Home
Class 11
CHEMISTRY
Pure PCl(5) is introduced into an evacu...

Pure `PCl_(5) ` is introduced into an evacuated chamber and to equilibrium at `247^(@)C` and 2.0 atm .The equilibrium gases mixture contains 40% chlorine by volume .
Calculate `K_(p)` at `247^(@)C` for the reaction
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`

A

(a) `0.625` atm

B

(b) `4 atm`

C

(c) `1.6 atm`

D

(d) none of these

Text Solution

Verified by Experts

The correct Answer is:
c
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

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

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

(K_p/K_c) for the given equilibrium is [PCl_5 (g) hArrPCl_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)

Find out the units of K_c and K_p for the following equilibrium reactions : PCl_5(g) hArr PCl_3 + Cl_2(g)

At 540 K, 0.10 mol of PCl_(5) is heated in a 8L flask. The pressure of equilibrium mixture is found to be 1.0 atm . Calculate K_(p) and K_(c ) for the reaction.

For PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g), write the expression of K_(c)

PCl_(5) is 10% dissociated at 1 atm. What is % dissociation at 4 atm . PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)