The equilibrium constant, Kc

Introduction:

Characteristics of the equilibrium state

 

1)   Equilibrium can only be established in a closed system.  Matter cannot be exchanged with the surroundings (this will affect the position of the equilibrium), but energy can be exchanged.

 

2)   Equilibrium can be approached from either direction.  The products can be used as the reactants to set up the equilibrium – reversible reactions. 

 

3)   Equilibrium is a dynamic state – At equilibrium the rate in both directions must be the same.

 

4)   Dynamic equilibrium is stable under fixed conditions but is sensitive to changes in temperature, pH, pressure.

The equilibrium law:

 

Kc

=

[PRODUCTS]p
[REACTANTS]r
  aA + bB D cC + dD
Kc

=

[C]c    [D]d
[A]a    [B]b

Approaching equilibrium

N2O4(g) D 2NO2(g)
  Initially:
  • N2O4 molecules decompose into 2 NO2 molecules,  the rate of the forward reaction is fast. 
  • Because there are very few molecules of NO2, the reverse reaction can only happen slowly.
  As the reaction proceeds:
  • There are now fewer N2O4 molecules available to decompose, the rate of the forward reaction decreases.
  • There are now more NO2 molecules present so the rate of the reverse reaction increases.
  At equilibrium:
  • Eventually the forward reaction takes place at exactly the same rate as the reverse reaction. 
  • Dynamic equilibrium is established and is denoted by  D
  • The concentrations of reactants and products now remain constant.

Writing expressions for Kc:

N2O4(g) D 2NO2(g)
Kc

=

[NO2]2
[N2O4]
2SO2(g) + O2(g) D 2SO3(g)
Kc

=

      [SO3]2
[SO2]2   [O2]

 

H2(g) + I2(g) D 2HI(g)
Kc

=

      [HI]2
[H2]   [I2]

Units of Kc

  Kc = [NO2]2     Kc = [SO3]2     Kc = [HI]2  
      [N2O4]1         [SO2]2   [O2]         [H2]   [I2]  
                             
  Kc = mol dm-3  x  mol dm-3   Kc = (mol dm-3)2   Kc = mol dm-3       mol dm-3 
      mol dm-3       (mol dm-3)2 mol dm-3       mol dm-3 mol dm-3
                             
  Kc = mol dm-3  x  mol dm-3   Kc = (mol dm-3)2   Kc = mol dm-3       mol dm-3 
      mol dm-3       (mol dm-3)2 mol dm-3       mol dm-3 mol dm-3
                             
  Kc = mol dm-3     Kc = 1     Kc = No units  
                mol dm-3            
                             
            Kc = dm3 mol-1            

Qu 1  P127

Calculations using Kc

Determining Kc from equilibrium concentrations:

1)  Hydrogen, Iodine, Hydrogen iodide equilibrium:

Reaction: H2(g) + I2(g) D 2HI(g)
Equilibrium concentrations: 0.140   0.040   0.320
  Kc = [HI]2       Kc = [HI]2  
      [H2]   [I2]           [H2]   [I2]  
                     
  Kc = (0.320)2     Kc = mol dm-3       mol dm-3 
      0.140  x  0.040         mol dm-3 mol dm-3
                     
  Kc = 18.3       Kc = mol dm-3       mol dm-3 
                mol dm-3 mol dm-3
                     
              Kc = No units  

2)  N2O4  /  NO2 equilibrium:  Given the number of moles at equilibrium in a volume of 2dm3.

Reaction: N2O4(g) D 2NO2(g)
Mole quantities at equilibrium: 0.400   3.20
Equilibrium concentrations: 0.400 / 2   3.20 / 2
Equilibrium concentrations: 0.20   1.60
  Kc = [NO2]2       Kc = [NO2]2  
      [N2O4]           [N2O4]  
                     
  Kc = (1.60)2     Kc = mol dm-3       mol dm-3 
      0.20         mol dm-3
                     
  Kc = 12.8 mol dm-3     Kc = mol dm-3       mol dm-3  
                mol dm-3
                     
              Kc = mol dm-3  

Calculating the quantities and concentrations present at equilibrium:

3)  Hydrogen, Iodine, Hydrogen iodide equilibrium:

Reaction: H2(g) + I2(g) D 2HI(g)
At start: 0.60   0.40   0.0
At equilibrium: 0.28   0.08   0.64
Reacted: 0.32   0.32   0.64
Equilibrium concentrations: 0.28   0.08   0.64
  Kc = [HI]2       Kc = [HI]2  
      [H2]   [I2]           [H2]   [I2]  
                     
  Kc = (0.64)2     Kc = mol dm-3       mol dm-3 
      0.28  x 0.08         mol dm-3 mol dm-3
                     
  Kc = 18.3       Kc = mol dm-3       mol dm-3 
                mol dm-3 mol dm-3
                     
              Kc = No units  

Practical          Workbook          Calculate Kc

Qu 1 - 2  P129

The equilibrium position and Kc

What is the significance of a Kc value?

 

Products favoured:    Kc > 1

Reactants favoured:    Kc < 1

How do changes in temperature affect Kc?

1)  Endothermic reactions:

 

  • The top number increases and the bottom number decreases.

  • This makes Kc a larger value.

                            Kc increases

                            and vice versa

2)  Exothermic reactions:

 

  • The top number decreases and the bottom number increases.

  • This makes Kc a lsmaller value.

                            Kc decreases

                            and vice versa

Qu 1  P131

The equilibrium constant, Kc, and the rate constant, k

How does a change in concentration and pressure affect Kc?

                                Kc is unaltered by changes in concentration and pressure

1)  Changes in concentration:

                                        A change in Concentration has no effect on the equilibrium constant.

Reaction: N2O4(g) D 2NO2(g)
[NO2] = 1.60 Mol dm-3   [NO2] = 1.60 Mol dm-3  
[N2O4] = 0.20 Mol dm-3 Double [  ] : [N2O4] = 0.40 Mol dm-3    
                   
  Kc = [NO2]2   Kc = [NO2]2    
      [N2O4]       [N2O4]    
                   
  Kc = (1.60)2   Kc = (3.20)2    
      0.20       0.40    
                   
  Kc = 12.8

Mol dm-3

Kc = 6.4 Mol dm-3

2)  Changes in Pressure:

                                        A change in Pressure has no effect on the equilibrium constant.

Reaction: N2O4(g) D 2NO2(g)
[NO2] = 1.60 Mol dm-3 Double [  ] : [NO2] = 3.20 Mol dm-3  
[N2O4] = 0.20 Mol dm-3   [N2O4] = 0.40 Mol dm-3    
                   
  Kc = [NO2]2   Kc = [NO2]2    
      [N2O4]       [N2O4]    
                   
  Kc = (1.60)2   Kc = (3.20)2    
      0.20       0.40    
                   
  Kc = 12.8

Mol dm-3

Kc = 25.6 Mol dm-3

How does the presence of a catalyst affect Kc?

                                        A catalyst has no effect on the equilibrium constant.

The equilibrium constant Kc, and the rate constant, k

a) Equilibrium How far
b) Rates How fast

 

a)  The equilibrium constant, Kc

  Large Kc Products predominate
  Small Kc Reactants predominate
  Endothermic Kc increases with an increase in temperature (increases products)
  Exothermic Kc decreases with an increase in temperature (decreases products)

                    Kc can be written from the balanced chemical equation

b)  The rate constant, k

  Large k Fast rate
  Small k Slow rate
  k increases with an increase in temperature - Rate increases with an increase in temperature
  k decreases with a decrease in temperature - Rate decreases with a decrease in temperature

                    k can only be determined experimentally from the rate equation

The importance of compromise:

a)  Increasing the rate:

Increasing temperature:  Increases the rate of production of product - desirable

Increasing temperature:  Decreases the amount of product made - undesirable

b)  Increasing the amount:

Decreasing the temperature:  Increases the amount of product made - desirable

Decreasing temperature:  Decreases the rate of production of product - undesirable

The compromise is:

Moderate temperature: k increases moderately - Rate increases by a moderate amount
Moderate temperature: Kc decreases by a moderate amount - allowing a moderate yield

Qu 1  P133  /  Qu 4 - 5  P159  /  Qu 3 - 4  P161