Applications of the Equilibrium Constant

Qualitative Interpretation of the Equilibrium Constant

Equilibrium constants can be very large or very small.  The magnitude of the constant provides us with important information about the equilibrium mixture.  For example consider the hypothetical reaction 

                                    [C]
 A(g)  +  B(g)  <==>  C(g)   Kc = ------- = 5.00 x 109
                                   [A][B]

In order for the  equilibrium constant to be so large, the numerator of the equilibrium constant expression must be much larger than the denominator.  This must mean that equilibrium concentration of the product must be much larger than that of the reactants.  It means, with this large a value of K, that when equilibrium is reached reactants will have been completely converted to products.  Keep in mind that the magnitude of the equilibrium constant has nothing to do with how long the mixture will take to reach equilibrium.  In general we can say

K >> 1: Equilibrium lies to the right; products favored.
K << 1: Equilibrium lies to the left; reactants favored.

Predicting the Direction of Reaction 

Suppose we place a mixture of 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 in a 1.00-L container at 472oC, would the reaction

N2(g)  +  3 H2(g)  <==>  2 NH3(g)  Kc = 0.105

proceed toward the right or the left? To answer this question, we determine the value of a quantity known as the reaction quotient, Qc and compare this to Kc.  The reaction quotient is an expression having the same form as the equilibrium constant expression but whose concentration values are not equilibrium concentrations, but usually initial concentrations.  In general

If Qc > Kc, the reaction will go to the left.

If Qc < Kc, the reaction will go to the right.

If Qc= Kc, the reaction is at equilibrium.

For the reaction above Qc =

          {NH3}2        (2.00)2
    Qc = --------- = -------------- = 0.500
         {N2}{H2}3    (1.00)(2.00)3 
Since Qc > Kc the reaction the reaction will proceed to the left to reach equilibrium.

Calculating Equilibrium Concentrations

Chemists frequently need to calculate equilibrium concentrations.  The approach we will use in solving problems of this type is similar to the one we used for calculating the equilibrium constant.  We tabulate the initial concentrations, the changes in these concentrations, and the final equilibrium concentrations.  Usually we end up using the equilibrium constant expression to derive an equation that must be solved, as we illustrate in the following examples.

Example 1

A 2.00-L flask contains an unknown amount of PCl5 and 0.040 mol each of  PCl3 and Cl at equilibrium at 600 K.  How many moles of  PCl5 are in the flask if Kc for this reaction is 0.0420 at this temperature.  

PCl5(g) <==> PCl3(g) + Cl2(g

     [PCl3][Cl2]             (0.020)(0.020)
Kc = ----------- = 0.0420 = ---------------
       [PCl5]                  [PCl5]
         (0.020)(0.020)
[PCl5] = ---------------   = 9.5 x 10-3
            0.0420
 

 The next example deals with the situation where we begin a reaction with known quantities of substances and wish to calculate the quantities at equilibrium.

Example 2

A mixture of 0.00623 mol H2, 0.00414 mol I2, and 0.0224 mol HI was placed in a 1.00-L flask at 700K.  The equilibrium constant Kc for the reaction

H2(g)  +  I2(g)<==>  2 HI(g)    

is 54.3 at this temperature.  Calculate the equilibrium concentration of these species.

Solution:

This is a good place to use the reaction quotient, to decide which direction the reaction will proceed.

          {HI}2        (0.0224)2
    Qc = --------- = ------------------ = 19.5, since Qc < Kc  reaction goes right
         {H2}{I2}3    (0.00623)(0.00414) 

Set up the usual table.  Let -x be the change in concentration of H2 and I2 and from the stoichiometry of the reaction the change in concentration of HI must be 2x.

Concentrations(M) H2(g) + I2(g) <==>
2 HI(g)
Starting 0.00623 0.00414 0.0224
Change -x -x +2x
Equilibrium 0.00623 - x 0.00414 - x 0.0224 + 2x 

Substitute these equilibrium values into the equilibrium constant expression.
              [HI]2         (0.0224 + 2x)2  
Kc  = 54.3 = -------- = --------------------------
             [H2][I2]    (0.00623 - x)(0.00414 - x)

Carry out the multiplications 
54.3(2.58 x 10-5 - 0.0104x + x2) = 5.02 x 10-4 + 0.0896x + 4x2 
Collecting terms we get 50.3x2 - 0.654x + 8.98 x 10-4 = 0  use the quadratic formula to solve.
    
     0.654  ± [(-0.654)2 - 4(50.3)(8.98 x 10-4)]1/2
x = ----------------------------------------------     the 1/2 power means square root   
                     2(50.3)
x = 0.0114 M  or x = 0.00156 M  The first solution is physically impossible since it is larger
than the amount of H2 and I2 originally present.
At equilibrium the concentrations are
[H2] = (0.00623 - 0.00156)M = 0.00467 M
[I2] = (0.00414 - 0.00156)M = 0.00258 M
[HI] = (0.0224 - 2 x 0.00156)M = 0.0255 M
 
Relations Between Equilibrium Constants

Below are some useful relationships of equilibrium constants.

1. The equilibrium constant of a reverse reaction is the reciprocal of the forward reaction.

     aA  +  bB  <==>  cC  +  dD    K1

     cC  +  dD  <==>   aA  +  bB   K2 = 1/K1   (This will be obvious if you write out the equilibrium constants for K1 and K2 and compare them.)

2. If an equation is multiplied by a constant the equilibrium constant of the new reaction is the equilibrium constant of the original reaction raised to the power of the constant.

 aA  +  bB  <==>  cC  +  dD    K1

 naA  +  nbB  <==>  ncC  +  ndD    K3 = K1 

2. If an equation can be obtained by taking the sum of other equations, the equilibrium constant of the new reaction equals the product of the equilibrium constants of the other equations.


aA  +  bB  <==>  cC  +  dD    K1 

 cC  +  dD  <==> eE  +  fF      K2 

--------------------------------------------

aA  +  bB  <==>  eE  +  fF      K3 = K1K2 

 

 

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