$X e F_{6} + H_{2} O ⇌ X e O F_{4} + 2 H F$ constant $= k_{1}, X e O_{4} + X e F_{6} ⇌ X e O F_{4} + X e O_{3} F_{2}$ constant $= K_{2}$ . Then equilibrium constant for the following reaction will be:
$X e O_{4} + 2 H F ⇌ X e O_{3} F_{2} + H_{2} O$ .

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

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K_{1} + K_{2}

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

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\frac{K_{1}}{(K_{2})^{2}}

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Solution

The correct option is A $\frac{K_{2}}{K_{1}}$
Given that
$X e F_{6} (g) + H_{2} O (g) ⇌ X e O F_{4} (g) + 2 H F (g) = K_{1}$
$X e O_{4} (g) + X e F_{6} (g) ⇌ X e O F_{4} (g) + X e O_{3} F_{2} (g)$ constant = $K_{2}$
equilibrium constant for $(K)$
$X e O_{4} + 2 H F ⇌ X e O_{3} F_{2} + H_{2} O$ (constant = $K$ )
In terms of $K_{1}$ and $K_{2}$
$K_{1} = \frac{[X e O F_{4}] [H F]^{2}}{[X e F_{6}] [H_{2} O]}$
$K_{2} = \frac{[X e O F_{4}] [X e O_{3} F_{2}}{[X e O_{4}] [X e F_{6}}$
$K = \frac{[X e O_{3} F_{2}] [H_{2} O]}{[X e O_{4}] [H F]^{2}}$
$\frac{X e O F_{4}}{X e F_{6}} = K_{1} \frac{[H_{2} O]}{[H F]^{2}}$
$K = \frac{K_{2}}{K_{1}}$
So, option $A$ is correct.

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Similar questions

X e F_{6} + H_{2} O ⇌ X e O F_{4} + 2 H F

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