Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.

Options

A. `0.530`
B. `0.266`
C. `0.130`
D. `0.170`

Answer

B

Explanation

`{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),(t=0,2,,0,,0),(t=t_("eq"),(2-0.08)/2,,0.8/2,,0.8/2):}`
`:. K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.8/2xx0.8/2)/(1.2/2)=0.64/2.4=0.266`