Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\text{NH}}_{\text{2}}{\text{COONH}}_{\text{4}}\text{(s)}\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\text{+}{\text{CO}}_{\text{2}}\text{(g)}$
Suppose P is pressure of$\mathrm{C}{\mathrm{O}}_2$at equilibrium than
${\mathrm{P}}_{\mathrm{N}{\mathrm{H}}_3}=2\mathrm{P}$
${\mathrm{P}}_{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=2\mathrm{P}+\mathrm{P}=3\mathrm{P}$
As${\mathrm{K}}_{\mathrm{P}}={\left({\mathrm{P}}_{\mathrm{N}{\mathrm{H}}_3}\right)}^2\left({\mathrm{P}}_{\mathrm{C}{\mathrm{O}}_2}\right)=4{\mathrm{P}}^3$
On adding$\mathrm{N}{\mathrm{H}}_3$at equilibrium
${\mathrm{P}}_{\mathrm{N}{\mathrm{H}}_3}={\mathrm{P}}_{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=3\mathrm{P}$
So${\mathrm{K}}_{\mathrm{P}}=4{\mathrm{P}}^3={\left({\mathrm{P}}_{\mathrm{N}{\mathrm{H}}_3}\right)}^2.\left(\mathrm{P}{\mathrm{\text{'}}}_{\mathrm{C}{\mathrm{O}}_2}\right)$
$4{\mathrm{P}}^3={\left(3\mathrm{P}\right)}^2\left(\mathrm{P}{\mathrm{\text{'}}}_{\mathrm{C}{\mathrm{O}}_2}\right)$
$\mathrm{P}{\mathrm{\text{'}}}_{\mathrm{C}{\mathrm{O}}_2}=\frac{4}{9}\mathrm{P}$
Now total$\mathrm{P}\mathrm{\text{'}}=3\mathrm{P}+\frac{4}{9}\mathrm{P}=\frac{31}{9}\mathrm{P}$
Hence pressure ratio$\left(\frac{\mathrm{P}\mathrm{\text{'}}}{\mathrm{P}}\right)=\frac{\frac{31}{9}\mathrm{P}}{3\mathrm{P}}$
$=\frac{31}{27}$