Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\mathrm{X}}_2(\mathrm{g})\underset{{\mathrm{K}}_{\mathrm{b}}=4{\sec }^{-1}{\mathrm{torr}}^{-1}}{\stackrel{{\mathrm{K}}_{\mathrm{f}}=4{\sec }^{-1}}{\rightleftharpoons }}2\mathrm{X}(\mathrm{g})$

Initial moles 1 0

moles at eq. $1-\alpha$ $2\alpha$ $\mathrm{P}=2$ torr $\begin{array}{rlr}& \frac{{\mathrm{r}}_{\mathrm{mix}}}{{\mathrm{r}}_{\mathrm{He}}}=\frac{1}{1.5}=\sqrt{\frac{4}{{\mathrm{M}}_{\mathrm{mix}}}}\Rightarrow {\mathrm{M}}_{\mathrm{mix}}=9& \because \mathrm{n}\propto \frac{1}{\mathrm{M}}\end{array}$

$\frac{1}{1+\alpha }=\frac{9}{M}⟹1+\alpha =\frac{M}{9}.\dots \left(1\right)$

$K_f{P}_{{\mathrm{X}}_2}=K_b{P}_{\mathrm{X}}^2$

or $\frac{K_f}{K_b}=\frac{P_{\mathrm{X}}^2}{P_{{\mathrm{X}}_2}}$.

So, $\frac{4}{4}=\frac{P_{\mathrm{X}}^2}{P_{{\mathrm{X}}_2}}$,

From (1), $1=\frac{{\left(\frac{2\alpha }{1+\alpha }\cdot 2\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }\cdot 2\right)}=\frac{8{\alpha }^2}{1-{\alpha }^2}$.

This leads to $1-{\alpha }^2=8{\alpha }^2$, so $1=9{\alpha }^2$, which means $\alpha =\frac{1}{3}$.

The molar mass $M=9\left(1+\frac{1}{3}\right)=9\cdot \frac{4}{3}=12$.