📚 Calculating Entropy Changes in Chemistry: Methods and Examples | 化学熵变的计算方法与实例
Entropy (S) is a measure of the disorder or randomness of a system. In A-Level chemistry, calculating entropy changes helps us predict whether a reaction is spontaneous and understand energy dispersal at the molecular level.
熵（S）是体系无序度或随机性的度量。在A-Level化学中，计算熵变有助于我们预测反应能否自发进行，并从分子层面理解能量的分散方式。
This article explains the core methods for calculating entropy changes, provides worked examples aligned with the CIE syllabus, and highlights common pitfalls to avoid in exams.
本文将解释熵变计算的核心方法，提供与CIE考纲同步的实例，并指出考试中常见的易错点。
1. What Is Entropy and Entropy Change? | 熵与熵变的基本概念
Entropy is a state function that quantifies the number of microscopic arrangements available to a system. A higher entropy value means greater disorder, such as when a solid melts or a gas expands.
熵是状态函数，用于量化体系可用的微观排列数。熵值越高意味着无序度越大，例如固体熔化或气体膨胀时熵增加。
The entropy change of a process, ΔS, is defined as the difference between the final and initial entropy of the system:
过程的熵变 ΔS 定义为系统终态熵与始态熵之差：
ΔS = Sfinal − Sinitial
If ΔS is positive, the system becomes more disordered; if negative, it becomes more ordered.
若 ΔS 为正值，体系变得更加无序；若为负值，则变得更加有序。
2. Standard Molar Entropy and Standard Entropy Change | 标准摩尔熵与标准熵变
Standard molar entropy (S°m) is the entropy of one mole of a pure substance at 298 K and 1 bar pressure. Units are J K⁻¹ mol⁻¹. Unlike enthalpy of formation, the entropy of an element in its standard state is not zero.
标准摩尔熵（S°m）是在298 K和1 bar压力下，1摩尔纯物质的熵，单位为 J K⁻¹ mol⁻¹。与标准生成焓不同，元素在其标准状态下的熵并不为零。
For a chemical reaction, the standard entropy change (ΔS°rxn) is calculated from the standard molar entropies of reactants and products:
对于化学反应，标准熵变（ΔS°rxn）由反应物和产物的标准摩尔熵计算：
ΔS°rxn = Σ S°m(products) − Σ S°m(reactants)
Remember to multiply each substance’s entropy by its stoichiometric coefficient in the balanced equation.
记住要用平衡方程中各物质的化学计量数乘以该物质的熵。
3. Worked Example: Combustion of Carbon | 实例计算：碳的燃烧
Calculate the standard entropy change for the reaction C(s) + O₂(g) → CO₂(g), given S°[C(s)] = 5.7 J K⁻¹ mol⁻¹, S°[O₂(g)] = 205.0 J K⁻¹ mol⁻¹, S°[CO₂(g)] = 213.7 J K⁻¹ mol⁻¹.
计算反应 C(s) + O₂(g) → CO₂(g) 的标准熵变，已知 S°[C(s)] = 5.7 J K⁻¹ mol⁻¹，S°[O₂(g)] = 205.0 J K⁻¹ mol⁻¹，S°[CO₂(g)] = 213.7 J K⁻¹ mol⁻¹。
Using the formula:
使用公式：
ΔS° = (1 × 213.7) − [(1 × 5.7) + (1 × 205.0)] = 213.7 − 210.7 = +3.0 J K⁻¹ mol⁻¹
The positive value is consistent with the fact that one gas molecule is produced from one solid and one gas molecule, but the change is small because the total number of gas molecules does not increase.
正值与事实一致：反应由一种固体和一种气体分子生成一种气体分子，但变化很小，因为气体分子总数没有增加。
4. Predicting the Sign of ΔS: Gas Moles Change | 预测ΔS的符号：气体摩尔数变化
For reactions involving gases, a reliable rule is: if the total number of moles of gaseous products is greater than that of gaseous reactants, ΔS is positive. If it decreases, ΔS is negative.
对于涉及气体的反应，一个可靠的规则是：如果气态产物的总摩尔数大于气态反应物的总摩尔数，则 ΔS 为正；如果减少，则 ΔS 为负。
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N₂(g) + 3H₂(g) → 2NH₃(g): gas moles decrease from 4 to 2, so ΔS is negative.
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N₂(g) + O₂(g) → 2NO(g): gas moles are equal (2 → 2), so ΔS is likely small.
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CaCO₃(s) → CaO(s) + CO₂(g): gas moles increase from 0 to 1, so ΔS is positive.
However, be careful: entropy also depends on the complexity of the molecules. For example, 2NO₂(g) → N₂O₄(g) has fewer gas moles (2 → 1), so ΔS is negative.
但需注意：熵还取决于分子的复杂性。例如，2NO₂(g) → N₂O₄(g) 气体摩尔数减少（2 → 1），所以 ΔS 为负。
5. Phase Change Entropy: ΔS = ΔH/T | 相变熵变：ΔS = ΔH/T
At a phase transition (melting, boiling, sublimation), the system is at equilibrium and the entropy change can be calculated from the enthalpy change and the transition temperature:
在相变（熔化、沸腾、升华）时，系统处于平衡态，熵变可由焓变和转变温度计算：
ΔStrans = ΔHtrans / Ttrans
Here, T must be in kelvin (K), and ΔH is the enthalpy change for the phase transition. For melting, use ΔHfus; for vaporisation, use ΔHvap.
此处 T 必须用开尔文（K），ΔH 是相变焓变。熔化用 ΔHfus，汽化用 ΔHvap。
Since melting and evaporation are endothermic processes that increase disorder, ΔH and T are positive, so ΔS is always positive for these transitions.
由于熔化和蒸发是吸热过程且增加无序度，ΔH 和 T 均为正值，因此这些相变的 ΔS 总是正值。
6. Worked Example: Melting of Ice | 实例计算：冰的熔化
Calculate the entropy change when 1.00 mol of ice melts at 0 °C. The enthalpy of fusion of water is 6.01 kJ mol⁻¹.
计算1.00 mol冰在0 °C熔化时的熵变。水的熔化焓为6.01 kJ mol⁻¹。
First convert temperature to kelvin: T = 0 + 273.15 = 273.15 K. Convert ΔH to J mol⁻¹: ΔHfus = 6010 J mol⁻¹.
首先将温度转换为开尔文：T = 0 + 273.15 = 273.15 K。将ΔH转换为 J mol⁻¹：ΔHfus = 6010 J mol⁻¹。
ΔSfus = 6010 J mol⁻¹ / 273.15 K = +22.0 J K⁻¹ mol⁻¹
Only a small entropy increase is observed because the ice and liquid water have similar intermolecular order.
观察到的熵增较小，因为冰和液态水的分子间有序度相似。
7. Worked Example: Vaporisation of Water | 实例计算：水的汽化
Estimate the entropy change for the vaporisation of water at 100 °C, given ΔHvap = 40.7 kJ mol⁻¹.
估算水在100 °C汽化的熵变，已知ΔHvap = 40.7 kJ mol⁻¹。
Convert values: T = 373.15 K, ΔHvap = 40700 J mol⁻¹.
转换数值：T = 373.15 K，ΔHvap = 40700 J mol⁻¹。
ΔSvap = 40700 / 373.15 = +109 J K⁻¹ mol⁻¹
The large positive value reflects the huge increase in disorder when liquid water becomes gaseous water vapor.
较大的正值反映了液态水变成气态水蒸气时无序度的大幅增加。
8. Complex Reaction Example: Thermal Decomposition of CaCO₃ | 复杂反应实例：CaCO₃的热分解
The decomposition of calcium carbonate is an important industrial reaction: CaCO₃(s) → CaO(s) + CO₂(g). Calculate ΔS° at 298 K using the following data:
碳酸钙的分解是重要的工业反应：CaCO₃(s) → CaO(s) + CO₂(g)。使用以下数据计算298 K下的ΔS°：
| Substance | S° / J K⁻¹ mol⁻¹ |
| CaCO₃(s) | 92.9 |
| CaO(s) | 39.8 |
| CO₂(g) | 213.7 |
Apply the standard entropy change formula:
应用标准熵变公式：
ΔS° = (39.8 + 213.7) − 92.9 = 253.5 − 92.9 = +160.6 J K⁻¹ mol⁻¹
The large positive entropy change arises because one mole of solid decomposes to produce one mole of gas, which has much higher entropy. This explains why the decomposition becomes spontaneous at high temperatures despite being endothermic.
较大的正熵变源于一摩尔固体分解产生一摩尔气体，气体的熵远高于固体。这解释了为什么该分解反应尽管吸热，但在高温下能自发进行。
9. Entropy Change and Spontaneity: Gibbs Free Energy | 熵变与自发性：吉布斯自由能
Entropy change alone does not determine whether a reaction is spontaneous; the total entropy of the universe must increase. For a process at constant temperature and pressure, the Gibbs free energy change combines enthalpy and entropy:
仅凭熵变不能决定反应是否自发；宇宙总熵必须增加。在恒温恒压下，吉布斯自由能变化结合了焓和熵：
ΔG = ΔH − TΔS
If ΔG < 0, the reaction is spontaneous. When ΔH is positive and ΔS is positive, increasing temperature makes TΔS larger, so ΔG may become negative at high temperatures — as seen in CaCO₃ decomposition.
若 ΔG < 0，则反应自发。当 ΔH 为正且 ΔS 为正时，升高温度使 TΔS 增大，因此 ΔG 在高温下可能变为负值——正如 CaCO₃ 的分解。
This relationship is essential in CIE A-Level questions, where you are asked to compare the sign of ΔS and explain the temperature dependence of spontaneity.
这一关系在CIE A-Level试题中至关重要，常要求你比较ΔS的符号并解释自发性对温度的依赖。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Here are frequent pitfalls when calculating entropy changes:
以下是计算熵变时常见的陷阱：
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Forgetting to multiply standard molar entropies by stoichiometric coefficients. Always balance the equation first.
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Using units incorrectly: standard entropies are usually given in J K⁻¹ mol⁻¹, while enthalpies are in kJ mol⁻¹. Convert to joules before calculating ΔS.
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Neglecting state symbols: the same substance in different states has different entropy (e.g., H₂O(g) vs H₂O(l)).
Also, remember that ΔS for a phase change uses the equilibrium temperature (melting point or boiling point) in kelvin. Do not use 298 K unless it is the transition temperature.
另外，相变熵变使用平衡温度（熔点或沸点）的开尔文值。除非转变温度就是298 K，否则不要用298 K。
In multiple-choice questions, quickly predict the sign of ΔS by counting gaseous moles. In calculation questions, show every step and include units.
在选择题中，通过数气体摩尔数快速判断ΔS的符号。在计算题中，写出每一步并带单位。
11. Summary of Key Formulas | 关键公式总结
The table below summarises the essential equations for entropy change calculations in CIE A-Level chemistry.
下表总结了CIE A-Level化学中熵变计算的基本方程。
| Process | Formula | Notes |
| Reaction entropy | ΔS°rxn = ΣS°(products) − ΣS°(reactants) | Use balanced equation coefficients |
| Phase transition | ΔStrans = ΔHtrans / Ttrans | T in kelvin; ΔH in J mol⁻¹ |
| Gibbs free energy | ΔG = ΔH − TΔS | Spontaneous if ΔG < 0 |
Master these formulas and practise with past exam questions. The ability to compute and interpret entropy changes is a high-scoring skill in physical chemistry.
掌握这些公式并练习历年试题。计算和解释熵变的能力是物理化学中得分率高的技能。
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