Question

For the equilibrium
$LiCl.3N{H}_3\left(s\right)\rightleftharpoons LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right);$
$K_p=9at{m}^2$ at ${37}^{\circ }C.$ A $5L$ vessel contains $0.1$ mole of $LiCl.N{H}_3$. How many moles of $N{H}_3$ should be added to the flask at this temperature to derive the backward reaction for completion?
Use : $R=0.082{\text{atm L mol}}^{-1}{\text{K}}^{-1}$

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

Answer 1

The correct option is D $0.79$

$LiCl.3N{H}_3\left(s\right)\rightleftharpoons LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right);$
$[K_p=9at{m}^2]$
Therefore,
$\begin{array}{cccccc}& LiCl.N{H}_3\left(s\right)& +& 2N{H}_3\left(g\right)& \rightleftharpoons & LiCl.3N{H}_3\left(s\right);\\ \text{Initial moles:}& 0.1& & \alpha & & 0\\ \text{Final moles at eqm.:}& 0& & (\alpha -0.2)& & 0.1\end{array}$
$[K_p^{\prime }=\frac{1}{9}(atm{)}^{-2}]$
Let, initial moles of $N{H}_3$ is $\alpha$ for completion of reaction.
At equilibrium, $K_p^{\prime }=\frac{1}{(P_{N{H}_3}^{\prime }{)}^2}or\frac{1}{9}=\frac{1}{(P_{N{H}_3}^{\prime }{)}^2}\therefore P_{N{H}_3}^{\prime }=3atm\because PV=nRT\Rightarrow 3\times 5=n\times 0.0820\times 310\therefore n=0.59i.e.,(\alpha -0.2)=0.59$
$\therefore$
$\text{Initial moles of}N{H}_3=0.79$