Prove that the pressure necessary to obtain 50% dissociation of `PCl_(5)` at 500 K is numerically three times the value of `K_(p)`.

Explanation

`{:(,PCl_5 hArr , PCl_3 + ,Cl_2),("Initial conc.", 1, 0,0),("Moles at equi.",1-0.5=0.5 , 0.5,0.5):}`
Total moles =0.5+0.5 +0.5 =1.5 mol
If p is the pressure required , then
`p(PCl_5)=0.5/1.5xxp=p/3`
`p(PCl_3)=0.5/1.5xxp=p/3`
`p(PCl_2)=0.5/1.5xxp =2/3`
`therefore K_p=(p(PCl_3)xxp(Cl_2))/(p(PCl_5))=(p/3xxp/3)/(p/3)=p/3`
`therefore p=3 K_p`