One mole of N2 and 3 moles of PCl5 are placed in a 100 litre vessel heated to 227ºC. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate degree of dissociation of PCl5 and KP

Write the eqn. first,

PCl5 ------->   PCl3   +  Cl2

Use ideal gas eqn. to get the value of n .i.e moles 

and use concenrtrn for findoong Kpinstead of using partial pressure.

Kp  = pcl3*cl2/pcl5

                 PCL5 PCL3 + CL2                                                                                                                                   3 mol          0 mol      0 mol  no.moles initially                                                                                                             3(3-x)            3x          3x  no.of moles at equi.                                                                                                   now,     

     on using 

                                 PV=nRT [P=2.05 , V=100 , R=0.0821 , T=227 DEGREE CECIOUS MEANS 500K]

                    =>   2.05*100=n*0.0821*500

                    =>  205=41.05n 

                      =>  n=4.9[ we can take approx 5]

                     total number of moles at equilirium 

           =>3-3x+3x+3x+1=5[we have taken 1  here because we have to add no. of moles of N2]

   => 3x+4=5

       => x=0.33[degree of dissociation] 

frome here we can find Kp also that will be 0.205atm