For a reaction $\mathrm{NH}_4 \mathrm{COONH}_{2(s)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}+\mathrm{

Question

For a reaction $\mathrm{NH}_4 \mathrm{COONH}_{2(s)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}+\mathrm{CO}_{2(\mathrm{~g})}$, the equilibrium pressure is $3 \mathrm{~atm} . K_p$ for the reaction will be

Professor PC Thomas Classes · Answer

(b) : $\mathrm{NH}_4 \mathrm{COONH}_{2(s)} ightleftharpoons \underset{2 p}{2 \mathrm{NH}_{3(g)}}+\underset{p}{\mathrm{CO}_{2(g)}}$ \When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure. So, $2 p+p=3$ 
$3 p=3, 
\quad 	herefore \quad p=1 \mathrm{~atm}$ 
$ K_p=(2 p)^2 	imes p=4 p^3=4 	imes(1)^3=4 \mathrm{~atm}^3 $