📚 Gibbs Free Energy Essentials for IGCSE Edexcel Chemistry | IGCSE Edexcel 化学：吉布斯自由能考点精讲
Gibbs free energy is one of the most powerful tools in chemistry for understanding why some reactions happen on their own while others need a constant push. In the Edexcel IGCSE specification, you are expected to calculate ΔG, interpret its sign, and grasp how enthalpy and entropy work together to determine feasibility. This article breaks down every key concept, equation, and common pitfall to help you master Gibbs free energy with confidence.
吉布斯自由能是化学中解释为什么有些反应能自发进行而另一些需要持续推动的最有力工具之一。在Edexcel IGCSE大纲中，你不仅要会计算ΔG、解读其正负，还要理解焓和熵如何共同决定反应的可行性。本文将拆解每一个核心概念、方程和常见易错点，帮助你稳稳拿下吉布斯自由能。
1. What Is Gibbs Free Energy? | 什么是吉布斯自由能？
Gibbs free energy (symbol G) is a thermodynamic potential that combines the enthalpy (heat content) and entropy (disorder) of a system. At constant temperature and pressure, the change in Gibbs free energy, ΔG, tells us whether a reaction is feasible (spontaneous) without needing external energy input. A negative ΔG means the reaction can occur on its own, while a positive ΔG means it cannot occur under those conditions.
吉布斯自由能（符号G）是一种结合了体系焓（热含量）和熵（混乱度）的热力学函数。在恒温恒压下，吉布斯自由能的变化ΔG可以告诉我们一个反应是否可行（自发），即无需外界输入能量就能发生。ΔG为负表示反应可自行发生，ΔG为正则表示在该条件下反应不能自发进行。
2. The Key Equation: ΔG = ΔH − TΔS | 关键方程：ΔG = ΔH − TΔS
The relationship is elegantly captured by a single equation:
ΔG = ΔH − TΔS
where ΔG is the Gibbs free energy change, ΔH is the enthalpy change, T is the absolute temperature in kelvin (K), and ΔS is the entropy change. It is essential to note the units: ΔH and ΔG are usually expressed in kJ mol⁻¹, while ΔS is often given in J K⁻¹ mol⁻¹. To use the equation correctly, you must convert ΔS into kJ K⁻¹ mol⁻¹ by dividing by 1000, or convert all values to joules. In IGCSE, using kJ and converting ΔS is the standard approach.
这个关系被一个简洁的方程所概括：ΔG = ΔH − TΔS，其中ΔG是吉布斯自由能变，ΔH是焓变，T是以开尔文(K)为单位的绝对温度，ΔS是熵变。必须注意单位：ΔH和ΔG通常用kJ mol⁻¹表示，而ΔS常以J K⁻¹ mol⁻¹给出。正确使用方程需要把ΔS除以1000转换成kJ K⁻¹ mol⁻¹，或者把所有数值都换算为焦耳。在IGCSE中，使用千焦并转换ΔS是标准做法。
3. Enthalpy Change (ΔH) and Its Role | 焓变(ΔH)及其作用
Enthalpy change, ΔH, is the heat absorbed or released during a reaction at constant pressure. An exothermic reaction has ΔH < 0 (negative), releasing heat and tending to make ΔG more negative, which favours spontaneity. An endothermic reaction has ΔH > 0 (positive), which alone works against spontaneity. However, a reaction with a positive ΔH can still be feasible if the entropy term −TΔS is negative enough to make ΔG negative overall.
焓变ΔH是恒压下反应吸收或放出的热量。放热反应ΔH为负值（< 0），释放热量，倾向于使ΔG更负，有利于自发进行。吸热反应ΔH为正值（> 0），仅从焓变角度看不利于自发。然而，如果熵项−TΔS足够负，使得总ΔG变为负值，吸热反应仍然可以可行。
4. Entropy Change (ΔS) and Disorder | 熵变(ΔS)与混乱度
Entropy measures the degree of disorder or randomness in a system. A positive ΔS means the products are more disordered than the reactants. Reactions that produce more gas molecules, or convert solids into liquids or gases, typically have positive ΔS. A negative ΔS indicates a more ordered system. Because the entropy term appears as −TΔS in the equation, a positive ΔS makes a negative contribution to ΔG, favouring feasibility. If ΔS is negative, then −TΔS becomes positive, hindering spontaneity.
熵衡量体系的混乱度。ΔS为正表示产物比反应物更混乱。生成更多气体分子、或将固体转化为液体或气体的反应通常具有正的ΔS。ΔS为负则表明体系变得更加有序。由于在方程中熵项以−TΔS形式出现，正的ΔS会对ΔG产生负贡献，有利于可行；负的ΔS则使−TΔS为正，阻碍自发过程。
5. The Importance of Absolute Temperature | 绝对温度的重要性
Temperature in the Gibbs free energy equation must be in kelvin (K), not degrees Celsius. To convert, use the relationship: T(K) = T(°C) + 273. (Some exam problems may use 298 K for 25°C, but always check the given data.) The magnitude of T directly influences the effect of ΔS on ΔG. At very high temperatures, the entropy term can dominate, while at low temperatures the enthalpy term is usually more significant.
吉布斯自由能方程中的温度必须使用开尔文(K)，而不是摄氏度。换算关系为：T(K) = T(°C) + 273。（有些题目中25°C常取298 K，但务必根据所给数据确认。）温度的大小直接影响熵对ΔG的贡献。在高温下，熵项可能起主导作用；在低温下，焓项通常更为重要。
6. Spontaneity Criteria: Interpreting ΔG | 自发性判据：解读ΔG
The sign of ΔG unambiguously predicts feasibility under the given conditions:
- ΔG < 0: The reaction is feasible (spontaneous) in the forward direction.
- ΔG = 0: The system is at equilibrium; no net change occurs.
- ΔG > 0: The reaction is not feasible in the forward direction. The reverse reaction would be feasible.
Importantly, a negative ΔG does not tell us anything about the rate of reaction. Some spontaneous reactions, like the conversion of diamond to graphite at room temperature, are extremely slow.
ΔG的正负可以明确预测反应在给定条件下的可行性：ΔG < 0时正向反应可行（自发）；ΔG = 0时体系处于平衡状态，无净变化；ΔG > 0时正向反应不可行，而逆反应将是可行的。需要强调的是，ΔG为负并不说明反应速率如何。有些自发反应（如室温下金刚石转化为石墨）极其缓慢，实际上几乎观察不到。
7. Worked Example: Calculating ΔG | 例题：计算ΔG
Calculate ΔG for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 298 K, given ΔH = −92 kJ mol⁻¹ and ΔS = −199 J K⁻¹ mol⁻¹. Determine whether the reaction is feasible under these conditions.
First, convert ΔS to kJ K⁻¹ mol⁻¹: ΔS = −199 J K⁻¹ mol⁻¹ = −0.199 kJ K⁻¹ mol⁻¹. Then substitute into ΔG = ΔH − TΔS:
ΔG = −92 − (298 × −0.199) = −92 + 59.3 = −32.7 kJ mol⁻¹
Since ΔG is negative (−32.7 kJ mol⁻¹), the reaction is feasible at 298 K. Notice that although ΔS is negative, the large exothermic ΔH makes the overall ΔG negative.
计算反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 在298 K下的ΔG，已知ΔH = −92 kJ mol⁻¹，ΔS = −199 J K⁻¹ mol⁻¹。判断该条件下反应是否可行。
首先将ΔS换算为kJ K⁻¹ mol⁻¹：ΔS = −199 J K⁻¹ mol⁻¹ = −0.199 kJ K⁻¹ mol⁻¹。然后代入方程ΔG = ΔH − TΔS：ΔG = −92 − (298 × −0.199) = −92 + 59.3 = −32.7 kJ mol⁻¹。由于ΔG为负值（−32.7 kJ mol⁻¹），该反应在298 K下是可行的。值得注意的是，虽然ΔS为负，但较大的放热焓值使得总ΔG为负。
8. Standard Gibbs Free Energy Change (ΔG°) | 标准吉布斯自由能变(ΔG°)
Standard conditions allow us to compare reactions fairly. For IGCSE, standard conditions are 100 kPa pressure, a temperature of 298 K (25°C), and solutions at a concentration of 1 mol dm⁻³. The symbol ΔG° refers to the Gibbs free energy change measured under these standard conditions. You can calculate ΔG° from standard ΔH° and standard ΔS° data using the same equation. Many exam questions provide tabulated ΔH° and S° values so you can work out ΔS° first.
标准条件使我们能公平地比较不同反应。IGCSE中标准条件为压强100 kPa、温度298 K（25°C）以及溶液浓度为1 mol dm⁻³。符号ΔG°表示在这些标准条件下测得的吉布斯自由能变。你可以利用标准ΔH°和标准S°数据，通过相同方程计算出ΔG°。很多考题会给出ΔH°和S°的表格，需要你先算出ΔS°。
9. Gibbs Free Energy and Chemical Equilibrium | 吉布斯自由能与化学平衡
When ΔG = 0, the system has reached equilibrium. This does not mean the reaction has stopped; it means the forward and reverse rates are equal. The relationship between ΔG and equilibrium is explored in more depth at A Level, but at IGCSE you need to know that a reversible reaction will proceed until ΔG becomes zero. The feasibility indicated by ΔG° refers to standard conditions, and altering concentrations, pressure, or temperature can shift the position of equilibrium, ultimately affecting the sign of ΔG under non‑standard conditions.
当ΔG = 0时，体系达到平衡。这并不表示反应停止了，而是正反应和逆反应速率相等。ΔG与平衡的深层关系会在A Level中进一步学习，但在IGCSE你需要知道可逆反应会持续进行直到ΔG变为零。ΔG°所表示的可行性是在标准条件之下的，改变浓度、压强或温度会使平衡位置移动，最终影响非标准条件下的ΔG符号。
10. Summary Table: Predicting Spontaneity from ΔH and ΔS | 总结表：由ΔH和ΔS预测自发性
The following table shows how the signs of ΔH and ΔS combine to influence ΔG and feasibility.
| ΔH sign | ΔS sign | ΔG = ΔH − TΔS | Feasibility |
|---|---|---|---|
| ΔH < 0 (exothermic) | ΔS > 0 | Always negative | Always feasible at any T |
| ΔH < 0 (exothermic) | ΔS < 0 | Negative only at low T | Feasible at low temperatures |
| ΔH > 0 (endothermic) | ΔS > 0 | Negative only at high T | Feasible at high temperatures |
| ΔH > 0 (endothermic) | ΔS < 0 | Always positive | Never feasible at any T |
这张表格展示了ΔH和ΔS的正负号如何共同影响ΔG和可行性。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
Mistake 1: Forgetting to convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹. Always check units before substituting into ΔG = ΔH − TΔS. If ΔS is given in J, convert to kJ by dividing by 1000.
错误1：忘记将ΔS从J K⁻¹ mol⁻¹转换为kJ K⁻¹ mol⁻¹。代入公式前一定要检查单位。如果ΔS以J为单位，必须除以1000转换成kJ。
Mistake 2: Using Celsius instead of Kelvin for T. Even if the question states 25°C, you must add 273 to get 298 K.
错误2：温度T使用摄氏度而不是开尔文。即使题目给出的是25°C，也必须加上273得到298 K。
Mistake 3: Assuming a feasible reaction will happen quickly. ΔG tells you about thermodynamics (feasibility), not kinetics (rate). Many reactions with negative ΔG require a catalyst or higher temperature to overcome activation energy.
错误3：认为可行的反应必定快速发生。ΔG反映的是热力学可行性，而不是动力学速率。许多ΔG为负的反应需要催化剂或更高温度以克服活化能才能明显进行。
Tip: In calculations, write down the equation first, list the values with correct units, convert ΔS if needed, and then substitute. Check the sign of TΔS carefully; a double negative gives a positive contribution. Practice with past paper questions to build speed.
技巧：计算时先写下方程，列出数值和单位，必要时转换ΔS，再代入。仔细处理−TΔS的符号——负负得正。用历年真题多练习，提升解题速度。
12. Quick Practice Question | 快速练习题
For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ mol⁻¹ and ΔS = −190 J K⁻¹ mol⁻¹. Determine whether the reaction is feasible at 500 K. Show your working and comment on the result.
Convert ΔS: −190 J K⁻¹ mol⁻¹ = −0.190 kJ K⁻¹ mol⁻¹. ΔG = −196 − (500 × −0.190) = −196 + 95 = −101 kJ mol⁻¹. ΔG is negative, so the reaction is feasible at 500 K. The exothermic ΔH outweighs the negative ΔS at this temperature. Try the same calculation at 1000 K and see how feasibility changes — this reinforces how temperature affects the entropy‑driven term.
对于反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，已知ΔH = −196 kJ mol⁻¹，ΔS = −190 J K⁻¹ mol⁻¹。判断该反应在500 K下是否可行。写出过程并评论结果。
转换ΔS：−190 J K⁻¹ mol⁻¹ = −0.190 kJ K⁻¹ mol⁻¹。ΔG = −196 − (500 × −0.190) = −196 + 95 = −101 kJ mol⁻¹。ΔG为负，因此该反应在500 K下是可行的。在此温度下放热的ΔH克服了负的ΔS。试在1000 K下重复计算，观察可行性如何变化——这能加深你对温度如何影响熵驱动项的理解。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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