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

Formula WHY the bridge exists — a 3-line derivation you can redo on this page

This page is self-contained: here is why the two constants are linked, so you never have to leave. Step 1 — one gas at a time. The ideal gas law $p V = n R T$ applied to just gas $X$ gives $p_{X} V = n_{X} R T \Rightarrow p_{X} = \frac{n _{X}}{V} R T = [X] R T .$ WHY: concentration $[X]$ is moles per volume $n_{X} / V$ , so pressure is literally "concentration times $R T$ ." Push $=$ crowd $\times R T$ . Step 2 — feed this into $K_{p}$ . Replace every $p_{X}$ in $K_{p}$ with $[X] R T$ : $K_{p} = \frac{([ C ] R T ) ^{c} ([ D ] R T ) ^{d}}{([ A ] R T ) ^{a} ([ B ] R T ) ^{b}} .$ WHY: we want $K_{p}$ written only in concentrations so we can compare it with $K_{c}$ . Step 3 — separate the two kinds of factor. Pull the concentrations together (they are $K_{c}$ ) and collect all the $R T$ 's: $K_{p} = = K_{c} \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} \times (R T)^{(c + d) - (a + b)} = K_{c} (R T)^{Δ n} .$ The leftover power of $R T$ is (product powers) − (reactant powers) $= Δ n$ . That is the whole formula — every $R T$ that doesn't cancel is one extra gas mole on the product side: $K_{p} = K_{c} (R T)^{Δ n}$

Recall The four things you need at your fingertips

Formula reveal ::: $K_{p} = K_{c} (R T)^{Δ n}$ Meaning of $Δ n$ ::: gaseous products minus gaseous reactants (count ONLY gases) Temperature unit ::: always kelvin, $T (K) = T (^{\circ} C) + 273$ $R$ for atm & mol L⁻¹ ::: $0.0821 L atm K^{- 1} mol^{- 1}$ $R$ for the energy equation $Δ G^{\circ} = - R T ln K$ ::: $8.314 J K^{- 1} mol^{- 1}$

Figure 1 — how the sign of $Δ n$ decides the $K_{p}$ vs $K_{c}$ comparison.

Alt-text / caption: A chalkboard bar chart. The horizontal axis lists five values of $Δ n$ from $- 2$ to $+ 2$ ; each bar is the resulting $K_{p}$ when $K_{c} = 1$ and $R T \approx 24.6$ (that is $R = 0.0821$ at $T = 300 K$ ). The vertical axis is on a log scale so the huge range fits. The white dashed line marks $K_{c} = 1$ : bars below it (pink, $Δ n < 0$ ) are cases where $K_{p} < K_{c}$ ; the middle bar (yellow, $Δ n = 0$ ) sits exactly on the line, $K_{p} = K_{c}$ ; bars above it (blue, $Δ n > 0$ ) are $K_{p} > K_{c}$ . How to use Figure 1 in an exercise: find your $Δ n$ on the axis, read whether that bar is below/on/above the dashed line, and you instantly know the direction of the comparison — provided $R T > 1$ , which every worked example on this page satisfies except the deliberately-constructed Exercise 5.5 (see the L1 mistake box for the $R T < 1$ caveat).

Level 1 — Recognition

(Can you read off $Δ n$ and pick the right relationship without a calculator?)

Recall Solution 1.1

WHAT we do: count gas moles on each side. Products: $2$ (that's the $2 S O_{3}$ ). Reactants: $2 + 1 = 3$ (the $2 S O_{2}$ and $1 O_{2}$ ). WHY: $Δ n$ is always products − reactants, gases only. $Δ n = 2 - 3 = - 1$ Since $Δ n < 0$ and $R T > 1$ , the factor $(R T)^{- 1} < 1$ , so $K_{p} < K_{c}$ . (On Figure 1 this is a pink bar below the dashed line.) Answer ::: $Δ n = - 1$ , and $K_{p} < K_{c}$ .

Recall Solution 1.2

Products: $2$ . Reactants: $1 + 1 = 2$ . $Δ n = 2 - 2 = 0 \Rightarrow (R T)^{0} = 1 \Rightarrow K_{p} = K_{c}$ WHY it matters: whenever the number of gas molecules doesn't change across the arrow, the two constants are numerically identical — the $R T$ 's cancel perfectly. Notice this conclusion needs no assumption about whether $R T$ is above or below $1$ , because anything to the power $0$ is $1$ . (Figure 1: the yellow bar sitting exactly on the dashed line.) Answer ::: $Δ n = 0$ , so $K_{p} = K_{c}$ .

Recall Solution 1.3

Compute $Δ n$ for each:

(a) $2 - 1 = + 1$ → $Δ n > 0$ → (with $R T > 1$ ) $K_{p} > K_{c}$ ✔
(b) $2 - 4 = - 2$ → $K_{p} < K_{c}$
(c) $2 - 2 = 0$ → $K_{p} = K_{c}$ Answer ::: only (a) gives $K_{p} > K_{c}$ .

Common mistake The L1 trap: assuming

$(R T)^{Δ n} > 1$ whenever $Δ n > 0$ Why the wrong path feels right: in every ordinary lab problem $R = 0.0821$ and $T$ is a few hundred kelvin, so $R T \approx 20 - 90$ , comfortably above $1$ . It becomes a reflex to say " $Δ n > 0$ $\Rightarrow K_{p} > K_{c}$ ." Why it can fail: the sign of $Δ n$ only fixes the direction if $R T > 1$ . If the pressure units made $R T < 1$ (e.g. an exotic unit system, or extremely low $T$ so that $R T < 1$ ), then for $Δ n > 0$ the factor $(R T)^{Δ n}$ would be less than $1$ and the inequality would reverse to $K_{p} < K_{c}$ . Only $Δ n = 0$ is safe in all cases (it always gives $K_{p} = K_{c}$ ). Exercise 5.5 makes you practice exactly this reversal. Fix: State the comparison as " $K_{p} > K_{c}$ provided $R T > 1$ ," and always check the actual value of $R T$ before trusting the shortcut. On this page every example except 5.5 has $R T > 1$ , so the shortcut is valid there.

Level 2 — Application

(Plug numbers in; watch units and the sign of the exponent.)

Recall Solution 2.1

Step 1 — $Δ n$ : products $2$ , reactant $1$ , so $Δ n = 2 - 1 = + 1$ . Step 2 — substitute: $K_{p} = K_{c} (R T)^{Δ n} = (4.0 \times 1 0^{- 2}) (0.0821 \times 300)^{1}$ Step 3 — evaluate: $R T = 0.0821 \times 300 = 24.63$ . $K_{p} = 4.0 \times 1 0^{- 2} \times 24.63 = 0.985$ WHY $K_{p} > K_{c}$ ? $Δ n = + 1 > 0$ and $R T = 24.63 > 1$ , so we multiply by something bigger than one. Answer ::: $K_{p} \approx 0.985$ .

Recall Solution 2.2

Step 1 — $Δ n$ : product $1$ , reactants $2$ , $Δ n = 1 - 2 = - 1$ . Step 2 — rearrange the master formula for $K_{c}$ : $K_{p} = K_{c} (R T)^{Δ n} \Rightarrow K_{c} = \frac{K _{p}}{( R T ) ^{Δ n}} = K_{p} (R T)^{- Δ n}$ With $Δ n = - 1$ , this is $K_{c} = K_{p} (R T)^{+ 1}$ . Step 3 — evaluate: $R T = 0.0821 \times 298 = 24.47$ . $K_{c} = 8.8 \times 24.47 = 215.3$ Answer ::: $K_{c} \approx 215$ .

Recall Solution 2.3

$R T = 0.0821 \times 500 = 41.05$ . $K_{p} = K_{c} (R T)^{2} = (1.0 \times 1 0^{- 3}) (41.05)^{2} = (1.0 \times 1 0^{- 3}) (1685.1) = 1.685$ Answer ::: $K_{p} \approx 1.69$ .

Level 3 — Analysis

(Now some species are solids/liquids, or the temperature is disguised.)

Recall Solution 3.1

Step 1 — count ONLY gases: the only gas is $C O_{2}$ (product). $C a C O_{3}$ and $C a O$ are solids and do not enter $K$ or $Δ n$ at all. $Δ n = 1 - 0 = + 1$ Step 2: $R T = 0.0821 \times 1000 = 82.1$ . $K_{p} = K_{c} (R T)^{1} = 0.05 \times 82.1 = 4.105$ Answer ::: $K_{p} \approx 4.11$ .

Recall Solution 3.2

Step 1 — temperature: $2 7^{\circ} C$ is Celsius. Convert: $T = 27 + 273 = 300 K$ . Step 2 — $Δ n$ : gases are $N H_{3}$ and $H C l$ (two products), solid $N H_{4} C l$ ignored. $Δ n = 2 - 0 = + 2$ Step 3: $R T = 0.0821 \times 300 = 24.63$ . $K_{p} = (2.0 \times 1 0^{- 4}) (24.63)^{2} = (2.0 \times 1 0^{- 4}) (606.6) = 0.1213$ Answer ::: $K_{p} \approx 0.121$ .

Recall Solution 3.3

Step 1 — $Δ n$ (gases only): products $= 2$ (the $2 C O$ ), reactant gas $= 1$ (the $C O_{2}$ ); solid carbon excluded. $Δ n = 2 - 1 = + 1$ Step 2 — solve for $K_{c}$ : $K_{c} = K_{p} (R T)^{- Δ n} = K_{p} (R T)^{- 1} = \frac{K _{p}}{R T}$ . $R T = 0.0821 \times 1100 = 90.31$ . $K_{c} = \frac{14.6}{90.31} = 0.1617$ Answer ::: $K_{c} \approx 0.162$ .

Level 4 — Synthesis

(Combine the formula with related equilibrium ideas.)

Recall Solution 4.1

(a) $Δ n = 2 - (1 + 3) = - 2$ . (b) Increasing pressure squeezes the system; by Le Chatelier's Principle equilibrium shifts toward the side with fewer gas moles to relieve the squeeze. That side is the products ( $2$ vs $4$ ), so yield of $N H_{3}$ rises. (c) The negative $Δ n$ is exactly the signature that products occupy fewer gas moles. The same $Δ n < 0$ that makes $(R T)^{Δ n} < 1$ (when $R T > 1$ , so $K_{p} < K_{c}$ ) is what tells Le Chatelier "products are the low-volume side." One number, two consequences. Answer ::: $Δ n = - 2$ ; higher pressure increases $N H_{3}$ yield; negative $Δ n$ marks the fewer-moles (product) side in both frameworks.

Recall Solution 4.2

WHY use $K_{p}$ (not $K_{c}$ ) here, and which $R$ ? The thermodynamic relation $Δ G^{\circ} = - R T ln K$ uses the dimensionless pressure-based constant for gases. To make $K_{p}$ dimensionless you divide each partial pressure by the standard pressure $p^{\circ} = 1 bar$ before forming the ratio:

= K_p\,(p^\circ)^{-\Delta n} \quad\text{with } p^\circ = 1\ \text{bar}.$$ Since $p^\circ = 1\ \text{bar}$ numerically, dividing by it does not change the *number* $8.8$ but it **does** strip the units, leaving a pure dimensionless $K^\circ = 8.8$. That is the value that goes inside $\ln$. Note the deliberate double switch versus earlier problems: those used $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$ to convert $K_p\!\leftrightarrow\!K_c$, but an **energy** result in joules needs $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ (the tiny $1\ \text{atm}$ vs $1\ \text{bar}$ standard-state difference is ignored at exam level). $$\ln(8.8) = 2.1748$$ $$\Delta G^\circ = -(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})(2.1748) = -5388\ \text{J mol}^{-1} \approx -5.39\ \text{kJ mol}^{-1}$$ Because $\Delta G^\circ < 0$, the forward reaction is **spontaneous** in the standard state. Answer ::: $K^\circ = 8.8$ (dimensionless); $\Delta G^\circ \approx -5.39\ \text{kJ mol}^{-1}$, spontaneous.

Level 5 — Mastery

(Multi-step; reverse-engineering; degenerate and limiting cases.)

Recall Solution 5.1

WHAT we need: solve $K_{p} = K_{c} (R T)^{Δ n}$ for the unknown exponent $Δ n$ . WHY logarithms? $Δ n$ sits in an exponent; the tool that pulls an exponent down into a multiplier is the logarithm. That is exactly the question " $Δ n$ is which power of $R T$ ?" Step 1 — isolate the power by dividing both sides by $K_{c}$ : $(R T)^{Δ n} = \frac{K _{p}}{K _{c}} = \frac{25.0}{0.0225} = 1111.1$ Step 2 — take the natural log of both sides. Using $ln ((R T)^{Δ n}) = Δ n ln (R T)$ : $Δ n ln (R T) = ln (1111.1)$ Step 3 — isolate $Δ n$ by dividing by $ln (R T)$ . With $R T = 0.0821 \times 400 = 32.84$ : $Δ n = \frac{l n ( 1111.1 )}{l n ( 32.84 )} = \frac{7.0131}{3.4917} = 2.008 \approx 2$ Step 4 — round to an integer. $Δ n$ must be a whole number (a difference of integer mole counts); $2.008$ rounds cleanly to $Δ n = 2$ . (The tiny excess over $2$ comes from the problem's rounded input $K_{c} = 0.0225$ ; the exact $K_{c}$ giving $Δ n = 2$ is $25.0/32.8 4^{2} \approx 0.02318$ .) Answer ::: $Δ n = + 2$ .

Recall Solution 5.2

Step 1 — $Δ n$ : products $2$ , reactants $2$ , $Δ n = 0$ . Step 2 — the degenerate case: $(R T)^{0} = 1$ for every value of $R T$ . Temperature drops out of the conversion entirely. $K_{p} = K_{c} \times 1 = 4.0 at 600 K and 900 K, and any T$ WHY this matters: for $Δ n = 0$ reactions, pressure- and concentration-based constants are the same number regardless of $T$ . (The value of $K_{c}$ itself can still change with $T$ through thermodynamics — but the $K_{p} = K_{c}$ equality is temperature-proof.) Answer ::: $K_{p} = 4.0$ at both temperatures (and all $T$ ), because $Δ n = 0$ .

Recall Solution 5.3

Step 1 — strip non-gases: $H_{2} O_{2} (l)$ and $H_{2} O (l)$ are liquids → excluded. The only gas is $O_{2}$ (product, coefficient $1$ ). $Δ n = 1 - 0 = + 1$ Step 2: $R T = 0.0821 \times 298 = 24.47$ . $K_{p} = K_{c} (R T)^{1} = 3.0 \times 24.47 = 73.4$ Answer ::: $K_{p} \approx 73.4$ .

Recall Solution 5.4

(a) $Δ n = 2 - 1 = + 1$ . $R T_{1} = 0.0821 \times 250 = 20.525$ . $K_{c} = K_{p} (R T_{1})^{- 1} = \frac{0.500}{20.525} = 0.02436$ (b) At $500 K$ : $R T_{2} = 0.0821 \times 500 = 41.05$ . $K_{p}^{'} = K_{c} (R T_{2})^{+ 1} = 0.02436 \times 41.05 = 1.000$ Comment: with $Δ n = + 1$ , doubling $T$ (from $250$ to $500 K$ ) doubles $R T$ , hence doubles the conversion factor $(R T)^{1}$ — so $K_{p}$ doubles from $0.500$ to $1.000$ , even if $K_{c}$ held fixed. This isolates the pure " $R T$ -conversion" effect from the deeper thermodynamic temperature dependence. Answer ::: (a) $K_{c} \approx 0.0244$ ; (b) $K_{p}^{'} \approx 1.00$ — the conversion factor scales linearly with $T$ when $Δ n = + 1$ .

Recall Solution 5.5

Step 1 — the factor. $Δ n = + 1$ , so $(R T)^{Δ n} = (0.40)^{1} = 0.40$ . Step 2 — evaluate. $K_{p} = K_{c} (R T)^{Δ n} = 5.0 \times 0.40 = 2.0$ Step 3 — compare. $K_{p} = 2.0 < K_{c} = 5.0$ . So $K_{p}$ is SMALLER than $K_{c}$ even though $Δ n > 0$ . WHY the shortcut fails here: the shortcut " $Δ n > 0 \Rightarrow K_{p} > K_{c}$ " secretly assumes $R T > 1$ (so that raising it to a positive power makes it bigger). Here $R T = 0.40 < 1$ , and raising a number below $1$ to a positive power makes it smaller — so the inequality flips. This is exactly the L1 trap made concrete. On Figure 1 the bars assume $R T \approx 24.6 > 1$ ; that picture would look upside-down for $R T < 1$ . Answer ::: $K_{p} = 2.0$ , which is smaller than $K_{c} = 5.0$ ; the shortcut does not hold because $R T < 1$ .

#recall check

Recall Rapid self-quiz (reveal after answering)

Solve for $K_{c}$ from $K_{p}$ : what exponent on $R T$ ? ::: $- Δ n$ (the negative of $Δ n$ ). $N H_{4} C l (s) ⇌ N H_{3} (g) + H C l (g)$ , $Δ n = ?$ ::: $+ 2$ (only the two product gases count). When is $K_{p} = K_{c}$ at all temperatures? ::: when $Δ n = 0$ , because $(R T)^{0} = 1$ for every $T$ . Which $R$ for $Δ G^{\circ} = - R T ln K$ ? ::: $8.314 J K^{- 1} mol^{- 1}$ . How do you pull $Δ n$ out of an exponent? ::: take logarithms: $Δ n = ln (K_{p} / K_{c}) / ln (R T)$ . When can $Δ n > 0$ still give $K_{p} < K_{c}$ ? ::: if $R T < 1$ , so the shortcut needs $R T > 1$ to hold.

Connections

Parent topic — the derivation
Law of Chemical Equilibrium — origin of $K_{c}$ .
Equilibrium Constant Kp — the partial-pressure constant.
Ideal Gas Law PV=nRT — the $p_{X} = [X] R T$ bridge every problem uses.
Partial Pressure and Dalton's Law — meaning of each $p_{X}$ .
Le Chatelier's Principle — Exercise 4.1.
Gibbs Free Energy and K — Exercise 4.2.