Question

For the equilibrium $LiCl.3N{H}_3\left(s\right)\rightleftharpoons LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right)$, $K_P$ is $9at{m}^2$ at ${27}^{o}C$. A $82.1L$ vessel contains $0.1$ mole $LiCl.N{H}_3$ at $3$ atm. How many moles of $N{H}_3$ would be required at the same temperature to drive the backward reaction for near completion?

• A. 8 moles
• B. 10 moles
• C. 12 moles
• D. 14 moles

Answer 1

The correct option is B 10 moles

The equilibrium pressure of ammonia is $\sqrt{K_P}=\sqrt{9at{m}^2}=3atm$

This is equal to the pressure of the system. Hence, the system is at equilibrium.

The number of moles of ammonia that must be added are

$n=\frac{PV}{RT}=\frac{3\times 82.1}{0.0821\times 300}=10$ moles

Thus, when 10 moles of ammonia is added, the reaction will proceed in the backward direction to completion.