Chemical Equilibrium Position Shift and Influencing Factors | 化学平衡位置的移动及影响因素

📚 Chemical Equilibrium Position Shift and Influencing Factors | 化学平衡位置的移动及影响因素

Chemical equilibrium is a cornerstone concept in A-Level physical chemistry. When a reversible reaction is confined in a closed system, the forward and reverse reactions proceed simultaneously. Equilibrium is reached when the rates of these opposing reactions become equal, and the macroscopic properties of the system remain constant. This article systematically explains how concentration, pressure, temperature, and catalysts influence the position of equilibrium, anchored in Le Chatelier’s Principle, and how these principles are applied in the Haber and Contact processes.

化学平衡是 A-Level 物理化学中的核心概念。当可逆反应被限制在封闭体系中时，正反应与逆反应同时进行。当正逆反应速率相等时，体系达到平衡，宏观性质保持不变。本文将系统讲解浓度、压强、温度和催化剂如何影响平衡位置，以勒夏特列原理为核心，并说明这些原理在哈伯法和接触法中的实际应用。

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1. Dynamic Equilibrium | 动态平衡

A reversible reaction reaches dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. The word “dynamic” is crucial: both reactions continue to occur at the microscopic level, but there is no net change in the concentrations of reactants or products. This state can only be achieved in a closed system.

当正反应速率等于逆反应速率时，可逆反应达到动态平衡。”动态”一词至关重要：在微观层面，两个反应仍在持续进行，但反应物和生成物的浓度没有净变化。这种状态只有在封闭体系中才能实现。

Key features of dynamic equilibrium:

动态平衡的要点如下：

• Equal forward and reverse rates — not equal concentrations of reactants and products.
• Constant macroscopic properties: colour, pressure, concentration, and density remain unchanged.
• Equilibrium can be approached from either direction — starting with only reactants or only products leads to the same equilibrium mixture.

• 正逆反应速率相等——并非反应物与生成物浓度相等。
• 宏观性质恒定：颜色、压强、浓度和密度保持不变。
• 平衡可以从任意方向接近——仅由反应物或仅由生成物出发，最终得到相同的平衡混合物。

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2. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s Principle states that when a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium position shifts in the direction that tends to counteract (oppose) the imposed change. This principle provides a qualitative tool for predicting how a system responds to external disturbances.

勒夏特列原理指出：当处于平衡的系统受到浓度、压强或温度变化时，平衡位置会向抵消（对抗）该变化的方向移动。该原理为预测系统对外界干扰的响应提供了定性工具。

For example, if a system experiences an increase in temperature, the equilibrium shifts in the endothermic direction to absorb the added heat energy. If pressure increases, the system shifts towards the side with fewer moles of gas to relieve the pressure. These predictions are essential for exam questions requiring qualitative reasoning about equilibrium shifts.

例如，若系统温度升高，平衡向吸热方向移动以吸收额外热量；若压强增大，系统向气体物质的量较少的一侧移动以减轻压强。这些预测对于考试中要求定性推理平衡移动方向的题目至关重要。

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3. Effect of Concentration | 浓度的影响

Changing the concentration of a reactant or product shifts the position of equilibrium, but it does not change the value of the equilibrium constant Kc (at constant temperature).

改变反应物或生成物的浓度会使平衡位置发生移动，但在恒温条件下不会改变平衡常数 Kc 的值。

Consider the well-known equilibrium involving iron(III) thiocyanate:

考虑以下常见的硫氰酸铁平衡体系：

Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) (deep red | 深红色)

• Adding Fe³⁺ or SCN⁻ ions increases the concentration of a reactant, so the equilibrium shifts to the right, producing more FeSCN²⁺; the deep red colour intensifies.
• Adding NaOH removes Fe³⁺ as Fe(OH)₃ precipitate, decreasing the Fe³⁺ concentration, so the equilibrium shifts to the left; the red colour fades.
• Diluting the solution decreases all concentrations. In this case, there is one ion on each side, so dilution produces no shift. However, in general, diluting shifts the equilibrium towards the side with more dissolved species.

• 加入 Fe³⁺ 或 SCN⁻ 离子，增加了反应物浓度，平衡向右移动，生成更多 FeSCN²⁺，深红色加深。
• 加入 NaOH 会以 Fe(OH)₃ 沉淀形式除去 Fe³⁺，使 Fe³⁺ 浓度降低，平衡向左移动，红色变浅。
• 稀释溶液会降低所有浓度。在此反应中，两侧各有一个离子，稀释不引起移动；但一般情况下，稀释使平衡向溶解物种数较多的一侧移动。

For a general gaseous reaction aA + bB ⇌ cC + dD, increasing the concentration of A or B shifts the equilibrium right; increasing the concentration of C or D shifts it left. Removing a product also shifts the equilibrium right, which is a common industrial strategy to maximise yield.

对于一般气体反应 aA + bB ⇌ cC + dD，增加 A 或 B 的浓度使平衡右移；增加 C 或 D 的浓度使平衡左移。移走生成物同样使平衡右移，这是工业中提高产率的常用策略。

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4. Effect of Pressure | 压强的影响

Pressure changes only affect equilibria that involve at least one gaseous species. An increase in pressure shifts the equilibrium towards the side with fewer moles of gas molecules, because this reduces the total pressure. Conversely, a decrease in pressure shifts the equilibrium towards the side with more moles of gas. If the total number of gas moles is identical on both sides, a pressure change has no effect on the equilibrium position.

压强变化只影响至少含有一种气态物质的平衡。增加压强使平衡向气体分子物质的量较少的一侧移动，因为这样可以降低总压强。相反，降低压强使平衡向气体分子物质的量较多的一侧移动。若两侧气体总物质的量相等，则压强变化对平衡位置没有影响。

A classic example is the synthesis of ammonia:

合成氨是经典例子：

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

There are 4 moles of gas on the left and 2 moles on the right. Increasing the pressure shifts the equilibrium to the right, favouring the production of ammonia. This is why the Haber process operates at extremely high pressures, typically around 200 atmospheres.

左侧有 4 mol 气体，右侧有 2 mol 气体。增加压强使平衡向右移动，有利于氨的生成。这就是哈伯法在约 200 个大气压的高压下操作的原因。

Note that pressure effects are analysed using partial pressures. Doubling the total pressure doubles the partial pressure of every gaseous component, which is how the shift is driven.

注意，压强效应应通过分压来分析。总压强加倍会使每种气体组分的分压都加倍，这正是推动平衡移动的原因。

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5. Effect of Temperature | 温度的影响

Temperature has a unique and crucial effect on equilibrium: it changes the value of the equilibrium constant itself. An increase in temperature shifts the equilibrium in the endothermic direction, while a decrease in temperature shifts it in the exothermic direction.

温度对平衡有独特且关键的影响：它改变平衡常数本身的值。升高温度使平衡向吸热方向移动，降低温度使平衡向放热方向移动。

Consider the exothermic Haber process:

考虑放热的哈伯法：

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

• Increasing temperature: the system absorbs heat by favouring the reverse (endothermic) reaction; NH₃ yield decreases.
• Decreasing temperature: the system releases heat by favouring the forward (exothermic) reaction; NH₃ yield increases.

• 升高温度：系统通过促进逆向（吸热）反应来吸收热量；NH₃ 产率降低。
• 降低温度：系统通过促进正向（放热）反应来释放热量；NH₃ 产率升高。

However, a lower temperature also lowers the rate of reaction. In industry, a compromise temperature (~450°C) is chosen to achieve a reasonable rate without sacrificing too much yield. This compromise between thermodynamics and kinetics is a classic exam theme.

然而，较低的温度也会降低反应速率。在工业中，选择折中温度（约 450°C），在保证合理速率的同时避免产率过低。热力学与动力学之间的折中方案是经典考点。

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6. Effect of Catalyst and Inert Gas | 催化剂与惰性气体的影响

A catalyst lowers the activation energy for both the forward and reverse reactions equally, increasing both rates by the same factor. Therefore, a catalyst does NOT change the position of equilibrium, nor does it change the value of Kc or Kp. It simply enables the system to reach equilibrium faster.

催化剂同等程度地降低正反应和逆反应的活化能，使两个反应的速率以相同倍数增大。因此，催化剂不会改变平衡位置，也不会改变 Kc 或 Kp 的值。它只是帮助体系更快地达到平衡。

The effect of adding an inert gas depends on the conditions:

加入惰性气体的影响取决于条件：

• Constant volume: adding an inert gas increases total pressure but does not change partial pressures of the reacting gases. No shift in equilibrium position.
• Constant pressure: the container volume expands, decreasing the partial pressures of all reacting gases. The equilibrium shifts towards the side with more moles of gas.

• 恒容条件：加入惰性气体使总压强增大，但不改变反应气体的分压。平衡位置不变。
• 恒压条件：容器体积增大，所有反应气体的分压降低。平衡向气体物质的量较多的一侧移动。

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7. Reaction Quotient Q vs Kc | 反应商 Q 与 Kc 的比较

The reaction quotient Q is calculated using the same expression as Kc, but with concentrations that may not be at equilibrium. Comparing Q to Kc predicts the direction of net movement:

反应商 Q 使用与 Kc 相同的表达式计算，但采用可能未达平衡的浓度。比较 Q 与 Kc 可以预测净移动方向：

If Q < Kc: reaction proceeds to the right (forward)　|　若 Q < Kc：反应向右（正向）进行

If Q > Kc: reaction proceeds to the left (reverse)　|　若 Q > Kc：反应向左（逆向）进行

If Q = Kc: the system is at equilibrium　|　若 Q = Kc：体系处于平衡

For example, for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if extra N₂ is injected so that Q becomes smaller than Kc, the system must shift right to restore equilibrium. The power of this method is that it provides a quantitative prediction that complements Le Chatelier’s qualitative reasoning.

例如，对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，若额外注入 N₂，使 Q 小于 Kc，则系统必须向右移动以恢复平衡。这种方法的价值在于提供定量预测，与勒夏特列原理的定性推理互为补充。

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8. Industrial Applications | 工业应用

The principles of equilibrium shift are directly applied in several large-scale industrial processes.

平衡移动原理在多个大规模工业过程中得到直接应用。

Haber Process | 哈伯法（合成氨）

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

• High pressure (≈200 atm) shifts equilibrium right because there are fewer gas moles on the product side.
• Low temperature favours the exothermic forward reaction, but a moderate temperature of ~450°C is used to maintain an adequate rate.
• An iron catalyst accelerates the attainment of equilibrium without altering its position.
• NH₃ is continuously liquefied and removed, shifting the equilibrium further to the right and increasing the overall yield.

• 高压（约 200 atm）使平衡右移，因为产物侧气体物质的量更少。
• 低温有利于放热的正反应，但为了保持足够的速率，实际采用约 450°C 的中等温度。
• 铁催化剂加速平衡的到达，但不改变平衡位置。
• 氨被持续液化并移走，使平衡进一步右移，提高总产率。

Contact Process | 接触法（制硫酸）

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)　ΔH = −197 kJ mol⁻¹

• V₂O₅ is used as a catalyst; it does not affect the equilibrium position.
• A temperature of ~450°C balances yield and rate.
• Excess oxygen (from air) is used to drive the equilibrium to the right by increasing the concentration of a reactant, thereby improving the conversion of SO₂ to SO₃.

• 使用 V₂O₅ 作催化剂，不影响平衡位置。
• 约 450°C 的温度兼顾产率与速率。
• 使用过量氧气（来自空气），通过增加反应物浓度推动平衡右移，提高 SO₂ 转化为 SO₃ 的转化率。

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9. Summary Table of Factors Affecting Equilibrium | 影响平衡的各因素总结表

Factor | 因素	Change | 变化	Direction of Shift | 移动方向	Effect on Kc | 对 Kc 的影响
Concentration | 浓度	Increase reactant | 增加反应物	Towards products | 向生成物	No change | 不变
Pressure | 压强	Increase | 增大	Towards fewer gas moles | 向气体物质的量少的一侧	No change | 不变
Temperature | 温度	Increase | 升高	Towards endothermic direction | 向吸热方向	Changes | 改变
Catalyst | 催化剂	Addition | 加入	No shift | 不移动	No change | 不变
Inert gas (constant V) | 惰性气体（恒容）	Addition | 加入	No shift | 不移动	No change | 不变

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10. Worked Example | 典型例题

Question | 题目： The reaction below is at equilibrium in a sealed flask.

Question | 题目： 以下反应在密封烧瓶中达到平衡。

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)　ΔH = −197 kJ mol⁻¹

(a) Predict the effect of increasing the pressure on the equilibrium position and explain your reasoning.

(a) 预测增大压强对平衡位置的影响，并解释原因。

ANSWER | 答案：Increasing the pressure shifts the equilibrium to the right. There are 3 moles of gas on the left (2SO₂ + 1O₂) and only 2 moles on the right (2SO₃). According to Le Chatelier’s Principle, the system shifts towards the side with fewer gas moles to reduce the pressure, which is the product side.

答案：增大压强使平衡向右移动。左侧共有 3 mol 气体（2SO₂ + 1O₂），右侧只有 2 mol 气体（2SO₃）。根据勒夏特列原理，系统向气体物质的量较少的一侧移动以减小压强，即生成物一侧。

(b) Predict the effect of increasing the temperature on the equilibrium position and on the value of Kc.

(b) 预测升高温度对平衡位置及 Kc 值的影响。

ANSWER | 答案：Since the forward reaction is exothermic (ΔH = −197 kJ mol⁻¹), the reverse reaction is endothermic. Increasing the temperature shifts the equilibrium to the left, towards the endothermic direction. Because the equilibrium constant depends on temperature, Kc will decrease as the position shifts towards reactants.

答案：由于正反应是放热的（ΔH = −197 kJ mol⁻¹），逆反应为吸热反应。升高温度使平衡向左（吸热方向）移动。由于平衡常数依赖于温度，随着平衡向反应物方向移动，Kc 将减小。

(c) A vanadium(V) oxide catalyst is added. State the effect on the equilibrium position and on the rate of attainment of equilibrium.

(c) 加入五氧化二钒催化剂。说明其对平衡位置和达到平衡速率的影响。

ANSWER | 答案：The catalyst has no effect on the equilibrium position; the same equilibrium mixture is reached. It lowers the activation energy for both forward and reverse reactions, so equilibrium is reached more quickly.

答案：催化剂对平衡位置没有影响；最终仍得到相同的平衡混合物。它同时降低正逆反应的活化能，因此使体系更快达到平衡。

(d) The volume of the flask is halved. Predict how Q compares with Kc immediately after the change.

(d) 将烧瓶体积减半。预测变化后瞬间 Q 与 Kc 的关系。

ANSWER | 答案：Halving the volume doubles the concentrations of all gases. For this reaction, Q increases: Q = [SO₃]² / ([SO₂]²[O₂]). Since concentrations double, [SO₃]² quadruples while [SO₂]²[O₂] octuples (2² × 2 = 8), so Q becomes smaller than Kc. The system must shift forward (right) to restore equilibrium.

答案：体积减半使所有气体浓度加倍。对本反应，Q 发生变化：Q = [SO₃]² / ([SO₂]²[O₂])。由于浓度加倍，分子 [SO₃]² 变为原来的 4 倍，分母 [SO₂]²[O₂] 变为原来的 8 倍（2² × 2 = 8），因此 Q 小于 Kc。系统必须正向（向右）移动以恢复平衡。

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