`PCl_5` decomposes as `PCl_5(g) hArr PCl_3(g)+Cl_2(g)`.If at equilibrium , total pressure is P and density of gaseous mixture is d at temperature T then degree of dissociation `(alpha)` is :
(Molecular wt. of `PCl_5=M`)

Phosphorus pentachloride dissociates as follows in a closed reaction vessel. PCl_5(g) hArr PCl_3(g) +Cl_2(g) If total pressure at equilibrium of the reactions mixture is P and degree of dissociation of PCl_5 is x, the partial pressure of PCl_3 will be:

PCl_(5) dissociation a closed container as : PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is alpha , the partial pressure of PCl_(3) will be:

PCl_(5) dissociates into PCl_(3) and Cl_(2) , thus PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) If the total pressure of the system in equilibrium is P at a density rho and temperature T , show that the degree of dissociation alpha=(PM)/(rhoRT)-1 , where M is the relative molar mass of PCl_(5) . If the vapour density of the gas mixture at equilibrium has the value of 62 when the temperature is 230^(@)C , what is the value of P//rho ?

For the reaction, PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g)

Phosphorous pentachloride when heated in a sealed tube at 700 K it undergoes decomposition as PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g), K_(p)=38 atm Vapour density of the mixture is 74.25 . Percentage dissociation of PCl_(5) may be given as

Delta n volue of the reaction, PCl_5(g)rarrPCl_3(g)+Cl_2(g) is