Question

Equilibrium constant can also be expressed in terms of $K_x$, when concentrations of the species are taken in mole fraction
$F_2(g)\rightleftharpoons F(g);K_x=\frac{X_F^2}{X_{F_2}}$
For the above equilibrium mixture, average molar mass at $1000K$ was $36.74gmo{l}^{-1}$. Thus, $K_x$ is

• A. $14.08$
• B. $2.124\times 10^2$
• C. $7.1\times 10^{-2}$
• D. $4.708\times 10^{-3}$

Answer 1

Answer: (D)

Let, $F_2(g)=x$ (mole fraction),
molar mass $=38.09mo{l}^{-1}$
$F(g)=(1-x),$ molar mass $=19.0gmo{l}^{-1}$
$\therefore$ Average molar mass $=\frac{M_1{x}_1+M_2{x}_2}{x_1+x_2}$
$36.74=38x+19(1-x)$
$\therefore x=0.9337$ (molar fraction of $F_2$)
$(1-x)=0.0633$ (mole fraction of $F$)
$\therefore K_x=\frac{x^{2}F}{x{F}_2}=\frac{{\left(0.0663\right)}^2}{0.6337}=4.708\times 10^{-3}$