A sample of pure `PCl_(5)` was introduced into aln evacuated vessel at `473` K. After equilibrium was attained, concentration of `PCl_(5)` was found to be `0.5 xx 10^(-1) L^(-1)`. If value of `K_(c)` is `8.3 xx 10^(-3)`. What are the concentration of `PCl_(3)` and `Cl_(2)` at equilibrium ?
`PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g)`

Explanation

Given: Concentration of `PCI_(5)=0.5xx10^(-1) "mol" L^(-1),K_(c)=8.3xx10^(-3)`.
Asked: Concentrations of `PCI_(3) "and" CI_(2) "at equilibrium"=?`
`PCI_(5)(g)hArrPCI_(3)(g)+CI_(2)(g)`
`"At eqm". 0.5xx10^(-1) x "mol"L^(-1) "(suppose)"`
Formulae: `K_(c)=([PCI_(3)][CI_(2)])/([PCI_(5)])`
Substitution & Calculation:
`therefore K_(c)=(X^(2))/(0.5xx10^(-1))=8.3xx10^(-3)`(Given)
or `x^(2)=(8.3xx10^(-3)(0.5xx10^(-1))=4.15xx10^(-4)`
or `x=sqrt4.15xx10^(-4)=2.04xx10^(-2)M=0.02M`