`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`, In above reaction, at equilibrium condition mole fraction of `PCl_(5)` is 0.4 and mole fraction of `Cl_(2)` is 0.3. Then find out mole fraction of `PCl_(3)`

PCl_(5)hArr PCl_(3)(g)+Cl_(2)(g) . For this reaction at the chemical equilibrium condition which of the following is correct-

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 ?

The system PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) attains equilibrium. If the equilibrium concentration of PCl_(3)(g) is doubled, the concentration of Cl_(2)(g) would become

The system PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) attains equilibrium. If the equilibrium concentration of PCl_(3)(g) is doubled, the concentration of Cl_(2)(g) would become

The system PCl_(5)(s)hArr PCl_(3)(g)+Cl_(2)(g) attains equilibrium. If the equilibrium concentration of PCl_(3)(g) is doubled, the concentration of Cl_(2)(g) would become :

Under certain conditions, the equilibrium constant for the decomposition of PCl_(5)(g) into PCl_(3)(g) and Cl_(2)(g) is 0.0211 "mol"L^(-1) . What are the equilibrium concentrations of PCl_(5),PCl_(3)d and Cl_(2) if the initial concentration of PCl_(5) was 1.00 M ?