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Reaction Quotient

The equilibri...

Question

The equilibrium constant $k_{p}$ for the reaction
$2 S O_{2} + O_{2} \Leftrightarrow 2 S O_{3}$ is 2.5 $a t m^{- 1}$ . What would be the partial pressure of $O_{2}$ , if the equilibrium pressures of $S O_{2}$ and $S O_{3}$ are equal?

A

304 mm

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B

30.4 mm

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C

0.04 mm

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D

760 mm

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Solution

The correct option is A 304 mm
The equilibrium reaction is $2 S O_{2} + O_{2} \Leftrightarrow 2 S O_{3}$
The expression for the equilibrium constant is $K_{p} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
But $[S O_{3}] = [S O_{2}]$ and $K_{p} = 2.5 a t m^{- 1}$
Substitute the values in the above expression.
$2.5 = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} = \frac{1}{[O_{2}]}$
Thus $[O_{2}] = 0.4 = 0.4 \times 760 = 304 m m$

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