Le Chatelier's Principle | 勒夏特列原理

📚 Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s Principle is a cornerstone of chemical equilibrium, allowing chemists to predict how a system at equilibrium responds to changes in concentration, pressure, or temperature. It is not only a theoretical tool but also the guiding logic behind many industrial processes, making it a high-frequency topic in AQA A-level Chemistry. Mastering it requires understanding the interplay between kinetics and thermodynamics, as well as the crucial distinction between the rate at which equilibrium is reached and the position of equilibrium itself.

勒夏特列原理是化学平衡的基石，能帮助化学家预测处于平衡态的系统如何应对浓度、压强或温度的变化。它不仅是理论工具，更是许多工业过程的指导逻辑，因而成为AQA A-level化学中的高频考点。要真正掌握该原理，需要理解动力学与热力学之间的相互作用，并能严格区分到达平衡的速率与平衡位置本身这两个概念。

1. Introduction to Le Chatelier’s Principle | 原理引言

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 re-establish equilibrium.

勒夏特列原理指出，如果动态平衡受到外界条件变化的干扰，平衡位置会朝着抵消该变化的方向移动，以便重新建立平衡。

It is important to remember that the principle only applies to systems that are already in dynamic equilibrium. Reactions that go to completion are not governed by this concept.

必须牢记，该原理仅适用于已经处于动态平衡的系统。对进行到底的反应，它是不适用的。

In every exam question, the key phrase “position of equilibrium shifts to the right/left” should be used rather than simply stating “more products are made”. The equilibrium moves to minimise the imposed change, but it rarely cancels it entirely.

在答题时，一定要用”平衡位置向右/左移动”这样的表述，而非简单地说”生成了更多产物”。平衡会去减弱外界施加的改变，但通常不会完全将其抵消。

2. Effect of Concentration Changes | 浓度变化的影响

When the concentration of a reactant is increased, the system shifts the equilibrium position to the right to use up the added substance, producing more product. Conversely, removing a product causes the equilibrium to shift to the right to replace the lost species.

增大反应物的浓度时，系统会将平衡位置向右移动，以消耗掉新增的物质，生成更多产物；反之，移走部分产物会使平衡向右移动，以补充被移走的物质。

If a product is added, the equilibrium shifts to the left to consume the extra product and reform reactants. These concentration adjustments do not change the value of the equilibrium constant Kc, provided the temperature stays constant.

如果加入更多产物，平衡则向左移动，消耗多余的产物并重新生成反应物。这些浓度调节并不会改变平衡常数Kc的值，前提是温度保持不变。

A common AQA example is the iron(III) thiocyanate equilibrium: Fe³⁺(aq) + NCS⁻(aq) ⇌ [FeNCS]²⁺(aq). Adding more Fe³⁺ deepens the blood-red colour as the equilibrium shifts right; adding NCS⁻ has the same effect. Diluting the mixture with water shifts the equilibrium back towards the left, lightening the colour.

AQA常考的例子是铁(III)离子与硫氰酸根离子的平衡：Fe³⁺(aq) + NCS⁻(aq) ⇌ [FeNCS]²⁺(aq)。加入更多Fe³⁺会使血红色加深，因为平衡右移；加入NCS⁻效果相同。用水稀释则使平衡左移，颜色变浅。

3. Effect of Pressure Changes | 压强变化的影响

Pressure changes only affect equilibria that involve gases, and even then only when there is a change in the total number of gas molecules (Δn ≠ 0). An increase in pressure forces the equilibrium to shift towards the side with fewer gaseous molecules, thereby reducing the pressure.

压强的变化只影响有气体参与的平衡，而且只有在气体分子总数发生变化（Δn ≠ 0）时才会产生作用。增大压强会迫使平衡向气体分子数较少的一侧移动，从而降低压强。

If both sides have the same number of gas molecules, altering the pressure has no effect on the position of equilibrium. This is frequently tested with the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g).

如果反应前后气体分子总数相等，改变压强就不会影响平衡位置。这一结论经常通过H₂(g) + I₂(g) ⇌ 2HI(g)平衡来进行考查。

For the reaction 2NO₂(g) ⇌ N₂O₄(g), an increase in pressure shifts the equilibrium to the right (fewer moles, from 2 to 1) and the colour of the mixture changes from dark brown to a paler shade as more colourless N₂O₄ is formed. This is a classic colour-change demonstration.

对于反应2NO₂(g) ⇌ N₂O₄(g)，增大压强会使平衡向右移动（气体分子数从2减至1），随着更多无色N₂O₄生成，混合气体的颜色从深棕色变浅。这是一个经典的变色实验。

It is vital to note that a change in pressure by adding an inert gas like argon at constant volume does not alter the partial pressures of the reacting gases, so the equilibrium position stays unchanged.

必须注意，如果在恒容条件下加入惰性气体（如氩气）来改变总压，并不会改变反应气体的分压，因此平衡位置保持不变。

4. Effect of Temperature Changes | 温度变化的影响

Temperature is the only external condition that changes the value of the equilibrium constant Kc. For an exothermic reaction (ΔH negative), increasing the temperature shifts the equilibrium to the left, favouring the endothermic reverse reaction, and Kc decreases.

温度是唯一能改变平衡常数Kc数值的外界条件。对于放热反应（ΔH为负），升高温度会使平衡向左移动，有利于吸热的逆反应，Kc值减小。

For an endothermic reaction (ΔH positive), raising the temperature moves the equilibrium to the right, increasing the yield of products and making Kc larger. AQA exams often ask you to predict the effect on Kc from a temperature change.

对于吸热反应（ΔH为正），升高温度会使平衡向右移动，提高产物的产率，Kc值增大。AQA试题经常要求考生根据温度变化预测对Kc的影响。

This can be summarised clearly in the table below, which focuses on the direction of shift and the Kc response:

下表清晰地总结了温度变化对平衡移动方向及Kc的影响：

Reaction type	Temperature increase	Temperature decrease
Exothermic (ΔH < 0)	Shifts left; Kc decreases	Shifts right; Kc increases
Endothermic (ΔH > 0)	Shifts right; Kc increases	Shifts left; Kc decreases

Using this table, the link between Le Chatelier’s Principle and the magnitude of Kc becomes clear: the equilibrium responds to temperature stress by favouring the reaction that absorbs or releases heat, exactly as the principle predicts.

借助此表，勒夏特列原理与Kc大小之间的关系就清晰了：平衡通过倾向于吸收或释放热量的方向来应对温度的压力，这完全符合原理的预测。

5. Effect of a Catalyst | 催化剂的影响

A catalyst provides an alternative reaction pathway with a lower activation energy. Crucially, it increases the rate of both the forward and reverse reactions by exactly the same amount.

催化剂提供了一条活化能更低的替代反应路径。关键之处在于，它使正反应和逆反应的速率以完全相同的幅度提高。

As a result, a catalyst has no effect on the position of equilibrium or on the value of Kc. It simply allows the system to reach equilibrium more quickly.

因此，催化剂不会影响平衡位置，也不会改变Kc值。它的作用仅仅是让系统更快达到平衡。

In AQA mark schemes, you must explicitly state that a catalyst does not alter the equilibrium yield but does reduce the time taken to attain equilibrium. This is often tested together with the Contact or Haber process.

在AQA评分标准中，必须明确写出催化剂不会改变平衡产率，但会缩短到达平衡所需的时间。这一点常常与接触法或哈伯法一起考查。

6. Relationship with Equilibrium Constant Kc | 与平衡常数Kc的关系

The equilibrium constant Kc for a given reaction is only affected by temperature. Changes in concentration or pressure can shift the equilibrium position, but the ratio of product concentrations to reactant concentrations at equilibrium (when raised to appropriate powers) stays constant as long as the temperature remains unchanged.

某一反应的平衡常数Kc只受温度影响。浓度或压强的变化可以移动平衡位置，但平衡时产物浓度与反应物浓度的比（取适当幂次）只要温度不变就始终保持恒定。

After a concentration disturbance, the system readjusts the individual concentrations so that the same Kc value is restored. This is a powerful way to distinguish the position of equilibrium (how far the reaction has gone) from Kc (a constant at fixed temperature).

在浓度扰动之后，系统会重新调整各组分的浓度，使得恢复原先相同的Kc值。这提供了一个强有力的区隔：平衡位置（反应进行到多远）与Kc（在固定温度下的常数）是不同的概念。

When a pressure change compresses a gas mixture, if the number of gas moles differs, the equilibrium shifts to maintain Kp (or Kc through a similar logic), even though the expression for Kc doesn’t always show an obvious pressure term; the connection is more subtle but the principle still holds.

当压强变化压缩气体混合物时，如果气体分子数不同，平衡会移动以维持Kp（或通过类似逻辑维持Kc），虽然Kc表达式中不一定明显包含压强项，但这种联系更为精妙，而原理仍然成立。

7. Industrial Application: The Haber Process | 工业应用：哈伯法

The Haber process for ammonia synthesis is the quintessential Le Chatelier application: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹.

哈伯法合成氨是勒夏特列原理的经典应用：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹。

Because the forward reaction is exothermic, a low temperature would favour a high equilibrium yield of ammonia. However, at low temperatures, the rate of reaction is unacceptably slow. A compromise temperature of around 400–450 °C is used to balance yield with economic rate.

由于正反应是放热的，低温有利于获得高的氨平衡产率。但低温下反应速率过慢，无法接受。实际选用约400–450 °C的折中温度，以兼顾产率和经济速率。

The reaction reduces the number of gas molecules from 4 to 2, so a high pressure drives the equilibrium to the right. A pressure of about 200 atm is employed, constrained by the cost and safety of building stronger vessels.

该反应使气体分子数从4减至2，因此高压能推动平衡向右移动。实际操作压力约为200 atm，这受到建造更高强度容器的成本和安全性的限制。

An iron catalyst is used to lower the activation energy, speeding up both forward and reverse reactions without affecting the equilibrium position. The gases are continuously recycled to improve overall atom economy.

使用铁催化剂来降低活化能，同时加速正、逆反应，而不影响平衡位置。气体被不断循环，以提高总体的原子经济性。

8. Industrial Application: The Contact Process | 工业应用：接触法

The Contact process produces sulfur trioxide, a key step in sulfuric acid manufacture: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹.

接触法生产三氧化硫，是硫酸制造中的关键步骤：2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹。

Once again the forward reaction is exothermic, so a low temperature shifts equilibrium right, but in practice a temperature of about 450 °C is used together with a vanadium(V) oxide catalyst (V₂O₅) to achieve a viable rate.

同样，正反应是放热的，因此低温使平衡右移，但实际操作中约450 °C的温度与五氧化二钒催化剂（V₂O₅）结合使用，以获得可接受的速率。

Since 3 moles of gas form 2 moles, high pressure favours SO₃ production, but the position is already well to the right at atmospheric pressure, so the process typically operates just above 1 atm to reduce capital costs.

由于3分子气体生成2分子，高压有利于SO₃的生成，但在常压下平衡位置已非常偏右，因此该过程通常在略高于1 atm的条件下操作，以降低设备成本。

9. Other Key Equilibrium Systems | 其他重要平衡体系

AQA frequently uses the dichromate(VI)–chromate(VI) equilibrium: 2CrO₄²⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l). Adding acid (H⁺) shifts the equilibrium right, turning the solution from yellow to orange; adding alkali removes H⁺ and shifts left, restoring yellow.

AQA经常考查重铬酸根–铬酸根的平衡：2CrO₄²⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l)。加入酸（H⁺）使平衡右移，溶液由黄色变为橙色；加入碱则消耗H⁺，平衡左移，恢复黄色。

The hydration of ethene to ethanol is also significant: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) ΔH = −45 kJ mol⁻¹. High pressure and relatively low temperature improve yield, but a phosphoric acid catalyst is used to increase rate at a compromise temperature of 300 °C and 60–70 atm.

乙烯水合制乙醇也颇具代表性：C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) ΔH = −45 kJ mol⁻¹。高压和较低温度能提高产率，但实际上使用磷酸催化剂，在折中温度300 °C和60–70 atm下提高反应速率。

Another exam-favourite is the formation of a blood-red complex: Fe³⁺(aq) + NCS⁻(aq) ⇌ [FeNCS]²⁺(aq). Any change in the concentration of the reactants or product can be followed by observing colour intensity.

另一个常考内容就是血红色配合物的形成：Fe³⁺(aq) + NCS⁻(aq) ⇌ [FeNCS]²⁺(aq)。反应物或产物浓度的任何变化都可以通过颜色深浅来追踪。

10. Common Misconceptions | 常见误区

Many students confuse the rate of reaction with the position of equilibrium. A catalyst increases the rate but does not change the yield at equilibrium. Similarly, adding more solid reactant (e.g., CaCO₃) does NOT shift the equilibrium because solids do not appear in the Kc expression and their concentration is effectively constant.

很多学生混淆了反应速率与平衡位置这两个概念。催化剂能提高速率，但不改变平衡时的产率。同样，加入更多固体反应物（如CaCO₃）并不会移动平衡，因为固体不出现在Kc表达式中，其浓度实际上是常数。

Another typical error is claiming that an increase in pressure always favours the forward reaction. This only happens if the forward reaction produces fewer gaseous molecules. If the numbers are equal, pressure has no effect.

另一个典型错误是声称增大压强总是有利于正反应。只有当正反应生成的气体分子数更少时才会如此。如果分子数相等，压强就不起作用。

Students also incorrectly believe that diluting an aqueous equilibrium will have no effect. In fact, adding water reduces all concentrations equally, but the system must respond according to the stoichiometry; for the FeNCS²⁺ equilibrium, dilution favours the side with more particles (fewer products), shifting left.

学生也会错误地认为稀释水溶液平衡不会产生任何影响。实际上，加水会等比例降低所有离子的浓度，但系统必须按照化学计量数作出响应；对于FeNCS²⁺平衡体系，稀释有利于总粒子数更多的方向（产物更少），平衡左移。

11. Exam Tips and Common Question Types | 应试技巧与常见题型

Always use precise language: “the position of equilibrium shifts to the right/left” rather than “more products are formed”. When asked to explain a shift, link it directly to the opposing change in the condition.

始终使用精确的语言：”平衡位置向右/左移动”，而不是”生成了更多产物”。在解释移动时，要将其与对条件变化的抗衡直接挂钩。

If a question provides an enthalpy change, you must identify whether the forward reaction is exothermic or endothermic before predicting the temperature effect. For pressure questions, count the gaseous moles on each side of the equation.

如果题目给出了焓变，必须先在预测温度效应前判断正反应是放热还是吸热。对于压强类问题，则要数出方程式两边气体分子的数目。

AQA often uses a multi-step question: (a) state the effect of a change on equilibrium position, (b) state what happens to Kc or Kp, and (c) explain the colour change. Practise linking the three parts logically.

AQA经常以多步骤设问：(a) 指出某一变化对平衡位置的影响，(b) 说明Kc或Kp将如何变化，(c) 解释颜色变化。要练习将这三部分有逻辑地串联起来。

When drawing graphs of concentration vs time after a disturbance, the concentration of the substance causing the disturbance spikes instantly, then the curves gradually level out to new equilibrium concentrations. The examiners look for a sharp change followed by a smooth relaxation.

在画出扰动后浓度-时间图像时，造成扰动的物质浓度会瞬间垂直突变，然后曲线缓慢平滑至新的平衡浓度。考官会留意是否画出了急剧变化和随后的平滑缓和过程。

For calculation of Kc, remember that only temperature can change its magnitude. If a question says “Kc for this reaction at 500 K is 0.12”, any shift induced by concentration or pressure will eventually restore that same Kc value at 500 K.

计算Kc时要记住，只有温度能改变其数值。如果题目说”该反应在500 K时的Kc值为0.12″，那么任何因浓度或压强引发的移动最终都会使系统在500 K恢复该Kc值。

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Le Chatelier’s Principle | 勒夏特列原理