If `K_(1)` and `K_(2)` are respective equilibrium constants for two reactions `:`
`XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF_(g)`
`XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)`
Then equilibrium constant for the reaction
`XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g)` will be

Options

A. `K_(1)K_(2)`
B. `K_(1)//(K_(2))^(2)`
C. `K_(1)(K_(2))^(-1)`
D. `K_(2)//K_(1)`

Answer

D

Explanation

`XeF_(6)(g)+H_(2)O(g) hArr XeOF_(4)(g)+2HF(g)`
`K_(1)=([XeOF_(4)][HF]^(2))/([XeF_(6)][H_(2)O])`
`XeO_(4)(g)+XeF_(6)(g) hArr XeOF_(4)(g)`
`K_(2)=([XeOF_(4)][XeO_(3)])/([XeO_(4)][XeF_(6)])`
`XeO_(4)(g)+2HF(g) hArr XeO_(3)(g)+H_(2)O(g)`
Let its eqn. constant `=K`
`:. K=([XeO_(3)F_(2)][H_(2)O])/([XeO_(4)][HF]^(2))=(K_(2))/(K_(1))`