You visited us 1 times! Enjoying our articles? Unlock Full Access!

Derivation of Kp and Kc

0.020 g of se...

Question

$0.020 g$ of selenium vapours at equilibrium occupying a volume of $2.463 mL$ at $1 atm$ and $27^{\circ}$ C.
The selenium is in a state of equilibrium according to the reaction
$3 S e_{2} (g) ⇌ S e_{6} (g)$
What is the degree of association of selenium?
(Atomic weight of Se = 79 amu)

0.205

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.315

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.51

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 0.315
$3 S e_{2} (g) ⇌ S e_{6} (g)$
a - a $α$ $\frac{a α}{3}$

$\frac{o b s e r v e d m o l e s a t e q u i l i b r i u m}{i n i t i a l m o l e s o f r e a c t a n t} = \frac{M_{i n i t}}{M_{e q}}$

$\frac{a (1 - α) + \frac{a α}{3}}{a}$ = $\frac{M_{i n i t}}{M_{e q}}$

$(1 - α) + \frac{α}{3}$ = $\frac{M_{i n i t}}{M_{e q}}$

$M_{i n i t} = M_{S e_{2}} = 2 \times 79 = 158$

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

Suggest Corrections

Similar questions

0.020

2.463 m L

1

27^{o} C

3 {S e}_{2} (g) ⇌ {S e}_{6} (g)

S e =

0.020 g

2.463 mL

1 atm

27^{\circ}

3 S e_{2} (g) ⇌ S e_{6} (g)

0.020

2.463

1

27^{o}

3 S e_{2} (g) ⇌ S e_{6} (g)

S e = 79

0.07755 g

114.2 m L

700^{\circ} C

185 m m

H g

S e_{6 (g)} ⇌ 3 S e_{e (g)}

S e

K_{p}

K_{C}

S e

79

90

S e = 78.96

Join BYJU'S Learning Program