At a certain temperature, the degree of dissociation of PCI_(5) was found to be 0.25 under a total pressure of 15 atm. The value of K_(P) for the dissociation of PCl_(5) is

At a certain temperature, the degree of dissociation of PCI_(5) was fo

1.	At a certain temperature, the degree of dissociation of PCI_(5) was found to be 0.25 under a total pressure of 15 atm. The value of K_(P) for the dissociation of PCl_(5) is
Answer» <P>1 0.25 0.5 0.75 Solution : Given `alpha = 0.25` Total no. of MOLES at equilibrium = `1+alpha=1.25` Total pressure = 15 atm `K_(P)=(P_(PCl_(3)) xx P_(Cl_(2)))/(P_(PCl_(5)))=(((0.25)/(1.25) xx 15)xx((0.25)/(1.25) xx 15))/(((0.75)/(1.25) xx 15))=1`