InterviewSolution
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InterviewSolution
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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)` isA. `14.08`B. `2.124xx10^(2)`C. `7.1xx10^(-2)`D. `4.708xx10^(-3)` |
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Answer» Correct Answer - D `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)` |
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