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The value of ...

Question

The value of $K_{p}$ for the reaction,
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is $5$ .
What will be the partial pressure of $O_{2}$ at equilibrium when equal moles of $S O_{2}$ and $S O_{2}$ are present at equilibrium?

A

0.5

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B

0.3

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C

0.2

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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

$K_{p} = \frac{[P_{S O_{3}}]}{[P_{S O_{2}}] [P_{O_{2}}]}$

as given
$[P_{S O_{3}}] = [P S O_{2}]$
so ,
$5 = \frac{1}{[O_{2}]}$

$⟹ [P_{O_{2}}] = \frac{1}{5}$

$⟹ [P_{O_{2}}] = 0.2$
so partial pressure of oxygen will be $0.2$ atm.

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