For the equilibrium $\mathrm{L}\mathrm{i}\mathrm{C}\mathrm{l}.3\mathrm{N}{\mathrm{H}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{L}\mathrm{i}\mathrm{C}\mathrm{l}.\mathrm{N}{\mathrm{H}}_3\left(\mathrm{s}\right)+2\mathrm{N}{\mathrm{H}}_3\left(\mathrm{g}\right);$ ${\mathrm{K}}_{\mathrm{P}}=9 \mathrm{a}\mathrm{t}{\mathrm{m}}^2$ at 310 K. A 5 L vessel contains 0.1 mole of $\mathrm{L}\mathrm{i}\mathrm{C}\mathrm{l}.\mathrm{N}{\mathrm{H}}_3$ . How many mole of $\mathrm{N}{\mathrm{H}}_3$ should be added to the flask at this temperature derive the backward reaction for completion? (use R = 0.082 atm-L/mol K)
[Report your answer upto two decimal places.]

Solution

${\mathrm{K}}_{\mathrm{P}}=9={\mathrm{P}}_{\mathrm{N}{\mathrm{H}}_3}^2$
${\mathrm{P}}_{\mathrm{N}{\mathrm{H}}_3}=3$
PV = nRT
$3\times 5=(\mathrm{a}-0.2)0.082\times 310$
$\mathrm{a}=0.79$

Hence, the correct answer is 0.79.