Question

$0.020\text{g}$ of selenium vapours at equilibrium occupying a volume of $2.463\text{mL}$ at $1\text{atm}$ and ${27}^{\circ }$C.
The selenium is in a state of equilibrium according to the reaction
$3S{e}_2\left(g\right)\rightleftharpoons S{e}_6\left(g\right)$
What is the degree of association of selenium?
(Atomic weight of Se = 79 amu)

• A. 0.205
• B. 0.315
• C. 0.14
• D. 0.51

Answer 1

The correct option is B 0.315

$3S{e}_2\left(g\right)\rightleftharpoons S{e}_6\left(g\right)$
a - a$\alpha$ $\frac{a\alpha }{3}$

$\frac{observedmolesatequilibrium}{initialmolesofreactant}=\frac{M_{init}}{M_{eq}}$

$\frac{a(1-\alpha )+\frac{a\alpha }{3}}{a}$ = $\frac{M_{init}}{M_{eq}}$

$(1-\alpha )+\frac{\alpha }{3}$ = $\frac{M_{init}}{M_{eq}}$

$M_{init}=M_{S{e}_2}=2\times 79=158$

Molar mass of mixture $M_{eq}=\frac{W\times R\times T}{P\times V}$
$M_{eq}=\frac{0.02\times 0.0821\times 300}{1\times 2.463\times {10}^{-3}}$
$M_{eq}=200$
$\therefore \alpha =0.315$