$X e F_{6} + H_{2} O ⇌ X e O F_{4} + 2 H F$ 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: $- X e O_{4} + 2 H F ⇌ X e O_{4} F_{2} + H_{2} O$ .

\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 C $\frac{K_{2}}{K_{1}}$
(1) $X e F_{6} + H_{2} O ⇌ {X e O F}_{4} + 2 H F K_{1}$
(2) ${X e O}_{4} + {X B F}_{6} ⇌ {X e O F}_{4} + {X e O}_{3} F_{2} K_{2}$
(3) ${X e O}_{4} + 2 H F ⇌ {X e O}_{3} F_{2} + H_{2} O K$
reaction (3) = reaction (2) - reaction (1)
Hence $K = \frac{K_{2}}{K_{1}}$

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Similar questions

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

= k_{1}, X e O_{4} + X e F_{6} ⇌ X e O F_{4} + X e O_{3} F_{2}

= K_{2}

X e O_{4} + 2 H F ⇌ X e O_{3} F_{2} + H_{2} O

X e F_{6} + H_{2} O ⇌ X e O F_{4} + 2 H F; K_{1}

X e O_{4} + X e F_{6} ⇌ X e O F_{4} + X e O_{3} F_{2}; K_{2}

X e O_{4} + 2 H F ⇋ X e O_{3} F_{2} + H_{2} O

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

Two gaseous equilibria $S O_{2} (g) + 0.5 O_{2} (g) ⇌ S O_{3} (g)$ and

$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ have equilibrium constants $K_{1}$ and $K_{2}$ respectively

at 298K. Which of the following relationships between $K_{1}$ and $K_{2}$ is correct

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