Gas-Phase and Heterogeneous Equilibria

What is it?

For reactions of gases, it is often easier to measure partial pressures than concentrations. The equilibrium constant written with partial pressures is $K_\text{p}$ . For

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}

K_\text{p} = \frac{(P_{\ce{NH3}})^2}{P_{\ce{N2}}\,(P_{\ce{H2}})^3}

It has exactly the same form as $K_\text{c}$ , with each partial pressure (in atm) in place of a concentration.

A heterogeneous equilibrium involves more than one phase. Pure solids and pure liquids are left out of $K$ , because their “concentration” does not change as they are used up. For the decomposition of limestone:

\begin{aligned} &\ce{CaCO3(s) <=> CaO(s) + CO2(g)} \\[4pt] &K_\text{p} = P_{\ce{CO2}} \end{aligned}

Key idea

$K_\text{c}$ and $K_\text{p}$ describe the same equilibrium in different units. Because $P = \frac{n}{V}RT = cRT$ for each gas, they are linked by

K_\text{p} = K_\text{c}\,(RT)^{\Delta n}

where $\Delta n$ = moles of gaseous products − moles of gaseous reactants. When $\Delta n = 0$ , $K_\text{p} = K_\text{c}$ .

Why does it matter?

Industry. Ammonia, sulfuric acid and methanol are made by gas-phase equilibria; engineers use $K_\text{p}$ to choose pressures and temperatures.
The atmosphere. Reactions such as $\ce{N2O4 <=> 2NO2}$ (a brown gas in smog) are gas equilibria.
Solids and gases together. Limestone kilns, rusting and the equilibrium vapour pressure of water are all heterogeneous equilibria.

How does it work?

1. Writing Kp

Products over reactants, each partial pressure raised to its coefficient; leave out solids and liquids.

2. Converting between Kc and Kp

Find $\Delta n$ from the gas coefficients only.
Use $R = 0.08206$ L·atm/(mol·K) and $T$ in kelvin.
$K_\text{p} = K_\text{c}(RT)^{\Delta n}$ .

3. ICE tables with pressures

Pressures behave like concentrations: set up Initial, Change and Equilibrium rows in atm, substitute into $K_\text{p}$ and solve.

Think of it like this

Kc and Kp are like the same price written in two currencies. The exchange rate is $(RT)^{\Delta n}$ . If the reaction has the same number of gas molecules on both sides, there is nothing to convert: the price is identical in both.

More precisely

Strictly, equilibrium constants use dimensionless activities: each partial pressure is divided by the standard pressure (1 bar, or 1 atm in older tables), which is why $K$ has no units. Values of $K_\text{p}$ therefore depend slightly on whether pressures are in bar or atm. The $K$ in $\Delta G° = -RT\ln K$ is $K_\text{p}$ for gas reactions.

Visualise it

Three reactions with their Kp expressions and delta n values. N2 plus 3 H2 to 2 NH3: delta n equals minus 2, so Kp equals Kc times RT to the minus 2. H2 plus I2 to 2 HI: delta n equals 0, so Kp equals Kc. CaCO3 solid to CaO solid plus CO2 gas: only the gas appears, Kp equals the pressure of CO2. — Δn counts gas molecules only; solids and liquids are left out of K.

Worked example

Worked example: Kp from partial pressures

Question: At 25 °C, an equilibrium mixture contains $\ce{N2O4}$ at 0.70 atm and $\ce{NO2}$ at 0.32 atm. Calculate $K_\text{p}$ for $\ce{N2O4(g) <=> 2NO2(g)}$ .

\begin{aligned} &K_\text{p} = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}} = \frac{(0.32)^2}{0.70} \\[4pt] &= 0.15 \end{aligned}

Worked example: From Kc to Kp

Question: For $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ , $K_\text{c} = 0.50$ at 400 °C. Find $K_\text{p}$ .

$\Delta n = 2 - (1 + 3) = -2$ ; $T = 400 + 273.15 = 673.15$ K.
$RT$ :
$\small\begin{aligned} &RT = 0.08206\ \tfrac{\text{L atm}}{\text{mol K}} \\[4pt] &\quad \times 673.15\ \text{K} \\[4pt] &= 55.24\ \text{L atm/mol} \end{aligned}$
Then, using the number 55.24 (K values are written without units):
$\begin{aligned} &K_\text{p} = 0.50 \times (55.24)^{-2} \\[4pt] &= 1.6 \times 10^{-4} \end{aligned}$

Worked example: A heterogeneous equilibrium

Question: Write $K_\text{c}$ and $K_\text{p}$ for $\ce{C(s) + H2O(g) <=> CO(g) + H2(g)}$ .

Carbon is a pure solid, so it is left out:

\begin{aligned} &K_\text{c} = \frac{[\ce{CO}][\ce{H2}]}{[\ce{H2O}]} \\[4pt] &K_\text{p} = \frac{P_{\ce{CO}}\,P_{\ce{H2}}}{P_{\ce{H2O}}} \end{aligned}

Worked example: An ICE table with pressures

Question: $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ has $K_\text{p} = 1.05$ at a certain temperature. Pure $\ce{PCl5}$ at 1.00 atm is allowed to reach equilibrium. Find the equilibrium partial pressures.

ICE (atm): $\ce{PCl5}$ : $1.00 - x$ ; $\ce{PCl3}$ : $x$ ; $\ce{Cl2}$ : $x$ .
Substitute:
$\begin{aligned} &\frac{x^2}{1.00 - x} = 1.05 \\[4pt] &x^2 + 1.05x - 1.05 = 0 \end{aligned}$
Quadratic formula, positive root, with $a = 1$ , $b = 1.05$ , $c = -1.05$ :
$\small\begin{aligned} &x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \\[4pt] &= \frac{-1.05 + \sqrt{5.3025}}{2} \\[4pt] &= 0.626 \end{aligned}$
$P_{\ce{PCl3}} = P_{\ce{Cl2}} = 0.626$ atm and $P_{\ce{PCl5}} = 1.00 - 0.626 = 0.374$ atm. Check: $\dfrac{0.626^2}{0.374} = 1.05$ ✓.

Common mistake

Common mistake: Including solids or liquids

In $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ , $K_\text{p} = P_{\ce{CO2}}$ only. Writing $\dfrac{P_{\ce{CaO}}P_{\ce{CO2}}}{P_{\ce{CaCO3}}}$ is wrong: solids have no partial pressure.

Common mistake: Counting all moles in Δn

Δn counts gas moles only. For $\ce{C(s) + H2O(g) <=> CO(g) + H2(g)}$ , $\Delta n = 2 - 1 = +1$ , not $2 - 2$ .

Common mistake: Using the wrong R or temperature

With pressures in atm, use $R = 0.08206$ L·atm/(mol·K), not 8.314, and always convert °C to K.

Notation note

$P_{\ce{A}}$ means the partial pressure of A at equilibrium (in atm here).
Kc uses mol/L; Kp uses partial pressures. Both are quoted without units.

Remember this

$K_\text{p}$ : same form as $K_\text{c}$ , with partial pressures.
$K_\text{p} = K_\text{c}(RT)^{\Delta n}$ ; Δn = gas moles (products − reactants); R = 0.08206; T in K.
Leave pure solids and liquids out of K.
ICE tables work with pressures just as with concentrations.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/gas-phase-equilibria/