Question

At 627∘C and 1 atm, SO3 is partially dissociated into SO2 and O2 . If the density of the equilibrium mixture is 0.925 g/L, what is the degree of dissociation?

Answer 1

V=nR900.......[1]

SO3 (g) <===> SO2 (g) + 1/2O2 (g)

n-x                    x               0.5x

Total mass=64x+16x+80(n-x)

Density=[64x+16x+80(n-x)]/V=0.925...........[2]

degree of dissociation=x/n can be obtained solving [1] & [2].

Answer 2

$\large\underline\mathrm\red{Solution}$

• Let the initial no. of moles of SO3 be 1 and its degree of dissociation, x.

$\large\mathrm\red{1 \:\:\:\:\:\:= \:\:\:\:\: 0 \:\:\:\: 0.}$ Initial no. of moles

$\large\mathrm\red{SO3 \:\:\:\: = \:\:SO2 \:\: + \:\:\:\: 1/2 O2}$

$\large\mathrm\red{1 \:\: - \:\:x \:\:\:\:\:\:\:\: x \:\:\:\:\:\:\:\: x/2}$

$\large\underline\mathrm\red{Total \: no. \: of \: moles \: at \: eqb.}$

$\implies$ 1 - x + x/2

$\implies$ 1 + x/2.

$\large\underline\mathrm\red{Thus, \: Applying \: pV \: = \: nRT}$

1 × V = ( 1 + x/2 ) × 0.0821 × (627 + 273)

V = ( 1 + x/2 ) × 73.89 litres.

$\large\underline\mathrm\red{Now}$ wt. of 1 moles of SO3 = 80 g and therefore, from the law of conservation of mass, we get, wt. of gases at eqb. 80 g.

Density = wt. in g / vol. in litres

$\implies$ 80/ (1 + x/2) × 73.89

$\implies$ 0.925

$\large\underline\mathrm\red{Or}$

$\large\mathrm{x \:\:\: = \:\:\: 0.34}$

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