Law of mass action and Kc, Kp: Visual walkthrough — Chemistry deep dive

Step 1 — What "reversible" and "equilibrium" even mean

WHAT. A reversible reaction runs both ways at once. Write it with a double harpoon: $a A + b B ⇌ c C + d D$

Read this out loud: " $a$ chunks of stuff $A$ plus $b$ chunks of stuff $B$ turn into $c$ chunks of $C$ plus $d$ chunks of $D$ , and the arrow points both ways."

$A, B$ — the reactants (the starting stuff), $C, D$ — the products (what forms).
$a, b, c, d$ — the stoichiometric coefficients: literally how many of each molecule take part in one reaction event. In $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ they are $1, 3, 2$ .

WHY. Nothing forces a reaction to go only forwards. As products pile up, they start crashing back into reactants. Equilibrium is the moment the forward traffic and backward traffic move at the same speed — the amounts stop changing even though molecules keep reacting. (This is the dynamic picture: busy, but balanced.)

PICTURE. Two crowds swapping across a river. Left bank turns into right bank; right bank turns back. When the flow across the bridge is equal both ways, the crowd sizes freeze.

Step 2 — Why the rate carries the coefficients as powers

WHAT. The forward rate is how fast reactants disappear: $forward speed r_{f} = forward rate constant k_{f} a copies of A [A]^{a} b copies of B [B]^{b}$

$[A]$ means "concentration of $A$ " — how crowded $A$ is, in moles per litre.
$k_{f}$ is a fixed number (at a given temperature) that bundles up how willing a collision is to succeed.
The exponents $a$ and $b$ are the coefficients from Step 1.

WHY the powers? A reaction event needs $a$ molecules of $A$ and $b$ of $B$ to be in the same spot at the same instant. The chance of finding one $A$ there scales with $[A]$ . To find $a$ of them at once, you multiply that chance $a$ times → $[A]^{a}$ . This is the exact same maths as "chance of $a$ coins all landing heads $= (one head)^{a}$ ". Independent lucky meetings multiply.

PICTURE. Denser crowd ⇒ collisions happen more often. Double $[A]$ and every $A$ -slot in the meeting is twice as easy to fill.

Step 3 — The backward rate, by the exact same logic

WHAT. Products collide too, reforming reactants: $backward speed r_{b} = backward rate constant k_{b} c copies of C [C]^{c} d copies of D [D]^{d}$

Same recipe as Step 2, just read right-to-left across the equation.

WHY. Symmetry. If forward collisions build products, backward collisions of those same products rebuild reactants. There is nothing special about "forward" — it's just the direction we happened to write first.

PICTURE. Same river, now watch the crowd flowing back left. Its speed grows as the right bank fills up.

Step 4 — Set the two rates equal (the definition of equilibrium)

WHAT. At equilibrium, forward speed $=$ backward speed: $r_{f} k_{f} [A]^{a} [B]^{b} = r_{b} k_{b} [C]^{c} [D]^{d}$

WHY. "No net change" is exactly " $r_{f} = r_{b}$ ". Molecules still cross the bridge — but for every one going right, one comes back. The concentrations stop moving.

PICTURE. Plot both rates against time. $r_{f}$ starts high (lots of reactants) and falls; $r_{b}$ starts at zero and climbs. They meet — that crossing point is equilibrium, and after it both stay level.

Step 5 — Rearrange into a constant: this is $K_{c}$

WHAT. Move both rate constants to one side, both concentration lumps to the other: $\frac{k _{f}}{k _{b}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

The left side is a ratio of two fixed numbers, so it is itself a fixed number. Name it: $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d} products, powered}{reactants, powered [ A ] ^{a} [ B ] ^{b}}$

Products on top, reactants on bottom.
Each concentration is raised to its coefficient.
The subscript $c$ says "built from concentrations".

WHY only temperature changes it. $K_{c} = k_{f} / k_{b}$ . Adding reactant speeds up both $r_{f}$ and $r_{b}$ until balance returns — the ratio survives. A catalyst boosts $k_{f}$ and $k_{b}$ by the same factor — the ratio is untouched. Only temperature nudges $k_{f}$ and $k_{b}$ by different amounts, so only temperature moves $K_{c}$ . (This is where Q vs K and Le Chatelier's Principle plug in.)

PICTURE. Two towers $k_{f}$ and $k_{b}$ : raise both equally (concentration/catalyst) and their height-ratio is unchanged; raise them unequally (temperature) and the ratio shifts.

Step 6 — Swap concentration for pressure (for gases)

WHAT. For gases we prefer partial pressure $p_{i}$ (easy to read on a gauge) over concentration. The ideal gas law $p V = n R T$ rearranges for one gas $i$ : $p_{i} = \frac{n _{i}}{V} R T = concentration = n_{i} / V [i] R T$

$n_{i}$ — moles of gas $i$ , $V$ — the shared container volume, so $n_{i} / V$ is the concentration $[i]$ .
$R$ — the gas constant, $T$ — absolute temperature (kelvin).
Punchline: for an ideal gas, pressure is just concentration multiplied by $R T$ .

WHY. At fixed $T$ , cramming more moles into the same box raises the pressure in lockstep with concentration. So a pressure ratio and a concentration ratio must be close cousins — they differ only by factors of $R T$ .

PICTURE. A piston box: same volume, more molecules ⇒ higher gauge reading. The gauge and the crowd-density rise together.

Now the pressure constant, built the same shape as $K_{c}$ : $K_{p} = \frac{p _{C}^{c} p _{D}^{d}}{p _{A}^{a} p _{B}^{b}}$

Step 7 — Substitute and collect the $R T$ factors

WHAT. Replace every $p_{i}$ with $[i] R T$ inside $K_{p}$ : $K_{p} = \frac{([ C ] R T ) ^{c} ([ D ] R T ) ^{d}}{([ A ] R T ) ^{a} ([ B ] R T ) ^{b}}$

Each bracket carries its own $R T$ . Pull every $R T$ out front. On top we get $R T$ raised to $(c + d)$ ; on the bottom $R T$ raised to $(a + b)$ . Dividing subtracts exponents: $K_{p} = = K_{c} \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} \cdot (R T)^{(c + d) - (a + b)}$

WHY separate them? Because the concentration lump is exactly $K_{c}$ from Step 5. Everything else is bookkeeping on the $R T$ tags — and the net power of $R T$ is just "how many gas moles the top has minus how many the bottom has".

PICTURE. Sort a pile of $R T$ tokens: pull the tokens from products to a "top" bin, from reactants to a "bottom" bin, then cancel matching pairs. Whatever's left over is the net exponent.

Name the leftover exponent: $Δ n_{g} = (c + d) - (a + b) = (gaseous moles of products) - (gaseous moles of reactants)$

Step 8 — The edge cases (every sign of $Δ n_{g}$ , and solids)

Never leave the reader in a scenario you didn't show. Here is every possibility for the sign of $Δ n_{g}$ , plus the degenerate solid case.

Case	Example	$Δ n_{g}$	Result
More gas made	$P C l_{5} ⇌ P C l_{3} + C l_{2}$	$2 - 1 = + 1$	$K_{p} = K_{c} (R T)^{+ 1} > K_{c}$ (if $R T > 1$ )
Gas moles equal	$H_{2} + I_{2} ⇌ 2 H I$	$2 - 2 = 0$	$(R T)^{0} = 1$ , so $K_{p} = K_{c}$
Fewer gas moles	$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$	$2 - 4 = - 2$	$K_{p} = K_{c} (R T)^{- 2} < K_{c}$ (if $R T > 1$ )
Degenerate: solids drop out	$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$	$1 - 0 = + 1$	only $C O_{2}$ counts: $K_{p} = p_{C O_{2}}$

WHY $Δ n_{g} = 0$ collapses the bridge: anything to the power zero is $1$ , so $K_{p}$ and $K_{c}$ become numerically identical (Example 2 in the parent, $K_{c} = 16$ ).
WHY solids leave no token: a pure solid's "concentration" is its fixed density, a constant. It never appeared as $[i]$ in Step 5, so it brings no $R T$ in Step 7. In $C a C O_{3}$ , both $C a C O_{3} (s)$ and $C a O (s)$ vanish; the equilibrium hinges on a single gas pressure.

The one-picture summary

$collisions \Rightarrow r_{f} = k_{f} [A]^{a} [B]^{b}, r_{b} = k_{b} [C]^{c} [D]^{d} r_{f} = r_{b} K_{c} = \frac{k _{f}}{k _{b}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} p = [] R T K_{p} = K_{c} (R T)^{Δ n_{g}}$

Recall Feynman retelling — the whole walkthrough in plain words

Picture a river with a bridge. On the left bank, reactant-people. On the right, product-people. They keep crossing both ways. How fast the left-to-right crossing goes depends on how crowded the left bank is — and because you need several of them to link arms and cross together, you multiply the crowdedness once for each person needed (that's the powers). Same story for the crowd crossing back. Wait long enough and both crossings move at the same speed: nobody's pile changes anymore, even though the bridge is busy. Take "product-crowd stuff on top, reactant-crowd stuff on bottom, each raised to its number" and you get one frozen number — call it $K_{c}$ . It only cares about temperature, because warming the river helps one direction cross more than the other; adding people or a helpful bridge-guard (catalyst) speeds both sides equally, so the ratio doesn't budge. For gases we'd rather read a pressure gauge than count crowds — and for a gas, pressure is just crowdedness times $R T$ . Swap every crowd for its gauge reading, tidy up the $R T$ bits, and the leftover power of $R T$ is simply "how many gas-people the top has minus the bottom." That count is $Δ n_{g}$ , and the whole story lands on $K_{p} = K_{c} (R T)^{Δ n_{g}}$ . Solids? They're boulders sitting on the bank — always the same, never counted.

Quick self-check

Fill from equilibrium

the two crossing rates

r_{f}

and

r_{b}

become equal, concentrations freeze

Why concentrations carry powers in the rate

a reaction event needs several molecules at once; independent meeting-chances multiply

K_{c}

from kinetics

K_{c} = k_{f} / k_{b} = [C]^{c} [D]^{d} / [A]^{a} [B]^{b}

Why only temperature moves

K_{c}

concentration and catalyst change

k_{f}, k_{b}

equally; only

T

changes them unequally

Pressure of an ideal gas in terms of concentration

p_{i} = [i] R T

Leftover

R T

power after substitution

Δ n_{g} =

gas moles of products

-

gas moles of reactants

Δ n_{g}

for

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

+ 1

(only

C O_{2}

; both solids ignored)

When

K_{p} = K_{c}

when

Δ n_{g} = 0

Connections

← Back to parent: Law of mass action and Kc, Kp
Chemical Equilibrium – Dynamic Nature
Reaction Quotient Q vs K
Le Chatelier's Principle
Rate Constants and Arrhenius Equation
Gibbs Free Energy and K
Ionic Equilibrium – Ka, Kb, Kw