Part C · How stuff changes

11 · Equilibrium

Most reactions don't run to completion. They run until the reverse reaction keeps pace with the forward one, and then the mixture stops changing even though the molecules never stop reacting.

That balanced state is chemical equilibrium. It's one of the hardest ideas in introductory chemistry, because what you see (nothing happening) is the opposite of what's going on underneath (constant, furious, balanced change). This chapter puts both on screen at once.

The same place from both sides

Here is chapter 9's box again, A2 + B2 ⇌ 2 AB, but now the reverse reaction is switched on: two AB molecules colliding hard enough can swap back. The double arrow ⇌ means "both ways at once".

Run it once starting from pure A2 + B2, and once starting from pure AB, at the same temperature. Where does each run end up?

Both end at the same mixture. The end point depends on the temperature and on how deep each molecule's wells are, not on the history. From the A2 + B2 side, forward reactions outnumber reverse ones at first. From the AB side it's the other way round. Either way the counts drift until forward and reverse happen equally often. Watch the two rate counters in the readout become equal while the flashes carry on.

80 molecules (plus any you add) at fixed temperature. Orange flashes are forward reactions (A2 + B2 → 2 AB), blue are reverse. The chart shows the count of each molecule over time; the faded line is the previous run. The dotted line is the Boltzmann prediction for AB. Each AB is 1.2 ε lower in energy, per molecule, than half an A2 plus half a B2.

Three things to try:

That last behaviour is Le Chatelier's principle: disturb an equilibrium and it shifts in the direction that partly undoes the disturbance. It isn't a law of its own. It falls straight out of the counting below.

Why the balance point is a ratio

Forward reactions need an A2 to meet a B2, so their rate is proportional to how many of each there are: kf[A2][B2]. Reverse reactions need two AB molecules to meet: kr[AB]2. At equilibrium the two rates are equal, which rearranges to:

K = [AB]2 / ([A2][B2]) = kf / kr

The equilibrium constant K depends only on temperature. Square brackets are amounts (concentrations), each raised to the power of how many of that molecule appear in the equation. This is the law of mass action. Put in a different mixture and the same ratio, called the reaction quotient Q, is simply wrong: if Q < K the reaction runs forward, if Q > K it runs backward, until Q = K. Adding A2 makes the bottom of the fraction bigger, so Q drops below K, and the system makes more AB to compensate. That's Le Chatelier.

Rate laws only follow the equation's coefficients like this for single-step reactions such as the toy one here. Real multi-step reactions often have quite different rate laws. But the equilibrium expression always follows the overall equation, because it comes from free energies (below), not from the mechanism.

One molecule at a time

Strip it to the bare bones: 200 molecules that can each flip between two shapes, A and B, independently. That's an isomerisation, like a molecule flexing between two shapes. Each molecule hops over the barrier at random, at a rate set by the Boltzmann factor for the pass (chapter 10). Set how much lower B is than A, and how high the pass is.

Left: the energy profile (A on the left, B on the right, the pass between them), in kJ/mol at 25 °C. Middle: the 200 molecules, blue = A, orange = B. Right: the fraction that is B over time, against the prediction K/(1 + K) with K = e−ΔG°/RT (dotted). Time is scaled so you can watch; real molecules with these barriers would flip thousands of times faster.

Move the pass slider and the approach to equilibrium speeds up or slows down, but the end point doesn't move. Move ΔG° and the end point moves. This is chapter 10's catalyst lesson again: rates depend on passes, equilibria depend only on valleys. And it's chapter 7's Boltzmann factor, exactly:

K = [B]/[A] = e−ΔG°/RT   ⇔   ΔG° = −RT ln K

At 25 °C, every 5.7 kJ/mol of ΔG° shifts K by a factor of 10. That's why chemists treat about 30 kJ/mol either way as "essentially complete" or "essentially doesn't happen": a factor of about 105.

Why reactions stop part way

If B is lower than A, why doesn't every molecule end up as B? Because of chapter 7's other half: entropy. A mixture of A and B has more ways to be arranged than pure B. Plot the total free energy of the system against how far the reaction has gone, and the mixing entropy bends the line into a curve with a minimum somewhere in the middle.

Free energy of 1 mol of A ⇌ B at 25 °C as a function of the fraction converted to B. Dashed: what it would be if A and B didn't mix (a straight line from pure A to pure B). Solid: with the entropy of mixing, RT[x ln x + (1−x) ln(1−x)]. The minimum is the equilibrium; the reaction rolls downhill to it from either side.

The slope of that curve at any point is ΔG = ΔG° + RT ln Q. Where it's negative, the reaction runs forward; where positive, backward; at the bottom, ΔG = 0 and Q = K. Near the pure ends the curve is infinitely steep, which is why no reaction ever goes to absolutely 100%. When ΔG° is large, the minimum is so close to the end that "100%" is a fine description.

Common trap

"At equilibrium the reaction has stopped", and "at equilibrium there are equal amounts of reactants and products". Neither is true. Both reactions carry on at equal rates, as the flashes show. The amounts sit wherever K puts them, which can be 99.999% products or 0.001%. Students also often picture the reactants and products as being in separate places. They're the same jumble of molecules, constantly trading atoms.

Physics / ML bridge

The flipping molecules are a two-state continuous-time Markov chain. Its stationary distribution satisfies detailed balance: πAkf = πBkr, with π ∝ e−G/RT. That's the same condition the Metropolis–Hastings acceptance rule is designed to enforce, so MCMC samples a Boltzmann distribution for the same reason a flask of molecules reaches equilibrium. The rate equation d[B]/dt = kf[A] − kr[B] is linear, so the approach is a pure exponential with rate kf + kr: the spectral gap of the generator. Lower the pass and both rates rise by the same factor: faster mixing, same stationary distribution.

For reactions between two molecules (like A2 + B2) the bulk rate equations are nonlinear. The exact description is the chemical master equation over molecule counts, and the deterministic equations are its large-number limit. And living cells are not at equilibrium: they burn fuel (ATP) to drive reaction cycles with a net flux, breaking detailed balance on purpose. A cell at equilibrium is a dead cell.

Real-world trade-off: making ammonia for fertiliser (N2 + 3 H2 ⇌ 2 NH3) gives out heat, so a low temperature gives the best equilibrium yield. But at low temperature it's hopelessly slow (chapter 10). The Haber–Bosch process compromises: around 400–450 °C for speed, an iron catalyst to speed it further, and very high pressure, which favours the side with fewer gas molecules. That's Le Chatelier again.

Tell it to a 6-year-old
Look
A glass of purple red-cabbage juice (a grown-up makes it by soaking chopped red cabbage in hot water, then straining it), a spoon, vinegar and baking soda.
Wonder
"What colour will it turn if we add vinegar? And if we then add baking soda? Can we make it go back?"
Try
A splash of vinegar: pink! A little baking soda: fizz, and it turns blue-green. More vinegar: pink again. Go back and forth a few times.
True sentence
"Some changes can go back and forth, again and again. The colour pieces in the cabbage juice change their shape when we add vinegar, and change back when we add baking soda. Each shape shows a different colour."

This is an equilibrium being pushed one way and then the other (Le Chatelier): vinegar adds acid, baking soda removes it, and the cabbage dye follows. Chapter 12 explains the colours properly. The fizz is carbon dioxide, a new gas. After several rounds the drink is salty and flat; don't drink it.

Play more