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Law of Mass Action

In the equili...

Question

In the equilibrium, $2 {S O}_{2} + O_{2} ⇌ 2 S O_{3}$ the partial pressure of ${S O}_{2}, O_{2}$ and $S O_{3}$ are 0.662,0.10 and 0.331 respectively.What should be the partial pressure of oxygen so that the equilibrium concentration of ${S O}_{2}$ and $S O_{3}$ are equal?

A

0.4atm

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B

1atm

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C

0.8atm

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D

0.25atm

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Solution

The correct option is B 0.4atm
The given reaction,

$2 S O_{2} + O_{2} ⇋ 2 S O_{3}$

Partial pressure,

$2 S O_{2} = 0.662, O_{2} = 0.10, 2 S O_{3} = 0.331$

$K_{p} = \frac{{p_{S O_{3}}}^{2}}{{p_{S O_{2}}}^{2} p_{O_{2}}}$

$= \frac{{[0.331]}^{2}}{{[0.662]}^{2} [0.10]}$

$= 2.5$

Now given that equilibrium condition,
$p_{S O_{3}} = p_{S O_{2}}$
$K_{p} = \frac{{p_{S O_{3}}}^{2}}{{p_{S O_{2}}}^{2} p_{O_{2}}}$

$2.5 = \frac{1}{p_{O_{2}}}$

$p_{o_{2}} = \frac{1}{2.5} = 0.4$

Hence,
The partial pressure of $O x y g e n = 0.4 a t m$

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