For the following reaction, `PCl_(5)hArrPCl_(3)(g)+Cl_(2)(g)`
0.4 mole of `PCl_(5)0.2` mole of `PCl_(3)` and 0.6 mole of `Cl_(2)` are taken in 1 litre flask if `K_(c)=0.2` then, predict the direction in which reaction proceeds.

Explanation

`PCl_(5)(g)hArrPCl_(3)(g)9+Cl_(2)(g)`
The given concentrations are
`[PCl_(5)]=(0.4)/(1)=0.4M.[PCl_(3)]=(0.2)/(1)=0.2M`
`[Cl_(2)]=(0.6)/(1)=0.6M`
hence, `Q=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.2xx0.6)/(0.4)=0.3gtK_(c)(=0.2)`
Since `Q ltk_(c),` the reaction proceeds in the backward direction, i.e., more `PCl_(50` is formed.