At a certain temperature, 0.5011 mol of N2 and 1.781 mol of H2 are placed in a 5.00-L container N,(g) + 3H2(g) 2NH3(g) At equilibrium, 0.1201 mol of N2 is present. Calculate the equilibrium constant, Kc. Number e146.7 Solution N2(g) + 3H2(g) <----> 2NH3(g) initial 0.5011/5 1.781/5 0 0.1 M 0.356 M 0 change -0.076 -0.228 +0.152 equilib 0.024 0.128 0.152 M at equilibrium, N2 = 0.1201/5 = 0.024 M Kc = [NH3]^2/[N2][H2]^3 Kc = 0.152^2/(0.128^3*0.024) = 459.03.