Chemical Equilibria, Le Chatelier's Principle and Kc | 化学平衡、勒夏特列原理与Kc

📚 Chemical Equilibria, Le Chatelier’s Principle and Kc | 化学平衡、勒夏特列原理与Kc

Chemical equilibria lie at the heart of many industrial and biological processes. Understanding how systems respond to change is essential for predicting yields and optimising conditions. This article explains the key concepts of dynamic equilibrium, Le Chatelier’s principle and the equilibrium constant Kc, with reference to the AQA A-level Chemistry specification.

化学平衡是许多工业和生物过程的核心。理解体系如何对外界变化作出响应，对于预测产率和优化条件至关重要。本文依据AQA A-level化学考纲，讲解动态平衡、勒夏特列原理以及平衡常数Kc的核心概念。

1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡

Many chemical reactions are reversible. In a closed system, as reactants form products, the products can also react to reform the reactants. When the rates of the forward and reverse reactions become equal, the system reaches dynamic equilibrium.

许多化学反应是可逆的。在密闭体系中，当反应物生成产物的同时，产物也可以重新反应生成反应物。当正反应速率和逆反应速率相等时，体系达到动态平衡。

The key features of dynamic equilibrium are:

动态平衡的关键特征如下：

Closed system: no substances can enter or leave the system.
Constant macroscopic properties: concentration, colour, pressure and pH remain constant.
Dynamic on the microscopic level: molecules continually interconvert even though bulk concentrations are constant.

密闭体系：没有物质能够进入或离开体系。
宏观性质恒定：浓度、颜色、压强和pH均保持不变。
微观上动态：尽管总体浓度不变，分子仍在不断相互转化。

The equilibrium position is the relative amounts of reactants and products present at equilibrium. It can be described as “lies to the right” (more products) or “lies to the left” (more reactants).

平衡位置是指平衡时反应物和产物相对的量。可以描述为“偏向右边”（产物较多）或“偏向左边”（反应物较多）。

2. Le Chatelier’s Principle: Predicting the Response to Change | 勒夏特列原理：预测体系对变化的响应

Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium position will shift to counteract the applied change and partially undo its effect.

勒夏特列原理指出：如果处于平衡的体系受到浓度、压强或温度的改变，平衡位置将向抵消外加改变的方向移动，从而部分削弱该改变的影响。

This principle is a powerful qualitative tool. It does not give numerical information but tells us whether the yield of products increases or decreases under new conditions.

该原理是强有力的定性工具。它不提供数值信息，但能告诉我们在新条件下产物的产率是增大还是减小。

3. Effect of Concentration Changes | 浓度改变的影响

If the concentration of a reactant is increased, the equilibrium position shifts to the right, consuming some of the added reactant and producing more products. Conversely, decreasing the concentration of a product shifts the equilibrium to the right; increasing a product concentration shifts it to the left.

如果增加反应物的浓度，平衡位置向右移动，消耗部分加入的反应物并生成更多产物。相反，降低产物浓度使平衡右移；增加产物浓度使平衡左移。

For the industrial synthesis of ammonia (Haber process):

以工业合成氨（哈伯法）为例：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹

If ammonia is continuously removed as it forms, the equilibrium position shifts to the right, increasing the overall yield of ammonia. This is why the Haber process uses condensation of ammonia to remove it from the recycled gases.

如果生成的氨被不断移走，平衡位置将向右移动，提高氨的整体产率。这就是哈伯法利用冷凝将氨从循环气中分离的原因。

4. Effect of Pressure Changes | 压强改变的影响

For gaseous equilibria, changing the total pressure affects the equilibrium position only if the number of moles of gaseous reactants differs from the number of moles of gaseous products. An increase in pressure shifts the equilibrium to the side with fewer moles of gas, because this reduces the pressure. A decrease in pressure shifts it to the side with more moles of gas.

对于气相平衡，只有当气态反应物的总物质的量与气态产物的总物质的量不同时，改变总压强才会影响平衡位置。增大压强使平衡向气体分子数较少的一侧移动，因为这样可以降低压强。减小压强使平衡向气体分子数较多的一侧移动。

Consider the reaction:

考虑反应：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

There are 4 moles of gas on the left and 2 moles on the right. Increasing the pressure drives the equilibrium to the right, producing more ammonia. If pressure is decreased, the equilibrium shifts to the left, producing more N₂ and H₂.

左边有4 mol气体，右边有2 mol气体。增大压强使平衡向右移动，生成更多氨。如果减小压强，平衡向左移动，生成更多N₂和H₂。

If the reaction has equal numbers of moles of gas on both sides, such as H₂(g) + I₂(g) ⇌ 2HI(g), pressure changes have no effect on the equilibrium position.

如果反应两边气体物质的量相等，例如H₂(g) + I₂(g) ⇌ 2HI(g)，压强改变对平衡位置没有影响。

5. Effect of Temperature Changes | 温度改变的影响

Temperature changes affect equilibrium because reactions have enthalpy changes. For an exothermic forward reaction, heat acts as a product; for an endothermic forward reaction, heat acts as a reactant.

温度改变影响平衡是因为反应具有焓变。对于放热正反应，热量相当于产物；对于吸热正反应，热量相当于反应物。

Increasing temperature shifts the equilibrium in the endothermic direction, absorbing the added heat. Decreasing temperature shifts the equilibrium in the exothermic direction, releasing heat to compensate.

升高温度使平衡向吸热方向移动，吸收加入的热量。降低温度使平衡向放热方向移动，释放热以补偿。

For the Haber process (exothermic forward reaction):

对于哈伯法（正反应放热）：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹

A low temperature would favour a high yield of ammonia, but the rate would be impractically slow. In practice, a compromise temperature of about 450 °C is used together with an iron catalyst to achieve a reasonable rate and a viable yield.

低温有利于氨的高产率，但反应速率会慢得不切实际。实际生产中，大约450 °C的折中温度配合铁催化剂，以获得合理的速率和可行的产率。

6. Effect of Catalysts | 催化剂的影响

A catalyst lowers the activation energy for both the forward and reverse reactions equally. It therefore increases the rate at which equilibrium is reached, but it does not alter the position of equilibrium or the value of Kc.

催化剂同等程度地降低正反应和逆反应的活化能。因此它加快了达到平衡的速率，但不会改变平衡位置，也不会改变Kc值。

Catalysts are commercially important because they allow equilibrium to be reached faster at lower temperatures, saving energy and reducing costs. However, they do not increase the equilibrium yield of products.

催化剂在商业上非常重要，因为它允许在较低温度下更快达到平衡，节省能源并降低成本。然而，它并不提高产物的平衡产率。

7. The Equilibrium Constant Kc | 平衡常数 Kc

For a general reaction at equilibrium:

对于一般反应在平衡状态时：

aA + bB ⇌ cC + dD

The equilibrium constant in terms of concentrations is:

基于浓度的平衡常数为：

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

Square brackets denote concentration in mol dm⁻³. Kc is a constant at a fixed temperature for a given reaction. It has no units unless the sum of the stoichiometric coefficients differs between products and reactants.

方括号表示浓度，单位为mol dm⁻³。Kc在温度固定时对给定反应是一个常数。除非产物和反应物的化学计量系数之和不相等，否则Kc没有单位。

Important points about Kc:

关于Kc的重要要点：

Kc only includes species in homogeneous equilibria; pure solids and pure liquids are omitted from the expression.
A large Kc (>> 1) means the equilibrium lies to the right, favouring products.
A small Kc (<< 1) means the equilibrium lies to the left, favouring reactants.
Kc does not depend on initial concentrations or on whether a catalyst is used.
Kc depends only on temperature.

Kc只包括均相平衡中的物种；纯固体和纯液体不出现在表达式中。
Kc很大（远大于1）表示平衡位于右侧，有利于产物。
Kc很小（远小于1）表示平衡位于左侧，有利于反应物。
Kc不依赖于初始浓度，也不依赖于是否使用催化剂。
Kc只取决于温度。

8. Writing Kc Expressions | 书写 Kc 表达式

When writing a Kc expression, you must use the stoichiometric coefficients as powers. Exclude solids and pure liquids. For example:

书写Kc表达式时，必须使用化学计量系数作为指数。排除固体和纯液体。例如：

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

对于反应N₂(g) + 3H₂(g) ⇌ 2NH₃(g)：

Kc = [NH₃]² / ([N₂][H₂]³)

For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g):

对于反应CaCO₃(s) ⇌ CaO(s) + CO₂(g)：

Kc = [CO₂]

The solids CaCO₃ and CaO are omitted because their concentrations are constant.

固体CaCO₃和CaO被省略，因为它们的浓度是常数。

9. Calculating Kc from Equilibrium Concentrations | 由平衡浓度计算 Kc

To calculate Kc, you need the equilibrium concentrations of all species. These are often determined from initial amounts and the known change in one species, using stoichiometry.

计算Kc需要所有物种的平衡浓度。这些通常由初始量和某一物种的已知变化量，通过化学计量关系来确定。

Worked example: 0.800 mol of SO₂ and 0.800 mol of O₂ are placed in a 1.00 dm³ vessel and allowed to reach equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). At equilibrium, 0.600 mol of SO₃ is present. Calculate Kc.

例题：0.800 mol SO₂和0.800 mol O₂置于1.00 dm³容器中，达到平衡：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)。平衡时存在0.600 mol SO₃。计算Kc。

Species	2SO₂	O₂	2SO₃
Initial / mol	0.800	0.800	0
Change / mol	−0.600	−0.300	+0.600
Equilibrium / mol	0.200	0.500	0.600
Equilibrium concentration / mol dm⁻³	0.200	0.500	0.600

Since the volume is 1.00 dm³, the number of moles equals the concentration.

因为体积为1.00 dm³，物质的量在数值上等于浓度。

Kc = [SO₃]² / ([SO₂]²[O₂]) = (0.600)² / ((0.200)²(0.500)) = 0.36 / 0.02 = 18

So Kc = 18 (no units because the sum of powers in the numerator is 2 and the sum in the denominator is 3; actually the overall units would be dm³ mol⁻¹, but many AQA questions treat units as required; here because Δn = 2 − 3 = −1, the units are dm³ mol⁻¹. Always calculate the units from the expression.)

所以Kc = 18（无单位？实际上因为分子指数和为2，分母指数和为3，总体单位应为dm³ mol⁻¹；这里Δn = 2 − 3 = −1，所以单位是dm³ mol⁻¹。始终根据表达式计算单位。）

Wait – the reaction is 2SO₂ + O₂ ⇌ 2SO₃. The numerator powers sum to 2, denominator powers sum to 3. Therefore Kc has units (mol dm⁻³)² / ((mol dm⁻³)² × (mol dm⁻³)) = (mol dm⁻³)⁻¹ = dm³ mol⁻¹. So Kc = 18 dm³ mol⁻¹.

注意——反应是2SO₂ + O₂ ⇌ 2SO₃。分子指数和为2，分母指数和为3。因此Kc的单位为(mol dm⁻³)² / ((mol dm⁻³)² × (mol dm⁻³)) = (mol dm⁻³)⁻¹ = dm³ mol⁻¹。所以Kc = 18 dm³ mol⁻¹。

10. Using Kc to Find Unknown Concentrations | 用 Kc 求未知浓度

Kc can be used to calculate an unknown equilibrium concentration if the other concentrations and Kc are known. For example, for the reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), Kc = 4.00 at a given temperature. At equilibrium, [CO₂] = 0.200 mol dm⁻³ and [H₂] = 0.200 mol dm⁻³, and [H₂O] = 0.100 mol dm⁻³. Find [CO].

Kc可以用来计算未知的平衡浓度，如果其他浓度和Kc已知。例如，对于反应CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)，某温度下Kc = 4.00。平衡时，[CO₂] = 0.200 mol dm⁻³，[H₂] = 0.200 mol dm⁻³，[H₂O] = 0.100 mol dm⁻³。求[CO]。

Kc = [CO₂][H₂] / ([CO][H₂O])

4.00 = (0.200 × 0.200) / ([CO] × 0.100)

Rearrange: [CO] = 0.0400 / (4.00 × 0.100) = 0.100 mol dm⁻³.

整理得：[CO] = 0.0400 / (4.00 × 0.100) = 0.100 mol dm⁻³。

11. Kc and the Extent of Reaction | Kc 与反应进行程度

The magnitude of Kc at a given temperature indicates how far the reaction proceeds. A very large Kc means that at equilibrium the numerator is much larger than the denominator, so products dominate. A very small Kc means reactants dominate.

Kc在给定温度下的数值大小表明反应进行的程度。非常大的Kc意味着平衡时分子远大于分母，因此产物占主导。非常小的Kc意味着反应物占主导。

This idea is used to predict whether a reaction is essentially complete or barely starts, and it guides the choice of reaction conditions for industrial processes.

这一概念用于预测反应是否基本完成或几乎不进行，也指导工业过程反应条件的选择。

However, Kc alone does not indicate the rate of reaction. A thermodynamically favourable reaction may be extremely slow if the activation energy is high.

然而，Kc本身并不表明反应速率。一个热力学上有利的反应，如果活化能很高，可能极其缓慢。

12. Homogeneous and Heterogeneous Equilibria | 均相与多相平衡

A homogeneous equilibrium is one in which all reactants and products are in the same physical state or same phase. For example, the reaction between hydrogen and iodine vapour:

均相平衡是指所有反应物和产物处于相同的物理状态或相同的相。例如，氢气和碘蒸气之间的反应：

H₂(g) + I₂(g) ⇌ 2HI(g)

Here Kc = [HI]² / ([H₂][I₂]).

这里Kc = [HI]² / ([H₂][I₂])。

A heterogeneous equilibrium involves species in different phases. For example, the thermal decomposition of calcium carbonate:

多相平衡涉及不同相的物种。例如，碳酸钙的热分解：

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Solids are omitted from the equilibrium constant expression. The concentration of a pure solid is effectively constant, so it does not appear in Kc. Thus Kc = [CO₂].

固体不出现在平衡常数表达式中。纯固体的浓度实际上是常数，所以它不出现在Kc中。因此Kc = [CO₂]。

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Chemical Equilibria, Le Chatelier’s Principle and Kc | 化学平衡、勒夏特列原理与Kc