A quantity of `PCI_(5)` was heated in a 10 llitre vessel at `250^(@)C, PCI_(5)(g)hArrPCI_(3)(g) + CI_(2)(g)`. At equilibrium the vessel contains 0.1 mole of `PCI_(5), 0.20` mole of `PCI_(3)` and `0.2` mole of `CI_(2)`. The equilibrium constant of the reaction is

Options

A. `0.02`
B. `0.05`
C. `0.04`
D. `0.025`

Answer

C

Explanation

`K_(c) = ([PCI_(3)][CI_(2)])/([PCI_(5)]) = ((0.2)/(10)xx(0.2)/(10))/([0.1//10]) = 0.04`