If K_1 and K_2 are respective equilibrium con | vedclass

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Question

(D) Let the given reactions be:$(1) XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)} ; K_1$$(2) XeO_{4(g)} + XeF_{6(g)} \rightleftha

Vedclass · Accepted Answer

$\frac{K_2}{K_1}$

Vedclass · Answer

(D) Let the given reactions be:$(1) XeF_{6(g)} + H_2O_{(g)} ightleftharpoons XeOF_{4(g)} + 2HF_{(g)} ; K_1$$(2) XeO_{4(g)} + XeF_{6(g)} ightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)} ; K_2$To obtain the target reaction $XeO_{4(g)} + 2HF_{(g)} ightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$,we perform the operation: $(2) - (1)$.This is equivalent to adding reaction $(2)$ and the reverse of reaction $(1)$.The equilibrium constant for the reverse of reaction $(1)$ is $\frac{1}{K_1}$.Therefore,the equilibrium constant $K$ for the target reaction is $K = K_2 	imes \frac{1}{K_1} = \frac{K_2}{K_1}$.

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