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Chemistry LibreTexts

13.4: The Equilibrium Constant in Terms of Pressure

  • Page ID
    49519
  • [ "article:topic", "ChemPrime", "Ideal Gas Law", "partial pressure", "pressure-equilibrium constant", "authorname:chemprime", "showtoc:no" ]

    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:

     

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

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

     

    \[K_{c}=\frac{P_{\text{H}_{\text{2}}\text{O}}}{RT}\]

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

     

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

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

     

    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:

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

     

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

     

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

    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:

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

     

    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.

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