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Dalton's Law of Partial Pressures

The equilibri...

Question

The equilibrium constant , $K_{p}$ for the reaction $2 S O_{2} (g)$ $⇌$ $2 S O_{3} (g) i s 4.0 a t m^{- 1}$ at $1000 K$ . What would be the partial pressure of $O_{2}$ at the equilibrium.

A

16.0 a t m

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B

0.25 a t m

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C

1 a t m

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D

0.75 a t m

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Solution

The correct option is B $0.25 a t m$
$2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}$
$K_{p} = \frac{P^{2} S O_{3}}{P^{2} S O_{2} . P O_{2}}$
Given data $K_{p} = 4 a t o m^{- 1}$
$[S O_{2}] = [S O_{3}]$
$P S O_{2} = P S O_{3}$
$4 = \frac{1}{P O_{2}}$
$P O_{2} = 0.25 a t m$

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