Question

For the reaction below (an undesired side reaction in the industrial production of nitric acid), if you begin by adding 0.0150 moles of both NH₃ and O₂ to a 1.00 L reaction vessel, and once equilibrium is established you have a nitrogen concentration of 0.00196 M, what is the value for K_c?

4 NH₃ (g)   + 3 O₂ (g)   <-->   2 N₂ (g) +   6 H₂O (g)

Answer 1

ICE Table

   4 NH3    + 3 O2   <===> 2 N2 + 6 H2O
I    0.0150 0.0150            0        0
C    -4x    -3x                +2x    +6x
E    0.0150-4x 0.0150-3x    2x        6x

[N2]equilibrium = 0.00196 = 2x
Therefore,    x = 0.00098 M

[NH3]eq = 0.0150-4x = 0.0111 M
[O2]eq = 0.0150-3x = 0.0121 M
[H2O]eq = 6x = 0.00588 M

Kc = [H2O]^6 * [N2]^2 / ( [NH3]^4 * [O2]^3 )
Kc = 4.76 x 10^-6