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Homogeneous equilibria

A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here.

C 2 H 2 ( a q ) + 2 Br 2 ( a q ) C 2 H 2 Br 4 ( a q ) K c = [ C 2 H 2 Br 4 ] [ C 2 H 2 ] [ Br 2 ] 2
I 2 ( a q ) + I ( a q ) I 3 ( a q ) K c = [ I 3 ] [ I 2 ] [ I ]
Hg 2 2+ ( a q ) + NO 3 ( a q ) + 3 H 3 O + ( a q ) 2 Hg 2+ ( a q ) + HNO 2 ( a q ) + 4 H 2 O ( l )
K c = [ Hg 2+ ] 2 [ HNO 2 ] [ Hg 2 2+ ] [ NO 3 ] [ H 3 O + ] 3
HF ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + F ( a q ) K c = [ H 3 O + ] [ F ] [ HF ]
NH 3 ( a q ) + H 2 O ( l ) NH 4 + ( a q ) + OH ( a q ) K c = [ NH 4 + ] [ OH ] [ NH 3 ]

In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H 2 O( l ) is the solvent for these solutions, its concentration does not appear as a term in the K c expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.

Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.

C 2 H 6 ( g ) C 2 H 4 ( g ) + H 2 ( g ) K c = [ C 2 H 4 ] [ H 2 ] [ C 2 H 6 ]
3 O 2 ( g ) 2 O 3 ( g ) K c = [ O 3 ] 2 [ O 2 ] 3
N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g ) K c = [ NH 3 ] 2 [ N 2 ] [ H 2 ] 3
C 3 H 8 ( g ) + 5 O 2 ( g ) 3 CO 2 ( g ) + 4 H 2 O ( g ) K c = [ CO 2 ] 3 [ H 2 O ] 4 [ C 3 H 8 ] [ O 2 ] 5

Note that the concentration of H 2 O( g ) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.

Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, n V .

P V = n R T
P = ( n V ) R T
= M R T

Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.

Using the partial pressures of the gases, we can write the reaction quotient for the system C 2 H 6 ( g ) C 2 H 4 ( g ) + H 2 ( g ) by following the same guidelines for deriving concentration-based expressions:

Q P = P C 2 H 4 P H 2 P C 2 H 6

In this equation we use Q P to indicate a reaction quotient written with partial pressures: P C 2 H 6 is the partial pressure of C 2 H 6 ; P H 2 , the partial pressure of H 2 ; and P C 2 H 6 , the partial pressure of C 2 H 4 . At equilibrium:

K P = Q P = P C 2 H 4 P H 2 P C 2 H 6

The subscript P in the symbol K P    designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, K P , is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.

Conversion between a value for K c    , an equilibrium constant expressed in terms of concentrations, and a value for K P , an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).

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Practice Key Terms 7

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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