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Arrhenius Acid

In the equili...

Question

In the equilibrium $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ , the partial pressures of $S O_{2}, O_{2}$ and $S O_{3}$ are $0.662, 0.101$ and $0.331 a t m$ respectively. What would be the partial pressure of oxygen so that the equilibrium concentration of $S O_{2}$ and $S O_{3}$ are equal?

A

0.4 a t m

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B

1.0 a t m

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C

0.8 a t m

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D

0.25 a t m

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Solution

The correct option is A $0.4 a t m$
Equilibrium constant $(K_{p})$ for the reaction is $\frac{[partial pressure of S O_{3}]^{2}}{[partial pressure of S O_{2}]^{2} \times [partial pressure of O_{2}]}$
$K_{p} = \frac{[0.331]^{2}}{[0.662]^{2} \times [0.101]} = 2.475$
$K_{p}$ for any reaction is constant and is independent of the partial pressures of the reacting species.
Let the partial pressures of $S O_{2}$ and $S O_{3}$ at equilibrium be $x$ (since it is given in the question that the equilibrium concentration of $S O_{2}$ and $S O_{3}$ are equal). Let the partial pressure of $O_{2}$ at equilibrium be $[p O_{2}]$ .
Now, $K_{p} = \frac{[x]^{2}}{[x]^{2} \times [p O_{2}]} = 2.475$
$p O_{2} = \frac{1}{2.475} = 0.4 a t m$

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