`3.00` mol of `PCl_(5)` kept in `1L` closed reaction vessel was allowed to attain equilibrium at `380K`. Calculate composition of the mixture at equlibrium `K_(c)=1.80`

Explanation

`PCl_(5) hArr Pcl_(5)+Cl_(2)`
`{:("Initial",*,,),("concentration:" ,3.0,0,*0):}`
Let x mol per litre of `PCl_(5)` be dissociated,
At equilibrium : `(3-x) " "x " " x`
`K_(6)[PCl_(3)][Cl_(2)]//[PCl_(5)]`
`1.8=x^(2)//(3-x)`
`x^(2)+1.8x-5.4=0`
`x=[-1.8+- sqrt((1.8)^(2))-4(-5-4)]//2`
`x=[-1.8+- sqrt(3.24)+21.6]//2`
`x=[-1.8+-4.98]//2`
`x=[-1.8+4.98]//2=159`
`[PCl_(5)]=3.0-x=3-1.59=1.41M`
`[PCl_(3)]=[Cl_(2)]=x=1.59M`