📚 Gibbs Free Energy | 吉布斯自由能
d temperature. In A-Level exams you must be able to use ΔG = ΔH − TΔS, interpret the sign of ΔG, and connect standard Gibbs free energy changes to equilibrium constants and electrode potentials.
吉布斯自由能将焓、熵和温度联系起来，用于预测恒压恒温下化学过程的热力学可行性。在 A-Level 考试中，你需要会使用 ΔG = ΔH − TΔS、解释 ΔG 的正负，并把标准吉布斯自由能变与平衡常数和电极电势联系起来。
1. What Is Gibbs Free Energy? | 什么是吉布斯自由能
Gibbs free energy, symbol G, is a thermodynamic state function that merges enthalpy and entropy into one single quantity. It is named after the American physicist Josiah Willard Gibbs, who developed it in the 1870s. Because almost every laboratory or industrial reaction is carried out at roughly constant temperature and pressure, G provides a direct test of whether a process can proceed without an external supply of useful work.
吉布斯自由能（符号 G）是一个热力学状态函数，它把焓和熵合并成一个量。它由美国物理学家吉布斯（Josiah Willard Gibbs）在 19 世纪 70 年代提出。由于几乎所有实验室或工业反应都在近似恒温恒压的条件下进行，G 可以直接判断一个过程能否在不外加有用功的情况下进行。
The absolute value of G cannot be measured or calculated, so chemists always work with changes in Gibbs free energy, written ΔG. A negative ΔG means the products have less free energy than the reactants, so energy is released to the surroundings as the reaction proceeds.
G 的绝对值无法测量或计算，因此化学家实际上总是使用吉布斯自由能的变化量，记作 ΔG。ΔG 为负，意味着产物的自由能低于反应物，反应进行时会向环境释放能量。
- State function: ΔG depends only on the initial and final states, not on the route taken.
状态函数：ΔG 只取决于始态与终态，与路径无关。 - Symbol and unit: G and ΔG are measured in kJ mol⁻¹ (sometimes J mol⁻¹).
符号与单位：G 和 ΔG 的单位是 kJ mol⁻¹（有时也用 J mol⁻¹）。 - Scope: the equation applies to a system at constant temperature and constant pressure.
适用范围：该式适用于恒温恒压的体系。
2. The Definition G = H − TS | 定义式 G = H − TS
S, where H is enthalpy, T is the absolute temperature in kelvin, and S is entropy. The term TS is sometimes called the "unavailable energy" the portion of the enthalpy that cannot be converted into useful work at that temperature.
吉布斯自由能定义为 G = H − TS，其中 H 为焓，T 为热力学温度（单位开尔文），S 为熵。TS 这一项有时称为"不可用能"，因为它代表在该温度下无法转化为有用功的那部分焓。
G = H − T S
ΔG = ΔH − T ΔS
ΔG = ΔH − TΔS. This is the version you must quote and apply in every A-Level calculation.
在恒温条件下，Δ(TS) 可以简化为 TΔS，因此定义式变为 ΔG = ΔH − TΔS。这就是 A-Level 所有计算中必须引用和使用的形式。
- ΔH: enthalpy change of the reaction, negative for exothermic processes.
ΔH：反应的焓变，放热过程为负值。 - T: temperature in kelvin, never in °C; K = °C + 273.
T：热力学温度，单位开尔文，绝不能用摄氏度；K = °C + 273。 - ΔS: entropy change of the system, usually in J K⁻¹ mol⁻¹.
ΔS：体系的熵变，常用单位 J K⁻¹ mol⁻¹。
3. ΔG at Constant Temperature and Pressure | 恒温恒压下的 ΔG
Most A-Level questions specify constant temperature and pressure, which is exactly the condition under which ΔG is the correct feasibility criterion. At constant T and P, a process with negative ΔG can proceed spontaneously in the forward direction, releasing free energy that can be used to do work.
大多数 A-Level 题目都给出恒温恒压的条件，而这正是 ΔG 可以作为可行性判据的前提。在恒温恒压下，ΔG 为负的过程可以自发正向进行，并释放出可用于做功的自由能。
A common misunderstanding is that "spontaneous" or "feasible" means the reaction happens quickly. Thermodynamics says nothing about rate. A reaction with a very negative ΔG can still be immeasurably slow if its activation energy is high, which is exactly why the Haber process needs an iron catalyst and why diamond is thermodynamically unstable yet persists for billions of years.
一个常见误解是把"自发"或"可行"理解为反应很快。热力学并不涉及速率问题。即使 ΔG 很负，如果活化能很高，反应仍然可能慢到几乎测不出来——这正是合成氨需要铁催化剂、而金刚石虽然热力学不稳定却可以稳定存在数十亿年的原因。
- Negative ΔG: forward reaction is thermodynamically feasible.
ΔG 为负：正反应在热力学上可行。 - Zero ΔG: the system is at equilibrium.
ΔG 为零：体系处于平衡状态。 - Positive ΔG: forward reaction is not feasible; the reverse reaction is feasible.
ΔG 为正：正反应不可行，逆反应可行。
4. Interpreting the Sign of ΔG | 判断 ΔG 的正负
The sign of ΔG is decided by a tug-of-war between the enthalpy term ΔH and the entropy term −TΔS. An exothermic reaction (negative ΔH) tends to make ΔG negative, while an increase in disorder (positive ΔS) also makes ΔG more negative. Temperature acts as the weighting factor on the entropy term.
ΔG 的正负取决于焓变 ΔH 与熵变项 −TΔS 之间的"拔河"。放热反应（ΔH 为负）倾向于使 ΔG 为负，而混乱度增大（ΔS 为正）同样使 ΔG 更负。温度则是熵变项的权重因子。
| ΔH | 焓变 | ΔS | 熵变 | Feasibility | 可行性 |
|---|---|---|
| negative | 负 | positive | 正 | feasible at all temperatures | 任何温度下都可行 |
| positive | 正 | negative | 负 | not feasible at any temperature | 任何温度下都不可行 |
| negative | 负 | negative | 负 | feasible only at low temperature | 仅低温下可行 |
| positive | 正 | positive | 正 | feasible only at high temperature | 仅高温下可行 |
Notice that only one combination is feasible at every temperature: exothermic with increasing entropy. Every other case needs a temperature condition, which is why exam questions often ask you to state the conditions under which a reaction becomes feasible.
注意只有一种组合在任何温度下都可行：放热且熵增。其余情况都需要附加温度条件，这也是考试中经常要求说明"反应在什么条件下变得可行"的原因。
5. How ΔH and TΔS Compete | ΔH 与 TΔS 的竞争
The enthalpy term reflects the strength of chemical bonding: forming stronger bonds releases energy and lowers the enthalpy. The entropy term reflects the number of ways energy can be distributed, so gases produced, disorder created and phase changes from solid to liquid to gas all push ΔS positive.
焓变项反映化学键的强弱：形成更强的键会释放能量、降低焓值。熵变项反映能量可能分布方式的数量，因此生成气体、混乱度增加以及从固→液→气的相变都会使 ΔS 增大（为正）。
Because T multiplies ΔS, raising the temperature strengthens the influence of entropy. A reaction with a positive ΔH and a positive ΔS therefore starts off unfavourable and becomes feasible only above a certain temperature. The reverse is true when ΔS is negative: the reaction is fine when cold and fails when hot.
由于 T 与 ΔS 相乘，升高温度会放大熵的影响。因此 ΔH 为正、ΔS 也为正的反应一开始不可行，只有超过某一温度后才变得可行。当 ΔS 为负时则相反：低温时可行，高温时反而不可行。
ΔG = ΔH − TΔS
This competition explains familiar industrial choices: the Haber process is exothermic but reduces the number of gas moles, so it is favoured by low temperature (though a compromise temperature is used for rate), while the cracking of alkanes and the thermal decomposition of carbonates need high temperatures because both are entropy-driven.
这种竞争可以解释我们熟悉的工业选择：合成氨放热且气体物质的量减少，因此低温有利（但为了速率采用折中温度）；而烷烃裂解和碳酸盐的热分解则因由熵驱动而需要高温。
6. Units and Simple Calculations | 单位与简单计算
The single most common arithmetic error in this topic is mixing units. ΔH is normally quoted in kJ mol⁻¹, but ΔS is almost always quoted in J K⁻¹ mol⁻¹. Before you subtract, you must divide the entropy term by 1000 to convert it into kJ K⁻¹ mol⁻¹.
这一主题最常见的算术错误就是单位不一致。ΔH 通常以 kJ mol⁻¹ 给出，而 ΔS 几乎总是以 J K⁻¹ mol⁻¹ 给出。在相减之前，必须把熵变项除以 1000，换算成 kJ K⁻¹ mol⁻¹。
ΔG (kJ mol⁻¹) = ΔH (kJ mol⁻¹) − T ΔS (J K⁻¹ mol⁻¹) ÷ 1000
Worked example: for the reaction N₂(g) + 3H₂(g) ⇌ . The negative value confirms the forward reaction is thermodynamically feasible at 298 K.
计算示例：反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，在 298 K 下 ΔH = −92.0 kJ mol⁻¹，ΔS = −199 J K⁻¹ mol⁻¹。则 TΔS = 298 × (−199) ÷ 1000 = −59.3 kJ mol⁻¹，ΔG = −92.0 − (−59.3) = −32.7 kJ mol⁻¹。负值说明正反应在 298 K 下热力学可行。
- Always convert T to kelvin before substituting.
代入前一定要把温度换成开尔文。 - Keep signs: a negative ΔS must be subtracted correctly.
注意符号：ΔS 为负时相减容易出错。 - Quote units in your final answer to gain the marking point.
最终答案要写出单位，才能拿到分数点。
7. Temperature and the Crossover Point | 温度与转变温度
ble to feasible or the other way round. Setting ΔG = 0 in ΔG = ΔH − TΔS gives T = ΔH ÷ ΔS.
转变温度是使 ΔG 恰好等于零的温度，也就是反应从不可行变为可行（或相反）的临界温度。令 ΔG = ΔH − TΔS 中的 ΔG = 0，可得 T = ΔH ÷ ΔS。
T = ΔH ÷ ΔS
− 47.8 = +130.2 kJ mol⁻¹ and the decomposition is not feasible. The crossover temperature is T = 178 ÷ 0.1605 = 1109 K (about 836 °C), above which the reaction becomes feasible.
计算示例：CaCO₃(s) → CaO(s) + CO₂(g)，ΔH = +178 kJ mol⁻¹，ΔS = +160.5 J K⁻¹ mol⁻¹。298 K 时 TΔS = 298 × 160.5 ÷ 1000 = +47.8 kJ mol⁻¹，故 ΔG = +178 − 47.8 = +130.2 kJ mol⁻¹，分解不可行。转变温度为 T = 178 ÷ 0.1605 = 1109 K（约 836 °C），高于此温度反应才变得可行。
- When ΔH is positive and ΔS is positive, use T > ΔH ÷ ΔS for feasibility.
当 ΔH 为正、ΔS 为正时，可行条件为 T > ΔH ÷ ΔS。 - When both are negative, use T < ΔH ÷ ΔS (the ratio is still positive).
当两者均为负时，可行条件为 T < ΔH ÷ ΔS（比值仍为正）。 - If ΔS = 0 the equation breaks down, because ΔG = ΔH at all temperatures.
若 ΔS = 0，该式失效，因为此时任何温度下均有 ΔG = ΔH。
8. Standard Gibbs Free Energy Change ΔG° | 标准吉布斯自由能变
The standard Gibbs free energy change, ΔG°, refers to a reaction carried out under standard conditions: 298 K, a pressure of 100 kPa for gases, and a concentration of 1 mol dm⁻³ for solutions. Standard values are tabulated as standard free energies of formation, ΔG°f, in kJ mol⁻¹.
标准吉布斯自由能变 ΔG° 指在标准条件下进行的反应：298 K，气体压力 100 kPa，溶液浓度 1 mol dm⁻³。标准数据以标准生成吉布斯自由能 ΔG°f 的形式列表给出，单位为 kJ mol⁻¹。
ΔG°reaction = Σ ΔG°f(products) − Σ ΔG°f(reactants)
By definition, ΔG°f of an element in its most stable standard state is zero, exactly as with standard enthalpy of formation. A reaction with a negative ΔG° therefore has products that are more thermodynamically stable than the reactants under standard conditions.
按照定义，元素在其最稳定标准状态下的 ΔG°f 为零，这一点与标准生成焓完全一致。因此 ΔG° 为负的反应，其产物在标准条件下比反应物在热力学上更稳定。
- Sign convention: ΔG° < 0 favours products, ΔG° > 0 favours reactants.
符号约定：ΔG° < 0 有利于产物，ΔG° > 0 有利于反应物。 - Temperature dependence: tabulated ΔG° values only apply at 298 K.
温度依赖：表中的 ΔG° 数值只适用于 298 K。 - Consistency: ΔG° tells you the position of equilibrium, not the rate.
一致性：ΔG° 说明平衡位置，而不是反应速率。
9. ΔG° and the Equilibrium Constant | ΔG° 与平衡常数
ln K, where R is the gas constant, 8.31 J K⁻¹ mol⁻¹, and T is the temperature in kelvin. This relation explains why a very negative ΔG° corresponds to a very large K.
标准吉布斯自由能变通过方程 ΔG° = −RT ln K 与平衡常数 K 直接联系，其中 R 为气体常数 8.31 J K⁻¹ mol⁻¹，T 为热力学温度。这一关系解释了为什么 ΔG° 越负，K 越大。
ΔG° = −R T ln K
K = e−ΔG° ÷ R T
The large value confirms that the equilibrium for ammonia synthesis lies well to the right under standard conditions.
计算示例：取 298 K 下 ΔG° = −32.7 kJ mol⁻¹ = −32 700 J mol⁻¹，则 ln K = 32 700 ÷ (8.31 × 298) = 13.2，故 K ≈ 5.4 × 10⁵。该值很大，说明标准条件下合成氨的平衡位置明显偏右。
- ΔG° = 0 ⟹ K = 1: products and reactants are equally favoured.
ΔG° = 0 ⟹ K = 1：产物与反应物同样有利。 - ΔG° < 0 ⟹ K > 1: the forward reaction is favoured at equilibrium.
ΔG° < 0 ⟹ K > 1：平衡时有利于正反应。 - Remember J vs kJ: convert ΔG° into J mol⁻¹ before dividing by RT.
注意 J 与 kJ：在除以 RT 之前要把 ΔG° 换成 J mol⁻¹。
10. ΔG° and Electrode Potential | ΔG° 与电极电势
s the number of moles of electrons transferred and F is the Faraday constant, 96 500 C mol⁻¹. A positive E°cell therefore automatically means a negative ΔG°, which is why cells with positive E°cell deliver electrical work.
在电化学中，电池反应的标准吉布斯自由能变与标准电池电动势的关系为 ΔG° = −nFE°cell，其中 n 为转移的电子的物质的量，F 为法拉第常数 96 500 C mol⁻¹。因此 E°cell 为正必然对应 ΔG° 为负，这正是 E°cell 为正的电池能对外做电功的原因。
ΔG° = −n F E°cell
ol⁻¹. If E°cell were negative, the same calculation would give a positive ΔG° and the cell reaction would not be feasible as written.
计算示例：丹尼尔电池 E°cell = +1.10 V，n = 2，则 ΔG° = −2 × 96 500 × 1.10 = −212 300 J mol⁻¹ ≈ −212 kJ mol⁻¹。若 E°cell 为负，同样的计算会得到正的 ΔG°，该电池反应按所写方向就不可行。
- n is the electron count after balancing the overall cell equation. 广告 / SponsoredChemistry Exam Preparation and Practice 剑桥国际AS & ALEVEL化学考试…化学 备考教辅 · 正版现货 · 与本文内容相关查看详情 / View on Taobao参考价 ¥253.0
