LearnChem · Part III — Why do some things react and others just sit there?
Stir milk into coffee and it never unstirs. Yet every collision in that cup would run backwards perfectly well, and the energy books balance either way. So what is actually stopping the milk?
“Why can't you unburn a match or unfry an egg?”
Last time you sealed a flame inside a can of water. You caught every joule the burn gave out. That ledger was strict. But it only answers one question. It tells you how much heat a crisp gives out. It never tells you why the smoke does not gather back into the crisp. The cold pack is worse. It swallows 25.7 kJ and turns icy by itself. Nothing pushes it. Module 12 asked how much. This module asks which way.
Strike a match. It burns down, the room warms a little, and the stick is left black. Now film it and run the film backwards. Smoke curls in from the corners of the room and gathers at the stick; the char pales back into wood; heat drains out of the air and into the match head. You have never seen this and you are certain you never will. Be careful about why.
A match burns. Film it, then run the film backwards: the smoke gathers, the flame draws in, the blackened stick is whole again. Commit — which law of nature does the backwards film break?
Commit before you look. First answers are counted anonymously, never named.
Count the joules in the backwards film. Going forward, the match released some quantity of heat into the room; running backwards, exactly that quantity leaves the room and goes into rebuilding the match. In, out, same number, opposite sign. Module 12's ledger does not merely permit this — it is built for it, which is why the kiln's +178 kJ and quicklime's −178 are one entry read twice. Energy is conserved in both films.
Go down a level and it gets worse. The backwards film is made of ordinary collisions, and a collision run backwards is still a collision: two things meet and go their ways. Nothing in the laws of motion distinguishes the two films. Whatever stops the match rebuilding, it is not a prohibition.
That leaves an awkward gap: the reverse is allowed, and it does not happen — not rarely, never. Something is doing the work of a law without being one. The same gap sits under the egg: frying one stiffens its proteins for good, and no amount of cooling unstiffens them, though nothing in the energy books says it could not. The rest of this module is the search for what is actually stopping it, and the answer turns out to be nothing cleverer than counting.
— Nothing in the energy books forbids the reverse. Something else is stopping it.
Start smaller than a match. Take a box, draw a line down the middle, and drop in a handful of numbered beads. Each bead is either on the left or on the right, and that list of sides — one entry per bead — is the entire description of where things are. Call one such list an arrangement. Now ask the only question this module ever really asks. Of all the arrangements those beads could be in, how many look sorted, everything piled on one side, and how many look mixed? Do not estimate. Count them.
Commit before the counting starts — twenty beads, each landing left or right at random. Roughly how often do all twenty land on the same side?
The shape on the bench is the whole argument. Twenty beads have 1,048,576 arrangements between them. Exactly one puts all twenty on the left. Exactly one puts all twenty on the right. The evenest split — ten and ten — accounts for 184,756 arrangements on its own, better than one in six. Sorted is not rare because something is keeping the beads out of it. Sorted is rare because there is hardly any sorted to be in.
Quick check — the bench has just counted every arrangement of 20 beads in two boxes: 1,048,576 of them, of which exactly one has all twenty on the left. Push it to a mole of beads. What happens to that count of one?
Chemists call the number of arrangements available to a state its multiplicity, written W — every tally on the bench is a multiplicity, counted. One awkwardness has to be fixed before W is usable. Multiplicities multiply. Stand two identical boxes side by side and the pair has W × W arrangements between them, though it is plainly just twice as much box.
Every other quantity you count in chemistry adds up instead. Two moles weigh twice as much and hold twice the energy; module 12's ledger added its entries. If this one is to sit alongside them it has to add up too, and turning multiplying into adding is exactly what a logarithm does. Take it, and the choice of base only changes the constant you put in front:
S = k ln W
S is the entropy: the count, logged. It comes out in joules per kelvin because of the constant in front. k is the Boltzmann constant, 1.380 649 × 10−23 J/K, and since the SI was rebuilt in 2019 that is not a measurement but a definition — exact, by international agreement.
Quick check — entropy in joules per kelvin, from S = k ln W, with k = 1.380 649 × 10−23 J/K. What entropy does a system with exactly one arrangement have?
Not quite whose it says. Ludwig Boltzmann had the idea — that entropy counts the ways a state can be realised — in an 1877 paper on the second law and probability. He never wrote it in this form. Max Planck did, in the winter of 1900–01, while working out the glow of hot objects, and it was Planck who put the constant in and argued that it was universal. The name “Boltzmann's constant” was attached later. Boltzmann himself died in 1906, and the equation he did not write was carved afterwards on his gravestone in Vienna.
— Mixed is not preferred. There is simply vastly more of it to land in.
Beads in a box sit where you put them; real mixing moves. So put the counting to work on something that runs. The bench below is a shallow tray of fixed pegs with two clouds of particles loose in it, one cloud starting on the left and one on the right, coloured only by where each began. Nothing in the model knows about colour. Nothing pushes the clouds together. Each particle simply goes where its last bounce sent it. Start it, and watch the one number that matters: how much of the left-born cloud is still on the left.
The trace falls from 1.00 to about 0.50 and stays there, wobbling. Leave it running all afternoon and it will still be wobbling about 0.50. The wobble repays a look. Push the particle slider up and the wobble shrinks — a hundred particles jitter visibly, eight hundred barely at all. The time the tray takes to mix hardly moves, though: each particle wanders at its own pace, whatever its neighbours do. Move the speed slider instead and the opposite happens: mixing gets faster, the wobble stays put. Numbers set how big the departures are. Speed sets how soon you get there.
Suppose you could reach into the mixed tray and reverse every particle's velocity exactly — every speed unchanged, every direction flipped. Commit — what happens next?
Commit before you look. First answers are counted anonymously, never named.
Every bounce in the tray is a mirror of itself. A particle arrives at a peg, turns, and leaves; reverse it and it arrives along the old outgoing path, turns, and leaves along the old incoming one. No step anywhere in the model knows which way time is pointing. So reversing every velocity must walk the whole tray back through its own history, to the sorted start it came from.
Press Reverse every velocity after a few hundred steps and it does exactly that. The colours climb apart. They unmix, and the number goes back to 1.00 — the readout will tell you what it reached.
So unmixing is more than a loophole in principle: there it is, happening, on your screen. Which sharpens the question rather than settling it — if separation is that easy to arrange, why has nobody ever seen a cup of coffee do it?
So try it late. Sort the tray again, let it run for a thousand steps this time, and then reverse it. Then do it again at two thousand. Watch the number each time.
Reversed inside the first few hundred steps it comes back to 1.00 on the nose. Reversed at a thousand it gets most of the way and stalls short. Reversed at two thousand it barely sets off before the trace flattens again. Nothing changed in the rules. What changed is that the computer carries about sixteen digits per number, and in a field of round pegs a difference in the sixteenth digit doubles every few bounces. Reverse late enough and the sixteenth digit has grown into the answer. You have just shuffled a deck backwards, and it worked only because the machine still knew every card's exact path. A cup of coffee carries no digits at all, and its pegs are other molecules, which move.
Two things happen in that cup and only one of them is mixing. The spoon does the first: it stretches the blob of milk into a long ribbon and folds it, over and over, until the ribbon is thin. That part is geometry, and in a slow enough liquid a careful enough spoon can largely undo it. The second part cannot be undone. Once the ribbon is thin enough, the molecules take over and wander off individually, as on Bench 2 — and after that no spoon helps, because there is no ribbon left to unfold. Stirring the other way only stirs a cup that is already mixed.
— Every collision runs backwards; the crowd of them does not.
A block of camphor left in a puja room dwindles away and never leaves a wet mark. It is not melting — that would take 178 °C — it is leaving as vapour, straight from the solid: C10H16O(s) → C10H16O(g). Where the molecules go afterwards is the part worth watching. They spread through the whole house and never come back. Strip that down to its bones and you have the cleanest case in the module: gas at one end of a box, vacuum at the other, a tap between them.
The tap opens and the gas rushes out into the vacuum. Commit — what does a thermometer sitting in the gas read once it has settled?
Commit before you look. First answers are counted anonymously, never named.
Work means pushing something. A gas expanding against a piston pushes the piston, and pays for that out of its own energy — which is why a deodorant can goes cold in your hand, and why a refrigerator works at all. Here there is nothing to push. The tap opens onto emptiness, and the molecules simply find themselves with further to fly before the next peg. Not one of them is slowed down by the expansion.
So no work leaves, and no heat enters. Watch the kinetic-energy line on the bench as the tap opens: it does not move.
The gas fills the box. Nothing surprising in that — the surprise is what it cost. Expanding into a vacuum the gas pushes against nothing, so no work is done; nothing is heated or cooled either, and the bench watches the total kinetic energy sit exactly where it was. Module 12's ledger records a flat zero for the whole event — and the event happens every time, and never once runs backwards.
Counting prices it in a line. Double the box and every molecule has twice as many places to be, so the multiplicity is multiplied by 2 for each molecule — by 2N altogether, where N is the number of molecules in the box. Feed that to S = k ln W and the powers come down as a multiplier:
ΔS = k ln 2N = N k ln 2 = n R ln 2
— where n is the number of moles and R is the gas constant, 8.314 J per kelvin per mole. R is nothing but k multiplied by module 03's Avogadro number, doing the job it always does: turning a price per particle into a price per mole. For one mole, doubling the box, ΔS = 5.76 J/K. The panel computes it live for whatever ratio you set — and the settling fraction on the trace is the same count read the other way round, one part in the ratio.
Quick check, and the bench cannot answer this one — its slider stops at four. One mole expanding from 1 litre into 2 litres cost ΔS = R ln 2 = 5.76 J/K. What would 1 litre into 8 litres come to?
Nobody has ever seen an arrangement. So why believe a quantity built out of counting them? Because it arrives at numbers that were already known from somewhere else entirely. That 5.76 J/K came out of an argument about places to be — no thermometer, no calorimeter, no experiment of any kind. The same figure is reachable from the laboratory end: let the gas expand slowly against a piston at a fixed temperature instead, measure the heat it draws in, and divide by the temperature. That route knows nothing about molecules, and was in working order before most chemists believed in them. Two roads, sharing no assumptions, one answer. That agreement is the evidence — and it is why chemists trust tables of entropies that nobody could ever count out by hand.
— A gas spreads into empty space for no reason but the count of places.
Two cases now that look like counterexamples. The first is module 12's unpaid debt. Squeeze an instant cold pack and the ammonium nitrate inside dissolves — NH4NO3(s) → NH4+(aq) + NO3-(aq) — swallowing about 25.7 kJ for every mole, taken out of whatever the pack is pressed against. On the energy ledger alone this should not happen at all. It is a climb, and module 12 was clear that the payouts come from falls.
Quick check — the cold pack goes uphill in energy, briskly, with nobody pushing it. Which reading settles it?
Uphill in energy, downhill in arrangements — and the second wins. Before the squeeze, every ion sat at a fixed address in a crystal — module 09's unit cell, one small box stamped over and over in all directions. After it, any ion may be anywhere in the water, with the water's own molecules rearranged around each one. That is a different order of number altogether, and it buys the 25.7 kJ easily.
The second case is harder, and it is the one that forces the rule into its real shape.
On a December night in Delhi a pan of water freezes solid, all by itself. Ice is the tidier arrangement — far fewer ways of being than the liquid had. Commit — how does that square with this module's rule?
Commit before you look. First answers are counted anonymously, never named.
The pan is not the whole system. Freezing gives out heat — H2O(l) → H2O(s) pays out exactly what melting will swallow again — and that heat goes into the December air, to be shared among the air's own molecules. Sharing energy among molecules is itself an arrangement problem: the more energy the air is holding, the more ways there are of splitting it up.
So two counts are moving at once, in opposite directions. The water's falls. The air's rises. Below 0 °C the air's rise is the bigger of the two, and the total goes up — which is the form of the rule that actually holds, and the form your board's textbook states:
ΔStotal = ΔSsystem + ΔSsurroundings > 0
Above 0 °C the same two counts come out the other way round, and the ice melts.
That inequality has a name. It is the second law of thermodynamics, and you have just counted your way to it. A process that goes on its own — a spontaneous one, which is the word your exam paper will use — is one that leaves more ways of being than it found, once the surroundings are on the books too.
Notice what you now need and still lack: two counts pointing opposite ways, and a verdict that flips at a particular temperature — 0 °C for water, some other number for everything else. Counting alone cannot deliver that verdict, because nothing so far says how much the surroundings' count moves for a given quantity of heat. That is one line of arithmetic, and it is the whole of the next module.
Two warnings before you go, and the first is about a word. Your textbook will gloss entropy as “randomness” or “disorder”. As a handle it is poor. The freezing pan is tidier at the end and the rule is satisfied anyway, and the exercises below will show you two identical gases “mixing” with the count not moving at all. Count, and you are right. Guess at tidiness, and you will not be.
The second is about scope. Oil and water still separate, and counting alone will not call it: the honest account needs both the grip the water keeps around a spread-out drop and a cost at the boundary itself — and chemists are still arguing over which of the two matters more. The egg from the hook is the same kind of half-case; frying it unfolds its proteins into far more arrangements, and those unfolded proteins then grip each other in ways they never could before. Counting is half a verdict. The next module supplies the other half.
One note on where this sits in your syllabus. NCERT folds the count and the verdict into a single chapter; this course has pulled them apart, and you have had the count first — so what reads there as one long argument should read here as two short ones.
— When a thing tidies itself, something outside it got untidier, by more.
These check themselves, and “New numbers” deals a fresh set — there is nothing to memorise. Each stem says how exact to be.
1. Module 09's counting trap, revisited: an atom sitting of a cubic unit cell is shared between how many cells? (A whole number — module 09 counted it for you.)
2. Bench 1's first number: beads in two boxes. How many arrangements altogether? (Exact.)
3. Of those arrangements, (Exact — it is N! ÷ (k! × (N−k)!), and the bench counts it by brute force if you would rather check than compute.)
4. The expansion price: (Use ΔS = nR ln(ratio), R = 8.314 J/K per mole. To ±0.2.)
A sealed box is divided by a tap. On the left, one mole of argon in 10 litres. On the right, one mole of neon in 20 litres. Same temperature throughout. Open the tap and the gases mix through all 30 litres.
Work out ΔS in J/K, to ±0.5. Mind the trap: this is not one gas expanding, it is two, and neither of them knows the other is there — so the two expansions are not the same size.
Now the second tap, and commit before you look at anything. Run it again with argon on both sides: one mole of argon in the 10 litres, two moles of argon in the 20 litres, same temperature. Every part of the box is at the same crowding as every other. Open the tap.
Argon on both sides, at the same crowding. What is ΔS when the tap opens, and what happens?
Argon and neon. Each gas ends up with the whole 30 litres, and neither is obstructed by the other, so each simply expands on its own. Argon goes from 10 litres to 30 — three times the room: ΔS = R ln 3 = 9.13 J/K. Neon goes from 20 to 30 — one and a half times: R ln 1.5 = 3.37 J/K. Total 12.51 J/K. Notice what never entered the arithmetic: the mixing. Only the two expansions did, and they are different sizes because the two gases started with different amounts of room.
Argon on both sides. Count the arrangements before and after. Before: ways of putting argon atoms in 10 litres, and argon atoms in 20 litres. After: ways of putting argon atoms in 30 litres. But the atoms are identical. There is no such thing as a left-hand argon atom, so “sorted” and “mixed” are not two states here at all — they are one state, described twice.
The count does not move. ΔS = 0, and when the tap opens, nothing happens: no entropy is made, and you could close the tap again with no trace anywhere that it was ever open. This is the sharpest test of whether you have been counting arrangements or merely counting particles. If “mixing raises entropy” were a rule about stirring things together, this case would break it. It is not. It is a rule about how many ways there are, and here there are exactly as many as before. (The puzzle of identical particles is an old one, usually called the Gibbs paradox, and this is as far as we take it.)
Your bedroom is sealed, and you are waiting for all the air in it to gather in one half, leaving you in the other. Estimate how long you would expect to wait. You will need to estimate three things before you can put a number on it; work out for yourself what they are.
Then do one more thing, and it matters more than the answer. Go back and change every assumption you made to something wildly more favourable — a smaller room, a faster shuffle, a longer life — and see how much the answer moves.
The discussion. The three quantities are the number of molecules, the share of arrangements with all of them on one side, and how often you are willing to say the air re-deals itself. A small bedroom is about 30 m³, and a cubic metre of air holds around 2.5 × 1025 molecules, so N ≈ 7.5 × 1026. The share is 2−N, which is 1 in 102.3 × 1026 — the exponent alone has 27 digits. Let the air re-deal itself a trillion times a second and you must still wait about 102.3 × 1026 seconds.
The second half of the exercise is the real one. Make the room a thousand times smaller and the exponent drops by a factor of a thousand — and the answer is still unimaginable. Let it re-deal a billion times faster and you subtract 9 from an exponent with 27 digits. Wait the whole age of the universe, about 1017 seconds, and you have subtracted 17 from a 27-digit exponent. The exponent does not notice.
That insensitivity is the lesson. Most estimates in science wobble when you lean on the assumptions; this one cannot be moved by anything you are prepared to do to it. Which is why a thing that is merely overwhelmingly unlikely ends up looking exactly like a thing that is forbidden. You are waiting for a deck to shuffle itself back into order — a deck with 1026 cards in it. (For a version you can almost picture: toss 100 coins once a second, and all heads or all tails turns up about once in 2 × 1022 years. A hundred coins is nothing at all beside a roomful of air.)
Things go the way that leaves more ways of being, surroundings counted in.