The Equilibrium Constant Kp

Drop a sealed flask of $\text{N}_2\text{O}_4$ into a warm bath and watch it turn brown as some converts to $\text{NO}_2$ . The reaction stops changing color at a fixed point, no matter how you got there. That fixed point is a ratio of partial pressures with a precise numerical value — the equilibrium constant $K_p$ . Every gas-phase industrial process, from ammonia synthesis to atmospheric ozone chemistry, is designed around knowing $K_p$ at the operating temperature. It tells engineers and chemists exactly how far a reaction can go before it stalls, and how pressure or temperature shifts that limit. This article builds $K_p$ from the chemical potential of an ideal gas. You will see how the reaction Gibbs energy vanishes at equilibrium. You will see why that condition forces a specific combination of pressures raised to stoichiometric powers. Then you will apply it to two worked numerical problems.

Key takeaways

$K_p$ is built from partial pressures each divided by a reference pressure $p^\circ$ (usually 1 bar), so it is dimensionless.
It follows directly from setting $\Delta_r G = 0$ at equilibrium, using $\mu = \mu^\circ + RT\ln(p/p^\circ)$ for each ideal gas.
$K_p = \exp(-\Delta_r G^\circ/RT)$ links the thermodynamic standard state to the measurable equilibrium composition.
Only gas-phase species with variable partial pressure appear in $K_p$ ; pure solids and liquids are omitted (activity = 1).
$K_p$ is a true constant only at fixed temperature; it does not depend on total pressure, volume, or catalyst.
Key idea: at equilibrium, the sum of stoichiometrically weighted chemical potentials is zero, and this single condition forces the partial pressures into a fixed dimensionless ratio, $K_p$ .

The core insight

$K_p$ is nothing more than the numerical fingerprint left behind when the condition $\Delta_r G = 0$ is translated, gas by gas, using the ideal-gas chemical potential $\mu = \mu^\circ + RT\ln(p/p^\circ)$ .

The physical picture

Consider a general gas-phase reaction,

aA + bB \rightleftharpoons cC + dD

Each species carries a chemical potential that depends on its own partial pressure, not on the total pressure of the mixture. As the reaction proceeds, $A$ and $B$ are consumed while $C$ and $D$ are produced. The Gibbs energy of the whole reacting system falls, because each species is moving to lower chemical potential as its partial pressure changes and the reaction mixture explores new compositions. The system stops evolving when the Gibbs energy of the mixture reaches a minimum with respect to the extent of reaction $\xi$ . At that point, adding a little more product costs exactly as much in potential as removing a little more reactant saves. This is the mechanical picture behind $\Delta_r G = 0$ : a balance of chemical potentials, weighted by stoichiometric coefficients. Because each ideal gas's chemical potential depends logarithmically on its partial pressure, the balance condition translates into a product of pressure ratios raised to powers — not a sum. That is the origin of the characteristic form of $K_p$ : a ratio of products, each partial pressure appearing raised to its stoichiometric coefficient. It is worth being clear about what varies and what does not. The individual partial pressures $p_A, p_B, p_C, p_D$ can each take a continuous range of values depending on how much gas you loaded into the vessel and at what total pressure. Yet the particular combination dictated by the reaction stoichiometry is pinned to one number, $K_p$ , at a given temperature — regardless of the starting amounts.

Figure

Gibbs energy of the reaction mixture falls to a minimum at the equilibrium extent of reaction

The core relations

K_p = \prod_i \left(\frac{p_i}{p^\circ}\right)^{\nu_i}

where: $p_i$ is the partial pressure of gas $i$ , $p^\circ$ is the standard pressure (1 bar), and $\nu_i$ is positive for products and negative for reactants.

K_p = \frac{\left(p_C/p^\circ\right)^{c}\left(p_D/p^\circ\right)^{d}}{\left(p_A/p^\circ\right)^{a}\left(p_B/p^\circ\right)^{b}}

where: $a, b, c, d$ are the stoichiometric coefficients from the balanced equation.

\Delta_r G^\circ = -RT\ln K_p

where: $\Delta_r G^\circ$ is the standard reaction Gibbs energy, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T$ in kelvin.

\mu_i = \mu_i^\circ + RT\ln\left(\frac{p_i}{p^\circ}\right)

where: this is the chemical potential of ideal gas $i$ , the building block from which $K_p$ is derived.

Derivation AFrom Δ_rG = 0 to the pressure-ratio expression for K_p

Goal. Show that the equilibrium condition $\Delta_r G = 0$ forces the partial pressures of a gas-phase reaction into the fixed combination $K_p = \exp(-\Delta_r G^\circ/RT)$ .

Step 1. Write the Gibbs energy of reaction in terms of chemical potentials. For the reaction $aA + bB \rightleftharpoons cC + dD$ , the reaction Gibbs energy at any composition is

\Delta_r G = c\mu_C + d\mu_D - a\mu_A - b\mu_B

This is the slope of total $G$ with respect to extent of reaction, $\left(\partial G/\partial \xi\right)_{T,p}$ , evaluated at the current composition.

Step 2. Insert the ideal-gas chemical potential for each species. Each $\mu_i = \mu_i^\circ + RT\ln(p_i/p^\circ)$ . Substituting,

\Delta_r G = \left[c\mu_C^\circ + d\mu_D^\circ - a\mu_A^\circ - b\mu_B^\circ\right] + RT\left[c\ln\frac{p_C}{p^\circ} + d\ln\frac{p_D}{p^\circ} - a\ln\frac{p_A}{p^\circ} - b\ln\frac{p_B}{p^\circ}\right]

The bracketed standard-state term is by definition $\Delta_r G^\circ$ , the reaction Gibbs energy when every gas is at $p^\circ$ .

Step 3. Collapse the logarithm sum into a single logarithm of a ratio. Using $n\ln x = \ln x^n$ and combining logs,

\Delta_r G = \Delta_r G^\circ + RT\ln\left[\frac{(p_C/p^\circ)^c (p_D/p^\circ)^d}{(p_A/p^\circ)^a (p_B/p^\circ)^b}\right]

Call the bracketed quantity the reaction quotient $Q_p$ . This equation, $\Delta_r G = \Delta_r G^\circ + RT\ln Q_p$ , holds at any composition, equilibrium or not.

Step 4. Impose the equilibrium condition. At equilibrium the Gibbs energy of the mixture is at a minimum along the reaction coordinate, so $\Delta_r G = 0$ . Substituting into Step 3's result,

0 = \Delta_r G^\circ + RT\ln Q_p^{\text{eq}}

Step 5. Solve for the equilibrium value of the quotient and name it $K_p$ . Rearranging,

\ln K_p = -\frac{\Delta_r G^\circ}{RT} \qquad\Longrightarrow\qquad \boxed{K_p = \exp\left(-\frac{\Delta_r G^\circ}{RT}\right) = \prod_i\left(\frac{p_i}{p^\circ}\right)^{\nu_i}\Bigg|_{\text{eq}}}

Because $\Delta_r G^\circ$ depends only on $T$ (it is built from standard chemical potentials, defined at fixed $p^\circ$ ), $K_p$ is a pure function of temperature alone — never of total pressure or the starting amounts of gas.

Worked examples

Example 1 — Foundational: the dissociation of dinitrogen tetroxide. The reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ has $\Delta_r G^\circ = +5.4\ \text{kJ mol}^{-1}$ at 298 K (standard tabulated value, using $p^\circ = 1$ bar). Step 1 — compute $K_p$ .

\ln K_p = -\frac{\Delta_r G^\circ}{RT} = -\frac{5400\ \text{J mol}^{-1}}{(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})} = -2.179

K_p = e^{-2.179} = 0.113

Step 2 — check the value makes physical sense. $K_p < 1$ signals that at 1 bar standard conditions, dissociation is not favored — most $\text{N}_2\text{O}_4$ remains intact, consistent with $\Delta_r G^\circ$ being positive (nonspontaneous at standard pressures). Step 3 — use $K_p$ to find the equilibrium partial pressures. Suppose 1.00 mol of pure $\text{N}_2\text{O}_4$ is placed in a vessel and the total pressure at equilibrium is held at $p_{\text{tot}} = 1.00$ bar. Let $\alpha$ be the fraction dissociated. Moles: $\text{N}_2\text{O}_4 = 1-\alpha$ , $\text{NO}_2 = 2\alpha$ , total $= 1+\alpha$ .

p_{\text{N}_2\text{O}_4} = \frac{1-\alpha}{1+\alpha}\,p_{\text{tot}}, \qquad p_{\text{NO}_2} = \frac{2\alpha}{1+\alpha}\,p_{\text{tot}}

K_p = \frac{(p_{\text{NO}_2}/p^\circ)^2}{p_{\text{N}_2\text{O}_4}/p^\circ} = \frac{4\alpha^2}{(1-\alpha)(1+\alpha)}\cdot\frac{p_{\text{tot}}}{p^\circ} = \frac{4\alpha^2}{1-\alpha^2}

Setting this equal to 0.113 and solving,

4\alpha^2 = 0.113(1-\alpha^2) \implies \alpha^2(4.113) = 0.113 \implies \alpha = \sqrt{0.02747} = 0.166

So about 16.6% of the $\text{N}_2\text{O}_4$ dissociates at 298 K and 1 bar total pressure — a modest but measurable extent, exactly as the small $K_p$ predicted.

Example 2 — Applied: ammonia synthesis and the effect of pressure. For $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$ at 700 K, tabulated data give $K_p = 1.45\times10^{-3}$ (dimensionless, referenced to $p^\circ = 1$ bar). Question: starting from a stoichiometric 1:3 mixture of $\text{N}_2$ and $\text{H}_2$ with no product, at a total equilibrium pressure of 300 bar, what fraction of the nitrogen has converted to ammonia? Step 1 — set up mole fractions. Let $z$ = mol $\text{N}_2$ reacted, starting from 1 mol $\text{N}_2$ + 3 mol $\text{H}_2$ = 4 mol total initially.

n_{\text{N}_2} = 1-z,\quad n_{\text{H}_2} = 3-3z,\quad n_{\text{NH}_3} = 2z,\quad n_{\text{tot}} = 4-2z

Step 2 — write $K_p$ in terms of $z$ and total pressure $P = 300$ bar (in units of $p^\circ$ ).

K_p = \frac{(p_{\text{NH}_3}/p^\circ)^2}{(p_{\text{N}_2}/p^\circ)(p_{\text{H}_2}/p^\circ)^3} = \frac{\left(\dfrac{2z}{4-2z}\right)^2 P^2}{\left(\dfrac{1-z}{4-2z}\right)\left(\dfrac{3-3z}{4-2z}\right)^3 P^4}

Simplify the pressure powers: numerator has $P^2$ , denominator has $P^{1+3}=P^4$ , giving an overall factor $P^{-2}$ .

K_p = \frac{4z^2 (4-2z)^2}{27(1-z)^4 P^2}

Step 3 — solve numerically. With $P = 300$ , $K_p = 1.45\times10^{-3}$ :

1.45\times10^{-3} = \frac{4z^2(4-2z)^2}{27(1-z)^4 (300)^2}

1.45\times10^{-3}\times27\times90000 = 4z^2(4-2z)^2/(1-z)^4

3523.5 = \frac{4z^2(4-2z)^2}{(1-z)^4}

Try $z = 0.70$ : numerator $= 4(0.49)(2.6)^2 = 4(0.49)(6.76) = 13.25$ ; denominator $= (0.30)^4 = 8.1\times10^{-3}$ ; ratio $= 1636$ — too small, need larger $z$ . Try $z = 0.80$ : numerator $= 4(0.64)(2.4)^2 = 4(0.64)(5.76) = 14.75$ ; denominator $= (0.20)^4 = 1.6\times10^{-4}$ ; ratio $= 92,200$ — too large. Try $z = 0.74$ : numerator $= 4(0.5476)(2.52)^2 = 4(0.5476)(6.350) = 13.91$ ; denominator $=(0.26)^4 = 4.570\times10^{-3}$ ; ratio $= 3045$ — close, slightly low. Try $z = 0.745$ : denominator $=(0.255)^4=4.228\times10^{-3}$ ; numerator $=4(0.555)(2.51)^2=4(0.555)(6.30)=13.99$ ; ratio $=3309$ — still a touch low. Try $z=0.75$ : denominator $=(0.25)^4=3.906\times10^{-3}$ ; numerator $=4(0.5625)(2.5)^2=4(0.5625)(6.25)=14.06$ ; ratio $=3600$ — matches 3523.5 closely. So $z \approx 0.748$ , meaning about 75% of the nitrogen converts to ammonia at 700 K and 300 bar. Sanity check: this is far higher than the ~15% typical of milder industrial conditions (450 K–500 K, moderate pressure), illustrating why the Haber process pushes to high pressure — Le Chatelier's principle in numbers, not just words, because $K_p$ itself never changes but the achievable conversion at fixed $K_p$ rises steeply with $P$ since the reaction reduces the mole number ( $\Delta \nu = 2-4=-2$ ).

Limiting cases & common traps

Situation	What happens to $K_p$
Pure solids or liquids in the reaction	Excluded from $K_p$ (activity fixed at 1); only gases appear
Total pressure changed at fixed $T$	$K_p$ unchanged; equilibrium composition shifts if $\Delta\nu \neq 0$
Temperature changed	$K_p$ changes — governed by the van 't Hoff relation
Inert gas added at constant volume	Partial pressures of reactive species unchanged, so $K_p$ and equilibrium position unchanged
Inert gas added at constant total pressure	Mole fractions of reactive gases drop, shifting equilibrium toward the side with more moles of gas

The chemist behind it

The equilibrium constant traces its conceptual roots to the Norwegian chemists Cato Guldberg and Peter Waage, who proposed the law of mass action in the 1860s from empirical reaction-rate reasoning. But the rigorous thermodynamic foundation — expressing $K$ through chemical potentials and the Gibbs energy — is the achievement of Josiah Willard Gibbs, whose 1876 monograph On the Equilibrium of Heterogeneous Substances supplied the mathematical machinery ( $\mu$ , $G$ , and the equilibrium condition $\sum \nu_i \mu_i = 0$ ) used in the derivation above. It was later chemists, notably applying Gibbs's framework to industrial gas reactions such as ammonia synthesis, who turned $K_p$ from abstract thermodynamics into a working engineering tool.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and defined the chemical potential underlying every equilibrium constant

Ask the tutor: How do I convert this $K_p$ into $K_c$ , and what does the van 't Hoff equation tell me about how $K_p$ shifts with temperature?

Why does K_p use a ratio p/p° instead of just the pressure p?

Thermodynamic functions like $\ln p$ are only meaningful for a dimensionless argument. The chemical potential expression $\mu = \mu^\circ + RT\ln(p/p^\circ)$ compares each gas's pressure to a fixed reference, $p^\circ = 1$ bar. Dividing by $p^\circ$ makes the ratio dimensionless, so $K_p$ itself is dimensionless and the logarithm in $\Delta_r G^\circ = -RT\ln K_p$ is well defined regardless of what pressure units you started with.

Does K_p change if I add a catalyst?

No. A catalyst speeds up both the forward and reverse reactions equally by lowering the activation energy, but it does not change $\Delta_r G^\circ$ or the standard chemical potentials of reactants and products. Since $K_p = \exp(-\Delta_r G^\circ/RT)$ depends only on those thermodynamic quantities, the equilibrium position and $K_p$ are identical with or without a catalyst — only the time to reach equilibrium changes.

Why do pure solids and liquids not appear in K_p?

Their chemical potential is essentially fixed at its standard value because their "concentration" — really their molar volume and structure — barely changes with the small amounts consumed or produced. Formally, their activity is defined as 1 relative to the pure standard state at that temperature. Since $\ln(1) = 0$ , they contribute nothing to the pressure-ratio expression, though they still matter for whether the reaction can proceed physically (there must be some present).

How is K_p related to K_c, the equilibrium constant in terms of concentrations?

For ideal gases, partial pressure and concentration are linked by $p_i = n_iRT/V = [i]RT$ , so a conversion factor involving $RT/p^\circ$ raised to the power $\Delta\nu$ (the change in moles of gas) connects the two constants. This conversion is explored in detail in equilibrium-constant-conversions, and is essential whenever you need to compare a spectroscopically measured concentration-based $K$ to a thermodynamically tabulated $K_p$ .

Does K_p depend on temperature, and how would I predict that dependence?

Yes — $K_p$ is a function of temperature only, because $\Delta_r G^\circ$ itself depends on $T$ through both the entalpic and entropic parts of the standard reaction Gibbs energy. The quantitative relationship between $\ln K_p$ and $1/T$ , derived from $\Delta_r H^\circ$ , is the subject of the van 't Hoff equation, which lets you extrapolate an equilibrium constant measured at one temperature to another.