GCSE WJEC Chemistry: Chemical Equilibrium - Key Points Explained | GCSE WJEC 化学：化学平衡

📚 GCSE WJEC Chemistry: Chemical Equilibrium – Key Points Explained | GCSE WJEC 化学：化学平衡 – 考点精讲

In GCSE WJEC Chemistry, understanding chemical equilibrium is essential for explaining how reversible reactions behave and for applying these ideas to real industrial processes like the Haber and Contact processes. This revision guide breaks down every key concept you need to know: dynamic equilibrium, Le Chatelier’s principle, the effects of concentration, temperature, pressure and catalysts, and how these factors are managed in industry. Each section presents the core content in both English and Chinese, helping you master the topic thoroughly.

在 GCSE WJEC 化学中，理解化学平衡对于解释可逆反应的行为以及将这些思想应用于哈伯法和接触法等实际工业过程至关重要。本复习指南分解了你需要掌握的每一个关键概念：动态平衡、勒夏特列原理、浓度、温度、压力和催化剂的影响，以及这些因素在工业中如何被调控。每个部分都以中英双语呈现核心内容，帮助你透彻掌握本专题。

1. Reversible Reactions and the Equilibrium Symbol | 可逆反应与平衡符号

Not all reactions go to completion. In a reversible reaction, the products can react together to re-form the original reactants. We use the ⇌ symbol to show this, for example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

并非所有反应都能进行到底。在可逆反应中，生成物可以再次反应变回原来的反应物。我们使用 ⇌ 符号来表示，例如：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。

A closed system is required for equilibrium to be established, so that no matter can escape. In an open container, gases may leave, preventing the reverse reaction from establishing a balance.

建立平衡需要一个封闭系统，这样物质无法逸出。在开口容器中，气体可能离开，从而阻止逆反应建立平衡。

2. Dynamic Equilibrium – What It Really Means | 动态平衡的真正含义

When a reversible reaction reaches equilibrium in a closed system, it does not stop; instead, it reaches a state of dynamic equilibrium. Both forward and backward reactions continue at exactly the same rate.

当可逆反应在封闭系统中达到平衡时，反应并不会停止；相反，它达到一种动态平衡状态。正反应和逆反应继续以完全相同的速率进行。

Because the rates are equal, the concentrations of reactants and products remain constant. This does not mean they are equal in amount – just that they no longer change over time.

由于速率相等，反应物和生成物的浓度保持恒定。这并不意味着它们的数量相等——只是它们不再随时间变化。

You should be able to identify equilibrium from a concentration-time graph: after a certain time, all concentrations level off horizontally, indicating equilibrium has been reached.

你应该能够从浓度-时间图中识别平衡：经过一段时间后，所有浓度都趋于水平，表明已达到平衡。

3. Le Chatelier’s Principle – Predicting the Shift | 勒夏特列原理——预测平衡移动

Le Chatelier’s principle states: if a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose the change.

勒夏特列原理指出：处于动态平衡的系统如果受到条件变化的影响，平衡位置将移动以对抗这种变化。

This principle allows us to predict how changing concentration, temperature or pressure will affect the yield of products. The system always tries to absorb or counteract the disturbance.

这一原理使我们能够预测改变浓度、温度或压力将如何影响产物的产率。系统总是试图吸收或抵消干扰。

It is important to remember that the shift only affects the position of equilibrium, not the equilibrium constant at that temperature (though no Kc calculations are required at GCSE).

重要的是要记住，这种移动只影响平衡位置，而不影响该温度下的平衡常数（尽管 GCSE 不要求 Kc 计算）。

4. Effect of Changing Concentration | 改变浓度的影响

If you increase the concentration of a reactant, the system tries to reduce it by favouring the forward reaction, shifting equilibrium to the right and producing more products.

如果你增加反应物的浓度，系统会试图通过加强正反应来减少它，使平衡向右移动，生成更多产物。

If you increase the concentration of a product, the system favours the backward reaction, shifting equilibrium to the left and forming more reactants. Removing a substance has the opposite effect.

如果你增加生成物的浓度，系统会加强逆反应，使平衡向左移动，生成更多反应物。移除一种物质则会产生相反的效果。

These shifts are predictable and clearly demonstrate Le Chatelier’s principle in action.

这些移动是可预测的，并且清楚地展示了勒夏特列原理的作用。

For example, in the reaction N₂ + 3H₂ ⇌ 2NH₃, adding more N₂ shifts equilibrium to the right, increasing the yield of ammonia.

例如，在反应 N₂ + 3H₂ ⇌ 2NH₃ 中，增加 N₂ 会使平衡向右移动，提高氨的产率。

5. Effect of Changing Temperature | 改变温度的影响

Temperature changes affect the equilibrium position depending on whether the forward reaction is exothermic or endothermic. If the forward reaction is exothermic (releases heat), increasing temperature favours the endothermic backward reaction, shifting equilibrium to the left.

温度变化对平衡位置的影响取决于正反应是放热还是吸热。如果正反应放热（释放热量），升高温度有利于吸热的逆反应，使平衡向左移动。

Decreasing temperature favours the exothermic forward reaction, shifting equilibrium to the right and increasing product yield. So lower temperatures give a higher equilibrium yield for exothermic reactions like the Haber process, but the rate becomes too slow – a compromise is needed.

降低温度有利于放热的正反应，使平衡向右移动，提高产物的产率。因此，对于像哈伯法这样的放热反应，较低的温度会给出更高的平衡产率，但速率会变得太慢——需要折中。

You must always link the shift to the enthalpy change. Remember: ΔH is negative for exothermic reactions. The system opposes an increase in temperature by absorbing heat.

你必须始终将移动与焓变联系起来。记住：放热反应的 ΔH 为负。系统通过吸收热量来对抗温度的升高。

6. Effect of Changing Pressure (for Gaseous Systems) | 改变压力的影响（针对气体系统）

Pressure changes only affect equilibria involving gases where there is a different number of gas molecules on each side. Increasing pressure favours the side with fewer gas molecules, reducing the overall pressure.

压力变化只影响涉及气体的平衡，并且各侧的气体分子总数不同。增大压力有利于气体分子数较少的一侧，从而降低总压力。

Decreasing pressure favours the side with more gas molecules. If the number of gas molecules is the same on both sides, pressure has no effect on the equilibrium position.

降低压力有利于气体分子数较多的一侧。如果两侧的气体分子总数相同，压力对平衡位置没有影响。

In the Haber reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), there are 4 gas molecules on the left and 2 on the right. So high pressure shifts equilibrium to the right, favouring ammonia production. However, very high pressures are expensive and dangerous, so a compromise of around 200 atm is used.

在哈伯反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 中，左侧有 4 个气体分子，右侧有 2 个。因此高压使平衡向右移动，有利于氨的生产。然而，极高的压力既昂贵又危险，所以采用了约 200 个大气压的折衷方案。

7. Effect of a Catalyst | 催化剂的影响

A catalyst speeds up both the forward and backward reactions equally. As a result, it does not change the position of equilibrium or the yield of products.

催化剂同等地加快正反应和逆反应。因此，它不会改变平衡位置或产物的产率。

Its role is to lower the activation energy, allowing equilibrium to be reached faster. Industrially, this means the same amount of product can be obtained in a shorter time, which is economically very important.

它的作用是降低活化能，使平衡更快达到。在工业上，这意味着可以在更短的时间内获得相同数量的产物，这在经济上非常重要。

It is a common exam mistake to say that a catalyst increases the yield; it does not. It only increases the rate at which equilibrium is established.

这是一个常见的考试错误：说催化剂能提高产率；实际上它不能。它只加快建立平衡的速率。

8. Industrial Case Study: The Haber Process | 工业实例：哈伯法合成氨

The Haber process manufactures ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂). The reversible reaction is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The forward reaction is exothermic (ΔH = –92 kJ mol⁻¹).

哈伯法利用氮气 (N₂) 和氢气 (H₂) 生产氨 (NH₃)。可逆反应为：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。正反应是放热的 (ΔH = –92 kJ mol⁻¹)。

Chosen conditions are a compromise: temperature about 450 °C, pressure about 200 atm, and an iron catalyst. Let’s examine why.

选定的条件是折中的结果：温度约 450 °C，压力约 200 atm，使用铁催化剂。我们来检视原因。

A lower temperature would give a higher equilibrium yield because the forward reaction is exothermic, but the rate would be too slow. 450 °C gives a reasonable rate while maintaining an acceptable yield.

较低温度会带来更高的平衡产率，因为正反应是放热的，但速率会太慢。450 °C 在给出合理速率的同时保持了可接受的产率。

High pressure shifts equilibrium to the right (fewer gas molecules), increasing yield. 200 atm is a compromise between yield, plant cost and safety. An iron catalyst speeds up the reaction so that equilibrium is reached quickly.

高压使平衡向右移动（气体分子数减少），提高产率。200 atm 是产率、设备成本和安全之间的折衷。铁催化剂加快反应，使平衡迅速达到。

The ammonia is continuously removed by liquefying it, which also helps to pull the equilibrium further to the right.

通过液化将氨不断移出，这也有助于将平衡进一步拉向右侧。

9. Industrial Case Study: The Contact Process | 工业实例：接触法制硫酸

The Contact process produces sulfur trioxide (SO₃), which is used to make sulfuric acid. The key reversible step is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The forward reaction is exothermic.

接触法生产三氧化硫 (SO₃)，用于制造硫酸。关键的可逆步骤为：2SO₂(g) + O₂(g) ⇌ 2SO₃(g)。正反应放热。

Typical conditions are: temperature around 450 °C, pressure close to atmospheric (1–2 atm), and a vanadium(V) oxide (V₂O₅) catalyst. Again, a compromise temperature is used to balance rate and yield.

典型条件为：温度约 450 °C，压力接近常压 (1–2 atm)，以及五氧化二钒 (V₂O₅) 催化剂。这里同样使用折中温度来平衡速率和产率。

The reaction has 3 gas molecules on the left (2SO₂ + O₂) and 2 on the right, so increasing pressure would favour SO₃. However, the equilibrium yield is already high at low pressure, so using a high pressure is not economically justified.

反应左侧有 3 个气体分子 (2SO₂ + O₂)，右侧有 2 个，因此增大压力有利于 SO₃ 的生成。然而，低压下平衡产率已经很高，因此使用高压在经济上并不合理。

The V₂O₅ catalyst ensures a fast rate without altering the equilibrium position. Unreacted gases are recycled to improve overall efficiency.

V₂O₅ 催化剂确保了高反应速率，但不改变平衡位置。未反应的气体被循环利用，以提高整体效率。

Comparing both processes, you can see how the same principles guide industrial design. The table below summarises the key details.

通过比较这两个过程，你可以看到相同的原理如何指导工业设计。下表总结了关键细节。

Process	Reaction	Temperature	Pressure	Catalyst
Haber	N₂ + 3H₂ ⇌ 2NH₃	450 °C	200 atm	Iron
Contact	2SO₂ + O₂ ⇌ 2SO₃	450 °C	1–2 atm	V₂O₅

10. Summary and Common Exam Pitfalls | 总结与常见考试陷阱

The key takeaways are: equilibrium is dynamic, not static; Le Chatelier’s principle explains the direction of shift; concentration, temperature and pressure changes have predictable effects; catalysts do not affect yield, only the speed of reaching equilibrium.

关键要点是：平衡是动态的，不是静止的；勒夏特列原理解释了移动的方向；浓度、温度和压力的变化有可预测的影响；催化剂不影响产率，只影响达到平衡的速度。

When answering exam questions, always specify whether you are discussing rate or equilibrium yield. A common mistake is to say that adding a catalyst increases yield or that a higher temperature always increases yield – both are false in the context of equilibrium.

在回答考试题目时，一定要指明你是在讨论速率还是平衡产率。一个常见错误是说添加催化剂能提高产率，或者说升高温度总是提高产率——在平衡语境下，这两者都是错误的。

Use precise language: ‘position of equilibrium shifts to the right’ rather than ‘more product is made’. The latter can be ambiguous because more product can be made faster, but the equilibrium amount might not change.

使用精确的语言：“平衡位置向右移动”，而不是“生成更多产物”。后者可能产生歧义，因为产物生成可以更快，但平衡时的量可能没有改变。

For industrial processes, be ready to explain why conditions are chosen as a compromise, and how catalysts, recycling unreacted gases and continuous removal of products improve efficiency.

对于工业过程，要准备好解释为什么所选条件是折衷方案，以及催化剂、未反应气体的循环利用和持续移出产物如何提高效率。

Remember that all these concepts link to the big idea of dynamic equilibrium. Mastering this will help you not only in Wales’ WJEC exams but also in understanding chemistry in the wider world.

请记住，所有这些概念都与动态平衡这一大思想相联系。掌握这些，不仅有助于你在威尔士 WJEC 的考试中取得好成绩，还能帮助你理解更广阔世界中的化学。

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GCSE WJEC Chemistry: Chemical Equilibrium – Key Points Explained | GCSE WJEC 化学：化学平衡 – 考点精讲