Disturb an equilibrium and it pushes back. For A + B ⇌ C, add reactant and the mixture is thrown off balance — then the reaction runs toward products to swallow the surplus, and settles at a new position. But the equilibrium constant never moved: at fixed temperature Q climbs right back to the same K. Down the center, data flows: the disturbance goes in, the mass-action ODE re-equilibrates, the new position and the unchanged K come out. The blue team builds and defends; the red team tries to break it.
source Le Chatelier, Sur un énoncé général des lois des équilibres chimiques, Comptes Rendus 99, 786–789 (1884) — English translation at web.lemoyne.edu/giunta/lechat.html (Classic Chemistry). Rendered, not quoted.
The reaction obeys one law: rate = kf[A][B] − kr[C]. Equilibrium is where that rate is zero, so K = kf/kr = [C]/([A][B]).
Add reactant: the forward term jumps, rate goes positive, C is made until the rate falls back to zero. The position moved; K did not. Live state of the current disturbance:
| species | at eq. | after re-eq. |
|---|
Q = [C]/([A][B]) — returns to K to <1e-9 whenever T is held fixed.
Le Chatelier (1884) is the words; the mass-action law (Guldberg & Waage, 1864) is the equation beneath them. "The equilibrium shifts to oppose the change" is exactly what falls out of setting kf[A][B] = kr[C].
The one deep fact the principle hides: at constant T the ratio K = kf/kr is a constant of the rate laws, not of the current mixture — so no concentration push can move it. Temperature is the sole lever that touches K, through van’t Hoff. Each sphere is the next one’s premise.
The blue team’s live check: re-integrate the ODE after a push and confirm K is invariant, that raising T lowers K (exothermic), and that inert gas moves nothing. If red tampers, this badge is where it shows.
The system starts at equilibrium for A + B ⇌ C with K = 4.00 at 298 K: [A] = [B] = 0.390, [C] = 0.610, Q = 4.000. You feed it one disturbance:
| disturbance | touches K? | predicted shift |
|---|---|---|
| add reactant A | no | → products |
| add product C | no | ← reactants |
| raise T (exothermic) | K falls | ← reactants |
| add inert gas (const. V) | no | no shift |
"Const. V" matters: at constant volume an inert gas changes total pressure but no partial pressure or concentration — so Q is untouched. That is what you feed the panel below.
Add reactant A: the forward rate jumps, so C is produced until Q falls back to K. The position moves toward products; K is unchanged.
Every number is integrated live from rate = kf[A][B] − kr[C] by RK4 to steady state — nothing is looked up.
What the machine proves: at fixed T a concentration push moves the position (here [C] rises 0.610 → 0.750 when 0.5 mol/L of A is added) while Q returns to K = 4.000000000 to <1e-9. Only temperature moves K — an exothermic reaction’s K falls as T rises (van’t Hoff, d ln K / d(1/T) = −ΔH/R). Inert gas at constant volume moves nothing.
The blue team’s witness (left) confirms these live; the red team (right) tries to make K appear to move.
The rigorous statement lives in the Gibbs energy and the reaction quotient, not in the slogan. This engine restricts to the case where the slogan is exact — a single push on one dilute species at constant T and V — and shows the invariant that actually holds: K.
"Adding reactant increases K." Cut. K is fixed by kf/kr at constant T; adding reactant changes only Q, which then relaxes back to the same K. The engine proves Q → K to <1e-9.
"A catalyst shifts the equilibrium." Cut. A catalyst raises kf and kr together, leaving K = kf/kr unchanged. It reaches the same position faster; it does not move it.
"Inert gas shifts things because pressure rose." Kept, corrected. Only at constant pressure (which forces volume up and dilutes) — at constant volume the partial pressures are untouched and nothing shifts.
The red team’s move: read K off the disturbed, non-equilibrium mixture the instant reactant is added — so K appears to change. The blue team’s witness (window 7) is watching.
This is the classic error: compute [C]/([A][B]) at the moment of the push, before re-equilibration. Q there is 1.75, not 4 — so "K changed." It never did; you measured the wrong instant. The witness recomputes, disagrees, and turns red.