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Dalton's Law of Partial Pressures

Kp value for ...

Question

$K_{p}$ value for $2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}$ is $5.0 a t m^{- 1}$ . What is the equilibrium partial pressure of $O_{2}$ if the equilibrium pressures of $S O_{2}$ and $S O_{3}$ are equal?

A

0.2

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B

0.3

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C

0.4

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D

0.1

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Solution

The correct option is C $0.2$
For the given reaction,
$2 S O_{2} + O_{2} \to 2 S O_{3}$
$K_{p}$ can be calculated as,
$K_{p} = \frac{[P_{S O_{3}}^{2}]}{[P_{S O_{2}}]^{2} \times [P_{O_{2}}]}$
As given,
$[P_{S O_{3}}] = [P_{S O_{2}}]$
So,
$5 = \frac{1}{[P_{O_{2}}]}$
$⟹ [P_{O_{2}}] = \frac{1}{5} = 0.2 a t m$

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