1 mol. $N_2$  and 3 mol $H_2$  are placed in a closed container at a pressure of 4 atm.  The pressure falls to 3 atm at the same temperature when the following equilibrium is attained.  $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$ .  The equilibrium constant Kp for dissociation of $N{H}_3$  is :

                               $2N{H}_{3\left(g\right)}\rightleftharpoons N_{2\left(g\right)}+3H_{2\left(g\right)}$

At equilibrium          2x                    1-x            3- 3x

                                    1-x+3-3x+2x= 3 (given)

                                      4 – 2x        =   3

                                            X          =   0.5

                                   $k_p=\frac{\left[N_2\right]{\left[H_2\right]}^3}{{\left[N{H}_3\right]}^2}=\frac{\left[0.5\right]{\left[1.5\right]}^3}{{\left[0.5\right]}^2}=\frac{{\left[1.5\right]}^3}{\left[0.5\right]}at{m}^{-2}$