1 mol `PCl_(2)(g)` is heated in a closed container of 2 litre capacity. If at equilibrium, the quantity of `PCl_(5)(g)` be 0.2 mol then calculate the value of equilibrium constant for the given reaction,
`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`.

Explanation

For the above reaction, `K_(c)=([PCl_(3)]xx[Cl_(2)])/([PCl_(5)])`
As per the given data, number of moles of `PCl_(5)(g)` at equilibrium=0.2. hence, numbre of moles of `PCl_(5)(g)` dissociated`=(1-0.2)=0.8`.
According to the equation, 1 mol `PCl_(5)(g)` dissociates to produce 1 mol of each of `PCl_(3)(g) and Cl_(2)(g)`.
Therefore, 0.8 mol `PCl_(5)(g) and Cl_(2)(g)`.
Therefore, 0.8 mol `PCl_(5)(g)` will dissociate to give 0.8 mol of each of `PCl_(3)(g) and Cl_(2)(g)`.
As given, volume of the container=2L. so, the equilibrium concentrations of different constituents are as follows:
`{:("Equilibrium",PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("conc."(mol*L^(-1)):,(0.2)/(2)=1,,(0.8)/(2)=0.4,,(0.8)/(2)=0.4):}`
`therefore K_(c)=([PCl_(3)]xx[Cl_(2)])/([PCl_(5)])=(0.4xx0.4)/(0.1)=1.6`