📚 Gibbs Free Energy for IGCSE CCEA Chemistry | IGCSE CCEA 化学：吉布斯自由能 考点精讲
ential for determining whether a reaction will occur spontaneously under given conditions.
吉布斯自由能（G）是一个综合了焓和熵的热力学函数，用于预测化学反应是否可行。在 IGCSE CCEA 化学中，掌握 ΔG = ΔH − TΔS 方程及其应用是判断反应能否自发进行的关键。
1. What is Gibbs Free Energy? | 什么是吉布斯自由能？
is enthalpy, T is absolute temperature in kelvin, and S is entropy. In a chemical reaction, the change in Gibbs free energy (ΔG) tells us whether a reaction is thermodynamically feasible without outside intervention.
吉布斯自由能（符号为 G）是一个状态函数，定义为 G = H − TS，其中 H 是焓，T 是开尔文绝对温度，S 是熵。在化学反应中，吉布斯自由能变化（ΔG）可以告诉我们反应是否能在不需要外界干预的条件下自发进行。
ΔG alone does not give information about the reaction rate — a reaction with a negative ΔG might still be extremely slow if the activation energy is high. For CCEA, you must be able to link ΔG to the idea of thermodynamic spontaneity, not speed.
单独的 ΔG 数值并不能说明反应速率——即使 ΔG 为负，如果活化能很高，反应依然可能极其缓慢。在 CCEA 考试中，你需要将 ΔG 与热力学自发性联系起来，而不是与反应快慢挂钩。
2. The Gibbs Free Energy Equation | 吉布斯自由能方程
nge of the reaction, T is the temperature in kelvin, and ΔS is the entropy change of the system. All quantities must use consistent units — mol⁻¹, so you must convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000 before using the formula.
基本关系式为：ΔG = ΔH − TΔS，其中 ΔH 是反应焓变，T 是开尔文温度，ΔS 是系统的熵变。所有量必须使用一致的单位——通常 ΔG 和 ΔH 以 kJ mol⁻¹ 表示，而 ΔS 常以 J K⁻¹ mol⁻¹ 给出，因此代公式前需将 ΔS 除以 1000 转换为 kJ K⁻¹ mol⁻¹。
When ΔS is given in J K⁻¹ mol⁻¹, the term TΔS will be in J mol⁻¹, so divide by 1000 to obtain kJ mol⁻¹ before subtracting from ΔH. CCEA exam questions frequently test this unit conversion, so always check the units given in the data.
当 ΔS 的单位为 J K⁻¹ mol⁻¹ 时，TΔS 项的单位将是 J mol⁻¹，因此在减去 ΔH 之前需除以 1000 得到 kJ mol⁻¹。CCEA 考题经常考查这种单位换算，因此务必核对数据中的单位。
ΔG = ΔH − TΔS
3. Enthalpy and Entropy Contributions | 焓变与熵变的贡献
ΔH represents the heat transferred during a reaction at constant pressure. A negative ΔH (exothermic) favours spontaneity because the system releases energy. A positive ΔH (endothermic) works against spontaneity. ΔS measures the dispersal of energy or disorder. A positive ΔS (more disorder) favours spontaneity, while a negative ΔS (less disorder) opposes it.
ΔH 表示恒压反应中的热量传递。负的 ΔH（放热）有利于自发进行，因为系统释放能量；正的 ΔH（吸热）则阻碍自发。ΔS 衡量能量或混乱度的分散程度。正的 ΔS（更混乱）有利于自发，而负的 ΔS（更有序）则不利于自发。
The TΔS term in the equation shows that the entropy contribution is temperature-dependent. At low temperatures, the TΔS term is small, so ΔH tends to dominate. At high temperatures, TΔS becomes important and may even reverse the sign of ΔG.
方程中的 TΔS 项表明熵的贡献依赖于温度。低温时 TΔS 项较小，因此 ΔH 往往起主导作用；高温时 TΔS 变得重要，甚至可能改变 ΔG 的正负号。
4. Spontaneity and the Sign of ΔG | 自发过程的 ΔG 符号判据
n ΔG = 0, the system is at equilibrium with no net tendency to change in either direction.
当 ΔG 为负时，反应在热力学上是自发的（可行的）。若 ΔG 为正，则该条件下反应不可行。当 ΔG = 0 时，体系处于平衡状态，没有向任一方向发生净变化的趋势。
It is vital to remember 'spontaneous' in this context does not mean fast. Many spontaneous reactions, such as the rusting of iron, are slow. CCEA mark schemes expect you to mention that ΔG only predicts thermodynamic feasibility, not rate.
必须牢记这里的“自发”并不意味着快速。许多自发反应（如铁的生锈）进行得很慢。CCEA 评分标准期望你能指出 ΔG 仅预测热力学可行性，而非反应速率。
| ΔG sign | Reaction feasibility |
| ΔG < 0 | Spontaneous / feasible |
| ΔG > 0 | Not feasible (reverse may be feasible) |
| ΔG = 0 | System at equilibrium |
5. Temperature Dependence of ΔG | ΔG 对温度的依赖关系
Because ΔG depends on TΔS, temperature can switch a reaction between feasible and non-feasible. The table below summarises the four combinations of ΔH and ΔS and how temperature influences spontaneity.
由于 ΔG 依赖于 TΔS，温度可以使反应在可行与不可行之间切换。下表总结了 ΔH 和 ΔS 的四种组合以及温度如何影响自发性。
| ΔH sign | ΔS sign | ΔG behaviour | Spontaneity |
| Negative (−) | Positive (+ | Always negative | Feasible at all temperatures |
| Negative (−) | Negative (−) | Negative only at low T | Feasible at low T, not at high T |
| Positive (+) | Positive (+) | Negative only at high T | Feasible at high T, not at low T |
| Positive (+) | Negative (−) | Always positive | Never feasible |
aO(s) + CO₂(g). This reaction is endothermic (ΔH > 0) and has a positive ΔS because a gas is produced. It becomes feasible only at high temperatures, which is why calcium carbonate decomposes at a high kiln temperature.
经典例子是碳酸钙分解：CaCO₃(s) → CaO(s) + CO₂(g)。该反应吸热（ΔH > 0）且因生成气体使 ΔS 为正，因此仅在高温下变得可行，这就是碳酸钙在窑中高温分解的原因。
6. Calculating ΔG at Standard Conditions | 标准条件下 ΔG 的计算
Standard Gibbs free energy change, ΔG°, is calculated at 298 K (25 °C) using standard enthalpy change ΔH° and standard entropy change ΔS°. For CCEA, you may be given ΔH° and ΔS° values and asked to calculate ΔG° and state whether the reaction is feasible at room temperature.
标准吉布斯自由能变化 ΔG° 是在 298 K（25 °C）下由标准焓变 ΔH° 和标准熵变 ΔS° 计算得出。CCEA 考试中，你可能会被给定 ΔH° 和 ΔS° 值，要求计算 ΔG° 并判断反应在室温下是否可行。
−92 + 59.3 = −32.7 kJ mol⁻¹. The negative ΔG° indicates the reaction is feasible at 298 K.
示例：ΔH° = −92 kJ mol⁻¹，ΔS° = −199 J K⁻¹ mol⁻¹。先将 ΔS° 转换：−199 J K⁻¹ mol⁻¹ = −0.199 kJ K⁻¹ mol⁻¹。则 ΔG° = −92 − (298 × −0.199) = −92 + 59.3 = −32.7 kJ mol⁻¹。负的 ΔG° 表明反应在 298 K 下可行。
Always show your working clearly and state the conversion step. When ΔS° is positive, the − TΔS° term becomes more negative as T increases, making ΔG° more negative.
务必清晰地写出计算过程并注明转换步骤。当 ΔS° 为正时，− TΔS° 项随温度升高变得更负，从而使 ΔG° 更负。
7. Using ΔG to Predict Feasibility | 利用 ΔG 预测反应可行性
ou can predict the feasibility of a reaction at any temperature if you know ΔH and ΔS. The 'crossover temperature' H / ΔS (with ΔH in J mol⁻¹ or ΔS in kJ K⁻¹ mol⁻¹). This is a common extension question in CCEA.
若已知 ΔH 和 ΔS，你就可以预测反应在任何温度下的可行性。设定 ΔG = 0 可求得反应变得可行的“转折温度”，T = ΔH / ΔS（此时 ΔH 的单位需为 J mol⁻¹，或 ΔS 的单位为 kJ K⁻¹ mol⁻¹）。这是 CCEA 常见的扩展题。
0.161 ≈ 1106 K. Above this temperature, ΔG becomes negative and decomposition is feasible.
示例：ΔH = +178 kJ mol⁻¹，ΔS = +161 J K⁻¹ mol⁻¹。转换：ΔS = 0.161 kJ K⁻¹ mol⁻¹。T = 178 / 0.161 ≈ 1106 K。当温度高于此值时，ΔG 变为负，分解反应可行。
Be prepared to discuss practical implications, such as why industrial processes run at high temperatures despite being endothermic — entropic benefits overcome the enthalpic cost at elevated temperatures.
要准备好讨论实际意义，例如为什么吸热的工业过程仍要在高温下进行——在高温下熵的有利效应克服了焓的不利效应。
8. Limitations of Gibbs Free Energy | 吉布斯自由能的局限性
Gibbs free energy does not tell us about activation energy or reaction rate. A reaction may be thermodynamically feasible but kinetically inert because of a high activation barrier. CCEA expects you to distinguish between thermodynamic and kinetic stability.
吉布斯自由能不能说明活化能或反应速率。由于存在高活化能屏障，反应可能在热力学上可行，却在动力学上是惰性的。CCEA 要求你能区分热力学稳定性与动力学稳定性。
Another limitation is that the equation applies to closed systems at constant temperature and pressure. In real scenarios, side reactions and non-standard conditions may alter the feasibility. Also, standard data are for 1 atm and 1 mol dm⁻³, so deviations occur under different concentrations or pressures.
另一个局限性是该方程适用于恒温恒压的封闭体系。在实际情况中，副反应及非标准条件可能改变可行性。此外，标准数据是在 1 atm 和 1 mol dm⁻³ 条件下测定的，在不同浓度或压力下会出现偏差。
Despite these limitations, ΔG remains an essential tool for predicting whether a reaction can, in principle, proceed under a given set of conditions. You may be asked to comment on limitations in higher-tier CCEA questions.
尽管有这些局限性，ΔG 仍是预测反应在给定条件下原则上能否进行的重要工具。CCEA 的高阶题目中可能会要求你评论这些局限性。
9. Common Exam Questions | 常见考题类型
- Calculating ΔG from ΔH and ΔS: Pay attention to unit conversion and sign. Often a mark is allocated for converting J to kJ.
- 从 ΔH 和 ΔS 计算 ΔG：注意单位换算和符号，常有一分值用于 J 到 kJ 的换算。
- Explaining the temperature effect: Use the table of ΔH/ΔS combinations to justify when a reaction is feasible.
- 解释温度影响：利用 ΔH/ΔS 组合表说明反应在何时可行。
- Determining the crossover temperature: Setting ΔG = 0 and solving for T = ΔH/ΔS, with correct units.
- 求算转折温度：设 ΔG = 0 并求解 T = ΔH/ΔS，注意单位正确。
- Interpreting the sign of ΔG: Linking negative ΔG to feasibility and stressing that rate is a separate issue.
- 解释 ΔG 的符号：将负的 ΔG 与可行性联系起来，并强调速率是另一个独立问题。
- Evaluating limitations: Mentioning activation energy, kinetic stability, and non-standard conditions.
- 评价局限性：提到活化能、动力学稳定性和非标准条件。
Many CCEA past papers include a structured question on Gibbs free energy, often embedded in a practical context such as industrial manufacturing or extraction of metals.
CCEA 的许多历年真题中都有一道关于吉布斯自由能的结构性题目，通常嵌入工业制造或金属提取等实际背景中。
10. Summary and Tips | 总结与备考建议
ΔH − TΔS. A negative ΔG indicates a feasible reaction. Temperature plays a crucial role through the TΔS term, and the four sign combinations must be memorised. Always convert ΔS to kJ K⁻¹ mol⁻¹ when ΔH is in kJ mol⁻¹, or convert ΔH to J mol⁻¹. Practice calculating crossover temperatures and explaining why some endothermic processes become feasible at high temperature.
牢记核心关系：ΔG = ΔH − TΔS。负的 ΔG 表示反应可行。温度通过 TΔS 项起关键作用，必须熟记四种符号组合。当 ΔH 用 kJ mol⁻¹ 时，务必将 ΔS 转换为 kJ K⁻¹ mol⁻¹，或把 ΔH 转换为 J mol⁻¹。多练习计算转折温度，并解释为什么某些吸热过程在高温下变得可行。
In the exam, read the question carefully: if you are given ΔS in J K⁻¹ mol⁻¹ and ΔH in kJ mol⁻¹, show the division by 1000 clearly. Use the unit-check method to avoid errors. Link your answers to the idea that thermodynamics sets the direction, while kinetics governs the speed.
考试时务必仔细读题：若给出的 ΔS 以 J K⁻¹ mol⁻¹ 为单位而 ΔH 以 kJ mol⁻¹ 为单位，要明确写出除以 1000 的步骤。用单位检查法避免错误。答题时要将热力学设定方向、动力学掌控速度的观点联系起来。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图