Calculate the % dissociation of N₂O₄(g):

         N₂O₄(g)   ---->    2 NO₂(g)
                         <----
Start with [N₂O₄] = 0.0500 M, and K_c= 4.6 x 10^-4 at 25^oC.
Find % dissociation.
                                  K_c = [NO₂]²/[N₂O₄]

                   N₂O₄(g)            NO₂(g)
Init []         0.0500             0
D[]              -x                     +2x
Equil []       0.0500-x         +2x

K_c = 4.6 x 10^-4 = (2x)²/0.0500-x = 4x²/0.0500-x
        2.30 x 10^-5 - 4.6 x 10^-4x - 4x² = 0
        4x² + 4.6 x 10^-4x - 2.30 x 10^-5 = 0  (use quadratic to
                    solve)

        x1 = 0.00228M  (correct solution)
        x2 = -0.00246M  (impossible because of - molarity!)

Therefore:  [N₂O₄] = 0.0500 - x = 0.0500 - 0.00228
                    [N₂O₄] = 0.0477M
                    [NO₂] = 2x = 2(0.00228) = 0.00456M

and % dissociation = x/[N₂O₄] x 100
   % dissociation = (0.00228/0.0500) x 100
                             = 4.56 % dissociated

NOTE:  % dissociated =
  (how much went to the right/initial concentration) x 100
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In the equilibrium mixture for the reaction:

        H₂(g) + Cl₂(g) ---->  2 HCl(g)
                                 <----
the following partial pressures were measured: 
    P_H2 = 0.387 atm,
    P_Cl2 = 0.152 atm, and P_HCl = 0.018 atm.  Calculate
        the value of the equilibrium constant, K_p.
      K_p = P²_HCl/(P_H2)(P_Cl2) = (0.018)²/(0.152)(0.180)
      K_p = 1.2 x 10^-2
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Relationship between K_p and K_c

From the ideal gas law: PV = nRT     
                    or   P/RT = n/V = moles/L = M = molar
                          P/RT = M 

For            N₂O₄(g)   ---->    2 NO₂(g)
                                 <----

                 K_c = [NO₂]²[N₂O₄]  =  (P_NO2/RT)²/P_N2O4/RT

                 K_c = [(P_NO2)²/P_N2O4] [(1/RT)²/(1/RT)]

                 K_c = K_p (1/RT)¹ = K_p (RT)^-1     
                    
         OR   K_p= K_c (RT)¹

Or, in general, K_p = K_c (RT)^Dn
 where Dn = #moles gas products - #moles gas reactants

Therefore, when equal # moles of gas on each side,
        K_p = K_c (RT)⁰    and K_p = K_c  because Dn = 0

CAUTION::!!!!!     R = 0.0821 L-atm/mol-K

Ex.: The equilibrium constant, K_p =1.92, at 252^oC for the
decomposition reaction of phosphorus pentachloride
according to the following equation:

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

a)  Calculate the partial pressures of all species present
      after 6.0 moles of PCl₅(g) are placed in an evacuated
      3.0L container and equilibrium is reached.

b)  Calculate the value of K_c.
 

SOLUTIONS:
a)  K_p = (P_PCl3)(P_Cl2)/(P_PCl5) = 1.92

     PV = nRT;  INITIALLY, P OF PCl₅(g) IS TOTAL P OF
                        SYSTEM
 P = nRT/V = (6.0mol)(0.0821L-atm/mol-K)(525K)/3.0L
 P = 86.2 atm

                  PCl₅(g)          PCl₃(g)            Cl₂(g)
Init P         86.2               0                      0
DP              -x                   +x                   +x
Equil P       86.2-x           +x                   +x

          K_p = (P_PCl3)(P_Cl2)/(P_PCl5) = 1.92 = (x)(x)/86.2-x
      1.92 = x²/86.2-x
                 x²  + 1.92x - 165.5 = 0   (use quadratic)

           x = 11.9 atm
Therefore, P_PCl5 = 86.2 - x = 86.2 - 11.9 = 74.3 atm
                   P_PCl3 = P_Cl2= x = 11.9 atm

Check:  K_p = (P_PCl3)(P_Cl2)/(P_PCl5) = (11.9)(11.9)/74.3
              K_p = 1.91

b)  K_p = K_c (RT)^Dn    or     K_c  = K_p (RT)^-Dn

   K_c = K_p (RT)^-Dn = 1.92 [(0.0821L-atm/mol-K)(525K)]^-1
   K_c = 1.92 /[(0.0821L-atm/mol-K)(525K)] = 0.0445
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Relationship between DG^o and Equilibrium Constant K:

    DG(reaction) = DG^o(reaction) + RT ln Q

    where DG^o is standard free energy for a reaction when
    reactants and products are in standard states.

     DG is standard free energy for the reaction for any
     other concentrations and pressures.

However, at equilibrium, DG = 0.
            DG(reaction) = 0 = DG^o(reaction) + RT ln Q

Therefore, DG^o(reaction) = - RT ln K  
                                                      (EQUILIBRIUM CONSTANT)

*When this is used with all gaseous reactants and
    products, then K = K_p

*When this is used with all solution reactants and
    products, then K = K_c

*When this is used with a mixture of gaseous and solution
    reactants and products, then K is the thermodynamic
    equilibrium constant and we do not make any distinction
    between K_p and K_c

When DG^o(reaction) < 0, reaction is spontaneous as it is
    written, and K>1

When DG^o(reaction) = 0, reaction is at equilibrium with    
    concentration of products exactly equal to concentration
    of reactants and K = 1     (this is very rare).

When DG^o(reaction) > 0, reaction is non-spontaneous as
    written, and K < 1.
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Ex.:  Find K_p for the reaction:  2 O₃(g) ---> 3 O₂(g)  at 25^oC
                                                                <---

DG^o(reaction) = 3 DG^o_f [O₂(g)] - 2 DG^o_f [O₃(g)]
                        = 3(0) - 2(163kJ/mol)
                        = -326 kJ/mol
Therefore, DG^o(reaction) = - RT ln K_p
                  DG^o = -(8.314 J/mol-K)(298K)(ln K_p)
     (-326000 J/mol)/-(8.314 J/mol-K)(298K) = ln K_p
                 131.58 = ln K
                        1.39 x 10⁵⁷ = K_p      Therefore, MUCH more
            products than reactants!

**************************************************************************
How about K at different T?         van't Hoff equation:

  ln (K₂/K₁) = (DH^o/R)[(1/T₁)-(1/T₂)].   This relates
    equilibrium constant at various temperatures to
    enthalpy.

Ex.:   K_p = 3.8 x 10⁴  at 25^oC for    
            N₂(g) + 3 H₂(g) ----> 2 NH₃(g)
                                       <----
If DH^o is known to be 150.0 kJ/mol, what is  K_p at 200^oC?

              ln K₂ - ln K₁ = (DH^o/R)[(1/T₁)-(1/T₂)]
  ln K₂ -ln(3.8 x 10⁴) =
    [(150.0 kJ/mol)/(8.314 J/mol-K)][(1/298)-1/473)]
  ln K₂ - 10.545 = (150000/8.314)(0.003356 - 0.002114)
   ln K₂ - 10.545 = 18042(0.001242) = 22.405
  Therefore,  ln K₂ = 32.950
And, therefore,  K_p  at 200^oC = 2.04 x 10¹⁴

As raise the temperature from room temp. (25^oC) to 200^oC,
the equilibrium shifts more to the right by a factor of about
10¹⁰.   Therefore, many more products!!

CAREFUL!!!  If K_c was given instead of K_p, you would have
    to convert K_c to K_p before using the van't Hoff equation!!

However, if the problem involved ONLY solutions and
    concentrations, then you could use K_c

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