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Degree of Dissociation

The equilibri...

Question

The equilibrium constant $K_{p}$ for the reaction $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ at $1000 K$ is 3.5. What would be the partial pressure of $O_{2}$ gas with equal moles of $S O_{2}$ and $S O_{3}$ at equilibrium?

A

0.29 atm

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B

3.50 atm

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C

0.53 atm

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D

1.87 atm

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Solution

The correct option is A 0.29 atm
The given reaction is $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{p} = \frac{(P_{S O_{3}})^{2}}{(P_{S O_{2}})^{2} P_{O_{2}}} = \frac{1}{P_{O_{2}}}$
Here, partial pressure of $S O_{3}$ and $S O_{2}$ are same because moles of $S O_{2}$ and $S O_{3}$ are equal at equilibrium.
$P_{O_{2}} = \frac{1}{K_{p}}$
$P_{O_{2}} = \frac{1}{3.5} = 0.285$
Hence, the correct answer is option (a).

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Degree of Dissociation

Standard XII Chemistry

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