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

Options

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

Answer

B

Explanation

Equilibrium conc. Of `PCl_(5) = (0.1)/(10) = 0.01M`
Equilibrium conc. Of `Cl_(2) = PCl_(3) = (0.2)/(10) = 0.02M`
`rArr K_(C) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]) = (0.02 xx 0.02)/(0.01) = (0.0004)/(0.01)`
`rArr K_(C) = 0.04`