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Dalton's Law of Partial Pressures

5 moles of ...

Question

$5$ moles of $S O_{2}$ and $5$ moles of $O_{2}^{-}$ are allowed to react. At equilibrium, it was found that $60$ % of $S O_{2}$ is used up. If the pressure of the equilibrium mixture is one atmosphere, the partial pressure of $O_{2}$ is :

A

0.52 a t m

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B

0.21 a t m

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C

0.41 a t m

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D

0.82 a t m

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Solution

The correct option is C $0.41 a t m$
${S O}_{2} + \frac{1}{2} O_{2} ⟶ {S O}_{3}$
Initial : $5$ mole $5$ mole $0$ mole
At eqn : $5 - 5 \times 0.6$ $5 - 5 \times 0.3$ $5 \times 0.6$
$= 2$ mole $= 3.5$ mole $= 3.0$ mole
$∴$ total moles at equilibrium $= 2 + 3.5 + 3.0 = 8.5$ mole
$∴ p_{O_{2}} = x_{O_{2}} \times P = \frac{3.5}{8.5} \times 1 a t m = 0.41 a t m$

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