Relating Kp, Kc, and Kx

Chemists measure equilibrium composition however is convenient: gas pressures, molar concentrations, or mole fractions. Yet the equilibrium constant itself is a single thermodynamic number, defined through activities. This article shows how $K_p$ , $K_c$ , and $K_x$ are three different bookkeeping systems for the same equilibrium. Converting between them is not optional trivia. It matters whenever you measure a reaction in one unit system but need a constant in another. It also matters when total pressure changes and you must predict how mole fractions shift. Getting the conversion factors wrong is a classic source of errors in gas-phase equilibrium problems. This article builds the conversions from the ideal gas law and the definition of mole fraction. You will see the general relations. Then you will see where the factors of $RT$ and $P^{\circ}$ come from. Finally, you will work two full numerical examples. One of these includes a change in total pressure.

Key takeaways

$K_p$ , $K_c$ , and $K_x$ describe the same equilibrium using pressures, concentrations, or mole fractions.
They are related through $K_p = K_c(RT)^{\Delta n}$ and $K_p = K_x(P/P^{\circ})^{\Delta n}$ .
$\Delta n$ is the change in moles of gas, products minus reactants, from the balanced equation.
Only $K_p$ (built from activities using $P^{\circ}=1\text{ bar}$ ) is truly dimensionless and temperature-only.
$K_c$ and $K_x$ carry implicit unit or pressure dependence hidden inside the conversion factor.
Key idea: the numerical value of an equilibrium constant depends on the concentration scale used to express it, but all scales trace back to one dimensionless thermodynamic constant via the ideal gas law.

The core insight

Because pressure, concentration, and mole fraction are related to each other through the ideal gas law ( $p_i = c_iRT = x_iP$ ), any equilibrium constant written in one variable can be rewritten in another by inserting these substitutions term by term and collecting the leftover factors into a power of $\Delta n$ , the net change in gas moles for the reaction.

The physical picture

Consider a gas-phase reaction at equilibrium, say the classic ammonia synthesis:

\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)

You can describe the equilibrium mixture in three equally valid languages. You can quote each gas's partial pressure $p_i$ in bar, its molar concentration $c_i = n_i/V$ in mol dm⁻³, or its mole fraction $x_i = n_i/n_{\text{total}}$ , a dimensionless number between 0 and 1. These three descriptions are not independent — for an ideal gas they are locked together by

p_i = c_i RT = x_i P

where $P$ is the total pressure of the mixture. Because the three composition variables are simply rescaled versions of one another, the equilibrium constants built from them, $K_p$ , $K_c$ , and $K_x$ , are also simply rescaled versions of one another. The rescaling factor depends only on temperature (and, for $K_x$ , on total pressure), and on how many moles of gas the reaction creates or destroys. The key structural feature is $\Delta n$ , the stoichiometric change in gas moles:

\Delta n = \sum_{\text{products}} \nu_i - \sum_{\text{reactants}} \nu_i

For ammonia synthesis, $\Delta n = 2 - (1+3) = -2$ . When $\Delta n = 0$ — equal moles of gas on each side, as in $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\,\text{HI}$ — all three constants become numerically identical, since every conversion factor collapses to unity.

Figure

Three composition scales for the same equilibrium mixture

The second figure shows how $K_x$ , unlike $K_p$ , changes with total pressure even at fixed temperature — a fact with direct practical consequences for reactions run under compression.

Figure

K_x drifts with total pressure while K_p stays fixed, for Delta n = -2

The core relations

p_i = c_i R T = x_i P

where $p_i$ is partial pressure, $c_i$ is molar concentration, $x_i$ is mole fraction, $P$ is total pressure, $R$ is the gas constant, $T$ is absolute temperature.

K_p = K_c\,(RT)^{\Delta n}

where $K_p$ uses partial pressures (in units matching $P^{\circ}$ ), $K_c$ uses molar concentrations, $\Delta n$ is moles gas product minus reactant.

K_p = K_x\left(\frac{P}{P^{\circ}}\right)^{\Delta n}

where $K_x$ uses mole fractions only, $P$ is total pressure, $P^{\circ}$ is the standard pressure (1 bar).

K_x = K_c\,\frac{(RT)^{\Delta n}}{(P/P^{\circ})^{\Delta n}}

where this follows by eliminating $K_p$ between the two relations above.

Derivation AFrom activities to Kp, Kc, and Kx

Goal. Starting from the thermodynamic (activity-based) equilibrium constant $K$ , derive how it relates to $K_p$ built from pressures, $K_c$ built from concentrations, and $K_x$ built from mole fractions.

Step 1 — Start from the thermodynamic constant. For an ideal gas, the activity of species $i$ is its partial pressure relative to the standard pressure:

a_i = \frac{p_i}{P^{\circ}}

The thermodynamic equilibrium constant for

aA + bB \rightleftharpoons cC + dD

K = \prod_i a_i^{\nu_i} = \frac{(p_C/P^{\circ})^{c}(p_D/P^{\circ})^{d}}{(p_A/P^{\circ})^{a}(p_B/P^{\circ})^{b}}

This $K$ is dimensionless by construction and depends only on temperature.

Step 2 — Define Kp as the pressure-quotient with the standard-pressure factors stripped off for convenience. Chemists commonly define

K_p = \frac{p_C^{\,c}\,p_D^{\,d}}{p_A^{\,a}\,p_B^{\,b}}

using pressures expressed in units of $P^{\circ}$ (bar), so that numerically $K_p = K$ when pressures are quoted in bar. This is the object introduced alongside the reaction quotient in earlier equilibrium work.

Step 3 — Substitute the ideal gas law to introduce concentrations. Since $p_i = c_iRT$ , substitute term by term:

K_p = \frac{(c_C RT)^{c}(c_D RT)^{d}}{(c_A RT)^{a}(c_B RT)^{b}}

Group the concentration terms and the $RT$ terms separately:

K_p = \frac{c_C^{\,c}c_D^{\,d}}{c_A^{\,a}c_B^{\,b}} \times (RT)^{(c+d)-(a+b)}

The concentration quotient is exactly $K_c$ , and the exponent is $\Delta n$ . Hence:

K_p = K_c (RT)^{\Delta n}

Step 4 — Substitute the ideal gas law to introduce mole fractions instead. Since $p_i = x_i P$ , substitute term by term:

K_p = \frac{(x_C P)^{c}(x_D P)^{d}}{(x_A P)^{a}(x_B P)^{b}}

Group the mole-fraction terms and the total-pressure terms separately:

K_p = \frac{x_C^{\,c}x_D^{\,d}}{x_A^{\,a}x_B^{\,b}} \times P^{(c+d)-(a+b)}

The mole-fraction quotient is $K_x$ , so, restoring the standard pressure to keep the exponent dimensionless:

K_p = K_x\left(\frac{P}{P^{\circ}}\right)^{\Delta n}

Step 5 — Combine. Both expressions equal $K_p$ , so they equal each other:

K_c (RT)^{\Delta n} = K_x \left(\frac{P}{P^{\circ}}\right)^{\Delta n}

\boxed{K_p = K_c(RT)^{\Delta n} = K_x\left(\frac{P}{P^{\circ}}\right)^{\Delta n}}

Worked examples

Example 1 — Foundational: converting Kc to Kp for ammonia synthesis. At $500\ \text{K}$ , the equilibrium constant for

\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)

expressed in concentrations (mol dm⁻³) is $K_c = 1.7 \times 10^{2}\ \text{mol}^{-2}\,\text{dm}^{6}$ . Find $K_p$ . Step 1: identify Δn.

\Delta n = 2 - (1+3) = -2

Step 2: compute RT in units matching Kc (dm³ bar mol⁻¹ K⁻¹ form of R). Use $R = 0.08314\ \text{dm}^3\,\text{bar}\,\text{mol}^{-1}\,\text{K}^{-1}$ :

RT = (0.08314)(500) = 41.57\ \text{dm}^3\,\text{bar}\,\text{mol}^{-1}

Step 3: apply the conversion.

K_p = K_c (RT)^{\Delta n} = (1.7\times10^{2})(41.57)^{-2}

(41.57)^{2} = 1728

K_p = \frac{1.7\times10^{2}}{1728} = 9.8\times10^{-2}

Since $\Delta n$ is negative here, $K_p$ came out with units of bar⁻² buried in the "dimensionless" activity-based number; when quoted per bar it is consistent with tabulated values of order $10^{-2}$ for this reaction near 500 K, matching the fact that the equilibrium lies well to the left (small $K_p$ ) at this temperature. Sanity check: raising the RT factor to a negative power because $\Delta n < 0$ correctly shrinks $K_p$ relative to $K_c$ , since more moles of gas exist on the reactant side and pressure units penalize that side.

Example 2 — Applied: predicting how Kx changes when a reactor is compressed. The gas-phase dimerization

2\,\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)

has $K_p = 6.75$ (dimensionless, pressures in bar, $P^{\circ}=1\ \text{bar}$ ) at $298\ \text{K}$ . A reactor initially runs at total pressure $P_1 = 1.00\ \text{bar}$ . Find $K_x$ at $P_1$ , then find $K_x$ if the reactor is compressed to $P_2 = 5.00\ \text{bar}$ at the same temperature. Step 1: identify Δn.

\Delta n = 1 - 2 = -1

Step 2: solve for Kx at P₁ = 1.00 bar.

K_p = K_x\left(\frac{P}{P^{\circ}}\right)^{\Delta n} \implies K_x = K_p\left(\frac{P}{P^{\circ}}\right)^{-\Delta n}

At $P_1 = 1.00\ \text{bar}$ :

K_x = 6.75 \times (1.00)^{1} = 6.75

Step 3: solve for Kx at P₂ = 5.00 bar.

K_x = K_p\left(\frac{P_2}{P^{\circ}}\right)^{1} = 6.75 \times 5.00 = 33.75

Step 4: interpret. $K_p$ stays fixed at $6.75$ regardless of pressure, because it is the true thermodynamic constant at this $T$ . But $K_x$ jumps from $6.75$ to $33.75$ on compression. This does not mean the reaction "shifted its thermodynamics" — it means the mole-fraction ratio of product to reactant favors $\text{N}_2\text{O}_4$ more strongly at higher pressure, exactly as Le Chatelier's principle predicts for a reaction that reduces gas moles. Sanity check: since $\Delta n = -1 < 0$ , increasing $P$ must increase $K_x$ , which is what we found — compression pushes equilibrium toward the side with fewer gas molecules.

Limiting cases & common traps

Situation	What happens
$\Delta n = 0$	$K_p = K_c = K_x$ numerically; no conversion factor needed
$\Delta n > 0$ (gas moles increase)	$K_p$ grows relative to $K_c$ as $(RT)^{\Delta n}$ ; increasing $P$ decreases $K_x$
$\Delta n < 0$ (gas moles decrease)	$K_p$ shrinks relative to $K_c$ ; increasing $P$ increases $K_x$
Reaction involves solids/liquids	Their activities are 1; only gas-phase species enter $\Delta n$ and the pressure/concentration quotients
Units of $R$	Must match the units used for $c_i$ (e.g., dm³ bar mol⁻¹ K⁻¹ paired with mol dm⁻³ and bar)

The chemist behind it

The conceptual foundation here — that equilibrium constants are ratios of activities, and that pressure, concentration, and mole fraction descriptions must be reconciled through the gas laws — rests on the broader thermodynamic framework built by Gilbert N. Lewis, who introduced the very concept of activity that makes $K_p$ , $K_c$ , and $K_x$ mutually consistent rather than three unrelated empirical numbers.

Origins · 1 minGilbert N. Lewis and how this was discovered

Gilbert N. Lewis

1875–1946 · American chemist who introduced fugacity and activity, giving equilibrium constants a rigorous, unit-independent thermodynamic foundation

Ask the tutor: How does adding an inert gas at constant pressure versus constant volume change these equilibrium constants and the position of equilibrium?

Why do we even need three different equilibrium constants

Different experiments naturally report different quantities. Gas mixtures are often characterized by total pressure and mole fractions from a manometer or gas chromatograph, while solution work uses molar concentrations. Rather than forcing every measurement into one variable before use, chemists convert between the constants after the fact. All three trace back to the same underlying activity-based constant $K$ , so no new physics is introduced — just a change of units and reference state, exactly as with the derivation from the ideal gas law shown above.

Does the value of Δn depend on how the equation is balanced

Yes, and this matters. If you double every stoichiometric coefficient, $\Delta n$ doubles too, and $K$ becomes squared, since $K$ for the doubled equation is $K^2$ of the original. Always state which balanced equation, and which $\Delta n$ , a quoted equilibrium constant corresponds to. This is the same subtlety that appears when comparing tabulated $K_p$ values across different sources.

Is Kx ever a genuinely useful, standalone constant

Yes, particularly in engineering contexts like reactor design, where you want to know directly what fraction of the mixture is product at a chosen operating pressure. But because $K_x$ depends on total pressure whenever $\Delta n \neq 0$ , it is not transferable between different pressure conditions without reapplying the conversion formula—unlike $K_p$ , which transfers directly across pressures at fixed temperature.

How does this connect to inert gas effects and Le Chatelier's principle

Adding an inert gas at constant volume does not change any partial pressure, so $K_p$ and the equilibrium position are unaffected. Adding inert gas at constant total pressure, however, dilutes the reactive species, changes their mole fractions, and shifts the mole-fraction equilibrium — this is exactly the pressure-inert-gas effect explored next, and it flows directly from the $K_x$ versus $P$ dependence derived here. Le Chatelier's principle is the qualitative shortcut for predicting these shifts.