Q8.one mole of N2 and 3 moles of PCl5 are placed in a 100l vessel heated to 227C the equilibrium pressure is 2.05 atm assuming ideal behaviour calculate the degree of dissociation for PCl5 and kp for the reaction PCl5--> PCl3 +Cl2 Share with your friends Share 0 Vartika Jain answered this Dear Student, PCl5(g) ↔ PCl3(g) + Cl2(g) 3 mol 0 03(1-α) 3α 3αTotal mol of gases in the vessel, n = nN2 + nPCl5 + nPCl3 +nCl2 =1+3(1-α)+3α+3α = 4+3αAlso, we know that PV=nRTSo, from the given equilibrium pressure, n=PVRT=2.05×1000.0821×500=5 molTherefore, 4+3α=5or, 3α=1or, α=13=0.333Therefore, nPCl5 = 3(1-0.333)= 2 mol nPCl3 =nCl2= 3(0.333)= 1 molNow, pPCl5 = nPCl5 npeq=25×2.05 = 0.82 atmpPCl3 =pCl2=15×2.05=0.41atmKp=(pPCl3 )(pCl2)(pPCl5 )=(0.41)(0.41)(0.82)= 0.205 atm -2 View Full Answer