Relationship Kp = Kc(RT)^Δn: Worked examples — Chemistry deep dive

The scenario matrix

Before solving anything, let us list every case class that this one formula can produce. Each row is a distinct trap or twist; each worked example below is labelled with the cell it covers.

Cell	Case class	What makes it different	Example
A	$Δ n > 0$	gas moles increase → $(R T)^{Δ n} > 1$ → $K_{p} > K_{c}$	Ex 1
B	$Δ n < 0$	gas moles decrease → $(R T)^{Δ n} < 1$ → $K_{p} < K_{c}$	Ex 2
C	$Δ n = 0$	$R T$ cancels → $K_{p} = K_{c}$ exactly	Ex 3
D	Solid/liquid present	only gases count in $Δ n$	Ex 4
E	Reverse direction	given $K_{p}$ , find $K_{c}$ → divide, not multiply	Ex 5
F	Units switch (bar / SI)	must pick the matching $R$	Ex 6
G	Real-world word problem	translate words → equation → $Δ n$	Ex 7
H	Exam twist: degrees Celsius + big $Δ n$	convert $T$ ; watch the exponent	Ex 8

Two symbols we lean on throughout, restated in plain words so nothing is assumed:

Below, a picture of the whole matrix so you can see how the sign of $Δ n$ pushes $K_{p}$ up or down relative to $K_{c}$ .

The red curve is $(R T)^{Δ n}$ as $Δ n$ sweeps from negative to positive at a fixed $R T > 1$ . Left of centre ( $Δ n < 0$ ) the factor is below the dashed line $y = 1$ , so $K_{p} < K_{c}$ (cell B). Exactly at $Δ n = 0$ it crosses $1$ , so $K_{p} = K_{c}$ (cell C). Right of centre ( $Δ n > 0$ ) it rises above $1$ , so $K_{p} > K_{c}$ (cell A). One picture, all three sign cases.

Worked examples

Cell A — $Δ n > 0$

Worked example Ex 1 — Dinitrogen tetroxide dissociating

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ at $T = 298 K$ , with $K_{c} = 5.7 \times 1 0^{- 3}$ . Find $K_{p}$ . Use $R = 0.0821 L atm K^{- 1} mol^{- 1}$ .

Forecast: one gas becomes two gases, so gas moles increase. Guess: will $K_{p}$ be bigger or smaller than $K_{c}$ ? (Bet on bigger.)

Count gas moles. Products: $2$ (the $2 N O_{2}$ ). Reactants: $1$ (the $N_{2} O_{4}$ ). So $Δ n = 2 - 1 = + 1$ . Why this step? The exponent of $R T$ is $Δ n$ ; nothing else can be computed until we have it.
Compute $R T$ . $R T = 0.0821 \times 298 = 24.47$ . Why this step? We need the numeric value of the bridge factor $R T$ before raising it to $Δ n$ .
Apply the formula. $K_{p} = K_{c} (R T)^{Δ n} = 5.7 \times 1 0^{- 3} \times (24.47)^{1} = 0.1395$ . Why this step? Exponent $+ 1$ means one factor of $R T$ , so we simply multiply once.

Verify: $Δ n > 0$ and $R T > 1$ , so $(R T)^{Δ n} = 24.47 > 1$ , hence $K_{p} > K_{c}$ . Indeed $0.1395 > 0.0057$ . Forecast confirmed. ✔

Cell B — $Δ n < 0$

Cell C — $Δ n = 0$

Cell D — solid/liquid present

Worked example Ex 4 — Limestone decomposition (don't count the solids!)

$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ at $T = 1000 K$ , $K_{c} = 0.05$ . Find $K_{p}$ . $R = 0.0821$ .

Forecast: it looks like $2$ products and $1$ reactant, but two of those are solids. Guess: only the $C O_{2}$ matters.

Strike out non-gases. $C a C O_{3} (s)$ and $C a O (s)$ are solids — they do not appear in $K$ at all, and they do not count in $Δ n$ . Why this step? The derivation replaced $p_{X}$ by $[X] R T$ only for gases; solids have no meaningful partial pressure to substitute.
Count only gases. Gaseous products $= 1$ ( $C O_{2}$ ); gaseous reactants $= 0$ . So $Δ n = 1 - 0 = + 1$ .
Compute $R T$ and apply. $R T = 0.0821 \times 1000 = 82.1$ ; $K_{p} = 0.05 \times (82.1)^{1} = 4.105$ .

Verify: If a student wrongly wrote $Δ n = 2 - 1 = 1$ they'd coincidentally get the same exponent here — but for the wrong reason. Change the reaction (e.g. two gaseous products) and the wrong method fails. Our count uses gases only. $4.105 > 0.05$ , consistent with $Δ n > 0$ . ✔

Cell E — reverse direction ( $K_{p} \to K_{c}$ )

Worked example Ex 5 — Given

$K_{p}$ , find $K_{c}$ For $2 N O (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$ at $T = 300 K$ , we measured $K_{p} = 1.0 \times 1 0^{6}$ . Find $K_{c}$ . $R = 0.0821$ .

Forecast: we run the formula backwards. Since gas moles decrease ( $Δ n < 0$ ), and going forward $K_{p} < K_{c}$ , going backward we expect $K_{c} > K_{p}$ — even bigger.

Rearrange the formula. From $K_{p} = K_{c} (R T)^{Δ n}$ , divide both sides: $K_{c} = \frac{K _{p}}{( R T ) ^{Δ n}} = K_{p} (R T)^{- Δ n}$ . Why this step? We are solving for the unknown; algebra flips the exponent's sign.
Find $Δ n$ . Products $= 2$ ; reactants $= 2 + 1 = 3$ ; $Δ n = 2 - 3 = - 1$ . So $- Δ n = + 1$ .
Compute. $R T = 0.0821 \times 300 = 24.63$ . Then $K_{c} = 1.0 \times 1 0^{6} \times (24.63)^{+ 1} = 2.463 \times 1 0^{7}$ . Why this step? With $- Δ n = + 1$ we multiply by $R T$ once.

Verify: Plug back forward: $K_{p} = K_{c} (R T)^{Δ n} = 2.463 \times 1 0^{7} \times (24.63)^{- 1} = 1.0 \times 1 0^{6}$ . ✔ Recovers the given $K_{p}$ , so the rearrangement is right, and $K_{c} > K_{p}$ as forecast.

Cell F — units switch (bar / SI)

Worked example Ex 6 — Same reaction, pressure in bar

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ at $T = 500 K$ , $K_{c} = 2.0 \times 1 0^{- 2}$ (mol L⁻¹). Find $K_{p}$ in bar. Use $R = 0.083 L bar K^{- 1} mol^{- 1}$ .

Forecast: identical reaction to the parent's Example 2, but now pressure is measured in bar, so we swap $R$ . Answer should be close to the atm value ( $0.821$ ) but slightly bigger, since $R_{bar} > R_{atm}$ .

Match $R$ to the units. $K_{p}$ is wanted in bar, so use $R = 0.083 L bar K^{- 1} mol^{- 1}$ . Why this step? $R$ is the exchange rate between concentration and pressure; it must carry the same pressure unit you want the answer in.
Find $Δ n$ . Products $= 2$ , reactants $= 1$ ; $Δ n = + 1$ .
Compute. $R T = 0.083 \times 500 = 41.5$ ; $K_{p} = 2.0 \times 1 0^{- 2} \times (41.5)^{1} = 0.830 bar$ .

Verify: The atm answer was $0.821$ ; the bar answer $0.830$ is slightly larger because $0.083 > 0.0821$ . The small ratio $0.083/0.0821 \approx 1.011$ matches $0.830/0.821 \approx 1.011$ . ✔ Physically the equilibrium is the same; only the yardstick (bar vs atm) changed.

Cell G — real-world word problem

Worked example Ex 7 — Industrial ammonia reactor (translate the words)

An engineer measures the equilibrium constant in concentration terms for the Haber process, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , as $K_{c} = 0.50$ at $T = 700 K$ . The plant's pressure gauges read in atm. She needs $K_{p}$ to predict the pressure balance. Find it. $R = 0.0821$ .

Forecast: four gas moles shrink to two — a big drop. Expect $K_{p}$ noticeably smaller than $K_{c}$ .

Turn the story into an equation and $Δ n$ . The balanced reaction gives products $= 2$ , reactants $= 1 + 3 = 4$ ; $Δ n = 2 - 4 = - 2$ . Why this step? Word problems hide the numbers inside phrases like "the Haber process"; you must recover the balanced equation to read $Δ n$ .
Compute $R T$ and its power. $R T = 0.0821 \times 700 = 57.47$ ; $(R T)^{- 2} = 1/57.4 7^{2} = 1/3302.8 = 3.028 \times 1 0^{- 4}$ . Why this step? Exponent $- 2$ means divide by $R T$ twice — a strongly shrinking factor.
Apply. $K_{p} = 0.50 \times 3.028 \times 1 0^{- 4} = 1.514 \times 1 0^{- 4} atm^{- 2}$ .

Verify: $Δ n = - 2 < 0$ so $K_{p} ≪ K_{c}$ : indeed $1.5 \times 1 0^{- 4} ≪ 0.50$ . ✔ This tells the engineer that in pressure terms the constant is tiny — consistent with why the plant runs at very high pressure to push the equilibrium toward $N H_{3}$ (see Le Chatelier's Principle).

Cell H — exam twist: Celsius input + large $Δ n$

Worked example Ex 8 — The classic "27 °C" trap with a large exponent

$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ at $27^{\circ} C$ , with $K_{c} = 8.0 \times 1 0^{- 3}$ . Find $K_{p}$ . $R = 0.0821$ .

Forecast: two traps stacked — (a) the temperature is in Celsius, and (b) two gas moles expand to four, so $Δ n$ is a large positive number. Expect $K_{p}$ much bigger than $K_{c}$ .

Convert temperature to kelvin. $T = 27 + 273 = 300 K$ . Why this step? The gas law and hence this formula are only valid in absolute temperature; plugging $27$ would understate $R T$ by a factor of $10$ and wreck the answer.
Find $Δ n$ . Products $= 1 + 3 = 4$ ; reactants $= 2$ ; $Δ n = 4 - 2 = + 2$ . Why this step? This is the reverse of ammonia synthesis, so the sign flips from $- 2$ to $+ 2$ — a useful cross-check against Ex 7's direction.
Compute. $R T = 0.0821 \times 300 = 24.63$ ; $(R T)^{2} = 606.6$ ; $K_{p} = 8.0 \times 1 0^{- 3} \times 606.6 = 4.85$ .

Verify: Two consistency checks. (i) $Δ n > 0$ so $K_{p} > K_{c}$ : $4.85 > 0.008$ . ✔ (ii) This reaction is exactly the reverse of Ex 7's; reversing a reaction inverts $K$ and flips the sign of $Δ n$ , and both effects appear (positive $Δ n$ here vs negative there). ✔

Recall Quick self-test on the matrix

Which cell does each belong to, and multiply or divide by $R T$ ?

$2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g)$ : which cell? ::: Cell A, $Δ n = + 1$ , multiply by $R T$ once. $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ : which cell? ::: Cell C, $Δ n = 0$ , $K_{p} = K_{c}$ . $C (s) + O_{2} (g) ⇌ C O_{2} (g)$ : what is $Δ n$ ? ::: $Δ n = 1 - 1 = 0$ (carbon is solid, not counted). Given $K_{p}$ and asked for $K_{c}$ with $Δ n = + 1$ : multiply or divide by $R T$ ? ::: Divide ( $K_{c} = K_{p} / (R T)$ ). $T = 25^{\circ} C$ — what value goes into $R T$ ? ::: $T = 298 K$ .

Connections

Parent: Kp = Kc(RT)^Δn — the formula these examples exercise.
Equilibrium Constant Kp — the pressure-based constant used throughout.
Law of Chemical Equilibrium — origin of $K_{c}$ .
Ideal Gas Law PV=nRT — supplies the $R T$ bridge and the choice of $R$ .
Partial Pressure and Dalton's Law — why gaseous species alone enter $Δ n$ .
Le Chatelier's Principle — the plant-pressure reasoning in Ex 7.
Gibbs Free Energy and K — deeper "why $K$ is dimensionless" thread.