Question

At $800°C$, the following equilibrium is established as:
$F_2\left(g\right)\rightleftharpoons 2F\left(g\right)$
The composition of the equilibrium may be determined by measuring the rate of effusion of the mixture through a pinhole. It is found that at $800°C$ and $1\text{atm}$ the mixture effuses $1.6$ times as fast as $S{O}_2$ effuses under similar conditions $(At.wt.ofF=19)$. The value of $K_P$ for the equilibrium is

Answer 1

We know, $r\propto \frac{1}{\sqrt{M}}$

$\frac{r_{mix}}{r_{S{O}_2}}=\sqrt{\frac{M_{S{O}_2}}{M_{mix}}}$

$1.6=\sqrt{\frac{64}{M_{mix}}}=M_{mix}=\frac{64}{(1.6{)}^2}$
$\Rightarrow M_{mix}=25$
For
$$\begin{array}{cccc}& F_2\left(g\right)& \rightleftharpoons & 2F\left(g\right)\\ \text{Initially}& 1& & 0\\ \text{At Equilibrium}& (1-\alpha )& & \left(2\alpha \right)\end{array}$$
$\text{Total moles at equilibrium}=1-\alpha +2\alpha =1+\alpha$

We know,
$\frac{\text{Normal molecular weight}}{\text{Experimental molecular weight}}=1+\alpha$
$\frac{38}{25}=1+\alpha$

$\alpha =0.52$

So, $\text{Total moles at equilibrium}=1+0.52=1.52$
$K_P=\frac{(P_F{)}^2}{P_{F_2}}=\frac{(\frac{2\alpha }{1+\alpha }{)}^2\times P^2}{\left(\frac{1-\alpha }{1+\alpha }\right)P}$

$K_P=\frac{4{\alpha }^{2}p}{(1-\alpha )(1+\alpha )}=\frac{4\times (0.52{)}^2\times 1}{1-(0.52{)}^2}=1.48$