Question

1)Given the following reaction, determine the equilibrium constant. Set up the equilibrium table.

3 NO ç==è N₂O + NO₂ (all gases)

1.95 moles of NO initially in a 3.78 liter container and 87.2 % reacts. Determine the equilibrium constant for the above reaction and the equilibrium constant for 3 N₂O + 3 NO₂ ===è 9 NO.

Answer 1

initial concentration of NO = n/V

            = 1.95/3.78

       = 0.516 M    

   3 NO(g) <===>    N2O(g) +   NO2(g)

initial   0.516 M          0 M        0 M

change 0.516*87.2/100     0.45/3     0.45/3

            = 0.45 M      = 0.15 M   = 0.15 M

equilibrium 0.516-0.45     0.15 M      0.15 M

              = 0.066 M

         Kc = [N2O][NO2]/[NO]^3

         = (0.15*0.15)/(0.066)^3

         = 78.262

9 NO(g) <===>   3 N2O(g) + 3 NO2(g) k1 = (Kc)^3

   and

3 N2O(g) + 3 NO2(g) <===> 9 NO(g)   K2 = 1/(Kc)^3

   k2 = (1/78.262^3) = 2.09*10^-6