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16 At 1.23atm...

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16 At 1.23atm &727^° C total equilibrium pressure partially dissociate in SO3,SO2 and O2 SO3(g) >SO2(g)+1/2O2(g) If the density of equilibrium mixture is 0.9gm/lit so what is the dissociation value 1)1/3 2)2/3 3)1/4 4)1/5

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727^{0} C

and 1.23 atm of total equilibrium pressure,

S O_{3}

is partially dissociated into

S O_{2}

O_{2}

S O_{3} (g) ⇌ S O_{2} (g) + 1 / 2 O_{2} (g)

. The density of equilibrium mixture is 0.9 gm/litre.The degree of dissociation is

627^{\circ} C

and one atmosphere pressure

S O_{3}

is partially dissociated into

S O_{2}

O_{2}

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?

727^{o}

1.23

atm of total equilibrium pressure,

S O_{3}

is partially dissociated into

S O_{2}

O_{2}

S O_{3} (g) ⇌ S O_{2} (g) + 1 / 2 O_{2} (g)

. The density of equilibrium mixture is

0.9

g/litre. The degree of dissociation is (assuming ideal behaviour of gases):

^{\circ} C

and one atmosphere,

S O_{3}

is partially dissociated into

S O_{2}

O_{2} .

S O_{3 (g)} ⇋ S O_{2 (g)} + \frac{1}{2} O_{2 (g)}

. If the density of the equilibrium mixture is 0.925 g/litre, then what is the degree of dissociation?

Q. The degree of dissociation of

S O_{3}

α

at equilibrium pressure P.

K_{P}

S O_{3} (g) ⇌ S O_{2} (g) + \frac{1}{2} O_{2} (g)

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