Question

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates:
4NH₃(g) + 3O₂(g) ⇔ 2N₂(g) + 6H₂O(g)
When 0.0200 mol gaseous NH₃ and 0.0100 mol gaseous O₂ are placed in a 1.00 L container at a certain temperature, the N₂ concentration at equilibrium is 1.20×10^-3 M.  
Calculate K_eq for the reaction at this temperature.

Answer 1

the reaction is

4 NH3 + 302 ---> 2N2 + 6 H20

we know that

concentration = moles / volume (L)

so

initially

[NH3] = 0.02 / 1 = 0.02 M

[02 ] = 0.01 / 1 = 0.01 M

now

consider the reaction

4 NH3 + 3 02 ---> 2 N2 + 6 H20

using ICE table

initial conc of NH3 , 02 , N2 , H20 are 0.02 , 0.01 , 0 , 0

change in conc of NH3 , 02 , N2 , H20 are -4x , -3x , 2x , 6x

equilibrium conc of NH3 , 02 ,N2 , H20 are 0.02 -4x , 0.01 -3x ,2x , 6x

given

equilibrium conc of N2 is 1.2 * 10-3

so

2x = 1.2 * 10-3

x = 6 * 10-4

now

[NH3]eq = 0.02 -4x = 0.02 - ( 4 * 6 * 10-4) = 0.0176

[02]eq = = 0.01 -3x = 0.01 - ( 3 * 6 * 10-4) = 8.2 * 10-3

[H20]eq = 6x = 6 * 6 * 10-4 = 3.6 * 10-3

now

Keq = [N2]^2 [H20]^6 / [NH3]^4 [02]^3

Keq = ( 1.2 * 10-3)^2 ( 3.6 * 10-3)^6 / ( 0.0176 )^4 ( 8.2 * 10-3)^3

Keq = 5.9 * 10-8