A-Level化学 化学平衡 Kc计算 勒夏特列原理

A-Level化学 化学平衡 Kc计算 勒夏特列原理

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 backward reaction. At this point, the concentrations of reactants and products remain constant over time ： not because the reaction has stopped, but because both directions are proceeding at equal rates. This is a dynamic equilibrium: molecules are continuously reacting, yet there is no net change in macroscopic properties such as colour, pressure, or concentration.

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。此时反应物和产物的浓度随时间保持不变，这不是因为反应停止了，而是因为两个方向的反应以相同的速率进行。这是一种动态平衡：分子在持续反应，但宏观性质（如颜色、压力或浓度）没有净变化。

A reversible reaction is denoted by the double arrow symbol ⇌. For a general reaction aA + bB ⇌ cC + dD, the system will reach equilibrium when the rates of the forward reaction (A + B → C + D) and the reverse reaction (C + D → A + B) become equal. Equilibrium can be approached from either direction: starting with only reactants, or starting with only products, will eventually reach the same equilibrium state.

可逆反应用双箭头符号⇌表示。对于一般反应 aA + bB ⇌ cC + dD，当正反应（A + B → C + D）和逆反应（C + D → A + B）的速率相等时，系统达到平衡。平衡可以从任一方向趋近：从纯反应物开始，或从纯产物开始，最终都会达到相同的平衡状态。

The Equilibrium Constant Kc / 平衡常数 Kc

The equilibrium constant Kc quantifies the position of equilibrium for a reaction at a given temperature. For the reaction aA + bB ⇌ cC + dD, the expression is: Kc = [C]^c × [D]^d / [A]^a × [B]^b, where [X] represents the equilibrium concentration of species X in mol/dm^3. The value of Kc is constant at a fixed temperature and is independent of initial concentrations, pressure, or the presence of a catalyst.

平衡常数 Kc 量化了在给定温度下反应平衡的位置。对于反应 aA + bB ⇌ cC + dD，表达式为：Kc = [C]^c × [D]^d / [A]^a × [B]^b，其中 [X] 表示物质 X 在平衡时的浓度，单位为 mol/dm^3。Kc 的值在固定温度下是常数，与初始浓度、压力或催化剂的存在无关。

There are several important rules for constructing Kc expressions. First, only gases and aqueous species appear in the expression ： pure solids and pure liquids have constant concentrations and are omitted. Second, stoichiometric coefficients become exponents in the expression. Third, Kc has units that depend on the specific reaction: if the total number of moles on both sides of the equation is equal, Kc is dimensionless (has no units).

构建 Kc 表达式有几条重要规则。第一，只有气体和水溶液中的物质出现在表达式中，纯固体和纯液体具有恒定的浓度，因此被省略。第二，化学计量系数成为表达式中的指数。第三，Kc 的单位取决于具体反应：如果方程式两边的总摩尔数相等，则 Kc 是无量纲的（无单位）。

A large Kc value (Kc >> 1) indicates that the equilibrium lies well to the right, favouring products. A small Kc value (Kc << 1) indicates that the equilibrium lies to the left, favouring reactants. A Kc value close to 1 suggests significant amounts of both reactants and products are present at equilibrium. Importantly, the magnitude of Kc tells you about the position of equilibrium, NOT the rate at which equilibrium is reached.

大的 Kc 值（Kc >> 1）表明平衡位置偏右，有利于产物。小的 Kc 值（Kc << 1）表明平衡位置偏左，有利于反应物。Kc 值接近 1 意味着平衡时反应物和产物都有显著的量。重要的是，Kc 的大小告诉你平衡的位置，而不是达到平衡的速率。

Calculating Kc: Worked Example / Kc计算：实例分析

Problem: Consider the reaction H₂(g) + I₂(g) ⇌ 2HI(g) at 700K. Initially, 1.00 mol of H₂ and 1.00 mol of I₂ are placed in a 2.00 dm³ container. At equilibrium, 1.56 mol of HI is present. Calculate Kc.

问题：考虑反应 H₂(g) + I₂(g) ⇌ 2HI(g) 在 700K 下进行。初始时将 1.00 mol H₂ 和 1.00 mol I₂ 放入 2.00 dm³ 容器中。平衡时，有 1.56 mol HI 存在。计算 Kc。

Step 1: Set up an ICE table (Initial, Change, Equilibrium). If 1.56 mol of HI is formed at equilibrium, from the stoichiometry 1 mol H₂ produces 2 mol HI, so 0.78 mol of H₂ and 0.78 mol of I₂ have been consumed. Equilbrium amounts: H₂ = 1.00 – 0.78 = 0.22 mol, I₂ = 1.00 – 0.78 = 0.22 mol, HI = 1.56 mol.

步骤1：建立 ICE 表格（初始、变化、平衡）。如果平衡时生成 1.56 mol HI，根据化学计量关系，1 mol H₂ 生成 2 mol HI，因此 0.78 mol H₂ 和 0.78 mol I₂ 已被消耗。平衡量：H₂ = 1.00 – 0.78 = 0.22 mol，I₂ = 1.00 – 0.78 = 0.22 mol，HI = 1.56 mol。

Step 2: Convert to concentrations. [H₂] = 0.22 / 2.00 = 0.11 mol/dm³, [I₂] = 0.11 mol/dm³, [HI] = 1.56 / 2.00 = 0.78 mol/dm³. Step 3: Apply Kc expression. Kc = [HI]² / ([H₂] × [I₂]) = (0.78)² / (0.11 × 0.11) = 0.6084 / 0.0121 = 50.3. Units: (mol/dm³)² / ((mol/dm³) × (mol/dm³)) = dimensionless. So Kc = 50.3 (no units). This relatively large Kc indicates the equilibrium strongly favours the formation of HI at 700K.

步骤2：转换为浓度。[H₂] = 0.22 / 2.00 = 0.11 mol/dm³，[I₂] = 0.11 mol/dm³，[HI] = 1.56 / 2.00 = 0.78 mol/dm³。步骤3：应用 Kc 表达式。Kc = [HI]² / ([H₂] × [I₂]) = (0.78)² / (0.11 × 0.11) = 0.6084 / 0.0121 = 50.3。单位：(mol/dm³)² / ((mol/dm³) × (mol/dm³)) = 无量纲。因此 Kc = 50.3（无单位）。这个相对较大的 Kc 值表明在 700K 时平衡强烈有利于 HI 的生成。

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 conditions, the position of equilibrium will shift to oppose that change. This principle allows chemists to predict how equilibrium responds to changes in concentration, pressure (for gases), and temperature. It is a qualitative tool that describes the direction of shift, not the new equilibrium concentrations.

勒夏特列原理指出，如果处于动态平衡的系统受到条件变化的影响，平衡位置将移动以抵消该变化。该原理使化学家能够预测平衡如何响应浓度、压力（对气体而言）和温度的变化。它是一种定性工具，描述移动的方向，而不是新的平衡浓度。

Effect of Concentration: Adding more reactant shifts equilibrium to the right (towards products) to consume the added reactant. Removing a product also shifts equilibrium to the right, as the system attempts to replace what was removed. Conversely, adding a product shifts equilibrium to the left. This is the principle behind industrial processes where one product is continuously removed to drive the reaction towards completion.

浓度的影响：增加反应物浓度使平衡向右移动（向产物方向），以消耗添加的反应物。移除产物也使平衡向右移动，因为系统试图补充被移除的物质。相反，添加产物使平衡向左移动。这就是工业过程中持续移除一种产物以推动反应完成的原理。

Effect of Pressure (gases only): Changing pressure only affects equilibria where the number of gas molecules differs between the two sides. Increasing pressure shifts equilibrium towards the side with fewer gas molecules, reducing the total pressure. Decreasing pressure shifts equilibrium towards the side with more gas molecules. If both sides have the same number of gas molecules, pressure changes have no effect on the position of equilibrium. Note: adding an inert gas at constant volume does NOT affect equilibrium position because it does not change the partial pressures of the reacting gases.

压力的影响（仅适用于气体）：改变压力只影响两侧气体分子数不同的平衡。增加压力使平衡向气体分子数较少的一侧移动，以降低总压力。降低压力使平衡向气体分子数较多的一侧移动。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。注意：在恒定体积下加入惰性气体不影响平衡位置，因为它不会改变反应气体的分压。

Effect of Temperature: Temperature is the ONLY change that alters the value of Kc. For an exothermic reaction (ΔH < 0), increasing temperature shifts equilibrium to the left (favouring reactants), and Kc decreases. For an endothermic reaction (ΔH > 0), increasing temperature shifts equilibrium to the right (favouring products), and Kc increases. Think of heat as a reactant (endothermic) or product (exothermic) when applying Le Chatelier’s Principle.

温度的影响：温度是唯一改变 Kc 值的变化。对于放热反应（ΔH < 0），升高温度使平衡向左移动（有利于反应物），Kc 减小。对于吸热反应（ΔH > 0），升高温度使平衡向右移动（有利于产物），Kc 增大。应用勒夏特列原理时，可以将热量视为反应物（吸热）或产物（放热）。

Effect of a Catalyst: A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the rate of BOTH the forward and reverse reactions equally. Therefore, a catalyst does NOT affect the position of equilibrium or the value of Kc ： it simply allows equilibrium to be reached faster. A catalyst cannot increase the yield beyond the equilibrium yield; it only changes the kinetics, not the thermodynamics.

催化剂的影响：催化剂提供具有较低活化能的替代反应路径，同等程度地提高正反应和逆反应的速率。因此，催化剂不影响平衡位置或 Kc 值，它只是使平衡更快达到。催化剂不能将产率提高到超过平衡产率；它只改变动力学，不改变热力学。

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

The Haber process for ammonia synthesis is the textbook example of applying Le Chatelier’s Principle industrially. The reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ/mol (exothermic). As written, there are 4 moles of gas on the left and 2 on the right, so high pressure favours ammonia production by shifting equilibrium to the right. However, since the forward reaction is exothermic, low temperature favours ammonia production thermodynamically.

哈伯法合成氨是勒夏特列原理工业应用的教科书范例。反应为 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = -92 kJ/mol（放热）。反应式左边有 4 摩尔气体，右边有 2 摩尔，因此高压有利于氨的生成，使平衡向右移动。然而，由于正反应是放热的，低温在热力学上有利于氨的生成。

In practice, a compromise is required. High pressure (typically 200 atm) shifts equilibrium favourably and increases the rate, but very high pressures are expensive and dangerous. Low temperature gives a higher equilibrium yield but makes the reaction too slow. The industrial compromise uses a temperature of about 450°C with an iron catalyst: the catalyst allows a reasonable rate at this moderate temperature, while the high pressure provides a favourable equilibrium shift. The result is an equilibrium yield of approximately 15-20% NH₃ gas, but the unreacted gases are continuously recycled, achieving an overall conversion of about 97%.

在实践中需要一个折衷方案。高压（通常 200 atm）有利于平衡移动并提高速率，但极高压力昂贵且危险。低温给出更高的平衡产率，但使反应太慢。工业折衷方案使用约 450°C 的温度和铁催化剂：催化剂允许在此中等温度下获得合理的速率，而高压提供了有利的平衡移动。结果是约 15-20% 的 NH₃ 平衡产率，但未反应的气体被持续循环利用，总转化率达到约 97%。

Common Misconceptions / 常见误区

Misconception 1: “Equilibrium means equal concentrations.” False. Equilibrium means equal forward and reverse rates, not equal concentrations. In the HI example above, [HI] = 0.78 mol/dm³ while [H₂] = [I₂] = 0.11 mol/dm³ ： clearly not equal concentrations.

误区1：”平衡意味着浓度相等。”错误。平衡意味着正逆反应速率相等，而非浓度相等。在上述 HI 例子中，[HI] = 0.78 mol/dm³，而 [H₂] = [I₂] = 0.11 mol/dm³，显然浓度不相等。

Misconception 2: “Adding a catalyst increases yield.” False. A catalyst affects only the rate, not the position of equilibrium. It cannot increase the equilibrium yield of a reaction. If the yield increases after adding a catalyst, it is because equilibrium was not yet reached before the addition.

误区2：”加入催化剂提高产率。”错误。催化剂只影响速率，不影响平衡位置。它不能提高反应的平衡产率。如果加入催化剂后产率提高，那是因为在加入之前尚未达到平衡。

Misconception 3: “Kc changes with concentration.” False. Kc is constant at a given temperature. Changing concentrations does shift the position of equilibrium, but the value of Kc remains the same ： the equilibrium concentrations adjust to satisfy the same Kc expression.

误区3：”Kc 随浓度变化。”错误。Kc 在给定温度下是常数。改变浓度确实会使平衡位置移动，但 Kc 值保持不变：平衡浓度会调整以满足相同的 Kc 表达式。

Exam Tips for A-Level Chemistry / A-Level化学考试技巧

When answering equilibrium questions, always show your ICE table clearly. State whether Kc is large, small, or moderate, and explain what this means in terms of product/reactant predominance. For Le Chatelier questions, always specify the DIRECTION of shift (left towards reactants or right towards products) and explain WHY using the principle’s language of “opposing the change.” Remember to mention that a catalyst has no effect on equilibrium position or yield ： this is a common exam trap.

在回答平衡问题时，始终清晰地展示你的 ICE 表格。说明 Kc 是大、小还是中等，并解释这在产物/反应物主导方面意味着什么。对于勒夏特列原理的题目，始终指定移动方向（向左移向反应物或向右移向产物），并使用原理解释”抵消变化”。记住提到催化剂对平衡位置或产率没有影响，这是常见的考试陷阱。

For Kc calculations, watch out for questions where the volume is not 1 dm³ ： you must divide moles by volume to get concentrations. Also, when the Kc expression involves a change in the number of moles, the units of Kc become a common source of error. Always derive the units from the expression and state them clearly. If your calculated Kc has unexpected units, re-check your stoichiometry and concentration conversions.

对于 Kc 计算，注意体积不是 1 dm³ 的题目：你必须将摩尔数除以体积得到浓度。此外，当 Kc 表达式涉及摩尔数的变化时，Kc 的单位成为常见的错误来源。始终从表达式推导单位并清晰地写出。如果计算出的 Kc 有出乎意料的单位，重新检查你的化学计量关系和浓度转换。

Temperature is the only factor that changes Kc ： this point appears in nearly every equilibrium exam. Be ready to explain why: temperature affects the relative rates of the forward and reverse reactions differently because they have different activation energies. The Arrhenius equation shows that a temperature change affects the rate constant of the endothermic direction more than the exothermic direction, shifting the equilibrium accordingly.

温度是唯一改变 Kc 的因素：这一点几乎出现在每道平衡考题中。准备好解释原因：温度对正反应和逆反应的相对速率有不同的影响，因为它们具有不同的活化能。阿伦尼乌斯方程表明，温度变化对吸热方向速率常数的影响大于对放热方向的影响，从而相应地移动平衡。