Question

At ${800}^{o}C$, the following equilibrium is established as:
$F_2\left(g\right)\rightleftharpoons 2F\left(g\right)$
The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pinhole. It is found that at ${800}^{o}C$ and $1$ atm mixture effuses $1.6$ times as fast as ${SO}_2$ 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 $1.49$

The rate of effusion is inversely proportional to square of the molar mass.
$\frac{r_{mix}}{r_{S{O}_2}}=\sqrt{\frac{MS{O}_2}{M_{mix}}}$

Here, $r_{mix}$ and $r_{S{O}_2}$ are the rates of effusion of mixture and sulphur dioxide respectively. $M_{S{O}_2}$ and $M_{mix}$ are the molar masses of $S{O}_2$ and mixture respectively.

Substitute values in the above expression-
$1.6=\sqrt{\frac{64}{M_{mix}}}$
$2.56=\frac{64}{M_{mix}}$
$M=25$

Hence, the molecular weight of the mixture is 25 g/mol respectively.

Let $x$ be the mole fraction of $F_2$ in the mixture. The mole fraction of F will be $1-x$.

The molar mass of the mixture is $38x+19-19x=25$
$19x=6$

The mole fraction of $F_2$ is $x=0.32$
The mole fraction of F is$1-x=0.68$

The partial pressure is the product of mole fraction and total pressure which is 1 atm.

The equilibrium constant expression is $K_p=\frac{P_F^2}{P_{F_2}}=\frac{(0.68{)}^2}{0.32}=1.49$