In the reaction, `PCl_(5)hArrPCl_(3)+Cl_(2)`, the amounts of `PCl_(5)`, `PCl_(3)` and `Cl_(2)`, `2 moles` each at equilibrium and the total pressure is `3 atm`. Find the equilibrium constant `K_(p)`.

In the reaction, PCl_5 hArr PCl_3 + Cl_2 , the amounts of PCl_5 PCl_3 and Cl_2 at equilibrium are 2 mole each and the total pressure is 3 atm. The equilibrium constant K_p is:

PCl_(5)hArrPCl_(3)+Cl_(2) in the reversible reaction the moles of PCl_(5).PCl_(3) and Cl_(2) are a,b and c respectively and total pressure is P then value of K_(p) will be

In the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) , the equilibrium concentrations of PCl_(5) and PCl_(3) are 0.4 and 0.2 mole // litre respectively. If the value of K_(c) is 0.5 , what is the concentration of Cl_(2) in moles // litre ?

For the reaction: PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) , if the initial concentration of PCl_(5)=1 mol/ l and x moles/litre of PCl_(5) are consumed at equilibrium, the correct expression of K_(p) if P is total pressure is

For the reaction: PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) , if the initial concentration of PCl_(5)=1 mol/ l and x moles/litre of PCl_(5) are consumed at equilibrium, the correct expression of K_(p) if P is total pressure is

In the equilibrium , PCl_(5)hArr PCl_(3)+Cl_(2) starting with 2 mol of PCl_(5) in 5 L flask at 350K , there is 80% dissociation. Hence the equilibrium pressure is

In the reaction PCl_(5)(g) rarr PCl_(3)(g) + Cl_(2)(g) PCl_(5),PCl_(3)"and"Cl_(2) are at equilibrium at 500 K. The concentration of PCl_(3) "and" Cl_(2) is 1.59M. K_(c) = 1.79 .Calculate the concentration of PCl_(5) .

In the reaction PCl_(5)(g) rarr PCl_(3)(g) + Cl_(2)(g) PCl_(5),PCl_(3)"and"Cl_(2) are at equilibrium at 500 K. The concentration of PCl_(3) "and" Cl_(2) is 1.59M. K_(c) = 1.79 .Calculate the concentration of PCl_(5) .