In the reaction
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`, the equilibrium concentrations of `PCl_(5)` and `PCl_(3)` are 0.4 and 0.2 mole `//` litre respectively. If the value of `K_(c)` is 0.5 , what is the concentration of `Cl_(2)` in moles `//` litre ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

Case (i) `PCl_(5) Leftrightarrow Pcl_(3)+cl_(2)`
`{:(,n_(eg),1,2,3),(,underset("molepct")(C),1,2,3):}`
`K_(c)=(2 xx 3)/(1)=6`
Case (ii)
`Pcl_(5) Leftrightarrow Pcl_(3)+cl_(2)`
`{:(,ni,x,2,3),(,n_(eq),x-a,2+a,3):}`
at equilibrium `n_(Pcl_(3))=3`
2+a=3
a=1
`6=(3 xx 4)/([Pcl_(5)])`
`[Pcl_(5)]=2`
`npcl_(5)=2`