2.0 mole of`PCl_(5)` were nttoducedd in a vessel of 5.0 L capacity of a particular temperature At equilibrium, `PCl_(5)` was found to be 35 % dissociated into `PCl_(3)and Cl_(2)` the value of `K_(c)` for the reaction
`PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)`

2.0 mole of PCl_(5) were introduced in a vessel of 5.0 L capacity of a particular temperature At equilibrium, PCl_(5) was found to be 35 % dissociated into PCl_(3)and Cl_(2) the value of K_(c) for the reaction PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)

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

Two moles of PCl_(5) were introduced in a 2L, flask and heated at 600 K to attain the equilibrium, PCl_(5) was found to be 40% dissociated into PCl_(3) and Cl_(2).K_(c ) for the reaction is :

Two moles of PCl_(5) were heated at 327^(@)C in a closed two litre vessel and when equilibrium was reached, PCl_(5) was found to be 40% dissociated into PCl_(3) and Cl_(2) The equilibrium constant K_(C ) for this reaction (approx) PCl_(5)(g)iffPCl_(3)(g)+Cl_(2)(g) for this reaction (approx)

1 mole of PCl_(5) taken at 5 atm, dissociates into PCl_(3) and Cl_(2) to the extent of 50% PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) Thus K_(p) is :

1 mole of PCl_(5) taken at 5 atm, dissociates into PCl_(3) and Cl_(2) to the extent of 50% PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) Thus K_(p) is :

1.5 mol of PCl_(5) are heated at constant temperature in a closed vessel of 4 L capacity. At the equilibrium point, PCl_(5) is 35% dissociated into PCl_(3) and Cl_(2) . Calculate the equilibrium constant.

1.5 mol of PCl_(5) are heated at constant temperature in a closed vessel of 4 L capacity. At the equilibrium point, PCl_(5) is 35% dissociated into PCl_(3) and Cl_(2) . Calculate the equilibrium constant.

2 "mole" of PCl_(5) were heated in a closed vessel of 2 litre capacity. At equilibrium 40% of PCl_(5) dissociated into PCl_(3) and Cl_(2) . The value of the equilibrium constant is: