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# If $K_1$ and $K_2$ are respective equilibrium constants of two reactions: $XeF_6 (g) + H_2O(g) \rightleftharpoons XeOF_4 (g) + 2HF(g)$ and $XeO_4 (g) + XeF_6(g) \rightleftharpoons XeOF_4 (g) + 2XeO_3F_2(g)$ Then equilibrium constant for the reaction $XeO_4(g) + 2HF(g) \rightleftharpoons XeO_3F_2(g) + H_2O(g)$ will be

(a) $K_1K_2$

(b) $\frac{K_1}{(K_2)^2}$

(c) $K_1(K_2)^{-1}$

(d) $\frac{K_2}{K_1}$

Answer: $\frac{K_2}{K_1}$

The First reaction,
$XeF_6(g) + H_2O(g) \rightleftharpoons XeOF_4(g) + 2HF(g)$

$K_1 = \frac{[XeOF_4][HF]^2}{[XeF_6][H_2O]}$

The Second reaction,
$XeO_4(g) + XeF_6(g) \rightleftharpoons XeOF_4(g) + XeO_3F_2(g)$

$K_2 = \frac{[XeOF_4][XeO_3F_2]}{[XeO_4][XeF_6]}$

The Third reaction,
$XeO_4(g) + 2HF(g) \rightleftharpoons XeO_3F_2(g) + H_2O(g)$

Let its equilibrium constant = K
$\therefore K = \frac{[XeO_3F_2][H_2O]}{[XeO_4][HF]^2} = \frac{K_2}{K_1}$