Reversible Reactions and Equilibrium | 可逆反应与化学平衡

📚 Reversible Reactions and Equilibrium | 可逆反应与化学平衡

Many chemical reactions do not go to completion. Instead, the forward and reverse reactions occur at the same rate under certain conditions, producing a dynamic equilibrium in which macroscopic properties remain constant. Understanding equilibrium is essential for predicting how reaction conditions affect yield in the laboratory and in industry.

许多化学反应不会进行到底。在一定条件下，正反应和逆反应以相同速率进行，形成动态平衡，此时宏观性质保持不变。理解平衡对于预测反应条件在实验室和工业中对产率的影响至关重要。

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

A reversible reaction is one in which the products can react to re-form the original reactants. For example, heating hydrated copper(II) sulfate removes water, but adding water to anhydrous copper(II) sulfate restores the hydrated form. At equilibrium, the forward and reverse reactions continue at the molecular level, but their rates are equal, so concentrations of all species remain constant.

可逆反应是指产物可以重新反应生成原来的反应物。例如，加热水合硫酸铜(II)会失去水，但向无水硫酸铜(II)中加水又恢复水合形式。达到平衡时，正、逆反应在分子水平上仍在进行，但速率相等，因此所有物质的浓度保持恒定。

2. Writing Equilibrium Expressions: Kc | 平衡常数表达式 Kc

For a general homogeneous reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is written as:

对于一般均相反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数写作：

K_c = [C]^c[D]^d / [A]^a[B]^b

Square brackets denote equilibrium concentrations in mol dm⁻³. Solids and pure liquids are omitted from the expression because their concentrations are essentially constant. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂].

方括号表示平衡浓度，单位为 mol dm⁻³。固体和纯液体不写入表达式，因为它们的浓度基本不变。例如，在 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 中，Kc = [CO₂]。

3. Equilibrium Constant Kp for Gaseous Systems | 气体体系平衡常数 Kp

For gas-phase equilibria, partial pressures are used instead of concentrations. If the total pressure is known, the partial pressure of a gas equals its mole fraction multiplied by the total pressure.

对于气相平衡，用分压代替浓度。如果已知总压，某气体的分压等于其摩尔分数乘以总压。

p(A) = x(A) × P_total    K_p = p(C)^c p(D)^d / p(A)^a p(B)^b

Kp has units that depend on the difference in the total number of gas moles between products and reactants. The exact units must be calculated from the expression for each equilibrium system.

Kp 的单位取决于产物与反应物气体总摩尔数之差。必须根据每个平衡体系的表达式计算具体单位。

4. Magnitude of Kc and Reaction Position | Kc 大小与反应位置

A large Kc value, much greater than 1, indicates that the equilibrium position lies to the right, favouring products. A small Kc value, much less than 1, indicates that the equilibrium position lies to the left, favouring reactants.

Kc 值很大（远大于 1）表明平衡位置偏向右边，有利于产物。Kc 值很小（远小于 1）表明平衡位置偏向左边，有利于反应物。

Do not confuse a large Kc with a fast reaction: equilibrium constants give no information about reaction rate.

不要将大的 Kc 与快速反应混淆：平衡常数不提供反应速率信息。

5. Le Chatelier’s Principle | 勒夏特列原理

Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium position shifts in the direction that opposes the change.

勒夏特列原理指出，如果处于平衡的体系受到浓度、压力或温度的变化，平衡位置会向削弱该变化的方向移动。

This principle helps predict qualitative changes, but it does not explain the rate at which a new equilibrium is reached. It must not be applied to changes in catalyst because a catalyst does not alter the position of equilibrium.

该原理有助于预测性质变化，但不能解释达到新平衡的速率。它不能用于催化剂变化，因为催化剂不改变平衡位置。

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

Increasing the concentration of a reactant shifts the equilibrium to the right, removing some of the added reactant. Decreasing the concentration of a product also shifts the equilibrium to the right. Adding water to an aqueous equilibrium can be treated as dilution, which shifts the position toward the side with more aqueous particles.

增大反应物浓度使平衡向右移动，消耗部分添加的反应物。降低产物浓度也使平衡向右移动。向水相平衡体系中加水可视为稀释作用，平衡会向水相粒子数较多的一侧移动。

• Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq) is red; adding Fe³⁺ or SCN⁻ makes the red colour more intense.
• Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq) 呈红色；加入 Fe³⁺ 或 SCN⁻ 会使红色加深。

Adding a solid or pure liquid does not shift the equilibrium because its concentration remains constant.

加入固体或纯液体不会移动平衡，因为其浓度保持不变。

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

Increasing pressure favours the side with fewer gas molecules. For the Haber reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), high pressure shifts the equilibrium toward ammonia because 4 gas moles become 2 gas moles.

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

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