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Solubility Product

The equilibri...

Question

The equilibrium constant of the following equilibrium are $K_{1}$ and $K_{2}$ respectively. The relationship between $K_{1}$ and $K_{2}$ is:
$S O_{2} + 1 / 2 O_{2} ⇌ S O_{3}$ and $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$

K_{2} = K_{1}

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K_{2} = K_{1}^{2}

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K_{2} = \frac{1}{K_{1}}

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K_{2} = \frac{1}{K_{1}^{2}}

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Solution

The correct option is A $K_{2} = K_{1}^{2}$
According to law of mass action:
$S O_{2} + 1 / 2 O_{2} ⇌ S O_{3}$

$K_{1} = \frac{[S O_{3}]}{[S O_{2}] [O_{2}]^{\frac{1}{2}}}$

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$
$K_{2} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
Hence, $K_{2} = K_{1}^{2}$

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