol dm⁻³)(mol dm⁻³) / (mol dm⁻³)² = no units. / 在 2.0 dm³ 容器中，0.80 mol HI 气体被加热。平衡 2HI(g) ⇌ H₂(g) + I₂(g) 建立。平衡时存在 0.10 mol I₂。建立 ICE 表格：Initial (mol)：HI = 0.80，H₂ = 0，I₂ = 0。Change：HI = -2x，H₂ = +x，I₂ = +x。Equilibrium：HI = 0.80 – 2x，H₂ = x，I₂ = x。由于 x = 0.10（来自平衡时 I₂），我们有 HI = 0.80 – 0.20 = 0.60 mol，H₂ = 0.10 mol，I₂ = 0.10 mol。浓度：[HI] = 0.60/2.0 = 0.30，[H₂] = 0.10/2.0 = 0.050，[I₂] = 0.10/2.0 = 0.050 mol dm⁻³。Kc = (0.050 × 0.050) / (0.30)² = 0.0025 / 0.090 = 0.0278。单位：(mol dm⁻³)(mol dm⁻³) / (mol dm⁻³)² = 无单位。

The Equilibrium Constant Kp / 平衡常数 Kp

For gaseous equilibria, the equilibrium constant in terms of partial pressure is Kp = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b, where P represents the partial pressure of each gas. Partial pressure is calculated as mole fraction multiplied by total pressure. Kp, like Kc, is constant at a given temperature and its units depend on the difference in the number of moles of gas between products and reactants, Δn = (c + d) – (a + b). For reactions where Δn ≠ 0, the units are typically atm^Δn or Pa^Δn. / 对于气相平衡，以分压表示的平衡常数为 Kp = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b，其中 P 表示每种气体的分压。分压等于摩尔分数乘以总压。与 Kc 一样，Kp 在给定温度下是常数，其单位取决于生成物和反应物之间的气体摩尔数差，Δn = (c + d) – (a + b)。对于 Δn ≠ 0 的反应，单位通常为 atm^Δn 或 Pa^Δn。

Kp Calculation Worked Example / Kp 计算例题

The dissociation of dinitrogen tetroxide: N₂O₄(g) ⇌ 2NO₂(g). At 60°C and a total pressure of 1.5 atm, N₂O₄ is 50% dissociated. Initial moles of N₂O₄ = 1.0. At equilibrium: N₂O₄ = 0.5 mol, NO₂ = 1.0 mol. Total moles = 1.5. Mole fraction of N₂O₄ = 0.5/1.5 = 1/3; mole fraction of NO₂ = 1.0/1.5 = 2/3. Partial pressures: P(N₂O₄) = (1/3) × 1.5 = 0.50 atm; P(NO₂) = (2/3) × 1.5 = 1.00 atm. Kp = (1.00)² / 0.50 = 1.00 / 0.50 = 2.00 atm. / 四氧化二氮的解离：N₂O₄(g) ⇌ 2NO₂(g)。在 60°C 和总压 1.5 atm 下，N₂O₄ 有 50% 解离。N₂O₄ 初始摩尔数 = 1.0。平衡时：N₂O₄ = 0.5 mol，NO₂ = 1.0 mol。总摩尔数 = 1.5。N₂O₄ 的摩尔分数 = 0.5/1.5 = 1/3；NO₂ 的摩尔分数 = 1.0/1.5 = 2/3。分压：P(N₂O₄) = (1/3) × 1.5 = 0.50 atm；P(NO₂) = (2/3) × 1.5 = 1.00 atm。Kp = (1.00)² / 0.50 = 1.00 / 0.50 = 2.00 atm。

Relationship Between Kc and Kp / Kc 与 Kp 的关系

Kc and Kp are related through the ideal gas equation. The relationship is Kp = Kc (RT)^Δn, where R is the gas constant (0.08206 dm³ atm mol⁻¹ K⁻¹), T is the absolute temperature in Kelvin, and Δn is the change in the number of moles of gas. This formula allows conversion between the two forms of the equilibrium constant. When Δn = 0, Kp = Kc because (RT)^0 = 1. This is a key quantitative relationship that appears frequently in A-Level examination questions. / Kc 和 Kp 通过理想气体方程相关联。关系式为 Kp = Kc (RT)^Δn，其中 R 是气体常数（0.08206 dm³ atm mol⁻¹ K⁻¹），T 是以开尔文为单位的绝对温度，Δn 是气体摩尔数的变化。此公式允许在两种形式的平衡常数之间进行转换。当 Δn = 0 时，Kp = Kc，因为 (RT)^0 = 1。这是一个关键的定量关系，经常出现在 A-Level 考试题目中。

Common Misconceptions and Exam Pitfalls / 常见误解和考试陷阱

A frequent error is confusing the position of equilibrium with the rate of reaction. Le Chatelier’s Principle predicts the direction of equilibrium shift, not the speed at which equilibrium is reached. Adding a catalyst increases the rate but does not affect the equilibrium position or Kc. Another common mistake is forgetting to omit solids and pure liquids from Kc and Kp expressions. Students also often confuse the effect of pressure on Kp ： pressure changes shift the equilibrium position but do NOT change Kp, because Kp is a constant at a given temperature. Only temperature changes alter the value of equilibrium constants. / 一个常见错误是混淆平衡位置与反应速率。勒夏特列原理预测的是平衡移动的方向，而不是达到平衡的速度。添加催化剂会提高速率但不影响平衡位置或 Kc。另一个常见错误是忘记从 Kc 和 Kp 表达式中省略固体和纯液体。学生也经常混淆压力对 Kp 的影响 ： 压力变化会移动平衡位置但不会改变 Kp，因为 Kp 在给定温度下是一个常数。只有温度变化才能改变平衡常数的值。

Exam Tips for Equilibrium Questions / 平衡题的考试技巧

When answering equilibrium questions, always state explicitly that the system is at equilibrium before applying Le Chatelier’s Principle. Write out the full Kc or Kp expression before substituting values, showing the stoichiometric coefficients as exponents. Check your units carefully ： this is a common source of mark loss. For Kp calculations, always calculate mole fractions before partial pressures, and ensure your mole fractions sum to 1.0. When a question asks for the effect of a change on Kc or Kp, remember that only temperature can alter these constants. / 在回答平衡题时，务必在应用勒夏特列原理之前明确说明系统处于平衡状态。先写出完整的 Kc 或 Kp 表达式再代入数值，将化学计量系数写为指数。仔细检查单位 ： 这是常见的失分点。对于 Kp 计算，始终在计算分压之前先计算摩尔分数，并确保摩尔分数之和为 1.0。当题目问及某种变化对 Kc 或 Kp 的影响时，请记住只有温度才能改变这些常数。