The chemical equation is
#"N"_2 + "3H"_2 ⇌ "2NH"_3#
The equilibrium moles of each component are:
#"N"_2 = "0.1 mol"#
#"H"_2 = "0.2 mol"#
#"NH"_3 = "(2 - 0.1 - 0.2) mol" = "1.7 mol"#
The volume of the container is #"0.500 dm"^3#, so the equilibrium concentrations of each component are
#["N"_2] = "0.1 mol"/"0.500 dm"^3 = "0.20 mol/dm"^3#
#["H"_2] = "0.2 mol"/"0.500 dm"^3 = "0.40 mol/dm"^3#
#["NH"_3] = "1.7 mol"/"0.500 dm"^3 = "3.4 mol/dm"^3#
We can set up part of an ICE table to solve this problem.
#color(white)(mmmmmmm)"N"_2 + "3H"_2 ⇌ "2NH"_3#
#"E/mol·dm"^"-3": color(white)(l)"0.20color(white)(ml)0.40color(white)(mml)3.4#
The equilibrium constant expression is
#K_text(c) = (["NH"_3]^2)/(["N"_2]["H"_2]^3)#
∴ #K_text(c) = 3.4^2/(0.20 × 0.40^3) = 900#
