If K₁ and K₂ arc the respective equilibrium constants for the two reactions,
$\begin{array}{l} {XeF}_{6} (g) + H_{2} O (g) ⇌ {XeOF}_{4} (g) + 2 HF (g) \\ {XeO}_{4} (g) + {XeF}_{6} (g) ⇌ {XeOF}_{4} (g) + {XeO}_{3} F_{2} (g) \end{array}$
the equilibrium constant of the reaction, ${XeO}_{4} (g) + 2 HF (g) \to {XeO}_{3} F_{2} (g) + H_{2} O (g)$ will be :

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K₁ / K₂

K₂ / K₁

K₁ . K₂

K₁ / (K₂)²

answer is B.

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Detailed Solution

${XeF}_{6} (g) + H_{2} O (g) ⟶ {XeOF}_{4} (g) + 2 HF (g)$
$K_{1} = \frac{[HF]^{2} [{XeOF}_{4}]}{[{XeF}_{6}] [H_{2} O]} . . . . (i)$
XeO₄(g) + XeF₆(g) + XeOF₄(g) + XeO₃F₂(g) ;
$K_{2} = \frac{[{XeOF}_{4}] [{XeO}_{3} F_{2}]}{[{XeO}_{4}] [{XeF}_{6}]} . . . . . (ii)$
For ${XeO}_{4} (g) + 2 HF (g) ⇌ {XeO}_{3} F_{2} (g) + H_{2} O (g)$
$K_{c} = \frac{[{XeO}_{3} F_{2}] [H_{2} O]}{[{XeO}_{4}] [HF]^{2}} = \frac{K_{2} \times K_{1}}{1}$