📚 Calculating Entropy Changes: System, Surroundings and Feasibility | 计算熵变：系统、环境与反应可行性
Entropy (S) is a fundamental thermodynamic quantity that measures the dispersal of energy and the degree of disorder or randomness in a system. In A-Level Chemistry, calculating entropy changes is vital for understanding why some reactions occur spontaneously while others do not. The total entropy change of the universe (system + surroundings) determines the feasibility of a process, linking together the concepts of enthalpy, temperature and the second law of thermodynamics.
熵（S）是一个基本的热力学量，衡量能量的分散以及系统的无序度或混乱度。在 A-Level 化学中，计算熵变对于理解为什么有些反应能自发发生而另一些不能至关重要。宇宙（系统 + 环境）的总熵变决定了一个过程是否可行，它把焓、温度和热力学第二定律的概念联系在一起。
1. What Is Entropy? | 什么是熵
Entropy is often described as a measure of disorder, but a more precise definition is the number of ways in which the energy of a system can be distributed among its particles. The more ways the energy can be arranged, the greater the entropy. This is why a gas has higher entropy than a liquid, which in turn has higher entropy than a solid: the particles in a gas have far greater freedom of motion and far more accessible energy levels.
熵常被描述为无序度的量度，但更精确的定义是：系统能量在粒子之间分配方式的数量。能量可能的分配方式越多，熵就越大。这就是为什么气体的熵高于液体，液体的熵又高于固体：气体粒子的运动自由度大得多，可占据的能级也多得多。
The symbol for entropy is S, and its units are joules per kelvin per mole (J K⁻¹ mol⁻¹). Entropy is a state function, which means its value depends only on the current state of the system, not on the route taken to reach that state.
熵的符号是 S，单位是焦耳每开尔文每摩尔（J K⁻¹ mol⁻¹）。熵是状态函数，这意味着它的数值只取决于系统当前的状态，而与达到该状态所经历的路径无关。
Three everyday observations show entropy increasing:
日常生活中可体现熵增的三个现象：
- Ice melting at room temperature - the ordered lattice collapses into a disordered liquid.
- 冰在室温下融化——有序的晶格变为无序的液体。
- A drop of ink spreading through water - particles become randomly distributed.
- 一滴墨水滴入水中扩散——粒子随机分布开来。
- A gas expanding to fill a larger container - the number of available positions increases.
- 气体膨胀充满更大的容器——可占据的位置数量增加。
2. The Third Law and Absolute Entropy | 热力学第三定律与绝对熵
The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero. At 0 K there is only one possible arrangement of the particles, so there is only one microstate and the entropy is exactly zero. This gives chemists a fixed reference point from which absolute entropy values can be measured.
热力学第三定律指出：完美晶体在绝对零度（0 K）时的熵为零。在 0 K 时粒子只有一种可能的排列方式，因此只有一个微观状态，熵恰好为零。这为化学家提供了一个固定参考点，从而可以测定绝对熵值。
Absolute entropies are obtained by measuring how the heat capacity of a substance changes as it is warmed from 0 K to the temperature of interest, including the entropy jumps at phase changes such as melting and boiling. The result is the standard molar entropy, S°, tabulated at 298 K and 100 kPa.
绝对熵的获得方法是：测量物质从 0 K 加热到目标温度的过程中热容如何变化，并把熔化、沸腾等相变时的熵增也计算进去。所得结果就是标准摩尔熵 S°，通常在 298 K 和 100 kPa 下列表给出。
Two important consequences follow from the Third Law:
由第三定律可得出两个重要结论：
- All substances have positive standard molar entropies at 298 K, because S° is measured from zero.
- 在 298 K 时所有物质的标准摩尔熵都是正值，因为 S° 是从零开始测定的。
- Entropy values are never negative, unlike enthalpy of formation, which can be negative.
- 与可以为负的生成焓不同，熵值永远不会是负值。
3. Predicting the Sign of the System Entropy Change | 判断系统熵变的符号
Before doing any calculation, you should be able to predict whether the entropy change of the system, ΔSsystem, is positive or negative. This is a common exam skill and it also acts as a check on your arithmetic.
在做任何计算之前，你应该能够预测系统的熵变 ΔSsystem 是正还是负。这既是常见的考点，也可以用来检验你的计算结果。
| Change | 变化 | Sign of ΔSsystem | ΔSsystem 的符号 | Reason | 原因 |
| Solid → liquid → gas | 固 → 液 → 气 | Positive | 正 | Particles gain freedom | 粒子自由度增大 |
| Fewer moles of gas | 气体摩尔数减少 | Negative | 负 | Gas molecules are combined | 气体分子被合并 |
| More moles of gas | 气体摩尔数增加 | Positive | 正 | More gaseous particles | 气态粒子变多 |
| Dissolving a solid in water | 固体溶于水 | Usually positive | 通常为正 | Ordered lattice is broken | 有序晶格被破坏 |
| Complex ion formed from ions | 由离子形成配离子 | Negative | 负 | Fewer independent particles | 独立粒子数减少 |
A useful shortcut in gaseous reactions is to count the change in the number of gas moles, Δngas. When Δngas is positive, ΔSsystem is positive; when Δngas is negative, ΔSsystem is negative.
在气相反应中一个实用的快捷判断法是计算气体摩尔数的变化 Δngas。当 Δngas 为正时，ΔSsystem 为正；当 Δngas 为负时，ΔSsystem 为负。
4. Calculating the Entropy Change of the System | 计算系统熵变
The entropy change of the system is found by subtracting the total standard molar entropy of the reactants from that of the products, taking stoichiometric coefficients into account. Absolute rather than formation values are used, because entropies are already measured from zero.
系统熵变的求法是用产物的标准摩尔熵总和减去反应物的标准摩尔熵总和，并考虑化学计量数。这里使用绝对熵而不是生成熵，因为熵本身就是从零开始测定的。
ΔS°system = Σ S°(products) − Σ S°(reactants)
Two practical points matter when using this equation:
使用这个方程时有两个实用要点：
- Multiply each S° value by its coefficient in the balanced equation.
- 把每个 S° 值乘以它在配平方程中的系数。
- Omit solids and liquids stated as pure phases only if the question gives no data; in practice A-Level questions provide S° for every species, so include all of them.
- 只有题目未给出数据时才可忽略纯固态或纯液态；实际上 A-Level 题目会给出所有物质的数据，因此应全部代入。
The answer comes out in J K⁻¹ mol⁻¹, and it refers to the reaction as written, so it is quoted per mole of reaction as written.
结果单位为 J K⁻¹ mol⁻¹，并且是针对所写的反应式而言，因此表示为"每摩尔所写反应"的熵变。
5. Calculating the Entropy Change of the Surroundings | 计算环境熵变
The surroundings gain exactly the heat that the system loses, and vice versa. At constant pressure, the heat exchanged per mole is the enthalpy change of the reaction, ΔH. Since entropy is heat divided by temperature, the entropy change of the surroundings is given by the expression below.
环境获得的热量正好等于系统失去的热量，反之亦然。在恒压下，每摩尔交换的热量就是反应的焓变 ΔH。由于熵等于热量除以温度，环境熵变由下式给出。
ΔSsurroundings = −ΔHreaction / T
The minus sign is essential and is the most common source of lost marks. An exothermic reaction (negative ΔH) releases heat to the surroundings, increasing their entropy, so ΔSsurroundings is positive. An endothermic reaction (positive ΔH) removes heat from the surroundings, decreasing their entropy.
负号至关重要，也是失分最多的地方。放热反应（ΔH 为负）向环境释放热量，使环境熵增大，因此 ΔSsurroundings 为正。吸热反应（ΔH 为正）从环境吸收热量，使环境熵减小。
Note that ΔSsurroundings depends strongly on temperature: at low temperature the same quantity of heat causes a much larger entropy change than at high temperature. Also remember to convert ΔH from kJ mol⁻¹ to J mol⁻¹ before dividing by T, otherwise the units will not match.
注意 ΔSsurroundings 与温度关系密切：在低温下，同样的热量引起的熵变比高温时大得多。同时要记得在除以 T 之前把 ΔH 由 kJ mol⁻¹ 换算为 J mol⁻¹，否则单位不匹配。
6. Total Entropy Change and the Second Law | 总熵变与热力学第二定律
The second law of thermodynamics states that the total entropy of the universe (system plus surroundings) must increase for a process to be feasible. The total entropy change is the sum of the two contributions calculated above.
热力学第二定律指出，对于一个可行的过程，宇宙（系统加环境）的总熵必须增加。总熵变就是上面两部分熵变之和。
ΔStotal = ΔSsystem + ΔSsurroundings = ΔSsystem − ΔH / T
The interpretation of the sign follows three simple rules:
符号的判断遵循三条简单规则：
- ΔStotal > 0: the reaction is thermodynamically feasible in the forward direction.
- ΔStotal > 0：正反应在热力学上是可行的。
- ΔStotal < 0: the reaction is not feasible in the forward direction; the reverse reaction is favoured.
- ΔStotal < 0：正反应不可行，逆反应更有利。
- ΔStotal = 0: the system is at equilibrium, and there is no net change.
- ΔStotal = 0：系统处于平衡状态，没有净变化。
Remember that thermodynamic feasibility says nothing about rate. A reaction with a positive ΔStotal may still be immeasurably slow if the activation energy is high, which is exactly why catalysts are needed.
要记住，热力学可行性并不涉及速率。一个 ΔStotal 为正的反应，如果活化能很高，仍可能慢到无法测量，这正是需要催化剂的原因。
7. Worked Example 1: Thermal Decomposition of Calcium Carbonate | 例题一：碳酸钙的热分解
Consider the decomposition of calcium carbonate and decide whether it is feasible at 298 K.
以碳酸钙的分解为例，判断它在 298 K 下是否可行。
CaCO₃(s) → CaO(s) + CO₂(g) ΔH = +178 kJ mol⁻¹
₂) = 213.7 J K⁻¹ mol⁻¹. Applying the system equation first gives the change in the system.
标准摩尔熵为 S°(CaCO₃) = 92.9，S°(CaO) = 39.8，S°(CO₂) = 213.7 J K⁻¹ mol⁻¹。先应用系统熵变公式求出系统熵变。
ΔSsystem = (39.8 + 213.7) − 92.9 = +160.6 J K⁻¹ mol⁻¹
The positive sign is expected, because a gas is produced from a solid. Next, calculate the entropy change of the surroundings at 298 K, remembering to convert kJ to J.
符号为正符合预期，因为由固体生成了气体。接下来计算 298 K 下的环境熵变，记得把 kJ 换算为 J。
ΔSsurroundings = −178000 / 298 = −597.3 J K⁻¹ mol⁻¹
Adding the two contributions gives a negative total entropy change, so the decomposition is not feasible at room temperature.
两者相加得到负的总熵变，因此该分解反应在室温下不可行。
ΔStotal = +160.6 − 597.3 = −436.7 J K⁻¹ mol⁻¹
The exam skill being tested here is the clear separation of the two entropy terms and the correct handling of the minus sign in the surroundings term.
本题考查的关键技能是把两个熵变项清楚分开，并正确处理环境项中的负号。
8. Worked Example 2: The Haber Process | 例题二：哈伯法合成氨
Now apply the same method to the synthesis of ammonia, an exothermic reaction in which the number of gas moles decreases.
现在用同样的方法分析合成氨反应，这是一个气体摩尔数减少的放热反应。
N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = −92 kJ mol⁻¹
) = 192.5 J K⁻¹ mol⁻¹, the system entropy change is calculated with coefficients included.
取 S°(N₂) = 191.6，S°(H₂) = 130.7，S°(NH₃) = 192.5 J K⁻¹ mol⁻¹，计算系统熵变时要代入系数。
ΔSsystem = 2(192.5) − (191.6 + 3 × 130.7) = −198.7 J K⁻¹ mol⁻¹
As predicted from the fall in gas moles from four to two, this term is negative. The surroundings term, however, is positive because the reaction is exothermic.
正如气体摩尔数由 4 降至 2 所预测的那样，这一项为负。但由于反应放热，环境项为正。
ΔSsurroundings = +92000 / 298 = +308.7 J K⁻¹ mol⁻¹
The positive surroundings term outweighs the negative system term, so the reaction is feasible at 298 K.
正的环境项超过了负的系统项，因此该反应在 298 K 下可行。
ΔStotal = −198.7 + 308.7 = +110.0 J K⁻¹ mol⁻¹
This example illustrates the important principle that a reaction can be feasible even when its own entropy change is negative, provided the enthalpy change is sufficiently exothermic and the temperature is low enough.
这个例子说明一个重要原则：只要放热足够多、温度足够低，即使系统自身的熵变是负的，反应仍可能可行。
9. The Crossover Temperature | 转变温度
At the temperature where ΔStotal becomes zero, the reaction changes from feasible to non-feasible. Setting ΔSsystem − ΔH / T equal to zero and rearranging gives the crossover temperature.
当 ΔStotal 变为零时，反应由可行转为不可行。令 ΔSsystem − ΔH / T 等于零并整理，即得转变温度。
T = ΔH / ΔSsystem
or the Haber process, T = 92000 / 198.7 = 463 K, meaning that above roughly 463 K the reaction becomes non-feasible on entropy grounds alone.
对碳酸钙的例子，T = 178000 / 160.6 = 1108 K（约 835 °C），这与工业上石灰窑采用高温相一致。对哈伯法，T = 92000 / 198.7 = 463 K，意味着仅从熵的角度看，温度高于约 463 K 时反应就不可行了。
This explains the industrial compromise in ammonia manufacture. A moderate temperature of around 400-450 °C and high pressure are used, because high temperature improves the rate (kinetics) even though it reduces the equilibrium yield (thermodynamics).
这就解释了合成氨工业中的折中方案：采用 400-450 °C 左右的适中温度和高压，因为高温虽然降低平衡产率（热力学），却能提高反应速率（动力学）。
10. Linking Total Entropy to Gibbs Free Energy | 总熵变与吉布斯自由能的联系
Multiplying the total entropy change by −T gives the Gibbs free energy change, which is the standard criterion used for spontaneity in many syllabuses.
把总熵变乘以 −T 就得到吉布斯自由能变，这是许多考纲中判断自发性的标准判据。
ΔG = −T ΔStotal and ΔG = ΔH − T ΔSsystem
The two statements are equivalent, but the entropy approach is often easier to reason about qualitatively. The table below summarises how enthalpy and entropy signs combine to control feasibility.
两种表述等价，但用熵来推理往往更直观。下表总结了焓变与熵变的符号如何共同决定反应可行性。
| ΔH | 焓变 | ΔSsystem | 系统熵变 | Feasibility | 可行性 |
| Negative | 负 | Positive | 正 | Feasible at all temperatures | 任何温度都可行 |
| Negative | 负 | Negative | 负 | Feasible at low temperature | 低温可行 |
| Positive | 正 | Positive | 正 | Feasible at high temperature | 高温可行 |
| Positive | 正 | Negative | 负 | Never feasible | 永不可行 |
11. Spontaneous Endothermic Dissolving | 自发的吸热溶解
Dissolving ammonium nitrate in water is endothermic, yet it happens readily. This apparent contradiction disappears once the surroundings are considered: the dissolving process increases the disorder of the system so much that the positive ΔSsystem outweighs the negative ΔSsurroundings.
把硝酸铵溶于水是吸热的，但它却能顺利发生。一旦考虑环境，这个表面上的矛盾就消失了：溶解过程使系统的无序度大幅增加，正的 ΔSsystem 超过了负的 ΔSsurroundings。
In an exam, if a question states that a reaction is endothermic and spontaneous, the expected reasoning is that ΔSsystem must be large and positive, and that the reaction becomes more feasible as the temperature rises.
在考试中，如果题目说明某反应吸热且能自发进行，预期的推理是：ΔSsystem 必须很大且为正，而且温度升高时反应会更加可行。
The same reasoning applies to the thermal decomposition of carbonates and to the cracking of long-chain hydrocarbons, both of which need high temperatures to become feasible.
同样的推理适用于碳酸盐的热分解和长链烃的裂解，这两类反应都需要高温才可行。
12. Common Exam Mistakes and Revision Checklist | 常见错误与复习清单
Most lost marks in entropy questions come from a small number of repeated errors. Working through the checklist below before an exam is a quick way to protect those marks.
熵变题目失分大多来自少数几个反复出现的错误。考前过一遍下面的清单，是保住这些分数的快捷方法。
- Forgetting the minus sign in ΔSsurroundings = −ΔH / T.
- 忘记 ΔSsurroundings = −ΔH / T 中的负号。
- Mixing units by leaving ΔH in kJ mol⁻¹ while S is in J K⁻¹ mol⁻¹.
- 单位混用：ΔH 仍用 kJ mol⁻¹，而 S 用 J K⁻¹ mol⁻¹。
- Ignoring stoichiometric coefficients in the Σ S°(products) − Σ S°(reactants) calculation.
- 在 Σ S°(产物) − Σ S°(反应物) 的计算中忽略化学计量数。
- Confusing feasibility with rate, and forgetting that a feasible reaction may be very slow.
- 把可行性与速率混为一谈，忘记可行的反应也可能非常慢。
- Stating that a reaction is spontaneous rather than thermodynamically feasible, which is the preferred A-Level wording.
- 用"自发"而非"热力学可行"来描述反应，后者才是 A-Level 更偏好的表述。
- Forgetting that ΔStotal = 0 corresponds to equilibrium, not to impossibility.
- 忘记 ΔStotal = 0 对应的是平衡状态，而不是不可能发生。
In summary, always work in three clear steps: calculate ΔSsystem from standard molar entropies, calculate ΔSsurroundings from −ΔH / T, then add the two to judge feasibility. Practising these three steps
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图