A sample of pure `PCl_(5)` was introduced into an evacuted vessel at `473 K`. After equilibrium was attained,concentration of `PCl_(5)` was found to be `0.5xx10^(-1)mol litre^(-1)`. If value of `K_(c)` is `8.3xx10^(-3) mol litre^(-1)`. What are the concentrations of `PCl_(3)` and `Cl_(2)` at equilibrium ?

Answer

`0.02"molL"^(-1)` for both.

Explanation

Let the concentrations of both `PCl_(3)` and `Cl_(2)` at equilibrium be x `"molL"^(-1)`. The given reaction is:
`{:(,PCl_(5(g)),harr,PCl_(3(g)),+,Cl_(2(g))),("At equilibrium",0.5xx10^(-1)"mol L"^(-1),,x"mol"L^(-1),,x "mol" L^(-1)):}`
It is given that the value of equilibrium constant. `K_(c)` is `8.3 xx 10^(-3)`.
now we can write the expression for equilibrium as :
`([PCl_(2)][Cl_(2)])/([PCl_(5)]) = K_(c)`
`rArr (x xx x)/(0.5 xx 10^(-1)) = 8.3 xx 10^(-3)`
`rArr x^(2) = 4.15 xx 10^(-4)`
`rArr x = 2.04 xx 10^(-2)`
`= 0.0204`
`= 0.2` (approximately)
Therefore, at equilibrium,
Therefore at equilibrium
`[PCl_(3)] = [Cl_(2)] = 0.02 "mol" L^(-1)`.