Effects of Pressure and Inert Gases on Equilibrium

Take the ammonia synthesis $N_2 + 3H_2 \rightleftharpoons 2NH_3$ . Squeeze the reaction vessel to twice the volume's pressure. Le Chatelier's principle says the equilibrium shifts toward fewer moles of gas. So more ammonia forms. Yet the equilibrium constant $K_p$ does not change at all. This apparent contradiction — composition moves, but $K$ stays fixed — is one of the most conceptually rich results in chemical thermodynamics. It trips up even careful students. The resolution hinges on distinguishing what $K$ actually is. It is a ratio built from a standard state and fugacities or activities. This differs from what mole fractions do. They respond to total pressure through the counting of gas moles. A second, equally subtle puzzle is inert-gas addition. Pumping argon into a fixed-volume reactor does nothing to equilibrium. But adding the same argon at constant total pressure shifts the composition. This happens exactly as if you had lowered the pressure. Both effects follow from one clean piece of algebra. This article derives the pressure dependence of $K_x$ from the pressure independence of $K_p$ . It works through the inert-gas cases at constant volume and constant pressure. It also gives two fully numeric examples. You will see why $\Delta G^\circ$ and $K$ ignore total pressure. You will see how $K_x$ absorbs all the pressure dependence. Finally, you will see how this sets up coupled multi-reaction equilibria.

Key takeaways

$K_p$ depends only on temperature; it is built from standard states (usually 1 bar), so total pressure never enters its definition.
$K_x$ (mole-fraction based) does depend on pressure whenever $\Delta n_{gas} \neq 0$ , via $K_x = K_p (p^\circ/p)^{\Delta n}$ .
Raising total pressure at constant $T$ shifts a gas equilibrium toward the side with fewer moles of gas, exactly as Le Chatelier predicts, without moving $K_p$ .
Adding inert gas at constant volume changes nothing: partial pressures of reactive species are untouched, so equilibrium composition is unchanged.
Adding inert gas at constant total pressure dilutes the reactive species (lowers each partial pressure), which is equivalent to expanding the system — it shifts equilibrium toward more moles of gas.
Key idea: $K$ is a function of $T$ alone; all pressure and inert-gas effects act on mole fractions and partial pressures, never on $K$ itself.

The core insight

The equilibrium constant $K$ is defined through the standard Gibbs energy change $\Delta G^\circ = -RT\ln K$ , and $\Delta G^\circ$ is evaluated at fixed standard-state pressure $p^\circ$ for every species — so $K$ is a pure function of temperature; total pressure and inert gases can only redistribute mole fractions among reactants and products at fixed $K$ , never change $K$ itself.

The physical picture

Consider a gas-phase reaction at temperature $T$ inside a container. Chemical potentials of ideal gases are written relative to a standard state at pressure $p^\circ = 1\ \text{bar}$ :

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

At equilibrium the condition $\sum_i \nu_i \mu_i = 0$ collapses to $\Delta G^\circ(T) = -RT\ln K_p$ , where $K_p$ is built entirely from the ratios $p_i/p^\circ$ . Because $\mu_i^\circ$ is defined at the fixed reference pressure $p^\circ$ , changing the total pressure of the real mixture never touches $\Delta G^\circ$ or $K_p$ . The dial you are turning — total pressure $p$ — simply is not a variable in the definition of $K_p$ . What does depend on total pressure is how you get from $K_p$ to actual mole fractions. Since $p_i = x_i p$ , the reaction quotient in mole fractions,

K_x = \prod_i x_i^{\nu_i},

is related to $K_p$ by a factor of $(p/p^\circ)^{\Delta n}$ , where $\Delta n = \sum_i \nu_i$ counts net gas moles produced by the reaction as written. If $\Delta n < 0$ (fewer gas moles as products), squeezing the system to higher $p$ forces $K_x$ to increase — meaning the reaction must shift forward to keep $K_p$ fixed. That is the entire mechanism behind Le Chatelier's pressure response, expressed as algebra instead of a qualitative rule. Inert gas addition is a special case of the same logic viewed through partial pressures instead of mole fractions. Held at constant volume, adding argon raises total pressure but leaves every reactive species' partial pressure, $p_i = n_iRT/V$ , completely unchanged — so $K_p$ is already satisfied and nothing shifts. Held at constant total pressure, adding argon must make room for itself by expanding the volume (or, equivalently, by diluting the reactive species' mole fractions at fixed $p$ ), lowering each $p_i$ exactly as if the vessel had expanded — and the equilibrium shifts as if the pressure had been reduced.

Figure

Same

K_p

, different mole fractions as total pressure changes for N₂ + 3H₂ ⇌ 2NH₃

Figure

Inert gas at constant V does nothing; at constant p it dilutes and shifts equilibrium

The core relations

\Delta G^\circ(T) = -RT\ln K_p

where: $K_p$ is built from standard-state ratios $p_i/p^\circ$ and depends only on $T$ .

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

where: $\nu_i$ is the stoichiometric coefficient (negative for reactants).

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

where: $K_x = \prod_i x_i^{\nu_i}$ , $p$ is total pressure, $\Delta n = \sum_i \nu_i$ .

\left(\frac{\partial \ln K_x}{\partial \ln p}\right)_T = -\Delta n

where: this quantifies how strongly mole-fraction composition responds to total pressure at fixed $T$ .

p_i = x_i p = \frac{n_i}{n_{\text{tot}} + n_{\text{inert}}}\, p

where: $n_{\text{inert}}$ is moles of added inert gas; note how it enters only through $n_{\text{tot}}$ when $p$ is fixed.

The core derivation

Derivation AFrom constant

K_p

to pressure-dependent

K_x

Goal. Show that $K_p$ is strictly pressure-independent, then derive how $K_x$ must depend on total pressure to compensate, for a generic gas reaction with net mole change $\Delta n$ .

Step 1 — Chemical potential of each ideal gas. Write the chemical potential of species $i$ at partial pressure $p_i$ referenced to standard pressure $p^\circ$ :

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

The reference term $\mu_i^\circ(T)$ is evaluated at $p^\circ$ only, a fixed number for given $T$ , regardless of the actual total pressure of the system.

Step 2 — Impose the equilibrium condition. At equilibrium, $\sum_i \nu_i \mu_i = 0$ . Substituting Step 1 and separating standard and pressure-dependent terms:

\sum_i \nu_i \mu_i^\circ(T) + RT\sum_i \nu_i \ln\left(\frac{p_i}{p^\circ}\right) = 0

The first sum is $\Delta G^\circ(T)$ , a function of temperature alone.

Step 3 — Define $K_p$ and isolate it. Rearranging,

\Delta G^\circ(T) = -RT \ln\left[\prod_i \left(\frac{p_i}{p^\circ}\right)^{\nu_i}\right] = -RT\ln K_p

Since $\Delta G^\circ$ depends only on $T$ (it is built from standard chemical potentials at fixed $p^\circ$ ), $K_p$ is a function of $T$ alone. No total-pressure variable appears anywhere on the right-hand side.

Step 4 — Introduce mole fractions. Using $p_i = x_i p$ with $p$ the total pressure,

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

Define $K_x \equiv \prod_i x_i^{\nu_i}$ and $\Delta n \equiv \sum_i \nu_i$ , giving

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

Step 5 — Solve for $K_x(p)$ . Since $K_p$ is fixed at given $T$ , rearrange for the pressure-dependent quantity:

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

Differentiating $\ln K_x$ with respect to $\ln p$ at constant $T$ (recall $K_p$ has no $p$ -dependence to differentiate):

\left(\frac{\partial \ln K_x}{\partial \ln p}\right)_T = -\Delta n

Step 6 — Interpret the sign. If $\Delta n < 0$ (moles of gas decrease going to products, as in $N_2+3H_2\rightleftharpoons 2NH_3$ , $\Delta n=-2$ ), then $-\Delta n>0$ , so $K_x$ increases with $p$ : the equilibrium mole fraction of products grows as pressure rises. If $\Delta n>0$ , raising $p$ decreases $K_x$ , favoring reactants. If $\Delta n=0$ , $K_x$ is pressure-independent and total pressure has no effect on composition at all.

\boxed{K_p(T) \text{ is fixed}; \quad K_x = K_p\left(\dfrac{p^\circ}{p}\right)^{\Delta n} \text{ carries all the pressure dependence}}

Worked examples

Example 1 — Ammonia synthesis, quantifying the pressure shift. Take $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ at 400 °C (673 K), where $K_p = 1.60\times 10^{-4}$ (dimensionless, using $p^\circ = 1$ bar and stoichiometric partial-pressure ratios; this is a standard literature value near this temperature). Here $\Delta n = 2 - 1 - 3 = -2$ . Suppose we start with a stoichiometric 1:3 mixture of $N_2$ and $H_2$ (no product yet) and ask how the equilibrium mole fraction of ammonia compares at $p = 10$ bar versus $p = 100$ bar. Using $K_x = K_p (p^\circ/p)^{\Delta n} = K_p (p/p^\circ)^{2}$ : At $p=10$ bar:

K_x = (1.60\times10^{-4})(10)^2 = 1.60\times10^{-2}

At $p=100$ bar:

K_x = (1.60\times10^{-4})(100)^2 = 1.60

$K_x$ rises by a factor of 100 as $p$ rises tenfold — exactly the $(p/p^\circ)^2$ scaling. To translate $K_x$ into a mole fraction of ammonia, let $\xi$ be the extent of reaction per mole of initial $N_2$ , starting from 1 mol $N_2$ + 3 mol $H_2$ (4 mol total, no dilution):

n_{N_2}=1-\xi,\quad n_{H_2}=3-3\xi,\quad n_{NH_3}=2\xi,\quad n_{tot}=4-2\xi

K_x = \frac{(2\xi/(4-2\xi))^2}{\big((1-\xi)/(4-2\xi)\big)\big((3-3\xi)/(4-2\xi)\big)^3}

Solving numerically at $p=10$ bar ( $K_x=1.60\times10^{-2}$ ) gives $\xi \approx 0.114$ , so $x_{NH_3} = 2(0.114)/(4-0.228) = 0.0605$ , about 6.1% ammonia. At $p=100$ bar ( $K_x=1.60$ ) solving gives $\xi\approx 0.457$ , so $x_{NH_3}=2(0.457)/(4-0.914)=0.296$ , about 29.6% ammonia.

Sanity check: $K_p$ itself never changed — we used the same $1.60\times10^{-4}$ at both pressures. Only the equilibrium mole fraction moved, from 6% to 30% ammonia, entirely consistent with Le Chatelier's qualitative prediction that higher pressure favors the side with fewer gas moles. This matches the real industrial practice of running the Haber process at 150–300 bar.

Example 2 — Inert gas at constant volume vs constant pressure. Consider the dissociation $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ at 298 K, where $K_p = 0.148$ (a standard textbook value, $\Delta n = +1$ ). Start with 1.00 mol of pure $N_2O_4$ in a 24.45 L vessel at 298 K (giving an initial pressure of 1.00 bar before any dissociation, since $RT/V = (0.08314)(298)/24.45 = 1.013$ bar·L/mol… using $V=24.45$ L gives approximately 1.00 bar for 1 mol ideal gas at 298 K). Case A — add 1.00 mol Ar at constant volume (24.45 L, rigid vessel). The partial pressures of $N_2O_4$ and $NO_2$ depend only on their own moles and $V$ , via $p_i = n_iRT/V$ . Adding argon raises total pressure (more molecules bouncing in the same box) but does not change $n_{N_2O_4}$ , $n_{NO_2}$ , or $V$ . Since $K_p$ depends only on the partial pressures of the reactive species, the equilibrium extent of dissociation is completely unchanged. Let $\alpha$ be the fraction of $N_2O_4$ dissociated; solving $K_p = 4\alpha^2 p_{N_2O_4,0}/(1-\alpha^2)$ with initial pressure 1.00 bar (no argon) gives:

0.148 = \frac{4\alpha^2(1.00)}{1-\alpha^2} \implies \alpha \approx 0.190

This value of $\alpha=0.190$ is exactly what remains true after argon is added at fixed $V$ — nothing shifts. Case B — add 1.00 mol Ar at constant total pressure (1.00 bar), letting the vessel expand. Now total moles increase, so the box must expand to keep $p=1.00$ bar fixed, which dilutes $N_2O_4$ and $NO_2$ exactly as if pure expansion had occurred. Total moles at equilibrium are $n_{tot} = 1-\alpha+2\alpha + 1_{Ar} = 2+\alpha$ (including the mole of argon), and mole fractions become $x_{N_2O_4}=(1-\alpha)/(2+\alpha)$ , $x_{NO_2}=2\alpha/(2+\alpha)$ . With $p_{N_2O_4}=x_{N_2O_4}p$ and $p_{NO_2}=x_{NO_2}p$ at $p=1.00$ bar:

K_p = \frac{(2\alpha/(2+\alpha))^2 (1.00)^2}{(1-\alpha)/(2+\alpha)} = \frac{4\alpha^2}{(2+\alpha)(1-\alpha)}

Setting this equal to 0.148 and solving: try $\alpha=0.24$ : $4(0.0576)/[(2.24)(0.76)] = 0.2304/1.7024=0.1353$ (too low). Try $\alpha=0.26$ : $4(0.0676)/[(2.26)(0.74)]=0.2704/1.6724=0.1617$ (slightly high). Interpolating, $\alpha \approx 0.253$ .

Sanity check: At constant $V$ , $\alpha$ stays at 0.190 — inert gas is a spectator. At constant $p$ , $\alpha$ rises to about 0.253 because diluting the reactive species with argon (while holding total $p$ fixed) is equivalent to lowering their partial pressures, pushing the equilibrium toward more moles of gas ( $NO_2$ ), just as $\Delta n=+1>0$ predicts for an effective pressure decrease.

Limiting cases & common traps

Situation	What happens to $K_p$	What happens to composition
$\Delta n = 0$ (e.g. $H_2+I_2\rightleftharpoons 2HI$ )	Unchanged (always)	Total pressure has no effect at all
$\Delta n \neq 0$ , raise total $p$	Unchanged	Shifts toward the side with fewer gas moles
Add inert gas at constant $V$	Unchanged	No shift — partial pressures of reactive species untouched
Add inert gas at constant $p$	Unchanged	Shifts toward more moles of gas (dilution mimics expansion)
Reaction has only condensed phases at fixed activity	Unchanged	Total gas pressure change has no effect if no gas-phase species involved

The chemist behind it

Henry Louis Le Chatelier formulated the qualitative principle bearing his name in 1884 while studying the industrial synthesis conditions for compounds like ammonia and steel-related equilibria. His insight — that a system at equilibrium responds to a disturbance by shifting to partially counteract it — long preceded the rigorous thermodynamic machinery of $K_p$ , $K_x$ , and standard states developed later by Gibbs, Lewis, and van't Hoff. The modern algebraic treatment shown here, separating the pressure-invariant $K_p$ from the pressure-sensitive $K_x$ , gives Le Chatelier's qualitative rule its full quantitative teeth and explains precisely why industrial ammonia plants operate at hundreds of bar.

Origins · 1 minHenry Louis Le Chatelier and how this was discovered

Henry Louis Le Chatelier

1850–1936 · French chemist who formulated the principle governing equilibrium shifts under pressure, temperature, and concentration perturbations, directly motivating high-pressure industrial synthesis

Ask the tutor: How do these pressure and dilution effects combine when two or more gas-phase equilibria share a common species?

Does Kp ever depend on pressure, even for non-ideal gases?

For ideal gases, never — $K_p$ is exactly a function of $T$ alone, since it is defined via standard chemical potentials at $p^\circ$ . For real gases, the analogous constant is built from fugacities, $K_f$ , which is still a pure function of $T$ ; however, converting between $K_f$ and pressure-based quotients introduces fugacity coefficients $\phi_i(T,p)$ that do depend on total pressure. So the fundamental constant remains pressure-independent, but the approximation $K_p \approx K_f$ breaks down at high pressure for real gases.

Why does Le Chatelier's principle work if K itself never moves?

Le Chatelier's principle is really a statement about how the reaction quotient $Q$ compares to the fixed $K$ after a perturbation. Squeezing the volume instantly changes each partial pressure, so $Q$ (computed with the new partial pressures) no longer equals $K$ . The system then reacts — shifting extent of reaction — until $Q$ returns to $K$ . The mole fractions move; the constant they must satisfy does not.

How is this different at constant volume versus constant pressure for the reaction quotient?

At constant volume, adding inert gas raises total pressure but each reactive species keeps its own $n_i$ and $V$ , so $p_i = n_iRT/V$ is untouched — $Q_p$ stays equal to $K_p$ and nothing shifts. At constant pressure, the vessel must expand to accommodate the extra gas, so $V$ increases while $n_i$ (initially) stays the same, lowering each $p_i$ . That drop makes $Q_p \ne K_p$ , and the system shifts to restore equality, exactly mimicking a pressure decrease.

Does this pressure independence of K extend to reactions in solution?

Yes, in the sense that $K$ (built from activities referenced to a standard state, typically 1 mol/L or unit mole fraction) is still a function of $T$ alone for ideal or ideally dilute solutions. However, real solution-phase $K$ values can show a modest pressure dependence through the reaction volume $\Delta V^\circ$ , via $(\partial \ln K/\partial p)_T = -\Delta V^\circ/RT$ . This effect is usually small at ordinary pressures but becomes significant in high-pressure or geochemical contexts, unlike the exactly zero pressure-dependence for ideal gas-phase $K_p$ .

How does this connect to systems with several simultaneous equilibria?

When multiple reactions occur at once — for example, in atmospheric or combustion chemistry — each reaction has its own pressure-independent $K_p^{(j)}(T)$ , but the mole fractions of shared species must simultaneously satisfy every reaction's $K_x^{(j)}(p)$ relation. Total pressure then couples the reactions together through shared species, even though no individual $K_p^{(j)}$ ever moves. This is the natural bridge to solving coupled multi-reaction equilibria.