At 550 K, the value of equilibrium constant `(K_(c))` is 0.08 for the given reaction: `PCl_(5)hArrPCl_(3)(g)+Cl_(2)(g)` occurring in a closed container. If the equilibrium concentration of `PCl_(5)(g)` and `Cl_(2)(g)` are 0.75 and 0.32 `mol*L^(-1)` respectively, then find the concentration of `PCl_(3)(g)`.

Explanation

For the given reaction, `K_(c)=([PCl_(3)]xx[Cl_(2)])/([PCl_(5)])`
As given, `[PCl_(5)]=0.75mol*L^(-1),[Cl_(2)]=0.32mol*L^(-1)`
and `K_(c)=0.08`
`therefore [PCl_(3)]=K_(c)xx([PCl_(5)])/([Cl_(2)])=0.08xx(0.75)/(0.32)=0.187mol*L^(-1)`
`therefore`Equilibrium concentration of `PCl_(3)=0.187mol*L^(-1)`