calculate the volume percent of chlorine gas at equilibrium in the dissociation of PCl5(g)under a total pressure of 1.5 atm . The Kp for its dissociation =0.3
PCl5 ↔ PCl3+Cl2
a-x x x (total moles=a+x)
Kp=[1.5x/(a+x)][1.5x/(a+x)]/[1.5(a-x)/(a+x)]=0.3
On solving vol. % of Cl2=100x/(a+x) can be obtained (vol. % is same as mol. % at const. pressure & temp.).