Question

At ${800}^o$C, the following equilibrium is established as:
$F_2\left(g\right)\to 2F\left(g\right)$
The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at ${800}^o$C and $1$ atm mixture effuses $1.6$ times as fast as $S{O}_2$ effuses under the similar conditions. (At. wt. of $F=19$). What is the value of $K_p$ (in atm)?

• A. $0.315$
• B. $0.685$
• C. $0.46$
• D. $1.49$

Answer 1

The correct option is D $1.49$

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

$\Rightarrow 1.6=\sqrt{\frac{64}{M_{mix}}}$

$\Rightarrow M_{mix}=\frac{64}{(1.6{)}^2}=\frac{64}{1.6\times 1.6}=25$

For $F_2\rightleftharpoons 2F$

$\Rightarrow \frac{\text{Normal molecular weight}}{\text{Exp. molecular weight}}=1+\alpha$

$\Rightarrow \frac{38}{25}=1+\alpha$

$\Rightarrow \alpha =0.52$

${F_2}_{\left(g\right)}⟶2F_{\left(g\right)}$

$1$ $0$ $I$

$1-\alpha$ $2\alpha$ $E$ Total moles at $eqm=1-\alpha +2\alpha =1+\alpha$

$=1+0.52=1.52$

$\therefore K_P=\frac{P_F^2}{P_{F_2}}$

$=\frac{(2\alpha /1+\alpha {)}^2{P}^2}{\frac{(1-\alpha )}{1+\alpha }P}$

$=\frac{4{\alpha }^{2}P}{(1+\alpha )(1-\alpha )}$

$=\frac{4(0.52{)}^2\times 1}{1-(0.52{)}^2}$

$=1.49$