📚 Gibbs Free Energy for IB & CIE Chemistry | IB CIE 化学：吉布斯自由能 考点精讲
Gibbs free energy is the ultimate arbiter of spontaneity in chemical reactions, linking enthalpy, entropy, and temperature into a single predictive tool. For IB and CIE Chemistry students, a firm grasp of ΔG is essential not only for determining whether a reaction will ‘go’ but also for understanding equilibrium, electrochemistry, and the thermodynamic boundaries of chemical systems. This article unpacks the key concepts, calculations, and applications of Gibbs free energy as required by both syllabi.
吉布斯自由能是判断化学反应自发性的终极标准，它将焓变、熵变和温度整合为一个预测工具。对于 IB 和 CIE 化学学生而言，牢固掌握 ΔG 不仅是判断反应是否“进行”的关键，更是理解化学平衡、电化学以及化学系统热力学边界的基础。本文深入解析吉布斯自由能的核心概念、计算方法和应用，紧扣两大考试局的考点。
1. The Gibbs Free Energy Concept | 吉布斯自由能的概念
Gibbs free energy (G) is a thermodynamic state function defined as the maximum amount of non-expansion work that can be extracted from a closed system at constant temperature and pressure. It combines the system’s enthalpy (H) and entropy (S) into a single value: G = H – TS. A change in Gibbs free energy (ΔG) indicates the direction of spontaneous change.
吉布斯自由能 (G) 是一种热力学状态函数，定义为在恒温恒压下，从一个封闭系统可获得的最大非膨胀功。它将系统的焓 (H) 和熵 (S) 结合为单一数值：G = H – TS。吉布斯自由能的变化 (ΔG) 指示了自发变化的方向。
For a chemical reaction, only the change in Gibbs free energy (ΔG) matters, not the absolute value of G. If ΔG < 0, the reaction can proceed spontaneously in the forward direction under the given conditions, releasing free energy (exergonic). If ΔG > 0, the forward reaction is non-spontaneous (endergonic), and energy must be supplied. At ΔG = 0, the system is at equilibrium.
对于化学反应，只有吉布斯自由能的变化 (ΔG) 是重要的，而不是 G 的绝对值。如果 ΔG < 0，反应在给定条件下可正向自发进行，释放自由能（放能反应）。如果 ΔG > 0，正向反应非自发（吸能反应），需要输入能量。当 ΔG = 0 时，系统处于平衡态。
It is crucial to distinguish spontaneity from rate: a negative ΔG says nothing about how fast a reaction occurs. Many spontaneous reactions (like combustion of diamond at room temperature) have a huge kinetic barrier and do not proceed at observable rates.
必须区分自发性和速率：负的 ΔG 并不说明反应发生的快慢。许多自发反应（如室温下金刚石的燃烧）由于巨大的动力学障碍，观察不到明显的反应速率。
2. The Gibbs Free Energy Equation | 吉布斯自由能方程
The fundamental equation linking Gibbs free energy change with enthalpy and entropy changes is:
ΔG = ΔH – TΔS
Where ΔG is the Gibbs free energy change (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 (J K⁻¹ mol⁻¹). Always convert ΔS to kJ K⁻¹ mol⁻¹ if ΔH and ΔG are in kJ, or convert all to consistent units. Common exam pitfall: using ΔS in J K⁻¹ mol⁻¹ while ΔH is in kJ mol⁻¹ — remember to divide ΔS by 1000.
关联吉布斯自由能变与焓变、熵变的基本方程为：
ΔG = ΔH – TΔS
式中 ΔG 为吉布斯自由能变 (kJ mol⁻¹)，ΔH 为焓变 (kJ mol⁻¹)，T 为绝对温度 (K)，ΔS 为熵变 (J K⁻¹ mol⁻¹)。如果 ΔH 和 ΔG 用 kJ，必须将 ΔS 转换为 kJ K⁻¹ mol⁻¹，或统一单位。典型考试陷阱：当 ΔS 以 J K⁻¹ mol⁻¹ 为单位，而 ΔH 以 kJ mol⁻¹ 为单位时，需将 ΔS 除以 1000。
This equation reveals the temperature-driven competition between enthalpy and entropy contributions to spontaneity. The term -TΔS becomes more significant at higher temperatures, often allowing endothermic reactions (ΔH > 0) to become spontaneous if the entropy increase is large enough.
此方程展示了焓与熵在温度驱动下对自发性的竞争贡献。项 -TΔS 在高温时变得更加显著，即使反应吸热 (ΔH > 0)，如果熵增足够大，反应也可能成为自发。
3. Sign of ΔG and Spontaneity | ΔG 的符号与自发性
For a reaction at constant temperature and pressure:
- ΔG < 0 : The forward reaction is spontaneous (feasible). Free energy is released.
- ΔG > 0 : The forward reaction is non-spontaneous. The reverse reaction is spontaneous.
- ΔG = 0 : The system is at equilibrium. There is no net change in composition.
在恒温恒压下：
- ΔG < 0 ：正向反应自发（可行），释放自由能。
- ΔG > 0 ：正向反应非自发，逆反应自发。
- ΔG = 0 ：系统处于平衡状态，组成不发生净变化。
Spontaneity depends both on the signs of ΔH and ΔS, and crucially on temperature. The four possible combinations are summarised below:
自发性取决于 ΔH 和 ΔS 的符号，以及至关重要的温度。四种可能的组合总结如下：
| ΔH | ΔS | ΔG = ΔH – TΔS | Spontaneity |
|---|---|---|---|
| Negative (-) | Positive (+) | Always negative | Spontaneous at all temperatures |
| Positive (+) | Negative (-) | Always positive | Never spontaneous (reverse reaction is always spontaneous) |
| Negative (-) | Negative (-) | Negative at low T, positive at high T | Spontaneous below a certain temperature |
| Positive (+) | Positive (+) | Positive at low T, negative at high T | Spontaneous above a certain temperature |
4. Temperature Dependence of Spontaneity | 温度对自发性的影响
The temperature at which a reaction changes from spontaneous to non-spontaneous (or vice versa) can be found by setting ΔG = 0 in the equation ΔG = ΔH – TΔS, which gives:
T = ΔH / ΔS
This is the ‘threshold temperature’ (often denoted Tcrossover). It applies only when ΔH and ΔS have the same sign (both positive or both negative). If ΔH and ΔS are both negative, the reaction is spontaneous at temperatures below T; if both positive, spontaneous above T.
通过在方程 ΔG = ΔH – TΔS 中令 ΔG = 0 可以求得反应由自发变为非自发（或相反）的温度：
T = ΔH / ΔS
此为“阈值温度”（常记作 Tcrossover）。仅当 ΔH 和 ΔS 同号（两者皆正或两者皆负）时适用。若 ΔH 和 ΔS 皆负，反应在温度低于 T 时自发；若皆正，则在温度高于 T 时自发。
It is assumed that ΔH and ΔS themselves do not change significantly with temperature for these calculations, which is a reasonable approximation over small temperature ranges. This assumption is frequently tested in both IB and CIE exam questions when students are asked to estimate ΔG at a new temperature using standard values at 298 K.
此类计算通常假设 ΔH 和 ΔS 本身不随温度显著变化，在较小温度范围内是合理的近似。IB 和 CIE 考试中常会要求学生使用 298 K 下的标准值估算另一温度下的 ΔG，这正是对此假设的考察。
5. Standard Gibbs Free Energy Change, ΔG° | 标准吉布斯自由能变，ΔG°
The standard Gibbs free energy change of a reaction (ΔG°) refers to the change when all reactants and products are in their standard states (1 bar pressure for gases, 1 mol dm⁻³ for solutions, pure solids or liquids, at a specified temperature, usually 298 K). ΔG° can be calculated in three main ways:
反应的标准吉布斯自由能变 (ΔG°) 是指所有反应物和产物处于标准状态（气体分压 1 bar，溶液浓度 1 mol dm⁻³，纯固体或液体，在指定温度下，通常为 298 K）时的变化。ΔG° 可通过三种主要方式计算：
- From standard enthalpies and entropies of formation: ΔG° = ΔH° – TΔS°, where ΔH° and ΔS° are calculated from standard formation data.
- From standard Gibbs free energies of formation (ΔGf°): ΔG° = Σ n ΔGf°(products) – Σ m ΔGf°(reactants).
- From the equilibrium constant: ΔG° = -RT ln K.
- 由标准生成焓和标准生成熵计算： ΔG° = ΔH° – TΔS°，其中 ΔH° 和 ΔS° 由标准生成数据求得。
- 由标准生成吉布斯自由能 (ΔGf°) 计算： ΔG° = Σ n ΔGf°(产物) – Σ m ΔGf°(反应物)。
- 由平衡常数计算： ΔG° = -RT ln K。
All three routes are examinable. For IB students, ΔGf° values are often provided in data booklets, while CIE learners may be required to retrieve them from tables or use the first two methods in combination. The reaction quotient Q must not be confused with the equilibrium constant K when discussing non-standard ΔG (see Section 7).
三种计算途径都可能出现在考试中。IB 学生常可在数据手册中查找 ΔGf° 值，而 CIE 学生可能需要自行查表，或结合前两种方法进行计算。在讨论非标准 ΔG 时，切勿将反应商 Q 与平衡常数 K 混淆（见第7节）。
6. Gibbs Free Energy and the Equilibrium Constant | 吉布斯自由能与平衡常数
One of the most powerful relationships in chemical thermodynamics links the standard Gibbs free energy change to the equilibrium constant K:
ΔG° = -RT ln K
Where R is the universal gas constant (8.31 J K⁻¹ mol⁻¹), T is temperature in kelvin, and K is the thermodynamic equilibrium constant (dimensionless, with activities or pressures relative to standard states). This equation shows that a large negative ΔG° corresponds to K >> 1, meaning the equilibrium position lies far to the right. A large positive ΔG° means K << 1.
化学热力学中最重要的关系式之一将标准吉布斯自由能变与平衡常数 K 联系起来：
ΔG° = -RT ln K
式中 R 为通用气体常数 (8.31 J K⁻¹ mol⁻¹)，T 为开氏温度，K 为热力学平衡常数（无量纲，使用相对于标准状态的活度或压力）。该方程表明，非常大的负 ΔG° 对应于 K >> 1，即平衡位置远偏向产物侧；非常大的正 ΔG° 意味着 K << 1。
For a reaction not at equilibrium, the sign of the actual ΔG (not ΔG°) determines the direction of change. The actual ΔG is related to ΔG° and the reaction quotient Q by:
ΔG = ΔG° + RT ln Q
At equilibrium, Q = K, and ΔG = 0, recovering the standard equation. If Q < K, then ΔG < 0 and the forward reaction is spontaneous. If Q > K, then ΔG > 0 and the reverse reaction occurs. This provides a quantitative bridge between thermodynamics and chemical equilibrium — a central theme in both IB and CIE higher-tier assessments.
对于非平衡态的反应，实际 ΔG（非 ΔG°）的符号决定变化方向。实际 ΔG 与 ΔG° 及反应商 Q 的关系为：
ΔG = ΔG° + RT ln Q
平衡时 Q = K，ΔG = 0，即可回到标准方程。若 Q < K，则 ΔG < 0，正向反应自发；若 Q > K，则 ΔG > 0，逆向反应发生。这便为热力学与化学平衡之间提供了定量桥梁——这是 IB 和 CIE 高难度试题中的核心主题。
7. Gibbs Free Energy and Electrochemical Cells | 吉布斯自由能与电化学电池
In electrochemistry, the electrical work done by a galvanic cell under reversible conditions is equal to the decrease in Gibbs free energy. The relationship is:
ΔG° = -nFE°cell
Where n is the number of moles of electrons transferred in the redox reaction, F is the Faraday constant (96 500 C mol⁻¹), and E°cell is the standard cell potential (in volts). A positive E°cell gives a negative ΔG°, confirming that a spontaneous redox reaction produces a positive emf.
在电化学中，原电池在可逆条件下所做的电功等于吉布斯自由能的减少。关系式为：
ΔG° = -nFE°cell
式中 n 为氧化还原反应中转移的电子摩尔数，F 为法拉第常数 (96 500 C mol⁻¹)，E°cell 为标准电池电动势（伏特）。正的 E°cell 给出负的 ΔG°，证实自发的氧化还原反应产生正电动势。
Combining the electrochemical relation with the equilibrium constant relation yields:
E°cell = (RT / nF) ln K
This is the Nernst equation in its standard form linking cell potential directly to the equilibrium constant. Students should be able to calculate ΔG° from E°cell, determine E°cell from ΔG°, and relate both to K. In CIE, calculations with the Faraday constant are common; in IB, data booklet values for F are given.
将电化学关系与平衡常数关系结合可得：
E°cell = (RT / nF) ln K
这是能斯特方程的标准形式，将电池电动势直接与平衡常数相关联。学生应能够从 E°cell 计算 ΔG°，从 ΔG° 确定 E°cell，并将两者与 K 关联。CIE 考试中常涉及法拉第常数的计算；IB 则在数据手册中提供 F 值。
8. Temperature Dependence of the Equilibrium Constant | 平衡常数与温度的关系
The Van ‘t Hoff equation describes how the equilibrium constant changes with temperature, directly linking the enthalpy change of a reaction to the shift in K:
ln(K₂/K₁) = -(ΔH°/R)(1/T₂ – 1/T₁)
or in differential form: d(ln K)/dT = ΔH°/(RT²). If ΔH° is negative (exothermic), K decreases with increasing temperature, shifting equilibrium towards reactants. If ΔH° is positive (endothermic), K increases with temperature, favouring products at higher temperatures.
范特霍夫方程描述了平衡常数如何随温度变化，将反应焓变直接与 K 的移动相联系：
ln(K₂/K₁) = -(ΔH°/R)(1/T₂ – 1/T₁)
或微分形式：d(ln K)/dT = ΔH°/(RT²)。若 ΔH° 为负（放热），K 随温度升高而减小，平衡向反应物方向移动；若 ΔH° 为正（吸热），K 随温度升高而增大，高温有利于生成产物。
This equation provides a way to determine ΔH° graphically by plotting ln K versus 1/T, where the slope equals -ΔH°/R. Such graphical analysis is common in practical assessments and data-based exam questions. Students must be careful to use K in correct dimensionless form and T in kelvin. The equation also explains why the ‘threshold temperature’ concept works: it’s the temperature where K becomes 1 and ΔG° = 0.
该方程提供了一种通过绘制 ln K 对 1/T 的图形来测定 ΔH° 的方法，斜率等于 -ΔH°/R。此类图形分析在实验评估和基于数据的考题中很常见。学生需注意使用正确无量纲形式的 K 并以开氏温度表示 T。该方程也解释了“阈值温度”概念的合理性：此温度下 K = 1 且 ΔG° = 0。
9. Coupled Reactions and Biochemical Applications | 耦合反应与生物化学应用
Gibbs free energy changes are additive for coupled reactions. A thermodynamically unfavourable reaction (ΔG > 0) can be driven forward by coupling it to a highly favourable reaction (ΔG << 0) if the net ΔG is negative. This is a foundational principle in biochemistry: the hydrolysis of adenosine triphosphate (ATP) to ADP has ΔG°' ≈ -30.5 kJ mol⁻¹, and it is used to power numerous endergonic cellular processes such as protein synthesis and active transport.
吉布斯自由能变化对耦合反应具有加和性。热力学不利的反应 (ΔG > 0) 可通过与一个非常有利的反应 (ΔG << 0) 相耦合而被驱动，只要净 ΔG 为负。这是生物化学的基本原理：ATP（三磷酸腺苷）水解为 ADP 的 ΔG°' 约为 -30.5 kJ mol⁻¹，用于驱动蛋白质合成、主动运输等众多吸能细胞过程。
For example, the synthesis of glutamine from glutamic acid and ammonia has ΔG° ≈ +14 kJ mol⁻¹, which is non-spontaneous. However, when coupled with ATP hydrolysis, the overall reaction becomes spontaneous. IB and CIE syllabi may include such qualitative reasoning about coupled reactions, often in the context of metabolism or industrial processes.
例如，由谷氨酸和氨合成谷氨酰胺的 ΔG° ≈ +14 kJ mol⁻¹，是非自发的。但与 ATP 水解耦合后，总体反应变得自发。IB 和 CIE 大纲可能涉及此类耦合反应的定性推理，通常在新陈代谢或工业过程的背景下考查。
10. Common Misconceptions and Exam Tips | 常见误区与考试技巧
Misconception 1: ‘A negative ΔG guarantees a fast reaction.’ Truth: Kinetics and thermodynamics are independent. Many spontaneous reactions (e.g., graphite → diamond at room conditions) are incredibly slow.
误区一：“负的 ΔG 保证反应快速进行。”事实：动力学与热力学是独立的。许多自发反应（如室温下石墨生成金刚石）极其缓慢。
Misconception 2: ‘Standard conditions imply 298 K.’ While 298 K is commonly used, standard state does not specify a temperature; it specifies pressure and concentration. ΔG° values are tabulated at 298 K, but the definition of standard state is broader.
误区二：“标准条件就意味着 298 K。”虽然 298 K 常用，但标准状态并不规定温度；它规定了压力和浓度。ΔG° 值在 298 K 下列表，但标准状态的定义更加宽泛。
Misconception 3: ‘If ΔG is positive, the reaction cannot happen at all.’ A positive ΔG simply means the forward reaction is not spontaneous under those conditions; the reaction may still proceed if conditions change or if coupled to another process.
误区三：“如果 ΔG 为正，反应完全不可能发生。”正的 ΔG 仅意味着在该条件下正向反应非自发；若条件改变或与其他过程耦合，反应仍可进行。
Exam tip: Always check unit consistency before plugging numbers into ΔG = ΔH – TΔS. In CIE papers, data may be given in kJ and J deliberately; convert ΔS to kJ K⁻¹ mol⁻¹. For IB, use data booklet values precisely, and remember that ΔG = -nFE° requires E° in volts and F in C mol⁻¹, giving ΔG in J mol⁻¹ — convert to kJ for comparison.
考试技巧：在将数字代入 ΔG = ΔH – TΔS 之前，务必检查单位一致性。CIE 试卷中，数据可能有意以 kJ 和 J 混合给出；须将 ΔS 转换为 kJ K⁻¹ mol⁻¹。IB 要准确使用数据手册的值，并记住 ΔG = -nFE° 要求 E° 以伏特、F 以 C mol⁻¹ 为单位，得出 ΔG 单位为 J mol⁻¹——需转换为 kJ 以进行比较。
Exam tip 2: When predicting spontaneity at a new temperature, state the assumption that ΔH and ΔS are temperature-independent. This shows understanding of the model’s limitations and can fetch marks in evaluation questions.
考试技巧二：在预测新温度下的自发性时，应说明假设 ΔH 和 ΔS 不随温度变化。这展示了对模型局限性的理解，可在评估类问题中得分。
Exam tip 3: For ΔG° = -RT ln K, if K is expressed in terms of concentration (Kc) or pressure (Kp), the standard state must be explicitly considered. Technically, one should divide each concentration by c° (1 mol dm⁻³) or pressure by p° (1 bar) to obtain the dimensionless equilibrium constant required. Many exam boards accept the direct use of Kc or Kp as numerical values, but understanding the dimensionless nature is important for university preparation and higher-level questions.
考试技巧三：对于 ΔG° = -RT ln K，若 K 以浓度 (Kc) 或压力 (Kp) 表示，必须明确考虑标准状态。技术上，应将每个浓度除以 c° (1 mol dm⁻³) 或压力除以 p° (1 bar) 以获得所需的无量纲平衡常数。许多考试局允许直接使用 Kc 或 Kp 的数值，但理解其无量纲性质对大学学习及更高级别问题至关重要。
11. Worked Example: Calculating ΔG and Threshold Temperature | 计算示例：计算 ΔG 和阈值温度
Question: For the reaction 2NO₂(g) → N₂O₄(g), given ΔH° = -57.2 kJ mol⁻¹ and ΔS° = -175.9 J K⁻¹ mol⁻¹ at 298 K. (a) Calculate ΔG° at 298 K. (b) Determine the temperature at which the reaction becomes non-spontaneous. (c) Predict the spontaneity at 400 K assuming ΔH° and ΔS° are constant.
问题：对于反应 2NO₂(g) → N₂O₄(g)，已知 ΔH° = -57.2 kJ mol⁻¹，ΔS° = -175.9 J K⁻¹ mol⁻¹（298 K）。(a) 计算 298 K 下的 ΔG°。(b) 求反应变为非自发时的温度。(c) 假设 ΔH° 和 ΔS° 不变，预测 400 K 下的自发性。
Solution (a):
Convert ΔS° to kJ K⁻¹ mol⁻¹: -175.9 J K⁻¹ mol⁻¹ = -0.1759 kJ K⁻¹ mol⁻¹.
ΔG° = ΔH° – TΔS° = -57.2 – (298)(-0.1759) = -57.2 + 52.4 = -4.8 kJ mol⁻¹
Since ΔG° is negative, the reaction is spontaneous at 298 K.
解答 (a)：
将 ΔS° 转换为 kJ K⁻¹ mol⁻¹：-175.9 J K⁻¹ mol⁻¹ = -0.1759 kJ K⁻¹ mol⁻¹。
ΔG° = ΔH° – TΔS° = -57.2 – (298)(-0.1759) = -57.2 + 52.4 = -4.8 kJ mol⁻¹
因 ΔG° 为负，反应在 298 K 下自发。
Solution (b):
Set ΔG° = 0: 0 = ΔH° – TΔS° → T = ΔH° / ΔS°.
Use ΔS° in kJ K⁻¹ mol⁻¹: T = (-57.2) / (-0.1759) ≈ 325 K. (51.8°C).
At temperatures above 325 K, TΔS outweighs ΔH, and ΔG becomes positive; the reaction is non-spontaneous. This matches the prediction for a process with negative ΔH and negative ΔS.
解答 (b)：
令 ΔG° = 0：0 = ΔH° – TΔS° → T = ΔH° / ΔS°。
使用 ΔS° 以 kJ K⁻¹ mol⁻¹ 为单位：T = (-57.2) / (-0.1759) ≈ 325 K (51.8°C)。
温度高于 325 K 时，TΔS 超过 ΔH，ΔG 变为正值；反应非自发。这与 ΔH 和 ΔS 均为负值的过程的预测相符。
Solution (c):
At 400 K: ΔG° = -57.2 – (400)(-0.1759) = -57.2 + 70.36 = +13.16 kJ mol⁻¹, so non-spontaneous. The reaction is spontaneous below 325 K and non-spontaneous above. This is why N₂O₄ dimerisation is favoured at low temperatures.
解答 (c)：
在 400 K 时：ΔG° = -57.2 – (400)(-0.1759) = -57.2 + 70.36 = +13.16 kJ mol⁻¹，非自发。反应在低于 325 K 时自发，高于 325 K 时非自发。这就是为什么 N₂O₄ 二聚在低温下有利。
12. Quick Reference Summary for Revision | 复习速查表
| Concept / Equation | Key points |
|---|---|
| ΔG = ΔH – TΔS | Unit check essential; T in K; ΔS often in J, must convert. |
| ΔG° = -RT ln K | R = 8.31 J K⁻¹ mol⁻¹; K dimensionless; use ln, not log. |
| ΔG° = -nFE°cell | n = mol e⁻; F = 96 500 C mol⁻¹; E° in V; ΔG in J. |
| ΔG = ΔG° + RT ln Q | Determines direction; at equilibrium Q = K and ΔG = 0. |
| Van ‘t Hoff eqn | Slope of ln K vs 1/T gives -ΔH°/R. |
| Coupled reactions | 更多咨询请联系16621398022（同微信）
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