At 627-C and one atmosphere pressure SO3 is partially dissociated into SO2 and O2 by SO3g-12 O2g- The density of the equilibrium mixture is 0-925 g-litre- What is the degree of dissociation-

Let the molecular mass of the mixture equilibrium be M_mix .Applying the relation, M _ mix = dRT P = 0.925× 0.0821× 900 1 =68.348 Molecular mass of ...

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Byju's Answer

Standard XII

Chemistry

Introduction and Degree of Hydrolysis

At 627∘C an...

Question

At $627^{\circ} C$ and one atmosphere pressure $S O_{3}$ is partially dissociated into $S O_{2}$ and $O_{2}$ by $S O_{3 (g)} ⇌ \frac{1}{2} O_{2 (g)}$ . The density of the equilibrium mixture is $0.925 g / l i t r e$ . What is the degree of dissociation?

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Solution

Let the molecular mass of the mixture equilibrium be $M_{m i x}$ .
Applying the relation,
$M_{m i x} = \frac{d R T}{P} = \frac{0.925 \times 0.0821 \times 900}{1} = 68.348$
Molecular mass of $S O_{3} = 80$
Vapour density of $O_{3}, D = \frac{80}{2} = 40$
vapour density of mixture, $d = \frac{68.348}{2} = 34.174$
Let the degree of dissociation be $x$
$x = \frac{D - d}{(n - 1) d} = \frac{40 - 34.174}{(\frac{3}{2} - 1) \times 34.174} = 0.34$
or $x = 34$ % dissociated
i.e., ${S O}_{3}$ is $34$ % dissociated.

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At 627∘C and one atmosphere pressure SO3 is partially dissociated into SO2 and O2 by SO3(g)⇌12O2(g). The density of the equilibrium mixture is 0.925 g/litre. What is the degree of dissociation?

At $627^{\circ} C$ and one atmosphere pressure $S O_{3}$ is partially dissociated into $S O_{2}$ and $O_{2}$ by $S O_{3 (g)} ⇌ \frac{1}{2} O_{2 (g)}$ . The density of the equilibrium mixture is $0.925 g / l i t r e$ . What is the degree of dissociation?