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.

Explanation

`PCI_5 hArr PCI_3+CI_2` Here concentration of `PCI_5=n/V`
=`2/2 =1`
`{:(1,0,0),(1-x,x,x):}`
x= 40% =0.4
1-x =1-0.4 =0.6
`K_c=([PCI_3][CI_2])/([PCI_5])`
`=(0.4xx0.4)/(0.6)`
`=(1.6)/6=(0.8)/3`
=0.267 moles /L