Question

0.020 gram of selenium vapour at equilibrium occupy a volume of 2.463 ml at 1 atm and 27ºC. The selenium is in a state of equilibrium according to reaction -

3Se₂ (g) $\rightleftharpoons$Se₆ (g)

What is the degree of association of selenium ?

(At. wt. of Se = 79 ;$\frac{1 \times 2.463 \times 10^{- 3}}{0.0821 \times 300}$ = 10^–4)

• A. 0.205
• B. 0.315
• C. 0.14
• D. None of these

Answer 1

Answer: (B)

0.315

Answer 2

3Se₂(g) $\rightleftharpoons$Se₆(g)

t = 0 n 0

̃ $n_{i} = n$

at equi. n(1–x) $\frac{xn}{3}$

̃ $n_{f} = \left( 1 - \frac{2x}{3} \right)n$ (x = degree of dissociation)

̃ $\frac{M_{i}}{M_{f}} = \frac{\left( 1 - \frac{2x}{3} \right)n}{n}$

̃ $\frac{79 \times 2}{M_{f}} = 1 - \frac{2x}{3}$ …(i)

Now n_f = $\frac{1 \times 2.463 \times 10^{- 3}}{0.0821 \times 300} = \frac{0.02}{M_{f}}$

̃ M_f = $\frac{0.02}{10^{- 4}}$ = 200 …(ii)

By (i) and (ii)

$\frac{2 \times 79}{200} = 1 - \frac{2x}{3} \Rightarrow \frac{2x}{3} = 1 - \frac{150}{200} = \frac{42}{200}$

̃ x = $\frac{3}{2} \times \frac{42}{200} = 0.315$