Reversible reactions and dynamic equilibrium: Visual walkthrough

We use the simplest possible reversible reaction so nothing hides: $A ⇌ B$

Read the double half-arrow $⇌$ as "goes forward AND comes back". The top arrow $⇀$ is A turning into B; the bottom arrow $↽$ is B turning back into A. Everything below is about those two arrows.

Step 1 — Draw the two doors

WHAT: picture one room with two one-way doors between an "A side" and a "B side".

WHY: before any maths, we must agree there are two separate journeys, each with its own crowd and its own speed. Mixing them up is the root of every mistake in the parent note's mistake list.

PICTURE: in the figure, the amber door carries molecules from A to B (forward); the cyan door carries them from B to A (backward). Nobody is standing still — every molecule is either waiting or crossing.

Figure — Reversible reactions and dynamic equilibrium

Step 2 — Turn "how fast" into a number: rate

WHAT: we attach a number, the rate, to each door.

WHY: "concentrations stay constant" is really a statement about speeds being equal. We cannot compare speeds until we can measure them, so we name them first. This is the rate idea — nothing more than "crossings per second".

PICTURE: each door in the figure now has a speedometer. The needle position is the rate.

Term by term, when we write $r_{f}$ : $forward speed r_{f} backward speed r_{b}$

Step 3 — Why rate depends on how much stuff there is

We write "amount of A per litre" as $[A]$ , called the concentration of A. Square brackets mean "concentration of whatever is inside", measured in mol per litre.

WHAT: we make each rate depend on the crowd size at its own door.

WHY: this is the one physical assumption of the whole derivation. Without it, the rates would be fixed and nothing could ever chase anything.

PICTURE: the figure shows a nearly-empty A side (slow forward door) versus a packed A side (fast forward door) — same door, different crowd, different needle.

$r_{f} \propto [A] r_{b} \propto [B]$

The symbol $\propto$ means "proportional to" — grows in lockstep.

Step 4 — Replace "proportional to" with an equals sign: the rate constants

WHAT: we upgrade the proportions into exact equations.

WHY: a " $\propto$ " cannot be solved; an " $=$ " can. And separating "how crowded" ( $[A]$ ) from "how leaky the door is" ( $k_{f}$ ) is exactly what lets us later cancel things cleanly.

PICTURE: the figure widens the amber door versus a narrow amber door for the same crowd — same $[A]$ , different $k_{f}$ , different rate. The door width is the constant $k$ .

$r_{f} = forward door width k_{f} A crowd [A] r_{b} = backward door width k_{b} B crowd [B]$

Step 5 — Press play: start with pure A

WHAT: we fill the room with A only and watch time run.

WHY: we need to see the two rates move, not just exist. Starting from a pure, one-sided state makes the chase as dramatic as possible.

At the very first instant:

$[A]$ is large $\Rightarrow r_{f} = k_{f} [A]$ is large — the amber door is packed and busy.
$[B] = 0 \Rightarrow r_{b} = k_{b} [B] = k_{b} \cdot 0 = 0$ — the cyan door is locked, because there is no B yet to come back.

So at $t = 0$ everything flows one way, A $\to$ B.

PICTURE: the figure freezes the first second: amber needle high, cyan needle at zero.

Step 6 — The scissor closes: rates chase each other

WHAT: we let time run and track both rates together.

WHY: this is the heart of the derivation. Because each rate depends on its own crowd, and the crowds change, the two rates must drift toward each other.

As B is made and A is used up:

$[A]$ falls $\Rightarrow r_{f} = k_{f} [A]$ falls (forward door quietens).
$[B]$ rises $\Rightarrow r_{b} = k_{b} [B]$ rises (backward door wakes up).

A falling curve and a rising curve, starting far apart, are guaranteed to meet. They do not pass through each other — the instant they touch, the crowds stop changing (Step 7), so the curves freeze in contact.

PICTURE: the figure plots $r_{f}$ (amber, descending) and $r_{b}$ (cyan, ascending) versus time. Look at the point where they touch — call it $t_{e q}$ . Left of it, $r_{f} > r_{b}$ (net forward); at it, they are equal.

Step 7 — The meeting point IS equilibrium

WHAT: we read off the condition at the touching point.

WHY: "concentrations constant" and "rates equal" are the same event seen two ways — a bookkeeping fact: the same molecule leaving A per second returns to A per second.

$A leaving per sec k_{f} [A]_{e q} = A returning per sec k_{b} [B]_{e q}$

The subscript $e q$ just tags "the value at that equilibrium instant".

PICTURE: the figure shows the concentration curves: $[A]$ falls then flattens, $[B]$ rises then flattens — and the flattening starts exactly under $t_{e q}$ from Step 6. Notice both flat lines sit at nonzero heights: reactants AND products remain.

Step 8 — Solve for the ratio: out pops $K_{c}$

WHAT: we rearrange the equal-rates equation to isolate the concentrations.

WHY: we want a number that describes where the balance sits, independent of how much we started with. Dividing both sides collects the constants on one side and the concentrations on the other.

Start from Step 7 and divide both sides by $k_{b}$ and by $[A]_{e q}$ : $k_{f} [A]_{e q} = k_{b} [B]_{e q} ⟹ \frac{[ B ] _{e q}}{[ A ] _{e q}} = \frac{k _{f}}{k _{b}}$

Because $k_{f}$ and $k_{b}$ are fixed at a given temperature, $K_{c}$ is fixed — so equilibrium always lands on the same ratio no matter where you begin. (Change the temperature and the door widths change — see Effect of temperature on equilibrium.)

PICTURE: the figure shows two runs — one starting from pure A, one from pure B — landing on the same flat ratio.

Step 9 — The edge cases (never leave a gap)

Three degenerate situations the reader must never hit unprepared.

(a) A leaky room (open system). If B is a gas and escapes, $[B]$ can never build up, so $r_{b} = k_{b} [B]$ stays tiny forever. The curves in Step 6 never meet — no equilibrium. This is why the parent insists on a closed system.

(b) A one-way door ( $k_{b} \approx 0$ ). If the backward door is essentially locked, $r_{b} \approx 0$ always, so $r_{f}$ never has anything to meet. The reaction runs to completion — this is an irreversible reaction. $K_{c}$ is huge.

(c) Adding a catalyst. A catalyst widens both doors — it multiplies $k_{f}$ and $k_{b}$ by the same factor. In the ratio $k_{f} / k_{b}$ that factor cancels, so $K_{c}$ is unchanged; only the meeting happens sooner. See also Le Chatelier's principle for what genuinely does move the balance.

PICTURE: the figure overlays a catalysed run (dashed, meets early) on the plain run (solid) — same final heights, earlier meeting.

The one-picture summary

Everything above compressed: two rates start far apart, chase, and lock; concentrations mirror this by sliding then freezing; the frozen ratio is $K_{c} = k_{f} / k_{b}$ .

Recall Feynman: the whole walkthrough in plain words

One room, two doors. The amber door sends molecules from the A pile to the B pile; the cyan door sends them back. A door's speed is just how many cross per second, and it depends on two things: how big the pile is at that door (the crowd) and how easy the door is (its width, the rate constant). Start with everything in the A pile: the amber door is slammed busy, the cyan door has nobody so it's dead quiet. As molecules pour over to B, the A pile shrinks (amber slows) and the B pile grows (cyan speeds up). A slowing thing and a speeding thing must meet. The instant they cross, every molecule leaving A comes straight back, so the pile sizes stop moving — even though both doors are still roaring. That frozen picture is dynamic equilibrium. Divide the two "equal speed" equations and the crowd-independent leftover is $K_{c} = k_{f} / k_{b}$ , the fixed number that says where the balance sits. If molecules could leak out of the room the doors could never match up (no equilibrium); if the cyan door were welded shut it'd never match either (irreversible); and a catalyst just oils both doors equally, so they meet sooner at the exact same place.

Recall Quick self-test

Why does $r_{b}$ start at zero when we begin with pure A? ::: Because $r_{b} = k_{b} [B]$ and $[B] = 0$ at the start, so $r_{b} = 0$ . At equilibrium, is $[A] = [B]$ ? ::: No — the rates are equal, $k_{f} [A] = k_{b} [B]$ ; the concentrations are equal only if $k_{f} = k_{b}$ . Why does a catalyst leave $K_{c}$ unchanged? ::: It multiplies both $k_{f}$ and $k_{b}$ by the same factor, which cancels in $k_{f} / k_{b}$ . If $k_{f} = 3 k_{b}$ , what is $[B]_{e q} / [A]_{e q}$ ? ::: $3$ .

Connections

Equilibrium constant Kc and Kp — the $K_{c}$ this walkthrough produces, quantified.
Rate of reaction and rate constants — the $k_{f}, k_{b}$ that set the door widths.
Collision theory — why rate $\propto$ concentration (Step 3).
Le Chatelier's principle — how the balance moves when disturbed.
Effect of temperature on equilibrium — why changing $T$ changes $k_{f}, k_{b}$ and hence $K_{c}$ .
Catalysis — the edge case in Step 9c.