find the kp for the reaction.\npcl3+cl2--pcl5

PCl5(g) <-----> PCl3(g) + Cl2(g)
initial pressure ..
0.48 <-----> 0 + 0
at equilibrium pressure will be ..
(0.48-x) <-----> x + x

given total pressure = 0.806
so partial pressure of PCl5 + partial pressure of PCl3 + partial pressure of Cl2 = 0.806
(0.48-x) + x + x = 0.806
0.48 + x = 0.806
x = 0.806-0.48 = 0.326

so at equilibrium partial pressure of PCl5 = 0.48-x = 0.48-0.326 = 0.154 atm
partial pressure of Cl2 = x = 0.326 atm
partial pressure of PCl3 = x = 0.326 atm
so Kp = P(PCl3) X P(Cl2)/P(PCl5) = 0.326 X 0.326/0.154 = 0.690