Question

Solid ammonium carbamate, $N{{H}_{4}}C{{O}_{2}}N{{H}_{2}}$ (s), dissociates into ammonia and carbon dioxide when it evaporates as shown by:
$N{{H}_{4}}C{{O}_{2}}N{{H}_{2}}(s)\Leftrightarrow 2N{{H}_{3}}(g)+C{{O}_{2}}(g)$
At ${{25}^{0}}C$, 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

Hint: This question can be solved only by the application of Le Chatelier’s principle. The pressure of gases on both the reactant and product side depend and change on the number of moles and volume.

Complete answer:
Let us see that what are the factors that bring about a change in the equilibria of reactions:
Effect of pressure change: If, in a state of equilibrium, the pressure of reaction is increased by keeping the temperature constant, then the volume is bound to decrease, as $PV=nRT$. This means that the number of moles per unit volume will increase. According to Le Chaitelier’s principle, equilibrium will shift in that direction that is accompanied by decrease in number of moles per unit volume.
Effect of temperature change: According to Le Chaitelier’s principle, when the temperature of the system is changed, then the equilibrium shifts in the opposite direction in order to neutralise the effect of the change. In an exothermic reaction, the equilibrium ets shifted to the left, on increasing temperature and vice-versa for endothermic reactions.
Effect of addition of inert gas:
At constant volume: When an inert gas is added to a reaction mixture in a state of equilibrium at constant volume, the total pressure of the system will increase, but partial pressure and molar concentrations will be unchanged. So there is no effect.
At constant pressure: In addition to gas at constant pressure, volume of mixture increases, moles per unit volume decrease. So, according to the principle, equilibrium shifts towards the side where the number of moles per unit volume decreases.
Considering all the above factors, we can say that the final pressure of $C{{O}_{2}}$ will be more than 0.1atm and the pressure of $N{{H}_{3}}$ will decrease, which gives options B and C as our answer.

So, Option B and C is the correct answer.

Note:
The condition of addition of inert gas in constant pressure can be remembered in an easier way. An inert gas shifts the equilibrium depending upon the nature of the reaction when added at a constant pressure( It actually shifts where the number of moles are more).