Introducing Entropy | 熵的引入

📚 Introducing Entropy | 熵的引入

Entropy is one of the most conceptually challenging yet fundamentally important ideas in A-Level Chemistry. It explains why reactions happen, why some processes are spontaneous, and why energy spreads out. This article introduces entropy in a clear, step-by-step way, aligned with the Cambridge A-Level syllabus.

熵是 A-Level 化学中最具概念挑战性但又至关重要的思想之一。它解释了反应为何发生、某些过程为何自发进行、以及能量为何会扩散。本文按照剑桥 A-Level 考纲的要求，循序渐进地介绍熵的概念。

1. What Is Entropy? | 什么是熵？

Entropy (symbol S) is a measure of the disorder or randomness of a system. The more disordered a system is, the higher its entropy. In chemical terms, we can think of entropy as a measure of how energy is spread out among the particles in a system.

熵（符号 S）是衡量系统无序程度或随机程度的物理量。系统越无序，其熵值越高。从化学角度来看，我们可以把熵理解为能量在系统粒子之间分布方式的度量。

A system with high entropy has particles moving randomly in many possible ways, with energy spread widely. A system with low entropy is highly ordered, with particles arranged in a regular pattern and energy concentrated in few arrangements.

高熵系统的粒子以多种可能的方式随机运动，能量分布广泛。低熵系统则高度有序，粒子按规则排列，能量集中在少数排列方式中。

2. Why Does Entropy Matter? | 熵为何重要？

In previous topics, you learned that exothermic reactions (ΔH negative) tend to be spontaneous. However, this is not the whole story. Some endothermic reactions occur spontaneously too, such as the dissolving of ammonium nitrate in water, which absorbs heat but still proceeds. Clearly, enthalpy change alone cannot predict spontaneity.

在之前的学习中，你了解到放热反应（ΔH 为负）往往是自发的。但这不是故事的全部。有些吸热反应也能自发进行，例如硝酸铵溶于水是吸热的，但该过程仍能发生。显然，仅凭焓变无法预测反应的自发性。

To determine whether a process is spontaneous, we must consider both enthalpy change and entropy change. Natural processes tend to proceed in the direction that increases the total entropy of the universe. This is a statement of the second law of thermodynamics.

要判断一个过程是否自发，我们必须同时考虑焓变和熵变。自然过程倾向于沿着使宇宙总熵增加的方向进行。这就是热力学第二定律的表述。

3. States of Matter and Entropy | 物质状态与熵

The entropy of a substance depends strongly on its physical state. For the same substance under the same conditions, the order of entropy is:

物质的熵与其物理状态密切相关。在相同条件下，同一种物质的熵值大小顺序为：

S(gas) > S(liquid) > S(solid)

A gas has the highest entropy because its particles move freely, occupy a large volume, and can adopt an enormous number of arrangements. A solid has the lowest entropy because its particles are fixed in a regular lattice with very few possible arrangements.

气体的熵最高，因为其粒子自由运动、占据较大体积，并且可以采用大量不同的排列方式。固体的熵最低，因为其粒子固定在规则的晶格中，可能的排列方式极少。

When a solid melts or a liquid boils, entropy increases significantly. Conversely, when a gas condenses or a liquid freezes, entropy decreases. This is why ΔS is positive for melting and boiling, and negative for freezing and condensing.

当固体熔化或液体沸腾时，熵显著增加。相反，当气体凝结或液体凝固时，熵减少。这就是为什么熔化过程 ΔS 为正、沸腾过程 ΔS 为正、而凝固和凝结过程 ΔS 为负的原因。

4. Factors That Increase Entropy | 使熵增加的因素

Several factors cause entropy to increase. You should be able to identify these in exam questions:

有多个因素会导致熵增加。你应该能够在考试题目中识别出这些因素：

Change of state from solid → liquid → gas: particles gain freedom of movement, increasing disorder.
Temperature increase: particles move faster and have more kinetic energy, spreading energy more widely.
Dissolving a solid in water: the solute particles separate and spread throughout the solution, greatly increasing disorder.
Increase in the number of gas molecules: more gas particles mean more possible arrangements and higher entropy.
Reaction of a solid to produce a gas: such as CaCO₃(s) → CaO(s) + CO₂(g), where gas is formed.

状态变化：固体 → 液体 → 气体：粒子获得运动自由度，无序程度增加。
温度升高：粒子运动加快，动能增大，能量分布更广泛。
固体溶于水：溶质粒子分离并在溶液中扩散，无序程度大大增加。
气体分子数增加：气体粒子越多，可能的排列方式越多，熵越高。
固体反应生成气体：例如 CaCO₃(s) → CaO(s) + CO₂(g)，生成了气体。

5. Standard Entropy Values | 标准熵值

The standard entropy of a substance, S°, is the entropy of one mole of the substance under standard conditions (298 K and 1 atm pressure). Its units are J K⁻¹ mol⁻¹.

物质的标准熵 S° 是指 1 摩尔该物质在标准条件下（298 K 和 1 atm 压力）的熵值。其单位是 J K⁻¹ mol⁻¹。

Here are some standard entropy values you should be familiar with:

以下是一些你应该熟悉的标准熵值：

Substance 物质	State 状态	S° / J K⁻¹ mol⁻¹
C (diamond)	s	2.4
C (graphite)	s	5.7
NaCl	s	72.1
H₂O	l	69.9
H₂O	g	188.8
CO₂	g	213.7
NH₃	g	192.3

Notice that gases have much higher S° values than solids, and that the same substance in the gas state has a much higher S° value than in the liquid state.

注意：气体的 S° 值远高于固体，同一物质气态的 S° 值也远高于液态。

6. Calculating the Entropy Change of a Reaction | 计算反应的熵变

For a chemical reaction, the standard entropy change of the system, ΔSᵢᵥₛₜₑₘ°, is calculated using the standard entropy values of the reactants and products:

对于化学反应，系统的标准熵变 ΔSᵢᵥₛₜₑₘ° 使用反应物和产物的标准熵值计算：

ΔSᵢᵥₛₜₑₘ° = ΣS°(products) − ΣS°(reactants)

In other words, you add up the standard entropies of all products (multiplied by their stoichiometric coefficients) and subtract the sum of the standard entropies of all reactants (also multiplied by their coefficients).

也就是说，将所有产物的标准熵相加（乘以各自的化学计量系数），然后减去所有反应物的标准熵之和（同样乘以各自的系数）。

Let us work through an example. For the reaction:

让我们通过一个例子来理解。对于反应：

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

Given S° values: NaHCO₃(s) = 102 J K⁻¹ mol⁻¹; Na₂CO₃(s) = 135 J K⁻¹ mol⁻¹; H₂O(g) = 189 J K⁻¹ mol⁻¹; CO₂(g) = 214 J K⁻¹ mol⁻¹.

已知 S° 值：NaHCO₃(s) = 102 J K⁻¹ mol⁻¹；Na₂CO₃(s) = 135 J K⁻¹ mol⁻¹；H₂O(g) = 189 J K⁻¹ mol⁻¹；CO₂(g) = 214 J K⁻¹ mol⁻¹。

ΔSᵢᵥₛₜₑₘ° = (135 + 189 + 214) − (2 × 102) = 538 − 204 = +334 J K⁻¹ mol⁻¹

The positive value makes sense: the reaction produces two moles of gas from a solid, so disorder increases greatly.

正值是合理的：该反应由固体生成了两摩尔气体，因此无序程度大大增加。

7. Dissolving and Entropy | 溶解与熵

When an ionic solid dissolves in water, the lattice structure breaks apart and the ions become dispersed throughout the solution. This greatly increases the disorder of the system, so the entropy change for dissolving is usually positive.

当离子固体溶于水时，晶格结构被破坏，离子分散到整个溶液中。这大大增加了系统的无序程度，因此溶解过程的熵变通常是正值。

However, there is an important subtlety: water molecules become ordered around the dissolved ions in a process called hydration. This ordering of water molecules decreases entropy. The overall entropy change depends on the balance between these two effects.

然而，这里有一个重要的细节：水分子会在溶解的离子周围形成有序排列，这一过程称为水合作用。水分子的这种有序化会降低熵。总熵变取决于这两种效应的平衡。

Despite this subtlety, for most ionic substances, dissolving in water leads to an increase in entropy overall. This is why many solids dissolve spontaneously even when the process is endothermic.

尽管有这些细节，对大多数离子化合物而言，溶于水总体上仍然导致熵增加。这就是为什么许多固体即使溶解过程是吸热的，也仍能自发溶解。

8. Predicting the Sign of ΔS | 判断 ΔS 的符号

In exams, you may be asked to predict whether the entropy change for a reaction is positive or negative without doing calculations. The key is to compare the number of moles of gas on the reactant side and the product side:

在考试中，你可能会被要求在不计算的情况下判断反应的熵变是正还是负。关键是比较反应物一侧和产物一侧的气体摩尔数：

More gas moles on the product side → ΔS is positive (disorder increases)
More gas moles on the reactant side → ΔS is negative (disorder decreases)
Equal gas moles on both sides → ΔS is small, sign uncertain (other factors matter)

产物侧气体摩尔数更多 → ΔS 为正（无序度增加）
反应物侧气体摩尔数更多 → ΔS 为负（无序度减少）
两侧气体摩尔数相等 → ΔS 较小，符号不确定（其他因素起作用）

For example, in the reaction N₂(g) + 3H₂(g) → 2NH₃(g), there are 4 moles of gas on the left and only 2 moles on the right. We would predict ΔS is negative, which is confirmed by calculation: ΔSᵢᵥₛₜₑₘ° = −199 J K⁻¹ mol⁻¹.

例如，在反应 N₂(g) + 3H₂(g) → 2NH₃(g) 中，左侧有 4 摩尔气体，右侧只有 2 摩尔。我们可以预测 ΔS 为负，计算结果证实了这一点：ΔSᵢᵥₛₜₑₘ° = −199 J K⁻¹ mol⁻¹。

9. Total Entropy Change and Spontaneity | 总熵变与自发性

The second law of thermodynamics states that for a process to be spontaneous, the total entropy of the universe must increase. The universe consists of the system (what we are studying) and the surroundings (everything else):

热力学第二定律指出：一个过程若要自发进行，宇宙的总熵必须增加。宇宙由系统（我们研究的对象）和环境（其他一切）组成：

ΔSᵢₒₜₐₗ = ΔSᵢᵥₛₜₑₘ + ΔSₛᵤᵣᵣₒᵤₙₔᵢₙ₉ₛ

For a spontaneous process, ΔSᵢₒₜₐₗ > 0. The entropy change of the surroundings is related to the enthalpy change of the system:

对于自发过程，ΔSᵢₒₜₐₗ > 0。环境的熵变与系统的焓变有关：

ΔSₛᵤᵣᵣₒᵤₙₔᵢₙ₉ₛ = −ΔHᵢᵥₛₜₑₘ / T

Here T is the temperature in kelvin. If the reaction is exothermic (ΔH negative), the surroundings gain heat, so their entropy increases. If the reaction is endothermic (ΔH positive), the surroundings lose heat, so their entropy decreases.

其中 T 是开尔文温度。如果反应放热（ΔH 为负），环境获得热量，因此环境的熵增加。如果反应吸热（ΔH 为正），环境失去热量，因此环境的熵减少。

10. Linking Entropy to Gibbs Free Energy | 熵与吉布斯自由能的联系

Predicting spontaneity by calculating ΔSᵢₒₜₐₗ directly can be cumbersome. Chemists prefer to use the Gibbs free energy change, ΔG, which combines both factors into a single equation:

通过直接计算 ΔSᵢₒₜₐₗ 来预测自发性可能很繁琐。化学家更倾向于使用吉布斯自由能变 ΔG，它将两个因素结合在一个等式中：

ΔG = ΔH − TΔS

This equation is derived from the total entropy criterion. A process is spontaneous when:

该方程源自总熵判据。一个过程自发时满足：

ΔG < 0

The relationship between ΔG and total entropy is:

ΔG 与总熵的关系是：

ΔSᵢₒₜₐₗ = −ΔG / T

When ΔG is negative, ΔSᵢₒₜₐₗ is positive, and the process is spontaneous. The sign of ΔG depends on the interplay between ΔH, T, and ΔS:

当 ΔG 为负时，ΔSᵢₒₜₐₗ 为正，过程自发。ΔG 的符号取决于 ΔH、T 和 ΔS 之间的相互作用：

ΔH	ΔS	Outcome 结果
Negative (exothermic)	Positive	Spontaneous at all temperatures 任何温度下都自发
Positive (endothermic)	Negative	Non-spontaneous at all temperatures 任何温度下都不自发
Negative (exothermic)	Negative	Spontaneous at low T only 仅在低温下自发
Positive (endothermic)	Positive	Spontaneous at high T only 仅在高温下自发

11. Worked Example | 例题解析

Consider the thermal decomposition of calcium carbonate:

考虑碳酸钙的热分解：

CaCO₃(s) → CaO(s) + CO₂(g)

Given: ΔH° = +178 kJ mol⁻¹; S°(CaCO₃) = 93 J K⁻¹ mol⁻¹; S°(CaO) = 40 J K⁻¹ mol⁻¹; S°(CO₂) = 214 J K⁻¹ mol⁻¹.

已知：ΔH° = +178 kJ mol⁻¹；S°(CaCO₃) = 93 J K⁻¹ mol⁻¹；S°(CaO) = 40 J K⁻¹ mol⁻¹；S°(CO₂) = 214 J K⁻¹ mol⁻¹。

Step 1: Calculate ΔSᵢᵥₛₜₑₘ°:

第一步：计算 ΔSᵢᵥₛₜₑₘ°：

ΔSᵢᵥₛₜₑₘ° = (40 + 214) − (93) = +161 J K⁻¹ mol⁻¹

Step 2: Calculate ΔG at 298 K. Remember to convert ΔS to kJ:

第二步：计算 298 K 时的 ΔG。注意将 ΔS 转换为 kJ：

ΔG = ΔH − TΔS = 178 − (298 × 0.161) = 178 − 48.0 = +130 kJ mol⁻¹

Since ΔG > 0 at 298 K, the reaction is non-spontaneous at room temperature. This is why calcium carbonate does not decompose on its own at room temperature.

由于在 298 K 时 ΔG > 0，该反应在室温下不能自发进行。这就是为什么碳酸钙在室温下不会自行分解的原因。

Step 3: Find the temperature at which the reaction becomes spontaneous. Set ΔG = 0:

第三步：求反应变为自发的温度。令 ΔG = 0：

0 = 178 − T(0.161)

T = 178 / 0.161 = 1106 K

Above approximately 1106 K (833 °C), the reaction becomes spontaneous. This matches the industrial conditions for lime production, which use temperatures around 900–1000 °C.

在大约 1106 K（833 °C）以上，反应变为自发。这与石灰生产的工业条件相符，工业上使用约 900–1000 °C 的温度。

12. Common Pitfalls in Exams | 考试常见陷阱

Students often make the following mistakes with entropy questions. Avoid them to secure full marks:

学生在熵相关的题目中经常犯以下错误。避免这些错误以拿到满分：

Forgetting units: S° and ΔS are measured in J K⁻¹ mol⁻¹, but ΔH is in kJ mol⁻¹. Always convert before using ΔG = ΔH − TΔS.
Ignoring stoichiometric coefficients: Multiply each S° value by the coefficient in the balanced equation.
Confusing ΔSᵢᵥₛₜₑₘ with ΔSᵢₒₜₐₗ: The system entropy change alone does not determine spontaneity; you need the total.
Saying “entropy always increases”: This is false. Entropy of the universe increases for spontaneous processes, but the entropy of a system can decrease (e.g., freezing water).
Predicting ΔS only by counting total moles: Only count moles of gas, not solids or liquids, when predicting the sign of ΔS.

忘记单位：S° 和 ΔS 的单位是 J K⁻¹ mol⁻¹，而 ΔH 的单位是 kJ mol⁻¹。使用 ΔG = ΔH − TΔS 之前务必换算。
忽略化学计量系数：每个 S° 值都要乘以平衡方程中的系数。
混淆 ΔSᵢᵥₛₜₑₘ 与 ΔSᵢₒₜₐₗ：仅凭系统熵变不能判断自发性；需要看总熵变。
认为”熵总是增加”：这是错误的。自发过程是宇宙的熵增加，但系统的熵可以减小（例如水结冰）。
仅通过总摩尔数判断 ΔS：预测 ΔS 的符号时只数气体摩尔数，固体和液体不算。

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Introducing Entropy | 熵的引入