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)^2`
B. `K_1.K_2`
C. `K_1//K_2`
D. `K_2//K_1`

Answer

D

Explanation

`XeF_6(g)+H_2O(g)hArr XeOF_4(g)+2HF(g)K_1=([XeOF_4][HF]^2)/([XeF_6][H_2O])`......(i)
`XeO_4(g)=XeF_6(g)hArrXeOF_4(g)+XeO_3F_2(g)K_2=([XeOF_4][XeO_3F_2])/([XeO_4][XeF_6])`......(ii)
For the reaction .
`XeO_4(g)+2HF(g)hArrXeO_3F_2(g)H_2O(g)K=([XeO_3F_2][H_2O])/([XeO_4][HF]^2)`......(iii)
By dividing eq.(ii) by (i) we get,
`K=(K_2)/(K_1)`