Chemical Equilibrium A Level Chemistry Guide

What Is Dynamic Equilibrium?

A reversible reaction is one that can proceed in both the forward and backward directions. At the start, the forward reaction dominates as reactants are plentiful. Over time, as products build up, the reverse reaction speeds up. Eventually, the rates of the forward and reverse reactions become equal — this is dynamic equilibrium. At equilibrium, the concentrations of all species remain constant, but the reaction has not stopped: both forward and reverse reactions continue at the same rate. This is why we call it “dynamic” rather than “static” equilibrium.

可逆反应是指可以同时在正方向和反方向进行的反应。最初，由于反应物充足，正反应占主导地位。随着时间的推移，生成物逐渐积累，逆反应速率加快。最终，正反应和逆反应的速率相等——这就是动态平衡。在平衡状态下，所有物质的浓度保持不变，但反应并未停止：正反应和逆反应仍在以相同速率进行。这就是为什么我们称之为”动态”平衡而非”静态”平衡。

For the general reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc is defined as:

对于一般反应 aA + bB ⇌ cC + dD，平衡常数 Kc 定义为：

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

A large Kc (much greater than 1) means the equilibrium lies far to the right — products are favoured. A small Kc (much less than 1) means the equilibrium lies to the left — reactants are favoured. Crucially, Kc is constant at a given temperature. Changing concentration or pressure may shift the equilibrium position, but it does not change the value of Kc. Only temperature can change Kc.

Kc 值很大（远大于 1）意味着平衡位置偏右——有利于生成物。Kc 值很小（远小于 1）意味着平衡位置偏左——有利于反应物。关键点是：在给定温度下，Kc 是常数。改变浓度或压力可能会改变平衡位置，但不会改变 Kc 的值。只有温度才能改变 Kc。

Le Chatelier’s Principle

Henry Louis Le Chatelier formulated one of the most useful principles in chemistry: if a system at dynamic equilibrium is subjected to a change in conditions, the equilibrium shifts to partially oppose the change. Think of it as the system “fighting back” against any disturbance. This simple idea allows chemists to predict and control the outcomes of reversible reactions — and it is absolutely fundamental to A-Level Chemistry examinations.

亨利·路易·勒夏特列提出了化学中最有用的原理之一：如果处于动态平衡的系统受到条件变化的干扰，平衡会移动以部分抵消这种变化。可以把这个过程理解为系统对任何干扰的”反抗”。这个简单的概念使化学家能够预测和控制可逆反应的结果——它在A-Level化学考试中是绝对的基础知识。

Effect of Concentration Changes

If you increase the concentration of a reactant, the equilibrium shifts to the right (towards products) to use up the added reactant. If you increase the concentration of a product, the equilibrium shifts to the left (towards reactants). Removing a product has the opposite effect: the equilibrium shifts to the right to replace it. This is the basis for many industrial processes — continuously removing the product drives the equilibrium forward, improving yield.

如果增加反应物的浓度，平衡会向右移动（朝向生成物），以消耗掉额外加入的反应物。如果增加生成物的浓度，平衡会向左移动（朝向反应物）。移除生成物则产生相反的效果：平衡会向右移动以补充被移除的物质。这是许多工业流程的基础——持续移除生成物可以推动平衡向正方向移动，提高产率。

Effect of Pressure Changes

Pressure changes only affect equilibria involving gases, and only when there is a different number of moles of gas on each side of the equation. Increasing pressure shifts the equilibrium to the side with fewer gas molecules (to reduce the pressure). Decreasing pressure shifts it to the side with more gas molecules. If the number of gas molecules is the same on both sides, a pressure change has no effect on the equilibrium position.

压力变化只影响涉及气体的平衡，并且只有在方程式两边气体分子数量不同时才有效。增加压力会使平衡移向气体分子较少的一侧（以降低压力）。降低压力则使其移向气体分子较多的一侧。如果两边气体分子数量相同，压力变化对平衡位置没有影响。

Consider the Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g). The left side has 4 moles of gas (1 + 3), the right side has 2 moles. Increasing pressure favours the forward reaction — making more ammonia. This is why the Haber Process operates at high pressure (around 200 atm).

以哈伯法为例：N2(g) + 3H2(g) ⇌ 2NH3(g)。左边有 4 摩尔气体（1+3），右边有 2 摩尔。增加压力有利于正反应——生成更多的氨。这就是哈伯法在高压下运行的原因（约 200 个大气压）。

Effect of Temperature Changes

Temperature is the only factor that changes the value of Kc. For an exothermic reaction (ΔH is negative), increasing temperature shifts the equilibrium to the left (towards reactants), and Kc decreases. For an endothermic reaction (ΔH is positive), increasing temperature shifts the equilibrium to the right (towards products), and Kc increases. Lowering temperature has the opposite effect in each case.

温度是唯一能改变 Kc 值的因素。对于放热反应（ΔH 为负），升高温度使平衡向左移动（朝向反应物），Kc 减小。对于吸热反应（ΔH 为正），升高温度使平衡向右移动（朝向生成物），Kc 增大。降低温度在每种情况下都会产生相反的效果。

Why does this happen? Think of heat as a “product” in an exothermic reaction and as a “reactant” in an endothermic reaction. Adding heat (raising temperature) favours the endothermic direction — the direction that absorbs heat. Removing heat (lowering temperature) favours the exothermic direction.

为什么会这样？可以把热量想象成放热反应中的”生成物”和吸热反应中的”反应物”。加入热量（升高温度）有利于吸热方向——即吸收热量的方向。移除热量（降低温度）有利于放热方向。

Effect of a Catalyst

A catalyst speeds up both the forward and reverse reactions equally. It provides an alternative reaction pathway with a lower activation energy. A catalyst does NOT affect the position of equilibrium — it simply allows equilibrium to be reached faster. It also does not affect the value of Kc. This is a common exam trap: students often write that a catalyst increases yield, but this is incorrect. Catalysts only increase the rate, not the yield.

催化剂同等程度地加速正反应和逆反应。它提供了一个活化能更低的替代反应途径。催化剂不会影响平衡位置——它只是让平衡更快达到。它也不会影响 Kc 的值。这是考试中常见的陷阱：学生经常写催化剂能提高产率，但这是错误的。催化剂只提高速率，不提高产率。

Equilibrium Constant Kp (Partial Pressures)

For gaseous equilibria, we often use Kp instead of Kc. Kp is expressed in terms of partial pressures rather than concentrations. The partial pressure of a gas in a mixture is the pressure that gas would exert if it alone occupied the entire volume. It is calculated as: partial pressure = mole fraction × total pressure.

对于气体平衡，我们通常使用 Kp 而不是 Kc。Kp 用分压而不是浓度来表示。混合气体中某气体的分压是指该气体单独占据整个体积时所施加的压力。计算公式为：分压 = 摩尔分数 × 总压强。

For the general reaction aA(g) + bB(g) ⇌ cC(g) + dD(g):

Kp = (PC)^c (PD)^d / (PA)^a (PB)^b

where PA, PB, PC, PD represent the partial pressures of each gas. Like Kc, Kp is constant at a given temperature and only changes with temperature. Kp has units, which depend on the difference between the total moles of gaseous products and reactants (Δn).

其中 PA、PB、PC、PD 代表每种气体的分压。与 Kc 一样，Kp 在给定温度下是常数，只随温度变化。Kp 有单位，取决于气态生成物和反应物的总摩尔数之差（Δn）。

Calculating Partial Pressures

Mole fraction of gas A = moles of A / total moles of all gases in the mixture. Then partial pressure of A = mole fraction of A × total pressure.

气体 A 的摩尔分数 = A 的摩尔数 / 混合物中所有气体的总摩尔数。然后 A 的分压 = A 的摩尔分数 × 总压强。

Example: In a container at 500 kPa total pressure, an equilibrium mixture contains 2.0 mol N2, 6.0 mol H2, and 4.0 mol NH3. Total moles = 12.0. Mole fraction of NH3 = 4.0/12.0 = 0.333. Partial pressure of NH3 = 0.333 × 500 = 166.7 kPa.

示例：在一个总压为 500 kPa 的容器中，平衡混合物含有 2.0 mol N2、6.0 mol H2 和 4.0 mol NH3。总摩尔数 = 12.0。NH3 的摩尔分数 = 4.0/12.0 = 0.333。NH3 的分压 = 0.333 × 500 = 166.7 kPa。

The Haber Process: A Case Study in Equilibrium Control

The Haber Process for ammonia synthesis is one of the most important industrial applications of equilibrium principles. Developed by Fritz Haber in the early 20th century, it produces ammonia from nitrogen and hydrogen: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92 kJ mol⁻¹.

哈伯法合成氨是平衡原理最重要的工业应用之一。由弗里茨·哈伯在 20 世纪初开发，它从氮气和氢气中生产氨：N2(g) + 3H2(g) ⇌ 2NH3(g)，ΔH = -92 kJ mol⁻¹。

Compromise Conditions

The forward reaction is exothermic, so Le Chatelier’s Principle tells us that a low temperature would favour ammonia production. However, at low temperatures the reaction is too slow to be economical. The Haber Process uses a compromise temperature of around 400-450°C — high enough for a reasonable rate, but not so high that yield is severely compromised.

正反应是放热的，所以勒夏特列原理告诉我们低温有利于氨的生成。但在低温下反应太慢，无法经济化生产。哈伯法采用了约 400-450°C 的折中温度——足够高以获得合理的速率，但又不会高到严重损害产率。

Similarly, high pressure favours the forward reaction (4 moles of gas → 2 moles), but very high pressures are expensive and dangerous due to the engineering challenges. The process operates at around 200 atm — high enough to achieve an acceptable yield (~15-20% per pass), but manageable from an engineering standpoint.

同样，高压有利于正反应（4 摩尔气体 → 2 摩尔），但非常高的压力由于工程挑战而昂贵且危险。该过程在约 200 个大气压下运行——足够高以获得可接受的产率（每次通过约 15-20%），但从工程角度看是可行的。

The Role of the Iron Catalyst

An iron catalyst is used to speed up the reaction. The catalyst does not affect the equilibrium position or yield — its sole purpose is to make equilibrium attainable in a reasonable time. Without the catalyst, the reaction would be far too slow at 400°C to be practical.

使用铁催化剂来加速反应。催化剂不会影响平衡位置或产率——其唯一目的是使平衡在合理的时间内达到。没有催化剂，在 400°C 下反应会太慢而无法实际应用。

Continuous Process Design

The Haber Process is designed as a continuous flow system. Unreacted N2 and H2 are separated from the ammonia product (which is liquefied by cooling) and recycled back into the reactor. This recycling achieves an overall conversion efficiency approaching 98%, even though the per-pass yield is only 15-20%.

哈伯法被设计为连续流动系统。未反应的 N2 和 H2 与氨产品（通过冷却液化）分离，并循环回反应器。这种循环使整体转化效率接近 98%，即使每次通过的产率仅为 15-20%。

ICE Tables: Systematic Equilibrium Calculations

ICE (Initial, Change, Equilibrium) tables are the systematic method for solving equilibrium problems. This method is tested in virtually every A-Level Chemistry examination.

ICE（初始、变化、平衡）表格是解决平衡问题的系统方法。几乎每次 A-Level 化学考试都会考查这种方法。

The ICE Method

1. I – Initial: Write down the initial amounts (moles or concentrations) of all species before any reaction occurs.
2. C – Change: Express the changes in terms of x, using the stoichiometric ratios from the balanced equation. Reactants decrease (-), products increase (+).
3. E – Equilibrium: Write expressions for the equilibrium amounts by combining I and C.

1. I – 初始：写出所有物质在反应发生前的初始量（摩尔数或浓度）。
2. C – 变化：用 x 表示变化量，使用平衡方程式中的化学计量比。反应物减少（-），生成物增加（+）。
3. E – 平衡：将 I 和 C 组合，写出平衡量的表达式。

Worked Example

A mixture of 1.00 mol of ethanoic acid and 1.00 mol of ethanol is allowed to reach equilibrium at 25°C. The equilibrium mixture contains 0.667 mol of ethyl ethanoate. Calculate Kc for: CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O.

将 1.00 mol 乙酸和 1.00 mol 乙醇的混合物在 25°C 下达到平衡。平衡混合物含有 0.667 mol 乙酸乙酯。计算以下反应的 Kc：CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O。

ICE Table (assuming 1 dm³ volume, so moles = concentration):

• Initial: CH3COOH = 1.00, C2H5OH = 1.00, CH3COOC2H5 = 0, H2O = 0
• Change: CH3COOH = -0.667, C2H5OH = -0.667, CH3COOC2H5 = +0.667, H2O = +0.667
• Equilibrium: CH3COOH = 0.333, C2H5OH = 0.333, CH3COOC2H5 = 0.667, H2O = 0.667

ICE 表格（假设体积为 1 dm³，因此摩尔数 = 浓度）：

• 初始：CH3COOH = 1.00, C2H5OH = 1.00, CH3COOC2H5 = 0, H2O = 0
• 变化：CH3COOH = -0.667, C2H5OH = -0.667, CH3COOC2H5 = +0.667, H2O = +0.667
• 平衡：CH3COOH = 0.333, C2H5OH = 0.333, CH3COOC2H5 = 0.667, H2O = 0.667

Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH] = (0.667 × 0.667) / (0.333 × 0.333) = 0.4449 / 0.1109 = 4.01 (no units, since equal moles on both sides).

Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH] = (0.667 × 0.667) / (0.333 × 0.333) = 0.4449 / 0.1109 = 4.01（无单位，因为两边摩尔数相等）。

Common Exam Pitfalls and Tips

Many A-Level students lose marks on equilibrium questions not because they do not understand the concepts, but because of avoidable mistakes. Here are the most common ones, along with how to avoid them:

许多 A-Level 学生在平衡问题上丢分，不是因为不理解概念，而是因为一些可以避免的错误。以下是最常见的错误及如何避免：

• Forgetting to include water in Kc expressions when water is a reactant or product in a non-aqueous system. In aqueous solutions, water is the solvent and its concentration is effectively constant, so it is omitted from Kc. But in esterification (as above), water is a product and must be included.
• 在水不是溶剂的系统中，忘记在 Kc 表达式中包含水。在水溶液中，水是溶剂，其浓度实际上恒定，因此从 Kc 中省略。但在酯化反应中（如上），水是生成物，必须包含在内。
• Confusing moles with concentrations. Kc uses equilibrium concentrations (mol dm⁻³), not amounts (mol). If the volume is not 1 dm³, you must divide moles by volume first.
• 混淆摩尔数与浓度。Kc 使用平衡浓度（mol dm⁻³），而不是物质的量（mol）。如果体积不是 1 dm³，必须先除以体积。
• Using the wrong stoichiometric ratio in change expressions. In the Haber Process, if x moles of N2 react, then 3x moles of H2 react, and 2x moles of NH3 form. Always check the balanced equation carefully.
• 在变化表达中使用错误的化学计量比。在哈伯法中，如果 x 摩尔 N2 反应，则 3x 摩尔 H2 反应，生成 2x 摩尔 NH3。始终仔细检查配平的方程式。
• Claiming a catalyst increases yield. Catalysts increase rate, not yield. They allow equilibrium to be reached faster but do not shift its position.
• 声称催化剂提高产率。催化剂提高速率，不提高产率。它们使平衡更快达到，但不改变其位置。
• Using Kc when Kp is required for gaseous systems. Read the question carefully — if partial pressures are given, use Kp.
• 在气体系统中需要 Kp 时使用了 Kc。仔细阅读题目——如果给出了分压，使用 Kp。

Past Paper Practice Questions

Question 1: Consider the equilibrium: 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = -197 kJ mol⁻¹. State and explain the effect of (a) increasing temperature, (b) increasing pressure, and (c) adding a catalyst on the equilibrium yield of SO3.

问题 1：考虑平衡：2SO2(g) + O2(g) ⇌ 2SO3(g)，ΔH = -197 kJ mol⁻¹。说明并解释 (a) 升高温度，(b) 增加压力，(c) 添加催化剂对 SO3 平衡产率的影响。

Answer 1: (a) The forward reaction is exothermic. Increasing temperature shifts equilibrium to the left (endothermic direction), decreasing the yield of SO3. (b) There are 3 moles of gas on the left and 2 on the right. Increasing pressure shifts equilibrium to the right (fewer gas molecules), increasing the yield of SO3. (c) A catalyst speeds up both forward and reverse reactions equally. It does not affect the equilibrium position or yield — only the rate at which equilibrium is reached.

答案 1：(a) 正反应放热。升高温度使平衡向左（吸热方向）移动，降低 SO3 的产率。(b) 左边 3 摩尔气体，右边 2 摩尔。增加压力使平衡向右（气体分子较少的一侧）移动，提高 SO3 的产率。(c) 催化剂同等加速正逆反应。不影响平衡位置或产率——只影响达到平衡的速率。

Question 2: At 500 K, an equilibrium mixture contains 0.80 mol N2, 2.4 mol H2, and 1.6 mol NH3 in a 2.0 dm³ vessel. Calculate Kc for N2(g) + 3H2(g) ⇌ 2NH3(g).

问题 2：在 500 K 时，一个 2.0 dm³ 容器中的平衡混合物含有 0.80 mol N2、2.4 mol H2 和 1.6 mol NH3。计算 N2(g) + 3H2(g) ⇌ 2NH3(g) 的 Kc。

Answer 2: [N2] = 0.80/2.0 = 0.40 mol dm⁻³, [H2] = 2.4/2.0 = 1.2 mol dm⁻³, [NH3] = 1.6/2.0 = 0.80 mol dm⁻³. Kc = [NH3]² / ([N2][H2]³) = (0.80)² / (0.40 × (1.2)³) = 0.64 / (0.40 × 1.728) = 0.64 / 0.6912 = 0.926. Units: (mol dm⁻³)² / ((mol dm⁻³)(mol dm⁻³)³) = mol⁻² dm⁶.

答案 2：[N2] = 0.80/2.0 = 0.40 mol dm⁻³，[H2] = 2.4/2.0 = 1.2 mol dm⁻³，[NH3] = 1.6/2.0 = 0.80 mol dm⁻³。Kc = [NH3]² / ([N2][H2]³) = (0.80)² / (0.40 × (1.2)³) = 0.64 / (0.40 × 1.728) = 0.64 / 0.6912 = 0.926。单位：mol⁻² dm⁶。

Summary: Key Facts to Memorise

• At dynamic equilibrium, rate of forward reaction = rate of reverse reaction.
• Le Chatelier’s Principle: a system at equilibrium shifts to oppose any imposed change.
• Concentration changes shift equilibrium; they do NOT change Kc.
• Pressure changes shift equilibrium only when Δn (gas) ≠ 0; Kc and Kp are unchanged.
• Temperature changes shift equilibrium AND change Kc and Kp.
• Catalysts increase the rate of reaching equilibrium but do NOT affect position, yield, or Kc/Kp.
• Kc uses concentrations (mol dm⁻³); Kp uses partial pressures (kPa or atm).
• The Haber Process uses compromise conditions: 400-450°C, 200 atm, iron catalyst.

• 在动态平衡中，正反应速率 = 逆反应速率。
• 勒夏特列原理：处于平衡的系统会移动以抵消任何施加的变化。
• 浓度变化会移动平衡，但不会改变 Kc。
• 压力变化仅在 Δn（气体）≠ 0 时移动平衡；Kc 和 Kp 不变。
• 温度变化会移动平衡并改变 Kc 和 Kp。
• 催化剂提高达到平衡的速率，但不影响位置、产率或 Kc/Kp。
• Kc 使用浓度（mol dm⁻³）；Kp 使用分压（kPa 或 atm）。
• 哈伯法使用折中条件：400-450°C，200 atm，铁催化剂。

Mastering chemical equilibrium is not just about memorising Le Chatelier’s Principle — it is about understanding how to apply it systematically, calculating Kc and Kp correctly, and avoiding the common pitfalls that examiners love to test. With the ICE table method and a solid grasp of the underlying principles, you can tackle any equilibrium question with confidence.

掌握化学平衡不仅仅是记住勒夏特列原理——而是要理解如何系统地应用它，正确计算 Kc 和 Kp，并避免出题人喜欢考察的常见陷阱。通过 ICE 表格方法和对基本原理的扎实掌握，你可以自信地应对任何平衡问题。