📚 Gibbs Free Energy: Key Exam Concepts | 吉布斯自由能：考点精讲
Gibbs free energy (G) is a thermodynamic potential that combines enthalpy and entropy to predict the spontaneity of a process at constant temperature and pressure. For IB and OCR Chemistry students, mastering Gibbs free energy calculations and interpreting the sign of ΔG is essential for tackling energetics and equilibrium questions. This article unpacks the core concepts, equations, and common pitfalls to help you secure those marks.
吉布斯自由能 (G) 是一个结合焓与熵的热力学势，用于判断恒温恒压下过程的自发性。对 IB 和 OCR 化学考生而言，掌握吉布斯自由能的计算、理解 ΔG 的符号含义是解答能量学与平衡类题目的关键。本文梳理核心概念、公式与常见误区，助你稳拿高分。
1. Introduction to Gibbs Free Energy | 吉布斯自由能简介
Gibbs free energy, denoted G, is defined as G = H – in Gibbs free energy (ΔG) indicates whether a reaction can occur spontaneously. A negative ΔG means the process is thermodynamically feasible, while a positive ΔG indicates a non‑spontaneous reaction under the given conditions.
吉布斯自由能记为 G，定义为 G = H – TS，其中 H 为焓，T 为绝对温度（单位开尔文），S 为熵。在恒温恒压的化学体系中，吉布斯自由能的变化 (ΔG) 可判断反应能否自发进行。ΔG 为负，表示过程在热力学上可行；ΔG 为正，表明该条件下反应不自发。
The concept links two driving forces: the tendency to minimize enthalpy (exothermic reactions) and the tendency to maximize entropy (disorder). When these two factors are in opposition, Gibbs free energy tells us which factor dominates at a particular temperature.
该概念整合了两大驱动力：倾向最低焓（放热反应）和倾向最大熵（混乱度增加）。当这两个因素对立时，吉布斯自由能可判断在特定温度下哪一个因素占主导。
2. The Gibbs Equation: ΔG = ΔH – TΔS | 吉布斯方程
The fundamental equation is:
核心方程如下：
ΔG = ΔH – TΔS
Here, ΔH is the enthalpy change (kJ mol⁻¹), T is the temperature in kelvin (K), and ΔS is the entropy change (J K⁻¹ mol⁻¹ or kJ K⁻¹ mol⁻¹). A critical exam tip is to ensure consistent units: ΔS is often given in J K⁻¹ mol⁻¹, so you must convert it to kJ K⁻¹ mol⁻¹ (by dividing by 1000) if ΔH is in kJ. In IB and OCR papers, forgetting this conversion is a common source of error.
式中 ΔH 为焓变 (kJ mol⁻¹)，T 为开尔文温度 (K)，ΔS 为熵变 (J K⁻¹ mol⁻¹ 或 kJ K⁻¹ mol⁻¹)。关键考点是单位统一：ΔS 常给为 J K⁻¹ mol⁻¹，若 ΔH 以 kJ 为单位，必须将 ΔS 转化为 kJ K⁻¹ mol⁻¹（除以 1000）再代入计算。IB 与 OCR 考题中，忘记单位换算是常见失分点。
You can calculate ΔG at any temperature using this relationship, provided ΔH and ΔS are known and assumed constant over the temperature range. The equation highlights that temperature can alter the feasibility of a reaction through the TΔS term.
只要已知 ΔH 和 ΔS 并假定在温度范围内不变，就可用此关系计算任意温度下的 ΔG。该方程表明，温度可通过 TΔS 项改变反应的自发性。
3. Units and Sign Conventions | 单位与符号约定
ion to kJ K⁻¹ mol⁻¹ is necessary before substituting. The sign of ΔG is the ultimate indicator: negative = spontaneous, positive = non‑ H, exothermic values are negative; for ΔS, an increase in disorder gives a positive value.
温度始终用开尔文 (K = °C + 273)。ΔG 和 ΔH 通常以 kJ mol⁻¹ 表示。ΔS 最常以 J K⁻¹ mol⁻¹ 给出，代入前须转换为 kJ K⁻¹ mol⁻¹。ΔG 的符号是最终判据：负值为自发，正值为非自发，零为平衡。ΔH 放热为负；ΔS 混乱度增加为正。
Be careful with the signs in the equation: ΔG = ΔH – ositive (never spontaneous). The interesting cases arise when ΔH and ΔS have the same sign — temperature then decides feasibility.
注意方程中的符号：ΔG = ΔH – TΔS。若 ΔH 为负且 ΔS 为正，ΔG 永为负（任何温度下自发）。若 ΔH 为正且 ΔS 为负，ΔG 永为正（绝不自发）。有趣的情况出现在 ΔH 与 ΔS 同号时——此时温度决定自发性。
4. Spontaneity Criteria | 自发性判据
The sign of ΔG alone determines thermodynamic spontaneity under constant T and P:
在恒温恒压下，仅凭 ΔG 的符号即可判定热力学自发性：
ΔG < 0 : The forward reaction is spontaneous.
ΔG < 0：正向反应自发。
ΔG > 0 : The forward reaction is non‑spontaneous (the reverse reaction is spontaneous).
ΔG > 0：正向反应非自发（逆反应自发）。
ΔG = 0 : The system is at equilibrium; there is no net change.
ΔG = 0：体系处于平衡态，无净变化。
Remember that spontaneity says nothing about the rate. A reaction with a negative ΔG may be immeasurably slow without a catalyst or suitable pathway. Thermodynamics tells us whether a process can happen, not whether it will happen quickly.
注意自发性与速率无关。ΔG 为负的反应若无催化剂或合适途径，速率可能极慢。热力学告诉我们过程“能否”发生，而非“快慢”。
5. Temperature Dependence | 温度依赖性
Because ΔG = ΔH – en ΔH and ΔS have identical signs. The table below summarises the four possible scenarios:
由于 ΔG = ΔH – TΔS，当 ΔH 与 ΔS 同号时温度具有决定性作用。下表总结了四种可能情景：
| ΔH sign | ΔS sign | ΔG sign & spontaneity | 中文说明 |
|---|---|---|---|
| Negative (–) | Positive (+) | Always negative; spontaneous at all T | 任何温度下自发 |
| Positive (+) | Negative (–) | Always positive; never spontaneous | 任何温度下非自发 |
| Negative (–) | Negative (–) | Spontaneous below T = ΔH/ΔS | 低温自发（T < ΔH/ΔS） |
| Positive (+) | Positive (+) | Spontaneous above T = ΔH/ΔS | 高温自发（T > ΔH/ΔS） |
vourite exam calculation: set ΔG = 0, rearrange, and substitute values with correct units.
ΔG = 0 的转折温度由 T = ΔH / ΔS 给出。这是常见考试计算：令 ΔG = 0，整理公式，代入数值（注意单位统一）。
6. Calculating ΔG° from ΔH° and ΔS° | 由标准焓变和熵变计算标准吉布斯自由能变
The standard Gibbs free energy change, ΔG°, refers to the change when all reactants and products are in their standard states (1 bar pressure, 1 mol dm⁻³ for solutions). ΔG° can be calculated directly from standard enthalpy and entropy changes:
标准吉布斯自由能变 ΔG° 指所有反应物与产物均处于标准态（气体分压 1 bar，溶液浓度 1 mol dm⁻³）时的变化值。ΔG° 可直接由标准焓变与熵变计算：
ΔG° = ΔH° – TΔS°
ted otherwise. Remember to convert ΔS° from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹. This method is particularly useful when you are given formation or combustion data to find ΔH° and absolute entropies to find ΔS°.
除非题目另有说明，一般使用 T = 298 K。务必先将 ΔS° 从 J K⁻¹ mol⁻¹ 转换为 kJ K⁻¹ mol⁻¹。若题目给出生成或燃烧数据求 ΔH°，并给出绝对熵值求 ΔS°，则可直接使用此法。
py values; OCR provides similar data. Be careful to apply the formula ΔS° = Σ S°(products) – Σ S°(reactants) and the analogous one for ΔH°.
IB 数据手册提供了标准生成焓与标准熵数据；OCR 同样提供类似数据。务必使用 ΔS° = Σ S°(产物) – Σ S°(反应物)，以及相似的 ΔH° 计算公式。
7. Standard Gibbs Free Energy of Formation, ΔG°f | 标准生成吉布斯自由能
The standard Gibbs free energy of formation, ΔG°f, is the change in Gibbs free energy when one mole of a compound is formed from its elements in their standard states. By definition, ΔG°f for any element in its standard state is zero.
标准生成吉布斯自由能 ΔG°f 是指由标准态下的元素生成 1 mol 化合物时的吉布斯自由能变。根据定义，任何标准态元素的 ΔG°f = 0。
For a reaction, ΔG° can be calculated from tabulated ΔG°f values:
对于某一反应，ΔG° 可由查表所得的 ΔG°f 计算：
ΔG° = Σ ΔG°f(products) – Σ ΔG°f(reactants)
This provides a direct route to ΔG° without needing ΔH° and ΔS°, though it is less commonly tested in IB than the enthalpy‑entropy method.
该法无需 ΔH° 和 ΔS° 即可直接求得 ΔG°，不过在 IB 考试中不如焓‑熵计算法常见。
8. Gibbs Free Energy and Equilibrium Constant | 吉布斯自由能与平衡常数
A pivotal relationship in chemical thermodynamics links ΔG° to the equilibrium constant K:
化学热力学中一个核心关系将 ΔG° 与平衡常数 K 联系起来：
ΔG° = –RT ln K
Where R is the gas constant (8.31 J K⁻¹ mol⁻¹) and T is the temperature in kelvin. This equation tells us:
式中 R 为气体常数 (8.31 J K⁻¹ mol⁻¹)，T 为开尔文温度。由该方程可知：
If K > 1, ln K is positive, so ΔG° is negative → products are favoured at equilibrium.
若 K > 1，ln K 为正，ΔG° 为负 → 平衡时产物占优。
If K < 1, ln K is negative, so ΔG° is positive → reactants are favoured.
若 K < 1，ln K 为负，ΔG° 为正 → 平衡时反应物占优。
If K = 1, ΔG° = 0.
t exam questions will ask you to calculate K from ΔG° or vice versa, or to explain the relationship qualitatively.
注意单位：使用 R = 8.31 J K⁻¹ mol⁻¹ 时，ΔG° 必须换算为 J mol⁻¹。多数考题会要求由 ΔG° 计算 K，或反之，或定性地解释两者关系。
9. Coupled Reactions | 耦合反应
In biochemistry and industrial processes, a thermodynamically unfavourable reaction (ΔG > 0) can be driven by coupling it with a highly favourable reaction (ΔG ≪ 0). The overall ΔG is the sum of the individual ΔG values. As long as the sum is negative, the coupled process is spontaneous.
在生物化学和工业过程中，热力学不利的反应 (ΔG > 0) 可通过与强有利反应 (ΔG ≪ 0) 耦合而得以驱动。总 ΔG 为各步 ΔG 之和。只要总和为负，耦合过程即可自发进行。
Classic examples include the hydrolysis of ATP (ΔG°' ≈ – s, you may be asked to calculate the overall ΔG and comment on feasibility after coupling.
经典例子包括 ATP 水解 (ΔG°' 约为 –30.5 kJ mol⁻¹) 驱动谷氨酸合成谷氨酰胺（单一过程 ΔG 为正）。考题可能要求计算耦合后的总 ΔG，并评价其自发性。
Remember that coupling does not imply both reactions occur simultaneously without a common intermediate; in practice, they share a chemical link, such as a phosphorylated intermediate.
注意耦合并不意味着两个反应无中间体而同时发生；实际过程中它们通过化学链路（如磷酸化中间体）相互关联。
10. Worked Examples | 例题详解
Example 1: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH° = –92 kJ mol⁻¹ and ΔS° = –199 J K⁻¹ mol⁻¹ at 298 K. Determine the temperature range for spontaneity.
例1： 已知反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH° = –92 kJ mol⁻¹，ΔS° = –199 J K⁻¹ mol⁻¹ (298 K)。求自发进行的温度范围。
Convert ΔS°: –199 J K⁻¹ mol⁻¹ = –0.199 kJ K⁻¹ mol⁻¹. Set ΔG° = 0, T = ΔH° / ΔS° = (–92) / (– ve, the reaction is spontaneous below 462 K. Above this temperature, ΔG° becomes positive.
转换 ΔS°：–199 J K⁻¹ mol⁻¹ = –0.199 kJ K⁻¹ mol⁻¹。设 ΔG° = 0，T = ΔH° / ΔS° = (–92) / (–0.199) ≈ 462 K。因 ΔH 和 ΔS 均为负值，反应在 462 K 以下自发；高于此温度 ΔG° 为正，不自发。
Example 2: At 298 K, the equilibrium constant for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is K = 4.0 × 10²⁴. Calculate ΔG° and comment on the position of equilibrium.
例2： 在 298 K 时，反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 的平衡常数 K = 4.0 × 10²⁴。计算 ΔG° 并评价平衡位置。
ΔG° = –RT ln K = –(8.31)(298) ln(4.0 × 10²⁴) = –(2476) × (56.6) ≈ –140 kJ mol⁻¹. The large negative ΔG° and K ≫ 1 indicate that the products are strongly favoured.
ΔG° = –RT ln K = –(8.31)(298) ln(4.0×10²⁴) = –(2476)×(56.6) ≈ –140 kJ mol⁻¹。ΔG° 极负且 K ≫ 1，说明平衡强烈倾向于产物。
11. Common Misconceptions | 常见误区
"If ΔG is negative, the reaction will be fast." Thermodynamics says nothing about kinetics. A negative ΔG only indicates feasibility, not rate. For instance, combustion of diamond has ΔG < 0 at room temperature but is immeasurably slow.
“ΔG 为负，反应就一定快。” 热力学与动力学无关。ΔG 为负仅表示反应可行，与速率无关。例如金刚石常温燃烧的 ΔG < 0，但速率极慢。
"Exothermic reactions are always spontaneous." Only when ΔS is positive or the temperature is sufficiently low for a negative ΔS. A reaction like freezing of water at 10 °C is exothermic but has a negative ΔS, and above 0 °C it is non‑spontaneous.
“放热反应总是自发的。” 仅当 ΔS 为正，或 ΔS 为负但温度足够低时才成立。比如 10 °C 时水结冰虽放热，但 ΔS 为负，高于 0 °C 即为非自发。
"ΔG° and ΔG are the same." ΔG° is the standard change under standard conditions; ΔG varies with actual concentration or partial pressure. The sign of ΔG, not ΔG°, determines spontaneity under specific conditions, though ΔG° can indicate whether products or reactants are favoured at equilibrium.
“ΔG° 与 ΔG 相同。” ΔG° 是标准态下的值；ΔG 随实际浓度或分压而变。判断具体条件自发性的应是 ΔG 的符号，而非 ΔG°。ΔG° 可表明平衡有利于产物还是反应物。
12. Summary and Key Takeaways | 总结与要点
enthalpy and entropy into a single criterion for spontaneity. Master the equation ΔG = ΔH – ng transition temperatures (ΔG = 0) and linking ΔG° to equilibrium constants through ΔG° = – stinct, and that coupling enables unfavourable reactions to proceed in biological systems.
吉布斯自由能将焓与熵完美统一为单一的自发性判据。熟练掌握 ΔG = ΔH – TΔS 方程、单位换算以及四种 ΔH/ΔS 符号组合情景。勤练转折温度计算 (ΔG = 0) 及通过 ΔG° = –RT ln K 关联平衡常数。牢记动力学与热力学相互独立，并理解生物体系中不利反应通过耦合得以进行的原理。
In exams, always show your unit conversions clearly, state assumptions (e.g., constant ΔH and ΔS), and link the sign of ΔG to the question context. With a solid grip on these concepts, you will confidently handle any Gibbs free energy question that appears on your IB or OCR paper.
考试中，清晰写出单位换算，注明假设（如假定 ΔH 与 ΔS 不变），并将 ΔG 符号与题目情境联系起来。扎实掌握这些概念后，你就能从容应对 IB 或 OCR 试卷中任何吉布斯自由能的考题。
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