Find out the equilibrium constant for the reaction, `XeO_(4)(g)+2HF(g)hArr XeO_(3)F_(2)(g)+H_(2)O(g)` consider `K_(1)` as the equilibrium constant for the reaction, `XeF_(6)(g)+H_(2)O(g)hArr XeOF_(4)(g)+2HF(g) and K_(2)` as the equilibrium constant for the reaction, `XeO_(4)(g)+XeF_(6)(g) hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)`.

Explanation

According to the given condition,
`K_(1)=([XeOF_(4)]xx[HF]^(2))/([XeF_(6)]xx[H_(2)O]) and K_(2)=([XeOF_(4)]xx[XeO_(3)F_(2)])/([XeO_(4)]xx[XeF_(6)])`
`therefore`For `XeO_(4)(g)+2HF(g)hArr XeO_(3)F_(2)(g)+H_(2)O(g)`
Equilibrium constant, `K=([XeO_(3)F_(2)]xx[H_(2)O])/([XeO_(4)]xx[HF]^(2))`
Now, `(K_(2))/(K_(1))=([XeO_(3)F_(2)]xx[H_(2)O])/([XeO_(4)]xx[HF]^(2))=K" "therefore K=(K_(2))/(K_(1))`.