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Entropy as a count of the arrangements of particles and energy, the second law as the total always rising, and entropy changes from tables.
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By the end of this lesson you will be able to explain entropy as a count of arrangements and calculate how fast that count grows, predict the sign of an entropy change from states and moles of gas, calculate a standard entropy change from tabulated entropies, and explain how a system's entropy can fall while the total entropy of system and surroundings rises.
From Chemistry 1 you have enthalpy changes: exothermic reactions give out heat and endothermic ones take it in. From the kinetics unit you have the particle model and the idea that energy is spread over particles in a distribution. The equilibrium unit showed reactions settling somewhere other than all products. None of that yet says why a change happens at all; this lesson begins to.
Entropy, $S$, measures the number of ways the particles of a system and their energy can be arranged; more ways means more entropy. Each such arrangement is a microstate. The standard molar entropy, $S^\circ$, is the entropy of one mole of a substance at $100$ kPa, usually at $25$ °C, in J/(K mol). The second law of thermodynamics says that in any change that happens of its own accord, the total entropy of the system and its surroundings increases. A spontaneous change is one that happens without continuing outside work, whether fast or slow.
Put four gas molecules, A, B, C and D, in a box with two halves. Each can be on the left or the right, independently, so the number of arrangements is $2 \times 2 \times 2 \times 2 = 16$:
| Molecules on the left | Arrangements |
|---|---|
| 4 | 1 |
| 3 | 4 |
| 2 | 6 |
| 1 | 4 |
| 0 | 1 |
Only one of the sixteen has every molecule on the left; six have them evenly split. If the molecules move at random, the box spends most of its time near the even split simply because there are more ways to be there. Nothing pushes the molecules apart.
Now scale up. With $10$ molecules there are $1024$ arrangements and one of them is all-left. With $100$ the share is about one in $10^{30}$. With a mole, $6 \times 10^{23}$ molecules, the evenly spread state is so overwhelmingly more numerous that the gas is never seen gathered in one half. That is why a gas spreads to fill its container, why perfume spreads through a room and why it never spreads back into the bottle.
Entropy is the measure of that number of arrangements — Boltzmann's formula makes it $S = k \ln W$, where $W$ is the count of microstates and $k$ a constant. The second law is the counting argument written as a rule: left to itself, a system of many particles moves towards the states with the most arrangements, because those are the states it is overwhelmingly likely to be found in.
Energy is arranged as well as particles. A hot object next to a cold one has fewer ways to share their energy than two objects at the same temperature, which is why heat flows from hot to cold and not back.
Another way: picture
Shuffle a pack of cards. There is one order that runs ace to king in every suit and there are about $8 \times 10^{67}$ others. Shuffling does not prefer mixed-up orders; there are just vastly more of them, so a shuffled pack is never found in order. Molecules shuffle themselves billions of times a second.
Another way: steps
To predict the sign of an entropy change:
Unlike enthalpy, entropy has an absolute zero: a perfect crystal at $0$ K has only one arrangement and zero entropy. So tables give every substance its own positive entropy, elements included:
| Substance | $S^\circ$ / J/(K mol) |
|---|---|
| diamond, $\mathrm{C(s)}$ | 2.4 |
| graphite, $\mathrm{C(s)}$ | 5.7 |
| liquid water | 69.9 |
| calcium carbonate | 92.9 |
| hydrogen gas | 130.7 |
| steam | 188.8 |
| oxygen gas | 205.2 |
| carbon dioxide gas | 213.8 |
Gases sit far above liquids and solids, and hard solids at the bottom. The entropy change of a reaction is products minus reactants, each multiplied by its coefficient:
$$\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$$
For $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$: $(38.1 + 213.8) - 92.9 = +159.0$ J/(K mol), positive because a gas is made from a solid. The units are joules, not kilojoules, which matters in the next lesson.
Water freezing in a freezer loses entropy: its molecules lock into a lattice. That is allowed, because the second law is about the total. Freezing releases heat, and that heat flows into the surroundings, giving their particles more ways to arrange their energy. The entropy the surroundings gain is the heat they receive divided by their temperature:
$$\Delta S_{surr} = \frac{-\Delta H}{T}$$
with $\Delta H$ in joules and $T$ in kelvin. The colder the surroundings, the bigger the gain from the same heat — a small gift means more to someone who has little. Below 0 °C the gain outweighs the water's loss and freezing is spontaneous; above 0 °C it does not, and ice melts instead. Living things, crystals growing and refrigerators all make local order the same way: by paying for it with a larger entropy increase somewhere else.
Entropy always increases in every part of a system. Only the total must increase. A system can lose entropy if its surroundings gain more.
Entropy is disorder. Disorder is a picture, and sometimes a misleading one. Entropy counts arrangements of particles and energy; a well-mixed but warmer system can have more entropy than a messier-looking cold one.
An element in its standard state has zero entropy. That is enthalpy of formation. Every substance above absolute zero has positive entropy, including oxygen and graphite.
Spontaneous means fast. Diamond turning into graphite is spontaneous and takes longer than the age of the universe. Spontaneity is about direction, not rate.
Entropy changes are in kJ. Standard entropies are tabulated in J/(K mol); forgetting to convert to match $\Delta H$ in kJ gives answers a thousand times out.
$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$, with $S^\circ$ values $191.6$, $130.7$ and $192.8$ J/(K mol).
Four moles of gas become two, so expect a decrease.
Products: $2 \times 192.8 = 385.6$. Reactants: $191.6 + 3 \times 130.7 = 583.7$.
Multiply each by its coefficient.
$\Delta S^\circ = 385.6 - 583.7 = -198.1$ J/(K mol).
Negative, as the gas count predicted.
At $500$ K a reaction has $\Delta H = -40$ kJ/mol and $\Delta S_{sys} = -50$ J/(K mol).
The system loses entropy; the surroundings gain heat.
$\Delta S_{surr} = 40\,000 \div 500 = +80$ J/(K mol).
Heat in joules over temperature in kelvin.
$\Delta S_{total} = -50 + 80 = +30$ J/(K mol): positive, so the reaction is spontaneous at $500$ K.
The total decides.
All gases; three moles on the left, two on the right.
Count the moles of gas.
Fewer moles of gas means fewer arrangements, so $\Delta S$ is $\ldots$
…negative: the entropy of the substances decreases, and a table confirms it.
Gas molecules move freely between the left and right halves of a box. Complete the table: how many arrangements there are of the molecules between the halves, and what fraction of them have every molecule on the left.
| arrangements | share with all on the left | |
|---|---|---|
| 1 molecule | ||
| 4 molecules | ||
| 7 molecules |
For each change, does the entropy of the substances increase or decrease?
| entropy of the substances | |
|---|---|
| ice melting to water | |
| steam condensing to water | |
| calcium carbonate decomposing to calcium oxide and carbon dioxide | |
| nitrogen and hydrogen forming ammonia | |
| sodium chloride dissolving in water |
Standard molar entropies at 25 °C, in J/(K mol): $\mathrm{C(s)}$ $5.7$, $\mathrm{O_2(g)}$ $205.2$, $\mathrm{CO_2(g)}$ $213.8$. What is the standard entropy change for $\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$?
Answer: J/(K mol)
Put these in order of their entropy per mole, lowest first.
Number the steps in order (write the number in the box):
Water left outside at $-5$ °C freezes of its own accord, although the ice has lower entropy than the liquid. How can that happen?
Water freezes in a freezer at $-18$ °C. Match each quantity to what happens to it.
| increases | decreases | |
|---|---|---|
| the entropy of the water | ||
| the entropy of the surroundings | ||
| the total entropy of water and surroundings | ||
| the energy of the water |
A flask is joined to an empty flask of the same size by an open tap, and $12$ gas molecules move freely between them. At any instant, what is the chance that all $12$ are back in the first flask? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
At $250$ K a process has $\Delta H = -58$ kJ/mol and an entropy change of the system of $-113$ J/(K mol). What is the total entropy change, system plus surroundings, in J/(K mol)?
Answer: J/(K mol)
You can count arrangements, read an entropy table and apply the second law to the total. Say out loud why water freezes below 0 °C although the ice has less entropy. Next: combining enthalpy, entropy and temperature into one quantity that decides spontaneity — Gibbs energy.
10. Your turn: predict the sign of $\Delta S$ for $\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}$ before using any table., step 3