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

0.024 mol `L^(-1)`

Explanation

`PCl_5 hArr PCl_3+ Cl_2`
Let concentration of `PCl_3` and `Cl_2` at equilibrium be x mol `L^(-1)` .
`[PCl_5]=0.05`
`K_c=8.3xx10^(-3)`
`K_c=([PCl_3][Cl_2])/([PCl_5])`
`8.3xx10^(-3)=(x xx x)/0.05`
`x=2.04xx10^(-2) "mol L"^(-1)`
`therefore [PCl_3]=2.04xx10^(-2) "mol L"^(-1)`
`[Cl_2]=2.04xx10^(-2) "mol L"^(-1)`