PC, (g)--PCI3 (g) t, Cl2 (g)In above reaction, at equilibrium conditio

1.	PC, (g)--PCI3 (g) t, Cl2 (g)In above reaction, at equilibrium condition molefraction of PCl, is 0.4 and mole fraction of Cl2 is0.3. Then find out mole fraction of PC(1) 0.3 (2) 0.7 (3) 0.4 (4) 0.6
Answer» Let total sum of moles be 1 mole and sum of mole fractions of all = 1 At equilibrium no. of moles of PCl3 will be same as that of Cl2. also given mole of PCl5 = 0.4/1 = 0.4 and mole of Cl2 = 0.3/1 = 0.3 moles so, mole fraction of PCl3 will also be 0.3 moles option A

PC, (g)--PCI3 (g) t, Cl2 (g)In above reaction, at equilibrium condition molefraction of PCl, is 0.4 and mole fraction of Cl2 is0.3. Then find out mole fraction of PC(1) 0.3 (2) 0.7 (3) 0.4 (4) 0.6

Answer»

Let total sum of moles be 1 mole

and sum of mole fractions of all = 1

At equilibrium no. of moles of PCl3 will be same as that of Cl2.

also given mole of PCl5 = 0.4/1 = 0.4 and mole of Cl2 = 0.3/1 = 0.3 moles

so, mole fraction of PCl3 will also be 0.3 moles

option A