Gibbs Free Energy | 吉布斯自由能 考点精讲

📚 Gibbs Free Energy | 吉布斯自由能 考点精讲

Gibbs free energy is the ultimate ‘boss level’ topic in thermochemistry for OCR IGCSE Chemistry. It gives us a single, elegant criterion to predict whether a chemical reaction will happen on its own – without needing to run the experiment first.

吉布斯自由能是 OCR IGCSE 化学热化学板块的终极考点。它为我们提供了一个简洁而强大的判据，让我们无需动手实验，就能提前判断一个化学反应是否能自发进行。

1. What is Gibbs Free Energy? | 什么是吉布斯自由能？

Gibbs free energy, symbol G, combines the two driving forces of any chemical change – the tendency to reach lower energy (enthalpy) and the tendency to become more disordered (entropy). The change in Gibs free energy, ΔG, tells us if a reaction is feasible at a given temperature.

吉布斯自由能，符号为 G，把化学变化背后的两大驱动力结合在了一起：一是体系趋向于达到更低的能量（焓），二是体系趋向于变得更加混乱（熵）。吉布斯自由能的变化值 ΔG 能够告诉我们，在既定温度下某一反应是否可行。

Think of ΔG as a balance account. If ΔG is negative, the universe has ‘approved’ the change; if positive, energy must be paid in for the reaction to happen.

你可以把 ΔG 想象成一个收支账单。如果 ΔG 为负值，代表宇宙给这个反应“开了绿灯”；如果 ΔG 为正，就需要外部提供能量，反应才能进行下去。

2. The Master Equation | 核心方程

The entire topic hangs on one equation that you must memorise perfectly:

整个考点的核心只有一个方程式，你必须牢牢记住：

ΔG = ΔH – TΔS

where ΔG is the change in Gibbs free energy (kJ mol⁻¹), ΔH is the enthalpy change (kJ mol⁻¹), T is the absolute temperature in kelvin (K), and ΔS is the entropy change (kJ K⁻¹ mol⁻¹). Note the units: ΔS is often given in kJ K⁻¹ mol⁻¹ to match ΔH, but you may also see it in J K⁻¹ mol⁻¹. Always convert to kJ before using the formula.

方程中，ΔG 是吉布斯自由能变（单位 kJ mol⁻¹），ΔH 是焓变（kJ mol⁻¹），T 是绝对温度（开尔文，K），ΔS 是熵变（kJ K⁻¹ mol⁻¹）。注意单位：ΔS 经常会以 J K⁻¹ mol⁻¹ 给出，使用前必须换算成 kJ，以保证与 ΔH 单位一致。

In OCR exams, you will be asked to plug in numbers, rearrange the equation to solve for T, or interpret the sign of ΔG given the signs of ΔH and ΔS. Practice converting °C to K by adding 273. This simple step is frequently tested.

在 OCR 试卷中，你会被要求直接代入数据计算，或者对方程进行变形求解温度 T，又或者根据 ΔH 和 ΔS 的符号解读 ΔG 的正负。摄氏温度转换为开尔文温度（加 273）是非常基础的步骤，却经常出现在考题里。

3. Enthalpy Change (ΔH) – The Energy Factor | 焓变 ΔH——能量因素

ΔH is the heat energy transferred during a reaction at constant pressure. A negative ΔH (exothermic) means the system loses energy to the surroundings; the products are more stable than the reactants. This contributes a negative term to ΔG, making a spontaneous reaction more likely.

ΔH 表示在恒压条件下反应体系转移的热量。ΔH 为负（放热反应）意味着体系向周围环境释放能量，生成物比反应物更稳定。这会给 ΔG 贡献一个负项，让反应更容易自发。

However, an exothermic reaction does not guarantee spontaneity. Many exothermic processes (like rusting) are spontaneous, but some require an initial energy input. The entropy term also has a vote.

然而，放热并不必然保证反应自发。不少放热过程（例如铁生锈）的确是自发的，但也有部分放热反应需要先投入能量才能启动。别忘了，熵项同样拥有投票权。

4. Entropy Change (ΔS) – The Disorder Factor | 熵变 ΔS——混乱度因素

Entropy, S, is a measure of the disorder or randomness of a system. Gases have the highest entropy, followed by liquids, and then solids. A reaction that produces more gas molecules than it consumes usually has a positive ΔS.

熵 S 是衡量体系混乱度或随机度的物理量。气体的熵最高，液体次之，固体最低。如果一个反应生成的气体分子数多于反应物消耗的气体分子数，那么该反应的 ΔS 通常为正值。

A positive ΔS favours spontaneity because –TΔS becomes negative, reducing ΔG. A negative ΔS (system becomes more ordered) makes ΔG more positive and works against spontaneity. Crystallisation from a solution is an example where ΔS is negative.

正值的 ΔS 有利于反应自发，因为 –TΔS 这一项会变成负数，从而拉低 ΔG。反之，如果 ΔS 为负（体系变得更加有序），就会使 ΔG 增大，不利于自发。从溶液中析出晶体就是 ΔS 为负的典型例子。

5. Temperature – The Decider | 温度 T——最终裁决者

The temperature T is a scaling factor in the TΔS term. Its effect is crucial: at low temperatures, the enthalpy term dominates; at high temperatures, the entropy term becomes magnified. This explains why some reactions only become spontaneous above a certain temperature.

温度 T 在 TΔS 项中扮演倍增因子的角色。它的影响至关重要：低温时焓变项占主导地位；高温时熵变项被放大。这就解释了为什么有些反应只有在高于某一温度时才会自发进行。

To find the temperature at which a reaction just becomes feasible (ΔG = 0), set ΔH – TΔS = 0 and solve: T = ΔH / ΔS. This is one of the most common calculation questions in the exam.

要找出反应刚好变得可行的温度（此时 ΔG = 0），只需令 ΔH – TΔS = 0，然后求解即可：T = ΔH / ΔS。这是考题中最常见的计算题型之一。

6. The Four Possibilities – Sign Analysis | 四种情况——符号分析法

By analysing the signs of ΔH and ΔS, we can predict how ΔG will behave with temperature. OCR examiners love testing this table:

通过分析 ΔH 和 ΔS 的正负号，我们可以预测 ΔG 随温度的变化规律。OCR 考官尤其偏爱考察这个表格：

ΔH	ΔS	ΔG = ΔH – TΔS	Spontaneous?
Negative (–)	Positive (+)	Always negative	Yes, at all temperatures
Positive (+)	Negative (–)	Always positive	No, at any temperature
Negative (–)	Negative (–)	Negative at low T, positive at high T	Yes, only below a certain temperature
Positive (+)	Positive (+)	Positive at low T, negative at high T	Yes, only above a certain temperature

This table summarises the entire qualitative aspect of Gibbs free energy. Memorise the reasoning: an exothermic reaction with increasing disorder is the ‘ideal combination’, while an endothermic reaction with decreasing order will never be spontaneous.

这张表概括了吉布斯自由能全部定性分析的内容。请牢记背后的推理逻辑：放热且混乱度增加的反应是“理想组合”，而吸热且混乱度降低的反应在任何温度下都不会自发。

7. Calculating ΔG Step-by-Step | ΔG 计算实战步骤

Follow this method for every numerical problem: (1) Write the equation ΔG = ΔH – TΔS. (2) Check units – convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000. (3) Convert °C to K by adding 273. (4) Substitute values and calculate. (5) State the conclusion: if ΔG < 0, the reaction is feasible.

对于每一道计算题，请按以下步骤操作：（1）写出方程 ΔG = ΔH – TΔS。（2）检查单位——将 ΔS 从 J K⁻¹ mol⁻¹ 转换为 kJ K⁻¹ mol⁻¹，除以 1000。（3）将摄氏温度转换为开尔文，加上 273。（4）代入数值计算结果。（5）根据结果得出结论：若 ΔG < 0，反应可行。

For example: A reaction has ΔH = –110 kJ mol⁻¹ and ΔS = –200 J K⁻¹ mol⁻¹. Is it spontaneous at 25 °C? Convert ΔS to –0.200 kJ K⁻¹ mol⁻¹; T = 298 K. ΔG = –110 – (298 × –0.200) = –110 + 59.6 = –50.4 kJ mol⁻¹. ΔG is negative, so the reaction is spontaneous at room temperature.

例如：某反应的 ΔH = –110 kJ mol⁻¹，ΔS = –200 J K⁻¹ mol⁻¹，问 25 °C 下反应是否自发？先将 ΔS 换算为 –0.200 kJ K⁻¹ mol⁻¹；T = 298 K。ΔG = –110 – (298 × –0.200) = –110 + 59.6 = –50.4 kJ mol⁻¹。ΔG 为负，因此反应在室温下自发进行。

8. Free Energy and Equilibrium | 自由能与化学平衡

While a negative ΔG predicts spontaneity, it says nothing about the speed of the reaction. A reaction may be thermodynamically favoured but kinetically hindered (like diamond turning into graphite). In the context of IGCSE OCR, you are not required to derive the linkage with equilibrium constant K, but you should know that ΔG = 0 corresponds to a system at equilibrium.

虽然 ΔG 为负意味着反应在热力学上可以自发，但这与反应速率无关。一个反应可能是热力学上有利的，却因为动力学障碍而几乎不进行（例如金刚石转化为石墨）。在 IGCSE OCR 的要求范围内，你不必推导 ΔG 与平衡常数 K 的关系，但需要了解 ΔG = 0 对应于体系处于平衡状态。

When ΔG becomes zero, the driving force disappears; the rates of forward and reverse reactions are equal, and there is no net change in composition.

当 ΔG 等于零时，化学反应的驱动力消失，正逆反应速率相等，体系的组成不再随时间发生净变化。

9. Common Exam Pitfalls | 常见失分点

• Unit mismatch: Forgetting to convert J to kJ for ΔS before multiplying by T. Always pause and check units.
• 温度单位错误：在代入 TΔS 计算前忘记将 ΔS 从 J 换算为 kJ。务必在落笔前停顿一下，检查单位。
• Sign errors: Misplacing a negative sign, especially when both ΔH and ΔS are negative. Write the equation as ΔG = ΔH + (–TΔS) if it helps.
• 符号混乱：处理双重负号时出错，特别是 ΔH 和 ΔS 同为负的情况。如果不够熟练，可以把方程写成 ΔG = ΔH + (–TΔS)，帮助理清符号。
• Using Celsius instead of Kelvin: T must always be in kelvin. This is the number one mistake at IGCSE.
• 使用摄氏温度而非开尔文温度：T 必须是以开尔文为单位的绝对温度。这是 IGCSE 阶段最严重的错误。

10. Linking to Real Reactions | 真实反应中的吉布斯自由能

Use familiar reactions to make sense of ΔG. Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O. ΔH is negative, and the number of gas molecules increases (ΔS positive) – this reaction is spontaneous at all reasonable temperatures.

用熟悉的反应来理解 ΔG 的判据。甲烷燃烧：CH₄ + 2O₂ → CO₂ + 2H₂O。该反应 ΔH 为负，气体分子数增加（ΔS 为正），因此在任何合理温度下都是自发的。

Dissolving ammonium nitrate in water is endothermic (ΔH > 0) but results in a large increase in disorder (ΔS > 0). At room temperature, the –TΔS term outweighs the positive ΔH, so ΔG becomes negative and the salt dissolves spontaneously. This elegant explanation is often worth marks in extended answers.

硝酸铵溶于水的过程是吸热的（ΔH > 0），但导致混乱度大幅增加（ΔS > 0）。在室温下，–TΔS 项足以压倒吸热项，使得 ΔG 变为负值，所以盐会自发溶解。这种精妙的解释在扩展作答中常常能拿到高分。

11. Practice Quickfire Questions | 快问快答精练

Test your understanding instantly: (a) If ΔH is negative and ΔS is positive, what is the sign of ΔG at any temperature? (b) A reaction has ΔH = +45 kJ mol⁻¹ and ΔS = 150 J K⁻¹ mol⁻¹. At what temperature does it become feasible? (c) True or false: A reaction with a negative ΔG will always be fast.

即时检验你的理解：（a）若 ΔH 为负、ΔS 为正，任意温度下 ΔG 的符号是什么？ （b）某反应的 ΔH = +45 kJ mol⁻¹，ΔS = 150 J K⁻¹ mol⁻¹，反应在什么温度变得可行？ （c）判断对错：ΔG 为负的反应一定进行得很快。

Answers: (a) Always negative; (b) ΔS = 0.150 kJ K⁻¹ mol⁻¹, T = ΔH/ΔS = 45/0.150 = 300 K (27 °C); (c) False – thermodynamic feasibility does not guarantee a fast rate.

参考答案：（a）恒为负；（b）ΔS = 0.150 kJ K⁻¹ mol⁻¹，T = ΔH/ΔS = 45/0.150 = 300 K（27 °C）；（c）错误——热力学上的可行并不保证反应速率快。

12. Summary – Your Gibbs Free Energy Checklist | 总结——你的吉布斯自由能自查清单

Before the exam, ensure you can: recite the equation ΔG = ΔH – TΔS; explain the meaning of each symbol and its unit; convert J to kJ and °C to K; predict spontaneity from the signs of ΔH and ΔS; calculate the temperature at which ΔG = 0; and state that a negative ΔG indicates a feasible reaction. Master these, and any Gibbs question becomes a simple puzzle.

考前务必自查是否能够：默写方程 ΔG = ΔH – TΔS；解释每个符号的含义及单位；进行 J 与 kJ、°C 与 K 的单位转换；根据 ΔH 和 ΔS 的符号预测反应的自发性；计算 ΔG = 0 时的温度；说明 ΔG 为负意味着反应可行。掌握了这些，任何关于吉布斯自由能的考题都将变成一道简单的拼图。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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