The equilibrium constant of the reaction $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)$ is $4 \times 10^{- 3} a t m^{- 1 / 2} .$ The equilibrium constant of the reaction $2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ would be:

250 atm

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4 \times 10^{3}

atm

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0.25 \times 10^{4}

atm

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6.25 \times 10^{4}

atm

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Solution

The correct option is D $6.25 \times 10^{4}$ atm
(1) $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g) K_{p} = 4 \times 10^{- 3} a t m^{- 1 / 2}$
$\Rightarrow$ $K_{P} = \frac{[P_{S O_{3}}]}{[P_{O_{2}}]^{1 / 2} [P_{S O_{2}}]}$
$\Rightarrow$ $\frac{1}{K_{P}} = \frac{[P_{O_{2}}]^{1 / 2} [P_{S O_{2}}]}{[P_{S O_{3}}]}$ ...(i)

(2) $2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
$\Rightarrow$ $K_{P}^{''} = \frac{[P_{O_{2}}] [P_{S O_{2}}]^{2}}{[P_{S O_{3}}]^{2}}$ ...(ii)
from equation (i) and (ii), we get

$K_{P}^{''} = (\frac{1}{K_{P}})^{2} = \frac{1}{(4 \times 10^{- 3})^{2}}$
$\Rightarrow K_{P}^{''} = \frac{1}{16 \times 10^{- 6}} a t m = 6.25 \times 10^{4} a t m .$
Therefore the correct option is (d)

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