📚 Changing the Position of Equilibrium | 改变化学平衡的位置
Chemical equilibrium is a dynamic state in which the forward and reverse reactions proceed at exactly the same rate, so the macroscopic concentrations of reactants and products remain constant. Understanding how to shift this balance is a cornerstone of physical chemistry at A-Level, with powerful applications from industrial synthesis to biological systems.
化学平衡是一种动态状态，此时正向反应和逆向反应的速率完全相同，因此反应物和产物的宏观浓度保持不变。理解如何移动这一平衡是A-Level物理化学的核心知识，在工业合成和生物体系中有着广泛的应用。
1. Dynamic Equilibrium and Le Chatelier’s Principle | 动态平衡与勒夏特列原理
A reversible reaction reaches dynamic equilibrium when the rates of the forward and reverse reactions become equal and the concentrations of all species stop changing over time. The system must be closed, and the equilibrium can be approached from either direction.
当正向反应和逆向反应的速率相等，且所有物种的浓度不再随时间变化时，可逆反应便达到了动态平衡。体系必须密闭，并且平衡可以从任意方向趋近。
Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure (for gases) or temperature, the position of equilibrium will shift in the direction that tends to counteract the imposed change.
勒夏特列原理指出，如果处于平衡的体系受到浓度、压力（对气体而言）或温度的改变，平衡位置就会朝着抵消该改变的方向移动。
2. Effect of Changing Concentration | 改变浓度的影响
Increasing the concentration of a reactant shifts the equilibrium to the right, favouring the forward reaction to produce more products until a new equilibrium is established. Conversely, increasing the concentration of a product shifts the equilibrium to the left, favouring the reverse reaction.
增加反应物的浓度会使平衡向右移动，有利于正向反应生成更多产物，直至达到新的平衡。相反，增加产物的浓度会使平衡向左移动，有利于逆向反应。
A classic colour-change demonstration uses the iron(III) thiocyanate equilibrium: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). Adding Fe³⁺ ions deepens the blood-red colour because the equilibrium shifts right, making more of the complex. Removing Fe³⁺ (e.g. by precipitation) shifts the equilibrium left and the solution becomes paler.
一个经典的颜色变化演示使用了铁(III)硫氰酸根平衡：Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)。加入Fe³⁺离子会使血红色加深，因为平衡向右移动，生成了更多的配合物。移除Fe³⁺（如通过沉淀）则会使平衡向左移动，溶液颜色变浅。
In industry, continuously removing a product (e.g. condensing ammonia in the Haber process) keeps the equilibrium shifted towards the product side, driving the reaction forward.
在工业中，持续移除产物（如哈柏法中冷凝氨）能够使平衡不断向产物一侧移动，推动反应正向进行。
3. Effect of Changing Pressure | 改变压力的影响
Pressure changes only affect equilibria involving gases where there is a difference in the total number of gas molecules on the two sides of the equation. Increasing pressure favours the side with fewer moles of gas, while decreasing pressure favours the side with more moles of gas.
压力的改变只影响有气体参与且反应方程式两边气体总分子数不同的平衡。增大压强有利于气体分子数较少的一侧，减小压强则有利于气体分子数较多的一侧。
For the dimerisation of nitrogen dioxide: 2NO₂(g) ⇌ N₂O₄(g), there are 2 moles of gas on the left and 1 mole on the right. Increasing the pressure shifts the equilibrium to the right, forming more N₂O₄ and the brown colour fades. If the pressure is reduced, the equilibrium shifts left and the brown colour intensifies.
对于二氧化氮的二聚反应：2NO₂(g) ⇌ N₂O₄(g)，左边有2摩尔气体，右边有1摩尔。增大压强，平衡向右移动，生成更多的N₂O₄，棕色变浅。如果减小压强，平衡向左移动，棕色加深。
When there is no change in the number of gas molecules, such as H₂(g) + I₂(g) ⇌ 2HI(g), altering the pressure has no effect on the position of equilibrium. The partial pressures of all components change in the same ratio, so the system remains undisturbed.
当气体分子总数没有变化时，如H₂(g) + I₂(g) ⇌ 2HI(g)，改变压强对平衡位置没有影响。所有组分的分压按相同比例变化，体系保持不受干扰。
4. Effect of Changing Temperature | 改变温度的影响
Temperature is the only variable that changes the value of the equilibrium constant, Kc. If the forward reaction is exothermic (ΔH negative), increasing the temperature shifts the equilibrium to the left, favouring the endothermic reverse reaction and decreasing the equilibrium yield of products. Lowering the temperature favours the exothermic forward reaction and shifts the equilibrium to the right.
温度是唯一能够改变平衡常数Kc数值的因素。如果正向反应放热（ΔH为负），升高温度会使平衡向左移动，有利于吸热的逆向反应，降低产物的平衡产率。降低温度则有利于放热的正向反应，平衡向右移动。
Consider the synthesis of ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹. Heating the mixture reduces the equilibrium concentration of NH₃ because the system absorbs some of the added heat by shifting to the left. Cooling the system increases the proportion of ammonia, though the rate becomes too slow for practical use.
以合成氨为例：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹。加热混合物会降低NH₃的平衡浓度，因为体系通过向左移动吸收了部分热量。冷却体系则会增加氨的比例，但反应速率变得太慢，无法实际应用。
For an endothermic forward reaction, the opposite is true: raising the temperature shifts the equilibrium to the right, increasing Kc. This knowledge is essential when choosing operating conditions for industrial processes.
对于正向吸热的反应，情况相反：升高温度使平衡向右移动，Kc增大。在选择工业过程的运行条件时，这一知识点至关重要。
5. Catalysts and Equilibrium | 催化剂与平衡
A catalyst provides an alternative reaction pathway with a lower activation energy, speeding up both the forward and reverse reactions to exactly the same extent. As a result, a catalyst does not change the position of equilibrium or the equilibrium composition – it only allows the system to reach equilibrium more quickly.
催化剂提供了一条活化能较低的反应途径，同等程度地加快了正向反应与逆向反应。因此，催化剂不会改变平衡位置，也不会改变平衡组成——它只是让体系更快地达到平衡。
In an energy profile diagram, the catalyst lowers the energy of the transition state for both directions equally, leaving the enthalpy change, ΔH, and the equilibrium constant unchanged. Industrially, this means a catalyst cannot increase the maximum possible yield, but it enables a lower temperature to be used without sacrificing the rate, saving energy.
在能级图中，催化剂同等地降低了两个方向过渡态的能量，焓变ΔH和平衡常数均保持不变。在工业上，这意味着催化剂不能提高最大可能的产率，但可以在不牺牲速率的前提下使用更低的温度，从而节省能源。
6. The Equilibrium Constant, Kc | 平衡常数 Kc
For a general homogeneous reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is written as:
Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ
where the square brackets represent equilibrium concentrations in mol dm⁻³. Pure solids and liquids are omitted from the expression because their concentrations are effectively constant.
对于一般的均相反应 aA + bB ⇌ cC + dD，用浓度表示的平衡常数写作：
Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ
方括号代表平衡浓度，单位为 mol dm⁻³。纯固体和纯液体不出现在表达式中，因为它们的浓度可视为常数。
Kc has a fixed value at a given temperature. Changing concentration or pressure may shift the position of equilibrium, but as long as the temperature stays constant, the system will re-establish the same Kc value. For example, if more A is added, the equilibrium shifts right, consuming some A and B to form more C and D. The new equilibrium concentrations still satisfy the original Kc expression.
在给定温度下，Kc具有唯一确定的值。改变浓度或压力可能会移动平衡位置，但只要温度保持不变，体系就会重新建立相同的Kc值。例如，加入更多的A，平衡向右移动，消耗部分A和B生成更多的C和D。新的平衡浓度仍然满足原来的Kc表达式。
Only a change in temperature alters Kc. For an exothermic reaction, increasing temperature decreases Kc; for an endothermic reaction, increasing temperature increases Kc. This is a very common examination point.
只有温度的变化才会改变Kc。对于放热反应，升高温度会减小Kc；对于吸热反应，升高温度会增大Kc。这是一个非常常见的考点。
7. Industrial Case Study: The Haber Process | 工业案例：哈柏法
The Haber process synthesises ammonia from nitrogen and hydrogen:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹
The equilibrium shifts to the right – producing more ammonia – at high pressure and low temperature. 4 moles of gas become 2 moles, so high pressure favours the forward reaction. Because the forward reaction is exothermic, low temperature increases the equilibrium yield.
哈柏法从氮气和氢气合成氨：
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ mol⁻¹
在高压和低温下，平衡向右移动——生成更多的氨。4摩尔气体变为2摩尔，因此高压有利于正向反应。由于正向放热，低温可以提高平衡产率。
In practice, a compromise temperature of about 450 °C is used. Although lower temperatures would give a higher equilibrium percentage of ammonia, the rate would be unacceptably slow. An iron catalyst is employed to speed up both forward and reverse reactions, allowing the equilibrium to be reached more quickly without changing its position. A pressure of around 200 atm is selected to push the equilibrium towards the products while balancing plant costs and safety.
实际上，采用约450 °C的折中温度。虽然更低的温度会带来更高的氨平衡百分含量，但反应速率会慢得无法接受。使用铁催化剂来加速正逆反应，让平衡更快达到而不改变其位置。选择约200 atm的压强以推动平衡向产物方向移动，同时兼顾设备成本和安全。
The continuous removal of ammonia by condensation also helps maintain a high net rate, as it reduces product concentration and prompts the equilibrium to shift right.
通过冷凝持续移走氨也有助于维持较高的净反应速率，因为这会降低产物浓度，促使平衡向右移动。
8. Industrial Case Study: The Contact Process | 工业案例：接触法
The Contact process produces sulfur trioxide, a key step in making sulfuric acid:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = –197 kJ mol⁻¹
There is a reduction in the number of gas molecules (3 moles → 2 moles), so high pressure would shift the equilibrium to the right. However, the forward reaction is already favoured at normal pressures to such an extent that the yield is very high – often above 99 % – so an expensive high-pressure vessel is not justified.
接触法生产三氧化硫，这是制造硫酸的关键步骤：
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = –197 kJ mol⁻¹
反应中气体分子数减少（3摩尔→2摩尔），因此高压会使平衡向右移动。但是，在常压下正向反应已经非常有利，产率非常高——通常高于99 %——因此没有必要使用昂贵的高压设备。
A temperature of around 450 °C is chosen as a compromise between equilibrium yield and rate, and a vanadium(V) oxide (V₂O₅) catalyst is used to increase the rate without affecting the position of equilibrium. The process operates at just above atmospheric pressure (1–2 atm) to maintain a sufficient flow through the reactor.
选择约450 °C的温度作为平衡产率和速率的折中，并使用五氧化二钒（V₂O₅）催化剂来提高速率而不影响平衡位置。该过程在略高于大气压（1–2 atm）下操作，以维持足够的反应器通量。
9. Key Summary for Predicting Equilibrium Shifts | 预测平衡移动的关键总结
To predict the direction in which the equilibrium will shift, always ask: ‘Which direction will absorb or oppose the applied change?’ For a concentration increase, the system shifts to consume some of the added substance. For a pressure increase (with unequal gas moles), the system shifts to the side with fewer gas molecules. For a temperature increase, the system shifts in the endothermic direction. Adding a catalyst produces no shift; it only shortens the time needed to reach equilibrium.
要预测平衡移动的方向，永远问自己：“哪个方向能够吸收或抵消施加的改变？”对于浓度增大，体系向消耗所加物质的方向移动。对于压强增大（且气体摩尔数不等），体系向气体分子数较少的一侧移动。对于温度升高，体系向吸热方向移动。加入催化剂不会引起任何移动，只会缩短达到平衡所需的时间。
It is essential to remember that Kc remains unchanged when the position of equilibrium shifts due to concentration or pressure changes, but does change when the temperature is altered. The magnitude of the shift is also limited: the system only partially counteracts the disturbance, it never fully reverses it.
必须记住，当平衡位置因浓度或压力变化而移动时，Kc保持不变；但当温度改变时，Kc也随之改变。移动的幅度也是有限的：体系只能部分抵消扰动，永远无法完全逆转。
10. Key Points for Exams | 考试要点
Be precise with terminology: ‘equilibrium shifts to the left/right’, not ‘reaction goes faster’. Always refer to the effect on the position of equilibrium, not just on the rate. When temperature changes, state how Kc is affected – for exothermic reactions, Kc decreases as temperature rises; for endothermic reactions, Kc increases.
术语要精准：“平衡向左/向右移动”，而不是“反应变快”。永远指出对平衡位置的影响，不单是速率。当温度改变时，要说明Kc如何变化——对于放热反应，温度升高Kc减小；对于吸热反应，Kc增大。
Pressure has no effect on the position of equilibrium if the number of moles of gas is the same on both sides. Catalysts increase the rate of both forward and reverse reactions equally, so they do not alter yield or Kc; they are often used to allow lower temperatures without slowing down production unacceptably.
如果反应两边气体摩尔数相同，压强对平衡位置没有影响。催化剂同等加快正逆反应速率，因此不会改变产率或Kc；工业上常使用催化剂以便采用较低温度而不致使生产速率慢得无法接受。
Finally, remember that Le Chatelier’s principle describes a response that partially counteracts a change, not one that reverses it entirely. Adding more reactant will increase its equilibrium concentration compared with the undisturbed state, even though some is consumed. This subtlety often appears in multiple-choice questions.
最后，记住勒夏特列原理所描述的响应只是部分抵消变化，而不是完全扭转。虽然
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