Question

Solid ammonium carbamate, $N{H}_{4}C{O}_{2}N{H}_2\left(s\right)$, dissociates into ammonia and carbon dioxide when it evaporates as shown by:
$N{H}_{4}C{O}_{2}N{H}_2\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
At ${25}^{o}C$, the total pressure of the gases in equilibrium with the solid is 0.116 atm. If 0.1 mole of $C{O}_2$ is introduced after equilibrium is reached then:

• A. final pressure of $C{O}_2$ will be less than 0.1 atm
• B. final pressure of $C{O}_2$ will be more than 0.1 atm
• C. pressure of $N{H}_3$ will decrease due to addition of $C{O}_2$
• D. pressure of $N{H}_3$ will increase due to addition of $C{O}_2$

Answer 1

Answer: (B), (C)

final pressure of $C{O}_2$ will be more than 0.1 atm
pressure of $N{H}_3$ will decrease due to addition of $C{O}_2$

The given reaction is:-

${NH}_4{CO}_{2}N{H}_2\left(s\right)\rightleftharpoons 2{NH}_3\left(g\right)+{CO}_2\left(g\right)$

Now, $K_P=\frac{{\left({pNH}_3\right)}^2.{pCO}_2}{p{NH}_4{CO}_3}$

$\because {NH}_4{CO}_3$ is pure solid. So,

$K_P={\left(p{NH}_3\right)}^2.p{CO}_2\to \left(I\right)$

Now, If amount of ${CO}_2$ is in creased at eqm then to keep $K_P$, a constant the pressure of ${NH}_3$ must decrease.

Also, since 0.1 moles of $C{O}_2$ is introduced after the equilibrium is reached thus its pressure will be more than 0.1 atm.