Chemical Equilibrium for IGCSE WJEC Chemistry | IGCSE WJEC 化学平衡考点精讲

📚 Chemical Equilibrium for IGCSE WJEC Chemistry | IGCSE WJEC 化学平衡考点精讲

In the IGCSE WJEC Chemistry syllabus, chemical equilibrium is a vital topic that links the concepts of reversible reactions and industrial process optimisation. A thorough understanding of dynamic equilibrium and Le Chatelier’s principle enables you to predict how changes in conditions affect the position of equilibrium, and to explain the operating conditions of the Haber and Contact processes.

在 IGCSE WJEC 化学考纲中，化学平衡是连接可逆反应概念与工业过程优化的重要主题。透彻理解动态平衡和勒夏特列原理，可以帮助你预测条件变化如何影响平衡位置，并解释哈伯法和接触法的操作条件。

1. What is a Reversible Reaction? | 什么是可逆反应？

A reversible reaction is one in which the products can react under the same or different conditions to re-form the original reactants. It is represented in equations by the symbol ⇌.

可逆反应是指生成物在相同或不同条件下可以重新反应，再次生成原来的反应物。在方程式中用符号 ⇌ 表示。

A classic laboratory example is the thermal decomposition of ammonium chloride. When heated, solid NH₄Cl decomposes into ammonia and hydrogen chloride gases, which recombine on cooling to form the solid again: NH₄Cl(s) ⇌ NH₃(g) + HCl(g).

一个经典的实验室例子是氯化铵的热分解。加热时，固态 NH₄Cl 分解成氨气和氯化氢气体；冷却时它们又重新化合为固体：NH₄Cl(s) ⇌ NH₃(g) + HCl(g)。

Another common example is hydrated cobalt(II) chloride, which changes colour with water content: CoCl₂·6H₂O(s) (pink) ⇌ CoCl₂(s) (blue) + 6H₂O(g).

另一个常见例子是水合氯化钴的变色反应：CoCl₂·6H₂O(s) (粉红色) ⇌ CoCl₂(s) (蓝色) + 6H₂O(g)。

2. Dynamic Equilibrium | 动态平衡

Dynamic equilibrium is reached in a closed system when the rate of the forward reaction becomes equal to the rate of the backward reaction. At this point the concentrations of reactants and products remain constant, but both reactions are still occurring at the molecular level.

动态平衡是在封闭体系中，当正反应速率与逆反应速率相等时达到的状态。此时，反应物和生成物的浓度保持恒定，但在分子水平上两个方向的反应仍在进行。

Equilibrium can only be established in a closed system, where no matter is allowed to enter or leave. If the system is open, gases may escape, so the backward reaction cannot proceed and equilibrium is never reached.

平衡只能在封闭体系中建立，即没有物质进入或离开。如果体系是开放的，气体可能逸出，逆反应无法进行，平衡也就无法达到。

It is important to note that equilibrium can be approached from either direction. Starting with only reactants or only products, the same equilibrium mixture will eventually be obtained under the same conditions.

需要特别注意的是，平衡可以从两个方向到达。无论从反应物开始还是从生成物开始，在相同条件下最终都会得到相同的平衡混合物。

3. Features of a System at Equilibrium | 平衡体系的特征

At equilibrium, the following macroscopic properties remain constant: colour (if coloured species are present), pressure (for gaseous systems), concentration of each species, and overall mass. However, the reaction is dynamic – particles continue to react in both directions.

处于平衡时，以下宏观性质保持恒定：颜色（若有颜色物种）、压强（对气体体系）、各物质浓度以及总质量。但反应是动态的——微粒仍在两个方向上持续反应。

For a system at equilibrium, the rate of the forward reaction equals the rate of the backward reaction. This does not mean that the concentrations of reactants and products are equal; they simply stay constant once equilibrium is established.

对于平衡体系，正反应速率等于逆反应速率。这并不意味着反应物和生成物的浓度相等；它们只是在平衡建立后保持恒定而已。

Temperature must remain constant for the equilibrium to persist. Any temperature change will disturb the equilibrium because the rates of the forward and backward reactions are affected differently.

温度必须保持不变，平衡才能维持。任何温度变化都会扰乱平衡，因为正逆反应的速率受到的影响不同。

4. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states that if a system at dynamic equilibrium is subjected to a change in concentration, temperature or pressure, the position of equilibrium will shift in such a way as to oppose the imposed change.

勒夏特列原理指出，如果处于动态平衡的体系受到浓度、温度或压强方面的变化，平衡位置将向着削弱这种变化的方向移动。

This principle provides a qualitative tool for predicting the direction in which the equilibrium will shift. It does not tell us how fast the new equilibrium will be reached, nor does it give quantitative values for the new composition.

这个原理提供了一个定性工具来预测平衡移动的方向。它不告诉我们达到新平衡需要多快，也不给出新组成的定量数值。

Using Le Chatelier’s principle, we can maximise the yield of a desired product in an industrial process by adjusting the reaction conditions, although economic and practical factors must also be considered.

利用勒夏特列原理，我们可以通过调整反应条件来提高工业过程中目标产物的产率，但同时也必须考虑经济性和实际因素。

5. The Effect of Concentration | 浓度变化的影响

If the concentration of a reactant is increased, the equilibrium shifts to the right (forward direction) to reduce the concentration of the added substance. This increases the yield of products.

如果增加某种反应物的浓度，平衡将向右（正向）移动以降低该加入物质的浓度，从而提高产物的产率。

Conversely, if a product is removed from the equilibrium mixture (for example, by continuous extraction in a flow process), the equilibrium shifts to the right to produce more of that product. This is exploited in the Haber process by condensing out the ammonia formed.

相反，如果从平衡混合物中移走某种产物（例如在连续流工艺中不断提取），平衡将向右移动，以生成更多该产物。哈伯法中就是通过将生成的氨冷凝分离来利用这一原理。

A visible example is the equilibrium between iron(III) ions and thiocyanate ions: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). The product is a deep red complex. Adding more Fe³⁺ or SCN⁻ deepens the colour as the equilibrium shifts to the right.

一个可视化的例子是铁(III)离子与硫氰酸根离子之间的平衡：Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)。产物是一种深红色配合物。加入更多的 Fe³⁺ 或 SCN⁻ 会使颜色加深，因为平衡向右移动。

6. The Effect of Temperature | 温度变化的影响

Temperature changes alter the equilibrium position because the forward and backward reactions have different enthalpy changes. If the forward reaction is exothermic (releases heat), then the backward reaction is endothermic (absorbs heat).

温度变化会改变平衡位置，因为正逆反应的焓变不同。如果正反应是放热的，那么逆反应就是吸热的。

According to Le Chatelier’s principle, increasing the temperature shifts the equilibrium in the endothermic direction – that is, the direction that absorbs heat. So, for an exothermic forward reaction, raising the temperature will shift the equilibrium to the left, decreasing the product yield.

根据勒夏特列原理，升高温度会使平衡向吸热方向移动，也就是向吸收热量的方向移动。因此，对于正向放热的反应，升温会使平衡向左移动，降低产物的产率。

A good illustrative system is the N₂O₄ / NO₂ equilibrium: N₂O₄(g, colourless) ⇌ 2NO₂(g, brown), where the forward reaction is endothermic. Warming the sealed tube intensifies the brown colour because the equilibrium shifts right to absorb the added heat.

一个很好的说明体系是 N₂O₄ / NO₂ 平衡：N₂O₄(g, 无色) ⇌ 2NO₂(g, 棕色)，其中正反应吸热。加热密封管会使棕色加深，因为平衡向右移动以吸收增加的热量。

In industrial processes such as the Haber process (exothermic forward reaction), a compromise temperature of about 450°C is used. Lower temperatures would give a higher equilibrium yield but the rate would be too slow to be economical.

在哈伯法这类工业过程中（正向放热），采用了大约 450 °C 的折中温度。更低的温度虽然能提供更高的平衡产率，但反应速率太慢，不具备经济性。

7. The Effect of Pressure (Gaseous Systems) | 压强变化的影响（气体反应）

Pressure changes only affect equilibria that involve gases, and only when there is a change in the total number of gas molecules between reactants and products. If the number of gas molecules is the same on both sides, pressure has no effect on the equilibrium position.

压强变化只影响有气体参与的平衡，且仅当反应物和生成物之间气体分子总数发生改变时才起作用。如果两边气体分子数相同，压强不影响平衡位置。

Increasing the pressure shifts the equilibrium towards the side with fewer gas molecules, as this helps to reduce the pressure. Decreasing the pressure has the opposite effect, favouring the side with more gas molecules.

增大压强会使平衡向气体分子数较少的方向移动，因为这有助于降低压强。减小压强则产生相反的效果，有利于气体分子数较多的一侧。

For example, in the Haber process reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), there are 4 moles of gas on the left and 2 moles on the right. High pressure therefore increases the yield of ammonia. An operating pressure of around 200 atm is used, as even higher pressures would require very costly equipment and pose safety risks.

例如，哈伯法反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 中，左边有 4 摩尔气体，右边有 2 摩尔气体。因此高压有利于提高氨的产率。实际操作中使用约 200 atm，因为更高的压强需要非常昂贵的设备并带来安全风险。

In the Contact process, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the left side has 3 moles of gas and the right side has 2 moles. A moderate pressure would theoretically favour SO₃ production, but at the temperature used the conversion is already high enough at atmospheric pressure, so operating at high pressure is not normally required.

在接触法中，2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，左边有 3 摩尔气体，右边有 2 摩尔。中等压强理论上有利于 SO₃ 的生成，但在操作温度下常压转化率已经足够高，因此通常不需要高压操作。

8. The Role of a Catalyst | 催化剂的作用

A catalyst speeds up both the forward and backward reactions equally by providing an alternative reaction pathway with a lower activation energy. It does not alter the position of equilibrium, so it does not increase the equilibrium yield.

催化剂通过提供一条活化能更低的新反应路径，同等程度地加快正反应和逆反应的速率。它不会改变平衡位置，因此不能提高平衡产率。

However, a catalyst enables a system to reach equilibrium more quickly. In industry this is extremely valuable because it allows a process to operate at a lower temperature while still achieving a reasonable rate, saving energy and making the process economically viable.

然而，催化剂可以使体系更迅速地达到平衡。在工业上，这极具价值，因为它允许工艺在较低温度下运行的同时仍能达到可观的速率，从而节省能源并使过程经济可行。

For example, iron in the Haber process and vanadium(V) oxide (V₂O₅) in the Contact process are used as catalysts. Neither changes the final percentage of ammonia or sulfur trioxide at equilibrium; they simply reduce the time needed to get there.

例如，哈伯法中的铁和接触法中的五氧化二钒（V₂O₅）都被用作催化剂。它们都不改变氨或三氧化硫的平衡百分含量，只是缩短了达到平衡所需的时间。

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

The Haber process synthesises ammonia from nitrogen and hydrogen: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The forward reaction is exothermic (ΔH = –92 kJ mol⁻¹). Nitrogen is obtained from the air, and hydrogen is usually derived from natural gas (methane) and steam.

哈伯法用氮气和氢气合成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。正反应放热（ΔH = –92 kJ mol⁻¹）。氮气来源于空气，氢气通常来自天然气（甲烷）与水蒸气。

The optimum conditions chosen for the process are a temperature of about 450 °C, a pressure of approximately 200 atm, and an iron catalyst. These conditions represent a compromise between yield, rate, and cost.

该工艺选择的最佳条件是：温度约 450 °C，压强约 200 atm，铁作为催化剂。这些条件代表了产率、速率和成本之间的折中。

Temperature: A lower temperature would give a higher equilibrium yield of ammonia (since the forward reaction is exothermic), but the rate would be unacceptably slow. 450 °C gives a reasonable yield with a moderate rate.

温度：较低的温度会提供更高的氨平衡产率（因为正反应放热），但速率会慢得无法接受。450 °C 在合理的速率下给出了可接受的产率。
Pressure: A higher pressure shifts the equilibrium to the right (fewer gas molecules) and also increases the rate. However, very high pressures demand very thick pipework and expensive compressors. 200 atm is a practical upper limit.

压强：高压使平衡向右移动（气体分子数减少的方向），还会加快速率。但极高的压强需要很厚的管道和昂贵的压缩机。200 atm 是一个实际的压强上限。
Catalyst: The iron catalyst allows the equilibrium to be reached quickly at the chosen temperature, without affecting the equilibrium mixture composition.

催化剂：铁催化剂使平衡在所选温度下迅速达到，但不影响平衡混合物的组成。

Ammonia is continuously removed from the reaction mixture by condensation, and unreacted N₂ and H₂ are recycled, which helps to shift the equilibrium further to the right.

氨通过冷凝从反应混合物中不断移走，未反应的 N₂ 和 H₂ 被循环利用，这有助于进一步使平衡向右移动。

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

The Contact process is used to manufacture sulfuric acid. The key equilibrium step is the oxidation of sulfur dioxide: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The forward reaction is exothermic (ΔH ≈ –98 kJ mol⁻¹). Sulfur dioxide is obtained by burning sulfur or roasting sulfide ores, and oxygen is supplied from air.

接触法用于生产硫酸。关键的平衡步骤是二氧化硫的氧化：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)。正反应放热（ΔH ≈ –98 kJ mol⁻¹）。二氧化硫通过燃烧硫磺或焙烧硫化物矿石获得，氧气由空气提供。

The typical operating conditions are a temperature around 450 °C, a pressure close to 1 atm (atmospheric), and a vanadium(V) oxide (V₂O₅) catalyst.

典型的操作条件是：温度约 450 °C，压强接近 1 atm（常压），采用五氧化二钒（V₂O₅）催化剂。

Temperature: As with the Haber process, a low temperature would give a higher yield of SO₃, but the rate decreases. 450 °C is a compromise that gives a sufficient rate and an economically viable conversion (over 99% when a catalyst is used).

温度：与哈伯法类似，低温能提供更高的 SO₃ 产率，但速率下降。450 °C 是折中选择，配合催化剂能达到足够的速率和经济上可行的转化率（使用催化剂时超过 99%）。
Pressure: In theory, increasing the pressure would favour the side with fewer gas molecules (SO₃). However, under the chosen conditions the equilibrium already lies heavily to the right, so using a high pressure is unnecessary and would add to equipment costs.

压强：理论上，增大压强将有利于气体分子数较少的 SO₃ 一侧。然而，在所选条件下平衡已经强烈偏向右边，因此高压操作没有必要，只会增加设备成本。
Catalyst: V₂O₅ provides an alternative pathway with lower activation energy, dramatically increasing the rate of attainment of equilibrium without changing the equilibrium composition.

催化剂：V₂O₅ 提供了活化能更低的反应路径，显著加快达到平衡的速率，而不改变平衡组成。

Excess oxygen is often used to drive the equilibrium further to the right, increasing the conversion of SO₂ to SO₃. The SO₃ produced is then absorbed in concentrated sulfuric acid to form oleum, which is later diluted.

实际操作中通常使用过量的氧气来推动平衡进一步向右移动，提高 SO₂ 转化为 SO₃ 的程度。生成的 SO₃ 随后被浓硫酸吸收形成发烟硫酸，再稀释得到所需浓度的硫酸。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Chemical Equilibrium for IGCSE WJEC Chemistry | IGCSE WJEC 化学平衡考点精讲