13.5: The Equilibrium Constant in Terms of Pressure


Some equilibria involve physical instead of chemical processes. One example is the equilibrium between liquid and vapor in a closed container. In other sections we stated that the vapor pressure of a liquid was always the same at a given temperature, regardless of how much liquid was present. This can be seen to be a consequence of the equilibrium law if we recognize that the pressure of a gas is related to its concentration through the ideal gas law. Rearranging PV = nRT we obtain

$P=\frac{n}{V}RT=cRT\label{1}$

since c = amount of substance/volume = n/V. Thus if the vapor pressure is constant at a given temperature, the concentration must be constant also. Equation $$\ref{1}$$ also allows us to relate the equilibrium constant to the vapor pressure. In the case of water, for example, the equilibrium reaction and Kc are given by:

$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g) \nonumber$ $\text{K}_c = [\text{H}_2\text{O}(g)] \nonumber$

Substituting for the concentration of water vapor from Equation $$\ref{1}$$, we obtain

$K_{c}=\frac{P_{\text{H}_{\text{2}}\text{O}}}{RT} \nonumber$

At 25°C for example, the vapor pressure of water is 17.5 mmHg (2.33 kPa), and so we can calculate

$K_{c}=\frac{\text{2}\text{.33 kPa}}{\text{(8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{)(298}\text{.15 K)}} \nonumber$

$=\text{9}\text{.40 }\times \text{ 10}^{-\text{4}}\text{ mol/L} \nonumber$

For some purposes it is actually more useful to express the equilibrium law for gases in terms of partial pressures rather than in terms of concentrations. In the general case:

$a\text{A}(g) + b\text{B}(g) \rightleftharpoons c\text{C}(g) + d\text{D}(g) \nonumber$

The pressure-equilibrium constant Kp is defined by the relationship:

$K_{p}=\frac{p_{\text{C}}^{c}p_{\text{D}}^{d}}{p_{\text{A}}^{a}p_{\text{B}}^{b}} \nonumber$

where pA is the partial pressure of component A, pB of component B, and so on. Since pA = [A] × RT, pB = [B] × RT, and so on, we can also write as follows:

\begin{align} K_{p} & =\frac{p_{\text{C}}^{c}p_{\text{D}}^{d}}{p_{\text{A}}^{a}p_{\text{B}}^{b}}=\frac{\text{( }\!\![\!\!\text{ C }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{c}\text{( }\!\![\!\!\text{ D }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{d}}{\text{( }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{a}\text{( }\!\![\!\!\text{ B }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{b}} \\ & =\frac{\text{ }\!\![\!\!\text{ C }\!\!]\!\!\text{ }^{c}\text{ }\!\![\!\!\text{ D }\!\!]\!\!\text{ }^{d}}{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }^{a}\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }^{b}}\text{ }\times \text{ }\frac{\text{(}RT\text{)}^{c}\text{(}RT\text{)}^{d}}{\text{(}RT\text{)}^{a}\text{(}RT\text{)}^{b}} \\ & =K_{c}\text{ }\times \text{ (}RT\text{)}^{\text{(}c\text{ + }d\text{ }-\text{ }a\text{ }-\text{ }b\text{ )}} \\ & =K_{c}\text{ }\times \text{ (}RT\text{)}^{\Delta n} \end{align} \nonumber

Again $$Δn$$ is the increase in the number of gaseous molecules represented in the equilibrium equation. If the number of gaseous molecules does not change, Δn = 0, Kp = Kc, and both equilibrium constants are dimensionless quantities.

Example $$\PageIndex{1}$$ : SI Units

In what SI units will the equilibrium constant Kc be measured for the following reactions? Also predict for which reactions Kc = Kp.

1. $$\text{2NOBr}(g) \rightleftharpoons \text{2NO}(g) + \text{Br}_2(g)$$
2. $$\text{H}_2\text{O}(g) + \text{C}(s) \rightleftharpoons \text{CO}(g) + \text{H}_2(g)$$
3. $$\text{N}_2(g) + \text{3H}_2(g) \rightleftharpoons \text{2NH}_3(g)$$
4. $$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons \text{2HI}(g)$$

Solution

We apply the rule that the units are given by (mol dm–3)Δn.

1. Since Δn = 1, units are moles per cubic decimeter.
2. Since Δn = 1, units are moles per cubic decimeter (the solid is ignored).
3. Here Δn = -2 since two gas molecules are produced from four. Accordingly the units are mol–2 dm6.
4. Since Δn = 0, Kc is a pure number. In this case also Kc = Kp.