A given quantity of `PCI_5` was heated in a 10 dm`""^3` vessel at `250^@C`
`PCl_(5(g)) Leftrightarrow PCl_(3(g)) +Cl_(2(g))`.
At equilibrium, the vessel contains 0.1 mole of `PCl_5` and 0.2 mole of `Cl_2.` The equilibrium constant of the reaction is

Options

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

Answer

B

Explanation

At equilibrium concentration of `PCl_3 and Cl_2` are equal
At eq. conc `("mol L"^(-1)) underset((0.1)/(10))(PCl_(5)) Leftrightarrow underset((0.2)/(10))(PCl_(3))+underset((0.2)/(10))(Cl_(2))`
`K=([PCl_3] [Cl_2])/([PCl_5]) =((0.2)/(10)) (0.2)/(10))/((0.1)/(10))=0.04`