AS Chemistry: Chemical Equilibrium Key Points | AS 化学：化学平衡 考点精讲

📚 AS Chemistry: Chemical Equilibrium Key Points | AS Chemical：化学平衡 考点精讲

Chemical equilibrium is one of the most fundamental and exam-critical topics in AS Chemistry. It brings together concepts of reversible reactions, dynamic balance, quantitative treatment via equilibrium constants, and the qualitative predictive power of Le Chatelier’s Principle. Mastery of equilibrium allows you to explain why industrial processes such as the Haber process are run under specific conditions, and how yields can be maximised. In this article we systematically break down every key idea, calculation method and misconception so that you can approach equilibrium questions with absolute confidence.

化学平衡是 AS 化学中最基础、考试出题率最高的主题之一。它将可逆反应、动态平衡、通过平衡常数进行的定量处理以及勒夏特列原理的定性预测能力融为一体。掌握平衡可以帮助你解释为什么哈伯法等工业过程要在特定条件下进行，以及如何实现产率最大化。本文会系统拆解每一个关键概念、计算方法和常见误区，让你在面对平衡题目时充满信心。

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1. Reversible Reactions and Dynamic Equilibrium | 可逆反应与动态平衡

A reversible reaction is one that can proceed in both forward and backward directions. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, so the concentrations of reactants and products remain constant – but the reactions continue to occur at the molecular level. This is called dynamic equilibrium.

可逆反应是指既可以正向进行也可以逆向进行的反应。达到平衡时，正反应速率等于逆反应速率，因此反应物和产物的浓度保持不变——但在分子水平上反应仍在持续进行。这就是动态平衡。

Equilibrium can only be established in a closed system where no matter can escape. In an open system, products may leave, preventing the backward reaction from reaching the same rate.

平衡只能在一个封闭系统中建立，物质无法逸出。在开放系统中，产物可能离开，导致逆反应无法达到相同速率。

At equilibrium, macroscopic properties such as colour, pressure and concentration are constant, but the forward and backward reactions have not stopped.

在平衡状态下，颜色、压力和浓度等宏观性质保持不变，但正向和逆向反应都没有停止。

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2. The Equilibrium Constant, Kc | 平衡常数 Kc

The equilibrium constant in terms of concentration, Kc, is a numerical value that expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of its stoichiometric coefficient.

以浓度表示的平衡常数 Kc 是一个数值，它表达了平衡时产物浓度与反应物浓度之比，每种物质的浓度以其化学计量系数为指数。

For a general reaction: aA + bB ⇌ cC + dD, the expression for Kc is written as:

对于一般反应：aA + bB ⇌ cC + dD，Kc 的表达式写作：

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Square brackets denote concentration in mol dm⁻³. Pure solids and pure liquids are omitted from the Kc expression because their concentrations remain essentially constant.

方括号表示浓度，单位为 mol dm⁻³。纯固体和纯液体不在 Kc 表达式中出现，因为它们的浓度基本保持不变。

Kc is temperature-dependent. A given equilibrium system has a specific Kc value at a fixed temperature; if the temperature changes, Kc changes.

Kc 取决于温度。一个给定的平衡体系在某一特定温度下有特定的 Kc 值；如果温度改变，Kc 也会改变。

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3. Writing Kc Expressions | 书写 Kc 表达式

When writing a Kc expression, only aqueous and gaseous species are included. If the reaction involves a solvent such as water and it appears as a reactant but is in large excess, it is often omitted as its concentration is effectively constant.

书写 Kc 表达式时，只包含溶液中的物种和气态物种。如果反应涉及溶剂如水，且水作为反应物但大量过量，常常被省略，因为其浓度实际上保持不变。

For example, in the esterification reaction: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l), all species are liquids. In this case dilute solution conditions are often assumed, and water may be treated as a product but the expression is simplified in different exam specifications. Always follow the guidance of your particular examination board.

例如，酯化反应：CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)，所有物种均为液体。在这种情况下，通常假定为稀溶液条件，水可能被视作产物，但表达式在不同的考试大纲中会简化处理。务必遵循具体考试局的要求。

For heterogeneous equilibria, only gases and aqueous ions feature in Kc. Solids and pure liquids are excluded.

对于多相平衡，只有气体和水溶液中的离子才会出现在 Kc 中。固体和纯液体排除在外。

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4. Calculating Kc Using ICE Tables | 使用 ICE 表格计算 Kc

The most reliable method for solving equilibrium calculations is the ICE table: Initial concentration, Change in concentration, Equilibrium concentration. This approach organises data clearly and reduces sign errors.

解决平衡计算最可靠的方法是 ICE 表格：初始浓度、浓度变化、平衡浓度。这种方法能够清晰组织数据，减少符号错误。

Worked example: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), 1.0 mol of H₂ and 1.0 mol of I₂ are placed in a vessel of volume V dm³. At equilibrium, 0.40 mol of H₂ remains. Calculate Kc assuming V = 1.0 dm³.

例题：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，将 1.0 mol H₂ 和 1.0 mol I₂ 放入体积为 V dm³ 的容器中。平衡时剩余 0.40 mol H₂。计算 Kc，假设 V = 1.0 dm³。

Initial concentrations: [H₂] = 1.0, [I₂] = 1.0, [HI] = 0. Change: H₂ decreased by 0.60, so [H₂] changes by –0.60, similarly [I₂] changes by –0.60, and [HI] increases by +1.20. Equilibrium: [H₂] = 0.40, [I₂] = 0.40, [HI] = 1.20.

初始浓度：[H₂] = 1.0，[I₂] = 1.0，[HI] = 0。变化：H₂ 减少了 0.60，因此 [H₂] 变化为 –0.60，同样 [I₂] 变化为 –0.60，[HI] 增加 +1.20。平衡浓度：[H₂] = 0.40，[I₂] = 0.40，[HI] = 1.20。

Kc = [HI]² / ([H₂][I₂]) = (1.20)² / (0.40 × 0.40) = 1.44 / 0.16 = 9.0. No unit for Kc here because Δn = 0.

Kc = [HI]² / ([H₂][I₂]) = (1.20)² / (0.40 × 0.40) = 1.44 / 0.16 = 9.0。此处 Kc 无单位，因为 Δn = 0。

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5. Units of Kc | Kc 的单位

The unit of Kc depends on the overall change in the number of moles of gas or aqueous ions in the balanced equation. Define Δn = (total moles of products) – (total moles of reactants) in the Kc expression. Then Kc has units (mol dm⁻³)^(Δn).

Kc 的单位取决于配平方程式中气体或水合离子的总摩尔数变化。定义 Δn = (产物总摩尔数) – (反应物总摩尔数)（在 Kc 表达式中）。那么 Kc 的单位为 (mol dm⁻³)^(Δn)。

If Δn = 0, Kc has no unit. If Δn = 1, unit is mol dm⁻³. If Δn = –1, unit is dm³ mol⁻¹. Always calculate the unit after evaluating the numerical value to avoid losing marks in exams.

如果 Δn = 0，则 Kc 无单位。如果 Δn = 1，单位为 mol dm⁻³。如果 Δn = –1，单位为 dm³ mol⁻¹。一定要在计算出数值后确定单位，以免考试丢分。

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6. 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, pressure or temperature, the position of equilibrium shifts to oppose the change. This principle allows qualitative prediction of the direction of shift.

勒夏特列原理指出，如果一个处于动态平衡的体系的浓度、压力或温度发生改变，平衡位置将向着减弱这种改变的方向移动。该原理可以定性预测平衡移动的方向。

It is important to remember that the principle does not explain why the shift occurs; it simply describes the observed response of the equilibrium system.

需要记住的是，该原理并不能解释移动发生的原因；它仅仅描述了平衡体系观察到的那种响应。

Only changes in concentration, pressure (for gaseous systems) and temperature affect the equilibrium position. Addition of a catalyst or changes in surface area do not alter the position of equilibrium.

只有浓度、压力（对于气体体系）和温度的变化会影响平衡位置。加入催化剂或改变表面积不会改变平衡位置。

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

If the concentration of a reactant is increased, the system shifts to use up the added substance, moving the equilibrium towards the products. Conversely, removing a product shifts the equilibrium to the right to replenish it.

如果增加反应物的浓度，体系会消耗掉加入的物质，平衡向产物方向移动。反之，移去产物会使平衡向右移动以补充产物。

This shift minimises the imposed change. However, Kc remains unchanged because concentration changes do not affect the value of the equilibrium constant at constant temperature.

这种移动会将外力影响降到最低。然而 Kc 保持不变，因为在温度恒定下浓度变化不会影响平衡常数的数值。

In industry, a cheap reactant is often used in excess to drive the reaction forward and improve the yield of an expensive product.

在工业中，经常使用过量廉价反应物来推动反应向正向进行，以提高昂贵产物的产率。

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8. Effect of Pressure Changes | 压力变化的影响

Pressure changes only affect equilibria involving gases where there is a change in the total number of gas molecules. If the pressure of the system is increased, the equilibrium shifts to the side with fewer moles of gas to reduce the pressure.

压力变化只影响有气体参与且气体总分子数发生变化的平衡。如果增大体系的压力，平衡将向气体摩尔数较少的一侧移动，以降低压力。

If both sides have an equal number of gas moles, changing pressure has no effect on the equilibrium position. For example, H₂(g) + I₂(g) ⇌ 2HI(g) has 2 moles on each side; pressure changes do not alter the composition at equilibrium, though rates may be affected.

如果两边气体摩尔数相等，改变压力不会影响平衡位置。例如 H₂(g) + I₂(g) ⇌ 2HI(g) 每侧均有 2 摩尔气体；压力变化不改变平衡组成，尽管速率可能受到影响。

Kc and Kp (the equilibrium constant in terms of pressure) are unaffected by pressure changes at constant temperature, because while partial pressures adjust, the ratio remains constant.

在温度恒定下，Kc 和 Kp（用压力表示的平衡常数）不受压力变化的影响，因为尽管分压发生调整，但其比值保持不变。

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9. 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. If the forward reaction is endothermic (ΔH > 0), increasing temperature favours the forward reaction – equilibrium shifts right and Kc increases.

温度是唯一能改变平衡常数值的外部条件。如果正向反应为吸热反应（ΔH > 0），升高温度有利于正向反应——平衡右移，Kc 增大。

If the forward reaction is exothermic (ΔH < 0), raising the temperature shifts the equilibrium to the left to absorb heat, so Kc decreases. Therefore, in the Haber process where the forward reaction is exothermic, lower temperatures give a higher equilibrium yield of ammonia – but a compromise temperature is used to maintain a reasonable rate.

如果正向反应为放热反应（ΔH < 0），升高温度会使平衡向左移动以吸收热量，因此 Kc 减小。所以在正向反应放热的哈伯法中，较低温度有利于氨的平衡产率更高——但工业上会使用折中温度以保持合理的反应速率。

Always express the shift using Le Chatelier’s Principle and explicitly state the effect on Kc in exam answers.

在考试作答时，务必运用勒夏特列原理说明移动方向，并明确指出对 Kc 的影响。

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10. Effect of Catalysts | 催化剂的影响

A catalyst speeds up both forward and backward reactions equally by providing an alternative pathway with lower activation energy. At equilibrium, a catalyst does not alter the position of equilibrium or the value of Kc.

催化剂通过提供一条活化能较低的替代反应途径，同等程度地加快正、逆反应的速率。在平衡状态下，催化剂不会改变平衡位置，也不会改变 Kc 的值。

The sole benefit of a catalyst is that equilibrium is reached more quickly. This is vital in industrial processes where achieving equilibrium rapidly means higher productivity, even though the maximum yield is unchanged.

催化剂的唯一好处是能够更快地达到平衡。这在工业过程中至关重要，因为快速达到平衡意味着更高的生产效率，尽管最大产率并未改变。

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11. Equilibrium Constant Kp for Gaseous Reactions | 气相反应的平衡常数 Kp

For homogeneous gas-phase reactions, the equilibrium constant can also be expressed in terms of partial pressure, denoted Kp. The partial pressure of a gas is the pressure it would exert if it alone occupied the whole volume.

对于均相气相反应，平衡常数也可以用分压表示，记作 Kp。某一气体的分压是指假设该气体单独占有整个体积时所施加的压力。

Partial pressure p(A) = mole fraction of A × total pressure P. Mole fraction = moles of A / total moles of gas present at equilibrium.

分压 p(A) = A 的摩尔分数 × 总压 P。摩尔分数 = A 的物质的量 / 平衡时气体总物质的量。

For a reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp is given by:

对于反应 aA(g) + bB(g) ⇌ cC(g) + dD(g)，Kp 表示为：

Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ

Kp has units of pressure raised to Δn, e.g., atm^(Δn) or Pa^(Δn), depending on the units used.

Kp 的单位为压力的 Δn 次方，例如 atm^(Δn) 或 Pa^(Δn)，取决于采用的单位。

Just like Kc, Kp is constant for a given system at a fixed temperature and is unaffected by pressure changes or catalysts.

与 Kc 一样，对于给定体系，Kp 在固定温度下为常数，不受压力变化或催化剂的影响。

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12. Industrial Processes and Equilibrium – The Haber Process | 工业过程与平衡——哈伯法

The Haber process for ammonia synthesis illustrates the interplay between equilibrium and kinetics: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = –92 kJ mol⁻¹. The forward reaction is exothermic and produces fewer gas molecules (4 mol → 2 mol).

合成氨的哈伯法很好地体现了平衡与动力学的相互作用：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = –92 kJ mol⁻¹。正向反应放热且气体分子数减少（4 mol → 2 mol）。

According to Le Chatelier, high pressure favours the forward reaction, increasing equilibrium yield. Low temperature also favours the exothermic forward reaction, giving a higher equilibrium yield. However, the actual industrial conditions are a compromise: pressure of about 200 atm and temperature around 400–450 °C, with an iron catalyst.

根据勒夏特列原理，高压有利于正向反应，提高平衡产率。低温也有利于放热正向反应，得到更高的平衡产率。但实际工业条件是折中方案：压力约 200 atm，温度约 400–450 °C，并使用铁催化剂。

The catalyst allows a lower temperature to achieve a fast enough rate without excessively sacrificing yield. Unreacted N₂ and H₂ are recycled to improve overall efficiency.

催化剂使得在较低温度下也能获得足够快的反应速率，同时不会过度牺牲产率。未反应的 N₂ 和 H₂ 会循环利用，以提高总效率。

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