Relationship Kp = Kc(RT)^Δn: Visual walkthrough — Chemistry deep dive

Step 1 — Two ways to describe one gas in a box

WHAT. Picture a sealed box. Inside float molecules of one gas we call $X$ . There are two honest ways to say "how much $X$ is here."

Concentration, written $[X]$ : how crowded the box is — molecules (in moles) per litre. Symbol $[X]$ means "moles of $X$ divided by the volume."
Partial pressure, written $p_{X}$ : how hard the $X$ molecules push on the walls, all on their own, ignoring every other gas.

WHY start here. Before we can relate two equilibrium constants, we must relate the two quantities they are built from: concentration and pressure. If we can't connect $[X]$ to $p_{X}$ for one gas, we have no hope of connecting $K_{c}$ to $K_{p}$ .

PICTURE. The same box, two labels: on the left we count dots-per-litre (crowd), on the right we count wall-hits (push).

Step 2 — The ideal gas law connects crowd to push

WHAT. Grab the ideal gas law and apply it to gas $X$ alone: $p_{X} V = n_{X} R T$

Term by term, right where it sits:

$p_{X}$ — the push of gas $X$ on the walls.
$V$ — the box volume the gas fills.
$n_{X}$ — moles of $X$ present.
$R$ — the gas constant, a fixed number that converts between the two languages.
$T$ — absolute temperature in kelvin (never Celsius).

WHY this tool and not another. We need a bridge whose two ends are pressure and (moles, volume) — because $n_{X} / V$ is exactly concentration. The ideal gas law is the one law that puts $p$ , $n$ , and $V$ in a single equation. No other relation ties push directly to crowd. That is precisely the connection we are missing.

PICTURE. Two vertical bars: the "push" bar ( $p_{X}$ ) rises in lock-step with the "crowd" bar ( $n_{X} / V$ ), the arrow between them labelled $\times R T$ .

Step 3 — Rewrite it as "push = crowd × $R T$ "

WHAT. Divide both sides of $p_{X} V = n_{X} R T$ by $V$ : $p_{X} = \frac{n _{X}}{V} R T = [X] R T$

$\frac{n _{X}}{V}$ becomes $[X]$ — we just renamed "moles per volume" as concentration.
$R T$ is the exact scale factor between push and crowd.

WHY. This is the sentence we will use over and over: every partial pressure is its concentration multiplied by $R T$ . It is the whole engine of the derivation compressed into one equation.

PICTURE. The clean identity $p_{X} = [X] R T$ shown as a conversion machine: feed in a concentration, turn the $R T$ crank, out comes a pressure.

Step 4 — Meet the two constants for a real reaction

WHAT. Take the general gas reaction $a A + b B ⇌ c C + d D$ The small letters $a, b, c, d$ are stoichiometric coefficients — how many moles of each gas take part. From the law of mass action we build two equilibrium constants: $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}, K_{p} = \frac{p _{C}^{c} p _{D}^{d}}{p _{A}^{a} p _{B}^{b}}$

Top row = products ( $C$ , $D$ ), each raised to its coefficient.
Bottom row = reactants ( $A$ , $B$ ), each raised to its coefficient.
Left uses concentrations $[]$ ; right uses pressures $p$ . See Equilibrium Constant Kp.

WHY. Both constants describe the same jar at the same equilibrium — one in crowd-language, one in push-language. Because they describe one physical state, they cannot be independent. Step 3 is exactly the dictionary that translates one into the other.

PICTURE. Two fraction-towers side by side, products stacked over reactants: the $K_{c}$ tower made of $[]$ blocks, the $K_{p}$ tower made of $p$ blocks.

Step 5 — Translate every pressure into concentration

WHAT. In $K_{p}$ , replace each $p$ with $[X] R T$ from Step 3: $K_{p} = \frac{( [ C ] R T ) ^{c} ( [ D ] R T ) ^{d}}{( [ A ] R T ) ^{a} ( [ B ] R T ) ^{b}}$

Look at one factor to see what happens: $([C] R T)^{c} = [C]^{c} (R T)^{c}$ . The power hits both the concentration and the $R T$ .

WHY. We are converting the entire $K_{p}$ tower into crowd-language so it can be compared, block for block, with the $K_{c}$ tower. Every $p$ carries one $R T$ into the expression.

PICTURE. The $K_{p}$ tower from Step 4 with each $p$ block "cracking open" into a $[X]$ block plus an $R T$ tag glued beside it.

Step 6 — Sort the concentrations from the $R T$ tags

WHAT. Pull all the concentration parts into one group and all the $R T$ tags into another: $K_{p} = = K_{c} \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} \times \frac{( R T ) ^{c} ( R T ) ^{d}}{( R T ) ^{a} ( R T ) ^{b}}$

The concentration group is exactly $K_{c}$ (compare with Step 4) — no coincidence, we built it that way.
The $R T$ group: multiply powers on top, divide on the bottom. Using the exponent rule $(R T)^{c} (R T)^{d} = (R T)^{c + d}$ and dividing subtracts exponents: $\frac{( R T ) ^{c + d}}{( R T ) ^{a + b}} = (R T)^{(c + d) - (a + b)}$

WHY. Separating the two groups makes the leftover visible. Everything that is $K_{c}$ collapses into the $K_{c}$ box; everything else is a single power of $R T$ . That surviving exponent is the entire story.

PICTURE. The mixed tower splits into two clean stacks: a $K_{c}$ box on the left, and an $R T$ ledger on the right with $+ c, + d$ credits on top and $- a, - b$ debits on the bottom.

Step 7 — The result

WHAT. Put the two groups back together: $K_{p} = K_{c} (R T)^{Δ n}$

$K_{c}$ — the concentration group we recognised.
$(R T)^{Δ n}$ — the leftover $R T$ tags, stacked to the net change in gas moles.

WHY it makes sense. The only difference between describing equilibrium by push versus by crowd is a factor of $R T$ per gas molecule, and molecules on the product and reactant sides partly cancel. What survives is one $R T$ for each net new gas molecule — that count is $Δ n$ .

PICTURE. The final one-liner with $K_{c}$ and the $(R T)^{Δ n}$ badge fused together.

Step 8 — The three cases (nothing is left uncovered)

WHAT. The sign of $Δ n$ decides how $K_{p}$ compares with $K_{c}$ , assuming $R T > 1$ (true for all ordinary temperatures, since $R T \approx 24$ – $40$ in atm·L units).

$Δ n$	$(R T)^{Δ n}$	Comparison	Reaction feel
$Δ n = 0$	$(R T)^{0} = 1$	$K_{p} = K_{c}$	equal gas moles both sides
$Δ n > 0$	$> 1$	$K_{p} > K_{c}$	net gas produced
$Δ n < 0$	$< 1$	$K_{p} < K_{c}$	net gas consumed

WHY the edge case $Δ n = 0$ matters. When products and reactants have the same number of gas molecules, every $R T$ tag cancels — top against bottom — and the two constants become literally equal. This is the degenerate case where push-language and crowd-language coincide exactly. Example: $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ has $Δ n = 2 - 2 = 0$ , so $K_{p} = K_{c}$ at every temperature.

Degenerate species. Pure solids and liquids are not counted in $Δ n$ — they have no partial pressure and no concentration term in $K$ . For $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ , only $C O_{2}$ is a gas, so $Δ n = 1 - 0 = + 1$ .

PICTURE. Three number-lines of $(R T)^{Δ n}$ : one landing exactly on $1$ (equal), one climbing above $1$ (bigger), one dropping below $1$ (smaller).

Worked mini-check (following the pictures)

The one-picture summary

Everything above, compressed: one gas gives $p_{X} = [X] R T$ ; feed that into the $K_{p}$ tower; the concentrations regroup into $K_{c}$ ; the $R T$ tags collapse into $(R T)^{Δ n}$ ; the sign of $Δ n$ sets the three cases.

Recall Feynman: the whole walk in plain words

Put a bunch of gases in a box. You can measure each gas two ways: how crowded it is (concentration) or how hard it pushes (pressure). The ideal gas law is a translator — push equals crowd times $R T$ , one clean sentence. The equilibrium constant is just "products over reactants." Write it in push-language ( $K_{p}$ ) and swap every push for crowd-times- $R T$ . All the crowds line up and rebuild the concentration version ( $K_{c}$ ) exactly. What's left over is a pile of $R T$ 's — one for every gas molecule. Molecules on the two sides of the arrow partly cancel, so the leftover pile is just the net change in gas molecules: that's $Δ n$ . Hence $K_{p} = K_{c} (R T)^{Δ n}$ . If both sides have the same number of gas molecules the pile is empty, $(R T)^{0} = 1$ , and the two constants are twins.

Recall Quick self-test

Why does only $Δ n$ (not the individual coefficients) appear in the final formula? ::: Because the $R T$ tags on top and bottom of the fraction subtract; only the net difference $(c + d) - (a + b)$ survives. In $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ , what is $Δ n$ ? ::: $+ 1$ — only the gas $C O_{2}$ counts; solids carry no $R T$ tag. When are $K_{p}$ and $K_{c}$ exactly equal? ::: When $Δ n = 0$ , so $(R T)^{0} = 1$ and every $R T$ cancels.

Connections

Parent topic — the summary version of this derivation.
Ideal Gas Law PV=nRT — the translator used in Steps 2–3.
Partial Pressure and Dalton's Law — what $p_{X}$ means.
Law of Chemical Equilibrium — where the $K_{c}$ tower comes from.
Equilibrium Constant Kp — the partial-pressure form.
Le Chatelier's Principle — why $Δ n \neq = 0$ reactions respond to pressure.
Gibbs Free Energy and K — $Δ G^{\circ} = - R T ln K$ .