Gibbs Free Energy for GCSE CIE Chemistry | GCSE CIE 化学：吉布斯自由能 考点精讲

📚 Gibbs Free Energy for GCSE CIE Chemistry | GCSE CIE 化学：吉布斯自由能 考点精讲

Gibbs free energy is a powerful concept that helps chemists predict whether a reaction will happen on its own. Although often taught at A level, a basic understanding of Gibbs free energy can greatly strengthen your grasp of energetics in GCSE CIE Chemistry. This article breaks down the key ideas into simple, exam-focused points so you can confidently tackle questions about feasibility and spontaneity.

吉布斯自由能是一个强大的概念，它帮助化学家预测一个反应是否会自发进行。虽然通常在 A level 阶段学习，但基础理解吉布斯自由能可以极大地加强你对 GCSE CIE 化学中能量学的掌握。本文将关键概念分解为简单、针对考试的要点，让你自信地应对有关反应可行性和自发性的问题。

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

Gibbs free energy, symbol G, combines enthalpy and entropy to give a single value that determines whether a process is feasible at a given temperature. It is named after the American scientist Josiah Willard Gibbs. A reaction is feasible if the overall Gibbs free energy change (ΔG) is negative.

吉布斯自由能，符号 G，结合了焓和熵，给出一个单一的值，用于判断在给定温度下过程是否可行。它以美国科学家约西亚·威拉德·吉布斯命名。如果总吉布斯自由能变化 (ΔG) 为负，反应就是可行的。

2. The Key Equation ΔG = ΔH – TΔS | 核心方程 ΔG = ΔH – TΔS

The famous equation is ΔG = ΔH – TΔS, where ΔH is the enthalpy change (in kJ mol⁻¹), T is the temperature in kelvin (K), and ΔS is the entropy change (in J K⁻¹ mol⁻¹, often converted to kJ for consistency). You must think carefully about units: often ΔS is given in J, so divide by 1000 to get kJ before using the equation.

著名的方程是 ΔG = ΔH – TΔS，其中 ΔH 是焓变（单位 kJ mol⁻¹），T 是开尔文温度（K），ΔS 是熵变（单位 J K⁻¹ mol⁻¹，通常转换为 kJ 以保持一致）。你必须仔细考虑单位：ΔS 常常以 J 给出，所以使用方程前要除以 1000 转换为 kJ。

ΔG = ΔH – TΔS

3. Spontaneous vs Feasible Reactions | 自发反应与可行反应

In chemistry, ‘spontaneous’ means a reaction will proceed on its own without needing a continuous input of energy. Feasible reactions are those with ΔG < 0. However, even if ΔG is negative, the reaction might be very slow if a high activation energy barrier exists, so kinetics also matter.

在化学中，“自发”意味着反应会自己进行，不需要持续的能量输入。可行的反应是那些 ΔG < 0 的反应。然而，即使 ΔG 为负，如果存在高活化能势垒，反应也可能非常缓慢，因此动力学也很重要。

4. Enthalpy Change ΔH and Its Role | 焓变 ΔH 及其作用

ΔH is the heat transferred under constant pressure. An exothermic reaction (ΔH negative) tends to make ΔG more negative, favouring feasibility. An endothermic reaction (ΔH positive) tends to make ΔG more positive, working against spontaneity unless entropy helps.

ΔH 是在恒压下传递的热量。放热反应（ΔH 为负）倾向于使 ΔG 更负，有利于可行性。吸热反应（ΔH 为正）倾向于使 ΔG 更正，不利于自发，除非熵有帮助。

5. Entropy Change ΔS and Disorder | 熵变 ΔS 与无序度

Entropy (S) measures disorder or randomness. A positive ΔS means the system becomes more disordered (e.g., solid → liquid, fewer moles → more moles of gas). A positive ΔS helps make -TΔS negative, which helps make ΔG negative, especially at high temperatures.

熵 (S) 衡量无序度或随机性。正的 ΔS 意味着系统变得更加无序（例如，固体→液体，较少摩尔数→较多摩尔数的气体）。正的 ΔS 有助于使 -TΔS 为负，从而帮助 ΔG 变为负，尤其在高温下。

6. Temperature Dependence and the TΔS Term | 温度依赖性与 TΔS 项

The TΔS term acts as a temperature multiplier for entropy. As T increases, the importance of entropy grows. Reactions with a large positive ΔS can become feasible at high temperatures even if ΔH is positive. For example, melting ice is endothermic (ΔH > 0) but becomes feasible above 273 K because TΔS outweighs ΔH.

TΔS 项作为熵的温度乘数。随着 T 增加，熵的重要性增加。具有较大正 ΔS 的反应即使在 ΔH 为正的情况下，在高温下也可能变得可行。例如，冰熔化是吸热的（ΔH > 0），但在 273 K 以上变得可行，因为 TΔS 超过了 ΔH。

7. Sign Conventions and Predicting Feasibility | 符号惯例与预测可行性

ΔH sign	ΔS sign	ΔG becomes negative?	Feasibility
– (exothermic)	+ (more disorder)	Always negative	Always feasible
– (exothermic)	– (less disorder)	Negative at low T	Feasible at low temperatures
+ (endothermic)	+ (more disorder)	Negative at high T	Feasible at high temperatures
+ (endothermic)	– (less disorder)	Never negative	Never feasible

The table above summarises how the signs of ΔH and ΔS combine to affect ΔG. Use this logic to quickly assess feasibility without calculations.

上表总结了 ΔH 和 ΔS 的符号如何结合影响 ΔG。使用这个逻辑可以在没有计算的情况下快速评估可行性。

8. Calculating Temperature of Feasibility | 计算可行性温度

Often, you are asked to find the temperature at which a reaction just becomes feasible, i.e., where ΔG = 0. Setting ΔG = 0 gives 0 = ΔH – TΔS, so T = ΔH / ΔS. Remember to use consistent units: ΔH in J mol⁻¹ and ΔS in J K⁻¹ mol⁻¹, or both in kJ. This temperature is sometimes called the ‘cross-over’ temperature.

经常会问到找到反应刚好变得可行的温度，即 ΔG = 0。令 ΔG = 0 得到 0 = ΔH – TΔS，所以 T = ΔH / ΔS。记住使用一致的单位：ΔH 用 J mol⁻¹ 和 ΔS 用 J K⁻¹ mol⁻¹，或都使用 kJ。这个温度有时称为“转变”温度。

T = ΔH ÷ ΔS (when ΔG = 0)

9. Worked Example: Decomposition of Calcium Carbonate | 计算示例：碳酸钙的分解

Consider CaCO₃(s) → CaO(s) + CO₂(g). Given ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. First convert ΔS to kJ: 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹. At 298 K, ΔG = 178 – (298 × 0.161) = 178 – 48.0 = +130 kJ mol⁻¹, so not feasible at room temperature. The temperature for ΔG = 0 is T = 178 / 0.161 = 1106 K. Above 1106 K, ΔG becomes negative and decomposition is feasible.

考虑 CaCO₃(s) → CaO(s) + CO₂(g)。给定 ΔH = +178 kJ mol⁻¹ 和 ΔS = +161 J K⁻¹ mol⁻¹。首先转换 ΔS 为 kJ：161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹。在 298 K 下，ΔG = 178 – (298 × 0.161) = 178 – 48.0 = +130 kJ mol⁻¹，因此在室温下不可行。ΔG = 0 的温度是 T = 178 / 0.161 = 1106 K。高于 1106 K 时，ΔG 变为负，分解反应可行。

10. Linking to CIE GCSE Exam Questions | 衔接 CIE GCSE 考试题

CIE GCSE Chemistry exam questions may ask you to interpret ΔG values directly or to explain why some reactions only happen at high temperatures. They might provide a table of ΔH and ΔS and ask you to state whether a reaction is feasible under certain conditions. Practice explaining using the ΔG = ΔH – TΔS idea in words, not just numbers.

CIE GCSE 化学考试题可能会要求你直接解释 ΔG 值，或解释为什么某些反应只在高温下发生。他们可能提供 ΔH 和 ΔS 的表格，并要求你说明在某条件下反应是否可行。练习用语言解释 ΔG = ΔH – TΔS 的概念，而不仅仅是数字。

11. Common Pitfalls and Misconceptions | 常见错误与误区

• A negative ΔH does not guarantee a reaction will occur. If ΔS is very negative, the -TΔS term becomes positive and large, making ΔG positive at higher temperatures — the reaction loses feasibility.

  负 ΔH 并不保证反应会发生。如果 ΔS 非常负，-TΔS 项变为正且很大，使 ΔG 在较高温度下为正——反应失去可行性。

• A positive ΔG means the reaction is not feasible under the stated conditions. But remember, a catalyst cannot change ΔG; it only lowers activation energy, speeding up a reaction that is already feasible (ΔG < 0).

  正 ΔG 意味着在该条件下反应不可行。但记住，催化剂不能改变 ΔG；它只降低活化能，加速已经可行（ΔG < 0）的反应。

• Always check the units of ΔS. If you forget to convert J to kJ, the TΔS term will be 1000 times too large or small, giving a wrong conclusion about feasibility.

  始终检查 ΔS 的单位。如果忘记将 J 转换为 kJ，TΔS 项会过大或过小 1000 倍，导致可行性判断错误。

12. Summary for Revision | 复习总结

Gibbs free energy combines the effects of enthalpy and entropy to predict reaction feasibility. The equation ΔG = ΔH – TΔS and the simple rule ‘ΔG < 0 means feasible' are your main tools. Always mind the units and temperature scale (kelvin). With practice, you can master this concept and apply it confidently in your CIE GCSE Chemistry exam.

吉布斯自由能结合了焓和熵的影响来预测反应的可行性。方程 ΔG = ΔH – TΔS 和简单的规则“ΔG < 0 代表可行”是你的主要工具。始终注意单位和温度标度（开尔文）。通过练习，你可以掌握这一概念，并在 CIE GCSE 化学考试中自信地应用它。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）