[" 12.How many moles per litre of "PCl_(5)" has to be taken to obtain "0.1" mole "Cl_(2)," if the value of equilitor "],[" constant "K_(c)" is "0.04?]

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:

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:

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:

How much PCI_5 must be taken in a 9.2 L vessel to get 0.5 moles of Cl_2 at a particular temperature? The value of equilibrium constant (K_c) at the given temperature is 0.0414.

0.6 moles of PCl_(5) , 0.3 mole of PCl_(3) and 0.5 mole of Cl_(2) are taken in a 1 L flask to obtain the following equilibrium , PCl_(5(g))rArrPCl_(3(g))+Cl_(2(g)) If the equilibrium constant K_(c) for the reaction is 0.2 Predict the direction of the reaction.

0.6 moles of PCl_(5) , 0.3 mole of PCl_(3) and 0.5 mole of Cl_(2) are taken in a 1 L flask to obtain the following equilibrium , PCl_(5(g))rArrPCl_(3(g))+Cl_(2(g)) If the equilibrium constant K_(c) for the reaction is 0.2 Predict the direction of the reaction.

One mole of PCl_(5) is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.