Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\rightleftharpoons 2{\mathrm{NH}}_3(\mathrm{g}){\mathrm{K}}_{\mathrm{P}}$

But the question is about the dissociation of ${\mathrm{NH}}_3$, so we have to reverse the reaction.

$\begin{array}{rl}& 2{\mathrm{NH}}_3(\mathrm{g})\rightleftharpoons {\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\frac{1}{{\mathrm{K}}_{\mathrm{P}}}\\ & \mathrm{t}=0aatm\\ & \mathrm{t}={\mathrm{t}}_{\mathrm{eq}}\underset{{\mathrm{P}}_{{\mathrm{NH}}_3}}{\underbrace{(\mathrm{a}-\mathrm{x})\mathrm{atm}}}\frac{\mathrm{x}}{2}\frac{3\mathrm{x}}{2}\\ & {\mathrm{P}}_{\text{total }}=\mathrm{P}={\mathrm{P}}_{{\mathrm{NH}}_3}+\frac{\mathrm{x}}{2}+\frac{3\mathrm{x}}{2}\\ & \mathrm{AsP}\gg {\mathrm{P}}_{{\mathrm{NH}}_3},\mathrm{P}=\frac{\mathrm{x}}{2}+\frac{3\mathrm{x}}{2}\\ & \mathrm{P}=2\mathrm{x}\\ & \mathrm{x}=\frac{\mathrm{P}}{2}\\ & {\mathrm{P}}_{{\mathrm{N}}_2}=\frac{\mathrm{x}}{2}=\frac{\mathrm{P}}{4}\\ & {\mathrm{P}}_{{\mathrm{H}}_2}=\frac{3\mathrm{x}}{2}=\frac{3\mathrm{P}}{4}\end{array}$

Writing the equilibrium constant for the reaction:

$$\begin{array}{rl}& \frac{1}{{\mathrm{K}}_{\mathrm{P}}}=\frac{{\mathrm{P}}_{{\mathrm{N}}_2}{\mathrm{P}}_{{\mathrm{H}}_2}^3}{{\mathrm{P}}_{{\mathrm{NH}}_3}^2}\\ & \Rightarrow \frac{1}{{\mathrm{K}}_{\mathrm{P}}}=\frac{\mathrm{P}/4(3\mathrm{P}/4{)}^3}{{\mathrm{P}}_{{\mathrm{NH}}_3}^2}\\ & {\mathrm{P}}_{{\mathrm{NH}}_3}^2=\frac{3^3{\mathrm{P}}^4}{256}{\mathrm{K}}_{\mathrm{P}}\\ & {\mathrm{P}}_{{\mathrm{NH}}_3}=\frac{3^{3/2}{\mathrm{P}}^2{\mathrm{K}}_{\mathrm{P}}^{1/2}}{16}\end{array}$$