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Derivation of Kp and Kc

What is K c f...

Question

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M$ and $[S O_{3}] = 1.90 M$ ?
$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$

12.24 M^{- 1}

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1.23 M^{- 1}

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22.23 M^{- 1}

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32.23 M^{- 1}

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Solution

The correct option is A $12.24 M^{- 1}$
The equilibrium constant $(K_{c})$ for the give reaction is:
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
$= \frac{(1.90)^{2} M^{2}}{(0.60)^{2} (0.821) M^{3}}$
$= 12.239 M^{- 1}$ (approximately)
Hence, $K_{c}$ for the equilibrium is $12.239 M^{- 1}$ .

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