The composition of the equilibrium mixture (Cl22C1)which is attained a

1.	The composition of the equilibrium mixture (Cl22C1)which is attained at 1200°C, is determined by measuringthe rate of effusion through a pin-hole. It is observed thatat 1.80 mm Hg pressure, the mixture effuses 1.16 times asfast as krypton effuses under the same conditions.Calculate the fraction of chlorine molecules dissociatedinto atoms (atomic weight of Kr 84)(1995)
Answer» Mixture Krypton rmix= 1.16rKr= 1 Mmix= ?MKr= 84 We know that rmix/rKr= √MKr/Mminor 1.16/1 = √84/Mmix or (1.16)284/Mmix=>Mmix= 84/(1.16)2= 62.426 Determination of the composition of the equilibrium mixture/Let the fraction of CI2molecules dissociated at equilibrium = x CI2⇌ 2CI Total Initially 1 0 1 At equilibrium 1 – x 2x 1 – x + 2x = 1 + x ∴ Total moles at equilibrium = 1 – x + 2x = 1 + 1 ∵ Normal molecular mass/Experimental molecular mass = 1+x ∴ 71/64.426 = 1+ ∝ ∴ ∝ = 0.137 =13.7%

The composition of the equilibrium mixture (Cl22C1)which is attained at 1200°C, is determined by measuringthe rate of effusion through a pin-hole. It is observed thatat 1.80 mm Hg pressure, the mixture effuses 1.16 times asfast as krypton effuses under the same conditions.Calculate the fraction of chlorine molecules dissociatedinto atoms (atomic weight of Kr 84)(1995)

Answer»

Mixture Krypton

rmix= 1.16rKr= 1

Mmix= ?MKr= 84

We know that

rmix/rKr= √MKr/Mminor 1.16/1 = √84/Mmix

or (1.16)284/Mmix=>Mmix= 84/(1.16)2= 62.426

Determination of the composition of the equilibrium mixture/Let the fraction of CI2molecules dissociated at equilibrium = x

CI2⇌ 2CI Total

Initially 1 0 1

At equilibrium 1 – x 2x 1 – x + 2x = 1 + x

∴ Total moles at equilibrium = 1 – x + 2x = 1 + 1

∵ Normal molecular mass/Experimental molecular mass = 1+x

∴ 71/64.426 = 1+ ∝

∴ ∝ = 0.137 =13.7%