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

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

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