Question

$0.07755g$ of selenium vapours occupying a volume of $114.2mL$ at ${700}^{\circ }C$ and $185mm$ of $Hg$. The vapour are in equilibrium as:
$S{e}_{6\left(g\right)}\rightleftharpoons 3S{e}_{e\left(g\right)}$
Calculate:
(i) Degree of dissociation of $Se$,
(ii) $K_p$,
(iii) $K_C$.
Atomic weight of $Se$ is $79$.

Answer 1

${se}_{6_{\left(g\right)}}\leftrightharpoons 3{se}_{2_{\left(g\right)}}$

$ot{t}_{2}OO$

$Ot$

$t{e}_q(a-x)$

Total numbers of moles $\Rightarrow a-x+3x=(a+2x)$

$\left[\right(a-x\left)\right(6\times 79)+(3\times 79\times 2\left)\right]=0.0755g$

$a\times 6\times 79=0.0755g$

$a=1.6\times {10}^{-4}moles....\left(i\right)$

$\because pv=nRT....ot$ equilibrium

$n=\frac{PV}{RT}$

$(a+2x)=\frac{\frac{185}{760}\times \frac{114.2}{1000}}{\frac{1}{12}\times \left(700\right)}$

$(a+2x)=4.76\times {10}^{-4}moles$

$2x=3.1\times {10}^{-4}$

$x=1.58\times {10}^{-4}moles$

$\therefore$ Degree of Dissociation $\left(\alpha \right)=\frac{x}{a}$

$=\frac{1.58\times {10}^{-4}}{1.6\times {10}^{-4}}$

$=\alpha =0.9875$

$KC=\frac{{\left(\frac{3\times 1.58\times {10}^{-4}\times 1000}{114.2}\right)}^3}{2\times {10}^{-6}}=3.6\times {10}^{-2}$