Home
Class 11
CHEMISTRY
16 mol of PCl(5)(g) is placed in 4 dm^(-...

16 mol of `PCl_(5)(g)` is placed in 4 `dm^(-3)` closed vessel. When the temperature is raised to 500 K, it decompses and at equilibrium, 1.2 mol of `PCl_(5)(g)` remains. What is `K_(c)` value for the decomposition of `PCl_(5)(g)` to `PCl_(3)(g)` and `Cl_(2)(g)` at 500K.

Promotional Banner

Similar Questions

Explore conceptually related problems

1.6 moles of PCl_(5)(g) is placed in 4dm^(3) closed vessel . When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 moles of PCl_(5)(g) remains . What is the K_(c) value for the decomposition of PCl_(5)(g) to PCl_(3)(g)andCl_(2)(g) at 500 K?

1.6 mol of PCl_(5) is placed in a 4 litre vessel. When the temperature is increased to 500 K, the PCl_(5) decomposes as PCl_(5)hArr PCl_(3)+Cl_(2) At equilibrium 1.20 mol of PCl_(5) remains' K_(c ) for the reaction is :

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

1 mole of PCl_(5) taken at 5 atm, dissociates into PCl_(3) and Cl_(2) to the extent of 50% PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) Thus K_(p) is :

1 mole of PCl_(5) taken at 5 atm, dissociates into PCl_(3) and Cl_(2) to the extent of 50% PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) Thus K_(p) is :

Initially, 0.8 mole of PCl_(5) and 0.2 mol of PCl_(3 are mixed in one litre vessel. At equilibrium, 0.4 mol of PCl_(3) is present. The value of K_(c ) for the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) would be