Equilibrium constant can also be expressed in terms of `K_(x)` , when concentrations of the species are taken in mole fraction `F_(2)(g)hArr2F(g),K_(x) = (X_(F)^(2))/(X_(F_(2)))`
For the above equilibrium mixture, aberage molar mass at 1000 K was `36.74` g `mol^(-1)` . Thus, `K_(x)` is

Options

A. `14.08`
B. `2.124xx10^(2)`
C. `7.1xx10^(-2)`
D. `4.708xx10^(-3)`

Answer

D

Explanation

`Let , f_(2)(g)=x("mole friction")`,
`"molar mass"=38.09 "mol"^(-1)`
`F(g)=(1-x),"molar mass"=19.0g "mol"^(-1)`
`therefore "molar mass"=(M_(1)x_(1)+M_(2)x_(2))/(x_(1)+x_(2))`
`36.74=38x+19(1-x)`
`therefore x=0.337("molar fraction of" F_(2))`
`(1-x)=0.0633("mole fraction of "F)`
`therefore K_(x)=x_(F)^(2)/x_(f_(2))=(0.0663)^(2)/(0.9337)=4.708xx10^(-3)`