1. Tardigrade
  2. Question
  3. Chemistry
  4. Equilibrium constant can also be expressed in terms of Kx, when concentrations of the species are taken in mole fraction F2(g) leftharpoons F(g); Kx =(X2F/XF2) For the above equilibrium mixture, average molar mass at 1000 K was 36.74 g mol-1. Thus, Kx is

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

Equilibrium

Solution:

Let, $F_{2}(g) = x$ (mole fraction),
molar mass $= 38.09\, mol^{-1}$
$F(g) = (1 - x),$ molar mass $= 19.0 \,g\, mol^{-1}$
$\therefore $ Average molar mass $= \frac{M_{1} x_{1} +M_{2} x_{2}}{x_{1} +x_{2}}$
$36 .74 = 38 x + 19(1- x)$
$\therefore x=0.9337$ (molar fraction of $F_{2}$)
$(1- x) = 0.0633$ (mole fraction of $F$)
$\therefore K_{x}=\frac{x^{2}F}{xF_{2}}=\frac{\left(0.0663\right)^{2}}{0.6337} =4.708 \times10^{-3}$