`1.5 mol` of `PCl_(5)` are heated at constant temperature in a closed vessel of `4 L` capacity. At the equilibrium point, `PCl_(5)` is `35%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant.

Answer

0.071

Explanation

Initial concentration of `PCl_5`=1.5 mol
No. of moles of `PCl_5` dissociated =`(1.5xx35)/100`=0.525
Now, 1 mole of `PCl_5` on dissociation gives 1 mol of `PCl_3` and 1 mol of `Cl_2`. 0.525 mol of `PCl_5` on dissociation will give 0.525 mol of `PCl_3` and 0.525 mol of `Cl_2`.
Volume of the flask =4L
Molar conc. at equilibrium are
`{:(,PCl_5(g)hArr , PCl_3(g)+ , Cl_2(g)),("Initial conc.",1.5//4,0,0),("Equi.conc.",(1.5-0.525)/4, 0.525/4, 0.525/4):}`
`K_c=([PCl_3][Cl_2])/([PCl_5])=((0.525//4)(0.525//4))/((0.975//4))`=0.071