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Dalton's Law of Partial Pressures

For the equil...

Question

For the equilibrium , ${2 S O}_{3} (g) ⇌ 2 {S O}_{2} (g) + O_{2} (g)$ ; the partial pressure of ${S O}_{3}$ , ${S O}_{2}$ and $O_{2}$ gases at $650 K$ are respectively $0.2 b a r, 0.6 b a r$ and $0.4 b a r$ . If the moles of both oxides of sulphur are so adjusted as equal, what will be the partial pressure of $O_{2}$ ?

A

4.4

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B

0.44

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C

3.6

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D

1.12

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Solution

The correct option is D $3.6$
$2 S O_{3} ⇌ 2 S O_{2} + O_{2}$
$K_{p} = \frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}} = \frac{(0.6)^{2} (0.4)}{(0.2)^{2}} = 3.6$
If $S O_{2}$ and $S O_{3}$ are made equal,
$3.6 = \frac{[S O_{2}]^{2}}{[S O_{3}]^{2}} \times [O_{2}] \Rightarrow [O_{2}] = 3.6$
Because moles of $S O_{2} & S O_{3}$ are equal, $P_{S O_{2}}, & P_{S O_{3}}$ came each other

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