Determination of the Equilibrium Constant

In the previous section we determined the equilibrium composition of reaction mixtures starting with a certain starting composition. If we start with a different initial composition, the mixture will again reach equilibrium but the equilibrium composition will be different this time. The condition of equilibrium is one in which the composition is not changing it is not some particular composition, in fact a particular reaction can have an infinite number of different compositions and be at equilibrium.

So how can we characterize the condition of chemical equilibrium if we cannot use composition.  It turns out that there is a certain ratio of concentrations which will be the same or very nearly the same for all of these infinite compositions at equilibrium.  This ratio is called the equilibrium constant. Keep in mind that this ratio will only be a constant at a particular temperature.  The equilibrium constant is a number for which units are customarily omitted.  The equilibrium constant is calculated from an equilibrium constant expression which is an equation.  The equilibrium constant expression can be derived from theoretical considerations, but we will have to postpone that for a while. What we are going to do here is present a set of rules for writing an equilibrium constant expression from a balanced chemical equation.

For the balanced reaction:

aA + bB <==> cC + dD     The equilibrium constant, Keq is defined as:

      [C]c [D]d
Keq = ---------
      [A]a [B]b
where the [ ] brackets indicate the equilibrium concentration of the chemical species

 

Rules for Writing Equilibrium Constant Expressions

  1. Products are always in the numerator.
  2. Reactants are always in the denominator.
  3. Express gas concentrations as partial pressure, P, and dissolved species in molar concentration, [ ].
  4. The partial pressures or concentrations are raised to the power of the stoichiometric coefficient for the balanced reaction.
  5. Leave out pure solids or liquids and any solvent.

Example 1

Zn (s) + 2 H+(aq) <==> Zn2+(aq) + H2 (g) 

     PH2 [Zn2+]
K = -----------
       [H+]2
where PH2 is the partial pressure of H2, [Zn2+] and [H+] are the molar concentrations of Zn2+ and H+, respectively, and Zn (s) is left out of the K expression because it is a pure solid. This example was of a heterogeneous equilibrium, that is one in which all substances are not in the same phase.

Example 2

What is the equilibrium constant expression for the reaction

4 NH3(g)  +  3 O2(g) <==> 2 N2(g)  +  6 H2O(g)

Notice that water is in the gaseous form. If it was in the liquid or solid form we would not put it in the equilibrium constant expression.

Using the rules above we get

     [N2]2 [H2O]6
Kc = -----------
     [NH3]4 [O2]3

The subscript 'c' on the K indicates that concentrations, as opposed to say partial pressures, were used in the equilibrium constant expression.

Calculating the Equilibrium Constant for Reactions

Example 3

What is the equilibrium constant for the reaction 

CO(g)  +  3 H2(g)  <==>  CH4(g)  +  H2O(g)

if at a temperature the equilibrium molar concentrations are [CO] = 0.613, [H2] = 1.839, [CH4] = 0.387, and [H2O] = 0.387?

Step 1: Write the equilibrium constant expression, which is:

     [CH4][H2O]
Kc = -----------
     [CO][H2]3

Step 2: Plug in the numbers and solve: 

     [0.387][0.387]
Kc = --------------   = 0.0393
     [0.613][1.839]3

Next we consider a slightly more complicated example.

Example 4

Obtain the value of Kc for the following reaction at a temperature 300 K using the following data.  A 1.355-mol sample of PCl5 was placed into a 5.00 L and allowed to reach equilibrium.  At equilibrium the mixture was found to contain 0.188 mol of Cl2.  

PCl5(g) <==> PCl3(g) + Cl2(g) 

We will set up a table similar to the ones we used in the previous topic, except this time we want molar concentrations instead of moles.  This is because we need molar concentrations in the equilibrium constant expression.  From the information given in the problem, the initial concentration of PCl5 is 1.355 mol / 5.00 L = 0.271 M and the equilibrium concentration of  Cl2 is 0.188 mol / 5.00 L = 0.0376 M.  Now we can build our table.

Concentrations(M) PCl5(g) <==> PCl3(g)
+
 Cl2(g)
Starting 0.271 0 0
Change -x +x +x
Equilibrium 0.271 - x x x = 0.0.0376 

Thus the equilibrium concentrations will be [PCl5] = (0.271 - x) M = (0.271 - 0.038) M = 0.233 M and [PCl3] = [Cl2] = 0.0376.  Next we will set up the equilibrium constant expression and plug in these values. 
     [PCl3][Cl2]
Kc = -----------
       [PCl5]

     [0.0376][0.0376]
Kc = ---------------   = 6.07 x 10-3
         [0.233]
 

 And now one final example.

Example 5

At 80oC 4.00 mol of nitrosyl bromide, NOBr, placed in a 2.00-L vessel dissociates to the extent of 8.9%; that is, for each mole of NOBr before reaction, (1.00 - 0.089) mol NOBr remains at equilibrium.  Calculate the value of Kc for the reaction.

2 NOBr(g)  <==>  2 NO(g)  +  Br2(g)

Set up the usual table.

Concentrations(M) 2 NOBr(g) <==> 2 NO(g)
+
 Br2(g)
Starting 2.00 0 0
Change -2x +2x +x
Equilibrium 2.00 - 2x 2x x 

The moles of NOBr at equilibrium is 4(1.00 - 0.089) = 3.644 mol, thus the equilibrium concentration is 3.644 mol / 2.00 L = 1.822 M.  Use this value to solve for x. 1.822 = 2.00 - 2x therefore x = 0.089 M.  Set up the equilibrium constant expression and plug in these values.

 

     [NO]2[Br2]
Kc  = -----------
       [NOBr]2



     [0.178]2[0.089]
Kc = ---------------   = 8.5 x 10-4
         [1.822]2

 

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