The composition of the equilibrium mixture for the equilibrium `Cl_(2) hArr 2 Cl` at 1400 K may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1400 K, the mixture diffuses 1.16 times as fast as krypton diffuses under the same conditions. Find the degree of dissociation of `Cl_(2)` equilibrium

Options

A. 0.1374
B. 0.2748
C. 0.4122
D. 0.637

Answer

A

Explanation

Equilibrium of dissociation of `Cl_(2)` may be represented as :
`{:(,Cl_(2) (g),hArr,2Cl (g),),(t = 0,a,,0,),(t_(eq),a(1- alpha),,2 a alpha,):}`
Total moles `= a (1 - alpha) + 2a alpha = a (1 + alpha)`
`M_("min") = (a M_(Cl_(2)))/(a (1 + alpha)) = (M_(Cl_(2)))/((1 + alpha))`
`(R_("mix"))/(R_(Kr)) = sqrt((M_(Kr))/(M_("mix")))`
`1.16 = (84 (1 + alpha))/(M_(Cl_(2)))`
`((1.16)^(2) xx 71)/(84) - 1 = alpha, alpha = 0.1374`