The composition of the equilibrium mixture for the equilibrium Cl_(2) hArr 2Cl at 1470 K may be determined by the rate of diffusion of the mixture through a pinhole. It is found that at 1470 K, the mixture diffuse 1.16 times as fast as krypton (83.8) diffuses under the same conditions. Find the degree of dissociation of Cl_(2) at equilibrium.

The composition of the equilibrium mixture for the equilibrium Cl_(2)

1.	The composition of the equilibrium mixture for the equilibrium Cl_(2) hArr 2Cl at 1470 K may be determined by the rate of diffusion of the mixture through a pinhole. It is found that at 1470 K, the mixture diffuse 1.16 times as fast as krypton (83.8) diffuses under the same conditions. Find the degree of dissociation of Cl_(2) at equilibrium.
Answer» Solution :For the EQUILIBRIUM, `{:(,Cl_(2),hArr,2Cl,,),("Initial moles : ",1,,0,,),("Moles at eqb. ",1-x,,2x,,"(x is the DEGREE of dissociation)"):}` Molecular weight of the MIXTURE of `Cl_(2)` and Cl at eqb., i.e., `M_("mix")` is calculated as `M_("mix") = ((1-x) 71 + (2x)35.5)/((1-x) + 2x) = (71)/(1+x)` Now we have, `(r_("mix"))/(r_(Kr)) = sqrt((M_(Kr))/(M_("mix"))) = sqrt((83.8)/(71(1+x))) = 1.16` `THEREFORE x = 0.14`