Question

The composition of the equilibrium mixture $(C{l}_2\rightleftharpoons 2Cl)$, which is attained at ${1200}^{\circ }C$ is determined by measuring the rate of effusion through a pinhole. It is observed that a 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms.

[atomic weight of Kr is 84].

Answer 1

$\frac{r}{r_{Kr}}=\sqrt{\frac{M_w\left(Kr\right)}{M}}$

$1.16=\sqrt{\frac{83.8}{M}}\therefore M=62.28$

(Molar weight of $C{l}_2$, and mixture $=$ 62.28)

$C{l}_2\rightleftharpoons 2Cl$
Initial 1 0
Final $1-x$ $2x$

$Totalmoles=1-x+2x=1-x$

$\therefore \frac{2\times \left(35.5\right)+(1-x)\times 71}{1+x}=62.28$

$\frac{710}{1+x}=62.28$

$x=0.14=14$% dissociated