In the following gaseous phase equilibrium at constant temperature.
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
The concentration of $[S O_{2}] = 3.0 \times 10^{- 3} M, [O_{2}] = 3.5 \times 10^{- 3} M, [S O_{3}] = 5.0 \times 10^{- 2} M$ . Calculate equilibrium constant.

K_{c} = 7.936 \times 10^{4}

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K_{c} = 17.936 \times 10^{4}

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K_{c} = 12.936 \times 10^{2}

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K_{c} = 2.936 \times 10^{4}

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Solution

The correct option is A $K_{c} = 7.936 \times 10^{4}$
Given, the concentration of $[S O_{2}] = 3.0 \times 10^{- 3} M; [O_{2}] = 3.5 \times 10^{- 3} M; [S O_{3}] = 5.0 \times 10^{- 2} M$
For the reaction,
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
Putting all the values,
$= \frac{(5 \times 10^{- 2})^{2}}{(3 \times 10^{- 3})^{2} (3.5 \times 10^{- 3})} = 7.936 \times 10^{4}$

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