Equilibrium

 

Consider the reaction

         N2O4 ® 2 NO2

After awhile, the reverse rxn start going.

         2 NO2 ® N2O4       

These two reactions are both thought to be elementary reactions

 

1)      N2O4 ® 2 NO2                 R1 = k1 [N2O4]

 

2)      2 NO2 ® N2O4                 R2 = k2 [NO2]2

 

The second reaction is the reverse of the first.  If we start will pure dinitrogen tetroxide at some temperature, this molecule will break apart to form nitrogen dioxide.  As more N2O4 decomposes, the first reaction rate decreases and the second reaction speeds up.  There will come a time when the two rates are equal and the two concentrations will not change.

 

At this time

 

         Rate1=R2

 

k1 [N2O4]  = k2 [NO2]2

 

         k2 / k1 = [NO2]2 /  N2O4

 

Both the forward and reverse reaction continue, but since the rates are equal, we don’t see a change in either concentration.  This state where the forward rate equals the backward rate, where the reactants and products concentration is unchanging is called dynamic equilibrium.  We discussed dynamic equilibrium when we discussed vapor pressure, when the rate of the condensed phase vaporizing equals the rate of the vapor condensing.  This quotient in bold above, is called the reaction quotient, Q.

 

At 100.°C, when this system reaches equilibrium, Q = .211

 

When the concentrations are no longer changing Q = Kc , which is called the equilibrium constant.  So for the above reaction, at 100.°C, Kc = .211.  We specify the T, because the rate constants k1  and k2 change different amounts with the change in temperature.  Thus a change in temperature will change Kc.

 

? if [NO2] = .0300 M and [N2O4] = .0100 M at 100.°C, is the system at equilibrium?

Calculate Q and see if it = .211

In this case Q < .211.

 

Since Q = [NO2]2 /  N2O4 and Q < .211, we need more product and less reactant to increase the value of Q.  We say the reaction will shift right to come to equilibrium.  This will make more products and consume reactants. 

 

If Q < Kc, the reaction shifts right to make more products and

If Q > Kc , the reaction shifts left to make more reactants.

 

In general, for any homogeneous reaction where all reactants and products are in the same phase,

 

         aA + bB = cC + dD

 

The mass action expression is

 

         Q = [C]c [D]d  /  [A]a [B]b 

 

         This = Kc  (@ equilibrium)

 

Examples

 

         Overall Reaction                                        Mass Action Expression

 

(1)     2 O3 = 3 O2                                                Q = [O2]3 / [O3]2

 

(2)     N2 + 3 H2 = 2 NH3                                        Q = [NH3]2 / [N2] [H2]3

 

(3)     SO2 + ½ O2 = SO3                                        Q = [SO3] / [SO2] [O2]1/2

 

(4)     2 SO2 + O2 = 2 SO3                                      Q = [SO3]2 / [SO2]2 [O2]

 

(5)     2 SO3 = 2 SO2 + O2                                      Q =  [SO2]2 [O2] / [SO3]2

 

Note that in reaction 4, we multiplied reactants and products by 2.  The effect on Q is that Q4= Q32

Reaction 5 is the reverse of equation 4, and Q5 = 1/Q4.

 

For gas phase reactions, it might be easier to measure partial pressures, so we could write expression in terms of P

 

         PV = nRT

 

         P = (n/V) RT = MRT

 

         Kp = PCc PDd / PAa PBb

 

Relate Kp to Kc  for the following reaction

 

         N2 + 3 H2 = 2 NH3   

 

         H2 + I2 = 2 HI

 

Consider

         N2 + O2 = 2 NO     at 25°C   Kc = 4.5 X 10-31

 

Compare the size of the numerator and denominator.

 

         at 2300 K Kc = .0017 for this rxn.

 

So, which way did the rxn shift at T increased?

 

         2 O3 = 3 O2   Kp = 8.5 X 1085

 

Compare the size of the numerator and denominator