📚 Combined 085: Free Energy and Equilibrium Constant | 综合08.5：自由能与平衡常数
ty to link thermodynamics and equilibrium is tested through a powerful relationship: ΔG° = – ibria (Topic 11). Understanding this link allows you to predict reaction feasibility, calculate equilibrium constants from thermodynamic data, and explain how temperature shifts the position of equilibrium. This article will walk you through the core theory, worked examples, and common exam traps, ensuring you can apply these concepts confidently.
在 Edexcel A-Level 化学中，连接热力学与平衡的能力通过一个强大的关系式来考查：ΔG° = –RT ln K。这一主题在复习框架中常被称为 Combined 085，它融合了能量学（Topic 8 与 Topic 13）和平衡（Topic 11）的思想。理解这一联系可让你预测反应可行性、由热力学数据计算平衡常数，并解释温度如何移动平衡位置。本文将带你梳理核心理论、典型例题和常见考试陷阱，确保你能够自信地应用这些概念。
1. What Is Combined 085? | 什么是综合08.5？
een free energy changes and equilibrium constants. In the Edexcel specification, the ΔG° = – I) but draws heavily on the equilibrium principles of Topic 11. By treating it as a single 'combined' ign multi-step questions requiring you to move seamlessly between ΔG°, K, and even E°cell.
综合08.5 是一个综合性复习标签，突出自由能变化与平衡常数之间的交汇点。在 Edexcel 大纲中，ΔG° = –RT ln K 方程出现在 Topic 13（能量学 II），但大量依赖于 Topic 11 的平衡原理。将其作为一个单独的“组合”概念来处理，你可以练习同时考查 ΔG、K 和温度效应的计算。考试命题常设计多步问题，要求你在 ΔG°、K 甚至 E°cell 之间自如切换。
2. Gibbs Free Energy Refresher | 吉布斯自由能要点回顾
action at constant temperature and pressure. The change in free energy is given by ΔG = ΔH – le reaction. At standard conditions (298 K, 100 kPa, 1 mol dm⁻³), we use the standard free energy change, ΔG°. This quantity can be calculated from standard enthalpies and entropies of formation or from ΔG° = ΔH° – TΔS°.
吉布斯自由能 G 综合了焓和熵，用以判断恒温恒压下反应的自发性。自由能变化由 ΔG = ΔH – TΔS 给出，其中 ΔH 为焓变，T 为热力学温度（开尔文），ΔS 为熵变。ΔG 为负值表明反应在热力学上可行。在标准条件（298 K、100 kPa、1 mol dm⁻³）下，我们使用标准自由能变化 ΔG°。该量可由标准生成焓与生成熵计算，或由 ΔG° = ΔH° – TΔS° 求得。
3. Equilibrium Constant K and Reaction Quotient Q | 平衡常数 K 与反应商 Q
For a general equilibrium aA + bB ⇌ s exactly the same mathematical form but uses non-equilibrium concentrations. If Q < K, the forward reaction is favoured; if Q > K, the reverse reaction is favoured. When Q = K, the system is at equilibrium and ΔG = 0.
对于一般平衡 aA + bB ⇌ cC + dD，平衡常数 Kc（或 Kp）由平衡时的浓度（或分压）定义。反应商 Q 的数学形式完全相同，但使用的是非平衡浓度。若 Q < K，正反应占优；若 Q > K，逆反应占优。当 Q = K 时，体系处于平衡，且 ΔG = 0。
4. Deriving the Key Link: ΔG° = –RT ln K | 推导核心联系式：ΔG° = –RT ln K
is in kelvin. At equilibrium, Q becomes K and ΔG = 0, so 0 = ΔG° + RT ln K. Rearranging yields the crucial equation:
ΔG° = –RT ln K
This shows that the more negative ΔG° is, the larger the equilibrium constant, meaning the equilibrium lies further to the right. Always remember that K must be dimensionless or expressed in terms of standard pressures (1 bar) if using Kp; for solution reactions, Kc values can be used directly provided they are numerically consistent with the standard state definition.
平衡的热力学条件是 ΔG = 0。反应混合物的自由能为 ΔG = ΔG° + RT ln Q，其中 R = 8.314 J mol⁻¹ K⁻¹，T 以开尔文为单位。平衡时，Q 变为 K 且 ΔG = 0，故 0 = ΔG° + RT ln K。移项得到关键方程：
ΔG° = –RT ln K
这表明 ΔG° 越负，平衡常数越大，平衡位置越偏右。务必记住，对于气相反应若使用 Kp，K 需为无量纲或与标准压力（1 bar）相关联；对于溶液反应，可直接使用 Kc，只要其数值与标准态定义一致。
5. Spontaneity and Magnitude of K | 自发性与 K 的大小
When ΔG° is very negative (typically less than –20 kJ mol⁻¹), K is huge (>10³), and the reaction is considered to go essentially to completion. When ΔG° is very positive (> +20 kJ mol⁻¹), K is tiny (<10⁻³), and the forward reaction barely occurs. For intermediate values of ΔG° (between –20 and +20 kJ mol⁻¹), K lies between 10⁻³ and 10³, and significant amounts of both reactants and products are present at equilibrium. This threshold helps you quickly predict equilibrium composition from a given ΔG°.
当 ΔG° 非常负（通常低于 –20 kJ mol⁻¹），K 极大（>10³），反应可视为基本进行完全。当 ΔG° 非常正（> +20 kJ mol⁻¹），K 极小（<10⁻³），正反应几乎不发生。对于介于 –20 至 +20 kJ mol⁻¹ 之间的 ΔG° 值，K 在 10⁻³ 至 10³ 之间，平衡时会有显著量的反应物和产物共存。这一阈值有助于你根据给定的 ΔG° 快速预测平衡组成。
6. Predicting Direction with ΔG = ΔG° + RT ln Q | 用 ΔG = ΔG° + RT ln Q 预测方向
s you which direction a non-equilibrium mixture will proceed. For example, if you know ΔG° and the initial concentrations, you can calculate ΔG. A negative ΔG means the forward reaction is spontaneous under those conditions; a positive ΔG favours the reverse reaction. This is particularly useful in industrial processes where initial concentrations are chosen far from equilibrium to drive product formation.
完整方程 ΔG = ΔG° + RT ln Q 能告诉你非平衡混合物将向哪个方向进行。例如，已知 ΔG° 和初始浓度，便可计算 ΔG。ΔG 为负意味着在该条件下正反应自发；ΔG 为正则倾向逆反应。这在工业过程中尤其有用，初始浓度往往选择远离平衡以推动产物的生成。
7. Worked Example: Calculating K from ΔG° | 例题：由 ΔG° 计算 K
the equilibrium constant Kp at 298 K for the decomposition of calcium carbonate: CaCO₃(s) ⇌ CaO(s) + CO₂(g), given ΔG° = +130.4 kJ mol⁻¹.
已知 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 的 ΔG° = +130.4 kJ mol⁻¹，计算 298 K 下的平衡常数 Kp。
First, convert ΔG° to J mol⁻¹: ΔG° = +130 400 J mol⁻¹. Use ΔG° = –RT ln K.
首先将 ΔG° 转换为 J mol⁻¹：ΔG° = +130 400 J mol⁻¹。运用 ΔG° = –RT ln K。
ln K = –ΔG° / (RT) = –130 400 / (8.314 × 298)
ln K = –130 400 / 2477.6 ≈ –52.63
K = e⁻⁵²·⁶³ ≈ 1.4 × 10⁻²³
artial pressure of CO₂ is produced. Note that since the solid activities are 1, Kp = pCO₂.
极小的 K 反映了 CaCO₃ 在室温下的热稳定性；仅产生极微小的 CO₂ 分压。注意由于固体活度为 1，Kp = pCO₂。
8. Temperature Dependence: The van't Hoff Equation | 温度依赖关系：范特霍夫方程
Because ΔG° = ΔH° – TΔS°, the equilibrium constant changes with temperature. By combining ΔG° = –RT ln K with ΔG° = ΔH° – TΔS°, we obtain:
ln K = –ΔH°/(R) × 1/T + ΔS°/R
This is the linear form of the van't Hoff equation, often written as:
ln(K₂/K₁) = –ΔH°/R × (1/T₂ – 1/T₁)
If ΔH° is positive (endothermic), increasing T increases K, favouring products. If ΔH° is negative (exothermic), increasing T decreases K.
由于 ΔG° = ΔH° – TΔS°，平衡常数随温度而变化。将 ΔG° = –RT ln K 与 ΔG° = ΔH° – TΔS° 结合可得：
ln K = –ΔH°/(R) × 1/T + ΔS°/R
这是范特霍夫方程的线性形式，常写作：
ln(K₂/K₁) = –ΔH°/R × (1/T₂ – 1/T₁)
若 ΔH° 为正值（吸热），升高 T 使 K 增大，有利于产物。若 ΔH° 为负值（放热），升高 T 则使 K 减小。
9. Relating K at Two Temperatures | 关联两个温度下的 K
The van' at T₂. Always use consistent units for ΔH° (J mol⁻¹) and R. For example, if K for 2NO₂(g) ⇌ N₂O₄(g) at 298 K is 8.8 and ΔH° = – emperature shifts equilibrium to the right, so K increases. The calculation confirms this.
范特霍夫方程在计算非 298 K 温度下的 K 时极为有用。给定 T₁ 时的 K₁ 和 ΔH°，便可求出 T₂ 时的 K₂。务必保证 ΔH°（J mol⁻¹）与 R 的单位一致。例如，2NO₂(g) ⇌ N₂O₄(g) 在 298 K 时 K = 8.8，ΔH° = –57.2 kJ mol⁻¹，可预测 273 K 时的 K。由于反应放热，降低温度使平衡右移，因此 K 增大。计算可证实这一点。
10. Linking to Electrochemistry: ΔG° = –nFE°cell | 联系电化学：ΔG° = –nFE°cell
ectrode potentials. The standard free energy change for a redox reaction is given by ΔG° = – ere n is the number of electrons transferred and F = 96 500 C mol⁻¹. By equating this with –RT ln K, you get:
nFE°cell = RT ln K
This means a positive E°cell corresponds to a large K, confirming that thermodynamically favourable redox reactions have high equilibrium constants. It also allows you to calculate K directly from standard cell potentials.
在 Edexcel Combined 085 题目中，你还可能需要将平衡与电极电势联系起来。氧化还原反应的标准自由能变由 ΔG° = –nFE°cell 给出，其中 n 为转移电子数，F = 96 500 C mol⁻¹。将其与 –RT ln K 相等联立，可得：
nFE°cell = RT ln K
这意味着正的 E°cell 对应大的 K，证实热力学上有利的氧化还原反应具有高平衡常数。由此也可直接从标准电池电势计算 K。
11. Common Pitfalls and Exam Tips | 常见误区与考试技巧
nother pitfall is using ΔG° values for standard temperature when the question involves a different temperature – always use ΔG° = ΔH° – mber that when calculating ΔG from ΔG° + RT ln Q, Q must be correctly formulated, including solids and pure liquids as 1.
典型错误是当使用 R = 8.314 J mol⁻¹ K⁻¹ 时忘记将 ΔG° 转换为 J mol⁻¹。若 ΔG° 单位仍为 kJ，计算出的 ln K 将偏差 1000 倍。另一误区是当题目涉及非标准温度时直接使用标准 ΔG° 值——应始终用 ΔG° = ΔH° – TΔS° 求出所需温度下的 ΔG°，再求 K。注意 Kp 与 Kc 的单位；Kp 问题通常要求你转换为分压。最后记住，在通过 ΔG° + RT ln Q 计算 ΔG 时，必须正确表达 Q，固体和纯液体按 1 处理。
12. Summary and Quick Reference | 总结与速查表
The Combined 085 topic encapsulates one of the most elegant relationships in physical chemistry. To master it, ensure you can fluently move between ΔG°, K, T, and E°cell. The following table summarises the key equations you should memorise.
综合08.5 主题概括了物理化学中最优雅的关系之一。要掌握它，需确保你能在 ΔG°、K、T 和 E°cell 之间流畅转换。下表总结了你应记住的关键方程。
| Equation / 方程 | Application / 用途 |
|---|---|
| ΔG = ΔH – TΔS | Calculate free energy change / 计算自由能变 |
| ΔG° = –RT ln K | Link standard free energy to equilibrium constant / 连接标准自由能与平衡常数 |
| ΔG = ΔG° + RT ln Q | Predict reaction direction under any conditions / 预测任意条件下的反应方向 |
| ln(K₂/K₁) = –ΔH°/R (1/T₂ – 1/T₁) | Find K at a new temperature / 求新温度下的 K |
| ΔG° = –nFE°cell | Relate free energy to cell potential / 关联自由能与电池电势 |
| nFE°cell = RT ln K | Direct route from cell potential to K / 从电池电势直接求 K |
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