In the dissociation of `PCl_(5)` as
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
If the degree of dissociation is `alpha` at equilibrium pressure P, then the equilibrium constant for the reaction is

The equilibrium established during thermal dissociation of PCl_(5)(g) in a closed vessel is as follows: PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) If the degree of dissociation of PCl_(5)(g) at equilibrium be 'x', then show that (1) 'x' is inversely proportional to the square root of the total pressure of reaction mixture at equilibrium and (2) 'x' increases with increase in volume of the system.

For the reaction PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , the equation connecting the degree of dissociation (alpha) of PCl_(5)(g) with the equilibrium constant K_(p) is

For the reaction PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , the equation connecting the degree of dissociation (alpha) of PCl_(5)(g) with the equilibrium constant K_(p) is

Prove alpha=sqrt((K_(p)/(P+K_(p)))) for PCl_(5) hArr PCl_(3)+Cl_(2) where alpha is the degree of dissociation at temperature when equilibrium constant is K_(p) .

Prove alpha=sqrt((K_(p)/(P+K_(p)))) for PCl_(5) hArr PCl_(3)+Cl_(2) where alpha is the degree of dissociation at temperature when equilibrium constant is K_(p) .

PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g),K_(p)=1.8 . At 250^(@)C if 50% of PCl_(5) dissociates at equilibrium then what should be the pressure of the reaction-system?