Question

Chem 142, Dr. Haefner Consider the following equilibrium for the formation of nitrogen dioxide from the reaction of NO (g) with oxygen gas 4. K-1.71 x 1012 2NO (g) O (g)2NO2(g) Calculate the equilibrium concentration all species if 0.00750 moles of NO2 are placed in a 25 mL glass bulb over

Answer 1

        conc of NO2    = no of moles/volume in L

                                = 0.0075/0.025   = 0.3M

     2NO(g)   + O2(g) -----------> 2NO2(g)

I               0 0    0.3

C    2x    x    -2x

E 2x x 0.3-2x

   K    = [NO2]^2/[NO]^2[O2]

1.71*10^12 = (0.3-2x)^2/(2x)^2 * x

1.71*10^12 *(2x)^2*x = (0.3-2x)^2

         x   = 2.36*10^-5

   [NO]    = 2x    = 2*2.36*10^-5    = 4.72*10^-5M

[O2]     = x        = 2.36*10^-5M

[NO2]    = 0.3-2x    = 0.3-2*2.36*10^-5

                              = 0.3-4.72*10^-5

                              = 0.3-0.0000472   = 0.299M