One mole of `PCl_(5)` is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.

Explanation

Let T be the equilibrium temperature and let `n_(1)` and `n_(2)` be the number of moles in vessels 1 and 2 respectively. As there is no loss of energy,
`n_(1)((3)/(2)RT_(1))+n_(2)((3)/(2)RT_(2))=(n_(1)+n_(2))((3)/(2)RT)`
`n_(1)T_(1)+n_(2)T_(2)=(n_(1)+n_(2))T`
`T=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`
`"Now, "n_(1)=(P_(1)V_(1))/(RT_(1)), n_(2)=(P_(2)V_(2))/(RT_(2))`
Substituting `n_(1) and n_(2)` values and solving, we get
`T=(T_(1)T_(2)(P_(1)V_(1)+P_(2)V_(2)))/(P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))`