Question

For the equilibrium,
$LiCl\cdot {3NH}_3\left(s\right)\rightleftharpoons LiCl\cdot {NH}_3\left(s\right)+2{NH}_3\left(g\right)$ $K_p=9{atm}^2$ at ${37}^{o}C$.

A $5$ litre vessel contains $0.1$ mole of $LiCl\cdot {3NH}_3$. How many moles of ${NH}_3$ should be added to the flask at this temperature to derive the backward reaction for completion?
$R=0.082atm.L/molK$

• A. $0.2$
• B. $0.59$
• C. $0.69$
• D. $0.79$

Answer 1

Answer: (D)

$0.79$

Ammonia is the only gaseous species in the reaction. Hence, its pressure will appear in the expression for the equilibrium constant.
$k_p=P_{N{H}_3}^2=9at{m}^2$
Hence, $P_{N{H}_3}=3atm$
Using ideal gas equation, calculate the number of moles corresponding to 3 atm pressure.
$n=\frac{3atm\times 5L}{0.082Latm/molK\times 310K}=0.59mole.$
The initial number of moles of $LiCl\cdot {3NH}_3\left(s\right),LiCl\cdot {NH}_3\left(s\right)and{NH}_3$ are 0.1, 0 and 0 respectively.
At equilibrium, the number of moles of $LiCl\cdot {3NH}_3\left(s\right),LiCl\cdot {NH}_3\left(s\right)and{NH}_3$ are $0.1-x,x$ and $2x$ respectively.
But $2x=0.59$

Since 0.1 mole of $N{H}_3$ of $Lic{l}_3.N{H}_3$ added .

$\therefore$ Twice $N{H}_3$ requires for backward reaction.

$\therefore$ $0.1\times 2=0.2$

total $N{H}_3=0.59+0.2=0.79mole$.