Question

The composition of the equilibrium mixture ( $C{l}_2\rightleftharpoons 2Cl$), which is attained at ${1200}^o\text{C}$, is detrmined by measuring the rate of effusion through a pin-hole. It is observed that at $1.80\text{mm}$ Hg pressure, the mixture effuses $1.183$ times as fast as krypton effuses under the same conditions. Calculate the fraction of the chlorine molecules dissociated into atoms. (Given : Relative atomic mass of $Kr=84$, $(1.183{)}^2=1.4$)

Answer 1

Mixture Krypton
$r_{mix}$ = 1.16 $r_{Kr}$ = 1
$M_{mix}$ = ?
We know that
$\frac{r_{mix}}{r_{Kr}}=\sqrt{\frac{M_{Kr}}{M_{min}}}\Rightarrow \frac{1.183}{1}=\sqrt{\frac{84}{M_{mix}}}\Rightarrow (1.183{)}^2=\frac{84}{M_{mix}}\Rightarrow M_{mix}=\frac{84}{(1.183{)}^2}=\frac{84}{1.4}=60$

Let the fraction of$C{l}_2$ molecuels dissociated equlibrium $=x$
$\begin{array}{ccccc}& & C{l}_2& \rightleftharpoons & 2Cl\\ \text{Initially}& & 1& & 0\\ \text{At equilibrium}& & 1-2x& & x\end{array}$
$\therefore$Total moles $=1+x$
$\frac{\text{Normal molecular mass}}{\text{Experimental molecular mass}}=1+\alpha$
$\Rightarrow \frac{71}{60}=1+\alpha$
$\therefore \alpha =0.183$
Percentage of $\alpha =18.3\%$