A-Level化学 化学平衡 勒夏特列原理 Kc计算

A-Level化学 化学平衡 勒夏特列原理 Kc计算

Introduction to Chemical Equilibrium

Chemical equilibrium is one of the most fundamental concepts in A-Level Chemistry. It describes the state in a reversible reaction where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of reactants and products. Understanding equilibrium is essential for predicting how chemical systems behave under different conditions and for optimising industrial processes.

化学平衡是A-Level化学中最基本的概念之一。它描述的是可逆反应中正反应和逆反应速率相等的状态，此时反应物和产物的浓度不再发生净变化。理解化学平衡对于预测化学体系在不同条件下的行为以及优化工业过程至关重要。

Dynamic Equilibrium: A Molecular Perspective

Many students mistakenly believe that at equilibrium, the reaction has stopped. In reality, equilibrium is dynamic: both the forward and reverse reactions continue at the molecular level, but because their rates are equal, the macroscopic concentrations remain constant. Consider the reversible decomposition of dinitrogen tetroxide: N2O4(g) ⇌ 2NO2(g). At equilibrium, N2O4 molecules continue to dissociate into NO2, while NO2 molecules simultaneously combine to form N2O4 at exactly the same rate.

许多学生误以为达到平衡时反应就停止了。实际上，平衡是动态的：在分子水平上，正反应和逆反应都在继续进行，但由于速率相等，宏观浓度保持不变。以四氧化二氮的可逆分解为例：N2O4(g) ⇌ 2NO2(g)。平衡时，N2O4分子继续分解为NO2，同时NO2分子也以完全相同的速率结合成N2O4。

A system at dynamic equilibrium has three key characteristics. First, it must be a closed system ： no matter can enter or leave. Second, the macroscopic properties such as colour, pressure, and concentration remain constant. Third, equilibrium can be approached from either direction: starting with pure reactants or pure products will eventually lead to the same equilibrium mixture, provided the conditions are identical.

处于动态平衡的体系有三个关键特征。第一，必须是封闭体系：物质不能进入或离开。第二，宏观性质如颜色、压强和浓度保持恒定。第三，平衡可以从任一方向达到：从纯反应物或纯产物开始，在相同条件下最终都会达到相同的平衡混合物。

Le Chatelier’s Principle

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 that change. This principle, formulated by Henri Louis Le Chatelier in 1884, is not a rigorous thermodynamic law but an extremely useful predictive tool that helps chemists anticipate how equilibrium systems respond to perturbations.

勒夏特列原理指出：如果处于动态平衡的体系受到浓度、压强或温度的变化，平衡位置将发生移动以抵消这种变化。这一原理由亨利·路易·勒夏特列于1884年提出，它不是一个严格的热力学定律，而是一个极为有用的预测工具，帮助化学家预测平衡体系如何响应外界干扰。

Effect of Concentration Changes

When the concentration of a reactant is increased, the equilibrium shifts to the right (towards products) to consume the added reactant. Conversely, when a product is removed from the equilibrium mixture, the forward reaction is favoured to replenish the lost product. For example, in the esterification reaction CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O, adding more ethanol shifts the equilibrium to the right, producing more ethyl ethanoate. Removing water (using a drying agent) has the same effect.

当反应物的浓度增加时，平衡向右移动（向产物方向）以消耗增加的反应物。相反，当从平衡混合物中移除产物时，正反应受到促进以补充失去的产物。例如，在酯化反应CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O中，增加乙醇会使平衡向右移动，生成更多的乙酸乙酯。移除水（使用干燥剂）会产生同样的效果。

Effect of Pressure Changes

Pressure changes only affect equilibria involving gases where there is a difference in the number of moles between reactants and products. If the pressure is increased, the equilibrium shifts towards the side with fewer moles of gas to reduce the pressure. In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), there are 4 moles of gas on the left and 2 moles on the right. Increasing the pressure therefore favours the forward reaction, increasing the yield of ammonia. Conversely, in a reaction where both sides have equal moles of gas ： such as H2(g) + I2(g) ⇌ 2HI(g) ： changing the pressure has no effect on the equilibrium position.

压强的变化只影响涉及气体且反应物和产物之间气体摩尔数存在差异的平衡。如果压强增加，平衡向气体摩尔数较少的一侧移动以降低压强。在哈伯法中，N2(g) + 3H2(g) ⇌ 2NH3(g)，左边有4摩尔气体，右边有2摩尔。因此，增加压强有利于正反应，提高氨的产率。相反，在两侧气体摩尔数相等的反应中：如H2(g) + I2(g) ⇌ 2HI(g)：改变压强对平衡位置没有影响。

It is important to note that adding an inert gas at constant volume does not affect the equilibrium position because the partial pressures of the reacting gases remain unchanged. However, adding an inert gas at constant pressure (allowing the volume to expand) does shift the equilibrium towards the side with more moles of gas ： the same effect as decreasing the pressure.

需要注意的是，在恒定体积下加入惰性气体不会影响平衡位置，因为反应气体的分压保持不变。然而，在恒定压强下加入惰性气体（允许体积膨胀）确实会使平衡向气体摩尔数更多的一侧移动：效果与降低压强相同。

Effect of Temperature Changes

Temperature is the only factor that changes the value of the equilibrium constant, Kc. For an exothermic reaction (ΔH is negative), increasing the temperature shifts the equilibrium to the left (towards reactants) because the reverse endothermic reaction absorbs the added heat. For an endothermic reaction (ΔH is positive), increasing the temperature shifts the equilibrium to the right (towards products). The Contact process for sulfuric acid production ： 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ mol⁻¹ ： illustrates this trade-off: lower temperatures give a higher equilibrium yield of SO3, but the rate is too slow, so a compromise temperature of around 450°C is used with a catalyst.

温度是唯一能改变平衡常数Kc值的因素。对于放热反应（ΔH为负），升高温度使平衡向左移动（向反应物方向），因为逆反应是吸热的，可以吸收增加的热量。对于吸热反应（ΔH为正），升高温度使平衡向右移动（向产物方向）。硫酸生产的接触法：2SO2(g) + O2(g) ⇌ 2SO3(g)，ΔH = −197 kJ mol⁻¹：说明了这种权衡：较低温度下SO3的平衡产率更高，但反应速率太慢，因此使用约450°C的折中温度并配合催化剂。

Effect of a Catalyst

A catalyst provides an alternative reaction pathway with a lower activation energy. It increases the rate of both the forward and reverse reactions equally, meaning it does NOT shift the position of equilibrium. A catalyst simply allows the system to reach equilibrium faster. It has no effect on the equilibrium constant Kc, nor on the equilibrium yield ： only on the speed at which equilibrium is attained. This is an extremely common exam question, and students often lose marks by claiming that catalysts increase the yield of products.

催化剂提供了活化能较低的替代反应路径。它同等程度地增加正反应和逆反应的速率，这意味着催化剂不会移动平衡位置。催化剂只是让体系更快地达到平衡。它对平衡常数Kc和平衡产率都没有影响：只影响达到平衡的速度。这是考试中非常常见的问题，学生常常因为声称催化剂能提高产物产率而丢分。

The Equilibrium Constant, Kc

The equilibrium constant Kc quantifies the position of equilibrium for a reversible reaction at a given temperature. For the general reaction aA + bB ⇌ cC + dD, the expression is: Kc = [C]^c[D]^d / [A]^a[B]^b, where square brackets denote equilibrium concentrations in mol dm⁻³. Kc is a constant at a fixed temperature: its value does not change with concentration or pressure, only with temperature. A large Kc (much greater than 1) indicates that the equilibrium lies far to the right, favouring products. A small Kc (much less than 1) indicates that the equilibrium favours reactants.

平衡常数Kc量化了可逆反应在给定温度下的平衡位置。对于一般反应aA + bB ⇌ cC + dD，其表达式为：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在固定温度下是常数：其值不随浓度或压强变化，只随温度变化。较大的Kc（远大于1）表明平衡位置偏右，有利于产物。较小的Kc（远小于1）表明平衡有利于反应物。

Homogeneous equilibria involve reactants and products all in the same phase ： typically all gases or all aqueous. In heterogeneous equilibria, the concentrations of pure solids and pure liquids are effectively constant and are therefore omitted from the Kc expression. For example, in CaCO3(s) ⇌ CaO(s) + CO2(g), Kc = [CO2], because the concentrations of the solid calcium carbonate and calcium oxide are constant and included in the value of Kc.

均相平衡涉及的反应物和产物都处于同一相：通常全是气体或全是水溶液。在多相平衡中，纯固体和纯液体的浓度实际上是恒定的，因此从Kc表达式中省略。例如，在CaCO3(s) ⇌ CaO(s) + CO2(g)中，Kc = [CO2]，因为固体碳酸钙和氧化钙的浓度是恒定的，已包含在Kc的值中。

Calculating Kc: Worked Examples

Calculating Kc is a core skill in A-Level Chemistry. The method involves constructing a table of initial amounts, changes in amounts, and equilibrium amounts ： commonly called an ICE table (Initial, Change, Equilibrium). Let us work through a typical example. Consider the reaction H2(g) + I2(g) ⇌ 2HI(g). 1.00 mol of H2 and 1.00 mol of I2 are placed in a 2.00 dm³ vessel and allowed to reach equilibrium at 700 K. At equilibrium, 0.40 mol of H2 remains. To find Kc, we first determine that 1.00 − 0.40 = 0.60 mol of H2 has reacted. By stoichiometry, 0.60 mol of I2 also reacts, leaving 0.40 mol of I2 at equilibrium, and 2 × 0.60 = 1.20 mol of HI is formed. Converting to concentrations: [H2] = 0.40 ÷ 2.00 = 0.20 mol dm⁻³, [I2] = 0.20 mol dm⁻³, [HI] = 0.60 mol dm⁻³. Then Kc = [HI]² ÷ ([H2][I2]) = (0.60)² ÷ (0.20 × 0.20) = 0.36 ÷ 0.04 = 9.0, with no units since the powers cancel.

计算Kc是A-Level化学的核心技能。方法包括构建一个初始量、变化量和平衡量的表格：通常称为ICE表（Initial, Change, Equilibrium）。我们来看一个典型的例子。考虑反应H2(g) + I2(g) ⇌ 2HI(g)。将1.00 mol H2和1.00 mol I2放入一个2.00 dm³的容器中，在700 K下达到平衡。平衡时剩余0.40 mol H2。为求Kc，我们首先确定1.00 − 0.40 = 0.60 mol H2已反应。根据化学计量比，0.60 mol I2也反应，平衡时剩余0.40 mol I2，生成2 × 0.60 = 1.20 mol HI。转换为浓度：[H2] = 0.40 ÷ 2.00 = 0.20 mol dm⁻³，[I2] = 0.20 mol dm⁻³，[HI] = 0.60 mol dm⁻³。则Kc = [HI]² ÷ ([H2][I2]) = (0.60)² ÷ (0.20 × 0.20) = 0.36 ÷ 0.04 = 9.0，由于幂次抵消，无单位。

For reactions where the total number of moles changes, Kc has units. Consider the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.80 mol of PCl5 is heated in a 2.00 dm³ vessel and at equilibrium 0.20 mol of PCl5 has dissociated, then the equilibrium amounts are: PCl5 = 0.80 − 0.20 = 0.60 mol, PCl3 = 0.20 mol, Cl2 = 0.20 mol. Concentrations: [PCl5] = 0.30, [PCl3] = 0.10, [Cl2] = 0.10 mol dm⁻³. Kc = (0.10 × 0.10) ÷ 0.30 = 0.0333 mol dm⁻³. The unit comes from mol dm⁻³ × mol dm⁻³ ÷ mol dm⁻³ = mol dm⁻³. Always derive units from the Kc expression itself ： do not memorise them.

对于总摩尔数发生变化的反应，Kc带有单位。考虑平衡PCl5(g) ⇌ PCl3(g) + Cl2(g)。将0.80 mol PCl5在2.00 dm³容器中加热，平衡时有0.20 mol PCl5解离，则平衡量为：PCl5 = 0.80 − 0.20 = 0.60 mol，PCl3 = 0.20 mol，Cl2 = 0.20 mol。浓度：[PCl5] = 0.30，[PCl3] = 0.10，[Cl2] = 0.10 mol dm⁻³。Kc = (0.10 × 0.10) ÷ 0.30 = 0.0333 mol dm⁻³。单位来自mol dm⁻³ × mol dm⁻³ ÷ mol dm⁻³ = mol dm⁻³。始终从Kc表达式本身推导单位：不要死记硬背。

Using Kc to Predict Direction

The reaction quotient Qc has the same form as the Kc expression but uses the current (non-equilibrium) concentrations. By comparing Qc with Kc, we can predict the direction in which a reaction will proceed to reach equilibrium. If Qc is less than Kc, the forward reaction is favoured to increase the concentration of products. If Qc is greater than Kc, the reverse reaction is favoured. If Qc equals Kc, the system is at equilibrium. This predictive power makes Kc an invaluable tool in both laboratory and industrial chemistry.

反应商Qc的形式与Kc表达式相同，但使用的是当前（非平衡）浓度。通过比较Qc与Kc，我们可以预测反应进行的方向以达到平衡。如果Qc小于Kc，正反应受到促进以增加产物浓度。如果Qc大于Kc，逆反应受到促进。如果Qc等于Kc，体系处于平衡状态。这种预测能力使Kc成为实验室和工业化学中不可或缺的工具。

Industrial Applications of Equilibrium Principles

The Haber process for ammonia synthesis ： N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol⁻¹ ： is perhaps the most famous industrial application of equilibrium principles. The forward reaction is exothermic and produces fewer moles of gas. Le Chatelier’s Principle predicts that high pressure and low temperature would maximise the equilibrium yield. In practice, the process operates at around 200 atm and 400-450°C, with an iron catalyst. The high pressure favours the forward reaction while the elevated temperature, though unfavourable for equilibrium yield, is necessary to achieve a commercially viable reaction rate.

哈伯法合成氨：N2(g) + 3H2(g) ⇌ 2NH3(g)，ΔH = −92 kJ mol⁻¹：可能是平衡原理最著名的工业应用。正反应是放热的且气体摩尔数减少。勒夏特列原理预测高压和低温可以使平衡产率最大化。实践中，该过程在约200 atm和400-450°C下运行，使用铁催化剂。高压有利于正反应，而较高温度虽然不利于平衡产率，但对实现商业上可行的反应速率是必要的。

Another important industrial process is the Contact process for sulfuric acid: 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ mol⁻¹. This exothermic reaction also produces fewer moles of gas (3 mol = 2 mol), so high pressure and low temperature favour the forward reaction. Industrially, the process uses a pressure of 1-2 atm (atmospheric pressure is sufficient because the equilibrium already lies far to the right at these conditions), a temperature of around 450°C, and a vanadium(V) oxide catalyst. The SO3 produced is then absorbed in concentrated sulfuric acid to form oleum, which is diluted to produce sulfuric acid.

另一个重要的工业过程是硫酸的接触法：2SO2(g) + O2(g) ⇌ 2SO3(g)，ΔH = −197 kJ mol⁻¹。这个放热反应气体摩尔数也减少（3 mol = 2 mol），所以高压和低温有利于正反应。工业上，该过程使用1-2 atm的压强（常压就足够了，因为在这些条件下平衡已经很偏右），约450°C的温度，以及五氧化二钒催化剂。生成的SO3随后被浓硫酸吸收形成发烟硫酸，再稀释得到硫酸。

Common Misconceptions and Exam Pitfalls

Several misconceptions about chemical equilibrium persist among A-Level students. The most common is confusing rate and equilibrium position. A catalyst increases the rate at which equilibrium is reached but does not affect the equilibrium position or yield. Similarly, increasing the concentration of a reactant increases the rate of the forward reaction, but the new equilibrium position is determined by Le Chatelier’s Principle ： not simply by which side is “faster”. Another common error is assuming that when Kc is large, the reaction goes to completion. Even with a very large Kc, an equilibrium mixture still contains some reactants, however small the concentration.

A-Level学生中存在几个关于化学平衡的常见误解。最常见的是混淆速率和平衡位置。催化剂增加达到平衡的速率，但不影响平衡位置或产率。同样，增加反应物浓度会增加正反应速率，但新的平衡位置由勒夏特列原理决定：而不是简单地看哪一侧”更快”。另一个常见错误是认为当Kc很大时，反应就进行完全了。即使Kc非常大，平衡混合物中仍然含有一些反应物，无论浓度多么小。

Students also frequently confuse the effects of temperature on rate and on equilibrium. Increasing temperature always increases the rate of reaction (it provides more molecules with energy exceeding the activation energy). However, the effect on equilibrium depends on whether the reaction is exothermic or endothermic. For an exothermic reaction, increasing temperature shifts the equilibrium to the left; for an endothermic reaction, it shifts to the right. These two effects ： rate and equilibrium ： act independently and must be analysed separately in exam questions on industrial process optimisation.

学生还经常混淆温度对速率和平衡的影响。升高温度总是增加反应速率（它使更多分子具有超过活化能的能量）。然而，对平衡的影响取决于反应是放热还是吸热。对于放热反应，升高温度使平衡向左移动；对于吸热反应，则向右移动。速率和平衡这两种效应独立作用，在关于工业过程优化的考试题中必须分开分析。

Summary and Exam Tips

Chemical equilibrium is a topic that rewards systematic thinking and careful application of principles. The key takeaways are: equilibrium is dynamic, not static; Le Chatelier’s Principle predicts qualitative responses to changes in concentration, pressure, and temperature; Kc quantifies the equilibrium position and changes only with temperature; catalysts do not affect equilibrium position; and ICE tables provide a structured method for calculating equilibrium concentrations and Kc. When answering exam questions, always state explicitly which principle you are applying, show your ICE table clearly, and include units for Kc where they exist. For industrial process questions, discuss the compromise between equilibrium yield and reaction rate, and explain why the chosen conditions represent an economically optimal balance.

化学平衡是一个需要通过系统思考和仔细应用原理才能掌握的课题。关键要点是：平衡是动态的，不是静态的；勒夏特列原理预测了对浓度、压强和温度变化的定性响应；Kc量化了平衡位置且只随温度变化；催化剂不影响平衡位置；ICE表为计算平衡浓度和Kc提供了结构化的方法。在回答考试问题时，始终明确说明你正在应用哪个原理，清楚地展示你的ICE表，并在Kc有单位时包含单位。对于工业过程问题，讨论平衡产率和反应速率之间的折中，并解释为什么所选条件代表了经济上的最优平衡。