For the equilibrium:
`LiC 1.3 NH(3)(s) hArr LiCl. NH_(3)(s)+2NH_(3)`
`K_(p)=9 atm^(2)` at `40^(@)C`. A `5-L` vessel contains `0.1` "mole" of `LiCl.NH_(3)`. How many moles of `NH_(3)` should be added to the flask at this temperature to derive the backward reaction for completion?

Answer

C

Explanation

`:' LiCl.3NH_(3)(s) hArr LiCi.NH_(3)(s)+2NH_(3)(g) [K_(p)=9("atm")^(2)]`
`:. LiCl.NH_(3)(s)+2NH_(3)(g) hArr LiCl.3NH_(3)(s) [K_(P_(1))=1/9 ("atm")^(-2)]`
`{:(0.1,(a+0.2),),(,,"Initial moles"),(0,a,0.1),(,,"Final moles at Equilibrium"):}`
`:.` Initial moles of `NH_(3)` should be `(a+0.02)` to bring in completion of reaction.
At equilibrium `K_(P_(1))=1/((P'_(NH_(3)))^(2))` or `1/9=1/((P'_(NH_(3)))^(2))`
`:. P'_(NH_(3))=3 "atm"`
`:' PV=nRT`
`3xx5=nxx0.0821xx313`
`:. n=0.5837 i.e. a=0.5837`
`:.` Initial moles of `NH_(3)=a+0.2=0.5837+0.2`
`=0.7837` moles.