Question

The composition of the equilibrium mixture for the equilibrium $C{l}_2\rightleftharpoons 2Cl$ at $1400K$ may be determined by the rate of diffusion of mixture through a pin hole. It is found that at $1400K$, the mixture diffuses $1.16$ times as fast as krypton diffuses under the same conditions. Find the degree of dissociation of $C{l}_2$ equilibrium.

Answer 1

Equilibrium of dissociation of $C{l}_2$ may be represented as:
$C{l}_2\left(g\right)\rightleftharpoons 2Cl\left(g\right)$
$\underset{t_{eq}}{t=0}\underset{a(1-\alpha )}{a}\underset{2a\alpha }{0}$
Total moles $=a(1-\alpha )+2a\alpha =a(1+\alpha )$
$M_{mix}=\frac{a{M}_{C{l}_2}}{a(1+\alpha )}=\frac{M_{C{l}_2}}{(1+\alpha )}$
$\frac{R_{mix}}{R_{Kr}}=\sqrt{\frac{M_{Kr}}{M_{Mix}}}$
$1.16=\frac{84(1+\alpha )}{M_{C{l}_2}}$
$\frac{(1.16{)}^2\times 71}{84}-1=\alpha$

$\alpha =0.1374$