The degree of dissociation of `PCl_(5) (alpha)` obeying the equilibrium, `PCl_(5)hArrPCl_(3)+Cl_(2)` is related to the pressure at equilibrium by :

The degree of dissociation of PCl_5 (alpha) obeying the equiliberium, PCl_5 ⇌ PCl_3 + Cl_2 is approximately realted to the pressure at equilibrium by :

At a particular temperature, PCl_(5)(g) undergoes 50% dissociation. The equilibrium constant for PCl_(5)(g) rarr PCl_(3)(g) + Cl_(2)(g) is 2atm. The pressure of the equilibrium mixture is

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 ?

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 ?