Home
Class 12
CHEMISTRY
K(p) for the reaction PCl(5)(g)ghArrPC...

`K_(p)` for the reaction
`PCl_(5)(g)ghArrPCl_(3)(g)+Cl_(2)(g)`
at `250^(@)C` is `0.82`. Calculate the degree of dissociation at given temperature under a total pressure of `5 atm`. What will be the degree of dissociation if the equilibrium pressure is `10 atm`, at same temperature.

A

`27.5^(@)`

B

`23%`

C

`35.5%`

D

`40%`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reaction PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g) the value of K_(p) at 250^(@)C is 0.61atm ^(-1) The value of K_(c) at this temperature will be

For the reaction PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , the equation connecting the degree of dissociation (alpha) of PCl_(5)(g) with the equilibrium constant K_(p) is

For the reaction PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) the forward reaction at constant temperature is favoured by

For the reaction PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g), the forward reaction at constant temperature favorrd by :

For reaction, PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g) the value of K_(c) at 250^(@)C is 26. The value of K_(p) at this temperature will be .

(K_p/K_c) for the given equilibrium is [PCl_5 (g) hArrPCl_3(g) + Cl_2 (g)]

A certain temperature, the degree of dissociationof PCI_(3) 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

{:("For the reaction",PCl_(5)hArrPCl_(3)+Cl_(2)),("Initial moles are"," a b c"):} If alpha is the degree of dissociation and P is the total pressure. The partial pressure of PCl_(3) is

Unit of equilibrium constant K_p for the reaction PCl_5(g) hArr PCl_3(g)+ Cl_2(g) is

The degree of dissociation of PCl_(5) at 1 atm pressure is 0.2 . Calculate the pressure at which PCl_(5) is dissociated to 50% ?