Gibbs Free Energy Calculations | 吉布斯自由能计算

📚 Gibbs Free Energy Calculations | 吉布斯自由能计算

Gibbs free energy, G, is a thermodynamic potential that combines enthalpy and entropy to determine the spontaneity of a reaction at constant temperature and pressure. For A-Level Chemistry students, mastering Gibbs free energy calculations is essential for predicting reaction feasibility and linking energetics to equilibrium.

吉布斯自由能（G）是结合了焓与熵的热力学函数，用于判断恒温恒压条件下反应的自发性。对 A-Level 化学学生而言，掌握吉布斯自由能计算是预测反应可行性并将能量学与平衡联系起来的关键。

1. The Fundamental Equation ΔG = ΔH – TΔS | 基本方程 ΔG = ΔH – TΔS

The change in Gibbs free energy for a reaction is given by the equation ΔG = ΔH – A negative ΔG indicates that the forward reaction is feasible under the given conditions.

反应的吉布斯自由能变由方程 ΔG = ΔH – TΔS 给出。其中 ΔH 是焓变，ΔS 是熵变，T 是热力学温度（开尔文，K）。ΔG 为负值表示该反应在给定条件下正向可行。

It is crucial to note that ΔH is usually expressed in kJ mol⁻¹, while ΔS is commonly given in J K⁻¹ mol⁻¹. The temperature term TΔS must therefore be converted to kJ mol⁻¹ before combining with ΔH. Many calculation errors arise from unit mismatches.

必须注意，ΔH 通常以 kJ mol⁻¹ 表示，而 ΔS 常以 J K⁻¹ mol⁻¹ 给出。因此 TΔS 项在与 ΔH 相加之前必须转换为 kJ mol⁻¹。许多计算错误均源于单位不匹配。

ΔG = ΔH – TΔS

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

The standard Gibbs free energy change, ΔG°, refers to the value when all reactants and products are in their standard states: 100 kPa pressure for gases, 1 mol dm⁻³ concentration for solutions, and a specified temperature, usually 298 K. ΔG° can be calculated from standard enthalpies and entropies of formation or from tabulated data.

标准吉布斯自由能变 ΔG° 指的是所有反应物和产物均处于标准状态下的数值：气体压强 100 kPa，溶液浓度 1 mol dm⁻³，温度通常指定为 298 K。ΔG° 可由标准生成焓和标准熵的数据表计算得出。

ΔG° is particularly useful because it allows chemists to predict the position of equilibrium and to compare the spontaneity of different reactions under the same reference conditions.

ΔG° 特别有用，因为它使化学家能够预测平衡位置，并在相同的参考条件下比较不同反应的自发性。

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3. Units and Conversions in ΔG Calculations | ΔG 计算中的单位与换算

When performing Gibbs free energy calculations, it is vital to ensure that the units of ΔH and TΔS are consistent. Typically, ΔH and ΔG are recorded in kJ mol⁻¹, whereas ΔS values are given in J K⁻¹ mol⁻¹. To combine them, convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000, or multiply ΔH by 1000 to work entirely in J mol⁻¹.

进行吉布斯自由能计算时，确保 ΔH 与 TΔS 的单位一致至关重要。通常 ΔH 和 ΔG 用 kJ mol⁻¹ 记录，而 ΔS 值以 J K⁻¹ mol⁻¹ 给出。合并时需将 ΔS 除以 1000 转换为 kJ K⁻¹ mol⁻¹，或将 ΔH 乘以 1000 全部采用 J mol⁻¹ 进行运算。

n J mol⁻¹ first, then convert to kJ mol⁻¹ before subtracting from ΔH. For example, if ΔH = –92 kJ mol⁻¹, ΔS = –199 J K⁻¹ mol⁻¹, and T = 298 K, then TΔS = 298 × (–199) = –59302 J mol⁻¹ = –59.3 kJ mol⁻¹. Then ΔG = –92 – (–59.3) = –32.7 kJ mol⁻¹.

考试中常见的方法是先以 J mol⁻¹ 为单位计算 TΔS，再转换为 kJ mol⁻¹ 与 ΔH 相减。例如，ΔH = –92 kJ mol⁻¹，ΔS = –199 J K⁻¹ mol⁻¹，T = 298 K，则 TΔS = 298 × (–199) = –59302 J mol⁻¹ = –59.3 kJ mol⁻¹，随后 ΔG = –92 – (–59.3) = –32.7 kJ mol⁻¹。

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4. Calculating ΔG from ΔH and ΔS Values | 由 ΔH 与 ΔS 计算 ΔG

ΔS: 160 J K⁻¹ mol⁻¹ = 0.160 kJ K⁻¹ mol⁻¹. Then TΔS = 298 × 0.160 = 47.7 kJ mol⁻¹. ΔG = 178 – 47.7 = +130.3 kJ mol⁻¹. Since ΔG is positive, the reaction is not feasible at 298 K.

以碳酸钙分解为例：CaCO₃(s) → CaO(s) + CO₂(g)。给定 ΔH = +178 kJ mol⁻¹，ΔS = +160 J K⁻¹ mol⁻¹，温度为 298 K。首先转换 ΔS：160 J K⁻¹ mol⁻¹ = 0.160 kJ K⁻¹ mol⁻¹。然后 TΔS = 298 × 0.160 = 47.7 kJ mol⁻¹。ΔG = 178 – 47.7 = +130.3 kJ mol⁻¹。由于 ΔG 为正，该反应在 298 K 下不可行。

0 = 1112.5 K. Above this temperature, ΔG becomes negative and the reaction is spontaneous.

要找到反应刚好变得可行的温度（即 ΔG = 0），令 ΔH = TΔS，得 T = ΔH / ΔS = 178 / 0.160 = 1112.5 K。高于此温度，ΔG 变为负值，反应自发进行。

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5. Effect of Temperature on Reaction Feasibility | 温度对反应可行性的影响

The sign of ΔH and ΔS determines how temperature affects spontaneity. Four cases are commonly tested: (i) ΔH negative, ΔS positive – reaction is feasible at all temperatures. (ii) ΔH positive, ΔS negative – reaction is never feasible. (iii) ΔH negative, ΔS negative – feasible only at low temperatures. (iv) ΔH positive, ΔS positive – feasible only at high temperatures.

ΔH 和 ΔS 的符号决定了温度如何影响自发性。常考四种情形：（一）ΔH 为负，ΔS 为正——所有温度下均可行；（二）ΔH 为正，ΔS 为负——任何温度下均不可行；（三）ΔH 为负，ΔS 为负——仅在低温下可行；（四）ΔH 为正，ΔS 为正——仅在高温下可行。

een feasible and non-feasible conditions. For case (iii), temperatures below this threshold yield a negative ΔG; for case (iv), temperatures above it are required.

转变温度 T = ΔH / ΔS（此时 ΔG = 0）是可行与不可行的分界线。对于情形（三），低于该阈值的温度产生负 ΔG；对于情形（四），则需要高于该阈值的温度。

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6. Relationship Between ΔG and Equilibrium Constant K | ΔG 与平衡常数 K 的关系

ndard Gibbs free energy change is linked to the equilibrium constant by the equation ΔG° = – m constant. This equation allows us to calculate K from thermodynamic data and vice versa.

标准吉布斯自由能变与平衡常数通过方程 ΔG° = –RT ln K 相联系。R 是摩尔气体常数（8.31 J K⁻¹ mol⁻¹），T 为热力学温度，K 为平衡常数。该方程使我们能够从热力学数据计算 K，反之亦然。

If ΔG° is very negative, ln K is large and positive, meaning K >> 1, so the equilibrium lies well to the right. If ΔG° is positive, K is small (< 1), indicating that the forward reaction is not favoured under standard conditions, though the reaction may still proceed under non-standard conditions.

若 ΔG° 非常负，则 ln K 为较大的正值，意味着 K >> 1，平衡强烈偏向右侧。若 ΔG° 为正，则 K 很小（< 1），表明在标准条件下正向反应不被看好，但在非标准条件下反应仍可能进行。

ΔG° = –RT ln K

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7. Using ΔG° to Calculate Equilibrium Constants | 利用 ΔG° 计算平衡常数

e ΔG° is expressed in J mol⁻¹ (multiply kJ by 1000). Rearranging the equation gives ln K = –ΔG° / (RT). Then K = e^{–ΔG° / RT}. For example, if ΔG° = –10 kJ mol⁻¹ at 298 K, then ΔG° = –10000 J mol⁻¹, RT = 8.31 × 298 ≈ 2476 J mol⁻¹, so ln K = –(–10000)/2476 ≈ 4.04, K ≈ e⁴ 7. This large K confirms the reaction goes almost to completion under standard conditions.

要从 ΔG° 计算 K，首先要确保 ΔG° 以 J mol⁻¹ 表示（将 kJ 乘以 1000）。整理方程得 ln K = –ΔG° / (RT)，然后 K = e^{–ΔG° / RT}。例如，若 ΔG° = –10 kJ mol⁻¹，T = 298 K，则 ΔG° = –10000 J mol⁻¹，RT = 8.31 × 298 ≈ 2476 J mol⁻¹，故 ln K = –(–10000)/2476 ≈ 4.04，K ≈ e⁴.04 ≈ 57。如此大的 K 表明在此标准条件下反应几乎进行完全。

Conversely, if K is known, ΔG° can be determined using the same formula. This is often applied in exam questions where K is obtained from experimental data or given directly.

反之，若已知 K，也可用同一公式求 ΔG°。考试中常以实验数据或直接给出 K 的形式命题。

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8. Two-Step Calculations: From ΔH and ΔS to K | 两步计算：由 ΔH 和 ΔS 求 K

n asks for the equilibrium constant at a particular temperature. First, calculate ΔG° = ΔH – TΔS (converting units as necessary). Next, substitute ΔG° into ΔG° = – . Remember that ΔH and ΔS are often assumed to be approximately constant with temperature.

典型的考试题目会给出 ΔH 和 ΔS 值，然后要求在某一温度下计算平衡常数。首先计算 ΔG° = ΔH – TΔS（必要时转换单位）。然后将 ΔG° 代入 ΔG° = –RT ln K 求解 K。需注意通常假设 ΔH 和 ΔS 在该温度范围内近似不变。

For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), given ΔH = –197 kJ mol⁻¹ and ΔS = –188 J K⁻¹ mol⁻¹ at 298 K. ΔS in kJ = –0.188 kJ K⁻¹ mol⁻¹. ΔG° = –197 – (298 × –0.188) = –197 + 56.0 = –141 kJ mol⁻¹. In J mol⁻¹, ΔG° = –141000 J mol⁻¹. Then ln K = –(–141000) / (8.31 × 298) ≈ 141000 / 2476 ≈ 56.9. K ≈ e⁵⁶.9, an extremely large number, indicating the equilibrium strongly favours products.

对于反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，给定 ΔH = –197 kJ mol⁻¹，ΔS = –188 J K⁻¹ mol⁻¹，T = 298 K。ΔS 转换为 kJ 为 –0.188 kJ K⁻¹ mol⁻¹。ΔG° = –197 – (298 × –0.188) = –197 + 56.0 = –141 kJ mol⁻¹。以 J mol⁻¹ 计，ΔG° = –141000 J mol⁻¹。则 ln K = –(–141000) / (8.31 × 298) ≈ 141000 / 2476 ≈ 56.9。K ≈ e⁵⁶.9，这是一个极其巨大的数值，表明平衡强烈偏向产物。

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9. Practical Examples with Detailed Steps | 带详细步骤的实际例题

Example: The Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At 298 K, ΔH° = –92 kJ mol⁻¹ and ΔS° = – at which the reaction ceases to be feasible. (c) Calculate K at 298 K. Solution: (a) ΔS° = –0.199 kJ K⁻¹ mol⁻¹; TΔS° = 298 × (–0.199) = –59.3 kJ mol⁻¹; ΔG° = –92 – (–59.3) = –32.7 kJ mol⁻¹. (b) Set ΔG = 0, T = ΔH°/ΔS° = –92 / (–0.199) ≈ 462 K. Above this temperature, ΔG becomes positive. (c) ΔG° = –32700 J mol⁻¹, RT = 2476 J mol⁻¹, ln K = 32700 / 2476 ≈ 13.2, K ≈ 5.4×10⁵.

例题：哈伯法合成氨 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。在 298 K 下，ΔH° = –92 kJ mol⁻¹，ΔS° = –199 J K⁻¹ mol⁻¹。(a) 计算 298 K 时的 ΔG°；(b) 求反应不再可行的温度；(c) 计算 298 K 时的 K。解答：(a) ΔS° = –0.199 kJ K⁻¹ mol⁻¹；TΔS° = 298 × (–0.199) = –59.3 kJ mol⁻¹；ΔG° = –92 – (–59.3) = –32.7 kJ mol⁻¹。(b) 令 ΔG = 0，T = ΔH°/ΔS° = –92 / (–0.199) ≈ 462 K。高于此温度，ΔG 变为正值。(c) ΔG° = –32700 J mol⁻¹，RT = 2476 J mol⁻¹，ln K = 32700 / 2476 ≈ 13.2，K ≈ 5.4×10⁵。

This example highlights the importance of unit conversion and the realistic interpretation of K values. Note that although the Haber process is feasible at low temperatures, a higher temperature is used industrially to increase rate, despite a lower equilibrium yield.

此例突显了单位换算的重要性以及对 K 值的实际解读。注意，虽然哈伯法在低温下可行，但工业上使用较高温度以提高反应速率，尽管这会降低平衡产率。

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10. Common Pitfalls and Exam Tips | 常见错误与考试技巧

One of the most frequent mistakes is forgetting to convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ before adding to ΔH. Always check the units and clearly write the conversion step. Also, never use Celsius temperatures; T must be in kelvin.

最常见的错误之一是在与 ΔH 相加之前忘记将 ΔS 从 J K⁻¹ mol⁻¹ 转换为 kJ K⁻¹ mol⁻¹。请始终检查单位并清晰写出换算步骤。同时，切勿使用摄氏温度，T 必须用开尔文。

When using ΔG° = – , remember that ln K is dimensionless; if you get a K with a unit, you have made an error.

使用 ΔG° = –RT ln K 时，确保 R 取 8.31 J K⁻¹ mol⁻¹，且 ΔG° 以 J mol⁻¹ 表示。许多考生混淆 kJ 与 J，得到错误的 K 值。另外，记住 ln K 是无量纲的；若算出的 K 带单位，说明有误。

In questions where both ΔG° and K are asked, you can cross-check: a negative ΔG° should correspond to K > 1, and a positive ΔG° to K < 1. This quick verification can save many marks.

在同时要求计算 ΔG° 和 K 的题目中，可以互相印证：负的 ΔG° 应对应 K > 1，正的 ΔG° 应对应 K < 1。这一快速验证能帮你保住不少分数。

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