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 = n RT 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 K c 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 K p 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 p A is the partial pressure of component A, p B of component B, and so on. Since p A = [A] × RT, p B = [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, K p = K c , and both equilibrium constants are dimensionless quantities. Example \(\PageIndex{1}\) : SI Units In what SI units will the equilibrium constant K c be measured for the following reactions? Also predict for which reactions K c = K p . \(\text{2NOBr}(g) \rightleftharpoons \text{2NO}(g) + \text{Br}_2(g)\) \(\text{H}_2\text{O}(g) + \text{C}(s) \rightleftharpoons \text{CO}(g) + \text{H}_2(g)\) \(\text{N}_2(g) + \text{3H}_2(g) \rightleftharpoons \text{2NH}_3(g)\) \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons \text{2HI}(g)\) Solution We apply the rule that the units are given by (mol dm –3 ) Δ n . Since Δn = 1, units are moles per cubic decimeter. Since Δn = 1, units are moles per cubic decimeter (the solid is ignored). Here Δn = -2 since two gas molecules are produced from four. Accordingly the units are mol –2 dm 6 . Since Δn = 0, K c is a pure number. In this case also K c = K p .