Home
Class 11
CHEMISTRY
The degree of dissociation of PCl(5)...

The degree of dissociation of `PCl_(5)`

Promotional Banner

Similar Questions

Explore conceptually related problems

For PCl_(5 (g)) hArr PCl_(3 (g)) + Cl_(2 (g)) at equilibrium , K_(P) = P//3 . Then degree of dissociation of PCl_(5) at that temperature is

2 mols of PCl_(5) were heated in a sealed 5L container to constant temperature. If the degree of dissociation of PCl_(5) at this temperature is 0.4, the K_(c) of the reaction PCl_(5)(g) iff PCl_(3)(g)+Cl_(2)(g) is very close to

The equilibrium established during thermal dissociation of PCl_(5)(g) in a closed vessel is as follows: PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) If the degree of dissociation of PCl_(5)(g) at equilibrium be 'x', then show that (1) 'x' is inversely proportional to the square root of the total pressure of reaction mixture at equilibrium and (2) 'x' increases with increase in volume of the system.

For PCl_5(g)⇌PCl_3(g)+Cl_2(g) at equilibrium, K_p = P/3 , where P is equilibrium pressure. Then degree of dissociation of PCl_5 at that temperature is ?

1 mol of Cl_(2) and 3 mol of PCl_(5) are placed in a 100 L vessel heated to 227^(@)C . The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for PCl_(5) and K_(p) for the reaction. PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)

1 mol of Cl_(2) and 3 mol of PCl_(5) are placed in a 100 L vessel heated to 227^(@)C . The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for PCl_(5) and K_(p) for the reaction. PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)

At a certain temperature, the degree of dissociation of PCI_(5) was found to be 0.25 under a total pressure of 15 atm. The value of K_(P) for the dissociation of PCl_(5) is

PCl_(5) dissociation a closed container as : PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is alpha , the partial pressure of PCl_(3) will be:

Phosphorus pentachloride dissociates as follows in a closed vesselm PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) . If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is x, the partial pressure of PCl_(3) will be :

PCl_(5) dissociates as follows in a closed reaction vessel PCI_(5(g)) harr PCl_(3(g)) + Cl_((g)) . If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is x, the partial pressure of PCI_(3) will be