📚 IB Chemistry: Entropy – Key Concepts and Exam Tips | IB 化学：熵 考点精讲
Entropy is one of the most conceptually demanding topics in IB Chemistry, yet it provides the crucial bridge between energetics and the direction of chemical change. This article unpacks every essential examination point—from the definition and standard molar entropy to calculations of ΔS° and the Gibbs free energy equation—helping you master spontaneity and equilibrium requirements for both SL and HL papers.
熵是 IB 化学中概念最具挑战性的主题之一，但它在能量变化与化学反应方向之间架起了关键的桥梁。本文逐一剖析所有必考的考点——从熵的定义和标准摩尔熵，到 ΔS° 的计算及吉布斯自由能方程——帮助你掌握自发过程和平衡判据，轻松应对 SL 与 HL 的考试要求。
1. What Is Entropy? | 什么是熵？
Entropy, symbol S, is a thermodynamic state function that measures the degree of disorder or randomness of a system. The more ways energy can be distributed among the particles in a system, the higher its entropy.
熵，符号 S，是一个热力学状态函数，用于衡量体系的混乱度或无序度。能量在体系微粒之间的分布方式越多，体系的熵值就越高。
At the microscopic level, entropy is related to the number of microstates, W, accessible to a system at a given energy: S = kB ln W, where kB is the Boltzmann constant. IB does not require quantitative use of this formula, but understanding the concept helps.
在微观层面，熵与体系在给定能量下可达到的微观状态数 W 相关：S = kB ln W，其中 kB 为玻尔兹曼常数。IB 课程不要求定量使用该公式，但理解这层含义有助于把握本质。
Entropy is measured in J K⁻¹ mol⁻¹. Unlike enthalpy, it is possible to determine absolute entropy values, thanks to the Third Law of Thermodynamics.
熵的单位是 J K⁻¹ mol⁻¹。与焓不同，借助热力学第三定律可以确定物质的绝对熵值。
2. The Second Law & Entropy Increase | 热力学第二定律与熵增
The Second Law of Thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases of reversible processes. For chemical and physical changes, the entropy change of the universe (ΔSuniverse) = ΔSsystem + ΔSsurroundings must be positive for a spontaneous process.
热力学第二定律指出，孤立体系的总熵总是随时间增加，在理想的可逆过程中则保持不变。对化学和物理变化而言，宇宙的熵变 ΔSuniverse = ΔSsystem + ΔSsurroundings 必须为正，过程才能自发进行。
If a cold pack dissolves ammonium nitrate and the solution becomes very cold, the system entropy increases as the solid dissociates into ions, and although the surroundings lose entropy, the overall increase in system entropy drives the process.
如果冷敷袋中硝酸铵溶解时溶液变得很冷，体系因固体解离为离子而使熵增大；尽管环境的熵减少，但体系熵增占主导，过程依然自发。
3. Standard Molar Entropy (S°) | 标准摩尔熵 (S°)
The standard molar entropy, S°, is the absolute entropy of one mole of a substance under standard conditions (100 kPa, 298 K). Values are tabulated for elements, compounds, and ions in J K⁻¹ mol⁻¹.
标准摩尔熵 S° 是指 1 mol 物质在标准条件（100 kPa、298 K）下的绝对熵，单位为 J K⁻¹ mol⁻¹。元素、化合物和离子都有相应的表列数据。
Key trends in S° include: gaseous substances have much larger S° than liquids, which in turn have larger S° than solids; more complex molecules (larger molar mass, more atoms) possess higher S°; and aqueous ions often have low or even negative S° owing to ordering of water molecules around them.
S° 的主要规律：气体的 S° 远大于液体，液体又大于固体；分子越复杂（摩尔质量越大、原子数越多），S° 越高；水合离子常因周围水分子有序排列而具有较低甚至负的 S° 值。
| Substance | State | S° (J K⁻¹ mol⁻¹) |
|---|---|---|
| C (diamond) | s | 2.4 |
| H₂O | l | 69.9 |
| H₂O | g | 188.8 |
| CO₂ | g | 213.7 |
| C₂H₆ | g | 229.6 |
4. Predicting Entropy Changes Qualitatively | 定性预测熵变
Before any calculation, you must be able to predict the sign of ΔS° for a reaction or process. Entropy increases when a substance melts or boils, when a solid dissolves forming ions, when the number of gaseous molecules increases, or when a complex molecule breaks into simpler ones.
在任何计算之前，你需要能够定性判断一个反应或过程的 ΔS° 符号。熔化、沸腾、固体溶解形成离子、气体分子数增加、复杂分子分解为简单分子等情况都会使熵增加。
For a reaction such as 2H₂O₂(l) → 2H₂O(l) + O₂(g), the system entropy increases dramatically because one mole of gas is produced from a liquid. Conversely, N₂(g) + 3H₂(g) → 2NH₃(g) involves a decrease in the number of gas molecules (4 → 2), so ΔS° is negative.
对于 2H₂O₂(l) → 2H₂O(l) + O₂(g) 这类反应，由于从液体生成了一摩尔气体，体系熵显著增大。反之，N₂(g) + 3H₂(g) → 2NH₃(g) 中气体分子数减少（4 → 2），因此 ΔS° 为负。
5. Entropy Changes in Physical Processes | 物理过程中的熵变
During melting and vaporisation, the entropy of a substance increases because the particles gain freedom of movement. The entropy change for a phase transition at constant temperature can be calculated from ΔS = ΔH / T, where ΔH is the enthalpy change of fusion or vaporisation and T is the transition temperature in kelvin.
熔化和汽化过程中，物质因粒子运动自由度增加而使熵增大。恒温相变的熵变可由 ΔS = ΔH / T 计算，其中 ΔH 为熔化或汽化的焓变，T 为相变温度的绝对温度（K）。
For water at 373 K, ΔHvap = 40.7 kJ mol⁻¹, so ΔSvap ≈ 40 700 J mol⁻¹ ÷ 373 K ≈ 109 J K⁻¹ mol⁻¹. This large positive value reflects the huge entropy increase when liquid water becomes steam.
在 373 K 下，水的 ΔHvap = 40.7 kJ mol⁻¹，因此 ΔSvap ≈ 40700 J mol⁻¹ ÷ 373 K ≈ 109 J K⁻¹ mol⁻¹。这一较大的正值反映了液态水变成水蒸气时巨大的熵增。
6. Calculating Entropy Change of a Reaction | 化学反应熵变的计算
The standard entropy change for a chemical reaction, ΔS°rxn, is determined using standard molar entropies of all reactants and products:
化学反应的标准化熵变 ΔS°rxn 利用所有反应物和产物的标准摩尔熵来计算：
ΔS°rxn = Σ S°(products) − Σ S°(reactants)
Stoichiometric coefficients must be multiplied by the corresponding S° values. For the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), you insert S° values: ΔS° = [S°(CO₂) + 2 × S°(H₂O,l)] − [S°(CH₄) + 2 × S°(O₂)] = [213.7 + 2 × 69.9] − [186.3 + 2 × 205.1] = (353.5) − (596.5) = −243.0 J K⁻¹ mol⁻¹. The negative sign is consistent with the reduction in moles of gas (3 mol gas → 1 mol gas).
化学计量系数必须与相应的 S° 值相乘。以甲烷的燃烧为例，CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)，代入 S° 值计算：ΔS° = [213.7 + 2×69.9] − [186.3 + 2×205.1] = 353.5 − 596.5 = −243.0 J K⁻¹ mol⁻¹，负号与气体摩尔数减少（3 mol 气体 → 1 mol 气体）相符。
Remember: many exam errors arise from forgetting to multiply S° of O₂ by its coefficient, or from using S° of H₂O(g) instead of H₂O(l). Always check the physical states given.
切记：许多考试中的失分都源于忘记将 O₂ 的 S° 乘以化学计量数，或错误地使用了 H₂O(g) 而非 H₂O(l) 的 S°。务必核对题目给出的物态。
7. Gibbs Free Energy Equation | 吉布斯自由能方程
The Gibbs free energy change, ΔG, combines enthalpy and entropy to determine reaction spontaneity at constant temperature and pressure:
吉布斯自由能变 ΔG 将焓和熵结合在一起，用于判断恒温恒压下反应的自发性：
ΔG° = ΔH° − TΔS°
Here T is the temperature in kelvin, ΔH° is in kJ mol⁻¹, and ΔS° must be converted to kJ K⁻¹ mol⁻¹ (÷1000). This unit conversion is a classic IB trap.
其中 T 为绝对温度（K），ΔH° 单位为 kJ mol⁻¹，而 ΔS° 需转换为 kJ K⁻¹ mol⁻¹（除以 1000）。这一单位换算是 IB 经典的失分陷阱。
A reaction is spontaneous (thermodynamically feasible) when ΔG < 0. If ΔG > 0, the reverse reaction is spontaneous. At equilibrium, ΔG = 0.
当 ΔG < 0 时，反应自发（热力学可行）；若 ΔG > 0，则逆反应自发；达到平衡时 ΔG = 0。
8. Spontaneity and Temperature Dependence | 自发性与温度依赖性
The sign of ΔG depends on the interplay between ΔH and ΔS. Four scenarios are possible:
ΔG 的符号取决于 ΔH 与 ΔS 之间的相互作用，共有四种可能的情形：
• ΔH < 0 and ΔS > 0: ΔG < 0 at all temperatures; always spontaneous (e.g., combustion).
• ΔH < 0 且 ΔS > 0：任意温度下 ΔG < 0，始终自发（如燃烧反应）。
• ΔH > 0 and ΔS < 0: ΔG > 0 at all temperatures; never spontaneous.
• ΔH > 0 且 ΔS < 0：任意温度下 ΔG > 0，永不自发。
• ΔH < 0 and ΔS < 0: ΔG negative only at low temperatures; reaction is spontaneous below a crossover temperature T = ΔH/ΔS.
• ΔH < 0 且 ΔS < 0：仅低温下 ΔG 为负；在转变温度 T = ΔH/ΔS 以下自发。
• ΔH > 0 and ΔS > 0: ΔG negative only at high temperatures; spontaneous above T = ΔH/ΔS.
• ΔH > 0 且 ΔS > 0：仅高温下 ΔG 为负；在 T = ΔH/ΔS 以上自发。
For example, the decomposition of calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), is endothermic (ΔH° = +178 kJ mol⁻¹) and has positive ΔS° (gas formed). It becomes spontaneous at around 1100 K.
例如，碳酸钙分解 CaCO₃(s) → CaO(s) + CO₂(g) 为吸热反应（ΔH° = +178 kJ mol⁻¹）且 ΔS° 为正（生成气体），它在大约 1100 K 时变得自发。
9. Entropy and Equilibrium | 熵与平衡
At equilibrium, the change in Gibbs free energy is zero, which links the equilibrium constant K to the standard free energy change: ΔG° = −RT ln K. Although the derivation is not required for SL, HL students must be able to perform calculations and interpret the relationship.
平衡时吉布斯自由能变为零，由此将平衡常数 K 与标准自由能变联系起来：ΔG° = −RT ln K。尽管 SL 不要求推导，HL 学生需要会计算并解释该关系。
A large negative ΔG° corresponds to K ≫ 1, meaning the equilibrium lies far to the right. A positive ΔG° gives K < 1, with the equilibrium favouring reactants. Entropy considerations are embedded in this framework: a reaction with a large positive ΔS° contributes to a more negative ΔG° and hence a larger K.
较负的 ΔG° 对应 K ≫ 1，平衡强烈偏向产物；ΔG° 为正则 K < 1，平衡偏向反应物。熵因素自然蕴含其中：较大的正 ΔS° 促使 ΔG° 更负，从而使 K 更大。
10. Worked Example & Exam Tips | 例题与考试技巧
Worked example: Determine the temperature at which the reaction 2SO₂(g) + O₂(g) → 2SO₃(g) becomes non-spontaneous given ΔH° = −196 kJ mol⁻¹ and ΔS° = −189 J K⁻¹ mol⁻¹.
例题：已知 2SO₂(g) + O₂(g) → 2SO₃(g) 的 ΔH° = −196 kJ mol⁻¹，ΔS° = −189 J K⁻¹ mol⁻¹，求反应变为非自发的温度。
Solution: Convert ΔS° to kJ K⁻¹ mol⁻¹: −189 J K⁻¹ mol⁻¹ = −0.189 kJ K⁻¹ mol⁻¹. At the crossover temperature, ΔG° = 0, so T = ΔH° / ΔS° = (−196) / (−0.189) ≈ 1037 K. For temperatures above 1037 K the negative TΔS° term outweighs the negative ΔH°, making ΔG° positive. Therefore, the reaction is spontaneous only below ~1040 K.
解答：将 ΔS° 转换成 kJ K⁻¹ mol⁻¹：−189 J K⁻¹ mol⁻¹ = −0.189 kJ K⁻¹ mol⁻¹。在转变温度处 ΔG° = 0，T = ΔH°/ΔS° = (−196) / (−0.189) ≈ 1037 K。温度高于 1037 K 时，负的 TΔS° 项压倒负的 ΔH°，使 ΔG° 为正。因此反应仅在约 1040 K 以下自发。
Top exam tips: Always convert ΔS° to kJ before inserting into ΔG° equation. Watch out for physical states—S°(H₂O,l) and S°(H₂O,g) are very different. Predict the sign of ΔS° even when data is not given: count gas moles. And recall that a reaction with ΔG° < 0 is not necessarily fast; kinetics is separate.
必考技巧：代入 ΔG° 方程前务必将 ΔS° 转换为 kJ。密切注意物态——S°(H₂O,l) 与 S°(H₂O,g) 差异很大。即使未给出数据，也要通过气体摩尔数预测 ΔS° 符号。还要记住，ΔG° < 0 的反应不一定迅速；动力学是另一回事。
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