IB Chemistry: The Relationship between Gibbs Free Energy and Equilibrium Constant | IB化学：吉布斯自由能与平衡常数的关联

📚 IB Chemistry: The Relationship between Gibbs Free Energy and Equilibrium Constant | IB化学：吉布斯自由能与平衡常数的关联

In chemical thermodynamics, the Gibbs free energy change (ΔG) and the equilibrium constant (K) are two sides of the same coin. Understanding their relationship is essential for predicting the direction of reactions and calculating equilibrium compositions. This article explores the fundamental connection between ΔG and K, with a focus on the IB Chemistry syllabus.

在化学热力学中，吉布斯自由能变化（ΔG）与平衡常数（K）如同一枚硬币的两面。理解二者之间的关系，对于判断反应方向以及计算平衡组成至关重要。本文将围绕IB化学教学大纲，深入探讨ΔG与K之间最基本的联系。

1. Gibbs Free Energy and Spontaneity | 吉布斯自由能与自发性

Gibbs free energy (G) is defined as the maximum useful work obtainable from a closed thermodynamic system at constant temperature and pressure. The change in Gibbs free energy, ΔG, determines whether a reaction is spontaneous (occurs without external input) under given conditions. The defining equation is:

吉布斯自由能（G）被定义为在恒温恒压的封闭系统中可以获得的最大有用功。吉布斯自由能的变化量ΔG决定了在给定条件下反应是否自发（即无需外界输入即可发生）。其定义式为：

ΔG = ΔH − TΔS

Here, ΔH is the enthalpy change, T is the absolute temperature (in kelvin), and ΔS is the entropy change. A negative ΔG indicates a spontaneous process, a positive ΔG indicates a non-spontaneous process, and ΔG = 0 indicates equilibrium.

其中ΔH为焓变，T为绝对温度（单位为开尔文），ΔS为熵变。ΔG为负值表示过程自发，为正值表示非自发，而ΔG = 0表示系统处于平衡状态。

2. Standard Gibbs Free Energy Change (ΔG°) | 标准吉布斯自由能变（ΔG°）

When all reactants and products are in their standard states (1 bar pressure for gases, 1 mol dm⁻³ for solutions, pure liquids and solids), the Gibbs free energy change is called the standard Gibbs free energy change, denoted ΔG°. For a reaction, ΔG° can be calculated from standard formation values:

当所有反应物和产物都处于标准状态时（气体为1 bar压强，溶液为1 mol dm⁻³浓度，纯液体和纯固体为各自标准相态），吉布斯自由能变化称为标准吉布斯自由能变，记作ΔG°。对于任一反应，可由标准生成吉布斯自由能数据计算：

ΔG° = Σ ΔG°_f(products) − Σ ΔG°_f(reactants)

ΔG° is a constant for a given reaction at a specific temperature, and it reflects the position of equilibrium under standard conditions.

ΔG°在给定温度下对特定反应是一个常数，它反映了标准状态下反应的平衡倾向。

3. The Equilibrium Constant K | 平衡常数K

The equilibrium constant, K, describes the ratio of product concentrations (or partial pressures) to reactant concentrations (or partial pressures) at equilibrium, each raised to the power of their stoichiometric coefficients. For a general reaction aA + bB ⇌ cC + dD:

平衡常数K描述了平衡时产物浓度（或分压）与反应物浓度（或分压）之比，且每项均以其化学计量系数为指数。对于一般反应 aA + bB ⇌ cC + dD：

K = [C]^c[D]^d / ([A]^a[B]^b)

K is dimensionless in IB Chemistry (activities are used in precise thermodynamics), and its value depends only on temperature, not on initial concentrations or reaction rates. For gases, K_p is used with partial pressures; for solutions, K_c is used with molar concentrations.

在IB化学中，K通常无量纲（精确热力学中使用活度），其值只取决于温度，与初始浓度或反应速率无关。对于气体，使用以分压表达的K_p；对于溶液，使用以物质的量浓度表达的K_c。

4. The Fundamental Equation: ΔG° = −RT ln K | 核心方程：ΔG° = −RT ln K

The bridge between thermodynamics and equilibrium is the equation:

连接热力学与平衡的桥梁是如下方程：

ΔG° = −RT ln K

where R is the universal gas constant (8.314 J mol⁻¹ K⁻¹), T is the absolute temperature, and ln K is the natural logarithm of the equilibrium constant. This equation is derived from the more general relation between ΔG and the reaction quotient Q.

其中R为普适气体常数（8.314 J mol⁻¹ K⁻¹），T为绝对温度，ln K为平衡常数的自然对数。该方程由更一般的ΔG与反应商Q的关系推导而来。

5. Derivation from ΔG = ΔG° + RT ln Q | 由ΔG = ΔG° + RT ln Q推导

At any point during a reaction, the instantaneous Gibbs free energy change is given by:

在反应进行到任意时刻时，瞬时吉布斯自由能变化由下式给出：

ΔG = ΔG° + RT ln Q

where Q is the reaction quotient, which has the same form as K but uses current concentrations or pressures, not equilibrium values. At equilibrium, the reaction reaches a state where no net change occurs, so ΔG = 0 and Q = K. Substituting these conditions gives:

其中Q为反应商，其表达式形式与K相同，但使用的是当前浓度或压强，而非平衡值。当反应达到平衡时，系统不再发生净变化，因此ΔG = 0且Q = K。代入这些条件可得：

0 = ΔG° + RT ln K → ΔG° = −RT ln K

This derivation shows that the standard free energy change is simply the free energy change when all species are converted from standard state to equilibrium. It also explains why K is a thermodynamic quantity that depends only on temperature for a given reaction.

这一推导表明，标准自由能变化实质上就是体系由标准状态到达平衡状态过程中的自由能变化。同时，它也解释了为什么K是只依赖于温度的纯热力学量。

6. Interpreting ΔG° and K Values | 解读ΔG°与K值

The sign and magnitude of ΔG° directly indicate the position of equilibrium:

ΔG°的符号和大小直接指示平衡的位置：

If ΔG° < 0, then ln K > 0, so K > 1. The equilibrium lies to the right (products are favored), and the reaction is exergonic under standard conditions.

若ΔG° < 0，则ln K > 0，即K > 1。平衡偏向右侧（有利于产物生成），反应在标准状态下为放能过程。
If ΔG° = 0, then ln K = 0, so K = 1. Reactants and products are equally favored at equilibrium.

若ΔG° = 0，则ln K = 0，即K = 1。平衡时反应物与产物比例相当。
If ΔG° > 0, then ln K < 0, so K < 1. The equilibrium lies to the left (reactants are favored), and the reaction is endergonic under standard conditions.

若ΔG° > 0，则ln K < 0，即K < 1。平衡偏向左侧（有利于反应物），反应在标准状态下为吸能过程。

The table below summarizes these relationships for a typical reaction at 298 K:

下表总结了298 K下典型反应中ΔG°与K的关系：

ΔG° (kJ mol⁻¹)	ln K	K	Equilibrium position
−20	8.07	~3.2 × 10³	Strongly favors products
−5	2.02	~7.5	Favors products
0	0	1	Equal
+5	−2.02	~0.13	Favors reactants
+20	−8.07	~3.1 × 10⁻⁴	Strongly favors reactants

7. Temperature Dependence: The van’t Hoff Equation | 温度依赖性：范特霍夫方程

Since ΔG° = ΔH° − TΔS° and ΔG° = −RT ln K, we can combine the two expressions:

由于 ΔG° = ΔH° − TΔS°，且 ΔG° = −RT ln K，将两个表达式合并可得：

−RT ln K = ΔH° − TΔS°

Dividing both sides by −RT and rearranging gives the van’t Hoff equation:

两边同除以−RT并整理，即得范特霍夫方程：

ln K = −ΔH°/(RT) + ΔS°/R

This equation shows that a plot of ln K versus 1/T is a straight line with slope −ΔH°/R and intercept ΔS°/R. It allows us to determine the standard enthalpy and entropy changes from equilibrium constants measured at different temperatures. If ΔH° > 0 (endothermic), increasing temperature increases K; if ΔH° < 0 (exothermic), increasing temperature decreases K, consistent with Le Chatelier's principle.

该方程表明，以ln K对1/T作图是一条直线，斜率为−ΔH°/R，截距为ΔS°/R。我们即可通过不同温度下测得的平衡常数来确定标准焓变和标准熵变。若ΔH° > 0（吸热），升温会使K增大；若ΔH° < 0（放热），升温会使K减小，这与勒夏特列原理完全一致。

8. Using ΔG and Q to Predict Reaction Direction | 利用ΔG与Q预测反应方向

For non-standard conditions, the equation ΔG = ΔG° + RT ln Q allows us to decide which way a reaction will proceed. By comparing Q and K, we can predict the direction without calculation:

对于非标准条件，方程ΔG = ΔG° + RT ln Q可用来判断反应进行的方向。通过比较Q与K，无需计算即可预测反应方向：

If Q < K, then ln(Q/K) < 0, so ΔG < 0. The forward reaction is spontaneous; the system moves toward products.

若Q < K，则ln(Q/K) < 0，即ΔG < 0。正反应自发进行，系统向产物方向移动。
If Q > K, then ln(Q/K) > 0, so ΔG > 0. The reverse reaction is spontaneous; the system moves toward reactants.

若Q > K，则ln(Q/K) > 0，即ΔG > 0。逆反应自发进行，系统向反应物方向移动。
If Q = K, then ΔG = 0; the system is at equilibrium.

若Q = K，则ΔG = 0；系统处于平衡状态。

This is a powerful tool for solving IB problems that ask you to determine whether a reaction mixture is at equilibrium and, if not, which direction it will shift.

这是解决IB化学问题的一个强大工具，尤其当题目要求判断某反应混合物是否已达平衡，以及若未达平衡将向哪个方向移动时。

9. Worked Example: Calculating K from ΔG° | 计算实例：由ΔG°求K

Consider the reaction N₂O₄(g) ⇌ 2NO₂(g) at 298 K, with ΔG° = +4.8 kJ mol⁻¹. To calculate K:

考虑反应 N₂O₄(g) ⇌ 2NO₂(g) 在298 K下，ΔG° = +4.8 kJ mol⁻¹。计算K：

Step 1: Convert ΔG° to J mol⁻¹: ΔG° = +4800 J mol⁻¹.

步骤1：将ΔG°换算为 J mol⁻¹：ΔG° = +4800 J mol⁻¹。

Step 2: Use ΔG° = −RT ln K:

步骤2：使用ΔG° = −RT ln K：

4800 = −(8.314)(298) ln K → ln K = −4800 / 2477.6 ≈ −1.94

Step 3: Take antilog: K = e⁻¹·⁹⁴ ≈ 0.14.

步骤3：取反对数：K = e⁻¹·⁹⁴ ≈ 0.14。

The positive ΔG° corresponds to K < 1, meaning reactants are favored, which agrees with our earlier discussion. Always remember to use consistent units (J for ΔG°, not kJ) when plugging into R = 8.314 J mol⁻¹ K⁻¹.

正的ΔG°对应K < 1，即反应物更稳定，这与前文讨论一致。注意代入R = 8.314 J mol⁻¹ K⁻¹时，ΔG°必须使用焦耳（J）而非千焦（kJ）。

10. Common Misconceptions and Exam Tips | 常见误区与考试要点

Students often confuse ΔG and ΔG°. ΔG° refers to standard conditions and is a constant for a given reaction at a fixed temperature; it does not change with concentrations. In contrast, ΔG is the actual free energy change at any instant and depends on the reaction quotient Q. At equilibrium, ΔG = 0, but ΔG° is not necessarily zero unless K = 1.

学生经常混淆ΔG与ΔG°。ΔG°指标准状态下的吉布斯自由能变，对给定温度和给定反应是一个常数，不随浓度改变。而ΔG是任意时刻的实际自由能变，取决于反应商Q。在平衡时ΔG = 0，但ΔG°并非必然为零，除非K = 1。

Another common error is forgetting to convert temperature to kelvin or using kPa with R = 8.314. In IB exams, always check units. Also, when using the equation ΔG° = −RT ln K, make sure ΔG° and R have the same unit of energy. Some calculators require you to know the relationship between ln and log₁₀: ln K = 2.303 log₁₀ K, so ΔG° = −2.303 RT log₁₀ K.

另一个常见错误是忘记将温度换成开尔文，或使用kPa却将R取为8.314。在IB考试中务必检查单位。此外，使用ΔG° = −RT ln K时，要确保ΔG°与R的能量单位一致。有些同学需要用到ln与log的换算关系：ln K = 2.303 log₁₀ K，因此ΔG° = −2.303 RT log₁₀ K。

Exam tip: Sketching a graph of ΔG versus Q can help visualize the equilibrium point. The curve crosses zero exactly when Q = K. The slope of a van’t Hoff plot (ln K vs 1/T) is a favourite data-analysis question in Paper 2 and Paper 3.

考试提示：画出ΔG随Q变化的示意图有助于形象化理解平衡点。曲线在Q = K时恰好穿过零点。在Paper 2和Paper 3中，利用范特霍夫图（ln K对1/T作直线）求ΔH°和ΔS°是常见的数据分析题。

11. Coupled Reactions and Biological Relevance | 偶联反应与生物学意义

In living systems, many thermodynamically unfavourable reactions (ΔG° > 0) are driven by coupling them to thermodynamically favourable reactions, most commonly the hydrolysis of ATP. For example, the synthesis of glucose-6-phosphate from glucose and phosphate has ΔG° ≈ +13.8 kJ mol⁻¹, but when coupled with ATP hydrolysis (ΔG° ≈ −30.5 kJ mol⁻¹), the overall ΔG° becomes −16.7 kJ mol⁻¹, making the reaction spontaneous.

在生物体内，许多热力学上不利的反应（ΔG° > 0）通过与热力学上有利的反应偶联而进行，其中最常见的是ATP的水解。例如，由葡萄糖和磷酸合成葡萄糖-6-磷酸的ΔG°约为+13.8 kJ mol⁻¹，但与ATP水解（ΔG°约为−30.5 kJ mol⁻¹）偶联后，总ΔG°变为−16.7 kJ mol⁻¹，从而使反应得以自发进行。

This coupling principle is a direct application of the additivity of Gibbs free energy changes: ΔG°_overall = ΔG°₁ + ΔG°₂. Since the conversion between ΔG° and K is logarithmic, even a small change in ΔG° can lead to a large change in K. A difference of 10 kJ mol⁻¹ at 298 K changes K by a factor of about e⁴⁰ ≈ 2.4 × 10¹⁷, which explains how biological systems can switch metabolic pathways on or off with fine thermodynamic control.

这一偶联原理正是吉布斯自由能变可加性的直接应用：ΔG°_总 = ΔG°₁ + ΔG°₂。由于ΔG°与K之间是对数关系，ΔG°即使微小变化也会导致K发生显著变化。在298 K下，ΔG°每相差10 kJ mol⁻¹，K约改变e⁴⁰ ≈ 2.4 × 10¹⁷倍。这也解释了生物体如何通过精细的热力学控制来开启或关闭代谢通路。

12. Summary and Key Equations | 总结与核心公式

Mastering the link between Gibbs free energy and equilibrium proceeds from three key equations:

掌握吉布斯自由能与平衡之间的联系，关键在于以下三个方程：

ΔG = ΔH − TΔS — determines spontaneity of any process at constant temperature and pressure.

ΔG = ΔH − TΔS — 判断恒温恒压过程中是否自发。
ΔG = ΔG° + RT ln Q — determines the actual free energy change under non-standard conditions.

ΔG = ΔG° + RT ln Q — 计算非标准状态下的实际自由能变。
ΔG° = −RT ln K — links the standard free energy change to the equilibrium constant.

ΔG° = −RT ln K — 将标准自由能变与平衡常数联系起来。

By combining these, you can calculate K from thermodynamic data, predict the direction of a reaction from Q, and analyse how temperature affects equilibrium via the van’t Hoff equation. For IB Chemistry, be confident in unit conversion, sign conventions, and the distinction between ΔG and ΔG°. Practice with past paper questions involving ΔG°, K, and van’t Hoff plots to solidify your understanding.

综合运用这些公式，你可以由热力学数据计算K，根据Q判断反应方向，并借助范特霍夫方程分析温度对平衡的影响。对于IB化学，请熟练掌握单位换算、符号规则，并清楚区分ΔG与ΔG°。通过练习涉及ΔG°、K和范特霍夫图的历年真题来巩固理解。

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IB Chemistry: The Relationship between Gibbs Free Energy and Equilibrium Constant | IB化学：吉布斯自由能与平衡常数的关联