`3.00` mol of `PCl_(5)` kept in `1L` closed reaction vessel was allowed to attain equilibrium at `3.80 K`. Calculate composition of the mixture at equilibrium `K_(c) = 1.80`

Answer

1.413,1.59,1.59 M

Explanation

`{:(,PCl_5 hArr , PCl_3 + , Cl_2),(,3.0,0,0),("Initial conc.",3-x,x,x):}`
`K_c=([PCl_3][Cl_2])/([PCl_5])`
`1.8=(x xx x)/(3-x) =x^2/(3-x)`
Solving for x , we get x =1.59
`[PCl_5]`=3-1.59 = 1.41 M, `[PCl_3]=[Cl_2]` =1.59 M