For the gaseous reaction,equilibrium constant | vedclass

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(D) Given reactions:$(i) \ XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$$(ii) \ XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$To obtain

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$\frac{K_2}{K_1}$

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(D) Given reactions:$(i) \ XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$$(ii) \ XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$To obtain the target reaction $XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$,we perform the operation $(ii) - (i)$:$(XeO_4 + XeF_6) - (XeF_6 + H_2O) \rightleftharpoons (XeOF_4 + XeO_3F_2) - (XeOF_4 + 2HF)$$XeO_4 + XeF_6 - XeF_6 - H_2O \rightleftharpoons XeOF_4 + XeO_3F_2 - XeOF_4 - 2HF$$XeO_4 - H_2O \rightleftharpoons XeO_3F_2 - 2HF$Rearranging gives: $XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$Since we subtracted reaction $(i)$ from reaction $(ii)$,the new equilibrium constant $K$ is given by $\frac{K_2}{K_1}$.

For the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be

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For the gaseous reaction,equilibrium constant is given:$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$The equilibrium constant for the reaction:$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be