28 g of ${\mathrm{N}}_2$ and 6 g of ${\mathrm{H}}_2$ were kept at ${400}^0\mathrm{C}$ in 1 litre vessel, the equilibrium mixture contained 27.54 g of ${\mathrm{NH}}_3$. The approximate value of ${\mathrm{K}}_{\mathrm{C}}$ for the above reaction can be

(in ${\mathrm{mole}}^{-2} {\mathrm{litre}}^2)$ 

                     ${\mathrm{N}}_2$     +     $3{\mathrm{H}}_2$    $\leftrightharpoons$    $2{\mathrm{NH}}_3$

Initial conc.    1                3                   0

at equilibrium  1 - 0.81    3 - 2.43       1.62

                          0.19        0.57

No. of moles of ${\mathrm{N}}_2$ = $\frac{28}{28}=1$ mole

No. of moles of ${\mathrm{H}}_2$ = $\frac{6}{2}=3$  mole

No. of moles of   ${\mathrm{NH}}_3$ = $\frac{27.54}{17}=1.62$ moles

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right] {\left[{\mathrm{H}}_2\right]}^3} = \frac{{\left[1.62\right]}^2}{\left[0.19\right] {\left[0.57\right]}^3}= 75$