A sample of pure `PC l_(5)` was introduced into an evacuated vessel at 473 K. after equilibrium was attained, concentration of `PCl_(5)` was found to be `0.5xx10^(-1)mol*L^(-1)`, if value of `K_(c)` is `8.3xx10^(-3)`, what are the concentration of `PCl_(3) and Cl_(2)` at equilibrium?

Explanation

Let the concentration of each of `PCl_(3) and Cl_(2)` be x `mol*L^(-1)`
`{:("Equilibrium",PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("concentration "(mol*L^(-1)), 0.5xx10^(-1),,x,,x):}`
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(x xx x)/(0.5xx10^(-1))=8.3xx10^(-3)`
or, `x^(2)=(8.3xx10^(-3))xx(0.5xx10^(-1))`
`therefore x=0.02(M)`
`therefore`The concentration of each of `PCl_(3) and Cl_(2)` at equilibrium is 0.02(M).