Answer to Question #155407 in General Chemistry for C

Question

1.  For the synthesis of ammonia, the equilibrium reaction is as follows:

N2(g)  + 3H2(g) <---->  2NH3(g)

At 474 K, Keq = 625.  The equilibrium mixture is analyzed and found to contain a nitrogen gas concentration of 3.2 mol/L and a hydrogen gas concentration of 2.13 mol/L.  What is the equilibrium concentration of ammonia gas in the mixture?    

Answer 1

Keq = [NH3]^2/[H2]^3/[N2] = [NH3]^2/(2.13)^3/3.2 = 625

[NH3] = (625*2.13^3*3.2)^(1/2) = 139.0 mol/L