9.3 Equilibrium Constant and its Units (Continued)

Problem 9.2

The equilibrium concentrations are

$[{\mathrm{N}}_2]=0.602\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{d}\mathrm{m}}^{-3};$$[\mathrm{N}_2] = 0.602 \mathrm{mol} \mathrm{dm}^{-3};$ $[{\mathrm{H}}_2]=0.420\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{d}\mathrm{m}}^{-3};$$[\mathrm{H}_2] = 0.420 \mathrm{mol} \mathrm{dm}^{-3};$ and $[{\mathrm{N}\mathrm{H}}_3]=0.113\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{d}\mathrm{m}}^{-3}\mathrm{.}$$[\mathrm{NH}_3] = 0.113 \mathrm{mol} \mathrm{dm}^{-3}.$

The equilibrium constant expression for this reaction is:

Now put the equilibrium concentration values in the equilibrium expression