The Equilibrium Constant in Terms of Pressure

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:42


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 pressureThe pressure (or partial pressure) exerted by the gaseous form of a substance in equilibrium with the liquid form. of a liquid was always the same at a given temperatureA physical property that indicates whether one object can transfer thermal energy to another object., 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 concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated. through the ideal gas law. Rearranging PV = nRT we obtain


      P=\frac{n}{V}RT=cRT\text{                                                      (1)}


since c = amount of substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula./volume = n/V. Thus if the vapor pressure is constant at a given temperature, the concentration must be constant also. Equation (1) also allows us to relate the equilibrium constantThe value of the equilibrium constant expression when equilibrium concentrations are substituted; a value greater than one indicates the position of equilibrium lies toward products (product-favored), and a value less than one indicates the position of equilibrium lies toward reactants (reactant-favored). to the vapor pressure. In the case of water, for example, the equilibrium reaction and Kc are given by


      H2O(l) \rightleftharpoons H2O(g)Kc = [H2O(g)]


Substituting for the concentration of water vapor from Eq. (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 dm}^{-\text{3}}


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


      aA(g) + bB(g) \rightleftharpoons cC(g) + dD(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}}      (2)


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:


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}


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 1 In what SI unitsThe international system of units (Système International d'Unité) based on seven fundamental units: meter, kilogram, second, ampere, kelvin, candela, mole. will the equilibrium constant Kc be measured for the following reactions? Also predict for which reactions Kc = Kp.


a) 2NOBr(g) \rightleftharpoons 2NO(g) + Br2(g)


b) H2O(g) + C(s) \rightleftharpoons CO(g) + H2(g)


c) N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)


d) H2(g) + I2(g) \rightleftharpoons 2HI(g)


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. We apply the rule that the units are given by (mol dm–3)Δn.

a) Since Δn = 1, units are moles per cubic decimeter.

b) Since Δn = 1, units are moles per cubic decimeter (the solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. is ignored).

c) Here Δn = -2 since two gas molecules are produced from four. Accordingly the units are

mol–2 dm6.

d) Since Δn = 0, Kc is a pure number. In this case also Kc = Kp.