Noesis
2.6.5 · D2Equilibrium

Visual walkthrough — Le Chatelier's principle — pressure, temperature, concentration, catalyst effects

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We will earn, in order: what "equilibrium" looks like → the ratio → the special number → the single sentence "compare to " → and then watch every stress (concentration, pressure, temperature, catalyst) as a picture of that comparison.

Step 1 — What "equilibrium" even looks like

WHAT. Picture a reaction as two crowds swapping sides of a room through one doorway. Reactants walk forward to become products; products walk back to become reactants. At the start the forward crowd is huge, so lots of people cross forward. As products build up, the back traffic grows. Eventually the number crossing forward each second equals the number crossing back.

WHY. When forward-per-second = backward-per-second, the amounts on each side stop changing — even though molecules never stop moving. That is what the word dynamic equilibrium means: still on the outside, busy on the inside. (See Chemical Equilibrium.)

PICTURE. The two arrows in the figure are the same length — that is the visual signature of equilibrium.

Step 2 — Turning "how far along" into a single number: the ratio

WHAT. We need one number that says how much product versus reactant is present right now. For a reaction

we build a fraction: products on top, reactants on the bottom. The little bar hanging off the fraction is our reminder "use the amounts at this instant":

Let us read every symbol as it sits:

  • — the square brackets mean "concentration of " (how crowded is, in moles per litre).
  • the little raised — the exponent, equal to how many molecules the balanced equation makes. More molecules made → that concentration counts more strongly (it is multiplied by itself times).
  • — the vertical bar with the word "current" underneath: it says evaluate this fraction with the concentrations you measure right now, before any settling.
  • top = products, so a big top means "lots of product made".
  • bottom = reactants, so a big bottom means "lots of reactant left over".

WHY a ratio and not a difference? Because the rates that drive the crossing depend on concentrations multiplied together (two molecules must meet). Multiplication and division track "meetings"; subtraction would not. A ratio also has the neat property that if you double every concentration the reactant and product parts fight, exactly the tug-of-war we want to see. (This is the Reaction Quotient Q.)

PICTURE. Watch the fraction climb as the reaction runs: starts near (no product yet) and rises.

Step 3 — The finish line: the constant

WHAT. Run the reaction long enough and stops climbing — it lands on one particular value and stays. That landing value is the equilibrium constant . It is the same fraction as , but now the bar underneath reads "equilibrium" instead of "current":

So is just "the value is trying to reach." (Solids and liquids are omitted here too, for exactly the reason in Step 2.)

WHY does it stop at a fixed number? Because forward rate and reverse rate . Setting them equal and rearranging gives exactly this ratio — a fixed number at a fixed temperature. (Full story in Equilibrium Constant.)

PICTURE. Draw as a horizontal finish line; 's curve flattens onto it.

Step 4 — The one sentence that runs everything: compare to

WHAT. Everything the parent note calls a "shift" is just being knocked off and then walking back. Three cases, no more:

  • → too little product → arrow leans forward (make product, rises).
  • → too much product → arrow leans backward (unmake product, falls).
  • → already home → no net motion.

WHY is this the deep reason, not just Le Chatelier's slogan? Because the true driver is free energy. The thermodynamic law is

term by term:

  • — the "downhill-ness" right now; negative means the reaction moves forward spontaneously.
  • — a fixed reference tilt, tied to by .
  • — the gas constant, a fixed conversion number.
  • — absolute temperature (kelvin); it scales how strongly pushes.
  • — the natural logarithm of our ratio; it turns " above or below " into "sign of ".

Substitute :

Now it is transparent: if then , its log is negative, → forward. If , log positive, → backward. The see-saw of Le Chatelier is literally the sign of . (See Gibs Free Energy.)

PICTURE. A number line for with marked; a ball placed left of rolls right, right of rolls left.

Step 5 — Stress #1: concentration (kick sideways, stays put)

WHAT. Take at equilibrium, so . Now pour in extra .

sits on the bottom. Making the bottom bigger makes the whole fraction smaller, so instantly .

WHY the system moves. is exactly Step 4's forward case: , so the reaction consumes some added (and ), builds , and climbs back up to the unchanged . Crucially never moved — only was disturbed. This is why the summary table says " changes? No."

PICTURE. The ball, resting on , is shoved left (to ) by the added reactant, then rolls back to .

Step 6 — Stress #2: pressure (squeeze the box, count the moles)

WHAT. Same reaction, gases only. Compress the container — halve the volume. Every concentration doubles at once. Does move up or down? The exponents decide.

Write with each concentration replaced by its old value:

=\frac{2^{2}}{2^{1}\cdot 2^{3}}\;Q_{\text{old}} =\frac{2^{2}}{2^{4}}\,Q_{\text{old}}=\tfrac14\,Q_{\text{old}}$$ Read the powers of $2$: the **top** collected $2^{2}$ (from the $2$ moles of product $NH_3$); the **bottom** collected $2^{1+3}=2^{4}$ (from $4$ moles of reactant gas). Bottom wins → $Q$ shrinks to a quarter → $Q<K$. **WHY moles are the whole game.** The exponent on each species is its mole count in the equation. Compression multiplies top and bottom by $2^{(\text{moles})}$; whichever side has *more* moles gets a *bigger* boost in the denominator-vs-numerator race. So compression always pushes toward the side with **fewer moles of gas** (here, the $2$-mole product side). If both sides have equal moles, the powers of $2$ cancel exactly and $Q=K$ still — **no shift**. That is the degenerate case. **PICTURE.** A piston squeezing; a mole-tally (4 balls vs 2 balls) with an arrow toward the 2-ball side. > [!example] Worked check (the parent's Example 2) > $N_2O_4\rightleftharpoons 2NO_2$: reactant $1$ mole, product $2$ moles. Compress → favor **fewer moles** → shift **backward** toward $N_2O_4$. ✓ > Degenerate: e.g. $H_2+I_2\rightleftharpoons 2HI$ has $2$ moles each side → pressure does **nothing**. --- ## Step 7 — Stress #3: temperature (the *only* stress that moves $K$ itself) **WHAT.** Concentration and pressure only nudged $Q$. Temperature is different: it slides the **finish line $K$** to a new place, and $Q$ chases it. How $K$ moves is set by the [[van 't Hoff Equation]]: $$\frac{d\ln K}{dT}=\frac{\Delta H^{\circ}}{R\,T^{2}}$$ symbol by symbol: - $\dfrac{d\ln K}{dT}$ — "how fast $\ln K$ changes as $T$ rises." Its **sign** tells us if $K$ grows or shrinks. - $\Delta H^{\circ}$ — the reaction's heat: **negative** if it releases heat (exothermic), **positive** if it absorbs heat (endothermic). This sign controls everything. - $R\,T^{2}$ — always positive, so it changes the *size* of the effect but never its sign. **WHY the shift direction.** The sign of the right side is the sign of $\Delta H^{\circ}$: - Exothermic ($\Delta H^{\circ}<0$): raising $T$ makes $\ln K$ **fall** → $K$ shrinks → the old $Q$ is now *above* the new smaller $K$ → $Q>K$ → **backward** shift. - Endothermic ($\Delta H^{\circ}>0$): raising $T$ makes $\ln K$ **rise** → $K$ grows → old $Q$ now *below* new $K$ → $Q<K$ → **forward** shift. This matches the "heat as a reagent" trick: for exothermic, heat is a *product*, so adding heat pushes backward. But now you see the real mechanism — the finish line itself slid. **PICTURE.** Two finish lines: heating slides $K$ down (exo) or up (endo); the stationary $Q$ is then caught on the wrong side and the ball rolls to the new $K$. > [!example] Worked check (the parent's Example 3) > $N_2+3H_2\rightleftharpoons 2NH_3,\ \Delta H=-92\text{ kJ/mol}$ (exothermic). Raise $T$: $K$ falls → $Q>K$ → **backward** → $[NH_3]$ drops. This is the [[Haber Process]] tension: heat speeds the reaction but lowers the yield. ✓ --- ## Step 8 — Stress #4: catalyst (moves *neither* $Q$ nor $K$) **WHAT.** Add a catalyst. It lowers the ==activation energy== (the hill both directions must climb — see [[Activation Energy]]) by the *same* amount $\Delta E_a$ for forward and reverse. Each rate constant follows the Arrhenius form $k = A\,e^{-E_a/RT}$, where: - $A$ — the ==pre-exponential (frequency) factor==: it counts *how often* molecules collide with the right orientation to react, before we even worry about energy. It is a fixed property of the reaction, unchanged by a catalyst. - $e^{-E_a/RT}$ — the fraction of those collisions that carry *enough* energy to clear the barrier $E_a$. Because $K=\dfrac{k_f}{k_r}$ (forward rate constant over reverse), we plug both in: $$K_{\text{new}}=\frac{A_f\,e^{-(E_{a,f}-\Delta E_a)/RT}}{A_r\,e^{-(E_{a,r}-\Delta E_a)/RT}} =\frac{A_f\,e^{-E_{a,f}/RT}}{A_r\,e^{-E_{a,r}/RT}}\cdot\frac{e^{\Delta E_a/RT}}{e^{\Delta E_a/RT}}=K$$ Here $A_f$ and $A_r$ are the forward and reverse frequency factors — same collision counts as before, since a catalyst changes energy hills, not collision rates. The two $e^{\Delta E_a/RT}$ factors — one from each direction — are **identical**, so they cancel. $K$ is untouched; and since equilibrium concentrations are unchanged, $Q$ ends up at the same $K$ too. **WHY it still matters.** It lowers *both* hills, so the system races to equilibrium far faster — years to seconds — without changing *where* equilibrium lies. **PICTURE.** An energy landscape: the barrier drops equally on both sides; the two valley floors (reactant, product) stay at the same heights, so their gap — hence $K$ — is unchanged. > [!mistake] Common trap > "A catalyst improves yield." **No** — it improves *speed only*. Yield is fixed by $K$, and a catalyst leaves $K$ exactly where it was. --- ## The one-picture summary Every stress in the chapter is one move of a ball on the $Q$–$K$ line: - **Concentration / pressure** → shove the **ball** ($Q$), finish line $K$ fixed. - **Temperature** → slide the **finish line** ($K$), ball chases it. - **Catalyst** → touch **neither**, only make the ball roll faster. > [!recall]- Feynman retelling — say it back in plain words > Imagine a ball resting exactly on a mark called $K$ on a track. That resting position is equilibrium. The ball's own spot on the track is a number called $Q$ — products over reactants (and we quietly leave out any pure solids or liquids, because their crowding never changes). > If I dump in extra reactant, the fraction shrinks, so the ball jumps to the left of the mark; being off the mark, it rolls right, back onto $K$. Squeeze the gas box and the same thing happens, but the direction is decided by counting molecules — the crowded side loses. > Temperature is the sneaky one: instead of moving the ball, it drags the *mark itself*. Heat a reaction that gives off heat and the mark slides toward reactants, so the ball, now stranded ahead of it, rolls backward. Heat one that soaks up heat and the mark slides the other way. > A catalyst? It doesn't touch the ball or the mark. It just polishes the track so the ball reaches the mark in seconds instead of years. That's the entire chapter — one ball, one mark, four ways to poke them. > [!recall]- Quick self-test > Q < K means the system shifts which way? ::: Forward (make more product). > Which single stress actually changes the value of K? ::: Temperature. > Compressing $N_2+3H_2\rightleftharpoons 2NH_3$ shifts toward which side and why? ::: Products — fewer moles of gas (2 vs 4). > Does a catalyst change the equilibrium yield? ::: No — it only speeds up reaching equilibrium; K is unchanged. > For an exothermic reaction, raising T does what to K? ::: K decreases (van 't Hoff, ΔH° < 0). > In $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$, what appears in K? ::: Only $[CO_2]$ — pure solids are omitted.