Entropy, Enthalpy and Free Energy Changes: How They Interconnect | 熵变、焓变与自由能的关联

📚 Entropy, Enthalpy and Free Energy Changes: How They Interconnect | 熵变、焓变与自由能的关联

In A-Level Chemistry, the thermodynamic trio — enthalpy change (ΔH), entropy change (ΔS) and Gibbs free energy change (ΔG) — forms the backbone of predicting whether a reaction is feasible. While enthalpy tells us about heat exchange, entropy tells us about disorder, and free energy combines both to answer the ultimate question: will this reaction happen spontaneously?

在A-Level化学中，热力学三兄弟——焓变（ΔH）、熵变（ΔS）和吉布斯自由能变（ΔG）——构成了判断反应是否可行的核心框架。焓变告诉我们热量交换的情况，熵变告诉我们混乱度的变化，而自由能则综合两者回答最关键的问题：这个反应能否自发进行？

1. Enthalpy Change (ΔH) — Heat at Constant Pressure | 焓变（ΔH）——恒压下的热量变化

Enthalpy change (ΔH) is the heat energy transferred in a reaction at constant pressure. A negative ΔH means the reaction releases heat to the surroundings (exothermic), while a positive ΔH means the reaction absorbs heat (endothermic). For many years, students assume that exothermic reactions are always spontaneous — but this is only half the story.

焓变（ΔH）是在恒压条件下反应中转移的热能。ΔH为负表示反应向环境释放热量（放热），ΔH为正表示反应吸收热量（吸热）。许多学生长期以为放热反应一定是自发的——但这只是故事的一半。

For example, the dissolving of ammonium nitrate in water is endothermic (ΔH is positive), yet it occurs spontaneously at room temperature. Clearly, enthalpy alone cannot predict spontaneity.

例如，硝酸铵溶于水是吸热的（ΔH为正），但在室温下却能自发进行。显然，单靠焓变无法预测反应的自发性。

2. Entropy Change (ΔS) — The Measure of Disorder | 熵变（ΔS）——混乱度的量度

Entropy (S) is a measure of the degree of disorder or randomness in a system. Gases have much higher entropy than liquids, which in turn have higher entropy than solids. When a substance changes from solid to liquid to gas, or when the number of gas molecules increases in a reaction, entropy increases.

熵（S）是系统混乱程度或随机性的量度。气体的熵远高于液体，液体的熵又高于固体。当物质从固态变为液态再变为气态，或反应中气体分子数增加时，熵增大。

Entropy change (ΔS) can be calculated using the formula:

熵变（ΔS）可以用以下公式计算：

ΔS = ΣS(products) − ΣS(reactants)

Standard entropy values (S°, measured in J K⁻¹ mol⁻¹) are provided in data tables. Note the units: entropy values are usually given in joules per kelvin per mole, not kilojoules.

标准熵值（S°，单位J K⁻¹ mol⁻¹）在数据表中给出。请注意单位：熵值通常以焦耳每开尔文每摩尔给出，而非千焦。

3. Gibbs Free Energy — The Master Predictor | 吉布斯自由能——终极预测者

Josiah Willard Gibbs introduced a single function that combines enthalpy and entropy to predict spontaneity. The Gibbs free energy change (ΔG) is defined as:

约西亚·威拉德·吉布斯提出了一个将焓和熵结合在一起的单一函数来预测自发性。吉布斯自由能变（ΔG）定义为：

ΔG = ΔH − TΔS

In this equation, ΔH is the enthalpy change (kJ mol⁻¹), T is the temperature in kelvin, and ΔS is the entropy change (converted to kJ K⁻¹ mol⁻¹ before multiplication).

在这个方程中，ΔH是焓变（kJ mol⁻¹），T是开尔文温度，ΔS是熵变（在相乘之前需转换为kJ K⁻¹ mol⁻¹）。

The beauty of this equation lies in its simplicity: it tells us the net driving force of a reaction. Enthalpy pushes the reaction toward lower energy (favourable), while entropy pushes it toward greater disorder (also favourable). Temperature controls the balance between them.

这个方程的优美之处在于它的简洁：它告诉我们反应的净驱动力。焓推动反应趋向更低能量（有利），熵推动反应趋向更大混乱度（同样有利）。温度控制着两者之间的平衡。

4. Decoding ΔG — Spontaneity and Equilibrium | 解读ΔG——自发与平衡

The sign of ΔG determines the feasibility of a reaction under standard conditions:

ΔG的符号决定了反应在标准条件下的可行性：

ΔG < 0 (negative): The reaction is spontaneous (feasible). Products are favoured at equilibrium.
ΔG = 0: The reaction is at equilibrium. No net change occurs.
ΔG > 0 (positive): The reaction is non-spontaneous. Reactants are favoured; the reverse reaction is spontaneous.

ΔG < 0（负值）：反应自发进行（可行），平衡时产物占优。
ΔG = 0：反应处于平衡状态，无净变化。
ΔG > 0（正值）：反应非自发，平衡时反应物占优；逆反应自发。

It is crucial to understand that ΔG only predicts feasibility — it says nothing about the rate of reaction. A reaction with a very negative ΔG can still be infinitely slow without a suitable catalyst, such as the conversion of diamond to graphite.

必须理解的是，ΔG只能预测可行性——它完全不涉及反应速率。一个ΔG非常负的反应如果没有合适的催化剂仍然可能极慢，例如金刚石转化为石墨。

5. The Four Combinations of ΔH and ΔS | ΔH与ΔS的四种组合

The relationship ΔG = ΔH − TΔS gives rise to four distinct scenarios depending on the signs of ΔH and ΔS. This is a favourite exam question theme:

ΔG = ΔH − TΔS的关系根据ΔH和ΔS的符号产生了四种不同的情形。这是考试题中最喜欢考查的主题之一：

ΔH	ΔS	Feasibility \| 可行性
Negative (exothermic)	Positive (more disorder)	Always spontaneous at all temperatures (ΔG always negative)
Positive (endothermic)	Negative (less disorder)	Never spontaneous at any temperature (ΔG always positive)
Negative (exothermic)	Negative (less disorder)	Spontaneous only at low temperatures
Positive (endothermic)	Positive (more disorder)	Spontaneous only at high temperatures

For the last two cases, the temperature at which the reaction switches from non-spontaneous to spontaneous is found by setting ΔG = 0:

对于最后两种情形，反应从非自发变为自发的转变温度可以通过令ΔG = 0求得：

T = ΔH / ΔS

At this temperature, the system is at equilibrium. Above it, the entropy term dominates; below it, the enthalpy term dominates.

在这个温度下，系统处于平衡。高于此温度，熵项占主导；低于此温度，焓项占主导。

6. Temperature-Dependent Feasibility — A Closer Look | 温度依赖性可行性——深入探讨

Consider the thermal decomposition of calcium carbonate:

以碳酸钙的热分解为例：

CaCO₃(s) → CaO(s) + CO₂(g)

This reaction is endothermic (ΔH is positive, about +178 kJ mol⁻¹) and involves the production of a gas, so ΔS is also positive (about +160 J K⁻¹ mol⁻¹, or +0.160 kJ K⁻¹ mol⁻¹). At room temperature (298 K):

这个反应是吸热的（ΔH为正，约+178 kJ mol⁻¹），并且产生了一种气体，因此ΔS也为正（约+160 J K⁻¹ mol⁻¹，即+0.160 kJ K⁻¹ mol⁻¹）。在室温（298 K）下：

ΔG = (+178) − (298 × 0.160) = +178 − 47.7 = +130.3 kJ mol⁻¹

Since ΔG is positive, CaCO₃ is stable at room temperature. But when we heat the system, the TΔS term grows. Using the transition temperature:

因为ΔG为正，CaCO₃在室温下是稳定的。但当我们加热系统时，TΔS项增大。使用转变温度公式：

T = 178 / 0.160 = 1113 K (approximately 840 °C)

Above this temperature, ΔG becomes negative and limestone decomposes spontaneously. This is exactly why lime kilns operate at high temperatures.

超过这个温度，ΔG变为负值，石灰石自发分解。这正是石灰窑在高温下运行的原因。

7. Calculating ΔS from Standard Entropies | 从标准熵计算ΔS

A typical exam question provides standard entropy values (S°) for reactants and products and asks for the entropy change of the reaction. The steps are straightforward:

典型的考题会给出反应物和产物的标准熵值（S°），要求计算反应的熵变。步骤非常直接：

Write the balanced chemical equation with state symbols.
Multiply each substance’s S° value by its stoichiometric coefficient.
Apply the formula: ΔS = ΣS°(products) − ΣS°(reactants).

写出带有状态符号的配平化学方程式。
将每种物质的S°乘以其化学计量系数。
套用公式：ΔS = ΣS°(产物) − ΣS°(反应物)。

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), given S°(N₂) = 192 J K⁻¹ mol⁻¹, S°(H₂) = 131 J K⁻¹ mol⁻¹, and S°(NH₃) = 193 J K⁻¹ mol⁻¹:

对于反应N₂(g) + 3H₂(g) → 2NH₃(g)，已知S°(N₂) = 192 J K⁻¹ mol⁻¹、S°(H₂) = 131 J K⁻¹ mol⁻¹、S°(NH₃) = 193 J K⁻¹ mol⁻¹：

ΔS = [2 × 193] − [192 + (3 × 131)] = 386 − (192 + 393) = 386 − 585 = −199 J K⁻¹ mol⁻¹

The negative sign makes chemical sense: three gas molecules and one gas molecule combine to form two gas molecules — the total number of gas particles decreases, so disorder decreases.

负号在化学上有明确意义：三个气体分子和一个气体分子结合生成两个气体分子——气体粒子总数减少，混乱度降低。

8. Units and Conversions — The Classic Trap | 单位与换算——经典陷阱

One of the most common mistakes in CIE exam papers is mixing up units. ΔH values are typically given in kJ mol⁻¹, while ΔS values are typically given in J K⁻¹ mol⁻¹. Before substituting into ΔG = ΔH − TΔS, you must convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000.

CIE考题中最常见的错误之一是混用单位。ΔH通常以kJ mol⁻¹给出，而ΔS通常以J K⁻¹ mol⁻¹给出。在代入ΔG = ΔH − TΔS之前，必须将ΔS除以1000换算为kJ K⁻¹ mol⁻¹。

Worked example: For a reaction, ΔH = −92 kJ mol⁻¹ and ΔS = −180 J K⁻¹ mol⁻¹. At what temperature does the reaction cease to be spontaneous?

例题：某反应ΔH = −92 kJ mol⁻¹，ΔS = −180 J K⁻¹ mol⁻¹。该反应在什么温度下不再自发进行？

ΔS = −180 J K⁻¹ mol⁻¹ = −0.180 kJ K⁻¹ mol⁻¹

At the boundary: ΔG = 0, so T = ΔH / ΔS = (−92) / (−0.180) = 511 K

Since ΔH is negative and ΔS is negative, the reaction is feasible only when TΔS is small, i.e. at temperatures below 511 K. Above this temperature, the entropy term (which opposes spontaneity here) overwhelms the enthalpy term.

由于ΔH为负、ΔS为负，反应只有在TΔS较小的时候才可行，即温度低于511 K。在此温度之上，熵项（这里阻碍自发性）压倒了焓项。

9. Feasibility vs Rate — A Crucial Distinction | 可行性与速率——一个关键区别

Students frequently confuse whether a reaction occurs with how fast it occurs. ΔG thermodynamically tells us whether a reaction can happen; the activation energy and the presence of a catalyst determine whether it will happen at an observable rate.

学生经常混淆反应是否发生和反应有多快发生这两个问题。ΔG从热力学角度告诉我们反应能否发生；而活化能和催化剂的存在决定反应能否以可观察的速率发生。

A classic example: the conversion of diamond to graphite is thermodynamically spontaneous (ΔG is negative), yet we are not worried about our diamonds disappearing. The activation energy for breaking diamond’s giant covalent lattice is so high that the process is kinetically inhibited at room temperature.

一个经典例子：金刚石转化为石墨在热力学上是自发的（ΔG为负），但我们并不担心钻石会消失。因为破坏金刚石巨大共价晶格所需的活化能太高，在室温下这一过程受到动力学抑制。

Exam questions often ask: “A reaction has ΔG < 0 but no observable reaction occurs. Explain why.” The answer must reference kinetic factors — high activation energy, insufficient energy for effective collisions, or the absence of a catalyst.

考试题经常问：”某反应ΔG < 0但没有观察到反应发生，请解释原因。”答案必须提及动力学因素——活化能过高、没有足够的能量实现有效碰撞，或缺少催化剂。

10. Predicting Entropy Changes Qualitatively | 定性预测熵变

For exam questions that do not provide data, you must be able to predict the sign of ΔS based on the reaction system:

对于不提供数据的考试题，你必须能够基于反应体系判断ΔS的符号：

Increase in the number of gas molecules: ΔS is positive.
Decrease in the number of gas molecules: ΔS is negative.
Change of state from solid to liquid or gas, or from liquid to gas: ΔS is positive.
Dissolving a solid in a solution (often): ΔS is positive, as the particles spread out.

气体分子数增加：ΔS为正。
气体分子数减少：ΔS为负。
从固态变为液态或气态，或从液态变为气态：ΔS为正。
固体溶解于溶液（通常）：ΔS为正，因为粒子分散开来。

Consider the reaction 2H₂O₂(l) → 2H₂O(l) + O₂(g). One liquid decomposes to give one liquid and one gas — the number of gas particles increases, so the entropy increases. This is consistent with the fact that this decomposition is spontaneous from a thermodynamic perspective.

考虑反应2H₂O₂(l) → 2H₂O(l) + O₂(g)。一种液体分解为一种液体和一种气体——气体粒子数增加，因此熵增大。这与该分解反应在热力学上自发的事实一致。

11. Standard Conditions and Exam Conventions | 标准条件与考试约定

In CIE A-Level Chemistry, standard free energy change (ΔG°) refers to conditions of 298 K (25 °C), 100 kPa and 1 mol dm⁻³ solutions. Examiners expect you to state these conditions when defining standard enthalpy of formation or other standard thermodynamic quantities.

在CIE A-Level化学中，标准自由能变（ΔG°）指的是298 K（25 °C）、100 kPa和1 mol dm⁻³溶液的条件。考官期望你在定义标准生成焓或其他标准热力学量时表述这些条件。

When ΔG° = 0, the reaction is at equilibrium under standard conditions. In this state, the equilibrium constant K equals 1. The relationship between ΔG° and the equilibrium constant is a deeper topic, but at A-Level it suffices to know that a large negative ΔG° corresponds to K > 1 (products favoured), while a large positive ΔG° corresponds to K < 1 (reactants favoured).

当ΔG° = 0时，反应在标准条件下处于平衡。在这一状态下，平衡常数K等于1。ΔG°与平衡常数之间关系的更深入内容超出本阶段，但在A-Level中只需知道ΔG°有很大负值时K > 1（产物占优），ΔG°有很大正值时K < 1（反应物占优）。

12. Exam Strategy Summary | 考试策略总结

When facing a thermodynamics question in the exam, follow these steps to maximise marks:

在考试中面对热力学问题时，按以下步骤操作以最大化得分：

Always convert ΔS from J to kJ before using ΔG = ΔH − TΔS.
Write down the equation with units at every step.
Distinguish between ΔS (reaction entropy change) and S° (standard molar entropy).
Check the feasibility: if ΔG < 0, the reaction is feasible, but not necessarily fast.
When asked for a “temperature at which a reaction becomes feasible”, set ΔG = 0 and solve T = ΔH / ΔS.

在使用ΔG = ΔH − TΔS前，务必将ΔS从J换算为kJ。
每一步都写出带单位的方程表达式。
区分ΔS（反应熵变）与S°（标准摩尔熵）。
判断可行性：如果ΔG < 0，反应可行，但不一定快。
当题目要求”反应可行的温度”时，令ΔG = 0，解T = ΔH / ΔS。

Mastering how enthalpy, entropy and temperature work together through the Gibbs equation is not just a box-ticking exercise — it explains phenomena ranging from why ice melts above 0 °C to why thermal decomposition reactions require high temperatures. Practice with past papers and always check your units — success in this topic comes down to meticulous habits.

掌握焓、熵和温度如何通过吉布斯方程协同作用，不仅仅是为了完成答题任务——它解释了从冰为何在0 °C以上融化到热分解反应为何需要高温等一系列现象。多练习历年真题，永远检查你的单位——在这个知识点上取得成功的秘诀就是一丝不苟的习惯。

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Entropy, Enthalpy and Free Energy Changes: How They Interconnect | 熵变、焓变与自由能的关联