The equilibrium constant K_c for the reaction N_2(g) + 3H_2(g) rightleftharpoons 2NH_3(g) at 450^oC is 0.159. What is the equilibrium composition when 1 mol N_2 is mixed with 3 mol H_2 in a 2.00-L vessel? Thank you so much!

1 Answer

2.07 mol of NH3

Explanation:

Kc = [NH3]^2/([N2]^1[H2]^3)

[NH3] = sqrt(Kc*[N2]^1 * [H2]^3

Plugging the value of N2 and H2 as 1 and 3 mol respectively, we get the equilibrium amount of NH3 as 2.07 mol.