📚 Gibbs Free Energy for IB & WJEC Chemistry | IB WJEC 化学：吉布斯自由能 考点精讲
rticle unpacks every essential concept, equation and examination tip you need for IB and WJEC chemistry, from ΔG = ΔH – TΔS to the relationship with equilibrium constants, all laid out in paired English–Chinese explanations for clarity and revision.
吉布斯自由能是判断恒温恒压下化学反应能否自发进行的终极判据。本文拆解 IB 与 WJEC 化学所需的每一个核心概念、公式与应试技巧，从 ΔG = ΔH – TΔS 到与平衡常数的关系，全部以中英配对讲解，便于理解与复习。
1. What is Gibbs Free Energy? | 什么是吉布斯自由能？
Gibbs free energy, G, is a thermodynamic state function defined as G = H – ture in kelvin and S is entropy. It combines the two driving forces of a chemical reaction – the tendency to minimise enthalpy (exothermic) and to maximise entropy (disorder) – into a single quantity.
吉布斯自由能 G 是一个热力学状态函数，定义为 G = H – TS，其中 H 是焓，T 是开尔文温标，S 是熵。它将化学反应的两种驱动力——焓最小化（放热）的趋势与熵最大化（混乱度）的趋势——结合为单一物理量。
The absolute value of G cannot be measured; only changes in Gibbs free energy, ΔG, are meaningful. A negative ΔG indicates a spontaneous process under the given conditions, while a positive ΔG means the process is non‑spontaneous.
G 的绝对值无法测量，只有吉布斯自由能的变化值 ΔG 才有意义。ΔG 为负表示该条件下过程自发，ΔG 为正表示过程非自发。
2. The Master Equation: ΔG = ΔH – TΔS | 核心方程：ΔG = ΔH – TΔS
temperature and pressure, the change in Gibbs free energy for a system is given by ΔG = ΔH – of the system, ΔS is the entropy change of the system, and T is the temperature in kelvin.
在恒温恒压下，系统的吉布斯自由能变由 ΔG = ΔH – TΔS 给出。其中 ΔH 是系统焓变，ΔS 是系统熵变，T 是开尔文温度。
This equation reveals that a reaction can be spontaneous (ΔG < 0) even if it is endothermic (ΔH > 0), provided the entropy increase is large enough for the –TΔS term to dominate. Conversely, a highly exothermic reaction may become non‑spontaneous if it involves a large decrease in entropy at high temperature.
该方程表明，即使反应吸热（ΔH > 0），只要熵增足够大，使得 –TΔS 项占主导，反应依然可以自发（ΔG < 0）。反之，强放热反应若伴随大幅熵减，在高温下也可能变为非自发。
- If ΔH < 0 and ΔS > 0, then ΔG is always negative at any temperature – the reaction is always spontaneous.
- 若 ΔH < 0 且 ΔS > 0，则任何温度下 ΔG 恒为负——反应始终自发。
- If ΔH > 0 and ΔS < 0, then ΔG is always positive – the reaction is never spontaneous at any temperature.
- 若 ΔH > 0 且 ΔS < 0，则 ΔG 恒为正——任何温度下反应均不自发。
- If ΔH < 0 and ΔS < 0, the reaction is spontaneous only at low temperatures where |ΔH| > |TΔS|.
- 若 ΔH < 0 且 ΔS < 0，仅当低温下 |ΔH| > |TΔS| 时反应自发。
- If ΔH > 0 and ΔS > 0, the reaction is spontaneous only at high temperatures where TΔS outweighs ΔH.
- 若 ΔH > 0 且 ΔS > 0，仅当高温下 TΔS 超过 ΔH 时反应自发。
3. Standard Gibbs Free Energy Change, ΔG° | 标准吉布斯自由能变 ΔG°
The standard Gibbs free energy change, ΔG°, refers to the Gibbs energy change when all reactants and products are in their standard states (1 bar for gases, 1 mol dm⁻³ for solutions, pure solids and liquids). It is calculated using standard free energies of formation, ΔGf°, in a manner analogous to ΔH°.
标准吉布斯自由能变 ΔG° 是指所有反应物和产物均处于标准状态（气体 1 bar，溶液 1 mol dm⁻³，纯固体和液体）时的吉布斯能量变化。它可利用标准生成自由能 ΔGf° 计算，方法与计算 ΔH° 类似。
ΔG° = Σ ΔGf°(products) – Σ ΔGf°(reactants)
For an element in its most stable form at 298 K and 1 bar, ΔGf° is defined as zero. This mirrors the definition used for standard enthalpy of formation.
对于 298 K、1 bar 下最稳定形态的单质，其 ΔGf° 定义为零，这与标准生成焓的定义一致。
4. Calculating ΔG° from ΔH° and S° | 由 ΔH° 与 S° 计算 ΔG°
an be obtained from standard enthalpy and standard absolute entropy values using ΔG° = ΔH° – TΔS°, where ΔS° = Σ S°(products) – o use entropy values in J K⁻¹ mol⁻¹ and convert to kJ when combining with ΔH° in kJ mol⁻¹.
由于吉布斯自由能是一个状态函数，ΔG° 可利用标准焓与标准绝对熵值由 ΔG° = ΔH° – TΔS° 求得，其中 ΔS° = Σ S°(产物) – Σ S°(反应物)。注意熵值单位是 J K⁻¹ mol⁻¹，与 ΔH° 的 kJ mol⁻¹ 组合时需换算为 kJ。
| Substance / 物质 | ΔHf° / kJ mol⁻¹ | S° / J K⁻¹ mol⁻¹ |
|---|---|---|
| CaCO₃(s) | –1207 | 92.9 |
| CaO(s) | –635 | 39.7 |
| CO₂(g) | –394 | 214 |
: ΔH° = +178 kJ mol⁻¹, ΔS° = +160.8 J K⁻¹ mol⁻¹ = 0.1608 kJ K⁻¹ mol⁻¹. At 298 K, ΔG° = 178 – room temperature. The temperature at which ΔG° = 0 can be found from T = ΔH°/ΔS° ≈ 1110 K.
对于 CaCO₃(s) → CaO(s) + CO₂(g) 的分解反应：ΔH° = +178 kJ mol⁻¹，ΔS° = +160.8 J K⁻¹ mol⁻¹ = 0.1608 kJ K⁻¹ mol⁻¹。298 K 时，ΔG° = 178 – 298 × 0.1608 ≈ +130 kJ mol⁻¹，因此室温下反应不自发。令 ΔG° = 0 可得 T = ΔH°/ΔS° ≈ 1110 K。
5. Spontaneity and the Sign of ΔG | 自发性与 ΔG 的符号
A negative ΔG is the thermodynamic condition for a spontaneous process at constant T and P. However, 'spontaneous' does not imply 'fast'; many spontaneous reactions, like the combustion of diamond, are kinetically hindered and require initiation or a catalyst.
ΔG 为负是在恒温恒压下过程自发的热力学条件。但“自发”并不意味着“快速”；许多自发反应（如金刚石的燃烧）受到动力学阻碍，需要引发或催化剂。
When ΔG = 0, the system is at equilibrium – are equal, and there is no net tendency to change. If ΔG > 0, the forward reaction is non‑spontaneous, but the reverse reaction must be spontaneous.
当 ΔG = 0 时，系统处于平衡——正逆反应速率相等，没有净变化的趋势。若 ΔG > 0，正向反应非自发，则逆向反应必定自发。
Always remember to assess spontaneity using ΔG for the system alone, not total entropy, because ΔG already incorporates the entropy change of the surroundings via the –TΔS term.
切记，判断自发性只需用系统的 ΔG，而非总熵，因为 ΔG 已通过 –TΔS 项包含了环境的熵变。
6. Relationship Between ΔG° and the Equilibrium Constant K | ΔG° 与平衡常数 K 的关系
uations in chemical thermodynamics links ΔG° directly to the equilibrium constant K: ΔG° = – namic equilibrium constant (with activities, or in terms of partial pressures for gases, Kp; and concentrations for solutions, Kc).
化学热力学中最强大的公式之一将 ΔG° 与平衡常数 K 直接关联：ΔG° = –RT ln K。其中 R = 8.314 J K⁻¹ mol⁻¹，T 为开尔文温度，K 为热力学平衡常数（使用活度，或对于气体用分压 Kp，溶液用浓度 Kc）。
< 1 (reactants favoured). When ΔG° = 0, K = 1, meaning reactants and products are equally favoured.
该式表明，强负的 ΔG° 对应 K >> 1（平衡有利于产物），而强正的 ΔG° 对应 K << 1（平衡有利于反应物）。ΔG° = 0 时 K = 1，反应物与产物同样受偏爱。
In calculations, ensure you use consistent units: ΔG° in J mol⁻¹, R in J K⁻¹ mol⁻¹, T in K, and that K is dimensionless or expressed relative to standard state. For Kc, the equilibrium constant is quoted without units in the thermodynamic expression, but examination boards often accept units in the context of Kc.
计算时务必统一单位：ΔG° 用 J mol⁻¹，R 用 J K⁻¹ mol⁻¹，T 用 K，且 K 在热力学表达式中为无量纲或相对于标准态。对于 Kc，平衡常数在热力学中不带单位，但考试常允许 Kc 带单位出现。
7. Temperature Dependence of ΔG and Le Chatelier's Principle | ΔG 的温度依赖性及勒夏特列原理
Since ΔG = ΔH – ΔS ≠ 0. This temperature dependence of ΔG is the thermodynamic reason behind Le Chatelier's principle for temperature changes.
因为 ΔG = ΔH – TΔS，当 ΔS ≠ 0 时反应的自发性强烈依赖于温度。ΔG 对温度的依赖正是勒夏特列原理在温度变化背后的热力学原因。
For an endothermic reaction (ΔH > 0) with a positive ΔS, raising the temperature makes –TΔS more negative, thus ΔG becomes more negative, favouring the forward reaction – exactly what Le Chatelier's principle predicts: heating favours the endothermic direction.
对于吸热反应（ΔH > 0）且 ΔS 为正，升高温度使 –TΔS 更负，于是 ΔG 变得更负，利于正向反应——这正是勒夏特列原理的预测：升温利于吸热方向。
Conversely, for an exothermic reaction (ΔH < 0) with a negative ΔS, increasing temperature makes –TΔS more positive, raising ΔG and eventually making the reaction non‑spontaneous, favouring the endothermic reverse reaction at high temperature.
反之，对于放热反应（ΔH < 0）且 ΔS 为负，升高温度使 –TΔS 更正，ΔG 升高，最终反应不再自发，高温时利于吸热的逆反应。
8. Gibbs Free Energy and Maximum Non‑Expansion Work | 吉布斯自由能与最大非体积功
The decrease in Gibbs free energy (–ΔG) for a spontaneous process at constant T and P equals the maximum useful (non‑n) work that the system can perform. This is a key concept in electrochemical cells, where –ΔG = nFEcell relates free energy to the cell potential Ecell.
恒温恒压下自发过程的吉布斯自由能减少（–ΔG）等于系统所能做的最大有用（非体积）功。这是电化学电池的核心概念，此时 –ΔG = nFEcell 将自由能与电池电动势 Ecell 联系起来。
When a reaction is not spontaneous (ΔG > 0), the magnitude ΔG represents the minimum work that must be supplied to drive the reaction, for example by electrolysis. This underlines why electrolysis of water requires a minimum voltage of 1.23 V under standard conditions.
当反应不自发（ΔG > 0）时，ΔG 的大小代表驱动该反应所需的最小功，例如电解。这解释了为什么标准条件下水的电解至少需要 1.23 V 的电压。
n endergonic reaction (ΔG > 0) to a strongly exergonic one (such as ATP hydrolysis, ΔG°′ ≈ –30.5 kJ mol⁻¹) is how cells perform otherwise unfavourable tasks.
在生物系统中，将吸能反应（ΔG > 0）与强放能反应（如 ATP 水解，ΔG°′ ≈ –30.5 kJ mol⁻¹）耦合起来，是细胞实现原本不利工作的途径。
9. Ellingham Diagrams – A WJEC Special | 埃林厄姆图——WJEC 特色考点
ainst temperature for the formation of metal oxides (and sometimes sulfides). The straight‑ ionship arises because ΔH° and ΔS° are roughly constant with temperature, giving ΔG° = ΔH° – TΔS° as a linear equation.
埃林厄姆图以 ΔG° 对温度作图，展示金属氧化物（有时包括硫化物）的生成反应。由于 ΔH° 和 ΔS° 随温度近似恒定，ΔG° = ΔH° – TΔS° 表现为线性关系。
, and to compare the reducing ability of different metals. A metal whose oxide line lies lower on the diagram can reduce the oxide of a metal above it, because its own oxidation has a more negative ΔG°.
通过埃林厄姆图可预测金属氧化物开始热分解的温度（ΔG° = 0），并可比较不同金属的还原能力。图中氧化物线位于更下方的金属可以还原位于上方的金属的氧化物，因为其自身的氧化反应具有更负的 ΔG°。
The classic example is aluminium reducing iron(III) oxide in the thermite reaction, which is clearly rationalised by the much lower position of the Al₂O₃ line compared to Fe₂O₃.
典型例子是铝热反应中铝还原氧化铁，Al₂O₃ 线远低于 Fe₂O₃ 线，清楚解释了这一过程。
10. Free Energy and Coupled Reactions | 自由能与耦合反应
In IB and WJEC contexts, you may be asked to show that an overall reaction can be made spontaneous by coupling a non‑spontaneous step to a highly spontaneous one. The overall ΔG° is the sum of the ΔG° values of the individual steps because G is a state function.
在 IB 与 WJEC 的试题中，可能要求你证明，通过将一个非自发步骤与一个高度自发的步骤耦合，可使总反应变为自发。总反应的 ΔG° 是各步骤 ΔG° 之和，因为 G 是状态函数。
TiCl₄ + 2CO involves two exergonic steps (carbon oxidation and TiCl₄ formation) that together provide a negative ΔG°. This underpins the industrial extraction of titanium.
例如，碳直接还原 TiO₂ 不可行，但耦合反应 TiO₂ + 2Cl₂ + 2C → TiCl₄ + 2CO 包含了两个放能步骤（碳的氧化和 TiCl₄ 的生成），共同给出负的 ΔG°，这成为工业提取钛的基础。
11. Common Pitfalls in ΔG Calculations | 计算 ΔG 常见陷阱
convert entropy from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000 before using ΔG = ΔH – TΔS, or convert ΔH to J. Be explicit about units in your working.
最常犯的错误是混淆 kJ 与 J。使用 ΔG = ΔH – TΔS 前务必先将熵从 J K⁻¹ mol⁻¹ 除以 1000 转为 kJ K⁻¹ mol⁻¹，或将 ΔH 转为 J。解题过程中要明确标注单位。
on must be in kelvin. Using Celsius will produce nonsensical results. And when using ΔG° = –RT ln K, the value of R must be 8.314 J K⁻¹ mol⁻¹ if ΔG° is in J mol⁻¹.
另一个常见错误是忘记吉布斯方程中的温度必须使用开尔文。使用摄氏度会导致荒谬的结果。此外，使用 ΔG° = –RT ln K 时，若 ΔG° 以 J mol⁻¹ 为单位，R 必须取 8.314 J K⁻¹ mol⁻¹。
Always check the sign of ΔS carefully. An increase in the number of gas moles generally leads to a positive ΔS, while a decrease leads to a negative ΔS. For phase changes, melting and vaporisation have ΔS > 0.
始终仔细检查 ΔS 的符号。气体分子数增加通常导致 ΔS 为正，减少则导致 ΔS 为负。对于相变，熔化和汽化的 ΔS > 0。
12. Exam Technique: Using ΔG to Predict and Explain | 答题技巧：用 ΔG 进行预测与解释
In data‑based questions, you will often be given ΔH and S° values and asked to calculate ΔG at a specific temperature, then predict spontaneity. Always show the conversion of units, set out the equation clearly, and state the conclusion with reference to the sign of ΔG.
在数据题中，常给出 ΔH 和 S° 值，要求计算某温度下的 ΔG 并预测自发性。务必展示单位换算，清晰列出方程，并依据 ΔG 的符号陈述结论。
When explaining why a reaction becomes spontaneous above a certain temperature, use the interplay of enthalpy and entropy: 'The positive ΔS makes the –TΔS term increasingly negative as temperature rises, and at sufficiently high T it overcomes the positive ΔH, giving ΔG < 0.'
解释为何反应在某温度以上变为自发时，要运用焓与熵的交互关系：“正的 ΔS 使得随温度升高 –TΔS 项越来越负，在足够高的温度下克服了正的 ΔH，使 ΔG < 0。”
For WJEC, practice reading Ellingham diagrams – rature from the intercept with the ΔG = 0 line, and justify which metal can reduce another's oxide by comparing the lines.
对 WJEC，要练习阅读埃林厄姆图——通过与 ΔG = 0 线的交会点确定热分解温度，并通过比较线位置论证哪种金属可还原另一种金属的氧化物。
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