Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\mathrm{K}}_{\mathrm{p}}\mathrm{ }=\frac{{{\mathrm{n}}^2}_{\mathrm{N}{\mathrm{H}}_3}}{{\mathrm{n}}_{{\mathrm{N}}_2}\times {\mathrm{n}}_{{\mathrm{H}}_2}^3}\times \mathrm{ }{\left(\frac{\mathrm{P}}{\mathrm{\Sigma }\mathrm{n}}\right)}^{\mathrm{\Delta }\mathrm{n}}$

At constant pressure, with the introduction inert gas$\mathrm{\Sigma }\mathrm{n}$increases. Since$\mathrm{\Delta }\mathrm{n}=–2$, therefore on increasing$\mathrm{\Sigma }\mathrm{n}\mathrm{ }{\left(\frac{\mathrm{P}}{\mathrm{\Sigma }\mathrm{n}}\right)}^{\mathrm{\Delta }\mathrm{n}}$increases, but${\mathrm{K}}_{\mathrm{p}}$remain constant, hence no. of

moles of${\mathrm{N}}_2\mathrm{ }\left(\mathrm{g}\right)$&${\mathrm{H}}_2\mathrm{ }\left(\mathrm{g}\right)$are increased and backward reaction is favoured.